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  <row Id="1" PostTypeId="1" AcceptedAnswerId="9" CreationDate="2010-07-20T19:09:27.200" Score="20" ViewCount="483" Body="&lt;p&gt;Can someone explain to me how there can be different kinds of infinities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading &quot;the man who loved only numbers&quot; and came across the concept of countable and uncountable infinities, but they're only words to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T19:27:38.740" LastActivityDate="2010-11-02T17:28:02.230" Title="Different kinds of infinities?" Tags="&lt;set-theory&gt;&lt;intuition&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="6" />
  <row Id="2" PostTypeId="1" AcceptedAnswerId="19" CreationDate="2010-07-20T19:09:44.937" Score="3" ViewCount="301" Body="&lt;p&gt;This &lt;code&gt;limit&lt;/code&gt; thing keeps coming up in my calculus textbook.  What is it?&lt;/p&gt;&#xA;" OwnerUserId="6" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-20T20:21:40.707" LastActivityDate="2010-08-21T20:25:07.867" Title="What is a limit?" Tags="&lt;calculus&gt;&lt;limit&gt;&lt;definition&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="1" />
  <row Id="3" PostTypeId="1" CreationDate="2010-07-20T19:12:14.353" Score="7" ViewCount="155" Body="&lt;p&gt;&lt;a href=&quot;http://mathfactor.uark.edu/&quot; rel=&quot;nofollow&quot;&gt;mathfactor&lt;/a&gt; is one I listen to.  Does anyone else have a recommendation?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:52:17.563" LastActivityDate="2010-07-28T17:39:29.080" CommunityOwnedDate="2010-07-20T20:35:23.263" Title="List of interesting math podcasts?" Tags="&lt;big-list&gt;&lt;soft-question&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="5" />
  <row Id="4" PostTypeId="2" ParentId="3" CreationDate="2010-07-20T19:14:10.603" Score="2" ViewCount="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.bbc.co.uk/podcasts/series/moreorless&quot; rel=&quot;nofollow&quot;&gt;More or Less&lt;/a&gt; is a BBC Radio 4 programme about maths and statistics in the news, and there is a free podcast. It's presented by &lt;a href=&quot;http://timharford.com/&quot; rel=&quot;nofollow&quot;&gt;Tim Harford&lt;/a&gt;, the Undercover Economist from the &lt;a href=&quot;http://www.ft.com/home/uk&quot; rel=&quot;nofollow&quot;&gt;Financial Times&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-20T19:14:10.603" CommunityOwnedDate="2010-07-20T20:35:23.263" CommentCount="1" />
  <row Id="5" PostTypeId="1" AcceptedAnswerId="7" CreationDate="2010-07-20T19:18:01.250" Score="9" ViewCount="452" Body="&lt;p&gt;I have read a few proofs that $\sqrt{2}$ is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never, however, been able to really grasp what they were talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simplified proof that $\sqrt{2}$ is irrational?&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T03:46:40.490" LastActivityDate="2011-01-06T18:48:04.747" Title="How can you prove that the square root of two is irrational?" Tags="&lt;algebra-precalculus&gt;&lt;irrational-numbers&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="6" PostTypeId="1" AcceptedAnswerId="281" CreationDate="2010-07-20T19:20:00.543" Score="4" ViewCount="87" Body="&lt;p&gt;I'm looking for a nice, quick online graphing tool. The ability to link to, or embed the output would be handy, too.&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T07:34:23.613" LastActivityDate="2010-07-21T07:34:23.613" CommunityOwnedDate="2010-07-20T19:20:00.543" Title="What is your favorite online graphing tool?" Tags="&lt;math-software&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7" PostTypeId="2" ParentId="5" CreationDate="2010-07-20T19:21:52.240" Score="26" ViewCount="0" Body="&lt;p&gt;You use a proof by contradiction. Basically, you suppose that $\sqrt{2}$ can be written as $p/q$. Then you know that $2q^2 = p^2$. However, both $q^2$ and $p^2$ have an even number of factors of two, so $2q^2$ has an odd number of factors of 2, which means it can't be equal to $p^2$.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-20T19:21:52.240" CommentCount="1" />
  <row Id="8" PostTypeId="1" AcceptedAnswerId="17" CreationDate="2010-07-20T19:22:12.190" Score="6" ViewCount="230" Body="&lt;p&gt;I was reading up on the Fibonacci Sequence when I've noticed some were able to calculate specific numbers. So far I've only figured out creating an array and counting to the value, which is incredibly simple, but I reckon I can't find any formula for calculating a Fibonacci number based on it's position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to do this? If so, how are we able to apply these formulas to arrays?&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T17:01:20.937" LastActivityDate="2010-07-28T09:52:46.637" Title="How are we able to calculate specific numbers in the Fibonacci Sequence?" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="6" FavoriteCount="1" />
  <row Id="9" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:22:20.193" Score="35" ViewCount="0" Body="&lt;p&gt;Suppose no one every taught you the names for ordinary numbers.  Then suppose that you and I agreed that we would trade one bushel of corn for each of my sheep.  But there's a problem, we don't know how to count the bushels or the sheep!  So what do we do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We form a &quot;bijection&quot; between the two sets.  That's just fancy language for saying you pair things up by putting one bushel next to each of the sheep.  When we're done we swap.  We've just proved that the number of sheep is the same as the number of bushels without actually counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can try doing the same thing with infinite sets.  So suppose you have the set of positive numbers and I have the set of rational numbers and you want to trade me one positive number for each of my rationals.  Can you do so in a way that gets all of my rational numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps surprisingly the answer is yes!  You make the rational numbers into a big square grid with the numerator and denominators as the two coordinates.  Then you start placing your &quot;bushels&quot; along diagonals of increasing size, &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aPairing_natural.svg&quot; rel=&quot;nofollow&quot;&gt;see wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that the rational numbers are &quot;countable&quot; that is you can find a clever way to count them off in the above fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remarkable fact is that for the real numbers there's &lt;em&gt;no way at all&lt;/em&gt; to count them off in this way.  No matter how clever you are you won't be able to scam me out of all of my real numbers by placing a natural number next to each of them.  The proof of that is Cantor's clever &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;diagonal argument&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T19:22:20.193" CommentCount="3" />
  <row Id="11" PostTypeId="1" AcceptedAnswerId="49" CreationDate="2010-07-20T19:23:09.117" Score="12" ViewCount="1244" Body="&lt;p&gt;I'm told by smart people that &lt;code&gt;0.999... = 1&lt;/code&gt; and I believe them but is there a proof that explains why?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T19:08:51.070" LastActivityDate="2010-08-17T18:58:40.717" CommunityOwnedDate="2010-07-28T06:53:43.987" Title="Does .99999... = 1?" Tags="&lt;algebra-precalculus&gt;&lt;decimal-expansion&gt;" AnswerCount="19" CommentCount="11" FavoriteCount="7" />
  <row Id="13" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:23:17.570" Score="4" ViewCount="0" Body="&lt;p&gt;The basic concept is thus:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A 'countable' infinity is one where you can give each item in the set an integer and 'count' them (even though there are an infinite number of them)&lt;/li&gt;&#xA;&lt;li&gt;An 'uncountable' infinity defies this. You cannot assign an integer to each item in the set because you will miss items.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The key to seeing this is using the 'diagonal slash' argument as originally put forward by Cantor. With a countable infinity, you can create a list of all the items in the set and assign each one a different natural number. This can be done with the naturals (obviously) and the complete range of integers (including negative numbers) and even the rational numbers (so including fractions). It cannot be done with the reals due to the diagonal slash argument:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Create your list of all real numbers and assign each one an integer&lt;/li&gt;&#xA;&lt;li&gt;Create a real number with the rule that the first digit after the decimal point is different from the first digit of your first number, the second digit is different from the second digit of your second number, and so on for all digits&lt;/li&gt;&#xA;&lt;li&gt;Try and place this number in your list of all numbers... it can't be the first number, or the second or the third... and so on down the list. &lt;/li&gt;&#xA;&lt;li&gt;Reductio Ad Absurdium, your number does not exist in your countable list of all real numbers and must be added on to create a new list. The same process can then be done again to show the list still isn't complete.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This shows a difference between two obviously infinite sets and leads to the somewhat scary conclusion that there are (at least) 2 different forms of infinity.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:23:17.570" />
  <row Id="14" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:23:20.317" Score="2" ViewCount="0" Body="&lt;p&gt;How many Natural (Integers) Numbers could you count?&#xA;There are infinitely many, yet you can count them.&#xA;It's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;Countable Set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many Real Numbers are there?&#xA;Infinitely as well (Since at least every Natural Number is a Real Number).&#xA;Yet you won't be able to count them (Intuitively, Let's say you name a number the first, then find the second, I can, for sure, find a number in between, their average which is Real Number as well)..&#xA;It's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncountable_set&quot; rel=&quot;nofollow&quot;&gt;Uncountable Set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are after is how we define how big is a given set.&#xA;Then you should look for &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinality&quot; rel=&quot;nofollow&quot;&gt;Cardinality&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditorDisplayName="" LastEditDate="2010-11-02T17:28:02.230" LastActivityDate="2010-11-02T17:28:02.230" CommentCount="3" />
  <row Id="15" PostTypeId="2" ParentId="6" CreationDate="2010-07-20T19:23:21.220" Score="5" ViewCount="0" Body="&lt;p&gt;Well, I am not sure where you want to embed the graphs, but &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sin%28x%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; is pretty handy for graphing. It has most of the features of Mathematica, can handle 3D functions, and fancy scaling and such. I highly recommend it.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="41" LastEditorDisplayName="" LastEditDate="2010-07-20T19:26:37.047" LastActivityDate="2010-07-20T19:26:37.047" CommunityOwnedDate="2010-07-20T19:23:21.220" />
  <row Id="16" PostTypeId="2" ParentId="5" CreationDate="2010-07-20T19:23:50.563" Score="9" ViewCount="0" Body="&lt;p&gt;Consider this proof by contradiction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\sqrt{2}$ is rational. Then there exists some rational $R=\sqrt{2}=\frac{Q}{D}$, where $Q$ and $D$ are positive integers and relatively prime (since $R$ can be expressed in simplified form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $R^2 = 2 = \frac{Q^2}{D^2}$. Since $Q$ and $D$ are relatively prime, this means that only $Q^2$ can have $2$ in its prime decomposition, and the exponent must be one. Thus, $Q^2 = 2^1 x$, for some odd integer $x$. But $Q^2$ is a square, and thus the exponents for all of its prime factors must be even. Here we have a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\sqrt{2}$ must be irrational.&lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2010-07-20T19:23:50.563" />
  <row Id="17" PostTypeId="2" ParentId="8" CreationDate="2010-07-20T19:23:54.230" Score="9" ViewCount="0" Body="&lt;p&gt;Wikipedia has a closed-form function called &quot;Binet's formula&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the Golden Ratio.&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-22T18:38:42.820" LastActivityDate="2010-07-22T18:38:42.820" CommentCount="5" />
  <row Id="18" PostTypeId="1" CreationDate="2010-07-20T19:24:34.823" Score="2" ViewCount="96" Body="&lt;p&gt;Given the semimajor axis and a flattening factor, is it possible to calculate the semiminor axis?&lt;/p&gt;&#xA;" OwnerUserId="17" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T16:08:54.593" LastActivityDate="2010-12-24T16:08:54.593" Title="How do you calculate the semiminor axis for an ellipsoid?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="19" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T19:24:52.200" Score="6" ViewCount="0" Body="&lt;p&gt;For many functions there are undefined values. For example in the function $$f(x) = \frac{x^2 - 1}{x - 1}$$ there is no value of $f(x)$ for $x = 1$, since that involves a division by 0.&#xA;In this instance though, as x gets closer and closer to 1, $f(x)$ gets closer and closer to 2, so we define that to be the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus limits come up when getting the slope of a linear function as I'm sure you've seen, the problem is that getting the slope at a certain point, using the slope formula we all know, involves dividing by zero, so we use limits to get the answer. More specifically we create a formula for the slope between the point $(x, f(x))$ and the point $(x+h, f(x+h))$ and find the slope as $h \to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limits as $x \to \infty$ are also used often, since even if a function is well defined for all $x$ it is still impossible to simply plug in infinite and work it out, in this way it can be used for approximation purposes. Calculating the runtime of computer programs on large inputs is often done by deriving a formula for the run-time given input of a certain length, then finding the limit as $x \to \infty$&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-21T20:25:07.867" LastActivityDate="2010-08-21T20:25:07.867" />
  <row Id="20" PostTypeId="1" CreationDate="2010-07-20T19:25:09.563" Score="3" ViewCount="427" Body="&lt;p&gt;In mathematics there seem to be a lot of different types of number. What exactly are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Real numbers&lt;/li&gt;&#xA;&lt;li&gt;Integers&lt;/li&gt;&#xA;&lt;li&gt;Rationals numbers&lt;/li&gt;&#xA;&lt;li&gt;Decimals&lt;/li&gt;&#xA;&lt;li&gt;Complex numbers&lt;/li&gt;&#xA;&lt;li&gt;Natural numbers&lt;/li&gt;&#xA;&lt;li&gt;Cardinals&lt;/li&gt;&#xA;&lt;li&gt;Ordinals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And as workmad3 points out, some more advanced categories (I'd never heard of)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hyper-reals&lt;/li&gt;&#xA;&lt;li&gt;Quaternions&lt;/li&gt;&#xA;&lt;li&gt;Imaginary numbers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or any other type of classification of a number I missed.&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T23:12:36.630" LastActivityDate="2010-10-13T06:01:26.353" Title="Can you explain what a real number is (also rational, decimal, integer, natural, cardinal, ordinal...)" Tags="&lt;terminology&gt;&lt;definition&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="1" />
  <row Id="22" PostTypeId="1" AcceptedAnswerId="53" CreationDate="2010-07-20T19:25:58.483" Score="6" ViewCount="209" Body="&lt;p&gt;By matrix-defined, I mean&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; x  =&#xA;&#xA;    | i j k |&#xA;det(| a b c |)&#xA;    | d e f |&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(instead of the definition of the product of the magnitudes multiplied by the sign of their angle, in the direction orthogonal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I try cross producting two vectors with no &lt;code&gt;k&lt;/code&gt; component, I get one with only &lt;code&gt;k&lt;/code&gt;, which is expected.  But why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been pointed out, I am asking why the algebraic definition lines up with the geometric definition.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:42:23.140" LastActivityDate="2010-07-28T02:03:15.647" Title="Why is the matrix-defined Cross Product of two 3D vectors always orthogonal?" Tags="&lt;linear-algebra&gt;" AnswerCount="4" />
  <row Id="23" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:26:13.353" Score="-2" ViewCount="0" Body="&lt;p&gt;well the simplest is &lt;code&gt;3x(1/3) = 1&lt;/code&gt; where &lt;code&gt;1/3 = 0.33·&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2010-07-20T19:26:13.353" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="1" />
  <row Id="24" PostTypeId="2" ParentId="8" CreationDate="2010-07-20T19:26:16.643" Score="2" ViewCount="0" Body="&lt;p&gt;The closed form calculation for Fibonacci sequences is known as &lt;a href=&quot;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&quot; rel=&quot;nofollow&quot;&gt;Binet's Formula&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="53" LastActivityDate="2010-07-20T19:26:16.643" />
  <row Id="25" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:26:36.643" Score="1" ViewCount="1" Body="&lt;p&gt;The proof I've always seen was that if 1/3 == 0.333... then 0.333... x 3 must be equal to 1, but at the same time, calculating it on a digit level gives 0.999...&lt;/p&gt;&#xA;" OwnerUserId="17" LastActivityDate="2010-07-20T19:26:36.643" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="8" />
  <row Id="26" PostTypeId="2" ParentId="8" CreationDate="2010-07-20T19:26:38.187" Score="1" ViewCount="1" Body="&lt;p&gt;You can use Binet's formula, described at &lt;a href=&quot;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see also Wikipedia for a proof: &lt;a href=&quot;http://en.wikipedia.org/wiki/Binet_formula#Closed_form_expression&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binet_formula#Closed_form_expression&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="55" LastActivityDate="2010-07-20T19:26:38.187" />
  <row Id="27" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T19:26:38.670" Score="0" ViewCount="0" Body="&lt;p&gt;A limit is the term used when you examine a function as it approaches some value. For example, the function f(x) = 1/x. You can take a limit of this function as x approaches infinity and as x approaches 0 and see what occurs, and get an idea of the behaviour of the function in these regions even though the function cannot be defined at the exact values you have taken limits of. (1/0 is undefined, and 1/infinity cannot be done without adding infinity to the number system - see the hyperreals for this)&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:26:38.670" />
  <row Id="29" PostTypeId="1" CreationDate="2010-07-20T19:27:09.700" Score="9" ViewCount="381" Body="&lt;p&gt;I'm looking for an online or software calculator that can show me the history of items I typed in, much like an expensive Ti calculator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you recommend any?&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T01:04:25.440" LastActivityDate="2010-10-08T02:42:51.233" CommunityOwnedDate="2010-07-28T01:32:30.370" Title="Can you recommend a decent online or software calculator?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;math-software&gt;&lt;computer-algebra-systems&gt;" AnswerCount="18" FavoriteCount="4" />
  <row Id="30" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:27:38.963" Score="-2" ViewCount="0" Body="&lt;p&gt;Given:&lt;br&gt;&#xA;$\frac{1}{3} = 0.\bar{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by 3:&lt;br&gt;&#xA;$3\times \left( \frac{1}{3} \right) = \left( 0.\bar{3} \right) \times 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;br&gt;&#xA;$\frac{3}{3} = 0.\bar{9}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-21T19:25:39.083" LastActivityDate="2010-07-21T19:25:39.083" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="31" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:27:39.433" Score="1" ViewCount="2" Body="&lt;p&gt;Indeed this is true. The underlying reason is that decimal numbers are not unique representations of the reals. (Technically, there does not exist a bijection between the set of all decimal numbers and the reals.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a very simple proof:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 / 3 = 0.333... (by long division)&#xA;&#xA;=&amp;gt; 0.333 * 3 = 0.999... (multiplying each digit by 3)&#xA;&#xA;But then we already know 0.333... * 3 = 1&#xA;&#xA;Therefore 0.999... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-21T07:21:11.583" LastActivityDate="2010-07-21T07:21:11.583" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="7" />
  <row Id="35" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:29:54.777" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Hotel&lt;/a&gt; is a classic demonstration.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2010-07-20T19:29:54.777" CommentCount="2" />
  <row Id="36" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T19:30:11.283" Score="2" ViewCount="4" Body="&lt;p&gt;What others have provided is a pretty good intuitive/more general definition of a limit. If you want something a bit more technical and rigorous, have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit&quot; rel=&quot;nofollow&quot;&gt;(ε, δ)-definition of limit&lt;/a&gt;, which is the one used in mathematical analysis.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:30:11.283" />
  <row Id="38" PostTypeId="2" ParentId="29" CreationDate="2010-07-20T19:30:50.313" Score="4" ViewCount="0" Body="&lt;p&gt;Lots of people like to use &lt;a href=&quot;http://instacalc.com/&quot; rel=&quot;nofollow&quot;&gt;Instacalc&lt;/a&gt; which lets you do unit conversions and store intermediate calculations in variables.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-20T19:30:50.313" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="40" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:32:21.590" Score="6" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Natural numbers&lt;br&gt;&#xA;The &quot;counting&quot; numbers.  (That is, all integers, that are one or greater).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whole numbers&lt;br&gt;&#xA;The Natural numbers, and zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Integers&lt;br&gt;&#xA;The Whole numbers, and the negatives of the Natural numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rational numbers&lt;br&gt;&#xA;Any number that may be expressed by any integer A divided by any integer B, where B is not zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Irrational numbers&lt;br&gt;&#xA;Any number that cannot be expressed as a rational number, but is not imaginary.  All irrational numbers have an infinite decimal representation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Real numbers&lt;br&gt;&#xA;All of the Rational and Irrational numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Imaginary numbers&lt;br&gt;&#xA;All Real numbers, multiplied by the square root of negative one.  Imaginary numbers are signified by the letter &lt;code&gt;i&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Complex numbers&lt;br&gt;&#xA;Numbers composed of the sum of a Real and an Imaginary number.  This includes all Real and all Imaginary numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="38" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-20T19:37:58.457" LastActivityDate="2010-07-20T19:37:58.457" CommentCount="15" />
  <row Id="41" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:32:44.553" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;code&gt;.999...  = 1&lt;/code&gt; because &lt;code&gt;.999...&lt;/code&gt; is a concise symbolic representation of &quot;the limit of some variable as it approaches one.&quot; Therefore, &lt;code&gt;.999... = 1&lt;/code&gt; for the same reason the limit of x as x approaches 1 equals 1.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-07-23T02:13:49.417" LastActivityDate="2010-07-23T02:13:49.417" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="42" PostTypeId="2" ParentId="22" CreationDate="2010-07-20T19:33:50.870" Score="-1" ViewCount="0" Body="&lt;p&gt;The obvious but slightly trite answer is &quot;because that's just how the cross-product works as an operation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for an intuitive reason, the cross-product by definition produces a vector that is orthogonal to the two operand (input) vectors. You know that the base vectors i, j, and k are all orthogonal, thus if your two input vectors lie on the (x, y) plane (i.e. only i and j components), you know that any orthogonal vector must have only a component in the z direction (multiple of k).&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:33:50.870" CommentCount="7" />
  <row Id="43" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:34:26.857" Score="0" ViewCount="0" Body="&lt;p&gt;You can visualise it by thinking about it in infinitesimals. The more 9's you have on the end of 0.999, the closer you get to 1. When you add an infinite number of 9's to the decimal expansion, you are infinitely close to 1 (or an infinitesimal distance away).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this isn't a rigorous proof, just an aid to visualisation of the result.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:34:26.857" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="44" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:35:09.227" Score="9" ViewCount="0" Body="&lt;p&gt;What I really don't like about all the above answers, is the underlying assumption that &lt;code&gt;1/3=0.3333...&lt;/code&gt;, How do you know that?. It seems to me like assuming the something which is already known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof I really like is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.9999... x 10 = 9.999999...&#xA;&#xA;0.9999... x 9 + 0.99999.... = 9.99999....&#xA;&#xA;0.9999... x 9 = 9.9999....-0.999999 = 9&#xA;&#xA;0.9999... x 9 = 9&#xA;&#xA;0.9999... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The only things I need to assume is, that &lt;code&gt;9.999.... - 0.9999... = 9&lt;/code&gt; and that &lt;code&gt;0.999... x 10 = 9.999...&lt;/code&gt;. These seems to me intuitive enough to take for granted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is from an old highschool level math book of the Open University in Israel.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2010-07-21T03:58:15.933" LastActivityDate="2010-07-21T03:58:15.933" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="2" />
  <row Id="46" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:36:43.670" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a collection of proofs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://premium.uploadit.org/AlphaNumeric/Recurring.pdf&quot; rel=&quot;nofollow&quot;&gt;http://premium.uploadit.org/AlphaNumeric/Recurring.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2010-07-20T19:36:43.670" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="47" PostTypeId="2" ParentId="29" CreationDate="2010-07-20T19:37:42.700" Score="6" ViewCount="1" Body="&lt;p&gt;Google's calculator is very powerful: &#xA;&lt;a href=&quot;http://www.googleguide.com/help/calculator.html&quot; rel=&quot;nofollow&quot;&gt;http://www.googleguide.com/help/calculator.html&lt;/a&gt; and your use history will be stored in your browser history.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-20T19:37:42.700" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="48" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:37:45.070" Score="1" ViewCount="1" Body="&lt;p&gt;You might as well start at the 'lowest':&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Integers are all the whole numbers,&lt;/li&gt;&#xA;&lt;li&gt;Natural numbers (&lt;code&gt;N&lt;/code&gt;) are the set of positive integers, &lt;code&gt;{1, 2, ...}&lt;/code&gt; (0 optional),&lt;/li&gt;&#xA;&lt;li&gt;Rational numbers (&lt;code&gt;Q&lt;/code&gt;) are any number that can be represented &lt;code&gt;a/b&lt;/code&gt; with a and b being Integers (&lt;code&gt;|b| &amp;lt; 0&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Real numbers are all the Rational numbers and all the others.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cardinality is the number of elements in a set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;An Ordinal is a well ordered set.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditorDisplayName="" LastEditDate="2010-07-29T12:16:54.263" LastActivityDate="2010-07-29T12:16:54.263" CommentCount="1" />
  <row Id="49" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:38:21.590" Score="37" ViewCount="1" Body="&lt;p&gt;What does it mean when you refer to .99999...?  Symbols don't mean anything in particular until you've &lt;em&gt;defined what you mean by them&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the definition is that you're taking the limit of .9, .99, .999, .9999, etc.  What does it mean to say that limit is 1?  Well, it means that no matter how small a number x you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance x of 1.  But certainly whatever number you chose your number is bigger than 10^-k for some k.  So I can just pick my point to be the kth spot in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more intuitive way of explaining the above argument is that the reason .99999... = 1 is that their difference is zero.  So let's subtract 1.0000... -.99999... = .00000... = 0.  That is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.0 -.9 = .1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.00-.99 = .01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.000-.999=.001, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.000... -.99999... = .000... = 0.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T19:38:21.590" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="10" />
  <row Id="50" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:38:49.920" Score="-4" ViewCount="0" Body="&lt;p&gt;If you want to go the calculus route, then basically you're gonna follow this pattern and use mathy magic and limits to follow it up to its logical conclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.9&#xA;0.99&#xA;0.999&#xA;0.9999&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. This pattern is a sum of 9*10^-i from i=1 to i=infinity. Every iteration, the difference between the pattern and 1 is a bit less. So at i=infinity, the difference is so small that it is literally zero.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-07-20T19:38:49.920" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="51" PostTypeId="1" AcceptedAnswerId="109" CreationDate="2010-07-20T19:39:11.557" Score="5" ViewCount="157" Body="&lt;p&gt;What length of rope should be used to tie a cow to an &lt;strong&gt;exterior fence post&lt;/strong&gt; of a &lt;em&gt;circular&lt;/em&gt; field so that the cow can only graze half of the grass within that field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;updated:&lt;/em&gt;&lt;/strong&gt; To be clear: the cow should be tied to a post on the exterior of the field, not a post at the center of the field. &lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:55:02.220" LastActivityDate="2010-08-05T19:02:22.920" Title="The cow in the field problem (intersecting circular areas)" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" />
  <row Id="52" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:40:38.467" Score="4" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;countably infinite&lt;/em&gt; set is a set for which you can list the elements a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,a&lt;sub&gt;3&lt;/sub&gt;,...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the set of all integers is countably infinite since I can list its elements as follows 0,1,-1,2,-2,3,-3,... So is the set of rational numbers, but this is more difficult to see. Let's start with the positive rationals. Can you see the pattern in this listing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/1,1/2,2/1,1/3,2/2,3/1,1/4,2/3,3/2,4/1,1/5,2/4,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hint: Add the numerator and denominator to see a different pattern.) This listing has lots of repeats, e.g. 1/1, 2/2 and 1/2, 2/4. That's ok since I can condense the listing by skipping over any repeats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/1,1/2,2/1,1/3,3/1,1/4,2/3,3/2,4/1,1/5,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write q&lt;sub&gt;n&lt;/sub&gt; for the n-th element of this list. Then 0,q&lt;sub&gt;1&lt;/sub&gt;,-q&lt;sub&gt;1&lt;/sub&gt;,q&lt;sub&gt;2&lt;/sub&gt;,-q&lt;sub&gt;2&lt;/sub&gt;,q&lt;sub&gt;3&lt;/sub&gt;,-q&lt;sub&gt;3&lt;/sub&gt;,... is a listing of all rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;countable set&lt;/em&gt; is a set which is either finite or countably infinite; an &lt;em&gt;uncountable set&lt;/em&gt; is a set which is not countable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, an uncountable set is an infinite set which has no listing of all of its elements (as in the definition of countably infinite set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of an uncountable set is the set of all real numbers. To see this, you can use the &lt;em&gt;diagonal method&lt;/em&gt;. Ask another question to see how this works...&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-20T19:40:38.467" />
  <row Id="53" PostTypeId="2" ParentId="22" CreationDate="2010-07-20T19:40:39.277" Score="4" ViewCount="0" Body="&lt;p&gt;Assuming you know the definition of orthogonal as &quot;a is orthogonal to b iff $a\cdot b=0$ then we could calculate $(a \times b)\cdot a = a_1(a_2b_3-a_3b_2)-a_2(a_1b_3-a_3b_1)-a_3(a_1b_2-a_2b_1)=0$ and  $(a \times b)\cdot b-0$, so the cross product is orthogonal to both. As Nold mentioned, if the two vectors a and b lie in the x,y plane, then the orthogonal vectors must be purely in the z direction.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T19:40:39.277" />
  <row Id="54" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:41:05.063" Score="4" ViewCount="4" Body="&lt;h1&gt;Real numbers&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Real numbers are any numbers you can locate (even approximately) on an infinite number line.  This is a theoretical number line with infinite &quot;resolution&quot; that extends infinitely in both positive and negative directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/0/09/Number-line.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One neat property about real numbers is that they are orderable -- that is, given any two real numbers, you can tell which one is &quot;higher&quot; and which one is &quot;lower&quot; than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real numbers are closed under multiplication, addition, and subtraction.  That is, if you perform any of these operations on two real numbers, their result will always be real as well.  They are &lt;em&gt;almost&lt;/em&gt; closed under division, except for the whole divide-by-zero issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not real numbers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;infinity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the square root of -1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1/0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Decimals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There isn't a rigorous definition of &quot;decimals&quot;, because depending on where you use it, you'll get different definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the elementary sense, it means any number that has a &quot;decimal part&quot;; or a part after the radix (decimal point, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more advance sense, it means any number written in &lt;a href=&quot;http://en.wikipedia.org/wiki/Base_10&quot; rel=&quot;nofollow&quot;&gt;Base 10&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Natural numbers&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Natural numbers are often also called &quot;counting numbers&quot;, because they are the numbers you count with.  (0,) 1, 2, 3, 4, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some disagreement in the mathematics community over whether or not 0 is a natural number.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Cardinals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In linguistics, this means the natural &quot;numbers&quot; themselves (1, 2, 3, etc.)  But you probably don't want to know about linguistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Set Theory, two sets have the same &lt;strong&gt;cardinality&lt;/strong&gt; if each element could be paired up with an element of the other set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1,2,3} and {4,5,6} share the same cardinality because you can pair up 1&amp;amp;4, 2&amp;amp;5, 3&amp;amp;6.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Ordinals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In linguistics, this means 1st, 2nd, 3rd, etc.  But you probably don't want to know about linguistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Set Theory, an ordinal is a well-ordered set.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-29T23:06:54.147" LastActivityDate="2010-07-29T23:06:54.147" CommentCount="5" />
  <row Id="55" PostTypeId="2" ParentId="29" CreationDate="2010-07-20T19:41:44.267" Score="18" ViewCount="0" Body="&lt;p&gt;Essentially the most helpful is &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;, as Ami said, you can use your browser history here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WolframAlpha can carry out complex equations can comparisons much like a TI Calculator. Additionally they have some areas where you can see the simplification of an equation paired with charts and graphs where possible.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2010-07-20T19:41:44.267" CommunityOwnedDate="2010-07-28T01:32:30.370" CommentCount="2" />
  <row Id="56" PostTypeId="1" CreationDate="2010-07-20T19:41:55.687" Score="6" ViewCount="152" Body="&lt;p&gt;As I've understood it, what I've learned as the dot product is just one of many possible &quot;inner product spaces&quot;. Can someone explain this concept? When is it useful to define it as something other than the dot product?&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:47:25.767" LastActivityDate="2010-07-23T06:43:35.283" Title="What is an inner product space?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="57" PostTypeId="1" AcceptedAnswerId="134" CreationDate="2010-07-20T19:42:33.500" Score="3" ViewCount="147" Body="&lt;p&gt;This is simply not clicking for me. I'm currently learning math during the summer vacation and I'm on the chapter for relations and functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are five properties for a relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflexive - R -&gt; R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetrical - R -&gt; S ; S -&gt; R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antisymmetrical - R -&gt; S &amp;amp;&amp;amp; (R -&gt; R || S -&gt; S)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asymmetrical - R -&gt; S &amp;amp;&amp;amp; !(R -&gt; R || S -&gt; S)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transitive - if R -&gt; S &amp;amp;&amp;amp; S -&gt; T, then R -&gt; T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not what you call the properties in English, then please let me know- I have to study it in German, unfortunately, and these are my rough translations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing on, I just don't know what to do with this information practically. The examples of the book are horrible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;Is the same age as&quot; is apparently reflexive, symmetrical and transitive. &#xA;2) &quot;Is related to&quot; is also apparently reflexive, symmetrical and transitive.&#xA;3) &quot;Is older than&quot; is asymmetric, antisymetric and transitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more useless examples like this. I have no idea how it comes to these conclusions because we're talking about a literal statement. I was hoping perhaps for some real mathematical examples, but the book falls short on those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would greatly appreciate it if somebody could explain the above example and perhaps give me a better use for Relations other than... that. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? &lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-25T23:39:24.247" LastActivityDate="2010-07-28T21:12:19.983" Title="How do the Properties of Relations work?" Tags="&lt;set-theory&gt;&lt;logic&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="58" PostTypeId="1" AcceptedAnswerId="70" CreationDate="2010-07-20T19:43:20.457" Score="8" ViewCount="358" Body="&lt;p&gt;I recently discovered Benford's Law. I find it very fascinating. I'm wondering what are some of the real life uses of Benford's law. Specific examples would be great.&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:41:55.237" LastActivityDate="2010-12-28T01:14:13.617" Title="Real life usage of Benford's Law" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;statistics&gt;&lt;applications&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="4" />
  <row Id="59" PostTypeId="1" AcceptedAnswerId="65" CreationDate="2010-07-20T19:43:30.097" Score="2" ViewCount="129" Body="&lt;p&gt;Given two points, around an origin (0,0), in 2D space how would you calculate an angle from p1 to p2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would this change in 3D space?&lt;/p&gt;&#xA;" OwnerUserId="24" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:54:27.277" LastActivityDate="2010-07-28T09:43:00.053" Title="Calculating an Angle from 2 points in space" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="60" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T19:45:14.633" Score="17" ViewCount="0" Body="&lt;p&gt;Suppose this was not the case, i.e. &lt;code&gt;0.9999... != 1&lt;/code&gt;. Then &lt;code&gt;0.9999... &amp;lt; 1&lt;/code&gt; (I hope we agree on that). But between two distinct real numbers, there's always another one (say &lt;code&gt;x&lt;/code&gt;) in between, hence &lt;code&gt;0.9999... &amp;lt; x &amp;lt; 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decimal representation of &lt;code&gt;x&lt;/code&gt; must have a digit somewhere that is not &lt;code&gt;9&lt;/code&gt; (otherwise &lt;code&gt;x == 0.9999...&lt;/code&gt;). But that means it's actually smaller &amp;ndash; &lt;code&gt;x &amp;lt; 0.9999...&lt;/code&gt;, contradicting the definition of &lt;code&gt;x&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the assumption that there's a number between &lt;code&gt;0.9999...&lt;/code&gt; and &lt;code&gt;1&lt;/code&gt; is false, hence they're equal.&lt;/p&gt;&#xA;" OwnerUserId="3" LastActivityDate="2010-07-20T19:45:14.633" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="1" />
  <row Id="62" PostTypeId="2" ParentId="51" CreationDate="2010-07-20T19:46:35.563" Score="0" ViewCount="1" Body="&lt;p&gt;So, the area of the field is pi*r^2 and you want the cow to be able to graze an area equal to half of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to do is set up the equation (r[1] is the radius of the field, r[2] is the length of the rope desired):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(pi*r[1]^2) / 2 = pi*r[2]^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then simplify it down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r[1]^2 / 2 = r[2]^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then taking roots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r[2] = r[1] / sqrt(2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need a rope that is equal to the radius divided by the square root of 2, and the post can be no closer than this distance to the edge of the field.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:46:35.563" />
  <row Id="63" PostTypeId="2" ParentId="58" CreationDate="2010-07-20T19:46:38.047" Score="1" ViewCount="0" Body="&lt;p&gt;I've heard of it being as a rough check to see if accounting numbers were being made up&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2010-07-20T19:46:38.047" CommentCount="1" />
  <row Id="64" PostTypeId="2" ParentId="51" CreationDate="2010-07-20T19:47:08.217" Score="0" ViewCount="0" Body="&lt;p&gt;Let the total area of the field = $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know $A = \pi R^2$ where $R$ = the radius of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want the cow to be able to graze half the area, so we solve for a length of rope $r$ such that $\pi r^2 = A / 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives: $\pi r^2 = \pi R^2 / 2$, hence $r = R / \sqrt(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words, the length of the cow's rope should be the radius of the field divided by sqrt(2).&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:47:08.217" CommentCount="4" />
  <row Id="65" PostTypeId="2" ParentId="59" CreationDate="2010-07-20T19:47:26.950" Score="5" ViewCount="0" Body="&lt;p&gt;Assuming this is relative to the origin (as John pointed out): Given two position vectors p&lt;sub&gt;1&lt;/sub&gt; and p&lt;sub&gt;2&lt;/sub&gt; the dot product is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;p&lt;sub&gt;1&lt;/sub&gt; * p&lt;sub&gt;2&lt;/sub&gt; = |p&lt;sub&gt;1&lt;/sub&gt;| *&#xA;  |p&lt;sub&gt;2&lt;/sub&gt;| * cos(v)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solving for v, we get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;v = arccos(p&lt;sub&gt;1&lt;/sub&gt; * p&lt;sub&gt;2&lt;/sub&gt; / (|p&lt;sub&gt;1&lt;/sub&gt;| * |p&lt;sub&gt;2&lt;/sub&gt;|))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For 2d space this equals:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;v = arccos((x&lt;sub&gt;1&lt;/sub&gt; * x&lt;sub&gt;2&lt;/sub&gt; + y&lt;sub&gt;1&lt;/sub&gt; * y&lt;sub&gt;2&lt;/sub&gt;) / sqrt((x&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;) * (x&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;)))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And extended for 3d space:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;v = arccos((x&lt;sub&gt;1&lt;/sub&gt; * x&lt;sub&gt;2&lt;/sub&gt; + y&lt;sub&gt;1&lt;/sub&gt; * y&lt;sub&gt;2&lt;/sub&gt; + z&lt;sub&gt;1&lt;/sub&gt; * z&lt;sub&gt;2&lt;/sub&gt;) / sqrt((x&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + z&lt;sub&gt;1&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;) * (x&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + y&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt; + z&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;2&lt;/sup&gt;)))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditorDisplayName="" LastEditDate="2010-07-20T19:52:47.110" LastActivityDate="2010-07-20T19:52:47.110" />
  <row Id="66" PostTypeId="2" ParentId="57" CreationDate="2010-07-20T19:49:03.750" Score="2" ViewCount="0" Body="&lt;p&gt;There are several important kinds of relations, each of which satisfy a different collection of properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalence relations:&#xA;These are reflexive, symmetric, and transitive.  Essentially they're relations that &quot;behave like equality.&quot;  The most important elementary one is &quot;equivalence modulo m,&quot; where say 1 = 6 = 11 modulo 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial orderings:&#xA;These are reflexive, transitive, and (anti-symmetric or maybe asymmetric, I'm having trouble parsing your logical statements).  Essentially they're relations that &quot;behave like less than or equal to.&quot;  An important elementary example is &quot;divides&quot; where we say a|b if the ratio b/a is an integer.  Note that 2|6 but 6 does not divide 2.  However if 2|4 and 4|8 certainly 2|8.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T19:49:03.750" CommentCount="1" />
  <row Id="67" PostTypeId="1" AcceptedAnswerId="72" CreationDate="2010-07-20T19:49:29.490" Score="17" ViewCount="266" Body="&lt;p&gt;I hear a lot about Elliptic Curve Cryptography these days, but I'm still not quite sure what they are or how they relate to crypto...&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T06:16:23.970" LastActivityDate="2010-09-05T10:22:57.960" Title="What is an elliptic curve, and how are they used in cryptography?" Tags="&lt;cryptography&gt;&lt;elliptic-curves&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="68" PostTypeId="2" ParentId="59" CreationDate="2010-07-20T19:49:50.033" Score="2" ViewCount="0" Body="&lt;p&gt;I will assume that you mean the angle of the line from p1 to p2 &lt;strong&gt;with respect to the x-axis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the best I can do given the information you have provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the official mathsy way would be to find the dot product between the two, and divide by the magnitude of p1-p2 and take the arccossine.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = (normalized vector from p1 to p2)&#xA;theta = arccos( v * &amp;lt;1,0&amp;gt; )    (dot product)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can normalize a vector by dividing every term by the magnitude (length) of the entire vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3D, the same thing applies:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theta = arccos( v * &amp;lt;1,0,0&amp;gt; )   (dot product)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You could also possibly mean the angle between the line from the origin to p1 and the line from the origin to p2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this with dot products, as well; but both vectors must be normalized.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theta = arccos( a * b ) (dot product)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;a&lt;/code&gt; is the normalized vector from the origin to p1 and &lt;code&gt;b&lt;/code&gt; is the normalized vector from the origin to p2.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T19:49:50.033" />
  <row Id="69" PostTypeId="2" ParentId="1" CreationDate="2010-07-20T19:49:51.500" Score="3" ViewCount="0" Body="&lt;p&gt;You can see that there are infinitely many natural numbers 1, 2, 3, ..., and infinitely many real numbers, such as 0, 1, pi, etc. But are these two infinities the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, suppose you have two sets of objects, e.g. people and horses, and you want to know if the number of objects in one set is the same as in the other. The simplest way is to find a way of corresponding the objects one-to-one. For instance, if you see a parade of people riding horses, you will know that there are as many people as there are horses, because there is such a one-to-one correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that an set with infinitely many things is &quot;countable,&quot; if we can find a one-to-one correspondence between the things in this set and the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., the integers are countable: 1 &amp;lt;-&gt; 0, 2 &amp;lt;-&gt; -1, 3 &amp;lt;-&gt; 1, 4 &amp;lt;-&gt; -2, 5 &amp;lt;-&gt; 2, etc, gives such a correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the set of real numbers is NOT countable! This was proven for the first time by Georg Cantor. Here is a proof using the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;diagonal argument&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2010-07-20T19:49:51.500" />
  <row Id="70" PostTypeId="2" ParentId="58" CreationDate="2010-07-20T19:51:24.680" Score="9" ViewCount="0" Body="&lt;p&gt;Forensic accountancy is a popular use, and is actually admissible as evidence in the USA.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:51:24.680" />
  <row Id="71" PostTypeId="1" AcceptedAnswerId="78" CreationDate="2010-07-20T19:52:14.927" Score="7" ViewCount="246" Body="&lt;p&gt;I've recently started reading about Quaternions, and I keep reading that for example they're used in computer graphics and mechanics calculations to calculate movement and rotation, but without real explanations of the benefits of using them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering what exactly can be done with Quaternions that can't be done as easily (or easier) using more tradition approaches, such as with Vectors?&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:49:40.137" LastActivityDate="2010-07-24T18:49:40.137" Title="Real world uses of Quaternions?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;linear-algebra&gt;&lt;applications&gt;&lt;quaternions&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="72" PostTypeId="2" ParentId="67" CreationDate="2010-07-20T19:52:31.587" Score="14" ViewCount="1" Body="&lt;p&gt;Here is a super nice powerpoint on the subject!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T19:52:31.587" CommentCount="1" />
  <row Id="73" PostTypeId="2" ParentId="56" CreationDate="2010-07-20T19:54:30.663" Score="5" ViewCount="0" Body="&lt;p&gt;An inner product space is a vector space for which the inner product is defined. The inner product is also known as the 'dot product' for 2D or 3D Euclidean space. An arbitrary number of inner products can be defined according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_product_space#Definition&quot; rel=&quot;nofollow&quot;&gt;three rules&lt;/a&gt;, though most are a lot less intuitive/practical than the Euclidean (dot) product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may seem slightly esoteric, but as a physicist the obvious application of inner product spaces are Hilbert spaces used in quantum mechanics. The inner product of an eigenfunction with a wavefunction in Hilbert space gives the corresponding eigenvalue.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T19:54:30.663" />
  <row Id="74" PostTypeId="1" CreationDate="2010-07-20T19:54:33.893" Score="2" ViewCount="252" Body="&lt;p&gt;Let's say I know 'X' is a Gaussian Variable.&lt;br&gt;&#xA;Moreover, I know 'Y' is a Gaussian Variable and Y=X+Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's X and Z are Independent.&#xA;How can I prove Y is a Gaussian Random Variable if and only if Z is a Gaussian R.V.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to show the other way around (X, Z Orthogonal and Normal hence create a Gaussian Vector hence any Linear Combination of the two is a Gaussian Variable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditorDisplayName="" LastEditDate="2010-07-21T05:28:28.740" LastActivityDate="2010-07-30T22:04:55.433" Title="Sum of Gaussian Variables" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="75" PostTypeId="1" AcceptedAnswerId="80" CreationDate="2010-07-20T19:55:05.293" Score="3" ViewCount="271" Body="&lt;p&gt;Rings, groups, and fields all feel similar. What are the differences between them, both in definition and in how they are used?&lt;/p&gt;&#xA;" OwnerUserId="37" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:50:08.467" LastActivityDate="2010-07-31T17:50:08.467" Title="differences between rings, groups, and fields" Tags="&lt;terminology&gt;&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="76" PostTypeId="2" ParentId="57" CreationDate="2010-07-20T19:55:53.373" Score="1" ViewCount="0" Body="&lt;p&gt;Asymmetric means simply &quot;not symmetric&quot;. So in the binary case, it is NOT the case that if a is related to b, b is related to a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antisymmetric means that if a is related to b, and b is related to a, a = b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain your third example:&#xA;&quot;is older than&quot; is asymmetric because if Alice is older than Bob, Bob is NOT older than Alice. &quot;is older than&quot; is antisymmetric since if Alice is older than Bob, and Bob is older than Alice, Alice must be Bob because someone must be older (and if this is not the case, Alice simply has two names..). &quot;is older than&quot; is transitive since if Alice is older than Bob, and Bob is older than Charlie, Alice is also older than Charlie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So asymmetric and antisymmetric don't cancel out because the first means it's sort of a one-way relation, whereas the second means, loosely, that if it you reverse the operands and both statements are true, the operands must be the same.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T19:55:53.373" CommentCount="2" />
  <row Id="77" PostTypeId="1" AcceptedAnswerId="85" CreationDate="2010-07-20T19:57:06.617" Score="10" ViewCount="204" Body="&lt;p&gt;What purposes do the Dot and Cross products serve? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any clear examples of when you would use them?&lt;/p&gt;&#xA;" OwnerUserId="24" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-31T15:15:30.327" LastActivityDate="2010-07-31T15:51:53.060" Title="Understanding Dot and Cross Product" Tags="&lt;linear-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="78" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T19:57:51.043" Score="4" ViewCount="0" Body="&lt;p&gt;I believe they are used in quantum physics as well, because rotation with quaternions models Spinors extremely well (due to the lovely property that you need to rotate a point in quaternionic space around 2 full revolutions to get back to your 'original', which is exactly what happens with spin-1/2 particles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are also, as you said, used in computer graphics a lot for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;they are much more space efficient to store than rotation matrices (4 floats rather than 16)&lt;/li&gt;&#xA;&lt;li&gt;They are much easier to interpolate than euler angle rotations (spherical interpolation or normalised liner interpolation)&lt;/li&gt;&#xA;&lt;li&gt;They avoid gimbal lock&lt;/li&gt;&#xA;&lt;li&gt;It's much cooler to say that your rotation is described as 'a great circle on the surface of a unit 4 dimensional hypersphere' :)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think there are other uses, but a lot of them have been superseded by more general Vectors.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T19:57:51.043" CommentCount="5" />
  <row Id="79" PostTypeId="2" ParentId="20" CreationDate="2010-07-20T19:58:12.743" Score="8" ViewCount="0" Body="&lt;p&gt;The &lt;strong&gt;natural numbers&lt;/strong&gt; can be defined by &lt;a href=&quot;http://mathworld.wolfram.com/PeanosAxioms.html&quot; rel=&quot;nofollow&quot;&gt;Peano's Axioms&lt;/a&gt; (sometimes called the Peano Postulates):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Zero is a number.&lt;/li&gt;&#xA;&lt;li&gt;If n is a number, the successor of n is a number.&lt;/li&gt;&#xA;&lt;li&gt;zero is not the successor of a number.&lt;/li&gt;&#xA;&lt;li&gt;Two numbers of which the successors are equal are themselves equal.&lt;/li&gt;&#xA;&lt;li&gt;(induction axiom.) If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(This definition includes 0 in the natural numbers; altering rules 1, 3, and 5 to refer to one instead of zero excludes 0 from the natural numbers.  Whether or not 0 is a natural number varies in various texts.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;whole numbers&lt;/strong&gt; are the natural numbers with the additive identity element called 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;integers&lt;/strong&gt; are the whole numbers and their additive inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;rational numbers&lt;/strong&gt; are numbers that can be expressed as a ratio of an integer to a non-zero integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;real numbers&lt;/strong&gt; are the set of numbers that are limits of &lt;a href=&quot;http://mathworld.wolfram.com/CauchySequence.html&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequences&lt;/a&gt; of rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;irrational numbers&lt;/strong&gt; are the real numbers that are not rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;complex numbers&lt;/strong&gt; are the numbers that can be expressed as &lt;em&gt;a&lt;/em&gt; + &lt;em&gt;b&lt;/em&gt; * &lt;em&gt;i&lt;/em&gt; where &lt;em&gt;a&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt; are real numbers and &lt;em&gt;i&lt;/em&gt; behaves like a real number under addition/multiplication/distribution/etc., with the added rule that &lt;em&gt;i&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; = -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;imaginary numbers&lt;/strong&gt; are sometimes defined to be the &quot;pure imaginary&quot; numbers--complex numbers for which the &quot;real part&quot; &lt;em&gt;a&lt;/em&gt; = 0, sometimes with the added restriction that &lt;em&gt;b&lt;/em&gt; is not zero--and are sometimes defined to be the non-real complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;algebraic numbers&lt;/strong&gt; are numbers that are solutions to polynomial equations with integer coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;transcendental numbers&lt;/strong&gt; are complex numbers (sometimes limited to real numbers) that are not algebraic.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-24T09:13:19.853" LastActivityDate="2010-07-24T09:13:19.853" CommentCount="3" />
  <row Id="80" PostTypeId="2" ParentId="75" CreationDate="2010-07-20T19:58:48.127" Score="3" ViewCount="0" Body="&lt;p&gt;They should feel similar! In fact, every ring is a group, and every field is a ring. A ring is a group with an additional operation, where the second operation is associative and the distributive properties make the two operations &quot;compatible&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A field is a ring such that the second operation also satisfies all the group properties (after throwing out the additive identity); i.e. it has multiplicative inverses, multiplicative identity, and is commutative.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-20T20:10:50.810" LastActivityDate="2010-07-20T20:10:50.810" CommentCount="2" />
  <row Id="81" PostTypeId="1" CreationDate="2010-07-20T20:01:55.013" Score="12" ViewCount="545" Body="&lt;p&gt;I have the one or other interesting Math blog in my feedreader that I follow. It would be interesting to compile a list of Math blogs that are interesting to read, and do not require research-level math skills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start with my entries:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Division By Zero, &lt;a href=&quot;http://divisbyzero.com/&quot; rel=&quot;nofollow&quot;&gt;http://divisbyzero.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Tanya Khovanova’s Math Blog, &lt;a href=&quot;http://blog.tanyakhovanova.com/&quot; rel=&quot;nofollow&quot;&gt;http://blog.tanyakhovanova.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="79" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T16:26:58.500" LastActivityDate="2010-08-03T05:30:35.467" CommunityOwnedDate="2010-07-20T20:01:55.013" Title="List of Interesting Math Blogs" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="11" CommentCount="4" FavoriteCount="9" />
  <row Id="82" PostTypeId="2" ParentId="81" CreationDate="2010-07-20T20:04:19.563" Score="4" ViewCount="0" Body="&lt;p&gt;Gil Kalai's &lt;a href=&quot;http://gilkalai.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;blog&lt;/a&gt; is pretty awesome. Mostly covers combinatorics.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T20:04:19.563" CommunityOwnedDate="2010-07-20T20:04:19.563" />
  <row Id="83" PostTypeId="1" AcceptedAnswerId="88" CreationDate="2010-07-20T20:04:20.733" Score="26" ViewCount="664" Body="&lt;p&gt;I'm talking in the range of 10-12 years old, but this question isn't limited to only that range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any advice on cool things to show kids that might spark their interest in spending more time with math? The difficulty for some to learn math can be pretty overwhelming. Do you have any teaching techniques that you find valuable?&lt;/p&gt;&#xA;" OwnerUserId="24" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T10:06:42.220" LastActivityDate="2010-10-16T20:14:28.650" CommunityOwnedDate="2010-08-07T22:32:44.343" Title="What are some good ways to get children excited about math?" Tags="&lt;big-list&gt;&lt;teaching&gt;" AnswerCount="16" CommentCount="4" FavoriteCount="13" />
  <row Id="84" PostTypeId="2" ParentId="77" CreationDate="2010-07-20T20:05:03.633" Score="1" ViewCount="0" Body="&lt;p&gt;The dot product can be used to find the length of a vector or the angle between two vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product is used to find a vector which is perpendicular to the plane spanned by two vectors.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-07-20T20:05:03.633" />
  <row Id="85" PostTypeId="2" ParentId="77" CreationDate="2010-07-20T20:05:31.633" Score="16" ViewCount="0" Body="&lt;p&gt;When you deal with vectors, sometimes you say to yourself, &quot;Darn I wish there was a function that...&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;was zero when two vectors are perpendicular, letting me test perpendicularness.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;would let me find the angle between two vectors.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt; (actually gives the cosine of the angle between two normalized vectors)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;would let me 'project' one vector onto another, or give the length of one vector in the direction of another.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much force is actually helping the object move, when pushing at an angle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much a vector field is 'spreading out'.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dot Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could give me a vector that is perpendicular to two other vectors.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much torque a force was applying to a rotating system.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;could tell me how much this vector field is 'curling' up.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Product&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are actually a lot more uses, but the more I study vectors, the more and more I run into a situation where I need a function to do &lt;em&gt;exactly&lt;/em&gt; something, and then realize that the cross/dot products already did exactly what I needed!&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T20:05:31.633" />
  <row Id="86" PostTypeId="2" ParentId="56" CreationDate="2010-07-20T20:09:05.667" Score="15" ViewCount="0" Body="&lt;p&gt;As for the utility of inner product spaces: They're vector spaces where notions like the length of a vector and the angle between two vectors are available. In this way, they generalize R^n but preserve some of its additional structure that comes on top of it being a vector space. Familiar friends like Cauchy-Schwartz, the parallelogram rule, and orthogonality all work in inner product spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that there is a more general class of spaces, normed spaces, where notions of length make sense always, but an inner product cannot necessarily be defined.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dot product is the standard inner product on R^n. In general, any symmetric, positive definite matrix will give you an inner product on C^n. And you can have inner products on infinite dimensional vector spaces, like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \langle \, f, \, g \, \rangle = \int_a^b \ f(x)\overline{g(x)} \, dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for f, g square-integrable functions on [a,b].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes useful, for example, in applications like Fourier series where you want a basis of orthonormal functions for some function space (it's not just the trigonometric functions that work).&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-23T06:43:35.283" LastActivityDate="2010-07-23T06:43:35.283" CommentCount="4" />
  <row Id="87" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T20:10:04.057" Score="1" ViewCount="0" Body="&lt;pre&gt;&lt;code&gt;x = 0.999...&#xA;10x = 9.999...&#xA;10x - x = 9.999... - 0.999... = 9&#xA;-&amp;gt; 9x = 9&#xA;-&amp;gt; x = 1 thus, 0.999... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="84" LastActivityDate="2010-07-20T20:10:04.057" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="3" />
  <row Id="88" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T20:10:38.563" Score="14" ViewCount="0" Body="&lt;p&gt;Graph theory! It's essentially connecting the dots, but with theorems working wonders behind the scenes for when they're old enough. Simple exercises like asking how many colors you need to color the faces or vertices of a graph are often fun (so I hear). (Also, most people won't believe the 4-color theorem.)&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T20:10:38.563" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="89" PostTypeId="2" ParentId="22" CreationDate="2010-07-20T20:11:12.370" Score="2" ViewCount="0" Body="&lt;p&gt;Note that if you replace i, j, and k with m, n, and p, the determinant becomes the dot-product of the vector (m, n, p) with the cross-product of the two original vectors.  If (m, n, p) = (a, b, c) or (m, n, p) = (d, e, f), the determinant is zero (any matrix with two identical rows has determinant zero), so the dot product of (a, b, c) or (d, e, f) with the cross-product is zero.  Hence, (a, b, c) and (d, e, f) are orthogonal to their cross-product.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T20:11:12.370" />
  <row Id="90" PostTypeId="1" CreationDate="2010-07-20T20:11:37.703" Score="11" ViewCount="201" Body="&lt;p&gt;Not sure if this is the place for it, but there are similar posts for podcasts and blogs, so I'll post this one. I'd be interested in seeing a list of online resources for mathematics learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone doing a non-maths degree in college I'd be interested in finding some resources for learning more maths online, most resources I know of tend to either assume a working knowledge of maths beyond secondary school level, or only provide a brief summary of the topic at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start off by posting &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/&quot; rel=&quot;nofollow&quot;&gt;MIT Open Courseware&lt;/a&gt;, which is a large collection of lecture notes, assignments and multimedia for the MIT mathematics courses, although in many places it's quite incomplete.&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="167" LastEditorDisplayName="" LastEditDate="2010-07-21T10:47:30.240" LastActivityDate="2010-07-21T10:47:30.240" CommunityOwnedDate="2010-07-20T20:11:37.703" Title="Online resources for learning Mathematics." Tags="&lt;books&gt;&lt;reference-request&gt;" AnswerCount="7" FavoriteCount="8" />
  <row Id="91" PostTypeId="2" ParentId="75" CreationDate="2010-07-20T20:13:11.210" Score="4" ViewCount="0" Body="&lt;p&gt;You're right to think that the definitions are very similar. The main difference between groups and rings is that rings have two binary operations (usually called addition and multiplication) instead of just one binary operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you forget about multiplication, then a ring becomes a group with respect to addition (the identity is 0 and inverses are negatives). This group is always commutative!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you forget about addition, then a ring does not become a group with respect to multiplication. The binary operation of multiplication is associative and it does have an identity 1, but some elements like 0 do not have inverses. (This structure is called a monoid.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commutative ring is a field when all nonzero elements have a multiplicative inverses. In this case, if you forget about addition and remove 0, the remaining elements do form a group under multiplication. This group is again commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A division ring is a (not necessarily commutative) ring in which all nonzero elements have multiplicative inverses. Again, if you forget about addition and remove 0, the remaining elements do form a group under multiplication. This group is not necessarily commutative. An example of a division ring which is not a field are the quaternions.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-20T20:13:11.210" />
  <row Id="92" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T20:15:09.723" Score="0" ViewCount="0" Body="&lt;p&gt;Two good general references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Portal%3aMathematics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathworld.wolfram.com/&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="37" LastActivityDate="2010-07-20T20:15:09.723" CommunityOwnedDate="2010-07-20T20:15:09.723" />
  <row Id="93" PostTypeId="2" ParentId="18" CreationDate="2010-07-20T20:16:01.453" Score="1" ViewCount="0" Body="&lt;p&gt;Possibly something like this. Correct me if I'm wrong. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;j = semi-major&lt;br /&gt;&#xA;n = semi-minor&lt;br /&gt;&#xA;e = eccentricity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = sqrt((j * sqrt(1 - e^2)) * (j * (1 - e^2)))&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-07-20T20:16:01.453" />
  <row Id="94" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T20:16:09.050" Score="3" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Kahn Academy, &lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You'll find tons of explanatory videos from various branches of mathematics; plus, each subject is explained pretty good, and the videos are easy to follow&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="79" LastEditorUserId="79" LastEditorDisplayName="" LastEditDate="2010-07-20T20:21:11.893" LastActivityDate="2010-07-20T20:21:11.893" CommunityOwnedDate="2010-07-20T20:16:09.050" CommentCount="2" />
  <row Id="95" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T20:17:42.790" Score="2" ViewCount="1" Body="&lt;p&gt;A useful one for undergraduate level maths is &lt;a href=&quot;http://www.mathscentre.ac.uk/&quot; rel=&quot;nofollow&quot;&gt;Mathcentre&lt;/a&gt;.  It has useful background material for people studying maths, or who need some maths background for other courses.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2010-07-20T20:17:42.790" CommunityOwnedDate="2010-07-20T20:17:42.790" />
  <row Id="96" PostTypeId="2" ParentId="81" CreationDate="2010-07-20T20:18:49.590" Score="7" ViewCount="0" Body="&lt;p&gt;Not always pure math, but I think John Baez' &lt;a href=&quot;http://math.ucr.edu/home/baez/this.week.html&quot; rel=&quot;nofollow&quot;&gt;This Week in Mathematical Physics&lt;/a&gt; contains a lot of really interesting math reads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;What's new&lt;/a&gt;. It's a very active math blog (both in posts and comments) and definitely covers some cutting edge math, even if it can be way over my head.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-07-21T15:37:45.827" LastActivityDate="2010-07-21T15:37:45.827" CommunityOwnedDate="2010-07-20T20:18:49.590" />
  <row Id="97" PostTypeId="1" AcceptedAnswerId="1224" CreationDate="2010-07-20T20:20:22.223" Score="2" ViewCount="228" Body="&lt;p&gt;I am looking for an accurate algorithm to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Error_function&quot; rel=&quot;nofollow&quot;&gt;error function&lt;/a&gt;. I have tried using &lt;a href=&quot;http://stackoverflow.com/questions/457408/is-there-an-easily-available-implementation-of-erf-for-python/457805#457805&quot;&gt;this formula&lt;/a&gt; (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486612724&quot; rel=&quot;nofollow&quot;&gt;Handbook of Mathematical Functions&lt;/a&gt;, formula 7.1.26), but the results are not accurate enough for the application.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T23:56:18.930" LastActivityDate="2010-08-27T23:56:18.930" Title="How to accurately calculate erf(x) with a computer?" Tags="&lt;algorithms&gt;&lt;statistics&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="98" PostTypeId="1" AcceptedAnswerId="106" CreationDate="2010-07-20T20:21:33.407" Score="7" ViewCount="117" Body="&lt;p&gt;I am looking for a reference (book or article) that poses a problem that seems to be a classic, in that I've heard it posed many times, but that I've never seen written anywhere: that of the possibility of a man in a circular pen with a lion, each with some maximum speed, avoiding capture by that lion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References to these sprts of pursuit problems in general would also be appreciated, and the original source of this problem.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-10-14T08:01:40.487" Title="Chased By a Lion and other Pursuit Problems" Tags="&lt;reference-request&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="99" PostTypeId="2" ParentId="81" CreationDate="2010-07-20T20:27:41.723" Score="1" ViewCount="0" Body="&lt;p&gt;Not a pure math blog, but it's one of the most fascinating blogs in my RSS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futility Closet&#xA;&lt;a href=&quot;http://www.futilitycloset.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.futilitycloset.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5" LastActivityDate="2010-07-20T20:27:41.723" CommunityOwnedDate="2010-07-20T20:27:41.723" />
  <row Id="100" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T20:29:10.767" Score="4" ViewCount="0" Body="&lt;p&gt;If they're fairly mathematically inclined anyway, then try to get them solving interesting problems with an obvious mathematical content, if they're less mathematically inclined try to find problems where the usage of maths isn't as explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems with a very mathematical bent can be found at places like &lt;a href=&quot;http://nrich.maths.org/public/&quot; rel=&quot;nofollow&quot;&gt;NRich&lt;/a&gt;, they update their problems monthly, the Stage 2 and 3 problems cover that age range. Other sources for problems could be video games (resource management based games require mathematical thinking), code breaking or programming a game (a simple driving game in Flash requires a lot of maths based problem solving).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they've convinced themselves that they can't do Maths then you basically need to smuggle the maths into your chosen activity without them realising that they're doing maths to solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2010-07-20T20:29:10.767" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="101" PostTypeId="1" AcceptedAnswerId="105" CreationDate="2010-07-20T20:30:39.280" Score="2" ViewCount="77" Body="&lt;p&gt;This is in relation to the Euler Problem 13 from www.ProjectEuler.net.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;37107287533902102798797998220837590246510135740250&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this was my thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can freely discard the last fourty digits and leave the last ten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0135740250&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then simply sum those. This would be large enough to be stored in a 64-bit data-type and a lot easier to compute. However, my answer isn't being accepted, so I'm forced to question my logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't see a problem. The last fourty digits will never make a difference because they are at least a magnitude of 10 larger than the preceding values and therefore never carry backwards into smaller areas. Is this not correct?&lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:42:42.350" LastActivityDate="2010-07-24T18:42:42.350" Title="Faulty logic when summing large integers?" Tags="&lt;arithmetic&gt;&lt;project-euler&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="102" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T20:34:32.283" Score="4" ViewCount="1" Body="&lt;p&gt;This really depend on how smart the kid is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I lean toward discrete math, elementary number theory related topics when talking to non-math people about math. They requires little background knowledge. There are some fun problems in discrete math, especially combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple probability is also nice. So are logic problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both topics can be used to formulate some simple puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple number theory puzzle&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;How many zeros are there in 20!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I assume a bright 10 year old can solve it.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T20:34:32.283" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="2" />
  <row Id="103" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T20:37:52.477" Score="4" ViewCount="0" Body="&lt;p&gt;One place they are frequently used is in computer games when you want to smoothly transition from one rotation to another. An artist might have said &quot;at this time I want the head oriented like this and at that time I want it like this&quot;. The computer needs to work out what happens in-between these poses. It's quite easy to find in-between poses using quaternions. If the two poses are reasonably similar, then you can get a half-way orientation simply by taking the average of the quaternions. You can find out more &lt;a href=&quot;http://en.wikipedia.org/wiki/Slerp&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-20T20:37:52.477" />
  <row Id="104" PostTypeId="2" ParentId="101" CreationDate="2010-07-20T20:37:56.830" Score="2" ViewCount="0" Body="&lt;p&gt;First you are doing it in the wrong end, second, the statement in general is still not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9999999999&#xA;1000000001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Say if you want the first 2 digits,&#xA;you will get 10 if you discard the last 2 digit and do the sum.&#xA;The right answer is 11&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T20:37:56.830" />
  <row Id="105" PostTypeId="2" ParentId="101" CreationDate="2010-07-20T20:38:04.910" Score="5" ViewCount="0" Body="&lt;p&gt;If you were supposed to find the &lt;em&gt;last&lt;/em&gt; ten digits, you could just ignore the first 40 digits of each number. However you're supposed to find the &lt;em&gt;first&lt;/em&gt; ten digits, so that doesn't work. And you can't just ignore the last digits of each number either because those &lt;em&gt;can&lt;/em&gt; carry over.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-07-20T20:38:04.910" />
  <row Id="106" PostTypeId="2" ParentId="98" CreationDate="2010-07-20T20:38:30.527" Score="3" ViewCount="0" Body="&lt;p&gt;Since you're asking for a reference, perhaps this will do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Mathworld says the problem was listed by Rado in 1925. The reference is on the problem description page, &lt;a href=&quot;http://mathworld.wolfram.com/LionandManProblem.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-20T20:38:30.527" />
  <row Id="107" PostTypeId="1" AcceptedAnswerId="1159" CreationDate="2010-07-20T20:38:57.700" Score="7" ViewCount="622" Body="&lt;p&gt;It is 10 o'clock, and I have a box. Inside the box is a ball marked 1.&lt;br&gt;&#xA;At 10:30, I will remove the ball marked 1, and add two balls, labeled 2 and 3. At 10:45, I will remove the balls labeled 2 and 3, and add 4 balls, marked 4, 5, 6, and 7. 7.5 minutes before 11, I will remove the balls labeled 4, 5, and 6, and add 8 balls, labeled 8, 9, 10, 11, 12, 13, 14, and 15. This pattern continues. Each time I reach the halfway point between my previous action and 11 o'clock, I add some balls, and remove some other balls. Each time I remove one more ball than I removed last time, but add twice as many balls as I added last time.&lt;br&gt;&#xA;The result is that as it gets closer and closer to 11, the number of balls in the box continues to increase. Yet every ball that I put in was eventually removed. So just how many balls will be in the box when the clock strikes 11? 0, or infinitely many? What's going on here? &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T14:01:59.213" LastActivityDate="2010-11-27T14:01:59.213" Title="paradox: increasing sequence that goes to 0?" Tags="&lt;soft-question&gt;&lt;paradox&gt;" AnswerCount="8" CommentCount="5" />
  <row Id="108" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T20:41:27.650" Score="1" ViewCount="0" Body="&lt;p&gt;You can view a real-world example of quaternions in computer graphics with the open source program known as NASA WorldWind (http://worldwind.arc.nasa.gov/java/).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses a Quaternion object to represent rotation of various geometries. The class definition itself is located in the src/gov/nasa/worldwind/geom/Quaternion.java file.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-07-20T20:41:27.650" />
  <row Id="109" PostTypeId="2" ParentId="51" CreationDate="2010-07-20T20:48:24.463" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/797940734.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field is the smaller/left circle, centered at A.  The cow is tied to the post at E.  The larger/right circle is the grazing radius.  Let the radius of the field be R and the length of the rope be L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grazable area is the union of a segment of the circular field and a segment of the circle defined by the rope length.  (A segment of a circle is a sector of a circle less the triangle defined by the center of the circle and the endpoints of the arc.)  The area of a segment of a circle of radius r with central angle &amp;theta; is $\frac{1}{2}R^2(\theta-\sin\theta)$, where &amp;theta; is measured in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to express the grazable area in terms of L and R, we need expressions for the measures of &amp;ang;CED and &amp;ang;CAD (which define the segments of the circles; call these &amp;alpha; and &amp;beta; for convenience).  By using the Law of Cosines on the blue triangle in the diagram, $\alpha=\mathrm{m}\angle CED=2\arccos\left(\frac{L}{2R}\right)$ and $\beta=\mathrm{m}\angle CAD=2\arccos\left(\frac{2R^2-L^2}{2R^2}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grazable area is $\frac{1}{2}L^2(\alpha-\sin\alpha)+\frac{1}{2}R^2(\beta-\sin\beta)$.  We want that to be equal to half the area of the field, $\frac{1}{2}\pi R^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the relationship between L and R is $$\begin{align}&amp;amp;\frac{L^2}{2}\left(2\arccos\left(\frac{L}{2R}\right)-\sin\left(2\arccos\left(\frac{L}{2R}\right)\right)\right) \\ &amp;amp;\quad\quad\quad+\frac{R^2}{2}\left(2\arccos\left(\frac{2R^2-L^2}{2R^2}\right)-\sin\left(2\arccos\left(\frac{2R^2-L^2}{2R^2}\right)\right)\right) \\ &amp;amp;=\frac{\pi R^2}{2}.\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no algebraic method of solving this equation for L in terms of R (Mathematica cannot &lt;code&gt;Solve[]&lt;/code&gt; or &lt;code&gt;Reduce[]&lt;/code&gt; it).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:02:22.920" LastActivityDate="2010-08-05T19:02:22.920" />
  <row Id="110" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T20:48:39.627" Score="-2" ViewCount="0" Body="&lt;p&gt;The argument against this is that 0.99999999... is &quot;somewhat&quot; less than 1. How much exactly?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      1 - 0.999999... = ε              (0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the above is true, the following also must be true:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 × (1 - 0.999999...) = ε × 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's calculate:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.999... ×&#xA;9        =&#xA;───────────&#xA;8.1&#xA;  81&#xA;   81&#xA;     .&#xA;      .&#xA;       .&#xA;&#xA;───────────&#xA;8.999...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     9 - 8.999999... = 9ε              (1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         8.999999... = 8 + 0.99999...  (2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Indeed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8.00000000... +&#xA;0.99999999... =&#xA;────────────────&#xA;8.99999999...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now let's see what we can deduce from &lt;code&gt;(0)&lt;/code&gt;, &lt;code&gt;(1)&lt;/code&gt; and &lt;code&gt;(2)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9 - 8.999999... = 9ε                      because of (2)&#xA;9 - 8.999999... = 9 - (8 + 0.99999...) =  because of (1)&#xA;                = 9 -  8 - (1 - ε)        because of (0)&#xA;                =   1    -  1 + ε         &#xA;                =               ε.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9ε = ε&#xA;&#xA;10ε = 0&#xA;&#xA;ε = 0&#xA;&#xA;1 - 0.999999... = ε = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Quod erat dimostrandum. Pardon my unicode.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-07-20T21:12:35.550" LastActivityDate="2010-07-20T21:12:35.550" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="4" />
  <row Id="111" PostTypeId="2" ParentId="107" CreationDate="2010-07-20T20:50:35.817" Score="0" ViewCount="0" Body="&lt;p&gt;This is a problem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeno%27s_paradoxes&quot; rel=&quot;nofollow&quot;&gt;Zeno's Paradox&lt;/a&gt;. If you continue to subdivide the time ad infinitum, you'll never reach 11 o'clock so the answer is undefined.&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2010-07-20T20:50:35.817" CommentCount="2" />
  <row Id="112" PostTypeId="1" AcceptedAnswerId="115" CreationDate="2010-07-20T20:51:07.517" Score="4" ViewCount="65" Body="&lt;p&gt;I keep seeing this symbol ∇ around and I know enough to understand that it represents the term &quot;gradient.&quot; But what is a gradient? When would I want to use one mathematically?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:44:02.677" LastActivityDate="2010-07-24T18:44:02.677" Title="What are gradients and how would I use them?" Tags="&lt;calculus&gt;&lt;terminology&gt;&lt;tag-removed&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="113" PostTypeId="1" CreationDate="2010-07-20T20:56:33.853" Score="4" ViewCount="133" Body="&lt;p&gt;If you know it, also try to include the precise reason why the proof is fallacious. To start this off, let me post the one that most people know already:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; Let a = b.&#xA;&lt;li&gt; Then a^2 = ab&#xA;&lt;li&gt; a^2 - b^2 = ab - b^2&#xA;&lt;li&gt; Factor to (a-b)(a+b) = b(a-b)&#xA;&lt;li&gt; Then divide out (a-b) to get a+b = b&#xA;&lt;li&gt; Since a = b, then b+b = b&#xA;&lt;li&gt; Therefore 2b = b&#xA;&lt;li&gt; Reduce to 2 = 1&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As @jan-gorzny pointed out, in this case, line 5 is wrong since a = b implies a-b = 0, and so you can't divide out (a-b).&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T05:09:26.823" LastActivityDate="2010-07-21T05:09:26.823" CommunityOwnedDate="2010-07-20T21:00:04.093" Title="What are some classic fallacious proofs?" Tags="&lt;big-list&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="114" PostTypeId="2" ParentId="107" CreationDate="2010-07-20T20:57:00.700" Score="2" ViewCount="0" Body="&lt;p&gt;During one math class I had the same problem. the professor told us the answer can be a arbitrary integer, depend on how you remove the balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you said every ball you put in was eventually removed, I will make it simple and assume you first number the balls to natural numbers, and then add and remove the balls with the smallest numbers first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the bin will be empty--every ball put in have been removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define the sequence in the way such that a_n = amount of balls when you do the nth move, then this sequence never get to zero, because there is no n correspond to 11 o'clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increasing sequence doesn't define how many balls in the bin at 11. Therefore there is no paradox.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T20:57:00.700" CommentCount="1" />
  <row Id="115" PostTypeId="2" ParentId="112" CreationDate="2010-07-20T21:02:32.183" Score="7" ViewCount="0" Body="&lt;p&gt;The ∇ (pronounced &quot;del&quot;) is an operator, more technically.  In 3D, it (more or less) means the vector&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt; df/dx, df/dy, df/dz &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, if &lt;code&gt;f(x,y,z) = x^2 + y^3*z + sin(z)&lt;/code&gt;, &lt;code&gt;∇f = &amp;lt; 2x, 3y^2*z, y^3 + cos(z) &amp;gt;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's actually a bit more subtle than that; technically it means&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt; d/dx, d/dy, d/dz &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And when you do ∇f, it's sort of like a &quot;multiplication&quot; of ∇ and f;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt; d/dx, d/dy, d/dz &amp;gt; f = &amp;lt; d/dx f, d/dy f, d/dz f &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Only, not multiplication, but &lt;em&gt;operation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some neat properties about the del operator.  Here are a couple:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The most famous is that &lt;code&gt;∇f&lt;/code&gt; yields the &lt;strong&gt;gradient&lt;/strong&gt; of &lt;code&gt;f&lt;/code&gt;.  That is, at any point &lt;code&gt;(x,y,z)&lt;/code&gt;, &lt;code&gt;∇f(x,y,z)&lt;/code&gt; is the vector pointing in the direction where it is &lt;strong&gt;most increasing&lt;/strong&gt;.  The magnitude of it is the magnitude of the increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easier to understand with, say, a 2D &lt;code&gt;f(x,y)&lt;/code&gt;.  If &lt;code&gt;f(x,y)&lt;/code&gt; represents the height of a point at &lt;code&gt;(x,y)&lt;/code&gt;, then &lt;code&gt;∇f(x,y)&lt;/code&gt; represents the steepest incline from that point.  Or rather, if you placed a ball on that point, it would &lt;em&gt;start rolling in the opposite direction of the gradient vector&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Normally, for multi-dimensional functions, it is easiest to find the derivative along an axis (x, y, z, etc.).  With ∇, you can find the derivative along any arbitrary direction by using ∇f * u, where * is the dot product and u is the unit vector along the direction you are calculating.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;∇ is also used to calculate divergence (amount that vectors are &quot;spreading out&quot;) and curl (amount that vectors are &quot;curling up&quot;) of a vector field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divergence is ∇ * f (dot product), and curl is ∇ x f (cross product)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They aren't truly &quot;products&quot; in the sense.  Rather, when you are calculating divergence and curl and you must do &lt;code&gt;d/dx * (something)&lt;/code&gt;, you are actually doing &lt;code&gt;d/dx (something)&lt;/code&gt; or &lt;code&gt;d(something)/dx&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-20T21:24:14.510" LastActivityDate="2010-07-20T21:24:14.510" />
  <row Id="116" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T21:02:52.433" Score="6" ViewCount="0" Body="&lt;p&gt;Assuming:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;infinite decimals are series where the terms are the digits divided by the proper power of the base&lt;/li&gt;&#xA;&lt;li&gt;the infinite geometric series a + a * r + a * r^2 + a * r^3 + ... has sum a/(1 - r) as long as |r|&amp;lt;1&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;0.99999... = 9/10 + 9/10^2 + 9/10^3 + ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the infinite geometric series with first term a = 9/10 and common ratio r = 1/10, so it has sum (9/10) / (1 - 1/10) = (9/10) / (9/10) = 1.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T21:02:52.433" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="1" />
  <row Id="117" PostTypeId="2" ParentId="98" CreationDate="2010-07-20T21:06:00.833" Score="4" ViewCount="0" Body="&lt;p&gt;Here is a book on this type of problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://press.princeton.edu/titles/8371.html&quot; rel=&quot;nofollow&quot;&gt;http://press.princeton.edu/titles/8371.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is also briefly mentioned in his other book &quot;Euler's Fabulous Formula&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T21:06:00.833" CommentCount="1" />
  <row Id="118" PostTypeId="1" AcceptedAnswerId="125" CreationDate="2010-07-20T21:06:33.610" Score="7" ViewCount="357" Body="&lt;p&gt;I keep hearing about this weird type of math called calculus. I only have experience with geometry and algebra. Can you try to explain what it is to me?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T03:05:43.040" LastActivityDate="2010-07-28T01:50:04.913" Title="How would you describe calculus in simple terms?" Tags="&lt;soft-question&gt;&lt;calculus&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="2" />
  <row Id="119" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T21:08:28.333" Score="4" ViewCount="0" Body="&lt;p&gt;The following reddit post has a decent list of math resources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.reddit.com/r/math/comments/bqbex/lets_list_all_the_useful_free_online_math/&quot; rel=&quot;nofollow&quot;&gt;http://www.reddit.com/r/math/comments/bqbex/lets_list_all_the_useful_free_online_math/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One site I did not see it their list that I've found very helpful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://betterexplained.com/&quot; rel=&quot;nofollow&quot;&gt;http://betterexplained.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="105" LastActivityDate="2010-07-20T21:08:28.333" CommunityOwnedDate="2010-07-20T21:08:28.333" />
  <row Id="120" PostTypeId="1" AcceptedAnswerId="122" CreationDate="2010-07-20T21:08:29.207" Score="14" ViewCount="631" Body="&lt;p&gt;Is 1 classified as a prime number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so, why? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:37:46.723" LastActivityDate="2010-12-07T05:25:50.567" Title="Is 1 a prime number?" Tags="&lt;number-theory&gt;&lt;terminology&gt;" AnswerCount="6" FavoriteCount="2" />
  <row Id="122" PostTypeId="2" ParentId="120" CreationDate="2010-07-20T21:11:24.393" Score="22" ViewCount="0" Body="&lt;p&gt;One of the whole &quot;points&quot; of defining primes is to be able to uniquely and finitely prime factorize every natural number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1 was prime, then this would be more or less impossible.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-25T21:59:33.263" LastActivityDate="2010-07-25T21:59:33.263" CommentCount="11" />
  <row Id="123" PostTypeId="1" CreationDate="2010-07-20T21:14:07.827" Score="7" ViewCount="712" Body="&lt;p&gt;I covered hyperbolic trigonometric functions in a recent maths course. However I was never presented with any reasons as to &lt;em&gt;why&lt;/em&gt; (or even if) they are useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good examples of their uses outside academia?&lt;/p&gt;&#xA;" OwnerUserId="108" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:52:17.790" LastActivityDate="2010-12-29T11:10:41.143" Title="Real world uses of hyperbolic trigonometric functions" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;applications&gt;&lt;trigonometry&gt;" AnswerCount="9" CommentCount="4" FavoriteCount="2" />
  <row Id="124" PostTypeId="2" ParentId="123" CreationDate="2010-07-20T21:18:58.203" Score="17" ViewCount="0" Body="&lt;p&gt;If you take a rope, fix the two ends, and let it hang under the force of gravity, it will naturally form a hyperbolic cosine curve.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-20T21:18:58.203" CommentCount="2" />
  <row Id="125" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:19:03.680" Score="10" ViewCount="0" Body="&lt;p&gt;There came a time in mathematics when people encountered situations where they had to deal with really, really, really small things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not just small like 0.01; but small as in &lt;em&gt;infinitesimally small&lt;/em&gt;.  Think of &quot;the smallest positive number that is still greater than zero&quot; and you'll realize what sort of problems mathematicians began encountering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon, this problem became more than just theoretical or abstract.  It became very, very real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, velocity.  We know that average velocity is the change in position per change in time (i.e., 5 miles per hour).  But what about velocity &lt;strong&gt;at a point in time&lt;/strong&gt;?  What does it mean to be going 5 mph &lt;strong&gt;at this moment&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution someone came up with was to say &quot;it's the change in position divided by the change in time, where the change in time is an infinitesimally small amount of time&quot;.  But how would you handle/calculate that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem came about trying to find the area under a curve.  The current accepted solution was to divide the curve into rectangles, and add together the area of the rectangles.  However, in order to find the &lt;strong&gt;exact&lt;/strong&gt; area under the curve, you'd need to divide it into rectangles that were &lt;strong&gt;infinitesimally tiny&lt;/strong&gt;, and, therefore, add up an infinite amount of tiny rectangles -- to something that was finite (area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus came about as the system of math dedicated to studying these infinitesimally small changes.  In fact, I do believe some people describe calculus as &quot;the study of continuous changes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T21:19:03.680" CommentCount="2" />
  <row Id="126" PostTypeId="2" ParentId="123" CreationDate="2010-07-20T21:19:10.890" Score="5" ViewCount="0" Body="&lt;p&gt;An equation for a catenary curve can be given in terms of hyperbolic cosine.  Catenary curves appear in many places, such as the Gateway Arch in St. Louis, MO.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T21:19:10.890" />
  <row Id="127" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:19:52.887" Score="3" ViewCount="0" Body="&lt;p&gt;Calculus is a field which deals with two seemingly unrelated things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) the area beneath a graph and the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) the slope (or gradient) of a curve at different points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (1) is also called 'integration' and 'anti-differentiation', and part (2) is called 'differentiation'.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-20T21:19:52.887" CommentCount="4" />
  <row Id="128" PostTypeId="1" AcceptedAnswerId="1163" CreationDate="2010-07-20T21:20:16.380" Score="6" ViewCount="218" Body="&lt;p&gt;The Weyl equidistribution theorem states that the sequence of fractional parts ${n \xi}$, $n = 0, 1, 2, \dots$ is uniformly distributed for $\xi$ irrational. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved using a bit of ergodic theory, specifically the fact that an irrational rotation is uniquely ergodic with respect to Lebesgue measure. It can also be proved by simply playing with trigonometric polynomials (i.e., polynomials in $e^{2\pi i k x}$ for $k$ an integer) and using the fact they are dense in the space of all continuous functions with period 1.  In particular, one shows that if $f(x)$ is a continuous function with period 1, then for any $t$,  $\int_0^1 f(x) dx = \lim \frac{1}{N} \sum_{i=0}^{N-1} f(t+i \xi)$. One shows this by checking this (directly) for trigonometric polynomials via the geometric series.  This is a very elementary and nice proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general form of Weyl's theorem states that if $p$ is a monic integer-valued polynomial, then the sequence ${p(n \xi)}$ for $\xi$ irrational is uniformly distributed modulo 1.  I believe this can be proved using extensions of these ergodic theory techniques -- it's an exercise in Katok and Hasselblatt.  I'd like to see an elementary proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the general form of Weyl's theorem be proved using the same elementary techniques as in the basic version?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:35:36.433" LastActivityDate="2010-07-29T16:10:13.567" Title="How do you prove that $p(n \xi)$ for $\xi$ irrational and $p$ a polynomial is uniformly distributed modulo 1?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="129" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:20:59.237" Score="2" ViewCount="0" Body="&lt;p&gt;To be very brief and succinct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculus is the study of how quantities change&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Slightly more technically, it a subject based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinitesimal&quot; rel=&quot;nofollow&quot;&gt;infinitesimals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be pointing out the obvious, but the &lt;a href=&quot;http://en.wikipedia.org/wiki/Calculus&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; does actually provide a pretty decent beginners introduction to the subject. You'll generally want to start with &lt;em&gt;differential calculus&lt;/em&gt; and move on quickly to &lt;em&gt;integral calculus&lt;/em&gt;, followed by linking up the two (fundamental theorem of calculus) and moving on from there.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-20T21:20:59.237" />
  <row Id="130" PostTypeId="2" ParentId="113" CreationDate="2010-07-20T21:25:30.033" Score="9" ViewCount="0" Body="&lt;p&gt;Wikipedia has a long list of these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_fallacy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathematical_fallacy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-20T21:25:30.033" CommunityOwnedDate="2010-07-20T21:25:30.033" />
  <row Id="132" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T21:31:08.557" Score="12" ViewCount="0" Body="&lt;p&gt;I recently taught a once-a-week geometry class to grades 7-10 where we did some graph theory, some surfaces, some spherical geometry, a lot of complex numbers, and basic ideas of homeomorphismm and homotopy. Most couldn't work with complex numbers at the beginning: showing their use in proofs of analytic geometry results had the benefit of not boring the kids who had already seen and worked with complex numbers, and giving the kids who hadn't a reason to learn the technicalities (getting under the hood of these pretty picures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler's formula and the relation of planar graphs to polyhedra showed them the basics of a connection between geometric and algebraic intuition, outside the coordinate geometry many were used to, and got them thinking about ways to define familiar things mathematically (space), and what we can learn about familiar things through mathematics (surprising things like orientability through the Mobius strip, or sphere eversion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general I've found geometry to be a very good place to start with people with a professed fear or disinterest in mathematics. For me, it's the quickest way to show the distance between what mathematicians play with when they're doing math and what was taught in high school, with the two column proof nonsense and the treatment of math as a branch of formal logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I suspect that I find this path the easiest because it is one of the ones I am most excited about in mathematics, and I further imagine that in many cases, the best thing to use to get people excited about math is something you yourself are really excited about, as long as you can think to translate it well.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-20T21:31:08.557" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="133" PostTypeId="2" ParentId="3" CreationDate="2010-07-20T21:31:44.030" Score="1" ViewCount="0" Body="&lt;p&gt;I listen to &lt;a href=&quot;http://www.aracnet.com/~eseligma/mm/&quot; rel=&quot;nofollow&quot;&gt;Math Mutation Podcast&lt;/a&gt;. The topics are interesting and understandable by a layman.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T21:31:44.030" CommunityOwnedDate="2010-07-20T21:31:44.030" />
  <row Id="134" PostTypeId="2" ParentId="57" CreationDate="2010-07-20T21:34:01.537" Score="5" ViewCount="0" Body="&lt;p&gt;I'd like to change the notation of your definitions, since R, S and T would usually be used to stand for the relations themselves (and x, y and z would be more commonly chosen for the objects that might bear the relation to each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reflexive&lt;/em&gt; - For all x: xRx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example reflexive relation: xRy stands for 'x is a factor of y' (in the set of natural numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Symmetric&lt;/em&gt; - For all x,y: if xRy then yRx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example symmetric relation: xRy stands for 'x and y are 2 metres apart' (in the set of all people in a particular room)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Antisymmetric&lt;/em&gt; - For all x,y: if xRy and yRx then x = y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example antisymmetric relation: xRy stands for 'x is a factor of y' (in the set of natural numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Asymmetric&lt;/em&gt; - For all x,y: if xRy then not yRx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example asymmetric relation: xRy stands for 'x is taller than y' (in the set of all people)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Transitive&lt;/em&gt; - For all x,y,z: if xRy and yRz then xRz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example transitive relation: xRy stands for 'x is taller than y' (in the set of all people)&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-20T21:34:01.537" />
  <row Id="135" PostTypeId="1" AcceptedAnswerId="138" CreationDate="2010-07-20T21:34:29.273" Score="10" ViewCount="324" Body="&lt;p&gt;Why is any number (other than zero) to the power of zero equal to one? Please include in your answer an explanation of why 0^0 should be undefined.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T16:59:46.510" LastActivityDate="2010-07-23T11:41:17.373" Title="Why is x^0 = 1 except when x = 0?" Tags="&lt;notation&gt;&lt;convention&gt;" AnswerCount="6" CommentCount="7" />
  <row Id="136" PostTypeId="1" AcceptedAnswerId="183" CreationDate="2010-07-20T21:36:40.990" Score="2" ViewCount="87" Body="&lt;p&gt;I was playing around with the squares and saw an interesting pattern in their differences.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;0^2 = 0&#xA;          + 1&#xA;1^2 = 1&#xA;          + 3&#xA;2^2 = 4&#xA;          + 5&#xA;3^2 = 9&#xA;          + 7&#xA;4^2 = 16&#xA;          + 9&#xA;5^2 = 25&#xA;          + 11&#xA;6^2 = 36&#xA;          etc.&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Also, in a very related question, which major Math Research Journal should I contact to publish my groundbreaking find in?)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:36:39.207" LastActivityDate="2010-07-21T01:36:39.207" Title="Why are the differences between consecutive squares equal to the sequence of odd numbers?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="137" PostTypeId="2" ParentId="67" CreationDate="2010-07-20T21:37:18.823" Score="7" ViewCount="0" Body="&lt;p&gt;The technical definition is a nonsingular projective curve of genus 1, which is an abelian variety under the group law: basially, this means that you   draw the line through two points on the curve -- which can be embedded in the projective plane -- and find where that line intersects the curve again (and call that the negative of the sum).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can always put elliptic curves in the (projectivization of the) form $y^2 = x^3 - Ax + B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the meaning of &quot;abelian variety&quot; is that you can add points on the elliptic curve, which is really useful; there isn't a way to do this for most objects in algebraic geometry.  Then one can study things like the torsion points on an elliptic curve, with respect to this abelian group structure: it's a theorem that there are $m^2$ torsion points of order $m$ for instance, if you 're working in an algebraically closed field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, one way to think of this is that an elliptic curve is really--algebraically and topologically--a torus if you are working over the complex numbers, and the torsion points in the torus are easily determined.  (Namely, a torus is algebraically the product of two copies of the unit circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also yields the theorem about torsion points for algebraically closed fields of characteristic zero via the &quot;Lefschetz principle.&quot;  (For characteristic p, you need a different argument.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things one can consider include the group of points with coordinates in, say, the rational numbers (assuming the curve is defined by rational coefficients). One of the central theorems is that this group is finitely generated.  The point is that the geometry of the elliptic curve leads to a rich algebraic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a bit about elliptic curves; I know nothing about cryptography and can't comment on that.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-20T21:57:10.940" LastActivityDate="2010-07-20T21:57:10.940" CommentCount="1" />
  <row Id="138" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:38:34.720" Score="3" ViewCount="0" Body="&lt;p&gt;For non-zero bases and exponents, the relation x^a * x^b = x^(a+b) holds. For this to make sense with an exponent of 0, x^0 needs to equal one. This gives you x^a * 1 = x^a * x^0 = x^(a+0) = x^a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the base is also zero, it's not possible to define a value for 0^0 because there is no value that is consistent with all the necessary constraints. For example, 0^x = 0 and x^0 = 1 for all positive x, and 0^0 can't be consistent with both of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to see that 0^0 can't have a reasonable definition is to look at the graph of f(x,y) = x^y which is discontinuous around (0,0). No chosen value for 0^0 will avoid this discontinuity.&lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="113" LastEditorDisplayName="" LastEditDate="2010-07-22T13:28:18.537" LastActivityDate="2010-07-22T13:28:18.537" CommentCount="7" />
  <row Id="139" PostTypeId="1" AcceptedAnswerId="176" CreationDate="2010-07-20T21:38:38.840" Score="2" ViewCount="125" Body="&lt;p&gt;I have a square that's 10m x 10m. I want to cut it in half so that I have a square with half the area. But if I cut it from top to bottom or left to right, I don't get a square, I get a rectangle!&lt;br&gt;&#xA;I know the area of the small square is supposed to be 50, so I can use my calculator to find out how long a side should be: it's 7.07106781. But my teacher said I should be able to do this without a calculator. How am I supposed to get that number by hand?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-20T23:55:32.657" Title="how do i cut a square in half?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="140" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:39:08.137" Score="5" ViewCount="0" Body="&lt;p&gt;0^x = 0, x^0 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;both are true when x &gt; 0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what happens when x=0? undefined, because there is no way to chose one definition over the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people define 0^0 = 1 in their books, like Knuth, because 0^x is less 'useful' than x^0.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-21T22:09:52.383" LastActivityDate="2010-07-21T22:09:52.383" CommentCount="3" />
  <row Id="141" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:40:17.747" Score="10" ViewCount="0" Body="&lt;p&gt;This is a question of definition, the question is &quot;why does it make sense to define x^0=1 except when x=0?&quot; or &quot;How is this definition better than other definitions?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that $x^a * x^b = x^{a+b}$ is an excellent formula that makes a lot of sense (multiplying a times and then multiplying b times is the same as multiplying a+b times) and which you can prove for a and b positive integers.  So any sensible definition of x^a for numbers a which aren't positive integers &lt;em&gt;should still satisfy this identity&lt;/em&gt;.  In particular, $x^0 * x^b = x^{0+b} = x^b$; now &lt;em&gt;if x is not zero&lt;/em&gt; then you can cancel x^b from both sides and get that x^0 = 1.  But if x=0 then x^b is zero and so this argument doesn't tell you anything about what you should define x^0 to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument should convince you that when x is not zero then x^-a should be defined as 1/x^a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argument using the related identity (x^a)^b = x^(ab) should convince you that x^{1/n} is taking the nth root.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T21:40:17.747" CommentCount="4" />
  <row Id="142" PostTypeId="2" ParentId="123" CreationDate="2010-07-20T21:40:22.333" Score="4" ViewCount="0" Body="&lt;p&gt;Apart from occuring naturally in the form of the catenary mentioned, hyperbolic trig functions are also fundamentally important in calculus. I hope you'll agree that calculus has many, many real world applications!&lt;/p&gt;&#xA;" OwnerUserId="67" LastActivityDate="2010-07-20T21:40:22.333" CommentCount="1" />
  <row Id="143" PostTypeId="2" ParentId="136" CreationDate="2010-07-20T21:41:21.523" Score="9" ViewCount="0" Body="&lt;p&gt;A nice thing to notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically because&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(n + 1) ^ 2 - n ^ 2 = 2 * n + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditorDisplayName="" LastEditDate="2010-07-20T21:50:48.713" LastActivityDate="2010-07-20T21:50:48.713" CommentCount="1" />
  <row Id="144" PostTypeId="2" ParentId="135" CreationDate="2010-07-20T21:42:02.833" Score="3" ViewCount="0" Body="&lt;p&gt;Exponents are only &quot;basically&quot; defined under the natural numbers above zero.  By this I mean, defined as &quot;iterated multiplication&quot; the same way multiplication is defined as iterated addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property &lt;code&gt;a^0 = 1&lt;/code&gt; only arises when we look at generalizing multiplication to the integers.  We do this by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a^4 / a^3 = (a*a*a*a)/(a*a*a) = a^1&#xA;a^4 / a^3 = a^(4-3)           = a^1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And using this, we can say:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a^2 / a^3 = a^(-1) = 1/a&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and also:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a^2 / a^2                 = 1&#xA;a^2 / a^2 = a^(2-2) = a^0 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So we say &lt;code&gt;a^0 = 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, notice that these proofs don't have any meaning when &lt;code&gt;a=0&lt;/code&gt;, because the whole concept/idea involves fractions, and you cannot have zero be in the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say &lt;code&gt;2^0 = 1&lt;/code&gt;, we really mean:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2^(1-1) = 2^1 / 2^1 = 2/2 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But we cannot say the same for &lt;code&gt;0^0&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0^(1-1) = 0^1/0^1 = 0/0 = UNDEFINED&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T21:42:02.833" />
  <row Id="145" PostTypeId="2" ParentId="139" CreationDate="2010-07-20T21:42:38.077" Score="0" ViewCount="0" Body="&lt;p&gt;A simple approximation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x^2 = 50&#xA;7^2 = 49&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so the number is around 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A even better approximation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(7+x)^2 = 50&#xA;14x + x^2 = 1 &#xA;(14+x)x = 1&#xA;14+x = 1/x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This basically shows x is pretty small, and around 1/14, without a calculator, it's not likely you can find what exactly 1/14 is, but at least you know it be smaller than 1/10. 7.1 is a good approximation. (off by less than 0.03)&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-20T21:48:27.607" LastActivityDate="2010-07-20T21:48:27.607" CommentCount="1" />
  <row Id="146" PostTypeId="2" ParentId="139" CreationDate="2010-07-20T21:44:00.230" Score="4" ViewCount="0" Body="&lt;p&gt;Take a pair of compasses and draw an arc between two opposite corners, centred at another corner, and then draw a diagonal across the arc. If you now draw two lines from the point of intersection, parallel to the sides of the square, the biggest of the resulting squares will have half the area of the original square.&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditorDisplayName="" LastEditDate="2010-07-20T23:55:32.657" LastActivityDate="2010-07-20T23:55:32.657" CommentCount="6" />
  <row Id="147" PostTypeId="2" ParentId="90" CreationDate="2010-07-20T21:45:47.093" Score="2" ViewCount="0" Body="&lt;p&gt;Andrea Feretti's &lt;a href=&quot;http://mathonline.andreaferretti.it/&quot; rel=&quot;nofollow&quot;&gt;MathOnline&lt;/a&gt; page.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-20T21:45:47.093" CommunityOwnedDate="2010-07-20T21:45:47.093" />
  <row Id="148" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T21:55:14.763" Score="4" ViewCount="0" Body="&lt;p&gt;One of the greatest achievements of human civilization is Newton's laws of motions.  The first law says that unless a force is acting then the &lt;em&gt;velocity&lt;/em&gt; (not the position!) of objects stay constant, while the second law says that forces act by causing an &lt;em&gt;acceleration&lt;/em&gt; (though heavy objects require more force to accellerate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However to make sense of those laws and to apply them to real life you need to understand how to move between the following three notions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Position&lt;/li&gt;&#xA;&lt;li&gt;Velocity (that is the rate of change in position)&lt;/li&gt;&#xA;&lt;li&gt;Acceleration (that is the rate of change of the velocity)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Moving down that list is called &quot;taking the derivative&quot; while moving up that list is called &quot;taking the integral.&quot;  Calculus is the study of derivatives and integerals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if you want to figure out how objects move under some force you need to be able to integrate twice.  This requires understanding a lot of calculus!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a first semester class you usually learn about derivatives and integrals of functions of one variable, that is what you need to understand physics in 1-dimension!  To understand the actual physics of the world you need to understand derivatives and integrals in 3-dimensions which requires several courses.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T21:55:14.763" />
  <row Id="149" PostTypeId="2" ParentId="120" CreationDate="2010-07-20T21:59:54.497" Score="15" ViewCount="0" Body="&lt;p&gt;The main point of talking about prime numbers is Euclid's theorem that every positive integer can be written uniquely as a product of primes.  As Justin remarks, this would break horribly if 1 were considered prime, for example we could factor 2 as 2*1*1*1*1*1.  Instead we say that 1 is not a prime, but it is the &lt;em&gt;product of zero primes&lt;/em&gt; (see &lt;a href=&quot;http://math.stackexchange.com/questions/135/why-is-x0-1-except-when-x-0&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt; to understand why any prime multiplied by itself 0 times is 1) so Euclid's theorem works out nicely!&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T21:59:54.497" CommentCount="1" />
  <row Id="150" PostTypeId="1" AcceptedAnswerId="151" CreationDate="2010-07-20T22:04:21.640" Score="3" ViewCount="314" Body="&lt;p&gt;It seems like functions that are continuous always seem to be differentiable, to me.  I can't imagine one that is not.  Are there any examples of functions that are continuous, yet not differentiable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way around seems a bit simpler -- a differentiable function is obviously always going to be continuous.  But are there any that do not satisfy this?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T04:42:28.373" LastActivityDate="2010-07-21T04:42:28.373" Title="Are there any functions that are (always) continuous yet not differentiable?  Or vice-versa?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="151" PostTypeId="2" ParentId="150" CreationDate="2010-07-20T22:07:02.073" Score="12" ViewCount="0" Body="&lt;p&gt;It's easy to find a function which is continuous but not differentiable at a single point, e.g. f(x) = |x| is continuous but not differentiable at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there are functions which are continuous but nowhere differentiable, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nowhere_differentiable&quot; rel=&quot;nofollow&quot;&gt;Weierstrass function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, continuity follows from differentiability, so there are no differentiable functions which aren't also continuous. If a function is differentiable at $x$, then the limit $(f(x+h)-f(x))/h$ must exist (and be finite) as $h$ tends to 0, which means $f(x+h)$ must tend to $f(x)$ as $h$ tends to 0, which means $f$ is continuous at $x$.&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="31" LastEditorDisplayName="" LastEditDate="2010-07-20T22:22:36.287" LastActivityDate="2010-07-20T22:22:36.287" CommentCount="4" />
  <row Id="152" PostTypeId="2" ParentId="71" CreationDate="2010-07-20T22:08:08.390" Score="3" ViewCount="0" Body="&lt;p&gt;To understand the benefits of using quaternions you have to consider different ways to represent rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are few ways with a summary of the pros and cons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Euler angles&lt;/li&gt;&#xA;&lt;li&gt;Rotation matrices&lt;/li&gt;&#xA;&lt;li&gt;Axis angle&lt;/li&gt;&#xA;&lt;li&gt;Quaternions  &lt;/li&gt;&#xA;&lt;li&gt;Rotors (normalized Spinors)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Euler angles are the best choice if you want a user to specify an orientation in a intuitive way. They are are also space efficient (three numbers). However, it is more difficult to linear interpolate values. Consider the case where you want to interpolate between 359 and 0 degrees. Linearly interpolating would cause a large rotation, even though the two orientations are almost the same. Writing shortest path interpolation, is easy for one axis, but non-trivial when considering the three Euler angles(for instance the shortest route between (240, 57, 145) and (35, -233, -270) is not immediately clear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotation matrices specify a new frame of reference using three normalized and orthogonal vectors (Right, Up, Out, which when multiplied become the new x, y, z). Rotation matrices are useful for operations like strafing (side ways movement), which only requires translating along the Right vector of the camera's rotation matrix. However, there is no clear method of interpolating between them. The are also expensive to normalize which is necessary to prevent scaling from being introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axis angle, as the name suggests, are a way of specifying a rotation axis and angle to rotate around that axis. You can think of Euler angles, as three axis angle rotations, where the axises are the x, y, z axis respectively. Linearly interpolating the angle in a axis angle is pretty straight forward (if you remember to take the shortest path), however linearly interpolating between different axises is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quaternions are a way of specifying a rotation through a axis and the cosine of half the angle. They main advantage is I can pick any two quaternions and smoothly interpolate between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotors are another way to perform rotations. Rotors are basically quaternions, but instead of thinking of them as 4D complex numbers, rotors are thought of as real 3D multivectors. This makes their visualization much more understandable (compared to quaternions), but requires fluency in geometric algebra to grasp their significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay with that as the background I can discuss a real world example. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say you are writing a computer game&#xA;  where the characters are animated in&#xA;  3ds Max. You need to export a&#xA;  animation of the character to play in&#xA;  your game, but cannot faithfully&#xA;  represent the interpolation used by&#xA;  the animation program, and thus have&#xA;  to sample. The animation is going to&#xA;  be represented as a list of rotations&#xA;  for each joint. How should we store&#xA;  the rotations? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If I am going to sample every frame,&#xA;  not interpolate, and space is not an&#xA;  issue, I would probably store the&#xA;  rotations as rotation matrices. If&#xA;  space was issue, then Euler angles.&#xA;  That would also let me do things like&#xA;  only store one angle for joints like&#xA;  the knee that have only one degree of&#xA;  freedom.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If I only sampled every 4 frames, and&#xA;  need to interpolate it depends on&#xA;  whether I am sure the frame-rate will&#xA;  hold. If I am positive that the&#xA;  frame-rate will hold I can use axis&#xA;  angle relative rotations to perform&#xA;  the interpolation. This is atypical.&#xA;  In most games the frame rate can drop&#xA;  past my sampling interval, which would&#xA;  require skipping an element in the&#xA;  list to maintain the correct playback&#xA;  speed. If I am unsure of what two&#xA;  orientations I need to interpolate&#xA;  between, then I would use quaternions&#xA;  or rotors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-20T23:34:33.507" LastActivityDate="2010-07-20T23:34:33.507" />
  <row Id="153" PostTypeId="1" AcceptedAnswerId="277" CreationDate="2010-07-20T22:17:25.933" Score="9" ViewCount="849" Body="&lt;p&gt;I have a non-professional interest in math and I would like to be able to be curious and play around with some math that I'm learning about (both here and from calculus-and-higher level math classes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica and MatLab seem like the best ways to play with math. I've seen demos of both and they both seem exceptionally powerful and complex. I'd like to start learning more about the syntax of one or the other so I can develop ideas and evaluate them, but I don't know which is better suited for me. Which one should I use?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-07-21T17:14:33.037" LastActivityDate="2010-09-13T18:56:55.120" Title="What should I learn first, Mathematica or MatLab?" Tags="&lt;math-software&gt;&lt;mathematica&gt;&lt;matlab&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="4" />
  <row Id="154" PostTypeId="1" AcceptedAnswerId="168" CreationDate="2010-07-20T22:17:36.010" Score="74" ViewCount="6896" Body="&lt;p&gt;Complex numbers involve the square root of negative one, and most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning. What's the best way to explain &lt;strong&gt;to a non-mathematician&lt;/strong&gt; that complex numbers are necessary and meaningful, in the same way that real numbers are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a Platonic question about the reality of mathematics, or whether abstractions are as real as physical entities, but an attempt to bridge a comprehension gap that many people experience when encountering complex numbers for the first time. The wording, although provocative, is deliberately designed to match the way that many people do in fact ask this question.&lt;/p&gt;&#xA;" OwnerUserId="113" LastActivityDate="2011-01-05T19:40:57.497" Title="Do complex numbers really exist?" Tags="&lt;soft-question&gt;&lt;teaching&gt;&lt;philosophy&gt;" AnswerCount="30" CommentCount="28" FavoriteCount="35" />
  <row Id="155" PostTypeId="1" AcceptedAnswerId="163" CreationDate="2010-07-20T22:17:38.477" Score="22" ViewCount="690" Body="&lt;p&gt;I've come across statements in the past along the lines of &quot;function f has no closed form integral&quot;, which I assume means that there is no combination of the operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;addition/subtraction&lt;/li&gt;&#xA;&lt;li&gt;multiplication/division&lt;/li&gt;&#xA;&lt;li&gt;raising to powers and roots&lt;/li&gt;&#xA;&lt;li&gt;trigonometric functions&lt;/li&gt;&#xA;&lt;li&gt;exponential functions&lt;/li&gt;&#xA;&lt;li&gt;logarithmic functions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which when differentiated gives the function f. I've heard this said about the function f(x) = x^x for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sort of techniques are used to prove statements like this? What is this branch of mathematics called?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Merged with &quot;&lt;a href=&quot;http://math.stackexchange.com/questions/2328/&quot;&gt;How to prove that some functions don’t have a primitive&lt;/a&gt;&quot; by &lt;a href=&quot;http://math.stackexchange.com/users/918/ismael&quot;&gt;Ismael&lt;/a&gt;:&lt;br&gt;&#xA;Sometimes we are told that some functions like $\sin(x)/x$ don't have a indefinite integral, or that it can't be expressed in term of other simple functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do wonder how we can prove that kind of assertions.&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T18:41:54.267" LastActivityDate="2010-11-25T23:10:30.527" Title="How can you prove that a function has no closed form integral?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;tag-removed&gt;&lt;differential-algebra&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="5" />
  <row Id="156" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:21:34.883" Score="8" ViewCount="0" Body="&lt;p&gt;Quantum mechanics, and hence physics and everything around us, fundamentally involves complex numbers.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-20T22:21:34.883" CommentCount="2" />
  <row Id="157" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:21:45.087" Score="3" ViewCount="0" Body="&lt;p&gt;The argument isn't worth having, as you disagree about what it means for something to 'exist'. There are many interesting mathematical objects which don't have an obvious physical counterpart. What does it mean for the Monster group to exist?&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-20T22:21:45.087" CommentCount="3" />
  <row Id="158" PostTypeId="2" ParentId="153" CreationDate="2010-07-20T22:22:04.197" Score="4" ViewCount="0" Body="&lt;p&gt;Conventional wisdom is that Matlab is the natural choice for numerical computation, whereas Mathematica is better for symbolic calculation and manipulation. Both can do both, and it's probably a matter of taste and the convention in the fields you're interested in. Matlab is much more widely used across the sciences and engineering, so if you've any interest in that direction, it might be more helpful for you to know.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-20T22:22:04.197" />
  <row Id="159" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:23:02.310" Score="7" ViewCount="0" Body="&lt;p&gt;I'll start by pointing out that a whole host of things that people think of as 'real' are on shakier ground than imaginary numbers. Given that quantum mechanics predicts a fundamental limit to how granular reality is, the whole concept of real numbers is on very shakey ground, yet people accept those as fine. I'd therefore suggest that it is merely a case of familiarity - people are less familiar with complex numbers than with some other mathematical constructs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for an actual existence outside the realms of pure maths... your best bet is to look at quantum mechanics again. This area has some fascinating results that are only possible through the use of imaginary numbers. Incidentally, fundamental particles are the place in nature that gave a 'physicality' to negative numbers (the charge of an electron is negative) well after they were accepted as normal by most people. &lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-20T22:23:02.310" CommentCount="8" />
  <row Id="160" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:24:43.197" Score="48" ViewCount="0" Body="&lt;p&gt;The concept of mathematical numbers and &quot;existing&quot; is a tricky one.  What actually &quot;exists&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do negative numbers exist?  Of course they do not.  You can't have a negative amount of apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, the beauty of negative numbers is that when we define them (rigorously), then all of a sudden we can use them to solve problems we were never ever been able to solve before, or to solve them in a much simpler way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine trying to do simple physics without the idea of negative numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are they &quot;real&quot;?  Do they &quot;exist&quot;?  No, they don't.  But they are just tools that help us solve real life problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go back to your question about complex numbers, I would say that the idea that they exist or not has no bearing onto whether or not they are actually useful in solving the problem of every day life, or making them many, many, many times more easy to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math that makes your computer run and involves the tool that is complex numbers, for instance.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T22:24:43.197" CommentCount="7" />
  <row Id="161" PostTypeId="2" ParentId="2" CreationDate="2010-07-20T22:25:54.740" Score="-3" ViewCount="0" Body="&lt;p&gt;Conceptually, to me at least, a limit is a way approximate something, by realizing the long run behavior is predictable and more important then the short term behavior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say your baby sitting a rambunctious kid. You could frustrate yourself trying to quite him, or just realize, in the limit he will tire himself out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of systems seem complicated in the short term, but are simple when viewed in the long term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit is way to take advantage of that, and relate a complex system to a simpler approximation. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-20T23:38:39.223" LastActivityDate="2010-07-20T23:38:39.223" CommentCount="2" />
  <row Id="162" PostTypeId="1" AcceptedAnswerId="171" CreationDate="2010-07-20T22:34:36.637" Score="8" ViewCount="552" Body="&lt;p&gt;I've heard of some other paradoxes involving sets (ie, &quot;the set of all sets that do not contain themselves&quot;) and I understand how paradoxes arise from them.  But this one I do not understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is &quot;the set of all sets&quot; a paradox?  It seems like it would be fine, to me.  There is nothing paradoxical about a set containing itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it something that arises from the &quot;rules of sets&quot; that are involved in more rigorous set theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I don't really have much of a background with Set Theory, so an answer that deals with common english would be much appreciated =)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-21T11:06:27.840" LastActivityDate="2010-08-19T19:57:08.697" Title="Why is &quot;the set of all sets&quot; a paradox?" Tags="&lt;paradox&gt;&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="4" />
  <row Id="163" PostTypeId="2" ParentId="155" CreationDate="2010-07-20T22:34:58.367" Score="12" ViewCount="0" Body="&lt;p&gt;It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions f and g, g nonconstant, the antiderivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int [f(x)\exp(g(x))] \mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be expressed in terms of elementary functions if and only if there exists some rational function h such that it is a solution to the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = h&amp;#39; + hg$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int e^{x^2} dx$ is another classic example of such a function with no elementary antiderivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how much math you've had, but some of this paper might be comprehensible in its broad strokes: &lt;a href=&quot;http://www.sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liouville's original paper: Liouville, J. &quot;&lt;em&gt;Sur la classification des Transcendantes et sur l'impossibilité d'exprimer les racines des certaines équations en fonction finie explicite des coefficients&lt;/em&gt;. Part 2.&quot; J. Math. Pure Appl. 3, 523-547, 1838. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Michael Spivak's book on Calculus also has a section with a discussion of this.&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T23:10:30.527" LastActivityDate="2010-11-25T23:10:30.527" CommentCount="1" />
  <row Id="164" PostTypeId="1" AcceptedAnswerId="181" CreationDate="2010-07-20T22:36:56.010" Score="17" ViewCount="771" Body="&lt;p&gt;I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-03T13:05:15.390" LastActivityDate="2010-09-03T16:59:07.623" Title="Why is the volume of a sphere $\frac{4}{3}\pi r^3$?" Tags="&lt;geometry&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="5" />
  <row Id="165" PostTypeId="2" ParentId="153" CreationDate="2010-07-20T22:37:23.013" Score="1" ViewCount="0" Body="&lt;p&gt;MatLab is a numerical computing environment. Mathematica is about symbolic manipulations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica helped me with calculus and some number theory problems. Almost everything I need is built into Mathematica, i can't say the same for MatLab. The math professors in my school uses maple(closer to mathematica than matlab). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have the good of both worlds, there are mathematica toolbox for matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find the problems on &lt;a href=&quot;http://projecteuler.net/&quot; rel=&quot;nofollow&quot;&gt;project euler&lt;/a&gt; interesting, mathematica are something you want to look into. &lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T22:37:23.013" CommentCount="3" />
  <row Id="166" PostTypeId="2" ParentId="97" CreationDate="2010-07-20T22:38:47.287" Score="4" ViewCount="0" Body="&lt;p&gt;You can use a Taylor polynomial of sufficient degree to guarantee the accuracy that you need.  (The Taylor series for erf(x) is given on the Wikipedia page to which you linked.)  The &lt;a href=&quot;http://mathworld.wolfram.com/LagrangeRemainder.html&quot; rel=&quot;nofollow&quot;&gt;Lagrange Remainder term&lt;/a&gt; can be used to bound the error in the Taylor series approximation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T22:38:47.287" />
  <row Id="167" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:40:39.890" Score="5" ViewCount="0" Body="&lt;p&gt;There are geometric interpretations of imaginary numbers where they are thought of as parallelograms with a front and back, or oriented parallelograms. That interpretation requires geometric algebra but only uses real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_algebra#Complex_numbers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geometric_algebra#Complex_numbers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't have any pictures so it is admittedly not intuitive, but the answer is yes. Whether you think of imaginary numbers as square root of negative 1 or as parallelogram with a front and back, they exist.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-20T22:52:52.997" LastActivityDate="2010-07-20T22:52:52.997" CommentCount="3" />
  <row Id="168" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:42:01.277" Score="87" ViewCount="0" Body="&lt;p&gt;There are a few good answers to this question, depending on the audience. I've used all of these on occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A way to solve polynomials&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We came up with equations like &lt;code&gt;x - 5 = 0, what is x?&lt;/code&gt;, and the naturals solved them (easily). Then we asked, &quot;wait, what about &lt;code&gt;x + 5 = 0&lt;/code&gt;?&quot; So we invented negative numbers. Then we asked &quot;wait, what about &lt;code&gt;2x = 1&lt;/code&gt;?&quot; So we invented rational numbers. Then we asked &quot;wait, what about &lt;code&gt;x^2 = 2&lt;/code&gt;?&quot; so we invented irrational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we asked, &quot;wait, what about &lt;code&gt;x^2 = -1&lt;/code&gt;?&quot; This is the only question that was left, so we decided to invent the &quot;imaginary&quot; numbers to solve it. All the other numbers, at some point, didn't exist and didn't seem &quot;real&quot;, but now they're fine. Now that we have imaginary numbers, we can solve every polynomial, so it makes sense that that's the last place to stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pairs of numbers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explanation goes the route of redefinition. Tell the listener to forget everything he knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good &quot;realistic&quot; use of pairs of numbers (many exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, show that in this system,  &lt;code&gt;(0,1) * (0,1) = (-1,0)&lt;/code&gt;, in other words, we've defined a new system, under which it makes sense to say that &lt;code&gt;sqrt(-1) = i&lt;/code&gt;, when &lt;code&gt;i=(0,1)&lt;/code&gt;. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there &lt;strong&gt;is&lt;/strong&gt; an answer to &lt;code&gt;sqrt(-1)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The historical explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain the history of the imaginary numbers. Showing that mathematicians &lt;em&gt;also&lt;/em&gt; fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little rusty, but I think there were certain equations that kept having parts of them which used &lt;code&gt;sqrt(-1)&lt;/code&gt;, and the mathematicians kept  throwing out the equations since there is no such thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, one mathematician decided to just &quot;roll with it&quot;, and kept working, and found out that all those square roots cancelled each other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazingly, the answer that was left was the &lt;strong&gt;correct answer&lt;/strong&gt; (he was working on finding roots of polynomials, I think). Which lead him to think that there &lt;em&gt;was&lt;/em&gt; a valid reason to use &lt;code&gt;sqrt(-1)&lt;/code&gt;, even if it took a long time to understand it.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-20T22:42:01.277" CommentCount="7" />
  <row Id="169" PostTypeId="2" ParentId="164" CreationDate="2010-07-20T22:45:12.100" Score="6" ViewCount="0" Body="&lt;p&gt;This was first &quot;guessed at&quot; by Archimedes, by what fraction of the volume of a cylinder a sphere is.  That is, a sphere that is contained within that cylinder.  From that, he figured out 4/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do it these days with the tools of calculus.  One way would to use the &lt;a href=&quot;http://library.thinkquest.org/3616/Calc/S3/TDM.html&quot; rel=&quot;nofollow&quot;&gt;Disk Method&lt;/a&gt;, over the graph of a semicircle.  Another would be to use spherical coordinates and calculate the 3D Integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these methods should give you 4/3rds pi * r^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it's not just r^3 because it's &quot;3D&quot;; but also because it's the only way the dimensional analysis would work; that is, the units would have to be in distance^3.  But that might be what you said, but just clarifying.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T22:45:12.100" />
  <row Id="170" PostTypeId="2" ParentId="120" CreationDate="2010-07-20T22:46:09.323" Score="13" ViewCount="0" Body="&lt;p&gt;It's important to understand that this is not something that can be &lt;strong&gt;proved&lt;/strong&gt;: it's a &lt;strong&gt;definition&lt;/strong&gt;. We &lt;em&gt;choose&lt;/em&gt; not to regard 1 as a prime number, simply because it makes writing lots of theorems much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noah gives the best example in his answer: Euclid's theorem that every positive integer can be written uniquely as a product of primes. If 1 &lt;em&gt;is&lt;/em&gt; defined to be a prime number, then we'd have to change that theorem to: &quot;every positive integer can be written uniquely as a product of primes, except for infinite multiplications by 1&quot;. So we choose to go with the easier path of defining 1 to not be a prime.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-20T22:46:09.323" />
  <row Id="171" PostTypeId="2" ParentId="162" CreationDate="2010-07-20T22:47:54.220" Score="19" ViewCount="0" Body="&lt;p&gt;Let |S| be the cardinality of S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S| &amp;lt; |2^S|$ can be proven with &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;generalized Cantor's diagonal argument&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of the set of all sets doesn't exist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let S be the set of all sets,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S| &amp;lt; |2^S|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $2^S$ is a subset of S, because every set in $2^S$ is in S. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|2^S| \leq |S|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contradiction. Therefore the set of all sets doesn't exist.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-04T03:37:04.360" LastActivityDate="2010-08-04T03:37:04.360" CommentCount="7" />
  <row Id="173" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T22:56:03.757" Score="8" ViewCount="0" Body="&lt;p&gt;You may be interested to read the MathOverflow question &quot;Demystifying Complex Numbers,&quot; &lt;a href=&quot;http://mathoverflow.net/questions/30156/demystifying-complex-numbers&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. A teacher is asking how to motivate complex numbers to students taking complex analysis.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-20T22:56:03.757" CommentCount="1" />
  <row Id="174" PostTypeId="2" ParentId="164" CreationDate="2010-07-20T22:57:01.727" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem&quot; rel=&quot;nofollow&quot;&gt;Pappus's centroid theorem&lt;/a&gt; (second theorem) says that the volume of a solid formed by revolving a region about an axis is the product of the area of the region and the distance traveled by the centroid of the region when it is revolved.  A sphere can be formed by revolving a semicircle about is diameter edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the semicircle is 1/2 &amp;pi; r^2.  The centroid of the semicircle can be found by intersecting two lines that both divide the area of the semicircle into two equal parts.  One such line is perpendicular to the diameter edge through the center of the semicircle (this is a line of symmetry of the semicircle).  Another such line is parallel to the diameter edge, (4r)/(3&amp;pi;) away from it (verification of this is left as an exercise for the reader).  When revolved about the diameter edge of the semicircle, the centroid travels 2&amp;pi; * (4r)/(3&amp;pi;) = 8/3 * r, so the volume of the sphere is 1/2 &amp;pi; r^2 * 8/3 * r = 4/3 &amp;pi; r^3.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-20T22:57:01.727" />
  <row Id="175" PostTypeId="2" ParentId="118" CreationDate="2010-07-20T22:58:42.847" Score="0" ViewCount="0" Body="&lt;p&gt;Calculus is the mathematics of change. In algebra, almost nothing ever changes. Here's a comparison of some algebra vs. calc problems:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;algebra: car A is driving at 50 kph. How far has it gone after 6 hours?&#xA;calc: car B starts at 10 mph and begins accelerating at the rate of 10 kph^2&#xA;(kilometers per hour per hour). How far has car B gone after 6 hours?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note how the algebra problem nothing changes, where in the calc problem, the speed of the car is constantly changing.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;calc: If a ball is rolling in a straight line at 10 fps with a diameter of 1 foot&#xA;and Q is a the point at the top of the ball when t=0, how fast is point Q moving &#xA;at time t=4 relative to the ground?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The speed of the point in relation to the ground is never the same (its zero when its at the bottom, 20fps when it's at the top. Calculus lets you figure out how fast it's going exactly at a specific moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main branches of calculus, differential and integral. These problems pertain to differential calculus as they concern how something is changing. Integral calculus deals with how much something has changed, the opposite of differential calculus. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The number of cubic feet of oxygen circulated by Jeff lungs per hour at time t &#xA;follows the equation f(t) = t^3 + sin(t) How many cubic feet of oxygen do &#xA;Jeff's lungs cycle per day?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To find out how much something has changed when its rate of change isn't constant requires integral calculus. (the equation is purely hypothetical unless Jeff happens to be the size of a beluga whale).&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2010-07-20T22:58:42.847" CommunityOwnedDate="2010-07-20T22:58:42.847" />
  <row Id="176" PostTypeId="2" ParentId="139" CreationDate="2010-07-20T23:09:57.103" Score="10" ViewCount="0" Body="&lt;p&gt;Does this give you any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://farm5.static.flickr.com/4119/4813793520_0d7fa53b99_o.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-20T23:09:57.103" CommentCount="3" />
  <row Id="177" PostTypeId="1" CreationDate="2010-07-20T23:10:36.897" Score="3" ViewCount="170" Body="&lt;p&gt;Consider the task of generating random points uniformly distributed within a circle of a given radius $r$ that is centered at the origin. Assume that we are given a random number generator $R$ that generates a floating point number uniformly distributed in the range $[0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following procedure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Generate a random point $p = (x, y)$ within a square of side $2r$ centered at the origin. This can be easily achieved by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Using the random number generator $R$ to generate two random numbers $x$ and $y$, where $x, y \in [0, 1)$, and then transforming $x$ and $y$ to the range $[0, r)$ (by multiplying each by $r$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Flipping a fair coin to decide whether to reflect $p$ around the $x$-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Flipping another fair coin to decide whether to reflect $p$ around the $y$-axis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, if $p$ happens to fall outside the given circle, discard $p$ and generate another point. Repeat the procedure until $p$ falls within the circle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is the previous procedure correct? That is, are the random points generated by it &lt;strong&gt;uniformly distributed&lt;/strong&gt; within the given circle? How can one formally [dis]prove it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background Info&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task was actually given in &lt;a href=&quot;http://rubyquiz.strd6.com/quizzes/234-random-points-within-a-circle&quot; rel=&quot;nofollow&quot;&gt;Ruby Quiz - Random Points within a Circle (#234)&lt;/a&gt;. If you're interested, you can &lt;a href=&quot;http://gist.github.com/447554&quot; rel=&quot;nofollow&quot;&gt;check my solution&lt;/a&gt; in which I've implemented the procedure described above. I would like to know whether the procedure is mathematically correct or not, but I couldn't figure out how to formally [dis]prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the actual task was to generate random points uniformly distributed within a circle of a given radius &lt;em&gt;and position&lt;/em&gt;, but I intentionally left that out in the question because the generated points can be easily translated to their correct positions relative to the given center.&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:42:00.333" LastActivityDate="2010-07-21T17:39:38.927" Title="Will this Procedure Generate Random Points Uniformly Distributed within a Given Circle? Proof?" Tags="&lt;algorithms&gt;&lt;probability-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="178" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T23:11:01.500" Score="1" ViewCount="0" Body="&lt;p&gt;We will will first consider the most common definition of i, as the square root of -1. When you first hear this, it sounds crazy. 0 squared is 0; a positive times a positive is positive and a negative times a negative is positive too. So there doesn't actually appear to be any number that we can square to get -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mathematician would collectively term 0, negative numbers and positive numbers as the real numbers. They would also define the term complex numbers as a group of numbers that includes these real numbers. So while we have shown that no real number can square to get -1, we haven't even defined complex numbers at this point, so we can't rule out that one might have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, it makes sense to ask what does a mathematician mean by a number? It certainly isn't what most people associate it with - as an abstract representation some kind of real world quantity. We need to understand that it isn't uncommon for one word to have different meanings for different groups of people - after all words mean whatever we make them mean. Most people only need to real world quantities, so they find it convenient to call those numbers. On the other hand, mathematicians explore a variety of different number systems. Indeed some, such as complex numbers, are useful for solving problems that are actually about real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So mathematicians define i as a number that obeys most of the normal algebraic laws. They also defined i*i to equal -1. From this we can derive all of the standard results about &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_number&quot; rel=&quot;nofollow&quot;&gt;complex number&lt;/a&gt;s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for whether they are real - it depends on what you want to know. Obviously, they don't correspond to quantities of physical objects. On the other hand, complex numbers can useful for representing &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_impedance&quot; rel=&quot;nofollow&quot;&gt;resistance in an electric circuit&lt;/a&gt;. Ultimately, they are an idea and while ideas don't exist physically, saying they don't exist at all is inaccurate.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-20T23:11:01.500" CommentCount="1" />
  <row Id="179" PostTypeId="2" ParentId="162" CreationDate="2010-07-20T23:13:32.773" Score="2" ViewCount="0" Body="&lt;p&gt;An informal explanation is Russel's Paradox. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Russel%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;wiki page&lt;/a&gt; is informative, here's the relevant quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let us call a set &quot;abnormal&quot; if it is&#xA;  a member of itself, and &quot;normal&quot;&#xA;  otherwise. For example, take the set&#xA;  of all squares. That set is not itself&#xA;  a square, and therefore is not a&#xA;  member of the set of all squares. So&#xA;  it is &quot;normal&quot;. On the other hand, if&#xA;  we take the complementary set that&#xA;  contains all non-squares, that set is&#xA;  itself not a square and so should be&#xA;  one of its own members. It is&#xA;  &quot;abnormal&quot;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now we consider the set of&#xA;  all normal sets, R. Attempting to&#xA;  determine whether R is normal or&#xA;  abnormal is impossible: If R were a&#xA;  normal set, it would be contained in&#xA;  the set of normal sets (itself), and&#xA;  therefore be abnormal; and if it were&#xA;  abnormal, it would not be contained in&#xA;  the set of normal sets (itself), and&#xA;  therefore be normal. This leads to the&#xA;  conclusion that R is both normal and&#xA;  abnormal: Russell's paradox.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-20T23:13:32.773" CommentCount="3" />
  <row Id="180" PostTypeId="2" ParentId="75" CreationDate="2010-07-20T23:14:16.800" Score="4" ViewCount="0" Body="&lt;p&gt;I won't explain what a ring or a group is, because that's already been done, but I'll add something else.  One reason groups and rings feel similar is that they are both &quot;algebraic structures&quot; in the sense of universal algebra. So for instance, the operation of quotienting via a normal subgroup (for a group) and a two-sided ideal (for a ring) are basically instances of quotienting via an invariant equivalence relation in universal algebra.  A field, by contrast, is not really a construction of universal algebra (because the operation $x \to x^{-1}$ is not everywhere defined) -- which is why free fields don't exist, for instance -- though they are a special case of rings.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-20T23:14:16.800" />
  <row Id="181" PostTypeId="2" ParentId="164" CreationDate="2010-07-20T23:15:55.393" Score="20" ViewCount="0" Body="&lt;p&gt;In addition to the methods of calculus, Pappus, and Archimedes already mentioned, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavalieri%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Cavalieri's Principle&lt;/a&gt; can be useful for these kinds of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have two solid figures lined up next to each other, each fitting between the same two parallel planes. (E.g., two stacks of pennies lying on the table, of the same height). Then, consider cutting the two solids by a plane parallel to the given two and in between them. If the cross-sectional area thus formed is the same for each of the solids for any such plane, the volumes of the solids are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're willing to accept that you know the volume of a cone is 1/3 that of the cylinder with the same base and height, you can use Cavalieri, comparing a hemisphere to a cylinder with an inscribed cone, to get the volume of the sphere. This diagram (from Wikipedia) illustrates the construction: &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/en/5/58/Sphere_cavalieri.svg&quot; rel=&quot;nofollow&quot;&gt;look here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a cylinder of radius R and height R, with, inside it, an inverted cone, with base of radius R coinciding with the top of the cylinder, and again height R. Put next to it a hemisphere of radius R. Now consider the cross section of each at height y above the base. For the cylinder/cone system, the area of the cross-section is pi*(R^2-y^2). It's the same for the hemisphere cross-section, as you can see by doing the Pythagorean theorem with any vector from the sphere center to a point on the sphere at height y to get the radius of the cross section (which is circular).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the cylinder/cone and hemisphere have the same height, by Cavalieri's Principle the volumes of the two are equal. The cylinder volume is Pi*R^3, the cone is a third that, so the hemisphere volume is (2/3)*pi*R^3. Thus the sphere of radius R has volume (4/3)*pi*R^3. &lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-21T02:20:34.283" LastActivityDate="2010-07-21T02:20:34.283" CommentCount="2" />
  <row Id="183" PostTypeId="2" ParentId="136" CreationDate="2010-07-20T23:27:10.833" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://www.education.vic.gov.au/images/content/studentlearning/mathscontinuum/SquareNos_sml.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/136/why-are-the-differences-between-consecutive-squares-equal-to-the-sequence-of-odd/143#143&quot; rel=&quot;nofollow&quot;&gt;Dan's&lt;/a&gt; algebraic justification is correct, but you may get more intuition about why this is happening from the above picture. Each time you want to enlarge the square by one unit, you have to add an extra row, an extra column, and one more square to fill in the corner. These correspond directly to the $n+n+1=2n+1$ that Dan mentioned. And of course, $2n+1$ is how odd numbers look.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it from the other direction, you can use the same idea to convince yourself of Noah's claim that $1+3+5+...+(2n-1) = n^2$. Imagine the left-hand side as representing the upper sequence of green and orange squares. You add on first 1 tile, then 3, then 5, and so on to construct larger and larger squares. At each step, you are adding one row (n) and one column (another n), then removing that one tile in the top right corner where they overlap (for a total of 2n-1). &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T01:10:36.993" LastActivityDate="2010-07-21T01:10:36.993" />
  <row Id="184" PostTypeId="2" ParentId="83" CreationDate="2010-07-20T23:30:03.247" Score="2" ViewCount="0" Body="&lt;p&gt;I have found most people liked math at some point, but something happened in their learning process that made them feel so stupid, they became disenfranchised with mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What tends to happen is students are presented with some mathematically result they are expected to memorize by route, which takes all the joy out of mathematics and prevents them from approaching mathematics intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would first try to zero in on what they don't like and what parts of mathematics they have had to take on faith. You might not have to excite them if you help them learn mathematics intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per the suggestions, I would show them how mathematically equations make pretty shapes in Processing.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-21T02:43:14.273" LastActivityDate="2010-07-21T02:43:14.273" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="185" PostTypeId="1" AcceptedAnswerId="193" CreationDate="2010-07-20T23:34:22.990" Score="1" ViewCount="47" Body="&lt;p&gt;I have a collection of 3D points in the standard x, y, z vector space. Now I pick one of the points &lt;code&gt;p&lt;/code&gt; as a new origin and two other points &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; such that &lt;code&gt;a - p&lt;/code&gt; and &lt;code&gt;b - p&lt;/code&gt; form two vectors of a new vector space. The third vector of the space I will call &lt;code&gt;x&lt;/code&gt; and calculate that as the cross product of the first two vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to recast or reevaluate each of the points in my collection in terms of the new vector space. How do I do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, if 'recasting' not the right term here, please correct me.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-20T23:51:38.390" LastActivityDate="2010-07-20T23:51:38.390" Title="Recasting points from one vector space to another" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="187" PostTypeId="1" AcceptedAnswerId="1492" CreationDate="2010-07-20T23:38:48.037" Score="4" ViewCount="166" Body="&lt;p&gt;Basic definitions: a tiling of d-dimensional Euclidean space is a decomposition of that space into polyhedra such that there is no overlap between their interiors, and every point in the space is contained in some one of the polyhedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vertex-uniform tiling is a tiling such that each vertex figure is the same: each vertex is contained in the same number of k-faces, etc: the view of the tiling is the same from every vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vertex-transitive tiling is one such that for every two vertices in the tiling, there exists an element of the symmetry group taking one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly all vertex-transitive tilings are vertex-uniform. For n=2, these notions coincide. However, Grunbaum, in his book on tilings, mentions but does not explain that for n &gt;= 3, there exist vertex uniform tilings that are not vertex transitive. Can someone provide an example of such a tiling, or a reference that explains this?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-28T20:57:48.987" Title="Euclidean Tilings that are Uniform but not Vertex-Transitive" Tags="&lt;discrete-geometry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="188" PostTypeId="2" ParentId="154" CreationDate="2010-07-20T23:39:39.317" Score="27" ViewCount="0" Body="&lt;p&gt;No number does &quot;really exist&quot; the way trees or atom exist.&#xA;In physics people however have found use for complex numbers just as they have found use for real numbers.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastActivityDate="2010-07-20T23:39:39.317" CommentCount="9" />
  <row Id="190" PostTypeId="2" ParentId="11" CreationDate="2010-07-20T23:44:38.403" Score="0" ViewCount="0" Body="&lt;p&gt;If you take two real numbers &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; then there per definition of the real number &lt;code&gt;z&lt;/code&gt; for which &lt;code&gt;x &amp;lt; z &amp;lt; y&lt;/code&gt; or &lt;code&gt;x &amp;gt; z &amp;gt; y&lt;/code&gt; is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;code&gt;x = 0.99999...&lt;/code&gt; and &lt;code&gt;y = 1&lt;/code&gt; you can't find a &lt;code&gt;z&lt;/code&gt; and therefore &lt;code&gt;0.99999... = 1&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastEditorUserId="2263" LastEditorDisplayName="" LastEditDate="2010-07-20T23:50:15.053" LastActivityDate="2010-07-20T23:50:15.053" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="191" PostTypeId="2" ParentId="177" CreationDate="2010-07-20T23:46:23.517" Score="0" ViewCount="0" Body="&lt;p&gt;I'm going to use $P(a,b)$ for probability of x = a, and y = b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(a,b) = P[x = |a|]P[y = |b|] * 1/4$ for any a, b inside the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you are picking a coordinate uniformly inside the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem reduced to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a element uniformly in a set A, then discard ones that are not B (B is a subset of A). Is it the same as pick a element uniformly in the set $A \cap B = B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is true, I don't know if the following is a formal proof(haven't done probability in a while). please point out my mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(B) is the event that a element in B is picked. P(A) is the event that a element in A is picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a x uniformly, we have b \in B is P(x=b|B), it's obviously uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditional probability P(B|A) = P(A \cap B)/P(A) = P(B)/P(A) = P(B) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(A) = 1, w/e you pick are in A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(x=b| (B|A)) = P(x=b|B), which is uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer to your question: It is uniformly distributed. &lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-20T23:46:23.517" />
  <row Id="193" PostTypeId="2" ParentId="185" CreationDate="2010-07-20T23:48:51.560" Score="2" ViewCount="0" Body="&lt;p&gt;You are changing your basis vectors, call your new ones i, j, and k where i is defined from a-p, j from b-p, and k the cross product. Now recall that your basis vectors should be unit, so take the length of your three vectors and divide the vectors by their length, making i,j, and k unit. Now a = d(a,p)*i, b =d(b,p)*j.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-20T23:48:51.560" />
  <row Id="194" PostTypeId="1" AcceptedAnswerId="199" CreationDate="2010-07-20T23:49:17.660" Score="5" ViewCount="89" Body="&lt;p&gt;Do primes become more or less frequent as you go further out on the number line?  That is, are there more or fewer primes between 1 and 1,000,000 than between 1,000,000 and 2,000,000?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof or pointer to a proof would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:37:11.750" LastActivityDate="2010-07-21T03:59:29.613" Title="Distribution of primes?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="195" PostTypeId="2" ParentId="185" CreationDate="2010-07-20T23:49:40.717" Score="1" ViewCount="0" Body="&lt;p&gt;What you are describing is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;Affine Transformation&lt;/a&gt;, which is a linear transformation followed by a translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this because any straight lines in your original vector space is also going to be a straight line in your transformed vector space.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T23:49:40.717" />
  <row Id="196" PostTypeId="1" AcceptedAnswerId="229" CreationDate="2010-07-20T23:50:41.403" Score="5" ViewCount="121" Body="&lt;p&gt;I keep looking at this picture and its driving me crazy. How can the smaller circle travel the same distance when its circumference is less than the entire wheel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathworld.wolfram.com/images/gifs/AristotlesWheel.gif&quot; alt=&quot;demonstration image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T01:40:25.533" LastActivityDate="2010-07-21T10:33:49.587" Title="How does the wheel paradox work?" Tags="&lt;geometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="197" PostTypeId="2" ParentId="196" CreationDate="2010-07-20T23:53:59.960" Score="7" ViewCount="0" Body="&lt;p&gt;If the two circles are fixed, then they will be traveling the same difference, but at different velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the ratio of the radii is equal to the ratio of the velocities a point on either circle will be traveling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tried to repeat this by putting two different-sized circles on a track and making them spin to come out to be the same distance with the same angular velocity, you will notice that one of the circles will have to slide/slip along the track in order to keep them at the same pace.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-20T23:53:59.960" />
  <row Id="198" PostTypeId="1" CreationDate="2010-07-20T23:54:20.210" Score="1" ViewCount="614" Body="&lt;p&gt;Since constructive mathematics allows us to avoid things like Russell's Paradox, then why don't they replace traditional proofs?  How do we know the &quot;regular&quot; kind of mathematics are free of paradox without a proof construction?&lt;/p&gt;&#xA;" OwnerUserId="117" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T05:16:00.523" LastActivityDate="2010-08-17T06:34:19.387" Title="Aren't constructive math proofs more &quot;sound&quot;?" Tags="&lt;proof-theory&gt;&lt;constructive-mathematics&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="199" PostTypeId="2" ParentId="194" CreationDate="2010-07-20T23:55:54.543" Score="9" ViewCount="0" Body="&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article about the prime number theorem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Roughly speaking, the prime number&#xA;  theorem states that if a random number&#xA;  nearby some large number N is&#xA;  selected, the chance of it being prime&#xA;  is about 1 / ln(N), where ln(N)&#xA;  denotes the natural logarithm of N.&#xA;  For example, near N = 10,000, about&#xA;  one in nine numbers is prime, whereas&#xA;  near N = 1,000,000,000, only one in&#xA;  every 21 numbers is prime. In other&#xA;  words, the average gap between prime&#xA;  numbers near N is roughly ln(N).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-20T23:55:54.543" />
  <row Id="200" PostTypeId="2" ParentId="162" CreationDate="2010-07-20T23:58:51.917" Score="4" ViewCount="0" Body="&lt;p&gt;The &quot;set of all sets&quot; is not so much a paradox in itself as something  that inevitably leads to a contradiction, namely the well-known (and referenced in the question) Russell's paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any set and a predicate applying to sets, the set of all things satisfying the predicate should be a subset of the original set.  If the &quot;set of all sets&quot; were to exist, because self-containment and non-self-containment are valid predicates, the set of all sets not containing themselves would have to exist as a set in order for our set theory to be consistent.  But this &quot;set of all sets&quot; cannot exist in a consistent set theory because of the Russel paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the non-existence of the &quot;set of all sets&quot; is a consequence of the fact that presuming it's existence would lead to the contradiction described by Russel's paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in fact the origin of Russel's paradox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his work &quot;The Basic Laws of Arithmetic&quot;, Gottlob Frege had taken as a postulate the existence of this &quot;set of all sets&quot;.  In a letter to Frege, Bertrand Russell essentially blew away the basis of Frege's entire work by describing the paradox and proving that this postulate could not be a part of a consistent set theory.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-20T23:58:51.917" CommentCount="2" />
  <row Id="201" PostTypeId="1" AcceptedAnswerId="211" CreationDate="2010-07-21T00:01:59.303" Score="8" ViewCount="1296" Body="&lt;p&gt;Prime numbers are numbers with no factors other than one and itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factors of a number are always lower or equal to than a given number; so, the larger the number is, the larger the pool of &quot;possible factors&quot; that number might have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the larger the number, it seems like the less likely the number is to be a prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely there must be a number where, simply, every number above it has some other factors.  A &quot;critical point&quot; where every number larger than it simply will always have some factors other than one and itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has there been any research as to finding this critical point, or has it been proven not to exist?  That for any &lt;code&gt;n&lt;/code&gt; there is always guaranteed to be a number higher than &lt;code&gt;n&lt;/code&gt; that has no factors other than one and itself?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T04:08:49.177" LastActivityDate="2010-11-04T04:08:49.177" Title="Is there possibly a largest prime number?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="8" CommentCount="8" FavoriteCount="0" />
  <row Id="202" PostTypeId="2" ParentId="155" CreationDate="2010-07-21T00:02:33.280" Score="5" ViewCount="0" Body="&lt;p&gt;Surprisingly, there is a procedure for determining this, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Risch_algorithm&quot; rel=&quot;nofollow&quot;&gt;Risch Algorithm&lt;/a&gt;. However, it is apparently very complicated to implement, and furthermore it is not a true algorithm: if the input &lt;em&gt;does&lt;/em&gt; have an anti-derivative, it can find it in a finite amount of time, but demonstrating there is no anti-derivative requires solving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Constant_problem&quot; rel=&quot;nofollow&quot;&gt;constant problem&lt;/a&gt;, which can fail to halt. (In that case, probabilistic methods can be used.)&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-07-21T00:02:33.280" CommentCount="1" />
  <row Id="203" PostTypeId="2" ParentId="198" CreationDate="2010-07-21T00:04:39.220" Score="3" ViewCount="0" Body="&lt;p&gt;The distinction between constructive mathematics and traditional mathematics has nothing to do with Russell's Paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructive mathematics simply requires working with one less basic postulate that many mathematicians have believed to be sensible and on which some proofs are based, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_choice&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T00:04:39.220" CommentCount="4" />
  <row Id="204" PostTypeId="2" ParentId="201" CreationDate="2010-07-21T00:06:12.167" Score="12" ViewCount="0" Body="&lt;p&gt;No there is not, here is a collection of proofs;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.mit.edu/~ssam/writings/primes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.mit.edu/~ssam/writings/primes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T00:06:12.167" CommentCount="5" />
  <row Id="205" PostTypeId="2" ParentId="29" CreationDate="2010-07-21T00:12:28.473" Score="7" ViewCount="0" Body="&lt;p&gt;Personally, I always use &lt;a href=&quot;http://www.python.org/download/&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt;'s console.&lt;br&gt;&#xA;It has history and allows all kinds of math operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img6.imageshack.us/img6/5497/pythony.png&quot; alt=&quot;Screenshot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is available for Linux, Windows, Mac, and others.&lt;/p&gt;&#xA;" OwnerUserId="135" LastEditorUserId="135" LastEditorDisplayName="" LastEditDate="2010-07-28T01:02:54.477" LastActivityDate="2010-07-28T01:02:54.477" CommunityOwnedDate="2010-07-28T01:32:30.370" CommentCount="5" />
  <row Id="206" PostTypeId="1" AcceptedAnswerId="248" CreationDate="2010-07-21T00:12:32.950" Score="9" ViewCount="576" Body="&lt;p&gt;Often in my studies (economics) the assumption of a &quot;well-behaved&quot; function will be invoked. I don't exactly know what that entails (I think twice continuously differentiability is one of the requirements), nor do I know why this is necessary (though I imagine the why will depend on each case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain it to me, and if there is an explanation of the why as well, I would be grateful. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: To give one example where the term appears, see this Wikipedia entry for utility functions, which says at one point:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to simplify calculations,&#xA;  various assumptions have been made of&#xA;  utility functions.     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;CES (constant elasticity of substitution, or&#xA;  isoelastic) utility&lt;br&gt;&#xA;  Exponential utility&lt;br&gt;&#xA;  Quasilinear utility&lt;br&gt;&#xA;  Homothetic preferences &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Most utility functions&#xA;  used in modeling or theory are&#xA;  &lt;strong&gt;well-behaved&lt;/strong&gt;. They are usually&#xA;  monotonic, quasi-concave, continuous&#xA;  and globally non-satiated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might be wrong, but I don't think &quot;well-behaved&quot; means monotonic, quasi-concave, continuous and globally non-satiated. What about twice differentiable?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T16:26:27.393" LastActivityDate="2010-08-10T21:44:15.657" Title="What exactly does it mean for a function to be &quot;well-behaved&quot;?" Tags="&lt;terminology&gt;&lt;analysis&gt;&lt;economics&gt;" AnswerCount="4" CommentCount="13" FavoriteCount="3" />
  <row Id="207" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T00:13:19.343" Score="10" ViewCount="0" Body="&lt;p&gt;As a free/libre alternative to Mathematica, you may want to have a look at &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt; for &lt;a href=&quot;http://en.wikipedia.org/wiki/Python_%28programming_language%29&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a free/libre alternative to MATLAB, which duplicates almost all of its functionality, have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/GNU_Octave&quot; rel=&quot;nofollow&quot;&gt;GNU Octave&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, &lt;a href=&quot;http://www.wolfram.com/products/mathematica/index.html&quot; rel=&quot;nofollow&quot;&gt;Mathematica&lt;/a&gt; is a beautiful system that has no equal.&lt;/p&gt;&#xA;" OwnerUserId="132" LastEditorUserId="132" LastEditorDisplayName="" LastEditDate="2010-08-15T21:20:11.747" LastActivityDate="2010-08-15T21:20:11.747" CommentCount="2" />
  <row Id="209" PostTypeId="2" ParentId="29" CreationDate="2010-07-21T00:15:05.363" Score="5" ViewCount="0" Body="&lt;p&gt;I use &lt;a href=&quot;http://www.r-project.org/&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt; these days. It was build to be a calculator and does the job well. It's syntax might be a bit strange but it allows you to do a lot with little typing.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastEditorUserId="2263" LastEditorDisplayName="" LastEditDate="2010-07-21T00:43:07.147" LastActivityDate="2010-07-21T00:43:07.147" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="210" PostTypeId="2" ParentId="198" CreationDate="2010-07-21T00:20:29.157" Score="7" ViewCount="0" Body="&lt;p&gt;A whole bunch of things in mathematics are inherently nonconstructive. For instance, invariant theory--recall the famous quote by Gordan that Hilbert's mathematics was &quot;theology.&quot; (A quote which, I believe, was in jest.) &#xA;The Hahn-Banach theorem, a fundamental tool in functional analysis (and a great tool for proving all sorts of results, like approximation results--Runge's theorem, the Stone-Weierstrass theorem, and more) relies on the axiom of choice, and is consequently nonconstructive.  The fact that any proper ideal in a ring is contained in a maximal ideal is frequently used in algebra, and yet it needs the axiom of choice.  The use of ultraproducts in logic (or the construction of hyperreal numbers) is inherently nonconstructive: you can't just &lt;em&gt;exhibit&lt;/em&gt; a nonprincipal ultrafilter on the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, a lot of mathematics just doesn't work without Zorn's lemma, and this is equivalent to the axiom of choice.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-17T06:34:19.387" LastActivityDate="2010-08-17T06:34:19.387" CommentCount="4" />
  <row Id="211" PostTypeId="2" ParentId="201" CreationDate="2010-07-21T00:24:29.570" Score="25" ViewCount="0" Body="&lt;p&gt;Euclid's famous proof is as follows: Suppose there is a finite number of primes.  Let X be the product of all of these primes.  Then look at X+1.  It is clear that X is coprime to X+1.  Therefore, no nontrivial factor of X is a factor of X+1, but every prime is a factor of X.  By the fundamental theorem of arithmetic, X+1 admits a prime factorization, and by the above remark, none of these prime factors can be a factor of X, but X is the product of all primes.  This is a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T00:24:29.570" CommentCount="6" />
  <row Id="212" PostTypeId="2" ParentId="201" CreationDate="2010-07-21T00:32:50.413" Score="3" ViewCount="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://xkcd.com/622/&quot; rel=&quot;nofollow&quot;&gt;XKCD&lt;/a&gt;, we have the following Haiku:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top Prime's Divisors'&lt;br&gt;&#xA;Product (Plus one)'s factors are...?&lt;br&gt;&#xA;Q.E.D B@%&amp;amp;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if we can edit it to make it correct&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-17T14:34:23.607" LastActivityDate="2010-08-17T14:34:23.607" CommentCount="14" />
  <row Id="213" PostTypeId="1" AcceptedAnswerId="498" CreationDate="2010-07-21T00:34:22.937" Score="9" ViewCount="752" Body="&lt;p&gt;I am learning geometric algebra, and it is incredible how much it helps me understand other branches of mathematics. I wish I had been exposed to it earlier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally I feel the same way about enumerative combinatorics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some less popular mathematical subjects that you think should be more popular?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-14T06:29:05.127" LastActivityDate="2010-09-26T00:02:24.270" CommunityOwnedDate="2010-07-21T00:34:22.937" Title="Mathematical subjects you wish you learned earlier" Tags="&lt;soft-question&gt;&lt;learning&gt;" AnswerCount="13" CommentCount="3" FavoriteCount="9" />
  <row Id="214" PostTypeId="2" ParentId="162" CreationDate="2010-07-21T00:36:42.743" Score="5" ViewCount="0" Body="&lt;p&gt;As others have explained, the existence of universal set is incompatible with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory&quot; rel=&quot;nofollow&quot;&gt;Zermelo&amp;ndash;Fraenkel axioms&lt;/a&gt; of set theory. However, there are alternative set theories which admit a universal set. One such theory is &lt;a href=&quot;http://en.wikipedia.org/wiki/New_Foundations&quot; rel=&quot;nofollow&quot;&gt;Quine's New Foundations&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-21T00:36:42.743" />
  <row Id="215" PostTypeId="2" ParentId="206" CreationDate="2010-07-21T00:40:08.307" Score="6" ViewCount="0" Body="&lt;p&gt;The short answer is that there is no &quot;exact&quot; meaning. Ideally, additional axioms are introduced to ensure that a certain function (or any mathematical object, for that matter) is &quot;well-behaved&quot; which, in effect, makes analysis easier. So, the meaning of &quot;well-behaved&quot; should be derived from those specific additional axioms.&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-21T00:40:08.307" CommentCount="6" />
  <row Id="216" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T00:41:36.323" Score="4" ViewCount="0" Body="&lt;p&gt;I think ideas like cardinality are suitable for getting kids interested in math -- the notion of a one-to-one correspondence is easy to grasp, and the proof of the uncountability of the reals is great fun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, elementary number theory can be useful.  This is the sort of thing that would be especially useful for, say, contest math, which is where a lot of kids get interested in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, popular math books can be a good source of ideas.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T00:41:36.323" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="1" />
  <row Id="217" PostTypeId="2" ParentId="155" CreationDate="2010-07-21T00:42:47.973" Score="0" ViewCount="0" Body="&lt;p&gt;Brian Conrad explains this in the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stanford.edu/~conrad/papers/finalint.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T00:42:47.973" CommentCount="1" />
  <row Id="218" PostTypeId="2" ParentId="177" CreationDate="2010-07-21T00:50:34.200" Score="7" ViewCount="0" Body="&lt;p&gt;Yes this will work; it's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Rejection_sampling&quot; rel=&quot;nofollow&quot;&gt;rejection sampling&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better is to generate a point in &lt;a href=&quot;http://mathworld.wolfram.com/DiskPointPicking.html&quot; rel=&quot;nofollow&quot;&gt;polar coordinates&lt;/a&gt; though:  pick θ from [0, 2π) and r&lt;sup&gt;2&lt;/sup&gt; from [0, R&lt;sup&gt;2&lt;/sup&gt;] (ie. multiply R by the square-root of a random number in [0, 1] - without the square-root it is non-uniform).&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-21T17:39:38.927" LastActivityDate="2010-07-21T17:39:38.927" CommentCount="1" />
  <row Id="219" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T00:53:34.957" Score="4" ViewCount="0" Body="&lt;p&gt;Besides all of these mathematics topics, I would contend that it is equally as important to show them the more human side of the subject.  Mathematics has a history and a culture that is accessible and exciting to some people.  Of course there are the exciting lives of Pascal and Galois, but there is also the story of Bourbaki (and the individual stories of their various members).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, introducing children to actual mathematicians/engineers/scientists is a great way to motivate their interest in the subject.  Let them see that mathematics has not been borne from the head of Zeus, that it is a living and breathing subject.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T00:53:34.957" CommunityOwnedDate="2010-08-07T22:32:44.343" CommentCount="4" />
  <row Id="220" PostTypeId="1" CreationDate="2010-07-21T00:56:14.780" Score="3" ViewCount="173" Body="&lt;p&gt;Okay, so hopefully this isn't too hard or off-topic. Let's say I have a very simple lowpass filter (something that smooths out a signal), and the filter object has a position variable and a cutoff variable (between 0 and 1). In every step, a value is put into the following bit of pseudocode as &quot;input&quot;: &lt;code&gt;position = position*(1-c)+input*c&lt;/code&gt;, or more mathematically, &lt;code&gt;f(n) = f(n-1)*(1-c)+x[n]*c&lt;/code&gt;. The output is the value of &quot;position.&quot; Basically, it moves a percentage of the distance between the current position and then input value, stores this value internally, and returns it as output. It's intentionally simplistic, since the project I'm using this for is going to have way too many of these in sequence processing audio in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the filter design, how do I construct a function that takes input frequency (where 1 means a sine wave with a wavelength of 2 samples, and .5 means a sine wave with wavelength 4 samples, and 0 is a flat line), and cutoff value (between 1 and 0, as shown above) and outputs the amplitude of the resulting sine wave? Sine wave comes in, sine wave comes out, I just want to be able to figure out how much quieter it is at any input and cutoff frequency combination.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-08-01T07:24:06.283" LastActivityDate="2010-08-01T07:24:06.283" Title="Simple lowpass frequency response" Tags="&lt;applications&gt;&lt;signal-processing&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="221" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T00:57:50.690" Score="6" ViewCount="0" Body="&lt;p&gt;Though I'm sure it's not unpopular, I don't think many people learn it early: Group Theory. It's a real nice area with a lot of cool math and some neat applications (like cryptography).&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-21T00:57:50.690" CommunityOwnedDate="2010-07-21T00:57:50.690" />
  <row Id="222" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T00:59:39.780" Score="2" ViewCount="0" Body="&lt;p&gt;I don't really think that &lt;strong&gt;graph theory&lt;/strong&gt; is a &quot;less popular mathematical subject,&quot; but I certainly wish I had been exposed to it earlier.&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-21T00:59:39.780" CommunityOwnedDate="2010-07-21T00:59:39.780" />
  <row Id="223" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T01:01:56.937" Score="3" ViewCount="0" Body="&lt;p&gt;Theory of computation, information theory and logic/foundation of mathematics are very interesting topics. I wish I knew them earlier. They are not unpopular(almost every university have a bunch of ToC people in CS depatment...) , but many math major I know have never touched them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They show you the limits of mathematics, computation and communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic shows there are things can't be proved from a set of axioms even if it's true--Godel's incompleteness theorem. There are other interesting theorems in foundation of mathematics. Like the independence of continuum hypothesis to ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theory of computation showed me things that's not computable. Problems that takes exponential time, exponential space, no matter what kind of algorithm you come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information theory proves the minimum amount of information required to reconstruct some other information. It pops up in unexpected places. There is a proof of there are infinite number of primes by information theory (Sorry I can't find it, I can only tell you it exists. I might find it later).&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-21T07:05:16.320" LastActivityDate="2010-07-21T07:05:16.320" CommunityOwnedDate="2010-07-21T01:01:56.937" CommentCount="4" />
  <row Id="224" PostTypeId="2" ParentId="154" CreationDate="2010-07-21T01:02:26.140" Score="2" ViewCount="0" Body="&lt;p&gt;Are real numbers &lt;em&gt;&quot;real&quot;&lt;/em&gt;?  It's not even computationally possible to compare two real numbers for equality!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, it is shown in Abstract Algebra courses that the idea of complex numbers arises naturally from the idea of real numbers - you could not say, for instance, that the real numbers are &lt;em&gt;valid&lt;/em&gt; but the complex numbers aren't (whatever your definition of &lt;em&gt;valid&lt;/em&gt; is...)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-21T01:02:26.140" CommentCount="1" />
  <row Id="225" PostTypeId="2" ParentId="213" CreationDate="2010-07-21T01:10:42.140" Score="3" ViewCount="0" Body="&lt;p&gt;I don't know what level of mathematics you are referring to, but here's my opinion after recently finishing my university's undergraduate curriculum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I would like to second Jan Gorzny's reply of Group Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, I wish that I had learned linear algebra earlier.  The topic usually has two semesters: matrix algebra and then an early proof-based introduction to vector spaces and linear transformations.  The real work in this topic can't begin until after both of these classes are completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wish I had been exposed to topology earlier than I was.  Of course there are two &quot;standard&quot; approaches here, and I suppose the approach that introduces general topology before advanced calculus would have better suited my tastes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good book that may give some more insight into the heavily debated area where your question lies: Thomas Garrity, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521792851&quot; rel=&quot;nofollow&quot;&gt;All the Mathematics You Missed But Need to Know for Graduate School&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T01:10:42.140" CommunityOwnedDate="2010-07-21T01:10:42.140" CommentCount="6" />
  <row Id="226" PostTypeId="2" ParentId="206" CreationDate="2010-07-21T01:23:22.517" Score="5" ViewCount="0" Body="&lt;p&gt;In general, we think of well-behaved functions as &lt;strong&gt;simpler&lt;/strong&gt;, somehow. In any field, we might want to limit ourselves to considering only well-behaved functions in order to avoid having to deal with nasty edge cases. And in each of these domains, the community is free to choose whatever definition of 'well-behaved' makes sense for them. A quick look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-behaved&quot; rel=&quot;nofollow&quot;&gt;wiki link&lt;/a&gt; that &lt;a href=&quot;http://math.stackexchange.com/questions/206/what-exactly-does-it-mean-for-a-function-to-be-well-behaved/215#215&quot; rel=&quot;nofollow&quot;&gt;e.James posted&lt;/a&gt; will show you the diversity in ideas of what it means to be well-behaved. I am no economist, so I will take for granted that the definition you put forth in your question is the one in common use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see &lt;strong&gt;twice-differentiable&lt;/strong&gt; as a reasonable requirement for a utility function to be 'well-behaved' is because the derivative of the utility function is marginal utility, and economists often care about the derivative of marginal utility. For example, if the second derivative of utility is negative, this means that the marginal utility has a negative derivative in other words, additional quantity of the good or service does not add utility as quickly. Also commonly referred to as diminishing returns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to be able to take the derivative even once, of course, we need the function to be &lt;strong&gt;continuous&lt;/strong&gt;. You probably don't need to worry about the formal definition here. Pencil test should work fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement for utility to be &lt;strong&gt;monotonic&lt;/strong&gt; means that it is either always increasing or always decreasing. In other words, that a particular good or service is either desirable or not. If 10 widgets were good, 20 must be better. Of course, as mentioned above, maybe not &lt;em&gt;that&lt;/em&gt; much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monotonic means that it will also be &lt;a href=&quot;http://en.wikipedia.org/wiki/Concave_function#Properties&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;quasi-concave&lt;/strong&gt;&lt;/a&gt;. (Except for weird stuff like a flat function) That is, they have at most one local maximum. We would prefer that functions be quasiconcave because we wish to avoid cases like the one below. It just makes it so much easier to optimize when you only have one possible maximum to worry about.&lt;br&gt;&#xA;&lt;img src=&quot;http://img.sparknotes.com/figures/E/eeb0fa64da5c2a1115cdf73c3ef2f0fc/derivapps1.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Globally non-satiated&lt;/strong&gt; someone else can talk about. I don't know enough to be sure I won't just be misleading you further.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T02:31:15.863" LastActivityDate="2010-07-21T02:31:15.863" CommentCount="5" />
  <row Id="227" PostTypeId="1" AcceptedAnswerId="231" CreationDate="2010-07-21T01:24:02.017" Score="5" ViewCount="563" Body="&lt;p&gt;I am doing a PhD in economics and I have the chance of taking one subject a semester in the maths department (I would like to do more, but &quot;unfortunately&quot; I have to work on my thesis). I want to have a plan so that I start with subjects I might know a bit but making sure I have all the basis, while at the same time ensuring I get the chance to sit in more advanced courses. My plan, for now, is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am skipping calculus I and II, because even though I don't know all of it, I know enough I think to be able to do the next course without trouble)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Linear Algebra  &lt;/li&gt;&#xA;&lt;li&gt;Real Analysis     &lt;/li&gt;&#xA;&lt;li&gt;Discrete Maths and Operations Research &lt;em&gt;or&lt;/em&gt;&lt;br&gt;&#xA;Engineering Maths &lt;em&gt;or&lt;/em&gt;&lt;br&gt;&#xA;Group Theory and Linear Algebra  &lt;/li&gt;&#xA;&lt;li&gt;Numerical &amp;amp; Symbolic Mathematics &lt;em&gt;or&lt;/em&gt;&lt;br&gt;&#xA;Vector Calculus&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you anyone give me a hand on making the decision or giving suggestions? In economics we use software such as matlab, lots of differentiation, Euler equations, difference equations, dynamic programming, statistical analysis (but for that I plan on taking Probability and Stochastic Modelling). &lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T02:00:53.190" LastActivityDate="2010-11-22T22:32:04.050" CommunityOwnedDate="2010-08-12T14:58:43.903" Title="Which 4 maths courses to take as an Economics PhD student?" Tags="&lt;economics&gt;&lt;soft-question&gt;&lt;education&gt;" AnswerCount="9" CommentCount="5" FavoriteCount="1" />
  <row Id="228" PostTypeId="1" CreationDate="2010-07-21T01:27:27.687" Score="1" ViewCount="182" Body="&lt;p&gt;&lt;i&gt;The sum of two Gaussian variables is another Gaussian.&lt;/i&gt;&lt;br&gt;&#xA;It seems natural, but I could not find a proof using Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a short way to prove this?&lt;br&gt;&#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Provided the two variables are independent.&lt;/p&gt;&#xA;" OwnerUserId="135" LastEditorUserId="135" LastEditorDisplayName="" LastEditDate="2010-07-28T00:53:59.567" LastActivityDate="2010-07-28T00:53:59.567" Title="Proof that the sum of two Gaussian variables is another Gaussian" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="229" PostTypeId="2" ParentId="196" CreationDate="2010-07-21T01:34:37.700" Score="8" ViewCount="0" Body="&lt;p&gt;That picture confuses things by making it look as though the red line is being &quot;unwound&quot; from the circle like paper towel being unwound from a roll. Our brains pick up on that, since it is a real-world example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both circles complete a single revolution, and both travel the same distance from left to right. If these really were rolls of paper towel, the smaller roll would have to spin faster (and therefore complete more than one full revolution) in order to lay out the same length of paper towel as the larger roll. Alternatively, if the two rolls &lt;em&gt;were&lt;/em&gt; spinning at the same rate, the free end of the strip of towel left behind by the smaller roll would also move to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the image is a kind of optical/mental illusion, and you're not going crazy &lt;code&gt;:)&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-07-21T01:34:37.700" />
  <row Id="230" PostTypeId="2" ParentId="227" CreationDate="2010-07-21T01:35:04.157" Score="3" ViewCount="0" Body="&lt;p&gt;You may consider focusing on Operations Research, and anything numerical.  I think you can pick up the tools of vector calculus as you go (which for your specialty will mostly be partial-differentiation, I think).  Maybe you can take two semesters of OR and a semester of something like Numerical Solutions to PDE, and one of Numerical Linear Algebra or Numerical Analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, I suppose all of this depends on what you plan to do with Econ.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T01:35:04.157" CommunityOwnedDate="2010-08-12T14:58:43.903" CommentCount="2" />
  <row Id="231" PostTypeId="2" ParentId="227" CreationDate="2010-07-21T01:37:29.953" Score="6" ViewCount="0" Body="&lt;p&gt;If you do take vector calculus, it's probably best taken before real analysis (where you'll learn many of the results from vector calc in a more general setting). If it were me, I'd take discrete math and operations research over the other two in (3), because they'll be useful for economics and because those subjects are the most different from linear algebra and real analysis. The group theory wouldn't, as far as I know, be tremendously applicable to economics except as it relates to analysis, and you'll get a sense of an algebraic way of thinking through linear. Numerical and symbolic math, if it includes numerical analysis, would be very relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A main criterion I am using here is that you'll want to learn more math as you go on in econ, and so the most important thing, for your ability to learn various things on your own later and for developing an appreciation for different kinds of math, is to take courses that cover many different kinds of mathematical thinking (including that which is more directly relevant to econ and other applied areas).&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T01:37:29.953" CommunityOwnedDate="2010-08-12T14:58:43.903" CommentCount="3" />
  <row Id="232" PostTypeId="2" ParentId="228" CreationDate="2010-07-21T01:55:04.623" Score="1" ViewCount="0" Body="&lt;p&gt;I don't know how I missed that one, indeed:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables&lt;/a&gt;&lt;br&gt;&#xA;Thanks Kaestur Hakarl!&lt;/p&gt;&#xA;" OwnerUserId="135" LastActivityDate="2010-07-21T01:55:04.623" CommentCount="1" />
  <row Id="233" PostTypeId="2" ParentId="123" CreationDate="2010-07-21T01:58:47.687" Score="9" ViewCount="0" Body="&lt;p&gt;On a map using the Mercator projection, the relationship between the latitude &lt;em&gt;L&lt;/em&gt; of a point and its &lt;em&gt;y&lt;/em&gt; coordinate on the map is given by &lt;em&gt;y&lt;/em&gt; = atanh(sin(&lt;em&gt;L&lt;/em&gt;)), where atanh is the inverse of the hyperbolic tangent function.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T01:58:47.687" />
  <row Id="234" PostTypeId="1" AcceptedAnswerId="510" CreationDate="2010-07-21T02:11:36.817" Score="5" ViewCount="220" Body="&lt;p&gt;Suppose there are two face down cards each with a positive real number and with one twice the other. Each card has value equal to its number. You are given one of the cards (with value x) and after you have seen it, the dealer offers you an opportunity to swap without anyone having looked at the other card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose to swap, your expected value should be the same, as you still have a 50% chance of getting the higher card and 50% of getting the lower card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the other card has a 50% chance of being 0.5x and a 50% chance of being 2x. If we keep the card, our expected value is x, while if we swap it, then our expected value if 05.*0.5x+0.5*2x=1.25x, so it seems like it is better to swap. Can anyone explain this apparent contradiction.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T04:29:25.300" LastActivityDate="2010-07-22T23:44:39.753" Title="Card doubling paradox" Tags="&lt;paradox&gt;&lt;probability-theory&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="235" PostTypeId="2" ParentId="154" CreationDate="2010-07-21T02:16:46.523" Score="13" ViewCount="0" Body="&lt;p&gt;To the degree that anything actually &quot;exists&quot; in math, yes complex number exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you accept that groups, rings and fields exist, and that isomorphism of rings makes sense, complex numbers can be recognized as (isomorphic to) the subring (which happens to be a field) of the ring of 2 by 2 real matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generators of this subring are the matrices&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; |1  0|&#xA; |0  1|&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; |0 -1|&#xA; |1  0|&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which correspond to 1 and i in the normal notation of complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As people tend to accept that matrices exist, this may be a convincing argument. ;-)&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T02:16:46.523" CommentCount="3" />
  <row Id="236" PostTypeId="2" ParentId="227" CreationDate="2010-07-21T02:24:29.700" Score="1" ViewCount="0" Body="&lt;p&gt;During my graduate studies, I had a friend who was extremely interested in differential topology and focused on that in his work toward an MS in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his PhD work, he switched to Economics, but was still essentially doing Differential Topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Economics can get just as abstract as math!&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T02:24:29.700" CommunityOwnedDate="2010-08-12T14:58:43.903" />
  <row Id="237" PostTypeId="1" AcceptedAnswerId="238" CreationDate="2010-07-21T02:31:40.650" Score="5" ViewCount="201" Body="&lt;p&gt;Given a point's coordinates (x,y), what is the procedure for determining if it lies within a polygon whose vertices are (x1,y1), (x2,y2), ..., (xn,yn)?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T02:33:05.423" LastActivityDate="2010-07-21T09:24:09.863" Title="How do you determine if a point sits inside a polygon?" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="238" PostTypeId="2" ParentId="237" CreationDate="2010-07-21T02:43:53.087" Score="5" ViewCount="0" Body="&lt;p&gt;I will assume the polygon has no intersections between the edges except at corners. Call the point (x0, y0). First we determine whether we are on a line - this is simple using substitution and range checking. For the range checking, suppose we have a segment (x1,y1), (x2,y2). We check that x1&amp;lt;=x0&amp;lt;=x2 or x2&amp;lt;=x0&amp;lt;=x1 and do the same for y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we aren't on a line, we draw a ray from the point and count how many lines you intersect. If the line doesn't intersect a corner or contain a non-degenerate segment of an edge, then an odd number of intersections should mean you are inside and an even number mean outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we intersect with a corner, it is more difficult. Clearly, the ray could leave with it only intersecting a corner, but it could also pass through the corner without entering the polygon. One solution is to pick the ray so that it doesn't intersect a corner or go along a line - this should always be possible. We can also use cross product to determine whether we are crossing the lines at their corner or not (just check if the adjacent vertices are on the same side of the line or not).&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-21T02:52:26.640" LastActivityDate="2010-07-21T02:52:26.640" CommentCount="2" />
  <row Id="239" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T03:07:37.587" Score="2" ViewCount="0" Body="&lt;p&gt;When I was that age, I discovered Raymond Smullyan's classic logic puzzle books in the library (such as &lt;em&gt;What is the name of this book?&lt;/em&gt;), and really got into it. I remember my amazement when I first understood how a complicated logic puzzle could become trivial, just symbolic manipulation really, with the right notation.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T03:07:37.587" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="240" PostTypeId="1" AcceptedAnswerId="1688" CreationDate="2010-07-21T03:19:46.880" Score="5" ViewCount="425" Body="&lt;p&gt;I remember hearing several times the advice that, we should avoid using a proof by contradiction, if it is simple to convert to a direct proof or a proof by contrapositive.  Could you explain the reason?  Do logicians think that proofs by contradiction are somewhat weaker than direct proofs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To clarify the question.  I am wondering, if there is any reason that one would still continue looking for a direct proof of some theorem, although a proof by contradiction has already been found.  I don't mean improvements in terms of elegance or exposition, I am asking about logical reasons.  For example, in the case of axiom of choice, there is obviously reason to look for a proof that does not use the axiom of choice.  Is there a similar case for proofs by contradiction?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T05:10:26.290" LastActivityDate="2010-08-06T01:54:45.447" Title="Are the &quot;proofs by contradiction&quot; weaker than other proofs?" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="7" CommentCount="2" />
  <row Id="241" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:22:18.083" Score="0" ViewCount="0" Body="&lt;p&gt;Nearly always the direct proof is easier to understand, shorter, and more helpful!&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-07-21T03:22:18.083" CommentCount="6" />
  <row Id="242" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:28:42.423" Score="6" ViewCount="0" Body="&lt;p&gt;Sometimes you might want to know not just that there &lt;em&gt;exists&lt;/em&gt; something, you might want to know how to actually go about finding it (and related questions like how quickly you can find it).  Proofs by contradiction are non-constructive, while direct proofs are typically constructive in the sense that they actually construct an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the proof that there are infinitely many primes usually proceeds by contradiction.  However, you can make it a direct proof which gives the stronger result that the nth prime is less than e^{e^n}.  (This is a good exercise to work out for yourself, but you can also find it as Prop 1.1.3 in my &lt;a href=&quot;http://math.columbia.edu/~nsnyder/thesismain.pdf&quot; rel=&quot;nofollow&quot;&gt;senior thesis&lt;/a&gt; and probably many other places as well.)&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T03:28:42.423" />
  <row Id="243" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:31:45.497" Score="7" ViewCount="0" Body="&lt;p&gt;Most logicians consider proofs by contradiction to be equally valid, however some people are &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructivism_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;constructivists/intuitionists&lt;/a&gt; and don't consider them valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible reason that the constructivist point of view makes a certain amount of sense is that statements like the continuum hypothesis are independent of the axioms, so it's a bit weird to claim that it's either true or false, in a certain sense it's neither.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless constructivism is a relatively uncommon position among mathematicians/logicians.  However, it's not considered totally nutty or beyond the pale.  Fortunately, in practice most proofs by contradiction can be translated into constructivist terms and actual constructivists are rather adept at doing so.  So the rest of us mostly don't bother worrying about this issue, figuring it's the constructivists problem.&lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="8" LastEditorDisplayName="" LastEditDate="2010-07-21T03:44:03.817" LastActivityDate="2010-07-21T03:44:03.817" CommentCount="3" />
  <row Id="244" PostTypeId="2" ParentId="83" CreationDate="2010-07-21T03:31:49.727" Score="0" ViewCount="0" Body="&lt;p&gt;Speaking from my own experience with elementary mathematics, yours is a very hard question to answer because there is little in the elementary math curriculum worth getting excited about. Before students are going to get excited about math, the math curriculum has to be changed so that the process of doing mathematics is made to be engaging and thought provoking - not tedious. I think every teacher of elementary mathematics needs to watch &lt;a href=&quot;http://www.ted.com/talks/lang/eng/dan_meyer_math_curriculum_makeover.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; video about how math education can and should be reformed.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T03:31:49.727" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="245" PostTypeId="2" ParentId="240" CreationDate="2010-07-21T03:39:26.747" Score="0" ViewCount="0" Body="&lt;p&gt;At first this seems like a silly question - after all isn't the point of a mathematical proof to be a proof and hence to be beyond question. But of course, to prove anything we need assumptions and &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructivism_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;some people&lt;/a&gt; do disagree with the axioms commonly used by mathematicians. I don't have much knowledge of this view, but I am sure they have theorems that show that contradiction like proofs are valid (given their axioms) within certain conditions. I would recommend going along with what everyone else does and treating &quot;proofs by contradiction&quot; as equally valid, unless you have investigated the Constructivism view and you decide that they are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether they are clearer, that will depend on the actual proof. Sometimes the clearest way to make a proof is to start from the assumptions and see what they are really saying and why that is going to lead to a contradiction. The most illustrative proof depends on the circumstances.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T03:39:26.747" />
  <row Id="246" PostTypeId="2" ParentId="150" CreationDate="2010-07-21T03:42:05.877" Score="2" ViewCount="0" Body="&lt;p&gt;Actually, in some sense, almost all of the continuous functions are nowhere differentiable:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function#Density_of_nowhere-differentiable_functions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Weierstrass_function#Density_of_nowhere-differentiable_functions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-07-21T03:42:05.877" />
  <row Id="247" PostTypeId="2" ParentId="194" CreationDate="2010-07-21T03:59:29.613" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Sieve of Eratosthenes&lt;/a&gt; is a very intuitive visual representation of why the frequency of prime numbers goes down as you go further out on the number line.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T03:59:29.613" CommentCount="3" />
  <row Id="248" PostTypeId="2" ParentId="206" CreationDate="2010-07-21T04:19:29.520" Score="15" ViewCount="0" Body="&lt;p&gt;In the sciences (as opposed to in mathematics) people are often a bit vague about exactly what assumptions they are making about how &quot;well-behaved&quot; things are.  The reason for this is that ultimately these theories are made to be put to the test, so why bother worrying about exactly which properties you're assuming when what you care about is functions coming up in real life which are probably going to satisfy all of your assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is particularly ubiquitous in physics where it is extremely common to make heuristic assumptions about well-behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in mathematics we do this sometimes.  When people say something is true for n sufficiently large, they often won't bother writing down exactly how large is sufficiently large as long as it's clear from context how to work it out.  Similarly, in an economics paper you could read through the argument and figure out exactly what assumptions they need, but it makes it easier to read to just say &quot;well-behaved.&quot;&lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="8" LastEditorDisplayName="" LastEditDate="2010-07-21T04:24:47.783" LastActivityDate="2010-07-21T04:24:47.783" />
  <row Id="249" PostTypeId="2" ParentId="234" CreationDate="2010-07-21T04:21:47.443" Score="-1" ViewCount="0" Body="&lt;p&gt;I understand your problem since Kaestur explained that you are talking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Two_envelopes_problem&quot; rel=&quot;nofollow&quot;&gt;two envelopes paradox&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a player is given a card and he knows the value of his, then the “expected value” reasoning makes sense. For example, if you are holding a card with a value 10 it makes sense to swap your card because there is a 50-50 chance you will either gain 10 or lose 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in your question, the player doesn't know the value on his or her card when the question of swapping is relevant. Therefore, in your question “expected value” reasoning cannot be applied. It is true that the player stands to either double the value of his or her card or half that value, but since the value of the card is unknown, the terms half and doubled are ill-defined. In particular, if the player is dealt the lower card of value x, a switch will increase his value by x. If the player is dealt the higher card of value 2x, a switch will decrease his value by x. Therefore, there is a fifty fifty chance that he will gain the value x or will lose the value x and therefore there is no incentive to switch cards.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T04:21:47.443" CommentCount="1" />
  <row Id="250" PostTypeId="1" CreationDate="2010-07-21T04:24:20.250" Score="34" ViewCount="1860" Body="&lt;p&gt;The following is a quote from &lt;em&gt;Surely you're joking, Mr. Feynman&lt;/em&gt; .  The question is: are there any interesting theorems that you think would be a good example to tell Richard Feynman, as an answer to his challenge?  Theorems should be totally counter-intuitive, and be easily translatable to everyday language.  (Apparently Banach-Tarski paradox was not a good example.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then I got an idea.  I challenged&#xA;  them: &quot;I bet there isn't a single&#xA;  theorem that you can tell me - what&#xA;  the assumptions are and what the&#xA;  theorem is in terms I can understand -&#xA;  where I can't tell you right away&#xA;  whether it's true or false.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It often went like this: They would&#xA;  explain to me, &quot;You've got an orange,&#xA;  OK? Now you cut the orange into a&#xA;  finite number of pieces, put it back&#xA;  together, and it's as big as the sun.&#xA;  True or false?&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;No holes.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Impossible!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;Ha! Everybody gather around! It's&#xA;  So-and-so's theorem of immeasurable&#xA;  measure!&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Just when they think they've got&#xA;  me, I remind them, &quot;But you said an&#xA;  orange! You can't cut the orange peel&#xA;  any thinner than the atoms.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;But we have the condition of&#xA;  continuity: We can keep on cutting!&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;No, you said an orange, so I&#xA;  assumed that you meant a real orange.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So I always won. If I guessed it&#xA;  right, great. If I guessed it wrong,&#xA;  there was always something I could&#xA;  find in their simplification that they&#xA;  left out. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="145" LastEditorUserId="122" LastEditorDisplayName="" LastEditDate="2010-07-21T06:58:27.397" LastActivityDate="2010-11-26T21:43:39.813" CommunityOwnedDate="2010-07-21T04:26:00.670" Title="A challenge by R. P.  Feynman: give counter-intuitive theorems that can be translated into everyday language" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;examples&gt;" AnswerCount="24" CommentCount="3" FavoriteCount="27" />
  <row Id="251" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:26:55.160" Score="5" ViewCount="0" Body="&lt;p&gt;Now that Wiles has done the job, I think that Fermat's Last Theorem may suffice. I find it a bit surprising still.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T04:26:55.160" CommunityOwnedDate="2010-07-21T04:26:55.160" CommentCount="3" />
  <row Id="252" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:32:06.670" Score="1" ViewCount="0" Body="&lt;p&gt;An infinite amount of coaches, each containing an infinite amount of people can be accommodated at &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Grand Hotel&lt;/a&gt;. Visual demonstration &lt;a href=&quot;http://demonstrations.wolfram.com/TheHilbertHotel/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T04:32:06.670" CommunityOwnedDate="2010-07-21T04:32:06.670" CommentCount="3" />
  <row Id="253" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:33:03.923" Score="6" ViewCount="0" Body="&lt;p&gt;All of three dimensional space can be filled up with an infinite &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_filling_curve&quot; rel=&quot;nofollow&quot;&gt;curve&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-07-21T05:02:27.473" LastActivityDate="2010-07-21T05:02:27.473" CommunityOwnedDate="2010-07-21T04:33:03.923" CommentCount="4" />
  <row Id="254" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T04:34:05.297" Score="5" ViewCount="0" Body="&lt;p&gt;My first thought is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ham_sandwich_theorem&quot; rel=&quot;nofollow&quot;&gt;ham sandwich theorem&lt;/a&gt;--given a sandwich formed by two pieces of bread and one piece of ham (these pieces can be of any reasonable/well-behaved shape) in any positions you choose, it is possible to cut this &quot;sandwich&quot; exactly in half, that is divide each of the three objects exactly in half by volume, with a single &quot;cut&quot; (meaning a single plane).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-21T04:34:05.297" CommunityOwnedDate="2010-07-21T04:34:05.297" />
  <row Id="255" PostTypeId="1" AcceptedAnswerId="259" CreationDate="2010-07-21T05:00:09.020" Score="14" ViewCount="213" Body="&lt;p&gt;Can someone give a simple explanation for why the series 1 + 1/2 + 1/3 + ... doesn't converge, but just grows very slowly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd prefer an easily comprehensible explanation rather than a rigorous proof of the type I could get from an undergraduate text book.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-21T05:23:55.037" LastActivityDate="2010-07-21T05:28:34.980" Title="Why does the series 1/1 + 1/2 + 1/3 + ... not converge?" Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="256" PostTypeId="2" ParentId="113" CreationDate="2010-07-21T05:06:14.363" Score="0" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Two_envelopes_problem&quot; rel=&quot;nofollow&quot;&gt;two envelopes problem&lt;/a&gt; is a good one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/234/card-doubling-paradox&quot; rel=&quot;nofollow&quot;&gt;http://math.stackexchange.com/questions/234/card-doubling-paradox&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/9037&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/9037&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-21T05:06:14.363" CommunityOwnedDate="2010-07-21T05:06:14.363" />
  <row Id="258" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T05:12:52.277" Score="19" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt; fits the bill pretty well. Almost everyone, including most mathematicians, answered it wrong on their first try, and some took a lot of convincing before they agreed with the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also very easy to explain it to people.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-21T05:12:52.277" CommunityOwnedDate="2010-07-21T05:12:52.277" CommentCount="4" />
  <row Id="259" PostTypeId="2" ParentId="255" CreationDate="2010-07-21T05:13:12.447" Score="17" ViewCount="0" Body="&lt;p&gt;Let's group the terms as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 1: 1/1                       (1 term) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 2: 1/2 + 1/3                 (2 terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 3: 1/4 + 1/5 + 1/6 + 1/7     (4 terms) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;group 4: 1/8 + 1/9 + ... + 1/15    (16 terms) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, group n contains 2^n terms.  But also, notice that the smallest element in group n is larger than 1/2^(n+1).  For example all elements in group 2 are larger than 1/2^2.  So the sum of the terms in each group is larger than 2^n * 1/2^(n+1) =  1/2.  Since there are infinitely many groups, and the sum in each group is larger than 1/2, it follows that the total sum is infinite.  &lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-07-21T05:13:12.447" CommentCount="4" />
  <row Id="260" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T05:23:28.717" Score="9" ViewCount="0" Body="&lt;p&gt;There are true statements in arithmetic which are unprovable.  Even more remarkably &lt;em&gt;there are explicit polynomial equations&lt;/em&gt; where it's unprovable whether or not they have integer solutions with ZFC! (We need ZFC + consistency of ZFC)&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-04T03:52:24.753" LastActivityDate="2010-08-04T03:52:24.753" CommunityOwnedDate="2010-07-21T05:23:28.717" CommentCount="6" />
  <row Id="261" PostTypeId="2" ParentId="255" CreationDate="2010-07-21T05:28:34.980" Score="6" ViewCount="0" Body="&lt;p&gt;This is not as good an answer as AgCl's, nonetheless people may find it interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're used to calculus then you might notice that the sum 1+1/2+1/3+...+1/n is very close to the integral from 1 to n of 1/x.  This definite integral is ln(n), so you should expect 1+1/2+1/3+...+1/n to grow like ln(n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of ln(x) is 1/x, which is probably harder than the original question.  Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T05:28:34.980" CommentCount="2" />
  <row Id="262" PostTypeId="1" CreationDate="2010-07-21T05:42:01.433" Score="11" ViewCount="1018" Body="&lt;p&gt;If you could go back in time and tell yourself to read a specific book at the beginning of your career as a mathematician, which book would it be?&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T06:22:24.537" LastActivityDate="2010-09-07T20:52:10.283" CommunityOwnedDate="2010-07-21T05:42:01.433" Title="What is the single most influential book every mathematician should read?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="19" FavoriteCount="11" />
  <row Id="263" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T05:47:22.827" Score="4" ViewCount="0" Body="&lt;p&gt;William Dunham's &quot;Journey through Genius.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, rather that is the book I read that made me want to be a mathematician.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T05:47:22.827" CommunityOwnedDate="2010-07-21T05:47:22.827" CommentCount="3" />
  <row Id="264" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T05:57:33.337" Score="2" ViewCount="0" Body="&lt;p&gt;Following Noah's lead I will mention;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Man Who Loved Only Numbers&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How to Read and Do Proofs&quot;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T05:57:33.337" CommunityOwnedDate="2010-07-21T05:57:33.337" />
  <row Id="265" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T05:57:57.783" Score="6" ViewCount="0" Body="&lt;p&gt;Polya's &quot;How To Solve It&quot;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-21T05:57:57.783" CommunityOwnedDate="2010-07-21T05:57:57.783" />
  <row Id="266" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T06:12:49.547" Score="45" ViewCount="0" Body="&lt;p&gt;Every simple closed curve that you can draw by hand will pass through the corners of some square.  The question was asked by Toeplitz in 1911, and has only been partially answered in 1989 by Stromquist.  As of now, the answer is only known to be positive, for the curves that can be drawn by hand. (i.e. the curves that are piecewise the graph of a continuous function) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the result beyond my intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gl7Ge.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For details, see &lt;a href=&quot;http://www.webpages.uidaho.edu/~markn/squares/&quot; rel=&quot;nofollow&quot;&gt;http://www.webpages.uidaho.edu/~markn/squares/&lt;/a&gt;  (the figure is also borrowed from this site)&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-11T01:39:38.820" LastActivityDate="2010-09-11T01:39:38.820" CommunityOwnedDate="2010-07-21T06:12:49.547" CommentCount="4" />
  <row Id="267" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T06:13:34.803" Score="1" ViewCount="0" Body="&lt;p&gt;Probability Theory the Logic of Science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or anything by Edwin T Jaynes.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-21T06:13:34.803" CommunityOwnedDate="2010-07-21T06:13:34.803" />
  <row Id="270" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T06:28:48.563" Score="3" ViewCount="0" Body="&lt;p&gt;This is an extremely broad question, especially given the wide variety of mathy people here, but I'll bite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HSM Coxeter's &lt;em&gt;Introduction to Geometry&lt;/em&gt; is a book that was very important to the development of my interest in mathematics and inclination towards its geometric aspects.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T06:28:48.563" CommunityOwnedDate="2010-07-21T06:28:48.563" CommentCount="1" />
  <row Id="271" PostTypeId="1" AcceptedAnswerId="273" CreationDate="2010-07-21T06:29:43.507" Score="12" ViewCount="167" Body="&lt;p&gt;Is there a way of taking a number known to limited precision (e.g. 1.644934) and finding out an &quot;interesting&quot; real number (e.g. $\pi^2/6$) that's close to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of something like Sloane's Online Encyclopedia of Integer Sequences, only for real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intended use would be: write a program to calculate an approximation to $\sum_{i=0}^\infty 1/n^2$, look up the answer (&quot;looks close to $\pi^2/6$&quot;) and then use the likely answer to help find a proof that the sum really is $\pi^2/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a thing exist?&lt;/p&gt;&#xA;" OwnerUserId="31" LastEditorUserId="31" LastEditorDisplayName="" LastEditDate="2010-07-26T19:57:37.337" LastActivityDate="2010-07-26T19:57:37.337" Title="Is there a real number lookup algorithm or service?" Tags="&lt;math-software&gt;&lt;experimental-mathematics&gt;" AnswerCount="3" />
  <row Id="272" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T06:32:05.717" Score="3" ViewCount="0" Body="&lt;p&gt;Nicolas Bourbaki's Éléments de mathématique (specifically Topologie Générale and Algèbre).  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T06:32:05.717" CommunityOwnedDate="2010-07-21T06:32:05.717" />
  <row Id="273" PostTypeId="2" ParentId="271" CreationDate="2010-07-21T06:34:58.427" Score="18" ViewCount="0" Body="&lt;p&gt;Try &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1.644934&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;. It actually does sequences as well.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T06:34:58.427" CommentCount="2" />
  <row Id="274" PostTypeId="2" ParentId="29" CreationDate="2010-07-21T06:57:37.633" Score="2" ViewCount="0" Body="&lt;p&gt;For desktop software, I use &lt;a href=&quot;http://www.speedcrunch.org/en_US/index.html&quot; rel=&quot;nofollow&quot;&gt;SpeedCrunch&lt;/a&gt;. Has a history, lots of mathematical functions, supports variable, etc.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-21T06:57:37.633" CommunityOwnedDate="2010-07-28T01:32:30.370" />
  <row Id="275" PostTypeId="1" CreationDate="2010-07-21T07:12:50.247" Score="3" ViewCount="692" Body="&lt;p&gt;I'm not a real Mathematician, just an enthusiast. I'm often in the situation where I want to learn some interesting Maths through a good book, but not through an actual Maths textbook. I'm also often trying to give people good Maths books to get them &quot;hooked&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question: What is a good book, for laymen, which teaches interesting Mathematics, but actually does it in a &quot;real&quot; way. For example, &quot;Fermat's Last Enigma&quot; doesn't count, since it doesn't actually feature any Maths, just a story, and most textbook don't count, since they don't feature a story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite example of this is &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/014014739X&quot; rel=&quot;nofollow&quot;&gt;Journey Through Genius&lt;/a&gt;&quot;, which is a brilliant combination of interesting storytelling and large amounts of actual Mathematics. It took my love of Maths to a whole other level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few more details on what I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The audience of &quot;laymen&quot; should be anyone who has the ability (and desire) to understand actual mathematics, but does not want to learn from a textbook. Obviously I'm thinking about myself here, as a programmer who loves mathematics, I love being exposed to real maths, but I'm not going to get into it seriously. That's why books that show actual maths, but give a lot more exposition (and much clearer explanations, especially of what the intuition should be) are great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say &quot;real maths&quot;, I'm talking about actual proofs, formulas, or other mathematical theories. Specifically, I'm not talking about philosophy, nor am I talking about books which &lt;strong&gt;only&lt;/strong&gt; talk about the history of mats (Simon Singh style), since they only talk &lt;em&gt;about&lt;/em&gt; maths, they don't actually show anything. William Dunham's books and Paul J. Nahin's books are good examples.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditorDisplayName="" LastEditDate="2010-07-21T22:25:14.570" LastActivityDate="2010-11-30T17:31:47.517" CommunityOwnedDate="2010-07-21T07:12:50.247" Title="Best Maths books for non-mathematicians" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="18" CommentCount="2" FavoriteCount="13" />
  <row Id="276" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T07:14:25.783" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/014014739X&quot; rel=&quot;nofollow&quot;&gt;Journey Through Genius&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://ecx.images-amazon.com/images/I/519AarC3x0L.jpg&quot; alt=&quot;cover&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brilliant combination of interesting storytelling and large amounts of actual Mathematics. It took my love of Maths to a whole other level.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T14:45:32.043" LastActivityDate="2010-11-07T14:45:32.043" CommunityOwnedDate="2010-07-21T07:14:25.783" CommentCount="1" />
  <row Id="277" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T07:20:11.040" Score="14" ViewCount="0" Body="&lt;p&gt;From &lt;a href=&quot;http://www.helium.com/items/895074-comparing-matlab-and-mathematica&quot; rel=&quot;nofollow&quot;&gt;helium.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The comparison between Matlab and&#xA;  Mathematica is NOT numeric versus&#xA;  symbolic as some claim. It is 'high&#xA;  level complete system' versus 'low&#xA;  level core with specialist toolboxes'.&#xA;  Mathematica can do all the numeric&#xA;  math and matrix work that Matlab can&#xA;  do, as fast and as accurately. Matlab&#xA;  is almost strictly a subset of the&#xA;  functionality of Mathematica.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Much of the wider functionality for&#xA;  Matlab is sold as additional&#xA;  toolboxes. e.g. Matlab has toolboxes&#xA;  for curve fitting, statistics,&#xA;  optimization, symbolic math, PDEs,&#xA;  Image acquisition etc but ALL of these&#xA;  areas of functionality are built in to&#xA;  Mathematica already. Aside from the&#xA;  cost and inconvenience of making you&#xA;  buy every function that you need&#xA;  separately, the Matlab design means&#xA;  that the core is inherently low level&#xA;  (since it cannot call on any of the&#xA;  toolbox functionality in case you have&#xA;  not purchased that toolbox), and many&#xA;  of these functions do not fit well&#xA;  together (e.g. no toolbox function can&#xA;  be used with symbolic parameters as&#xA;  they are all written using the&#xA;  numeric-only Matlab core).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Matlab's only comparative strength&#xA;  lies in some of its specialist&#xA;  toolboxes (e.g. control theory where&#xA;  it is dominant) but if you are not in&#xA;  that specialism, those are of no use.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some specific differences between the&#xA;  core systems:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Matlab is machine precision numeric only with some support for complex&#xA;  arithmetic - all Mathematica functions&#xA;  support machine precision and&#xA;  arbitrary precision real and complex&#xA;  arithmetic, exact arithmetic and&#xA;  symbolic computation.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab basic data model is 2D (matrices) with a pointer mechanism to&#xA;  emulate higher dimensions -&#xA;  Mathematica basic data model handles&#xA;  any dimensionality without extra work.&lt;/li&gt;&#xA;  &lt;li&gt;Many of Mathematica's functions switch automatically between different&#xA;  algorithms (with a manual overide) -&#xA;  in Matlab you must always choose the&#xA;  algorithm.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab requires you to manually linearize ODEs to a system of first&#xA;  order equations before it can solve&#xA;  them - Mathematica accepts any order&#xA;  ODE.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab can do integration up to only three dimensions (with&#xA;  triplequad) Mathematica handles&#xA;  n-dimensional integration for any n.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab's language is procedural only (rather like higher level&#xA;  FORTRAN) - Mathematica allows&#xA;  programming in procedural, functional,&#xA;  object oriented and rule based styles.&lt;/li&gt;&#xA;  &lt;li&gt;Matlab has text interface and a basic home grown IDE - Mathematica has&#xA;  text interface, document interface,&#xA;  GUI tools, and full Eclipse based IDE.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-08-16T07:10:58.270" LastActivityDate="2010-08-16T07:10:58.270" CommentCount="5" />
  <row Id="278" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T07:20:16.393" Score="5" ViewCount="0" Body="&lt;p&gt;I've been successful using Courant and Robbins, &lt;em&gt;What is Mathematics&lt;/em&gt;, on adults who have not had a math class fora few decades, but are open to the idea of learning more about math. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sections are too advanced for someone with only high school math, and many more will appear that way to the person at first, but do not actually rely on anything beyond high school math.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T07:20:16.393" CommunityOwnedDate="2010-07-21T07:20:16.393" />
  <row Id="279" PostTypeId="2" ParentId="271" CreationDate="2010-07-21T07:26:04.723" Score="2" ViewCount="0" Body="&lt;p&gt;Sometimes the decimal digits of numbers will appear in Sloane's On-Line Encyclopedia of Integer Sequences &lt;a href=&quot;http://www.research.att.com/~njas/sequences/&quot; rel=&quot;nofollow&quot;&gt;OIES&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A000796&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is the decimal expansion of pi.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-07-21T07:26:04.723" />
  <row Id="280" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T07:32:14.413" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_Elements&quot; rel=&quot;nofollow&quot;&gt;Euclid's Elements&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Principia_mathematica&quot; rel=&quot;nofollow&quot;&gt;Newton's Principia Mathematica&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally in the original languages of Ancient Greek and Latin respectively! No, just kidding. But they are &lt;em&gt;true&lt;/em&gt; classics that any accomplished mathematician should read at some point during their career. Not because they'll teach you something you don't already know, but they provide a unique insight into the mind of these giants.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-28T15:50:09.213" LastActivityDate="2010-07-28T15:50:09.213" CommunityOwnedDate="2010-07-21T07:32:14.413" CommentCount="6" />
  <row Id="281" PostTypeId="2" ParentId="6" CreationDate="2010-07-21T07:32:58.407" Score="5" ViewCount="0" Body="&lt;p&gt;Some good options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://fooplot.com/&quot; rel=&quot;nofollow&quot;&gt;FooPlot&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.onlinecharttool.com/graph.php&quot; rel=&quot;nofollow&quot;&gt;Online Chart Tool&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://graph.seriesmathstudy.com/&quot; rel=&quot;nofollow&quot;&gt;GraphFunc Tool&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-07-21T07:32:58.407" CommunityOwnedDate="2010-07-21T07:32:58.407" />
  <row Id="282" PostTypeId="1" CreationDate="2010-07-21T07:34:33.270" Score="0" ViewCount="151" Body="&lt;p&gt;What are the latest major discoveries in math? What areas and problems are considered important for overall advance of math?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given that most of those topics are most likely very advanced - what is the best learning path to approach them?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T07:36:43.060" LastActivityDate="2010-07-21T07:54:15.483" CommunityOwnedDate="2010-07-21T07:55:49.940" ClosedDate="2010-07-21T07:55:54.620" Title="What are the cutting-edge topics in modern math?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="283" PostTypeId="1" AcceptedAnswerId="293" CreationDate="2010-07-21T07:42:33.560" Score="2" ViewCount="258" Body="&lt;p&gt;Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems as though formerly 0 was considered in the set of natural numbers but now it seems more common to see definitions saying that the natural numbers are precisely the positive integers.  &lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-07-21T12:15:31.773" LastActivityDate="2010-07-23T14:53:20.897" Title="Is 0 a natural number?" Tags="&lt;terminology&gt;&lt;natural-number&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="284" PostTypeId="2" ParentId="90" CreationDate="2010-07-21T07:42:48.740" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.purplemath.com/lessons.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Purplemath&lt;/strong&gt;&lt;/a&gt; has a list for math lessons and tutoring. It's a list with various links and short reviews referring to tutoring and instructional resources.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-07-21T07:42:48.740" CommunityOwnedDate="2010-07-21T07:42:48.740" />
  <row Id="285" PostTypeId="2" ParentId="282" CreationDate="2010-07-21T07:45:02.323" Score="2" ViewCount="0" Body="&lt;p&gt;Higher category theory and Brave new algebraic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to approach these subjects, you should take a look at (roughly in this order):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebra by Lang&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Categories for the working mathematician by Mac Lane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Categories and Sheaves by Kashiwara-Schapira&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic Concepts of Enriched Category Theory by Kelly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model Categories by Hovey&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model Categories and their Localizations by Hirschhorn&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplicial Homotopy theory by Goerss-Jardine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes on Quasicategories by Joyal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher Topos Theory by Lurie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher Operads, Higher Categories by Leinster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to commutative algebra by Atiyah-MacDonald&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math 614 notes by Hochster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math 615 notes (2010) by Hochster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derived Algebraic Geometry (thesis) by Lurie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homotopical Algebraic Geometry (I&amp;amp;II) by Toen-Vezzosi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all seriousness though, this question is stupid.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T07:45:02.323" CommunityOwnedDate="2010-07-21T07:45:02.323" />
  <row Id="286" PostTypeId="2" ParentId="81" CreationDate="2010-07-21T07:47:29.243" Score="4" ViewCount="0" Body="&lt;p&gt;It isn't quite a blog, but &lt;a href=&quot;http://opinionator.blogs.nytimes.com/category/steven-strogatz/&quot; rel=&quot;nofollow&quot;&gt;Steven Strogatz's 15 part series&lt;/a&gt; for the New York Times was excellent.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T07:47:29.243" CommunityOwnedDate="2010-07-21T07:47:29.243" />
  <row Id="287" PostTypeId="1" AcceptedAnswerId="290" CreationDate="2010-07-21T07:53:17.747" Score="12" ViewCount="416" Body="&lt;p&gt;Below is a visual proof (!) that 32.5 = 31.5. How could that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://farm1.static.flickr.com/48/152036443_ca28c8d2a1_o.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-21T12:48:18.710" LastActivityDate="2010-07-21T12:48:18.710" Title="How come 32.5 = 31.5?" Tags="&lt;geometry&gt;&lt;fake-paradoxes&gt;" AnswerCount="3" CommentCount="20" FavoriteCount="3" />
  <row Id="288" PostTypeId="2" ParentId="282" CreationDate="2010-07-21T07:54:15.483" Score="2" ViewCount="0" Body="&lt;p&gt;There are a few places on the web that have lists of open problems in specific subfields. For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://garden.irmacs.sfu.ca/&quot; rel=&quot;nofollow&quot;&gt;The Open Problems Garden&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is not an endorsement, I don't know much about the site besides its existence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://maven.smith.edu/~orourke/TOPP/&quot; rel=&quot;nofollow&quot;&gt;The Open Problems Project&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is specific to discrete and computational geometry. The latter is a favorite of some people looking for nice research projects for undergraduates or high school students, as the problems tend to have statements, at least, accessible without too much background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The broader question you ask is one I won't address, by dint of intractability.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T07:54:15.483" />
  <row Id="289" PostTypeId="2" ParentId="287" CreationDate="2010-07-21T07:57:42.277" Score="9" ViewCount="0" Body="&lt;p&gt;The red and blue triangles are not similar (the ratios of the sides are 3/8 = 0.375 and 2/5 = 0.4 respectively), so the &quot;hypotenuse&quot; of your big triangle is not a straight line.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-21T07:57:42.277" />
  <row Id="290" PostTypeId="2" ParentId="287" CreationDate="2010-07-21T07:58:04.337" Score="26" ViewCount="0" Body="&lt;p&gt;It's an optical illusion - neither the first nor second set of blocks actually describes a triangle. The diagonal edge of the first is slightly concave and that of the second is slightly convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see clearly, look at the gradients of the hypotenuses of the red and blue triangles - they're not 'similar'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gradient of blue triangle hypotenuse = 2/5&lt;br&gt;&#xA;gradient of red triangle hypotenuse = 3/8  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these gradients are different, combining them in the ways shown in the diagram does not produce an overall straight (diagonal) line.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-21T12:36:30.200" LastActivityDate="2010-07-21T12:36:30.200" CommentCount="1" />
  <row Id="291" PostTypeId="2" ParentId="81" CreationDate="2010-07-21T07:58:35.410" Score="2" ViewCount="0" Body="&lt;p&gt;John D Cook writes &lt;a href=&quot;http://www.johndcook.com/blog/&quot; rel=&quot;nofollow&quot;&gt;The Endeavor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the MathWorks blogs: &lt;a href=&quot;http://blogs.mathworks.com/loren/&quot; rel=&quot;nofollow&quot;&gt;Loren on the Art of Matlab&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... a few more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://unimodular.net/blog/?p=185&quot; rel=&quot;nofollow&quot;&gt;eon&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cameroncounts.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Peter Cameron's Blog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.walkingrandomly.com/&quot; rel=&quot;nofollow&quot;&gt;Walking Randomly&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://topologicalmusings.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Todd and Vishal's Blog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Check their blogrolls for more)&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T07:58:35.410" CommunityOwnedDate="2010-07-21T07:58:35.410" />
  <row Id="292" PostTypeId="2" ParentId="287" CreationDate="2010-07-21T07:59:12.960" Score="6" ViewCount="0" Body="&lt;p&gt;No, this is not a proof of your statement. If you look very closely, you will see that the hypotenuse of the triangles aren't straight. You can also verify this algebraically by calculating the internal angles of the triangle using trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way to demonstrate the non-straightness of the hypotenuse is in the fact that the red and blue triangles aren't similar, which can be seen by the difference in the ratios of their width and height. For the blue triangle, this is 2/5 and for the red triangle this is 3/8. As they are both right angled triangles, these ratios would be the same for similar triangles and the fact that they aren't means the interior angles are different. &lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-21T07:59:12.960" />
  <row Id="293" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T08:02:01.160" Score="4" ViewCount="0" Body="&lt;p&gt;Simple answer: sometimes yes, sometimes no, it's usually stated (or impled by notation). From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_number&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematics, there are two&#xA;  conventions for the set of natural&#xA;  numbers: it is either the set of&#xA;  positive integers  {1, 2, 3, ...}&#xA;  according to the traditional&#xA;  definition; or the set of non-negative&#xA;  integers {0, 1, 2, ...} according to a&#xA;  definition first appearing in the&#xA;  nineteenth century.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Saying that, more often than not I've seen the natural numbers only representing the 'counting numbers' (i.e. excluding zero). This was the traditional historical definition, and makes more sense to me. Zero is in many ways the 'odd one out' - indeed, historically it was not discovered (described?) until some time after the natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-21T08:02:01.160" CommentCount="4" />
  <row Id="294" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T08:03:58.067" Score="2" ViewCount="0" Body="&lt;p&gt;There are the two definitions, as you say. However the set of strictly positive numbers being the natural numbers is actually the older definition. Inclusion of 0 in the natural numbers is a definition for them that first occurred in the 19th century. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Peano Axioms for natural numbers take 0 to be one though, so if you are working with these axioms (and a lot of natural number theory does) then you take 0 to be a natural number.&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-21T11:58:57.843" LastActivityDate="2010-07-21T11:58:57.843" CommentCount="2" />
  <row Id="295" PostTypeId="2" ParentId="11" CreationDate="2010-07-21T08:20:33.500" Score="-2" ViewCount="0" Body="&lt;p&gt;For any digit n = 0,1,2,3,4,5,6,7,8,9 the fraction&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n/9 = 0.nnnnnnnn... (periodically).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore 9/9 is equal to 0.999999999.... but also since 9 divided by 9 equals one,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9/9 = 0.9999999.... = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-21T08:20:33.500" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="2" />
  <row Id="296" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T08:25:25.423" Score="7" ViewCount="0" Body="&lt;p&gt;There is no &quot;official rule&quot;, it depends from what you want to do with natural numbers. Originally they started from 1 because 0 was not given the status of number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays if you see N&lt;sup&gt;+&lt;/sup&gt; (it's a superscript) you may be assured we are talking about numbers from 1 above; N is usually for numbers from 0 above.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T08:25:25.423" />
  <row Id="297" PostTypeId="1" AcceptedAnswerId="315" CreationDate="2010-07-21T08:41:58.600" Score="3" ViewCount="426" Body="&lt;p&gt;Are there any simple methods for calculating the digits of Pi? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T19:37:32.223" LastActivityDate="2010-09-04T01:48:27.217" Title="Simple numerical methods for calculating the digits of Pi" Tags="&lt;approximation&gt;&lt;numerical-methods&gt;&lt;pi&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="299" PostTypeId="2" ParentId="120" CreationDate="2010-07-21T08:47:03.843" Score="8" ViewCount="0" Body="&lt;p&gt;actually 1 was considered a prime number until the beginning of 20th century. Unique factorization was a driving force beneath its changing of status, since it's formulation is quickier if 1 is not considered a prime; but I think that group theory was the other force. Indeed I prefer to describe numbers as primes, composites and &lt;strong&gt;unities&lt;/strong&gt;, that is numbers whose inverse exists (so if we take the set of integer numbers Z, we have that 1 and -1 are unities and we still have unique factorization up to unities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can always amend the defition of a prime number and say it is a number with exactly two divisors: in this way 1 is not a prime by definition :-)&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-21T10:15:01.427" LastActivityDate="2010-07-21T10:15:01.427" />
  <row Id="300" PostTypeId="2" ParentId="198" CreationDate="2010-07-21T08:54:52.967" Score="1" ViewCount="0" Body="&lt;p&gt;when dealing with infinite sets, many propositions cannot be proved mithout using non-constructive proofs. Axiom of Choice (AC, in brief) or an equivalent proposition is required. Russell's paradox is not a problem per se, you just rule out certain collections of things as sets; Banach-Tarski paradox (you may take a ball, &quot;divide&quot; it in a finite number of parts, translate and rotate them and obtain &lt;strong&gt;two&lt;/strong&gt; balls equal to the first) may be worse indeed. But few mathematicians would prefer not to do a lot of maths because AC is not allowed!&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T08:54:52.967" />
  <row Id="301" PostTypeId="1" AcceptedAnswerId="312" CreationDate="2010-07-21T08:59:10.337" Score="3" ViewCount="96" Body="&lt;p&gt;The arithmetic hierarchy defines the &amp;Pi;_1 formulae of arithmetic to be formulae that are provably equivalent to a formula in &lt;a href=&quot;http://en.wikipedia.org/wiki/Prenex_normal_form&quot; rel=&quot;nofollow&quot;&gt;prenex normal form&lt;/a&gt; that only has universal quantifiers, and &amp;Sigma;_1 if it is provably equivalent to a prenex normal form with only existential quantifiers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A formula is &amp;Delta;_1 if it is both &amp;Pi;_1 and &amp;Sigma;_1.  These formulae are often called recursive: why?&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-21T11:10:04.480" LastActivityDate="2010-08-04T01:23:26.147" Title="Why are Delta-1 sentences of arithmetic called recursive?" Tags="&lt;logic&gt;&lt;proof-theory&gt;&lt;arithmetic-hierarchy&gt;&lt;recursive&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="302" PostTypeId="2" ParentId="237" CreationDate="2010-07-21T09:24:09.863" Score="1" ViewCount="0" Body="&lt;p&gt;Representing a polygon by its edge path might not be the most useful, especially if you want to ask about inclusion for many points.  Consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Polygon_triangulation&quot; rel=&quot;nofollow&quot;&gt;triangulating the polygon&lt;/a&gt;, which is trivial for convex polygons, and not difficult to find O(n log(n)) for hairier cases.  Then determining whether the point is in the polygon reduces to whether it is in anyone of the triangles, which is easy.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-21T09:24:09.863" CommentCount="1" />
  <row Id="303" PostTypeId="1" AcceptedAnswerId="310" CreationDate="2010-07-21T09:33:20.273" Score="1" ViewCount="95" Body="&lt;p&gt;Any homomorphism φ between the rings Z_18 and Z_15 is completely defined by φ(1). So from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 = φ(0) = φ(18) = φ(18 * 1) = 18 * φ(1) = 15 * φ(1) + 3 * φ(1) = 3 * φ(1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get that φ(1) is either 5 or 10. But how can I prove or disprove that these 2 are valid homomorphisms?&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-23T18:28:06.873" LastActivityDate="2010-07-23T18:28:06.873" Title="What are all the homomorphisms between the rings Z_18 and Z_15?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="304" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T09:52:00.850" Score="0" ViewCount="0" Body="&lt;p&gt;I remember all of my courses at University using only positive integers (not including 0) for the Natural Numbers. It's possible that they had come to an agreement amongst the Maths Faculty, but during at least two courses we generated the set of natural numbers in ways that wouldn't make sense if 0 was included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One involved the cardinality of Sets of Sets, the other defined the natural numbers in terms of the number 1 and addition only (0 and Negative Integers come into the picture later when you define an inverse to addition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result when teaching the difference between Integers and Natural Numbers I always define 0 as an integer that isn't a Natural Number.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2010-07-21T09:52:00.850" CommentCount="1" />
  <row Id="305" PostTypeId="2" ParentId="90" CreationDate="2010-07-21T10:20:17.880" Score="2" ViewCount="0" Body="&lt;p&gt;For basic mathematics &lt;a href=&quot;http://www.themathpage.com/Arith/arithmetic.htm&quot; rel=&quot;nofollow&quot;&gt;mathpage&lt;/a&gt; has a long course, including the list below. Don't be put off by the elementary school style lesson names, it does go into some depth with each one.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Lesson 1    Reading and Writing Whole Numbers&#xA;Lesson 2    The Meaning of Decimals&#xA;Lesson 3    Multiplying and Dividing&#xA;Lesson 4    More Elementary Addition&#xA;Lesson 5    Adding Whole Numbers and Decimals&#xA;Lesson 6    The Meaning of Subtraction&#xA;Lesson 7    Subtracting Whole Numbers and&#xA;Lesson 8    The Meaning of Multiplication&#xA;Lesson 9     Multiplying Whole Numbers&#xA;Lesson 10    The Meaning of Division&#xA;Lesson 11    Short Division&#xA;Lesson 12    Dividing Decimals  &#xA;Lesson 13    Percent with a Calculator  &#xA;Lesson 14    Parts of Natural Numbers 1  &#xA;Lesson 15    Parts of Natural Numbers 2  &#xA;Lesson 16    Ratio and Proportion  1  &#xA;Lesson 17    Ratio and Proportion  2  &#xA;Lesson 18    Proportionality  &#xA;Lesson 19    Proper Fractions, Mixed Numbers&#xA;Lesson 20    Unit Fractions  &#xA;Lesson 21    Equivalent Fractions  &#xA;Lesson 22    Lowest Common Multiple  &#xA;Lesson 23    Fractions into Decimals  &#xA;Lesson 24    Adding and Subtracting Fractions &#xA;Lesson 25    Multiplying Fractions &#xA;Lesson 26    The Meaning of Multiplying Fractions&#xA;Lesson 27    Percents are Ratios  &#xA;Lesson 28    Percent of a Number  &#xA;Lesson 29    What Percent?&#xA;Lesson 30    Percent Increase or Decrease  &#xA;Lesson 31    Prime Numbers  &#xA;Lesson 32    Greatest Common Divisor&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="54" LastActivityDate="2010-07-21T10:20:17.880" CommunityOwnedDate="2010-07-21T10:20:17.880" />
  <row Id="306" PostTypeId="2" ParentId="196" CreationDate="2010-07-21T10:33:49.587" Score="5" ViewCount="0" Body="&lt;p&gt;the smaller wheel does not just rotate, but also slides. If you had cogwheels instead of smooth wheels, you'd notice that movement is not possible.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T10:33:49.587" />
  <row Id="307" PostTypeId="1" AcceptedAnswerId="308" CreationDate="2010-07-21T11:23:38.983" Score="3" ViewCount="520" Body="&lt;p&gt;How can I show that (n-1)! is congruent to -1 (mod n) iff n is prime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-10-15T15:51:28.870" Title="Prove that (n-1)! = -1(mod n) iff n is prime" Tags="&lt;number-theory&gt;&lt;group&gt;" AnswerCount="4" />
  <row Id="308" PostTypeId="2" ParentId="307" CreationDate="2010-07-21T11:28:20.957" Score="10" ViewCount="0" Body="&lt;p&gt;$$n\text{ is prime if }(n-1)! \equiv -1 \pmod n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This direction is easy. If $n$ is composite, then there exists $k|n$ and $k\lt n$. So $k|(n-1)!$ and $k \equiv 1 \pmod n$. This means $k$ needs to divide $1$. So $n$ must be prime (or $1$, but we can eliminate this by substitution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n-1)! \equiv -1\text{ if }n\text{ is prime}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia contains &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson%27s_theorem#Proofs&quot; rel=&quot;nofollow&quot;&gt;two proofs&lt;/a&gt; of this result known as Wilson's theorem. The first proof only uses basic abstract algebra and so should be understandable with a good knowledge of modular arithmetic. Just in case, I prove below that each element &lt;code&gt;1, 2, ... n-1&lt;/code&gt; has a unique inverse &lt;code&gt;mod n&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They use the fact that integers &lt;code&gt;mod p&lt;/code&gt; form a group and hence that each element &lt;code&gt;x not congruent 0&lt;/code&gt; has a multiplicative inverse (a number y such that &lt;code&gt;xy congruent 1 (mod n)&lt;/code&gt;.&#xA;We show this as follows. Suppose &lt;code&gt;n not | x&lt;/code&gt;, for &lt;code&gt;n&lt;/code&gt; prime. From the uniqueness of prime factorisations, &lt;code&gt;xn&lt;/code&gt; is the first product of &lt;code&gt;x&lt;/code&gt;, after &lt;code&gt;0x&lt;/code&gt;, divisible by &lt;code&gt;n&lt;/code&gt; (use prime factorisation theorem). If we look at the series &lt;code&gt;kn&lt;/code&gt; in &lt;code&gt;mod n&lt;/code&gt;, this cycles and must have cycle length &lt;code&gt;n&lt;/code&gt;. Therefore, each element &lt;code&gt;x, 2x,... nx&lt;/code&gt; must be different modulo &lt;code&gt;n&lt;/code&gt;, including one, &lt;code&gt;y&lt;/code&gt;, with &lt;code&gt;xy congruent 1 mod n&lt;/code&gt;. Furthermore, due to the cycle length being &lt;code&gt;n&lt;/code&gt;, each only one of those elements will be an inverse. So every element has a unique inverse (although 1 and -1 are their own inverses).&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-10-15T15:51:28.870" CommentCount="1" />
  <row Id="309" PostTypeId="1" AcceptedAnswerId="327" CreationDate="2010-07-21T12:16:11.880" Score="4" ViewCount="129" Body="&lt;p&gt;In category theory, a subobject of X is defined as an object Y with a monomorphism, from Y to X. If A is a subobject of B, and B a subobject of A, are they isomorphic? It is not true in general that having monomorphisms going both ways between two objects is sufficient for isomorphy, so it would seem the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because I'm working through the exercises in Geroch's Mathematical Physics, and one of them asks you to prove that the relation &quot;is a subobject of&quot; is reflexive, transitive and antisymmetric. But it can't be antisymmetric if I'm right...&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-12-11T15:32:59.133" Title="If A is a subobject of B, and B a subobject of A, are they isomorphic?" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="310" PostTypeId="2" ParentId="303" CreationDate="2010-07-21T12:27:04.667" Score="7" ViewCount="0" Body="&lt;p&gt;If one has a homomorphism of two rings $R, S$, and $R$ has an identity, then the identity must be mapped to an idempotent element of $S$, because the equation $x^2=x$ is preserved under homomorphisms.  Now $5$ is not an idempotent element in $Z_{15}$, so the map generated by $1 \to 5$ is not a homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, 10 is an idempotent element of $Z_{15}$. In particular, the subring $T \subset Z_{15}$ generated by 10 has unit 10.  Since it is annihilated by 3, and consequently by 18, there is a unital homomorphism $Z_{18} \to T$ (i.e., mapping 1 to 10).  So your second map is a legitimate homomorphism of rings (composing with the injection $T \to Z_{15}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the point of this answer is to check that one of your maps preserves the relations of the two rings, while the other doesn't.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-21T13:29:30.147" LastActivityDate="2010-07-21T13:29:30.147" CommentCount="7" />
  <row Id="311" PostTypeId="1" AcceptedAnswerId="325" CreationDate="2010-07-21T12:30:36.140" Score="5" ViewCount="121" Body="&lt;p&gt;I've read about about &lt;a href=&quot;http://en.wikipedia.org/wiki/Higher_order_logic&quot; rel=&quot;nofollow&quot;&gt;higher-order logics&lt;/a&gt; (i.e. those that build on first-order predicate logic) but am not too clear on their applications. While they are capable of expressing a greater range of proofs (though never &lt;em&gt;all&lt;/em&gt;, by Godel's Incompleteness theorem), they are often said to be less &quot;well-behaved&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians generally seem to stay clear of such logics when possible, yet they are certainly necessary for prooving some more complicated concepts/theorems, as I understand. (For example, it seems the reals can only be constructed using at least 2nd order logic.) Why is this, what makes them less well-behaved or less useful with respect to logic/proof theory/other fields?&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-27T23:31:34.267" Title="Higher Order Logics" Tags="&lt;logic&gt;&lt;higher-order-logic&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="312" PostTypeId="2" ParentId="301" CreationDate="2010-07-21T12:31:56.217" Score="2" ViewCount="0" Body="&lt;p&gt;If a formula $\phi(x_1, ... x_n)$ is in both $\Sigma_1, \Prod_1$, then one can define a Turing machine to determine whether it is true or false. Namely, in parallel, search for a collection of parameters that makes true the existential formula, and search for a collection of formulas that makes false the universal formula. If the first happens, return true; if the second happens, return false.  One of these must exist, so the Turing machine always halts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The set of $x_1,...x_n$ such that $\phi(x_1, ... , x_n)$ is valid if $\phi$ belongs to  $\Sigma_1$ is, by contrast, is only recursively enumerable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, since the action of any Turing machine is simulable by existential formulas in first-order logic (i.e. there exists a number $k$ such that $M$ halts in $k$ steps), any language which is recursively enumerable can be expressed by existential formulas. Any language whose &lt;em&gt;complement&lt;/em&gt; is recursively enumerable can similarly be expressed by universal formulas (by the analog of deMorgan's laws). So any recursive language (i.e., one which is both recursively enumerable and whose complement is r.e.), can be expressed in both ways.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-22T11:34:02.793" LastActivityDate="2010-07-22T11:34:02.793" CommentCount="2" />
  <row Id="313" PostTypeId="2" ParentId="311" CreationDate="2010-07-21T12:35:23.993" Score="2" ViewCount="0" Body="&lt;p&gt;Second-order logic does not satisfy the completeness and compactness theorems. Here is a proof that the compactness theorem fails (which itself implies that the completeness theorem fails, because completeness implies compactness).  Namely, there is a second-order way (call it $F$) of expressing that a set is finite: every injective function on the set is surjective. This means that if $P$ is a relation on the product set $S \times S$ such that $P$ satisfies the conditions to be a function, and $P(x,y), P(z,y)$ imply $x=z$, then for all $w$ there is $q$ with $P(w,q)$. This statement is quantified over $P$ as well as the variables $x,y,z,w,q$ so is second-order and clearly expresses finiteness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we also have the statement $T_n$ &quot;there exist distinct $x_0, \dots x_n$ in the set,&quot; which is even first order, and states that the set has cardinality at least $n$.&#xA;So the conjunction of $F$ and any finite collection of the $T_n$ is satisfiable, but all of them together cannot be satisfied.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T12:35:23.993" CommentCount="2" />
  <row Id="314" PostTypeId="2" ParentId="271" CreationDate="2010-07-21T12:42:23.590" Score="13" ViewCount="0" Body="&lt;p&gt;I've long used Simon Plouffe's &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html&quot; rel=&quot;nofollow&quot;&gt;inverse symbolic calculator&lt;/a&gt; for this purpose.  It is essentially a searchable list of &quot;interesting&quot; numbers.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T12:42:23.590" />
  <row Id="315" PostTypeId="2" ParentId="297" CreationDate="2010-07-21T12:46:15.347" Score="7" ViewCount="0" Body="&lt;p&gt;One of the simplest is the series for arctangent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tan&lt;sup&gt;-1&lt;/sup&gt; x = x - x&lt;sup&gt;3&lt;/sup&gt;/3 + x&lt;sup&gt;5&lt;/sup&gt;/5 - x&lt;sup&gt;7&lt;/sup&gt;/7 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;pi;/4 = tan&lt;sup&gt;-1&lt;/sup&gt; 1, but that converges slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html&quot; rel=&quot;nofollow&quot;&gt;addition formula for the tangent&lt;/a&gt; (tan (a+b) = (tan a + tan b) / (1 - tan a tan b)) to break down &amp;pi;/4  to the sum of two angles and repeat; this can be used to come up with values for the arctangent that are smaller than 1 and therefore converge faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;exercise for the reader: show that &amp;pi;/4 = 2 tan&lt;sup&gt;-1&lt;/sup&gt; 2/11 + 3 tan&lt;sup&gt;-1&lt;/sup&gt; 1/7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then means that  &amp;pi; = 8 tan&lt;sup&gt;-1&lt;/sup&gt; 2/11 + 12 tan&lt;sup&gt;-1&lt;/sup&gt; 1/7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the arctangent formula with only a few terms and you will have pi to quite a few decimal places. (3 terms yields 3.14160, 4 terms yields 3.1415924, 5 terms yields 3.1415926592)&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-21T12:46:15.347" CommentCount="1" />
  <row Id="316" PostTypeId="1" AcceptedAnswerId="345" CreationDate="2010-07-21T12:47:23.143" Score="8" ViewCount="1867" Body="&lt;p&gt;When I tried to approximate $\int_0^1 (1-x^7)^{1/5} - (1-x^5)^{1/7} dx$, I kept getting answers that were really close to 0, so I think it might be true. But why? When I &lt;a href=&quot;http://integrals.wolfram.com/index.jsp?expr=%281-x%5E7%29%5E%281%2F5%29+-+%281-x%5E5%29%5E%281%2F7%29&amp;amp;random=false&quot; rel=&quot;nofollow&quot;&gt;ask Mathematica&lt;/a&gt;, I get a bunch of symbols I don't understand! &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T11:06:49.250" LastActivityDate="2010-08-03T09:08:01.803" Title="Why is $\int_0^1 (1-x^7)^{1/5} - (1-x^5)^{1/7} dx=0$?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="318" PostTypeId="2" ParentId="283" CreationDate="2010-07-21T12:56:10.100" Score="2" ViewCount="0" Body="&lt;p&gt;Unambiguous names for the set of nonnegative integers (0, 1, 2, 3, ...) are &lt;strong&gt;whole numbers&lt;/strong&gt;&#xA;Unambiguous names for the set of positive integers (1, 2, 3, ...) are &lt;strong&gt;counting numbers&lt;/strong&gt; and $\mathbb{N}^+$.&lt;br&gt;&#xA;As other answers (and your own question) indicate, &lt;strong&gt;natural numbers&lt;/strong&gt;, or $\mathbb{N}$, can have different meanings, depending on context. If which usage you mean is not clear from context, you are better off using one of the other terms.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-23T14:53:20.897" LastActivityDate="2010-07-23T14:53:20.897" CommentCount="3" />
  <row Id="319" PostTypeId="2" ParentId="8" CreationDate="2010-07-21T12:59:08.067" Score="8" ViewCount="0" Body="&lt;p&gt;A lot of people have mentioned Binet's formula.  But I suspect this is not the most practical way to compute the nth Fibonacci number for large n, because it requires either having a very accurate value of sqrt(5) and carrying around lots of decimal places (if you want to do floating-point arithmetic) or expanding large powers of (1+sqrt(5)) using the binomial formula.  The latter comes out to writing the Fibonacci number as a sum of binomial coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following formulas hold, though:&#xA;F(2n-1) = F(n)^2 + F(n-1)^2&#xA;F(2n) = (2*F(n-1) + F(n)) F(n)&#xA;which you can find derivations of in &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on Fibonacci numbers&lt;/a&gt;.  This lets you find F(k), for any k even or odd, in terms of two Fibonacci numbers with approximately half the index.  The result is faster than Binet's formula.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T12:59:08.067" />
  <row Id="320" PostTypeId="2" ParentId="307" CreationDate="2010-07-21T13:00:49.560" Score="2" ViewCount="0" Body="&lt;p&gt;to show that (p-1)! &amp;equiv; -1 (mod p) without explicitly use group theory, maybe the simplest path is: (the following assumes p is odd, but if p=2 then the result is immediate)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;given n &amp;ne; 0, all values n, 2n, ... (p-1)n are different mod p. Otherwise, if hn &amp;equiv; kn (mod p) then (h-k)n &amp;equiv; 0 (mod p) against the hyphotesis that p is prime.&lt;/li&gt;&#xA;&lt;li&gt;this means that each n has an inverse mod p, that is for each n there is a m such that mn &amp;equiv; 1 (mod p).&lt;/li&gt;&#xA;&lt;li&gt;the equation x&lt;sup&gt;2&lt;/sup&gt; &amp;equiv; 1 (mod p) may be written as (x+1)(x-1) &amp;equiv; 0 (mod p); therefore its only solutions are x &amp;equiv; 1 (mod p) and x &amp;equiv; -1 (mod p). For each other number n, an inverse m must exist (because of the pigeonhole principle) but m &amp;ne; n.&lt;/li&gt;&#xA;&lt;li&gt;we are nearly done. Let's couple every number from 2 to p-2 with its own inverse. Their product is 1 (mod p), so they don't count in the overall total. 1 does not count either; it remains just p-1, that is -1 (mod p) as requested.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T13:00:49.560" CommentCount="2" />
  <row Id="321" PostTypeId="2" ParentId="303" CreationDate="2010-07-21T13:01:37.873" Score="0" ViewCount="0" Body="&lt;p&gt;I will quote &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_homomorphism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; on the definition of a ring homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;More precisely, if R and S are rings,&#xA;  then a ring homomorphism is a function&#xA;  f : R → S such that&lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_homomorphism&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;f(a + b) = f(a) + f(b) for all a and b in R&lt;/li&gt;&#xA;  &lt;li&gt;f(ab) = f(a) f(b) for all a and b in R&lt;/li&gt;&#xA;  &lt;li&gt;f(1) = 1&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last requirement is being relaxed. We can then sub in our &lt;code&gt;f&lt;/code&gt;, in this case &lt;code&gt;f(a1)=5a&lt;/code&gt; or &lt;code&gt;f(a1)=10a&lt;/code&gt; and see if these relations hold. Given that these are small, finite and well understood groups, the problem is easy to solve from here.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T13:01:37.873" />
  <row Id="322" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T13:03:12.423" Score="1" ViewCount="0" Body="&lt;p&gt;T.W.Körner, &lt;em&gt;The Pleasures of Counting&lt;/em&gt;. It shows how mathematics is alive.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T13:03:12.423" CommunityOwnedDate="2010-07-21T13:03:12.423" />
  <row Id="323" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T13:07:21.883" Score="7" ViewCount="0" Body="&lt;p&gt;I think that a non-mathematician could appreciate T.W.Körner's book &lt;em&gt;The Pleasures of Counting&lt;/em&gt;; but I still believe that the collection of &quot;Mathematical Games&quot; columns from Martin Gardner are the very best thing.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="4079" LastEditorDisplayName="" LastEditDate="2010-11-30T17:31:47.517" LastActivityDate="2010-11-30T17:31:47.517" CommunityOwnedDate="2010-07-21T13:07:21.883" CommentCount="1" />
  <row Id="324" PostTypeId="1" AcceptedAnswerId="330" CreationDate="2010-07-21T13:09:57.197" Score="1" ViewCount="58" Body="&lt;p&gt;What is a unital homomorphism? Why are they important?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-23T18:28:26.733" LastActivityDate="2010-07-23T18:28:26.733" Title="Unital homomorphism" Tags="&lt;terminology&gt;&lt;definition&gt;&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="325" PostTypeId="2" ParentId="311" CreationDate="2010-07-21T13:10:04.170" Score="3" ViewCount="0" Body="&lt;p&gt;Unfortunately the term is ambiguous: there two kinds of semantics of higher-order languages, and only one is problematic.  Consider the language of second-order arithmetic, where there are quantifiers over both natural numbers and sets of natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is what Quine called &quot;set-theory in sheep's clothing&quot;: this is where  quantification over sets of natural numbers is defined to be over all the sets of number that can be posited.  It's the theory we use when we prove that there can be only one complete, totally ordered field.  It isn't really a logic, there's no complete notion of proof formalisation for it.  Wikipedia calls this &quot;standard semantics&quot;; I'm not sure if there is an authority for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's Henkin semantics, which uses the rule analogous to the lambda-calculus to define a  semantics for second-order quantifiers.  This can be seen as still in the realm of first-order logic, in that sense that the second-order system can be translated into a first-order system conserving provability.  This is how second-order arithmetic is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the theorems of the Henkin semantics will be &quot;theorems&quot; in the first.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-21T13:10:04.170" CommentCount="5" />
  <row Id="326" PostTypeId="1" AcceptedAnswerId="328" CreationDate="2010-07-21T13:17:16.040" Score="2" ViewCount="187" Body="&lt;p&gt;So you have 2 dice and you want to get at least a 1 or a 5 (on the dice not added).  How do you go about calculating the answer for this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question comes from the game farkle.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditorDisplayName="" LastEditDate="2010-07-21T18:43:07.317" LastActivityDate="2010-07-21T18:43:07.317" Title="Calculating the probability of two dice getting at least a 1 or a 5" Tags="&lt;probability-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="327" PostTypeId="2" ParentId="309" CreationDate="2010-07-21T13:20:59.917" Score="5" ViewCount="0" Body="&lt;p&gt;I don't think this can be true in general.  What if we just take the category consisting of two objects A, B and take maps f:A --&gt; B, g: B--&gt; A with no relations between the maps, but forcing associativity.  Then certainly f and g are monomorphisms but A and B are not isomorphic (since there are no relations between maps). &lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-21T13:20:59.917" CommentCount="8" />
  <row Id="328" PostTypeId="2" ParentId="326" CreationDate="2010-07-21T13:23:13.453" Score="11" ViewCount="0" Body="&lt;p&gt;Go backwards: Calculate the probability that &lt;em&gt;neither&lt;/em&gt; of them shows a 1 or 5. That means both show a 2, 3, 4, or 6. Thats $(4/6)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that &lt;em&gt;at least one&lt;/em&gt; shows a 1 or 5 is $1-(2/3)^2=5/9$.&lt;/p&gt;&#xA;" OwnerUserId="3" LastActivityDate="2010-07-21T13:23:13.453" CommentCount="5" />
  <row Id="329" PostTypeId="1" AcceptedAnswerId="483" CreationDate="2010-07-21T13:28:20.443" Score="12" ViewCount="1042" Body="&lt;p&gt;Which is the single best book for &lt;a href=&quot;http://en.wikipedia.org/wiki/Number_theory&quot; rel=&quot;nofollow&quot;&gt;Number Theory&lt;/a&gt; that everyone who loves Mathematics should read?&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T16:53:08.063" LastActivityDate="2011-01-05T20:25:02.063" CommunityOwnedDate="2010-07-21T14:53:06.763" Title="Best ever book on Number Theory" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;number-theory&gt;&lt;books&gt;&lt;reference-request&gt;" AnswerCount="14" CommentCount="4" FavoriteCount="9" />
  <row Id="330" PostTypeId="2" ParentId="324" CreationDate="2010-07-21T13:28:39.320" Score="0" ViewCount="0" Body="&lt;p&gt;A unital homomorphism between rings R and S is a ring homomorphism that sends the identity element of R to the identity element of S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Homomorphisms (between objects in any algebraic category like groups, rings, vector spaces, etc.) preserve the algebraic structure, and if you want a map between rings with an identity element, it is natural to require this to preserve this element (since it satisfies unique properties).&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-21T13:28:39.320" />
  <row Id="331" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T13:31:47.427" Score="9" ViewCount="0" Body="&lt;p&gt;I would still stick with Hardy and Wright, even if it is quite old.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T13:31:47.427" />
  <row Id="332" PostTypeId="2" ParentId="324" CreationDate="2010-07-21T13:32:48.360" Score="2" ViewCount="0" Body="&lt;p&gt;A lot of results about rings just won't work otherwise: for instance, a unital homomorphism of rings sends units to units.  A nonunital homomorphism doesn't have to do that.  Nonunital homomorphisms can be very degenerate, e.g. the zero homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason you want homomorphisms to preserve the unit is that this is how you get a map $\spec S \to \spec R$ from a ring-homomorphism $R \to S$.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T13:32:48.360" />
  <row Id="333" PostTypeId="2" ParentId="326" CreationDate="2010-07-21T13:38:29.280" Score="3" ViewCount="0" Body="&lt;p&gt;To visually see the answer given by balpha above, you could write out the entire set of dice rolls&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]&#xA;[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6]&#xA;[3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [3, 6]&#xA;[4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [4, 6]&#xA;[5, 1], [5, 2], [5, 3], [5, 4], [5, 5], [5, 6]&#xA;[6, 1], [6, 2], [6, 3], [6, 4], [6, 5], [6, 6]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Total number of possible dice rolls: 36&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dice rolls that contain 1 or a 5: 20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20/36 = 5/9&lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2010-07-21T13:38:29.280" />
  <row Id="334" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T13:49:51.470" Score="0" ViewCount="0" Body="&lt;p&gt;I think any book by John Allen Paulos would be something any Math enthusiast could enjoy and learn from.&lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2010-07-21T13:49:51.470" CommunityOwnedDate="2010-07-21T13:49:51.470" />
  <row Id="335" PostTypeId="2" ParentId="326" CreationDate="2010-07-21T13:57:25.867" Score="6" ViewCount="0" Body="&lt;p&gt;The other way to visualise this would be to draw a probability tree like so:&#xA;&lt;img src=&quot;http://img.skitch.com/20100721-xwruwx7qnntx1pjmkjq8gxpifs.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(apologies for my poor standard of drawing :) )&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-21T16:36:15.153" LastActivityDate="2010-07-21T16:36:15.153" />
  <row Id="336" PostTypeId="1" AcceptedAnswerId="338" CreationDate="2010-07-21T13:58:48.657" Score="6" ViewCount="227" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; Many (if not all) of the transformation matrices used in 3D computer graphics are 4x4, including the three values for &lt;code&gt;x&lt;/code&gt;, &lt;code&gt;y&lt;/code&gt; and &lt;code&gt;z&lt;/code&gt;, plus an additional term which usually has a value of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the extra computing effort required to multiply 4x4 matrices instead of 3x3 matrices, there must be a substantial benefit to including that extra fourth term, even though 3x3 matrices &lt;em&gt;should&lt;/em&gt; (?) be sufficient to describe points and transformations in 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why is the inclusion of a fourth term beneficial? I can guess that it makes the computations easier in some manner, but I would really like to know &lt;em&gt;why&lt;/em&gt; that is the case.&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T16:56:59.480" LastActivityDate="2010-07-21T16:56:59.480" Title="Why are 3D transformation matrices [4]x[4] instead of [3]x[3]?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="337" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T14:03:38.800" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;A concise introduction to the theory of numbers&lt;/em&gt; by Alan Baker (1970 Fields medalist) covers a lot of ground in less than 100 pages, and does so in a fluid way that never feels rushed. I love this little book.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T14:03:38.800" />
  <row Id="338" PostTypeId="2" ParentId="336" CreationDate="2010-07-21T14:04:20.653" Score="10" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;even though 3x3 matrices should (?) be sufficient to describe points and transformations in 3D space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, they aren't enough! Suppose you represent points in space using 3D vectors. You can transform these using 3x3 matrices. But if you examine the definition of matrix multiplication you should see immediately that multiplying a zero 3D vector by a 3x3 matrix gives you another zero vector. So simply multiplying by a 3x3 matrix can never move the origin. But translations and rotations do need to move the origin. So 3x3 matrices are not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tried to explain exactly how 4x4 matrices are used. But I hope I've convinced you that 3x3 matrices aren't up to the task and that something more is needed.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T14:04:20.653" CommentCount="2" />
  <row Id="339" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T14:12:45.597" Score="6" ViewCount="0" Body="&lt;p&gt;John Derbyshire's &lt;em&gt;Prime Obsession&lt;/em&gt; is about Riemann's hypothesis. One of the stated goals of the author is to explain what &quot;all non-trivial zeros of the zeta function have real part one-half&quot; means to readers who have no background in calculus. Odd-numbered chapters tell the story of how Riemann came to his hypothesis, and even-numbered chapters are more mathematical in nature.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T14:12:45.597" CommunityOwnedDate="2010-07-21T14:12:45.597" CommentCount="2" />
  <row Id="340" PostTypeId="2" ParentId="336" CreationDate="2010-07-21T14:20:30.210" Score="7" ViewCount="0" Body="&lt;p&gt;To follow up user80's answer, you want to get transformations of the form v --&gt; Av + b, where A is a 3 by 3 matrix (the linear part of transformation) and b is a 3-vector.  We can encode this transformation in a 4 x 4 matrix by putting A in the top left with three 0's below it and making the last column be (b,1).  Multiplying the 4-vector (v,1) with this matrix will give you (Av + b, 1).&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-21T16:12:21.147" LastActivityDate="2010-07-21T16:12:21.147" CommentCount="1" />
  <row Id="341" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T14:24:58.623" Score="1" ViewCount="0" Body="&lt;p&gt;On a slightly different angle, Matlab is widely used in engineering, while Mathematica is more used among mathematicians. If I had to learn just one, I'd go with the one most used in my field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica is also much cheaper for personal use (I think they have a licence for a few hundred dollars for home use, and even cheaper versions for students).&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2010-07-21T14:24:58.623" CommentCount="1" />
  <row Id="342" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T14:25:23.817" Score="4" ViewCount="0" Body="&lt;p&gt;I like Niven and Zuckerman, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471625469&quot; rel=&quot;nofollow&quot;&gt;Introduction to the Theory of Numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-21T14:25:23.817" />
  <row Id="343" PostTypeId="2" ParentId="336" CreationDate="2010-07-21T14:31:54.567" Score="7" ViewCount="0" Body="&lt;p&gt;I'm going to copy &lt;a href=&quot;http://stackoverflow.com/questions/2465116/understanding-opengl-matrices/2465290#2465290&quot;&gt;my answer from Stack Overflow&lt;/a&gt;, which also shows why 4-component vectors (and hence 4&amp;times;4 matrices) are used instead of 3-component ones.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In most 3D graphics a point is represented by a 4-component vector (x, y, z, w), where w = 1. Usual operations applied on a point include translation, scaling, rotation, reflection, skewing and combination of these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These transformations can be represented by a mathematical object called &quot;matrix&quot;. A matrix applies on a vector like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ a b c tx ] [ x ]   [ a*x + b*y + c*z + tx*w ]&#xA;| d e f ty | | y | = | d*x + e*y + f*z + ty*w |&#xA;| g h i tz | | z |   | g*x + h*y + i*z + tz*w |&#xA;[ p q r s  ] [ w ]   [ p*x + q*y + r*z +  s*w ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example, scaling is represented as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 2 . . . ] [ x ]   [ 2x ]&#xA;| . 2 . . | | y | = | 2y |&#xA;| . . 2 . | | z |   | 2z |&#xA;[ . . . 1 ] [ 1 ]   [ 1  ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and translation as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 1 . . dx ] [ x ]   [ x + dx ]&#xA;| . 1 . dy | | y | = | y + dy |&#xA;| . . 1 dz | | z |   | z + dz |&#xA;[ . . . 1  ] [ 1 ]   [   1    ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;One of the reason for the 4th component is to make a translation representable by a matrix.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of using a matrix is that multiple transformations can be combined into one via matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the purpose is simply to bring translation on the table, then I'd say (x, y, z, 1) instead of (x, y, z, w) and make the last row of the matrix always &lt;code&gt;[0 0 0 1]&lt;/code&gt;, as done usually for 2D graphics. In fact, the 4-component vector will be mapped back to the normal 3-vector vector via this formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ x(3D) ]   [ x / w ]&#xA;| y(3D) ] = | y / w |&#xA;[ z(3D) ]   [ z / w ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Homogeneous_coordinates#Use_in_computer_graphics&quot; rel=&quot;nofollow&quot;&gt;homogeneous coordinates&lt;/a&gt;. &lt;strong&gt;&lt;em&gt;Allowing this makes the perspective projection expressible with a matrix too,&lt;/em&gt;&lt;/strong&gt; which can again combine with all other transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, since objects farther away should be smaller on screen, we transform the 3D coordinates into 2D using formula&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(2D) = x(3D) / (10 * z(3D))&#xA;y(2D) = y(3D) / (10 * z(3D))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now if we apply the projection matrix&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 1 . .  . ] [ x ]   [  x   ]&#xA;| . 1 .  . | | y | = |  y   |&#xA;| . . 1  . | | z |   |  z   |&#xA;[ . . 10 . ] [ 1 ]   [ 10*z ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then the real 3D coordinates would become&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(3D) := x/w = x/10z&#xA;y(3D) := y/w = y/10z&#xA;z(3D) := z/w = 0.1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so we just need to chop the z-coordinate out to project to 2D.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-21T14:31:54.567" CommunityOwnedDate="2010-07-21T14:31:54.567" />
  <row Id="344" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T14:35:54.197" Score="9" ViewCount="0" Body="&lt;p&gt;Serre's &quot;A course in Arithmetic&quot; is pretty phenomenal.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T14:35:54.197" />
  <row Id="345" PostTypeId="2" ParentId="316" CreationDate="2010-07-21T14:51:01.790" Score="25" ViewCount="0" Body="&lt;p&gt;Note that if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y = \left(1 - x^7\right)^{1/5} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(1 - y^5\right)^{1/7} = x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means $(1-x^7)^{1/5}$ is the inverse function of $(1-x^5)^{1/7}$. In the graph, one will be the same as the other when reflected along the diagonal line y = x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, both functions &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;share the same range [0, 1] and domain [0, 1] and&lt;/li&gt;&#xA;&lt;li&gt;monotonically decreasing,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, the area under the graph in [0, 1] will be the same for both functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^1 \left(1-x^7\right)^{1/5} dx = \int_0^1 \left(1-y^5\right)^{1/7} dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grouping the two integrals yield the equation in the title.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-03T09:08:01.803" LastActivityDate="2010-08-03T09:08:01.803" CommentCount="1" />
  <row Id="346" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T14:52:45.437" Score="8" ViewCount="0" Body="&lt;p&gt;Similar to the Monty Hall problem, but trickier: at the latest Gathering 4 Gardner, Gary Foshee asked &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are assuming that births are equally distributed during the week, that every child is a boy or girl with probability 1/2, and that there is no dependence relation between sex and day of birth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: 13/27. This was in the news a lot recently, see for instance &lt;a href=&quot;http://news.bbc.co.uk/2/hi/programmes/more_or_less/8735812.stm&quot; rel=&quot;nofollow&quot;&gt;BBC News&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T14:52:45.437" CommunityOwnedDate="2010-07-21T14:52:45.437" CommentCount="4" />
  <row Id="347" PostTypeId="2" ParentId="250" CreationDate="2010-07-21T15:33:16.183" Score="14" ViewCount="0" Body="&lt;p&gt;You have two identical pieces of paper with the same picture printed on them. You put one flat on a table and the other one you crumple up (without tearing it) and place it on top of the first one. &lt;a href=&quot;http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Brouwer's fixed point theorem&lt;/a&gt; states that there is some point in the picture on the crumpled-up page that is directly above the same point on the bottom page. It doesn't matter how you place the pages, or how you deform the top one.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-07-21T15:33:16.183" CommunityOwnedDate="2010-07-21T15:33:16.183" CommentCount="5" />
  <row Id="348" PostTypeId="2" ParentId="57" CreationDate="2010-07-21T15:41:46.653" Score="1" ViewCount="0" Body="&lt;p&gt;It's not clear what the question is asking. Is it asking for logical relations among the properties? e.g. a relation that is transitive and symmetric MUST be reflexive (unless the relation holds between no two objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it asking for mathematical relations that have these properties? &quot;&gt;&quot; is transitive, asymmetric and total or complete (for all distinct x,y either xRy or yRx). It is a strict total order. &quot;greater or equal&quot; is transitive, complete, symmetric, reflexive. This is a weak total order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;x is a subset of y&quot; is reflexive and antisymmetric but not necessarily transitive, symmetric (although it can be either of those things if you restrict yourself to the right sort of sets). &quot;x is a proper subset of y&quot; is asymmetric and irreflexive (for all x, it is not the case that xRx), and can be transitive if you restrict yourself to the right sort of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in both these cases, one can define one of each pair of relations in terms of the other (plus a notion of identity or non-identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In logic the notion of &quot;x entails y&quot; is transitive, reflexive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation &quot;x has the same integer part as y&quot; over the real numbers is transitive, reflexive, symmetric. This is called an equivalence relation.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-21T15:41:46.653" />
  <row Id="349" PostTypeId="2" ParentId="81" CreationDate="2010-07-21T15:49:18.933" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://math-blog.com/&quot; rel=&quot;nofollow&quot;&gt;Math-blog&lt;/a&gt; is one I have in my google reader and I just found this one through google reader but it's a little over my head:&lt;a href=&quot;http://unapologetic.wordpress.com/2010/07/20/upper-and-lower-ordinate-sets/&quot; rel=&quot;nofollow&quot;&gt;The Unapologetic Mathematician&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="180" LastEditorDisplayName="" LastEditDate="2010-07-22T14:00:08.583" LastActivityDate="2010-07-22T14:00:08.583" CommunityOwnedDate="2010-07-21T15:49:18.933" CommentCount="1" />
  <row Id="350" PostTypeId="1" AcceptedAnswerId="354" CreationDate="2010-07-21T15:58:51.407" Score="2" ViewCount="148" Body="&lt;p&gt;I'm looking for course notes and assignments  and hopefully some example exams for Discrete Math, I'm taking a placement exam in the subject after having taken it 4 years ago. &lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-21T16:56:08.763" LastActivityDate="2010-09-13T18:42:45.433" CommunityOwnedDate="2010-09-13T20:09:02.853" Title="Where can I find a review of discrete math " Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;education&gt;&lt;discrete-mathematics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="351" PostTypeId="2" ParentId="135" CreationDate="2010-07-21T16:05:57.817" Score="3" ViewCount="0" Body="&lt;p&gt;If $a$ and $b$ are natural numbers, then $a^b$ is the number of ways you can make a sequence of length $b$ where each element in the sequence is chosen from a set of size $a$. You're allowed replacements. For example $2^3$ is the number of 3 digit sequences where each digit is zero or 1: 000, 001, 010, ..., 111.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is precisely one way to make a zero length sequence: one. So you'd expect $0^0=1$.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T16:05:57.817" CommentCount="1" />
  <row Id="352" PostTypeId="2" ParentId="57" CreationDate="2010-07-21T16:09:07.313" Score="1" ViewCount="0" Body="&lt;p&gt;I think you're thinking about this in the wrong way. Properties don't &quot;work&quot;, properties are things that are true for the given relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, you say that the a relation has the reflexive property if it satisfies the condition that all elements are related to themselves. Similarly, it has the symmetric property if for all a and b, if a is related to b then b is related to a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could simply say that &quot;∀a : a R a&quot; each time, but because this is something that happens often, this particular property has been given a name, i.e. reflexivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain types of specific relations are also given names. For instance, a partial order is reflexive, antisymmetric and transitive, and equivalence relations are reflexive, symmetric and transitive. Again, these are just names that aren't strictly needed, but they make it more convenient talking about these kinds of things.&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="157" LastEditorDisplayName="" LastEditDate="2010-07-21T16:22:11.827" LastActivityDate="2010-07-21T16:22:11.827" />
  <row Id="353" PostTypeId="2" ParentId="297" CreationDate="2010-07-21T16:13:22.677" Score="6" ViewCount="0" Body="&lt;p&gt;Some of the easiest algorithms to use are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bailey-Borwein-Plouffe_formula&quot; rel=&quot;nofollow&quot;&gt;spigot&lt;/a&gt; algorithms. They allow you to jump rapidly to the nth digit without computing the n-1 digits before. The catch is that they only really work in binary. &lt;a href=&quot;http://en.literateprograms.org/Pi_with_the_BBP_formula_%28Python%29&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; are some implementations in Python. You can see how short they are.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T16:13:22.677" CommentCount="1" />
  <row Id="354" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T16:14:24.860" Score="1" ViewCount="0" Body="&lt;p&gt;This is the discrete math course one in my school. It contain lecture notes, homework and previous exams. &lt;a href=&quot;http://www.cs.sunysb.edu/~cse547/&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.sunysb.edu/~cse547/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-21T16:57:39.213" LastActivityDate="2010-07-21T16:57:39.213" CommunityOwnedDate="2010-09-13T20:09:02.853" />
  <row Id="355" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T16:18:15.213" Score="6" ViewCount="0" Body="&lt;p&gt;One's that were suggested to me by my Calculus teacher in High School. Even my wife liked them and she hates math now:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1589880331&quot; rel=&quot;nofollow&quot;&gt;The Education of T.C. Mits: What&#xA;modern mathematics means to you&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1589880366&quot; rel=&quot;nofollow&quot;&gt;Infinity: Beyond the Beyond the&#xA;Beyond&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Written and illustrated(Pictures are great ;p) by a couple: Lillian R. Lieber, and Hugh Gray Lieber. These books were hard to find before because they went out of print but I have this new version and like it a lot. The books explains profound topics in a way that is graspable by anyone without being dumbed down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1589880331&quot; rel=&quot;nofollow&quot;&gt;Godel's proof&lt;/a&gt; is one I enjoyed. It's was a little hard to understand but there is nothing in this book that makes it inaccessible to someone without a strong math background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping with Godel in the title, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465026567&quot; rel=&quot;nofollow&quot;&gt;Godel, Escher, Bach: An Eternal Golden Braid&lt;/a&gt; while not just about math was a good read (a bit long ;p).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0060935588&quot; rel=&quot;nofollow&quot;&gt;The Music of the Primes: Searching to Solve the Greatest Mystery in Mathematics&lt;/a&gt;  It describes the Riemann Hypothesis and people who were involved with it somehow. My favorite part was learning about the people who attempted to solve it. Many I never heard off before this book. (Side not: I'll have to read pguertin suggestion, sounds in like a similar but more profound book).&lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="4122" LastEditorDisplayName="" LastEditDate="2010-11-30T16:31:08.247" LastActivityDate="2010-11-30T16:31:08.247" CommunityOwnedDate="2010-07-21T16:18:15.213" CommentCount="1" />
  <row Id="356" PostTypeId="1" CreationDate="2010-07-21T16:42:55.237" Score="3" ViewCount="252" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very interested in Computer Science (computational complexity, etc.). I've already finished a University course in the subject (using Sipser's &quot;Introduction to the Theory of Computation&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the basics, i.e. Turing Machines, Computability (Halting problem and related reductions), Complexity classes (time and space, P/NP, L/NL, a little about BPP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm looking for a good book to learn about some more advanced concepts. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T07:52:26.070" LastActivityDate="2010-12-04T19:02:05.233" CommunityOwnedDate="2010-07-21T19:11:11.363" Title="Best Intermediate/Advanced Computer Science book" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;&lt;reference-request&gt;" AnswerCount="4" CommentCount="11" FavoriteCount="3" />
  <row Id="357" PostTypeId="2" ParentId="329" CreationDate="2010-07-21T16:49:45.870" Score="2" ViewCount="0" Body="&lt;p&gt;Stewart&amp;amp;Tall's &quot;Algebraic Number Theory&quot; is great.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T16:49:45.870" CommunityOwnedDate="2010-07-21T16:49:45.870" />
  <row Id="358" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T16:50:55.913" Score="0" ViewCount="0" Body="&lt;p&gt;Possibly this may not really qualify as presenting much interesting maths, but I think Hardy's &lt;a href=&quot;http://math.boisestate.edu/~holmes/holmes/A%20Mathematician%27s%20Apology.pdf&quot; rel=&quot;nofollow&quot;&gt;A Mathematician's Apology&lt;/a&gt; should be on the must-read list.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-07-21T16:50:55.913" CommunityOwnedDate="2010-07-21T16:50:55.913" CommentCount="2" />
  <row Id="359" PostTypeId="2" ParentId="356" CreationDate="2010-07-21T16:56:49.387" Score="2" ViewCount="0" Body="&lt;p&gt;Papadimitriou's &lt;em&gt;Computational Complexity&lt;/em&gt; covers complexity theory at a higher level than Sipser, but has essentially no prerequisites.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T16:56:49.387" CommunityOwnedDate="2010-07-21T19:50:37.107" />
  <row Id="360" PostTypeId="2" ParentId="262" CreationDate="2010-07-21T17:07:54.750" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/A_Mathematician%27s_Apology&quot; rel=&quot;nofollow&quot;&gt;A Mathematician's Apology&lt;/a&gt; by G H Hardy. I did in fact read this in high school, and it raised my view of mathematics from a thing of utility to a thing of beauty and wonder. It inspired me to go on to study mathematics at Cambridge myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a pity that the &quot;introduction&quot; by C P Snow is longer than the original and contains a rather depressing view of Hardy's later life. I would recommend readers to skip the introduction altogether and concentrate on Hardy's own words.&lt;/p&gt;&#xA;" OwnerUserId="113" LastActivityDate="2010-07-21T17:07:54.750" CommunityOwnedDate="2010-07-21T17:07:54.750" CommentCount="2" />
  <row Id="361" PostTypeId="2" ParentId="153" CreationDate="2010-07-21T17:09:43.997" Score="1" ViewCount="0" Body="&lt;p&gt;If you have prior programming experience, Matlab will probably seem more intuitive. Mathematica's notions about structure, layout and order of execution can often be vexing to the coder mindset -- as can its set of operators, which we might charitably describe as &lt;em&gt;rich&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the notion that it is helpful for programs to resemble what a mathematician might scrawl on a piece of parchment, with a quill pen, is one you will either love or hate...&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-07-21T17:09:43.997" />
  <row Id="362" PostTypeId="1" AcceptedAnswerId="363" CreationDate="2010-07-21T17:12:50.780" Score="13" ViewCount="363" Body="&lt;p&gt;I am vaguely familiar with the broad strokes of the development of group theory, first when ideas of geometric symmetries were studied in concrete settings without the abstract notion of a group available, and later as it was formalized by Cayley, Lagrange, etc (and later, infinite groups being well-developed). In any case, it's intuitively easy for me to imagine that there was substantial lay, scientific, and artistic interest in several of the concepts well-encoded by a theory of groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a few of the corresponding names for who developed the abstract formulation of rings initially (Wedderburn etc.), but I'm less aware of the ideas and problems that might have given rise to interest in ring structures. Of course, now they're terribly useful in lots of math, and $\mathbb{Z}$ is a natural model for elementary properties of commutative rings, and I'll wager number theorists had an interest in developing the concept. And if I wanted noncommutative models, matrices are a good place to start looking. But I'm not even familiar with what the state of knowledge and formalization of things like matrices/linear operators was at the time rings were developed, so maybe these aren't actually good examples for how rings might have been motivated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone outline or point me to some basics on the history of the development of basic algebraic structures besides groups? &lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-07T20:34:33.857" Title="History of the Concept of a Ring" Tags="&lt;math-history&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="363" PostTypeId="2" ParentId="362" CreationDate="2010-07-21T17:22:25.990" Score="8" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Bill Dubuque has pointed out that much of this answer (specifically, the part about FLT) is essentially a mathematical urban legend, albeit a pervasive one. I cannot delete an accepted answer, so here is a &lt;a href=&quot;http://mathoverflow.net/questions/26613/papers-that-debunk-common-myths-in-the-history-of-mathematics/30272#30272&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to an answer of his on MO explaining it.&lt;br&gt;&#xA;Here is also a link to a related &lt;a href=&quot;http://mathoverflow.net/questions/34806/what-was-the-relative-importance-of-flt-vs-higher-reciprocity-laws-in-kummers-i&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's some of the history here in Bourbaki's &lt;em&gt;Commutative Algebra,&lt;/em&gt; in the appendix.  Basically, a fair bit of ring theory was developed for algebraic number theory.  This in turn was because people were trying to prove Fermat's last theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why's this? Let $p$ be a prime. Then the equation $x^p + y^p = z^p$ can be written as $\prod (x+\zeta_p^iy) = z^p$ for $\zeta_p$ a primitive $p$th root of unity.  All these quantities are elements of the ring $Z[\zeta_p]$.  So if $p&gt;3$ and there is unique factorization in the ring $Z[\zeta_p]$, it isn't terribly hard to show that this is impossible at least in the basic case where $p $ does not divide $xyz$ (and can be found, for instance, in Borevich-Shafarevich's book on number theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lame actually thought he had a proof of FLT via this argument. But he was wrong: these rings generally don't admit unique factorization.  So, it became a problem to study these &quot;generalized integers&quot; $Z[\zeta_p]$, which of course are basic examples of rings.  It wasn't until Dedekind that the right notion of unique factorization -- namely, factorization of ideals -- was found. In fact, the case of FLT I just mentioned generalizes to the case where $p$ does not divide the class number of $Z[\zeta_p]$ (the class number is the invariant that measures how far it is from being a UFD).  And, according to this &lt;a href=&quot;http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Ring_theory.html&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;, Dedekind was the first to define a ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article I linked to, incidentally, has a fair bit of additional interesting history.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-07T20:34:33.857" LastActivityDate="2010-08-07T20:34:33.857" CommentCount="11" />
  <row Id="364" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T17:51:22.493" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/Paul-J.-Nahin/e/B001HCS1XI/ref=sr_ntt_srch_lnk_1?_encoding=UTF8&amp;amp;qid=1279734093&amp;amp;sr=8-1&quot; rel=&quot;nofollow&quot;&gt;Paul Nahin&lt;/a&gt; has a number of accessible mathematics books written for non-mathematicians, the most famous being&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691146004&quot; rel=&quot;nofollow&quot;&gt;An Imaginary Tale: The Story of $\sqrt{-1}$&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691118221&quot; rel=&quot;nofollow&quot;&gt;Dr. Euler's Fabulous Formula (Cures Many Mathematical Ills!)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/Ian-Stewart/e/B000APQ9NM/ref=ntt_athr_dp_pel_1&quot; rel=&quot;nofollow&quot;&gt;Professor Ian Stewart&lt;/a&gt; also has many books which each give laymen overviews of various fields or surprising mathematical results&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465013023&quot; rel=&quot;nofollow&quot;&gt;Professor Stewart's Cabinet of Mathematical Curiosities&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465017754&quot; rel=&quot;nofollow&quot;&gt;Professor Stewart's Hoard of Mathematical Treasures&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0199562075&quot; rel=&quot;nofollow&quot;&gt;Cows in the Maze: And Other Mathematical Explorations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0631232516&quot; rel=&quot;nofollow&quot;&gt;Does God Play Dice? The New Mathematics of Chaos&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-21T19:17:49.260" LastActivityDate="2010-07-21T19:17:49.260" CommunityOwnedDate="2010-07-21T17:51:22.493" CommentCount="2" />
  <row Id="365" PostTypeId="1" CreationDate="2010-07-21T18:11:33.560" Score="5" ViewCount="108" Body="&lt;p&gt;I'm looking to find out if there's any easy way to calculate the number of ways to tile a $3 \times 2n$ rectangle with dominoes. I was able to do it with the two codependent recurrences&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(0) = g(0) = 1&#xA;f(n) = f(n-1) + 2g(n-1)&#xA;g(n) = f(n) + g(n-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where $f(n)$ is the actual answer and $g(n)$ is a helper function that represents the number of ways to tile a $3 \times 2n$ rectangle with two extra squares on the end (the same as a $3 \times 2n+1$ rectangle missing one square).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By combining these and doing some algebra, I was able to reduce this to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(n) = 4f(n-1) - f(n-2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which shows up as sequence &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001835&quot; rel=&quot;nofollow&quot;&gt;A001835&lt;/a&gt;, confirming that this is the correct recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways to tile a $2 \times n$ rectangle is the Fibonacci numbers because every rectangle ends with either a verticle domino or two horizontal ones, which gives the exact recurrence that Fibonacci numbers do. My question is, &lt;strong&gt;is there a similar simple explanation for this recurrence for tiling a $3 \times 2n$ rectangle&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-21T18:22:13.133" LastActivityDate="2010-07-21T18:40:01.863" Title="Tiling a 3 by 2n rectangle with dominoes" Tags="&lt;combinatorics&gt;&lt;tiling&gt;&lt;dominoes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="366" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T18:12:48.237" Score="2" ViewCount="0" Body="&lt;p&gt;When wanting to know about a particular mathematics subject, I often find that starting with the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_mathematics#Further_reading&quot; rel=&quot;nofollow&quot;&gt;further reading&lt;/a&gt;&quot; section of the relevant wikipedia page is a good way in.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-21T18:12:48.237" CommunityOwnedDate="2010-09-13T20:09:02.853" />
  <row Id="367" PostTypeId="2" ParentId="365" CreationDate="2010-07-21T18:40:01.863" Score="3" ViewCount="0" Body="&lt;p&gt;Here's my best shot at the sort of explanation you're asking for, although it's not nearly as clear as the 2-by-n case.  The negative sign makes combinatorial proofs difficult, so let's rearrange this as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(n) + f(n-2) = 4f(n-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you want to show that the number of n-tilings, plus the number of (n-2)-tilings, is four times the number of (n-1)-tilings. (An &quot;n-tiling&quot; is a tiling of a 3 by 2n rectangle by dominoes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bijective terms, then, we want a bijection between the set of n-tilings and (n-2)-tilings and the set of (n-1)-tilings, where the (n-1)-tilings are each tagged with the number 1, 2, 3, or 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an (n-1)-tiling, there are three &quot;obvious&quot; ways to obtain an n-tiling from it, namely by adding one of the three 1-tilings on the right end.  These generate tilings which have a vertical line passing all the way through,  two units from the right end; call these &quot;faulted&quot; tilings, and those which don't have a vertical line in that position &quot;faultless&quot;.&lt;br&gt;&#xA;So it suffices to show that the number of faultless n-tilings, plus the number of (n-2)-tilings, is the number of (n-1)-tilings.  It's easy to see that the number of faultless n-tilings is 2g(n-2); a faultless tilings must have a horizontal domino covering the second and third squares from the right in some row, and this assumption forces the placement of some other dominoes.  So we need 2g(n-2) + f(n-2) = f(n-1).  Shifting the indices, we need 2g(n-1) + f(n-1) = f(n), which you have already said is true.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T18:40:01.863" />
  <row Id="368" PostTypeId="1" AcceptedAnswerId="374" CreationDate="2010-07-21T19:18:25.310" Score="5" ViewCount="385" Body="&lt;p&gt;I know that the Fibonacci sequence can be described via the Binet's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was wondering if there was a similar formula for $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this possible?  If not, why not?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2010-09-11T05:43:04.290" Title="Is there a closed-form equation for $n!$?  If not, why not?" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="369" PostTypeId="2" ParentId="368" CreationDate="2010-07-21T19:24:55.437" Score="6" ViewCount="0" Body="&lt;p&gt;The relative error of &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt; gets arbitrarily small as n gets larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/4/6/c/46c33d496f99c39d9886c66fa12e655d.png&quot; alt=&quot;(picture)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is only an approximation, not a closed-form of $n!$&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-21T20:01:30.527" LastActivityDate="2010-07-21T20:01:30.527" CommentCount="3" />
  <row Id="370" PostTypeId="1" AcceptedAnswerId="377" CreationDate="2010-07-21T19:27:09.313" Score="11" ViewCount="950" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the best book or lecture notes on category theory?&lt;/p&gt;&#xA;" OwnerUserId="179" LastEditorUserId="179" LastEditorDisplayName="" LastEditDate="2010-07-24T10:25:04.913" LastActivityDate="2010-12-31T05:36:55.767" CommunityOwnedDate="2010-07-24T10:25:04.913" Title="Good book/lecture notes about category theory" Tags="&lt;soft-question&gt;&lt;books&gt;&lt;reference-request&gt;&lt;category-theory&gt;" AnswerCount="16" CommentCount="4" FavoriteCount="10" />
  <row Id="371" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T19:34:46.737" Score="1" ViewCount="0" Body="&lt;p&gt;if you don't mind shell out a good amount of money, &lt;em&gt;Concrete Mathematics&lt;/em&gt; by Graham, Knuth and Patashnik could be nice (I own the first edition)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T19:34:46.737" CommunityOwnedDate="2010-09-13T20:09:02.853" CommentCount="2" />
  <row Id="372" PostTypeId="2" ParentId="297" CreationDate="2010-07-21T19:35:23.600" Score="4" ViewCount="0" Body="&lt;p&gt;There is also the option of approximating $\pi$ using &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_integration&quot; rel=&quot;nofollow&quot;&gt;Monte Carlo integration&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is this: If we agree that the area of a circle is $\pi r^2$, for simplicity we build a circle of area $\pi$ by setting $r=1$.  Placing this circle wholly inside of another region of known area, preferably by inscribing it in a square of side length 2, then we have a ratio of the circle's area to the total area of the square  (in this example that ratio is $\frac{\pi}{4}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Monte Carlo method works by approximating areas based on the ratio of the number of sample points lying within our region of interest and the total number of sample points we choose to try.  If we spread a uniformly distributed sequence of $N$ points over our square of area 4, and call the number of points that land inside of the inscribed circle $p$, then we can say $\frac{\pi}{4} = \frac{p}{N}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My implementation of this in Matlab requires tens of thousands of test points to achieve 3.14159xxxx, but I have not tried it for low-discrepancy sequences, or any other uniformly distributed point sets.    &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-21T19:35:23.600" />
  <row Id="373" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T19:35:57.110" Score="8" ViewCount="0" Body="&lt;p&gt;Categories for the Working mathematician by Mac Lane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Categories and Sheaves by Kashiwara and Schapira&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-21T19:35:57.110" CommunityOwnedDate="2010-08-04T15:59:49.803" CommentCount="1" />
  <row Id="374" PostTypeId="2" ParentId="368" CreationDate="2010-07-21T19:36:10.433" Score="6" ViewCount="0" Body="&lt;p&gt;If you're willing to accept an integral as an answer, then $n! = \int_0^\infty t^n e^{-t} \: dt$.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-21T19:36:10.433" CommentCount="13" />
  <row Id="375" PostTypeId="1" AcceptedAnswerId="376" CreationDate="2010-07-21T19:43:27.840" Score="2" ViewCount="220" Body="&lt;p&gt;Context: Rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguably three questions in one, but since they all are properties of the multiplication, I'll &#xA;try my luck...&lt;/p&gt;&#xA;" OwnerUserId="174" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2010-07-22T22:38:08.670" LastActivityDate="2010-07-22T22:38:08.670" Title="Are x*0 = 0 and x*1 = x and -(-x) = x axioms?" Tags="&lt;algebra&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="376" PostTypeId="2" ParentId="375" CreationDate="2010-07-21T19:48:29.717" Score="9" ViewCount="0" Body="&lt;p&gt;I will assume this is in the context of rings (e.g. real numbers, integers, etc).  In this case, the axiom defining 0 is that x + 0 = x for all x.  x*0 = 0 is a result of this since we have x*0 = x*(0+0) = x*0 + x*0 which implies x*0 = 0 (canceling one of the x*0's).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing that for the second one you mean x*1 = x.  This is a definition (axiom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third one is a consequence of the definition of -x being the element such that x + (-x) = 0.  For then we have (-x) + x is also zero so that x is the negative of -x.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-21T19:48:29.717" CommentCount="4" />
  <row Id="377" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T20:18:39.363" Score="8" ViewCount="0" Body="&lt;p&gt;Lang's &lt;em&gt;Algebra&lt;/em&gt; contains a lot of introductory material on categories, which is really nice since it's done with constant motivation from algebra (e.g. coproducts are introduced right before the free product of groups is discussed). &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T20:18:39.363" CommunityOwnedDate="2010-08-04T15:59:49.803" CommentCount="2" />
  <row Id="378" PostTypeId="2" ParentId="350" CreationDate="2010-07-21T20:24:35.730" Score="1" ViewCount="0" Body="&lt;p&gt;When it comes to textbooks, the Kenneth Rosen text &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073229725&quot; rel=&quot;nofollow&quot;&gt;Discrete Mathematics and its Applications&lt;/a&gt; is highly recommended. I was first introduced to it at my university, but I've seen it cited in several places.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-07-21T20:24:35.730" CommunityOwnedDate="2010-09-13T20:09:02.853" CommentCount="2" />
  <row Id="379" PostTypeId="2" ParentId="375" CreationDate="2010-07-21T20:28:20.760" Score="5" ViewCount="0" Body="&lt;p&gt;The question is more profound than is initially seems, and is really about algebraic structures. The first question you have to ask yourself is &lt;em&gt;where you're working&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, addition and multiplication are defined on a &lt;em&gt;structure&lt;/em&gt;, which in this case  is a &lt;em&gt;set&lt;/em&gt; (basically a collection of &quot;things&quot;) with two &lt;em&gt;operators&lt;/em&gt; we call &lt;em&gt;addition&lt;/em&gt; (marked +) and &lt;em&gt;multiplication&lt;/em&gt; (marked · or × or * or whatever). If this structure holds some properties, which are sometimes called &lt;em&gt;axioms&lt;/em&gt;, then it is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Unit_ring&quot; rel=&quot;nofollow&quot;&gt;unit ring&lt;/a&gt;. The properties are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The set is closed under the operator +. That is, if a and b are in R, then a+b is also in R.&lt;/li&gt;&#xA;&lt;li&gt;The set has a member which we mark as 0. It has the properties that for every a in R, a+0 = 0 and 0+a = 0.&lt;/li&gt;&#xA;&lt;li&gt;The operation + is commutative: a+b = b+a.&lt;/li&gt;&#xA;&lt;li&gt;The operation + is associative: (a+b)+c = a+(b+c).&lt;/li&gt;&#xA;&lt;li&gt;Every member has an additive inverse: for every a in R there is some b in R such that a+b = 0 (we mark b as -a).&lt;/li&gt;&#xA;&lt;li&gt;The set is closed under the operator *. That is, if a and b are in R, then a*b is also in R.&lt;/li&gt;&#xA;&lt;li&gt;The set has a member which we mark as 1. It has the properties that for every a in R, a*1 = a and 1*a = a.&lt;/li&gt;&#xA;&lt;li&gt;The operation * is associative: (a*b) * c = a * (b*c).&lt;/li&gt;&#xA;&lt;li&gt;Multiplication is distributive over addition: a * (b+c) = a*b + a*c and (a+b) * c = a*b + a*c. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While this is a long list, and introduces the operator + which is not even explicitly mentioned in the question, these properties are quite natural. For example, the integers {..., -2, -1, 0, 1, 2, ...} we all know and love indeed form a ring. The real numbers also form a ring  (in fact they form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;field&lt;/a&gt;, which means they hold even more properties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regard to your question, the identity x * 1 = x (I assume that's what you meant) is in fact an axiom - it is axiom 7. However, the other two identities are results of the other axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First identity:&lt;/strong&gt; We use axioms 2 and 9 to get  0 * x = (0+0) * x = 0*x + 0*x  and then by adding -(0*x) (the additive inverse of 0*x, from axiom 5) to both sides,  0 = 0*x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second identity:&lt;/strong&gt; As stated in axiom 5, -(-x) is just a notation used which means &quot;the additive inverse of -x&quot;. To show that -(-x) = x we need to show that x is in fact the additive inverse of -x, or in other words that x + -x = 0 and -x + x = 0. But that's just what axiom 5 says, so we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last point: You might be wondering why did we have to go and introduce addition to answer a question about multiplication? Well, it so happens that without addition the other two identities are simply not true. For example, if we look at the positive integers {1, 2, 3, ...} with only multiplication, then there is no 0 there! Simply put, this is because the positive integers do not form a ring.&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-07-22T10:17:50.120" LastActivityDate="2010-07-22T10:17:50.120" />
  <row Id="380" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T20:35:21.180" Score="2" ViewCount="0" Body="&lt;p&gt;Arbib, &lt;em&gt;Arrows, Structures, and Functors: The Categorical Imperative&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More elementary than MacLane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know very much about this, but some stripes of computer scientist have taken an interest in category theory recently, and there are lecture notes floating around with that orientation.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-21T20:35:21.180" CommunityOwnedDate="2010-08-04T15:59:49.803" />
  <row Id="381" PostTypeId="1" AcceptedAnswerId="449" CreationDate="2010-07-21T20:40:41.727" Score="8" ViewCount="316" Body="&lt;p&gt;The fibonacci sequence is very well known, and is often explained with a story about how many rabbits there are after &lt;code&gt;n&lt;/code&gt; generations if they each produce a new pair every generation. Is there any &lt;em&gt;other&lt;/em&gt; reason you would care about the fibonacci sequence?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:16:39.697" LastActivityDate="2010-07-29T13:55:45.977" CommunityOwnedDate="2010-08-03T18:46:03.740" Title="Applications of the fibonacci sequence" Tags="&lt;big-list&gt;&lt;combinatorics&gt;&lt;applications&gt;" AnswerCount="15" CommentCount="2" FavoriteCount="1" />
  <row Id="382" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:50:30.707" Score="5" ViewCount="0" Body="&lt;p&gt;Yep, the solution by Matiyasevich to Hilbert's tenth problem relies heavily on showing exponentiation is Diophantine.  In other words, the relation R given by $R(x,y,z)$ if and only if $x^y = z$ can be expressed by a polynomial equation $T(x,y,z,w)=0$, such that $R(x,y,z)$ holds if and only if $T(x,y,z,w)=0$ has a solution with $w$ integral.  This is a very difficult and technical result, as well as rather counterintuitive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the Fibonacci numbers grow essentially exponentially. One of the lemmas in showing that exponentiation is Diophantine is to show that the Fibonacci sequence (and its variants) are Diophantine.  See the book by Matiyasevich on Hilbert's 10th problem.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-21T20:50:30.707" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="383" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:50:50.440" Score="0" ViewCount="0" Body="&lt;p&gt;They are sometimes used or occur in financial applications, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliot_wave&quot; rel=&quot;nofollow&quot;&gt;Elliot Wave Theory&lt;/a&gt;.  This seems more like magic than math to me, but I am not a trader.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2010-07-21T20:50:50.440" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="384" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:51:59.050" Score="4" ViewCount="0" Body="&lt;p&gt;It isn't exactly an &lt;em&gt;application&lt;/em&gt; as such, but the upper bound of the size of a subtree in a Fibonacci heap whose root is a node with degree k is F_(k+2) where F_n is the n'th Fibonacci number.&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="157" LastEditorDisplayName="" LastEditDate="2010-07-21T22:15:15.967" LastActivityDate="2010-07-21T22:15:15.967" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="1" />
  <row Id="385" PostTypeId="2" ParentId="375" CreationDate="2010-07-21T20:55:51.783" Score="0" ViewCount="0" Body="&lt;p&gt;Answering what you asked, the question is poorly formed because you didn't specify the theory you are talking about. To do that, well, you must define symbols, deduction rules... and axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering what you probably meant to ask, no. Those are properties that - and × have in the Z set, not axioms. They can be deduced from the operations themselves.&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2010-07-21T20:55:51.783" />
  <row Id="386" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T20:56:34.280" Score="1" ViewCount="0" Body="&lt;p&gt;They're useful in determining the amortized run time of the appropriately named Fibonacci Heap in computer science.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-21T20:56:34.280" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="387" PostTypeId="1" AcceptedAnswerId="388" CreationDate="2010-07-21T20:57:35.510" Score="6" ViewCount="156" Body="&lt;p&gt;I know that the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... diverges. I also know that the sum of the inverse of prime numbers 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... diverges too, even if really slowly since it's O(log log n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think I read that if we consider the numbers whose decimal representation do not have a certain digit (say, 7) and sum the inverse of these numbers, the sum is finite (usually between 19 and 20, it depends from the missing digit). Does anybody know the result, and some way to prove that the sum is finite?&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-21T21:08:50.333" LastActivityDate="2010-07-21T21:35:25.143" Title="sum of inverse of  numbers with certain terms omitted" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="388" PostTypeId="2" ParentId="387" CreationDate="2010-07-21T21:06:16.283" Score="4" ViewCount="0" Body="&lt;p&gt;EDIT: &lt;a href=&quot;http://mathworld.wolfram.com/KempnerSeries.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; might be what you're looking for. Found it from looking at the source below. They're called Kempner series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An article &lt;a href=&quot;http://www.jstor.org/stable/2974352&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (and cited below) says that one Dr. Kempner proved in 1914 that the series 1+ 1/2 + 1/3 + ..., with any term that has a 9 in the denominator removed, is convergent (though he doesn't say what it converges to in the introductory paragraph). The article goes on to generalize the result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Curious Convergent Series&#xA;Frank Irwin&#xA;The American Mathematical Monthly, Vol. 23, No. 5 (May, 1916), pp. 149-152&#xA;Published by: Mathematical Association of America&#xA;Stable URL: &lt;a href=&quot;http://www.jstor.org/stable/2974352&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2974352&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-07-21T21:06:16.283" CommentCount="2" />
  <row Id="389" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T21:07:51.460" Score="4" ViewCount="0" Body="&lt;p&gt;it pops up every now and then. For example, positions (F&lt;sub&gt;2n&lt;/sub&gt;, F&lt;sub&gt;2n+1&lt;/sub&gt;) are winning ones in &lt;a href=&quot;http://en.wikipedia.org/wiki/Wythoff%27s_game&quot; rel=&quot;nofollow&quot;&gt;Wythoff's Game&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-21T21:07:51.460" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="392" PostTypeId="1" AcceptedAnswerId="1093" CreationDate="2010-07-21T21:18:40.193" Score="11" ViewCount="1088" Body="&lt;p&gt;One of the first things ever taught in a differential calculus class:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The derivative of sin(x) is cos(x)&lt;/li&gt;&#xA;&lt;li&gt;The derivative of cos(x) is -sin(x)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leads to a rather neat (and convenient?) chain of derivatives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;sin(x)&#xA;cos(x)&#xA;-sin(x)&#xA;-cos(x)&#xA;sin(x)&#xA;...&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An analysis of the shape of their graphs confirms &lt;em&gt;some&lt;/em&gt; points; for example, when sin(x) is at a maximum, cos(x) is zero and moving downwards; when cos(x) is at a maximum, sin(x) is zero and moving upwards.  But these &quot;matching points&quot; only work for multiples of pi/4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us move back towards the original definition(s) of sine and cosine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the most basic level, sin(x) is defined as -- for a right triangle with internal angle x -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To generalize this to the domain of all real numbers, sin(x) was then defined as the y coordinate of a point on the unit circle that is an angle x from the positive x axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of cos(x) was then made the same way, but with adj/hyp and the x coordinate, as we all know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything about this &lt;strong&gt;basic&lt;/strong&gt; definition that allows someone to look at these definitions, alone, and think, &quot;Hey, the derivative of the sine function with respect to angle is the cosine function!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, from &lt;strong&gt;the unit circle definition alone&lt;/strong&gt;.  Or, even more amazingly, the &lt;strong&gt;right triangle definition alone&lt;/strong&gt;.  Ignoring graphical analysis of their plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, I am asking, essentially, &quot;Intuitively &lt;em&gt;why&lt;/em&gt; is the derivative of the sine the cosine?&quot;&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-21T22:19:59.910" LastActivityDate="2010-11-01T03:49:15.053" Title="Intuitive understanding of the derivatives of sin(x) and cos(x)" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="13" CommentCount="12" FavoriteCount="8" />
  <row Id="393" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T21:27:01.883" Score="0" ViewCount="0" Body="&lt;p&gt;This isn't exactly what you asked, but look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;Taylor series&lt;/a&gt; for the polynomials:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\sin x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\text{ for all } x\!&#xA;&#xA;\cos x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\text{ for all } x\! &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The relationships between the derivatives are clear from this.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-21T21:27:01.883" CommentCount="6" />
  <row Id="394" PostTypeId="2" ParentId="387" CreationDate="2010-07-21T21:35:25.143" Score="6" ViewCount="0" Body="&lt;p&gt;It is not very surprising that the sum is finite, since numbers without a 7 (or any other digit) get rarer and rarer as the number of digits increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;em&gt;S&lt;/em&gt; be the harmonic series with all terms whose denominator contains the digit &lt;em&gt;k&lt;/em&gt; removed. We can write &lt;em&gt;S&lt;/em&gt; =&lt;em&gt;S&lt;/em&gt;_1 + &lt;em&gt;S&lt;/em&gt;_2 + &lt;em&gt;S&lt;/em&gt;_3 + ..., where &lt;em&gt;S&lt;/em&gt;_i is the sum of all terms whose denominator contains exactly &lt;em&gt;i&lt;/em&gt; digits, all different from &lt;em&gt;k&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the number of &lt;em&gt;i&lt;/em&gt;-digit numbers that do not contain the digit k is 8 x 9^(&lt;em&gt;i&lt;/em&gt;-1) (there are 8 choices for the first digit, excluding 0 and k, and 9 choices for the other digits). [Well, if k=0 there are 9 choices for the first digit, but the proof still works.] So there are 8 x 9^(&lt;em&gt;i&lt;/em&gt;-1) numbers in the sum &lt;em&gt;S&lt;/em&gt;_i.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now each number in &lt;em&gt;S&lt;/em&gt;_i is of the form 1/&lt;em&gt;a&lt;/em&gt;, where &lt;em&gt;a&lt;/em&gt; is an &lt;em&gt;i&lt;/em&gt;-digit number. So &lt;em&gt;a&lt;/em&gt; &gt;= 10^(&lt;em&gt;i&lt;/em&gt;-1), which implies 1/&lt;em&gt;a&lt;/em&gt; &amp;lt;= 1/10^(&lt;em&gt;i&lt;/em&gt;-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;em&gt;S&lt;/em&gt;_i &amp;lt;= 8 x 9^(&lt;em&gt;i&lt;/em&gt;-1) / 10^(&lt;em&gt;i&lt;/em&gt;-1) = 8 x (9/10)^(&lt;em&gt;i&lt;/em&gt;-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;S&lt;/em&gt; = sum of &lt;em&gt;S&lt;/em&gt;_i &amp;lt;= sum of 8 x (9/10)^(&lt;em&gt;i&lt;/em&gt;-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a  geometric series of ratio 9/10 &amp;lt; 1, which converges. Since &lt;em&gt;S&lt;/em&gt; is a positive series bounded above by a converging series, &lt;em&gt;S&lt;/em&gt; converges. &lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-21T21:35:25.143" />
  <row Id="395" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T21:56:46.540" Score="3" ViewCount="0" Body="&lt;p&gt;As a Physics Major, I would like to propose an answer that comes from my understanding of seeing sine and cosine in the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In doing this, I will examine uniform circular motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the point-on-a-unit-circle definition of sine and cosine, we can say that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r(t) = &amp;lt; cos(t), sin(t) &amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is a proper parametric function to describe a point moving along the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider what the first derivate, in a physical context, should be.  The first derivative of position &lt;em&gt;should&lt;/em&gt; represent, ideally, the &lt;em&gt;velocity&lt;/em&gt; of the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a physical context, we would expect the velocity to be the line tangent to the direction of motion at a given time &lt;code&gt;t&lt;/code&gt;.  Following from this, it would be tangent to the circle at angle &lt;code&gt;t&lt;/code&gt;.  Also, because the angular velocity is constant, the magnitude of the velocity should be constant as well.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r'(t)     = &amp;lt; -sin(t), cos(t) &amp;gt;&#xA;|r'(t)|^2 = (-sin(t))^2 + cos(t)^2&#xA;|r'(t)|^2 = sin(t)^2 + cos(t)^2&#xA;|r'(t)|^2 = 1&#xA;|r'(t)|   = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As expected, the velocity is constant, so the derivatives of sine and cosine are behaving as they should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also think about what the &lt;strong&gt;direction&lt;/strong&gt; of the velocity would be, as well, compared to the position vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is &quot;cheating&quot; by the bounds of the question, but by visualizing the graph we can see that the velocity, by nature of being tangent to the circle, must be perpendicular to the position vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, then position * velocity = 0 (dot product).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                              r(t) * r'(t) = 0&#xA;  &amp;lt; cos(t), sin(t) &amp;gt; * &amp;lt; -sin(t), cos(t) &amp;gt; = 0&#xA;( cos(t) * -sin(t) ) + ( sin(t) * cos(t) ) = 0&#xA;              -sin(t)cos(t) + sin(t)cos(t) = 0&#xA;                                         0 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Life is good.  If we assume that the definition of cos(t) is -sin(t) and that the definition of sin(t) is cos(t), we find physical behavior &lt;strong&gt;exactly&lt;/strong&gt; like expected: a constant velocity that is always perpendicular to the position vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can take this further and look at the acceleration.  In Physics, we would call this the restoring force.  In a circle, what acceleration would have to exist in order to keep a point moving in a circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, in what direction would this acceleration have to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes little thought to arrive at the idea that acceleration would have to be center-seeking, and pointing towards the center.  So, if we can find that acceleration is &lt;strong&gt;in the opposite direction&lt;/strong&gt; as the position vector, the we can be almost certain about the derivatives of sine and cosine.  That is, their internal angle should be &lt;code&gt;pi&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                            r(t) * r''(t) = |r(t)| * |r''(t)| * cos(pi)&#xA;                            r(t) * r''(t) = |r(t)| * |r''(t)| * -1&#xA;&amp;lt; cos(t), sin(t) &amp;gt; * &amp;lt; -cos(t), -sin(t) &amp;gt; = |&amp;lt;cos(t),sin(t)&amp;gt;| * |&amp;lt;-cos(t),-sin(t)&amp;gt;| * -1&#xA;                    -cos(t)^2 + -sin(t)^2 = 1 * 1 * -1&#xA;               -1 * (cos(t)^2 + sin(t)^2) = -1&#xA;                                   -1 * 1 = -1&#xA;                                       -1 = -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-21T22:06:58.390" LastActivityDate="2010-07-21T22:06:58.390" />
  <row Id="396" PostTypeId="2" ParentId="381" CreationDate="2010-07-21T22:05:25.537" Score="4" ViewCount="0" Body="&lt;p&gt;Suppose you're writing a computer program to search a sorted array for a particular value. Usually the best method to use is a binary search. But binary search assumes it's the same cost to read from anywhere in the array. If it costs something to move from one array element to another, and the cost is proportional to how many array elements you need to skip over to get from one element you read to the next, then &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_search_technique&quot; rel=&quot;nofollow&quot;&gt;Fibonacci search&lt;/a&gt; works better. This can apply to situations like searching through arrays that don't fit entirely in your computer's cache so it's generally cheaper to read nearby elements that distant ones.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2010-07-21T22:05:25.537" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="397" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T22:11:32.083" Score="0" ViewCount="0" Body="&lt;p&gt;I don't think you can get an intuitive feel for the derivatives without looking at the plots personally. When you consider that a derivative is a rate of change, you need to be looking at a function that is varying, which implies you are looking at the plot/graph of the function. When you further consider that a derivative (by definition of it being a rate of change) is a gradient function, the intuitive answer is that cos is the gradient function of sin, and -sin is the gradient function of cos (and so on). So if you calculate the gradient of the sin curve at any point, the value you get will be the cosine value for that point.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-21T22:11:32.083" />
  <row Id="398" PostTypeId="1" AcceptedAnswerId="406" CreationDate="2010-07-21T22:16:34.997" Score="4" ViewCount="290" Body="&lt;p&gt;We will call the set of all positive even numbers &lt;code&gt;E&lt;/code&gt; and the set of all positive integers &lt;code&gt;N&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, it seems obvious that &lt;code&gt;E&lt;/code&gt; is smaller than &lt;code&gt;N&lt;/code&gt;, because for &lt;code&gt;E&lt;/code&gt; is basically &lt;code&gt;N&lt;/code&gt; with half of its terms taken out.  The size of &lt;code&gt;E&lt;/code&gt; is the size of &lt;code&gt;N&lt;/code&gt; divided by two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could see this as, for every item in &lt;code&gt;E&lt;/code&gt;, two items in &lt;code&gt;N&lt;/code&gt; could be matched (the item x and x-1).  This implies that &lt;code&gt;N&lt;/code&gt; is &lt;strong&gt;twice as large as &lt;code&gt;E&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second glance though, it seems less obvious.  Each item in &lt;code&gt;N&lt;/code&gt; could be mapped with &lt;em&gt;one&lt;/em&gt; item in &lt;code&gt;E&lt;/code&gt; (the item x*2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is larger, then?  Or are they both equal in size?  Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My background in Set theory is quite extremely scant)&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-08-02T17:06:16.650" Title="What is larger -- the set of all positive even numbers, or the set of all positive integers?" Tags="&lt;set-theory&gt;&lt;infinity&gt;" AnswerCount="8" CommentCount="2" />
  <row Id="399" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T22:19:51.873" Score="5" ViewCount="0" Body="&lt;p&gt;They are both the same size, the size being 'countable infinity' or 'aleph-null'. The reasoning behind it is exactly that which you have already identified - you can assign each item in E to a single value in N. This is true for the Natural numbers, the Integers, the Rationals but not the Reals (see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Diagonal Slash&lt;/a&gt; argument for details on this result).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Added explanation from comment --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reasoning is invalid because the cardinality of infinite sets doesn't follow 'normal' multiplication rules. If you multiply a set with cardinality of aleph-0 by 2, you still have aleph-0. The same is true if you divide it, add to it, subtract from it by any finite amount. &lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-21T22:36:17.057" LastActivityDate="2010-07-21T22:36:17.057" CommentCount="6" />
  <row Id="400" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T22:20:00.300" Score="1" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Each item in N could be mapped with one item in E  (the item x*2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Both sets have &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinality&quot; rel=&quot;nofollow&quot;&gt;cardinality&lt;/a&gt; aleph-0.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-21T22:20:00.300" CommentCount="4" />
  <row Id="401" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T22:40:29.437" Score="5" ViewCount="0" Body="&lt;p&gt;The word 'size' doesn't have a intuitive meaning for set of infinite items. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians defined cardinality by one-to-one correspondence(bijection), and it's generally what it means by 'size'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there &lt;strong&gt;exist&lt;/strong&gt; a bijection between A and B, then the two sets have the same cardinality. You have shown the existence of a bijection, therefore E and N have the same cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might mean the 'density' of N is twice as large as E. density of A(a subset of natural number) is limit of |{a&amp;lt;=n|a\in A}|/n as n goes to infinity.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-21T22:40:29.437" />
  <row Id="402" PostTypeId="2" ParentId="275" CreationDate="2010-07-21T22:47:58.727" Score="1" ViewCount="0" Body="&lt;p&gt;As an undergrad, I read a fair number of pop math books. The best by far was Ash and Gross' &quot;Fearless Symmetry&quot;. This book is very beautiful. It sustains a nice level of rigor while being approachable by those who aren't professionals. Additionally, it weaves the tale of one of the most beautiful recent stories in mathematics. Everyone I know of who have read the book have found it wonderful.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-21T22:47:58.727" CommunityOwnedDate="2010-07-21T22:47:58.727" CommentCount="1" />
  <row Id="403" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T22:48:13.517" Score="4" ViewCount="0" Body="&lt;p&gt;This interesting pattern of derivatives involving sine and cosine is related to the fact that e^x is its own derivative and that e^(ix) = cos(x) + i*sin(x) (&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's Formula&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two facts are in some sense the math hiding behind Justin L's more physical explanation, which you might well find more intuitive.  &lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-07-21T22:48:13.517" CommentCount="3" />
  <row Id="404" PostTypeId="2" ParentId="392" CreationDate="2010-07-21T23:03:36.737" Score="4" ViewCount="0" Body="&lt;p&gt;One of the main ways that sine and cosine come up is as two solutions to the differential equation y'' = -y.  Why is this an important differential equation?  Well, by Newton's second law it's the differential equation that says &quot;the force is proportional and opposite to the position.&quot;  For example, this is what happens with a spring!  Now that's a degree 2 diff eq so it has a 2-dimensional space of solutions.  How to pick a nice basis of that space?  Well one way would be to pick f and g such that f' = i f and g' = -i g.  However, that involves too many imaginary numbers, so another option is f' = -g, and g' = -f.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if you're trying to find two functions which explain oscillatory motion you're naturally lead to picking functions that have f' = g, g' = -f, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(On the other hand it's totally unclear from this point of view why Sine and Cosine should have anything to do with triangles...)&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T23:03:36.737" />
  <row Id="405" PostTypeId="1" AcceptedAnswerId="408" CreationDate="2010-07-21T23:25:43.533" Score="15" ViewCount="441" Body="&lt;p&gt;I have been learning some functional programming recently and I so I have come across monads. I understand what they are in programming terms, but I would like to understand what they are mathematically. Can anyone explain what a monad is using as little category theory as possible?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-31T10:14:29.723" LastActivityDate="2010-08-19T12:28:32.053" Title="Simple explanation of a monad" Tags="&lt;intuition&gt;&lt;category-theory&gt;&lt;universal-algebra&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="6" />
  <row Id="406" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T23:32:07.050" Score="19" ViewCount="0" Body="&lt;p&gt;Mathematics is the art of clever forgetting.  The first mathematical breakthrough, numbers, came about when people realized that if you just forgot about whether it was 5 cows + 3 cows, or 5 rocks + 3 rocks or whatever you always got 8.  Numbers are what you get when you look at collections of objects and &lt;em&gt;forget what kind of object they are&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;as sets&quot; you mean you're forgetting a lot of information, in particular you don't care about what the names of the elements in that set are or what properties those elements have.  As sets the positive numbers and the positive even numbers are &quot;the same&quot; (that is are in bijection) because you can take 1,2,3,...  and just &lt;em&gt;rename&lt;/em&gt; 1 to 2, and 2 to 4, and n to 2n, and you've just renamed all the elements and got the even numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you want to remember more about these sets, for example that they're not arbitrary sets they're both subsets of the natural numbers, then they become distinguishable.  Depending on how you want to measure &quot;size of a subset of the natural numbers&quot; they might be different sizes.  For example, a common way to measure &quot;size of a subset of the natural numbers&quot; is by its &quot;density.&quot;  That is you look at the first N numbers and calculate what fraction of them are in your set, and then take the limit as N goes to infinity (warning for sufficiently complicated sets this limit might not exist).  So for your two examples, one has density 1 and the other has density 1/2, which is one way to make precise the intuition that the former is bigger &lt;em&gt;as a subset of the natural numbers&lt;/em&gt; (though not &lt;em&gt;as a set&lt;/em&gt;) than the latter.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-21T23:32:07.050" CommentCount="1" />
  <row Id="407" PostTypeId="2" ParentId="398" CreationDate="2010-07-21T23:36:51.333" Score="1" ViewCount="0" Body="&lt;p&gt;Your example is an example of a countable sets: &lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Countable_set&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have explained, if you find a bijection between these two sets then they are of the same cardinality or have the same number of elements.(I hope that statement is true :)) But in this case you have found one with N which is what makes it a countable set. f(x)=2x would be an example of a bijection between N and 2N. Since for every x in the set of natural numbers there is a corresponding number in the set of 2N (only one) then you can say that they have the same cardinality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cooler example would be the interval of the real numbers (0, 1) and R itself, those are not countable but their cardinality is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other notes, if you are wondering what aleph is: &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Aleph_number&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="180" LastEditorDisplayName="" LastEditDate="2010-07-22T01:51:50.317" LastActivityDate="2010-07-22T01:51:50.317" />
  <row Id="408" PostTypeId="2" ParentId="405" CreationDate="2010-07-21T23:42:48.640" Score="2" ViewCount="0" Body="&lt;p&gt;If you want to avoid too much category theory, you can first read this link to understand the definition of monads in &lt;a href=&quot;http://www.haskell.org/tutorial/monads.html&quot; rel=&quot;nofollow&quot;&gt;Haskell&lt;/a&gt;. Then look at &lt;a href=&quot;http://en.wikibooks.org/wiki/Haskell/Category_theory&quot; rel=&quot;nofollow&quot;&gt;Wikibooks&lt;/a&gt; for a more mathematical look (thanks Jonathan Fischoff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two descriptions that I know of. The first can easily be found by looking at wiki under Monad or consulting Harry's nice summary. The second is more interesting in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that you don't know the definition of a monoidal action, if you do, just skip ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A monoidal action is a functor from a monoid to the category of endofunctors on a category satisfying two coherence relations. These two coherence relations simply verify that your monoidal product is the same as composition in the target, and that the identity object behaves with the action. The relations are normally written as diagrams, but without latex implement, I wont type them here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea of a monoidal action, consider a group action, and formulate it a little more categorically, by writing the two axioms as diagrams. These diagrams, when converted to the language of monoidal categories, are exactly those of a monoidal action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the best part is once you have monoidal action, monads on a category are simply the category of monoidal actions from the trivial monoidal category to your category. Note here that the trivial monoidal category will be the monoidal category with one object one morphism and all the other monoidal data is trivially determined. The monadic coherence relations come for free from your monoidal action coherence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my simple explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In this way, we can formulate monads functorially as &quot;representations&quot; of the trivial monoidal category.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can readily show the two definitions are the same.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-19T12:28:32.053" LastActivityDate="2010-08-19T12:28:32.053" CommentCount="8" />
  <row Id="409" PostTypeId="2" ParentId="370" CreationDate="2010-07-21T23:45:18.633" Score="5" ViewCount="0" Body="&lt;p&gt;Another book that is more elementary, not requiring any algebraic topology for motivation, and formulating the basics through a question and answer approach is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/052171916X&quot; rel=&quot;nofollow&quot;&gt;Conceptual Mathematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An added benefit is that it is written by an expert!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-03T15:06:07.790" LastActivityDate="2010-08-03T15:06:07.790" CommunityOwnedDate="2010-08-04T15:59:49.803" CommentCount="5" />
  <row Id="411" PostTypeId="1" AcceptedAnswerId="413" CreationDate="2010-07-22T00:02:34.840" Score="4" ViewCount="245" Body="&lt;p&gt;Say we have a function $\displaystyle f(z)=\sum_{n=0}^\infty a_n z^n$ with radius of convergence R&gt;0. Why is the radius of convergence only R? Can we conclude that there must be a pole, branch cut or discontinuity for some z_0 with |z_0|=R? What does that mean for functions like&lt;br&gt;&#xA;$$f(z)=\begin{cases}&#xA;  0 &amp;amp; \text{for $z=0$} \\&#xA;  e^{-\frac{1}{z^2}} &amp;amp; \text{for $z \neq 0$} \end{cases}$$&lt;br&gt;&#xA;that have a radius of convergence 0?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-12T13:08:32.587" LastActivityDate="2010-08-12T13:08:32.587" Title="Why do complex functions have a finite radius of convergence?" Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="412" PostTypeId="2" ParentId="405" CreationDate="2010-07-22T00:09:50.567" Score="3" ViewCount="0" Body="&lt;p&gt;Let $C$ be a category.  Then a monad &lt;em&gt;based at $C$&lt;/em&gt; is a &lt;em&gt;monoid&lt;/em&gt; in the strict monoidal category &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{End}(C)=\mathcal{Hom}_{Cat}(C,C),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the natural monoidal product is given by composition of endofunctors, and the monoidal unit is the identity functor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A monoid in a monoidal category is defined &lt;a href=&quot;http://en.wikipedia.org/wiki/Monoid_object&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more explanation, just give me a call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notation: The category of functors $C\to D$ is also written as $Fun(C,D)$, but this notation is nonstandard.  The standard notations are $\mathcal{Hom}_{Cat}(C,D)$ or simply $Cat(C,D)$.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T00:17:31.097" LastActivityDate="2010-07-22T00:17:31.097" />
  <row Id="413" PostTypeId="2" ParentId="411" CreationDate="2010-07-22T00:42:49.470" Score="12" ViewCount="0" Body="&lt;p&gt;If the radius of convergence is $R$, that means there is a singular point on the circle $|z| = R$. In other words, there is a point $\xi$ on the circle of radius $R$ such that the function cannot be extended via &quot;analytic continuation&quot; in a neighborhood of $\xi$.  This is a straightforward application of compactness of the circle and can be found in books on complex analysis, e.g. Rudin's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it does not mean that there is a pole, branch cut, or discontinuity, though those would cause singular values. Indeed, a &quot;pole&quot; on the boundary would only make sense if you can analytically continue the power series to some proper domain containing the disk $D_R(0)$, and this is generally impossible. For instance, the power series $\sum z^{2^j}$ cannot be continued in any way outside the unit disk, because it is unbounded along any ray whose angle is a dyadic fraction.  The unit circle is its natural boundary, though it does not make sense to say that the function has a branch point or pole there.  (More generally, one can show that given any domain in the plane, there is a holomorphic function in that domain which cannot be extended any further, essentially using variations on the same theme.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $\sum_j \frac{z^j}{j^2}$, incidentally, is continuous on the closed unit disk, but even though there is a singular point there. So continuity may happen at singular points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last function you mention does not have a power series expansion in a neighborhood of zero. In fact, it is not continuous at zero, because it blows up if you approach zero along the imaginary axis.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T00:42:49.470" />
  <row Id="414" PostTypeId="1" AcceptedAnswerId="417" CreationDate="2010-07-22T00:44:57.890" Score="1" ViewCount="150" Body="&lt;p&gt;We define Lie algebra's abstractly as algebras whose multiplication satisfies anti-commutativity and Jacobi's Identity. A particular instance of this is an associative algebra equipped with the commutator bracket [a,b]=ab-ba. However, the notation suggests that this bracket is the one we think about. Additionally, the right adjoint to the functor I just mentioned creates the universal enveloping algebra by quotient-ing the tensor algebra by the tensor version of this bracket; but we could always start with some arbitrary Lie algebra with  some other satisfactory bracket and apply this functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;why the commutator bracket?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it purely from a historical standpoint(and if so could you explain why)? Or is there a result that says any Lie algebra is essentially one with the commutator bracket(maybe something about the faithfulness of the functor from above)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of(a colleague told me) a proof that the Jacobi identity is also an artifact of the right adjoint to the universal enveloping algebra. He can show that it is the necessary identity for the universal enveloping algebra to be associative(if someone knows of this in the literature I would also appreciate the link to this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question is clear, if not, I can revise and try to make it a bit more specific.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-22T01:03:08.603" Title="Why do we use the commutator bracket for Lie algebra's" Tags="&lt;notation&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="415" PostTypeId="1" AcceptedAnswerId="416" CreationDate="2010-07-22T00:47:12.957" Score="6" ViewCount="134" Body="&lt;p&gt;It appears to me that only Triangles, Squares, and Pentagons are able to &quot;tessellate&quot; (is that the proper word in this context?) to become regular 3D convex polytopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What property of those regular polygons themselves allow them to faces of regular convex polyhedron?  Is it something in their angles?  Their number of sides?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why are there more Triangle-based Platonic Solids (three) than Square- and Pentagon- based ones? (one each)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, is this the same property that allows certain Platonic Solids to be used as &quot;faces&quot; of regular polychoron (4D polytopes)?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-25T07:39:02.780" LastActivityDate="2010-08-05T18:33:17.190" Title="What property of certain regular polygons allows them to be faces of the Platonic Solids?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="416" PostTypeId="2" ParentId="415" CreationDate="2010-07-22T01:00:59.263" Score="12" ViewCount="0" Body="&lt;p&gt;The regular polygons that form the Platonic solids are those for which the measure of the interior angles, say &amp;alpha; for convenience, is such that $3\alpha&amp;lt;2\pi$ (360&amp;deg;) so that three (or more) of the polygons can be assembled around a vertex of the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular (equilateral) triangles have interior angles of measure $\frac{\pi}{3}$ (60&amp;deg;), so they can be assembled 3, 4, or 5 at a vertex ($3\cdot\frac{\pi}{3}&amp;lt;2\pi$, $4\cdot\frac{\pi}{3}&amp;lt;2\pi$, $5\cdot\frac{\pi}{3}&amp;lt;2\pi$), but not 6 ($6\cdot\frac{\pi}{3}=2\pi$--they tesselate the plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular quadrilaterals (squares) have interior angles of measure $\frac{\pi}{2}$ (90&amp;deg;), so they can be assembled 3 at a vertex ($3\cdot\frac{\pi}{2}&amp;lt;2\pi$), but not 4 ($4\cdot\frac{\pi}{2}=2\pi$--they tesselate the plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular pentagons have interior angles of measure $\frac{3\pi}{5}$ (108&amp;deg;), so they can be assembled 3 at a vertex ($3\cdot\frac{3\pi}{5}&amp;lt;2\pi$), but not 4 ($4\cdot\frac{3\pi}{5}&gt;2\pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular hexagons have interior angles of measure $\frac{2\pi}{3}$ (120&amp;deg;), so they cannot be assembled 3 at a vertex ($3\cdot\frac{2\pi}{3}=2\pi$--they tesselate the plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other regular polygon will have larger interior angles, so cannot be assembled into a regular solid.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T18:33:17.190" LastActivityDate="2010-08-05T18:33:17.190" CommentCount="5" />
  <row Id="417" PostTypeId="2" ParentId="414" CreationDate="2010-07-22T01:03:08.603" Score="2" ViewCount="0" Body="&lt;p&gt;Well Lie algebras naturally arise from the Lie bracket of vector fields and from taking the Lie algebra of a Lie group.  If we look at a the Lie algebra of a matrix subgroup, then the Lie bracket is the commutator of matrices.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T01:03:08.603" CommentCount="2" />
  <row Id="418" PostTypeId="1" AcceptedAnswerId="420" CreationDate="2010-07-22T01:04:43.687" Score="4" ViewCount="304" Body="&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find a bijective function from (0,1) to [0,1]. I haven't find the solution to this since I saw it a few days ago. It strike me as odd--mapping a open set into a closed set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;S is countable. It's trivial to find a bijective function f:N -&gt; N-S, (when |N| = |N-S|) f(n) = the nth smallest number in N-S. Are they any analogous trivial solutions to f: R-&gt;R-S?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="96" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-26T15:24:26.753" LastActivityDate="2010-11-26T15:24:26.753" Title="Construct a bijection from R to R-S, where S is countable" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="419" PostTypeId="1" AcceptedAnswerId="421" CreationDate="2010-07-22T01:07:25.070" Score="2" ViewCount="165" Body="&lt;p&gt;I just got my statistics test back and I am totally confused about one of the questions!  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A study was done that took a simple random sample of 40 people and measured whether&#xA;  the subjects were right-handed or&#xA;  left-handed, as well as their ages.&#xA;  The study showed that the proportion&#xA;  of left-handed people and the ages had&#xA;  a strong negative correlation. What&#xA;  can we conclude? Explain your answer. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that we can't conclude that getting older causes people to become right-handed. Something else might be causing it, not the age. If two things are correlated, we can only conclude association, not causation. So I wrote: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can conclude that many people&#xA;  become right-handed as they grow&#xA;  older, but we cannot tell why.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's exactly what association means, but my teacher marked me wrong! What mistake did I make? Is 40 too small of a sample size to make any conclusions?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T02:27:46.657" LastActivityDate="2010-07-23T01:53:22.163" Title="What can we conclude from correlation?" Tags="&lt;statistics&gt;&lt;homework&gt;&lt;high-school&gt;" AnswerCount="3" CommentCount="16" FavoriteCount="1" />
  <row Id="420" PostTypeId="2" ParentId="418" CreationDate="2010-07-22T01:13:00.127" Score="7" ViewCount="0" Body="&lt;p&gt;The proof of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Schroeder-Bernstein theorem&lt;/a&gt; allows you to get a bijection for 1, since we have an injection (0,1) \to [0,1] and a bijection $f: [0,1] \to [1/4, 3/4] \subset (0,1)$ (say $x \to x/2 +1/4$).  The function's definition will be somewhat messy (basically, it depends on how many times you can lift a point under these to injections already defined, and specifically the parity of the number of times), but it'll do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2, iterate this construction to get a bijection $R \to R - N$.  Then given any countable set $S$, define the map of $R$ that interchanges $N$ and $S$ and leaves every other point fixed.  Then the composition of the first bijection with this second map is your bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuity considerations imply that the map can't be continuous: in 1, for instance, we'd otherwise have that (0,1) is compact, which it's not.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T01:13:00.127" CommentCount="3" />
  <row Id="421" PostTypeId="2" ParentId="419" CreationDate="2010-07-22T01:52:19.083" Score="2" ViewCount="0" Body="&lt;p&gt;This is wrong: &quot;We can conclude that many people become right-handed as they grow older.&quot; We cannot conclude this at all from the given data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one, the study only takes a sample at one point in time, rather than selecting a sample and monitoring their progress through many decades. This is what would be needed for us to even entertain the possibility that aging causes a change in handedness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible causes include that left handed people might have a shorter life expectancy, or perhaps there was a spike in the birth rate of right handed people in the past. There are many other possibilities that have been mentioned in others answers which would also account for the skewed proportions &lt;em&gt;without&lt;/em&gt; requiring people to change handedness with age, which is what you falsely concluded in the test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, just an observation, but it appears the &quot;study&quot; was conducted under false pretenses. Handedness is a false dichotomy, people can also be ambidextrous.&lt;/p&gt;&#xA;" OwnerUserId="156" LastEditorUserId="156" LastEditorDisplayName="" LastEditDate="2010-07-23T01:53:22.163" LastActivityDate="2010-07-23T01:53:22.163" CommentCount="7" />
  <row Id="422" PostTypeId="2" ParentId="240" CreationDate="2010-07-22T01:57:38.540" Score="2" ViewCount="0" Body="&lt;p&gt;In mathematics you can construct a mathematical theory with different sets of axioms. This can be really useful. When mathematicians ignored the parallel line axiom in Euclidian geometry it gave raise to non-Euclidian geometries, which became really important in Einsteinian physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An axiom of logic is the law of the excluded third which basically says that one statement is either true or false. This means that any theorem that depends completely in this axiom is not valid on mathematical theories that decide to ignore the axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof by contradiction is using the axiom directly; if the consequent is false then the antecent is false, then the converse of the consequent is true (because it must be either true or false). If the theorem can be proved in a constructive way, then it does not depend on the Law of Excluded Third and is valid in theories that does not use the law.&lt;/p&gt;&#xA;" OwnerUserId="187" LastActivityDate="2010-07-22T01:57:38.540" />
  <row Id="423" PostTypeId="1" AcceptedAnswerId="429" CreationDate="2010-07-22T02:12:22.140" Score="4" ViewCount="201" Body="&lt;p&gt;I've sort of gotten a grasp on the Chain rule with one variable.  If you hike up a mountain at 2 feet an hour, and the temperature decreases at 2 degrees per feet, the temperature would be decreasing for you at 2*2 = 4 degrees per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm having a bit more trouble understanding the Chain Rule as applied to multiple variables.  Even the case of 2 dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = f(x,y) where x = g(t) and y = h(t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dz/dt = dz/dx dx/dt + dz/dy dy/dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is easy enough to &quot;calculate&quot; (and figure out what goes where).  My teacher taught me a neat tree-based graphical method for figuring out partial derivatives using chain rule.  All-in-all, it was rather hand-wavey.  However, I'm not sure exactly how this works, intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, intuitively, is the equation above true?  Why &lt;strong&gt;addition&lt;/strong&gt;?  Why not multiplication, like the other chain rule?  Why are some multiplied and some added?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T07:32:01.610" LastActivityDate="2010-07-22T11:43:11.577" Title="Intuitive reasoning behind the Chain Rule in multiple variables?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="424" PostTypeId="2" ParentId="419" CreationDate="2010-07-22T02:13:36.677" Score="0" ViewCount="0" Body="&lt;p&gt;We can only conclude that the result is interesting and it deserves more research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without further study we can not say if right-handedness causes age (ie. being left handed causes biological alterations that shorten the life span), age causes right-handedness (ie. as people age they become right handed), they are correlated because they are caused by another variable (ie. older people became educated in a different system that discouraged left-handedness), the study had bad luck selecting its sample, the study sample was bad designed, the study was bad designed, etc.&lt;/p&gt;&#xA;" OwnerUserId="187" LastActivityDate="2010-07-22T02:13:36.677" CommentCount="3" />
  <row Id="425" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:17:38.100" Score="0" ViewCount="0" Body="&lt;p&gt;Changing time is the same as changing x and changing y. If the changes each one caused in z didn't interact, then the total change would be the sum of both changes. If the function is well behaved (differentiatable) then the interaction caused by an infinitesimal change in x with an infinitesimal change in y will be doubly infinitesimal. The proof of the double chain rule just shows this formally.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-22T02:24:27.210" LastActivityDate="2010-07-22T02:24:27.210" />
  <row Id="426" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:19:45.070" Score="4" ViewCount="0" Body="&lt;p&gt;The correct context of the chain rule is that taking the tangent bundle is functorial.  A more down-to-earth answer is provided by working coordinate free using linear algebra.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f:X\to Y$ and $g:Y\to Z$ are functions between Banach spaces (these are a generalized version of R^n) such that $f$ is differentiable at $\vec{v}$ and $g$ is differentiable at $f(\vec{v})$ (note that in the general case we must require that their derivatives are toplinear (continuous and linear), since not all linear maps are continuous in the context of infinite dimensional spaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then taking the total differential, we see that the chain rule is equivalent to saying that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_\vec{v}(g\circ f)=T_{f(\vec{v})}(g) \circ T_\vec{v}(f).$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description you get with coordinates comes from this very much simpler presentation (Where T denotes the total differential) as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To derive the formula with coordinates (say, for example, in three dimensions), we present the total differentials (which are linear transformations) as their Jacobian matrices and test along the column vector ${}^t[x,y,z]$. (Where the leftthand exponent t denotes the matrix transpose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: When we present linear operators by their matrices, composition of linear transformations becomes matrix multiplication, and evaluation at a vector $\vec{w}$ becomes righthand multiplication by a column matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main point is that coordinates &lt;em&gt;obscure&lt;/em&gt; what's actually going on here.  The beauty of the coordinate free definition is destroyed by the complicated description of matrix multiplication.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-22T07:31:29.990" LastActivityDate="2010-07-22T07:31:29.990" CommentCount="1" />
  <row Id="427" PostTypeId="2" ParentId="240" CreationDate="2010-07-22T02:20:12.463" Score="1" ViewCount="0" Body="&lt;p&gt;Proof by contradiction is just as logically valid as any other type of proof. If you are unsure, I think it might help to consider exactly what a proof by contradiction entails.&lt;br&gt;&#xA;Say we have a set of statements $\Gamma$, and that $\Gamma\cup{(\neg\phi)}$ is not consistent. That is, the statement $\neg\phi$ contradicts something in $\Gamma$. (In other words, we supposed $\phi$ was false, and reached a contradiction.) Say that statement was $\psi$. Then $\Gamma\cup{(\neg\phi)}\vdash\psi$ and $\Gamma\cup{(\neg\phi)}\vdash \neg\psi$.&#xA;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_explosion&quot; rel=&quot;nofollow&quot;&gt;principle of explosion&lt;/a&gt;, we conclude that $\Gamma\cup{(\neg\phi)}\vdash\phi$. (We can prove any statement, so we prove $\phi$.&lt;br&gt;&#xA;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_deduction#Proofs_and_type-theory&quot; rel=&quot;nofollow&quot;&gt;deduction&lt;/a&gt;, we know that $\Gamma\vdash(\neg\phi\rarrow\phi)$.&lt;br&gt;&#xA;Most first-order logic systems have an axiom that gives us $((\neg\phi\rarrow\phi)\rarrow\phi$. I hope you can convince yourself that this is true without trouble.&lt;br&gt;&#xA;This yields $\Gamma\vdash\phi$.&lt;br&gt;&#xA;We started with the idea that the negation of the statement $\phi$ was incompatible with your working set of axioms and theorems $\Gamma$, and concluded that therefore $\Gamma$ proves $\phi$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is more than one way to prove anything. Other methods can often be more intuitive, more elegant, or may lead to some other useful results. However, that is distinct from &quot;weaker.&quot; Proof by contradiction is perfectly sound.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T02:44:44.980" LastActivityDate="2010-07-22T02:44:44.980" />
  <row Id="428" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:21:57.997" Score="1" ViewCount="0" Body="&lt;p&gt;Think of it in terms of causality &amp;amp; superposition.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;z = f(x,y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you keep x fixed then dz/dt = df/dx * dx/dt, and if you keep y fixed then dz/dt = df/fy * dy/dt. Superposition says you can just add the two together.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-07-22T11:43:11.577" LastActivityDate="2010-07-22T11:43:11.577" />
  <row Id="429" PostTypeId="2" ParentId="423" CreationDate="2010-07-22T02:26:47.253" Score="6" ViewCount="0" Body="&lt;p&gt;The basic reason is that one is simply composing the derivatives just as one composes the functions. Derivatives are linear approximations to functions. When you compose the functions, you compose the linear approximations---not a surprise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to try to expand on Harry Gindi's answer, because that was the only way I could grok it, but in somewhat simpler terms.&#xA;The way to think of a derivative in multiple variables is as a linear approximation.   In particular, let $f: R^m \to R^n$ and $q=f(p)$.  Then near $p$, we can write $f$ as $q$ basically something linear plus some &quot;noise&quot; which &quot;doesn't matter&quot; (i.e. is little oh of the distance to $p$).  Call this linear map $L: R^m \to R^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $g: R^n \to R^s$ is some map and $r = g(q)$.  We can approximate $g$ near $q$ by $r$ plus some linear map $N$ plus some &quot;garbage&quot; which is, again, small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, I'm going to assume that $p,q,r$ are all zero. This is ok, because one can just move one's origin around a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as before, applying $f$ to a point near zero corresponds loosely to applying the linear transformation $L$. Applying $g$ to a point near zero corresponds loosely to applying $N$.  Hence applying $g \circ f$ corresponds up to some ignorable &quot;garbage&quot; to the map $N \circ L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $N \circ L$ is the linear approximation to $g \circ f$ at zero, in particular this composition is the derivative of $g \circ f$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T02:26:47.253" />
  <row Id="431" PostTypeId="2" ParentId="419" CreationDate="2010-07-22T02:52:07.877" Score="1" ViewCount="0" Body="&lt;p&gt;Here is one example of a plausible explanation that disagrees with your analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cultural expectations for left- and right-handedness have changed over time. Older people may have gone to school at a time where left-handedness was discouraged and students were forced to write with their right hands, training children never to use the left hand instead of the right. Younger participants in the study were in school more recently and learned to write at a time where left-handedness was &lt;em&gt;not&lt;/em&gt; discouraged, creating a positive correlation between left-handedness and youth.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-22T02:52:07.877" />
  <row Id="432" PostTypeId="1" AcceptedAnswerId="670" CreationDate="2010-07-22T03:03:44.997" Score="6" ViewCount="116" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant&quot; rel=&quot;nofollow&quot;&gt;discriminant&lt;/a&gt; &lt;a href=&quot;http://mathurl.com/2dqz7gw&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2dqz7gw.png&quot;&gt;&lt;/a&gt; of the cubic polynomial &lt;a href=&quot;http://mathurl.com/24gdf5r&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/24gdf5r.png&quot;&gt;&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_equation#The_nature_of_the_roots&quot; rel=&quot;nofollow&quot;&gt;indicates not only if there are repeated roots when Δ vanishes&lt;/a&gt;, but also that there are three distinct, real roots if &lt;a href=&quot;http://mathurl.com/2wooxlb&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2wooxlb.png&quot;&gt;&lt;/a&gt;, and that there is one real root and two complex roots (complex conjugates) if &lt;a href=&quot;http://mathurl.com/2cuwdfa&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2cuwdfa.png&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does &lt;img src=&quot;http://mathurl.com/2cuwdfa.png&quot;&gt; indicate complex roots? I understand that because of the way that the discriminant is defined, it indicates that there is a repeated root if it vanishes, but why does Δ greater than 0 or less than 0 have special meaning, too?&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:30:47.977" LastActivityDate="2010-07-26T08:30:47.977" Title="Why does the discriminant of a cubic polynomial being less than 0 indicate complex roots?" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;&lt;zeros&gt;&lt;discriminant&gt;&lt;roots&gt;" AnswerCount="2" />
  <row Id="433" PostTypeId="2" ParentId="432" CreationDate="2010-07-22T03:11:57.567" Score="5" ViewCount="0" Body="&lt;p&gt;The discriminant of any monic polynomial is the product $\prod_{i \neq j} (x_i - x_j)^2$ of the squares of the differences of the roots (in an algebraic closure, e.g. $C$). Cf. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant#Discriminant_of_a_polynomial&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; on this. Consequently, if the roots are all real and distinct, this must be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the polynomial is not monic, the factor $a_0^{2n-2}$ is thrown in, for $a_0$ the leading coefficient and $n$ the degree; this is positive for a real polynomial.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T03:11:57.567" CommentCount="2" />
  <row Id="434" PostTypeId="1" AcceptedAnswerId="437" CreationDate="2010-07-22T03:22:51.490" Score="7" ViewCount="176" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/250/a-challenge-by-r-p-feynman-give-counter-intuitive-theorems-that-can-be-transl/260#260&quot; rel=&quot;nofollow&quot;&gt;This answer&lt;/a&gt; suggests that there are explicit polynomial equations for which the existence &#xA;(or nonexistence) of integer solutions is unprovable. How can this be?&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="38" LastEditorDisplayName="" LastEditDate="2010-07-25T23:39:47.427" LastActivityDate="2010-07-25T23:39:47.427" Title="How can there be explicit polynomial equations for which the existence of integer solutions is unprovable?" Tags="&lt;algebra&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="435" PostTypeId="2" ParentId="250" CreationDate="2010-07-22T03:24:31.227" Score="-1" ViewCount="0" Body="&lt;p&gt;Theorem: Any predicate which can be evaluated in polynomial time on a nondeterministic turing machine can be evaluated in polynomial time on a deterministic turing machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is asking an open question cheating? :)&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="48" LastEditorDisplayName="" LastEditDate="2010-07-22T03:30:16.190" LastActivityDate="2010-07-22T03:30:16.190" CommunityOwnedDate="2010-07-22T03:24:31.227" CommentCount="1" />
  <row Id="437" PostTypeId="2" ParentId="434" CreationDate="2010-07-22T03:32:36.017" Score="8" ViewCount="0" Body="&lt;p&gt;Edit: This has been explained better by Joel David Hamkins; cf. &lt;a href=&quot;http://mathoverflow.net/questions/32892/does-anyone-know-a-polynomial-whose-lack-of-roots-cant-be-proved&quot; rel=&quot;nofollow&quot;&gt;this MO thread&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because given any first-order formula $\phi(x_1, \dots, x_n)$ with $n$ parameters in the integers, there is a polynomial $P(x_1, \dots, x_n, z_1, \dots, z_m)$ which can be proved to have a root in $z_1, \dots, z_m$ in the integers if and only if $\phi(x_1, \dots, x_n)$---this is a version of Matiyasevich's solution to Hilbert's tenth problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's possible to construct within any formal system explicit first-order formulas which can't be proved or disproved (for instance, the statement &quot;this formal system is consistent&quot;).  This is  the second incompleteness theorem.  These first-order formulas correspond to polynomials by the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, one can show that you can construct a polynomial $P(z_1, \dots, z_m)$ corresponding to the statement &quot;mathematics* is inconsistent&quot; translated to a formula.  Thus, if mathematics is consistent, there's no mathematical proof that this polynomial doesn't have a root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Mathematics=ZFC here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on these lines, see Ebbinghaus-Frum-Thomas's very accessible book on mathematical logic.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-22T16:37:55.700" LastActivityDate="2010-07-22T16:37:55.700" CommentCount="10" />
  <row Id="438" PostTypeId="1" AcceptedAnswerId="439" CreationDate="2010-07-22T04:05:54.007" Score="3" ViewCount="307" Body="&lt;p&gt;I know there must be something unmathematical in the following but I don't know where it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sqrt{-1} &amp;amp;= i \\ \\&#xA;\frac1{\sqrt{-1}} &amp;amp;= \frac1i \\ \\&#xA;\frac{\sqrt1}{\sqrt{-1}} &amp;amp;= \frac1i \\ \\&#xA;\sqrt{\frac1{-1}} &amp;amp;= \frac1i \\ \\&#xA;\sqrt{\frac{-1}1} &amp;amp;= \frac1i \\ \\&#xA;\sqrt{-1} &amp;amp;= \frac1i \\ \\&#xA;i &amp;amp;= \frac1i \\ \\&#xA;i^2 &amp;amp;= 1 \\ \\&#xA;-1 &amp;amp;= 1 \quad !!!&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="187" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T18:11:18.780" LastActivityDate="2010-08-24T18:11:18.780" Title="-1 is not 1, so where is the mistake?" Tags="&lt;arithmetic&gt;&lt;complex-numbers&gt;&lt;fake-proofs&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="439" PostTypeId="2" ParentId="438" CreationDate="2010-07-22T04:11:45.640" Score="11" ViewCount="0" Body="&lt;p&gt;Between your third and fourth lines, you use sqrt(a)/sqrt(b) = sqrt(a/b).  This is only true when a &gt;= 0 and b &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  As pointed out in the comments, what I meant was that the identity sqrt(a)/sqrt(b) = sqrt(a/b) has domain a &gt;= 0 and b &gt; 0.  Outside that domain, applying the identity is inappropriate, whether or not it &quot;works.&quot;  In general, sqrt(x) where x is a negative real number must first be rewritten as i * sqrt(|x|) before any other algebraic manipulations can be applied (because the identities relating to manipulation of square roots [perhaps exponentiation with non-integer exponents in general] require nonnegative numbers).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-22T05:04:12.227" LastActivityDate="2010-07-22T05:04:12.227" CommentCount="7" />
  <row Id="440" PostTypeId="1" AcceptedAnswerId="441" CreationDate="2010-07-22T04:22:56.800" Score="7" ViewCount="197" Body="&lt;p&gt;An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any a and b,&#xA;a R b implies b R a by symmetry. Using transitivity, we have a R a.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-22T11:28:41.117" Title="Why isn't reflexivity redundant in the definition of equivalence relation?" Tags="&lt;relations&gt;" AnswerCount="3" />
  <row Id="441" PostTypeId="2" ParentId="440" CreationDate="2010-07-22T04:28:18.623" Score="14" ViewCount="0" Body="&lt;p&gt;Actually, without the reflexivity condition,   the empty relation would count as an equivalence relation, which is non-ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument used the hypothesis that for each $a$, there exists $b$ such that $aRb$ holds. If this is true, then symmetry and transitivity imply reflexivity, but this is not true in general.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T04:28:18.623" />
  <row Id="442" PostTypeId="2" ParentId="440" CreationDate="2010-07-22T04:36:02.390" Score="6" ViewCount="0" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing condition is sometimes called  'seriality' -- for any x there must be a y such that x R y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add seriality to the symmetry and transitivity you get a reflexive relation again.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-22T04:36:02.390" CommentCount="2" />
  <row Id="443" PostTypeId="1" AcceptedAnswerId="445" CreationDate="2010-07-22T05:03:43.867" Score="9" ViewCount="176" Body="&lt;p&gt;1/7 = 0.(142857)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the digits in the parentheses repeating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the reason it's a repeating fraction is because 7 and 10 are coprime.  But this...cyclical nature is something that is not observed by any other reciprocal of any natural number that I know of (besides multiples of 7). (if I am wrong, I hope that I may find others through this question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;cyclical,&quot; i mean:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1/7 = 0.(142857)...&#xA;2/7 = 0.(285714)...&#xA;3/7 = 0.(428571)...&#xA;4/7 = 0.(571428)...&#xA;5/7 = 0.(714285)...&#xA;6/7 = 0.(857142)...&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where all of the repeating digits are the same string of digits, but shifted.  Not just a simple &quot;they are all the same digits re-arranged&quot;, but the same digits &lt;strong&gt;in the same order&lt;/strong&gt;, but shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps more strikingly, from the &lt;a href=&quot;http://en.wikipedia.org/wiki/142857_%28number%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1 × 142,857 = 142,857&#xA;2 × 142,857 = 285,714&#xA;3 × 142,857 = 428,571&#xA;4 × 142,857 = 571,428&#xA;5 × 142,857 = 714,285&#xA;6 × 142,857 = 857,142&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What is it about the number 7 in relation to the base 10 (and its prime factorization 2*5?) that allows its reciprocal to behave this way?  Is it (and its multiples) unique in having this property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_number&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has an article on this subject, and gives a form for deriving them and constructing arbitrary ones, but does little to show the &quot;why&quot;, and finding what numbers have cyclic inverses.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-22T07:22:23.383" LastActivityDate="2010-08-05T20:05:24.247" Title="Why is the decimal representation of 1/7 &quot;cyclical&quot;?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="444" PostTypeId="2" ParentId="443" CreationDate="2010-07-22T05:30:54.930" Score="6" ViewCount="0" Body="&lt;p&gt;It works with 1/19 = 0.(052631578947368421) too, while n/13 has two cycles: 1/13 = 0.(076923), 2/13 = 0.(153846), 3/13 = 0.(230769), 4/13 = 0.(307692), 5/13 = 0.(384615), and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That a cycle must appear when you have a prime number p different from the base in which we work (so in base 10 different from 2 and 5) is clear: if you perform the long division 1/p, sooner or later partial quotients must be repeated, and from that point on the quotients repeat themselves. The length of the cycle must be a divisor of p-1: it may be short (think at 1/11 = 0.(09) ) or have the maximum possible lenght like the cases of 7 and 19.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_number&quot; rel=&quot;nofollow&quot;&gt;an article on Cyclic numbers&lt;/a&gt;, and some other example is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Repeating_decimal&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;; &#xA;unfortunately no sufficient rule is given for a number to have its inverse cyclical.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T05:30:54.930" CommentCount="3" />
  <row Id="445" PostTypeId="2" ParentId="443" CreationDate="2010-07-22T05:37:42.593" Score="3" ViewCount="0" Body="&lt;p&gt;For a prime p, the length of the repeating block of $\frac{1}{p}$ is the least positive integer k for which $p|(10^k-1)$.  As in mau's answer, $k|(p-1)$, so $k\leq p-1$.  When $k=p-1$, then $\frac{1}{p}$ and its multiples behave as discussed in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the first 100 primes, this is true for 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(List generated in Mathematica using &lt;code&gt;Select[Table[Prime[n], {n, 1, 100}], # - 1 == Length[RealDigits[1/#][[1]][[1]]]&amp;amp;]&lt;/code&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T20:05:24.247" LastActivityDate="2010-08-05T20:05:24.247" CommentCount="1" />
  <row Id="446" PostTypeId="2" ParentId="418" CreationDate="2010-07-22T06:13:15.787" Score="5" ViewCount="0" Body="&lt;p&gt;An explicit bijection (0,1) \to [0,1] for part 1 is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(1/2) = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(1/3) = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(1/n) = 1/(n-2) for n &gt; 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x) = x for x not equal to a reciprocal of an integer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bijection R to R-S, we can number the elements of S (because S is countable) as s_1, s_2, s_3., .... Choose an infinite sequence t_1, t_2, ... of distinct elements of R, none of which are in S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(s_i) = t_{2i}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(t_i) = t_{2i+1}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(x) = x (otherwise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a bijection from R to R-S.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-22T06:13:15.787" />
  <row Id="448" PostTypeId="1" AcceptedAnswerId="452" CreationDate="2010-07-22T07:25:11.347" Score="4" ViewCount="72" Body="&lt;p&gt;Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I  can place them? And what if I also have 3 red balls?&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="143" LastEditorDisplayName="" LastEditDate="2010-07-22T07:53:18.487" LastActivityDate="2010-07-22T09:01:04.033" Title="In how many different ways can I sort balls of two different colors" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="449" PostTypeId="2" ParentId="381" CreationDate="2010-07-22T07:25:38.507" Score="10" ViewCount="0" Body="&lt;p&gt;Perhaps it's not an entirely practical application, but Fibonacci numbers can be used to &lt;a href=&quot;http://www.catonmat.net/blog/using-fibonacci-numbers-to-convert-from-miles-to-kilometers&quot; rel=&quot;nofollow&quot;&gt;convert from miles to kilometers and vice versa&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take two consecutive Fibonacci&#xA;  numbers, for example 5 and 8. And&#xA;  you're done converting. No kidding –&#xA;  there are 8 kilometers in 5 miles. To&#xA;  convert back just read the result from&#xA;  the other end - there are 5 miles in 8&#xA;  km!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But why does it work?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fibonacci numbers have a property that&#xA;  the ratio of two consecutive numbers&#xA;  tends to the &lt;a href=&quot;http://mathworld.wolfram.com/GoldenRatio.html&quot; rel=&quot;nofollow&quot;&gt;Golden ratio&lt;/a&gt; as&#xA;  numbers get bigger and bigger. The&#xA;  Golden ratio is a number and it&#xA;  happens to be approximately 1.618.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Coincidentally, there are 1.609&#xA;  kilometers in a mile, which is within&#xA;  0.5% of the Golden ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-07-22T07:25:38.507" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="2" />
  <row Id="450" PostTypeId="1" CreationDate="2010-07-22T07:35:26.143" Score="2" ViewCount="307" Body="&lt;p&gt;I just came back from an intense linear algebra lecture which showed that linear transformations could be represented by transformation matrices; with more generalization, it was later shown that affine transformations (linear + translation) could be represented by matrix multiplication as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me to thinking about all those other transformations I've picked up over the past years I've been studying mathematics.  For example, polar transformations -- transforming &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; to two new variables &lt;code&gt;r&lt;/code&gt; and &lt;code&gt;theta&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mapped &lt;code&gt;r&lt;/code&gt; to the &lt;code&gt;x&lt;/code&gt; axis and &lt;code&gt;theta&lt;/code&gt; to the &lt;code&gt;y&lt;/code&gt; axis, you'd basically have a coordinate transformation.  A rather warped one, at that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to represent this using a transformation matrix?  I've tried fiddling around with the numbers but everything I've tried to work with has fallen apart quite embarrassingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, is there a way to, given a specific non-linear transformation, construct a transformation matrix from it?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-09-09T23:00:16.617" Title="Can non-linear transformations be represented as Transformation Matrices?" Tags="&lt;linear-algebra&gt;&lt;transformation&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="451" PostTypeId="2" ParentId="448" CreationDate="2010-07-22T07:40:50.280" Score="2" ViewCount="0" Body="&lt;p&gt;The case of two colors is simple: if you have m yellow balls and n blue ones you only need to choose m positions among (m+n) possibilities, that is (m+n)!/(m!&amp;middot;n!). The other balls' positions are automatically set up.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T07:40:50.280" />
  <row Id="452" PostTypeId="2" ParentId="448" CreationDate="2010-07-22T07:48:08.623" Score="7" ViewCount="0" Body="&lt;p&gt;This is a standard problem involving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination&quot; rel=&quot;nofollow&quot;&gt;combinations of sets&lt;/a&gt;, though perhaps not very obvious intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly consider the number of ways you can rearrange the entire set of balls, counting each ball as indepndent (effectively ignoring colours for now). This is simply &lt;code&gt;(4 + 5)! = 9!&lt;/code&gt;, since the 1st ball can be any of the 9, the 2nd can be any of the remaining 8, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we calculate how many different ways the yellow balls can be arranged within themselves, since for the purpose of this problem they are considered equivalent. The number of combinations is of course &lt;code&gt;4!&lt;/code&gt;; similarly, for the blue balls the number is &lt;code&gt;5!&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, overall we find:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;total no. of arrangements = number of arrangements of all balls / (number of arrangements of yellow balls * number of arrangements of blue balls)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore in our case we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;total no. of arrangements = 9! / (5! * 4!) = 126&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can see how this can be easily extended if we also have 3 red balls too. (Hint: the total changes and we have another multiple of identical arrangements to account for.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps...&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-22T07:48:08.623" />
  <row Id="453" PostTypeId="2" ParentId="450" CreationDate="2010-07-22T07:52:24.933" Score="0" ViewCount="0" Body="&lt;p&gt;No.  Everything is determined by a choice of basis.  For a more in-depth answer, I would need to explain the first two weeks of linear algebra and draw some commutative diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like a better explanation, see pages 12-14 of Emil Artin's monograph &lt;em&gt;Geometric Algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-08T20:21:00.297" LastActivityDate="2010-08-08T20:21:00.297" CommentCount="4" />
  <row Id="454" PostTypeId="2" ParentId="356" CreationDate="2010-07-22T07:59:14.977" Score="6" ViewCount="0" Body="&lt;h3&gt;The Art of Computer Programming&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;(Donald Knuth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The legendary book (of multiple volumes, still incomplete) can't go without mention. For learning about algorithms and their complexities, there is no rival. It's written with practicality in mind, though from a largely theoretical perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/The_art_of_computer_programming&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/en/6/62/ArtOfComputerProgramming.jpg&quot; alt=&quot;The Art of Computer Programming&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-22T07:59:14.977" CommunityOwnedDate="2010-07-22T07:59:14.977" CommentCount="2" />
  <row Id="455" PostTypeId="2" ParentId="450" CreationDate="2010-07-22T08:00:36.677" Score="1" ViewCount="0" Body="&lt;p&gt;You can't represent a non linear transformation with a matrix, however there are some tricks (for want of a better word) available if you use homogenous co-ordinates. For example, 3d translation is a non-linear transformation in a 3x3 3d transformation matrix, but is a linear transformation in 3d homogenous co-ordinates using a 4x4 transformation matrix. The same is true of other things like perspective projections. This is why 4x4 matrices are used in 3d graphics as the homogenous co-ordinate system simplifies things a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify - using homogenous co-ordinates increases the range of transformations representable using matrices from plain linear transformations to affine transformations and some projections, but it doesn't make all non-linear transformations representable using matrices. The non-linear transformation provided as an example is still beyond representation as an affine transformation (Thanks to @Harry for prompting this clarification in the comments)&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-22T10:39:48.367" LastActivityDate="2010-07-22T10:39:48.367" CommentCount="3" />
  <row Id="456" PostTypeId="2" ParentId="450" CreationDate="2010-07-22T08:19:00.767" Score="3" ViewCount="0" Body="&lt;p&gt;As Harry says, you can't (the example of affine transformations can be tweaked to work because they're just linear ones with the origin translated). However, approximating a nonlinear function by a linear one is something we do all the time in calculus through the derivative, and is what we often have to do to make a mathematical model of some real-world phenomenon tractable.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-22T08:19:00.767" CommentCount="2" />
  <row Id="457" PostTypeId="1" CreationDate="2010-07-22T08:34:29.107" Score="28" ViewCount="833" Body="&lt;p&gt;I like to use physical demonstrations when teaching mathematics (putting physics in the service of mathematics, for once, instead of the other way around), and it'd be great to get some more ideas to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for nontrivial ideas in abstract mathematics that can be demonstrated with some contraption, construction or physical intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one can restate Euler's proof that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ in terms of the flow of an incompressible fluid with sources at the integer points in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, consider the problem of showing that, for a convex polyhedron whose $i^{th}$ face has area A_i and outward facing normal vector $n_i$, $\sum A_i \cdot n_i = 0$. One can intuitively show this by pretending the polyhedron is filled with gas at uniform pressure. The force the gas exerts on the i_th face is proportional to $A_i \cdot n_i$, with the same proportionality for every face. But the sum of all the forces must be zero; otherwise this polyhedron (considered as a solid) could achieve perpetual motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example showing less basic mathematics, consider &quot;showing&quot; the double cover of SO(3) by SU(2) by needing to rotate your hand 720 degrees to get it back to the same orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have more demonstrations of this kind?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-11-11T16:59:54.103" CommunityOwnedDate="2010-07-22T08:34:29.107" Title="Good Physical Demonstrations of Abstract Mathematics" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;physics&gt;&lt;education&gt;&lt;teaching&gt;" AnswerCount="14" CommentCount="2" FavoriteCount="12" />
  <row Id="458" PostTypeId="2" ParentId="381" CreationDate="2010-07-22T08:41:38.497" Score="2" ViewCount="0" Body="&lt;p&gt;They appear in &lt;strong&gt;pattern formation&lt;/strong&gt;, for instance in the numbers of petals on a plant, the segements on a pineapple or pinecone, and the structure of nautilus shells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://library.thinkquest.org/27890/applications5.html&quot; rel=&quot;nofollow&quot;&gt;http://library.thinkquest.org/27890/applications5.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.popmath.org.uk/rpamaths/rpampages/sunflower.html&quot; rel=&quot;nofollow&quot;&gt;http://www.popmath.org.uk/rpamaths/rpampages/sunflower.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/&quot; rel=&quot;nofollow&quot;&gt;http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-07-22T08:41:38.497" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="1" />
  <row Id="460" PostTypeId="1" AcceptedAnswerId="467" CreationDate="2010-07-22T08:54:37.340" Score="11" ViewCount="242" Body="&lt;p&gt;Are there any interesting and natural examples of semigroups that are not monoids (that is, they don't have an identity element)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a bit more precise, I guess I should ask if there any interesting examples of semigroups (X, *) for which there is not a monoid (X, *, e) where e is in X. I don't consider an example like the set of real numbers greater than 10 (considered under addition) to be a sufficiently 'natural' semigroup for my purposes; if the domain can be extended in an obvious way to include an identity element then that's not what I'm after.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:49:24.800" LastActivityDate="2010-12-11T01:58:42.147" CommunityOwnedDate="2010-07-22T12:28:57.663" Title="Are there any interesting semigroups that aren't monoids?" Tags="&lt;big-list&gt;&lt;abstract-algebra&gt;&lt;semigroup&gt;&lt;monoid&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="2" />
  <row Id="461" PostTypeId="2" ParentId="448" CreationDate="2010-07-22T09:01:04.033" Score="1" ViewCount="0" Body="&lt;p&gt;For some reason I find it easier to think in terms of letters of a word being rearranged, and your problem is equivalent to asking how many permutations there are of the word YYYYBBBBB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for counting permutations of words with repeated letters (whose reasoning has been described by Noldorin) gives us the correct answer of 9!/(4!5!) = 126.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-22T09:01:04.033" />
  <row Id="462" PostTypeId="1" AcceptedAnswerId="463" CreationDate="2010-07-22T09:06:01.573" Score="6" ViewCount="175" Body="&lt;p&gt;I'm currently trying to understand the concepts and theory behind some of the common proof verifiers out there, but am not quite sure on the exact nature and construction of the sort of systems/proof calculi they use. Are they essentially based on higher-order logics that use Henkin semantics, or is there something more to it? As I understand, extending Henkin semantics to higher-order logic does not render the formal system any less sound, though I am not too clear on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I'm mainly looking for a general answer with useful examples, here are a few specific questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What exactly is the role of type theory in creating higher-order logics? Same goes with category theory/model theory, which I believe is an alternative.&lt;/li&gt;&#xA;&lt;li&gt;Is extending a) natural deduction, b) sequent calculus, or c) some other formal system the best way to go for creating higher order logics?&lt;/li&gt;&#xA;&lt;li&gt;Where does typed lambda calculus come into proof verification?&lt;/li&gt;&#xA;&lt;li&gt;Are there any other approaches than higher order logic to proof verification?&lt;/li&gt;&#xA;&lt;li&gt;What are the limitations/shortcomings of existing proof verification systems (see below)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia pages on proof verification programs such as &lt;a href=&quot;http://en.wikipedia.org/wiki/HOL_Light&quot; rel=&quot;nofollow&quot;&gt;HOL Light&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Calculus_of_constructions&quot; rel=&quot;nofollow&quot;&gt;Coq&lt;/a&gt;, and &lt;a href=&quot;http://us.metamath.org/&quot; rel=&quot;nofollow&quot;&gt;Metamath&lt;/a&gt; give some idea, but these pages contain limited/unclear information, and there are rather few specific high-level resources elsewhere. There are so many variations on formal logics/systems used in proof theory that I'm not sure quite what the base ideas of these systems are - what is required or optimal and what is open to experimentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a good way of answering this, certainly one I would appreciate, would be a brief guide (albeit with some technical detail/specifics) on how one might go about generating a complete proof calculus (proof verification system) from scratch? Any other information in the form of explanations and examples would be great too, however.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-26T08:50:09.473" LastActivityDate="2010-07-28T06:12:18.627" Title="How do proof verifiers work?" Tags="&lt;math-software&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" />
  <row Id="463" PostTypeId="2" ParentId="462" CreationDate="2010-07-22T09:29:23.427" Score="5" ViewCount="0" Body="&lt;p&gt;I'll answer just part of your question: I think the other parts will become clearer based on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof verifier is essentially a program that takes one argument, a proof representation, and checks that this is properly constructed, and says OK if it is, and either fails silently otherwise, or highlights what is invalid otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the proof representation could just be a sequence of formulae in a Hilbert system: all logics (at least, first-orderisable logics) can be represented in such a way.  You don't even need to say which rule is specified at each step, since it is decidable whether any formula follows by a rule application from earlier formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, though, the proof representations are more complex.  Metamath is rather close to Hilbert systems, but has a rich set of rules.  Coq and LF use (different) typed lambda calculi with definitions to represent the steps, which are computationally quite expensive to check (IIRC, both are PSPACE hard).  And the proof verifier can do much more: Coq allows ML programs to be extracted from proofs.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T11:34:40.453" LastActivityDate="2010-07-22T11:34:40.453" CommentCount="4" />
  <row Id="464" PostTypeId="2" ParentId="381" CreationDate="2010-07-22T09:41:32.900" Score="0" ViewCount="0" Body="&lt;p&gt;It's not really a mathematical application, but the Fibonacci sequence is closely related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Golden_ratio#Relationship_to_Fibonacci_sequence&quot; rel=&quot;nofollow&quot;&gt;Golden Ratio&lt;/a&gt;, used in architecture (amongst others).&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-07-22T09:41:32.900" CommunityOwnedDate="2010-08-03T18:46:03.740" CommentCount="0" />
  <row Id="465" PostTypeId="2" ParentId="83" CreationDate="2010-07-22T10:06:14.170" Score="1" ViewCount="0" Body="&lt;p&gt;Seymour Papert's ideas about &quot;maths you can build&quot;.  Look at &lt;a href=&quot;http://scratch.mit.edu/&quot; rel=&quot;nofollow&quot;&gt;Scratch&lt;/a&gt; (much inspired by Papert), and read &lt;a href=&quot;http://en.wikipedia.org/wiki/Mindstorms%3a_Children,_Computers,_and_Powerful_Ideas&quot; rel=&quot;nofollow&quot;&gt;Mindstorms&lt;/a&gt;; note that quite a few of his ideas there are contentious, but there is a lot of maths you can make exciting this way.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-22T10:06:14.170" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="466" PostTypeId="2" ParentId="440" CreationDate="2010-07-22T10:11:04.430" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;del&gt;It can be, if you use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_relation&quot; rel=&quot;nofollow&quot;&gt;Euclideanity&lt;/a&gt; instead of transitivity: &lt;em&gt;a R c&lt;/em&gt; and &lt;em&gt;b R c&lt;/em&gt; implies &lt;em&gt;a R b&lt;/em&gt;.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, Euclideanity plus seriality characterises the equivalence relations.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-22T11:28:41.117" LastActivityDate="2010-07-22T11:28:41.117" CommentCount="6" />
  <row Id="467" PostTypeId="2" ParentId="460" CreationDate="2010-07-22T10:16:54.910" Score="13" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/Folded.html&quot; rel=&quot;nofollow&quot;&gt;Convolution&lt;/a&gt; of functions/distributions is useful in a variety of fields, and the identity element, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;dirac delta&lt;/a&gt;, is not strictly a function.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-22T10:16:54.910" CommentCount="2" />
  <row Id="468" PostTypeId="2" ParentId="460" CreationDate="2010-07-22T10:29:51.280" Score="0" ViewCount="0" Body="&lt;p&gt;Let G be the set of (continuous) functions f: R -&gt; R where f(x) tends to 0 as x tends to infinity: $lim_{x\to \infty}f(x) = 0$. The operator is the usual point-wise multiplication of functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G is closed under * since lim(f(x)) = 0 and lim(g(x)) = 0 imply lim(f(x)*g(x)) = 0.  G is a subgroup of {f: R-&gt;R}, so the identity must be the same - the function which is constantly 1. But this identity is not in G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; As Harry correctly points out, {f: R-&gt;R} is not a group. Therefore the following correction is needed: consider only functions such that $f(x) \neq 0$ everywhere.&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-07-22T11:36:31.260" LastActivityDate="2010-07-22T11:36:31.260" CommentCount="3" />
  <row Id="469" PostTypeId="1" AcceptedAnswerId="478" CreationDate="2010-07-22T10:52:29.193" Score="3" ViewCount="430" Body="&lt;p&gt;What's the meaning of this symbol in mathematical notation? :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;⊧&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-10-27T10:58:57.823" LastActivityDate="2010-10-27T10:58:57.823" ClosedDate="2010-07-26T06:25:07.010" Title="What is the meaning of this symbol?" Tags="&lt;logic&gt;&lt;notation&gt;&lt;latex&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" />
  <row Id="470" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T10:56:04.867" Score="4" ViewCount="0" Body="&lt;p&gt;It is a symbol from model theory denoting &lt;a href=&quot;http://en.wikipedia.org/wiki/Entailment&quot; rel=&quot;nofollow&quot;&gt;entailment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ⊧ B is read as &quot;A entails B&quot;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-22T10:56:04.867" />
  <row Id="472" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T11:07:52.947" Score="3" ViewCount="0" Body="&lt;p&gt;It is called a 'double turnstile': &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_turnstile&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Double_turnstile&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-22T11:07:52.947" CommentCount="6" />
  <row Id="473" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T11:22:43.013" Score="1" ViewCount="0" Body="&lt;p&gt;$\models$ is also known as the &lt;a href=&quot;http://planetmath.org/encyclopedia/SatisfactionRelation.html&quot; rel=&quot;nofollow&quot;&gt;satisfication relation&lt;/a&gt;. For a structure $\mathcal{M}=(M,I)$ and an $\mathcal{M}$-assignment $\nu$, $(\mathcal{M},\nu)\models \varphi$ means that the formula $\varphi$ is true with the particular assignment $\nu$.&lt;br&gt;&#xA;See &lt;a href=&quot;http://www.trinity.edu/cbrown/topics_in_logic/struct/node2.html&quot; rel=&quot;nofollow&quot;&gt;http://www.trinity.edu/cbrown/topics_in_logic/struct/node2.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-22T11:22:43.013" />
  <row Id="474" PostTypeId="2" ParentId="460" CreationDate="2010-07-22T11:52:19.533" Score="1" ViewCount="0" Body="&lt;p&gt;Finite sets of matrices of varying dimensions, where the product A*B={PQ|P in A &amp;amp; Q in B &amp;amp; dim(Q)=codim(P)}, and dim &amp;amp; codim are the dimensions of the source &amp;amp; target spaces of a matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinite case has an obvious unit.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-22T11:52:19.533" />
  <row Id="475" PostTypeId="2" ParentId="457" CreationDate="2010-07-22T11:57:18.243" Score="10" ViewCount="0" Body="&lt;p&gt;The book &quot;The Mathematical Mechanic&quot; by Mark Levi is a very good source of such examples, which Levi has been collecting for some time. The first two here are in the book, if I recall correctly.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-22T11:57:18.243" CommunityOwnedDate="2010-07-22T11:57:18.243" />
  <row Id="476" PostTypeId="1" CreationDate="2010-07-22T12:03:10.743" Score="0" ViewCount="86" Body="&lt;p&gt;Many years ago, I had read a book entitled &quot;Think of a Number&quot; by Malcolm E. Lines, and it was an eminently readable and thought provoking book. In the book, there were topics like Fibonacci numbers (along with the live examples from the nature) and Golden Section. Now I'm looking for a similar book. Can anyone recommend me one?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T20:08:51.293" LastActivityDate="2010-07-22T20:08:51.293" Title="Looking for a book similar to &quot;Think of a Number&quot;" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="4" />
  <row Id="477" PostTypeId="1" AcceptedAnswerId="479" CreationDate="2010-07-22T12:07:38.830" Score="5" ViewCount="197" Body="&lt;p&gt;The set of all ℝ → ℝ continuous functions is &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinality_of_the_continuum&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;c&lt;/strong&gt;&lt;/a&gt;. How to show that? Is there are bijection between ℝ&lt;sup&gt;n&lt;/sup&gt; and the set of continuous functions?&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-22T12:19:00.337" Title="Cardinality of set of real continuous functions" Tags="&lt;calculus&gt;&lt;set-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="478" PostTypeId="2" ParentId="469" CreationDate="2010-07-22T12:11:48.120" Score="7" ViewCount="0" Body="&lt;p&gt;Just to enlarge on Harry's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your symbol denotes one of two specified notions of implication in formal logic &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdash$ -the &lt;strong&gt;turnstile&lt;/strong&gt; symbol denotes &lt;strong&gt;syntactic&lt;/strong&gt; implication (syntactic here means  related to syntax, the structure of a sentence), where the 'algebra' of the logical system in play (for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Sentential_calculus&quot; rel=&quot;nofollow&quot;&gt;sentential calculus&lt;/a&gt;) allows us to 'rearrange and cancel' the stuff we know on the left into the thing we want to prove on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example might be the classic &quot;all men are mortal $\wedge$ socrates is a man $\vdash$ socrates is mortal&quot; ('$\wedge$' of course here just means 'and'). You can almost imagine cancelling out the 'man bit' on the left to just give the sentence on the right (although the truth may be more complex...).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\models$ -the &lt;strong&gt;double turnstile&lt;/strong&gt;, on the other hand, is not so much about algebra as meaning (formally it denotes &lt;strong&gt;semantic&lt;/strong&gt; implication)- it means that any interpretation of the stuff we know on the left must have the corresponding interpretation of the thing we want to prove on the right true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be if we had an infinite set of sentences: $\Gamma$:= {&quot;1 is lovely&quot;, &quot;2 is lovely&quot;, ...} in which all numbers appear, and the sentence A= &quot; the natural numbers are precisely {1,2,...}&quot; listing all numbers. Any interpretation would give us B=&quot;all natural numbers are lovely&quot;. So $\Gamma$, A $\models$ B. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, the goal of any logician trying to set up a formal system is to have $\Gamma \vdash A \iff \Gamma \models B$, meaning that the 'algebra' must line up with the interpretation, and this is not something we can take as given. Take the second example above- can we be sure that algebraic operations can 'parse' those infinitely many sentences and make the simple sentence on the right (this is to do with a property called &lt;strong&gt;compactness&lt;/strong&gt;)??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal can be split into two distict subgoals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Soundness:&lt;/strong&gt; $A \vdash B \Rightarrow A \models B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Completeness:&lt;/strong&gt; $A \models B \Rightarrow A \vdash B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the first stops you proving things that aren't true when we interpret them and the second means that everything we know to be true on interpretation, we must be able to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sentential calculus, for example, can be proved complete (and was in Godel's lesser known, but celebrated completeness theorem), but other for other systems Godel's incompleteness theorem, give us a terrible choice between the two.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary:&lt;/strong&gt; The interplay of meaning and axiomatic machine mathematics, captured by the difference between $\models$ and $\vdash$, is a subtle and interesting thing.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T21:42:11.713" LastActivityDate="2010-09-16T21:42:11.713" CommentCount="2" />
  <row Id="479" PostTypeId="2" ParentId="477" CreationDate="2010-07-22T12:19:00.337" Score="11" ViewCount="0" Body="&lt;p&gt;The cardinality is at least that of the continuum because every real number corresponds to a constant function.  The cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $R^{N}$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Schroeder-Bernstein theorem&lt;/a&gt; now implies the cardinality is precisely that of the continuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that  then the set of sequences of reals is also of the same cardinality as the same cardinality as the reals.  This is because if we have a sequence of binary representations $.a_1a_2..., .b_1b_2..., .c_1c_2...$, we can splice them together via $.a_1 b_1 a_2 c_1 b_2 a_3...$ so that a sequence of reals can be encoded by one real number.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-22T12:19:00.337" CommentCount="5" />
  <row Id="480" PostTypeId="2" ParentId="457" CreationDate="2010-07-22T13:05:26.903" Score="6" ViewCount="0" Body="&lt;p&gt;There is the hairy ball theorem, which states that no even dimensional sphere admits a nowherevanishing continuous vector field.  In the case of $S^2$, the physical demonstration of this is that one cannot comb the hair of a ball without getting a cowlick.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T13:05:26.903" CommunityOwnedDate="2010-07-22T13:05:26.903" CommentCount="3" />
  <row Id="481" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T13:15:36.210" Score="1" ViewCount="0" Body="&lt;p&gt;did you try the books of Eastaway and Wyndham? &lt;em&gt;Why Do Buses Come in Threes?&lt;/em&gt; and &lt;em&gt;How Long Is a Piece of String?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blurb for the first one says &quot;An amusing explanation of how maths is relevant to almost everything in life. Citing many examples of the way mathematics can explain common phenomena&quot;; for the second, &quot;This title is for anyone wanting to remind themselves - or discover for the first time - that maths is relevant to almost everything we do. Dating, cooking, travelling by car, gambling and ranking sportsmen all have links with intriguing mathematical problems that are explained in this book&quot;. &lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-22T14:49:47.403" LastActivityDate="2010-07-22T14:49:47.403" />
  <row Id="482" PostTypeId="2" ParentId="438" CreationDate="2010-07-22T13:19:27.403" Score="2" ViewCount="0" Body="&lt;p&gt;Isaac's &lt;a href=&quot;http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake/439#439&quot; rel=&quot;nofollow&quot;&gt;answer&lt;/a&gt; is correct, but it can be hard to see if you don't have a strong knowledge of your laws. These problems are generally easy to solve if you examine it line by line and simplify both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt(-1) = î&#xA;LHS=i, RHS=i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/sqrt(-1) = 1/î&#xA;LHS=1/i=-i, RHS=-i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt(1)/sqrt(-1) = 1/î&#xA;LHS=1/i=-i, RHS=-i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt(1/-1) = 1/î&#xA;LHS=sqrt(-1)=i, RHS=-i&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then see that the error must be assuming sqrt(1)/sqrt(-1)=sqrt(1/-1).&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-22T13:19:27.403" CommentCount="2" />
  <row Id="483" PostTypeId="2" ParentId="329" CreationDate="2010-07-22T13:20:23.173" Score="16" ViewCount="0" Body="&lt;p&gt;Ireland and Rosen  hands down!&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T13:20:23.173" CommunityOwnedDate="2010-07-22T13:20:23.173" CommentCount="1" />
  <row Id="484" PostTypeId="2" ParentId="164" CreationDate="2010-07-22T13:22:16.210" Score="3" ViewCount="0" Body="&lt;p&gt;Integration. \int 4pi*r^2 dr = 4/3 pi r^3. &lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T13:22:16.210" CommentCount="4" />
  <row Id="486" PostTypeId="1" AcceptedAnswerId="495" CreationDate="2010-07-22T13:28:58.457" Score="7" ViewCount="176" Body="&lt;p&gt;When I used to compete in Olympiad Competitions back in high school, a decent number of the easier geometry questions were solvable by what we called a geometry bash. Basically, you'd label every angle in the diagram with the variable then use a limited set of basic geometry operations to find relations between the elements, eliminate equations and then you'd eventually get the result. It seems like the kind of thing you could program a computer to do. So, I'm curious, does there exist any software to do this? I know there is lots of software for solving equations, but is there anything that lets you actually input a geometry problem without manually converting to equations? I'm not looking for anything too advance, even seeing just an attempt would be interesting. If there is anything decent, I think it'd be rather interesting to run the results on various competitions and see how many of the questions it solves.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-11T06:46:03.067" LastActivityDate="2010-08-11T06:46:03.067" Title="Software for solving geometry questions" Tags="&lt;algebraic-geometry&gt;&lt;logic&gt;&lt;euclidean-geometry&gt;&lt;math-software&gt;&lt;quantifier-elimination&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="487" PostTypeId="2" ParentId="2" CreationDate="2010-07-22T13:29:04.040" Score="0" ViewCount="0" Body="&lt;p&gt;A rigorous mathematical treatment follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say the limit of f(x) as x approaches c is \tau iff for all epsilon in R+ there exists delta in R+ such that |x-c|&amp;lt; \delta implies |f(x)-\tau| &amp;lt; epsilon.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T13:29:04.040" CommentCount="1" />
  <row Id="488" PostTypeId="2" ParentId="75" CreationDate="2010-07-22T13:34:17.930" Score="1" ViewCount="0" Body="&lt;p&gt;In the tactile sense, groups and rings do not &quot;feel&quot; very similar. Indeed, groups of grapes feel very plush and juicy, while rings are metallic and hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HAHAHHA Just kidding! xD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically look. The intro abstract algebra idea is the following (something you might feel is the purpose of the field {nice pun, eh?} if you read as silly a book as Fraleigh): Take a set $S.$ Create an operation and impose some structural conditions. A ring is just more structure on a group (an abelian group - abelian grape hahah!-, in fact) where you introduce a second operation. In this way, a field is the ultimate beast because it has two operations that are invertible.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T13:34:17.930" />
  <row Id="491" PostTypeId="2" ParentId="2" CreationDate="2010-07-22T13:45:06.123" Score="0" ViewCount="0" Body="&lt;p&gt;There are two slightly different types of limits (at least, in one real variable, as in single-variable calculus): limits as a variable goes to positive or negative infinity, and limits as a variable goes to some real value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit of f(x) as x goes to infinity (or as x goes to negative infinity) is asking what happens to f(x) at the right (left) end of its graph.  Does f(x) get arbitrarily close to a single real number as x increases (or decreases) without bound--that is, as x gets really big and positive (or really big and negative)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit of f(x) as x goes to some real value, say c, is asking what happens to f(x) near, but not at, x = c.  Does f(x) get arbitrarily close to a single real number as x gets arbitrarily close to c?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Arbitrarily close&quot; is used somewhat informally above, but these ideas can be translated into more rigorous definitions (as in other people's answers) by reformulating &quot;arbitrarily close&quot; in terms of inequalities with &amp;epsilon; and/or &amp;delta;.  Also, while a limit is said to not exist if there isn't the single real number as described above, it is somewhat common to talk about the limit being infinity (or negative infinity) if the limit does not exist, but the values of f(x) increase without bound (or decrease without bound).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-22T13:45:06.123" />
  <row Id="493" PostTypeId="2" ParentId="81" CreationDate="2010-07-22T13:46:56.540" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://amathew.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;http://amathew.wordpress.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He climbs Mount Bourbaki with the grace of a leaping gazelle. &lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditorDisplayName="" LastEditDate="2010-07-22T17:25:02.870" LastActivityDate="2010-07-22T17:25:02.870" CommunityOwnedDate="2010-07-22T13:46:56.540" />
  <row Id="494" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T14:27:22.967" Score="1" ViewCount="0" Body="&lt;p&gt;I don't know the book you are speaking of, but for some reason I am reminded of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198516231&quot; rel=&quot;nofollow&quot;&gt;1089, and all that&lt;/a&gt; by David Acheson.  This is a charming book about some mathematical curiosities and can be appreciated by the professional and enjoyed by the layperson.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-22T14:27:22.967" />
  <row Id="495" PostTypeId="2" ParentId="486" CreationDate="2010-07-22T14:30:51.677" Score="7" ViewCount="0" Body="&lt;p&gt;This is just a special case of automated theorem proving. A nice thing is some geometry problem can indeed be solved by an algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are theories show such system can be realized algorithmically. &lt;strike&gt;I don't know if anyone have really wrote the program to do it.&lt;/strike&gt; &lt;a href=&quot;http://woody.cs.wichita.edu/&quot; rel=&quot;nofollow&quot;&gt;JGEX&lt;/a&gt; seems to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mechanical geometry proof technique was popularized in China by &lt;a href=&quot;http://woody.cs.wichita.edu/&quot; rel=&quot;nofollow&quot;&gt;Jingzhong Zhang&lt;/a&gt;. He first introduced it as a way for machines to solve geometric problems relating the proportions between areas, lengths or angles. Then some Olympiad people I know start using it to bash that kind of problem. I don't know what the name is in English, but a literal translation of the method is &quot;point removal method&quot;. Although it's not exactly same as what you are talking about, because &quot;input a geometry problem&quot; requires you to provide the construction of the problem from a straight edge and a compass, which is almost like &quot;manually converting to equations&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the basic idea:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Construct the original problem by compass and straightedge, make a list of every constructed point ordered by the order of construction, let it be L. Record which points are used in the construction of every point.(Only the points used to construct point P are used in the substitution steps 3 and 4)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Translate the theorem into a equivalent form. Usually a/b = 1, where a and b are functions of length and area of certain segments or triangles. Let's call this equation E.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let P be the last point of L. For every P appears in E, substitute it with another relation using other points from L(P itself is also allowed), usually if we are proving about lengths, we might use area. A list of possible operations are required for this step. It can branches off as a proof tree when the program decide to use different substitutions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do another substitution that eliminates the point P. For example, if in the step before, we substitute length to area, then we want to find something involve the length.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do 3 to 4 over and over until we have 1=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;An example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle bisector theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: $AD$ is the angle bisector of $\angle BAC$ of triangle $\triangle ABC$. Let $XYZ$ be the area of triangle $\triangle XYZ$, and $XY$ be the length of segment $XY$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove: $\frac{AB}{AC} = \frac{BD}{DC}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First construct $ABC$. Then construct $AD$. The points in the list $L$ are $A,B,C,D$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The equivalent equation to the theorem is&#xA;$\frac{AB}{AC} \frac{DC}{BD} = 1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{AB}{AC} \frac{DC}{BD} = \frac{AB}{AC} \frac{ACD}{ABD}$ (substitute length with area)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$=\frac{AB}{AC} \frac{\frac{1}{2} AC\cdot AD \sin \angle CAD}{\frac{1}{2} AB\cdot AD \sin \angle BAD}$, this step successfully remove point $D$ by cancellation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$=\frac{AB}{AC} \frac{AC}{AB} =1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is only a non-formal explanation of how such automated system would work. I think the following book from Zhang will tell you more about it: Machine proofs in geometry: automated production of readable proofs for geometry theorems. I did not read the book, but the description of it seems like what you are seeking. &lt;a href=&quot;http://woody.cs.wichita.edu/help/pdf/&quot; rel=&quot;nofollow&quot;&gt;A few paper by Zhang and his colleague&lt;/a&gt; can be found in the JGEX's website. &lt;a href=&quot;http://woody.cs.wichita.edu/help/atp.html&quot; rel=&quot;nofollow&quot;&gt;The JGEX documentation on it's automated theorem prover&lt;/a&gt; is also a great resource.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-22T14:39:28.130" LastActivityDate="2010-07-22T14:39:28.130" CommentCount="2" />
  <row Id="496" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T14:30:59.960" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805062998&quot; rel=&quot;nofollow&quot;&gt;The Number Devil&lt;/a&gt; may be something like what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Young Robert's dreams have taken a decided turn for the weird. Instead of falling down holes and such, he's visiting a bizarre magical land of number tricks with the number devil as his host. Starting at one and adding zero and all the rest of the numbers, Robert and the number devil use giant furry calculators, piles of coconuts, and endlessly scrolling paper to introduce basic concepts of numeracy, from interesting number sequences to exponents to matrices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(It's not a watered-down kids' book, even though the description might suggest it.)&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-22T14:30:59.960" />
  <row Id="497" PostTypeId="2" ParentId="81" CreationDate="2010-07-22T14:34:07.113" Score="5" ViewCount="0" Body="&lt;p&gt;I find &lt;a href=&quot;http://qchu.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Annoying Precision&lt;/a&gt; to be wonderfully readable, and has many many interesting topics. Additionally, &lt;a href=&quot;http://rigtriv.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Rigorous Trivialities&lt;/a&gt; is a bit higher level, but has a really useful intro to Algebraic Geometry.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-22T14:34:07.113" CommunityOwnedDate="2010-07-22T14:34:07.113" CommentCount="2" />
  <row Id="498" PostTypeId="2" ParentId="213" CreationDate="2010-07-22T14:45:31.017" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Category theory and algebraic geometry&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a lot of time in undergrad studying things that were kinda nifty, but way to classical to be of any use/interest beyond &quot;fun math&quot;. When I got to grad school, &lt;strong&gt;Category theory&lt;/strong&gt; was assumed and made some of my courses much harder than they should of been. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the words of Ravi Vakil &quot;&lt;strong&gt;Algebraic Geometry&lt;/strong&gt; should be learned slowly over a number of years&quot;. I currently &lt;em&gt;NEED&lt;/em&gt; algebraic geometry, so I don't have this number of years. I wish I would of started that a long time ago. Additionally, both of these topics would of helped me learn the things I was thinking about anyways, in particular &lt;em&gt;commutative algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-16T18:45:40.760" LastActivityDate="2010-08-16T18:45:40.760" CommunityOwnedDate="2010-07-22T14:45:31.017" CommentCount="4" />
  <row Id="499" PostTypeId="2" ParentId="476" CreationDate="2010-07-22T14:54:21.263" Score="2" ViewCount="0" Body="&lt;p&gt;Another more general pop math book covering mathematical curiosities is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393059456&quot; rel=&quot;nofollow&quot;&gt;Coincidences, Chaos, and All that Math Jazz&lt;/a&gt;. It is at the level you're talking about. A more focused book on the golden ratio is that by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0767908163&quot; rel=&quot;nofollow&quot;&gt;Mario Livio&lt;/a&gt;. One nice feature of this second book is that he does a nice job of pointing out places where people believe in coincidences that aren't really there. He stays nicely objective, where many others fail to.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-22T14:54:21.263" CommentCount="1" />
  <row Id="500" PostTypeId="2" ParentId="213" CreationDate="2010-07-22T15:52:58.950" Score="2" ViewCount="0" Body="&lt;p&gt;I did mathematics as an undergrad, and I thought that differential equations were boring and pointless. Type of diff eq -&gt; existence and uniqueness proofs for solutions -&gt; rinse and repeat. Yawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I find my lack of knowledge of differential equations is hampering my learning some interesting parts of physics that I'd like to know more about...&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-22T15:52:58.950" CommunityOwnedDate="2010-07-22T15:52:58.950" />
  <row Id="501" PostTypeId="1" AcceptedAnswerId="1878" CreationDate="2010-07-22T15:58:45.687" Score="7" ViewCount="459" Body="&lt;p&gt;An extreme form of constructivism is called &lt;em&gt;finitisim&lt;/em&gt;.  In this form, unlike the standard axiom system, infinite sets are not allowed.  There are important mathematicians, such as Kronecker, who supported such a system.  I can see that the natural numbers and rational numbers can easily defined in a finitist system, by easy adaptations of the standard definitions.   But in order to do any significant mathematics, we need to have definitions for the irrational numbers that one is likely to encounter in practice, such as  e  or sqrt(2).  In standard constructions, real numbers are defined as Dedekind cuts, or Cauchy sequences, which are actually sets of infinite cardinality, so they are of no use here. My question is, how would a real number like those be defined in a finitist axiom system.  (Of course we have no hope to construct the entire set of real numbers, since that set is uncountably infinite.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing a little research  I found a constructivist definition in Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Constructivism_(mathematics)#Example_from_real_analysis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Constructivism_(mathematics)#Example_from_real_analysis&lt;/a&gt; , but we need a finitist definition of a function for this definition to work. (Because in the standard system, a function over the set of natural numbers is actually an infinite set.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question boils down to this:  How can we define a function f over the natural numbers in a finitist axiom system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Original version of this question, &lt;a href=&quot;http://meta.math.stackexchange.com/questions/172/why-did-you-close-my-question-if-all-sets-were-finite&quot;&gt;which had been closed during private beta&lt;/a&gt;, is as follows:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;If all sets were finite, how would mahtematics be like?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If we replace the axiom that 'there&#xA;  exists an infinite set' with 'all sets&#xA;  are finite', how would mathematics be&#xA;  like? My guess is that, all the theory&#xA;  that has practical importance would&#xA;  still show up, but everything would be&#xA;  very very unreadable for humans. Is&#xA;  that true?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We would have the natural numbers,&#xA;  athough the class of all natural&#xA;  numbers would not be a set. In the&#xA;  same sense, we could have the rational&#xA;  numbers. But could we have the real&#xA;  numbers? Can the standard&#xA;  constructions be adapted to this&#xA;  setting?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-09T16:19:49.210" LastActivityDate="2010-08-09T16:24:44.743" Title="If all sets were finite, how could the real numbers be defined?" Tags="&lt;logic&gt;&lt;set-theory&gt;&lt;proof-theory&gt;" AnswerCount="6" CommentCount="9" FavoriteCount="1" />
  <row Id="502" PostTypeId="2" ParentId="501" CreationDate="2010-07-22T16:14:27.127" Score="-1" ViewCount="0" Body="&lt;p&gt;You wouldn't even have the natural numbers (in totality), let alone the real numbers (or complex numbers) at all. The reals form an &lt;em&gt;uncountably infinite&lt;/em&gt; set and it is the arbitrarily small spacing between them that really gives them their unique properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, you certainly wouldn't have anything like calculus and thus no mathematical analysis whatsoever! Mathematics would not be mathematics in effect. It's almost an inherent contradiction to ask this question. Does that not say enough?&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-07-22T16:30:15.240" LastActivityDate="2010-07-22T16:30:15.240" CommentCount="11" />
  <row Id="503" PostTypeId="2" ParentId="501" CreationDate="2010-07-22T16:14:48.077" Score="0" ViewCount="0" Body="&lt;p&gt;well: sqrt(2), e and \pi were defined before set theory was even conceived ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that algebraic numbers could have been defined without resorting to the standard theory of real numbers, just looking at them as roots of polynomials; certain classes of transcendent numbers could also, but I wonder if there is &quot;finitistic&quot; way to define e. I am at loss with this.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-22T16:26:06.820" LastActivityDate="2010-07-22T16:26:06.820" CommentCount="6" />
  <row Id="504" PostTypeId="2" ParentId="501" CreationDate="2010-07-22T16:28:14.133" Score="3" ViewCount="0" Body="&lt;p&gt;There is a fragment of mathematics that is given by a set of axioms known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Peano_axioms&quot; rel=&quot;nofollow&quot;&gt;Peano axioms&lt;/a&gt;. Using these rules you can carry out a vast amount of mathematics relating to natural numbers. For example you can prove lots of theorems in number theory using these axioms. The Peano axioms make no reference to sets at all, whether finite or infinite. The only things that exist in this theory are naturals. You can't even form the set of all integers. You can only talk about the naturals themselves. So a vast amount of mathematics would work absolutely fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though Peano's axioms are about naturals, you can already use them to talk about finite sets. The idea is that any finite set could be encoded as a finite sequence of symbols which in turn could be represented as naturals using &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del_numbering&quot; rel=&quot;nofollow&quot;&gt;Godel numbering&lt;/a&gt;. So questions like &quot;is this set a subset of that one?&quot; could be turned into purely arithmetical statements about Godel numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm pretty sure that declaring that there is no infinite set would make little difference to people working within the system defined by Peano's axioms. We'd still have all of the natural numbers to work with, we just wouldn't be able to assemble them into a single entity, the set of all natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are theorems that make essential use of an infinite set. Like &lt;a href=&quot;http://en.wikipedia.org/wiki/Goodstein%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Goodstein's theorem&lt;/a&gt;. Without infinite sets (or a substitute of some sort) it would be impossible to prove this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the overall result would be, I think, that you could still do lots of mathematics fine. The mathematics you could do wouldn't be all that weird. And you'd simply be depriving yourself of a useful proof technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, you'd still be able to say many things about real numbers. A real number can be thought of as a Cauchy sequence. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_sequence&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequence&lt;/a&gt; is a certain type of sequence of rational numbers. So many statements about real numbers, when unpacked, are really statements about rational, and hence naturals, but in disguise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Uncovering precisely what parts of mathematics you need in order to prove things is a field known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_mathematics&quot; rel=&quot;nofollow&quot;&gt;reverse mathematics&lt;/a&gt;. Hilbert, and others mathematicians, were interested in trying to prove as much mathematics as possible using finite methods. Although it was ultimately shown that you can't carry out all mathematics using finite methods, it's surprising how much you can. &lt;a href=&quot;http://www.andrew.cmu.edu/user/avigad/Papers/elementary.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a paper that talks about a system called EA which has no infinite sets. Amazingly we can use results from &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_number_theory&quot; rel=&quot;nofollow&quot;&gt;analytic number theory&lt;/a&gt; in EA. This is because propositions about analytic functions can be interpreted as statements about natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="80" LastEditorUserId="80" LastEditorDisplayName="" LastEditDate="2010-07-22T16:50:17.980" LastActivityDate="2010-07-22T16:50:17.980" CommentCount="5" />
  <row Id="505" PostTypeId="1" AcceptedAnswerId="507" CreationDate="2010-07-22T16:52:36.447" Score="8" ViewCount="376" Body="&lt;p&gt;Akhil showed that the &lt;a href=&quot;http://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions/479#479&quot;&gt;Cardinality of set of real continuous functions&lt;/a&gt; is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree everywhere, since the rationals are dense in the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't an obvious step, so why is it true?&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-08-09T16:01:43.040" LastActivityDate="2010-08-09T16:01:43.040" Title="Can there be two distinct, continuous functions that are equal at all rationals?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="506" PostTypeId="2" ParentId="505" CreationDate="2010-07-22T16:57:33.693" Score="7" ViewCount="0" Body="&lt;p&gt;If there were two continuous functions $f(x)$ and $g(x)$ that were equal at all rationals, then (because the rationals are dense) we can show that $\lim_{x \to a} f(x) - g(x) = 0$ for all values of $a$ using a delta-epsilon proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the difference of two continuous functions is continuous, we know $\lim_{x \to a} f(x) - g(x) = f(a) - g(a)$ for all $a$, and therefore $f(a) - g(a) = 0$ and $f(x) = g(x)$, proving that $f$ and $g$ must be identical.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-22T17:04:22.980" LastActivityDate="2010-07-22T17:04:22.980" CommentCount="2" />
  <row Id="507" PostTypeId="2" ParentId="505" CreationDate="2010-07-22T17:06:50.840" Score="10" ViewCount="0" Body="&lt;p&gt;Without resorting to &amp;epsilon;-&amp;delta; arguments: Let f and g be continuous real functions and f(x) = g(x) for all rational x.  For any real number c (in particular, an irrational c), there exists a Cauchy sequence of rational numbers  such that $\lim_{n \to \infty}x_{n}=c$.  Since f and g are continuous, $\lim_{n \to \infty}f({x_{n}})=f({c})$ and $\lim_{n \to \infty}g({x_{n}})=g({c})$.  Since $x_n$ is rational, $f(x_n) = g(x_n)$ for all n, so the two limits must be equal and so f(c) = g(c) for all real c.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-03T18:01:20.823" LastActivityDate="2010-08-03T18:01:20.823" CommentCount="2" />
  <row Id="508" PostTypeId="2" ParentId="486" CreationDate="2010-07-22T17:24:19.657" Score="7" ViewCount="0" Body="&lt;p&gt;You might be interested in Doron Zeilberger's website. He has a page entitled &quot;Plane Geometry: An Elementary Textbook (Circa 2050)&quot; where he envisioned a world in which computers can derive all of plane geometry without human intervention or interference. The accompanying Maple package proves many statements by computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page exists at &lt;a href=&quot;http://www.math.rutgers.edu/~zeilberg/PG/gt.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.rutgers.edu/~zeilberg/PG/gt.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-22T17:24:19.657" CommentCount="1" />
  <row Id="509" PostTypeId="2" ParentId="234" CreationDate="2010-07-22T18:06:36.787" Score="3" ViewCount="0" Body="&lt;p&gt;This paradox has always interested me.  Something to think about is that there does not exist a uniform probability distribution over the positive real numbers (since they are infinite).  In arriving at your paradox, it seems you are assuming that any real number is equally likely, but this cannot be the case.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T18:06:36.787" />
  <row Id="510" PostTypeId="2" ParentId="234" CreationDate="2010-07-22T18:33:41.163" Score="2" ViewCount="0" Body="&lt;p&gt;This puzzle is known as the two envelope paradox. &lt;a href=&quot;http://www.franzdietrich.net/Papers/DietrichList-TwoEnvelopeParadox.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; contains a nice explanation of the two envelope paradox, and some references to further literature regarding the puzzle.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-22T18:33:41.163" CommentCount="1" />
  <row Id="511" PostTypeId="2" ParentId="81" CreationDate="2010-07-22T18:37:39.953" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://wildaboutmath.com/&quot; rel=&quot;nofollow&quot;&gt;Wild about Math!&lt;/a&gt; and &lt;a href=&quot;http://satmathblog.com/&quot; rel=&quot;nofollow&quot;&gt;SatMathBlog&lt;/a&gt; (more for amateurs)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T18:37:39.953" CommunityOwnedDate="2010-07-22T18:37:39.953" />
  <row Id="512" PostTypeId="2" ParentId="234" CreationDate="2010-07-22T20:01:49.750" Score="2" ViewCount="0" Body="&lt;p&gt;Edit: My original answer was slightly misleading. Here is a better explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the number on the card, and see that it is x, where x is a real number, you can deduce that your expected return is x if you stay, or 1.25x if you switch. So obviously you should switch, and &lt;strong&gt;there is no paradox the way you've stated the question&lt;/strong&gt;. It's better to switch because there's a fifty-fifty chance of gaining/losing, but if you gain, the difference is greater than if you lose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'paradox' comes from applying that expected return to the situation &lt;strong&gt;without looking at the card first&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If I were to look at the first card, I'd see that it's better to switch.&lt;/li&gt;&#xA;&lt;li&gt;But then if I look at the second card, I'd see it's better to switch.&lt;/li&gt;&#xA;&lt;li&gt;But if it would be better to switch from the second card, I should just take the first card.&lt;/li&gt;&#xA;&lt;li&gt;But then I'd see that I should take the second card...&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Which card do I pick, if 1.25x is always m y expected return?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answer is that you should pick either card, and then switch, because it's not &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;expected return = 1.25x, where x is the number on either card&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The correct expression is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;expected return = [first card chosen]*1.25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So no matter which card you choose, that's your expected return. And if your expected return is always the same, there is no point in switching.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;original answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion you've reached is incorrect - x is a completely random variable, and multiplying it won't help us reach any conclusions. Read the &lt;a href=&quot;http://en.wikipedia.org/wiki/Two_envelopes_problem&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; about the two envelopes problem for more on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is some correct math to demonstrate that switching won't help matters at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might have the 'half' envelope H, or we might have the 'double' envelope D. Let E denote the expected value of our earnings if we open the envelope now:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E = 0.5H + 0.5D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;D is 2H, so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E = 0.5H + 0.5*(2H)&#xA;E = 1.5H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we switch, our expected earnings are still 1.5H because the same math applies, so switching will not help us. QED.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditorDisplayName="" LastEditDate="2010-07-22T23:44:39.753" LastActivityDate="2010-07-22T23:44:39.753" CommentCount="12" />
  <row Id="513" PostTypeId="2" ParentId="1" CreationDate="2010-07-22T20:45:36.930" Score="3" ViewCount="0" Body="&lt;p&gt;Infinity is an overloaded term that can mean many things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common non-mathematical use of infinity is to refer to everything in the universe. This is &lt;strong&gt;not&lt;/strong&gt; what mathematicians mean when they say infinity. That would be a kin to the set of all sets, which is a paradoxical concept that is not part of mathematical discourse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians will use infinity as a way to represent a process that continues indefinitely. This is a kin to saying &quot;take the limit as n goes to infinity&quot;, which is close to saying &quot;continue this process indefinitely.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinity is also use infinity to talk about size. All sets are either infinite or finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story doesn't stop there. There is something fundamentally different about sets like the points on a line, where there are no holes, and sets like the integers where there are holes. They are both infinite but one seems denser then the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where whole countable uncountable thing comes in. Infinite sets have a size, but it is not a number in the traditional sense. Its more like &quot;relative size&quot;. Bijections are how we determine size for infinite sets, which are explained well on this page, so I won't repeat the explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more in-depth, but still understandable explanation is given in Computability and Logic by George Boolos.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-22T20:45:36.930" />
  <row Id="514" PostTypeId="1" AcceptedAnswerId="515" CreationDate="2010-07-22T20:49:28.590" Score="17" ViewCount="547" Body="&lt;p&gt;I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that everyone here is familiar with it; it describes an operation on a natural number -- n/2 if it is even, 3n+1 if it is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjecture states that if this operation is repeated, all numbers will eventually wind up at 1 (or rather, in an infinite loop of 1-4-2-1-4-2-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fired up Python and ran a quick test on this for all numbers up to 5.76 × 10^18 (using the powers of cloud computing and dynamic programming magic).  Which is millions of millions of millions.  And all of them eventually ended up at 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely I am close to testing every natural number?  How many natural numbers could there be?  Surely not much more than millions of millions of millions. (I kid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I explained this to my friend, who told me, &quot;Why would numbers suddenly get different at a certain point?  Wouldn't they all be expected to behave the same?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To which I said, &quot;No, you are wrong!  In fact I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And he said, &quot;It is my conjecture that there are none! (and if any, they are rare)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me, smart math people.  Can you provide a counterexample to his conjecture?  Perhaps, more convincingly, several?  I've only managed to find one! (Polya's)  One, out of the many thousands (I presume) of conjectures.  It's also one that is hard to explain the finer points to the layman.  Are there any more famous or accessible examples?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-22T22:18:57.237" LastActivityDate="2010-08-08T16:21:28.287" Title="Conjectures that have been disproved with extremely large counterexamples?" Tags="&lt;big-list&gt;&lt;conjectures&gt;&lt;contradiction&gt;" AnswerCount="9" CommentCount="1" FavoriteCount="4" />
  <row Id="515" PostTypeId="2" ParentId="514" CreationDate="2010-07-22T20:59:58.363" Score="12" ViewCount="0" Body="&lt;p&gt;The wikipedia article on the Collatz conjecture gives these three examples of conjectures that were disproved with large numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/P%C3%B3lya_conjecture&quot; rel=&quot;nofollow&quot;&gt;Polya conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mertens_conjecture&quot; rel=&quot;nofollow&quot;&gt;Mertens conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Skewes%27_number&quot; rel=&quot;nofollow&quot;&gt;Skewes number&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T20:59:58.363" />
  <row Id="516" PostTypeId="2" ParentId="514" CreationDate="2010-07-22T21:03:05.330" Score="10" ViewCount="0" Body="&lt;p&gt;The first example which came to my mind is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Skewes%27_number&quot; rel=&quot;nofollow&quot;&gt;Skewes' number&lt;/a&gt;, that is the smallest natural number n for which &amp;pi;(n) &gt; li(n). Wikipedia states that now the limit is near e&lt;sup&gt;727.952&lt;/sup&gt;, but the first estimation was much higher.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-22T21:03:05.330" />
  <row Id="517" PostTypeId="2" ParentId="118" CreationDate="2010-07-22T21:11:04.467" Score="3" ViewCount="0" Body="&lt;p&gt;Calculus is basically a way of calculating rates of changes (similar to slopes, but called derivatives in calculus), and areas, volumes, and surface areas (for starters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to calculate these kinds of things with algebra and geometry if the shapes you're interested in are simple.  For example, if you have a straight line you can calculate the slope easily.  But if you want to know the slope at an arbitrary point (any random point) on the graph of some function like x-squared or some other polynomial, then you would need to use calculus.  In this case, calculus gives you a way of &quot;zooming in&quot; on the point you're interested in to find the slope EXACTLY at that point.  This is called a derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a cube or a sphere, you can calculate the volume and surface area easily.  If you have an odd shape, you need to use calculus.  You use calculus to make an infinite number of really small slices of the object you're interested in, determine the sizes of the slices, and then add all those sizes up.  This process is called integration.  It turns out that integration is the reverse of derivation (finding a derivative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, calculus is a tool that lets you do calculations with complicated curves, shapes, etc., that you would normally not be able to do with just algebra and geometry.&lt;/p&gt;&#xA;" OwnerUserId="195" LastActivityDate="2010-07-22T21:11:04.467" />
  <row Id="518" PostTypeId="1" CreationDate="2010-07-22T21:43:15.497" Score="9" ViewCount="335" Body="&lt;p&gt;Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent.  Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be proved without using localization?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-30T21:44:02.087" LastActivityDate="2010-08-06T23:33:46.173" Title="Why does a minimal prime ideal consist of zerodivisors?" Tags="&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="519" PostTypeId="1" AcceptedAnswerId="597" CreationDate="2010-07-22T21:49:58.990" Score="11" ViewCount="256" Body="&lt;p&gt;The group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = &amp;lt; x, y | x^2 = y^3 = (xy)^7 = 1 &gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is infinite, or so I've been told. How would I go about proving this? (To prove finiteness of a finitely presented group, I could do a coset enumeration, but I don't see how this helps if I want to prove that it's infinite.)&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-30T16:33:34.130" Title="How do you prove that a group specified by a presentation is infinite?" Tags="&lt;group-theory&gt;&lt;presentation&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="6" />
  <row Id="520" PostTypeId="2" ParentId="518" CreationDate="2010-07-22T21:58:37.160" Score="0" ViewCount="0" Body="&lt;p&gt;If A has an identity then what about this argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take p in P and look at pP.  This is an ideal of A and a subset of P so we must have that pP = P by minimality.  Thus pa = p for some a in P.  a cannot be one for otherwise P = A is not a prime ideal.  Thus p(a-1) = 0 so that p is a zero divisor.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-22T21:58:37.160" CommentCount="4" />
  <row Id="521" PostTypeId="2" ParentId="11" CreationDate="2010-07-23T00:33:32.400" Score="3" ViewCount="0" Body="&lt;p&gt;Okay I burned a lot of reputation points (at least for me) on MathOverflow to gain clarity on how to give some intuition into this problem, so hopefully this answer will be at least be somewhat illuminating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To gain a deeper understanding of what is going on, first we need to answer the question, &quot;What is a number?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of ways to define numbers, but in general numbers are thought of as symbols that represent sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy for things like the natural numbers. So 10 would correspond to the set with ten things -- like a bag of ten stones. Pretty straight forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part is that when we consider ten a subset of the real numbers, we actually redefine it. This is not emphasized even in higher mathematics classes, like real analysis; it just happens when we define the real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is 10 when constructed in the real numbers? Well, at least with the Dedekind cut version of the real numbers, &lt;strong&gt;all&lt;/strong&gt; real numbers correspond to a set with an infinite amount of elements. This makes 10 under the hood look drastically different, although in practice it operates exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's return to the question: Why is 10 the same as 9.99999? Because the real numbers have this completely surprising quality, where there is no next real number. So when you have two real numbers that are as close together as possible, they are the same. I can't think of any physical object that has this quality, but it's how the real numbers work (makes &quot;real&quot; seem ironic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With integers (bag of stones version) this is not the same. When you have two integers as close to each other as possible they are still different, and they are distance one apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, 10 bag of stones are not the same as 9.9999999 but 10 the natural number, where natural numbers are a subset of the real numbers is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that the real numbers have these tricky edge cases that are hard to understand intuitively. Don't worry, your intuition is not really failing you. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't feel confident answering until I got this Terence Tao link: &lt;a href=&quot;http://www.google.com/buzz/114134834346472219368/RarPutThCJv/In-the-foundations-of-mathematics-the-standard&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/buzz/114134834346472219368/RarPutThCJv/In-the-foundations-of-mathematics-the-standard&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-23T00:56:33.743" LastActivityDate="2010-07-23T00:56:33.743" CommunityOwnedDate="2010-07-28T06:53:43.987" />
  <row Id="522" PostTypeId="2" ParentId="514" CreationDate="2010-07-23T01:12:25.927" Score="4" ViewCount="0" Body="&lt;p&gt;A famous example that is not quite as large as these others is the &lt;a href=&quot;http://mathworld.wolfram.com/ChebyshevBias.html&quot; rel=&quot;nofollow&quot;&gt;prime race&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjecture states, roughly: Consider the first n primes, not counting 2 or 3. Divide them into two groups: A contains all of those primes congruent to 1 modulo 3 and B contains those primes congruent to 2 modulo 3. A will never contain more numbers than B. The smallest value of n for which this is false is 23338590792.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T01:15:07.540" LastActivityDate="2010-07-29T01:15:07.540" CommentCount="1" />
  <row Id="523" PostTypeId="1" AcceptedAnswerId="526" CreationDate="2010-07-23T02:09:04.340" Score="0" ViewCount="56" Body="&lt;p&gt;Polygons are, in this question, defined as non-unique if they similar to another (by rotation, reflection, translation, or scaling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this answer be any different if similar but non-identical polygons were allowed?  And if only if rotated/translated by rational coefficients?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this answer be any different if we constrained the length and internal angles of all polygons to rational numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the number of sides is finite but unbounded, and greater than two.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-23T06:41:43.400" LastActivityDate="2010-07-23T06:41:43.400" Title="Is the set of all unique (convex) polygons countable?  If so, by what bijection to the natural numbers?" Tags="&lt;geometry&gt;&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="524" PostTypeId="1" AcceptedAnswerId="1206" CreationDate="2010-07-23T02:10:15.867" Score="1" ViewCount="85" Body="&lt;p&gt;I want to find the least squares solution to &lt;strong&gt;Ax&lt;/strong&gt;=&lt;b&gt;b&lt;/b&gt; where &lt;strong&gt;A&lt;/strong&gt; is a highly sparse square matrix.&lt;br&gt;&#xA;I found two methods that look like they might lead me to a solution: &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_decomposition&quot; rel=&quot;nofollow&quot;&gt;QR factorization&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_value_decomposition#Pseudoinverse&quot; rel=&quot;nofollow&quot;&gt;singular value decomposition&lt;/a&gt;. Unfortunately, I haven't taken linear algebra yet, so I can't really understand most of what those pages are saying. I can calculate both in Matlab though, and it looks like the SVD gave me a smaller squared error. Why did that happen? How can I know which one I should be using in the future?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-03T06:14:23.543" Title="How can I tell which matrix decomposition to use for OLS?" Tags="&lt;linear-algebra&gt;&lt;math-software&gt;&lt;numerical-methods&gt;" AnswerCount="1" />
  <row Id="525" PostTypeId="2" ParentId="11" CreationDate="2010-07-23T02:10:22.280" Score="-11" ViewCount="0" Body="&lt;p&gt;I do not think that 0.999... = 1 unless we're defining the equality operator in some strange way.  Assuming you allow N decimals in the 0.999... term, you can take N as large as you want and it will still be less than 1.  There is no N for which 0.999... is not less than 1.  Sure, if you take the limit as N approaches infinity then you get 1, but &quot;lim 0.999... as N approaches infinity&quot; is not the same as &quot;0.999...&quot; by itself.  By itself, there is always some infinitesimal difference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the whole point of writing 0.999... is to indicate that it is NOT equal to 1 and to indicate that there is some infinitesimal difference.  Otherwise you would just write 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My opinion is that most people say 0.999... = 1 because that's what they were taught in college and they accepted it without really thinking about it.  All the arguments I've heard or read so far are just contrived ways of making the equality hold, but there is always some subtle flaw that takes a long time to point out, that the proponents will not accept anyway (for reasons I will not get into here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a comment I made:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10x - x != 9.  10x - x would be 8.9999...1.  However infinite the extent of 9s is in x, if we multiply it by 10, the nines are shifted left by one position and a zero inserted at the &quot;last&quot; place, and then when you subtract the other number there is a nine subtracted from a zero at the far right.  Otherwise we'd have to give 0.999.. some unusual properties like automatically increasing the number of nines when it is multiplied.  It would not be just an ordinary number.  Maybe that's the problem.  0.999... might just not be an ordinary type number as some people are using it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current winning response as of 7/22/10 treats 0.999... as being equivalent to lim (0.999...) as the number of decimals goes to infinity.  But the sequence and the limit of the sequence are not the same thing.&lt;/p&gt;&#xA;" OwnerUserId="195" LastEditorUserId="195" LastEditorDisplayName="" LastEditDate="2010-07-23T02:20:35.140" LastActivityDate="2010-07-23T02:20:35.140" CommunityOwnedDate="2010-07-28T06:53:43.987" CommentCount="4" />
  <row Id="526" PostTypeId="2" ParentId="523" CreationDate="2010-07-23T02:13:47.357" Score="7" ViewCount="0" Body="&lt;p&gt;There are uncountably many, because for example one can have rectangles with arbitrary side ratios.  For your second question, if everything is constrained to be rational, there will be countably many, because a polygon is uniquely determined by its ordered collection of sides and angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of a general fact: if $A$ is a countable set, then the collection of ordered $n$-tuples of elements of $A$ for all $n$ is still countable.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T02:13:47.357" CommentCount="6" />
  <row Id="527" PostTypeId="2" ParentId="514" CreationDate="2010-07-23T02:15:52.577" Score="2" ViewCount="0" Body="&lt;p&gt;I heard this story from &lt;a href=&quot;http://www.ams.sunysb.edu/~estie/estie.html&quot; rel=&quot;nofollow&quot;&gt;Professor Estie Arkin&lt;/a&gt; at Stony Brook &lt;em&gt;(sorry, I don't know what conjecture she was talking about)&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For weeks we tried to prove the conjecture (without success) while we left a computer running looking for counter-examples.  One morning we came in to find the computer screen flashing: &lt;em&gt;&quot;Counter-example found&quot;&lt;/em&gt;.  We all thought for sure that there must have been a bug in the algorithm, but sure enough, it was a valid counter-example.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I tell this story to my students to emphasize that &lt;em&gt;&quot;proof by lack of counter-example&quot;&lt;/em&gt; is not a proof at all!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T02:15:52.577" CommentCount="1" />
  <row Id="528" PostTypeId="2" ParentId="356" CreationDate="2010-07-23T02:45:15.487" Score="0" ViewCount="0" Body="&lt;p&gt;I think you should determine what kinds of applications you want to apply CS to and then learn general theory relevant to those applications.  Not all theory is equally applicable everywhere.&lt;/p&gt;&#xA;" OwnerUserId="195" LastActivityDate="2010-07-23T02:45:15.487" CommunityOwnedDate="2010-07-23T02:45:15.487" CommentCount="1" />
  <row Id="529" PostTypeId="1" AcceptedAnswerId="547" CreationDate="2010-07-23T03:11:15.673" Score="12" ViewCount="249" Body="&lt;p&gt;Why are the only (associative) division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a division algebra is an associative algebra where every nonzero number is invertible (like a field, but without assuming commutativity of multiplication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an old result proved by Frobenius, but I can't remember how the argument goes.  Anyone have a quick proof?&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-08-04T01:19:55.590" Title="Why are the only division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?" Tags="&lt;quaternions&gt;&lt;ring-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="530" PostTypeId="1" CreationDate="2010-07-23T03:17:09.670" Score="9" ViewCount="236" Body="&lt;p&gt;In intro number theory a key lemma is that if a and b are relatively prime integers, then there exist integers x and y such that ax+by=1.  In a more advanced course instead you would use the theorem that the integers are a PID, i.e. that all ideals are principal.  Then the old lemma can be used to prove that &quot;any ideal generated by two elements is actually principal.&quot;  Induction then says that any finitely generated ideal is principal.  But what if all finitely generated ideals are principal but there are some ideals that aren't finitely generated?  Can that happen?&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-09-08T08:49:41.177" Title="Can you find a domain where ax+by=1 has a solution for all a and b relatively prime, but which is not a PID?" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="531" PostTypeId="1" CreationDate="2010-07-23T03:20:13.713" Score="23" ViewCount="537" Body="&lt;p&gt;I know there's something called &quot;ultrafinitism&quot; which is a very radical form of constructivism that I've heard said means people don't believe that really large integers actually exist.  Could someone make this a little bit more precise?  Are there good reasons for taking this point of view?  Can you actually get math done from that perspective?&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-10-13T18:04:19.743" Title="What is &quot;ultrafinitism&quot; and why do people believe it?" Tags="&lt;soft-question&gt;&lt;philosophy&gt;&lt;constructive-mathematics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="532" PostTypeId="2" ParentId="531" CreationDate="2010-07-23T03:30:20.270" Score="15" ViewCount="0" Body="&lt;p&gt;The philosophy is explained in Doron Zeilberger's &lt;a href=&quot;http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/real.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;.&#xA; Basically, it's the belief that there is a largest natural number! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard a funny story (on Scott Aaronson's blog) about someone who was an ultrafinitist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do you believe in 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Yes, he responded immediately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do you believe in 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Yes, he responded after a brief pause&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do you believe in 3? &#xA;-Yes, he responded after a slightly longer pause&#xA;-Do you believe in 4?&#xA;-Yes, after several seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It soon become clear that he would take twice as long to answer the next question as the previous one. (I believe &lt;a href=&quot;http://en.wikipedia.org/wiki/Alexander_Esenin-Volpin&quot; rel=&quot;nofollow&quot;&gt;Alexander Esesin-Volpin&lt;/a&gt; was the person.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T03:30:20.270" CommentCount="8" />
  <row Id="533" PostTypeId="1" AcceptedAnswerId="561" CreationDate="2010-07-23T03:36:00.833" Score="3" ViewCount="126" Body="&lt;p&gt;I know some references where I can find this, but they seem tedious(Both Hartshorne and Ueno cover this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there is an elegant way to describe these. If this task is too difficult in general, how about just $\mathbb{P}^n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-23T11:53:21.343" Title="Classifying Quasi-coherent Sheaves on Projective Schemes" Tags="&lt;algebraic-geometry&gt;&lt;projective-schemes&gt;&lt;quasicoherent-sheaves&gt;&lt;projective-space&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="534" PostTypeId="1" AcceptedAnswerId="565" CreationDate="2010-07-23T03:56:46.910" Score="5" ViewCount="69" Body="&lt;p&gt;There is a little triviality that has been referred to as the &lt;a href=&quot;http://www.johndcook.com/blog/2010/01/13/soft-maximum&quot; rel=&quot;nofollow&quot;&gt;&quot;soft maximum&quot;&lt;/a&gt; over on &lt;a href=&quot;http://www.johndcook.com/blog/&quot; rel=&quot;nofollow&quot;&gt;John Cook's Blog&lt;/a&gt; that I find to be fun, at the very least.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is this:  given a list of values, say $x_1,x_2,\ldots,x_n$ , the function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x_1,x_2,\ldots,x_n) = \log(\exp(x_1) + \exp(x_2) + \cdots + \exp(x_n))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;returns a value very near the maximum in the list.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens because that exponentiation exaggerates the differences between the $x_i$ values.  For the largest $x_i$, $\exp(x_i)$ will be $really$ large.  This largest exponential will significantly outweigh all of the others combined. Taking the logarithm, i.e. undoing the exponentiation,  we essentially recover the largest of the $x_i$'s.  (Of course, if two of the values were very near one another, we aren't guaranteed to get the true maximum, but it won't be far off!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About this, John Cook says: &quot;The soft maximum approximates the hard maximum but it also rounds off the corners.&quot;  This couldn't really be said any better.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall trying to cleverly construct sequences for proofs in advanced calculus where not-everywhere-differentiable operations would have been great to use if they didn't have that pesky non-differentiable trait.  I can't recall a specific incidence where I was tempted to use $max(x_i)$, but this seems at least plausible that it would have come up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone used this before or have a scenario off hand where it would be useful?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-23T04:58:15.617" LastActivityDate="2010-07-23T12:33:52.603" Title="applications of the &quot;soft maximum&quot;" Tags="&lt;numerical-methods&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="535" PostTypeId="1" CreationDate="2010-07-23T03:59:36.487" Score="8" ViewCount="314" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pWnq4.png&quot; alt=&quot;complex plot of the zeros&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Diagram and setup from UCSMP &lt;em&gt;Precaluclus and Discrete Mathematics&lt;/em&gt;, 3rd ed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is a partial plot of the zeros of $p_c(x)=4x^4+8x^3-3x^2-9x+c$.  The text stops at showing the diagram and does not discuss the shape of the locus of the zeros or describe the resulting curves.  Are the curves in the locus some specific (named) type of curve?  Is there a simple way to describe the curves (equations)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question need not be limited to the specific polynomial given--a similar sort of locus is generated by the zeros of nearly any quartic polynomial as the constant term is varied.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-11T06:51:54.967" LastActivityDate="2010-12-11T06:51:54.967" Title="Describe the locus in the complex plane of the zeros of a quartic polynomial as the constant term varies" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;&lt;zeros&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="536" PostTypeId="1" CreationDate="2010-07-23T04:05:29.323" Score="14" ViewCount="386" Body="&lt;p&gt;How does one prove that a simple (steps of length 1 in directions parallel to the axes) symmetric (each possible direction is equally likely) random walk in 1 or 2 dimensions returns to the origin with probability 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: note that while returning to the origin is guaranteed (p = 1) in 1 and 2 dimensions, it is not guaranteed in higher dimensions; this means that something in a correct justification for the 1- or 2-d case must fail to extend to 3-d (or fail when the probability for each direction drops from 1/4 to 1/6).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-23T05:08:37.597" LastActivityDate="2010-08-04T12:42:49.603" Title="Proving that 1- and 2-d simple symmetric random walks return to the origin with probability 1" Tags="&lt;probability-theory&gt;" AnswerCount="5" CommentCount="12" FavoriteCount="3" />
  <row Id="537" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T04:35:13.213" Score="0" ViewCount="0" Body="&lt;p&gt;I'll do 1D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1D walks are building binary strings, 010101, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say take six steps. Then 111111 is just as likely as 101010. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how many of the possible sequences have six ones? 1. How many of the possibly sequences have three ones and three zeros? Much more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That number is called multiplicity, and it grows mighty fast. In the limit its log becomes Shannon entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequences are equally likely, but combinations are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit the combinations with maximum entropy are going dominate all the rest. So the walk is going to have gone an equal number of right and left steps...almost surely.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-23T04:54:20.617" LastActivityDate="2010-07-23T04:54:20.617" CommentCount="5" />
  <row Id="538" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T04:51:17.470" Score="1" ViewCount="0" Body="&lt;p&gt;I can prove the 1 dimensional case a bit more formally than &lt;a href=&quot;http://math.stackexchange.com/questions/536/proving-that-1-and-2-d-simple-symmetric-random-walks-return-to-the-origin-with-p/537#537&quot; rel=&quot;nofollow&quot;&gt;Jonathan&lt;/a&gt;. First we only look at the absolute values. Let us try to calculate the probability of this never exceeding x. The probability of 2x+1 consecutive moves all being the same is some p&gt;0. If this ever occurs, then the absolute value will exceed x. Consider n groups of 2x+1 moves, the probability that at least one of these is all the same is 1-(1-p)^n, which approaches 1. So the probability of reaching each absolute value x (other than 0) is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets consider the probability of reaching 0 again. Without loss of generality, suppose our first move is +1. We have a 100% chance of reaching a point a distance of 1 from this. There is a 50% that the first such point is 0 and 50% that it is 2. From 2, we have a 100% chance of reaching a point two away from this. 50% chance that this is 0, 50% that this is 4. Repeating, it is easy to see that we have to reach 0 again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, after we have reached an absolute value x, the probability of reaching it again is 1. So we reach each absolute value an infinite number of times. Due to symmetry, we can expect to reach each x an infinite number of times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I originally tried to apply it to the 2d case as you can see below. This is incorrect, as even though there will be no maximal absolute value, there is no reason we will necessarily come back down.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Similar reasoning will work in the 2d&#xA;  case. Instead of using symmetry, we&#xA;  simply note that each point with the&#xA;  same Manhattan distance from 0 has an&#xA;  expected value that is a non-zero&#xA;  proportion of the expected value all&#xA;  points a particular Manhattan distance&#xA;  away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-23T06:16:39.330" LastActivityDate="2010-07-23T06:16:39.330" CommentCount="6" />
  <row Id="539" PostTypeId="2" ParentId="2" CreationDate="2010-07-23T06:34:30.527" Score="0" ViewCount="0" Body="&lt;p&gt;If you were an ant crawling along a curve, at a really small (almost zero) distance away from the point you're taking the limit of, where are you?&lt;/p&gt;&#xA;" OwnerUserId="198" LastActivityDate="2010-07-23T06:34:30.527" />
  <row Id="540" PostTypeId="1" CreationDate="2010-07-23T06:42:52.100" Score="22" ViewCount="856" Body="&lt;p&gt;In laymen's terms, as much as possible: What is the Riemann-Zeta function, and why does it come up so often with relation to prime numbers?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-14T06:28:36.233" LastActivityDate="2010-08-14T06:28:36.233" Title="What is the Riemann-Zeta function?" Tags="&lt;terminology&gt;&lt;complex-analysis&gt;&lt;prime-numbers&gt;&lt;riemann-zeta&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="11" />
  <row Id="541" PostTypeId="1" AcceptedAnswerId="611" CreationDate="2010-07-23T07:37:42.770" Score="1" ViewCount="165" Body="&lt;p&gt;By this I mean books with math puzzles and problems similar to the ones you would find in mathematical olympiads.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:31:18.443" LastActivityDate="2010-07-29T23:31:18.443" CommunityOwnedDate="2010-07-23T16:12:30.563" Title="Which books would you recommend about Recreational Mathematics?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;&lt;recreational-mathematics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="542" PostTypeId="2" ParentId="540" CreationDate="2010-07-23T07:52:32.097" Score="12" ViewCount="0" Body="&lt;p&gt;Giving an explanation in layman's terms is always going to be challenging, given that the Riemann-Zeta function (and related hypothesis) inevitably lies in the domain of abstract mathematics, but I shall do my best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Riemann-Zeta function&lt;/a&gt; is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_function#Complex_functions&quot; rel=&quot;nofollow&quot;&gt;complex function&lt;/a&gt; that tells us many things about the theory of numbers. Its mystery is increased by the fact it has no &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed-form_expression&quot; rel=&quot;nofollow&quot;&gt;closed form&lt;/a&gt; - i.e. it can't be expressed a single formula that contains other standard (elementary) functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there are many different ways of expressing the Riemann-Zeta function (the Wikipedia article gives several), it can ultimately be derived from the following simple series of real numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/4/f/e/4fe5e85d36c09cc13a973a39d6387caa.png&quot; alt=&quot;Eq1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by extending it into the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_plane&quot; rel=&quot;nofollow&quot;&gt;complex plane&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this strange and esoteric function is so famous and actively discussed in mathematics is due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Riemann hypothesis&lt;/a&gt; - proposedi n 1859 by the by the great Bernhard Riemann and still unsolved. The Wiki article states the problem in quite simple terms:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Riemann zeta-function ζ(s) is&#xA;  defined for all complex numbers s ≠ 1.&#xA;  It has zeros at the negative even&#xA;  integers (i.e. at s = −2, −4, −6,&#xA;  ...). These are called the trivial&#xA;  zeros. The Riemann hypothesis is&#xA;  concerned with the non-trivial zeros,&#xA;  and states that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The real part of any non-trivial zero of the Riemann zeta function is 1/2.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Thus the non-trivial zeros should lie&#xA;  on the critical line, 1/2 + it, where&#xA;  t is a real number and i is the&#xA;  imaginary unit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although the conjecture (it is only that at the moment) has many consequences for mathematics (number theory in particular), the primary one, at least the one Riemann originally proposed, is about the &lt;em&gt;distribution of prime numbers&lt;/em&gt;. In other words, it tells us with great precision what the average gaps between primes are as we move to greater and greater numbers. Many of the other implications are rather more esoteric, though perhaps equally important for pure mathematicians.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-23T07:52:32.097" />
  <row Id="544" PostTypeId="1" AcceptedAnswerId="545" CreationDate="2010-07-23T08:16:24.300" Score="7" ViewCount="301" Body="&lt;p&gt;What is a intuitive explanation of a Markov Chain, and how they work? Please provide at least one practical example.&lt;/p&gt;&#xA;" OwnerUserId="156" LastEditorUserId="156" LastEditorDisplayName="" LastEditDate="2010-07-25T15:12:42.147" LastActivityDate="2010-07-31T05:45:16.580" Title="What is a  Markov Chain?" Tags="&lt;terminology&gt;&lt;probability-theory&gt;&lt;intuition&gt;&lt;stochastic-processes&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="545" PostTypeId="2" ParentId="544" CreationDate="2010-07-23T08:28:36.050" Score="11" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A Markov chain is a discrete random&#xA;  process with the property that the&#xA;  next state depends only on the current&#xA;  state (&lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So P(X_n | X_1, X_2, ... X_n-1) = P(X_n | X_n-1). An example could be when you are modelling the weather. You then can take the assumption that the weather of today can be predicted by only using the knowledge of yesterday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have Rainy and Sunny. When it is rainy on one day the next day is Sunny with probability 0.3. When it is Sunny, the probability for Rain next day is 0.4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when it is today Sunny we can predict the weather of the day after tomorrow, by simply calculating the probability for Rain tomorrow, multiplying that with the probablity for Sun after rain plus the probability of Sun tomorrow times the probability of Sun after sun. In total the probability of Sunny of the day after tomorrow is P(R|S)*P(S|R) + P(S|S)*P(S|S) = 0.3*0.4+0.6*0.6 = 0.48&lt;/p&gt;&#xA;" OwnerUserId="143" LastActivityDate="2010-07-23T08:28:36.050" />
  <row Id="546" PostTypeId="2" ParentId="544" CreationDate="2010-07-23T08:30:06.530" Score="5" ViewCount="0" Body="&lt;p&gt;In a nutshell, a Markov chain is (the behavior of) a random process which may only find itself in a (not necessarily finite) number of different states. The process moves from a state to another in discrete times (that is, you define a sequence S(t) of states at time t=0,1,2,...), and for which the probability of going from state S to state R depends just from S and R; that is, there is no &quot;memory of the past&quot; and the process is &quot;timeless&quot;. This means that the Markov chain may be modeled as a n*n matrix, where n is the number of possible states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a process which may be modeled by a Markov chain is the sequence of faces of a die showing up, if you are allowed to rotate the die wrt an edge. The corresponding matrix is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    1   2   3   4   5   6&#xA;   ------------------------ &#xA;1 | 0   1/4 1/4 1/4 1/4 0&#xA;2 | 1/4 0   1/4 1/4 0   1/4&#xA;3 | 1/4 1/4 0   0   1/4 1/4&#xA;4 | 1/4 1/4 0   0   1/4 1/4&#xA;5 | 1/4 0   1/4 1/4 0   1/4&#xA;1 | 0   1/4 1/4 1/4 1/4 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As usual, &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/MarkovChain.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; are your friends.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-23T08:30:06.530" />
  <row Id="547" PostTypeId="2" ParentId="529" CreationDate="2010-07-23T08:32:59.613" Score="4" ViewCount="0" Body="&lt;p&gt;Essentially one first proves that any real division algebra D is a Clifford algebra (i.e. it's generated by elements of some inner product vector space I subject to relations v&lt;sup&gt;2&lt;/sup&gt;=&amp;lt;v,v&amp;gt;): first one splits D as $\mathbb R\oplus D_0$ where D&lt;sub&gt;0&lt;/sub&gt; is the space of elements with Tr=0 and then one observes that minimal polynomial of a traceless element has the form x&lt;sup&gt;2&lt;/sup&gt;-a=0 (it's quadratic because it's irreducible and the coefficient of x is zero because it is the trace). Now it remains to find out which Clifford algebras are division algebras which is pretty straightforward (well, and it follows from the classification of Clifford algebras).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)&quot; rel=&quot;nofollow&quot;&gt;is written&lt;/a&gt; in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-30T07:33:29.343" LastActivityDate="2010-07-30T07:33:29.343" />
  <row Id="548" PostTypeId="1" CreationDate="2010-07-23T08:51:02.597" Score="7" ViewCount="347" Body="&lt;p&gt;(This was asked due to the comments and downvotes on this &lt;a href=&quot;http://stackoverflow.com/questions/3236489/why-is-0-divided-by-0-an-error/3236541#3236541&quot;&gt;Stackoverflow&lt;/a&gt; answer. I am not that good at maths, so was wondering if I had made any basic mistakes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring limits, I would like to know if this is a valid explanation for why 0/0 is undefined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x = 0/0&lt;br&gt;&#xA;   x * 0 = 0  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hence There are an infinite number of values for x as anything multiplied by 0 is 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it seems to have got comments, with two general themes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once is that you lose the values of x by multiplying by 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is that the last line is &lt;code&gt;x * 0 = 0/0 * 0&lt;/code&gt; as it involves a division by 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any merit to either argument? More to the point, are there any major flaws in my explanation and is there a better way of showing why 0/0 is undefined?&lt;/p&gt;&#xA;" OwnerUserId="108" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:31:38.957" LastActivityDate="2010-08-02T11:38:02.740" Title="Explanation of method for showing that 0 / 0 is undefined" Tags="&lt;algebra&gt;&lt;algebra-precalculus&gt;&lt;zero-divisor&gt;" AnswerCount="7" CommentCount="10" FavoriteCount="2" />
  <row Id="549" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T09:07:55.223" Score="0" ViewCount="0" Body="&lt;p&gt;It's undefined because $0/0 = 0$ is against the identity $a/a = 1$. Or at least, that's how I understand it, although there are probably deeper explanations that involve the consistency of algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also approach it in a simple manner from the direction that what you have is $0^1 * 0^-1$ which would give you $0^0$ which is a very disturbing expression (and could be taken as either 0 or 1 depending on your definition of exponents).&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2010-07-23T09:07:55.223" CommentCount="4" />
  <row Id="550" PostTypeId="2" ParentId="392" CreationDate="2010-07-23T09:34:33.213" Score="1" ViewCount="0" Body="&lt;p&gt;From first principles,using trig identities and small-angle approximations:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;sin'(x) = lim(h-&gt;0){(sin(x+h)-sin(x))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(x+h) = sin(x)cos(h)+cos(x)sin(h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; sin'(x) = lim(h-&gt;0){(sin(x)(cos(h)-1) + cos(x)sin(h))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For x small, sin(x)~=x, so lim(h-&gt;0){sin(h)/h}=1, and cos(x)~= 1 -x^2/2, so (h-&gt;0){(cos(h)-1)/h}=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; sin'(x) = cos(x)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;cos'(x) = lim(h-&gt;0){(cos(x+h)-cos(x))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cos(x+h) = cos(x)cos(h) - sin(x)sin(h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; cos'(x) = lim(h-&gt;0){(cos(x)(cos(h)-1) - sin(x)sin(h))/h}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= -sin(x) by the same reasoning above.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-07-23T09:34:33.213" CommentCount="1" />
  <row Id="551" PostTypeId="2" ParentId="531" CreationDate="2010-07-23T09:39:52.700" Score="19" ViewCount="0" Body="&lt;p&gt;Ultrafinitism is basically resource-bounded  constructivism: proofs have constructive content, and what you get out of these constructions isn't much more than you put in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the universal and existential quantifier should help clarify things.  Constructively, a universally quantified sentence means that if I am given a parameter, I can construct something that satisfies the quantified predicate.  Ultrafinitistically, the thing you give back won't be much bigger: typically there will be a polynomial bound on the size of what you get back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For existentially quantified statements, the constructive content is a pair of the value of the parameter, and the construction that satisfies the predicate.  Here the resource is the size of the proof: the size of the parameter and construction will be related to the size of the proof of the existential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, addition and mutliplication are total functions, but exponentiation is not.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-verifying_theories&quot; rel=&quot;nofollow&quot;&gt;Self-verifying theories&lt;/a&gt; are more extreme: addition is total in the strongest of these theories, but multiplication cannot be.  So the resource bound is linear for these theories, not polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A foundational problem with ultrafinitism is that there aren't nice ultrafinitist logics that support an attractive formulae-as-types correspondence in the way that intuitionistic logic does.  This makes ultrafinitism as less comfy kind of contructivism than intuitionism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do people believe it?&lt;/strong&gt;  For the same kinds of reasons people believe in constructivism: they want mathematical claims to be backed up by something they can regard as concrete.  Just as an intuitionist might be bothered by the idea of cutting a ball into 10^50 non-measurable pieces and putting it back together into two balls, so too  an ultrafinitist might be concerned about the idea that towers of exponentials are meaningful ways of constructing numbers.  Wittgenstein argued this point in his &quot;Lectures on the Foundations of Mathematics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you actually get math done from that perspective?&lt;/strong&gt; Yes.  If intuitionism is the mathematics of the computable, ultrafinitism is the mathematics of the feasibly computable.  But the difference in ease of working with between ultrafinitism and intuitionism is much bigger than that between intuitionism and classical mathematics.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-10-02T13:30:23.827" LastActivityDate="2010-10-02T13:30:23.827" CommentCount="1" />
  <row Id="552" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T09:42:11.283" Score="3" ViewCount="0" Body="&lt;p&gt;I think that ignoring limits is problematic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was a limit of the function $f(x,y)=x/y for x,y-&gt;0$ regardless of how the limit is performed, then one would define that value to be $f(0,0)$, even if everything else is strange. Since the limiting value depends on the way the limit is done, choosing a value for $f(0,0)$ is counterproductive as it gives a non-continuous function. Better to have a continuous function over a slightly smaller domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also forces the point that if you do have a limit process that results in the evaluation of $f(0,0)$, you realize early on that you should examine the limit carefully rather than use $\lim g(x) = g(\lim x)$ (which is only true for continuous functions, of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this might be too trivial, but I'll give an example of how the limiting value depends on the limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow0} \frac{\sin(x)}{x} = 1$, (both $\sin(x)$ and $x$ goto 0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow0} \frac{cos x - 1}{x} = 0$ (again, both num and denum goto 0, but num goes &quot;faster&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow0} \frac{\sqrt{x}}{x} =+\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS, sorry about the latex notation, hopefully, we could have it working nicely like in MO...&lt;/p&gt;&#xA;" OwnerUserId="165" LastActivityDate="2010-07-23T09:42:11.283" CommentCount="3" />
  <row Id="553" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T09:53:19.190" Score="7" ViewCount="0" Body="&lt;p&gt;For &quot;all&quot; x,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0/x = 0 ⇒ 0/0 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For &quot;all&quot; x,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x/x = 1 ⇒ 0/0 = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Moreover, if one could say &lt;code&gt;0/0 = k&lt;/code&gt;, any k, we could then say &lt;code&gt;2 = 3&lt;/code&gt; -- just divide both sides by 0 and get &lt;code&gt;k = k&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no reasonable value &lt;code&gt;0/0&lt;/code&gt; can have, &lt;code&gt;0/0&lt;/code&gt; must be undefined.&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-07-30T20:24:53.447" LastActivityDate="2010-07-30T20:24:53.447" CommentCount="3" />
  <row Id="554" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T09:53:30.360" Score="4" ViewCount="0" Body="&lt;p&gt;I'll try it for 2D and then we can get 1D as a corollary [excercise!]...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the only proof I know of, there may be a more intuitive (and less messy without tex!) proof out there, but I like this one- it uses generating functions in a really nifty way.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the probability of being at the origin after 2n steps (notice we cannot return in an odd number of steps):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$u_0=1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$u_{2n} = p(n,n,0,0)+ p(n-1,n-1,1,1)+...+p(n-k,n-k,k,k)+...+p(0,0,n,n)$ (when $n \neq 0) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $p(u,d,l,r)$ is the probability of the first 2n steps being u up, d down, l left and r right in any order. Each order has probability $\frac{1}{4^{2n}}$, and there are $\frac{(2n)!}{u!d!l!r!}$ distinct orders, giving $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \frac{(2n)!}{(n-k)!(n-k)!k!k!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $(2n)!=n! n! \binom{2n}{n}$ we have $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \binom{2n}{n} \binom{n}{k}^2$ giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{2n}= \frac{1}{4^{2n}}  \Sigma_k \binom{2n}{n} \binom{n}{k}^2$ which, by one of those silly binomial results, can be contracted to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$u_{2n}= \frac{1}{4^{2n}} \binom{2n}{n}^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us put that in our back pocket for now and consider in stead the probability of &lt;strong&gt;first&lt;/strong&gt; returning after 2n steps $f_{2n}$- this is rather difficult to tackle directly but we can make ourselves a cunning little formula involving it, jazz out some &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_function&quot; rel=&quot;nofollow&quot;&gt;generating function&lt;/a&gt; fun and seal the deal with some .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula in question is:&#xA;$u_{2n}= f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which we shall not prove so much as explain: to return to the origin after 2n steps (LHS) you must either first return after 2 steps and do a 'return to the origin in 2n-2 steps walk' (first term RHS) or first return after 4 steps and do a 'return to the origin in 2n-4 steps walk' (2nd term) or... or first return after 2n-2 steps and do a 'return to the origin in 2 steps walk or first return to the origin after 2n steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall now tweak our formula just a tiny bit so it has the right symmetry properties for what is to come, we do this by adding $f_0=0$ and $u_0=1$ to give &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$u_{2n}= f_0 u_{2n}+ f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}u_0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is secretly a statement about generating functions (this is the sweet bit!), see look at the generating functions of u_n, f_n:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(x)= \Sigma_m u_{2m} x^{2m}$, $F(x)= \Sigma_m f_{2m} x^{2m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$U(x)= 1+ U(x)F(x)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(where the '1+' is to compensate for the fact that $u_0$ does not appear in the product) Rearranging to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)= \frac{U(x)-1}{U(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that the probability of return is $\Sigma f_n= F(1)$, which comes out as 1  because $U(1)$ diverges by some tedious stirlings formula bounds that I forget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Until Tex comes online, this is pretty unreadable, so &lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter12.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;'s a link to some lecture notes I found with the same proof (and, fortunately, the same notation!). Enjoy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit$^2$:&lt;/strong&gt; Hooray, Tex has come online!!! Enjoy.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-04T12:42:49.603" LastActivityDate="2010-08-04T12:42:49.603" />
  <row Id="555" PostTypeId="2" ParentId="514" CreationDate="2010-07-23T10:00:35.643" Score="4" ViewCount="0" Body="&lt;p&gt;For an old example, Mersenne made the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Mersenne_conjectures&quot; rel=&quot;nofollow&quot;&gt;conjecture&lt;/a&gt; in 1644:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;The Mersenne numbers, $M_n=2^n  − 1$, are prime for n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127 and 257, and no others.&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler observed that the Mersenne number at M_61 is prime, so refuting the conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M_61 is quite large by the standards of the day: 2 305 843 009 213 693 951.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, there are 47 known Mersenne primes.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-08-08T07:59:24.667" LastActivityDate="2010-08-08T07:59:24.667" CommentCount="4" />
  <row Id="556" PostTypeId="2" ParentId="540" CreationDate="2010-07-23T10:00:46.830" Score="4" ViewCount="0" Body="&lt;p&gt;Here there is another attempt at an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the sum of the inverse of the positive numbers, 1 + 1/2 + 1/3 + ..., diverges. Euler shown that the sum of the inverse of the squares,  1/(1&lt;sup&gt;2&lt;/sup&gt;) + 1/(2&lt;sup&gt;2&lt;/sup&gt;) + 1/(3&lt;sup&gt;2&lt;/sup&gt;) + ..., has a finite sum, namely &amp;pi;&lt;sup&gt;2&lt;/sup&gt;/6. Mathematicians love to generalize things, so they thought at the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathurl.com/35scyna.png&quot; alt=&quot;f(x) = sum for n=1 to infinity of (1/n^x)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is defined for x &gt; 1. But this was not enough: they decided that the variable could be a complex number and not a real one. There is a standard tecnique (&lt;a href=&quot;http://mathworld.wolfram.com/AnalyticContinuation.html&quot; rel=&quot;nofollow&quot;&gt;Analytic continuation&lt;/a&gt;) which allows us to extend the function to nearly all the complex plane. So we now have a function which formally is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathurl.com/5euwuy.png&quot; alt=&quot;zeta(s) = sum for n=1 to infinity of (1/n^s)]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the variable being s and not x to show that we are dealing with complex numbers) but is not computed in this way. Just to make an example, &amp;zeta;(0)=1/2, and sum of an infinity of ones is not 1/2 :-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be shown that for s = -2n (n positive integer) &amp;zeta;(s) = 0. But there are infinite other point s'=(x,y) where &amp;zeta;(s') = 0. For all of these points, 0 &amp;lt; x &amp;lt; 1; Riemann's hypothesis says that for all such points x = 1/2. If it were true, we could have the best asymptotic expression to count &amp;pi;(n), that is the number of primes below n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the function pop up when we talk about primes? I don't know, but in the case of integer values Euler proved that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathurl.com/34s664v.png&quot; alt=&quot;sum of 1/(n^s) = product over all primes of 1/(1-p^s)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this could be a good start.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-23T12:07:54.467" LastActivityDate="2010-07-23T12:07:54.467" CommentCount="1" />
  <row Id="558" PostTypeId="2" ParentId="213" CreationDate="2010-07-23T10:22:47.437" Score="2" ViewCount="0" Body="&lt;p&gt;Statistics is the topic in which I am still poor and it is still useful to me which I learned so late and that's why I am poor in Statistics.&lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2010-07-23T10:22:47.437" CommunityOwnedDate="2010-07-23T10:22:47.437" />
  <row Id="559" PostTypeId="2" ParentId="548" CreationDate="2010-07-23T11:16:56.147" Score="0" ViewCount="0" Body="&lt;p&gt;It seems as though this is a question of conventions (like the question about why 0^0 is undefined rather than just being 1 &lt;a href=&quot;http://math.stackexchange.com/questions/135/why-is-x0-1-except-when-x-0&quot; rel=&quot;nofollow&quot;&gt;as was asked here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could define 0/0 = 0, and I don't think it would break Mathematics. I'm happy to defend my view (or revise it) based on the comments this answer receives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not be keen to define it to be anything else, since if we have 0/0 = x then (I'd like the axioms to allow me to say) we have x = 0/0 = (2*0)/0 = 2*(0/0) = 2*x. But it's unclear that we should require (a*b)/c = a*(b/c) to hold even when c = 0.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-23T11:16:56.147" CommentCount="1" />
  <row Id="560" PostTypeId="2" ParentId="135" CreationDate="2010-07-23T11:41:17.373" Score="1" ViewCount="0" Body="&lt;p&gt;If we use the idea of set exponentiation to define exponentiation of cardinals, we have the following natural idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$$A^B:=\{f:B\to A\}$$&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the exponential of cardinals as follows: $|A|^|B|:=|A^B|$.  It's easy to check that this agrees with our intuition for exponentiation of natural numbers when $B$ is nonempty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one set representing the cardinal $0$, namely the empty set.  Then we may look at $0^0$ as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$$0^0=|\emptyset|^{|\emptyset|}=|\emptyset^\emptyset|&lt;br&gt;&#xA;=|\{f:\emptyset\to\emptyset\}|=1$$&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-23T11:41:17.373" />
  <row Id="561" PostTypeId="2" ParentId="533" CreationDate="2010-07-23T11:53:21.343" Score="2" ViewCount="0" Body="&lt;p&gt;Quasi-coherent sheaves on affine schemes (say $Spec(A)$) are obtained by taking an $A$-module $M$ and the associated sheaf (by localizing $M$).  This gives an equivalence of categories between $A$-modules and q-c sheaves on $Spec(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a graded ring, $R = R_0 + R_1 + \dots$ (direct sum).  Then we can, given a &lt;em&gt;graded&lt;/em&gt; $R$-module $M$, consider its associated sheaf $\tilde{M}$.  The stalk of this at  a homogeneous prime ideal $P$ is defined to be the localization $M_{(P)}$, which is defined as generated by quotients $m/s$ for $s$ homogeneous of the same degree as $m$ and not in $P$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, we get sheaves of modules on the affine scheme just as we get the normal  sheaves of rings.  We get sheaves of modules on the projective scheme in the same homogeneous localization way as we get the sheaf of rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's no longer an equivalence of categories. Why? Say you had a graded module $M= M_0 + M_1 + \dots$ (in general, we allow negative gradings as well).  Then it is easy to check that the sheaves associated to $M$ and $M' = M_1 + M_2 + \dots$ are exactly the same.&#xA;Nevertheless, it is possible to get every sheaf on $Proj(R)$ for $R$ a graded ring in this way.  See Proposition II.5.15 in Hartshorne.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T11:53:21.343" CommentCount="5" />
  <row Id="562" PostTypeId="2" ParentId="530" CreationDate="2010-07-23T11:58:10.147" Score="5" ViewCount="0" Body="&lt;p&gt;If I'm not mistaken, the integral domain of holomorphic functions on a connected open set $U &#xA;\subset \mathbb{C}$ works. It is a theorem (in Chapter 15 of Rudin's &lt;em&gt;Real and Complex Analysis,&lt;/em&gt; and essentially a corollary of the Weierstrass factorization theorem), that every finitely generated ideal in this domain is principal. This implies that if $a,b$ have no common factor, they generate the unit ideal.  However, for instance, the ideal of holomorphic functions in the unit disk that vanish on all but finitely many of ${1-\frac{1}{n}}$ is nonprincipal.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-07T12:59:48.080" LastActivityDate="2010-09-07T12:59:48.080" CommentCount="0" />
  <row Id="563" PostTypeId="2" ParentId="536" CreationDate="2010-07-23T12:01:33.107" Score="7" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://www.math.cornell.edu/~durrett/PTE/PTE4_Jan2010.pdf&quot; rel=&quot;nofollow&quot;&gt;Durrett, &lt;i&gt;Probability and Examples&lt;/i&gt;&lt;/a&gt; (link goes to online copy of the fourth edition).  On p. 164 Durrett gives a proof that simple random walk is recurrent in two dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First find the probability that simple random walk in &lt;i&gt;one&lt;/i&gt; dimension is at 0 after 2n steps; this is clearly rho_1(2n) = (2n choose n)/2^(2n), since 2n choose n is the number of paths with n right steps and n left steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the probability that simple random walk in &lt;i&gt;two&lt;/i&gt; dimensions -- call this rho_2(2n) -- is at 0 after 2n steps is the square of the previous probability.  Consider the simple random walk which makes steps to the northeast, northwest, southeast, and southwest with equal probability. The projections of this walk onto the x- and y-axes are independent simple random walks in one dimension.  Roating and rescaling gives the &quot;usual&quot; SRW in two dimensions (with steps north, east, south and west) and doesn't change the probability of being at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So rho_2(2n) = ((2n choose n)/2^(2n))^2.  This is asymptotic to 1/(pi*n), and the expected number of returns to the origin is the sum of rho_2(2n) over positive integers n, so the expected number of returns to the origin is infinite. It's not hard to show (and is in fact true, but the proof is unenlightening so I'll leave it to Durrett) that in this case the probability of eventually returning to the origin is 1.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-23T12:01:33.107" CommentCount="1" />
  <row Id="564" PostTypeId="2" ParentId="544" CreationDate="2010-07-23T12:31:25.573" Score="3" ViewCount="0" Body="&lt;p&gt;Markov chains, especially hidden Markov models are huge in computation linguistics. A hidden Markov model is one where we can't directly view the state, but we do have some information about what the state might be. For example, consider breaking down a sentence in parts of speech such as verbs, adjectives, ect. We don't know what the parts of speech are, but we can attempt to deduce them from the word. For example, the word &lt;a href=&quot;http://www.thefreedictionary.com/run&quot; rel=&quot;nofollow&quot;&gt;run&lt;/a&gt; might be used 80% as a verb, 18% of the time as a noun and 2% of the time as an adjective. We also have Markov relations between the parts of speech, so for example an adjective might be followed by a noun 70% of the time and another adjective 30% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Viterbi_algorithm&quot; rel=&quot;nofollow&quot;&gt;Viterbi algorithm&lt;/a&gt; to decide which sequence is most likely to have generated the observed output (the algorithm takes into account both the probability of such a sequence of parts of speech occurring together in a sentence and the relative chance that such parts of speech would be responsible for us observing the given words).&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-23T12:31:25.573" CommentCount="1" />
  <row Id="565" PostTypeId="2" ParentId="534" CreationDate="2010-07-23T12:33:52.603" Score="2" ViewCount="0" Body="&lt;p&gt;This is close to being the flipside of the geometric mean, which is the nth root of the product of the numbers, and can be expressed as the exponential of the sum of the logarithms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another pair of dual mean measures is the regular mean and the harmonic mean (n divided by the sum of the reciprocals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say the soft maximum is close to being the flipside of the geometric mean, but it lacks the good property that all of the others have of taking a list of the same value to that value (for definedness, let all values be positive).  Let's call the hyperbolic mean the &quot;soft maximum&quot; of the nth roots of the terms in the list: then this has that good property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyperbolic mean the emphasises large values in a roughly symmetric manner to the way that the geometric mean emphasises small values (which is always smaller than the regular mean), and is, of course, much smaller for long list of large values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I say, consider it an amplified version of a useful addition to the family of of mean operators.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-23T12:33:52.603" />
  <row Id="566" PostTypeId="1" CreationDate="2010-07-23T12:44:16.403" Score="15" ViewCount="757" Body="&lt;p&gt;Okay, so this question was bound to come up sooner or later- the hope was to ask it well before someone asked it badly...&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;We all love a good puzzle&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of sudoku for example is hidden in &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_squares&quot; rel=&quot;nofollow&quot;&gt;latin squares&lt;/a&gt;). Mathematicians and puzzles get on, it seems, rather well.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;But what is a good puzzle?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Okay, so in order to make this question worthwhile (and not a ten-page wadeathon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify for answerhood- to do so it must&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Not be widely known:&lt;/strong&gt; If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that &lt;em&gt;hilarious&lt;/em&gt; scene in the film 21, where kevin spacey explains the monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disemvowelled.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Be mathematical:&lt;/strong&gt; as much as possible- true: logic &lt;em&gt;is&lt;/em&gt; mathematics, but puzzles beginning 'There is a street where everyone has a different coloured house...' are much of a muchness and tedious as hell. Note: there is a happy medium between this and trig substitutions.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Not be too hard:&lt;/strong&gt; any level is cool but if the answer requires more than two sublemmas, you are misreading your audience&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Actually have an answer:&lt;/strong&gt; crank questions will not be appreciated! You can post the answers/hints in &lt;a href=&quot;http://en.wikipedia.org/wiki/Rot_13&quot; rel=&quot;nofollow&quot;&gt;Rot-13&lt;/a&gt; underneath as comments as on MO if you fancy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And should  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ideally include where you found it:&lt;/strong&gt; so we can find more cool stuff like it  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Have that indefinable spark that makes a puzzle awesome:&lt;/strong&gt; a situation that seems familiar, requiring unfamiliar thought...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For ease of voting- one puzzle per post is bestest.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Some examples to set the ball rolling&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Simplify $\sqrt{2+\sqrt{3}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From:&lt;/strong&gt; problem solving magazine &lt;strong&gt;Hint:&lt;/strong&gt; GEL N GJB GREZ FBYHGVBA&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can one make an equilateral triangle with all vertices at integer coordinates?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From:&lt;/strong&gt; Durham distance maths challenge 2010 &lt;strong&gt;Hint:&lt;/strong&gt; GUVF VF RDHVINYRAG GB GUR ENGVBANY PNFR&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;nxn &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_squares&quot; rel=&quot;nofollow&quot;&gt;Magic squares&lt;/a&gt; form a vector space over $\mathbb{R}$ prove this, and by way of a linear transformation, derive the dimension of this vector space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From:&lt;/strong&gt; Me, I made this up (you can tell, can't you!) &lt;strong&gt;Hint:&lt;/strong&gt;  NCCYL GUR ENAX AHYYVGL GURBERZ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy puzzling!&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-10-27T12:34:22.917" LastActivityDate="2010-10-27T12:34:22.917" CommunityOwnedDate="2010-07-23T12:44:16.403" Title="Your favourite maths puzzles" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;puzzle&gt;" AnswerCount="15" CommentCount="6" FavoriteCount="14" />
  <row Id="567" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T13:02:28.817" Score="3" ViewCount="0" Body="&lt;p&gt;From &lt;em&gt;Mathematical Puzzles&lt;/em&gt; by Peter Winkler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide an hexagon in equilateral triangles, like in the figure. Now fill all the hexagon with the three kinds of diamonds made from two triangles, also shown in the figure. Prove that the number of each kind of diamond is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i27.tinypic.com/33nkqrr.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-23T13:02:28.817" CommunityOwnedDate="2010-07-23T13:02:28.817" CommentCount="3" />
  <row Id="568" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T13:04:13.867" Score="3" ViewCount="0" Body="&lt;p&gt;unknown source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the plane be colored with two different colors (say, red and blue) so that there is no equilateral triangle whose vertices are all of the same color?&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-23T13:04:13.867" CommunityOwnedDate="2010-07-23T13:04:13.867" CommentCount="4" />
  <row Id="571" PostTypeId="1" AcceptedAnswerId="578" CreationDate="2010-07-23T13:36:43.717" Score="7" ViewCount="233" Body="&lt;p&gt;You are standing on a cliff at a height &lt;em&gt;h&lt;/em&gt; above the sea. You are capable of throwing a stone with velocity &lt;em&gt;v&lt;/em&gt; at any angle &lt;em&gt;a&lt;/em&gt; between horizontal and vertical. What is the value of &lt;em&gt;a&lt;/em&gt; when the horizontal distance travelled &lt;em&gt;d&lt;/em&gt; is at a maximum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On level ground, when &lt;em&gt;h&lt;/em&gt; is zero, it's easy to show that &lt;em&gt;a&lt;/em&gt; needs to be midway between horizontal and vertical, and thus π/4 or 45°. As &lt;em&gt;h&lt;/em&gt; increases, however, we can see by heuristic reasoning that &lt;em&gt;a&lt;/em&gt; decreases to zero, because you can put more of the velocity into the horizontal component as the height of the cliff begins to make up for the loss in the vertical component. For small negative values of &lt;em&gt;h&lt;/em&gt; (throwing up onto a platform), &lt;em&gt;a&lt;/em&gt; will actually be greater than 45°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a fully-solved, closed-form expression for the value of &lt;em&gt;a&lt;/em&gt; when &lt;em&gt;h&lt;/em&gt; is not zero?&lt;/p&gt;&#xA;" OwnerUserId="113" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-23T14:50:53.553" LastActivityDate="2010-08-03T09:18:31.527" Title="What is the optimum angle of projection when throwing a stone off a cliff?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="572" PostTypeId="2" ParentId="541" CreationDate="2010-07-23T13:40:21.850" Score="2" ViewCount="0" Body="&lt;p&gt;Art of Problem Solving: &lt;a href=&quot;http://www.artofproblemsolving.com/Store/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Store/index.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything by Martin Gardner.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-23T13:40:21.850" CommunityOwnedDate="2010-07-23T16:12:30.563" CommentCount="1" />
  <row Id="573" PostTypeId="1" AcceptedAnswerId="579" CreationDate="2010-07-23T13:40:43.287" Score="4" ViewCount="105" Body="&lt;p&gt;So I know that given a chain complex we can define the d-th cohomology by taking ker{d}/im_{d+1}. But I don't know how this corresponds to the idea of holes in topological spaces (maybe this is homology, I'm a tad confused).&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:40:46.240" LastActivityDate="2010-07-24T18:40:46.240" Title="varying definitions of cohomology" Tags="&lt;cohomology&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="574" PostTypeId="1" AcceptedAnswerId="878" CreationDate="2010-07-23T13:43:25.603" Score="1" ViewCount="192" Body="&lt;p&gt;What does it mean (in a mathematically rigorous way) to claim something is &quot;generic?&quot; How does this coincide with the Zariski topology?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:29:49.287" LastActivityDate="2010-07-31T01:07:09.890" Title="genericness and the zariski topology" Tags="&lt;algebraic-geometry&gt;&lt;topology&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="575" PostTypeId="1" AcceptedAnswerId="830" CreationDate="2010-07-23T13:46:38.513" Score="4" ViewCount="94" Body="&lt;p&gt;Given a curve how do you intuitively construct the picture of its projective dual? I know points --&gt; lines, lines--&gt; points but for something like the swallowtail this is not really obvious.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-28T01:37:45.227" Title="projective duality" Tags="&lt;projective-space&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="576" PostTypeId="2" ParentId="571" CreationDate="2010-07-23T13:52:28.750" Score="0" ViewCount="0" Body="&lt;p&gt;I don't have a complete solution, but I attempted to solve this problem using calculus.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x'=v cos a&#xA;y''= -g and y' (at t=0) = v sin a&#xA;So y'= v sin a -gt&#xA;x0=0, so x=vt cos a&#xA;y0=h, so y=vt sin a - 0.5gt^2+c (integrating with respect to t)&#xA;Subbing in h, y=vt sin a - 0.5gt^2+h&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The ball will hit the ground when y=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as far as I got, but it appears that you can find a &lt;a href=&quot;http://math.stackexchange.com/questions/571/what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff/578#578&quot; rel=&quot;nofollow&quot;&gt;closed solution&lt;/a&gt; after all. I originally tried solving the quadratic for t and subbing that into x, but it seems to work much better to do the substitution the other way round. I will leave this solution here in case anyone wants to see how to derive the basic equations for x and y.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-23T23:20:49.457" LastActivityDate="2010-07-23T23:20:49.457" CommentCount="4" />
  <row Id="577" PostTypeId="1" AcceptedAnswerId="1760" CreationDate="2010-07-23T14:07:57.573" Score="3" ViewCount="236" Body="&lt;p&gt;According to Wikipedia, Godel's incompleteness theorem states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No consistent system of axioms whose&#xA;  theorems can be listed by an&#xA;  &quot;effective procedure&quot; (essentially, a&#xA;  computer program) is capable of&#xA;  proving all facts about the natural&#xA;  numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This obviously includes our current system. So has anyone proposed any additional axioms that seem credible?&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-06T21:00:58.133" Title="Has anyone ever proposed additional axioms?" Tags="&lt;axioms&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="578" PostTypeId="2" ParentId="571" CreationDate="2010-07-23T14:29:17.023" Score="10" ViewCount="0" Body="&lt;p&gt;Assume no friction and uniform gravity &lt;em&gt;g&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you throw a stone at point (0, &lt;em&gt;h&lt;/em&gt;), with velocity (&lt;em&gt;v&lt;/em&gt; cos &lt;em&gt;&amp;theta;&lt;/em&gt;, &lt;em&gt;v&lt;/em&gt; sin &lt;em&gt;&amp;theta;&lt;/em&gt;), then we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;d &amp;amp;= vt\cos\theta &amp;amp;&amp;amp; (1) \\&#xA;0 &amp;amp;= h + vt\sin\theta - \frac12 gt^2 &amp;amp;&amp;amp; (2)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only unknown to be solved is &lt;em&gt;t&lt;/em&gt; (total travel time). We could eliminate it by using $t = \frac d{v\cos\theta}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 = h + d\tan\theta - \frac{gd^2\sec^2\theta}{2v^2}\qquad(3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we compute the total derivative with respect to &lt;em&gt;θ&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;0 &amp;amp;= \frac d{d\theta}\left(d\tan\theta\right) - \frac g{2v^2}\frac d{d\theta}\left(d^2\sec^2\theta\right) \\&#xA;&amp;amp;= \ldots&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then set $\frac{dd}{d\theta}=0$ (because it is maximum) to solve &lt;em&gt;d&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d = \frac{v^2}{g\tan\theta} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute this back to (3) gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;h &amp;amp;= \frac{v^2}g \left( \frac1{2\sin^2\theta} - 1\right) \\&#xA;\Rightarrow \sin\theta &amp;amp;= \left( 2 \left(\frac{gh}{v^2} + 1\right) \right)^{-1/2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the closed form of &lt;em&gt;θ&lt;/em&gt; in terms of &lt;em&gt;h&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-03T09:18:31.527" LastActivityDate="2010-08-03T09:18:31.527" CommentCount="2" />
  <row Id="579" PostTypeId="2" ParentId="573" CreationDate="2010-07-23T14:52:06.373" Score="5" ViewCount="0" Body="&lt;p&gt;Edited to clear some things up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplicial and singular (co)homology were invented to detect holes in spaces.  To get an intuitive idea of how this works, consider subspaces of the plane.  Here the 2-chains are formal sums of things homeomorphic to the closed disk, and 1-chains are formal sums of things homeomorphic to a line segment.  The operator d takes the boundary of a chain.  For example, the boundary of the closed disk is a circle.  If we take d of the circle we get 0 since a circle has no boundary.  And in general it happens that d^2 = 0, that is boundaries always have no boundaries themselves.  Now suppose we remove the origin from the plane and take a circle around the origin.  This circle is in the kernel of d since it has no boundary.  However, it does not bound any 2-chain in the space (since the origin is removed) and so it is not in the image of the boundary operator on two-dimensions.  Thus the circle represents a non-trivial element in the quotient space ker d / im d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I have defined things makes the above a homology theory simply because the d operator decreases dimension.  Cohomology is the same thing only the operator increases dimension (for example the exterior derivative on differential forms).  Thus algebraically there really is no difference between cohomology and homology since we can just change the grading from i to -i.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a homology we can get a corresponding cohomology theory by dualizing, that is by looking at maps from the group of chains to the underlying group (e.g. Z or R).  Then d on the cohomology theory becomes the adjoint of the previous boundary operator and thus increases degrees.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-23T15:05:02.630" LastActivityDate="2010-07-23T15:05:02.630" />
  <row Id="580" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:38:37.710" Score="3" ViewCount="0" Body="&lt;p&gt;Most of us know that, being deterministic, computers cannot generate true &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudorandom_number_generator&quot; rel=&quot;nofollow&quot;&gt;random numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let's say you have a box which generates truly random binary numbers, but is biased:  it's more likely to generate either a &lt;code&gt;1&lt;/code&gt; or a &lt;code&gt;0&lt;/code&gt;, but you don't know the exact probabilities, or even which is more likely &lt;em&gt;(both probabilities are &gt; 0 and sum to 1, obviously)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you use this box to create an unbiased random generator of binary numbers?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="114" LastEditorDisplayName="" LastEditDate="2010-09-07T23:29:29.537" LastActivityDate="2010-09-07T23:29:29.537" CommunityOwnedDate="2010-07-23T15:38:37.710" CommentCount="6" />
  <row Id="582" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:39:57.557" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming you have unlimited time and cash, is there a strategy that's guaranteed to win at roulette?&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T15:39:57.557" CommunityOwnedDate="2010-07-23T15:39:57.557" CommentCount="5" />
  <row Id="584" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:40:36.243" Score="-2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Frk n th rd 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y'r n pth n n slnd, cme t  frk n th rd.  Bth pths ld t vllgs f ntvs; th ntr vllg thr lwys tlls th trth r lwys ls &lt;em&gt;(bth villgs cld b trth-tllng r lyng vllgs, r n f ch)&lt;/em&gt;.  Thr r tw ntvs t th frk - thy cld bth b frm th sm vllg, r frm dffrnt vllgs &lt;em&gt;(s bth cld b trth-tllrs, both lrs, r ne f ch)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n pth lds t sfty, th thr t dm.  Y'r llwd t sk nly n qstn t ch ntv t fgr t whch pth s whch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wht d y sk?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-23T17:43:23.640" LastActivityDate="2010-07-23T17:43:23.640" CommunityOwnedDate="2010-07-23T15:40:36.243" CommentCount="4" />
  <row Id="585" PostTypeId="2" ParentId="566" CreationDate="2010-07-23T15:40:51.703" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Fork in the road 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're once again at a fork in the road, and again, one path leads to safety, the other to doom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three natives at the fork.  One is from a village of truth-tellers, one from a village of liars, one from a village of random answerers.  Of course you don't know which is which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the natives answer &quot;pish&quot; and &quot;posh&quot; for yes and no, but you don't know which means &quot;yes&quot; and which means &quot;no.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're allowed to ask only two yes-or-no questions, each question being directed at one native.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you ask?&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T15:40:51.703" CommunityOwnedDate="2010-07-23T15:40:51.703" CommentCount="1" />
  <row Id="586" PostTypeId="2" ParentId="107" CreationDate="2010-07-23T15:50:56.943" Score="2" ViewCount="0" Body="&lt;p&gt;I've always thought this sort of puzzle is a wonderful example of the difference between behaviour as you approach a limit, versus behaviour AT the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make the puzzle more revealing like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have an infinite number of banknotes, each with a unique serial number. I give you bill number 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you have two options, either you keep that bill and the game ends, or I give you 10 more banknotes, but you have to burn the lowest numbered bill. It seems obvious that option two is much better.&#xA;Now, we repeat this game over and over. You keep getting more and more money.&#xA;But at the limit, every banknote has been burned and you're worse off than you would have been had you just taken the one banknote and left.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-23T15:50:56.943" />
  <row Id="587" PostTypeId="2" ParentId="577" CreationDate="2010-07-23T15:54:17.940" Score="4" ViewCount="0" Body="&lt;p&gt;Adding additional axioms would make more truths provable. But it wouldn't make all truths provable (unless the axiom was inconsistent with the already given ones, in which case all falsehoods would also be provable too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So adding additional axioms isn't going to help make all the truths provable. I guess you could just add, say, Goldbach's conjecture and the Riemann hypothesis as extra axioms and carry on doing mathematics-PLUS! but why would you want to?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-07-23T15:54:17.940" CommentCount="5" />
  <row Id="588" PostTypeId="1" CreationDate="2010-07-23T16:00:43.943" Score="6" ViewCount="302" Body="&lt;p&gt;How do we know if a particular function can be represented as a power series? And once we have come up with a power series representation, how does one figure out its radius of convergence ?&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:35:02.243" LastActivityDate="2010-07-28T19:35:02.243" Title="What functions can be represented as power series?" Tags="&lt;calculus&gt;&lt;complex-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="589" PostTypeId="2" ParentId="588" CreationDate="2010-07-23T16:04:55.110" Score="0" ViewCount="0" Body="&lt;p&gt;This is a very general question, as one can create all sorts of power series for different functions. (e.g. Taylor series, Laurent series, Fourier series).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give the obvious example of Taylor series: a power series representation of a function can be found if the function is infinitely differentiable in the neighbourhood of the given point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all power series, you will need to find the recursion relation (formula giving a successive term from the current term) and then use the ratio test to solve for the value of the input variable that gives a ratio of convergence of 1.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-23T16:04:55.110" CommentCount="10" />
  <row Id="590" PostTypeId="2" ParentId="540" CreationDate="2010-07-23T16:13:17.913" Score="3" ViewCount="0" Body="&lt;p&gt;The key point is that the Riemann zeta function is a function whose properties encode properties about the prime numbers.  As mentioned by Noldorin, in order to fully understand the Riemann zeta function you need to &quot;analytically continue it to the complex plane&quot; which is a tricky process which takes serious study.  Fortunately for some easier properties of the primes you can just use the definition of the zeta function for real s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim (due to Euler): The fact that \zeta(s) goes to infinity as s-&gt;1 tells you that there are infinitely many primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sketch of proof: Use the &quot;Euler factorization&quot; mentioned by mau (expand the RHS as a geometric series and then multiply it out using unique factorization into primes):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^\infty \frac{1} {n^s} = \prod_{p prime} \frac{1} {1-p^-s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take log of both sides to get:&#xA;log \zeta(s) = \sum_{p prime} \log \frac{1} {1-p^-s}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the taylor series for \log and send s to zero.  You'll get that the left hand side goes to infinity (http://math.stackexchange.com/questions/255/), while the right hand side looks like \sum 1/p + bounded terms.  So there must be infinitely many primes.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-23T16:13:17.913" />
  <row Id="591" PostTypeId="2" ParentId="588" CreationDate="2010-07-23T16:22:15.813" Score="13" ViewCount="0" Body="&lt;p&gt;A function can be represented as a power series if and only if it is complex differentiable in an open set. This follows from the general form of Taylor's theorem for complex functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being real differentiable--even infinitely many times--is not enough, as the function $e^{-1/x^2}$ on the real line (equal to 0 at 0) is $C^\infty$ yet does not equal its power series expansion since all its derivatives at zero vanish. The reason is that the complexified version of the function is not even continuous at the origin.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T16:22:15.813" CommentCount="2" />
  <row Id="592" PostTypeId="2" ParentId="577" CreationDate="2010-07-23T16:29:37.447" Score="3" ViewCount="0" Body="&lt;p&gt;Sure, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;continuum hypothesis&lt;/a&gt;.  We know, thanks to the work of Godel and Paul Cohen, that this is a legitimate axiom, in that it doesn't destroy consistency: you can't disprove it (or prove it) using Zermelo-Frank set theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People do actually use the continuum hypothesis in model theory, for instance, where there are several results that depend on it.  For an article about its uses, cf. the &lt;a href=&quot;http://www.tricki.org/article/How_to_use_the_continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Tricki page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T19:59:50.097" LastActivityDate="2010-08-06T19:59:50.097" CommentCount="3" />
  <row Id="593" PostTypeId="1" AcceptedAnswerId="682" CreationDate="2010-07-23T17:21:44.563" Score="7" ViewCount="107" Body="&lt;p&gt;When I look at the Taylor series for e^x and the volume formula for oriented simplexes, it makes e^x look like it is, at least almost, the sum of simplexes volumes from n to infinity. Does anyone know of a stronger relationship beyond, &quot;they sort of look similar&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some links:&lt;br&gt;&#xA;Volume formula&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Simplex#Geometric_properties&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Simplex#Geometric_properties&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taylor Series&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Complex_numbers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Complex_numbers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-30T01:10:28.307" LastActivityDate="2010-07-30T07:28:06.127" Title="Is there a relationship between e and the sum of n-simplexes volumes?" Tags="&lt;geometry&gt;&lt;intuition&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="594" PostTypeId="1" CreationDate="2010-07-23T17:25:13.187" Score="11" ViewCount="653" Body="&lt;p&gt;It is a theorem in elementary number theory that if $p$ is a prime and congruent to 1 mod 4, then it is the sum of two squares. Apparently there is a trick involving arithmetic in the gaussian integers that lets you prove this quickly. Can anyone explain it?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-22T01:34:17.307" Title="How do you prove that a prime is the sum of two squares iff it is congruent to 1 mod 4?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="4" />
  <row Id="595" PostTypeId="1" AcceptedAnswerId="596" CreationDate="2010-07-23T17:29:53.710" Score="3" ViewCount="267" Body="&lt;p&gt;I'm learning Linear Algebra using MIT's Open Courseware &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2005/&quot; rel=&quot;nofollow&quot;&gt;Course 18.06&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite often, the professor says &quot;... assuming that the matrix is invertible ...&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere in the lecture he says that using a determinant on an $n \times n$ matrix is on the order of $O(n!)$ operations, where an operation is a multiplication and a subtraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a more efficient way?&lt;/strong&gt;  If the aim is to get the inverse, rather than just determine the invertibility, what is the most effecient way to do this?&lt;/p&gt;&#xA;" OwnerUserId="38" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T00:57:20.783" LastActivityDate="2010-08-04T00:57:20.783" Title="What is the most efficient way to determine if a matrix is invertible?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="596" PostTypeId="2" ParentId="595" CreationDate="2010-07-23T17:38:22.477" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Jordan_elimination&quot; rel=&quot;nofollow&quot;&gt;Gauss-Jordan elimination&lt;/a&gt; can be used to determine when a matrix is invertible and can be done in polynomial (in fact, cubic) time.  The same method (when you apply the opposite row operation to identity matrix) works to calculate the inverse in polynomial time as wel.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T17:38:22.477" CommentCount="5" />
  <row Id="597" PostTypeId="2" ParentId="519" CreationDate="2010-07-23T17:41:22.817" Score="10" ViewCount="0" Body="&lt;p&gt;&amp;lt;x,y|x^2=y^3=1&amp;gt;&amp;cong;PSL_2(Z) and this isomorphism identifies G with $PSL_2/T^7=1$ (where $T:z\mapsto z+1$). Result is the symmetry group of &lt;a href=&quot;http://en.wikipedia.org/wiki/(2,3,7)_triangle_group&quot; rel=&quot;nofollow&quot;&gt;the tiling&lt;/a&gt; of the hyperbolic plane. From this description one can see that G is infinite (e.g. because there are infinitely many triangles in the tiling and G acts on them transitively).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-30T07:31:43.497" LastActivityDate="2010-07-30T07:31:43.497" CommentCount="2" />
  <row Id="598" PostTypeId="2" ParentId="594" CreationDate="2010-07-23T17:46:06.803" Score="7" ViewCount="0" Body="&lt;p&gt;Let $p$ be a prime congruent to 1 mod 4. Then to write $p = x^2 + y^2$ for $x,y$ integers is the same as writing $p = (x+iy)(x-iy) = N(x+iy)$ for $N$ the norm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well-known that the ring of Gaussian integers $\mathbb{Z}[i]$ is a principal ideal domain, even a euclidean domain.   Now I claim that $p$ is not prime in $\mathbb{Z}[i]$. To determine how a prime $p$ of $\mathbb{Z}$ splits in $\mathbb{Z}[i]$ is equivalent to determining how the polynomial $X^2+1$ splits modulo $p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, $-1$ is a quadratic residue modulo $p$ because $p \equiv 1 \mod 4$.  Consequently, there is $t \in \mathbb{Z}$ with $t^2 \equiv -1 \mod p$, so $X^2+1$ splits modulo $p$, and $p$ does not remain prime in $\mathbb{Z}[i]$.  (Another way of seeing this is to note that&#xA;if $p$ remained prime, then we'd have $p \mid (t+i)(t-i)$, which means that $p \mid t+i$ or $t \mid t-i$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, as a result there is a non-unit $x+iy$ of $\mathbb{Z}[i]$ that properly divides $p$.  This means that the norms properly divide as well.  In particular, $N(x+iy) = x^2+y^2$ properly divides $p^2$, so is $p$ or $1$. It cannot be the latter since otherwise $x+iy$ would be a unit.  So $x^2+y^2 = p$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-23T17:46:06.803" CommunityOwnedDate="2010-07-23T17:46:32.027" CommentCount="2" />
  <row Id="599" PostTypeId="2" ParentId="595" CreationDate="2010-07-23T17:49:08.667" Score="6" ViewCount="0" Body="&lt;p&gt;A matrix is invertible iff its determinant is non-zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are algorithms which find the determinant in &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant#Algorithmic_implementation&quot; rel=&quot;nofollow&quot;&gt;slightly worse&lt;/a&gt; than O(n&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T17:49:08.667" CommentCount="5" />
  <row Id="600" PostTypeId="1" AcceptedAnswerId="825" CreationDate="2010-07-23T18:14:56.057" Score="5" ViewCount="195" Body="&lt;p&gt;Given n distinct objects, there are n! permutations of the objects and n!/n &quot;circular permutations&quot; of the objects (orientation of the circle matters, but there is no starting point, so 1234 and 2341 are the same, but 4321 is different).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given n objects of k types (where the objects within each type are indistinguishable), $r_i$ of the ith type, there are $\frac{n!}{r_1!r_2!\cdots r_k!}$ permutations.  How many circular permutations are there of such a set?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-27T22:02:50.807" Title="Circular permutations with indistinguishable objects" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="601" PostTypeId="1" AcceptedAnswerId="602" CreationDate="2010-07-23T18:21:15.327" Score="5" ViewCount="143" Body="&lt;p&gt;From a standard 52-card deck, how many ways are there to pick a hand of $k$ cards that includes one card from all four suits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for any specific $k$, it's possible to break it up into cases based on the partitions of $k$ into four parts. For example, if I want to choose a hand of six cards, I can break it up into two cases based on whether there are (1) three cards from one suit and one card from each of the other three or (2) two cards from each of two suits and one card from each of the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simpler, more general solution that doesn't require splitting the problem into many different cases?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-24T06:08:44.547" LastActivityDate="2010-08-05T19:44:31.013" Title="Counting how many hands of cards use all four suits" Tags="&lt;combinatorics&gt;&lt;card-games&gt;" AnswerCount="1" />
  <row Id="602" PostTypeId="2" ParentId="601" CreationDate="2010-07-23T18:38:18.040" Score="6" ViewCount="0" Body="&lt;p&gt;Count the number of hands that do not contain at least one card from every suit and subtract from the total number of k-card hands.  To count the number of hands that do not contain at least one card from every suit, use inclusion-exclusion considering what suits are not in a given hand.  That is, letting $N(\dots)$ mean the number of hands meeting the given criteria, $$\begin{align}&#xA;&amp;amp;N(\mathrm{no\ }\heartsuit)+N(\mathrm{no\ }\spadesuit)+N(\mathrm{no\ }\clubsuit)+N(\mathrm{no\ }\diamondsuit)&#xA;\\&#xA;&amp;amp;\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit)-N(\mathrm{no\ }\heartsuit\clubsuit)-N(\mathrm{no\ }\heartsuit\diamondsuit)-N(\mathrm{no\ }\spadesuit\clubsuit)-N(\mathrm{no\ }\spadesuit\diamondsuit)-N(\mathrm{no\ }\clubsuit\diamondsuit)&#xA;\\&#xA;&amp;amp;\quad\quad+N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit)+N(\mathrm{no\ }\heartsuit\spadesuit\diamondsuit)+N(\mathrm{no\ }\heartsuit\clubsuit\diamondsuit)+N(\mathrm{no\ }\spadesuit\clubsuit\diamondsuit)&#xA;\\&#xA;&amp;amp;\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit\diamondsuit)&#xA;\\&#xA;&amp;amp;=4{39 \choose k}-6{26 \choose k}+4{13 \choose k}-{0 \choose k}.&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the number of hands of k cards that include at least one card from every suit is $${52 \choose k}-4{39 \choose k}+6{26 \choose k}-4{13 \choose k}+{0 \choose k}.$$  [Drop terms as appropriate for larger values of k.]&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:44:31.013" LastActivityDate="2010-08-05T19:44:31.013" CommentCount="2" />
  <row Id="603" PostTypeId="1" AcceptedAnswerId="634" CreationDate="2010-07-23T18:40:32.417" Score="1" ViewCount="112" Body="&lt;p&gt;Congruence transformations (isometries) and similarity transformations (isometries + dilations) should be constructable.  What about other affine transformations?  Other conformal mappings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: by constructable, I mean given the defining information for the transformation in a geometric way (e.g. a dilation requires a center and a ratio, so the given could be a point and two segments), can you construct the image of a point under the transformation from its preimage?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:35:17.123" LastActivityDate="2010-07-26T08:35:17.123" Title="What transformations of the plane are geometrically constructable (compass &amp; straight edge)?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;compass-and-straightedge&gt;&lt;geometric-construction&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="604" PostTypeId="1" AcceptedAnswerId="605" CreationDate="2010-07-23T19:15:05.187" Score="7" ViewCount="65" Body="&lt;p&gt;For a long time I've eschewed bulky and inelegant calculators for the use of my trusty trig/log-log slide rule.  For those unfamiliar, here is a simple &lt;a href=&quot;http://www.antiquark.com/sliderule/sim/virtual-slide-rule.html&quot; rel=&quot;nofollow&quot;&gt;slide rule simulator using Javascript&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate, find the LL_3 scale, which is on the back of the virtual one.  Let's say we want to solve  &lt;code&gt;3^n&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you would move the cursor (the red line) over where &lt;code&gt;3&lt;/code&gt; is on the LL_3 scale.  Then, you would &lt;em&gt;slide&lt;/em&gt; the middle slider until the &lt;code&gt;1&lt;/code&gt; on the C scale is lined up to the cursor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And voila, your slide rule is set up to find &lt;code&gt;3^n&lt;/code&gt; for any arbitrary &lt;code&gt;n&lt;/code&gt;!  For example, to find &lt;code&gt;3^2&lt;/code&gt;, move the cursor to &lt;code&gt;2&lt;/code&gt; on the C scale, and your answer is what the cursor is on on the LL_3 scale (&lt;code&gt;9&lt;/code&gt;).  Move your cursor to &lt;code&gt;3&lt;/code&gt; on C, and it should be lined up with &lt;code&gt;27&lt;/code&gt; on LL_3.  To &lt;code&gt;4&lt;/code&gt; on C, it is on &lt;code&gt;81&lt;/code&gt; on LL_3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even do this for non-integer exponents (1.3, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do this for &lt;strong&gt;exponents less than one&lt;/strong&gt;, by using the LL_2 scale.  For example, to do &lt;code&gt;3^0.5&lt;/code&gt;, you would find &lt;code&gt;5&lt;/code&gt; on the C scale, and look where the cursor is lined up at on the LL_2 scale (which is about &lt;code&gt;1.732&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, I was wondering if anyone could explain to me how this all works?  It works, but...why?  What property of logarithms and exponents (and logarithms of logarithms?) allows this to work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already understand how the basics of the Slide Rule works (&lt;code&gt;ln(m) + ln(n) = ln(mn)&lt;/code&gt;), with only multiplication, but this exponentiation eludes me.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-23T19:49:20.130" Title="Why does the log-log scale on my Slide Rule work?" Tags="&lt;explanation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="605" PostTypeId="2" ParentId="604" CreationDate="2010-07-23T19:49:20.130" Score="4" ViewCount="0" Body="&lt;p&gt;If x = 3&lt;sup&gt;n&lt;/sup&gt;, then log x = n log 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The C scale is logarithmic, which means if the reading is p, then the distance is proportional to &amp;nbsp; log p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in the LLx scale the distance is proportional to &amp;nbsp; log log p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, when you align 1 to &quot;3&quot; in LL3, you introduce an &lt;em&gt;offset&lt;/em&gt; of (log log 3). Suppose you get a reading of &lt;em&gt;n&lt;/em&gt; in the C scale, then the corresponding value in LL3 would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;log log p = log log 3 + log n&#xA;  (LL3)      (offset)    (C)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;eliminating one level of log gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    log p = log 3 * n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;eliminating one more level of log gives&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        p = 3^n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;LL2 is the same as LL3 except it covers a different range.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-23T19:49:20.130" CommentCount="2" />
  <row Id="606" PostTypeId="2" ParentId="519" CreationDate="2010-07-23T20:15:16.727" Score="10" ViewCount="0" Body="&lt;p&gt;Grigory has already answered your particular question.  However, I wanted to point out that your question &quot;How do you prove that a group specified by a presentation is infinite?&quot; has no good answer &lt;em&gt;in general&lt;/em&gt;.  Indeed, in general the question of whether a group presentation defines &lt;em&gt;the trivial group&lt;/em&gt; is undecidable.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-23T20:15:16.727" CommentCount="2" />
  <row Id="607" PostTypeId="1" AcceptedAnswerId="608" CreationDate="2010-07-23T20:56:22.497" Score="3" ViewCount="205" Body="&lt;p&gt;Assuming I can play forever, what are my chances of coming out ahead in a coin flipping series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I want &quot;heads&quot;...then if I flip once, and get heads, then I win, because I've reached a point where I have more heads than tails (1-0).  If it was tails, I can flip again.  If I'm lucky, and I get two heads in a row after this, this is another way for me to win (2-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if I can play forever, my chances are probably pretty decent.  They are at least greater than 50%, since I can get that from the first flip.  After that, though, it starts getting sticky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drawn a tree graph to try to get to the point where I could start see the formula hopefully dropping out, but so far it's eluding me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your chances of coming out ahead after 1 flip are 50%.  Fine.  Assuming you don't win, you have to flip at least twice more.  This step gives you 1 chance out of 4.  The next level would be after 5 flips, where you have an addtional 2 chances out of 12, followed by 7 flips, giving you 4 out of 40.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I may be able to work through this given some time, but I'd like to see what other people think...is there an easy way to approach this?  Is this a known problem?&lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2010-08-07T12:29:15.587" Title="Given enough time, what are the chances I can come out ahead in a coin toss contest?" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="608" PostTypeId="2" ParentId="607" CreationDate="2010-07-23T21:08:17.693" Score="14" ViewCount="0" Body="&lt;p&gt;100%, for the same reason as the &lt;a href=&quot;http://math.stackexchange.com/questions/536/proving-that-1-and-2-d-simple-symmetric-random-walks-return-to-the-origin-with-p&quot; rel=&quot;nofollow&quot;&gt;1-D walk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact (again for the same reason), your chances are 100% of eventually reaching X-greater heads than tails (or tails than heads), where X is any non-negative integer.&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-23T21:08:17.693" CommentCount="0" />
  <row Id="610" PostTypeId="2" ParentId="600" CreationDate="2010-07-23T22:57:44.907" Score="2" ViewCount="0" Body="&lt;p&gt;This problem is best solved with Pólya's enumeration theorem, which follows from Burnside's lemma.  See the first section of &lt;a href=&quot;http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-23T22:57:44.907" CommentCount="1" />
  <row Id="611" PostTypeId="2" ParentId="541" CreationDate="2010-07-23T23:25:35.780" Score="4" ViewCount="0" Body="&lt;p&gt;If you're after Olmpiad-level books, get &lt;a href=&quot;http://books.google.com/books?id=rVvr2p76cKQC&amp;amp;printsec=frontcover&amp;amp;dq=mathematical+olympiad&amp;amp;hl=en&amp;amp;ei=tiJKTKnAIcL-8AbAgs0z&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CEAQ6AEwAw#v=onepage&amp;amp;q=mathematical%20olympiad&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The IMO Compendium&lt;/a&gt; which is a collection of problems from the International Math Olympiad, 1959-2004. You can find similar books with national Olympiad problems by going to Amazon and searching for &quot;mathematical Olympiad&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two books that offer collections of techniques useful for olympiad-level contests are Paul Zeitz's &lt;a href=&quot;http://books.google.com/books?id=Go_iAAAACAAJ&amp;amp;dq=paul+zeitz&amp;amp;hl=en&amp;amp;ei=0SNKTO_MFIL48AaZ8PE3&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;The Art and Craft of Problem Solving&lt;/a&gt; and Arthur Engel's &lt;a href=&quot;http://books.google.com/books?id=B3EYPeKViAwC&amp;amp;printsec=frontcover&amp;amp;dq=paul%20zeitz&amp;amp;source=gbs_slider_thumb#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problem Solving Strategies&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of other books with similar titles and descriptions. Just follow Google Books's suggestions.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-23T23:25:35.780" CommunityOwnedDate="2010-07-23T23:25:35.780" />
  <row Id="612" PostTypeId="1" AcceptedAnswerId="615" CreationDate="2010-07-23T23:38:24.513" Score="2" ViewCount="81" Body="&lt;p&gt;Are there any results from algebraic geometry that have led to an interesting &quot;real world&quot; application? &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-24T00:52:55.863" CommunityOwnedDate="2010-07-24T01:15:56.183" Title="What are some applications outside of mathematics for algebraic geometry?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="613" PostTypeId="2" ParentId="612" CreationDate="2010-07-23T23:47:06.493" Score="2" ViewCount="0" Body="&lt;p&gt;Broadly speaking, algebraic geometry is used a lot in some areas of robotics and mechanical engineering. Real algebraic geometry, for example, is important to the development of CAD systems (think NURBS, computing intersections of primitives, etc.) And AG comes up in robotics when it is important to figure out, say, what motions a robotic arm in a given configuration is capable of, or to construct some kind of linkage that draws a prescribed curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something specific in that vein: Kempe's Universality Theorem gives that any bounded algebraic curve in $\mathbb{R}^2$ is the locus of some linkage. The &quot;locus of a linkage&quot; being the path drawn out by all the vertices of a graph, where the edge lengths are all specified and one or more vertices remains still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, Kempe's orginal proof of the theorem was flawed, and more recent proofs have been more involved. However, &lt;a href=&quot;http://erikdemaine.org/theses/tabbott.pdf&quot; rel=&quot;nofollow&quot;&gt;Timothy Abbott's MIT masters thesis&lt;/a&gt; gives a simpler proof that gives a working linkage for a given curve, and makes for interesting reading concerning the problem in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The NURBS connection is, in part, that can construct a B-spline that approximates a given real algebraic curve, which is crucial in displaying intersection curves, for example. See &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.51.3585&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more details (I'm afraid I don't know many on this.)&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-24T00:52:55.863" LastActivityDate="2010-07-24T00:52:55.863" CommentCount="1" />
  <row Id="614" PostTypeId="1" AcceptedAnswerId="616" CreationDate="2010-07-24T00:06:42.180" Score="1" ViewCount="55" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A 10-year loan of $500 is repaid with&#xA;  payments at the end of each year. The lender charges interest at an&#xA;  annual effective rate of 10%. &#xA;  Each of the first ten payments is&#xA;  150% of the amount of interest due. &#xA;  Each of the last ten payments is X. &#xA;  Calculate X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came across this practice question while studying for my actuary exam. I tried it on my own and got stuck:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$500 will earn $50 interest each year, so each of the first 10 payments must be $75.&lt;br&gt;&#xA;Then after 10 years, a total of 750 has been repaid.&lt;br&gt;&#xA;In 10 years, I can find the accumulated debt by saying &lt;code&gt;PV&lt;/code&gt;=500, &lt;code&gt;I/Y&lt;/code&gt;=10, &lt;code&gt;N&lt;/code&gt;=10, giving me &lt;code&gt;FV&lt;/code&gt;=$1296.87. So the balance would be $1296.87-$750=$546.97.&lt;br&gt;&#xA;Now I am stuck! How do I find out what the last payments should be? I know I can't just divide $546.97/10=$54.697, because the lender is still charging interest while the borrower pays off this remaining debt, so there would still be the interest left over.&lt;br&gt;&#xA;This situation isn't mentioned anywhere in my calculator manual! Can one of you give me some explanation about what is going on so that I can do it by hand?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I tried working on it some more, and came up with a really great idea! Since the payments are 1.5x the 10% interest, it's just like paying off 5% of the principal each year! This saves me a lot of time, because I can just set &lt;code&gt;I/Y=-5&lt;/code&gt; and get &lt;code&gt;FV=598.74&lt;/code&gt; on my calculator. I did it the long way by calculating the future value of each interest payment (turns out they were not all $75, because the outstanding principal got smaller), and they were the same. Is this always going to work, or did I just get lucky here?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another update!&lt;br&gt;&#xA;I think I solved it. All I needed to do was to set &lt;code&gt;FV=0&lt;/code&gt;, &lt;code&gt;I/Y=10&lt;/code&gt;, &lt;code&gt;N=10&lt;/code&gt;, &lt;code&gt;PV=598.74&lt;/code&gt;, and then I got &lt;code&gt;PMT=97.44&lt;/code&gt;. I never used the &lt;code&gt;PMT&lt;/code&gt; button before, though, so is there some other way I can check the answer is right?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T03:54:11.520" LastActivityDate="2010-07-25T09:59:56.960" Title="How to determine annual payments on a partially repaid loan?" Tags="&lt;finance&gt;&lt;actuarial-science&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="615" PostTypeId="2" ParentId="612" CreationDate="2010-07-24T00:08:18.137" Score="1" ViewCount="0" Body="&lt;p&gt;The following &lt;a href=&quot;http://www.dms.uaf.edu/~eallman/mega.pdf&quot; rel=&quot;nofollow&quot;&gt;slideshow&lt;/a&gt; gives an explanation of how algebraic geometry can be used in phylogenetics.&#xA;See also &lt;a href=&quot;http://rigtriv.wordpress.com/2008/08/07/phylogenetics-and-algebraic-geometry/&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; of Charles Siegel on Rigorous Trivialties. This is not an area I've looked at in much detail at all, but it appears that the idea is to use a graph to model evolutionary processes, and such that the &quot;transition function&quot; for these processes is given by a polynomial map.  In particular, it'd be of interest to look at the potential outcomes, namely the image of the transition function; that corresponds to the image of a polynomial map (which is not necessarily an algebraic variety, but it is a constructible set, so not that badly behaved either).  (In practice, though, it seems that one studies the closure, which is a legitimate algebraic set.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-24T00:08:18.137" CommentCount="2" />
  <row Id="616" PostTypeId="2" ParentId="614" CreationDate="2010-07-24T00:24:01.380" Score="1" ViewCount="0" Body="&lt;p&gt;This problem is in two stages. For the first stage, notice that you are paying 150% interest, but ending up owing more. This is because you subtracted $750 from the future value, when in fact each $75 amount was paid at a time in the past and needs be converted to a future value too. The payment in the nth year has a future value of 75*1.1^(10-n). The total future value of the repayment is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;75*1.1^9+75*1.1*8+...+75*1.1^0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I have assumed that the interest is charged before the repayments are made. This sequence is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_progression&quot; rel=&quot;nofollow&quot;&gt;geometric progression&lt;/a&gt;. We consider it as a geometric sequence in reverse to make the maths easier. It has first term (a) 75, each term 1.1 times the previous (r) and 10 terms (n). The sum is given by the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a(r^n-1)/(r-1)=75*(1.1^10-1)/0.1=$1195.31&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After we have solved this first part, then it is just a standard &lt;a href=&quot;http://math.stackexchange.com/questions/618/repayments-of-a-loan-with-compound-interest/685#685&quot; rel=&quot;nofollow&quot;&gt;interest with repayments problem.&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-25T09:59:56.960" LastActivityDate="2010-07-25T09:59:56.960" CommentCount="6" />
  <row Id="617" PostTypeId="2" ParentId="607" CreationDate="2010-07-24T00:38:09.387" Score="5" ViewCount="0" Body="&lt;p&gt;The question can be answered using &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;Catalan numbers&lt;/a&gt;. Let C_n denote the number of sequences of 2n coin tosses in which you are never ahead. Formally, we count sequences in which every prefix has no less &lt;code&gt;T&lt;/code&gt;'s than &lt;code&gt;H&lt;/code&gt;'s. We call this &lt;em&gt;property A&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of total sequences of length 2n is $2^{2n}$. We then show that as n→∞, the ratio $C_n / 2^{2n}$ tends to 0. This means that in almost every sequence you will eventually be ahead (the chances of a random sequence having property A tend to 0 as the sequence gets longer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n = \frac{(2n)!}{(n+1)!n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n / 2^{2n} = \frac{(2n)!}{2^{2n}} \cdot \frac{1}{(n+1)!n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it can be shown that this tends to 0 by &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt; (multiply and divide by $(2n/e)^{2n}$).&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-08-07T12:29:15.587" LastActivityDate="2010-08-07T12:29:15.587" />
  <row Id="618" PostTypeId="1" AcceptedAnswerId="685" CreationDate="2010-07-24T01:10:30.970" Score="1" ViewCount="35" Body="&lt;p&gt;Suppose I have a loan of M dollars. At the end of each year, I am charged interest at rate R and make a repayment of P. The loan is repaid after n years.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How long (n) does it take to repay the loan if I am given the other variables?&lt;/li&gt;&#xA;&lt;li&gt;How much are the repayments of P if I am given the other variable?&lt;/li&gt;&#xA;&lt;li&gt;Suppose that the payments were at the start of the year. How would this change the problem?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T18:37:49.753" LastActivityDate="2010-07-25T09:58:14.807" Title="Repayments of a loan with compound interest" Tags="&lt;finance&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="619" PostTypeId="1" AcceptedAnswerId="620" CreationDate="2010-07-24T01:22:34.420" Score="6" ViewCount="115" Body="&lt;p&gt;&lt;em&gt;The Problem:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two trains travel on the same track towards each other, each going at a speed of 50 kph. They start out 50km apart. A fly starts at the front of one train and flies at 75 kph to the front of the other; when it gets there, it turns around and flies back towards the first. It continues flying back and forth til the two trains meet and it gets squashed (the least of our worries, perhaps).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far did the fly travel before it got squashed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Attempt at a solution:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can do this by summing the infinite series of the fly's distance for each leg. I get an answer of 37.5 km: but that's so nice! There must be a more intuitive way...is there?&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-24T01:34:44.343" LastActivityDate="2010-07-24T01:34:44.343" Title="Problem: Two Trains and a Fly" Tags="&lt;high-school&gt;&lt;sequences-and-series&gt;&lt;fun-problem&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="620" PostTypeId="2" ParentId="619" CreationDate="2010-07-24T01:30:46.100" Score="9" ViewCount="0" Body="&lt;p&gt;The trains take half an hour to collide, which, at a rate of 75kph, leads to the fly travelling 37.5km.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-24T01:30:46.100" CommentCount="1" />
  <row Id="621" PostTypeId="2" ParentId="618" CreationDate="2010-07-24T02:20:09.947" Score="2" ViewCount="0" Body="&lt;p&gt;Suggest you look into &lt;a href=&quot;http://en.wikipedia.org/wiki/Amortization_calculator&quot; rel=&quot;nofollow&quot;&gt;amortization&lt;/a&gt; calculators.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-24T02:20:09.947" CommentCount="1" />
  <row Id="622" PostTypeId="1" CreationDate="2010-07-24T02:42:29.813" Score="14" ViewCount="507" Body="&lt;p&gt;Representation theory is a subject I want to like (it can be fun finding the representations of a group), but it's hard for me to see it as a subject that arises naturally or why it is important.  I can think of two mathematical reasons for studying it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The character table of a group is packs a lot of information about the group and is concise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It is practically/computationally nice to have explicit matrices that model a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are for sure deeper things that I am missing.  I can understand why one would want to study group actions (the axioms for a group beg you to think of elements as operators), but why look at group actions on vector spaces?  Is it because linear algebra is so easy/well-known (when compared to just modules, say)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also told that representation theory is important in quantum mechanics.  For example, physics should be SO(3) invariant and when we represent this on a Hilbert space of wave-functions, we are led to information about angular momentum.  But this seems to only trivially invoke representation theory since we already start with a subgroup of GL(n) and then extend it to act on wave functions by $\psi(x,t) \mapsto \psi(Ax,t)$ for A in SO(n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_physics_and_representation_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Particle_physics_and_representation_theory&lt;/a&gt; wikipedia article claims that if our physical system has G as a symmetry group, then there is a correspondence between particles and representations of G.  I'm not sure if I understand this correspondence since it seems to be saying that if we act an element of G on a state that corresponds to some particle, then this new state also corresponds to the same particle.  So a particle is an orbit of the G action?  Anyone know of good sources that talk about this?&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-21T00:46:52.693" Title="Importance of Representation Theory" Tags="&lt;representation-theory&gt;&lt;physics&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="8" />
  <row Id="623" PostTypeId="1" AcceptedAnswerId="644" CreationDate="2010-07-24T02:54:51.733" Score="5" ViewCount="315" Body="&lt;p&gt;The volume of a cone with height $h$ and radius $r$ is $\frac{1}{3} \pi r^2 h$, which is exactly one third the volume of the smallest cylinder that it fits inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved easily by considering a cone as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_of_revolution&quot; rel=&quot;nofollow&quot;&gt;solid of revolution&lt;/a&gt;, but I would like to know if it can be proved or at least visual demonstrated without using calculus.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-24T17:32:17.447" Title="Why is the volume of a cone one third of the volume of a cylinder?" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="624" PostTypeId="1" AcceptedAnswerId="629" CreationDate="2010-07-24T03:00:59.880" Score="7" ViewCount="100" Body="&lt;p&gt;In my AP chem class, I often have to balance chemical equations like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Al + O&lt;sub&gt;2&lt;/sub&gt; ---&gt; Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to make both side of the arrow have the same amount of atoms by adding compounds in the equation to each side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4Al + 3O&lt;sub&gt;2&lt;/sub&gt; ---&gt; 2Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the subscripts become really large, or there are lot of atoms involved, trial and error is impossible unless performed by a computer. What if some chemical equation can not be balanced(does such equations exist)? I tried one for a long time only to realize the problem was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher said trial and error is the only way. Are there other methods?&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-07-25T02:30:21.940" LastActivityDate="2010-07-25T03:11:29.843" Title="Balance chemical equations without trial and error?" Tags="&lt;chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="625" PostTypeId="1" CreationDate="2010-07-24T03:01:09.037" Score="7" ViewCount="242" Body="&lt;p&gt;When differentiated with respect to $r$, the derivative of $\pi r^2$ is $2 \pi r$, which is the circumference of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when the formula for a sphere's volume $\frac{4}{3} \pi r^3$ is differentiated with respect to $r$, we get $4 \pi r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just a coincidence, or is there some deep explanation for why we should expect this?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T03:03:29.950" LastActivityDate="2010-07-24T16:59:38.733" Title="Why is the derivative of a circle's area its perimeter (and similarly for spheres)?" Tags="&lt;geometry&gt;&lt;calculus&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="626" PostTypeId="2" ParentId="622" CreationDate="2010-07-24T03:16:11.290" Score="11" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The representation theory of finite groups can be used to prove results about finite groups themselves that are otherwise much harder to prove by &quot;elementary&quot; means. For instance, the proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Burnside's theorem&lt;/a&gt; (that a group of order $p^a q^b$ is solvable).  A lot of the classification proof of finite simple groups relies on representation theory (or so I'm told, I haven't read the proof...).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematical physics. Lie algebras and Lie groups definitely come up here, but I'm not familiar enough to explain anything. In addition, the classification of complex simple Lie algebras relies on the root space decomposition, which is a significant (and nontrivial) fact about the representation theory of semisimple Lie algebras.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Number theory.  The nonabelian version of L-functions (Artin L-functions) rely on the representations of the Galois group (in the abelian case, these just correspond to sums of 1-dimensional characters).  For instance, the proof that Artin L-functions are meromorphic in the whole plane relies on (I think) &lt;del&gt;Artin&lt;/del&gt; Brauer's theorem (i.e., a corollary of the usual statement) that any irreducible character is an &lt;del&gt;rational&lt;/del&gt; integer combination of induced characters from cyclic subgroups -- this is in Serre's &lt;em&gt;Linear Representations of Finite Groups.&lt;/em&gt; Also, the Langlands program studies  representations of groups $GL_n(\mathbb{A}_K)$ for $\mathbb{A}_K$ the adele ring of a global field. This is a generalization of standard &quot;abelian&quot; class field theory (when $n=1$ and one is determining the character group of the ideles).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Combinatorics. The representation theory of the symmetric group has a lot of connections to combinatorics, because you can parametrize the irreducibles explicitly (via Young diagrams), and this leads to the problem of determining how these Young diagrams interact. For instance, what does the tensor product of two Young diagrams look like when decomposed as a sum of Young diagrams? What is the dimension of the irreducible representation associated to a Young diagram? These problems have a combinatorial flavor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I should add the disclaimer that I have not formally studied representation theory, and these are likely to be an unrepresentative sample of topics (some of which I have only vaguely heard about).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-18T11:34:01.253" LastActivityDate="2010-08-18T11:34:01.253" CommentCount="4" />
  <row Id="627" PostTypeId="2" ParentId="625" CreationDate="2010-07-24T03:23:06.523" Score="1" ViewCount="0" Body="&lt;p&gt;Because you use the integral (read: anti-derivative) to find the area under the curve - even a curve in polar coordinates.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-24T16:59:38.733" LastActivityDate="2010-07-24T16:59:38.733" CommentCount="4" />
  <row Id="628" PostTypeId="2" ParentId="625" CreationDate="2010-07-24T03:32:36.620" Score="10" ViewCount="0" Body="&lt;p&gt;Consider increasing the radius of a circle by an infinitesimally small amount, dx. This increases the area by an &lt;a href=&quot;http://en.wikipedia.org/wiki/Annulus_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;annulus&lt;/a&gt; (or ring) with inner radius 2pi*r and outer radius 2pi*(r+dx). At this ring is extremely thin, we can imagine cutting the ring and then flattening it out to form a rectangle with width 2pi*r and height dx (the side of length 2pi*(r+dx) is close enough to 2pi*r that we can ignore that). So the area gain is 2pi*r*dx and to determine the rate of change with respect to x, we divide by dx and so we get 2pi*r. This is of course, just an informative, intuitive explanation and not a formal proof. The same reasoning works with a sphere, we just flatten it out to a rectangular prism instead.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-24T03:32:36.620" />
  <row Id="629" PostTypeId="2" ParentId="624" CreationDate="2010-07-24T03:34:46.620" Score="8" ViewCount="0" Body="&lt;p&gt;Yes; it's possible to write a system of equations that can be solved to find the correct coefficients. Here's an example for the given formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're trying to find coefficients A, B, and C such that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A (Al) + B (O_2) -&amp;gt; C (Al_2 O_3)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In order to do this, we can write an equation for each element based on how many atoms are on each side of the equation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for Al:  A = 2C&#xA;for O:  2B = 3C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is an uninteresting example, but these will always be linear equations in terms of the coefficients. Note that we have fewer equations than variables. This means that there's more than one way to correctly balance the equation (and there is, because any set of coefficients can be scaled by any factor). We just need to find one integral solution to these equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve, we can arbitrarily set one of the variables to 1 and we'll get a solution with (probably fractional) coefficients. If we add $A=1$, the solution is $(A,B,C) = (1,3/4,1/2)$. To get the smallest solution with integer coefficients, just multiply by the least common multiple of the denominators (4 in this case), giving us $(4,3,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the set of equations has no solution where the coefficients are nonzero, then you know that the equation cannot be balanced.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-25T03:11:29.843" LastActivityDate="2010-07-25T03:11:29.843" CommentCount="4" />
  <row Id="631" PostTypeId="1" CreationDate="2010-07-24T04:46:02.207" Score="24" ViewCount="722" Body="&lt;p&gt;The 24 game is as follows.  Four numbers are drawn; the player's objective is to make 24 from the four numbers using the four basic arithmetic operations (in any order) and parentheses however one pleases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following generalization. Given n+1 numbers, determine whether the last one can be obtained from the first n using elementary arithmetical operations as above. This problem admits succint certificates so is in NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it NP-complete?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-24T13:10:37.653" LastActivityDate="2010-08-04T00:14:22.177" Title="Is the 24 game NP-complete?" Tags="&lt;computer-science&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="12" />
  <row Id="632" PostTypeId="1" AcceptedAnswerId="1245" CreationDate="2010-07-24T05:07:01.957" Score="5" ViewCount="176" Body="&lt;p&gt;I am working on computing phase diagrams for alloys.  These are&#xA;blueprints for a material that show what phase, or combination of&#xA;phases, a material will exist in for a range of concentrations and&#xA;temperatures (see &lt;a&#xA;href=&quot;http://web.cos.gmu.edu/~tstephe3/talks/SIAMMaterialsScience2010.pdf&quot; rel=&quot;nofollow&quot;&gt;this&#xA;pdf presentation&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial step in drawing the boundaries that separate one phase&#xA;from another on these diagrams involves minimizing a free energy&#xA;function subject to basic physical conservation constraints.  I am&#xA;going to leave out the chemistry/physics and hope that we can move forward&#xA;with the minimization using Lagrange multipliers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free energy that is to be minimized is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\widetilde{G}(x_1, x_2) = f^{(1)}G_{1}(x_1) + f^{(2)}G_{2}(x_2),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(1)}x_1 + f^{(2)}x_2 = c_1,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(1)} + f^{(2)} = 1. $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and also that the $x_{i} &gt; 0$ and $f^{(i)} &gt; 0$, for $i=1,2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrange formulation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(x_1,x_2,f^{(1)},f^{(2)},\lambda_1, \lambda_2, \lambda_3) =&#xA;f^{(1)}G_{1}(x_1) + f^{(2)}G_{2}(x_2)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$- \lambda_{1}(f^{(1)}x_1 + f^{(2)}x_2 - c_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$- \lambda_{2}(f^{(1)} + f^{(2)} - 1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimization of $\widetilde{G}$ follows from finding the $x_{i}$'s  that satisfy $\nabla L = 0:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial x_{1}}   = f^{(1)}G_{1}'(x_1) - \lambda_{1}f^{(1)} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial x_2}     = f^{(2)}G_{2}'(x_2) - \lambda_{1}f^{(2)} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial f^{(1)}} = G_{1}(x_1) - \lambda_{1}x_{1} - \lambda_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial f^{(2)}} = G_{2}(x_2) - \lambda_{1}x_{2} - \lambda_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(*) f^{(1)}\left[G_{1}'(x_1) - \lambda_1 \right] = 0$       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(**) f^{(2)}\left[G_{2}'(x_2) - \lambda_1 \right]= 0 $       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(***) G_{1}(x_1) - G_{2}(x_2) = \lambda_1 \left[ x_1 - x_2\right]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $f^{(1)}$ and $f^{(2)}$ are not to be zero, from (*) and (**) we have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_{1}'(x_1) = G_{2}'(x_2) = \lambda_{1}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, a manipulation of equation (***) looks like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{G_{1}(x_1) -G_{2}(x_2)}{x_1 - x_2} = \lambda_{1}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, think of $G_{i}$ as an even degree polynomial (which it isn't, but&#xA;it's graph sometimes resembles one) in the plane.  Let the points $x_1$&#xA;and $x_2$ be locations along the x-axis that lie roughly below the&#xA;minima of this curve.  The constraints (*),(*&lt;em&gt;), and (*&lt;/em&gt;*) describe the&#xA;condition that the line drawn between $(x_1,G_{1}(x_1))$ and $(x_2,G_{2}(x_2))$ form a common tangent&#xA;to the &quot;wells&quot; of the curve.  It is these points $x_1$ and $x_2$,&#xA;which represent concentrations of pure components in our alloy, that&#xA;become mapped onto a phase diagram.  It is essentially by repeating this procedure for many&#xA;temperatures that we can trace out the boundaries in the desired phase diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question is:&lt;/strong&gt;  Looking at this from a purely analytic geometry&#xA;perspective, &lt;strong&gt;how would one derive the &quot;variational&quot; approach to find a common tangent line that we seem to have found using the above Lagrangian?&lt;/strong&gt;  (warning: I don't really know how to&#xA;model things using variational methods.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And, secondly:&lt;/strong&gt; I have presented a model of a binary alloy, meaning&#xA;two variables to keep track of representing concentrations.  I have&#xA;been working on ternary alloys, where this free energy $\widetilde{G}$&#xA;is a function of three variables (two independent: $x_1,x_2,x_3$,&#xA;where $x_3 = 1- x_1 - x_2$) and is therefore a surface over a Gibbs&#xA;triangle.  Then $\nabla L = 0$ produces partial derivatives that no&#xA;longer &quot;speak geometry&quot; to me, although the solution is a common tangent&#xA;plane.  (I have attempted to characterize a common tangent plane&#xA;based purely in analytic geometry - completely disregarding the&#xA;Lagrangian - and have come up with several relations between&#xA;directional derivatives... &lt;strong&gt;How might directional derivatives relate&#xA;to the optimality conditions set forth by the Lagrangian?&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;  Thank you Greg Graviton for wading through this sub-optimal notation and pointing out several mistakes in the statement of the problem.  (Also, thank you for the &lt;a href=&quot;http://math.stackexchange.com/questions/632/validating-a-mathematical-model-lagrange-formulation-and-geometry/1245#1245&quot;&gt;excellent discussion below&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-01T18:29:44.957" LastActivityDate="2010-08-01T18:29:44.957" Title="validating a mathematical model (Lagrange formulation and geometry)" Tags="&lt;calculus&gt;&lt;applications&gt;&lt;materials-science&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="633" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T05:55:46.600" Score="1" ViewCount="0" Body="&lt;p&gt;You can use Pappus's centroid theorem &lt;a href=&quot;http://math.stackexchange.com/questions/164/why-is-the-volume-of-a-sphere-frac43-pi-r3/174#174&quot; rel=&quot;nofollow&quot;&gt;as in my answer here&lt;/a&gt;, but it does not provide much insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead of a cylinder and a cone, you consider a cube and a square-based pyramid where the &quot;top&quot; vertex of the pyramid (the one opposite the square base) is shifted to be directly above one vertex of the base, you can fit three such pyramids together to form the complete cube.  (I've seen this as physical toy/puzzle with three pyramidal pieces and a cubic container.)  This may give some insight into the 1/3 &quot;pointy thing rule&quot; (for pointy things with similar, linearly-related cross-sections) that Katie Banks discussed in her comment.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-24T05:55:46.600" />
  <row Id="634" PostTypeId="2" ParentId="603" CreationDate="2010-07-24T06:14:57.360" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: The question is much more involved than I'd originally thought.  As implied in the question, I knew that congruence and similarity transformations are constructible.  Immediately below this section is my original answer, which only demonstrates congruence transformations and was intended more to give an idea of what an answer might look like (since, at the time, there was another answer that was not particularly helpful).  In the last section of this answer is my justification that all affine transformations of the plane are constructible.  In re-reading that now, I realize that I'd assumed the ability to construct a point, say P', on a line, say RP, such that RP'/RP is equal to some known ratio.  This is equivalent to being able to construct the dilation of P by the known ratio about center R.  I've added the construction of such a dilation below the congruence transformation section.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a partial answer, here are constructions of the basic congruence transformations, assuming basic construction techniques like constructing a line parallel or perpendicular to a given line through a given point and angle-copying:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;reflection&lt;/strong&gt;  Given a point P and a line l, construct the line perpendicular to l through P, and construct the circle centered at the intersection of this new line and l and passing through P.  The image of P under a reflection over the line l is the point of intersection of the circle and the new line (the one not at P).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;translation&lt;/strong&gt;  Given a point P and a vector AB (from A to B), construct the line through A and P, the line through B parallel to line PA, and the line through P parallel to AB.  The image of P under translation by vector AB is the intersection of the two constructed parallels.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;rotation&lt;/strong&gt;  Given a point P, a center of rotation R, and an angle ABC (from A to C), construct the line through P and R, copy angle ABC such that the copy of B coincides with R and the copy of A is on ray RP and let the copy of C be called D, construct the circle with center at R and passing through P.  The image of P under rotation by angle ABC about point R is the intersection of the circle with ray RD.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;dilation&lt;/strong&gt; Given a point P, a center of dilation R, and a ratio AC/AB (where point C lies on ray AB), translate B to B' and C to C' by the translation that takes A to R, construct line B'P, construct the line through C' parallel to B'P, and construct ray RP.  The image of P under a dilation about R by a factor of AC/AB is the intersection of ray RP and the line through C' parallel to B'P.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All &lt;strong&gt;affine transformations&lt;/strong&gt; are constructible.  Per &lt;a href=&quot;http://mathworld.wolfram.com/AffineTransformation.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation#Affine_transformation_of_the_plane&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, an affine transformation of the plane is a transformation of the plane that preserves collinearity and preserves ratios of distances on any given line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, to show that affine transformations preserve parallelism, suppose that two lines MN and PQ are parallel, and that their images, lines M'N' and P'Q', intersect at X', the image of X.  Since affine transformations preserve collinearity, X must be on MN and on PQ, which is a contradiction, so M'N' and P'Q' cannot intersect.  Thus, affine transformations preserve parallelism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An affine transformation is determined by a triangle ABC and its image, triangle A'B'C' (per &lt;a href=&quot;http://mathworld.wolfram.com/AffineTransformation.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;; &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation#Affine_transformation_of_the_plane&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; talks about defining an affine transformation by a parallelogram and its image, which is equivalent since affine transformations preserve parallelism).  Given point P and triangles ABC and A'B'C', construct line l_1 through P parallel to AB and line l_2 through P and parallel to AC, call the intersection of l_1 with AC I_1 and call the intersection of l_2 with AB I_2, extend A'B' past B' to a point I'_2 such that AB/AI_2 = A'B'/A'I'_2, extend A'C' past C' to a point I'_1 such that AC/AI_1 = A'C'/A'I'_1, construct line l'_1 through I'_1 parallel to A'B' and line l'_2 through I'_2 parallel to A'C'.  The image of P under the affine transformation mapping ABC onto A'B'C' is the intersection of lines l'_1 and l'_2.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T01:34:29.577" LastActivityDate="2010-07-26T01:34:29.577" CommentCount="4" />
  <row Id="635" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T07:30:06.133" Score="5" ViewCount="0" Body="&lt;p&gt;One can cut a cube into 3 pyramids with square bases -- so for such pyramids the volume is indeed 1/3 hS. And then one uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavalieris_principle&quot; rel=&quot;nofollow&quot;&gt;Cavalieri's principle&lt;/a&gt; to prove that the volume of any cone is 1/3 hS.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-07-24T07:30:06.133" />
  <row Id="636" PostTypeId="1" AcceptedAnswerId="919" CreationDate="2010-07-24T07:36:50.767" Score="4" ViewCount="121" Body="&lt;p&gt;The knight's tour is a sequence of 64 squares on a chess board, where each square is visted once, and each subsequent square can be reached from the previous by a knight's move.  Tours can be cyclic, if the last square is a knight's move away from the first, and acyclic otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several symmetries among knight's tours.  Both acyclic and cyclic tours have eight reflectional symmetries, and cyclic tours additionally have symmetries arising from starting at any square in the cycle, and from running the sequence backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it known how many knight's tours there are, up to all the symmetries?&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-28T05:57:09.560" Title="How many knight's tours are there?" Tags="&lt;combinatorics&gt;&lt;chess&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="637" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T08:34:05.543" Score="0" ViewCount="0" Body="&lt;p&gt;This is certainly a non-trivial proof if you have to avoid calculus. See the answer over on the &lt;a href=&quot;http://mathforum.org/library/drmath/view/55263.html&quot; rel=&quot;nofollow&quot;&gt;Dr. Math forums&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-24T08:34:05.543" CommentCount="1" />
  <row Id="639" PostTypeId="1" AcceptedAnswerId="1106" CreationDate="2010-07-24T11:03:59.873" Score="15" ViewCount="1344" Body="&lt;p&gt;I just got out from my Math and Logic class with my friend.  During the lecture, a well-known math/logic puzzle was presented:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The King has $1000$ wines, $1$ of which is poisoned.  He needs to identify the poisoned wine as soon as possible, and with the least resources, so he hires the protagonist, a Mathematician.  The king offers you his expendable servants to help you test which wine is poisoned.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The poisoned wine is very potent, so much that one molecule of the wine will cause anyone who drinks it to die.  However, it is slow-acting.  The nature of the slow-acting poison means that there is only time to test one &quot;drink&quot; per servant. (A drink may be a mixture of any number of wines)  (Assume that the King needs to know within an hour, and that any poison in the drink takes an hour to show any symptoms)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the minimum amount of servants you would need to identify the poisoned wine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With enough time and reasoning, one can eventually see that this requires at most &lt;strong&gt;ten&lt;/strong&gt; ($10$) servants (in fact, you could test 24 more wines on top of that 1000 before requiring an eleventh servant).  The proof/procedure is left to the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend and I, however, was not content with resting upon this answer.  My friend added the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be different if there were $2$ wines that were poisoned out of the 1000?  What is the new minimum then?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We eventually generalized the problem to this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $N$ bottles of wine ($N \gt 1$) and, of those, $k$ poisoned wines  ($0 \lt k \lt N$), what is the optimum method to identify the all of the poisoned wines, and how many servants are required ($s(N,k)$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After some mathsing, my friend and I managed to find some (possibly unhelpful) lower and upper bounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ log_2 {N \choose k} \le s(N,k) \le N-1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because $log_2 {N \choose k}$ is the minimum number of servants to uniquely identify the $N \choose k$ possible configurations of $k$ poisoned wines in $N$ total wines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help us find an optimum strategy?  Besides the trivial one requiring $N-1$ servants.  How about a possible approach to start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this problem be any different if you were only required to find a strategy that would for sure find a wine that is &lt;strong&gt;not&lt;/strong&gt; poisoned, instead of identifying all poisoned wines? (other than the slightly trivial solution of $k$ servants)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:28:51.460" LastActivityDate="2010-12-23T19:26:16.583" Title="Logic problem: Identifying poisoned wines out of a sample, minimizing test subjects with constraints" Tags="&lt;puzzle&gt;&lt;recreational-mathematics&gt;" AnswerCount="3" CommentCount="16" FavoriteCount="8" />
  <row Id="640" PostTypeId="1" AcceptedAnswerId="820" CreationDate="2010-07-24T11:24:56.920" Score="8" ViewCount="398" Body="&lt;p&gt;When I studied complex analysis, I could never understand how once-differentiable complex functions could be possibly be infinitely differentiable. After all, this doesn't hold for functions from R2 to R2. Can anyone explain what is different about complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-28T17:09:07.587" Title="Why are differentiable complex functions infinitely differentiable?" Tags="&lt;complex-analysis&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="641" PostTypeId="2" ParentId="640" CreationDate="2010-07-24T11:51:04.973" Score="3" ViewCount="0" Body="&lt;p&gt;The proofs I have seen derive this as a corollary of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula&quot; rel=&quot;nofollow&quot;&gt;Cauchy's integral formula&lt;/a&gt;.  Look at the difference quotient as an integral, play around with it, and you get that it converges to what you'd get if you differentiated under the integral sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that since harmonic functions also satisfy a similar integral equation, they are also infinitely differentiable in the same way (this also follows since they are real and imaginary parts of holomorphic functions).&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-24T12:01:33.627" LastActivityDate="2010-07-24T12:01:33.627" CommentCount="3" />
  <row Id="642" PostTypeId="2" ParentId="622" CreationDate="2010-07-24T13:08:06.907" Score="4" ViewCount="0" Body="&lt;p&gt;Particles correspond to specific vectors in a representation, not to G-orbits!  The reason has to do with &quot;symmetry breaking.&quot;  The 8 particles in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eightfold_Way_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;meson octet&lt;/a&gt; correspond to a basis of a certain 8-dimensional representation of the group SU(3) called the &quot;adjoint representation.&quot;  At high enough energies these particles would be indistinguishable.  But at low energies the &quot;SU(3) symmetry has been broken&quot; and the particles become distinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good physics example that's easier to understand is that the orbital states of electrons in atoms correspond to representations of the group SO(3) of symmetries of space (well, really SU(2) if you want to incorporate spin).  Try reading a standard quantum mechanics textbook for a little bit of this picture and then try thinking about it in terms of representation theory.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-07-24T13:08:06.907" CommentCount="2" />
  <row Id="644" PostTypeId="2" ParentId="623" CreationDate="2010-07-24T14:52:32.947" Score="14" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://www.mathematische-basteleien.de/wuerfel16.gif&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;A visual demonstration for the case of a pyramid with a square base. As &lt;a href=&quot;http://math.stackexchange.com/questions/623/why-is-the-volume-of-a-cone-one-third-of-the-volume-of-a-cylinder/635#635&quot; rel=&quot;nofollow&quot;&gt;Grigory states&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavalieris_principle&quot; rel=&quot;nofollow&quot;&gt;Cavalieri's principle&lt;/a&gt; can be used to get the formula for the volume of a cone. We just need the base of the square pyramid to have side length r√pi. Such a pyramid has volume 1/3 * h * pi * r&lt;sup&gt;2&lt;/sup&gt;.&lt;br&gt;&#xA;&lt;img src=&quot;http://www.csrjjsmp.com/images/offsite/cone.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Then the area of the base is clearly the same. The cross-sectional area at distance a from the peak is a simple matter of similar triangles: The radius of the cone's cross section will be a/h * r. The side length of the square pyramid's cross section will be a/h *   r√pi.&lt;br&gt;&#xA;Once again, we see that the areas must be equal. So by Cavalieri's principle, the cone and square pyramid must have the same volume: 1/3 * h * pi * r&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T17:32:17.447" LastActivityDate="2010-07-24T17:32:17.447" />
  <row Id="645" PostTypeId="1" AcceptedAnswerId="647" CreationDate="2010-07-24T15:24:08.620" Score="8" ViewCount="162" Body="&lt;p&gt;Chicken is a famous game where two people drive on a collision course straight towards each other. Whoever swerves is considered a 'chicken' and loses, but if nobody swerves, they will both crash. So the payoff matrix looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  B swerves               B straight&#xA;A swerves            tie               A loses, B wins&#xA;A straight     B loses, A wins            both lose &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I have heard of another situation called the prisoner's dilemma, where two prisoners are each given the choice to testify against the other, or remain silent. The payoff matrix for prisoner's dilemma also looks like  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  B silent               B testify&#xA;A silent            tie               A loses, B wins&#xA;A testify     B loses, A wins            both lose &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I remember hearing that in the prisoner's dilemma, it was always best for both prisoners to testify. But that makes no sense if you try to apply it to chicken: both drivers would crash every time, and in real life, almost always someone ends up swerving. What's the difference between the two situations?&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-24T15:44:51.053" Title="How is prisoner's dilemma different from chicken?" Tags="&lt;game-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="646" PostTypeId="2" ParentId="645" CreationDate="2010-07-24T15:36:48.387" Score="7" ViewCount="0" Body="&lt;p&gt;The games aren't just about winning or losing, but also about utility. Here is a more accurate table for chicken:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  B swerves                B straight&#xA;A swerves    No gain for either          A loses, B wins&#xA;A straight     B loses, A wins         Both have large loss&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is one for prisoners dilemma:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    B silent                        B testify&#xA;A silent       Both have small loss         A large loss, B loses nothing&#xA;A testify  B loses nothing, A large loss,        both medium lose &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the prisoners dilemma, an individual prisoner will always do better by testifying (look at the table), however, by both testifying they end up in a worse position than if both were silent. In contrast, in chicken, going straight will be better if the other swerves and swerving will be better if the other goes straight. Your tables represent some strategies as being equally good for players when they are not.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-24T15:44:51.053" LastActivityDate="2010-07-24T15:44:51.053" />
  <row Id="647" PostTypeId="2" ParentId="645" CreationDate="2010-07-24T15:38:40.977" Score="12" ViewCount="0" Body="&lt;p&gt;(See &lt;a href=&quot;http://en.wikipedia.org/wiki/Chicken_%28game%29#Prisoner.27s_dilemma&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chicken_%28game%29#Prisoner.27s_dilemma&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is in the payoff. In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chicken_%28game%29&quot; rel=&quot;nofollow&quot;&gt;&quot;chicken&quot; game&lt;/a&gt;, the payoff matrix is like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Sw       St&#xA;   Sw  0,  0   -1, +1&#xA;   St +1, -1  -10, -10&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;While in the PD game, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Si       Te&#xA;   Si  3, 3     0, 5&#xA;   Te  5, 0     1, 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Both games have this structure in the payoff table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         C       D&#xA;   C   R, R     S, T&#xA;   D   T, S     P, P&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In PD game, the order is T &gt; R &gt; P &gt; S&lt;/li&gt;&#xA;&lt;li&gt;In chicken game, the order is T &gt; R &gt; S &gt; P&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leads a different Nash equilibria. In the PD game, if A remains silent, B chooses to testify because T &gt; R, while if A testifies, B also testify because S &gt; P. So testifying is the most rational choice of B after considering all possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the chicken game, as P &gt; S, if A goes straight, B should swerve. This leads to two Nash equilibria in the pure game: (St, Sw) and (Sw, St). &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-07-24T15:44:42.443" LastActivityDate="2010-07-24T15:44:42.443" />
  <row Id="648" PostTypeId="2" ParentId="434" CreationDate="2010-07-24T16:16:57.587" Score="0" ViewCount="0" Body="&lt;p&gt;The answer to this question depends on how the problem is defined, but the answer is no, at least without defining the problem in a misleading way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since my first solution was completely off the mark, I have deleted it and posted this new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider polynomial p. If it has an integer solution, then the solution will eventually be found by random guessing. So if it is impossible to prove the existential status, there must be no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know from this link that there is a polynomial, q, that is unsolvable in the integers iff ZFC is consistent. It is well known that ZFC cannot prove its own consistency. So if it ZFC is consistent, then q is unsolvable, but we cannot prove this as then we could prove ZFC. So it seems like it is accurate to say that if mathematics is consistent, we have a polynomial with no integer roots, but we can't prove it. However, if we are assuming maths is consistent, we can use this to prove that the equation is unsolvable (indeed that is what we have done). So, it really isn't accurate an accurate statement at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further clarify, when considering mathematical truth, there are two basic ways of viewing it. The first is where we are assuming that are axioms are true, which necessarily means assuming consistency. If we show any problem is equivalent to consistency, then we consider it to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is where we are considering a formal set of statements, of which the axioms have been defined to be true and seeing which statements can be derived to be true. From this viewpoint, we don't actually know whether the axioms are consistent or not. In fact, Godel's second incompleteness theorem shows that no &quot;non-trivial&quot; atomic system can prove its own consistency. So showing a problem is equivalent to consistency is actually the same as showing that the problem is unprovable by the atomic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusion comes from assuming ZFC is consistent to eliminate one possibility in a choice, yet not allowing this assumption to be used as an axiom in the proofs.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-24T19:39:04.343" LastActivityDate="2010-07-24T19:39:04.343" CommentCount="3" />
  <row Id="649" PostTypeId="2" ParentId="566" CreationDate="2010-07-24T18:13:44.017" Score="5" ViewCount="0" Body="&lt;p&gt;A probability problem I love.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a shuffled deck of cards. Deal off the cards one by one until you reach any Ace. Turn over the next card, and note what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;: which card has a higher probability of being turned over, the Ace of Spades or the Two of Hearts?&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2010-07-24T18:13:44.017" CommunityOwnedDate="2010-07-24T18:13:44.017" CommentCount="3" />
  <row Id="650" PostTypeId="2" ParentId="123" CreationDate="2010-07-24T19:40:33.433" Score="7" ViewCount="0" Body="&lt;p&gt;I don't know if you consider General Relativity &quot;outside acadamia&quot;(and I don't care to argue the point!) but if you do,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the group of symmetries with respect to the Lorentzian Metric can be written as Matrices containing hyperbolic trig functions as elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note Kenny's comment.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-24T20:20:49.873" LastActivityDate="2010-07-24T20:20:49.873" CommentCount="3" />
  <row Id="651" PostTypeId="2" ParentId="640" CreationDate="2010-07-24T20:31:14.640" Score="-1" ViewCount="0" Body="&lt;p&gt;It isn't true.  Take any singly but not doubly differentiable function f, with the property that f(x)=f(-x).  Then obtain F from f by rotating f around the origin: F(r e^it)=f(r).  This function is then singly but not doubly differentiable over the whole complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are confusing differentiable with holomorphic.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-24T20:31:14.640" CommentCount="4" />
  <row Id="652" PostTypeId="2" ParentId="370" CreationDate="2010-07-24T20:40:30.187" Score="2" ViewCount="0" Body="&lt;p&gt;I've read a fair amount of Sets for Mathematics and found it to be a gentle introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521010608&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Sets-Mathematics-F-William-Lawvere/dp/0521010608/ref=pd_sim_b_5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-24T20:40:30.187" CommunityOwnedDate="2010-07-24T20:40:30.187" />
  <row Id="653" PostTypeId="1" CreationDate="2010-07-24T20:59:19.173" Score="4" ViewCount="396" Body="&lt;p&gt;I covered homotopy theory in a recent maths course. However I was never presented with any reasons as to why (or even if) it is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good examples of its use outside academia?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T21:06:29.763" LastActivityDate="2010-11-19T06:01:04.227" Title="Real world uses of homotopy theory" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;applications&gt;&lt;homotopy-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="654" PostTypeId="2" ParentId="653" CreationDate="2010-07-24T21:02:37.217" Score="2" ViewCount="0" Body="&lt;p&gt;Robert Ghrist is an amazing applied mathematician who uses a lot of interesting algebraic topology for engineering applications. He uses homology and sheaf theory. I claim this answers your question since homology is a generalization of homotopy theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant link:&#xA;&lt;a href=&quot;http://www.math.uiuc.edu/~ghrist/index_files/research.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/~ghrist/index_files/research.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-24T21:02:37.217" CommentCount="3" />
  <row Id="655" PostTypeId="1" AcceptedAnswerId="658" CreationDate="2010-07-24T21:30:52.573" Score="3" ViewCount="51" Body="&lt;p&gt;I've learned that multiplying two generating functions f(x) and g(x) gives you the result $\sum_{k=0}^\infty\left(\sum_{j=0}^k a_j\,b_{k-j}\right)x^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used the result, but it was presented in my class without proof and I'm having some trouble tracking one down.  Weak google-foo today, I suppose.  Can anyone give me a pointer to a proof?  If this is a question better answered in book form, that is fine as well.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-24T22:30:23.300" LastActivityDate="2010-07-25T00:47:31.660" Title="Proof for multiplying generating functions" Tags="&lt;generating-functions&gt;&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="656" PostTypeId="1" AcceptedAnswerId="657" CreationDate="2010-07-24T21:32:04.680" Score="8" ViewCount="288" Body="&lt;p&gt;One thing I have been wondering about lately is how to write an equation that describes a pattern of numbers. What I mean is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   0    1    2 &#xA;y   1    5    9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I have this, I can tell that an equation that describes this would be y=4x+1. In fact, I don't even need the 3rd pair of numbers. It's very easy when the equation is a straight line. But when the equation is a parabola, its not always that easy. For example:   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   0    1    2 &#xA;y   1    2    5  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can tell this is y=x&lt;sup&gt;2&lt;/sup&gt;+1, because I recognize the pattern. But I can't always tell just by looking at the numbers what the right equation should be. Is there some way to always know the right equation? I know that if you get the x=0 term you can get the c in y=ax&lt;sup&gt;2&lt;/sup&gt;+bx+c, but that's not enough to let me solve it like I can when the equation is just a line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, can someone show me how you would do it for  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x    0     1     2&#xA;y    5     4     7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's not a homework question, I promise!&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:29:10.943" LastActivityDate="2010-07-26T08:29:10.943" Title="How can I write an equation that matches any sequence?" Tags="&lt;middle-school&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="657" PostTypeId="2" ParentId="656" CreationDate="2010-07-24T21:47:45.883" Score="9" ViewCount="0" Body="&lt;p&gt;If you know your relationship is going to be a &lt;strong&gt;polynomial&lt;/strong&gt;, then there are some pretty (conceptually) simple ways you can do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know what degree your polynomial is (line, parabola, cubic, etc.) then your job will be much easier.  But if not, then you simply need to look at the number of points you have.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you are given one point, the best you can do is of degree 0  ( y = k )&lt;/li&gt;&#xA;&lt;li&gt;If you are given two points, the best you can do is of degree 1 ( y = A x + B )&lt;/li&gt;&#xA;&lt;li&gt;If you are given three points, the best you can do is of degree 2 ( y = A x&lt;sup&gt;2&lt;/sup&gt; + B x + C )&lt;/li&gt;&#xA;&lt;li&gt;If you are given four points, the best you can do is of degree 3 ( y = A x&lt;sup&gt;3&lt;/sup&gt; + B x&lt;sup&gt;2&lt;/sup&gt; + C x + D )&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When I say &quot;the best you can do&quot;, what I mean is -- if you have a parabola, but are only given two points, then you really can't identify the parabola.  But you can say that it's a simple line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume you have three points.  The &quot;best you can do&quot; is assume that it is degree 2.  If it is actually of degree one, your answer will magically turn into a line ( your &lt;code&gt;x^2&lt;/code&gt; coefficient will be &lt;code&gt;0&lt;/code&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea of solving relationships/equations is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you have &lt;code&gt;n&lt;/code&gt; unknowns, you need &lt;code&gt;n&lt;/code&gt; equations/points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice how, in the form of the Parabola ( y = A x&lt;sup&gt;2&lt;/sup&gt; + B x + C ), you have three unknowns?  And also three equations! (points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's pick three arbitrary points&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;x   1    2    4&#xA;y   6    7    3&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You would set up three equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 = 1&lt;sup&gt;2&lt;/sup&gt; * A + 1 * B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 = 2&lt;sup&gt;2&lt;/sup&gt; * A + 2 * B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 = 4&lt;sup&gt;2&lt;/sup&gt; * A + 4 * B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three equations, three unknowns.  You should be able to solve this with a combination of most system-of-equation-solving rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, we find:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;A = -1&#xA;B = 4&#xA;C = 3&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So our equation is y = -x&lt;sup&gt;2&lt;/sup&gt; + 4 x + 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, if your original three points formed a line, your $A$ would $= 0$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, if your equation is &lt;strong&gt;NOT&lt;/strong&gt; a polynomial, then you are left with little more than guess and check, plugging in various coefficients and trials (exponential?  trigonometric?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beauty of the polynomial approach is that a polynomial of high enough degree will always fit &lt;strong&gt;any&lt;/strong&gt; list of points. (provided that the points form a function)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-24T23:46:58.333" LastActivityDate="2010-07-24T23:46:58.333" CommentCount="2" />
  <row Id="658" PostTypeId="2" ParentId="655" CreationDate="2010-07-24T21:50:20.643" Score="2" ViewCount="0" Body="&lt;p&gt;It is actually the other way round. A generating function is generally defined to have an addition operation where the components are added and a multiplication operation like that you mentioned. Once we have made these definitions, we observe that polynomials obey the same laws and so that it is convenient to represent generating functions as infinite polynomials rather than just an infinite tuple.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-25T00:47:31.660" LastActivityDate="2010-07-25T00:47:31.660" />
  <row Id="659" PostTypeId="1" AcceptedAnswerId="663" CreationDate="2010-07-24T22:11:12.957" Score="0" ViewCount="105" Body="&lt;h3&gt;Motivatation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am presently looking at a structure that I am trying to pin down- my strategy being to pull the thing up into the greatest possible generality (based on the bits I'm sure about) and narrow it down from there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation I have is somewhat similar to the dual space structure in vector spaces, though almost certainly less well-behaved and vector spaces alone will not cut it.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Construction&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Consider a vector space $V$ over a field $k$- its dual space $V^*$ appears naturally as the set linear of maps   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w^* : V \to k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, coincidentally, form a vector space in themselves. We can generalise this to arbitrary categories $A$, $B$, $C$ by setting $B=hom(A,C). Then, at least in some sense, $B=A^&lt;em&gt;$. So far so standard, but I want more: a nice property of dual spaces is that an element of $V \otimes V^&lt;/em&gt;$ can be canonically seen as an element of $hom(V,V)$- this is because of the way that $k$ acts on $V$ by multiplication. We can mimic this by letting $C$ be a monoid acting on $A$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;In summary:&lt;/strong&gt; A category $A$ acted on by a monoid $C$ and a dual A $A^*:=hom(A,C)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am particularly interested in when $A$ is also a monoid, especially so when $A$ is a space of stochastic matrices.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Questions:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So this isn't too unlikely a construction, in fact it's probably forehead-slappingly well known, so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is it called, if anything?&lt;/li&gt;&#xA;&lt;li&gt;In what cases can we have $A=B=C$? Is that necessarily a permutation group for example?&lt;/li&gt;&#xA;&lt;li&gt;Are there any useful canonical examples, besides vector spaces?&lt;/li&gt;&#xA;&lt;li&gt;Better yet, theorems??? Papers???&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can probably tell, I am no category theorist, so any help would be awesome.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-06T20:21:01.613" Title="Is this the right categorical generalisation of dual space" Tags="&lt;terminology&gt;&lt;reference-request&gt;&lt;category-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="660" PostTypeId="2" ParentId="655" CreationDate="2010-07-24T22:15:09.880" Score="3" ViewCount="0" Body="&lt;p&gt;Casebash is correct that this is a definition and not a theorem. But the motivation from 3.48 (Defintion of product of series) of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/007054235X&quot; rel=&quot;nofollow&quot;&gt;little Rudin&lt;/a&gt; may convince you that this is a good definition:&lt;br&gt;&#xA;$\sum_{n=0}^{\inf} a_n z^n \cdot \sum_{n=0}^{\inf} b_n z^n = (a_0+a_1z+a_2z^2+ \cdots)(b_0+b_1z+b_2z^2+ \cdots)$&lt;br&gt;&#xA;$=a_0b_0+(a_0b_1 + a_1b_0)z + (a_0b_2+a_1b_1+a_2b_0)z^2 + \cdots$&lt;br&gt;&#xA;$=c_0+c_1z+c_2z^2+ \cdots $&lt;br&gt;&#xA;where $c_n=\sum_{k=0}^n a_k b_{n-k}$  &lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-24T22:15:09.880" CommentCount="1" />
  <row Id="661" PostTypeId="2" ParentId="656" CreationDate="2010-07-24T22:38:04.883" Score="3" ViewCount="0" Body="&lt;p&gt;Given a list of terms of a sequence as you describe, one technique that may be of use (supplementary to Justin's answer) is finite differences.  Calculate the differences between successive terms.  If these first differences are constant, then a linear equation fits the terms you have.  If not, compute the differences of the differences.  If these second differences are constant, then a quadratic equation fits the terms you have.  If not, you can continue to compute differences until you reach a constant difference (in the nth differences means an nth degree polynomial), differences that are a constant multiple of the previous differences (exponential of some sort), or you run out of terms.  In any case, what you find is limited to matching the terms you know, as without some kind of general rule for the sequence, the first unknown term could be anything and completely alter the pattern (and with n known terms, a polynomial of degree n-1 will always perfectly fit).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-24T22:38:04.883" CommentCount="4" />
  <row Id="662" PostTypeId="2" ParentId="656" CreationDate="2010-07-24T22:50:07.723" Score="2" ViewCount="0" Body="&lt;p&gt;In the general case of trying to predict some infinite sequence of integers, there is no formula. This is because there is no reason to expect a pattern to continue, since all sequences are possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can say a certain number is more likely given a set of functions. For instance if you considered all Turing machines, you could say that given n elements of a sequence look at all the Turing machines that predict the current sequence, and then find the most predicted next number. There still isn't a efficient way to compute what the most likely next number is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ray Solomonoff called this &quot;Universal Probabilistic Induction&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained in more depth here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_inference&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inductive_inference&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-24T23:19:39.207" LastActivityDate="2010-07-24T23:19:39.207" />
  <row Id="663" PostTypeId="2" ParentId="659" CreationDate="2010-07-24T23:06:21.473" Score="2" ViewCount="0" Body="&lt;p&gt;I believe the right generalization of dual spaces is that of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_object&quot; rel=&quot;nofollow&quot;&gt;dual object&lt;/a&gt; in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_category&quot; rel=&quot;nofollow&quot;&gt;tensor category&lt;/a&gt;, which I will assume symmetric for convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall what makes a dual space of a vector space work: We have a map $V \times V^* \to k$ (for $k$ the ground field). The problem is, this isn't a homomorphism in the category of vector spaces; it is rather a bilinear map. So you can think of it as a map $V \otimes V^* \to k$ instead.  This is why you need a tensor structure to think of duals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't enough, though, because we need to know that the pairing is nondegenerate.  One way to express this is that there's a map $k \to V \times V^*$ mapping 1 to the &quot;Casimir element&quot; (which is the sum $\sum e_i \otimes e_i^{\vee}$ where $e_i$ ranges over a basis of $V$ and $e_i^{\vee}$ the dual basis; it is independent of the choice of $e_i$ as a quick computation shows).  The Casimir morphism satisfies the condition that $$V \to (k) \otimes V \to (V \otimes V^&lt;em&gt;) \otimes V \to V \otimes (V^&lt;/em&gt; \otimes V)$ is just the identity.&lt;br&gt;&#xA;Conversely, this is enough to show that the pairing is nondegenerate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, anyway, how does this make sense in a symmetric tensor category?  Basically, $V$ is the object, $V^*$ the putative dual, and $k$ replaced by the unital object. This definition is entirely arrow-theoretic, and it all goes through as usual.&#xA;It is an exercise to check that the dual is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This coincides with the usual dual in the category of vector spaces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This coincides with the dual sheaf if one is working in the category of locally free sheaves on a scheme&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This corresponds to the dual (contragredient) representation in the (tensor) category of representations of any Hopf algebra (so this includes representations of finite groups and Lie algebras)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Oh, and what happens if you don't have a symmetric tensor category? Then you have to worry about &quot;left&quot; and &quot;right&quot; duals, respectively.  For more about all this, I recommend the &lt;a href=&quot;http://www-math.mit.edu/~etingof/tenscat1.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; of Pavel Etingof on tensor categories.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-24T23:06:21.473" CommentCount="4" />
  <row Id="664" PostTypeId="1" AcceptedAnswerId="665" CreationDate="2010-07-25T00:19:52.090" Score="8" ViewCount="119" Body="&lt;p&gt;Hi, I just came back from my Introduction to Rotational Kinematics class, and one of the important concepts they described was &lt;em&gt;Rotational Inertia&lt;/em&gt;, or &lt;em&gt;Moment of Inertia&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically the equivalent of mass in Netwon's $F = m a$ in linear motion.  The equivalent rotational equation is $\tau = I \alpha$, where $\tau$ is rotational force, $\alpha$ is rotational acceleration, and $I$ is rotational inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a point about an axis, $I$ is $m r^2$, where $r$ is the distance from the point to the axis of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a continuous body, this is an integral -- $I = \int r^2 \,dm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really doesn't make any sense to me...you have two independent variables?  I am only used to having one independent variable and one constant.  So I would solve this, using my experience with calculus (which encompasses a read through the Sparks Notes packet) as $ I = m r^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But obviously, this is wrong?  $r$ is not a constant!  How do I deal with it?  Do I need to replace $r$ with an expression that varies with $m$?  But how could $r$ possibly vary with $m$?  Isn't it more likely the other way around?  But how can $m$ vary with $r$?  It's all rather confusing me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me figure out what to do with all these substitutions for, example, figuring out the Moment of Inertia of a hoop with no thickness and width $w$, with the axis of rotation running through its center orthogonal to its plane?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-11-09T08:17:27.780" Title="Finding Moment of Inertia (rotional inertia?) $I$ using integration?" Tags="&lt;calculus&gt;&lt;physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="665" PostTypeId="2" ParentId="664" CreationDate="2010-07-25T01:15:03.773" Score="5" ViewCount="0" Body="&lt;p&gt;Seeing an expression like $I = \int r^2 dm$ is certainly confusing the first time you see it. When you see an expression like $I = \int x^2 dx$ we effectively iterating over a range on the x axis and adding up the area of an infinite amount of infinitesimally small strips. Note that the area of each strip is approximated as the function value (x^2) times the strip length (dx). When you see an expression like dm, we are iterating over masses instead. We still sum up the function value (r^2), but this time we multiply it by the strip mass (dm). To solve the problem, we usually put m in terms of another variable which we can iterate over more easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the moment of inertia of a rod of length L around its center with total mass of L. Each bit of length (dx) has mass (dm) and r=|x|. Solving for $I = \int r^2 dm = \int |x|^2 dx = \int x^2 dx = (x^3)/3+c$. Now, we have definite values for x to sub in (-L/2 and L/2), so we write &lt;code&gt;I=(L/8)/3-(-L/8)/3=L/12&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets calculate the moment of inertia of the hoop case you described. We break the hoop up into infinitesimally small rings the same distance from the center. Let the hoop have inner thickness r and outer thickness R. The area is &lt;code&gt;R^2-r^2&lt;/code&gt;. The area density, d, is therefore &lt;code&gt;m/(R^2-r^2)&lt;/code&gt;. A ring at radius k with thickness dk has area &lt;code&gt;2pi k dk&lt;/code&gt;, mass is &lt;code&gt;2pi k dk * m/(R^2-r^2)&lt;/code&gt; and moment of inertia around the central axis &lt;code&gt;2pi k^3 dk * m/(R^2-r^2)&lt;/code&gt;. The integral is &lt;code&gt;pi/2*k^4 *m/(R^2-r^2)+c&lt;/code&gt;. Subbing in the exact values, we get, &lt;code&gt;pi/2*(R^4-r^4) *m/(R^2-r^2)=pi/2*(R^2+r^2)*m&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-11-09T08:17:27.780" LastActivityDate="2010-11-09T08:17:27.780" CommentCount="1" />
  <row Id="667" PostTypeId="1" CreationDate="2010-07-25T02:14:47.343" Score="7" ViewCount="200" Body="&lt;p&gt;I have heard that one can think of stacks on a site as taking sheaves but instead of the restrictions being equal, we just loosen it to isomorphic, and treat the sheaf conditions with the &quot;obvious&quot; coherence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How seriously can one take this analogy? Note that my background is stacks is feeble at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question isn't too vague. One may choose to respond to this question with &quot;That analogy is stupid because of &lt;strong&gt;__&lt;/strong&gt; gotcha&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T00:57:10.353" LastActivityDate="2010-07-29T00:57:10.353" Title="Stacks are just sheaves up to Isomorphism" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="668" PostTypeId="1" AcceptedAnswerId="669" CreationDate="2010-07-25T02:25:11.593" Score="26" ViewCount="681" Body="&lt;p&gt;In my linear algebra class we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a 2 by 2 matrix by the formula. Our teacher showed us how to compute the determinant of an N by N matrix by breaking it up into the determinants of smaller matrices, and apparently there’s a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is intuitively, and how all those definitions of it are related?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-11-14T18:37:07.847" Title="What's an intuitive way to think about the determinant?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="10" />
  <row Id="669" PostTypeId="2" ParentId="668" CreationDate="2010-07-25T02:25:45.523" Score="15" ViewCount="0" Body="&lt;p&gt;Your trouble with determinants is pretty common. They’re a hard thing to teach well, too, for two main reasons that I can see: the formulas you learn for computing them are messy and complicated, and there’s no “natural” way to interpret the value of the determinant, the way it’s easy to interpret the derivatives you do in calculus at first as the slope of the tangent line. It’s hard to believe things like the invertibility condition you’ve stated when it’s not even clear what the numbers mean and where they come from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than show that the many usual definitions are all the same by comparing them to each other, I’m going to state some general properties of the determinant that I claim are enough to specify uniquely what number you should get when you put in a given matrix. Then it’s not too bad to check that all of the definitions for determinant that you’ve seen satisfy those properties I’ll state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to think about if you want an “abstract” definition of the determinant to unify all those others is that it’s not an array of numbers with bars on the side. What we’re really looking for is a function that takes N vectors (the N columns of the matrix) and returns a number. Let’s assume we’re working with real numbers for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember how those operations you mentioned change the value of the determinant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Switching two rows or columns changes the sign. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Multiplying one row by a constant multiplies the whole determinant by that constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The general fact that number two draws from: the determinant is &lt;em&gt;linear in each row&lt;/em&gt;. That is, if you think of it as a function $\det: \mathbb{R}^{n^2} \rightarrow \mathbb{R}$, then $ \det(a \vec{v_1} +b \vec{w_1}, \vec{v_2},...,\vec{v_n}) = a \det(\vec{v_1},\vec{v_2},...,\vec{v_n}) + b \det(\vec{w_1}, \vec{v_2}, ...,\vec{v_n})$, and the corresponding condition in each other slot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that these facts, together with the fact that the determinant of the identity matrix is one, is enough to define a &lt;em&gt;unique function&lt;/em&gt; that takes in N vectors (each of length N) and returns a real number, the determinant of the matrix given by those vectors. I won’t prove that, but I’ll show you how it helps with some other interpretations of the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, there’s a nice geometric way to think of a determinant. Consider the unit cube in N dimensional space: the set of vectors of length N with coordinates 0 or 1 in each spot. The determinant of the linear transformation (matrix) T is the &lt;em&gt;signed volume of the region gotten by applying T to the unit cube&lt;/em&gt;. (Don’t worry too much if you don’t know what the “signed” part means, for now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that follow from our abstract definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you apply the identity to the unit cube, you get back the unit cube. And the volume of the unit cube is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stretch the cube by a constant factor in one direction only, the new volume is that constant. And if you stack two blocks together aligned on the same direction, their combined volume is the sum of their volumes: this all shows that the signed volume we have is linear in each coordinate when considered as a function of the input vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when you switch two of the vectors that define the unit cube, you flip the orientation. (Again, this is something to come back to later if you don’t know what that means).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are ways to think about the determinant that aren’t symbol-pushing. If you’ve studied multivariable calculus, you could think about, with this geometric definition of determinant, why determinants (the Jacobian) pop up when we change coordinates doing integration. Hint: a derivative is a linear approximations of the associated function, and consider a “differential volume element” in your starting coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s not too much work to check that the area of the parallelogram formed by vectors $(a,b)$ and $(c,d)$ is $\det((a,b),(c,d))$, either: you might try that to get a sense for things.&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-14T18:37:07.847" LastActivityDate="2010-11-14T18:37:07.847" CommunityOwnedDate="2010-07-25T06:05:47.733" CommentCount="3" />
  <row Id="670" PostTypeId="2" ParentId="432" CreationDate="2010-07-25T03:22:36.777" Score="2" ViewCount="0" Body="&lt;p&gt;These implications are reached by considering the three, different cases for the roots &lt;a href=&quot;http://mathurl.com/33klydl&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/33klydl.png&quot;&gt;&lt;/a&gt; of the polynomial: repeated root, all distinct real roots, or two complex roots and one real root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one of the roots is repeated, say &lt;a href=&quot;http://mathurl.com/2emr9hn&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2emr9hn.png&quot;&gt;&lt;/a&gt; and &lt;a href=&quot;http://mathurl.com/3yf4gya&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/3yf4gya.png&quot;&gt;&lt;/a&gt;, then it is clear that the discriminant is 0 because the &lt;a href=&quot;http://mathurl.com/2uboj2q&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2uboj2q.png&quot;&gt;&lt;/a&gt; term of the product is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one root is a complex number &lt;a href=&quot;http://mathurl.com/3xbav5m&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/3xbav5m.png&quot;&gt;&lt;/a&gt;, then by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem&quot; rel=&quot;nofollow&quot;&gt;complex conjugate root theorem&lt;/a&gt;, &lt;a href=&quot;http://mathurl.com/247trlo&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/247trlo.png&quot;&gt;&lt;/a&gt; is also a root. By the same theorem, the remaining third root must be real. Evaluating the product in the discriminant for this case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathurl.com/2wcefqd&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://mathurl.com/2wcefqd.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is less than or equal to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when all roots are real, the product is clearly positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, Δ:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;less than 0 implies that one root is complex&lt;/li&gt;&#xA;&lt;li&gt;equal to 0 implies that one root is repeated&lt;/li&gt;&#xA;&lt;li&gt;greater than 0 implies that all roots are distinct and real&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-07-25T03:22:36.777" />
  <row Id="671" PostTypeId="1" AcceptedAnswerId="672" CreationDate="2010-07-25T04:45:55.390" Score="3" ViewCount="139" Body="&lt;p&gt;I just came back from a class on Probability in Game Theory, and was musing over something in my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming, for the sake of the question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Playing cards in their current state have been around for approximately eight centuries&lt;/li&gt;&#xA;&lt;li&gt;A deck of playing cards is shuffled to a random configuration one billion times per day&lt;/li&gt;&#xA;&lt;li&gt;Every shuffle ever is completely (theoretically) random and unaffected by biases caused by human shuffling and the games the cards are used for&lt;/li&gt;&#xA;&lt;li&gt;By &quot;deck of cards&quot;, I refer to a stack of unordered 52 unique cards, with a composition that is identical from deck to deck.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This would, approximately, be on the order of 3 x 10&lt;sup&gt;14&lt;/sup&gt; random shuffles in the history of playing cards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to shuffle a new deck today, completely randomly, what are the probabilistic odds (out of 1) that you create a new unique permutation of the playing cards that has never before been achieved in the history of 3 x 10&lt;sup&gt;14&lt;/sup&gt; similarly random shuffles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was to think that it was a simple matter of $\frac{1}{52!} * 3 * 10^{14}$, but then I ran into things like &lt;a href=&quot;http://en.wikipedia.org/wiki/Birthday_Paradox&quot; rel=&quot;nofollow&quot;&gt;Birthday Paradox&lt;/a&gt;.  While it is not analogous (I would have to be asking about the odds that any two shuffled decks in the history of shuffled decks ever matched), it has caused me to question my intuitive notions of Probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong in my initial approach, if it is wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the true probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if the probability is less than 0.5, if we how many more years (centuries?) must we wait, assuming the current rate of one billion shuffles per day, until we reach a state where the probability is 0.5+?   0.9+?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Out of curiosity, it would be neat to know the analogous birthday paradox answer, as well)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-25T04:58:07.650" LastActivityDate="2010-07-25T05:12:44.320" Title="When you randomly shuffle a deck of cards, what is the probability that it is a unique permutation never before configured?" Tags="&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="672" PostTypeId="2" ParentId="671" CreationDate="2010-07-25T05:04:51.993" Score="2" ViewCount="0" Body="&lt;p&gt;Your original answer of 3 * 10^14 / 52! is not far from being right. That is in fact the expected number of times any ordering of the cards has occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that any particular ordering of the cards has not occurred, given your initial assumptions, is (1-1/52!)^(3*10^14), and the probability that it has occurred is 1 minus this value.  But for small values of n&amp;epsilon;, (1+&amp;epsilon;)^n is nearly 1+n&amp;epsilon;.  In particular  1-(1-1/52!)^(3*10^14) is nearly 1/52! * 3*10^14.  &lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-25T05:04:51.993" CommentCount="3" />
  <row Id="673" PostTypeId="2" ParentId="671" CreationDate="2010-07-25T05:12:44.320" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose we shuffle a deck and get a permutation p. For each previous shuffling there is a 1-1/52! chance that p doesn't match it. Each previous shuffling is independent, in that regardless of what p and the other permutations are, the chance of p matching the shuffling is 1-1/52! When probabilities are independent we can simply multiple them to find the chance of all the events happening. In this case, the each event is actually a match &lt;em&gt;not&lt;/em&gt; happening, so the chance of no matches given n previous shuffles is (1-1/52!)^n. We can then complete the calculations as Michael did.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-25T05:12:44.320" />
  <row Id="674" PostTypeId="1" AcceptedAnswerId="2405" CreationDate="2010-07-25T05:14:24.020" Score="6" ViewCount="312" Body="&lt;p&gt;I have been using Lagrange multipliers in constrained optimization problems, but I don't see &lt;em&gt;how&lt;/em&gt; they actually work to simultaneously satisfy the constraint &lt;em&gt;and&lt;/em&gt; find the lowest possible value of an objective function.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-25T06:15:16.997" LastActivityDate="2010-08-14T14:51:28.360" CommunityOwnedDate="2010-07-25T05:14:24.020" Title="How do Lagrange multipliers work to find the lowest value of a function subject to a constraint?" Tags="&lt;calculus&gt;&lt;optimization&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="675" PostTypeId="2" ParentId="674" CreationDate="2010-07-25T05:18:20.160" Score="4" ViewCount="0" Body="&lt;p&gt;This type of problem is generally referred to as &lt;a href=&quot;http://en.wikipedia.org/wiki/Constraint_optimization&quot; rel=&quot;nofollow&quot;&gt;constrained optimization&lt;/a&gt;.  A general technique to solve many of these types of problems is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;method of Lagrange multipliers&lt;/a&gt;, here is an example of such a problem using Lagrange multipliers and a short justification as to why the technique works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the parabaloid given by $f(x,y) = x^2 + y^2$.  The global minimum of this surface lies at the origin (at $x=0$, $y=0$).  If we are given the constraint, a requirement on the relationship between $x$ and $y$, that $3x+y=6$, then the origin can no longer be our solution (since $3\cdot 0 + y \cdot 0 \neq 6$).  Yet, there is a lowest point on this function satisfying the given constraint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What we have so far:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objective function: $f(x,y) = x^2 + y^2$, &lt;br&gt;&#xA;subject to: $3x+y=6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we can derive the Lagrange formulation of our constrained minimization problem.  This will be a function $L$ of $x$, $y$, and a single Lagrange multiplier $\lambda$ (since we have only a single constraint). &lt;em&gt;It will be this new function that we minimize.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(x,y,\lambda) = x^2 + y^2 + \lambda(3x+y-6)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrange formulation incorporates our original function along with our constraint(s).  On the way toward minimizing $L$, we will have to minimize the objective function $x^2 + y^2$, as well as minimize the contribution from the constraint, which is now weighted by a factor of $\lambda$.  If the constraint is met, then the expression $3x+y-6$ will necessarily be zero, and will not contribute anything the value of $L$. This is the &lt;em&gt;trick&lt;/em&gt; of the technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Minimizing the Lagrange formulation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To minimize $L$ we simply find the $x,y$, and $\lambda$ values that make its gradient zero.  (This is exactly analogous to setting the first derivative to zero in calculus.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla L = 0:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial x} = 2x + 3 \lambda = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial y} = 2y + \lambda = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial L}{\partial \lambda} = 3x + y - 6 = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our example we have arrived at a system of simultaneous linear equations which can (and should) be solved with matrix algebra.  The solution will be a vector holding values for $x, y$ and $\lambda$.  The lowest value of the objective function, subject to the given constraint, sits at $(x,y,f(x,y))$, and the Lagrange multiplier does not have an immediate physical interpretation.  (The multipliers have meaning when appearing in certain contexts, more info on that &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers#Interpretation_of_the_Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-07-25T06:07:09.480" LastActivityDate="2010-07-25T06:07:09.480" CommunityOwnedDate="2010-07-25T05:18:20.160" />
  <row Id="676" PostTypeId="1" AcceptedAnswerId="679" CreationDate="2010-07-25T05:31:54.210" Score="4" ViewCount="188" Body="&lt;p&gt;Randomly break a stick (or a piece of dry spaghetti, etc.) in two places, forming three pieces.  The probability that these three pieces can form a triangle is 1/4 (coordinatize the stick form 0 to 1, call the breaking points x and y, consider the unit square of the coordinate plane, shade the areas that satisfy the triangle inequality &lt;em&gt;edit&lt;/em&gt;: see comments on the question, below, for a better explanation of this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other day in class&lt;sup&gt;*&lt;/sup&gt;, my professor was demonstrating how to do a Monte Carlo simulation of this problem on a calculator and wrote a program that, for each trial did the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick a random number x between 0 and 1.  This is the first side length.&lt;/li&gt;&#xA;&lt;li&gt;Pick a random number y between 0 and 1 - x (the remaning part of the stick).  This is the second side length.&lt;/li&gt;&#xA;&lt;li&gt;The third side length is 1 - x - y.&lt;/li&gt;&#xA;&lt;li&gt;Test if the three side lengths satisfy the triangle inequality (in all three permutations).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;He ran around 1000 trials and was getting 0.19, which he said was probably just random-chance error off 0.25, but every time the program was run, no matter who's calculator we used, the result was around 0.19.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with the simulation method?  What is the theoretical answer to the problem actually being simulated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;sup&gt;*&lt;/sup&gt; the other day was more than 10 years ago)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-25T07:03:56.587" LastActivityDate="2010-07-25T07:03:56.587" Title="Probability that a stick randomly broken in two places can form a triangle" Tags="&lt;probability&gt;&lt;monte-carlo&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="677" PostTypeId="1" AcceptedAnswerId="1105" CreationDate="2010-07-25T06:00:35.030" Score="10" ViewCount="155" Body="&lt;p&gt;About two years ago, I started thinking about the following problem: You're given an N and an S, positive integers. You start with an N-gon that has positive integer labels at each vertex, such that the labels sum to S. A &quot;move&quot; is replacing the label at some vertex with the positive difference of the labels at its two adjacent vertices. You win if you can exhibit a sequence of moves such that all of the vertex labels are zero at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization of 2003 USAMO problem B3, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A positive integer is written at each vertex of a hexagon. A move is to replace a number by the (non-negative) difference between the two numbers at the adjacent vertices. If the starting numbers sum to 2003^2003, show that it is always possible to make a sequence of moves ending with zeros at every vertex.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The USAMO problem itself can be answered in the affirmative by doing a modified greedy algorithm for most of the moves, showing there's always a move where you can reduce the problem to a smaller sum while changing the parities systematically, then doing casework for the &quot;end game&quot; to show that each configuration of 1s and 0s (or k's and zeros, the cases are of course equivalent) can be gotten down to all zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to check that for n=4 and S odd, there's a winning strategy. I have lots of simulation/numerical evidence and some half baked ideas to suggest that, in general, there's a winning strategy iff N and S have opposite parities. But in two years of coming back to it on and off, I haven't really made progress. It's open, as far as I know, though I knew that more surely two years ago than I do now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-29T03:30:58.303" Title="Proving that this Game on Polygons Ends" Tags="&lt;open-problem&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="679" PostTypeId="2" ParentId="676" CreationDate="2010-07-25T06:29:45.050" Score="10" ViewCount="0" Body="&lt;p&gt;The three triangle inequalities are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x + y &amp;gt; 1-x-y&#xA;x + (1-x-y) &amp;gt; y&#xA;y + (1-x-y) &amp;gt; x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your problem is that in picking the smaller number first from a uniform distribution, it's going to end up being bigger than it would if you had just picked two random numbers and taken the smaller one. (You'll end up with an average value of $1/2$ for the smaller instead of $1/3$ like you actually want.) Now when you pick $y$ on $[0, 1-x]$, you're making it smaller than it should be (ending up with average value of $1/4$). To understand this unequal distribution, we can substitute $y (1-x)$ for $y$ in the original inequalities and we'll see the proper distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CARdj.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the $y$-axis of the graph doesn't really go from $0$ to $1$; instead the top represents the line $y=1-x$. I'm showing it as a square because that's how the probabilities you were calculating were being generated.) Now the probability you're measuring is the area of the strangely-shaped region on the left, which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{1/2} \frac{1}{2-2x} - \frac{2x-1}{2x-2} \,dx = \ln(2) - 1/2 \approx 0.19314. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that's the answer you calculated.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-25T06:58:50.747" LastActivityDate="2010-07-25T06:58:50.747" CommentCount="4" />
  <row Id="680" PostTypeId="1" AcceptedAnswerId="681" CreationDate="2010-07-25T06:57:11.093" Score="6" ViewCount="178" Body="&lt;p&gt;Two mathematicians each come into a coffee shop at a random time between 8:00 a.m. and 9:00 a.m. each day. Each orders a cup of coffee then sits at a table, reading a newspaper for 20 minutes before leaving to go to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On any day, what is the probability that both mathematicians are at the coffee shop at the same time (that is, their arrival times are within 20 minutes of each other)?&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:29:27.107" LastActivityDate="2010-09-29T04:19:45.697" Title="Probability that two people see each other at the coffee shop" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="681" PostTypeId="2" ParentId="680" CreationDate="2010-07-25T07:13:22.623" Score="13" ViewCount="0" Body="&lt;p&gt;Working in hours and letting 8:00 a.m. be t=0, each mathematician's arrival time is a number between 0 and 1.  The sample space can be represented by the unit square in the coordinate plane with one professor's arrival time as x and the other's as y, where regions with equal areas are equally likely.  We want x - 1/3 &amp;lt; y &amp;lt; x + 1/3 -- that is, the second professor arrives earlier than the first by no more than 1/3 of an hour or later than the first by no more than 1/3 of an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Z2M3h.png&quot; alt=&quot;plot of the inequalities&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the desired region is 5/9.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:19:45.697" LastActivityDate="2010-09-29T04:19:45.697" CommentCount="7" />
  <row Id="682" PostTypeId="2" ParentId="593" CreationDate="2010-07-25T07:48:10.590" Score="4" ViewCount="0" Body="&lt;p&gt;The answer is, it's just a fact &amp;ldquo;cone over a simplex is a simplex&amp;rdquo; rewritten in terms of the generating function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;observe that because n-simplex is a cone over (n-1)-simplex $\frac{\partial}{\partial x}vol(\text{n-simplex w. edge x}) = vol(\text{(n-1)-simplex w. edge x})$; in other words $e(x):=\sum_n vol\text{(n-simplex w. edge x)}$ satisfies an equvation $e'(x)=e(x)$. So $e(x)=Ce^x$ -- and C=1 because e(0)=1.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-30T07:28:06.127" LastActivityDate="2010-07-30T07:28:06.127" CommentCount="2" />
  <row Id="683" PostTypeId="1" CreationDate="2010-07-25T07:56:11.163" Score="8" ViewCount="84" Body="&lt;p&gt;I want to randomly generate trees, i.e. undirected acyclic graphs with a single root, making sure that all possible trees with a fixed number of nodes &lt;em&gt;n&lt;/em&gt; are equally likely.&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2010-07-25T19:40:11.843" Title="How can I randomly generate trees?" Tags="&lt;probability-theory&gt;&lt;random&gt;" AnswerCount="1" />
  <row Id="684" PostTypeId="2" ParentId="683" CreationDate="2010-07-25T08:24:10.440" Score="4" ViewCount="0" Body="&lt;p&gt;Knuth says to look at it as generating all nested parentheses in lexicographic order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look here for the details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-cs-faculty.stanford.edu/~uno/fasc4a.ps&quot; rel=&quot;nofollow&quot;&gt;http://www-cs-faculty.stanford.edu/~uno/fasc4a.ps&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-25T19:40:11.843" LastActivityDate="2010-07-25T19:40:11.843" CommentCount="6" />
  <row Id="685" PostTypeId="2" ParentId="618" CreationDate="2010-07-25T09:58:14.807" Score="2" ViewCount="0" Body="&lt;p&gt;Okay, I think that I've given other people a fair chance to answer this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to solve this problem is to calculate how much each payment reduces your debt after you have been repaying your loan for n years. Let r=1+R/100, ie. this converts the interest rate from a percentage to a value you can multiply your debt by to calculate how much you owe after adding one time period's interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I make a payment of $P at the end of the kth year, then we avoid paying interest on this money n-k times and so we reduce our debt by P*r^(n-k). We sum up the future values of all our payments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sum over k from 1 to n of: P*r^(n-k)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reverse this, it is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sum over k from 0 to n-1 of: P*r^k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series&quot; rel=&quot;nofollow&quot;&gt;geometric series&lt;/a&gt;, which can be solved using the formula a(r^n-1)/(r-1), where a is the first term, r is the factor and n is the number of terms being summed. P(r^n-1)/(r-1). We then attempt to equate this will the debt owed after n years, which is Mr^n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know equate the two locations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(r^n-1)/(r-1)=Mr^n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating n&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We group the r^n terms:&#xA;P/(r-1)=r^n(M-P/(r-1))&#xA;r^n=p/(M(r-1)-P)&#xA;So we just take the nth log of the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating repayments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P=Mr^n*(r-1)/(r^n-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Payments made at the start of the year&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the future values of our interest payment simply become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sum over k from 1 to n of: P*r^k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We proceed as we did before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also solve this problem using present value instead of future value.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-25T09:58:14.807" />
  <row Id="686" PostTypeId="1" AcceptedAnswerId="691" CreationDate="2010-07-25T10:54:29.373" Score="4" ViewCount="97" Body="&lt;p&gt;Suppose that I am buying cakes for a party. There are k different types and I intend to buy a total of n cakes. How many different combinations of cakes could I possibly bring to the party?&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-25T15:52:28.867" Title="Combinations of selecting n objects with k different types" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="687" PostTypeId="1" AcceptedAnswerId="1050" CreationDate="2010-07-25T11:08:30.817" Score="5" ViewCount="90" Body="&lt;p&gt;Suppose we model traffic flow between two points with a directed graph. Each route has either a constant travel time or one that linearly increases with traffic. We assume that each driver wishes to minimise their own travel time and we assume that the drivers form a Nash equilibria. Can removing a route ever decrease the average travelling time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the existence of multiple Nash equilibria makes this question a bit complicated. To clarify, I am looking for a route removal that will guarantee a decrease in the average traveling time regardless of the Nash equilibria that are chosen before and after.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T21:02:19.803" LastActivityDate="2010-07-28T21:02:19.803" Title="Increasing network throughput by cutting routes" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="690" PostTypeId="2" ParentId="686" CreationDate="2010-07-25T14:37:35.910" Score="1" ViewCount="0" Body="&lt;p&gt;Let g(n,k) = # combinations of cakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;g(n,1) = 1. (all the cakes are the same)&lt;/li&gt;&#xA;&lt;li&gt;g(n,2) = n+1. (e.g. for 5 cakes, the # of cakes of type 1 can be 0, 1, 2, 3, 4, 5)&lt;/li&gt;&#xA;&lt;li&gt;g(1,k) = k.&lt;/li&gt;&#xA;&lt;li&gt;g(2,k) = k*(k-1)/2 + k (the first term is two different cakes; the second term is when both cakes are the same), as long as k &gt; 1. (otherwise g(2,1) = 1)&lt;/li&gt;&#xA;&lt;li&gt;g(3,k) = k * (k-1) * (k-2)/6 + k*(k-1)/2 * 2 + k (the first term is 3 different cakes; the second term is 2 different cakes, with a *2 since there are two choices for which one to duplicate, the third term is when all 3 cakes are the same), as long as k &gt; 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we think of k as a radix rather than the # of cakes, then this problem is equivalent to expressing the # of distinct n-digit numbers in base k whose digits are in sorted order. (e.g. 1122399 is equivalent to 9921231)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; I can express it as a nonrecursive sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g(n,k) = sum from j=1 to max(n,k) of { (k choose j) * h(n,j) } &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where h(n,j) is the # of ways to partition N cakes using j different types. (the term in the sum is when there are j distinct cakes actually chosen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's about as far as I can get... :/&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;edit: looks like it's &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;combinations with repetitions&lt;/a&gt; = ((k+n-1) choose n). (same as the wikipedia article with n and k swapped)&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-25T14:37:35.910" />
  <row Id="691" PostTypeId="2" ParentId="686" CreationDate="2010-07-25T15:52:28.867" Score="8" ViewCount="0" Body="&lt;p&gt;Using a method that's often called &quot;stars and bars&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We draw $n$ stars in a row to represent the cakes, and $k-1$ bars to divide them up. All of the stars to the left of the first bar are cakes of the first type; stars between the first two bars are of the second type; . . . .&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;**|***||*|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's an example with $n=6$ and $k=5$. We're getting 2 of the first type, 3 of the second type, 0 of the third type, 1 of the fourth type, and 0 of the fifth type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to solve the problem, we just need to reorder the stars and bars by choosing the $k-1$ spots for the bars out of the $n+k-1$ total spots, so our answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{n+k-1}{k-1}. $$&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-25T15:52:28.867" CommentCount="1" />
  <row Id="692" PostTypeId="2" ParentId="154" CreationDate="2010-07-25T15:57:22.297" Score="2" ViewCount="0" Body="&lt;p&gt;Complex numbers are the final step in a sequence of increasingly &quot;unreal&quot; extensions to the number system that humans have found it necessary to add over the centuries in order to express significant numerical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first such &quot;unreal number&quot; was zero, back in the mists of time. It seems obvious to us now, but it must have seemed strange at first. How can the number of sheep I have be zero, when I don't actually have any sheep?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negative numbers are the next most obvious addition to the family of numbers. But what does it mean to have -2 apples? If I have 3 apples and you have 5, it's convenient to be able to say that I have -2 more apples than you. Even so, during the middle ages many mathematicians were very uncomfortable with the idea of negative numbers and tried to arrange their equations so that they didn't occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rationals (fractions) seem real enough, since I'm happy to have 2/5 of a pizza. However, this is not the number of pizzas that I have, just a ratio between 0 and 1 pizzas, and so is further removed from the concept of counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More significant philosophical problems problems arose when irrational numbers were first discovered by the classical Greeks. They were astonished when Euclid (or one of his predecessors) proved that the diagonal of a unit square could not be represented by ratio of two integers. This was such an outrageous idea to them that they called these numbers &quot;irrational&quot;. However, they couldn't easily deny their existence since they have a direct geometric representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irrational numbers were first understood as the solutions to algebraic equations such as x^2 = 2 but this still doesn't cover all the possible numbers that we need. For example, the ratio of a circle's diameter to its circumference is π and so has a direct geometric representation. However, in 1882 π was proved to be transcendental, meaning that it can't be defined as the solution to a specific algebraic equation. This begins to seem a lot less &quot;real&quot;, especially when you consider that there are many important transcendental numbers that, unlike π, don't have a geometric interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course many algebraic equations that don't have a solution even among the irrational numbers, and in some ways there's no reason why they should. However, when 16th century mathematicians like Gerolamo Cardano began working on solutions to cubic equations they found that the square roots of negative numbers began cropping up very naturally in their procedures, even though the solutions themselves were purely real. This eventually led people to explore the arithmetic of complex numbers and they were surprised to find that it produced a consistent and beautiful theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, complex numbers don't have such an intuitively obvious geometrical meaning as the numbers that came before. They are typically represented graphically as points in the 2D plane, and the rules of addition and multiplication are equivalent to certain operations on lengths and angles, but those operations aren't driven by geometrical necessity in quite the way that squares and circles are. Even so, complex numbers are a perfect representation for various physical phenomena such as the state of particles in quantum mechanics and the behaviour of varying currents in electrical circuits. They are also very useful for reducing the cost of computation in 3D computer graphics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really special thing about complex numbers, though, is that they are the end of this journey that has been going on for millenia. There is no need to invent further number systems to provide solutions to problems expressed in terms of complex numbers because now every non-contradictory equation, algebraic or transcendental, has a solution within the complex numbers. They are a self-sufficient, consistent system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physicists have been breaking matter down into smaller and smaller particles over the years. Molecules, atoms, nuclei, protons, quarks. When will the process stop? For several decades it was felt that quarks could be the final, indivisible particle. However, new theoretical frameworks such as string theory are suggesting that there may be more fundamental entities than quarks. Because physics ultimately relies on experimental verification, we can never be sure that there aren't going to be more steps in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, however, we can prove things for all time. Complex numbers are the final step in the sequence, the numbers that we have been reaching for since before the beginning of recorded history. Every other number system is just a subset of the complex numbers, just a part of the true picture. Complex numbers are the real thing.&lt;/p&gt;&#xA;" OwnerUserId="113" LastActivityDate="2010-07-25T15:57:22.297" CommentCount="3" />
  <row Id="693" PostTypeId="1" AcceptedAnswerId="694" CreationDate="2010-07-25T16:09:21.277" Score="2" ViewCount="65" Body="&lt;p&gt;I need n cakes for a party. I go to the cake shop and there are k different kinds of cake. For variety, I'd like to get at least one of every cake. How many ways can I do this?&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-08-05T19:13:10.093" Title="Picking cakes if we need at least one of each type" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="694" PostTypeId="2" ParentId="693" CreationDate="2010-07-25T16:18:22.500" Score="6" ViewCount="0" Body="&lt;p&gt;Similar to &lt;a href=&quot;http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types/691#691&quot;&gt;the stars and bars technique&lt;/a&gt;, consider the n cakes as a row of n stars *.  Instead of permuting them with k-1 bars | (which allows two bars next to each other, giving 0 of a type, place the k-1 bars (needed to split the n stars into k types) into the n-1 spaces between the stars, allowing at most one bar per space.  The number of ways to do this is ${n-1 \choose k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, since you need one of each type, there are only n-k cakes for which you are choosing types.  Using &lt;a href=&quot;http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types/691#691&quot;&gt;the stars and bars technique&lt;/a&gt;, there are ${(n-k)+k-1 \choose k-1} = {n-1 \choose k-1}$ ways to do it.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:13:10.093" LastActivityDate="2010-08-05T19:13:10.093" CommentCount="3" />
  <row Id="696" PostTypeId="2" ParentId="656" CreationDate="2010-07-25T20:41:20.947" Score="2" ViewCount="0" Body="&lt;p&gt;Of course there are infinite equation (even if you require them to be infinitely differentiable...) that satisfy the given constraints. &#xA;As Isaac and Justin already wrote, you may always find a polynomial of degree at most n-1 (where n is the number of points given) which satisfies the given data; but you cannot be sure that this is the right answer. &#xA;Moreover, if data is not exact but approximate the resulting polynomial may be quite different from the correct function, since it would likely hade huge peaks and falls. In such cases, an approximate method like &lt;a href=&quot;http://en.wikipedia.org/wiki/Least_squares&quot; rel=&quot;nofollow&quot;&gt;least squares&lt;/a&gt; could be more useful.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-25T20:41:20.947" CommentCount="1" />
  <row Id="697" PostTypeId="1" CreationDate="2010-07-25T22:19:40.330" Score="5" ViewCount="170" Body="&lt;p&gt;In least-squares approximations the normal equations act to project a vector existing in N-dimensional space onto a lower dimensional space, where our problem actually lies, thus providing the &quot;best&quot; solution we can hope for (the orthogonal projection of the N-vector onto our solution space).  The &quot;best&quot; solution is the one that minimizes the &lt;strong&gt;Euclidean distance (two-norm)&lt;/strong&gt; between the N-dimensional vector and our lower dimensional space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist other norms and other spaces besides $\mathbb{R}^d$, what are the analogues of least-squares under a different norm, or in a different space?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T00:56:38.330" LastActivityDate="2010-11-19T00:56:38.330" CommunityOwnedDate="2010-08-10T18:09:45.803" Title="Are there variations on least-squares approximations?" Tags="&lt;linear-algebra&gt;&lt;big-list&gt;&lt;approximation&gt;&lt;regression&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="698" PostTypeId="1" AcceptedAnswerId="1472" CreationDate="2010-07-25T22:55:49.980" Score="0" ViewCount="169" Body="&lt;p&gt;Recently I discussed an experiment with a friend. Assume we start a random experiment. At first there is an array with size 100 000, all set to 0. We calculate at each round a random number modulo 2 and select one random position in that array. If the number in the array is 1, nothing is changed and otherwise the pre-computed value is set. The question is: how many distinct hash values would we have added in 1%, 5%, 50%, 95%, 99% of all cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: 4 rounds with array of size 10:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Array                     Position   random number&#xA;[0,...,0]                    5              0&#xA;[0,...,0]                    7              1&#xA;[0,...0,1,0,0,0]             6              1&#xA;[0,..0,.1,1,0,0,0]           6              0&#xA;[0,..0,.1,1,0,0,0]           2              0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;First we considered this a somehow simple problem. But after thinking about for some hours, searching the web and asking some math students we couldn't find a solution.So do you know a probability distribution for this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Was also posted on &lt;a href=&quot;http://mathoverflow.net/questions/33335/looking-for-a-probability-distribution&quot; rel=&quot;nofollow&quot;&gt;Math Overflow&lt;/a&gt; and got its answer there.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditorDisplayName="" LastEditDate="2010-08-02T11:56:18.977" LastActivityDate="2010-08-03T20:32:23.647" Title="Looking for a probability distribution" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="699" PostTypeId="2" ParentId="697" CreationDate="2010-07-26T00:43:42.073" Score="0" ViewCount="0" Body="&lt;p&gt;Sure, there are variations on least-squares approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an engineering answer (not really a pure math answer): On one project I did at my company, we had a thermal model that was approximating the thermal transfer function from measured power to device temperature. The cost of error in the positive direction (underestimating temperature) was worse than the cost of the same error in the negative direction (overestimating temperature) -- so we used a weighting function that was (K * error ^ 2) where K was 1 for negative temperature error and greater than 1 (e.g. 1.5 or 2) for positive temperature error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thought about using more complicated mappings (10 degrees underestimate much worse than 1 degree underestimate) but didn't want to go there... I assume this has some analog to a utility function (e.g. expected monetary gain of a system with random outcome has nonlinear mapping to &quot;happiness&quot; or &quot;utility&quot;) where nonlinearity is intentionally introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could do something similar for approximating functions with polynomials: a least-squares fit treats error in a linear way, but there may be places in the function (e.g. at the ends or at the center) where minimizing error is more or less important.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-26T00:43:42.073" />
  <row Id="700" PostTypeId="2" ParentId="57" CreationDate="2010-07-26T01:05:00.097" Score="1" ViewCount="0" Body="&lt;p&gt;The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) &amp;isin; R.  This allows all of the properties of relations to be expressed in equational form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eq = {(x,x) | for each individual x}.  This is the diagonal or equality relation that holds only between something and itself.  It is reflexive, symmetric and transitive, which is to say it is an equivalence relation.&lt;/li&gt;&#xA;&lt;li&gt;All = {(x,y) | for all individuals x,y}.  The whenever relation, it is also an equivalence relation.&lt;/li&gt;&#xA;&lt;li&gt;Neq = {(x,y) | for all individuals x,y for which x &amp;ne; y}: the inequality relation that holds between anything different.  It is symmetric and irreflexive, but not transitive.&lt;/li&gt;&#xA;&lt;li&gt;Never is the empty set.  It is symmetric and transitive, and irreflexive.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Three basic binary operations on relations, and one unary relation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R&amp;cap;S = {(x,y) | (x,y) &amp;isin; R and (x,y) &amp;isin; S};&lt;/li&gt;&#xA;&lt;li&gt;R&amp;ndash;S = {(x,y) | (x,y) &amp;isin; R and (x,y) &amp;notin; S};&lt;/li&gt;&#xA;&lt;li&gt;R&amp;sdot;S = {(x,z) | there is some y such that (x,y) &amp;isin; R and (y,z) &amp;isin; S} (composition);&lt;/li&gt;&#xA;&lt;li&gt;tr(R) = {(y,x) | (x,y) &amp;isin; R} (transpose).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Observe that Never = All&amp;ndash;All, and Neq = All&amp;ndash;Eq.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say R -&gt; S when S holds whenever R holds.  This is the same as saying either that  R is a subset of S, or that R&amp;cap;S=R, or that R-S=Never.  So Never -&gt; Eq, and Eq -&gt; All.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can express your five relations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;R is reflexive when Eq -&gt; R; dually we have an additional property of a relation, where R is anti-reflexive when R -&gt; Neq;&lt;/li&gt;&#xA;&lt;li&gt;R is symmetric when R -&gt; tr(R);&lt;/li&gt;&#xA;&lt;li&gt;R is anti-symmetric when R &amp;cap; tr(R) -&gt; Eq, or equivalently when R &amp;cap; tr(R) &amp;cap; Neq = 0.&lt;/li&gt;&#xA;&lt;li&gt;R is asymmetric when R &amp;cap; tr(R)=Never: a relation is asymmetric when it is anti-symmetric and anti-reflexive;&lt;/li&gt;&#xA;&lt;li&gt;R is transitive when R&amp;sdot;R -&gt; R.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out?&lt;/em&gt;  &amp;mdash; Look at Eq: it is both symmetric and anti-symmetric, although it is not asymmetric.  In fact, all asymmetric relations are anti-symmetric, but not vice versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: anti-symmetric doesn't care about what pairs there might be along the diagonal, whilst asymmetric insists that there are no pairs along the diagonal, which is irreflexivity.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-28T21:12:19.983" LastActivityDate="2010-07-28T21:12:19.983" CommentCount="2" />
  <row Id="701" PostTypeId="1" CreationDate="2010-07-26T04:55:18.153" Score="5" ViewCount="120" Body="&lt;p&gt;Hi, I've just came back from my Mathematics of Packing and Shipping lecture, and I've run into a problem I've been trying to figure out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a rectangle of length $l$ and width $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple equation that can be used to show me how many circles of radius $r$ can be packed into the rectangle, in &lt;strong&gt;the optimal way&lt;/strong&gt;?  So that no circles overlap. ($r$ is less than both $l$ and $w$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm rather in the dark as to what the optimum method of packing circles together in the least amount of space is, for a given shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equation with a non-integer output is useful to me as long as the truncated (rounded down) value is the true answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not that interested in &lt;strong&gt;how&lt;/strong&gt; the circles would be packed, as I am going to go into business and only want to know how much I can demand from the packers I hire to pack my product)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-26T06:22:12.773" LastActivityDate="2010-07-26T12:42:57.450" Title="How many circles of a given radius can be packed into a given rectangular box?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="702" PostTypeId="2" ParentId="603" CreationDate="2010-07-26T05:26:53.267" Score="0" ViewCount="0" Body="&lt;p&gt;Couldn't all transformation which send each point (x,y) to another point (x',y') which can be computed from the first one by performing only the four operations and extraction of square root?&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-26T05:26:53.267" CommentCount="1" />
  <row Id="704" PostTypeId="1" CreationDate="2010-07-26T07:06:24.977" Score="3" ViewCount="127" Body="&lt;p&gt;Is there symbolism (or even a name) for groups of statements in which any fixed-number of them imply all the rest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in linear algebra, a basis is sometimes defined as a set of n-dimensional vectors which:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is linearly independent&lt;/li&gt;&#xA;&lt;li&gt;Spans the vector space&lt;/li&gt;&#xA;&lt;li&gt;Contains exactly n elements&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, it is then shown that any TWO of these statements being true implies the third is true.  How would you write this symbolically? (other than $A \wedge B \Rightarrow C, A \wedge C \Rightarrow B, B \wedge C \Rightarrow A$, which does not scale well...)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-11-07T03:13:54.083" Title="n out of m theorems (some imply the rest)" Tags="&lt;terminology&gt;&lt;logic&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="705" PostTypeId="1" CreationDate="2010-07-26T07:13:48.847" Score="2" ViewCount="75" Body="&lt;p&gt;Given&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A straight line of arbitrary length&lt;/li&gt;&#xA;&lt;li&gt;The ability to construct a straight line in any direction from any starting point with the &quot;unit length&quot;, or the length whose square root of its magnitude yields its own magnitude.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there a way to geometrically construct (using only a compass and straightedge) the a line with the length of the square root of the arbitrary-lengthed line?  What is the mathematical basis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why can't this be done without the unit line length?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T08:25:37.177" LastActivityDate="2010-07-26T08:52:06.800" Title="Compass-and-straightedge construction of the square root of a given line?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;geometric-construction&gt;&lt;compass-and-straightedge&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="706" PostTypeId="2" ParentId="705" CreationDate="2010-07-26T07:29:47.623" Score="4" ViewCount="0" Body="&lt;p&gt;Without the unit-length segment--that is, without something to compare the first segment to--its length is entirely arbitrary, so can't be valued, so there's no value of which to take the square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the given segment (with length x) be AB and let point C be on ray AB such that BC = 1.  Construct the midpoint M of segment AC, construct the circle with center M passing through A, construct the line perpendicular to AB through B, and let D be one of the intersections of that line with the circle centered at M (call the other intersection E).  BD = sqrt(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AC and DE are chords of the circle intersecting at B, so by the power of a point theorem, AB * BC = DB * BE, so x * 1 = x = DB * BE.  Since DE is perpendicular to AC and AC is a diameter of the circle, AC bisects DE and DB = BE, so x = DB^2 or DB = sqrt(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: this is a special case of the more general geometric-mean construction.  Given two lengths AB and BC (arranged as above), the above construction produces the length BD = sqrt(AB * BC), which is the geometric mean of AB and BC.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T07:41:47.890" LastActivityDate="2010-07-26T07:41:47.890" CommentCount="2" />
  <row Id="707" PostTypeId="1" AcceptedAnswerId="789" CreationDate="2010-07-26T07:34:21.030" Score="1" ViewCount="147" Body="&lt;p&gt;Say there are three jars, $j_1, j_2, j_3$ filled with different binary sequences of length two.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution of the binary sequences in each of the jars is given by the $p_i^k(1-p_i)^{n-k}$, where &#xA;$p_i = \frac{i}{m + 1}$ where m is the number of jars, i is the jar index, k is number of 1's and n is the length of the string.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for three jars we have $p_1 = 0.25, p_2 = 0.5$, and $p_3 = 0.75$ for $j_1, j_2, j_3$ respectively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the sequences and their probabilities for $j_1$ with $p_1 = 0.25$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(00) = 9 / 16&lt;br&gt;&#xA;P(10) = 3 / 16&lt;br&gt;&#xA;P(01) = 3 / 16&lt;br&gt;&#xA;P(11) = 1 / 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I tell you that I have selected a binary sequence and the first element is 1 what is the E($p_i$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this can be calculated by looking at each of the jars and adding up the probability of candidate sequences times the value of $p_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I wasn't normalizing this conditionally space properly. I'm skipping a step which I'll explain, someone wants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E($p_i$) = (4/24 * 1/4) + (8/24 * 1/2) + (12/24 * 3/4) = 14 / 24 = 0.58&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is ... what is E($p_i$) when the numbers of jars goes to infinity (or alternatively, when p can take on values between 0 and 1)? Also what happens when the size of the binary strings goes to infinity? Does it have an effect on the outcome? If it does, does the order we take the limits change the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And most importantly what is the general case for when I have $s$ 1's and $r$ 0's?, with a continuous p from 0 to 1 and infinite sequences?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-27T19:38:43.117" LastActivityDate="2010-08-03T02:22:09.930" Title="The limiting case of a discrete probability problem" Tags="&lt;probability-theory&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="708" PostTypeId="2" ParentId="705" CreationDate="2010-07-26T07:40:53.620" Score="3" ViewCount="0" Body="&lt;p&gt;If you have a segment AB, place the unit lenght segment on the line where AB lies, starting with A and in the direction opposite to B; let C be the other point of the segment. Now draw a semicircle with diameter AC and the perpendicular to B; this line crosses the semicircle in a point D. BD is then the square root of AB. &#xA;Triangle BCD is a right triangle, like ACD and ABD; all of these are similar, so you find out that AC/AD = AD/AB. But AC=1, so AD = sqrt(AB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the drawing below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i26.tinypic.com/2ivgp03.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-26T08:52:06.800" LastActivityDate="2010-07-26T08:52:06.800" CommentCount="1" />
  <row Id="709" PostTypeId="1" AcceptedAnswerId="6842" CreationDate="2010-07-26T07:49:39.480" Score="4" ViewCount="550" Body="&lt;p&gt;Hi, I'm taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = a x&lt;sup&gt;2&lt;/sup&gt; + b x + c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher gave me the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -\frac{b}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the x coordinate of the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked her why, and she told me not to tell her how to do her job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My smart friend mumbled something about it involving calculus, but I've always found him a rather odd fellow and I doubt I'd be able to understand a solution involving calculus, because I have no background in it.  If you use something you know from calculus, explain it to someone who has no background in it.  Because I sure don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a purely algebraic or geometrical yet &lt;strong&gt;elegant&lt;/strong&gt; derivation for the x coordinate of a parabola of the above form?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-16T04:36:44.793" LastActivityDate="2010-10-16T04:36:44.793" Title="Derivation of the formula for the vertex of a Parabola" Tags="&lt;algebra-precalculus&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;&lt;parabola&gt;" AnswerCount="10" CommentCount="3" FavoriteCount="1" />
  <row Id="710" PostTypeId="2" ParentId="709" CreationDate="2010-07-26T08:02:56.187" Score="12" ViewCount="0" Body="&lt;p&gt;By the vertex I assume you mean the minimum/maximum point of the parabola. Indeed, this result can be discovered easily through a bit of calculus, but there is also a simple purely algebraic way, which I will present here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a generic quadratic expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = ax^2 + bx + c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now &lt;a href=&quot;http://en.wikipedia.org/wiki/Completing_the_square&quot; rel=&quot;nofollow&quot;&gt;complete the square&lt;/a&gt; on this formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a[x^2 + b/a + c/a]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a[(x + b/2a)^2 - (b/2a)^2 + c/a]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression $- (b/2a)^2 + c/a$ is a constant (it does not depend on x), so we can replace it with &lt;code&gt;k&lt;/code&gt; for the matter of discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a[(x + b/2a)^2 + k]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, depending on whether a is positive or negative, the parabola given by y will either have a maximum or minimum. Since a and k are fixed, this must occur when $(x + b/2a)^2$ is zero (we know it cannot be less than zero, and it can extend to infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we know that for $(x + b/2a)^2$ to be zero, $x = -b/2a$. This in turn implies that the function y is at a minimum or a maximum when this is true. Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-07-26T08:02:56.187" />
  <row Id="711" PostTypeId="2" ParentId="709" CreationDate="2010-07-26T08:05:03.357" Score="5" ViewCount="0" Body="&lt;p&gt;Parabolas of the form you described (y = ...) are symmetric over a vertical line through their vertex.  Let's call that line x = k.  This means that if the graph crosses the x-axis (meaning that $ax^2+bx+c=0$ has real solution(s)), they must be equidistant from x = k, so (k,0) must be the midpoint of the segment with endpoints at the zeros of the quadratic or k is the average of the zeros.  From the quadratic formula, the two zeros of the quadratic are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, so their sum is $-\frac{b}{a}$ and their average is $k=-\frac{b}{2a}$.  So, the x-coordinate of the vertex must be $-\frac{b}{2a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the parabola does not cross the x-axis (no real solutions), there is another parabola with equation $y=ax^2+bx+c'$ for some $c'$ for which the graph is a vertical translation of the graph of the original quadratic, but crosses the x-axis.  Its axis of symmetry is $x=-\frac{b}{2a}$ and because it is a vertical translation of the original, the axis of symmetry of the original is also $x=-\frac{b}{2a}$, so the vertices of both have x-coordinate $-\frac{b}{2a}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T20:11:56.420" LastActivityDate="2010-08-05T20:11:56.420" />
  <row Id="712" PostTypeId="1" AcceptedAnswerId="715" CreationDate="2010-07-26T08:11:56.960" Score="10" ViewCount="211" Body="&lt;p&gt;I read &lt;a href=&quot;http://math.stackexchange.com/questions/625/why-is-the-derivative-of-a-circles-area-its-perimeter-and-similarly-for-spheres&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt; the other day and it got me thinking: the area of a circle is $\pi r^2$, which differentiates to $2 \pi r$, which is just the perimeter of the circle. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why doesn't the same thing happen for squares? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we start with the area formula for squares, $l^2$, this differentiates to $2l$ which is sort of right but only &lt;em&gt;half&lt;/em&gt; the perimeter. I asked my calculus teacher and he couldn't tell me why. Can anyone explain???&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-07-26T09:41:01.373" Title="Why does area differentiate to perimeter for circles and not for squares?" Tags="&lt;geometry&gt;&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="713" PostTypeId="1" CreationDate="2010-07-26T08:12:43.027" Score="8" ViewCount="304" Body="&lt;p&gt;The notion (rank-2) &quot;tensor&quot; appears in many different parts of physics, e.g. stress tensor, moment of inertia tensor, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know mathematically a tensor can be represented by a 3x3 matrix. But I can't grasp its geometrical picture — unlike scalar (a number) and vector (an arrow with direction and magnitude) which I can easily see what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to visualize a tensor?&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T23:35:58.330" LastActivityDate="2010-08-22T01:44:03.140" Title="How to visualize a rank-2 tensor?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;intuition&gt;&lt;tensors&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="714" PostTypeId="2" ParentId="712" CreationDate="2010-07-26T08:17:48.273" Score="6" ViewCount="0" Body="&lt;p&gt;If you use the formula you describe for squares, your measurements are coming from one corner of the square.  Imagine those measurements growing slowly.  The square will grow, but only along the two sides opposite the corner from which you measured, so the derivative of the area formula is only the perimeter on those two sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, consider measuring the size of a square by the distance d from its center to the midpoint of a side.  This would make the side length 2d, the perimeter 8d and the area 4d^2.  Now, the derivative of the area is the perimeter.  (Also, if you imagine growing the square slowly with this measurement, it grows from the center outward, growing on all four sides.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-26T08:17:48.273" />
  <row Id="715" PostTypeId="2" ParentId="712" CreationDate="2010-07-26T08:17:57.990" Score="13" ViewCount="0" Body="&lt;p&gt;Actually, it is also true for squares (and for regular polygons in general!), the problem you ran into is what the &lt;a href=&quot;http://en.wikipedia.org/wiki/Apothem&quot; rel=&quot;nofollow&quot;&gt;equivalent of &quot;r&quot;&lt;/a&gt; is. The side length of a square is actually more comparable to the circle's diameter.&lt;br&gt;&#xA;Instead, the correct analogue of the circle's radius is the distance from the center of the square to the midpoint of one side, which is only half as long as the square's side.&lt;br&gt;&#xA;&lt;img src=&quot;http://www.csrjjsmp.com/images/offsite/square.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Here, we have A=(2r)&lt;sup&gt;2&lt;/sup&gt;=4r&lt;sup&gt;2&lt;/sup&gt; and P=4*(2r)=8r.&lt;br&gt;&#xA;The perimeter is derivative of the area with respect to r, just as in the case of a circle.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-26T09:41:01.373" LastActivityDate="2010-07-26T09:41:01.373" CommentCount="0" />
  <row Id="716" PostTypeId="1" AcceptedAnswerId="718" CreationDate="2010-07-26T08:38:16.320" Score="4" ViewCount="202" Body="&lt;p&gt;I know that the harmonic series 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/n + ... diverges, but what about the alternating harmonic series 1/1 - 1/2 + 1/3 - 1/4 + 1/5 - ... + (-1)^n/n + ...?  Does it converge?  If so, what is its sum?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-26T12:28:15.030" Title="Sum of the alternating harmonic series" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="717" PostTypeId="2" ParentId="716" CreationDate="2010-07-26T08:50:52.000" Score="5" ViewCount="0" Body="&lt;p&gt;it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, it converges to log(2) (log being the natural logarithm). See &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercator_series&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-26T08:50:52.000" />
  <row Id="718" PostTypeId="2" ParentId="716" CreationDate="2010-07-26T09:14:15.523" Score="6" ViewCount="0" Body="&lt;p&gt;Complementary to Mau's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call a series $a_n$ &lt;strong&gt;absolutely convergent&lt;/strong&gt; if $\Sum|a_n|$ converges. If $a_n$ converges but is not absolutely convergent we call $a_n$ &lt;strong&gt;conditionally convergent&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;The Riemann series theorem&lt;/a&gt; states that any conditionally convergent series can be reordered to converge to any real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morally this is because both the positive and negative parts of your series diverge but the divergences cancel each other out, one or other's cancelling the other can be staggered by adding on, say,  the negative bits every third term in stead of every other term. This means that in the race for the two divergences to cancel each other out, we give the positive bit something of a head-start and will get a larger positive outcome. Notice how, even in this rearranged version of the series, every term will still come up exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting, on the wikipedia link Mau provided, that the convergence to $ln2$ of your series is at the edge of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Radius_of_convergence&quot; rel=&quot;nofollow&quot;&gt;radius of convergence&lt;/a&gt; for the series expansion of $ln(1-x)$- this is a fairly typical occurrence: at the boundary of a domain of convergence of a taylor series, the series is &lt;em&gt;only just&lt;/em&gt; converging- which is why you see this conditional convergence type behavior. &lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-07-26T09:14:15.523" />
  <row Id="719" PostTypeId="2" ParentId="704" CreationDate="2010-07-26T10:29:09.160" Score="2" ViewCount="0" Body="&lt;p&gt;Often the most obvious solution is the best solution. I would write T(s) be 1 if a statement is true and 0 if it is false. I would then call the statements A, B and C and let S={A, B, C}. Then is would write sum T(s) over S &gt;=2 =&gt; all s are true. You can make these statement more symbolic if you really like&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-26T12:01:45.160" LastActivityDate="2010-07-26T12:01:45.160" CommentCount="2" />
  <row Id="720" PostTypeId="1" AcceptedAnswerId="770" CreationDate="2010-07-26T11:22:17.353" Score="4" ViewCount="149" Body="&lt;p&gt;Given an initial position and a subsequent position, each given by latitude and longitude in the WGS-84 system. How do you determine the heading in degrees clockwise from true north of movement?&lt;/p&gt;&#xA;" OwnerUserId="193" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-27T09:03:35.653" LastActivityDate="2010-09-29T03:45:13.390" Title="How to calculate a heading on the earths surface?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="721" PostTypeId="1" AcceptedAnswerId="722" CreationDate="2010-07-26T11:59:51.513" Score="9" ViewCount="161" Body="&lt;p&gt;Is there an algorithm for working out the best way (i.e. fewest multiplications) of calculating A^n in a structure where multiplication is associative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose A is a square matrix. Matrix multiplication is associative, and I can compute A^9 with 4 multiplications:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A^2 = A*A&#xA;A^3 = A^2*A&#xA;A^6 = A^3*A^3&#xA;A^9 = A^6*A^3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;One method which works is to compute A^(2^i) and use the binary representation of n, but this is not always optimal, e.g. with n=23, we can do it in 6 multiplications:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A^2 = A*A&#xA;A^3 = A^2*A&#xA;A^5 = A^3*A^2&#xA;A^10 = A^5*A^5&#xA;A^20 = A^10*A^10&#xA;A^23 = A^23*A^3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;rather than 7:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A^2 = A*A&#xA;A^4 = A^2*A^2&#xA;A^8 = A^4*A^4&#xA;A^16 = A^8*A^8&#xA;A^20 = A^16*A^4&#xA;A^22 = A^20*A^2&#xA;A^23 = A^22*A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is there an algorithm which gives the quickest way?&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-26T21:37:34.503" Title="Algorithm for calculating A^n with as few multiplications as possible" Tags="&lt;algorithms&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="722" PostTypeId="2" ParentId="721" CreationDate="2010-07-26T12:15:26.850" Score="8" ViewCount="0" Body="&lt;p&gt;There seems to be no efficient algorithm for this. Quoting Wikipedia on &lt;a href=&quot;http://en.wikipedia.org/wiki/Addition-chain_exponentiation&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Addition-chain exponentiation&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;… the addition-chain method is much more complicated, since the determination of a shortest addition chain seems quite difficult: &lt;strong&gt;no efficient optimal methods are currently known for arbitrary exponents&lt;/strong&gt;, and the related problem of finding a shortest addition chain for a given set of exponents has been proven NP-complete.  Even given a shortest chain, addition-chain exponentiation requires more memory than the binary method, because it must potentially store many previous exponents from the chain simultaneously. In practice, therefore, shortest addition-chain exponentiation is primarily used for small fixed exponents for which a shortest chain can be precomputed and is not too large.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the same page, however, it does link to a reference* of some methods better than binary exponentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple example is to use base-N number instead of base-2. e.g. for A&lt;sup&gt;510&lt;/sup&gt;,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;with binary:   2 = 1 + 1&#xA;               3 = 2 + 1&#xA;               6 = 3 + 3&#xA;               7 = 6 + 1&#xA;              14 = 7 + 7&#xA;              15 = 14 + 1&#xA;              30 = 15 + 15&#xA;              31 = 30 + 1&#xA;              62 = 31 + 31&#xA;              63 = 62 + 1&#xA;             126 = 63 + 63&#xA;             127 = 126 + 1&#xA;             254 = 127 + 127&#xA;             255 = 254 + 1&#xA;             510 = 255 + 255   (15 multiplications)&#xA;&#xA;with base-8:   2 = 1 + 1&#xA;               3 = 2 + 1&#xA;               4 = 3 + 1&#xA;               5 = 4 + 1&#xA;               6 = 5 + 1&#xA;               7 = 6 + 1&#xA;              14 = 7 + 7&#xA;              28 = 14 + 14&#xA;              56 = 28 + 28&#xA;              63 = 56 + 7&#xA;             126 = 63 + 63&#xA;             252 = 126 + 126&#xA;             504 = 252 + 252&#xA;             510 = 504 + 6      (14 multiplications)&#xA;&#xA;with base-4:   2 = 1 + 1&#xA;               3 = 2 + 1&#xA;               4 = 2 + 2&#xA;               7 = 4 + 3&#xA;              14 = 7 + 7&#xA;              28 = 14 + 14&#xA;              31 = 28 + 3&#xA;              62 = 31 + 31&#xA;             124 = 62 + 62&#xA;             127 = 124 + 3&#xA;             254 = 127 + 127&#xA;             508 = 254 + 254&#xA;             510 = 508 + 2      (13 multiplications)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(I don't know how to choose the optimal N.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*: &lt;a href=&quot;http://eref.uqu.edu.sa/files/a_survey_of_fast_exponentiation_methods__64651.pdf&quot; rel=&quot;nofollow&quot;&gt;Daniel M. Gordon, &lt;em&gt;A survey of fast exponentiation methods&lt;/em&gt;, Journal of Algorithms &lt;strong&gt;27&lt;/strong&gt; (1998), pp 129–146&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-07-26T13:17:36.607" LastActivityDate="2010-07-26T13:17:36.607" CommentCount="3" />
  <row Id="723" PostTypeId="2" ParentId="716" CreationDate="2010-07-26T12:28:15.030" Score="1" ViewCount="0" Body="&lt;p&gt;Let's  say you have a sequence of nonnegative numbers $a_1 \geq a_2 \geq \dots$ tending to zero. Then it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternating_series_test&quot; rel=&quot;nofollow&quot;&gt;theorem&lt;/a&gt; that the alternating sum $\sum (-1)^i a_i$ converges (not necessarily absolutely, of course).&#xA;This in particular applies to your series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if your curious why it converges to $\log(2)$ (which seems somewhat random), it's because of the Taylor series of $\log(1+x)$ while letting $x \to 1$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-26T12:28:15.030" />
  <row Id="724" PostTypeId="2" ParentId="713" CreationDate="2010-07-26T12:39:20.013" Score="2" ViewCount="0" Body="&lt;p&gt;While tensors are generalizations of vectors, I don't think you can really generalize the way you visualize them.  This is because you really want to think of tensors as multi-linear functions and you usually don't think of a vector as a linear function from the dual space to the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if you can't get as nice a geometric picture of a tensor, you do get a nice grasp on what they are if you view them as multi-linear functions (as opposed to just a collection of numbers) from some copies of your vector space V (and/or its dual V*) into R.  For example, a metric is a type of rank 2 tensor and has a nice geometric meaning-- applying it to two copies of a vector gives the vectors squared length, you can apply to to two vectors to get the angle between them, etc.  The moment of inertia tensor is a 2-tensor I such that I(u,u) is the moment of inertia about the u-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also usually different ways of looking at the same tensor, which may make it easier to get a grasp on it.  For example, a (1,1) tensor is a multilinear map from V x V* into R.  However, this can naturally (i.e. basis independently) be identified with a linear map from V to itself: if T is a (1,1) tensor then contracting it with a vector gives a linear map V* to R, which is just another vector.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-26T12:39:20.013" />
  <row Id="725" PostTypeId="2" ParentId="701" CreationDate="2010-07-26T12:42:57.450" Score="3" ViewCount="0" Body="&lt;p&gt;I had an answer before, but I looked into it a bit more and my answer was incorrect so I removed it. This link may be of interest: &lt;a href=&quot;http://en.wikipedia.org/wiki/Circle_packing_in_a_square&quot; rel=&quot;nofollow&quot;&gt;Circle Packing in a Square (wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was suggested by KennyTM that there may not be an optimal solution yet to this problem in general. Further digging into this has shown me that this is probably correct. Check out this page: &lt;a href=&quot;http://hydra.nat.uni-magdeburg.de/packing/csq/csq.html&quot; rel=&quot;nofollow&quot;&gt;Circle Packing - Best Known Packings&lt;/a&gt;. As you can see, solutions up to only 30 circles have been found and proven optimal. (Other higher numbers of circles have been proven optimal, but 31 hasn't)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that although problem defined on the wikipedia page and the other link is superficially different than the question asked here, the same fundamental question is being asked, which is &quot;what is the most efficient way to pack circles in a square/rectangle container?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...And it seems the answer is &quot;we don't really know&quot; :)&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-07-26T12:42:57.450" />
  <row Id="726" PostTypeId="1" AcceptedAnswerId="727" CreationDate="2010-07-26T12:51:35.810" Score="8" ViewCount="344" Body="&lt;p&gt;I keep hearing questions about whether something is NP-complete, but they never really mention what it is. Why do people care so much about NP-complete problems?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-26T13:41:05.957" LastActivityDate="2010-07-30T16:27:13.650" Title="What are NP-complete problems and why are they so important?" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="3" FavoriteCount="6" />
  <row Id="727" PostTypeId="2" ParentId="726" CreationDate="2010-07-26T12:58:10.257" Score="10" ViewCount="0" Body="&lt;p&gt;A problem is in class NP if its solution may be &lt;strong&gt;verified&lt;/strong&gt; in polynomial time, that is if the dimension of the problem is n you may be sure that for large enough n you need less than r&amp;middot;n&lt;sup&gt;k&lt;/sup&gt; operations to verify the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem is in class P if its solution may be &lt;strong&gt;found&lt;/strong&gt; in polynomial time, instead. A problem in P is in NP by definition, but the converse may not be the case; probably the most important open question in computer science is whether classes P and NP are the same, that is P=NP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NP-complete is a family of NP problems for which you know that if one of them had a polynomial solution then &lt;strong&gt;everyone of them&lt;/strong&gt; has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDITED) For the time being, only known algorithms for NP-&lt;strong&gt;complete&lt;/strong&gt; problems are exponential in number of operations, so they are not practically solvable for n large.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-07-26T17:20:32.843" LastActivityDate="2010-07-26T17:20:32.843" CommentCount="1" />
  <row Id="728" PostTypeId="2" ParentId="726" CreationDate="2010-07-26T13:32:27.070" Score="10" ViewCount="0" Body="&lt;p&gt;To expand on Mau's answer, you should care about NP-complete problems because there is an &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_NP-complete_problems&quot; rel=&quot;nofollow&quot;&gt;entire family&lt;/a&gt; of them that spans a large number of seemingly basic algorithms across a wide range of disciplines. These aren't obscure problems, but extremely important and highly practical questions. For examples, consider the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Travelling_salesman_problem&quot; rel=&quot;nofollow&quot;&gt;Travelling salesman problem&lt;/a&gt; - finding the shortest path (on a graph) that allows you to visit every city exactly once.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bin_packing_problem&quot; rel=&quot;nofollow&quot;&gt;Bin packing problem&lt;/a&gt; - there are a number of fixed (integer) size bins and objects of varying sizes. Minimise the number of bins required to hold all of the objects&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;Knapsack problem&lt;/a&gt; - given objects of various sizes and values and a knapsack with a fixed integer size, choose the objects that can fit inside with the most value&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vertex_cover&quot; rel=&quot;nofollow&quot;&gt;Minimal Vertex cover&lt;/a&gt; - finding the smallest set of vertices such that every edge contains at least one chosen vertice&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Clique_problem&quot; rel=&quot;nofollow&quot;&gt;Clique&lt;/a&gt; - finding that largest group of people who all know each other&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem&quot; rel=&quot;nofollow&quot;&gt;Subgraph isomorphism&lt;/a&gt; - does one graph contain a subgraph isomorphic to another?&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Set_packing&quot; rel=&quot;nofollow&quot;&gt;Set packing&lt;/a&gt; - given a number of sets, what is the maximum number of disjoint sets that can be selected? This is related to set cover, where we are trying to choose sets so that every element is within at least one set&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;Subset sum&lt;/a&gt; - Given a set of integers, does some subset sum to 0?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although many of these problems may seem abstract, many more complicated problems can't be efficiently solved with current techniques, as they are equivalent to one of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of NP completeness has received a huge amount of attention. Once you've reduced a problem to NP-complete, you know to give up on an efficient fast algorithm and to start looking at approximations.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-26T21:19:05.073" LastActivityDate="2010-07-26T21:19:05.073" CommentCount="4" />
  <row Id="729" PostTypeId="2" ParentId="726" CreationDate="2010-07-26T16:05:34.357" Score="3" ViewCount="0" Body="&lt;p&gt;Any problem for which a solution (once found) can be quickly &lt;em&gt;verified&lt;/em&gt; as a solution is said to be &quot;in &lt;strong&gt;NP&lt;/strong&gt;&quot;  (Here, &quot;quickly&quot; means in polynomial-time).  Any problem for which a solution can be &lt;em&gt;found&lt;/em&gt; quickly is said to be &quot;in &lt;strong&gt;P&lt;/strong&gt;.&quot;  &lt;strong&gt;P&lt;/strong&gt; is a subset of &lt;strong&gt;NP&lt;/strong&gt; - that is, any problem for which a solution can be quickly &lt;em&gt;found&lt;/em&gt; can also be quickly &lt;em&gt;verified&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem is &lt;strong&gt;NP-complete&lt;/strong&gt; if it is the hardest problem in &lt;strong&gt;NP&lt;/strong&gt;.  Surprisingly, there are many &lt;strong&gt;NP-complete&lt;/strong&gt; problems, which are all equivalent - here, &lt;em&gt;equivalent&lt;/em&gt; means that a quick (polynomial-time) solution to any one of them would give you a quick solution to all the rest.  Also somewhat surprisingly, a quick solution to any &lt;strong&gt;NP-complete&lt;/strong&gt; problem would also give you a quick solution to &lt;em&gt;any&lt;/em&gt; problem in &lt;strong&gt;NP&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a quick (polynomial-time) algorithm to solve &lt;strong&gt;NP-complete&lt;/strong&gt; problems?  That is the &lt;a href=&quot;http://en.wikipedia.org/wiki/P%3DNP&quot; rel=&quot;nofollow&quot;&gt;P=NP problem&lt;/a&gt;, one of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Millennium_Prize_Problems&quot; rel=&quot;nofollow&quot;&gt;greatest unsolved problems&lt;/a&gt; of our time.  However, most sane mathemeticians believe (and hope!) that &lt;strong&gt;P≠NP&lt;/strong&gt;, because proving math-theorems is &lt;strong&gt;NP-Complete&lt;/strong&gt;; so if &lt;strong&gt;P=NP&lt;/strong&gt;, we'd be out of a job! :)&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-26T21:56:09.773" LastActivityDate="2010-07-26T21:56:09.773" />
  <row Id="730" PostTypeId="1" CreationDate="2010-07-26T17:29:10.320" Score="1" ViewCount="63" Body="&lt;p&gt;In an example I have worked out for my work, I have constructed a category whose objects are graded R-modules(R is a graded ring), and with morphisms the usual morphisms quotient(by quotient I mean simply that this class of morphisms are Isomorphisms, thus creating an eq relation) the following class of morphisms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma=\left\lbrace f\in Hom_{grR-mod}\left(A,B\right)|ker\left(f\right)_0\neq 0,  cok\left(f\right)_0\neq 0\right\rbrace$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if this category has a better(more canonical) description, or if I can show it is equivalent to some other interesting category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-26T17:29:10.320" Title="An elegant description for graded-module morphisms with non-zero zero component." Tags="&lt;modules&gt;&lt;graded-modules&gt;" CommentCount="4" />
  <row Id="731" PostTypeId="2" ParentId="721" CreationDate="2010-07-26T18:11:26.853" Score="1" ViewCount="0" Body="&lt;p&gt;Another option would be to use eigendecomposition . It allows you to raise the eigenvalues in the diagonal of the decomposition A = VDV^-1 to a power. It changes the problem from matrix multiplication to the multiplication of the eigenvalues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once in eigendecomposition form you could perform the same addition-chain exponentiation technique but it would be with scalars, not matrices. Much more efficient, because each matrix multiplication has n^3 multiplies, but with ed you would only have n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More is explained here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_decomposition#Eigendecomposition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Matrix_decomposition#Eigendecomposition&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-26T21:37:34.503" LastActivityDate="2010-07-26T21:37:34.503" CommentCount="3" />
  <row Id="732" PostTypeId="1" AcceptedAnswerId="733" CreationDate="2010-07-26T20:08:48.103" Score="5" ViewCount="444" Body="&lt;p&gt;Can someone point me to a proof that the set of irrational numbers is uncountable. I know how to show that the set &lt;strong&gt;Q&lt;/strong&gt; is countable but how would you then show that the irrationals are uncountable? &lt;/p&gt;&#xA;" OwnerUserId="180" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-04T17:43:57.480" LastActivityDate="2011-01-04T17:43:57.480" Title="Proof that the irrational numbers are uncountable" Tags="&lt;elementary-set-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="1" />
  <row Id="733" PostTypeId="2" ParentId="732" CreationDate="2010-07-26T20:12:04.727" Score="20" ViewCount="0" Body="&lt;p&gt;Given that the reals are uncountable (can be shown via &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_diagonalization&quot; rel=&quot;nofollow&quot;&gt;Cantor diagonalization&lt;/a&gt;) and the rationals are countable, the irrationals are the reals with the rationals removed, which is uncountable.  (Or, since the reals are the union of the rationals and the irrationals, if the irrationals were countable, the reals would be the union of countable sets and would have to be countable, so the irrationals must be uncountable.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-26T20:38:10.780" LastActivityDate="2010-07-26T20:38:10.780" />
  <row Id="734" PostTypeId="1" CreationDate="2010-07-26T20:13:26.987" Score="6" ViewCount="214" Body="&lt;p&gt;Are there any real online mathematics (applied math, statistics, ...) degree programs out there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm full-time employed, thus not having the flexibility of attending an on campus program. I also already have a MSc in Computer Science. My motivation for a math degree is that I like learning and am interested in the subject. I've studied through number of OCW courses on my own, but it would be nice if I could actually be able to have my studying count towards something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done my share of Googling for this, but searching for online degrees seems to bring up a lot of institutions that (at least superficially) seem a bit shady (diploma mills?).&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T01:14:52.763" LastActivityDate="2010-10-24T16:24:38.453" CommunityOwnedDate="2010-07-26T22:29:23.650" Title="Online Math Degree Programs" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;online-degree&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="735" PostTypeId="1" AcceptedAnswerId="739" CreationDate="2010-07-26T20:23:17.057" Score="6" ViewCount="200" Body="&lt;p&gt;I was coming back from my Driver's Education class, and something mathsy really stuck out to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the essential properties of a car is its current speed.  Or speed at a current time.  For example, at a given point in time in my drive, I could be traveling 40 mph.  But what does that &lt;em&gt;mean&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my basic algebra classes, I've learned that speed = distance/time.  So if I travel ten miles in half an hour, my average speed would be 20 mph ($\frac{10 mi}{25 h}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instantaneous velocity...you aren't measuring average speed for a given amount of time.  You're measuring instantaneous speed over an...instantaneous amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would be something like (miles) / (time), where time = 0?  Isn't that infinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And perhaps, in a difference of time = 0, then I'd be traveling 0 miles.  So would I be said to be going 0 mph at an instantaneous moment in time?  I'd like to be able to tell that to any cops pull me over for &quot;speeding&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then if miles = 0 and time = 0, then you have 0/0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all rather confusing.  What does it &lt;strong&gt;mean&lt;/strong&gt; to be going 40 mph at a given moment in time, exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard this explained using this strange art called &quot;calculus&quot; before, and it's all gone over my head.  Can anyone explain this using terms I (a High School Algebra and Geometry and Driving student) will understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I figured that my problem had numbers in it, and therefore has to do with Maths.)&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:40:48.933" LastActivityDate="2010-07-28T00:52:00.427" Title="What does it mean to be going 40 mph (or 64 kph, etc.) at a given moment?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="737" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T20:36:28.617" Score="6" ViewCount="0" Body="&lt;p&gt;If you were to try to measure instantaneous speed as you described, you would in fact have traveled 0 miles in 0 time and 0/0 is undefined  If, however, you look at your average speed over smaller and smaller periods of time around the instant you care about--that is, (distance traveled from $t=t_0$ to $t=t_0+\epsilon$ for various small values of &amp;epsilon;--and these average speeds &quot;converge&quot; (they all get closer to a single value as &amp;epsilon; gets closer to 0), then we say that the instantaneous speed is that single value upon which the average speeds around that point converge.  This is, of course, a somewhat informal explanation; to be more precise requires getting into differential calculus.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-26T20:36:28.617" />
  <row Id="738" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T20:59:48.653" Score="1" ViewCount="0" Body="&lt;p&gt;in the real world you need a bit of time to brake or accelerate. This means that in a small amount of time (let's say 1/10 of a second) your speed is more or less constant. So, instead of dividing by 0 seconds, just take a small elapsed time and do the division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus you will learn how to be more precise, noticing that using a smaller and smaller elapsed time the ratio between distance and time changes less and less; at that point you postulate that &lt;strong&gt;in this scenario only&lt;/strong&gt; you may say that the limit (which is 0/0 and therefore indefinite) is the instantaneous speed. But the odometers don't care for instantaneous speed, so for the moment you shouldn't worry.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-26T20:59:48.653" CommentCount="2" />
  <row Id="739" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T21:22:38.663" Score="8" ViewCount="0" Body="&lt;p&gt;I think there is a very clear meaning in the physical world: If, at some moment, you were going 40 mph, if you were to stop de/accelerating and just hold that velocity, you would cover 40 miles in 1 hour.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-26T21:22:38.663" />
  <row Id="740" PostTypeId="1" CreationDate="2010-07-26T21:35:48.560" Score="11" ViewCount="617" Body="&lt;p&gt;What are some particularly well-known functions that exhibit pathological behavior at or near at least one value and are particularly useful as examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if f'(a) = b, then f(a) exists, f is continuous at a, f is differentiable at a, but f' need not be continuous at a.  A function for which this is true is f(x) = x^2 sin(1/x) at x=0.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-25T22:10:29.297" CommunityOwnedDate="2010-07-26T21:38:01.230" Title="Useful examples of pathological functions" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="14" CommentCount="2" FavoriteCount="8" />
  <row Id="741" PostTypeId="1" AcceptedAnswerId="743" CreationDate="2010-07-26T21:40:11.743" Score="4" ViewCount="399" Body="&lt;p&gt;I asked this question on Stack Overflow but it was closed as &quot;not programming related&quot;. So I think this is probably the best place for it...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I read over the wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;, but it seems to be beyond my comprehension. It says it's for optimization, but how is it different than any other method for optimizing things?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An answer that introduces me to linear programming so I can begin diving into some less beginner-accessible material would be most helpful.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:39:28.670" LastActivityDate="2010-07-28T04:30:38.603" Title="What is linear programming?" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="742" PostTypeId="1" AcceptedAnswerId="746" CreationDate="2010-07-26T21:48:35.177" Score="3" ViewCount="183" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have sheets of 42-cent stamps and&#xA;  29-cent stamps, but you need at least&#xA;  $3.20 to mail a package.  What is the&#xA;  least amount you can make with the 42-&#xA;  and 29-cent stamps that is sufficient&#xA;  to mail the package?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A contest problem such as this is probably most easily solved by tabulating the possible combinations, using 0 through ceiling(total/greater value) of the greater-value stamp and computing the necessary number of the smaller stamp and the total postage involved.  The particular example above would be solved with a 9-row table, showing the minimum to be $3.23, made with 7 42-cent stamps and 1 29-cent stamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better algorithm for solving this kind of problem?  What if you have more than two values of stamps?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:40:02.070" LastActivityDate="2010-08-11T22:01:46.380" Title="Least wasteful use of stamps to achieve a given postage" Tags="&lt;number-theory&gt;&lt;algebra-precalculus&gt;&lt;algorithms&gt;&lt;contest-math&gt;" AnswerCount="5" />
  <row Id="743" PostTypeId="2" ParentId="741" CreationDate="2010-07-26T21:56:42.473" Score="4" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming#Standard_form&quot; rel=&quot;nofollow&quot;&gt;standard form&lt;/a&gt; (and &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming#Example&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;) sections pretty well describe what it is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it different than any other method for optimizing things?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's, well, just another method.  However, it is somewhat special in that many other optimization algorithms either &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming_relaxation&quot; rel=&quot;nofollow&quot;&gt;use linear programming&lt;/a&gt; as part of their solution, or are in reality a &lt;a href=&quot;http://en.wikipedia.org/wiki/Flow_network&quot; rel=&quot;nofollow&quot;&gt;specialized solution to a linear programming problem&lt;/a&gt;.  &lt;em&gt;In fact&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_programming#Integer_unknowns&quot; rel=&quot;nofollow&quot;&gt;integer linear programming&lt;/a&gt; is &lt;strong&gt;NP-complete&lt;/strong&gt;, meaning that &lt;em&gt;any&lt;/em&gt; problem in &lt;strong&gt;NP&lt;/strong&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/726/what-are-np-complete-problems-and-why-are-they-so-important/729#729&quot; rel=&quot;nofollow&quot;&gt;can be stated as an (integer) linear programming problem&lt;/a&gt;.&lt;br&gt;&#xA;&lt;em&gt;(this also means solving your typical integer linear programming problem is &lt;strong&gt;much&lt;/strong&gt; more difficult than if we didn't restrict ourselves to integers..)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-26T22:51:00.397" LastActivityDate="2010-07-26T22:51:00.397" CommentCount="1" />
  <row Id="744" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T22:08:42.607" Score="0" ViewCount="0" Body="&lt;p&gt;A state-variable kind of approach: (represents the physical states of a system)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's first law of motion&lt;/a&gt; says that without external forces, masses will move at constant velocity. Every change in velocity (for a car, whether a change in speed due to braking, or a change in direction due to friction of the tires and the steering mechanism) is due to an external force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;state&quot; of the car at any instant in time is its velocity; distance traveled can be derived from velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you are going 60mi/hr = 26.8m/sec (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Miles_per_hour#Conversions&quot; rel=&quot;nofollow&quot;&gt;conversion factors&lt;/a&gt;) in a particular direction at constant velocity, you can multiply a change in time times velocity and get distance: 26.8m/sec = 26.8 meters in one second. Or 26.8mm in one millisecond. Or 26.8 micrometers in one microsecond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of going from velocity to (directed) distance is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral&quot; rel=&quot;nofollow&quot;&gt;integration&lt;/a&gt;, and for those two variables (velocity and directed distance) is just the summing up of distance increments (= velocity * time increments) over infinitesimally short periods of time. For smoothly varying velocities, coarse time increments are a good approximation to reality -- differential equation solvers use this or similar approaches (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Simpson%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Simpson's Rule&lt;/a&gt;) to perform integration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But instantaneous velocity...you aren't measuring average speed for a given amount of time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, instantaneous velocity is just the limit of average velocity as the time period becomes infinitely small.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-26T22:08:42.607" />
  <row Id="745" PostTypeId="2" ParentId="741" CreationDate="2010-07-26T22:09:22.463" Score="3" ViewCount="0" Body="&lt;p&gt;BlueRaja's answer is certainly more complete than this one (and gives good references), but here's a rough overview of linear programming.  Suppose that you have a linear function (in high school courses, it's typically a function of two variables) that you want to optimize on a convex &quot;feasible&quot; region bounded by linear equations (again, in high school, the bounds are typically described using linear inequalities in two variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the function to optimize is linear, the set of points for which the function has a particular value, say c, is a line (and all such lines are parallel) and the value of the function anywhere to one side of the line is greater than c and anywhere to the other side of the line is less than c.  So, you can think about moving through this set of constant-value lines to increase/decrease the value of the target function as appropriate to optimize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the feasible region is bounded by linear equations, as the constant-value line moves through and out of the feasible region, it last touches the feasible region at a vertex (or possibly at all points on an edge connecting two vertices), so the optimal solution must occur at a vertex of the feasible region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all that, linear programming comes down to evaluating the target function at all the vertices of the feasible region to find the optimal value.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-26T22:09:22.463" CommentCount="5" />
  <row Id="746" PostTypeId="2" ParentId="742" CreationDate="2010-07-26T22:13:10.403" Score="1" ViewCount="0" Body="&lt;p&gt;With two stamps, you can do it in &lt;strike&gt;linear&lt;/strike&gt; pseudo-linear time - &lt;strong&gt;O(totalCost/costOfLargerStamp)&lt;/strong&gt; - by simply enumerating every possibility (there is only one possible count of the smaller stamp for each count of the larger stamp).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, however, solving this is equivalent to solving a general &lt;a href=&quot;http://en.wikipedia.org/wiki/Integer_linear_programming#Integer_unknowns&quot; rel=&quot;nofollow&quot;&gt;integer linear programming problem&lt;/a&gt; written in standard form, which is NP-complete.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-07-26T23:52:29.077" LastActivityDate="2010-07-26T23:52:29.077" CommentCount="1" />
  <row Id="747" PostTypeId="2" ParentId="740" CreationDate="2010-07-26T22:15:40.447" Score="2" ViewCount="0" Body="&lt;p&gt;sin(x)/x is useful; it has a singularity at x=0, but if you take the union of y=sin(x)/x with the point (x=0,y=1) then you get sinc(x/&amp;pi;). The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sinc_function&quot; rel=&quot;nofollow&quot;&gt;sinc function&lt;/a&gt; has a lot of applications in signal processing and diffraction.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-26T22:15:40.447" CommunityOwnedDate="2010-07-26T22:15:40.447" />
  <row Id="748" PostTypeId="2" ParentId="740" CreationDate="2010-07-26T22:27:12.387" Score="4" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;Dirac delta &quot;function.&quot;&lt;/a&gt; It's not a &quot;function,&quot; strictly speaking, but rather a very simple example of a distribution that isn't a function.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-26T22:27:12.387" CommunityOwnedDate="2010-07-26T22:27:12.387" />
  <row Id="749" PostTypeId="2" ParentId="742" CreationDate="2010-07-26T22:40:45.853" Score="1" ViewCount="0" Body="&lt;p&gt;This is a simple variation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;Knapsack problem&lt;/a&gt;. Which is a NP-Complete problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let n to be the target value, it can be solve using knapsack's dynamic programming solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a finite amount of stamps, then you can add them all up, subtract the target value. Run 0-1 knapsack on the stamps with the resulting number. The stamps outside the knapsack is the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a infinite amount of stamps. You can either make them finite: $\lceil \frac{n}{t_i}\rceil$ stamps for stamps with value $t_i$ and run the above process, or you can compute the result by running knapsacks O(log n) times similar to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search&quot; rel=&quot;nofollow&quot;&gt;binary search&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's no better than a naive &lt;a href=&quot;http://en.wikipedia.org/wiki/Backtracking&quot; rel=&quot;nofollow&quot;&gt;backtracking&lt;/a&gt; solution when n is large. I assume you are not going to pay $100 for stamps. ;)&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-26T22:40:45.853" />
  <row Id="750" PostTypeId="1" AcceptedAnswerId="754" CreationDate="2010-07-26T22:54:07.037" Score="4" ViewCount="62" Body="&lt;p&gt;The absolute value of a 2 by 2 matrix determinant is the area of a corresponding parallelogram with the 2 row vectors as sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute value of a 3 by 3 matrix determinant is the volume of a corresponding parallelepiped with the 3 row vectors as sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be generalized to n-D? The absolute value of a n by n matrix determinant is the volume of a corresponding n-parallelotope?&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-26T23:16:16.113" Title="Determinants and volume of parallelotopes" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="751" PostTypeId="2" ParentId="735" CreationDate="2010-07-26T23:02:12.000" Score="3" ViewCount="0" Body="&lt;p&gt;This is similar to other answers. Imagine another car beside yours. That car is covering a distance of 40 miles over the next hour at a constant velocity. At the point at which you keep pace with that car (relative velocity = 0), you are traveling at 40 mph.&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2010-07-26T23:02:12.000" />
  <row Id="752" PostTypeId="2" ParentId="750" CreationDate="2010-07-26T23:11:18.070" Score="3" ViewCount="0" Body="&lt;p&gt;Yes--see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the change of variables formula with the Jacobian. However, it is probably more accurate to say that the change-of-variables formula is a (nontrivial) consequence of this.  One argument for this fact, which is given in Rudin's &lt;em&gt;Real and Complex Analysis,&lt;/em&gt; is the fact that both the determinant and the volume function up to sign behave very nicely satisfying certain specific properties (i.e., multilinear, alternating, normalized) hence are the same.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-26T23:11:18.070" />
  <row Id="753" PostTypeId="2" ParentId="750" CreationDate="2010-07-26T23:14:36.613" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, that is because in its most general form the determinate is &lt;strong&gt;defined&lt;/strong&gt; that way, except without the absolute value part. The determinant is a way to compare volumes of subspaces that have the same grade, which is the multi-vector equivalent of the size of matrix.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-26T23:14:36.613" />
  <row Id="754" PostTypeId="2" ParentId="750" CreationDate="2010-07-26T23:16:16.113" Score="4" ViewCount="0" Body="&lt;p&gt;Yes it can. In fact, as Katie Banks noted, a determinant is an &lt;a href=&quot;http://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant/669#669&quot; rel=&quot;nofollow&quot;&gt;intuitive way of thinking about volumes&lt;/a&gt;. To summarise her argument, if we consider the vectors as a matrix, switching two rows, multiplying one by a constant or adding a linear combination will have the same effect on the volume as on the determinate. We can use these operations to transform any n-parallelotope to cube and note that the determinate matches the signed volume here, so it will match it everywhere as well.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-26T23:16:16.113" />
  <row Id="755" PostTypeId="2" ParentId="742" CreationDate="2010-07-26T23:27:18.700" Score="1" ViewCount="0" Body="&lt;p&gt;It is important to understand that NP-completeness is defined in terms of input size, not in terms of either cost to be paid, C, or the number of types of stamps, S, when we are dealing with integers rather than real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algorithm can actually be solved in time CS. We take the first stamp value an mark all multiple of it as obtainable (up to the least number greater than the cost). Then, for each subsequent stamp value, we mark as obtainable all numbers that can be obtained by adding a multiple of its value with an obtainable number (up to the least number greater than the cost). We can keep track of the lowest cost postage amount that we have found so far.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-26T23:27:18.700" />
  <row Id="756" PostTypeId="1" CreationDate="2010-07-27T00:26:20.523" Score="8" ViewCount="344" Body="&lt;p&gt;Here's one that popped into my mind when I was thinking about binary search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of an integer between 1 and &lt;em&gt;n&lt;/em&gt;. You have to guess my number. You win as soon as you guess the correct number. If your guess is not correct, I'll give you a hint by saying &quot;too high&quot; or &quot;too low&quot;. What's your best strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an easy problem if I always tell the truth: by guessing the (rounded) mean of the lower bound and the upper bound, you can find my number in roughly log&lt;sub&gt;2&lt;/sub&gt; &lt;em&gt;n&lt;/em&gt; guesses at most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I'm allowed to cheat once? What is your best strategy then? To clarify: if you guess my number, you always win instantly. But I'm allowed, at most once, to tell you &quot;too high&quot; when your guess is actually too low, or the opposite. I can also decide not to lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a rough upper bound: you can ask each number twice to make sure I'm not cheating, and if I ever give two different answers, just ask a third time and from then on, play the regular game. In this way, you can win with about 2 log&lt;sub&gt;2&lt;/sub&gt; &lt;em&gt;n&lt;/em&gt; guesses at most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that bound can be improved. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-27T00:38:44.943" LastActivityDate="2010-11-02T13:11:07.107" Title="Twenty questions against a liar" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;discrete-mathematics&gt;&lt;searching&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="757" PostTypeId="2" ParentId="740" CreationDate="2010-07-27T00:49:43.627" Score="11" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;Weierstrass function&lt;/a&gt; is continuous everywhere and differentiable nowhere.&lt;br&gt;&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nowhere_continuous_function&quot; rel=&quot;nofollow&quot;&gt;Dirichlet function&lt;/a&gt; (the indicator function for the rationals) is continuous nowhere.&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;A modification&lt;/a&gt; of the Dirichlet function is continuous at all irrational values and discontinuous at rational values.&lt;br&gt;&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_function&quot; rel=&quot;nofollow&quot;&gt;Devil's Staircase&lt;/a&gt; is uniformly continuous but not absolutely. It increases from 0 to 1, but the derivative is 0 almost everywhere.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-25T22:10:29.297" LastActivityDate="2010-08-25T22:10:29.297" CommunityOwnedDate="2010-07-27T00:49:43.627" CommentCount="4" />
  <row Id="758" PostTypeId="2" ParentId="734" CreationDate="2010-07-27T00:56:57.370" Score="3" ViewCount="0" Body="&lt;p&gt;I would recommend against getting a degree from any online-only university. Even if you happen to find one that's not shady, everyone else who hasn't heard of it will assume it is some kind of diploma mill without bothering to do much research. Instead, I think you'd be better off going to a nearby university you're interested in, and ask them if they would be willing to make some kind of special arrangement for you. Many universities allow reduced course loads for students that have families or work full time, and aren't very good about advertising it.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-27T00:56:57.370" CommunityOwnedDate="2010-07-27T00:56:57.370" />
  <row Id="759" PostTypeId="2" ParentId="740" CreationDate="2010-07-27T00:57:18.650" Score="0" ViewCount="0" Body="&lt;p&gt;(1/x)^(1/x). Try graphing it if you dare (including over the negatives)&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-27T00:57:18.650" CommunityOwnedDate="2010-07-27T00:57:18.650" CommentCount="4" />
  <row Id="760" PostTypeId="1" AcceptedAnswerId="762" CreationDate="2010-07-27T01:09:24.273" Score="2" ViewCount="160" Body="&lt;p&gt;I have a 3-D sphere of radius R, centered at the origin.&lt;br&gt;&#xA;(x&lt;sub&gt;1&lt;/sub&gt;,y&lt;sub&gt;1&lt;/sub&gt;,z&lt;sub&gt;1&lt;/sub&gt;) and&#xA;(x&lt;sub&gt;2&lt;/sub&gt;,y&lt;sub&gt;2&lt;/sub&gt;,z&lt;sub&gt;2&lt;/sub&gt;) are two points on the sphere. The Euclidean distance is easy to calculate, but what if I were to restrict myself to traveling on the surface of the sphere?&lt;br&gt;&#xA;Two approaches come to mind: use &lt;a href=&quot;http://en.wikipedia.org/wiki/Arc_length&quot; rel=&quot;nofollow&quot;&gt;arc-length&lt;/a&gt; in some way, or simply use trigonometry: calculate the angle between the two points and get a distance from that.&lt;br&gt;&#xA;Will both/either of these methods work? Which would be easier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat related to &lt;a href=&quot;http://math.stackexchange.com/questions/720/how-to-calculate-a-heading-vector-on-the-earths-surface&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;. Maybe it will inspire someone to go answer it!&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-11T22:01:26.910" Title="How do I measure distance on a globe?" Tags="&lt;geometry&gt;&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="761" PostTypeId="1" AcceptedAnswerId="763" CreationDate="2010-07-27T01:18:03.910" Score="0" ViewCount="115" Body="&lt;p&gt;Wind resistance -- upwards acceleration, typically varies either linearly or quadratically by the current velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a constant downward acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we model the velocity over time of a falling object, subject only to wind resistance and downwards gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have much experience with differential equations, but I do know that this answer necessarily involves it, so could you possibly explain every step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-07-27T06:20:24.090" LastActivityDate="2010-07-27T06:20:24.090" Title="Using differential equations to graph velocity over time of a falling object subject to wind resistance" Tags="&lt;calculus&gt;&lt;physics&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="762" PostTypeId="2" ParentId="760" CreationDate="2010-07-27T01:33:51.347" Score="7" ViewCount="0" Body="&lt;p&gt;What you are looking for is what is called the great circle distance: the shortest curve connecting two points on the surface of a sphere is given by traveling along the (there will be exactly one unless the points are polar opposites)  arc of the great circle (that is, the circle of radius R) connecting them.  So you do need to find an arc length but it is easy to do this without invoking calculus if you know about the dot product.  Suppose the two points on the circle are represented by the vectors v and w.  If v*w denotes the dot product of these two vectors, then the angle between them will be arccos(v*w/R^2) (we divide by R^2 since v and w have length R).  Assuming this is in radians, to get the length of the arc connecting them we just multiply this angle by R to get R*arcos(v*w/R^2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are quite lucky that there is such a simple formula.  For most manifolds, the curves that minimize distances are not very easy to find since it involves solving a non-linear differential equation (the geodesic equation).  The fact that the sphere is so symmetric helps in this case, and you can maybe convince yourself that an arc of a great circle minimizes distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find this wikipedia article interesting: &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geodesic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-27T01:39:57.477" LastActivityDate="2010-07-27T01:39:57.477" />
  <row Id="763" PostTypeId="2" ParentId="761" CreationDate="2010-07-27T01:42:46.160" Score="2" ViewCount="0" Body="&lt;p&gt;F = ma  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; dv/dt = a = F/m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = mg - kv &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; dv/dt = (g - k/m*v)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a differential equation, solution = v = A + B * exp(-k/m*t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dv/dt = - B * k/m * exp(-k/m*t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;match terms and initial conditions (v = 0 at time t = 0) and you get g - k/m*A - k/m*B*exp(-k/m*t) = -B * k/m*exp(-k/m*t) --&gt; A = g * m / k, B = -g * m / k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; v = mg/k * (1 - exp(-k/m*t))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a linear differential equation (dv/dt is a linear function of v); the F = -kv&lt;sup&gt;2&lt;/sup&gt; is a nonlinear differential equation (can't remember off the top of my head how to deal with that one; it may not have a closed form solution). It's a bit difficult to summarize the techniques in general, but any good book on differential equations would cover them.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-27T01:42:46.160" />
  <row Id="764" PostTypeId="1" AcceptedAnswerId="765" CreationDate="2010-07-27T03:16:25.763" Score="3" ViewCount="196" Body="&lt;p&gt;From Rudin's PMA:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6.2 Definition&lt;/strong&gt;  Let $\alpha$ be a monotonically increasing function on $[a,b]$. ... Corresponding to each partition $P$ of $[a,b]$, we write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \alpha_i = \alpha(x_i) - \alpha(x_{i-1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then goes on to define the Riemann Stieltjes integral of $f$ with respect to $\alpha$, over the interval $[a,b]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Riemann integral is then pointed out to be a special case of this when $\alpha(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\alpha(x)=x$, I understand $\Delta x = x_i - x_{i-1}$ to represent the directed magnitude of the &quot;base of the approximating rectangle&quot; that we then multiply by the value of $f$ taken somewhere within this interval, thus obtaining the area of an approximating rectangle.  I don't know where to begin to interpret the case where $\alpha(x) \not\equiv x$.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-05T17:10:06.220" Title="Riemann vs. Stieltjes Integral" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="765" PostTypeId="2" ParentId="764" CreationDate="2010-07-27T04:01:40.240" Score="4" ViewCount="0" Body="&lt;p&gt;The idea is that we define a different way of measuring the width of the rectangle. Instead of using the difference of the two coordinates, we take the difference of the &lt;em&gt;images&lt;/em&gt; of those two coordinates under the map $\alpha$.  Basically, the idea is that the $(x_i-x_{i-1})$ is to be interpreted as the width or measure of the interval, a construction which leads more generally to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;Lebesgue measure&lt;/a&gt; of a subset of the reals.  The $\alpha(x_i)-\alpha(x_{i-1})$ leads to a different measure, starting with a different assignment of &quot;size&quot; to an interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $\alpha$ is continuously differentiable, then $\alpha(x_i) -\alpha(x_{i-1})$ is equal to $\alpha'(x) (x_i - x_{i-1})$ for $x \in [x_{i-1}, x_i]$ up to higher order terms. So in this case the Stieltjes integral with respect to $\alpha$ is the Riemann integral of the same function times $\alpha'$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-27T04:01:40.240" CommentCount="2" />
  <row Id="766" PostTypeId="1" AcceptedAnswerId="767" CreationDate="2010-07-27T06:25:37.183" Score="3" ViewCount="92" Body="&lt;p&gt;RHS is a well known test for determining the &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_%28geometry%29#Determining_congruence&quot; rel=&quot;nofollow&quot;&gt;congruency of triangles&lt;/a&gt;. It is easy enough to prove it works, simply use Pythagorus' theorem to reduce to SSS. I thought that it seems strange that this only works for an angle being 90 degrees - or does it? What if I tried changed the given angle to 89 degrees or 91 degrees, would it still be uniquely identified up to congruence?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-27T06:47:59.760" LastActivityDate="2010-08-05T19:59:38.613" Title="RHS Congruency test - What makes 90 degrees different?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;&lt;high-school&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="767" PostTypeId="2" ParentId="766" CreationDate="2010-07-27T06:46:59.903" Score="5" ViewCount="0" Body="&lt;p&gt;Supposing that we knew two triangles had one angle congruent, a side adjacent to the angle congruent, and the side opposite the angle congruent.  This is sometimes referred to as SSA, which is not a congruence theorem (and I've heard it said that it is &quot;ass-backwards&quot;).  With a little more information, it is possible to determine congruence in some instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you'd said about RHS, in that case you can use right-triangle trigonometry to determine that the unknown sides are congruent, then use SSS to establish congruence.  Without the right angle, the technique for determining the length of the third side of the triangle is to use the Law of Sines to determine the measure of the unknown angle opposite the known side, use that to find the measure of the third angle, then use the Law of Cosines to determine the length of the unknown side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call one of the triangles ABC with &amp;ang;A and AB and BC being known.  From the Law of Sines, $\frac{\sin A}{BC}=\frac{\sin C}{AB}$ or $\sin C=\frac{AB\cdot\sin A}{BC}$.  There will be two values of C in the range 0&amp;deg; to 180&amp;deg; that satisfy this equation, unless sin &lt;em&gt;C&lt;/em&gt; = 1.  So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if sin &lt;em&gt;C&lt;/em&gt; = 1, then C is a right angle, the triangle is uniquely determined, so congruence can be established;&lt;/li&gt;&#xA;&lt;li&gt;if A &amp;ge; 90&amp;deg;, C &amp;lt; 90&amp;deg;, so there is only one solution for C that makes sense in this triangle, the triangle is uniquely determined, and congruence can be established (one might call this SSobtuseA);&lt;/li&gt;&#xA;&lt;li&gt;if A &amp;lt; 90&amp;deg;, but BC &amp;ge; AB, then B &amp;ge; C (in a triangle, the largest/smallest side is opposite the largest/smallest angle), so the only solution for C that makes sense in this triangle is the one with C&amp;nbsp;&amp;lt;&amp;nbsp;90&amp;deg; (if B&amp;nbsp;&amp;ge;&amp;nbsp;C&amp;nbsp;&gt;&amp;nbsp;90&amp;deg;, then A + B + C &gt; 180&amp;deg;), the triangle is uniquely determined, and congruence can be established (one might call this SsA, with the relative sizes of the S/s indicating the relative lengths);&lt;/li&gt;&#xA;&lt;li&gt;otherwise (when A &amp;lt; 90&amp;deg; and BC &amp;lt; AB), there are two possible values for C, both of which lead to triangles, so there are two possible triangles satisfying the given information, and congruence cannot be established.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, to your specific question, if the angle were 91&amp;deg; (case 2), congruence would follow; if the angle were 89&amp;deg;, congruence may or may not follow, depending on what you can determine about the other sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, RHS is also commonly referred to (at least in the midwestern U.S. in contemporary high school geometry) as hypotenuse-leg or HL.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T19:59:38.613" LastActivityDate="2010-08-05T19:59:38.613" CommentCount="3" />
  <row Id="768" PostTypeId="1" AcceptedAnswerId="773" CreationDate="2010-07-27T08:02:16.660" Score="11" ViewCount="160" Body="&lt;p&gt;Pick's theorem says that given a square grid (that is, all points in the plane with integer coordinates) and a polygon without holes and non selt-intersecting whose vertices are grid points, its area is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i + b/2 - 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where i is the number of interior lattice points and b is the number of points on its boundary. Theorem and proof may be found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us suppose that the grid is not square but triangular (or hexagonal). Does a similar theorem hold?&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-30T08:06:23.087" Title="Pick's Theorem on a triangular (or hex) grid" Tags="&lt;geometry&gt;&lt;discrete-geometry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" />
  <row Id="769" PostTypeId="2" ParentId="664" CreationDate="2010-07-27T08:26:39.377" Score="1" ViewCount="0" Body="&lt;p&gt;$dm$ takes density fluctuations into account, it's just $dm = \rho(\vec r) d^3r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. for a homogeneous cylinder with the rotational axis align parallel to the z-axis it is $\rho = \frac{m}{V}\cdot\theta(R-r)\theta(z^2-a^2)$, where $\theta(x) = \begin{cases} 0 for x&amp;lt;0 and\ 1 for x&gt;0 \end{cases}$ and $r = \sqrt{x^2+y^2}$ is the radial coordinate in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;cylindrical coordinates&lt;/a&gt;. The result is that your integration over $r$ is from 0 to R and for z from -a to a. $V=\pi R^2 \cdot 2a$ is the volume of the cylinder, $m$ its weight. In cylindrical coordinates, $d^3r = r dr d\phi dz$, so you got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{r=0}^{R} \int_{z=-a}^a \int_{\phi=0}^{2\pi} r^2 \frac{m}{V} \cdot r d\phi dz dr = 4a\pi \frac{m}{V} \int_{r=0}^R r^3 dr = \frac{am\pi}{V}R^4 = \frac{1}{2}mR^2$.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-27T08:26:39.377" />
  <row Id="770" PostTypeId="2" ParentId="720" CreationDate="2010-07-27T08:39:51.483" Score="5" ViewCount="0" Body="&lt;p&gt;First, some background.  A &quot;line&quot; (the object containing the path of least distance between two points) on a sphere is a great circle; a great circle is the intersection of a sphere with a plane passing through the center of the sphere.  A spherical triangle is the triangle formed by the spherical line segments connecting three non-collinear points on a sphere.  Below is the Spherical Law of Cosines as it appears in UCSMP &lt;em&gt;Functions, Statistics, and Trigonometry&lt;/em&gt;, 3rd ed., copied here because the diagram is good and helps with clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KAOhC.png&quot; alt=&quot;Spherical Law of Cosines (from UCSMP Functions, Statistics, and Trigonometry, 3rd ed.)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ABC is a spherical triangle with arcs a, b, and c (meaning the measures of the arcs, not the lengths), then cos c = cos a * cos b + sin a * sin b * cos C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to the specific problem at hand.  Let's use the diagram below, also from UCSMP &lt;em&gt;Functions, Statistics, and Trigonometry&lt;/em&gt;, 3rd ed., for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DRkJ3.png&quot; alt=&quot;globe (from UCSMP Functions, Statistics, and Trigonometry, 3rd ed.)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let B be the initial point and A be the final point of the movement (and for simplicity, I'll assume they are both in the northern hemisphere, and leave extending the solution to any points as an exercise); N and S are the north and south poles, respectively; C and D are the points on the equator that are on the same line of longitude as A and B, respectively.  Consider spherical triangle ABN.  a = (90&amp;deg; - latitude of point A); b = (90&amp;deg; - latitude of point B). N = positive difference in longitude between points A and B.  Use the Spherical Law of Cosines (cos n = ... form) to determine n, then use the Spherical Law of Cosines (cos b = ... form) to determine B.  If the points are arranged as pictured, the heading of travel from B to A is 360&amp;deg; - B; for other arrangements of the points, the exact method of determining the heading based on the measure of angle B may vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(graphics from Lesson 5-10 of UCSMP Functions, Statistics, and Trigonometry, 3rd ed., © 2010 Wright Group/McGraw Hill)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T03:45:13.390" LastActivityDate="2010-09-29T03:45:13.390" CommentCount="1" />
  <row Id="771" PostTypeId="1" AcceptedAnswerId="775" CreationDate="2010-07-27T09:13:21.393" Score="2" ViewCount="262" Body="&lt;p&gt;I am trying to show the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1885087/Expectation.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I really can't remember what I am supposed to do to get from the LHS to the RHS. I have tried using integration this way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1885087/integ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then use integration by parts, but even though what I get resembles it, it can't be correct (because $e^{-\gamma W}$ is not the distribution of W).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also tried using Taylor series expansion, but I think I am way off, and I don't think an approximation here is what I need, because the equality above is exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, this is not homework, I am working through a &lt;a href=&quot;http://www.princeton.edu/~markus/research/papers/liquidity.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (page 10) and I would really like to know how every step was derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone at least point me to the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: This expectation on the RHS is very similar to the moment generating function formula (with a negative exponent). If you check &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function#Examples&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, you will see that the moment generating function for the normal distribution is like the LHS (but with a positive sign). So in a way I have my answer, but I still would like to know how to derive it, if there is a way. I know little if anything at all about moment generating functions, so maybe I shouldn't try and derive it but rather just use the result? Does it even make sense to try and derive it?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="85" LastEditorDisplayName="" LastEditDate="2010-08-04T09:53:01.523" LastActivityDate="2010-08-04T09:53:01.523" Title="Expected value of a function of a random variable: help!" Tags="&lt;probability-theory&gt;&lt;random-variable&gt;&lt;taylor-expansion&gt;&lt;expectation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="772" PostTypeId="2" ParentId="768" CreationDate="2010-07-27T09:22:36.813" Score="2" ViewCount="0" Body="&lt;p&gt;This is a very interesting question. I don't have a complete solution yet, but I did get some results. Consider an arbitrary distribution of points. Let P(i,b)=i+b/2-1 where i is the number of internal points and P is the number of boundary points. Let P(A)=P(i_A,b_A), where A is a simple polygon with all its vertices on the points. Wikipedia shows that P(C)=P(A)+P(T) where T is a triangle that shares a single edge with A and C a simple polygon formed by the union of A and T. Since all simple polygons can be triangulated, P(C)=sum P(t) for all t in the triangulation of C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof then has a second part that shows P(t) equals the area for any triangle. So if we want to generalise it for other grids, we have to find a property equal to P(T) for any triangle.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-27T09:22:36.813" CommentCount="1" />
  <row Id="773" PostTypeId="2" ParentId="768" CreationDate="2010-07-27T09:27:33.757" Score="8" ViewCount="0" Body="&lt;p&gt;The short answer is that, no, there can be no formula for polygons with vertices in the hexagonal lattice in terms of just boundary and interior points. This is based on the fact that primitive triangles on this lattice--ones with no lattice points on their boundary (besides the vertices) or in the interior--can have different areas, whereas for the square lattice all primitive triangles have area 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Casebash has partly gotten at in his answer, you can approximate things well if you compute what, in the below paper, is called the &quot;boundary characteristic&quot; of the polygon, a number that is somewhat complicated to think to compute, but which gives a decent proxy for how many of each type of primitive triangle the polygon contains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kolodziejczyk has been the main one doing work on hexagonal lattice results of this type that I know; he's worth looking up for similar results. Ding Ren is another, and the older work of Grunbaum, etc., still bears on the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A Fast Pick-Type Approximation for the Area of H-Polygons,&quot; Ren, Kolodziejczyk, et al., American Mathematical Monthly, 1993.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Available at &lt;a href=&quot;http://www.jstor.org/pss/2323889&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt; if you have access.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-27T09:27:33.757" />
  <row Id="774" PostTypeId="2" ParentId="764" CreationDate="2010-07-27T09:33:27.300" Score="2" ViewCount="0" Body="&lt;p&gt;To add to Akhil's answer, the Stieltjes integral made a lot more sense to me when I started thinking about it in terms of physics. Think of taking something like the integral you get to determine moment of inertia, where it's mass--a function of position--you're integrating over, and not position itself.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-27T09:33:27.300" />
  <row Id="775" PostTypeId="2" ParentId="771" CreationDate="2010-07-27T09:51:33.053" Score="2" ViewCount="0" Body="&lt;p&gt;If W is randomly chosen with the PDF P(x), then the expectation value should be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;http://mathcache.appspot.com/?tex=%5cpng%5c%5bE%5Be%5E%7B-%5Cgamma%20W%7D%5D%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20P%28x%29%20e%5E%7B-%5Cgamma%20x%7D%20dx%5c%5d&quot; alt=&quot;$E[e^{-\gamma W}]=\int_{-\infty}^\infty P(x) e^{-\gamma x} dx$&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I think that equation (E[e&lt;sup&gt;-&amp;gamma;W&lt;/sup&gt;] = e&lt;sup&gt;-&amp;gamma;(E[W] - &amp;frac12;&amp;gamma;Var[W])&lt;/sup&gt;) is correct only when W is a normal distribution.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-07-27T10:22:37.633" LastActivityDate="2010-07-27T10:22:37.633" CommentCount="3" />
  <row Id="776" PostTypeId="1" AcceptedAnswerId="777" CreationDate="2010-07-27T09:52:02.100" Score="3" ViewCount="561" Body="&lt;p&gt;There is a well known theorem often stated as the angle in a semi-circle being 90 degrees. To be more accurate, any triangle with one of its sides being a diameter and all vertices on the circle has its angle opposite the diameter being 90 degrees. The standard proof uses isosceles triangles and is worth having as an answer, but there is also a much more intuitive proof as well (this proof is more complicated though).&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-27T20:43:11.277" LastActivityDate="2010-08-06T14:15:55.930" Title="Proof of Angle in a Semi-Circle is 90 degrees" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="777" PostTypeId="2" ParentId="776" CreationDate="2010-07-27T10:02:15.607" Score="12" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Nonstandard proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the semi-circle with endpoints A and C and center O and the inscribed angle &amp;ang;ABC (B on the semi-circle) together with the rotation image of both about O by 180&amp;deg;.  The image of A is C and vice versa; let B' be the image of B.  The image of a line under a 180&amp;deg; rotation is parallel to the original line, AB is parallel to CB' and BC is parallel to B'A, so ABCB' is a parallelogram.  BO and its image must be parallel, but the image of O is itself, since it is the center of rotation, and if BO and B'O are parallel and contain a point in common, they must lie on the same line, so BB' passes through O.  AC and BB' (the diagonals of ABCB') are both diameters of the circle, so they are congruent.  A parallelogram with congruent diagonals is a rectangle.  Thus, &amp;ang;ABC is a right angle (and has measure 90&amp;deg;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/762828246.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Standard proof&lt;/strong&gt; (or, at least, my guess at it based on the description in the question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As above, consider the semi-circle with endpoints A and C and center O and the inscribed angle &amp;ang;ABC (B on the semi-circle).  Draw in radius OB.  OA = OB, so &amp;#x25B3;AOB is isosceles and &amp;ang;OAB&amp;cong;&amp;ang;OBA.  OB = OC, so &amp;#x25B3;BOC is isosceles and &amp;ang;OBC&amp;cong;&amp;ang;OCB.  Let &amp;alpha;=m&amp;ang;OAB=m&amp;ang;OBA and &amp;beta;=m&amp;ang;OBC=m&amp;ang;OCB.  In &#xA;&amp;#x25B3;ABC, the measures of the angles are &amp;alpha;, &amp;alpha;+&amp;beta;, and &amp;beta;, so &amp;alpha;+(&amp;alpha;+&amp;beta;)+&amp;beta;=180&amp;deg; or 2(&amp;alpha;+&amp;beta;)=180&amp;deg; or &amp;alpha;+&amp;beta;=90&amp;deg;, so &amp;ang;ABC has measure 90&amp;deg; and is a right angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/319527897.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit:&lt;/em&gt; &lt;strong&gt;Another Nonstandard proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the labeling as above and apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Stewart%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Stewart's Theorem&lt;/a&gt; to &amp;#x25B3;ABC: $$(AB)^2(OC) + (BC)^2(AO) = (AC)((BO)^2 + (AO)(OC))$$  Substituting the length r of the radius of the semicircle as appropriate: $$(AB)^2r + (BC)^2r = 2r(r^2 + r^2)=4r^3$$  Dividing both sides by r: $$(AB)^2+(BC)^2=(2r)^2=(AC)^2$$&#xA;So, by the converse of the Pythagorean Theorem, &amp;ang;ABC is a right angle.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T06:22:02.320" LastActivityDate="2010-08-04T06:22:02.320" CommentCount="13" />
  <row Id="778" PostTypeId="2" ParentId="756" CreationDate="2010-07-27T10:20:44.983" Score="6" ViewCount="0" Body="&lt;p&gt;Look at &lt;a href=&quot;http://mathoverflow.net/questions/32269&quot; rel=&quot;nofollow&quot;&gt;this answer&lt;/a&gt; on MathOverflow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, there is a way to guess a number asking 14 questions in worst case. To do it you need a linear code with length 14, dimension 10 and distance at least 3. One such code can be built based on Hamming code (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamming_code&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hamming_code&lt;/a&gt;).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here is the strategy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let us denote bits of first player's number as &lt;em&gt;a&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;i&lt;/em&gt; ∈ [1..10]. We start with asking values of all those bits. That is we ask the following questions: &quot;is it true that &lt;em&gt;i&lt;/em&gt;-th bit of your number is zero?&quot; Let us denote answers on those questions as &lt;em&gt;b&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;i&lt;/em&gt; ∈ [1..10].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now we ask 4 additional questions:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;5&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; is equal to zero? (⊗ is sumation modulo 2).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;6&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; is equal to zero?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;8&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; is equal to zero?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Is it true that &lt;em&gt;a&lt;sub&gt;5&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;6&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;8&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;a&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; is equal to zero?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let &lt;em&gt;q&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;q&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;q&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;q&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; be answers on those additional questions. Now second player calculates &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; (&lt;em&gt;i&lt;/em&gt; ∈ [1..4]) --- answers on those questions based on bits &lt;em&gt;b&lt;sub&gt;j&lt;/sub&gt;&lt;/em&gt; which he previously got from first player.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now there are 16 ways how bits &lt;em&gt;q&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; can differ from &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;. Let &lt;em&gt;d&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; = &lt;em&gt;q&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; ⊗ &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; (hence &lt;em&gt;d&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; = 1 iff &lt;em&gt;q&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; ≠ &lt;em&gt;t&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; ).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let us make table of all possible errors and corresponding values of &lt;em&gt;d&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;:&#xA;  position of error -&gt; (&lt;em&gt;d&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt;)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;no error -&gt; (0, 0, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 1, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 1, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;5&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;6&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;7&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 1, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;8&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 0, 1, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;9&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 1, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;b&lt;sub&gt;10&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 1, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; -&gt; (1, 0, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 1, 0, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 0, 1, 0)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;error in &lt;em&gt;q&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt; -&gt; (0, 0, 0, 1)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All the values of (&lt;em&gt;d&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;3&lt;/sub&gt;&lt;/em&gt;, &lt;em&gt;d&lt;sub&gt;4&lt;/sub&gt;&lt;/em&gt;) are different. Hence we can find where were an error and hence find all &lt;em&gt;a&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Answered by &lt;a href=&quot;http://mathoverflow.net/users/7079/falagar&quot; rel=&quot;nofollow&quot;&gt;falagar&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="227" LastEditorDisplayName="" LastEditDate="2010-07-27T21:49:17.880" LastActivityDate="2010-07-27T21:49:17.880" CommunityOwnedDate="2010-07-27T10:20:44.983" CommentCount="2" />
  <row Id="779" PostTypeId="2" ParentId="742" CreationDate="2010-07-27T11:02:18.070" Score="0" ViewCount="0" Body="&lt;p&gt;Let &lt;code&gt;t&lt;/code&gt; be your target and &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b &amp;lt; a&lt;/code&gt; be the values of the stamps you have. Let &lt;code&gt;n = ⌈t/a⌉&lt;/code&gt; be the maximum number of &lt;code&gt;a&lt;/code&gt; stamps it makes sense to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each number &lt;code&gt;i&lt;/code&gt; from &lt;code&gt;n&lt;/code&gt; to &lt;code&gt;0&lt;/code&gt;, consider the cost of the solution with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;i&lt;/code&gt; stamps of value &lt;code&gt;a&lt;/code&gt; and&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;j = ⌈(t - i*a) / b⌉&lt;/code&gt; stamps of value &lt;code&gt;b&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The cost of each solution would obviously be &lt;code&gt;i*a + j*b&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your solution is the pair &lt;code&gt;(i, ⌈(t - i*a) / b⌉ )&lt;/code&gt; that minimizes said cost. (Note there's only one unknown in that pair, &lt;code&gt;i&lt;/code&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional optimization would be to stop immediately as soon as a division without rest is performed, either when determining &lt;code&gt;n&lt;/code&gt; or &lt;code&gt;j&lt;/code&gt;. A division without rest means that we have found a solution that perfectly pays the due amount and cannot thus be improved on. (There may be others like it, but none that is actually better, so there's no need to keep searching).&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2010-07-27T11:02:18.070" CommentCount="1" />
  <row Id="780" PostTypeId="2" ParentId="631" CreationDate="2010-07-27T11:21:22.967" Score="2" ViewCount="0" Body="&lt;p&gt;This is still WIP. There are a few missing details, still I think it's better than nothing. Feel free to edit in the missing details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a problem of &lt;code&gt;SUBSET-SUM&lt;/code&gt;. We have a set of &lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,...,a&lt;sub&gt;n&lt;/sub&gt;} numbers, and another number &lt;code&gt;s&lt;/code&gt;. The question we're seeking answer to is, whether or not there's a subset of &lt;code&gt;A&lt;/code&gt; whose sum is &lt;code&gt;s&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the 24-game allows you to use rational numbers. Even if it doesn't, I think that it is possible to emulate rational numbers up to denominator of size &lt;code&gt;p&lt;/code&gt; with integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that &lt;code&gt;SUBSET-SUM&lt;/code&gt; is NP-complete even for integers only. I think the &lt;code&gt;SUBSET-SUM&lt;/code&gt; problem is &lt;code&gt;NP&lt;/code&gt;-hard even if you allow treating each a&lt;sub&gt;i&lt;/sub&gt; as a negative number. That is even if &lt;code&gt;A&lt;/code&gt; is of the form &lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;,-a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,-a&lt;sub&gt;2&lt;/sub&gt;,...,a&lt;sub&gt;n&lt;/sub&gt;,-a&lt;sub&gt;n&lt;/sub&gt;}. This is still a wrinkle I need to iron out in this reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if there's a subset of &lt;code&gt;A&lt;/code&gt; with sum &lt;code&gt;s&lt;/code&gt;, then there's a solution to the &lt;code&gt;24&lt;/code&gt;-problem for how to reach using &lt;code&gt;A&lt;/code&gt; to &lt;code&gt;s&lt;/code&gt;. The solution is only using the &lt;code&gt;+&lt;/code&gt; sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, what happens if there's no solution which only uses the &lt;code&gt;+&lt;/code&gt; sign, but there is a solution which uses other arithmetic operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider the following problem. Let's take a prime &lt;code&gt;p&lt;/code&gt; which is larger than &lt;code&gt;n&lt;/code&gt;, the total number of elements in &lt;code&gt;A&lt;/code&gt;. Given an oracle which solves the &lt;code&gt;24&lt;/code&gt;-problem, and a &lt;code&gt;SUBSET-SUM&lt;/code&gt; problem of &lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;,a&lt;sub&gt;2&lt;/sub&gt;,...,a&lt;sub&gt;n&lt;/sub&gt;} and &lt;code&gt;s&lt;/code&gt;. We'll ask the oracle to solve the &lt;code&gt;24&lt;/code&gt;-problem on&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;A&lt;/code&gt;={a&lt;sub&gt;1&lt;/sub&gt;+(1/p),a&lt;sub&gt;2&lt;/sub&gt;+(1/p),...,a&lt;sub&gt;n&lt;/sub&gt;+(1/p)}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for the following values:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt;=s+1/p,s&lt;sub&gt;2&lt;/sub&gt;=s+2/p,...,s&lt;sub&gt;n&lt;/sub&gt;=s+n/p.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the solution includes multiplication, we will have a denominator larger than &lt;code&gt;p&lt;/code&gt; in the end result, and thus we will not be able to reach any s&lt;sub&gt;i&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an arithmetric expression that contains a&lt;sub&gt;i&lt;/sub&gt;a&lt;sub&gt;j&lt;/sub&gt;=x+(1/p&lt;sup&gt;2&lt;/sup&gt;), It is impossible that the denominator p&lt;sup&gt;2&lt;/sup&gt; would &quot;cancel&quot; out, since there are at most &lt;code&gt;n&lt;/code&gt; elements in the summation, and thus the numerator would never reach &lt;code&gt;p&lt;/code&gt;, since &lt;code&gt;p&amp;gt;n&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THIS IS NOT QUIT RIGHT! The expression a&lt;sub&gt;i&lt;/sub&gt;a&lt;sub&gt;j&lt;/sub&gt;-a&lt;sub&gt;k&lt;/sub&gt;a&lt;sub&gt;l&lt;/sub&gt; will be an integer, and therefor our oracle might return answer which includes two multiplications one negative and one positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about division? How can we be sure no division will occur. Find another prime &lt;code&gt;q&lt;/code&gt; which is different than p, and larger than the largest a&lt;sub&gt;i&lt;/sub&gt; times &lt;code&gt;n&lt;/code&gt;. Multiply all answers by &lt;code&gt;q&lt;/code&gt;. The set &lt;code&gt;A&lt;/code&gt; will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;A&lt;/code&gt;={qa&lt;sub&gt;1&lt;/sub&gt;+(1/p),qa&lt;sub&gt;2&lt;/sub&gt;+(1/p),...,qa&lt;sub&gt;n&lt;/sub&gt;+(1/p)}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We will look for the following values:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt;=qs+1/p,s&lt;sub&gt;2&lt;/sub&gt;=qs+2/p,...,s&lt;sub&gt;n&lt;/sub&gt;=qs+n/p.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that case, a&lt;sub&gt;i&lt;/sub&gt;/a&lt;sub&gt;j&lt;/sub&gt; will be smaller than the minimal element in &lt;code&gt;A&lt;/code&gt;, and therefor the end result which will contains a&lt;sub&gt;i&lt;/sub&gt;/a&lt;sub&gt;j&lt;/sub&gt; will never be one of the s&lt;sub&gt;i&lt;/sub&gt; we're looking for.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2010-07-28T15:25:00.803" LastActivityDate="2010-07-28T15:25:00.803" CommentCount="8" />
  <row Id="781" PostTypeId="1" AcceptedAnswerId="863" CreationDate="2010-07-27T13:14:14.543" Score="4" ViewCount="226" Body="&lt;p&gt;Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Benford%27s_law&quot; rel=&quot;nofollow&quot;&gt;Benford's Law&lt;/a&gt; (if you take a list of values, the distribution of the most significant digit is rougly proportional to the logarithm of the digit) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Zipf%27s_law&quot; rel=&quot;nofollow&quot;&gt;Zipf's Law&lt;/a&gt; (given a corpus of natural language utterances, the frequency of any word is roughly inversely proportional to its rank in the frequency table) are not theorems in a mathematical sense, but they work quite good in the real life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have an idea why this happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see also &lt;a href=&quot;http://math.stackexchange.com/questions/58&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-10-13T18:03:48.933" Title="Why does Benford's Law (or Zipf's Law) hold?" Tags="&lt;soft-question&gt;&lt;statistics&gt;&lt;applications&gt;&lt;philosophy&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="782" PostTypeId="2" ParentId="781" CreationDate="2010-07-27T13:16:41.230" Score="4" ViewCount="0" Body="&lt;p&gt;For the case of Bendford's Law, of course scale invariance is a necessary condition; it the law must be true either if we measure things in meters or in feet or in furlongs, thus multiplying given data for a constant, the only distribution which allows this is the logarithmic one. But its being necessary does not mean that it is the answer, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scale invariance is not relevant for Zipf's Law, however, since we have an absolute rank.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-27T13:16:41.230" CommentCount="2" />
  <row Id="783" PostTypeId="1" AcceptedAnswerId="784" CreationDate="2010-07-27T17:19:25.283" Score="1" ViewCount="105" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in a category i have two objects $a$ and $b$ and a morphism $m$ from $a$ to $b$ and one $n$ from $b$ to $a$. Is this always an isomorphism? Why is it emphasized that this has to be true, too: $m \circ n$ = $id_b$ and $n \circ m$ = $id_a$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an example in which the id-part is not true and therefore $m$ and $n$ are not isomorph.&lt;/p&gt;&#xA;" OwnerUserId="179" LastActivityDate="2010-09-01T12:40:27.527" Title="isomorphism and $id$" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="784" PostTypeId="2" ParentId="783" CreationDate="2010-07-27T17:41:32.347" Score="4" ViewCount="0" Body="&lt;p&gt;If you have no restrictions on m and n, then clearly they cannot be isomorphisms in general.  For instance, take any two groups G and H and let m: G --&gt; H, n: H --&gt; G be the zero homomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you say that n and m are monomorphisms, then it is still not true in general that they are isomorphisms.  I believe it is true however if your category is one whose objects are sets with additional structure. See this question: &lt;a href=&quot;http://math.stackexchange.com/questions/309/if-a-is-a-subobject-of-b-and-b-a-subobject-of-a-are-they-isomorphic/327#327&quot; rel=&quot;nofollow&quot;&gt;http://math.stackexchange.com/questions/309/if-a-is-a-subobject-of-b-and-b-a-subobject-of-a-are-they-isomorphic/327#327&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-27T17:41:32.347" CommentCount="3" />
  <row Id="785" PostTypeId="1" AcceptedAnswerId="786" CreationDate="2010-07-27T18:31:44.133" Score="2" ViewCount="477" Body="&lt;p&gt;There is a general formula for solving quadratic equations, namely the Quadratic Formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For third degree equations of the form $ax^3+bx^2+cx+d=0$, there is a set of thee equations: one for each root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a general formula for solving equations of the form&#xA;$ax^4+bx^3+cx^2+dx+e=0$&#xA;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about for higher degrees? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2010-08-05T23:00:44.320" Title="Is there a general formula for solving 4th degree equations?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="786" PostTypeId="2" ParentId="785" CreationDate="2010-07-27T18:44:11.520" Score="9" ViewCount="0" Body="&lt;p&gt;There is, in fact, a general formula for solving quartic (4th degree polynomial) equations.  As the cubic formula is significantly more complex than the quadratic formula, the quartic formula is significantly more complex than the cubic formula.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_formula#Solving_a_quartic_equation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on quartic functions&lt;/a&gt; has a lengthy process by which to get the solutions, but does not give an explicit formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex numbers and there are two different ways to define such a principal root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There cannot be explicit algebraic formulas for the general solutions to higher-degree polynomials, but proving this requires mathematics beyond precalculus (it is typically proved with Galois Theory now, though it was originally proved with other methods).  This fact is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;the Abel-Ruffini theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of note, Wolfram &lt;a href=&quot;http://library.wolfram.com/examples/quintic/&quot; rel=&quot;nofollow&quot;&gt;sells a poster&lt;/a&gt; that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation.  This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  I believe that the formula given below gives the correct solutions for x to $ax^4+bx^3+cx^2+d+e=0$ for all complex a, b, c, d, and e, under the assumption that $w=\sqrt{z}$ is the complex number such that $w^2=z$ and $\arg(w)\in(-\frac{\pi}{2},\frac{\pi}{2}]$ and $w=\sqrt[3]{z}$ is the complex number such that $w^3=z$ and $\arg(w)\in(-\frac{\pi}{3},\frac{\pi}{3}]$ (these are typically how computer algebra systems and calculators define the principal roots).  Some intermediate parameters $p_k$ are defined to keep the formula simple and to help in keeping the choices of roots consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let: $$\begin{align*}&#xA;p_1&amp;amp;=2c^3-9bcd+27ad^2+27b^2e-72ace&#xA;\\&#xA;p_2&amp;amp;=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2}&#xA;\\&#xA;p_3&amp;amp;=\frac{c^2-3bd+12ae}{3a\sqrt[3]{\frac{p_2}{2}}}+\frac{\sqrt[3]{\frac{p_2}{2}}}{3a}&#xA;\end{align*}\quad\quad\quad\quad\begin{align*}&#xA;p_4&amp;amp;=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3}&#xA;\\&#xA;p_5&amp;amp;=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3&#xA;\\&#xA;p_6&amp;amp;=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4}&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: $$\begin{align}&#xA;x&amp;amp;=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2}&#xA;\\&#xA;\mathrm{or\ }x&amp;amp;=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2}&#xA;\\&#xA;\mathrm{or\ }x&amp;amp;=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2}&#xA;\\&#xA;\mathrm{or\ }x&amp;amp;=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These came from having Mathematica explicitly solve the quartic, then seeing what common bits could be pulled from the horrifically-messy formula into parameters to make it readable/useable.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T23:00:44.320" LastActivityDate="2010-08-05T23:00:44.320" CommentCount="4" />
  <row Id="787" PostTypeId="2" ParentId="785" CreationDate="2010-07-27T18:44:52.423" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_formula#The_general_case.2C_along_Ferrari.27s_lines&quot; rel=&quot;nofollow&quot;&gt;Yes&lt;/a&gt;, there is a quartic formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such solution by radicals for higher degrees. This is a result of Galois theory, and follows from the fact that the symmetric group $S_5$ is not solvable. It is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Insolubility_of_the_quintic&quot; rel=&quot;nofollow&quot;&gt;Abel's theorem&lt;/a&gt;. In fact, there are specific fifth-degree polynomials whose roots cannot be obtained by using radicals from $\mathbb{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-27T18:44:52.423" />
  <row Id="788" PostTypeId="1" AcceptedAnswerId="792" CreationDate="2010-07-27T19:28:44.280" Score="3" ViewCount="261" Body="&lt;p&gt;The question that follows was inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic formula is much more simple than the quartic formula. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;That the general solutions to various polynomial equations are so complex and difficult to derive seems to suggest a fundamental limitation in the problem solving capabilities of the mathematical machinery. Does this intuition of mine make any sense? What should I make of it?&lt;/li&gt;&#xA;&lt;li&gt;Why is it that with each successive degree in a polynomial equation, the solution becomes so much more complex? Can I gain some intuition about what makes finding the roots so hard?&lt;/li&gt;&#xA;&lt;li&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Ruffini theorem&lt;/a&gt;: &quot;there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher.&quot; What is so special about the quintic that makes it the cut-off for finding a general algebraic solution?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-07-27T19:37:28.737" LastActivityDate="2010-07-27T20:33:23.530" Title="Why is it so hard to find the roots of polynomial equations?" Tags="&lt;intuition&gt;&lt;polynomials&gt;&lt;roots&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="789" PostTypeId="2" ParentId="707" CreationDate="2010-07-27T20:21:09.660" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;First, sticking to strings of length n=2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(10)=p_i (1-p_i)=\frac{i}{m+1}\left (1-\frac{i}{m+1}  \right )=\frac{i(m+1-i)}{(m+1)^2}$ and $P(11)=p_i^2=\left (\frac{i}{m+1}  \right )^2=\frac{i^2}{(m+1)^2}$; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{P(10)+P(11)}{\sum_{i=1}^{m}(P(10)+P(11))} \right )$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}{\sum_{i=1}^{m}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(i(m+1-i)+i^2\right )}{\sum_{i=1}^{m}\left(i(m+1-i)+i^2\right )}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}$ $=\frac{\left(\frac{m(m+1)(2m+1)}{6} \right )}{(m+1)\frac{m(m+1)}{2}}$ $=\frac{2m+1}{3(m+1)}$; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For general n:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(1...)=p_i\sum_{k=0}^{n-1}\left({n-1 \choose k}p_i^k(1-p_i)^{n-1-k} \right )=p_i$ (yes, P(1...) does not depend on n at all!);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}p_i\left(\frac{P(1...)}{\sum_{i=1}^{m}P(1...)} \right )=\sum_{i=1}^{m}p_i\left(\frac{p_i}{\sum_{i=1}^{m}p_i} \right )$ $=\frac{\sum_{i=1}^{m}p_i^2}{\sum_{i=1}^{m}p_i}=\frac{\sum_{i=1}^{m}\left(\frac{i}{m+1} \right )^2}{\sum_{i=1}^{m}\frac{i}{m+1}}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}=\frac{2m+1}{3(m+1)}$ (no surprise there, given that $P(1...)$ doesn't depend on n);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-03T02:22:09.930" LastActivityDate="2010-08-03T02:22:09.930" CommentCount="0" />
  <row Id="790" PostTypeId="1" AcceptedAnswerId="824" CreationDate="2010-07-27T20:26:21.083" Score="1" ViewCount="123" Body="&lt;p&gt;For $n &gt; 1$ an integer, there are well-known formulas for volume of the balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the analogous statement in a Banach space/Hilbert Space?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T13:04:01.453" LastActivityDate="2010-08-05T13:28:22.340" Title="Volume of the sphere in a Banach space" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="791" PostTypeId="1" AcceptedAnswerId="913" CreationDate="2010-07-27T20:32:42.910" Score="3" ViewCount="87" Body="&lt;p&gt;What is the volume of intersection of the three cylinders with axes of length $1$ in $x, y, z$ directions starting from the origin, and with radius $1$?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T04:29:55.297" Title="How to compute the volume of this object via integration?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="792" PostTypeId="2" ParentId="788" CreationDate="2010-07-27T20:33:23.530" Score="4" ViewCount="0" Body="&lt;p&gt;When you try to solve a degree n equation, there are n roots you have to find (in principal) and none of them is favoured over any of the others, which (in some metaphorical sense) means that you have to break an n-fold symmetry in order to write down the roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the symmetry group of the n roots becomes more and more complicated the larger n is.&#xA;For n = 2 it is abelian (and very small!); for n = 3 and 4 it is still solvable (in the technical sense of group theory), which explains the existence of an explicit formula involving radicals (this is due to Galois, and is a part of so-called Galois theory); for n = 5 or more this group is non-solvable (in the technical sense of group theory), and this corresponds to the fact that there is no explicit formula involving radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: The complexity of the symmetry group of the n roots leads to a corresponding complexity in explicitly solving the equation.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-27T20:33:23.530" />
  <row Id="793" PostTypeId="1" AcceptedAnswerId="813" CreationDate="2010-07-27T20:36:51.667" Score="0" ViewCount="100" Body="&lt;p&gt;How to prove that the exponential function and the logarithm function are the inverses of each other? I want it the following way. We must use the definition as power series, and must verify that all the terms of the composition except the coefficent of $z$ vanish, and that the first degree term is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write down the proof for the coefficient of $x^n$ for arbitrary but fixed $n$ by explicit verification. But how to settle this for all $n$ at one go?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-27T21:14:16.983" Title="Exponential and log functions compose to identity" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="794" PostTypeId="2" ParentId="462" CreationDate="2010-07-27T20:38:58.120" Score="1" ViewCount="0" Body="&lt;p&gt;I don't think the people working in higher-order theorem proving really care about Henkin semantics or models in general, they mostly work with their proof calculi. As long as there are no contradictions or other counterintuitive theorems they are happy. The most important and most difficult theorem they prove is usually that their proof terms terminate, which IIRC can be viewed as a form of soundness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Henkin semantics is most interesting for people trying to extend their first-order methods to higher-order logic, because it behaves essentially like models of first-order logic. Henkin semantics is somewhat weaker than what you would get with standard set-theoretic semantics, which by Gödels incompleteness theorem can't have a complete proof calculus. I think type theories should lie somewhere in between Henkin and standard semantics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where does typed lambda calculus come into proof verification?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To prove some implication &lt;code&gt;P(x) --&amp;gt; Q(x)&lt;/code&gt; with some free variables &lt;code&gt;x&lt;/code&gt; you need to map any proof of &lt;code&gt;P(x)&lt;/code&gt; to a proof of &lt;code&gt;Q(x)&lt;/code&gt;. Syntactically a map (a function) can be represented as a lambda term.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any other approaches than higher order logic to proof verification?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can also verify proofs in first-order or any other logic, but then you would loose much of the power of the logic. First-order logic is mostly interesting because it is possible to automatically find proofs, if they are not too complicated. The same applies even more to propositional logic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the limitations/shortcomings of existing proof verification systems (see below)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more powerful the logic becomes the harder it becomes to construct proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the systems are freely available I suggest you play with them, e.g. Isabelle and Coq for a start.&lt;/p&gt;&#xA;" OwnerUserId="210" LastEditorUserId="210" LastEditorDisplayName="" LastEditDate="2010-07-28T06:12:18.627" LastActivityDate="2010-07-28T06:12:18.627" CommentCount="2" />
  <row Id="795" PostTypeId="1" AcceptedAnswerId="867" CreationDate="2010-07-27T20:43:00.637" Score="3" ViewCount="89" Body="&lt;p&gt;There is a  theorem of Riemann to that effect. How to prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This was asked by Kenny in the beta for &quot;calculus&quot;.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T00:23:58.393" Title="How to prove that a conditionally convergent series can be rearranged to sum to any real number?" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="796" PostTypeId="1" CreationDate="2010-07-27T20:44:24.130" Score="3" ViewCount="88" Body="&lt;p&gt;What class of Partial Differential Equations can be solved using the method of separation of variables?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T04:27:18.310" Title="Separation of variables for diff eqns" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="797" PostTypeId="1" AcceptedAnswerId="817" CreationDate="2010-07-27T20:46:30.707" Score="14" ViewCount="375" Body="&lt;p&gt;There are so many available bases. Why is the strange number 'e' preferred over all else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one could integrate 1/x and see this. But is there more to the story?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-11-09T08:05:09.417" Title="What's so &quot;natural&quot; about the base of natural logarithms?" Tags="&lt;calculus&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="798" PostTypeId="1" CreationDate="2010-07-27T20:48:40.110" Score="0" ViewCount="57" Body="&lt;p&gt;There are quite simple, intuitive and straightforward expressions for evaluating the area or volume of a figure. But why is the expression for the length of a curve so complicated?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-27T21:15:23.240" Title="Why is the integral expression for the length of a curve more complicated than the expression for area or volume of a figure rotated?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="799" PostTypeId="1" AcceptedAnswerId="821" CreationDate="2010-07-27T20:51:34.517" Score="1" ViewCount="101" Body="&lt;p&gt;Ok, Ok, I know that in fact the discriminant is defined (up to sign) as a product of differences of the roots of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does it then have integral coefficients, if the polynomial you started with had integer coefficients?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T13:25:57.633" LastActivityDate="2010-07-28T13:25:57.633" Title="Why does the discriminant of an integral polynomial have integral coefficients?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="800" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T20:57:59.073" Score="19" ViewCount="0" Body="&lt;p&gt;Differentiation and integration is precisely why it is considered natural, but not just because $\int 1/x dx=ln x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^x$ has the two following nice properties&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ d/dx e^x=e^x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \int e^x dx=e^x+c $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we looked at a^x instead, we would get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d/dx a^x= d/dx e^(ln a*x)=ln a * a^x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int a^x dx= integral e^(ln a*x) dx=a^x/(ln x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So e is vital to the integration and differentiation of exponentials.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-11-09T08:05:09.417" LastActivityDate="2010-11-09T08:05:09.417" />
  <row Id="801" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T20:58:18.043" Score="2" ViewCount="0" Body="&lt;p&gt;The wikipedia page on e, &lt;a href=&quot;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29&lt;/a&gt;, has many properties of the constant that make it naturally occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the biggest reason it is natural when it comes to exponentiation/logarithms is that it is the only number that satisfies d/dt e^t = e^t, while every other number satisfies d/dt a^t = c a^t, where c is some constant (different than 1).  This makes it &quot;normalized&quot; in a sense.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-27T20:58:18.043" />
  <row Id="802" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T20:58:19.620" Score="1" ViewCount="0" Body="&lt;p&gt;A few months ago when I was learning LaTeX, I wrote a quick page to test its math rendering. I never though it'd help anyone, but here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dl.dropbox.com/u/977994/Stack/Logarithmic%20Growth%20Model.pdf&quot; rel=&quot;nofollow&quot;&gt;Logarithmic Growth Model&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first sentence is key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The logarithmic growth model is defined by its proportionality of rate of growth to population size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-07-27T20:58:19.620" CommentCount="2" />
  <row Id="803" PostTypeId="1" CreationDate="2010-07-27T20:58:47.027" Score="12" ViewCount="699" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow&quot;&gt;Pythagorean Theorem&lt;/a&gt; is one of the most popular to prove by mathematicians, and there are &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;many proofs available&lt;/a&gt; (including one from &lt;a href=&quot;http://en.wikipedia.org/wiki/James_A._Garfield&quot; rel=&quot;nofollow&quot;&gt;James Garfield&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the most elegant proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite is this graphical one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/1/16/Pythagorean_Proof_%283%29.PNG/800px-Pythagorean_Proof_%283%29.PNG&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to cut-the-knot:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Loomis (pp. 49-50) mentions that the&#xA;  proof &quot;was devised by Maurice Laisnez,&#xA;  a high school boy, in the&#xA;  Junior-Senior High School of South&#xA;  Bend, Ind., and sent to me, May 16,&#xA;  1939, by his class teacher, Wilson&#xA;  Thornton.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The proof has been published by Rufus&#xA;  Isaac in Mathematics Magazine, Vol. 48&#xA;  (1975), p. 198.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="225" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:47:53.663" LastActivityDate="2010-11-18T18:51:15.880" CommunityOwnedDate="2010-07-27T21:20:09.110" Title="What is the most elegant proof of the Pythagorean theorem?" Tags="&lt;geometry&gt;&lt;big-list&gt;&lt;algebra-precalculus&gt;&lt;euclidean-geometry&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="5" />
  <row Id="804" PostTypeId="1" CreationDate="2010-07-27T21:00:07.103" Score="2" ViewCount="207" Body="&lt;p&gt;I have seen and studied this class equation for a finite group acting on itself by conjugations. The only applications I know are Cauchys' theorm and Sylow's theorem. Are there more?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:46:59.970" LastActivityDate="2010-08-04T00:44:09.747" Title="Applications for the class equation from group theory" Tags="&lt;big-list&gt;&lt;abstract-algebra&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="805" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T21:00:42.513" Score="9" ViewCount="0" Body="&lt;p&gt;The wikipedia article on &lt;a href=&quot;http://bit.ly/FtPFd&quot; rel=&quot;nofollow&quot;&gt;e&lt;/a&gt; tells a bit of the story.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One example is an account that starts with 1.00 and pays 100% interest per year. If the interest is credited once, at the end of the year, the value is 2.00; but if the interest is computed and added twice in the year, the 1 is multiplied by 1.5 twice, yielding 1.00×1.5² = $2.25. Compounding quarterly yields 1.00×1.25&lt;sup&gt;4&lt;/sup&gt; = 2.4414…, and compounding monthly yields 1.00×(1.0833…)&lt;sup&gt;12&lt;/sup&gt; = 2.613035….&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Bernoulli noticed that this sequence approaches a limit (the force of interest) for more and smaller compounding intervals. Compounding weekly yields 2.692597…, while compounding daily yields 2.714567…, just two cents more. Using n as the number of compounding intervals, with interest of 100%/n in each interval, the limit for large n is the number that came to be known as e; with continuous compounding, the account value will reach 2.7182818…. More generally, an account that starts at $1, and yields (1+R) dollars at simple interest, will yield e&lt;sup&gt;R&lt;/sup&gt; dollars with continuous compounding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally, it is the base of the exponential function &lt;code&gt;y = k^x&lt;/code&gt;, finding a specific value for &lt;code&gt;k&lt;/code&gt; where &lt;code&gt;d/dx k^x = k^x&lt;/code&gt;.  That is, the rate of change of the exponential curve at any point is equal to the y value of the curve at that point.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-07-27T21:00:42.513" />
  <row Id="806" PostTypeId="1" AcceptedAnswerId="933" CreationDate="2010-07-27T21:01:27.537" Score="3" ViewCount="243" Body="&lt;p&gt;There is the notion of class number from algebraic number theory. Why is such a notion defined and what good comes out of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is nice if it is $1$; we have unique factorization of all ideals; but otherwise?&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T15:26:24.330" LastActivityDate="2010-08-04T00:19:10.063" Title="Applications of class number " Tags="&lt;algebraic-number-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="807" PostTypeId="2" ParentId="799" CreationDate="2010-07-27T21:02:54.693" Score="4" ViewCount="0" Body="&lt;p&gt;The square of the discriminant is &lt;em&gt;symmetric&lt;/em&gt; in the roots of the polynomial; if you permute the roots, it stays the same.  By the &lt;a href=&quot;http://mathworld.wolfram.com/FundamentalTheoremofSymmetricFunctions.html&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of symmetric functions&lt;/a&gt;, that means it can be expressed as a polynomial, with integer coefficients, in the coefficients of the original polynomial.  (I am using both &quot;coefficients&quot; and &quot;polynomial&quot; in two senses here, so let me know if this doesn't make sense.)  This is a basic observation which will become important if you ever study Galois theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example will probably make this clearer.  Suppose I have a quadratic polynomial $x^2 + bx + c$ with two roots $r_1, r_2$.  Then $x^2 + bx + c = (x - r_1)(x - r_2)$, so $b = - r_1 - r_2$ and $c = r_1 r_2$.  These are the &lt;em&gt;elementary symmetric functions&lt;/em&gt; in two variables, and the theorem above implies that every polynomial function of $r_1$ and $r_2$ which is invariant under switching the two is actually a polynomial in $b$ and $c$.  For example, the discriminant is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(r_1 - r_2)^2 = r_1^2 - 2r_1 r_2 + r_2^2 = (r_1 + r_2)^2 - 4r_1 r_2 = b^2 - 4c.$$&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-27T21:17:44.847" LastActivityDate="2010-07-27T21:17:44.847" CommentCount="3" />
  <row Id="808" PostTypeId="2" ParentId="795" CreationDate="2010-07-27T21:04:59.073" Score="7" ViewCount="0" Body="&lt;p&gt;Let the number be r.  The idea is that you first add all the positive elements of the sequence in order, until you get over r, then you add all the negative elements until you get under r, and then all the positive until you get over r, and so forth.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-27T21:04:59.073" />
  <row Id="809" PostTypeId="2" ParentId="798" CreationDate="2010-07-27T21:05:07.433" Score="0" ViewCount="0" Body="&lt;p&gt;Obviously this depends on the shape, as you implied. I would say for a circle the circumference formula is simpler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the length of the curve &lt;strong&gt;is a volume&lt;/strong&gt; in some sense, its just 1 dimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually in calculus we are calculating the area relative to the the x-axis, and your formula is in terms of x, so that makes things simpler. I imagine that if the curve was parameterized in terms of its arc length finding its length would be quite trivial, and finding the area between the curve and the x-axis would seem very complicated.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-27T21:15:23.240" LastActivityDate="2010-07-27T21:15:23.240" />
  <row Id="811" PostTypeId="2" ParentId="803" CreationDate="2010-07-27T21:06:36.277" Score="5" ViewCount="0" Body="&lt;p&gt;The proof that uses the fact that shearing a parallelogram parallel to one of its sides preserves area is my favorite.  Here's an animation from &lt;a href=&quot;http://www.arseweb.com/rupe/pythag/proof.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/W9r8D.gif&quot; alt=&quot;animated proof&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:17:50.523" LastActivityDate="2010-09-29T04:17:50.523" CommunityOwnedDate="2010-09-30T10:29:54.723" CommentCount="1" />
  <row Id="812" PostTypeId="2" ParentId="798" CreationDate="2010-07-27T21:08:31.293" Score="3" ViewCount="0" Body="&lt;p&gt;The length of a curve doesn't transform in a nice way under scaling of one of the variables the same way that the area of a 2d figure or the volume of a 3d figure does.  That puts a hard lower bound on how complicated it has to be.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:08:31.293" />
  <row Id="813" PostTypeId="2" ParentId="793" CreationDate="2010-07-27T21:14:16.983" Score="3" ViewCount="0" Body="&lt;p&gt;There is a combinatorial proof.  If $A(x) = \sum_{n \ge 0} a_n \frac{x^n}{n!}$ is an exponential generating function for the number of ways $a_n$ to put a certain structure on a set of size $n$, where $a_0 = 0$, then $\exp A(x) = \sum_{n \ge 0} b_n \frac{x^n}{n!}$ has coefficients $b_n$ which count the number of ways to split a set of size $n$ into subsets, each of which is given an $A$-structure.  For example, $A(x) = e^x - 1$ is the structure of &quot;being a nonempty set,&quot; and $\exp A(x)$ counts the number of ways to partition a set into nonempty subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\log \frac{1}{1 - x} = \sum_{n \ge 1} \frac{x^n}{n}$ can be thought of as the structure of &quot;being a cycle,&quot; which is to say there are $(n-1)!$ ways to arrange $n$ objects into a cycle $s_1 \to s_2 \to ... \to s_n \to s_1$.  So $\exp \log \frac{1}{1 - x}$ counts the number of ways to split $n$ objects up into disjoint cycles, which is the same as the number of permutations of $n$ by cycle decomposition.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp \log \frac{1}{1 - x} = \frac{1}{1 - x}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A generalization of this argument lets you compute the &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;cycle index polynomials of the symmetric groups&lt;/a&gt;.  Note that all of the above manipulations take place in the ring of formal power series over $\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:14:16.983" CommentCount="1" />
  <row Id="814" PostTypeId="2" ParentId="804" CreationDate="2010-07-27T21:18:22.197" Score="6" ViewCount="0" Body="&lt;p&gt;A nice application is Wedderburn's theorem: every finite skewfield is necessarily commutative. Here a skewfield is something which satisfies the same axioms as field, except that multiplication is not required to be commutative; the typical example are quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, let $F$ be a finite skewfield, $Z$ be its center. It is easy to see that $Z$ is a field, hence it must be $\mathbb{F}_q$ for some prime power $q = p^k$. $F$ will then be a vector space of finite dimension $n$ over $\mathbb{F}_q$, hence it will have $q^n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write the class equation for the multiplicative group of $F$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q^n - 1 = q - 1 + \sum_i \frac{q^n - 1}{q^{t_i} - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $q - 1$ appears as the cardinality of the center, while the sum extends over a set of representatives of the non-trivial conjugacy classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for $q^{t_i} - 1$ to divide $q^n - 1$, $t_i$ must divide $n$. Indeed the order of $q$ modulo $q^{t_i} - 1$ is $t_i$, and $q^n = 1 \pmod{q^{t_i} - 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $f_n$ be the $n$-th cyclotomic polynomial. Then $f_n(q)$ divides $q^n - 1$ and it also divides each term in the sum, so $f_n(q)$ divides $q - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is impossible unless $n = 1$ and the sum is empty, in which case $F$ is commutative. Indeed $f_n(q)$ is a product of terms of the form $q - \omega$, where $\omega$ is a root of unity, and this product will have bigger absolute value than $q - 1$ as soon as $n &gt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditorDisplayName="" LastEditDate="2010-08-03T12:49:13.113" LastActivityDate="2010-08-03T12:49:13.113" CommentCount="3" />
  <row Id="815" PostTypeId="1" CreationDate="2010-07-27T21:18:45.677" Score="4" ViewCount="434" Body="&lt;p&gt;This question is lifted from &lt;a href=&quot;http://mathoverflow.net/questions/5499/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow.&lt;/a&gt;. I feel it belongs here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are mathematicians whose creativity, insight and taste have the power of driving anyone into a world of beautiful ideas, which can inspire the desire, even the need for doing mathematics, or can make one to confront some kind of problems, dedicate his life to a branch of math, or choose an specific research topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this kind of force must not be underestimated; on the contrary, we have the duty to take advantage of it in order to improve the mathematical education of those who may come after us, using the work of those gifted mathematicians (and even their own words) to inspire them as they inspired ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm interested on knowing who (the mathematician), when (in which moment of your career), where (which specific work) and why this person had an impact on your way of looking at math. Like this, we will have an statistic about which mathematicians are more akin to appeal to our students at any moment of their development. Please, keep one mathematician for post, so that votes are really representative.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2011-01-07T14:48:06.590" CommunityOwnedDate="2010-07-27T21:24:12.777" Title="Which mathematicians have influenced you the most?" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="1" />
  <row Id="816" PostTypeId="2" ParentId="740" CreationDate="2010-07-27T21:24:11.607" Score="2" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_distribution&quot; rel=&quot;nofollow&quot;&gt;Cauchy distribution&lt;/a&gt; is another example.  This distribution pops up frequently in statistical reasoning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the ratio of two independent standard normal random variables is a standard Cauchy variable.  We calculate ratios all the time, but often forget to consider that their distribution does not follow a simple normal distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is interesting for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The mean and variance for the Cauchy distribution are undefined (--&gt; infinity).  If one is trying to estimate these parameters for the ratio of two normal variables, the results may blow up to rather large values that are hard to interpret.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In such a situation, one would therefore prefer to use more robust estimators of the central tendency (such as the median) and scale (median absolute deviation).  When designing and testing robust estimators (en.wikipedia.org/wiki/Robust_statistics), one way to test them is to try them out on a Cauchy distribution, and make sure that they don't blow up like the usual formulas for mean and variance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="234" LastActivityDate="2010-07-27T21:24:11.607" CommunityOwnedDate="2010-07-27T21:24:11.607" CommentCount="1" />
  <row Id="817" PostTypeId="2" ParentId="797" CreationDate="2010-07-27T21:28:11.640" Score="9" ViewCount="0" Body="&lt;p&gt;If you know some linear algebra, then here is an abstract reason: $e^x$ is the unique eigenvector of eigenvalue $1$ of the derivative $D$ acting on, say, the space of smooth functions on $\mathbb{R}$.  Why is this important?  The study of solutions of linear differential equations with constant coefficients is equivalent to the study of nullspaces of operators which are polynomials in $D$, e.g. operators of the form $\sum a_k D^k$.  Any such operator automatically commutes with $D$, so this nullspace splits up into eigenspaces of $D$.  That's why solutions to linear differential equations with constant coefficients can be expressed as sums of complex exponentials.  The choice of $e$ makes it particularly easy to see what the eigenvalue is: the eigenvalue of the eigenvector $e^{\lambda x}$ is $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:28:11.640" CommentCount="7" />
  <row Id="818" PostTypeId="2" ParentId="806" CreationDate="2010-07-27T21:33:35.587" Score="5" ViewCount="0" Body="&lt;p&gt;We say that a prime p is regular if it does not divide the class number of the p-th cyclotomic field. For regular primes, it is easy to prove Fermat's last theorem, as outlined for instance in &lt;a href=&quot;http://www.jmilne.org/math/CourseNotes/ant.html&quot; rel=&quot;nofollow&quot;&gt;Milne's notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, everything would be easy if the class number was 1, in which case one could use unique factorization. If the class number is prime to p, you use the fact that every ideal whose p-th power is principal must itself be principal. This, together with unique factorization for ideals, turns out to be all that you need in the naif proof.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-07-27T21:33:35.587" />
  <row Id="819" PostTypeId="2" ParentId="803" CreationDate="2010-07-27T21:35:14.013" Score="4" ViewCount="0" Body="&lt;p&gt;This is not really a proof, in the sense that it depends on certain physical assumptions, but it forces you to think very hard about what those physical assumptions are.  I learned it from Mark Levi's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691140200&quot; rel=&quot;nofollow&quot;&gt;The Mathematical Mechanic&lt;/a&gt;, although unfortunately I don't have a diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a fish tank in the shape of a triangular prism, where the triangle on top has vertices $A, B, C$ ($C$ is the right angle) and the triangle on the bottom has opposite vertices $A', B', C'$.  Let $a, b, c$ denote the lengths of the opposite sides.  Run a rod through the edge $A A'$ on which the fish tank is allowed to pivot, and fill it with water.  By conservation of energy, this system is in mechanical equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I claim (and it is not hard to see) that the total torque about the rod is proportional to $c^2 - b^2 - a^2$, each contribution coming from the pressure of the water in the tank against the corresponding side.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:35:14.013" CommunityOwnedDate="2010-07-27T21:35:14.013" CommentCount="2" />
  <row Id="820" PostTypeId="2" ParentId="640" CreationDate="2010-07-27T21:48:44.120" Score="12" ViewCount="0" Body="&lt;p&gt;As Akhil mentions, the keyword is elliptic regularity.  Since I don't know anything about this, let me just say some low-level things and maybe they'll make sense to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A differentiable function $f : \mathbb{R} \to \mathbb{R}$ can be thought of as a function which behaves locally like a linear function $f(x) = ax + b$.  So, very roughly, it is a collection of tiny vectors which fit together.  These tiny vectors can, however, fit together in a very erratic manner.  That's because since you only have to fit one vector to the two vectors that are its neighbors, there is a lot of room for bad behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A differentiable function $f : \mathbb{C} \to \mathbb{C}$ has to satisfy a much more stringent requirement: locally, it has to behave like a linear function $f(z) = az + b$ where $z, a, b$ are complex, which is a &lt;strong&gt;rotation&lt;/strong&gt; (and scale, and translation).  So, very roughly, it is a collection of tiny rotations which fit together.  Now one rotation has a continuum of neighbors to worry about, and it becomes much harder for erratic behavior to persist.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T21:48:44.120" CommentCount="1" />
  <row Id="821" PostTypeId="2" ParentId="799" CreationDate="2010-07-27T21:53:18.830" Score="3" ViewCount="0" Body="&lt;p&gt;Another definition of discriminant is as the resultant of the polynomial with it's first derivative (up to a scalar), and the resultant of two polynomials vanishes if and only if they have a common root.  So when does a polynomial have a common root with it's derivative? That's when a zero is also a local max or min, precisely a multiple root.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T21:53:18.830" />
  <row Id="822" PostTypeId="2" ParentId="785" CreationDate="2010-07-27T21:58:48.933" Score="0" ViewCount="0" Body="&lt;p&gt;Regarding the inability to solve the quintic, this is sort-of true and sort-of false.  No, there is no general solution in terms of $+$, $-$, $\times$ and $\div$, along with $\sqrt[n]{}$.  However, if you allow special theta values (a new operation, not among the standard ones!) then yes, you can actually write down the solutions of arbitrary polynomials this way.  Also, you can do construct lengths equal to the solutions by intersecting lower degree curves (for a quintic, you can do so with a &lt;a href=&quot;http://mathworld.wolfram.com/TridentofDescartes.html&quot; rel=&quot;nofollow&quot;&gt;trident&lt;/a&gt; and a circle.)&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T21:58:48.933" CommentCount="1" />
  <row Id="823" PostTypeId="2" ParentId="640" CreationDate="2010-07-27T22:00:11.930" Score="7" ViewCount="0" Body="&lt;p&gt;The existence of a complex derivative means that locally a function can only rotate and expand.  That is, in the limit, disks are mapped to disks.  This rigidity is what makes a complex differentiable function infinitely differentiable, and even more, analytic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a complex derivative to exist, it has to exist and have the same value for all ways the &quot;h&quot; term can go to zero in (f(z+h) - f(z))/h.  In particular, h could approach 0 along any radial path, and that's why an analytic function must map disks to disks in the limits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, an infinitely differentiable function of two real variables could map a disk to an ellipse, stretching more in one direction than another.  An analytic function can't do that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A smooth function of two variables could also flip a disk over, such as f(x, y) = (x, -y).  An analytic function can't do that either.  That's why complex conjugation is not an analytic function.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-07-28T17:09:07.587" LastActivityDate="2010-07-28T17:09:07.587" />
  <row Id="824" PostTypeId="2" ParentId="790" CreationDate="2010-07-27T22:02:00.617" Score="1" ViewCount="0" Body="&lt;p&gt;Maybe this is helpful, a formula for the &lt;a href=&quot;http://www.johndcook.com/blog/2010/07/02/volumes-of-generalized-unit-balls/&quot; rel=&quot;nofollow&quot;&gt;volume of Lp balls in R^n&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-27T22:02:00.617" CommentCount="2" />
  <row Id="825" PostTypeId="2" ParentId="600" CreationDate="2010-07-27T22:02:50.807" Score="4" ViewCount="0" Body="&lt;p&gt;I wrote a series of blog posts which explains how to solve questions like this; the relevant one is &lt;a href=&quot;http://qchu.wordpress.com/2009/06/21/gila-v-the-polya-enumeration-theorem-and-applications/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  The generating function you want is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n} \sum_{d | n} (x_1^{n/d} + ... + x_k^{n/d})^d \varphi \left( \frac{n}{d} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficient of $x_1^{r_1} ... x_k^{r_k}$ is the number you want.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T22:02:50.807" />
  <row Id="826" PostTypeId="2" ParentId="760" CreationDate="2010-07-27T22:04:12.723" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an &lt;a href=&quot;http://www.johndcook.com/lat_long_distance.html&quot; rel=&quot;nofollow&quot;&gt;online calculator&lt;/a&gt; for great circle distance and here is a &lt;a href=&quot;http://www.johndcook.com/lat_long_details.html&quot; rel=&quot;nofollow&quot;&gt;detailed explanation&lt;/a&gt; of the calculations behind it.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-27T22:04:12.723" />
  <row Id="827" PostTypeId="2" ParentId="815" CreationDate="2010-07-27T22:05:59.197" Score="1" ViewCount="0" Body="&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Isaac_Newton&quot; rel=&quot;nofollow&quot;&gt;Isaac Newton&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'm not going to give a complete biography, but for those who don't know...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He single-handedly invented The Calculus. Of course while at the same time, Leibniz &quot;single-handedly&quot; invented it too. :)&lt;/li&gt;&#xA;&lt;li&gt;He theorized the connection between an object's tendency to fall to the earth and the motion of the heavenly bodies.&lt;/li&gt;&#xA;&lt;li&gt;Studied the laws of motion and developed formulas we commonly use today&lt;/li&gt;&#xA;&lt;li&gt;Investigated light refraction&lt;/li&gt;&#xA;&lt;li&gt;Calculated the speed of sound to less than 1% the experimented value&lt;/li&gt;&#xA;&lt;li&gt;He used the &quot;dot notation&quot; to signify time derivatives, which I prefer far over prime notation. :)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-07-27T22:05:59.197" CommunityOwnedDate="2010-07-27T22:05:59.197" CommentCount="4" />
  <row Id="828" PostTypeId="2" ParentId="535" CreationDate="2010-07-27T22:06:16.293" Score="1" ViewCount="0" Body="&lt;p&gt;Well, you're looking at $p_c(x) = 4x^4 + 8x^3 - 3x^2 - 9x + c=0$, which is just $c=-4x^4-8x^3+3x^2+9x$, so if you graph in the $xc$-plane, you'll get a quartic that opens down with the horizontal cross sections the zeros for given $c$.  In fact, because we can solve for $c$ like this to make it a graph in the affine plane, it is a rational quartic curve, so you might want to start looking there.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T22:06:16.293" CommentCount="2" />
  <row Id="829" PostTypeId="2" ParentId="756" CreationDate="2010-07-27T22:13:54.087" Score="0" ViewCount="0" Body="&lt;p&gt;Falagar's answer doesn't seem to be for this question. In this question we only can ask higher/lower, not questions about arbitrary bits. I would think only checking if he lied after a certain number of questions would be best. When playing the game, if he lied at one point, you'd start to get all the answers being either higher or lower (because you're searching in the wrong area) So the point is to maximize the number of times he must change his answer to be able to detect his lie more easily. So I think it would skew your guesses to either the upper or lower end of range. [Sorry for this long post, I can't comment yet and I didn't want to just say people aren't reading the question correctly]&lt;/p&gt;&#xA;" OwnerUserId="248" LastActivityDate="2010-07-27T22:13:54.087" />
  <row Id="830" PostTypeId="2" ParentId="575" CreationDate="2010-07-27T22:17:36.657" Score="2" ViewCount="0" Body="&lt;p&gt;Answer edited, in response to the comment and a second wind for explaining mathematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F(x_0,x_1,x_2)=0$ be the equation for your curve, and take $(y_0,y_1,y_2)$ to be coordinates on $(\mathbb{P}^2)^*$.  Also, assume that $F$ is irreducible and has no linear factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $y_0 x_0+y_1 x_1+y_2 x_2=0$ is the equation of a general line in $\mathbb{P}^2$ (recall, here $y_0,y_1,y_2$ are fixed, and the $x_i$ are the coordinates on the plane) and we look at the open set of $(\mathbb{P}^2)^*$ where $y_2\neq 0$.  On this open set, we can solve the equation of the line for $x_2$, and look at $g(x_0,x_1)=y_2^n F(x_0,x_1,-\frac{1}{y_2}(y_0x_0+y_1x_1))$, a homogeneous polynomial of degree $n$ in $x_0,x_1$ with coefficients homogeneous polynomials in the $y_i$.  This polynomial has zeros the intersections of our curve $C$ with the line $L$ we're looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want to find points of multiplicity at least two.  So how do we find multiple roots of a polynomial? We take the discriminant! Specifically, we do it for an affinization, and we get a homogeneous polynomial of degree $2n^2-n$ in the $y_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that's left is to factor the polynomial, and kill all the linear factors, just throw them away, the reasons are explained in more computational detail on my &lt;a href=&quot;http://rigtriv.wordpress.com/2008/08/04/dual-curves/&quot; rel=&quot;nofollow&quot;&gt;blogpost&lt;/a&gt;, and there I also do explicit examples, but the method for calculating the equation of the dual curve is as above.&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2010-07-28T01:37:45.227" LastActivityDate="2010-07-28T01:37:45.227" CommentCount="1" />
  <row Id="831" PostTypeId="2" ParentId="815" CreationDate="2010-07-27T22:19:05.907" Score="1" ViewCount="0" Body="&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Beno%C3%AEt_Mandelbrot&quot; rel=&quot;nofollow&quot;&gt;Benoît Mandelbrot&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Although he provided many valuable contributions to the field, I am most in love with his work on Fractals. I find math to be quite beautiful, and the Mandelbrot Set (magnified portion shown below) is a perfect example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/3/37/Mandel_zoom_14_satellite_julia_island.jpg&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/3/37/Mandel_zoom_14_satellite_julia_island.jpg/800px-Mandel_zoom_14_satellite_julia_island.jpg&quot; alt=&quot;mandelbrot set zoom 14: julia island&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2011-01-06T16:26:20.040" LastActivityDate="2011-01-06T16:26:20.040" CommunityOwnedDate="2010-07-27T22:19:05.907" CommentCount="5" />
  <row Id="832" PostTypeId="2" ParentId="540" CreationDate="2010-07-27T22:22:26.763" Score="9" ViewCount="0" Body="&lt;p&gt;Suppose you want to put a probability distribution on the natural numbers for the purpose of doing number theory.  What properties might you want such a distribution to have?  Well, if you're doing number theory then you want to think of the prime numbers as acting &quot;independently&quot;: knowing that a number is divisible by $p$ should give you no information about whether it's divisible by $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That quickly leads you to the following realization: you should choose the exponent of each prime in the prime factorization independently.  So how should you choose these?  It turns out that the probability distribution on the non-negative integers with maximum entropy and a given mean is a geometric distribution, as explained for example by Keith Conrad &lt;a href=&quot;https://docs.google.com/viewer?url=http://www.math.uconn.edu/~kconrad/blurbs/analysis/entropypost.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  So let's take the probability that the exponent of $p$ is $k$ to be equal to $(1 - r_p) r_p^k$ for some constant $r_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the probability that a positive integer $n = p_1^{e_1} ... p_k^{e_k}$ occurs as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle C \prod_{i=1}^{k} r_p^{e_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C = \prod_p (1 - r_p)$.  So we need to choose $r_p$ such that this product converges.  Now, we'd like the probability that $n$ occurs to be monotonically decreasing as a function of $n$.  It turns out (and this is a nice exercise) that this is true if and only if $r_p = p^{-s}$ for some $s &gt; 1$ (since $C$ has to converge), which gives the probability that $n$ occurs as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{ \frac{1}{n^s} }{ \zeta(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta(s)$ is the zeta function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of thinking about this argument is that $\zeta(s)$ is the partition function of a statistical-mechanical system called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Primon_gas&quot; rel=&quot;nofollow&quot;&gt;Riemann gas&lt;/a&gt;.  As $s$ gets closer to $1$, the temperature of this system increases until it would require infinite energy to make $s$ equal to $1$.  But this limit is extremely important to understand: it is the limit in which the probability distribution above gets closer and closer to uniform.  So it's not surprising that you can deduce statistical information about the primes by studying the behavior as $s \to 1$ of this distribution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me mention two other reasons to care about the limit as $s \to 1$ of the above distribution.  First, the basic reason to think of the primes as acting independently is the Chinese Remainder Theorem.  Second, a natural reason to look at a distribution where the probability that a number is divisible by $k$ is $(1 - p^{-1}) p^{-k}$ is that this is precisely the distribution you get on the residues $\bmod p^n$ for $k &amp;lt; n$.  In fact, I believe this can be upgraded to the corresponding statement about Haar measure on the $p$-adic integers.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-01T21:26:25.423" LastActivityDate="2010-08-01T21:26:25.423" CommentCount="4" />
  <row Id="833" PostTypeId="2" ParentId="519" CreationDate="2010-07-27T22:30:35.367" Score="0" ViewCount="0" Body="&lt;p&gt;Your group is the alternating subgroup of a Coxeter group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x, y, z | x^2 = y^2 = z^2 = (xy)^2 = (yz)^3 = (zx)^7 = 1 \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not on the (well-understood) list of &lt;a href=&quot;http://en.wikipedia.org/wiki/Coxeter_group#Finite_Coxeter_groups&quot; rel=&quot;nofollow&quot;&gt;finite Coxeter groups&lt;/a&gt;, and the alternating subgroup always has index $2$.  (The connection to the hyperbolic plane is that Coxeter groups of rank $3$ always act as symmetries of a tiling of either the sphere, the Euclidean plane, or the hyperbolic plane by triangles.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T22:30:35.367" />
  <row Id="834" PostTypeId="2" ParentId="815" CreationDate="2010-07-27T22:51:14.237" Score="8" ViewCount="0" Body="&lt;h1&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Leonhard_Euler&quot; rel=&quot;nofollow&quot;&gt;Leonhard Euler&lt;/a&gt;&lt;/h1&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;He made important discoveries in pretty much &lt;em&gt;every&lt;/em&gt; mathematical field there was at his time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He invented graph theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He is responsible for much of the current mathematical notation we use today, including Σ, &lt;em&gt;i&lt;/em&gt;, &lt;em&gt;e&lt;/em&gt;, f(x), π, and sin/cos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_topics_named_after_Leonhard_Euler&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;EVERYTHING&lt;/strong&gt; is named after him&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;His &lt;a href=&quot;http://www.math.dartmouth.edu/~euler/tour/tour_08.html&quot; rel=&quot;nofollow&quot;&gt;combined works&lt;/a&gt; fill 80 (!) volumes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And last but not least,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/c/6/6/c669a6c5e0faf3a8ba0befed0f517ae5.png&quot; alt=&quot;(Euler's formula)&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-27T22:51:14.237" CommunityOwnedDate="2010-07-27T22:51:14.237" />
  <row Id="835" PostTypeId="1" CreationDate="2010-07-27T22:55:17.460" Score="2" ViewCount="251" Body="&lt;p&gt;When I have watched Deal or No Deal (I try not to make a habit of it) I always do little sums in my head to work out if the banker is offering a good deal. Where odds drop below &quot;evens&quot; it's easy to see it's a bad deal, but what would be the correct mathematical way to decide if you're getting a good deal?&lt;/p&gt;&#xA;" OwnerUserId="270" LastActivityDate="2010-12-08T04:17:24.557" Title="Optimal Strategy for Deal or No Deal" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="836" PostTypeId="2" ParentId="806" CreationDate="2010-07-27T23:01:14.113" Score="8" ViewCount="0" Body="&lt;p&gt;Well, class field theory states that the class number is the degree of the largest everywhere-unramified abelian extension of a number field (namely, the Hilbert class field).  But class field theory really says a lot more: it says that there's an isomorphism between the Galois group and the ideal class group. And in general, for any abelian extension $L/K$, there's an isomorphism between $G(L/K)$ and a certain &quot;generalized ideal class group&quot; of $K$ where you quotient by ideals that are norms from $L$.  This can be stated somewhat more elegantly using ideles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, you asked about the class number. In class field theory, it's difficult (to my knowledge) to prove directly that the Artin reciprocity map defined from ideals to the Galois group is an isomorphism. You can show that it is surjective and injective, however; to do this, you have to estimate the order of these generalized ideal class groups, which you can do using either analytic (L-function) methods or algebraic (e.g., using the Herbrand quotient) methods. So this is where the size and seemingly &quot;decategorified&quot; properties (like the order) become more important than the groups themselves: the proofs.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T14:54:32.163" LastActivityDate="2010-07-28T14:54:32.163" CommentCount="6" />
  <row Id="837" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:01:37.857" Score="3" ViewCount="0" Body="&lt;p&gt;The rational numbers are both a continuum (between any two rationals you can find another rational) and countable (they can be lined up in correspondence with the positive integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematician missed it for hundreds (thousands?) of years, until Cantor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the proof of that works both ways, and is equally surprising the other way -- there is a way to order the integers (or any countable set) that makes it into a continuum.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-07-27T23:01:37.857" CommunityOwnedDate="2010-07-27T23:01:37.857" CommentCount="4" />
  <row Id="838" PostTypeId="2" ParentId="640" CreationDate="2010-07-27T23:09:13.553" Score="2" ViewCount="0" Body="&lt;p&gt;On an intuition level:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Complex numbers are somewhat equivalent to R^2 (homeomorphic to a 2-dimentional real space), on the other hand they are also a 1-dimentional complex space. This gives them an extra structure with consequences like this theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another view on this is that differentiable complex functions must also satisfy the Cauchy-Reimann equations and this extra hypothesis is what makes them holomorphic&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditorDisplayName="" LastEditDate="2010-07-27T23:28:02.027" LastActivityDate="2010-07-27T23:28:02.027" />
  <row Id="839" PostTypeId="2" ParentId="835" CreationDate="2010-07-27T23:12:25.170" Score="0" ViewCount="0" Body="&lt;p&gt;In maths, the expected value is the average of how much you'd win if you used the same strategy from the same position a large (approaching infinity) number of times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should first note that expected value alone doesn't decide the best option. We also have to take into account risk. Most people (other than gamblers) would prefer a certain dollar rather than a fifty percent chance of two and fifty percent of nothing. To deal with this, we typically define &lt;a href=&quot;http://en.wikipedia.org/wiki/Utility&quot; rel=&quot;nofollow&quot;&gt;utility&lt;/a&gt; instead. Utility varies between individuals and is determined by their risk profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the dollar amount are quite large for contestants, it would be logical (in theory) to find someone who can afford to take the risk to insure you. This would allow you to receive the gains from a risker strategy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For deal or no deal, they encourage you to play by making their offers worse than the expected value at the start of the game. Later, (&lt;a href=&quot;http://en.wikipedia.org/wiki/Deal_or_No_Deal&quot; rel=&quot;nofollow&quot;&gt;according to Wikipedia&lt;/a&gt;) the offers may even exceed the expected value. Without knowing how exactly the offers are calculated (or forming a model), we can't answer this question accurately, but only make general statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore risk and offers being greater than expected utility, you'd always want to go as far as possible in the game. But, as is, the game is extremely difficult to analyse mathematically.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-27T23:20:17.560" LastActivityDate="2010-07-27T23:20:17.560" />
  <row Id="840" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:15:09.097" Score="16" ViewCount="0" Body="&lt;p&gt;Suppose you have a large collection of books, all of the same size.  &lt;strike&gt;Balance one of them on the edge of a table so that one end of the book is as far from the table as possible.  Balance another book on top of that one, and again try to get as far from the table as possible.&lt;/strike&gt;  Take $n$ of them and try to balance them on top of each other so that the top book is as far as possible away from the edge of the table horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt;  With enough books, you can get arbitrarily far from the table.  If you are &lt;em&gt;really&lt;/em&gt; careful.  This is a consequence of the divergence of the harmonic series.  I think if you haven't heard this one before it's very hard to tell whether it's true or false.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-04T03:15:24.027" LastActivityDate="2010-08-04T03:15:24.027" CommunityOwnedDate="2010-07-27T23:15:09.097" CommentCount="14" />
  <row Id="841" PostTypeId="1" CreationDate="2010-07-27T23:16:48.580" Score="1" ViewCount="127" Body="&lt;p&gt;If I remember rightly there are some integrals of real functions which are easier to compute by using complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because of properties of the particular function or because of a lack of a known real analysis technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there functions which would require hypercomplex analysis to integrate?&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2010-07-27T23:31:45.967" Title="Integral via complex analysis. Integral via hypercomplex analysis" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="842" PostTypeId="1" AcceptedAnswerId="864" CreationDate="2010-07-27T23:19:07.500" Score="11" ViewCount="350" Body="&lt;p&gt;In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most logic texts either don't explain the terminology, or allude to the topological property of compactness. I see an analogy as, given a topological space X and a subset of it S, S is compact iff for every open cover of S, there is a finite subcover of S. But, it doesn't seem strong enough to justify the terminology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there more to the choice of the terminology in logic than this analogy?&lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-08-11T22:03:15.910" Title="Why is compactness in logic called compactness?" Tags="&lt;terminology&gt;&lt;logic&gt;&lt;topology&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="843" PostTypeId="2" ParentId="841" CreationDate="2010-07-27T23:23:08.977" Score="3" ViewCount="0" Body="&lt;p&gt;I'm pretty sure you mean the evaluation of &lt;em&gt;definite&lt;/em&gt; integrals. For instance, $e^{-x^2}$ has no elementary antiderivative, but it's definite integral over the real line can be computed explicitly (it's $\sqrt{\pi}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask what complex analysis has to do with this. Well, the idea is that complex integrals around closed curves (of holomorphic functions) are usually comparatively easy to evaluate: it's because we have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Residue_theorem&quot; rel=&quot;nofollow&quot;&gt;residue theorem&lt;/a&gt; as a tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these real integrals are usually over the whole real line, which doesn't satisfy the hypothesis of a path under the residue theorem. For instance, it's not a finite closed path.  However, we can replace the real number by a large semicircle (to pick one common example) that goes from $-R$ to $R$, then around the upper-half-plane from $R$ to $-R$. This is a closed path and the residue theorem can apply to this. When you let $R$ to $\infty$, the integral around semicircular path often tends to zero (by direct bounding arguments). So what you're left with is the integral along the real line, and it's equal to the residues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of this will be in any complex analysis textbook, e.g. Ahlfors's.  The Wikipedia article on the residue theorem (link above) also has examples.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-27T23:23:08.977" />
  <row Id="844" PostTypeId="2" ParentId="842" CreationDate="2010-07-27T23:26:19.687" Score="0" ViewCount="0" Body="&lt;p&gt;This isn't a complete answer, partly because I've had discussions with other grad students and we weren't able to work it out satisfactorily, but I've been told that you can actually put a topology on logical statements such that the compactness theorem translates to &quot;The set of true statements is a compact subset of the set of all statements&quot; or something similar.  But as I said, a few of my friends and I weren't able to work out the topology.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-27T23:26:19.687" />
  <row Id="845" PostTypeId="2" ParentId="842" CreationDate="2010-07-27T23:27:21.497" Score="-1" ViewCount="0" Body="&lt;p&gt;As far as I know, the link comes from the syntactic theory. You are given a set of symbols of sentence F = {f_i} and you are allowed allowed to form complex statements using them. You can combine the elementary statements with AND, OR, NOT operators and parentheses, in the usual way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get a set X of composed sentences, like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f AND g) OR (NOT h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A syntactic version of the compactness theorem states the following.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume that for every finite subset Y of X you can assign truth values to the f_i in such a way that all sentences in Y are true. Than you can do the same for X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Proof&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the topological space A obtained by taking the product of {0, 1} over the set F. The topology on A is the product topology. By Tychonoff's theorem, A is compact. For every composed statement s, the set of truth values which make s true is a finite intersection of cylinders, hence it is a closed set of A. The hypothesis say that every finite intersection of such closed sets, for s ranging in X, is not empty. Hence the intersection of all such closed set is not empty, which means one can make all statements in X true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditorDisplayName="" LastEditDate="2010-07-27T23:32:26.137" LastActivityDate="2010-07-27T23:32:26.137" CommentCount="1" />
  <row Id="846" PostTypeId="2" ParentId="842" CreationDate="2010-07-27T23:30:16.513" Score="1" ViewCount="0" Body="&lt;p&gt;The analogy for the compactness theorem for propositional calculus is as follows. Let $p_i $ be propositional variables; together, they take values in the product space $2^{\mathbb{N}}$.  Suppose we have a collection of statements $S_t$ in these boolean variables such that every finite subset is satisfiable. Then I claim that we can prove that they are all simultaneously satisfiable by using a compactness argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F$ be a finite set. Then the set of all truth assignments (this is a subset of $2^{\mathbb{N}}$) which satisfy $S_t$ for $t \in F$ is a closed set $V_F$ of assignments satisfying the sentences in $F$.  The intersection of any finitely many of the $V_F$ is nonempty, so by the finite intersection property, the intersection of all of them is nonempty (since the product space is compact), whence any truth in this intersection satisfies all the statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how this works in predicate logic.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-27T23:48:53.003" LastActivityDate="2010-07-27T23:48:53.003" />
  <row Id="847" PostTypeId="1" AcceptedAnswerId="1570" CreationDate="2010-07-27T23:30:47.480" Score="3" ViewCount="281" Body="&lt;p&gt;Is there any mathematical significance to the fact that the law of cosines...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cos(angle between a and b) = (a^2 + b^2 - c^2) / (2ab)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... for an impossible triangle yields a cosine &amp;lt; -1 (when c &gt; a+b), or &gt; 1 (when c &amp;lt; |a-b|)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a = 3, b = 4, c = 8 yields cos(angle ab) = -39/24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or a = 3, b = 5, c = 1 yields cos(angle ab) = 33/30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something to do with hyperbolic geometry/cosines?&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-04T21:40:43.287" Title="Law of cosines with impossible triangles" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="3" />
  <row Id="848" PostTypeId="2" ParentId="311" CreationDate="2010-07-27T23:31:34.267" Score="1" ViewCount="0" Body="&lt;p&gt;A further sense in which Higher Order Logics with standard or saturated semantics (HOL, hereafter) are less well-behaved than First Order Logic (FOL, hereafter) is a direct consequences of the failure of Completeness (and thus, as explained in other answers, of Compactness). The set of logical truths and the set of correct claims of semantic consequence for these logics are not recursively enumerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FOL is Complete, yet not Decidable. So, we determine of an arbitrary sentences and sets of sentences of the language of FOL if those sentences are logical truths or if a set has a given sentence as a consequence. But, since FOL is Complete and proofs are finitely long, we can (in the mathematicians sense of &quot;can&quot;) enumerate the proofs and inspect one by one, checking what sentence the proof shows as a theorem or what sentence the proof derives from what set of assumptions. This gets us a recursive enumeration of the truths and the sentence/set pairs that stand in the consequence relation. (This does not contradict the failure of decidability as we cannot conclude that since we've yet to come across a proof in out enumeration, there isn't one if only we kept looking.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since HOL's are not Complete, this means of showing them recursively enumerable is not available. Indeed, there can be no means; were there such a means, it could be exploited to induce a Complete proof system, and there cannot be such a Complete proof system as the HOL's are not Compact. &lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-07-27T23:31:34.267" CommentCount="2" />
  <row Id="849" PostTypeId="2" ParentId="841" CreationDate="2010-07-27T23:31:45.967" Score="2" ViewCount="0" Body="&lt;p&gt;A generic real-valued function carries no additional information other than its values.  But many of the common functions encountered in analysis are just the restriction to $\mathbb{R}$ of holomorphic functions, and these functions carry a lot of additional information, namely, their values in $\mathbb{C}$.  Unlike real-analytic functions, holomorphic functions are extremely rigid: for example, if two holomorphic functions agree on a set of complex numbers with a limit point, they must be equal everywhere.  That means it's possible to deduce information about how a holomorphic function behaves on part of its domain (say, $\mathbb{R}$) from information about how it behaves on other parts of its domain, since the two have to determine each other.  This information is extracted using contour integrals.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-27T23:31:45.967" />
  <row Id="851" PostTypeId="2" ParentId="847" CreationDate="2010-07-27T23:38:38.607" Score="2" ViewCount="0" Body="&lt;p&gt;One can prove the triangle inequality in any abstract inner product space, such as a Hilbert space; it is a consequence of the Cauchy-Schwarz inequality $\langle a, b \rangle^2 \le ||a||^2 ||b||^2$.  In order for the triangle inequality to fail, the Cauchy-Schwarz inequality has to fail, and in order for Cauchy-Schwarz to fail (which corresponds to the &quot;cosine&quot; no longer being between $1$ and $-1$), some axiom of an abstract inner product space has to be given up.  One choice is to give up positive-definiteness; in other words, lengths of vectors are no longer always non-negative.  This leads to geometries like &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski spacetime&lt;/a&gt; which are relevant to relativity.  In Minkowski spacetime, there is a &lt;em&gt;reverse&lt;/em&gt; triangle inequality instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I should also mention that the &quot;unit sphere&quot; in Minkowski spacetime is a model for hyperbolic geometry called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperboloid_model&quot; rel=&quot;nofollow&quot;&gt;hyperboloid model&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-27T23:45:20.123" LastActivityDate="2010-07-27T23:45:20.123" CommentCount="6" />
  <row Id="852" PostTypeId="2" ParentId="847" CreationDate="2010-07-27T23:41:57.573" Score="0" ViewCount="0" Body="&lt;p&gt;For some a, b, and c that form a triangle: increasing the length of c increases the measure of angle C and as m&amp;ang;C approaches 180&amp;deg;, cos C approaches -1; decreasing the length of c increases the measure of angle C and as m&amp;ang;C approaches 0&amp;deg;, cos C approaches 1.  Extending this pattern, it makes sense that if c &gt; a + b, c has gotten bigger past making a triangle with a and b, so cos C &amp;lt; -1, and if c &amp;lt; |a-b|, c has gotten smaller past making a triangle with a and b, so cos C &gt; 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hyperbolic geometry, the definition of lines (and hence triangles) is different and the sum of the measures of the angles in a triangle is less than 180&amp;deg;.  There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines&quot; rel=&quot;nofollow&quot;&gt;Hyperbolic Law of Cosines&lt;/a&gt;, but it's not quite the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there's a sensible way to relate hyperbolic cosine to the Law of Cosines in Euclidean geometry.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-27T23:41:57.573" />
  <row Id="853" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:52:55.273" Score="5" ViewCount="0" Body="&lt;p&gt;Position-based Cryptography. This is a fun example since it seems very &quot;out of left field&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The setup:&lt;/strong&gt;&#xA;Three servers are positioned in known locations on the globe (their positions can be arbitrary, provided they aren't on top of each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single computer wants to prove its location to the servers. In other words, if the computer is actually located where it claims, then the protocol will accept. However, if the computer is located anywhere else, then the protocol will reject, no matter how the computer cheats (it is even allowed to recruit friends to help it cheat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All communication is subject to the laws of physics -- information travels at speed c, and quantum mechanics holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 1:&lt;/strong&gt; This is impossible if all communication is classical. Cheating is always possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 2:&lt;/strong&gt; This is possible if quantum communication is possible.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-07-27T23:52:55.273" CommunityOwnedDate="2010-07-27T23:52:55.273" CommentCount="2" />
  <row Id="854" PostTypeId="2" ParentId="250" CreationDate="2010-07-27T23:56:58.277" Score="2" ViewCount="0" Body="&lt;p&gt;Goldbach's Conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted this is open, so it may be cheating a bit. However, this seems like a very hard problem to intuit your way to the &quot;conventional wisdom&quot;. By contrast, &lt;code&gt;P != NP&lt;/code&gt; is &quot;obviously&quot; true. Indeed, Feynman had trouble believing &lt;code&gt;P != NP&lt;/code&gt; was even open.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-07-27T23:56:58.277" CommunityOwnedDate="2010-07-27T23:56:58.277" CommentCount="3" />
  <row Id="855" PostTypeId="1" AcceptedAnswerId="856" CreationDate="2010-07-27T23:59:09.817" Score="11" ViewCount="926" Body="&lt;p&gt;I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T00:17:44.240" LastActivityDate="2011-01-08T03:36:50.213" Title="For any prime $p &gt; 3$, why is $p^2-1$ always divisible by 24?" Tags="&lt;prime-numbers&gt;&lt;number-theory&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="1" />
  <row Id="856" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T00:02:04.647" Score="14" ViewCount="0" Body="&lt;p&gt;P^2-1  = (P+1)(P-1). P must be either 1 or (2 mod 3), so we have a factor of 3 in the product. And P is also either 1 or 3 mod 4. Hence either 2|(P+1) and 4|(P-1) or 2|(P-1) and 4|(P+1). Thus 8*3= 24 divides the product.&lt;/p&gt;&#xA;" OwnerUserId="296" LastActivityDate="2010-07-28T00:02:04.647" />
  <row Id="857" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T00:03:48.777" Score="8" ViewCount="0" Body="&lt;p&gt;$p$ must be congruent either to 1,3,5,7 modulo 8.  Then $p^2$ is congruent to $1$ modulo $8$ in either case.  So $8$ divides $p^2-1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $p$ is not a multiple of 3, so either $p-1$ or $p+1$ is a multiple of three. So $3$ divides $p^2-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together, it follows that 24 divides $p^2 -1 $.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-08T03:08:28.423" LastActivityDate="2011-01-08T03:08:28.423" CommentCount="1" />
  <row Id="858" PostTypeId="2" ParentId="742" CreationDate="2010-07-28T00:04:11.413" Score="0" ViewCount="0" Body="&lt;p&gt;A similar problem is known in mathematics.  It is called the &quot;Postage-Stamp&quot; problem, and usually asks which postal values can be realized and which cannot.  Dynamic programming is a common but not polynomial-time solution for this.  A polynomial time solution using continued fractions in the case of two stamp denominations exists, and more complex algorithms exist for three or more stamp denominations.  Look up &quot;Frobenius Number&quot; for more technical information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given problem, I would roughly estimate the Frobenius number, and based on the estimate make an easy guess as to a near-solution and then use a dynamic programming/tabular solution, depending on the amount of time I had to solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:01:46.380" LastActivityDate="2010-08-11T22:01:46.380" CommentCount="0" />
  <row Id="859" PostTypeId="2" ParentId="81" CreationDate="2010-07-28T00:07:35.650" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://sbseminar.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Secret Blogging Seminar&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-07-28T00:07:35.650" CommunityOwnedDate="2010-07-28T00:07:35.650" CommentCount="1" />
  <row Id="860" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T00:08:53.867" Score="25" ViewCount="0" Body="&lt;p&gt;The most elementary proof I can think of without explicitly mentioning any number theory: out of the three consecutive numbers p-1, p, p+1, one of them must be divisible by 3, also since the neighbours of p are consecutive even numbers, one of them must be divisible by 2 and the other by 4, so their product is divisible by &lt;code&gt;3*2*4=24&lt;/code&gt; (and of course, we can throw p out since it's prime and those factors cannot come from p).&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-28T00:08:53.867" CommentCount="2" />
  <row Id="861" PostTypeId="1" AcceptedAnswerId="865" CreationDate="2010-07-28T00:13:25.637" Score="3" ViewCount="109" Body="&lt;p&gt;This may be a poorly phrased question - please let me know of it - but what is the correct way to think of the cotangent bundle? It seems odd to think of it as the dual of the tangent bundle (I am finding it odd to reconcile the notions of &quot;maps to the ground field&quot; with this object).&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T00:15:36.867" LastActivityDate="2010-07-28T00:53:38.753" Title="Cotangent bundle" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="3" />
  <row Id="862" PostTypeId="2" ParentId="842" CreationDate="2010-07-28T00:15:14.417" Score="2" ViewCount="0" Body="&lt;p&gt;A motto which is related (and sometimes true) is &quot;proofs are finite&quot;.  In most systems of logic under consideration, the statement and a proof of the statement are finite strings of symbols.  One can think of them as &quot;compactly&quot; represented.  How nice then that certain systems will always allow a proof to be found if there is one.  While this does not give a tight connection to topology, it suggests (and you need to work this through on your own to be convinced) that certain infinite abnormalities (like an infinite proof) will not occur.  A similar topological claim is that, for compact sets, an infinite inclusive chain of certain subsets will have nonempty intersection, such a claim easily seen to be false for some non-compact sets, such as the collection { C_n | C_n = {x | x &gt;= n and x is a real number } for n a non-negative integer } .  &lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:03:15.910" LastActivityDate="2010-08-11T22:03:15.910" CommentCount="0" />
  <row Id="863" PostTypeId="2" ParentId="781" CreationDate="2010-07-28T00:15:29.330" Score="1" ViewCount="0" Body="&lt;p&gt;As a rough/somewhat-intuitive explanation of why Benford's Law makes sense, consider it with respect to amounts of money.  The amount of time(/effort/work) needed to get from $1000 to $2000 (100% increase) is a lot greater than the amount of time needed to get from $8000 to $9000 (12.5% increase)--increasing money is usually done in proportion to the money one has.  Thought about in the other direction, it should take a fixed amount of time to, say, double one's money, so going from $1000 to $2000 takes as long as from $2000 to $4000 and $4000 to $8000, so the leading digit spends more time at lower values than at higher values.  Because the value growth is exponential, the time spent at each leading digit is roughly logarithmic.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T00:15:29.330" />
  <row Id="864" PostTypeId="2" ParentId="842" CreationDate="2010-07-28T00:17:34.070" Score="8" ViewCount="0" Body="&lt;p&gt;The Compactness Theorem is equivalent to the compactness of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras&quot; rel=&quot;nofollow&quot;&gt;Stone space&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindenbaum%E2%80%93Tarski_algebra&quot; rel=&quot;nofollow&quot;&gt;Lindenbaum&amp;ndash;Tarski algebra&lt;/a&gt; of the first-order language &lt;em&gt;L&lt;/em&gt;. (This is also the space of &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_%28model_theory%29&quot; rel=&quot;nofollow&quot;&gt;0-types&lt;/a&gt; over the empty theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point in the Stone space &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt; is a complete theory &lt;em&gt;T&lt;/em&gt; in the language &lt;em&gt;L&lt;/em&gt;. That is, &lt;em&gt;T&lt;/em&gt; is a set of sentences of &lt;em&gt;L&lt;/em&gt; which is closed under logical deduction and contains exactly one of &amp;sigma; or &amp;not;&amp;sigma; for every sentence &amp;sigma; of the language. The topology on the set of types has for basis the open sets &lt;em&gt;U&lt;/em&gt;(&amp;sigma;) = {&lt;em&gt;T&lt;/em&gt;: &amp;sigma; &amp;isin; &lt;em&gt;T&lt;/em&gt;} for every sentence &amp;sigma; of L. Note that these are all clopen sets since &lt;em&gt;U&lt;/em&gt;(&amp;not;&amp;sigma;) is complementary to &lt;em&gt;U&lt;/em&gt;(&amp;sigma;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how the Compactness Theorem implies the compactness of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt;, suppose the basic open sets &lt;em&gt;U&lt;/em&gt;(&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), i &amp;isin; I, form a cover of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt;. This means that every complete theory T contains at least one of the sentences &amp;sigma;&lt;sub&gt;i&lt;/sub&gt;. I claim that this cover has a finite subcover. If not, then the set {&amp;not;&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;: i &amp;isin; I} is finitely consistent. By the Compactness Theorem, the set consistent and hence (by Zorn's Lemma) is contained in a maximally consistent set &lt;em&gt;T&lt;/em&gt;. This theory &lt;em&gt;T&lt;/em&gt; is a point of the Stone space which is not contained in any &lt;em&gt;U&lt;/em&gt;(&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), which contradicts our hypothesis that the &lt;em&gt;U&lt;/em&gt;(&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), i &amp;isin; I, form a cover of the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how the compactness of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt; implies the Compactness Theorem, suppose that {&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;: i &amp;isin; I} is an inconsistent set of sentences in &lt;em&gt;L&lt;/em&gt;. Then &lt;em&gt;U&lt;/em&gt;(&amp;not;&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;), i &amp;isin; I, forms a cover of &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;&lt;em&gt;L&lt;/em&gt;&lt;/sub&gt;. This cover has a finite subcover, which corresponds to a finite inconsistent subset of {&amp;sigma;&lt;sub&gt;i&lt;/sub&gt;: i &amp;isin; I}. Therefore, every inconsistent set has a finite inconsistent subset, which is the contrapositive of the Compactness Theorem.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditorDisplayName="" LastEditDate="2010-07-30T01:03:24.220" LastActivityDate="2010-07-30T01:03:24.220" />
  <row Id="865" PostTypeId="2" ParentId="861" CreationDate="2010-07-28T00:18:21.587" Score="3" ViewCount="0" Body="&lt;p&gt;One fruitful way to think about it, if you have physics background, is as phase space.  Your manifold is the configuration space for some system of particles, and the cotangent bundle is then the phases, so the cotangent directions are velocities.  This is helpful also with the symplectic structure on $T^*M$.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T00:18:21.587" CommentCount="4" />
  <row Id="866" PostTypeId="2" ParentId="815" CreationDate="2010-07-28T00:18:44.877" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nicolas_Bourbaki&quot; rel=&quot;nofollow&quot;&gt;&lt;h1&gt;Nicolas Bourbaki&lt;/h1&gt;&lt;/a&gt;&#xA;The story had me from the moment I realized it is true. And as I come to understand more mathematics, I find that I frequently ask myself &quot;What would Bourbaki do?&quot; Also, the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3764326506&quot; rel=&quot;nofollow&quot;&gt;individuals&lt;/a&gt; have &lt;a href=&quot;http://en.wikipedia.org/wiki/Jean_Dieudonn%C3%A9&quot; rel=&quot;nofollow&quot;&gt;done&lt;/a&gt; much to expose the &lt;a href=&quot;http://www.fermentmagazine.org/home5.html&quot; rel=&quot;nofollow&quot;&gt;human&lt;/a&gt; side of mathematics.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-28T00:18:44.877" CommunityOwnedDate="2010-07-28T00:18:44.877" />
  <row Id="867" PostTypeId="2" ParentId="795" CreationDate="2010-07-28T00:23:58.393" Score="1" ViewCount="0" Body="&lt;p&gt;More generally, if $\sum_{i=1}^{\infty} c_n$ is a conditionally convergent series of &lt;em&gt;complex&lt;/em&gt; numbers, then the collection of sums of all convergent rearrangements is an affine subspace of the plane $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sketch of why this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A$ and $B$ are two rearrangements that converge to the distinct complex numbers $a,b$, respectively. Focus on a partial sum of $A$ that's pretty close to $a$. A sufficiently long partial sum of $B$ will include every term of the former partial sum of $A$, but it will also tend to $b$ as you take more and more terms. It's &quot;clear&quot; then that it's possible to obtain any combination $ta + (1 - t)b$ for $t \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in addition there is a point $c$ which is the sum of some rearrangement but is not on the line through $a$ and $b$, then the previous result can be reapplied, replacing the numbers $a$ and $b$ with the numbers $c$ and $d$, where $d$ is any point on the line through $a$ and $b$. It follows that in this case the entire plane $\mathbb{C}$ is obtained by rearranging the original series.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-07-28T00:23:58.393" />
  <row Id="868" PostTypeId="1" AcceptedAnswerId="904" CreationDate="2010-07-28T00:30:59.580" Score="3" ViewCount="164" Body="&lt;p&gt;I am currently working on a Computer Algebra System and was wondering for suggestions on methods of finding roots/factors of polynomials. I am currently using the Numerical Durand-Kerner method but was wondering if there are any good non-numerical methods (primarily for simplifying fractions etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally this should work for equations in multiple variables.&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="302" LastEditorDisplayName="" LastEditDate="2010-07-28T02:03:51.857" LastActivityDate="2010-07-28T03:44:15.640" Title="What is the best way to factor arbitrary polynomials" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="869" PostTypeId="2" ParentId="861" CreationDate="2010-07-28T00:41:25.627" Score="2" ViewCount="0" Body="&lt;p&gt;You might be interested in this MathOverflow post: &lt;a href=&quot;http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, I'd leave this as a comment but I just joined this site and don't have enough reputation.)&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T00:41:25.627" />
  <row Id="870" PostTypeId="2" ParentId="868" CreationDate="2010-07-28T00:42:24.800" Score="2" ViewCount="0" Body="&lt;p&gt;If you're looking to factor exactly, then you'll need to use something that's not one of the fundamental operations of addition, subtraction, multiplication, division and extraction of roots.  The Abel-Ruffini theorem says so for degree five and above.  However, there are numerous other methods to find roots exactly, using more general functions, my favorite being theta functions, as explained in the appendix to Mumford's &quot;Tata Lectures on Theta II&quot;&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T00:42:24.800" CommentCount="1" />
  <row Id="871" PostTypeId="2" ParentId="392" CreationDate="2010-07-28T00:45:23.483" Score="5" ViewCount="0" Body="&lt;p&gt;If you look carefully and geometrically at the quotient limit that defines sin'(x) in the unit circle, and take the chord and tangent as approximations to the arc (that is the angle; this is the essence of sin(x)/x approaches 1), you will see that limit of the derivative quotient tends exactly to cos(x), that is, it's adjacent/hypotenuse. In other words, it's built into right triangle geometry, like so many phenomena in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in that geometry, you'll see lurking the proof for the sin(x+y) formula, which, along with the limit of sin(x)/x, is how the standard proof that sin'(x) = cos(x) goes.  But skipping that algebra and going directly to the geometry is the most straightforward way I know to answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I don't have time or tools to draw the pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this saying the same thing as the physics answer above, but perhaps more directly.  I do think all the answers referring to series expansions miss the point.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-07-28T01:46:01.523" LastActivityDate="2010-07-28T01:46:01.523" CommentCount="2" />
  <row Id="872" PostTypeId="1" AcceptedAnswerId="881" CreationDate="2010-07-28T00:45:46.913" Score="2" ViewCount="113" Body="&lt;p&gt;I understand what the Hodge dual is, but I can't quite wrap my head around the dual space of vector space. They seem very similar, almost the same, but perhaps they are unrelated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in R^3, the blade a^b gives you a subspace that's like a plane, and the dual is roughly the normal to the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similarly simple example for the dual space of a vector space, or is there a way to describe the vector space dual in terms of the Hodge dual?&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-11-22T22:08:07.980" Title="What is the relationship between the Hodge dual of p-vectors and the dual space of an ordinary vector space?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="873" PostTypeId="2" ParentId="806" CreationDate="2010-07-28T00:45:58.847" Score="5" ViewCount="0" Body="&lt;p&gt;Frequently in studying diophantine equations, or related problems, one is forced to look at rings of integers of algebraic number fields, and it may be that the class number of the ring your are forced to deal with is &gt; 1.  In this case, there is nothing you can do to change that fact; you have to live with it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a response, it is natural to try and develop a theory of the class number and the class group, as a means of finding ways to deal with the failure of unique factorization.  Kummer was the one who invented the notion of class number, and he did in the course of his work on Fermat's Last Theorem; this is what is described in Andrea Ferretti's answer, and is an excellent example of what I am talking about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, quite different in nature, occurs in the proof of Dirichlet's theorem on primes in arithmetic progressions.    This is the theorem that if a and d are coprime natural numbers, then there are infinitely many primes p such that p is congruent to a mod d.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his proof, he reduces everything to showing that a certain number (the value at 1 of a certain so-called L-function) is non-zero.  He then is able to give a precise formula for this L-function, and shows that it is equal to some non-zero constants times a certain class number.  Since the class number is positive (and so in particular, is non-zero!), he is able to complete his proof.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T00:45:58.847" CommentCount="2" />
  <row Id="874" PostTypeId="1" CreationDate="2010-07-28T00:50:36.187" Score="0" ViewCount="99" Body="&lt;p&gt;When drawing an angle line (45 degrees) in a rectangle from a general point &lt;code&gt;p(x,y)&lt;/code&gt; that located on the right or the top line of the rectangle. How can I find the intersection point &lt;code&gt;p2&lt;/code&gt; of this line with the rectangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I want to write the target point, &lt;code&gt;p2&lt;/code&gt;, with my current information: &lt;code&gt;x, y, w, h&lt;/code&gt;. (This variables are described in the picture below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point (0,0) is in the top-right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img690.imageshack.us/img690/9819/rectangleline.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T03:32:15.807" LastActivityDate="2010-07-28T04:58:52.313" Title="How to get an equation that output the end point of an angle line in rectangle?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="875" PostTypeId="2" ParentId="574" CreationDate="2010-07-28T00:51:57.447" Score="4" ViewCount="0" Body="&lt;p&gt;Generic has different meanings in different branches of geometry.  In algebraic geometry,&#xA;it usually means that the property in question holds on a Zariski dense open set.  In other geometric contexts, it could also that the property holds on a dense open set (in whatever is the natural topology under consideration), but can also mean that it holds on a countable intersection of such sets.  (The reason for considering such intersections as being big is&#xA;motivated by the Baire category theorem, which says that in reasonable contexts the complement of such a set will be very &quot;thin&quot;, in some sense.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T00:51:57.447" />
  <row Id="876" PostTypeId="2" ParentId="735" CreationDate="2010-07-28T00:52:00.427" Score="1" ViewCount="0" Body="&lt;p&gt;Hmm, apparently the other answerers' algebra classes were a lot more intense than mine...My answer's based on what I would have been comfortable with after basic algebra and geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you think graphically, instantaneous velocity is the slope of a line at a single point rather than over an interval. At least that's what you want. So, if you happen to be going at a constant velocity (any straight line on a position graph), you can just use the slope formula. Where calculus comes in is if you're dealing with an inconstant velocity (a curved line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not making sense right off, just try to realistically graph the movement of a car as it accelerates/decelerates, where the y-axis is position and the x-axis is time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using algebra, you can't take the slope of a curved line. What you can do is take the slope between two points on the curve. The closer these points get to each other, the closer the slope between them will approximate the actual slope of the curve. So, if the slope between t=5.001 min. and t=5.01 min. is 40, then that approximates the actual slope, and instantaneous velocity, at t=5.0055 min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I can get much more specific/accurate without going into (pre-)calculus.&lt;/p&gt;&#xA;" OwnerUserId="220" LastActivityDate="2010-07-28T00:52:00.427" />
  <row Id="877" PostTypeId="2" ParentId="861" CreationDate="2010-07-28T00:53:38.753" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not completely sure what you mean by this: &quot;It seems odd to think of it as the dual of the tangent bundle (I am finding it odd to reconcile the notions of &quot;maps to the ground field&quot; with this object),&quot; but maybe the following will help you see why it is natural to consider the dual space of the tangent bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a function f on our manifold, we want to associate something like the gradient of f.  Well, in calculus, what characterized the gradient of a function?  Its the vector field such that when we take its dot product with a vector v at some point p, we get the directional derivative, at p, of f along v.  In a general manifold we don't have a dot product (which is a metric) but we can form a covector field (something which gives an element of the cotangent bundle at any point) such that, when applied to a vector v, we get the directional derivative of f along v.  This covector field is denoted df and is called the exterior derivative of f.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-28T00:53:38.753" />
  <row Id="878" PostTypeId="2" ParentId="574" CreationDate="2010-07-28T00:54:39.220" Score="5" ViewCount="0" Body="&lt;p&gt;The term &lt;em&gt;generic&lt;/em&gt; usually applies to something which happens in an open dense set of some space. The idea is that open dense sets are large subsets of the space. Indeed, they are closed under finite intersections and thus form a base for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Filter_%28mathematics%29#Filter_on_a_set&quot; rel=&quot;nofollow&quot;&gt;filter of subsets&lt;/a&gt; of the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, the term is applied more generally to something which happens in a countable intersection of open dense sets (a dense G&lt;sub&gt;&amp;delta;&lt;/sub&gt; set) of some space. Usually, this is in the context of a complete metric space or a locally compact Hausdorff space for which the &lt;a href=&quot;http://en.wikipedia.org/wiki/Baire_category_theorem&quot; rel=&quot;nofollow&quot;&gt;Baire Category Theorem&lt;/a&gt; applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;generic point&lt;/em&gt; is a (sometimes fictitious) point which lies in every open dense set of the space. Irreducible &lt;a href=&quot;http://en.wikipedia.org/wiki/Sober_space&quot; rel=&quot;nofollow&quot;&gt;sober spaces&lt;/a&gt; always have a generic point, it is the unique point whose closure is the whole space. The only Hausdorff space with a generic point is the one-point space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fictitious generic points have a variety of uses. Usually one means a point which lies in all open dense sets which are considered in the current argument, without specifying the open dense sets in question. This is fine because the intersection of finitely many (or even countably many in the case of Baire spaces) open dense sets is guaranteed to be nonempty.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditorDisplayName="" LastEditDate="2010-07-28T01:38:46.330" LastActivityDate="2010-07-28T01:38:46.330" />
  <row Id="879" PostTypeId="2" ParentId="874" CreationDate="2010-07-28T00:56:09.203" Score="0" ViewCount="0" Body="&lt;p&gt;Doesn't this depend on the slope of the segment that begins at P = (x,y)? Perhaps assuming some sort of proportionality we can solve this problem. Please provide more information.--&lt;/p&gt;&#xA;" OwnerUserId="298" LastActivityDate="2010-07-28T00:56:09.203" CommentCount="1" />
  <row Id="881" PostTypeId="2" ParentId="872" CreationDate="2010-07-28T01:01:41.547" Score="3" ViewCount="0" Body="&lt;p&gt;(Edit:  I have edited this answer several times because my understanding of the situation has been improving.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is always profitable to understand these kind of constructions by understanding exactly what information they depend on.  The Hodge dual depends on a surprising amount of information: you need a vector space $V$ which is equipped with both an inner product and an orientation, which is essentially a choice of which bases of $V$ are &quot;right-handed.&quot;  So let's see what we can say ignoring all this information first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any abstract vector space $V$ of finite dimension $n$ has exterior powers $\Lambda^2 V, \Lambda^3 V, ... \Lambda^n V$, the last of which is one-dimensional.  The vector spaces $\Lambda^k V$ and $\Lambda^{n-k} V$ always have the same dimension, so we would like to be able to define some sort of &quot;canonical&quot; map between them.  What can we say?  Well, they are always dual: the wedge product defines a natural bilinear map $\Lambda^k V \times \Lambda^{n-k} V \to \Lambda^n V$, and since the latter is one-dimensional this means (once you've proven nondegeneracy) that the two vector spaces are in fact dual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But duality does not give you a map between them.  When two vector spaces $V, W$ are dual, meaning there is a nondegenerate bilinear map $V \times W \to F$ (where $F$ is the ground field), all you get is an isomorphism $V \simeq W^{\ast}$.  Here you get an isomorphism $\Lambda^k V \simeq \Lambda^{n-k} V^{\ast}$, &lt;strong&gt;once you have specified an isomorphism&lt;/strong&gt; $\Lambda^n V \simeq F$.  This is equivalent to picking out a distinguished vector in $\Lambda^n V$, which there is no way to do in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is to introduce extra data.  To identify $\Lambda^{n-k} V^{\ast}$ with $\Lambda^{n-k} V$, we need an inner product.  An inner product gives you &lt;strong&gt;two&lt;/strong&gt; distinguished vectors in $\Lambda^n V$, as follows: take any orthonormal basis $b_1, ... b_n$.  Then wedging together the $b_i$ in any order gets you one of two elements of $\Lambda^n V$, depending on whether the corresponding permutation is even or odd.  But without any extra data, there is no way to identify one of these elements with $1$ and one of these elements with $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra data that does this is an orientation on $V$, which tells you which bases are &quot;right-handed&quot; and which are &quot;left-handed.&quot;  So an oriented orthonormal basis gives you a distinguished element of $\Lambda^n V$, which gives you a distinguished isomorphism $\Lambda^k V \simeq \Lambda^{n-k} V^{\ast}$, which composed with the isomorphism $\Lambda^{n-k} V^{\ast} \simeq \Lambda^{n-k} V$ is the Hodge dual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained in &lt;a href=&quot;http://www-users.math.umd.edu/~toni/hodge.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; I just found on Google.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-28T01:19:14.083" LastActivityDate="2010-07-28T01:19:14.083" CommentCount="8" />
  <row Id="882" PostTypeId="2" ParentId="835" CreationDate="2010-07-28T01:02:01.250" Score="0" ViewCount="0" Body="&lt;p&gt;From what I hear about game-shows in general, if your performance does not make it to air, then you don't get anything.  Hence you cannot just accept the first amount offered (if it turns out to be a better choice) and expect to get it, since it won't make an interesting show.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-07-28T01:02:01.250" CommentCount="4" />
  <row Id="886" PostTypeId="1" AcceptedAnswerId="888" CreationDate="2010-07-28T01:20:31.943" Score="2" ViewCount="244" Body="&lt;p&gt;Okay so &lt;a href=&quot;http://math.stackexchange.com/questions/835/optimal-strategy-for-deal-or-no-deal&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt; reminded me of one my brother asked me a while back about the hit day-time novelty-worn-off-now snoozathon &lt;a href=&quot;http://en.wikipedia.org/wiki/Deal_or_no_deal&quot; rel=&quot;nofollow&quot;&gt;Deal or no deal&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;For the uninitiated:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In playing deal or no deal, the player is presented with one of 22 boxes (randomly selected) each containing different sums of money, he then asks in turn for each of the 21 remaining boxes to be opened, occasionally receiving an offer (from a wholly unconvincing 'banker' figure) for the mystery amount in his box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he rejects all of the offers along the way, the player is allowed to work his way through several (for some unfathomable reason, emotionally charged) box openings until there remain only two unopened boxes: one of which is his own, the other not. He is then given a choice to stick or switch (take the contents of his own box or the other), something he then agonises pointlessly over for the next 10 minutes.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Monty hall&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;[If you have not seen the monty hall 'paradox' check out this &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;wikipedia link&lt;/a&gt; and prepare to be baffled, then enlightened, then disappointed that the whole thing is so trivial. After which feel free to read on.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a certain similarity, you will agree, between the situation a deal or no deal player finds himself in having rejected all offers and the dilemma of Monty's contestant in the classic problem: several 'bad choices' have been eliminated and he is left with a choice between a better and worse choice with no way of knowing between them.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;So???&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; The solution to the monty hall problem is that it is, in fact, better to switch- does the same apply here? Does this depend upon the money in the boxes? Should every player opt for 'switch', cutting the 10 minutes of agonising away???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-12-21T19:51:14.277" Title="Deal or no deal: does one switch (to avoid a goat)?/ Should deal or no deal be 10 minutes shorter?" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="887" PostTypeId="2" ParentId="22" CreationDate="2010-07-28T01:34:53.483" Score="5" ViewCount="0" Body="&lt;p&gt;Here's an explanation in terms of the Hodge dual and the exterior (wedge) product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ${e_1, e_2, e_3}$ be the standard orthonormal basis for $\mathbb{R}^3$. Consider the two vectors $a = a_1 e_1 + a_2 e_2 + a_3 e_3$ and $b = b_1 e_1 + b_2 e_2 + b_3 e_3$. From the matrix computation we obtain the familiar formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a\times b = (a_2 b_3 - a_3 b_2) e_1 + (a_3 b_1 - a_1 b_3) e_2 + (a_1 b_2 - a_2 b_1) e_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (see note at the bottom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \wedge b = (a_1 b_2 - a_2 b_1) e_1 \wedge e_2 + (a_2 b_3 - a_3 b_2) e_2 \wedge e_3 + (a_3 b_1 - a_1 b_3) e_3 \wedge e_1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the wedge $\wedge$ represents the exterior product. One can now compute the dual of this latter expression using that the left contraction of $(e_1 \wedge e_2)$ onto $(e_3 \wedge e_2 \wedge e_1)$ is $e_3$ (and similar relations). The result is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \times b = (a \wedge b)^*$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, the cross product of $a$ and $b$ is the dual of their exterior product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, this is an incredible picture. The exterior product is the plane element spanned by both $a$ and $b$, and the dual is the vector orthogonal to that plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my favorite interpretation of the cross product, but it's only helpful, of course, if you're familiar with exterior algebra and the Hodge dual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The wedge product can be found by formally computing &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a_1 e_1 + a_2 e_2 + a_3 e_3) \wedge (b_1 e_1 + b_2 e_2 + b_3 e_3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the distributivity and anticommutation relations of the exterior product.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-07-28T02:03:15.647" LastActivityDate="2010-07-28T02:03:15.647" CommentCount="1" />
  <row Id="888" PostTypeId="2" ParentId="886" CreationDate="2010-07-28T01:41:01.723" Score="5" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would ceratinly plump for saying they were opened at random&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then no, the Monty Hall solution doesn't apply.  The whole point is that the door isn't &lt;em&gt;randomly&lt;/em&gt; opened, it's always a goat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way of seeing this is imagining there are 100 doors, with 99 goats.  If, after you pick a door, the host always opens 98 doors of goats, then switching is very intuitively favorable.  However, if he had just opened 98 doors at random, then most of the time (98 out of 100) he would open the door with the car behind it; and even on the rare occasions he didn't, you still wouldn't be any better off switching than staying.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-12-21T19:51:14.277" LastActivityDate="2010-12-21T19:51:14.277" CommentCount="2" />
  <row Id="889" PostTypeId="2" ParentId="874" CreationDate="2010-07-28T01:43:46.567" Score="1" ViewCount="0" Body="&lt;p&gt;If it is just 45 degrees, then the answer is not very difficult. Center a coordinate system at the bottom left hand corner of the rectangle. Hence the coordinates of the (???) point are (q,0) for some q&#xA;&#xA;&lt;p&gt;Note that the because theta is 45 degrees, y = w-q . (Isosceles right triangle). Hence q=w-y, and our point is simply (w-y, 0)...&lt;/p&gt;&#xA;" OwnerUserId="298" LastActivityDate="2010-07-28T01:43:46.567" CommentCount="1" />
  <row Id="890" PostTypeId="2" ParentId="118" CreationDate="2010-07-28T01:50:04.913" Score="0" ViewCount="0" Body="&lt;p&gt;Have a look at this explanation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://betterexplained.com/articles/a-gentle-introduction-to-learning-calculus/&quot; rel=&quot;nofollow&quot;&gt;http://betterexplained.com/articles/a-gentle-introduction-to-learning-calculus/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you love it :)&lt;/p&gt;&#xA;" OwnerUserId="313" LastActivityDate="2010-07-28T01:50:04.913" />
  <row Id="891" PostTypeId="2" ParentId="886" CreationDate="2010-07-28T01:51:46.220" Score="2" ViewCount="0" Body="&lt;p&gt;This doesn't have the key features of the Monty Hall problem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The contestant makes a choice&lt;/li&gt;&#xA;&lt;li&gt;The contestant is given information about other choices they could have made&lt;/li&gt;&#xA;&lt;li&gt;The contestant can change their choice based on this new information&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Without being given extra information, their is no point in changing your choice.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-28T01:51:46.220" />
  <row Id="892" PostTypeId="1" AcceptedAnswerId="893" CreationDate="2010-07-28T01:55:26.570" Score="0" ViewCount="90" Body="&lt;p&gt;So the projective plane RP^2 is not a vector space.  Is it still isomorphic to its dual?  If not, is there at least an invertible map that takes RP^2 to its dual?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-07-28T02:23:49.593" Title="Projective plane and its dual." Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="893" PostTypeId="2" ParentId="892" CreationDate="2010-07-28T01:59:42.910" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, though the word &quot;dual&quot; is somewhat questionable.  If you mean is $\mathbb{P}(\mathbb{R}^3)$, the standard projective plane, isomorphic to $\mathbb{P}((\mathbb{R}^3)^*)$, the projectivization of the dual, then yes, it follows from the isomorphism of vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more interestingly, the duality allows you to switch points and lines in theorems, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pascal%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Mystic Hexagon&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Brianchon%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Brianchon's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T01:59:42.910" CommentCount="2" />
  <row Id="894" PostTypeId="1" AcceptedAnswerId="928" CreationDate="2010-07-28T02:06:45.263" Score="3" ViewCount="157" Body="&lt;p&gt;In First Order Logic with Identity (FOL+I), one can express the proposition that there are exactly 3 items that have the property P. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it not possible to express the proposition that there is a finite number of items that have the property P (in FOL+I)?&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-07-29T12:37:26.077" Title="Why is the &quot;finitely many&quot; quantifier not definable in First Order Logic?" Tags="&lt;logic&gt;&lt;quantifiers&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="895" PostTypeId="2" ParentId="894" CreationDate="2010-07-28T02:10:44.943" Score="3" ViewCount="0" Body="&lt;p&gt;Any property expressible under first-order logic is closed under ultraproducts. The property of finite sets is not, however, closed under ultraproducts.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-28T02:10:44.943" CommentCount="1" />
  <row Id="896" PostTypeId="2" ParentId="894" CreationDate="2010-07-28T02:11:56.220" Score="0" ViewCount="0" Body="&lt;p&gt;Well, it would mean that you have the statement $P_1\vee P_2\vee\ldots$ where $P_i$ stands for &quot;there are $i$ objects with property $P$&quot;, and infinite disjunctions aren't allowed.  As for proving that this isn't equivalent to anything else you can write that IS allowed, I don't know how to do that.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-07-28T02:11:56.220" CommentCount="1" />
  <row Id="897" PostTypeId="2" ParentId="835" CreationDate="2010-07-28T02:20:53.437" Score="2" ViewCount="0" Body="&lt;p&gt;There are (at least) two factors that mean that simply calculating the average of the remaining options is not enough to describe how someone should play.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Risk_aversion&quot; rel=&quot;nofollow&quot;&gt;Risk aversion&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Someone's utility is not a predictable function of the amount of money that they win. For instance my utility from winning \$5 is more than 100 times my utility from winning 5 cents. However, my utility from winning \$100 million is less than 100 times my utility from winning \$1 million.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditorDisplayName="" LastEditDate="2010-12-08T04:17:24.557" LastActivityDate="2010-12-08T04:17:24.557" />
  <row Id="898" PostTypeId="2" ParentId="892" CreationDate="2010-07-28T02:23:49.593" Score="2" ViewCount="0" Body="&lt;p&gt;This is a comment more than an answer, mainly in response to Qiaochu's question, but I don't have sufficient rep. to comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual to a projective plane is the set of all lines in the plane, which itself is a&#xA;projective plane (as hinted at in Charles Siegel's answer).  This is an important concept in classical projective geometry.  (Concretely, the equation for a line has the form&#xA;a x + b y + c z = 0, where a,b, and c are some parameters, not all zero, and x,y,z are homogeneous coords. for the points in the proj. plane.  The set of all such lines can thus&#xA;be thought of as the set of all (a,b,c) not all zero; but note that simultaneously multiplying a, b, and c by a non-zero scalar doesn't change the solution set, i.e. doesn't change the line, so the line should really be thought of as corresponding to the homogeneous coordinates (a:b:c); thus the set of all lines is again a projective plane.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T02:23:49.593" />
  <row Id="899" PostTypeId="2" ParentId="154" CreationDate="2010-07-28T02:30:18.093" Score="12" ViewCount="0" Body="&lt;p&gt;In my opinion, the most natural way to view complex number is as a class of maps from the plane to itself. Specifically, lets define $(R, \theta)$ to be the map which multiplies every point in the plane by the number $R$, and then rotates it by the angle $\theta$. We may call these maps &quot;dilations with rotations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such maps can be added and composed (multiplied) in the obvious way, and its not hard to work out that the sum and product of two such mappings is another dilation with rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also identify the real number $x$ with the map $(x,0)$, i.e. the map which multiplies every point in the plane by $x$.  Then we see that these maps have the magical property that $-1$ has a square root! Namely, if $P$ is the mapping $(1,\pi/2)$ (i.e. rotate every point by angle $\pi/2$), then applying $P$ twice is the same as multiplying every number by $-1$, i.e. $P^2=-1$! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As should be obvious by now, these maps are just complex numbers in disguise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unsurprisingly, they are &#xA;singularly useful for solving polynomial equations. Indeed, the real number $x'$ is a root of the polynomial equations $a_0 x^n + a_1 x^{n-1} + \cdots + a_n =0$ if and only if the mapping $(x',0)$ satisfies the same equation. So viewing polynomial equations over the set of these mappings loses no solutions, while at the same time giving us additional freedom to do operations such as taking the square roots of negative numbers. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-07-28T02:30:18.093" CommentCount="1" />
  <row Id="900" PostTypeId="2" ParentId="667" CreationDate="2010-07-28T02:30:26.223" Score="5" ViewCount="0" Body="&lt;p&gt;What you are referring to is the &quot;stacks as sheaves of groupoids&quot; point of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate where it comes from, imagine for example that we are talking about the moduli stack of elliptic curves (on the category of schemes).  To give an elliptic curve over a scheme, it is not just enough to specify the elliptic curve over the members of the open cover; we have to explain how we glue the restrictions of the curves on the various opens on their overlaps, and this gluing has to be coherent over triple overlaps.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that elliptic curves can have non-trivial automorphisms, so that there is no &lt;em&gt;a priori&lt;/em&gt; determined way to make the identifications on the overlaps (because having non-trivial automorphisms is the same as saying that when two curves are isomorphic, they can be isomorphic in more than one way), so it is your job to choose these identifications, and to make sure that you do it in a coherent way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here elliptic curves can be replaced by any other moduli problem you can think of, of course.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-07-29T00:13:37.180" LastActivityDate="2010-07-29T00:13:37.180" />
  <row Id="901" PostTypeId="2" ParentId="631" CreationDate="2010-07-28T02:53:48.710" Score="1" ViewCount="0" Body="&lt;p&gt;There are a few subtleties that will probably effect the final answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are we required to find the solution, or merely establish existence? By analogy, determining if a number has a prime factorization is trivial, but finding its prime factorization is hard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the runtime being measured in terms of {a_1,...,a_n,s} or {log(a_i),...,log(a_n),log(s)}? By analogy, SUBSET-SUM is in P in the first case, but NP-complete in the second case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="275" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2010-07-28T02:56:07.330" LastActivityDate="2010-07-28T02:56:07.330" CommentCount="6" />
  <row Id="902" PostTypeId="2" ParentId="518" CreationDate="2010-07-28T02:56:36.067" Score="1" ViewCount="0" Body="&lt;p&gt;This is a comment on the proof sketch in the question (I don't have sufficient rep. to actually leave comments): the localization A_P is local of dimension zero (its unique maximal ideal is also a minimal prime ideal, and hence is the unique prime ideal of A_P), but need not be Artinian, as far as I can tell.  E.g. if A = C[[x^{1/2},x^{1/3},...]]/(x), then A is local&#xA;with a unique prime ideal (namely the ideal generated by all the x^{1/n}), but is not Artinian (equivalently, not Noetherian), since if (a_i) is any strictly descending sequence of rational numbers in the interval (0,1), then the ideals x^{a_i} A form a strictly descending sequence&#xA;of ideals in A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully I'm not blundering here; if I am, someone please let me know!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, I should add that it is still the case that since PA_P is the unique minimal&#xA;prime of A_P, every element of PA_P is nilpotent, and hence every element of P is&#xA;a zero divisor, so my comment is very nitpicky: it is just about the use of the terminology Artinian.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T02:56:36.067" CommentCount="3" />
  <row Id="903" PostTypeId="1" CreationDate="2010-07-28T03:10:02.303" Score="5" ViewCount="379" Body="&lt;p&gt;Which is a better textbook - Dugundji or Munkres? I'm concerned with clarity of exposition and explanation of motivation, etc.&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T03:26:01.203" LastActivityDate="2010-08-06T12:53:00.723" CommunityOwnedDate="2010-07-28T03:17:30.603" Title="choosing a topology text" Tags="&lt;soft-question&gt;&lt;books&gt;&lt;topology&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="2" />
  <row Id="904" PostTypeId="2" ParentId="868" CreationDate="2010-07-28T03:38:52.363" Score="5" ViewCount="0" Body="&lt;p&gt;If you are interested in the factorization algorithms employed in modern computer algebra systems such as Macsyma, Maple, or Mathematica, then see any of the standard introductions to computer algebra , e.g. Geddes et.al. &quot;Algorithms for Computer Algebra&quot;; Knuth, &quot;TAOCP&quot; v.2; von zur Gathen &quot;Modern Computer Algebra&quot;; Zippel &quot;Effective Polynomial Computation&quot;. See also&#xA;Kaltofen's surveys on polynomial factorization [116,68,56,7] in his publications &lt;a href=&quot;http://www4.ncsu.edu/~kaltofen/bibliography/kaltofen.html&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt;, which contains plenty of theory, history and literature references. Note: Kaltofen's home page appears to be temporarily down so instead see his &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt; to get started (see comments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; Kaltofen, E. Factorization of Polynomials, pp. 95-113 in:&lt;br&gt;&#xA;Computer Algebra, B. Buchberger, R. Loos, G. Collins, editors, Vienna, Austria,  (1982).&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.7916&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-07-28T03:44:15.640" LastActivityDate="2010-07-28T03:44:15.640" CommentCount="1" />
  <row Id="905" PostTypeId="2" ParentId="29" CreationDate="2010-07-28T03:39:45.887" Score="3" ViewCount="0" Body="&lt;p&gt;Try sage.  It's like python (it is actually written in python, and uses something similar to a python shell), but with a vastly larger set of built-in mathematical operations.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-07-28T03:39:45.887" CommunityOwnedDate="2010-07-28T03:39:45.887" />
  <row Id="906" PostTypeId="2" ParentId="29" CreationDate="2010-07-28T03:53:58.273" Score="1" ViewCount="0" Body="&lt;p&gt;Some people have suggested Sage or Python. Sage can do the symbolic manipulation that TI calculators can do using Maxima. Python with the right packages can do this as well (SymPy package I believe). Plus these things will be programmable. Sage however is fairly heavy at a few GB if I recall correctly. I'm not sure if there is anyway to interface with Maxima through Python like in Sage without actually having Sage.&lt;/p&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-07-28T04:01:11.157" LastActivityDate="2010-07-28T04:01:11.157" CommunityOwnedDate="2010-07-28T03:53:58.273" />
  <row Id="907" PostTypeId="1" AcceptedAnswerId="924" CreationDate="2010-07-28T03:56:11.673" Score="3" ViewCount="463" Body="&lt;p&gt;While I feel quite confident that I've inferred the correct meaning of &quot;In the sequel&quot; from context, I've never heard anyone explicitly tell me, so first off, to remove my niggling doubts: What does this phrase mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Someone recently argued to me that &quot;sequel&quot; was actually supposed to refer to a forthcoming second part of a paper, which I found highly unlikely, but I'd just like to make sure. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main questions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what points in the text, and for what kinds of X, is it appropriate to use the phrase &quot;In the sequel, X&quot; in a paper? In a book? Is it ever acceptable to introduce definitions via &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In the sequel, we introduce the concept of a &quot;blah&quot;, which is a thing satisfying ...&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the start of a paper or book &lt;strong&gt;without&lt;/strong&gt; a formal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Definition. A &quot;blah&quot; is a thing, satsifying ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the main text of the paper or book? Finally, out of curiosity, I'm wondering how long this phrase has been around, if it's considered out of date or if it's still a popular phrase, and what some good alternatives are.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T01:16:54.680" LastActivityDate="2010-09-08T08:36:29.900" Title="Correct usage of the phrase &quot;In the sequel&quot;? History? Alternatives?" Tags="&lt;soft-question&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="908" PostTypeId="2" ParentId="907" CreationDate="2010-07-28T04:01:41.187" Score="2" ViewCount="0" Body="&lt;p&gt;I would never write that (mostly because it really sounds like the definition will be in another paper...). I'm pretty sure I'd write «In what follows, ...».&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T04:01:41.187" CommentCount="2" />
  <row Id="910" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T04:07:58.903" Score="6" ViewCount="0" Body="&lt;p&gt;In fact the result holds a bit more generally, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm \quad 24\ |\ M^2 - N^2 \;$ if $\rm \; M,N \perp 6, \;$ i.e. coprime to $6.\;$ The proof is easy:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad N\perp 2 \;\Rightarrow\; N = \pm 1, \pm 3 \pmod 8 \;\Rightarrow\; N^2 = 1 \pmod 8$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad N\perp 3 \;\Rightarrow\; N \;\;= \;\;\;\pm 1\:\quad \pmod 3 \;\Rightarrow\; N^2 = 1 \pmod 3 \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm \quad\; 3, 8\ |\ N^2 - 1 \;\Rightarrow\;  24\ |\ N^2 - 1 \ $ since $\ {\rm lcm}(3,8) = 24.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case $\rm\ n = 24\ $ of this much more general result  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $  For naturals $\rm\ a,e,n $ with $\rm\ e,n&amp;gt;1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad n\ |\ a^e-1$ for all $\rm a\perp n \ \iff\  \phi&amp;#39;(p^k)\:|\:e\ $ for all $\rm\ p^k\:|\:n\:,\ \ p\:$ prime&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\rm \;\;\; \phi&amp;#39;(p^k) = \phi(p^k)\ $ for odd primes $\rm p\:,\ $ where $\phi$ is Euler's totient function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and   $\rm\ \quad \phi&amp;#39;(2^k) = 2^{k-2}\ $ if $\rm  k&amp;gt;2\:,\ $ else $2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter exception is due to $\rm \mathbb Z/2^k$ having multiplicative group $\rm C(2) \times C(2^{k-2})$ for $\rm k&amp;gt;2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the least such exponent $\rm e$ is given by $\rm \;\lambda(n)\; = \;{\rm lcm}\;\{\phi&amp;#39;(\;{p_i}^{k_i})\}\;$ where $\rm \; n = \prod {p_i}^{k_i}\;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\lambda(n)$ is called the (universal) &lt;a href=&quot;http://planetmath.org/encyclopedia/Exponent.html&quot; rel=&quot;nofollow&quot;&gt;exponent&lt;/a&gt; of the group $\rm \mathbb Z/n^*,\;$ a.k.a. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_function&quot; rel=&quot;nofollow&quot;&gt;Carmichael function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the case at hand is simply $\rm\ \lambda(24) = lcm(\phi&amp;#39;(2^3),\phi&amp;#39;(3)) = lcm(2,2) = 2\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zk55tqcie.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for proofs and further discussion.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-08T03:36:50.213" LastActivityDate="2011-01-08T03:36:50.213" />
  <row Id="911" PostTypeId="2" ParentId="250" CreationDate="2010-07-28T04:22:44.383" Score="13" ViewCount="0" Body="&lt;p&gt;Scott Aaronson once basically did this for a bunch of theorems in computer science &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://scottaaronson.com/blog/?p=153&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  I particularly like this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose a baby is given some random examples of grammatical and ungrammatical sentences, and based on that, it wants to infer the general rule for whether or not a given sentence is grammatical. If the baby can do this with reasonable accuracy and in a reasonable amount of time, for any “regular grammar” (the very simplest type of grammar studied by Noam Chomsky), then that baby can also break the RSA cryptosystem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="232" LastEditorUserId="794" LastEditorDisplayName="" LastEditDate="2010-11-26T16:44:21.833" LastActivityDate="2010-11-26T16:44:21.833" CommunityOwnedDate="2010-07-28T04:22:44.383" CommentCount="1" />
  <row Id="912" PostTypeId="2" ParentId="796" CreationDate="2010-07-28T04:27:18.310" Score="2" ViewCount="0" Body="&lt;p&gt;There is an extremely beautiful Lie-theoretic approach to separation of variables,&#xA;e.g. see Willard Miller's &lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;book [1]&lt;/a&gt;. I quote from his introduction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This book is concerned with the&#xA;  relationship between symmetries of a &#xA;  linear second-order partial&#xA;  differential equation of mathematical&#xA;  physics,  the coordinate systems in&#xA;  which the equation admits solutions&#xA;  via separation  of variables, and the&#xA;  properties of the special functions&#xA;  that arise in  this manner. It is an&#xA;  introduction intended for anyone with&#xA;  experience in  partial differential&#xA;  equations, special functions, or Lie&#xA;  group theory, such  as group&#xA;  theorists, applied mathematicians,&#xA;  theoretical physicists and  chemists,&#xA;  and electrical engineers. We will&#xA;  exhibit some modem group-theoretic &#xA;  twists in the ancient method of&#xA;  separation of variables that can be &#xA;  used to provide a foundation for much&#xA;  of special function theory. In &#xA;  particular, we will show explicitly&#xA;  that all special functions that arise &#xA;  via separation of variables in the&#xA;  equations of mathematical physics can &#xA;  be studied using group theory. These&#xA;  include the functions of Lam6, Ince, &#xA;  Mathieu, and others, as well as those&#xA;  of hypergeometric type. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is a very critical time in the&#xA;  kistory of group-theoretic methods in &#xA;  special Iunction theory. The basic&#xA;  relations between Lie groups, special &#xA;  functions, and the method of&#xA;  separation of variables have recently&#xA;  been  clarified. One can now construct&#xA;  a group-theoretic machine that, when &#xA;  applied to a given differential&#xA;  equation of mathematical physms,&#xA;  describes  in a rational manner the&#xA;  possible coordinate systems in which&#xA;  the equation  admits solutions via&#xA;  separation of variables and the&#xA;  various expansion  theorems relating&#xA;  the separable (special function)&#xA;  solutions in distinct  coordinate&#xA;  systems. Indeed for themost important&#xA;  linear equations, the  separated&#xA;  solutions are characterized as common&#xA;  eigenfunctions of sets of &#xA;  second-order commuting elements in the&#xA;  umversal enveloping algebra of  the&#xA;  Lie symmetry algebra corresponding to&#xA;  the equation. The problem of &#xA;  expanding one set of separable&#xA;  solutions in terms of another reduces&#xA;  to a  problem in the representation&#xA;  theory of the Lie symmetry algebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; Willard Miller. Symmetry and Separation of Variables.&lt;br&gt;&#xA;Addison-Wesley, Reading, Massachusetts, 1977 (out of print)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ima.umn.edu/~miller/separationofvariables.html&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://gigapedia.com/items:links?id=64401&quot; rel=&quot;nofollow&quot;&gt;http://gigapedia.com/items:links?id=64401&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-28T04:27:18.310" />
  <row Id="913" PostTypeId="2" ParentId="791" CreationDate="2010-07-28T04:29:55.297" Score="2" ViewCount="0" Body="&lt;p&gt;The answer can be found on pages 27-28 of the following link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="309" LastActivityDate="2010-07-28T04:29:55.297" CommentCount="0" />
  <row Id="914" PostTypeId="2" ParentId="741" CreationDate="2010-07-28T04:30:38.603" Score="1" ViewCount="0" Body="&lt;p&gt;I suggest you &lt;a href=&quot;http://rjlipton.wordpress.com/2010/06/26/stating-pnp-without-turing-machines/&quot; rel=&quot;nofollow&quot; title=&quot;this&quot;&gt;this&lt;/a&gt; article, which talks about Linear Equations, Linear Programming, Integer Programming and P=NP. It's easy to understand and talks about the differences among these things&lt;/p&gt;&#xA;" OwnerUserId="229" LastActivityDate="2010-07-28T04:30:38.603" />
  <row Id="915" PostTypeId="2" ParentId="362" CreationDate="2010-07-28T04:44:48.480" Score="8" ViewCount="0" Body="&lt;p&gt;For a nice introduction to the history of ring theory see the following paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. Kleiner. &lt;strong&gt;From numbers to rings: the early history of ring theory&lt;/strong&gt;.&lt;br&gt;&#xA;Elemente der Mathematik 53 (1998) 18-35.&lt;br&gt;&#xA;&lt;a href=&quot;http://retro.seals.ch/cntmng?type=pdf&amp;amp;rid=elemat-001:1998:53::22&amp;amp;subp=hires&quot; rel=&quot;nofollow&quot;&gt;http://retro.seals.ch/cntmng?type=pdf&amp;amp;rid=elemat-001:1998:53::22&amp;amp;subp=hires&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.springerlink.com/content/e7p7kxk2y3a71j0d/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/e7p7kxk2y3a71j0d/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-28T04:44:48.480" CommentCount="2" />
  <row Id="916" PostTypeId="2" ParentId="874" CreationDate="2010-07-28T04:58:52.313" Score="0" ViewCount="0" Body="&lt;p&gt;Alright, I'm not 100% sure I'm understanding this correctly. You say that p can be located on the right or top line and that p2 can be located on the bottom or left line. Do you mean the rectangle can be rotated? If that's the case, the question should say that p can be on the right or bottom line of the rectangle. Also, are you looking for two separate answers or one that works both when p2 is on the bottom and on the left?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do mean that the rectangle can be rotated, and want two different answers, it's pretty simple. First I'll deal with when p2 is on the bottom and p is on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since p2 is on the bottom line we know the y-coordinate is h, according to the diagram. We also know that p is (0,y). Because of the 45 degree angle, we know that the distance between p's y-coordinate and the lower right corner is the same as the distance between the lower right corner and p2's x-coordinate, which in this case is p2's x-coordinate. Therefore, the coordinates of p2 are (h-y, h).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p2 is on the left and p is on the bottom, it's very similar. Since p2 is on the left, it's x-coordinate is h. Because p is on the x-axis, it's (x,0). Because of the 45 degree angle, the distance between the lower left corner and p is the same as the distance between the lower left corner and p2, which this time gives us p2's y-coordinate. Therefore the coordinates of p2 are (h,h-x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully I understood your intentions correctly. If not, I hope you can use my misunderstandings to further improve your question.&lt;/p&gt;&#xA;" OwnerUserId="220" LastActivityDate="2010-07-28T04:58:52.313" CommentCount="1" />
  <row Id="917" PostTypeId="2" ParentId="250" CreationDate="2010-07-28T05:06:14.937" Score="21" ViewCount="0" Body="&lt;p&gt;My favorite would probably be Goodstein's theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with your favorite number (mine is 37) and express it in hereditary base 2 notation.  That is, write it as a power of 2 with exponents powers of 2, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 37 = 2^(2^2 + 1) + 2^2 + 1.  This is the first element of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, change all the 2's to 3's, and subtract one from what's remaining and express in hereditary base 3 notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get 3^(3^3 + 1) + 3^3 + 1 - 1= 3^(3^3 + 1) + 3^3 (which is roughly 2 x 10^13).  This is the second element of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, change all 3's to 4's, subtract one, and express in hereditary base 4 notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get 4^(4^4 + 1) + 4^4 - 1 = 4^(4^4 + 1) + 3*4^3 + 3*4^2 + 3*4 + 3 (which is roughly 5 x 10^154) .  This is the third element of the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rinse, repeat:  at the nth stage, change all the &quot;n+1&quot; to &quot;n+2&quot;, subtract 1, and reexpress in hereditary base n+2 notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem is: no matter which number you start with, eventually, your sequence hits 0, despite the fact that it grows VERY quickly at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if instead of starting with 37, we started with 4, then (according to the wikipedia page), it takes 3*2^(402653211) - 2 steps ( VERY roughly 10^(100,000,000), or a 1 followed by a hundred million 0s).  37 takes vastly longer to drop to 0.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-07-28T05:06:14.937" CommunityOwnedDate="2010-07-28T05:06:14.937" CommentCount="2" />
  <row Id="918" PostTypeId="1" AcceptedAnswerId="929" CreationDate="2010-07-28T05:32:29.170" Score="12" ViewCount="741" Body="&lt;p&gt;Let $S$ be a set of size $n$.  There is an easy way to count the number of subsets with an even number of elements.  Algebraically, it comes from the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} {n \choose k} = (1 + 1)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} (-1)^k {n \choose k} = (1 - 1)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n/2} {n \choose 2k} = 2^{n-1}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A direct combinatorial proof is as follows: fix an element $s \in S$.  If a given subset has $s$ in it, add it in; otherwise, take it out.  This defines a bijection between the number of subsets with an even number of elements and the number of subsets with an odd number of elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous formulas for the subsets with a number of elements divisible by $3$ or $4$ are more complicated, and divide into cases depending on the residue of $n \bmod 6$ and $n \bmod 8$, respectively.  The algebraic derivations of these formulas are as follows (with $\omega$ a primitive third root of unity):  observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} \omega^k {n \choose k} = (1 + \omega)^n = (-\omega^2)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n} \omega^{2k} {n \choose k} = (1 + \omega^2)^n = (-\omega)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that $1 + \omega^k + \omega^{2k} = 0$ if $k$ is not divisible by $3$ and equals $3$ otherwise.  (This is a special case of the discrete Fourier transform.)  It follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n/3} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega)^{2n}}{3}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\omega$ and $-\omega^2$ are sixth roots of unity, so this formula splits into six cases (or maybe three).  Similar observations about fourth roots of unity show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{n/4} {n \choose 4k} = \frac{2^n + (1+i)^n + (1-i)^n}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $1+i = \sqrt{2} e^{ \frac{\pi i}{4} }$ is a scalar multiple of an eighth root of unity, so this formula splits into eight cases (or maybe four).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  Does anyone know a direct combinatorial proof of these identities? &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-03T08:05:40.423" LastActivityDate="2010-08-03T08:05:40.423" Title="How do I count the subsets of a set whose number of elements is divisible by 3? 4?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" />
  <row Id="919" PostTypeId="2" ParentId="636" CreationDate="2010-07-28T05:57:09.560" Score="4" ViewCount="0" Body="&lt;p&gt;I was recently surprised to discover that it's actually not known. The number of &lt;em&gt;closed&lt;/em&gt; knight's tours (cyclic) was computed in the 1990s, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_decision_diagram&quot; rel=&quot;nofollow&quot;&gt;Binary decision diagrams&lt;/a&gt;. There are 26,534,728,821,064 closed directed knight's tours, and the number of undirected ones is half that or 13,267,364,410,532. If you count equivalence classes under rotation and reflection, there are 1/8th of that: 1,658,420,855,433.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Loebbing and  Wegener (1996) wrote a paper &lt;a href=&quot;http://www.combinatorics.org/Volume_3/Abstracts/v3i1r5.html&quot; rel=&quot;nofollow&quot;&gt;&quot;The Number of Knight's Tours Equals 33,439,123,484,294 — Counting with Binary Decision Diagrams&quot;&lt;/a&gt;; the number in the title in the mistake, as they pointed out in &lt;a href=&quot;http://www.combinatorics.org/Volume_3/Comments/v3i1r5.html&quot; rel=&quot;nofollow&quot;&gt;a comment&lt;/a&gt; to their paper. Brendon McKay independently &lt;a href=&quot;http://www.combinatorics.org/Volume_3/Comments/v3i1r5.01.ps&quot; rel=&quot;nofollow&quot;&gt;computed&lt;/a&gt; the correct number with another method, and the original authors seem to have later &lt;a href=&quot;http://books.google.com/books?id=-DZjVz9E4f8C&amp;amp;pg=PA369&amp;amp;dq=532&quot; rel=&quot;nofollow&quot;&gt;found&lt;/a&gt; the same answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the exact number of open tours (not cyclic/reentrant) remains &lt;a href=&quot;http://www.ktn.freeuk.com/pa.htm&quot; rel=&quot;nofollow&quot;&gt;open, but it is estimated&lt;/a&gt; to be about 10&lt;sup&gt;15&lt;/sup&gt; or 2×10&lt;sup&gt;16&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-28T05:57:09.560" />
  <row Id="920" PostTypeId="2" ParentId="20" CreationDate="2010-07-28T06:04:47.113" Score="4" ViewCount="0" Body="&lt;p&gt;I think you were being a little too hard on Isaac.  The truth is that the real numbers are a sophisticated mathematical construction and that any explanation of what they &quot;are&quot; which pretends otherwise is a convenient fiction.  Mathematicians need these kind of sophisticated constructions because they are what is required for rigorous proofs.  Before people explicitly constructed the real numbers and used them to define and prove things about other concepts, it was never totally clear what was true or what was false, and everybody was very confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Cantor proved that the number of points in the plane is the same as the number of  points on a line.  Many people thought that this was impossible before he did it; they had an intuition that you couldn't possibly &quot;fit&quot; the plane into the line.  More generally, people were pretty sure you couldn't fit $\mathbb{R}^n$ into $\mathbb{R}^m$ if $n$ was greater than $m$.  It wasn't until quite a bit later that mathematicians formalized and proved a rigorous mathematical statement which justified this intuition called &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariance_of_domain&quot; rel=&quot;nofollow&quot;&gt;invariance of domain&lt;/a&gt;, which says you can't do this in a &lt;em&gt;continuous&lt;/em&gt; way.  One of the many mathematical constructions you need to even state this theorem is the construction of the real numbers.  (Another is a formal definition of what &quot;continuous&quot; means, but one thing at a time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what are the real numbers?  They are a formal way to fill &quot;holes&quot; in the rational numbers, which is necessary for all sorts of things.  The most basic thing they are necessary for is doing geometry.  You probably know that the square root of 2 is irrational.  What this means is that it is impossible to think about the diagonal of a square as being the same kind of object as the sides of a square while only using rational numbers.  But you can rotate a diagonal, and it looks just like the side of a square, only a bit longer.  So you'd like a number system in which you can sensibly talk about any number you can construct geometrically.  You'd also like to be able to talk about rotation!  You can't do that with just rational numbers, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you fill in enough holes to do geometry?  &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_cut&quot; rel=&quot;nofollow&quot;&gt;Dedekind&lt;/a&gt; came up with a very clever way to do this.  It starts by observing that a rational number $q$ is completely determined by the set of rational numbers greater than it and the set of rational numbers less than it.  For example, 1/2 is completely determined by the fact that it's always between 1/2 + 1/n and 1/2 - 1/n.  (For the initiated, this is a special case of the Yoneda lemma.)  But there are &quot;numbers,&quot; such as the square root of 2, which aren't rational, and yet have the property that we can always tell what rational numbers are greater than it and what rational numbers are less than it.  For the square root of 2, these are precisely the fractions p/q such that 2q^2 &amp;lt; p^2 and such that 2q^2 &gt; p^2, respectively.  Dedekind's brilliant idea was the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Define&lt;/em&gt; a real number to be a partition of the rational numbers!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Dedekind's construction, the square root of 2 quite literally &lt;em&gt;is&lt;/em&gt; the set of rational numbers that are greater than it, and the set of rational numbers that are less than it. You can define all the usual arithmetic operations on these &quot;numbers,&quot; called Dedekind cuts, and prove all the wonderful theorems you'll find in a standard book on real analysis.  In particular, the property that guarantees that all the holes are filled is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_number#Completeness&quot; rel=&quot;nofollow&quot;&gt;completeness&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Figured I might as well add something about the complex numbers.  The story here is beautiful, and if you're really interested you should check out Tristan Needham's &lt;a href=&quot;http://usf.usfca.edu/vca//&quot; rel=&quot;nofollow&quot;&gt;Visual Complex Analysis&lt;/a&gt;.  Some people say that the point of the complex numbers is to let you solve polynomials, but this is really selling them short.  The complex numbers are an inherently &lt;em&gt;geometric&lt;/em&gt; construction, and should be understood as such.  Their geometry and topology just happens to be responsible for the fact that you can solve polynomials with them, but it's also responsible for much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a quick sketch.  Now that you've got the real numbers on your hands, you can rigorously talk about plane geometry.  In plane geometry, an important notion is that of &lt;em&gt;similarity&lt;/em&gt;.  Informally, two figures are similar if they have the same shape.  More formally, two figures are similar if you can rotate, translate, and scale one figure so that it matches up with the other.  So similarity is all about a certain collection of transformations of the points in the plane.  It was &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlangen_program&quot; rel=&quot;nofollow&quot;&gt;Klein&lt;/a&gt; who first realized that the important features of different flavors of &quot;geometry&quot; are captured in what kind of transformations are allowed.  So to do geometry the modern way we should focus our attention on these transformations, which form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_theory&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this easier, let's ignore the translations for now.  We'll pick an origin for our plane, and we'll only allow rotations and scalings about this origin.  Rotations and scalings have the property that they are both linear transformations; this means that if you know what the transformation does to two points $u, v$, you also know what it does to the vector sum $u + v$.  In particular, a linear transformation is determined by what it does to the point $(1, 0)$ and to the point $(0, 1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, rotations and scalings satisfy an extra property: they are, in fact, determined by what they do to the point $(1, 0)$.  This is because $(0, 1)$ can be obtained from $(1, 0)$ by a rotation by 90 degrees, and rotations and scalings &lt;em&gt;commute&lt;/em&gt; with each other: if you rotate x degrees then y degrees, that's the same as rotating y degrees then x degrees, which is the same as rotating x+y degrees.  Similarly, if you rotate x degrees then scale by 2, that's the same as scaling by 2, then rotating x degrees.  So if you know what a rotation-and-scaling does to $(1, 0)$, you just rotate that vector by 90 degrees, and you know what it did to $(0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to every rotation-and-scaling, we can assign two real numbers: the coordinates of the image of the point $(1, 0)$.  In general, a rotation by $\theta$ angles followed by a scaling by $r$ sends $(1, 0)$ to $(r \cos \theta, r \sin \theta)$.  A different transformation, say a rotation by $\phi$ angles followed by a scaling by $s$, sends $(1, 0)$ to $(s \cos \phi, s \sin \phi)$.  And their &lt;em&gt;composition&lt;/em&gt; sends $(1, 0)$ to $(rs \cos (\theta + \phi), rs \sin (\theta + \phi))$.  In other words, composition of rotations-and-scalings defines a multiplication law on pairs of real numbers.  What is this law, exactly?  Well, by the angle addition formulas, it's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a, b) * (c, d) = (ac - bd, ad + bc).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is precisely the rule for multiplication in the complex numbers, where $(a, b)$ corresponds to $a + bi$.  You get the rule for addition by observing that not only can you compose two rotations-and-scalings, you can also add their results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together, the real numbers and the complex numbers provide a foundation for much of modern mathematics and physics.  For example, the complex numbers turn out (for reasons which are still not well understood) to be fundamental in the description of quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-28T06:35:01.467" LastActivityDate="2010-07-28T06:35:01.467" />
  <row Id="921" PostTypeId="1" AcceptedAnswerId="926" CreationDate="2010-07-28T06:58:36.537" Score="-2" ViewCount="398" Body="&lt;p&gt;Mathematics is beautiful. Many numbers are interesting such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1729 = 1&lt;sup&gt;3&lt;/sup&gt; + 12&lt;sup&gt;3&lt;/sup&gt; = 9&lt;sup&gt;3&lt;/sup&gt; + 10&lt;sup&gt;3&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please list out the interesting numbers and reasons to make number interesting&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-28T17:20:37.613" LastActivityDate="2010-07-29T20:38:30.733" CommunityOwnedDate="2010-07-28T06:58:53.930" ClosedDate="2010-07-30T02:30:08.647" Title="Interesting numbers in Maths" Tags="&lt;big-list&gt;&lt;recreational-mathematics&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="923" PostTypeId="2" ParentId="20" CreationDate="2010-07-28T07:04:41.780" Score="4" ViewCount="0" Body="&lt;p&gt;The &lt;em&gt;natural numbers&lt;/em&gt; are defined by the Peano axioms, as in the &lt;a href=&quot;http://math.stackexchange.com/questions/20/can-you-explain-what-a-real-number-is-also-rational-decimal-integer-natural/79#79&quot;&gt;answer of Isaac&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also view the natural numbers as the cardinalities of finite sets, which implies that zero is a natural number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the other number domains arise because mathematicians want to give values for certain operations which otherwise are only defined partially. Alternatively, they want to give solutions to equations of certain forms, which cannot always be solved in the smaller domain.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;integers&lt;/em&gt; arise from subtraction (solutions of &lt;code&gt;a + x = b&lt;/code&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;rational numbers&lt;/em&gt; arise from division (solutions of &lt;code&gt;a * x = b&lt;/code&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;real numbers&lt;/em&gt; arise from taking limits or upper bounds&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;complex numbers&lt;/em&gt; arise from taking roots (solutions of polynomial equations)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;cardinal numbers&lt;/em&gt; arise as the sizes of sets. They can be constructed as isomorphism classes of sets, e.g. the number one would then be the class of all sets containing one element. They are useful for talking about the different sizes of infinite sets, which can get extremely large. Some set theorists compete over who can define the largest cardinal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;ordinal numbers&lt;/em&gt; arise as the sizes of well-orderings. They can be constructed as isomorphism classes of well-orderings, e.g. the number two would then be the class of all well-orderings containing two elements (with the first smaller than the second in the ordering). Interestingly, the ordinals are themselves well-ordered in a single well-ordering, so they come in a well-defined sequence, and an ordinal number can be viewed as describing its position in this sequence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="210" LastEditorUserId="210" LastEditorDisplayName="" LastEditDate="2010-07-29T19:29:08.340" LastActivityDate="2010-07-29T19:29:08.340" CommentCount="2" />
  <row Id="924" PostTypeId="2" ParentId="907" CreationDate="2010-07-28T07:12:25.707" Score="12" ViewCount="0" Body="&lt;p&gt;Dear Zev,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your interpretation of how &quot;in the sequel&quot; is used in mathematical literature is correct: it means &quot;in what follows&quot;, &quot;in the remainder of the present text&quot;, &quot;from now on&quot;...As you can see, there are many other such expressions, and I don't think that &quot;in the sequel&quot; has any particular nuance of meaning that these other phrasings lack.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to your question of whether this is acceptable usage in a math paper: acceptable, yes, but not completely recommended.  In more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ths phrase is, as you point out, rather common, so any experienced reader of math will have seen it before.  Also its meaning should be relatively transparent to a literate native speaker of English: this is after all what the dictionary says that &quot;sequel&quot; means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are two drawbacks.  First, as Mariano says, nowadays we hear &quot;sequel&quot; used most commonly for the next movie in a franchise, so its use in a math paper will inevitably make some readers think that you are referring to &lt;strong&gt;[TITLE OF YOUR PAPER] Part II: This Time, It's Personal&lt;/strong&gt;, or whatever.  Second, in my opinion it is somewhat hackneyed language and an idiomatic usage that doesn't add any meaning.  If you read enough math papers you'll find that there are certain linguistic ticks that people pick up from each other (and also sometimes, from writing in other languages), e.g. beginning a sentence with &quot;Remark that...&quot; sounds a little stilted and is probably a semiconscious translation of the french phrase &quot;Remarquer que&quot; (which I would translate as &quot;Notice that&quot;).  Most people agree that good mathematical writing is as stylistically unobstrusive as possible, so using phrases that make people ponder and ask questions about them on websites is, arguably, slightly too distracting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it's  no big deal.  If I were a referee or even a copyeditor, I would almost certainly let the author use &quot;in the sequel&quot; if she wants.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-07-28T07:12:25.707" CommentCount="2" />
  <row Id="925" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T07:18:03.370" Score="2" ViewCount="0" Body="&lt;p&gt;$163$ is interesting because it's the largest &lt;a href=&quot;http://en.wikipedia.org/wiki/Heegner_number&quot; rel=&quot;nofollow&quot;&gt;Heegner number&lt;/a&gt;.  This means that it's the largest positive integer $d$ such that the ring of integers in $\mathbb{Q}(\sqrt{-d})$ has unique factorization.  If you don't know what that means, two other elementary consequences of $163$ being a Heegner number are that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n^2 + n + 41$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is prime for $n = 0, 1, 2, ... 39$ and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{\pi \sqrt{163}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &lt;em&gt;extremely&lt;/em&gt; close to an integer.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-29T20:38:30.733" LastActivityDate="2010-07-29T20:38:30.733" CommunityOwnedDate="2010-07-28T07:18:03.370" />
  <row Id="926" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T07:26:24.550" Score="10" ViewCount="0" Body="&lt;p&gt;All positive integers are interesting. Suppose they weren't. Let k be the lowest non-interesting positive integer. That is an interesting property. Therefore contradiction. Generalisation is left as an exercise to the reader.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T08:22:33.930" LastActivityDate="2010-07-28T08:22:33.930" CommunityOwnedDate="2010-07-28T07:26:24.550" CommentCount="3" />
  <row Id="927" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T07:27:17.107" Score="3" ViewCount="0" Body="&lt;p&gt;$196883$ is interesting because it's the smallest dimension of a faithful complex representation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monster_group&quot; rel=&quot;nofollow&quot;&gt;Monster group&lt;/a&gt;.  Roughly speaking, a group is a collection of symmetries of something.  A simple group is a group which can't be decomposed further into smaller groups; to borrow a phrase, it's an &quot;atom of symmetry.&quot;  There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups&quot; rel=&quot;nofollow&quot;&gt;classification&lt;/a&gt; of these atoms with several infinite families and 26 exceptions, and the Monster group is the largest of these exceptions.  And the simplest way to write it down is using $196883$-by-$196883$ matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$196883$ was crucial to McKay's discovery of &lt;a href=&quot;http://en.wikipedia.org/wiki/Monster_group#Moonshine&quot; rel=&quot;nofollow&quot;&gt;Monstrous moonshine&lt;/a&gt;, which began with the observation that the &lt;a href=&quot;http://en.wikipedia.org/wiki/J-invariant&quot; rel=&quot;nofollow&quot;&gt;j-invariant&lt;/a&gt;, an important function in number theory, has a Taylor series in which $196884$ appears.  The story of this discovery is explained in Mark Ronan's wonderful book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0192807226&quot; rel=&quot;nofollow&quot;&gt;Symmetry and the Monster&lt;/a&gt;, which I highly recommend.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T07:27:17.107" CommunityOwnedDate="2010-07-28T07:27:17.107" CommentCount="1" />
  <row Id="928" PostTypeId="2" ParentId="894" CreationDate="2010-07-28T08:01:11.347" Score="5" ViewCount="0" Body="&lt;p&gt;We can define formula $P_i$ that says &quot;there are at most $i$ elements satisfying $P$&quot;. Now, if the infinite disjunction of the $P_i$ was definable in FO, it would (by compactness) imply a conjunction of some finite subset of the $P_i$, hence it would imply $P_i$ for some $i$. That is not true, if $P$ can have (say) $i+1$ elements satisfying it.&lt;/p&gt;&#xA;" OwnerUserId="236" LastEditorUserId="236" LastEditorDisplayName="" LastEditDate="2010-07-29T12:37:26.077" LastActivityDate="2010-07-29T12:37:26.077" CommentCount="1" />
  <row Id="929" PostTypeId="2" ParentId="918" CreationDate="2010-07-28T08:25:13.880" Score="6" ViewCount="0" Body="&lt;p&gt;Fix two elements s&lt;sub&gt;1&lt;/sub&gt;,s&lt;sub&gt;2&lt;/sub&gt;&amp;isin;S and divide subsets of S into two parts (subsets of S containing only s&lt;sub&gt;2&lt;/sub&gt;)&amp;cup;(subsets of S which contains s&lt;sub&gt;1&lt;/sub&gt; if they contain s&lt;sub&gt;2&lt;/sub&gt;). The second part contains equal number of sets for all reminders mod 3 (because Z/3 acts there adding s&lt;sub&gt;1&lt;/sub&gt;, then s&lt;sub&gt;2&lt;/sub&gt;, then removing both of them) -- namely, 2&lt;sup&gt;n-2&lt;/sup&gt;. And for the first part we have a bijection with subsets &lt;em&gt;(edit: with 2 mod 3 elements)&lt;/em&gt; of a set with (n-2) elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get a recurrence relation that gives an answer 2&lt;sup&gt;n-2&lt;/sup&gt;+2&lt;sup&gt;n-4&lt;/sup&gt;+... -- i.e. (2&lt;sup&gt;n&lt;/sup&gt;-1):3 for even and (2&lt;sup&gt;n&lt;/sup&gt;-2):3 for odd n.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Errata.&lt;/strong&gt; For n=0,1,5 mod 6 one should add &quot;+1&quot; to the answer from the previous paragraph (e.g. for n=6 the correct answer is 1+20+1=22 and not 21).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to rephrase the solution to make it obvious. For n=2k divide S on k pairs and consider an action of a group Z/3Z on each pair described in a first paragraph. We get an action of (Z/3Z)&lt;sup&gt;k&lt;/sup&gt; on subsets of S, and after removal of it's only fixed point (k-element set consisting of second points from each pair) we get a bijection between subsets which have 0, 1 or 2 elements mod 3. So there are (2&lt;sup&gt;n&lt;/sup&gt;-1):3 sets with i mod 3 elements excluding the fixed point &lt;em&gt;and&lt;/em&gt; to count that point one should add &quot;+1&quot; for i=k mod 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for n=2k+1 there are 2 fixed points&amp;nbsp;&amp;mdash; including or not (2k+1)-th element of S&amp;nbsp;&amp;mdash; with k+1 and k elements respectively.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-07-31T13:46:57.907" LastActivityDate="2010-07-31T13:46:57.907" CommentCount="4" />
  <row Id="930" PostTypeId="2" ParentId="855" CreationDate="2010-07-28T09:09:28.857" Score="11" ViewCount="0" Body="&lt;p&gt;This is somewhere between an answer and commentary.  As others have said, the question is equivalent to showing: for any prime $p &gt; 3$, $p^2 \equiv 1 \pmod 3$ and $p^2 \equiv 1 \pmod 8$.  Both of these statements are straightforward to show by just looking at the $\varphi(3) = 2$ reduced residue classes modulo $3$ and the $\varphi(8) = 4$ reduced residue classes modulo $8$.  But what is their significance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a positive integer $n$, let $U(n) = (\mathbb{Z}/n\mathbb{Z})^{\times}$ be the multiplicative group of units (&quot;reduced residues&quot;) modulo $n$.  Like any abelian group $G$, we have a squaring map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[2]: G \rightarrow G$, $g \mapsto g^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the image of which is the set of squares in $G$.  So, the question is equivalent to: for $n = 3$ and also $n = 8$, the subgroup of squares in $U(n)$ is the trivial group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $U(3) = { \pm 1}$ has order $2$; since $(-1)^2 = 1$, the fact that the subgroup of squares is equal to $1$ is pretty clear.  But more generally, for any odd prime $p$, the squaring map $[2]$ on $U(p)$ is two-to-one onto its image -- an element of a field has no more than two square roots -- so that precisely half of the elements of $U(p)$ are squares.  It turns out that when $p = 3$, half of $p-1$ is $1$, but of course this is somewhat unusual: it doesn't happen for any other odd prime $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $U(8) = {1,3,5,7}$ has order $4$.  By analogy to the case of $U(p)$, one might expect the squaring map to be two-to-one onto its image so that exactly half of the elements are squares.  But that is not what is happening here: indeed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \pmod 8$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the subgroup of squares is again trivial.  What's different?  Since $\mathbb{Z}/8\mathbb{Z}$ is not a field, it is legal for a given element to have more than two square roots, but a more insightful answer comes from the structure of the groups $U(n)$.  For any odd prime $p$, the group $U(p)$ is &lt;em&gt;cyclic&lt;/em&gt; of order $p-1$ (&quot;existence of primitive roots&quot;).  It is easy to see that in any cyclic group of even order, exactly half of the elements are squares.  So $U(8)$ must not be cyclic, so it must be the other abelian group of order $4$, i.e., isomorphic to the Klein $4$-group $C_2 \times C_2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $p$ is an odd prime number and $a$ is a positive integer, then &#xA;$U(p^a)$ is cyclic of order $p^{a-1}(p-1)$ hence isomorphic to $C_{p^{a-1}} \times C_{p-1}$, whereas for any $a geq 2$, the group $U(2^a)$ is isomorphic to $C_{2^{a-2}} \times C_2$.  This is one of the first signs in number theory &quot;there is something odd about the prime $2$&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Note that the above considerations allow us to answer the more general question: &quot;What is the largest positive integer $N$ such that for all primes $p$ with $\operatorname{gcd}(p,N) = 1$, $N$ divides $(p^2-1)$?&quot;  (Answer: $N = 24$.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-07-28T14:00:25.960" LastActivityDate="2010-07-28T14:00:25.960" />
  <row Id="931" PostTypeId="2" ParentId="8" CreationDate="2010-07-28T09:21:32.260" Score="4" ViewCount="0" Body="&lt;p&gt;Also you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Matrix%20form&quot; rel=&quot;nofollow&quot;&gt;matrix equation&lt;/a&gt; for fibonacci numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;       n &#xA;[ 1 1 ] = [ F_{n+1} F_{n}   ]&#xA;[ 1 0 ] = [ F_{n}   F_{n-1} ]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To calculate $n$-th power of the matrix you can use exponentiation by squaring algorithm:&#xA;_http://en.wikipedia.org/wiki/Exponentiation_by_squaring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach could also be generalized on the case of arbitrary sequence with linear recurrence relation.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T09:21:32.260" CommentCount="1" />
  <row Id="932" PostTypeId="2" ParentId="59" CreationDate="2010-07-28T09:43:00.053" Score="1" ViewCount="0" Body="&lt;p&gt;In case you want to implement it on some programming language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually there is a function called like &lt;b&gt;atan2(y, x)&lt;/b&gt; which returns oriented angle between points (x, y) and (1, 0). In that case use could use&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;atan2(vector product, scalar product).&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is usually more stable than using just &lt;b&gt;arccos&lt;/b&gt; or &lt;b&gt;arcsin&lt;/b&gt;.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T09:43:00.053" />
  <row Id="933" PostTypeId="2" ParentId="806" CreationDate="2010-07-28T09:45:24.453" Score="6" ViewCount="0" Body="&lt;p&gt;As others have said, often what you want for a particular Diophantine application is that the class number of a certain number field be relatively prime to a certain number.  The famous example of this (as already noted by others) is Kummer's Theorem that for an odd prime $p$, the Fermat equation $x^p + y^p = z^p$ has no integer solutions with $xyz \neq 0$ if the ring of integers of $\mathbb{Q}[e^{\frac{2 \pi i}{p}}]$ has class number prime to $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another -- simpler -- nice example is the &lt;strong&gt;Mordell equation&lt;/strong&gt; $y^2 + k = x^3$.  If $k \equiv 1,2 \pmod 4$ and the ring $\mathbb{Z}[\sqrt{-k}]$ has class number prime to $3$, then all of the integer solutions to the Mordell equation can be found.  See Section 4 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400MordellEquation.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400MordellEquation.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for an exposition of this which is (I hope) reasonably elementary and accessible to undergraduates. &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-07-28T09:45:24.453" />
  <row Id="934" PostTypeId="2" ParentId="8" CreationDate="2010-07-28T09:52:46.637" Score="1" ViewCount="0" Body="&lt;p&gt;To expand on falagar's answer, my favourite proof of Binet's formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Which I was going to post a summary of here, but remembered that everything was awful without Tex, so here is a link to &lt;a href=&quot;http://livetoad.org/Courses/Documents/132d/Notes/fibonacci_matrix.pdf&quot; rel=&quot;nofollow&quot;&gt;some notes on it I found on google&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is to treat pairs of fibonnacci numbers, adjacent in the sequence, as vectors. Moving on to the next adjacent pair induces a linear transformation not unlike that of the matrix falagar posted. Calculating eigenvalues and eigenvectors can give a complete prediction of where an initial vector will find itself, predicting the whole sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite a lot of work but I think it's rather illuminating.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-07-28T09:52:46.637" />
  <row Id="935" PostTypeId="2" ParentId="594" CreationDate="2010-07-28T10:21:41.717" Score="2" ViewCount="0" Body="&lt;p&gt;Here is another proof without complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with proving that there exists $z \in \mathbb{N}$ such that $z^2 + 1 \equiv 0 \pmod p$. We do this in the same way as Akhil Mathew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let we have $a^2 + b^2 = pm$. Take $x$ and $y$ such that&lt;br /&gt;&#xA;$x \equiv a \pmod m$ and&lt;br /&gt;&#xA;$y \equiv b \pmod m$ and&lt;br /&gt;&#xA;$x, y \in [-m/2, m/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $u = ax + by$ and $v = ay - bx$. Then $u^2 + v^2 = (a^2 + b^2)(x^2 + y^2)$.&#xA;Moreover, $u$ and $v$ are multiples of $m$.&#xA;Hence $(u/m)^2 + (v/m)^2 = p (x^2 + y^2)/m$. $(x^2 + y^2)/m$ is an integer because of the definition of $x$ and $y$ and that $a^2 + b^2 = pm$.&lt;br /&gt;&#xA;Also $(x^2 + y^2)/m$ is less than $m/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we change $a$ by $u$ and $b$ by $v$ and continue this process until we get $m=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this is quite efficient way to find representation of $p$ as a sum of two squares - it takes $O(\log p)$ steps to find it provided we have found $z$ such that $z^2 + 1$ is multiple of $p$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-09T15:55:17.210" LastActivityDate="2010-08-09T15:55:17.210" CommentCount="1" />
  <row Id="936" PostTypeId="1" AcceptedAnswerId="937" CreationDate="2010-07-28T10:41:41.603" Score="1" ViewCount="173" Body="&lt;p&gt;I am looking for functions and/or constants that when being integrated from minus infinity to infinity produce 1. I think the Dirac delta function is one example but perhaps there are some more? References on useful material is also greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T14:37:18.693" LastActivityDate="2010-07-28T15:03:54.620" CommunityOwnedDate="2010-08-03T18:47:52.520" Title="Improper Integrals from -oo to oo with solution 1" Tags="&lt;big-list&gt;&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="937" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T10:47:25.527" Score="6" ViewCount="0" Body="&lt;p&gt;Any integrable functions that gives a finite nonzero answer can be modified to suit your need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose ∫ f(x) dx = A, then let g(x) = f(x)/A, automatically we have ∫ g(x) dx = A/A = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Actually, all continuous &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_probability_distributions#Continuous_distributions&quot; rel=&quot;nofollow&quot;&gt;probability distribution function&lt;/a&gt; must have this property.)&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-07-28T10:47:25.527" CommunityOwnedDate="2010-08-03T18:47:52.520" CommentCount="2" />
  <row Id="938" PostTypeId="2" ParentId="653" CreationDate="2010-07-28T10:47:27.460" Score="1" ViewCount="0" Body="&lt;p&gt;Homotopy theory is useful in quantum mechanics when (for instance) talking about manifolds of Hamiltonians.  You might have a collection of Hamiltonians that depend continuously on some parameters $p_1, p_2, \ldots, p_n$ such that the matrix representation of the Hamiltonian is periodic in some subset of the parameters $p_{r_1}, p_{r_2}, \ldots, p_{r_m}$.  From this, the fundamental group of the manifold of Hamiltonians can be computed, which has some physical ramifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example of this would be the Hamiltonian that describes the quantum hall effect.  The quantum hall effect is the phenomena that the resistance in a 2-dimensional substrate exposed to a perpendicular electric field at close to zero temperature is quantized.  In condensed matter physics there is a notion called quasimomentum that can be thought of as being related to momentum but is a bit different.  We need something different from the classical definition of momentum because the classical definition depends on translation invariance, and in a crystal there is only discrete translation invariance.  In the quantum hall effect, we have two quasimomenta: $k_x, and k_y$, corresponding to quasimomentum in the $x$ and $y$ directions.  The Hamiltonian is periodic in both of these parameters, leading to a fundamental group of the manifold of Hamiltonians of $\mathbb{Z}^2$, i.e. the fundamental group of the torus $T^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot more to the topic than this.  This paper by Avron, Seiler, and Simon has more details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v51/i1/p51_1&quot; rel=&quot;nofollow&quot;&gt;Homotopy and Quantization in Condensed Matter Physics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="352" LastActivityDate="2010-07-28T10:47:27.460" CommentCount="4" />
  <row Id="939" PostTypeId="1" AcceptedAnswerId="956" CreationDate="2010-07-28T10:59:44.590" Score="2" ViewCount="247" Body="&lt;p&gt;Let $pi(x)$ be the number of primes not greater than $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Prime-counting_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; says that $pi(10^{23}) = 1,925,320,391,606,803,968,923$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how to calculate $pi(x)$ for large $x$ in a reasonable time? What algorithms do exist for that?&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T04:33:06.937" LastActivityDate="2010-09-02T04:33:06.937" Title="Counting primes" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="940" PostTypeId="2" ParentId="939" CreationDate="2010-07-28T11:04:33.520" Score="1" ViewCount="0" Body="&lt;p&gt;You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;inclusion exclusion principle&lt;/a&gt; to get a boost over the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Eratosthenes sieve&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T11:19:11.453" LastActivityDate="2010-07-28T11:19:11.453" CommentCount="1" />
  <row Id="941" PostTypeId="1" AcceptedAnswerId="945" CreationDate="2010-07-28T11:23:15.767" Score="0" ViewCount="222" Body="&lt;p&gt;This could be a trivial question, but what is exactly the difference of between these two expressions? Am I correct to state the both interchangeably whenever I need to express the approximation of pi? I'm bit confused as &lt;a href=&quot;http://en.wikipedia.org/wiki/Approximation#Mathematics&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; it states pi can be express by ≒ as it's not a rational number, but pi can also be expressed by a series (asymptotic), so it should be ≈ as well.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;π ≈ 3.14..&#xA;π ≒ 3.14..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="102" LastEditorUserId="102" LastEditorDisplayName="" LastEditDate="2010-07-28T11:48:02.467" LastActivityDate="2010-07-29T05:43:06.793" Title="Approximation symbol: Is π ≈ 3.14.. equal to π ≒ 3.14.. ?" Tags="&lt;definition&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="942" PostTypeId="1" AcceptedAnswerId="984" CreationDate="2010-07-28T11:41:32.120" Score="4" ViewCount="221" Body="&lt;p&gt;This is a question in Liu's book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a quasi-compact scheme. Show that $X$ contains a closed point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I'm unable to do this question, so any help would be appreciated. This question also makes me curious to know about the meaning/use of closed points of a scheme in general - by that I mean a scheme which is not an algebraic variety/local scheme over a field, which has a geometric meaning. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-28T19:47:33.367" LastActivityDate="2010-12-21T15:38:30.933" Title="Meaning of closed points of a scheme" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="943" PostTypeId="2" ParentId="594" CreationDate="2010-07-28T11:43:23.520" Score="2" ViewCount="0" Body="&lt;p&gt;Sorry that I don't have enough rep yet, I just want to point out the fact Casebash's asking about in Akhil's comment goes by the name Kummer-Dedekind theorem.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-07-28T11:43:23.520" CommentCount="1" />
  <row Id="944" PostTypeId="1" AcceptedAnswerId="950" CreationDate="2010-07-28T11:49:37.720" Score="3" ViewCount="360" Body="&lt;p&gt;Consider the following assertion from &lt;a href=&quot;http://scottaaronson.com/blog/?p=152#&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's blog&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Supposing you do prove the Riemann&#xA;  Hypothesis, it’s possible to convince&#xA;  someone of that fact, without&#xA;  revealing anything other than the fact&#xA;  that you proved it. It’s also possible&#xA;  to write the proof down in such a way&#xA;  that someone else could verify it,&#xA;  with very high confidence, having only&#xA;  seen 10 or 20 bits of the proof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain where this result comes from?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T11:59:11.490" LastActivityDate="2010-08-03T03:54:04.270" Title="Proving the Riemann Hypothesis without revealing anything other than you proved it" Tags="&lt;computer-science&gt;&lt;riemann-hypothesis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="945" PostTypeId="2" ParentId="941" CreationDate="2010-07-28T11:54:49.487" Score="9" ViewCount="0" Body="&lt;p&gt;Any mathematical notation is ok as long as it is common knowledge in your community. For instance, I believe I fully understand the meaning of the $\approx$ symbol. However, I haven't ever seen the second symbol you provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be on the sure side you should provide a definition of any relation symbol you don't consider to be common knowledge. This may happen as a short remark (&quot;..., where $\approx$ denotes ...&quot;) or maybe as a table of the used symbols in the front matter of your work. As with any definition in mathematics, there is no right or wrong in the symbol/notion/etc. you use, only proper or unsound definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: When in doubt, use the symbol that is used more commonly in the standard textbooks of your field. There is no benefit in being avant-garde at notation.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-07-28T11:54:49.487" />
  <row Id="946" PostTypeId="1" CreationDate="2010-07-28T11:59:11.490" Score="10" ViewCount="206" Body="&lt;p&gt;Here is another result from &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's blog&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If every second or so your computer’s&#xA;  memory were wiped completely clean,&#xA;  except for the input data; the clock;&#xA;  a static, unchanging program; and a&#xA;  counter that could only be set to 1,&#xA;  2, 3, 4, or 5, it would still be&#xA;  possible (given enough time) to carry&#xA;  out an arbitrarily long computation —&#xA;  just as if the memory weren’t being&#xA;  wiped clean each second. This is&#xA;  almost certainly not true if the&#xA;  counter could only be set to 1, 2, 3,&#xA;  or 4. The reason 5 is special here is&#xA;  pretty much the same reason it’s&#xA;  special in Galois’ proof of the&#xA;  unsolvability of the quintic equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does anyone have idea of how to show this?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-28T21:06:51.040" LastActivityDate="2010-12-06T23:24:29.673" Title="Computation with a memory wiped computer" Tags="&lt;computer-science&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="6" />
  <row Id="947" PostTypeId="2" ParentId="944" CreationDate="2010-07-28T12:03:33.543" Score="1" ViewCount="0" Body="&lt;p&gt;The concept behind it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-knowledge_proof&quot; rel=&quot;nofollow&quot;&gt;Zero knowledge proof&lt;/a&gt;. Wikipedia has a good article about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The similar question was asked in paper &quot;How to Prove a Theorem So No One Else Can Claim It&quot; by M. Blum. Also &lt;a href=&quot;http://euler.nmt.edu/~brian/students/pope.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article discusses the question.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T12:03:33.543" />
  <row Id="948" PostTypeId="2" ParentId="942" CreationDate="2010-07-28T12:06:45.673" Score="7" ViewCount="0" Body="&lt;p&gt;Zorn's lemma implies there is a minimal nonempty closed set $F \subset X$ with no proper closed subsets (because the closed sets have the finite intersection property in view of quasi-compactness).  It is sufficient to find a closed point in $F$. Now $F$ is itself a scheme, and it has an open subset $U = \spec A$ for $A$ a ring. This must be all of $F$ by minimality.  A maximal ideal in $A$ gives a closed point in $F$, hence in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are people here who could give a much better answer to your other question, so I'll leave it.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-28T12:06:45.673" CommentCount="3" />
  <row Id="949" PostTypeId="1" CreationDate="2010-07-28T12:15:49.507" Score="3" ViewCount="87" Body="&lt;p&gt;Say I have an image, with pixels that can be either 0 or 1. For simplicity, assume it's a 2D image (though I'd be interested in a 3D solution as well). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pixel has 8 neighbors (if that's too complicated, we can drop to 4-connectedness). Two neighbouring pixels with value 1 are considered to be connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I know the probability &lt;code&gt;p&lt;/code&gt; that an individual pixel is 1, and if I can assume that all pixels are independent, how many groups of at least &lt;code&gt;k&lt;/code&gt; connected pixels should I expect to find in an image of size &lt;code&gt;n-by-n&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really need is a good way of calculating the probability of &lt;code&gt;k&lt;/code&gt; pixels being connected given the individual pixel probabilities. I have started to write down a tree to cover all the possibilities up to &lt;code&gt;k=3&lt;/code&gt;, but even then, it becomes really ugly really fast. Is there a more clever way to go about this?&lt;/p&gt;&#xA;" OwnerUserId="356" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T13:38:01.037" LastActivityDate="2010-07-28T13:46:55.227" Title="Probability to find connected pixels" Tags="&lt;probability&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="950" PostTypeId="2" ParentId="944" CreationDate="2010-07-28T12:26:56.757" Score="6" ViewCount="0" Body="&lt;p&gt;The correct answer has already been given by Akhil Mathew in the comments above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topic belongs to the field of complexity theory in computer science. In complexity theory, there exists an intriguing concept for the problem of deciding whether a given word belongs to a given formal language or not: &lt;a href=&quot;http://en.wikipedia.org/wiki/Interactive_proof_system&quot; rel=&quot;nofollow&quot;&gt;interactive proof systems&lt;/a&gt;. These systems model the interaction between resource-limited verifier (say, you or me) and an almighty prover (say, your much, much smarter older sister). The goal of the interaction is that the prover convinces the verifier from the fact that the given word is or is not an element of the language such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;almost surely, the verfier can only be convinced from the true answer and&lt;/li&gt;&#xA;&lt;li&gt;the verifier can only be fooled to believe the opposite within a very small margin of error.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are a large number of theoretical results with respect to these interactive systems. These resuts include the follwoing two statements (given informally):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;everything that can be proven by such an interactive protocoll can also be proven using an interactive protocoll that convinces the verifier (within small margin of error) but it does not reveal &lt;em&gt;any&lt;/em&gt; information about the proof itself. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-knowledge_proof&quot; rel=&quot;nofollow&quot;&gt;Zero-Knowledge-Proof&lt;/a&gt;, &quot;Everything provable is provable in zero-knowledge&quot; by Ben-Or et al, 1988)&lt;/li&gt;&#xA;&lt;li&gt;every proof can be rewritten such that inspection of just a constant number of bits from this proof convinces the verifier within only a small margin of error (&lt;a href=&quot;http://en.wikipedia.org/wiki/PCP_theorem&quot; rel=&quot;nofollow&quot;&gt;PCP-Theorem&lt;/a&gt;, &quot;Proof verification and the hardness of approximation problems&quot; by Arora et al, 1992, and a number of other papers)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both of these concepts and results are highly non-trivial and beyond the scope of this forum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in the quote above Scott Aaronson is just using some everyday's problem to illustrate these concepts. To use these results formally, would have to convert the task of &quot;proving the Riemann Hypothesis&quot; to a decision problem of formal languages, as is standard in complexity theory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: There is in fact a small modification in the model between both results stated above which I omitted. First, the interaction between one prover and the verifier can be generalized to multiple provers and a verifier. Next, there is a result that finds that the case of multiple provers can be equivalently reformulated as follows: The provers are removed from the protocol, rather there is a single (possibly very long) string which acts as a written proof for the word problem. Interaction is now looking at an arbitrarily chosen bit of this proof, and the verifier may choose the location of these bits randomly. This is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistically_checkable_proof&quot; rel=&quot;nofollow&quot;&gt;Probabilistic Checkable Proof&lt;/a&gt; (hence, PCP). So ,in this scenario, the &quot;written proof&quot; of the Riemann Hypothesis would be interpreted as the proof string.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-07-28T12:52:06.407" LastActivityDate="2010-07-28T12:52:06.407" CommentCount="2" />
  <row Id="951" PostTypeId="1" AcceptedAnswerId="952" CreationDate="2010-07-28T12:28:00.623" Score="4" ViewCount="172" Body="&lt;p&gt;I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T15:41:58.513" LastActivityDate="2010-07-28T17:02:13.010" CommunityOwnedDate="2010-07-28T12:28:00.623" Title="Visualising functions from complex numbers to complex numbers" Tags="&lt;big-list&gt;&lt;math-software&gt;&lt;complex-analysis&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="952" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T12:37:01.900" Score="8" ViewCount="0" Body="&lt;p&gt;One way that functions from C to C can be represented is to show the image of a grid.  That is, plot the images of the lines x = constant and y = constant under your function, where z = x + yi.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is what Wikipedia calls &lt;a href=&quot;http://en.wikipedia.org/wiki/Domain_coloring&quot; rel=&quot;nofollow&quot;&gt;domain coloring&lt;/a&gt; (see &lt;a href=&quot;http://www.mai.liu.se/~halun/complex/domain_coloring-unicode.html&quot; rel=&quot;nofollow&quot;&gt;this article by Hans Lundmark&lt;/a&gt; for a more detailed exposition.  The idea here is to color the range of the complex function -- since color space is three-dimensional there are multiple ways to do this -- and then color each point in the domain by the color of the corresponding point in the range.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-07-28T12:37:01.900" CommunityOwnedDate="2010-07-28T12:37:01.900" CommentCount="1" />
  <row Id="953" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T12:50:06.147" Score="2" ViewCount="0" Body="&lt;p&gt;One good example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;standard Gaussian distribution&lt;/a&gt;, $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$.  This is the most straightforward example of a continuous probability distribution function as mentioned by KennyTM above.&lt;/p&gt;&#xA;" OwnerUserId="349" LastActivityDate="2010-07-28T12:50:06.147" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="954" PostTypeId="1" CreationDate="2010-07-28T12:51:19.983" Score="0" ViewCount="101" Body="&lt;p&gt;There is a very simple expression for the inverse of Fourier transform. What is the easiest known expression for the inverse Laplace transform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, what is the easiest way to prove it?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-29T14:22:08.317" Title="Inverse of Laplace  transform" Tags="&lt;analysis&gt;&lt;integral-transforms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="955" PostTypeId="1" AcceptedAnswerId="958" CreationDate="2010-07-28T12:52:39.747" Score="0" ViewCount="52" Body="&lt;p&gt;Is there any connection between exact differential equations and forms, or is the similarity in name just an accident?&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T12:56:36.820" Title="Exact differential eqns and exact differential forms" Tags="&lt;differential-equations&gt;&lt;global-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="956" PostTypeId="2" ParentId="939" CreationDate="2010-07-28T12:52:46.220" Score="5" ViewCount="0" Body="&lt;p&gt;The most efficient prime counting algorithms currently known are all essentially optimizations of the method developed by Meissel in 1870, e.g. see the discussion here &lt;a href=&quot;http://primes.utm.edu/howmany.shtml&quot; rel=&quot;nofollow&quot;&gt;http://primes.utm.edu/howmany.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-28T12:52:46.220" />
  <row Id="957" PostTypeId="1" AcceptedAnswerId="960" CreationDate="2010-07-28T12:54:07.043" Score="5" ViewCount="206" Body="&lt;p&gt;Why are noncommutative nonassociative Hopf algebras are called quantum groups? This seems to be a purely mathematical notion and there is no quantum anywhere in it prima facie.&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-29T18:34:53.743" LastActivityDate="2010-07-29T18:34:53.743" Title="Why are noncommutative &quot;algebraic&quot; groups called quantum groups?" Tags="&lt;terminology&gt;&lt;noncommutative-geometry&gt;&lt;quantum-groups&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="958" PostTypeId="2" ParentId="955" CreationDate="2010-07-28T12:56:36.820" Score="1" ViewCount="0" Body="&lt;p&gt;The differential equation $I(x,y)dx + J(x,y)dy = 0$ is exact is by definition the same as this differential being an exact form.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-07-28T12:56:36.820" />
  <row Id="959" PostTypeId="1" AcceptedAnswerId="962" CreationDate="2010-07-28T13:09:12.457" Score="-2" ViewCount="147" Body="&lt;p&gt;Lets say we have a sphere, cube, and cylinder of same dimensions, that is&#xA;The diameter of the sphere is same as one side of a cube and same as circumference of the cylinder.&#xA;Of these three containers, how can one prove that the sphere has largest capacity?&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T13:34:52.137" LastActivityDate="2010-07-29T21:46:44.523" Title="How to prove which amongst a Sphere, Cube and Cylinder of same dimensions has more capacity" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="960" PostTypeId="2" ParentId="957" CreationDate="2010-07-28T13:18:41.627" Score="2" ViewCount="0" Body="&lt;p&gt;One way that Hopf algebras come up is as the algebra of (real or complex) functions on a topological group.  The multiplication is commutative since it is just pointwise multiplication of functions.  However, in non-commutative geometry you want to replace the algebra of functions on a space with a non-commutative algebra, giving a non-commutative Hopf algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relates to quantum mechanics because there the analog of the classical coordinate functions of position and momentum do not commute.  Therefore we think of the algebra of functions on a quantum &quot;space&quot; as being non-commutative.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-29T18:29:02.197" LastActivityDate="2010-07-29T18:29:02.197" CommentCount="9" />
  <row Id="961" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T13:20:53.150" Score="3" ViewCount="0" Body="&lt;p&gt;Open two browser windows side-by-side and use &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;wolfram alpha&lt;/a&gt;. Recast your function f(z) as f(x+iy) and plot the &lt;a href=&quot;http://bit.ly/bEZgsA&quot; rel=&quot;nofollow&quot;&gt;real part&lt;/a&gt; in one window and the &lt;a href=&quot;http://bit.ly/bfeLyS&quot; rel=&quot;nofollow&quot;&gt;imaginary part&lt;/a&gt; in another. My example links provide plots for z&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="167" LastEditorDisplayName="" LastEditDate="2010-07-28T13:37:32.940" LastActivityDate="2010-07-28T13:37:32.940" CommunityOwnedDate="2010-07-28T13:20:53.150" />
  <row Id="962" PostTypeId="2" ParentId="959" CreationDate="2010-07-28T13:33:33.073" Score="3" ViewCount="0" Body="&lt;p&gt;It is easy to show which has the largest volume (once you've provided the height of the cylinder as per my comment) as long as you accept proofs built on top of the existing verified equations of the volumes of each solid. Then you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sphere = \frac{4\pi}{3}(\frac{d}{2})^3 = \frac{\pi d^3}{6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cube = d^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cylinder = (\frac{d}{2\pi})^2 \pi h = \frac{d^2}{4\pi^2}h\pi = \frac{hd^2}{4\pi}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final equation would be:&#xA;$\frac{d^3}{4\pi}$ &#xA;if you make the height of the cylinder equal to it's circumference and therefore equal to the diameter of the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually shows a different result to the one you suggested too, namely that the cube has the greatest volume in this setup. This can easily be seen geometrically by envisioning the shapes. The sphere with diameter $d$ will fit inside a cube with side of length $d$. The cylinder with circumference $d$ is the smallest (assuming the height is equal to the circumference) because it is a cylinder with diameter $\frac{d}{2\pi}$ which is a much smaller circle than that described by a cross-section of the sphere taken through a great circle drawn on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you may have gotten your assumption from a related but different statement - namely that a sphere has the largest volume given that the &lt;em&gt;surface areas&lt;/em&gt; are equal in the shapes being considered. You can show this in a similar way by simply comparing the equations for volumes again, but calculating them from the surface area equations instead of directly from sides/diameters.&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="35" LastEditorDisplayName="" LastEditDate="2010-07-29T21:46:44.523" LastActivityDate="2010-07-29T21:46:44.523" CommentCount="2" />
  <row Id="963" PostTypeId="2" ParentId="949" CreationDate="2010-07-28T13:46:55.227" Score="2" ViewCount="0" Body="&lt;p&gt;This looks a bit like percolation theory to me. In the 4-neighbour case, if you look at the dual of the image, the chance that an edge is connected (runs between two pixels of the same colour) is &lt;code&gt;1-2p+2p^2&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can get nice closed-form answer for your question, but maybe a computer can help with some Monte Carlo simulation?&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-28T13:46:55.227" />
  <row Id="964" PostTypeId="2" ParentId="939" CreationDate="2010-07-28T13:50:00.850" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Atkin&quot; rel=&quot;nofollow&quot;&gt;Sieve of Atkin&lt;/a&gt; is one of the fastest algorithm used to calculate $pi(x)$. The Wikipedia page says that its complexity is O(N/ log log N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;http://numbers.computation.free.fr/Constants/Primes/Pix/pixproject.html&quot; rel=&quot;nofollow&quot;&gt;distributed computation project&lt;/a&gt; which was able to calculate $\pi(4\times 10^{22})$, maybe it could be useful.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2010-07-28T16:28:02.443" LastActivityDate="2010-07-28T16:28:02.443" CommentCount="3" />
  <row Id="965" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:08:21.867" Score="0" ViewCount="0" Body="&lt;p&gt;The function which is 1 on the interval &lt;code&gt;[0;1]&lt;/code&gt;, and 0 elsewhere, is a non-continuous probability distribution function. The function which is 3 on &lt;code&gt;[0;1]&lt;/code&gt; and -1 on (1;3], and so on and on. What kind of answer do you want? What kind of properties do you want your functions to have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really are too many functions to list, since multiplying &lt;strong&gt;any&lt;/strong&gt; function by a C^oo function with compact support and then applying Kenny's trick gives you an answer.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-28T14:08:21.867" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="966" PostTypeId="2" ParentId="957" CreationDate="2010-07-28T14:37:05.403" Score="0" ViewCount="0" Body="&lt;p&gt;I cannot comment, and this should be a comment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that the question in your title and the question in the body of your question are quite different! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-commutative non-cocommutative Hopf algebra is not the same thing as a non-commutative group, and quantum groups &lt;em&gt;are&lt;/em&gt; usually associative.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-07-28T14:45:05.320" LastActivityDate="2010-07-28T14:45:05.320" CommentCount="9" />
  <row Id="967" PostTypeId="1" CreationDate="2010-07-28T14:42:21.270" Score="8" ViewCount="301" Body="&lt;p&gt;What are some good online/free resources (tutorials, guides, exercises, and the like) for learning Lambda Calculus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I am interested in the following areas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Untyped lambda calculus&lt;/li&gt;&#xA;&lt;li&gt;Simply-typed lambda calculus&lt;/li&gt;&#xA;&lt;li&gt;Other typed lambda calculi&lt;/li&gt;&#xA;&lt;li&gt;Church's Theory of Types (I'm not sure where this fits in).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(As I understand, this should provide a solid basis for the understanding of type theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice and suggestions would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-07-30T08:19:20.123" LastActivityDate="2010-08-04T13:06:13.490" CommunityOwnedDate="2010-08-07T16:55:42.303" Title="Learning Lambda Calculus" Tags="&lt;logic&gt;&lt;lambda-calculus&gt;&lt;type-theory&gt;" AnswerCount="4" FavoriteCount="6" />
  <row Id="968" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:45:22.340" Score="1" ViewCount="0" Body="&lt;p&gt;If you take any odd function F differentiable on R and such that F(x)-&gt;l (with l a nonzero real) for x-&gt;infinity, then f(x) = 1/(2l)*F'(x) statisfies your request. For example F(x) = arctan(x)&lt;/p&gt;&#xA;" OwnerUserId="365" LastActivityDate="2010-07-28T14:45:22.340" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="969" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:49:35.327" Score="2" ViewCount="0" Body="&lt;p&gt;This should be a comment but I cannot comment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Dirac delta function is &lt;em&gt;not&lt;/em&gt; a function!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T14:49:35.327" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="970" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T14:53:46.267" Score="1" ViewCount="0" Body="&lt;p&gt;Most practical &lt;a href=&quot;http://en.wikipedia.org/wiki/Wavelet&quot; rel=&quot;nofollow&quot;&gt;mother wavelet&lt;/a&gt; have square norm 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{-\infty}^\infty |\psi(t)|^2 dt = 1$&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-07-28T14:53:46.267" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="971" PostTypeId="2" ParentId="936" CreationDate="2010-07-28T15:03:54.620" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Any&lt;/strong&gt; function f(x) which integrates to 1 over &lt;strong&gt;any&lt;/strong&gt; range [a,b] fits this bill, since we can define g(x) = f(x) on [a,b], and 0 everywhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you only want continuous functions, restricting ourselves above to f(x) where f(a) = f(b) = 0 still satisfies this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want continuous functions strictly &gt; 0 everywhere, these are known as probability distributions (continuous on [-∞, ∞]).  A large list of such functions can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_probability_distributions#Supported_on_the_whole_real_line&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  A few more notable examples are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;normal distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Skew_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;skew-normal distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Student%27s_t-distribution&quot; rel=&quot;nofollow&quot;&gt;t-distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_distribution&quot; rel=&quot;nofollow&quot;&gt;cauchy distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Fisher-Tippett_distribution&quot; rel=&quot;nofollow&quot;&gt;extreme-value distribution&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-28T15:03:54.620" CommunityOwnedDate="2010-08-03T18:47:52.520" />
  <row Id="972" PostTypeId="2" ParentId="946" CreationDate="2010-07-28T15:16:34.497" Score="3" ViewCount="0" Body="&lt;p&gt;As Scott himself states in comments section of &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;the post in question&lt;/a&gt; (comment #9):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(4) Width-5 branching programs can compute NC1 (Barrington 1986); corollary pointed out by Ogihara 1994 that width-5 bottleneck Turing machines can compute PSPACE&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't have any ideas how this is proved.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-07-28T15:16:34.497" CommunityOwnedDate="2010-07-28T15:16:34.497" CommentCount="1" />
  <row Id="973" PostTypeId="2" ParentId="967" CreationDate="2010-07-28T15:18:53.593" Score="2" ViewCount="0" Body="&lt;p&gt;It might be nice to work through &lt;em&gt;Structure and Interpretation of Computer Programs&lt;/em&gt;, which is &lt;a href=&quot;http://mitpress.mit.edu/sicp/&quot; rel=&quot;nofollow&quot;&gt;available online&lt;/a&gt; for free. This book is an introduction to computer science and the programming language Scheme, which is a flavor of the programming language Lisp, which is based on the lambda calculus. Although it is not strictly a book about the lambda calculus, it might be fun or useful to gain some hands-on and &quot;practical&quot; experience with the lambda calculus by reading some of this book and working through some of its exercises.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T15:18:53.593" CommentCount="2" />
  <row Id="974" PostTypeId="2" ParentId="262" CreationDate="2010-07-28T15:28:10.947" Score="1" ViewCount="0" Body="&lt;p&gt;I would read a book about Perelman's proof of the Poincaré conjecture (or even the papers themselves). Oh, you mean the book had to be written when I was starting? &lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-28T15:28:10.947" CommunityOwnedDate="2010-07-28T15:28:10.947" />
  <row Id="975" PostTypeId="1" CreationDate="2010-07-28T15:46:10.150" Score="1" ViewCount="235" Body="&lt;p&gt;Hi guys, &#xA;In light of the &lt;a href=&quot;http://math.stackexchange.com/questions/81/list-of-interesting-math-blogs&quot;&gt;List of Interesting Math Blogs&lt;/a&gt; I thought it would benefit the community if we have a list of textbooks for specific topics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trusted users and/or I will edit this post depending on the answers. Please try to include the textbook name, the field it relates too, the author and if you want, give it a rating. I am not sure about how the rating works but lets say 5 is something a first year mathematics student needs while 1 is for high school and 10 is for advanced levels. If you can think of a better rating system, I am all ears.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will have the categories setup as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculus&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear Algebra:&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T15:48:10.693" LastActivityDate="2010-07-28T15:48:10.693" CommunityOwnedDate="2010-07-30T05:57:16.610" Title="List of interesting and well-explained textbooks." Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;books&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="976" PostTypeId="2" ParentId="262" CreationDate="2010-07-28T15:46:35.237" Score="2" ViewCount="0" Body="&lt;p&gt;I am not a mathematician but &lt;a href=&quot;http://books.google.com/books?id=R6E0AAAAMAAJ&amp;amp;dq=flatland&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=eE9QTLKVF5G5rAex79zVDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=11&amp;amp;ved=0CEQQ6AEwCg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Flatland: A Romance of Many Dimensions&lt;/a&gt; blew my mind. I read it when I was a college student in a class on Special Relativity and wish I had read it way earlier.&lt;/p&gt;&#xA;" OwnerUserId="333" LastActivityDate="2010-07-28T15:46:35.237" CommunityOwnedDate="2010-07-28T15:46:35.237" />
  <row Id="977" PostTypeId="2" ParentId="942" CreationDate="2010-07-28T15:54:51.207" Score="4" ViewCount="0" Body="&lt;p&gt;Closed points should be thought of as being &quot;actual points&quot;, whereas non-closed points can correspond to all sorts of different things: subvarieties, &quot;fat points&quot;, generic points, etc. You might be interested in reading &lt;a href=&quot;http://www.neverendingbooks.org/index.php/mumfords-treasure-map.html&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; about Mumford's drawing of $\operatorname{Spec} \mathbb{Z}[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible way to justify the claim that closed points are the &quot;actual points&quot; is the fact that if we have, for instance, a smooth variety over $\mathbb{C}$, then its &lt;a href=&quot;http://books.google.com/books?id=ZhzXJHUgcRUC&amp;amp;lpg=PA67&amp;amp;ots=aVQoeMkBwc&amp;amp;dq=analytification&amp;amp;pg=PA61#v=onepage&amp;amp;q=analytification&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;analytification&lt;/a&gt; will be a complex manifold. The closed points of the former will then correspond exactly to the points of the latter.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-07-28T19:44:00.380" LastActivityDate="2010-07-28T19:44:00.380" CommentCount="1" />
  <row Id="978" PostTypeId="1" AcceptedAnswerId="981" CreationDate="2010-07-28T16:02:17.240" Score="0" ViewCount="557" Body="&lt;p&gt;Let A and B be two matrices which can be multiplied.&lt;br&gt;&#xA;Then &lt;strong&gt;$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proved &lt;strong&gt;$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$&lt;/strong&gt; interpreting AB as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of &lt;em&gt;loss of information&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to manage &lt;strong&gt;$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(A))$&lt;/strong&gt;? Is there a nice interpretation like the previous one?&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-02T10:04:17.533" LastActivityDate="2010-09-02T10:04:17.533" Title="How to prove and interpret rank(AB) &lt;= min(rank(A), rank(B)) ?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="979" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T16:02:37.630" Score="1" ViewCount="0" Body="&lt;p&gt;For Moebius transformations, check out &lt;a href=&quot;http://www.youtube.com/watch?v=JX3VmDgiFnY&quot; rel=&quot;nofollow&quot;&gt;this nice YouTube video&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T16:02:37.630" CommunityOwnedDate="2010-07-28T16:02:37.630" CommentCount="2" />
  <row Id="980" PostTypeId="1" CreationDate="2010-07-28T16:04:45.773" Score="0" ViewCount="601" Body="&lt;p&gt;Especially with relation to topology - rigorous definitions are appreciated, but just as important is the intuition!&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-28T19:48:10.760" LastActivityDate="2010-07-28T19:57:50.113" Title="What's the difference between open and closed sets?" Tags="&lt;terminology&gt;&lt;set-theory&gt;&lt;intuition&gt;&lt;topology&gt;" AnswerCount="5" CommentCount="8" />
  <row Id="981" PostTypeId="2" ParentId="978" CreationDate="2010-07-28T16:11:23.520" Score="3" ViewCount="0" Body="&lt;p&gt;Yes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get &quot;more stuff&quot; in the former case, as the former is bigger than the latter.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T16:11:23.520" CommentCount="1" />
  <row Id="982" PostTypeId="2" ParentId="978" CreationDate="2010-07-28T16:11:25.330" Score="1" ViewCount="0" Body="&lt;p&gt;Prove first that if $f:X\to Y$ and $g:Y\to Z$ are functions between finite sets, then &#xA;&lt;code&gt;$|g(f(X))| \leq \min \{ |f(X)|, |g(Y)| \}.$&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use the same idea.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-07-29T20:49:23.670" LastActivityDate="2010-07-29T20:49:23.670" CommentCount="1" />
  <row Id="983" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T16:27:04.723" Score="3" ViewCount="0" Body="&lt;p&gt;The rigorous definition of open and closed sets is fundamental to topology: you define a topology by saying what its open sets are.  From this perspective, open and closed sets are axiomatic, like points and lines in geometry.  In any case, closed sets are the complements of open sets and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most familiar example of open sets would be open intervals on the real line, intervals of the form {x : a &amp;lt; x &amp;lt; b}. Such sets and their arbitrary unions define the standard topology on the real line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the paragraph above describes the &lt;em&gt;standard&lt;/em&gt; topology, but not the only one.  You could put a very different topology on the real line.  That's because a topology is determined by what you call open sets and not by the underlying space &lt;em&gt;per se&lt;/em&gt;.  For example, another topology on the real line defines a set to be open if its complement has only a finite number of points. &lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-07-28T18:22:11.493" LastActivityDate="2010-07-28T18:22:11.493" />
  <row Id="984" PostTypeId="2" ParentId="942" CreationDate="2010-07-28T16:30:13.550" Score="6" ViewCount="0" Body="&lt;p&gt;Kevin Lin's answer regarding the meaning of closed points is quite reasonable, especialy in the case when the scheme in question underlies a classical variety.  I want to add some additional remarks and examples for thinking about more general schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some tautological remarks: recall that a point x in a scheme X is called a specialization of y if x lies in the Zariski closure of y (and y is called a generalization of x).  So tautologically, a closed points is one that cannot be specialized any further (just as a generic point cannot be generalized any further).  What does specialization really mean:&#xA;ring theoretically, it means taking the image under a homomorphism; so if p and q are prime&#xA;ideals of a ring A, then q is a specialization of p in Spec A if and only if q contains p, i.e. if A/p surjects onto A/q.  It is perhaps best to think of an example: say A is C[x,y],&#xA;p is the prime ideal gen'd by (x-1) and q is the prime (actually maximal ideal) gen'd by (x-1,y).  Then in A/p, we have &quot;specialized&quot; the value of x to equal 1 (because we have &#xA;declared x-1 = 0) but y is still a free variable.  When we pass to the further quotient A/q, we have specialized both x and y: x is specialized to 1 and y is specialized to 0.   At this point, we can't specialize any more; technically, this is because q is a maximal ideal of A,&#xA;so a closed point of Spec A; intuitively, it is because both x and y have now both been &quot;specialized&quot; to actual numbers, and so we can't specialize any further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose now we set B = Z[x,y], and take p and q to be the same, i.e. gen'd by x-1 and by&#xA;(x-1,y) respectively.  Then q is &lt;em&gt;not&lt;/em&gt; maximal; there is more capacity for specialization.&#xA;How is this?  Well, x and y are now taking values in Z (rather than the field C) and so we can also reduce both x and y modulo some prime, say 5; this gives a prime ideal r = (x-1,y,5) in B containing q.  Now r &lt;em&gt;is&lt;/em&gt; maximal, and so we are done specializing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a scheme that is finite type over Z, the closed points will correspond to &#xA;&quot;actual points&quot;, in Kevin's terminology, but defined over finite fields.  The points of the scheme whose coordinates are integers, say, will &lt;em&gt;not&lt;/em&gt; be closed.  One has the choice of thinking them of them as &quot;actual points&quot; which nevertheless can be specialized further by reducing modulo primes, or as subvarieties rather than &quot;actual points&quot;, by identifying them&#xA;with their Zariski closures (for a picture of this, see the drawing of Mumford that Kevin links to).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-07-28T19:56:54.470" LastActivityDate="2010-07-28T19:56:54.470" CommentCount="0" />
  <row Id="985" PostTypeId="2" ParentId="921" CreationDate="2010-07-28T16:37:59.720" Score="1" ViewCount="0" Body="&lt;p&gt;I always liked 23 (or more precisely, -23) because Q adjoin the square-root of -23 is the first quadratic imaginary field whose class number is not a power of 2; in fact, it's class group is of order 3. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-28T16:37:59.720" CommunityOwnedDate="2010-07-28T16:37:59.720" />
  <row Id="986" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T16:40:31.667" Score="4" ViewCount="0" Body="&lt;p&gt;Intuitively speaking, an open set is a set without a border: every element of the set has, in its neighborhood, other elements of the set. If, starting from a point of the open set, you move away a little, you never exit the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed set is the complement of an open set (i.e. what stays &quot;outside&quot; from the open set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some set exists, that are neither open nor closed.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2010-07-28T19:47:01.683" LastActivityDate="2010-07-28T19:47:01.683" CommentCount="2" />
  <row Id="987" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T16:43:01.447" Score="2" ViewCount="0" Body="&lt;p&gt;A set X is open if for every point p in X, there exists a neighborhood (open ball) N of p such that N is a subset of X. We call the point p of the set X a limit point if every neighborhood of p has another point q which is also in X. The set X is closed if every limit point of X is a point of X. A set can be both open and closed, and such sets are occasionally termed &quot;clopen.&quot; Trivial examples of clopen sets are the empty set (since it has no points, both the above definitions are vacuously true) and the set of all real numbers. You can in turn visualize an open set in R as an open interval on the real line, and a closed set as a closed interval on the real line.&lt;/p&gt;&#xA;" OwnerUserId="378" LastEditorUserId="378" LastEditorDisplayName="" LastEditDate="2010-07-28T17:10:35.430" LastActivityDate="2010-07-28T17:10:35.430" CommentCount="3" />
  <row Id="988" PostTypeId="2" ParentId="951" CreationDate="2010-07-28T17:02:13.010" Score="1" ViewCount="0" Body="&lt;p&gt;The graphs in the middle of &lt;a href=&quot;http://mathworld.wolfram.com/ConformalMapping.html&quot; rel=&quot;nofollow&quot;&gt;the MathWorld page on Conformal Mapping&lt;/a&gt; show examples of the first method in Michael Lugo's answer as well as something somewhat similar to the second method in that answer.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T17:02:13.010" CommunityOwnedDate="2010-07-28T17:02:13.010" />
  <row Id="989" PostTypeId="2" ParentId="250" CreationDate="2010-07-28T17:04:42.087" Score="7" ViewCount="0" Body="&lt;p&gt;What is the smallest area of a parking lot in which a car (that is, a segment of) can perform a complete turn (that is, rotate 360 degrees)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is obviously the Kakeya Needle Problem. Fairly easy to explain, models an almost reasonable real-life scenario, and has a very surprising answer as you probably know - the lot can have as small an area as you'd like).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kakeya_set&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry: Kakeya Set&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-07-28T17:04:42.087" CommunityOwnedDate="2010-07-28T17:04:42.087" CommentCount="2" />
  <row Id="990" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T17:05:58.433" Score="2" ViewCount="0" Body="&lt;p&gt;An open set is a set S for which, given any of its element A, you can find a ball centered in A and whose points are all in S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed set is a set S for which, if you have a sequence of points in S who tend to a limit point B, B is also in S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, a closed set is a set which contains its own boundary, while an open set is a set where you are able not to leave it if you move just a little bit.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-28T17:05:58.433" CommentCount="1" />
  <row Id="992" PostTypeId="2" ParentId="3" CreationDate="2010-07-28T17:30:41.997" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://www.scienceofbetter.org/podcast/&quot; rel=&quot;nofollow&quot;&gt;Science of Better&lt;/a&gt; is a good applied math podcast.  Features applications of operations research to practical problems.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-07-28T17:39:29.080" LastActivityDate="2010-07-28T17:39:29.080" CommunityOwnedDate="2010-07-28T17:30:41.997" />
  <row Id="993" PostTypeId="2" ParentId="3" CreationDate="2010-07-28T17:38:48.037" Score="1" ViewCount="0" Body="&lt;p&gt;Peter Rowlett has a couple mathematical podcasts.  &lt;a href=&quot;http://www.travelsinamathematicalworld.co.uk/&quot; rel=&quot;nofollow&quot;&gt;Travels in a Mathematical World&lt;/a&gt; produced 64 episodes, but recently stopped.  He has a new podcast &lt;a href=&quot;http://www.pulse-project.org/pulsemathsmaths/&quot; rel=&quot;nofollow&quot;&gt;Math / Maths&lt;/a&gt; that he co-hosts with Samuel Hansen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Samuel Hansen also has a couple other podcasts: &lt;a href=&quot;http://itunes.apple.com/us/podcast/strongly-connected-components/id334982081&quot; rel=&quot;nofollow&quot;&gt;Strongly Connected Components&lt;/a&gt; and &lt;a href=&quot;http://cppodcast.libsyn.com/&quot; rel=&quot;nofollow&quot;&gt;Permutations and Combinations&lt;/a&gt;. Strongly Connected Components is more mathematical and often features interviews.  I've only listened to Permutations and Combinations once or twice.  I believe it's more of a comedy show.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T17:38:48.037" CommunityOwnedDate="2010-07-28T17:38:48.037" />
  <row Id="994" PostTypeId="2" ParentId="903" CreationDate="2010-07-28T17:46:08.790" Score="5" ViewCount="0" Body="&lt;p&gt;Try Simmons, Introduction to Topology and Modern Analysis.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T17:46:08.790" CommunityOwnedDate="2010-07-28T17:46:08.790" CommentCount="2" />
  <row Id="995" PostTypeId="2" ParentId="370" CreationDate="2010-07-28T17:47:40.813" Score="1" ViewCount="0" Body="&lt;p&gt;First Chapter of Jacobson's Basic Algebra -II.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T17:47:40.813" CommunityOwnedDate="2010-07-28T17:47:40.813" />
  <row Id="996" PostTypeId="1" CreationDate="2010-07-28T17:52:07.437" Score="0" ViewCount="117" Body="&lt;p&gt;I have lifted this from &lt;a href=&quot;http://mathoverflow.net/questions/11669/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt; since it belongs here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I'm taking matrix theory, and our textbook is Strang's Linear Algebra. Besides matrix theory, which all engineers must take, there exists linear algebra I and II for math majors. What is the difference, if any, between matrix theory and linear algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;kolistivra&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T18:50:50.077" LastActivityDate="2010-07-28T18:50:50.077" Title="What is the difference between matrix theory and linear algebra?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="997" PostTypeId="2" ParentId="329" CreationDate="2010-07-28T17:53:44.890" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Problems in Algebraic Number Theory&lt;/em&gt; is written in a style I'd like to see in more textbooks&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-07-28T17:53:44.890" CommunityOwnedDate="2010-07-28T17:53:44.890" />
  <row Id="998" PostTypeId="1" CreationDate="2010-07-28T18:00:21.333" Score="7" ViewCount="575" Body="&lt;p&gt;Lifted from &lt;a href=&quot;http://mathoverflow.net/questions/2446&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think (almost) everyone agrees that Hartshorne's Algebraic Geometry is still the best.&#xA;Then what might be the 2nd best? It can be a book, preprint, online lecture note, webpage, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One suggestion per answer please. Also, please include an explanation of why you like the book, or what makes it unique or useful.&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:39:14.497" LastActivityDate="2010-08-22T09:36:03.033" CommunityOwnedDate="2010-07-28T18:00:32.910" Title="Best Algebraic Geometry text book? (other than Hartshorne)" Tags="&lt;big-list&gt;&lt;books&gt;&lt;algebraic-geometry&gt;" AnswerCount="8" FavoriteCount="4" />
  <row Id="1002" PostTypeId="1" CreationDate="2010-07-28T18:07:57.030" Score="5" ViewCount="499" Body="&lt;p&gt;A vague question of Kevin Lin which didn't quite fit at &lt;a href=&quot;http://mathoverflow.net/questions/446&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ... what is the Fourier transform? What does it do? Why is it useful (both in math and in engineering, physics, etc)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Answers at any level of sophistication are welcome.)&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-28T20:56:38.927" LastActivityDate="2010-11-20T18:11:51.367" Title="Fourier transform for dummies" Tags="&lt;fourier-analysis&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="5" />
  <row Id="1008" PostTypeId="2" ParentId="996" CreationDate="2010-07-28T18:17:43.873" Score="2" ViewCount="0" Body="&lt;p&gt;My answer from the MO thread:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. When you talk about matrices, you're allowed to talk about things like the entry in the 3rd row and 4th column, and so forth. In this setting, matrices are useful for representing things like transition probabilities in a Markov chain, where each entry indicates the probability of transitioning from one state to another. You can do lots of interesting numerical things with matrices, and these interesting numerical things are very important because matrices show up a lot in engineering and the sciences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In linear algebra, however, you instead talk about linear transformations, which are &lt;strong&gt;not&lt;/strong&gt; (I cannot emphasize this enough) a list of numbers, although sometimes it is convenient to use a particular matrix to write down a linear transformation. The difference between a linear transformation and a matrix is not easy to grasp the first time you see it, and most people would be fine with conflating the two points of view. However, when you're given a linear transformation, you're not allowed to ask for things like the entry in its 3rd row and 4th column because questions like these depend on a choice of basis. Instead, you're only allowed to ask for things that don't depend on the basis, such as the rank, the trace, the determinant, or the set of eigenvalues. This point of view may seem unnecessarily restrictive, but it is fundamental to a deeper understanding of pure mathematics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T18:17:43.873" />
  <row Id="1010" PostTypeId="2" ParentId="996" CreationDate="2010-07-28T18:21:49.857" Score="0" ViewCount="0" Body="&lt;p&gt;I think of Linear Algebra is a semi-applied Matrix theory. I can't summarize it better then Qiaochu. &lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-07-28T18:21:49.857" />
  <row Id="1012" PostTypeId="1" CreationDate="2010-07-28T18:24:38.307" Score="2" ViewCount="49" Body="&lt;p&gt;An off-topic question posed at &lt;a href=&quot;http://mathoverflow.net/questions/5095/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt; by Andrew Stacey, but one which fits here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that came out of Terry Tao's recent blog posts on this matter (&lt;a href=&quot;http://terrytao.wordpress.com/2009/10/29/displaying-mathematics-on-the-web/&quot; rel=&quot;nofollow&quot;&gt;first post&lt;/a&gt; and &lt;a href=&quot;http://terrytao.wordpress.com/2009/11/04/displaying-maths-online-ii/&quot; rel=&quot;nofollow&quot;&gt;follow up&lt;/a&gt;) is that it's hard to get an overview of all the different ways of getting one's amazing mathematics onto the web.  I thought it'd be useful to gather together a list of such.  This meant to be a list of ways to do it, not examples of where it's already being done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard community wiki rules: one thing per answer and feel free to edit other's answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional rules: it'd be useful to have a little more than just links.  A brief description, pros and cons (be objective), platforms (does it only work on Linux, sort of thing) - things that might help someone decide which things to examine further.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T18:32:36.403" CommunityOwnedDate="2010-07-28T18:24:38.307" Title="Resources for getting maths on to the web." Tags="&lt;soft-question&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="1016" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T18:25:58.740" Score="1" ViewCount="0" Body="&lt;p&gt;I'll give an engineering answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a time series that you think is the result of a additive collection of periodic function, the Fourier transform will help you determine what the dominant frequencies are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the way guitar tuners work. The perform and FFT on the sound data and pick out the frequency with the greatest power (squares of the real and imaginary parts) and consider that the &quot;note.&quot; This is called the fundamental frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other uses, so you might want to add big list as a tag.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-28T18:25:58.740" CommentCount="1" />
  <row Id="1017" PostTypeId="1" AcceptedAnswerId="1262" CreationDate="2010-07-28T18:26:41.203" Score="3" ViewCount="170" Body="&lt;p&gt;Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, (x1,y1), and (x2,y2) is half of the absolute value of the determinant of the 2x2 matrix of the vertices (x1,y1) and (x2,y2), 1/2 * |x1*y2 - x2*y1|) how can one prove that &lt;a href=&quot;http://en.wikipedia.org/wiki/Shoelace_Method&quot; rel=&quot;nofollow&quot;&gt;the shoelace method&lt;/a&gt; works for all non-self-intersecting polygons?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-30T20:44:23.697" Title="Proving the Shoelace Method at the Precalculus Level" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;contest-math&gt;&lt;analytic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1018" PostTypeId="1" CreationDate="2010-07-28T18:26:53.573" Score="2" ViewCount="150" Body="&lt;p&gt;An off-topic question asked at &lt;a href=&quot;http://mathoverflow.net/questions/3036/&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt; by Andrew Stacey; but one which fits here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in hearing of examples of mathematical (or, at a pinch, scientific) websites with serious content where the design of the website actually makes it easy to read and absorb the material. To be absolutely clear, the mathematical content of the website should be on the website itself and not in an electronic article (so meta-sites that make it easy to find material, like MathSciNet or the arXiv, don't count).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm extending this to non-internet material. I want examples where the design of the document/website/whatever actually helped when reading the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a little background, I know that LaTeX is meant to help us separate content from context and concentrate on each one in turn, but I often feel when reading an article that the author has concentrated solely on the content and left all of the context to TeX. This is most obvious with websites where there are some really well-designed websites to compare with, but holds as well with articles.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-07-28T21:03:46.010" CommunityOwnedDate="2010-07-28T18:26:53.573" Title="Examples of well-displayed mathematics on the internet" Tags="&lt;soft-question&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1019" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T18:27:08.130" Score="2" ViewCount="0" Body="&lt;p&gt;You could think of a Fourier series expanding a function as a sum of sines and cosines analogous to the way a Taylor series expands a function as a sum of powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could think of the Fourier series as a change of variables.  A fundamental skill in engineering and physics is to pick the coordinate system that makes your problem simplest.  Since the derivatives of sines and cosines are more sines and cosines, Fourier series are the right &quot;coordinate system&quot; for many problems involving derivatives.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:27:08.130" CommentCount="0" />
  <row Id="1022" PostTypeId="2" ParentId="1012" CreationDate="2010-07-28T18:32:36.403" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://mathurl.com&quot; rel=&quot;nofollow&quot;&gt;http://mathurl.com&lt;/a&gt; is handy if you want to send someone a quick link to a mathematical expression.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:32:36.403" CommunityOwnedDate="2010-07-28T18:32:36.403" />
  <row Id="1024" PostTypeId="2" ParentId="1018" CreationDate="2010-07-28T18:33:22.843" Score="1" ViewCount="0" Body="&lt;p&gt;For students in lower level university courses, there are two amazing resources.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://tutorial.math.lamar.edu/&quot; rel=&quot;nofollow&quot;&gt;Paul's online math notes&lt;/a&gt;. My linear algebra book was complete garbage. I pretty much used these notes to get through the class. It explains everything very well and does not assume that you already know graduate level mathematics. Presentation wise, its very simple. A plain HTML website, however content is king. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And ofcourse, Khan Academy. I think everyone knows about Khan. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Last but not least, I would like to add the MIT online courses. I didn't like them that much but they did help me. Maybe its just me but I can not learn anything while staring at a computer screen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-07-28T18:33:22.843" CommunityOwnedDate="2010-07-28T18:33:22.843" CommentCount="5" />
  <row Id="1025" PostTypeId="1" AcceptedAnswerId="1027" CreationDate="2010-07-28T18:36:05.147" Score="1" ViewCount="160" Body="&lt;p&gt;Conic sections are a frequent target for dropping when attempting to make room for other topics in advanced algebra and precalculus courses.  A common argument in favor of dropping them is that typical first-year calculus doesn't use conic sections at all.  Do conic sections come up in typical intro-level undergraduate courses?  In typical prelim grad-level courses?  If so, where?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T21:32:34.943" Title="Usefulness of Conic Sections" Tags="&lt;algebra-precalculus&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="4" />
  <row Id="1026" PostTypeId="1" AcceptedAnswerId="1028" CreationDate="2010-07-28T18:39:19.617" Score="3" ViewCount="85" Body="&lt;p&gt;How many ways are there to color faces of a cube with N colors if &#xA;two colorings are the same if it's possible to rotate the cube such that&#xA;one coloring goes to another?&lt;/p&gt;&#xA;" OwnerUserId="382" LastActivityDate="2010-07-28T18:41:56.710" Title="Number of colorings of cube's faces" Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1027" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T18:40:51.893" Score="4" ViewCount="0" Body="&lt;p&gt;Conic sections are basic examples in algebraic geometry, since they are (the real forms of) curves of genus zero.  As such, they are also basic examples in number theory, since it is easy to determine the rational points on a conic section, and this is a good warm-up for studying more complicated Diophantine equations.  In fact, curves of genus zero are the &lt;em&gt;only&lt;/em&gt; class of variety for which an algorithm provably exists to determine the rational points!  Even for the next hardest case, elliptic curves, there are no algorithms which provably always work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great survey of these topics is Bjorn Poonen's &lt;a href=&quot;http://www-math.mit.edu/~poonen/papers/millennial.pdf&quot; rel=&quot;nofollow&quot;&gt;Computing rational points on curves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  There is also Franz Lemmermeyer's &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/publ/conics.pdf&quot; rel=&quot;nofollow&quot;&gt;Conics - a poor man's elliptic curves&lt;/a&gt;, which explains how certain conics can be thought of as &quot;degenerate&quot; elliptic curves.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-28T19:51:03.437" LastActivityDate="2010-07-28T19:51:03.437" />
  <row Id="1028" PostTypeId="2" ParentId="1026" CreationDate="2010-07-28T18:41:56.710" Score="3" ViewCount="0" Body="&lt;p&gt;The number of different colorings is equal to $\frac{n^6 + 3n^4 + 12n^3 + 8n^2}{24}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get this number using &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside lemma&lt;/a&gt;.&#xA;The wikipedia article contains solution of your problem as well.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-28T18:41:56.710" />
  <row Id="1029" PostTypeId="2" ParentId="740" CreationDate="2010-07-28T18:52:19.560" Score="0" ViewCount="0" Body="&lt;p&gt;Three examples suffice to show why some modes of convergence don't imply other modes of convergence: pointwise convergence, Lp norm convergence, convergence in measure, etc.  See the counterexamples section &lt;a href=&quot;http://www.johndcook.com/modes_of_convergence.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:52:19.560" CommunityOwnedDate="2010-07-28T18:52:19.560" />
  <row Id="1030" PostTypeId="2" ParentId="740" CreationDate="2010-07-28T18:55:51.020" Score="1" ViewCount="0" Body="&lt;p&gt;When I was first learning calculus, the fact that $sin(1/x)$ is continuous on the set $(0,\infty)$ gave me a headache. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-07-28T18:55:51.020" CommunityOwnedDate="2010-07-28T18:55:51.020" CommentCount="1" />
  <row Id="1031" PostTypeId="2" ParentId="544" CreationDate="2010-07-28T18:58:18.923" Score="1" ViewCount="0" Body="&lt;p&gt;Markov chains are used in Markov Chain Monte Carlo (MCMC).  This computational technique is extremely common in Bayesian statistics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Bayesian statistics, you want to compute properties of a posterior distribution.  You'd like to draw independent samples from this distribution, but often this is impractical.  So you construct a Markov chain that has as its limiting distribution the the distribution you want.  So, for example, to get the mean of your posterior distribution you could take the mean of the states of your Markov chain.  (Ergodic theory blesses this process.)&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T18:58:18.923" />
  <row Id="1032" PostTypeId="2" ParentId="668" CreationDate="2010-07-28T19:05:34.603" Score="10" ViewCount="0" Body="&lt;p&gt;You could think of a determinant as a volume.  Think of the columns of the matrix as vectors at the origin forming the edges of a skewed box.  The determinant gives the volume of that box.  For example, in 2 dimensions, the columns of the matrix are the edges of a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can derive the algebraic properties from this geometrical interpretation.  For example, if two of the columns are linearly dependent, you're box is missing a dimension and so it's been flattened to have zero volume.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:05:34.603" CommentCount="1" />
  <row Id="1033" PostTypeId="2" ParentId="381" CreationDate="2010-07-28T19:08:36.673" Score="1" ViewCount="0" Body="&lt;p&gt;Here's an example of Fibonacci numbers applied to &lt;a href=&quot;http://www.johndcook.com/blog/2008/04/23/fibonacci-numbers-at-work/&quot; rel=&quot;nofollow&quot;&gt;numerical integration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:08:36.673" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1034" PostTypeId="2" ParentId="381" CreationDate="2010-07-28T19:10:27.917" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a humorous application of Fibonacci numbers to &lt;a href=&quot;http://www.johndcook.com/blog/2009/05/19/golden-ratio-rational-approximation/&quot; rel=&quot;nofollow&quot;&gt;breastfeeding twins&lt;/a&gt;.  Even though the post is somewhat of a joke, it makes a serious point about what are called &quot;almost periodic functions.&quot; &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:10:27.917" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1035" PostTypeId="2" ParentId="213" CreationDate="2010-07-28T19:14:48.937" Score="3" ViewCount="0" Body="&lt;p&gt;I wish I'd understood the importance of inequalities earlier.  I wish I'd carefully gone through the classic book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521358809&quot; rel=&quot;nofollow&quot;&gt;Inequalities&lt;/a&gt; by Hardy, Littlewood, and Poyla early on.  Another good book is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/052154677X&quot; rel=&quot;nofollow&quot;&gt;The Cauchy-Schwarz Masterclass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can study inequalities as a subject in their own right, often without using advanced math.  But they're critical techniques for advanced math.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:14:48.937" CommunityOwnedDate="2010-07-28T19:14:48.937" CommentCount="2" />
  <row Id="1036" PostTypeId="2" ParentId="588" CreationDate="2010-07-28T19:17:44.657" Score="1" ViewCount="0" Body="&lt;p&gt;To your question regarding radius of convergence, &lt;a href=&quot;http://en.wikipedia.org/wiki/Radius_of_convergence&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives a good answer.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-07-28T19:17:44.657" />
  <row Id="1037" PostTypeId="1" AcceptedAnswerId="1047" CreationDate="2010-07-28T19:19:40.470" Score="5" ViewCount="396" Body="&lt;p&gt;Is the purpose of the derivative notation d/dx strictly for symbolic manipulation purposes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember being confused when I first saw the notation for derivatives - it looks vaguely like there's some division going on and there are some fancy 'd' characters that are added in... I recall thinking that it was a lot of characters to represent an action with respect to one variable. Of course, once you start moving the dx around it makes a little more sense as to why they exist - but is this the only reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any history lesson or examples where this notation is helpful or unhelpful is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T19:35:35.550" LastActivityDate="2010-12-18T22:36:49.877" Title="Why are derivatives specified as d/dx?" Tags="&lt;reference-request&gt;&lt;notation&gt;&lt;math-history&gt;" AnswerCount="8" FavoriteCount="1" />
  <row Id="1038" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T19:24:26.903" Score="7" ViewCount="0" Body="&lt;p&gt;If you have access to it, the book &lt;em&gt;A History of Mathematical Notations&lt;/em&gt;, by Florian Cajori, has a pretty detailed description of the history of notations for derivatives in its second volume.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T19:24:26.903" />
  <row Id="1039" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T19:38:49.083" Score="0" ViewCount="0" Body="&lt;p&gt;Adding to Mariano's answer, its worthwhile to look at the modern notation and compare it with the older one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link that explains it very well. &#xA;&lt;a href=&quot;http://www.vendian.org/mncharity/dir3/dxdoc/&quot; rel=&quot;nofollow&quot;&gt;http://www.vendian.org/mncharity/dir3/dxdoc/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-07-28T19:38:49.083" />
  <row Id="1040" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T19:40:11.203" Score="-3" ViewCount="0" Body="&lt;p&gt;A more complicated answer (yet it's going to be imprecise, because I haven't touched this in 15 years...) is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 3-dimentional space (for example) you can represent a vector &lt;strong&gt;v&lt;/strong&gt; by its end point coordinates, x, y, z, in a very simple way. You choose three vectors which are of unit length and orthogonal with each other (a base), say &lt;strong&gt;i&lt;/strong&gt;, &lt;strong&gt;j&lt;/strong&gt; and &lt;strong&gt;k&lt;/strong&gt;, and calculate the coordinates as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = &lt;strong&gt;v&lt;/strong&gt; ∙ &lt;strong&gt;i&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = &lt;strong&gt;v&lt;/strong&gt; ∙ &lt;strong&gt;j&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = &lt;strong&gt;v&lt;/strong&gt; ∙ &lt;strong&gt;k&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In multidimentional space, the equations still hold. In a discrete infinite space, the coordinates and the base vectors become a sequence. The dot product becomes an infinite sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a continuous infinite space (like the space of good functions) the coordinates and the bases become functions and the dot product an infinite integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Fourier transform is exactly this kind of operation (based on a set of base functions which are basically a set of sines and cosines). In other words, it is a different representation of the same function in relation to a particular set of base functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, for example, functions of time, represented against functions of time and space (in other words integrated over time multiplied by functions of space and time), become functions of space, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps!&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T19:40:11.203" CommentCount="6" />
  <row Id="1041" PostTypeId="2" ParentId="998" CreationDate="2010-07-28T19:53:44.257" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387977163&quot; rel=&quot;nofollow&quot;&gt;Algebraic Geometry: A First Course&lt;/a&gt; by Joe Harris is a very good book that sits in that region between undergraduate treatments and the prerequisites of Hartshorne. In particular, one does not need to know much commutative algebra to get a lot out of Harris's book. Harris himself recommends reading Hartshorne after his book for the theory of schemes.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-07-28T19:53:44.257" CommunityOwnedDate="2010-07-28T19:53:44.257" CommentCount="1" />
  <row Id="1042" PostTypeId="2" ParentId="980" CreationDate="2010-07-28T19:57:50.113" Score="1" ViewCount="0" Body="&lt;p&gt;I will not reiterate the very nice definitions found in the other answers, however I think that these &quot;practical&quot; definitions might help you as well on an intuitive level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open sets are typically used as domains for functions, as they are more useful for analysing &quot;continuous&quot; properties like differentiability. Also they don't have nasty borders (hence you don't have to deal with functions which are well behaved only on one side of the edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closed sets are useful because, if they are limited, they are compact.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T19:57:50.113" />
  <row Id="1043" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T19:57:57.367" Score="0" ViewCount="0" Body="&lt;p&gt;The study of parabolas (with axis parallel to y-axis) is useful when you have to solve 2nd degree inequations.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-28T19:57:57.367" CommentCount="1" />
  <row Id="1044" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T20:04:14.967" Score="1" ViewCount="0" Body="&lt;p&gt;This is the Leibniz notation, which is based on the ratio of &quot;infinitesimals&quot;. dy and dx are, respectively, the infinitesimal increment of the dependent variable y and the infinitesimal increment of the variable x. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other notations: Newton notation, which puts a dot over the variable name, and Cauchy notation, which uses the operator D, as in D(sin(x))=cos(x).&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-28T20:04:14.967" CommentCount="2" />
  <row Id="1045" PostTypeId="1" AcceptedAnswerId="1055" CreationDate="2010-07-28T20:06:54.867" Score="1" ViewCount="58" Body="&lt;p&gt;I have some polygons I would like to map onto the face of a cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see from &lt;a href=&quot;http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/cylindrical/body.htm#skip3&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; that I can convert the points of the polygon to cylinderical coordinates, which is almost what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about modifying the formulas to work for conical coordinates?&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-07-28T20:45:10.763" LastActivityDate="2010-07-28T20:56:53.403" Title="How do I convert from cartesian to conical coordinates?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1046" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T20:08:43.270" Score="4" ViewCount="0" Body="&lt;p&gt;If you're a physics kind of person, then a good reason to like this notation is that it gives the correct units for the derivative: whatever units $f(x)$ is in, the units for $\frac{d}{dx} f(x)$ are obtained by dividing by the units for $x$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T20:08:43.270" />
  <row Id="1047" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T20:10:20.723" Score="10" ViewCount="0" Body="&lt;p&gt;Because of their definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a function, calculate the difference in value between two points and divide by the size of the interval between the two. You can represent this as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x&lt;sub&gt;2&lt;/sub&gt;) - f(x&lt;sub&gt;1&lt;/sub&gt;) / x&lt;sub&gt;2&lt;/sub&gt; - x&lt;sub&gt;1&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;∆f(x) / ∆x (where ∆, delta, is the greek capital D and indicates an interval)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, take the limit as ∆x goes to zero, and you have the differential. This is indicated by using a lower case d instead of the ∆.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;df(x) / dx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if this operation is treated as an operator applied to a function, it is usually represented as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d / dx (applied to f(x) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that (typically in physics), you can also use the letter δ to indicate very small intervals and in general you would use the symbol ∂ to represent partial differentials. They are all variations of the letter D.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T20:10:20.723" CommentCount="1" />
  <row Id="1048" PostTypeId="1" CreationDate="2010-07-28T20:21:51.977" Score="6" ViewCount="317" Body="&lt;p&gt;In school, we learn that sin is &quot;opposite over hypotenuse&quot; and cos is &quot;adjacent over hypotenuse&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on, we learn the power series definitions of sin and cos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one prove that these two definitions are equivalent?&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-08-04T04:31:50.797" Title="Different definitions of trigonometric functions" Tags="&lt;geometry&gt;&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="1049" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T20:27:01.353" Score="0" ViewCount="0" Body="&lt;p&gt;Here's some simple &lt;a href=&quot;http://www.gnu.org/software/octave/&quot; rel=&quot;nofollow&quot;&gt;Matlab&lt;/a&gt; code to play around with if you like.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&lt;code&gt;&#xA;% This code will approxmmate the function f using the DFT&#xA;&#xA;clear all &#xA;close all &#xA;&#xA;a=0;b=2*pi; % define interval, i.e. endpoints of domain(f) &#xA;&#xA;N=20; % number of sample points to take from f&#xA;&#xA;% build vector of points in domain(f) to sample from&#xA;for j=1:N+1&#xA;   x(j) = (b-a)*(j-1)/N;&#xA;end&#xA;&#xA;f= cos(x); % approximate cos(x) with resulting Fourier series&#xA;&#xA;&#xA;% build matrix of powers of roots of unity&#xA;for m=1:N+1&#xA;   for n=1:N+1&#xA;      F(m,n) = exp((m-1)*(n-1)*(2*pi*i)/N);&#xA;   end&#xA;end&#xA;&#xA;% solve f = Fc by domng F\f&#xA;c = F\f'; % c is vector of Fourier coefficients&#xA;&#xA;% plot discrete cos(x) using N points &#xA;&#xA;xx=0:0.01:2*pi;&#xA;plot(xx,cos(xx),'g')&#xA;hold on&#xA;&#xA;% build the Fourier series using coefficients from c&#xA;summ=0;&#xA;for k=1:length(c)&#xA;   summ = summ + c(k)*exp(i*(k-1)*x); &#xA;end&#xA;&#xA;% plot the fourier series against the discrete sin function &#xA;plot(x,summ)&#xA;legend('actual','approx.')&#xA;&#xA;&#xA;&lt;/code&gt;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As written you will have the first N=20 terms of the Fourier approximation to the cosine on the interval [a,b]=[0,2*pi].  Not very interesting as is...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0961408804&quot; rel=&quot;nofollow&quot;&gt;Gilbert Strang&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-07-28T20:27:01.353" />
  <row Id="1050" PostTypeId="2" ParentId="687" CreationDate="2010-07-28T20:28:40.367" Score="2" ViewCount="0" Body="&lt;p&gt;The form this question is usually asked is whether adding a route can increase the average traveling time, and this is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Braess_paradox&quot; rel=&quot;nofollow&quot;&gt;Braess's paradox&lt;/a&gt;.  The Wiki article gives an explicit example in which the travel time on some of the routes depends on the traffic.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-28T20:28:40.367" />
  <row Id="1051" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T20:43:13.187" Score="4" ViewCount="0" Body="&lt;p&gt;If you're a physics sort of person, conic sections clearly come up when you study how Kepler figured out what the shapes of orbits are, and some of their synthetic properties give useful shortcuts to things like proving &quot;equal area swept out in equal time&quot; that need not involve calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other skills you typically learn while studying conic sections in analytic geometry - polar parametrization of curves, basic facts about various invariants related to triangles and conics, rotations and changing coordinate systems (so as to recognize the equation of a conic in general form as some sort of transformation of a standard done), are all extremely useful in physics. I'd say that plane analytic geometry was the single most useful math tool for me in solving physics problems until I got to fluid dynamics stuff (where that is replaced by complex analysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relatedly, independent of their use in physics, I think they're a great way to show the connections between analytic and synthetic thinking in math, which will come up over and over again for people who go on to study math (coordinate-based versus intrinsic perspectives, respectively).&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-28T20:43:13.187" />
  <row Id="1052" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T20:43:45.980" Score="5" ViewCount="0" Body="&lt;p&gt;Most of the proofs in elementary calculus textbooks use the definition of $\sin x$ via geometry to prove that the derivative of $\sin x$ is $\cos x$ (namely, the fact that $\lim_{x \to 0} \frac{ \sin x}{x} = 1$). Consequently, it follows that $\sin x$ and $\cos x$ are the two linearly independent solutions of $y'' = -y$.  The power series equations are also two linearly independent solutions of this differential equation.  Moreover, $\sin x$ and its derivative coincide with the derivative of the power series for $ \sin x$ at zero (no surprise, it's a Taylor series).  Same for $\cos x$. By uniqueness of solutions to ordinary differential equations, this proves that $\sin x$ and $\cos x$ as defined in school are equal to their power series.  (This is an expansion of Qiaochu's comment.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-28T20:43:45.980" />
  <row Id="1053" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T20:44:50.827" Score="1" ViewCount="0" Body="&lt;p&gt;As a rough outline, the circular definitions of sine and cosine (the y- and x-coordinates of the image of (1,0) under a rotation about the origin) lead to being able to differentiate sine and cosine, and once you know how to differentiate them (infinitely), &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Taylor's Theorem&lt;/a&gt; justifies that the power series is equal to the function.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-28T20:44:50.827" />
  <row Id="1054" PostTypeId="2" ParentId="1002" CreationDate="2010-07-28T20:54:58.573" Score="2" ViewCount="0" Body="&lt;p&gt;Let me partially steal from the accepted answer on MO, and illustrate it with examples I understand:&lt;br&gt;&#xA;&lt;em&gt;The Fourier transform is a different representation that makes convolutions easy.&lt;/em&gt;&lt;br&gt;&#xA;Or, to quote directly from there: &quot;the Fourier transform is a unitary change of basis for functions (or distributions) that diagonalizes all convolution operators.&quot; This often involves expressing an arbitrary function as a superposition of &quot;symmetric&quot; functions of some sort, say functions of the form e&lt;sup&gt;itx&lt;/sup&gt; — in the common signal-processing applications, an arbitrary &quot;signal&quot; is decomposed as a superposition of &quot;waves&quot; (or &quot;frequencies&quot;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example 1: Polynomial multiplication&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is the use of the discrete Fourier transform I'm most familiar with. Suppose you want to multiply two polynomials of degree n, given by their coefficients (a&lt;sub&gt;0&lt;/sub&gt;, …, a&lt;sub&gt;n&lt;/sub&gt;) and (b&lt;sub&gt;0&lt;/sub&gt;, …, b&lt;sub&gt;n&lt;/sub&gt;). In their product, the coefficient of x&lt;sup&gt;k&lt;/sup&gt; is c&lt;sub&gt;k&lt;/sub&gt; = &amp;sum;a&lt;sub&gt;i&lt;/sub&gt;b&lt;sub&gt;k-i&lt;/sub&gt;. This is a convolution, and doing it naively would take O(n&lt;sup&gt;2&lt;/sup&gt;) time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, suppose we represent the polynomials by their &lt;em&gt;values&lt;/em&gt; at 2n points. Then the value of the product polynomial (the one we want) at any point is simply the product of the values of our original two polynomials. Thus we have reduced convolution to pointwise multiplication. The Fourier transform and its inverse correspond to polynomial evaluation and interpolation respectively, for certain well-chosen points (roots of unity). The Fast Fourier Transform (FFT) is a way of doing both of these in O(n log n) time.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example 2: Convolution of probability distributions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Suppose we have two independent (continuous) random variables X and Y, with probability densities f and g respectively. In other words, P(X ≤ x) = ∫&lt;sup&gt;x&lt;/sup&gt;&lt;sub&gt;-∞&lt;/sub&gt; f(t)dt and P(Y ≤ y) = ∫&lt;sup&gt;y&lt;/sup&gt;&lt;sub&gt;-∞&lt;/sub&gt; f(t)dt. We often want the distribution of their sum X+Y, and this is given by a convolution: P(X+Y ≤ z) = ∫f(t)g(z-t)dt. This integration may be hard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instead of representing the random variables by their densities, we can also represent them by their &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;characteristic functions&lt;/a&gt; &amp;phi;&lt;sub&gt;X&lt;/sub&gt;(t) =  E[e&lt;sup&gt;itX&lt;/sup&gt;] and &amp;phi;&lt;sub&gt;Y&lt;/sub&gt;(t) = E[e&lt;sup&gt;itY&lt;/sup&gt;]. Then the characteristic function of X+Y is just:&lt;br&gt;&#xA;&amp;phi;&lt;sub&gt;X+Y&lt;/sub&gt;(t) = E[e&lt;sup&gt;it(X+Y)&lt;/sup&gt;] = &amp;phi;&lt;sub&gt;X&lt;/sub&gt;(t)&amp;phi;&lt;sub&gt;Y&lt;/sub&gt;(t) since they're independent. The characteristic function is the continuous Fourier transform of the density function; it is a change of representation in which convolution becomes pointwise multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote again the answer on MO, many transformations we want to study (translation, differentiation, integration, …) are actually convolutions, so the Fourier transform helps in a wide number of instances. &lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-28T20:54:58.573" />
  <row Id="1055" PostTypeId="2" ParentId="1045" CreationDate="2010-07-28T20:56:53.403" Score="1" ViewCount="0" Body="&lt;p&gt;Presumably you already have the formulas for converting from conical to rectangular coordinates as listed on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conical_coordinates&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page for conical coordinates&lt;/a&gt;. You'll need to solve for r, μ, and ν in terms of x, y, and z to get your answer. I can't see offhand the easiest way to find a general formula, but if you're trying to find it for particular values of r, μ, and ν, it shouldn't be too hard.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-28T20:56:53.403" />
  <row Id="1056" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T20:57:24.090" Score="0" ViewCount="0" Body="&lt;p&gt;Call the highschool functions (defined by the &lt;em&gt;right triangle inscribed in a unit circle&lt;/em&gt;, the angle being equal to the &lt;em&gt;length of the arc of the circle&lt;/em&gt;) $\sin_h$ and $\cos_h$, and let $\sin_p$ and $\cos_p$ be the power series definitions. (Note that these functions are continuous and agree at the end points $0$ and $2\pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sin_p^2(\theta)+\cos_p^2(\theta)=1$ the power series definitions also form a right triangle. Hence $\sin_h = \sin_p \circ \gamma$ and $\cos_h = \cos_p \circ \gamma$ for some parameterization $\gamma$. We know the power series definitions satisfy the arc length criteria so $\gamma$ must be the identity function.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-03T21:45:56.810" CommentCount="2" />
  <row Id="1057" PostTypeId="1" AcceptedAnswerId="1165" CreationDate="2010-07-28T21:01:19.230" Score="1" ViewCount="225" Body="&lt;p&gt;Suppose a finite group has the property that for every $x, y$, it follows that $(xy)^3 = x^3 y^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you prove that it is abelian?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: I recall that the correct exercise needed in addition that the order of the group is not divisible by 3.&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="218" LastEditorDisplayName="" LastEditDate="2010-07-29T18:06:58.200" LastActivityDate="2010-08-04T08:17:33.700" Title="Group with an endomorphism that is &quot;almost&quot; abelian is abelian." Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="1058" PostTypeId="2" ParentId="1018" CreationDate="2010-07-28T21:03:46.010" Score="0" ViewCount="0" Body="&lt;p&gt;Paul Bourke's &lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt;, in particular his &lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/geometry/&quot; rel=&quot;nofollow&quot;&gt;geometry&lt;/a&gt; section fueled my passion for geometry at a young age, infact I learned the idea of analytic geometry from it. For this reason it holds a special place in my heart. It is a very rich site full of graphics and derivations, very easy to get sucked in and lose a few hours!&lt;/p&gt;&#xA;" LastActivityDate="2010-07-28T21:03:46.010" CommunityOwnedDate="2010-07-28T21:03:46.010" />
  <row Id="1059" PostTypeId="2" ParentId="967" CreationDate="2010-07-28T21:05:58.750" Score="4" ViewCount="0" Body="&lt;p&gt;Recommendations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Barendregt &amp;amp; Barendsen, 1984, &lt;a href=&quot;ftp://ftp.cs.kun.nl/pub/CompMath.Found/lambda.pdf&quot; rel=&quot;nofollow&quot;&gt;Introduction to lambda-calculus&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;Girard, Lafont &amp;amp; Taylor, 1987, &lt;a href=&quot;http://www.paultaylor.eu/stable/Proofs+Types.html&quot; rel=&quot;nofollow&quot;&gt;Proofs and Types&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;Sørenson &amp;amp; Urzyczyn, 1999, &lt;a href=&quot;http://folli.loria.fr/cds/1999/esslli99/courses/sorensen-urzyczyn.html&quot; rel=&quot;nofollow&quot;&gt;Lectures on the Curry-Howard Isomorphism&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of these are mentioned in &lt;a href=&quot;http://lambda-the-ultimate.org/node/492&quot; rel=&quot;nofollow&quot;&gt;the LtU Getting Started thread&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-07-28T21:05:58.750" CommentCount="2" />
  <row Id="1061" PostTypeId="2" ParentId="1037" CreationDate="2010-07-28T21:11:54.930" Score="2" ViewCount="0" Body="&lt;p&gt;hysterical raisins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus used to be done with infinitesimals (Archimedes &lt;em&gt;The Method of Mechanical Theorems&lt;/em&gt;, Newtons &lt;em&gt;Fluxions&lt;/em&gt;, ...) but there was some controversy about these &lt;a href=&quot;http://www.maths.tcd.ie/pub/HistMath/People/Berkeley/AnalCont.html&quot; rel=&quot;nofollow&quot;&gt;ghostly quantities&lt;/a&gt; eventually the whole foundation of analysis was rebuilt using limits but the old notations have been kept. So there is (as you noticed) a strange gap between reality (epsilonics) and intuition (infinitesimal quantities) but there are a few more recent redevelopments of the foundation of analysis for example &lt;a href=&quot;http://www.math.wisc.edu/~keisler/calc.html&quot; rel=&quot;nofollow&quot;&gt;Keisler&lt;/a&gt; or &lt;a href=&quot;http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf&quot; rel=&quot;nofollow&quot;&gt;Bell&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-07-28T21:11:54.930" CommentCount="2" />
  <row Id="1062" PostTypeId="2" ParentId="275" CreationDate="2010-07-28T21:19:25.870" Score="5" ViewCount="0" Body="&lt;p&gt;I liked the &lt;a href=&quot;http://press.princeton.edu/titles/8350.html&quot; rel=&quot;nofollow&quot;&gt;Princeton Companion to Mathematics&lt;/a&gt;, though it is a heavy book and not always light reading. I'm a computer scientist with an interest in mathematics.&lt;/p&gt;&#xA;" OwnerUserId="210" LastActivityDate="2010-07-28T21:19:25.870" CommunityOwnedDate="2010-07-28T21:19:25.870" CommentCount="2" />
  <row Id="1063" PostTypeId="2" ParentId="370" CreationDate="2010-07-28T21:23:26.200" Score="2" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://ncatlab.org/&quot; rel=&quot;nofollow&quot;&gt;nLab&lt;/a&gt; is a great resource for category theory.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T21:23:26.200" CommunityOwnedDate="2010-07-28T21:23:26.200" CommentCount="1" />
  <row Id="1064" PostTypeId="1" CreationDate="2010-07-28T21:30:16.330" Score="5" ViewCount="288" Body="&lt;p&gt;Give an example of a perfect set in $\mathbb R^n$ that does not contain any of the rationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or prove that it does not exist).&lt;/p&gt;&#xA;" OwnerUserId="218" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-28T22:20:48.400" LastActivityDate="2010-12-12T21:57:02.770" Title="Perfect set without rationals" Tags="&lt;topology&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="1065" PostTypeId="2" ParentId="1025" CreationDate="2010-07-28T21:32:34.943" Score="0" ViewCount="0" Body="&lt;p&gt;Conic sections should definitely be retained. If you don't cover conic sections, then what other examples can you cover?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lines? Too simple. General curves? Insufficiently concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples are very important for illustrating the general theory and techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in a multivariable calculus course, typical examples will involve quadric surfaces. Here conic sections will come into play, since hyperplane sections (or &quot;level curves&quot;) of quadric surfaces are conic sections.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-28T21:32:34.943" />
  <row Id="1066" PostTypeId="1" AcceptedAnswerId="1071" CreationDate="2010-07-28T21:58:08.247" Score="2" ViewCount="135" Body="&lt;p&gt;From &lt;a href=&quot;http://scottaaronson.com/blog/?p=152&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's blog&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There’s a finite (and not&#xA;  unimaginably-large) set of boxes, such&#xA;  that if we knew how to pack those&#xA;  boxes into the trunk of your car, then&#xA;  we’d also know a proof of the Riemann&#xA;  Hypothesis. Indeed, every formal proof&#xA;  of the Riemann Hypothesis with at most&#xA;  (say) a million symbols corresponds to&#xA;  some way of packing the boxes into&#xA;  your trunk, and vice versa.&#xA;  Furthermore, a list of the boxes and&#xA;  their dimensions can be feasibly&#xA;  written down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;His later commented to explain where he get this from: &quot;3-dimensional bin-packing is NP-complete.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how these two are related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question inspired by the same article is &lt;a href=&quot;http://math.stackexchange.com/questions/946/computation-with-a-memory-wiped-computer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T00:58:01.943" LastActivityDate="2010-07-29T00:58:01.943" Title="Packing boxes and proof of Riemann Hypothesis" Tags="&lt;computer-science&gt;&lt;proof-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1067" PostTypeId="2" ParentId="1064" CreationDate="2010-07-28T22:01:12.833" Score="14" ViewCount="0" Body="&lt;p&gt;An easy example comes from the fact that a number with an infinite &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; expansion is irrational (and conversely). The set of all irrationals with continued fractions consisting only of 1's and 2's in any arrangement is a perfect set of irrational numbers.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-07-28T22:01:12.833" CommentCount="2" />
  <row Id="1068" PostTypeId="1" AcceptedAnswerId="2280" CreationDate="2010-07-28T22:19:32.567" Score="6" ViewCount="552" Body="&lt;p&gt;All the integrals I'm familiar with have the form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int f(x)\mathrm{d}x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I understand these as the sum of infinite tiny rectangles with an area of: f(x_i) * dx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it valid to have integrals that do not have a differential, such as dx, or that have the differential elsewhere than as a factor ? Let me give couple of examples on what I'm thinking of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is valid notation, I'd expect it to sum infinite ones together, thus to go inifinity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int e^{\mathrm{d}x}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, I'd expect this to go to infinity as e^0 = 1, assuming the notation is valid.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\int (e^{\mathrm{d}x} - 1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This I could potentially imagine to have a finite value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are any such integrals valid ? If so, are there any interesting / enlightening examples of such integrals?&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditorDisplayName="" LastEditDate="2010-08-12T15:17:22.003" LastActivityDate="2010-08-12T20:17:43.890" Title="Usage of dx in Integrals" Tags="&lt;calculus&gt;&lt;notation&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="1069" PostTypeId="1" CreationDate="2010-07-28T22:21:41.067" Score="2" ViewCount="117" Body="&lt;p&gt;I would be glad if someone could explain in intuitive terms what these different derivatives are, and possibly give some practical, understandable examples of how they would produce different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, I would like to understand the geometrical or physical meaning of these operators more than the mathematical or topological subtleties that lead to them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-29T00:10:31.523" LastActivityDate="2010-07-29T00:10:31.523" Title="Intuitive explanation of covariant, contravariant and Lie derivatives" Tags="&lt;intuition&gt;&lt;differential-topology&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" />
  <row Id="1070" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T22:23:11.300" Score="11" ViewCount="0" Body="&lt;p&gt;When you write $\int f(x) dx$, the whole of $\int ... dx$ is an indivisible symbol, just as the $d/dx$ is an indivisible symbol when you write $df/dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there &lt;em&gt;are&lt;/em&gt; reasons why the notation is as it is, but trying to manipulate it like you suggest in $\int e^{dx}$, for example, is simply meaningless. &lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-03T03:38:00.153" LastActivityDate="2010-08-03T03:38:00.153" CommentCount="7" />
  <row Id="1071" PostTypeId="2" ParentId="1066" CreationDate="2010-07-28T22:23:36.590" Score="5" ViewCount="0" Body="&lt;p&gt;The question of whether a formal proof of the Riemann Hypothesis exists (with at most a million symbols) is a problem in NP: given such a proof, it can be verified to be correct in polynomial time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bin-packing is NP-complete: this means that every problem in NP can be reduced to bin packing. In particular, the problem mentioned in the previous paragraph can. (This is a reduction that can be made explicit, so once we specify the proof verifier etc., we can carry out the steps of the reduction to get an instance of bin packing. We also need the reduction to be &quot;parsimonious&quot; i.e. solutions correspond one-to-one; I believe it is.)&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-28T22:23:36.590" CommentCount="5" />
  <row Id="1072" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T22:26:38.787" Score="1" ViewCount="0" Body="&lt;p&gt;No, it's not valid. The dx in the integral is a representation of the fact that the integral is obtained as an area, so multiplying the &quot;average&quot; of the function value at each point by an infinitesimal interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the manner in which we don't calculate the area does not change, the notation does not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different notations that are used when the integral is over a curve, or over more than variable (thus leading for example to volumes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The d(variable) notation is also used as a reminder that the integral is against a specific variable and not another, e.g. that int x/y dx differs from int x/y dy.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T22:26:38.787" />
  <row Id="1073" PostTypeId="2" ParentId="381" CreationDate="2010-07-28T22:52:06.230" Score="0" ViewCount="0" Body="&lt;p&gt;It is useful when making estimates (for examples for programming tasks) as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;∀ n &gt; 1 ∈ &lt;strong&gt;N&lt;/strong&gt;, F&lt;sub&gt;n+1&lt;/sub&gt; ≠ (F&lt;sub&gt;n+2&lt;/sub&gt; + F&lt;sub&gt;n&lt;/sub&gt;) / 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore it discourages bad practices like assuming that two equal-sized tasks necessarily take twice as much as one of them (it's false because the larger the scope, the less accurate the estimate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scrummethodology.com/scrum-effort-estimation-and-story-points/&quot; rel=&quot;nofollow&quot;&gt;Example&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-28T22:52:06.230" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1074" PostTypeId="2" ParentId="1069" CreationDate="2010-07-28T23:10:38.137" Score="2" ViewCount="0" Body="&lt;p&gt;The Lie derivative is a derivative of a vector &lt;em&gt;field&lt;/em&gt; V along another vector &lt;em&gt;field&lt;/em&gt; W.  It is defined at a point p as follows: flow the point p along W for some time t and look at the value of V at this point.  Then push this forward along the flow of W to a vector at p.  Subtract V_p from this, divide by t, and take the limit as t-&gt;0.  So this is a measure of how V changes as it gets pushed around by the flow of W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The covariant derivative is a derivative of a vector &lt;em&gt;field&lt;/em&gt; V along a vector W.  Unlike the Lie derivative, this does not come for free: we need a connection, which is a way of identifying tangent spaces.  The reason we need this extra data is because if we wanted to take the directional derivative of V along the vector W how we do in Euclidean space, we would be taking something like V_(p+tW) - V_p, which is the difference of vectors living in different tangent spaces.  If we have a metric, then we can impose reasonable conditions that give us a unique connection (the Levi-Civita connection). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what a contravariant derivative is.  I'd guess it has to do with applying a covariant derivative and lowering indices.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-28T23:10:38.137" />
  <row Id="1075" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T23:33:30.033" Score="17" ViewCount="0" Body="&lt;p&gt;I think your question here shows that, while you have been using these symbols, you haven't really been given a proper motivation for where they came from.&lt;br&gt;&#xA;Let's go back and consider how we came up with the idea of an integral. In a typical class, you will see a lot of pictures like this:&lt;br&gt;&#xA;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/9/91/Integral_approximations.svg/420px-Integral_approximations.svg.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;We find the area under the curve by summing up the area of all these little rectangles. If we wanted to write an expression for the area, it would look like:&lt;br&gt;&#xA;&lt;img src=&quot;http://mathurl.com/27gla3n.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;The Σ means that we are computing a sum. We are adding the areas of the rectangles, which we have numbered 1 through n, to get the complete area under the curve. The area of each rectangle is given by multiplying the height by the width. The height is given by f(x&lt;sub&gt;i&lt;/sub&gt;) because the base of the rectangle is at 0, and the top of the rectangle is where it meets the function f. The Δx represents the width of each rectangle.&lt;br&gt;&#xA;When we find the integral, we are taking the limit of this sum as the number of rectangles goes to infinity, and each individual rectangle becomes infinitesimally tiny. You can think of the dx as the equivalent of Δx: it represents the infinitesimally small width of each rectangle that we added up to get the area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you realize this, we can see why integrals only make sense when written ∫f(x)dx. Because we are adding up the areas of rectangles that have height f(x) and width dx. If you try to interpret the expressions you wrote in this way, you will see that they do not really make sense as integrals: you are not summing up rectangles, so you are not finding an area under a curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, of course, define your own notation in which those expressions behave the way you expect them to, but all mathematical notation is driven based on what people find useful, and what people can agree on and easily understand. Your reuse of the integral sign and dx that people are used to seeing in a particular context will probably result in few people adopting your definition.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-28T23:33:30.033" CommentCount="4" />
  <row Id="1076" PostTypeId="2" ParentId="1048" CreationDate="2010-07-28T23:34:06.787" Score="1" ViewCount="0" Body="&lt;p&gt;There is another proof that the derivative of sine is cosine that doesn't use the sandwich theorem mentioned by Qiaochu and Akhil above.  Instead, one can use the definition of arcsine and the standard calculus formula for arc length in terms of an integral to show that arcsine = the integral of (1 - x^2)^(-.5).  It follows that the derivative of arcsine is (1 - x^2)^(-.5), and (by the chain rule) one can use this fact to prove that the derivative of sine is cosine.&#xA;In fact, I'm not sure why this proof is presented less frequently then the one via the sandwich theorem.  The unit circle definition of sine is based on arc length, and in calculus we learn a formula for arc length based on integration.  Why not connect these two concepts for a natural proof that the derivative of sine is cosine?&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2010-07-28T23:34:06.787" CommentCount="1" />
  <row Id="1077" PostTypeId="2" ParentId="713" CreationDate="2010-07-28T23:42:24.700" Score="0" ViewCount="0" Body="&lt;p&gt;One way to &quot;visualize&quot; a 2-tensor (in the presence of an inner product) is as follows: a vector $X$ can be regarded as a function mapping each direction $D$ to a scalar in a linear and homogeneous fashion, namely, the inner product $X\dot D$. In the same vein, a 2-tensor is a function mapping directions to vectors linearly and homogeneously.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-28T23:42:24.700" />
  <row Id="1078" PostTypeId="2" ParentId="946" CreationDate="2010-07-28T23:44:41.060" Score="4" ViewCount="0" Body="&lt;p&gt;1) Why does a small number of states suffice?&#xA;Regardless of whether the constant is 5 or 500, its still very surprising. Thankfully, it's fairly to prove if you allow the counter to be &lt;code&gt;{1...8}&lt;/code&gt; instead of &lt;code&gt;{1...5}&lt;/code&gt;. [This proof is by Ben-Or and Cleve.]. Start by representing the computation as a circuit, and ignore the whole wiping-clean thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a register machine as follows: It has 3 registers &lt;code&gt;(R_1,R_2,R_3)&lt;/code&gt;, each of which holds a single bit. At each step, the program performs some computation on the registers of the form &lt;code&gt;R_i &amp;lt;-- R_a + x_b R_c&lt;/code&gt; or &lt;code&gt;R_i &amp;lt;-- R_a + x_b R_c + R_d&lt;/code&gt; (where &lt;code&gt;x_1...x_n&lt;/code&gt; is the input).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, set &lt;code&gt;(R_1,R_2,R_3) = (1,0,0)&lt;/code&gt;. The machine should end in the state &lt;code&gt;R_3 + f R_1&lt;/code&gt;. We'll simulate the circuit using a register machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now proceed by induction on the depth of the circuit. If the circuit has depth 0, then we just copy the appropriate bit: &lt;code&gt;R_3 &amp;lt;-- R_3 + x_i R_1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the induction, we have 3 cases, according to whether the final gate is NOT, AND, or OR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the circuit is &lt;code&gt;NOT(f)&lt;/code&gt;. By induction, we can compute &lt;code&gt;f&lt;/code&gt;, yielding the state &lt;code&gt;(R_1,R_2,R_3 + f R_1)&lt;/code&gt;. We can therefore perform the instruction &lt;code&gt;R_3 &amp;lt;-- R_3 + R_1&lt;/code&gt; to get the desired output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the circuit is &lt;code&gt;AND(f_1,f_2)&lt;/code&gt;, then life is a tad more complicated. By induction, we then execute the following 4 instructions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;R_2 &amp;lt;-- R_2 + f_1 R_1&lt;/code&gt; &lt;br&gt;&#xA;&lt;code&gt;R_3 &amp;lt;-- R_3 + f_2 R_2&lt;/code&gt; &lt;br&gt;&#xA;&lt;code&gt;R_2 &amp;lt;-- R_2 + f_1 R_1&lt;/code&gt; &lt;br&gt;&#xA;&lt;code&gt;R_3 &amp;lt;-- R_3 + f_2 R_2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I haven't made any typos, we are left with the state &lt;code&gt;(R_1,R_2,R_3+f_1f_2R_1)&lt;/code&gt;, as desired. OR works similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a moment to process what just happened. It's a slick proof that you have to read 2 or 3 times before it begins to sink in. What we've shown is that we can simulate a circuit by applying a fixed program that stores only 3 bits of information at any time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert this into Aaronson's version, we encode the three registers into the counter (that's why we needed the extra 3 spaces). The simple program uses the input and the clock to determine how far we've made it through the computation and then applies the appropriate change to the counter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) But what's the deal with 5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get from 8 states down to 5, you use a similar argument, but are much more careful about exactly how much information needs to be propagated between stages and how it can be encoded. A formal proof requires lots of advanced group theory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit to answer Casebash's questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Correct. Any computation can be expressed as a circuit composed solely of &quot;NOT&quot;, binary-&quot;AND&quot;, and binary-&quot;OR&quot; gates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The notation &lt;code&gt;f R_1&lt;/code&gt; means (boolean) multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The program for computing &lt;code&gt;f&lt;/code&gt; should take input &lt;code&gt;(R_1,R_2,R_3)&lt;/code&gt; to &lt;code&gt;(R_1,R_2,R_3 + f R_1)&lt;/code&gt;. We insist that the first two registers are unchanged since we use those as temporary storage in the induction. For example, when computing &lt;code&gt;AND&lt;/code&gt;, we compute the first branch and store the result in &lt;code&gt;R_2&lt;/code&gt; while computing the second branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The single bit of output is the final value of &lt;code&gt;R_3&lt;/code&gt;. Since we started with &lt;code&gt;(1,0,0)&lt;/code&gt;, we end with &lt;code&gt;(1,0,f)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditorDisplayName="" LastEditDate="2010-12-06T23:24:29.673" LastActivityDate="2010-12-06T23:24:29.673" CommentCount="3" />
  <row Id="1079" PostTypeId="1" AcceptedAnswerId="1083" CreationDate="2010-07-28T23:44:55.770" Score="11" ViewCount="452" Body="&lt;p&gt;Let f  and g be two periodic functions over R with the following property: If T is a period of f, and S is a period of g, then T/S is irrational. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture&lt;/strong&gt;: f+g is &lt;em&gt;not&lt;/em&gt; periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you give a proof or a counter example?  It is easier if we assume continuity.  But is it true for arbitrary real valued functions?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-07-29T17:51:22.023" LastActivityDate="2010-08-11T22:05:01.260" Title="Sum of two periodic functions" Tags="&lt;analysis&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="5" />
  <row Id="1080" PostTypeId="2" ParentId="1068" CreationDate="2010-07-28T23:53:16.580" Score="2" ViewCount="0" Body="&lt;p&gt;In the context of calculus, $dx$ simply means 'integrate with respect to $x$'. Som books even omit $dx$ entirely because 'of course we're integrating with respect to $x$'. The $dx$-bit does not get a proper meaning before differential forms.&lt;/p&gt;&#xA;" OwnerUserId="401" LastActivityDate="2010-07-28T23:53:16.580" />
  <row Id="1081" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T00:00:49.343" Score="3" ViewCount="0" Body="&lt;p&gt;Pick a basis $B$ of $\mathbb R$ as a $\mathbb Q$ vector space, and split it into two non-empty disjoint parts $B_1$ and $B_2$. Define $\mathbb Q$-linear maps $f,g:\mathbb R\to\mathbb R$ such that $f(x)=x$ and $g(x)=0$ if $x\in B_1$, $f(x)=0$ and $g(x)=x$ if $x\in B_2$. Then $f(x)+g(x)=x$ for all $x\in B$, so that in fact $f+g=\operatorname{id}_{\mathbb R}$, which is not a periodic function. Morever $f$ and $g$ are periodic, and their sets of periods are precisely $B_1$ and $B_2$. Since $B_1\cup B_2$ is linearly independent over $\mathbb Q$, it is easy to see that $x/y\not\in\mathbb Q$ whenever $x\in B_1$ and $y\in B_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is then an example where the sum is not periodic.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-07-29T00:00:49.343" />
  <row Id="1082" PostTypeId="2" ParentId="713" CreationDate="2010-07-29T00:26:03.490" Score="0" ViewCount="0" Body="&lt;p&gt;The part of me that tells me to learn about tensors good and proper tells me that a place to begin is to think about it like this: A tensor (i.e., an element of the tensor product $V\otimes V$) is an equivalence class of elements $(u,v)\in V\times V$ which maps to the same element under some bilinear map. If we think of $(u,v)$ as the square spanned by $u,v$, we may think of $u\otimes v\in V\otimes V$ as a family of squares such that 'through the eyes of a bilinear map' they are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, tell me if I'm off so I don't misguide anyone.&lt;/p&gt;&#xA;" OwnerUserId="401" LastActivityDate="2010-07-29T00:26:03.490" CommentCount="2" />
  <row Id="1083" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T00:32:21.480" Score="9" ViewCount="0" Body="&lt;p&gt;Here is a counterexample.  Let $a, b, c \in \mathbb{R}$ be linearly independent over $\mathbb{Q}$.  Let $\text{span}(x, y, z, ...)$ be the $\mathbb{Q}$-vector space in $\mathbb{R}$ spanned by $x, y, z, ...$.  Let $AB = \text{span}(a, b), BC = \text{span}(b, c), AC = \text{span}(a, c)$.  And for a subset $S$ of $\mathbb{R}$, let $\chi_S$ denote the characteristic function of $S$.  Now define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x) = \chi_{AB} - 2 \chi_{BC}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle g(x) = 3 \chi_{AC} + 2 \chi_{BC}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f$ has period set $\text{span}(b)$, $g$ has period set $\text{span}(c)$, and $f + g$ has period set $\text{span}(a)$.  (I am not sure if the coefficients are necessary; they're just precautions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you still interested in the continuous case?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Old answer below.  I slightly misunderstood the question when I wrote this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simpler example.  I claim that the function $h(x) = \sin x + \sin \pi x$ cannot possibly be periodic.  Why?  Suppose an equation of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin x + \sin \pi x = \sin (x+T) + \sin \pi (x+T)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;held for all $x$ and some $T &gt; 0$.  Take the second derivative of both sides with respect to $x$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin x + \pi^2 \sin \pi x = \sin (x+T) + \pi^2 \sin \pi(x+T).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that $\sin x = \sin (x+T)$ and that $\sin \pi x = \sin \pi(x+T)$, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or is the question whether the sum &lt;em&gt;can&lt;/em&gt; be periodic?)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-31T04:39:10.863" LastActivityDate="2010-07-31T04:39:10.863" CommentCount="3" />
  <row Id="1084" PostTypeId="2" ParentId="668" CreationDate="2010-07-29T00:36:44.637" Score="4" ViewCount="0" Body="&lt;p&gt;The top exterior power of an $n$-dimensional vector space $V$ is one-dimensional. Its elements are sometimes called pseudoscalars, and they represent oriented $n$-dimensional volume elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linear operator $f$ on $V$ can be extended to a linear map on the exterior algebra according to the rules $f(\alpha) = \alpha$ for $\alpha$ a scalar and $f(A \wedge B) = f(A) \wedge f(B), f(A + B) = f(A) + f(B)$ for $A$ and $B$ blades of arbitrary grade. Trivia: some authors call this extension an &lt;em&gt;outermorphism&lt;/em&gt;. The extended map will be grade-preserving; that is, if $A$ is a homogeneous element of the exterior algebra of grade $m$, then $f(A)$ will also have grade $m$. (This can be verified from the properties of the extended map I just listed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this implies that a linear map on the exterior algebra of $V$ once restricted to the top exterior power reduces to multiplication by a constant: the &lt;strong&gt;determinant&lt;/strong&gt; of the original linear transformation. Since pseudoscalars represent oriented volume elements, this means that the determinant is precisely the factor by which the map scales oriented volumes.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-07-29T03:16:51.460" LastActivityDate="2010-07-29T03:16:51.460" CommentCount="1" />
  <row Id="1085" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T00:45:18.067" Score="6" ViewCount="0" Body="&lt;p&gt;Before Hartshorne's book there was &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/354063293X&quot; rel=&quot;nofollow&quot;&gt;Mumford's Red Book of Varietie&lt;/a&gt;s.  I think it is a great introductory textbook to modern algebraic geometry (scheme theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that Mumford is quite good at motivating new concepts; in particular I really enjoy his development of nonsingularity and the sheaf of differentials.  I think another great aspect about this book is that it emphasizes how to define things intrinsically (i.e. without reference to a closed or open immersion into affine space) but also explains how to make local arguments (i.e. using immersion into affine space).  A classic example of the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(non intrinsic tangent space): Say X is a variety and p is a point of X.  Choose an affine neighborhood so that p corresponds to the origin.  Then this affine neighborhood is spec k[x1, ..., xn]/I for some ideal.  Let I' be all the linear terms of I (i.e. if I = (x,y^2), then I' = (x)).  Then the tangent space at p is spec k[x1,...,xn]/I'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(intrinsic tangent space): Let m be the maximal ideal of the local ring of the structure sheaf at p, then the tangent space is the dual of the vector space m/m^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking spec of the symmetric algebra of the latter gives you the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some drawbacks.  This book doesn't cover nearly as much as Hartshorne's book.  It doesn't have that many exercises.  The notation is slightly different; integral finite type schemes are called pre-varieties and you can remove the `pre' if it's also separated.  Nevertheless I think its a great compliment to reading Hartshorne.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-07-29T00:45:18.067" CommunityOwnedDate="2010-07-29T00:45:18.067" />
  <row Id="1086" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T00:55:28.563" Score="0" ViewCount="0" Body="&lt;p&gt;Another book I wish I had known about when I was first reading Hartshorne is Miranda's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521423538&quot; rel=&quot;nofollow&quot;&gt;Complex Algebraic Curves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again this book covers much less then Hartshorne and only discusses curves over the complex numbers (and their Jacobians).  But it gives a lot more details and examples of concepts which I found particularly difficult when I first started learning algebraic geometry (sheafs, divisors, cohomology).  It also has a bunch of exercises which I think are often not as challenging as the the exercises in Hartshorne.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also covers a lot more of the 'classical' theory of curves than Hartshrone does; e.g. &#xA;Weierstrass points. &lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-07-29T00:55:28.563" CommunityOwnedDate="2010-07-29T00:55:28.563" />
  <row Id="1087" PostTypeId="2" ParentId="381" CreationDate="2010-07-29T00:56:04.130" Score="0" ViewCount="0" Body="&lt;p&gt;They're a much easier way to evaluate the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x) = (ø^x -(-ø)^x) / sqrt(5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by hand, where ø is the golden ratio and x is an integer :)&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2010-07-29T00:56:04.130" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1088" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T01:07:22.173" Score="5" ViewCount="0" Body="&lt;p&gt;My last suggestion would be Ravi Vakil's &lt;a href=&quot;http://math.stanford.edu/~vakil/0910-216/&quot; rel=&quot;nofollow&quot;&gt;online notes on the foundations of algebraic geometry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these notes might be made into a full on textbook someday.  I haven't looked through all of them but these notes seem to cover as much as Hartshorne does (if not more).  Only rarely do Hartshorne and Vakil define things differently (`projective morphisms' is the only example that comes to mind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard it said that Hartshorne's book is a `baby' version of EGA.  I think Vakil's notes are somewhere between Hartshorne and EGA (probably not the midpoint though).  At least Vakil discusses much more the theory of representable functors, and Noetherian hypothesis are less prevalent in Vakil's notes.  Also Vakil's notes are more complete in that they also include proofs of many of the commutative algebra results that are just stated in Hartshorne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Vakil spends a lot more time motivating the material and often the notes are a bit conversational.  Also there are tons of exercises and most of the them are appended with useful qualifiers like (easy but important exercises, unimportant exercise, tedious but useful exercise, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One drawback is that they are very long and they are online notes so there are many typos.  But most of them are grammatical and easy to spot.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-07-29T01:07:22.173" CommunityOwnedDate="2010-07-29T01:07:22.173" CommentCount="4" />
  <row Id="1089" PostTypeId="2" ParentId="907" CreationDate="2010-07-29T01:09:48.737" Score="-2" ViewCount="0" Body="&lt;p&gt;If you think that's bad, you should see how mathematicians use the words &quot;group&quot; and &quot;field&quot;.  Totally non-standard English.&lt;/p&gt;&#xA;" OwnerUserId="407" LastActivityDate="2010-07-29T01:09:48.737" />
  <row Id="1090" PostTypeId="2" ParentId="250" CreationDate="2010-07-29T01:10:24.537" Score="5" ViewCount="0" Body="&lt;p&gt;For any five points on the globe, there is an angle in outer space from which you could see at least 4 of the 5 points (assuming the moon or anything isn't in the way).  The proof is pretty simple, too...&lt;/p&gt;&#xA;" OwnerUserId="209" LastActivityDate="2010-07-29T01:10:24.537" CommunityOwnedDate="2010-07-29T01:10:24.537" CommentCount="1" />
  <row Id="1091" PostTypeId="1" AcceptedAnswerId="1095" CreationDate="2010-07-29T01:19:48.370" Score="5" ViewCount="422" Body="&lt;p&gt;The wiki &lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalue,_eigenvector_and_eigenspace&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; on eigenvectors offers the following geometrical interpretation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each application of the matrix to an arbitrary vector yields a result which will have rotated towards the eigenvector with the largest eigenvalue.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Qn 1: If there is any other geometrical interpretation particularly in the context of a covariance matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki also discusses the difference between left and right eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qn 2: Do the above geometrical interpretations hold irrespective of whether they are left or right eigenvectors?&lt;/p&gt;&#xA;" OwnerUserId="116" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T01:27:38.477" LastActivityDate="2010-07-30T04:05:04.443" Title="Is there a geometrical interpretation to the notion of eigenvector and eigenvalues?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="1092" PostTypeId="2" ParentId="1091" CreationDate="2010-07-29T01:29:06.133" Score="0" ViewCount="0" Body="&lt;p&gt;Of course! Consider a coordinate transformation of rotation and/or scaling (but not translation):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = Au&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where v and u are vectors, and A is a transformation matrix. Then the eigenvectors, if they have real components, are the axes which are left unrotated (scaling only) by the transformation. (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalue,_eigenvector_and_eigenspace&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A covariance matrix is a symmetric, positive definite matrix, so it has orthonormal eigenvectors, and these form a tuple of axes; I am fairly sure the eigenvectors form a new basis of linear combinations of the input variables where the basis variables are uncorrelated, but I can't remember how to show this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;For example, if w1 = [x;y] is a pair of independent unit-variance zero-mean Gaussian random variables, consider w2 = [u;v] = [1 1; 2 1][x;y] = (x+y,2x+y), so that w1 = [-1 1;2 -1][u;v] = [v-u;2u-v]. Then cov(w2) = [2 3; 3 5]. This has eigenvectors which have sqrt(5) in them, hmmmm...&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for question 2, I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-07-29T01:55:14.603" LastActivityDate="2010-07-29T01:55:14.603" CommentCount="4" />
  <row Id="1093" PostTypeId="2" ParentId="392" CreationDate="2010-07-29T02:04:40.680" Score="10" ViewCount="0" Body="&lt;p&gt;Perhaps the following diagram will provide insight:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/dsindcos.jpg&quot; alt=&quot;(Non)Proof without Words: Derivatives of Sine and Cosine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to look at the sine and cosine curves as projections of a helix drawn on a cylinder. If you look at the cylinder itself as a curled planar square of length 2pi, then helix is a curled version of the square's diagonal. A tangent vector along the flat square's diagonal always lies at 45 degrees to the square's sides, say with length-&quot;1&quot; shadows in each direction; after smoothly curling the square into the cylinder, the tangent vector lies at 45 degrees to the cylinder's (z-)axis and the perpendicular (xy-)plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Projecting the helix into the zy- and zx-planes gives graphs of sine and cosine. Projecting the helix's tangent vector gives tangent vectors to those graphs. The &quot;dz&quot;s for these projected tangents are always 1 (the &quot;vertical&quot; shadow of the helix's tangent vector). To get at &quot;dy&quot; and &quot;dx&quot; (&quot;v_x&quot; and &quot;v_y&quot; in the diagram) we project &lt;em&gt;down&lt;/em&gt; into the xy-plane where we see a circle, and yet another projected tangent vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic geometry tells us that a tangent to a circle is perpendicular to the radius at the point of tangency. In our circle, the point of tangency --and the radius vector it-- is parameterized as &quot;&amp;lt; cos, sin, 0 &gt;&quot;. The perpendicular tangent line must therefore have a &quot;negative-reciprocal&quot; direction vector: &quot;&amp;lt; -sin, cos, 0 &gt;&quot;, which gives us our &quot;dx&quot; and &quot;dy&quot; for the helix tangent ... and the projected graph tangents as well, so that we may make the following conclusions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of cosine --by its &lt;em&gt;conceptual definition&lt;/em&gt; as &quot;slope of the tangent line&quot;-- is change-in-x-over-change-in-z = dx/dz = -sin/1 = -sin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the derivative of sine is dy/dz = cos/1 = cos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like this approach because the conceptual &quot;slope of tangent line&quot; definition of the derivative is used throughout; there are no (obvious) appeals to digressive computational tricks involving trig identities and limits of difference quotients. I also like that the curious negative sign in the derivative of cosine traces back to an elementary property of circle geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this approach doesn't constitute &lt;em&gt;proof&lt;/em&gt; of the formulas. The process of curling the planar square into a cylinder and claiming that the tangent vector behaves as claimed actually &lt;em&gt;assumes&lt;/em&gt; the computational machinery covered by the traditional limit arguments. Nevertheless, on an &lt;em&gt;intuitive&lt;/em&gt; level, I think this argument explains the &quot;why&quot; of the derivatives quite beautifully. Then, knowing what the formulas &lt;em&gt;are&lt;/em&gt; (or &quot;should be&quot;) helps motivate the investigation of the computational tricks needed to provide a rigorous proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://daylateanddollarshort.com/math/pdfs/dsindcos.pdf&quot; rel=&quot;nofollow&quot;&gt;Here's a PDF&lt;/a&gt; with a variant of the above discussion (but the same image). &lt;a href=&quot;http://demonstrations.wolfram.com/CalculusFreeDerivativesOfSineAndCosine/&quot; rel=&quot;nofollow&quot;&gt;Here's a Mathematica Demonstration&lt;/a&gt; that animates the various elements, including the square curling into the cylinder.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-07-29T14:25:30.383" LastActivityDate="2010-07-29T14:25:30.383" CommentCount="2" />
  <row Id="1094" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T02:14:13.390" Score="3" ViewCount="0" Body="&lt;p&gt;for &lt;em&gt;Undergraduate&lt;/em&gt; algebraic geometry (significantly below the level of Hartshorne), Cox, Little and O'Shea's &lt;em&gt;Ideals, Varieties, and Algorithms&lt;/em&gt; is a pleasant treatment.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-07-29T02:14:13.390" CommunityOwnedDate="2010-07-29T02:14:13.390" CommentCount="1" />
  <row Id="1095" PostTypeId="2" ParentId="1091" CreationDate="2010-07-29T02:32:27.370" Score="6" ViewCount="0" Body="&lt;p&gt;Here is a partial answer in the case where M is a real symmetric matrix.  This is to ensure, by the real spectral theorem, that M has real eigenvectors with real eigenvalues, so there is a chance for a genuine geometric interpretation which stays in R^n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M acts on the unit sphere in R^n in the following way: it sends the unit sphere v^T v = 1 to v^T (M^T M) v = 1.  This modified shape is not generally a sphere, but is generally an ellipsoid.  The axes of this ellipsoid are the eigenvectors of M, and the sizes of each axis are given by the squares of the corresponding eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-29T02:32:27.370" CommentCount="3" />
  <row Id="1096" PostTypeId="1" AcceptedAnswerId="1097" CreationDate="2010-07-29T02:38:41.303" Score="6" ViewCount="471" Body="&lt;p&gt;Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;1 = 1^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2 + 2 = 2^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3 + 3 + 3 = 3^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;x + x + x + ... (repeated x times) = x^2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Take the derivative of lhs and rhs and we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 + 1 + 1 (repeated x times) = 2x&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;x = 2x&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and hence &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 = 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clearly something is wrong but I am unable pinpoint my mistake.&lt;/p&gt;&#xA;" OwnerUserId="116" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T06:25:55.463" LastActivityDate="2010-08-11T04:24:21.853" Title="Where is the flaw in this argument of a proof that 1=2?" Tags="&lt;calculus&gt;&lt;recreational-mathematics&gt;&lt;fake-proofs&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="1097" PostTypeId="2" ParentId="1096" CreationDate="2010-07-29T02:47:45.227" Score="15" ViewCount="0" Body="&lt;p&gt;You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even beyond that, if you can express x&lt;sup&gt;2&lt;/sup&gt; as x + x + x + ... (repeated x times), then x must be an integer and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;edit&lt;/em&gt;: I gave my first reason first because the second reason can be smoothed over by taking &quot;repeated x times&quot; to mean something like $\underset{\lfloor x\rfloor\mathrm{\ addends}}{\underbrace{x+x+\cdots+x}}+(x-\lfloor x\rfloor)\cdot x$.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-05T18:40:51.450" LastActivityDate="2010-08-05T18:40:51.450" />
  <row Id="1098" PostTypeId="2" ParentId="1096" CreationDate="2010-07-29T02:50:51.570" Score="3" ViewCount="0" Body="&lt;p&gt;You cannot differentiate the LHS of your equation&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x + x + x + ... (repeated x times) = x^2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because the LHS is not a continuous function; the number of terms depends on x so the LHS is not well defined when x is not an integer. We can only differentiate continuous functions, so this is not valid.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-29T02:50:51.570" CommentCount="1" />
  <row Id="1099" PostTypeId="1" AcceptedAnswerId="1110" CreationDate="2010-07-29T02:57:50.193" Score="6" ViewCount="314" Body="&lt;p&gt;Are there any well-studied analogs to the Mandelbrot set using functions other than f(z)=z^2+c?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-30T09:20:05.043" LastActivityDate="2010-11-16T19:08:05.717" Title="Mandelbrot-like sets for functions other than f(z)=z^2+c" Tags="&lt;fractals&gt;&lt;dynamical-systems&gt;" AnswerCount="6" FavoriteCount="3" />
  <row Id="1100" PostTypeId="2" ParentId="381" CreationDate="2010-07-29T03:05:47.383" Score="2" ViewCount="0" Body="&lt;p&gt;How many ways are there to tile a 2 x n grid with 1 x 2 dominos? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem can be solved using Fibonacci numbers. Let S&lt;sub&gt;n&lt;/sub&gt; be the number of valid tilings of the 2 x n grid. Each such tiling has either a vertical 1 x 2 domino or two horizontal dominos on the right. Therefore, each tiling of a 2 x (n-1) grid or a 2 x (n-2) grid generates a tiling of the 2 x n grid, and hence we have a recurrence relation S&lt;sub&gt;n&lt;/sub&gt; = S&lt;sub&gt;n-1&lt;/sub&gt; + S&lt;sub&gt;n-2&lt;/sub&gt;. This is precisely the recurrence relation of the Fibonacci numbers. Checking our base cases, we see that there is one way to tile a 1 x 2 grid and two ways to tile a 2 x 2 grid, so S&lt;sub&gt;1&lt;/sub&gt; = 1 and S&lt;sub&gt;2&lt;/sub&gt; = 2. Therefore, the number of tilings is precisely the Fibonacci sequence.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-29T03:05:47.383" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1101" PostTypeId="2" ParentId="514" CreationDate="2010-07-29T03:11:46.200" Score="5" ViewCount="0" Body="&lt;p&gt;Another example: &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture&quot; rel=&quot;nofollow&quot;&gt;Euler's sum of powers conjecture&lt;/a&gt;, a generalization of Fermat's Last Theorem. It states:&lt;br&gt;&#xA;If the equation &amp;sum;&lt;sub&gt;i=0&lt;/sub&gt;&lt;sup&gt;k&lt;/sup&gt; x&lt;sub&gt;i&lt;/sub&gt;&lt;sup&gt;n&lt;/sup&gt; = z&lt;sup&gt;n&lt;/sup&gt; has a solution in positive integers, then n ≤ k (unless k=1). Fermat's Last Theorem is the k=2 case of this conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counterexample for n=5 was found in 1966: it's&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 61917364224 = 27&lt;sup&gt;5&lt;/sup&gt; + 84&lt;sup&gt;5&lt;/sup&gt; + 110&lt;sup&gt;5&lt;/sup&gt; + 133&lt;sup&gt;5&lt;/sup&gt; = 144&lt;sup&gt;5&lt;/sup&gt;.&lt;br&gt;&#xA;The smallest counterexample for n=4 was found in 1988:&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 31858749840007945920321 = 95800&lt;sup&gt;4&lt;/sup&gt; + 217519&lt;sup&gt;4&lt;/sup&gt; + 414560&lt;sup&gt;4&lt;/sup&gt; = 422481&lt;sup&gt;4&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example used to be even more useful in the days before FLT was proved, as an answer to the question &quot;Why do we need to prove FLT if it has been verified for thousands of numbers?&quot; :-)&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-29T03:11:46.200" CommentCount="1" />
  <row Id="1102" PostTypeId="2" ParentId="1099" CreationDate="2010-07-29T03:17:46.047" Score="3" ViewCount="0" Body="&lt;p&gt;There is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Julia_set&quot; rel=&quot;nofollow&quot;&gt;Julia set&lt;/a&gt;, which can be defined for any complex rational map f(z) = P(z)/Q(z) where P(z) and Q(z) are polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Julia set for the map f&lt;sub&gt;c&lt;/sub&gt;(z) = z^2 + c is related to the Mandelbrot set in that a point z is in the Mandelbrot set if the Julia set of f&lt;sub&gt;c&lt;/sub&gt;(z) is connected.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-29T03:17:46.047" CommentCount="1" />
  <row Id="1103" PostTypeId="2" ParentId="1048" CreationDate="2010-07-29T03:25:59.173" Score="1" ViewCount="0" Body="&lt;p&gt;If you allow yourself a tiny bit of calculus ( &quot;$\sin x / x \to 1$&quot; as &quot;$x \to 0$&quot; ) and apply some combinatorics, there's a really nice geometric interpretation of the terms of the power series for the functions. Consider this diagram and the polygonal &quot;spiral&quot; that starts at $P_0$ and closes in on the point $P$ (where $|P_0 P| = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/sincosinvolutes.jpg&quot; alt=&quot;The Sine and Cosine Involute Pinwheel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal segments $P_{2n} P_{2n+1}$ alternately overshoot and undershoot the length of the cosine segment; the vertical segments $P_{2n+1} P_{2n+2}$ do the same for the sine segment. So, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\cos \theta = \sum_{n=0}^{\infty}(-1)^n | P_{2n} P_{2n+1} |$ &lt;br /&gt;&lt;br /&gt;&#xA;  $\sin \theta = \sum_{n=0}^{\infty} (-1)^n | P_{2n+1} P_{2n+2} |$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the lengths $|P_{k} P_{k+1}|$ are equal to the lengths of the curves $|I_k|$, which constitute a series of successive &lt;a href=&quot;http://en.wikipedia.org/wiki/Involute&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;involutes&lt;/em&gt;&lt;/a&gt; (with $I_0$ defined to be a segment, and $I_1$ defined to be an arc of the unit circle). Combinatorics and the calculus result I mentioned show that the involute lengths satisfy ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$|I_k| = \theta^k / k!$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... so that the above are, in fact, &lt;em&gt;power series&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the same thing can be done with secant and tangent, using an involute zig-zag:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/sectaninvolutes.jpg&quot; alt=&quot;The Secant and Tangent Involute Zig-Zag&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\sec \theta = \sum_{n=0}^{\infty} | P_{2n} P_{2n+1} | = \sum_{n=0}^{\infty} | I_{2n} |$&lt;br /&gt;&lt;br /&gt;&#xA;  $\tan \theta = \sum_{n=0}^{\infty} | P_{2n+1} P_{2n+2} | = \sum_{n=0}^{\infty} | I_{2n+1} |$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the lengths $|I_k|$ turn out to be the appropriate multiples of powers of $\theta^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference to the argument for sine and cosine (attributed to Y. S. Chaikovsky, as reported by Leo Gurin), a complete discussion of the trickier argument for secant and tangent, and then a refinement of the argument for sine and cosine, are in my note &lt;a href=&quot;http://daylateanddollarshort.com/math/pdfs/sectan.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Zig-Zag Involutes, Up-Down Permutations, and Secant and Tangent&quot; (PDF)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: I have not (yet) cracked the case for cosecant and cotangent.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-08-03T16:23:35.267" LastActivityDate="2010-08-03T16:23:35.267" />
  <row Id="1104" PostTypeId="2" ParentId="1096" CreationDate="2010-07-29T03:28:59.503" Score="4" ViewCount="0" Body="&lt;p&gt;Here's my explanation from an old sci.math post:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zachary Turner  wrote on 26 Jul 2002:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let D = d/dx = derivative wrt x. Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D[x^2] = D[x  +   x  + ... +   x  (x times)]&#xA;       = D[x] + D[x] + ... + D[x] (x times)&#xA;       =   1  +   1  + ... +   1  (x times)&#xA;       =   x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An obvious analogous fallacious argument proves both&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  D[x f(x)]  =  Df(x) (x  times) = x Df(x)&#xA;&#xA;  D[x f(x)]  =   Dx (f(x) times) = f(x), via  Dx = 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;vs.  the correct result: their sum  f(x) + x Df(x)&#xA;as given by the Leibniz product rule (= chain rule for times).&#xA;The error arises from overlooking the dependence upon x in &lt;em&gt;both&lt;/em&gt;&#xA;arguments of the product  x * f(x)  when applying the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of the error becomes clearer if we consider a&#xA;discrete analog. This will also eliminate any tangential&#xA;concerns on the meaning of &quot;(x times)&quot; for non-integer x.&#xA;Namely, we consider the shift operator  S: n -&gt; n+1  on polynomials  p(n)&#xA;with integer coef's, where  S p(n) = p(n+1). Here's a similar fallacy&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  S[n^2] =  S[n  +   n  + ... +   n  (n times)]&#xA;         =  S[n] + S[n] + ... + S[n] (n times)&#xA;         =  1+n  + 1+n  + ... + 1+n  (n times)&#xA;         = (1+n)n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But correct is  S[n^2] = (n+1)^2. Here the &quot;product rule&quot; is&#xA;simply  S[fg] = S[f] S[g], not  S[f] g  as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fallacy actually boils down to operator non-commutativity.&#xA;On the space of functions f(x), consider &quot;x&quot; as the linear&#xA;operator of multiplication by  x, so  x: f(x) -&gt; x f(x). Then&#xA;the linear operators  D  and  x  generate an operator algebra&#xA;of polynomials  p(x,D) in NON-commutative indeterminates x,D&#xA;since we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  (Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so  Dx = xD+1&#xA;&#xA;  (Sn)[f] = S[nf] = (n+1)S[f], so  Sn = (n+1)S != nS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This viewpoint reveals the error simply as mistakenly&#xA;assuming commutativity of the operators  x,D  or  n,S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps something to ponder on boring &lt;em&gt;commutes&lt;/em&gt; !&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-07-29T03:28:59.503" />
  <row Id="1105" PostTypeId="2" ParentId="677" CreationDate="2010-07-29T03:30:58.303" Score="2" ViewCount="0" Body="&lt;p&gt;I like your problem. But your opposite parity conjecture doesn't seem to handle the case N=3 and S=2 (even when corrected as in Mau's comment by gcd), with a triangle labeled as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;        1&#xA;&#xA;     1     0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This position appears to be invariant under any move, so you cannot reach all 0s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, a similar obstacle works for any 3n-gon, with S=2n. If you repeat the pattern 1 1 0 all the way around, it is stable under any move. If $n$ is odd, these will have opposite parities.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-07-29T03:30:58.303" CommentCount="1" />
  <row Id="1106" PostTypeId="2" ParentId="639" CreationDate="2010-07-29T03:42:46.310" Score="7" ViewCount="0" Body="&lt;p&gt;For $k = 2$ I can do it with ${\lceil \log_2 N \rceil + 2 \choose 2} - 1$ servants.  In particular for $N = 1000$ I can do it with $65$ servants.  The proof is somewhat long, so I don't want to post it until I've thought about the problem more.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to improve on the above result.  Here's how it works.  Let $n = \lceil \log_2 N \rceil$.  Let me go through the algorithm for $k = 1$ so we're all on the same page.  Number the wines and assign each of them the binary expansion of their number, which consists of $n$ bits.  Find $n$ servants, and have servant $i$ drink all the wines whose $i^{th}$ bit is $1$.  Then the set of servants that die tells you the binary expansion of the poisoned wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k = 2$ we need to find $n$ butlers, $n$ maids, and ${n \choose 2}$ cooks.  The cooks will be named $(i, j)$ for some positive integers $1 \le i &amp;lt; j \le n$.  Have butler $i$ drink all the wines whose $i^{th}$ bit is $1$, have maid $i$ drink all the wines whose $i^{th}$ bit is $0$, and have cook $(i, j)$ drink all the wines such that the sum of the $i^{th}$ bit through the $j^{th}$ bit, inclusive, mod 2, is $1$.  This is how the casework breaks down for butlers and maids.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If both butler $i$ and maid $i$ die, then one of the poisoned wines has $i^{th}$ bit $0$ and the other has $i^{th}$ bit $1$.&lt;/li&gt;&#xA;&lt;li&gt;If only butler $i$ dies, then both of the poisoned wines have $i^{th}$ bit $1$.&lt;/li&gt;&#xA;&lt;li&gt;If only maid $i$ dies, then both of the poisoned wines have $i^{th}$ bit $0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second two cases are great.  The problem with case 1 is that if it occurs more than once, there's still ambiguity about which wine has which bit.  (The worst scenario is if all the butlers and maids die.)  To fix the issue with case 1, we use the cooks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $i_1 &amp;lt; ... &amp;lt; i_m$ be the set of bits where case 1 occurs.  We'll say that the poisoned wine whose $(i_1)^{th}$ bit is $1$ is wine A, and the other one is wine B.  Notice that the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine A mod 2 is the same as the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine B mod 2, and we can determine what this sum is by looking at whether cook $(i_1, i_2)$ died.  The value of this sum determines whether the $(i_2)^{th}$ bit of wine A is 1 or 0 (and the same for wine B).  Similarly, looking at whether cook $(i_j, i_{j+1})$ died tells us the remaining bits of wine A, hence of wine B.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One last comment for now.  The lower bound is not best possible when $k$ is large compared to $N$; for example, when $k = N-1$ it takes $N-1$ servants.  The reason is that any servant who drinks more than one wine automatically dies, hence gives you no information. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-29T21:55:57.263" LastActivityDate="2010-07-29T21:55:57.263" CommentCount="7" />
  <row Id="1107" PostTypeId="2" ParentId="1099" CreationDate="2010-07-29T04:15:05.790" Score="11" ViewCount="0" Body="&lt;p&gt;I have no idea about well-studied, but here's what I've gotten from a bit of playing around (&lt;a href=&quot;http://gist.github.com/497210&quot; rel=&quot;nofollow&quot;&gt;here's the relevant Mathematica code&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\boldsymbol{f_c(z)=\cos(z)+c}$ where $\cos(z)$ is defined however Mathematica defines it, using the escape time algorithm with escape radius $10\pi$ and 100 iterations maximum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2Ddw7.png&quot; alt=&quot;cosine1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pL8nx.png&quot; alt=&quot;cosine2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\boldsymbol{f_c(z)=\sin(z)+c}$ appears to be the same as cosine, but horizontally translated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\boldsymbol{f_c(z)=e^z+c}$, using the escape time algorithm with escape radius 50 and 100 iterations maximum (it's vertically periodic with period $2\pi$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1cJU6.png&quot; alt=&quot;exp1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dq0cl.png&quot; alt=&quot;exp2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; (Mathematica code edited, too): $\boldsymbol{f_c(z)=cz(1-z)}$ (since camccann's answer mentions logistic maps) using the escape time algorithm with escape radius 50 and 200 iterations maximum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/V3csJ.png&quot; alt=&quot;logistic1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/21gSR.png&quot; alt=&quot;logistic2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:33:19.810" LastActivityDate="2010-09-29T04:33:19.810" CommentCount="4" />
  <row Id="1108" PostTypeId="1" AcceptedAnswerId="1114" CreationDate="2010-07-29T04:32:57.897" Score="0" ViewCount="108" Body="&lt;p&gt;I understand the mechanics of the proof of Ostrowski's Theorem, but I'm a little unclear on why one should expect valuations to be related to primes. Is this a special property of number fields and function fields, or do primes of K[x,y] correspond to valuations on K(x,y) in the same way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping for an answer that can explain what exactly are the algebraic analogs of archimedian valuations, and how to use them - for example, I've heard that the infinite place on K(x) corresponds to the &quot;prime (1/x)&quot; - how does one take a polynomial in K[x] &quot;mod (1/x)&quot; rigorously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-29T07:36:36.973" LastActivityDate="2010-07-29T20:22:39.630" Title="Why should one expect valuations to be related to primes? How to treat an infinite place algebraically?" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="1109" PostTypeId="2" ParentId="1064" CreationDate="2010-07-29T04:49:16.473" Score="3" ViewCount="0" Body="&lt;p&gt;It can be proven that the Cantor set is perfect. Certainly, this contains infinitely many rationals. How about modifying the construction of the Cantor set by defining:&#xA;$I_1 = [\sqrt{2},\sqrt{2}+1/3] \cup [\sqrt{2}+2/3,\sqrt{2}+1]$, $I_2 = [\sqrt{2},\sqrt{2}+1/9] \cup [\sqrt{2}+2/9,\sqrt{2}+1/3]\cup[\sqrt{2}+2/3,\sqrt{2}+7/9]\cup[\sqrt{2}+8/9,\sqrt{2}+1]$, etc and setting $P = \cap_{i=1}^\infty I_i$? Each of end points of any interval that appears in the construction is a member of $P$ and is irrational. However, is it true that all the members of $P$ must be an end point of a certain interval? I am tempted to think so because we can prove that $P$ does not contain any interval. &lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2010-07-29T04:49:16.473" CommentCount="1" />
  <row Id="1110" PostTypeId="2" ParentId="1099" CreationDate="2010-07-29T05:17:57.723" Score="7" ViewCount="0" Body="&lt;p&gt;A variant on the M-set can be defined in straightforward fashion for any iterated function in the complex plane parameterized by a single initial value. For instance, slight modifications produce the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tricorn_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;tricorn&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Burning_Ship_fractal&quot; rel=&quot;nofollow&quot;&gt;burning ship&lt;/a&gt; fractals. However, most such variations tend to be either boring, incoherent, or obviously derived from the Mandelbrot set--nothing particularly novel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some obvious patterns also emerge quickly from many variations: Real exponents alter symmetry, imaginary exponents cause asymmetric twisting, disreputable functions produce misshapen lumps like the burning ship, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it's more difficult than you might expect to avoid the M-set in the first place: For a well known example, consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;the Newton-Raphson method for approximating roots&lt;/a&gt;, which can be generalized to the complex plane in straightforward fashion. For some polynomial, a point in the complex plane may or may not converge to a particular root after some number of Newton-Raphson iterations. In most cases, plotting the regions of divergence and convergence per root produces a fractal. Modifying the polynomial and the iteration formula produces effects analogous to modifying the constant and iteration function, respectively, of a standard Julia set, and in fact &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/newton/&quot; rel=&quot;nofollow&quot;&gt;it turns out that Julia fractals can be considered a special case of Newton-Raphson fractals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous structure can also be found elsewhere than the complex plane. The period-doubling behavior of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logistic_map&quot; rel=&quot;nofollow&quot;&gt;logistic map&lt;/a&gt; relates to the behavior of points on the real line of the M-set, and islands of stability for the logistic map correspond to the positions of mini-Mandelbrots. Elaborate Julia-like structures can also be found in the quaternions, although unfortunately visualization of 4-dimensional fractals is somewhat challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspicion is that Julia-like structure will arise for &lt;em&gt;any&lt;/em&gt; fractal defined by similar means, e.g., classifying points into sets based on their behavior under repeated iteration of a position-sensitive transformation, but I'm not sure how to define &quot;similar means&quot; precisely enough to formalize that.&lt;/p&gt;&#xA;" OwnerUserId="211" LastActivityDate="2010-07-29T05:17:57.723" CommentCount="1" />
  <row Id="1111" PostTypeId="2" ParentId="1091" CreationDate="2010-07-29T05:25:31.097" Score="4" ViewCount="0" Body="&lt;p&gt;If you are interested in covariance matrices, then the eigenvectors of the covariance matrix tell you how to change variables to make your random variables uncorrelated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, let $M$ be a covariance matrix of the random variables $X_1,...,X_n$. For simplicity, lets assume that all of these random variables are zero mean. Lets also define $X$ to be the random vector whose $i$&quot;th component is the random variable $X_i$. Let the eigenvectors of $M$ be $v_1, v_2, ..., v_n$. We can assume that these are orthogonal since $M$ is symmetric. Consider the random variables $Y_i$ obtained by taking the dot product of $v_i$ and $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the random variables $Y_1, ..., Y_n$ are uncorrelated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[ Y_i Y_j] = E[ v_i^T X X^T v_j] = v_i^T M v_j = \lambda_j v_i^T v_j = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-07-29T05:25:31.097" CommentCount="1" />
  <row Id="1112" PostTypeId="2" ParentId="941" CreationDate="2010-07-29T05:43:06.793" Score="1" ViewCount="0" Body="&lt;p&gt;While it is certainly true that with the proper definition there is now 'wrong' notation, perhaps it should be mentioned that some notation is more suggestive and/or easier to work with than others, e.g. Arabic numeral vs. Roman numerals, the various symbols for the derivative, and countless others. The actual symbols are arbitrary, but good notation can certainly promote the flow of ideas more easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, do I remember correctly that Feynman gave up trying to invent more efficient notation for simple math when he was quite young because nobody could understand what he was doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good notation has a subtlety and suggestiveness which at times make it almost seem like a live teacher.&#xA;    --Bertrand Russell&lt;/p&gt;&#xA;" OwnerUserId="416" LastActivityDate="2010-07-29T05:43:06.793" CommentCount="1" />
  <row Id="1113" PostTypeId="1" AcceptedAnswerId="1387" CreationDate="2010-07-29T06:34:12.370" Score="2" ViewCount="157" Body="&lt;p&gt;Is it true that there are infinitely many nonprime integers $n$ such that $3^{n-1} - 2^{n-1}$ is a multiple of $n$?&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-02T12:18:54.860" Title="Nonprimes with $3^{n-1} \equiv 2^{n-1} \pmod n$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1114" PostTypeId="2" ParentId="1108" CreationDate="2010-07-29T07:03:02.577" Score="2" ViewCount="0" Body="&lt;p&gt;I couldn't divine much information on your background (e.g. undergraduate, master's level, PhD student...) from the question, but I recently taught an intermediate level graduate course which had a unit on valuation theory.  Sections 1.6 through 1.8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/8410Chapter1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;address your questions.  In particular, if your field $K$ is the fraction field of a Dedekind domain $R$, then you can always use each prime ideal $\mathfrak{p}$ of $R$ to define a valuation on $K$, essentially the &quot;order at $\mathfrak{p}$&quot;.  There is also a converse result, Theorem 13: if you have a valuation on $K$ which has the additional property that it is non-negative at every element of the Dedekind domain $R$, then it has to be (up to equivalence) the $\mathfrak{p}$-adic valuation for some $\mathfrak{p}$.   I felt the need to give this additional condition a name, so I called such a valuation &lt;strong&gt;R-regular&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that (as Qiaochu says in his comments), in case $K$ is a number field and $R$ is its ring of integers, every valuation on $K$ is $R$-regular.  However, in the function field setting this is not true and this leads to a discussion of &quot;infinite places&quot;.  Note that I do describe the analogues of Ostrowski's Theorem for finite extensions both of $\mathbb{Q}$ and of $F(t)$ for any field $F$ (in the latter case, one restricts to valuations which are trivial on $F$; when $F$ is finite, this condition is automatic).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested to know whether you find the notes helpful.  If not, I or someone else can probably recommend an alternative reference.    &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-07-29T07:03:02.577" />
  <row Id="1115" PostTypeId="2" ParentId="1108" CreationDate="2010-07-29T07:06:28.077" Score="1" ViewCount="0" Body="&lt;p&gt;Discrete valuations &amp;lt;-&gt; points on a curve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a nonsingular projective curve over an algebraically closed field, there is a one-one correspondence between the points on it, and the discrete valuations of the function field (i.e. all the meromorphic functions of the curve). The correspondence is point P -&gt; the valuation that sends a function f, to the order of zero/pole of f at P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximal ideals &amp;lt;-&gt; points on a curve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for varieties (common zeros of several polynomials) over an algebraically closed field, there is a one-one correspondence between points on it, and the maximal ideal in $k[x_1,\cdots,x_n]$. The correspondence is point $P = (a_1,\cdots,a_n)$ -&gt; the polynomials vanishing at P, which turns out to be $(x_1-a_1,\cdots,x_n-a_n)$. This is something true not only for curves, but for varieties. (Hilbert's Nullstellensatz) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting these together, for nonsingular projective curves over an algebraically closed field, you know that there is a one-one correspondence between the maximal ideals (think them as points) and the discrete valuations of the function field. Now the situation here is analogous. You consider a &quot;curve&quot;, whose coordinate ring is $\mathbb{Z}$, with function field $\mathbb{Q}$. The nonarchimedean valuations correspond to discrete valuations in this case. So they should capture order of zeros/poles at some &quot;points&quot;. What are the points? They should correspond to the maximal ideals of $\mathbb{Z}$, which are exactly the primes here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $K(x)$, look at it as the function field of $K\mathbb{P}^1$. Just like the usual real/complex projective spaces, you should have two pieces here. Let's say $K[x]$ corresponds to the piece where the second coordinate is nonzero. So the corresponding homogeneous coordinates here is like $[x,1]$. We know there is one point missing, which is $[1,0]$. For this, we change our coordinates $[x,1] \to [1,1/x]$, so the piece where the first coordinate is nonzero should be $K[1/x]$. The missing point corresponds to the ideal $(1/x - 0) = (1/x)$, so this is why the infinite place corresponds to (1/x). Of course, a more straight forward interpretation is that for a rational function, you divide both numerator and denominator sufficiently high power of $x$ so that they both become polynomials in 1/x, have nonzero constant term, with an extra term (x to the some power). The infinite place measures this power.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2010-07-29T20:22:39.630" LastActivityDate="2010-07-29T20:22:39.630" />
  <row Id="1116" PostTypeId="1" AcceptedAnswerId="8447" CreationDate="2010-07-29T07:07:04.470" Score="9" ViewCount="458" Body="&lt;p&gt;How many ways can a rectangle be partitioned by either vertical or horizontal lines into &lt;code&gt;n&lt;/code&gt; sub-rectangles? At first I thought it would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      f(n) = 4f(n-1) - 2f(n-2)  &#xA;where f(0) = 1  &#xA;  and f(1) = 1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but the recurrence relation only counts the cases in which at least one side (either top, bottom, left or right of the original rectangle) is not split into sub-rectangles. There are many other partitions that don't belong to those simple cases like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img228.imageshack.us/img228/6572/partitions.png&quot; alt=&quot;sample partitions&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other related problem suggestions are welcome. Also it is nice to know how to traverse this partition efficiently.&lt;/p&gt;&#xA;" OwnerUserId="333" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-29T07:17:13.603" LastActivityDate="2010-11-02T09:35:40.290" Title="Number of ways to partition a rectangle into n sub-rectangles" Tags="&lt;combinatorics&gt;&lt;computer-science&gt;&lt;recursive&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="5" />
  <row Id="1117" PostTypeId="1" AcceptedAnswerId="1119" CreationDate="2010-07-29T07:13:45.857" Score="5" ViewCount="124" Body="&lt;p&gt;How to find all natural X for that X^2 + (X+1)^2 is a perfect square?&lt;/p&gt;&#xA;" OwnerUserId="382" LastActivityDate="2010-07-29T07:19:22.397" Title="Find all X for that X^2 + (X+1)^2 is a square" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="1118" PostTypeId="1" AcceptedAnswerId="1122" CreationDate="2010-07-29T07:15:46.650" Score="9" ViewCount="192" Body="&lt;p&gt;I'd like to characterise the functions that 'have square roots' in the function composition sense. That is, can a given function f be written as f = g.g (where . is function composition)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the function f(x) = x+10 has a square root g(x) = x+5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the function f(x) = 9x has a square root g(x) = 3x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the function f(x) = x^2 + 1 has a square root, but I couldn't think of any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to determine which functions have square roots? To keep things simpler, I'd be happy just to consider functions f: R -&gt; R.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T11:06:06.090" LastActivityDate="2010-07-29T11:06:06.090" Title="Characterising functions f that can be written as f = g.g?" Tags="&lt;algebra&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="1119" PostTypeId="2" ParentId="1117" CreationDate="2010-07-29T07:19:22.397" Score="9" ViewCount="0" Body="&lt;p&gt;Suppose $x^2 + (x+1)^2 = y^2$. We can rewrite it as $(2x+1)^2 + 1 = 2y^2$ or $(2x+1)^2 - 2y^2 = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $z=2x+1$ then we have $z^2 - 2y^2 = -1$. This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Pell_equation&quot; rel=&quot;nofollow&quot;&gt;Pell's equation&lt;/a&gt;.&#xA;Wikipedia article shows how to solve it.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T07:19:22.397" CommentCount="1" />
  <row Id="1120" PostTypeId="1" AcceptedAnswerId="1124" CreationDate="2010-07-29T07:49:22.323" Score="3" ViewCount="141" Body="&lt;p&gt;Are there 2 subsets, say, $A$ and $B$, of the naturals such that&lt;br&gt;&#xA;$$\sum_{n\in A}   f(n)   =  \sum_{n\in B}   f(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(n)=1/n^2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(n)=1/n$ then there are many counterexamples, which is probably a consequence of the fact that the harmonic series diverges:&#xA;$$\frac23 = \frac12 + \frac16 = \frac14+\frac13+\frac1{12}$$&lt;br&gt;&#xA;And if $f(n)=b^{-n}$ for some base b then it is true because for all $M$, $\sum_{n&gt;M} f(n) &amp;lt; f(M)$. (This is just the base-b representation of a real number.The case b=2 gives a &lt;s&gt;bijection&lt;/s&gt; surjection 2^{\N} \to [0,1]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have sort of an in-between case here.&lt;br&gt;&#xA;Also, what if $A$,$B$:&lt;br&gt;&#xA;-are required to be finite sets?&lt;br&gt;&#xA;-are required to be infinite and disjoint?&lt;/p&gt;&#xA;" OwnerUserId="414" LastEditorUserId="414" LastEditorDisplayName="" LastEditDate="2010-07-30T02:15:15.297" LastActivityDate="2010-07-30T02:15:15.297" Title="Summing 1/n^2 over subsets of N." Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1121" PostTypeId="2" ParentId="370" CreationDate="2010-07-29T08:02:00.330" Score="2" ViewCount="0" Body="&lt;p&gt;And when you get bored of reading, let the &lt;a href=&quot;http://www.youtube.com/TheCatsters#p/u/68/xqLgGB7Hv7g&quot; rel=&quot;nofollow&quot;&gt;Catsters&lt;/a&gt; take over. (78 videos on Category theory!)&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-07-29T08:02:00.330" CommunityOwnedDate="2010-07-29T08:02:00.330" />
  <row Id="1122" PostTypeId="2" ParentId="1118" CreationDate="2010-07-29T08:07:08.430" Score="8" ViewCount="0" Body="&lt;p&gt;I showed you the link to the MO question mostly to convince you that this is a hard question.  I will &quot;answer&quot; it in the special case that $f$ is a bijection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that given a bijection $f : S \to S$, where $S$ is a set, a cycle of $f$ length $n$ is a set of distinct points $x, f(x), ... f^{n-1}(x)$ such that $f^n(x) = x$.  A cycle of infinite length is a set of distinct points $x, f(x), f^2(x), ...$.  It is not hard to see that $S$ is a disjoint union of cycles of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;  A bijection $f : S \to S$ has a square root if and only if there are an even number of cycles of $f$ of any given even length.  (For the purposes of this result, infinity is an even number; so there can be an infinite number of cycles, and you need to consider cycles of infinite length.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  First we show that any bijection with a square root has this property.  Let $g : S \to S$ be a bijection such that $g(g(x)) = f(x)$.  Then each cycle of $g$ corresponds to either one or two cycles of $f$, as follows.  If the cycle has odd length, it corresponds to one cycle of $f$.  For example, the cycle $1 \to 2 \to 3 \to 1$ of $g$ would correspond to the cycle $1 \to 3 \to 2 \to 1$ of $f$.  If the cycle has even length, it corresponds to two cycles of $f$.  For example, the cycle $1 \to 2 \to 1$ of $g$ would correspond to the pair of cycles $1 \to 1$ and $2 \to 2$, and the cycle $1 \to 2 \to 3 \to ... $ would correspond to the pair of cycles $1 \to 3 \to ... $ and $2 \to 4 \to ...$.  In particular, cycles of $f$ of odd length can come from cycles of $g$ one at a time or two at a time, but cycles of $f$ of even length can only come from cycles of $g$ two at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we show the reverse implication.  Given a cycle of $f$ of odd length $2k+1$, consider the corresponding cycle of $f^{k+1}$ of odd length.  Since $f^{2k+2} = f$ when restricted to this cycle, make this a cycle of $g$.  Given a pair of cycles of $f$ of the same even length, just weave them together to get a cycle of $g$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say &quot;answer&quot; instead of answer because it's not obvious if you can always find the cycle decomposition of some complicated bijection on an infinite set.  In any case, if $f$ isn't assumed to be a bijection this question becomes much harder; the analogue of cycle decomposition is much more difficult to work with.  I suggest you look at some examples where $S$ is finite if you really want to get a grip on this case; best of luck.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-29T08:07:08.430" CommentCount="1" />
  <row Id="1123" PostTypeId="2" ParentId="855" CreationDate="2010-07-29T08:07:09.613" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a very simplistic proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n^2 = 1 (mod 24) for n=1,5,7 and 11, by checking each case individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(n+12)^2 = n^2 + 24n + 144 = n^2 (mod 24).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, n^2 = 1 (mod 24) when n is odd and not divisible by 3, and so n^2-1 is divisible by 24 for these n. You don't need primality of p here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slight modification would be to use 1 and 5 as &quot;base cases&quot;, and use the fact that (n+6)^2 = n^2 + 12n + 36 = n^2 + 12(n+3), which is equal to n^2 mod 24 when (n+3) is even, i.e. n is odd.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-07-29T08:07:09.613" />
  <row Id="1124" PostTypeId="2" ParentId="1120" CreationDate="2010-07-29T08:38:10.443" Score="5" ViewCount="0" Body="&lt;p&gt;Yes to both cases:&#xA;1) $\frac{1}{15^2}+\frac{1}{20^2}=\frac{1}{12^2}$&#xA;2) $\frac{1}{15^2}+\frac{1}{150^2}+\frac{1}{1500^2}+...+\frac{1}{20^2}+\frac{1}{200^2}+\frac{1}{2000^2}+...=\frac{1}{12^2}+\frac{1}{120^2}+\frac{1}{1200^2}...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for first case - if we have pythagorean triple (a,b,c), such that $a^2+b^2=c^2$, then:&#xA;$\frac{1}{a^2 b^2}=\frac{1}{a^2 c^2}+\frac{1}{b^2 c^2}$&lt;/p&gt;&#xA;" OwnerUserId="425" LastActivityDate="2010-07-29T08:38:10.443" />
  <row Id="1125" PostTypeId="1" AcceptedAnswerId="1161" CreationDate="2010-07-29T08:43:19.217" Score="1" ViewCount="84" Body="&lt;p&gt;I know volume preserving diffeomorphisms of a sphere^2 make a group sdiff(S2). I would to know if it is a Lie group, which I assume if it is that makes interpolation easier (like with rotations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that is one question, is it a Lie group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is the group path connected? If so, how can I interpolate between two elements in the group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not subjects I know very little about. I apologize if Im phrasing it in some way that sounds ridiculous.  &lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="99" LastEditorDisplayName="" LastEditDate="2010-07-29T09:15:23.347" LastActivityDate="2010-07-29T18:38:40.070" Title="Interpolating between volume preserving diffeomorphisms of sphere" Tags="&lt;geometry&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1126" PostTypeId="2" ParentId="1113" CreationDate="2010-07-29T08:43:34.643" Score="0" ViewCount="0" Body="&lt;p&gt;as Qiaochu Yuan pointed out, take a &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_number&quot; rel=&quot;nofollow&quot;&gt;Carmichael number&lt;/a&gt; &lt;em&gt;q&lt;/em&gt;; by definition, 3&lt;sup&gt;&lt;em&gt;q&lt;/em&gt;-1&lt;/sup&gt; and 2&lt;sup&gt;&lt;em&gt;q&lt;/em&gt;-1&lt;/sup&gt; are both congruent to 1 mod &lt;em&gt;q&lt;/em&gt;, so their difference is a multiple of &lt;em&gt;q&lt;/em&gt;. Since Carmichael numbers are infinite, you are done.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-29T08:43:34.643" CommentCount="1" />
  <row Id="1127" PostTypeId="1" AcceptedAnswerId="1128" CreationDate="2010-07-29T08:59:19.303" Score="6" ViewCount="212" Body="&lt;p&gt;I just came up with this problem yesterday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;:&#xA;Assume there is an important segment of straight line &lt;code&gt;AB&lt;/code&gt; that needs to be watched at all time. A watchdog can see in one direction in front of itself and must walk at a constant &lt;em&gt;non-zero&lt;/em&gt; speed at all time. (All watchdogs don't need to have the same speed.) When it reaches the end of the segment, it must turn back (at no time) and keep watching the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many watchdogs does it need to guarantee that the line segment is watched at all time? And how (initial positions and speeds of the dogs)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: &#xA;It's clear that two dogs are not enough. I conjecture that four will suffice and three will not. For example, the below configuration doesn't work from 7.5 second if &lt;code&gt;AB&lt;/code&gt;'s length is 10 meters.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Dog 1 at A               walks to the right with speed 1.0 m/s&#xA;Dog 2 at between A and B walks to the right with speed 1.0 m/s&#xA;Dog 3 at B               walks to the left  with speed 1.0 m/s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or it can be illustrated as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         A ---------------------------------------- B&#xA;0.0 sec    1 --&amp;gt;               2 --&amp;gt;          &amp;lt;-- 3&#xA;&#xA;2.5 sec              1 --&amp;gt;          &amp;lt;-- 32 --&amp;gt;&#xA;&#xA;5.0 sec                   &amp;lt;-- 31 --&amp;gt;          &amp;lt;-- 2&#xA;&#xA;7.5 sec         &amp;lt;-- 3               &amp;lt;-- 21 --&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Please provide your solutions, hints, or related problems especially in higher dimensions or looser conditions (watchdogs can walk with acceleration, etc.)&lt;/p&gt;&#xA;" OwnerUserId="333" LastEditorUserId="333" LastEditorDisplayName="" LastEditDate="2010-07-30T04:33:42.937" LastActivityDate="2010-07-30T04:33:42.937" Title="Watchdog Problem" Tags="&lt;discrete-mathematics&gt;&lt;recreational-mathematics&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="1128" PostTypeId="2" ParentId="1127" CreationDate="2010-07-29T09:29:50.983" Score="8" ViewCount="0" Body="&lt;p&gt;Three dogs is enought I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the length of line segment be equal to 1 (with coordinates from 0 to 1).&lt;br &gt;&#xA;First dog: start position = 0, speed = +1/3&lt;br /&gt;&#xA;Second dog: start position = 2/3, speed = +1/3&lt;br /&gt;&#xA;Third dog: start position = 2/3, speed = -1/3&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 1 second the position becomes similar.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T09:29:50.983" CommentCount="3" />
  <row Id="1130" PostTypeId="1" AcceptedAnswerId="1137" CreationDate="2010-07-29T09:53:03.543" Score="1" ViewCount="54" Body="&lt;p&gt;Suppose we have some function $f(x)$ with local extrema at $x_1, x_2, \dots$, and a second function $g(x)$ which is continuous, strictly increasing and non-zero everywhere over the range of the $x_i$. Will $g(f(x))$ have its local extrema at the same $x_i$ and no others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, are there any obvious loosenings of the constraints on $g$ for which this will remain true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm really thinking of this in the context of signal processing, looking at transformations that preserve the visual structure of an image, but it seems like a general question that must have been trivially proved by someone 250 years ago...)&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T11:00:37.177" LastActivityDate="2010-07-29T11:00:37.177" Title="Preserving the extrema of one function after applying another" Tags="&lt;calculus&gt;&lt;signal-processing&gt;" AnswerCount="2" />
  <row Id="1131" PostTypeId="2" ParentId="1130" CreationDate="2010-07-29T10:02:33.837" Score="3" ViewCount="0" Body="&lt;p&gt;Since g is continuous and strictly increasing, its inverse $g^{-1}$ is a function and strictly increasing.  Since both are strictly increasing, $a&amp;lt;b\Leftrightarrow g(a)&amp;lt;g(b)\Leftrightarrow g^{-1}(a)&amp;lt;g^{-1}(b)$.  From this, it follows that $x_i$ is a local max (min) of g(f(x)) iff it is a local max (min) of f(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If g were continuous and strictly decreasing, it would exchange local maximums and minimums (because $a&amp;lt;b\Leftrightarrow g(a)&amp;gt;g(b)$ and $a&amp;lt;b\Leftrightarrow g^{-1}(a)&amp;gt;g^{-1}(b)$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-29T10:16:47.663" LastActivityDate="2010-07-29T10:16:47.663" />
  <row Id="1132" PostTypeId="1" AcceptedAnswerId="13930" CreationDate="2010-07-29T10:12:29.380" Score="6" ViewCount="138" Body="&lt;p&gt;How can I show with a heuristic argument based on a Taylor expansion that for Stratonovich&#xA;stochastic calculus the chain rule takes the form of the classical (Newtonian) one? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning Ito calculus the fact that dX^2 = dt results via a Taylor expansion in Ito's lemma - this fact should stay the same with Stratonovich but it should somehow cancel out in there - I just don't know how...&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T11:03:40.633" LastActivityDate="2010-12-11T17:47:37.827" Title="Taylor expansion to show that for Stratonovich stochastic calculus the chain rule takes the form of the classical one" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1133" PostTypeId="1" AcceptedAnswerId="1168" CreationDate="2010-07-29T10:17:31.483" Score="9" ViewCount="344" Body="&lt;p&gt;Bourbaki shows in  a very natural way that every continuous group isomorphism of the additive reals to the positive multiplicative reals is determined by its value at $1$, and in fact, that every such isomorphism is of the form $f_a(x)=a^x$ for $a&gt;0$ and $a\neq 1$.  We get the standard real exponential (where $a=e$) when we notice that for any $f_a$, $(f_a)'=g(a)f_a$ where $g$ is a continuous group isomorphism from the positive multiplicative reals to the additive reals.  By the intermediate value theorem, there exists some positive real $e$ such that $g(e)=1$ (by our earlier classification of continuous group homomorphisms, we notice that $g$ is in fact the natural log).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that every deduction above follows from a natural question.  We never need to guess anything to proceed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any natural way like the above to derive the complex exponential?  The only way I've seen it derived is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derive the real exponential by some method (inverse function to the natural log, which is the integral of $1/t$ on the interval $[1,x)$, Bourbaki's method, or some other derivation), then show that it is analytic with infinite radius of convergence (where it converges uniformly and absolutely), which means that it is equal to its Taylor series at 0, which means that we can, by a general result of complex analysis, extend it to an entire function on the complex plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This derivation doesn't seem natural to me in the same sense as Bourbaki's derivation of the real exponential, since it requires that we notice some analytic properties of the function, instead of relying on its unique algebraic and topological properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a derivation similar to Bourbaki's for the complex exponential?  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-07-29T17:02:00.233" Title="Natural derivation of the complex exponential function?" Tags="&lt;complex-analysis&gt;&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="2" />
  <row Id="1134" PostTypeId="2" ParentId="1125" CreationDate="2010-07-29T10:36:24.243" Score="0" ViewCount="0" Body="&lt;p&gt;The term &quot;volume preserving&quot; sounds a bit ambiguous to me: do you mean that your map preserves the total volume or do you mean that its differential at every point preserves volume (i.e. has determinant 1)? The former is weaker than the latter, and gives you more room for interpolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there is a famous invariant of continuous maps $S^2\to S^2$ called the &lt;em&gt;degree&lt;/em&gt;. Any two maps with the same degree are homotopic to each other. Being volume preserving (in the former sense) implies that the degree is $1$ (taking orientation into account!), so you can interpolate between any two volume preserving maps. &lt;em&gt;However&lt;/em&gt;, the intermediate maps in this line of reasoning are only continuous, not necessarily diffeomorphisms. I'm confident that with a standard argument &quot;approximate continuous functions by differentiable ones&quot; you can get them to be differentiable, but I don't about &quot;is diffeomorphism&quot; and &quot;is locally volume preserving&quot; parts.&lt;/p&gt;&#xA;" OwnerUserId="430" LastActivityDate="2010-07-29T10:36:24.243" CommentCount="1" />
  <row Id="1135" PostTypeId="1" AcceptedAnswerId="1136" CreationDate="2010-07-29T10:47:23.870" Score="5" ViewCount="269" Body="&lt;p&gt;I can never figure out (because the English language is imprecise) which part of &quot;if and only if&quot; means which implication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A if and only if B) = (A &amp;lt;=&gt; B), but is the following correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A only if B) = (A =&gt; B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A if B) = (A &amp;lt;= B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, one never comes into contact with &quot;A if B&quot; or &quot;A only if B&quot; using those constructions in everyday common speech.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-07-29T10:52:37.360" LastActivityDate="2010-07-29T23:10:35.437" Title="If and only if, which direction is which?" Tags="&lt;terminology&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="1136" PostTypeId="2" ParentId="1135" CreationDate="2010-07-29T10:52:33.460" Score="6" ViewCount="0" Body="&lt;p&gt;I wouldn't say that I &lt;em&gt;never&lt;/em&gt; come into contact with those phrasings--they are certainly rare in technical use, but perhaps more common in plain language.  Below is a table of equivalent phrasings of p=&gt;q, from UCSMP &lt;em&gt;Precalculus and Discrete Mathematics&lt;/em&gt;, 3rd ed., &amp;copy; 2010 Wright Group/McGraw Hill (Lesson 1-5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/171340741.png&quot; alt=&quot;table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-29T10:52:33.460" CommentCount="9" />
  <row Id="1137" PostTypeId="2" ParentId="1130" CreationDate="2010-07-29T10:52:44.257" Score="4" ViewCount="0" Body="&lt;p&gt;You don't need to assume that $g$ is non-zero, and it could be strictly decreasing as well. Furthermore, the conditions on $g$ only need to hold on the image of $f$ (which doesn't need to the be whole of $\mathbb{R}$, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $g$ has a local extremum at $y=f(z)$ and $f$ is strictly increasing around $z$, then you're obviously in trouble, because $g\circ f$ will have a local extremum at $z$. But having no local extrema is equivalent to being strictly monotonous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only that might relax the conition on $g$ is that it has local extrema exactly where $f$ does and they &quot;cancel each other out&quot; or &quot;amplify each other&quot;.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-07-29T10:52:44.257" />
  <row Id="1138" PostTypeId="2" ParentId="1135" CreationDate="2010-07-29T10:53:15.173" Score="6" ViewCount="0" Body="&lt;p&gt;This example may be more clear, because apples ⊂ fruits is more obvious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This is an apple if it is a fruit&quot; is false.&lt;br&gt;&#xA;&quot;This is an apple only if it is a fruit&quot; is true.&lt;br&gt;&#xA;&quot;This is a fruit if it is an apple&quot; is true.&lt;br&gt;&#xA;&quot;This is a fruit only if it is an apple&quot; is false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is an apple =&gt; A is a fruit&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:10:35.437" LastActivityDate="2010-07-29T23:10:35.437" />
  <row Id="1139" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T11:11:47.873" Score="2" ViewCount="0" Body="&lt;p&gt;The function f(x) = x over the rationals and 2x over the irrationals is locally increasing in 0 but it is neither increasing nor decreasing.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-29T11:11:47.873" CommunityOwnedDate="2010-07-29T11:11:47.873" CommentCount="1" />
  <row Id="1140" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T11:16:22.330" Score="4" ViewCount="0" Body="&lt;p&gt;Also &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway_base_13_function&quot; rel=&quot;nofollow&quot;&gt;Conway base 13 function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function has the following properties:&lt;br&gt;&#xA;1. On every closed interval $[a, b]$ take every real value.&lt;br&gt;&#xA;2. Is continuous nowhere.  &lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T11:16:22.330" CommunityOwnedDate="2010-07-29T11:16:22.330" />
  <row Id="1143" PostTypeId="1" CreationDate="2010-07-29T11:58:10.617" Score="4" ViewCount="175" Body="&lt;p&gt;Can you give a definition of the Conway base-13 function better than the one actually present on wikipedia (&lt;a href=&quot;http://www.google.it/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBgQFjAA&amp;amp;url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FConway_base_13_function&amp;amp;ei=empRTNK-OsqWON2_uPUE&amp;amp;usg=AFQjCNFqIS-UhBV9Miw1QnAZJaxnswI3Yg&amp;amp;sig2=AlFPiLqMkh3gd4VPEHJI-Q&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), which isn't clear? Maybe with some examples?&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-29T20:58:09.280" Title="A definition of Conway base-13 function" Tags="&lt;calculus&gt;&lt;number-theory&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="1144" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T12:28:46.330" Score="3" ViewCount="0" Body="&lt;p&gt;Have also a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-07-29T12:28:46.330" CommunityOwnedDate="2010-07-29T12:28:46.330" />
  <row Id="1145" PostTypeId="2" ParentId="514" CreationDate="2010-07-29T12:44:40.587" Score="3" ViewCount="0" Body="&lt;p&gt;Further counterexamples can be found here: &lt;a href=&quot;http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="273" LastActivityDate="2010-07-29T12:44:40.587" />
  <row Id="1146" PostTypeId="1" AcceptedAnswerId="1188" CreationDate="2010-07-29T13:12:32.010" Score="4" ViewCount="193" Body="&lt;p&gt;I know that this is meant to explain variance butthe description on Wikpiedia stinks and it is not clear how you can explain variance using this technique&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain it in a simple way?&lt;/p&gt;&#xA;" OwnerUserId="437" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:44:52.727" LastActivityDate="2010-07-29T23:44:52.727" Title="Intuitive Way To Understand Principal Component Analysis" Tags="&lt;terminology&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="1147" PostTypeId="1" AcceptedAnswerId="1148" CreationDate="2010-07-29T13:21:02.440" Score="3" ViewCount="128" Body="&lt;p&gt;How to calculate volume of tetrahedron given lengths of all it's edges?&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T13:24:25.193" Title="Tetrahedron volume" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1148" PostTypeId="2" ParentId="1147" CreationDate="2010-07-29T13:24:25.193" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Cayley-MengerDeterminant.html&quot; rel=&quot;nofollow&quot;&gt;Cayley-Menger Determinant&lt;/a&gt; - A generalization of Herons Formula.&lt;/p&gt;&#xA;" LastActivityDate="2010-07-29T13:24:25.193" />
  <row Id="1149" PostTypeId="2" ParentId="1143" CreationDate="2010-07-29T13:36:26.367" Score="4" ViewCount="0" Body="&lt;p&gt;You just need to switch back and forth from the lexicographic meaning of the base-13 expansion of the number (think of having ABC instead of .-+) and the loaded meaning you give to the well-formed string as a base-10 number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a number for which Conway base-13 function is 0 is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.12-34++1+2-34..11111111111...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the leftmost . is the threedecimal point (that is, it has a semantic meaning), the three rightmost . mean that the base-13 representation has an infinite number of 1 (that is, they have a metameaning), and the other two . are &quot;digits&quot; of the number (that is, they have a syntactic meaning).&lt;br&gt;&#xA;An example of a number for which Conway base-13 function is not 0 is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.12-34++1+2-34.11111111111...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that number, the function has value -34.11111111111... (in base 10)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-29T13:36:26.367" />
  <row Id="1150" PostTypeId="2" ParentId="1064" CreationDate="2010-07-29T13:39:56.827" Score="7" ViewCount="0" Body="&lt;p&gt;Consider the set of reals x whose binary expansion, if you look only at the even digit places, is some fixed non-eventually-repeating pattern z. This is perfect, since we have branching at the odd digits, but they are all irrational, since z is not eventually repeating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can draw a picture of this set, and it looks something like the Cantor middle third set, except that you divide into four pieces, and take either first+third or second+fourth, depending on the digits of z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution: Begin with an interval having irrational endpoints, and perform the usual Cantor middle-third construction, except that at stage n, be sure to exclude the n-th rational number (with respect to some fixed enumeration), using a subinterval having irrational endpoints. By systematically excluding all rational numbers, you have the desired perfect set of irrationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hi François!)&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-07-29T21:09:19.270" LastActivityDate="2010-07-29T21:09:19.270" CommentCount="1" />
  <row Id="1151" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T13:44:40.870" Score="4" ViewCount="0" Body="&lt;p&gt;So, what's unnatural about the complex differential equation...&#xA;f : C -&gt; C satisfying f'(z) = f(z) and f(0)=1 ?&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-07-29T13:44:40.870" CommentCount="1" />
  <row Id="1152" PostTypeId="2" ParentId="381" CreationDate="2010-07-29T13:55:45.977" Score="6" ViewCount="0" Body="&lt;p&gt;Consecutive Fibonacci numbers are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm#Algorithmic%20efficiency&quot; rel=&quot;nofollow&quot;&gt;worst-case&lt;/a&gt; (maximum number of steps) numbers for Euclid's gcd algorithm.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-07-29T13:55:45.977" CommunityOwnedDate="2010-08-03T18:46:03.740" />
  <row Id="1153" PostTypeId="2" ParentId="954" CreationDate="2010-07-29T14:22:08.317" Score="1" ViewCount="0" Body="&lt;p&gt;The Laplace transform can be simply interpreted as a Wick rotated Fourier transform of a function $f(t)$ which vanishes for $t&amp;lt;0$. Wick rotation means (in this case) changing the frequency $\omega$ of the Fourier transform into an imaginary parameter $s=-i\omega$ of the Laplace transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the inversion&#xA;    $\mathcal{L}^{-1} {F(s)} = f(t) = \frac{1}{2\pi i}\lim_{T\to\infty}\int\limits_{\gamma-iT}^{\gamma+iT}e^{st}F(s)\,ds, \qquad s=\Re(\gamma)$&#xA;to be a bit more complicated than the inverse Fourier transform (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse_Laplace_transform&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, to quote @Akhil) is this imaginary frequency which, if left purely imaginary, would lead to a non-convergent integral. The formula is still similar to the inverse Fourier transform. In fact I think (but have not verified) you could use it for that purpose to.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-29T14:22:08.317" />
  <row Id="1154" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T14:26:04.720" Score="-1" ViewCount="0" Body="&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;f(x) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;df/dx = -sin(x) + i*cos(x)&#xA;      = i*f(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;∫(1/f(x)) df = ∫i dx&#xA;ln(f(x)) = ix  + C&#xA;f(x) = e^(ix + C) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since &lt;code&gt;f(0) = 1&lt;/code&gt;, &lt;code&gt;C = 0&lt;/code&gt;, so&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;e^(ix) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I will update this with LaTeX when that functionality becomes available)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-29T14:26:04.720" CommentCount="7" />
  <row Id="1155" PostTypeId="2" ParentId="1143" CreationDate="2010-07-29T14:26:35.530" Score="1" ViewCount="0" Body="&lt;p&gt;The idea of the Conway base-13 number is to find a function that is not continuous, yet if &lt;code&gt;f(a)&amp;lt;x&amp;lt;f(b)&lt;/code&gt;, then there is some &lt;code&gt;c&lt;/code&gt; between &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; with &lt;code&gt;f(c)=x&lt;/code&gt;. This a counterexample to the converse of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate_value_theorem&quot; rel=&quot;nofollow&quot;&gt;intermediate value theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is defined by encoding base-10 values in the tail (the digits left after skipping a finite number). We use &lt;code&gt;+&lt;/code&gt;, &lt;code&gt;-&lt;/code&gt;, &lt;code&gt;.&lt;/code&gt; and the digits to represent an encoded number in the tail and require the encoded number to start with a &lt;code&gt;+&lt;/code&gt; or &lt;code&gt;-&lt;/code&gt;. In base 10, every number ending in an infinite number of &lt;code&gt;9&lt;/code&gt;s can be rewritten to end in an infinite number of &lt;code&gt;0&lt;/code&gt;s instead (ie. 0.999...=1.0). Similarly, we decide we will rewrite each Conway number ending in an infinite number of &lt;code&gt;+&lt;/code&gt;, to ensure that each real number has a unique decimal representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each number can have up to one base-10 encoded value, which is the result of applying Conway's Base 13 function if it exists. If no such encoding exists for x(ie. &lt;code&gt;+&lt;/code&gt; occurring infinite times in the expansion), then we define f(x)=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then show that for each &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; that we can find a &lt;code&gt;c&lt;/code&gt; in between with an arbitrary encoded value. We first ensure the number being constructed is between &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt; by copying enough digits from &lt;code&gt;a&lt;/code&gt; and incrementing a digit that won't matter. This is easier because we have disallowed ending in an infinite number of &lt;code&gt;+&lt;/code&gt;. We then concatenate the digits of the signed base-10 representation of the value we wish the function to have to those digits  we have already fixed.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-29T20:58:09.280" LastActivityDate="2010-07-29T20:58:09.280" CommentCount="2" />
  <row Id="1156" PostTypeId="2" ParentId="1143" CreationDate="2010-07-29T14:29:47.020" Score="3" ViewCount="0" Body="&lt;p&gt;I understand why the Wikipedia article uses the notation it does, but I find it annoying. Here is a transliteration, with some elaboration.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Expand &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;(0,1) in base 13, using digits {0, 1, ...&amp;nbsp;, 9, &lt;em&gt;d&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, &lt;em&gt;p&lt;/em&gt;} --- using the convention &lt;em&gt;d&lt;/em&gt;&amp;nbsp;=&amp;nbsp;10, m&amp;nbsp;=&amp;nbsp;11, &lt;em&gt;p&lt;/em&gt;&amp;nbsp;=&amp;nbsp;12. &lt;strong&gt;N.B.&lt;/strong&gt; for rational numbers whose most reduced expression &lt;em&gt;a/b&lt;/em&gt; is such that &lt;em&gt;b&lt;/em&gt; is a power of 13, there are two such expansions: a terminating expansion, and a non-terminating one ending in repeated &lt;em&gt;p&lt;/em&gt; digits. In such a case, use the terminating expansion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let &lt;em&gt;S&lt;/em&gt;&amp;nbsp;&amp;sub;&amp;nbsp;(0,1) be the set of reals whose expansion involves finitely many &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, and &lt;em&gt;d&lt;/em&gt; digits, such that the final &lt;em&gt;d&lt;/em&gt; digit occurs after the final &lt;em&gt;p&lt;/em&gt; digit and the final &lt;em&gt;m&lt;/em&gt; digit. (We may require that there be at least one digit 0--9 between the final &lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt; digit and the final &lt;em&gt;d&lt;/em&gt; digit, but this does not seem to be necessary.) Then, every &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt; has a base 13 expansion of the form&lt;br&gt;&lt;br&gt;&#xA;0.&lt;em&gt;x&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;x&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;x&lt;/em&gt;&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;[&amp;nbsp;&lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt;&amp;nbsp;]&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;[&amp;nbsp;&lt;em&gt;d&lt;/em&gt;&amp;nbsp;]&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&lt;br&gt;&lt;br&gt;&#xA;for some digits &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;j&lt;/sub&gt;&amp;nbsp;&amp;isin;&amp;nbsp;{0, ...&amp;nbsp;, &lt;em&gt;p&lt;/em&gt;} and where the digits &lt;em&gt;a&lt;/em&gt;&lt;sub&gt;j&lt;/sub&gt; and &lt;em&gt;b&lt;/em&gt;&lt;sub&gt;j&lt;/sub&gt; are limited to {0, ...&amp;nbsp;, 9} for all j. The square brackets above are only intended for emphasis; and in particular the n+1&lt;sup&gt;st&lt;/sup&gt; base-13 digit of &lt;em&gt;x&lt;/em&gt; is the final occurance of either &lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt; in the expansion of &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;, we define f(&lt;em&gt;x&lt;/em&gt;) by transliterating the string format above. We ignore the digits &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; through &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;, transliterate the &lt;em&gt;p&lt;/em&gt; or &lt;em&gt;m&lt;/em&gt; as a plus-sign or minus-sign, and the &lt;em&gt;d&lt;/em&gt; as a decimal point. This yields a decimal expansion for a real number, either&lt;br&gt;&lt;br&gt; +&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;.&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&lt;br&gt;&lt;br&gt; or&lt;br&gt;&lt;br&gt; &amp;minus;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;.&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;...&lt;br&gt;&lt;br&gt;according to whether the n+1&lt;sup&gt;st&lt;/sup&gt; base-13 digit of &lt;em&gt;x&lt;/em&gt; is a &lt;em&gt;p&lt;/em&gt; or an &lt;em&gt;m&lt;/em&gt; respectively. For &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;, we set f(&lt;em&gt;x&lt;/em&gt;) to this number; for &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;notin;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;, we set f(&lt;em&gt;x&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: this function is not computable, as there is no way that you can determine in advance whether the base-13 expansion of &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;(0,1) has only finitely many occurances of any of the digits &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, or &lt;em&gt;d&lt;/em&gt;; even if you are provided with a number which is promised to have only finitely many, in general you cannot know when you have found the last one. However, if you are provided with a number &lt;em&gt;x&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;(0,1) for which you know the location of the final &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt;, and &lt;em&gt;d&lt;/em&gt; digits, you can compute f(&lt;em&gt;x&lt;/em&gt;) very straightforwardly.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-07-29T14:29:47.020" CommentCount="2" />
  <row Id="1157" PostTypeId="2" ParentId="1057" CreationDate="2010-07-29T14:48:37.347" Score="6" ViewCount="0" Body="&lt;p&gt;You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field &#xA;$\mathbb {Z}/3\mathbb{Z}$ has exponent three, so your equation holds, but it is not abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of examples: the most famous ones are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_problem#Bounded_Burnside_problem&quot; rel=&quot;nofollow&quot;&gt;Burnside groups&lt;/a&gt; $B(m,3)$: the group I described above is $B(2,3)$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-07-29T20:54:40.320" LastActivityDate="2010-07-29T20:54:40.320" CommentCount="5" />
  <row Id="1158" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T15:05:15.577" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Some Assumptions&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that you are ok with power series being used, just not Taylor's theorem. I will also assume you will allow us to observe a solution to a DE since you used it in your derivation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Defn&lt;/strong&gt; A series of the form $\sum_{n=0}^{\infty}c_n\left(z-z_0\right)^n$ for $c_n,z,z_0\in\mathbb{C}$ is called a &lt;em&gt;power series&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thm&lt;/strong&gt; There is some $R\in[0,\infty]$ such that the power series above converges absolutely for all $z\in\mathbb{C}$ with $\mid z-z_0\mid &amp;lt; R$ and uniformly in $D\left(z_0,\rho\right)$ for all $\rho &amp;lt; R$. Further, the terms are unbounded for all $z$ with $\mid z-z_0\mid &gt; R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;pf&lt;/strong&gt; Use the geometric series' convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; Inside the disk of convergence $\sum_{n=1}^{\infty}nc_n\left(z-z_0\right)^{n-1}$ is the derivative of the power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Construction&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For power functions $y(z)=\sum c_nz^n$ can we find a unique solution to $y'(z)=y(z)$ in $\mathbb{C}$? We can observe that this implies $nc_n=c_{n-1}$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Defn&lt;/strong&gt; Let $E\left(z\right):=\sum_{n=0}^{\infty}\frac{1}{n!}z^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thm&lt;/strong&gt; 1) $E'=E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $E\left(z_1+z_2\right)=E\left(z_1\right)E\left(z_2\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)$E_{\mid_{\mathbb{R}}}$ is strictly increasing and $E(\mathbb{R})=(0,\infty)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) $x\mapsto E(ix)$ sends $\mathbb{R}$ onto $\mathbb{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) There is some real $\pi&gt;0$ such that $E\left(\frac{\pi}{2}i\right)=i$ and for all $z_1, z_2\in\mathbb{C}$, then $E\left(z_1\right)=E\left(z_2\right)$ iff $\frac{z_1-z_2}{2\pi i}\in\mathbb{Z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) $E\left(\mathbb{C}\right)=\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all of these can be shown purely at the level of power series using no complex analysis. Further the proofs are not hard, if you want more details here let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary&lt;/strong&gt; $E$ is a homomorphism of the additive group $\mathbb{C}$ onto the multiplicative group with $\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Application&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you care about loops(which I think you do!) lets observe that $\gamma(t)=e^{it}$ for $t\in[0,2\pi]$, is $\mathbb{T}$ (Note here that $\pi$ is simply the real number we found before, not some existential thing!). We could now push this to get winding numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could use this definition of exponential, and the power series definition of inverse to get a (branchless) logarithm. In particular, you can show that the derivative of that fella is $\frac{1}{x}$. So we don't have to &lt;em&gt;define&lt;/em&gt; it that way. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Comments&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with you that the definitions of complex exponentials feel contrived and the logarithm is even worse. The branched logarithms is the only part of Palka that I dislike as a complex book. These definitions I can stomach, as they require no Dues ex Machina.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-07-29T15:12:39.570" LastActivityDate="2010-07-29T15:12:39.570" CommentCount="5" />
  <row Id="1159" PostTypeId="2" ParentId="107" CreationDate="2010-07-29T15:21:26.037" Score="3" ViewCount="0" Body="&lt;p&gt;The problem is that although the set of balls grows larger and larger, their &lt;em&gt;intersection&lt;/em&gt; does not. What is important is not the size of the set of balls at any particular time, but rather how the set of balls changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Different example.&lt;/strong&gt; Let's consider an alternative game, wherein at time t=0 you have the ball &quot;1&quot; in the box, and at each time t&gt;0 you add the balls {2&lt;sup&gt;t&lt;/sup&gt;, 2&lt;sup&gt;t&lt;/sup&gt;+1, 2&lt;sup&gt;t&lt;/sup&gt;+2, ...&amp;nbsp;, 2&lt;sup&gt;t+1&lt;/sup&gt;&amp;nbsp;&amp;minus;&amp;nbsp;1}, removing only the odd-numbered ball 2t&amp;minus;1. Let &lt;em&gt;S&lt;/em&gt;(t) denote the set of balls in the box after the t&lt;sup&gt;th&lt;/sup&gt; step. Then, you would have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;S&lt;/em&gt;(0) = {1}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(1) = {2,3}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(2) = {2,4,5,6,7}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(3) = {2,4,6,7,8,9,10,11,12,13,14,15}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(4) = {2,4,6,8,9,10,11,...,31}&lt;br&gt;&#xA;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sets are increasing in size at a tremendous rate, but in the end any odd numbered-ball will be removed. What matters is the set of balls which are at some point added, and then &lt;strong&gt;never&lt;/strong&gt; removed.* We define the set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I&lt;/em&gt;(n)&amp;nbsp;=&amp;nbsp;&lt;em&gt;S&lt;/em&gt;(n)&amp;nbsp;&amp;cap;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;(n+1)&amp;nbsp;&amp;cap;&amp;nbsp;...&amp;nbsp;&amp;cap;&amp;nbsp;&lt;em&gt;S&lt;/em&gt;(2n)&amp;nbsp;&amp;cap;&amp;nbsp;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is an intersection of all of the sets from step n onwards; it's just the set of balls which are never removed after step n. Then, what matters is how the set I(n) grows. In the example I give above, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I&lt;/em&gt;(0) = &amp;empty;&lt;br&gt;&#xA;&lt;em&gt;I&lt;/em&gt;(1) = {2}&lt;br&gt;&#xA;&lt;em&gt;I&lt;/em&gt;(2) = {2,4,6}&lt;br&gt;&#xA;&lt;em&gt;I&lt;/em&gt;(3) = {2,4,6,8,10,12,14}&lt;br&gt;&#xA;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we may see that in general, the size of &lt;em&gt;I&lt;/em&gt;(t) will be 2&lt;sup&gt;t&lt;/sup&gt;&amp;nbsp;&amp;minus;&amp;nbsp;1; then the limit of this process will be a box with infinitely many balls, &lt;em&gt;i.e.&lt;/em&gt; the even numbered balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your original question.&lt;/strong&gt; Let's consider how the intersections of the box-sets work in the game you present. At step n, you add 2&lt;sup&gt;n&lt;/sup&gt; balls, and remove the n lowest-numbered balls: this yeilds a box-set function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;S&lt;/em&gt;(0) = {1}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(1) = {2,3}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(2) = {4,5,6,7}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(3) = {7,8,9,10,11,12,13,14,15}&lt;br&gt;&#xA;&lt;em&gt;S&lt;/em&gt;(4) = {11,12,...,31}&lt;br&gt;&#xA;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because each ball is eventually removed, never to be added again; this means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I&lt;/em&gt;(0) = &lt;em&gt;I&lt;/em&gt;(1) = &lt;em&gt;I&lt;/em&gt;(2) = ... = &amp;empty;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which amounts to saying that no particular ball is left in the box; and so finally there are zero balls left in the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Footnotes:&#xA;&lt;strong&gt;1.&lt;/strong&gt; More generally, we can talk about games in which balls are added and removed several times, but in which some balls are eventually re-added and then left there for the rest of the game; we might say that these balls are &quot;eventually never&quot; removed.   &lt;strong&gt;2.&lt;/strong&gt; This process of taking limits only makes sense if every ball is added and removed finitely many times. For example, in a game where all you do is repeatedly add the ball 1 and remove the ball 1, the process does not have any sort of meaningful limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Edit&lt;/strong&gt;: corrected the discussion of the original ball-adding-removing scheme to correctly correspond to the original problem.]&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-14T15:09:32.170" LastActivityDate="2010-08-14T15:09:32.170" CommentCount="4" />
  <row Id="1160" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T15:30:02.437" Score="5" ViewCount="0" Body="&lt;p&gt;Here's an example of a strictly increasing function on &amp;#8477; which is continuous exactly at the irrationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick your favorite absolutely convergent series &amp;sum;a&lt;sub&gt;n&lt;/sub&gt; in which all the terms are positive (mine is &amp;sum;1/2&lt;sup&gt;n&lt;/sup&gt;) and your favorite enumeration of the rationals: &amp;#8474;={q&lt;sub&gt;1&lt;/sub&gt;,q&lt;sub&gt;2&lt;/sub&gt;,...}. For a real number x, define f(x) to be the sum of all the a&lt;sub&gt;n&lt;/sub&gt; for which q&lt;sub&gt;n&lt;/sub&gt; &amp;le; x.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2010-07-29T15:30:02.437" CommunityOwnedDate="2010-07-29T15:30:02.437" />
  <row Id="1161" PostTypeId="2" ParentId="1125" CreationDate="2010-07-29T15:43:50.367" Score="3" ViewCount="0" Body="&lt;p&gt;First, as others have pointed out, the group of volume preserving diffeomorphisms will be infinite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, there is a beautiful technique known as Moser's trick which answers it.  Moser's trick, in fancy language, says that if (M,w) and (M,w') are two symplectic structures on the same manifold, and if [w] = [w'] in H^2(M;R) (de Rham cohomology), then there is a family of diffeomorphism f_t:M-&gt;M with f_0 = Id and such that f_1 pulls w' back to w.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 2-dimensional compact, oriented manifold (like the sphere), we have H^2(M;R) = R (the real numbers), and a symplectic form is nothing but a nonzero element in R (which can be interpreted as the total volume).  Since in this setting, [w] = [w'] iff they both give the same signed volume, it follows from Moser's trick that the group of (signed) volume preserving maps is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider unsigned volume, there will be 2 components to the group diffeomorphisms preserving the unsigned volume.  This is because, as others have pointed out, one has the notion of &quot;degree&quot; which shows the map x-&gt; -x is not homotopic to Id, even through just continuous maps (not neccesarily volume preserving).  This shows there are AT LEAST two componenets.  There are at most two components because every volume preserving diffeo can be connected to Id or (x -&gt; -x) by Moser's trick again.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:  I misspoke a little bit.  Moser's trick says that if you have a family w_t of symplectic forms, then there is a family of diffeomorphisms as I described above.  It's not clear to me that what I said (that it's enough to have [w] = [w'] in H^2) is enough to guarantee that there is a family of symplectic forms connecting them.  Further, it seems that Moser's trick only guarantees you have a path of diffeos which starts and ends at a volume preserving diffeo, but may not preserve volume for all time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the case of S^2 (or any closed, orientable 2-manifold), I can patch things up. Given w and w', volume forms, with [w] = [w'] (i.e, they have the same volume), then the form w_t = tw + (1-t)w' is a path of symplectic forms which connects them.  For a fixed t, the form w_t is closed since it's a sum of closed forms (or, even easier, because it has top degree), and is nondegenerate because it's a volume form (integration shows the volume given is that of w).  The fact that the volume is constant for each w_t implies that the path of diffeos preserves volume for all time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I wasn't able to immediately convince myself that in general, the convex sum of symplectic forms was nondegenerate, hence my initial hesitation.  In fact, I think that it need not be nondegenerate.)&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-07-29T18:38:40.070" LastActivityDate="2010-07-29T18:38:40.070" CommentCount="1" />
  <row Id="1162" PostTypeId="2" ParentId="107" CreationDate="2010-07-29T16:03:19.423" Score="-1" ViewCount="0" Body="&lt;p&gt;At every &quot;tick&quot; you have one more ball. So at tick 23, you have 23 balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that at 11 you have N balls in the basket, where N is any positive integer or zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is impossible because you would have had N+1 balls at tick N+1, which would have happened before 11, and the number of balls only increases in time, so surely you must have more than N balls in the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, then, the only possible answer is infinite balls, because an infinite number of ticks would have occurred at 11.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-07-29T16:03:19.423" CommentCount="1" />
  <row Id="1163" PostTypeId="2" ParentId="128" CreationDate="2010-07-29T16:10:13.567" Score="5" ViewCount="0" Body="&lt;p&gt;There is a fairly good exposition in &lt;a href=&quot;http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/&quot; rel=&quot;nofollow&quot;&gt;Terry Tao's post&lt;/a&gt;, see Corollaries 4-6. Here is a sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We prove the more general statement: Let $p(n)= \chi n^d + a_{d-1} n^{d-1} + \cdots + a_1 n + a_0$ be any polynomial, with $\chi$ irrational. Then $p(n) \mod 1$ is equidistributed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $e(x) = e^{2 \pi i x}$. By the standard trickery with exponential polynomials, it is enough to show &#xA;$$\sum_{n=0}^{N-1} e(p(n)) = o(N).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a positive integer $h$. With a small error, we can replace the sum by&#xA;$$\sum_{n=0}^{N-1} (1/h) \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right).$$&#xA;By Cauchy-Schwartz, this bounded by&#xA;$$\sqrt{N}/h \left[ \sum_{n=0}^{N-1}  \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right) \overline{ \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right)} \right]^{1/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding the inner sum, we get $h^2$ terms of the form $e(p(n) - p(n+k))$. There are $h$ terms where $k=0$; these each sum up to $N$. For the other $h^2-h$ terms, the sum is of the form $\sum_{n=0}^{N-1} e(q(n))$, where $q$ has leading term $\chi d n^{d-1}$. By induction, each of these sums is $o(N)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the quantity in the square root is&#xA;$$hN+o(N)$$&#xA;where the constant in the $o$ depends on $h$ and $\chi$. Putting it all together, we get a bound of&#xA;$$N/\sqrt{h} + o(N).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ was arbitrary, this proves the result.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-07-29T16:10:13.567" CommentCount="1" />
  <row Id="1164" PostTypeId="1" CreationDate="2010-07-29T16:18:09.690" Score="1" ViewCount="151" Body="&lt;p&gt;Does combinatorial mean discrete ?&lt;/p&gt;&#xA;" OwnerUserId="450" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-07-29T16:35:30.860" LastActivityDate="2010-07-29T16:44:44.360" ClosedDate="2010-07-29T21:06:18.763" Title="difference between combinatorics and discrete mathematics" Tags="&lt;soft-question&gt;&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1165" PostTypeId="2" ParentId="1057" CreationDate="2010-07-29T16:21:20.203" Score="3" ViewCount="0" Body="&lt;p&gt;On the other hand, if the order of your group is not a multiple of 3 then it must be abelian!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a proof &lt;a href=&quot;http://groupprops.subwiki.org/wiki/Cube_map_is_endomorphism_iff_abelian_%28if_order_is_not_a_multiple_of_3%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-07-29T16:21:20.203" CommentCount="2" />
  <row Id="1166" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T16:39:51.603" Score="1" ViewCount="0" Body="&lt;p&gt;Also, if you are OK with power series, the Prologue to Walter Rudin's &lt;em&gt;Real and Complex Analysis&lt;/em&gt; seems like exactly what you want.  It's a beautiful development of &lt;em&gt;exp&lt;/em&gt;, as well as &lt;em&gt;sin&lt;/em&gt;, &lt;em&gt;cos&lt;/em&gt;, &lt;em&gt;e&lt;/em&gt;, and even &lt;em&gt;&amp;pi;&lt;/em&gt;, all quite organically.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-07-29T16:39:51.603" />
  <row Id="1167" PostTypeId="2" ParentId="1164" CreationDate="2010-07-29T16:44:44.360" Score="5" ViewCount="0" Body="&lt;p&gt;Combinatorics is a subset of discrete mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combinatorics is the study of the ways discrete structures can be arranged/combined. Discrete mathematics is the study of all aspects of discrete structures (thus including combinatorics).&lt;/p&gt;&#xA;" OwnerUserId="393" LastActivityDate="2010-07-29T16:44:44.360" />
  <row Id="1168" PostTypeId="2" ParentId="1133" CreationDate="2010-07-29T17:02:00.233" Score="9" ViewCount="0" Body="&lt;p&gt;I think essentially the same characterization holds. The complex exponential is the unique Lie group homomorphism from $\mathbb{C}$ to $\mathbb{C}^*$ such that the (real) derivative at the identity is the identity matrix.&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2010-07-29T17:02:00.233" CommentCount="1" />
  <row Id="1169" PostTypeId="2" ParentId="740" CreationDate="2010-07-29T17:05:49.243" Score="2" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy functionals&lt;/a&gt;, which satisfy the very simple equation &lt;em&gt;f(a+b) = f(a)+f(b)&lt;/em&gt; for all real &lt;em&gt;a,b&lt;/em&gt;.  These are either a line through the origin (the &quot;nice&quot; ones) or really &quot;ugly&quot; functions that are discontinuous and unbounded in every interval.  The latter are possible because the Axiom of Choice implies (actually is equivalent to) that infinite dimensional vector spaces have bases; i.e. the reals over the rationals have a Hamel basis.  A great explanation of all this (including the nice/ugly terminology) is in Horst Herrlich's monograph &lt;em&gt;The Axiom of Choice&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="452" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-07T18:28:23.390" LastActivityDate="2010-08-07T18:28:23.390" CommunityOwnedDate="2010-07-29T17:05:49.243" />
  <row Id="1170" PostTypeId="2" ParentId="1146" CreationDate="2010-07-29T18:04:04.790" Score="0" ViewCount="0" Body="&lt;p&gt;Have a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.symmys.com/AttilioMeucci/Teaching/Talks/Meucci_ModelingEstimationTechniquesForPortfolioManagement.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.symmys.com/AttilioMeucci/Teaching/Talks/Meucci_ModelingEstimationTechniquesForPortfolioManagement.pdf&lt;/a&gt; esp. p. 22f &lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-07-29T18:09:19.100" LastActivityDate="2010-07-29T18:09:19.100" />
  <row Id="1171" PostTypeId="2" ParentId="1146" CreationDate="2010-07-29T18:10:06.680" Score="0" ViewCount="0" Body="&lt;p&gt;Unfortunately I can also post one hyperlink per answer - so here is another one:&#xA;Also have a look at:&lt;br&gt;&#xA;&lt;a href=&quot;http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1358533&quot; rel=&quot;nofollow&quot;&gt;http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1358533&lt;/a&gt; esp. p. 3f&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-07-29T18:10:06.680" />
  <row Id="1174" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T20:14:31.867" Score="2" ViewCount="0" Body="&lt;p&gt;If each function has a smallest period, and otherwise fits the conditions, then a proof may be forthcoming by attempting to compute the smallest period of the sum and failing.  However, things become unclear if there is no smallest period, as in the case of the characteristic function of the rationals.  Progress might be made in this case by decomposing such a function as an infinite sum of periodic functions, or at least give more&#xA;counterexamples to study.  (e.g. Write the characteristic function of the rationals as&#xA;an infinite sum of functions of smallest period 1. )&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:05:01.260" LastActivityDate="2010-08-11T22:05:01.260" CommentCount="4" />
  <row Id="1175" PostTypeId="2" ParentId="1079" CreationDate="2010-07-29T20:22:41.597" Score="2" ViewCount="0" Body="&lt;p&gt;as a first start, if f+g is periodic the period cannot be rational wrt to the period of f  and g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us suppose that T is the smallest period of f(x), that is &amp;forall;x, f(x+T) = f(x); similarly S is the smallest period of g(x), that is &amp;forall;x, g(x+S) = g(x). If f+g had a period Q, and Q/T = m/n, we have that &amp;forall;x, f(x+nQ)+g(x+nQ) = f(x)+g(x). But f(x+nQ)=f(x+mT)=f(x), thus &amp;forall;x, g(x+nQ)=g(x) and therefore nQ is a period of g, which is impossible since it would mean that T/S is rational.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-29T20:22:41.597" CommentCount="1" />
  <row Id="1176" PostTypeId="2" ParentId="998" CreationDate="2010-07-29T20:29:17.463" Score="0" ViewCount="0" Body="&lt;p&gt;I'm really enjoying &lt;a href=&quot;http://www.mathematik.uni-kl.de/~gathmann/alggeom.php&quot; rel=&quot;nofollow&quot;&gt;Andreas Gathmann's lecture notes&lt;/a&gt;.  They are pretty elementary and surprisingly complete (for lecture notes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reid also has a really nice text on algebraic geometry («Undergraduate algebraic geometry»).&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-07-29T20:29:17.463" CommunityOwnedDate="2010-07-29T20:29:17.463" />
  <row Id="1177" PostTypeId="1" AcceptedAnswerId="1205" CreationDate="2010-07-29T20:30:03.983" Score="1" ViewCount="91" Body="&lt;p&gt;What are the differences between a (discrete) cosine transform and a (discrete) Fourier transform? I know the former is used in JPEG encoding while the latter plays a big part is signal and image processing- how related are they?&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-07-30T03:44:04.673" Title="Comparing/Contrasting Cosine and Fourier Transforms" Tags="&lt;transformation&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1178" PostTypeId="2" ParentId="1127" CreationDate="2010-07-29T20:51:21.433" Score="3" ViewCount="0" Body="&lt;p&gt;I'll make the trivial answer: 1 dog at point A, facing point B, walking with a velocity of 0.  Presumably, you should really highlight that the dogs' velocities must be non-zero...this is the kind of side case that math people love to exploit. &lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2010-07-29T20:51:21.433" CommentCount="1" />
  <row Id="1179" PostTypeId="2" ParentId="123" CreationDate="2010-07-29T21:07:02.397" Score="4" ViewCount="0" Body="&lt;p&gt;Velocity addition in (special) relativity is not linear, but becomes linear when expressed in terms of hyperbolic tangent functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, if you add two motions in the same direction, such as a man walking at velocity $v_1$ on a train that moves at $v_2$ relative to the ground, the velocity $v$ of the man relative to ground is not $v_1 + v_2$; velocities don't add (otherwise by adding enough of them you could exceed the speed of light).  What does add is the &lt;em&gt;inverse hyperbolic tangent&lt;/em&gt; of the velocities (in speed-of-light units, i.e., $v/c$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tanh^{-1}(v/c)=\tanh^{-1}(v_1/c) + \tanh^{-1}(v_2/c)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one way of deriving special relativity: assume that a velocity addition formula holds, respecting a maximum speed of light and some other assumptions, and show that it has to be the above.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T11:10:41.143" LastActivityDate="2010-12-29T11:10:41.143" />
  <row Id="1180" PostTypeId="2" ParentId="501" CreationDate="2010-07-29T21:39:50.027" Score="5" ViewCount="0" Body="&lt;p&gt;Disclaimer: I am not a finitist --- but as a theoretical computer scientist, I have a certain sympathy for finitism. The following is the result of me openly speculating what an &quot;official&quot; finitist response would be, based on grounds of computability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short version is this: &lt;strong&gt;(a)&lt;/strong&gt;&amp;nbsp;It depends on what you mean by a 'number', but there's a reasonable approach which makes it reasonable to talk about finitistic approaches to real numbers; &lt;strong&gt;(b)&lt;/strong&gt;&amp;nbsp;What you can do finitisitically with numbers, real, rational, or otherwise, depends on how you represent those numbers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What is a number?&lt;/strong&gt; Is &amp;minus;1 a number? Is sqrt(2) a number? Is &lt;em&gt;i&lt;/em&gt;&amp;nbsp;=&amp;nbsp;sqrt(&amp;minus;1) a number? What about quaternions? --- I'm going to completely ignore this question and suggest a pragmatic, formalist approach: a &quot;number&quot; is an element of a &quot;number system&quot;; and a &quot;number system&quot; is a collection of expressions which you can transform or describe properties of in some given ways (&lt;em&gt;i.e.&lt;/em&gt; certain given arithmetic operations) and test certain properties (&lt;em&gt;e.g.&lt;/em&gt; tests for equality, ordering, &lt;em&gt;etc.&lt;/em&gt;) These expressions don't have to have a meaningful interpretation in terms of quantities or magnitudes as far as I'm concerned; &lt;em&gt;you&lt;/em&gt; get to choose which operations/tests you care about.&lt;br&gt;&lt;br&gt; A finitist would demand that any operation or property be described by an algorithm which provably terminates. That is, it isn't sufficient to prove existence or universality &lt;em&gt;a la&lt;/em&gt; classical logic; existence proofs must be finite constructions --- of a &quot;number&quot;, that is a representation in some &quot;number system&quot; --- and univserality must be shown by a computable test. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Representation of numbers:&lt;/strong&gt; How we represent the numbers matters. A finitist should have no qualms about rational numbers: ratios which ultimately boil down to ordered pairs. Despite this, the decimal expansions of these numbers may be infinitely long: 1/3 = 0.33333... what's going on here?&lt;br&gt;&lt;br&gt; Well, the issue is that we have two representations for the same number, one of which is finite in length (and allows us to perform computations) and another which is not finite in length. However, the decimal expansion can be easily expressed as a function: for all &lt;em&gt;k&lt;/em&gt;, the &lt;em&gt;k&lt;/em&gt;&lt;sup&gt;th&lt;/sup&gt; decimal place after the point is '3'; so you can still characterize it precisely in terms of a finite rule.&lt;br&gt;&lt;br&gt; What's important is that there exists &lt;strong&gt;some&lt;/strong&gt; finite way to express the number. But the way in which we choose to &lt;em&gt;define&lt;/em&gt; the number (as a part of system or numbers, using some way of expressing numbers) will affect what we can do with it...&#xA;there is now a question about what operations we can perform.&lt;br&gt;&lt;br&gt;--- For rationals-as-ratios, we can add/subtract, multiply/divide, and test order/equality. So this representation is a very good one for rationals. &lt;br&gt;&lt;br&gt;--- For rationals-as-decimal-expansions, we can still add/subtract and multiply/divide, by defining a new digit-function which describes how to compute the result from the decimal expansions; these will be messier than the representations as ratios. Order comparisons are still possible for &lt;em&gt;distinct&lt;/em&gt; rationals; but you cannot test equality for arbitrary decimal-expansion representations, because you cannot necessarily verify that all decimal places of the difference |&lt;em&gt;a&lt;/em&gt;&amp;minus;&lt;em&gt;b&lt;/em&gt;| are 0. The best you can do in general is testing &quot;equality up to precision &amp;epsilon;&quot;, wherein you show that |&lt;em&gt;a&lt;/em&gt;&amp;minus;&lt;em&gt;b&lt;/em&gt;|&amp;nbsp;&amp;lt;&amp;nbsp;&amp;epsilon;, for some desired precision &amp;epsilon;. This is a number system which informally we may say has certain amount of &quot;vagueness&quot;; but it is in principle completely specified --- there's nothing wrong with this in principle. It's just a matter of how you wish to define your system of arithmetic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What representation of reals?&lt;/strong&gt; Obviously, because there are uncountably many real numbers, you cannot represent all real numbers even if you &lt;em&gt;aren't&lt;/em&gt; a finitist. But we can still express some of them. The same is true if you're a finitist: you just don't have access to as many, and/or you're restricted in what you can do with them, according to what your representation can handle.&lt;br&gt;&lt;br&gt;&#xA;--- Algebraic irrational numbers such as sqrt(2) can be expressed simply like that: &quot;sqrt(2)&quot;. There's nothing wrong with the expressions &quot;sqrt(2)&amp;nbsp;&amp;minus;&amp;nbsp;1&quot; or &quot;[1&amp;nbsp;+&amp;nbsp;sqrt(5)]/2&quot; --- they express quantities perfectly well. You can perform arithmetic operations on them perfectly well; and you can also perform ordering/equality tests by transforming them into a normal form of the type &quot;[sum of integers and roots of integers]/[positive integer]&quot;; if the difference of two quantities is zero, the normal form of the difference will just end up being '0'. For order comparisons, we can compute enough decimal places of each term in the sum to determine whether the result is positive or negative, a process which is guaranteed to terminate.&lt;br&gt;&lt;br&gt;&#xA;--- Numbers such as &amp;pi; and e can be represented by decimal expansions, and computed with in this form, as with the rational numbers. The decimal expansions can be gotten from classical equalities (&lt;em&gt;e.g.&lt;/em&gt; &quot;infinite&quot; series, except computing only &lt;em&gt;partial&lt;/em&gt; sums; a number such as e may be expressed by some finite representation of such an 'exact' formula, together with a computable function which describes how many terms of the series are required to get a correct evaluation of the first &lt;em&gt;k&lt;/em&gt; decimal places.) Of course, what you can do finitistically with these representations is limited in the same way as described above with the rationals; specifically, you cannot always test equality.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-07-29T21:54:11.680" LastActivityDate="2010-07-29T21:54:11.680" CommentCount="2" />
  <row Id="1181" PostTypeId="1" AcceptedAnswerId="1182" CreationDate="2010-07-29T22:11:12.637" Score="4" ViewCount="206" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside Lemma&lt;/a&gt; looks like it should have an intuitive explanation. Does anyone have one?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-11-18T18:47:12.027" LastActivityDate="2010-11-19T07:22:51.490" Title="Intuitive explanation of the Burnside Lemma" Tags="&lt;group-theory&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="1182" PostTypeId="2" ParentId="1181" CreationDate="2010-07-29T22:11:45.070" Score="2" ViewCount="0" Body="&lt;p&gt;As an example, we consider the number of ways of colouring a cube with n colours with uniqueness up to rotation. We call each unique colouring where rotations are not allowed a static colouring and each unique one where they are allowed a dynamic colouring. We define the the set of orbits to be the (disjoint) static colourings that correspond to each dynamic colouring. We will use rotations to mean a rotation that makes the cube occupy the same space, and as being unique if it is a unique function from the cube to the cube. This includes the identity rotation. Intuitively, the lemma says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition 1. #Orbits * #Rotations = sum for each rotation r of #static colourings unchanged by this rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now consider each orbit O separately. Pick a static colouring c inside O. Suppose two (possibly equal) rotations p, q give the same static colouring, d, when applied on c. Then p^-1*q fixes d. Additionally, suppose r (possibly the identity) fixes d. p^-1*(p*r) will also fix d. So p*r will take c to d. Since p*r is different for each r, and p^-1*q is different for each q, the mapping functions are injective in both directions and there is a bijection between the q and r values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each O, the number of rotations is the sum over each static colorings x in O times the number of rotations producing x. This can be rewritten as the sum over each rotation r of the number of static colourings in O fixed by r (due to the bijection in the previous paragraph). We get proposition 1 by adding over all O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general proof is quite similar to this, except that it uses group theory.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-29T22:11:45.070" CommentCount="1" />
  <row Id="1183" PostTypeId="1" AcceptedAnswerId="1184" CreationDate="2010-07-29T22:12:28.233" Score="8" ViewCount="382" Body="&lt;p&gt;I ran into a problem dividing by imaginary numbers recently. I was trying to simplify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \over i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with two methods, which produced different results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 1: ${2 \over i} = {2i \over i^2} = {2i \over -1} = -2i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 2: ${2 \over i} = {2 \over \sqrt{-1}} = {\sqrt{4} \over \sqrt{-1}} = \sqrt{4 \over -1} = \sqrt{-4} = 2i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from using the formula from &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_numbers#Operations&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; that method 1 produces the correct result. My question is: &lt;strong&gt;why does method 2 give the incorrect result&lt;/strong&gt;? What is the invalid step?&lt;/p&gt;&#xA;" OwnerUserId="477" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:50:52.350" LastActivityDate="2010-07-31T17:50:52.350" Title="Division by imaginary number" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="1184" PostTypeId="2" ParentId="1183" CreationDate="2010-07-29T22:15:08.030" Score="17" ViewCount="0" Body="&lt;p&gt;The incorrect step is saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{4}/\sqrt{-1} = \sqrt{4/-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{a}/\sqrt{b} = \sqrt{a/b}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is only justified when $a$ and $b$ are positive.&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-07-29T22:15:08.030" CommentCount="5" />
  <row Id="1185" PostTypeId="2" ParentId="1183" CreationDate="2010-07-29T22:24:16.773" Score="0" ViewCount="0" Body="&lt;p&gt;This is exactly the same issue as in &lt;a href=&quot;http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake/482#482&quot;&gt;this question&lt;/a&gt;. Each non-zero complex number has two numbers that square to give it, with the same magnitude, but with opposite sign. When we define the square root function, we have to decide which of the roots we want. For positive numbers, it is obvious to choose the positive root. For negative number, we choose to have the positive imaginary values, although because of symmetry the choice doesn't mean much anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to see if the standard multiplication and division laws apply, then we have to consider domain the numbers are in. We already know they apply for non-negative real numbers. It is easy enough to verify that for negative numbers sqrt(a)*sqrt(b)=-sqrt(ab) and sqrt(a)/sqrt(b)=sqrt(a/b). We also see that, if a is positive and b negative, then sqrt(ab)=sqrt(a)sqrt(b), sqrt(a/b)=-sqrt(a)/sqrt(b) and sqrt(b/a)=sqrt(b)/sqrt(a).&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-29T22:45:08.907" LastActivityDate="2010-07-29T22:45:08.907" CommentCount="2" />
  <row Id="1186" PostTypeId="1" CreationDate="2010-07-29T22:58:13.583" Score="6" ViewCount="278" Body="&lt;p&gt;Anything to speed up or simplify calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example might be to get a multiple of 19, for instance, 38 * 19 = 38 * 20 - 38.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(This is hard to tag with so few tags in play!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mental-calculations tips tricks shortcut cheats time-saver&lt;/p&gt;&#xA;" OwnerUserId="483" LastEditorUserId="483" LastEditorDisplayName="" LastEditDate="2010-07-30T12:32:19.310" LastActivityDate="2010-07-30T13:38:00.860" CommunityOwnedDate="2010-07-30T12:32:19.310" Title="What calculation shortcuts exist to help or speed-up mental (or paper) calculations?" Tags="&lt;big-list&gt;&lt;arithmetic&gt;&lt;elementary&gt;&lt;mental-math&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="1187" PostTypeId="1" CreationDate="2010-07-29T22:58:59.543" Score="0" ViewCount="216" Body="&lt;p&gt;I'd like to help a 6 year old who already has a pretty good grasp of 2, 5, and 10 times tables.&lt;/p&gt;&#xA;" OwnerUserId="483" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-07-29T23:00:58.633" LastActivityDate="2010-07-31T00:16:31.383" Title="What's the most effective ways of teaching kids - times tables?" Tags="&lt;soft-question&gt;&lt;teaching&gt;&lt;elementary&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="1188" PostTypeId="2" ParentId="1146" CreationDate="2010-07-29T23:41:41.033" Score="3" ViewCount="0" Body="&lt;p&gt;Principal component analysis is a useful technique when dealing with large datasets. In some fields, (bioinformatics, internet marketing, etc) we end up collecting data which has many thousands or tens of thousands of dimensions. Manipulating the data in this form is not desirable, because of practical considerations like memory and CPU time. However, we can't just arbitrarily ignore dimensions either. We might lose some of the information we are trying to capture!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Principal component analysis is a common method used to manage this tradeoff. The idea is that we can somehow select the 'most important' directions, and keep those, while throwing away the ones that contribute mostly noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this picture shows a 2D dataset being mapped to one dimension: &#xA; &lt;img src=&quot;http://cmp.felk.cvut.cz/cmp/software/stprtool/examples/pca_example.gif&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Note that the dimension chosen was not one of the original two: in general, it won't be, because that would mean your variables were uncorrelated to begin with.&lt;br&gt;&#xA;We can also see that the direction of the principal component is the one that maximizes the variance of the projected data. This is what we mean by 'keeping as much information as possible.'&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-07-29T23:41:41.033" />
  <row Id="1189" PostTypeId="2" ParentId="1186" CreationDate="2010-07-29T23:51:00.967" Score="3" ViewCount="0" Body="&lt;p&gt;Art Benjamin is your man! He has many tricks to speed up mental calculation and other fun mathemagical tricks. He also wrote two books on the subject!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a video of him in action: &lt;a href=&quot;http://www.youtube.com/watch?v=M4vqr3_ROIk&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=M4vqr3_ROIk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is his new book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0307338401&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Secrets-Mental-Math-Mathemagicians-Calculation/dp/0307338401&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-29T23:51:00.967" CommunityOwnedDate="2010-08-03T18:44:06.083" />
  <row Id="1190" PostTypeId="2" ParentId="1187" CreationDate="2010-07-30T00:05:16.260" Score="10" ViewCount="0" Body="&lt;p&gt;I'm guessing you are looking for a hard data research answer, but unfortunately, the studies I know of are dubious, partly because they rely heavily on &lt;a href=&quot;http://numberwarrior.wordpress.com/2010/04/15/are-learning-styles-dead/&quot; rel=&quot;nofollow&quot;&gt;learning style myths&lt;/a&gt;, partly because some are funded by commercial parties pushing particular systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can give some general advice --&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Give the system for the 9s that the digits add to 9, let your pupil fiddle with the patterns, and they tend to get this one fairly quickly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Asian countries do fairly well with a &quot;musical&quot; or &quot;chant&quot; method. I've heard of a couple commercial products, including &lt;a href=&quot;http://www.abbeytape.com/ttrap/&quot; rel=&quot;nofollow&quot;&gt;Times Table Rap&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Play with the visual representation using marbles or tokens, especially so the pupil can understand why 3 x 7 is the same as 7 x 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;circle&quot; method of visualizing skip counting (samples of what I mean &lt;a href=&quot;http://overthecrescentmoon.blogspot.com/2009/09/skip-counting.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) can be brought in to help with understanding the &quot;rotating&quot; nature of counting on the times tables.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are a fair number of games that help multiplication table prowess, either intentionally or unintentionally. Yahtzee in particular naturally covers the 1-6 portion of the table. &lt;a href=&quot;http://numberwarrior.wordpress.com/2010/03/11/multiplication-table-game/&quot; rel=&quot;nofollow&quot;&gt;I've written a game myself&lt;/a&gt;, although I don't know how well it would work with 6 year olds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While I wouldn't necessarily recommend the DS game itself, check out how &lt;a href=&quot;http://bit.ly/9SBY5U&quot; rel=&quot;nofollow&quot;&gt;Professor Kageyama's Maths Training: The Hundred Cell Calculation Method&lt;/a&gt; mixes up the columns and rows of the times table to force students to think about the arrangement. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="295" LastActivityDate="2010-07-30T00:05:16.260" />
  <row Id="1192" PostTypeId="2" ParentId="1186" CreationDate="2010-07-30T00:15:34.210" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;To square a number ending in 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove the ending 5. Let the resulting number be n, and compute n(n+1). Append 25 to the end of n(n+1) and that's your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&#xA;85&lt;sup&gt;2&lt;/sup&gt;. Here, we drop the last digit to get 8, compute 8*9 = 72, so 85&lt;sup&gt;2&lt;/sup&gt; = 7225. Similarly, we can compute &#xA;115&lt;sup&gt;2&lt;/sup&gt;. Here, we drop the last digit to get 11, compute 11*12 = 132, so 115&lt;sup&gt;2&lt;/sup&gt; = 13225.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does this work?:&lt;/strong&gt;&#xA;Note that&#xA;(10n + 5)&lt;sup&gt;2&lt;/sup&gt; = 100n&lt;sup&gt;2&lt;/sup&gt; + 100n + 25 = 100 * n(n+1) + 25.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T00:15:34.210" CommunityOwnedDate="2010-08-03T18:43:51.840" />
  <row Id="1193" PostTypeId="2" ParentId="566" CreationDate="2010-07-30T00:21:41.217" Score="6" ViewCount="0" Body="&lt;p&gt;Here are three of my favorite variations on the hats and prisoners puzzle that I've collected over time:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Fifteen prisoners sit in a line, and hats are placed on their heads. Each hat can be one of two colors: white or black. They can see the colors of the people in front of them but not behind them, and they can’t see their own hat colors. Starting from the back of the line (with the person who can see every hat except his own), each prisoner must try to guess the color of his own hat. If he guesses correctly, he escapes. Otherwise, he is fed to cannibals (because that’s the canonical punishment for failing at hat problems). Each prisoner can hear the guess of each person behind him. By listening for painful screaming and the cheering of cannibals, he can also deduce if each of those guesses was accurate. Of course, this takes place in some magical mathematical universe where people don’t cheat. Assuming that they do not want to be eaten, find the optimal guessing strategy for the prisoners. (The cannibals should eat no more than one prisoner.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the year 3141, Earth’s population has exploded. A countably infinite number of prisoners sit in a line (there exists a back of the line, but the other end extends forever). As in the previous problem, white and black hats are placed on their heads. Due to modern technology, each person can see the hat colors of all infinitely many people in front of them. However, they cannot hear what the people behind them say, and they do not know if those people survive. Assuming that they do not want to be eaten, find the optimal guessing strategy for the prisoners. Assume that there are enough cannibals to eat everyone who fails. (The cannibals should eat no more than finitely many prisoners. Assume the Axiom of Choice.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are seven prisoners, and colored hats will be placed on their heads. The hats have seven possible colors (red, orange, yellow, green, blue, indigo, violet), and may be placed in any arrangement: all the same color, all different colors, or some other arrangement. Each person can see everyone else’s hat color but cannot see his own hat color. They may not communicate after getting their hats, and as in the previous problems, they remain in a magical universe where no one cheats. They must guess their hat colors all at the same time. If at least one person guesses correctly, they are all released. If no one guesses correctly, however, the entire group is fed to cannibals. Assuming that they don’t want to be eaten, find the optimal guessing strategy for the prisoners. (By this point, the cannibals have probably eaten far too much. It would be cruel to force them to eat any more, so spare the cannibals and find a way to guarantee that the seven prisoners survive.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T00:21:41.217" CommunityOwnedDate="2010-07-30T00:21:41.217" CommentCount="0" />
  <row Id="1194" PostTypeId="2" ParentId="187" CreationDate="2010-07-30T01:26:01.123" Score="0" ViewCount="0" Body="&lt;p&gt;The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Honeycomb_(geometry)#Space-filling_polyhedra.5B2.5D&quot; rel=&quot;nofollow&quot;&gt;honeycombs&lt;/a&gt; has several examples of 3d tilings by a single polyhedron which aren't vertex-transitive.  I am not positive if they are vertex-uniform, though, since (again) I don't really understand your definition and can't find a good one online.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T01:56:26.340" LastActivityDate="2010-07-30T01:56:26.340" />
  <row Id="1195" PostTypeId="2" ParentId="1187" CreationDate="2010-07-30T01:32:41.653" Score="5" ViewCount="0" Body="&lt;p&gt;My personal experience with my 7/8 year old is that Math War works very well for practice. You can play with a normal deck of cards---each player plays two cards, and announces the product, with the higher number winning all the cards. In case of tie, then there is &quot;War&quot;, which means you put two (or more) cards face down and then have aother battle on top, with the winner taking the whole pile. You can also play the game with any set of ordinary flash cards. My son is interested to go for a long time with this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also add, however, that although I have heard many people say that one must master the basic math facts before learning more advanced math concepts, I think that this is  nonsense. Go ahead and talk about any kind of mathematics (at the right level) with your child.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-07-30T01:46:30.967" LastActivityDate="2010-07-30T01:46:30.967" />
  <row Id="1196" PostTypeId="1" AcceptedAnswerId="1199" CreationDate="2010-07-30T02:07:43.037" Score="2" ViewCount="563" Body="&lt;p&gt;I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Problem:&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;For any natural number n , n&lt;sup&gt;3&lt;/sup&gt; + 2n is divisible by 3.&#xA;  &lt;code&gt;This makes sense&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;h2&gt;Proof:&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Basis Step:&lt;/strong&gt; If n = 0, then n&lt;sup&gt;3&lt;/sup&gt; + 2n = 0&lt;sup&gt;3&lt;/sup&gt; +&#xA;  2*0 = 0. So it is divisible by 3.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Induction:&lt;/strong&gt; Assume that for an arbitrary natural number n,&#xA;  n&lt;sup&gt;3&lt;/sup&gt; + 2n is divisible by 3. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Induction Hypothesis:&lt;/strong&gt; To prove this for n+1, first try to express ( n&#xA;  + 1 )&lt;sup&gt;3&lt;/sup&gt; + 2( n + 1 )   in terms of n&lt;sup&gt;3&lt;/sup&gt; + 2n   and use&#xA;  the induction hypothesis. &lt;code&gt;Got it&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;( n + 1 )&lt;sup&gt;3&lt;/sup&gt; + 2( n + 1 ) = (  n&lt;sup&gt;3&lt;/sup&gt; + 3n&lt;sup&gt;2&lt;/sup&gt; + 3n + 1 ) + ( 2n + 2 )  &lt;code&gt;Just some simplifying&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;= ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + ( 3n&lt;sup&gt;2&lt;/sup&gt; + 3n + 3 ) &lt;code&gt;simplifying&#xA;  and regrouping&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;li&gt;= ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + 3( n&lt;sup&gt;2&lt;/sup&gt; + n + 1 ) &lt;code&gt;factored out&#xA;  the 3&lt;/code&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;which is divisible by 3, because (&#xA;  n&lt;sup&gt;3&lt;/sup&gt; + 2n ) is divisible by 3&#xA;  by the induction hypothesis. &lt;code&gt;What?&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone explain that last part? I don't see how you can claim ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + 3( n&lt;sup&gt;2&lt;/sup&gt; + n + 1 ) is divisible by 3. &lt;/p&gt;&#xA;" OwnerUserId="494" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-07-30T04:45:45.777" LastActivityDate="2010-08-09T00:07:09.457" Title="Proof that n^3+2n is divisble by 3" Tags="&lt;number-theory&gt;&lt;high-school&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" />
  <row Id="1197" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:13:44.880" Score="1" ViewCount="0" Body="&lt;p&gt;In a proof by induction, we try to prove that a statement is true for all integers n. To do this, we first check the base case, which is the &quot;Basic Step&quot; above. Then, we have the induction hypothesis, where we assume that the statement is true for an integer k. Using this fact, we prove that the statement is also true for the next integer k+1. This produces a bootstrapping ladder where we use the base case (n=1) to show that the statement is true for n=2 via the inductive hypothesis, and then for n=3, etc, off to infinity; this shows that the statement is true for all integers n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we claimed that ( n&lt;sup&gt;3&lt;/sup&gt; + 2n ) + 3( n&lt;sup&gt;2&lt;/sup&gt; + n + 1 ) is divisible by 3 because this was the inductive hypothesis; we were using this to show that&#xA;( (n+1)&lt;sup&gt;3&lt;/sup&gt; + 2(n+1) ) + 3( (n+1)&lt;sup&gt;2&lt;/sup&gt; + (n+1) + 1 ).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T02:13:44.880" />
  <row Id="1198" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:15:17.760" Score="1" ViewCount="0" Body="&lt;p&gt;Why don't you just test the validity of this using modular arithmetic? I.e. take n congruent to 1 mod 3 and we easily get 3=0. If you try n congruent to 2 mod 3, you get 8+4 =12 =0, so you're done. No ugly inductions (although this particular case is not so dirty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful idea when thinking of induction is to think of dominos. If you know something is true for one fixed tile and if you know that it being true for one tile means that it's true for the neighbour on the right, then it's like knocking one over knocks them all over.&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="298" LastEditorDisplayName="" LastEditDate="2010-07-30T02:20:51.290" LastActivityDate="2010-07-30T02:20:51.290" CommentCount="3" />
  <row Id="1199" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:26:09.203" Score="6" ViewCount="0" Body="&lt;p&gt;In the inductive hypothesis, you assumed that $n^3 + 2n$ was divisible by 3 for some $n$, and now you're proving the same for $n+1$. It's like knocking down dominoes: if you can prove that the first domino falls over (base case) and each domino knocks over the next (inductive step), then that means that all of the dominoes get knocked down eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that $(n^3 + 2n) + 3(n^2 + n + 1)$ is divisible by 3 because $n^3 + 2n$ is (because of the inductive hypothesis) and $3(n^2 + n + 1)$ is (because it's 3 times an integer). So, their sum is as well.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-30T12:20:44.747" LastActivityDate="2010-07-30T12:20:44.747" CommentCount="1" />
  <row Id="1200" PostTypeId="2" ParentId="1196" CreationDate="2010-07-30T02:34:59.507" Score="2" ViewCount="0" Body="&lt;p&gt;Presumably you're only looking for a way to understand the induction problem, but you can note that $n^3+2n = n^3 - 3n^2 + 2n - 3n^2 = (n-2)(n-1)(n) + 3n^2$. Since any three consecutive integers has a multiple of three, we're adding two multiples of three and so get another multiple of 3.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2010-07-30T02:34:59.507" CommentCount="2" />
  <row Id="1201" PostTypeId="2" ParentId="1186" CreationDate="2010-07-30T02:35:19.460" Score="3" ViewCount="0" Body="&lt;p&gt;One simple one that you're probably familiar with already: When you multiply a one-digit number, n, by 9, the result has n-1 in the 10s place, and then the ones digit is such that the sum of the digits is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: 9*6 = 54 because 5 is 6-1, and then 5+4 is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really cool is that you can use this trick, plus your fingers, to get the answer instantly. Hold out your hands with all 10 fingers up, then put down the n'th finger (which might be on either hand. Then the number is just (how many fingers there are to the left of the finger you put down)*10 + (how many fingers there are to the right of the finger you put down)&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-30T02:35:19.460" CommunityOwnedDate="2010-08-03T18:45:03.010" />
  <row Id="1202" PostTypeId="1" AcceptedAnswerId="1203" CreationDate="2010-07-30T02:37:40.377" Score="2" ViewCount="80" Body="&lt;p&gt;I was reading &quot;Functional Equations and How to Solve Them&quot; by Small and the following comment pops up without much justification on p. 13: If a(x) is an involution, then f(a(x))=f(x) has as solutions f(x) = T[x,a(x)], where T is an arbitrary symmetric function of u and v.  I was wondering why this was true (it works for examples I've tried, but I am not sure how to prove this -and if there's anything obvious staring at me in the face here).&lt;/p&gt;&#xA;" OwnerUserId="298" LastActivityDate="2010-07-30T02:43:39.330" Title="f(a(x))=f(x) - functional eq" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1203" PostTypeId="2" ParentId="1202" CreationDate="2010-07-30T02:43:39.330" Score="4" ViewCount="0" Body="&lt;p&gt;Any function f(x) that is a solution to your functional equation f(a(x)) = f(x) must satisfy the property that it is unchanged when you plug in a(x) instead of x. In addition, f(x) clearly must be a function f(x) = T[x, a(x)] depending on the x and a(x); the question is to see why T must be symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, T[x, a(x)] = f(x) = f(a(x)) = T[a(x), a(a(x))] = T[a(x), x] since a(x) is an involution. In particular, this means that T must be symmetric in its two variables.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T02:43:39.330" CommentCount="2" />
  <row Id="1204" PostTypeId="1" AcceptedAnswerId="1208" CreationDate="2010-07-30T02:58:21.027" Score="4" ViewCount="182" Body="&lt;p&gt;I know that a superalgebra is a Z/2Z graded algebra and that it behaves nicely. I know very little physics though, so even though I know &lt;em&gt;that&lt;/em&gt; the super- prefix is related to supersymmetry, I don't know what that means; is there a compelling &lt;em&gt;mathematical&lt;/em&gt; reason to consider superalgebras? &lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T17:46:19.847" LastActivityDate="2010-08-02T22:16:17.183" Title="Why are superalgebras so important?" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="1205" PostTypeId="2" ParentId="1177" CreationDate="2010-07-30T03:44:04.673" Score="1" ViewCount="0" Body="&lt;p&gt;Cosine transforms are nothing more than shortcuts for computing the Fourier transform of a sequence with special symmetry (e.g. if the sequence represents samples from an even function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good book on how FFT and the DCTs/DSTs are related is Van Loan's &quot;Computational Frameworks for the Fast Fourier Transform&quot;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-07-30T03:44:04.673" />
  <row Id="1206" PostTypeId="2" ParentId="524" CreationDate="2010-07-30T03:53:59.347" Score="2" ViewCount="0" Body="&lt;p&gt;It's all dependent on the 2-norm condition number of your matrix (the ratio of the largest singular value to the smallest); as a nice rule of thumb, if the base-10 logarithm of the reciprocal of the condition number is much less than the number of digits your computer uses to store numbers, QR (and maybe even the normal equations) might be sufficient. Otherwise, SVD is a &quot;safer bet&quot;: it always works, but is much slower than the other methods for solving least squares problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good references would be the classic &quot;Solving Least Squares Problems&quot; by Lawson and Hanson, and the newer &quot;Numerical Methods for Least Squares Problems&quot; by Björck.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To add to the answer I gave previously, one way you can proceed for a matrix A whose conditioning you don't know would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Compute the QR decomposition of A.&lt;/li&gt;&#xA;&lt;li&gt;Estimate the condition number of R.&lt;/li&gt;&#xA;&lt;li&gt;If R is well conditioned (condition number is &quot;small enough&quot;), stop; else&lt;/li&gt;&#xA;&lt;li&gt;Compute the singular value decomposition of R=U∑V&lt;sup&gt;T&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;Multiply Q and U to get the SVD of A.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The advantage of proceeding in this manner is if your matrix is badly conditioned enough to necessitate the use of SVD, the algorithm for computing the SVD has to handle only a triangular matrix instead of the original matrix (which may have more rows than columns as is usual in least-squares applications).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T06:14:23.543" LastActivityDate="2010-08-03T06:14:23.543" CommentCount="2" />
  <row Id="1207" PostTypeId="2" ParentId="1091" CreationDate="2010-07-30T04:05:04.443" Score="1" ViewCount="0" Body="&lt;p&gt;Instead of giving an answer, let me point out to you this &lt;a href=&quot;http://www.mathworks.com/moler/eigs.pdf&quot; rel=&quot;nofollow&quot;&gt;chapter&lt;/a&gt; in Cleve Moler's book &quot;Numerical Computing with MATLAB&quot;, there is a nice geometric demonstration in MATLAB on how eigenvalues/eigenvectors (as well as singular values/vectors) of an order-2 square matrix are involved in how a circle is transformed into an ellipse after a linear transformation represented by the matrix.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-07-30T04:05:04.443" />
  <row Id="1208" PostTypeId="2" ParentId="1204" CreationDate="2010-07-30T04:05:14.647" Score="4" ViewCount="0" Body="&lt;p&gt;Sure, supermodules over a superalgebra form a nice and nontrivial example of a tensor category. In fact, it is a theorem that Deligne that all tensor categories over $\mathbb{C}$ with suitable growth conditions can be obtained as categories of super-representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is given by Qiaochu, that supercommutativity arises naturally in a lot of places.&lt;br&gt;&#xA;For instance, the exterior algebra is supercommutative; thus the wedge product on differential forms acts the same way.  Interestingly, the exterior algebra of a direct sum corresponds to taking the super tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example: With singular cohomology on a topological space, there is a way to define a cup product, which satisfy the supercommutative law $ab = (-1)^{deg b deg a} ba$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that given a $\mathbb{Z}$-graded super-commutative algebra, you can get a $\mathbb{Z}/2$-graded superalgebra as you indicate in your question by taking the sum of the odd parts and the sum of the even parts.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T04:05:14.647" />
  <row Id="1209" PostTypeId="2" ParentId="1204" CreationDate="2010-07-30T04:34:50.267" Score="3" ViewCount="0" Body="&lt;p&gt;I can summarize one really basic reason, which is actually the reason I originally got interested in the definition.  Take a finite-dimensional vector space $V$ of dimension $n$, and let $\left( {n \choose k} \right) = {n+k-1 \choose k}$ denote the number of multisets of size $k$ on a set of size $n$.  (Multisets are like subsets except that more than one copy of a given element is possible.)  Then the symmetric powers of $V$ have dimensions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left( {n \choose 1} \right), \left( {n \choose 2} \right), ... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas the exterior powers of $V$ have dimensions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle {n \choose 1}, {n \choose 2}, ....$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is a funny identity: it is not hard to see that ${n \choose k} = (-1)^k \left( {-n \choose k} \right)$.  One way we might interpret this identity is that the $k^{th}$ exterior power of a vector space of dimension $n$ is like the $k^{th}$ symmetric power of a vector space of dimension $-n$, whatever that means.  So what could that possibly mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer (and I'll let you work this out for yourself, because it's fun) is to work in the category of supervector spaces!  A supervector space $V$ is a direct sum $V_0 \oplus V_1$, and while one notion of dimension is to take $\dim V_0 + \dim V_1$, another (which can be motivated by thinking of $\mathbb{Z}/2\mathbb{Z}$-graded vector spaces as the category of representations of $\mathbb{Z}/2\mathbb{Z}$) is to take $\dim V_0 - \dim V_1$.  So while a purely even vector space has positive dimension, a purely odd vector space has negative dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: graded-commutativity implies that the symmetric power of a purely odd vector space is the exterior power of the corresponding purely even vector space.  More generally, the $k^{th}$ symmetric power of a vector space of dimension $n$ (for all integers $n$) has dimension $\left( {n \choose k} \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And, of course, the symmetric algebra of a supervector space is naturally a graded-commutative superalgebra.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The physics connection is that symmetric powers = bosons, exterior powers = fermions, and there is a duality between the two.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-30T04:34:50.267" CommentCount="3" />
  <row Id="1210" PostTypeId="2" ParentId="631" CreationDate="2010-07-30T05:51:11.413" Score="1" ViewCount="0" Body="&lt;p&gt;It's worth considering a few ways of showing that the problem is neither in P, nor NPC. I've marked this answer &quot;community wiki&quot;, so please feel free to add suggestions and flesh out ideas here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my experience playing the 24 game, it seems that most combinations of numbers are solvable. If we could formalize this, we could show that the 24 game is not NPC. Formally, consider the &lt;code&gt;2^n&lt;/code&gt; inputs of length &lt;code&gt;n&lt;/code&gt;. If all but polynomially-many of them solvable, then the language is sparse and cannot be NPC (unless P=NP).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-07-30T05:51:11.413" CommunityOwnedDate="2010-07-30T05:51:11.413" />
  <row Id="1211" PostTypeId="1" AcceptedAnswerId="1269" CreationDate="2010-07-30T07:46:11.700" Score="4" ViewCount="460" Body="&lt;p&gt;Roots behave strangely over &lt;a href=&quot;http://math.stackexchange.com/questions/1183/division-by-imaginary-number/1185#1185&quot;&gt;complex numbers&lt;/a&gt;. Given this, how do non-integer powers behave over negative numbers? More specifically:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we define fractional powers such as (-2)^-1.5?&lt;/li&gt;&#xA;&lt;li&gt;Can we define irrational powers (-2)^pi?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-21T07:13:37.997" LastActivityDate="2010-08-21T07:13:37.997" Title="Non-integer powers of negative numbers" Tags="&lt;complex-numbers&gt;&lt;powers&gt;" AnswerCount="4" />
  <row Id="1212" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T08:02:04.923" Score="1" ViewCount="0" Body="&lt;p&gt;We can define complex powers as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^w := e^{(l(z)*w)}$ for $z,w \in \mathbb{C}$ and a complex logarithm function $l$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of course you need to make sure there actually exists such a logarithm $l:U\to\mathbb{C}$ function ($U$ needs to be a simply-connected subset of $\mathbb{C}\setminus\{0\}$&lt;/p&gt;&#xA;" OwnerUserId="340" LastActivityDate="2010-07-30T08:02:04.923" />
  <row Id="1213" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T08:03:21.410" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says &quot;Complex powers of positive reals are defined via e^x as in section Complex powers of positive real numbers above. These are continuous functions.&#xA;Trying to extend these functions to the general case of noninteger powers of complex numbers that are not positive reals leads to difficulties. Either we define discontinuous functions or multivalued functions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let me say a few more words. Essentially what we are looking for is $z^a=e^{a log(t)}$. But as you may or may not know, &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_logarithm#Problems_with_inverting_the_complex_exponential_function&quot; rel=&quot;nofollow&quot;&gt;complex logarithms&lt;/a&gt; are problematic depending on your viewpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an &lt;a href=&quot;http://math.stackexchange.com/questions/1133/natural-derivation-of-the-complex-exponential-function/1158#1158&quot;&gt;answer&lt;/a&gt; to Harry's question, I describe a more careful way to define the complex exponential function, and suggest that one should just inverse this to get your logarithm. This way is safe and will yield your logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are less fickle, and are willing to take on the branched logarithms, may I suggest $Log(z)=ln\mid z\mid +iArg(z)$ where Arg is the principle argument. This will lead you to branched exponents. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-30T08:03:21.410" />
  <row Id="1214" PostTypeId="2" ParentId="768" CreationDate="2010-07-30T08:06:23.087" Score="1" ViewCount="0" Body="&lt;p&gt;(For an integer &amp;lambda; and a lattice polygon P) denote by &amp;lambda;P polygon P stretched by &amp;lambda; times. Then number N(&amp;lambda;P) of points inside the polygon &amp;lambda; is a quadratic polynomial in &amp;lambda; with leading coefficient S(P). This is the form of Pick's theorem that holds for any lattice (and obvious analogue works in any dimension&amp;nbsp;&amp;mdash; unlike usual Pick's formula that has no analogue in 3d even for the cubic lattice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the square lattice it yields ordinary Pick's theorem since for a parallelogram P on the square lattice $N(\lambda P)\approx\lambda^2(i+\frac{b}{2}-1)S(P)$ as &amp;lambda;&amp;rarr;&amp;infin; and general theorem follows from the theorem for parallelograms by additivity.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-07-30T08:06:23.087" />
  <row Id="1215" PostTypeId="1" CreationDate="2010-07-30T08:23:48.053" Score="3" ViewCount="296" Body="&lt;p&gt;Can we find f(x) given that 1-f(x) = f(-x) for all real x?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start by rearranging to: f(-x) + f(x) = 1. I can find an example such as f(x) = abs(x) that works for some values of x, but not all. Is there a method here? Is this possible?&lt;/p&gt;&#xA;" OwnerUserId="503" LastActivityDate="2010-07-30T17:13:44.180" Title="Solution to 1-f(x) = f(-x)" Tags="&lt;algebra-precalculus&gt;" AnswerCount="6" CommentCount="5" />
  <row Id="1216" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:32:04.523" Score="10" ViewCount="0" Body="&lt;p&gt;f(x)=1/2+(any odd function) (e.g. f(x)=1/2+x or, say, f(x)=1/2+99x^3+7x^5).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-07-30T08:32:04.523" CommentCount="1" />
  <row Id="1217" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:32:10.933" Score="2" ViewCount="0" Body="&lt;p&gt;Let f(x) = 1 if x&gt;0,   1/2 if x = 0,  and 0 if x&amp;lt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If x &gt; 0, then -x &amp;lt; 0 so that f(x)+f(-x)=1+0=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise with x&amp;lt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If x=0, then f(x)+f(-x)=(1/2)+(1/2)=1.&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2010-07-30T08:32:10.933" />
  <row Id="1218" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:56:49.720" Score="5" ViewCount="0" Body="&lt;p&gt;Clearly, we only have relations between f(x) and f(-x). The relation means that 0 has to have value 1/2. We can divide all non-zero real numbers into disjoint pairs of x and -x and define the function f on each pair separately. For each pair, f(x) can be given any value and then f(-x) has a single valid value. As mentioned by &lt;a href=&quot;http://math.stackexchange.com/questions/1215/solution-to-1-fx-f-x/1216#1216&quot;&gt;Grigory&lt;/a&gt;, the valid functions can be characterised as any odd function plus 1/2.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-30T09:47:26.877" LastActivityDate="2010-07-30T09:47:26.877" CommentCount="2" />
  <row Id="1219" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T08:59:03.443" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Usually&lt;/em&gt; simple problems like this ask you to find &lt;strong&gt;a&lt;/strong&gt; function that respects the condition, not &lt;strong&gt;all&lt;/strong&gt; of them. And (again) &lt;em&gt;usually&lt;/em&gt; you start by checking if a simple polynomial function of the first degree could be a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) = a * x + b&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) + f(-x) = 1 =&gt; a * x + b + a * (-x) + b = 1 =&gt; 2 * b = 1 =&gt; b = 1/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the condition is satisfied by any function of the type:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f(x) = a * x + 1/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditorDisplayName="" LastEditDate="2010-07-30T11:24:47.393" LastActivityDate="2010-07-30T11:24:47.393" />
  <row Id="1220" PostTypeId="1" AcceptedAnswerId="1232" CreationDate="2010-07-30T09:05:52.880" Score="0" ViewCount="107" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, a characteristic function completely determines the properties of a probability distribution. This means it must be unique. However, the definition give is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Char of X (t)=E[e^itX]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now e^iz repeats for every 2 pi increase in z. So how can it be unique?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-30T23:10:52.143" LastActivityDate="2010-07-30T23:10:52.143" Title="Uniqueness of Characterstic Functions in Probability" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="1221" PostTypeId="2" ParentId="1183" CreationDate="2010-07-30T09:35:56.080" Score="3" ViewCount="0" Body="&lt;p&gt;The only soundproof way to be sure to find the right result while dividing two complex numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a+bi)/(c+di)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is reducing it to a multiplication. The answer is of the form x+yi; therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c+di)(x+yi) = a+bi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you will end up with two linear equations, one for the real coefficient and another for the imaginary one. As Simon and Casebash already wrote, taking a square root leads to problems, since you cannot be sure which value must be chosen.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-30T09:35:56.080" />
  <row Id="1222" PostTypeId="1" AcceptedAnswerId="1243" CreationDate="2010-07-30T10:10:29.233" Score="4" ViewCount="117" Body="&lt;p&gt;I'm trying to read a document that applies Riemann-Roch left, right and center. I don't know this theorem or the theory it comes from so I need to build up a bit more background before I can tackle this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please recommend good books or (online) lecture notes which cover &quot;multi-valued functions&quot;, Riemann Surfaces and similar up to (at least) Riemann-Roch? (I also want to pick up a bit about modular forms and the relation between lattices and elliptic curves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've done a bit of complex calculus but it was all with analogy to real analysis so I am not sure that it will really give me any head start here. Also apologies for being so vauge with this but I don't know enough about this subject to be any more precise)&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-30T10:16:31.023" LastActivityDate="2010-08-14T09:53:42.810" Title="Books/Notes recommendation request: Multivalued functions/Riemann surfaces" Tags="&lt;books&gt;&lt;reference-request&gt;&lt;learning&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="1223" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T10:53:45.523" Score="3" ViewCount="0" Body="&lt;p&gt;Consider instead the functions g that satisfy the identity -g(x)=g(-x) for all x.  If (x,g(x)) is a point of the function g, then (-x,-g(x)) is also a point (since g(-x)=-g(x)).  Therefore, every function g is symmetric when rotated by 180 degrees about the point (0,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do things change for the identity 1-f(x)=f(-x)?  We merely shift the point of symmetry to (0,1/2).  Here the point (x,f(x)) implies the point (-x,1-f(x))&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The function f satisfies the identity 1-f(x)=f(-x) for all real numbers x if and only if it is symmetric when rotated about the point (0,1/2) by 180 degrees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's going to be many of these functions; some of which will be polynomials, some of which will not.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-07-30T10:53:45.523" />
  <row Id="1224" PostTypeId="2" ParentId="97" CreationDate="2010-07-30T11:08:58.043" Score="6" ViewCount="0" Body="&lt;p&gt;I am assuming that you need the error function only for real values. For complex arguments there are other approaches, more complicated than what I will be suggesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going the Taylor series route, the best series to use is formula &lt;a href=&quot;http://www.math.ucla.edu/~cbm/aands/page_297.htm&quot; rel=&quot;nofollow&quot;&gt;7.1.6&lt;/a&gt; in Abramowitz and Stegun. It is not as prone to subtractive cancellation as the series derived from integrating the power series for e&lt;sup&gt;-x²&lt;/sup&gt;. This is good only for &quot;small&quot; arguments. For large arguments, you can use either the asymptotic series or the continued fraction representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, may I direct you to &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.90.6481&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://homepages.physik.uni-muenchen.de/~Winitzki/erf-approx.pdf&quot; rel=&quot;nofollow&quot;&gt;papers&lt;/a&gt; by S. Winitzki that give nice approximations to the error function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-07-30T11:08:58.043" CommentCount="3" />
  <row Id="1225" PostTypeId="2" ParentId="697" CreationDate="2010-07-30T11:18:20.117" Score="3" ViewCount="0" Body="&lt;p&gt;The usual cases treated apart from least squares are the one-norm and infinity-norm (Chebyshev) cases; they crop up in function approximation for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually both of these are &lt;a href=&quot;http://dx.doi.org/10.1007/BF02162565&quot; rel=&quot;nofollow&quot;&gt;solved via linear programming techniques&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-07-30T11:18:20.117" />
  <row Id="1227" PostTypeId="2" ParentId="220" CreationDate="2010-07-30T12:47:44.910" Score="2" ViewCount="0" Body="&lt;p&gt;As I understand it, you are given a sequence $(x_n)_{n\in\mathbb{N}}$ of input values from which you calculate a sequence $f_n$ that is given by the following recurrence relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{n+1} = (1-c)f_n + c\cdot x_{n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &lt;strong&gt;question&lt;/strong&gt; is: given a sine wave $x_n=\sin(\omega n)$, you assert that $f_n$ is also a sine wave and you want to know its amplitude in dependence on the frequency $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;: It's easier to calculate the frequency response with exponential functions instead of sine waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n = Ae^{i\omega n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_n = e^{i\omega n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f_{n+1} = e^{i\omega (n+1)} = e^{i\omega} e^{i\omega n} = e^{i\omega} f_n$, the recurrence relation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{i\omega} A e^{i\omega n} = (1-c)Ae^{i\omega n} + c e^{i\omega} e^{i\omega n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(\omega) = \frac{c}{1 - (1-c)e^{-i\omega}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the response for sine waves, you can represent the sine function as a linear combination of two exponential functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_n = \sin(\omega n) = \frac1{2i}(e^{i\omega n}-e^{-i\omega n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n = \frac1{2i}(A(\omega)e^{i\omega n}-\bar A(\omega)e^{-i\omega n}) = |A(\omega)| \sin(\omega n + \phi)$ where $A(\omega) =: |A(\omega)| e^{i\phi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the complex frequency response $A(\omega)$ encodes both the change in amplitude and the phase shift. I'll leave the exact calculation of $|A(\omega)|$ to you, so that you can get familiar with this method of calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without exponential functions, you'd have to use the addition and subtraction theorems for sine and cosine. Put differently, exponential functions provide a very convenient formulation of the trigonometric identities.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-07-30T15:43:29.187" LastActivityDate="2010-07-30T15:43:29.187" CommentCount="2" />
  <row Id="1228" PostTypeId="2" ParentId="83" CreationDate="2010-07-30T12:48:39.713" Score="2" ViewCount="0" Body="&lt;p&gt;Show them large buildings, tallest buildings, pyramids, visualize them about different structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explain importance of maths&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brain remember things easily when comparison is done.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2010-07-30T12:48:39.713" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="1229" PostTypeId="2" ParentId="1220" CreationDate="2010-07-30T13:15:03.083" Score="1" ViewCount="0" Body="&lt;p&gt;It's just a Fourier transform. E(x) is an integral over the probability distribution. The function is unique; if you focus on the value inside the expectation integral, that's not, but so what?&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-07-30T13:15:03.083" />
  <row Id="1230" PostTypeId="2" ParentId="1186" CreationDate="2010-07-30T13:38:00.860" Score="4" ViewCount="0" Body="&lt;p&gt;When squaring a number, break the calculation into three smaller calculations and add them using the &lt;a href=&quot;http://en.wikipedia.org/wiki/FOIL_method&quot; rel=&quot;nofollow&quot;&gt;FOIL method&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;302^2 = 300^2 + 2*(300*2) + 2^2.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-07-30T13:38:00.860" CommunityOwnedDate="2010-07-30T13:38:00.860" CommentCount="1" />
  <row Id="1231" PostTypeId="2" ParentId="220" CreationDate="2010-07-30T13:55:24.143" Score="2" ViewCount="0" Body="&lt;p&gt;I don't have enough mojo to comment on Greg's answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Greg made a silly calculational mistake: The transfer function $A(\omega)$ should be $c/(1-(1-c)e^{-i\omega})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What you want is the modulus of $A(\omega)$. Note that $\sin \omega n$ is precisely the imaginary part of $e^{i\omega n}$. Because the relation between input and output is linear, the response to $\sin\omega n$ will be the imaginary part of $A(\omega)e^{i\omega n}$. That's going to be a sinusoid with some shifting and the amplitude $|A(\omega)|$. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+|.5/%281-.5+exp%28-ix%29%29|&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a plot for $c=1/2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To read more about this sort of things, google &quot;IIR filter&quot; or &quot;infinite impulse response&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-30T13:55:24.143" CommentCount="3" />
  <row Id="1232" PostTypeId="2" ParentId="1220" CreationDate="2010-07-30T15:05:51.883" Score="3" ViewCount="0" Body="&lt;p&gt;I thought at first you were asking how the characteristic function can be unique. There's no issue here, because e^ix is a well-defined function: for a given value of x, there is a unique value of e^ix. And so the expectation (an integral) has a unique value as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, it appears you're asking how it's possible, considering that e^ix is not &lt;em&gt;injective&lt;/em&gt; (i.e., multiple x can have the same value of e^ix), that the original distribution can be completely recovered from the characteristic function. The answer to &lt;em&gt;that&lt;/em&gt; is that you're probably missing the &quot;t&quot; in the expression: the characteristic function is a function of t, and although for a &lt;em&gt;given&lt;/em&gt; value of t (e.g. t=1), the random variables $X$ and $X+2k\pi/t$ &lt;em&gt;would&lt;/em&gt; have the same value of the characteristic function at that point, they would have different values at other t. So the characteristic &lt;em&gt;functions&lt;/em&gt; are different, and yes, the distribution can be completely recovered from the characteristic function.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-30T15:05:51.883" CommentCount="3" />
  <row Id="1233" PostTypeId="2" ParentId="967" CreationDate="2010-07-30T15:07:00.430" Score="2" ViewCount="0" Body="&lt;p&gt;I like &lt;a href=&quot;http://www.cs.kent.ac.uk/people/staff/sjt/TTFP/&quot; rel=&quot;nofollow&quot;&gt;Type Theory and Functional Programming&lt;/a&gt; by Simon Thompson.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-30T15:07:00.430" />
  <row Id="1234" PostTypeId="1" CreationDate="2010-07-30T16:06:46.553" Score="1" ViewCount="265" Body="&lt;p&gt;This was really inspired by Solitaire, but a few people reacted with ``oh, it's like the towers of Hanoi, isn't it?'' so I'll try to pose the problem in terms of discs here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start. There are n disks on the real line, one of size 1 at position $x_1$, one of size 2 at position $x_2$, ..., and one of size n at position $x_n$. Your goal is to make a tower with all n discs, consuming as little energy as possible in the process. You are allowed to move a tower whose base is a disk of size k only on top of the disk with size k+1 (which may be the top of another mini-tower). The energy you consume to perform such a move is the distance traveled by the moved mini-tower. For example, the energy consumed by the first move is $|x_k-x_{k+1}|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you'd like to write a program that tells you whether the energy you have is enough to perform the task. It just needs to say Y or N. (If the answer is Y, then clearly the list of moves is proof enough that the answer is correct, so the problem is NP. If the answer is N, there's no point in even attempting the task---you are too tired.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the fastest such program you can find? Is the problem NP-complete? If it helps, consider a simplifications that restrict $x_k$'s to be rational, integers, integers in a certain range, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an upper bound: $O(n2^n)$. Represent the initial state by the list $x_1$, $x_2$ ,... ,$x_n$. A move affects the state by deleting a number in this list and the energy consumed by the move is the absolute difference between the deleted number and the number coming after. Clearly there are $2^n$ states with at most $n$ moves each. (See &lt;a href=&quot;http://rgrig.blogspot.com/2010/07/solitaire.html&quot; rel=&quot;nofollow&quot;&gt;my blog post&lt;/a&gt; for an example run of this algorithm if it's not clear. The description there is in terms of cards.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Some of the comments below refer to older versions of the questions. See the history of edits if they seem confusing.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-07-31T11:47:02.263" LastActivityDate="2010-07-31T12:15:45.357" Title="Is this version of the Hanoi towers problem NP-complete?" Tags="&lt;algorithms&gt;&lt;np-complete&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="1235" PostTypeId="2" ParentId="83" CreationDate="2010-07-30T16:09:24.400" Score="1" ViewCount="0" Body="&lt;p&gt;Check out the book «&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821804308&quot; rel=&quot;nofollow&quot;&gt;Mathematical circles&lt;/a&gt;».  It's written by some Russian mathematicians who used to run &quot;math circles&quot; for middle and high school students and has a lot of interesting material.&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-07-30T16:09:24.400" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="1236" PostTypeId="1" CreationDate="2010-07-30T16:15:38.950" Score="9" ViewCount="217" Body="&lt;p&gt;How have experts estimated the amount of oil that was shooting out of that pipe in the Gulf? I bet there's some neat math or physics involved here, and some interesting assumptions considering how little concrete data are available.&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-07-30T17:17:46.717" LastActivityDate="2010-08-25T18:11:37.750" Title="How do you estimate the flow rate of one fluid into another like the Deep Horizon Oil leak?" Tags="&lt;applications&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="1237" PostTypeId="1" CreationDate="2010-07-30T16:21:12.173" Score="0" ViewCount="221" Body="&lt;p&gt;A fun question I ask students or interviewees (in engineering) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;This is not my question, this is an example:&lt;/strong&gt;&lt;br&gt;&#xA;  Using only what you know now, how many&#xA;  cans of soda would you estimate are&#xA;  produced per day (on average) in the&#xA;  United States?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For this question, the result doesn't matter so much as the process you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this theme of estimation, what's your favorite question?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T19:05:37.770" LastActivityDate="2010-08-16T19:05:37.770" CommunityOwnedDate="2010-07-30T16:21:12.173" ClosedDate="2010-07-31T17:35:23.297" Title="What is your favorite estimation exercise? " Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="1238" PostTypeId="2" ParentId="519" CreationDate="2010-07-30T16:24:10.423" Score="1" ViewCount="0" Body="&lt;p&gt;Another approach, which can't work in general (see Noah's answer), but will surely work in this case, is to find a normal form for each element of the group, and then see whether there are finitely or infinitely many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, that means imagining a word in x and y, and then applying the relations as much as possible to simplify it, and then trying to figure out (and prove!) what the possible different &quot;irreducible words&quot; (i.e. words that can no longer be simplified) are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the first thing one would note is that x can only appear to the 1st power&#xA;(since any higher power can be simplified using x^2 = 1), while y can only appear to the powers +1 or -1 (for the same reason).  Also, we can't have too many expressions of the&#xA;form  xy  or  yx  in a row, because of the third relation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can keep going like this.  I didn't, but what I imagine is that one can have &#xA;expressions of the form  x y x y^{-1} x y x y^{-1}   ...   that are arbitrarily long,&#xA;and inequivalent, explaining the infinite order of the group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shouldn't be so hard&#xA;to settle the question from this point of view by sitting down with pencil and paper and just playing around with different words to get a feel for what kinds of reductions can&#xA;take place.  (In geometric arguments like Grigory's, one uses a geometric context, such as&#xA;an action on a hyperbolic tiling, as a more conceptual way of understanding the relations and distinguishing inequivalent words.  But in this case I'm sure it won't be hard to see&#xA;everything directly from the presentation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added after rereading the question: what I am suggesting is precisely that even when the answer might be infinite, you can still hope to find a coset enumeration, at least in a case&#xA;like this with relatively simple relations.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-07-30T16:33:34.130" LastActivityDate="2010-07-30T16:33:34.130" CommentCount="2" />
  <row Id="1240" PostTypeId="2" ParentId="1237" CreationDate="2010-07-30T16:39:31.703" Score="15" ViewCount="0" Body="&lt;p&gt;&quot;How many estimation questions are asked in interviews across the world during a typical 24h period?&quot;&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-30T16:39:31.703" CommunityOwnedDate="2010-07-30T16:39:31.703" CommentCount="2" />
  <row Id="1241" PostTypeId="1" AcceptedAnswerId="1248" CreationDate="2010-07-30T17:09:32.753" Score="3" ViewCount="196" Body="&lt;p&gt;The generalized mean (power mean) with exponent $p$ of $n$ numbers $x_1, x_2, \ldots, x_n$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar x = (\frac{1}{n} \sum x_i^p)^{1/p}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the harmonic mean, arithmetic mean, and root mean square for $p = -1$, $p = 1$, and $p = 2$, respectively. Also its limit at $p = 0$ is equal to the geometric mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When should the different means be used? I know harmonic mean is useful when averaging speeds and the plain arithmetic mean is certainly used most often, but I've never seen any uses explained for the geometric mean or root mean square. (Although standard deviation is the root mean square of the deviations from the arithmetic mean for a list of numbers.)&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditorDisplayName="" LastEditDate="2010-07-30T21:34:40.073" LastActivityDate="2010-07-30T23:46:52.207" Title="Which average to use? (RMS vs. AM vs. GM vs. HM)" Tags="&lt;average&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="1242" PostTypeId="2" ParentId="1215" CreationDate="2010-07-30T17:13:44.180" Score="0" ViewCount="0" Body="&lt;p&gt;WolframAlpha provides a solution to this (and many other) recurrence equations:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=1-f(x)+%3D+f(-x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=1-f(x)+%3D+f(-x&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-07-30T17:13:44.180" CommentCount="1" />
  <row Id="1243" PostTypeId="2" ParentId="1222" CreationDate="2010-07-30T17:14:23.383" Score="1" ViewCount="0" Body="&lt;p&gt;The book by Otto Forster on Riemann Surfaces is pretty good. I never finished reading it myself, but it covers things like Riemann-Roch and Abel's theorem from a sheafish viewpoint. In particular, the proof of Riemann-Roch is analogous to the one in Hartshorne; it follows from the Serre duality theorem and an inductive argument. Learning about sheaves is definitely a plus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there's a book by Springer, though the level is a bit more elementary.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T17:14:23.383" />
  <row Id="1244" PostTypeId="2" ParentId="1241" CreationDate="2010-07-30T17:24:12.590" Score="2" ViewCount="0" Body="&lt;p&gt;I admit I don't really know what type of answer your looking for. So, I'll say something that might very well be entirely irrelevant for your purposes but which I enjoy. At least, it'll provide some context for the power means you asked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These generalized power means are basically the discrete (finitary) analogs of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_norm#lp_spaces&quot; rel=&quot;nofollow&quot;&gt;L^p norms&lt;/a&gt;. So, for instance, it's with these norms that you prove (using, say, elementary calculus) the finitary version of &lt;a href=&quot;http://en.wikipedia.org/wiki/Holder%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Holder's inequality&lt;/a&gt;, which is really important in analysis, because it leads (via a limiting argument) to the more important fact that $L^p$ and $L^q$ spaces (which are continuous analogs of these finitary $l^p$ spaces) are dual for $p,q$ conjugate exponents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This duality is really important: one example is that if you are trying to prove something about the $L^p$ spaces that is preserved under duality, you just have to restrict yourself to the case $1 \leq p \leq 2$. The theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_integral&quot; rel=&quot;nofollow&quot;&gt;singular integral operators&lt;/a&gt; provides examples of this: basically, it's easy to prove they are bounded (i.e., reasonably well-behaved) for $p=2$ by Fourier analysis; you prove that they're &quot;weak-bounded&quot; on $L^1$ (in some sense which I won't make precise); then you apply to general results on interpolation to get boundedness in the range $1-2$; finally, this duality operation gives it for $p&gt;2$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;http://en.wikipedia.org/wiki/Root_mean_square_speed&quot; rel=&quot;nofollow&quot;&gt;root-mean-square speed&lt;/a&gt; is used to define temperature in physics.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T17:24:12.590" />
  <row Id="1245" PostTypeId="2" ParentId="632" CreationDate="2010-07-30T17:26:01.447" Score="4" ViewCount="0" Body="&lt;p&gt;Concerning the physical meaning, I take it that $f_1$ and $f_2$ represent the fractions of the two phases in the alloy (this implies $f_1 + f_2 = 1$). I imagine $x_1$ and $x_2$ to correspond to an intensional variable like pressure, whose average $f_1x_1 + f_2x_2 =: \bar x$ is held constant in the experiment. Now, the alloy minimizes the free energy under these constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &lt;strong&gt;answer&lt;/strong&gt; your first question: there is a geometric reason why the solution is the common tangent to $G_1$ and $G_2$ in the case of two dimensions. Namely, the fractions $f_1$ and $f_2$ are exactly the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barycentric_coordinates_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Barycentric coordinates&lt;/a&gt; of the average $\bar x$ sitting between $x_1$ and $x_2$. In particular, the value of the total energy $\tilde G$ is the height of the line drawn between $(x_1,G_1(x_1))$ and $(x_2,G_2(x_2))$ evaluated at $\bar x$. Here's a sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i30.tinypic.com/2l2fk7.jpg&quot; alt=&quot;Free energy of an alloy of two compounds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this picture, it is clear that if this line is &lt;em&gt;not&lt;/em&gt; tangent to both $G_1$ and $G_2$, then you can move it a bit so that the value at $\bar x$ will decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your &lt;strong&gt;second question&lt;/strong&gt;, the geometry readily extends to higher dimensions. For instance, for an alloy of three compounds, one has to consider the triangle enclosed by the three points $(x_1,G_1(x_1))$, $(x_2,G_2(x_2))$ and $(x_3,G_3(x_3))$. The situation is a bit degenerate here, any point inside this triangle whose first coordinate is $\bar x$ represents a valid value of $\tilde G$. Of these, nature will choose the smallest one. Consequently, the lower side of the triangle has to be tangent to two of the individual free energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i28.tinypic.com/335dwrl.jpg&quot; alt=&quot;alloy of three compounds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar reasoning applies when the variable $x$ is not just a number, but, say, a pair of numbers, then we're dealing with a plane tangent to three individual Gibbs functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While not terribly useful in this case, there is also a very general geometric interpretation of the method of Lagrange multipliers. Namely, consider a goal function $f$ and a holonomic constraint $g$. Then, the Euler-Lagrange-equations give $\nabla f = \lambda \nabla g$ which means that $f$ changes only in directions orthogonal to the surface $g$. But since we're confined to the surface $g$, this must be an extremum. Wikipedia has a &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aLagrangeMultipliers2D.svg&quot; rel=&quot;nofollow&quot;&gt;picture&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-08-01T12:53:29.397" LastActivityDate="2010-08-01T12:53:29.397" CommentCount="4" />
  <row Id="1246" PostTypeId="2" ParentId="595" CreationDate="2010-07-30T17:31:07.103" Score="3" ViewCount="0" Body="&lt;p&gt;Computing the determinant and Gaussian elimination are both fine if you are using exact computations, for instance if the entries of your matrix are rational numbers and you are using only rational numbers during the computations. The disadvantage is that the numerator and denominator can get very large indeed. So the number of operations may indeed be O(n&lt;sup&gt;2.376&lt;/sup&gt;) or O(n&lt;sup&gt;3&lt;/sup&gt;), but the cost of every addition and multiplication gets bigger as n grows because the numbers get bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an issue if you are using floating point numbers, but then you have the problem that floating point computations are not exact. Some methods are more sensitive to this than others. In particular, checking invertibility by computing the determinant is a bad idea in this setting. Gaussian elimination is better. Even better is to use the singular value decomposition, which will be treated towards the end of the MIT course.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-07-30T17:31:07.103" />
  <row Id="1247" PostTypeId="2" ParentId="1237" CreationDate="2010-07-30T17:33:54.413" Score="0" ViewCount="0" Body="&lt;p&gt;I like &quot;how many licks does it take to get to the center of a tootsie roll pop&quot;. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-30T17:33:54.413" CommunityOwnedDate="2010-07-30T17:33:54.413" />
  <row Id="1248" PostTypeId="2" ParentId="1241" CreationDate="2010-07-30T17:52:59.590" Score="2" ViewCount="0" Body="&lt;p&gt;One possible answer is for defining unbiased estimators of probability distributions. Often times you want some transformation of the data that gets you closer to, or exactly to, a normal distribution. For example, products of lognormal variables are again lognormal, so the geometric mean is appropriate here (or equivalently, the additive mean on the natural log of the data). Similarly, there are cases where the data are naturally reciprocals or ratios of random variables, and then the harmonic mean can be used to get unbiased estimators. These show up in actuarial applications, for example.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-07-30T17:52:59.590" />
  <row Id="1249" PostTypeId="1" CreationDate="2010-07-30T17:55:21.223" Score="6" ViewCount="258" Body="&lt;p&gt;I am in the process of trying to learn algebraic geometry via schemes and am wondering if there are simple motivating examples of why you would want to consider these structures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my biggest issue is the following: I understand (and really like) the idea of passing from a space to functions on a space.  In passing from $k^n$ to $R:=k[x_1,\ldots,x_n]$, we may recover the points by looking at the maximal ideas of $R$.  But why consider $\operatorname{Spec} R$ instead of $\operatorname{MaxSpec} R$?  Why is it helpful to have non-closed points that don't have an analog to points in $k^n$?  On a wikipedia article, it mentioned that the Italian school used a (vague) notion of a generic point to prove things.  Is there a (relatively) simple example where we can see the utility of non-closed points?&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-07-30T23:51:05.600" Title="Motivating Example for Algebraic Geometry/Scheme Theory" Tags="&lt;algebraic-geometry&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="1250" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T18:07:36.283" Score="5" ViewCount="0" Body="&lt;p&gt;To start off,  the &lt;a href=&quot;http://sbseminar.wordpress.com/2009/08/06/algebraic-geometry-without-prime-ideals/&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; at Sbseminar has comments from lots of people who actually know algebraic geometry, and if anything I say contradicts something they say, please trust them and not me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason is that you lose the functoriality of $Spec$ if you stick to $MaxSpec$: the inverse image of a maximal ideal is not necessarily maximal. Nevertheless, if you stick to schemes of finite type over a field, this is true (it's basically a version of the Nullstellensatz).  In particular, in Serre's FAC paper he defines a &quot;variety&quot; by gluing together regular affine algebraic sets in the sense of classical algebraic geometry.&#xA;But this is less general.  One natural example of a scheme which is not of finite type over a field is simply $Spec \mathbb{Z}$. Then given a scheme $X$ over this (well, admittedly every scheme $X$ is a scheme over $Spec \mathbb{Z}$ in a canonical way), the fibers at the non-closed points of $Spec \mathbb{Z}$ are still interesting and basically amount to studying polynomial equations over finite fields (when $X \to \mathbb{Z}$ is of finite type).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a (simple) example of how generic points can be used, one can prove that a coherent sheaf on a noetherian integral scheme is free on a dense open subset. Why? Because it must be free at the generic point (since the local ring there is a field), and it is a general fact that two coherent sheaves whose stalks are isomorphic are isomorphic in a neighborhood. (This is true actually for sheaves of finite presentation over a ringed space.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T18:07:36.283" CommentCount="2" />
  <row Id="1252" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T19:09:42.600" Score="4" ViewCount="0" Body="&lt;p&gt;I think I can at least convince you that it's a good idea to work with non-reduced rings, which aren't really captured by their maximal ideals.  The idea is that you get more functors.  Just as $\text{Hom}(-, \mathbb{C})$ is the functor which sends a finitely-generated domain over $\mathbb{C}$ to its set of $\mathbb{C}$-points, it turns out that $\text{Hom}(-, \mathbb{C}[x]/x^2)$ sends a finitely-generated domain over $\mathbb{C}$ to its set of $\mathbb{C}$-points &lt;em&gt;together with a choice of tangent vector.&lt;/em&gt;   In fact this is one way to define the Zariski tangent space.  So, on the scheme side, it's a good idea to study morphisms from $\text{Spec } \mathbb{C}[x]/x^2$ to your scheme, and you can't do this if you identify $\mathbb{C}[x]/x^2$ with its set of $\mathbb{C}$-points.  (You can't even do this if you identify $\mathbb{C}[x]/x^2$ with its prime ideals; you really need the entire structure sheaf.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A really nice application: an algebraic definition of the Lie algebra of an algebraic group.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-30T19:09:42.600" />
  <row Id="1253" PostTypeId="1" AcceptedAnswerId="1255" CreationDate="2010-07-30T19:20:49.810" Score="0" ViewCount="67" Body="&lt;p&gt;In my geometry class last year I remember putting down the statement in a column proof &quot;That all isosceles are always and only similar to other isosceles&quot;. I do not remember what I was trying to prove. But, I do remember that I was stressed and that was the only thing I could think of and made a guess thinking I would probably get the proof wrong on my test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Funny enough though, I didn't get the proof wrong and I was wondering if anyone could show a proof as to why this would be true. I mean I makes sense but, I do not see any way to prove it. Could you please explain how this is true?&lt;/p&gt;&#xA;" LastActivityDate="2010-07-30T19:41:28.683" Title="Are isosceles always and only similar to other isosceles?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1254" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T19:24:44.747" Score="3" ViewCount="0" Body="&lt;p&gt;See my answer &lt;a href=&quot;http://math.stackexchange.com/questions/942/meaning-of-closed-points-of-a-scheme/984#984&quot;&gt;here&lt;/a&gt; for a brief discussion of how points that are closed in one optic (rational solutions to a Diophantine equation, which are closed points on the variety over Q attached to the Diophantine equation) become non-closed in another optic (when we clear denominators and think of the Diophantine equation as defining a scheme over Z).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of rings (and connecting to Qiaochu's answer), under the natural map&#xA;Z[x_1,...,x_n] --&gt; Q[x_1,...,x_n], the preimage of maximal ideals are prime, but not maximal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These examples may give impression that non-closed points are most important in arithmetic situations, but actually that is not the case.  The ring C[t] behaves much like Z, and so one can have the same discussion with Z and Q replaced by C[t] and C(t).  Why would one do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, suppose you have an equation (like y^2 = x^3 + t) which you want to study, where you think of t as a parameter.  To study the generic behaviour of this equation, you can think of it as a variety over C(t).  But suppose you want to study the geometry for one particular value of t_0 of t.  Then you need to pass from C(t) to C[t], so that you can apply the &#xA;homomorphism C[t] --&gt; C given by t |--&gt; t_0  (specialization at t_0).  This is completely&#xA;analogous to the situation considered in my linked answer, of taking integral solutions to a a Diophantine equation and then reducing them mod p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the upshot?  Basically, any serious study of varieties in families (whether arithmetic families, i.e. schemes over Z, or geometric families, i.e. parameterized families&#xA;of varieties) requires scheme-theoretic techniques and the consideration of non-closed points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, serious such studies were made by the Italian geometers, by Lefschetz, by Igusa, by Shimura, and by many others before Grothendieck's invention of schemes, but the whole point of schemes is to clarify what came before and to give a precise and workable theory that encompasses all of the contexts considered in the &quot;old days&quot;, and is also more systematic and more powerful than the older techniques.) &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-07-30T23:51:05.600" LastActivityDate="2010-07-30T23:51:05.600" />
  <row Id="1255" PostTypeId="2" ParentId="1253" CreationDate="2010-07-30T19:25:36.833" Score="4" ViewCount="0" Body="&lt;p&gt;Re-reading your question, I see two possible interpretations of your statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First (and my original answer), &quot;If &amp;#x25B3;ABC is isosceles and &amp;#x25B3;ABC~&amp;#x25B3;DEF, then &amp;#x25B3;DEF is isosceles.&quot;  Two triangles are similar if and only if the three angles of one are congruent to the three angles of the other.  Since a triangle is isosceles if and only if two of its angles are congruent, if a triangle is similar to an isosceles triangle, then it will also have two congruent angles and must be isosceles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, &quot;If &amp;#x25B3;ABC and &amp;#x25B3;DEF are isosceles, then they are similar.&quot;  This is not true.  Suppose one triangle has angles with measures 20&amp;deg;, 20&amp;deg;, and 140&amp;deg; and another other triangle has angles with measures 85&amp;deg;, 85&amp;deg;, and 10&amp;deg;.  Both triangles are isosceles (since within each triangle, there is a pair of congruent angles), but the triangles are not similar (because the angles of one are not congruent to the angles of the other).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-30T19:41:28.683" LastActivityDate="2010-07-30T19:41:28.683" CommentCount="4" />
  <row Id="1256" PostTypeId="2" ParentId="20" CreationDate="2010-07-30T19:42:21.053" Score="0" ViewCount="0" Body="&lt;p&gt;There are many types of numbers, though the natural numbers, the integers, the rational, the decimal, the real, and the complex form a nice self-complete expository whole. Hopefully, the following block(s) of text aren't too poorly formatted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The set of &lt;strong&gt;natural numbers&lt;/strong&gt; consists of numbers with which we count {0,1,2,3,...}. As noted in some of the other answers, some people think that 0 is not a natural number (see &lt;a href=&quot;http://www.marriedtothesea.com/031107/zero.gif&quot; rel=&quot;nofollow&quot;&gt;one of my desktop backgrounds&lt;/a&gt;). Whether or not it is, it's a matter of taste. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not a matter of taste are the defining properties of the natural numbers. In particular, there is a (binary) operation called +, which takes two numbers a and b and spits out a third number a+b, which is the sum of a and b. It satisfies the usual properties that you would expect from counting: it is commutative (the order of the summands doesn't matter, i.e. a+b=b+a) and associative (the order in which you add summands to each other doesn't matter, i.e. (a+b)+c=a+(b+c), so you can always write a+b+c for a specific number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an order relation &amp;lt; such that for any two different numbers a and b either a&#xA;&#xA;&lt;p&gt;The order plays nice with addition in that a &amp;lt; b implies a+c &amp;lt; b+c. We also have a cancellation property that if a &amp;lt; c, then there exists a number b such that a+b=c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order also satisfies one unobvious property called the well-ordering principle and states that if you take any collection of natural numbers, there is a smallest one, i.e. a number that is smaller than all the other numbers in the collection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The well-ordering principle, together with cancellation, implies that there is a smallest number, and that any two consecutive numbers differ by the same number. From this it follows that either the smallest number is 0 (in the sense that 0 is the number such that 0+a=a+0=a) or that it is 1, where 1 is the common difference between any two consecutive numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing one or the other of the two possibilities defines the natural numbers uniquely as either {0,1,2,...} or {1,2,...}. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fun exercise to show that the above axioms are equivalent to the axioms of Peano (note that the axioms of Peano state nothing about addition, so in fact BOTH {0,1,2,...} and {1,2,...} satisfy the axioms; how we define addition with respect to the smallest element is what specifies one of the two sets)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The set of &lt;strong&gt;integers&lt;/strong&gt;, also known as whole numbers, is {..., -3, -2, -1, 0, 1, 2, 3, ...}.&#xA;It arises as we try to &quot;complete&quot; the arithmetic of the integers in the following sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any two numbers a &amp;lt; b we have a number b-a such that (b-a)+a=b, thus we have subtraction of a smaller number from a bigger number. We do not have such numbers b-a if a&gt;b (i.e. there is no natural number which when you add a to you get b if a&gt;b), but from experience we see that numbers such as ... -3, -2, -1 &#xA;seem to be sensical to add and subtract as we try to solve equations and whatnot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while numbers such as b-a when a &amp;lt; b are defined, the same number could be represented in different ways. For example, 2 is both 5-3 and 8-6. So we need to answer the question of when b-a=d-c for a &amp;lt; b, c &amp;lt; d, and by algebra the answer is that they're equal whenever b+c=d+a. Hence, we can define all the integers, positive and negative, by taking them to be pairs of natural numbers (a,b) with the caveat that (a,b)=(c,d) whenever a+c=b+c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the integer -2 consists of (among others) the pairs (5,3), (6,8), (0,2). This is all seems like much formal ado about intuitive nothings, but that is only because we can represent any integer as either the pair (a,0), which we write simply as a, which we write as -a, and think of as (a,0) is the integer which when you add a to you get 0, or as (0,a), which we think of as the integer to which which you add 0 you get the natural number a. Hence what we actually do arithmetic with are the pairs {... (3,0), (2,0), (1,0), (0,0), (0,1), (0,2), (0,3),...}, but the fact that we only need to throw in a negative sign to make the arithmetic meaningful is precisely because there exists these special representative pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: define the addition and order on the pairs of natural numbers for the above to work (we lose of course the property that any set of integers has a smallest integers [e.g. the set of negative integers], but the principle still holds for sets of positive integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The &lt;strong&gt;rational numbers (or fractions)&lt;/strong&gt; are obtained by the exact same process of completion as above, except with respect to multiplication. the fraction a/b, where a and b are integers corresponds to the number which when multiplied by b gives you a. Note that a/b=c/d if and only if ad=bc, which mirrors the fact that a-b=c-d if and only if a+b=b+c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The real numbers are best understood as coming from &lt;strong&gt;decimals (or decimal expansions)&lt;/strong&gt;, so what are decimals and where do they come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rational numbers as fractions are hard to compare to one another. Is 7/5 bigger or smaller than 4/3? The answer is yes, becayse 7*3&gt;3*5. In general if you have a/b and c/d, we have a/b &amp;lt; c/d, a/b=c=d or a/b&gt;c/d if correspondingly adbc. But that's a tedious process, is there some way of writing them down so that it is clearer who's bigger than who? Better, can the process of writing them down be easier than the process of checking who's bigger than who by cross-multiplication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes, and we only need consider the usual way in which we write natural numbers.  We can express the number 1729 compactly based on the distributive properties of addition and exponentiation: 1729=1000+700+20+9=10^3+7*10^2+2*10+9, where the digits 1,2,3,4,5,6,7,8,9,10 are the first ten successors of zero, and zero is denoted by 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can write rational numbers with denominator 10^n as a.b=a+b/10^n with a and b are natural numbers and b is smaller than 10^n. For example, 1729/100 can be written as 17.29 where 17 and 29 are natural numbers with 29&amp;lt;100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for any rational number c/d there exists a rational number with denominator closest 10^n that's closest to and smaller than c/d (this is intuitively obvious and not that hard to formalize). Then we can say that c/d ~ a.b where a+b/10^n is that rational number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the closest number with denominator 10 to 1/3 is 3/10=.3, with denominator 100 is 33/100=.33, and so on. Then that the infinite string 0.3333... equals 1/3 means that the closest rational number less than or equal to 1/3 with denominator 10^n is 333...3/10^n where you have n 3's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, one can prove that rational numbers correspond have periodic expansions, i.e. that the string a.b has a repeating sequence of digits from some point onward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(even better, we can compute decimal expansions by the usual process of long division by just adding a decimal point, and we can efficiently compare numbers by looking at the expansion until the first point of difference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Real numbers come from taking arbitrary infinite sequences of digits as decimals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From above, we know that decimals actually encode sequences of rational numbers with denominators 1,10, 10^2, 10^3, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When should two such sequences be equivalent? Consider the standard question of whether .999...=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decimal expansion of 1 is just 1.000... because the sequence of rational numbers with denominators 10, 10^2, 10^3, ... has to be a sequence of rational numbers less than or equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we drop the requirement that the numbers be less than or equal to 1, keep the requirement that the sequence of rational numbers (approximations really) is non-decreasing, and say allow the rational number to be either the closest number less than 1 or 1 itself, then we obtain the other expansion 1=.999...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a measuring instrument that was accurate to n decimal places, then upon measuring .999... and 1.000... you would always get .999...9 with n-2 9s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it turns out that with finite precision measuring, you can't tell the difference between certain decimal expansions, so we might as way say those two decimal expansions are equal. This is the process of completion by Cauchy sequences put into words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And those expansions then are the real numbers, and they have addition, multiplication, subtraction and division and ordering just like the rational numbers, except that they're complete in that every sequence of real numbers converges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Complex numbers&#xA;It turns out that every polynomial with real coefficients of degree &gt;2 has a square root. -1 does not have a a real square root, so we wish to throw a root of -1 in to algebraically complete the real numbers, and thus get the complex numbers. The algebraic construction is best done by explaining concepts such as rings and ideals and quotients and homomorphisms, which is too much machinery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, consider Euclidean geometry. It is a(n advanced) fact that you can coordinize it with real numbers, i.e. you can represent points by pairs of real numbers (a,b), lines by the solutions of the two variable equation ax+by=c, lengths given by square root of ((a-c)^2+(b-d)^2), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding pairs (a,b) and (c,d) gives you (a,b)+(c,d) so addition of pairs corresponds to translations. Another way of interpreting a pair of (a,b) is by interpreting it in polar form (r, theta) where r is the distance of (a,b) form the origin and theta is the angle from the x-axis to (a,b). Then we can think of (a,b)~(r,theta) as a dilation and rotation with center (0,0) and sending (1,0) to (a,b), i.e. a dilation by a factor of r and  a rotation by an angle of theta. But this then defines a multiplication of vectors, which turns out to distribute over addition. Clearly we have inverse rotations/dilations and inverse translations, giving us subtraction and division. Finally, we can identify the real numbers with pairs (r,0) which are only a dilation without any rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These pairs are then our complex numbers and you can deduce all their properties from the above description. Note that the vector (0,1) multiplied by (0,1) gives (-1,0) since it corresponds to a rotation of 90 followed by a rotation of 90, which is a rotation of 180 of (1,0) which equals (-1,0). See Qiaouchu's answer for more details.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-07-30T19:42:21.053" CommentCount="1" />
  <row Id="1257" PostTypeId="1" AcceptedAnswerId="1259" CreationDate="2010-07-30T19:59:03.430" Score="6" ViewCount="452" Body="&lt;p&gt;I've solved for it making a computer program, but was wondering there was a mathematical equation that you could use to solve for the nth prime?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T17:39:07.933" Title="Is there a known mathematical equation to find the nth prime?" Tags="&lt;prime-numbers&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="1258" PostTypeId="2" ParentId="1257" CreationDate="2010-07-30T20:06:31.370" Score="2" ViewCount="0" Body="&lt;p&gt;No such formula is known, but there are a few that give impressive results. A famous one is Euler's:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(n) = n^2 − n + 41&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which yields a prime for every natural number lower than 41, though not necessarily the &lt;code&gt;n&lt;/code&gt;th prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See more &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-07-30T20:06:31.370" />
  <row Id="1259" PostTypeId="2" ParentId="1257" CreationDate="2010-07-30T20:10:21.033" Score="12" ViewCount="0" Body="&lt;p&gt;No, there is no formula that gives the nth prime, except artificial ones you can write that are basically equivalent to &quot;the nth prime&quot;. But if you only want an approximation, the nth prime is roughly around $n \ln n$ (or more precisely, near the number m such that $m/\ln m = n$) by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;prime number theorem&lt;/a&gt;. In fact, we have the following asymptotic bound on the nth prime $p_n$:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$n \ln n + n(\ln\ln n - 1) &amp;lt; p_n &amp;lt;  n \ln n + n \ln \ln n$ for n≥6  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_primes#Prime_sieves&quot; rel=&quot;nofollow&quot;&gt;sieve&lt;/a&gt; within this range if you want the $n$th prime. [Edit: There are better ideas than a sieve, see &lt;a href=&quot;#4314&quot; rel=&quot;nofollow&quot;&gt;the answer by Charles&lt;/a&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entirely unrelated: if you want to see formulae that generate a lot of primes (not the nth prime) up to some extent, like the famous f(n)=n&lt;sup&gt;2&lt;/sup&gt;-n+41, look at the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;formula for primes&lt;/a&gt;, or Mathworld for &lt;a href=&quot;http://mathworld.wolfram.com/PrimeFormulas.html&quot; rel=&quot;nofollow&quot;&gt;Prime Formulas&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-09-09T07:30:36.230" LastActivityDate="2010-09-09T07:30:36.230" CommentCount="2" />
  <row Id="1260" PostTypeId="1" AcceptedAnswerId="1284" CreationDate="2010-07-30T20:17:02.063" Score="-1" ViewCount="202" Body="&lt;p&gt;I'm looking for a tool for drawing vectors (without using Microsoft paint, Paint.net etc.) to paste into a Google docs document.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest an easy solution? I need up to 2 column/row vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update:&lt;/em&gt; This is the format I'm after (not arrows, but just the items):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/b/6/4/b6459d183a924dad2e5067a15fd45f2c.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T19:05:01.237" LastActivityDate="2010-08-16T19:05:01.237" ClosedDate="2010-08-02T14:27:52.290" Title="Is there a tool for drawing column and row vectors?" Tags="&lt;math-software&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="1261" PostTypeId="2" ParentId="548" CreationDate="2010-07-30T20:19:41.947" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT:I botched my original answer, here's what I actually meant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fractions are equivalence classes of pairs of integers subject to (a,b)=(c,d) iff ad=bc, with addition and multiplication extended from {(a,1)} as a copy of the integers by multiplication defined component-wise in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we allow pairs of the form (a,0) for some integer a, we have with the multiplication axioms (a,0)=(b,0) for any a and b, and in particular all of them coincide with (0,1)(a,0)=(0,0) as multiplication is component-wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (0,0)=(a,b) for every pair, so our construction gives us the trivial ring. Hence, to get a non-trivial structure, we must disallow pairs of the form (a,0), including (0,0) and that gives us the rationals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Anyway, the above is the standard reason why 0/0 is undefined. You can actually define it, though to do that you must change some of the fundamental arithmetic properties. For a fun read, check out the following paper on the topic:&#xA;&lt;a href=&quot;http://www2.math.su.se/~jesper/research/wheels/wheels.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www2.math.su.se/~jesper/research/wheels/wheels.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Original answer:&#xA;By definition (or by construction) a/b, where a and b are natural numbers, is defined to be the rational number which when multiplied by b gives you a. From this you have that 0/0=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different fractions a/b and c/d are &lt;strong&gt;defined&lt;/strong&gt; to be equal if ad=bc. This allows us to extend the arithmetic of the integers to an arithmetic of fractions. From this you have that 0/0=c/d for every rational number c/d since x0=0 for all x in any system where multiplication distributes over addition (proof: x0=x(0-0)=x0-x0=0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you allow 0/0 among your fractions, and you attempt to extend the arithmetic of the integers, all your fractions must be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: the reason you cannot allow a/0 among your fractions is when a is not equal to 0 is because if you extend the arithmetic to where you have multiplication distributing over addition, you would have 0x=0 for all x and so the existence of a/0 is only consistent if a=0, which is the case above).&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-07-31T17:48:00.530" LastActivityDate="2010-07-31T17:48:00.530" CommentCount="2" />
  <row Id="1262" PostTypeId="2" ParentId="1017" CreationDate="2010-07-30T20:39:10.357" Score="3" ViewCount="0" Body="&lt;p&gt;One way is to note that $x_1y_2 - x_2y_1$ is a &lt;em&gt;signed&lt;/em&gt; area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i27.tinypic.com/2hyz7mb.jpg&quot; alt=&quot;Adding signed areas for the sides of a polygon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $\frac12 \oint_{polygon} x\,dy - y\,dx$ here.) By induction, you then only have to verify the formula for a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i32.tinypic.com/2d9wm7o.jpg&quot; alt=&quot;Shoelace sums of two polygons&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-07-30T20:44:23.697" LastActivityDate="2010-07-30T20:44:23.697" />
  <row Id="1263" PostTypeId="2" ParentId="548" CreationDate="2010-07-30T20:51:10.327" Score="0" ViewCount="0" Body="&lt;p&gt;Everyone knows 0/0 is &lt;a href=&quot;http://www.bbc.co.uk/berkshire/content/articles/2006/12/06/divide_zero_feature.shtml&quot; rel=&quot;nofollow&quot;&gt;nullity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(yes, this is a joke post, but that was a serious article)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-07-30T20:51:10.327" CommentCount="1" />
  <row Id="1264" PostTypeId="2" ParentId="1260" CreationDate="2010-07-30T20:51:48.377" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming you want arrows, you can try using Mathematica. In particular, you probably want to use the &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/Arrow.html&quot; rel=&quot;nofollow&quot;&gt;Arrow&lt;/a&gt; command.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Graphics[{Arrow[{{0, 0}, {1, 0}}], Arrow[{{0, 0}, {1, 2}}]}] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;gives the two vectors (1, 0) and (1, 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you want 1xn matrices, try LaTeX.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-07-30T20:51:48.377" CommentCount="1" />
  <row Id="1265" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T21:18:07.600" Score="2" ViewCount="0" Body="&lt;p&gt;Below is my geometric understanding of why irrational powers of negatives are difficult to define. As such it is probably not rigorous and may be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way irrational powers of the real numbers are usually defined is by limits of fractional powers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For complex numbers, the same is true, except the limiting process is more complicated. We can of course coast as usual on our real numbers result and see that we only need to define irrational powers on the unit circle, since every other points is some positive real multiple of a point on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in general z -&gt; z^n wraps the circle around itself n times. What does z-&gt;z^(1/n) do? Well, it's not clear since each point has n possible points it could have come from, in particular if you partition the circle into n arcs of length 2pi/n, each of those gets mapped to the full circle. Once you choose a starting arc though, z -&gt; z^m maps the starting arc to other arcs in the following way. Partition your starting arc of length 2pi/n into arcs of length 2pi/(nm) and then each of the little arcs gets mapped to a big arc that is 2pim/n away from the previous big arc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you choose arcs as opposed to weirdly distributed discrete sets, is because you want exponentiation to be continuous and hence you want the inverse image sets to be as the whole circle is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no problem with fractional powers of -1, you have n choices for starting arc. But if you want your exponentiation to have some semblance of continuity with respect to the exponent, then you have to be choosing branches (arcs) for z-&gt;z^(1/n) that cohere, i.e. such that for very large n the roots get closer and closer together. This is done by requiring that the nth root of 1 is always 1, which makes all of the arcs into neighborhoods of 1. But this means that all the points with argument between 0 and pi get mapped in the half-arc above 1, and all the points with argument between pi and 2pi get mapped in the half-arc below 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if you approach -1 from above and from below, the two limits of the nth root will be different, and hence you cannot have continuous exponentiation at -1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result (you can do some more visualization if you wish), the irrational powers of -1 cannot even be defined as limits of rational powers.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-07-30T21:18:07.600" CommentCount="1" />
  <row Id="1266" PostTypeId="2" ParentId="1257" CreationDate="2010-07-30T21:26:18.713" Score="6" ViewCount="0" Body="&lt;p&gt;There are &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;formulas&lt;/a&gt; on Wikipedia, though they are messy. No polynomial $p(x)$ can output the $n$th prime for all $n$, as is explained in the first section of the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, a &lt;a href=&quot;http://primes.utm.edu/glossary/xpage/MatijasevicPoly.html&quot; rel=&quot;nofollow&quot;&gt;polynomial in 26 variables&lt;/a&gt; whose &lt;em&gt;nonnegative&lt;/em&gt; values are precisely the primes. (This is fairly useless as far as computation is concerned.) This comes from the fact that the property of being a prime is decidable, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_set&quot; rel=&quot;nofollow&quot;&gt;theorem of Matiyasevich&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-07-30T21:26:18.713" CommentCount="2" />
  <row Id="1268" PostTypeId="2" ParentId="756" CreationDate="2010-07-30T21:39:34.520" Score="10" ViewCount="0" Body="&lt;p&gt;There is a fairly simple strategy which requires (1&amp;nbsp;+&amp;nbsp;&amp;epsilon;)log(n)&amp;nbsp;+&amp;nbsp;1/&amp;epsilon;&amp;nbsp;+&amp;nbsp;O(1) queries, for any constant &amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0. I illustrate this strategy below.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, I ask you whether or not X, the secret number, is n/2. Without loss of generality, suppose you answer &quot;less&quot;. I learn nothing at first, because you may be lying; but for the moment I give you the benefit of the doubt.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I next ask you whether X is n/4. If you say &quot;less&quot;, I don't know whether or not 0&amp;nbsp;&amp;lt;&amp;nbsp;X&amp;nbsp;&amp;lt;&amp;nbsp;n/4, but I &lt;strong&gt;do&lt;/strong&gt; know that 0&amp;nbsp;&amp;lt;&amp;nbsp;X&amp;nbsp;&amp;lt;&amp;nbsp;n/2, because you can't lie twice. Similarly, if I go about a normal binary search, so long as you continue to answer &quot;less&quot;, I know that your answer to the &lt;em&gt;preceding&lt;/em&gt; query was honest. So we may reduce to the case where you say &quot;more&quot; to my query of whether X is n/4.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I continue to take you at your word, persue a binary search, and enquire about whether X is 3n/8, you may say either &quot;more&quot; or &quot;less&quot;. If you say &quot;more&quot;, then I don't know whether X&amp;nbsp;&amp;gt;&amp;nbsp;n/2, but I do know that X&amp;nbsp;&amp;gt;&amp;nbsp;n/4, again because you can't lie twice. So again so long as you continue to answer &quot;more&quot; in my normal binary search, I know that your answer to the preceding query was honest.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More generally:&lt;/strong&gt; if I consistently guess under the hypothesis that you are being honest, in any &quot;monotonic&quot; sequence of responses, I know that all but (possibly) the last of them are honest. So it might seem as though the worst case scenario is where your responses alternate a lot, as would occur for instance if you had chosen something like X&amp;nbsp;=&amp;nbsp;&lt;sub&gt;&amp;lfloor;&lt;/sub&gt;&amp;thinsp;n/3&amp;thinsp;&lt;sub&gt;&amp;rfloor;&lt;/sub&gt;. But in the alternating case too, I can be confident of the honesty of some of your answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you say (non-monotonically) that X is less than 3n/8, I don't know whether X is less than 3n/8 for sure; but I do know that X&amp;nbsp;&amp;lt;&amp;nbsp;n/2, because again you can't have lied about both.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More generally:&lt;/strong&gt; not only do I know that monotonic subsequences of answers are mostly honest, I know that any time that you answer &quot;more&quot; or &quot;less&quot;, your previous answer of the same kind was also honest. So in fact I should be most suspicious, when I encounter long monotonic subsequences in your answers, that the answer previous to that monotonic subsequence was a lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is a strategy that will tell me when to revisit an old question, depending on how large n is. Ideally, revisiting old questions would require very little overhead. If I encounter a monotonic sequence of f(n) responses, I revisit the last question before that monotonic sequence started.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For instance, if your responses to queries about n/4, 3n/8, 7n/16, etc. are all monotonically &quot;more&quot;, I eventually ask about the last time you said &quot;less&quot;, which is for n/2, just in case you lied back then. This allows me to avoid the scenario where you lie about n/2, and keep me guessing at (2&lt;sup&gt;&lt;em&gt;t&lt;/em&gt;&amp;minus;1&lt;/sup&gt;&amp;nbsp;&amp;minus;&amp;nbsp;1)/2&lt;sup&gt;&lt;em&gt;t&lt;/em&gt;&lt;/sup&gt; until I eliminate all possibilities, catch you in your lie, and then redo all of my binary search for queries greater than n/2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If I do a double-check like this every time I encounter a monotonic sequence of length r, I will in the worst case query you about (r+1)log(n)/r = (1&amp;nbsp;+&amp;nbsp;1/r)log(n) times. If I catch you in a lie, I will have only &quot;wasted&quot; r queries, and my strategy afterwards can be just a simple binary search without double-checks; so your optimal strategy for maximizing the number of queries I make is actually not to lie at all, or to save your lie for nearly the end of the game to cost me about r additional queries. Here r can be an arbitrarily large integer; thus for any &amp;epsilon;, I can achieve a query rate of (1&amp;nbsp;+&amp;nbsp;&amp;epsilon;)log(n)&amp;nbsp;+&amp;nbsp;1/&amp;epsilon;&amp;nbsp;+&amp;nbsp;O(1) by setting r&amp;nbsp;&amp;gt;&amp;nbsp;1/&amp;epsilon;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus problem #1.&lt;/strong&gt; Without too much extra work, I think you can improve this to a strategy requiring only log(n)&amp;nbsp;+&amp;nbsp;O(log&amp;nbsp;log(n)) queries, but I'm too lazy to work out the details right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus problem #2.&lt;/strong&gt; Generalize this strategy to the regime where you are allowed to lie to me some fixed number of times L&amp;nbsp;&gt;&amp;nbsp;0.  &lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-11-02T13:11:07.107" LastActivityDate="2010-11-02T13:11:07.107" />
  <row Id="1269" PostTypeId="2" ParentId="1211" CreationDate="2010-07-30T21:44:55.380" Score="4" ViewCount="0" Body="&lt;p&gt;As other posters have indicated, the problem is that the complex logarithm isn't well-defined on $\mathbb{C}$.  This is related to my comments in &lt;a href=&quot;http://math.stackexchange.com/questions/1183/division-by-imaginary-number&quot;&gt;a recent question&lt;/a&gt; about the square root not being well-defined (since of course $\sqrt{z} = e^{ \frac{\log z}{2} }$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point of view is that the complex exponential $e^z : \mathbb{C} \to \mathbb{C}$ does not really have domain $\mathbb{C}$.  Due to periodicity it really has domain $\mathbb{C}/2\pi i \mathbb{Z}$.  So one way to define the complex logarithm is not as a function with range $\mathbb{C}$, but as a function with range $\mathbb{C}/2\pi i \mathbb{Z}$.  Thus for example $\log 1 = 0, 2 \pi i, - 2 \pi i, ...$ and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are we doing when we don't do this?  Well, let us suppose that for the time being we have decided that $\log 1 = 0$.  This is how we get other values of the logarithm: using power series, we can define $\log (1 + z)$ for any $z$ with $|z| &amp;lt; 1$.  We can now pick any number in this circle and take a power series expansion about &lt;em&gt;that&lt;/em&gt; number to get a different power series whose circle of convergence is somewhere else.  And by repeatedly changing the center of our power series, we can compute different values of the logarithm.  This is called &lt;em&gt;analytic continuation&lt;/em&gt;, and typically it proceeds by choosing a (say, smooth) path from $1$ to some other complex number and taking power series around different points in that path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem you quickly run into is that the value of $\log z$ depends on the choice of path from $1$ to $z$.  For example, the path $z = e^{2 \pi i t}, 0 \le t \le 1$ is a path from $1$ to $1$, and if you analytically continue the logarithm on it you will get $\log 1 = 2 \pi i$.  And that is not what you wanted.  (This is essentially the same as the contour integral of $\frac{1}{z}$ along this contour.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way around this problem is to arbitrarily choose a ray from the origin and declare that you are not allowed to analytically continue the logarithm through this ray.  This is called choosing a branch cut, and it is not canonical, so I don't like it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another way to resolve this situation, which is to consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_surface&quot; rel=&quot;nofollow&quot;&gt;Riemann surface&lt;/a&gt; $(z, e^z) \subset \mathbb{C}^2$ and to think of the logarithm as the projection to the first coordinate from this surface to $\mathbb{C}$.  So all the difficulties we have encountered above have been due to the fact that we have been trying to pretend that this projection has certain properties that it doesn't have.  A closed path like $z = e^{2\pi i t}$ in which the logarithm starts and ends with different values corresponds to a path on this surface which starts and ends at different &lt;em&gt;points&lt;/em&gt;, so there is no contradiction.  This was Riemann's original motivation for defining Riemann surfaces, and it is this particular Riemann surface that powers things like the residue theorem.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T21:55:52.240" LastActivityDate="2010-07-30T21:55:52.240" CommentCount="1" />
  <row Id="1271" PostTypeId="1" AcceptedAnswerId="1272" CreationDate="2010-07-30T21:59:14.623" Score="0" ViewCount="89" Body="&lt;p&gt;In part of my research, the following problem has come up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the system of equations (in complex numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^b w^c = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^d w^e = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the solution set when we restrict both z and w to be ath roots of unity, for some positive integer a.  Of course, one immediately sees that (z,w) = (1,1) is a solution.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are some nice necessary and sufficient conditions on a,b,c,d, and e which guarantee that (z,w) = (1,1) is the ONLY solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To give an idea of the flavor of answer I'd be most happy with, one must have gcd(a,b,d) = gcd(c,e) = 1, because if z is any gcd(a,b,d)th root of 1 (which is neccesarily an ath root of 1), then (z,1) is a solution to both equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also turns out that z and w must both be gcd(a, be-cd)th roots of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd love to have an answer like &quot;gcd(a, be-cd) = gcd(a,b,d) = gcd(c,e) = 1 is necessary and sufficient&quot;, with, perhaps, a few more estimates on gcd terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem can also been generalized (and I am interested in that case as well).  Suppose you are given 3 equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^a w^b = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^c w^d = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z^e w^f = 1, with z and w complex numbers of modulus 1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are necessary and sufficient conditions on a,b,c,d,e and f which guarantee that the only simultaneous solution is (z,w) = (1,1)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The previous problem is a special case of this (which comes from setting b = 0.  Clearly then, z must be an ath root of unity.  It turns that if b is 0, using the fact that gcd(d,f)=1 one can show w must also be an ath root of unity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And please feel free to retag as appropriate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T22:06:02.513" LastActivityDate="2010-07-30T22:18:30.490" Title="Solving systems of equations in roots of unity." Tags="&lt;linear-algebra&gt;&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="1272" PostTypeId="2" ParentId="1271" CreationDate="2010-07-30T22:04:14.857" Score="2" ViewCount="0" Body="&lt;p&gt;You are computing the nullspace of a $2 \times 2$ matrix over $\mathbb{Z}/a\mathbb{Z}$, so a necessary and sufficient condition is that the matrix in question is invertible, hence $\gcd(a, be-dc) = 1$ should be necessary and sufficient.  In the second case you should compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;Smith normal form&lt;/a&gt; of your matrix over $\mathbb{Z}$.  I believe a necessary and sufficient condition is that the Smith normal form is [[1 0][0 1][0 0]] (looks like matrices aren't working yet).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-30T22:18:30.490" LastActivityDate="2010-07-30T22:18:30.490" CommentCount="3" />
  <row Id="1273" PostTypeId="2" ParentId="74" CreationDate="2010-07-30T22:04:55.433" Score="1" ViewCount="0" Body="&lt;p&gt;Your question: given that X and Z are independent, X is Gaussian (I'll use &quot;normal&quot;), and Y = X+Z, prove that Y is normal iff Z is normal. Right? As you observed, one direction is easy: &lt;em&gt;if&lt;/em&gt; Z is normal, then so is Y=X+Z. So for the other direction, assume that Y is normal. We need to prove that is Z normal too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps there's an even easier way, but it's straightforward to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;characteristic functions&lt;/a&gt;, which completely characterise distributions. Because X and Z are independent,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \varphi_Y(t) = E[e^{itY}] = E[e^{it(X+Z)}] = E[e^{itX}]E[e^{itZ}]$, and so,&lt;br&gt;&#xA;$ \varphi_Z(t) = E[e^{itZ}] = E[e^{itY}]/E[e^{itX}] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that Z has exactly the right characteristic function for a normal variable, and hence it's normal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More interestingly and much more generally, there is a theorem of Cramer (e.g. see &lt;a href=&quot;http://books.google.com/books?id=B7Ch-c2G21MC&amp;amp;pg=PA205&amp;amp;dq=cramer+theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) which says that if X and Z are independent and X+Z is normally distributed, then both X and Z are!&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-30T22:04:55.433" CommentCount="2" />
  <row Id="1275" PostTypeId="2" ParentId="1249" CreationDate="2010-07-30T23:39:46.963" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a simple intersection theoretic example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the intersection of the line $y=0$ and the parabola $y=x^2$. Classically, the intersection is a point. But note that there is more to the intersection than just the point; there is the fact that the two curves are tangent at that point. Scheme-theoretically, the intersection is $\operatorname{Spec} k[x,y]/(y,y-x^2) \cong \operatorname{Spec} k[x]/(x^2)$. This reflects the tangency. If the intersection were transverse, then the scheme-theoretic intersection would have been just $\operatorname{Spec} k$.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-30T23:39:46.963" CommentCount="0" />
  <row Id="1276" PostTypeId="2" ParentId="1241" CreationDate="2010-07-30T23:46:52.207" Score="0" ViewCount="0" Body="&lt;p&gt;An important special case of the AM-GM inequality is that the product of two (positive) numbers with a constant sum is at a maximum when they are equal. This comes up a lot.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-07-30T23:46:52.207" />
  <row Id="1278" PostTypeId="2" ParentId="574" CreationDate="2010-07-31T00:43:33.840" Score="4" ViewCount="0" Body="&lt;p&gt;In general*, if something is &quot;generic&quot;, it means it happens or is true &quot;almost all of the time&quot; or &quot;almost everywhere&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In measure theory, for example, when you say &quot;$P(x)$ is true for almost all $x$&quot;, this has the precise meaning that the set of $x$'s for which $P(x)$ does not hold has measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can relate this to the Zariski topology via the fact that Zariski closed subsets of $\mathbb{C}^n$ have Lebesgue measure zero: &lt;a href=&quot;http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also these MO posts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/19688/what-does-generic-mean-in-algebraic-geometry&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/19688/what-does-generic-mean-in-algebraic-geometry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/2162/what-are-the-most-important-instances-of-the-yoga-of-generic-points&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/2162/what-are-the-most-important-instances-of-the-yoga-of-generic-points&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*or perhaps I should say... generically ;-)&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-07-31T01:07:09.890" LastActivityDate="2010-07-31T01:07:09.890" />
  <row Id="1279" PostTypeId="1" AcceptedAnswerId="1282" CreationDate="2010-07-31T01:02:48.323" Score="3" ViewCount="104" Body="&lt;p&gt;I'm looking for the name of a shape that is like a capsule, but where each circle can have different radii. The shape could be described using two circles (two centers and two radii). Something like this (this is a 2D shape, not a lame attempt at 3D):&#xA;&lt;img src=&quot;http://img694.imageshack.us/img694/3212/shape.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="529" LastEditorUserId="529" LastEditorDisplayName="" LastEditDate="2010-07-31T01:38:30.830" LastActivityDate="2010-07-31T06:27:29.753" Title="What is the name for a shape that is like a capsule, but with two different radii?" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="2" />
  <row Id="1280" PostTypeId="2" ParentId="405" CreationDate="2010-07-31T01:03:34.703" Score="3" ViewCount="0" Body="&lt;p&gt;TheCatsters have short YouTube videos explaining monads, and other parts of category theory: &lt;a href=&quot;http://www.youtube.com/user/TheCatsters&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/user/TheCatsters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-07-31T01:03:34.703" />
  <row Id="1282" PostTypeId="2" ParentId="1279" CreationDate="2010-07-31T01:23:26.823" Score="6" ViewCount="0" Body="&lt;p&gt;You can just say it is the convex hull of two circles.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-07-31T01:23:26.823" CommentCount="5" />
  <row Id="1283" PostTypeId="1" CreationDate="2010-07-31T04:11:00.823" Score="3" ViewCount="307" Body="&lt;p&gt;I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$.  I am already aware of expressing it in terms of the gamma function.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Background / Motivation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$.  Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to do the same for a generating function in the form&#xA;$g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Attempted Solutions&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Differential equation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$.  Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$.  Our initial condition $y(0) = 0$ so $c=0$.  By &lt;a href=&quot;http://mathworld.wolfram.com/IncompleteGammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;(2)&lt;/a&gt;, $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Recurrence Relation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$.  The point $z = 1$ is a pole of order one with residue $e^x$.  To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$.  The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ by the positively oriented unit circle centered at the origin.  By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Finite Calculus&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I've tried to evaluate the sum using finite calculus, but can't seem to make much progress. &lt;/p&gt;&#xA;" OwnerUserId="381" LastEditorUserId="381" LastEditorDisplayName="" LastEditDate="2010-08-06T20:55:56.303" LastActivityDate="2010-08-06T20:55:56.303" Title="Closed form of a partial sum of the power series of exp(x)" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="1284" PostTypeId="2" ParentId="1260" CreationDate="2010-07-31T04:54:21.547" Score="2" ViewCount="0" Body="&lt;p&gt;I'm assuming you want to typeset vectors or matrices in a Google Docs document, not actually draw the arrow. A simple solution is to use the online TeXify tool at &lt;a href=&quot;http://www.texify.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.texify.com/&lt;/a&gt;. You type some LaTeX code, and it gives you some code that you can paste into your Google Docs document.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, give it the following input:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\begin{pmatrix} x &amp;amp; y &amp;amp; z \\ 1 &amp;amp; 2 &amp;amp; 3 \end{pmatrix}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and click the big button. You'll see a 2 x 3 matrix. If you're not familiar with LaTeX notation, just google &quot;pmatrix&quot; and you'll find lots of documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A word of caution: what you're pasting into your Google Docs document is actually a link to the Texify site. If it ever goes offline, your equations will disappear as well. &lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-07-31T04:54:21.547" CommentCount="1" />
  <row Id="1285" PostTypeId="1" AcceptedAnswerId="1288" CreationDate="2010-07-31T05:14:01.170" Score="5" ViewCount="84" Body="&lt;p&gt;I found this question a while ago on a SAT practice exam or something, can't quite remember. So given an acute triangle $ABC$ with $P$ a point inside it and $AP$, $BP$, and $CP$ meeting the opposite sides at $D$, $E$, and $F$ respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/tnZew.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you find the area of triangle $ABC$ given the areas of triangles $x$, $y$, and $z$?&lt;/p&gt;&#xA;" OwnerUserId="215" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T06:22:37.313" LastActivityDate="2010-07-31T06:40:19.400" Title="Area of a triangle from some of its parts" Tags="&lt;geometry&gt;&lt;contest-math&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1286" PostTypeId="2" ParentId="544" CreationDate="2010-07-31T05:45:16.580" Score="2" ViewCount="0" Body="&lt;p&gt;I had a programming project in college where we generated large amounts of psuedo-English text using Markov chains. The &lt;a href=&quot;http://scf.usc.edu/~csci102b/curr/assgts/PA3/pa3.html&quot; rel=&quot;nofollow&quot;&gt;assignment is here&lt;/a&gt;, although I don't know if that link will be good forever. From that page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, suppose that [our Markov chains are of length] 2 and the sample file&#xA;  contains&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I like the big blue dog better than the big elephant with the big blue hat on his tusk. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;Here is how the first three words&#xA;  might be chosen:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;A two-word sequence is chosen at random to become the initial prefix.&#xA;  Let's suppose that &quot;the big&quot; is&#xA;  chosen.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The first word must be chosen based on the probability that it&#xA;  follows the prefix (currently &quot;the&#xA;  big&quot;) in the source. The source&#xA;  contains three occurrences of &quot;the&#xA;  big&quot;. Two times it is followed by&#xA;  &quot;blue&quot;, and once it is followed by&#xA;  &quot;elephant&quot;. Thus, the next word must&#xA;  be chosen so that there is a 2/3&#xA;  chance that &quot;blue&quot; will be chosen, and&#xA;  a 1/3 chance that &quot;elephant&quot; will be&#xA;  chosen. Let's suppose that we choose&#xA;  &quot;blue&quot; this time.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The next word must be chosen based on the probability that it&#xA;  follows the prefix (currently &quot;big&#xA;  blue&quot;) in the source. The source&#xA;  contains two occurrences of &quot;big&#xA;  blue&quot;. Once it is followed by &quot;dog&quot;,&#xA;  and the other time it is followed by&#xA;  &quot;hat&quot;. Thus, the next word must be&#xA;  chosen so that there is a 50-50&#xA;  probability of choosing &quot;dog&quot; vs.&#xA;  &quot;hat&quot;. Let's suppose that we choose&#xA;  &quot;hat&quot; this time.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The next word must be chosen based on the probability that it&#xA;  follows the prefix (currently &quot;blue&#xA;  hat&quot;) in the source. The source&#xA;  contains only one occurrence of &quot;blue&#xA;  hat&quot;, and it is followed by &quot;on&quot;.&#xA;  Thus, the next word must be &quot;on&quot; (100%&#xA;  probability).&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Thus, the first three words in the output text would be &quot;blue hat&#xA;  on&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You keep going like that, generating text that is completely nonsensical, but ends up having sort of the same &quot;tone&quot; as the original text. For example, if your sample file is the complete text of Alice In Wonderland (one of the texts we tried it on) then your nonsense comes out kind of whimsical and Carrollian (if that's a word). If your sample file is The Telltale Heart, you get somewhat dark, morbid nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, while not a rigorous, formal, definition, I hope this helps give you a sense of what a Markov chain is.&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-31T05:45:16.580" CommentCount="1" />
  <row Id="1287" PostTypeId="2" ParentId="1285" CreationDate="2010-07-31T06:04:23.997" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Method 1&lt;/strong&gt;: We can perform an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;affine transformation&lt;/a&gt; to make ADC any triangle we wish, while preserving colinearality and the ratios of areas. In particular, this allows us to assign exact coordinates to A, D and C. We can then use the ratio of x and y to find E and the ratio of z and x+y+z to find P. We can then use these coordinates to find B. Once we know B, we know the ratio of BC to DC and hence the ratio of the area of the whole triangle to the known areas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 2&lt;/strong&gt;: Let Q be the area of the whole triangle&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AE:EC=x:y=|AEB|:|BEC|&#xA;|APB|=xQ/(x+y)-x&#xA;|BPD|=yQ/(x+y)-y-z&#xA;Now AP:PD=x+y:z&#xA;z(xQ/(x+y)-x)=(x+y)(yQ/(x+y)-y-z)&#xA;Q(zx/(x+y)-y)=xz-xy-y^2-xz-yz=-xy-y^2-xz&#xA;Q(zx-xy-y^2)/(x+y)=(-xy-y^2-xz)&#xA;Q=(-xy-y^2-xz)*(x+y)/(zx-xy-y^2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-07-31T06:39:12.070" LastActivityDate="2010-07-31T06:39:12.070" CommentCount="4" />
  <row Id="1288" PostTypeId="2" ParentId="1285" CreationDate="2010-07-31T06:21:32.323" Score="4" ViewCount="0" Body="&lt;p&gt;This was not likely to have been an SAT practice problem, though it is a typical contest problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AE:EC = x:y (since those two triangles have the same altitude to AC, the ratios of their areas is the ratios of their bases with respect to that altitude) and AP:PD = (x+y):z (same idea as AE:EC).  Knowing these two ratios, apply the technique of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_points&quot; rel=&quot;nofollow&quot;&gt;mass points&lt;/a&gt;, putting masses zy at A, zx at C (gives the ratio x:y for AE:EC), and y(x+y) at D (gives (x+y):z for AP:PD).  This results in a mass of y(x+y)-zx at B, so the ratio BD:DC = zx:(y(x+y)-zx).  This must also be the ratio of the areas of &amp;#x25B3;ABD to &amp;#x25B3;ADC (common altitude again), so (area of &amp;#x25B3;ABD):(x+y+z) = zx:(y(x+y)-zx).  Solving from there is a matter of bashing out the algebra.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T06:40:19.400" LastActivityDate="2010-07-31T06:40:19.400" CommentCount="2" />
  <row Id="1289" PostTypeId="2" ParentId="1279" CreationDate="2010-07-31T06:27:29.753" Score="2" ViewCount="0" Body="&lt;p&gt;Here is an article at Mathworld on &lt;a href=&quot;http://mathworld.wolfram.com/Circle-CircleTangents.html&quot; rel=&quot;nofollow&quot;&gt;Circle-Circle Tangents&lt;/a&gt;. Perhaps &quot;CirclesWithTangents&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathworld.wolfram.com/images/eps-gif/CircleCircleTangentGeneral_1000.gif&quot; alt=&quot;Circle-Circle Tangents&quot; title=&quot;Circle-Circle Tangents&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="208" LastActivityDate="2010-07-31T06:27:29.753" CommentCount="1" />
  <row Id="1290" PostTypeId="2" ParentId="164" CreationDate="2010-07-31T06:38:13.307" Score="2" ViewCount="0" Body="&lt;p&gt;To expand on what anosov_diffeomorphism said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we agree that the surface area of a sphere is &lt;strong&gt;4 &amp;pi; r&lt;sup&gt;2&lt;/sup&gt;&lt;/strong&gt; (If we do not agree on this, that would be a great question to ask on this site, and then someone can edit this answer to link to it!). Now if I have a sphere of radius &lt;em&gt;r&lt;/em&gt;, and I increase the radius by a tiny amount, &lt;em&gt;dr&lt;/em&gt;, then the new, expanded sphere has a volume that is bigger, by the volume of the thin spherical shell that was just added. If &lt;em&gt;dr&lt;/em&gt; is small enough, then it doesn't much matter whether we look at the inner or outer radius of that shell, and we can just say that the shell's volume is approximately &lt;strong&gt;4 &amp;pi; r&lt;sup&gt;2&lt;/sup&gt; dr&lt;/strong&gt;. So the difference in volume between two spheres of radius $r_1$ and $r_2$ is $\int_{r_1}^{r_2}$ 4 &amp;pi; r&lt;sup&gt;2&lt;/sup&gt; dr, which is 4/3 &amp;pi; r^3&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-31T06:38:13.307" />
  <row Id="1291" PostTypeId="2" ParentId="1237" CreationDate="2010-07-31T06:48:48.637" Score="4" ViewCount="0" Body="&lt;p&gt;One that I remember from some book (I think it was Innumeracy by John Allen Paulos) was &quot;How fast does your hair grow, in miles per hour?&quot;&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-07-31T06:48:48.637" CommunityOwnedDate="2010-07-31T06:48:48.637" />
  <row Id="1292" PostTypeId="1" AcceptedAnswerId="1293" CreationDate="2010-07-31T07:09:16.437" Score="13" ViewCount="632" Body="&lt;p&gt;The &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Angle_addition_formula#Angle_sum_and_difference_identities&quot; rel=&quot;nofollow&quot;&gt;sum and difference&lt;/a&gt;&quot; formulas often come in handy, but it's not immediately obvious that they would be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sin(\alpha \pm \beta) &amp;amp;= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\&#xA;\cos(\alpha \pm \beta) &amp;amp;= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I want to know is, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I prove that these formulas are correct?&lt;/li&gt;&#xA;&lt;li&gt;More importantly, how can I understand these formulas intuitively?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ideally, I'm looking for answers that make no reference to Calculus, or to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;, although such answers are still encouraged, for completeness.&lt;/p&gt;&#xA;" OwnerUserId="496" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-13T12:35:28.923" LastActivityDate="2010-08-13T12:35:28.923" Title="How can I understand and prove the &quot;sum and difference formulas&quot; in trigonometry? (cos(a ± b) = ..., etc.)?" Tags="&lt;intuition&gt;&lt;trigonometry&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="8" />
  <row Id="1293" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T07:19:26.117" Score="9" ViewCount="0" Body="&lt;p&gt;The key fact here is that rotation is a linear transformation, e.g. the rotation of $u + v$ is the rotation of $u$ plus the rotation of $v$.  You should draw a diagram that shows this carefully if you don't believe it.  That means a rotation is determined by what it does to $(1, 0)$ and to $(0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $(1, 0)$ rotated by $\theta$ degrees counterclockwise is just $(\cos \theta, \sin \theta)$, whereas $(0, 1)$ rotated by $\theta$ degrees counterclockwise is just $(-\sin \theta, \cos \theta)$.  (Again, draw a diagram.)  That means a rotation by $\theta$ is given by a $2 \times 2$ matrix with those entries.  (Matrices don't work here yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take a rotation by $\theta$ and another one by $\theta'$, and multiply the corresponding matrices.  What you get is the sine and cosine angle addition formulas.  (The connection to complex numbers is that one can represent complex numbers as $2 \times 2$ real matrices.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you believe that $a \cdot b = |a| |b| \cos \theta$, this implies the cosine angle difference formula when $a$ and $b$ are unit vectors.  Ditto for the cross product and the sine angle difference formula.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-07-31T07:19:26.117" />
  <row Id="1294" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T07:51:38.223" Score="7" ViewCount="0" Body="&lt;p&gt;There are several typical derivations used in high school texts.  Here's one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/400545892.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take two points on the unit circle, one a rotation of (1,0) by &amp;alpha;, the other a rotation of (1,0) by &amp;beta;.  Their coordinates are as shown in the diagram.  Let c be the length of the segment joining those two points.  By the Law of Cosines (on the blue triangle), $c^2=1^2+1^2-2\cdot1\cdot1\cdot\cos(\alpha-\beta)$.  Using the distance formula, $c=\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}$.  Squaring the latter and setting the two equal, $1^2+1^2-2\cdot1\cdot1\cdot\cos(\alpha-\beta)=(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2$.  Simplifying both sides, $2-2\cos(\alpha-\beta)=\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta$ $=2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$ (using the Pythagorean identity).  Solving for $\cos(\alpha-\beta)$, $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this identity, the other three can be derived by substituting $\frac{\pi}{2}-\alpha$ for &amp;alpha; (gives sin(&amp;alpha;+&amp;beta;)), then -&amp;beta; for &amp;beta; (gives the remaining two).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to understanding the formulas intuitively, if you accept that multiplying by a complex number $z_\theta$ for which |z|=1 rotates by &amp;theta;, then you can think about what happens when you multiply $z_\alpha=\cos\alpha+i\sin\alpha$ and $z_\beta=\cos\beta+i\sin\beta$ (by expanding the binomial product), which should result in $\cos(\alpha+\beta)+i\sin(\alpha+\beta)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-07-31T07:51:38.223" CommentCount="6" />
  <row Id="1295" PostTypeId="2" ParentId="1237" CreationDate="2010-07-31T08:16:32.233" Score="0" ViewCount="0" Body="&lt;p&gt;What is the mass of the atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="453" LastActivityDate="2010-07-31T08:16:32.233" CommunityOwnedDate="2010-07-31T08:16:32.233" />
  <row Id="1296" PostTypeId="2" ParentId="405" CreationDate="2010-07-31T08:55:15.557" Score="3" ViewCount="0" Body="&lt;p&gt;(Sorry for the LaTeX, the symbols &lt;code&gt;\mapsto&lt;/code&gt; and &lt;code&gt;\{&lt;/code&gt; and &lt;code&gt;\}&lt;/code&gt; don't work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monads in Haskell and monads in category theory are very much the same: A &lt;em&gt;monad&lt;/em&gt; consists of a functor $T: C \to C$ and two natural transformations $\eta_X : X \to T(X)$ (&lt;code&gt;return&lt;/code&gt; in Haskell) and $\mu_X : T(T(X)) \to T(X)$ (&lt;code&gt;join&lt;/code&gt; in Haskell) subject to the following laws&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_X \circ T(\eta_X) = \mu_X \circ \eta_{T(X)} = 1_{T(X)}$ (left and right unit laws)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_X \circ \mu_{T(X)} = \mu_X \circ T(\mu_X)$ (associativity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, compared to Haskell, the monad is defined in terms of &lt;code&gt;return&lt;/code&gt;, &lt;code&gt;join&lt;/code&gt; and &lt;code&gt;fmap&lt;/code&gt; instead of &lt;code&gt;return&lt;/code&gt; and &lt;code&gt;(&amp;gt;&amp;gt;=)&lt;/code&gt;. For more details on this, see also the &lt;a href=&quot;http://en.wikibooks.org/wiki/Haskell/Category_theory#Monads&quot; rel=&quot;nofollow&quot;&gt;Haskell wikibook&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two examples may illuminate this definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;powerset functor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathcal{P} = X \mapsto \mathcal{P}(X)$ maps a set to the set of its subsets.&lt;/li&gt;&#xA;&lt;li&gt;Functions $f:X \to Y$ are extended point-wise to $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$&lt;/li&gt;&#xA;&lt;li&gt;$\eta_X : X \to \mathcal{P}(X)$ is the function $x \mapsto {x}$&lt;/li&gt;&#xA;&lt;li&gt;$\mu_X : \mathcal{P}(\mathcal{P}(X)) \to \mathcal{P}(X)$ flattens the inner layer of subsets: $\mu_X(A) = { b | a \in A, b \in a }$.&lt;/li&gt;&#xA;&lt;li&gt;In Haskell, this is known as the &lt;em&gt;list monad&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;closure&lt;/strong&gt; operation on the subsets of a topological space $S$ is a monad, too.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The objects of the category $C$ are the subsets of a given topological space $S$.&lt;/li&gt;&#xA;&lt;li&gt;There is an unique arrow $X \to Y$ between to objects $X$ and $Y$ exactly when $X \subseteq Y$.&lt;/li&gt;&#xA;&lt;li&gt;The monad is given by the functor that maps each object $X$ to its topological closure $\bar X$ and the arrow $X \subseteq Y$ to the arrow $\bar{X}\subseteq \bar{Y}$.&lt;/li&gt;&#xA;&lt;li&gt;Clearly, we have $X \subseteq \bar X$; this is $\eta_X$.&lt;/li&gt;&#xA;&lt;li&gt;Also, we know that $\bar{\bar X} = \bar X$, in particular $\bar{\bar X} \subseteq \bar X$; this is $\mu_X$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-07-31T09:01:47.167" LastActivityDate="2010-07-31T09:01:47.167" />
  <row Id="1297" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T09:24:09.130" Score="8" ViewCount="0" Body="&lt;p&gt;You can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosine#Relationship_to_exponential_function_and_complex_numbers&quot; rel=&quot;nofollow&quot;&gt;complex representation,&lt;/a&gt;&lt;br&gt;&#xA;$\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$&lt;br&gt;&#xA;$\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$&lt;br&gt;&#xA;and the rules for powers ($a^{x+y}=a^x a^y$)&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-07-31T09:24:09.130" CommentCount="1" />
  <row Id="1298" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T11:52:37.290" Score="5" ViewCount="0" Body="&lt;p&gt;Though the standard high-school derivations are not the most useful way to remember it in the long run, here's another one which I like because you can &quot;see&quot; it directly without much algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/QnIhq.png&quot; alt=&quot;Angle sum formulae&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let P be the point on the unit circle got by rotating (1,0) by angle α+β. Drop a perpendicular N to the α-rotated line, and R to the x-axis. So from the right triangle ONP, you see &lt;strong&gt;ON = cos &amp;beta;&lt;/strong&gt;. You can see that the angle RPN is &amp;alpha; too: it's the complement of ∠PNQ, and so is ∠QNO = &amp;alpha;. Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(\alpha + \beta) = \mbox{PR} = \mbox{PQ} + \mbox{QR} = \sin(\beta)\cos(\alpha) + \cos(\beta)\sin(\alpha)$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos(\alpha + \beta) = \mbox{OR} = \mbox{OM} - \mbox{RM} = \cos(\beta)\cos(\alpha) - \sin(\beta)\sin(\alpha)$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-31T11:52:37.290" />
  <row Id="1299" PostTypeId="2" ParentId="1292" CreationDate="2010-07-31T11:56:33.290" Score="4" ViewCount="0" Body="&lt;p&gt;I remember that $e^{i\alpha}=\cos\alpha+i\sin\alpha$ and that $i^2=-1$. Both these relations are useful in many other situations and pretty fundamental to understanding complex numbers. Then your equalities are the real and, respectively, the imaginary part of $e^{i(\alpha+\beta)}=e^{i\alpha}e^{i\beta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not very different from the other answers, but I actually prefer the algebra perspective. The only place where I think geometrically is in interpreting $e^{i\alpha}=\cos\alpha+i\sin\alpha$ by thinking of the unit circle in the complex plane.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-07-31T20:37:06.617" LastActivityDate="2010-07-31T20:37:06.617" CommentCount="3" />
  <row Id="1300" PostTypeId="2" ParentId="1234" CreationDate="2010-07-31T12:15:45.357" Score="1" ViewCount="0" Body="&lt;p&gt;Here's a polynomial time solution given to me by Javi Gomez. Let (i,j) with $i\le j$ represent the situation where disks i, i+1, ..., j are on top of each other in position $x_j$, and let E(i,j) represent the energy needed to obtain that configuration. Clearly E(i,i) is zero for all i. Also, $E(i,j)=\min_k E(i,k)+E(k,j)+|x_k-x_j|$. What we want is E(1,n).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I made this a community wiki since the answer wasn't really found by me. Feel free to edit.)&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-07-31T12:15:45.357" CommunityOwnedDate="2010-07-31T12:15:45.357" />
  <row Id="1302" PostTypeId="1" AcceptedAnswerId="1305" CreationDate="2010-07-31T13:33:51.307" Score="3" ViewCount="118" Body="&lt;p&gt;Following my &lt;a href=&quot;http://math.stackexchange.com/questions/1279/what-is-the-name-for-a-shape-that-is-like-a-capsule-but-with-two-different-radii&quot;&gt;previous question&lt;/a&gt;, I'm wondering how I can determine if a point is within the convex hull of two circles (a boolean value). There's no problem testing if the point is in either of the two circles, but it can also be &quot;between&quot; them and I don't know how to test for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing Wolfram MathWorld's article on &lt;a href=&quot;http://mathworld.wolfram.com/Circle-CircleTangents.html&quot; rel=&quot;nofollow&quot;&gt;Circle-Circle Tangeants&lt;/a&gt;, it seems that an inequation that tests if the point is on the internal side the two external circle tangeants would do the trick, but I'm afraid my solving skills are too far away to be able to turn the tangeant equations into a fitting inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm defining the convex hull of two circles using both centers and radii.&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2010-08-01T09:56:16.397" Title="How to test if a point is inside the convex hull of two circles?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="1303" PostTypeId="2" ParentId="1302" CreationDate="2010-07-31T13:49:21.940" Score="1" ViewCount="0" Body="&lt;p&gt;The solution to this problem is indeed to check whether the point is in one of the circles or in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isosceles_trapezoid&quot; rel=&quot;nofollow&quot;&gt;isosceles trapezoid&lt;/a&gt; determined by the points were circles touch the tangent lines. However, without some care the equations get messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with a circle at (x1', y1') of radius r1', a circle at (x2', y2') with radius r2', and a point (x', y') to test. We may assume that r1' is at least r2'. By shifting followed by rotation followed by scaling this can be transformed into a simpler instance: a circle at (0,0) of radius r1, a circle at (1,0) of radius r2, and a point (x,y) to test. We have: r1=r1'/D, r2=r2'/D, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=((y'-y'_1)\sin\alpha+(x'-x'_1)\cos\alpha)/D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=((y'-y'_1)\sin\alpha-(x'-x'_1)\cos\alpha)/D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D^2=(x'_1-x'_2)^2+(y'_1-y'_2)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\alpha$ is the angle where (x2'-x1',y2'-y1') lies. (In C, there's a function atan2 that takes the two coordinates of a point and gives the angle. In math, atan doesn't really distinguish between points symmetric wrt (0,0).)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the point is in one of the circles say YES, otherwise continue.&lt;/li&gt;&#xA;&lt;li&gt;If r1=r2 then return (0 &amp;lt; x &amp;lt; 1 and |y| &amp;lt; r1), otherwise continue. (At this point the difference between radii Dr=r1-r2 is strictly between 0 and 1.)&lt;/li&gt;&#xA;&lt;li&gt;If x &amp;lt; r1 Dr, say NO.&lt;/li&gt;&#xA;&lt;li&gt;If x &gt; 1 + r2 Dr, say NO.&lt;/li&gt;&#xA;&lt;li&gt;If $x\Delta r + |y|\sqrt{1-{\Delta r}^2}&gt;r_1$, say NO.&lt;/li&gt;&#xA;&lt;li&gt;Say YES.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There's only one thing I used, really, namely that certain triangles are similar. Draw the tangent that touches the two circles 'above'. Let's say that it intersects the big circle at A=(x1,y1), the small circle at B=(x2,y2), and the horizontal axis at C=(d,0). Let's also write D=(0,0) and E=(x1,0) and F=(1,0) and G=(x2,0). The similar triangles are ADC, AED, BFC, and BGD. From ADC similar to BFC you find d = r1/Dr. From ADC similar to AED you find x1 = r1 Dr and from BFD similar to BGD you find x2 - 1 = r2 Dr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check that a point is on the right side of a line write its equation as ax+by+c=0 and replace equality by inequality. (It's good to remember that the vector $(a,b)^T$ is perpendicular to the line and that the equation can be interpreted as defining all vectors/points whose scalar/dot product with a fixed vector $(a,b)^T$ gives the constant c.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if I messed up again :P&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;This is the old answer, which is wrong as pointed by the first comment below. It starts with the same transformations (without giving details) but then it uses some line that is clearly &lt;em&gt;not&lt;/em&gt; tangent to the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the first circle has the center at (0,0) (if not, shift the figure) with radius r_1, and the second at (1,0) (if not, rotate the figure and scale) with radius r_2. If your point is (x,y) and is not in one of the two circles, then x must be between 0 and 1 and |y| must be at most r_1+(r_2-r_1)x.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-08-01T09:56:16.397" LastActivityDate="2010-08-01T09:56:16.397" CommentCount="3" />
  <row Id="1304" PostTypeId="2" ParentId="1260" CreationDate="2010-07-31T14:04:11.890" Score="1" ViewCount="0" Body="&lt;p&gt;You can try with &lt;a href=&quot;http://mathurl.com/&quot; rel=&quot;nofollow&quot;&gt;mathurl&lt;/a&gt;, which allows you to write in LaTeX and copy&amp;amp;paste everywhere.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-07-31T14:04:11.890" />
  <row Id="1305" PostTypeId="2" ParentId="1302" CreationDate="2010-07-31T14:40:43.003" Score="8" ViewCount="0" Body="&lt;p&gt;You can first calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_lines_to_circles&quot; rel=&quot;nofollow&quot;&gt;four points of tangency&lt;/a&gt;, then use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_in_polygon&quot; rel=&quot;nofollow&quot;&gt;Point in Polygon&lt;/a&gt; algorithm to determine if your point is inside the quadrilateral (this is assuming you have a programming related problem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, once you have found the four tangent points, you can form four lines and get four simultaneous inequalities.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T14:40:43.003" CommentCount="1" />
  <row Id="1306" PostTypeId="2" ParentId="77" CreationDate="2010-07-31T15:51:53.060" Score="2" ViewCount="0" Body="&lt;p&gt;Two more uses that I haven't seen mentioned yet:  if you want to find the area of the parallelogram formed by two vectors (each vector gives a pair of parallel sides), then you would use the magnitude of the cross product of the two vectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One use of this is to aid in defining a surface integral.  Let x(u,v) be a parameterization of a surface.  Then at each point, we can find tangent vectors T&lt;sub&gt;u&lt;/sub&gt; =  &amp;part;x/&amp;part;u and T&lt;sub&gt;v&lt;/sub&gt; = &amp;part;x/&amp;part;v.  From the idea of a linear approximation, T&lt;sub&gt;u&lt;/sub&gt; and T&lt;sub&gt;v&lt;/sub&gt; will define a tangent plane at that particular point.  Consider the parallelogram formed with T&lt;sub&gt;u&lt;/sub&gt; and T&lt;sub&gt;v&lt;/sub&gt; as sides.  Informally, we can see that each area element will be |T&lt;sub&gt;u&lt;/sub&gt; x T&lt;sub&gt;v&lt;/sub&gt;| du dv.  Then a function f(u,v) integrated over this surface is &amp;int;&amp;int; f(u,v) |T&lt;sub&gt;u&lt;/sub&gt; x T&lt;sub&gt;v&lt;/sub&gt;| du dv.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second use, if you wanted to find the volume of the parallelepiped having the three vectors a, b, c as sides, then you would use the magnitude of the scalar triple product |a &amp;sdot; (b x c)|. &lt;/p&gt;&#xA;" OwnerUserId="381" LastActivityDate="2010-07-31T15:51:53.060" />
  <row Id="1307" PostTypeId="1" AcceptedAnswerId="1309" CreationDate="2010-07-31T16:51:50.433" Score="4" ViewCount="98" Body="&lt;p&gt;Let $A$ be a valuation ring of rank two. Then $A$ gives an example of a commutative ring such that $Spec(A)$ is a noetherian topological space, but $A$ is nonnoetherian. (Indeed, otherwise $A$ would be a discrete valuation ring.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a concrete example of such a ring $A$? &lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-07-31T18:17:31.513" LastActivityDate="2010-11-04T00:37:46.830" Title="Valuation rings of rank two" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="1308" PostTypeId="2" ParentId="1307" CreationDate="2010-07-31T17:28:30.750" Score="1" ViewCount="0" Body="&lt;p&gt;Take the ring of formal power series (over $\mathbb{C}$, say) with exponents in $\mathbb{Z}^2$ under lex order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  As Robin Chapman mentions, one must be careful about what this means.  The precise construction for any totally ordered abelian group is described &lt;a href=&quot;http://en.wikipedia.org/wiki/Valuation_ring#Construction&quot; rel=&quot;nofollow&quot;&gt;in the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-07-31T20:25:21.323" LastActivityDate="2010-07-31T20:25:21.323" />
  <row Id="1309" PostTypeId="2" ParentId="1307" CreationDate="2010-07-31T18:14:22.037" Score="7" ViewCount="0" Body="&lt;p&gt;Qiaochu's answer is sound in principle, but in practice one needs to be more careful&#xA;with the definition of the ring. The quotient field $K$ of $A$ consists of&#xA;formal Laurent series of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=\sum_{r=-r_0}^\infty x^r\sum_{s=-s_0(r)}^\infty a_{r,s}y^s.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $r_0$ is an integer and for each integer $r$, $s_0(r)$ is an integer&#xA;(depending on $r$).&#xA;So these are the power series where the powers of $x$ are bounded below&#xA;and for each integer $r$ the coefficient of $x^r y^s$ is zero for all&#xA;$s$ below a bound depending on $r$. This complicated-looking condition&#xA;ensures that the product of two elements of $K$ is also an element&#xA;of $K$ (note that one cannot multiply two general Laurent series).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $A$ will consist of all such series with the additional conditions&#xA;that $r_0=0$ and $s_0(0)=0$. The valuation of an element $f$ is&#xA;the least $(r,s)$ under lexicographic ordering with $a_{r,s}\ne0$.&#xA;Here the ordering is $(r,s)&amp;lt;(r',s')$ if $r &amp;lt; r'$ or $r=r'$ and $s &amp;lt; s'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more high-brow interpretation of the condition for memebership of&#xA;$K$ is that the &lt;em&gt;support&lt;/em&gt; of $f$, the setof $(r,s)$ for which $a_{r,s}\ne0$,&#xA;should be well-ordered, that is each subset of the support has a least&#xA;element. (With repsct to this lexicographic ordering of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By considering a version of this construction in $n$ variables&#xA;one can construct explicitly a ring with a valuation of rank $n$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-07-31T18:14:22.037" CommentCount="1" />
  <row Id="1310" PostTypeId="2" ParentId="803" CreationDate="2010-07-31T18:38:40.893" Score="9" ViewCount="0" Body="&lt;p&gt;More than dissection proofs, I find the proof using similarity most enlightening and intuitive. See &lt;a href=&quot;http://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/&quot; rel=&quot;nofollow&quot;&gt;post by Terence Tao&lt;/a&gt;: drop a perpendicular from C to the hypotenuse AB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.math.ucla.edu/~tao/pythagoras.jpeg&quot; alt=&quot;Image from Tao&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure, the triangles whose are areas are marked x and y are similar to the original triangle (which has area x+y). So accepting that areas of similar right-angled triangles are proportional to the squares of the hypotenuse, x:y:x+y are in ratio a&lt;sup&gt;2&lt;/sup&gt;:b&lt;sup&gt;2&lt;/sup&gt;:c&lt;sup&gt;2&lt;/sup&gt;, which is Pythagoras's theorem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See also the linked discussion &lt;a href=&quot;http://math.berkeley.edu/~giventh/papers/eu.pdf&quot; rel=&quot;nofollow&quot;&gt;by Alexander Giventhal&lt;/a&gt; where he remarks that this proof is more general than tiling or dissection proofs, and is even proved by Euclid. If you think of a&lt;sup&gt;2&lt;/sup&gt;+b&lt;sup&gt;2&lt;/sup&gt;=c&lt;sup&gt;2&lt;/sup&gt; as the geometrical result that the sum of areas of squares constructed with sides a and b is the area of a square placed on c, then the Pythagorean theorem is true not just for constructing squares on the sides, but &lt;em&gt;any&lt;/em&gt; similar figures. For instance, as Euclid himself proves, something like the following is true (though it's still just saying a&lt;sup&gt;2&lt;/sup&gt;+b&lt;sup&gt;2&lt;/sup&gt;=c&lt;sup&gt;2&lt;/sup&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/OpDUA.png&quot; alt=&quot;From Giventhal&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-07-31T18:38:40.893" CommunityOwnedDate="2010-07-31T18:38:40.893" CommentCount="3" />
  <row Id="1311" PostTypeId="1" AcceptedAnswerId="1312" CreationDate="2010-07-31T19:49:24.733" Score="4" ViewCount="423" Body="&lt;p&gt;I've been told that there are precisely the same number of rationals as there are of integers. The set of rationals is countably infinite, therefore every rational can be associated with a positive integer, therefore there are the same number of rationals as integers. I've ignored sign-related issues, but these are easily handled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count the rationals, consider sets of rationals where the denominator and numerator are positive and sum to some constant. If the constant is 2 there's 1/1. If the constant is 3, there's 1/2 and 2/1. If the constant is 4 there's 1/3, 2/2 and 3/1. So far we have counted out 6 rationals, and if we continue long enough, we will eventually count to any specific rational you care to mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, I find this very hard to accept. I have two reasons. First, this logic seems to assume that infinity is a finite number. You can count to and number any rational, but you cannot number all rationals. You can't even count all positive integers. Infinity is code for &quot;no matter how far you count, you have never counted enough&quot;. If it were possible to count to infinity, it would be possible to count one step less and stop at count infinity-1 which must be different to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason is that it's very easy to construct alternative mappings. Between zero and one there are infinitely many rational numbers, between one and two there are infinitely many rational numbers, and so on. To me, this seems a much more reasonable approach, implying that there are infinite rational numbers for every integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even then, this is just one of many alternative ways to map between ranges of rationals and ranges of integers. Since you can count the rationals, you can equally count stepping by any amount for each rational. You can use 1..10 for the first rational and 11..20 for the second etc. Or 1..100 and 101..200 etc, or 1..1000 and 1001..2000 etc. You can map finite range of integers of any size to each rational this way and, since there is no finite upper bound to the stepping amount, you could argue there are potentially infinite integers for every single rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... can anyone convince me that there is a single unambiguous correct answer to this question? Are there more rational numbers than integers, or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I've already accepted an answer, I'll just add some extra context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reason for questioning this relates to the Hilbert space-filling curve. I find this interesting because of applications to multi-dimensional indexing data structures in software. However, I found Hilberts claim that the Hilbert curve literally filled a multi-dimensional space hard to accept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in a comment below, a one meter line segment and a two meter line segment can both be seen as sets of points and, but (by the logic in answers below), those two sets are both the same size (cardinality). Yet we would not claim the two line segments are both the same size. The lengths are finite and different. Going beyond this, we most certainly wouldn't claim that the size of &lt;em&gt;any&lt;/em&gt; finite straight line segment is equal to the size of a one-meter-by-one-meter square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hilbert curve reasoning makes sense now - the set of points in the curve is equal to the set of points in the space it fills. Previously, I was thinking too much about basic geometry, and couldn't accept the size of a curve as being equal to the size of a space. However, this isn't based on a fallacious counting-to-infinity argument - it's a necessary consequence of an alternative line of reasoning. The two constructs are equal because they both represent the same set of points. The area/volume/etc of the curve follows from that.&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-07-31T21:22:29.987" LastActivityDate="2010-07-31T21:22:29.987" Title="Are there more rational numbers than integers?" Tags="&lt;set-theory&gt;&lt;infinity&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="1312" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:01:41.193" Score="10" ViewCount="0" Body="&lt;p&gt;Mathematicians have very precise definitions for terms like &quot;infinite&quot; and &quot;same size&quot;.  The single unambiguous correct answer to this question is that using the standard mathematical definitions, the rationals have the &quot;same size&quot; as the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, here are the definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Define &quot;0&quot; = emptyset, &quot;1&quot; = {0}, &quot;2&quot; = {0,1}, &quot;3&quot; = {0,1,2}, etc.  So, the number &quot;n&quot; is really a set with &quot;n&quot; elements in it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set A is called &quot;finite&quot; iff there is some n and a function f:A-&gt;n which is bijective.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set A is called &quot;infinite&quot; iff it is not finite.  (Note that this notion says nothing about &quot;counting never stops&quot; or anything like that.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two sets A and B are said to have the &quot;same size&quot; if there is a some function f:A-&gt; B which is a bijection.  Note that we do NOT require that ALL functions be bijections, just that there is SOME bijection.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once one accepts these definitions, one can prove that the rationals and integers have the same size.  One just needs to find a particular bijection between the two sets.  If you don't like the one you mentioned in your post, may I suggest that Calkin-Wilf enumeration of the rationals?  (Simply google search Calkin Wilf counting rationals.  The first .pdf has what I'm talking about).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these give bijections between the naturals (with out 0) and the rationals, but once you have a bijection like this, it's easy to construct a bijection from the integers to the rationals by composing with a bijection from the naturals to the integers.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-07-31T20:01:41.193" CommentCount="6" />
  <row Id="1313" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:04:39.830" Score="0" ViewCount="0" Body="&lt;p&gt;You can think about it a different way. Consider the set of real numbers between 0 and 1, and then the set of real numbers between 0 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By intuition, it seems that the set of real numbers between 0 and 2 has double the size of the set between 0 and 1. However, this is not the case, because the two sets have the same &lt;em&gt;cardinality&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $f(x) = 2x$. Every real between 0 and 1 is &lt;em&gt;bijected&lt;/em&gt; to a real between 0 and 2. Therefore the sets are of the same size.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T20:04:39.830" CommentCount="1" />
  <row Id="1314" PostTypeId="2" ParentId="803" CreationDate="2010-07-31T20:07:46.967" Score="0" ViewCount="0" Body="&lt;p&gt;Not a proof in itself, but the book &lt;a href=&quot;http://www.amazon.com/exec/obidos/ISBN=0873530365/ctksoftwareincA/&quot; rel=&quot;nofollow&quot;&gt;The Pythagorean Proposition&lt;/a&gt; by Loomis has probably the most comprehensive collection of proofs of the Pythagorean theorem.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T20:07:46.967" CommunityOwnedDate="2010-07-31T20:07:46.967" />
  <row Id="1315" PostTypeId="1" AcceptedAnswerId="1320" CreationDate="2010-07-31T20:08:01.177" Score="1" ViewCount="152" Body="&lt;p&gt;Suppose you have a normal distribution with mean=0, and stdev=1.  So the expected value is 0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose you limit the outcomes, such that no values can be below 0.  So 50% of values now equal 0, and rest of distribution is still normal.  Running 1000000 trials, I come out with an expected value of .4  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how can I get this expected value through calculation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2010-08-01T13:53:17.917" LastActivityDate="2010-08-01T13:53:17.917" Title="How do I calculate expected value of partial normal distribution?" Tags="&lt;calculus&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="1316" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:09:02.033" Score="1" ViewCount="0" Body="&lt;p&gt;The cardinality of the set of rationals is the same as the cardinality of the integers is the same as the cardinality of the natural numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we count a finite set of elements, we are constructing a one-one map from the set onto a finite initial segment of the natural numbers. If we want to know if two finite sets have the same cardinality (are equi-cardinal) we can either: 1) count both sets and see if we get the same number, or 2) attempt to construct a one-one map from one set onto the other. If we can construct the map aimed at in (2), then the sets are equi-cardinal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalizing that procedure from the finite sets to arbitrary sets, we get that for any two sets, the sets have the same cardinality (are equi-cardinal) if there exists a bijection (a one-one map between the sets that is onto the target rather than merely into). For the finite case, if there is a one-one map that is a bijection, all one-one maps are bijective. That is not the case for infinite sets, which is the root of your second concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address that second concern, consider the map from the negative integers to the positive integers which maps each negative integer to its absolute value. The existence of that map shows that the two sets are equi-cardinal. We can, of course, construct one-one maps from the negative integers to the positive integers that are into rather than on to. (Consider the map that takes each negative integer to its product with -2.) But, the existence of these alternative maps doesn't affect the fact that there is at least one bijection between the sets, and that is all it takes for those sets to be equi-cardinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your first concern, I don't see why you think the procedure assumes that &quot;infinity is a finite number&quot;. What it involves is specifying a mapping function from one set to the other that is one-one and onto. That attempt can certainly fail, as Cantor's Diagonalization Argument that the cardinality of a set is always strictly less than the cardinality of its power set shows. (A relevant application of that technique is the well-known proof that the cardinality of the reals is greater than the cardinality of the natural numbers.)&lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-07-31T20:09:02.033" CommentCount="2" />
  <row Id="1317" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:11:34.837" Score="3" ViewCount="0" Body="&lt;p&gt;In mathematics a set is called &lt;em&gt;infinite&lt;/em&gt; if it can be put into a 1-1 correspondence with a proper subset of it, and &lt;em&gt;finite&lt;/em&gt; it is not infinite. (I know it seems crazy to have the concept of infinite as primitive and finite as a derivate, but it's simpler to do this, since otherwise you must assume that the integers exist before saying that a set is finite)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your remarks: &#xA;- with your method (if you don't forget to throw out fractions like 4/6 which is equal to 2/3) you actually &lt;strong&gt;counted&lt;/strong&gt; the rationals, since for each number you have a function which associates it to a natural number. It's true that you cannot count ALL rationals, or all integers; but you cannot either draw a whole straight line, can you?&#xA;- with infinite sets you may build infinite mappings, but you just need a single 1-1 mapping to show that two sets are equal.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-07-31T20:11:34.837" CommentCount="3" />
  <row Id="1318" PostTypeId="2" ParentId="329" CreationDate="2010-07-31T20:18:16.647" Score="2" ViewCount="0" Body="&lt;p&gt;My favorite is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321500318&quot; rel=&quot;nofollow&quot;&gt;Elementary Number Theory&lt;/a&gt; by Rosen, which combines computer programming with number theory, and is accessible at a high school level.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-07-31T20:18:16.647" CommunityOwnedDate="2010-07-31T20:18:16.647" />
  <row Id="1319" PostTypeId="2" ParentId="1311" CreationDate="2010-07-31T20:26:04.660" Score="2" ViewCount="0" Body="&lt;p&gt;You may not be very satisfied with this answer, but I'll try to explain anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Countability.&lt;/strong&gt; We're not really talking about whether you can &quot;count all of the rationals&quot;, using some finite process. Obviously, if there is an infinite number of elements, you cannot count them in a finite amount of time using any reasonable process. The question is whether there is &lt;em&gt;the same number&lt;/em&gt; of rationals as there are positive integers; this is what it means for a set to be &quot;countable&quot; --- for there to &lt;em&gt;exist&lt;/em&gt; a one-to-one mapping from the positive integers to the set in question. You have described such a mapping, and therefore the rationals are &quot;countable&quot;. (You may disagree with the terminology, but this does not affect whether the concept that it labels is coherent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative mappings.&lt;/strong&gt; You seem to be dissatisfied with the fact that, unlike the case of a finite set, you can define an injection from the natural numbers to the rationals which is not surjective --- that you can in fact define a more general &lt;em&gt;relation&lt;/em&gt; in which each integer is related to infinitely many rationals, but no two integers are related to the same rational numbers. Well, two can play at that game: you can define a relation in which every rational number is related to infinitely many integers, and no two rationals are related to the same integers! Just define the relation that each positive rational a/b is related to all numbers which are divisible by 2&lt;sup&gt;a&lt;/sup&gt; but not 2&lt;sup&gt;a+1&lt;/sup&gt;, and by 3&lt;sup&gt;b&lt;/sup&gt; but not 3&lt;sup&gt;b+1&lt;/sup&gt;; or more generally respectively 2&lt;sup&gt;ka&lt;/sup&gt; and 3&lt;sup&gt;kb&lt;/sup&gt; for any positive integer k. (There are, as you say, sign issues, but these can be smoothed away.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might complain that the relation I've defined isn't &quot;natural&quot;. Perhaps you have in mind the fact that the integers are a &lt;em&gt;subset&lt;/em&gt; of the rationals --- a subgroup, in fact, taking both of them as additive groups --- and that the factor group &amp;#8474;/&amp;#8484; is infinite. Well, this is definitely interesting, and it's a natural sort of structure to be interested in. But it's more than what the issue of &quot;mere cardinality&quot; is trying to get at: set theory is interested in size regardless of structure, and so we don't restrict to maps which have one or another kind of &quot;naturalness&quot; about them. Of course, if you are interested in mappings which respect some sort of structure, you can build theories of size based on that: this is what is done in measure theory (with measure), linear algebra (with dimension), and indeed group theory (with index). So if you don't like cardinality as set theorists conceive it, you can look at more structured measures of size that you find more interesting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Immediate predecessors.&lt;/strong&gt; A somewhat unrelated (but still important) complaint that you make is this: &quot;&lt;em&gt;If it were possible to count to infinity, it would be possible to count one step less and stop at count infinity-1 which must be different to infinity.&lt;/em&gt;&quot; The question is: &lt;strong&gt;why&lt;/strong&gt; would you necessarily be able to stop at 'infinity minus one'? This is true for finite collections, but it does not necessarily hold that anything which is true of finite collections is true also for infinite ones. (In fact, obviously, some things necessarily will fail.) --- This is important if you study ordinals, which mirrors the process of counting itself in some ways (labelling things as being &quot;first&quot;, &quot;second&quot;, &quot;third&quot;, and so forth), because of the concept of a limit ordinal: the first &quot;infinitieth&quot; element of a well-ordering doesn't have any immediate predecessors! Again, you are free to say that these are concepts that you are not interested in exploring personally, but this does not mean that they are necessarily incoherent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To summarize:&lt;/strong&gt; the set theorists measure &quot;the size of a set&quot; using a simple definition which doesn't care about structure, and which may violate your intuitions if you like to take the structure of the integers (and the rational numbers) very seriously, and also want to preserve your intuitions about finite sets. There are two solutions to this: try to stretch your intuition to accomodate the ideas of the set theorists, or study a different branch of math which you find more interesting!&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-07-31T20:26:04.660" CommentCount="3" />
  <row Id="1320" PostTypeId="2" ParentId="1315" CreationDate="2010-07-31T20:26:47.403" Score="4" ViewCount="0" Body="&lt;p&gt;The normal distribution has density function $f(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$; your new distribution has that density function on the positive reals, $P(0)=\frac{1}{2}$, and $P(x)=0$ for the negative reals.  The expected value is $0\cdot\frac{1}{2}+\int_{0}^{\infty}x\cdot f(x)dx=\frac{1}{\sqrt{2\pi}}\approx0.398942$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: If you were to cut off at $x=c$ (assigning all the probability from below c to c itself) instead of $x=0$, your density function would be $f(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$ for $x&amp;gt;c$, $P(c)=\int_{-\infty}^{c}\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx$, and $P(x)=0$ for $x&amp;lt;c$, so the expected value is $c\cdot P(c) + \int_{c}^{\infty}x\cdot \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: note that the exponent on e in all of the above is $-\frac{x^2}{2}$ (the exponent 2 on the x is, in the current TeX rendering, positioned and sized such as to be somewhat ambiguous)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 3&lt;/em&gt;: my explanation incorrectly mixed probability density functions and literal probabilities--this was solely an issue of terminology and the analytic results still stand, but I have attempted to clarify the language above.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-01T05:45:33.857" LastActivityDate="2010-08-01T05:45:33.857" CommentCount="4" />
  <row Id="1321" PostTypeId="2" ParentId="460" CreationDate="2010-07-31T20:53:06.657" Score="7" ViewCount="0" Body="&lt;p&gt;One source of monoids is given by taking rings with identity, and forgetting about addition.&#xA;So similarly, one source of semigroups that are not monoids is taking rings without identity, and forgetting about addition.  With this in mind, let me explain one basic source of rings without identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic source of rings is given by taking functions satisfying some reasonable condition on a space, e.g. continuous real or complex valued functions on a space, with pointwise addition and multiplication.  Of course, the constant function 1 is continuous, and so this gives a ring with identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose now that we impose some condition, such as &quot;all functions that are continuous, and which furthermore vanish at some specified point&quot;.  This throws out the constant function 1, and so gives a ring without identity.  Now you could naturally object that this is artificial&#xA;(as per the requirement in the question that there not be an obvious extension to a monoid),&#xA;so let me add more explanation as to why it need not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of a point to consider is &quot;the point at infinity&quot;, i.e. we could look at all functions which vanish at infinity, i.e. which on the complement of larger and larger compact sets, grow smaller and smaller.  This is a natural condition to impose in many analytic contexts, and so gives a natural example.   (The reason that this kind of growth&#xA;condition is natural in analysis is that, on a non-compact space, e.g. the real line, a random continuous function may not be integrable (just as an example), and imposing some decay at infinity (perhaps of the kind I specified, or perhaps something more quantitive) becomes a way to rescue the situation.)  (Note also that the example that Tomer Vromen gives is exactly of this form.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that if your semigroup doesn't have an identity, then you can always formally adjoin one, just by throwing in an extra element e and declaring that ex = x for all x.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can do a similar thing for rings without identity.  If A is a C-algebra (say) without an identity, then one can form A + C e  (the direct sum), and declare that e acts as a multiplicative identity.  This is a frequently-used technique in the theory of rings-without-identity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  I don't know much literature about semigroups without identity, but for rings without identity, the best literature I know of is in functional analysis books; e.g. Naimark's classic &lt;em&gt;Normed Rings&lt;/em&gt; often treats the case of Banach algebras (and the like) without identity in addition to the case when they do have identity, exactly so as to be able to handle examples such as the ring of continuous functions on a locally compact space that vanish at infinity. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-07-31T20:53:06.657" CommunityOwnedDate="2010-07-31T20:53:06.657" />
  <row Id="1323" PostTypeId="1" CreationDate="2010-07-31T21:28:34.173" Score="6" ViewCount="162" Body="&lt;p&gt;I was wondering if there was a resource that listed known algebraic groups and their corresponding coordinate rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The previous wording was terrible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an algebraic group $G$, with Borel subgroup $B$ we can form the Flag Variety $G/B$ which is projective. I am hoping for a list of the graded ring $R$ such that $Proj(R)$ corresponds to this Flag Variety.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:41:29.873" LastActivityDate="2010-08-01T17:41:29.873" Title="Is there an atlas of Algebraic Groups and corresponding Coordinate rings?" Tags="&lt;reference-request&gt;&lt;algebraic-geometry&gt;&lt;algebraic-groups&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="1324" PostTypeId="1" AcceptedAnswerId="1325" CreationDate="2010-07-31T21:53:33.887" Score="0" ViewCount="198" Body="&lt;p&gt;What is a primitive polynomial? I was looking into some random number generation algorithms and 'primitive polynomial' came up a sufficient number of times that I decided to look into it in more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure of what a primitive polynomial is, and why it is useful for these random number generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd find it particularly helpful if an example of a primitive polynomial could be provided.&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-07-31T23:40:59.560" Title="What is a primitive polynomial?" Tags="&lt;polynomials&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="1325" PostTypeId="2" ParentId="1324" CreationDate="2010-07-31T22:05:52.727" Score="8" ViewCount="0" Body="&lt;p&gt;Consider a finite field $F_p$, then we know that it is cyclic. We call an element primitive if it generates this field. Further, given a field and some polynomial over that field(all the coefficients are in the field), we can form a field extension by any of its roots. This is adjoining on that root and making a field of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a simple result of Galois Theory that if we take a field and extend by some root of some polynomial and get a finite extension(one who's degree as a vector space over the original field is finite), that we can find a polynomial $m$ over our ground field such that $m$ vanishes at this root and is minimal(smallest degree, i.e. it divides all other polys which vanish at this root).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a primitive element and its minimal polynomial, that poly is call primitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;more details on &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_polynomial&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-07-31T22:05:52.727" CommentCount="1" />
  <row Id="1326" PostTypeId="1" AcceptedAnswerId="1328" CreationDate="2010-07-31T23:37:24.167" Score="3" ViewCount="97" Body="&lt;p&gt;The title really say's it all, but once again is a curve's curvature invariant under rotation and uniform scaling?&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-07-31T23:48:16.190" Title="Is a curve's curvature invariant under rotation and uniform scaling?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="1327" PostTypeId="2" ParentId="1324" CreationDate="2010-07-31T23:40:59.560" Score="7" ViewCount="0" Body="&lt;p&gt;BBischof's answer is correct, but unfortunately there's another, quite different possible meaning of the same term: that is a polynomial whose coefficients have no common prime factor (this makes sense over the integers, or other UFDs as well).&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-07-31T23:40:59.560" CommentCount="3" />
  <row Id="1328" PostTypeId="2" ParentId="1326" CreationDate="2010-07-31T23:43:08.917" Score="8" ViewCount="0" Body="&lt;p&gt;A curve's curvature is invariant under rotation. Intuitively, a curve turns just as much no matter how it is oriented. More formally, for a curve $\gamma(s)$ that is parametrized by arc length, the curvature is $\kappa(s) = ||\gamma''(s)||$. Rotation does not change the length of the $\gamma''(s)$ vector, only the direction; therefore, rotation does not affect curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curve's curvature is not invariant under uniform scaling, however. Consider the example of a circle. All circles are the same up to scaling, but they don't all have the same curvature; in general, a circle of radius r has curvature 1/r.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditorDisplayName="" LastEditDate="2010-07-31T23:48:16.190" LastActivityDate="2010-07-31T23:48:16.190" CommentCount="1" />
  <row Id="1329" PostTypeId="1" AcceptedAnswerId="1332" CreationDate="2010-07-31T23:56:42.353" Score="3" ViewCount="99" Body="&lt;p&gt;This is a followup to &lt;a href=&quot;http://math.stackexchange.com/questions/1326/is-a-curves-curvature-invariant-under-rotation-and-uniform-scaling&quot;&gt;this question&lt;/a&gt;, where I learned that curvature is invariant to rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have learned of a version of curvature that is &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_curvature#Affine_curvature&quot; rel=&quot;nofollow&quot;&gt;invariant under affine transformations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there a is a form of curvature between the two. Invariant under uniform scaling and rotation but not all affine transformations?&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-01T03:33:21.160" LastActivityDate="2010-08-01T03:33:21.160" Title="What is the form of curvature that is invariant under rotations and uniform scaling" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="1332" PostTypeId="2" ParentId="1329" CreationDate="2010-08-01T02:58:53.723" Score="2" ViewCount="0" Body="&lt;p&gt;I don't know if this would suit you, but one thing you can consider (much more naive than&#xA;the notion of affine curvature) is to fix a point P_0 on your curve, and then consider the&#xA;function on the curve given by sending a point P to the quantity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;curvature(P)/curvature(P_0)  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a kind of relative curvature, where you measure how much everything is curving&#xA;in comparison to the curvature at P_0, and is invariant under scaling and rotation.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-01T02:58:53.723" CommentCount="5" />
  <row Id="1333" PostTypeId="1" AcceptedAnswerId="1335" CreationDate="2010-08-01T07:45:27.030" Score="10" ViewCount="233" Body="&lt;p&gt;Me and my wife are often not knowing which DVD to watch. If we have two options we have a simple solution, I put one DVD in one hand behind my back and the other DVD in the other hand. She will randomly choose a hand and the DVD I have in that hand will be the one to watch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure is easy to expand to any power of 2. If we have 4 DVD's I hold 2 in one hand, 2 in the other. When a pair of DVD's is chosen, I split them out to two hands and she choses again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what can we do when we have 3 DVD's. The assumptions we make are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am not impartial. If I can influence the result somehow I will try to do that&lt;/li&gt;&#xA;&lt;li&gt;My wife really choses randomly a side every-time, independent of what she chose earlier.&lt;/li&gt;&#xA;&lt;li&gt;I don't have any other place to hide the DVD's, so every DVD is either visible or in one of the two hands.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As requirement we have that it must be a procedure with a predetermined number of steps. Not more, not less. If this is not possible, a solution that guarantees to finish with an upperbound, this is a good second choice. Off course the DVD we choose must be really random!&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:56:02.873" LastActivityDate="2010-08-01T21:22:26.807" Title="Is there a possibility to choose fairly from three items when every choice can only have 2 options" Tags="&lt;probability-theory&gt;&lt;algorithms&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="1334" PostTypeId="2" ParentId="1333" CreationDate="2010-08-01T08:40:07.303" Score="5" ViewCount="0" Body="&lt;p&gt;There is no such procedure which has an upperbound on number of steps. Here is proof.&#xA;Let there is such procedure with no more then $N$ steps. On each step you basicly generate random integer between $1$ and $2$. Consider all possible sequences of no more than $N$ generated numbers. Every such sequence has probability of form $\frac{x}{2^N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider sequences for which the procedure says &quot;1&quot; (the result is equal to $1$). The sum of theirs probabilities is $\frac{x_1}{2_N}$. It also must be equal to $\frac{1}{3}$. But $\frac{x_1}{2_N}$ can't be equal to $\frac{1}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-01T08:40:07.303" />
  <row Id="1335" PostTypeId="2" ParentId="1333" CreationDate="2010-08-01T08:49:35.987" Score="10" ViewCount="0" Body="&lt;p&gt;Nice question. Firstly, there's no solution which will always take a fixed number of steps each time: after n independent random trials (choosing a hand), there are 2&lt;sup&gt;n&lt;/sup&gt; possible choices, and 2&lt;sup&gt;n&lt;/sup&gt; is never divisible by 3. (Equivalently: 1/3 does not have a terminating representation in base 2.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can come up with solutions that terminate within a fixed number of steps with high probability. Note that you need at least 2 trials, since log&lt;sub&gt;2&lt;/sub&gt;(3) &gt; 1. Here's one simple method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hold 2 DVDs in one hand and 1 DVD in other (your wife doesn't know which is which).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the hand with 2 DVDs is chosen: play them against each other.&lt;br&gt;&#xA;Else: If the 1 DVD is chosen: play it against an empty hand.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each of the 3 DVDs has equal probability 1/4 of being chosen, and with probability 1/4, you have to repeat from scratch. This solution takes 2 trials with probabilty 3/4, 4 trials with probability (1/4)(3/4), 6 trials with probability (1/4)&lt;sup&gt;2&lt;/sup&gt;(3/4) and so on, so takes an expected number of 8/3 (≈ 2.66) trials. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_distribution&quot; rel=&quot;nofollow&quot;&gt;geometric distribution&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is optimal (famous last words?) if you just have to choose out of 3 DVDs once. Probably, if you're going to choose from 3 DVDs many times, you can do better in the long run (amortised) and achieve the lower bound of log&lt;sub&gt;2&lt;/sub&gt; 3 ≈ 1.58 trials on average, by &quot;caching&quot; some random choices from last time. (But will you remember them? :-)) At least you can do something similar when generating random numbers (see  also &lt;a href=&quot;http://stackoverflow.com/questions/137783/given-a-function-which-produces-a-random-integer-in-the-range-1-to-5-write-a-fun&quot;&gt;this mildly related Stack Overflow thread&lt;/a&gt;), but in this case with game-theoretic fairness complications I'm not so sure.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-01T08:55:24.973" LastActivityDate="2010-08-01T08:55:24.973" CommentCount="2" />
  <row Id="1336" PostTypeId="2" ParentId="803" CreationDate="2010-08-01T10:03:35.580" Score="5" ViewCount="0" Body="&lt;p&gt;My favorite proof of the Pythagorean Theorem is a special case of this picture-proof of the Law of Cosines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/lawofcos.jpg&quot; alt=&quot;Proof Without Words: The Law of Cosines&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drop three perpendiculars and let the definition of cosine give the lengths of the sub-divided segments. Then, observe that like-colored rectangles have the same area (computed in slightly different ways) and the result follows immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When C is a right angle, the blue rectangles vanish and we have the Pythagorean Theorem via what amounts to Proof #5 on &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;Cut-the-Knot's Pythagorean Theorem page&lt;/a&gt;. (Note that, as mentioned on CtK, the use of cosine here doesn't amount to an invalid &quot;trigonometric proof&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture works for obtuse C as well. (Proof left as an exercise for the reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final note ... Because the same-colored rectangles have the same area, they're &quot;equidecomposable&quot; (aka &quot;scissors congruent&quot;): it's possible to cut one into a finite number of polygonal pieces that reassemble to make the other. While there's at least one standard procedure for determining how to make the cuts, the resulting pieces aren't necessarily pretty. Some popular dissection proofs of the Pythagorean Theorem --such as Proof #36 on &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;Cut-the-Knot&lt;/a&gt;-- demonstrate a specific, clear pattern for cutting up the figure's three squares, a pattern that applies to all right triangles. I have yet to find a similarly straightforward cutting pattern that would apply to all triangles and show that my same-colored rectangles &quot;obviously&quot; have the same area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another exercise for the reader, perhaps? :)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-01T10:03:35.580" CommunityOwnedDate="2010-08-01T10:03:35.580" CommentCount="1" />
  <row Id="1337" PostTypeId="1" AcceptedAnswerId="1339" CreationDate="2010-08-01T11:22:30.790" Score="2" ViewCount="171" Body="&lt;p&gt;Dot product of two vectors on plane could be defined as product of lengths&#xA;of those vectors and cosine of angle between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cartesian coordinates dot product of vectors with coordinates (x1, y1) and&#xA;(x2, y2) is equal to x1*x2 + y1*y2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove it?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:29:06.960" LastActivityDate="2010-08-01T17:29:06.960" Title="Dot product in coordinates" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="1338" PostTypeId="1" AcceptedAnswerId="1345" CreationDate="2010-08-01T11:26:29.567" Score="1" ViewCount="108" Body="&lt;p&gt;Given a list of coordinates of a coplanar plane (&lt;code&gt;pt1&lt;/code&gt;, &lt;code&gt;pt2&lt;/code&gt;, &lt;code&gt;pt3&lt;/code&gt; and so on), how to compute the centroid of the coplanar plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do it is to project the plane onto XY and YZ plane, but I don't really favor this approach as you have to check the orientation of the coplanar plane first before doing the projection and computing the centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I'm looking for a natural extension of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon&quot; rel=&quot;nofollow&quot;&gt;2D centroid plane algorithm&lt;/a&gt; in 3D:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/e/e/1/ee14cbb2b170c4bb435f1d84e78f6d66.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://upload.wikimedia.org/math/a/4/c/a4cee81a1d18e4d067f66d4d40a8a1fe.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/0/2/a/02aecb75f67f8c7b2fc11fdcbcb6ea80.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-08-01T15:21:05.333" Title="Compute the Centroid of a 3D Planar Polygon Without Projecting It To Specific  Planes" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1339" PostTypeId="2" ParentId="1337" CreationDate="2010-08-01T11:39:17.987" Score="6" ViewCount="0" Body="&lt;p&gt;I suppose you want to prove that two your definitions of dot product are the same.&#xA;We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$. We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$ and prove that it also satisfies $(\vec{u}, \vec{v}) = x_1 x_2 + y_1 y_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first you can prove that dot product is linear: $(\vec{v_1}, \vec{v_2} + \alpha \vec{v_3}) = (\vec{v_1}, \vec{v_2}) + \alpha (\vec{v_1}, \vec{v_3})$. This is true because $(\vec{v_1}, \vec{v_2})$ is equal to the product of $|\vec{v_1}|$ and projection of $\vec{v_2}$ on $\vec{v_1}$. Projection of sum of vectors is equal to sum of projections. Hence dot product is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\vec{e_1}$ and $\vec{e_2}$ be vectors with coordinates $(1, 0)$ and $(0, 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that if $\vec{v_1} = x_1 \vec{e_1} + y_1 \vec{e_2}$ and $\vec{v_2} = x_2 \vec{e_2} + y_2 \vec{e_2}$ then by linearity of dot product we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 (\vec{e_1}, \vec{e_1}) + x_1 y_2 (\vec{e_1}, \vec{e_2}) + x_2 y_1 (\vec{e_2}, \vec{e_1}) + x_2 y_2 (\vec{e_2}, \vec{e_2})$.&#xA;Since $(\vec{e_1}, \vec{e_1}) = (\vec{e_2}, \vec{e_2}) = 1$ and $(\vec{e_1}, \vec{e_2}) = (\vec{e_2}, \vec{e_1}) = 0$ we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 + y_1 y_2$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-01T15:22:23.023" LastActivityDate="2010-08-01T15:22:23.023" CommentCount="9" />
  <row Id="1340" PostTypeId="1" AcceptedAnswerId="1343" CreationDate="2010-08-01T11:42:22.300" Score="0" ViewCount="101" Body="&lt;p&gt;Let's say we have a graph, with a list of edges and vertexes &lt;code&gt;(E,V)&lt;/code&gt;, all the vertexes are connected to at least one edge at one end. There are many ways a complete set of &lt;a href=&quot;http://www.mathreference.com/gph,basis.html&quot; rel=&quot;nofollow&quot;&gt;cycle basis&lt;/a&gt; can be found out from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the issue is, is it &lt;strong&gt;always&lt;/strong&gt; possible to find a complete set of cycle basis that each edge is shared by at most 2 cycles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Edit: There is a &lt;a href=&quot;http://math.stackexchange.com/questions/1340/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-basis-with-each-an/1343#1343&quot;&gt;mathematical argument&lt;/a&gt; proving why it is not possible. But admittedly such a highly abstract reasoning is a bit hard for me to grasp. I would appreciate if someone can provide a graphical example of such a graph.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-08-05T05:47:31.717" LastActivityDate="2010-08-05T05:47:31.717" Title="In a graph, is it always possible to construct a set of cycle basis, with each and every edge Is shared by at most 2 cycle bases?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1341" PostTypeId="2" ParentId="1337" CreationDate="2010-08-01T12:39:02.813" Score="2" ViewCount="0" Body="&lt;p&gt;The dot product is invariant under rotations, we may therefore rotate our coordinate system so that v is along the x-axis.  In this case, $v = (|v|, 0)$.  Letting $w = (x,y)$ we have (using the definition of dot product in Cartesian coordinates) $v \cdot w = |v| x$.  But what is $x$?  Well, if you draw the picture and let $\theta$ be the angle between v and w, then we see that $\cos \theta = x/|w||$ so that $x = |w| \cos \theta$.  Thus $v\cdot w = |v||w| \cos \theta$.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-01T12:39:02.813" CommentCount="2" />
  <row Id="1342" PostTypeId="2" ParentId="1292" CreationDate="2010-08-01T13:25:38.840" Score="6" ViewCount="0" Body="&lt;p&gt;Here are my favorite diagrams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://daylateanddollarshort.com/misc/angaddsub.jpg&quot; alt=&quot;Proof Without Words: Angle Sum and Difference for Sine and Cosine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As given, the diagrams put certain restrictions on the angles involved: neither angle, nor their sum, can be larger than 90 degrees; and neither angle, nor their difference, can be negative. The diagrams can be adjusted, however, to push beyond these limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a bonus mnemonic cheer (which probably isn't as exciting to read as to hear):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sine, Cosine, &lt;em&gt;Sign&lt;/em&gt;, Cosine, Sine!&lt;br /&gt;&#xA;  Cosine, Cosine, &lt;em&gt;Co-Sign&lt;/em&gt;, Sine, Sine!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first line encapsulates the sine formulas; the second, cosine. Just drop the angles in (in order $\alpha$, $\beta$, $\alpha$, $\beta$ in each line), and know that &quot;Sign&quot; means to use the &lt;em&gt;same sign&lt;/em&gt; as in the compound argument (&quot;+&quot; for angle sum, &quot;-&quot; for angle difference), while &quot;Co-Sign&quot; means to use the &lt;em&gt;opposite sign&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-01T13:25:38.840" CommentCount="1" />
  <row Id="1343" PostTypeId="2" ParentId="1340" CreationDate="2010-08-01T14:03:03.260" Score="3" ViewCount="0" Body="&lt;p&gt;The question was answered in the negative on Math Overflow.  See &lt;a href=&quot;http://mathoverflow.net/questions/30759/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-bases-with-each-an&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/30759/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-bases-with-each-an&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get a counterexample from any non-planar graph where every edge is part of at least one cycle. Here's a reference: &lt;a href=&quot;http://www.ams.org/journals/proc/1973-037-02/S0002-9939-1973-0313098-X/S0002-9939-1973-0313098-X.pdf&quot; rel=&quot;nofollow&quot;&gt;P. V. O'Neil, Proc. AMS, 37 (2), Feb. 1973, 617-8&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll repeat the argument here.  Take a nonplanar graph with every edge in at least one cycle. If that had a cycle basis like the one you want, then we could generate one for $K_{3,3}$ or $K_{5}$. Suppose we had a basis for $K_{3,3}$. There are 4 cycles in that basis. (A cycle basis has m-n+1 elements, where m is the number of vertices and n is the number of edges.) Also, take the binary sum of the four cycles. The five cycles include every edge exactly twice, so there are a total of 2 $\cdot$ (number of edges) = 18 edges in those five cycles. But each cycle has at least 4 edges, so the five cycles must have at least 20 edges. Contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at $K_{5}$. A cycle basis has 6 cycles. Also take the binary sum. The 7 cycles include each edge exactly twice, so there are 2 $\cdot$ (number of edges) = 20 edges in the collection. But each cycle has at least three edges, so the 7 cycles have at least 7*3=21 edges. Again, a contradiction. &lt;/p&gt;&#xA;" OwnerUserId="273" LastEditorUserId="273" LastEditorDisplayName="" LastEditDate="2010-08-02T14:51:48.877" LastActivityDate="2010-08-02T14:51:48.877" CommentCount="12" />
  <row Id="1344" PostTypeId="1" AcceptedAnswerId="1373" CreationDate="2010-08-01T14:04:15.177" Score="4" ViewCount="261" Body="&lt;p&gt;Proposition 0.4.5.4 in EGA appears to be a general representability theorem. It reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $F$ is a contravariant functor from the category of locally ringed spaces over $S$ to the category of sets. Suppose given representable sub-functors $F_i$ of $F$, such that the morphisms $F_i \to F$ are representable by open immersions. Suppose furthermore that if $Hom(-, X) \to F$ is a morphism and the functors $F_i \times_F Hom(-,X)$ are representable by $Hom(-,X_i)$, the family $X_i$ forms an open covering of $X$. (That $X_i \to X$ is an open immersion follows from the definitions.)  Finally, suppose that if $U$ ranges over the open subsets of a locally ringed space $X$, the functor $U \to F(U)$ is a sheaf.   Then, $F$ is representable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't yet been able to grok the proof, but it appears to be some sort of extended gluing construction. This result appears to be used in proving that fibered products exist in the category of schemes. However, it's fairly easy to directly construct fibered products by gluing open affines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there examples where this result actually makes the life of algebraic geometers easier?  Also, I'd appreciate any links to examples (outside of EGA) where this result is used.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-01T17:43:27.437" LastActivityDate="2010-09-02T08:01:22.617" Title="Examples/other references for EGA 0.4.5.4" Tags="&lt;reference-request&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="1345" PostTypeId="2" ParentId="1338" CreationDate="2010-08-01T15:21:05.333" Score="2" ViewCount="0" Body="&lt;p&gt;You can take any two orthogonal vectors $\vec{e_1}$ and $\vec{e_2}$ on the plane and use them as a basis. You also need some point $(x_0, y_0, z_0)$ on the plane as origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given point with coordinates $(x_1, y_1, z_1)$ on your plane you calculate it's coordinates with respect to new basis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = (x_1 - x_0) e_{1x} + (y_1 - y_0) e_{1y} + (z_1 - z_0) e_{1z}$&lt;br /&gt;&#xA;$y = (x_1 - x_0) e_{2x} + (y_1 - y_0) e_{2y} + (z_1 - z_0) e_{2z}$&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after that you can apply your formulae to get $C_x$ and $C_y$. Those coordinates are easyly transformed back into original 3d coordinates:&lt;br /&gt;&#xA;$x = x_0 + e_{1x} C_x + e_{2x} C_y$&lt;br /&gt;&#xA;$y = y_0 + e_{1y} C_x + e_{2y} C_y$&lt;br /&gt;&#xA;$z = z_0 + e_{1z} C_x + e_{2z} C_y$&lt;br /&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-01T15:21:05.333" CommentCount="5" />
  <row Id="1346" PostTypeId="2" ParentId="164" CreationDate="2010-08-01T18:22:34.037" Score="9" ViewCount="0" Body="&lt;p&gt;The volume of a sphere with radius $a$ may be found by evaluating the triple integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle V = \int\int\int_{\ S}\ dx\ dy\ dz,$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S$ is the volume enclosed by the sphere $x^2+y^2+z^2=a^2$. Changing variables to spherical polar coordinates, we obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle V = \int_{0}^{2\pi}d\phi\int_{0}^{\ \pi}d\theta\int_{0}^{\ a}r^2\sin\theta \ dr = \int_{0}^{2\pi}d\phi\int_{0}^{\pi}\sin\theta \ d\theta\int_{0}^{\ a}r^2\ dr = \frac{4\pi a^3}{3},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as expected.&lt;/p&gt;&#xA;" OwnerUserId="207" LastEditorUserId="207" LastEditorDisplayName="" LastEditDate="2010-08-01T18:28:21.637" LastActivityDate="2010-08-01T18:28:21.637" />
  <row Id="1347" PostTypeId="1" AcceptedAnswerId="1383" CreationDate="2010-08-01T18:31:43.533" Score="35" ViewCount="510" Body="&lt;p&gt;You've spent your whole life in the hyperbolic plane. It's second nature to you that the area of a triangle depends only on its angles, and it seems absurd to suggest that it could ever be otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But recently a good friend named Euclid has raised doubts about the fifth postulate of Poincaré's &lt;em&gt;Elements&lt;/em&gt;. This postulate is the obvious statement that given a line $L$ and a point $p$ not on $L$ there are at least two lines through $p$ that do not meet $L$. Your friend wonders what it would be like if this assertion were replaced with the following: given a line $L$ and a point $p$ not on $L$, there is exactly one line through $p$ that does not meet $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You begin investigating this Euclidean geometry, but you find it utterly impossible to visualize intrinsically. You decide your only hope is to find a model of this geometry within your familiar hyperbolic plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What model do you build?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if there's a satisfying answer to this question, but maybe it's entertaining to try to imagine. For clarity, we Euclidean creatures have built models like the upper-half plane model or the unit-disc model to visualize hyperbolic geometry within a Euclidean domain. I'm wondering what the reverse would be.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-20T22:48:37.860" LastActivityDate="2010-08-21T03:18:02.510" Title="Hyperbolic critters studying Euclidean geometry" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="6" />
  <row Id="1348" PostTypeId="1" CreationDate="2010-08-01T18:47:30.570" Score="5" ViewCount="263" Body="&lt;p&gt;I have a Ph.D. in computational and theoretical chemistry with advanced but field-oriented knowledge of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fascinated by fractals, but I am unable to understand them from the formal point of view. To my level of understanding, they look like a graphical rendering of an ill-conditioned iterative problem, where small variations of the initial condition lead to huge changes in the final result, but that's just what I got out of it with my current knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you explain fractals (such as the Mandelbrot set) to a layperson with basic mathematics knowledge from high school, and how would you instead explain it to someone which has more math training, but not formal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is collateral to a post on the &lt;a href=&quot;http://forthescience.org/blog/2010/07/12/the-mandelbrot-set-in-python/&quot; rel=&quot;nofollow&quot;&gt;Mandelbrot set&lt;/a&gt; I did on my blog some time ago. If you have any comments on what I was doing with my tinkering of the parameters (to get some keywords for further exploration), it's greatly appreciated. I would like to explain it better to my readers, but I am unable to do it. Thanks&lt;/p&gt;&#xA;" OwnerUserId="559" LastActivityDate="2010-08-03T00:47:55.047" Title="How to explain fractals to a layperson and to someone with more math training?" Tags="&lt;fractals&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="1349" PostTypeId="2" ParentId="1348" CreationDate="2010-08-01T19:09:50.977" Score="4" ViewCount="0" Body="&lt;p&gt;There are a lot of examples that are not hard to define rigorously, e.g. the Cantor set. This can be defined rigorously simply by taking a suitable intersection of nested intervals when the middle thirds are successively removed. It is easy to make all sorts of variants, constructing &quot;Cantor-like&quot; sets when you replace &quot;middle third&quot; with &quot;middle fourth&quot; or even with a construction where the ratio of the interval removed changes with the iteration. (This is important because you can then get a set of positive measure.) Leaving aside the remark about measure, all this uses nothing more than elementary mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of fractals is as of subsets of the phase space of certain dynamical systems. For instance, consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smale_horseshoe&quot; rel=&quot;nofollow&quot;&gt;Smale horseshoe&lt;/a&gt;. This is a two-dimensional map that can be visualized geometrically; this is explained in the Wikipedia article. Then the collection of points that stay in the square when you iterate this procedure both forwards and backwards is a fractal set, basically a two-dimensional Cantor set, which is very interesting from the perspective of dynamical systems (it's chaotic---one consequence of this is that knowing a point to arbitrary precision will not tell you what its orbit looks like later on with much precision).  So much of the interesting behavior of dynamical systems is on fractals. You can define a map of the interval into itself such that the points in the interval that stay in the interval no matter how many times you iterate the map is just the ordinary Cantor set, as well.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-01T19:09:50.977" />
  <row Id="1350" PostTypeId="1" CreationDate="2010-08-01T19:29:28.793" Score="-2" ViewCount="287" Body="&lt;p&gt;Is the set of polynomials of degree n a vector space? If not, are there any similar sets that are vector spaces?&lt;/p&gt;&#xA;" OwnerUserId="560" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-03T07:30:35.280" LastActivityDate="2010-08-03T07:30:35.280" ClosedDate="2010-08-02T08:37:12.670" Title="Are the polynomials of degree n a vector space?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="1351" PostTypeId="2" ParentId="1348" CreationDate="2010-08-01T19:53:29.330" Score="8" ViewCount="0" Body="&lt;p&gt;For a &quot;high-level&quot; explanation, I would say this: Fractals are surprisingly complex patterns that result from the repeated application of relatively simple operations/rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the easiest to visualize examples is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Koch_snowflake&quot; rel=&quot;nofollow&quot;&gt;Koch snowflake&lt;/a&gt;, constructed by adding smaller triangles to each face of the figure at each iteration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/d/d9/KochFlake.svg/280px-KochFlake.svg.png&quot; alt=&quot;Koch snowflake&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more real-world example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barnsley%27s_fern&quot; rel=&quot;nofollow&quot;&gt;fern leaf&lt;/a&gt;. The DNA in a single plant cell encodes enough information to describe the structure of an entire leaf (and the entire plant, for that matter) &lt;em&gt;without explicitly describing the location of each cell&lt;/em&gt;. Instead, the cells grow according to a set of simple rules that result in the self-similar appearance of the fern, even at smaller and smaller levels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/7/76/Barnsley_fern_plotted_with_VisSim.PNG&quot; alt=&quot;Barnsley fern&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more complex mathematical explanation that still remains tied to the real world, have a look at the basic &lt;a href=&quot;http://en.wikipedia.org/wiki/Ricker_model&quot; rel=&quot;nofollow&quot;&gt;Ricker model&lt;/a&gt; of population growth and the resulting &lt;a href=&quot;http://www.phaser.com/modules/ecology/ricker/index.html&quot; rel=&quot;nofollow&quot;&gt;bifurcation diagrams&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.phaser.com/modules/ecology/ricker/bif.jpg&quot; alt=&quot;bifurcation diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The x-axis on this graph is population growth rate and the y-axis is population density. Although it looks complex, All it takes is a handful of iterations of the basic formula on a hand calculator to see how the results can oscillate between seemingly random population levels.&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-08-03T00:47:55.047" LastActivityDate="2010-08-03T00:47:55.047" CommentCount="1" />
  <row Id="1352" PostTypeId="1" CreationDate="2010-08-01T20:11:01.703" Score="8" ViewCount="443" Body="&lt;p&gt;I just learned (thanks to &lt;a href=&quot;http://mathoverflow.net/questions/10014/applications-of-the-chinese-remainder-theorem/10017#10017&quot; rel=&quot;nofollow&quot;&gt;Harry Gindi&lt;/a&gt; and on &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/blog.php?b=10595&quot; rel=&quot;nofollow&quot;&gt;Qiaochu Yuan&lt;/a&gt; MO) that the chinese remainder theorem and lagrange interpolation are really just two instances of the same thing.  Similarly the method of partial fractions can be applied to rationals rather than polynomials. I find that seeing a method applied in different contexts, or just learning a connection that wasn't apparent helps me appreciate a deeper understanding of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask, can you help me find more examples of this? Especially ones which you personally found inspiring. &lt;/p&gt;&#xA;" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-01T20:35:53.970" LastActivityDate="2010-08-11T22:02:30.823" CommunityOwnedDate="2010-08-02T09:17:13.293" Title="Surprising Generalizations" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;intuition&gt;" AnswerCount="8" CommentCount="8" FavoriteCount="4" />
  <row Id="1353" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T20:22:20.867" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Localization&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I learned that you could localize categories(and not just abelian!) I was floored. The general idea that we take a class of morphisms in a category and send them functorially to another category where they are isos is awesome. It is also very important in my work, which is generalizing some ideas of Algebraic Geometry to a more categorical setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://ncatlab.org/nlab/show/localization&quot; rel=&quot;nofollow&quot;&gt;link!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-01T20:22:20.867" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1354" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T20:26:23.947" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Galois Connections&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's be honest, the correspondence between Galois groups and field extension is pretty hott. The first time I saw this I was duly impressed. However, about two years ago, I learned about universal covering spaces. Wow! I swear my understanding of covering spaces doubled when the prof told me that this was a &quot;Galois correspondence for fundamental groups and covering spaces&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again here is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_connection#Algebraic_topology%3a_covering_spaces&quot; rel=&quot;nofollow&quot;&gt;link!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-01T20:26:23.947" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="3" />
  <row Id="1355" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T20:27:16.377" Score="5" ViewCount="0" Body="&lt;p&gt;Classification of finitely-generated abelian groups and Jordan normal form are two instances of &lt;a href=&quot;http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain&quot; rel=&quot;nofollow&quot;&gt;the structure theorem for finitely generated modules over a principal ideal domain&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-01T20:27:16.377" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="1" />
  <row Id="1356" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T21:00:47.437" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Model categories&lt;/strong&gt; as a framework for both complexes of R-modules and topological spaces (making precise, for example, analogy between taking projective resolution and replacing a space with weakly homotopy equivalent CW-complex).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-01T21:00:47.437" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1357" PostTypeId="2" ParentId="1333" CreationDate="2010-08-01T21:22:26.807" Score="3" ViewCount="0" Body="&lt;p&gt;In 1976, &lt;a href=&quot;http://www.prismmodelchecker.org/bibitem.php?key=KY76&quot; rel=&quot;nofollow&quot;&gt;Knuth and Yao&lt;/a&gt; proved a result that may help you choose a DVD out of 5, 6, etc.  (I couldn't find a good reference online.) Consider the following problem: You have a fair coin and you must write an algorithm that outputs 1 with probability p1, outputs 2 with probability p2, ..., and outputs n with probability pn. (p1+p2+...+pn=1) Note that &lt;em&gt;any&lt;/em&gt; possible algorithm can be described in terms of certain (possibly infinite) binary trees: A node that has two children means &quot;throw a coin and pick one of the children according to the result&quot;; a leaf is tagged with one of the numbers 1, 2, ..., n and means &quot;output that number&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;An optimal tree that solves the problem has one leaf k on level m if and only if the m-th bit of pk is 1. Otherwise there are zero such leafs.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Optimal&quot; here means not only that the average runtime is as good as possible, but the  stronger guarantee that the probability of running more than m steps is not worse than that of any other algorithm that solves the problem correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, p1=p2=p3=0.01010101... The digit just before dot corresponds to the root (level 0) and and the subsequent digits to subsequent levels. So a tree that has no leaf on levels 0 and 1, has leafs (1, 2, and 3) on level 2, has no leaf on level 3, ... is optimal. This is exactly the solution given by ShreevatsaR. Such a tree is fairly easy to find once you know what leafs you need to put on each level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, if p1=...=p6=1/6=0.00101010101..., then you should have leafs (one of each 1, 2, 3, 4, 5, 6) on levels 3, 5, 7, and so on. If you draw a symmetric tree that obeys this you'll find how to choose one DVD out of 6: First grab three with each hand, then do what ShreevatsaR said for three.  You are also guaranteed that it's optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can mock me and my friends for &lt;a href=&quot;http://www.youtube.com/watch?v=3JBE5ykC1No#t=1m&quot; rel=&quot;nofollow&quot;&gt;using a suboptimal algorithm for allocating hotel rooms&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-01T21:22:26.807" CommentCount="1" />
  <row Id="1358" PostTypeId="2" ParentId="457" CreationDate="2010-08-01T21:26:32.957" Score="4" ViewCount="0" Body="&lt;p&gt;Some principles of inversive geometry can be demonstrated using &lt;a href=&quot;http://mathworld.wolfram.com/PeaucellierInversor.html&quot; rel=&quot;nofollow&quot;&gt;Peaucellier's Inversor&lt;/a&gt;, a linkage device designed to transform circular motion into motion in a straight line and vice versa. There are various online demonstrations however, you may wish to build a &quot;real&quot; one for yourself. The book &quot;Mathematical Models&quot; by Cundy and Rollett has an entire chapter devoted to making mechanical models, including the above and related likages. &lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-08-01T21:26:32.957" CommunityOwnedDate="2010-08-01T21:26:32.957" CommentCount="1" />
  <row Id="1359" PostTypeId="2" ParentId="457" CreationDate="2010-08-01T21:46:41.357" Score="3" ViewCount="0" Body="&lt;p&gt;I'm not sure how good of an example this is since the original problem already has a natural physical interpretation. Nonetheless this is an example of something that can be proved using Euclidean geometry or mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting on one of the edges of a plane polygon $P$ one can set into motion a point-mass which moves along a straight trajectory except on collision with another edge, when it is subjected to an elastic collision response (i.e., its velocity vector is reflected in the edge it has collided with). The trajectory traced out by such a point-mass is its orbit, and an orbit is periodic if the point-mass eventually returns to its starting spot with its starting velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most basic questions one can ask is whether periodic orbits exist in acute triangles. The answer is yes, and a particularly nice periodic orbit in an acute triangle can be constructed by drawing the three altitudes and connecting their bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved using geometry, but it can also be seen in the following physical manner. Assume the triangle's edges are thin wires. Place around each wire a small ring which is free to move along its edge. Now thread a rubber band through the three rings. Tighten this rubber band. This system reaches equilibrium when the rings occupy the positions of the bases of the altitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, in the equilibrium state the triangle whose vertices are the rings will be the inscribed triangle of minimal perimeter. There's no such minimum in an obtuse triangle, so the construction doesn't work in that case. As of today it is unknown whether every obtuse triangle admits a periodic orbit, although much partial progress has been achieved.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-11-08T18:32:35.527" LastActivityDate="2010-11-08T18:32:35.527" CommunityOwnedDate="2010-08-01T21:46:41.357" />
  <row Id="1360" PostTypeId="2" ParentId="1350" CreationDate="2010-08-01T21:53:52.680" Score="1" ViewCount="0" Body="&lt;p&gt;It is not a vector space because there is no zero element.&lt;/p&gt;&#xA;" OwnerUserId="563" LastActivityDate="2010-08-01T21:53:52.680" CommentCount="0" />
  <row Id="1361" PostTypeId="2" ParentId="1347" CreationDate="2010-08-01T22:01:51.833" Score="6" ViewCount="0" Body="&lt;p&gt;Look up &quot;horosphere&quot; (for example, in page 90 of the Princeton Companion to Mathematics).  Wikipedia describes it on its &lt;a href=&quot;http://en.wikipedia.org/wiki/horoball&quot; rel=&quot;nofollow&quot;&gt;Horoball&lt;/a&gt; page. &lt;/p&gt;&#xA;" OwnerUserId="273" LastEditorUserId="273" LastEditorDisplayName="" LastEditDate="2010-08-01T22:15:33.460" LastActivityDate="2010-08-01T22:15:33.460" CommentCount="4" />
  <row Id="1362" PostTypeId="2" ParentId="1348" CreationDate="2010-08-01T22:07:53.733" Score="6" ViewCount="0" Body="&lt;p&gt;That's easy.  I would buy them some of this stuff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://farm2.static.flickr.com/1340/614003477_2d7fd27ca5.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Then I would tell them to stare at it for awhile.)  It's called Romanesco broccoli, and it grows like this for the same reasons, as explained in e.James' answer, that ferns do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, I would tell them to check out &lt;em&gt;Indra's Pearls&lt;/em&gt; by Mumford, Series, and Wright.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-01T22:22:55.683" LastActivityDate="2010-08-01T22:22:55.683" CommentCount="3" />
  <row Id="1363" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T22:16:59.357" Score="6" ViewCount="0" Body="&lt;p&gt;I agree!  I spend much of my mathematical free time exploring such connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a basic one that I constantly ponder.  The rules of matrix multiplication encode two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How to compose a linear transformation $A$ with another linear transformation $B$, with respect to a fixed basis.&lt;/li&gt;&#xA;&lt;li&gt;How to follow an edge of type $A$ on a graph, and then follow an edge of type $B$ (where $A$ and $B$ are just a disjoint partition of the set of edges).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means that one can study walks on graphs by studying how a matrix called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjacency_matrix&quot; rel=&quot;nofollow&quot;&gt;adjacency matrix&lt;/a&gt; behaves.  This leads into all sorts of &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_graph_theory&quot; rel=&quot;nofollow&quot;&gt;beautiful mathematics&lt;/a&gt;; for example, this is the basic tool behind Google's PageRank algorithm, and it also in some sense motivated Heisenberg's &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_mechanics&quot; rel=&quot;nofollow&quot;&gt;matrix mechanics&lt;/a&gt; formulation of quantum mechanics.   I often try to recast results in linear algebra in terms of some combinatorial statement about walks on graphs.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-01T22:16:59.357" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="4" />
  <row Id="1364" PostTypeId="1" AcceptedAnswerId="1365" CreationDate="2010-08-01T22:33:45.520" Score="1" ViewCount="40" Body="&lt;p&gt;I guarantee there is an easy reference on this, but for some reason I cannot find it. If you can point me to a reference or just write a short proof for me, I would be appreciative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a graded ring $R_{\bullet}$ and a localization $R_{\bullet}^{*}$. We also have a graded $R_{\bullet}$-mod, $M_{\bullet}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I want to know; is &lt;code&gt;$\left(R_{\bullet}^{*}\otimes M_{\bullet}\right)_0=\left(R_{\bullet}^{*}\right)_0\otimes \left(M_{\bullet}\right)_0$&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-01T22:55:55.400" LastActivityDate="2010-08-01T23:13:19.207" Title="The zero component of tensor products of graded mods" Tags="&lt;reference-request&gt;&lt;graded-modules&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="1365" PostTypeId="2" ParentId="1364" CreationDate="2010-08-01T23:13:19.207" Score="1" ViewCount="0" Body="&lt;p&gt;For a counterexample, take $R=k[t]$ with its usual grading and $M=R(1)$, the free module of rank one generated in degree $1$. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-01T23:13:19.207" CommentCount="1" />
  <row Id="1366" PostTypeId="2" ParentId="1352" CreationDate="2010-08-01T23:37:04.630" Score="2" ViewCount="0" Body="&lt;p&gt;The law of cosines and the equation for the variance of a sum of (possibly correlated) random variables are both consequences of basic inner product space properties. Details &lt;a href=&quot;http://www.johndcook.com/blog/2010/06/17/covariance-and-law-of-cosines/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-01T23:37:04.630" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1367" PostTypeId="2" ParentId="1352" CreationDate="2010-08-02T00:06:44.387" Score="5" ViewCount="0" Body="&lt;p&gt;I loved learning about how differential forms and the exterior derivative generalize 3-d vector calculus (div, grad, curl).  Differential forms are so elegant in comparison, work in arbitrary dimensions, and give rise to beautiful mathematics (e.g. de Rham cohomology, Hodge theory).  And of course, the generalized Stoke's theorem is one of the prettiest equations: $\int_{\partial R} \phi = \int_R d\phi$.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-02T00:06:44.387" CommunityOwnedDate="2010-08-02T15:52:41.847" />
  <row Id="1368" PostTypeId="1" AcceptedAnswerId="1370" CreationDate="2010-08-02T00:54:21.387" Score="2" ViewCount="590" Body="&lt;p&gt;Struggling with yet another proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that, for any positive integer n: (a + b)&lt;sup&gt;n&lt;/sup&gt; &gt;= a&lt;sup&gt;n&lt;/sup&gt; + b&lt;sup&gt;n&lt;/sup&gt; for all a, b &gt; 0:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wasted 3 pages of notebook paper on this problem, and I'm getting nowhere slowly. So I need some hints.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What technique would you use to prove this (e.g. induction, direct, counter example)&lt;/li&gt;&#xA;&lt;li&gt;Are there any tricks to the proof? I've seen some crazy stuff pulled out of nowhere when it comes to proofs...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="494" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-02T01:00:50.487" LastActivityDate="2010-08-11T22:05:41.697" ClosedDate="2010-08-03T00:12:53.650" Title="Prove: (a + b)^n &gt;= a^n + b^n" Tags="&lt;algebra-precalculus&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="1369" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T01:00:04.590" Score="6" ViewCount="0" Body="&lt;p&gt;This follows directly from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;binomial theorem&lt;/a&gt;.&#xA;Alternatively, you can prove it inductively (which is probably more fun): suppose the inequality true for $n-1$. Then $(a+b)^n = (a+b)(a+b)^{n-1} \geq (a+b)(a^{n-1} + b^{n-1})$ by the inductive hypothesis. So $(a+b)^n \geq  a(a^{n-1}+ b^{n-1}) + b(b^{n-1} + a^{n-1})$, and this is at least $a^n + b^n$.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-02T01:00:04.590" />
  <row Id="1370" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T01:00:14.090" Score="7" ViewCount="0" Body="&lt;p&gt;Hint: Use the binomial theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This states that &#xA;$(a + b)^n = \sum_{k = 0}^n {n \choose k} a^{n-k} b^k = a^n + b^n + \sum_{k=1}^{n-1} {n \choose k} a^{n-k} b^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that every term in the second sum is positive; this is because a, b, and the binomial coefficients are all positive. Therefore, (a+b)&lt;sup&gt;n&lt;/sup&gt; = a&lt;sup&gt;n&lt;/sup&gt; + b&lt;sup&gt;n&lt;/sup&gt; + (sum of positive terms) &gt;= a&lt;sup&gt;n&lt;/sup&gt; + b&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2010-08-02T23:55:06.150" LastActivityDate="2010-08-02T23:55:06.150" CommentCount="5" />
  <row Id="1371" PostTypeId="2" ParentId="760" CreationDate="2010-08-02T01:48:26.383" Score="0" ViewCount="0" Body="&lt;p&gt;I only wish to add the point made in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Great-circle_distance&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; for the great circle distance: namely, the classical formula making use of the arccosine can suffer from numerical instability. The formula using the arctangent (which when implemented on a computer should be done using the two-argument arctangent) is numerically robust.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-02T01:48:26.383" />
  <row Id="1372" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T02:10:27.063" Score="6" ViewCount="0" Body="&lt;p&gt;It might also be helpful for you to think a little about the geometry of the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2$, find a way to put an $a \times a$ square and a $b \times b$ square into a $(a+b) \times (a+b)$ square without any &#xA;overlaps. For $n=3$, see if you can fit an $a \times a \times a$ cube and a $b \times b \times b$ cube within an $(a+b) \times (a+b) \times (a+b)$ cube without overlaps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the notion of having more than three dimensions might seem a little weird, but &#xA;think of the box in $n$ dimensions whose sides have length $a+b$. Can you fit two boxes within it, one with side length $a$ and one with side length $b$? &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-08-02T02:10:27.063" />
  <row Id="1373" PostTypeId="2" ParentId="1344" CreationDate="2010-08-02T04:30:58.563" Score="5" ViewCount="0" Body="&lt;p&gt;Dear Akhil,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one typical situation when this would be used (which you probably already know)&#xA;is when you want to reduce the construction of some object which is supposed to lie over&#xA;a base to the case when the base is affine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, imagine you want to build a scheme T over a base S, and you cover S by open affines S_i; then if you construct the schemes T_i (pull-back of T over S_i) with appropriate gluing data, you will be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in practice, you won't have T but will instead have the functor it represents on S-schemes, T_i will be the fibre product of that functor and S_i (i.e. just restrict the functor to S_i-schemes), and now if T is a sheaf, and the T_i are representable, you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could imagine applying this to build projective spaces or Grassmanians over T. E.g. suppose you had a locally free sheaf from which you wanted to form the associated projective space bundle.  One could do this directly, with some Proj construction, but one could instead figure out the universal property, and then replace S by an open cover S_i over which the locally free sheaf is actually free, in which case the usual projective space of the appropriate dimension&#xA;(taken over S_i) will obviously represent your functor (provided you figured the functor out correctly!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should say that, outside of the context of EGA itself, I don't imagine that people cite this result (or analogous ones) very often.  They are more likely just to write something like &quot;since F is a sheaf, we can reduce our construction to the case when S is affine&quot;.  It is just one of the standard techniques that float around for trying to represent moduli problems.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-02T04:30:58.563" CommentCount="1" />
  <row Id="1374" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T06:02:08.590" Score="2" ViewCount="0" Body="&lt;p&gt;Let's have a precalculus answer: Consider the function of $a$ depending on the parameter $b$ that is $f_b(a)=(a+b)^n-a^n-b^n$. Its derivative relative to $a$ is $f'_b(a)=n((a+b)^{n-1}-a^{n-1})$ because $b&gt;0$ $f'_b(a,b)$ is nonnegative and $f_b$ is an increasing function of $a$ and you can conclude.&lt;/p&gt;&#xA;" OwnerUserId="441" LastEditorUserId="441" LastEditorDisplayName="" LastEditDate="2010-08-02T10:43:39.783" LastActivityDate="2010-08-02T10:43:39.783" CommentCount="2" />
  <row Id="1375" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T07:54:15.427" Score="3" ViewCount="0" Body="&lt;p&gt;Here is another way to look at the inequality.  Pick the larger of a^n and b^n and divide through by that quantity.  This reduces the problem to showing that (1+r)^n &gt; = 1 + r^n for&#xA;some positive real number r &amp;lt;= 1.  If r &amp;lt; 1, we have r^n &amp;lt; r, so 1 + r^n &amp;lt; 1 + r &amp;lt; (1+r)^n.&#xA;I leave the case r = 1 (and n rational and &gt;= 1)  to others.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:05:41.697" LastActivityDate="2010-08-11T22:05:41.697" CommentCount="0" />
  <row Id="1376" PostTypeId="2" ParentId="760" CreationDate="2010-08-02T08:05:28.067" Score="1" ViewCount="0" Body="&lt;p&gt;If you have the Euclidean distance d between the two points, and set D = the diameter of the sphere (so D = 2R), then the great circle distance is D*InvSin(d/D), where InvSin is the inverse sine function.  If you know that d/D is usually very small, then you can  approximate the great circle distance using a few terms of the Taylor series for inverse sine.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:01:26.910" LastActivityDate="2010-08-11T22:01:26.910" CommentCount="0" />
  <row Id="1377" PostTypeId="2" ParentId="1352" CreationDate="2010-08-02T08:12:36.520" Score="2" ViewCount="0" Body="&lt;p&gt;Galois connections are further reaching than one first realizes.  One small application is in model theory, where the relation R between sentences of a theory and models given by&#xA;t R M if sentence t is true in model M, leads to deductively closed theories versus classes of models closed under certain operations, many of which are algebraic in nature.&#xA;They also arise in algebraic geometry and computer science, among other fields.  There&#xA;is a book on Galois connections edited by Marcel Erne; for the strongly inquisitive I recommend checking it out.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:02:30.823" LastActivityDate="2010-08-11T22:02:30.823" CommunityOwnedDate="2010-08-02T15:52:41.847" CommentCount="1" />
  <row Id="1378" PostTypeId="1" CreationDate="2010-08-02T08:34:58.713" Score="0" ViewCount="177" Body="&lt;p&gt;While reading one aptitude book, i came across this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;35% alcohol mixed with 60% alcohol to get a 50% alcohol. In what ratio were they mixed?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent 20 minutes on this question, but I couldn't even figure out what it means! Could anyone explain this to me?&lt;/p&gt;&#xA;" OwnerUserId="569" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-02T09:23:05.357" LastActivityDate="2010-08-04T13:52:32.663" Title="Mixing solutions of specified strength" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="1379" PostTypeId="2" ParentId="1378" CreationDate="2010-08-02T08:54:42.007" Score="1" ViewCount="0" Body="&lt;p&gt;Call the ratio you want to find r=(volume of 35% alcohol):(volume of 60% alcohol).  Suppose that there are r liters of the 35% alcohol (so there are 0.35 L of alcohol and 0.65 L of water), 1 liter of the 60% alcohol (so the ratio is r), and 1+r liters of the 50% alcohol.  From there, you can construct an equation about the alcohol (or about the water) and solve for r.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-02T08:54:42.007" />
  <row Id="1380" PostTypeId="2" ParentId="1378" CreationDate="2010-08-02T08:55:45.233" Score="-1" ViewCount="0" Body="&lt;p&gt;Given &lt;code&gt;a&lt;/code&gt; liters of &lt;code&gt;35%&lt;/code&gt; alcohol solution, and &lt;code&gt;b&lt;/code&gt; liters of &lt;code&gt;60%&lt;/code&gt; alcohol solution, what would be the alcohol ratio after we'll mix this two solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;a&lt;/code&gt; liters are actually &lt;code&gt;a*35/100&lt;/code&gt; alcohol, and &lt;code&gt;a*(1-(35/100))&lt;/code&gt; water. Likewise the &lt;code&gt;b&lt;/code&gt; liters contains &lt;code&gt;b*60/100&lt;/code&gt; alcohol and &lt;code&gt;b*(1-(60/100))&lt;/code&gt; water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing them together would give us &lt;code&gt;a*35/100+b*60/100&lt;/code&gt; alcohol and &lt;code&gt;a*(1-(35/100))+b*(1-(60/100))&lt;/code&gt; water. The question is for which &lt;code&gt;a,b&lt;/code&gt; we have the ratio&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;((a*35+b*60)/100)/((a*65+b*40)/100) = 50% = 1/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(a*35+b*60)/(a*65+b*40) = 1/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I believe you can take it from here...&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2010-08-04T06:32:22.297" LastActivityDate="2010-08-04T06:32:22.297" CommentCount="3" />
  <row Id="1381" PostTypeId="2" ParentId="1368" CreationDate="2010-08-02T09:21:38.727" Score="0" ViewCount="0" Body="&lt;p&gt;Induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n=1 it is trivially true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume true for n=k i.e. (a + b)^k &gt;= a^k + b^k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider case n=k+1 (a+b)^(k+1)=(a+b)(a+b)^k&gt;=(a+b)(a^k+b^k)=a^(k+1)+b^(k+1)+ab^k+ba^k&gt;=a^(k+1)+b^(k+1) as required&lt;/p&gt;&#xA;" OwnerUserId="570" LastActivityDate="2010-08-02T09:21:38.727" />
  <row Id="1382" PostTypeId="1" AcceptedAnswerId="1746" CreationDate="2010-08-02T10:41:04.887" Score="7" ViewCount="279" Body="&lt;p&gt;Some time ago &lt;a href=&quot;http://math.stackexchange.com/users/232/qiaochu-yuan&quot;&gt;Qiaochu Yuan&lt;/a&gt; asked about &lt;a href=&quot;http://math.stackexchange.com/questions/918/how-do-i-count-the-subsets-of-a-set-whose-number-of-elements-is-divisible-by-3-4/&quot;&gt;counting subsets of a set whose number of elements is divisible by 3 (or 4)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story becomes even more interesting if one asks about number of subsets of n-element set with r mod 5 elements. Denote this number by, say, P&lt;sub&gt;n&lt;/sub&gt;(r mod 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experiment shows that for small n P&lt;sub&gt;n&lt;/sub&gt;(r mod 5)-P&lt;sub&gt;n&lt;/sub&gt;(r' mod 5) is always a Fibonacci number (recall that for &quot;r mod 3&quot; corresponding difference is always 0 or 1 and for &quot;r mod 2&quot; they are all 0). It's not hard to prove this statement by induction but as always inductive proof explains nothing. &lt;em&gt;Does anybody have a combinatorial proof?&lt;/em&gt; (Or maybe some homological proof&amp;nbsp;&amp;mdash; I've heard one for &quot;r mod 3&quot;-case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is there some theorem about P&lt;sub&gt;n&lt;/sub&gt;(r mod l) for arbitrary l (besides that it satisfies some recurrence relation of degree growing with l)?&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-03T08:05:09.893" LastActivityDate="2010-08-07T07:24:39.447" Title="Counting subsets with r mod 5 elements" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="1383" PostTypeId="2" ParentId="1347" CreationDate="2010-08-02T11:19:44.533" Score="29" ViewCount="0" Body="&lt;p&gt;Here's another version of Doug Chatham's answer, but with details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you lived in Hyperbolic space, then Euclidean geometry would be natural to you as well.  The reason is that you can take what is called a horosphere (in the half-space model for us, this is just a hyperplane which is parallel to our limiting hyperplane) and this surface actually has a Euclidean geometry on it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unlike for us, where the hyperbolic plane cannot be embedded into Euclidean 3-space, the opposite is true: the Euclidean plane can be embedded into hyperbolic 3-space! So this is analogous to our understanding of spherical geometry.  It's no surprise the spherical geometry is slightly different, however, it fits nicely into our Euclidean view of things, because spherical geometry is somewhat contained in three-dimensional geometry because of the embedding.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-02T11:19:44.533" CommentCount="2" />
  <row Id="1384" PostTypeId="2" ParentId="1260" CreationDate="2010-08-02T11:27:05.767" Score="0" ViewCount="0" Body="&lt;p&gt;This was the specific code I needed:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\begin{bmatrix} &#xA;x_{1} &amp;amp; x_{4} &amp;amp; x_{7} \\&#xA;x_{2} &amp;amp; x_{5} &amp;amp; x_{8} \\&#xA;x_{3} &amp;amp; x_{6} &amp;amp; x_{9} &#xA;\end{bmatrix}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} &#xA;x_{1} &amp;amp; x_{4} &amp;amp; x_{7} \&#xA;x_{2} &amp;amp; x_{5} &amp;amp; x_{8} \&#xA;x_{3} &amp;amp; x_{6} &amp;amp; x_{9} &#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Looks like the library the site is using doesn't support bmatrix?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Texify's version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.texify.com/img/%5CLARGE%5C%21%5Cbegin%7Bbmatrix%7D%20x_%7B1%7D%20%26%20x_%7B4%7D%20%26%20x_%7B7%7D%20%5C%5Cx_%7B2%7D%20%26%20x_%7B5%7D%20%26%20x_%7B8%7D%20%5C%5Cx_%7B3%7D%20%26%20x_%7B6%7D%20%26%20x_%7B9%7D%20%5Cend%7Bbmatrix%7D.gif&quot; alt=&quot;column vector&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="54" LastActivityDate="2010-08-02T11:27:05.767" />
  <row Id="1385" PostTypeId="2" ParentId="548" CreationDate="2010-08-02T11:38:02.740" Score="2" ViewCount="0" Body="&lt;p&gt;The existence of a (multiplicative) inverse for the (additive) zero is &lt;a href=&quot;http://en.wikipedia.org/wiki/Consistency&quot; rel=&quot;nofollow&quot;&gt;inconsistent&lt;/a&gt; with the other &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29#Definition_and_illustration&quot; rel=&quot;nofollow&quot;&gt;field axioms&lt;/a&gt;. As you noticed, the crux is that anything multiplied by zero gives zero. Let us then establish this fact first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 is identity for addition: $0+0=0$ &lt;br/&gt;&#xA;multiply by some x: $(0+0)\cdot x=0\cdot x$&lt;br/&gt;&#xA;distributivity: $0\cdot x + 0\cdot x = 0\cdot x$&lt;br/&gt;&#xA;add the additive inverse of $0\cdot x$ to both sides: $(0\cdot x + 0\cdot x) + (-(0\cdot x)) = 0\cdot x + (-(0\cdot x))$&lt;br/&gt;&#xA;associativity of addition: $0\cdot x + (0\cdot x + (-(0\cdot x))) = 0\cdot x + (-(0\cdot x))$&lt;br/&gt;&#xA;definition of &quot;additive inverse&quot;: $0\cdot x + 0 = 0$&lt;br/&gt;&#xA;zero is additive identity: $0\cdot x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume now that there is a multiplicative inverse of 0, denoted by Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) $0\times Z=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the last two relations, 1=0, which contradicts another field axiom (often forgotten), which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ne0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you either accept that 0 has no inverse or you change at least one of the field axioms---you can't have both at the same time. In a sense, it is a matter of convention which axioms you choose. In practice, some sets of axioms lead to more useful consequences. For example, if you want 0 to have an inverse and drop the axiom saying that 1 does not equal 0, then the 'arithmetic' you end up doing won't be very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, you are right.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-02T11:38:02.740" />
  <row Id="1386" PostTypeId="2" ParentId="536" CreationDate="2010-08-02T12:06:59.440" Score="0" ViewCount="0" Body="&lt;p&gt;If you'd need only the 1d case, you can get this statement from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm&quot; rel=&quot;nofollow&quot;&gt;Law of the iterated logarithm&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not give much combinatorial insight, and well... requires said law, which is a little more work to proof.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-02T12:06:59.440" />
  <row Id="1387" PostTypeId="2" ParentId="1113" CreationDate="2010-08-02T12:18:54.860" Score="7" ViewCount="0" Body="&lt;p&gt;I think I know how to prove it without using of Carmichael numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $n = 3^{2^k} - 2^{2^k}$. Then it's not hard to prove by induction on $k$ that $2^k | n - 1$. Hence $3^{2^k} - 2^{2^k} | 3^{n-1} - 2^{n-1}$ (that's true because $a - b | a^r - b^r$ for any natural $a, b$ and $k$).&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-02T12:18:54.860" />
  <row Id="1388" PostTypeId="1" AcceptedAnswerId="1399" CreationDate="2010-08-02T12:42:27.637" Score="7" ViewCount="322" Body="&lt;p&gt;Galois theory tells us that $\frac{z^{11}-1}{z-1} = z^{10} + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the roots be $\zeta^1,\zeta^2,\ldots,\zeta^{10}$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group $[2,4,8,5,10,9,7,3,6,1]$ we can partition into the five subgroups of conjugate pairs $[2,9]$,$[4,7]$,$[8,3]$,$[5,6]$,$[10,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $q_1 = \zeta^2+\zeta^9$,$q_2 = \zeta^4+\zeta^7$,$q_3 = \zeta^8+\zeta^3$,$q_4 = \zeta^5+\zeta^6$,$q_5 = \zeta^{10}+\zeta^1$. So if we can solve the quintic $(q - q_1)(q - q_2)(q - q_3)(q - q_4)(q - q_5) = q^5 + q^4 - 4q^3 - 3q^2 + 3q + 1 = 0$ we would just be left to solve a few quadratic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pari/gp tells me this quintic has the cyclic group C(5):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? polgalois(x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1)&#xA;%1 = [5, 1, 1, &quot;C(5) = 5&quot;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've worked through examples of solving quadratic and cubic equations based on the galois group but when it comes to this quintic I'm completely stumped so any advice would be tremendously helpful! Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Thanks to Robin Chapman the problem is reduced significantly. Let $\omega$ be a primitive 5th root of unity (which is easy to express in radicals), it only remains to express $(q_1 + \omega q_2  + \omega^2 q_3  + \omega^3 q_4  + \omega^4 q_5)^5$ in terms of rationals and powers of $\omega$ (and then everything can be substituted back and solved easily). We know this is possible because the term is fixed by the quintics galois group, how to actually perform this evades me but I will try to find a way.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-05T19:34:55.197" Title="How to solve a cyclic quintic in radicals?" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="1389" PostTypeId="1" AcceptedAnswerId="1391" CreationDate="2010-08-02T13:06:19.443" Score="7" ViewCount="304" Body="&lt;p&gt;Prove that the equation &lt;i&gt;n&lt;sup&gt;a&lt;/sup&gt; + n&lt;sup&gt;b&lt;/sup&gt; = n&lt;sup&gt;c&lt;/sup&gt;&lt;/i&gt;, with &lt;i&gt;a,b,c,n &lt;/i&gt; positive integers, has infinite solutions if &lt;i&gt;n&lt;/i&gt;=2, and no solution if &lt;i&gt;n&lt;/i&gt;&amp;ge;3.&lt;/p&gt;&#xA;" OwnerUserId="89" LastEditorUserId="89" LastEditorDisplayName="" LastEditDate="2010-08-02T13:17:52.630" LastActivityDate="2010-08-02T20:24:33.080" Title="Not quite Fermat's Last Theorem" Tags="&lt;number-theory&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="4" />
  <row Id="1390" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T13:08:19.157" Score="2" ViewCount="0" Body="&lt;p&gt;(assume b&amp;ge;a) &lt;code&gt;$n^b&amp;lt;n^a+n^b&amp;lt;n^{b+1}$&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-02T13:08:19.157" CommentCount="3" />
  <row Id="1391" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T13:11:50.193" Score="17" ViewCount="0" Body="&lt;p&gt;So this is fermats last theorem upside down? It occurs to me if we have two binary numbers we may add them to get another power of two,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   1000000&#xA;   1000000&#xA;+ --------&#xA;  10000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but if we had two numbers in base 3, say&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  1000000&#xA;  1000000&#xA;+ -------&#xA;  2000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;we would not have so much luck.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-02T13:11:50.193" CommentCount="14" />
  <row Id="1392" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T13:13:21.440" Score="5" ViewCount="0" Body="&lt;p&gt;If $n=2$ we can take $a=k, b=k, c=k+1$ for any $k \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n \ge 3$. We can assume that $a, b, c \ge 0$ because if not we could multiply left and right side by $n^k$ to make them positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's clear that $c \ge a$ and $c \ge b$. Then we have $n^a | n^c$, hence $n^a | n^a + n^b$ and $a \le b$. In the same way $b \le a$. So $a = b$. Hence $2n^a = n^c$ and $n=2$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-02T13:13:21.440" CommentCount="1" />
  <row Id="1393" PostTypeId="1" AcceptedAnswerId="1404" CreationDate="2010-08-02T13:14:03.730" Score="1" ViewCount="115" Body="&lt;p&gt;I already asked if there are more rationals than integers here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1311/are-there-more-rational-numbers-than-integers&quot;&gt;http://math.stackexchange.com/questions/1311/are-there-more-rational-numbers-than-integers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is one particular argument that I didn't give before which I still find compelling...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every integer is also a rational. There exist (many) rationals that are not integers. Therefore there are more rationals than integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, in a sense, I am simply choosing one particular bijection, so by the definition of set cardinality this argument is irrelevant. But it's still a compelling argument for &quot;size&quot; because it's based on a trivial/identity bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; please note that the above paragraph indicates that I know about set cardinality and how it is defined, and accept it as a valid &quot;size&quot; definition, but am asking here about something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, the set of integers is a proper subset of the set of rationals. It seems strange to claim that the two sets are equal in size when one is a proper subset of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there, for example, some alternative named &quot;size&quot; definition consistent with the partial ordering given by the is-a-proper-subset-of operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; clearly it is reasonable to define such a partial order and evaluate it. And while I've use geometric analogies, clearly this is pure set theory - it depends only on the relevant sets sharing members, not on what the sets represent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helpful answers might include a name (if one exists), perhaps for some abstraction that is consistent with this partial order but defined in cases where the partial order is not. Even an answer like &quot;yes, that's valid, but it isn't named and doesn't lead to any interesting results&quot; may well be correct - but it doesn't make the idea unreasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if some of my comments aren't appropriate, but this is pretty frustrating. As I said, it feels like I'm violating some kind of taboo.&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-08-02T14:38:15.900" LastActivityDate="2010-08-02T15:20:10.567" Title="Relative sizes of sets of integers and rationals revisited - how do I make sense of this?" Tags="&lt;set-theory&gt;" AnswerCount="7" CommentCount="4" />
  <row Id="1394" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:23:49.073" Score="0" ViewCount="0" Body="&lt;p&gt;Actually to elaborate a bit more on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems strange to claim that the two sets are equal in size&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us consider binary representations, every natural number can be written in binary. For example $13 = 1101_2$ but we can define two functions $N(q) = 1+q$, $D(q) = 1/(1+1/q)$ and interpret a binary sequence as a composition of these functions applied to 1, for example $1101_{\mathbb{Q}} = (N \circ N \circ D \circ N) 1 = 8/3$ and by Euclids algorithm this defines every (positive) rational number exactly once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all they are, are different interpretations of binary sequences, it would be weird not to claim have equal size!&lt;/p&gt;&#xA;" LastActivityDate="2010-08-02T13:23:49.073" CommentCount="2" />
  <row Id="1395" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:24:46.730" Score="0" ViewCount="0" Body="&lt;p&gt;It may seem strange that the set of integers is a proper subset of the set of rationals, but this is exactly the definition of an infinite set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it could be easier for you to see that there cannot be a size-based definition when talking about infinite set if you consider this: with two finite sets, in whichever way you choose elements to be removed one at a time from both sets at one, you'll end up with the larger set having still some element while the smaller remained empty. This does not happen with infinite sets; even if you find a way to remove all elements (for example, remove he first at T=0, the second a T=1/2, the third at T=3/4, and so on) you may always choose an ordering in which the &quot;larger&quot; set becomes empty, and the &quot;smaller&quot; has still some element.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-02T13:24:46.730" CommentCount="6" />
  <row Id="1396" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:32:01.190" Score="1" ViewCount="0" Body="&lt;p&gt;I've found &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Hotel&lt;/a&gt; a useful example to understand (or fail to understand, but on a higher level), how much &quot;infinity&quot; really is, and how much the naive view on things fails when confronted with infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It deals with the easier case, comparing the integers with even integers, but maybe it will help. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The wikipedia article linked is not great, but google will surely turn out more useful.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-02T13:32:01.190" CommentCount="4" />
  <row Id="1397" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:38:42.877" Score="1" ViewCount="0" Body="&lt;p&gt;You could consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_relation&quot; rel=&quot;nofollow&quot;&gt;order relation&lt;/a&gt; generated by the operation of inclusion between sets. It is a partial order, and it is, in some way, related to the &quot;size&quot; of the sets.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-02T13:38:42.877" CommentCount="1" />
  <row Id="1398" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T13:44:04.250" Score="0" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there, for example, some alternative named &quot;size&quot; definition consistent with the partial ordering given by the is-a-proper-subset-of operator?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it is important to keep in mind the context you are looking at something when you want to talk about sizes.  In a topological or geometric context, if $A \subset B$ then we may want to think of $A$ as smaller than $B$.  However, when talking about cardinality of a set, we need to think of the structure of the set as a set (and not part of another set like the real numbers or in its geometric/topological context).  In this way, the only reasonable definition for sets to be the same (isomorphic) is if there is a bijection between them.  Therefore if we want to assign some cardinality to something as a set, it should be the same for all sets that are in bijection with it.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-02T13:44:04.250" CommentCount="2" />
  <row Id="1399" PostTypeId="2" ParentId="1388" CreationDate="2010-08-02T13:49:38.980" Score="8" ViewCount="0" Body="&lt;p&gt;For full details of this and more, the best place to look is the&#xA;following paper (if you have access to it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. S. Dummit, Solving solvable quintics. &lt;em&gt;Math. Comp.&lt;/em&gt; &lt;strong&gt;57&lt;/strong&gt; (1991), 387-401.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea (which extends to any equation with a cyclic Galois group)&#xA;is to consider Lagrange resolvents. Let the equation have roots $x_1,\ldots,x_5$&#xA;with an element $\tau$ of the Galois group permuting them as $x_i\mapsto x_{i+1}$.&#xA;Let $\zeta=\exp(2\pi i/5)$ be the standard fifth root of unity. Then the Lagrange&#xA;resolvents are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_0=x_1+x_2+x_3+x_4+x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_1=x_1+\zeta x_2+\zeta^2 x_3+\zeta^3 x_4+\zeta^4 x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_2=x_1+\zeta^2 x_2+\zeta^4 x_3+\zeta x_4+\zeta^3 x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_3=x_1+\zeta^3 x_2+\zeta x_3+\zeta^4 x_4+\zeta^2 x_5,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_4=x_1+\zeta^4 x_2+\zeta^3 x_3+\zeta^2 x_4+\zeta x_5.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once one has $A_0,\ldots,A_4$ one easily gets $x_1,\ldots,x_5$.&#xA;It's easy to find $A_0$ :-)&#xA;The point is that $\tau$ takes $A_j$ to $\zeta^{-j}A_j$&#xA;and so takes $A_j^5$ to $A_j^5$. Thus $A_j^5$ can be written down in terms&#xA;of rationals (if that's your starting field) and powers of $\zeta$.&#xA;Alas here is where the algebra becomes difficult. The coefficients&#xA;of powers of $\zeta$ in $A_1^5$ are complicated. They can be expressed&#xA;in terms of a root of a &quot;resolvent polynomial&quot; which will have a rational&#xA;root as the equation is cyclic. Once one has done this, you have $A_1$&#xA;as a fifth root of a certain explicit complex number. Then one can&#xA;express the other $A_j$ in terms of $A_1$. The details are not very pleasant,&#xA;but Dummit skilfully navigates through the complexities, and produces&#xA;formulas which are not as complicated as they might be. Alas, I don;t have&#xA;the time nor the energy to provide more details.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-02T13:49:38.980" CommentCount="1" />
  <row Id="1400" PostTypeId="1" AcceptedAnswerId="1407" CreationDate="2010-08-02T13:59:04.213" Score="7" ViewCount="165" Body="&lt;p&gt;What is the probability that the convex hull of n+2 random points on n-dimensional sphere contains sphere's center?&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-02T14:16:35.327" LastActivityDate="2010-08-03T17:08:01.413" Title="Probability that the convex hull of random points contains sphere's center" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="1401" PostTypeId="1" AcceptedAnswerId="1498" CreationDate="2010-08-02T14:19:57.380" Score="4" ViewCount="348" Body="&lt;p&gt;Why are groups $PSL_3(\mathbb{F}_2)$ and $PSL_2(\mathbb{F}_7)$ isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update.&lt;/em&gt; There is a group-theoretic proof (see answer). But is there any geometric proof? Or some proof using octonions, maybe?&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-02T17:01:00.947" LastActivityDate="2010-08-04T07:29:19.783" Title="Why PSL_3(F_2)=PSL_2(F_7)" Tags="&lt;group-theory&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="3" />
  <row Id="1402" PostTypeId="2" ParentId="1401" CreationDate="2010-08-02T14:37:05.560" Score="7" ViewCount="0" Body="&lt;p&gt;Both are simple groups of order 168, and each simple group of order $168$&#xA;is isomorphic to $PSL_2(7)$. An extended exercise, with hints. Prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a simple group of order $168$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has $8$ Sylow $7$-subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be identified with a subgroup of $A_8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Labelling the objects it acts on as $\infty,0,1,\ldots,6$&#xA;one Sylow $7$-subgroup is generated by $g=(0\ 1\ 2\ 3\ 4\ 5\ 6)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $G$ is $2$-transitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normalizer of $\langle g\rangle$ is generated by $g$ and&#xA;$h=(1\ 2\ 4)(3\ 6\ 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setwise stabilizer $H$ of &lt;code&gt;$\{\infty,0\}$&lt;/code&gt; is generated by $h$&#xA;and another element $k$ which is the product of $(\infty\ 0)$ and&#xA;three other disjoint transpositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $H$ is cyclic, then the Sylow $2$-subgroup of $G$ would be unique,&#xA;leading to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $H$ is nonabelian and we can take $k=(\infty\ 0)(1\ 6)(2\ 3)(4\ 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally $G$ is $PSL_2(7)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-02T14:37:05.560" CommentCount="2" />
  <row Id="1403" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T14:47:49.870" Score="1" ViewCount="0" Body="&lt;p&gt;Purely set theoretically, cardinality is the right way to think of the &quot;size&quot; of a set. A bijection $f:A\to B$ simply renames each element $x$ in $A$ to $f(x)$, and one reasonably wants the size of a set not to depend on the names given to its elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other notions of size if you let your sets have more structure. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; (if it exists) of a subset $A$ of the natural numbers $\mathbb N$ can be thought as the relative size of $A$ to $\mathbb N$. The natural density of the even numbers is $1/2$, for example, so one might say there are half as many even natural numbers as there are natural numbers altogether. If $A$ and $B$ have natural densities $d(A)$ and $d(B)$, and $A\subseteq B\subseteq \mathbb N$, then $d(A)\leq d(B)$. Not all subsets of $\mathbb N$ have a natural density though, so in particular we can't compare the &quot;sizes&quot; of all sets of naturals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is considering the (measurable) subsets of a set $X$ equipped with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;measure&lt;/a&gt; $m$. If $A$ and $B$ are measurable subsets of $X$, and $A\subseteq B$, then $m(A)\leq m(B)$. For example, we can use the Lebesgue measure $m$ on $X=\mathbb R$, which gives measure 1 to the interval $[0,1]$ and measure $1/2$ to the interval $[0,1/2]$. But again, not all subsets of $X$ are measurable, so not all sets can be compared size-wise this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in both the approaches above, we can only compare the size of a set relative to some other fixed set ($\mathbb N$ or $X$). Any finite set and the set of rational numbers both have measure 0 with respect to the Lebesgue measure on $\mathbb R$, for example, so we would be forced to admit them to have the same size in this setting.&lt;/p&gt;&#xA;" OwnerUserId="405" LastEditorUserId="405" LastEditorDisplayName="" LastEditDate="2010-08-02T14:56:14.810" LastActivityDate="2010-08-02T14:56:14.810" CommentCount="2" />
  <row Id="1404" PostTypeId="2" ParentId="1393" CreationDate="2010-08-02T14:48:40.757" Score="3" ViewCount="0" Body="&lt;p&gt;Of course, there are other notions of size.  In particular, your notion of &quot;a partial order based on inclusion of sets&quot; is a very fruitful concept which has been used frequently.  As a quick example, there is a technique in mathematical logic/set theory called &quot;forcing&quot; which is used to show that certain mathematical statements are unprovable.  Forcing often starts with a partial ordered set where the order is given by inclusion of subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the everyday world interpretation of the word &quot;size&quot;, there are (at least) two problems with the using the partial order given by inclusion of subsets.  The first is, as you said, a &lt;em&gt;partial&lt;/em&gt; order: there are two sets which cannot be compared, i.e., there are 2 sets where you cannot say one is bigger than the other.  The second is that two things will have the same size precisely when the two things are absolutely the same.  There is no notion of different things which happen to be the same size - that can't happen in this partial order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, lets say we're looking at subsets of the integers.  You pull out your favorite subset: all the odd integers and I pull out mine: all the even integers.  Using the partial order definition of size, these two sets are incomparable.  Mine is neither bigger than, smaller than, or the same size as yours.  To contrast that, using the cardinality notion of size, they have the same size.  This is evidenced by simply taking everything in your set and adding 1 to it to get everything in my set.  For an even more absurd example, consider the set {0} and the set {1}.  One would expect these two sets to have the same notion of &quot;size&quot; (for any notion of &quot;size&quot;!), but using the partial order notion, one cannot compare these two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, cardinality (or, the way I used &quot;size&quot; in the previous link) is defined on ALL sets (assuming the axiom of choice), even those which a priori have no subset relation.  And there are many examples of sets which have the same cardinality, but which are not equal.  (For example, the set of evens and odds, or the sets {0} and {1}).&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-02T15:20:10.567" LastActivityDate="2010-08-02T15:20:10.567" CommentCount="10" />
  <row Id="1405" PostTypeId="2" ParentId="398" CreationDate="2010-08-02T15:10:37.740" Score="0" ViewCount="0" Body="&lt;p&gt;Almost all of this misses the idea that you can't do traditional arithmetic with infinity that way that most of us think of it.  10/2 = 5, five is half of 10, no problem.  If Judy has five apples and Joe has 10 apples, then it is proper to say &quot;Judy has half as many apples as Joe.&quot;  But to say &quot;there are half as many even positive integers as there are positive intergers&quot; is to infer something about infinity/2, or half of infinity, which is not entirely kosher.&lt;/p&gt;&#xA;" OwnerUserId="575" LastActivityDate="2010-08-02T15:10:37.740" />
  <row Id="1406" PostTypeId="2" ParentId="1400" CreationDate="2010-08-02T15:34:25.363" Score="5" ViewCount="0" Body="&lt;p&gt;This problem is discussed in J. G. Wendel; A Problem in Geometric Probability, Mathematica Scandinavica 11 (1962) 109-111. Wendel showed, that the probability of $N$ random points lying on the surface of the unit sphere in dimension $n$ all lie on one hemisphere is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{-N+1}\sum_{k=0}^{n-1} {{n-1}\choose k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found this &lt;a href=&quot;http://www.math.uni-bielefeld.de/~sillke/PUZZLES/ranpoint.txt&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-08-03T06:13:11.830" LastActivityDate="2010-08-03T06:13:11.830" CommentCount="1" />
  <row Id="1407" PostTypeId="2" ParentId="1400" CreationDate="2010-08-02T15:44:39.370" Score="4" ViewCount="0" Body="&lt;p&gt;This is one of those old chestnuts that come up again and again.&#xA;To be precise, the probability that the convex hull of $n+2$ points in $S^n$&#xA;(the unit sphere in $\mathbb{R}^{n+1}$) contains the origin is $2^{-n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a brief argument at &lt;a href=&quot;http://mathworld.wolfram.com/SphereTetrahedronPicking.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram's mathworld&lt;/a&gt; which I don't find&#xA;entirely convincing but which certainly can be patched to form&#xA;a convincing argument. In brief, show that for random points $P_1,\ldots,P_{n+2}$&#xA;on the sphere, then with probability one, exactly one choice of signs&#xA;will put the centre in the convex hull of $\pm P_1,\pm P_2,\ldots,\pm P_{n+1}$&#xA;and $P_{n+2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (3/8/2010)&#xA;Thanks to Grigory for his comment. Changing the notation slightly,&#xA;one can show that under some fairy weak hypotheses, if we choose $m+1$&#xA;points randomly and indepedently in $\mathbb{R}^m$ the probability their convex&#xA;hull contains the origin is $2^{-m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a probability distribution on $\mathbb{R}^m$ and choose a sequence&#xA;of points (which we identify with vectors) independently from that distribution.&#xA;Our first condition on this distribution is that $m$ vectors $v_1,\ldots,v_m$&#xA;chosen independently from it are linearly independent with probability one.&#xA;This can fail if say some point occurs with nonzero probability or the&#xA;distribution lies in a hyperplane through the origin. Assume this condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a sequence $v_0,v_1,\ldots,v_m$ of random points chosen according to&#xA;our distribution are linearly independent: there are reals $a_i$ not all zero with&#xA;$\sum_i a_i v_i=0$. By our condition, with probability one, the sequence&#xA;$(a_0,\ldots,a_m)$ is unique up to constant multiple, and moreover all the&#xA;$a_i$ are nonzero. So we may assume $a_0=1$ and $a_1,\ldots,a_m$&#xA;are nonzero and uniquely determined. Then the convex hull of the $v_i$&#xA;contains the origin if an only if all the $a_i$ are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we introduce another condition: that the distribution is centrally symmetric;&#xA;in detail the probability that a random vector $v$ lies in a set $A$ equals&#xA;the probability that $-v$ lies in $A$. A condition like this is clearly necessary;&#xA;it stops the distribution being supported on a small region far from the origin.&#xA;This condition shows that all the $2^m$ possibilities of signs&#xA;for $a_1,\ldots,a_m$ are equiprobable; since changing the sign of some&#xA;$v_i$ changes the sign of $a_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, if our probability distribution on $\mathbb{R}^m$ satisfies&#xA;these two condition, the probability that the convex hull of $m+1$&#xA;indepdendently chosen points contains the origin is $2^{-m}$.&#xA;These conditions are satisfied by the uniform distribution on a sphere&#xA;with centre at the origin, but also by many others.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-03T17:08:01.413" LastActivityDate="2010-08-03T17:08:01.413" CommentCount="2" />
  <row Id="1408" PostTypeId="1" AcceptedAnswerId="1411" CreationDate="2010-08-02T16:20:02.310" Score="6" ViewCount="232" Body="&lt;p&gt;Suppose that $X$ is a topological space with a sheaf of rings $\mathcal{O}_X$. In general, the stalk at a point $p \in X$ is the direct limit of the rings $\mathcal{O}_X(U)$ for all open sets $U$ containing $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two questions on computing stalks - I think both should be true, since a direct limit should be some sort of &quot;limiting process&quot;, but that's far from convincing for me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can I compute the stalk of $\mathcal{O}_X$ at a point $p \in X$ by only limiting over basic open sets of $X$ containing $p$?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Can I compute the stalk of $\mathcal{O}_X$ at a point $p \in X$ by excluding some finite number of &quot;large&quot; open sets around $p$, and then limiting over the remaining open sets around $p$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-08-04T17:12:21.593" Title="Computing stalks: do direct limits behave like limits?" Tags="&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="4" />
  <row Id="1409" PostTypeId="2" ParentId="1388" CreationDate="2010-08-02T16:43:34.530" Score="1" ViewCount="0" Body="&lt;p&gt;I recommend you go and look at this reference. www.passhema.org/proceedings/2007/2007DavisGupta.pdf&lt;/p&gt;&#xA;" OwnerUserId="441" LastActivityDate="2010-08-02T16:43:34.530" CommentCount="1" />
  <row Id="1410" PostTypeId="2" ParentId="398" CreationDate="2010-08-02T16:54:02.130" Score="0" ViewCount="0" Body="&lt;p&gt;You can also get it the other way around -- N is (kind of) a subset of E. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Map each integer n into 4n. Then there's a bijection (1-1, onto function) between N and (let's call it) N4, the set of all integers divisible by 4, so N and N4 are, in your intuition, of the same &quot;size&quot;.  But N4 is a proper subset of E !  Of course, you could just as well have mapped N directly to E with bijection f(n) = 2n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, that shows you how slippery it is to apply concepts of finite size to infinite sets, and why bijection is the way to go.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-08-02T17:06:16.650" LastActivityDate="2010-08-02T17:06:16.650" />
  <row Id="1411" PostTypeId="2" ParentId="1408" CreationDate="2010-08-02T16:55:26.730" Score="4" ViewCount="0" Body="&lt;p&gt;Yes. The general statement is the following: limit over a poset is equal to limit over its any &lt;a href=&quot;http://en.wikipedia.org/wiki/Cofinal&quot; rel=&quot;nofollow&quot;&gt;coinitial&lt;/a&gt; subset. Formal proof is easy (hint: construct maps in both directions) and informally it's an analogue of &quot;subsequence has the same limit as a sequence&quot; theorem.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-02T16:55:26.730" CommentCount="3" />
  <row Id="1412" PostTypeId="2" ParentId="398" CreationDate="2010-08-02T17:01:30.167" Score="0" ViewCount="0" Body="&lt;p&gt;The density argument above is not quite to the point, since it involves not only a set but an ordering.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if you play with orderings, you can get even stranger results, such as the integers and rationals can both be ordered as a continuum (according to the ordering, between any two members in the set there is another member) and a non-continuum (there are two members with no other member in between). Or worse, between any two members is at most a finite set of other members (non-continuum) or an infinite set of other members (continuum).  That's kind of an extreme form of density.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-02T17:01:30.167" />
  <row Id="1413" PostTypeId="1" AcceptedAnswerId="1414" CreationDate="2010-08-02T17:10:25.987" Score="11" ViewCount="424" Body="&lt;p&gt;This is what I know about why someone might care about fields of positive characteristic:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;they are useful for number theory&lt;/li&gt;&#xA;&lt;li&gt;in algebraic geometry, a theory of &quot;geometry&quot; can be developed over them, and it's fun to see how this geometry works out&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some people might read this and think, &quot;What more could you need?&quot; But I've never been able to make myself care about number theory, so (1) doesn't help me. (2) is nice for what it is, but I'm hoping there's something more. My understanding of (2) is that this is only geometry in a rather abstract sense and, for instance, there's no generally useful way to directly visually represent these fields or varieties over them the way we can over the reals or complex numbers. (Drawing a curve in R^2 and saying it's the curve over some other field may be helpful for some purposes, but it's not what I'm after here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything else? The ideal (surely impossible) answer for me would be &quot;Yes, such fields are very good models for these common and easy to understand physical systems: A, B, C. Also, we can visualize them and varieties over them quite easily by method D. Finally, here's a bunch of surprising and helpful applications to 500 other areas of mathematics.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: to answer Qiaochu's comment about what I do care about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I care about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;algebraic &amp;amp; geometric topology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;differential geometry &amp;amp; topology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;applications to physics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I certainly care about algebraic geometry over C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is to say I understand the motivations behind these subjects and the general idea, not necessarily that I know them in depth)&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2010-08-02T18:26:49.887" LastActivityDate="2010-08-17T22:50:03.043" Title="Why should I care about fields of positive characteristic?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="3" />
  <row Id="1414" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T17:43:12.907" Score="13" ViewCount="0" Body="&lt;p&gt;Let's suppose you care about finite groups, one way or another.  Then you probably care about the classification of finite simple groups.  The bulk of this classification is the groups of Lie type, which were discovered by finding analogues of Lie groups over finite fields.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finite fields are also (as one might guess) very important in computer science.  I'm certainly no expert, but here are some applications I know of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cryptographic protocols like &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange&quot; rel=&quot;nofollow&quot;&gt;Diffie-Hellman&lt;/a&gt; have as their basis the simple fact that it is difficult to invert exponentiation in finite fields.&lt;/li&gt;&#xA;&lt;li&gt;The standard way that one factors polynomials over the integers is to apply something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Berlekamp_algorithm&quot; rel=&quot;nofollow&quot;&gt;Berlekamp's algorithm&lt;/a&gt; to factor them over several finite fields first, then combine the factorizations.  &lt;/li&gt;&#xA;&lt;li&gt;The classic theorem that &lt;a href=&quot;http://en.wikipedia.org/wiki/IP_(complexity)#Proof_that_IP_.3D_PSPACE&quot; rel=&quot;nofollow&quot;&gt;IP=PSPACE&lt;/a&gt; requires some work over finite fields.  &lt;/li&gt;&#xA;&lt;li&gt;Elliptic curves over finite fields are used for &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve_cryptography&quot; rel=&quot;nofollow&quot;&gt;elliptic curve cryptography&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;I have also been told that vector spaces and varieties over finite fields can be used to construct error-correcting codes.  I don't know anything about this, but here is a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521771943&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; on the subject.  For the special case of linear codes this leads to a beautiful analogy between lattice sphere packings and error-correcting codes which is described, for example, by Noam Elkies &lt;a href=&quot;http://www.ams.org/notices/200010/fea-elkies-1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, even if you are only interested in varieties over $\mathbb{C}$ (say), if your variety happens to also be nice and defined over $\mathbb{Z}$ then it can be nice and defined over $\mathbb{F}_p$ for all but finitely many $p$ and you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_conjectures#Statement_of_the_Weil_conjectures&quot; rel=&quot;nofollow&quot;&gt;Weil conjectures&lt;/a&gt; to compute its Betti numbers by &lt;em&gt;counting&lt;/em&gt;.  This is particularly easy to do for varieties with nice moduli interpretations like flag varieties.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:  You might also be interested in reading Serre's expository article &lt;a href=&quot;http://arxiv.org/abs/0903.0517&quot; rel=&quot;nofollow&quot;&gt;How to use finite fields for problems concerning infinite fields&lt;/a&gt;, as well as Manin's &lt;a href=&quot;http://books.google.de/books?id=ffV07pGT6soC&amp;amp;pg=PA518&amp;amp;lpg=PA518&amp;amp;dq=Manin+%22arithmetical+physics%22&amp;amp;source=bl&amp;amp;ots=Bl0wfHR7iG&amp;amp;sig=kC9sfstSnBdY4eXtZZvWmHBAQgY&amp;amp;hl=de&amp;amp;ei=bA9PS4vOA4PqnAOu38GiCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CAcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Reflections on arithmetical physics&lt;/a&gt;.  I got the latter link from an excellent &lt;a href=&quot;http://mathoverflow.net/questions/11716/mirror-symmetry-mod-p-physics-mod-p/11739#11739&quot; rel=&quot;nofollow&quot;&gt;answer to an MO question&lt;/a&gt; on mirror symmetry over finite fields.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-02T19:20:56.460" LastActivityDate="2010-08-02T19:20:56.460" CommentCount="3" />
  <row Id="1415" PostTypeId="2" ParentId="1408" CreationDate="2010-08-02T18:09:13.323" Score="6" ViewCount="0" Body="&lt;p&gt;Basically, here's a way to think of the stalk that is more &quot;down-to-earth&quot; than direct limits. An element of the stalk $O_x$ is given by a pair $(f, U)$ where $f$ is a section over the open set $U$ and $U$ contains $x$. Two pairs $(f,U), (g,V)$ are considered equivalent if $f=g$ on a neighborhood $W$ of $x$ (contained in $U \cap V$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this definition, it's easy to see that what happens at $x$ doesn't depend on what happens on $F$, where $F$ is any closed set disjoint from $x$. The stalk is a purely local construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why this is equivalent to the direct limit: that's a direct corollary of how the construction works in most familiar categories with which one might define a sheaf (sets, groups, rings, etc.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-02T18:09:13.323" CommentCount="2" />
  <row Id="1416" PostTypeId="1" AcceptedAnswerId="1424" CreationDate="2010-08-02T18:19:20.480" Score="3" ViewCount="211" Body="&lt;p&gt;I've been looking online (and in teaching journals) for a good introduction to Algebras of Random Variables (on an undergraduate level) and their usage, and have come up short. I know I can find the probability distribution of $h(z)$ where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = x + y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ and $y$ are from known independent probability distributions (the solution is simply a convolution). Two other operations $z=xy$ and $z=y/x$ can be solved for quite easily as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any other, more complicated, uses for treating random variables as objects to be manipulated? &lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-08-02T21:13:51.843" Title="Algebra of Random Variables?" Tags="&lt;probability-theory&gt;&lt;random-variable&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1417" PostTypeId="2" ParentId="1401" CreationDate="2010-08-02T18:27:42.897" Score="2" ViewCount="0" Body="&lt;p&gt;There is a reasonably geometric proof in Section 1.4 of Noam Elkies' &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.163.1487&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;The Klein Quartic in number theory&lt;/a&gt;.  A very rough summary is as follows.  The unique simple group $G$ of order $168$ has a $3$-dimensional irreducible representation $V$ which is defined over a number field $k$.  By reducing this representation modulo a prime over $2$ we obtain a $3$-dimensional representation of $G$ over $\mathbb{F}_2$ which identifies $G$ with $\text{GL}_3(\mathbb{F}_2)$ by simplicity (and a counting argument).  By reducing this representation modulo a prime over $7$ we obtain a $3$-dimensional representation of $G$ over $\mathbb{F}_7$ which is, as it turns out, the symmetric square of the defining representation of $\text{PSL}_2(\mathbb{F}_7)$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-02T18:35:07.403" LastActivityDate="2010-08-02T18:35:07.403" CommentCount="1" />
  <row Id="1418" PostTypeId="2" ParentId="1416" CreationDate="2010-08-02T19:03:01.013" Score="2" ViewCount="0" Body="&lt;p&gt;There are nice algebraic relationships for special families.  The sum of normal (Cauchy, Levy) random variables is normal (Cauchy, Levy). The product of log-normal random variables is log-normal.  The sum of gamma random variables is gamma if the distributions have a common scale parameter, etc. See more &lt;a href=&quot;http://www.johndcook.com/distribution_chart.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-02T19:03:01.013" CommentCount="1" />
  <row Id="1419" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T20:10:32.380" Score="8" ViewCount="0" Body="&lt;p&gt;If you are interested in algebraic geometry over &lt;strong&gt;C&lt;/strong&gt;, here is another reason. A basic technique in birational geometry is the bend and break, which roughly amounts to taking any curve on a projective variety and deforming it until it becomes a union of curves of lower genus. Eventually this technique can be used to produce rational curves (under appropriate hypothesis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a major tool, and has been used for instance to study the geometry of Fano varieties, in particular to prove that they are uniruled, or to prove Hartshorne conjecture on the characterization of projective space by the positivity of its tangent bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is, to make the trick work you have to have a curve whose space of embedded deformations is big enough. This is easy to achieve in positive characteristic: the curve itself may not deform, but if you compose the inclusion with a sufficiently high power of the Frobenius morphism, the map that you obtain will have enough deformations. To prove the result in characteristic 0, a technique of reduction to characteristic p is used.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-02T20:10:32.380" />
  <row Id="1420" PostTypeId="2" ParentId="1389" CreationDate="2010-08-02T20:24:33.080" Score="2" ViewCount="0" Body="&lt;p&gt;Wlog assume $a \le b$. Dividing by $n^a$ yields $1 + n^{b-a} = n^{c-a}$ =&gt; $a=b$ (else $n|1$) =&gt; $ n = 2, c = a+1$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-02T20:24:33.080" />
  <row Id="1421" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T20:34:21.840" Score="8" ViewCount="0" Body="&lt;p&gt;There is a whole &lt;a href=&quot;http://arxiv.org/abs/0903.0517&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; of Serre on this, which I haven't read; I will explain briefly one of my favorite applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the really cool applications of this is to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ax%E2%80%93Grothendieck_theorem&quot; rel=&quot;nofollow&quot;&gt;Ax-Grothendieck theorem&lt;/a&gt;.  Which is a 100% analytic statement: if $P: \mathbb{C}^n \to \mathbb{C}^n$ is an injective polynomial map, then it's surjective. And yet the proof uses finite fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How? Well, first note that the theorem is trivial when $\mathbb{C}$ is replaced by a finite field: an injective map on finite sets is surjective. Using a little algebra (and a simple direct limit argument), it follows that the theorem is true for the algebraically closed field of characteristic $p$ which is the algebraic closure of the integers mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply to a bit of model theory; the theorem of Robinson says that any statement of first-order logic that is true in an alg. closed field of char. $p$ for each $p$ is true in any algebraically closed field of char 0, in particular the complex numbers. The trick is that the Ax-Grothendieck theorem can be phrased as a collection of (infinitely many) statements in first-order logic. And each is true in the algebraic closure of the finite fields. Whence, by Robinson's theorem, it's true in $\mathbb{C}$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be cheating if Robinson's theorem were something obscure. In fact, however, it is a direct application of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactness_theorem&quot; rel=&quot;nofollow&quot;&gt;compactness theorem for first-order logic&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another proof via the Nullstellensatz (but based on finite fields, still), see Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-02T20:34:21.840" />
  <row Id="1422" PostTypeId="1" CreationDate="2010-08-02T20:34:54.600" Score="3" ViewCount="214" Body="&lt;p&gt;Suppose $f \in C(\mathbb{R}^n)$, the space of continuous $\mathbb{R}$-valued functions on $\mathbb{R}^n$.  Are there conditions on $f$ that guarantee it is the pullback of a polynomial under some homeomorphism?  That is, when can I find $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$?  I have tried playing around with the implicit function theorem but haven't gotten far.  It feels like I may be missing something very obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some related questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A necessary condition in the case of $n = 1$ is that $f$ cannot attain the same value infinitely many times (since a polynomial has only finitely many roots).  Is this sufficient?&lt;/li&gt;&#xA;&lt;li&gt;What if we replace $\mathbb{R}$ by $\mathbb{C}$?  &lt;/li&gt;&#xA;&lt;li&gt;What if we look at smooth functions instead?&lt;/li&gt;&#xA;&lt;li&gt;What about the complex analytic case?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-15T09:34:21.720" Title="Which continuous functions are polynomials?" Tags="&lt;algebraic-geometry&gt;&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="1423" PostTypeId="1" CreationDate="2010-08-02T20:58:02.360" Score="6" ViewCount="241" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_finite_simple_groups&quot; rel=&quot;nofollow&quot;&gt;This Wikipedia article&lt;/a&gt; states that the isomorphism type of a finite simple group is determined by its order, except that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;L&lt;sub&gt;4&lt;/sub&gt;(2) and L&lt;sub&gt;3&lt;/sub&gt;(4) both have order 20160&lt;/li&gt;&#xA;&lt;li&gt;O&lt;sub&gt;2n+1&lt;/sub&gt;(q) and S&lt;sub&gt;2n&lt;/sub&gt;(q) have the same order for q odd, n &gt; 2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think this means that for each integer g, there are 0, 1 or 2 simple groups of order g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we need the full strength of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups&quot; rel=&quot;nofollow&quot;&gt;Classification of Finite Simple Groups&lt;/a&gt; to prove this, or is there a simpler way of proving it?&lt;/p&gt;&#xA;" OwnerUserId="31" LastActivityDate="2010-11-29T15:42:13.990" Title="Number of finite simple groups of given order is at most 2 - is a classification-free proof possible?" Tags="&lt;group-theory&gt;&lt;finite-simple-groups&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="1424" PostTypeId="2" ParentId="1416" CreationDate="2010-08-02T21:02:09.167" Score="5" ViewCount="0" Body="&lt;p&gt;In the special case that the sample space is the non-negative integers (or a subset thereof), one can think of a probability distribution as a generating function $f(x) = \sum_{n \ge 0} a_n x^n$ where $a_n \ge 0$ and $f(1) = 1$.  Then the sum of random variables corresponds to the product of generating functions, so one can bring generating function techniques (see, for example, &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;Wilf&lt;/a&gt;) to bear on such random variables.  For example, it is particularly easy to compute expected values this way: the expected value is $f'(1)$, and the product rule expresses the fact that expected value is additive.  Similarly, the variance is $f''(1) + f'(1) - f'(1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know a place where these issues are discussed in detail, but one spectacular family of examples is the computation of the expected values and variances of certain statistics on permutations.  For example, suppose we want to compute the expected number of fixed points that a permutation of $n$ elements has.  By Burnside's lemma, the answer is $1$.  But another way to do this computation is to construct the family of polynomials &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P_n(x) = \frac{1}{n!} \sum_{\pi \in S_n} x^{c_1(\pi)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_1(\pi)$ is the number of fixed points.  Then the number we want is $P_n'(1)$.  It turns out we can compute all these numbers at the same time because the bivariate generating function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P(x, y) = \sum_{n \ge 0} P_n(x) y^n = \frac{1}{1 - y} \exp \left( xy - y \right).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the family of numbers we want is $\frac{\partial}{\partial x} P(x, y)$ evaluated at $x = 1$, which (as it is not hard to verify) is $\frac{1}{1 - y}$.  In fact this is true for the second derivative and all higher derivatives as well; in particular, the variance of the number of fixed points is also $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we want to know the expected number and variance of, say, the total number of cycles?  Now we want to look at the family of polynomials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle Q_n(x) = \frac{1}{n!} \sum_{\pi \in S_n} x^{c(\pi)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c(\pi)$ is the total number of cycles.  Then the number we want is $Q_n'(1)$.  Now it turns out that the bivariate generating function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle Q(x, y) = \sum_{n \ge 0} Q_n(x) y^n = \frac{1}{(1 - y)^x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which should be interpreted as $\exp \left( x \log \frac{1}{1 - y} \right)$).  The partial derivative $\frac{\partial}{\partial x} Q(x, y)$ evaluated at $x = 1$ is now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{1 - y} \log \frac{1}{1 - y} = \sum_{n \ge 1} H_n y^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_n$ is the $n^{th}$ harmonic number.  Thus the expected number of cycles of a permutation of $n$ elements is about $\log n$.  (It actually turns out that the expected number of cycles of length $r$ is $\frac{1}{r}$, from which this result immediately follows.)  The second partial derivative evaluated at $x = 1$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{1 - y} \log^2 \frac{1}{1 - y} = \sum_{n \ge 1} G_n y^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle G_n = \sum_{k=1}^{n-1} \frac{1}{k} H_{n-k}$; I'm not sure of the asymptotic growth of this sequence, though, but whatever it is, the variance of the total number of cycles is $G_n + H_n - H_n^2$.  (In any case $G_n \le H_n^2$, so the variance is less than or equal to $H_n$, and this is probably about right asymptotically.)  One can deduce asymptotics for these kind of sequences using methods such as those in Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt;, which is my best guess for more examples of using generating functions in this way.  There are probably examples there related to statistics of trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the generating function identities I used above are a consequence of the exponential formula, one version of which is proven and discussed &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;in this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-02T21:13:51.843" LastActivityDate="2010-08-02T21:13:51.843" CommentCount="2" />
  <row Id="1425" PostTypeId="2" ParentId="1413" CreationDate="2010-08-02T22:03:53.247" Score="3" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can't avoid finite fields if you do complex algebraic geometry, just as you can't avoid (and it is an advantage to utilize) variables that square to zero.  Some major techniques work by reduction mod $p$ and there aren't always substitutes that accomplish the same directly in characteristic 0. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A lot of algorithms work by reducing modulo many primes and then patching together the results (Chinese remainder theorem).  For the analysis of what at happens the primes you need to understand finite fields, and in some cases, p-adic fields.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Error-correcting codes have a fundamental significance as constructions of &quot;uniformly distributed&quot; or &quot;well spaced&quot; objects in many dimensions.  This is true irrespective of their additional commercial, cryptographic or computational significance.  And it happens that the comprehensible part of error-correcting code theory is the part related to linear algebra, number theory, and algebraic geometry over finite fields.  To the extent that error-correcting codes are seen to be a model for phenomena in nature or its mathematical idealizations (crystals, packings, coverings, spin glasses, combinatorial phase transitions, etc) then routine use of finite fields is unavoidable, even if you don't care about number theory per se. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-02T22:03:53.247" />
  <row Id="1426" PostTypeId="2" ParentId="1204" CreationDate="2010-08-02T22:16:17.183" Score="1" ViewCount="0" Body="&lt;p&gt;If you are willing to build a formalism of supergeometry then some constructions are easier to state in this language, notably differential forms and the De Rham complex.  The first one is just functions from the odd line ($R^{(0 | 1)}$) to your manifold, and the De Rham differential comes from super-diffeomorphisms acting on that line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is a symplectic/orthogonal duality in representation theory that some people (most famously Kontsevich) have advocated as best understood in terms of Lie superalgebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructions using a formally negative-dimensional object (or calculations that look like they might come from such an object) sometimes can be interpreted in terms of bona fide $Z/2$ graded objects whose superdimension (even dimension minus odd dimension) is the negative dimension in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: you will get more knowledgeable answers if you post to Mathoverflow, where some of the contributors have written papers on super- or noncommutative geometry.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-02T22:16:17.183" CommentCount="5" />
  <row Id="1427" PostTypeId="2" ParentId="530" CreationDate="2010-08-02T22:16:47.150" Score="2" ViewCount="0" Body="&lt;p&gt;How about this construction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a domain $R_0$ as follows. Take a field $K$, adjoin an indeterminate $x_0$, and localize at $(x_0)$ (that is, adjoin inverses to everything not a multiple of $x_0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_0$ has all its ideals principal and linearly ordered: $(x_0)$ contains $(x_0^2)$ contains $(x_0^3)$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given $R_i$, define $R_{i+1}$ inductively: Adjoin an indeterminate $x_{i+1}$, so we have $R_i[x_{i+1}]$. Quotient by $(x_{i+1}^2 - x_i)$. Finally, localize at the prime ideal $(x_{i+1})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effectively just gives us one more principal ideal containing all the principal ideals from $R_i : (x_{i+1})$ contains $(x_{i+1}^2)=(x_i)$ contains $(x_i^2)$... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $R$ be the union of all the $R_i$, and it's obvious that any finitely generated ideal is principal, but there's a non-fg one generated by all the $x_i$. &lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-07T13:18:30.800" LastActivityDate="2010-09-07T13:18:30.800" CommentCount="1" />
  <row Id="1429" PostTypeId="2" ParentId="1422" CreationDate="2010-08-02T23:03:07.563" Score="1" ViewCount="0" Body="&lt;p&gt;Since I can't leave comments I'm writing this here.  I think this question is made difficult by the condition that $\phi$ is just required homeomorphism versus say a diffeomorphism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case n = 1 you can certainly come up with continuous functions that are not differentiable on a discrete set but can be pulled back to yield a polynomial.  As a baby example consider the function $f$ that is $\sqrt{x}$ on the positive reals and x on the negative reals.  Consider the homeomorphism that is $x^2$ on the positive reals and x on the negative reals, then $f$ pulls back to the polynomial $x$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think its sufficient that $f$ doesn't attain the same value infinitely many times.  I don't have a counterexample but I think a candidate might be contained in this &lt;a href=&quot;http://projecteuclid.org/DPubS?verb=Display&amp;amp;version=1.0&amp;amp;service=UI&amp;amp;handle=euclid.pja/1200672011&amp;amp;page=record&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;. The gist is that there are functions everywhere continuous and strictly monotonic but with derivative 0 almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you'd have more luck using the implicit function theorem if you required $\phi$ to be a diffeomorphism.  Also I believe its true that 'most' continuous function from $\mathbb{R} \to \mathbb{R}$ are not very nice (nonwhere differential) so a more tractable question might be the same question but requiring $f$ to be smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace $\mathbb{R}$ with $\mathbb{C}$ and impose $f$ and $\phi$ both be holomorphic then I think it suffices that $f^{(n)}$ vanish for all sufficiently large $n$ because you can recover $f$ from its taylor series.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-02T23:03:07.563" CommentCount="1" />
  <row Id="1431" PostTypeId="1" AcceptedAnswerId="1470" CreationDate="2010-08-03T00:13:56.220" Score="10" ViewCount="931" Body="&lt;p&gt;This is an exercise from &lt;em&gt;Problems from the Book&lt;/em&gt; by Andreescu and Dospinescu.  When it was &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?t=290872&quot; rel=&quot;nofollow&quot;&gt;posted on AoPS&lt;/a&gt; a year ago I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;  Prove that the function $f : [0, 1) \to \mathbb{R}$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x) = \log_2 (1 - x) + x + x^2 + x^4 + x^8 + ...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A preliminary observation is that $f$ satisfies $f(x^2) = f(x) + \log_2 (1 + x) - x$.  I played around with using this functional equation for awhile, but couldn't quite make it work.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-04T13:06:33.503" LastActivityDate="2010-08-04T13:06:33.503" Title="Prove that this function is bounded" Tags="&lt;real-analysis&gt;&lt;functional-equations&gt;&lt;asymptotics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="1432" PostTypeId="2" ParentId="1401" CreationDate="2010-08-03T00:14:36.033" Score="1" ViewCount="0" Body="&lt;p&gt;At the last joint meetings in SF there was a talk about exactly this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/amsmtgs/2124_abstracts/1056-z1-379.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/amsmtgs/2124_abstracts/1056-z1-379.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to the talk but I don't remember the explicit isomorphism; but at the very least a very concrete answer exists.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-03T00:14:36.033" CommentCount="1" />
  <row Id="1433" PostTypeId="2" ParentId="530" CreationDate="2010-08-03T00:34:29.783" Score="3" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezout_domain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bezout_domain&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-08T08:49:41.177" LastActivityDate="2010-09-08T08:49:41.177" CommentCount="3" />
  <row Id="1434" PostTypeId="2" ParentId="1423" CreationDate="2010-08-03T01:34:41.807" Score="10" ViewCount="0" Body="&lt;p&gt;Although I am not (by any stretch) an expert on finite simple groups, let me flesh out my above comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following &lt;strong&gt;QCFSG&lt;/strong&gt; (i.e., &quot;qualitative&quot; CFSG): with only finitely many exceptions, every finite simple group has prime order, is alternating, or is one of the finitely many known infinite families of Lie type.  QCFSG must have been conjectured rather early on, whereas the exact statement of CFSG was much harder to come by, as much of the early work on the classification problem resulted in discovery of new sporadic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that early on someone must have looked at the nonsporadic finite simple groups and noticed that, except for the two exceptions listed above, they have distinct orders. [Assuming this is actually true, that is.  I have no reason to doubt it, but I haven't checked it myself.]  Once you notice that, if you believe QCFSG, then you certainly think that the order of a simple group determines the group &lt;em&gt;up to finitely many exceptions&lt;/em&gt;.  It is very hard for me to imagine how you could prove that the number of exceptions is precisely two without knowing the full CFSG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot resist conveying a story of Jim Milne, whose moral is that you shouldn't feel too bad when you say something absolutely stupid in public: better mathematicians than you or I have said stupider things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally, a story to keep in mind the next time you ask a totally stupid question at a major lecture. During a Bourbaki seminar on the status of the classification problem for simple finite groups, the speaker mentioned that it was not known whether a simple group (the monster) existed of a certain order. &quot;Could there be more than one simple group of that order?&quot; asked Weil from the audience. &quot;Yes, there could&quot; replied the speaker. &quot;Well, could there be infinitely many?&quot; asked Weil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the source, and for some further fun stories, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jmilne.org/math/apocrypha.html&quot; rel=&quot;nofollow&quot;&gt;http://www.jmilne.org/math/apocrypha.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-29T15:42:13.990" LastActivityDate="2010-11-29T15:42:13.990" CommentCount="2" />
  <row Id="1435" PostTypeId="2" ParentId="804" CreationDate="2010-08-03T01:35:44.643" Score="0" ViewCount="0" Body="&lt;p&gt;Sorry, I don't have enough reputation to comment on Andrea Ferretti's answer, but that proof of Wedderburn's theorem is also given in detail in Herstein's &lt;em&gt;Topics in Algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-03T01:35:44.643" CommentCount="1" />
  <row Id="1436" PostTypeId="2" ParentId="1388" CreationDate="2010-08-03T01:42:09.077" Score="4" ViewCount="0" Body="&lt;p&gt;Just for fun: Gap+RadiRoot tells me that if $\zeta_n$ is a primitive $n$th root of unity, and we set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_1 = \sqrt[5]{\left(\frac{66}{3125}\zeta_{5} + \frac{451}{3125}\zeta_{5}^{2} + \frac{176}{3125}\zeta_{5}^{3} + \frac{286}{3125}\zeta_{5}^{4}\right)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_2 = \sqrt[2]{ - \frac{11}{20} + \left(\frac{1}{4}\zeta_{5}^{4}\right)\omega_1 + \left( - \frac{5}{44}\zeta_{5} + \frac{15}{44}\zeta_{5}^{2} + \frac{5}{44}\zeta_{5}^{3} + \frac{5}{44}\zeta_{5}^{4}\right)\omega_1^2 + \left(\frac{25}{121}\zeta_{5} - \frac{75}{242}\zeta_{5}^{2} - \frac{75}{484}\zeta_{5}^{3} + \frac{75}{242}\zeta_{5}^{4}\right)\omega_1^3 + \left( - \frac{375}{2662}\zeta_{5} + \frac{625}{1331}\zeta_{5}^{2} + \frac{625}{2662}\zeta_{5}^{3} + \frac{4375}{5324}\zeta_{5}^{4}\right)\omega_1^4}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the roots of the polynomial are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ - \frac{1}{10} + \frac{1}{2}\omega_1 + \left( - \frac{5}{22}\zeta_{5}^{2} - \frac{5}{11}\zeta_{5}^{3} + \frac{5}{11}\zeta_{5}^{4}\right)\omega_1^2 + \left(\frac{75}{242}\zeta_{5} + \frac{150}{121}\zeta_{5}^{2} + \frac{75}{121}\zeta_{5}^{3} + \frac{125}{121}\zeta_{5}^{4}\right)\omega_1^3 + \left(\frac{1625}{1331}\zeta_{5} + \frac{1000}{1331}\zeta_{5}^{2} + \frac{5125}{2662}\zeta_{5}^{3} + \frac{375}{1331}\zeta_{5}^{4}\right)\omega_1^4-\omega_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; I generated this (in a gap4 session, with RadiRoot properly installed) as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; LoadPackage(&quot;radiroot&quot;);&#xA;---------------------------------------------------------------------------------&#xA;Loading  RadiRoot 2.4 (Roots of a Polynomial as Radicals)&#xA;by Andreas Distler (a.distler@tu-bs.de).&#xA;---------------------------------------------------------------------------------&#xA;true&#xA;gap&amp;gt; g := UnivariatePolynomial( Rationals, [1,1,1,1,1,1,1,1,1,1,1]);&#xA;x_1^10+x_1^9+x_1^8+x_1^7+x_1^6+x_1^5+x_1^4+x_1^3+x_1^2+x_1+1&#xA;gap&amp;gt;  RootsOfPolynomialAsRadicals(g, &quot;latex&quot;);&#xA;&quot;/tmp/tmp.sfoZ6C/Nst.tex&quot;&#xA;gap&amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That created a $\LaTeX$ file, /tmp/tmp.sfoZ6C/Nst.tex, containing the formulas&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-04T16:38:07.303" LastActivityDate="2010-08-04T16:38:07.303" />
  <row Id="1437" PostTypeId="1" AcceptedAnswerId="1440" CreationDate="2010-08-03T02:44:53.480" Score="5" ViewCount="391" Body="&lt;p&gt;How to solve this equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = 10&lt;sup&gt;x/10&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="595" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T16:56:15.487" LastActivityDate="2010-08-05T15:23:43.270" Title="How to solve this exponential equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="1438" PostTypeId="2" ParentId="1431" CreationDate="2010-08-03T02:47:49.043" Score="6" ViewCount="0" Body="&lt;p&gt;Starting from (the natural logarithm of) $(1-x)^{-1} = (1+x)(1+x^2)(1+x^4) \dots$, it becomes clearer where the $\log(2)$ factor comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to show that $\Sigma (x^{2^k} - C\log(1 + x^{2^k}))$ is bounded sum of positive terms.  The sum of the first $n$ terms approaches $n - Cn\log(2)$ as $x \to 1-$, so we need $C = 1/\log(2)$ if there is to be boundedness.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T02:47:49.043" CommentCount="2" />
  <row Id="1440" PostTypeId="2" ParentId="1437" CreationDate="2010-08-03T03:18:33.500" Score="13" ViewCount="0" Body="&lt;p&gt;There is an obvious solution $x = 10$.  For $x &gt; 10$ the derivative of the RHS is at least $\log 10 &gt; 1$ so there are no solutions.  For $x \le 0$ there are obviously no solutions.  By the IVT there is a solution in $(0, 10)$, and by convexity this solution is unique.  In fact this solution is in $(1, 2)$.  It can be expressed using the Lambert W-function, but it is really not worth writing down explicitly.  Numerically it is about $1.37$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-03T03:18:33.500" CommentCount="2" />
  <row Id="1441" PostTypeId="2" ParentId="944" CreationDate="2010-08-03T03:54:04.270" Score="4" ViewCount="0" Body="&lt;p&gt;The point is that SHORTPROOF is an NP-complete problem: given a sentence of length $n$ in the language of some formal proof system (ZFC, Peano arithmetic, etc), does it have a proof of length at most some fixed polynomial in $n$, such as $(2n)^{100}$?  It's in NP because for a reasonable formal system you can check a given proof fairly quickly.  This problem was considered in Goedel's letter to von Neumann that implicitly stated what we now call the $P \neq NP$ question (the heart of the problem, the universality of concrete NP-complete problems, wasn't known until much later). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any NP-complete problem has a zero-knowledge proof protocol for demonstrating solutions of instances, e.g. that we have a SHORTPROOF of the Riemann hypothesis.  These are &quot;proofs that reveal nothing other than their own validity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The role of the PCP theorem is to show that the proof protocols (interactive challenge/response games) can be very efficient for any stipulated level of confidence.  The probability that the prover really does have a SHORTPROOF of Riemann, given that we follow the protocol and the prover wins, is at least 99 percent, or whatever specified degree of certainty.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T03:54:04.270" />
  <row Id="1442" PostTypeId="1" AcceptedAnswerId="1655" CreationDate="2010-08-03T04:18:36.063" Score="10" ViewCount="597" Body="&lt;p&gt;The identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (n+1) \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose n} \right) = \text{lcm}(1, 2, ... n+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is probably not well-known.  The only way I know how to prove it is by using &lt;a href=&quot;http://planetmath.org/encyclopedia/KummersTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Kummer's theorem&lt;/a&gt; that the power of $p$ dividing ${a+b \choose a}$ is the number of carries needed to add $a$ and $b$ in base $p$.  Is there a more direct proof, e.g. by showing that each side divides the other?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-03T08:04:27.773" LastActivityDate="2010-08-11T22:01:07.300" Title="Is there a direct proof of this lcm identity?" Tags="&lt;number-theory&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" FavoriteCount="4" />
  <row Id="1443" PostTypeId="1" AcceptedAnswerId="1444" CreationDate="2010-08-03T04:52:05.810" Score="0" ViewCount="76" Body="&lt;p&gt;In mathematics, geometry is the topic I like. Just out of curiosity I would like to know what computational geometry is about. Who studies this subject and why. What are its applications, or is it confined to a theoretical side of things? How does it relate to computer science?&lt;/p&gt;&#xA;" OwnerUserId="328" LastActivityDate="2010-08-03T05:00:21.960" ClosedDate="2010-08-03T05:18:59.090" Title="What is computational geometry and who should take a course on this subject?" Tags="&lt;geometry&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1444" PostTypeId="2" ParentId="1443" CreationDate="2010-08-03T05:00:21.960" Score="0" ViewCount="0" Body="&lt;p&gt;Computational geometry is a major part of the machinery necessary for creating computer graphics, either in games or movies. There are many uses, but if you are interested in computer graphics, it will be a very helpful class.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-03T05:00:21.960" />
  <row Id="1445" PostTypeId="2" ParentId="81" CreationDate="2010-08-03T05:30:35.467" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://gowers.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Timothy Gowers'&lt;/a&gt; blog is excellent.  Like Terence Tao, he is both a Fields medalist and an excellent writer.  Together their blogs were my first real introduction into how professional mathematicians think, and their writing has taught me a lot, both about mathematics and about mathematical writing.  If you are a serious student of mathematics you will find all the blogs you need by scrolling through their blogrolls.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-03T05:30:35.467" CommunityOwnedDate="2010-08-03T05:30:35.467" />
  <row Id="1446" PostTypeId="1" CreationDate="2010-08-03T05:41:57.103" Score="2" ViewCount="171" Body="&lt;p&gt;Many people are familiar with some properties of binary relations, such as reflexivity, symmetry and transitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the commonly studied properties of ternary (3-ary) relations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could provide a motivating example of why the property is interesting that would also be helpful. &lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2010-08-04T07:14:22.030" Title="Interesting properties of ternary relations?" Tags="&lt;logic&gt;&lt;relations&gt;&lt;tag-removed&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="1447" PostTypeId="2" ParentId="1446" CreationDate="2010-08-03T07:49:47.823" Score="2" ViewCount="0" Body="&lt;p&gt;One interesting kind of ternary relation is the &quot;betweenness&quot; relation characterised by the Axioms of Order in Hilbert's Foundations of Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect ternary relations are typically studied less because identifying an interesting one requires much more involved definitions than is normally the case for binary relations...&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-03T07:49:47.823" />
  <row Id="1448" PostTypeId="2" ParentId="1446" CreationDate="2010-08-03T08:38:36.867" Score="2" ViewCount="0" Body="&lt;p&gt;One interesting example is &quot;being &lt;a href=&quot;http://en.wikipedia.org/wiki/Steiner_system&quot; rel=&quot;nofollow&quot;&gt;Steiner triple system&lt;/a&gt;&quot; (and this is has a connection with Qiaochu Yuan's comment: any Steiner triple system defines commutative quasigroup).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-03T08:38:36.867" CommunityOwnedDate="2010-08-10T17:16:40.313" />
  <row Id="1449" PostTypeId="2" ParentId="1442" CreationDate="2010-08-03T08:47:17.063" Score="0" ViewCount="0" Body="&lt;p&gt;It will need something clever.  For n + 1 = 6, you need 6 times the lcm of 1 5 and 10 in&#xA;order to get enough powers of 2 ( and 3 ).  Is there a characterization of those n + 1 where the&#xA;entire factor of n + 1 is needed, and not the lcm, for the left hand side?&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:01:07.300" LastActivityDate="2010-08-11T22:01:07.300" CommentCount="0" />
  <row Id="1450" PostTypeId="2" ParentId="1401" CreationDate="2010-08-03T08:56:30.503" Score="2" ViewCount="0" Body="&lt;p&gt;(Idea of a proof from V.Dotsenko's paper in &lt;a href=&quot;http://www.mccme.ru/free-books/matprosd.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mccme.ru/free-books/matprosd.html&lt;/a&gt; (in Russian))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of 4-tuples of points on $\mathbb{P}^1(\mathbb{F}_7)$ with cross-ratio equal to 3. This set contains exactly 28 elements. Identifying 4-tuple with its complement (recall that there are exactly 8 points on $\mathbb{P}^1(\mathbb{F}_7)$) one gets 14-element set — exactly the number of points + the number of lines on $\mathbb{P}^2(\mathbb{F}_2)$. Define two tuples to be incident if they intersect. Claim: 1) result is indeed $\mathbb{P}^2(\mathbb{F}_2)$; 2) induced homomorphism $PSL_2(\mathbb{F}_7)\to PSL_3(F_2)$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hope to update the answer with more details when I understand them. Anyway, the first step already looks somewhat mysterious — does anybody have any comments?)&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-03T08:56:30.503" />
  <row Id="1451" PostTypeId="2" ParentId="1442" CreationDate="2010-08-03T09:12:48.513" Score="-1" ViewCount="0" Body="&lt;p&gt;Couldn't it be proved by induction? For n=1 or 2 it's clear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose the result is true for n-1; now, ${ n \choose k } = {n\over k} { n-1 \choose k }$ and ${ n \choose n } = 1$, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;lcm $({ n \choose 0, }, { n \choose 1 }, \ldots, { n \choose n }$ = n &amp;middot; lcm $({ n-1 \choose 0, }, { n-1 \choose 1 }, \ldots, { n-1 \choose n-1 }$ = lcm (1,2,...n) for induction hypothesis. n+1 does not divide neither $({ n \choose 0, }, { n \choose 1 }, \ldots, { n \choose n }$ nor {1, 2, ... n}, so the thesis is true also for n.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-03T09:12:48.513" CommentCount="2" />
  <row Id="1452" PostTypeId="2" ParentId="1437" CreationDate="2010-08-03T09:18:29.967" Score="8" ViewCount="0" Body="&lt;p&gt;You can study and graph the two functions &lt;em&gt;y&lt;/em&gt; = &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; = 10&lt;sup&gt;&lt;em&gt;x&lt;/em&gt;/10&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img594.imageshack.us/img594/3081/function.png&quot; alt=&quot;Graph of y=x and y=10^(x/10)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which you can see that there are only two solutions.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-03T09:18:29.967" CommentCount="2" />
  <row Id="1453" PostTypeId="1" AcceptedAnswerId="1606" CreationDate="2010-08-03T09:21:02.880" Score="1" ViewCount="110" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/PCP_theorem&quot; rel=&quot;nofollow&quot;&gt;PCP theorem&lt;/a&gt; states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every decision problem in &lt;a href=&quot;http://en.wikipedia.org/wiki/NP_%28complexity%29&quot; rel=&quot;nofollow&quot;&gt;NP&lt;/a&gt; has&#xA;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistically_checkable_proof&quot; rel=&quot;nofollow&quot;&gt;probabilistically checkable proofs&lt;/a&gt; of&#xA;  constant &lt;a href=&quot;http://en.wikipedia.org/wiki/Query_complexity&quot; rel=&quot;nofollow&quot;&gt;query complexity&lt;/a&gt; and&#xA;  logarithmic randomness complexity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone give an intuitive explanation of how this can be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Links&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is a follow up from &lt;a href=&quot;http://math.stackexchange.com/questions/944/proving-the-riemann-hypothesis-without-revealing-anything-other-than-you-proved-i&quot;&gt;this&#xA;question&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;There is a proof &lt;a href=&quot;http://eccc.uni-trier.de/report/2005/046/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-05T12:55:01.437" Title="Intutive explanation of the PCP Theorem" Tags="&lt;proof-theory&gt;" AnswerCount="1" />
  <row Id="1454" PostTypeId="1" AcceptedAnswerId="1544" CreationDate="2010-08-03T11:01:03.730" Score="1" ViewCount="94" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://people.ischool.berkeley.edu/~hal/Papers/2006/position.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; on position auctions for web ads. Basically, there are N slots each with an expected number of clicks (in a particular time period) $x$. Each agent makes a bid $B_i$ of how much they are willing to pay per click. The bids are put in decreasing order and agent who makes the $i^\text{th}$ highest bid receives receives the slot with the $i^\text{th}$ highest click through rate for a price $P_i$ equal to $B_{i+1}$ (except for the last agent who pays nothing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the least price to win the $i^\text{th}$ slot, we note that we have to beat the $i^\text{th}$ agent's &lt;em&gt;bid&lt;/em&gt; if we are moving up, but that we only have to beat the $i^\text{th}$ agents &lt;em&gt;price&lt;/em&gt; if we are moving down. We easily obtain the following equations for Nash Equilibria:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$(v_s-p_s)x_s\ge(v_s-p_t)x_t$ for $t&gt;s$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$(v_s-p_s)x_s\ge(v_s-P_{t-1})x_t$ for $t &amp;lt; s$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The paper then defines the symmetric Nash equilibrium to be a set of prices with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(v_s-p_s)x_s\ge(v_s-p_t)x_t$ for all $t$ and $s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, instead of having the second part of the previous conditions, the first part of the previous equations is valid anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the symmetric Nash equilibrium defined more generally? In particular, is it the same as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_equilibrium&quot; rel=&quot;nofollow&quot;&gt;symmetric equilibrium in this Wikipedia article&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T20:24:34.067" LastActivityDate="2010-08-04T22:33:52.443" Title="Symmetric nash equilibrium" Tags="&lt;game-theory&gt;&lt;nash-equilibrium&gt;&lt;auction&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="1456" PostTypeId="2" ParentId="1446" CreationDate="2010-08-03T12:27:15.567" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_triple&quot; rel=&quot;nofollow&quot;&gt;Pythagorean triples&lt;/a&gt; induce a ternary relation that has many interesting properties.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-03T12:27:15.567" CommentCount="4" />
  <row Id="1457" PostTypeId="2" ParentId="457" CreationDate="2010-08-03T12:30:56.930" Score="2" ViewCount="0" Body="&lt;p&gt;How about probability &amp;amp; statistics?  Not exactly physics, but lots of applications which can be demonstrated with empirical data.  Any example where &quot;taking an average&quot; seems reasonable is amenable to finding a distribution.  Many examples: frequencies of arrival (traffic, say) as Poisson or negative binomial; arrival times as geometric; insurance claims as lognormal or gamma (or other more complex skewed distributions, but no need to get that complicated); percentiles as beta; human physical characteristics as normal.  Depending upon your course, you could even take empirical data and try fitting distributions using various techniques, which employ calculus, numerical methods, power series (e.g. moments), etc.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-03T12:30:56.930" CommunityOwnedDate="2010-08-03T12:30:56.930" />
  <row Id="1458" PostTypeId="2" ParentId="457" CreationDate="2010-08-03T12:58:50.547" Score="11" ViewCount="0" Body="&lt;p&gt;I cannot resist mentioning the waiter's trick as a physical demonstration of the fact that $SO(3)$ is not simply connected. For those who don't know it, it is the following: you can hold a dish on your hand and perform two turns (one over the elbow, one below) in the same direction and come back in the original position. I guess one can find it on youtube if it is not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why the two things are related, I borrow the following explanation by Harald Hanche-Olsen on MathOverflow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw a curve through your body from a stationary point, like your foot, up the leg and torso and out the arm, ending at the dish. Each point along the curve traces out a curve in SO(3), thus defining a homotopy. After you have completed the trick and ended back in the original position, you now have a homotopy from the double rotation of the dish with a constant curve at the identity of SO(3). You can't stop at the halfway point, lock the dish and hand in place, now at the original position, and untwist your arm: This reflects the fact that the single loop in SO(3) is not null homotopic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-03T12:58:50.547" CommunityOwnedDate="2010-08-03T12:58:50.547" CommentCount="1" />
  <row Id="1459" PostTypeId="1" AcceptedAnswerId="1491" CreationDate="2010-08-03T13:04:36.257" Score="1" ViewCount="218" Body="&lt;p&gt;Exercise 10 in Geroch's Mathematical Physics asks whether direct products distribute over direct sums in arbitrary categories. (They do in the category of sets, which is what motivates the question). That is, (using AxB to mean the direct product and A+B to be the direct sum), whether there is an isomorphism between Ax(B+C) and (AxB)+(AxC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An earlier question asked you to show that associativity of products holds. (AxB)xC is isomorphic to Ax(BxC). This is true, even when not all objects have products (i.e. it is true that &lt;em&gt;if&lt;/em&gt; both Ax(BxC) and (AxB)xC exist, then they are isomorphic...). So I take this question to not require that all objects have products...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe there is the relevant isomorphism, but I'm not sure what I've done proves it. So my question is: Is the following argument a good way of approaching category theory questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wrote out a A,B,C,AxB,AxC,B+C,Ax(B+C),(AxB)+(AxC) and drew in all the arrows I know exist from the definitions of products and sums (e.g. AxB guarantees that there is an arrow from AxB to A and one to B...). Then I observed that this diagram has no arrows that go from Ax(B+C) to (AxB)+(AxC) or the other way. That is, even allowing for composition of arrows, there need not be a morphism between the two above mentioned objects. So, I said, this means that this diagram is a diagram of a category where Ax(B+C) is not isomorphic to (AxB)+(AxC), since if there are no morphisms between them, there can't be any isomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a good way of constructing counterexamples in category theory? Is there any rigorous discussion of &quot;diagrams&quot; used this way? (I've heard it mentioned, but I don't know where to look).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question about whether the strategy I am using is a good one, not really about the actual truth value of the statement in question.&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-08-04T16:41:47.360" LastActivityDate="2010-08-04T16:41:47.360" Title="Constructing a counterexample in category theory" Tags="&lt;category-theory&gt;&lt;universal-algebra&gt;&lt;proof-strategy&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="1460" PostTypeId="2" ParentId="1459" CreationDate="2010-08-03T13:45:35.753" Score="0" ViewCount="0" Body="&lt;p&gt;I am not an expert on this topic, so forgive me if this is obvious.  I see what you have done, but isn't this a rather &quot;vacuous&quot; category?  Products and coproducts satisfy universal properties in that we can factor through them with respect to other morphisms.  For example, if there are arrows from B and C each to A, then suddenly you have a new arrow from B+C to A which is not in your diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's where I lose the thread: if you assume products and coproducts exist for all objects A, B, and C, must not some other arrows exist between A, B, and C?  If not, then you may be OK, but I just don't know if you can assume that.  The definitions of (co)products reference such other arrows, so if there are none, then aren't all the (co)products vacuous?&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-03T13:45:35.753" CommentCount="1" />
  <row Id="1461" PostTypeId="1" AcceptedAnswerId="1469" CreationDate="2010-08-03T14:13:02.740" Score="1" ViewCount="174" Body="&lt;p&gt;Let $O(f(n))$ be the minimum time complexity for output a real sequence ${a_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input of the algorithm is a integer $n$. It output the finite sequence $a_1, a_2, \ldots, a_n$. (Clearly $f(n)\geq n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we say about the time complexity of $\max(a_1, a_2, \ldots, a_n)$? Is there always an algorithm to solve this problem in time complexity less than $O(f(n))$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the algorithm that output the sequence of natural numbers is a $O(n)$ algorithm. Finding the maximum value between 1 and n is $O(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have some more information, like Kolmogorov complexity of the sequence, how will it change the time bounds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that if a sequence can be described easily, we can find a way to find the maximum value easier than naively compute the sequence and compare.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T19:24:20.967" LastActivityDate="2010-08-03T19:24:20.967" Title="Time complexity to find the extremal value of a function" Tags="&lt;computer-science&gt;&lt;algorithms&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1463" PostTypeId="2" ParentId="1459" CreationDate="2010-08-03T16:25:47.593" Score="1" ViewCount="0" Body="&lt;p&gt;The dual to the statement that $A\times(B+C)$ is isomorphic to&#xA;$(A\times B)+(A\times C)$ is that $A+(B\times C)$ is isomorphic&#xA;to $(A+B)\times(A+C)$. Now if this statement fails in some&#xA;category $\mathcal{C}$ then the original fails in the opposite&#xA;category $\mathcal{C}^{\mathrm{op}}$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-03T16:25:47.593" CommentCount="2" />
  <row Id="1464" PostTypeId="2" ParentId="1347" CreationDate="2010-08-03T16:59:22.630" Score="4" ViewCount="0" Body="&lt;p&gt;An alternative to the horosphere model ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.jstor.org/pss/2323257&quot; rel=&quot;nofollow&quot;&gt;&quot;A Euclidean Model for Euclidean Geometry&quot;&lt;/a&gt;, Adolf Madur discusses a Disk model of the Euclidean plane. (Madur says that David Gans has priority for discussing this model, so I'll call it the &quot;Gans Disk&quot;.) The &quot;lines&quot; consist of diameters of the Disk, and half-ellipses that have a diameter as a major axis; the measure of the angle between two &quot;lines&quot; is defined as the traditional measure of the angle between their respective major axes. With an appropriate metric (which I have forgotten, and which is &lt;em&gt;just&lt;/em&gt; missing in the document preview linked), we get all of the Euclidean plane crammed into the Disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overlaying the Gans Disk on the Poincare Disk provides another way for Hyperbolians to study Euclidean geometry. They just have to agree to treat these half-ellipse paths (which I don't think are &lt;em&gt;ellipses&lt;/em&gt; to them) as &quot;lines&quot;, and to alter their concept of angle measure and length accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model might be considerably harder for Hyperbolians to wrap their minds around than the horosphere model, though.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-03T16:59:22.630" />
  <row Id="1465" PostTypeId="1" AcceptedAnswerId="1468" CreationDate="2010-08-03T17:04:24.330" Score="2" ViewCount="97" Body="&lt;p&gt;...and encoding it as a probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a sequence of non-negative integers that is periodic with period N:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{1},A_{2},...,A_{N},A_{1}...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $A_{k}$ takes on a value no greater than some constant B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq A_{k} \leq B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then take this sequence and do a simple convolution, for some constant $L &gt; 0$ and $1 \leq n \leq N$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{L}(n) = A_{n} + A_{n+1} +...+ A_{n+L-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $S_{L}(n)$ we then form a probability distribution $P(n)$ which gives the frequency of each of its values. Let $e_{j}(k) = 1$ if $j = k$ and $0$ otherwise. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(n) = (e_{n}(S_{L}(1)) + e_{n}(S_{L}(2)) +...+ e_{n}(S_{L}(N))) / N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to find out is the extent to which this process can be reversed. I have two data points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I know (pretty much) everything about the probability distribution $P(n)$: the distribution itself, its mean, range, variance, skewness, kurtosis, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I can tell you the frequency of values of $A_{k}$ in one period, so that if the sequence is 1,0,2,3,1,0, I can tell you there are two 0's, two 1's, one 2, and one 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent am I able to reconstruct the sequence $A_{k}$ from these two data points?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-08-03T18:10:59.330" Title="Recreating an Integer Sequence After Convolution..." Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1466" PostTypeId="2" ParentId="1461" CreationDate="2010-08-03T17:15:43.063" Score="1" ViewCount="0" Body="&lt;p&gt;Not exactly what you're looking for, but probably related: global optimization of one-variable real functions is NP-complete.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-03T17:15:43.063" />
  <row Id="1467" PostTypeId="1" AcceptedAnswerId="1471" CreationDate="2010-08-03T18:07:41.393" Score="2" ViewCount="400" Body="&lt;p&gt;Let &lt;em&gt;C&lt;/em&gt; = {0, 1, 2, …, &amp;alefsym;&lt;sub&gt;0&lt;/sub&gt;, &amp;alefsym;&lt;sub&gt;1&lt;/sub&gt;, &amp;alefsym;&lt;sub&gt;2&lt;/sub&gt;, …}. What is |&lt;em&gt;C&lt;/em&gt;|? Or is it even well-defined?&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-04T15:45:45.260" Title="Cardinality of all cardinalities" Tags="&lt;set-theory&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="3" />
  <row Id="1468" PostTypeId="2" ParentId="1465" CreationDate="2010-08-03T18:10:59.330" Score="2" ViewCount="0" Body="&lt;p&gt;No, you cannot recover the sequence $A_k$. As a trivial example, note that any cyclic permutation of $(A_1, A_2, \dots, A_N)$ would result in the same distribution (and frequencies). But since this is a periodic sequence, you probably don't care about distinguishing between cyclic permutations of the same sequence, so here's another example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider L=1. Then your probability distribution is just equivalent to the frequencies, so &lt;em&gt;any&lt;/em&gt; permutation of the $A_k$s would give the same distribution. If L=1 is too degenerate, here's another example with L=2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $L=2$, and $N=10$. Then, both sequences $(1, 1, 0, 0, 0, 1, 1, 0, 0, 0)$ and $(1, 1, 0, 0, 0, 0, 1, 1, 0, 0)$ would have the same distribution of sums $S_L$: 2 twice, 1 four times, and 0 four times. You can easily extend this example to any $L$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-03T18:10:59.330" CommentCount="1" />
  <row Id="1469" PostTypeId="2" ParentId="1461" CreationDate="2010-08-03T18:46:51.930" Score="2" ViewCount="0" Body="&lt;p&gt;Let $a_n = $ number of 1's in the first $n$ terms of another given 0-1 sequence, $b_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $a_n$ could equal the number of odd perfect numbers between 1 and (2n-1), or the number of Fermat primes $2^{2^k} + 1$ with $k \leq n$, or the number of the first $n$ LISP programs that terminate within 4000*(program-length)^4 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n$ is non-decreasing, but to find $a_n$ (which is the maximum of the first $n$ terms) is as difficult as calculating all the previous $a_i$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T18:46:51.930" />
  <row Id="1470" PostTypeId="2" ParentId="1431" CreationDate="2010-08-03T19:02:05.137" Score="10" ViewCount="0" Body="&lt;p&gt;OK, a second trick is needed (but it actually finishes the problem).  It is nice and simple enough that it's probably what the authors intended by a &quot;Book&quot; solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = x \log(2) - \log(1+x)$.  We want to show that $S(x) = f(x) + f(x^2) + f(x^4) + \dots$ is bounded.  Because $f(0)=f(1)=0$ and $f$ is differentiable, we can find a constant $A$ such that $|f(x)| \leq Ax(1-x) = Ax - Ax^2$.  The sum of this bound over the powers $x^{2^k}$ is telescopic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the role of $\log(2)$ was to ensure that $f(1)=0$.   &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-03T19:02:05.137" CommentCount="3" />
  <row Id="1471" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T20:01:12.813" Score="10" ViewCount="0" Body="&lt;p&gt;C is not a set - it is infact a proper class. If C were a set, then |C| would be defined. It then follows that |C| would be the largest cardinality, since there is a total order between all the cardinalities, and $|C| &gt; \kappa$ for every cardinality $\kappa$ (every cardinality is equivalent to the set of all smaller cardinalities). But $2^{|C|} &gt; |C|$ and so there cannot be a largest cardinal.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-03T20:01:12.813" CommentCount="6" />
  <row Id="1472" PostTypeId="2" ParentId="698" CreationDate="2010-08-03T20:05:13.227" Score="2" ViewCount="0" Body="&lt;p&gt;As answered by &lt;a href=&quot;http://mathoverflow.net/users/6579/t&quot; rel=&quot;nofollow&quot;&gt;T.&lt;/a&gt; on &lt;a href=&quot;http://mathoverflow.net/questions/33335/looking-for-a-probability-distribution/33342#33342&quot; rel=&quot;nofollow&quot;&gt;MathOverflow.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is equivalent to (among other&#xA;  names) the Coupon Collector problem.&#xA;  Your are asking about the distribution&#xA;  of the number of coupons collected&#xA;  after t steps, when the total number&#xA;  of possible coupons is n. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;ADDED: this and related distributions&#xA;  are also studied under other names&#xA;  such as Birthday Problem, random&#xA;  mappings, and random hashing.&#xA;  Kolchin-Sevastyanov-Chistyakov &lt;em&gt;Random&#xA;  Allocations&lt;/em&gt;, Knuth &lt;em&gt;The Art Of Computer&#xA;  Programming, vol. 2&lt;/em&gt;, and Flajolet &amp;amp;&#xA;  Sedgwick &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/book.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Analytic Combinatorics&lt;/em&gt;&lt;/a&gt; all&#xA;  discuss these problems and may contain&#xA;  the precise asymptotics of the&#xA;  distribution you are looking for.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;III.10 in Flajolet and Sedgewick gives&#xA;  the Poisson answer $1−\exp(−t/n)$ when&#xA;  the ratio is held constant, but other&#xA;  asymptotic regimes are also of&#xA;  interest especially in hashing&#xA;  problems. Birthday problem is when&#xA;  $t=O(n^{1/2})$ and one gets statistics&#xA;  of the number of collisions. For t=n^k&#xA;  with 1/2&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T20:32:23.647" LastActivityDate="2010-08-03T20:32:23.647" CommunityOwnedDate="2010-08-03T20:05:13.227" />
  <row Id="1473" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T21:09:51.713" Score="1" ViewCount="0" Body="&lt;p&gt;It is not possible to answer the question as such. The best I can do is to redirect you to the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Grothendieck_universe&quot; rel=&quot;nofollow&quot;&gt;Grothendieck universes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more detailed explanation, someone more knowledgeable in logic has to come forth.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-03T21:09:51.713" />
  <row Id="1474" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T21:33:36.150" Score="3" ViewCount="0" Body="&lt;p&gt;The most common way to define the cardinal number $|X|$ of a set $X$ is as the least ordinal which is in bijection with $X$. Then $C$ is an unbounded class of ordinals, and any such is necessarily a proper class. Since $C$ is not a set, it does not have a cardinality.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-03T21:33:36.150" />
  <row Id="1475" PostTypeId="1" CreationDate="2010-08-03T22:09:46.883" Score="31" ViewCount="763" Body="&lt;p&gt;&lt;a href=&quot;http://math.gmu.edu/~eobrien/Venn4.html&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; gives a few examples of Venn diagrams for 4 sets. Some examples:&lt;br&gt;&#xA;&lt;img src=&quot;http://math.gmu.edu/~eobrien/image/Vennflower.gif&quot; alt=&quot;alt text&quot;&gt; &#xA;&lt;img src=&quot;http://math.gmu.edu/~eobrien/image/Venn4r.gif&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;Thinking about it for a little, it is impossible to partition the plane into the 16 segments required for a complete 4-set Venn diagram using only circles as we could do for &amp;lt;4 sets. Yet it is doable with ellipses or rectangles, so we don't require non-convex shapes as &lt;a href=&quot;http://en.wikipedia.org/wiki/Venn_diagram#Edwards.27_Venn_diagrams&quot; rel=&quot;nofollow&quot;&gt;Edwards&lt;/a&gt; uses.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what properties of a shape determine its suitability for n-set Venn diagrams? Specifically, why are circles not good enough for the case n=4?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-03T22:15:57.890" LastActivityDate="2010-08-04T09:19:53.013" Title="Why can a venn diagram for 4+ sets not be constructed using circles?" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;logic&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="5" />
  <row Id="1476" PostTypeId="2" ParentId="1467" CreationDate="2010-08-03T23:20:37.197" Score="6" ViewCount="0" Body="&lt;p&gt;This is &quot;Fact 20&quot; on page 10 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are notes on infinite sets from the most &quot;naive&quot; perspective (e.g., one of the facts is that every infinite set has a countable subset, so experts will see that some weak form of the Axiom of Choice is being assumed without comment.  But this is consistent with the way sets are used in mainstream mathematics).  It is meant to be accessible to undergraduates.  In particular, ordinals are not mentioned, although there are some further documents -- replace &quot;1&quot; in the link above with &quot;2&quot;, &quot;3&quot; or &quot;4&quot; -- which describe such things a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see why it is necessary or helpful to speak of ordinals (or universes!) to answer this question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: to be clear, I wish to recast the question in the following way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no set $C$ such that for every set $X$, there exists $Y \in C$ and a bijection from $X$ to $Y$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is easy to prove via Cantor's diagonalization and it sidesteps the &quot;reification problem for cardinalities&quot;, i.e., we do not need to say what a cardinality of a set &lt;em&gt;is&lt;/em&gt;, only to know when two sets have the same cardinality.  I believe this is appropriate for a general mathematical audience.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-04T10:46:05.350" LastActivityDate="2010-08-04T10:46:05.350" />
  <row Id="1477" PostTypeId="2" ParentId="804" CreationDate="2010-08-03T23:56:37.217" Score="2" ViewCount="0" Body="&lt;p&gt;The class equation implies a nontrivial finite $p$-group has a nontrivial center, or more generally that a nontrivial normal subgroup of a nontrivial finite $p$-group $G$ contains a nontrivial element of the center of $G$.  (Note: there are infinite groups where every element has $p$-power order and the center is trivial, so the finiteness assumption on the group is important.) This has standard further consequences for finite $p$-groups, although they are no longer direct consequences of the class equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The class equation itself is a special case of a more general result: the orbit-stabilizer formula for group actions.  &lt;em&gt;That&lt;/em&gt; has lots of uses.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-03T23:56:37.217" />
  <row Id="1479" PostTypeId="2" ParentId="806" CreationDate="2010-08-04T00:19:10.063" Score="6" ViewCount="0" Body="&lt;p&gt;The class group of a number field $K$ can be used to &lt;em&gt;parametrize&lt;/em&gt; other objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $[L:K] = n$, the possible $O_K$-module structure of $O_L$ is described by the ideal classes of $K$, although it is still an open question in general to show for each $n &gt; 1$ and each ideal class of $K$ that there's an extension $L/K$ with degree $n$ such that $O_L$ as an $O_K$-module corr. to that ideal class.  (This is known for small $n$, but not for general $n$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The orbits of the action of $\text{SL}_2(O_K)$ on ${\mathbf P}^1(K)$ are in bijection with ideals classes in $K$.  For instance, the action is transitive iff $K$ has class number 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) When $O$ is a quadratic order with discriminant $d$, the (narrow) class group of $O$ describes the primitive quadratic forms of discriminant $d$ up to proper equivalence.  Here we need a slightly more general concept than the usual ideal class group (unless $O = O_K$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Weierstrass equations for an elliptic curve over $K$ up to a standard change of variables are related to ideal classes in $K$ (see Silverman's first book on ell. curves, Chap. VIII).&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-04T00:19:10.063" />
  <row Id="1480" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T00:36:28.577" Score="2" ViewCount="0" Body="&lt;p&gt;I'm also a fan of Tom Leinster's lecture notes, available on his webpage &lt;a href=&quot;http://www.maths.gla.ac.uk/~tl/msci/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. In difficulty level, I would say these are harder than Conceptual Mathematics but easier than Categories and Sheaves, and at a similar level as Categories for the Working Mathematician.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2010-08-04T00:36:28.577" CommunityOwnedDate="2010-08-04T00:36:28.577" />
  <row Id="1481" PostTypeId="2" ParentId="804" CreationDate="2010-08-04T00:44:09.747" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Burnside&lt;/a&gt;'s_lemma&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burnside's lemma can be easily deduced from the class equation and is useful in combinatorics&lt;/p&gt;&#xA;" OwnerUserId="621" LastActivityDate="2010-08-04T00:44:09.747" />
  <row Id="1482" PostTypeId="1" CreationDate="2010-08-04T01:00:29.497" Score="0" ViewCount="236" Body="&lt;p&gt;I want to solve $0&amp;lt;1−an/(mb^2)e^{−r(T−t)}&amp;lt;1$, where $r, a, b, T, t&gt;0$. The solution is that either $an\leq mb^2$ or $mb^2\leq an\leq mb^2e^{rT}$ and $t&amp;lt; T − (\ln(an) − \ln(mb^2 ))/r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts are that the first part $0&amp;lt;1−\frac{an}{mb^2}e^{−r(T−t)}$ gives me $an \leq mb^2$ because $e^{−r(T−t)}&gt;0$, so I have the first part. The second part  $1−\frac{an}{mb^2}e^{−r(T−t)}&amp;lt;1$ does not give me useful information since $\frac{an}{mb^2}e^{−r(T−t)}&gt;0$ always. How do I get the other half of the solution ( $mb^2\leq an\leq mb^2e^{rT}$ and $t&amp;lt; T − (\ln(an) − \ln(mb^2 ))/r$)? I also realise that the problem I have to solve reduces to solving $xy&amp;lt;1$ where both $x,y&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Merged from: &lt;a href=&quot;http://math.stackexchange.com/questions/1543/tricky-inequality&quot;&gt;http://math.stackexchange.com/questions/1543/tricky-inequality&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about solving $0&amp;lt;1−\frac{an}{mb^2}e^{−r(T−t)}&amp;lt;1$, where a,b,T&gt;t&gt;0? I have been stuck here for some time now.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-08-05T00:57:28.763" LastActivityDate="2010-08-05T10:41:55.200" Title="problem with inequality" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="0" />
  <row Id="1483" PostTypeId="2" ParentId="529" CreationDate="2010-08-04T01:19:55.590" Score="2" ViewCount="0" Body="&lt;p&gt;There is a simple proof of Frobenius's theorem in Lam's book on noncommutative rings, pp. 208--209.  He attributes the argument to Palais. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should consider this theorem to be two theorems: (1) C is the only C-central division algebra and (2) R and H are the only R-central division algebras. The reason there are so few choices is that C is alg. closed and R is nearly so. Division algebras with center equal to a particular field can be created using cyclic Galois extensions* and since Q has such extensions of arb. high degree there are Q-central division alg. of arb. high dimension.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*There are further technical conditions to be satisfied on the cyclic extension in order for the construction of a division algebra to work, e.g., a finite field has a cyclic extension of each degree but there are no central div. alg. of dim. &gt; 1 over a finite field.  The relevant technical conditions are satisfied when the base field is the rational numbers. &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-04T01:19:55.590" />
  <row Id="1484" PostTypeId="2" ParentId="301" CreationDate="2010-08-04T01:23:26.147" Score="0" ViewCount="0" Body="&lt;p&gt;The general theorem giving the relationship between the arithmetical hierarchy and computability is known as Post's theorem [1]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part, Post's theorem states that a set of natural numbers is $\Sigma^0_1$ if and only if it is recursively enumerable. If the set is also $\Pi^0_1$ then its complement is recursively enumerable too. So a $\Delta^0_1$ set of natural numbers must be computable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more general consequence of Post's theorem is that a set of natural numbers is $\Delta^0_{n+1}$ if and only if the set is computable from the $n$th Turing jump of the empty set, $\emptyset^{(n)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Post%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Post%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-04T01:23:26.147" />
  <row Id="1485" PostTypeId="2" ParentId="1283" CreationDate="2010-08-04T01:31:28.547" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure you'll like this, but in terms of the incomplete $\Gamma$ function, one can get a closed form as&#xA;$$\frac{e^{x}\Gamma(n+1,x)}{\Gamma(n+1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Incomplete_gamma_function&quot; rel=&quot;nofollow&quot;&gt;incomplete $\Gamma$&lt;/a&gt; function is defined as&#xA;$$\Gamma(s,x) = \int_x^{\infty} t^{s-1}e^{-t}dt$$. &lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-08-04T01:31:28.547" CommentCount="4" />
  <row Id="1486" PostTypeId="2" ParentId="198" CreationDate="2010-08-04T01:45:39.907" Score="4" ViewCount="0" Body="&lt;p&gt;Proof theorists have obtained several &quot;relative consistency&quot; proofs between classical and constructive theories. These show that if certain theories of classical mathematics are inconsistent, then corresponding theories of constructive mathematics are also inconsistent. These relative consistency results are proved constructively. They show that the consistency problem does not simply disappear if we switch to constructive mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the more famous relative consistency techniques uses a &quot;double negation translation&quot;. This method assigns each formula $\phi$ of a system a corresponding formula $\phi^N$ (the &quot;translation&quot; of $\phi$). The exact definition of the translation varies from author to author, depending on the system at hand. But the name is somewhat accurate: the definition of $\phi^N$ involves adding additional negation symbols to $\phi$ in the right places. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1933, Gödel proved there is a translation $N$ of formulas of Peano arithmetic so that whenever a formula $\phi$ is provable in Peano arithmetic, the corresponding formula $\phi^N$ is provable in the constructive system of Heyting arithmetic. Moreover, if $\phi$ is of the form $A \land \lnot A$ then Gödel's translation assigns it the formula $\phi^N = A^N \land \lnot A^N$, which is still contradictory. This means that if Peano arithmetic is inconsistent, so is its constructive counterpart Heyting arithmetic. Gödel's proof is constructive, like you would hope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we were only worried about consistency, there would be no advantage to working in Heyting arithmetic instead of Peano arithmetic. But people who work in constructive mathematics do not do it only for the sake of consistency. Constructive proofs carry more information than classical proofs, so constructive provability is interesting even to classical mathematicians. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a little more about Gödel's result at &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%E2%80%93Gentzen_negative_translation&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-04T01:45:39.907" />
  <row Id="1487" PostTypeId="1" AcceptedAnswerId="1520" CreationDate="2010-08-04T01:59:21.300" Score="0" ViewCount="119" Body="&lt;p&gt;I am trying to understand this problem and yes this is from my assignment and I should be doing it myself, but I have been staring at it for 2 hours and not getting anywhere, so decided to post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G be a connected cubic simple graph that contains 2 edge-disjoint spanning trees show that |G| = 4.&lt;/p&gt;&#xA;" OwnerUserId="634" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-04T02:10:08.813" LastActivityDate="2010-08-04T12:03:10.313" Title="Connected simple cubic graph" Tags="&lt;graph-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1488" PostTypeId="2" ParentId="1487" CreationDate="2010-08-04T02:20:30.137" Score="3" ViewCount="0" Body="&lt;p&gt;Express the number of edges in a connected cubic simple graph in terms of the number of vertices. Then do the same for a graph that contains 2 edge-disjoint spanning trees. Solve this equation for the number of vertices.&lt;/p&gt;&#xA;" OwnerUserId="635" LastActivityDate="2010-08-04T02:20:30.137" CommentCount="1" />
  <row Id="1489" PostTypeId="2" ParentId="1048" CreationDate="2010-08-04T04:21:00.310" Score="0" ViewCount="0" Body="&lt;p&gt;To do this rigorously you have essentially no choice as to the logic of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first definition constructs (sin,cos) as a pair of functions on the unit circle X^2 + Y^2 = 1.  The functions are just Y and X respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second definition constructs another pair (Sin,Cos), of functions on the real line.  The functions are specific power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that these pairs are &quot;equal&quot; has a unique meaning: to construct some local identification of the two spaces (a locally invertible parametrization of the circle by the line, and vice versa, i.e., a covering map) such that under this identification (sin,cos) corresponds to (Sin,Cos).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This identification is itself unique: the rotationally invariant angle measure in the plane (Haar measure on SO(2,R) in its conventional normalization to have total length 2*Pi).  If angle measure is already available by other means, such as length measurement on the circle, one gets a second construction of (sin,cos) through their inverse functions (using integrals) and one then has to check that arcsin^{-1} = Sin where the function on the left is an integral and the right is a power series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise we have only the map in one direction, (Cos(t),Sin(t)) from the line to the circle using power series, and one has to check that  Cos^2(t) + Sin^2(t) = 1  and the local invertibility (everywhere nonzero velocity vector) of the map.  This would &lt;em&gt;define&lt;/em&gt; angle measure on the circle as &quot;t&quot;, ie., the inverse of the power-series parametrization of the circle by the line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edited to remove Tex, click on the edit-history to see it with the TeX.) &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-04T04:31:50.797" LastActivityDate="2010-08-04T04:31:50.797" />
  <row Id="1490" PostTypeId="1" AcceptedAnswerId="1494" CreationDate="2010-08-04T04:29:44.467" Score="2" ViewCount="129" Body="&lt;p&gt;Is there a collective term for the XY, YZ and ZX planes in 3D co-ordinate geometry? I was thinking &quot;principal planes&quot; but I'm not sure where I heard that.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-04T04:33:12.560" LastActivityDate="2010-08-04T05:29:36.147" Title="Collective Term for XY, YZ and ZX planes" Tags="&lt;geometry&gt;&lt;terminology&gt;&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="1491" PostTypeId="2" ParentId="1459" CreationDate="2010-08-04T04:32:56.350" Score="1" ViewCount="0" Body="&lt;p&gt;What you have done is a sensible start. You have tried proving the result and you suspect it isn't true. In this situation you want to look for a counter example (and just one is enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion is the category of non-commutative rings. You still have direct sum. However the product is the free product. (For commutative rings it is the tensor product.)&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-04T04:32:56.350" CommentCount="2" />
  <row Id="1492" PostTypeId="2" ParentId="187" CreationDate="2010-08-04T04:59:26.443" Score="0" ViewCount="0" Body="&lt;p&gt;Transitive action on the vertices is usually the definition of &quot;looking the same from each vertex&quot;.   So maybe you have two different groups in mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The automorphism group of the combinatorics of the tiling; the abstract structure of vertices, faces, edges, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The group of Euclidean motions that leave the tiling invariant. Motions meaning isometries of space, where one should also specify whether orientation-reversal is allowed or not.  Any instance of this is also an automorphism of the tiling combinatorics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If tilings that are vertex-transitive under group #1 are &quot;vertex-uniform&quot; and those that are vertex-transitive in the stricter sense of group #2 are &quot;vertex transitive&quot;, then it is easy to give examples where some of the polyhedra are deformations of others (so not isometrically transitive in sense #2).  For example, a one-dimensional periodic tiling with several different sizes of interval will have all vertices equivalent combinatorially but a finite number of distinct vertex types under geometric equivalence.&#xA;This is maybe too simple to be what Grunbaum had in mind, can you quote the book?&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T04:59:26.443" />
  <row Id="1493" PostTypeId="2" ParentId="1401" CreationDate="2010-08-04T05:01:44.113" Score="0" ViewCount="0" Body="&lt;p&gt;Can't leave comments yet, but the details of there being only one simple group of order 168, and why PSL(2,7) and PSL(3,2) are order 168 and simple, are spelled out on pages 141-147 in Smith and Tabachnikova's &quot;Topics in Group Theory&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-04T05:01:44.113" CommentCount="1" />
  <row Id="1494" PostTypeId="2" ParentId="1490" CreationDate="2010-08-04T05:29:36.147" Score="10" ViewCount="0" Body="&lt;p&gt;&quot;Principal planes&quot; (and principal axes, principal projections) are terms used in engineering, architecture and technical drawing when using orthogonal projections (usually several of them) to represent a 3-dimensional image.  Names of other fields where the same words could arise: descriptive geometry, perspective drawing, computer graphics, axonometry, surveying, geodesy, artistic perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, &quot;coordinate planes&quot; is completely standard, as can be seen from the first comment and its multiple upvotes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from being non-standard, use of &quot;principal&quot; to refer to the coordinate axes or planes is imprecise, and it conflicts with a web of other (useful because mutually consistent) terms such as principal axes of an ellipsoid or of a stress, principal component analysis in statistics, principal diagonal of a matrix.  One wants to keep the latter terms as they are because they all are expressing the same circle of concepts related to diagonalization of quadratic forms.  So &quot;coordinate planes&quot; it is.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T05:29:36.147" />
  <row Id="1495" PostTypeId="1" AcceptedAnswerId="1524" CreationDate="2010-08-04T05:48:12.653" Score="0" ViewCount="117" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very weak in solving problem related to probability and permutation and combination. Can you please guide me the good content by which I can understand it better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="644" LastActivityDate="2010-08-04T12:42:58.247" ClosedDate="2010-08-04T12:50:35.423" Title="Learn Probability and Permutation and combination" Tags="&lt;probability&gt;&lt;learning&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1496" PostTypeId="2" ParentId="250" CreationDate="2010-08-04T05:51:50.697" Score="1" ViewCount="0" Body="&lt;p&gt;Every set can be well ordered, I once sat on the bar of my favourite place and ran into this girl I haven't seen for years. I explained a bit about AC, Zorn's Lemma and the Well-Ordering principle and that they are all equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In another time, my friend told me that if every set can be well-ordered I should tell him the successor of 0 in the real numbers, I answered that 1 is. He then argued that he means the real numbers, and not the natural numbers. I told him that my well-order puts the natural numbers with the usual ordering first, then the rationals and then the irrationals. But I can shift a finite number of positions if he wants me to.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-04T05:51:50.697" CommunityOwnedDate="2010-08-04T05:51:50.697" />
  <row Id="1497" PostTypeId="2" ParentId="1446" CreationDate="2010-08-04T06:19:17.207" Score="0" ViewCount="0" Body="&lt;p&gt;One class of example arises in Lie theory. Take $L$ a simple Lie algebra. Then there is a particular $\mathfrak{sl}(2)\subset L$, name take $E$ to be a highest root, $F$ a lowest root, and $H=[E,F]$. Then decompose $L$ as a representation of this subalgebra. You get $L=L_0\oplus L_1\otimes T \oplus \mathfrak{sl}(2)$. Then $L_1$ is a ternary system. This satisfies a (complicated) identity. You can reconstruct $L$ from the ternary system and you need this identity for the Jacobi identity.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-04T06:19:17.207" />
  <row Id="1498" PostTypeId="2" ParentId="1401" CreationDate="2010-08-04T06:26:47.567" Score="2" ViewCount="0" Body="&lt;p&gt;The group $G=\operatorname{PSL}_2(7)$ acts on $X=P^1(\mathbb{F}_7)$. Fix $p\in X$, and consider the action of the stabilizer subgroup $G_p=\{g\in G:g\cdot p=p\}$ on the set $\binom{X\setminus\{p\}}{3}$ of $3$-element subsets of $X\setminus\{p\}$. It has three orbits, of sizes $7$, $7$ and $21$; this can be checked by considering cross-ratios. Pick one of the small ones: one can check that it is a triple Steiner system $S(2,3,7)$, so it is isomorphic as a design, to $P^2(\mathbb{F}_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing a bit with this construction can be used to realize the isomorphism explicitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later.&lt;/strong&gt; An observation, which helps explainwhy this works, is that if $o\in\binom{X}{4}$ is one of the $G$-orbits of size $14$, then the automorphism group of $o$ (that is, the set of permutations of $X$ which map elements of $o$ to elements of $o$) is $\operatorname{GL}_3(\mathbb{F}_2)\rtimes\mathbb{F}_2^3$, the group of affine maps of $\mathbb{F}_2^3$ (here  &lt;code&gt;\rtimes&lt;/code&gt; is supposed to mean &lt;em&gt;crossed product&lt;/em&gt;). Fixing an element $p\in X$, as I did above, amounts to picking a 'zero' in $\mathbb{F}_2^3$ viewed as an affine space, that is, looking at it as a vector space. (This puts a structure of affine $3$-space over $\mathbb{F}_2$ on $X$, and if we start with the other $14$-element orbit $o'\in\binom{X}{4}$ we get &lt;em&gt;another&lt;/em&gt; structure of affine $3$-space over $\mathbb{F}_2$; $\operatorname{PGL}_2(7)$ is precisely the set of permutations of $X$ which preserves both affine structures)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-04T07:29:19.783" LastActivityDate="2010-08-04T07:29:19.783" CommentCount="3" />
  <row Id="1499" PostTypeId="1" AcceptedAnswerId="1517" CreationDate="2010-08-04T06:30:18.277" Score="0" ViewCount="143" Body="&lt;p&gt;If $(\mathbb{R},+)$ is a group and $H$ is a proper subgroup of $\mathbb{R}$ then prove that $H$ is of measure zero.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-04T13:24:13.907" Title="Proper Measurable subgroups of R" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1501" PostTypeId="2" ParentId="1499" CreationDate="2010-08-04T06:57:33.497" Score="6" ViewCount="0" Body="&lt;p&gt;Your title talks about &quot;proper measurable subgroups of R&quot;, but the body of your post doesn't require that H be measurable.  The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0.  I'm not sure if every subgroup must be measurable or not....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough outline of the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1:  If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0.&#xA;Proof:  See &lt;a href=&quot;http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/&quot; rel=&quot;nofollow&quot;&gt;http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2:  If H is a subgroup of G, then H-H=H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 3:  If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-04T06:57:33.497" />
  <row Id="1502" PostTypeId="1" AcceptedAnswerId="1505" CreationDate="2010-08-04T07:00:33.320" Score="6" ViewCount="311" Body="&lt;p&gt;What is a set? I know that results such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Russell%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;Russell's paradox&lt;/a&gt; mean that the definition isn't as straight forward as one might expect.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-04T15:58:12.053" Title="Definition of a set" Tags="&lt;set-theory&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="1503" PostTypeId="2" ParentId="1446" CreationDate="2010-08-04T07:14:22.030" Score="1" ViewCount="0" Body="&lt;p&gt;So far nobody is actually giving properties, but just examples.  I'll continue that theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Gauss defined composition of quadratic forms, on the level of quadratic forms what he defined was not really a law of composition but a ternary relation (three quadratic forms $Q_1$, $Q_2$, and $Q_3$ are &quot;in composition&quot; if $Q_1(x,y)Q_2(x',y') = Q_3(B,B')$ where $B$ and $B'$ are linear in $xx', xy', yx', yy'$). At the level of proper equivalence classes of quadratic forms this becomes a group law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could say any group law is defined by a ternary relation $ghk = 1$ on the group. &#xA;This fits the geometric description and addition of points on an elliptic curve or Bhargava's interpretation of Gauss's composition.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-04T07:14:22.030" />
  <row Id="1505" PostTypeId="2" ParentId="1502" CreationDate="2010-08-04T07:26:07.573" Score="7" ViewCount="0" Body="&lt;p&gt;The intuitive notion of a set is of a collection whose identity is entirely determined by its members. Russell's paradox resulted, in effect, from taking any condition at all to determine a collection of members and thus a set. (Russell formulated his paradox against &lt;a href=&quot;http://plato.stanford.edu/entries/frege-logic/#S2&quot; rel=&quot;nofollow&quot;&gt;Gottlob Frege's Basic Law V&lt;/a&gt; which has, as a consequence, that every condition determines a set.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, there is no &quot;one notion of a set.&quot; In response to Russell's Paradox (and others, such as the &lt;a href=&quot;http://mathworld.wolfram.com/Burali-FortiParadox.html&quot; rel=&quot;nofollow&quot;&gt;Burali-Forti Paradox&lt;/a&gt;), various axiomatizations of set theory have been developed to try to capture and render precise the notion of a set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results due to Goedel and Cohen show that widely accepted principles governing sets as captured by the &lt;a href=&quot;http://mathworld.wolfram.com/Zermelo-FraenkelSetTheory.html&quot; rel=&quot;nofollow&quot;&gt;Zermelo–Fraenkel axiomatization&lt;/a&gt; fail to decide some interesting and contentious &quot;higher&quot; claims about sets (the &lt;a href=&quot;http://mathworld.wolfram.com/AxiomofChoice.html&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/ContinuumHypothesis.html&quot; rel=&quot;nofollow&quot;&gt;Continuum Hypothesis&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in a real way, there is not consensus in the mathematical community about what exactly a set is nor about what principles give a full description of sets. (In practice, most mathematicians use Zermelo–Fraenkel set theory and make appeal to the Axiom of Choice, though many prefer to avoid the later when possible. Some refuse to accept it outright.)&lt;/p&gt;&#xA;" OwnerUserId="277" LastEditorUserId="277" LastEditorDisplayName="" LastEditDate="2010-08-04T15:58:12.053" LastActivityDate="2010-08-04T15:58:12.053" />
  <row Id="1506" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T07:39:54.490" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Category_theory#External_links&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has some nice free texts linked at the bottom. There's an online version of &lt;a href=&quot;http://katmat.math.uni-bremen.de/acc/acc.pdf&quot; rel=&quot;nofollow&quot;&gt;Abstract and Concrete Categories&lt;/a&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve Awodey has some &lt;a href=&quot;http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/Awodey.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; available online too. (Awodey's newish book is expensive, but probably rather good)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Patrick Schultz's answer, and BBischoff's comment on an earlier answer also have good links to freely available resources.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-04T07:39:54.490" CommunityOwnedDate="2010-08-04T07:39:54.490" />
  <row Id="1507" PostTypeId="2" ParentId="1057" CreationDate="2010-08-04T08:17:33.700" Score="2" ViewCount="0" Body="&lt;p&gt;I wrote a short proof &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&amp;amp;t=346228&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-04T08:17:33.700" />
  <row Id="1508" PostTypeId="2" ParentId="1475" CreationDate="2010-08-04T08:19:49.513" Score="19" ViewCount="0" Body="&lt;p&gt;The short answer, from a &lt;a href=&quot;http://www.ams.org/notices/200611/ea-wagon.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Frank Ruskey, Carla D. Savage, and Stan Wagon is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... it is impossible to draw a Venn diagram with circles that will represent all the possible intersections of four (or more) sets. This is a simple consequence of the fact that circles can finitely intersect in at most two points and &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph#Euler.27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler’s relation&lt;/a&gt; F − E + V = 2 for the number of faces, edges, and vertices in a plane graph.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same paper goes on in quite some detail about the process of creating Venn diagrams for higher values of &lt;em&gt;n&lt;/em&gt;, especially for simple diagrams with rotational symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple summary, the best answer I could find was on &lt;a href=&quot;http://wiki.answers.com/Q/How_do_you_solve_a_four_circle_venn_diagram&quot; rel=&quot;nofollow&quot;&gt;WikiAnswers&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two circles intersect in at most two points, and each intersection creates one new region. (Going clockwise around the circle, the curve from each intersection to the next divides an existing region into two.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since the fourth circle intersects the first three in at most 6 places, it creates at most 6 new regions; that's 14 total, but you need 2^4 = 16 regions to represent all possible relationships between four sets.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But you can create a Venn diagram for four sets with four ellipses, because two ellipses can intersect in more than two points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both of these sources indicate that the critical property of a shape that would make it suitable or unsuitable for higher-order Venn diagrams is the number of possible intersections (and therefore, sub-regions) that can be made using two of the same shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate further, consider some of the complex shapes used for &lt;em&gt;n&lt;/em&gt;=5, &lt;em&gt;n&lt;/em&gt;=7 and &lt;em&gt;n&lt;/em&gt;=11 (from &lt;a href=&quot;http://mathworld.wolfram.com/VennDiagram.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram Mathworld&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://mathworld.wolfram.com/images/gifs/VennDiagrams.jpg&quot; alt=&quot;Venn diagrams for n=5, 7 and 11&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of these shapes is chosen such that they can intersect with each-other in as many different ways as required to produce the number of unique regions required for a given &lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://www.brynmawr.edu/math/people/anmyers/PAPERS/Venn.pdf&quot; rel=&quot;nofollow&quot;&gt;Are Venn Diagrams Limited to Three or Fewer Sets?&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-08-04T09:19:53.013" LastActivityDate="2010-08-04T09:19:53.013" CommentCount="1" />
  <row Id="1509" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T08:23:26.307" Score="3" ViewCount="0" Body="&lt;p&gt;Awodey's new book, while pricey, is a really pleasant read and a good tour of Category Theory from a logician's perspective all the way up to topos theory, with a more up to date view on categories than Mac Lane.&lt;/p&gt;&#xA;" OwnerUserId="666" LastActivityDate="2010-08-04T08:23:26.307" CommunityOwnedDate="2010-08-04T08:23:26.307" CommentCount="2" />
  <row Id="1510" PostTypeId="2" ParentId="370" CreationDate="2010-08-04T08:28:13.943" Score="0" ViewCount="0" Body="&lt;p&gt;Barr and Wells, in addition to &lt;em&gt;Toposes, Triples and Theories&lt;/em&gt;, have written &lt;em&gt;Category Theory for the Computing Sciences&lt;/em&gt;, a comprehensive tome which goes through most of the interesting aspects of category theory with a constant explicit drive to relate everything to computer science whenever possible.&lt;/p&gt;&#xA;" OwnerUserId="666" LastActivityDate="2010-08-04T08:28:13.943" CommunityOwnedDate="2010-08-04T08:28:13.943" />
  <row Id="1512" PostTypeId="1" AcceptedAnswerId="1556" CreationDate="2010-08-04T09:40:40.957" Score="6" ViewCount="255" Body="&lt;p&gt;This is Problem 1.7 from &lt;a href=&quot;http://www.colby.edu/personal/fqgouvea/deform.dvi&quot; rel=&quot;nofollow&quot;&gt;Gouvea's lecture notes on deformations of Galois representations&lt;/a&gt;. In particular, he asks you to show that it has many subgroups of finite index which are not closed. So here's what I've got so far, which may be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write the compositum as F = &lt;b&gt;Q&lt;/b&gt;[&amp;radic;–1, &amp;radic;2, &amp;radic;3, &amp;radic;5, &amp;radic;7 ...] (can I?) and then the Galois group G = Gal(F/&lt;b&gt;Q&lt;/b&gt;)  is isomorphic to a direct product &amp;Pi;&lt;sub&gt;p&lt;/sub&gt; (&lt;b&gt;Z&lt;/b&gt;/2&lt;b&gt;Z&lt;/b&gt;) where the product is taken over all primes p, as well as p=-1, and the pth component is generated by the conjugation &amp;sigma;&lt;sub&gt;p&lt;/sub&gt; defined by &amp;radic;p -&gt; –&amp;radic;p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a subgroup which isn't closed would be the subgroup H consisting of &lt;i&gt;finite&lt;/i&gt; products of conjugations, since for example, H contains the sequence &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;, &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;&amp;sigma;&lt;sub&gt;3&lt;/sub&gt;, &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;&amp;sigma;&lt;sub&gt;3&lt;/sub&gt;&amp;sigma;&lt;sub&gt;5&lt;/sub&gt;, &amp;sigma;&lt;sub&gt;2&lt;/sub&gt;&amp;sigma;&lt;sub&gt;3&lt;/sub&gt;&amp;sigma;&lt;sub&gt;5&lt;/sub&gt;&amp;sigma;&lt;sub&gt;7&lt;/sub&gt;... which converges to the automorphism &quot;conjugate everything&quot;, and this automorphism is not contained in H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this subgroup is nowhere near being finite index--it has the cardinality of the natural numbers, whereas G has the cardinality of the reals. The only finite index subgroups I can think of take are of the form Gal(F/K) where K is a finite extension of &lt;b&gt;Q&lt;/b&gt;, but of course these are by definition all closed. So I guess I've stuffed up somewhere, and I'd be really grateful for any help?! In know this may seem a bit &quot;homework questiony&quot; but it's not, it's just something that's really bugging me!&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2010-08-05T22:02:31.707" Title="How to describe the Galois group of the compositum of all quadratic extensions of Q?" Tags="&lt;number-theory&gt;&lt;group-theory&gt;&lt;galois-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="1513" PostTypeId="2" ParentId="1502" CreationDate="2010-08-04T11:30:07.073" Score="5" ViewCount="0" Body="&lt;p&gt;In very naive set theory (say in the late 19th century), a set was taken to be an arbitrary collection of objects. The difficulty is in telling which things that seem like they should be collections actually are well-defined collections.  The paradoxes show that it's unclear whether this concept of set is coherent, although it &lt;em&gt;is&lt;/em&gt; the natural-language meaning of the word &quot;set&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the concept seems poorly defined, almost all contemporary set theory deals with a more restrictive notion: pure, well-founded sets. These are the sets that can be constructed starting with the empty set and taking powersets and subsets. The only elements of these sets are other sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sets are defined in stages. At the first stage, you only have the empty set. At every larger stage, you add the powerset of every set that has already been constructed. There is one stage for every ordinal number, and the collection of all sets available after stage $\alpha$ is named $V_\alpha$. Symbolically, we have $V_0 = \emptyset$ and, in general, &#xA;$$&#xA;V_\alpha = \bigcup_{\beta &amp;lt; \alpha} P(V_\beta)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each ordinal $\alpha$, $V_\alpha$ is a set. The union $V = \bigcup_\alpha V_\alpha$ is a proper class. The sequence $( V_\alpha )$, indexed by ordinals, is known as the &lt;em&gt;cumulative hierarchy&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;sets&quot; that mathematicians study and that are formalized in Zermelo-Fraenkel set theory are exactly the sets in $V$.  Moreover, the formalization of mathematics into set theory does not require any other sets than those in $V$ (which is why essentially all mathematics can be formalized into ZFC). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for all practical mathematical purposes (outside of set theory), the answer to &quot;what is a set&quot; is &quot;a set is an element of $V$&quot;. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-04T11:30:07.073" CommentCount="2" />
  <row Id="1514" PostTypeId="1" AcceptedAnswerId="1516" CreationDate="2010-08-04T11:38:47.437" Score="3" ViewCount="233" Body="&lt;p&gt;Given the sum of first &lt;code&gt;N&lt;/code&gt; elements of sequence $A$: $S_{n} = n^{2} + 3n + 4$.&lt;br&gt;&#xA;Compute $A_{1} + A_{3} + \ldots + A_{21}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this problem can be tackled by carefully calculating each value of the sequence. But I wonder what are the better ways to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="696" LastEditorUserId="696" LastEditorDisplayName="" LastEditDate="2010-08-04T11:43:53.730" LastActivityDate="2010-08-11T22:00:44.973" Title="A high school competition-level problem concerning sum and sequence" Tags="&lt;sequences-and-series&gt;&lt;high-school&gt;&lt;contest-math&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="1515" PostTypeId="2" ParentId="162" CreationDate="2010-08-04T11:44:44.743" Score="0" ViewCount="0" Body="&lt;p&gt;how about a set that contains everything with the exception of himself o_O&lt;/p&gt;&#xA;" OwnerUserId="699" LastActivityDate="2010-08-04T11:44:44.743" />
  <row Id="1516" PostTypeId="2" ParentId="1514" CreationDate="2010-08-04T11:45:10.577" Score="5" ViewCount="0" Body="&lt;p&gt;$A_n=S_n-S_{n-1}=2n+2$. Now one has to find a sum of an arithmetic progression $4+8+\dots+44$.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-04T11:45:10.577" CommentCount="0" />
  <row Id="1517" PostTypeId="2" ParentId="1499" CreationDate="2010-08-04T11:46:58.903" Score="2" ViewCount="0" Body="&lt;p&gt;As was noted by Jason DeVito if H is measurable then measure of H is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $\mathbb{R}$, generated by others. Then H is a subgroup of ($\mathbb{R}$, +).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: H is not measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, h\in H}$, where $q\in Q$, has measure 0 (because it is just a shift of H). But $\mathbb{R}$ is equal to union of countable many sets with measure 0: $\mathbb{R} = \cup_{q\in\mathbb{Q}}(H+qe)$. Therefore $m(\mathbb{R})=0$. We have come to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: If H is measurable proper subgroup of $\mathbb{R}$, then m(H)=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = H\cap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=\delta&gt;0$. Let's take integer N, such that $\delta N &gt; z+1$ (we will see later, why).&#xA;Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x&amp;lt;1, such that $x\notin H$. Suppose y=x/N!. Then for every positive integer n, number ny obeys the following properties:&lt;br&gt;&#xA;1. 1&gt;ny&gt;0.&lt;br&gt;&#xA;2. ny is not in H.&lt;br&gt;&#xA;Then sets H_0, H_0+y, ..., H_0+(N-1)y do not intersect. Therefore we can write&#xA;$1+z = m( [0, 1+z) ) \leq m(\cup_{n=0}^{N-1} (H_0 + n y) = N \delta.$ Here we have a contradiction with definition of N.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-08-04T13:24:13.907" LastActivityDate="2010-08-04T13:24:13.907" CommentCount="1" />
  <row Id="1518" PostTypeId="2" ParentId="1514" CreationDate="2010-08-04T11:57:40.667" Score="-1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_progression&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arithmetic_progression&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sn = n/2(A1+An)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="699" LastEditorUserId="699" LastEditorDisplayName="" LastEditDate="2010-08-04T12:14:12.310" LastActivityDate="2010-08-04T12:14:12.310" />
  <row Id="1519" PostTypeId="1" CreationDate="2010-08-04T11:58:10.697" Score="4" ViewCount="187" Body="&lt;h3&gt;For the uninitiated&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_theory&quot; rel=&quot;nofollow&quot;&gt;Morse theory&lt;/a&gt;, as many other early alebraic-topology widgets, leads to a picture of smooth manifolds as being built up from 'cells', copies of $\mathbb{D}^n$ for varying $n$, 'glued' to each other by the usual topological tools; giving rise to (in some sense) a more natural picture of &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;homology&lt;/a&gt; as 'coming from' &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellular_homology&quot; rel=&quot;nofollow&quot;&gt;cellular homology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As an example, consider the torus $\mathbb{T}^2$: we begin with empty space, attach a 0-cell ($\mathbb{D}^0=$ a point), attach a 1-cell ($\mathbb{D}^1=$ a line) to your point (both ends of the line are attached to the point, creating a circle), attach another 1-cell (in the same way, to the same point, creating a sort of figure 8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardest bit to visualise is next: attaching a 2-cell ($\mathbb{D}^2=$ a disk, which we will think of as its homeomorphism equivalent, a square). Begin by twisting your figure 8 so that one circle is in the xy-plane, the other in the xz-plane, now attach the top and bottom of your square (coloured red in picture below) to the xy circle (creating a 'curling round' tube) and the left and right edges (coloured blue below) of your square (now a tube) either side of the xz circle, completing the torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://inperc.com/wiki/images/thumb/8/8c/Torus.JPG/150px-Torus.JPG&quot; alt=&quot;Torus cell decomposition&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The above takes some thinking, but a little reading around shows that this is fairly easy to see. What makes it so easy is that the cells we are attaching are of &lt;strong&gt;adjacent dimensions&lt;/strong&gt;, that is; we may easily identify the boundary of one with the entirety of another. Where it gets harder to visualise is when the dimensions of the cells we are attaching to one another differ by &gt;1- the canonical example of this is the complex projective plane $\mathbb{CP}^2$, a 4-manifold built by attaching a disk to a point (making a sphere) and then attaching a 4-ball to that sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter attaching map (wherein points are identified with their images), I know, may be thought of as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf_map&quot; rel=&quot;nofollow&quot;&gt;Hopf fibration&lt;/a&gt; $\partial \mathbb{D}^4=S^3 \to \mathbb{CP}^1=S^2 $, but I have no way of visualising this, particularly with regard to the interior of the 4-disk. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does a 4 cell wrap around a 2 cell without producing a singularity of some kind? Is this analagous in some sense to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dehn_surgery&quot; rel=&quot;nofollow&quot;&gt;Dehn surgery&lt;/a&gt; in which one uses a thickening? Is there a right way to think about this or can it only really be thought of 'intellectually'?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-04T12:23:49.190" LastActivityDate="2010-09-27T16:30:14.270" Title="Visualising CP^2- a problem of attaching cells with a dimension gap &gt;1" Tags="&lt;intuition&gt;&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="3" />
  <row Id="1520" PostTypeId="2" ParentId="1487" CreationDate="2010-08-04T12:03:10.313" Score="1" ViewCount="0" Body="&lt;p&gt;Let $G$ be a simple cubic graph, with two edge-disjoint spanning trees, that has $n$ vertices and $m$ edges. Since $G$ is regular of degree 3, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3n = 2m.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now any spanning tree has $n-1$ edges, and $G$ has two disjoint spanning trees. So we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \ge 2n - 2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the first equation and the second inequality, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \ge 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are no cubic simple graphs on 1, 2 or 3 vertices, so the result follows. (It is debatable whether there is one on 0 vertices.)&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2010-08-04T12:03:10.313" CommentCount="1" />
  <row Id="1521" PostTypeId="2" ParentId="1519" CreationDate="2010-08-04T12:20:56.650" Score="5" ViewCount="0" Body="&lt;p&gt;1) Near a point of $S^2$ the picture looks like $\mathbb{C}$ sitting inside, naturally $\mathbb{C}^2$ (because $\mathbb{C}P^2\cong(\mathbb{C}^3\setminus{0})/\mathbb{C}^{\times}$ and $S^2=\mathbb{C}P^1$ is the image of a hyperplane in $\mathbb{C}^3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think first about $S^2$: one attaches a disk to a point, contracting the circle on the disk's boundary. Now, to make $\mathbb{C}P^2$ one takes 4-disk and do pretty much the same but not for one circle but for all, well, fibers of the Hopf fibration at once — or, in other words, for all points of $S^2$ at once. So, since there were no singularities after gluing $S^2$, there will be no singularities here either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not sure if it's an answer, but only hope it helps.)&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-04T12:20:56.650" CommentCount="1" />
  <row Id="1522" PostTypeId="2" ParentId="1459" CreationDate="2010-08-04T12:36:17.137" Score="0" ViewCount="0" Body="&lt;p&gt;The category of of all groups is also a counterexample. Letting $A,B,C:=\mathbb{Z}$ , the LHS is $\mathbb{Z}\times (\mathbb{Z} * \mathbb{Z})$ and the RHS is $(\mathbb{Z}\times \mathbb{Z})*(\mathbb{Z}\times \mathbb{Z})$, which are not isomorphic, as their Abelianizations are $\mathbb{Z}^3$, resp. $\mathbb{Z}^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the category of Vector spaces. If $A,B,C$ are just one dimensional vector spaces the dimension of the LHS and the RHS don't agree (Note $\times\neq \otimes$).&lt;/p&gt;&#xA;" OwnerUserId="707" LastEditorUserId="707" LastEditorDisplayName="" LastEditDate="2010-08-04T12:41:44.363" LastActivityDate="2010-08-04T12:41:44.363" />
  <row Id="1523" PostTypeId="2" ParentId="1459" CreationDate="2010-08-04T12:40:08.627" Score="0" ViewCount="0" Body="&lt;p&gt;Note that in your example (as far as I understand it, you have a category with only 8 objects) your category does not have products and coproducts (you have them only for some special pairs of objects (for example, you don't have product $(A\times B)\times(A\times C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if your exercise was about categories with products and coproducts, then your example is not an answer on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to construct a counterexample is to check some known categories. It will be quite useful for studying category theory, because you will check, what do abstract notions mean in concrete examples. In this case this works, you can check for example category of vector spaces over fixed field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is to use some method to construct new category by those you know (for example, as was it was shown by Robin Chapman).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is just create some simple category --- it is the method you have tried.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-08-04T12:40:08.627" />
  <row Id="1524" PostTypeId="2" ParentId="1495" CreationDate="2010-08-04T12:42:58.247" Score="1" ViewCount="0" Body="&lt;p&gt;You learn to swim by swimming, learn to talk by talking and &lt;strong&gt;you learn combinatorics by solving problems.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get problems at &lt;a href=&quot;http://www.artofproblemsolving.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Art of Problem Solving&lt;/strong&gt;&lt;/a&gt;, try to solve as many &quot;different&quot; problems as you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need help you know &lt;a href=&quot;http://math.stackexchange.com/&quot;&gt;&lt;strong&gt;where to get it&lt;/strong&gt;&lt;/a&gt; ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May the force be with you! &lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T12:42:58.247" />
  <row Id="1525" PostTypeId="2" ParentId="967" CreationDate="2010-08-04T13:06:13.490" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://worrydream.com/AlligatorEggs/pieces_1.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://worrydream.com/AlligatorEggs/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Alligator Eggs&lt;/strong&gt;&lt;/a&gt; is a cool way to learn lambda calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also learning functional programming languages like Scheme, Haskell etc. will be added fun.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T13:06:13.490" CommentCount="2" />
  <row Id="1526" PostTypeId="1" AcceptedAnswerId="1536" CreationDate="2010-08-04T13:33:43.277" Score="3" ViewCount="521" Body="&lt;p&gt;I have to divide 2860 by 3186. The question gives only 2 minutes and that division is only half part of question. Now I can't possibly make that division in or less than 2 minutes by applying traditional methods, which I can't apply on that division anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anyone can perform below division using faster technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2860/3186&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading, hoping to get some answers. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a multiple choice question, with answers 6/7, 7/8, 8/9 and 9/10.&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-04T17:51:29.800" LastActivityDate="2010-08-06T15:01:58.817" Title="Need faster division technique for 4 digit numbers" Tags="&lt;number-theory&gt;&lt;arithmetic&gt;&lt;numerical-methods&gt;" AnswerCount="8" CommentCount="7" FavoriteCount="2" />
  <row Id="1527" PostTypeId="2" ParentId="1378" CreationDate="2010-08-04T13:44:23.413" Score="0" ViewCount="0" Body="&lt;p&gt;Mixture1 : 60%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixture2 : 35%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say you take mixture using a pot of 100 units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also you take Mixture1 'a' times and Mixture2 'b' times using the pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow$&#xA;$\frac{(a  60 + b  35)}{(a 100 + b  100)} = \frac{50}{100}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solving this you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a}{b} = \frac{3}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&gt; 300 units of Mixture1 and 200 units of Mixture2 will give you 500 units of mixture with alcohol concentration of 50% because 3 x 60 + 2 x 35 = 250.  &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-04T13:52:32.663" LastActivityDate="2010-08-04T13:52:32.663" />
  <row Id="1528" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T13:44:29.670" Score="0" ViewCount="0" Body="&lt;p&gt;You could maybe use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Euclid's Algorithm&lt;/a&gt; to find the GCD of these two numbers, and use this to reduce the fraction. With luck, the result will be easier to manage.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-04T13:44:29.670" CommentCount="2" />
  <row Id="1529" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T14:05:00.470" Score="2" ViewCount="0" Body="&lt;p&gt;If this problem appears on a standardized test you may want to take into account some broader considerations...  How many of these digits are even significant (relative to other quantities you will be computing with in &quot;second half&quot; of the problem)? What further operations will you be performing with this quantity - i.e. will you be significantly amplifying your error with nonlinear operations?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you may find that you can simply approximate the quantity as $\frac{2.9}{3.2}$, which can quickly be converted to the decimal 0.906, or so.  Seeing that the original number is approximately 0.898, you should be good to go.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-04T14:05:00.470" CommentCount="1" />
  <row Id="1530" PostTypeId="2" ParentId="1514" CreationDate="2010-08-04T14:25:03.850" Score="0" ViewCount="0" Body="&lt;p&gt;Determining a nice form for A(n) is a good start.  This is S(n) - S(n-1) = 2n + 2 .&#xA;Now the desired sum is sum {k= 1 to 10} A(2k-1) = sum {k=1 to 10} 4k = 4 * 10 * 11 / 2 = 220 .&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:00:44.973" LastActivityDate="2010-08-11T22:00:44.973" CommentCount="0" />
  <row Id="1531" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T14:32:05.737" Score="3" ViewCount="0" Body="&lt;p&gt;If you are given options like those in your comment (&lt;strong&gt;ie. where the numerators/denominators are of comparable size&lt;/strong&gt;) you can use the fact that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2860}{3186}=\frac{a}{b} \iff 2860b= 3186a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see which $\frac{a}{b}$ is best, calculate $2860b- 3186a$ for each starting from one of the two midsize fractions and repeating for larger or smaller fractions depending on the sign of the result (if your answer is positive a larger fraction would approximate better, if negative: a smaller one). The choice of a and b giving the smallest answer is the correct one.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-04T17:33:29.313" LastActivityDate="2010-08-04T17:33:29.313" CommentCount="3" />
  <row Id="1532" PostTypeId="2" ParentId="201" CreationDate="2010-08-04T14:35:50.003" Score="1" ViewCount="0" Body="&lt;p&gt;I found this proof on this &lt;a href=&quot;http://www.matrix67.com/blog/archives/2700&quot; rel=&quot;nofollow&quot;&gt;Chinese website&lt;/a&gt;. Here is the translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a proof with information theory (the proof can be done without information theory, but this is just cool). $\log(n)$ always means $\log_2(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniformly pick a integer N between 1 and n. We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) = \log(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $H(x)$ is the entropy function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N$ can be represented by $m$ integers $X_1,\ldots,X_m$ uniquely as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N = p_i^{X_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the amount of primes no larger than $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) = H(X_1, X_2, \ldots, X_m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) \leq H(X_1)+H(X_2)+\ldots+H(X_m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $X_i \leq \log(N)$ ($2^{\log_2(N)} = N$), we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(N) \leq m H(log(N))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\log(n)}{\log(\log(n)+1)} \leq m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lhs can increase indefinitely. This even give a bound for the amount of primes smaller than $n$&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-08-04T14:35:50.003" CommentCount="1" />
  <row Id="1533" PostTypeId="1" CreationDate="2010-08-04T14:41:49.183" Score="5" ViewCount="1054" Body="&lt;p&gt;As a programmer, mathematics is important basic knowledge to study some topics, especially Algorithms. Many websites, and my fellows suggest me to study Discrete Mathematics before going to Algorithms, so I want to know which Discrete Mathematics book is suitable for my needs?&lt;/p&gt;&#xA;" OwnerUserId="725" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-14T07:10:56.690" LastActivityDate="2010-08-14T07:19:25.220" CommunityOwnedDate="2010-08-04T20:33:55.820" Title="What is the best book for studying discrete mathematics?" Tags="&lt;big-list&gt;&lt;books&gt;&lt;discrete-mathematics&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="8" />
  <row Id="1534" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T14:44:22.797" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Men_of_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Men of Mathematics by E T Bell&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img.fkcdn.com/img/185/9780671628185.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T14:44:22.797" CommunityOwnedDate="2010-08-04T14:44:22.797" CommentCount="2" />
  <row Id="1535" PostTypeId="2" ParentId="1533" CreationDate="2010-08-04T14:46:13.760" Score="15" ViewCount="0" Body="&lt;p&gt;By Donald Knuth himself! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://ecx.images-amazon.com/images/I/518GBVWEBYL._SL500_AA300_.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-04T14:46:13.760" CommunityOwnedDate="2010-08-04T20:33:55.820" CommentCount="9" />
  <row Id="1536" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T14:46:56.800" Score="7" ViewCount="0" Body="&lt;p&gt;if an approximate answer is enough, you may just throw out digits at the right of the two numbers; so 2860/3186 is more or less 28/31 (it's better to leave two digits, especially because 2860 is much more than 2000 while 3186 is not so much more than 3000)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second kind of approximation consists in adding or subtracting from both sides two numbers whose ratio is more or less what you expect the answer is. If you start from 2860/3186, which is near 1/1, you may subtract 186 from both sides, ending up with 2674/3000 ~ .891; if you are quick in additions and subtractions you may first add 14 to both sides, obtaining 2874/3200, then subtracting 180 and 200 (whose ratio is .9) obtaining 2694/3000 ~ .896.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-04T14:46:56.800" CommentCount="1" />
  <row Id="1537" PostTypeId="1" AcceptedAnswerId="1539" CreationDate="2010-08-04T15:01:54.853" Score="21" ViewCount="636" Body="&lt;p&gt;There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers.  Is there a simple reason why Euler's Gamma function $\Gamma (z) = \int_0^\infty t^{z-1} e^t dt$ is &quot;best&quot;?  In particular, I'm looking for reasons that I can explain to first-year calculus students.&lt;/p&gt;&#xA;" OwnerUserId="726" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-28T09:52:18.710" LastActivityDate="2010-08-28T09:52:18.710" Title="Why is Euler's Gamma function the &quot;best&quot; extension of the factorial function to the reals?" Tags="&lt;real-analysis&gt;&lt;education&gt;&lt;gamma-function&gt;&lt;special-functions&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="7" />
  <row Id="1539" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T15:41:57.133" Score="27" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%25E2%2580%2593Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Bohr–Mollerup theorem&lt;/a&gt; shows that the gamma function is the &lt;em&gt;only&lt;/em&gt; function that satisfies the properties &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(1)=1$;&lt;/li&gt;&#xA;&lt;li&gt;$f(x+1)=xf(x)$ for every $x\geq 0$;&lt;/li&gt;&#xA;&lt;li&gt;$\log f$ is a convex function. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By the way, the gamma function is &lt;em&gt;not the only meromorphic function satisfying&lt;/em&gt;&#xA;$$f(z+1)=z f(z),\qquad f(1)=1,$$&#xA;&lt;em&gt;with no zeroes and no poles other than the points&lt;/em&gt; $z=-n$, $n=0,1,2\dots$. There is a whole family of such functions, which, in general, have the form&#xA;$$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$&#xA;where $g(z)$ is an entire function such that&#xA;$$g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z, $$&#xA;($\gamma$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%25E2%2580%2593Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler's constant&lt;/a&gt;). The gamma function corresponds to the simplest choice&#xA;$g(z)=\gamma z$.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-08-04T22:08:58.947" LastActivityDate="2010-08-04T22:08:58.947" CommentCount="2" />
  <row Id="1541" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T15:42:02.950" Score="0" ViewCount="0" Body="&lt;p&gt;Doesn't one choose the Gamma Function because it satisfies the functional equation $\Gamma(x) = (x-1)*\Gamma(x-1)$ for all positive $x$ and therefore looks a lot like the factorial?&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-04T15:42:02.950" CommentCount="2" />
  <row Id="1542" PostTypeId="2" ParentId="1467" CreationDate="2010-08-04T15:45:45.260" Score="0" ViewCount="0" Body="&lt;p&gt;I may be wrong but it seems that C is not a set of all sets or anything similar. It has IMHO a trivial bijection into $\mathbb{Z}$ and therefore into $\mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't met any paradox to forbid $|C| \in C$ ($\{1\}$ is such set).&lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-04T15:45:45.260" CommentCount="4" />
  <row Id="1543" PostTypeId="1" CreationDate="2010-08-04T15:55:08.090" Score="0" ViewCount="173" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/1482/problem-with-inequality&quot;&gt;problem with inequality&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;How do I go about solving $0&amp;lt;1−\frac{an}{mb^2}e^{−r(T−t)}&amp;lt;1$, where a,b,T&gt;t&gt;0? I have been stuck here for some time now.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-08-04T16:38:08.100" LastActivityDate="2010-08-04T17:54:26.630" ClosedDate="2010-08-04T20:37:43.003" Title="tricky inequality " Tags="&lt;algebra-precalculus&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="1544" PostTypeId="2" ParentId="1454" CreationDate="2010-08-04T16:08:31.747" Score="1" ViewCount="0" Body="&lt;p&gt;The notion &quot;symmetric equilibrium&quot; (the one from Wikipedia article) is not applicable here, because the game is not symmetric (different players have different &quot;profits per click&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've a look at the paper and I think, that &quot;symmetric Nash equilibrium&quot; in your case is nothing but technically convenient case of Nash equilibrium (the latter is not unique in your case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I've noted a probable misprint in the proof of the &quot;Fact 1&quot;.&#xA;It should be $(v_s-p_s)x_s \geq (v_s-p_{S+1})x_{S+1}$ instead of $(v_s-p_s)x_s \geq (v_{S+1}-p_{S+1})x_{S+1}$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-08-04T22:33:52.443" LastActivityDate="2010-08-04T22:33:52.443" />
  <row Id="1545" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T16:11:34.100" Score="16" ViewCount="0" Body="&lt;p&gt;Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.luschny.de/math/factorial/hadamard/f1.gif&quot; alt=&quot;Euler's Gamma Function&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Euler's Gamma Function&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.luschny.de/math/factorial/hadamard/hadgamma1.gif&quot; alt=&quot;Hadamard's Gamma function&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Hadamard's Gamma function&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.luschny.de/math/factorial/hadamard/LuschnyFactorial5.gif&quot; alt=&quot;Luschny's factorial function&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Luschny's factorial function&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-08-04T16:11:34.100" />
  <row Id="1546" PostTypeId="2" ParentId="1519" CreationDate="2010-08-04T16:25:23.167" Score="0" ViewCount="0" Body="&lt;p&gt;Concerning the Hopf fibration, the film &quot;&lt;a href=&quot;http://www.dimensions-math.org/Dim_E.htm&quot; rel=&quot;nofollow&quot;&gt;Dimensions&lt;/a&gt;&quot; does a superb job of visualizing it (chapters 7-8 in the &lt;a href=&quot;http://www.dimensions-math.org/Dim_tour_E.htm&quot; rel=&quot;nofollow&quot;&gt;table of contents&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the visualization of the glueing, maybe it helps to visualize other, similar constructions to get at least some intuition. For instance, you could try to get a picture of how the torus looks like after every &quot;meridian&quot; (blue circle in your picture) has been squashed to a point.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-08-04T16:25:23.167" />
  <row Id="1547" PostTypeId="2" ParentId="198" CreationDate="2010-08-04T16:40:50.263" Score="0" ViewCount="0" Body="&lt;p&gt;Here is one elementary result you can't prove without the Axiom of Choice (that I used to think was not part of constructive mathematics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everybody knows how to prove that, if two sets $X$ and $Y$ are non-empty, then their cartesian product $X \times Y$ is non-empty too: since $X$ is non-empty, you can choose one element from it: $x \in X$. For the same reason, you can choose also one $y \in Y$. So you have an element $(x,y) \in X\times Y$ and therefore it is non-empty. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a finite family of non-empty sets $X_1, \dots , X_n$ this procedure still works: you choose one element in each set starting from the first: $x_1 \in X_1$. And then keep on doing the same for the rest: choose $x_2$ in the second set $X_2$, $x_3$ in the third one $X_3$,... And so on, till you reach $x_n \in X_n$. You have thus produced an element of the cartesian product: $(x_1, \dots , x_n) \in X_1 \times \dots \times X_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you had an infinite &lt;em&gt;countable&lt;/em&gt; family of non-empty sets $X_1, X_2, \dots , X_n, \dots$, you could produce an element of their product $\prod_{n=1}^{\infty} X_n$ this way: choose successively elements $x_1 \in X_1$, $x_2 \in X_2$,..., $x_n \in X_n$... And you would obtain your element $(x_n)$ of your infinite cartesian product. Hence $\prod_{n=1}^{\infty} X_n$ is non-empty too if all the $X_n$, $n=1, 2, \dots $ are non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, try to do the same with an &lt;em&gt;arbitrary&lt;/em&gt; family of sets $X_i$, with the indexes $i$ belonging to an &lt;em&gt;arbitrary&lt;/em&gt; indexing set $I$. That is, set $I$ can be infinite and &lt;em&gt;not&lt;/em&gt;  countable. This kind of cartesian products do exist in &quot;nature&quot;. For instance: $I$ could be the set of real numbers $\mathbb{R}$. Take $X_i = \mathbb{R}$ for all $i \in \mathbb{R}$ too. This cartesian product $\prod_{i\in \mathbb{R}} X_i = \mathbb{R}^\mathbb{R}$ is the same as the set of all (not necessarily continuous) functions from $\mathbb{R}$ to $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can prove that $\mathbb{R}^\mathbb{R}$ is non-empty by showing an element of it (for instance, $f(x) = x$ for all $x\in \mathbb{R}$), but how can you prove, in general, that the product of an arbitrary family of non-empty sets $\prod_{i\in I}X_i$ is non-empty too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to imitate what you have just done in the finite or countable cases, you'll find asking yourself: &lt;em&gt;where&lt;/em&gt; do I start? Which is the &lt;em&gt;first&lt;/em&gt;  $\ i \in I$? Assuming there is a &lt;em&gt;first&lt;/em&gt; $i$, which is the &lt;em&gt;next&lt;/em&gt; one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that, for an infinite, non-countable, set of indexes $I$ there is in general no such a thing as a &lt;em&gt;first&lt;/em&gt; index $i$, nor a &lt;em&gt;next&lt;/em&gt; one. So you can't use the previous algorithm in order to produce an element of the product $\prod_{i\in I} X_i$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except you have something that allows you to &lt;em&gt;choose&lt;/em&gt;, one particular element of each set $x_i \in X_i$ -no matter how, even with no precise constructive algorithm as in the finite or countable cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were able to do that for no matter which set of indexes $I$, then you would have your element $(x_i)_{i\in I}$ and you would have proved that, if every $X_i \neq \emptyset$, so is their cartesian product $\prod_{i\in I} X_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do &lt;em&gt;allow&lt;/em&gt; you for an infinite, non-countable, set of indexes $I$ to &lt;em&gt;choose&lt;/em&gt; those $x_i \in X_i$ for every $I$? Well, this is exactly what the &lt;em&gt;Axiom of Choice&lt;/em&gt; says.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-17T01:27:45.070" LastActivityDate="2010-08-17T01:27:45.070" CommentCount="5" />
  <row Id="1548" PostTypeId="1" AcceptedAnswerId="1625" CreationDate="2010-08-04T16:53:25.383" Score="8" ViewCount="193" Body="&lt;p&gt;Here is a classic theorem of Herstein:&#xA;$G$ is a finite group, $M$ a maximal subgroup, which is abelian.  Then $G$ is solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is pretty easy, but it uses character theory (specifically, Frobenius' theorem on Frobenius groups).  Is there a character-theory-free proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get things going, note that we can reduce to the case where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $M$ is core-free and a Hall subgroup of $G$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $Z(G)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-05T14:19:01.960" Title="G a finite group, M a maximal subgroup; M abelian implies G solvable?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="1549" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T16:55:16.410" Score="8" ViewCount="0" Body="&lt;p&gt;For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the &quot;correct&quot; substitute for the factorial in various integrals, which seems to come more or less from its integral definition.  For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{0}^1 t^a (1 - t)^b \, dt = \frac{a! b!}{(a+b+1)!}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (non-negative?) real values of $a$ and $b$ the correct generalization is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_0^1 t^a (1 - t)^b \, dt = \frac{\Gamma(a-1) \Gamma(b-1)}{\Gamma(a+b)}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, integrals are important, so the Gamma function must also be important.  For example, the Gamma function appears in the general formula for the &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere#Volume_and_surface_area&quot; rel=&quot;nofollow&quot;&gt;volume of an n-sphere&lt;/a&gt;.  But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation&quot; rel=&quot;nofollow&quot;&gt;functional equation for the Riemann zeta function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-04T16:55:16.410" CommentCount="3" />
  <row Id="1550" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T17:06:00.823" Score="11" ViewCount="0" Body="&lt;p&gt;Gödel, Escher, Bach: An Eternal Golden Braid  by Douglas Hofstadter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/en/thumb/f/f1/GEBcover.jpg/200px-GEBcover.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2263" LastActivityDate="2010-08-04T17:06:00.823" CommunityOwnedDate="2010-08-04T17:06:00.823" />
  <row Id="1551" PostTypeId="2" ParentId="1408" CreationDate="2010-08-04T17:12:21.593" Score="3" ViewCount="0" Body="&lt;p&gt;I think there is a missing word in Akhil Mathew's answer: and it's &quot;filtered&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do that because stalks are &lt;em&gt;filtered&lt;/em&gt; colimits (aka &quot;direct limits&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For filtered colimits, $\varinjlim_i X_i$, you can take representatives of elements $x \in \varinjlim_i X_i$ for some $i$ belonging to the set of indexes $I$ (in our case, the open sets $U$). That is, you can find some $i$ and $x_i \in X_i$ that goes to your $x$ through the universal arrow $X_i \longrightarrow \varinjlim_i X_i$. For instance, every element of the stalk $O_{X,p}$ can be represented by a section $f \in O_X(U)$ for some open set $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not true for other kinds of colimits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take the push-out of two arrows $f: A \longrightarrow B$ and $g: A \longrightarrow C$ in the category of, say, abelian groups. Elements of this push out $B \oplus_A C$ are classes of pairs $(b,c) \in B\oplus C$, where you quotient out elements of the form $(f(a), 0) - (0, g(a))$, for all $a\in A$. That is, $(f(a),0) = (0,g(a))$ in $B\oplus_A C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements of $B\oplus_A C$ cannot be represented, in general, by elements coming from just $B$ or $C$, which are of the form $(b,0)$ or $(0,c)$, respectively: so, for a general $(b,c) \in B\oplus_A C$ there is no $b \in B$, nor $c\in C$, that represents it.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-04T17:12:21.593" CommentCount="1" />
  <row Id="1552" PostTypeId="2" ParentId="1512" CreationDate="2010-08-04T17:48:00.320" Score="2" ViewCount="0" Body="&lt;p&gt;The finite index subgroups that you say that you can think of are precisely&#xA;the closed (equivalently open) finite index subgroups of $Gal(F/\mathbb{Q})$ &#xA;with respect to the Krull topology on the group. These are really the only relevant&#xA;ones in terms of Galois theory. But the group has non-closed subgroups too,&#xA;but their existence requires some non-constructive principle such as Zorn's&#xA;lemma. These subgroups won't be open or closed in the Krull topology.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-04T18:46:40.320" LastActivityDate="2010-08-04T18:46:40.320" />
  <row Id="1553" PostTypeId="2" ParentId="1482" CreationDate="2010-08-04T17:54:26.630" Score="4" ViewCount="0" Body="&lt;p&gt;There are many unnecessary variables. Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \frac{an}{mb^2}.\qquad(1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the inequality becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; 1 - \alpha e^{-r (T - t)} &amp;lt; 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first obvious step is perform “1 &amp;minus;” on every parts,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 &gt; \alpha e^{-r (T-t)} &gt; 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the exponential function's range is positive and &amp;lt; 1 (since &lt;em&gt;r&lt;/em&gt; &gt; 0 and &lt;em&gt;T&lt;/em&gt; &gt; &lt;em&gt;t&lt;/em&gt; &gt; 0), we can ensure &lt;em&gt;&amp;alpha;&lt;/em&gt; is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 0 &amp;lt; &lt;em&gt;&amp;alpha;&lt;/em&gt; ≤ 1, then every &lt;em&gt;t&lt;/em&gt; will satisfy the inequality (the first solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume &lt;em&gt;&amp;alpha;&lt;/em&gt; &gt; 1. Now it's pretty obvious on how to solve &lt;em&gt;t&lt;/em&gt; in terms of &lt;em&gt;r&lt;/em&gt;, &lt;em&gt;T&lt;/em&gt; and &lt;em&gt;&amp;alpha;&lt;/em&gt;. Substitute (1) again to get back &lt;em&gt;a&lt;/em&gt;, &lt;em&gt;n&lt;/em&gt;, &lt;em&gt;m&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things you may consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;e&lt;sup&gt;x&lt;/sup&gt;&lt;/em&gt; is strictly increasing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-05T10:41:55.200" LastActivityDate="2010-08-05T10:41:55.200" CommentCount="6" />
  <row Id="1554" PostTypeId="2" ParentId="1537" CreationDate="2010-08-04T17:54:55.770" Score="2" ViewCount="0" Body="&lt;p&gt;Comment on Jens: You can take any function on $[0,1)$ and extend to $\mathbb{R}$ using this functional equation. You have three examples above. These can obviously be made continuous. More surprising is that there are several ways of getting a convex function. However if you insist on log-convex there is a unique function.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-04T17:54:55.770" CommentCount="2" />
  <row Id="1555" PostTypeId="1" AcceptedAnswerId="1560" CreationDate="2010-08-04T18:14:41.537" Score="4" ViewCount="287" Body="&lt;p&gt;I'm trying to understand better the following addition formula: $$\int_0^a \frac{\mathrm{d}x}{\sqrt{1-x^2}} + \int_0^b \frac{\mathrm{d}x}{\sqrt{1-x^2}} = \int_0^{a\sqrt{1-b^2}+b\sqrt{1-a^2}} \frac{\mathrm{d}x}{\sqrt{1-x^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term $a\sqrt{1-b^2}+b\sqrt{1-a^2}$ can be derived from trigonometry (since $\sin(t) = \sqrt{1 - \cos^2(t)}$) but I have not been able to find any way to derive this formula without trigonometry, how could it be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; fixed a mistake in my formula.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-06T20:34:09.730" Title="Finding an addition formula without trigonometry" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="0" />
  <row Id="1556" PostTypeId="2" ParentId="1512" CreationDate="2010-08-04T18:14:50.697" Score="7" ViewCount="0" Body="&lt;p&gt;Let's forget the Galois group structure: it is enough to consider the infinite product $G = \mathbb{Z}/(2)^\mathbb{N}$ with its profinite topology and exhibit non-closed subgroups of finite index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An element of $G$ is naturally identified with a subset of $\mathbb{N}$: the string $(a_i)$ corresponds to the subset $A$ such that $n \in A$ if and only if $a_n = 1$. With this interpretation, the operation on $G$ just becomes the symmetric difference of sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the filter $F'$ of all cofinite subsets of $\mathbb{N}$, and let $F$ be any ultrafilter containing it. I show that we can associate to $F$ a subgroup of index $2$. The $F$ will be non closed since $F$ is not a principal ultrafilter (indeed any non-principal ultrafilter will do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the characteristic function of $F$: $$f = \chi_F \colon G = \mathcal{P}(\mathbb{N}) \to \mathbb{Z}/(2)$$. I claim that this is a homomorphism, and its kernel is the desired subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy to verify directly. Let $A, B \in G$ and assume $f(A) = f(B) = 0$. This means that neither $A$ nor $B$ belong to $F$, so their complementary sets do. It follows that $A^c \cap B^c \in F$, so that $A \cup B \notin F$. A fortiori the symmetric difference $A \Delta B \notin F$, whic means $f(A \Delta B) = 0$. The other cases are similar.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-04T18:14:50.697" />
  <row Id="1557" PostTypeId="2" ParentId="1526" CreationDate="2010-08-04T18:36:03.180" Score="8" ViewCount="0" Body="&lt;p&gt;Another very specific trick, based on the multiple choices: since all the choices are of the form $\displaystyle \frac{n-1}{n} = 1 - \frac{1}{n}$, you're trying to find $n$ for which $1/n$ is closest to 1&amp;minus;(your fraction). So you'd consider (denominator&amp;minus;numerator)/denominator = (3186&amp;minus;2860)/3186 which is (around 320)/3186, clearly closest to 1/10.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-04T18:36:03.180" />
  <row Id="1560" PostTypeId="2" ParentId="1555" CreationDate="2010-08-04T19:40:04.787" Score="3" ViewCount="0" Body="&lt;p&gt;Replace the first integral by the same thing from $-a$ to $0$, and consider the points W,X,Y,Z on the unit circle above $-a,0,b$ and $c = a\sqrt{1-b^2} + b \sqrt{1-a^2}$.  Draw the family of lines parallel to XY (and WZ). This family sets up a map from the circle to itself; through each point, draw a parallel and take the other intersection of that line with the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your formula says that this map [edit: or rather the map it induces on the $x$-coordinates of points on the circle] is a change of variables converting the integral on $[-a,0]$ to the same integral on $[b,c]$.  Whatever differentiation you perform in the process of proving this, will be the verification that $dx/y$ is a rotation-invariant differential on the circle $x^2 + y^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[The induced map on x-coordinates is: $x \to$ point on semicircle above $x \to$ corresponding point on line parallel to XY $\to x$-coordinate of the second point.  Here were are just identifying $[-1,1]$ with the semicircle above it.]&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T20:34:09.730" LastActivityDate="2010-08-06T20:34:09.730" CommentCount="4" />
  <row Id="1562" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T20:12:40.433" Score="0" ViewCount="0" Body="&lt;p&gt;I'll recommend two, which are similar in that they take fairly elementary mathematical problems and give very thorough and careful &quot;talking out loud&quot; illustrations of how a proper mathematician would go about thinking them through - what's really going on, what's a good example, what's a definitive counterexample, how to generalise, how to realise you've reached a dead end, and so on.  &quot;Proofs and refutations&quot; by Imre Lakatos (just one, geometrical, problem, in glorious detail).  &quot;Mathematics and plausible reasoning Vol 1&quot; by G. Polya (a little more advanced, and much more satisfying, than &quot;How to solve it&quot;).&lt;/p&gt;&#xA;" OwnerUserId="770" LastActivityDate="2010-08-04T20:12:40.433" CommunityOwnedDate="2010-08-04T20:12:40.433" />
  <row Id="1563" PostTypeId="1" CreationDate="2010-08-04T21:01:22.910" Score="4" ViewCount="274" Body="&lt;p&gt;I just started learning model theory on my own, and I was wondering if there are any interesting examples of two structures of a language L which are not isomorphic, but are elementarily equivalent (this means that any L-sentence satisfied by one of them is satisfied by the second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the notaion of David Marker's book &quot;Model theory: an introduction&quot;.&lt;/p&gt;&#xA;" OwnerUserId="777" LastActivityDate="2010-09-14T05:40:03.343" Title="Example of non-isomorphic structures which are elementarily equivalent" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="5" FavoriteCount="0" />
  <row Id="1564" PostTypeId="2" ParentId="1563" CreationDate="2010-08-04T21:13:07.820" Score="0" ViewCount="0" Body="&lt;p&gt;Take a look at the wiki article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_closed_field&quot; rel=&quot;nofollow&quot;&gt;Real closed fields&lt;/a&gt;. Briefly, they are fields that are first-order equivalent to the field of the real numbers. An example is the field of real numbers that are roots of polynomials with rational coefficients. That's pretty much all I know about them, though.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-04T21:13:07.820" />
  <row Id="1565" PostTypeId="2" ParentId="1563" CreationDate="2010-08-04T21:13:08.320" Score="2" ViewCount="0" Body="&lt;p&gt;There are a real numbers worth of countable models of Peano arithmetic, all elementarily equivalent to the usual model, but all pairwise nonisomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every nonstandard example (nonstandard means not isomorphic to the usual model) is characterized by the fact that it contains &quot;infinitely large&quot; numbers.  That is, each model M has a canonical copy of N = usual naturals inside.  However, each nonstandard model M has an element p (in fact, countably many such elements) which the model things is bigger than everything in its copy of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are created by standard compactness arguments paired with the downward Lowenheim-Skolem theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let PA denote the axioms for Peano Arithmetic and set T = Th(N), the theory of N (i.e., the set of all first order statements which are true in the usual interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, add a constant c to the language and let phi_n be the statement c &gt; n, or more appropriately, c &gt; SSSS.....S(0), where S is the successor function of PA and there are n S's in the expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the set of axioms PA union T union phi_n for every n.  By a standard compactness argument, this set of axioms is consistent, and so by the Godel completeness theorem, there is a model M of all the axioms simultaneously.  By downward Lowenheim-Skolem, we can assume M is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, ask yourself, what can c be?  Well, since M satisfies phi_n for every n, we must have c &gt; n for each &quot;usual&quot; natural number in M, i.e., c is infinite!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really cool is that by playing around with different kinds of phi_n, one can find models that have elements which are divisible by NO &quot;standard&quot; prime number.  One can also find models that have elements which are simultaneously divisible by EVERY &quot;standard&quot; prime number!&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-04T21:13:08.320" />
  <row Id="1566" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T21:20:39.347" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471608394&quot; rel=&quot;nofollow&quot;&gt;Geometric Algebra&lt;/a&gt; by Emil Artin. Though not for the beginner, it can do wonders for an intermediate undergraduate in terms of expanding their horizons and helping them appreciate the beauty and interconnectedness of mathematics. It did for me and I think convinced me that I'm a geometer at heart.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-04T21:20:39.347" CommunityOwnedDate="2010-08-04T21:20:39.347" CommentCount="1" />
  <row Id="1567" PostTypeId="2" ParentId="1563" CreationDate="2010-08-04T21:23:33.447" Score="6" ViewCount="0" Body="&lt;p&gt;Laws, yes&lt;sup&gt;[1]&lt;/sup&gt;: if there weren't such examples, there wouldn't be a subject called model theory and therefore Marker's book would not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, however, not very many explicit examples which can be given, with proof, right after the definition of elementary equivalence, a pedagogical problem that I encounered recently when I taught a short summer course on model theory.  When I first gave the definition, all I was able to come up with was the following argument: for any language $L$, the class of $L$-structures is a proper class [any nonempty set can be made into the &quot;universe&quot;, or underlying set of, an $L$-structure] whereas since there are at most &#xA;&lt;code&gt;$2^{\max|L|,\aleph_0}$&lt;/code&gt; different theories in the language $L$, the $L$-structures up to elementary equivalence must form a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you just want examples without proof, sure, here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In the empty language, two sets are elementarily equivalent iff they are both infinite or both finite of the same cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Any two dense linear orders without endpoints are elementarily equivalent.  [The same holds for DLOs &lt;em&gt;with&lt;/em&gt; endpoints, but the two classes of structures are not elementarily equivalent to each other.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Any two algebraically closed fields of the same characteristic are elementarily equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Any two real-closed fields are elementarily equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;5) Any two infinite models of the theory of finite fields are elementarily equivalent.&lt;/s&gt; (Nope! See Dave Marker's answer.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, such structures exist of every infinite cardinality, so the class of isomorphism classes of such structures is a proper class, not a set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &quot;M-O-O-N, that spells model theory!&quot;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-06T09:24:27.740" LastActivityDate="2010-08-06T09:24:27.740" CommentCount="1" />
  <row Id="1568" PostTypeId="1" AcceptedAnswerId="1574" CreationDate="2010-08-04T21:23:48.410" Score="2" ViewCount="135" Body="&lt;p&gt;As a follow up to &lt;a href=&quot;http://math.stackexchange.com/questions/1416/algebra-of-random-variables&quot;&gt;algebra-of-random-variables&lt;/a&gt;, is it possible to compute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = \frac{dx}{dy}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $x$ and $y$ are drawn from a known, independent probability distribution $x \in f(x)$, $y \in g(y)$? Geometrically the problem makes sense, one has a function $h(x,y) = f(x)g(y)$ that is the joint probability distribution and we are essentially asking for the gradient along one direction. I can get this to the expression ($z \in q(z)$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q(z) = \int f(x) g( \int{[z dx]} ) ( \frac{d}{dz} \int{[z dx]} ) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not quite sure what to make of this, or even how to compute it given specific PDF's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: For simplicity, assume all the PDF's considered are continuous and smooth.&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-08-04T22:00:03.460" LastActivityDate="2010-08-04T22:19:26.333" Title="Derivatives of random variables" Tags="&lt;probability-theory&gt;&lt;random-variable&gt;" AnswerCount="2" />
  <row Id="1569" PostTypeId="2" ParentId="1533" CreationDate="2010-08-04T21:34:26.313" Score="-1" ViewCount="0" Body="&lt;p&gt;You don't need to study discrete mathematics before for you start with algorithms, that will probably only mean that you never do it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need some basic understanding about graphs (which you will get anyhow from your algorithm book) and you need to be able to do induction proofs (same here). That's it basically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a good book for algorithmic study you should check Jon Kleinberg's book, I think its called &quot;Algorithm Design&quot;. Good luck! :)&lt;/p&gt;&#xA;" OwnerUserId="779" LastActivityDate="2010-08-04T21:34:26.313" CommunityOwnedDate="2010-08-04T21:34:26.313" CommentCount="1" />
  <row Id="1570" PostTypeId="2" ParentId="847" CreationDate="2010-08-04T21:40:43.287" Score="3" ViewCount="0" Body="&lt;p&gt;This is not a directly a matter of hyperbolic geometry but of &lt;em&gt;complex Euclidean geometry&lt;/em&gt;.  The construction of &quot;impossible&quot; triangles is the same as the construction of square roots of negative numbers, when considering the coordinates the vertices of those triangles must have.  If you calculate the coordinates of a triangle with sides 1,3,5 or 3,4,8 you get complex numbers.  In ordinary real-coordinate Euclidean geometry this means there is no such triangle.  If complex coordinates are permitted, the triangle exists, but not all its points are visible in drawings that only represent the real points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plane analytic geometry where the Cartesian coordinates are allowed to be complex, concepts of point,line, circle, squared distance, dot-product, and (with suitable definitions) angle and cosine can be interpreted using the same formulas.  This semantics extends the visible (real-coordinate) Euclidean geometry to one where any two circles intersect, but possibly at points with complex coordinates.  We &quot;see&quot; only the subset of points with real coordinates, but the construction that builds a triangle with given distances between the sides continues to work smoothly, and some formulations of the law of Cosines will continue to hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly relations of this picture to hyperbolic geometry.  One is that $cos(z)=cosh(iz)$ so you can see the hyperbolic cosine and cosine as the same once complex coordinates are permitted.  Another is that the Pythagorean metric on the complex plane, considered as a 4-dimensional real space, is of the form $x^2 + y^2 - w^2 - u^2$, so that the locus of complex points at distance $0$ from the origin contains copies of the hyperboloid model of hyperbolic geometry.  But there is no embedding of the hyperbolic plane as a linear subspace of the complex Euclidean plane, so we don't get from this an easier way of thinking about hyperbolic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help visualize what is going on it is illuminating to calculate the coordinates of a triangle with sides 3,4,8 or other impossible case, and the dot-products of the vectors involved.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T21:40:43.287" CommentCount="2" />
  <row Id="1571" PostTypeId="2" ParentId="1568" CreationDate="2010-08-04T21:46:00.977" Score="1" ViewCount="0" Body="&lt;p&gt;You can take a derivative of a function $x$ with respect to one of its variables and the function has to be smooth w.r.t. this parameter.   If $x$ is being drawn from a set of smooth functions then it is certainly possible to consider the derivative of a random variable $x$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm really not sure what $\frac{dx}{dy}$ would even mean in your case.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-08-04T21:46:00.977" />
  <row Id="1572" PostTypeId="2" ParentId="29" CreationDate="2010-08-04T22:05:50.883" Score="3" ViewCount="0" Body="&lt;p&gt;SpeQ is really nice. If you close it's left/right areas it's just like Notepad, but a Notepad which can compute stuff. You can change any line displayed, press Enter at it's end and the line will be recomputed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like in Notepad, you can select any part of what's displayed and delete it. You can also paste non-mathematical text into it. It will also remember what's displayed inside it the next time you start it. It has an option to clear it's content on Exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people recommended Python. It's is my main programming language and I love it, but SpeQ it's MUCH more powerful and easy to use than Python as a calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/190212/public/speq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.speqmath.com/index.php?id=1&quot; rel=&quot;nofollow&quot;&gt;http://www.speqmath.com/index.php?id=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="775" LastEditorDisplayName="" LastEditDate="2010-08-04T22:13:24.340" LastActivityDate="2010-08-04T22:13:24.340" CommunityOwnedDate="2010-08-04T22:05:50.883" />
  <row Id="1573" PostTypeId="2" ParentId="262" CreationDate="2010-08-04T22:08:04.640" Score="2" ViewCount="0" Body="&lt;p&gt;When I was in my fourth year of high school I got a copy of &lt;em&gt;What is Mathematics?&lt;/em&gt; by Courant and Robbins. That book showed to me that Mathematics is far more than a &quot;boring tool&quot; to do Physics and opened up new worlds. I would recommend it to any bright high school kid with an interest in math and sciences.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-04T22:08:04.640" CommunityOwnedDate="2010-08-04T22:08:04.640" />
  <row Id="1574" PostTypeId="2" ParentId="1568" CreationDate="2010-08-04T22:08:56.793" Score="4" ViewCount="0" Body="&lt;p&gt;Radon-Nikodym derivative of the measures represented by variables $x$ and $y$ is one answer to your question, and is somewhat similar to the reasoning you sketched about a product density $h(x,y)=f(x)g(y)$.   For &quot;change in $x$ per change in $y$&quot; you perhaps want the correlation coefficient (multiplied by standard deviation of $x$). &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-04T22:08:56.793" CommentCount="0" />
  <row Id="1576" PostTypeId="1" CreationDate="2010-08-04T23:05:09.973" Score="2" ViewCount="162" Body="&lt;p&gt;I have an application that wants controllable random functions from $\mathbb{Z}^2$ and $\mathbb{Z}^3$ to $2^{32}$ , where by controllable I basically mean seedable by some parameters (say, on the order of 3 to 5 32-bit integers) such that the same seeds will always produce the same functions.  The most obvious way of doing this (for the two-dimensional case, say) would seem to be computing the value at some point $(x,y)$ by using $x$, $y$, and the seed parameters as seeds for something like an LFSR generator or a Mersenne Twister, then running the RNG for some fixed number of steps and taking the resultant value as the value of the function at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I be certain that this procedure won't keep too much correlation between adjacent 'seed points', and is there either a straightforward analysis or even just some general guideline for how many iterations would be necessary to eliminate that correlation?  My first back-of-the-envelope guess would be that each iteration roughly doubles the decorrelation between given seed values, so that 32 iterations would be necessary to achieve the requisite decorrelation over a range of $2^{32}$ values (and in practice I'd probably double it to 64 iterations), but that's strictly a guess and any proper analysis would be welcome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited for clarification:&lt;/strong&gt; To further outline the issue, I may be sampling this random function $f$ (for some given seed parameters) at arbitrary values, and need those samples to be identical between passes; so for instance, if a first application computes $f(0, 0)$, $f(437, 61)$, $f(-23, 129)$, and then $f(5,3)$, and a second (potentially concurrent) application computes $f(1,0)$ and then $f(5,3)$, both passes need to find the same value of $f$ at $(5,3)$.  I may also be sampling $f$ at arbitrary points, so I'd like the evaluation to take constant time (and in particular, evaluating $f(x,y)$ shouldn't take time linear in $x+y$).&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2010-08-05T18:02:49.817" LastActivityDate="2010-08-05T18:02:49.817" Title="'(Pseudo)-random functions' by seeding of PRNGs?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="1577" PostTypeId="1" AcceptedAnswerId="1578" CreationDate="2010-08-05T00:36:13.250" Score="0" ViewCount="45" Body="&lt;p&gt;Say we have a family {$A_\alpha$} of disjoint Baire spaces. Also suppose that each $A_\alpha$ is disjoint from the closure of the union of the other sets. Show that $\bigcup_{\alpha} A_\alpha$ is a Baire space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can prove this by transfinite induction. Suppose the property holds for all $\beta$ &amp;lt; $\alpha$, show it holds for $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha$ is a limit ordinal: we want to show that for every sequence of open dense subsets of $\bigcup_{\beta&amp;lt;\alpha} A_\alpha$ their intersection is open dense. This sequence of open dense sets in the union is defined as the union of all open dense sets in each of the sets since they all are Baire spaces and by induction hypothesis the union up to $\beta$ is a Baire space. Suppose this is not the case. Then there exists an open set $U$ such that the intersection of these open dense sets misses $U$. There is some ordinal $\beta$ such that $U \in A_\beta$, but this is a contradiction since we supposed that the union up to $\beta$ was a Baire space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do I handle the successor case? it seems trickier.&lt;/p&gt;&#xA;" OwnerUserId="786" LastActivityDate="2010-08-05T00:47:57.033" Title="disjoint union of Baire spaces which is a Baire space" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="1578" PostTypeId="2" ParentId="1577" CreationDate="2010-08-05T00:47:57.033" Score="2" ViewCount="0" Body="&lt;p&gt;I'm pretty sure it's simpler and doesn't require transfinite induction or anything nearly as difficult. In this case the $A_\alpha$ are open sets (as the complement of the closure of the union of the others). So suppose $U_1, \dots,$ are open dense sets in this union. Then each intersection $U_i \cap A_\alpha$ is open and dense in $A_\alpha$. The countable collection has intersection which is dense in each $A_\alpha$ by the Baire property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the countable intersection, call it $F$, is dense in the union $\bigcup A_\alpha$. (Any point in the union belongs one of the $A_\alpha$, and there is a point of $F \cap A_\alpha$ nearby, hence a point of $F$ nearby.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-05T00:47:57.033" />
  <row Id="1579" PostTypeId="1" AcceptedAnswerId="1581" CreationDate="2010-08-05T00:49:39.170" Score="4" ViewCount="295" Body="&lt;p&gt;This question was motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/1442/is-there-a-direct-proof-of-this-lcm-identity&quot;&gt;pondering this lcm identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that gcd(1,6,15) = 1, but lcm(1,6,15)=30, but 1*6*15 = 90.  (2,6,15) shows a similar phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the correct identity for n-ary gcd/lcm?&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-08-18T02:11:55.883" Title="n-ary version of gcd(a,b)lcm(a,b) = ab " Tags="&lt;number-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="1580" PostTypeId="2" ParentId="29" CreationDate="2010-08-05T00:53:40.923" Score="2" ViewCount="0" Body="&lt;p&gt;If I am at school (and in my lab) I will just have Maple running for symbolic stuff. It's pretty good at it. Other then that, for simple calculations I use windows calculator (its a hotkey on my keyboard) and for more advanced stuff, I'll use WolframAlpha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my plan is to learn mathematica. &lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-08-05T00:53:40.923" CommunityOwnedDate="2010-08-05T00:53:40.923" />
  <row Id="1581" PostTypeId="2" ParentId="1579" CreationDate="2010-08-05T01:06:02.443" Score="7" ViewCount="0" Body="&lt;p&gt;For $n \ge 2$ there isn't just one identity; there are actually $n+1$ identities, half of which are dual to the other half, and two of which are the trivial identity $a_1 ... a_n = a_1 ... a_n$.  The first two are as follows.  Let $P = a_1 ... a_n$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a_1 ... a_n = \text{gcd}(a_1, a_2, ... a_n) \text{lcm} \left( \frac{P}{a_1}, \frac{P}{a_2}, ... \right).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a_1 ... a_n = \text{gcd}(a_1 a_2, a_1 a_3, ..., a_{n-1} a_n) \text{lcm} \left( \frac{P}{a_1 a_2}, \frac{P}{a_1 a_3}, ..., \frac{P}{a_{n-1} a_n} \right).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other identities are similar.  This is because &quot;pointwise&quot; (e.g. at a particular prime $p$) the two-variable identity is just the identity $\text{min}(a, b) + \text{max}(a, b) = a + b$ for non-negative integers $a, b$, and the above identities are the natural generalization of this one. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-05T01:11:48.877" LastActivityDate="2010-08-05T01:11:48.877" CommentCount="2" />
  <row Id="1582" PostTypeId="2" ParentId="1533" CreationDate="2010-08-05T01:56:21.643" Score="3" ViewCount="0" Body="&lt;p&gt;Discrete Math knowledge is needed to become adept in proving the correctness and deriving the complexity of algorithms and data structures. You will be taught those in Algo/DS books, but you can only get the mathematical proficiency by practicing just discrete math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knuth book is very good for that. But IMHO, you will only need it if you for doing advanced proofs in DS/Algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a beginner, it would be great to go over &quot;Grimaldi&quot; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201199122&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Discrete-Combinatorial-Mathematics-Applied-Introduction/dp/0201199122&lt;/a&gt;&#xA;and then quickly move to Algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you will continue going deep in Discrete Math and never get to Algorithms/DS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, Discrete Math does not teach you how to design algorithms or Data structures. That will come only by practicing Algorithm problems @ topcoder, acm icpc , spoj etc and reading books on Algos/DS or courses on those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My 2 cents.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2010-08-05T01:56:21.643" CommunityOwnedDate="2010-08-05T01:56:21.643" CommentCount="1" />
  <row Id="1583" PostTypeId="2" ParentId="29" CreationDate="2010-08-05T02:02:54.517" Score="6" ViewCount="0" Body="&lt;p&gt;A few people have mentioned Sage, but I think deserves a bit more of a plug. Firstly someone mentioned that Sage is quite large to install. This is true, the binary downolad comes in at around 1 gig, however you can use it for free online, without having to install a thing! Head on over to  &lt;a href=&quot;http://www.sagenb.org&quot; rel=&quot;nofollow&quot;&gt; sagenb.org &lt;/a&gt;, and sign up for a free account, which will allow you to create and save your own worksheets. I think this is really nice--it feels like it's your &quot;gmail&quot; but for maths, so you can just log in from anywhere and &quot;check your maths&quot; if you want :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very popular option--I believe that in the last 9  months around 30,000 people have created accounts! It's perfectly suited for high school students and undergraduates, since no installation is necessary, and hence it can be used straight away in any computer lab with internet access. Here's a screenshot of what the notebook looks like. (I just googled &quot;sage notebook screenshot.&quot;) It's been designed to look and work like the Mathematica notebook system, and it's surprisingly slick and easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://sidk.info/wp-content/uploads/2007/11/sage_screenshot2.jpeg&quot; alt=&quot;Sage screenshot&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really is a lot to recommend it. I used the online notebook for around 6 months before I installed a copy on my own computer, and this was because I became interested in developing Sage. Although Sage is based on Python (so in particular any Python syntax will work), you really don't need to know Python to get started using Sage--I didn't when I started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sage is quite powerful too. Aside from using Maxima for symbolic integration, I belive Sage uses Pari for highly optimised polynomial arithmatic, and I have used Sage for some pretty intense linear algebra computations (over the integers as well as over finite fields), although for this kind of thing you'll probably not want to rely on the online server which can be a little slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's not forget that you also get the moral boost of being part of something really positive--an movement to create an open source alternative to costly and &quot;closed&quot; alternatives like Mathematica, Maple and Matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Disclaimer: I just came back from the Sage Days 23 sage developers workshop in Leiden, and I'm a bit pumped!)&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2010-08-05T02:02:54.517" CommunityOwnedDate="2010-08-05T02:02:54.517" />
  <row Id="1584" PostTypeId="2" ParentId="1563" CreationDate="2010-08-05T02:19:19.103" Score="3" ViewCount="0" Body="&lt;p&gt;Take any complete theory $T$ that has two models of of different cardinalities. Then the models are elementary equivalent (they both model $T$) but they cannot be isomorphic because isomorphisms preserve cardinality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Lowenheim-Skolem theorem, every theory with infinite models has models of different cardinalities, so really any complete theory with infinite models will work for $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personal note: Whenever I teach students about cardinality, I always point out this sort of application. Because cardinality is preserved by bijections, showing two objects (groups, rings, models, etc) have different cardinalities is an easy way to show they are not isomorphic.  I find this a very compelling reason to care about cardinality outside of set theory. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-08-05T11:18:54.957" LastActivityDate="2010-08-05T11:18:54.957" CommentCount="6" />
  <row Id="1585" PostTypeId="2" ParentId="1533" CreationDate="2010-08-05T04:10:42.853" Score="2" ViewCount="0" Body="&lt;p&gt;Theres many different areas to discrete math, and many good books. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;theres Graph Theory by Diestel, which has a free pdf version available at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://diestel-graph-theory.com/&quot; rel=&quot;nofollow&quot;&gt;diestel-graph-theory.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;theres generatingfunctionology by wilf, free pdf version at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;math.upenn.edu/~wilf/DownldGF.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other books that are good include Enumerative combinatorics 1 and 2 by Richard P Stanley (a book which is sufficiently dense that having at least 1 analysis and algebra course each will help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that being said, for more introductory expositions in terms of expected mathematical maturity, I'd suggest googling around and looking at various lecture notes of the &quot;intro to combinatorics&quot; or &quot;mathematics for computer scientists&quot; sorts. I found that MIT OCW's &quot;mathematics for Computer Scientists&quot; notes were quite nice when I looked at them several years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2005/readings/&quot; rel=&quot;nofollow&quot;&gt;ocw.mit.edu/courses/electrical-engineering-and-computer-science...&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;has a link to the lecture notes. There are some really funny asides in it. One of my favorites &quot;... anyone who says that is wrong, and you should make fun of them until they cry&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, If you want to dig even deeper into discrete math/ combinatorics, the value of building up a wee bit of mathematical basics in other areas of math. Complex Analysis, real analysis (at the level of at least baby rudin, and perhaps even up to functional analysis), maybe some probability up to its measure theory formulation level, and at least a smidge of abstract algebra.  Then you can do stuff like look at the combinatorics of random processes (great for analyzing randomized algorithms) and look at cool problems like percolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;theres probably other things I should suggest, but the point is discrete math is accessible without that much of a background, but is also rewards you for enriching that mathematics background with some amazingly beautiful stuff thats 1) awesome and fun 2) useful.&lt;/p&gt;&#xA;" OwnerUserId="798" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-05T04:56:48.560" LastActivityDate="2010-08-05T04:56:48.560" CommunityOwnedDate="2010-08-05T04:10:42.853" />
  <row Id="1586" PostTypeId="2" ParentId="1512" CreationDate="2010-08-05T04:52:56.257" Score="5" ViewCount="0" Body="&lt;p&gt;One way to think of the situation is that the countable product of $\mathbf Z/2\mathbf Z$s that you are considering&#xA;is a vector space $V$ over $\mathbf F_2$ that is uncountable, and thus has an (even more) uncountable dual space.  If $f:V \to \mathbf F_2$ is a non-zero element of the dual space, its kernel is an index 2 subgroup of $V$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the open subgroups that you know about correspond to (i.e. are the kernels of) the projections onto the various factors&#xA;in the product, and these are countable in number.  So there is a massive host of other index 2 subgroups, corresponding to all the other non-zero elements $f$ of the dual space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have noted, you won't be able to write these down explicitly though.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-05T04:52:56.257" CommentCount="3" />
  <row Id="1587" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T05:41:21.980" Score="1" ViewCount="0" Body="&lt;p&gt;As BBischof says: Willard is nice. I also learned a lot from Mendelson, a little dover book. I think that you should use whatever gets you through the subject &quot;quickest&quot;. Point Set topology is now just basic language that is very necessary, but it seems there is not much research in the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My dad did not go into math because of a course out of munkres, he had no intuition for the topic the way it was taught. I dont see this as a criticism of Munkres because I think one of the major goals of any early course is to untrain the student. There are spaces that are pathological that you have to deal with. You have to be rigorous. I think that is one of the main points of such a course, that and preparing the student for what comes next. Which is why mariano asks what your goals are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing about willard though is that you should make sure you do the exercises, they contain key facts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I felt it pertinent to add that I did go through a course from munkres (by the time I did it was pointless for me, I had already been using the relevant stuff for my field daily) and it was quite well done I thought. It is the classic, I don't think you can go wrong with it.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-08-05T15:07:36.513" LastActivityDate="2010-08-05T15:07:36.513" CommunityOwnedDate="2010-08-05T05:41:21.980" CommentCount="4" />
  <row Id="1588" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T05:57:26.957" Score="5" ViewCount="0" Body="&lt;p&gt;Good and Free &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://uob-community.ballarat.edu.au/~smorris/topology.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Topology Without Tears by Sidney A. Morris&lt;/strong&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-05T05:57:26.957" CommunityOwnedDate="2010-08-05T05:57:26.957" CommentCount="3" />
  <row Id="1589" PostTypeId="1" AcceptedAnswerId="1591" CreationDate="2010-08-05T06:00:14.783" Score="5" ViewCount="233" Body="&lt;p&gt;I asked this question a while back on MO : &lt;a href=&quot;http://mathoverflow.net/questions/32689/how-should-a-homotopy-theorist-think-about-sheaf-cohomology&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/32689/how-should-a-homotopy-theorist-think-about-sheaf-cohomology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that really helped in learning the Serre SS was doing particular computations (like $H^*(CP^{\infty})$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious, as a sort of followup if anyone can suggest:&#xA;1. a reference where small computations are carried out? or&#xA;2. a specific computation to do with a small enough sheaf an some simple topological space that would be able to give one a feel for sheaf cohomology. So this space that we are working over need not be a scheme, in fact it would probably be best if it were not a scheme since i dont understand them quite yet. And are there tricks of the trade to computing these things? or do people just hammer away ate injective resolutions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in short, please suggest a space and a sheaf on it that i should work on computing the sheaf cohomology of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for your time,&#xA;Sean&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS I of course welcome any other suggestions for understanding how to compute sheaf cohomology.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-08-26T16:05:08.073" LastActivityDate="2010-08-26T16:05:08.073" Title="toy Sheaf Cohomology Computation" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="1590" PostTypeId="2" ParentId="1519" CreationDate="2010-08-05T06:08:32.050" Score="0" ViewCount="0" Body="&lt;p&gt;I think you understand attaching cells that have dimension gap better than you think! Consider any $S^n$ for $n&gt;&gt;0$, we can think of this as being the attachment of $D^n$ to a point which is 0-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would recommend that you think of this attaching cells thing a bit more carefully. if you understand the hopf fibration &lt;em&gt;then you do&lt;/em&gt; understand the attaching of the cell. The hopf fibration is &lt;em&gt;not&lt;/em&gt; easy to picture (as far as I know). I would recommend thinking about the map in terms of what happens when you work with coordinates. I will be back later to add more... I promise.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-08-05T06:08:32.050" />
  <row Id="1591" PostTypeId="2" ParentId="1589" CreationDate="2010-08-05T06:39:00.947" Score="2" ViewCount="0" Body="&lt;p&gt;Any de Rham cohomology (or Dolbeault cohomology) computation is a computation in sheaf cohomology. Actually --- any computation in singular cohomology is a computation in sheaf cohomology!! ;-) We're just taking different resolutions of the appropriate constant sheaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IIRC, there are some good Cech cohomology computations and examples in Bott-Tu. Also, have you read section 3.H of Hatcher's algebraic topology book, on &quot;local coefficients&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple example from algebraic geometry, compute the cohomology of the structure sheaf of $\mathbb{A}^2$ minus a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to recall an exercise or an example in Hartshorne in which the genus of a degree $d$ curve in $\mathbb{P}^2$ is computed using Cech cohomology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The section in Hartshorne on the cohomology of $\mathbb{P}^n$ uses Cech cohomology, and I remember finding it pretty instructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eisenbud's commutative algebra book probably has lots of good examples. However, he might not use words like &quot;sheaf cohomology&quot; (since I don't think the book introduces sheaves), although that's what it is.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-08-05T06:48:53.283" LastActivityDate="2010-08-05T06:48:53.283" />
  <row Id="1592" PostTypeId="2" ParentId="653" CreationDate="2010-08-05T06:43:35.837" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an answer to your meta question since i can not yet post there:&#xA;I disagree that such questions should be closed without thought. Some people may genuinely be interested in real world applications and some may come up and be particularly helpful in ways I might not have conceived of. Perhaps the &quot;questioner&quot; meant something broader like what comes out of homtopy theory, then certainly the Ghrist work would be relevant (despite the fact the that homoglogy is definitely not a generalization of homotopy, not evenof $\pi_1$).The work that Carlsson is doing is sort of a real world application, although not specifically of any model categorical stuff (... maybe it is, i dunno).&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-08-05T06:43:35.837" />
  <row Id="1593" PostTypeId="2" ParentId="1576" CreationDate="2010-08-05T07:10:07.140" Score="0" ViewCount="0" Body="&lt;p&gt;This does not strictly answer your question, but might be too long for a comment. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe you should rethink your approach. Most randon number generators are designed to be seeded once, and than return a sequence of &quot;random&quot; numbers, as close to independent and uniformly distributed as possible. I don't think that there are many results about the correlation of the 64th sequence element for different seeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, (at least in the implementations I know) the period of the algorithm is a lot longer than the number of possible seeds. If you have to reuse seeds in your case, you could introduce correlations that are avoidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of precomputing the entire grid, you could get a random value just-in-time and store it for future reference, saving memory if you don't access every node.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-05T07:10:07.140" CommentCount="1" />
  <row Id="1594" PostTypeId="2" ParentId="709" CreationDate="2010-08-05T07:33:05.207" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://hotmath.com/images/gt/lessons/genericalg1/parabs.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = ax^2 + bx + c$&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for parabola written in &lt;strong&gt;Vertex Form&lt;/strong&gt; vertex is (h,k).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = m(x-h)^2 + k = \underline{m}x^2 + \underline{(-2mh)}x+ \underline{mh^2 +k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b = -2mh = -2ah \rightarrow h = -b/2a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = mh^2 + k = ah^2 + k \rightarrow k = c - b^2/4a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;source&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-05T07:33:05.207" />
  <row Id="1595" PostTypeId="2" ParentId="1512" CreationDate="2010-08-05T07:35:10.837" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an easy point which is implicit in the other answers, but which is part of the question (as I read it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $(p_1,p_2,p_3,p_4,\dots):=(-1,2,3,5,\dots)$, and define the fields $K_0,K_1,\dots$ by $K_0:=\mathbb Q$ and $K_n:=K_{n-1}(\sqrt{p_n})$ for $n\ge 1$. [It's clear that any quadratic field is contained into some $K_n$.] To make sure that the Galois group in question is the indicated one, we need to check that $\sqrt{p_n}$ is &lt;strong&gt;not&lt;/strong&gt; in $K_{n-1}$. To do that we prove the following apparently stronger claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be a positive integer and $a$ the product of the $p_k^{e(k)}$, where $k$ runs over the integers $\ge n$ and $e(k)$ is an integer equal to $0$ for almost all $k$. Assume $\sqrt a\in K_{n-1}$. Then all the $e(k)$ are even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily proved by induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; Things can be described in this way: The obvious morphism from $\mathbb Q^{\times}/\mathbb Q^{\times 2}$ into our Galois group induces an isomorphism between the profinite completions. [It's nice to obtain a Galois group by applying a simple construction to the base field.]&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-05T09:27:07.193" LastActivityDate="2010-08-05T09:27:07.193" />
  <row Id="1596" PostTypeId="1" CreationDate="2010-08-05T08:56:51.800" Score="0" ViewCount="214" Body="&lt;p&gt;Can someone explain to me in simple terms what a non-linear transformation is in maths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know some single-variable calculus, but I read it has to do with multi-variable calculus, which I'm not familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could explain it in simple words, that would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2010-08-05T11:41:51.863" Title="Non-Linear Transformation" Tags="&lt;calculus&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="1597" PostTypeId="2" ParentId="1537" CreationDate="2010-08-05T09:45:56.930" Score="1" ViewCount="0" Body="&lt;p&gt;Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(1)=1$&lt;/li&gt;&#xA;&lt;li&gt;$f(z+1)=zf(z)$ for all $z&gt;0$&lt;/li&gt;&#xA;&lt;li&gt;$f(z)$ is analytic for $\operatorname{Re}z&gt;0$&lt;/li&gt;&#xA;&lt;li&gt;$f(z)$ is bounded for $1\leq \operatorname{Re}z\leq 2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(See also the related MathOverflow thread &lt;a href=&quot;http://mathoverflow.net/questions/23229/importance-of-log-convexity-of-the-gamma-function&quot; rel=&quot;nofollow&quot;&gt;Importance of Log Convexity of the Gamma Function&lt;/a&gt;, where I learned about the above theorem.)&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-05T09:45:56.930" />
  <row Id="1598" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T10:29:55.607" Score="1" ViewCount="0" Body="&lt;p&gt;Sutherland's &lt;em&gt;Metric and topological spaces&lt;/em&gt; is nice. I know a guy who swears by Kelley's &lt;em&gt;General topology&lt;/em&gt; but I've not read it myself.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-05T10:29:55.607" CommunityOwnedDate="2010-08-05T10:29:55.607" />
  <row Id="1599" PostTypeId="2" ParentId="1596" CreationDate="2010-08-05T10:33:57.537" Score="2" ViewCount="0" Body="&lt;p&gt;Let $V_1, V_2$ be two &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;vector spaces&lt;/a&gt; over the &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;field&lt;/a&gt; $F$. A transformation $T: V_1 \to V_2$ is linear if for every $x, y \in V_1$ and every $\alpha \in F$ it is true that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) $T(x + \alpha y) = T(x) + \alpha T(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T is not a linear transformation if there are some $x, y, \alpha$ such that (*) is not true.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-05T10:33:57.537" />
  <row Id="1600" PostTypeId="2" ParentId="1537" CreationDate="2010-08-05T10:44:36.710" Score="3" ViewCount="0" Body="&lt;p&gt;This is a comment posted as an answer for lack of reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the &quot;prime at infinity&quot;, and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is at least as old as Tate's thesis, and a possible reference is Weil's &lt;strong&gt;Basic Number Theory&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-05T11:13:14.637" LastActivityDate="2010-08-05T11:13:14.637" CommentCount="1" />
  <row Id="1601" PostTypeId="1" AcceptedAnswerId="1702" CreationDate="2010-08-05T11:14:05.370" Score="2" ViewCount="324" Body="&lt;p&gt;In a recent discussion, I came across the idea of proving lower bound for number of comparisons required to find the largest element in an array. The bound is $n - 1$. This is so because, the set of comparisons performed by every such algorithm looks like a tournament tree, which always has $n - 1$ internal nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious question is then, What should be the lower bound for finding 2nd largest elelment from an array.&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T23:10:33.557" LastActivityDate="2010-08-06T08:47:12.300" Title="lower bound for finding second largest element" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;searching&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="1602" PostTypeId="2" ParentId="1576" CreationDate="2010-08-05T11:36:06.083" Score="1" ViewCount="0" Body="&lt;p&gt;Note that when an &lt;a href=&quot;http://www.newwaveinstruments.com/resources/articles/m_sequence_linear_feedback_shift_register_lfsr.htm&quot; rel=&quot;nofollow&quot;&gt;LFSR&lt;/a&gt; is used to generate a sequence of +1 / -1 values, the correlation between that sequence and any time-shifted version of that sequence is near zero: (sum from i = 0 to N-1 of a[i]a[i+k]) is either N if k is a multiple of N, or -1 otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial state of an LFSR has a one-to-one mapping to the time shift -- this is analogous to a discrete logarithm problem in Galois fields; given a time shift k, it's easy to find the initial state of an LFSR (O(log(k), I think, as it's just computing exponentiation in the appropriate Galois field), but given the initial state of an LFSR, it's hard to find the time shift k in anything shorter than O(k). So your PRNG could be an LFSR with very long period e.g. 64 bits or greater, and the &quot;seed&quot; could be a time shift used to derive a different initial state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as 2- or 3- or k-dimensional mapping, interleave the bits of the coordinates, and use a hash function to map those bits to an integer for use in deriving LFSR seeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what kind of correlation you're looking to avoid between functions. (e.g. correlation in a statistical sense, or cryptographic independence e.g. one PRNG cannot be used to predict future output of another).&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-08-05T11:36:06.083" CommentCount="4" />
  <row Id="1603" PostTypeId="2" ParentId="1596" CreationDate="2010-08-05T11:41:51.863" Score="0" ViewCount="0" Body="&lt;p&gt;In addition to the definition of linear map that Tomer remind you, here are two examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, $f(x,y) = x^2y$ is &lt;em&gt;not&lt;/em&gt; a linear map $f: \mathbb{R}^2 \longrightarrow \mathbb{R}$ because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(2x,2y) = 4x^22y \neq 2x^2y = 2f(x,y) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, the linear maps $f: \mathbb{R}^m \longrightarrow \mathbb{R}^n$ are &lt;em&gt;necesseraly&lt;/em&gt; of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(x_1, \dots , x_m) = (a_1^1 x_1 + \dots + a_1^m x_m , \dots , a_n^1 x_1 + \dots + a_n^m x_m)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $a^i_j$ constant coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, two more examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x,y) = x + 2y$ is a linear map.&lt;/li&gt;&#xA;&lt;li&gt;$f(x,y,z) = 3x + 1$ is a non-linear map&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-05T11:41:51.863" />
  <row Id="1604" PostTypeId="2" ParentId="1589" CreationDate="2010-08-05T11:42:04.530" Score="0" ViewCount="0" Body="&lt;p&gt;Rotman does some very elementary explicit computations of Cech cohomology in his book Homological Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly he does these computations using resolutions, spectral sequences, and by just starting with some sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a complete beginner to this material, I was able to understand his treatment and compute some specific examples on my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I just looked at the first edition, and it seems to be different slightly. For your information I used the second edition.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-05T11:42:04.530" CommentCount="3" />
  <row Id="1605" PostTypeId="2" ParentId="198" CreationDate="2010-08-05T11:42:51.033" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;Taking the principle of excluded middle from the mathematician would be the same, say, as proscribing the telescope to the astronomer or to the boxer the use of his fists. To prohibit existence statements and the principle of excluded middle is tantamount to relinquishing the science of mathematics altogether.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-David Hilbert&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-08-05T11:42:51.033" CommunityOwnedDate="2010-08-05T11:42:51.033" />
  <row Id="1606" PostTypeId="2" ParentId="1453" CreationDate="2010-08-05T11:57:21.947" Score="3" ViewCount="0" Body="&lt;p&gt;A detailed explanation can be found in many places. I'll to provide an intuitive one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem&quot; rel=&quot;nofollow&quot;&gt;Cook-Levin theorem&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_satisfiability_problem&quot; rel=&quot;nofollow&quot;&gt;Boolean satisfiability problem&lt;/a&gt; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Np_complete&quot; rel=&quot;nofollow&quot;&gt;NP-complete&lt;/a&gt; - ie. every decision problem in NP can be reduced to it. We will ask of our prover to supply the input and output of every gate in the circuit and consider this as the proof that $x\in L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to stop there, then the prover could cheat by changing a single bit in his proof, which would require us to check a large (non-constant) number of bits in it. So we must ask something more from the prover. This something more will be the encoding of his proof using some (special) error-correcting code. Intuitively, this &quot;smudges&quot; the false bits onto a large number of bits in the code-word. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The verifier receives a word from the prover, there are 2 ways in which the prover might try and cheat, this word might not be a code-word in the chosen code, or it might be the encoding of something which isn't a proof. We'll examine the (slightly different) separation of cases of the word either being far from every codeword (a large number of bits must be changed to reach a codeword) or that it is close to codeword which isn't an encoding of a proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be able to detect these we demand that our code has the following 2 properties:&#xA;1) locally checkable - we can, by reading a constant number of bits, detect w.h.p if a word is far from any codeword.&#xA;2) locally decodable - we can, by reading a constant number of bits, decode w.h.p a bit from the encoded word.&#xA;(finding such codes is hard, hadamard has this properties but the code-word's size is exponential, RM codes have these properties as well (to a lesser degree) and they are the ones generally used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the verifier checks if the word is close to being a code-word (using 1), and if the test succeeds, picks a random gate and decodes it's outputs and output (using 2). this is sufficient to achieve constant query complexity and logarithmic randomness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be mentioned that  (&lt;a href=&quot;http://www.wisdom.weizmann.ac.il/~dinuri/mypapers/combpcp.pdf&quot; rel=&quot;nofollow&quot;&gt;Dinur&lt;/a&gt;) has a combinatorial proof of PCP which is quite different from what i've discribed, but her version is (for me, at least) less intuitive.  &lt;/p&gt;&#xA;" OwnerUserId="816" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T12:55:01.437" LastActivityDate="2010-08-05T12:55:01.437" CommentCount="3" />
  <row Id="1607" PostTypeId="2" ParentId="1601" CreationDate="2010-08-05T11:57:32.260" Score="2" ViewCount="0" Body="&lt;p&gt;An observation: it will always require at least n-1. Draw a graph. If we test two nodes, then we draw an arrow from the smaller to the larger. We know a is greater than b if and only if there is a path from a to b along the arrows. We will always need at least n-1 arrows to connect the graph and hence n-1 comparisons to find a lower bound. It is easy enough to see that this isn't the minimal upper bound though - we will only know the second highest element if our arrows form a line (ie. we can go from a &quot;start&quot; node to an &quot;end&quot; node by passing through each node and each arrow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fact - there are algorithms for finding the &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_algorithm&quot; rel=&quot;nofollow&quot;&gt;kth largest number&lt;/a&gt; in guaranteed linear time. That could give you some information, but there could always be a special algorithm for finding the second largest element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a solution for the problem yet. But one thing we can try is breaking the problem into two groups and combining them. Suppose we have two groups, A and B, where we know both the largest and second largest elements, but no knowledge of any relationships between the groups. Then it takes two arrows to find the second highest element. First we compare the highest elements in A and B. Without loss of generality assume A has the highest element. Then we would compare B with the highest element in A. Showing that the groups can't be combined with a single comparison is trivial. This gives us a recursion relation for the time taken to solve by combining groups which have been solved separately: $f(n)=min f(k)+f(n-k)+2$ where $ 0 &amp;lt; k &amp;lt; n $.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T23:08:25.010" LastActivityDate="2010-08-05T23:08:25.010" CommentCount="3" />
  <row Id="1608" PostTypeId="1" AcceptedAnswerId="1616" CreationDate="2010-08-05T11:58:30.107" Score="0" ViewCount="62" Body="&lt;p&gt;How can I calculate the length of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathetus&quot; rel=&quot;nofollow&quot;&gt;cathetus&lt;/a&gt; in a triangle if I know the length of the hypotenuse and the ratio between the two catheti?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypotenuse is 5cm long, and the ratio between the catheti is 4:3 - how long are the catheti or either cathetus?&lt;/p&gt;&#xA;" OwnerUserId="470" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-05T12:39:05.777" LastActivityDate="2010-08-05T12:45:53.860" Title="Get length of cathetus/i from length of hypothenuse and ratio between catheti?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1609" PostTypeId="1" AcceptedAnswerId="1635" CreationDate="2010-08-05T12:05:17.973" Score="8" ViewCount="267" Body="&lt;p&gt;when trying to explain AG-codes to computer scientists, the major points of contention i am faced with are the concepts of divisors, Riemann-Roch space and the genus of a function field. are there any intuitive explanations for these concepts? or at least ones that are less dependent on knowledge of algebraic-geometry/topology? &lt;/p&gt;&#xA;" OwnerUserId="816" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-05T19:47:57.917" LastActivityDate="2010-08-09T20:47:20.013" Title="intuitive explantions for the concepts of divisor and genus" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;&lt;algebraic-curves&gt;" AnswerCount="4" FavoriteCount="8" />
  <row Id="1610" PostTypeId="2" ParentId="1608" CreationDate="2010-08-05T12:07:38.157" Score="1" ViewCount="0" Body="&lt;p&gt;You can call one cathetus 4*x*, the other 3*x*, and apply Pythagora's theorem: (4*x*)&lt;sup&gt;2&lt;/sup&gt;+(3*x*)&lt;sup&gt;2&lt;/sup&gt;=5&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will obtain 25*x*&lt;sup&gt;2&lt;/sup&gt;=25, which yields &lt;em&gt;x&lt;/em&gt;=1. So one cathetus is 4cm, and the other 3cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Remember that 3,4,5 is a Pythagorean triple)&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-05T12:07:38.157" CommentCount="2" />
  <row Id="1611" PostTypeId="2" ParentId="1609" CreationDate="2010-08-05T12:11:59.867" Score="2" ViewCount="0" Body="&lt;p&gt;Well, the concepts of genus and divisors come from geometry/topology intuition originally, so those explanations are generally going to be the fastest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can give very rough versions without needing to say &quot;genus is the number of holes in the surface that the curve looks like.&quot;  You can think of genus as a measure of how complicated the function field is.  Naturally $k(x)$ is going to be the simplest possible one, and it happens to be genus zero.  Genus one function fields look like $k(x,y)$ but where $y^2$ is a cubic in $x$.  Now, this isn't very precise, but it's a rough thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for divisors, the topology/geometry of them is &quot;linear combinations of points, modulo zero-$\infty$ loci of functions,&quot; but to handle it purely algebraically, you'll want to change the word &quot;point&quot; to &quot;discrete valuations&quot;, so these are integer linear combinations of the various ways to describe the order of an element of the function field, modulo some set of trivial ones.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-05T12:11:59.867" />
  <row Id="1612" PostTypeId="2" ParentId="1609" CreationDate="2010-08-05T12:28:46.967" Score="3" ViewCount="0" Body="&lt;p&gt;Charles has already explained the notion of genus (which comes from topology--there's also a purely algebraic notion of genus as a first cohomology group, but I at least would find it less intuitive). So I'll talk about divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the analytic notion of a nonsingular projective curve is a compact Riemann surface. &quot;Nonsingular&quot; translates to &quot;complex manifold,&quot; &quot;curve&quot; translates to &quot;dimension 1,&quot; and &quot;projective&quot; translates to &quot;compact.&quot; So I will talk about Riemann surfaces.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a Riemann surface, it's locally the same as the complex plane. And in the complex plane, you have a way to measure the order of the zero of a holomorphic function. More generally, you can measure (in $\mathbb{Z}$) the order of a zero of a meromorphic function (which is negative if it has a pole, zero if it is nonvanishing and analytic, positive if it has a zero).  So to any meromorphic function (of course, we don't restrict to holomorphic ones---they're all constant) on a compact Riemann surface, we can assign a finite set of points with multiplicities consisting of the places where the order is nonzero (i.e., the zeros and poles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A divisor is more general. It's just a formal sum of points on the Riemann surfaces with multiplicities.  A divisor may come from a function as above; then it's called principal. In general, however, it does not. Yet given a divisor $D$, we can associate a vector space of meromorphic functions such that $div(f)+ D$ is a divisor with nonnegative coefficients. This is denoted $L(D)$ and its dimension by $l(D)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann-Roch_Theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann-Roch theorem&lt;/a&gt; is then a statement about the dimensions $l(D)$. One of its consequences, in particular, is that if $|D|$ (the sum of the multiplicities) is large, then you can always find a (nonzero) element of $L(D)$. This is intuitively easy to understand.  If $D$ is large, then you are allowing lots of leeway for $f$, potentially having poles at many places. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I'm being informal here, but there is a general &quot;GAGA&quot; theory about these equivalences that goes into much more detail, which I know next to nothing about.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-05T12:28:46.967" />
  <row Id="1613" PostTypeId="2" ParentId="1589" CreationDate="2010-08-05T12:33:54.223" Score="1" ViewCount="0" Body="&lt;p&gt;This is rather scheme-y, but there's  a really nice paper by &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.rmjm/1250128841&quot; rel=&quot;nofollow&quot;&gt;Kempf&lt;/a&gt; (hopefully you have institutional access :() that gives a very basic and elementary proof that the higher cohomology of a quasi-coherent sheaf on an affine scheme is trivial. The first part of the paper uses nothing more than the basic properties  (e.g. long exact sequence) of cohomology, and might be fun. I thought it was fun, anyway; it's also nice because it shows that Hartshorne is unnecessarily restrictive in sticking to noetherian affine schemes in chapter III (even if one wants to avoid anything fancy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, update: here is the proof &lt;a href=&quot;https://amathew.wordpress.com/2010/08/05/the-paper-of-kempf/&quot; rel=&quot;nofollow&quot;&gt;explained&lt;/a&gt; (admittedly by a beginner :)).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T02:51:39.390" LastActivityDate="2010-08-06T02:51:39.390" CommentCount="2" />
  <row Id="1614" PostTypeId="1" AcceptedAnswerId="1622" CreationDate="2010-08-05T12:36:01.677" Score="1" ViewCount="115" Body="&lt;p&gt;&lt;strong&gt;Stromquist's Theorem&lt;/strong&gt;:  If the simple closed curve J is &quot;nice enough&quot; then it has an inscribed square.&lt;br&gt;&#xA;&quot;Nice enough&quot; includes polygons.&lt;br&gt;&#xA;Read more about it here:  www.webpages.uidaho.edu/~markn/squares&lt;br&gt;&#xA;An &quot;inscribed square&quot; means that the &lt;em&gt;corners&lt;/em&gt; of a square overlap with the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to suggest a counter-example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curve connected by the points:&#xA;(.2,0)&#xA;    (1,0)&#xA;    (1,1)&#xA;    (0,1)&#xA;    (0, .2)&#xA;    (-.2, -.2)&#xA;    (.2,0)&lt;br&gt;&#xA;Link to plot: &lt;a href=&quot;http://www.freeimagehosting.net/uploads/5b289e6824.png&quot; rel=&quot;nofollow&quot;&gt;http://www.freeimagehosting.net/uploads/5b289e6824.png&lt;/a&gt;&lt;br&gt;&#xA;Can this curve be incribed by a square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(An older version of this question had another example: a triangle on top of a square (without their mutual side.) )&lt;/p&gt;&#xA;" OwnerUserId="821" LastEditorUserId="821" LastEditorDisplayName="" LastEditDate="2010-08-05T13:49:20.537" LastActivityDate="2010-08-05T14:01:55.283" Title="Is this a counter example to Stromquist's Theorem?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="1615" PostTypeId="2" ParentId="1614" CreationDate="2010-08-05T12:42:49.153" Score="8" ViewCount="0" Body="&lt;p&gt;What about the square on the bottom- with it and the triangle's mutual side returned?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Btw. Don't be disheartened that your counterexample didn't come through- looking for counterexamples to proved theorems will never give a genuine counterexample (unless ZFC turns out to be inconsistent!!;)), but it can provide valuable insight into what the theorem means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can see from experience that 'most of the square' may be contained within the curve- this is something that may not be obvious looking at the examples of Stormquist in action in books on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to find counterexamples, IMHO, is a great way to develop as a mathematician- just do not be afraid to be wrong ;)&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-05T12:54:36.057" LastActivityDate="2010-08-05T12:54:36.057" CommentCount="1" />
  <row Id="1616" PostTypeId="2" ParentId="1608" CreationDate="2010-08-05T12:45:53.860" Score="1" ViewCount="0" Body="&lt;p&gt;Let the ratio be 1:r and the hypotenuse be h. Then the sides are then x and rx for some x. By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow&quot;&gt;Pythagorean Theorem&lt;/a&gt; we get $x^2+r^2*x^2=h^2$. So $x=\sqrt{{h^2}/(1+r^2)}$. We can then calculate rx, the other side.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-05T12:45:53.860" />
  <row Id="1617" PostTypeId="1" CreationDate="2010-08-05T13:09:38.787" Score="0" ViewCount="252" Body="&lt;p&gt;Pardon me for not knowing LaTeX code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3x3 matrix A:&lt;br&gt;&#xA;[  2  -1   1  ]&lt;br&gt;&#xA;| -2   3  -2  |&lt;br&gt;&#xA;[ -1   1   0  ]&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...produces this characteristic equation:&lt;br&gt;&#xA;&amp;lambda;&lt;sup&gt;2&lt;/sup&gt; - 4&amp;lambda; + 3 = 0&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...these eigenvalues:&lt;br&gt;&#xA;&amp;lambda;&lt;sub&gt;1&lt;/sub&gt; = 1&lt;br&gt;&#xA;&amp;lambda;&lt;sub&gt;2&lt;/sub&gt; = $\dfrac{-1 + \sqrt{13}}{2}$&lt;br&gt;&#xA;&amp;lambda;&lt;sub&gt;3&lt;/sub&gt; = $\dfrac{-1 - \sqrt{13}}{2}$&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and these eigenvectors:&lt;br&gt;&#xA;@ &amp;lambda;&lt;sub&gt;1&lt;/sub&gt;, &lt;b&gt;x&lt;/b&gt; = s&amp;lt;1, 1, 0&gt; + t&amp;lt;-1, 0, 1&gt; &lt;b&gt;or&lt;/b&gt; span{&amp;lt;1, 1, 0&gt;, &amp;lt;-1, 0, 1&gt;}&lt;br&gt;&#xA;&lt;b&gt;x&lt;sub&gt;1&lt;/sub&gt;&lt;/b&gt; = &amp;lt;1, 1, 0&gt;&lt;br&gt;&#xA;&lt;b&gt;x&lt;sub&gt;2&lt;/sub&gt;&lt;/b&gt; = &amp;lt;-1, 0, 1&gt;&lt;br&gt;&#xA;@ &amp;lambda;&lt;sub&gt;3&lt;/sub&gt;, &lt;b&gt;x&lt;sub&gt;3&lt;/sub&gt;&lt;/b&gt; = &amp;lt;0, 0, 0&gt;&lt;br&gt;&#xA;@ &amp;lambda;&lt;sub&gt;4&lt;/sub&gt;, &lt;b&gt;x&lt;sub&gt;4&lt;/sub&gt;&lt;/b&gt; = &amp;lt;0, 0, 0&gt;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After normalizing the already orthogonal &lt;b&gt;x&lt;sub&gt;1&lt;/sub&gt;&lt;/b&gt;, I get span(, &amp;lt; -($\sqrt{2}$/2), 0, $\sqrt{2}$/2&gt;}.&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting up my P matrix for diagonalization, it appears to be a 3x4 matrix that is singular:&lt;br&gt;&#xA;$[\left(&#xA;\begin{array}{ccc}&#xA;\sqrt{2}/2 &amp;amp; - \sqrt{2}/2 &amp;amp; 0 &amp;amp; 0\&#xA;\sqrt{2}/2 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\&#xA;0 &amp;amp; \sqrt{2}/2 &amp;amp; 0 &amp;amp; 0\end{array}&#xA;\right)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: Why do I have four vectors, and why are two of them zero vectors? Symmetric matrices are &lt;b&gt;always&lt;/b&gt; diagonalizable, where there's a P such that P&lt;sup&gt;-1&lt;/sup&gt; = P&lt;sup&gt;T&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="819" LastActivityDate="2010-08-05T14:02:41.820" Title="Symmetric matrices and orthogonal diagonalization. " Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="1618" PostTypeId="1" AcceptedAnswerId="1628" CreationDate="2010-08-05T13:15:27.857" Score="7" ViewCount="345" Body="&lt;p&gt;Let $X,Y$ be $\sigma$-finite measure spaces, and let $L^p(X) \otimes L^p(Y)$ be the algebraic tensor product.  The product has a natural map into $L^p(X \times Y)$ which takes $\sum a_{ij} f_i \otimes g_j$ to the function $F(x,y) = \sum a_{ij} f_i(x) g_j(y)$.  A moment's thought shows that this map is well-defined.  Is it also injective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that this should be true, but I can't see how to prove it.  Intuitively, one needs to show that if $\sum a_{ij} f_i(x) g_j(y) = 0$ a.e., then one should be able to cancel all the terms in the sum using bilinearity.  It is not quite clear how to do this without knowing anything about the terms.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-08-05T13:41:13.250" LastActivityDate="2010-08-10T21:06:56.040" Title="Is the natural map $L^p(X) \otimes L^p(Y) \to L^p(X \times Y)$ injective?" Tags="&lt;linear-algebra&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="1619" PostTypeId="2" ParentId="1617" CreationDate="2010-08-05T13:16:13.050" Score="2" ViewCount="0" Body="&lt;p&gt;Well, first off, a vector isn't orthogonal, a set of vectors is.  Remember that for two vectors to be orthogonal, it means that they are at right angles to each other, and orthonormal means that plus they're unit vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I had to say where I think your first error is, you took a 3 x 3 matrix and got a quadratic equation somehow, but you should have a cubic.  And also, as for your eigenvectors, where did $\lambda_4$ come from? What is it? The symbol just appears from nowhere.  Perhaps you have typos and it's $\lambda_2,\lambda_3$ rather than 3 and 4, and each of them must have a nonzero eigenvector, because they are eigenvalues of multiplicity 1 (though with the equation error, they might not be eigenvalues), and so you would have to be incorrect about having a two dimensional eigenspace in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, both of the eigenvectors for 1 check out, which means that you've incorrectly calculated the eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(instead of just an answer, I put in all the thinking I did to get there, because I thought it might help clarify how to check your work in the future)&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-05T13:16:13.050" CommentCount="6" />
  <row Id="1620" PostTypeId="2" ParentId="790" CreationDate="2010-08-05T13:28:22.340" Score="2" ViewCount="0" Body="&lt;p&gt;In an infinite-dimensional normed space with a translation-invariant measure, the measure of a ball must be either $0$ or $\infty$.  This fact is sometimes summarized as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;there is no infinite-dimensional Lebesgue measure&lt;/a&gt;.&quot;  So unless you have some other notion of &quot;volume&quot; in mind, only the finite-dimensional case (i.e. $\mathbb{R}^n$ with some other norm) has any content.  And with regard to Hilbert spaces, the only finite-dimensional Hilbert spaces are $\mathbb{R}^n$ with the Euclidean inner product, so of course we know about this.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-05T13:28:22.340" />
  <row Id="1622" PostTypeId="2" ParentId="1614" CreationDate="2010-08-05T14:01:55.283" Score="4" ViewCount="0" Body="&lt;p&gt;Regarding your edit: (0.2, 0) — (1, 0.2) — (0.8, 1) — (0, 0.8) (and many others)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.imgftw.net/img/326639277.png&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://www.imgftw.net/img/326639277.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-05T14:01:55.283" />
  <row Id="1623" PostTypeId="2" ParentId="1526" CreationDate="2010-08-05T14:01:56.563" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;Continued fractions&lt;/a&gt; give the best approximations to a number using smaller terms in the fraction. This may take a bit of practice to calculate at speed, but for the number in question there are only 8 terms:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[0;1,8,1,3,2,2,7],&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with convergents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 8/9, 9/10, 35/39, 79/88, 193/215, 1430/1593&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can clearly see that the option 9/10 is the best one from the choices on offer. Note the convergents get progressively more accurate and oscillate between being over-estimates and under-estimates. The final term is, of course, the original fraction itself in its simplest form.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-08-05T14:01:56.563" />
  <row Id="1624" PostTypeId="2" ParentId="1617" CreationDate="2010-08-05T14:02:41.820" Score="0" ViewCount="0" Body="&lt;p&gt;Your characteristic polynomial is incorrect.&#xA;Go to wolframalpha.com and enter: {{2, -1, 1},{-2, 3, 2}, {-1, 1, 0}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.wolframalpha.com/input/?i={{2,+-1,+1},{-2,+3,+2},+{-1,+1,+0}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see that the characteristic polynomial is of 3rd degree.&#xA;It must be of 3rd degree because your matrix is 3x3.&#xA;Here is what you'll get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(x)=-x^3+5 x^2-3 x-1&lt;br&gt;&#xA;lambda_1 = 4.23607&lt;br&gt;&#xA;lambda_2 = 1&lt;br&gt;&#xA;lambda_3 = -0.236068  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you will also get 3 eigenvectors. (Never 4 for a 3x3 matrix).&lt;/p&gt;&#xA;" OwnerUserId="821" LastActivityDate="2010-08-05T14:02:41.820" CommentCount="1" />
  <row Id="1625" PostTypeId="2" ParentId="1548" CreationDate="2010-08-05T14:19:01.960" Score="4" ViewCount="0" Body="&lt;p&gt;You can apply &lt;a href=&quot;http://groupprops.subwiki.org/wiki/Burnside%27s_normal_p-complement_theorem&quot; rel=&quot;nofollow&quot;&gt;Burnside's normal $p$-complement theorem&lt;/a&gt; to get a normal complement $N$ of $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then take an element $m$ of $M$ with prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: The centralizer $C_N(m)$ of $m$ in $N$ is nontrivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $M$ centralizes $m$, it acts on the fixed points of $m$ (in the action by conjugation on $N$), i.e., $C_N(m)$ is an $M$-invariant subgroup of $N$. If $C_N(m) = N$, then $m \in Z(G)$ contradicts ii). Otherwise $M &amp;lt; C_N(m)M &amp;lt; G$ contradicting the maximality of $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $N$ has the fixed point free automorphism $m$ of prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Thompson's thesis $N$ is nilpotent, hence $G$ solvable.&lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2010-08-05T14:19:01.960" />
  <row Id="1626" PostTypeId="1" AcceptedAnswerId="1634" CreationDate="2010-08-05T14:24:36.597" Score="4" ViewCount="358" Body="&lt;p&gt;How can I calculate $27^{41}\ \mathrm{mod}\ 77$ as simple as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know that $27^{60}\ \mathrm{mod}\ 77 = 1$ because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler’s theorem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a^{\phi(n)}\ \mathrm{mod}\ n = 1 $$&#xA;and&#xA;$$ \phi(77) = \phi(7 \cdot 11) = (7-1) \cdot (11-1) = 60 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know from using modular exponentiation that $27^{10} \mathrm{mod}\ 77 = 1$ and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 27^{41}\ \mathrm{mod}\ 77 = 27^{10} \cdot 27^{10} \cdot 27^{10} \cdot 27^{10} \cdot 27^{1}\ \mathrm{mod}\ 77 = 1 \cdot 1 \cdot 1 \cdot 1 \cdot 27 = 27 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can I derive the result of $27^{41}\ \mathrm{mod}\ 77$ using $27^{60}\ \mathrm{mod}\ 77 = 1$ somehow?&lt;/p&gt;&#xA;" OwnerUserId="659" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-05T15:30:14.937" LastActivityDate="2010-08-15T21:28:05.590" Title="Modular exponentiation using Euler’s theorem" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="1627" PostTypeId="1" AcceptedAnswerId="1636" CreationDate="2010-08-05T14:52:14.937" Score="1" ViewCount="46" Body="&lt;p&gt;I try to learn the theory of category from &lt;a href=&quot;http://katmat.math.uni-bremen.de/acc/acc.pdf&quot; rel=&quot;nofollow&quot;&gt;The Joy of Cats&lt;/a&gt;. I got stacked with the first exercise (3A a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a simple graph with one vertex and 2 nodes all we know is that in category there is 2 objects and one homomorphism (beside identities). Hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[C = (\{\{x, y\}, \{1, 2\}\}, \{f, id\}, id, \cdot), f = \{(x, 1), (y, 2)\}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[C = (\{\{x\}, \{1\}, \{f, id\}, id, \cdot), f = \{(x, 1)\}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should have the same graph but are they isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note about notation:&lt;/strong&gt; I treat category as quadruple $(O, H, id, \cdot)$ where $O$ is class of objects (here is set), $H$ is class of homomorphisms (in example set). $id$ is identity and $\cdot$ is composition of homomorphisms.&lt;/p&gt;&#xA;" OwnerUserId="728" LastEditorUserId="728" LastEditorDisplayName="" LastEditDate="2010-08-05T16:02:29.750" LastActivityDate="2010-08-05T16:39:26.027" Title="Thin categories and graphs (isomorphism of categories)" Tags="&lt;category-theory&gt;" AnswerCount="1" />
  <row Id="1628" PostTypeId="2" ParentId="1618" CreationDate="2010-08-05T14:52:25.310" Score="2" ViewCount="0" Body="&lt;p&gt;The previous version of this answer was incorrect. It has generously been accepted by Nate Eldredge on the ground that &quot;it definitely contains the right idea&quot;. This previous version has better remain forgotten forever. The reader should consult &lt;a href=&quot;http://math.stackexchange.com/questions/1618/is-the-natural-map-lpx-otimes-lpy-to-lpx-times-y-injective/1678#1678&quot;&gt;Nate's answer&lt;/a&gt;, which is correct and complete. I'll just try here to spell out the underlying linear algebra lemma (which is trivial, but which I failed to see). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;UNDERLYING LEMMA.&lt;/strong&gt; Let $U,V,W$ vector spaces over some field $K$, and $\alpha:U\otimes V\to W$ a linear map. Assume that for each nonzero vector $u_0$ in $U$ there are linear maps $\beta:W\to V$ and $\gamma:U\to K$ such that $\gamma(u_0)\not=0$ and &#xA;  $$\beta(\alpha(u\otimes v))=\gamma(u)v\ \forall\ u,v.$$ Then $\alpha$ is injective. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof can be very easily extracted from &lt;a href=&quot;http://math.stackexchange.com/questions/1618/is-the-natural-map-lpx-otimes-lpy-to-lpx-times-y-injective/1678#1678&quot;&gt;Nate's answer&lt;/a&gt;. Here is the most standard application: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;APPLICATION.&lt;/strong&gt; Let $U,V,W$ be respectively the spaces of $K$-valued functions on the sets $X,Y,X\times Y$, and $\beta,\gamma$ are given by $(\beta(w))(y)=w(x_0,y)$, $\gamma(u)=u(x_0)$ where $x_0$ is chosen so that $u_0(x_0)\not=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; Here is a mild generalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a measure space, let $\mathcal V(X)$ be the vector space of all measurable complex valued functions on $X$, let $\mathcal N$ be the subspace of functions vanishing almost everywhere, and let $V(X)$ be the quotient. [Is there a standard notation for these spaces?]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ be another measure space. The bilinear map from $\mathcal V(X)\times\mathcal V(Y)$ to $\mathcal V(X\times Y)$ sending $(f,g)$ to $(x,y)\mapsto f(x)g(y)$ induces a linear map $\Phi$ from $V(X)\otimes V(Y)$ to $V(X\times Y)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; Assume $X$ and $Y$ are $\sigma$-finite. Then $\Phi$ is injective.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This slightly more general than the statement proved in &lt;a href=&quot;http://math.stackexchange.com/questions/1618/is-the-natural-map-lpx-otimes-lpy-to-lpx-times-y-injective/1678#1678&quot;&gt;Nate's answer&lt;/a&gt; because no integrability assumptions are made on the functions.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;LEMMA.&lt;/strong&gt; Let $f_i,\dots,f_n$ be in $\mathcal V(X)$, and $S\subset X^n$ the set of $x\in X^n$ such that $\det(f_i(x_j))\not=0$. Then the $f_i$ are linearly independent in $V(X)$ if and only if the measure of $S$ is positive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Denote by $|A|$ the measure of $A$, and by $S_g$ the support of the function $g$, that is the set of points where it is nonzero. Put $d(x):=\det(f_i(x_j))$ for $x\in X^n$. The above set $S$ is now $S_d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the if part of the lemma, assume the $f_i$ are linearly dependent in $V(X)$ and check $|S_d|=0$ as follows. Let $\lambda$ be a nonzero vector of $\mathbb C^n$ such that $g:=\sum \lambda_i f_i$ satisfies $|S_g|=0$. Denote again by $\lambda$ the 1 by $n$ matrix formed by the $\lambda_i$, and by $A(x)$ the matrix $(f_i(x_j))$. For $x$ in $(S_g^c)^n$, where the superscript c mean &quot;complement&quot;, we have $\lambda A(x)=0$. Multiplying on the right by the adjugate of $A(x)$ we get $d(x)=0$. This shows that $S_d$ is contained in the measure zero subset $((S_g^c)^n)^c$ of $X^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the only if part of the lemma, assume $|S_d|=0$ and check that the $f_i$ are linearly dependent in $V(X)$ as follows. Expand $d(x)$ as $$d(x)=\sum\ d_i(x')\ f_i(x_n)$$ with $x':=(x_1,\dots,x_{n-1})$. Arguing by induction, we can assume $|S_{d_i}|&gt;0$ for all $i$. Let $U$ be the union of the $S_{d_i}$. Denoting by $d(x',\bullet)$ the function $x_n\mapsto d(x)$, we have&#xA;$$S_d=\bigcup_{x'\in X^{n-1}}\ {x'}\times S_{d(x',\bullet)}.$$ Fubini yields&#xA;$$0=|S_d|=\int_{X^{n-1}}\ |S_{d(x',\bullet)}|\ dx'.$$&#xA;If the $f_i$ were linearly independent in $V(X)$, we would have $|S_{d(x',\bullet)}|&gt;0$ for all $x'$ in $U$, a contradiction. The lemma is proved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's prove the theorem. In the next lines, $i,j,k$ will lie between 1 and $m$, whereas $p,q,r$ will lie between 1 and $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_1,\dots,f_m$ be in $\mathcal V(X)$; let $g_1,\dots,g_n$ be in $\mathcal V(Y)$; and put $F_{ip}(x,y):=f_i(x)g_p(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem is equivalent to the statement that the linear independence of the $f_i$ and $g_p$ in $V(X)$ and $V(Y)$ implies that of the $F_{ip}$ in $V(X\times Y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S\subset X^m$ and $T\subset Y^n$ be defined by the conditions $\det(f_i(x_j))\not=0$ and $\det(g_p(y_q))\not=0$. Embedd $U:=S^n\times T^m$ into $(X\times Y)^{mn}$ in the obvious way, and denote the $(i,p)$-th coordinate of $z\in(X\times Y)^{mn}$ by $z_{ip}=(x_{pi},y_{ip})$. Then $z$ is in $U$ iff $x_p\in S$ and $y_i\in T$ for all $(i,p)$, that is iff&#xA;$$\det\Big(f_j(x_{pk})\Big)_{jk}\not=0\not&#xA;=\det\Big(g_q(y_{ir})\Big)_{qr}$$&#xA;for all $(i,p)$. We claim that the validity of this double nonequality for all $(i,p)$ implies the nonequality&#xA;$$\det\Big(f_i(x_{qj})\ g_p(y_{jq})\Big)_{(i,p)(j,q)}\not=0.$$&#xA;In view of the lemma, the theorem is equivalent to the claim. But the claim doesn't depend on the measures defined on $X$ and $Y$. As the theorem holds when $X$ and $Y$ are equipped with the counting measure (see the application of the &quot;underlying lemma&quot; at the beginning of the post), we are done.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-10T08:41:44.417" LastActivityDate="2010-08-10T08:41:44.417" CommunityOwnedDate="2010-08-10T08:43:03.587" />
  <row Id="1629" PostTypeId="2" ParentId="653" CreationDate="2010-08-05T15:11:55.163" Score="3" ViewCount="0" Body="&lt;p&gt;Homotopy groups and homotopy classes of maps are used in physics to study topological defects.  Roughly speaking, $\pi_k(X)$ classifies codimension $k+1$ defects in textures modeled on $X$ (maps from $R^d$ to $X$ continuous except at the loci of defects).  Other homotopy classes of maps can be used to study linked defects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following beautiful review paper by N.D. Mermin for an introduction: &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v51/i3/p591_1&quot; rel=&quot;nofollow&quot;&gt;http://rmp.aps.org/abstract/RMP/v51/i3/p591_1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="826" LastEditorUserId="826" LastEditorDisplayName="" LastEditDate="2010-08-06T19:47:28.357" LastActivityDate="2010-08-06T19:47:28.357" />
  <row Id="1630" PostTypeId="2" ParentId="1626" CreationDate="2010-08-05T15:13:57.283" Score="2" ViewCount="0" Body="&lt;p&gt;You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;exponentiation by squaring&lt;/a&gt; (all operations are modulo 77):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27^{41} = 27^{32+8+1} = 27 \cdot 27^8 \cdot (27^8)^4 = (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\big[ 27^8 = ((27^2)^2)^2 = (36^2)^2 = 64^2 = 15 \big]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(*) = 27 \cdot 15 \cdot 15^4 = 27 \cdot 15 \cdot (15^2)^2 = 27 \cdot 15 \cdot 71^2 = 27 \cdot 15 \cdot 36 = 27$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses only 7 multiplications instead of 41.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-05T15:13:57.283" />
  <row Id="1631" PostTypeId="2" ParentId="709" CreationDate="2010-08-05T15:23:07.250" Score="2" ViewCount="0" Body="&lt;p&gt;The &lt;em&gt;vertex&lt;/em&gt; of a parabola is the point on the axis of symmetry that intersects it (&lt;a href=&quot;http://en.wikipedia.org/wiki/Parabola&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point $(x,y)=\left( 0,0\right) $ is the vertex of the parabola given by $y=ax^{2}$.&#xA;By making the change of coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=x+\dfrac{b}{2a},Y=y+\dfrac{b^{2}-4ac}{4a}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;equation $y=ax^{2}+bx+c$ is transformed into $Y=aX^{2}$, the vertex of which is the point $(X,Y)=(0,0)$, i.e. $(x,y)=\left(-\dfrac{b}{2a},-\dfrac{b^{2}-4ac}{4a}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This change of coordinates is a &lt;em&gt;translation&lt;/em&gt; of both axes $x$, $y$, respectively, by $-\dfrac{b}{2a},-\dfrac{b^{2}-4ac}{4a}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the $x$-coordinate of the vertex of the parabola is $-\dfrac{b}{2a}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; &lt;em&gt;Example&lt;/em&gt;. Plot of $y=2x^2-x+4$; blue axes: $X=x-1/4,Y=y-31/8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://problemasteoremas.files.wordpress.com/2010/08/parabola.jpg?w=490&amp;amp;h=392&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T10:30:47.447" LastActivityDate="2010-08-29T10:30:47.447" />
  <row Id="1632" PostTypeId="1" CreationDate="2010-08-05T15:23:30.370" Score="1" ViewCount="55" Body="&lt;p&gt;A discrete one-dimensional model of optical imaging looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r) = \sum_i e_i P(r - r_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the $e_i$ are point light sources at locations $r_i$ in the object and $P$ is a point spread function that blurs each point. We can assume that $P$ is even, non-negative and has a finite extent, ie $P(x) = 0$ for $|x| &gt; p$. The $e_i$ are all positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complex imaging process instead produces an image like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r)^2 = \sum_{i,j} e_i e_j P(r - r_i) P(r - r_j) \cos (r_i - r_j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The $^2$ is a consequence of the reconstruction method. In practice we normally take the square root, but per &lt;a href=&quot;http://math.stackexchange.com/questions/1130/preserving-the-extrema-of-one-function-after-applying-another&quot;&gt;my earlier question here&lt;/a&gt;, I think the salient structural features of the resulting image should be unchanged by this transformation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical simulations suggest that the latter method allows better resolution of the points when $p \sim \pi$, and is not materially different when $p \ll \pi$. (We ignore the case where $p \gg \pi$ as not physically tenable.) Intuitively, this is because the &quot;trough&quot; in the $\cos (r_i - r_j)$ term reduces the interaction between points at some intermediate separations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to be more analytical about this. We can readily extend the above to a continuous model with an object function $O$ in place of the $e_i$. The ordinary image becomes a convolution integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r) = \int O(s) \: P(r - s) \; \mathrm{d}s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which there are standard analytical approaches available. However, the more complex reconstruction looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(r)^2 = \int \int  O(s) \; O(t) \; P(r-s) \; P(r-t) \; \cos (s - t) \mathrm{d}s \mathrm{d}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks fairly intractable to me, and I'm not really sure where to begin with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my first question is: is there any point in trying to look at the continuous model, or should I just concentrate on the discrete? On the one hand, every single practical use case for the technique will actually be using discrete measurements, so in that sense the discrete model is more realistic. On the other, it may be that there are things one could demonstrate using the continuous model that are not available otherwise. (But even if that's the case, I'm probably not capable of doing so without at least being nudged in the right direction!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I guess my second question is: continuous or not, does anyone have any other suggestions of useful ways to analyse this model?&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-08-05T15:23:30.370" Title="Analysing an optics model in discrete and continuous forms" Tags="&lt;analysis&gt;&lt;signal-processing&gt;&lt;physics&gt;" />
  <row Id="1633" PostTypeId="2" ParentId="1437" CreationDate="2010-08-05T15:23:43.270" Score="0" ViewCount="0" Body="&lt;p&gt;Open a graphic calculator such as fooplot.com&lt;br&gt;&#xA;Enter a function:&lt;br&gt;&#xA;y(x) = 10^(x/10)&lt;br&gt;&#xA;and another function:&lt;br&gt;&#xA;y(x) = x&lt;br&gt;&#xA;Click the intersection button and then the intersections on the image.&lt;/p&gt;&#xA;" OwnerUserId="821" LastActivityDate="2010-08-05T15:23:43.270" CommentCount="1" />
  <row Id="1634" PostTypeId="2" ParentId="1626" CreationDate="2010-08-05T15:26:21.820" Score="6" ViewCount="0" Body="&lt;p&gt;As suggested in the comment above, you can use the Chinese Remainder Theorem, by using Euler's theorem / Fermat's theorem on each of the primes separately. You know that $27^{10} \equiv 1 \mod 11$, and you can also see that modulo 7, $27 \equiv -1 \mod 7$, so $27^{10} \equiv (-1)^{10} \equiv 1\mod 7$ as well. So $27^{10} \equiv 1 \mod 77$, and $27^{41} = 27^{40+1} \equiv 27 \mod 77$. (We've effectively found the order of 27 as 10, but a more mechanical procedure may be to use that $27^{41} \equiv 27 \equiv 5 \mod 11$ and $27^{6} \equiv 1 \mod 7$ to see that $27^{41} = 27^{42-1} \equiv 27^{-1} \equiv -1 \mod 7$, and put 5 mod 11 and -1 mod 7 together to get 27.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is if you're doing it by hand, for this case. In general, algorithmically, you would just use repeated squaring to exponentiate numbers. You don't gain much by using Euler's theorem, since finding the order of a number mod $n$ is as hard as factoring $n$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-05T15:32:29.663" LastActivityDate="2010-08-05T15:32:29.663" CommentCount="3" />
  <row Id="1635" PostTypeId="2" ParentId="1609" CreationDate="2010-08-05T15:42:56.583" Score="14" ViewCount="0" Body="&lt;p&gt;If you want to avoid algebraic geometry and topology, then you will probably be forced to use algebra.  This is okay, because the things you ask about have algebraic interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it probably helps to write the function field as $k(x,y),$ where $x$ and $y$ are related by some equation $f(x,y) = 0$.  One problem is going to be that $f$ can't be chosen to be smooth in general (because not every curve can be embedded as a smooth plane curve), but it's probably best to ignore this; if it becomes important, then probably your colleagues will be delving deeper into the theory anyway, and so the whole level of explanation can be ramped up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to explain divisors, you can have them imagine intersecting $f(x,y) = 0$ with some other curve $g(x,y) = 0$; the intersections will be a bunch of points, possibly with multiplicity.&#xA;This is a divisor.   So divisors are just natural ways of encoding how curves intersect one another.  (Again, I am ignoring here the issue of points at infinity; you will have to decide whether that is appropriate, or whether you need a higher level of precision in your explanations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the genus, it's a little more complicated to explain algebraically, but possible; here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the equation $f(x,y) = 0$ has degree $d$, so your function field corresponds to a degree $d$ curve in the plane.  Now let $V_n$ be the vector space of all polynomials of degree $\leq n$ in $x$ and $y$.  Let's suppose that $n \geq d$ (and in general, we should&#xA;think that $n$ is large).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple computation shows that $V_n$ has dimension $(n+2)(n+1)/2$.  Inside $V_n$, we have&#xA;a subspace consisting of all the multiples of $f$.  This subspace is obtained by taking&#xA;the elements of $V_{n-d}$ (i.e. polynomials of degree at most $n-d$) and multiplying them by $f$, i.e. it is the subspace $f V_{n-d}$ of $V_n$, and so has dimension $(n-d+2)(n-d+1)/2$.  If we look at the quotient $V_n/fV_{n-d},$ this then has dimension &#xA;$n d + 1 - (d-1)(d-2)/2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the meaning of this quotient?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g \in V_n$ represents a non-zero element of $V_n/fV_{n-d}$, then it is a degree $n$&#xA;equation that is &lt;em&gt;not&lt;/em&gt; divisible by $f$, so it &lt;em&gt;doesn't&lt;/em&gt; vanish identically on $f(x,y) = 0$, so by Bezout's theorem, the intersection of $g(x,y) = 0$ and $f(x,y) = 0$ is $n d$ &#xA;points, i.e. a divisor of degree $n d$.  So we see that non-zero elements of &#xA;$V_n/f V_{n-d}$ correspond to those divisors of degree $n d$ that are obtained by &#xA;intersecting $f(x,y) = 0$ with a degree $n$ curve.  (In fact, we should think about non-zero elements up to scaling, because if we multiply $g$ by a non-zero scalar,&#xA;the curve $g(x,y) = 0$ doesn't change.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus among all the degree $n d$ divisors on $f(x,y) = 0$, those that come by intersecting with a degree $n$ curve form a space of dimension $n d - (d-1)(d-2)/2.$  (Here I have&#xA;subtracted 1, because rescaling accounts for 1 of the dimensions in the above formula.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is the dimension of the space of all divisors of degree $n d$ (with non-negative coefficients,&#xA;which are the only kind that can possibly arise as intersections; divisors with non-negative coefficients are called effective)?  Well, we just have to choose $n d$ points and add them together.  We are choosing the points from a curve, which is one-dimensional, so that means there is an $n d$-dimensional space of effective divisors of degree $n d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we see that, in the $n d$-dimensional space of all effective divisors of degree $n d$,&#xA;those that arise by intersecting with a degree $n$ curve are a subspace of dimension&#xA;$n d -(d-1)(d-2)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity $(d-1)(d-2)/2$ is precisely the genus.&#xA;So what we see is that the bigger the genus is, the harder it is for a degree $n d$&#xA;divisor on $f(x,y) = 0$ to be obtained by intersecting with another curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the degree $d$ is $1$ or $2$, then the genus vanishes, and &#xA;every degree $n d$ divisor comes from intersecting with a degree $n$ curve.&lt;br&gt;&#xA;E.g. on a conic (i.e. when $d = 2$) any two points come from intersecting with a line&#xA;(the line that passes through those two points), any four points come from intersecting &#xA;with a conic, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the degree $d = 3$, then not all triples of points on $f(x,y) = 0$&#xA;come from intersecting with a line: in fact if you give yourself two points, then&#xA;they determine a line, which in turn determines the 3rd point of intersection.&#xA;Similarly (going up to the case $n =3$), a general set of 9 points on the cubic doesn't come from&#xA;intersecting with another cubic; instead, if you give yourself 8 points on the cubic&#xA;curve, you can find another cubic passing through these 8 points, and its 9th point of intersection is uniquely determined by the given 8 points.  (Since 9 = 8 + 1, this &#xA;is a particular manifestation of the fact that our cubic curve has genus 1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, I've lapsed into a more geometrical way of thinking (using notions such as dimension), but I don't think one can avoid this completely: the concept of genus arose historically from essentially the kind of computations that I've just been making, and at some point you have to think about spaces of divisors and their dimensions if you want to understand it.  Still, I hope that this gives you an avenue to explaining the concept which is more algebraic, and so more accessible to your colleagues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more technical remark: The space $V_{n}/f V_{n-d}$ is an example of a Riemann--Roch space, and the formula for its dimension ($n d + 1 - $ the genus) is a special case of the Riemann-Roch formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Technical remark added in response to a question of T. in the comments below; feel free to ignore it if it is at too high a level:]  Note that the above discussion works even if $f$ is allowed to be singular (and irreducible, say, so that&#xA;we can't have polynomials $g$ which vanish on one component of $f(x,y) = 0$ without vanishing &#xA;on the whole curve).  The reason is that $(d-1)(d-2)/2$ is always the &lt;em&gt;arithmetic genus&lt;/em&gt; of a plane curve of degree $d$, whether or not the curve is smooth, and it is the arithmetic genus that intervenes in the Riemann--Roch formula.  (I believe that this is the origin of the adjective &lt;em&gt;arithmetic&lt;/em&gt; in arithmetic genus: it is this version of the genus for singular curves that comes up when you make Riemann--Roch-type calculations of the dimensions of various spaces of divisors.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-09T03:26:02.570" LastActivityDate="2010-08-09T03:26:02.570" CommentCount="6" />
  <row Id="1636" PostTypeId="2" ParentId="1627" CreationDate="2010-08-05T15:53:59.867" Score="1" ViewCount="0" Body="&lt;p&gt;It is a bit hard to understand your notation. If understand correctly, the isomorphism is realized by the functor $F$ which sends $\{x, y\}$ to $\{ x \}$ and $\{ 1, 2 \}$ to $\{ 1 \}$ (on objects) and send the unique non trivial morphism $f$ to $f$.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="233" LastEditorDisplayName="" LastEditDate="2010-08-05T16:39:26.027" LastActivityDate="2010-08-05T16:39:26.027" CommentCount="3" />
  <row Id="1637" PostTypeId="2" ParentId="764" CreationDate="2010-08-05T17:10:06.220" Score="0" ViewCount="0" Body="&lt;p&gt;In Stieltjes integral you assign different importance to different parts of the set you are integrating over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usefulness of this would become clearer when you know the theory of Lebesgue measure, which generalizes this even further. For example there is the Dirac measure which when used for integrating functions cares only about the value of a function at a point(typically the origin). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding it in the Stieljes form is not any harder; moreover it will pave the way for measure theory.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:10:06.220" />
  <row Id="1638" PostTypeId="2" ParentId="998" CreationDate="2010-08-05T17:11:06.403" Score="2" ViewCount="0" Body="&lt;p&gt;Nobody has mentioned the book of Shafarevich; so I mention myself.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:11:06.403" CommunityOwnedDate="2010-08-05T17:11:06.403" />
  <row Id="1639" PostTypeId="2" ParentId="262" CreationDate="2010-08-05T17:13:27.977" Score="2" ViewCount="0" Body="&lt;p&gt;Anybody who wants to be a serious mathematician better read W. Rudin's &quot;Principles of mathematical Analysis&quot;. It gives a rigorous foundation to the basic notions analysis and introduces the reader to the world of rigor, after the sloppy days of calculus courses. One must learn the notion of rigor properly if one wants to be a mathematician. More than anything else, it is an exercise in the rectitude of thought. No other book is so universally used that would teach this notion, than Rudin. &lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:13:27.977" CommunityOwnedDate="2010-08-05T17:13:27.977" />
  <row Id="1640" PostTypeId="2" ParentId="1413" CreationDate="2010-08-05T17:15:31.403" Score="1" ViewCount="0" Body="&lt;p&gt;I do not want to be mean or rude here; but my personal belief is that if you do not appreciate number theory, you are missing the most important topic in mathematics. It is literally the heart of mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need a concrete example, I can give one from the point of view of algebra. To have an example of a purely non-separable polynomial, you go to char p.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:15:31.403" CommentCount="3" />
  <row Id="1641" PostTypeId="2" ParentId="457" CreationDate="2010-08-05T17:16:54.430" Score="2" ViewCount="0" Body="&lt;p&gt;The Fourier transform of the Dirac delta function is a constant function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore whatever frequency your radio is turned to during a storm, or your TV, you will receive a noise.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T17:16:54.430" CommunityOwnedDate="2010-08-05T17:16:54.430" />
  <row Id="1642" PostTypeId="2" ParentId="450" CreationDate="2010-08-05T17:53:53.310" Score="1" ViewCount="0" Body="&lt;p&gt;As others have already mentioned, the Jacobian determinant transforms one coordinate system to another by relating infinitesimal areas (or volumes) from one system to another. Consider going from Cartesian to Polar coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;J = \det\frac{\partial(x,y)}{\partial(r,\theta)}&#xA;=\begin{vmatrix}&#xA; \frac{\partial x}{\partial r} &amp;amp; \frac{\partial x}{\partial \theta} \\&#xA; \frac{\partial y}{\partial r} &amp;amp; \frac{\partial y}{\partial \theta} \\&#xA;\end{vmatrix}&#xA;$$&#xA;$$&#xA;=\begin{vmatrix}&#xA; \cos\theta &amp;amp; -r\sin\theta \\&#xA; \sin\theta &amp;amp; r\cos\theta \\&#xA;\end{vmatrix}&#xA;=r\;\cos^2\theta + r\;\sin^2\theta = r&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is useful because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}A = J\;\mathrm{d}r\mathrm{d}\theta = r\mathrm{d}r\mathrm{d}\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\iint_\mathbf{R} f(r,\theta)\mathrm{d}A = \int_a^b \int_0^{r(\theta)} f(r,\theta) r\mathrm{d}r\mathrm{d}\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which tells you that if you have a function $f(r, \theta)$ you can compute the integral as long as you remember to add a factor of $r$. The common transformations have all been worked out and can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_canonical_coordinate_transformations&quot; rel=&quot;nofollow&quot;&gt;here on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-09T23:00:16.617" LastActivityDate="2010-09-09T23:00:16.617" CommentCount="3" />
  <row Id="1643" PostTypeId="2" ParentId="903" CreationDate="2010-08-05T17:58:54.490" Score="0" ViewCount="0" Body="&lt;p&gt;I would like to recommend &lt;em&gt;&lt;a href=&quot;http://www.pearson.ch/HigherEducation/MathematicsStatistics/AdvancedMathematics/1471/9780131848696/Introduction-to-Topology-Pure-and.aspx&quot; rel=&quot;nofollow&quot;&gt;Topology: Pure and Applied&lt;/a&gt;&lt;/em&gt;, by Colin Adams and Robert Franzosa.&#xA;Anyone familiar with Colin Adams's &lt;em&gt;The Knot Book&lt;/em&gt; will expect this to be equally accessible.&#xA;And they will not be disappointed.  In terms of your criteria&amp;mdash;clarity and motivation&amp;mdash;I&#xA;find it unsurpassed.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-08-05T17:58:54.490" CommunityOwnedDate="2010-08-05T17:58:54.490" />
  <row Id="1644" PostTypeId="2" ParentId="1626" CreationDate="2010-08-05T18:17:21.653" Score="4" ViewCount="0" Body="&lt;p&gt;If you are serious about &quot;as simple as possible&quot; then observe that $27^{41} = 3^{123}$ and use  &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_function&quot; rel=&quot;nofollow&quot;&gt;Carmichael's theorem&lt;/a&gt; (a strengthening of Euler's theorem which actually gives a tight bound) to deduce that $3^{30} \equiv 1 \bmod 77$ and hence $3^{123} \equiv 3^3 \equiv 27 \bmod 77$.  But I do not think this is the right question to ask; you should really be asking &quot;as general as possible,&quot; and then the answer you have accepted is most appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course it suffices to use Euler's theorem for the above computation, but few people seem to learn Carmichael's theorem and I always like to point it out when I can.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-05T18:17:21.653" CommentCount="2" />
  <row Id="1645" PostTypeId="2" ParentId="262" CreationDate="2010-08-05T18:18:27.687" Score="1" ViewCount="0" Body="&lt;p&gt;I've been rereading &lt;i&gt;Littlewood's Miscellany&lt;/i&gt; recently. It's a very readable collection of the writings of J. E. Littlewood, carefully edited by Béla Bollobás. Any budding mathematician will draw much inspiration from it. I like &lt;i&gt;A Mathematician's Apology&lt;/i&gt;, but if I was forced into choosing only one book, it would be &lt;i&gt;Littlewood's Miscellany&lt;/i&gt;.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2010-08-05T18:18:27.687" CommunityOwnedDate="2010-08-05T18:18:27.687" />
  <row Id="1646" PostTypeId="2" ParentId="450" CreationDate="2010-08-05T18:33:14.460" Score="0" ViewCount="0" Body="&lt;p&gt;You can represent some non-linear transforms (like translation) of a n dimensional vector with a n+1 dimensional matrix. However, converting the vector to its n+1 dimensional homogeneous version and back is not a linear transformation and also not representable as a matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More is explained here: &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Transformation_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Transformation_matrix&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-05T18:33:14.460" />
  <row Id="1647" PostTypeId="1" CreationDate="2010-08-05T18:57:39.233" Score="3" ViewCount="62" Body="&lt;p&gt;I am writing a Gaussian blur filter in graphics shader code.  I want to make the blur parameterized by radius from the users perspective.  The best method I can figure to do this is to pick a suitable stopping point for y, say .001, and solve for the variance to plug into the normal distribution function that will achieve that value of y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I cannot for the life of me solve this equation for v...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 10$ (blur radius)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$.001 = \frac{1}{2 \pi v^2}e^{-\frac{x^2}{2v^2}}$$.&lt;/p&gt;&#xA;" OwnerUserId="836" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-04T19:41:48.183" LastActivityDate="2010-09-05T23:27:59.197" Title="How do I calculate a specific variation for a known value of the normal distribution function?" Tags="&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1648" PostTypeId="1" CreationDate="2010-08-05T19:05:28.907" Score="4" ViewCount="512" Body="&lt;p&gt;i've sampled a real world process, network ping times. The &quot;round-trip-time&quot; is measured in milliseconds. Results are plotted in a histogram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i36.tinypic.com/2qtfp6f.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ping times have a minimum value, but a long upper tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i want to know what statistical distribution this is, and how to estimate its parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the distribution is not a normal distribution, i can still show what i am trying to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal distribution uses the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/9/e/1/9e1e4a3af93c9680ba75669a0b69fbf6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the two parameters&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;μ (mean)&lt;/li&gt;&#xA;&lt;li&gt;σ&lt;sup&gt;2 &lt;/sup&gt; (variance)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Parameter estimation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The formulas for estimating the two parameters are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/3/2/4/324081bc08a66fbeba64f43a3dbd5e1a.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying these formulas against the data i have in Excel, i get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;μ = 10.9558 (mean)&lt;/li&gt;&#xA;&lt;li&gt;σ&lt;sup&gt;2 &lt;/sup&gt; = 67.4578 (variance)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With these parameters i can plot the &quot;&lt;em&gt;normal&lt;/em&gt;&quot; distribution over top my sampled data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i34.tinypic.com/2mqu3wl.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously it's not a normal distribution. A normal distribution has an infinite top and bottom tail, and is symmetrical. This distribution is not symmetrical.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What principles would i apply, what flowchart, would i apply to determine what kind of distribution this is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And cutting to the chase, what is the formula for that distribution, and what are the formulas to estimate its parameters?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;i want to get the distribution so i can get the &quot;average&quot; value, as well as the &quot;spread&quot;:&#xA;&lt;img src=&quot;http://i38.tinypic.com/2i27sp4.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am actually plotting the histrogram in software, and i want to overlay the theoretical distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i38.tinypic.com/nqpnck.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tags&lt;/strong&gt;: sampling, statistics, parameter-estimation, normal-distribution&lt;/p&gt;&#xA;" OwnerUserId="835" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T19:48:35.717" LastActivityDate="2010-08-10T02:40:44.470" Title="How do I figure out what kind of distribution this is?" Tags="&lt;discrete-mathematics&gt;&lt;random-variable&gt;&lt;average&gt;&lt;gamma-function&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="1649" PostTypeId="1" AcceptedAnswerId="1657" CreationDate="2010-08-05T19:06:08.530" Score="0" ViewCount="210" Body="&lt;p&gt;Are there any examples of solving for the global maximum of a non-differentiable function where you:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Construct a series of differentiable functions that approach the non-differentiable function in the limit&lt;/li&gt;&#xA;&lt;li&gt;Show the maximum of each differentiable function converges to some value, which is thus your answer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For all I know, the procedure above is fatally flawed (or there are trivial examples, I would be most interested in non-trivial examples) in some way, if that is the case let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically interested in examples involving absolute values.&lt;/p&gt;&#xA;" OwnerUserId="99" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-08-05T22:05:53.933" LastActivityDate="2010-08-06T11:46:42.377" Title="Finding the Extrema Non-differentiable Functions. " Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="1650" PostTypeId="2" ParentId="1648" CreationDate="2010-08-05T19:13:22.353" Score="1" ViewCount="0" Body="&lt;p&gt;The go-to distribution for things like wait times is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;Exponential&lt;/a&gt;. Yours doesn't look exactly the same because of the tiny lower tail, but I would be inclined to attribute that to noise/measurement error. (The assumption of independence of events is almost certainly wrong for ping times, but it's probably still your best choice.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you would probably be better off asking this sort of question on the &lt;a href=&quot;http://stats.stackexchange.com&quot;&gt;stats site&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As pointed out by &lt;a href=&quot;http://stats.stackexchange.com/questions/1315/how-do-i-figure-out-what-kind-of-distribution-this-is/1316#1316&quot;&gt;Srikant Vadali&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution&quot; rel=&quot;nofollow&quot;&gt;Gamma distribution&lt;/a&gt; is more general and can account for a non-negligible short tail, so may be a more appropriate choice. It's easier to estimate the parameter for the exponential, though.&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="144" LastEditorDisplayName="" LastEditDate="2010-08-05T22:04:55.963" LastActivityDate="2010-08-05T22:04:55.963" CommentCount="3" />
  <row Id="1651" PostTypeId="1" CreationDate="2010-08-05T19:18:33.297" Score="1" ViewCount="100" Body="&lt;p&gt;I am seeking a listing of the distinct &lt;a href=&quot;http://mathworld.wolfram.com/HamiltonianCycle.html&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian cycles&lt;/a&gt; following the edges of the icosahedron and the dodecahedron.  By &lt;em&gt;distinct&lt;/em&gt; I mean they are not congruent by some symmetry&#xA;of the icosahedron or dodecahedron (respectively). So they do not make the same sequence of angular turns. For example (as Gerhard corrected me in the comments), there is just one distinct Hamiltonian cycle on the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hamiltonian cycles of the Platonic solids are all over the web, but I am not finding a definitive list of the number and a description of each.  Thanks to anyone who can point me in the right direction!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-05T22:38:20.607" LastActivityDate="2010-08-05T22:38:20.607" Title="Distinct Hamiltonian cycles of the icosahedron and dodecahedron" Tags="&lt;geometry&gt;&lt;reference-request&gt;&lt;euclidean-geometry&gt;&lt;graph-theory&gt;&lt;discrete-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1652" PostTypeId="2" ParentId="1647" CreationDate="2010-08-05T19:20:20.717" Score="0" ViewCount="0" Body="&lt;p&gt;I don't think the radius has to involve the variance. You could just use the radius to scale the x-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could sample the distribution from -1 to 1, always. If the radius is 4, would sample 9 regions. For 10 you sample 21 times between -1 and 1. The variance can stay the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to try this question here too &lt;a href=&quot;http://gamedev.stackexchange.com/&quot;&gt;http://gamedev.stackexchange.com/&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-05T19:20:20.717" />
  <row Id="1653" PostTypeId="2" ParentId="1651" CreationDate="2010-08-05T19:28:05.820" Score="2" ViewCount="0" Body="&lt;p&gt;There is a listing of a hamilton cycle count of 2560 at this website: &lt;a href=&quot;http://mathworld.wolfram.com/IcosahedralGraph.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/IcosahedralGraph.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-08-05T19:28:05.820" CommentCount="1" />
  <row Id="1654" PostTypeId="2" ParentId="1388" CreationDate="2010-08-05T19:34:55.197" Score="2" ViewCount="0" Body="&lt;p&gt;Solving $x^{11} = 1$ in radicals was first accomplished by Vandermonde, whose method is described on p.24-26 of Edwards' book on Galois theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=0bH6SUHSvloC&amp;amp;pg=PA24&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=0bH6SUHSvloC&amp;amp;pg=PA24&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-05T19:34:55.197" />
  <row Id="1655" PostTypeId="2" ParentId="1442" CreationDate="2010-08-05T19:44:33.743" Score="7" ViewCount="0" Body="&lt;p&gt;Consider &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_harmonic_triangle&quot; rel=&quot;nofollow&quot;&gt;Leibnitz harmonic triangle&lt;/a&gt;&lt;/em&gt; — a table that is like &amp;laquo;Pascal triangle reversed&amp;raquo;: on it's sides lie numbers $\frac{1}{n}$ and each number is the sum of two beneath it (see the &lt;a href=&quot;http://upload.wikimedia.org/math/5/8/1/581098eb5e9213bf6c66e932ed218e08.png&quot; rel=&quot;nofollow&quot;&gt;picture&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can easily proove by induction that m-th number in n-th row of Leibnitz triangle is $\frac{1}{(n+1)\binom{n}{m}}$. So LHS of our identity is just lcd of fractions in n-th row of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's not hard to see that any such number is an integer linear combination of fractions on triangle's sides (i.e. $1/1,1/2,\dots,1/n$) — and vice versa. So LHS is equal to $lcd(1/1,\dots,1/n)$ — and that is exactly RHS.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-05T19:44:33.743" CommentCount="2" />
  <row Id="1656" PostTypeId="1" AcceptedAnswerId="1671" CreationDate="2010-08-05T19:51:01.563" Score="1" ViewCount="42" Body="&lt;p&gt;Say P ~ Q (P and Q are «projectively equivalent») iff there is a projective transformation f such that f(P) = Q.  Then ~ is an equivalence relation.  I read that the space of inscribed n-gons modulo projective equivalence has dimension n-3.  Why is this?  Also, are there any related results?&lt;/p&gt;&#xA;" OwnerUserId="314" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-05T23:08:49.410" LastActivityDate="2010-08-05T23:08:49.410" Title="Space of inscribed n-gons modulo projective transformations." Tags="&lt;geometry&gt;&lt;projective-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1657" PostTypeId="2" ParentId="1649" CreationDate="2010-08-05T19:53:53.507" Score="1" ViewCount="0" Body="&lt;p&gt;A simple example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F_n(x) = \sqrt{x^2+2^{-n}}$. It is not hard to show that $F_n(x) \to \sqrt{x^2} = |x|$.&#xA;Every $F_n$ is differentiable and has a local minimum at 0, and indeed so does |x|.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this is what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-05T19:53:53.507" CommentCount="5" />
  <row Id="1658" PostTypeId="1" AcceptedAnswerId="1661" CreationDate="2010-08-05T20:07:03.340" Score="4" ViewCount="70" Body="&lt;p&gt;Whenever I think about how a function behaves, I always try to identify a general pattern of behavior with some common numbers (somewhere between 5 and 100 maybe) and then I try to see if anything interesting happens around 1, 0 and into negative numbers if applicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that all works out, I essentially assume that I know that the function is going to behave similarly for very large numbers as it does for those relatively small numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there notable (famous, clever or common) functions where very large numbers would cause them to behave significantly differently than would initially be thought if I followed my regular experimental pattern? If so, are there any warning signs I should be aware of?&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-08-05T20:56:26.957" Title="Function behavior with very large variables" Tags="&lt;functions&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="1659" PostTypeId="1" AcceptedAnswerId="1663" CreationDate="2010-08-05T20:09:41.823" Score="2" ViewCount="107" Body="&lt;p&gt;Suppose I have a point $x \in \mathbb{R}^n$ on an &lt;a href=&quot;http://en.wikipedia.org/wiki/N-sphere&quot; rel=&quot;nofollow&quot;&gt;n-sphere&lt;/a&gt;. Suppose I divide the n-sphere into 4 sections (I think this makes sense in $n$ dimensions), how do I know which section $x$ lies on? &lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2010-08-05T22:50:30.577" LastActivityDate="2010-08-05T22:50:30.577" Title="Locating the quadrant containing a point on an n-sphere" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="1660" PostTypeId="1" AcceptedAnswerId="1719" CreationDate="2010-08-05T20:18:00.743" Score="5" ViewCount="178" Body="&lt;p&gt;Denote by $D(a_1,\dots,a_n)$ the product $\prod_{j&gt;i}(a_j-a_i)$. Assuming that $a_i$ are integers s.t. $a_1\le a_2\le\dots\le a_n$, proove that $D(a_1,...,a_n)/D(1,...,n)$ is the number of &lt;em&gt;Gelfand-Zetlin triangles&lt;/em&gt; (that is, &lt;a href=&quot;http://imrn.oxfordjournals.org/content/vol2010/issue13/images/medium/rnp223ueq3.gif&quot; rel=&quot;nofollow&quot;&gt;triangles&lt;/a&gt; consisting of $\frac{n(n+1)}2$ integers, s.t. each number is greater it's lower-left neighbor but not greater than lower-right neighbor) with the base $a_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for n=3 one needs to prove that number of b&lt;sub&gt;1&lt;/sub&gt;, b&lt;sub&gt;2&lt;/sub&gt;, b' s.t. $a_1\le b_1&amp;lt;a_2\le b_2&amp;lt;a_3$, $b_1\le b'&amp;lt;b_2$ is exactly $\frac{(a_3-a_2)(a_3-a_1)(a_2-a_1)}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one can guess from the name &amp;ldquo;Gelfand-Zetlin&amp;rdquo;, this fact is well-known in representation theory (namely, in LHS we count dimension of a $gl_n$-representation by Weyl formula, and in RHS we count elements in Gelfand-Zetlin basis of the same representation). But maybe someone can come with more or less direct proof? (Some kind of bijective proof, maybe.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Informal probabilistic argument&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For simplicity, consider the case $n=3$: D(a_1,a_2,a_3) counts the number of triangles s.t. $a_1\le b_1&amp;lt;a_2\le b_2&amp;lt;a_3$, $a_1\le b'&amp;lt;a_3$ — and we're interested only in G-Z triangles, i.e. in triangles s.t. $b_1\le b'&amp;lt; b_2$. Now, mathematical expectation of the length of the interval $(b_1,b_2)$ is exactly one half of the length of the interval from which $b'$ is chosen. So one may &lt;em&gt;expect&lt;/em&gt; that the probability that random triangle is G-Z is $1/2$ — and the answer is indeed $D(a_1,a_2,a_3)/D(1,2,3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The main problem with this computation is that we're multiplying probabilities for events that are clearly not independent. And although for n=3 it's not hard to transform this heuristic argument into a formal proof, even for n=4 I failed to do such thing.)&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-06T07:54:36.650" LastActivityDate="2010-08-19T19:02:35.283" Title="Direct proof of Gelfand-Zetlin identity" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" />
  <row Id="1661" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:19:02.367" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://www.geatbx.com/docu/fcnindex-01.html#P160_7291&quot; rel=&quot;nofollow&quot;&gt;Griewank function&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(\mathbf x) = \frac1{4000}\sum_{i=1}^n x_i^2 - \prod_{i=1}^n \cos\left(\frac{x_i}{\sqrt i}\right) + 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is one of the objective functions used in testing optimization algorithms, looks completely different in large scale (dominated by &lt;strong&gt;x&lt;/strong&gt;&lt;sup&gt;2&lt;/sup&gt;) and small scale (dominated by cos &lt;em&gt;x&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.geatbx.com/docu/fcnindex-msh_f8_500-19.gif&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;http://www.geatbx.com/docu/fcnindex-msh_f8_50-20.gif&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;http://www.geatbx.com/docu/fcnindex-msh_f8_8-21.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-05T20:19:02.367" />
  <row Id="1662" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:19:13.113" Score="2" ViewCount="0" Body="&lt;p&gt;Many rational functions $f(x)=\frac{n(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)}$ (where deg(r) &amp;lt; deg(d)) behave very differently in the general vicinity of the zeros of d(x) than for large (positive or negative) values x.  Near the zeros of d(x), the values of $\frac{r(x)}{d(x)}$ dominate the values of q(x) (that is, f(x) behaves like $\frac{r(x)}{d(x)}$), whereas for large (positive or negative) values of x, the values of q(x) dominate the values of $\frac{r(x)}{d(x)}$ (that is, f(x) behaves like q(x)).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-05T20:19:13.113" />
  <row Id="1663" PostTypeId="2" ParentId="1659" CreationDate="2010-08-05T20:30:29.270" Score="3" ViewCount="0" Body="&lt;p&gt;This is just a rephrasing of ShreevastasR's answer; no credit to me.  It does make sense to divide an $n$-sphere into quadrants, as you explain in $\mathbb{R}^3$: partition by two coordinate planes.&#xA;But then deciding which quadrant is, as ShreevastasR says, simply looking at the signs of the coordinates of $x$.  If $x_1$ and $x_2$ are both positive, you are in the first, $++$, quadrant;&#xA;if $x_1$ is negative and $x_2$ positive, you are in the second, $-+$, quadrant. And so on.  If instead you partition the sphere into $2^n$ orthants, then you consider all the signs of the coordinates of $x$.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-05T22:40:30.963" LastActivityDate="2010-08-05T22:40:30.963" CommentCount="1" />
  <row Id="1664" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:33:45.130" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/ChebyshevBias.html&quot; rel=&quot;nofollow&quot;&gt;Chebyshev bias&lt;/a&gt; behavior is properly understood only by examining up to very very large numbers.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-05T20:33:45.130" />
  <row Id="1665" PostTypeId="1" AcceptedAnswerId="1669" CreationDate="2010-08-05T20:39:29.283" Score="2" ViewCount="250" Body="&lt;p&gt;Let $p$ be a prime. If $\frac{p-1}{4}$ and $\frac{p+1}{2}$ are also primes then prove that $p=13$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-06T15:40:28.367" Title="Prime satisfying a given condition" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="1666" PostTypeId="2" ParentId="1658" CreationDate="2010-08-05T20:44:45.497" Score="1" ViewCount="0" Body="&lt;p&gt;The values of a function $f(x)$ for relatively small values of its argument $x$ is &lt;em&gt;typically&lt;/em&gt; a very bad predictor of asymptotic behavior of $f(x)$ for large $x$. This is true even when $f(x)$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_function&quot; rel=&quot;nofollow&quot;&gt;analytic function&lt;/a&gt; which is uniquely determined by its values on any small interval $x\in[-\epsilon,\epsilon]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at this excerpt from &lt;a href=&quot;http://books.google.co.uk/books?id=pntQAAAAMAAJ&amp;amp;q=concrete+mathematics+Ronald+L.+Graham,+Donald+Ervin+Knuth,+Oren+Patashnik&amp;amp;dq=concrete+mathematics+Ronald+L.+Graham,+Donald+Ervin+Knuth,+Oren+Patashnik&amp;amp;hl=en&amp;amp;ei=pSBbTJ7nHcj64Aapq-38AQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CC4Q6AEwAA&quot; rel=&quot;nofollow&quot;&gt;&quot;Concrete Mathematics&quot;&lt;/a&gt; for a (not the worst possible) example of how deceptive the &quot;small argument values&quot;  intuition could be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It helps to cultivate an expansive attitude when we're doing asymptotic analysis: We should think big, when imagining a variable that approaches infinity. For example, the hierarchy says that $\log n\prec n^{0.0001}$; this might seem wrong if we limit our horizons to teeny-tiny numbers like one googol, $n = 10^{100}$. For in that case, $\log n = 100$, while $n^{0.0001}$ is only $10^{0.01}\approx 1.0233$. But if we go up to a googolplex, $n = 10^{10^{100}}$, then $\log n = 10^{100}$ pales in comparison with $n^{0.0001} = 10^{10^{96}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-08-05T20:56:26.957" LastActivityDate="2010-08-05T20:56:26.957" />
  <row Id="1667" PostTypeId="2" ParentId="1649" CreationDate="2010-08-05T20:56:58.657" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a stab at why the problem is hard for highly non-differentiable but continuous functions. So say $f$ is nowhere differentiable on $[a,b]$. I claim that there are either infinitely or no local extrema of $f$. (By local extrema, I mean extrema that occur in the interior of the interval $[a,b]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, suppose there were finitely many, say $c_1 &amp;lt; c_2&amp;lt;\dots f(c_2)$. Then we can take the global maximum of $f$ on $[c_1, c_2]$, which occurs at an &lt;em&gt;interior&lt;/em&gt; point; it is thus a local extremum of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is a monotone function, then it is a theorem of Lebesgue that it is differentiable a.e. In particular, the above reasoning shows that the existence of finitely many local extrema implies that $f$ is differentiable a.e.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-06T11:46:42.377" LastActivityDate="2010-08-06T11:46:42.377" CommentCount="4" />
  <row Id="1668" PostTypeId="2" ParentId="1649" CreationDate="2010-08-05T21:21:07.437" Score="2" ViewCount="0" Body="&lt;p&gt;The approach you outlined is commonly used in practice.  If your original problem has some nice properties, such as convexity, the approach will work well.  For example, the &lt;a href=&quot;http://www.johndcook.com/blog/2010/01/13/soft-maximum/&quot; rel=&quot;nofollow&quot;&gt;soft maximum&lt;/a&gt; is a common way to construct a series of smooth, convex approximations to the maximum function. &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-05T21:21:07.437" CommentCount="1" />
  <row Id="1669" PostTypeId="2" ParentId="1665" CreationDate="2010-08-05T21:22:35.970" Score="4" ViewCount="0" Body="&lt;p&gt;nice one.&#xA;For (p-1)/4 to be odd, p must be of the form 8k+5, so (p-1)/4 is of the form 2k+1 and (p+1)/2 of the form 4k+3.&lt;br&gt;&#xA;If k is = 0,1,2 (mod 3) then the three numbers are respectively congruent to&lt;br&gt;&#xA;k=0(mod 3): 2,1,0 (mod 3)&lt;br&gt;&#xA;k=1(mod 3): 1,0,1 (mod 3)&lt;br&gt;&#xA;k=2(mod 3): 0,2,2 (mod 3)&lt;br&gt;&#xA;This means that the only way all three of them are prime numbers is that one of them is 3. &#xA;For k=0 we have 5,1,3 which is ruled out; for k=1 we have 13,3,7 which satisfies hypotheses; for k&gt;1 all numbers are &gt; 3.&#xA;The only other case to be checked is (p-1)/4 = 2; in this case however p = 9, so this could not be a solution.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-05T21:22:35.970" />
  <row Id="1670" PostTypeId="2" ParentId="1660" CreationDate="2010-08-05T22:03:21.940" Score="0" ViewCount="0" Body="&lt;p&gt;Just to make sure: you know that what you are asking for is actually Weyl's dimension formula for the highest weight representation in the case of SL(n,C) resp. SU(n)? I suppose there is no direct combinatorial proof that does not make use of this fact.&lt;/p&gt;&#xA;" OwnerUserId="843" LastActivityDate="2010-08-05T22:03:21.940" CommentCount="2" />
  <row Id="1671" PostTypeId="2" ParentId="1656" CreationDate="2010-08-05T22:54:57.777" Score="2" ViewCount="0" Body="&lt;p&gt;Of the projective transformations fixing a conic, there is a unique one sending any given ordered triple of points to any other given ordered triple of points.  So you are free to determine the location of the first 3 vertices of the n-gon, modulo projective equivalence, but any two placements of the remaining (n-3) points are projectively inequivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the space of n-gons up to projective equivalence can be thought of as the space of (n-3) points on a projective line (or conic, it is isomorphic) punctured at three given points.  This has dimension (n-3).&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-05T22:54:57.777" />
  <row Id="1672" PostTypeId="2" ParentId="1601" CreationDate="2010-08-05T22:56:14.577" Score="3" ViewCount="0" Body="&lt;p&gt;The (tight) lower bound is $n + \lceil \lg n \rceil - 2$ (where $\lg n$ means $\log_2 n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll prove tightness first: that this can be achieved (apparently the idea is due to Lewis Carroll!). First find the maximum using a &quot;tennis tournament&quot; structure: first compare the $n$ elements in pairs, then compare the $n/2$ &quot;winners&quot; in pairs, and so on. Since every element except the winner loses exactly once, this takes $n-1$ comparisons. But now note that the second largest element must be one which lost to the winner, as it couldn't have been defeated by any other element. So you need to find the maximum among all the $\lg n$ elements that were defeated by the winner, and finding this maximum can be done in $\lg n - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can prove this is a lower bound as well. Firstly, you need to find the maximum, since  cannot be sure some element is the second maximum without knowing which element is the maximum. Further, for each element that lost to the maximum ($m$ of them, say), this comparison was useless in determining whether it was or not the second maximum. In other words, all elements other than the maximum must lose at least once, &lt;em&gt;and&lt;/em&gt; all but one ($m-1$) of the elements that directly lost to the maximum must lose at least once more, no matter in which order the comparisons were made. So we need at least $n-1 + m-1$ comparisons. (To restate: $n-2$ of them must be found less than the second-largest element — comparisons with the largest element do not help here — plus $m$ of them must lose directly to the maximum by definition.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An adversary can make sure you always have at least $\lg n$ elements that lost to the winner (proved &lt;a href=&quot;http://www.cse.unsw.edu.au/~cs2011/lect/2711_Adversary.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://euler.slu.edu/~goldwasser/class/loyola/comp363/2003_Spring/handouts/selectionproofs.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), so this proves the $n + \lg n  - 2$ lower bound. The adversarial argument for the lower bound: the adversary keeps for each element a list of elements known to be less than or equal to it. Whenever a query's result is already &quot;known&quot;, the adversary gives that answer, else it declares the one which has a larger list the winner (breaking ties arbitrarily). The maximum is determined when the size of its list grows to n, and with each comparison the size can only double (can prove this…), so the number of comparisons was at least $\lg n$.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-06T08:47:12.300" LastActivityDate="2010-08-06T08:47:12.300" CommentCount="5" />
  <row Id="1673" PostTypeId="2" ParentId="1648" CreationDate="2010-08-05T23:17:39.257" Score="1" ViewCount="0" Body="&lt;p&gt;I'd vote for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_distribution&quot; rel=&quot;nofollow&quot;&gt;Poisson distribution&lt;/a&gt; with a constant offset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hand-wavy reasoning might be that the round trip time is due to a constant offset being the best-case round-trip time assuming no delays due to router queues (= wave propagation velocity over physical distance, + minimum processing time), with &quot;rare events&quot; (see wikipedia page) corresponding to queueing delays in one or more routers that make up the network path(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as parameter estimation goes, I'm not familiar with how to do it for samples taken from a (suspected) Poisson distribution, but I'm sure you could find something on the Internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aha, here we go: &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_distribution#Parameter_estimation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poisson_distribution#Parameter_estimation&lt;/a&gt; -- you could use this after subtracting off the minimum of a large number of samples.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;drat, stupid me, I glossed over the fact that Poisson = discrete probability distribution. &lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-08-05T23:17:39.257" CommentCount="2" />
  <row Id="1674" PostTypeId="1" AcceptedAnswerId="1679" CreationDate="2010-08-05T23:30:46.287" Score="1" ViewCount="85" Body="&lt;p&gt;If X and Y are normed spaces I know that the space of a bounded linear functions from X to Y , B(X,Y),  is complete if Y is complete.  Is there an example of a normed spaces X,Y s.t. B(X,Y) is not complete?  &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-06T00:19:56.443" Title="Bounded Linear Operators" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="1675" PostTypeId="1" AcceptedAnswerId="1690" CreationDate="2010-08-05T23:34:27.810" Score="2" ViewCount="70" Body="&lt;p&gt;In DotA, there is a character called &quot;&lt;a href=&quot;http://www.dotastrategy.com/hero-71-MogulKhanAxe.html&quot; rel=&quot;nofollow&quot;&gt;Axe&lt;/a&gt;&quot;. Every time he is attacked, he has a chance to spin (17%) his blade and deal damage based on what level the skill is, 100 / 125 / 150 / 175 damage for levels 1, 2, 3, and 4. When the spin activates, it triggers the cooldown of 0.7 / 0.65 / 0.6 / 0.55 seconds, so that attacking Axe does not generate a chance to spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to calculate an average damage per second that this skill generates, given that most of the time you will find Axe taking around 5 attacks per second (average 1.667 attacks per second from creeps, 3 creeps per camp), so I figured that would mean the probability of him spinning in any second is 1 - 0.83&lt;sup&gt;5&lt;/sup&gt;, so given the damage from earlier, we should expect him to deal 60.61 / 75.76 / 90.91 / 106.07 damage per second?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this calculation, but then I realized that I have to factor in the cooldown somewhere, but I have no idea where to start. &lt;/p&gt;&#xA;" OwnerUserId="850" LastEditorUserId="850" LastEditorDisplayName="" LastEditDate="2010-08-06T03:44:01.317" LastActivityDate="2010-08-06T03:44:01.317" Title="How can I calculate this expected rate?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1676" PostTypeId="1" CreationDate="2010-08-05T23:38:10.800" Score="7" ViewCount="261" Body="&lt;p&gt;Somewhat related to &lt;a href=&quot;http://math.stackexchange.com/questions/1249/motivating-example-for-algebraic-geometry-scheme-theory&quot;&gt;this question&lt;/a&gt;, but almost infinitely more basic.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A Confession&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am, should classification prove essential, a differential geometer and a topologist by inclination and by training: as an undergraduate I shunned any ring that wasn't $\mathbb{Z}_n$ or a ring of differential operators and held close the differentiable and the non-singular. It did not seem to matter then that these exotic 'schemes' and their exciting projective morphisms were beyond me, and to a certain extent it does not seem to matter now; but increasingly my old uni friends, fellow MOers (and, hey, even math.stack exchangers) are talking about nothing else but schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In recent months (after a &lt;a href=&quot;http://mathoverflow.net/questions/27971/why-is-there-no-cayleys-theorem-for-rings&quot; rel=&quot;nofollow&quot;&gt;frighteningly eye-opening MO question&lt;/a&gt;) I have found myself becoming more amenable to rings, and am less daunted by my paucity of understanding than previously. In spite of this, though, I remain &lt;strong&gt;entirely&lt;/strong&gt; in the dark about schemes.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Where I Sit&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am not a complete novice. I completed an undergraduate course in algebraic geometry: illuminating, interesting, but all classical beyond belief. I have read and re-read the &lt;a href=&quot;http://en.wikipedia.org/wiki/Scheme_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; on schemes several times- taking in all of the neccessary components: &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrum_of_a_ring&quot; rel=&quot;nofollow&quot;&gt;the spectrum of a ring&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Locally_ringed_space&quot; rel=&quot;nofollow&quot;&gt;a locally ringed space&lt;/a&gt; et al, but have no idea how these fit together to make the objects I fiddled with over a semester two years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made numerous guesses about generalised nulstellensatze and structure sheaves, but to explain any would probably be to complicate matters further unneccessarily. I am aware there are probably brilliant texts that do exactly what I am asking for, but I am not currently affiliated to a university and my current library would require ordering in, which for the sort of toe-dipping excercise I intend here would be overkill. So I ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can anyone provide me with a canonical example of a scheme, pointing along the way the topology and the spectra associated to each open set. Perhaps deeper, if it pleases: what I am looking for is a sort of 'scheme jargon safari'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am aware this is silly, and perhaps asking for a verbatim quotation of page 2 of any decent algebraic geometry text, but I would be ever so grateful. Can anyone help? &lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-05T23:50:37.597" LastActivityDate="2010-12-06T18:26:48.483" Title="An example of a scheme in the language of schemes" Tags="&lt;intuition&gt;&lt;algebraic-geometry&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="5" />
  <row Id="1678" PostTypeId="2" ParentId="1618" CreationDate="2010-08-06T00:14:48.557" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT: Here is a cleaned-up and corrected version of this answer, based on Pierre-Yves' suggestion (thanks!).  His answer above contains a much more complete version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sum_{i=1}^n a_{i} f_i \otimes g_i$ is not the zero element of $L^p(X) \otimes L^p(Y)$, we may assume without loss of generality that the $f_i$ are linearly independent.  We can also assume that $a_1 \ne 0$ and $g_1 \ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the corresponding function $F(x,y) = \sum_{i=1}^n a_{i} f_i(x) g_i(y) = 0$ a.e.  Since $g_1 \ne 0$, there is a measurable $B \subset Y$ of positive finite measure such that $\int_B g_1 \ne 0$ (the integral is finite by H&amp;ouml;lder).    Then by Fubini's theorem, for a.e. $x$ we have&#xA;$$&#xA;0 = \int_{B} F(x,y)dy = \sum_{i=1}^n a_i \left(\int_{B} g_i\right) f_i(x).&#xA;$$&#xA;This contradicts the assumed linear independence of the $f_i$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-08-10T19:56:55.087" LastActivityDate="2010-08-10T19:56:55.087" CommentCount="1" />
  <row Id="1679" PostTypeId="2" ParentId="1674" CreationDate="2010-08-06T00:19:56.443" Score="2" ViewCount="0" Body="&lt;p&gt;Let $X = \mathbb{R}$ with the Euclidean norm and let $Y$ be a normed space which is not complete.  You should find that $B(X, Y) \simeq Y$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T00:19:56.443" CommentCount="1" />
  <row Id="1680" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T00:24:09.603" Score="9" ViewCount="0" Body="&lt;p&gt;Dear Tom,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to consider everything here over the field $\mathbb C$.  You can replace $\mathbb C$ by any algebraically closed field (or even any field) with essentially no changes, but working over $\mathbb C$ is the natural starting point, and has the advantage that one can connect with the kind of geometry/topololgy with which you are more familiar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a classical variety $V$, you can consider all the closed subvarieties.  These satisfy the axioms of a topology, called the Zariski topology.   For definiteness, let's say that our&#xA;variety is an affine variety, so it is being cut out by polynomials in $\mathbb C^n$, for some $n$.  The usual topology on $\mathbb C^n$ induces a topology on $V$, which has many more open sets than the Zariski topology (unless $V$ is $0$-dimensional).  The point is that to be closed in the Zariski topology, you have to really be the zero locus of some polynomial, i.e. another variety, so it is hard to be Zariski closed, and hence similarly hard to be Zariski open.    (Just to be absolutely clear, let's look at an example: the real line is closed in $\mathbb C$, but is not Zariski closed; there is no polynomial in one variable over $\mathbb C$ that vanishes precisely on the points of the real line; indeed,&#xA;such a polynomial either vanishes at only finitely many points, or else is identically zero, &#xA;and so vanishes everywhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have the notion of rational function on the variety (just think of the restriction of a ratio of polynomials in $n$-variables to $V$, such that the denominator does not vanish&#xA;identically on $V$); a rational function is called &lt;em&gt;regular&lt;/em&gt; at a point $P$ of the variety if it has no singularity at that point.  Being a singularity is a Zariski closed condition (singularities occur where the denominator of the rational function, which is a polynomial, vanish), so being regular at a point is a Zariski open condition.  If we fix a Zariski open&#xA;set in advance, we can look at the ring of all rational functions that are regular on that&#xA;open set.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These form a sheaf on $V$ (with its Zariski topology).  It is much &quot;smaller&quot; than the sheaves you are used to, like smooth or continuous functions.  Not only are there many&#xA;fewer open sets to think about (just the Zariski open ones), but on a given open set,&#xA;there will be incredibly more continuous or smooth functions than regular functions, just because being the ratio of polynomials is a very restrictive condition on a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the global sections of this sheaf, i.e. the functions that are regular on the whole of $V$, we in particular get a ring which is called the affine ring of $V$.  If I just hand you this ring (as a $\mathbb C$-algebra), it turns out that you can recover $V$, namely&#xA;$V$ is the maximal spectrum of this ring (i.e. point of $V$ are in natural bijection with&#xA;maximal ideals of $V$).  The map one way is easy: given a point, we can look at all regular functions on $V$ that vanish at the point; this gives a maximal ideal in the ring of all&#xA;regular functions.  That this is a bijection is harder, and is essentially equivalent to the Nullstellensatz.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the role of the entire spectrum of the ring (i.e. the prime ideals as well as the maximal ideals) one has to say and think about more, but this is probably enough for now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To learn more, you should google &quot;affine ring of a variety&quot; or similar expressions, and you should find troves of information, at a great range of levels.  Once you understand this &#xA;basic connection, it makes sense to look at schemes in more detail.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-06T16:38:23.850" LastActivityDate="2010-12-06T16:38:23.850" CommentCount="1" />
  <row Id="1681" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T00:24:12.847" Score="5" ViewCount="0" Body="&lt;p&gt;$\text{Spec } \mathbb{C}[x]$ is probably as basic as it gets.  As a set, this is the collection of maximal ideals $(x - a), a \in \mathbb{C}$ together with the prime ideal $(0)$.  As a topological space, this is $\mathbb{C}$ in the cofinite topology (the closed sets are the finite ones) together with a point $(0)$ whose closure is the entire space (and which is in every open set except the empty set).  The local ring at $(x - a)$ is the subring of $\mathbb{C}(x)$ of rational functions which are defined at $a$, and the local ring at $(0)$ is all of $\mathbb{C}(x)$; more generally, the section of the structure sheaf over a Zariski-open set $U$ is the subring of $\mathbb{C}(x)$ of rational functions which are defined at every $a \in U$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T00:24:12.847" CommentCount="1" />
  <row Id="1682" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T00:33:40.130" Score="9" ViewCount="0" Body="&lt;p&gt;You should read David Eisenbud and Joe Harris's &lt;em&gt;The Geometry of Schemes&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Really.&lt;/em&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-06T01:08:58.673" LastActivityDate="2010-08-06T01:08:58.673" CommentCount="3" />
  <row Id="1683" PostTypeId="2" ParentId="709" CreationDate="2010-08-06T00:53:35.193" Score="2" ViewCount="0" Body="&lt;p&gt;Echoing Americo's and Isaac's answers, but without appealing to the quadratic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First treat the special case f(x)=ax²+d (i.e. no &quot;x&quot; term); by remembering that the line that divides the parabola symmetrically passes through the vertex, and is in fact the y-axis (x=0), we see that 0 is the abscissa of the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To treat the more general f(x)=ax²+bx+c, we try a substitution of the form x=x'-h (geometrically, this corresponds to translating your coordinate system horizontally by h units). Expanding this, you will get something like a(x')²+(2ah+b)x'+(constant term). To go back to the special case we first treated, we find the value of h such that the coefficient of x' is 0. Thus you obtain the expression for the vertex's x-coordinate.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-06T01:28:54.390" LastActivityDate="2010-08-06T01:28:54.390" />
  <row Id="1684" PostTypeId="2" ParentId="1676" CreationDate="2010-08-06T01:01:55.487" Score="3" ViewCount="0" Body="&lt;p&gt;Well, Qiaochu lists the  affine line, so I'm going to do the projective line. (Everything here is over the complex numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projective line $P^1$, or equivalently the Riemann sphere.  This is the space of all lines through the origin in $\mathbb{C}^2$.  Recall that the Riemann sphere can be obtained by gluing two copies of the complex plane. Namely, if $S^2 = \mathbb{C} \cup { \infty }$, then one copy is just the subset $\mathbb{C}$. The other copy is $\mathbb{C}^* \cup {\infty}$, which is identified with the complex numbers by taking reciprocals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously a Riemann surface.  In fact, the maps defining this &quot;gluing&quot; can be checked to be algebraic (they're just reciprocals), so it is in fact a scheme, and a toy example which isn't affine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a regular function over an open subset of $S^2$? Well, one defines a holomorphic function over an open subset of $S^2$ by saying that the pull-back to each chart is holomorphic. Now here we say that the pull-back to each chart is regular in the sense of its being a rational function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, there are only constant functions which are regular on all of $P^1$. Indeed, such a function (considered as a map $P^1 \to \mathbb{C}$) would be a holomorphic map from a compact Riemann surface, hence constant. Here's an algebraic argument. Consider the open set $\mathbb{C}^*$. On this (affine) open set, it can be checked that the only regular functions are polynomials in $z $ and $1/z$ (or otherwise the denominator would blow up).  However, if the function is regular everywhere, then $1/z$ can't occur (that would blow up at the origin) and $z$ can't occur (that isn't defined at ${\infty}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projective line is important because it is compact in the complex topology. The algebraic version of this is that it is &lt;em&gt;proper&lt;/em&gt; over $\mathbb{C}$. In particular, any map out of it into a complex variety is a &lt;em&gt;emph&lt;/em&gt; closed map. However, with affine varieties, you don't have this closed property anymore. For instance, the hyperbola $xy=1$ in the affine plane (clearly a closed set in the Zariski topology) projects to the non-open complement of the origin of $A^1$ via $(x,y) \to x$. This doesn't happen for the projective line.  (&quot;Proper&quot; is an algebraic analog of &quot;compact,&quot; just as &quot;separated&quot; is an analog of &quot;Hausdorff.&quot;)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-06T01:01:55.487" CommentCount="1" />
  <row Id="1685" PostTypeId="1" AcceptedAnswerId="1697" CreationDate="2010-08-06T01:04:40.570" Score="5" ViewCount="165" Body="&lt;p&gt;Are there books or article that develop (or sketch the main points) of Euclidean geometry without fudging the hard parts such as angle measure, but might at times use coordinates, calculus or other means so as to maintain rigor or avoid the detail involved in Hilbert-type axiomatizations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of Hilbert's foundations and the book by Moise.   I was wondering if there is anything more modern that tries to stay (mostly) in the tradition of synthetic geometry.   &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T01:22:47.757" LastActivityDate="2010-08-06T06:59:41.937" Title="Rigorous synthetic geometry without Hilbert axiomatics" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="1686" PostTypeId="2" ParentId="1563" CreationDate="2010-08-06T01:46:44.667" Score="8" ViewCount="0" Body="&lt;p&gt;First, I'm glad you are reading my book! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me make a couple of comments on Pete's answer--this is my first time here and I don't see how to leave comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Any two dense linear orders without endpoints are elementarily equivalent.&#xA;In particular $(Q,&amp;lt;)$ and $(R,&amp;lt;)$ are elementarily equivalent.  So there is no first order way of expressing the completeness of the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Any two algebraically closed fields of the same characteristic are elementarily equivalent.  So the algebraic numbers&#xA;is elementarily equivalent to the complex numbers.  This means you can prove first order things about the algebraic numbers using complex analysis or the complex numbers using Galois Theory or countability.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Similarly the reals field is elementarily equivalent to the real algebraic numbers or to the field of real Puiseux series.&#xA;One can for example use the Puiseux series to prove asymptotic properties of semialgebraic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Pete's comment 5) about infinite models of the theory of finite fields being elementarily equivalent isn't quite right.  This is only true if the relative algebraic closure of the prime fields are isomorphic.&#xA;For example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) take an  ultrapower of finite fields $F_p$ where the ultrafilter containing {2,4,8,...} then the resulting model has &#xA;characteristic 2, while if the ultrafilter contains the set of primes, then the ultraproduct has characteristic 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) if the ultrapower contains the set of primes congruent to 1 mod 4, in the ultraproduct -1 is a square, while if&#xA;the ultraproduct contains the set of primes congruent to 3 mod 4 then in the ultraproduct -1 is not a square..&lt;/p&gt;&#xA;" OwnerUserId="851" LastEditorUserId="851" LastEditorDisplayName="" LastEditDate="2010-08-08T13:44:06.557" LastActivityDate="2010-08-08T13:44:06.557" CommentCount="5" />
  <row Id="1687" PostTypeId="1" AcceptedAnswerId="1689" CreationDate="2010-08-06T01:52:18.913" Score="2" ViewCount="48" Body="&lt;p&gt;I want to try to emulate what this application can do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img530.imageshack.us/img530/5756/bezier.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the red round dots (from the mouse) it is able to solve for the bezier handles given that the tension of the curve is set to 0.6. How could I solve for the bezier handles? Is there some way I could do this with the slope or perpendicular of the previous and next curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I should also add that each point is equadistant meaning point i + 1 is the distance to point i or point i + 2.&lt;/p&gt;&#xA;" OwnerUserId="852" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-26T01:32:00.180" LastActivityDate="2010-08-26T01:32:00.180" Title="Generating bezier handles based on constraints" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;bezier-curve&gt;" AnswerCount="1" />
  <row Id="1688" PostTypeId="2" ParentId="240" CreationDate="2010-08-06T01:54:45.447" Score="9" ViewCount="0" Body="&lt;p&gt;Over at &lt;a href=&quot;http://mathoverflow.net/questions/12342/reductio-ad-absurdum-or-the-contrapositive/12400#12400&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow question&lt;/a&gt;, I posted the following answer to a similar question (and there are several other interesting answers there):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;With good reason&lt;/em&gt;, we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the reason? The reason is the &lt;em&gt;fecundity&lt;/em&gt; of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r&lt;sub&gt;1&lt;/sub&gt; and r&lt;sub&gt;2&lt;/sub&gt; and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when we prove the contrapositive (&amp;not;q implies &amp;not;p) directly, we assume &amp;not;q, make intermediary conclusions  r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, and then finally conclude &amp;not;p. Thus, we have also established not only that &amp;not;q implies &amp;not;p, but also, that it implies r&lt;sub&gt;1&lt;/sub&gt; and r&lt;sub&gt;2&lt;/sub&gt; and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (&amp;not;r&lt;sub&gt;1&lt;/sub&gt; implies q), we learn about many different hypotheses that all imply q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and &amp;not;q, and argue  r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, and so on, before arriving at a contradiction. The statements r&lt;sub&gt;1&lt;/sub&gt; and r&lt;sub&gt;2&lt;/sub&gt; are all deduced under the contradictory hypothesis that p and &amp;not;q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistant, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; are &lt;em&gt;all&lt;/em&gt; the primes. It follows that since none of them divide the product-plus-one p&lt;sub&gt;1&lt;/sub&gt;...p&lt;sub&gt;n&lt;/sub&gt;+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced becasue this is a proof by contradiction, and that conclusion relied on the assumption that p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; were &lt;em&gt;all&lt;/em&gt; the primes. If one organizes the proof, however, as a direct argument showing that whenever p&lt;sub&gt;1&lt;/sub&gt;, ..., p&lt;sub&gt;n&lt;/sub&gt; are prime, then there is yet another prime not on the list, then one is lead to the true conclusion, that p&lt;sub&gt;1&lt;/sub&gt;...p&lt;sub&gt;n&lt;/sub&gt;+1 has merely a prime divisor not on the original list.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-06T01:54:45.447" CommentCount="2" />
  <row Id="1689" PostTypeId="2" ParentId="1687" CreationDate="2010-08-06T02:00:53.407" Score="2" ViewCount="0" Body="&lt;p&gt;From the graphic, it appears that the handles at a given point $p_k$ are parallel to the line through $p_{k-1}$ and $p_{k+1}$ and that the lengths of the handles at every point are equal.  I suspect that the length of the handles is determined by the &quot;tension&quot; that you mention.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-06T02:00:53.407" CommentCount="1" />
  <row Id="1690" PostTypeId="2" ParentId="1675" CreationDate="2010-08-06T02:56:12.237" Score="1" ViewCount="0" Body="&lt;p&gt;Let r be the chance an attack caused axe to spin and d be the damage it does. Since we know the number of attacks on Axe per second, we just need to find the expected spin damage per attack, call this x. If someone attacks Axe once, we expect the retaliation damage to be rd. However, each attack blocks a certain number of attacks depending on cooldown, call this number b. This reduces our expected value by brx.  So we have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x=rd-brx&#xA;x(1+br)=rd&#xA;x=rd/(1+br)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that this assumes the combat goes for an infinite amount of attacks. When the combat is shorter, the expected damage will be higher. Given the speed that combat happens in Dota, this effect will often be significant&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-06T02:56:12.237" />
  <row Id="1691" PostTypeId="1" AcceptedAnswerId="1695" CreationDate="2010-08-06T04:36:39.063" Score="2" ViewCount="136" Body="&lt;p&gt;If we have a curve $C:\{ P(x,y) = 0 \}$ and define $\omega=\frac{\mathrm{d}x}{y}$ then is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^A \omega + \int_0^B \omega = \int_0^{A \oplus B} \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $\oplus$ being addition on a group on the curve) a theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite sure this is what's happening when C is an elliptic curve, but I have failed to make this work out when C is defined by $P(x,y) = x^2 + y^2 - 1$ (unit circle) and the group law is defined by firing a ray through $(3/5,4/5)$ (chosen arbitrarily) parallel to $AB$ and taking it's intersection with the circle as $A \oplus B$.&lt;/p&gt;&#xA;" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T06:19:12.360" LastActivityDate="2010-08-06T12:17:16.223" Title="Trying to piece together an integral addition theorem" Tags="&lt;calculus&gt;&lt;algebraic-geometry&gt;&lt;conic-sections&gt;&lt;elliptic-curves&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1692" PostTypeId="2" ParentId="709" CreationDate="2010-08-06T04:51:20.560" Score="3" ViewCount="0" Body="&lt;p&gt;Here is the simplest way I know, using a bit of algebra.  Starting with Isaac's observation that the vertex lies on the axis of symmetry, the line x = k.  Question is, what's k? That's your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if f(x) = ax&lt;sup&gt;2&lt;/sup&gt; + bx + c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the symmetry around x = k means that, for any h &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(k+h) = f(k-h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that, just multiply it all out and some simple additive cancellation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-2akh - bh = 2akh + bh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h cancels out, and a bit more algebra gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k = -$\frac{b}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really complete this, however, we should show that x = k = -$\frac{b}{2a}$ is a maximum or minimum point. Again, with more algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(k+h) - f(k) boils down to ah&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when a &gt; 0, then f(k+h) &gt; f(k) for any h != 0, establishing that x = k is a minimum point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And likewise when a &amp;lt; 0, x = k is a maximum. This also shows the symmetry around x = k, and how the coefficient a determines whether the curve is concave upward or downward.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-08-06T12:20:35.610" LastActivityDate="2010-08-06T12:20:35.610" />
  <row Id="1694" PostTypeId="1" CreationDate="2010-08-06T05:56:39.303" Score="7" ViewCount="626" Body="&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithm&quot; rel=&quot;nofollow&quot;&gt;sez&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The natural logarithm of x is often written &quot;ln(x)&quot;, instead of log_e(x) especially in disciplines where it isn't written &quot;log(x)&quot;. However, some mathematicians disapprove of this notation. In his 1985 autobiography, Paul Halmos criticized what he considered the &quot;childish ln notation,&quot; which he said no mathematician had ever used. In fact, the notation was invented by a mathematician, Irving Stringham, professor of mathematics at University of California, Berkeley, in 1893.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently the notation &quot;ln&quot; first appears in Stringham's book &lt;em&gt;Uniplanar algebra: being part I of a propædeutic to the higher mathematical analysis&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't explain why &quot;ln&quot; has become so pervasive. I'm pretty sure that most high schools in the US at least still use the notation &quot;ln&quot; today, since all of the calculus students I come into contact with at Berkeley seem to universally use &quot;ln&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did this happen?&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-08-07T21:47:12.533" Title="How did the notation &quot;ln&quot; for &quot;log base e&quot; become so pervasive?" Tags="&lt;notation&gt;&lt;math-history&gt;" AnswerCount="7" CommentCount="7" />
  <row Id="1695" PostTypeId="2" ParentId="1691" CreationDate="2010-08-06T06:01:55.567" Score="2" ViewCount="0" Body="&lt;p&gt;$0$, the lower limit for the integrals, should be whatever point on the curve is $0$ for the group law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $dx/y$ should be rotation-invariant for any circle. It will not be true for a rotated ellipse; there is an invariant differential but it is not $dx/y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an elliptic curve, $dx/y$ being the invariant differential relies on the curve being written as $y^2=f(x)$ with $\deg(f)=3$. If you move the curve the invariant differential will be a different one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher genus the curve does not have an addition law.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-06T06:31:16.967" LastActivityDate="2010-08-06T06:31:16.967" CommentCount="1" />
  <row Id="1697" PostTypeId="2" ParentId="1685" CreationDate="2010-08-06T06:59:41.937" Score="2" ViewCount="0" Body="&lt;p&gt;You might look at Hartshorne's&#xA;&lt;a href=&quot;http://books.google.co.uk/books?id=EJCSL9S6la0C&amp;amp;lpg=PP1&amp;amp;dq=hartshorne&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Geometry: Euclid and Beyond&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-06T06:59:41.937" CommentCount="0" />
  <row Id="1698" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T07:06:06.650" Score="1" ViewCount="0" Body="&lt;p&gt;One simple possible reason&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is very difficult to come up with a&#xA;  notation that is concise, correct and&#xA;  understandable by at least two people.&#xA;  - P A M Dirac / Richard Feynmann&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one is concise, correct and is understandable at least by you and me! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, notations used in popular textbooks spread like virus. Books become popular because of good notation and notation becomes widespread because book is popular.  &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-06T07:14:23.187" LastActivityDate="2010-08-06T07:14:23.187" CommentCount="1" />
  <row Id="1699" PostTypeId="2" ParentId="67" CreationDate="2010-08-06T07:16:04.540" Score="4" ViewCount="0" Body="&lt;p&gt;Elliptic curve cryptography (ECC), or at least some version of it, essentially follows the same idea as RSA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulo some details, here's how RSA works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;public key&quot; $N$ which is a product of two large prime numbers $p$ and $q$. Now pick two numbers $e$ and $d$ such that $ed = 1$ mod $(p-1)(q-1)$. The number $e$ is a &quot;public key&quot;. The number $d$ is a &quot;secret key&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be our secret message (a number). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the encrypted message is $B := A^e$ mod $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fact that $A$ is equal to $B^d$ mod $N$. (It follows by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem#Generalizations&quot; rel=&quot;nofollow&quot;&gt;a generalization of Fermat's little theorem&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if we know $d$ and $B$, we can decrypt the message (that is, we can recover $A$ from $B$) by computing $B^d$ mod $N$. The point is that, even if an adversary knows $B$, $N$, and $e$, presumably he also needs $d$ to recover the secret message $A$. Moreover, to find $d$ he presumably needs to know $p$ and $q$, and finding $p$ and $q$ is presumably difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RSA works by exploiting the properties of $\mathbb{Z}/N\mathbb{Z}$ (or actually its group of units). ECC is analogous, except we use an elliptic curve (over a finite field) instead of $\mathbb{Z}/N\mathbb{Z}$. An elliptic curve is, in fancy terms, an example of an &quot;abelian variety&quot;. In less fancy terms, you can think of an elliptic curve as simply a set of &quot;things&quot; with a funny multiplication law $\ast$: that is, given two &quot;things&quot; $a$ and $b$, we can &quot;multiply&quot; them together to get a new &quot;thing&quot; $a \ast b$. Now in ECC, our secret message $A$ is such a &quot;thing&quot;, and the way we encrypt it is by multiplying $A$ with itself $e$ times (using the funny multiplication) to get a new &quot;thing&quot; $B$. The way we decrypt $B$ and recover $A$ is by multiplying $B$ with itself $d$ times (for some appropriate $d$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope I am getting this right. I am working mainly from memory. My memory of this stuff might be inaccurate, or flat out wrong...)&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-08-06T07:34:18.187" LastActivityDate="2010-08-06T07:34:18.187" CommentCount="1" />
  <row Id="1700" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T07:39:09.190" Score="5" ViewCount="0" Body="&lt;p&gt;It is probably at least in part an instance of consumer lock-in, the phenomenon where the advantages of the majority of the market using the same or compatible products outweigh sufficiently minor differences of one version of the product versus another.  (When I learned this term, the canonical example was VHS versus Beta.  I am just about old enough to remember that when VCR terminology first came out, vendors would carry both.  Already when the first video stores opened up there was more product available in VHS, so a lot of stores would have just one rack with all the Beta products.  And eventually of course Beta died.  However those who remember and care seem to largely agree that Beta was the superior technology.  Apologies for giving such an old-fogey example.  Can someone suggest something more current?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, when it comes to electronic calculators, having everyone agree what's going to happen when you press a certain button is a good thing.  (In fact, another lock-in phenomenon is that when I was in high school in the early 90's, most students had Texas Instruments calculators of one kind or other.  Among the real geeks it was known that the &quot;cadillac of calculators&quot; was actually the Hewlett-Packard, which used &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_Polish_notation&quot; rel=&quot;nofollow&quot;&gt;reverse polish notation&lt;/a&gt;.  Serious CS people appreciate RPN, but the problem is that if you're a high school kid and pick up such a calculator for the first time, it's very hard to figure out what's going on.  I haven't seen an HP calculator for many years.)  The notation $\ln$ is simple and unambiguous: you don't have to like it (and I don't, especially), but you know what it means, and it's easier to fit on a small calculator key than $\log_e$.  I think if you're first learning about logarithms, then base ten is probably the simplest (to have any clue what $e$ is other than &quot;about $2.71828...$&quot; requires some calculus, and is in my experience one of the more subtle concepts of first year calculus), so it's reasonable to have that be the standard base for the logarithm for a general audience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm sure everyone here knows this but I wish my calculus students had a better appreciation of it: exactly what base you take for the logarithm only matters up to a multiplicative constant anyway, since $\log_c x = \frac{\log_a x}{\log_a c}$.  So it's no big deal either way.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-06T08:51:54.617" LastActivityDate="2010-08-06T08:51:54.617" CommentCount="5" />
  <row Id="1701" PostTypeId="1" AcceptedAnswerId="1708" CreationDate="2010-08-06T07:43:00.133" Score="6" ViewCount="246" Body="&lt;p&gt;I came across the statement, that Yoneda-lemma is a generalization of Cayley`s theorem which states, that every group is isomorphic to a group of permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly does generalizes Yoneda-lemma Cayley`s theorem? Can Cayley's theorem be deduced from Yoneda lemma, is it a generalization of a special case of Yoneda, or is this rather a philosiophical statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems that Yoneda embedding is more canonical than Cayleys theorem, because in the latter you have to choose whether the group acts from the left or from the right on itself. But maybe this is an optical illusion.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-08-06T12:37:14.343" Title="Yoneda-Lemma as generalization of Cayley`s theorem?" Tags="&lt;category-theory&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="5" />
  <row Id="1702" PostTypeId="2" ParentId="1601" CreationDate="2010-08-06T07:53:39.567" Score="2" ViewCount="0" Body="&lt;p&gt;The definitive answer for any $k$ (not just the second largest) is discussed in &lt;a href=&quot;http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;. In short, if $V(n,k)$ is the number of comparisons needed to determine the kth largest element in a set of size $n$, then &#xA;$$ V(n,k) \ge n + R(n,k) - 2\sqrt{R(n,k)} $$&#xA;where&#xA;$$ R(n,k) = \log \binom{n}{k} - \log (n-k+1) + 3$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-08-06T07:53:39.567" CommentCount="1" />
  <row Id="1703" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T08:01:17.010" Score="0" ViewCount="0" Body="&lt;p&gt;The &lt;code&gt;log&lt;/code&gt; abbreviation was more understandable then the &lt;code&gt;lg&lt;/code&gt; abbreviation, and also was more natural compared to the &lt;code&gt;sin&lt;/code&gt;, &lt;code&gt;cos&lt;/code&gt;, &lt;code&gt;tan&lt;/code&gt; abbreviations. Compare the &lt;code&gt;tg&lt;/code&gt; notation versus &lt;code&gt;tan&lt;/code&gt;. They are both in use, however &lt;code&gt;tan&lt;/code&gt; is more common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;code&gt;log_a(x)&lt;/code&gt; we have two important special cases: &lt;code&gt;log_10&lt;/code&gt; and &lt;code&gt;log_e&lt;/code&gt;. &lt;code&gt;Log_10&lt;/code&gt; was so common, especially in high school, that it became abbreviated to &lt;code&gt;log&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Log_e&lt;/code&gt; is/was more difficult to understand at high school level  - I can remember that. The notation &lt;code&gt;ln&lt;/code&gt; to distinguish this special case is helpful especially for beginners. Two different set of rules for two different scenarios (&lt;code&gt;log=log_10&lt;/code&gt; and &lt;code&gt;ln=log_e&lt;/code&gt;). Later on the can &lt;em&gt;perhaps&lt;/em&gt; understand the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add to the confusion: my teachers at the university used the notation &lt;code&gt;log&lt;/code&gt; to mean &lt;code&gt;ln&lt;/code&gt;!! From there point of view any extra notation should be avoided and &lt;code&gt;log_10&lt;/code&gt; had no particular importance to them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &lt;code&gt;ln&lt;/code&gt; notation turns out to be useful from a didactical perspective.&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="706" LastEditorDisplayName="" LastEditDate="2010-08-06T08:06:38.560" LastActivityDate="2010-08-06T08:06:38.560" />
  <row Id="1704" PostTypeId="1" AcceptedAnswerId="1707" CreationDate="2010-08-06T08:19:05.400" Score="3" ViewCount="68" Body="&lt;p&gt;In group cohomology theory, we know that the cohomological dimension cd(G) for a profinite group is a fundamental numerical invariant. We say G is of virtual cohomological dimension n (denote it by vcd(G)) if there exists an open subgroup H such that cd(H)=n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that vcd(G) is not always a fixed integer and my question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any criterion when vcd(G) become fixed ,or say, when cd(G)=cd(H) for all open subgroup H of G ? If there are any related references please let me know, thx !&lt;/p&gt;&#xA;" OwnerUserId="866" LastActivityDate="2010-08-06T09:09:30.197" Title="When does the virtual cohomological dimension become a numerical invariant?" Tags="&lt;cohomology&gt;" AnswerCount="1" />
  <row Id="1705" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T08:27:29.157" Score="6" ViewCount="0" Body="&lt;p&gt;Cajori, in his &lt;em&gt;History of mathematical notations&lt;/em&gt;, Vol. II, as far as I can see, mentions the notation «$\operatorname{ln}$» once when he is talking about notations for logarithm. He refers to [Irving Stringham, &lt;em&gt;Uniplanar Algebra&lt;/em&gt;, (San Francisco 1893), p. xiii] as someone who used that notation; the way this guy is mentioned makes me doubt he was the first or alone in this, though (I'd love to know what 'uniplanar algebra' is (was?)!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mention of «$\operatorname{ln}$» is quite minor, and I would guess that at the time Cajori was writing (the volume was completed in August 1925) the notation was essentially not used, for otherwise he would have been more interested in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; I think I got the link from someone here, or on MO... &lt;a href=&quot;http://spikedmath.com/043.html&quot; rel=&quot;nofollow&quot;&gt;http://spikedmath.com/043.html&lt;/a&gt; is pretty relevant!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-06T08:33:04.000" LastActivityDate="2010-08-06T08:33:04.000" CommentCount="1" />
  <row Id="1706" PostTypeId="1" AcceptedAnswerId="1731" CreationDate="2010-08-06T09:03:33.143" Score="5" ViewCount="520" Body="&lt;p&gt;I've been trying to find answer to this question for some time but in every document I've found so far it's taken for granted that reader know what $\mathbf ℝ^+$ is.&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-03T01:05:00.710" LastActivityDate="2010-11-03T01:05:00.710" Title="Does set  $\mathbb{R}^+$ include zero?" Tags="&lt;notation&gt;&lt;number-systems&gt;" AnswerCount="5" CommentCount="13" FavoriteCount="1" />
  <row Id="1707" PostTypeId="2" ParentId="1704" CreationDate="2010-08-06T09:09:30.197" Score="3" ViewCount="0" Body="&lt;p&gt;Although it is not obvious from the definition, in fact -- assuming that there exists at least one open subgroup of finite cohomological dimension, otherwise there is nothing to say -- the virtual cohomological dimension is always a fixed integer.  In other words, if there exists a single open subgroup $H_0$ of $G$ such that $\operatorname{cd}(H_0) &amp;lt; \infty$, then for all open subgroups $H$ of $G$ with $\operatorname{cd}(H) &amp;lt; \infty$, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{cd}(H) = \operatorname{cd}(H_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Moreover, this also holds for the $p$-cohomological dimension at any prime $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this result was first proved by Serre.  In any case, it follows easily from Proposition I.14 in Serre's &lt;em&gt;Galois Cohomology&lt;/em&gt;.  In fact, a slightly stronger result is given as Proposition I.14$'$: again assuming that there exists at least one open subgroup $H_0$ of finite cohomological dimension, an open subgroup $H$ has infinite cohomological dimension if $H$ has nontrivial elements of finite order and cohomological dimension equal to $\operatorname{cd}(H_0)$ otherwise.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-06T09:09:30.197" CommentCount="2" />
  <row Id="1708" PostTypeId="2" ParentId="1701" CreationDate="2010-08-06T10:38:14.537" Score="7" ViewCount="0" Body="&lt;p&gt;Just as Cayley's theorem states that every group is a subgroup of a symmetric group, Yoneda's lemma states that every (locally small) category $C$ embeds into a category of functors defined on $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, Yoneda's lemma states that if $F:C\to Set$ is an arbitrary set-valued functor, then $F(A) = Nat(\hom(A,-), F)$, so that natural transformations from a hom-set functor are in bijective correspondence with the elements in the functor image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this to be a straight generalization, we may consider a group $G$ as a category $C_G$ (actually a groupoid) with a single object $\ast$, and each group element a morphism. Then, there is only a single object, so a set-valued functor is the same thing as a set $S$ with a map from $G$ to $Bij(S,S)$, the set of bijections from $S$ to $S$. We can see this by unrolling the definition of set-valued functor: $\ast$ goes to $S$, and each element of $G$ goes to some map $S\to S$; all of which maps have to be bijections since otherwise the group properties suffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now that we have some such $G$-set $S$, Yoneda's lemma tells us that its elements are bijective with natural transformations from $G$ to $S$; so what is a natural transformation here? Our two functors are $S$, that takes $\ast$ to $S$, and $\hom(\ast,-)$ that takes $\ast$ to $G$; both as sets. A natural transformation of these is a set-valued map from one image to the other, such that the 'obvious' square of induced maps from morphisms in $C_G$ commutes - thus, $Nat(\hom(\ast,-),S)$ is the collection of $G$-set maps from $G$ as a left $G$-representation to $S$ itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the things Yoneda brings is a bijection $Nat(\hom(a,-),\hom(b,-)) = \hom(b,a)$: the Yoneda embedding. Applied to the group situation, this tells us that $\hom_G(G,G)=G$. Certainly, left multiplication by an element is a group endomorphism; and what this tells us is that these are all there is. This bijection is exactly what is used in the proof of Cayley's theorem on the wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="666" LastEditorUserId="666" LastEditorDisplayName="" LastEditDate="2010-08-06T11:27:56.573" LastActivityDate="2010-08-06T11:27:56.573" CommentCount="1" />
  <row Id="1709" PostTypeId="2" ParentId="1648" CreationDate="2010-08-06T10:45:04.263" Score="2" ViewCount="0" Body="&lt;p&gt;Sounds an awful lot like the conditions you would expect for &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlang_distribution&quot; rel=&quot;nofollow&quot;&gt;an Erlang distribution&lt;/a&gt; to me- it also looks a lot like one... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erlang distributions model the times between occurrences in poisson processes and are frequently used as parts of models of internet traffic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is this: as a site sending back a signal, one processes and sends stuff for a given user in an approximately poisson process (the approximate 'limit' of bernoilli trials p-&gt; stuff for user 1-p -&gt; stuff for a different user) and the time spent waiting for one to occur is therefore distributed Erlangwise, with a shift to the right (to account for the user's sending of the signal). This gives the shape you have above :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This should be Erlang-2 if that is not already clear, since receiving and sending are two poisson occurrences from the same distribution depending (as indicated above) on traffic [That is: occurrence 1- server has free bit to process receiving, occurrence 2- computer has free bit to process sending]&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-06T12:21:45.543" LastActivityDate="2010-08-06T12:21:45.543" CommentCount="2" />
  <row Id="1710" PostTypeId="1" AcceptedAnswerId="1712" CreationDate="2010-08-06T11:35:44.143" Score="5" ViewCount="189" Body="&lt;p&gt;I've been experimenting with the summation of polynomials. My line of attack is to treat the subject the way I would for calculus, but not using limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of a very simple example, suppose I wish to add the all numbers between 10 and 20 inclusive, and find a polynomial which I can plug the numbers into to get my answer.  I suspect its some form of polynomial with degree 2. So I do a integer 'differentiation':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;diff(x^2) = x^2-(x-1)^2 = 2x-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see from this that I nearly have my answer, so assuming an inverse 'integration' operation and re-arranging:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1/2)diff(x^2+int(1)) = x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know that the 'indefinite integral' of 1 is just x, from 'differentiating' x-(x-1) = 1. So ultimately:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1/2)(x^2+x) = int(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get my answer I take the 'definite' integral: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;int(x):10,20 = (1/2)(20^2+20)-(1/2)(9^2+9) = 165 (the lower bound needs decreasing by one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there a general way I can 'integrate' any polynomial, in this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse my lack of rigour and odd notation.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-08-06T16:17:25.780" Title="Indefinite Summation of Polynomials" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="1711" PostTypeId="2" ParentId="540" CreationDate="2010-08-06T11:45:39.807" Score="3" ViewCount="0" Body="&lt;p&gt;The above answers give excellent explanations about why the zeta function has close connections to number theory, but I thought I'd mention something about why the Riemann Hypothesis should matter so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking the logarithm and then differentiating the zeta function, one gets the formula&#xA;$$\frac{\zeta'(s)}{\zeta(s)}=\sum_{n=1}^\infty\frac{\Lambda(n)}{n^s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\Re(s)&gt;1$, where $\Lambda(n)$ is the von Mangoldt function which takes the value $\log p$ at powers of primes $p$, and is 0 everywhere else. Think of it as a weighted way of counting the primes (the prime number theorem tells us that $\log p$ is the natural weight to choose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of analytic number theory proceeds by choosing a weight of the set we wish to consider (often the primes), and then encoding this weighting in a so-called Dirichlet series (an infinite sum of the form above). We can then use analysis to study this series and get lots of useful information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, then, the function we need to study to get information about the primes is $\frac{\zeta'(s)}{\zeta(s)}$, which we can study using complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In complex analysis, a good slogan is 'the only things that matter are zeros and poles' (effectively points where the function shoots off to infinity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence to understand the prime numbers, we just need to understand the zeros and poles of $\frac{\zeta'(s)}{\zeta(s)}$ - we know about the simple pole at $s=1$, we know there aren't any other zeros where it counts, and we also know that the only other poles are at zeros of $\zeta(s)$ (roughly because dividing by zero causes infinity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if we knew where these zeros are (i.e. the Riemann hypothesis) we can work with $\frac{\zeta'(s)}{\zeta(s)}$ in all kinds of clever ways to get good results on the prime numbers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More specifically, in the usual contour proof of the prime number theorem, knowing that there aren't any other zeros in $\Re(s)&gt;1/2$ would allow us to shift the contour further to the left, reducing the error term in the result to (roughly) $O(\sqrt{x})$.&lt;/p&gt;&#xA;" OwnerUserId="870" LastActivityDate="2010-08-06T11:45:39.807" />
  <row Id="1712" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T11:58:31.347" Score="6" ViewCount="0" Body="&lt;p&gt;Your &quot;diff&quot; is actually called &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;(backward) finite difference&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\nabla_1 [ P ](x) &amp;amp;= P(x) - P(x-1) \\&#xA;&amp;amp;= P(x-1+1) - P(x-1) \\&#xA;&amp;amp;= \Delta_1[ P ](x-1)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse the forward finite difference is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Indefinite_sum&quot; rel=&quot;nofollow&quot;&gt;indefinite sum&lt;/a&gt;. Extracted from Wikipedia, the useful formulae for polynomials is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\Delta^{-1}_1 x^n &amp;amp;= \frac{B_{n+1}(x)}{n+1} + C \\&#xA;\Delta^{-1}_1 af(x) &amp;amp;= a \Delta^{-1}_1 f(x)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B&lt;sub&gt;n+1&lt;/sub&gt;(x) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_polynomials&quot; rel=&quot;nofollow&quot;&gt;Bernoulli polynomial&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &amp;Delta;&lt;sup&gt;-1&lt;/sup&gt; can be converted back to your &quot;int&quot; by substituting $x \mapsto x + 1$.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-06T12:06:41.143" LastActivityDate="2010-08-06T12:06:41.143" CommentCount="1" />
  <row Id="1713" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T12:01:14.540" Score="5" ViewCount="0" Body="&lt;p&gt;You seem to be reaching for the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;calculus of finite differences&lt;/a&gt;, once a well-known topic but rather&#xA;unfashionable these days. The answer to your question is yes: given a polynomial&#xA;$f(x)$ there is a polynomial $g(x)$ (of degree one greater than $f$) such that&#xA;$$f(x)=g(x)-g(x-1).$$&#xA;This polynomial $g$ (like the integral of $f$) is unique save for&#xA;its constant term.&#xA;Once one has $g$ then of course&#xA;$$f(a)+f(a+1)+\cdots+f(b)=g(b)-g(a-1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $f(x)=x^n$ is a monomial, the coefficients&#xA;of $g$ involve the endlessly fascinating &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-06T12:12:04.237" LastActivityDate="2010-08-06T12:12:04.237" />
  <row Id="1714" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T12:09:46.660" Score="2" ViewCount="0" Body="&lt;p&gt;As other answers have noted, you are about to discover the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;calculus of finite differences&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For practical calculations, here a most useful fact: the rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx} x^n = n x^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;corresponds to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_1 x(x-1)(x-2)\cdots(x-(n-1)) = n x(x-1)(x-2)\cdots(x-(n-2))$&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-08-06T12:09:46.660" CommentCount="2" />
  <row Id="1715" PostTypeId="2" ParentId="1691" CreationDate="2010-08-06T12:17:16.223" Score="1" ViewCount="0" Body="&lt;p&gt;muad,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right! This is what's happening on elliptic curves.  And, if you look at higher genus, as T.. said, there is no group law on the curve.  But something silly like there not being a group law on curves of high genus wasn't enough to stop the 19th century mathematicians, they discovered that it's the group law on an abelian variety related to the curve that matters, and the curve and the abelian variety happen to be the same for cubics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find all this worked out in any good book that talks about elliptic integrals or abelian integrals, and I'm personally a big fan of the exposition in Stillwell's &quot;Mathematics and its History&quot;&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-06T12:17:16.223" CommentCount="2" />
  <row Id="1716" PostTypeId="2" ParentId="709" CreationDate="2010-08-06T12:25:40.793" Score="1" ViewCount="0" Body="&lt;p&gt;I like Americo's answer using translation of y = ax^2. But to finish it, we should at least note that ax^2 has a maximum or minimum at &amp;lt;0,0&gt;, which is quite obvious, and that translation preserves order and therefore maxima and minima, which is pretty easy.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-06T12:25:40.793" />
  <row Id="1717" PostTypeId="2" ParentId="1701" CreationDate="2010-08-06T12:37:14.343" Score="3" ViewCount="0" Body="&lt;p&gt;I can also add that both Yoneda and Cayley are results which follow from the general philosophy of investigating algebraic structures by letting them act on themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If you let a group $G$ act on itself, you realize it as a subgroup of $\mathfrak{S}_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If you let a ring with unit act on itself, you realize it as a subring of $E$, where $E$ is the underlying additive group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Similarly if you let a $k$-algebra act on itself, you realize it as a subalgebra of $\mathcal{M}_n(k)$. In particular this gives the classical realization of $\mathbb{C}$ as a matrix algebra over $\mathbb{R}$ and of the quaternions as a matrix algebra over $\mathbb{C}$ or $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) You can let a Lie algebra act on itself, but unfortunately this action need not be faithful (Lie algebra don't have units...). So you only obtain the easy first step of Ado's theorem about embedding Lie algebras into mtraix algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) If you let a category $\mathcal{C}$ act on itself, you obtain an embedding into $Fun(\mathcal{C}^{op}, Set)$, which is the content of Yoneda's lemma.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-06T12:37:14.343" CommentCount="3" />
  <row Id="1718" PostTypeId="2" ParentId="1648" CreationDate="2010-08-06T12:42:56.610" Score="1" ViewCount="0" Body="&lt;p&gt;From the &lt;a href=&quot;http://stats.stackexchange.com/questions/1315&quot;&gt;comments on stats.stackexchange&lt;/a&gt;, it seems like you may not care too much about the distribution, but just a pretty curve to overlay on your graph. In which case, some kind of spline is your best bet. Use some kind of curves with asymptotes at y=0 for your upper- and lower-most segments, and whatever fits best in between.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do actually care about the underlying distribution:&lt;br&gt;&#xA;The first step would be to use whatever outside knowledge you have to characterize the distribution. For example:&lt;br&gt;&#xA;Network ping is a sum of independent wait times (the individual nodes in the network). This would suggest a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution&quot; rel=&quot;nofollow&quot;&gt;Gamma&lt;/a&gt;/Erlang distribution if each of these steps is identical, and a &lt;a href=&quot;http://www.math.bme.hu/~balazs/sumexp.pdf&quot; rel=&quot;nofollow&quot;&gt;more complex distribution&lt;/a&gt; if they are not.&lt;br&gt;&#xA;Ping is a measure of time until the computer at the other end responds to your request, the likelihood of which is proportional to the time elapsed. This would suggest a &lt;a href=&quot;http://en.wikipedia.org/wiki/Weibull_distribution&quot; rel=&quot;nofollow&quot;&gt;Weibull&lt;/a&gt; distribution.&lt;br&gt;&#xA;Ping time is the accumulation of a large number of factors that all have a multiplicative effect on the result. Then a &lt;a href=&quot;http://en.wikipedia.org/wiki/Log-normal_distribution&quot; rel=&quot;nofollow&quot;&gt;log-normal&lt;/a&gt; distribution would be best. &#xA;I don't know enough about networking to say anything about the accuracy of any of the above models, and it's also perfectly likely that ping time follows some other model which I haven't thought of. I just wanted to demonstrate the idea: that you should think about what factors contribute to the thing you are trying to model, and how they interact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, the distribution does not necessarily have to be a known one! In which case the above won't get you very far! In this case you might want to come up with your own empirical distribution, for which a variety of methods exist. The most common are to take your measurements as the distribution (as long as you have a sufficiently large number) or to take each of those data points and treat it as the center of some uniform/normal/other distribution, and sum everything with appropriate scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you know the type of distribution, you may also be able to use domain knowledge to estimate some of its parameters. For example, you might guess at the number of exponentials being summed based on the shape of the network. You can also use your measured mean and variance to form estimates of the distribution parameters. For example, if you thought that your distribution was a Gamma(3,θ), then you could use your measured variance to estimate θ=4.74182454 based on our known formula for variance of a Gamma Distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have your guess at a distribution, you will want to test its goodness of fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, the standard method would be to apply the one-sample &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov-Smirnov_test&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov-Smirnov test&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other potentially applicable tests are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer-von-Mises_test&quot; rel=&quot;nofollow&quot;&gt;Cramer-von-Mises&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Anderson-Darling_test&quot; rel=&quot;nofollow&quot;&gt;Anderson-Darling&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Pearson%27s_chi-square_test&quot; rel=&quot;nofollow&quot;&gt;chi-square&lt;/a&gt; tests.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is incomplete, I will add more later.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-10T02:40:44.470" LastActivityDate="2010-08-10T02:40:44.470" />
  <row Id="1719" PostTypeId="2" ParentId="1660" CreationDate="2010-08-06T12:46:36.323" Score="3" ViewCount="0" Body="&lt;p&gt;As Qiaochu points out the formula given by Gjergji Zaimi is certainly relevant.&#xA;It realizes $D(a_1,\ldots,a_n)/D(1,\ldots,n)$ as the&#xA;determinant of the $n$-by-$n$ matrix $M$ with $(i,j)$-entry&#xA;$${a_i\choose j-1}.$$&#xA;By row operations this equals the determinant of the $(n-1)$-by-$(n-1)$  matrix $N$&#xA;with $(i,j)$-entry&#xA;$${a_{i+1}\choose j}-{a_i\choose j}.$$&#xA;The $i$-th row of $n$ is the sum of vectors $v_{a_i},v_{a_i+1},\ldots,v_{a_{i+1}-1}$&#xA;where the $j$-th entry of $v_c$ is&#xA;$${c\choose j-1}.$$&#xA;By the multilinearity of the determinant as a function of its rows, $N$&#xA;is the sum the determinants with rows $v_{b_1},\ldots,v_{b_{n-1}}$&#xA;where $a_1\le b_i &amp;lt; a_2\le b_2 &amp;lt; a_3\le\cdots$. These tuples&#xA;$(b_1,\ldots,b_{n-1})$ are precisely the admissible penultimate rows&#xA;in GZ triangles with bottom row $(a_1,\ldots,a_n)$. Hence both sides of the&#xA;sought equality satisfy the same recurrence.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-06T12:46:36.323" CommentCount="1" />
  <row Id="1720" PostTypeId="2" ParentId="1526" CreationDate="2010-08-06T12:46:39.427" Score="0" ViewCount="0" Body="&lt;p&gt;Well 3186-2860 = 326.  That is very nearly a tenth of 3186, but 3186/9 = 354. But 326 is closer to 318.6 than to 354, so I'd go for 9/10 instead of 8/9.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-08-06T12:46:39.427" />
  <row Id="1721" PostTypeId="2" ParentId="903" CreationDate="2010-08-06T12:53:00.723" Score="2" ViewCount="0" Body="&lt;p&gt;I think Munkres' is a great book, though you can safely skip some sophisticated counter-examples in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Janick's &quot;Topology&quot; is more intuitive, less technical or just less narrowly focused on the specific subject of Topology -so it links to other parts of mathematics. But I've never dared to base my lectures on it, precisely because elementary Topology can be explained on its own, out of the blue, without assuming any previous knowledge at all, whereas mixing it with Differential Geometry, for instance, can be more confusing than useful for undergraduate students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, Janick's book has some masterpieces; for instance, the proof of Urysohn's lemma: the first time I read it I realised that there are people in this world who &lt;em&gt;actually&lt;/em&gt; understand maths, and others, well... we just repeat, more or less succesfully, what we have learnt (or so we hope). The same proof in Munkres' simply doesn't compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Dugundji, it's a great book, but nowadays it seems to me much too difficult for undergraduate students.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-06T12:53:00.723" CommunityOwnedDate="2010-08-06T12:53:00.723" />
  <row Id="1722" PostTypeId="2" ParentId="776" CreationDate="2010-08-06T13:00:43.033" Score="3" ViewCount="0" Body="&lt;p&gt;In fact it is a corollary of the theorem stating that the angle subtended at the centre of the circle (here 180 degrees) is double the angle subtended at the centre. If this is not satisfactory use the contrapositive instead: if angle ABC is not 90 degrees, the angle AOB cannot be 180 degrees, i.e. A,O,B are not collinear.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-06T13:00:43.033" />
  <row Id="1723" PostTypeId="2" ParentId="1665" CreationDate="2010-08-06T13:29:49.513" Score="3" ViewCount="0" Body="&lt;p&gt;The cases $p=2,3$ can be trivially checked. I'll assume $p\ge 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note $\left( \displaystyle\frac{p-1}{4} \right) \left( \displaystyle\frac{p+1}{2}\right)=\displaystyle\frac{p^2-1}{8t}=a$ (say). Then $a$ has only two prime divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is well known that $24|p^2-1$ for $p&gt;3$. Let $p^2-1=24t$. Then $a=3t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $3$ is a prime divisor of $a$, implying $3$ equals one of $\displaystyle\frac{p-1}{4},\frac{p+1}{2}$. Direct substitution shows that $p=13$ is the only solution.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-06T13:29:49.513" />
  <row Id="1724" PostTypeId="2" ParentId="1555" CreationDate="2010-08-06T13:51:27.147" Score="0" ViewCount="0" Body="&lt;p&gt;I will try to summarize here what I've found so far on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the form being integrated is $\frac{\mathrm{d}x}{\sqrt{1 - x^2}} =: \omega$ let $y = \sqrt{1 - x^2}$, we can also write this as $x^2 + y^2 - 1 = 0$ which makes it more obvious that this is a unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N$ be the point $(1,0)$ and we can define a group structure on the curve as follows. To add $A$ and $B$, fire a ray from $N$ parallel to $AB$ and pick the point it intersects the curve as $A \oplus B$. Symbolically, this means $A \oplus B = N - k(B - A)$ for some nonzero $k$, since $A \oplus B$ lies on the circle we can expand it into the equation of the curve to solve for $k$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;(- k (x_b - x_a))^2 + (1 - k (y_b - y_a))^2 - 1 &amp;amp;= 0 \\&#xA;k ((x_b - x_a)^2 + (y_b - y_a)^2) - 2 (y_b - y_a) &amp;amp;= 0 \\&#xA;\frac{2 (y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} &amp;amp;= k&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence $x_{a \oplus b} = - 2 \frac{(x_b - x_a)(y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} = x_a \sqrt{1-x_b^2} + x_b \sqrt{1-x_a^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I think by &lt;a href=&quot;http://math.stackexchange.com/questions/1691/trying-to-piece-together-an-integral-addition-theorem&quot;&gt;the theorem&lt;/a&gt;, if &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\omega$ is invariant along the curve&lt;/li&gt;&#xA;&lt;li&gt;$\oplus$ is the group law on points of the curve&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;we can conclude now that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^a \omega + \int_0^b \omega = \int_0^{a \oplus b} \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, the previous equation holds - but for this post to count as a proof the theorem is needed. Furthermore I believe this same idea works for any conic section and also for cubic curves and perhaps not any others curves.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-06T13:51:27.147" CommentCount="1" />
  <row Id="1725" PostTypeId="2" ParentId="776" CreationDate="2010-08-06T14:15:55.930" Score="3" ViewCount="0" Body="&lt;p&gt;Really short vector proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Center the circle at the origin, and scale to have radius 1. Let the vertex of the right triangle be at vector $v$, and let the diameter be the segment from the vector $w$ to $-w$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$(v-w) \cdot (v-(-w)) = (v-w) \cdot (v+w) = (v \cdot v) - (w \cdot w) = 1 - 1 = 0$, so the angle formed by $vw$ and $v(-w)$ is a right angle.&lt;/p&gt;&#xA;" OwnerUserId="831" LastActivityDate="2010-08-06T14:15:55.930" CommentCount="1" />
  <row Id="1726" PostTypeId="2" ParentId="1626" CreationDate="2010-08-06T14:44:42.393" Score="1" ViewCount="0" Body="&lt;p&gt;By little Fermat: $\; 6,10|120 \Rightarrow 3^{120} \equiv 1 \pmod{7, 11} \Rightarrow 3^{123} \equiv 3^3 \pmod{77}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also these &lt;strong&gt;Fermat-Euler-Carmichael&lt;/strong&gt; generalizations of little Fermat-Euler from my post &lt;a href=&quot;http://google.com/groups?selm=y8zk55tqcie.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 1&lt;/strong&gt;.  For natural numbers $ a,e,n $ with $ e,n&gt;1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\;\; n|a^e-a$ for all $a \iff n$ is squarefree and prime $ p|n \Rightarrow p-1|e-1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; The special case  $e = n$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_number#Korselt.27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Korselt's criterion&lt;/a&gt; for Carmichael numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 2&lt;/strong&gt;.  For natural numbers $ a,e,n $ with $ e,n&gt;1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\;\; n|a^e-1$ for all $a$ coprime to $n \iff  p$ prime, $\; p^k||n \Rightarrow E(p^k)|e $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\;\;\; E(p^k) = \phi(p^k) $ for odd primes $p$, or $p=2,\; k \le 2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and   $\quad E(2^k) = 2^{k-2} $ for $ k&gt;2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter exception is due to $\mathbb Z/2^k$ having multiplicative group $C(2) \times C(2^{k-2})$ for $k&gt;2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the least such exponent $e$ is given by $\;\lambda(n)\; = \;{\rm lcm}\;\{E(\;{p_i}^{k_i})\}\;$ where $\; n = \prod {p_i}^{k_i}\;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda(n)$ is called the (universal) &lt;a href=&quot;http://planetmath.org/encyclopedia/Exponent.html&quot; rel=&quot;nofollow&quot;&gt;exponent&lt;/a&gt; of the group $\mathbb Z/n^*,\;$ a.k.a. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carmichael_function&quot; rel=&quot;nofollow&quot;&gt;Carmichael function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my post &lt;a href=&quot;http://google.com/groups?selm=y8zk55tqcie.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for proofs and further discussion.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-15T21:28:05.590" LastActivityDate="2010-08-15T21:28:05.590" />
  <row Id="1727" PostTypeId="1" AcceptedAnswerId="1729" CreationDate="2010-08-06T14:58:42.553" Score="4" ViewCount="175" Body="&lt;p&gt;Related to &lt;a href=&quot;http://math.stackexchange.com/questions/1659/locating-the-quadrant-containing-a-point-on-an-n-sphere&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm using the &lt;a href=&quot;http://mathworld.wolfram.com/Hypersphere.html&quot; rel=&quot;nofollow&quot;&gt;geometer definition&lt;/a&gt; of an $n$-sphere of radius $r$, i.e.$ \{ x \in \mathbb{R}^n : \|x\|_2 = r  \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have an $n$-sphere centered at $\bf 0$ in $\mathbb{R}^n$ with radius $r$ which has been divided into $2^k$ orthants by $k$ axis-aligned hyperplanes (note, $k \le n$) in $\mathbb{R}^{n-1}$ passing through $\bf 0$. For e.g., if $k=1$, we have $2$ $n$-hemispheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the &lt;strong&gt;question&lt;/strong&gt;: how do I find the center of mass of such an orthant? Or (since I'm still working on it), how would you find the center of mass of an $n$-hemisphere?&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2010-08-06T16:27:32.287" LastActivityDate="2010-08-06T19:21:31.747" Title="Center of mass of an $n$-hemisphere" Tags="&lt;geometry&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="1728" PostTypeId="2" ParentId="1526" CreationDate="2010-08-06T15:01:58.817" Score="1" ViewCount="0" Body="&lt;p&gt;This isn't so much a math answer as a test-taking answer: &lt;strong&gt;You don't have to compute the fraction, you just have to determine which is the right answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2860/3186&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is a multiple choice question, with answers 6/7, 7/8, 8/9 and 9/10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you reduce the fraction to get a/b, then b must be a divisor of 3186. This allows you to immediately eliminate some choices. It can't be 9/10 because 10 doesn't divide 3186. You can quickly check that 7 and 8 don't divide 3186, but 9 does, so the only one of the choices that &lt;em&gt;has a shot&lt;/em&gt; at being the correct answer is 8/9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, &lt;strong&gt;none of those answer is correct&lt;/strong&gt;; it's not 8/9 either. The furthest you can reduce the fraction is 1430/1593, so you must have an error in your question. Either the fraction is wrong or you're supposed to find the &lt;em&gt;best approximation&lt;/em&gt; rather than the actual value.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2010-08-06T15:01:58.817" />
  <row Id="1729" PostTypeId="2" ParentId="1727" CreationDate="2010-08-06T15:14:15.180" Score="2" ViewCount="0" Body="&lt;p&gt;One finds the centre of mass for these bodies just as one&#xA;does for any body, by integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $x_i$-coordinate of the centre of mass of a region $A$ in $\mathbb{R}^n$&#xA;is&#xA;$$\frac{\int_A x_i\ dx_1\cdots dx_n}{\int_A\ dx_1\cdots dx_n}.$$&#xA;Here the region might as well be that between the planes $x_n=a$ and $x_n=b$&#xA;in the sphere of  radius $r$ centred at the origin. Using the symmetry&#xA;of $A$ this ratio of integrals equals&#xA;$$\frac{\gamma_{n-1}\int_a^b x_n(r^2-x_n^2)^{(n-1)/2} dx_n}&#xA;{\gamma_{n-1}\int_a^b (r^2-x_n^2)^{(n-1)/2} dx_n}$$&#xA;(in the $x_n$ direction) where $\gamma_{n-1}$ is the volume of the $(n-1)$-dimensional&#xA;unit ball (and obligingly cancels). These integrals can be attacked by&#xA;trig substitutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited&lt;/strong&gt;&#xA;It's now clear that my original interpretation of your question&#xA;was wrong. However it's still not clear whether your centre of mass&#xA;is for a solid orthant or its curved surface. In any case if your orthant&#xA;is defined by the conditions $x_1,\ldots,x_k\ge0$ then its centre&#xA;of mass has the form $(a,\ldots,a,0\ldots,0)$ where $a$ depends on $r$&#xA;and $n$ but &lt;strong&gt;not&lt;/strong&gt; on $k$. One sees this from the symmetry of the problem.&#xA;Thus the problem reduces to the hemispheric case. In the solid case the&#xA;answer is&#xA;$$a=\frac{\int_0^r x(r^2-x^2)^{(n-1)/2} dx}&#xA;{\int_0^r (r^2-x^2)^{(n-1)/2} dx}$$&#xA;while in the &quot;shell&quot; case it is&#xA;$$a=\frac{\int_0^r x(r^2-x^2)^{(n-3)/2} dx}&#xA;{\int_0^r (r^2-x^2)^{(n-3)/2} dx}$$&#xA;(if I've done my sums right).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-06T19:21:31.747" LastActivityDate="2010-08-06T19:21:31.747" CommentCount="3" />
  <row Id="1730" PostTypeId="1" CreationDate="2010-08-06T15:30:44.160" Score="12" ViewCount="659" Body="&lt;p&gt;I'm teaching a College Algebra class in the upcoming semester, and only a small portion of the students will be moving on to further mathematics. The class is built around functions, so I need to start with the definition of one, yet many &quot;official&quot; definitions I have found too convoluted (or poorly written) for general use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one of the better &quot;light&quot; definitions I've found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A function is a relationship which&#xA;  assigns to each input (or domain)&#xA;  value, a unique output (or range)&#xA;  value.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sounds simple enough on the surface, but putting myself &quot;in the head&quot; of a student makes me pause. It's almost &lt;em&gt;too&lt;/em&gt; compact with potentially ambiguous words for the student (relationship? assigns? unique?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my personal best attempt, in 3 parts. Each part of the definition would include a discussion and examples before moving to the next part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A relation is a set of links between&#xA;  two sets.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Each link of a relation has an input&#xA;  (in the starting set) and an output&#xA;  (in the ending set).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A function is a relation where every&#xA;  input has one and only one possible&#xA;  output.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm somewhat happier here: starting with a relation gives some natural examples and makes it easier to impart the special importance of a function (which is &quot;better behaved&quot; than a relation in practical circusmtances).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm also still uneasy (&quot;links&quot;? A set between sets?) and I was wanting to see if anyone had a better solution.&lt;/p&gt;&#xA;" OwnerUserId="295" LastActivityDate="2010-12-22T07:13:24.917" Title="How do you define functions for non-mathematicians?" Tags="&lt;algebra-precalculus&gt;&lt;teaching&gt;&lt;definition&gt;&lt;functions&gt;" AnswerCount="9" FavoriteCount="4" />
  <row Id="1731" PostTypeId="2" ParentId="1706" CreationDate="2010-08-06T15:33:47.170" Score="10" ViewCount="0" Body="&lt;p&gt;It depends on the choice of the person using the notation: sometimes it does, sometimes it doesn't. It is just a variant of the situation with $\mathbb N$, which half the world (the mistaken half!) considers to include zero.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-06T15:33:47.170" />
  <row Id="1732" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T15:40:20.550" Score="8" ViewCount="0" Body="&lt;p&gt;The way you've restated the definition is fairly common in contemporary high school books in the U.S. (perhaps changing &quot;links between two sets&quot; to &quot;ordered pairs&quot;).  What I've seen a lot of in middle school and earlier algebra settings is the idea of a &quot;function machine.&quot;  The function machine graphic below is from &lt;a href=&quot;http://etc.usf.edu/clipart/41800/41849/function_41849.htm&quot; rel=&quot;nofollow&quot;&gt;FCIT (&amp;copy;2009)&lt;/a&gt;, but &lt;a href=&quot;http://www.google.com/images?q=function+machine&quot; rel=&quot;nofollow&quot;&gt;a google image search for function machine&lt;/a&gt; will show you many different ways the concept can be visualized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BgduF.gif&quot; alt=&quot;function machine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this probably pushes the idea that a function has a formula, I'd claim that &quot;rule&quot; could be as general as a specific listing of which inputs map to which outputs, as in your definition.  To me, the prevalence of this machine metaphor in middle school contexts suggests that it works well for students who do not necessarily yet have a sense of symbolic algebra.  I've seen function machines used as low as 3rd grade.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-29T04:16:05.283" LastActivityDate="2010-09-29T04:16:05.283" CommentCount="4" />
  <row Id="1733" PostTypeId="2" ParentId="1665" CreationDate="2010-08-06T15:40:28.367" Score="5" ViewCount="0" Body="&lt;p&gt;The primes have product (p^3 - p)/8 which is divisible by 3. So one prime = 3. &#xA;The rest is trivial.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-06T15:40:28.367" />
  <row Id="1734" PostTypeId="2" ParentId="1694" CreationDate="2010-08-06T15:43:58.553" Score="1" ViewCount="0" Body="&lt;p&gt;I suggest that everybody from now on uses ln, lg, and lb respectively for the natural, decimal, and binary logarithmic functions, reserving log for when the base is displayed explicitly or is some fixed arbitrary number if the scale doesn't matter. I do (with explanation if necessary at the first instance) . What could be simpler?&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-08-06T15:43:58.553" CommentCount="5" />
  <row Id="1735" PostTypeId="1" CreationDate="2010-08-06T15:44:16.930" Score="2" ViewCount="124" Body="&lt;p&gt;Let $f$ be a function defined on an open interval $(a,b)$ which has continuous derivatives of order $1,2, \dots , n-1$, and there is a point $c \in (a,b)$ such that $f(c), f^{{1}}(c), \dots , f^{(n-1)}(c)$ are all $0$.  However $f^{n}(c)$ is not $0$ and we can assume that $f^{n}(x) &gt; 0$ for $x \in (a,b)$, however &lt;strong&gt;no assumptions about the continuity of&lt;/strong&gt; $f^{(n)}(x)$ &lt;strong&gt;are made&lt;/strong&gt;. &#xA;In this case is it true that $\lim_{x\to c}\frac{f(x)}{(x-c)^n} = \frac{f^{(n)}(c)}{n!}$?&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-08-06T15:58:28.373" Title="A particular case of L'Hospital" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1736" PostTypeId="2" ParentId="1735" CreationDate="2010-08-06T15:53:02.417" Score="2" ViewCount="0" Body="&lt;p&gt;L'Hospital's rule does not depend on the continuity of the derivative.  Under your conditions, $\lim_{x\to c}f^{(k)}(x)=0$ for $k\le n-1$, so L'Hospital's rule can be applied to give:&#xA;$$\lim_{x\to c}\frac{f(x)}{(x-c)^n}&#xA;=\lim_{x\to c}\frac{f'(x)}{n(x-c)^{n-1}}&#xA;=\cdots&#xA;=\lim_{x\to c}\frac{f^{(n-1)}(x)}{n!(x-c)}&#xA;=\lim_{x\to c}\frac{f^{(n)}(x)}{n!}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to assert that $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ requires that $f^{(n)}$ be continuous at c. (n! is independent of x, so $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ and $\lim_{x\to c}f^{(n)}(x)=f^{(n)}(c)$ are equivalent and the latter is the definition of continuity for $f^{(n)}$ at c.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-06T15:58:28.373" LastActivityDate="2010-08-06T15:58:28.373" CommentCount="0" />
  <row Id="1737" PostTypeId="2" ParentId="1706" CreationDate="2010-08-06T16:11:30.543" Score="3" ViewCount="0" Body="&lt;p&gt;As a rule of thumb most mathematicians of the anglo saxon school consider that positive numbers (be it $\mathbb{N}$ or $\mathbb{R}^{+}$) do not include while the latin (French, Italian) and russian schools make a difference between positive and strictly positive and between negative and strictly negative. This means by the way that $0$ is the intersection of positive and negative numbers. One needs to know upfront the convention.&lt;/p&gt;&#xA;" OwnerUserId="441" LastActivityDate="2010-08-06T16:11:30.543" CommentCount="6" />
  <row Id="1738" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T16:14:41.257" Score="3" ViewCount="0" Body="&lt;p&gt;Isaac's answer is almost exactly the first definition that I give. But what comes after is similar to what you are describing, I discuss each ambiguous term in the definition at great length, replacing the word with synonyms. Hopefully this lets the students take in the slight abstraction of the definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I do the defining and explaining of words I do about 9 examples. Three from each of three classes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;little point-set diagrams, i.e. ovals with points inside them and lines going between them.&#xA;algebraically with f(x)=&#xA;graphs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two from each class of example are functions and I point out all the parts of the definition and what they correspond to. The third example in each class is a non-example, i.e. not a function. I point out where the issue is, thus showing that one point has two arrows coming out of it, one number can be plugged in to get two, and the vertical line test respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find that this is very successful. &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-06T16:14:41.257" />
  <row Id="1739" PostTypeId="2" ParentId="1710" CreationDate="2010-08-06T16:17:25.780" Score="4" ViewCount="0" Body="&lt;p&gt;For any particular polynomial there is an easier way to do indefinite summation than using the Bernoulli numbers, going off of Greg Graviton's answer.  Here we'll use the forward difference $\Delta f(x) = f(x+1) - f(x)$.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \Delta {x \choose n} = {x \choose n-1}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that we can perform a &quot;Taylor expansion&quot; on any polynomial to write it in the form $f(x) = \sum a_n {x \choose n}$ by evaluating the finite differences $\Delta^n f(0)$ at zero.  For any particular polynomial $f$ it is easy to write these finite differences down by constructing a table.  In general, the formula is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a_n = \Delta^n f(0) = \sum_{k=0}^{n} (-1)^{n-k} {n \choose k} f(k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as one can readily prove by writing $\Delta = S - I$ where $S$ is the shift operator $S f(x) = f(x+1)$ and $I$ is the identity operator $I f(x) = f(x)$.  Then the indefinite sum of $f$ is just $\sum a_n {x \choose n+1}$.  This is the easiest way I know how to do such computations by hand, and it also leads to a fairly easy method for polynomial interpolation given the values of a polynomial at consecutive integers.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T16:17:25.780" CommentCount="3" />
  <row Id="1740" PostTypeId="2" ParentId="1706" CreationDate="2010-08-06T17:08:32.327" Score="10" ViewCount="0" Body="&lt;p&gt;You will often find $\mathbf R^+$ for the positive reals, and $\mathbf R^+_0$ for the positive reals and the zero.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-06T17:08:32.327" />
  <row Id="1741" PostTypeId="1" CreationDate="2010-08-06T18:09:59.140" Score="-1" ViewCount="196" Body="&lt;p&gt;Classes are larger than sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's larger than classes ?&lt;/p&gt;&#xA;" OwnerUserId="882" LastActivityDate="2010-08-08T20:42:53.020" Title="Sets. Classes. ... ?" Tags="&lt;set-theory&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="1742" PostTypeId="2" ParentId="514" CreationDate="2010-08-06T18:16:48.003" Score="2" ViewCount="0" Body="&lt;p&gt;Another class of examples arise from diophantine equations with huge minimal solutions. Thus the conjecture that such an equation is unsolvable in integers has only huge counterexamples. Well-known examples arise from Pell equations, e.g. the smallest solution to the classic &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedes%27_cattle_problem&quot; rel=&quot;nofollow&quot;&gt;Archimedes Cattle problem&lt;/a&gt; has &lt;strong&gt;206545 decimal digits&lt;/strong&gt;, namely 77602714 ... 55081800.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-06T18:16:48.003" />
  <row Id="1743" PostTypeId="2" ParentId="1741" CreationDate="2010-08-06T18:20:53.190" Score="0" ViewCount="0" Body="&lt;p&gt;A class is a collection of things. Some classes are sets (in some sense, only &quot;smaller&quot; classes can be sets). Since &lt;em&gt;any&lt;/em&gt; collection is a class, there can't be anything larger.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-06T18:20:53.190" CommentCount="1" />
  <row Id="1744" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T18:36:32.957" Score="23" ViewCount="0" Body="&lt;p&gt;For fun, I like to liven-up the &quot;black box&quot;/machine view of a function by putting a monkey into the box. (I got pretty good at chalkboard-sketching a monkey that looked a little bit like &lt;a href=&quot;http://pbskids.org/curiousgeorge/&quot; rel=&quot;nofollow&quot;&gt;Curious George&lt;/a&gt;, but with a tail.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give the Function Monkey an input and he'll cheerfully give you an output. The Function Monkey is smart enough to read and follow rules, and make computations, but he's not qualified to make decisions: his rules &lt;em&gt;must&lt;/em&gt; provide for exactly one output for a given input. &lt;em&gt;(Never let a Monkey choose!)&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can continue the metaphor by discussing the monkey's &quot;domain&quot; as the inputs he understands (what he can control); giving him an input outside his domain just confuses and frightens him ... or, depending upon the nature of the audience, kills him. &lt;em&gt;(What? You gave the Reciprocal Monkey a Zero? You killed the Function Monkey!)&lt;/em&gt; Of course, it's probably more appropriate to say that the Function Monkey simply &lt;em&gt;ignores&lt;/em&gt; such inputs, but students seem to like the drama. (As warnings go, &lt;em&gt;&quot;Don't kill the Function Monkey!&quot;&lt;/em&gt; gets more attention than &lt;em&gt;&quot;Don't bore the Function Monkey!&quot;&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Function Monkey comes in handy later when you start graphing functions: imagine that the x-axis is covered with coconuts (one coconut per &quot;x&quot; value). The Function Monkey strolls along the axis, picks up a &quot;x&quot; coconut, computes the associated &quot;y&quot; value (because that's what he &lt;em&gt;does&lt;/em&gt;), and then throws the coconut up (or down) the appropriate height above (or below) the axis, where it magically &lt;em&gt;sticks&lt;/em&gt; (or &lt;em&gt;hovers&lt;/em&gt; or whatever). So, if you ever want to plot a function, just &quot;Be a Function Monkey and throw some coconuts around&quot;. (Warning: Students may insist that that's not a &lt;em&gt;coconut&lt;/em&gt; the Monkey is throwing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further on, you can make the case that we're &lt;em&gt;smarter&lt;/em&gt; than monkeys (at least, we should strive to be): We don't always have to mindlessly plot points to know what the graph of an equation looks like; we can sometimes anticipate the outcome by studying the equation. This motivates manipulating an equation to tease out clues about the shape of its graph, explaining, for instance, our interest in the slope-intercept form of a line equation (and the almost-never-taught intercept-intercept form, which I personally like a lot), the special forms of conic section equations (which aren't all &lt;em&gt;functions&lt;/em&gt;, of course), and all that stuff related to translations and scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parametric equations can be presented as a way to let the Function Monkey plot elaborate curves ... both in the plane and in space (and beyond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, I find that the Function Monkey can make the course material more engaging &lt;em&gt;without dumbing it down&lt;/em&gt;; he provides a fun way to interpret the definitions and behaviors of functions, not a way to avoid them. Now, is the Function Monkey too cutesy for a College Algebra class? My high school students loved him, even at the Calculus level. One former student told me that he would often invoke the Function Monkey when tutoring his college peers. If it's clear to the students that the instructor isn't trying to patronize them, the Function Monkey may prove quite helpful.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-06T18:36:32.957" CommentCount="3" />
  <row Id="1745" PostTypeId="1" AcceptedAnswerId="1757" CreationDate="2010-08-06T19:02:27.273" Score="2" ViewCount="104" Body="&lt;p&gt;I needed (for my research) to solve a Diophantine equation, in particular,&#xA;$$ 2 a + 3 b + 4 c + 5 d = 12 .$$&#xA;And I could easily solve it&#xA;(for example, on solution is $a=2, b=1, c=0, d=1$).&#xA;But this made me wonder if such equations, with their coefficients increasing sequences of&#xA;natural numbers, are a special case of Diophantine equations that are &lt;em&gt;always&lt;/em&gt; explicitly solvable, &#xA;despite the negative solution to Hilbert's 10th problem.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T03:30:34.870" LastActivityDate="2010-09-21T03:30:34.870" Title="Diophantine equations whose coefficients are increasing sequences of integers" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="1746" PostTypeId="2" ParentId="1382" CreationDate="2010-08-06T19:09:37.260" Score="2" ViewCount="0" Body="&lt;p&gt;(Sketch of a bijective solution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that binomial coefficients count number of walks with steps (+1,+1) and (+1,-1) from the origin to different points (e.g. the number of walks to the point (2n,0) is $\binom{2n}{n}$). Consider the following involution on the set of all such walks: if the path intersects with the line y=l-1 or the line y=-1, reflect its part starting from the first intersection point (w.r.t. corresponding line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This involution almost gives a bijection between P&lt;sub&gt;n&lt;/sub&gt;(r mod l) and P&lt;sub&gt;n&lt;/sub&gt;(-r-2 mod l) (and moving the strip we can get other correspondences of this kind). But it has some fixed points — namely, paths that lie inside the strip 0&amp;le;y&amp;le;l-2 (aka walks on the path graph of length l-1 mentioned by &lt;a href=&quot;http://math.stackexchange.com/users/232/qiaochu-yuan&quot;&gt;Qiaochu Yuan&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to answer original question one only needs to recall that numbers of such paths for l=5 are exactly Fibonacci numbers.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-07T07:24:39.447" LastActivityDate="2010-08-07T07:24:39.447" CommentCount="2" />
  <row Id="1747" PostTypeId="1" CreationDate="2010-08-06T19:12:23.137" Score="16" ViewCount="810" Body="&lt;p&gt;Not a math student, so forgive me if the question seems trivial or if I pose it &quot;wrong&quot;. Here goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I'm flipping a coin a &lt;em&gt;n&lt;/em&gt; times. I am not sure if it's a &quot;fair&quot; coin, meaning I am not sure if it will come up heads and tails each with a propability of exactly 0.5. Now, if after &lt;em&gt;n&lt;/em&gt; throws it has come up heads exactly as many times as it has come up tails, then obviously there's nothing to indicate that the coin is not fair. But my intuition tells me that it would be improbable even for a completely fair coin to come up with heads and tails an exact even number of times given a large amount of tosses. My question is this: How &quot;off&quot; should the result be for it to be probable that the coin is not fair? IOW, how many more tosses should come up heads rather than tails in a series of &lt;em&gt;n&lt;/em&gt; throws before I should assume the coin is weighted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone mentioned Pearson's chi-square test but then for some reason deleted their answer. Can someone confirm if that is indeed the right place to look for the answer?&lt;/p&gt;&#xA;" OwnerUserId="886" LastEditorUserId="886" LastEditorDisplayName="" LastEditDate="2010-08-07T09:07:19.153" LastActivityDate="2010-08-28T08:48:11.187" Title="How to determine if coin comes up heads more often than tails?" Tags="&lt;probability&gt;" AnswerCount="6" CommentCount="10" FavoriteCount="3" />
  <row Id="1748" PostTypeId="1" CreationDate="2010-08-06T19:40:32.847" Score="2" ViewCount="344" Body="&lt;p&gt;What are the best algebraic geometry textbooks for undergraduate students?&lt;/p&gt;&#xA;" OwnerUserId="888" LastEditorUserId="888" LastEditorDisplayName="" LastEditDate="2010-08-06T19:59:10.517" LastActivityDate="2010-09-01T03:48:41.877" CommunityOwnedDate="2010-08-06T19:48:54.623" Title="(undergraduate) Algebraic Geometry Textbook Recomendations" Tags="&lt;algebraic-geometry&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="5" />
  <row Id="1749" PostTypeId="2" ParentId="1741" CreationDate="2010-08-06T19:41:44.047" Score="2" ViewCount="0" Body="&lt;p&gt;Classes are defined differently in systems, but essentially--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sets are defined from bottom-up (adding elements starting from nothing) while classes are defined from top-down (starting with the universal class &lt;em&gt;V&lt;/em&gt; and then deciding from there which classes are sets). So it doesn't make sense to talk about things larger than the universal class, and if notions of comparing sizes are possible, you're simply dealing with classes that are also sets. (&lt;em&gt;V&lt;/em&gt; itself is not a set.)&lt;/p&gt;&#xA;" OwnerUserId="295" LastActivityDate="2010-08-06T19:41:44.047" CommentCount="1" />
  <row Id="1750" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T19:49:21.067" Score="6" ViewCount="0" Body="&lt;p&gt;See the MathOverflow question about exactly this: &lt;a href=&quot;http://mathoverflow.net/questions/2446/best-algebraic-geometry-text-book-other-than-hartshorne&quot; rel=&quot;nofollow&quot;&gt;Best Algebraic Geometry text book?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, from what little I've seen, &lt;a href=&quot;http://www.cs.amherst.edu/~dac/iva.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Ideals, Varieties and Algorithms&lt;/em&gt;&lt;/a&gt; by Cox, Litle and O'Shea is a very nice book for undergraduates. It requires little background, and has a computational perspective that's especially good for those with concrete or computational interests. I don't know how good it is as a preparation for further &quot;serious&quot; study in algebraic geometry, but it's certainly engaging.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-06T20:09:47.640" LastActivityDate="2010-08-06T20:09:47.640" CommunityOwnedDate="2010-08-06T19:49:21.067" CommentCount="4" />
  <row Id="1752" PostTypeId="1" AcceptedAnswerId="1754" CreationDate="2010-08-06T19:57:01.113" Score="0" ViewCount="89" Body="&lt;p&gt;It seems like an extremely easy question, but I can't solve it. All I know is that it is a $\frac{1}{2^n}$ chance that all tosses will result in heads and $\sum_{k=0}^{n} f(k, n)$ is equal to 1 given the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though school has just started for me, this is not a homework question.&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-06T20:03:04.660" Title="What is the probability that exactly $i$ heads will result from $n$ coin tosses?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1754" PostTypeId="2" ParentId="1752" CreationDate="2010-08-06T20:03:04.660" Score="4" ViewCount="0" Body="&lt;p&gt;This probability is equal to $f(i, n) = \binom{n}{i}2^{-n}$, where $\binom{n}{i} = \frac{n!}{i! (n-i)!}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow&quot;&gt;binomial coefficient&lt;/a&gt;. This is true because there are $\binom{n}{i}$ ways to choose which coins will be heads up and each such combinaton of $i$ heads up coins has the same probability $2^{-n}$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-06T20:03:04.660" CommentCount="1" />
  <row Id="1755" PostTypeId="2" ParentId="1747" CreationDate="2010-08-06T20:12:48.480" Score="3" ViewCount="0" Body="&lt;p&gt;Given your prefatory comment, I'm going to avoid talking about the normal curve and the associated variables and use as much straight probability as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do a side problem first. If on a A-D multiple choice test you guess randomly, what's the probability you get 8 out of 10 questions right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each problem you have a 25% (.25) chance of getting right and a 75% (.75) chance of getting wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to first choose which eight problems you get right. That can be done in &lt;a href=&quot;http://www.wolframalpha.com/input/?i=10+choose+8&quot; rel=&quot;nofollow&quot;&gt;10 choose 8&lt;/a&gt; ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want .25 to happen eight times [$(.25)^8$] and .75 to happen twice [$(.75)^2$]. This needs to be multiplied by the possible number of ways to arrange the eight correct problems, hence your odds of getting 8 out of 10 right is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${10 \choose{8}}(.25)^8(.75)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so let's say you throw a coin 3000 times. What's the probability that it comes up heads only 300 times? By the same logic as the above problem that would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${3000 \choose{300}}(.5)^{300}(.5)^{2700}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or a rather unlikely 6.92379... x 10^-482.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given throwing the coin n times, the probability it comes up heads x times is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${n \choose{x}}(.5)^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or if you want to ask the probability it comes up heads x times or less&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{x}{{n \choose{i}}(.5)^n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so all you have to do is decide now how unlikely are you willing to accept?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This was a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_probability&quot; rel=&quot;nofollow&quot;&gt;Binomial Probability&lt;/a&gt; if you want to read more and all the fancier methods involving an integral under the normal curve and whatnot start with this concept.)&lt;/p&gt;&#xA;" OwnerUserId="295" LastEditorUserId="295" LastEditorDisplayName="" LastEditDate="2010-08-06T20:18:58.093" LastActivityDate="2010-08-06T20:18:58.093" />
  <row Id="1756" PostTypeId="2" ParentId="659" CreationDate="2010-08-06T20:21:01.613" Score="0" ViewCount="0" Body="&lt;p&gt;You can discuss duals in a monoidal category (which may not be symmetric).&#xA;This has been mentioned by Akhil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ and $W$ be objects (in your monoidal category) with $K$ the identity for tensor product. Then you require morphisms $K\rightarrow V\otimes W$ and $W\otimes V\rightarrow K$&#xA;which satisfy the zig-zag identities (so-called because this becomes clear if you draw string diagrams). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally the zig-zag identities are &#xA;$$V=K\otimes V\rightarrow V\otimes W\otimes V\rightarrow V\otimes K=V$$&#xA;is the identity map and&#xA;$$W=W\otimes K\rightarrow W\otimes V\otimes W\rightarrow K\otimes W=k$$&#xA;is the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to $Hom(W\otimes X,Y)=Hom(X,V\otimes Y)$ and $Hom(X\otimes V,Y)=Hom(V,Y\otimes W)$ (both natural in $X$ and in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you say $V$ is left/right dual to $W$ and $W$ is right/left dual to $V$&#xA;(I can never remember which). Then we can define $V$ to be dual to $W$ if it is left dual and right dual.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-06T20:21:01.613" />
  <row Id="1757" PostTypeId="2" ParentId="1745" CreationDate="2010-08-06T20:38:46.367" Score="7" ViewCount="0" Body="&lt;p&gt;Linear Diophantine equations are always &lt;strike&gt;solvable&lt;/strike&gt; decidable (in linear time).  If the coefficients are $a_1, a_2, ... a_n$ then the numbers that can appear on the RHS are precisely the multiples of $\text{gcd}(a_1, ... a_n)$, and one can find solutions using the extended Euclidean algorithm.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-06T21:15:48.970" LastActivityDate="2010-08-06T21:15:48.970" CommentCount="3" />
  <row Id="1758" PostTypeId="1" AcceptedAnswerId="1792" CreationDate="2010-08-06T20:40:58.017" Score="5" ViewCount="326" Body="&lt;p&gt;Can you divide a disk into 7 pieces of equal area, with 3 line segments? (You can surely divide it into 7 pieces, but could those have equal areas?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question was left unanswered at another forum. I can see with some visual arguments that the answer should be &lt;em&gt;no&lt;/em&gt;, but I couldn't find a nice way to write it down.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-06T21:01:39.757" LastActivityDate="2010-08-09T21:23:41.117" Title="Dividing a disk into 7 equal pieces with 3 line segments" Tags="&lt;euclidean-geometry&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="1759" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T20:51:34.690" Score="2" ViewCount="0" Body="&lt;p&gt;Everyone is linking to that MO question, so allow me to link to another MO &lt;a href=&quot;http://mathoverflow.net/questions/14371/looking-for-an-introductory-textbook-on-algebraic-geometry-for-an-undergraduate-l&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt; which is more relevant to this specific request.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I should mention that it partially depends on what you are looking for. Should this be for self study with little help from professors? Then your best bet would be to follow Ravi Vakil's &lt;a href=&quot;http://math216.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;course next semester&lt;/a&gt; or use his old notes &lt;a href=&quot;http://math.stanford.edu/~vakil/0708-216/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you intend on having a professor over see the self study then I think that Ueno's books: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821808621&quot; rel=&quot;nofollow&quot;&gt;Alg Geom 1&lt;/a&gt; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821813579&quot; rel=&quot;nofollow&quot;&gt;and 2&lt;/a&gt;, which I think are great for self study.(I should mention that his lower level &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821811444&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; is not nearly as good and I would NOT recommend it.) Another strong candidate for led self study would be &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/144193099X&quot; rel=&quot;nofollow&quot;&gt;Harris' nice intro&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are going to offer a course or take a course then I think &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1441928073&quot; rel=&quot;nofollow&quot;&gt;everybody's buddy&lt;/a&gt; would be nice augmented by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387986375&quot; rel=&quot;nofollow&quot;&gt;this masterpiece&lt;/a&gt; would be a sweet course, but likely too challenging for undergrad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that covers the basics. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Two more comments, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471050598&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; book is frustrating and I do not think it is appropriate to look at for undergrads trying to learn algebraic geometry, especially if they are moving to modern scheme theory. And another book with a very different feeling, but a really fun way to start thinking about algebraic geometry is Katz' &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821836870&quot; rel=&quot;nofollow&quot;&gt;Enumerative Geometry book&lt;/a&gt;. It was written especially for undergrads, and covers some really neat material that can give some good motivation to think about algebraic geometry and topology.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-06T20:51:34.690" CommunityOwnedDate="2010-08-06T20:51:34.690" CommentCount="14" />
  <row Id="1760" PostTypeId="2" ParentId="577" CreationDate="2010-08-06T21:00:58.133" Score="6" ViewCount="0" Body="&lt;p&gt;Set theory is completely full of new axioms, some of them expressing fundamental set principles, and many of them having consequences even in natural number arithmetic that are not provable without them (via consistency strength statements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past fifty years of research in set theory, a major lesson that has been learned is that many many fundamental questions about set theory are simply independent of the usual ZFC axioms. This includes most all of the questions about infinite cardinal arithmetic, but also subtle questons about infinite combinatorics.  For example, Is the exponential function (size of power set function) injective on infinite cardinalities? Independent of ZFC. Do Souslin trees exist?  Independent of ZFC. Is the size of the smallest dominating family of functions $f:\omega\to\omega$ necessarily $\aleph_1$?  Independent of ZFC. Can Lebesgue measure be more than countably additive, when CH fails?  Independent of ZFC. &#xA;There are hundreds of examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The response to this phenomenon was naturally to investigate the various hypotheses that go beyond ZFC. So we now know a great deal about what happens when CH holds, or when it fails, or when CH fails but the dominating number is low, and so on. The effect of this is to treat these hypotheses as semi-temporary axioms within a domain of research, which may not be the same usage of axiom that you inquired about, but the effect is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The powerful method of forcing, invented by Cohen in order to prove the consistency of $ZFC+\neg CH$, allows us to show that many other theories are also consistent. This method has given rise to a number of forcing axioms, such as Martin's Axiom, the Proper Forcing Axiom, Martin's Maximum and many others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much larger and very interesting class of axioms are provided by the large cardinal hierarchy, involving such notions as inaccessible cardinals, measurable cardinals, Woodin cardinals and so on. A curious feature of these strong axioms of infinity is that they imply certain highly regular features to occur among the projective sets of reals. Thus, although the axioms themselves seem to have nothing to do with sets of reals, they imply very nice properties for the sets of reals that we can easily define. Many set theorists take this as some kind of positive evidence for their truth. Meanwhile, the large cardinal hypotheses are intensely studied as a fundamental research effort to understand the nature of mathematical infinity. But you cannot do this without assuming that these cardinals exist, since this is provably not provable in ZFC, if consistent, and so these large cardinal hypotheses constitute new axioms. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-06T21:00:58.133" CommentCount="1" />
  <row Id="1761" PostTypeId="2" ParentId="1730" CreationDate="2010-08-06T21:06:19.310" Score="1" ViewCount="0" Body="&lt;p&gt;I start with the notion of an expression. An &lt;em&gt;expression&lt;/em&gt; is a grammatically meaningful combination of variables and constants. I don't need to tell &lt;em&gt;this&lt;/em&gt; audience what I mean by that. An &lt;em&gt;equation&lt;/em&gt; relate two expressions. To solve an equation for a variable one manipulates the equation according to rules until one variable is written unambiguously in terms of the others. This may not be always be done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;function&lt;/em&gt; is an equation in two variables in which one variable (y) can be solved as an unambiguous expression in the other variable (x). Thus y can be written as an expression in x. &#xA;Then I let the students know (a) this definition is not quite good enough for mathematicians, (b) it will work pretty well for all of the applications that we have in mind. Throughout the process examples are given.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-08-06T21:06:19.310" />
  <row Id="1762" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T21:12:05.710" Score="1" ViewCount="0" Body="&lt;p&gt;My favorite book at undegraduate level is Harris' Algebraic Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something more advanced, with a strongly geometric point of view, I advise Griffiths and Harris Principles of algebraic geometry. It is far more demanding, though.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-06T21:12:05.710" CommunityOwnedDate="2010-08-06T21:12:05.710" />
  <row Id="1763" PostTypeId="2" ParentId="1741" CreationDate="2010-08-06T21:17:17.673" Score="3" ViewCount="0" Body="&lt;p&gt;From one possible point of view, conglomerate come next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a model of set theory $\mathcal{U}$ and assume you have an inaccesible cardinal; let $S \in \mathcal{U}$ be a set of that cardinality. Then $S$ itself is another model of set theory. So you can (re)-define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Sets as elements of $S$. That is, $A \in \mathcal{U}$ is a set if and only if $A \in S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Classes as subsets of $S$. That is, $A \in \mathcal{U}$ is a class if and only if $A \subset S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Conglomerates as elements of $\mathcal{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way sets and classes will satisfy the usual axioms, and you have conglomerates as even bigger collections.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-06T21:17:17.673" CommentCount="3" />
  <row Id="1764" PostTypeId="2" ParentId="1748" CreationDate="2010-08-06T21:21:05.740" Score="7" ViewCount="0" Body="&lt;p&gt;Cox, Little, and O'Shea is an excellent book.  I'd also like to recommend Miles Reid's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521356628&quot; rel=&quot;nofollow&quot;&gt;Undergraduate Algebraic Geometry&lt;/a&gt;, which unlike Hartshorne or many other standard references was written specifically with undergraduates in mind.  You should also check out his book on commutative algebra; I found both very enlightening.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-06T21:21:05.740" CommunityOwnedDate="2010-08-06T21:21:05.740" />
  <row Id="1765" PostTypeId="2" ParentId="1758" CreationDate="2010-08-06T21:31:15.733" Score="1" ViewCount="0" Body="&lt;p&gt;Each line segment is a chord of the circle, with 3 of the 7 regions in the segment of the circle determined by the chord (the segment is the region between the chord and the minor arc between its endpoints).  The area of a segment with arc measure &amp;alpha; radians of a circle with radius r is $\frac{r^2}{2}(\alpha-\sin\alpha)$&amp;mdash;for simplicity, let $r=\sqrt{2}$ so the segment area is $\alpha-\sin\alpha$ and the area of the circle is $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each of the 7 regions has equal area, then the area of the segment determined by each chord must be $\frac{3}{7}$ of the area of the circle, so for each chord, $\alpha-\sin\alpha=\frac{6\pi}{7}\Rightarrow\alpha\approx2.91624$.  With 3 such chords, there is no way to get the central triangular region to have area more than about $\frac{1}{10}$ of the circle, so no way to get 7 regions of equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below left is just about the largest the central triangular region can be; below right is a more symmetric configuration, which actually doesn't make the central triangle much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/225458569.png&quot; alt=&quot;largest central triangle&quot;&gt; &lt;img src=&quot;http://www.imgftw.net/img/193455867.png&quot; alt=&quot;somewhat symmetric configuration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  When I said &quot;empirically&quot; in a comment below, that was a bit imprecise: given an angle of measure 2.91624 radians, I constructed (compass and straightedge) a circle with one chord meeting the given criteria.  I then constructed a few dozen possible locations for the second chord and for each of those, a few hundred locations for the third chord, measuring the area of the central triangle, when it existed.  There are 4 distinct configurations where it exists, corresponding to the 4 regions in the contour plot in my other answer.  The area function is continuous on these regions, so a well-distributed large sample of values across each of the regions gives an accurate depiction of the behavior of the function.  When a triangle has some aspects free and some aspects constrained, the maximum area is typically at an extreme value of the free aspects or at a configuration of maximal symmetry.  The multitude of constructions indicated local extrema for the configuration above left (narrowest but tallest) and for a configuration where the side of the triangle on the other chord is nearly the whole chord (shortest but widest).  The configuration shown above left had the largest area (this is verified computationally in my other answer).  Neither of these extreme configurations, however, are anywhere near 7 equal-area regions&amp;mdash;in each case, at least one region is reduced to the degenerate case of a point.  Above right is the most symmetrical configuration, which is intuitively the most likely to generate 7 equal-area regions.  This corresponds to the middle of the lighter-colored region in the middle of the contour plot in my other answer and is at or near a local minimum of the area function.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-07T17:52:15.070" LastActivityDate="2010-08-07T17:52:15.070" CommentCount="6" />
  <row Id="1766" PostTypeId="1" AcceptedAnswerId="1783" CreationDate="2010-08-06T22:28:06.517" Score="10" ViewCount="562" Body="&lt;p&gt;Suppose $a_{n}&gt;0$ and the following series converges&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^{\infty} a_{n}^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this imply that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^{\infty} \frac{a_{n}}{n}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;converges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to prove that the second series also converges by using the limit comparision test.   Is there another way to show the second series converges (e.g. root or ratio test)?&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-07T02:35:37.280" LastActivityDate="2010-08-08T22:07:07.723" Title="If $\sum_{n=1}^{\infty} a_{n}^{3}$ converges does $\sum_{n=1}^{\infty} \frac{a_{n}}{n}$  converge?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="1767" PostTypeId="1" AcceptedAnswerId="1770" CreationDate="2010-08-06T22:59:41.843" Score="5" ViewCount="157" Body="&lt;p&gt;Hi All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one has been bugging me for a while - I have a number of definitions for a symmetric group, none of which I understand. Clearly, $S_n$ is a group, which I follow. Now, the wikipedia definition (most readily to hand) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The symmetric group on a set X is the&#xA;  group whose underlying set is the&#xA;  collection of all bijections from X to&#xA;  X and whose group operation is that of&#xA;  function composition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of that, I can see (clearly) we have an underlying set which combined with an operation follows set rules as per the definition of a group. I understand a bijection from X to X implies that $\forall x \in X$, $x \mapsto x$. So, just as in geometric symmetry a set is symmetrical if it becomes itself again after some map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All well and good, but how do composite functions relate to this? Surely taking the bijective requirement into consideration the only operation permitted is one that effectively leaves the set unchanged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can anyone please either explain how composite functions fit in my definition or (equally fine) provide an example of a function that works so I can try to fit the idea to the definitions I've come across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-08-07T16:22:59.393" Title="What is a symmetric group?" Tags="&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="1768" PostTypeId="1" AcceptedAnswerId="1769" CreationDate="2010-08-06T23:07:31.847" Score="4" ViewCount="136" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1766/convergent-series&quot;&gt;This question&lt;/a&gt; just reminded me of a conundrum I posed myself in my first year of university. I never did get a satisfactory answer...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $a_n$ be a &lt;a href=&quot;http://www.answers.com/topic/null-sequence&quot; rel=&quot;nofollow&quot;&gt;null sequence&lt;/a&gt;. Does it follow that $\sum \frac{a_n}{n}$ converges?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-07T01:06:31.277" LastActivityDate="2010-08-07T01:06:31.277" Title="More convergent series..." Tags="&lt;sequences-and-series&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1769" PostTypeId="2" ParentId="1768" CreationDate="2010-08-06T23:20:13.870" Score="12" ViewCount="0" Body="&lt;p&gt;If by null sequence you mean a sequence that converges to 0, then no. Try $a_n=1/\log n.$ By integral comparison, the series diverges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_2^\infty\dfrac1{n\log n}\geq\int_2^\infty\dfrac{dx}{x\log x}=\int_{\log 2}^\infty\dfrac{du}u=\infty,$$&#xA;where I've used the change of variables $u=\log x$.&lt;/p&gt;&#xA;" OwnerUserId="405" LastEditorUserId="405" LastEditorDisplayName="" LastEditDate="2010-08-07T00:03:50.947" LastActivityDate="2010-08-07T00:03:50.947" CommentCount="3" />
  <row Id="1770" PostTypeId="2" ParentId="1767" CreationDate="2010-08-06T23:25:07.577" Score="4" ViewCount="0" Body="&lt;p&gt;I think you misunderstood the definition/use of &lt;em&gt;bijection&lt;/em&gt;. I'll use an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $X = \{1, 2, 3\}$, the appropriate symmetric group is $S_3$, and it is the set of all possible permutations of X:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f_1: 1\to 1, 2\to 2, 3\to 3$&lt;/li&gt;&#xA;&lt;li&gt;$f_2: 1\to 1, 2\to 3, 3\to 2$&lt;/li&gt;&#xA;&lt;li&gt;$f_3: 1\to 3, 2\to 2, 3\to 1$&lt;/li&gt;&#xA;&lt;li&gt;$f_4: 1\to 2, 2\to 1, 3\to 3$&lt;/li&gt;&#xA;&lt;li&gt;$f_5: 1\to 2, 2\to 3, 3\to 1$&lt;/li&gt;&#xA;&lt;li&gt;$f_6: 1\to 3, 2\to 1, 3\to 2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Generally, bijections from a set X to itself are &lt;em&gt;permutations&lt;/em&gt;, and intuitively they are &quot;re-orderings&quot; of the members of the set. It then follows that for a set of size n there are exactly n! permutations, and thus $|S_n| = n!$.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-08-06T23:25:07.577" CommentCount="2" />
  <row Id="1771" PostTypeId="2" ParentId="1767" CreationDate="2010-08-06T23:27:06.823" Score="1" ViewCount="0" Body="&lt;p&gt;Each member of $S_n$ is a bijection and the composition of two bijections is another bijection. Therefore $S_n$ is closed under composition.&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-08-06T23:27:06.823" />
  <row Id="1772" PostTypeId="2" ParentId="518" CreationDate="2010-08-06T23:33:46.173" Score="0" ViewCount="0" Body="&lt;p&gt;In the early days of commutative algebra people did not use the language of localization, so probably such proof exists. In particular, I have been told Kaplansky's early book (around 1950?) might be such reference, but I don't have it right now.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, &lt;a href=&quot;http://books.google.com/books?id=Fm_yPgZBucMC&amp;amp;lpg=PA79&amp;amp;ots=7Ap1zxS3Ig&amp;amp;dq=localization%20without%20admitting%20it&amp;amp;pg=PA79#v=onepage&amp;amp;q=localization%20without%20admitting%20it&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Exercise 2.3&lt;/a&gt; of Eisenbud tells you how to cheat and localize without admitting it. I might be wrong here, but what you want to prove is about some multiplicative properties of elements in $R$, so perhaps any proof you can find is just localization in disguise. &lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-06T23:33:46.173" />
  <row Id="1774" PostTypeId="1" CreationDate="2010-08-07T00:52:37.047" Score="0" ViewCount="398" Body="&lt;p&gt;I don't know whether the books metioned in &lt;a href=&quot;http://math.stackexchange.com/questions/329/best-ever-book-on-number-theory&quot;&gt;http://math.stackexchange.com/questions/329/best-ever-book-on-number-theory&lt;/a&gt; are beyond undergraduate/high-school-olympiad level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please recommend your favourite.&lt;/p&gt;&#xA;" OwnerUserId="696" LastActivityDate="2010-08-17T21:44:34.547" CommunityOwnedDate="2010-08-07T00:55:26.913" Title="Undergraduate/High-School-Olympiad Level Introductory Number Theory Books For Self-Learning" Tags="&lt;number-theory&gt;&lt;books&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="1775" PostTypeId="2" ParentId="362" CreationDate="2010-08-07T00:57:13.837" Score="0" ViewCount="0" Body="&lt;p&gt;There's also the books &lt;a href=&quot;http://www.springer.com/birkhauser/historyofscience/book/978-0-8176-4684-4&quot; rel=&quot;nofollow&quot;&gt;A History of Abstract Algebra&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=hmath-32&quot; rel=&quot;nofollow&quot;&gt;Episodes in the History of Modern Algebra (1800-1950)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-07T00:57:13.837" />
  <row Id="1776" PostTypeId="2" ParentId="740" CreationDate="2010-08-07T02:31:06.337" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bump_function1.1.&quot; rel=&quot;nofollow&quot;&gt;bump function&lt;/a&gt; $\psi\colon \mathbb{R} \to \mathbb{R}$ given by $\psi(x) = e^{-1/(1-x^2)}$ for $|x|\leq 1$ and $\psi(x) = 0$ for $|x| &gt; 1$ is not exactly pathological per se, but it's very useful for (at least) two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's an example of a function that is smooth (all derivatives exist) but not analytic (not equal to its Taylor series in a neighbourhood of every point) -- look at the points $x=\pm 1$.&lt;/li&gt;&#xA;&lt;li&gt;You can use it to build partitions of unity on manifolds, which is important when you want to build a global object (like a Riemannian metric) out of local data (defined in each chart).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-07T02:31:06.337" CommunityOwnedDate="2010-08-07T02:31:06.337" CommentCount="2" />
  <row Id="1777" PostTypeId="2" ParentId="1747" CreationDate="2010-08-07T02:31:23.823" Score="-1" ViewCount="0" Body="&lt;p&gt;There is a very simple probability formula called the Binomial Distribution.&#xA;The Binomial Distribution works as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two start off with, I need a game with only two possible outcomes: win or lose. This is where the &quot;bi&quot; (meaning two) of the &quot;binomial distribution&quot; comes from. Next, each possible outcome needs to have an associated probability. In this case, both the heads and tails have a probability of 50%, which we write as $0.5$. The binomial distribution will tell us the probability of getting $k$ heads in $n$ trials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B(k,n,p) = \frac{n!}{k!(n-k)!}  {p^{k}}  {{(1-p)}^{n-k}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, k is the number of heads, n is the number of trials, and p the probability of getting a heads. The probability of getting a tails is thus 1-p, which is also 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we roll the dice 100 times and get only 35 heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B(35,100,0.5) = \frac{100!}{35!(100-35)!}  {0.5^{35}}  {{(1-0.5)}^{100-35}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equal to 0.000863855665741652, which is a 0.0863855666% probability (out of 100). That's more than enough to show that this dice is probably loaded. You can now use this formula with any number of heads in any number of flips to calculate the probability. It's up to you to decide how low a percentage you're willing to take, but I'd generally advise below two standard deviations from the mean: below 4.54%.&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-08-07T02:31:23.823" CommentCount="1" />
  <row Id="1778" PostTypeId="1" CreationDate="2010-08-07T02:49:12.603" Score="6" ViewCount="338" Body="&lt;p&gt;Motivation: A friend asked me this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Problem: Suppose you start off with a dollar. You flip a fair coin, if it lands on heads you win 50 cents otherwise you lose 50 cents. If after n flips you have a nonzero amount of money, you win. Whats the probability you win? What about the limiting case as n tends to infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: In this game you are not allowed to have negative money. Thanks, Jonathan Fischoff, the linked helped greatly.&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-08-07T09:10:11.407" LastActivityDate="2010-09-05T10:17:19.317" Title="A probability game" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="1779" PostTypeId="2" ParentId="1774" CreationDate="2010-08-07T03:27:42.397" Score="3" ViewCount="0" Body="&lt;p&gt;There are several elementary number theory books which you could use and which do not assume a level of knowledge beyond high school math. I would strongly recommend Underwood Dudley's Elementary Number Theory and Harold Stark's An Introduction to Number Theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're both beautifully written and cover most of the things that are usually covered in any introductory number theory course (at a basic level of course). Some of the other books that have already been suggested are excelent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly Katie Banks' suggestion of Ireland and Rosen, although this book makes some use of the language of groups, rings and fields, so it may be more advanced for a high school student.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-08-07T03:27:42.397" CommunityOwnedDate="2010-08-07T03:27:42.397" CommentCount="1" />
  <row Id="1781" PostTypeId="1" AcceptedAnswerId="1784" CreationDate="2010-08-07T04:11:28.180" Score="6" ViewCount="197" Body="&lt;p&gt;This is the equation I'm having trouble with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G \frac{M m}{r^2} = m \frac{d^2 r}{dt^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the non-vector form of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow&quot;&gt;universal law of gravitation&lt;/a&gt; on the left and &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's second law of motion&lt;/a&gt; on the right. I assume that upon correctly modeling and solving this, I will have a function of time that outputs the distance from a spherical mass in space (ie. distance from the Earth given an initial condition of $r(0) = 10,000 \mathrm{km}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, WolframAlpha gives a &lt;a href=&quot;http://tinyurl.com/35vuvoe&quot; rel=&quot;nofollow&quot;&gt;hell of an answer&lt;/a&gt;, which leads me to believe that I'm modeling this equation completely wrong. Can anyone shed some light on this problem?&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-07T15:55:51.887" Title="Can this gravitational field differential equation be solved, or does it not show what I intended?" Tags="&lt;physics&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="1782" PostTypeId="2" ParentId="1747" CreationDate="2010-08-07T04:44:33.790" Score="3" ViewCount="0" Body="&lt;p&gt;This question is one of statistics (esp. statistical inference), not probability per se.  Keywords: &quot;binomial sampling&quot;; &quot;confidence interval for a proportion&quot;.  Asking this at &lt;a href=&quot;http://stats.stackexchange.com&quot;&gt;http://stats.stackexchange.com&lt;/a&gt; will get more complete answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The related probability fact is: if a coin has probability $p$ of coming up heads and is tossed $n$ times, then the observed number of heads will on average be $np$, but we expect the observed number to fluctuate (if the experiment with $n$ tosses is repeated many times) around the average by an amount on the order of $\sqrt{np(1-p)}$.  Keywords: normal distribution, bell curve, Central Limit Theorem, binomial distribution, convergence of binomial distribution to normal (Gaussian) distribution.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-07T06:41:51.903" LastActivityDate="2010-08-07T06:41:51.903" CommentCount="3" />
  <row Id="1783" PostTypeId="2" ParentId="1766" CreationDate="2010-08-07T05:10:35.953" Score="6" ViewCount="0" Body="&lt;p&gt;$3|a_n|/n \leq (|a_n|^3 + (2/n^{3/2})$ by a well known inequality.  Sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also shows the range of values that the second infinite sum can assume, given the value of the first, and assuming all terms are non-negative.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-07T08:09:22.480" LastActivityDate="2010-08-07T08:09:22.480" />
  <row Id="1784" PostTypeId="2" ParentId="1781" CreationDate="2010-08-07T06:21:08.480" Score="10" ViewCount="0" Body="&lt;p&gt;Yes you're modeling it wrong. Gravity is an attractive force, so if the mass is at r = 0, and r is nonnegative, then the gravity force should point towards zero, i.e. negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - \frac{GMm}{r^2} = m\ddot r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now Alpha even refuses to solve it. That's fine, because Mathematica can't solve many things. The usual approach is to notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d^2r}{dt^2} = \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to convert that 2nd-order ODE to a 1st-order ODE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -GM \frac{dr}{r^2} = v \,dv $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the solution v(r).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{GM}r + \text{constant} = \frac{v^2}2 \qquad (*) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But recall $v = dr/dt$. So we have another 1st-order ODE to solve, which gives r(t). The solution should look like that “hell of answer” because of the square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If you rearrange the terms you should see that Equation (*) is just conservation of energy. &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-07T15:55:51.887" LastActivityDate="2010-08-07T15:55:51.887" />
  <row Id="1786" PostTypeId="2" ParentId="1778" CreationDate="2010-08-07T07:21:51.247" Score="3" ViewCount="0" Body="&lt;p&gt;This type problem on random walk is usually solved with the Reflection Principle, with the walk visualized as a lattice path.  Strangely, I can't find an online reference to the solution but it is given in Feller's book on probability theory, volume 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, measuring the money in units of 0.5 dollars,  the walk, drawn in the $(x,y)$ plane, starts at (2,0), moves by  $(+1,\pm 1)$ at each step, and the question is what is the probability that over $n$ moves the walk is always $\geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-07T07:21:51.247" CommentCount="1" />
  <row Id="1788" PostTypeId="2" ParentId="1747" CreationDate="2010-08-07T08:20:59.570" Score="1" ViewCount="0" Body="&lt;p&gt;In typical math problems involving coins we know the probability ahead of time. If you know the probability of heads and tails with absolute certainty the sequence of heads and tails you get from flipping the coin tells you &lt;strong&gt;nothing&lt;/strong&gt; about the what the probability should be (because you already know it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if you have absolutely no idea what the probability is, and all you know is that the there are two sides you will end up with something called the &quot;Rule of Succession&quot; or $\frac{Heads + 1}{Total + 2}$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_succession&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rule_of_succession&lt;/a&gt; for assigning the probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case it sounds like you are in between the two. You have some information that leads you to believe that the coin is balanced but you are not positive, and the thus the frequencies from flipping will affect your probability assignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why this question is tricky. You have to take everything you know about the coin and encode mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One why to think about doing that is assume that you have already flipped the coin x number of times. So in the case of knowing the probability for certain, it is as though we have flipped the coin an infinite number of times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe you would start by assuming that your prior knowledge is equivalent to having flipped the coin 10,000 times with half heads and half tails. That way the frequency will still will affect it, but not so much initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't of any mathematically texts that have attempted at solving anything but the two extreme cases (complete information, zero information). A physicist, Jaynes attempted to broach this problem here &lt;a href=&quot;http://www-biba.inrialpes.fr/Jaynes/cc18i.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www-biba.inrialpes.fr/Jaynes/cc18i.pdf&lt;/a&gt; (I can only partially follow his reasoning), but I don't know if there is an accepted mathematically rigorous way to solve this type of problem.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-07T08:20:59.570" CommentCount="6" />
  <row Id="1789" PostTypeId="1" AcceptedAnswerId="1790" CreationDate="2010-08-07T08:41:44.397" Score="2" ViewCount="258" Body="&lt;p&gt;I just finished reading the Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_condensation_test&quot; rel=&quot;nofollow&quot;&gt;Cauchy condensation test&lt;/a&gt;. I understand the trapezoidal view, but apparently &quot;the 'condensation' of terms is analogous to a substitution of an exponential function&quot;. Can anyone explain what is meant by this?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-07T18:02:50.093" LastActivityDate="2010-08-07T18:02:50.093" Title="Cauchy condensation test - Exponential function explanation" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="1790" PostTypeId="2" ParentId="1789" CreationDate="2010-08-07T08:46:27.163" Score="4" ViewCount="0" Body="&lt;p&gt;Explanation is quite straightforward. Consider integral $\int f(x)dx$. After change of variable $x=2^t$ it becomes $const\cdot\int 2^t f(2^t)dt$ — compare with $\sum 2^n f(2^n)$ from Cauchy condensation test.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-07T08:46:27.163" />
  <row Id="1792" PostTypeId="2" ParentId="1758" CreationDate="2010-08-07T08:57:11.433" Score="4" ViewCount="0" Body="&lt;p&gt;Let's focus on two of the chords. These have to split the circle in&#xA;the ratio $4:3$ so the minor arc subtended by each is equal. Call&#xA;one $AB$ and regard it as &quot;fixed&quot; and the other $CD$ which we think&#xA;of as moving. As I can't draw diagrams, look at Isaac's left diagram.&#xA;Let the two chords start in the position given by Isaac's two chords&#xA;meeting on the circumference. Let $AB$ be the one on the left and $CD$ be&#xA;the one on the right with $B=C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rotate $CD$ clockwise. Let $P$ be the point of intersection.&#xA;Then the angle $APD$ increases, the point $P$ moves monotonely from $B$ towards&#xA;$A$ on the line $AB$, and $C$ moves monotonely from $B$ towards $A$ on the minor&#xA;arc. Thus the area of the &quot;sector&quot; (bounded by two&#xA;lines and an arc of the circle) $PBC$ steadily increases, since at a later&#xA;time it will contain the sector at an earlier time. There is a unique&#xA;time at which the &quot;sector&quot; has area $1/7$ of the circle. Let $\theta_0$&#xA;be the angle $APD$ at this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sought configuration, all three angles of the central triangle have&#xA;area $\pi/3$ and so it exists if and only if $\theta_0=\pi/3$. So&#xA;deciding whether the configuration exists reduces to a single calculation:&#xA;what is the area of &quot;sector&quot; $PBC$ when angle $APD$ equals $\pi/3$;&#xA;this is a calculation that I'm not going to do :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-07T08:57:11.433" CommentCount="8" />
  <row Id="1793" PostTypeId="2" ParentId="1442" CreationDate="2010-08-07T09:10:43.950" Score="3" ViewCount="0" Body="&lt;p&gt;More generally, for $0 \leq k \leq n$, there is an identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1) {\rm lcm} ({n \choose 0}, {n \choose 1}, \dots {n \choose k}) = {\rm lcm} (n+1,n,n-1, \dots n+1-k)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is simply the fact that any integer linear combination of $f(x), \Delta f(x), \Delta^2 f(x), \dots \Delta^k f(x)$ is an integer linear combination of $f(x), f(x-1), f(x-2), \dots f(x-k)$ where $\Delta$ is the difference operator, $f(x) = 1/x$, and $x = (n+1)$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-07T09:10:43.950" CommentCount="3" />
  <row Id="1794" PostTypeId="2" ParentId="1789" CreationDate="2010-08-07T09:37:57.243" Score="3" ViewCount="0" Body="&lt;p&gt;&quot;Condensation&quot; because the series $a_n$ is compressed to a subseries $a_{2^n}$ carrying all the same information about convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the condensation test is easy, as Grigory M's answer indicates and whenever it works so does a comparison of sum with integral, which is a more fundamental idea.  So you might naturally wonder why this seemingly trivial convergence test deserves its own name, other than the fame of its supposed inventor.  The reason it is retained in textbooks is that it is often a convenient criterion to use for series when the terms or the partial sums have some dependence on $\log (n)$.  The proof of divergence of the harmonic series is an example.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-07T09:47:49.810" LastActivityDate="2010-08-07T09:47:49.810" CommentCount="3" />
  <row Id="1795" PostTypeId="1" AcceptedAnswerId="1796" CreationDate="2010-08-07T10:24:47.263" Score="1" ViewCount="79" Body="&lt;p&gt;Can anyone explain the syntax from the following Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Definition&quot; rel=&quot;nofollow&quot;&gt;convergence of random variables&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/math/1/b/d/1bd88f4785429841e1d24c2996cbb6e0.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the squiggly line?&lt;/li&gt;&#xA;&lt;li&gt;What are $d$, $D$, $L$, $L_X$?&lt;/li&gt;&#xA;&lt;li&gt;Why does it say $X_n$ implies X? They don't seem to be statements&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-07T10:49:47.747" LastActivityDate="2010-08-07T11:43:46.200" Title="Syntax for the convergence of random variables" Tags="&lt;probability-theory&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1796" PostTypeId="2" ParentId="1795" CreationDate="2010-08-07T11:25:25.680" Score="4" ViewCount="0" Body="&lt;p&gt;All of these expressions are just notations for &quot;X_n converges to $X$ in distribution&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The squiggly line is really just a notation (a rather strange one) for convergence in distribution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The symbols $d$ and $\mathcal D$ stand for &lt;em&gt;distribution&lt;/em&gt;, $\mathcal L$ and $\mathcal L_X$ stand for &lt;em&gt;law&lt;/em&gt;. (One also speaks of &quot;convergence in law&quot; instead of &quot;convergence in distribution&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is also just a notation for convergence in distribution (also a rather strange one); the arrow does not symbolize an implication of statements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-07T11:43:46.200" LastActivityDate="2010-08-07T11:43:46.200" CommentCount="1" />
  <row Id="1797" PostTypeId="2" ParentId="1781" CreationDate="2010-08-07T11:34:49.583" Score="2" ViewCount="0" Body="&lt;p&gt;I think you are doing it all wrong. Your Newtonian gravitational force equation looks wrong to me in spherical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://en.wikiversity.org/wiki/Advanced_Classical_Mechanics/Central_Forces&quot; rel=&quot;nofollow&quot;&gt;a different approach&lt;/a&gt; using Lagrangian and Hamiltonian approach. The solution is &#xA;&lt;img src=&quot;http://upload.wikimedia.org/math/0/4/6/04604f267d01b2296779593b47ca1d71.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\V(r)\ &amp;amp;= -\frac1r\&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-08-07T11:34:49.583" />
  <row Id="1798" PostTypeId="2" ParentId="1694" CreationDate="2010-08-07T15:06:07.213" Score="0" ViewCount="0" Body="&lt;p&gt;Well, we call $e^x$ natural exponentation and therefore &lt;code&gt;\log_e(x)&lt;/code&gt; natural logarithm. This one is abbreviated as &lt;code&gt;\ln&lt;/code&gt; (for logarithmus naturalis).&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-07T15:06:07.213" />
  <row Id="1799" PostTypeId="1" AcceptedAnswerId="1803" CreationDate="2010-08-07T15:44:42.270" Score="2" ViewCount="220" Body="&lt;p&gt;For a given $k&gt;0$ constant, assuming that $x,y&gt;0$. Also this equation has a particular name, or some mathematician associated with it?&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2010-08-07T20:05:20.733" Title="How to solve the equation $x^y=y^x=k$?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="1800" PostTypeId="2" ParentId="1767" CreationDate="2010-08-07T16:22:59.393" Score="1" ViewCount="0" Body="&lt;p&gt;The original question claims that the author doesn't yet understand the definitions of the symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I think it's best to start with (what I think) is the most accessible definition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$S_n$, the symmetric group on the $n$ numbers $1,\ldots,n$ is the set of all possible ways to permute the set of numbers $1,\ldots,n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, one possible permutation (element of $S_n$) is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$g = (1,2,\ldots,n) \Rightarrow (2,3,\ldots,n,1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This set of permutations forms a group: the multiplication is simply performing one permutation, and then the next. So &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$g\cdot g = g^2: (1,2,\ldots,n) \Rightarrow (3,4,\ldots,n,1,2)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you should get comfortable with this definition first. Once you think about it for a while, you'll realize that each permutation can be represented as a function on the set ${1,\ldots,n}$. For example, the permutation $g$ above would be the function that sends $1 \mapsto 2$, $2 \mapsto 3$, and so on (in general, $n \mapsto n + 1$, where $n$ maps back to $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think a little further, you'll realize that any function you get from a permutation in such a fashion must be bijective (that is, a permutation can't, say, send all the numbers $1,\ldots,n$ to $1$, since that's clearly not a permutation).&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-08-07T16:22:59.393" />
  <row Id="1801" PostTypeId="2" ParentId="740" CreationDate="2010-08-07T16:28:12.663" Score="2" ViewCount="0" Body="&lt;p&gt;I like $f(x,y) = \frac{xy}{x^{2}+y{2}}$ for $(x,y) \ne (0,0)$ and $f(x,y)=0$ for $(x,y)=(0,0)$.  Here $f$ has partial derivatives at (0,0) but it is not differentiable at $(0,0)$&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-07T16:34:16.977" LastActivityDate="2010-08-07T16:34:16.977" CommunityOwnedDate="2010-08-07T16:28:12.663" CommentCount="1" />
  <row Id="1802" PostTypeId="1" AcceptedAnswerId="1805" CreationDate="2010-08-07T16:53:06.537" Score="0" ViewCount="127" Body="&lt;p&gt;Let $A$, $X$, and $Y$ be arbitrary topological spaces. Let $F:X\times A\rightarrow Y$ be a continuous function. Let $P$ be the map from $X\times A$ to $Y\times A$ taking $(x,a)$ to $(F(x,a),a)$. Does it follow from continuity of $F$ that $P$ is continuous?&lt;/p&gt;&#xA;" OwnerUserId="921" LastActivityDate="2010-08-07T17:01:14.323" Title="Is this function continuous?" Tags="&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="1803" PostTypeId="2" ParentId="1799" CreationDate="2010-08-07T16:56:17.637" Score="7" ViewCount="0" Body="&lt;p&gt;WLOG suppose that $x \ge y$ and let $x = y^n$ for some $n \ge 1$.  Then (ignoring $k$) the equation becomes $y^{ny} = y^{y^n}$, or $ny = y^n$, or $y^{n-1} = n$, which gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y = n^{ \frac{1}{n-1} }, x = n^{ \frac{n}{n-1} }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it only remains to find the values of $n$ such that $x^y = k$, which you should just do numerically.  (I guess one must also account for the solutions where $x = y$; the graph $x^y = y^x$ has two components which intersect at $(e, e)$, which one gets from the above by taking the limit as $n \to 1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am almost positive this equation does not have a name because, in all honesty, it is not interesting.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-07T17:34:18.657" LastActivityDate="2010-08-07T17:34:18.657" CommentCount="2" />
  <row Id="1804" PostTypeId="2" ParentId="1758" CreationDate="2010-08-07T16:59:37.510" Score="0" ViewCount="0" Body="&lt;p&gt;Each line segment is a chord of the circle, with 3 of the 7 regions in the segment of the circle determined by the chord (the segment is the region between the chord and the minor arc between its endpoints).  The area of a segment with arc measure $\theta$ radians of a circle with radius r is $\frac{r^2}{2}(\theta-\sin\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each of the 7 regions has equal area, then the area of the segment determined by each chord must be $\frac{3}{7}$ of the area of the circle, so for each chord, $\frac{r^2}{2}(\theta-\sin\theta)=\frac{3\pi r^2}{7}\Rightarrow\theta\approx 2.91624$.  Going forward, I'll use $\omega$ to refer to this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's work with the unit circle centered at the origin and place one chord so that its minor arc would be swept out by the rotation images of (1,0) about the origin by angles in the interval $[0,\omega]$.  Let the other two chords be placed such that their minor arcs would be swept out by the rotation images of (1,0) about the origin by angles in the intervals $[\alpha,\alpha+\omega]$ and $[\beta,\beta+\omega]$ where $0&amp;lt;\alpha&amp;lt;\beta&amp;lt;2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any arrangement of chords that yields 7 regions, there will be a central triangular region.  The lines containing the chords have equations $y=\frac{\sin\omega}{\cos\omega-1}(x-1)$, $y-\sin\alpha=\frac{\sin\alpha-\sin(\alpha+\omega)}{\cos\alpha-\cos(\alpha+\omega)}(x-\cos\alpha)$, and $y-\sin\beta=\frac{\sin\beta-\sin(\beta+\omega)}{\cos\beta-\cos(\beta+\omega)}(x-\cos\beta)$.  The vertices of the central triangular region are at the pairwise points of intersection of those three chords:&#xA;$\left(\cos\frac{\omega}{2}\cos\frac{\alpha+\omega}{2}\sec\frac{\alpha}{2},&#xA;\cos\frac{\omega}{2}\sin\frac{\alpha+\omega}{2}\sec\frac{\alpha}{2}\right)$, &#xA;$\left(\cos\frac{\omega}{2}\cos\frac{\beta+\omega}{2}\sec\frac{\beta}{2},&#xA;\cos\frac{\omega}{2}\sin\frac{\beta+\omega}{2}\sec\frac{\beta}{2}\right)$, &#xA;and $\left(\cos\frac{\omega}{2}\cos\frac{\alpha+\beta+\omega}{2}\sec\frac{\alpha-\beta}{2},\frac{\csc(\alpha-\beta)}{2}(-\cos\alpha+\cos\beta-\cos(\alpha+\omega)+\cos(\alpha-\beta))\right)$.  The area of this triangle is $\frac{1}{2}\left|\left(\cos\left(\alpha-\frac{\beta}{2}\right)-\cos\frac{\beta}{2}\right)\cos^2\frac{\omega}{2}\sec\frac{\alpha}{2}\sec\frac{\alpha-\beta}{2}\tan\frac{\beta}{2}\right|$ (restricted to only values of $\alpha$ and $\beta$ for which the three points of intersection are within the circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contour plot of the area is shown below (white areas do not satisfy the constraints described above; darker colors represent smaller values of the area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/455062883.png&quot; alt=&quot;contour plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest possible area for the triangle is approximately 0.226346 when $\alpha\approx 2.91624$ and $\beta\approx 3.36694$ (numerical approximation from Mathematica), which is well short of one seventh the area of the circle, $\frac{\pi}{7}\approx 0.448799$.  Therefore, it is not possible to divide the circle into 7 regions of equal area using 3 chords.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-07T16:59:37.510" />
  <row Id="1805" PostTypeId="2" ParentId="1802" CreationDate="2010-08-07T17:01:14.323" Score="6" ViewCount="0" Body="&lt;p&gt;Yes.  This follows from the fact that a function $U \to V \times W$ is continuous if and only if its component functions $U \to V, U \to W$ are, and from the fact that the projection maps $V \times W \to V$ and $V \times W \to W$ are continuous.  Both of these facts in turn follow  from the universal property of the product topology.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-07T17:01:14.323" CommentCount="1" />
  <row Id="1806" PostTypeId="2" ParentId="1116" CreationDate="2010-08-07T17:46:55.937" Score="0" ViewCount="0" Body="&lt;p&gt;I don't have enough rep to comment yet, so I'll make this CW since it's not a complete answer, just a remark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it overcome some of the difficulty by using some sort of graph representation of the subdivided rectangle? Each rectangle is a vertex and an edge represents sharing a side, i.e.&#xA;the first rectangle above would be represented by the edges ${(1,2),(1,3),(2,4),(3,4)}$ and the second by ${(1,2),(1,3),(1,4),(2,4),(2,5),(3,4),(3,5)}$, numbering the rectangles from top-left to bottom-right. The tricky part here would then be to implement the operations for subdividing by a vertical or horizontal line segment. Starting instead with lines instead of segments could be promising.&lt;/p&gt;&#xA;" OwnerUserId="854" LastActivityDate="2010-08-07T17:46:55.937" CommunityOwnedDate="2010-08-07T17:46:55.937" CommentCount="1" />
  <row Id="1807" PostTypeId="2" ParentId="1778" CreationDate="2010-08-07T18:05:43.473" Score="0" ViewCount="0" Body="&lt;p&gt;As n goes to infinity, you run out of money with probability 1.  This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Gambler%27s_ruin&quot; rel=&quot;nofollow&quot;&gt;gambler's ruin&lt;/a&gt; or the drunk walk--if you are winning/losing money or walking side to side such that you win/lose/go left/go right with probability 1/2 (a 1-d simple symmetric &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;random walk&lt;/a&gt;), and there is a boundary like going broke or ending up in the gutter, as the number of games/steps goes to infinity, the probability of hitting that boundary goes to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to prove it, but per Joshua Zucker's comment on &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001405&quot; rel=&quot;nofollow&quot;&gt;A001405 in the OEIS&lt;/a&gt;, and agreeing with T..'s answer, the probability of ending up with a positive amount of money after n flips if you started with no money would be $\frac{1}{2^n}{n\choose \left\lfloor\frac{n}{2}\right\rfloor}$.  Because you start with one dollar instead of no money, this is offset so that the probability of having money after n flips is $\frac{1}{2^{n-1}}{n-1\choose \left\lfloor\frac{n-1}{2}\right\rfloor}$ for n&amp;ge;1.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-07T18:05:43.473" />
  <row Id="1808" PostTypeId="1" CreationDate="2010-08-07T18:47:03.710" Score="5" ViewCount="414" Body="&lt;p&gt;I'm mainly a CS major, but I also want to learn more advanced  mathematics. I see lot of cross over between CS and applied math, same can't be said about pure math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of 'advanced': Something a pure math major learns during and after late undergraduate years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the fields that combines advanced mathematical topics with computer science? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical computer science and computational group theory are good examples.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-08-09T08:23:58.353" CommunityOwnedDate="2010-08-07T18:47:03.710" Title="Fields that require both CS and pure math" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="10" CommentCount="2" FavoriteCount="2" />
  <row Id="1809" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T19:13:39.587" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Number Theory&lt;/strong&gt;: many large numbers were tested by using computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear programming&lt;/strong&gt;: many algorithms have been developed.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-08-07T19:13:39.587" CommunityOwnedDate="2010-08-07T19:13:39.587" />
  <row Id="1810" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T19:22:17.247" Score="4" ViewCount="0" Body="&lt;p&gt;Computational algebraic geometry is the first example that comes to mind. It's used in phylogenetics (http://siam.org/pdf/news/1146.pdf can also look up the book Algebraic Statistics for Computational Biology). It also shows up in robotics and control theory, but I know a lot less about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polytopes (see Coxeter - Regular Polytopes or Grunbaum - Convex Polytopes) also appear in a lot of applications. The only specific one that comes to mind at the moment is the problem of figuring out whether an object (like a robot) can traverse a given maze. The algorithms use the convex hull of the robot to test whether or not it could traverse the maze. I'm sure there are a lot of people more knowledgeable than me on the subject.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-07T19:22:17.247" CommunityOwnedDate="2010-08-07T19:22:17.247" />
  <row Id="1812" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T20:30:21.323" Score="2" ViewCount="0" Body="&lt;p&gt;Robotics.  I don't know much about this field but from a theoretical point of view a lot of lie group theory is used (to describe joint motion and position).  In particular the lie theory of rigid motions i.e. of the Euclidean group $SE(3)$ is relevant.  I would imagine CS would come in more from the computational side when larger calculations are involved.  Also its robots which need to be built so plenty of mechanical and electrical engineering going on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mostly based on a quick glance at &lt;a href=&quot;http://www.springer.com/computer/ai/book/978-0-387-20874-9&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/computer/ai/book/978-0-387-20874-9&lt;/a&gt;.  Which is written for an undergraduate math or engineering major.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-07T20:30:21.323" CommunityOwnedDate="2010-08-07T20:30:21.323" />
  <row Id="1813" PostTypeId="2" ParentId="1808" CreationDate="2010-08-07T21:40:36.610" Score="0" ViewCount="0" Body="&lt;p&gt;Today is an open question to find an efficient algorithm/data structure that can be used to index/search information in a high dimensional space (ie d &gt; 20).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen approaches using computational geometry, space filling curves, principal component analysis, etc. &lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2010-08-07T21:40:36.610" CommunityOwnedDate="2010-08-07T21:40:36.610" />
  <row Id="1814" PostTypeId="1" CreationDate="2010-08-07T21:44:25.857" Score="2" ViewCount="107" Body="&lt;p&gt;Are there any books or online notes that cover the basics of lie groups using nonstandard analysis? Another thing I would like is a to see these things set in category theory (along the lines of &lt;em&gt;Algebra: Chapter 0&lt;/em&gt; except for differential geometry?)&lt;/p&gt;&#xA;" LastActivityDate="2010-08-08T06:13:16.587" CommunityOwnedDate="2010-08-08T13:00:02.230" Title="Books on Lie Groups via nonstandard analysis?" Tags="&lt;differential-geometry&gt;&lt;lie-algebra&gt;&lt;nonstandard-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="1815" PostTypeId="2" ParentId="1694" CreationDate="2010-08-07T21:47:12.533" Score="2" ViewCount="0" Body="&lt;p&gt;I don't have enough rep. on this site to post a comment on Gerard's answer, so I'll write this as an answer. Those who say &quot;tan&quot; is more common than tg should keep in mind where they live. In Russian the standard abbreviation is tg, not tan. They write sin and cos, but the trig functions tan, cot, and csc have alternate abbreviations. &#xA;See the first line at &lt;a href=&quot;http://ru.wikipedia.org/wiki&quot; rel=&quot;nofollow&quot;&gt;http://ru.wikipedia.org/wiki&lt;/a&gt;/Тригонометрические_функции, for instance, where the trigonometric functions are named and then abbreviated in parentheses. (You may need to cut and paste, rather than click, on that link since the cyrillic part of the link is not working properly.) &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-07T21:47:12.533" CommentCount="1" />
  <row Id="1816" PostTypeId="2" ParentId="83" CreationDate="2010-08-07T21:49:50.613" Score="2" ViewCount="0" Body="&lt;p&gt;My biggest hobby throughout junior high was video game development. I taught myself how to program in C/C++ and first wrote pong. This introduced me to vectors and I saw first-hand how convenient they could be for geometric computations. I then gradually made more and more complex programs. By high school I was mainly interested in computer graphics, which led to my learning linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having lots of programming experience is in itself extremely beneficial. In today's computer age, you can never know enough about programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is obviously a quite indirect approach and it's not exactly a teaching technique, but I would &lt;em&gt;highly&lt;/em&gt; recommend any young child at least look into game development. It's very intellectually stimulating, it will expose them to a variety of fields that they might otherwise have to wait years and years to be exposed to, it will condition them to think logically and rigorously from a young age, and above all it's lots of fun to see computer science and mathematics work together to produce a video game--which, by the way, you can then play!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days there are simply incredible resources for teaching oneself the techniques of game development. Anyone interested should visit &lt;a href=&quot;http://www.gamedev.net&quot; rel=&quot;nofollow&quot;&gt;http://www.gamedev.net&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-08-07T21:49:50.613" CommunityOwnedDate="2010-08-07T22:32:44.343" />
  <row Id="1817" PostTypeId="1" AcceptedAnswerId="1825" CreationDate="2010-08-07T23:49:06.073" Score="2" ViewCount="113" Body="&lt;p&gt;Let $X$ be a compact metric space.  Given a sequence $x_n \in X$ and an ultrafilter $\mathcal{U}$ on $\mathbb{N}$, we can define the &quot;Banach limit&quot; of ${x_n}$ with respect to $\mathcal{U}$. This limit is the unique element of $X$ such that every neighborhood of it contains a $\mathcal{U}$-large collection of the sequence ${x_n}$.  However, this process is not necessarily translation-invariant. When $X$ is a closed interval in the reals, then one can define a Banach limit by using an ultrafilter and taking the ultrafilter-limit of the Cesaro means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to define a translation-invariant &quot;limit&quot; with reasonable properties for sequences taking values in an arbitrary compact metric space?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-08T03:45:38.183" Title="Banach limits in metric spaces" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="1818" PostTypeId="2" ParentId="1116" CreationDate="2010-08-08T00:25:55.657" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a non-answer which may prove interesting and tractable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a configuration of rectangles that properly tile a surrounding rectangle, one can&#xA;represent the tiling in a number of ways.  Two that come to mind I will call C and LL.&#xA;C represents each rectangle by giving its dimensions (x-span and y-span) and the coordinates for the center of the rectangle, and LL does the same, except it specifies the coordinate for the lower left vertex (instead of the center) of each rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With plenty of rectangles in a tiling, there is some redundancy in the LL form of a tiling.  For example, it should be possible to deduce the dimensions of the rectangle in the extreme lower left corner of the tiling by noting stored coordinates of the rectangle above and the rectangle to the right of this corner rectangle.  No such boundaries exist for the upper rightmost rectangle in a tiling, but perhaps an &quot;end point&quot;&#xA;could be added to indicate the maximum x and maximum y values in the tiled rectangle.  Suppose we add such an end point, and let LL' be a representation like LL that contains the lower left coordinates and the end point, and does NOT contain the dimension information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1: LL' is an equivalent representation to LL.  That is, given a representation in LL form, one can construct a representation in LL' form and use it to recreate the tiling; conversely, one can build an LL form from an LL' form to represent the same tiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1 appears to be true; it is clear how to make an LL' form from an LL form, while for the converse, it should be possible to check all the lower left points to see which would help determine the dimensions for a given rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is statement 1 true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider C and make C' from C by again forgetting the dimensions, except that you can&#xA;provide two endpoints representing the upper right and lower left corners of the tiled rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 2: C and C' are equivalent in the sense used above for LL and LL' .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear how to get a C' form from a C form.  It is NOT clear that it is possible to get a C form from a C' form where you have no dimension information at all, except for the&#xA;entire rectangle as implied by the two endpoints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Statement 2 true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statements 1 and 2 have a bearing on the problem above: it may be possible to enumerate geometrically inequivalent tilings with n rectangles by looking at normalized forms of either C, C', LL, or LL'.  Also, there may be advantages to switching between the forms of the tiling in determining equivalence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are interesting variations that can occur.  For example, C'' might have the center of the tiled rectangle instead of the two endpoints, or even give just the dimensions of&#xA;the tiled rectangle instead of the coordinates for its upper right and lower left corners.  What information is lost by such representations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meta-question:  Is there a geometric property at work here that might explain why LL' uses less information but is still equivalent to C (assuming Statement 1 is true), while C' does not (assuming Statement 2 is false)?  Can something be said in general about what properties of a form for tiling are needed in order to represent the tiling both minimally and faithfully?&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:02:54.053" LastActivityDate="2010-08-11T22:02:54.053" CommentCount="0" />
  <row Id="1819" PostTypeId="1" CreationDate="2010-08-08T00:32:12.913" Score="2" ViewCount="190" Body="&lt;p&gt;What is the difference between convergence in distribution, convergence in probability and almost sure convergence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;Convergence of random variables&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-09-11T07:20:19.580" Title="Convergence of random variables" Tags="&lt;probability-theory&gt;&lt;real-analysis&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1820" PostTypeId="2" ParentId="1819" CreationDate="2010-08-08T02:01:56.513" Score="4" ViewCount="0" Body="&lt;p&gt;Convergence in distribution means pointwise convergence of the CDFs.  It's not a form of convergence of the random variables per se. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convergence in probability says that for every epsilon &gt; 0, the probability of the sequence of random variables being more than epsilon from the limiting random variable goes to zero. Convergence in probability is also called convergence in measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convergence almost everywhere = almost sure convergence = pointwise convergence of the random variables except possibly on a set of measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.johndcook.com/modes_of_convergence.html&quot; rel=&quot;nofollow&quot;&gt;diagram&lt;/a&gt; showing which modes of convergence imply which other modes.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-08T02:01:56.513" />
  <row Id="1821" PostTypeId="2" ParentId="1814" CreationDate="2010-08-08T02:26:21.643" Score="1" ViewCount="0" Body="&lt;p&gt;I never looked at that part of the book in any detail at all, but Abraham Robinson's &lt;em&gt;Nonstandard Analysis&lt;/em&gt; includes some material on Lie groups via nonstandard analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little bit of googling found a &lt;a href=&quot;http://arxiv.org/abs/funct-an/9205003&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; on the arXiv that might be relevant.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-08T02:26:21.643" />
  <row Id="1822" PostTypeId="1" AcceptedAnswerId="1823" CreationDate="2010-08-08T02:34:27.177" Score="6" ViewCount="871" Body="&lt;p&gt;I have just started auditing Linear Algebra, and the first thing we learned was how to solve a system of linear equations by reducing it to row echelon form (or reduced row echelon form). I know how to solve a system of linear equations using determinants, and so the using row echelon form seems to be very inefficient and an easy way to make mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lecturer seemed to indicate that this will be useful in the future for something else, but a friend who has a PhD in mathematics said it is totally useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? Is this technique totally useless or is there some &quot;higher mathematics&quot; field/ technique where row echelon form will be useful?&lt;/p&gt;&#xA;" OwnerUserId="85" LastActivityDate="2010-08-08T10:27:53.353" Title="Is reducing a matrix to row echelon form useful at all?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" />
  <row Id="1823" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T02:56:55.597" Score="14" ViewCount="0" Body="&lt;p&gt;Go beyond 2 x 2 and row echelon form is &lt;em&gt;the&lt;/em&gt; way to solve systems of linear equations.&#xA;This stuff is not done by hand, but by a computer. Row echelon form is much, much faster than &#xA;using determinants (Cramer's formula).  Ask someone who works in numerical analysis about this.  Your friend with the Ph.D. is clearly unaware of the needs of applied math. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pure math, Cramer's formula is very important for some proofs (and I personally prefer it for conceptual reasons over the algorithmic row echelon business, but then I don't work in numerical analysis). In applied math problems that give rise to systems of linear equations, you typically face equations in hundreds of variables.  You are not going to solve that with a determinant.  Moreover, if a computer needed to find a determinant of a 100 x 100 matrix it would &lt;em&gt;not&lt;/em&gt; use the sum of 100! terms as in the Laplace expansion formula for determinants, but instead apply faster algorithms (the QR algorithm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your profile says you are studying economics.  At some point when you may need to deal with some system of linear equations in economics and have the computer solve it for you, the method of solution could be treated as a black box in the same way most people drive  without knowing how a car really works. But the reality is that if it were not for the algorithm of reducing to row echelon form, the computer couldn't solve systems of hundreds of equations rapidly. &lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="619" LastEditorDisplayName="" LastEditDate="2010-08-08T03:06:06.027" LastActivityDate="2010-08-08T03:06:06.027" CommentCount="3" />
  <row Id="1824" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T03:10:50.993" Score="10" ViewCount="0" Body="&lt;p&gt;Gaussian elimination on an $n \times n$ matrix takes takes on the order of $n^3$ operations, where an operation is an addition, subtraction, multiplication, or a division. By contrast, computing determinants using the cofactor expansion takes $n!$ operations, which is much, much longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider a $100 \times 100$ matrix. Then $100^3$ is a million, and a million operations can be done by a typical laptop in under a second; but $100!$ is approximately $1$ followed by $158$ zeros, which would take even a supercomputer more time than the age of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I believe the opposite of your claim is true: if you want to compute determinants, the best way is to to reduce to echelon form. &lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-08-08T06:21:36.890" LastActivityDate="2010-08-08T06:21:36.890" CommentCount="6" />
  <row Id="1825" PostTypeId="2" ParentId="1817" CreationDate="2010-08-08T03:45:38.183" Score="3" ViewCount="0" Body="&lt;p&gt;Let $X = \lbrace 0,1\rbrace$ be the compact metric space with exactly two points.  Then the space of sequences is just $\Sigma_2^+ = \lbrace 0,1 \rbrace^\mathbb{N}$, and an ultrafilter $\mathcal{U}$ defines a Banach limit in a fairly straightforward way:  for $i=0,1$, write $A_i = \lbrace n \in \mathbb{N} \mid x_n = i \rbrace$, and $\lim_\mathcal{U} x_n = i$ if and only if $A_i \in \mathcal{U}$.  This limit is translation invariant if and only if the ultrafilter itself is, so your question reduces to the existence of a translation invariant ultrafilter on $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that no such ultrafilter exists, and hence the answer to your question is &quot;no&quot;; however, I'm not particularly knowledgeable about ultrafilters, so I welcome correction or confirmation on this last point.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-08T03:45:38.183" CommentCount="1" />
  <row Id="1826" PostTypeId="2" ParentId="1747" CreationDate="2010-08-08T05:07:50.303" Score="0" ViewCount="0" Body="&lt;p&gt;Others have provided fine answers for the standard method of calculating the probability of getting &lt;em&gt;k&lt;/em&gt; or more heads in &lt;em&gt;n&lt;/em&gt; tosses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think you are approaching the question from the wrong direction. Few people have tossed a single coin enough times. Pearson did it around 1900: He tossed a coin 24,000 times and came up with 12,012 heads. Count Buffon tossed one 4040 times and got 2048 heads and Kerrich did it 10,000 times in a German POW camp and got 5067 heads. (Moore et al. &quot;Introduction to the Practice of Statistics&quot;, 6th ed, p.240).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to make sure a given coin is fair, you should use physical measurements and analyze the surface and composition of the coin to make sure its weight is uniformly distributed, it's not bent etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone offers you a bet where you pay 10 dollars if the coin comes up heads, and he pays 10,000 dollars if it comes up tails, you can be assured that the coin is not fair ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, we assume a generic coin picked without any special procedure is fair. Without that assumption, life as we know it would not be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://www.state.nj.us/casinos/actreg/reg/docs_chapter46/c46s01sec15to19b.pdf&quot; rel=&quot;nofollow&quot;&gt;casinos are not required to roll their dice thousands of times&lt;/a&gt; to ascertain their fairness. In fact, doing so may wear out the corners or edges. Rather, there are specific manufacturing process requirements that need to be met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, no one wastes time trying to ensure that the coin used to determine who gets the ball at the start of a football game is fair. &lt;a href=&quot;http://www.nfl.com/rulebook/cointoss&quot; rel=&quot;nofollow&quot;&gt;NFL's rules on the coin flip are simple&lt;/a&gt; and they do not involve ascertaining the fairness of the coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability will tell you that if 1,000 people each toss their &lt;em&gt;fair&lt;/em&gt; coins 30 times, most of the percentages will be very close to 50%. Determining whether an individual coin is fair is not a task for Statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf&quot; rel=&quot;nofollow&quot;&gt;Dynamical Bias in Coin Tossing&lt;/a&gt; and &lt;a href=&quot;http://www.dartmouth.edu/~chance/chance_news/recent_news/chance_news_11.02.html#item2&quot; rel=&quot;nofollow&quot;&gt;Chance News 11.02&lt;/a&gt; referenced in that article.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2010-08-08T15:18:34.913" LastActivityDate="2010-08-08T15:18:34.913" CommentCount="7" />
  <row Id="1827" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T05:53:27.510" Score="5" ViewCount="0" Body="&lt;p&gt;Aside from its incredible utility in practical computations, echelon form is also a special case of a general construction called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bruhat_decomposition&quot; rel=&quot;nofollow&quot;&gt;Bruhat decomposition&lt;/a&gt; which is important in the theory of Lie groups and algebraic groups.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-08T05:53:27.510" CommentCount="1" />
  <row Id="1828" PostTypeId="2" ParentId="1814" CreationDate="2010-08-08T06:13:16.587" Score="0" ViewCount="0" Body="&lt;p&gt;You can find references to such literature in Isaac Goldbring's &lt;a href=&quot;http://www.math.uiuc.edu/~igoldbr2/researchstatement.pdf&quot; rel=&quot;nofollow&quot;&gt;statement of research&lt;/a&gt;. In particular see the work of Robinson [13], Bate [1], Singer [14] and Hirschfeld [6] referenced there.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-08T06:13:16.587" />
  <row Id="1829" PostTypeId="1" CreationDate="2010-08-08T08:02:29.217" Score="2" ViewCount="187" Body="&lt;p&gt;I am working a bit on a collection of Linear Algebra examples, &#xA;as well as some examples on induction. This is what is taught freshman year at our university.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intend to release this to the public, either by selling printed copies or releasing it online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I do not have experience using such material myself, there are some questions I would like some opinions on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How much theory should I include? Is references to course litterature enough?&lt;/li&gt;&#xA;&lt;li&gt;Is there a format preference? Small text, so that the collection is more enviromental-friendly, or with big marginals for notes?&lt;/li&gt;&#xA;&lt;li&gt;Best way to deal with misprints?&lt;/li&gt;&#xA;&lt;li&gt;Should induction and Linear algeba be separate pieces? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please share your experience if you have done something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;An answer I seek is something along the lines of:&#xA;&quot;I am a &quot;something&quot; stident, and I prefer &quot;something&quot;, and would like to see more of &quot;something&quot;.&lt;/p&gt;&#xA;" OwnerUserId="934" LastEditorUserId="934" LastEditorDisplayName="" LastEditDate="2010-08-15T20:33:19.260" LastActivityDate="2010-08-20T21:06:15.857" CommunityOwnedDate="2010-08-08T12:30:22.677" Title="Tips for writing math solutions for others" Tags="&lt;linear-algebra&gt;&lt;examples&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1830" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T08:11:36.853" Score="0" ViewCount="0" Body="&lt;p&gt;I am working with the Schroedinger equation,&#xA;which suprisinly leads to graph theory (via cell decompositions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a good grip on these graphs, I did a lot of programming,&#xA;since doing it by hand takes way too much time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I generated about 100 examples, and from there, I could easily see what needed to be proved,&#xA;something which was quite hard prior to that discovery.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-08-08T08:11:36.853" CommunityOwnedDate="2010-08-08T08:11:36.853" CommentCount="2" />
  <row Id="1831" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T08:12:02.610" Score="3" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;I think the only &lt;em&gt;real&lt;/em&gt; utility of Cramer's rule -besides resolving 2x2 systems by hand- is that, for a general system of linear equations $AX = B$ with $\det A \neq 0$, it shows blatantly &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer%27s_rule&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cramer%27s_rule&lt;/a&gt; the continuous dependence of the solution with respect both $A$ and $B$.&lt;/li&gt;&#xA;&lt;li&gt;But, for the same reason, Cramer's rule is very dangerous because that $\det A$ appearing in the denominator: using it with a computer for small values of $\det A$ makes errors grow quickly.&lt;/li&gt;&#xA;&lt;li&gt;Nevertheless, I must confess that I feel somewhat unconfortable telling these kind of things to my students, since they could (and sometimes do) easely argue: (a) Are we going to &lt;em&gt;actually&lt;/em&gt; solve some, say, 5x5 system of linear equations &lt;em&gt;by hand&lt;/em&gt;? (b) Who cares what the computer really does in order to solve the system? -The only thing I need is the solution.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For these embarrassing objections, I have the following problem: try to solve some &lt;em&gt;simultaneuous&lt;/em&gt; systems of linear equations. That is to say, for instance, two systems $AX = B_1$ and $AX = B_2$, with the same $A$. Of course, you can apply Cramer's rule two times, but the reducing row echelon algorithm allows you to solve it with one go, since the operations for reducing the system depend &lt;em&gt;only&lt;/em&gt; on $A$. In manual computations, you can take advantage of this putting both matrices $B_1$ and $B_2$ together like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(A \vert B_1 B_2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this can be exploited further for any number of simultaneous systems $AX = B_1, \dots , AX=B_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(A \vert B_1 B_2 \dots B_n) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, inverting a matrix is a problem of solving a simultaneous system of linear equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;AX = I  \quad \Longleftrightarrow \quad (A \vert e_1 \dots e_n)  \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the identity matrix and $e_1, \dots , e_n$ is the standard basis of $\mathbb{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-08T08:44:56.617" LastActivityDate="2010-08-08T08:44:56.617" CommentCount="2" />
  <row Id="1832" PostTypeId="1" AcceptedAnswerId="1843" CreationDate="2010-08-08T09:55:43.383" Score="3" ViewCount="124" Body="&lt;p&gt;Let $m,n \geq 2$ be integers. Find all functions $f:[0,\infty) \to \mathbb{R}$ continuous at at least one point in $[0,\infty)$ and such that $$f\biggl(\frac{1}{n} \sum\limits_{i=1}^{n} x_{i}^{m} \biggr)=\frac{1}{n} \sum\limits_{i=1}^{n}(f(x_i))^{m} \quad \text{for} \ x_{i} \geq 0, i=1,2,...,n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not getting any idea to proceed.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-08T13:37:43.530" LastActivityDate="2010-08-08T14:04:07.957" Title="Functions satisfying a given condition" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1833" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T09:58:14.643" Score="2" ViewCount="0" Body="&lt;p&gt;Cryptography is a field which requires lot of Number theory. Formal languages and automata theory requires some Graph theory, and these automata things can be connected to computer science as well.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-08T09:58:14.643" CommunityOwnedDate="2010-08-08T09:58:14.643" />
  <row Id="1835" PostTypeId="1" AcceptedAnswerId="1837" CreationDate="2010-08-08T10:17:48.177" Score="-3" ViewCount="201" Body="&lt;p&gt;Given a Non abelian group $G$ with $|G|=pq$, where $p,q$ are primes $p &amp;lt; q$. How do you find the class equation of this group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Attempts: Since $G$ is non abelian we can assume that $p \mid (q-1)$ for if $ p \nmid (q-1)$ then $G$ would become cyclic and hence abelian. Now i looked at the $q$ sylow subgroups $n_{q} = 1+kq$ which says the $q-$ sylow subgroup would be normal.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-08T10:37:05.113" Title="Class equation for a Non Abelian Group of order $|G|=pq$" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1836" PostTypeId="2" ParentId="1822" CreationDate="2010-08-08T10:27:53.353" Score="4" ViewCount="0" Body="&lt;p&gt;alex already covered much of what I had wanted to say, so allow me to share something that had been written by Forman Acton in his (excellent!) &quot;Numerical Methods that Work&quot; on the matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;...the fact that this sequence of operations takes (much more) labor on the part of the computer as a direct solution of the problem by (Gaussian elimination) has never penetrated (the student's) consciousness. The glibness of the formal mathematical notation has obscured the realities of the arithmetic process. The cure is simple enough--the student should be asked to solve a 4-by-4 system, by hand, both ways. But this sort of realism is currently unfashionable--arithmetic labor being deemed ony suitable for machines.&quot;&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-08T10:27:53.353" CommentCount="1" />
  <row Id="1837" PostTypeId="2" ParentId="1835" CreationDate="2010-08-08T10:37:05.113" Score="4" ViewCount="0" Body="&lt;p&gt;In such a group, the Sylow subgroup of order $q$ is unique and normal.&#xA;The $q-1$ elements of order $q$ split into conjugacy classes of&#xA;size $p$. However there are $q$ Sylow $p$-subgroups&#xA;and so $q(p-1)$ elements of order $p$. These split into conjugacy&#xA;classes of size $q$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-08T10:37:05.113" />
  <row Id="1838" PostTypeId="1" AcceptedAnswerId="1858" CreationDate="2010-08-08T12:15:59.717" Score="5" ViewCount="214" Body="&lt;p&gt;I understand what the problem with Gimbal Lock is, such that at the North Pole, all directions are south, there's no concept of east and west. But what I don't understand is why this is such an issue for navigation systems? Surely if you find you're in Gimbal Lock, you can simply move a small amount in any direction, and then all directions are right again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this cause such a problem for navigation?&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditorDisplayName="" LastEditDate="2010-08-08T12:25:55.697" LastActivityDate="2010-08-10T02:52:13.350" Title="Why is Gimbal Lock an issue?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="1839" PostTypeId="1" AcceptedAnswerId="1841" CreationDate="2010-08-08T12:33:18.440" Score="6" ViewCount="128" Body="&lt;p&gt;Let $a_n\geq 0$ be a sequence of non-negative numbers. Consider the following two statements:&#xA;$$&#xA;\text{(I)}\qquad\qquad \lim_{n\to\infty} \frac{1}{n^2}\sum_{i=1}^n a_i =0,&#xA;$$&#xA;$$&#xA;\text{(II)}\qquad\qquad\qquad \sum_{n=1}^\infty \frac{a_n}{n^2}&amp;lt;\infty.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; Does (I) imply (II)? Does (II) imply (I)? Otherwise, please provide counterexamples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Motivation: Both statements occur in the context of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;law of large numbers&lt;/a&gt; for non-identically distributed random variables. With $a_n=\mathrm{Var}(X_n)$, one can conclude the weak LLN if the $X_n$ are pairwise uncorrelated and condition (I) holds. The strong LLN can be concluded if the $X_n$ are stochastically independent and condition (II) holds. Therefore, one might expect that (II) implies (I).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-08T16:27:41.513" LastActivityDate="2010-08-08T16:27:41.513" Title="Comparison of two convergence conditions for sequences of non-negative numbers" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="1840" PostTypeId="1" CreationDate="2010-08-08T12:52:58.083" Score="0" ViewCount="114" Body="&lt;p&gt;Given the following:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum_{i=1}^{n} a_i = 1, a_i \geq 0 \forall i$&lt;/li&gt;&#xA;&lt;li&gt;$\sum_{i=1}^{n} b_i = 1, b_i \geq 0 \forall i$&lt;/li&gt;&#xA;&lt;li&gt;$\sum_{i=1}^{n} |a_i - b_i| \leq e$ where $e \ll 1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the upper bound on:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i\cdot b_j|$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am almost sure it is bounded by $2e$ but am not able to prove it.&lt;/p&gt;&#xA;" OwnerUserId="938" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-08T13:56:21.143" LastActivityDate="2010-08-08T17:36:59.260" Title="a proof on the upperbound of this expression:" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="1841" PostTypeId="2" ParentId="1839" CreationDate="2010-08-08T12:57:02.083" Score="10" ViewCount="0" Body="&lt;p&gt;II implies I:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(we deal with the partial sums here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One applies the Cauchy-Schwarz inequality to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left( \sum_{i=1}^n \frac{a_i}{i^2} \right)\left(\sum_{i=1}^n i^2\right) \ge (\sum_{i=1}^n a_i)^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that $ \sum_{i=1}^n i^2 $ is of the magnitude $n^3$ as $n \rightarrow \infty$. So when we divide both sides by $n^4$, LHS will converge to 0 as $n \rightarrow 0$, which implies that $\displaystyle  \frac{\sum a_n}{n^2}$ converges to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I does NOT imply II:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. take $a_1,a_2$ to be anything, $\displaystyle a_n = \frac{n}{\log n}$ for $n \ge 3$. This does not satisfy II (well known). But it does satisfy I, because asymptotically, &#xA;$\displaystyle \frac{1}{n^2} \sum_{i=1}^n a_i \sim \frac{1}{n^2} \int_3^n \frac{x}{\log x} dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_integral_function&quot; rel=&quot;nofollow&quot;&gt;logarithmic integral&lt;/a&gt; has a well known approximation&#xA;$\displaystyle \int_3^n \frac{1}{\log x}dx = O\left(\frac{n}{\log n}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\displaystyle \frac{1}{n^2} \int_3^n \frac{x}{\log x} dx \leq \frac{n}{n^2} \int_3^n \frac{1}{\log x} dx = O\left(\frac{1}{\log n}\right) \rightarrow 0$ as $n \rightarrow \infty$.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2010-08-08T13:04:50.720" LastActivityDate="2010-08-08T13:04:50.720" CommentCount="1" />
  <row Id="1842" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T13:24:37.680" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Semantics of programming languages&lt;/strong&gt;: the modern version is a mixture of type theory and category theory.  Also, there is a lot of subtle issues in logic (linear logic in particular) that crops up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Design of programming languages&lt;/strong&gt;: Things like &lt;em&gt;linear logic&lt;/em&gt; have influenced the addition of a number of features to languages (or reasonable restrictions).  &lt;em&gt;monads&lt;/em&gt; in Haskell come straight out of category theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Analysis of programming languages&lt;/strong&gt;: static analysis is nowadays all based on &lt;em&gt;abstract interpretation&lt;/em&gt;, which is fundamentally an application of lattices and of Galois connections.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-08-08T13:24:37.680" CommunityOwnedDate="2010-08-08T13:24:37.680" CommentCount="1" />
  <row Id="1843" PostTypeId="2" ParentId="1832" CreationDate="2010-08-08T13:28:49.353" Score="2" ViewCount="0" Body="&lt;p&gt;Let $x_1=x_2=\ldots=x_n=x$, then $f(x^m)=f(x)^m$, and $f(0)=f(0)^m$, so $f(0)$ is either 0,1 or -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $y_i=x_i^m$, then we have $f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$ for any $y_i\in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(0)=0$, then $y_1=y$, $y_2=\ldots=y_n=0$ implies that $f(\frac{1}{n} y)=\frac{1}{n}f(y)$.&#xA;This implies $\frac{1}{n}f(\sum_{i=1}^n y_i)=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $y_3=\ldots=y_n=0$, we get that $f(y_1+y_2)=f(y_1)+f(y_2)$, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy's Functional Equation&lt;/a&gt;. Since $f$ is continuous at one point, it's linear, so $f(x)=ax$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get $a=a^m$, so the solutions in this case are $f(x)=0$, $f(x)=x$ and $f(x)=-x$ (if $m$ is odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(0)\neq 0$, WLOG let $f(0)=1$ (or else replace $f$ by $-f$ if $m$ is odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously $f(\frac{1}{n}y)=\frac{f(y)+n-1}{n}$, and $\frac{f(\sum_{i=1}^n y_i)+n-1}{n}=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$, so that $f(\sum_{i=1}^n y_i)+n-1=\sum_{i=1}^n f(y_i)\Rightarrow g(\sum_{i=1}^n y_i)=\sum_{i=1}^n g(y_i)$ where $g(y)=f(y)-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we have that $g$ satisfies Cauchy's functional equation, so that $f(y)=ay+1$. But $ay^m+1=f(y^m)=f(y)^m=(ay+1)^m$, which implies $a=0$ by comparing coefficients. So in this case the solutions are $f(x)=1$ and $f(x)=-1$ (if $m$ is odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that the solutions are $f(x)=0$, $f(x)=x$, $f(x)=1$ and if $m$ is odd $f(x)=-x$ and $f(x)=-1$.&lt;/p&gt;&#xA;" OwnerUserId="446" LastEditorUserId="446" LastEditorDisplayName="" LastEditDate="2010-08-08T14:04:07.957" LastActivityDate="2010-08-08T14:04:07.957" CommentCount="3" />
  <row Id="1844" PostTypeId="1" AcceptedAnswerId="1849" CreationDate="2010-08-08T14:52:41.617" Score="2" ViewCount="425" Body="&lt;p&gt;On a 2D plane, line X is at x radians angle, an incoming light travels at y radians angle, how to calculate the angle of the outgoing light reflected off line X? How to do this in a way to cover all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was trying to figure out &lt;a href=&quot;http://projecteuler.net/index.php?section=forum&amp;amp;id=144&quot; rel=&quot;nofollow&quot;&gt;Project Euler problem 144&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-08-09T03:35:27.663" LastActivityDate="2010-08-09T03:35:27.663" Title="How to calculate reflected light angle?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;project-euler&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="1845" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T15:47:01.057" Score="1" ViewCount="0" Body="&lt;p&gt;Let the line has equation $ax + by + c = 0$, point where light comes on the line has coordinates $(x_0, y_0)$ and direction of the light is $(v_x, v_y)$. Suppose also that line and light direction are normed: $a^2 + b^2 = 1$ and $v_x^2 + v_y^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vector $(-b, a)$ is a vector parallel to the line. Let $d = -bv_x + av_y$ --- projection of light's direction to the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then new direction of the light is $((-b)\cdot(2d) - v_x, (a)\cdot(2d) - v_y)$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-08T15:47:01.057" CommentCount="1" />
  <row Id="1846" PostTypeId="2" ParentId="709" CreationDate="2010-08-08T15:51:33.067" Score="1" ViewCount="0" Body="&lt;p&gt;There are nice answers already but I would like to add another way to look at this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadratic equation $a(x-r_1)(x-r_2) = ax^2 - a(r_1 + r_2)x + r_1 r_2 = 0$ defines a parabola, $r_1, r_2$ are the roots (intersections with the x-axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle point is between the two roots, $\frac{r_1 + r_2}{2} = -\frac{- a(r_1 + r_2)}{2a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to an interesting question about what happens when the parabola does not have real roots.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-08T15:51:33.067" />
  <row Id="1848" PostTypeId="2" ParentId="1838" CreationDate="2010-08-08T16:08:00.047" Score="0" ViewCount="0" Body="&lt;p&gt;You seem to have a very idealised idea of what is likely to happen. In the vicinity of a pole the magnetic field lines are nearly vertical. Any compass that needs gimbals is not going to be reliable over a wide area.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-08-08T16:08:00.047" CommentCount="2" />
  <row Id="1849" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T16:14:33.067" Score="3" ViewCount="0" Body="&lt;p&gt;Since you've stated all three angles in similar terms, and want a formula that works in all cases, lets use the angle with the x axis, in 360 degree terms, which accomplishes both purposes and is good for computation.  So here's the picture, with z the angle you are seeking, and a the angle of reflection...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;&lt;img src=&quot;http://djlewis.smugmug.com/photos/962600635_sUzmn-O.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, using the two triangles with the x axis as base and the fact that an exterior angle of a triangle is the sum of the other interior angles, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = x + a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = $\pi$ - 2a + z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And solving those for z in terms of x and y gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = $\pi$ + 2x - y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, this is actually far from a general analysis -- angles x, y and z could occur in any order on the x axis, but we may assume that y is to the right of x without losing generality, so there are only two other cases. Also angle x could &gt; $\frac{\pi}{2}$, but since the lines are undirected, we may assume that x &amp;lt; $\pi$. Finally, angle x could be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmmm... thinking about this some more.  When z falls to the right of y, the same formula follows because the roles of z and y are interchanged. But when z falls to the left of x, it's because the line of the reflected light intersects the x axis &quot;backwards&quot;.  And then the geometry yields z = 2x - y, or the prior formula if you take the angle of the reflected light as the supplement of z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we really need vectors for the light rays, not undirected lines, and/or the original problem is not altogether well-formed, that is the notion &quot;angle&quot; of a light ray needs to be defined as the direction of its vector. If you do that, then the angle labeled z in my diagram is not the correct vector directional angle. It should be $\pi$ + z, so the true formula, in vector direction is z = 2x - y, and that works for all cases. (Haven't checked the degenerate cases, but no reason to expect them to fail).&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-08-08T22:10:26.597" LastActivityDate="2010-08-08T22:10:26.597" CommentCount="3" />
  <row Id="1850" PostTypeId="2" ParentId="514" CreationDate="2010-08-08T16:21:28.287" Score="1" ViewCount="0" Body="&lt;p&gt;I don't know if I would consider this accessible or 'large', but the counterexample of Adyan to the famous General Burnside Problem (http://en.wikipedia.org/wiki/Burnside%27s_problem) in group theory requires an odd exponent greater than or equal to 665. The &quot;shorter&quot; counterexample (proof) due to Olshanskii requires an exponent greater than 10^10. The reason for the large number in the latter proof is essentially due to 'large scale' consequences of Gauss-Bonnet theorem for certain planar graphs expressing relations in groups. It may be that a finer analysis can show that a counterexample can occur at exponent as low as 5, but this is still not known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably essentially different than what you are asking, since we aren't forced to consider 665 because the cases 1-664 are known to be true. I thought it may be fun to point out, here, though! &lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-08-08T16:21:28.287" />
  <row Id="1851" PostTypeId="2" ParentId="709" CreationDate="2010-08-08T16:47:32.927" Score="4" ViewCount="0" Body="&lt;p&gt;An alternative approach, which isn't necessarily the &lt;em&gt;best&lt;/em&gt; in and of itself, but stick with me ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of using the fact that the vertex is the point where a &lt;em&gt;vertical&lt;/em&gt; line cuts the parabola exactly in half, we'll use the fact that it's also the point where a &lt;em&gt;horizontal&lt;/em&gt; line &lt;em&gt;just&lt;/em&gt; brushes the curve, meeting it in a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider how the horizontal line $y=d$ crosses the parabola as you adjust $d$. If $d$ places the line above or below the vertex, there are either 2 or 0 points of intersection (which is which depends upon whether the parabola points up or down); when $d$ places the line at the same height as the vertex, the parabola seems to sit comfortably on the line. (The line is &lt;em&gt;tangent&lt;/em&gt; to the parabola the point they have in common, in the same way that lines can be tangent to circles.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find the point of tangency (the vertex), then, by using algebra to determine when the intersection of our parabola and line equations have a single solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We begin with the standard find-the-intersection strategy, setting our formulas for $y$ equal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$d = a x^2 + b x + c$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When does this equation have a &lt;em&gt;single&lt;/em&gt; solution for $x$? Let's see. Rearranging and invoking the Quadratic Formula, we have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a x^2 + b x + c - d = 0$&lt;br /&gt;&lt;br /&gt;&#xA;  $x = \frac{1}{2a}\left( - b \pm \sqrt{ b^2 - 4 a (c-d) } \right)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, when the stuff under the radical &lt;em&gt;matters&lt;/em&gt;, then we have either 2 or 0 values of $x$: two values if the stuff is positive (the square root gives a quantity to add and subtract from $-b$); no roots if the stuff is negative (the square root gives an imaginary number we can't use here). We'll have our sought-after single value of $x$ when $d$ is whatever-it-has-to-be (its exact value doesn't matter) to make the stuff under the radical becomes zero, making the radical term vanish; when that happens, our value of $x$ falls out as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x = -\frac{b}{2a}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just like with the other approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;em&gt;this&lt;/em&gt; approach, however, we can consider a more sophisticated situation. It may be somewhat advanced for your class, but --if you can make sense of my rambling-- you can use it to blow your teacher's mind. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Before we begin: I wouldn't necessarily recommend this exploration for every student just beginning to learn about parabola equations. However, I think having it &quot;in the archive&quot;, as it were, can be worthwhile for people who come across the question to see that the answer can lead to some interesting places.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ask ourselves this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if we &lt;em&gt;tilt the line&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is: What can we say about the $x$-coordinate of the point of tangency where the parabola is touched by a line with slope $m$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the slanted line's equation be $y = m x + d$, where $d$ is adjustable. We can make the same observations as in the horizontal case: if $d$ places the line above or below the point of tangency, the line will meet the parabola in either 2 or 0 places; for a &lt;em&gt;special&lt;/em&gt; value of $d$, however, the line meets the parabola at the single point, the point of tangency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, we set our formulas for $y$ equal to each other ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$m x + d = a x^2 + b x + c$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... then rearrange and invoke the Quadratic Formula ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a x^2 + ( b - m ) x + ( c - d ) = 0$&lt;br /&gt;&lt;br /&gt;&#xA;  $x = \frac{1}{2a}\left( -(b-m) \pm \sqrt{stuff} \right) = \frac{1}{2a}\left(m-b\pm \sqrt{stuff}\right)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as before, we don't really care about the complicated mess that &quot;$stuff$&quot; might be, since we're only interested in the situation where it goes away anyhow.[*] Once the radical term vanishes and the dust settles, we have this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the parabola described by the equation $y = a x^2 + b x + c$, the $x$-coordinate of the point where a line of slope $m$ lies tangent to the curve is given by &lt;br /&gt;&lt;br /&gt;&#xA;  $x = \frac{1}{2a}\left( m - b \right)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the kicker that your smart friend[**] can appreciate: That statement &lt;em&gt;really is&lt;/em&gt; a fact that most people don't see until Calculus. Except there, the issue tends to be phrased a little differently: It's not about finding the $x$-coordinate of the point-of-tangency based on a given slope, but about finding the slope of the line based on the $x$-coordinate for the point-of-tangency. So, we re-phrase the above equation to yield $m$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the parabola described by the equation $y = a x^2 + b x + c$, the slope $m$ of the line tangent to the curve at a point with $x$-coordinate $x$ is given by&lt;br /&gt;&lt;br /&gt;&#xA;  $m = 2 a x + b$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll be way ahead of the game if you can see a neat way to get directly from the parabola equation to the slope formula. I'll give you a hint by showing how the formula fits in with a couple of facts you already know about the slopes of lines, and show you one step further, with a few &lt;em&gt;unnecessary&lt;/em&gt; zeros thrown in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$y = a$ &amp;emsp; slope = $0$.&lt;br /&gt;&#xA;  $y = a x + b$ &amp;emsp; slope = $a + 0$.&lt;br /&gt;&#xA;  $y = a x^2 + b x + c$ &amp;emsp; slope = $2 a x + b + 0$&lt;br /&gt;&#xA;  $y = a x^3 + b x^2 + c x + d$ &amp;emsp; slope = $3 a x^2 + 2 b x + c + 0$&lt;br /&gt;&#xA;  $y = a x^4 + b x^3 + c x^2 + d x + e$ &amp;emsp; slope = &lt;em&gt;you tell me&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Detect a pattern?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] Actually, you should care enough to convince yourself that you can find a value of $d$ that &lt;em&gt;makes&lt;/em&gt; &quot;$stuff$&quot; go away, but for now you can trust me that such a $d$ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[**] Keep in mind that &quot;smart&quot; doesn't always mean &quot;having the right answers&quot;; just as often, it means &quot;asking the right questions&quot;. The fact that you chose not to accept a formula blindly and asked your teacher &quot;why&quot; --and then steadfastly continued your quest for an answer when she wouldn't provide one-- suggests to me that your smart friend has a smart friend, too.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-08T16:47:32.927" CommentCount="1" />
  <row Id="1852" PostTypeId="1" CreationDate="2010-08-08T16:48:13.563" Score="2" ViewCount="101" Body="&lt;p&gt;Imagine a process with two variables &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt;, and two counters &lt;em&gt;hi&lt;/em&gt; and &lt;em&gt;lo&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We initialize &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; by selecting two random numbers (assume a uniform (0,1) distribution for convenience), and sorting them.  We also set the &lt;em&gt;hi&lt;/em&gt; and &lt;em&gt;lo&lt;/em&gt; counters to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each step we select a new random number &lt;em&gt;r&lt;/em&gt; from our distribution and do the following&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (r &amp;lt; min) {&#xA;  lo++;&#xA;} else if (r &amp;gt; max) {&#xA;  hi++;&#xA;} else if (lo &amp;gt; hi) {&#xA;  max = r;&#xA;  hi++;&#xA;} else if (hi &amp;gt; lo) {&#xA;  min = r;&#xA;  lo++;&#xA;} else {&#xA;  if (rand(0,1) &amp;gt; 0.5 ){&#xA;    max = r;&#xA;    hi++;&#xA;  } else {&#xA;    min = r;&#xA;    lo++;&#xA;  }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the random number is outside our range, we increment the appropriate counter, and if it lies inside our range we increment the lower counter and adjust our range appropriately.  The question is, after &lt;em&gt;n&lt;/em&gt; iterations, where do we expect min and max to end up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be seen that the difference between min and max (assuming a uniform(0, 1) distribution) is distributed as the minimum of &lt;em&gt;n + 2&lt;/em&gt; uniform (0, 1) random variables.  However, it should be possible to say considerably more about the location of &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;  This algorithm is only vaguely related to median finding.  The process given here just takes a sequence of unifomly distributed random numbers and attempts to guess the median by remembering only two values.  This, of course is not an effective method, but similar algorithms are used (remembering more than two variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process that I want to analyze has two values and two counters, and takes a sequence of random numbers, if the random number is larger than both, then increase the &lt;em&gt;hi&lt;/em&gt; counter, if the number is smaller than both then increase the &lt;em&gt;lo&lt;/em&gt; counter, and if the number lies between the two, either replace &lt;em&gt;max&lt;/em&gt; with the current number and increase &lt;em&gt;hi&lt;/em&gt; or else replace &lt;em&gt;min&lt;/em&gt; with the current number and increase &lt;em&gt;lo&lt;/em&gt;, whichever makes the &lt;em&gt;hi&lt;/em&gt; or &lt;em&gt;lo&lt;/em&gt; counters closer (in cases where either choice could be made, flip a coin).  &lt;em&gt;hi&lt;/em&gt; and &lt;em&gt;lo&lt;/em&gt; are counting the numbers that we have seen that are larger than &lt;em&gt;max&lt;/em&gt; or smaller that &lt;em&gt;min&lt;/em&gt; respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know, is how to figure out the distribution of &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; as the number of iterations becomes large, also of interest is the distribution of &lt;em&gt;hi - lo&lt;/em&gt;.  I can find the distribution of &lt;em&gt;max - min&lt;/em&gt; using elementary order statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;min = max&lt;/em&gt;, then &lt;em&gt;hi - lo&lt;/em&gt; is an ordinary one dimensional random walk, and is easy to analyze.  When they are different the walk is subtly biased towards 0.  Similarly the values of &lt;em&gt;min&lt;/em&gt; and &lt;em&gt;max&lt;/em&gt; are biased towards 0.5, I want to know how to find out by how much they are biased.&lt;/p&gt;&#xA;" OwnerUserId="943" LastEditorUserId="943" LastEditorDisplayName="" LastEditDate="2010-08-08T21:20:01.863" LastActivityDate="2010-08-08T21:20:01.863" Title="Analysis of a biased random walk related to median finding" Tags="&lt;algorithms&gt;&lt;probability&gt;&lt;stochastic-processes&gt;" CommentCount="0" FavoriteCount="0" />
  <row Id="1854" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T17:12:42.310" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/412079240.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that regardless of what line the angles are measured from, there is a parallel line through the point of reflection on the mirror for which the angle measures will be the same.  My diagram is not fully general as drawn (in terms of the angles), but the argument below is fully general, provided that all angle measures are directed&amp;mdash;that is, counterclockwise angles are positive and clockwise angles are negative.  x is the angle from a reference line to (one half of) the mirror, y is the angle from that same reference line to the incoming light ray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle from the measured half of the mirror to the incoming light ray is $y-x$ (note that if the diagram is as I'd drawn, except that x were to the other half of the mirror, $x&gt;y$ so $y-x&amp;lt;0$, which makes sense since $y-x$ would be going counterclockwise from the other half of the mirror to the incoming light ray).  Because the magnitude of the angles of incidence and reflection are the same, but the angle of reflection is the opposite direction from the other half of the mirror, the angle from the other half of the mirror to the outgoing light ray is $-(y-x)=x-y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes the directed angle from the measured half of the mirror to the outgoing light ray $\pi+(x-y)$ (start at the measured half of the mirror, rotate $\pi$ counterclockwise, then rotate $x-y$, which is directed back clockwise).  From the angle reference line to the outgoing light ray is then $x+\pi+x-y=\pi+2x-y$ (as in David Lewis's answer).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-08T17:12:42.310" />
  <row Id="1855" PostTypeId="2" ParentId="1840" CreationDate="2010-08-08T17:36:59.260" Score="3" ViewCount="0" Body="&lt;p&gt;$\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i\cdot b_j|$ = $\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i.a_j + b_i.a_j - b_i\cdot b_j|$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \sum_{i,j=1}^{n} |(a_i - b_i).a_j + (a_j - b_j).b_i|$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq \sum_{i,j=1}^{n} [|(a_i - b_i).a_j| + |(a_j - b_j).b_i|]$ ........ (using the triangle inequality ,i.e., $|x+y| \leq |x|+|y|$)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= (\sum_{i,j=1}^{n} |(a_i - b_i).a_j|) + (\sum_{i,j=1}^{n} |(a_j - b_j).b_i|)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq (\sum_{i}^{n} |(a_i - b_i).1|) + (\sum_{i}^{n} |e.b_i|)$ .... (By solving the inner summation over j and using the given (in)equations).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq e + e$&lt;br&gt;&#xA;$= 2.e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence Proved.&lt;/p&gt;&#xA;" OwnerUserId="938" LastActivityDate="2010-08-08T17:36:59.260" />
  <row Id="1856" PostTypeId="2" ParentId="1808" CreationDate="2010-08-08T19:55:32.963" Score="1" ViewCount="0" Body="&lt;p&gt;Computer graphics is a prime example of a field that needs sophisticated CS and math.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-08T19:55:32.963" CommunityOwnedDate="2010-08-08T19:55:32.963" />
  <row Id="1857" PostTypeId="2" ParentId="1778" CreationDate="2010-08-08T20:00:06.867" Score="0" ViewCount="0" Body="&lt;p&gt;As Issac pointed out, this is known as the Gamler's ruin problem. I recently wrote a couple of blog posts (&lt;a href=&quot;http://randomdeterminism.wordpress.com/2010/07/09/gamblers-ruin-2/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and the post linked from there) explaining how to calculate the ruin probability. I'll repeat part of one of the proofs here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem Formulation&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A gambler enters a casino with $n$ dollars in cash and starts playing a game where he wins with probability  $p$ and looses with probability $q = 1-p$  The gampler plays the game repeatedly, betting $1$ dollar in each round. He leaves the gave it his total fortune reaches $N$ or he runs out of money (he is &lt;em&gt;ruined&lt;/em&gt;), whichever happens first. What is the probability that the gambler is ruined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A gambler's ruin can be modeled as a one-dimensional random walk in which we are interested in the hitting probability of the absorbing states. Calculating these probabilities is fairly straightforward. Let $P_N(n)$ denote the probability that the gambler's fortune reaches $N$ dollars before he is ruined on the condition that his current fortune is $n$. Then,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$P_N(n) = p P_N(n+1) + q P_N(n-1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle [P_N(n+1) - P_N(n)] = \left(\frac q p \right)[ P_N(n) - P_N(n-1)]$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since $P_N(0) = 0$, we have that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(2) - P_N(1) = \left(\frac qp \right) P_N(1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and similarly&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(3) - P_N(2) = \left(\frac qp \right) [P_N(2) - P_N(1)] = \left( \frac qp \right)^2 P_N(1)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continuing this way, we get that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) - P_N(n-1) = \left( \frac qp \right)^{n-1} P_N(1) $.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and therefore, by adding the first $n$ such terms, we get&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) = \sum_{k=0}^{n-1} \left( \frac qp \right)^k P_N(1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moreover, we know that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(N) = \sum_{k=0}^{N-1} \left( \frac qp \right)^k P_N(1) = 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(1) = \frac 1{\sum_{k=0}^{N-1} \left( \frac qp \right)^k} = \frac { 1 - (q/p)}{\strut 1 - (q/p)^N}, \quad  p \neq q $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$P_N(1) = \frac 1N, \quad p = q $.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combining with the previous expression for $P_N(n)$ we get,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) = \begin{cases} \frac{ 1 - (q/p)^n} {\strut 1 - (q/p)^N}, &amp;amp; p \neq 1/2 \ \frac{n}{N}, &amp;amp; p = 1/2 \end{cases}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For ease of representation let $\lambda = q/p$. Then, the probability of &lt;em&gt;winning&lt;/em&gt; are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\displaystyle P_N(n) = \frac{ 1 - \lambda^n} {\strut 1 - \lambda^N}, \quad \lambda \neq 1 $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$P_N(n) = \frac{n}{N}, \quad \lambda = 1 $.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="763" LastActivityDate="2010-08-08T20:00:06.867" CommentCount="2" />
  <row Id="1858" PostTypeId="2" ParentId="1838" CreationDate="2010-08-08T20:20:18.283" Score="5" ViewCount="0" Body="&lt;p&gt;I don't imagine that this is a practical issue for navigation any longer, given the advent of GPS technology. However, it is of practical concern in 3-d animation and robotics. To get back to your navigation example, suppose that I have a mechanical gyro mounted in an airplane flying over the North Pole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gyro is only mounted on three gimbals, one of the gimbals will freeze because moving smoothly to the proper orientation would require at least one of the gimbals to flip 180 degrees instantaneously. The gimbal lock problem can be countered by adding a redundant degree of freedom in the form of an extra gimbal, an extra joint in a robotic arm, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out, it's the singularity at the poles of the representation that's the problem. Having a redundant degree of freedom helps because you can have enough information at the pole to move the correct direction. In 3-d graphics, if an axis-angle representation (Euler axis and angle) or quaternions are used instead of a triple-axis representation (Euler angles), then weird rotation artifacts due to gimbal lock are eliminated (performing a YouTube search for &quot;gimbal lock&quot; yields several visual demonstrations of the problem).&lt;/p&gt;&#xA;" OwnerUserId="950" LastActivityDate="2010-08-08T20:20:18.283" />
  <row Id="1859" PostTypeId="2" ParentId="1741" CreationDate="2010-08-08T20:42:53.020" Score="1" ViewCount="0" Body="&lt;p&gt;Sets are first-order objects in a system like ZFC, although there are systems like second-order arithmetic where the first-order objects are non-set individuals and sets of individuals are the second-order objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernays-Gödel set theory (BG set theory), which I see both WP and Wolfram prefer to call by the unwieldy term &lt;a href=&quot;http://mathworld.wolfram.com/vonNeumann-Bernays-GoedelSetTheory.html&quot; rel=&quot;nofollow&quot;&gt;Von Neumann–Bernays–Gödel set theory&lt;/a&gt;, adds second-order quantifiers over classes.  The result is conservative over ZFC (i.e., its first-order theorems are the same), but allows much shorter proofs of theorems.  You can formalise talk of classes in first-order set theory by using &lt;em&gt;class terms&lt;/em&gt;, where you treat one-place predicates as defining a set by extension, but this is weaker than using second-order quantifiers, and an important issue in second-order set theories is what classes a particular set theory is committed to.  This question can be addressed axiomatically in third-order set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen the usage &quot;second-order set&quot; and &quot;third-order set&quot; to talk about classes and their third-order generalisation.  &lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-08-08T20:42:53.020" />
  <row Id="1860" PostTypeId="1" CreationDate="2010-08-08T20:43:43.737" Score="4" ViewCount="327" Body="&lt;p&gt;Prove (or disprove) the following statement:&#xA;For any positive integers $x,y,t$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{i=1}^{t(y+1)-1} \frac{1}{t(xy+x-1)-x+i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an increasing function of $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts:&#xA;The statement appears to be true numerically.&#xA;Tried some obvious bounds to compare the sums for consecutive values of $t$ but didn't find one that was strong enough to prove the statement. &lt;/p&gt;&#xA;" OwnerUserId="951" LastActivityDate="2010-08-12T17:36:50.857" Title="Comparing sums of reciprocals" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="1861" PostTypeId="2" ParentId="1766" CreationDate="2010-08-08T20:53:10.920" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a more conceptual answer that doesn't resort to inequalities ad hoc.  Assume $a_n &gt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can be rephrased, by writing $(a_n)^3 = 1/(n^{3/2 + p})$ (with $p$ a function of $n &gt; 1$, and ignoring $a_1$), as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If $\Sigma 1/(n^{3/2 + p})$ converges then $\Sigma 1/(n^{3/2 + (p/3)})$ converges&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation moves the series toward the (convergent) one with $p=0$.  This preserves convergence, because the series can be split into the terms with $p \leq 0$ and $p&gt;0$.  For the first set, convergence is improved, and for the second set, the sum is dominated by $\Sigma 1/n^{3/2}$. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-08T22:07:07.723" LastActivityDate="2010-08-08T22:07:07.723" />
  <row Id="1862" PostTypeId="1" AcceptedAnswerId="1863" CreationDate="2010-08-08T20:53:50.873" Score="3" ViewCount="125" Body="&lt;p&gt;I've been reading about tropical geometry and many papers reference polyhedral fans. I feel like I have a decent intuitive picture of what they are from reading articles but I still haven't been able to guess the general definition. All the ones I've encountered have been systems of linear inequalities, so that is my best guess at a general definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments on where they appeared first historically or links/books to general resources on learning about them would be appreciated. Also, I'm curious to know what other areas of math these show up in?&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-21T14:39:53.640" Title="What is the rigorous definition of polyhedral fan? What are some good resources to learn about them? What context do they arise naturally in?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="1863" PostTypeId="2" ParentId="1862" CreationDate="2010-08-08T21:12:44.340" Score="5" ViewCount="0" Body="&lt;p&gt;A polyhedral cone is a subset of a real vector space which is the intersection of finitely many closed half spaces. (The defining planes of these half spaces must pass through $0$.) A fan is a finite set $F$ of polyhedral cones, all living in the same vector space, such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) if $\sigma$ is a cone in $F$, and $\tau$ is a face of $\sigma$, then $\tau$ is in $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) if $\sigma$ and $\sigma'$ are in $F$, then $\sigma \cap \sigma'$ is a face of both $\sigma$ and of $\sigma'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://sbseminar.wordpress.com/2009/02/18/toric-varieties-and-fans/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; of mine might help you visualize these definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most mathematicians I know learned fans from Fulton's &lt;em&gt;Toric Varieties&lt;/em&gt;. This would involve learning a lot of algebraic geometry on top of your combinatorics, although it is algebraic geometry that is very relevant to tropical geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a pure combinatorics reference, have you tried Chapter 2 of &lt;a href=&quot;http://www.math.ucdavis.edu/~deloera/BOOK/final.pdf&quot; rel=&quot;nofollow&quot;&gt;De Loera, Rambau and Santos&lt;/a&gt;? They focus on polyhedral complexes, which is the more general setup where you don't require that the half spaces pass through $0$, but they talk about fans as well. I haven't had a chance to look at it yet but, based on my knowledge of the authors, I expect it is very good.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-08-21T14:39:53.640" LastActivityDate="2010-08-21T14:39:53.640" CommentCount="3" />
  <row Id="1864" PostTypeId="2" ParentId="1116" CreationDate="2010-08-08T21:14:51.033" Score="1" ViewCount="0" Body="&lt;p&gt;This problem is tantalizing. It may be a &quot;toy&quot;-version of some substantial ongoing research in topology, so it's valuable and interesting.  It has immediate analogs in areas of classical topology, knot theory, universal algebra, and topological field theory.  By toy-version, I don't mean it's necessarily easy to solve, but that it can be used to present some advanced ideas in a non-technical way.  Forgive me because what follows is not an answer to your question, but a rambling list of related problems and possible applications. I only hope it inspires someone here to investigate the connections in algebraic topology. (Thus, community wiki.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are asking to count equivalence classes of parametrized rectangle framings (where a parametrized framing is a framing with precise line positions).  In some sense, this type of enumeration problem comes up often in algebraic topology when one wants to understand a cellular decomposition of a complicated space.  Take for instance, the space of n-planes in R^d.  These spaces are called Grassmannians and their topology can be analyzed by breaking them down into subsets of various dimensions.  These subsets are called Schubert cells.  Counting these Schubert cells and understanding how they fit together was an important advancement.  See the wikipedia entries on &lt;strong&gt;Grassmannian&lt;/strong&gt; and &lt;strong&gt;Enumerative Geometry&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your parametrized rectangle framing spaces (denoted perhaps Fi^{n} where i indexes the equivalence classes of framings having n sub-rectangles) are analogous to the cells of the Grassmannian, but the resulting stratified space is some universal space of parametrized rectangular framings which includes framings with an arbitrarily large but finite number of sub-rectangles.  That is, each of your framing spaces F (the equivalence classes you want to count)  represents a &quot;cell&quot; of some fixed dimension which degenerates into lower degree framings at the boundaries.  By carefully studying the simple shape of each cell and how they degenerate into other cells, we can create a universal space $F^{\infty}$ = $\bigcup$ $F^{n}_{i}$ (&lt;em&gt;modulo the boundary case equivalence relations&lt;/em&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes one can approach the topology of this universal space by other means, and that global knowledge can be used to count the number of cells needed in each stratum.  Finally, as you point out, the spaces have certain symmetries which may pass up to the full space in an interesting way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stratification trick comes up all the time  If you want to understand some large space, find a way to break it down into strata which fit together in an analyzable way.  Or, find a related space which has a natural stratification, analyze this stratified space, and make inferences about the original space.  A great, non-trivial use of this in the 90's was when Vassiliev realized that knots could be studied by analyzing the space of non-knots which admitted a very clear stratification (essentially stratify the non-knots by the number of points where the circle map is non-injective)  Vassiliev found clear topological structure in this space and this allowed him to make claims about the structure of the set of knots.  This led others like Kontsevich and Bar Natan to provide real computational tools using tricks for counting and integrating over cells.  For instance, Mathworld has two good introductory articles on &lt;strong&gt;Vassiliev Invariants&lt;/strong&gt; and &lt;strong&gt;Kontsevich Integrals&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, your rectangle framing space is evocative of the little-disks operad which encodes an algebraic structure.  See the wikipedia page:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Operad_theory#.22Little_something.22_operads&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Operad_theory#.22Little_something.22_operads&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever we have a set of topological spaces such that elements of the spaces can be geometrically combined to get higher degree spaces, it hints that the geometry can be studied using techniques from algebra.  If you imagine putting variables in each of your frame's sub-rectangles, you've sort of described an algebraic operation.  Imagine that every one of your parametrized n-framings defined a map from n inputs to 1 output, an &quot;n-ary operation.&quot;  Further assume that the geometry forces some consistency condition that demands that when you place the result of a p-ary operation into one of the sub-rectangles in a q-ary frame, you get just what you'd get from the corresponding (p+q-1)-ary frame operation.  This means your algebra must satisfy certain standard relations (maybe up to homotopy) and then these operations may descend to operations on an appropriately defined cohomology theory.  That is, the topology of your universal framing space may index various operations which satisfy certain relations exactly.  See for instance the ncatlab.org entries on &lt;strong&gt;operad&lt;/strong&gt;, &lt;strong&gt;L-infinity-algebra&lt;/strong&gt;, and &lt;strong&gt;A-infinity-algebra&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-08-08T21:14:51.033" CommunityOwnedDate="2010-08-08T21:16:11.643" />
  <row Id="1865" PostTypeId="2" ParentId="307" CreationDate="2010-08-08T21:27:37.610" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ (p-1)!\ mod\ p\:$ is the product of all elts of $\rm\: {\mathbb F}_p^*$. The map $\rm n \mapsto n^{-1}$ is a permutation of $\rm\:{\mathbb F}_p^*\:$ of order 2 so it decomposes into cycles of length 1 or 2, which partition the product. The 2-cycles $\rm (n, n^{-1})$ have product 1 so they can be deleted from the product, leaving only the product of 1-cycles $\rm (n)$. They satisfy $\rm\: n^{-1} = n \Rightarrow n^2 = 1 \Rightarrow n = \pm 1\:$ via $\rm{\mathbb F}_p$ a field. So the product reduces to $\:-1*1 = -1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes: if a finite abelian group has a unique elt of order 2 then it is equal to the product of all the elts; otherwise the product is 1, e.g. see this &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/784a5eb7087387eb&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; for hints.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how we've exploited the existence of a symmetry - here an involution that induces a natural pairing of elts. Frequently involution and reflection symmetries lie at the heart of elegant proofs, e.g. see the elegant proof by Liouville, &lt;a href=&quot;http://eprints.maths.ox.ac.uk/677/1/invariant.pdf&quot; rel=&quot;nofollow&quot;&gt;Heath-Brown&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/31113&quot; rel=&quot;nofollow&quot;&gt;Zagier&lt;/a&gt;, that every prime $\equiv 1 \pmod 4$ is a sum of 2 squares, or the little-known beautiful &lt;a href=&quot;http://mathoverflow.net/questions/33726&quot; rel=&quot;nofollow&quot;&gt;reflective&lt;/a&gt; generation of the ternary tree of primitive Pythagorean triples due to Aubry.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-14T22:30:00.247" LastActivityDate="2010-10-14T22:30:00.247" />
  <row Id="1866" PostTypeId="2" ParentId="307" CreationDate="2010-08-08T22:02:25.603" Score="2" ViewCount="0" Body="&lt;p&gt;Here are a couple possible proofs of Wilson's theorem for $p&gt;2$ ($p=2$ is easily checked):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We have that $x^{p-1}-1$ has roots $1,2,\ldots,p-1$ over $\mathbb{Z}/p\mathbb{Z}$ (by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's Little Theorem&lt;/a&gt;). But as $\mathbb{Z}/p\mathbb{Z}$ is a field, we have unique factorization of polynomials, so that $x^{p-1}-1=(x-1)(x-2)\ldots(x-(p-1))$. Comparing constant terms wields Wilson's theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $g$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive root&lt;/a&gt; modulo $p$. Then $(p-1)!\equiv g\times g^2\times \ldots \times g^{p-1}=g^{p\frac{p-1}{2}}\equiv g^{\frac{p-1}{2}}\bmod{p}$ by Fermat's Little Theorem, and $g^{\frac{p-1}{2}}\equiv -1 \bmod{p}$ because if $(g^{\frac{p-1}{2}})^2=g^{p-1}\equiv 1 \bmod{p}$ and $g^{\frac{p-1}{2}}\not \equiv 1 \bmod{p}$ by the definition of primitive root.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-08T22:02:25.603" />
  <row Id="1867" PostTypeId="2" ParentId="1844" CreationDate="2010-08-08T23:08:50.270" Score="0" ViewCount="0" Body="&lt;p&gt;Ok, here's another way to look at it, using vector directions and kinda intuitive, but I think it is pretty close to a real proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the mirror line X horizontal, that is, its angle x = 0. The it's clear that the vector angle, z, of the reflected light is the negative of the vector angle, y, of the light itself: z = -y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://djlewis.smugmug.com/photos/962899093_ZPQ96-O.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rotate line X by d degrees around the point of reflection, either way, leaving the original light ray (angle y) fixed and letting the reflected ray (angle z) rotate around the point of reflection to maintain the angle of reflection equal to the angle of incidence. Assuming counterclockwise rotation, d &gt; 0, this &quot;pushes&quot; the reflected line by 2d, one d for a direct push to maintain the angle fo reflection, and another d because the angle of incidence also increases by d, so the reflected light must rotate that much more to keep the two angles of the light equal. Likewise when d &amp;lt; 0 for clockwise rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are increasing (or decreasing, for counterclockwise rotation) angle x by d, but angle z (vector angle) by 2d. Hence...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = 2d - y = 2x - y&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-08-08T23:08:50.270" />
  <row Id="1869" PostTypeId="2" ParentId="1196" CreationDate="2010-08-09T00:07:09.457" Score="1" ViewCount="0" Body="&lt;p&gt;Alternatively, $f:{\mathbb N} \rightarrow {\mathbb Z}/3{\mathbb Z} \;$ is constant since $f(n+1) = f(n)$, so $f(n) = f(0) = 0$. This reduces the induction to a trivial lemma: $f(n)$ is constant if $f(n+1) = f(n)$ for all $n\in \mathbb N$, a trivial telescopy.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-09T00:07:09.457" />
  <row Id="1870" PostTypeId="1" AcceptedAnswerId="2781" CreationDate="2010-08-09T00:29:13.923" Score="4" ViewCount="408" Body="&lt;p&gt;What is the the most &lt;em&gt;educational&lt;/em&gt; software for high school and college math?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not the one that just gives you the answer, but has &lt;strike&gt;any of&lt;/strike&gt; the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: Software mentioned in answer added in brackets&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;provides step-by-step solutions [&lt;em&gt;WolframAlpha&lt;/em&gt;];&lt;/li&gt;&#xA;&lt;li&gt;can verify your step-by-step solution [&lt;em&gt;empty&lt;/em&gt;];&lt;/li&gt;&#xA;&lt;li&gt;can give you hints [&lt;em&gt;empty&lt;/em&gt;];&lt;/li&gt;&#xA;&lt;li&gt;can generate problems similar to given one [&lt;em&gt;empty&lt;/em&gt;];&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or something along those lines.&lt;/li&gt;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="956" LastEditorDisplayName="" LastEditDate="2010-08-09T17:12:26.533" LastActivityDate="2010-09-07T20:35:49.157" CommunityOwnedDate="2010-08-09T00:57:28.670" Title="Educational Math Software" Tags="&lt;math-software&gt;&lt;education&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" />
  <row Id="1871" PostTypeId="2" ParentId="1870" CreationDate="2010-08-09T00:44:04.687" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;. Try, for instance &quot;solve x^2-x-1=0&quot; and click &quot;show steps&quot;.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-09T00:44:04.687" CommunityOwnedDate="2010-08-09T02:57:41.033" />
  <row Id="1872" PostTypeId="1" CreationDate="2010-08-09T01:52:39.957" Score="4" ViewCount="189" Body="&lt;p&gt;Both matrix multiplication and quaternion multiplication are non-commutative; hence the use of terms like &quot;premultiplication&quot; and &quot;postmultiplication&quot;. After encountering the concept of &quot;quaternion matrices&quot;, I am a bit puzzled as to how one may multiply two of these things, since there are at least four ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some searching has netted &lt;a href=&quot;http://en.cnki.com.cn/Article_en/CJFDTOTAL-LXXB198402006.htm&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, but not having any access to it, I have no way towards enlightenment except to ask this question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are indeed these four ways to multiply quaternion matrices, how does one figure out which one to use in a situation, and what shorthand might be used to talk about a particular version of a multiplication?&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-12T04:11:07.923" Title="On multiplying quaternion matrices" Tags="&lt;linear-algebra&gt;&lt;quaternions&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="1" />
  <row Id="1873" PostTypeId="1" CreationDate="2010-08-09T03:46:53.487" Score="0" ViewCount="130" Body="&lt;p&gt;I need to find eigenvalues/eigenvectors of different kinds of nxn matrices. For example as listed below. What is the typical process? Should I always go by the route of finding eigenvalues by finding roots of characteristic polynomial and then getting eigenvectors by solving (A-lambda*I)*x = 0?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     [,1] [,2] [,3]&#xA;[1,]    2    0    0&#xA;[2,]    1    2    0&#xA;[3,]    0    1    2&#xA;&#xA;      [,1] [,2] [,3] [,4]&#xA; [1,]    4    1    1    1&#xA; [2,]    1    4    1    1&#xA; [3,]    1    1    4    1&#xA; [4,]    1    1    1    4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These are just examples. Typically I want to find eigenvectors of nxn matrices. If you can show me the process of finding solution of one of these matrices, that would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-09T07:38:35.497" LastActivityDate="2010-08-09T11:10:52.327" Title="process of finding eigenvalues and eigenvectors" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="1874" PostTypeId="1" AcceptedAnswerId="1875" CreationDate="2010-08-09T03:55:53.467" Score="6" ViewCount="98" Body="&lt;p&gt;I recently noticed a few things in some recent questions on MO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the fundamental group of $S^2$ minus, say, 4 points, is $\langle a,b,c,d\ |\ abcd=1\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The fundamental group of a torus minus a point is $\langle a,b,c\ |\ [a,b]c=1\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just wondering if you have a manifold $M$, and you know a presentation of its fundamental group, can you quickly get a presentation of the fundamental group of $M$ minus $n$ points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this could merely be a coincidence.  These are both surfaces, and both their fundamental groups only have one relation.  But perhaps there is more to it than that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any and all helpful comments appreciated,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-09T04:21:41.677" Title="Presentation of the fundamental group of a manifold minus some points" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1875" PostTypeId="2" ParentId="1874" CreationDate="2010-08-09T04:21:41.677" Score="7" ViewCount="0" Body="&lt;p&gt;The basic tool here is Van Kampen's theorem.  Let $M$ be the manifold, and let $M' = M\setminus {x}.$  Let $D$ be a small ball around $x$.&#xA;Then $M = M' \cup D,$ and $M \cap D = D\setminus {x}$ is homotopic to $S^{n-1}$ (here $n$ &#xA;is the dimension of $M$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Van Kampen's theorem says that $\pi_1(M) = \pi_1(M')*_{\pi_1(S^{n-1})} \pi_1(D)$.&#xA;If $n &gt; 2,$ then $S^{n-1}$ is simply connected, and of course $\pi_1(D)$ is simply connected,&#xA;and so this reduces to $\pi_1(M) = \pi_1(M')$; in other words, deleting points doesn't change&#xA;$\pi_1$ in dimensions $&gt; 2$.  (You can probably convince yourself of this directly: imagine you have some loop contracting in a 3-dimensional space; then if you remove a point, you can always just perturb the contraction slightly so that it misses the point.  Of course,&#xA;this is not the case in two dimensions.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 2$, then $\pi_1(S^1) = \mathbb Z$ is infinite cyclic, while $\pi_1(D)$ is trivial&#xA;again.  So we see that $\pi_1(M)$ is obtained from $\pi_1(M')$ by killing a loop.   One can be more precise, of course (and I will restrict myself to the orientable case, so as to make life easier):  if you have a genus $g$ surface, with $r&gt;0$ punctures, and also&#xA;$g + r &gt; 1$, then $\pi_1$ is a free group on $2 g + r - 1$ generators. (Every puncture after the first adds another independent loop, namely the loop around that puncture.) If $g \geq 1,$ and $r = 0$, then $\pi_1$ is obtained via Van Kampen as above: you&#xA;begin with a free group on $2g$ generators for $M'$, and then you kill off the loop&#xA;around the puncture when you fill it in (so you get the standard presentation for the&#xA;$\pi_1$ of a compact orientable surface of genus $g \geq 1$).  (Conversely, going from&#xA;the compact surface $M$ to the once-punctured surface $M'$ does not add any generators&#xA;to $\pi_1$, but gets rid of a relation.)&#xA;If $g = 0$ and $r \leq 1,$ then you have either a disk ($r = 1$) or a sphere ($r = 0$)&#xA;and so $\pi_1$ is trivial.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-09T04:21:41.677" CommentCount="1" />
  <row Id="1876" PostTypeId="1" AcceptedAnswerId="1911" CreationDate="2010-08-09T05:11:09.750" Score="1" ViewCount="139" Body="&lt;p&gt;I am not sure if this question is appropriate here but would appreciate any help. Let $n&gt;2$ and $x_1,\cdots,x_n$ be real numbers. What is the infimum of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \sum_{i=1}^n \frac{1}{(1+x_i)^2} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to the constraint $\prod_{i=1}^nx_i=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar question for: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B = \sum_{i=1}^n \frac{1}{(1-x_i)^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2010-08-09T06:19:44.423" LastActivityDate="2010-08-09T14:28:37.840" Title="Infimum of some quantities" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1877" PostTypeId="1" AcceptedAnswerId="1883" CreationDate="2010-08-09T05:37:43.387" Score="3" ViewCount="92" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is a standard and important fact in basic calculus/real analysis that a continuous function on a compact metric space is in fact uniformly continuous.  That is, suppose (X,d) is a compact metric space and f: X &amp;rarr; &amp;#8477; is such that for every x&amp;isin;X and &amp;epsilon;&gt;0 there exists &amp;delta;&gt;0 such that d(x,y)&amp;lt;&amp;delta; implies |f(x)-f(y)|&amp;lt;&amp;epsilon;.  Then in fact, such a &amp;delta; can be chosen independently of x.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Does a similar statement hold regarding &lt;em&gt;semi&lt;/em&gt;-continuous functions?  For concreteness, let's consider upper semi-continuous functions, so suppose (X,d) is compact and f: X &amp;rarr; &amp;#8477; has the property that for every x&amp;isin;X and &amp;epsilon;&gt;0 there exists &amp;delta;&gt;0 such that d(x,y)&amp;lt;&amp;delta; implies f(y) &amp;lt; f(x)+&amp;epsilon;.  (Note the asymmetry of x and y in this definition.)  Then is it true that &amp;delta;=&amp;delta;(&amp;epsilon;) can be chosen independently of x?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reformulation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given $\delta, \epsilon &gt; 0$, consider the set&#xA;$$&#xA;X_\delta^\epsilon := \lbrace x\in X \mid f(y) &amp;lt; f(x) + \epsilon \text{ for every } y\in B(x,\delta) \rbrace.&#xA;$$&#xA;Then $f$ is upper semi-continuous if and only if $\bigcup_{\delta&gt;0} X_\delta^\epsilon = X$ for every $\epsilon &gt; 0$, and $f$ is &lt;em&gt;uniformly&lt;/em&gt; upper semi-continuous if and only if this union stabilises -- that is, if for every $\epsilon &gt; 0$ there exists $\delta&gt;0$ such that $X_\delta^\epsilon = X$.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditorDisplayName="" LastEditDate="2010-08-09T15:11:17.877" LastActivityDate="2010-08-09T15:11:17.877" Title="Uniform semi-continuity" Tags="&lt;real-analysis&gt;" AnswerCount="2" />
  <row Id="1878" PostTypeId="2" ParentId="501" CreationDate="2010-08-09T05:59:45.900" Score="6" ViewCount="0" Body="&lt;p&gt;Set theory with all sets finite has been studied, is a familiar theory in disguise, and is enough for most/all concrete real analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, Zermelo-Fraenkel set theory with the Axiom of Infinity replaced by its negation (informally, &quot;there is no infinite set&quot;) is equivalent to first-order Peano Arithmetic.  Call this system &lt;em&gt;finite ZF&lt;/em&gt;, the theory of hereditarily finite sets.  Then under the Goedel arithmetic encoding of finite sets, Peano Arithmetic can prove all the theorems of Finite ZF, and under any of the standard constructions of integers from finite sets, Finite ZF proves all the theorems of Peano Arithmetic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implication is that theorems unprovable in PA involve intrinsically infinitary reasoning.  Notably, finite ZF was used as an equivalent of PA in the Paris-Harrington paper &quot;A Mathematical Incompleteness in Peano Arithmetic&quot; which proved that their modification of the finite Ramsey theorem can't be proved in PA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real numbers and infinite sequences are not directly objects of the finite ZF universe, but there is a clear sense in which real (and complex, and functional) analysis can be performed in finite ZF or in PA.  One can make statements about $\pi$ or any other explicitly defined real number, as theorems about a specific sequence of rational approximations ($\forall n P(n)$) and these can be formulated and proved using a theory of finite sets.  PA can perform very complicated induction proofs, i.e., transfinite induction below $\epsilon_0$.   In practice this means any concrete real number calculation in ordinary mathematics.  For the example of the prime number theorem, using complex analysis and the Riemann zeta function, see Gaisi Takeuti's &lt;em&gt;Two Applications of Logic to Mathematics&lt;/em&gt;.    More discussion of this in a MO thread and my posting there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence/31942#31942&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence/31942#31942&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof theory in general and reverse mathematics in particular contain analyses of the logical strength of various theorems in mathematics (when made suitably concrete as statements about sequences of integers), and from this point of view PA, and its avatar finite set theory, are very powerful systems.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-09T06:22:45.603" LastActivityDate="2010-08-09T06:22:45.603" CommentCount="0" />
  <row Id="1879" PostTypeId="2" ParentId="1876" CreationDate="2010-08-09T06:12:51.163" Score="0" ViewCount="0" Body="&lt;p&gt;As it is, this question doesn't quite make sense.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Infimum&quot; rel=&quot;nofollow&quot;&gt;infimum&lt;/a&gt; of a set X of real numbers is its greatest lower bound.  Currently you don't actually have a set of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your after inf(X) of the set $$X:=\{\sum_{i=1}^n \frac{1}{(1+x_i)^2}:x_1 \cdot x_2 \cdots x_n=1\},$$ then the two obvious tries are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When $x_1=x_2=\cdots=x_n=1$ which identifies that the element $n/2^2 \in X$.&lt;/li&gt;&#xA;&lt;li&gt;When $x_1 \rightarrow 0$ from the positive direction and $x_2=x_3=\cdots=x_n=(1/x_1)^{n-1} \rightarrow \infty$ (if n is fixed).  Hence inf(X) is at most 1.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first case is better when $n \in \{1,2,3\}$.  This suggests to me that finding inf(X) is not going to be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-08-09T09:04:19.363" LastActivityDate="2010-08-09T09:04:19.363" CommentCount="6" />
  <row Id="1880" PostTypeId="1" AcceptedAnswerId="1882" CreationDate="2010-08-09T06:21:19.367" Score="2" ViewCount="671" Body="&lt;p&gt;This seems counterintuitive, but 22/7 is closer to pi than 3.14=314/100 which has a significantly greater denominator.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is 22/7 a better approximation for pi than 3.14?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This has important implications:  e.g. Should &quot;pi day&quot; by the 14th of March or 22nd of July?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-04T16:11:35.890" Title="Why is 22/7 a better approximation for pi than 3.14?" Tags="&lt;approximation&gt;&lt;pi&gt;" AnswerCount="4" CommentCount="11" FavoriteCount="3" />
  <row Id="1881" PostTypeId="2" ParentId="1880" CreationDate="2010-08-09T06:39:44.057" Score="10" ViewCount="0" Body="&lt;p&gt;Well, just measure |pi - 22/7| and |pi-3.14| ...&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-08-09T06:39:44.057" CommentCount="2" />
  <row Id="1882" PostTypeId="2" ParentId="1880" CreationDate="2010-08-09T06:42:08.903" Score="12" ViewCount="0" Body="&lt;p&gt;It has to do with the continued fraction expansion of $\pi$.  Suppose $[a_1, a_2, \ldots]$ is the continued fraction of an irrational number $\alpha$ -- that is, if $a_n$ is the (essentially unique) sequence of natural numbers such that if we define partial convergents by $x_1 = a_1$, $x_2 = a_1 + 1/a_2$, $x_3 = a_1 + 1/(a_2 + 1/a_3)$, $x_4 = a_1 + 1/(a_2 + 1/(a_3 + 1/a_4))$, and so on, then $\alpha = \lim_{n\to\infty} x_n$.  Then the partial convergents $x_n$ are rational numbers that approximate $\alpha$ better than anything that is not a partial convergent, in the following sense:  a rational number $\frac pq$ satisfies the inequality $|\alpha - \frac pq| &amp;lt; \frac 1{2q^2}$ if and only if $\frac pq$ is one of the convergents $x_n$.  (One could, of course, come up with different notions of what constitutes a &quot;good&quot; approximation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continued fraction expansion of $\pi$ is $[3,7,15,1,292,1,1,\dots]$, so the first few convergents are $3$, $\frac{22}{7}$, $\frac{333}{106}$, $\frac{355}{113}$, etc.  Thus $\frac{22}{7}$ is a better approximation than $\frac{314}{100}$ (in the above sense) because it appears in the list of partial convergents, while $\frac{314}{100}$ does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the approximation $x_n$ is best when the coefficient $a_{n+1}$ is quite large, so the size of $a_5 = 292$ means that $x_4 = \frac{355}{113}$ is a particularly good approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the risk of self-promotion, I wrote a brief exposition of all this in a bit more detail -- you can find it on my website if you're interested, at &lt;a href=&quot;http://www.math.psu.edu/climenha/contfrac.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.psu.edu/climenha/contfrac.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditorDisplayName="" LastEditDate="2010-08-21T03:59:26.697" LastActivityDate="2010-08-21T03:59:26.697" CommentCount="17" />
  <row Id="1883" PostTypeId="2" ParentId="1877" CreationDate="2010-08-09T07:15:14.900" Score="3" ViewCount="0" Body="&lt;p&gt;f(x)=0 (x&amp;le;0), f(x)=-1/x (x&amp;gt;0) is upper semi-continuous on [-1;1] — but not uniformly.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-09T07:15:14.900" CommentCount="2" />
  <row Id="1884" PostTypeId="1" AcceptedAnswerId="1885" CreationDate="2010-08-09T07:27:08.177" Score="0" ViewCount="116" Body="&lt;p&gt;How would you find eigenvalues/eigenvectors of a nxn matrix where each diagonal entry is scalar 'd' and all other entries are '1' ? I am looking for a decomposition but cannot find anything for this. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; a&#xA;[,1] [,2] [,3] [,4]&#xA;[1,]    2    1    1    1&#xA;[2,]    1    2    1    1&#xA;[3,]    1    1    2    1&#xA;[4,]    1    1    1    2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-08-09T07:45:18.133" Title="find eigenvectors/eigenvalues of this matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1885" PostTypeId="2" ParentId="1884" CreationDate="2010-08-09T07:45:18.133" Score="5" ViewCount="0" Body="&lt;p&gt;The matrix is $(d-1)I + J$ where $I$ is the identity matrix and $J$ is the all-ones matrix, so once you have the eigenvectors and eigenvalues of $J$ the eigenvectors of $(d-1)I + J$ are the same and the eigenvalues are each $d-1$ greater.  (Convince yourself that this works.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $J$ has rank $1$, so it has eigenvalue $0$ with multiplicity $n-1$.  The last eigenvalue is $n$, and it's quite easy to write down all the eigenvectors.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-09T07:45:18.133" CommentCount="1" />
  <row Id="1886" PostTypeId="2" ParentId="594" CreationDate="2010-08-09T07:50:15.627" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Taken From I.N. Herstein&lt;/strong&gt; There are 2 results which i am going to use.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $p$ be a prime integer and suppose that for some integer $c$ relatively prime to $p$ we can find integers $x$ and $y$ such that $x^{2}+y^{2}=cp$. Then $p$ can be written as the sum of 2 squares.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $p$ is a prime of the form $4n+1$, then we can solve the congruence $x^{2} \equiv \ -1 \  (mod)  \ p$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now the main result. If $p$ is a prime of the form $4n+1$ then $p$ is the sum of 2 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. By 2 there exists and $x$ such that $x^{2} \equiv -1 \text{mod} \ p$. So $x$ can be chose such that $0 \leq x \leq (p-1)$. We can restrict the size of $p$ even further, namely to satisfy $|x| \leq \frac{p}{2}$. For if $x &gt; p/2$ then, $y=p-x$ satisfies $y^{2} \equiv -1 \text{mod} \ p$ but $|y| \leq p/2$. Thus we may assume that we have an integer $x$ such that $|x| \leq p/2$ and $x^{2}+1$ is a multiple of $p$ say $cp$. Now $cp=x^{2}+1 \leq p^{2}/4 +1 &amp;lt; p^{2}$, hence $c &amp;lt; p$ and hence $(c,p)=1$. Invoking (1) we have $p=a^{2}+b^{2}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-15T18:34:46.983" LastActivityDate="2010-08-15T18:34:46.983" CommentCount="2" />
  <row Id="1887" PostTypeId="2" ParentId="1808" CreationDate="2010-08-09T07:56:10.043" Score="2" ViewCount="0" Body="&lt;p&gt;Much of data mining requires an understanding of high dimensional geometry. Lots of machine learning also requires an understandind of advanced probability theory.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-08-09T07:56:10.043" CommunityOwnedDate="2010-08-09T07:56:10.043" CommentCount="1" />
  <row Id="1888" PostTypeId="1" AcceptedAnswerId="1929" CreationDate="2010-08-09T08:04:47.353" Score="-1" ViewCount="138" Body="&lt;p&gt;Let $\mathbb{Q}$ be the field of rational numbers. What are the necessary and sufficient conditions on $a$ and $b$ so that the splitting field of $x^{3}+ax+b$ has degree 3 exactly over $\mathbb{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-09T16:21:45.183" Title="Necessary Conditions For a Polynomial" Tags="&lt;galois-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1889" PostTypeId="2" ParentId="1808" CreationDate="2010-08-09T08:23:58.353" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Computable_analysis&quot; rel=&quot;nofollow&quot;&gt;Computable analysis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-08-09T08:23:58.353" CommunityOwnedDate="2010-08-09T08:23:58.353" />
  <row Id="1890" PostTypeId="1" AcceptedAnswerId="1891" CreationDate="2010-08-09T08:27:40.857" Score="-4" ViewCount="480" Body="&lt;p&gt;Let $f:[0,1] \to [0,1]$ be an injection such that $f(2x-f(x))=x$ for $x \in [0,1]$. Prove that $f(x)=x$ for $x \in [0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-28T12:48:26.433" LastActivityDate="2010-11-09T17:55:45.697" Title="A characterization of the identity map of the interval [0,1]" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="41" />
  <row Id="1891" PostTypeId="2" ParentId="1890" CreationDate="2010-08-09T10:09:51.730" Score="5" ViewCount="0" Body="&lt;p&gt;Since other people think this could be a homework question, I'll only give you a hint. The first thing you should try is just substituting in random values.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Suppose f(0.5)=0.4&#xA;Then 0.5=f(1-f(0.5))=f(0.6)&#xA;Then 0.6=f(1.2-f(0.6))=f(0.7)&#xA;...&#xA;Until we reach: 1=f(1.1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maybe you should see if a similar fact holds algebraically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Now that Chandru has updated the question, I'll add more info on generalising it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a &lt;code&gt;m&lt;/code&gt; and let &lt;code&gt;f(m)=m+n&lt;/code&gt;. Then we get&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m=f(2m-f(m))=f(m-n)&#xA;m-n=f(2m-2n-f(m-n))=f(m-2n)&#xA;...&#xA;m-kn=f(m-(k-1)n)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Clearly, if &lt;code&gt;n&lt;/code&gt; is non-zero, then we end up with something being out of bounds&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-28T12:44:32.150" LastActivityDate="2010-08-28T12:44:32.150" CommentCount="6" />
  <row Id="1892" PostTypeId="1" AcceptedAnswerId="1895" CreationDate="2010-08-09T10:47:27.760" Score="6" ViewCount="692" Body="&lt;p&gt;Although whether $$ P = NP $$ is important from theoretical computer science point of view, but I fail to see any practical implication of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we can prove all questions that can be verified in polynomial time have polynomial time solutions, it won't help us in finding the actual solutions. Conversely, if we can prove that $$ P \ne NP,$$ then this doesn't mean that our current NP-hard problems have no polynomial time solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From practical point of view ( practical as in the sense that we can immediately use the solution to real world scenario), it shouldn't bother me whether P vs NP is proved or disproved any more than whether my &lt;strong&gt;&lt;em&gt;current problem&lt;/em&gt;&lt;/strong&gt; has a polynomial time solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-09T11:32:17.017" LastActivityDate="2010-08-10T01:04:04.810" Title="The Practical Implication of P vs NP Problem" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="3" />
  <row Id="1893" PostTypeId="2" ParentId="1873" CreationDate="2010-08-09T10:48:31.377" Score="1" ViewCount="0" Body="&lt;p&gt;Just a little warning about the first example: let's call that matrix $J$. It is a (non-diagonal) Jordan matrix, so it is not diagonalizable. That is, there is no base of eigenvectors for this matrix, although, as J. Mangaldan observed, 2 is the only eigenvalue and the last vector of the standard $\mathbb{R}^3$ basis is an eigenvector, since, as you can see, $Je_3 = 2e_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can convince yourself that $J$ is not diagonalizable computing its characteristic polynomial: $Q(t) = (2-t)^3$ and observing that $2$ has multiplicity $3$, but the eigenspace $\ker (J-2I)$ has dimension $1$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-09T11:10:52.327" LastActivityDate="2010-08-09T11:10:52.327" CommentCount="1" />
  <row Id="1894" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T11:11:54.320" Score="2" ViewCount="0" Body="&lt;p&gt;Actually $P \ne NP$ &lt;em&gt;does&lt;/em&gt; mean that our current NP-hard problems have no polynomials time solutions. NP-complete problems are the hardest problems in NP and NP-hard problems are at least as hard as this. So if $P \ne NP$, then all these NP-hard problems must be harder than P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the proof helps us find solutions will of course depend on the proof. If $P \ne NP$, then we know not to waste time looking for polynomial solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P=NP$, then the real practical benefits would of course come from the solution, rather than the proof. That is fine - there is no reason why all theoretical computer science needs to be &lt;em&gt;directly&lt;/em&gt; practical.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-09T21:45:46.140" LastActivityDate="2010-08-09T21:45:46.140" CommentCount="2" />
  <row Id="1895" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T11:11:57.830" Score="8" ViewCount="0" Body="&lt;p&gt;Many of the problems we know to be in NP or NP-complete are problems that we actually want to solve, problems that arise, say, in circuit design or in other industrial design applications. Furthermore, since the diverse NP-complete problems are all polynomial time related to one another, if we should ever learn a feasible means of solving any of them, we would have feasible means for all of them. The result of this would be extraordinary, something like a second industrial revolution. It would be as though we suddenly had a huge permanent increase in computational power, allowing us to solve an enormous array of practical problems heretofore out of our computational reach. The P vs. NP question is important in part because of this tantalizing possibility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were proved that P = NP and the proof provided a specific polynomial time algorithm for an NP-complete problem, then because of the existing reduction proofs, we could immediately produce polynomial time algorithms for all our other favorite NP problems. Of course, a proof may be indirect, and not provide a specific polynomial time algorithm, but you can be sure that if we have a proof of P=NP, then enormous resources will be put into extracting from the proof a speciffic algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if someone were to prove $P \neq NP$, then it would mean that there could be no polynomial time solution for any NP complete problem. (In particular, the last sentence of your second paragraph is not correct.) &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-09T11:11:57.830" CommentCount="7" />
  <row Id="1896" PostTypeId="1" AcceptedAnswerId="1920" CreationDate="2010-08-09T11:22:02.447" Score="0" ViewCount="68" Body="&lt;p&gt;If $k_1,k_2,...,k_n$ are principal curvatures of a hypersurface in $\mathbb{R}^{n+1}$ then one can apparently locally parametrize the manifold as $(x_1,x_2,...,x_n,y)$ such that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = \frac{1}{2}(k_1 x_1^2 +...+k_n x_n^2) + O(\vert x \vert ^3) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how this is proven. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the coordinates are obtained by &quot;integrating&quot; the eigen-vectors of the shape operator whose eigen-values are the principal curvatures anyway. I suppose one has to be able to find out curves passing through a point which have as velocity at that point the principal directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have no experience of doing such a calculation on arbitrary manifolds and hence I can't see how to do this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further I would like to understand how for dimension 2 the principal curvatures (as defined above) become the minimum and the maximum of geodesic curvatures and also minimum and maximum of curvatures of curves obtained by intersecting the manifold with planes orthogonal to the tangent space. (in elementary surface theory the principal curvatures get defined as the maximum and the minimum of the above quantities)&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-10-04T04:19:18.463" Title="Local equation for the manifold from its principal curvatures" Tags="&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1897" PostTypeId="1" CreationDate="2010-08-09T11:46:26.987" Score="0" ViewCount="88" Body="&lt;p&gt;Suppose we have a functional question, such as the &lt;a href=&quot;http://math.stackexchange.com/questions/1890/real-analysis-problem&quot;&gt;one here&lt;/a&gt;. Often a key step in these problems is to try to equate or chain results together (&lt;a href=&quot;http://math.stackexchange.com/questions/1890/real-analysis-problem&quot;&gt;see my hint&lt;/a&gt;). From looking at these questions, it seems that there is often only a finite number of ways that we can attempt to join them together. Can this fact be examined more formally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am not asking you to turn my answer from that question into something algebraic. It is that way purposely as I was only trying to give a hint, not a complete solution. What I am asking, is whether there are any results that show that the functions can only be &quot;combined&quot; or &quot;equated&quot; in a certain number of ways on some reasonably general class of problems.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-09T12:44:41.940" Title="Solving functional questions" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="1898" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T12:40:34.537" Score="1" ViewCount="0" Body="&lt;p&gt;Currently, if a manager asks their software engineering team to look at implementing some utility, and the team says that requirements are NP hard, that's a reason that the project requirements need to be changed before work on implementation can begin.  That's because no-one knows how to give feasible solutions to such problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plurality of complexity theorists furthermore believe P =/= NP, so that means that there's a widespread belief among experts that feasible solutions to these problems will never be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone shows P=NP, then if the team says the requirements are NP-complete, then the manager and the team will start to move from talk of theory to possible realisations and their efficiency.  &lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-08-09T12:40:34.537" />
  <row Id="1899" PostTypeId="2" ParentId="1897" CreationDate="2010-08-09T12:44:41.940" Score="1" ViewCount="0" Body="&lt;p&gt;One example is that there's an easy way to get uniqueness of functional equations in certain cases. Consider the &quot;cohomological&quot; equation $f(Tx) - f(x) = g(x)$ (which is important in dynamical systems, e.g. it comes up when you are trying to find an invariant $n$-form). The goal, of course, is to prove that there exists $f$ given $g$. If $T: X \to X$ is topologically transitive, then any two solutions differ  by a constant. This is because if $x_0$ has a dense orbit, then the value of $f(x_0)$ determines $f(Tx_0), f(T^2x_0)$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, there is the same uniqueness for the &quot;twisted&quot; cohomological equation $f(Tx) - \alpha f(x) = g(x)$, where $\alpha$ is a nonzero real number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is not, I think, relevant to the question, but I'll mention that one can get existence of a Holder solution in the untwisted case for transitive Anosov diffeomorphisms of compact manifolds if the sum of $g$ over every finite orbit is zero, though it requires a clever trick. This is a result of Livsic.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-09T12:44:41.940" CommentCount="1" />
  <row Id="1901" PostTypeId="1" CreationDate="2010-08-09T13:22:20.880" Score="4" ViewCount="344" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that some issues in group theory prevent classifying all manifolds upto homotopy using the fundamental group invariant. Does anyone know what are those issues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;K.&lt;/p&gt;&#xA;" OwnerUserId="975" LastActivityDate="2010-11-29T20:51:56.813" Title="Why is fundamental group insufficient to classify manifolds?" Tags="&lt;group-theory&gt;&lt;topology&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="1902" PostTypeId="2" ParentId="1872" CreationDate="2010-08-09T13:24:25.917" Score="0" ViewCount="0" Body="&lt;p&gt;This gives quite an intuitive idea about what is going on:&lt;br&gt;&#xA;&lt;a href=&quot;http://plus.maths.org/content/curious-quaternions&quot; rel=&quot;nofollow&quot;&gt;http://plus.maths.org/content/curious-quaternions&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-09T13:24:25.917" CommentCount="1" />
  <row Id="1905" PostTypeId="2" ParentId="1872" CreationDate="2010-08-09T13:27:30.013" Score="0" ViewCount="0" Body="&lt;p&gt;...unfortunately I am only allowed to post one link per answer - so here is the sequel&#xA;&lt;a href=&quot;http://plus.maths.org/content/ubiquitous-octonions&quot; rel=&quot;nofollow&quot;&gt;http://plus.maths.org/content/ubiquitous-octonions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-09T13:27:30.013" CommentCount="1" />
  <row Id="1907" PostTypeId="2" ParentId="1901" CreationDate="2010-08-09T13:31:35.700" Score="2" ViewCount="0" Body="&lt;p&gt;Well, there are no reasons to expect that manifolds with equal &amp;pi;&lt;sub&gt;1&lt;/sub&gt; are homotopy equivalent, and in general, they indeed aren't. Even for closed manifolds of fixed dimension: say, both CP&lt;sup&gt;2&lt;/sup&gt; and S&lt;sup&gt;4&lt;/sup&gt; have trivial fundamental group.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-10T16:53:10.693" LastActivityDate="2010-08-10T16:53:10.693" />
  <row Id="1909" PostTypeId="1" AcceptedAnswerId="1915" CreationDate="2010-08-09T13:44:35.387" Score="3" ViewCount="150" Body="&lt;p&gt;Let $(\mathbb{R},+)$ be the group of Real Numbers under addition. Let $H$ be a proper subgroup of $\mathbb{R}$. Prove that either $H$ is dense in $\mathbb{R}$ or there is an $a \in \mathbb{R}$ such that $H={ na : n=0, \pm{1},\pm{2},...}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to proceed. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T06:26:38.223" LastActivityDate="2010-08-10T06:26:38.223" Title="Characterizing Dense Subgroups of the Reals" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1910" PostTypeId="1" AcceptedAnswerId="1912" CreationDate="2010-08-09T13:46:07.163" Score="1" ViewCount="122" Body="&lt;p&gt;School starts in 1.5 weeks and I'm taking a course in Symbolic Logic. Here is a quote from the syllabus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You need to keep a notebook with all your homework in it. This notebook can take one of two forms. 1. A looseleaf (or some kind of removable/replaceable page notebook) or 2. A folder (aka subdirectory) on your computer. The latter is only recommended for true computer geeks who like ﬁguring out how to work with symbols (e.g., ∃, ∀, ⊃, ⊢, →, ↔). I will periodically collect homework.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to try and do homework on my computer, if only to see if I can do it, even though in the long run I'll probably get frustrated and just use paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you recommend for this type of document? I assume it'll be a mix of words and equations with the above symbols. Is there a good special program for this, or instructions for doing this in MS Office 2010 or OpenOffice? Is this a good application for L&lt;sup&gt;A&lt;/sup&gt;T&lt;sub&gt;E&lt;/sub&gt;X, and if so, how does someone with no L&lt;sup&gt;A&lt;/sup&gt;T&lt;sub&gt;E&lt;/sub&gt;X knowledge (me) learn enough to do my homework? Is it even worth the trouble or should I just stick with deadtree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My primary OS is Windows 7, if it makes a difference.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2010-08-12T17:58:59.880" Title="How to do homework for a class in Symbolic Logic on my computer?" Tags="&lt;logic&gt;&lt;math-software&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1911" PostTypeId="2" ParentId="1876" CreationDate="2010-08-09T13:46:38.487" Score="3" ViewCount="0" Body="&lt;p&gt;I assume you intend for all the $x_i$ to be positive. The infimum is $\min(n/4, 1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good example of how to use convexity to minimize an expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the change of variables $x_i = e^{t_i}$, so $\sum t_i=0$. Define $f(t) = 1/(1+e^t)^2$. We want to minimize $\sum f(t_i)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $f(t)$ is convex for $t&gt; - \log 2$ and concave for $t &amp;lt; - \log 2$. This is an easy calculation with the double derivative of $f$. (You might find this &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Plot%5B1%2F%281%2Be%5Et%29%5E2%2C%20%7Bt%2C-Infinity%2C%20Infinity%7D%5D&quot; rel=&quot;nofollow&quot;&gt;graph&lt;/a&gt; helpful.) So, if all of the $t_i$ lie in $[- \log 2, \infty)$, then &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt; shows that the minimum occurs when all the $t_i$ are $0$. In that case, we get $n/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim that we can restrict to the case that at most one of the $t_i$ lies in $(- \infty, - \log 2)$. Proof: Consider any solution $(t_i)$, with $t_1$, $t_2$, ..., $t_k &amp;lt; - \log 2$. Replace these values by $k-1$ copies of $- \log 2$ and $t_1+t_2+\cdots+t_k + (k-1) \log 2$. Since $f$ is concave in this range, we have decreased the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we may assume that there is one $t_i$, call it $u$, in $(- \infty, - \log 2)$, and all the others are in $[- \log 2, \infty)$. For a fixed value of $u$, by convexity, we will obtain the minimum when all the other values are equal to each other, say $v$. So $u+(n-1)v=0$. Our sum is&#xA;$$g(v) :=(1+e^{-(n-1)v})^{-2} + (n-1) (1+e^v)^{-2}.$$&#xA;We want to minimize this quantity, subject to the condition that $v \geq -\log 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $v \to \infty$, the function $g$ goes to $1$. We'd like to know whether there is any point in $[\log 2, \infty)$ where $g$ dips below $1$. Such a point must be a local minimum. Here are some plots for $n$ &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Plot%5B%281%2Be%5E%28-%282%29v%29%29%5E%28-2%29%20%2B%20%282%29%20%281%2Be%5Ev%29%5E%28-2%29%2C%20%7Bt%2C-0.5%2C%200.5%7D%5D&quot; rel=&quot;nofollow&quot;&gt;equal to 3&lt;/a&gt; and &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Plot%5B%281%2Be%5E%28-%289%29v%29%29%5E%28-2%29%20%2B%20%289%29%20%281%2Be%5Ev%29%5E%28-2%29%2C%20%7Bt%2C-0.5%2C%200.5%7D%5D&quot; rel=&quot;nofollow&quot;&gt;equal to 10&lt;/a&gt;. Numerically, it looks like there is a local minimum at $0$, where $g$ is $n/4$ and a local maximum at some positive value. If we can confirm this, the infimum is $\min(1, n/4)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's put the cherry on this sundae and prove that there is no other local minimum. The derivative $g'$ is&#xA;$$\frac{-2(n-1)e^{-(n-1)v}}{(1+e^{-(n-1)v})^3} + \frac{(n-1) e^v}{(1+e^v)^3}$$&#xA;Set this equal to zero and clear out denominators to get&#xA;$$(1+e^v)^3 e^{2 (n-1)v} = (1+e^{(n-1)v})^3 e^v.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $w=e^v$. For $n \geq 4$, this is a polynomial in $w$ with two sign changes so, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs&quot; rel=&quot;nofollow&quot;&gt;Descartes' rule of signs&lt;/a&gt;, it has at most two positive roots. There also turn out to be two real roots for $n=2$ and $3$, by computer check.  These are the local minimum we have already noticed at $v=0$ ($w=1$), and the local maximum nearby. So there are no other local minima, and we are done.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-09T14:28:37.840" LastActivityDate="2010-08-09T14:28:37.840" CommentCount="3" />
  <row Id="1912" PostTypeId="2" ParentId="1910" CreationDate="2010-08-09T14:02:10.557" Score="2" ViewCount="0" Body="&lt;p&gt;For &quot;a mix of words and equations with symbols&quot;  I'd clearly recommend using latex. &lt;a href=&quot;http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is a good introduction.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-09T14:02:10.557" />
  <row Id="1913" PostTypeId="1" CreationDate="2010-08-09T14:07:06.677" Score="3" ViewCount="52" Body="&lt;p&gt;Let $X $ be a set, $|X|=n$ and $G$ be a group with a 2-transitive action on $X$.&#xA;what can be said about the size of $G$?&lt;/p&gt;&#xA;" OwnerUserId="816" LastEditorUserId="816" LastEditorDisplayName="" LastEditDate="2010-08-09T14:20:05.083" LastActivityDate="2010-08-09T15:08:58.757" Title="minimal size of 2-transitive groups" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="1914" PostTypeId="2" ParentId="1913" CreationDate="2010-08-09T14:12:53.167" Score="3" ViewCount="0" Body="&lt;p&gt;When $n$ is a prime (or indeed a prime power), there are two-transitive&#xA;subgroups of Sym$(X)$ (where $|X|=n$) of order $n(n-1)$. In any case $n(n-1)$&#xA;is a lower bound for the order of $G$, since this is the number of ordered pairs&#xA;$(x,x')$ of distinct elements of $X$. Alas I can't see where your $n^2+n$ comes from.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-09T14:12:53.167" CommentCount="1" />
  <row Id="1915" PostTypeId="2" ParentId="1909" CreationDate="2010-08-09T14:14:28.950" Score="7" ViewCount="0" Body="&lt;p&gt;If there is a smallest positive element, then we are done, since any positive element must be an integer multiple of it, or otherwise we could use a euclidean-type-algorithm to get a positive element with smaller value. (I.e., suppose $a$ is the smallest positive element, and $b$ a   positive element which is not an integer multiple of $a$---keep subtracting copies of $a$ until you get something that is strictly between $0 $ and $a$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume there is a sequence $a_n$ contained in the group that consists of positive numbers tending to zero. Then the group contains each ${\mathbb{Z} a_n}$. This means that for each $n$, any number in $\mathbb{R}$ is within $|a_n|$ of an element of the group. Since the $|a_n|$ can be small, we find that the group is dense.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-09T14:14:28.950" CommentCount="2" />
  <row Id="1916" PostTypeId="2" ParentId="1901" CreationDate="2010-08-09T14:34:00.183" Score="6" ViewCount="0" Body="&lt;p&gt;I think that what you are thinking of is the insolvability of the so-called word problem in group theory.  What this means is  (among other things) that it is not possible to write an algorithm (i.e. a computer program) which can tell whether two groups, each presented by finitely many generators and relations, are or aren't isomorphic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are given a manifold (presented say by gluing together various open balls), you can get&#xA;such a presentation of its fundamental group, but (at least when the dimension is 4 or higher --- and I think that 4 is the right bound here) in fact you can get an arbitrary presentation in this way, and so the insolvability of the word problem means that, in practice, you don't have any way to figure out what the fundamental group of your manifold is (even whether or not it is trivial!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;As a consequence, you can't find an algorithm to determine whether two manifolds (of dimension 4 or higher) are homotopic (let along homeomorphic or diffeomorphic), since homotopic manifolds will have isomorphic&#xA;$\pi_1$s, and so we could use such an algorithm to solve the word problem, by encoding two finitely presented groups as the $\pi_1$ of some manifolds, and then applying our hypothetical manifold algorithm. &#xA;&lt;/strike&gt;&#xA;Following Pete Clark's comment below, I'm no longer sure about the correctness of the preceding statement.  What is true is that there is no general algorithm to determine the fundamental group of a manifold (in dimension $\geq 4$), or even to determine if a manifold &#xA;is simply connected.  (Whether this actually prohibits a classification is not clear to me; with luck, someone with more expertise will weigh in.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, a lot (although not all) investigations of higher dimensional manifolds restrict attention to simply connected manifolds.  Then there is no obstruction to classification coming from issues of the uncomputatibility of $\pi_1$.  Of course,&#xA;as others have noted in their answers, there can be many non-diffeomorphic, non-homeomorphic manifolds, and non-homotopic manifolds of a given dimension, so the problem with $\pi_1$ is far from the only problem when it comes to classifying manifolds.  But it is certainly one problem, and you are correct that (at least from the point of view I'm explaining here) it is a problem of group theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: See Damiano's answer to this question for a precise explanation of the relationship between the word problem for groups and the classification problem for manifolds.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-10T20:35:28.583" LastActivityDate="2010-08-10T20:35:28.583" CommentCount="3" />
  <row Id="1917" PostTypeId="2" ParentId="1901" CreationDate="2010-08-09T14:34:42.367" Score="3" ViewCount="0" Body="&lt;p&gt;For a better pair of examples take the once punctured torus and a pair of pants. Both have fundamental groups free of rank 2. They are both homotopy equivalent to a figure 8, but as manifolds they are not homeomorphic since one has 3 boundary components and the other has one boundary component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are simply connected closed manifolds of dimension 4 and higher that are not homeomorphic. In even dimensions they can be distinguished by their middle dimensional homology intersection forms.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-08-09T14:34:42.367" />
  <row Id="1918" PostTypeId="1" CreationDate="2010-08-09T14:42:46.870" Score="2" ViewCount="251" Body="&lt;p&gt;What are some interesting mathematical paradoxes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have in mind are things like the Banach-Tarski paradox, Paradox of Zeno of Elea, Russel's paradox, etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As an additional restriction, let us focus on paradoxes that are not already in the list at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Category:Mathematics_paradoxes&quot; rel=&quot;nofollow&quot;&gt;https://secure.wikimedia.org/wikipedia/en/wiki/Category:Mathematics_paradoxes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-10T01:05:13.373" LastActivityDate="2010-08-10T01:05:13.373" CommunityOwnedDate="2010-08-09T14:44:53.277" Title="Mathematical paradoxes?" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="1919" PostTypeId="1" CreationDate="2010-08-09T14:54:03.613" Score="8" ViewCount="236" Body="&lt;p&gt;The simple group of order 60 can be generated by the permutations (1,2)(3,4) and (1,3,5), but all you need to do is square the first one and it becomes the identity.  Can't we find a version of the simple group where the elements of small order can be ignored?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a group H, define Ω&lt;sub&gt;n&lt;/sub&gt;(H) to be the subgroup generated by elements of order less than n.  For instance, if n=3 and H=SL(2,5) is the perfect group of order 120, then Ω&lt;sub&gt;3&lt;/sub&gt;(SL(2,5)) has order 2, and H/Ω&lt;sub&gt;n&lt;/sub&gt;(H) is the simple group of order 60.  If n=4 and H=SL(2,5)⋅3&lt;sup&gt;4&lt;/sup&gt; is the perfect group of order (60)⋅(162) whose 3-core is not complemented, then Ω&lt;sub&gt;n&lt;/sub&gt;(H) has order 162 and H/Ω&lt;sub&gt;n&lt;/sub&gt;(H) is again the simple group of order 60.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is if there are smaller examples for n=4, since the jump 1, 2, 162 seems a bit drastic for n=2, 3, 4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a group H of order less than (60)⋅(162) such that H/Ω&lt;sub&gt;4&lt;/sub&gt;(H) is the simple group of order 60?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably, for each positive integer n, there is a finite group H such that H/Ω&lt;sub&gt;n&lt;/sub&gt;(H) is the simple group of order 60.  I am interested in whether such H can be chosen to be &quot;small&quot; somehow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a sequence of finite groups H&lt;sub&gt;n&lt;/sub&gt; and a constant C such that H&lt;sub&gt;n&lt;/sub&gt;/Ω&lt;sub&gt;n&lt;/sub&gt;(H&lt;sub&gt;n&lt;/sub&gt;) is the simple group of order 60 and such that |H&lt;sub&gt;n&lt;/sub&gt;| ≤ C⋅n?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would also be fine with some references to where such a problem is discussed.  It would be nice if there was some sort of analogue to the Schur multiplier describing the largest non-silly kernel, and a clear definition of what a silly kernel is (I think it is too much to ask for a non-silly kernel to be contained in the Frattini subgroup, and I think it might be unreasonable to ask for the maximum amongst minimal kernels).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In case it helps, here are some reduced cases that I know can be handled:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler example: if instead of the simple group of order 60, we concentrate on the simple group of order 2, then we can choose H&lt;sub&gt;n&lt;/sub&gt; to be the cyclic group of order 2&lt;sup&gt;1+lg(n−1)&lt;/sup&gt; when n≥2, and the order of H&lt;sub&gt;n&lt;/sub&gt; is bounded above and below by multiples of n.  We can create much larger H&lt;sub&gt;n&lt;/sub&gt; for n≥3 by taking the direct product of our small H&lt;sub&gt;n&lt;/sub&gt; with an elementary abelian 2-group of large order, but then Ω&lt;sub&gt;n&lt;/sub&gt;(H&lt;sub&gt;n&lt;/sub&gt;×2&lt;sup&gt;n&lt;/sup&gt;) =  Ω&lt;sub&gt;n&lt;/sub&gt;(H&lt;sub&gt;n&lt;/sub&gt;)×2&lt;sup&gt;n&lt;/sup&gt; has just become silly since the entire elementary abelian 2-group part, 2&lt;sup&gt;n&lt;/sup&gt;, is unrelated and uses a lot of extra generators, that is, it is not contained within the Frattini subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moderate example: if instead of the simple group of order 60, we take the non-abelian group of order 6, then I can find a natural choice of H&lt;sub&gt;n&lt;/sub&gt; with |H&lt;sub&gt;n&lt;/sub&gt;| ≤ C⋅n, but I am not sure if there are other reasonable choices.  My choice of H&lt;sub&gt;n&lt;/sub&gt; has Φ(H&lt;sub&gt;n&lt;/sub&gt;)=1, which suggests to me that Frattini extensions may not be the right idea.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-09T14:54:03.613" Title="Ignoring elements of small order in the simple group of order 60" Tags="&lt;group-theory&gt;" CommentCount="1" />
  <row Id="1920" PostTypeId="2" ParentId="1896" CreationDate="2010-08-09T15:01:08.527" Score="1" ViewCount="0" Body="&lt;p&gt;This won't be a complete derivation, but I will post it here on the chance that it is useful.&#xA;If no-one else comes along with a more complete explanation, I can try to expand on it at some later point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be a point on your hypersurace in $\mathbb R^{n+1}$, and let $\Pi$ be the tangent plane to your hypersurface at $P$, interpreted as an actual plane in $\mathbb R^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change coordinates so that $P$ is the origin, and the tangent plane is given by $y = 0$.&#xA;(Here $y$ is the $n+1$st coordinate of $\mathbb R^{n+1}$.)  Now choose the first $n$-coordinates&#xA;$x_1, \ldots, x_n$ so that they are orthogonal and point in the direction of the various principal curvatures, with principal curvatures $k_1,\ldots,k_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim, then, is that your hypersurface is given locally at $P$ by the equation&#xA;$y = (1/2)(k_1^2 x_1^2 + \cdots + k_n x_n^2) + O(x^3),$ or in other words, that the quadric&#xA;$y = (1/2)(k_1^2 x_1^2 + \cdots + k_n x_n^2)$ has contact order 2 with the hypersurface at the point $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will follow more-or-less from the definition of the principal curvatures as eigenvalues of the second fundamental form: they encode the shape of your manifold at the point $P$ up to second order, and you can check that quadric we are writing down has the same principal curvatures at the origin, and so has the same shape as your manifold at the point $P$.  Thus they coincide up to second order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't actually checked this, that would be a useful exercise: take the quadric&#xA;$y = (1/2)(k_1x_1^2 + \cdots + k_n x_n^2),$ and compute its second fundamental form at the origin; you should find that it is diagonal, with eigenvalues precisely $k_1,\ldots,k_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your questions about the various different interpretations of the principal curvatures when $n = 2$, these are discussed extensively in one of the volumes of Spivak&#xA;(I think the second volume; in any case, if you look in the contents you will find the&#xA;place where he first starts to discuss curvature, and he talks about principal curvatures&#xA;of surfaces in a lot of detail.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a brief hint: a 2x2 symmetrix matrix $A$ can be thought of as defining a conic section centered at the origin (via the equation $x^T A x = 1$), and the two eigendirections&#xA;are the major and minor axes of the conic.  (This is just linear algebra and basic analytic geometry.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now apply this statement in the context of the second fundamental form for a surface in space, you will be able to derive that the two principal curvatures at a point correspond to maximizing and minizing the curvature of curves passing through that point.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-04T04:19:18.463" LastActivityDate="2010-10-04T04:19:18.463" />
  <row Id="1921" PostTypeId="2" ParentId="1913" CreationDate="2010-08-09T15:02:50.927" Score="4" ViewCount="0" Body="&lt;p&gt;As a complement to Robin Chapman's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since G is 2-transitive, its order is divisible by n*(n-1), not just bounded below by, and so it would be interesting to bound |G|/(n*(n-1)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If n is not a prime power, then it is quite possible for the lower bound to be huge.  The smallest reasonable non-prime, n=6, has its smallest 2-transitive group with order 60 = 6*5*2.  The next, n=10, has its smallest 2-transitive group with order 10*9*4.  For most n, the smallest multiple is (n-2)!/2, that is, the alternating group on n points is the smallest 2-transitive group.  This happens already at n=22, 33, 34, 35, and asymptotically takes over.  Cameron–Neumann–Teague showed this in their 1982 paper &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=661693&quot; rel=&quot;nofollow&quot;&gt;MR661693&lt;/a&gt;, and I believe it is covered in Dixon–Mortimer's textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on the one hand the lower bound for a 2-transitive group on n points is n*(n-1) for prime powers n, but for most n the lower bound is n!/2.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-09T15:08:58.757" LastActivityDate="2010-08-09T15:08:58.757" />
  <row Id="1922" PostTypeId="2" ParentId="1877" CreationDate="2010-08-09T15:08:12.190" Score="2" ViewCount="0" Body="&lt;p&gt;A little further thought reveals the following:  &lt;em&gt;uniform semi-continuity implies uniform continuity&lt;/em&gt;.  Thus the answer to my question is a resounding &quot;no&quot;, since any function that is upper semi-continuous but not continuous &lt;em&gt;cannot&lt;/em&gt; be uniformly upper semi-continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;.  Let $f$ be uniformly upper semi-continuous.  Then for every $\epsilon&gt;0$ there exists $\delta&gt;0$ such that for every $x\in X$, we have $f(y) &amp;lt; f(x) + \epsilon$ whenever $y\in B(x,\delta)$.  However, since this statement holds for every $x$, it also holds with $x$ and $y$ reversed; in the language of the original post, both $x$ and $y$ are contained in the set $X_\delta^\epsilon = X$.  Since $y$ is in this set and $x\in B(y,\delta)$, we also have $f(x) &amp;lt; f(y) + \epsilon$, and thus $|f(x) - f(y)| &amp;lt; \epsilon$.  But this is just the definition of uniform continuity.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-09T15:08:12.190" CommentCount="1" />
  <row Id="1923" PostTypeId="1" CreationDate="2010-08-09T15:21:11.143" Score="1" ViewCount="60" Body="&lt;p&gt;Noncommutative algebraic geometry is a developing field. Things have not yet got the final form as in commutative geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one might wonder whether things are any better in the case of skew-fields, ie division rings, ie possibly noncommutative rings in which each nonzero element has a multiplicative inverse. Algebraic geometry is much simpler in the case of fields. So are things better in the case of skew fields? For instance, is there any particularly nice geometry over the quaternions, like there is one over the complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-09T15:21:11.143" Title="Noncommutative algebraic geometry in the case of skew fields" Tags="&lt;algebraic-geometry&gt;&lt;noncommutative-geometry&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="1924" PostTypeId="2" ParentId="1918" CreationDate="2010-08-09T15:43:59.210" Score="3" ViewCount="0" Body="&lt;p&gt;One example not in the above list is &lt;a href=&quot;http://google.com/groups?selm=WGD.95Dec11023450%40martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;Goodstein's theorem&lt;/a&gt;, a highly nonintuitive concrete number theoretic theorem which is unprovable in Peano arithmetic (or, similarly, the Hercules vs. Hydra game). They essentially encode induction up to the ordinal $\epsilon_0 = \omega^{\omega^{\omega^{\cdot^{\cdot^{\cdot}}}}}$ - something that is not at all intuitive to those who are not familiar with such ordinals - especially their Cantor normal form.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-09T15:43:59.210" CommunityOwnedDate="2010-08-09T15:43:59.210" CommentCount="6" />
  <row Id="1925" PostTypeId="2" ParentId="594" CreationDate="2010-08-09T15:46:07.027" Score="4" ViewCount="0" Body="&lt;p&gt;Perhaps my favorite argument (other than any arguably &quot;correct&quot; arguments, such as the one Akhil has given, or arguments starting from the fact that $x^2 + y^2$ is the unique binary quadratic form of discriminant $-4$ up to equivalence) uses continued fractions.  Suppose that $u^2 \equiv -1 \pmod{p}$, and consider the continued fraction expansion of the rational number $u/p$.  Let $r/s$ be the last convergent to $u/p$ with the property that $s &amp;lt; \sqrt{p}$.  Then, setting $x=s$ and $y = rp-us$, one has $x^2 + y^2 = p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the argument: let $r'/s'$ be the convergent following $r/s$.  Then the basic theory of continued fractions gives the estimate $|r/s - u/p| &amp;lt; 1/ss'$, and the right-hand side is less than $1/s\sqrt{p}$ by hypothesis.  Clearing denominators gives $y &amp;lt; \sqrt{p}$, so that $0 &amp;lt; x^2 + y^2 &amp;lt; 2p$.  On the other hand $x^2 + y^2$ is checked to be divisible by $p$ (by choice of $u$), so must be equal to $p$.&lt;/p&gt;&#xA;" OwnerUserId="631" LastEditorUserId="631" LastEditorDisplayName="" LastEditDate="2010-08-09T16:13:29.633" LastActivityDate="2010-08-09T16:13:29.633" CommentCount="3" />
  <row Id="1926" PostTypeId="1" AcceptedAnswerId="1943" CreationDate="2010-08-09T15:47:19.053" Score="1" ViewCount="107" Body="&lt;p&gt;I know a little bit of the theory of compact Riemann surfaces, wherein there is a very nice divisor -- line bundle correspondence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I take up the book of Hartshorne, the notion of &lt;a href=&quot;http://planetmath.org/encyclopedia/CartierDivisor.html&quot; rel=&quot;nofollow&quot;&gt;Cartier divisor&lt;/a&gt; there is very confusing. It is certainly not a direct sum of points; perhaps it is better to understand it in terms of line bundles. But Cartier divisors do not seem to be quite the same thing as line bundles. The definition is hard to figure out. Can someone clear the misunderstanding for me and explain to me how best to understand Cartier divisors?&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-09T15:54:12.560" LastActivityDate="2010-08-09T19:00:55.773" Title="Divisor -- line bundle correspondence in algebraic geometry" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="1927" PostTypeId="1" AcceptedAnswerId="1930" CreationDate="2010-08-09T16:18:20.740" Score="0" ViewCount="89" Body="&lt;p&gt;Let's take field $(K, +,\times,0,1)$. Let's take set $S(K) \subset K$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ \forall_{e \in K}\exists_{a \in S(K)}\exists_{p \in K} e = a \times p \times p \]&#xA;\[ \lnot \exists_{a, b \in S(K)}\exists_{p \in K} (a \times p \times p = b \land a \neq b) \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example $S(\mathbb{R}) = \{-1, 1\}$, $S(\mathbb{C}) = \{1\}$, $S(\mathbb{Z}_3) = \{1, 2\}$, $S(\mathbb{Z}_5) = \{1, 2\}$. Of course sets are not unique like $S(\mathbb{R}) = \{-e,\pi\}$ is also possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry if I misuse terms - I haven't found english translation - but I found it useful to generalize the signature of quadratic polynomials).&lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-09T16:24:36.990" Title="What's the name of this set " Tags="&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="1928" PostTypeId="2" ParentId="1910" CreationDate="2010-08-09T16:21:34.123" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=atIhEDQSI1U&quot; rel=&quot;nofollow&quot;&gt;GNU Emacs + AUCTeX&lt;/a&gt; works fine for me.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-09T16:21:34.123" />
  <row Id="1929" PostTypeId="2" ParentId="1888" CreationDate="2010-08-09T16:21:45.183" Score="2" ViewCount="0" Body="&lt;p&gt;First and foremost, it must be irreducible over $\mathbb{Q}$. Second, the splitting field of an irreducible cubic polynomial over $\mathbb{Q}$ is either of degree $3$ or of degree $6$ over $\mathbb{Q}$. If it is degree $6$, then the Galois group must be either isomorphic to $C_6$ (the cyclic group of order $6$) or to $S_3$; either way, it contains a unique subgroup of index $2$, corresponding to a unique subextension of degree 2; this is given by adjoining the square root of the discriminant. So the degree is $6$ if and only if the polynomial is irreducible and the discriminant is &lt;em&gt;not&lt;/em&gt; a rational square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a necessary and sufficient condition on the polynomial is that it be irreducible and that its discriminant be a square. I'll let you figure out how to translate that into conditions on $a$ and $b$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-09T16:21:45.183" CommentCount="3" />
  <row Id="1930" PostTypeId="2" ParentId="1927" CreationDate="2010-08-09T16:24:36.990" Score="2" ViewCount="0" Body="&lt;p&gt;If I am deciphering your notation correctly, I would call $S$ &quot;a set of coset representatives for $K^{\times}/(K^{\times})^2$.&quot;&lt;/p&gt;&#xA;" OwnerUserId="631" LastActivityDate="2010-08-09T16:24:36.990" CommentCount="1" />
  <row Id="1931" PostTypeId="2" ParentId="501" CreationDate="2010-08-09T16:24:44.743" Score="2" ViewCount="0" Body="&lt;p&gt;Finitism still allows you to use infinitary definitions of real numbers, because a finitist is content with finite &lt;em&gt;proofs&lt;/em&gt; even if the concepts mentioned by those proofs would seem to require infinite sets. For example, a finitist would still recognize that &quot;ZFC proves that every bounded nonempty set of reals has a least upper bound&quot; even if the finitist does not accept that infinite sets exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proofs in various infinitary systems are of interest to finitists because of conservation results. In this setting, a conservation result would show that if a sentence about the natural numbers of a certain form is provable in some infinitary system, the sentence is actually provable in a finitistic system.  For example, there are finitistic proofs that if any $\Pi^0_2$ sentence about the natural numbers is provable in the infinitary system $\text{WKL}_0$ of second order arithmetic, that sentence is also provable in the finitistic system $\text{PRA}$ of primitive-recursive arithmetic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many consistency results are proven finitistically. For example, there is a finitistic proof that if ZF set theory without the axiom of choice is consistent, then ZFC set theory with the axiom of choice is also consistent. This proof studies infinitary systems of set theory, but the objects actually handled are finite formal proofs rather than infinite sets.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-09T16:24:44.743" CommentCount="0" />
  <row Id="1932" PostTypeId="1" CreationDate="2010-08-09T16:46:05.870" Score="3" ViewCount="121" Body="&lt;p&gt;In &lt;a href=&quot;http://meta.math.stackexchange.com/questions/172/why-did-you-close-my-question-if-all-sets-were-finite/178#178&quot;&gt;the comment thread of an answer&lt;/a&gt;, I said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The computable numbers are based on the intuitionistic continuum, and are not finitary. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To which T.. replied:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Computable numbers are not based on the intuitionistic continuum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This disagreement contains, I think, a good example of a philosophical question: are the computable reals within the scope of finitistic mathematics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bendegem, 2010, &lt;a href=&quot;http://plato.stanford.edu/entries/geometry-finitism/&quot; rel=&quot;nofollow&quot;&gt;Finitism in Geometry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Edalat, 2009, &lt;a href=&quot;http://www.doc.ic.ac.uk/~ae/papers/longo-f.pdf&quot; rel=&quot;nofollow&quot;&gt;A computable approach to measure and integration theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Zach, 2001, &lt;a href=&quot;http://www.ucalgary.ca/rzach/papers/hilbert.html&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Finitism&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Zach, 2003, &lt;a href=&quot;http://plato.stanford.edu/entries/hilbert-program&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Program&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-10-13T18:03:36.857" Title="Are the computable reals finitary?" Tags="&lt;soft-question&gt;&lt;constructive-mathematics&gt;&lt;philosophy&gt;&lt;real-number-computation&gt;&lt;finitism&gt;" AnswerCount="2" />
  <row Id="1933" PostTypeId="1" AcceptedAnswerId="1945" CreationDate="2010-08-09T16:57:54.247" Score="3" ViewCount="109" Body="&lt;p&gt;I have seen a lot of places where the group cohomology when a group acts on a module, is extensively used. But beyond seeing the definition and some claims of partial results, I havent seen any uses of the case when we replace action on a module by action on noncommutative group. It seems hard to believe that much can be made out of it as H_1 is just a set, not even a group. Can somebody explain some uses of bringing up and studying this notion?&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-09T23:02:01.287" LastActivityDate="2010-08-09T23:02:01.287" Title="Use of noncommutative group cohomology" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1934" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T17:14:56.107" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Microsoft_Math&quot; rel=&quot;nofollow&quot;&gt;Microsoft Math&lt;/a&gt; looks like TI and shows the history of items&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-09T17:14:56.107" CommunityOwnedDate="2010-08-09T17:14:56.107" />
  <row Id="1935" PostTypeId="1" CreationDate="2010-08-09T17:22:03.380" Score="2" ViewCount="157" Body="&lt;p&gt;Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-09T23:01:05.987" LastActivityDate="2010-08-10T02:01:21.663" Title="Methods to see if a polynomial is irreducible" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="1936" PostTypeId="2" ParentId="1918" CreationDate="2010-08-09T17:24:02.893" Score="2" ViewCount="0" Body="&lt;p&gt;A nice paradox (in the sense of going against the common opinion) which is not in that list is &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrow%27s_impossibility_theorem&quot; rel=&quot;nofollow&quot;&gt;Arrow's theorem&lt;/a&gt;. More or less, it states the following. Let be given a set of people who vote on some issue, and have a finite number of alternatives (at least 3). Each person orders the alternatives according to her preferences; the outcome of the vote is an order on the set of alternatives which is supposed to reflect the common consensus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, a preference is a total order on the set of alternatives, and a voting system is a function which associates to each $n$-uple of preferences another preference. It turns out that the only voting system which satisfies some innocent-looking hypothesis is the projection on some factor, that is, the dictatorship of one of the people.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-09T18:22:05.807" LastActivityDate="2010-08-09T18:22:05.807" CommunityOwnedDate="2010-08-09T17:24:02.893" />
  <row Id="1937" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T17:25:02.970" Score="1" ViewCount="0" Body="&lt;p&gt;Not directly related to the question, but definitely relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three days ago &lt;a href=&quot;http://www.hpl.hp.com/personal/Vinay_Deolalikar/Papers/pnp_preliminary.pdf&quot; rel=&quot;nofollow&quot;&gt;proof&lt;/a&gt; for P != NP is published. Community thinks it looks serious.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="956" LastEditorDisplayName="" LastEditDate="2010-08-09T23:11:20.820" LastActivityDate="2010-08-09T23:11:20.820" CommentCount="5" />
  <row Id="1938" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T17:30:41.760" Score="1" ViewCount="0" Body="&lt;p&gt;Here are three links you may find useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wims.unice.fr/wims/wims.cgi?lang=en&amp;amp;+session=YV6390EF2C.1&amp;amp;+module=home&quot; rel=&quot;nofollow&quot;&gt;http://wims.unice.fr/wims/wims.cgi?lang=en&amp;amp;+session=YV6390EF2C.1&amp;amp;+module=home&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.numbertheory.org/php/php.html&quot; rel=&quot;nofollow&quot;&gt;http://www.numbertheory.org/php/php.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.alpertron.com.ar/NUMBERT.HTM&quot; rel=&quot;nofollow&quot;&gt;http://www.alpertron.com.ar/NUMBERT.HTM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-08-09T17:30:41.760" CommunityOwnedDate="2010-08-09T17:30:41.760" />
  <row Id="1939" PostTypeId="2" ParentId="1935" CreationDate="2010-08-09T17:32:50.803" Score="4" ViewCount="0" Body="&lt;p&gt;To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the &lt;strong&gt;master theorem&lt;/strong&gt; behind all these related results. A good place to start is Filaseta's notes - see the links below. Note: you may need to write to Filaseta to get access to his interesting book [1] on this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Newton Polygon Applet &#xA;&lt;a href=&quot;http://www.math.sc.edu/~filaseta/newton/newton.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.sc.edu/~filaseta/newton/newton.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[4] Abhyankar, Shreeram S.&lt;br&gt;&#xA;Historical ramblings in algebraic geometry and related algebra.&lt;br&gt;&#xA;Amer. Math. Monthly 83 (1976), no. 6, 409-448.&lt;br&gt;&#xA;&lt;a href=&quot;http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG&quot; rel=&quot;nofollow&quot;&gt;http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAG&lt;/a&gt;... &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-09T17:32:50.803" CommentCount="2" />
  <row Id="1940" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T17:35:54.930" Score="2" ViewCount="0" Body="&lt;p&gt;For those of us on a Mac, there's &lt;a href=&quot;http://www.acqualia.com/soulver/&quot; rel=&quot;nofollow&quot;&gt;Soulver&lt;/a&gt;, which is more like a notepad with calculating in mind. Lots of interesting features like conversions and instant stock prices, but no graphing if you're reliant on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in having a history of disconnected equations over the course of a day, you could just open a new document, use it for the day and then discard it when you quit. However, it's more designed to create a set of calculations which you can save as a document and then return to later. You can link calculations together as well, letting you change a single value and recomputing all your answers instantly.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-08-09T17:35:54.930" CommunityOwnedDate="2010-08-09T17:35:54.930" />
  <row Id="1941" PostTypeId="2" ParentId="1932" CreationDate="2010-08-09T17:53:26.567" Score="3" ViewCount="0" Body="&lt;p&gt;A real number $r$ is defined to be computable if there is an algorithm that, given any natural number $n$, will produce a rational number within distance $1/n$ of $r$. Thus each individual computable number is defined using a finite amount of information (the algorithm). On the other hand, there is a well-known intuitive sense in which an arbitrary real number can contain an infinite amount of information. In this sense, computable real numbers are finitary in a way that arbitrary real numbers are not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a computable real number can be stated in either classical mathematics or in intuitionistic mathematics. Nothing in the concept of a computable real number ties you to one logic or the other. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-09T17:53:26.567" CommentCount="2" />
  <row Id="1942" PostTypeId="1" CreationDate="2010-08-09T17:58:32.297" Score="2" ViewCount="183" Body="&lt;p&gt;We have a pool of items.  Each item has a set of characteristics.  Each characteristic is a string, integer, or floating-point number.  Each item is also of a particular type.  Types are hierarchical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have a set of bins.  Each bin has a requirement, which is a boolean expression comprised of atomic expressions regarding item characteristics.  These atomic expressions can be string, integer, or floating-point comparisons and type comparisons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wish to put the items in the bins such that each item placed in a bin satisfies the requirement of that bin.  Each bin also must contain a specified number of items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we might have the following type hierarchy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;shape &#xA;&lt;ul&gt;&#xA;&lt;li&gt;triangle &#xA;&lt;ul&gt;&#xA;&lt;li&gt;equilateral triangle&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;rectangle &#xA;&lt;ul&gt;&#xA;&lt;li&gt;square&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We could have the following items:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;         &lt;li&gt;LargeRedTriangle&lt;/li&gt;&#xA;    &lt;ul&gt;&#xA;         &lt;li&gt;type = triangle&lt;/li&gt;&#xA;         &lt;li&gt; color = &quot;red&quot;&lt;/li&gt;&#xA;         &lt;li&gt; area = 1000&lt;/li&gt;&#xA;    &lt;/ul&gt;&#xA; &lt;li&gt;SmallShape&lt;/li&gt;&#xA;    &lt;ul&gt;&#xA; &lt;li&gt;type = shape&lt;/li&gt;&#xA; &lt;li&gt;area = 5&lt;/li&gt;&#xA;    &lt;/ul&gt;&#xA;&lt;li&gt;LargeBlueSquare&#xA;    &lt;ul&gt;&#xA; &lt;li&gt;type = square&lt;/li&gt;&#xA; &lt;li&gt;color = &quot;blue&quot;&lt;/li&gt;&#xA; &lt;li&gt;area = 2000&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is an example bin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LargeRectangles&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(type == rectangle) and (area &gt; 750)&lt;/li&gt;&#xA;&lt;li&gt;item count = 5&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We could put LargeBlueSquare in the LargeRectangles bin.  We want exactly 5 items in this bin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example bin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;NonBlueTriangles&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(type == triangle) and (not (color == &quot;blue&quot;))&lt;/li&gt;&#xA;&lt;li&gt;item count = 10&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;LargeRedTriangle can go in the NonBlueTriangles bin.  We want exactly 10 items in this bin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have a set of items and a set of bins.  The question is to find out how many ways there are to put the items in the bins while satisfying the constraints (i.e., the requirements for items in bins and the number of items in each bin).  (We don't need to enumerate all the ways to do this; we only need to know how many different ways of doing this there are.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an efficient way of determining this number?  Or do we just have to 'brute-force' it and try every possible combination?&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="984" LastEditorDisplayName="" LastEditDate="2010-08-09T18:23:25.147" LastActivityDate="2010-09-16T21:58:48.003" Title="Counting placing items in bins according to boolean restrictions" Tags="&lt;combinatorics&gt;" AnswerCount="3" />
  <row Id="1943" PostTypeId="2" ParentId="1926" CreationDate="2010-08-09T18:10:04.580" Score="6" ViewCount="0" Body="&lt;p&gt;When discussing divisors, a helpful distinction to make at the beginning is effective divisors vs. all divisors.  Normally effective divisors have a more geometric description; all divisors can then be obtained from the effective ones by allowing some minus signs to come into the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An irreducible effective Weil divisor on a variety $X$ is the same thing as an irreducible codimension one subvariety, which in turn is the same thing as a height one point $\eta$ of $X$.  (We get $\eta$ as the generic point of the irred. codim'n one subvariety, and we recover the subvariety as the closure of $\eta$.)&lt;br&gt;&#xA; An effective Weil divisor is a non-negative integral linear combination of irreducible ones, so you can think of it as a non-negative integral linear combination of height one points $\eta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, one restricts to normal varieties, so that all the local rings at height one points  are DVRs.    Then, given any pure codimension one subscheme $Z$ of $X$, you can attach a Weil&#xA;divisor to $Z$, in the following way:&#xA;because the local rings at height one points are DVRs, if $Z$ is any codimension one subscheme of $X$, cut out by an ideal sheaf $\mathcal I_Z$,  and $\eta$ is a height one point, then the stalk $\mathcal I_{Z,\eta}$ is&#xA;an ideal in the DVR $\mathcal O_{X,\eta}$, thus is just some power of the maximal ideal&#xA;$\mathfrak m_{\eta}$ (using the DVR property), say $\mathcal I_{Z,\eta} = \mathfrak m_{\eta}^{m_{Z,\eta}},$ and so the multiplicity $m$ of $Z$ at $\eta$ is well-defined.&lt;br&gt;&#xA;Thus the effective Weil divisor $$div(Z) := \sum_{\eta \text{ of height one}} m_{Z,\eta}\cdot  \eta$$&#xA;is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this recipe is only goes one way: starting with the Weil divisor, we can't recover $Z$, because the Weil divisor does not remember all the scheme structure (i.e. the whole&#xA;structure sheaf, or equivalently, the whole ideal sheaf) of $Z$, but only its behaviour at its generic points (which amounts to the same thing as remembering the irreducible components and their multiplicities). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An effective Cartier divisor is actually a more directly geometric objects, namely, &#xA;it is a &lt;em&gt;locally principal&lt;/em&gt; pure codimension one subscheme, that is,&#xA;a subscheme, each component of which is codimension one, and which, locally around each point, is the zero locus of a section of the structure sheaf.   Now in order to cut out&#xA;a pure codimension one subscheme as its zero locus, a section of the structure&#xA;sheaf should be regular (in the commutative algebra sense), i.e. a non-zero divisor.&#xA;Also, two regular sections will cut out the same zero locus if their ratio is a unit&#xA;in the structure sheaf.  So if we let $\mathcal O_X^{reg}$ denote the subsheaf of&#xA;$\mathcal O_X$ whose sections are regular elements (i.e. non-zero divisors in each stalk),&#xA;then the equation of a Cartier divisor is a well-defined global section of the quotient&#xA;sheaf&#xA;$\mathcal O_X^{reg}/\mathcal O_X^{\times}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that we are on a smooth variety.  Then any irreducible codimension one subvariety&#xA;is in fact locally principal, and so given a Weil divisor &#xA;$$D = \sum_{\eta \text{ of height one}} m_{\eta} \cdot\eta,$$&#xA;we can actually find attach a canonical Cartier divisor to it, in the following way:&#xA;in a n.h. of some point $x$, let $f_{\eta}$ be a local equation for the Zariski closure&#xA;of $\eta$; then if $Z(D)$ is cut out locally by $\prod_{\eta} f_{\eta}^{m_{\eta}} = 0,$&#xA;then $Z(D)$ is locally principal by construction, and, again by construction,&#xA;$div(Z(D)) = D.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the smooth setting,&#xA;we see that $Z \mapsto div(Z)$ and $D \mapsto Z(D)$ establish a bijection between&#xA;effective Cartier divisors and effective Weil divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, on a singular variety, it can happen that an irreducible codimension one subvariety need not be locally principal in the neighbourhood of a singular point (e.g. a generating line on the cone $x^2 +y^2 + z^2 = 0$&#xA;in $\mathbb A^3$ is not locally&#xA;principal in any neighbourhood of the cone point).  Thus there can be Weil divisors that&#xA;are not of the form $div(Z)$ for any Cartier divisor $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from effective Weil divisor to all Weil divisors, you just allow negative coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from effective Cartier divisors to all Cartier divisors, you have to allow yourself &#xA;to invert the functions $f$ that cut out the effective Cartier divisors, or equivalently, to go from the sheaf of monoids $\mathcal O_X^{reg}/\mathcal O_X^{\times}$ to the associated sheaf of groups, which is $\mathcal K_X^{reg}/\mathcal O_X^{\times}.$&#xA;(Here, it helps to remember that $\mathcal K_X$ is obtained from $\mathcal O_X$&#xA;by inverting non-zero divisors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for the connection with line bundles: if $\mathcal L$ is a line bundle,&#xA;and $s$ is a regular section (i.e. a section whose zero locus is pure codimension one,&#xA;or equivalently, a section which, when we choose a local isomorphism $\mathcal L_{| U}&#xA;\cong \mathcal O_U$, is not a zero divisor), then the zero locus $Z(s)$ of $s$&#xA;is an effective Cartier divisor, essentially by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have a map $(\mathcal L,s) \mapsto Z(s)$ which sends line bundles with regular sections to effective Cartier divisors.  This is in fact an isomorphism of monoids&#xA;(where on the left we consider pairs $(\mathcal L,s)$ up to isomorphism of pairs):&#xA;given an effective Cartier divisor $D$, we can define $\mathcal O(D)$ to be the&#xA;subsheaf of $\mathcal K_X$ consisting (locally) of sections $f$ such that the locus of&#xA;poles of $f$ (a well-defined Cartier divisor) is contained (as a subscheme)in the Cartier&#xA;divisor $D$ (perhaps less intuitively, but more concretely: if $D$ is locally cut out&#xA;by the equation $g = 0$, then $\mathcal O(D)$ consists (locally) of sections $f$&#xA;of $\mathcal K_X$ such that $fg$ is in fact a section of $\mathcal O_X$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant function $1$ certainly lies in $\mathcal O(D)$, and (thought of as a section&#xA;of $\mathcal O(D)$ -- not as a function!) its zero locus is exactly $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $D \mapsto (\mathcal O(D), 1)$ is an inverse to the above map $(\mathcal L,s) \mapsto&#xA;Z(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if we choose two different regular sections of the same line bundle,&#xA;the corresponding Cartier divisors are linearly equivalent.  Thus we are led to the&#xA;isomorphism &quot;line bundles up to isomorphism = Cartier divisors up to linear equivalence&quot;.&#xA;But, just to emphasize, to understand this it is best to restrict first to line bundles&#xA;which admit a regular section, and then think of the corresponding Cartier divisor as being&#xA;the zero locus of that section.   This brings out the geometric nature of the Cartier divisor quite clearly.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-09T19:00:55.773" LastActivityDate="2010-08-09T19:00:55.773" CommentCount="1" />
  <row Id="1944" PostTypeId="2" ParentId="1935" CreationDate="2010-08-09T18:23:40.420" Score="5" ViewCount="0" Body="&lt;p&gt;One method for polynomials over $\mathbb{Z}$ is to use complex analysis to say something about the location of the roots.  Often Rouche's theorem is useful; this is how Perron's criterion is proven, which says that a monic polynomial $x^n + a_{n-1} x^{n-1} + ... + a_0$ with integer coefficients is irreducible if $|a_{n-1}| &gt; 1 + |a_{n-2}| + ... + |a_0|$ and $a_0 \neq 0$.  A basic observation is that knowing a polynomial is reducible places constraints on where its roots can be; for example, if a monic polynomial with prime constant coefficient $p$ is reducible, one of its irreducible factors has constant term $\pm p$ and the rest have constant term $\pm 1$.  It follows that the polynomial has at least one root inside the unit circle and at least one root outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important thing to keep in mind here is that there exist irreducible polynomials over $\mathbb{Z}$ which are reducible modulo every prime.  For example, $x^4 + 16$ is such a polynomial.  So the modular technique is not enough in general.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-09T18:23:40.420" CommentCount="1" />
  <row Id="1945" PostTypeId="2" ParentId="1933" CreationDate="2010-08-09T18:29:26.133" Score="8" ViewCount="0" Body="&lt;p&gt;Often elements of $H^1$ are used to classify objects of a certain kind.  E.g. if $k$ is a field and $k^s$ its separable closure, and $G_k = Gal(k^s/k)$, then if $X$ is a variety&#xA;over $k$, and $Aut(X)$ is the group of automorphisms of $X_{/k^s}$,&#xA;then the set $H^1(G_k,Aut(X))$ classifies varieties over $k$ that become isomorphic to&#xA;$X$ when both are regarded as varieties over $k^s$.   The base point of $H^1$ corresponds &#xA;to the original variety $X$ itself. (Such varieties are called &lt;em&gt;twists&lt;/em&gt; of $X$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an important construction, which (along with variants) is used all the time in&#xA;arithmetic geometry, including (perhaps especially) the theory of algebraic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one concrete example, let me mention&#xA;a famous example, namely the result that $H^1(G_k, GL_n(k^s)) = 1$.  (A generalized form of Hilbert's Thm. 90.)   If we now consider the short exact sequence&#xA;$1 \to (k^s)^{\times} \to GL_n(k^s) \to PGL_n(k^s) \to 1,$&#xA;we then obtain an injection $H^1(G_k, PGL_n(k^s))\hookrightarrow H^2(G_k,(k^s)^{\times}).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $PGL_n(k^s)$ acts as automorphisms of $\mathbb P^{n-1}$, so we see that the twists of&#xA;projective spaces (such twists are called Brauer--Severi varieties) are classified by certain elements of the Brauer group.  For example, smooth conics in the plane (which are twists&#xA;of $\mathbb P^1$, because over $k^s$ they acquire rational points, and so are isomorphic&#xA;to $\mathbb P^1$) correspond to quaternion algebras.  (To see this concretely --- or rather,&#xA;to go the other way ---&#xA;if $D$ is a quaternion algebra, then $D^{tr = 0}$, the subspace of elements of reduced trace zero, is 3-dimensional, and the reduced norm gives a quadratic form on this space.&#xA;Passing to the associated two-dimensional projective space, we get a conic in a projective&#xA;plane.   This is more concrete than the cohomological story, but the cohomological story has the advantage of being very general, and relies just on a general fact, namely Hilbert's Thm. 90, rather than specific knowledge of the geometric situation.  This is one typical advantage of cohomological arguments and constructions, when you can find them/make them.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-09T18:29:26.133" />
  <row Id="1946" PostTypeId="1" AcceptedAnswerId="2049" CreationDate="2010-08-09T18:29:59.503" Score="0" ViewCount="240" Body="&lt;p&gt;Was reading &quot;Taylor's Expansion&quot; when i came across this interesting problem, interesting for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f^{n+2}$ is continuous on $[0,x]$. Prove that there is a $\theta \in (0,1)$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;f(x) &amp;amp;= f(0) + \frac{f'(0)}{1!}x + \cdots + \frac{f^{n-1}(0)}{(n-1)!}x^{n-1} \\&#xA;&amp;amp; \quad + \frac{f^{n}\left(\frac{x}{n+1}\right)}{n!}x^{n} + \frac{n}{2(n+1)}f^{n+2}(\theta x)x^{n+2}{(n+2)!}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-08-10T20:41:20.487" LastActivityDate="2010-08-10T20:41:20.487" Title="Existence of a $\theta$ - Taylor Expansion Problem" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="1947" PostTypeId="1" AcceptedAnswerId="1950" CreationDate="2010-08-09T18:36:38.833" Score="0" ViewCount="66" Body="&lt;p&gt;Before starting algebraic geometry, I got some understanding of Compact Riemann Surfaces which is more or less rigorous; and then I attended a couple of lectures on analytic geometry. I did not quite fully grasp the analytic geometry rigorously; but it stuck in my mind and this previous experiences helped me greatly when learning algebraic geometry. However I am not able to imagine what would be the precise analogue of Cartan's Theorem A for algebraic varieties over a field. There is the wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cartan&lt;/a&gt;'s_theorems_A_and_B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which tells me the precise analogue of Cartan Theorem B in the book of Hartshorne. And indeed it is really analogous. But it does not mention Theorem A. Note that here I am restricting the situation to varieties out of fear of any pathologies schemes might have.&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-09T18:43:21.677" LastActivityDate="2010-08-09T19:00:17.023" Title="Analogue of Cartan theorem A in algebraic geometry" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="1948" PostTypeId="1" AcceptedAnswerId="1997" CreationDate="2010-08-09T18:55:11.073" Score="2" ViewCount="308" Body="&lt;p&gt;These are two famous identities of &lt;strong&gt;Ramanujan&lt;/strong&gt;.  Where can I find the proofs of them:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \displaystyle \frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum\limits_{k=0}^{\infty} \frac{(4k!)(1103 + 26390k)}{(k!)^{2} (396)^{4k}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int\limits_{0}^{\infty} \frac{1 + x^{2}/(b+1)^{2}}{1+x^{2}/a^{2}} \times \frac{1+ x^{2}/(b+2)^{2}}{1 + x^{2}/(a+1)^{2}} \times  \cdots dx= \frac{\sqrt{\pi}}{2} \times \frac{\Gamma(a+1) \Gamma(b+\frac{1}{2}) \Gamma(b-a+\frac{1}{2}}{\Gamma(a)\Gamma(b+\frac{1}{2} \Gamma(b-a+1)}$ for $0 &amp;lt; a &amp;lt; b+\frac{1}{2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-09T19:09:20.497" LastActivityDate="2010-08-10T00:57:12.550" Title="Where can I find the proofs of the following" Tags="&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1949" PostTypeId="1" CreationDate="2010-08-09T18:56:26.907" Score="2" ViewCount="135" Body="&lt;p&gt;I would like to know how to explicitly prove that Riemann Curvature,Ricci Curvature, Sectional Curvature and Scalar Curvature are left invariant under an isometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see this explained in most books I have looked at. &#xA;They atmost explain preservation of the connection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess doing an explicit proof for the sectional curvature should be enough (and easiest?) since all the rest can be written in terms of it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given Akhil's reply I think I should try to understand the connection invariance proof better and here goes my partial attempt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\nabla$ be the connection on the manifold $(M,g)$ and $\nabla '$ be the Riemann connection on the manifold $(M',g')$ and between these two let $\phi$ be the isometry. Then one wants to show two things,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$D\phi [\nabla _ X Y] = \nabla ' _{D\phi[X]} D\phi [Y]$&lt;/li&gt;&#xA;&lt;li&gt;$R(X,Y)Z = R'(D\phi [X],D\phi [Y]) D\phi [Z]$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where $R$ and $R'$ are the Riemann connection on $(M,g)$ and $(M',g')$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One defines the map $\nabla ''$ on M which maps two vector fields on M to another vector field using $\nabla '' _X Y = D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y]$. By the uniqueness of the Riemann connection the proof is complete if one can show that this $\nabla ''$ satisfies all the conditions of being a Riemann connection on M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am getting stuck after a few steps while trying to show the Lebnitz property of $\nabla ''$. Let $f$ be some smooth function on M and then one would like to show that,&#xA;$\nabla '' _X fY = X(f)Y + f\nabla '' _X Y$ which is equivalent to showing that,&#xA;$D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [fY]) = X(f) + f D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y])$ knowing that $\nabla '$ satisfies the Leibnitz property on $M'$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some how I am not being able to unwrap the above to prove this.&#xA;I can get the second term of the equation but not the first one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving symmetry of $\nabla ''$ is easy but again proving metric compatibility is stuck for me. If $X,Y,Z$ are 3 vector fields on M then one would want to show that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Xg(Y,Z) = g(\nabla ''_X Y,Z) + g(Y, \nabla '' _X Z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to showing that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Xg(Y,Z) = g(D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y]),Z) + g(Y,D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Z]) )$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;knowing that $\nabla'$ satisfies metric compatibility equation on $M'$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be helpful if someone can help me fill in the steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one is left with proving the curvature endomorphism equation. &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-08-12T16:04:32.893" LastActivityDate="2010-08-16T17:16:07.940" Title="Explicitly proving invariance of curvatures under isometry" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="3" />
  <row Id="1950" PostTypeId="2" ParentId="1947" CreationDate="2010-08-09T19:00:17.023" Score="3" ViewCount="0" Body="&lt;p&gt;Over an affine scheme any quasi-coherent sheaf is generated by its global sections, so I guess Theorem A is always true.&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-09T19:00:17.023" CommentCount="2" />
  <row Id="1951" PostTypeId="2" ParentId="1942" CreationDate="2010-08-09T19:02:48.920" Score="0" ViewCount="0" Body="&lt;p&gt;If you only have one bin, it's actually quite easy.  Let N be how many items qualify to go into the bin, and K be the number of items the bin is required to have.  Your answer would be nCr(N, K).  The nCr (a.k.a. &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient&quot; rel=&quot;nofollow&quot;&gt;binomial coefficient&lt;/a&gt;) function is defined as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;               n!&#xA;nCr n k = ------------&#xA;           k! * (n-k)!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(&lt;code&gt;n!&lt;/code&gt; means n &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial&quot; rel=&quot;nofollow&quot;&gt;factorial&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for multiple bins, I doubt that knowledge of the type hierarchy or the predicates in detail will be needed to make it efficient (but I may be wrong).  The approach I'd take is to, for each item, list all the bins that item qualifies for.  Then, just find out how many combinations of filled-up bins you get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I don't have the answer to having multiple bins (yet), but I hope this gets you started.  &lt;/p&gt;&#xA;" OwnerUserId="985" LastEditorUserId="985" LastEditorDisplayName="" LastEditDate="2010-08-09T19:08:20.437" LastActivityDate="2010-08-09T19:08:20.437" />
  <row Id="1952" PostTypeId="1" CreationDate="2010-08-09T19:28:49.687" Score="1" ViewCount="265" Body="&lt;p&gt;The correspondence between algebraic and analytic geometry is thought-provoking. The GAGA makes this precise to some extent. But there is more of this analogy. As I checked an article on &lt;a href=&quot;http://eom.springer.de/a/a012430.htm&quot; rel=&quot;nofollow&quot;&gt;analytic space&lt;/a&gt;, most of the notions there seems to have an algebraic analogue. I had been wondering why people studied analytic geometry separately to the extent it has been. It would be nice to know about some theorems in analytic geometry that have no analogue in algebraic geometry, and this would justify the study of this subject separately. Here I mean theorems that do need to use the full power of analytic geometry; not things like Hodge decomposition for which Kahler is sufficient.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-11T15:07:38.980" Title="Non-algebraic theorems in analytic geometry" Tags="&lt;algebraic-geometry&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="1953" PostTypeId="1" CreationDate="2010-08-09T19:40:43.120" Score="0" ViewCount="107" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out what the correct notation for a multiplicative algebra should be.  I've seen a bunch of weird and conflicting ways for writing these things in the literature, but they are all either too specialized or bring in extra baggage.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific, given a ring R and an R-module, M, with a finite basis B, we can identify the elements of M with functions of the form $f, g : B \to R$.  Then we construct the &lt;em&gt;multiplicative algebra&lt;/em&gt; on M according to the following rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (f \, g)(x) = f(x) g(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is simple:  What is the standard notation for this construction?  I've seen a bunch of weird forms.  For example, in the case where R is the field of complex numbers, we can write something like [; C^k(B) ;] for this algebra.  Or if we are dealing with matrices, we can talk about Hadamard products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can we write something simpler?  I would really just like to do something along the lines of say; define R(B) to be this multiplicative algebra and be done with it.  The reason for this is that if we are working with something like Pontryagin duality or representation theory, the set B may be a group/monoid/whatever, and there could be multiple algebras we defined over it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate why this is an issue, let $B = \mathbb R$ be the group of reals under addition, then we can define both a convolution algebra, $R[ \mathbb R ]$ and a multiplicative algebra $ R( \mathbb R )$ which are identified with the same R-module, but have totally different structures.  (Though they are related by Pontryagin duality in the case where $R = \mathbb C$.)&lt;/p&gt;&#xA;" OwnerUserId="986" LastEditorUserId="986" LastEditorDisplayName="" LastEditDate="2010-08-09T23:51:54.267" LastActivityDate="2010-08-09T23:51:54.267" Title="Notation for a multiplicative algebra" Tags="&lt;abstract-algebra&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="1954" PostTypeId="2" ParentId="1952" CreationDate="2010-08-09T19:55:59.013" Score="-4" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yes they do have some connections. Please check this link: &lt;a href=&quot;http://www.cmi.ac.in/~vipul/studenttalks/coaxialcirclestrianglecentersandthenullstellensatz.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cmi.ac.in/~vipul/studenttalks/coaxialcirclestrianglecentersandthenullstellensatz.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-09T19:55:59.013" CommentCount="4" />
  <row Id="1955" PostTypeId="1" CreationDate="2010-08-09T19:59:49.377" Score="1" ViewCount="133" Body="&lt;p&gt;I have seen at many places the notions that Lie Algebras are infinitesimal objects and they look really close at a point. But I never understood this. They are abstract algebraic objects different from rings in that they are equipped with a weird sort of product and a weird Jacobi identity. Any hints on how to make this connection to infinitesimals?&lt;/p&gt;&#xA;" OwnerUserId="987" LastActivityDate="2010-08-10T04:47:54.727" Title="Lie algebras and infinitesimals" Tags="&lt;lie-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1956" PostTypeId="1" AcceptedAnswerId="1959" CreationDate="2010-08-09T20:06:44.197" Score="11" ViewCount="540" Body="&lt;p&gt;The following integral,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is clearly positive, which proves that $\pi &amp;lt; 22/7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar integral which proves $\pi &gt; 333/106$?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-09T21:07:19.760" Title="Is there an integral that proves $\pi &gt; 333/106$?" Tags="&lt;calculus&gt;&lt;approximation&gt;&lt;pi&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" />
  <row Id="1957" PostTypeId="2" ParentId="1955" CreationDate="2010-08-09T20:11:16.883" Score="1" ViewCount="0" Body="&lt;p&gt;Lie algebras are infinitesimal in the sense that they naturally arise (over $\mathbb{R}$) as tangent space at the identity of a Lie group, which is also the same as the left-invariant vector fields (infinitesimal flows) on the Lie group.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-09T20:11:16.883" CommentCount="2" />
  <row Id="1958" PostTypeId="2" ParentId="1609" CreationDate="2010-08-09T20:47:20.013" Score="0" ViewCount="0" Body="&lt;p&gt;If your CS friends are like me, they might still find the answers above a little overwhelming. So you could start as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, show them the &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisor&quot; rel=&quot;nofollow&quot;&gt;Wiki's divisor&lt;/a&gt; page (it always works!). Then explain to them that by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt;, any divisor is just a bunch of prime numbers with multiplicities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, tell them that completely similar to $\mathbb Z$, the FTA works over $\mathbb C[x]$ (which you can think of as a line). Except now that each &quot;divisor&quot; (polynomial) can be thought of as a bunch of points (roots of the polynomial) on that line, with multiplicities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why stop as a straight line? One can do the same thing for a (reasonably nice) curve on the plane, and a natural way to get a bunch of points is to intersect with another curve. By the way, a bunch of points &quot;divide&quot; the curve, further justifying the terminology! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to $\mathbb C[x]$, you can point out that the total numbers of points (with mult.) is the degree of your polynomial $p(x)$, or the dimension of the vector space $\mathbb C[x]/(p(x))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, if they still follow you, show them Matt E's splendid answer (-: &lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-09T20:47:20.013" />
  <row Id="1959" PostTypeId="2" ParentId="1956" CreationDate="2010-08-09T20:58:48.610" Score="13" ViewCount="0" Body="&lt;p&gt;I think this integral would do the job&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int\limits_{0}^{1} \frac{x^{5}(1-x)^{6}(197+462x^{2})}{530(1+x^{2})}= \pi -\frac{333}{106}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can refer to S.K. Lucas Integral proofs that $335/113 &gt; \pi$ Gazette Aust.math.soc  &lt;strong&gt;32&lt;/strong&gt; (2005), 263-266.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-09T20:58:48.610" CommentCount="3" />
  <row Id="1960" PostTypeId="2" ParentId="1956" CreationDate="2010-08-09T21:07:19.760" Score="6" ViewCount="0" Body="&lt;p&gt;Although this is not exactly an answer to the question, it seems sufficiently related to mention: there are some direct generalizations, given on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page about this integral&lt;/a&gt;. For instance, $$0 &amp;lt; \frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\ dx=\pi -\frac{47171}{15015}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, &#xA;$$\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\ dx&#xA;&amp;lt;\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\ dx&#xA;&amp;lt;\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which for n=1 (the integral in the question) gives slightly better bounds than just $\pi &amp;lt; 22/7$:&#xA;$$ \frac{1}{1260} &amp;lt; \frac{22}{7} - \pi &amp;lt; \frac{1}{630}$$&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-09T21:07:19.760" CommentCount="2" />
  <row Id="1961" PostTypeId="2" ParentId="1933" CreationDate="2010-08-09T21:17:32.873" Score="3" ViewCount="0" Body="&lt;p&gt;Matt E's response is the canonical first answer.  As someone (else) who works with nonabelian group (in particular, Galois) cohomology frequently, let me give a second answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain interesting maps between &lt;em&gt;commutative&lt;/em&gt; cohomology groups are defined via a non-commutative intermediary.  The justification for this is that, though one does not in general have anything like a &quot;long exact sequence&quot; in non-commutative cohomology, if one has an extension of $\mathfrak{g}$-modules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \rightarrow Z \rightarrow G \rightarrow A \rightarrow 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is commutative and $Z$ is &lt;strong&gt;central&lt;/strong&gt;, then one gets a connecting map in cohomology &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta: H^1(\mathfrak{g},A) \rightarrow H^2(\mathfrak{g},Z).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often elements in an $H^2$ may be viewed as &quot;obstructions&quot; to something desirable happening at the $H^1$-level.  In particular, this is the case for the &lt;strong&gt;period-index obstruction map&lt;/strong&gt; in the Galois cohomology of abelian varieties.  See for instance &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/wc1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/wc1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also publications [12], [14], [16] on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/papers.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/papers.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-09T21:33:50.297" LastActivityDate="2010-08-09T21:33:50.297" />
  <row Id="1962" PostTypeId="2" ParentId="1758" CreationDate="2010-08-09T21:23:41.117" Score="0" ViewCount="0" Body="&lt;p&gt;(One more attempt at a simple and complete solution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each chord must divide the circle into the ratio 4:3, so each chord must correspond to an arc of measure approximately 2.91624.  Consider two such chords, one in a fixed location, and the other moving through all possible configurations in which it intersects the first chord.  For simplicity, let the area of the circle be 7, so that the fixed chord divides the circle into regions with area 3 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/114911250.gif&quot; alt=&quot;animation of moving chord&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that if the chord were to continue rotating counterclockwise, what is currently the &quot;bottom&quot; of the chord would intersect the fixed chord, but any such configuration is the reflection image over the perpendicular bisector of the fixed chord of a configuration in the animation shown.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the chord rotates counterclockwise, the point of intersection of the two chords and the endpoint of the moving chord on the upper half of the circle both move solely leftward, so the area of the blue region is monotonically increasing and the area of the red region is monotonically decreasing as the chord rotates counterclockwise.  The areas of the colored regions vary between 0 and 3, and are continuous functions of the rotation of the chord.  There are 2 positions of the rotating chord for which one of the two colored regions has area 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/284449919.png&quot; alt=&quot;1:2:1:3 configuration&quot;&gt;&#xA;&lt;img src=&quot;http://www.imgftw.net/img/843189976.png&quot; alt=&quot;2:1:2:2 configuration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each configuration, we can determine the areas of the regions by symmetry (exchange the moving chord and the fixed chord).  In the left diagram, the areas are (counterclockwise, starting with the blue region) 1:2:1:3.  In the right diagram, the areas are 2:1:2:2.  Since the final chord must divide 3 of the existing 4 regions in order to have 7 regions, the configuration with areas 1:2:1:3 (above left) won't work, since dividing either region with area 1 won't give us 7 regions with area 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, working with the configuration with areas 2:1:2:2 (above right), add a third chord, rotating through all possible positions that divide the lower right region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/936642181.gif&quot; alt=&quot;adding the third chord&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, the area of the purple region is monotone and continuous as the chord rotates and is between 0 and 2, so there is a unique configuration where the area of the purple region is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.imgftw.net/img/948982901.png&quot; alt=&quot;purple region has area 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the position of the purple chord relative to the original fixed chord, the small wedge region at the right cannot have area 1, and this can be verified computationally.  (It is also the case that the area of the triangle in the center cannot have are 1 in this configuration.)  So, it is not possible to divide the circle into 7 regions of equal area using 3 chords.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-09T21:23:41.117" CommentCount="3" />
  <row Id="1965" PostTypeId="2" ParentId="1955" CreationDate="2010-08-09T21:47:59.633" Score="2" ViewCount="0" Body="&lt;p&gt;This is all carefully explained in Chapter 8 of Fulton and Harris.  The key fact is that if $G$ is a connected Lie group, then $G$ is generated by the elements in any neighborhood of the identity.  This implies that a morphism out of $G$ is determined by what it does to elements arbitrarily close to the identity (and it is a general principle in category theory that an object is determined by the morphisms out of it).  Since we have a tangent space at the identity, we can say even more: it turns out that a morphism $f : G \to H$ is determined by its differential $df : T_e G \to T_e H$, where $T_e$ is the tangent space at the identity, otherwise known as the Lie algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differential is just a linear map between finite-dimensional vector spaces, so it's much easier to handle than the original map $f$.  The problem is then to characterize which linear maps can occur.  As a necessary condition, $f$ must preserve the Lie bracket on $T_e G$, and if $G$ is simply connected this is both necessary and sufficient.  So essentially the whole point of the wacky definition of a Lie algebra is to make this theorem true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, roughly speaking, that Lie algebras capture the local, or infinitesimal, structure of a Lie group.  The Lie algebra of a Lie group can't capture the global topological structure, but being able to separate out the easy part and the hard part of understanding a Lie group is very valuable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The connection to Charles' answer is that a tangent vector at the identity determines, by translation, a left-invariant vector field on $G$.  You can think of such a vector field as specifying, at each element of $G$, a direction in which something can flow.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-09T21:47:59.633" CommentCount="1" />
  <row Id="1966" PostTypeId="1" AcceptedAnswerId="1968" CreationDate="2010-08-09T21:49:35.953" Score="0" ViewCount="87" Body="&lt;p&gt;Can every compact metric space be realized as the continuous image of a cantor set?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-09T21:54:56.707" Title="Cantor Set and Compact Metric Spaces" Tags="&lt;real-analysis&gt;" AnswerCount="2" />
  <row Id="1967" PostTypeId="2" ParentId="1955" CreationDate="2010-08-09T21:51:18.123" Score="4" ViewCount="0" Body="&lt;p&gt;The direct infinitesimal analogue of a given Lie group is not the Lie algebra with its bracket operation and Jacobi identity, but the Lie algebra (thought of as just the space of tangent vectors at the identity, without the added structure of a bracket operation) with addition of elements being the group operation.  Addition is commutative but the group generally is not. To capture the noncommutativity you need to squeeze additional information from the group down to the infinitesimal level of the Lie algebra (by taking second-order information in the series expansion of group elements near the identity; the tangent space is first-order).&#xA;The Lie multiplication $[x,y]$ is the second-order infinitesimal analogue of the commutator $x y x^{-1} y^{-1}$, and the Jacobi identity is an analogue of an identity for the commutator.  Historically, the Jacobi identity for algebras (that is, for Lie algebras whose bracket is $XY - YX$ in an associative algebra) must have come first, and is used mainly in algebras, but you can think of it as coming from the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locally, second-order information is enough: the Lie algebra determines the structure of the group, up to some questions of a different, &quot;global&quot; nature (topology) about connectivity and covering spaces. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-10T04:47:54.727" LastActivityDate="2010-08-10T04:47:54.727" />
  <row Id="1968" PostTypeId="2" ParentId="1966" CreationDate="2010-08-09T21:54:41.607" Score="5" ViewCount="0" Body="&lt;p&gt;Yes (assuming it's nonempty, of course). Moreover if you google &quot;continuous image of the Cantor set&quot;, the first hit takes you to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cantor_set&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where you can read that this theorem is true and a reference is given to Willard's &lt;em&gt;General Topology&lt;/em&gt;.  (The article does not say this and perhaps it should, but it is specifically Theorem 30.7 on p. 217 of the Dover edition.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-09T21:54:41.607" CommentCount="1" />
  <row Id="1969" PostTypeId="1" AcceptedAnswerId="1976" CreationDate="2010-08-09T21:54:55.473" Score="3" ViewCount="550" Body="&lt;p&gt;If i were to ask this: Which quote by a &quot;Mathematician&quot; do you like, then what would be your answer!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-09T23:01:30.400" LastActivityDate="2010-08-10T11:20:54.510" CommunityOwnedDate="2010-08-09T21:56:50.373" ClosedDate="2010-08-10T14:53:14.440" Title="Best Quotes by a Mathematician" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="12" CommentCount="8" FavoriteCount="3" />
  <row Id="1970" PostTypeId="2" ParentId="1966" CreationDate="2010-08-09T21:54:56.707" Score="5" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/5357/theorems-for-nothing-and-the-proofs-for-free/5388#5388&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/5357/theorems-for-nothing-and-the-proofs-for-free/5388#5388&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the comment of Harald-Hanche Olsen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surprising, yes, but once you know about it, it seems easy enough to cook up a proof. Just write the set as a union of two closed subsets, decide to map the left half of the Cantor set onto one and the right half to the other, then do the same to each of these two sets, and so on. In the limit you have the map you want, provided you have arranged for the diameters of the parts to go to zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-08-09T21:54:56.707" />
  <row Id="1971" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T21:56:26.007" Score="2" ViewCount="0" Body="&lt;p&gt;For me this is one which i like a lot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Perfect numbers like perfect men are very rare.&quot; Rene Descartes&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-09T21:57:03.183" LastActivityDate="2010-08-09T21:57:03.183" CommunityOwnedDate="2010-08-09T21:57:03.183" />
  <row Id="1972" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T22:06:01.067" Score="4" ViewCount="0" Body="&lt;blockquote&gt;Mathematics is one of the few disciplines that teaches us about the power of thought as distinct from the power of authority.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This appears on my university's &lt;a href=&quot;http://math.gmu.edu/&quot; rel=&quot;nofollow&quot;&gt;department website&lt;/a&gt;, many of us attribute it to our late chair Klaus Fischer.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-09T22:06:01.067" CommunityOwnedDate="2010-08-09T22:06:01.067" CommentCount="1" />
  <row Id="1973" PostTypeId="2" ParentId="1932" CreationDate="2010-08-09T22:23:55.967" Score="2" ViewCount="0" Body="&lt;p&gt;The following are unrelated concepts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Finitism&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;intuitionistic continuum&quot; (does that mean Weyl's book? the real numbers as defined in Bishop's book on constructive analysis?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Computable real numbers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, you can read the definition of computable real numbers under a classical or an &quot;intuitionistic&quot; interpretation of the words, and they will differ as to whether &quot;1 if Goldbach conjecture is true and 2 if it's false&quot; is a computable real number.  But the theory of computable reals will make sense in both environments.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-09T22:23:55.967" />
  <row Id="1974" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T22:23:58.870" Score="6" ViewCount="0" Body="&lt;p&gt;I'd probably go with this nice one from Richard Askey:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If things are nice there is probably a&#xA;  good reason why they are nice: and if&#xA;  you do not know at least one reason&#xA;  for this good fortune, then you still&#xA;  have work to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-09T22:23:58.870" CommunityOwnedDate="2010-08-09T22:23:58.870" />
  <row Id="1975" PostTypeId="2" ParentId="1946" CreationDate="2010-08-09T22:39:04.460" Score="2" ViewCount="0" Body="&lt;p&gt;Something is wrong with this formula. It leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e = 1 + 1 + \frac{e^{\frac{1}{3}}}{2} + e^\theta 8 &gt; e$.&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2010-08-09T22:39:04.460" CommentCount="3" />
  <row Id="1976" PostTypeId="2" ParentId="1969" CreationDate="2010-08-09T22:39:30.577" Score="7" ViewCount="0" Body="&lt;p&gt;&quot;If only I had the theorems! Then I should find the proofs easily enough.&quot; - Riemann&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-09T22:39:30.577" CommunityOwnedDate="2010-08-09T22:39:30.577" CommentCount="1" />
  <row Id="1977" PostTypeId="2" ParentId="29" CreationDate="2010-08-09T22:48:00.983" Score="1" ViewCount="0" Body="&lt;p&gt;I tend to use &lt;em&gt;Mathematica&lt;/em&gt; heavily for its symbolic capabilities, but I have a soft spot for &lt;a href=&quot;http://www.gnu.org/software/octave/&quot; rel=&quot;nofollow&quot;&gt;Octave&lt;/a&gt;, an open-source MATLAB clone. For simple number crunching, it works nicely.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-09T22:48:00.983" CommunityOwnedDate="2010-08-09T22:48:00.983" />
  <row Id="1978" PostTypeId="2" ParentId="1935" CreationDate="2010-08-09T22:52:09.540" Score="1" ViewCount="0" Body="&lt;p&gt;Below is another method for irreducibility testing - excerpted from one of my old sci.math posts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1918 Stackel published the following simple observation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  If $ p(x) $ is a composite integer coefficient polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then  $ p(n) $ is  composite for all $|n| &gt; B $, for some bound $B$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in fact $ p(n) $ has at most $ 2d $ prime values, where  $ d = {\rm deg}(p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple proof can be found online in &lt;a href=&quot;http://web.math.fsu.edu/~aluffi/archive/paper134.ps&quot; rel=&quot;nofollow&quot;&gt;Mott &amp;amp; Rose [3]&lt;/a&gt;, p. 8.&#xA;I highly recommend this delightful and stimulating 27 page paper&#xA;which discusses prime-producing polynomials and related topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value&#xA;for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)&#xA;that prime $ p(x) $ assume infinitely many prime values (except in trivial&#xA;cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which&#xA;states that $ p(x) \in {\mathbb Z}[x]$ is prime if  $ p(b) $&#xA;yields a prime in radix $b$ representation (so necessarily $0 \le p_i &amp;lt; b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example $f(x) = x^4 + 6 x^2 + 1 \pmod p$ factors for all primes $p$,&#xA;yet  $f(x)$ is prime since  $f(8) = 10601$ octal $= 4481$ is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.&#xA;$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further discussion see my prior &lt;a href=&quot;http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post [1]&lt;/a&gt;, along with Murty's online &lt;a href=&quot;http://www.mast.queensu.ca/~murty/polya4.dvi&quot; rel=&quot;nofollow&quot;&gt;paper [2]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials&lt;br&gt;&#xA;&lt;a href=&quot;http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Murty, Ram.  Prime numbers and irreducible polynomials.&lt;br&gt;&#xA;Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.mast.queensu.ca/~murty/polya4.dvi&quot; rel=&quot;nofollow&quot;&gt;http://www.mast.queensu.ca/~murty/polya4.dvi&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Mott, Joe L.; Rose, Kermit.&#xA;Prime producing cubic polynomials&lt;br&gt;&#xA;Ideal theoretic methods in commutative algebra, 281-317.&lt;br&gt;&#xA;Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.&lt;br&gt;&#xA;&lt;a href=&quot;http://web.math.fsu.edu/~aluffi/archive/paper134.ps&quot; rel=&quot;nofollow&quot;&gt;http://web.math.fsu.edu/~aluffi/archive/paper134.ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-09T22:52:09.540" />
  <row Id="1979" PostTypeId="2" ParentId="1918" CreationDate="2010-08-09T22:58:18.237" Score="1" ViewCount="0" Body="&lt;p&gt;One of the consequences of &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems&quot; rel=&quot;nofollow&quot;&gt;Goedel's incompleteness theorem&lt;/a&gt; is that if $T$ is a finitely axiomatizable theory of arithmetic, then &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$T$ proves that $T$ is consistent&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;if and only if &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$T$ is inconsistent!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reason is that an inconsistent theory proves anything, and a consistent theory never proves its own consistency.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-08-09T23:03:19.977" LastActivityDate="2010-08-09T23:03:19.977" CommunityOwnedDate="2010-08-09T22:58:18.237" />
  <row Id="1980" PostTypeId="2" ParentId="1942" CreationDate="2010-08-09T23:22:24.327" Score="0" ViewCount="0" Body="&lt;p&gt;This is an example of a Constraint Satisfaction Problem.  If you have more objects&#xA;than bins, you can sometimes treat the bins individually (as a poster did with the case of one bin) and multiply the counts together, or use the individual counts for each bin in some similar fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to do some form of divide and conquer; for example, suppose, among all the other constraints and attributes that are&#xA;present, all the objects belong to exactly one of five different classes.  Then an attempt at counting can begin with enumerating all the possibilities for partially filling bins with class 1 items.  Each possibility leads to ways to enumerating the wayss to fill the remaining space fully or partially with class 2 items, and so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to process things is to bundle certain items together: fries, a burger, and a drink can (setting certain things like nutrition aside) be called a meal, and the constraints may be that enumeration is simplified if an attempt is made to pack the bins with bundled groups of items.  However, you will have a hard time getting nice lower and upper bounds that are close if you have any reasonable complexity in your constraints.&#xA;If your constraints are such that there are very few ways to pack some of the&#xA;bins, then a depth first search approach which packs those bins first may give you a&#xA;good handle on approximate counts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides Constraint Satisfaction Problem, you might try Polya Enumeration, reduction to one of a number of basic counting problems, Inclusion-Exclusion formula, and topics in enumerative combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:00:11.167" LastActivityDate="2010-08-11T22:00:11.167" CommentCount="0" />
  <row Id="1981" PostTypeId="2" ParentId="1892" CreationDate="2010-08-09T23:27:08.967" Score="3" ViewCount="0" Body="&lt;p&gt;If $P = NP$, computational revolution (once a specific algorithm is identified for an NP-hard problem, with explicit asymptotic runtime bounds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P &amp;lt; NP$ &lt;em&gt;and one can prove it&lt;/em&gt;, secure (classical) cryptography provably exists, and a huge missing piece in our understanding of computation is filled in.  The first already has significant implications for daily life, and developing the second would have much larger implications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also understand that after 40 years of research, today P=NP carries a host of related ideas like: easy-to-hard phase transition in combinatorial problems; quantifiable boundaries between easy and hard approximate versions of specific NP-complete problems (so getting within 7/8 of the optimal solution is easy but anything closer is NP-complete); counting and randomly sampling combinatorial objects are the same problem; zero-knowledge proofs &quot;that reveal nothing but their own validity&quot; (unforgeable ID cards).  It's a very rich universe of ideas and it doesn't run out of questions once you know the answer to P=NP.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-09T23:41:38.657" LastActivityDate="2010-08-09T23:41:38.657" CommentCount="2" />
  <row Id="1982" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T00:09:56.260" Score="5" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pierre de Fermat&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;I have discovered a truly marvelous proof that it is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second into two like powers. This margin is too narrow to contain it.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ref: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="918" LastActivityDate="2010-08-10T00:09:56.260" CommunityOwnedDate="2010-08-10T00:09:56.260" CommentCount="1" />
  <row Id="1983" PostTypeId="1" CreationDate="2010-08-10T00:17:32.873" Score="3" ViewCount="96" Body="&lt;p&gt;The &lt;a href=&quot;http://eom.springer.de/p/p072630.htm&quot; rel=&quot;nofollow&quot;&gt;Phragmen-Lindelöf theorem&lt;/a&gt; gives a consequence of the Riemann hypothesis, viz, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindel%C3%B6f_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Lindelöf hypothesis&lt;/a&gt;. As such this is weaker than Riemann hypothesis; but it is still considered that even a proof of this weaker result will be a breakthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the strongest known result yet on the Lindelöf hypothesis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T07:54:41.083" Title="Latest known result on Lindelöf hypothesis" Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1984" PostTypeId="1" CreationDate="2010-08-10T00:21:00.137" Score="7" ViewCount="729" Body="&lt;p&gt;Hello. I am A Neophyte Amateur Mathematician. I have beed reading a lot about Journals and the topic of peer-review in Mathematics Journals. Does one have to have Professional crediendtials or have a Doctorate in order to Publish in Peer-review Mathematics Jourals or just the desire to competenyly solve Mathematical problems?  &lt;/p&gt;&#xA;" OwnerUserId="993" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2011-01-09T19:19:21.650" LastActivityDate="2011-01-09T19:19:21.650" Title="Does a Person Need a  Mathematics Degree in order to Publish in a Mathematics jounal?" Tags="&lt;soft-question&gt;&lt;journals&gt;" AnswerCount="9" CommentCount="2" FavoriteCount="8" />
  <row Id="1985" PostTypeId="1" CreationDate="2010-08-10T00:24:04.763" Score="8" ViewCount="476" Body="&lt;p&gt;What software do you use to &lt;strong&gt;accurately draw geometry diagrams&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2010-08-10T15:59:10.417" CommunityOwnedDate="2010-08-10T00:24:04.763" Title="Software for drawing geometry diagrams" Tags="&lt;geometry&gt;&lt;big-list&gt;&lt;math-software&gt;" AnswerCount="13" FavoriteCount="6" />
  <row Id="1986" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T00:31:50.300" Score="23" ViewCount="0" Body="&lt;p&gt;No, no sort of degree or specific credentials are required.  It is far from unheard of that someone publishes one or more papers while still an undergraduate.  There are even a few instances of important work being published by high school students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are an uncredentialled mathematician you should take extra pains to make sure that your work passes superficial tests for looking serious and competent.  Off the top of my head, this would include the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Make sure that your grammar and spelling is virtually flawless (in particular, you have some spelling mistakes in your post, which is probably not such a big deal for this site, but would create a bad impression in a journal submission).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use some form of TeX to format your document.  (This is freely available online and just takes a little practice to become reasonably proficient.  Not to do so is a big warning sign of outsider status.)&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make sure that your work contains a bibliography which gives due acknowledgment to previous work in the field.  Wherever possible, you should try to cite primary sources (i.e., journal papers) or standard texts, and you should try to avoid citing things like wikipedia and popular books.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;[1] I am aware that there are a few leading mathematicians who do not use TeX.  The geometer Cliff Taubes is a popular example.  However, Professor Taubes began his mathematical career before TeX became widespread and, in any event, does not need any help in getting his work taken seriously.  I am not saying that someone who does not follow the above precepts is not a serious mathematician, just that they look less like one at a first impression.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-09T08:33:54.067" LastActivityDate="2011-01-09T08:33:54.067" CommentCount="1" />
  <row Id="1987" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:36:20.663" Score="2" ViewCount="0" Body="&lt;p&gt;One of the best is &lt;a href=&quot;http://poincare.matf.bg.ac.rs/~janicic//gclc/&quot; rel=&quot;nofollow&quot;&gt;WinGCLC&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graph quality is really exceptional, it's easy to use, can be easily exported to SVG/EPS/BMP/LaTeX, supports command line for scripting, supports animation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is really useful is that once you define all the points/lines/intersections/circles, you can easily change the location of one of the starting objects and the whole diagram will be accurately redrawn.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-10T00:36:20.663" CommunityOwnedDate="2010-08-10T00:36:20.663" />
  <row Id="1988" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T00:38:35.027" Score="6" ViewCount="0" Body="&lt;p&gt;Pete Clark has given an excellent answer. I just want to add two more things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Try to discuss with trustworthy professional mathematicians before publication who will be sympathetic and honorable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;First publish in a website like the arXiv. It is free-for-all, and you will get to know in short order if your paper is really interesting.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T00:38:35.027" />
  <row Id="1989" PostTypeId="1" CreationDate="2010-08-10T00:38:45.760" Score="0" ViewCount="172" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Morera%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Morera's theorem&lt;/a&gt; is an underappreciated theorem in complex analysis. I have been struck by the simplicity of its proof and some clever applications of it and I had been interested in finding out more of such. Please contribute examples. One example is the Weierstrass theorem that if a sequence of holomorphic functions converge absolutely and uniformly on every compact subset in a domain, then the limit is also holomorphic. And there are numerous applications of this latter fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, please come ahead and contribute clever and slick applications of Morera's theorem that will impress people!&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T12:55:38.543" Title="Striking applications of Morera's theorem" Tags="&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="1990" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T00:39:57.677" Score="2" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;There is no branch of mathematics, however abstract, which may not some&#xA;  day be applied to phenomena of the&#xA;  real world.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Quotations/Lobachevsky.html&quot; rel=&quot;nofollow&quot;&gt;Nikolai Ivanovich Lobachevsky&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-10T00:49:33.940" LastActivityDate="2010-08-10T00:49:33.940" CommunityOwnedDate="2010-08-10T00:39:57.677" />
  <row Id="1991" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:42:42.320" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://asymptote.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Asymptote&lt;/a&gt; is my tool of choice for virtually any mathematics diagram, geometry included.  &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-10T00:42:42.320" CommunityOwnedDate="2010-08-10T00:42:42.320" />
  <row Id="1992" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T00:43:54.017" Score="5" ViewCount="0" Body="&lt;p&gt;&quot;It's like if you want to be a good pianist, &#xA;you have to do a lot of scales and a lot of practice, &#xA;and a lot of that is kind of boring, it's work.&#xA;But you need to do that before you can really be very expressive and really play beautiful music.&#xA;You have to go through that phase of practice and drill.&quot;&#xA;- Terry Tao&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-10T00:43:54.017" CommunityOwnedDate="2010-08-10T00:43:54.017" />
  <row Id="1993" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:46:58.173" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.dynamicgeometry.com/&quot; rel=&quot;nofollow&quot;&gt;Geometer's Sketchpad&lt;/a&gt;&amp;mdash; Sketchpad 5 exports/copy-pastes diagrams much better than 4 did.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T01:55:27.060" LastActivityDate="2010-08-10T01:55:27.060" CommunityOwnedDate="2010-08-10T00:46:58.173" />
  <row Id="1994" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T00:46:58.783" Score="14" ViewCount="0" Body="&lt;p&gt;There is usually no technical requirement for any particular credentials, and it seems unlikely to me that editors and referees will check for this. The usual publication process is that you submit a paper to a journal editor, who then forwards it one or more competent referees, who make a recommendations on the suitability for publication. Often, the referees make suggestions for improving the paper, whether or not it is found acceptable for publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, an amateur mathematician will find him or herself at a disadvantage in several respects of this process. First, for someone with less experience in the professional mathematical community, it may be more difficult to recognize which accomplishments actually merit publication, and editors sometimes receive articles submitted in good faith from amateurs, which are trivial or seriously wrong in some respect. Second, even when the result is correct and worthwhile, if the presentation of the paper deviates from the accepted norms, it may be found wanting. For example, it is more difficult for an amateur to know which topics need more careful explaining in a paper and which do not, and inappropriate decisions in such cases can hurt the reception of the paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see (as I write this) that others have now given some concrete advice. The most important advice I can give is to take the suggestions of the editors and referees seriously. If someone claimes that part of your paper is unclear or wrong, then you should try to understand exactly why they thought so; doing so will inevitably lead you to a better paper.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-08-10T00:52:13.483" LastActivityDate="2010-08-10T00:52:13.483" CommentCount="2" />
  <row Id="1995" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:48:03.897" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.adobe.com/products/illustrator/&quot; rel=&quot;nofollow&quot;&gt;Adobe Illustrator&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T00:48:03.897" CommunityOwnedDate="2010-08-10T00:48:03.897" />
  <row Id="1996" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T00:54:08.483" Score="1" ViewCount="0" Body="&lt;p&gt;If you are a mathematician and wants to use such pictures for your notes or presentations using TeX files, use the pstricks package enhanced for use of easy by the package &lt;a href=&quot;http://latexdraw.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;LaTeXdraw&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a very useful linux program called &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBwQFjAA&amp;amp;url=http%3A%2F%2Fwww.xfig.org%2F&amp;amp;ei=JqNgTLDDO5WHOK-gwb0J&amp;amp;usg=AFQjCNHK2HkYQJAR_5YssWZqz9QguYesng&quot; rel=&quot;nofollow&quot;&gt;xfig&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T00:54:08.483" CommunityOwnedDate="2010-08-10T00:54:08.483" />
  <row Id="1997" PostTypeId="2" ParentId="1948" CreationDate="2010-08-10T00:57:12.550" Score="1" ViewCount="0" Body="&lt;p&gt;Since you seem to be in Chennai, walk into any mathematics library and pick up the collected works of Ramanujan. That is the best. Or look into the notes edited by Bruce Berndt.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T00:57:12.550" CommentCount="0" />
  <row Id="1998" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T01:00:54.523" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://latexdraw.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;LaTeXDraw&lt;/a&gt; is a graphical front end for &lt;a href=&quot;http://en.wikipedia.org/wiki/PSTricks&quot; rel=&quot;nofollow&quot;&gt;PSTricks&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-08-10T01:00:54.523" CommunityOwnedDate="2010-08-10T01:00:54.523" CommentCount="0" />
  <row Id="1999" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T01:01:45.330" Score="4" ViewCount="0" Body="&lt;p&gt;I've never published a paper, so I'm the wrong person to answer your question; this answer is really more of a long comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few years ago, I had worked on something independently that I thought I'd try to get published. Looking back, it wasn't very interesting--basically an attempt to generalize Riemann integration to an abstract setting---but I tried submitting it to the Proceedings of the AMS. It was rejected, but I got about two pages of helpful and interesting comments from a referee (suggesting that I try a non-researchish journal; I never actually had the time to do that though) as well as kind words of advice from the person sending the email. So it was a good experience in the long run. I also got practice in writing concise mathematics and with LaTeX (though I was fairly fluent by that point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the point is that even if it gets rejected (which it won't necessarily), you'll probably learn something about mathematical life from the experience. And, besides, you can put it on the arXiv if someone &quot;endorses&quot; you.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-10T01:01:45.330" />
  <row Id="2000" PostTypeId="2" ParentId="1892" CreationDate="2010-08-10T01:04:04.810" Score="0" ViewCount="0" Body="&lt;p&gt;There is an interesting heuristic to suggest that P is actually not NP. It is that, roughly, the task of finding out a proof of a statement is an NP task, but that of verifying it is a P task. From our actual experience that verifying a proof is far easier than finding one, we can intuitively expect P != NP to hold true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The practical application is that a result which would agree with our intuition would be a great thing, and psychologically satisfying moreover.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T01:04:04.810" CommentCount="1" />
  <row Id="2001" PostTypeId="1" CreationDate="2010-08-10T01:10:21.193" Score="2" ViewCount="157" Body="&lt;p&gt;What is a simple example, without getting into the mess of triangulated categories, of an additive category that is not abelian?&lt;/p&gt;&#xA;" OwnerUserId="997" LastActivityDate="2010-08-10T20:42:55.163" Title="Additive category that is not abelian" Tags="&lt;category-theory&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="2002" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T01:13:06.020" Score="3" ViewCount="0" Body="&lt;p&gt;For geometry I've always used &lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;Geogebra&lt;/a&gt;, and I think it's pretty good.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-10T01:13:06.020" CommunityOwnedDate="2010-08-10T01:13:06.020" />
  <row Id="2003" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T01:14:56.330" Score="3" ViewCount="0" Body="&lt;p&gt;The category of finitely generated modules over a non-Noetherian ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The category of filtered modules over a ring is an example given in Gelfand-Manin.&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="977" LastEditorDisplayName="" LastEditDate="2010-08-10T10:54:52.823" LastActivityDate="2010-08-10T10:54:52.823" CommentCount="6" />
  <row Id="2004" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T01:16:06.310" Score="7" ViewCount="0" Body="&lt;p&gt;Implicit in the previous answers: &lt;em&gt;do your best to pick an appropriate journal&lt;/em&gt;. Even professionals don't always know whether a particular journal will publish their paper (even if the results are perfectly correct). There are too many factors to count that can affect the editors' decision. For an amateur, the choice can be even more opaque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graduate students often rely on advisors to guide them in picking the right journals for their first papers. If you have academic contacts, they may be willing to give you advice about possible journals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been a heavy emphasis on undergraduate mathematics research recently, at least in the United States. Several journals have sprung up to publish papers written by students. If this applies to you, search for &quot;undergraduate mathematics journal&quot; on Google to find several possibilities. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-10T01:16:06.310" />
  <row Id="2005" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T01:23:00.960" Score="2" ViewCount="0" Body="&lt;p&gt;In translation from German:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We must know, we will know.&quot; - David Hilbert&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-10T01:23:00.960" CommunityOwnedDate="2010-08-10T01:23:00.960" />
  <row Id="2006" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T01:24:34.887" Score="2" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The only advice we offer to new researchers is read! read! read! many papers with complete proofs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-- Iwaniec &amp;amp; Kowalski.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T01:24:34.887" CommunityOwnedDate="2010-08-10T01:24:34.887" />
  <row Id="2007" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T01:44:55.617" Score="0" ViewCount="0" Body="&lt;p&gt;(A bit too long for comments, so...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I may: /. recently discussed something of this sort: &lt;a href=&quot;http://ask.slashdot.org/story/10/06/24/1739247&quot; rel=&quot;nofollow&quot;&gt;http://ask.slashdot.org/story/10/06/24/1739247&lt;/a&gt; . Apparently, most journals don't reject submissions just because the submitter has no degree to speak of. Good luck! Make sure however that you have already done a thorough review of the related literature to make sure you are not merely duplicating anybody's results.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-10T01:44:55.617" />
  <row Id="2008" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T01:48:31.973" Score="0" ViewCount="0" Body="&lt;p&gt;I've used &lt;a href=&quot;http://www.satimage.fr/software/en/downloads/index.html&quot; rel=&quot;nofollow&quot;&gt;Smile by Satimage&lt;/a&gt; on my Mac. (Not SmileLab, just the free Smile. And not to be confused with PDF utility company &lt;a href=&quot;http://www.smileonmymac.com/&quot; rel=&quot;nofollow&quot;&gt;&quot;Smile on My Mac&quot;&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smile uses AppleScript for drawing. Some &lt;a href=&quot;http://www.satimage.fr/software/en/smile/graphics/graphics_samples.html&quot; rel=&quot;nofollow&quot;&gt;example files&lt;/a&gt; show an amazing facility for geometric drawings: named points, marked angles, primitives for things like barycenter and circumcenter, TeX-styled labels, etc. A fully-labeled illustration of the &lt;a href=&quot;http://www.satimage.fr/software/en/smile/graphics/graphics_sample_euler.html&quot; rel=&quot;nofollow&quot;&gt;Euler Circle&lt;/a&gt; can be created with a few lines of script. Outputs to PDF, JPG, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always have lots of trouble navigating the online documentation, so I know I'm not getting the full benefit of the program. Nevertheless, the ability to describe a drawing precisely with a script is what I need, and it's what the app delivers.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-08-10T02:40:09.650" LastActivityDate="2010-08-10T02:40:09.650" CommunityOwnedDate="2010-08-10T01:48:31.973" />
  <row Id="2009" PostTypeId="2" ParentId="1935" CreationDate="2010-08-10T02:01:21.663" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540407146&quot; rel=&quot;nofollow&quot;&gt;Polynomials&lt;/a&gt; by Prasolov covers amoung others:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eisenstein's criterion&lt;/li&gt;&#xA;&lt;li&gt;Duma's criterion&lt;/li&gt;&#xA;&lt;li&gt;Irreducibility of polynomials attaining small values&lt;/li&gt;&#xA;&lt;li&gt;Hilbert's criterion&lt;/li&gt;&#xA;&lt;li&gt;Irreducibility of trinomials and fournomials&lt;/li&gt;&#xA;&lt;li&gt;A few algorithms for factorization&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-10T02:01:21.663" />
  <row Id="2010" PostTypeId="1" AcceptedAnswerId="2017" CreationDate="2010-08-10T02:34:40.443" Score="4" ViewCount="159" Body="&lt;p&gt;I'm optimizing a family of objectives over x1,...,xn with a single constraint that x's add up to 1, and when using method of Lagrange multipliers, the single multiplier always ends up 0 for members of this family...does this tell us anything interesting about these objectives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the answer was -- it means the constraint it inactive, ie, removing the constraint doesn't change the answer of optimization problem&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-08-11T20:00:41.400" LastActivityDate="2010-08-11T20:00:41.400" Title="Meaning of Lagrange multiplier always being 0?" Tags="&lt;calculus&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2011" PostTypeId="2" ParentId="1952" CreationDate="2010-08-10T02:42:47.647" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not sure that the main motivation for studying analytic geometry is that there are theorems true there that are not true in algebraic geometry; if anything, the reverse is true, since any complex variety is also a complex analytic space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are analytic varieties that are not algebraic, so analytic geometry is a broader subject than complex algebraic geometry, which has its own appeal.  Thinking analytically also&#xA;gives a source of techniques that are not available by purely algebraic methods (and maybe this is what you mean by theorems that are true in analytic geometry but not algebraic geometry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me also mention that &#xA;very interesting phenomena can occur when you study the interaction between the algebraic and analytic worlds, for example of the following kind:  the moduli space of analytic K3 surfaces is a connected 20 dimensional space.  The moduli space of algebraic K3 surfaces is a closed subspace, which is the union of countably many 19 dimensional connected components.   This shows that some problems, like the study of moduli spaces, can become&#xA;simpler in the analytic regime than they are if you restrict to the algebro-geometric setting.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-10T02:42:47.647" CommentCount="5" />
  <row Id="2012" PostTypeId="2" ParentId="1838" CreationDate="2010-08-10T02:52:13.350" Score="2" ViewCount="0" Body="&lt;p&gt;One problem that I have come across is when roller coasters are being designed. If you have the pitch at +/-90 degrees (pointing straight up/down) then using normal Euler-based orientation you can't easily specify an angle of banking as you have no reference to 'up'. To solve this Quaternions are often used.&lt;/p&gt;&#xA;" OwnerUserId="999" LastActivityDate="2010-08-10T02:52:13.350" />
  <row Id="2013" PostTypeId="1" AcceptedAnswerId="2019" CreationDate="2010-08-10T03:03:40.500" Score="1" ViewCount="247" Body="&lt;p&gt;How are operations such as the sum, the product, the quotient, exponentiation, etc. of random variables solved or approached?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1416/algebra-of-random-variables&quot;&gt;This question&lt;/a&gt; addresses a similar problem but starts one step further: knowing how to find the distribution of a function of random variables. I'm asking for the previous step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, my question is how, in general, all the operations over random variables are handled? &lt;/p&gt;&#xA;" OwnerUserId="1000" LastEditorUserId="1000" LastEditorDisplayName="" LastEditDate="2010-11-16T15:16:35.000" LastActivityDate="2010-11-16T15:16:35.000" Title="How to work with operations/algebra of random variables?" Tags="&lt;probability-theory&gt;&lt;random-variable&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="2015" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T03:10:10.187" Score="4" ViewCount="0" Body="&lt;p&gt;No professional credentials or degree is required. However, lacking such will subject your paper to &lt;em&gt;much&lt;/em&gt; higher scrutiny so it is essential to adequately prepare for such. Since many uncredentialed submissions are by cranks it is important to differentiate your paper from such submissions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, be sure to choose an appropriate journal. Browse the recent publications of various journals to see if they accept papers on your subject at the level of your work. Seek advice from professors at local universities or in online math forums. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, make it clear that you are aware of prior work by properly referencing such. Consult mathscinet and Zentralblatt as well as web searches (e.g. Google Scholar) to find related prior work. You may be able to access mathscinet at a local university library.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crank submissions usually fail on both accounts. Namely, they often submit to journals whose prestige far exceeds their results and they usually fail to adequately cite prior work. So if you succeed on both these points you're off to a good start differentiating your work from a crank submission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck!&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-10T03:10:10.187" />
  <row Id="2017" PostTypeId="2" ParentId="2010" CreationDate="2010-08-10T03:18:49.253" Score="2" ViewCount="0" Body="&lt;p&gt;One of two things that I can think of are happening here.  The global minimum (or max, whatever your goal is here) of your function may actually lie within the simplex defined by $\sum_{i=1}^{4} q_{i} = 1$.  In this case your algorithm is finding the correct solution and the constraint is trivially satisfied.  (Consider minimizing $x^2+y^2$ constrained to the line $x = 0$, so the Lagrange formulation is $L(x,y,\nu) = x^2 + y^2 + \nu x$.  Solving $\nabla L=0$ demands that $\nu=0$. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, your algorithm could be sampling only points that already satisfy the constraint - so you are only checking those points (intentionally or otherwise).  Then the algorithm is hopefully finding a local minimum (or max) along this constraint and your multiplier will always be zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you identify where a problem may be in your model - or clears up a conceptual issue and allows you to move forward with the solution you have.  As far as interpreting the meaning of Lagrange multipliers, &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers#Interpretation_of_the_Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;look here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-10T03:37:22.337" LastActivityDate="2010-08-10T03:37:22.337" CommentCount="1" />
  <row Id="2018" PostTypeId="2" ParentId="1952" CreationDate="2010-08-10T03:56:33.960" Score="1" ViewCount="0" Body="&lt;p&gt;How about Yau's proof of the Calabi conjecture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keywords: Geometric analysis; geometric PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Matt E says, we have many (beautiful, interesting, and useful) tools in the analytic category which don't exist in the algebraic category; my favorite example is Dolbeault resolutions. There are also various PDE methods that can be used. Conversely, analytic geometry also gives PDE-ists interesting examples to study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many things that are now studied in the algebraic category were first studied or discovered in the analytic category. Two examples off the top of my head: Hodge theory, Kodaira-Spencer deformation theory. At least when these theories first came out, they had no analogues in the algebraic category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat more &quot;modern&quot; example is the Bogomolov-Tian-Todorov theorem: deformations of Calabi-Yau manifolds are unobstructed. Until &lt;a href=&quot;http://arxiv.org/abs/0902.0732&quot; rel=&quot;nofollow&quot;&gt;very recently&lt;/a&gt;, I think this had no algebraic proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working in the analytic category also allows us to use the theory of (real) Lie groups and Lie algebras. See for example &lt;a href=&quot;http://mathoverflow.net/questions/16207/why-can-the-dolbeault-operators-be-realised-as-lie-algebra-actions&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/16207/why-can-the-dolbeault-operators-be-realised-as-lie-algebra-actions&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/16880/dolbeault-operators-for-cp1-as-mathfraksu2-actions&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/16880/dolbeault-operators-for-cp1-as-mathfraksu2-actions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the analytic theory is of independent interest in mathematical physics and in string theory. See for instance the famous paper &quot;Kodaira-Spencer Theory of Gravity and Exact Results for Quantum String Amplitudes&quot; by Bershadsky-Cecotti-Ooguri-Vafa.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-08-10T20:36:18.940" LastActivityDate="2010-08-10T20:36:18.940" CommentCount="1" />
  <row Id="2019" PostTypeId="2" ParentId="2013" CreationDate="2010-08-10T05:47:41.463" Score="1" ViewCount="0" Body="&lt;p&gt;In abstract probability based on measure theory, you are asking about the proof that any function from your list is measurable, such as $f(X,Y)=X+Y$ or $XY$ or $|X|^Y$.  You can see proofs of this for specific $f$ in any textbook that discusses Lebesgue measure theory.&#xA;Any continuous multivariable function is measurable, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice $f$ is some concrete function and you need only that for any reasonable subset $S$ of the range of $f$ (an open set, a compact set, countable or &quot;tame&quot; unions and intersections of these), $f^{-1}(S)$ is reasonable in a similar sense and therefore &quot;obviously measurable&quot;.   Explicitly proving the measurability of a specific function would be unusual.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-10T20:08:10.503" LastActivityDate="2010-08-10T20:08:10.503" />
  <row Id="2021" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T06:28:53.213" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.livegeometry.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Live Geometry&lt;/strong&gt;&lt;/a&gt; for Windows users.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-10T06:28:53.213" CommunityOwnedDate="2010-08-10T06:28:53.213" />
  <row Id="2022" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T06:32:27.540" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolfram.com/products/mathematica/index.html&quot; rel=&quot;nofollow&quot;&gt;Mathematica&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T06:32:27.540" CommunityOwnedDate="2010-08-10T06:32:27.540" />
  <row Id="2025" PostTypeId="2" ParentId="1983" CreationDate="2010-08-10T07:54:41.083" Score="3" ViewCount="0" Body="&lt;p&gt;The best result currently known along these lines, due to Hermann Weyl, is that&#xA; $$|\zeta(\frac{1}{2}+it)|=\mathcal{O}(\tau^{16+\epsilon})$$ for any $\epsilon &gt; 0$ as $\tau \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can refer more about this on: &lt;a href=&quot;http://www.openquestions.com/oq-ma014.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.openquestions.com/oq-ma014.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-10T07:54:41.083" CommentCount="3" />
  <row Id="2026" PostTypeId="2" ParentId="1989" CreationDate="2010-08-10T08:02:41.627" Score="0" ViewCount="0" Body="&lt;p&gt;For example, suppose that $f_{1},f_{2},\cdots,f_{n}$ is a sequence of holomorphic functions converging uniformly to a function $f$ on an open disc. By cauchy's theorem we know that $$\oint\limits_{C} f(z) \ dz = 0 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for every $n$, along any closed curve $C$ in the disc. Then the uniform convergence implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint\limits_{C} f(z) \ dz = \lim_{n \to \infty} \oint\limits_{C} f_{n}(z) \ dz =0$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-10T08:02:41.627" CommentCount="2" />
  <row Id="2027" PostTypeId="2" ParentId="1236" CreationDate="2010-08-10T08:31:31.807" Score="0" ViewCount="0" Body="&lt;p&gt;I think this may be of some help: &lt;a href=&quot;http://iopscience.iop.org/1873-7005/15/3/A03/pdf/1873-7005_15_3_A03.pdf&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1873-7005/15/3/A03/pdf/1873-7005_15_3_A03.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-10T08:31:31.807" />
  <row Id="2028" PostTypeId="1" AcceptedAnswerId="2039" CreationDate="2010-08-10T08:36:44.617" Score="2" ViewCount="155" Body="&lt;p&gt;I recently read a very good inequality concerning the no of primes $\pi(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi(n)&gt;\frac{1}{6}\frac{n}{\log n}\mathrm{\ for\ }n\ge 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are any other such elementary inequalities concerning the primes?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T16:39:53.177" LastActivityDate="2010-08-10T16:39:53.177" ClosedDate="2010-08-10T16:49:37.720" Title="Inequalities Concerning Primes" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2029" PostTypeId="1" AcceptedAnswerId="2030" CreationDate="2010-08-10T09:08:10.110" Score="4" ViewCount="277" Body="&lt;p&gt;A prime number not equal to 2 and 5 can't have last digit equal to 2,4,5,6 and 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that this is the only restriction on last digits of prime numbers?&#xA;I mean if its true that for any sequence of digits with last digit not equal to 2,4,5,6 and 8&#xA;there exists a prime number with given sequence of digits?&lt;/p&gt;&#xA;" OwnerUserId="1006" LastActivityDate="2010-08-11T06:40:59.707" Title="Last digits of primes" Tags="&lt;prime-numbers&gt;" AnswerCount="5" />
  <row Id="2030" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T09:14:33.193" Score="13" ViewCount="0" Body="&lt;p&gt;I think that you forgot about zero. It's clear that prime number can't have zero as last digit. And if you add zero to your list then your statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_1, x_2, \cdots x_n$ be your sequence. Take $x = x_1 10^{n-1} + \cdots + x_n$ ---&#xA;number formed by this sequence. Then $x$ and $10^n$ are coprime since last digit of $x$ is odd and is not divisible by 5.&#xA;Hence by &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirichlet theorem&lt;/a&gt; there exists&#xA;infinitely many primes $p$ with $p \equiv x \pmod {10^n}$. Those primes have required last digits.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-10T09:14:33.193" CommentCount="1" />
  <row Id="2031" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T09:16:00.590" Score="4" ViewCount="0" Body="&lt;p&gt;Don't forget that the last digit could also be $0$, but there are no such prime numbers (since anything ending in $0$ is divisible by $10$).   Otherwise, yes, this is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a string of final digits $a_n \cdots a_1$ with $a_1 \neq 0,2,4,5,6,8$, then the number $k = 10^{n-1} a_n + \ldots + a_1$ is not divisible by $2$ or $5$ so is prime to $10^n$.  By a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_primes_in_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;celebrated theorem of Dirichlet&lt;/a&gt;, the arithmetic progression $k, k + 10^n, k + 2 \cdot 10^n,\ldots$ then contains infinitely many prime numbers.  Any one of these is &#xA;congruent to $k$ modulo $10^n$, which is equivalent to saying that the sequence of the last $n$ digits is $a_n \cdots a_1$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-10T09:16:00.590" />
  <row Id="2032" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T09:21:59.440" Score="2" ViewCount="0" Body="&lt;p&gt;There have already been a lot of good advices. Let me stress the importance of chosing the journal right. There are too difficulties here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The journal is interested in your subject.&lt;/li&gt;&#xA;&lt;li&gt;Your paper fits the quality standards of the journal.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for (1) there is an easy algorithm: you sure cite other works in you bibliography, right? Look at the journals in which they were published. If an specific journal does appear more often than others, there is a good chance that your subject interests that journal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue is more difficult to grasp for beginners without the advise of a senior colleague. It takes some time to develop the sense of how &lt;em&gt;good&lt;/em&gt; is your paper and, consulting the JCR list of mathematical journals, find the appropriate journal according to its quality. Of course, you could try another easy &quot;algorithm&quot; for this: start with the highest quality journal in your field and, as they reject your paper, go down the list. There is nothing wrong in doing this, but some experience could help you in not trying Annals of Maths as your first option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a final advice: I'm sure that almost &lt;em&gt;every&lt;/em&gt; professional mathematician has had at least one of his/her papers rejected (in my case, I think that 30% of my papers were rejected by the first journal where I've sent them).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-10T09:21:59.440" />
  <row Id="2033" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T09:35:55.887" Score="2" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A mathematician is a device for&#xA;  turning coffee into theorems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Paul Erdős (or Alfréd Rényi)&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-08-10T09:35:55.887" CommunityOwnedDate="2010-08-10T09:35:55.887" />
  <row Id="2034" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T10:42:08.673" Score="5" ViewCount="0" Body="&lt;p&gt;I like Newton's:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have seen a little further it is by standing on the shoulders of Giants&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If only because of its lesser known interpretation. The above quote is from a letter by Newton to Robert Hooke, a man born with a severe stoop who criticized Newton for having stolen his theory of optics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's response, below the waterline, can be seen as saying &quot;I didn't steal optics from you, shorty!&quot; Hiding such snideness in a concession to modesty, even if terribly mean and apocryphal as an interpretation, is pretty stunning...&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-10T10:42:08.673" CommunityOwnedDate="2010-08-10T10:42:08.673" />
  <row Id="2035" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T10:53:04.290" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.inkscape.org/&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt; is a popular open source drawing package.  One feature that makes it particularly nice for drawing mathematical diagrams is that you can easily &lt;a href=&quot;http://www.johndcook.com/blog/2009/12/22/including-latex-in-inkscape/&quot; rel=&quot;nofollow&quot;&gt;insert LaTeX output&lt;/a&gt; in your drawings.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-10T10:53:04.290" CommunityOwnedDate="2010-08-10T10:53:04.290" />
  <row Id="2036" PostTypeId="2" ParentId="1969" CreationDate="2010-08-10T11:20:54.510" Score="4" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Young man, in mathematics you don't understand things. You just get used to them.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;John von Neumann&lt;/p&gt;&#xA;" OwnerUserId="1011" LastActivityDate="2010-08-10T11:20:54.510" CommunityOwnedDate="2010-08-10T11:20:54.510" />
  <row Id="2037" PostTypeId="1" AcceptedAnswerId="2069" CreationDate="2010-08-10T11:22:22.450" Score="6" ViewCount="327" Body="&lt;p&gt;What is the probability of getting &lt;a href=&quot;http://en.wikipedia.org/wiki/Yahtzee#Lower_section&quot; rel=&quot;nofollow&quot;&gt;yahtzee&lt;/a&gt; using N dices with X sides in Y throws in a single round?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which side of the dice one gets yahtzee with doesn't matter (e.g. it doesn't matter if it is ones or twos etc.). I also assume perfect strategy is used, that is, after each throw one saves the number it is the most of (and if equal, just pick one at random).&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="893" LastEditorDisplayName="" LastEditDate="2010-08-11T08:53:17.867" LastActivityDate="2010-10-11T14:03:30.083" Title="What is the probability of getting yahtzee?" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="2038" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T11:26:13.287" Score="3" ViewCount="0" Body="&lt;p&gt;There are at least two kinds of (interesting) examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. When we can get an abelian category, but have to add more (co)kernels: 1) the category of free modules over a ring; 2) the category of projective modules over a ring; 3) the category of vector bundles on a topological space (if fact, 3 is a particular case of 2).&#xA;(From 1 or 2 one gets abelian category of all modules over the ring, from 3 — abelian category of sheafs of vector spaces on X.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II. When we already have (co)kernels (&quot;category is pre-abelian&quot;) but not all mono-/epimorphisms are normal. As explained in another answer, an example is the category of filtered modules.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-10T16:51:14.300" LastActivityDate="2010-08-10T16:51:14.300" CommunityOwnedDate="2010-08-11T11:29:36.717" />
  <row Id="2039" PostTypeId="2" ParentId="2028" CreationDate="2010-08-10T11:42:22.387" Score="1" ViewCount="0" Body="&lt;p&gt;What is your question? This is well-known stuff. It appears for instance in section 4.5 of Apostol's &lt;em&gt;Introduction to Analytic Number Theory&lt;/em&gt;, which also contains an upper bound. You could at least say where you got yours from. (This should have been a comment, not an answer, but I don't have enough reputation for adding comments.)&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-10T11:42:22.387" CommentCount="1" />
  <row Id="2040" PostTypeId="1" AcceptedAnswerId="2041" CreationDate="2010-08-10T11:47:45.387" Score="3" ViewCount="116" Body="&lt;p&gt;We should have $\mathbb{Q} = \lim_\rightarrow \frac 1n \mathbb{Z}$ but a few things are confusing me. Since the index category is a set, we should get the coproduct: $\bigsqcup \frac 1n \mathbb{Z}$. But in this coproduct we have $1/2 \neq 2/4$ which clearly is wrong. This make me suspect that the index category is not an index set. What we probably should get is a union (and not a coproduct) in the category of rings (in which $1/2 = 2/4$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Help me understand what the index category is (eg, what are its maps) and am I correct when believing we should get an ordinary union? &lt;/p&gt;&#xA;" OwnerUserId="1013" LastActivityDate="2010-08-10T12:57:11.487" Title="Colimit of $\frac{1}{n} \mathbb{Z}$" Tags="&lt;category-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2041" PostTypeId="2" ParentId="2040" CreationDate="2010-08-10T11:52:28.670" Score="6" ViewCount="0" Body="&lt;p&gt;Index category is the category of positive integers with a morphism from n to m iff n|m. The limit is not coproduct but union of $\frac{1}{n}\mathbb{Z}$ with natural embedings (which is the same as union of $\frac{1}{n}\mathbb{Z}$ inside $\mathbb{Q}$ — hence the equality).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-10T12:57:11.487" LastActivityDate="2010-08-10T12:57:11.487" />
  <row Id="2042" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T12:25:19.743" Score="0" ViewCount="0" Body="&lt;p&gt;I use the programme CaRMetal. Here are a few screenshots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i34.tinypic.com/2vwcits.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i34.tinypic.com/2vwcits.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i35.tinypic.com/oa42ts.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i35.tinypic.com/oa42ts.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i35.tinypic.com/2lxfin8.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i35.tinypic.com/2lxfin8.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i35.tinypic.com/rble9l.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i35.tinypic.com/rble9l.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess most geometric constructions programmes are good, although I wouldn't recommend Kig.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-10T12:25:19.743" CommunityOwnedDate="2010-08-10T12:25:19.743" />
  <row Id="2043" PostTypeId="1" AcceptedAnswerId="2047" CreationDate="2010-08-10T12:36:57.627" Score="5" ViewCount="217" Body="&lt;p&gt;It's easy for scalars, $(\exp(a(x)))' = a' e^a$. But can anything be said about matrices? Do $A(x)$ and $A'(x)$ commute such that $(\exp(A(x)))' = A' e^A = e^A A'$ or is this only a special case?&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-08-11T13:50:17.213" Title="Is there a general formula for the derivative of $\exp(A(x))$ when $A(x)$ is a matrix?" Tags="&lt;calculus&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2044" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T12:39:56.780" Score="0" ViewCount="0" Body="&lt;p&gt;When you write down numbers in base $10 = 2\cdot 5$, the last digit only allows to determine divisibility by the base's factors 2 and 5, indicated by $0,2,4,5,6,8$. Would you pick another base, e.g. $30=2\cdot 3\cdot 5$ you could read the divisibility by 3 in addition.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-08-10T12:39:56.780" />
  <row Id="2045" PostTypeId="1" AcceptedAnswerId="2053" CreationDate="2010-08-10T12:53:04.397" Score="3" ViewCount="298" Body="&lt;p&gt;I'm working on a problem from a past exam and I'm stuck, so I'm asking for help.&#xA;Here it is:  $A = \frac12 \left[\begin{array}{rrrr}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{array}\right]$ find  $\mathbf A^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem isn't the inverse matrix itself. We just get the determinant, see if it's zero or not, get the adjoint matrix and divide it by determinant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is space. As you can see, it's a 4x4 matrix meaning that I'd have to do 4x4 3x3 determinants to get the adjoint matrix plus 2 3x3 determinants to get determinant of the matrix. Now we get one A3 piece of paper for 6 problems. The problems are printed on one side and the other side is blank. This and the fact that inverse matrix is $A = \frac12 \left[\begin{array}{rrrr}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{array}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;led me to believe that there's some catch that I do not see. Any ideas what could it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;STRIKE&gt;Also if someone could edit these matrices from MATLAB format into something that this site will parse would be great!&lt;/STRIKE&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;STRIKE&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Unfortunately it seem that TeX code for matrices doesn't work here. Here's the matrix in MATLAB form, if anyone wants it A=(1/2)*[1,1,1,1;1,1,-1,-1;1,-1,1,-1;1,-1,-1,1];&lt;/STRIKE&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt; Answer by Jack Schmidt contains code for matrices. &lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-08-10T13:56:05.043" LastActivityDate="2010-08-11T00:42:38.457" Title="How to get inverse of this matrix using least amount of space?" Tags="&lt;matrices&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="2046" PostTypeId="2" ParentId="1989" CreationDate="2010-08-10T12:55:38.543" Score="1" ViewCount="0" Body="&lt;p&gt;Ok, please see this link, i hope this helps: &lt;a href=&quot;http://www.jstor.org/page/termsConfirm.jsp?redirectUri=/stable/pdfplus/2309592.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/page/termsConfirm.jsp?redirectUri=/stable/pdfplus/2309592.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-10T12:55:38.543" CommentCount="2" />
  <row Id="2047" PostTypeId="2" ParentId="2043" CreationDate="2010-08-10T13:09:40.663" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/users/121/tom-stephens&quot;&gt;Tom Stephens&lt;/a&gt; linked to &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_exponential#The_exponential_map&quot; rel=&quot;nofollow&quot;&gt;the exponential map&lt;/a&gt;, which states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ \frac{d}{dt}e^{X(t)} = \int\limits_0^1 e^{\alpha X(t)} \frac{dX(t)}{dt} e^{(1-\alpha) X(t)} d\alpha $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $X(t)$ and $\frac{d}{dt}X(t)$ commute, the latter also commutes with $\exp(X(t))$ and &lt;em&gt;then&lt;/em&gt; it simplifies into $ \frac{d}{dt}e^{X(t)} = \frac{d X(t)}{dt} e^{X(t)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counter-example is&#xA;$$X(t) = \begin{pmatrix} \cos(t) &amp;amp; \sin(t) \\ \sin(t) &amp;amp; -\cos(t) \end{pmatrix}$$&#xA;at $t=0$, i.e. $X(0) = \sigma_3, \dot X(0) = \sigma_1$ ($\sigma_i$ are the non-commuting &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_matrices&quot; rel=&quot;nofollow&quot;&gt;Pauli Matrices&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-08-10T14:59:38.213" LastActivityDate="2010-08-10T14:59:38.213" />
  <row Id="2048" PostTypeId="1" CreationDate="2010-08-10T13:13:35.100" Score="2" ViewCount="104" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure a 'power equation' is the right name for the equation I'd like to know more about (and, specifically, about its solutions), but I don't know the 'proper' way to name it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest 'power polynomial' is : $P(x) = x^x $. The simplest 'power equation is: (1) $x^x = c$ for some $c \in \mathbb{R} $. What are the &lt;em&gt;exact&lt;/em&gt; solutions to (1) of x (in $\mathbb{C}$) with regards to c, apart from the 'obvious' solutions (x=0, c=0), (x=1, c=1) and (x=2, c=4) and all the other solutions of the form $ x^x = n^n$ for $n \in \mathbb{N}$ ?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could extend the power polynomial: $P_{2}(x) = x^{{ax}^{bx}} + x^{cx} $. What are the &lt;em&gt;exact&lt;/em&gt; solutions of $P_{2}(x) = d$ for $a,b,c,d \in \mathbb{R}, x\in \mathbb{C}$ ? Or perhaps I should ask what the exact form of the solutions is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could generalize the power polynomial even further to $P_{3}(x)$ and $P_{n}(x)$, but I don't know how to write down the latter, general polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I References are always welcome. II If you think this question belongs to MO, please tell me).&lt;/p&gt;&#xA;" OwnerUserId="1014" LastActivityDate="2010-08-10T13:21:19.947" Title="Solution(s) to 'power equations'" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="2049" PostTypeId="2" ParentId="1946" CreationDate="2010-08-10T13:15:33.817" Score="2" ViewCount="0" Body="&lt;p&gt;@ Julio Cesar: I don't know about the problem, but this is why i think that the result could be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Let $f$ and $g$ be continuous on $[a,b]$, and let $g$ have constant sign on $[a,b]$. Then there is a $c \in (a,b)$ such that $$\int\limits_{a}^{b} f(x)g(x) \ dx = f(c) \int\limits_{a}^{b} g(x) \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer &quot;E.I. Poffald, Amer.Math.Monthly 97 (1990), 205-213&quot; for a proof of this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this result, by Taylor's formula with integral remainder we have $$f^{n}\Bigl(\frac{x}{n+1}\Bigr)=f^{n}(0) + f^{n+1}(0)\frac{x}{n+1} + \int\limits_{0}^{\frac{x}{n+1}} f^{n+2}(t) \Bigl(\frac{x}{n+1}-t\Bigr) \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $$f(0)+ \frac{f'(0)}{1!}x + \cdots + \frac{f^{n-1}(0)}{(n-1)!}x^{n-1} + \frac{f^{n}(\frac{x}{n+1})}{n!}x^{n} = \text{ Our sum with the Integral Remainder}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one can define $g$ in this manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(t)= \frac{(x-t)^{n+1}}{n+1} - x^{n}\Bigl(\frac{x}{n+1}-t\Bigr) \quad ; t \in [0,x]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $g'(t)&gt;0$ for $t \in (0,x)$ and $g(0)=0$. Thus $g$ is positive on the open interval $(0,x)$. Then using our theroem we have $$\int\limits_{0}^{\frac{x}{n+1}} f^{n+2}(t)(g(t))=f^{n+2}(c) \int\limits_{0}^{\frac{x}{n+1}} g(t) \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where i am stuck.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-10T13:15:33.817" />
  <row Id="2050" PostTypeId="2" ParentId="2048" CreationDate="2010-08-10T13:21:19.947" Score="2" ViewCount="0" Body="&lt;p&gt;Maybe you should read about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W-function&lt;/a&gt; which gives the solutions to expressions like $z=x^x$. However, I am not sure what to do in case of such a &quot;power tower&quot; like $P_2$.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-08-10T13:21:19.947" />
  <row Id="2051" PostTypeId="1" AcceptedAnswerId="2063" CreationDate="2010-08-10T13:24:08.287" Score="0" ViewCount="177" Body="&lt;p&gt;Give an example of sequence of Measurable functions defined on some measurable subset $E$ of $\mathbb{R}$  such that $f_{n} \to f$ pointwise almost everywhere on $E$ but $$\int\limits_{E} f \ dm \not\leq \lim_{n} \inf \int\limits_{E} f_{n} \ dm$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-11T19:51:08.943" LastActivityDate="2010-08-11T19:51:08.943" Title="Fatou's Lemma Counterexample" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2052" PostTypeId="2" ParentId="2051" CreationDate="2010-08-10T13:39:18.907" Score="0" ViewCount="0" Body="&lt;p&gt;For $E=[0,1]$ set: $f_n(x)=n- n^2x$ for $x\in [0,\frac1n]$ $f_n(x)=0$ else.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-10T13:39:18.907" CommentCount="3" />
  <row Id="2053" PostTypeId="2" ParentId="2045" CreationDate="2010-08-10T13:44:13.857" Score="2" ViewCount="0" Body="&lt;p&gt;I guess A is &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_matrix&quot; rel=&quot;nofollow&quot;&gt;orthogonal&lt;/a&gt; and symmetric, so that tells you A&lt;sup&gt;−1&lt;/sup&gt; = A&lt;sup&gt;T&lt;/sup&gt; = A, but uh, that's not a very common situation to my mind.  Maybe someone else has a better &quot;test-taking strategy&quot; explanation, but me personally, I would just row reduce or whatever method you use in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orthogonal matrix is defined to be a matrix whose transpose is its inverse.  However, for us the better (almost) definition is a matrix whose rows (or columns) are orthogonal, as in, perpendicular.  So (1,1,1,1) is orthogonal to (1,-1,1,-1) since their dot product is (1)(1)+(1)(-1)+(1)(1)+(1)(-1) = 1 - 1 + 1 - 1 is zero.  You also should check that the length of the vector in each row is 1, $\sqrt{(1/2)^2 + (1/2)^2 + (1/2)^2 + (1/2)^2} = 1$ so good, but even if not, that part is easily fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes you can tell just by looking that a matrix is orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As far as parsing goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the matrix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$A = \frac12 \begin{pmatrix}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\\\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{pmatrix}$&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$A = \frac12 \begin{pmatrix}1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;-1&amp;amp;-1\\1&amp;amp;-1&amp;amp;1&amp;amp;-1\\1&amp;amp;-1&amp;amp;-1&amp;amp;1\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is using the array environment:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$A = \frac12 \left(\begin{array}{rrrr}&#xA;1 &amp;amp;  1 &amp;amp;  1 &amp;amp;  1 \\\\&#xA;1 &amp;amp;  1 &amp;amp; -1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp;  1 &amp;amp; -1 \\\\&#xA;1 &amp;amp; -1 &amp;amp; -1 &amp;amp;  1&#xA;\end{array}\right)$&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$A = \frac12 \left(\begin{array}{rrrr}1&amp;amp;1&amp;amp;1&amp;amp;1\\1&amp;amp;1&amp;amp;-1&amp;amp;-1\\1&amp;amp;-1&amp;amp;1&amp;amp;-1\\1&amp;amp;-1&amp;amp;-1&amp;amp;1\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The backslashes get eaten by the markdown software, so you just double them.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-10T14:00:12.193" LastActivityDate="2010-08-10T14:00:12.193" CommentCount="6" />
  <row Id="2054" PostTypeId="2" ParentId="2029" CreationDate="2010-08-10T13:52:45.317" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, including 0. Generally  $\pmod m$ the number of elts coprime to $m$ is counted by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_totient_function&quot; rel=&quot;nofollow&quot;&gt;Euler totient function&lt;/a&gt; $\phi(m)$. These elts are precisely the units, i.e. invertible elts $\pmod m$. Here $\phi(10) = 4$ with units $\{1,3,7,9\}$ Generally each such unit may be represented by a prime - this is the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;Dirichlet theorem&lt;/a&gt; on arithmetic progressions. These units form a group and the study of such unit groups plays a fundamental role in number theory and algebra.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-10T14:05:29.407" LastActivityDate="2010-08-10T14:05:29.407" />
  <row Id="2055" PostTypeId="1" CreationDate="2010-08-10T14:06:03.147" Score="0" ViewCount="207" Body="&lt;p&gt;I'd like to write down proofs. I use first-order logic and natural deduction. Formulas tend to be long, too long for LaTeX. Writing formulas with the computer is also a slow process. Handwriting is a lot faster and more natural, but I can't trust paper. Time, water, dirt, fire are its enemies. I could write down something and then immediately scan it, but I'm an eco-guy and I'm worried about using a whole page for only two lines. How do you write math? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1018" LastActivityDate="2010-08-10T16:14:13.213" CommunityOwnedDate="2010-08-10T14:49:09.253" Title="How to write down proofs?" Tags="&lt;soft-question&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="2" />
  <row Id="2056" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:09:10.473" Score="1" ViewCount="0" Body="&lt;p&gt;On a blackboard.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-10T14:09:10.473" />
  <row Id="2057" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:22:24.327" Score="2" ViewCount="0" Body="&lt;p&gt;There are a number of possible approaches. One can become quite proficient even with a text-based editor if it is user extensible - e.g. see the (La)TeX extensions for Emacs. These are easily user customizable if you have basic programming knowledge. Alternatively see the WYSIWYG LaTeX front-ends, e.g. Lyx. Another possibility is to employ a graphics / digitizing tablet (e.g. Wacom) for input, but this may require writing your own software to obtain optimal productivity.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-10T14:22:24.327" CommentCount="2" />
  <row Id="2058" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:38:30.917" Score="0" ViewCount="0" Body="&lt;p&gt;If you are a user of Windows 7 and have a graphics tablet , you could use it in combination with hand-writing recognition using the Math Input Panel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take some time to train your computer to correctly recognize your handwriting, but if you are only going to use it for formulas, then it shouldn't take more that few days. You can also use it with mouse, but it's less natural then with a real pen and tablet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor that can impact your experience is the choice of the tablet. Good ones are mainly from Wacom and are expensive (about 50€ here for cheapest models!) but come with good pens and good drivers. They are also powered using EM induction, so the mass distribution in the pen is more like in real pens. (BTW, I really don't want to advertise Wacom, but I'm not aware of any other manufacturer whose tablets have good drivers and whose pens are small and light enough.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also used a tablet made by Trust and had problems with drivers and weight distribution. Pen is powered by 2 AAA batteries which are in top part of the pen making it very heavy. Also pen is larger that a real pen so it feels clumsy.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-08-10T14:38:30.917" CommunityOwnedDate="2010-08-10T15:01:02.720" />
  <row Id="2059" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T14:58:40.807" Score="1" ViewCount="0" Body="&lt;p&gt;If you plan to enter the scientific community you should begin to learn LaTeX at precisely this moment.  I recommend text-based editors such as emacs or vim for the sole reason that they are available on &lt;em&gt;every&lt;/em&gt; unix-based machine, whether accessed remotely or on your own personal shiny macbook. &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-10T14:58:40.807" CommunityOwnedDate="2010-08-10T14:58:40.807" />
  <row Id="2060" PostTypeId="2" ParentId="2043" CreationDate="2010-08-10T15:19:39.843" Score="1" ViewCount="0" Body="&lt;p&gt;The question comes down to computing what's called the &quot;derivative of&quot;, the &quot;differential of&quot;, or the &quot;tangent map to&quot;, the exponential map from $M_n(\mathbb R)$ into itself at a given matrix $A$ (not necessarily the zero matrix). There is a classical formula for this. Here is the first reference I found: pages 1 and 2 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.columbia.edu/~woit/notes4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.columbia.edu/~woit/notes4.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Peter Woit. Here it is (with Peter Woit's notation) &#xA;$$\exp_*(X)\ Y=\exp(X)\ \frac{1-e^{-ad(X)}}{ad(X)}\ Y.$$ &#xA;Here is a reference for the Chain Rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule#The_fundamental_chain_rule&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chain_rule#The_fundamental_chain_rule&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reads, in Peter Woit's notation and under appropriate assumptions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f\circ g)_*(x)=f_*(g(x))\circ g_*(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Thank you to &lt;a href=&quot;http://math.stackexchange.com/users/171/kennytm&quot;&gt;KennyTM&lt;/a&gt; for having edited this formula.]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1.&lt;/strong&gt; Here are the two formulas written in another notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp'(X)=\exp(X)\ \frac{1-e^{-ad(X)}}{ad(X)}\quad,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f\circ g)'(x)=f'(g(x))\circ g'(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2.&lt;/strong&gt; Here is another reference. This is a post by &lt;a href=&quot;http://math.stackexchange.com/users/536/akhil-mathew&quot;&gt;Akhil Mathew&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://deltaepsilons.wordpress.com/2009/11/07/helgasons-formula-ii/&quot; rel=&quot;nofollow&quot;&gt;http://deltaepsilons.wordpress.com/2009/11/07/helgasons-formula-ii/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-11T13:50:17.213" LastActivityDate="2010-08-11T13:50:17.213" CommentCount="4" />
  <row Id="2062" PostTypeId="1" AcceptedAnswerId="2407" CreationDate="2010-08-10T15:45:55.957" Score="2" ViewCount="198" Body="&lt;p&gt;We know that if $ \displaystyle d(n)= \sum\limits_{d \mid n} 1$, then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum\limits_{n \leq x} d(n)= x\log{x} + (2C-1)x + \mathcal{O}(\sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have referred Apostol's &quot;Analytic Number theory&quot; and i understood the first half of the proof where the error term is $\mathcal{O}(x)$, but please tell me as to how to improve the error term to $\sqrt{x}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-27T18:05:27.470" LastActivityDate="2010-10-27T18:05:27.470" Title="Dirichlet's Divisor Problem" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;" AnswerCount="3" />
  <row Id="2063" PostTypeId="2" ParentId="2051" CreationDate="2010-08-10T15:49:54.207" Score="2" ViewCount="0" Body="&lt;p&gt;Since Fatou’s Lemma &lt;em&gt;holds&lt;/em&gt; for non-negative measurable functions, I suppose you are looking for an example involving not necessarily non-negative functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take, for instance, $f_n(x)=-1_{n\leq x\leq n+1}$ (just $-1$ times the indicator function on the interval $[n,n+1]$). Then $f=\lim\limits_{n\to\infty}\inf f_n=0$. Hence,&#xA;$$&#xA;0=\int f dm \not\leq \lim_{n\to\infty}\inf\int f_n dm = -1.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-10T15:58:30.197" LastActivityDate="2010-08-10T15:58:30.197" CommentCount="2" />
  <row Id="2064" PostTypeId="2" ParentId="1985" CreationDate="2010-08-10T15:59:10.417" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://tug.org/metapost.html&quot; rel=&quot;nofollow&quot;&gt;Metapost&lt;/a&gt;. There are a few packages to simplify drawing geometry drawings.&lt;/p&gt;&#xA;" OwnerUserId="763" LastActivityDate="2010-08-10T15:59:10.417" CommunityOwnedDate="2010-08-10T15:59:10.417" />
  <row Id="2065" PostTypeId="2" ParentId="2037" CreationDate="2010-08-10T16:04:41.960" Score="2" ViewCount="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Yahtzee#Probabilities&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; &quot;the probability&#xA;of a Yahtzee for any three-roll turn is about 0.04603 (or $\frac{347897}{7558272}$), or roughly 1 in 22 attempts.&quot;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-10T16:15:30.687" LastActivityDate="2010-08-10T16:15:30.687" />
  <row Id="2066" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T16:07:04.687" Score="-1" ViewCount="0" Body="&lt;p&gt;On book margins.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-10T16:07:04.687" CommunityOwnedDate="2010-08-10T16:07:04.687" CommentCount="1" />
  <row Id="2067" PostTypeId="2" ParentId="2055" CreationDate="2010-08-10T16:14:13.213" Score="1" ViewCount="0" Body="&lt;p&gt;Judging from your question, I presume you want to write down &lt;em&gt;formal&lt;/em&gt; proofs in natural deduction. If so, a proof assistant (check Wikipedia for examples) will help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't care that your proofs are formally checked, you can use $\LaTeX$ with the &lt;a href=&quot;http://math.ucsd.edu/~sbuss/ResearchWeb/bussproofs/&quot; rel=&quot;nofollow&quot;&gt;bussproofs package&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1024" LastActivityDate="2010-08-10T16:14:13.213" CommunityOwnedDate="2010-08-10T16:14:13.213" />
  <row Id="2068" PostTypeId="2" ParentId="2062" CreationDate="2010-08-10T16:20:26.110" Score="4" ViewCount="0" Body="&lt;p&gt;I'd post this as a comment, but I don't have the reputation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There usual route is via the Hyperbola Method: &lt;a href=&quot;http://myyn.org/m/article/dirichlet-hyperbola-method/&quot; rel=&quot;nofollow&quot;&gt;http://myyn.org/m/article/dirichlet-hyperbola-method/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-08-10T16:20:26.110" />
  <row Id="2069" PostTypeId="2" ParentId="2037" CreationDate="2010-08-10T16:25:04.423" Score="4" ViewCount="0" Body="&lt;p&gt;What follows is a solution to a smaller problem; I don't have a good way to determine the probabilities of exactly k of some kind being the most of any kind on the first roll, but the work below could be used from that point to get an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(N,X,Y)$ be the probability of getting all 1s with N dice, each with X sides, in at most Y rolls, where after each roll, 1s are retained and only non-1 dice are re-rolled.  Working 1 roll at a time, the first roll could have exactly $k=0,\dots,N$ 1s.  The probability of exactly k 1s in a single roll is ${N\choose k}\left(\frac{1}{X}\right)^k\left(\frac{x-1}{x}\right)^{N-k}$ and having k 1s in the first roll, we then need $N-k$ 1s in the remaining $Y-1$ rolls.  So, &#xA;$$f(N,X,Y)=\sum_{k=0}^{N}{N\choose k}\left(\frac{1}{X}\right)^k\left(\frac{x-1}{x}\right)^{N-k}f(N-k,X,Y-1).$$&#xA;Also, the probability of all 1s on 0 dice is 1 (0 of the 0 dice are guaranteed to be 1), so $f(0,X,Y)=1$, and the probability of all 1s on $N\ge 0$ dice in 0 rolls is 0 (can't get any 1s without rolling some dice), so $f(N,X,0)=0$ for $N\ge 0$.  This is a complete definition of $f(N,X,Y)$, though it does not lend itself to easy computation.  (However, software like Mathematica may be able to compute from this definition.  In Mathematica: &lt;code&gt;f[n_, x_, y_] := If[n == 0, 1, If[y == 0, 0, &#xA;       Sum[&#xA;    Binomial[n, k]*(1/x)^k*(1 - 1/x)^(n - k)*f[n - k, x, y - 1], &#xA;         {k, 0, n}]]]&lt;/code&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T16:25:04.423" CommentCount="1" />
  <row Id="2070" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T16:25:11.100" Score="1" ViewCount="0" Body="&lt;p&gt;Another nice example is the category of finite-dimensional vector bundles over a fixed base space (with bundle maps over the identity as morphisms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the base space is not too simple (the interval suffices), then this category is not (pre-)Abelian because there are, in general, neither kernels nor cokernels. Intuitively speaking, the obvious candidate for the kernel of a bundle map need not form a vector bundle because its dimension need not be locally constant.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-10T18:07:57.147" LastActivityDate="2010-08-10T18:07:57.147" CommentCount="3" />
  <row Id="2071" PostTypeId="1" CreationDate="2010-08-10T16:52:26.263" Score="6" ViewCount="156" Body="&lt;p&gt;Recently I found &lt;a href=&quot;http://www.reddit.com/r/programming/comments/cy4m6/and_e_appears_from_nowhere_quick_numeric/&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; on Reddit. It describes the following algorithm to find e: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here is an example of e turning up&#xA;  unexpectedly. Select a random number&#xA;  between 0 and 1. Now select another&#xA;  and add it to the first. Keep doing&#xA;  this, piling on random numbers. How&#xA;  many random numbers, on average, do&#xA;  you need to make the total greater&#xA;  than 1? The answer is e.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that you need on average ~2.7 random real numbers to make the sum greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a random number between 0 and 1 would on average be equal to 0.5. So intuitively I would think that, on average, only 2 random numbers would be required to have a sum &gt; 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where did I go wrong in my thinking? &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I just figured it out myself: You need at least two numbers to have a sum &gt; 1, but often you'll need three, sometimes you'll need four, sometimes five, etc... So it only natural that the average required numbers is above 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the replies!&lt;/p&gt;&#xA;" OwnerUserId="1025" LastEditorUserId="1025" LastEditorDisplayName="" LastEditDate="2010-08-10T17:44:48.100" LastActivityDate="2010-08-11T05:52:37.793" Title="Explanation of numeric experiment that approximates e?" Tags="&lt;number-theory&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="2072" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T17:01:39.853" Score="4" ViewCount="0" Body="&lt;p&gt;It is never possible to get greater than 1 on the first try, so the average must be strictly greater than 2.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-10T17:01:39.853" />
  <row Id="2073" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T17:16:22.543" Score="0" ViewCount="0" Body="&lt;p&gt;&quot;How many..., on average, ...&quot; means the expected number of random numbers.  That is, the sum of each possible number of random numbers multiplied by its probability.  As Qiaochu Yuan's answer says, it is impossible to be greater than 1 on the first try, so the expected number is&#xA;$$2\cdot P(\mathrm{it\ takes\ exactly\ 2\ numbers})+3\cdot P(\mathrm{it\ takes\ exactly\ 3\ numbers})$$ $$+\;4\cdot P(\mathrm{it\ takes\ exactly\ 4\ numbers})+\cdots$$&#xA;Since each of those probabilities is nonzero, but the sum of those probabilities is 1, the expected number must be strictly greater than 2.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-10T17:16:22.543" />
  <row Id="2074" PostTypeId="2" ParentId="1942" CreationDate="2010-08-10T17:28:36.283" Score="1" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://stackoverflow.com/questions/3442770/counting-placing-items-in-bins-according-to-boolean-restrictions&quot;&gt;Moron's answer&lt;/a&gt; to this question on StackOverflow.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I understood correctly your problem boils down to the problem of finding the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Matching_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;Perfect Matchings&lt;/a&gt; in a bipartite graph.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Items form the left set. Bins form the right. If a bin needs k items, you make k copies of that bin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now you form an edge between an item and a bin if the item is a possible candidate for the bin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now you need to find number of perfect matchings in this graph.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unfortunately, this is a hard problem. Calculating the number of perfect matchings in a graph is equivalent to finding the &lt;a href=&quot;http://en.wikipedia.org/wiki/Permanent&quot; rel=&quot;nofollow&quot;&gt;Permanent&lt;/a&gt; of an associated matrix, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sharp-P&quot; rel=&quot;nofollow&quot;&gt;#P-Complete&lt;/a&gt;. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Computation_of_the_permanent_of_a_matrix&quot; rel=&quot;nofollow&quot;&gt;Computation of the Permanent&lt;/a&gt;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Your best bet might be randomized/approximation algorithms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="984" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T21:58:48.003" LastActivityDate="2010-09-16T21:58:48.003" CommentCount="1" />
  <row Id="2075" PostTypeId="2" ParentId="674" CreationDate="2010-08-10T17:42:12.703" Score="3" ViewCount="0" Body="&lt;p&gt;You probably figured it out already, but for posterity -- this &lt;a href=&quot;http://www-diglib.stanford.edu/~klein/lagrange-multipliers.pdf&quot; rel=&quot;nofollow&quot;&gt;document&lt;/a&gt; gives intuition which I didn't see in Boyd book, esp. Figure 3 on page 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve equality constrained problem you look for points where gradient of objective is in the same direction as gradient of the constraint function. Why those points? Well, suppose we have a feasible point that's a maximum of objective and the gradients of objective/constraint functions are not collinear. Then you can take an infinitesimal step in the direction orthogonal to the gradient of constraint function to increase the objective. And because you are moving in the direction orthogonal to gradient of the constraint function, your constraint is still satisfied. Hence you get a feasible point with larger objective value, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-08-10T17:42:12.703" CommunityOwnedDate="2010-08-10T17:42:12.703" />
  <row Id="2076" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T17:45:59.780" Score="-1" ViewCount="0" Body="&lt;p&gt;Sorry, can't explain it either. It's obvious you need at least two random numbers to go greater than 1. And if the random numbers are on average .5 you'll need (edit) on average (/edit) maximum three numbers. So we end up in the range [2..3], but that's as close as I can get to e.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the method sure looks curious, it must be about the least efficient way ever to calculate e digits. A quick-and-dirty test program I wrote only gave 4 correct significant digits after a billion iterations. Or maybe Delphi's PRNG isn't perfect.&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-11T05:52:37.793" LastActivityDate="2010-08-11T05:52:37.793" CommentCount="3" />
  <row Id="2077" PostTypeId="2" ParentId="5" CreationDate="2010-08-10T17:49:48.213" Score="2" ViewCount="0" Body="&lt;p&gt;If $\sqrt 2$ were rational, we could write it as a fraction $a/b$ in&#xA;&lt;strong&gt;lowest terms&lt;/strong&gt;. Then&#xA;$$a^2 = 2 b^2.$$&#xA;Look at the last digit of $a^2$. It has to be $0$, $1$, $4$, $5$, $6$ or $9$.&#xA;Now look at the last digit of $2b^2$. It has to be $0$, $2$ or $8$.&#xA;As $a^2$ and $2b^2$ are the same number, its last digit must be $0$.&#xA;But that's only possible if $a$ ends in $0$ and $b$ ends in $0$ or $5$.&#xA;Either way both $a$ and $b$ are multiples of $5$ contradicting $a/b$&#xA;being in lowest terms.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-10T17:49:48.213" CommentCount="2" />
  <row Id="2078" PostTypeId="2" ParentId="697" CreationDate="2010-08-10T17:51:50.707" Score="2" ViewCount="0" Body="&lt;p&gt;For regression, least sum of squares criterion tried to fit your function through the mean of the data. In other words, with enough degrees of freedom, value of your fitted function for x will be the average of all observed values at x. Least sum of absolute values instead produces a function that'll go through the median of observed values. Some &lt;a href=&quot;http://hunch.net/?p=269&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; on this&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-08-10T17:51:50.707" />
  <row Id="2079" PostTypeId="1" AcceptedAnswerId="2081" CreationDate="2010-08-10T18:05:11.567" Score="2" ViewCount="293" Body="&lt;p&gt;Is there a isomorphism between the additive group of real algebraic numbers and the multiplicative group of positive real algebraic numbers, which is order preserving.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T04:55:45.747" LastActivityDate="2010-09-02T16:16:44.620" Title="Isomorphism between algebraic Numbers." Tags="&lt;abstract-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2080" PostTypeId="2" ParentId="1901" CreationDate="2010-08-10T18:08:08.363" Score="7" ViewCount="0" Body="&lt;p&gt;This would have belonged as a comment to MattE answer, but I do not have the reputation to leave comments. The reason it is impossible to algorithmically classify topological manifolds (up to homotopy) is, as MattE very clearly explained, that the word problem for groups can be embedded in the question of classifying manifolds. Indeed, suppose that you want to decide if a finitely presented group $G$ is trivial. First, there are several constructions that yield a compact 4-manifold whose fundamental group is isomorphic to $G$; choose your favorite construction and let $M$ be the resulting manifold. Observe that the intersection pairing in homology is computable, so we may just as well assume that we know what this is for our manifold $M$. It turns out that the homotopy type of a simply connected manifold is completely determined by the intersection form on the second integral homology group (and that any intersection form can be realised). So if on the side we construct the simply connected manifold $N$ with intersection form coinciding with the one of $M$, then deciding whether the homotopy types of $M$ and $N$ coincide will decide if the fundamental group of $M$ is trivial or not, i.e. if $G$ is the trivial group or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some issues on how you compute the intersection pairing and how you find the manifold $N$ given the intersection form, and there are tricks to simplify all these questions, but I will not go into this, in the hope that what I wrote will suffice! Of course, if you want more details, feel free to ask, and I will try to answer!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-10T18:08:08.363" CommentCount="2" />
  <row Id="2081" PostTypeId="2" ParentId="2079" CreationDate="2010-08-10T18:24:44.647" Score="20" ViewCount="0" Body="&lt;p&gt;Any such isomorphism would have to extend to a continuous morphism from the additive group of the real numbers to the multiplicative group of the positive real numbers because the order-preserving condition means convergent sequences get sent to convergent sequences.  Any such morphism is $a^x$ for some positive real $a$, and this is impossible e.g. by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfond-Schneider theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-10T18:24:44.647" CommentCount="2" />
  <row Id="2082" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T18:30:37.350" Score="1" ViewCount="0" Body="&lt;p&gt;I believe the answer is related to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that, by choosing random numbers, you're approximating a sampling from that series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your amusement, here's a little Python program that does a brute force calculation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;import random&#xA;iterations = 100000000&#xA;count = 0.0&#xA;for i in range(iterations):&#xA;    sum = 0&#xA;    while sum &amp;lt; 1:&#xA;            sum += random.random()&#xA;            count += 1&#xA;print count / iterations&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="208" LastActivityDate="2010-08-10T18:30:37.350" CommentCount="4" />
  <row Id="2083" PostTypeId="2" ParentId="2045" CreationDate="2010-08-10T18:59:50.740" Score="1" ViewCount="0" Body="&lt;p&gt;Have you tried reducing the matrix to row echelon form? Maybe my answer to this question:&#xA; &lt;a href=&quot;http://math.stackexchange.com/questions/1822/is-reducing-a-matrix-to-row-echelon-form-useful-at-all/1831#1831&quot;&gt;http://math.stackexchange.com/questions/1822/is-reducing-a-matrix-to-row-echelon-form-useful-at-all/1831#1831&lt;/a&gt; could be helpful.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-10T18:59:50.740" CommentCount="6" />
  <row Id="2084" PostTypeId="2" ParentId="2071" CreationDate="2010-08-10T19:21:35.857" Score="4" ViewCount="0" Body="&lt;p&gt;The other answers should have resolved why the number is &gt;2, but for the exact reason why it is &lt;em&gt;e&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For having N random numbers whose sum is &amp;lt;1, the probability is exactly the same as the volume of a unit &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplex&quot; rel=&quot;nofollow&quot;&gt;standard N-simplex&lt;/a&gt;, i.e. 1/N!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the probability that it takes exactly N random numbers for a sum of ≥1, the probability is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac1{(N-1)!} - \frac1{N!} = \frac1{N(N-2)!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the expected number is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{N=1}^\infty \frac N{N(N-2)!} = e $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-10T19:35:54.823" LastActivityDate="2010-08-10T19:35:54.823" CommentCount="3" />
  <row Id="2085" PostTypeId="1" CreationDate="2010-08-10T20:22:17.710" Score="0" ViewCount="293" Body="&lt;p&gt;Please explain Ito's Lemma in layman's terms if possible and how this relates to Taylor series&lt;/p&gt;&#xA;" OwnerUserId="1032" LastActivityDate="2010-08-10T20:22:17.710" ClosedDate="2010-08-11T00:34:47.973" Title="What is Ito's Lemma?" Tags="&lt;stochastic-processes&gt;" AnswerCount="0" CommentCount="6" />
  <row Id="2086" PostTypeId="2" ParentId="1984" CreationDate="2010-08-10T20:33:38.477" Score="3" ViewCount="0" Body="&lt;p&gt;No, but reputation counts for a lot and not having an academic track record means that you are missing many of the easy credibility markers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, when I was working towards my PhD back in the mid 1990s my supervisor made an interesting observation that some of &lt;em&gt;his&lt;/em&gt; best work in fact had been the hardest to get published.  He felt that really innovative, ground breaking ideas were often the most difficult to get accepted and were subjected to far greater scrutiny than normal.&lt;/p&gt;&#xA;" OwnerUserId="1032" LastActivityDate="2010-08-10T20:33:38.477" />
  <row Id="2087" PostTypeId="2" ParentId="2001" CreationDate="2010-08-10T20:42:55.163" Score="2" ViewCount="0" Body="&lt;p&gt;In infinite dimensions, all hell breaks loose. For example, neither the category of Banach spaces nor the category of Hilbert spaces, although additive, are abelian.&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-10T20:42:55.163" />
  <row Id="2088" PostTypeId="1" CreationDate="2010-08-10T20:57:01.197" Score="-3" ViewCount="119" Body="&lt;p&gt;my techer now gave us harderproblems man. :(((&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can u solve dis problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;26*8y = 4x - 27/8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;u solved before on other qestion but this is a lot different from b4 yeah? thx bros :)))&lt;/p&gt;&#xA;" OwnerUserId="1001" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-10T21:17:03.350" LastActivityDate="2010-08-10T21:51:09.850" ClosedDate="2010-08-10T21:52:00.257" Title="how to solve division on variabels and other hard stuff? :(" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2089" PostTypeId="2" ParentId="2088" CreationDate="2010-08-10T21:02:36.597" Score="-3" ViewCount="0" Body="&lt;p&gt;plz help!!! i have to finish this rly fastor i will fail :((((&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2010-08-10T21:02:36.597" CommentCount="3" />
  <row Id="2091" PostTypeId="2" ParentId="1890" CreationDate="2010-08-10T21:13:45.500" Score="1" ViewCount="0" Body="&lt;p&gt;Comment posted as an answer for lack of reputation: I think it would be better to assume explicitly $2x-f(x)\in[0,1]$ for all $x\in[0,1]$, and I believe that the title of the question is not very well chosen. &lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-08-10T21:13:45.500" CommentCount="5" />
  <row Id="2092" PostTypeId="2" ParentId="2088" CreationDate="2010-08-10T21:16:19.713" Score="1" ViewCount="0" Body="&lt;p&gt;Well, the problem seems to be same as the last one, just with numbers switched. I'd get right side under same denominator. After that you can get rid of denominator by multiplying both sides of equation with it.  This way you'll get exactly same problem as last time. After that you can use division to express what just y is.  You'll get something like a*y=b*x-c, where a,b,c are some numbers. Surely you can see what you need to use to get y?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and if on your last question, someone solved the problem, click on the tick mark next to the question.&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-08-10T21:37:27.120" LastActivityDate="2010-08-10T21:37:27.120" CommentCount="3" />
  <row Id="2093" PostTypeId="1" CreationDate="2010-08-10T21:32:53.563" Score="5" ViewCount="191" Body="&lt;p&gt;How can I determine the size of the largest collection of k-element subsets of an n-element set such that each pair of subsets has at most m elements in common?&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="984" LastEditorDisplayName="" LastEditDate="2010-08-10T21:57:37.837" LastActivityDate="2010-10-12T07:19:33.643" Title="Choosing subsets of a set with a specified amount of maximum overlap between them" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="2094" PostTypeId="2" ParentId="2013" CreationDate="2010-08-10T21:35:57.213" Score="1" ViewCount="0" Body="&lt;p&gt;See Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;Functions of random variables&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-10T21:35:57.213" CommentCount="1" />
  <row Id="2095" PostTypeId="2" ParentId="206" CreationDate="2010-08-10T21:44:15.657" Score="0" ViewCount="0" Body="&lt;p&gt;In practical application of mathematical analysis, the most useful &quot;well-behavedness&quot; condition on a function would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Lipschitz_continuity&quot; rel=&quot;nofollow&quot;&gt;Lipshitz continuity&lt;/a&gt;. This ensures that the variations of a function are not too wild. The most important consequence is that differential equations will have a unique solution. All sorts of modelling of natural and physical situations uses differential equations, and it is valuable to know abstractly that they would have solutions under some tame conditions. Given such a theorem, one can concentrate exclusively on finding the solutions.&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T21:44:15.657" />
  <row Id="2096" PostTypeId="2" ParentId="227" CreationDate="2010-08-10T21:50:30.897" Score="0" ViewCount="0" Body="&lt;p&gt;You should certainly know ordinary differential equations, partial differential equations, probability theory and statistics. &lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-08-10T21:50:30.897" CommunityOwnedDate="2010-08-12T14:58:43.903" />
  <row Id="2098" PostTypeId="2" ParentId="1096" CreationDate="2010-08-10T22:53:09.750" Score="0" ViewCount="0" Body="&lt;p&gt;Lets define what is &lt;code&gt;x+x+x+... x times for x - real&lt;/code&gt;. Natural definition is &lt;code&gt;x+x+x.. := x*x&lt;/code&gt; (note - just the same as Isaac has wrote in his edit).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we want left our initial definition as is. We don't know what is &lt;code&gt;x+x+.. repeat x times for x - real&lt;/code&gt; (and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. &lt;code&gt;f(x):=x+x+x.. repeat x times&lt;/code&gt;, &lt;code&gt;Df(x)=(f(x+h)-f(x))/h, h-&amp;gt;0&lt;/code&gt;. &lt;code&gt;Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h-&amp;gt;0&lt;/code&gt;. Suppose &lt;code&gt;x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times)&lt;/code&gt; we have &lt;code&gt;Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h-&amp;gt;0&lt;/code&gt;, &lt;code&gt;Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h-&amp;gt;0&lt;/code&gt;, or &lt;code&gt;Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h-&amp;gt;0&lt;/code&gt; and at last &lt;code&gt;Df(x)=x + x+h, h-&amp;gt;0 = 2x&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1035" LastActivityDate="2010-08-10T22:53:09.750" />
  <row Id="2099" PostTypeId="1" CreationDate="2010-08-10T23:08:39.053" Score="4" ViewCount="173" Body="&lt;p&gt;From:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moduli_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Moduli_space&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that moduli of curves is a very algebro-geometric topic. It is easy to understand its relevance and importance in algebraic geometry. But the mind boggles when we try to imagine how on earth such a topic from pure and abstruse mathematics is relevant in physics. I will be thankful if somebody can give some explanation.&lt;/p&gt;&#xA;" OwnerUserId="1036" LastActivityDate="2010-08-11T14:15:58.043" Title="How is moduli of curves relevant in physics?" Tags="&lt;algebraic-geometry&gt;&lt;physics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="2100" PostTypeId="1" CreationDate="2010-08-10T23:31:38.860" Score="1" ViewCount="218" Body="&lt;p&gt;I am trying to brush up my graph theory skills. I have not done any in over 4 years and i am rusty...If someone could help me out with this simple proof i would appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that for any graph $G$ of order at least 2, the degree sequence has at least one pair of repeated entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the degree sequence if a list of the degrees of each vertex (usually written in descending order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the sum of the degrees of the vertices of a graph is equal to $2|E|$ and that the number of vertices of odd degree is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could help me out and point me in the right direction i would appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-08-11T22:04:10.543" Title="Degree Sequence of a Graph" Tags="&lt;graph-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2101" PostTypeId="2" ParentId="2099" CreationDate="2010-08-10T23:39:27.287" Score="6" ViewCount="0" Body="&lt;p&gt;In string theory, a particle is actually a string.  In closed string theory, it is a closed string, i.e. a loop.  As time progresses, this loop will trace out a surface (its &lt;em&gt;worldsheet&lt;/em&gt;).  If you follow the particle from its creation to its destruction, this surface will be closed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For certain computations of physical quantities attached to the worldsheet (and perhaps at this point someone with more expertise can add details) one has to choose a conformal structure on the worldsheet; but to get an answer independent of the choice, one then integrates over all possible conformal structures; or sometimes one can show that the quantity computed is independent of the conformal structure (perhaps because it is computed by a horizontal section of a bundle with connection over the space of conformal structures). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we note that the space of all conformal structures on a closed topological surface of genus $g$ is the same as the moduli space of algebraic curves of genus $g$.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-10T23:39:27.287" />
  <row Id="2102" PostTypeId="2" ParentId="2100" CreationDate="2010-08-10T23:50:24.413" Score="2" ViewCount="0" Body="&lt;p&gt;If there were no repeats, what would the degree sequence look like?  Why could it not look like that?&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:04:10.543" LastActivityDate="2010-08-11T22:04:10.543" CommentCount="4" />
  <row Id="2103" PostTypeId="2" ParentId="2045" CreationDate="2010-08-11T00:11:33.450" Score="2" ViewCount="0" Body="&lt;p&gt;If you have a $n\times n$ matrix $A$ , the inverse matrix $A^{-1}$ can be&#xA;computed by using a &lt;em&gt;compact&lt;/em&gt; method, that consists of&#xA;the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Augment your matrix with the identity matrix $I$: $\left( A|I\right) $&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_elimination&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Gaussian elimination&lt;/em&gt;&lt;/a&gt; to get an upper triangular matrix on the left&#xA;and a matrix $J$ on the right: $\left( U|J\right) $, where $J=\left(&#xA;J_{1}|J_{2}|\ldots |J_{n}\right) $. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To get the $n$ vector columns of the inverse matrix: $A^{-1}=\left(&#xA;x_{1}|x_{2}|\ldots |x_{n}\right) $, solve for $x_{k}$, with $k=1,2,\ldots ,n$, $n$ systems of equations $U\cdot x_{k}=J_{k}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note 1: in your problem $n=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: there is a variant of this method called &quot;Gaussian elimination with&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_elimination&quot; rel=&quot;nofollow&quot;&gt;partial pivoting&lt;/a&gt;&quot; that gives a better accuracy, when the entries are not&#xA;rational numbers, which is not the present situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now I have not yet computed matrix $A^{-1}$ in this case. Instead I have used&#xA;The Scientific Notebook (included in the Scientific Work Place) to find that your matrix satisfies $A=A^{-1}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=\begin{pmatrix}1/2&amp;amp;1/2&amp;amp;1/2&amp;amp;1/2 \\ 1/2&amp;amp;1/2&amp;amp;-1/2&amp;amp;-1/2\\1/2&amp;amp;-1/2&amp;amp;1/2&amp;amp;-1/2\\1/2&amp;amp;-1/2&amp;amp;-1/2&amp;amp;1/2\end{pmatrix}=A^{-1}$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-11T00:42:38.457" LastActivityDate="2010-08-11T00:42:38.457" CommentCount="4" />
  <row Id="2104" PostTypeId="2" ParentId="153" CreationDate="2010-08-11T00:51:11.700" Score="11" ViewCount="0" Body="&lt;p&gt;As someone who uses both Matlab and Mathematica regularly, I'd like to give some strengths of Matlab over Mathematica (since Mehper C. Palavuzlar gave such a good overview of Mathematica's strengths over Matlab)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You can make Matlab scripts fully stand-alone. Yes, it will be a 500 MB executable for a program that prints &quot;Hello World&quot;, but it can be run by someone who doesn't have Matlab. This also makes it possible to run your calculations in parallel on a server farm like Amazon's EC2 (is it even possible to use EC2 for Mathematica's computations?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Matlab has a much larger and more active community. There are more Matlab blogs out there to browse for tips and you can get answers to questions on Matlab Central within 10 minutes sometimes. Mathematica has a moderated newsgroup which is high quality, but slow -- because of moderation it typically takes one day for your question to be seen by other people&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bigger community means more ready-to-use code out there. For instance, I started using Matlab for some visual recognition tasks, and I could get pretty much all the pieces for segmentation, low level feature extraction, adaptive thresholding etc from the web, whereas in Mathematica I would have to implement them from scratch. It's common for researchers to publish Matlab code they used for their computations on their homepages, whereas I'm yet to see anyone (at least in vision/statistics/optimization) publishing Mathematica code to go with along with their papers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Matlab's language is easier to learn (and to debug!). Mathematica offers an &quot;rewriting system&quot; paradigm unfamiliar to many, and even after years of programming in Mathematica I occasionally run into unexpected behaviors due to some intricacy of the evaluation sequence&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Overall, I'd recommend to go with Mathematica if you want to play around by yourself, and Matlab if you want to share your code or use code of others.&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-08-11T21:01:40.567" LastActivityDate="2010-08-11T21:01:40.567" />
  <row Id="2105" PostTypeId="2" ParentId="2100" CreationDate="2010-08-11T01:43:59.523" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, it just came to me after i posted the comment to the previous answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graph with at least 2 vertices and no edges the 0 degree is repeated.&#xA;A graph with 2 vertices and 1 edge the degree 1 is repeated.&#xA;Each vertex has at least one entry in the degree sequence, so there are a total of $N$ entries. But each degree can have a maximum of degree = $N-1$. Therefore according to the pigeonhole principle at least one of the degrees has to repeat.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditorDisplayName="" LastEditDate="2010-08-11T01:57:23.813" LastActivityDate="2010-08-11T01:57:23.813" CommentCount="2" />
  <row Id="2107" PostTypeId="2" ParentId="2099" CreationDate="2010-08-11T03:27:04.563" Score="4" ViewCount="0" Body="&lt;p&gt;In quantum field theory many quantities are calculated as (formal) Feynman path integrals, that is, integrals &quot;over the space of all paths&quot;.  To make sense of this in dimensions higher than 2, one uses a perturbation expansion (in powers of Planck's constant) of the integral which leads to a sequence of finite dimensional integrals described by Feynman diagrams: finite graphs with (among other decorations) some &quot;input&quot; and &quot;output&quot; vertices --- think of a process where some particles collide and some are emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string theory instead of a sum over paths of a point there is a sum over trajectories of a closed loop (a string), i.e., integration over some space of surfaces.  The analogue of the finite-dimensional perturbative Feynman diagrams are surfaces connecting some oriented loops. Loops with positive orientation relative to the surface are the &quot;inputs&quot; and the others are the &quot;outputs&quot;.  The surface can also have empty input or output.  Integration over all surfaces of this type is still an infinite-dimensional problem.  To reduce to a finite-dimensional integral, one integrates over the moduli space of conformal structures on each topological type of bordered surface.  This is finite-dimensional and there are standard measures on it (Weil-Petersson).  For problems with conformal invariance (&quot;conformal field theory&quot;) this recipe was given by Polyakov and is basic for string theory.   As far as I understand it, the worldsheet is always governed by a conformal field theory, so the Polyakov recipe is &quot;the&quot; method for defining the observables in perturbative string theory.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for open strings one might want the boundaries to be open circles, or punctures, thus cutting out closed discs or points from the surface.   So the general moduli space is that of  conformal structures on surfaces of a fixed finite genus with a given finite type of boundary and one can at least define a measure on this space.  Rigorous calculations can be extremely complicated as seen in the arxiv papers by D'Hoker and Phong where they went through a huge tour-de-force to construct the perturbative superstring measure through 2 and 3 loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Matt explained, the relation to algebraic geometry is that (at least for compact surfaces without boundary) the moduli space of conformal structures is the same as that of algebraic structures; in each conformal class there is an algebraic curve of that genus.  I am less sure of this in the non-compact case of punctured surfaces, and for bordered Riemann surfaces where the boundary contains loops, the moduli space is of geometric/symplectic/complex-analytic nature and I don't know if it has an algebraic analogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: I think that there is a basic difference between QFT and string theory in this analogy, because in string theory only the perturbative measure is known at present, while in QFT path-space integrals are thought of as an underlying non-perturbative theory and one can make sense of that to a large extent even if Feynman's &quot;measure&quot; on path space doesn't exist as an integration measure in the ordinary mathematical sense.   In string theory the non-perturbative definition of the theory is presently unknown.  There are objects in string theory that are considered non-perturbative and thus hints of the underlying theory, but even conjecturally or heuristically there is not a consensus as to what perturbative string theory is a perturbation of.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added: &#xA;According to this paper, for bosonic string theory in Polyakov's prescription, the moduli spaces in the perturbative integrals are those of smooth closed surfaces, without punctures or boundary loops.  This is interesting because conformal field theory does use the more complicated surfaces, but I don't know the details of how CFT is used in string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1104116138&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1104116138&lt;/a&gt;   )&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-11T14:15:58.043" LastActivityDate="2010-08-11T14:15:58.043" />
  <row Id="2108" PostTypeId="2" ParentId="1096" CreationDate="2010-08-11T04:24:21.853" Score="4" ViewCount="0" Body="&lt;p&gt;I think the discrete/continuous issue is sort of a red herring.  To me, the problem is forgetting to use the chain rule!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + \dots + u$, $v$ times.  Then $x^2 = F(x,x)$.  Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true.  In the fallacious example, the problem is essentially that the $F_v$ term has been omitted.  In some sense, one has forgotten to differentiate the operation &quot;$x$ times&quot; with respect to $x$!  Of course, the notation makes this easier to do.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-11T04:24:21.853" CommentCount="1" />
  <row Id="2110" PostTypeId="1" CreationDate="2010-08-11T06:29:51.057" Score="2" ViewCount="136" Body="&lt;p&gt;Here is a time sample: $Q = \{(t_i, x_i) | 0 \leq x_i \leq x_{i+1}, 1 \leq i \leq n\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $T_1 \leq t_{i+1} - t_i &amp;lt; T_2$ where  $T_1, T_2 &gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $x_i$ comes with error:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i = \lfloor x(t_i) \rfloor$ where $\lfloor x \rfloor$ - whole part of $x$, $x(t_i)$ - voluntary unknown law of object motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How we can find (approximately but sufficiently accurate) speed of the object at the given time?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="1041" LastEditorDisplayName="" LastEditDate="2010-11-21T21:00:22.003" LastActivityDate="2010-11-21T21:00:22.003" Title="Determine speed of the object at the current time by the non-uniform time sample" Tags="&lt;sequences-and-series&gt;&lt;statistics&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="2111" PostTypeId="2" ParentId="2029" CreationDate="2010-08-11T06:40:59.707" Score="0" ViewCount="0" Body="&lt;p&gt;There is infinite may constrains on digits of the prime number. Each one may be deduced as follows ( from Euclid method): suppose You have number $S =  \sum 10^i *a_i$ which is divisible by $p$. Then You have $S= ( \sum 10^j * b_j )*p = \sum 10^j b_j*p $ . You may perform multiplication and look for additional constrains. So as You have criteria for divisibility by 2, 4, or 5, which gives You constraints on last digits, You may construct similar criteria for divisibility by 13 etc.  as well. It is possible, that some of them gives You constrains of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for S&gt;p last digit has to be in a set M, where M is proper subset of {1,3,7}&lt;/p&gt;&#xA;" OwnerUserId="1043" LastActivityDate="2010-08-11T06:40:59.707" />
  <row Id="2112" PostTypeId="1" AcceptedAnswerId="2150" CreationDate="2010-08-11T06:55:31.287" Score="3" ViewCount="608" Body="&lt;p&gt;The reason I ask this is because difficult mathematical proofs are just not plain self-evident. You would need a few years of intensive study before you can get to the point of understanding the topics and the proofs. The problem is that for a difficult math theorem, maybe only a handful of mathematicians have studied and agreed on the validity of the proof. There is a chance, given that the topic is highly esoteric, that there are bugs in the proof that render it invalid. We just don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would give us a conundrum; we thought that mathematics is the most secure branch of knowledge, but it isn't; because a lot of the proofs are convoluted and esoteric to the point that only a handful a people who can understand them. So we can never be 100% sure that a proof is always valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-11-05T11:26:06.570" CommunityOwnedDate="2010-08-12T13:15:22.263" Title="Is It True that We Can Never Be Sure of Validity of a Mathematical Proof?" Tags="&lt;meta-math&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="2" />
  <row Id="2113" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T06:59:51.727" Score="0" ViewCount="0" Body="&lt;p&gt;You'll be sure when you'll understand it.&lt;/p&gt;&#xA;" OwnerUserId="480" LastActivityDate="2010-08-11T06:59:51.727" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="3" />
  <row Id="2114" PostTypeId="2" ParentId="2093" CreationDate="2010-08-11T07:16:03.263" Score="0" ViewCount="0" Body="&lt;p&gt;Hi Please refer this link: &lt;a href=&quot;http://gilkalai.wordpress.com/2008/10/06/extremal-combinatorics-iv-shifting/&quot; rel=&quot;nofollow&quot;&gt;http://gilkalai.wordpress.com/2008/10/06/extremal-combinatorics-iv-shifting/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also look here as well to get more idea for this type of problems: &lt;a href=&quot;http://mathoverflow.net/questions/25067/given-n-k-element-subsets-of-n-is-there-a-small-subset-a-of-n-which-intersects-t&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/25067/given-n-k-element-subsets-of-n-is-there-a-small-subset-a-of-n-which-intersects-t&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-11T07:16:03.263" />
  <row Id="2115" PostTypeId="1" AcceptedAnswerId="2170" CreationDate="2010-08-11T07:32:20.277" Score="9" ViewCount="578" Body="&lt;p&gt;We know that there exists real functions which are continuous at each irrational and dis-&#xA;continuous at each rational number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does there exists functions $f: \mathbb{R} \to \mathbb{R}$ which is differentiable at every irrational and discontinuous at every rational.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-14T22:51:25.323" Title="Functions discontinuous at $\mathbb{Q}$" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="6" />
  <row Id="2116" PostTypeId="1" AcceptedAnswerId="2177" CreationDate="2010-08-11T07:41:57.820" Score="0" ViewCount="194" Body="&lt;p&gt;For any natural number $n &gt; 1$, define $E(n)$,to be the highest exponent to&#xA;which a prime divides it. For instance, $E(12)=E(36)=2$. Show that $$\lim_{N \to \infty} \frac{1}{N} \sum\limits_{n=2}^{N} E(n)$$ exists and find its value &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-11T15:13:30.920" LastActivityDate="2010-08-11T22:02:04.303" Title="Mean of highest exponent in prime factorization of all integers ≥ 2" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2120" PostTypeId="1" AcceptedAnswerId="2122" CreationDate="2010-08-11T08:30:49.793" Score="7" ViewCount="2970" Body="&lt;p&gt;We know that $\displaystyle\zeta(2)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and it converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now does there exists a bijective map $f:\mathbb{N} \to \mathbb{N}$ such that the sum $$\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^2}$$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our $s=2$ was not fixed, then can we have a function such that $\displaystyle \zeta(s)=\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^s}$ converges&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-07T18:44:22.227" LastActivityDate="2010-11-09T18:25:33.667" Title="Does there exist a bijective $f:\mathbb{N} \to \mathbb{N}$ such that $\sum f(n)/n^2$ converges?" Tags="&lt;number-theory&gt;&lt;convergence&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="2121" PostTypeId="1" AcceptedAnswerId="2153" CreationDate="2010-08-11T08:41:14.963" Score="7" ViewCount="344" Body="&lt;p&gt;Does there exists a finite group $G$ and a normal subgroup $H$ of $G$ such that $|Aut(H)|&gt;|Aut(G)|$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-29T02:05:01.930" Title="Existence of a normal subgroup with $|Aut(H)|&gt;|Aut(G)|$" Tags="&lt;group-theory&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="2122" PostTypeId="2" ParentId="2120" CreationDate="2010-08-11T08:41:34.213" Score="27" ViewCount="0" Body="&lt;p&gt;For $s&gt;2$ you can take $f(n)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $s=2$ if you have $m &amp;lt; n$ and $a=f(m) &gt; b=f(n)$ then&#xA;$$\frac{a}{m^2}+\frac{b}{n^2}&gt;\frac{b}{m^2}+\frac{a}{n^2}$$&#xA;(proved either naively or as a case of the &quot;rearrangement inequality&quot;)&#xA;so the sum $\sum f(n)/n^2$ can be reduced by swapping the values $f(m)$ and $f(n)$.&#xA;Hence&#xA;$$\sum\frac{f(n)}{n^2}\ge\sum\frac{n}{n^2}$$&#xA;which is divergent.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T08:41:34.213" />
  <row Id="2123" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T08:48:44.993" Score="5" ViewCount="0" Body="&lt;p&gt;An elementary Abelian group can have a large automorphism group.&#xA;Let $H$ be a largish elementary Abelian group, and $G$&#xA;be an extension of $H$, say be a semidirect product with&#xA;a cyclic group acting faithfully on $H$. Then it is possible&#xA;for Aut$(G)$ to be smaller than Aut$(H)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T08:48:44.993" CommentCount="4" />
  <row Id="2124" PostTypeId="2" ParentId="2120" CreationDate="2010-08-11T08:50:44.927" Score="8" ViewCount="0" Body="&lt;p&gt;For $s=2$ the answer is negative. This series doesn't converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this we can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel_transformation&quot; rel=&quot;nofollow&quot;&gt;Abel transformation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_{n=1}^{n=N} \frac{f(n)}{n^2} = \sum_{n=1}^{n=N} (\sum_{k=1}^{k=n} f(k)) (\frac{1}{n^2} - \frac{1}{(n + 1)^2}) + (\sum_{n=1}^{N} \frac{1}{n^2})\frac{1}{(N+1)^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is bijection, $\sum_{k=1}^{k=n} f(k) \ge \frac{n^2}{2}$ hence the first sum is greater than $\sum_{n=1}^{N}\frac{c}{n}$ for some $c &gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-11T08:50:44.927" />
  <row Id="2125" PostTypeId="1" AcceptedAnswerId="2127" CreationDate="2010-08-11T09:05:06.373" Score="7" ViewCount="160" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;: Let $n$ be an integer and $p$ a prime dividing $5(n^2-n+\frac{3}{2})^2-\frac{1}{4}$. Prove that $p \equiv 1 \pmod{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomial can be re-written as $(\sqrt{5}(n^2-n+\frac{3}{2})-\frac{1}{2})(\sqrt{5}(n^2-n+\frac{3}{2})+\frac{1}{2})$. If this vanishes mod $p$ then $5$ is a quadratic residue mod $p$, which shows that $p \equiv \pm 1 \pmod{5}$ (the primes 2 and 5 are easily ruled out). It feels like the problem should be solvable by understanding the splitting of primes in the splitting field of this polynomial, but I can't find an appropriate &quot;reciprocity law&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The things I'm not sure about are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How does one rule out the primes congruent to -1 mod 5?&lt;/li&gt;&#xA;&lt;li&gt;Under what circumstances is it the case that the set {rational primes that split in the ring of integers of some number field} is the union of arithmetic progressions? This a kind of generalized reciprocity law but I don't know in what generality they are known to hold.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-11T16:52:12.510" Title="Primes dividing the values of integer polynomials" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="2126" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T09:14:31.163" Score="7" ViewCount="0" Body="&lt;p&gt;The quasidihedral group of order 16, QD16, has automorphism group of order 16, but it has Q8 (the quaternion group) as a normal subgroup, and | Aut(Q8) | = 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-11T09:14:31.163" />
  <row Id="2127" PostTypeId="2" ParentId="2125" CreationDate="2010-08-11T09:16:17.280" Score="5" ViewCount="0" Body="&lt;p&gt;The only way that this situation could possibly arise&#xA;is if the splitting field of $f(x)=5(x^2-x+3/2)^2-1/4$&#xA;is the cyclotomic field generated by the fifth roots of unity.&#xA;To show this, it helps to get your hands dirty and actually&#xA;solve the equation $f(x)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, number fields are determined up to isomorphism&#xA;by their splitting laws. The splitting law will be determined by a&#xA;congruence condition iff the extension is an Abelian extension of&#xA;$\mathbb{Q}$ (by class field theory).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T09:16:17.280" CommentCount="5" />
  <row Id="2128" PostTypeId="2" ParentId="2120" CreationDate="2010-08-11T10:32:00.607" Score="-1" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  I got a new solution now.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be a permutation of $\mathbb{N}$ and let $N \in \mathbb{N}$. We shall claim that $$\sum\limits_{N+1}^{3N} \frac{f(n)}{n^2} &gt; \frac{1}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact of the $2N$ numbers $f(N+1), \cdots, f(3N)$ only $N$ can be $\leq N$ so that atleast $N$ of them are $&gt;N$ Hence $$\sum\limits_{N+1}^{3N} \frac{f(n)}{n^{2}} \geq \frac{1}{3N^{2}} \sum\limits_{N+1}^{3N} f(n) &gt; \frac{1}{9N^{2}} \times N.N =\frac{1}{9}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-11T10:32:00.607" CommentCount="4" />
  <row Id="2129" PostTypeId="2" ParentId="392" CreationDate="2010-08-11T10:36:44.690" Score="0" ViewCount="0" Body="&lt;p&gt;In the spirit of the question, this answer addresses the remark: &quot;Or, even more amazingly, the right triangle definition alone&quot;. In essence the same &lt;a href=&quot;http://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sinx-and-cosx/871#871&quot;&gt;answer as David Lewis'&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically &lt;code&gt;d(sinθ)/dθ&lt;/code&gt; can be derived in a right triangle by enlarging the right triangle from &lt;code&gt;θ-&amp;gt;θ+dθ&lt;/code&gt; while keeping &lt;code&gt;a:=adj&lt;/code&gt; and the right angle fixed. In first order &lt;code&gt;d(sinθ)=(o+do)/(h+dh)-o/h≈do/h&lt;/code&gt;, where &lt;code&gt;o:=opp&lt;/code&gt;, &lt;code&gt;h:=hyp&lt;/code&gt;.  The small part of the circle with radius &lt;code&gt;h&lt;/code&gt; that defines &lt;code&gt;dθ&lt;/code&gt; is, again in first order, equal to the opposite side of a triangle with the perpendicular projection of h on h+dh, so that &#xA;&lt;code&gt;dθ=do┴/h&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that the derivative of &lt;code&gt;sinθ&lt;/code&gt; equals the proportion between &lt;code&gt;do&lt;/code&gt; and &lt;code&gt;do┴&lt;/code&gt;.&#xA;One can immediately see that this proportion equals &lt;code&gt;sin(π/2-θ)=cos(θ)&lt;/code&gt; in the small triangle in the upper right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &quot;the proportion between &lt;code&gt;do&lt;/code&gt; and &lt;code&gt;do┴&lt;/code&gt; equals the proportion of the two adjoining sides of the angle &lt;code&gt;θ&lt;/code&gt;&quot; would be the intuitive, geometrical meaning of &lt;code&gt;sin'(θ)=cos(θ)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="706" LastActivityDate="2010-08-11T10:36:44.690" />
  <row Id="2130" PostTypeId="1" CreationDate="2010-08-11T11:01:04.907" Score="2" ViewCount="223" Body="&lt;p&gt;Sometimes one encounter the requirement that the objects of a category needs to be a set. What if it was not, could you provide examples of what could go wrong? (One example per answer).&lt;/p&gt;&#xA;" OwnerUserId="1013" LastEditorUserId="1013" LastEditorDisplayName="" LastEditDate="2010-08-11T15:02:30.550" LastActivityDate="2010-11-20T02:51:11.153" Title="What's so special with small categories?" Tags="&lt;big-list&gt;&lt;set-theory&gt;&lt;category-theory&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="2" />
  <row Id="2131" PostTypeId="1" CreationDate="2010-08-11T11:06:30.357" Score="3" ViewCount="143" Body="&lt;p&gt;I have a data set &lt;code&gt;2 3 3 4 4 4 4&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the number of unique numbers of 3 digit numbers that can be formed using this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of doing &lt;code&gt;7 P 3 /(4!*2!)&lt;/code&gt;, but this doesn't seem right.&lt;/p&gt;&#xA;" OwnerUserId="1048" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-11T13:11:36.807" LastActivityDate="2010-08-11T15:07:35.173" Title="Permutations with duplicates" Tags="&lt;combinatorics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="2132" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T11:07:33.227" Score="3" ViewCount="0" Body="&lt;p&gt;I guess the point that mathematicians are trying to make is that a theorem is &lt;em&gt;always&lt;/em&gt; valid.  However, there may be human flaws in the process, which end up labelling non-theorems as theorems.  A scientific model, on the other hand, could be flawed in some (possibly untestable) instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, I'm 100% convinced, for example, that Euclid's proof of an infinite number of primes is correct.  But for more complicated proofs, errors are reduced by publishing (and in particular, peer review).  Some theorems (particularly important ones) have multiple published proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Computer proofs are just as invalid on the philosophical level since humans make computers, etc.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-11T11:07:33.227" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="2" />
  <row Id="2133" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T11:22:30.067" Score="1" ViewCount="0" Body="&lt;p&gt;If you start off with the 7P3 ways of choosing three distinct ordered elements from the multiset {2,3,3,4,4,4,4} then you will need to compensate for the multiply-counted 3-digit numbers in different ways depending on the structure of the number itself.  Consequently, you're likely to be stuck using an almost brute-force attack on the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you could generate all the unordered 3-digit numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;233, 234, 244, 334, 344, 444&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then consider one-by-one how many ordered 3-digit numbers have those particular digits.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3!/2! + 3! + 3!/2! + 3!/2! + 3!/2! + 3!/3!&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Once you're done, you could check your result using &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt; by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Arrangements([2,3,3,4,4,4,4],3);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-11T11:22:30.067" />
  <row Id="2134" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T11:27:33.710" Score="3" ViewCount="0" Body="&lt;p&gt;If you have &lt;code&gt;2&lt;/code&gt; as the first digit, you can have &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;4&lt;/code&gt; as the second and third digits, so this gives &lt;code&gt;233 234 243 244&lt;/code&gt;, &lt;strong&gt;four&lt;/strong&gt; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;code&gt;3&lt;/code&gt; as the first digit, you can have &lt;code&gt;2&lt;/code&gt; and &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;4&lt;/code&gt; as second and third digits (but &lt;code&gt;2&lt;/code&gt; only once), giving &lt;code&gt;323 332 324 342 334 343 344&lt;/code&gt;, &lt;strong&gt;seven&lt;/strong&gt; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;code&gt;4&lt;/code&gt; as the first digit, you can have &lt;code&gt;2&lt;/code&gt;, &lt;code&gt;3&lt;/code&gt; and &lt;code&gt;4&lt;/code&gt; as second and third digits, giving &lt;code&gt;424 442 423 432 433 434 443 444&lt;/code&gt;, &lt;strong&gt;eight&lt;/strong&gt; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 + 7 + 8 = 19 unique 3 digit numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather brute force, but it works pretty easily for this problem. &lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-08-11T11:27:33.710" />
  <row Id="2135" PostTypeId="2" ParentId="2093" CreationDate="2010-08-11T11:59:40.373" Score="0" ViewCount="0" Body="&lt;p&gt;In some instances, such a collection of sets would be a t-design (a generalised &lt;a href=&quot;http://en.wikipedia.org/wiki/Block_design&quot; rel=&quot;nofollow&quot;&gt;block design&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-11T11:59:40.373" />
  <row Id="2136" PostTypeId="1" AcceptedAnswerId="2142" CreationDate="2010-08-11T12:07:35.583" Score="3" ViewCount="94" Body="&lt;p&gt;Is possible to construct infinitely differentiable functions that interpolate through arbitrary points, the way polynomial splines do? If so, do they have a name and is there an algorithm for constructing the curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose a infinite degree polynomial spline might qualify but I was hoping there might be something more like a bump function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I definitely am a novice when it comes to this stuff, so please do not hesitate to correct me if I misusing any terminology.&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-08-11T14:26:11.637" Title="Are there smooth analogs to polynomial splines" Tags="&lt;numerical-methods&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="2137" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T12:21:52.233" Score="1" ViewCount="0" Body="&lt;p&gt;Alternatively, start from all 3-digit permutations of {2,3,4}, with repeats. That's $3^3=27$. Then remove numbers that have too many 2's or 3's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take out numbers with exactly 2 2's: 2 choices for the remaining digit and 3 ways to permute the 3 digits is 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take out numbers with exactly 3 2's, of which there's only 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take out numbers with 3 3's, of which there's only 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what's left is 27-6-1-1 = 19&lt;/p&gt;&#xA;" OwnerUserId="541" LastEditorUserId="541" LastEditorDisplayName="" LastEditDate="2010-08-11T14:58:17.253" LastActivityDate="2010-08-11T14:58:17.253" />
  <row Id="2138" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T12:36:29.550" Score="4" ViewCount="0" Body="&lt;p&gt;I think there are two issues floating around here. Very complicated informal proofs (i.e. what most mathematicians write, using ordinary languages and verbal reasoning) may indeed contain hard to spot errors. However, a correct proof could, in theory, be translated into a formal language of set theory, say ZFC. Such a proof would be enormous, and might look a lot like machine language (being a combination of symbolic first order logic and the membership symbol of ZFC set theory).  But every step of reasoning in such a proof is trivial to verify.  Thus, we can in theory check such proofs with computers (or by hand) and be convinced they are correct with very high certainty.  Of course, most proofs are not done this way, but there are projects (e.g. Metamath, Mizer, etc.) to formalize large parts of mathematics.  I think this is what you are getting at with your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue, however, is the more complex question having to do with Godel incompleteness.  In brief, to construct a formal proof, we need a foundation, which is typically ZFC set theory.  But by incompleteness, we cannot know that ZFC is consistent &lt;em&gt;inside of ZFC&lt;/em&gt;!  Most set theorist strongly believe ZFC is consistent, however, as no one has ever exhibited an inconsistency.  So mathematicians tacitly assume that ZFC is consistent, and from that assumption write proofs.  In set theory itself, you often see proofs that read &quot;Assuming ZFC is consistent, X is true.&quot;  But by contraposition, this yields the tantalizing possibility that we might find, via other means, that X is &lt;em&gt;false&lt;/em&gt;, and thus that ZFC is &lt;em&gt;inconsistent&lt;/em&gt;!  This would be an absolutely explosive event in mathematics, of course.  I don't think this is what your question was getting at, but I thought it would be useful to know that there is indeed a &quot;fundamental&quot; incompleteness issue floating around here.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-11T12:36:29.550" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="1" />
  <row Id="2139" PostTypeId="2" ParentId="2136" CreationDate="2010-08-11T12:38:54.977" Score="3" ViewCount="0" Body="&lt;p&gt;Yes - if you have a finite list of points on the plane that pass the vertical line test, you can make a smooth (infinitely differentiable) function that goes through all of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can actually be done somewhat explicitly, using the formulas at [1]. Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(x) = &#xA;\begin{cases}&#xA;0 &amp;amp; x \leq 0 \\&#xA;e^{-1/x}&amp;amp; x &gt; 0 \\&#xA;\end{cases}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be proved that this is a smooth function. Then define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;g(x) = \frac{f(x)}{f(x)+f(1-x)}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $g(x)$ is smooth, defined for all real $x$, and has the property that $g(x) = 0$ for all $x \leq 0$ while $g(x) = 1$ for all $x \geq 1$. By adding together shifted and scaled copies of $g(x)$, you can make a smooth function that goes through any finite list of points that pass the vertical line test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-11T14:26:11.637" LastActivityDate="2010-08-11T14:26:11.637" CommentCount="7" />
  <row Id="2140" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T12:40:42.320" Score="3" ViewCount="0" Body="&lt;p&gt;I have been convinced for years of the following claim: &lt;em&gt;Every compact topological space is sequentially compact.&lt;/em&gt; (The converse is trivially false.) I'm positive I have even been taught this in my topology course as an undergraduate, and I can provide at least two wrong, but very subtly so, &quot;proofs&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is one thing simple enough for me to understand 100% what is going on, still I did not spot the mistake. By the way, I'm not alone in this respect: a long conversation about the issue &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1217&quot; rel=&quot;nofollow&quot;&gt;has been going&lt;/a&gt; on at the NForum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, before this episode I was much more confident in the absolute certainty of mathematical proofs.&lt;/p&gt;&#xA;" OwnerUserId="233" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-05T09:00:55.780" LastActivityDate="2010-11-05T09:00:55.780" CommunityOwnedDate="2010-08-12T13:15:22.263" CommentCount="3" />
  <row Id="2142" PostTypeId="2" ParentId="2136" CreationDate="2010-08-11T12:52:34.993" Score="3" ViewCount="0" Body="&lt;p&gt;For &lt;em&gt;n&lt;/em&gt; data points, there is a unique polynomial of degree &lt;em&gt;n&lt;/em&gt;−1 that passes through the points; one way to get it is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange polynomial&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another class of interpolating functions is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thin_plate_spline&quot; rel=&quot;nofollow&quot;&gt;thin plate spline&lt;/a&gt;, which is typically used for 2D image warping, but can easily be applied to 1D interpolation as well, which is a strictly simpler problem. The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyharmonic_spline&quot; rel=&quot;nofollow&quot;&gt;polyharmonic splines&lt;/a&gt;, of which the thin plate spline is a special case, has examples of other infinitely differentiable splines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that in principle, you can choose any basis function whose translations are linearly independent, and find an interpolating function as a sum of translations and scalings of the basis. The above are essentially examples of particular basis functions people have used because of their other nice properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Sorry, thin plate splines and other examples in the article, apart from the Gaussian, are infinitely differentiable &lt;em&gt;almost&lt;/em&gt; everywhere. However, if you really want infinite differentiability everywhere, you can use any infinitely differentiable function as basis instead, such as the Gaussian, or $x^2/\sqrt{x^2+1}$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-08-11T13:59:12.107" LastActivityDate="2010-08-11T13:59:12.107" />
  <row Id="2145" PostTypeId="1" AcceptedAnswerId="2148" CreationDate="2010-08-11T13:42:30.323" Score="-1" ViewCount="152" Body="&lt;p&gt;if a&amp;lt; b which of the following are always correct&lt;br&gt;&#xA;1) a^3&amp;lt; a^2b&lt;br&gt;&#xA;the answer to this question would be true because there is theorm that &quot;if a&amp;lt; b and we add the same postive number to both sides of the inequalities , this would not effect the inequality&quot;&lt;br&gt;&#xA;Now the second problem is interesting&lt;br&gt;&#xA;2) a^2 &amp;lt; ab&lt;br&gt;&#xA;Now there is also another theorm that says &quot;if a &amp;lt; b and we multiply the same positive number with both sides this would not effect the inequality&quot;&lt;br&gt;&#xA;now as it is specified on the top of the question that a&amp;lt; b , now in this question we are multiplying a on both sides which makes it a^2 &amp;lt; ab so this should be correct. but in the answers at the end of the book the question number of this question is not given. please help.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2010-08-11T15:31:28.653" Title="i am confused with this question whoes answer would be either yes or no" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="2146" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T13:50:24.920" Score="0" ViewCount="0" Body="&lt;p&gt;1) a^3&amp;lt; a^2b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is always true&#xA;because even if 'a' is negative &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^3 will always be less than a^2b (positive) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2, if you have for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = -10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&#xA;a &amp;lt; b = true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^2 = 20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ab = -20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then a^2 &amp;lt; ab becomes false&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so 1 is always true, 2 is not.&lt;/p&gt;&#xA;" OwnerUserId="1053" LastActivityDate="2010-08-11T13:50:24.920" CommentCount="1" />
  <row Id="2148" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T13:57:41.923" Score="2" ViewCount="0" Body="&lt;p&gt;I think the easiest way to see this is to bring everything to the left side.  For inequality 1 you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^3 - a^2 b &amp;lt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now factor out a^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a^2( a - b) &amp;lt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a true statement - provided a != 0 - since a^2 is positive and a-b is negative.  Recall that a &amp;lt; b.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For inequality 2 you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a (a - b) &amp;lt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While (a - b) is negative we can't say whether or not a is positive or negative.  The inequality is true only if a is positive and is false if a is negative.&lt;/p&gt;&#xA;" OwnerUserId="1054" LastActivityDate="2010-08-11T13:57:41.923" CommentCount="1" />
  <row Id="2149" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T14:06:42.090" Score="0" ViewCount="0" Body="&lt;p&gt;Both are false; try for instance for a = 0 &amp;lt; 1 = b.&lt;/p&gt;&#xA;" OwnerUserId="1056" LastActivityDate="2010-08-11T14:06:42.090" />
  <row Id="2150" PostTypeId="2" ParentId="2112" CreationDate="2010-08-11T14:12:45.460" Score="11" ViewCount="0" Body="&lt;p&gt;Most philosophers have agreed for centuries that we can be &lt;em&gt;completely sure&lt;/em&gt; of very little, if anything.  Descartes famously argued &lt;a href=&quot;http://en.wikipedia.org/wiki/Cogito_ergo_sum&quot; rel=&quot;nofollow&quot;&gt;cogito ergo sum&lt;/a&gt;, i.e., that the action of contemplating the possibility of his nonexistence confirms his existence (a nice early example of a &quot;self-defeating proof by contradiction&quot;).  This particular argument is no longer thought to be unassailable: see the linked article. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the idea that you cannot be &lt;em&gt;completely sure&lt;/em&gt; that a mathematical proof is correct is true, but in the same way that you cannot be completely sure of virtually anything: i.e., in a not very interesting way and one which has nothing to do with mathematical knowledge in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be arguing that a special feature of mathematical verifications is that they can be long and complicated.  I disagree that this is particular to mathematics.  Suppose I wanted to create a list of all days between September 1, 2010 and September 1, 2020 in which the temperature in my condo never exceeds 80 degrees Fahrenheit.  What a pain in the butt it would be to be even reasonably certain about this: possibly I would want someone to remain in my condo at all times, probably I would want to buy some scientific equipment (I don't trust my thermostat that much, especially to distinguish between 79.5 and 80 degrees); certainly the task will take ten years to complete.  This is a long, difficult, esoteric verification that has nothing to do with mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have tried to point out, nothing that you say argues against &quot;we thought that mathematics is the most secure branch of knowledge&quot;.  You also don't see who &quot;we&quot; are or why we thought that, but as it happens &lt;em&gt;I&lt;/em&gt; do think that mathematics is, relatively speaking, an especially secure branch of knowledge.  If I spent ten years thinking about the proof of the Poincare Conjecture, then at the end I would be much more secure in my opinion that it was correct [or, possibly, incorrect!] than I would be at the end of my condo temperature experiment.  For PC I would have a coherent, well-understood argument that I could turn around, adapt to other situations, explore consequences of, etc.  I could burn all the original documents I read during those 10 years and not be much worse off.  For my condo temperature experiment I would just have reams of data and greater or lesser amounts of experimental doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complicated things are indeed complicated, and I think we must be honest that there is always the possibility (and very often, the reality) of human error.  Coming back to Descartes, he believed that he had discovered a &lt;em&gt;method&lt;/em&gt; by which it was possible to reason perfectly without error, and he applied his method to develop the new field of analytic geometry.  This was fantastic, pioneering work.  But entirely without error?  Of course not; as is well documented, he made some mistakes.  So do almost all of us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I would claim that a relatively simple mathematical proof has one of the highest degrees of certainty that we can muster.  In another response, someone mentioned the infinitude of prime numbers and indeed there was a previous question on this site with the title &quot;Is it possible that there are only finitely many primes?&quot;  When we answer &lt;strong&gt;No!&lt;/strong&gt; we are more sure of this than a researcher in any other field I can think of could be.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-11T14:12:45.460" CommunityOwnedDate="2010-08-12T13:15:22.263" />
  <row Id="2151" PostTypeId="2" ParentId="2136" CreationDate="2010-08-11T14:22:40.427" Score="2" ViewCount="0" Body="&lt;p&gt;It's a bit disingenuous to say polynomial splines are infinitely differentiable; a spline with nth order pieces can only have at most C^{n-1} continuity. That being said, a different basis set (e.g. the Fourier basis or rational functions) would in principle be infinitely differentiable, but might be a very &quot;wiggly&quot; way to &quot;connect the dots&quot; so to speak. Polynomial interpolation, whatever method you use, will also suffer from this &quot;wiggliness&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would you need your approximant to be infinitely differentiable anyway? Unless you're going to be doing something special I don't see it being a necessity.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-11T14:22:40.427" CommentCount="3" />
  <row Id="2152" PostTypeId="2" ParentId="227" CreationDate="2010-08-11T14:35:20.960" Score="1" ViewCount="0" Body="&lt;p&gt;You have a choice of a course called &quot;Group Theory and Linear Algebra&quot;?  You want linear algebra but not group theory if you have limited time.  If you have an interest in understanding stochastic models in a serious way then you want to understand stochastic differential equations, which requires serious doses of measure-theoretic probability (not lower-level probability using no measure theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you looked at Greg Mankiw's blog?  He has a section of links called Advice for Students and one is about math courses.  The link is &lt;a href=&quot;http://gregmankiw.blogspot.com/2006/05/which-math-courses.html&quot; rel=&quot;nofollow&quot;&gt;http://gregmankiw.blogspot.com/2006/05/which-math-courses.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-11T14:35:20.960" CommunityOwnedDate="2010-08-12T14:58:43.903" />
  <row Id="2153" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T14:39:12.480" Score="8" ViewCount="0" Body="&lt;p&gt;A version of Robin Chapman's answer that might be easier to verify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additive group H of a field of order 2&lt;sup&gt;n&lt;/sup&gt; has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism.  The semi-direct product G is called AΓL(1,2&lt;sup&gt;n&lt;/sup&gt;).  It has order (2&lt;sup&gt;n&lt;/sup&gt;)⋅(2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(n) ≤ 2&lt;sup&gt;3n&lt;/sup&gt;, and for n ≥ 2 it is its own automorphism group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has the additive group H of the field as a normal subgroup.  However, as just an additive group, H has automorphism group all n×n matrices over the field with 2 elements, which has size (2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(2&lt;sup&gt;n&lt;/sup&gt;−2)⋅⋅⋅(2&lt;sup&gt;n&lt;/sup&gt;−2&lt;sup&gt;n−1&lt;/sup&gt;) ≥ 2&lt;sup&gt;(n−1)&lt;sup&gt;2&lt;/sup&gt;&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large n, |Aut(G)| = (2&lt;sup&gt;n&lt;/sup&gt;)⋅(2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(n) ≤ 2&lt;sup&gt;3n&lt;/sup&gt; ≤ 2&lt;sup&gt;(n−1)&lt;sup&gt;2&lt;/sup&gt;&lt;/sup&gt; ≤ (2&lt;sup&gt;n&lt;/sup&gt;−1)⋅(2&lt;sup&gt;n&lt;/sup&gt;−2)⋅⋅⋅(2&lt;sup&gt;n&lt;/sup&gt;−2&lt;sup&gt;n−1&lt;/sup&gt;) = |Aut(H)|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 4, G = AΓL(1,16), Aut(G) = G has order 960, and Aut(H) = GL(4,2) has order 20160.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 10, G = AΓL(1,1024), Aut(G) = G has order 10475520, and Aut(H) = GL(10,2) has order 366440137299948128422802227200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, Aut(H) can be enormously bigger than Aut(G).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is reasonably important in finite group theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G is a very rigid group with lots of structure.  Because it contains all the &quot;automorphisms&quot; of the field, the group itself determines the field.  An automorphism of the group will have to be an automorphism of the field, and we've already listed them all.  Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H is a very floppy group to which you can do nearly anything.  Without the maps encoding scalar multiplication, H no longer remembers the field that defined it.  It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism.  H has lost its structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think H is the canonical example of a horribly structureless group.  Groups like GL and AΓL are pretty standard examples of groups with very clear structure.  The symmetric group is very similar.  Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-11T15:07:07.497" LastActivityDate="2010-08-11T15:07:07.497" />
  <row Id="2154" PostTypeId="2" ParentId="2125" CreationDate="2010-08-11T14:43:38.303" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; Your polynomial $p(n)$ splits over ${\mathbb Q}(w), w = \zeta_5$, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 125 \; p(x) = 125 \; (5 x^4-10 x^3+20 x^2-15 x+11) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\; =  \;\; (5 x+3 w^3-4 w^2-w-3) (5 x+4 w^3+3 w^2+7 w+1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad\; * \; (5 x-3 w^3+4 w^2+w-2)  (5 x-4 w^3-3 w^2-7 w-6) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the other questions in the query and the comments: there has been much research on various ways of characterizing number fields by splitting behavior, norm sets, etc - going all the way back back to Kronecker. Searching on the terms &quot;Kronecker equivalent&quot; or &quot;arithmetically equivalent&quot; will find pertinent literature. E.g. below is one enlightening review&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=485790&quot; rel=&quot;nofollow&quot;&gt;MR0485790 (58 #5595)&lt;/a&gt; 12A65 (12A75)&lt;br&gt;&#xA;Gauthier, François&lt;br&gt;&#xA;Ensembles de Kronecker et représentation des nombres premiers par une forme quadratique binaire.&lt;br&gt;&#xA;Bull. Sci. Math. (2) 102 (1978), no. 2, 129--143.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;L. Kronecker [Berlin Monatsber. 1880, 155--162; Jbuch 12, 65] first tried to characterize algebraic number fields by the decomposition behavior of primes. Recently, the Kronecker classes of algebraic number fields have been studied by W. Jehne [J. Number Theory 9 (1977), no. 2, 279--320; MR0447184 (56 #5499)] and others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article deals with the following types of questions:&lt;br&gt;&#xA;(a) When does the set of primes having a given splitting type in an algebraic number field contain (up to a finite set) an arithmetic progression?&lt;br&gt;&#xA;(b) When is this set a union of arithmetic progressions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K$ is an algebraic number field, let $\text{spl}(K)$ denote the set of rational primes which split completely in $K$ and let $\text{spl}^1(K)$ denote the set of rational primes which have at least one linear factor in $K$. Moreover, if $K/Q$ is a Galois extension with Galosis group $G$, let ${\text Art}_{K/Q}$ denote the Artin map which assigns a conjugacy class of $G$ to almost all rational primes $p$. If $C$ is a conjugacy class of $G$ then $\text{Art}_{K/Q}^{-1}(C)$ is the set of primes having Artin symbol $C$. Finally a set $S$ of rational primes is said to contain an arithmetic progression or to be the union of arithmetic progressions if the set of primes in the arithmetic progression(s) differs from $S$ by at most a finite set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G'$ denote the commutator subgroup of the Galois group $G$. Two results proved in the article are:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem A.&lt;/strong&gt; The following statements are equivalent:&lt;br&gt;&#xA;(a) $|C|=|G'|$;&lt;br&gt;&#xA;(b) $\text{Art}_{K/Q}^{-1}(C)$ is the union of arithmetic progressions;&lt;br&gt;&#xA;(c) $\text{Art}_{K/Q}^{-1}(C)$ contains an arithmetic progression.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem B.&lt;/strong&gt; The following statements are equivalent:&lt;br&gt;&#xA;(a) $K/Q$ is abelian;&lt;br&gt;&#xA;(b) $\text{spl}(K)$ contains an arithmetic progression;&lt;br&gt;&#xA;(c) $\text{spl}(K)$ is the union of arithmetic progressions;&lt;br&gt;&#xA;(d) there exist a modulus $m$ and a subgroup ${r_1,\cdots,r_t}$ of the multiplicative group modulo $m$ such that $\text{spl}(K)$ is the union of the arithmetic progressions $mx+r_i\ (i=1,\cdots,t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $K/Q$ is a non-Galois extension it is well known that $\text{spl}(K)=\text{spl}(\overline K)$ where $\overline K$ denotes the normal closure of $K$. It follows from Theorem B that $\text{spl}(K)$ cannot contain an arithmetic progression. However, the author gives two conditions, one necessary and the other sufficient, for $\text{spl}^1(K)$ to be the union of arithmetic progressions when $K/Q$ is non-Galois. As a final application of his result the author gives a necessary and sufficient condition for the set of primes represented by a quadratic form to be the union of arithmetic progressions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proofs use class field theory, properties of the Artin map and the Čebotarev density theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reviewed by Charles J. Parry&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-11T16:52:12.510" LastActivityDate="2010-08-11T16:52:12.510" CommentCount="1" />
  <row Id="2155" PostTypeId="2" ParentId="2145" CreationDate="2010-08-11T15:05:12.900" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; If $b-a &gt; 0 $ then $c(b-a) &gt; 0 \iff c &gt; 0$. Now specialize $c = a, \; a^2$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-11T15:05:12.900" />
  <row Id="2156" PostTypeId="2" ParentId="2131" CreationDate="2010-08-11T15:07:35.173" Score="0" ViewCount="0" Body="&lt;p&gt;It seems to me that a general formula for this simple but irregular question would be an overkill. Perhaps it is more efficient to simply count the possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for each of the possible 3 locations of 2 there are 4 such numbers, for a total of 12;&lt;/li&gt;&#xA;&lt;li&gt;there are a total of 8 different 3-digit numbers which have digits 3 or 4; however exactly one of them, namely 333, is not permissible; thus there are 7 different allowable 3-digit numbers, which have 3 and 4 as their only digits;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We see that all together there are 12+7 = 19 different solutions (3-digit numbers, which satisfy the stated assumptions).&lt;/p&gt;&#xA;" OwnerUserId="1056" LastActivityDate="2010-08-11T15:07:35.173" />
  <row Id="2157" PostTypeId="2" ParentId="1952" CreationDate="2010-08-11T15:07:38.980" Score="1" ViewCount="0" Body="&lt;p&gt;How about the Mumford conjecture, now proven by Madsen &amp;amp; Weiss? The statement can be stated rather simply in terms of cohomology of &lt;em&gt;M&lt;sub&gt;g&lt;/sub&gt;&lt;/em&gt;, but their proof is highly non-algebraic and uses some quite involved homotopy theory that I won't claim to understand. Very roughly, they consider the mapping class groups $\Gamma_{g}$ (the fundamental group of the moduli space of curves of genus &lt;em&gt;g&lt;/em&gt;) and form the limit $\Gamma_\infty$ as &lt;em&gt;g&lt;/em&gt; becomes large. Now each moduli space is homotopic to the classifying space for $\Gamma_{g}$, and the stable homology of the moduli spaces is the homology of the classifying space for $\Gamma_\infty$. But this latter space is a weird infinite-dimensional CW-complex, and there are certainly no algebro-geometric analogues of such a beast.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-08-11T15:07:38.980" />
  <row Id="2158" PostTypeId="1" AcceptedAnswerId="2171" CreationDate="2010-08-11T15:25:13.767" Score="8" ViewCount="451" Body="&lt;p&gt;I have a partition of a positive integer (p). How can I prove that the factorial of p can always be divided by the product of the factorials of the parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a quick example 9!/(2!3!4!) = 1260 (no remainder), where 9=2+3+4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can &lt;em&gt;nearly&lt;/em&gt; see it by looking at factors, but I can't see a way to guarantee it.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-10-02T07:09:57.843" Title="Division of Factorials" Tags="&lt;number-theory&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="2" />
  <row Id="2159" PostTypeId="1" AcceptedAnswerId="2163" CreationDate="2010-08-11T15:25:58.790" Score="3" ViewCount="94" Body="&lt;p&gt;Now, I may have only slept two hours last night and would currently struggle to discern a 'proof' by induction of FLT from a piece of genuine mathematics, but that doesn't stop mathematics from bugging me. At present I am puzzled by &lt;a href=&quot;http://mathoverflow.net/questions/34982/are-there-space-filling-curves-for-the-hilbert-cube&quot; rel=&quot;nofollow&quot;&gt;something I saw on MO this morning&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked question concerns the hilbert cube $[0,1]^\mathbb{N}$ (an infinite product of intervals) and the exitence of space filling curves thereof- that is: continuous images of the unit circle that are surjections on the hilbert cube. &lt;a href=&quot;http://mathoverflow.net/questions/34982/are-there-space-filling-curves-for-the-hilbert-cube/34983#34983&quot; rel=&quot;nofollow&quot;&gt;The accepted answer&lt;/a&gt;, together with another answer (which actually constructs such a map) and various comments, seems to allude toward an answer in the affirmative. However, the linked theorem (the 'Hahn–Mazurkiewicz theorem') which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A nonempty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;seems in direct contradiction to this since (and I may be mistaken for reasons explained above):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The hilbert cube is a subset of a normed space and hence a metric space&lt;/li&gt;&#xA;&lt;li&gt;The sequence (1,0,0...), (0,1,0...), (0,0,1,...) has no convergent subsequence&lt;/li&gt;&#xA;&lt;li&gt;So the hilbert cube is not sequentially compact, therefore non-compact (the two are equivalent in metric spaces)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which seems at odds with the only if portion of the theorem's statement. Maybe this is wikipedia taking me for a ride. Maybe I am just hallucinating a portion of this argument. Either way, this is annoying me. Thanks in advance for clearing this up...&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-11T15:42:51.023" Title="A curious compactness confusion: space filling curves in the hilbert cube that contradict a bona fide theorem?!?!?" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="2161" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T15:34:25.657" Score="0" ViewCount="0" Body="&lt;p&gt;Hi--&#xA; Please look at this link:&#xA;mathdl.maa.org/images/upload.../00029890.di021346.02p0064l.pdf&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-11T15:34:25.657" CommentCount="2" />
  <row Id="2162" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T15:35:45.000" Score="13" ViewCount="0" Body="&lt;p&gt;These quotients are integers since they solve counting problems.&#xA;For instance, how many nine-letter words are there with 2 As, 3 Bs and 4 Cs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the full story see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;multinomial theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T15:35:45.000" CommentCount="2" />
  <row Id="2163" PostTypeId="2" ParentId="2159" CreationDate="2010-08-11T15:41:26.763" Score="4" ViewCount="0" Body="&lt;p&gt;$(1,0,0,\ldots), (0,1,0,\ldots)$ converges to $(0,0,0,\ldots)$, so your example doesn't contradict the compactness of the Hilbert cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[0,1]^{\mathbb N}$ is homeomorphic to $[0,1]\times[0,1/2]\times[0,1/3]\times\cdots$ with the $\ell_2$-norm. So the norm of $(a_1,a_2,a_3,\ldots)$ is $\sum\left(\dfrac{a_n}{n}\right)^2$, and therefore your sequence converges to $(0,0,0,\ldots)$ since the norm of $(0,\ldots,0,1,0,\ldots)$ (with the 1 in the $n$th position) is $1/n^2$, which tends to 0.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-08-11T15:41:26.763" />
  <row Id="2164" PostTypeId="2" ParentId="2159" CreationDate="2010-08-11T15:42:51.023" Score="5" ViewCount="0" Body="&lt;p&gt;The sequence you give converges to $0$ in the product topology.  For instance, convergence in the product topology is equivalent to pointwise (or coordinatewise) convergence, and your sequence converges coordinatewise to $0$.  See e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pointwise_convergence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pointwise_convergence&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;b&gt;Added&lt;/b&gt;: Samuel's answer proceeds differently from mine -- via an explicit metric on the &quot;Hilbert cube&quot; -- and is also correct.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-11T15:42:51.023" />
  <row Id="2165" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T15:46:39.837" Score="7" ViewCount="0" Body="&lt;p&gt;If you believe (:-) in the two-part Newton case, then the rest is easily obtained by induction. For instance (to motivate you to write a full proof):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9!/(2!3!4!) = (9!/(5!\cdot 4!)) \cdot (5!/(2!\cdot 3!))$$&lt;/p&gt;&#xA;" OwnerUserId="1056" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-11T17:02:51.137" LastActivityDate="2010-08-11T17:02:51.137" CommentCount="4" />
  <row Id="2166" PostTypeId="1" AcceptedAnswerId="2173" CreationDate="2010-08-11T15:59:34.917" Score="1" ViewCount="199" Body="&lt;p&gt;&lt;strong&gt;Statement&lt;/strong&gt;  For every $n &gt; 1$ there is always at least one prime $p$ such that $n &amp;lt; p &amp;lt; 2n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious to know that if I replace that $2n$ by $2n-\epsilon$, ($\epsilon&gt;0$) then what is the  $\inf (\epsilon)$ so that the inequality still holds, meaning there is always a prime between $n$ and $2n-\epsilon$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-11T16:20:22.543" LastActivityDate="2010-08-13T20:39:56.267" Title="Bertrand's Postulate" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="2167" PostTypeId="2" ParentId="2130" CreationDate="2010-08-11T16:05:58.640" Score="1" ViewCount="0" Body="&lt;p&gt;In the theory of abelian categories, the problem with non-small categories (or at least categories without some smallness assumption) is that the collection of inclusions might not be a set.  This can prevent you from taking constructions you might want, such as taking the union over all submodules with a given property.  This happens, for instance, when trying to construct essential extensions.  This is usually fixed by requiring that the collection of inclusions into an object forms a set, rather than requiring smallness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that even outside the realm of abelian categories, non-smallness wreaks havoc with categorical constructions which involve a 'for all' such as equalizers, fibered products and more general limits.  However, I don't know of any specific examples of this.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-08-11T16:05:58.640" />
  <row Id="2168" PostTypeId="2" ParentId="2166" CreationDate="2010-08-11T16:10:23.003" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand%27s_postulate&quot; rel=&quot;nofollow&quot;&gt;Bertrand's postulate&lt;/a&gt; is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if n &gt; 3 is an integer, then there always exists at least one prime number p with n &amp;lt; p &amp;lt; 2n − 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus &amp;epsilon; &amp;lt; 2 for n &gt; 3. What if n ≤ 3?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For n = 3, 3 &amp;lt; 5 &amp;lt; 6 - &amp;epsilon; &amp;rArr; &amp;epsilon; &amp;lt; 1&lt;/li&gt;&#xA;&lt;li&gt;For n = 2, 2 &amp;lt; 3 &amp;lt; 4 - &amp;epsilon; &amp;rArr; &amp;epsilon; &amp;lt; 1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence we have 0 &amp;lt; &amp;epsilon; &amp;lt;  1, if &amp;epsilon; is a constant.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-11T17:39:21.873" LastActivityDate="2010-08-11T17:39:21.873" CommentCount="3" />
  <row Id="2169" PostTypeId="2" ParentId="392" CreationDate="2010-08-11T16:16:16.417" Score="6" ViewCount="0" Body="&lt;p&gt;I agree with David (+1), this is the pertinent graph, and it works for me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.animations.physics.unsw.edu.au/jw/graphics/calculus8%28circle%29.gif&quot; alt=&quot;Sin and cosine on the unit circle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation &lt;a href=&quot;http://www.animations.physics.unsw.edu.au/jw/calculus.htm#Trigonometric&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-08-11T16:16:16.417" />
  <row Id="2170" PostTypeId="2" ParentId="2115" CreationDate="2010-08-11T16:48:13.467" Score="-2" ViewCount="0" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think i have an answer. I found it &lt;a href=&quot;http://www.isibang.ac.in/~statmath/problems/soljan09.pdf&quot; rel=&quot;nofollow&quot;&gt;here on the internet&lt;/a&gt;. Its very interesting.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We clarify what is meant by the slightly more general concept of a $k$-times &lt;strong&gt;Peano&lt;/strong&gt; differentiable function $f$ at a point $x$. This simply requires that there is a polynomial $P_{x}$ of degree at most $k$ such that $$\frac{f(x+t)-P_{x}(t)}{t^k} \to 0$$ as $t \to 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, if possible, $f$ is a function which is discontinuous at all rationals and differentiable at all irrationals. Consider the set $\displaystyle A = \bigcap_{n} A_{n}$ where $$A_{n} = \Bigl[ p : \exists \ x,y ; |x-p|&amp;lt;\frac{1}{n},|y-p|&amp;lt;\frac{1}{n},\frac{f(x)-f(p)}{x-p} - \frac{f(y)-f(p)}{y-p}&gt;1 \Bigr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the very definition of $A$, $f$ is not differentiable at any of the points $p$&#xA;in A. Thus, $A$ is a subset of the rationals. However, we shall observe that&#xA;for each $n$, the interior points of $A_{n}$ form a dense set. This&#xA;would be a contradiction to $A$ being contained in the set of rationals. Start&#xA;with any open interval $I$ and with some rational number $q \in I$. Choose $h,k$ such that $$ \max ( f(q), \limsup_{t \to q} f(t) ) &gt; h &gt; k &gt; \min (f(q), \liminf_{t \to q} f(t))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle p \in I \cap \Bigl(q-\frac{1}{n},q\Bigr)$ and $p &gt; q-(h-k)$, then $$\frac{h-f(p)}{q-p}- \frac{k-f(p)}{q-p}=\frac{h-k}{q-p} &gt; 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, choosing $x,y$ very close to $q$ so that $f(x)&gt;h$ and $f(y) &amp;lt; k$, we get $$\frac{f(x)-f(p)}{x-p} - \frac{f(y)-f(p)}{y-p} &gt; 1$$ Hence $p \in A_{n}$. As the set $$ \Bigl[ t \in I \cap \Bigl(q-\frac{1}{n},q\Bigr) : t &gt; q-(h-k) \Bigr]$$ is open, the above $p$ is an interior point of $A_{n}$. Hence interior of $A_{n}$ is everywhere dense.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-14T22:51:25.323" LastActivityDate="2010-08-14T22:51:25.323" CommentCount="13" />
  <row Id="2171" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T17:08:55.947" Score="3" ViewCount="0" Body="&lt;p&gt;The key observation is that the product of $n$ consecutive integers is divisible by $n!$. This can be proved by induction.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-11T17:08:55.947" CommentCount="2" />
  <row Id="2172" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T17:27:28.460" Score="4" ViewCount="0" Body="&lt;p&gt;Besides the obvious combinatorial interpretations, one also has the has the following reduction from multinomial coefficient to products of binomial coefficients. Namely for $n = i+j+k+\cdots + m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{n!}{i!j!k!\cdots m!} = \binom{n}{i} \frac{(n-i)!}{j!k!\cdots m!} = \binom{n}{i}\binom{n-i}{j} \frac{(n-i-j)!}{k!\cdots m!} = \;\cdots$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-11T17:27:28.460" CommentCount="1" />
  <row Id="2173" PostTypeId="2" ParentId="2166" CreationDate="2010-08-11T17:29:31.483" Score="2" ViewCount="0" Body="&lt;p&gt;Three related points are worthy of mention, showing that epsilon can be close to n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a result of Finsler that approximates how many primes lie between n and 2n, which is of order o(n/log(n)) as is to be expected by the Prime Number Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literature on prime gaps will tell you the exponent delta such that there is (for sufficiently large n) at least one prime in the interval (n , n + n^delta).  I think delta is less than 11/20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observed data suggests that n^delta can be replaced by something much smaller: for n between something like 3 and 10^14 , some function like 2(log(n))^2 works.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:03:45.550" LastActivityDate="2010-08-11T22:03:45.550" CommentCount="0" />
  <row Id="2174" PostTypeId="1" AcceptedAnswerId="2179" CreationDate="2010-08-11T18:17:45.760" Score="1" ViewCount="116" Body="&lt;p&gt;As a teenager I was given this problem which took me a few years to solve.  I'd like to know if this hae ever been published.  When I presented my solution I was told that it was similar to one of several he had seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an &lt;code&gt;n&lt;/code&gt; dimensional space, develop a formula that evaluates the maximum number of &lt;code&gt;n&lt;/code&gt; dimensional regions when divided by &lt;code&gt;k&lt;/code&gt; &lt;code&gt;n-1&lt;/code&gt; dimensional (hyper)planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:  A line is partitioned by points: 1 point, 2 line segments. 10 points, 11 line segments, and so one.  &lt;/p&gt;&#xA;" OwnerUserId="1067" LastEditorUserId="1067" LastEditorDisplayName="" LastEditDate="2010-08-11T23:17:25.603" LastActivityDate="2010-08-11T23:17:25.603" Title="Are there publications or references describing N dimensional space divided by N-1 dimensional (hyper)planes?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="2175" PostTypeId="2" ParentId="2013" CreationDate="2010-08-11T18:23:29.867" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a (smaller) concrete answer for the sum, ie. what is the PDF of $h(z)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ z = x + y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $x$ and $y$ are drawn from independent probability distributions $f(x)$ and $g(y)$. It is very useful to realize that if $x,y$ are independent, then the function $p(x,y)=f(x)g(y)$ is the joint PDF, that is when you have two independent events the chance of both of them occurring is simply their product. Now think of the 2D plane where the point $(x,y)$ is the value $f(x)g(y)$. Integrating some area in this plane is equal to the probability of the event occouring over the area (just like in 1D).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for $z=x+y$ consider all the points where $z$ is constant - this defines a line with slope $-1$ and a y-intercept of $z$. We need to get an area for this, so we can integrate the area defined by a line that is very close to it. That is, we need to find the are between the two lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y_1 = -x + z $$&#xA;$$ y_2 = -x + z + dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h(z) = \int_{x=-\infty}^{\infty} \int_{y=-x+z}^{y=-x+z+dx} f(x)g(y) dy dx $$&#xA;$$ h(z) = \int_{x=-\infty}^{\infty} f(x) \int_{y=-x+z}^{y=-x+z+dx} g(y) dy dx $$&#xA;$$ h(z) = \int_{x=-\infty}^{\infty} f(x)g(z-x) dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot swept under the rug here and for that I'd refer you to the &lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf&quot; rel=&quot;nofollow&quot;&gt;reference given by Kaestur Hakarl&lt;/a&gt; in the comments. Also, if you're physics minded (like myself) I got introduced to the subject in Arfken &amp;amp; Weber's Mathematical Methods for Physicists vol 6, pg 1126. Here they introduce the change in area through the Jacobian, which helps you generalize to other functions such as the product and ratio.&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-08-11T18:23:29.867" CommentCount="0" />
  <row Id="2176" PostTypeId="2" ParentId="2174" CreationDate="2010-08-11T18:27:47.330" Score="2" ViewCount="0" Body="&lt;p&gt;Though not discussed in the full generality you describe, the problem is discussed at length for n = 1, 2, and 3 in &lt;em&gt;Mathematics for High School Teachers: An Advanced Persepective&lt;/em&gt; by Usiskin, Peressini, Machisotto, and Stanley (section 5.1.4; &amp;copy; 2003; published by Pearson Education).  Specifically, the text uses that problem as an example for discussing how induction can be applied to prove the formulae using the geometry of the problem in the inductive step (when you add the next (n-1)-dimensional boundary object, how many additional regions are created?).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-11T18:27:47.330" CommentCount="3" />
  <row Id="2177" PostTypeId="2" ParentId="2116" CreationDate="2010-08-11T18:28:41.760" Score="2" ViewCount="0" Body="&lt;p&gt;Here are some suggestions to approximate the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider F(n) to be the greater of 1 and the highest power of 2 that appears in the prime factorization of n.  Show what the limit of the average over n of F(n) is as n gets large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G_p(n) be defined like F(n), except replace 2 by an arbitrary prime p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that F(n) &amp;lt;= E(n), and that E(n) = max(G_p(n) over all primes p), so that if limit&#xA;of the average over n of E(n) exists, you can bound it using a sum of limits involving G_p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more accurate estimates, consider inclusion-exclusion.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T22:02:04.303" LastActivityDate="2010-08-11T22:02:04.303" CommentCount="1" />
  <row Id="2178" PostTypeId="1" AcceptedAnswerId="2226" CreationDate="2010-08-11T18:32:51.797" Score="0" ViewCount="122" Body="&lt;p&gt;Let $G$ be a graph such that $\delta(G) \geq k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Prove that $G$ has a path of length at least $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solution: We know that &#xA;$\delta(G) = min\lbrace deg(v) | v \in V(G) \rbrace$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $\delta(G) = k$ then there exists some $v \in V(G)$ such that $deg(v) = k$. This means all other vertices $u \in V(G)$ have $deg(u) \geq k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i know this must be part of the proof. How would i prove that there exists a path of at least k?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)if $k \geq 2$, prove that $G$ has a cycle of length at least $k+1$.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditorDisplayName="" LastEditDate="2010-08-12T11:45:19.260" LastActivityDate="2010-08-12T17:10:49.260" Title="Graph Path Length Problem" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2179" PostTypeId="2" ParentId="2174" CreationDate="2010-08-11T18:43:50.527" Score="4" ViewCount="0" Body="&lt;p&gt;Section 3 of &lt;a href=&quot;http://www.macalester.edu/~bressoud/pub/Art_of_Counting/Art_of_Counting.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.macalester.edu/~bressoud/pub/Art_of_Counting/Art_of_Counting.pdf&lt;/a&gt; discusses the problem for $n=3$ and poses the problem of generalizing to higher dimensions.  This article states that the first published solution was by Jakob Steiner in 1826.  Other references are cited.&lt;/p&gt;&#xA;" OwnerUserId="273" LastActivityDate="2010-08-11T18:43:50.527" />
  <row Id="2180" PostTypeId="1" AcceptedAnswerId="2185" CreationDate="2010-08-11T18:50:05.977" Score="0" ViewCount="89" Body="&lt;p&gt;&lt;code&gt;Ax = b&lt;/code&gt;.  I need a way to analyze a square matrix &lt;code&gt;A&lt;/code&gt; to see if its solution vector &lt;code&gt;x&lt;/code&gt; will always be positive when &lt;code&gt;b&lt;/code&gt; is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question arises from solving the radiosity equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.siggraph.org/education/materials/HyperGraph/radiosity/images/slide11.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested to know when A is &lt;em&gt;incorrect&lt;/em&gt;, which would be when &lt;code&gt;x&lt;/code&gt; has negative values even though &lt;code&gt;b&lt;/code&gt; is positive.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-08-11T20:17:21.750" Title="How can I characterize the type of solution vector that comes out of a matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="2181" PostTypeId="2" ParentId="2178" CreationDate="2010-08-11T19:20:10.953" Score="1" ViewCount="0" Body="&lt;p&gt;You can construct path and cycle inductively. Start at an arbitrary vertex $v_0$, and move along an arbitrary edge to $v_1$. Since vertex $v_1$ has degree k or greater, it has at least $k-1$ neighbours other than $v_0$. Move along an edge to one of these neighbours $v_2$. Since vertex $v_2$ has degree $k$ or greater, it has at least $k-2$ neighbours other than $v_0$ and $v_1$. Move along an edge to one of these neighbours $v_3$. And so on... until you reach vertex $v_k$. At this stage, you already have a path of length $k$. There may then be no unvisited vertices left to move to, in which case v_0 must be one of them, and you can return to it for a cycle of length k+1. Otherwise, just keep on moving to new vertices. If when you get to vertex v_n you find that it is connected to one of the vertices $v_j$ for $0 \leq j \leq (n-k)$ then you can go back to that vertex for a cycle of length k+1 or greater. If not, then v_n will have a neighbour that you have not yet visited, and the journey can continue. Assuming that G has only finitely many vertices, you must eventually complete a cycle.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-11T19:20:10.953" CommentCount="1" />
  <row Id="2182" PostTypeId="1" AcceptedAnswerId="2184" CreationDate="2010-08-11T19:45:00.923" Score="0" ViewCount="104" Body="&lt;p&gt;Suppose $G$ is a locally compact abelian group, with Haar Measure $\mu$, then is $\mu(E)=\mu(E^{-1})$? for all subsets $E$ of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen that this is true for all Borel subsets of $G$, but i am in need of a proof and also an example as to why it doesn't hold for other subsets.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T02:02:56.197" Title="Locally Compact Group with Haar Measure" Tags="&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2183" PostTypeId="2" ParentId="2174" CreationDate="2010-08-11T19:49:27.857" Score="1" ViewCount="0" Body="&lt;p&gt;I presume you are asking for the number of regions into which $n$-space&#xA;is divided into by $k$ hyperplanes &quot;in general position&quot;. &#xA;See the notes of Richard Stanley on &lt;a href=&quot;http://math.mit.edu/~rstan/arrangements/arr.html&quot; rel=&quot;nofollow&quot;&gt;hyperplane arrangements&lt;/a&gt;.&#xA;The answer to your problem is Proposition 2.4, and there are lots&#xA;more goodies too!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T19:49:27.857" CommentCount="1" />
  <row Id="2184" PostTypeId="2" ParentId="2182" CreationDate="2010-08-11T19:53:20.593" Score="3" ViewCount="0" Body="&lt;p&gt;Not necessarily. This condition is equivalent to each left-invariant Haar measure&#xA;being right-invariant. A group having this property is called &lt;em&gt;unimodular&lt;/em&gt;.&#xA;Since the concept has a name, then there must be groups where it fails.&#xA;See the Wikipedia article on Haar measure for an example.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-11T19:53:20.593" CommentCount="3" />
  <row Id="2185" PostTypeId="2" ParentId="2180" CreationDate="2010-08-11T19:53:52.773" Score="4" ViewCount="0" Body="&lt;p&gt;If all entries of $A^{-1}$ are positive numbers, then $A$ has the property you desire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edit:&lt;/strong&gt; This condition is both &lt;em&gt;sufficient&lt;/em&gt; and &lt;em&gt;necessary&lt;/em&gt;. If one entry $A^{-1}_{ij}$ is negative, then the choice $b=e_j$ will make the $i$-th component of $x=A^{-1}b$ equal to $A^{-1}_{ij}$, but this is negative.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following argument shows that your particular example matrix does have this property: your matrix $A$ has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = 1 - R F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R=\mathrm{diag}(\rho_1,\rho_2,\dots,\rho_n)$ is the diagonal matrix of reflectivities and $F$ the matrix of form factors. The physical interpretation of $R$ and $F$ makes sure that the norm of $RF$ will be smaller than 1, so that $A^{-1}$ can be expressed as a geometric series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^{-1} = (1- RF)^{-1} = 1 + RF + (RF)^2 + (RF)^3 + \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since all entries of $RF$ are positive numbers, this formula implies that the entries of $A^{-1}$ are positive numbers as well.&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-08-11T20:17:21.750" LastActivityDate="2010-08-11T20:17:21.750" />
  <row Id="2186" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T20:02:36.403" Score="9" ViewCount="0" Body="&lt;p&gt;The &quot;high-level&quot; way to see this is to recall that whenever a finite group $G$ has a subgroup $H$, we know that $|H|$ divides $|G|$.  Then note that $S_n$ clearly contains $S_{n_1} \times ... \times S_{n_k}$ as a subgroup for any partition $n_1 + ... + n_k = n$.  (This is actually the same as the combinatorial interpretation in Robin Chapman's answer, since what we are counting is the number of cosets $G/H$, and these cosets are precisely what the multinomial theorem is counting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basic lemma is surprisingly useful.  For example, it is not hard to use it to show that $m! (n!)^m$ divides $(mn)!$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-11T20:02:36.403" CommentCount="1" />
  <row Id="2187" PostTypeId="1" AcceptedAnswerId="2201" CreationDate="2010-08-11T20:13:42.307" Score="0" ViewCount="121" Body="&lt;p&gt;Supose we have a set S of natural numbers. We define: &#xA;f(x) = x / 2 if x is even and f(x) = (x - 1) / 2 if x is odd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let a[0] = a[1] = 1 and a[n] = a[n - 1] + a[n - 2] if n does not belong to S and a[n] = f(a[n - 1] + a[n - 2]) otherwise, n &gt; 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the remainder of a[n] modulus a prime p. The problem is that the numbers become quite large so I can not store then. I must work with the reminders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fist idea was to work with the modular inverse of 2 and e kept the remainder mod p at each step and instead of diving by 2 I multiply by 2^(-1). The problem is that is does not work if S has more than one element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the solution is to keep the values mod 2^y*p where y is the numbers of elements in S (#S). And in the end take do remainder of a[n] mod p. It seems to work but I don't understand why...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if I was not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1072" LastActivityDate="2010-08-11T21:59:04.537" Title="Calculate the remainder when there are division" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2188" PostTypeId="1" AcceptedAnswerId="2217" CreationDate="2010-08-11T20:16:26.480" Score="-3" ViewCount="283" Body="&lt;p&gt;&lt;em&gt;This problem is taken from &lt;a href=&quot;http://imc2009.elte.hu/Problems/imc2009-day1.pdf&quot; rel=&quot;nofollow&quot;&gt;International Mathematics Competition for University Students 2009&lt;/a&gt; (IMC 2009), Day 1, Problem 4.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $p(z)= a_{0}+a_{1}z + \cdots + a_{n}z^{n}$ be a complex polynomial. Suppose that $1=c_{0} \geq c_{1} \geq c_{2} \geq \cdots \geq c_{n} \geq 0$ is a sequence of real numbers which is convex (i.e $2c_{k} \leq c_{k-1} + c_{k+1}$, for every $k=1,2,...,n-1$), and consider the polynomial $$q(z)= c_{0}a_{0} + c_{1}a_{1}z + \cdots + c_{n}a_{n}z^{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$ \substack{\max_{|z|\leq 1}} |q(z)|  \leq \substack{\max_{|z|\leq 1}} |p(z)|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts: The polynomials $p$ and $q$ are regular on the complex plane, so by the Maximum Principle we have $\max_{|z|\leq 1} |q(z)| = \max_{|z|=1} |q(z)|$, and similarly for $p$ also. Let us denote $M_{f}= \max_{|z|=1} f(z)$, then we have to show that $M_{q} \leq M_{p}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T19:51:01.873" LastActivityDate="2010-08-15T19:51:01.873" Title="Complex Analysis Question" Tags="&lt;complex-analysis&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2189" PostTypeId="1" AcceptedAnswerId="2195" CreationDate="2010-08-11T20:22:06.607" Score="2" ViewCount="197" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... octonions are the crazy old uncle nobody lets out of the attic: they are nonassociative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Comes from a a quote by John Baez. Clearly, the sucessor to quaterions from the Caley-Dickson process is a numerical beast, but has anybody found any real-world uses for them? For example, quaterions have a nice connection to computer graphics through the connection to SO(4), and that alone makes them worth studying. What can be done with a nonassociative algebra like the octonions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: simply mentioning that they&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;have applications in fields such as string theory, special relativity, and quantum logic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not what I'm looking for (I can read wikipedia too). A specific example, especially one that is geared to someone who is &lt;strong&gt;not&lt;/strong&gt; a mathematician by trade would be nice!&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="1028" LastEditorDisplayName="" LastEditDate="2010-08-23T15:17:39.730" LastActivityDate="2010-08-23T15:17:39.730" CommunityOwnedDate="2010-08-11T20:55:10.957" Title="What are some real-world uses of Octonions?" Tags="&lt;abstract-algebra&gt;&lt;octonions&gt;&lt;division-algebra&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="2190" PostTypeId="1" AcceptedAnswerId="2197" CreationDate="2010-08-11T20:25:53.260" Score="0" ViewCount="176" Body="&lt;p&gt;&lt;em&gt;This problem is taken from &lt;a href=&quot;http://imc2009.elte.hu/Problems/imc2009-day2.pdf&quot; rel=&quot;nofollow&quot;&gt;International Mathematics Competition for University Students 2009&lt;/a&gt; (IMC 2009), Day 2, Problem 5.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb{M}$ be the vector space of $m \times p$ real matrices. For a vector subspace $S \subset \mathbb{M}$, denote $\delta(S)$ the dimesion of the vector space generated by all columns of all matrices in $S$. Say that a vector space $T \subset \mathbb{M}$ is a covering space, if $$ \bigcup\limits_{A \in T, A \neq 0} ker(A) = \mathbb{R}^{p}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a $T$ is minimal if it does not contain a proper vector subspace $S \subset T$ which is also a covering matrix space. Let $T$ be a minimal covering matrix space and let $n= \text{dim}(T)$. Prove that $$\delta(T) \leq { n \choose 2 } $$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-08-15T22:22:50.653" LastActivityDate="2010-08-15T22:22:50.653" Title="Dimension of Vector space" Tags="&lt;linear-algebra&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2191" PostTypeId="2" ParentId="2189" CreationDate="2010-08-11T20:29:33.297" Score="0" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer this link: &lt;a href=&quot;http://www.citeulike.org/user/Scis0000002/article/6831691&quot; rel=&quot;nofollow&quot;&gt;http://www.citeulike.org/user/Scis0000002/article/6831691&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of applications, and one of them is this.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-11T20:29:33.297" CommunityOwnedDate="2010-08-11T20:55:10.957" />
  <row Id="2192" PostTypeId="2" ParentId="2158" CreationDate="2010-08-11T20:39:58.810" Score="8" ViewCount="0" Body="&lt;p&gt;Below is a sketch of a little-known purely arithmetical proof that binomial coefficients are integral. I purposely constructed the proof so that it would be comprehensible to an educated layperson. The proof gives an algorithm to rewrite a binomial coefficient as a product of fractions whose denominators are coprime to any given prime. The method of proof is  best comprehended by applying the algorithm to a specific example.&lt;br&gt;&#xA;[Note: It may prove helpful to first read &lt;a href=&quot;http://math.stackexchange.com/questions/5829&quot;&gt;this simpler example&lt;/a&gt; before proceeding to the exposition below].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. consider $\ \ \binom{39}{17}\: =\: \frac{39!}{22! \; 17!}\: =\: \frac{23 \cdot 24 \cdots\; 39}{1 \cdot 2 \cdots\; 17}\:.\ $ When this fraction is reduced to lowest terms $\rm\:a/b\:,\ $ no prime $\rm\ p &gt; 17\ $ can divide its denominator $\rm\: b\:,\: $ since $\rm\ b\:|\:17\:!\ \:$ Hence, to show that $\rm\ a/b\ $ is an integer, it suffices to show that no prime $\rm\ p \le 17\ $ divides its denominator $\rm\: b\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. we show that $2$ doesn't divide $\rm b$. The highest power of $2$ in the &#xA;denominator terms is $16 &amp;lt; 17$. Now align the numerator and denominator terms $\rm (mod\ 16)$&#xA;by shifting the 1st numerator term so it lies above its value $\rm (mod\ 16)$,&#xA;viz. $23 \equiv 7 \pmod{16}$ so right-shift the numerator terms until $23\:$ lies above $7$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{}{1}\frac{}{2}\frac{}{3}\frac{}{4}\frac{}{5}\frac{}{6}\frac{23}{7}\frac{24}{8}\frac{25}{9}\frac{26}{10}\frac{27}{11}\frac{28}{12}\frac{29}{13}\frac{30}{14}\frac{31}{15}\frac{32}{16}\frac{33}{17}\frac{34}{}\frac{35}{}\frac{36}{}\frac{37}{}\frac{38}{}\frac{39}{}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I claim that $2$ doesn't divide the reduced denominator of each aligned fraction. Indeed $\ 24/8 = 3$, $\: 26/10 = 13/5 $,  $\:\ 28/12 = 7/3 $,  $\:\ 30/14 = 15/7 $,  $\:\ 32/16 = 2$.&#xA;This holds because these fractions $\rm\; c/d \;$  satisfy  $\rm c \equiv d\ (mod\ 16)\:$&#xA;i.e.  $\rm c = d + 16\: n \;$  so  $\rm 2|d \Rightarrow 2|c$, $\rm\; 4|d \Rightarrow 4|c$, $\:\cdots $, $\rm\; 16|d \Rightarrow 16|c$,&#xA;i.e. any power of $2\:$ below $16$ that divides $\rm d$ must also divide $\rm c$, so it cancels out upon reduction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to prove that $2$ doesn't divide the reduced denominator of $\binom{39}{17}$ &#xA;it suffices to prove the same for the &quot;leftover&quot; fraction composed &#xA;of the above non-aligned terms $(34 \cdots 39)/(1 \cdots 6) = \binom{39}{6}$.&#xA;Being an $\rm\binom{n}{k}$ with smaller $\rm k = 6 &amp;lt; 17\:,\:$ this follows by induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the same proof works for any prime $\rm p$, we conclude that no prime divides the &#xA;reduced denominator of $\binom{39}{17}$, therefore it is an integer.$\quad$ &lt;strong&gt;QED&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally, the reason that this works is because the denominator sequence starts at $1$, which is coprime to every prime $\rm p$. This ensures that it is the &quot;greediest&quot; possible contiguous sequence of integers, in the sense that its product contains the least power of $\rm\:p\:$ compared to other contiguous sequences of equal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm extends to multinomials by using the simple reduction of multinomials to products of binomials mentioned in my prior post here.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-02T07:09:57.843" LastActivityDate="2010-10-02T07:09:57.843" CommentCount="1" />
  <row Id="2193" PostTypeId="1" AcceptedAnswerId="2196" CreationDate="2010-08-11T20:42:22.503" Score="0" ViewCount="146" Body="&lt;p&gt;Given a finite group $G$ and its subgroups $H,K$ such that $$G \times H \cong G \times K$$ does it imply that $H=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, one can see that this doesn't work out for all subgroups. Is there any condition by which this can remain true.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-11T23:41:29.340" Title="Cancellation of Direct Products" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2194" PostTypeId="1" AcceptedAnswerId="2216" CreationDate="2010-08-11T20:44:57.440" Score="13" ViewCount="182" Body="&lt;p&gt;The Hilbert Cube $H$ is defined to be $[0,1]^{\mathbb{N}}$, i.e., a countable product of unit intervals, topologized with the product topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've read that the Hilbert Cube is homogeneous.  That is, given two points $p, q\in H$, there is a homeomorphism $f:H\rightarrow H$ with $f(p)=q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's confusing to me is that it seems like there seems to be a stratification of points.  That is, there are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Points contained in $(0,1)^{\mathbb{N}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Points which have precisely n coordinate a 0 or 1 for n a fixed natural number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Point which have countably many coordinates equaling 0 or 1 and countably many not and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Points which have n many coordinates NOT equal to 0 or 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, for fixed $p$ and $q$ both in class 1 or 3 (or fix an n and use class 2 or 4), it's clear to me that there is a homeomorphism taking $p$ to $q$, simply by swapping around factors and using the fact that $(0,1)$ is clearly homogeneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what are the homeomorphisms which mix the classes?  In particular, what homemorphism takes (0,0,0,....) to (1/2, 1/2,1/2,....)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said another way, for any natural number $n&gt;1$, $[0,1]^n$ is NOT homogeneous, precisely because of these boundary points.  What allows you to deal with the boundary points in the infinite product case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, feel free to retag, and thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;  In the off chance that someone stumbles across this question, I just wanted to provide a rough idea of the answer, as garnered from the link Pete provided in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one has a point of the form $(1,p)$ in $[0,1] \times [0,1]$, then there is a self homeomorphism of $[0,1]\times[0,1]$ taking $(1,p)$ to $(q,1)$ with $q\neq 0, 1$.  For example, one can use a &quot;square rotation&quot;.  From here, the idea is simple:  given a point in $H$ of the form $(1, p_2, p_3, p_4,...)$, apply the square rotation on the first two factors to get a new point of the form $(q_1, 1, p_2, p_3,...)$.  Now, apply the square rotation on the second two factors to get a new point of the form $(q_1, q_2, 1, p_3,...)$.  The point is that after k iterations, the first k coordinates are all in the interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one proves a techinical lemma that states that the infinite composition of these homeomorphisms is a well defined homeomorphism.  The infinite composition maps the point $(1, p_2, ....)$ to a point of the form $(q_1, q_2, ....)$ which lies on the &quot;interior&quot; of $H$.  Finally, using the fact that $(0,1)$ is clearly homogeneous, one can easily map $(q_1, q_2, ...)$ to $(1/2,1/2,...)$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-12T06:17:56.357" LastActivityDate="2010-08-12T06:17:56.357" Title="Why is the Hilbert Cube homogeneous?" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="2195" PostTypeId="2" ParentId="2189" CreationDate="2010-08-11T20:50:27.850" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/octonions/octonions.html&quot; rel=&quot;nofollow&quot;&gt;John Baez&lt;/a&gt; has a long online article about uses of the octonions, at least some of which is concerned with their relationship to physics.  You might also want to read his papers with Huerta, &lt;a href=&quot;http://arxiv.org/abs/0909.0551&quot; rel=&quot;nofollow&quot;&gt;Division Algebras and Supersymmetry I&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/1003.3436&quot; rel=&quot;nofollow&quot;&gt;Division Algebras and Supersymmetry II&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you'll be able to find an easy application to explain to a layman, since the octonions are naturally connected to geometry in higher dimensions than most people can be bothered to care about.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-11T20:50:27.850" CommunityOwnedDate="2010-08-11T20:55:10.957" />
  <row Id="2196" PostTypeId="2" ParentId="2193" CreationDate="2010-08-11T21:03:38.630" Score="4" ViewCount="0" Body="&lt;p&gt;Yes.  This is an easy consequence of the Krull-Schmidt theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/KrullRemakSchmidtTheorem.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/KrullRemakSchmidtTheorem.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: To be clear, what I am claiming is that if one has finite groups $H,K,G$ such that $G \times H \cong G \times K$, then $H \cong K$.  As Steve D points out below, the hypothesis and conclusion of the OP's literal question are a bit different than this.  But &lt;em&gt;because&lt;/em&gt; of the Krull-Schmidt theorem, the OP's literal question becomes: let $H$ and $K$ be isomorphic subgroups of a finite group $G$.  When do we have $H = K$?  It is quite clear that the answer is &quot;not always&quot;, and I find implausible that there would be a clean necessary and sufficient condition for this.  But let's see what transpires...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-11T22:39:00.277" LastActivityDate="2010-08-11T22:39:00.277" CommentCount="4" />
  <row Id="2197" PostTypeId="2" ParentId="2190" CreationDate="2010-08-11T21:06:40.073" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.imc-math.org.uk/imc2009/imc2009-day2-solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;Problem 5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-11T21:06:40.073" CommentCount="1" />
  <row Id="2198" PostTypeId="2" ParentId="2121" CreationDate="2010-08-11T21:12:25.850" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.imc-math.org.uk/imc2008/day1_solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;Problem 5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-11T21:12:25.850" CommentCount="1" />
  <row Id="2199" PostTypeId="2" ParentId="2193" CreationDate="2010-08-11T21:13:01.870" Score="2" ViewCount="0" Body="&lt;p&gt;Here's an entry point into the literature, from the introduction to Lam's &lt;a href=&quot;http://math.berkeley.edu/~lam/ohio.ps&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the study of any algebraic system in which there is a notion of a direct sum, the  theme of cancellation arises very naturally: if $A \oplus B \cong A\oplus C$ in the given system, can we conclude that $B \cong  C$? (For an early treatment of this problem, see the work of Jonsson  and Tarski [JT] in 1947.) The answer is, perhaps not surprisingly, sometimes &quot;yes&quot; and  sometimes &quot;no&quot;: it all depends on the algebraic system, and it depends heavily on the  choice of A as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with a simple example, we all know that, by the Fundamental Theorem of  Abelian Groups, the category of finitely generated abelian groups satisfies cancellation.  But a little more is true, which solved what would have been the &quot;Third Test Problem&quot;  for §6 in Kaplansky's book [Ka 1] (see the Notes in [Ka_1:§20]): if A is a f.g. (finitely  generated) abelian group, then for any abelian groups $B$ and $C$, $A\oplus B \cong  A\oplus C$ still implies $B \cong C$. Thus, f.g. abelian groups A remain &quot;cancellable&quot; (with respect  to direct sums) in the category of all abelian groups. This takes a proof, which was  first given, independently, by P. M.Cohn [Co] and E. A. Walker [W]. And yet, there exist  many torsionfree abelian groups of rank 1 (that is, nonzero subgroups of the rational  numbers Q ) that are not cancellable in the category of torsionfree abelian groups of  finite rank, according to B. Jonsson [Jo].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] T.Y. Lam. A Crash Course on Stable Range, Cancellation, Substitution, and Exchange&lt;br&gt;&#xA;University of California, Berkeley, Ca 94720&lt;br&gt;&#xA;&lt;a href=&quot;http://math.berkeley.edu/~lam/ohio.ps&quot; rel=&quot;nofollow&quot;&gt;http://math.berkeley.edu/~lam/ohio.ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-11T22:03:17.360" LastActivityDate="2010-08-11T22:03:17.360" CommentCount="1" />
  <row Id="2200" PostTypeId="2" ParentId="2194" CreationDate="2010-08-11T21:18:54.387" Score="4" ViewCount="0" Body="&lt;p&gt;This appears to be a difficult result.  It was first proven in 1931 by O. H. Keller (&lt;a href=&quot;http://www.springerlink.com/content/r8nqu8v469110152/&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;), but the paper is in German.  Keller actually proves a stronger result: any infinite-dimensional compact convex set in $\ell^2$ is homeomorphic to the Hilbert cube.  (I guess there is an obvious choice of such a set which is clearly homogeneous, but I can't think of one at the moment.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another proof of the theorem in &lt;a href=&quot;http://iopscience.iop.org/1468-4810/42/3/A08/pdf/1468-4810_42_3_A08.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; of Ageev, which is in English.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-11T21:18:54.387" CommentCount="8" />
  <row Id="2201" PostTypeId="2" ParentId="2187" CreationDate="2010-08-11T21:47:44.570" Score="1" ViewCount="0" Body="&lt;p&gt;This sounds similar to computing the last nonzero base-10 digit of factorial(n).  The fact that you are, according to some scheme, dropping the last bit of what would otherwise be a Fibonacci sequence expressed in binary makes me think there is no hope for finding any periodicity in the values mod p, which is what happens in the Fibonacci sequence itself.  If you drop the last bit only finitely many times, then continue after that, your sequence modulo p will become periodic.&lt;/p&gt;&#xA;" OwnerUserId="297" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-11T21:59:04.537" LastActivityDate="2010-08-11T21:59:04.537" CommentCount="2" />
  <row Id="2202" PostTypeId="2" ParentId="2193" CreationDate="2010-08-11T23:41:29.340" Score="1" ViewCount="0" Body="&lt;p&gt;Concerning Steve D's interpretation of the question, here is a partial answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following property (P) of a group: for any two subgroups $H$ and $K$ of $G$, if $H \cong K$, then $H = K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: For a finite group $G$, the following are equivalent:&lt;br&gt;&#xA;(i) $G$ has property (P).&lt;br&gt;&#xA;(ii) $G$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclic groups are characterized among finite groups by having at most one subgroup of any given order, so certainly (ii) $\implies$ (i).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, assume $G$ has property (P).  Then it is a &lt;strong&gt;Dedekind group&lt;/strong&gt;: all of its subgroups are normal (for otherwise it has two subgroups which are conjugate -- hence isomorphic -- but unequal).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: $G$ is abelian, of exponent $N$.  Then $G$ is isomorphic to $C_N \times G'$ for some subgroup $G'$ (this is a step in the classification of finite abelian groups; it also follows easily from the theorem).  If $G'$ is not cyclic, then $G'$ is not trivial hence contains an element of order $p | N$ and thus over all $G$ contains at least two subgroups of order $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $G$ is a nonabelian Dedekind group, i.e., a &lt;a href=&quot;http://planetmath.org/encyclopedia/QuasiHamiltonianGroup.html&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian group&lt;/a&gt;.  As Dedekind showed, a Hamiltonian group (finite or otherwise!) must contain a subgroup isomorphic to the quaternion group $Q_8$.  But $Q_8$ contains three cyclic subgroups of order $4$, namely those generated by the elements $i$, $j$ and $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some noncyclic infinite abelian groups with property (P), e.g. $\mathbb{Q}/\mathbb{Z}$, but probably one can classify them as well: I haven't thought much about it.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-11T23:41:29.340" CommentCount="4" />
  <row Id="2204" PostTypeId="1" AcceptedAnswerId="2207" CreationDate="2010-08-12T00:16:43.700" Score="12" ViewCount="315" Body="&lt;p&gt;There are $N$ bugs in a plane. All bugs are moving at the same constant (nonzero) speed, but no two bugs are moving in the same direction (velocity vectors are of the same speed, but no two are parallel). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that at some point in time $N$ bugs will form &lt;strong&gt;convex&lt;/strong&gt; polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Can you loosen up any of the conditions so that the statement still holds?&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-08-17T05:50:32.830" LastActivityDate="2010-08-22T08:00:11.617" Title="Bugs walking in a plane" Tags="&lt;geometry&gt;&lt;intuition&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="2205" PostTypeId="2" ParentId="2204" CreationDate="2010-08-12T00:24:59.923" Score="9" ViewCount="0" Body="&lt;p&gt;The counterexample I thought I had here doesn't work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proof.  Since none of the bugs are moving in the same direction, any pair of lines determined by the velocity vectors intersect.  Let $C$ denote the convex hull of these intersection points.  Since after waiting a sufficiently long period of time the bugs will be arbitrarily far away from $C$, if we &quot;zoom out&quot; far enough $C$ will become arbitrarily small with respect to the convex hull of the location of the bugs.  It follows that we can assume that $C$ is arbitrarily small to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now claim that the bugs eventually form a convex polygon in which the angle at each vertex is strictly less than $\pi$.  To do this it suffices to examine a configuration of three bugs $a, b, c$ in consecutive counterclockwise order.  Pick a coordinate system in which the centroid of $C$ is the origin and $b$ travels in the positive $y$-direction (hence $a$ travels to the right and $c$ travels to the left).  Then it is easy to see that regardless of where $a, b, c$ initially begin along their routes, $b$ will eventually have $y$-coordinate greater than either $a$ or $c$, so angle $abc$ will eventually be strictly less than $\pi$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that by waiting sufficiently long the bugs will always form a convex polygon.  In fact, the bugs are approximating the convex polygon whose vertices are the unit velocity vectors of the bugs.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-12T22:10:51.063" LastActivityDate="2010-08-12T22:10:51.063" CommentCount="10" />
  <row Id="2206" PostTypeId="2" ParentId="2204" CreationDate="2010-08-12T00:47:41.033" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose the bugs all started at the origin, then they would form a convex polygon at all times after $t = 0$. It should be noted we have some wiggle room - we may move any bug a small distance (proportional to time) and the polygon will remain convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since our bugs lie initially within a circle, let them walk far enough that they are within the wiggle room of the convex polygon that would have been formed had they started at the origin. Hence the bugs form a convex polygon.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T00:47:41.033" CommentCount="0" />
  <row Id="2207" PostTypeId="2" ParentId="2204" CreationDate="2010-08-12T01:02:43.157" Score="9" ViewCount="0" Body="&lt;p&gt;Assuming no bugs get squashed in the process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $\lim_{t\to\infty}$ when $t$ is time, the bugs' beginning points shrink to $\frac{P}{t} = 0$ as observed when zoomed out. This means the final position of each bug lies on $\sqrt{r^2 + (vt)^2}$, or on the edge of a huge circle. Thus, you can see that all the bugs make a convex polygon (or $N$-gon) since a circle can be thought of as a regular (and convex) $\infty$-gon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this isn't the most vigorous proof in the world, but it's written in the style &lt;a href=&quot;http://www.smbc-comics.com/index.php?db=comics&amp;amp;id=1914&quot; rel=&quot;nofollow&quot;&gt;that most humans can understand&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="227" LastEditorUserId="227" LastEditorDisplayName="" LastEditDate="2010-08-12T01:35:35.513" LastActivityDate="2010-08-12T01:35:35.513" CommentCount="4" />
  <row Id="2209" PostTypeId="1" AcceptedAnswerId="2213" CreationDate="2010-08-12T01:30:02.050" Score="8" ViewCount="165" Body="&lt;p&gt;There are 100 ropes in a bag. In each step, two rope ends are picked at random, tied together and put back into a bag. The process is repeated until there are no free ends.&lt;br&gt;&#xA;What is the expected number of loops at the end of the process?&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T02:51:06.847" Title="Expected number of loops" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2210" PostTypeId="1" AcceptedAnswerId="2214" CreationDate="2010-08-12T01:49:36.043" Score="4" ViewCount="149" Body="&lt;p&gt;I was wondering if anyone had any experience with an undergraduate course that emphasized the building of mathematical theories or if they'd ever heard of this being done? How did the class work (did the professor list axioms and definitions each week and see what the students were able to prove, did you discuss a specific mathematical theory and the various problems that arose during its creation?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this sort of thing work well for undergraduates? I'd be very interested in taking a class like this because I've never really had any experience with that aspect of mathematics. So if there are any classes that are commonly taught in this sort of style, I'd be interested to know that as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-12T01:54:25.923" LastActivityDate="2010-08-12T02:26:19.027" CommunityOwnedDate="2010-08-12T01:54:25.923" Title="undergraduate courses emphasizing theory building?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2211" PostTypeId="1" AcceptedAnswerId="2215" CreationDate="2010-08-12T02:00:21.523" Score="1" ViewCount="127" Body="&lt;p&gt;Some points on (integer, rectangular) grid in a plane are colored white, some black and some are not colored.&lt;br&gt;&#xA;In each step, one vertical or horizontal line can be selected, and all colored points on that line would toggle their color.&lt;br&gt;&#xA;Prove there is a finite sequence of steps after which on every vertical or horizontal line number of black and number of white points differs by 1 at most.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T04:14:10.027" Title="Black and White points on a grid" Tags="&lt;discrete-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2212" PostTypeId="2" ParentId="2182" CreationDate="2010-08-12T02:02:56.197" Score="2" ViewCount="0" Body="&lt;p&gt;Of course, a measure is in general only defined on some $\sigma$-algebra of sets, and for Haar measure that's the Borel sets.  As Pete mentioned you can complete the measure if you like, but that still won't yield all sets.  So for an arbitrary set $E$, $\mu(E)$ may not even be defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this is necessary with an example you already know: let $G = S^1$ thought of as $[0,1]$ mod endpoints, and let $E$ be the usual example of a non-measurable set (the Vitali set).  There's no value for $\mu(E)$ that will let $\mu$ be translation invariant, finite, and nontrivial, which are required for a Haar measure, and so there's no &quot;Haar measure&quot; defined for all subsets.  To be able to have a Haar measure, we need to define it only on &quot;reasonable&quot;, i.e. Borel, sets.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-12T02:02:56.197" />
  <row Id="2213" PostTypeId="2" ParentId="2209" CreationDate="2010-08-12T02:24:00.063" Score="9" ViewCount="0" Body="&lt;p&gt;Suppose you start with $n$ ropes. You pick two free ends and tie them together:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you happened to pick two ends of the same rope, you've added one additional loop (which you can set aside, since you'll never pick it now on), and have $n-1$ ropes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you happened to pick ends of different ropes, you've added no loop, and effectively replaced the two ropes with a longer rope, so you have $n-1$ ropes in this case too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of the $\binom{2n}{2}$ ways of choosing two ends, $n$ of them result in the first case, so the first case has probability $\frac{n}{2n(2n-1)/2} = 1/(2n-1)$. So the expected number of loops you &lt;strong&gt;add in the first step&lt;/strong&gt;, when you start with $n$ ropes, is $$\left(\frac{1}{2n-1}\right)1 + \left(1-\frac{1}{2n-1}\right)0 = \frac{1}{2n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, you start over with $n-1$ ropes. Since what happens in the first step and later are independent (and expectation is linear anyway), the expected number of loops is $$ \frac{1}{2n-1} + \frac{1}{2n-3} + \dots + \frac{1}{3} + 1 = H_{2n} - \frac{H_n}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for $n=100$, the answer is roughly $3.28$, which come to think of it seems surprisingly small for the number of loops!&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-12T02:51:06.847" LastActivityDate="2010-08-12T02:51:06.847" CommentCount="6" />
  <row Id="2214" PostTypeId="2" ParentId="2210" CreationDate="2010-08-12T02:26:19.027" Score="5" ViewCount="0" Body="&lt;p&gt;I have benefited from more than one professor who taught more or less according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore_method&quot; rel=&quot;nofollow&quot;&gt;Moore Method&lt;/a&gt;.  These professors happened to come out of SUNY Binghamton in the early 1980's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The courses required lots of input from the students.  One professor in particular would ask the class to make a conjecture about some construction he would put on the board and then name the conjecture after the student - we were then personally responsible for success or failure of our ideas.  After the first few humiliations, this became wonderfully empowering and fun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest impact I have noticed from taking these courses has been on my independent studying in other classes.  I notice that I am able to replicate the types of attacks on my own reasoning that my professor used to - hence leading me closer toward the boundaries of my own understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know how to respond to your idea of theory building... it seems to me that mathematics is not created upward from luckily chosen axioms, but in a highly nonlinear and back-and-forth fashion.  The only example I know of to illustrate this clearly is the development of topology, starting in the late 1800's with Cantor and more or less culminating in the early 1900's with Hausdorff (excellent discussions with references appear in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0800202090&quot; rel=&quot;nofollow&quot;&gt;Engelking, General Topology&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that it would be a great benefit to you to find courses which at least approach what you have in mind.  Good luck to you. &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-12T02:26:19.027" CommunityOwnedDate="2010-08-12T02:26:19.027" CommentCount="1" />
  <row Id="2215" PostTypeId="2" ParentId="2211" CreationDate="2010-08-12T03:21:46.837" Score="5" ViewCount="0" Body="&lt;p&gt;What sequence of horizontal or vertical steps would lead&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to have the number of black and white points almost even?&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditorDisplayName="" LastEditDate="2010-08-12T04:14:10.027" LastActivityDate="2010-08-12T04:14:10.027" CommentCount="1" />
  <row Id="2216" PostTypeId="2" ParentId="2194" CreationDate="2010-08-12T03:49:35.893" Score="3" ViewCount="0" Body="&lt;p&gt;When I first saw the question, I intended to post the following link as an answer, but then I saw that Qiaochu had come first with an essentially equivalent answer.  I still don't see anything lacking in his answer (I was the first to upvote it), but if someone is looking for a different reference, here is one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.narcis.info/publication/RecordID/oai:cwi.nl:7603&quot; rel=&quot;nofollow&quot;&gt;http://www.narcis.info/publication/RecordID/oai:cwi.nl:7603&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-12T03:49:35.893" CommentCount="1" />
  <row Id="2217" PostTypeId="2" ParentId="2188" CreationDate="2010-08-12T04:04:41.593" Score="3" ViewCount="0" Body="&lt;p&gt;Hi Chandru, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can reduce your question, considering the following argument. &#xA;Suppose that $c_k$ is a sequence satisfying the Carathéodory–Toeplitz Theorem, then&#xA;there exist a Borel Probability Measure $\mu$ on $\partial\mathbb D$, where $\mathbb D$ is the unity disc, such that &#xA;$$&#xA;\sum_{k=0}^{\infty}c_kz^k=\int_{\partial \mathbb D}\frac{1}{1-\zeta z}d\mu(\zeta).&#xA;$$&#xA;For $z\in\mathbb D$ fixed, we can use geometric series to obtain the following identity&#xA;$$&#xA;\sum_{k=0}^{\infty} c_kz^k =\int_{\partial \mathbb D} \sum_{k=0}^{\infty}(\zeta z)^k d\mu(\zeta)\ \ \ \ (1).&#xA;$$&#xA;Therefore $c_k=\int_{\partial \mathbb D}\zeta^k d\mu(\zeta)$. Let $\|f(z)\|$ to denote the maximum modulus of the function $f$ over the boundary of the unit disk. By (1) it follows that&#xA;$$&#xA;\|\sum_{k=0}^{n} a_kc_kz^k \|=\|\sum_{k=0}^n a_k\left(\int_{\partial\mathbb D}\zeta^k d\mu(\zeta)\right) z^k\|.&#xA;$$&#xA;By the linear properties of the integral, we obtain the following equality&#xA;$$&#xA;\|\sum_{k=0}^n a_k\left(\int_{\partial\mathbb D}\zeta^k d\mu(\zeta)\right) z^k\|=\|\int_{\partial\mathbb D}\sum_{k=0}^n a_k(\zeta z)^k d\mu(\zeta)\|.&#xA;$$&#xA;From the definition of $q(z)$ and $p(z)$, it follows that&#xA;$$&#xA;\|q(z)\|\leq\|\int_{\partial\mathbb D}\sum_{k=0}^n a_k(\zeta z)^k d\mu(\zeta)\|=&#xA;\|\int_{\partial\mathbb D}p(\zeta z) d\mu(\zeta)\|\leq \|p(z)\|\mu(\partial \mathbb D)=\|p(z)\|,&#xA;$$ &#xA;where the inequalities above are obtained by the Modulus Maximum Theorem and a suitable limit process. &#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;So your inequality is fully proved, if we are able to justify that $c_k$ satisfy  Carathéodory–Toeplitz.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditorDisplayName="" LastEditDate="2010-08-12T04:32:27.263" LastActivityDate="2010-08-12T04:32:27.263" />
  <row Id="2218" PostTypeId="2" ParentId="1872" CreationDate="2010-08-12T04:11:07.923" Score="2" ViewCount="0" Body="&lt;p&gt;I guess I should expand my comment into an answer.  Given two matrices $a_{ij}$ and $b_{ij}$ with entries in any (associative) ring $R$, the natural definition of the product has entries&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c_{ij} = \sum_k a_{ik} b_{kj}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This multiplication is associative, and it also agrees with the multiplication one obtains from any finite-dimensional matrix representation of $R$ by replacing each entry by the corresponding matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not see any particular reason to consider a different notion of multiplication.  Changing the order of some of the multiplications seems nonsensical to me, and multiplying in the opposite order gives you essentially the same multiplication.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition does &lt;em&gt;not&lt;/em&gt; agree with the definition in my first comment; multiplication by one of the above matrices does not define an $R$-module homomorphism when $R$ is noncommutative.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-12T04:11:07.923" CommentCount="8" />
  <row Id="2219" PostTypeId="2" ParentId="2188" CreationDate="2010-08-12T04:39:42.123" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.imc-math.org.uk/imc2009/imc2009-day1-solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;Problem 4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Chandru1 - this is at least the third question you asked that is identical to a question posed and answered at the &lt;a href=&quot;http://www.imc-math.org.uk/&quot; rel=&quot;nofollow&quot;&gt;IMC&lt;/a&gt;: it seems quite unlikely that you are not aware of the IMC website. Are you looking for solutions other than the one given there? If so, it would be better to explicitly mention this in your questions.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-12T04:39:42.123" CommentCount="2" />
  <row Id="2220" PostTypeId="2" ParentId="2188" CreationDate="2010-08-12T04:54:50.640" Score="1" ViewCount="0" Body="&lt;p&gt;Hi damiano, &#xA;thank you for post the link. I do not knew such exercise and the solution is more simple because it not invoke the Carathéodory-Toeplitz Theorem, that in this case I don't know if it can be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am sorry, for post this as an answer, but I have not enough reputation to post comments)&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-08-12T04:54:50.640" CommentCount="1" />
  <row Id="2221" PostTypeId="1" CreationDate="2010-08-12T05:30:45.557" Score="6" ViewCount="255" Body="&lt;p&gt;Suppose that $c_k$ is an decreasing sequence of non-negative real numbers, such that $c_0=1$&#xA;and $c_{k}\leq \frac{1}{2}(c_{k-1}+c_{k+1})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that the generated function of $c_k$ admit an integral representation as below&#xA;$$&#xA;\sum_{k=0}^{\infty}c_kz^k=\int_{\partial\mathbb D} \frac{1}{1-\zeta z}d\mu(\zeta),&#xA;$$&#xA;where $\mu$ is a Borel Probability Measure in $\partial\mathbb D$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Motivation:&lt;/b&gt; This question is related a possible slight different solution for the question &#xA;asked in &lt;a href=&quot;http://math.stackexchange.com/questions/2188/complex-analysis-question&quot;&gt;http://math.stackexchange.com/questions/2188/complex-analysis-question&lt;/a&gt; whose the answer, as pointed out by damiano, can be found at the IMC website.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="633" LastEditorDisplayName="" LastEditDate="2010-08-12T11:56:39.613" LastActivityDate="2010-08-24T21:46:14.487" Title="Convex sequences and Integral representation for the generating function " Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2222" PostTypeId="2" ParentId="154" CreationDate="2010-08-12T07:14:29.090" Score="4" ViewCount="0" Body="&lt;p&gt;As phrased, your question invites a philosophical answer but you said that's not what you want. One way to approach this question is to ask if complex numbers correspond to anything in the real world. What can you count or measure with them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, complex numbers are most directly useful for measuring things that rotate or oscillate. They're used by electrical engineers because voltages and currents can oscillate, and could be used for measuring springs or pendulums, or for anything that behaves like a wave. There are not that many situations in day-to-day life where you'd use complex numbers, but they're used extensively in physics because waves and oscillations show up everywhere.&lt;/p&gt;&#xA;" OwnerUserId="1086" LastActivityDate="2010-08-12T07:14:29.090" />
  <row Id="2223" PostTypeId="2" ParentId="535" CreationDate="2010-08-12T09:39:01.460" Score="0" ViewCount="0" Body="&lt;p&gt;One way to look at your problem is that you are trying to find the graph of an equation p(x,c)=0. In other words, you are trying to solve an implicit function equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the solutions to implicit polynomial equations in two variables are called Algebraic Varieties, and can be very complicated (http://en.wikipedia.org/wiki/Algebraic_variety). So you can't really say necessarily what kind of curve you will get. However, maybe for this specific polynomial it is a simple curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be said is that if you know p(x, c) = 0 for some given (x,c), then for all points d near enough to c one can find a root x(d) such that p(x(d), d) = 0. This is called the implicit function theorem, and another result of the theorem is that x(d) is even differentiable with respect to d.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-12T09:39:01.460" />
  <row Id="2224" PostTypeId="2" ParentId="1236" CreationDate="2010-08-12T09:44:03.210" Score="0" ViewCount="0" Body="&lt;p&gt;The following page should be of some help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diffusion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Diffusion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-12T09:44:03.210" CommentCount="1" />
  <row Id="2225" PostTypeId="1" AcceptedAnswerId="2228" CreationDate="2010-08-12T10:36:39.460" Score="1" ViewCount="283" Body="&lt;p&gt;If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function and satisfies $f(x)=f(2x+1)$, then its not to hard to show that $f$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is suppose $f$ is continuous and it satisfies $f(x)=f(2x+1)$, then can the domain of $f$ be restricted so that $f$ doesn't remain a constant. If yes, then give an example of such a function.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T12:35:48.427" Title="Function Satisfying $f(x)=f(2x+1)$" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2226" PostTypeId="2" ParentId="2178" CreationDate="2010-08-12T10:42:13.323" Score="1" ViewCount="0" Body="&lt;p&gt;b) consider a path $v_0\dots v_k$ of length $k$ (existing for part a) and for Chandru's construction) such that $v_i \neq v_j$ $\forall i\neq j$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $v_0$ and $v_k$ are adjacent, you have a cycle of lenght $k+1$.&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, since $deg(v_k)\geq k$, $v_k$ has at least one adjacent vertex $v_l$ with $l&gt;k$ and you can extend the path with this new vertex. If $v_l$ is adjacent with $v_0$ or $v_1$ you're done, otherwise you can extend the path again with a vertex that doesn't belong to the path yet. Since $G$ has only a finite number of vertices, at some point you will find a vertex $v_n$ whose adjacent vertices are all part of the path. $deg(v_n)\geq k$, so $v_n$ is adjacent to $v_i$ for some $i&amp;lt; n-k$ and you can finally close the path with a cycle of lenght $n-i \geq k+1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1065" LastEditorDisplayName="" LastEditDate="2010-08-12T12:58:45.760" LastActivityDate="2010-08-12T12:58:45.760" CommentCount="1" />
  <row Id="2227" PostTypeId="1" AcceptedAnswerId="2348" CreationDate="2010-08-12T10:46:33.783" Score="0" ViewCount="252" Body="&lt;p&gt;Can one think of an example of a continuous $2\pi$ periodic function whose Fourier series fails to converge on $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I referred this in the wikipedia page but no avail: &lt;strong&gt;It might be interesting to note that Jean-Pierre Kahane and Yitzhak Katznelson proved that for any given set E of measure zero, there exists a continuous function ƒ such that the Fourier series of ƒ fails to converge on any point of E.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-13T14:16:42.610" LastActivityDate="2010-08-13T14:16:42.610" Title="Example of a function whose Fourier Series fails  to converge at One point" Tags="&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2228" PostTypeId="2" ParentId="2225" CreationDate="2010-08-12T10:56:07.593" Score="5" ViewCount="0" Body="&lt;p&gt;Let $f$ have value $1$ on $[0,\infty)$ and value $0$ on $(-\infty,-1]$. This function is not constant (although it is locally constant), and satisfies $f(x)=f(2x+1)$ whenever $x$ is in its domain. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-12T10:56:07.593" CommentCount="4" />
  <row Id="2229" PostTypeId="1" AcceptedAnswerId="2246" CreationDate="2010-08-12T10:56:17.673" Score="6" ViewCount="183" Body="&lt;p&gt;Using the duality between locally compact Hausdorff spaces and commutative $C^*$-algebras one can write down a vocabulary list translating topological notions regarding a locally compact Hausdorff space $X$ into algebraic notions ragarding its ring of functions $C_0(X)$ (see Wegge-Olsen's book, for instance). For example, we have the following correspondences:&#xA;\begin{align*}&#xA;\text{open subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{ideal in $C_0(X)$}\newline&#xA;\text{dense open subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{essential ideal in $C_0(X)$}\newline&#xA;\text{closed subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{quotient of $C_0(X)$}\newline&#xA;\text{locally closed subset of $X$}\quad &amp;amp;\longleftrightarrow\quad\text{subquotient of $C_0(X)$}\newline&#xA;\text{???}\quad &amp;amp;\longleftrightarrow\quad\text{$C^*$-subalgebra in $C_0(X)$}&#xA;\end{align*}&#xA;By ideal I always mean a two-sided closed (and hence self-adjoint) ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I can't quite see &lt;strong&gt;how to reconvert a $C^*$-subalgebra in $C_0(X)$ into something topological involving only the space $X$.&lt;/strong&gt; Can you come up with something handy?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; A simple example of a subalgebra of a commutative $C^*$-algebra not being an ideal is&#xA;$$&#xA;\mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Alternatively, we could think about this question within the duality of affine algebraic varieties and finitely generated commutative reduced algebras or even within the duality between affine schemes and commutative rings.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Since I was not completely satisfied by the response I got here, I reposted this question on &lt;a href=&quot;http://mathoverflow.net/questions/35507/what-is-the-commutative-analogue-of-a-c-subalgebra&quot; rel=&quot;nofollow&quot;&gt;MO&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-15T13:17:56.627" LastActivityDate="2010-08-15T13:17:56.627" Title="What is the commutative analogue of a $C^*$-subalgebra?" Tags="&lt;functional-analysis&gt;&lt;noncommutative-geometry&gt;&lt;c-star-algebras&gt;&lt;noncommutative-topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2231" PostTypeId="2" ParentId="2130" CreationDate="2010-08-12T11:29:04.123" Score="1" ViewCount="0" Body="&lt;p&gt;If a category is not small, its presheaf category is not locally small and this can lead to all sorts of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not despair though, as there are several work-arounds. The best of all may simply be to pretend the problem does not exist and proceed naively; 99% percent of the time this works like a charm as one can always clean up the act through the use of more or less contorted devices / axioms.&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-12T11:29:04.123" />
  <row Id="2232" PostTypeId="2" ParentId="2227" CreationDate="2010-08-12T11:55:20.037" Score="6" ViewCount="0" Body="&lt;p&gt;No. Any continuous function is in $L^2$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Carleson-Hunt theorem&lt;/a&gt; states that an $L^2$ function's Fourier series converges almost everywhere to the function.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-12T11:55:20.037" CommentCount="5" />
  <row Id="2233" PostTypeId="1" AcceptedAnswerId="2245" CreationDate="2010-08-12T11:59:35.550" Score="-1" ViewCount="149" Body="&lt;p&gt;This question has been in my head for a long time, but i think now is the time to ask some great people who are here, doing a wonderful job of making Mathematics more pleasurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways of proving a Mathematical problem: One is direct way of proving the problem and the other one is through contradiction. I would like to know as to which one is better in getting results quicker. Is this dependent on the type of problem which we are solving or is it independent of what we do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;G.H. Hardy once remarked : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reductio ad absurdum, which Euclid loved so much, is one of a mathematician's finest weapons. It is a far finer gambit than any chess play: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a student i don't really find any differences between the two methods of solving a problem, but i am curious to know other people's opinion.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="76" LastEditorDisplayName="" LastEditDate="2010-08-12T16:48:13.383" LastActivityDate="2010-08-12T16:48:13.383" CommunityOwnedDate="2010-08-12T16:48:13.383" Title="Methodology of Mathematical Proofs" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="2234" PostTypeId="1" AcceptedAnswerId="2236" CreationDate="2010-08-12T12:31:33.753" Score="2" ViewCount="595" Body="&lt;p&gt;some guy asked for formula for sum of harmonic progression.&#xA;So i have done some calculation and give him some approx formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;integration of 1/x from 1 to n = 1/2(y1+y2)+1/2(y2+y3)..........1/2(yn-1+yn)               where yi is ith term of HP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ln(n) = (y1)/2+y2+y3..............(yn)/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sum of HP = ln(n)+(first term+last&#xA;  term)/2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;e.g. 1+1/2+..............+1/10 = 2.8525                  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;actual result = 2.9289&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is how to correct this formula&lt;/p&gt;&#xA;" OwnerUserId="1093" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T03:25:28.067" LastActivityDate="2010-10-27T21:17:37.567" Title="sum of harmonic progression?" Tags="&lt;sequences-and-series&gt;" AnswerCount="4" />
  <row Id="2235" PostTypeId="2" ParentId="2225" CreationDate="2010-08-12T12:35:48.427" Score="4" ViewCount="0" Body="&lt;p&gt;As in &lt;a href=&quot;http://mathoverflow.net/questions/31990/continuous-functions-remaining-constant&quot; rel=&quot;nofollow&quot;&gt;the previous proof&lt;/a&gt; of $f$ being constant on $\mathbb{R}$, define $g(x) = f(x-1)$, so that $g(x) = g(2x)$; the domains of $f$ and $g$ are just shifted versions of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, if the domain of $g$ is small enough, say $[2,3]$, then $g$ can be any continuous function, because the domain contains no $x$ and $2x$ at the same time. A more interesting question is: how large can we make the domain so that $g$ will still not be constant? The answer to this is suggested by &lt;a href=&quot;http://math.stackexchange.com/questions/2225/function-satisfying-fxf2x1/2228#2228&quot;&gt;JDH's answer&lt;/a&gt;: if we remove only the single point 0, making the domain $\mathbb{R} \setminus \{0\}$, it is disconnected into two components which can independently have constant values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big can a domain be on which $g$ is not even locally constant? Remove an arbitrarily small interval around $0$. Take any non-constant continuous function $h$ which is periodic with unit period, and let $g(x) = h(\log_2 x)$. Then $g(x) = g(2x)$ for all $x$, and is continuous everywhere.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-12T12:35:48.427" />
  <row Id="2236" PostTypeId="2" ParentId="2234" CreationDate="2010-08-12T12:35:49.753" Score="8" ViewCount="0" Body="&lt;p&gt;The partial sums of the harmonic series are called &quot;harmonic numbers.&quot; The difference between the nth harmonic number and ln(n) tends to a limit as n increases, and that limit is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_constant&quot; rel=&quot;nofollow&quot;&gt;Euler's constant&lt;/a&gt; or gamma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a great book about all this called &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691141339&quot; rel=&quot;nofollow&quot;&gt;Gamma: Exploring Euler's Constant&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-08-13T00:41:43.427" LastActivityDate="2010-08-13T00:41:43.427" CommentCount="2" />
  <row Id="2237" PostTypeId="1" AcceptedAnswerId="2259" CreationDate="2010-08-12T13:19:17.043" Score="3" ViewCount="149" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the answer to the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am playing poker and am dealt two cards neither of which are spades, what are the chances of two or more of the flop cards (the flop is the first three community cards) being spades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answered this question in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have no spades, there are still 13 spades unseen and a total of 50 cards remaining unseen (52 cards in the pack less my two cards which I've been dealt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the probability of the first flop card being a spade is p(A) and the second p(B) and the third p(C) then I need to find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p((A&amp;cap;B)&amp;cup;(A&amp;cap;C)&amp;cup;(B&amp;cap;C))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so given the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(X&amp;cup;Y&amp;cup;Z) = p(X) + p(Y) + p(Z) - p(X&amp;cap;Y) - p(X&amp;cap;Z) - p(Y&amp;cap;Z) + p(X&amp;cap;Y&amp;cap;Z)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(2 or more spades on flop) = 3(13/50 x 12/49) - 3((13/50 x 12/49) x 11/48) + (13/50 x 12/49 x 11/48)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives &lt;strong&gt;16.18%&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I wanted to do next was solve the question using combinations. Through a process of trial and error I got the same answer, but am a bit confused as to why it works this way. My method was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13C2 * 48 = 3744&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3744 - (3 * 13C3) + 13C3 = 3172&lt;/p&gt;&#xA;&#xA;&lt;p&gt;total number possible 3 card combinations on flop = 50C3 = 19600&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3172 / 19600 = &lt;strong&gt;16.18%&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I get the total number of 2 card combination of spades by 13C2 = 78. Then there are 48 other cards which can form combinations with each of these 2 card combination so 13C2 * 48.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm losing my grasp of whats happening. Why does 13C2 = p(A) + p(B) + p(C)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I have done to follow is find the number of combos for p(A&amp;cap;B), p(A&amp;cap;C), p(B&amp;cap;C) and p(A&amp;cap;B&amp;cap;C) and then crunched the numbers as in the calculation shown above. But even though I (think I've) got the right answer, I've come to it in a try-it-and-see way and can't understand for instance why 13C2 * 48 doesn't give the right answer straight off. Surely that gives the number of 3 card combinations on the flop with 2 or more spades which is what I'm looking for, doesn't it? But I don't get the same answer as with the original calculation if I use that figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the waffle. I hope someone can see where my thinking is going wrong and can help explain the combinations part a bit better for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1094" LastEditorUserId="1094" LastEditorDisplayName="" LastEditDate="2010-08-12T16:29:32.857" LastActivityDate="2010-08-12T17:05:38.147" Title="I've got the same answer to my question via probability and combinations, but I don't know why!" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2238" PostTypeId="1" AcceptedAnswerId="2244" CreationDate="2010-08-12T13:20:59.503" Score="2" ViewCount="84" Body="&lt;p&gt;There are $N&gt;1$ points in a plane. Consider the set of all distances between pairs of the points. Let $n$ be the number of elements of this set.&#xA;We know that $$n \leq {N \choose 2}.$$&#xA;What is the lower bound estimate on $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of clearity, if points are $A_1, A_2, \dots, A_N$, and $S = \{d(A_i, A_j)| 1\leq i &amp;lt; j \leq N \}$, then $n = |S|$.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-13T06:53:59.167" LastActivityDate="2010-08-13T06:53:59.167" Title="Lower bound for a set of distances between pairs of points in a plane" Tags="&lt;discrete-geometry&gt;&lt;erdos-problem&gt;&lt;extremal-combinatorics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2239" PostTypeId="1" AcceptedAnswerId="2247" CreationDate="2010-08-12T13:23:30.853" Score="4" ViewCount="229" Body="&lt;p&gt;Despite appearances, this is not a question on computational aspects of number theory. The background is as follows. I once asked a number theorist about what he considered to be the most important unsolved problems in arithmetic geometry. He told me about a few, but along with some well-known problems he told me the following one also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to determine conceptually when a number is squarefree or not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I protested that this sounded like a computational question, he told me that no, this is not so, and demonstrated that this has a rather nice solution for the ring of polynomials over a field, which is in many senses analogous to the ring of integers. Take the polynomial, take its derivative and compute the gcd using the Euclidean algorithm. But for the ring of integers there is nothing analogous to the derivative, and he wanted a solution of the problem by constructing a good notion of a differential in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What are the known investigations along this line? What well-known topics in arithmetic geometry are related to this? And what would be some other interesting consequences of a successful development of such a method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other comments that might enlighten me further would be received with gratitude.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-12T14:07:17.677" LastActivityDate="2010-08-12T16:48:22.523" Title="Importance of determining whether a number is squarefree, using geometry." Tags="&lt;arithmetic-geometry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2240" PostTypeId="1" AcceptedAnswerId="2241" CreationDate="2010-08-12T13:30:08.577" Score="1" ViewCount="130" Body="&lt;p&gt;We know that $(\mathbb{Q},+,\times)$ is a subfield of $(\mathbb{R},+,\times)$. It is easy to see that the automorphism of $\mathbb{Q}$ is only the identity. For a quick proof, lets go through the main steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(1)=1 \Longrightarrow f(n)=n$ for all $n \in \mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(-1)=-1$ which says that $f(x)=x$ for all $x \in \mathbb{Z}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle f \Bigl(\frac{p}{q}\Bigr) = \frac{p}{q}$, where $q \neq 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One, then uses the continuity of $f$ and the denseness of $\mathbb{Q}$ to prove that the Automorphism of $\mathbb{R}$ is also trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question&lt;/strong&gt; Given a subfield $K$ of $\mathbb{C}$ is it and an automorphism of $K$ can it be extended to the whole of $\mathbb{C}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-12T13:39:20.630" LastActivityDate="2010-08-12T19:51:29.600" Title="Extending Automorphism of a Field." Tags="&lt;field-theory&gt;" AnswerCount="3" />
  <row Id="2241" PostTypeId="2" ParentId="2240" CreationDate="2010-08-12T13:35:57.453" Score="5" ViewCount="0" Body="&lt;p&gt;Firstly, you need a little more to show that the only automorphism of $\mathbb R$ is the identity. You have to prove that any such automorphism is continuous. For a proof, let $x$ be a positive number; so it is the square of some number; so it is taken to another positive number under the automorphism. Since positive numbers go to positive numbers, the automorphism is order-preserving, and hence the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, this result is not true if you replace $\mathbb R$ by $\mathbb C$. The automorphism group of $\mathbb C$ over $\mathbb Q$ is uncountable. This is because $\mathbb C$ can be constructed by attaching uncountably many algebraicaly-independent transcendentals to $\mathbb Q$ and then taking the algebraic closure. Any automorphism of $\mathbb Q$ can be extended to an automorphism of $\mathbb C$, by first extending to the algebraic closure, and then to $\mathbb C$. The same argument would work for any algebraic number field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More general subfields are obtained as follows: You attach some number of transcendentals to $\mathbb Q$, and then take an algebraic extension of it. An automorphism of this field can be similarly extended to the whole of $\mathbb C$ as above.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-12T13:45:01.363" LastActivityDate="2010-08-12T13:45:01.363" CommentCount="2" />
  <row Id="2243" PostTypeId="2" ParentId="2239" CreationDate="2010-08-12T13:49:05.317" Score="4" ViewCount="0" Body="&lt;p&gt;The problem of finding some sort of analogue of differentiation of polynomials, in the context of integers, is a celebrated and difficult one.&#xA;I don't have anything useful to say about it, except to remark that the standard context in which this problem arises is that of the ABC conjecture: namely, the polynomial analogue of the ABC conjecture is a theorem, and one of the tools in its proof involves differentiating polynomials.  This has led people to try to construct some sort of analogous tool over the integers, with the goal of proving the ABC conjecture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a short article of Faltings (called &quot;Does there exist an arithmetic Kodaira--Spencder class&quot;) in which he writes down some reasonable axioms such a notion of differentiation would satisfy (in scheme theoretic language over Spec $\mathbb Z$) and shows that it is impossible to satisfy them.  Nevertheless, one might (and many do!) hope for some more subtle structure, more complicated than the direct analogue that Faltings rules out, but still having the same basic nature as a derivative.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-12T14:04:46.263" LastActivityDate="2010-08-12T14:04:46.263" CommentCount="5" />
  <row Id="2244" PostTypeId="2" ParentId="2238" CreationDate="2010-08-12T13:54:34.057" Score="5" ViewCount="0" Body="&lt;p&gt;You might want to read &lt;a href=&quot;http://www.math.ucsd.edu/~fan/mypaps/fanpap/67distances.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract of this article says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classical problem in combinatorial geometry is that of determining&#xA;the minimum number $f(n)$ of different distances determined by $n$ points in&#xA;the Euclidean plane. In 1952, L. Moser proved that $f(n) &gt; \frac{n^{2/3}}{2\cdot3^{1/3}} - 1$ and this has remained for 30 years as the best lower bound known for f(n). It is shown that $f(n) &gt; cn^{5/7}$ for some fixed constant c.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-12T13:54:34.057" />
  <row Id="2245" PostTypeId="2" ParentId="2233" CreationDate="2010-08-12T14:03:42.257" Score="3" ViewCount="0" Body="&lt;p&gt;Making an argument is a complicated business, and you do what you have to do, using whatever tools you can bring to hand.  One advantage of arguing by contradiction is that you give yourself something to apply your tools to.  E.g. in the proof of the odd order theorem by Feit and Thompson (to take a famous example), you give yourself a minimal counterexample, i.e. a non-abelian simple group of odd order, every proper subgroup of which is solvable.  (And you then try to prove that this group can't exist!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you something to work with: you have this minimal counterexample, with quite specific properties at hand, and you can try to prove things about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, this is one reason why mathematicians phrase proofs as proofs by contradiction, even in cases where they don't need to: for psychological reasons, it helps to have something like the minimal counterexample at hand; you now have an object in front of you that you can reason with.  (Even though in the end, you will prove that this object can't exist.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not all arguments proceed this way: for example, the proof of the Shimura--Taniyama conjecture by Wiles and Taylor--Wiles (the result which implies FLT) isn't based around a proof by contradiction.  (Of course, proofs of big results like this are extremely long and involved, with many local arguments, some of which may be by contradiction.  But the overall thrust of the argument is direct, unlike the Feit and Thompson argument.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-12T14:03:42.257" />
  <row Id="2246" PostTypeId="2" ParentId="2229" CreationDate="2010-08-12T14:08:25.817" Score="5" ViewCount="0" Body="&lt;p&gt;Roughly, the answer will be that closed C*-subalgebras will correspond to quotient spaces&#xA;(via pull-back of functions).  In your example, the quotient map is one which identifies&#xA;the two points into a single point.   I haven't thought through, though, whether this is a completely correct statement as it stands, or whether one has to add additional caveats.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added to answer a question in comments:]  The idea is that if $X$ surjects onto $Y$&#xA;then we get an injection $C_0(Y) \to C_0(X)$, and conversely.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Additional discussion added after more thought:] Let me say something about the analogous situation in algebraic geometry, where I am more comfortable with the technical issues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Affine algebraic sets over $\mathbb C$ correspond to finite type reduced $\mathbb C$-algebras.&#xA;Giving an inclusion $A \hookrightarrow B$ of finite type reduced $\mathbb C$-algebras&#xA;corresponds to giving a map $X \to Y$ of algebraic sets which is dominant, i.e. the image is dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in your set-up: if $X \to Y$ is a map of locally compact Hausdorff spaces with dense image, then again the map $C_0(X) \to C_0(Y)$ will be injective; so I might have been too hasty when I asserted that we get a surjective map.  On the other hand, perhaps the&#xA;image of $C_0(X) \to C_0(Y)$ will not be closed in this level of generality; it's a while since I've thought carefully about these sorts of things, so I don't think I can say more&#xA;right now with any certainty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm not so used to working in the case of rings without unit, so my suggestion has more chance to be correct in the case when the spaces are compact.&#xA;So perhaps it would be easiest to think about the case when $X$ and $Y$ are compact first; note then a map with dense image will automatically be surjective, and so this case might be simpler to understand for this reason too.  (In fact, thinking about your example of an ideal that you mention in comments, it might be easier to pass to one-point compactifications --- and thus add a unit --- before proceeding.  Because indeed I think&#xA;that in the ideal case, what will happen is that we will get a map from the 1 point compactification of the space to the one point compactification of the open set which crushes the complement of the open set down to the point at infinity, exactly as you suggest in your comment.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-12T14:36:26.687" LastActivityDate="2010-08-12T14:36:26.687" CommentCount="3" />
  <row Id="2247" PostTypeId="2" ParentId="2239" CreationDate="2010-08-12T14:11:09.117" Score="3" ViewCount="0" Body="&lt;p&gt;The standard example is the proof, by differentiation, of Fermat's last theorem for polynomials.  More generally, the proof of the ABC conjecture for polynomials (Mason's theorem) by differentiation.   In light of the analogies between algebraic number theory and algebraic geometry this suggests the hope that some kind of arithmetic differentiation, if it exists, could be a missing structure leading to breakthroughs such as a proof of the ABC conjecture or a simple proof of Fermat.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As somebody once said, &quot;define PDE's over a number field and you'll be a rich man&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know the furthest-developed approach to this problem at present is in a series of works by Alexander Buium on arithmetic differential operators: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.unm.edu/~buium/prebook.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.unm.edu/~buium/prebook.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-12T14:11:09.117" CommentCount="2" />
  <row Id="2248" PostTypeId="1" AcceptedAnswerId="2301" CreationDate="2010-08-12T14:36:27.107" Score="3" ViewCount="252" Body="&lt;p&gt;Bernhard Elsner, alias MathOMan, posted this exercise in plane Geometry, &lt;a href=&quot;http://www.mathoman.com/en/index.php/1529-theorem-about-a-circle-three-chords-and-a-midpoint&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Theorem about a circle, three chords and a midpoint&lt;/em&gt;&lt;/a&gt; on January 29th, 2010.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let $\mathcal{C}$ be a circle, $A,B$ two distinct points on $\mathcal{C}$ and $M$ be the midpoint of the chord $[AB]$. Take two other chords,$[PQ]$ and $[SR]$, that pass through $M$. Let $C$ (resp. $D$) be the intersection of $[AB]$ with $[PS]$ (resp. $[RQ]$).&#xA;Prove that $M$ is the midpoint of the chord $[CD]$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.latroika.com/mathoman/pix/exercice-cercle-cordes-milieu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove it I've written the following (failed) argument, in the &lt;a href=&quot;http://www.mathoman.com/de/index.php/1529-verschiedene-sehnen-in-einem-kreis&quot; rel=&quot;nofollow&quot;&gt;German version&lt;/a&gt; of this post (translation of mine):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://calcauxprobteor.files.wordpress.com/2010/04/cordas.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure is symmetric with respect to $M$: $\overline{AM}=\overline{MB}$, $\overline{PM}=\overline{MU}$, $\overline{RM}=\overline{MW}$, $\overline{QM}=\overline{MV}$, $\overline{QR}=\overline{VW}$, $\overline{SM}=\overline{MT}$. From $\dfrac{\overline{SC}}{\overline{DT}}=\dfrac{\overline{CM}}{\overline{MD}}=1$ follows that $\overline{CM}=\overline{MD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an extract of the author's reply (translation of mine): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is not clear that  $\overline{QM}=\overline{MV}$. Is the point $V$ on the line $(QM)$ defined by this equality or is $V$ defined as the intersection point of the lines $(QM)$ and $(WC)$? Why are both definitions to give the same point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C^{\prime }$ be the intersection of $(SP)$ and $(VW)$, and $D^{\prime }$&#xA;the intersection of $(TU)$ and $(QR)$. (...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One still has to show that $C^{\prime }=C$ and $D^{\prime }=D$.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have agreed with these objections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now no proof has been posted. The author considers that the &quot;proof is not quite simple&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q. What is the theorem this exercise refers to? Or how does one prove it?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-12T23:47:43.147" LastActivityDate="2010-08-12T23:47:43.147" Title="How to prove the midpoint of a chord is also the midpoint of the line segment defined by the points of intersection of other two chords with it?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="0" />
  <row Id="2249" PostTypeId="2" ParentId="2239" CreationDate="2010-08-12T15:06:11.030" Score="9" ViewCount="0" Body="&lt;p&gt;No feasible (polynomial time) algorithm is currently known for &#xA;recognizing squarefree integers or for computing the squarefree &#xA;part of an integer. In fact it may be the case that this problem &#xA;is no easier than the general problem of integer factorization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is important because one of the main tasks &#xA;of computational algebraic number theory reduces to it (in &#xA;deterministic polynomial time). Namely the problem of computing &#xA;the ring of integers of an algebraic number field depends upon &#xA;the square-free decomposition of the polynomial discriminant &#xA;when computing an integral basis, e.g. [2] S.7.3 p.429 or [1] &#xA;This is due to Chistov [0]. See also Problems 7,8, p.9 in [3], &#xA;which lists 36 open problems in number theoretic complexity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that such problems are simpler in function fields versus number fields&#xA;is due to the availability of derivatives. This opens up a powerful&#xA;toolbox that is not available in the number field case. For example&#xA;once derivatives are available so are Wronskians - which provide powerful&#xA;measures of dependence in transcendence theory and diophantine approximation.&#xA;A simple yet stunning example is the elementary proof of the polynomial case of Mason's ABC theorem, which yields as a very special case FLT for polynomials, cf.&#xA;my recent &lt;a href=&quot;http://mathoverflow.net/questions/35147&quot; rel=&quot;nofollow&quot;&gt;MO post&lt;/a&gt; and my old &lt;a href=&quot;http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post [4]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[0] A. L. Chistov. The complexity of constructing the ring of integers&lt;br&gt;&#xA;of a global field. Dokl. Akad. Nauk. SSSR, 306:1063--1067, 1989.&lt;br&gt;&#xA;English Translation: Soviet Math. Dokl., 39:597--600, 1989. 90g:11170&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/showciting?cid=854849&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/showciting?cid=854849&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Lenstra, H. W., Jr.  Algorithms in algebraic number theory.&lt;br&gt;&#xA;Bull. Amer. Math. Soc. (N.S.) 26 (1992), no. 2, 211--244.  93g:11131&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.105.8382&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.105.8382&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Pohst, M.; Zassenhaus, H.  Algorithmic algebraic number theory.&lt;br&gt;&#xA;Cambridge University Press, Cambridge, 1997.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Adleman, Leonard M.; McCurley, Kevin S.&lt;br&gt;&#xA;Open problems in number-theoretic complexity. II.&lt;br&gt;&#xA;Algorithmic number theory (Ithaca, NY, 1994), 291--322,&lt;br&gt;&#xA;Lecture Notes in Comput. Sci., 877, Springer, Berlin, 1994. 95m:11142&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.48.4877&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.48.4877&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[4] sci.math.research, 1996/07/17&lt;br&gt;&#xA; poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1]&lt;br&gt;&#xA;&lt;a href=&quot;http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed&quot; rel=&quot;nofollow&quot;&gt;http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T16:48:22.523" LastActivityDate="2010-08-12T16:48:22.523" />
  <row Id="2250" PostTypeId="1" AcceptedAnswerId="2380" CreationDate="2010-08-12T15:24:39.323" Score="0" ViewCount="175" Body="&lt;p&gt;Given the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;an $(n \times z)$ matrix $A = {(a_1,a_2, ... ,a_n)}^{T}$ where $z \geq n$ and every $a_i$ is a $z$-dimensional row vector.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$a_i = [a_{i1}  a_{i2} ...  a_{iz}]$ where $a_{ij} \geq 0,  \forall j$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sum_{i=1}^{z}a_{ri} = 1, \forall r \in ${$1,2,...,n$}.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sum_{i=1}^{z}|a_{pi} - a_{qi}| \leq \epsilon, \forall p,q$ where $\epsilon &amp;lt;&amp;lt; 1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Find a provable upper bound on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum_{i,j=1}^{z}|(1/n)*\sum_{k=1}^{n}[a_{ki}.(a_{f(k)j} - a_{g(k)j})]|$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where f and g are permutations over the set {$1,2,...,n$} such that $f(i) \neq g(i) \forall i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am expecting the bound to be $\epsilon^2$ but I have no idea how to prove it.&lt;/p&gt;&#xA;" OwnerUserId="938" LastEditorUserId="938" LastEditorDisplayName="" LastEditDate="2010-08-14T08:49:27.763" LastActivityDate="2010-08-14T08:49:27.763" Title="a provable upper bound on the summation" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2251" PostTypeId="1" AcceptedAnswerId="2253" CreationDate="2010-08-12T15:51:04.227" Score="6" ViewCount="144" Body="&lt;p&gt;Is it guaranteed that there will be some p s.t. p|2^n-1 but not p|2^m-1 for any m&#xA;&#xA;&lt;p&gt;In other words, does each 2^x-1 introduce a &lt;em&gt;new&lt;/em&gt; prime factor?&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-12T16:35:58.173" Title="a prime factor of 2^n-1 which is not a factor of 2^m-1 for m &lt; n" Tags="&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2252" PostTypeId="2" ParentId="2251" CreationDate="2010-08-12T16:12:37.367" Score="1" ViewCount="0" Body="&lt;p&gt;No. $2^6-1 = 3^2 \cdot 7$. But we have that $3|2^2-1$ and $7|2^3-1$&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2010-08-12T16:12:37.367" />
  <row Id="2253" PostTypeId="2" ParentId="2251" CreationDate="2010-08-12T16:16:30.310" Score="5" ViewCount="0" Body="&lt;p&gt;Yes, it's true (except for $2^6-1=7\times 9$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is known as Bang's theorem, and is a corollary of &lt;a href=&quot;http://en.wikipedia.org/wiki/Zsigmondy%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Zsigmondy's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a proof &lt;a href=&quot;http://www.ams.org/journals/proc/1997-125-07/S0002-9939-97-03981-6/S0002-9939-97-03981-6.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (Theorem 3).&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-12T16:16:30.310" />
  <row Id="2254" PostTypeId="1" AcceptedAnswerId="2257" CreationDate="2010-08-12T16:20:04.363" Score="2" ViewCount="159" Body="&lt;p&gt;Let $H,K$ be proper subgroups of a group $G$ having a complete set $S$ of representatives of left cosets in common, that is,&#xA;$$&#xA;G = \bigsqcup_{s \in S} sH = \bigsqcup_{s \in S} sK&#xA;$$&#xA;It seems in general one cannot expect any serious relation on $H,K.$ But I am afraid of overlooking some general result here. Any information on the subject will be warmly accepted. Regards, Olod&lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-08-12T18:25:36.237" LastActivityDate="2010-08-12T18:25:36.237" Title="If two subgroups have a complete set of left coset representatives in common, then ..." Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2255" PostTypeId="2" ParentId="2251" CreationDate="2010-08-12T16:29:45.867" Score="3" ViewCount="0" Body="&lt;p&gt;This result is due to Zsigmondy (1892), with special cases (b=1) &#xA;(re)discovered by Bang (1886), Birkhoff &amp;amp; Vandiver (1904) ... see &#xA;Ribenboim, The New Book of Prime Number Records, p. 43, 67-68, 338, 437. &#xA;Vandiver, a &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Vandiver.html&quot; rel=&quot;nofollow&quot;&gt;prolific researcher&lt;/a&gt; on FLT,&#xA;applied it to Fermat's Last Theorem (first case) and related diophantine equations, &#xA;e.g. see Ribenboim's book 13 Lectures on FLT, pages 52,&lt;em&gt;161&lt;/em&gt;,206,234,236. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such results have many applications: a MathSciNet search on &#xA;Anywhere=(Zsigmondy or (Birkhoff and Vandiver))  will find &#xA;over 35 related Math Reviews. Schinzel (1974, MR 93k:11107) &#xA;extended the theorem to arbitrary algebraic number fields $K$: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $A$ and $B$ are algebraic integers in $K$, $(A,B)=1$, and $A/B$ is of degree $d$ &#xA;  and not a root of unity, then there exists  $n_0 = n_0(d)$ such that &#xA;  for all  $n &gt; n_0,  A^n - B^n $ has a prime ideal factor $P$&lt;br&gt;&#xA;  not dividing   $A^m - B^m $ for all $ m &amp;lt; n$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later (1993, same MR) &quot;he generalized this to show that every &#xA;algebraic number which is not a root of unity satisfies only a &#xA;finite number of independent generalized cyclotomic equations &#xA;considered by the reviewer [in Structural properties of &#xA;polylogarithms, Chapter 11, see p. 236, Amer. Math. Soc., &#xA;Providence, RI, 1991; see MR 93b:11158]&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;http://arxiv.org/abs/math/0409540&quot; rel=&quot;nofollow&quot;&gt;elliptic&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/1002.4829&quot; rel=&quot;nofollow&quot;&gt;polynomial&lt;/a&gt; generalizations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T16:35:58.173" LastActivityDate="2010-08-12T16:35:58.173" />
  <row Id="2256" PostTypeId="2" ParentId="2254" CreationDate="2010-08-12T16:41:39.040" Score="2" ViewCount="0" Body="&lt;p&gt;If $S$ is a complete set of (left) coset representatives for $H$, then for every $x\in G$ there exists $s\in S$ such that $xH=sH$, and moreover, if $s_1,s_2\in S$ are such that $s_1H=s_2H$, then $s_1=s_2$. That is: there is one and exactly one representative from each coset of $H$ in $G$ in the set $S$. As such, your condition is trivially satisfied under the assumption that $S$ is a complete set of coset representatives for both $H$ and $K$, since for $s_1,s_2\in S$, you have&#xA;$$s_1H=s_2H \Longleftrightarrow s_1=s_2\Longleftrightarrow s_1K=s_2K.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either you meant something else, or you are just asking for condition under which two subgroups $H$ and $K$ can have the same complete set of coset representatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack Schmidt already gave an example with $H$ and $K$ not conjugate, and asked if there is an example in which $H$ and $K$ are not isomorphic. I think this does it: take $G=\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_4$, the product of two cyclic groups of order two and one of order four. Take $H={0}\times{0}\times\mathbb{Z}_4$, and $K={0}\times\mathbb{Z}_2\times\langle 2\rangle$ (so $H$ is cyclic of order $4$, and $K$ is the Klein $4$-group). Let $S={(0,0,1), (1,0,1), (1,1,0), (0,1,0)}$. If I did not make some silly mistake, then this is a complete set of coset representatives for both $H$ and $K$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-12T17:00:55.610" LastActivityDate="2010-08-12T17:00:55.610" CommentCount="6" />
  <row Id="2257" PostTypeId="2" ParentId="2254" CreationDate="2010-08-12T16:50:50.353" Score="3" ViewCount="0" Body="&lt;p&gt;|S| = [G:H] = [G:K], so at least their indexes are equal. I don't think much else is true, because of the following example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If G = 2 × 2 is the Klein four group, then H = 2 × 1 and K = 1 × 2 are two subgroups with a common set of coset representatives: S = { (0,0), (1,1) }.  H and K are not conjugate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general if G is a semi-direct product H⋉N, then H has S=N as a set of coset representatives.  It is very possible for N to have more than one &quot;complement&quot; H, that is, another subgroup K such that G=K⋉N.  In nice situations like G nonabelian of order 6, all complements are conjugate, but in general they need not be as the dihedral 2-groups (including the Klein four group) show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice to have an example where H and K are not even isomorphic, but semi-direct products won't do that.  I don't believe H, K need to be complemented to have a common transversal, but I guess that is another reasonable guess to rule out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Well, the dihedral group G of order 8 has a cyclic normal subgroup H and Klein four normal subgroup K, the union of which is not all of G.  Since a set of coset representatives S has only 2 elements, we just need to take the identity and an element neither in H nor K.  In particular, H need not be isomorphic to K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the dihedral group of order 16 has a similar pair (H cyclic order 4, K a four-group), and so neither H nor K need be complemented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also use Abelian examples as Arturo points out, and one can extend the D8 example to S4 as Steve points out.  A fun problem.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-12T17:13:55.200" LastActivityDate="2010-08-12T17:13:55.200" CommentCount="4" />
  <row Id="2259" PostTypeId="2" ParentId="2237" CreationDate="2010-08-12T17:05:38.147" Score="1" ViewCount="0" Body="&lt;p&gt;OK, I think I've got the answer. I tripped myself up initially because I didn't really understand what I was doing when I was multplying ${13\choose2}$ by 48. I'd reasoned that if I had Ace, King of spades then I could times by 48 because there were 48 other cards to make 48 combinations with AK spades in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I now realise is that multiplying the whole of ${13\choose2}$ by 48 causes an overlap in counting. For instance if I have A7 of spades then one of the other 48 cards is the King of spades. And in fact I need to apply the Inclusion/Exclusion rules just the same as if I'd been using probability to come to the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ${13\choose2}\times48$ does equal p(A) + p(B) + p(C) and I need to use the inclusion exclusion rules to take away any combinations that have been counted twice. SO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(A&amp;cap;B) = p(A&amp;cap;C) = p(B&amp;cap;C) = ${13\choose3}$ = 286&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(A&amp;cap;B&amp;cap;C) = ${13\choose3}$ = 286&#xA;and as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(A&amp;cup;B&amp;cup;C) = p(A) + p(B) + p(C) - p(A&amp;cap;B) - p(A&amp;cap;C)- p(B&amp;cap;C) + p(A&amp;cap;B&amp;cap;C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= (${13\choose2}\times48$) - (3 x ${13\choose3}$) + ${13\choose3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 3744 - 858 + 286&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 3172&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3172 / 19,600 (total number of possible flops) = 0.1618 = 16.18%&lt;/p&gt;&#xA;" OwnerUserId="1094" LastActivityDate="2010-08-12T17:05:38.147" />
  <row Id="2260" PostTypeId="1" AcceptedAnswerId="2263" CreationDate="2010-08-12T17:15:42.560" Score="5" ViewCount="1265" Body="&lt;p&gt;Apparently 1+2+3+4+...+n = 1/2*n*(n+1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How? What's the proof? Or maybe it is self apparent just looking at the above? Does this problem have a name and maybe a presence on the net? And what is this math called so I can add the right tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thx if you can help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;This problem is know as &quot;The sum of the first n positive integers&quot;. Do a search. &lt;/p&gt;&#xA;" OwnerUserId="1103" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-12T22:00:24.160" LastActivityDate="2010-08-12T22:57:36.210" Title="Proof for formula for sum of sequence 1+2+3+...+n?" Tags="&lt;algebra-precalculus&gt;&lt;sequences-and-series&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="3" />
  <row Id="2261" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:18:31.213" Score="4" ViewCount="0" Body="&lt;p&gt;Once you have a formula like this, you can prove it by induction.&#xA;But that begs the question as to how you get such a formula.&#xA;In this case you might ask: (a) what's the &quot;average&quot; term?&#xA;and (b) how many terms are there?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-12T17:18:31.213" />
  <row Id="2262" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:19:45.953" Score="2" ViewCount="0" Body="&lt;p&gt;The most general case of this is called an &lt;em&gt;arithmetic progression&lt;/em&gt; or &lt;em&gt;(finite) arithmetic series&lt;/em&gt;. There are many, many, many proofs. An easy one: write all the summands in a row; write them again just below, but from right to left now (so $1$ is under $n$, $2$ is under $n-1$, etc). Add them up, and figure out how it relates to the quantity you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-12T18:07:02.437" LastActivityDate="2010-08-12T18:07:02.437" CommentCount="1" />
  <row Id="2263" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:23:29.567" Score="11" ViewCount="0" Body="&lt;p&gt;Let S = 1 + 2 + ... + (n-1) + n.&#xA;Write it backwards: S = n + (n-1) + ... + 2 + 1.&#xA;Add the two equations, term by term; each term is n+1, so&#xA;2S = (n+1) + (n+1) + ... + (n+1) = n(n+1).&#xA;Divide by 2: S = n(n+1)/2&lt;/p&gt;&#xA;" OwnerUserId="1105" LastActivityDate="2010-08-12T17:23:29.567" CommentCount="2" />
  <row Id="2264" PostTypeId="2" ParentId="566" CreationDate="2010-08-12T17:31:18.823" Score="6" ViewCount="0" Body="&lt;p&gt;The Blue-Eyed Islander problem is one of my favorites.  You can read about it &lt;a href=&quot;http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; on Terry Tao's website, along with some discussion.  I'll copy the problem here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[For the purposes of this logic puzzle, &quot;highly logical&quot; means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One evening, he addresses the entire tribe to thank them for their hospitality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What effect, if anything, does this faux pas have on the tribe?&lt;/p&gt;&#xA;" OwnerUserId="1101" LastActivityDate="2010-08-12T17:31:18.823" CommunityOwnedDate="2010-08-12T17:31:18.823" CommentCount="1" />
  <row Id="2265" PostTypeId="2" ParentId="1860" CreationDate="2010-08-12T17:36:50.857" Score="2" ViewCount="0" Body="&lt;p&gt;You should be able to use the fact that the $n^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;Harmonic Number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_n = \ln n + \gamma + \frac{1}{2n} - O(\frac{1}{n^2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sum is a difference of two such numbers and so is approximately of the form $\ln\frac{at+b}{ct+d}$ where $a &gt; c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, haven't done the complete math, but this approach looks promising.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-12T17:36:50.857" CommentCount="4" />
  <row Id="2266" PostTypeId="2" ParentId="2240" CreationDate="2010-08-12T17:52:19.690" Score="0" ViewCount="0" Body="&lt;p&gt;First of all thanks a lot to George S. Secondly i recently found a paper entitled, which discusses something on Automorphisms of Complex Numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link for that paper: mathdl.maa.org/images/upload_library/22/Ford/PaulBYale.pdf&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T17:52:19.690" />
  <row Id="2267" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T17:56:23.783" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; Pair each summand $k$ with it's &quot;reflection&quot; $n+1-k$. This is simply a discrete analog of the method of computing the area of the triangle under the diagonal of a square by reflecting a subtriangle through the midpoint of the diagonal to form an $n$ by $n/2$ rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the analogous proof of &lt;a href=&quot;http://math.stackexchange.com/questions/1865&quot;&gt;Wilson's theorem&lt;/a&gt;, the method exploits the existence of a nontrivial symmetry. In Wilson's theorem we exploit the symmetry $n \mapsto n^{-1}$ which exists due to the fact that ${\mathbb F}_p^*$ forms a group. Here we exploit a reflection through a line - a symmetry that exists due to the &lt;em&gt;linear&lt;/em&gt; nature of the problem (which doesn't work for nonlinear sums, e.g. $\sum n^2$).  Symmetries often lead to elegant proofs. One should always look for innate symmetries when first pondering problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally there are (Galois) theories and algorithms for &lt;strong&gt;summation in closed form&lt;/strong&gt;, in analogy to the differential case (Ritt, Kolchin, Risch et al.). A &lt;em&gt;very&lt;/em&gt; nice motivated introduction can be found in the introductory chapter of Carsten Schneider's thesis &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.67.9442&quot; rel=&quot;nofollow&quot;&gt;Symbolic Summation in Difference Fields&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T19:32:30.147" LastActivityDate="2010-08-12T19:32:30.147" CommentCount="1" />
  <row Id="2268" PostTypeId="2" ParentId="1910" CreationDate="2010-08-12T17:58:59.880" Score="1" ViewCount="0" Body="&lt;p&gt;I would concur with the suggestion that you should use LaTeX. I'm bothering with a separate answer to &lt;em&gt;strongly discourage&lt;/em&gt; the use of MS Word with symbol fonts for logical notation. Before I learned LaTeX, I'd tried to write up logic in word processors (both Word and, later, OpenOffice). It was deeply frustrating when my documents got quite messed up when I moved from one computer to another and found myself working without the symbol fonts I had at my disposal when I wrote the documents. (Yes, there are ways around that problem, but it nevertheless seems worth the warning.)&lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2010-08-12T17:58:59.880" CommentCount="1" />
  <row Id="2269" PostTypeId="1" AcceptedAnswerId="2271" CreationDate="2010-08-12T18:10:12.557" Score="3" ViewCount="254" Body="&lt;p&gt;I was just reading a paper of Ramanujan entitled &quot; On question 330 of Professor Sanjana&quot; when i got stuck up with a Proposition which i am unable to answer. The proposition is if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle f(p) = \frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{ \Bigl[ \Gamma(\frac{P+2}{2}) \Bigr]^{2}}$, then $$\log{f(p)}= \log(\frac{\pi}{2}) - p\log{2}  + \frac{p^2}{2} \Bigl(1-\frac{1}{2}\Bigr)S_{2} - \frac{p^3}{3} \Bigl(1-\frac{1}{2^2}\Bigr)S_{3} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle S_{n} = \frac{1}{1^n} + \frac{1}{2^n} + \cdots \text{ad inf.}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain me as to how can i deduce this.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T02:53:37.263" LastActivityDate="2010-08-27T02:53:37.263" Title="Question Relating Gamma Function to Riemann Zeta function evaluated at integers" Tags="&lt;gamma-function&gt;&lt;riemann-zeta&gt;&lt;special-functions&gt;&lt;classical-analysis&gt;&lt;ramanujan&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="2270" PostTypeId="1" AcceptedAnswerId="2275" CreationDate="2010-08-12T18:19:22.863" Score="10" ViewCount="706" Body="&lt;p&gt;Does $S_k= \sum_{n=1}^{\infty}\sin(n^k)/n$ converge for all $k&gt;0$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: I recently learned that $S_1$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_test&quot; rel=&quot;nofollow&quot;&gt;converges&lt;/a&gt;. I think $S_2$ converges by the integral test. Was the question known in general?   &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-19T22:00:09.730" LastActivityDate="2010-11-12T13:24:02.667" Title="Convergence of $\sum_{n=1}^{\infty}\sin(n^k)/n$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" />
  <row Id="2271" PostTypeId="2" ParentId="2269" CreationDate="2010-08-12T18:26:59.620" Score="5" ViewCount="0" Body="&lt;p&gt;I presume this is a Maclaurin series. I don't know how it comes&#xA;about, but my instinct would be to expand out the infinite product&#xA;for $f(p)$ (which becomes an infinite sum for $\log f(p)$) in powers&#xA;of $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (13/8/2010)&#xA;As a hint: recall that&#xA;$$\Gamma(p+1)=p\Gamma(p)&#xA;=e^{-\gamma p}\prod_{k=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}$$&#xA;by the infinite product for the gamma function. Applying&#xA;this to $\Gamma(p+1)/\Gamma(p/2+1)^2$ one gets some nice&#xA;cancellation. Then take logs and apply the series for $\log(1+x)$.&#xA;You get a double sum; change the order of summation and hope that&#xA;the inner sum comes to something like a zeta series.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-13T17:13:44.230" LastActivityDate="2010-08-13T17:13:44.230" CommentCount="6" />
  <row Id="2272" PostTypeId="2" ParentId="2270" CreationDate="2010-08-12T18:58:12.037" Score="-1" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this help: $ \displaystyle \int\limits_{0}^{1} (\sin{x})^{k} \ dx = \lim_{n \to \infty} \sum\limits_{r=0}^{n} \frac{1}{n} \Bigl[ \sin(r/n)\Bigr]^{k}$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T18:58:12.037" CommentCount="6" />
  <row Id="2273" PostTypeId="1" CreationDate="2010-08-12T19:26:07.727" Score="1" ViewCount="111" Body="&lt;p&gt;Sometimes when I am reading a mathematical text I will be unable to decipher the steps necessary to translate one mathematical statement into the next. Usually I have to ask for help, but I was wondering if there is a software package that would let me enter two mathematical statements and the it would find the steps necessary to get from the start to the finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this not tractable in general, however, it would seem with the right heuristics it could still be possible many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a response both to my own experience and this question:&#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/2269/question-relating-gamma-function&quot;&gt;http://math.stackexchange.com/questions/2269/question-relating-gamma-function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="99" LastActivityDate="2010-10-05T17:25:21.780" Title="Is are there software package that can help one find a path between two mathematical statements?" Tags="&lt;math-software&gt;&lt;learning&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" />
  <row Id="2274" PostTypeId="2" ParentId="2273" CreationDate="2010-08-12T19:35:37.237" Score="0" ViewCount="0" Body="&lt;p&gt;It is probably not exactly on things You are looking for, but prover9-mace4 tries to find prof in FOL for given statements starting from given axioms ( prover9), or opposite, tries to find counterexample (mace4). I can imagine that for simple statements, like in group theory, it gives what You are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Prover9&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Prover9&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1043" LastActivityDate="2010-08-12T19:35:37.237" CommentCount="1" />
  <row Id="2275" PostTypeId="2" ParentId="2270" CreationDate="2010-08-12T19:41:11.530" Score="20" ViewCount="0" Body="&lt;p&gt;This is a replacement for my previous answer. The sum converges, and this fact needs even more math than I believed before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin by using summation by parts. This gives &#xA;$$\sum_{n=1}^N \left(\sum_{m=1}^N \sin(m^k) \right) \left( \frac{1}{n}-\frac{1}{n+1}\right) + \frac{1}{N+1} \left(\sum_{m=1}^N \sin(m^k) \right).$$ &#xA;Write $S_n:= \left(\sum_{m=1}^n \sin(m^k) \right)$. So this is&#xA;$$\sum_{n=1}^N S_n/(n(n+1)) + s_N/(N+1).$$&#xD;&#xA;The second term goes to zero by Weyl's &lt;a href=&quot;http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/&quot; rel=&quot;nofollow&quot;&gt;polynomial equidistribution theorem&lt;/a&gt;. So your question is equivalent to the question of whether $\sum s_n/(n(n+1))$ converges. We may as well clean this up a little: Since $|S_n| \leq n$, we know that $\sum S_n \left( 1/n(n+1) - 1/n^2 \right)$ converges. So the question is whether&#xA;$$\sum \frac{S_n}{n^2}$$&#xD;&#xA;converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show that $S_n$ is small enough that $\sum S_n/n^2$ converges absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I want to prove this is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Weyl's inequality&lt;/a&gt;. Let $p_i/q_i$ be an infinite sequence of rational numbers such that $|1/(2 \pi) - p_i/q_i| &amp;lt; 1/q_i^2$. Such a sequence exists by a standard lemma. Weyl inequality gives that&#xA;$$S_N = O\left(N^{1+\epsilon} (q_i^{-1} + N^{-1} + q_i N^{-k})^{1/2^{k-1}} \right)$$&#xD;&#xA;for any $\epsilon&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to George Lowther for pointing out the next step: According to &lt;a href=&quot;http://mathworld.wolfram.com/IrrationalityMeasure.html&quot; rel=&quot;nofollow&quot;&gt;Salikhov&lt;/a&gt;, for $q$ sufficiently large, we have&#xA;$$|\pi - p/q| &amp;gt; 1/q^{7.6304+\epsilon}.$$&#xD;&#xA;Since $x \mapsto 1/(2x)$ is Lipschitz near $\pi$, and since $p/q$ near $\pi$ implies that $p$ and $q$ are nearly proportional, we also have the lower bound $|1/(2 \pi) - p/q|&amp;gt; 1/q^{7.6304+\epsilon}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_i/q_i$ be the convergents of the continued fraction of $1/(2 \pi)$. By a standard result, $|1/(2 \pi) - p_i/q_i| \leq 1/(q_i q_{i+1})$. Thus, $q_{i+1} \leq q_i^{6.6304 + \epsilon}$ for $i$ sufficiently large. Thus, the intervals $[q_i, q_i^{7}]$ contain all sufficiently large integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any large enough $N$, choose $q_i$ such that $N^{k-1} \in [q_i, q_i^7]$. Then Weyl's inequality gives the bound&#xA;$$S_N = O \left( N^{1+\epsilon} (N^{-1/(7(k-1))} + N^{-1} + N^{-1} \right)^{1/2^{k-1}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $S_N = O(N^{1-1/(7(k-1)2^{k-1}) + \epsilon})$, which is enough to make sure the sum converges.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-11-12T13:24:02.667" LastActivityDate="2010-11-12T13:24:02.667" CommentCount="11" />
  <row Id="2276" PostTypeId="2" ParentId="2240" CreationDate="2010-08-12T19:51:29.600" Score="0" ViewCount="0" Body="&lt;p&gt;This and related questions about automorphism groups of algebraically closed fields (a topic I find interesting and have spent some time thinking about) are discussed in Section 9.1 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/FieldTheory.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, Theorem 77 answers the OP's question affirmatively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These notes are still very rough.  In particular there is not yet a bibliography.  When this gets remedied, a citation to Paul Yale's paper will be in order: it was definitely something I read when writing these notes.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-12T19:51:29.600" CommentCount="1" />
  <row Id="2277" PostTypeId="2" ParentId="566" CreationDate="2010-08-12T20:05:42.640" Score="3" ViewCount="0" Body="&lt;p&gt;The odd town puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a town with $m$ clubs formed by $n$ citizens of the town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clubs are so formed that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each club has an odd number of members.&lt;/li&gt;&#xA;&lt;li&gt;Any two clubs have an even number of common members. (Could be zero too).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Show that $m \le n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="555" LastEditorDisplayName="" LastEditDate="2010-09-06T19:12:27.333" LastActivityDate="2010-09-06T19:12:27.333" CommunityOwnedDate="2010-08-12T20:05:42.640" CommentCount="1" />
  <row Id="2278" PostTypeId="1" AcceptedAnswerId="2302" CreationDate="2010-08-12T20:06:37.630" Score="4" ViewCount="140" Body="&lt;p&gt;Let us be given a $n\times n$ matrix containing only zeros and ones.Now, the goal is to remove some 'ones' from the matrix (i.e. replace them with zeros) so that in each row and each column there is at most one 'one'. Of course, it is preferable that we leave as much 'ones' as it is possible (maximum is obviously $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us define the following algorithm: if there is a 'one' which is all alone in a column or a row, leave it there, and replace all 'ones' in its column and row with zeros. We repeat the process as long as there are single 'ones' in rows or columns. If there are no single 'ones' left, we take the first row or column (first is counted from up (for rows) and from left (for columns)) containing two 'ones' (if there are no pairs, we go for the first triplet and so on). In that pair (or triplet, or ...) we leave the first 'one' intact (first is counted from the left for elements of row, from the top for elements of a column) and change all other 'ones' in the row and column of that first 'one' into zeros.&#xA;Now, we check are there now any single 'ones' as defined above - if yes, we repeat that part of the algorithm, if no, we look for the first pair, triplet, etc. The algorithm ends when all 'ones' are singles both in their column and row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this algorithm always give the maximum covering, i.e. does it always terminate with maximal number of 'ones' in the final matrix? In other words, can there be a matrix with a possible exclusion of some ones such that it ends up with $n$ ones using some other exclusion scheme, while ending up with less than $n$ 'ones' using the algorithm described above?&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-12T20:20:08.753" LastActivityDate="2010-08-13T08:00:47.697" Title="Can this algorithm on removing 1's from a (0,1)-matrix fail?" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;" AnswerCount="3" />
  <row Id="2279" PostTypeId="1" AcceptedAnswerId="2282" CreationDate="2010-08-12T20:09:58.917" Score="0" ViewCount="88" Body="&lt;p&gt;Prove that a number in the sequence $2,3,4,...,n \ (n&gt;2$, is relatively prime to all other numbers if and only if it is a prime that exceeds $\displaystyle\frac{n}{2}$. Does such a prime always exist?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T20:18:51.050" Title="Proving that a Number is Relatively Prime in a sequence." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" />
  <row Id="2280" PostTypeId="2" ParentId="1068" CreationDate="2010-08-12T20:17:43.890" Score="3" ViewCount="0" Body="&lt;p&gt;First of all &lt;a href=&quot;http://www.math.wisc.edu/~keisler/calc.html&quot; rel=&quot;nofollow&quot;&gt;Non-standard Analysis&lt;/a&gt; makes nonsense like $\frac{\mathrm{d}y}{\mathrm{d}x}$ into a meaningful fraction rather than just a &quot;symbol&quot; (whatever that means).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and simplest example of differentials being used beyond the $\int f(x)\mathrm{d}x$ format is would be multidimensional integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_V \mathrm{d}x \mathrm{d}y \mathrm{d}z = \int \left(\int \left(\int \mathrm{d}x \right) \mathrm{d}y \right) \mathrm{d}z $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives the volume of the solid $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next example is Greens Theorem from vector calculus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int A(x,y) \mathrm{d}x - B(x,y) \mathrm{d}y = \int (\partial_1 B -  \partial_2 A) \mathrm{d}x \mathrm{d}y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to forget integration completely and just use &lt;a href=&quot;http://www.math.oregonstate.edu/bridge/papers/CMJdifferentials.pdf&quot; rel=&quot;nofollow&quot;&gt;equations with differentials in them&lt;/a&gt; to solve calculus problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see the standard format is not even close to the whole picture, If you want to integrate terms like $\int (e^{\mathrm{d}x} - 1)$ &lt;em&gt;please do&lt;/em&gt;! There is absolutely nothing to stop you figuring out from scratch how to solve this sort of integral and making the theory rigorous.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-12T20:17:43.890" CommentCount="5" />
  <row Id="2281" PostTypeId="2" ParentId="2278" CreationDate="2010-08-12T20:18:00.207" Score="5" ViewCount="0" Body="&lt;p&gt;I'm not sure about whether this works, but I'd be doubtful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to express this problem is finding maximum&#xA;partial matchings in a bipartite graph. (Just interpret&#xA;the matrix an adjacency matrix for the graph).&#xA;By a maximum partial matching I mean a maximal-sized&#xA;set of edges no two having a vertex in common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for my scepticism is that if your algorithm worked,&#xA;then it would give an algorithm for finding the&#xA;solution in the situation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hall%27s_marriage_theorem&quot; rel=&quot;nofollow&quot;&gt;Hall's marriage theorem&lt;/a&gt;.&#xA;Your algorithm is &quot;monotone&quot; in the sense that it adds&#xA;edges but never removes them, but all the algorithms I have&#xA;seen for Hall's theorem, one needs to remove edges in certain&#xA;circumstances in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem is hand, an algorithm comes from any that solves&#xA;the maximum flow problem in a network. Attach a source to each&#xA;vertex on one side of the bipartition and a sink to each vertex&#xA;on the other side. Give all edges unit capacity and find&#xA;a maximum flow.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-12T20:18:00.207" />
  <row Id="2282" PostTypeId="2" ParentId="2279" CreationDate="2010-08-12T20:18:51.050" Score="2" ViewCount="0" Body="&lt;p&gt;If it's not a prime, all its divisors are there.&lt;br&gt;&#xA;If it's not greater than n/2, than 2p is not greater than n, and it's in the list.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every prime greater than n/2 trivially satisfies the condition.&#xA;It always exists by Bertrand's postulate ($n/2 &amp;lt; p &amp;lt; 2\cdot n/2$).&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T20:18:51.050" />
  <row Id="2283" PostTypeId="2" ParentId="2273" CreationDate="2010-08-12T20:21:50.293" Score="0" ViewCount="0" Body="&lt;p&gt;There's only a handful of math software that deals with human solving/thinking process. Most apply some kind of normalization to more abstract concepts or more convenient data structures, run an algorithm on that, and then try to simplify the answer and present it in more human form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to get more into this, I suspect that you could use any powerful CAS to normalize both expressions and get the same canonical form, and from there you could trace back to both expressions.&lt;br&gt;&#xA;This of course would be very painful process.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-12T20:21:50.293" />
  <row Id="2284" PostTypeId="1" AcceptedAnswerId="2285" CreationDate="2010-08-12T20:41:22.647" Score="4" ViewCount="286" Body="&lt;p&gt;Everyone knows that $\pi$ is an irrational number, and one can refer to this page &lt;a href=&quot;http://planetmath.org/encyclopedia/PiAndPi2AreIrrational.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/PiAndPi2AreIrrational.html&lt;/a&gt; for the proof that $\pi^{2}$ is also irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the Highers powers of $\pi$, meaning is $\pi^{n}$ irrational for all $n \in \mathbb{N}$ or does there exists a $m \in \mathbb{N}$ when $\pi^{m}$ is rational.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T22:40:31.647" Title="Irrationality of powers of $\pi$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2285" PostTypeId="2" ParentId="2284" CreationDate="2010-08-12T20:50:36.010" Score="13" ViewCount="0" Body="&lt;p&gt;What Robin hinted at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\pi^{n}$ was rational, then $\pi$ would not be transcendental, as it would be the root of $ax^{n}-b = 0$ for some integers $a,b$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-12T20:50:36.010" CommentCount="4" />
  <row Id="2286" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T20:51:39.690" Score="0" ViewCount="0" Body="&lt;p&gt;You know, this method of writing backwards and finding the sum of an arithmetic series was invented by Karl Friedrich Gauss in a matter of minutes when he was a schoolboy. He is one of the greatest mathematicians that ever lived!&lt;/p&gt;&#xA;" OwnerUserId="1113" LastActivityDate="2010-08-12T20:51:39.690" CommentCount="3" />
  <row Id="2287" PostTypeId="2" ParentId="2273" CreationDate="2010-08-12T20:53:20.950" Score="2" ViewCount="0" Body="&lt;p&gt;The likely outcome of applying software to the reading of ordinary proofs would be to discover that most proofs contain steps that are not formally correct.  For example, in classical geometry many steps in the arguments are &quot;generically true&quot; but false for special degenerate configurations.  Sometimes these configurations, such as a triangle with all three vertices the same point, are implicitly excluded from the cases considered in the theorem, or can be handled in the proof using additional steps that might be considered trivial and are not articulated in the written argument.  In such cases the theorem would be correct in substance but some steps could not be filled in by a computerized proof assistant, because the steps are wrong.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less often, there are more difficult cases, where the non-generic configurations and components can't be ignored, or the intuitive idea of what is generic behavior is incorrect.  In this case a theorem may be correct, but a proof that doesn't deal with the special cases errs not only in the details but in its overall argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar thing happens in computational linguistics.  When trying to process natural language with formal grammars, it turns out that simple sentences are multiply ambiguous, with a large number of unintended but correct interpretations.  People constantly use additional knowledge to disambiguate the language they hear, but software does not yet have that capability. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-12T20:58:25.960" LastActivityDate="2010-08-12T20:58:25.960" />
  <row Id="2288" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T21:19:17.387" Score="10" ViewCount="0" Body="&lt;p&gt;My favourite proof is the one given &lt;a href=&quot;http://mathoverflow.net/questions/8846/proofs-without-words/8847#8847&quot; rel=&quot;nofollow&quot;&gt;here on MathOverflow&lt;/a&gt;.  I'm copying the picture here for easy reference, but full credit goes to &lt;a href=&quot;http://mathoverflow.net/users/1409/mariano-suarez-alvarez&quot; rel=&quot;nofollow&quot;&gt;Mariano Su&amp;aacute;rez-Alvarez&lt;/a&gt; for this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img31.imageshack.us/img31/919/triangle1y.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Takes a little bit of looking at it to see what's going on, but it's nice once you get it.  Observe that if there are n rows of yellow discs, then:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there are a total of 1 + 2 + ... + n yellow discs;  &lt;/li&gt;&#xA;&lt;li&gt;every yellow disc corresponds to a unique pair of blue discs, and vice versa;&lt;/li&gt;&#xA;&lt;li&gt;there are ${n+1 \choose 2} = \frac 12 n(n+1)$ such pairs.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-08-12T21:19:17.387" CommunityOwnedDate="2010-08-12T21:19:17.387" CommentCount="1" />
  <row Id="2290" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T21:31:15.130" Score="3" ViewCount="0" Body="&lt;p&gt;Another &quot;picture proof&quot; I just thought of... but without a picture, since I can't draw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to add up all the integers from 1 to n. Then draw n rows on a board, put 1 unit in the first row, 2 in the second, and so on. If you draw a right triangle of height and length n to try and contain this shape, you will cut off half of each unit on the diagonal. So let's add all of these up; you have n^2/2 units inside the triangle, and you have n/2 unites cut off on the diagonal (there are n squares on the diagonal, and half of each one is cut off). Adding these gives n^2/2 + n/2 = n(n+1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw it out it makes more sense, and I think it's geometrically a bit more straightforward than the other picture proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side note: I have no idea how to write math on this site... I'll go consult the meta, I'm sure there's something there about it.)&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-12T21:31:15.130" CommentCount="2" />
  <row Id="2291" PostTypeId="1" CreationDate="2010-08-12T21:34:56.583" Score="1" ViewCount="181" Body="&lt;p&gt;How is the power series expansion of an analytic function at a point constructed, without using Cauchy's theorem(or formula)?&lt;/p&gt;&#xA;" OwnerUserId="1117" LastActivityDate="2010-08-12T22:21:25.127" Title="Power series expansion without Cauchy theorem" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="2292" PostTypeId="1" AcceptedAnswerId="2298" CreationDate="2010-08-12T21:48:13.887" Score="0" ViewCount="240" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Problem 2.4.31 (page 84) from &lt;a href=&quot;http://books.google.com/books?id=OBK7D1qcY0oC&amp;amp;pg=PA84&amp;amp;lpg=PA84&amp;amp;dq=%22Give+an+example+of+a+bounded+function%22+%22which+is+not+Riemann+Integrable,+but+is+a+derivative+of+some+function%22&amp;amp;source=bl&amp;amp;ots=JgTTEAPY3o&amp;amp;sig=4qcimfqsmlE-5EWKVuEJFlapn7A&amp;amp;hl=en&amp;amp;ei=1UNoTM_3I4OYuAOMvKD-Aw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CBYQ6AEwAQ#v=onepage&amp;amp;q=%22Give%20an%20example%20of%20a%20bounded%20function%22%20%22which%20is%20not%20Riemann%20Integrable%2C%20but%20is%20a%20derivative%20of%20some%20function%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problems in Mathematical Analysis: Integration&lt;/a&gt; by W. J. Kaczor, Wiesława J. Kaczor and Maria T. Nowak.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Give an example of a bounded function $f:[0,1] \to \mathbb{R}$ which is not Riemann Integrable, but is a derivative of some function $g$ on $[0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T19:46:48.653" LastActivityDate="2010-08-15T19:46:48.653" Title="Bounded Function Which is Not Riemann Integrable." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2293" PostTypeId="1" CreationDate="2010-08-12T21:48:21.733" Score="4" ViewCount="368" Body="&lt;p&gt;What are some nice applications of algebraic topology that can be presented to beginning students? To give examples of what I have in mind: Brouwer's fixed point theorem, Borsuk-Ulam theorem, Hairy Ball Theorem, any subgroup of a free group is free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper the methods used, the better. All the above can be proved with just the fundamental group. More involved applications would be nice.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T00:19:08.030" LastActivityDate="2010-12-18T07:52:59.350" CommunityOwnedDate="2010-08-12T21:55:25.927" Title="Applications of algebraic topology" Tags="&lt;big-list&gt;&lt;applications&gt;&lt;algebraic-topology&gt;" AnswerCount="11" CommentCount="0" FavoriteCount="4" />
  <row Id="2294" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T21:50:57.630" Score="3" ViewCount="0" Body="&lt;p&gt;The Jordan-Brouwer separation theorem is a nice application of singular homology.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-12T21:50:57.630" CommunityOwnedDate="2010-08-12T21:50:57.630" />
  <row Id="2295" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T21:52:33.523" Score="1" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one interesting application of Algebraic Topology in Molecular Biology: www.springerlink.com/index/h55gv40w17306n57.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a computer related application please see this link: &lt;a href=&quot;http://www.lix.polytechnique.fr/~sanjeevi/atmcs/&quot; rel=&quot;nofollow&quot;&gt;http://www.lix.polytechnique.fr/~sanjeevi/atmcs/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T21:52:33.523" CommunityOwnedDate="2010-08-12T22:28:12.873" />
  <row Id="2296" PostTypeId="2" ParentId="2291" CreationDate="2010-08-12T21:53:37.893" Score="1" ViewCount="0" Body="&lt;p&gt;Below are a couple of possible approaches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR0123687 (23 #A1010)  30.20&lt;br&gt;&#xA;Porcelli, P.; Connell, E. H.&lt;br&gt;&#xA;&lt;strong&gt;A proof of the power series expansion without Cauchy's formula.&lt;/strong&gt;&lt;br&gt;&#xA;Bull. Amer. Math. Soc. 67 1961 177--181.&lt;br&gt;&#xA;&lt;a href=&quot;http://projecteuclid.org/euclid.bams/1183524076&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/euclid.bams/1183524076&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the basic result of topological analysis that a differentiable function of a complex variable generates an open mapping, the authors succeed in establishing in remarkably simple fashion, independent of any integration theory, the validity of the power-series expansion of such a function. A series of lemmas leads up to the theorem that differentiability of a function $f(z)$ on $0&amp;lt;|z-z_0| &amp;lt; r $ plus continuity at $z_0$ implies differentiability also at $z_0$. Using this, along with existence of higher-ordered derivatives, it is then shown that if $f(z)$ is continuous and $|f(z)|\leq 1$ for $|z|\leq 1$ and differentiable for $|z|&amp;lt;1$, then $|f^{(n)}(0)|\leq n!2^n$, for all $n$, and the Taylor series for $f$ converges to $f$ for $|z|&amp;lt;1/2$.&lt;br&gt;&#xA;Reviewed by G. T. Whyburn&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR0993637 (90m:30002a)   30A99 (26A39 30B10 42A20)&lt;br&gt;&#xA;Shisha, Oved(1-RI)&lt;br&gt;&#xA;&lt;strong&gt;Proof of power series and Laurent expansions of complex differentiable functions without use of Cauchy's integral formula or Cauchy's integral theorem.&lt;/strong&gt;&lt;br&gt;&#xA;J. Approx. Theory 57 (1989), no. 2, 117--135.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR1006337 (90m:30002b)  30A99 (26A39 30B10 42A20)&lt;br&gt;&#xA;Shisha, Oved&lt;br&gt;&#xA;&lt;strong&gt;Erratum: ``Proof of power series and Laurent expansions of complex differentiable functions without use of Cauchy's integral formula or Cauchy's integral theorem''.&lt;/strong&gt;&lt;br&gt;&#xA;J. Approx. Theory 58 (1989), no. 2, 246.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author shows how to establish the basic results of complex function theory in the context of real variable Fourier analysis. His basic tool is a generalization of the classical Riemann integral, which he proposes should become the standard integral of the working analyst in place of currently used integrals (including the Lebesgue integral), all of which it contains as a special case. A topological development, eliminating all use of integrals, is due to \n E. Connell\en, \n R. L. Plunket\en, \n P. Porcelli\en, \n A. H. Read\en and \n G. T. Whyburn\en [Whyburn, Topological analysis, revised edition, Princeton Univ. Press, Princeton, NJ, 1964; MR0165476 (29 #2758)]. The present development generalizes the work of \n P. R. Beesack\en [Canad. Math. Bull. 15 (1972), 473--480; MR0310199 (46 #9301)], who placed various conditions on the derivative to achieve his results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant property of the new integral is the fact that if $f$ is an arbitrary differentiable function on an interval $[a, b]$, then $\int^b_a f'(x)\,dx=f(b)-f(a)$. For any function $f$ on $[a,b]$, the new integral is defined as the number $I$ such that for every $\varepsilon &gt;0$, there exists a positive function $\delta_\varepsilon(x)$ on $[a,b]$, such that for every partition $a=x_0 &amp;lt; x_1 &amp;lt; \cdots &amp;lt; x_n=b$ and sequence $s_1,\cdots, s_n$, $x_{k-1} \le s_k\le x_k$, for $k=1,\cdots,n$, such that $x_k-x_{k-1}&amp;lt;\delta_{\epsilon}(k)$ for $k=1,\cdots,n$, one has $|I-\sum ^n_{k=1}f(s_k)(x_k-x_{k-1})|&amp;lt;\varepsilon$. Let $f$ be a complex differentiable function on an annulus $0\le R' &amp;lt; |z| &amp;lt; R''&amp;lt;\infty$. Then by Dini's test, since $f$ is everywhere differentiable, for each $r$, $R' &amp;lt; r &amp;lt; R''$, the function $f(re^{i\phi})$, $0\le \phi\le 2\pi$, can be expanded in a Fourier series $f(re^{i\phi}) =\sum^{+\infty}_{n=-\infty}c_k(r)e^{i\phi}$. The objective of the paper is achieved if it can be shown that the coefficients $a_k(r)$ are independent of $r$. The basic lemma used states that $$ \iint_{\begin{array}{} R_1\le r\le R \\ 0\le \varphi\le 2\pi \end{array}} f'(re^{i\varphi})drd\varphi =\int_0^{2\pi}e^{-i\varphi}[f(Re^{i\varphi})-F(R_1e^{-i\varphi})]d\varphi,$$ $$ \widehat{\iint}_{\begin{array}{} R_1\le r\le R \\ 0\le \varphi\le 2\pi \end{array}} f'(re^{i\varphi})drd\varphi =\int_{R_1}^{R}r^{-1}\int_0^{2\pi}e^{-i\varphi}f(re^{i\varphi})d\varphi dr,$$ where the integrals on the left are two-dimensional generalized Riemann integrals. The exposition is quite detailed and fully self-contained, including the development of all relevant characteristics of the generalized Riemann integral.&lt;br&gt;&#xA;\edref {Two minor errors are corrected in the erratum.}&lt;br&gt;&#xA;Reviewed by Kenneth O. Leland&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-12T22:21:25.127" LastActivityDate="2010-08-12T22:21:25.127" />
  <row Id="2297" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T21:58:09.763" Score="5" ViewCount="0" Body="&lt;p&gt;Brouwer's fixed point theorem can be illustrated using the game &lt;a href=&quot;http://en.wikipedia.org/wiki/Hex_(board_game)#Theory_and_Proofs&quot; rel=&quot;nofollow&quot;&gt;Hex&lt;/a&gt;.  In fact, the determinacy of Hex is equivalent to the 2-dimensional version.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-12T21:58:09.763" CommunityOwnedDate="2010-08-12T21:58:09.763" />
  <row Id="2298" PostTypeId="2" ParentId="2292" CreationDate="2010-08-12T22:01:12.430" Score="4" ViewCount="0" Body="&lt;p&gt;I gave an answer to this question on Math Overflow some months ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/6711/integrability-of-derivatives/6716#6716&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6711/integrability-of-derivatives/6716#6716&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, in particular, the following paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/Goffman77.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/Goffman77.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-12T22:01:12.430" />
  <row Id="2299" PostTypeId="2" ParentId="2293" CreationDate="2010-08-12T22:02:26.017" Score="2" ViewCount="0" Body="&lt;p&gt;You can prove the Grushko decomposition theorem, as well as the Kurosh subgroup theorem.  See for example the last two sections of Massey's &quot;Algebraic Topology: an Introduction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-12T22:02:26.017" CommunityOwnedDate="2010-08-12T22:02:26.017" />
  <row Id="2300" PostTypeId="1" AcceptedAnswerId="2307" CreationDate="2010-08-12T22:18:51.603" Score="1" ViewCount="198" Body="&lt;p&gt;It is said that there is a proof of fundamental theorem of algebra using Lie Theory. I have seen this claim at various places. But I could never find such a proof. Can anybody help me out?&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:41:05.400" LastActivityDate="2010-08-14T09:41:05.400" Title="Fundamental theorem of algebra using Lie Theory" Tags="&lt;complex-variables&gt;&lt;lie-groups&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="2301" PostTypeId="2" ParentId="2248" CreationDate="2010-08-12T22:24:32.410" Score="3" ViewCount="0" Body="&lt;p&gt;The theorem is known as the &quot;&lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/Butterfly.shtml&quot; rel=&quot;nofollow&quot;&gt;Butterfly Theorem&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fairly simple proof is the following (from &lt;em&gt;Selected Problems and Theorems of Elementary Mathematics&lt;/em&gt; by D. O. Shklyarsky, N. N. Chentsov and I. M. Yaglom).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let O be the center of the given circle. Since OM ⊥ CD, in order to show that CM = MD, we have to prove that ∠COM = ∠DOM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drop perpendiculars OK and ON from O onto PS and QR, respectively. Obviously, K is the midpoint of PS and N is the midpoint of QR. Further,&#xA;    ∠PSR = ∠PQR and ∠QPS = ∠QRS,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as angles subtending equal arcs. Triangles SPM and QRM are therefore similar, and SP/SM = QR/QM, or SK/SM = QN/QM. In other words, in triangles SKM and QNM two pairs of sides are proportional. Also, the angles between the corresponding sides are equal. We infer that the triangles SKM and QNM are similar. Hence, ∠SKM = ∠QNM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, have a look at the quadrilaterals OKCM and ONDM. Both have a pair of opposite straight angles, which implies that both are inscribable in a circle. In OKCM, ∠SKM = ∠COM. In ONDM, ∠QNM = ∠DOM. From which we get what we've been looking for: ∠COM = ∠DOM.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-12T22:24:32.410" CommentCount="3" />
  <row Id="2302" PostTypeId="2" ParentId="2278" CreationDate="2010-08-12T22:31:14.480" Score="5" ViewCount="0" Body="&lt;p&gt;Yes. Consider a large square matrix with two large square blocks filled with ones and one empty row and one empty column.  Delete one one from each of the blocks off the diagonal, and place a corresponding one in the empty row and column.  You now have one row with two elements, the rest having many elements, and one column having only two elements, the rest having many.  Your procedure will deal with the sparesely populated row and column first.  Permute the rows and columns so that you procedure links the two choices with different blocks.  Now the remainder of each block has one more row than it has columns or vice versa, as a perfect matching cannot exist.  It is easy to see however that a perfect matching existed before your algorithm matched the sparse row and column.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-08-12T22:31:14.480" CommentCount="0" />
  <row Id="2303" PostTypeId="2" ParentId="2278" CreationDate="2010-08-12T22:35:05.880" Score="3" ViewCount="0" Body="&lt;p&gt;Diagram to give an example for deinst's answer. I tried putting this in a comment, but it stuffed up the formatting. The Xs marks ones that give a complete match &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X01100001&#xA;1X1100000&#xA;11X100000&#xA;111X00000&#xA;00001011X&#xA;0000X1110&#xA;000011X10&#xA;0000111X0&#xA;01000X000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we take the two squares marked by Y's&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*0**0000Y&#xA;111100000&#xA;111100000&#xA;111100000&#xA;00001011*&#xA;00001*110&#xA;00001*110&#xA;00001*110&#xA;0*000Y000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We have two 3 by 4 squares left, so we can take 6 more items. This give a total of 8 instead of 9. We can enforce those two blocks to be taken by using a permutation.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-13T08:00:47.697" LastActivityDate="2010-08-13T08:00:47.697" CommentCount="0" />
  <row Id="2304" PostTypeId="1" AcceptedAnswerId="3445" CreationDate="2010-08-12T22:39:10.180" Score="3" ViewCount="173" Body="&lt;p&gt;Suppose $M_1$ and $M_2$ are invariant subspaces of the unilateral shift U such that $M_1$ subset $M_2$ and $M_1$ is of codimension strictly larger than $1$ in $M_2$. Show that there exists $M$ invariant under $U$ satisfying $M_1 \subset M \subset M_2$ where the inclusions are strict. All subspaces are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is from the Springer GTM: &quot;An introduction to operators on the Hardy-Hilbert space&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Perhaps I can take $M := U M_2$? Maybe I should give that some more thought.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-18T23:50:07.387" LastActivityDate="2010-08-27T08:03:53.797" Title="Invariant Subspaces (Hardy Space)" Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="2305" PostTypeId="1" AcceptedAnswerId="2337" CreationDate="2010-08-12T22:40:24.547" Score="6" ViewCount="182" Body="&lt;p&gt;I need an algorithm to produce all strings with the following property. Here capital letter refer to strings, and small letter refer to characters. $XY$ means the concatenation of string $X$ and $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Sigma = \{a_0, a_1,\ldots,a_n,a_0^{-1},a_1^{-1},\ldots,a_n^{-1}\}$ be the set of usable characters. Every string is made up of these symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out put any set $S_n$ with the following property achieves the goal.($n\geq 2$)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\in S_n$, then any cyclic shift of $W$ is not in $S_n$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\in S_n$, then $|W| = n$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\in S_n$, then $W \neq Xa_ia_i^{-1}Y$, $W \neq Xa_i^{-1}a_iY$, $W \neq a_iXa_i^{-1}$ and $W \neq a_i^{-1}Xa_i$ for any string $X$ and $Y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $W\not \in S_n$, $S_n \cup \{W\}$ will violate at least one of the above 3 properties. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Clearly any algorithm one can come up with is an exponential algorithm. but I'm still searching for a fast algorithm because this have some practical uses. At least for $\Sigma=\{a_0,a_1,a_0^{-1},a_1^{-1}\}$ and $n&amp;lt;25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive approach for my practical application requires $O(4^n)$ time. It generate all strings of length n. When ever a new string is generated, the program create all cyclic permutations of the string and check if it have been generated before though a hash table. If not, add to the list of the result strings. Total amount of operation are $O(n4^n)$, and that's assuming perfect hashing. 12 is the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there better approaches? clearly a lot of useless strings were generate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The practical usage is to find the maximum of minimum self intersection of a curve on a torus with a hole. Every curve can be characterized by a string described above. Therefore I have to generate every string and feed it to a program that calculate the minimum self intersection.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-08-14T00:08:33.477" LastActivityDate="2010-08-22T12:49:47.297" Title="A fast way to generate cyclic strings with few restrictions?" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" />
  <row Id="2306" PostTypeId="2" ParentId="2284" CreationDate="2010-08-12T22:40:31.647" Score="0" ViewCount="0" Body="&lt;p&gt;To nitpick Moron's answer - for any polynomial over the rational numbers, $\pi$ won't be a root of the said polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, any linear combination of positive powers of $\pi$ with rational coefficients is non-zero.&#xA;In particular, $\pi^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My problem with Moron's answer is that for example the unit roots of order 3 won't be the root of any polynomial $ax^n + b$ for rational $a,b$.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-12T22:40:31.647" CommentCount="3" />
  <row Id="2307" PostTypeId="2" ParentId="2300" CreationDate="2010-08-12T22:40:42.393" Score="2" ViewCount="0" Body="&lt;p&gt;The references can be found in the comments here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/34699/approaches-to-riemann-hypothesis-using-methods-outside-number-theory/34718#34718&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/34699/approaches-to-riemann-hypothesis-using-methods-outside-number-theory/34718#34718&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For those who don't know it, the (well, a) Lie-theoretic proof of Fund. Thm. of Algbera due to Witt is on p. 245 of the book &quot;Numbers&quot; by Ebbinghaus et al. – KConrad Aug 6 at 4:44&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Witt's Lie-theoretic proof of Fund. Thm. of Algbera seems to be Witt (Ernst), Über einen Satz von Ostrowski, Arch. Math. (Basel) 3, (1952). 334. – Chandan Singh Dalawat Aug 6 at 6:11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-08-12T22:40:42.393" CommentCount="6" />
  <row Id="2308" PostTypeId="2" ParentId="2300" CreationDate="2010-08-12T22:41:57.773" Score="0" ViewCount="0" Body="&lt;p&gt;Muad: Are you talking about this www.fc.up.pt/mp/jcsantos/PDF/artigos/FTA.pdf&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-12T22:41:57.773" CommentCount="3" />
  <row Id="2310" PostTypeId="2" ParentId="2260" CreationDate="2010-08-12T22:57:36.210" Score="3" ViewCount="0" Body="&lt;p&gt;How many ways are there to choose a $2$-element subset out of an $n$-element set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, you can choose the first element of the set in $n$ ways, then the second element of the set in $n-1$ ways, then divide by $2$ because it doesn't matter which you choose first and which you choose second.  This gives $\frac{n(n-1)}{2}$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose the $n$ elements are $1, 2, 3, ... n$, and suppose the larger of the two elements you choose is $j$.  Then for every $j$ between $2$ and $n$ there are $j-1$ possible choices of the smaller of the two elements, which can be any of $1, 2, ... j-1$.  This gives $1 + 2 + ... + (n-1)$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the two expressions above count the same thing, they must be equal.  This is known as the principle of double counting, and it is one of a combinatorialist's favorite weapons.  A generalization of this argument allows one to deduce the sum of the first $n$ squares, cubes, fourth powers...&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-12T22:57:36.210" />
  <row Id="2312" PostTypeId="2" ParentId="1533" CreationDate="2010-08-12T23:50:23.967" Score="2" ViewCount="0" Body="&lt;p&gt;A very good textbook for discrete mathematics at an undergraduate level is the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073229725&quot; rel=&quot;nofollow&quot;&gt;Kenneth Rosen book titled Discrete Mathematics and Its Applications&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book provides solutions to half of the problems. You can also buy the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073107794&quot; rel=&quot;nofollow&quot;&gt;Student's Solutions Guide&lt;/a&gt;. I don't own it, but I would suspect that it either provides the answers to the other half of the questions or provides a step-by-step guide to solving the problems (the book only provides final answers with minimal explanations of those answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's used for the two-quarter sequence in Discrete Mathematics that is taken by computer science and software engineering majors, as well as a number of mathematics programs at my university. I kept this book around even after I took the course, and I'm currently using it to brush up on my discrete math skills for my Certified Software Development Associate exam.&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-13T07:49:11.017" LastActivityDate="2010-08-13T07:49:11.017" CommunityOwnedDate="2010-08-12T23:50:23.967" />
  <row Id="2314" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T00:16:50.343" Score="6" ViewCount="0" Body="&lt;p&gt;One can prove the fundamental theorem of algebra using the fundamental group of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a generalization of the Hairy Ball Theorem, one may compute, for all spheres, the maximum number of linearly independent vector fields that may live on that sphere. (the hairy ball theorem says that this number is 0 for even dimensional spheres, and at least 1 for odd-dimensional spheres). For dimensions less than 15 (I think), one can compute this number using only cohomology operations. The general result was proven by Adams, but I'm not sure how he did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note: I don't think you can prove the general versions of Borsuk-Ulam, Brouwer, etc. with just the fundamental group... You need either higher homotopy groups or higher homology groups. &lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-13T00:16:50.343" CommunityOwnedDate="2010-08-13T00:16:50.343" CommentCount="1" />
  <row Id="2315" PostTypeId="1" AcceptedAnswerId="2316" CreationDate="2010-08-13T00:24:38.533" Score="3" ViewCount="479" Body="&lt;p&gt;I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The rank of a matrix A is the number&#xA;  of non-zero rows in the reduced&#xA;  row-echelon form of A&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lecturer then explained that if the matrix $A$ has size $m * n$, then $rank(A) \leq m$ and $rank(A) \leq n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I had been taught about rank was that it was the smallest of &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the number of rows bringing new information&lt;/li&gt;&#xA;&lt;li&gt;the number of columns bringing new information. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't see how that would change if we transposed the matrix, so I said in the lecture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;then the rank of a matrix is the same of its transpose, right?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the lecturer said: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;oh, not so fast! Hang on, I have to think about it&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the class has about 100 students and the lecturer was just substituting for the &quot;normal&quot; lecturer, he was probably a bit nervous, so he just went on with the lecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tested &quot;my theory&quot; with one matrix and it works, but even if I tried with 100 matrices and it worked, I wouldn't have proven that it always works because there might be a case where it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is first whether I am right, that is, whether the rank of a matrix is the same as the rank of its transpose, and second, if that is true, how can I prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  :)&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="85" LastEditorDisplayName="" LastEditDate="2010-08-13T00:42:24.177" LastActivityDate="2010-08-13T00:42:24.177" Title="Is the rank of a matrix the same of its transpose? If yes, how can I prove it?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2316" PostTypeId="2" ParentId="2315" CreationDate="2010-08-13T00:31:40.423" Score="7" ViewCount="0" Body="&lt;p&gt;The answer is yes.  This statement often goes under the name &quot;row rank equals column rank&quot;.  Knowing that, it is easy to to search the internet for proofs, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://homepage.mac.com/ehgoins/ma265/lecture_28.pdf&quot; rel=&quot;nofollow&quot;&gt;http://homepage.mac.com/ehgoins/ma265/lecture_28.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also any reputable linear algebra text should prove this: it is indeed a rather important result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since you said that you had only a substitute lecturer, I won't castigate him, but this would be a distressing lacuna of knowledge for someone who is a regular linear algebra lecturer.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-13T00:31:40.423" CommentCount="1" />
  <row Id="2317" PostTypeId="2" ParentId="2315" CreationDate="2010-08-13T00:33:11.120" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, it is a fact. This is true over any commutative field. See for instance the first chapter of Emil Artin, Galois Theory for a very elementary argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to phrase that argument in more conceptual terms, consider the matrices as linear transformations. If A is the matrix, then let $A^t$ be the transpose, and then $A^tA$ and $A$ have the same domain, and use the fact that they have the same null space, and use the dimension theorem rank + nullity = dimension of the space.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T00:33:11.120" CommentCount="5" />
  <row Id="2318" PostTypeId="2" ParentId="2269" CreationDate="2010-08-13T00:34:31.103" Score="1" ViewCount="0" Body="&lt;p&gt;Take a look at equation 34 on &lt;a href=&quot;http://mathworld.wolfram.com/GammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/GammaFunction.html&lt;/a&gt; After taking the log of both sides, the zeta function appears in essentially the same form as in your equation. They reference a 1968 paper by Wrench for more information. I haven't worked through the details myself, but the similarities are striking, and it looks like a promising approach.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2010-08-13T00:34:31.103" />
  <row Id="2319" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T00:55:15.570" Score="7" ViewCount="0" Body="&lt;p&gt;The only finite-dimensional division algebras over $\mathbb{R}$ have dimensions 1, 2, 4, and 8.  Note that we do not require that there be an identity element. For weaker versions of this result, see pages 173 and 222 in Hatcher's book &lt;em&gt;Algebraic Topology&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: An earlier version of this answer stated, mistakenly, that the only finite-dimensional division algebras are $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, and $\mathbb{O}$.  For this to be true, one needs the additional hypothesis that there is an identity element.&lt;/p&gt;&#xA;" OwnerUserId="455" LastEditorUserId="455" LastEditorDisplayName="" LastEditDate="2010-08-13T13:58:34.090" LastActivityDate="2010-08-13T13:58:34.090" CommunityOwnedDate="2010-08-13T00:55:15.570" CommentCount="1" />
  <row Id="2320" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T01:18:36.973" Score="3" ViewCount="0" Body="&lt;p&gt;This is my favorite.  One can show that for any continuous map from $S^{1}$ to $R^{3}$ there is a direction along which the map has at least 4 extrema (in particular, at least 2 global minima and 2 global maxima.)  More colloquially, one can show that every potato chip can be placed on a table so its edge touches the table in at least two points and its edge simultaneously has two points of maximum height. &lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="948" LastEditorDisplayName="" LastEditDate="2010-08-13T02:51:31.960" LastActivityDate="2010-08-13T02:51:31.960" CommunityOwnedDate="2010-08-13T01:18:36.973" CommentCount="1" />
  <row Id="2321" PostTypeId="1" AcceptedAnswerId="2327" CreationDate="2010-08-13T02:27:14.560" Score="3" ViewCount="215" Body="&lt;p&gt;Let $p_N/q_N$ be the $N^\text{th}$ convergent of the continued fraction for some irrational number $\alpha$. It turns out that for any other approximation $p/q$ (with $q \le q_N$) which isn't a convergent $|\alpha q - p| &gt; |\alpha q_{N-1} - p_{N-1}|$. I'm wondering if there are any nice proofs for this result?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In my book this is proved by picking $x,y$ that solves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix} p_N &amp;amp; p_{N-1} \\ q_N &amp;amp; q_{N-1} \end{pmatrix}&#xA;\begin{pmatrix} x \\ y \end{pmatrix} =&#xA;\begin{pmatrix} p \\ q \end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $x$ and $y$ have opposite sign, as well as $\alpha q_N - p_N$ and $\alpha q_{N-1} - p_{N-1}$ have opposite sign we can conclude that $|\alpha q - p| = |x (\alpha q_N - p_N) + y (\alpha q_{N-1} - p_{N-1})| = |x| |\alpha q_N - p_N| + |y| |\alpha q_{N-1} - p_{N-1}|$ which proves the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for different proofs than this one.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-23T13:16:01.510" Title="A nicer proof of Lagrange's 'best approximations' law?" Tags="&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2322" PostTypeId="2" ParentId="457" CreationDate="2010-08-13T03:04:55.733" Score="2" ViewCount="0" Body="&lt;p&gt;Katie Banks wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one can restate Euler's proof that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ in terms of the flow of an incompressible fluid with sources at the integer points in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody please elaborate or provide a reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1127" LastActivityDate="2010-08-13T03:21:37.753" CommunityOwnedDate="2010-08-13T03:04:55.733" CommentCount="2" />
  <row Id="2323" PostTypeId="2" ParentId="2093" CreationDate="2010-08-13T04:53:40.637" Score="1" ViewCount="0" Body="&lt;p&gt;I think this problem is still open, but the following might be useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ray-Chaudhuri-Wilson Theorem&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L$ be a set of $m$ integers and $F$ be an $L$-intersecting $k$-uniform family of subsets of a set of $n$ elements, where $m \le k$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|F| \le {n \choose m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$-uniform family is a set of subsets, each subset being of size k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $L$-intersecting family is such that the intersection size of any two distinct sets in the family is in $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following result of Frankl gives us a lower bound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Frankl's Result:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $k \ge m \ge 1$ and $n \ge 2k^{2}$ there exists a $k$-uniform family $F$ of size $&gt; (\frac{n}{2k})^{m}$ on $n$ points such that $|A \cap B| \le m-1$ for any two distinct sets $A,B \in F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an algorithm for constructing such sets (based on Frankl's result) refer: &lt;a href=&quot;http://stackoverflow.com/questions/2955318/creating-combinations-that-have-no-more-one-intersecting-element/2955527#2955527&quot;&gt;http://stackoverflow.com/questions/2955318/creating-combinations-that-have-no-more-one-intersecting-element/2955527#2955527&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-13T06:00:40.287" LastActivityDate="2010-08-13T06:00:40.287" />
  <row Id="2324" PostTypeId="1" CreationDate="2010-08-13T05:32:28.170" Score="2" ViewCount="104" Body="&lt;p&gt;What might be some potential methods to encode a 100-point signal (curve) for input to a Artificial Neural Network?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: we have a large number of 100-pt 'curves' ranging from flat-line to approximately a half-sine wave with a wide range of max amplitude.  The curves have some noise and occasional anomalous 'bumps'.  These training set curves can be broken down into known positives or negatives for 'truthing'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wanted to use an Artificial Neural Network to give a result for untrained data curves, what are potential method to encode the input data sets?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would transforming each of the 100 points in a curve to a [0.0000 .... 1.0000] range be useful (and thus we'd have 100 inputs to the ANN?  Or another method might produce better results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I understand that FFT and Power Spectrum analysis might be a completely different alternative here as well, but am focussing right now on a possible neural network method.&lt;/p&gt;&#xA;" OwnerUserId="1131" LastActivityDate="2010-12-14T18:44:25.953" Title="Encoding a Discrete Signal to an Artificial Neural Network?" Tags="&lt;signal-processing&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2325" PostTypeId="1" CreationDate="2010-08-13T05:55:57.990" Score="2" ViewCount="79" Body="&lt;p&gt;I have recently been thnking some about algebraic groups and reading parts of Humphreys book on them, and I was wondering if there is a general process to showing they are simply connected. In particular I was wondering over other fields than C but if the answer only works in C I will settle for that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea I had was that may e using borel-weil-bott one could make a slick arguement for when the fundamental group is trivial. I would just like to get away from the ad-hoc thinking process I am using.  &lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-13T09:00:56.663" Title="Checking if Algebraic Groups are simply connected. " Tags="&lt;algebraic-geometry&gt;&lt;algebraic-groups&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2326" PostTypeId="2" ParentId="457" CreationDate="2010-08-13T06:27:59.577" Score="7" ViewCount="0" Body="&lt;p&gt;From wikipedia article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorema_Egregium&quot; rel=&quot;nofollow&quot;&gt;Gauss Theorema Egregium&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An application of the Theorema Egregium is seen in a common pizza-eating strategy: A slice of pizza can be seen as a surface with constant Gaussian curvature 0. Gently bending a slice must then roughly maintain this curvature (assuming the bend is roughly a local isometry). If one bends a slice horizontally along a radius, non-zero principal curvatures are created along the bend, dictating that the other principal curvature at these points must be zero. This creates rigidity in the direction perpendicular to the fold, an attribute desirable when eating pizza (since it prevents the pizza toppings from falling off).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1043" LastActivityDate="2010-08-13T06:27:59.577" CommunityOwnedDate="2010-08-13T06:27:59.577" />
  <row Id="2327" PostTypeId="2" ParentId="2321" CreationDate="2010-08-13T06:32:09.227" Score="3" ViewCount="0" Body="&lt;p&gt;Of course there is a nicer proof! In fact, it's almost obvious if one thinks about geometric interpretation of continued fraction: consider the line $y=\alpha x$; then the best approximation (i.e. approximation that minimizes $|\alpha q-p|=q|\alpha-\frac{p}{q}|$) is the point of integer lattice nearest to this line; finally observe that convergents with even/odd numbers of the continued fraction give coordinates of [verices of] the convex hull of [points of] lattice lying over/under the line.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One can also (as &lt;a href=&quot;http://math.stackexchange.com/users/498/j-mangaldan&quot;&gt;J. Mangaldan&lt;/a&gt; suggests) take a look at &lt;em&gt;Lorentzen, Waadeland. Continued Fractions with Applications&lt;/em&gt;, I.2.1 (esp. figure 1 and the text near it; there are no words &quot;convex hull&quot; there but implicitly everything is explained, more or less).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upd. One more reference is Davenport's &quot;The Higher Arithmetics&quot; (section IV.12).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, an illustration (from Arnold's book).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bold line $y=\alpha x$ (on the picture $\alpha=\frac{10}7$) is approximated by vectors $e_i$ corresponding to convergents (namely, $e_i$ ends at $(p_i,q_i)$);&lt;/li&gt;&#xA;&lt;li&gt;each vector (starting from $e_3$) is a linear combination of two preciding ones: $e_{i+2}=e_i+a_{i-1}e_{i+1}$, where $a$ is the maximal integer s.t. the new vector still doesn't intersect the line;&lt;/li&gt;&#xA;&lt;li&gt;this is exactly the algorithm for representing $\alpha$ by a continued fraction: $\alpha=[a_0,a_1,\dots]$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/coRwt.png&quot; alt=&quot;geometry of continued fractions&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-23T13:16:01.510" LastActivityDate="2010-08-23T13:16:01.510" CommentCount="9" />
  <row Id="2329" PostTypeId="2" ParentId="155" CreationDate="2010-08-13T07:34:38.773" Score="9" ViewCount="0" Body="&lt;p&gt;Have you ever heard of Galois theory? It is a theory that studies the solutions of equations over fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, there is a special type of Galois theory called Differential Galois theory, which studies fields with a differential operator on them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_galois_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Differential_galois_theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this theory, one can prove that functions like sin(x)/x and x^x don't have an indefinite integral.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-13T07:34:38.773" CommentCount="1" />
  <row Id="2330" PostTypeId="1" CreationDate="2010-08-13T07:58:30.463" Score="1" ViewCount="97" Body="&lt;p&gt;I need to find the minimum of a function f(t) = int g(t,x) dx over [0,1]. What I did in mathematica is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[t_] = NIntegrate[g[t,x],{x,-1,1}]&#xA;FindMinimum[f[t],{t,t0}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However mathematica halts at the first try, because NIntegrate does not work with the symbolic t. Although Plot[f[t],{t,0,1}] works perferctly, FindMinimum stops at the initial point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way to get around it? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1132" LastActivityDate="2010-08-13T18:49:53.220" Title="find minimum of a function defined by integration in Mathematica" Tags="&lt;math-software&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2331" PostTypeId="2" ParentId="2325" CreationDate="2010-08-13T09:00:56.663" Score="1" ViewCount="0" Body="&lt;p&gt;Please see &lt;a href=&quot;http://mathoverflow.net/questions/19262/simply-connectedness-of-algebraic-group&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/19262/simply-connectedness-of-algebraic-group&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-13T09:00:56.663" CommentCount="1" />
  <row Id="2332" PostTypeId="1" CreationDate="2010-08-13T10:18:53.667" Score="0" ViewCount="135" Body="&lt;p&gt;Is the transpose of (A-B) where A and B are matrices equal to A'-B'?&lt;/p&gt;&#xA;" OwnerUserId="1134" LastActivityDate="2010-08-13T10:48:20.080" ClosedDate="2010-08-13T12:29:43.067" Title="Is the transpose of (A-B) where A and B are matrices equal to A'-B'?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2333" PostTypeId="2" ParentId="2332" CreationDate="2010-08-13T10:30:31.803" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, transposition of matrices is linear. You can check this directly using the definition of the transpose.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-13T10:48:20.080" LastActivityDate="2010-08-13T10:48:20.080" />
  <row Id="2334" PostTypeId="2" ParentId="155" CreationDate="2010-08-13T10:37:36.773" Score="4" ViewCount="0" Body="&lt;p&gt;A key result in this area is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28differential_algebra%29#Basic_theorem&quot; rel=&quot;nofollow&quot;&gt;theorem by Liouville&lt;/a&gt;. You'll find easy expositions of this result in the references below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Antoine Chambert-Loir. A Field Guide to Algebra, last chapter, Springer, 2005.&lt;/li&gt;&#xA;&lt;li&gt;M Rosenlicht. Integration in finite terms, Amer.    Math. Monthly,&#xA;    (1972), pp 963-972.&lt;/li&gt;&#xA;&lt;li&gt;Toni Kasper. Integration in ﬁnite&#xA;terms: the Liouville theory. Math.&#xA;Mag.,  53(4):195–201, 1980.&lt;/li&gt;&#xA;&lt;li&gt;D.    H. Potts. Elementary&#xA;integrals. Amer.    Math. Monthly,&#xA;63:545–554, 1956.&lt;/li&gt;&#xA;&lt;li&gt;A. D. Fitt and G. T. Q. Hoare. The closed-form integration of arbitrary functions,&#xA;The Mathematical Gazette, Vol. 77, No. 479 (Jul., 1993), pp. 227-236&lt;/li&gt;&#xA;&lt;li&gt;Elena Anne Marchisotto and Gholam-Ali Zakeri. An invitation to integration in finite terms, The College Mathematics Journal, Vol. 25, No. 4 (Sep., 1994), pp. 295-308&lt;/li&gt;&#xA;&lt;li&gt;D. G. Mead. Integration, The American Mathematical Monthly, Vol. 68, No. 2 (Feb., 1961), pp. 152-156 &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-13T10:37:36.773" CommentCount="1" />
  <row Id="2335" PostTypeId="1" AcceptedAnswerId="2442" CreationDate="2010-08-13T10:49:30.720" Score="-3" ViewCount="261" Body="&lt;p&gt;Let ${f_n}$ be a sequence of functions which are either (strictly increasing or strictly decreasing) on $\mathbb{R}$ and uniformly bounded on $\mathbb{R}$. Prove that ${f_{n}}$ contains a subsequence which is pointwise convergent on $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-13T13:39:57.393" LastActivityDate="2010-08-14T07:44:14.100" Title="Subsequence which is pointwise convergent " Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="2336" PostTypeId="2" ParentId="2130" CreationDate="2010-08-13T11:53:49.693" Score="0" ViewCount="0" Body="&lt;p&gt;As a rule, &quot;small&quot; things live most naturally in the &quot;thing of things.&quot; In this case, the category Cat of small categories is locally small, but the category CAT of all categories is not. As most categorical constructions (adjunctions, universals, ...) assume hom(A,B) is a set, CAT can be problematic to reason about.&lt;/p&gt;&#xA;" OwnerUserId="1136" LastActivityDate="2010-08-13T11:53:49.693" />
  <row Id="2337" PostTypeId="2" ParentId="2305" CreationDate="2010-08-13T12:37:22.273" Score="2" ViewCount="0" Body="&lt;p&gt;Making explicit what is implicit in Qiaochu Yuan's comment, and demonstrating that &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.97.4244&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;someone else's work&lt;/a&gt; has failed to evade my eyes. (It is a neat article, read it.)  I present this adaptation of Duval's algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assign an order to your symbols, say $a_1, a_2, a_1^{-1}, a_2^{-1}$ let first_symbol and _last_symbol be the first and last symbols in the set.  Let next be a function that gives the next symbol in sequence. The function conflict checks to see if the two symbols are inverses of each other.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;w[1] &amp;lt;- first_symbol&#xA;i &amp;lt;- 1&#xA;repeat&#xA;  for j = 1 to n–i&#xA;    do w[i+j] &amp;lt;- w[j]&#xA;  if i = n and not conflict(w[1], w[n])&#xA;    then output w[1] ... w[n]&#xA;  i &amp;lt;- n&#xA;  while i &amp;gt; 0 and w[i] = last_symbol&#xA;    do i &amp;lt;- i–1&#xA;  if i &amp;gt; o  &#xA;     then w[i] &amp;lt;- next(w[i])&#xA;  if i &amp;gt; 1 and conflict(w[i-1], w[i]) &#xA;     then w[i] &amp;lt;- next(w[i])&#xA;until i = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is just Duval's algorithm for generating a list of the lexicographically minimal cyclic shifts with extra checks to step over the cases where a conflict should occur. I have neither bothered to work out either a formal proof that this works, or implemented it in actual code.  Caveat Emptor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; As expected, I missed a corner case.  The following python code appears to work. It takes the length of the cycle and a list of integers (I use integers for the group)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def cycles(n,l):&#xA;    w = range(n+1)&#xA;    m = len(l) - 1&#xA;    w[1] = 0&#xA;    i = 1&#xA;    while i &amp;gt; 0:&#xA;        for j in range(n-i):&#xA;            w[j + i + 1] = w[j + 1]&#xA;        if i == n and l[w[1]] + l[w[n]] != 0:&#xA;            print [l[w[i]] for i in xrange(1,n+1)]&#xA;        i = n&#xA;        while i &amp;gt; 0 and w[i] == m:&#xA;            i = i - 1&#xA;        while i &amp;gt; 0:&#xA;            if i &amp;gt; 0:&#xA;                w[i] = w[i] + 1&#xA;            if i &amp;gt; 1 and l[w[i-1]] + l[w[i]] == 0:&#xA;                w[i] = w[i] + 1&#xA;            if w[i] &amp;lt;= m:&#xA;                break&#xA;            i = i - 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to get the length four cycles for {-2, -1, 1, 2} call&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cycles(4, [-2, -1, 1, 2])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;resulting in&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[-2, -2, -2, -1]&#xA;[-2, -2, -2, 1]&#xA;[-2, -2, -1, -1]&#xA;[-2, -2, 1, 1]&#xA;[-2, -1, -2, 1]&#xA;[-2, -1, -1, -1]&#xA;[-2, -1, 2, -1]&#xA;[-2, -1, 2, 1]&#xA;[-2, 1, 1, 1]&#xA;[-2, 1, 2, -1]&#xA;[-2, 1, 2, 1]&#xA;[-1, -1, -1, 2]&#xA;[-1, -1, 2, 2]&#xA;[-1, 2, 1, 2]&#xA;[-1, 2, 2, 2]&#xA;[1, 1, 1, 2]&#xA;[1, 1, 2, 2]&#xA;[1, 2, 2, 2]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ahem&lt;/strong&gt; Didn't I say&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def cycles(n,l):&#xA;    w = range(n+1)&#xA;    m = len(l) - 1&#xA;    w[1] = 0&#xA;    i = 1&#xA;    while i &amp;gt; 0:&#xA;        for j in range(n-i):&#xA;            w[j + i + 1] = w[j + 1]&#xA;        if (i == n) and ((l[w[1]] + l[w[n]]) != 0):&#xA;            print [l[w[i]] for i in xrange(1,n+1)]&#xA;        i = n&#xA;        while i &amp;gt; 0 and w[i] == m:&#xA;            i = i - 1&#xA;        while i &amp;gt; 0:&#xA;            if i &amp;gt; 0:&#xA;                w[i] = w[i] + 1&#xA;            if (i &amp;gt; 1) and ((l[w[i-1]] + l[w[i]]) == 0):&#xA;                w[i] = w[i] + 1&#xA;            if w[i] &amp;lt;= m:&#xA;                break&#xA;            i = i - 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That's what I should have said if I took my own advice.  Sorry.&lt;/p&gt;&#xA;" OwnerUserId="943" LastEditorUserId="943" LastEditorDisplayName="" LastEditDate="2010-08-15T01:44:37.713" LastActivityDate="2010-08-15T01:44:37.713" CommentCount="5" />
  <row Id="2338" PostTypeId="1" AcceptedAnswerId="2342" CreationDate="2010-08-13T12:45:26.733" Score="6" ViewCount="179" Body="&lt;p&gt;Is it true that sin(n&lt;sup&gt;k&lt;/sup&gt;) ≠ (sin n)&lt;sup&gt;k&lt;/sup&gt; for any positive integers n and integers k ≠ 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if n (&gt; 0), k (≠ 1) are rational?&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-13T13:12:44.917" Title="Is sin(n^k) ≠ (sin n)^k in general?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="2339" PostTypeId="1" AcceptedAnswerId="2343" CreationDate="2010-08-13T12:54:33.847" Score="4" ViewCount="273" Body="&lt;p&gt;How can you prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for every integer $n\geq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: no, it's not a homework... :-)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-13T13:27:13.773" LastActivityDate="2010-08-13T13:34:58.093" Title="Prove that  $\sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2340" PostTypeId="2" ParentId="2293" CreationDate="2010-08-13T13:04:14.997" Score="2" ViewCount="0" Body="&lt;p&gt;Beginning students might appreciate applications outside mathematics. Here are a couple of current (undergraduate) projects: &lt;a href=&quot;http://inperc.com/wiki/index.php?title=The_topology_of_data&quot; rel=&quot;nofollow&quot; title=&quot;The topology of data&quot;&gt;The topology of data&lt;/a&gt; and  3D image analysis. The tool used is called &quot;persistent homology&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1138" LastEditorUserId="1138" LastEditorDisplayName="" LastEditDate="2010-08-13T16:20:49.523" LastActivityDate="2010-08-13T16:20:49.523" CommunityOwnedDate="2010-08-13T13:04:14.997" />
  <row Id="2341" PostTypeId="1" CreationDate="2010-08-13T13:11:30.677" Score="0" ViewCount="212" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new at learning boolean algebra and understand the basic laws. But now I'm trying to proof:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x'z+ xyz + xy'z = z&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can someone help me with this please ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1139" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T13:37:17.447" LastActivityDate="2010-08-14T03:38:08.367" Title="boolean algebra proof" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="2342" PostTypeId="2" ParentId="2338" CreationDate="2010-08-13T13:12:44.917" Score="11" ViewCount="0" Body="&lt;p&gt;If $\sin(n^k)=\sin(n)^k$, then $\frac{e^{in^k}-e^{-in^k}}{2i}=(\frac{e^{in}-e^{-in}}{2i})^k$, so that $e^i$ would be algebraic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that directly contradicts the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Lindemann-Weierstrass theorem&lt;/a&gt; because $i$ is algebraic.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-13T13:12:44.917" CommentCount="1" />
  <row Id="2343" PostTypeId="2" ParentId="2339" CreationDate="2010-08-13T13:21:50.230" Score="8" ViewCount="0" Body="&lt;p&gt;By well know formula we have&#xA;$$&#xA;(\cos{\frac{k\pi}{2n}} + i\sin{\frac{k\pi}{2n}})^{2n}=(-1)^{k}&#xA;$$&#xA;Hence by &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_formula&quot; rel=&quot;nofollow&quot;&gt;Binomial theorem&lt;/a&gt; we have ($[x]$ is not an integer part, brackets are added for clarity)&#xA;$$&#xA;\sum_{t=0}^{2n}\binom{2n}{t}[\cos{\frac{k\pi}{2n}}]^t \cdot [i\sin{\frac{k\pi}{2n}}]^{2n-t}=(-1)^{k}&#xA;$$&#xA;Now we consider only imaginary part of this:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[\cos{\frac{k\pi}{2n}}]^{2r+1} \cdot [i\sin{\frac{k\pi}{2n}}]^{2n-2r-1}=0&#xA;$$&#xA;Divide it by $[\cos{\frac{k\pi}{2n}}]^{2n}$:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[i\tan{\frac{k\pi}{2n}}]^{2n-2r-1}=0&#xA;$$&#xA;Now multiply by $i\tan{\frac{k\pi}{2n}}$:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[i\tan{\frac{k\pi}{2n}}]^{2n-2r}=0&#xA;$$&#xA;So $[\tan{\frac{k\pi}{2n}}]^2$ are roots of the following polynomial:&#xA;$$&#xA;\sum_{r=0}^{n-1}\binom{2n}{2r+1}[i\cdot x]^{n-r}=0&#xA;$$&#xA;Hence by &lt;a href=&quot;http://en.wikipedia.org/wiki/Vieta%27s_formulas&quot; rel=&quot;nofollow&quot;&gt;Vieta's_formulas&lt;/a&gt; sum of it roots is equal to&#xA;$$&#xA;\frac{\binom{2n}{3}}{\binom{2n}{1}} =\frac{(2n-1)(n-1)}{3}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-13T13:21:50.230" CommentCount="1" />
  <row Id="2344" PostTypeId="2" ParentId="2339" CreationDate="2010-08-13T13:24:16.607" Score="0" ViewCount="0" Body="&lt;p&gt;I am doing a similar thing for $\cot$ i hope you can reciprocate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $$(\cot(\theta)-i)^{n} = \frac{ \cos{n\theta} - \sin{n \theta}}{\sin^{n}(\theta)}$$&#xA;Equating the real and imaginary parts on both sides we have $$ \frac{\sin(n\theta)}{\sin{\theta}} = \sum\limits_{s} {n \choose 2s+1}  (-1)^{s} \cot^{n-2s-1}(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $2n+1$ instead of $n$ we have, $$\sin{(2n+1)\theta} = \sin^{2n+1}(\theta)P_{n}\cot^{2}(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\displaystyle 0 &amp;lt; \theta &amp;lt; \frac{\pi}{2}$ and where $P_{n}$ is the polynomial given by $${ 2n+1 \choose 1}T^{n} - {2n+1 \choose 3}T^{n-1} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that the zeros of $P_{n}$ are precisely $\displaystyle \frac{r\pi}{2n+1},\ n =1,2,..$, we have the first identity from the sum of the roots formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the inequality $\sin{x} &amp;lt; x &amp;lt; \tan{x}$ we have $$\cot^{2m}(x) &amp;lt; \frac{1}{x^{2m}} &amp;lt; (1 + \cot^{2}(x))^{m}$$ we have $$ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1}  &amp;lt; \frac{(2n+1)^2m}{\pi^{2m}} \sum\limits_{r=1}^{n} \frac{1}{r^{2m}} &amp;lt; \sum\limits_{r=1}^{n} \Bigl( 1 + \cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $$\sum\limits_{r=1}^{n} \Bigl(1+\cot^{2} \frac{r\pi}{2n+1} \Bigr)^{m} =  \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} + \mathcal{O}(n^{2m-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words to find $c_{2m}$ where $ \sum\limits_{r=1}^{n} \cot^{2m} \frac{r\pi}{2n+1} = c_{2m}n^{2m} + \mathcal{O}(n^{2m-1})$ it suffices to look at the sum, $$ \cot^{2m} \frac{\pi}{2n+1} + \cot^{2m} \frac{2\pi}{2n+1} + \cdots + cot^{2m} \frac{n\pi}{2n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the sum $s_{m}$ of $m-th$ powers of the roots of the polynomial $P_{n}$. Then you can complete it using newtons formula for the sum of the roots.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-13T13:34:58.093" LastActivityDate="2010-08-13T13:34:58.093" CommentCount="0" />
  <row Id="2345" PostTypeId="2" ParentId="2341" CreationDate="2010-08-13T13:29:00.290" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming you mean what is more commonly written as $(\lnot x\land z)\lor(x\land y\land z)\lor(x\land\lnot y\land z)=z$, then it is equivalent, as Tom Stephens points out, to proving that $\lnot x\lor(x\land(y\lor\lnot y))$ is valid (that is, evaluates to true for all x and y). But that's clearly the same as $\lnot x \lor x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;distributivity: $x\lor(y\land z)=(x\lor y)\land(x\lor z)$ and also $x\land(y\lor z)=(x\land y)\lor(x\land z)$&lt;/li&gt;&#xA;&lt;li&gt;excluded middle: $x\lor\lnot x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-13T13:29:00.290" CommentCount="3" />
  <row Id="2346" PostTypeId="1" AcceptedAnswerId="2388" CreationDate="2010-08-13T13:33:55.397" Score="6" ViewCount="360" Body="&lt;p&gt;How does one set out to find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ which satisfy $f^{k}(x)=f(x^k)$ , where $k \in \mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation : &lt;a href=&quot;http://math.stackexchange.com/questions/2338/is-sinnk-sin-nk-in-general&quot;&gt;http://math.stackexchange.com/questions/2338/is-sinnk-sin-nk-in-general&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T08:01:05.733" Title="Continuous function Satisfying $f^{k}(x)=f(x^k)$" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="3" />
  <row Id="2347" PostTypeId="2" ParentId="2346" CreationDate="2010-08-13T13:47:42.883" Score="3" ViewCount="0" Body="&lt;p&gt;For a continuous $f:[0,1] \rightarrow (0, \infty)$ it is easy to show that $f(x)=1$ for $\forall x \in [0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-13T13:47:42.883" CommentCount="3" />
  <row Id="2348" PostTypeId="2" ParentId="2227" CreationDate="2010-08-13T14:15:46.170" Score="0" ViewCount="0" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All, please see this example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G_{n}$ denote the grouping of this $2n$ numbers, $$\frac{1}{2n-1},\frac{1}{2n-3},...,\frac{1}{3},1,-1,-\frac{1}{3},\cdots,-\frac{1}{2n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take a strictly increasing sequence of positive integers ${\lambda_n}$ and consider the groups $G_{\lambda_1},G_{\lambda_2},\cdots,$. We multiply each number of the group $G_{\lambda_n}$ by $n^{-2}$ and obtain the sequence $$\frac{1}{1^{2}(2\lambda_{1}-1)}, \cdots,-\frac{1}{1^{2}(2\lambda_{1}-1)}, \frac{1}{2^{2}(2\lambda_{2}-1)},...,-\frac{1}{2^2(2\lambda_{2}-1)},....,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;say $\alpha_{1},\alpha_{2},\cdots$. Our aim is to show that $$\sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$&#xA;is the fourier series of a continuous function. We group the terms in the following way $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2\lambda_{1}+1}^{2\lambda_{1}+2\lambda_{2}} \alpha_{n}\cos{nx} + \sum\limits_{n=2\lambda_{1}+2\lambda_{2}+1}^{2\lambda_{1}+2\lambda_{2}+2\lambda_{3}} \alpha_{n} \cos{nx}\cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last series can be written as $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $$\phi(n,r,x)= \frac{\cos{(r+1)x}}{2n-1} + \frac{\cos{(r+2)x}}{2n-3} + \cdots + \frac{\cos{(r+n)x}}{1} - \frac{\cos{(r+n+1)x}}{1} - \cdots - \frac{\cos{(r+2n)x}}{2n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one can show that there is a constant $M$ (independent of $n,r$ and $x$) such that $|\phi(n,r,x)|\leq M$. From this it follows that the grouped series $$\sum\limits_{n=1}^{2\lambda_{1}} \alpha_{n} \cos{nx} + \sum\limits_{n=2}^{\infty} \frac{\phi(\lambda_{n},2\lambda_{1}+2\lambda_{2} + \cdots + 2\lambda_{n-1},x)}{n^2}$$&#xA;converges absolutely on $\mathbb{R}$, say to $f(x)$, and $f$ is continuous on $\mathbb{R}$. It is also easy to check that $$f(x) \sim \sum\limits_{n=1}^{\infty} \alpha_{n} \cos{nx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall finally show that ${\lambda_n}$ can be chose so that the above series diverges at zero, that is $S_{n} = \alpha_{1} + \alpha_{2} +  \cdots + \alpha_{n}$ diverges to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $$S_{2\lambda_{1}+2\lambda_{2}+ \cdots + 2 \lambda_{n-1} + \lambda_{n}} = \frac{1}{n^2} \Bigl( \frac{1}{2\lambda_{n}-1} + \cdots + \frac{1}{3} + 1 \Bigr)$$ behaves as $\frac{\ln{\lambda_{n}}}{{2n^{2}}}$ as $n \to \infty$, it is enough to take $\lambda_{n}=n^{n^2}$. Then the fourier series does not converge to $f$ at $x=2k\pi, \ k\in \mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-13T14:15:46.170" />
  <row Id="2349" PostTypeId="1" AcceptedAnswerId="2350" CreationDate="2010-08-13T14:33:08.500" Score="2" ViewCount="439" Body="&lt;p&gt;I'm doing exercise on discrete mathematics and I'm stuck with question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $f$ is an invertible function from Y to Z and $g$ is an invertible function from X to Y&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Show that composition of $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've no idea how to prove this, please help me by give me some reference or hint to its solution.&lt;/p&gt;&#xA;" OwnerUserId="725" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-14T04:23:46.953" LastActivityDate="2010-08-16T13:05:17.240" Title="How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$ ?" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;&lt;homework&gt;" AnswerCount="5" CommentCount="6" />
  <row Id="2350" PostTypeId="2" ParentId="2349" CreationDate="2010-08-13T14:37:19.533" Score="5" ViewCount="0" Body="&lt;p&gt;Use the definition of an inverse and associativity of composition to show that the right hand side is the inverse of $(f \circ g)$.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-13T14:37:19.533" CommentCount="2" />
  <row Id="2351" PostTypeId="2" ParentId="2349" CreationDate="2010-08-13T14:38:14.617" Score="1" ViewCount="0" Body="&lt;p&gt;This is straightforward. Take x in the domain of f. It goes to f(x) = y. And g takes y to z = g(y). Therefore $g^{-1}$ takes z to y and $f^{-1}$ takes y to x. Both sides of your equation takes $z$ to $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please try to think more before asking. This was not hard, was it? :) &lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-13T14:48:23.940" LastActivityDate="2010-08-13T14:48:23.940" CommentCount="3" />
  <row Id="2352" PostTypeId="1" CreationDate="2010-08-13T14:54:26.093" Score="5" ViewCount="191" Body="&lt;p&gt;This is a strengthening of Chandru's question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/2227/&quot;&gt;http://math.stackexchange.com/questions/2227/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a nice and concrete example of a Fourier series that fails to converge on some &quot;big&quot; set of measure zero, for instance on the Cantor ternary set?&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:39:49.957" LastActivityDate="2010-10-13T11:01:25.923" Title="A Fourier series failing to converge on the Cantor Set" Tags="&lt;real-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="2353" PostTypeId="2" ParentId="2130" CreationDate="2010-08-13T15:00:30.383" Score="1" ViewCount="0" Body="&lt;p&gt;I will give one example of a statement that will go wrong if you are not careful with set theoretical foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal C$ be a category. Then the Yoneda lemma gives an embedding $\mathcal C \hookrightarrow Funct(\mathcal{C}, Sets)$ into the category of all functors from $\mathcal C$ to the category $Sets$ of suitably &quot;small&quot; sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $CAT$ be the category of all &quot;small&quot; categories. Again by the naturality of the Yoneda Construction, you get an embedding $CAT \hookrightarrow Funct(CAT, Sets)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not careful, various set-theoretical and logical problems might arise with the above statement. Here to really make sense of things without any problems anywhere, you have to fix a Grothendieck Universe and work within it, as far as I understand things.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:00:30.383" />
  <row Id="2354" PostTypeId="2" ParentId="2352" CreationDate="2010-08-13T15:04:30.670" Score="1" ViewCount="0" Body="&lt;p&gt;Hi--&#xA; I don't have an exact answer for this but i think this might help: &lt;a href=&quot;http://www.springerlink.com/content/p5t566u325582613/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/p5t566u325582613/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One more link: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this paper the authors show that for any set of zero measure there exists a continuous function on the circle whose Fourier series diverges on that set. In French.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sergei Vladimirovich Konyagin, On divergence of trigonometrique Fourier series everywhere, C. R. Acad. Sci. Paris 329 (1999), 693-697&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-13T15:04:30.670" CommentCount="3" />
  <row Id="2355" PostTypeId="1" AcceptedAnswerId="2531" CreationDate="2010-08-13T15:14:33.010" Score="19" ViewCount="565" Body="&lt;p&gt;The number of elements of order 2 in a group is fairly restricted: 0, odd, or infinity.  All such possibilities occur already in the trivial group and in dihedral groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of elements of order 3 in a group can be shown to be similarly restricted: 0, 2 mod 6, or infinity.  However, something strange happens: not all possibilities can be realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even worse, there is a fairly small number that I cannot decide whether it is or is not the number of elements of order 3 in a finite group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a group with exactly 92 elements of order 3?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More boldly, I would like to know (but feel free to answer only the first question):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Exactly which numbers occur as the number of elements of order 3 in a group?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;: Such questions were studied a bit by Sylow and more heavily by Frobenius.  The theorem that the number of elements of order p is equal to −1 mod p is contained in one of Frobenius's 1903 papers.  Since elements of order 3 come in pairs, this doubles to give 2 mod 6 for p=3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Frobenius's results were improved some 30 years later by P. Hall who showed that if the Sylow p-subgroups are not cyclic, then the number of elements of order p is −1 mod p&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Sylows are cyclic of order p&lt;sup&gt;n&lt;/sup&gt;, then the number of subgroups of order p is congruent to 1 mod p&lt;sup&gt;n&lt;/sup&gt; by the standard counting method.  If the Sylow itself is order p, then the subgroup generated by the elements of order p acts faithfully and transitively on the Sylow subgroups, so for small enough numbers, the subgroup can just be looked up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases, we can assume the group is finite since the subgroup generated by the elements of a fixed order is finite (assuming there are only finitely many elements of that fixed order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Easier example&lt;/strong&gt;: For instance there is no group with exactly 68 elements of order 3, since such a group would have cyclic Sylow 3-subgroups by Hall, order 3 Sylows by the counting, but then would have 34 Sylow 3-subgroups, and so (the subgroup generated by the elements of order 3 would) be a primitive group of degree 34.  One checks the list of primitive groups of degree 34 (that is, A&lt;sub&gt;34&lt;/sub&gt; and S&lt;sub&gt;34&lt;/sub&gt;, both with ginormous Sylow 3-subgroups) to see no such group exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could also try 140, but the action need not be primitive so the table lookup is harder.  Such a group has Sylows of order 3, but is not solvable, so is somewhat restricted.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-17T04:17:46.750" Title="Is there a group with exactly 92 elements of order 3?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="2356" PostTypeId="1" AcceptedAnswerId="2357" CreationDate="2010-08-13T15:16:50.157" Score="11" ViewCount="515" Body="&lt;p&gt;When flipping a coin to make important decisions in life you can flip once to choose between 2 possible outcomes. (Heads I eat cake, Tails I eat chocolate!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also flip twice to choose between 4 outcomes. (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you use a coin to choose evenly between three possible choices? If so how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ignoring slight abnormalities in weight)&lt;/p&gt;&#xA;" OwnerUserId="1099" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-14T17:59:44.437" LastActivityDate="2010-08-25T20:07:17.217" Title="Is it possible to 'split' coin flipping 3 ways?" Tags="&lt;probability&gt;&lt;fair-division&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="2357" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T15:21:12.017" Score="15" ViewCount="0" Body="&lt;p&gt;If you throw your coin $n$ times you have $2^n$ outcomes, the probability of each of which is $\frac{1}{2^n}$. The larger $n$ is, the better you can divide $2^n$ into three approximately equal parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just define $a_n=[2^n/3]$ and $b_n=[2\cdot 2^n/3]$, where $[\cdot]$ denotes rounding off (or on). Since $\frac{a_n}{2^n}\to\frac{1}{3}$ and $\frac{b_n}{2^n}\to\frac{2}{3}$ as $n\to\infty$, each of the three outcomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the number of Heads is between $0$ and $a_n$&quot;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the number of Heads is between $a_n$ and $b_n$&quot;, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the number of Heads is between $b_n$ and $2^n$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has approximately the probability $\frac{1}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Alternatively&lt;/em&gt;, you could apply your procedure to get four outcomes with the same probability (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads) to your problem in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associate the three outcomes Heads-Heads, Tails-Tails, Heads-Tails with your three possible choices. In the case that Tails-Heads occurs, just repeat the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sooner or later you will find an outcome different from Tails-Heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, by symmetry, the probability for first Heads-Heads, first Tails-Tails, or first Heads-Tails is $\frac{1}{3}$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Alternatively&lt;/em&gt;, you could of course throw a die and select your first choice if the outcome is 1 or 2, select your second choice if the outcome is 3 or 4, and select your third choice if the outcome is 5 or 6.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-13T22:19:18.823" LastActivityDate="2010-08-13T22:19:18.823" CommentCount="4" />
  <row Id="2358" PostTypeId="2" ParentId="1533" CreationDate="2010-08-13T15:28:08.370" Score="0" ViewCount="0" Body="&lt;p&gt;Mathematical Thinking: Problem-Solving and Proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John P. D'Angelo, Douglas B. West.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Available at &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130144126&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is supposed to be an introduction to mathematical proofs. As such it is not restricted to discrete mathematics. But it does a very good job for discrete mathematics. You would also see some proof in real analysis; but you can focus only on the discrete part ignoring this.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:28:08.370" CommunityOwnedDate="2010-08-13T15:28:08.370" />
  <row Id="2360" PostTypeId="1" AcceptedAnswerId="2362" CreationDate="2010-08-13T15:40:04.273" Score="3" ViewCount="336" Body="&lt;p&gt;I'm looking for the equation of a family of functions that roughly resembles the sketch below (with apologies for the crudeness of said sketch):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img202.imageshack.us/img202/6907/functionsketch.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Properties I'm looking for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\lim_{x\to-\infty}f(x)=y_1$ (i.e. approaches asymptote y=y&lt;sub&gt;1&lt;/sub&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{x\to+\infty}f(x)=y_2$ (i.e. approaches asymptote y=y&lt;sub&gt;2&lt;/sub&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$|f'(x)|$ is at a maximum at &lt;em&gt;x&lt;/em&gt;=0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I remember something involving e&lt;sup&gt;&lt;em&gt;x&lt;/em&gt;&lt;/sup&gt;/(??), but I can't remember exactly and can't seem to come up with the formula. I want to use it to weight tags in a tag cloud (i.e. the more recently a tag was used (smaller &lt;em&gt;x&lt;/em&gt;), the more weight (larger &lt;em&gt;y&lt;/em&gt;) the tag gets). The whole graph would be shifted up and to the right of what is shown in the sketch, but I know how to do that part once I have a formula.&lt;/p&gt;&#xA;" OwnerUserId="477" LastEditorUserId="477" LastEditorDisplayName="" LastEditDate="2010-08-13T15:45:40.687" LastActivityDate="2010-08-13T17:03:56.273" Title="Family of functions with two horizontal asymptotes" Tags="&lt;functions&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="2361" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T15:43:15.317" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Oh dear, Rasmus has extended his answer and rendered this one obsolete!  In fact, stop reading this right now and go see what he has done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpret what you are asking as trying to find a way to decide without introducing bias (as would be introduced by counting tails-heads as the same as heads-tails).  &lt;a href=&quot;http://math.stackexchange.com/questions/2356/is-it-possible-to-split-coin-flipping-3-ways/2357#2357&quot;&gt;Rasmus's suggestion&lt;/a&gt; of repeating the experiment for a certain configuration seems the best choice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew a little tree and tried to group outcomes into three &quot;equally-likely&quot; sets and then realized this is impossible because $3$ does not divide $2^n$ for any $n$.  (The quantity $2^n$ being the number of unique outcomes after $n$ &quot;flips&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-13T15:43:15.317" />
  <row Id="2362" PostTypeId="2" ParentId="2360" CreationDate="2010-08-13T15:47:35.490" Score="7" ViewCount="0" Body="&lt;p&gt;I think what you're looking for is a form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Logistic_function&quot; rel=&quot;nofollow&quot;&gt;logistic function&lt;/a&gt;, such as $$f(x)=\frac{2}{1+e^x}-1.$$&#xA;&lt;img src=&quot;http://i.imgur.com/gyurj.png&quot; alt=&quot;graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: For your specific criteria with $y_1$ and $y_2$: $$f(x)=\frac{y_1-y_2}{1+e^x}+y_2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: For comparison of my answer to the other two answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/WHMcb.png&quot; alt=&quot;comparison graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dashed/black: $\frac{2}{1+e^x}-1$; blue: $-\frac{2}{\pi}\arctan x$; red: $-\frac{x}{\sqrt{x^2+1}}$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-13T17:03:56.273" LastActivityDate="2010-08-13T17:03:56.273" />
  <row Id="2363" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T15:50:05.550" Score="1" ViewCount="0" Body="&lt;p&gt;If you are at a crossroads in life requiring you to choose one of three options, then coin tossing is not the correct thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, pick up a dice and roll it. Go for first option if the diece turns up 1 or 2. Go for second option  if the dice turns up 3 or 4. Go for the third option if the dice turns up 5 or 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Oops, I saw that Rasmus already gave this option. Anyway I strongly suggest that you prefer this one to the other two methods given by him.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:50:05.550" CommentCount="5" />
  <row Id="2364" PostTypeId="2" ParentId="2360" CreationDate="2010-08-13T15:54:38.450" Score="5" ViewCount="0" Body="&lt;p&gt;Just take $- tan^{-1} x$. Vary it appropriately for additional constraints.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T15:54:38.450" CommentCount="1" />
  <row Id="2365" PostTypeId="2" ParentId="594" CreationDate="2010-08-13T16:16:07.250" Score="-3" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have got another interesting proof, for your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $S = \{(x,y,z)\ in \mathbb{N} \times \mathbb{N} \times \mathbb{N} \ | x^{2}+4yz=p \} $. Define the map $\sigma : S \to S$ as follows: $ (x,y,z ) \mapsto (x+2z,z,y-x-z), \ \text{if} \ x \leq y-z$, $(x,y,z) \mapsto (2y-x,y,x+z-y), \ \text{if} \ y-z &amp;lt; x &amp;lt;2y$  and $(x,y,z) \mapsto (x-2y,x+z-y,y) \ \text{if} \ x \geq 2y$. Then one can show that $\sigma$ is a permutation of order 2 and has a unique fixed point. Hence conclude that any prime number $p$ is of the form $4n+1$ is the sum of 2 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof was made famous by &lt;a href=&quot;http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares%23Zagier%27s_%22one-sentence_proof%22&quot; rel=&quot;nofollow&quot;&gt;Don Zagier&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-22T01:34:17.307" LastActivityDate="2010-11-22T01:34:17.307" CommentCount="18" />
  <row Id="2366" PostTypeId="2" ParentId="2360" CreationDate="2010-08-13T16:36:17.043" Score="4" ViewCount="0" Body="&lt;p&gt;Another famous family of functions that behave as you describe is those of form $y=\dfrac{x}{\sqrt{x^2+1}}$. (This function is actually the sine of the arctan function &lt;a href=&quot;http://math.stackexchange.com/questions/2360/family-of-functions-with-two-horizontal-asymptotes/2364#2364&quot;&gt;George suggested&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graph of $y=-\dfrac{x}{\sqrt{x^2+1}}$:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/JBNSo.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general y&lt;sub&gt;1&lt;/sub&gt; and y&lt;sub&gt;2&lt;/sub&gt;, the formula would be $y=-\dfrac{y_1-y_2}{2}*\dfrac{x}{\sqrt{x^2+1}}+\dfrac{y_1+y_2}{2}$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-13T16:36:17.043" />
  <row Id="2367" PostTypeId="1" AcceptedAnswerId="2410" CreationDate="2010-08-13T16:44:35.100" Score="2" ViewCount="202" Body="&lt;p&gt;How does one prove that ring of Hamilton Quaternions with coefficients coming from the field $\mathbb{Z}/p\mathbb{Z}$ is not a divison ring.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-13T16:57:55.580" LastActivityDate="2010-08-13T20:58:40.043" Title="A question on Hamilton Quaternions." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2368" PostTypeId="2" ParentId="2367" CreationDate="2010-08-13T16:53:58.287" Score="3" ViewCount="0" Body="&lt;p&gt;The sledgehammer is to invoke &lt;a href=&quot;http://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Wedderburn's theorem&lt;/a&gt;&#xA;to the effect that every finite division ring is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a slightly lighter sledgehammer invoke &lt;a href=&quot;http://en.wikipedia.org/wiki/Four_square_theorem&quot; rel=&quot;nofollow&quot;&gt;Legendre's four&#xA;square theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-13T17:15:27.457" LastActivityDate="2010-08-13T17:15:27.457" CommentCount="3" />
  <row Id="2369" PostTypeId="1" AcceptedAnswerId="2371" CreationDate="2010-08-13T16:57:34.710" Score="0" ViewCount="91" Body="&lt;p&gt;I heard that one can construct field based on $p^n$ elements where $p$ is prime. I tried with $p = n = 2$. It seemed easy as there are 2 groups which $\{0,1,2,3\}$ can form and 1 group formed by $\{1,2,3\}$. However each time I get into contradiction. Is there anything I missed or it is not possible to form $p^n$ element group?&lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-13T17:08:52.320" Title="Construction of $p^n$ field" Tags="&lt;set-theory&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2370" PostTypeId="2" ParentId="2369" CreationDate="2010-08-13T17:02:01.893" Score="0" ViewCount="0" Body="&lt;p&gt;Did you refer the wikipedia article for constructing fields. Please refer the section of explicitly constructing finite fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_field&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Finite_field&lt;/a&gt;. Or if you want to be more advance you can refer this also: www.emis.de/journals/IJOPCM/Vol/09/IJOPCM(vol.2.3.1.S.9).pdf&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-13T17:02:01.893" />
  <row Id="2371" PostTypeId="2" ParentId="2369" CreationDate="2010-08-13T17:04:42.777" Score="3" ViewCount="0" Body="&lt;p&gt;There are finite fields of order $p^n$ for all prime powers $p^n$. They&#xA;are often called &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_field&quot; rel=&quot;nofollow&quot;&gt;Galois fields&lt;/a&gt;. But their additive groups&#xA;are not cyclic (when $n\ne 1$). So for order $4$ one can have an&#xA;addition table&#xA;$$\begin{array}{cccc}&#xA;0&amp;amp;1&amp;amp;a&amp;amp;b\\&#xA;1&amp;amp;0&amp;amp;b&amp;amp;a\\&#xA;a&amp;amp;b&amp;amp;0&amp;amp;1\\&#xA;b&amp;amp;a&amp;amp;1&amp;amp;0&#xA;\end{array}$$&#xA;but not an addition table&#xA;$$\begin{array}{cccc}&#xA;0&amp;amp;1&amp;amp;2&amp;amp;3\\&#xA;1&amp;amp;2&amp;amp;3&amp;amp;0\\&#xA;2&amp;amp;3&amp;amp;0&amp;amp;1\\&#xA;3&amp;amp;0&amp;amp;1&amp;amp;2&#xA;\end{array}.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-13T17:04:42.777" />
  <row Id="2372" PostTypeId="1" AcceptedAnswerId="2500" CreationDate="2010-08-13T17:07:54.007" Score="6" ViewCount="412" Body="&lt;p&gt;How can I find the number of &lt;em&gt;k&lt;/em&gt;-permutations of &lt;em&gt;n&lt;/em&gt; objects, where there are &lt;em&gt;x&lt;/em&gt; types of objects, and r&lt;sub&gt;1&lt;/sub&gt;, r&lt;sub&gt;2&lt;/sub&gt;, r&lt;sub&gt;3&lt;/sub&gt; ... r&lt;sub&gt;x&lt;/sub&gt; give the number of each type of object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have 20 letters from the alphabet. There are some duplicates - 4 of them are &lt;em&gt;a&lt;/em&gt;, 5 of them are &lt;em&gt;b&lt;/em&gt;, 8 of them are &lt;em&gt;c&lt;/em&gt;, and 3 are &lt;em&gt;d&lt;/em&gt;. How many unique 15-letter permutations can I make?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 20&lt;br&gt;&#xA;k = 15&lt;br&gt;&#xA;x = 4&lt;br&gt;&#xA;r&lt;sub&gt;1&lt;/sub&gt; = 4 | r&lt;sub&gt;2&lt;/sub&gt; = 5 | r&lt;sub&gt;3&lt;/sub&gt; = 8 | r&lt;sub&gt;4&lt;/sub&gt; = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was originally going to solve this problem in order to solve a simpler problem, but I managed to find a simpler solution to that problem instead. Now I'm still looking for the solution to this more general problem out of interest.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done some more work on this problem but haven't really come up with anything useful. Intuition tells me that as Douglas suggests below there will probably not be an easy solution. However, I haven't been able to prove that for sure - does anyone else have any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've now re-asked this question (&lt;a href=&quot;http://mathoverflow.net/questions/37211/permutations-with-identical-objects&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) on MO.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditorDisplayName="" LastEditDate="2010-08-31T03:20:40.913" LastActivityDate="2010-08-31T06:32:02.473" Title="Permutations with identical objects" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="2373" PostTypeId="2" ParentId="2369" CreationDate="2010-08-13T17:08:52.320" Score="2" ViewCount="0" Body="&lt;p&gt;First, any field of order $p^n$ will have characteristic $p$, so the underlying &lt;em&gt;additive&lt;/em&gt; structure of the group is necessarily $(\mathbb{Z}_p)^n$; for order $2^2$, it follows that the additive structure has to be the Klein $4$-group (it cannot be cyclic of order $4$). Second, the multiplicative group of nonzero elements of any finite field is necessarily &lt;em&gt;cyclic&lt;/em&gt;, so the multiplicative structure will always be isomorphic to $\mathbb{Z}_{p^n-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, trying to fit together the two structures abstractly is likely to prove very hard. Instead, the easiest way to construct such a field is as an extension of degree $n$ of the field $\mathbb{F}_p$ of $p$ elements. This is achieved by finding a monic irreducible polynomial of degree $n$ with coefficients in $\mathbb{F}_p$, call it $f(x)$, and considering the quotient $\mathbb{F}_p[x]/(f(x))$, where $\mathbb{F}_p[x]$ is the ring of polynomials with coefficients in $\mathbb{F}_p$, and $(f(x))$ is the ideal of all multiples of $f(x)$. Arithmetic of this field is easily computed using polynomial multiplication and long division (inverses can be found using the Euclidean algorithm). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may not be immediately obvious that there is always a monic irreducible polynomial of degree $n$ over $\mathbb{F}_p$ for all primes $p$ and all positive integers $n$; but you can deduce that this is the case by showing that there in fact is a (unique up to isomorphism) field of order $p^n$ (consider the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$), and then showing that if $\alpha$ is a generator for the multiplicative group of this field, then the field is given by $\mathbb{F}_p(\alpha)$. General theory of field extensions then tells you the monic irreducible of $\alpha$ will do.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-13T17:08:52.320" />
  <row Id="2374" PostTypeId="1" CreationDate="2010-08-13T17:33:04.730" Score="5" ViewCount="118" Body="&lt;p&gt;I am learning the axioms of Zermelo-Fraenkel (ZF) set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One axiom schema basically says that given any set S and any formula phi(x), there is a set T consisting of all those elements x of S such that phi(x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this axiom schema unsatisfying because it only guarantees that subsets of S &lt;em&gt;definable by a formula&lt;/em&gt; are really sets.  It's like saying that a function from X to Y only exists if you can write down a formula for it rather than just allowing arbitrary single-valued subsets of X cross Y.  Is there a way in the language of ZF to say &quot;given a set S, if T is a subset of S then T is a set?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related is the power set axiom.  Given a set S, there is a set P(S) consisting of exactly &lt;em&gt;all&lt;/em&gt; the subsets of S.  But in ZF the objects of the theory are sets (no urelements).  Everything is a set.  So, the elements of P(S) are all sets.  Doesn't this mean that &lt;em&gt;every&lt;/em&gt; subset of a given set S is a set?  If so, why the need for the subset axioms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can anticipate some difficulty.  I want to say that &quot;if x is a member of P(S) then x is a set,&quot; but I cannot express the predicate &quot;is a set&quot; in the language of set theory, and strangely enough there is no need to since everything is a set!  I attempted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;(for all S)(for all x)[x subset S --&gt; (there exists y)(x = y)]&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &quot;x subset S&quot; is an abbreviation for &quot;(for all z)[z in x --&gt; z in S].&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this attempt is silly because when I say &quot;for all x,&quot; x is automatically a set and there is no need to say it is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are the subset axioms necessary?&lt;/li&gt;&#xA;&lt;li&gt;Is there a way to remove reference to a defining formula so that &lt;em&gt;every&lt;/em&gt; subset of a given set is a set?&lt;/li&gt;&#xA;&lt;li&gt;If the answer to 1 is &quot;yes&quot; and the answer to 2 is &quot;no&quot; then why is set theory so weird as to allow only definable subsets on the one hand and yet allow for a set of all subsets on the other?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1146" LastActivityDate="2010-08-13T19:15:25.333" Title="Is the Subset Axiom Schema in ZF necessary?" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="2375" PostTypeId="1" AcceptedAnswerId="2377" CreationDate="2010-08-13T17:47:57.870" Score="0" ViewCount="50" Body="&lt;p&gt;Let U be a unitary transformation from Hilbert space H to Hilbert space K.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you call a *-homomorphism f from B(H) to B(K), defined by $f(a) = U a U^{-1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested both in a symbol, which can be used in formulas, and a name for it, which can be used in texts or in speech.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T04:53:28.340" LastActivityDate="2010-08-18T04:53:28.340" Title="Notation of homeomorphism from B(H) to B(K), corresponding to unitary transformation of Hilbert spaces." Tags="&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;notation&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2376" PostTypeId="2" ParentId="2346" CreationDate="2010-08-13T17:48:06.590" Score="2" ViewCount="0" Body="&lt;p&gt;Define $g$ on $[2,2^{k}]$ by $(g(2))^{k}=g(2^k)$.  Then using the functional equation you can extend to an $f$ on all numbers $\geq 1$. We can then use the similar trick for $(0,1)$ and for the negative numbers as well.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-13T17:48:06.590" />
  <row Id="2377" PostTypeId="2" ParentId="2375" CreationDate="2010-08-13T17:52:01.777" Score="1" ViewCount="0" Body="&lt;p&gt;If $H=K$ and $U$ is a unitary, then $f$ is called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_automorphism#Ring_case&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;inner automorphism&lt;/em&gt;&lt;/a&gt; of $\mathbb B(H)$,&#xA;because it is given by conjugation with a unitary element of your algebra $\mathbb B(H)$ (these are the &quot;obviously existing automorphism&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notation (for the group case) is proposed &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_automorphism#Notation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. At least for rings this notation isn't standard, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above naming doesn't nicely generalise to your more general situation and I don't think there's a common name for isomorphisms given by conjugation with a unitary transformation.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-15T14:23:40.163" LastActivityDate="2010-08-15T14:23:40.163" CommentCount="2" />
  <row Id="2378" PostTypeId="1" AcceptedAnswerId="2379" CreationDate="2010-08-13T18:03:12.893" Score="0" ViewCount="122" Body="&lt;p&gt;How did the definition of Algebraic objects like group, ring and field come up? When groups were first introduced, were they given the 4 axioms as we give now. And what made Mathematicians to think of something like this.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-21T05:44:06.157" CommunityOwnedDate="2010-08-13T18:42:59.240" Title="Definition of an Algebraic Objects" Tags="&lt;soft-question&gt;&lt;math-history&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2379" PostTypeId="2" ParentId="2378" CreationDate="2010-08-13T18:18:34.250" Score="4" ViewCount="0" Body="&lt;p&gt;I'm certainly no historian, so let me give my impression of the history in a CW answer and others can improve on this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups were understood as permutations of objects long before they were understood as abstract groups, just as manifolds were understood as nice subsets of $\mathbb{R}^n$ long before they were understood as abstract manifolds.  Most notably, Galois worked with permutations of the roots of a polynomial in order to understand the solvability of polynomials by radicals.  Similarly, number theorists like Kummer worked with concrete examples of number rings and fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the people responsible for the spread of the abstract point of view are (in no particular order) Hilbert, Noether, Artin, van der Waerden, (Dedekind?), .... together they more or less determined the abstract algebra curriculum as we now understand it.  Their reasons for doing everything axiomatically include that the proofs are shorter and more powerful this way; in particular, algebraic geometry needed rigorous tools and commutative algebra provided them (and Zariski used them).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One big payoff of the abstract point of view is that people now recognized fundamental groups and homology groups in topology as groups.  Before the abstract study of groups these objects, while defined in some sense, weren't recognized as groups because they didn't permute anything.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-13T18:18:34.250" CommunityOwnedDate="2010-08-13T18:18:34.250" CommentCount="2" />
  <row Id="2380" PostTypeId="2" ParentId="2250" CreationDate="2010-08-13T18:38:49.407" Score="2" ViewCount="0" Body="&lt;p&gt;The sum in question is at most ε&lt;sup&gt;2&lt;/sup&gt;.  (We do not need the condition that the row sum equals 1 or the condition f(i)≠g(i) to obtain this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;.  Since&#xA;$$\sum_{k=1}^n(a_{f(k)j}-a_{g(k)j})=\sum_{k=1}^na_{f(k)j}-\sum_{k=1}^na_{g(k)j}=0,$$&#xA;we have&#xA;$$\left|\sum_{k=1}^na_{ki}(a_{f(k)j}-a_{g(k)j})\right|&#xA;=\left|\sum_{k=1}^n(a_{ki}-a_{1i})(a_{f(k)j}-a_{g(k)j})\right|$$&#xA;$$\le\sum_{k=1}^n|a_{ki}-a_{1i}||a_{f(k)j}-a_{g(k)j}|.$$&#xA;Therefore, the sum in question is at most&#xA;$$\frac1n\sum_{i,j=1}^z\sum_{k=1}^n|a_{ki}-a_{1i}||a_{f(k)j}-a_{g(k)j}|&#xA;=\frac1n\sum_{k=1}^n\left(\sum_{i=1}^z|a_{ki}-a_{1i}|\right)\left(\sum_{j=1}^z|a_{f(k)j}-a_{g(k)j}|\right)$$&#xA;$$\le\frac1n\sum_{k=1}^n\epsilon^2=\epsilon^2.$$&lt;/p&gt;&#xA;" OwnerUserId="926" LastEditorUserId="926" LastEditorDisplayName="" LastEditDate="2010-08-13T19:09:27.280" LastActivityDate="2010-08-13T19:09:27.280" CommentCount="1" />
  <row Id="2381" PostTypeId="2" ParentId="2330" CreationDate="2010-08-13T18:49:53.220" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;OP's &lt;a href=&quot;http://stackoverflow.com/questions/3474961/find-minimum-of-a-function-defined-by-integration-in-mathematica&quot;&gt;equivalent question on Stack Overflow&lt;/a&gt; was already answered. This is &lt;a href=&quot;http://stackoverflow.com/questions/3474961/find-minimum-of-a-function-defined-by-integration-in-mathematica/3475061#3475061&quot;&gt;the accepted answer by Andrew Moylan&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Try this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[58]:= g[t_, x_] := t^3 - t + x^2&#xA;&#xA;In[59]:= f[t_?NumericQ] := NIntegrate[g[t, x], {x, -1, 1}]&#xA;&#xA;In[60]:= FindMinimum[f[t], {t, 1}]&#xA;&#xA;Out[60]= {-0.103134, {t -&amp;gt; 0.57735}}&#xA;&#xA;In[61]:= Plot[f[t], {t, 0, 1}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Two relevant changes I made to your code:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Define f with &lt;code&gt;:=&lt;/code&gt; instead of with &lt;code&gt;=&lt;/code&gt;. This effectively gives a definition for f &quot;later&quot;, when the user of f has supplied the values of the arguments. See &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/SetDelayed.html&quot; rel=&quot;nofollow&quot;&gt;SetDelayed&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Define f with &lt;code&gt;t_?NumericQ&lt;/code&gt; instead of &lt;code&gt;t_&lt;/code&gt;. This says, t can be anything numeric (Pi, 7, 0, etc). But not anything non-numeric (t, x, &quot;foo&quot;, etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-13T18:49:53.220" CommunityOwnedDate="2010-08-13T18:49:53.220" />
  <row Id="2382" PostTypeId="2" ParentId="2374" CreationDate="2010-08-13T18:51:15.450" Score="4" ViewCount="0" Body="&lt;p&gt;The subtle point you are speaking about is exactly the difference between first order logic and second order logic. In first order logic, we quantify only over objects in the universe. In second order logic, we can quantify also over subcollections of the universe. ZFC is a first order theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the ZFC subset axiom is &lt;em&gt;not&lt;/em&gt; saying that every subset of a set is describable in the way you mention. Rather, it is making the much milder claim that if we can describe a subset, then it is a set. And this seems totally unobjectionable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You evidently want to consider a stronger version of the axiom, which says somehow that all the subcollections that exist anywhere are also sets. This is a second order assertion, since it is quantifying over subsets of the universe. This second order subset axiom, however, is not expressible in first order logic. The reason is that any model with an infinite set, satisfying the second-order subset axiom would have to uncountable, since it would truly contain all the subsets of its sets. Thus, there could be no countable models of your theory, which would violate the Lowenheim Skolem theorem if the axiom were expressible in first order logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it is natural to look into the possibilities of second order ZFC, and this is studied in set theory. It turns out that the models of your second order ZFC (using second order Replacement also) are exactly the same as the universes known as $H_\kappa$ for an inaccessible cardinal $\kappa$, one of the large cardinal notions. These models are also known as the Grothendieck universes, and are used throughout  category theory as a convenient way of formalizing the large/small distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a good exercise for a beginner in this area to prove that if $\kappa$ is a (strongly) inaccessible cardinal, then the collection $H_\kappa$ of sets whose transitive closure has size less than $\kappa$ is a model of second order ZFC. When $\kappa$ is inaccessible, then this universe $H_\kappa$ is the same as $V_\kappa$ in the Levy cumulative hierarchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me remark that some might find it to be somewhat incoherent to propose a second-order theory as the main axiomatization of set theory. The purpose of the axiomatization, after all, is to set forth certain minimal bedrock principles that the set concept should obey, but if we make these principles second-order, then we would seem to need a prior concept of set to interpret them. That is, a second order logic only makes sense in the context of a larger set-theoretic background. So it wouldn't seem to work foundationally to provide such an axiomatization as the principal axiomatization of set theory.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-08-13T19:15:25.333" LastActivityDate="2010-08-13T19:15:25.333" CommentCount="2" />
  <row Id="2383" PostTypeId="2" ParentId="2374" CreationDate="2010-08-13T18:52:50.937" Score="1" ViewCount="0" Body="&lt;p&gt;So, maybe I should write this as a response. So, let's take the notions in order. First, I will reserve the word &quot;set&quot; for objects in the theory, and use &quot;collection&quot; for the more intuitive notion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &quot;$A$ is a subset of $B$&quot; is an abbreviation for two statements: (i) $A$ is a set; and (ii) for every $x$, if $x\in A$ then $x\in B$. So &quot;if $A$ is a subset of $B$ then $A$ is a set&quot; is a tautology of the form $P\wedge Q\Rightarrow P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As user80 points out, you are misinterpreting the Subset Axiom schema slightly; it does not state that &lt;em&gt;only&lt;/em&gt; those subcollections which you express as ${a\in A: \phi(a)}$ are sets (it is not an &quot;if and only if&quot; statement). It 'merely' states that all those are sets, leaving open that perhaps there may be &quot;other&quot; subcollections that may also be sets. So in your analogy, it would be like saying that if you can write down a formula for a function from $X$ to $Y$ then the result will &lt;em&gt;certainly&lt;/em&gt; be a function, but it does not tell you that you get a function &lt;em&gt;only&lt;/em&gt; if you can write down a formula: just that writing down a formula is &lt;em&gt;a&lt;/em&gt; way of getting a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It it &quot;necessary&quot;? Well, it is necessary if you wish to be able to construct new sets by specifying certain elements from something that you already know to be a set. Otherwise, you'll have no warrant for saying that such objects are sets, only that they are collections. The Axiom of Separation/Subset is about one way to take a set, pare it down, and get a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the axiom of the power set, you are again misinterpreting it. It's not that the axiom of the power set says that subsets of S are sets; that's already in the notion of &quot;subset&quot;. Rather, the Power Set Axiom tells you that you have a set whose elements &lt;em&gt;are&lt;/em&gt; the subsets of your given set $S$. It guarantees the existence of a set which is &quot;one level up&quot; from $S$, in the sense that its elements are subsets of $S$. From the Axiom of Pairs and other issues it is not hard to show that if you have a finite collection of things that you know are sets, then there is a set whose elements are exactly those in your original collection (that is, it allows you to pull together those things and put them in a set). But what if you have an infinite number of things that you know are subsets of $A$? In order to make sure you have a set that contains all of them as elements, you need something beyond knowing that they are sets; that's where the Axiom of the Power Set comes in: it gives you an overset (which is a set) in which all these objects will be elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does Power Set interact with Separation/Subset? Anything you obtain using Separation will necessarily be an element of the Power Set. It does not say that those are the only things in the power set, just that those are &lt;em&gt;guaranteed&lt;/em&gt; to be in the Power Set.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-13T18:52:50.937" CommentCount="3" />
  <row Id="2384" PostTypeId="1" CreationDate="2010-08-13T18:58:45.980" Score="8" ViewCount="267" Body="&lt;p&gt;I am tutoring several talented students, middle school level and early high school level, in mathematics. I am always looking for new sources from which to draw questions. Can anyone recommend books, web-sites, etc. with a interesting questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of the ArtofProblemSolving.com. I am currently using the books &quot;Challenging Problems in Algebra&quot; and &quot;Challenging Problems in Geometry&quot; by Posamentier and Salkind.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T01:33:43.623" LastActivityDate="2010-11-16T09:32:51.477" CommunityOwnedDate="2010-08-13T19:38:28.320" Title="Sources of problems for teaching/tutoring young mathematicians." Tags="&lt;education&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="9" FavoriteCount="7" />
  <row Id="2385" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T19:00:54.273" Score="1" ViewCount="0" Body="&lt;p&gt;Try Yakov Perelman, &quot;Mathematics can be Fun&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T19:00:54.273" CommunityOwnedDate="2010-08-13T19:38:28.320" CommentCount="1" />
  <row Id="2386" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T19:04:25.593" Score="0" ViewCount="0" Body="&lt;p&gt;There is this &quot;Math Help Forum&quot; which discusses about high school level and undergraduate level mathematics. You can find lots and lots of things. Here is one problem website : &lt;a href=&quot;http://amc.maa.org/e-exams/e8-usamo/usamo.shtml&quot; rel=&quot;nofollow&quot;&gt;http://amc.maa.org/e-exams/e8-usamo/usamo.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus by Thomas finney is a Classic book for teaching Calculus.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-13T19:04:25.593" CommunityOwnedDate="2010-08-13T19:38:28.320" CommentCount="1" />
  <row Id="2388" PostTypeId="2" ParentId="2346" CreationDate="2010-08-13T19:14:21.933" Score="3" ViewCount="0" Body="&lt;p&gt;Case $k=1$ is trivial. So, suppose $k&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose k is odd. For even k one should correct this solution a bit.&#xA;First note, that $a^k=a$ is equivalent to $a\in{-1,0,1}$. We have this equation for a=f(-1), a=f(0), a=f(1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for $x\in (-1,1)$ sequence $x,x^k,(x^k)^k,\dots$ goes to 0. Denote this sequence by $b_n$: $b_n = x^{(k^n)}$. From continuity of f we know, that $f(b_n)\rightarrow f(0)$ and $(f(x))^{(k^n)}\rightarrow f(0)$. Therefore for $x\in(-1,1)$ we know, that $f(x) \in (-1,1)$ or f is constant on this interval and equal to 1 or -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose x is in $(0,\infty)$. By the same reason the sequence $(f(x))^{(k^{-n})}$ goes to f(1). It is true iff  (f(x)=0 and f(1)=0) or (f(x) is in $(-\infty,0)$ and f(1)=-1) or (f(x) is in $(0,\infty)$ and f(1)=1). The same is for x in $(-\infty,0)$ and f(-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have the following cases&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;f(0)=1, then f(x)=1 for $x \in&#xA;[-1,1]$. Other values of f can be&#xA;uniquely determined by values on&#xA;$[-2^k,-2]\cup[2,2^k]$. This values can be chosen arbitrary to form continuous function&lt;br&gt;&#xA;to $(0,\infty)$ such that&#xA;$f(-2^k)=(f(-2))^k$ and&#xA;$f(2^k)=(f(2))^k$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now I will just list all other cases&#xA;    without going to details, which are&#xA;    similar to shown in the first case.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=-1, f(1)=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=-1, f(1)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=-1, f(1)=0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=-1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;f(0)=0,f(-1)=1, f(1)=0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="695" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-02T08:01:05.733" LastActivityDate="2010-09-02T08:01:05.733" />
  <row Id="2389" PostTypeId="1" CreationDate="2010-08-13T19:28:36.330" Score="1" ViewCount="79" Body="&lt;p&gt;How do you calculate the following summation? $\sum_{i=k}^n i$&lt;/p&gt;&#xA;" OwnerUserId="1153" LastActivityDate="2010-08-14T19:08:20.490" Title="Calculate integer summation when lower bound is a variable" Tags="&lt;sequences-and-series&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="3" />
  <row Id="2390" PostTypeId="2" ParentId="2389" CreationDate="2010-08-13T19:31:10.833" Score="4" ViewCount="0" Body="&lt;p&gt;This is a special case for the formula for the sum of n terms in an arithmetic progression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_progression&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arithmetic_progression&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T19:31:10.833" CommentCount="1" />
  <row Id="2391" PostTypeId="2" ParentId="2389" CreationDate="2010-08-13T19:38:36.243" Score="1" ViewCount="0" Body="&lt;p&gt;Okay. I was able to find it here, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Summation#Some_summations_of_polynomial_expressions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Summation#Some_summations_of_polynomial_expressions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1153" LastActivityDate="2010-08-13T19:38:36.243" />
  <row Id="2392" PostTypeId="1" AcceptedAnswerId="2394" CreationDate="2010-08-13T19:41:56.033" Score="1" ViewCount="85" Body="&lt;p&gt;Familiar Exercise: Suppose $G$ is a finite group and $T$ is an automorphism of $G$ which sends more than three quarters of elements of $G$ onto their inverses, then prove that $G$ is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group of Quarternions is an example of a group which sends exactly $\displaystyle \frac{3}{4}$ elements onto their inverses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a finite group with an automorphism $T$ which sends exactly $\displaystyle\frac{4}{5}$ of elements of $G$ onto their inverses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly can we find groups with Automorphism $T$ which sends exactly $\displaystyle \frac{n}{n+1}$ of elements of $G$ onto their inverses.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-13T20:20:53.380" LastActivityDate="2010-08-13T20:20:53.380" Title="Finding groups with given Automorphism" Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2393" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T19:50:43.110" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathcounts.org/&quot; rel=&quot;nofollow&quot;&gt;Mathcounts&lt;/a&gt; is aimed at talented middle school students. If you are affiliated with a school, you can register as a coach with them, and they can provide you with many helpful resources. If not, they still have problem sets available for sale from their catalog, as well as coaching handbooks, which include solutions and teaching advice.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-08-13T19:50:43.110" CommunityOwnedDate="2010-08-13T19:50:43.110" CommentCount="1" />
  <row Id="2394" PostTypeId="2" ParentId="2392" CreationDate="2010-08-13T19:55:32.613" Score="5" ViewCount="0" Body="&lt;p&gt;Prove that&#xA;if $G$ is abelian and $\phi$ is an automorphism of $G$&#xA;then $\{g\in G : \phi(g)=g^{-1} \}$&#xA;is a subgroup of $G$. &lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-13T19:55:32.613" CommentCount="2" />
  <row Id="2395" PostTypeId="1" CreationDate="2010-08-13T19:57:38.023" Score="3" ViewCount="442" Body="&lt;p&gt;I've already taken my calculus sequence and I'm interested in brushing up and staying sharp on the basics. So far, my calculus background is limited to single-variable calculus, which I applied in my physics sequence and engineering statistics. I learned a little bit of vector math in physics as well, but I'm not strong in it. My mathematical strengths lie in the discrete mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm particularly interested in studying for my IEEE Certified Software Development Associate exam - 10% of the exam is mathematics (calculus, differential equations, and statistics). Calculus is one of the mathematics knowledge areas that a software engineer should be competent in, so I'm looking for not only solidifying the courses that I've already taken, but moving forward with slightly more advanced topics, at an undergraduate mathematics level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently using the textbook from my calculus sequence, but I'm interested to know what books other people used to learn calculus and if the book is potentially worth checking out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What books are available to help me?&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="91" LastEditorDisplayName="" LastEditDate="2010-08-13T22:40:17.210" LastActivityDate="2010-09-11T00:32:10.590" CommunityOwnedDate="2010-08-13T19:57:45.683" Title="What are the recommended textbooks for introductory calculus?" Tags="&lt;calculus&gt;&lt;books&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="3" />
  <row Id="2396" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T19:59:50.700" Score="2" ViewCount="0" Body="&lt;p&gt;Thomas &amp;amp; Finney, Calculus and Analytic Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spivak, Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apostol, Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend Apostol the most.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-13T19:59:50.700" CommunityOwnedDate="2010-08-13T19:59:50.700" CommentCount="4" />
  <row Id="2397" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T20:03:22.160" Score="1" ViewCount="0" Body="&lt;p&gt;My university uses &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0495011665&quot; rel=&quot;nofollow&quot;&gt;Calculus: Early Transcendentals by James Stewart&lt;/a&gt; in the standard calculus 3 quarter sequence, the prolonged 4 quarter calculus sequence, as well as in multivariable calculus. This book covers topics such as functions and models, limits and derivatives, differentiation and applications of differentiation, integrals, techniques of integration, vectors, vector functions, partial derivatives, multiple integrals, vector calculus, and second-order differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also used a different James Stewart book in high school during pre-calculus and calculus. I don't remember the name of that book - it might have even been an earlier edition of this book.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-13T20:03:22.160" CommunityOwnedDate="2010-08-13T20:03:22.160" CommentCount="2" />
  <row Id="2398" PostTypeId="2" ParentId="2335" CreationDate="2010-08-13T20:06:03.493" Score="3" ViewCount="0" Body="&lt;p&gt;Sorry but I cannot post comments yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm getting confused. Fix x. Then the sequence $f_n(x)$ is monotone and bounded, hence it converges.&#xA;Define a function $f$ by the rule $f(x) = \lim_nf_n(x)$.&#xA;Then the sequence of functions $f_n$ pointwise converges to $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I misunderstood the question?&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2010-08-13T20:06:03.493" CommentCount="5" />
  <row Id="2399" PostTypeId="2" ParentId="2384" CreationDate="2010-08-13T20:06:48.377" Score="0" ViewCount="0" Body="&lt;p&gt;Check out the &lt;a href=&quot;http://blossoms.mit.edu/&quot; rel=&quot;nofollow&quot;&gt;Blossoms&lt;/a&gt; videos.  Although aimed more at a classroom setting there may be interesting snippets to grab.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-08-13T20:06:48.377" CommunityOwnedDate="2010-08-13T20:06:48.377" />
  <row Id="2400" PostTypeId="1" AcceptedAnswerId="2408" CreationDate="2010-08-13T20:11:16.260" Score="5" ViewCount="196" Body="&lt;p&gt;Having picked up a rudimentary understanding of tensors from reading mechanics papers and Wikipedia, I tend to think of rank-2 tensors simply as square matrices (along with appropriate transformation rules). Certainly, if the distinction between vectors and dual vectors is ignored, a rank 2 tensor $T$ seems to be simply a multilinear map $V \times V \rightarrow \mathbb{R}$, and (I think) any such map can be represented by a matrix $\mathbf{A}$ using the mapping $(\mathbf{v},\mathbf{w}) \mapsto \mathbf{v}^T\mathbf{Aw}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: Is this a reasonable way of thinking about things, at least as long as you're working in $\mathbb{R}^n$? Are there any obvious problems or subtle misunderstandings that this naive approach can cause? Does it break down when you deal with something other than $\mathbb{R}^n$? In short, is it &quot;morally wrong&quot;?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-13T20:54:14.643" Title="Is it misleading to think of rank-2 tensors as matrices?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;&lt;tensors&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="2401" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T20:14:02.383" Score="1" ViewCount="0" Body="&lt;p&gt;Elementary Analysis : The theory of calculus by Kenneth Ross is a beautiful written book. Link is here : &lt;a href=&quot;http://www.springer.com/mathematics/analysis/book/978-0-387-90459-7&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/mathematics/analysis/book/978-0-387-90459-7&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diferential and Integral Calculus by R.Courant is again a very good book which one can learn.&#xA;&lt;a href=&quot;http://books.google.co.in/books?id=eyC1nk9-YjkC&amp;amp;printsec=frontcover&amp;amp;dq=Calculus+R.Courant&amp;amp;source=bl&amp;amp;ots=v0rLcDvVW_&amp;amp;sig=VPB6ZBXfMm6yj0vLG4YalVkpV7I&amp;amp;hl=en&amp;amp;ei=_6ZlTLXAN47IvQOhtZHyDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCEQ6AEwAQ#v=onepage&amp;amp;q=Calculus%20R.Courant&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.co.in/books?id=eyC1nk9-YjkC&amp;amp;printsec=frontcover&amp;amp;dq=Calculus+R.Courant&amp;amp;source=bl&amp;amp;ots=v0rLcDvVW_&amp;amp;sig=VPB6ZBXfMm6yj0vLG4YalVkpV7I&amp;amp;hl=en&amp;amp;ei=_6ZlTLXAN47IvQOhtZHyDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCEQ6AEwAQ#v=onepage&amp;amp;q=Calculus%20R.Courant&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advanced Calculus by J.M.H Ohmsted is also a very good book. Although title says advanced its not really advanced.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-13T20:14:02.383" CommunityOwnedDate="2010-08-13T20:14:02.383" />
  <row Id="2402" PostTypeId="2" ParentId="2400" CreationDate="2010-08-13T20:22:32.910" Score="8" ViewCount="0" Body="&lt;p&gt;It's not misleading as long as you change your notion of equivalence.  When a matrix represents a linear transformation $V \to V$, the correct notion of equivalence is similarity: $M \simeq B^{-1} MB$ where $B$ is invertible.  When a matrix represents a bilinear form $V \times V \to \mathbb{R}$, the correct notion of equivalence is &lt;em&gt;congruence&lt;/em&gt;: $M \simeq B^TMB$ where $B$ is invertible.  As long as you keep this distinction in mind, you're fine.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-13T20:22:32.910" CommentCount="1" />
  <row Id="2403" PostTypeId="1" AcceptedAnswerId="2412" CreationDate="2010-08-13T20:24:29.923" Score="4" ViewCount="188" Body="&lt;p&gt;The statements of Poincare duality for manifolds and Serre Duality for coherent sheaves on algebraic varieties or analytic spaces look tantalizingly similar. I have heard tangential statements from some people that there is indeed some connection between the two. But I was never able to figure it out on myself. For instance for a naive attempt on a smooth complex manifold, the dimensions don't match. Can somebody help me out? &lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:40:34.137" LastActivityDate="2010-08-14T09:40:34.137" Title="Precise connection between Poincare Duality and Serre Duality. " Tags="&lt;algebraic-geometry&gt;&lt;differential-topology&gt;&lt;coherent-sheaves&gt;&lt;duality-theorems&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="2404" PostTypeId="2" ParentId="2395" CreationDate="2010-08-13T20:28:58.213" Score="1" ViewCount="0" Body="&lt;p&gt;Even though I took a course out of it more than 10 years ago, I still refer back to Spivak's Calculus all the time.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2010-08-13T20:28:58.213" CommunityOwnedDate="2010-08-13T20:28:58.213" CommentCount="4" />
  <row Id="2405" PostTypeId="2" ParentId="674" CreationDate="2010-08-13T20:39:55.737" Score="2" ViewCount="0" Body="&lt;p&gt;This elaborates on Mr. Bulatov's answer (note: Mr. Bulatov's link is broken, but I infer it is supposed to link to Dan Klein's notes &lt;a href=&quot;http://www.cs.berkeley.edu/~klein/papers/lagrange-multipliers.pdf&quot; rel=&quot;nofollow&quot;&gt;Lagrange Multipliers without Permanent Scarring&lt;/a&gt;, which are indeed excellent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, for the sake of simplicity let's work with functions in 2 variables, i.e. we want to find critical points of f(x,y) where the constraint is g(x,y)=c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take any point (a,b) in the xy-plane such that g(a,b)=c. Suppose that there was a (unit) vector $\vec v$ tangent to g at (a,b) whose dot product with the gradient vector of f at (a,b) was non-zero. We can take $\vec v$ such that the dot product with $\vec v$ is positive since $\nabla f(a,b)\cdot \vec v=-(\nabla f(a,b)\cdot -\vec v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what this means is that the the partial derivative of f with respect to $\vec v\quad$ is positive and that points along $\vec v\quad$ close enough to (a,b) have higher f-value. I (and Mr. Bulatov) claim that this means that there are points on g(x,y)=c close to (a,b) with higher f-value, meaning that f is not a maximum. Mr. Bulatov phrases the argument well in terms of infinitesimals, but I feel that a little bit more rigor is illuminating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a sequence of points (a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;) on g(x,y)=c that converge to (a,b). Each point (a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;) comes with a unit vector $\vec u_i\quad$ that corresponds to the direction (is a scalar multiple of) (a,b)-(a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;). Because $\vec v\quad$ is a tangent vector, we can choose the (a&lt;sub&gt;i&lt;/sub&gt;,b&lt;sub&gt;i&lt;/sub&gt;) such that the $\vec u_i$ converge to $\vec v$ . But because f is differentiable in more than one variable, its partial derivatives are continuous, and so we have that the partial derivatives $\nabla f(a,b)\cdot \vec u_i$ converge to $\nabla f(a,b) \cdot \vec v$, and so for all sufficiently large i the partial derivative in the u&lt;sub&gt;i&lt;/sub&gt; direction is positive and so $f(a_i,b_i)&gt;f(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if f(a,b) was a maximum (or a minimum) of f on g(x,y)=c, gradient vector of f at (a,b) would be have dot product zero with every vector in the line tangent to g at (a,b). But that means that the gradient vector of f is orthogonal to the tangent line of g, which means that the gradient of f must be parallel to (i.e. a scalar multiple of) the gradient of g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we obtain the equations $\nabla f(x,y)=\lambda\nabla g(x,y)$ and $g(x,y)=c$, which are satisfied if and only if $(x,y)$ is a minimum of $f(x,y)+\lambda (g(x,y)-c)$ for positive $\lambda$ and a maximum for negative $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-13T20:39:55.737" CommunityOwnedDate="2010-08-13T20:39:55.737" />
  <row Id="2406" PostTypeId="2" ParentId="2166" CreationDate="2010-08-13T20:39:56.267" Score="0" ViewCount="0" Body="&lt;p&gt;The answer depends if you want an answer that is true &quot;for all &lt;em&gt;n&lt;/em&gt;&quot; or an answer that is true &quot;for all sufficiently large &lt;em&gt;n&lt;/em&gt;.&quot; For instance, there is not always a prime in an interval of the form (n, 3n/2). Take n=7, for instance. There is always a prime in such an interval &quot;for sufficiently large &lt;em&gt;n&lt;/em&gt;,&quot; however.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastActivityDate="2010-08-13T20:39:56.267" />
  <row Id="2407" PostTypeId="2" ParentId="2062" CreationDate="2010-08-13T20:51:30.950" Score="6" ViewCount="0" Body="&lt;p&gt;You can use inclusion/exclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n\leq x} d(n) = \sum_{mn\le x} 1 = \sum_{m\le \sqrt{x}} \ \sum_{n\le x/m} 1 + \sum_{n\le \sqrt{x}} \ \sum_{m\le x/n} 1 - \sum_{m\le\sqrt{x}} 1 \sum_{n\le\sqrt{x}} 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first two double sums are the same (with the roles of $m$ and $n$ interchanged). Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n\leq x} d(n) = 2 \sum_{n\le \sqrt{x}} \Big\lfloor\frac{x}{n}\Big\rfloor - \Big( \big\lfloor\sqrt{x}\big\rfloor \Big)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$. Now use the fact that $\lfloor x \rfloor=x+O(1)$ to finish the proof. You will have to use the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n\leq x} \frac{1}{n} = \log x +\gamma + O\Big(\frac{1}{x}\Big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is Euler's constant.&lt;/p&gt;&#xA;" OwnerUserId="1151" LastEditorUserId="1151" LastEditorDisplayName="" LastEditDate="2010-09-30T19:53:52.943" LastActivityDate="2010-09-30T19:53:52.943" />
  <row Id="2408" PostTypeId="2" ParentId="2400" CreationDate="2010-08-13T20:54:14.643" Score="2" ViewCount="0" Body="&lt;p&gt;You're absolutely right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone will find useful a couple of remarks telling the same story in coordinate-free way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;what happens here is indeed identification of space with it's dual: so a bilinear map $T\colon V\times V\to\mathbb{R}$ is rewritten as $V\times V^*\to\mathbb{R}$ — which is exactly the same thing as a linear operator $A\colon V\to V$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;an identification of $V$ and $V^*$ is exactly the same thing as a scalar product on $V$, and using this scalar product one can write $T(v,w)=(v,Aw)$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;so orthogonal change of basis preserves this identification — in terms of Qiaochu Yuan's answer one can see this from the fact that for orthogonal matrix $B^T=B^{-1}$ (moral of the story: if you have a canonical scalar product, there is no difference between $T$ and $A$ whatsoever; and if you don't have one — see Qiaochu Yuan's answer.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-13T20:54:14.643" />
  <row Id="2409" PostTypeId="1" AcceptedAnswerId="2413" CreationDate="2010-08-13T20:57:23.450" Score="-5" ViewCount="357" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Problem N-88 (page 36) of &lt;a href=&quot;http://books.google.com/books?id=Snecq4MHl8YC&amp;amp;pg=PA36#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Group Theory: Selected Problems&lt;/a&gt; by B. Sury.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $\theta : G \to M$ be a surjective homomorphism of finite groups. Prove that there exists a subgroup $H$ of $G$ such that $\theta(H)=M$ and the set of primes dividing $|H|$ is same as the set of primes dividing $|M|$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T19:36:35.043" LastActivityDate="2010-08-15T19:36:35.043" Title="Surjective Homomorphism of Finite Groups" Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="2410" PostTypeId="2" ParentId="2367" CreationDate="2010-08-13T20:58:40.043" Score="4" ViewCount="0" Body="&lt;p&gt;This answer builds on damiano's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of Wedderburn's theorem, we know that there are no finite-dimensional non-commutative division algebras over $\mathbb{F}_p$.  In particular, every quaternion algebra over $\mathbb{F}_p$ is split, i.e., isomorphic to the algebra $M_2(\mathbb{F}_p)$ of $2 \times 2$ matrices over $\mathbb{F}_p$.  Can we see this directly by arguments particular to quaternion algebras?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes.  For instance, a quaternion algebra is split iff its associated norm form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(x) = x \overline{x} = q(x_1,x_2,x_3,x_4)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is isotropic -- i.e., there exists $(x_1,\ldots,x_4) \neq (0,\ldots,0)$ such that $q(x_1,\ldots,x_4) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Hamiltonian quaternion algebra,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Robin points out, one way to see that this form is isotropic over $\mathbb{F}_p$ is to &#xA;realize that by Lagrange's Four Squares Theorem, $p$ is a sum of four integral squares, and then reduce modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is overkill.  The fact that any quadratic form in at least three variables over a finite field is isotropic is a special case of the (easy to prove) Chevalley-Warning theorem: see e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400ChevalleyWarning.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400ChevalleyWarning.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even this is more than is necessary: it follows from a simple counting argument that any nondegenerate quadratic form in at least two variables over $\mathbb{F}_p$ is &lt;em&gt;universal&lt;/em&gt;, i.e., represents every nonzero element of the field.  For the details, see e.g. p. 5 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/quadraticforms2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/quadraticforms2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we get that every nondegenerate quadratic form in at least three variables over a finite field is isotropic, and in particular the norm form of any quaternion algebra over a finite field is split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the universality of binary quadratic forms over finite fields is the key to this result.  Assuming the characteristic is not $2$, the quaternion algebra may be represented as $\langle a, b \rangle$.  For a quaternion algebra $Q = \langle a, b \rangle$ over any field $F$ (of characteristic not $2$), a necessary and sufficient criterion for $Q$ to be split is that $b$ is a norm in the extension $F(\sqrt{a})/F$.  Thus the universality of the norm form of $\mathbb{F}_p(\sqrt{a})/\mathbb{F}_p$ implies that all quaternion algebras over $\mathbb{F}_p$ are split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, from the perspective of Galois cohomology, Wedderburn's theorem is not especially deep or difficult to prove.  Using standard properties of cohomology of cyclic groups, it comes down to showing that for any finite degree extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ of finite fields, the norm map is surjective, and this may also be proved by an elementary counting argument (or simply by a straightforward calculation).   &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-13T20:58:40.043" />
  <row Id="2411" PostTypeId="2" ParentId="2062" CreationDate="2010-08-13T21:04:49.873" Score="1" ViewCount="0" Body="&lt;p&gt;@Jim: You can represent the floor function with \lfloor and \rfloor (e.g. $\lfloor x \rfloor$). Would post this as a comment, but I don't have the reputation.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-08-13T21:04:49.873" />
  <row Id="2412" PostTypeId="2" ParentId="2403" CreationDate="2010-08-13T21:13:46.427" Score="3" ViewCount="0" Body="&lt;p&gt;As far as I know, to make a precise connection, one has to invoke Hodge theory.  Suppose&#xA;that $X$ is a compact smooth projective variety of dimension $d$.  Then Poincare duality pairs $H^n(X,\mathbb C)$ with $H^{2d-n}(X,\mathbb C),$ for any $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Hodge decomposition gives&#xA;$$H^n(X,\mathbb C) = \oplus_{p+q = n}&#xA;H^q(X,\Omega^p)$$&#xA;and &#xA;$$H^{2d-n}(X,\mathbb C) = \oplus_{p'+q' = 2 d - n} H^{q'}(X,\Omega^{p'})&#xA;= \oplus_{p + q = n}H^{d-q}(X,\Omega^{d - p}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Serre duality gives a duality between&#xA;$H^q(X,\Omega^p)$ and $H^{d - q}(X,\Omega^{d-p}),$&#xA;and the compatibility statement is that Poincare duality between $H^n$ and $H^{2 d - n}$&#xA;is induced by the direct sum of the pairings on the various summands in the Hodge decomposition given by Serre duality.  (Perhaps up to signs and powers of $2 \pi i$,&#xA;which I'm not brave enough to work out right now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: A good case to think about for a newcomer to Hodge theory is the case when $X$ is&#xA;a compact Riemann surface (or equivalently, an algebraic curve).  If the genus of $X$ is $g$,&#xA;then $H^1(X,\mathbb C)$ is $2g$-dimensional, and is endowed with a symplectic pairing via&#xA;Poincare duality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hodge theory breaks $H^1(X,\mathbb C)$ up into the sum of two $g$-dimensional subspaces, namely $H^0(X,\Omega^1)$ and $H^1(X,\mathcal O)$.  These are isotropic under Poincare duality (i.e. the Poincare duality pairing vanishes when restricted to either of them),&#xA;but the become dual to one another under Poincare duality, and that pairing agrees with&#xA;the Serre duality pairing (up to a factor of $2\pi i$, perhaps).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest part of this to understand is the inclusion $H^0(X,\Omega^1) \subset&#xA;H^1(X,\mathbb C)$: a holomorphic differential gives a cohomology class just via de Rham&#xA;theory (i.e. we integrate the holomorphic one form over 1-cycles); note that holomorphic&#xA;1-forms are automatically exact, because if you apply the exterior derivative, you get a holomorphic 2-form, which must vanish (because $X$ is a curve, i.e. of complex dimension &#xA;one).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why $H^0(X,\Omega^1)$ is isotropic under Poincare duality, note that in the de Rham&#xA;picture, the Poincare duality pairing corresponds to wedging forms. But wedging two holomorphic 1-forms again&#xA;gives a holomorphic 2-form, which must vanish (as we already noted).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-14T00:18:48.353" LastActivityDate="2010-08-14T00:18:48.353" CommentCount="4" />
  <row Id="2413" PostTypeId="2" ParentId="2409" CreationDate="2010-08-13T21:25:51.943" Score="3" ViewCount="0" Body="&lt;p&gt;This a comment more than an answer, but is a bit long for the comment box, and may point you in a useful direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that my interpretation of the question (see my comment above) is correct, and I will consider the special case when the kernel of $\theta$ (call it $K$) has order coprime to $|M|$.  Then if the $H$ you are looking for exists, we would have $H\bigcap K&#xA;= \{1\}$ (because the order of this subgroup has to divide both $|H|$ and $|K|$, and these&#xA;are coprime by our assumptions on $|K|$, $|M|$, and $|H|$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\theta$ would map $H$ isomorphically to $M$, and you would have written $G$ as a semi-direct product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the statement is true, then: if $\theta:G \to M$, with kernel $K$, has $|K|$ and $|M|$&#xA;coprime, then $G$ is isomorphic to a semi-direct product of $M$ by $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is true: it is the Schur--Zassenhaus theorem.  It is not so trivial to prove, but googling will lead you to some write-ups of it.  Perhaps you can adapt one of those arguments to your more general situation.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-13T21:25:51.943" />
  <row Id="2415" PostTypeId="2" ParentId="2378" CreationDate="2010-08-13T21:43:04.133" Score="3" ViewCount="0" Body="&lt;p&gt;Let me address the question: &quot;what made Mathematicians ... think of something like this&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: Galois, in studying the problem of factorization of polynomials, realized that&#xA;reasoning about the symmetries of the roots could be more powerful a tool than studying explicit formulas (which, loosely speaking, had been the basic method in the theory of equations up to that time).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this structural/conceptual point of view began to reveal its power in solving difficult concrete problems, more and more mathematicians began to think in this way.  The structural concepts were then isolated from their concrete settings, and this is how they are taught today.  But the motivation was, and for many remains, the applications of these abstract notions to concrete problems.  (A standard but helpful example is the pivotal role that group theory, Galois theory, cohomology, and ring theory played in the proof of Fermat's Last Theorem.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, nowadays we train ourselves and our students to recognize any hint of structure in a problem, and to exploit such structure to the hilt.  So these concepts have become basic tools in the problem-solving toolkits of contemporary mathematicians.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-13T21:43:04.133" CommunityOwnedDate="2010-08-13T21:43:04.133" />
  <row Id="2416" PostTypeId="2" ParentId="2378" CreationDate="2010-08-13T21:52:07.157" Score="3" ViewCount="0" Body="&lt;p&gt;See the books &lt;a href=&quot;http://www.springer.com/birkhauser/historyofscience/book/978-0-8176-4684-4&quot; rel=&quot;nofollow&quot;&gt;A History of Abstract Algebra&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=hmath-32&quot; rel=&quot;nofollow&quot;&gt;Episodes in the History of Modern Algebra (1800-1950)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-13T21:52:07.157" CommunityOwnedDate="2010-08-13T21:52:07.157" />
  <row Id="2417" PostTypeId="2" ParentId="2349" CreationDate="2010-08-13T21:56:35.393" Score="6" ViewCount="0" Body="&lt;p&gt;You put your socks first and then your shoes but you take off your shoes before taking off your socks.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-13T21:56:35.393" CommentCount="3" />
  <row Id="2418" PostTypeId="1" CreationDate="2010-08-13T22:37:03.223" Score="16" ViewCount="803" Body="&lt;p&gt;I had seen this problem a long time back and wasn't able to solve it. For some reason I was reminded of it and thought it might be interesting to the visitors here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, this problem is from a mathematics magazine of some university in the United States (sorry, no idea about either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $S \subset \mathbb{Z}$ (set of integers) such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $|S| = 15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\forall s \in S, \exists a,b \in S$ such that $s = a+b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that for every such $S$, there is a non-empty subset $T$ of $S$ such that the sum of elements of $T$ is zero and $|T| \leq 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; (Sep 13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an approach which seems promising and others might be able to take it ahead perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the set as a vector $s$, then there is a matrix $A$ with the main diagonal being all $1$, each row containing exactly one $1$ and one $-1$ (or a single $2$) in the non-diagonal position such that $As = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem becomes equivalent to proving that for any such matrix $A$ the row space of $A$ contains a vector with all zeroes except for a $1$ and $-1$ or a vector with all zeroes except $\leq 7$ ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the numbers in the set $S$ themselves don't matter and we can perhaps replace them with elements from a different field (like say reals, or complex numbers).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-13T17:56:43.500" LastActivityDate="2010-09-13T17:56:43.500" Title="A zero sum subset of a sum-full set." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="7" />
  <row Id="2419" PostTypeId="1" CreationDate="2010-08-13T22:40:19.287" Score="4" ViewCount="170" Body="&lt;p&gt;A friend of mine is subediting a book. The author of the book died in 1999. The author, at some point, uses the word &quot;trillion&quot; which is, unfortunately, an ambiguous word in the UK: when I was at school it used to mean $10^{18}$ but nowadays it means $10^{12}$. My friend is faced with the following paragraph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'According to Asimov, the amino acids of the proteins behave in a much freer way than our words do: they can be rearranged in any manner and always retain some meaning. A simple protein is made up of eight amino acids, which can be classified putting the numbers one to eight in a series, changing the order of sequence by one digit each time. Out of the same number of &quot;words&quot; we can construct a little over 40,000 organised &quot;biological phrases&quot; from the same genetic code, each one with its own meaning, which is the mission of every protein. But if the chains become longer, as in the case of more complicated molecules such as insulin, which consists of 30 amino acids, the tally rises to a staggering eight million trillion possibilities.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The editor doesn't like &quot;eight million trillion&quot; and wants to replace it with &quot;800000..000&quot;. The question, of course, is &quot;how many zeros&quot;? More precisely, the question is: does someone with a better understanding of chemistry than me able to reconstruct Asimov's calculation and see whether the answer is approximately $8\times 10^{24}$ or $8\times 10^{18}$?&lt;/p&gt;&#xA;" OwnerUserId="1162" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-14T08:17:16.233" LastActivityDate="2010-08-14T08:17:16.233" Title="Asimov quote about &quot;eight million trillion&quot; arrangements of amino acids" Tags="&lt;combinatorics&gt;&lt;chemistry&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2420" PostTypeId="2" ParentId="713" CreationDate="2010-08-13T23:06:53.523" Score="0" ViewCount="0" Body="&lt;p&gt;A common approach for displaying symmetric tensors for scientific data visualization is the use of what are called &quot;tensor glyphs&quot; (see for example &quot;&lt;a href=&quot;http://www.cs.utah.edu/~gk/papers/vissym04/&quot; rel=&quot;nofollow&quot;&gt;Superquadric Tensor Glyphs&lt;/a&gt;&quot; by Kindlmann, 2004). The basic idea helps for visualizing tensors intuitively, too. It really only makes sense for &lt;em&gt;symmetric&lt;/em&gt; tensors, but then, many tensors that arise in physics are symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since my familiarity is only with linear algebra and not tensor analysis, I'm going to equate &quot;tensor&quot; and &quot;matrix&quot; in the following. According to the answers to &lt;a href=&quot;http://math.stackexchange.com/questions/2400/is-it-misleading-to-think-of-rank-2-tensors-as-matrices&quot;&gt;this question&lt;/a&gt;, this is not a horribly wrong thing to do.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the tensor $T$ is symmetric and positive definite. Treat it as a matrix, and transform the unit sphere $S = \{x : \|x\| = 1\}$ with it to get an ellipsoid $TS = \{x : \|T^{-1}x\| = 1\}$. The ellipsoid's axes are parallel to $T$'s eigenvectors, with lengths proportional to the corresponding eigenvalues. This has some other nice properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The identity matrix is represented by a unit sphere.&lt;/li&gt;&#xA;&lt;li&gt;Scalar multiplication $\alpha T$ uniformly scales the ellipsoid by $\alpha$.&lt;/li&gt;&#xA;&lt;li&gt;A tensor with one large and two small eigenvalues looks like a line segment pointing along the large eigenvector.&lt;/li&gt;&#xA;&lt;li&gt;A tensor with two large and one small eigenvalues looks like a disk normal to the small eigenvector.&lt;/li&gt;&#xA;&lt;li&gt;The mapping between symmetric positive definite matrices/tensors and ellipsoids centered on the origin is bijective.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If your tensors are symmetric but &lt;em&gt;not&lt;/em&gt; positive definite, you can still visualize them in terms of deformations of the unit sphere, corresponding to $I + \epsilon T$. Now a negative eigenvalue corresponds to shrinking, and a positive one to stretching.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-13T23:06:53.523" CommentCount="1" />
  <row Id="2421" PostTypeId="2" ParentId="2419" CreationDate="2010-08-13T23:11:13.547" Score="2" ViewCount="0" Body="&lt;p&gt;What I think the author intends in the first part is that some piece of genetic code gives you $8$ distinct amino acids and you want to count the number of ways to rearrange them to get a protein, which is where the $8! \approx 40000$ number comes from.  Since there are $20$ amino acids, the best you can do for $30$ amino acids is to use each of them once and ten of them twice, giving an answer of $\frac{30!}{2^{10}} \approx 2.59 \times 10^{29}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to find the actual list of amino acids in insulin (which is ridiculous to me; why isn't this information on a wiki somewhere?), but the above calculation suggests that the $10^{24}$ number is closer.  If anyone wants to help me out, the actual polypeptide chain in question is the B chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  According to &lt;a href=&quot;http://www.greatcom.org/resources/reasons_skeptics/ch_15/default.htm&quot; rel=&quot;nofollow&quot;&gt;this citation&lt;/a&gt;, Asimov's actual estimate is $8 \times 10^{27}$.  So it looks like the author misquoted him.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-13T23:48:10.583" LastActivityDate="2010-08-13T23:48:10.583" CommentCount="4" />
  <row Id="2423" PostTypeId="2" ParentId="2356" CreationDate="2010-08-13T23:24:17.020" Score="2" ViewCount="0" Body="&lt;p&gt;Here's another method: consider H to be 0 and T to be 1. Consider the results that you get from your coins as successive bits (after the binary point) in a binary number.  Assign numbers in the interval [0, 1/3) to the first choice; [1/3, 2/3) to the second choice; [2/3, 1) to the third choice.  Flip coins until you know, no matter what happens on the remaining flips, in which of these intervals the resulting real number in [0, 1) will lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concretely, this means that if you flip two coins and observe HH, make the first choice; no matter what happens the result will be between .0000... = 0 and .001111... = 1/4.  Similarly if you observe TT, make the third choice; the result will be between 3/4 and 1.  If you observe HT, you can't commit to a choice yet; the result will be between 1/4 and 1/2, which overlaps two of the intervals. Something similar is true for TH.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case flip a third time.  HTT corresponds to the interval [3/8, 1/2) which lies entirely in [1/3, 2/3), so HTT corresponds to the second choice; similarly for THH.  If you see HTH you're still undecided between the first and second choices; THT is still undecided between the second and third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing in this way, HTHH corresponds to the first choice, HTHT and THTH are still undecided, THTT corresponds to the third choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for the cut-points 1/3 and 2/3, you flip coins until you get either two heads or two tails in a row. If you first get two heads in a row, this corresponds to the first choice if the initial flip was a head, the second if the initial flip was a tail. If you first get two tails in a row, this corresponds to the second choice if the initial flip was a head, the third choice if the initial flip was a tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might wonder how many flips this takes, on average, to give a result. With probability 1/2 you get a result on the second flip; with probability 1/4, on the third flip; with probability 1/8, on the fourth flip, and in general with probability $1/2^{n-1}$ for $n \ge 2$.  The sum $\sum_{n \ge 2} n/2^{n-1}$ has value 3, so on average this method takes three flips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a bit slower on average than the method Rasmus suggested, which takes on average 8/3 flips.  However, the same method works even if the three choices are not equally likely! (It won't have such a clean way of being expressed as &quot;flip until you get two in a row&quot;, though.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-13T23:24:17.020" />
  <row Id="2424" PostTypeId="2" ParentId="2419" CreationDate="2010-08-13T23:43:46.443" Score="2" ViewCount="0" Body="&lt;p&gt;The second sentence of that paragraph is meaningless, I think. The number of eight-amino-acid polypeptides (those are chains of amino acids) is 20^8, which is much larger than 40,000.  And insulin isn't really big enough to be a good example of a typical protein; typical proteins have hundreds of amino acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, Qiaochu's interpretation seems correct.  I was able to track down &lt;a href=&quot;http://www.indulgedfurries.com/petdiabetes/insulin_sequence.htm&quot; rel=&quot;nofollow&quot;&gt;the structure of human insulin&lt;/a&gt; -- note that the image is for bovine insulin, and human insulin differs by substituting Thr for Ala in the final position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the B-chain of human insulin has the sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phe Val Asn Gln His Leu Cys Gly Ser His Leu Val Glu Ala Leu Tyr Leu Val Cys Gly Glu Arg Gly Phe Phe Tyr Thr Pro Lys Ala&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which contains &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 Leu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 each Phe, Val, Gly,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 each His, Cys, Glu, Ala, Tyr, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 each Asn, Gln, Ser, Arg, Thr, Pro , Lys&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the number of ways that all of these can be rearranged is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{30!}{4! 3!^3 2!^5} \approx 1.6 \times 10^{27} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not all that close to either of the proposed replacements.  Then again, it's possible that whoever did the calculation was working from the sequence of some insulin other than human, or just made a mistake.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-13T23:53:45.237" LastActivityDate="2010-08-13T23:53:45.237" CommentCount="4" />
  <row Id="2425" PostTypeId="1" CreationDate="2010-08-14T00:16:55.830" Score="4" ViewCount="273" Body="&lt;p&gt;Experimentally, this limit appears to converge to 1/e, but I can't figure out how to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it's the limit as x goes to infinity of (x^(x+1) / (x+1)^x) - ((x-1)^x / x^(x-1))&lt;/p&gt;&#xA;" OwnerUserId="1164" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-14T00:25:02.350" LastActivityDate="2010-08-14T14:38:01.353" Title="What is the limit as x goes to infinity of x^(x+1)/(x+1)^x - (x-1)^x/x^(x-1)" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2426" PostTypeId="2" ParentId="2349" CreationDate="2010-08-14T00:42:41.660" Score="3" ViewCount="0" Body="&lt;p&gt;$$\begin{align}&#xA;&amp;amp; \text{id} \\&#xA;=&amp;amp; f \circ f^\circ \\&#xA;=&amp;amp; f  \circ \text{id} \circ f^\circ \\&#xA;=&amp;amp; f \circ (g \circ g^\circ) \circ f^\circ \\&#xA;=&amp;amp; f \circ g \circ g^\circ \circ f^\circ \\&#xA;=&amp;amp; (f \circ g) \circ (g^\circ \circ f^\circ)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $(f \circ g)^\circ = g^\circ \circ f^\circ$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-14T00:42:41.660" />
  <row Id="2427" PostTypeId="1" AcceptedAnswerId="2498" CreationDate="2010-08-14T00:45:44.867" Score="9" ViewCount="358" Body="&lt;p&gt;Is there any overarching reason why, after excluding the infinite classes of finite simple groups&#xA;(cyclic, alternating, Lie-type),&#xA;what remains---the &lt;em&gt;sporadic&lt;/em&gt;, exceptional finite simple groups, is in fact a finite list (just 26)?  In some sense, the prime numbers can be viewed as &quot;sporadic,&quot; but there is an infinite supply.  Is there some principle that indicates that there must be only a finite number of these&#xA;exceptional groups, and the &quot;only issue&quot; (to minimize a huge, multi-year community effort) was to identify them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask in relative ignorance of modern group theory, and apologize in advance for the naiveness of my question.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-14T11:32:01.850" Title="Why are there only a finite number of sporadic simple groups?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="2428" PostTypeId="2" ParentId="2341" CreationDate="2010-08-14T00:48:42.210" Score="3" ViewCount="0" Body="&lt;p&gt;Hi User. Are you still stuck? I'll use your terminology, which seems to be used in electrical engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) x'z + xyz + xy'z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) x'z + xz(y+y')       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since y+y' = 1, and xz1 = xz then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) x'z + xz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) (x'+x)z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="592" LastEditorDisplayName="" LastEditDate="2010-08-14T03:38:08.367" LastActivityDate="2010-08-14T03:38:08.367" />
  <row Id="2429" PostTypeId="1" CreationDate="2010-08-14T01:08:02.280" Score="0" ViewCount="181" Body="&lt;p&gt;How should I rotate a point (x,y) to location (a,b) on the co-ordinate by any angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LQiiA.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1167" LastActivityDate="2010-08-14T07:05:03.973" Title="Rotate a point in circle about an angle" Tags="&lt;trigonometry&gt;" AnswerCount="3" />
  <row Id="2430" PostTypeId="2" ParentId="2429" CreationDate="2010-08-14T01:11:47.250" Score="1" ViewCount="0" Body="&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rotation_matrix#Two_dimensions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rotation_matrix#Two_dimensions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here since the rotation is clockwise rather than counter clockwise, you would have to substitute $- \theta$ in place of $\theta$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T01:11:47.250" CommentCount="3" />
  <row Id="2431" PostTypeId="2" ParentId="2429" CreationDate="2010-08-14T01:13:38.663" Score="2" ViewCount="0" Body="&lt;p&gt;The rotation transformation of a point, treated as a vector, would be (you seem to be going clockwise so I'll swing with that):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigl(\begin{smallmatrix}&#xA;x'\\&#xA;y'&#xA;\end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix}&#xA;\cos(\theta)&amp;amp;\sin(\theta)\\ &#xA;-\sin(\theta)&amp;amp;\cos(\theta)&#xA;\end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix}&#xA;x\\ &#xA;y&#xA;\end{smallmatrix}\bigr)$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-14T06:24:27.070" LastActivityDate="2010-08-14T06:24:27.070" CommentCount="1" />
  <row Id="2432" PostTypeId="2" ParentId="2425" CreationDate="2010-08-14T01:19:12.943" Score="3" ViewCount="0" Body="&lt;p&gt;Making the change of variables $\; z = 1/x \;$ and collecting together like exponents yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad \displaystyle &#xA;\lim_{z\to 0^+} {\frac{(1+z)^{-1/z} - (1-z)^{1/z}}{z}}\; = \;\lim_{z\to 0^+} {\frac{f(-z)-f(z)}{z}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the well-known $\rm exp$ and $\rm log$ taylor series to $f(z) =  (1-z)^{1/z}\; = \; e^{{\rm log}(1-z)/z}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = e^{-1-\frac{z}{2}+\;\cdots} = e^{-1} (1 + (-\frac{z}{2} +\;\cdots) + (-\frac{z}{2} +\cdots)^2 + \;\cdots) =  \frac{1}{e} - \frac{z}{2e} + O(z^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore  $\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle&#xA;\frac{f(-z)-f(z)}{z}\; = \;\frac{1}{e} + O(z)\quad\quad\quad$  as confirmed by Macsyma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i33.tinypic.com/bi1m6h.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-14T14:38:01.353" LastActivityDate="2010-08-14T14:38:01.353" CommentCount="1" />
  <row Id="2433" PostTypeId="2" ParentId="2341" CreationDate="2010-08-14T01:19:50.180" Score="2" ViewCount="0" Body="&lt;p&gt;If you think of the Boolean operations as complement, intersection and union of sets, which is equivalent since every Boolean algebra is isomorphic to a field of sets, then the equation is obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, on the right side we have the set $z$. On the left side, we have the elements of $z$ separated into three categories: those that are not in $x$, those that are in $x$ and also in $y$, and those that are in $x$ but not in $y$. Since this accounts for all possible elements of $z$, we may deduce the equality.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-08-14T01:19:50.180" />
  <row Id="2434" PostTypeId="2" ParentId="2425" CreationDate="2010-08-14T01:49:29.687" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;b&gt;HINT&lt;/b&gt; You can write the difference as $x \left( y^{-x} - z^x \right)=x\left(1-(yz)^x\right)/y^z$, where $y$ and $z$ are rational functions of $x$ such that the limiting value of $y^x$ (and $z^x$) is well-known, finite, and non-zero. Now focus on finding the limit of $x\left(1-(yz)^x\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Restricting $x$ to integers and invoking the Binomial Theorem helps, but that only shows what the limit --&lt;em&gt;if&lt;/em&gt; it exists-- would have to be, leaving open the possibility that the limit might not exist when considering non-integer $x$.)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-14T01:49:29.687" CommentCount="1" />
  <row Id="2435" PostTypeId="1" AcceptedAnswerId="2674" CreationDate="2010-08-14T04:11:46.293" Score="2" ViewCount="174" Body="&lt;p&gt;I have a function that (more or less) is supposed to select a small number m of random number from the range [1,n] (for some n &gt;&gt; m) and I need to test that it work. Is there an easy to implement test that will give good confidence that it is working correctly? (For reference the full on chi squared test is not simple enough.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;m in [5,500], n is small enough I can use normal int/float types for most math.&lt;/li&gt;&#xA;&lt;li&gt;I can get as many set as I'm willing to wait for (&gt;100s per second)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reason I'm looking for something simple is that the code I'm checking isn't that complicated. If the test code is more complex than the code under test, then the potential for the test being wrong becomes a real problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took another look at Chi-squared and it's not as bad as I remembered... as long as I have a fixed number of buckets and a fixed significance threshold. (Or a canned CDF, and I don't think I do.)&lt;/p&gt;&#xA;" OwnerUserId="1171" LastEditorUserId="1171" LastEditorDisplayName="" LastEditDate="2010-08-18T18:59:13.183" LastActivityDate="2010-08-20T15:11:14.547" Title="Is there a simple test for uniform distributions?" Tags="&lt;statistics&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="2436" PostTypeId="2" ParentId="2425" CreationDate="2010-08-14T06:47:28.473" Score="4" ViewCount="0" Body="&lt;p&gt;A fairly mechanical approach is to write the limit as&#xA;$$\lim_{x\to\infty}(f(x)-f(x-1))$$&#xA;where&#xA;$$f(x)=\frac{x^{x+1}}{(x+1)^x}.$$&#xA;Then&#xA;$$\log f(x)=(x+1)\log x - x\log(x+1)&#xA;= \log x - x \log(1+1/x)$$&#xA;and so&#xA;$$\log f(x)= \log x - 1 + 1/(2x) + O(x^{-2})$$&#xA;as $x\to\infty$ (using the Maclaurin series for $\log(1+t)$).&#xA;Therefore&#xA;$$f(x) = (x/e)(1+1/(2x)+O(x^{-2}))=x/e+1/(2e)+O(x^{-1})$$&#xA;and so&#xA;$$f(x-1) =(x-1)/e+1/(2e) +O((x-1)^{-1}).$$&#xA;Subtracting,&#xA;$$f(x)-f(x-1)= 1/e+O(x^{-1})).$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-14T06:47:28.473" />
  <row Id="2437" PostTypeId="2" ParentId="2429" CreationDate="2010-08-14T06:50:23.117" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming that you ask where will &lt;em&gt;x&lt;/em&gt; end up after being rotated clockwise for an angle $\theta$ around the origin, then the answer is as given by George S. and J. Mangaldan. I'd like to give a brief explanation of &lt;em&gt;why&lt;/em&gt; that is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The key is that rotation is a linear transformation.&lt;/em&gt; If we write &lt;em&gt;Ax&lt;/em&gt; for the point &lt;em&gt;x&lt;/em&gt; rotated, then then $A(ax+bx) = a(Ax) + b(Ax)$, where &lt;em&gt;a&lt;/em&gt; and &lt;em&gt;b&lt;/em&gt; are some real numbers. Here &lt;em&gt;ax&lt;/em&gt; is a point obtained by stretching both coordinates, and lies on the line that contains x and the origin. It's a scaling operation. Also, $x+y$ is obtained by adding the coordinates, which you can imagine as the diagonal of a parallelogram whose edges are &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt;. So the equation $A(ax+bx) = a(Ax) + b(Ax)$ says that if you scale a figure with two points, then `add' the points by taking the parallelogram diagonal, and then rotate you get the same result you get if you start by rotating, and then you scale and add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/c5TbD.png&quot; alt=&quot;&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming that equation is OK when &lt;em&gt;A&lt;/em&gt; represents a rotation, note that every vector $x=(a,b)$ can be written as $a(1,0)+b(0,1)$, so its rotation will be $a(A(1,0))+b(A(0,1))$. In other words, all you need to know is where do the points (1,0) and (0,1) end up after being rotated (namely, you need the points A(1,0) and A(0,1), each having two coordinates). So four numbers are enough to figure out where &lt;em&gt;all&lt;/em&gt; points end up after being rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear transformations are &lt;em&gt;the&lt;/em&gt; reason matrices are important.&lt;/p&gt;&#xA;" OwnerUserId="512" LastEditorUserId="512" LastEditorDisplayName="" LastEditDate="2010-08-14T07:05:03.973" LastActivityDate="2010-08-14T07:05:03.973" CommentCount="1" />
  <row Id="2439" PostTypeId="1" CreationDate="2010-08-14T07:20:02.330" Score="7" ViewCount="174" Body="&lt;p&gt;Here's a question.  Have there been any theoretical results showing that if P != NP, there must exist some problems in NP which are not NP-complete and which are not in P either?  Just curious because I've never seen this question addressed.&lt;/p&gt;&#xA;" OwnerUserId="1173" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-14T07:36:53.197" LastActivityDate="2010-08-14T07:36:53.197" Title="Assuming P != NP, do we know whether there are problems which are in NP, not in P and are not NP complete ?" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="2441" PostTypeId="2" ParentId="2439" CreationDate="2010-08-14T07:29:19.173" Score="10" ViewCount="0" Body="&lt;p&gt;Yes, this is known as Ladner's theorem, proved by Richard Ladner in 1975. The class of such problems are called &lt;a href=&quot;http://en.wikipedia.org/wiki/NP-Intermediate&quot; rel=&quot;nofollow&quot;&gt;NP-intermediate&lt;/a&gt;. &lt;a href=&quot;http://books.google.com/books?id=47GwSJfb1mQC&amp;amp;pg=PA38&quot; rel=&quot;nofollow&quot;&gt;Here's one proof&lt;/a&gt; that does &quot;cut and paste&quot; between some problem in P and some NP-complete problem, &lt;a href=&quot;http://blog.computationalcomplexity.org/2003/03/foundations-of-complexity-lesson-16.html&quot; rel=&quot;nofollow&quot;&gt;here are two&lt;/a&gt;, and there are &lt;a href=&quot;http://www.google.com/search?q=ladner%27s+theorem&quot; rel=&quot;nofollow&quot;&gt;others&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also problems that are already &lt;em&gt;believed&lt;/em&gt; to be neither NP-complete nor in P, like integer factorisation and graph isomorphism, as well as &lt;a href=&quot;http://mathoverflow.net/questions/31821/problems-known-to-be-in-both-np-and-conp-but-not-known-to-be-in-p&quot; rel=&quot;nofollow&quot;&gt;problems known to be in NP ∩ coNP&lt;/a&gt; but not known to be in P.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-14T07:29:19.173" CommentCount="2" />
  <row Id="2442" PostTypeId="2" ParentId="2335" CreationDate="2010-08-14T07:44:14.100" Score="-2" ViewCount="0" Body="&lt;p&gt;Please look at this answer and then decide whether the question is wrong or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will first show that there is a subsequence $(f_{n_k})$ convergent on the set of rationals $\mathbb{Q}$. Since $\mathbb{Q}$ is countable, we can write $\mathbb{Q}= (r_{1},r_{2},...,)$. The sequence $(f_{n,1(r_{1})})$ is bounded so it has a convergent subsequence. Similarly the sequence $(f_{n,2(r_{2})})$ is bounded so it has a convergent subsequence. Clearly, $(f_{n,2(r_1)})$ is also convergent. Repeating this process, we obtain the sequence of sequences $(f_{n,1}),(f_{n,2}),\cdots,$ with the following properties:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$(f_{n,k+1})$ is a subsequence of $(f_{n,k})$ for $k=1,2,...$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sequence $(f_{n,k(r_i)})$ is convergent for $k \in \mathbb{N}$ and $i= 1,2,...,k$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the diagonal sequence $(f_{n,n})$ is convergent on $\mathbb{Q}$. In this way we have constructed a pointwise convergent sequence on $\mathbb{Q}$ say to $f$. Clearly $f$ is increasing on $\mathbb{Q}$. Now, we extend $f$ to $\mathbb{R}$ by setting $$f(x)= \sup [f(x) : r \in \mathbb{Q}, \ r \leq x ]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extended functuon $f$ is also increasing on $\mathbb{R}$. Now we show that if $f$ is continuous at $x$, then $\displaystyle \lim_{k \to \infty} f_{n_k}(x)=f(x)$. Consider two sequence of rational $(p_n)$ and $(q_n)$ converging to $x$ such that $p_{n} &amp;lt; x &amp;lt; q_{n}$. The monotonicity of $f_{n_k}$ implies that $f_{n_k}(p_n) \leq f_{n_k}(x) \leq f_{n_k}(q_n)$. Now letting $k \to \infty$, we get $$ f(p_n) \leq \lim_{k \to \infty} \inf f_{n_k}(x) \leq \lim_{k \to \infty} \sup f_{n_k}(x) \leq f(q_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also obtain $$ f(x^{-}) \leq \lim_{k \to \infty} \inf f_{n_k}(x) \leq \lim_{k \to \infty} \sup f_{n_k}(x) \leq f(x^{+})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then follows that $\displaystyle f(x)= \lim_{k \to \infty} f_{n_k}(x)$ at each point of $x$ of conitnuity of $f$. We know that the set $C$ of discontinuity of a monotonic function is always countable. Thus we have $f(x) = \displaystyle \lim_{k \to \infty} f_{n_k}(x)$ on the set $\mathbb{R} \setminus C$, and since $f_{n_k}$ is bounded n the countable set $C$ we can use the diagonal method again to chose a subsequence of $(f_{n_k})$ which is pointwise convergent on $C$. Clearly this subsequence will be convergent on all of $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-14T07:44:14.100" CommentCount="4" />
  <row Id="2443" PostTypeId="2" ParentId="1870" CreationDate="2010-08-14T07:53:39.613" Score="1" ViewCount="0" Body="&lt;p&gt;There is &lt;a href=&quot;http://www.microsoft.com/learningspace/&quot; rel=&quot;nofollow&quot;&gt;Microsoft Math&lt;/a&gt;. I haven't actually used it, but I know that it can provide step by step solutions. &lt;a href=&quot;http://www.squarecirclez.com/blog/microsoft-math-30-review/713&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s a nice review. Also it's $20.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-08-14T07:53:39.613" CommunityOwnedDate="2010-08-14T07:53:39.613" />
  <row Id="2444" PostTypeId="1" AcceptedAnswerId="2456" CreationDate="2010-08-14T08:28:27.470" Score="0" ViewCount="182" Body="&lt;p&gt;I'd like to prove that $\forall x\left(x\not\in x\right)$ in the context of Morse-Kelley set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call $A=\left\{y:y\not\in y\right\}$. I can easily prove that $A\not\in A$. In fact, if you suppose that $A\in A$, it follows that $A\not\in A$ by definition of class comprehension. Since this is a contradiction, then the hypothesis is wrong, that is, $A\not\in A$. But I cannot do the same for an arbitrary class $k$, since I know nothing about $k$ and I cannot apply the definition of class comprehension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2010-08-15T21:21:09.563" Title="How to prove that nothing is a member of itself?" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="2445" PostTypeId="2" ParentId="2395" CreationDate="2010-08-14T08:31:56.090" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://college.cengage.com/mathematics/ostebee/calculus_graphical/2e/students/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Calculus from Graphical, Numerical, and&#xA;Symbolic Points of View&lt;/em&gt; by Ostebee and Zorn&lt;/a&gt; is my favorite book from which to teach (er, actually, I have a slight preference for the 1st edition over the 2nd, but they're both good).  It's aimed a little above average (i.e. harder than the Harvard Consortium book [Hughes-Hallett, Gleason, et al]), but it's much more of a modern reform-calculus than Spivak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I think the conversational tone of the book makes it more readable and the variety of problems (especially the fact that it has challenging problems rather than just repetitive exercised) makes it a good book from which to study.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-14T08:37:59.163" LastActivityDate="2010-08-14T08:37:59.163" CommunityOwnedDate="2010-08-14T08:31:56.090" />
  <row Id="2446" PostTypeId="2" ParentId="2395" CreationDate="2010-08-14T08:32:23.250" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://bcs.wiley.com/he-bcs/Books?action=index&amp;amp;itemId=0471448761&amp;amp;itemTypeId=BKS&amp;amp;bcsId=1404&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Calculus: Single Variable, Course Advantage Edition&lt;/em&gt; by Hughes-Hallett, Gleason, et al&lt;/a&gt;, aka the &quot;Harvard Consortium&quot; book, is perhaps the standard &quot;reform calculus&quot; book.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-14T08:32:23.250" CommunityOwnedDate="2010-08-14T08:32:23.250" CommentCount="2" />
  <row Id="2447" PostTypeId="2" ParentId="2418" CreationDate="2010-08-14T08:35:13.163" Score="7" ViewCount="0" Body="&lt;p&gt;A weaker statement, where we allow elements in $T$ to be repeated, can be proved as below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we can look at the set $\{-s | s\in S \}$ we may assume there are at most $7$ positive numbers in $S$. Let each positive number be a vertex, from each vertex $s$ we draw an arrow to any vertex $a$ such that $s=a+b$. Since if $s&gt;0$, one of $a,b$ must be positive, there is at least one arrow from any vertex. So there must be a cycle $s_1,\cdots,s_n=s_1$ with $n\leq 8$. We can let $T$ consists of $s_i-s_{i+1}$, $1\leq i\leq n-1$.   &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-09T10:06:33.777" LastActivityDate="2010-09-09T10:06:33.777" />
  <row Id="2448" PostTypeId="1" AcceptedAnswerId="2449" CreationDate="2010-08-14T08:45:14.577" Score="0" ViewCount="188" Body="&lt;p&gt;The Sum of 2 sets of Measure zero might well be very large for example, the sum of $x$-axis and $y$-axis, is nothing but the whole plane. Similarly one can ask this question about Cantor sets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $C$ is the cantor set, then what is the measure of $C+C$?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-11T09:05:12.607" LastActivityDate="2010-12-11T09:05:12.607" Title="Measure of the Cantor set plus the Cantor set" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" />
  <row Id="2449" PostTypeId="2" ParentId="2448" CreationDate="2010-08-14T08:52:47.383" Score="9" ViewCount="0" Body="&lt;p&gt;If you are asking the case where $C$ is the Cantor ternary set, then you can show that $C+ C$ is actually $[0,2]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more general cantor sets, you can find a description in the paper: On the topological structure of the arithmetic sum of two Cantor sets, P Mendes and F Oliveira, available at &lt;a href=&quot;http://iopscience.iop.org/0951-7715/7/2/002&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0951-7715/7/2/002&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T09:20:40.417" LastActivityDate="2010-08-14T09:20:40.417" CommentCount="2" />
  <row Id="2450" PostTypeId="1" AcceptedAnswerId="2457" CreationDate="2010-08-14T09:14:31.940" Score="2" ViewCount="404" Body="&lt;p&gt;The tradition way for two children to divide a piece of cake fairly between them is &quot;you cut, I choose&quot;, What process accomplishes this is intuitively clear. Is there a similar process which works for $3$ children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that three children $(1),(2)$ and $(3)$ have associated measures (ways of evaluating the pieces of cake) $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$. Assume for the simplicity of the notation that the sum of total pieces of the cake is equal to $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question : Does the cake have three subsets $C_{1},C_{2}$ and $C_{3}$  that are pairwise disjoint and their union is equal to $C$, and that have moreover the property that $$\alpha(C_{1}) \geq \frac{1}{3}, \quad \alpha(C_{2} )\geq \frac{1}{3} , \quad \alpha(C_{3}) \geq \frac{1}{3}?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt; Let one of the three children say $(1)$, divide the cake into what he regards three equal pieces $C_{1},C_{2}$ and $C_{3}$: in other words $\alpha(C_{1})= \alpha(C_2)=\alpha(C_{3})=\frac{1}{3}$. Let each of $(2)$ and $(3)$ claim dibs on one of those three pieces, the one he prefers the most, the one he would be quite satisfied to have. It is possible that $(2)$ and $(3)$ will claim different pieces, and it is possible that they will claim the same one, but no matter, atleast one of the three pieces, $C_{1},C_{2},C_{3}$ will be left unclaimed. Let $(1)$ remove that piece (one of those pieces) and eat it. So far everyone is satisfied. Child (1) got what he believes to be exactly a third, and in the judgement of the children $(2)$ and $(3)$ the best piece (the claimed piece) is still available. The problem reduces therefore, to dividing whats fairly left, - and that's the classical problem of dividing a piece of cake between two claimants. Let that problem be solved by &quot;you cut, i choose&quot; and everybody is happy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt; The Proposed solution is wrong; whats wrong with it?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-14T17:55:31.793" LastActivityDate="2010-08-15T00:16:12.787" Title="Cutting the Cake Problem." Tags="&lt;set-theory&gt;&lt;fair-division&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2451" PostTypeId="2" ParentId="1222" CreationDate="2010-08-14T09:53:42.810" Score="0" ViewCount="0" Body="&lt;p&gt;You might like Raghavan Narasimhan's &quot;Compact Riemann Surfaces&quot;. It does things like the construction of the Jacobian. It seems to suit your interests best.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T09:53:42.810" CommentCount="1" />
  <row Id="2452" PostTypeId="2" ParentId="1037" CreationDate="2010-08-14T10:02:40.683" Score="0" ViewCount="0" Body="&lt;p&gt;It is possible to attach a precise meaning to &quot;dx&quot; as an infinitesimal using nonstandard analysis. There using model theory it is possible to rigorously make intuitive statements in calculus that we use all the time. This is a theory created by Abraham Robinson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a brief informal discussion, see the appendix(by Stewart) of &quot;What is mathematics&quot; by Courant, Robbins and Stewart. Or see Robinson's book &quot;Nonstandard Analysis&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-08-14T10:13:58.117" LastActivityDate="2010-08-14T10:13:58.117" />
  <row Id="2453" PostTypeId="2" ParentId="1" CreationDate="2010-08-14T10:40:09.497" Score="3" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How there can be different kinds of infinities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is very simple to see. This is because of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: A given set $X$ and its power set $P(X)$ can never be in bijection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: By contradiction. Let $f$ be any function from $X$ to $P(X)$. It suffices to prove $f$ cannot be surjective. That means that some member of $P(X)$ i.e., some subset of $S$, is not in the image of $f$.  Consider the set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T={ x\in X: x\not\in f(x) }.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $x$ in $X$ , either $x$ is in $T$ or not. If $x$ is in $T$, then by definition of $T$, $x$ is not in $f(x)$, so $T$ is not equal to $f(x)$. On the other hand,  if $s$ is not in $T$, then by definition of $T$, $x$ is in $f(x)$, so again $T$ is not equal to $f(x)$. Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus take any infinite set you like. Then take its power set, its power set, and so on. You get an infinite sequence of sets of increasing cardinality(Here I am skipping a little; but a use of the Schroeder-Bernstein theorem will fix things).&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T10:40:09.497" />
  <row Id="2454" PostTypeId="2" ParentId="154" CreationDate="2010-08-14T10:54:17.793" Score="1" ViewCount="0" Body="&lt;p&gt;Even natural numbers are abstract notions. Through abstract constructions the complex numbers can be demonstrated. So asking whether complex numbers is something like asking whether the natural numbers exist, whether the real numbers exist, or not. They are all things in the abstract domain of thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to reconcile it with your physical intuition, try the following. I suppose that your mental picture of the real numbers is by associating a line to it. Similarly, you can associate a plane to the set of complex numbers. Points are complex numbers. The real numbers will be the x-axis and imaginary numbers would be the y-axis. Addition is straightforward to visualize. For visualizing multiplication, write a complex number in the form $re^{i\theta}$. Then multiplication is a combination of stretching by a magnitude of $r$ and rotation by an angle of $\theta$. I hope this helps you in mental visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electrical networks, complex numbers are very good for analysis of alternating current networks. If some current has an imaginary component, it means that the current is leading or lagging the voltage by some &quot;phase difference&quot;, or &quot;angle&quot;. A resistance having imaginary components mean that it has capacitance or inductance. Only the &quot;real&quot; parts contribute to energy loss. But for imepdance calculations you have to consider capacitance and inductance too. A similar analysis can be made for every kind of control systems. So complex number are all very much there in nature and all sorts of engineering problems. I cannot stress their importance enough.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-14T10:54:17.793" CommentCount="1" />
  <row Id="2455" PostTypeId="1" AcceptedAnswerId="2618" CreationDate="2010-08-14T11:22:11.197" Score="5" ViewCount="254" Body="&lt;p&gt;I was quite impressed by a &lt;a href=&quot;http://mathoverflow.net/questions/8846/proofs-without-words/8847#8847&quot; rel=&quot;nofollow&quot;&gt;beautiful proof&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt;  of the formula for n'th triangular number — it's kind of a bijective proof built on the top of geometric summation idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: does this argument generalize to higher dimensions? i.e. can someone prove the formula for n-th &amp;laquo;k-dimensional simplex number&amp;raquo; this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/km6YV.png&quot; alt=&quot;the proof&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the statement for k=3 is that n-th &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedral_number&quot; rel=&quot;nofollow&quot;&gt;pyramidal number&lt;/a&gt; is $\binom{n+2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One obvious fallback would be observing that any ball in the pyramid is characterized by its coordinates&lt;sup&gt;2&lt;/sup&gt;, i.e. balls are counted by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiset#Multiset_coefficients&quot; rel=&quot;nofollow&quot;&gt;negative binomial coefficient&lt;/a&gt; — which is exactly the desired result. That's a good proof, but needless to say, that's not the proof I'm looking for.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; original post by &lt;a href=&quot;http://math.stackexchange.com/users/274/mariano-suarez-alvarez&quot;&gt;Mariano Suárez-Alvarez&lt;/a&gt;, &lt;a href=&quot;http://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-n/2288#2288&quot;&gt;cited&lt;/a&gt; by &lt;a href=&quot;http://math.stackexchange.com/users/876/vaughn-climenhaga&quot;&gt;Vaughn Climenhaga&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; k-simplex is the subset $x_0+\dots+x_k=1$, $x_i\ge0$ in $\mathbb{R}^{k+1}$, and n-th k-simplex number is the number of non-negative integer solutions of $x_0+\dots+x_k=n$&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-15T12:09:28.520" LastActivityDate="2010-08-16T19:40:37.027" Title="Geometric Proof of the Formula for Simplex Numbers" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="2456" PostTypeId="2" ParentId="2444" CreationDate="2010-08-14T11:38:27.783" Score="4" ViewCount="0" Body="&lt;p&gt;This can be proved using the foundation axiom. Suppose there exists a class $A$ such that $A \in A$, then $A$ is a set and we can consider the set $\left\{A\right\}$. The foundation axiom says that $\exists B \in \left\{A\right\}$ such that $B \cap \left\{A\right\} = \emptyset$. Since $B \in \left\{A\right\}$, $B$ must be $A$. It follows that $A \in A=B$, thus $A \in B \cap \left\{A\right\} \neq \emptyset$. And you have a contraddiction.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1065" LastEditorDisplayName="" LastEditDate="2010-08-15T21:21:09.563" LastActivityDate="2010-08-15T21:21:09.563" CommentCount="4" />
  <row Id="2457" PostTypeId="2" ParentId="2450" CreationDate="2010-08-14T12:02:33.697" Score="3" ViewCount="0" Body="&lt;p&gt;After giving to (1) one of unclaimed pieces, less than 2/3 of cake can be left (in measure of (2) or (3)), so one of them will be unsatisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The right solution is the following.&lt;br&gt;&#xA;1. Let (1) divide cake into three equal pieces.&lt;br&gt;&#xA;2. Ask (2) and (3) &quot;What is the best one?&quot;. If their answers are different, we are done: each one of (2) and (3) gets what he (or she) wants, and (1) takes unclaimed piece. Otherwise go to step 3.&lt;br&gt;&#xA;3. Ask (2) and (3) &quot;Is there any other piece, you would satisfied with?&quot;. If at least one of the answers is &quot;Yes&quot; we are done similar to step 2. Otherwise go to step 4.&lt;br&gt;&#xA;4. (1) takes any of two unclaimed pieces.&lt;br&gt;&#xA;5. Thanks to the answer &quot;No&quot; at the step 3 we know, that at least 2/3 of the cake left in both measures (the one of (2) and the one of (3)). Therefore we can apply &quot;you cut, I choose&quot; method to the leaving 2 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This problem can be solved with arbitrary number of children using the algorithm for finding maximum matching in a bipartite graph. Suppose, number of children is n. Cutting the cake algorithm is the following.&lt;br&gt;&#xA;1. Let (1) divide cake into three equal pieces.&lt;br&gt;&#xA;2. Ask others the following: &quot;Please, list all the pieces, you will be satisfied with.&quot;&lt;br&gt;&#xA;3. Consider bipartite graph with 2n-1 vertices: n pieces of cake and all children except (1). Children is connected with a cake iff it will be satisfied with it.&#xA;4. Find maximum matching in this graph. Lets call vertex &quot;free&quot; iff it is not matched in this matching with another one.&lt;br&gt;&#xA;5. If there are no free children, give pieces of the cake according to the matching (and free piece of cake is for (1)). In this case we are done. Otherwise go to step 6.&lt;br&gt;&#xA;6. In the algorithm of maximum matching we are able to find its certificate, i.e. pair of sets (U,V) with U containing all free children and some non-free, and V containing some pieces of cake, such that 1) amount of elements in V is equal to amount of non-free elements in U; 2) every vertex in U has edges only to vertexes form V 3) there are no edges from chosen matching, connecting vertex from V to some vertex outside U. In our case it means that every piece of cake outside V is less than 1/n in all measures of children from U.&lt;br&gt;&#xA;7. Give pieces of cake to children outside U (except (1) ) according to chosen matching.&lt;br&gt;&#xA;8. Give any free piece of cake outside V to (1).&lt;br&gt;&#xA;9. Note, that according to 6, all pieces of cake given at 7 and 8 are unimportant for elements of U. Therefore there is enough cake left for them. Divide it according to this algorithm between elements of U (note, that (1) is not in U, and, therefore size of U is less than n).&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-08-14T20:52:36.297" LastActivityDate="2010-08-14T20:52:36.297" CommentCount="0" />
  <row Id="2458" PostTypeId="2" ParentId="2450" CreationDate="2010-08-14T12:21:25.233" Score="0" ViewCount="0" Body="&lt;p&gt;Two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You are assuming that (1) cut the cake into three pieces of equal value to him, and then took one piece. Now there are two pieces left for (2) and (3) to choose between, and you claimed that this problem could be solved by the original &quot;you cut, I choose&quot; method. But, since the cake is already cut, they cannot use that method.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(1) cut the cake into three pieces of equal value to himself and then took a piece for himself, but this piece may have been more valuable to (2) or (3) than any of the other pieces, so they would not be satisfied.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have an idea of how to do it, but its not complete:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let (1) take three pieces of string to mark off where he intends to cut the cake into three slices of what he believes are equally valued. Then there are two cases for people (2) and (3). If (2) and (3)'s optimal slices are different, then they should just take those slices and (1) can take the last one- everyone is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both (2) and (3) want the same slice, then remove the string and start again with (2) setting the string on the cake this time instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By doing this eventually you reduce the problem to the case where when any person cuts the cake into equal slices, the other two people will want the same piece.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-14T12:21:25.233" />
  <row Id="2459" PostTypeId="1" AcceptedAnswerId="2460" CreationDate="2010-08-14T12:57:43.883" Score="1" ViewCount="47" Body="&lt;p&gt;Hello, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;perhaps a somewhat silly question, but is there a generally accepted formal notation for denoting the number of columns a matrix has? Something I could use in the description of an algorithm like: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if horizontaldim(V) &amp;gt; x then&#xA;end if&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if size(V,2) &amp;gt; x then &#xA;end if&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or should I just use a description like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if number of columns in V &amp;gt; x then&#xA;end if&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-08-14T13:16:04.470" Title="formal notation for number of rows/columns in a matrix (e.g to use in pseudocode/algorithm environment in latex)" Tags="&lt;notation&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="2460" PostTypeId="2" ParentId="2459" CreationDate="2010-08-14T13:16:04.470" Score="3" ViewCount="0" Body="&lt;p&gt;None that I know of, but I've seen numerical linear algebra books (e.g. Golub and Van Loan) just say something like $V\in\mathbb{R}^{m\times n}$ for a matrix V with m rows and n columns, and then use m and n in the following algorithm description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MATLAB notation, which some other people use, just says rows(V) and columns(V).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-14T13:16:04.470" CommentCount="1" />
  <row Id="2462" PostTypeId="1" AcceptedAnswerId="2466" CreationDate="2010-08-14T13:22:01.570" Score="0" ViewCount="96" Body="&lt;p&gt;Is it true that for $n \in \mathbb{N}$ we can have $4n = x^{2} + y^{2}$ or $4n = x^{2} - y^{2}$ for $x,y \in \mathbb{N} \cup (0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just working out and this came out to be true from $n=1$ to $n=20$. After that i didn't try, but i would like to see as what will be a counter example.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-14T14:13:38.433" Title="Multiples of 4 as sum or difference of 2 squares" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" />
  <row Id="2463" PostTypeId="2" ParentId="2450" CreationDate="2010-08-14T13:44:16.883" Score="0" ViewCount="0" Body="&lt;p&gt;Fiktor is correct. &#xA;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Proportional_%28fair_division%29&quot; rel=&quot;nofollow&quot;&gt;proportional division&lt;/a&gt;, you want to achieve the result, that each of the n player will end up with at lest 1/n cake in her own valuation. The procedure above does not achieve this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counter example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total valuation of cake&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Value P1 P2 P3&#xA;       3  5  8&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So each player should end up with at least&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Value P1 P2 P3&#xA;       1  5/3  8/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Player 1 makes first cut, valuation after first cut:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Value P1 P2 P3&#xA;C1    1   3  1&#xA;C2    1   2  4&#xA;C3    1   1  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;P3 always chooses C2, P2 chooses C1.  So P1 is chooses C3. &#xA;So it remains:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   Value P2 P3&#xA;    C1    3   1&#xA;    C2    2   4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Say P3 now makes the second cut, so that C1 and C2 have same valuation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   Value P2 P3&#xA;    C1    &amp;gt;3   5/2&#xA;    C2    &amp;lt;2   5/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so P2 chooses C1. Now P3 is left with a piece that she values at 5/2 which is less than the proportional 8/3. &lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-08-15T00:16:12.787" LastActivityDate="2010-08-15T00:16:12.787" CommentCount="1" />
  <row Id="2465" PostTypeId="2" ParentId="2462" CreationDate="2010-08-14T13:52:14.960" Score="3" ViewCount="0" Body="&lt;p&gt;It is true because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (n+1)^2 - (n-1)^2 = 4n $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-14T13:52:14.960" />
  <row Id="2466" PostTypeId="2" ParentId="2462" CreationDate="2010-08-14T13:55:09.913" Score="2" ViewCount="0" Body="&lt;p&gt;Here is a hint: the form $x^2-y^2$ factors as $(x+y)(x-y)$.  Therefore, if you want to represent an integer $N$ as $x^2-y^2$, you can attempt to do so by choosing a factorization $N = ab$ and solving the linear system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+y = a$&lt;br&gt;&#xA;$x-y = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system has the unique rational solution $x = \frac{a+b}{2}$, $y = \frac{a-b}{2}$.  This gives an integral solution iff $a$ and $b$ have the same parity.  Use this to show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positive integer $N$ is of the form $x^2 - y^2$ for $x,y \in \mathbb{Z}$ (possibly $0$) iff $N$ is odd or $N$ is divisible by $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $4n$ is always of the form $x^2-y^2$.  You want a little more: that $x$ and $y$ are both nonzero.  Clearly $x$ cannot be zero, so you need to analyze the case $y = 0$ and show that whenever all possible solutions to $n = x^2 - y^2$ have $y = 0$, then there are nonzero $X$ and $Y$ such that $4n = X^2 + Y^2$.  This is not so hard...&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-14T14:13:38.433" LastActivityDate="2010-08-14T14:13:38.433" CommentCount="5" />
  <row Id="2467" PostTypeId="2" ParentId="2409" CreationDate="2010-08-14T14:07:11.117" Score="3" ViewCount="0" Body="&lt;p&gt;You can do this via induction, and the Schur-Zassenhaus theorem, as Matt alludes to above.  The base case $n=1$ is obvious, so assume the statement is true for all gruops $G$ with order less than $n$.  Let $\rho:\ G\rightarrow L$ be a surj. group homomorphism, and $K=\text{ker}(\rho)$.  Now if there is a maximal subgroup $M$ of $G$ such that $K$ is not contained in $M$, then $\rho(M)=L$, and by induction we can find such an $H$ in $M$.  Otherwise, $K\le\Phi(G)$, the Frattini subgroup of $G$.  Now $\Phi(G)$ is nilpotent, so all the Sylow subgroups of $K$ are normal in $G$. For every prime $p$ that divides $K$ and divides $L$, let $S_p$ be the Sylow $p$-subgroup of $K$. Taking their product, we get a normal subgroup $N\le K$ of $G$, and $G/N$ still maps onto $L$.  We are thus reduced to the case where $\text{gcd}(|K|,|L|)=1$, which is just the S-Z theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastEditorUserId="641" LastEditorDisplayName="" LastEditDate="2010-08-14T14:17:10.610" LastActivityDate="2010-08-14T14:17:10.610" CommentCount="2" />
  <row Id="2468" PostTypeId="1" CreationDate="2010-08-14T14:09:32.343" Score="6" ViewCount="199" Body="&lt;p&gt;I discovered while teaching Calc 2 that if you apply L'Hopital's rule to $\frac{x}{\sqrt{x^2+1}}$  you get $\frac{\sqrt{x^2+1}}{x}$, and if you apply L'Hopital &lt;em&gt;again&lt;/em&gt; you get $\frac{x}{\sqrt{x^2+1}}$ back.  In other words the L'Hopital &lt;em&gt;operator&lt;/em&gt; has a cycle of order two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (Thanks KennyTM): &quot;I suppose the L'Hopital operator should be defined on equivalence classes of pairs $(f(x),g(x)$ of differentiable functions with the fractional equivalence:  $(f(x),g(x))\equiv(h(x),k(x))$ if and only if $fk=gh$.&quot;  &lt;em&gt;This does not work&lt;/em&gt;.  But it doesn't really take pairs of functions to pairs of function, either.  So the first problem is to find out &lt;em&gt;how&lt;/em&gt; it is an operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever studied this operator?  Wikipedia tells me nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I know it is easier to find the limit by dividing through by $x$, but some students want to apply L'Hopital to &lt;em&gt;everything&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="318" LastEditorUserId="318" LastEditorDisplayName="" LastEditDate="2010-08-14T15:27:54.437" LastActivityDate="2010-08-14T17:18:30.607" Title="Has L'Hopital's Rule been studied as an operator?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="2469" PostTypeId="1" AcceptedAnswerId="2491" CreationDate="2010-08-14T14:46:48.510" Score="2" ViewCount="95" Body="&lt;p&gt;Let $G$ be a group. A normal subgroup $N$ of $G$ is called &lt;i&gt;ultracharacteristic&lt;/i&gt; if for every normal subgroup $U$ of $G$ the condition&#xA;$$&#xA;G/U \cong G/N&#xA;$$&#xA;implies that $U \ge N.$ What are instances of natural conditions on $G/N$ that make $N$ ultracharacteristic? For example, suppose $G/N$ has only inner automorphisms; is then $N$ ultracharacteristic? &lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-08-15T00:10:38.180" Title="Sufficient conditions for ultracharactericity " Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2470" PostTypeId="2" ParentId="674" CreationDate="2010-08-14T14:51:28.360" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Two variables and one constraint&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have an objective function of two real variables $f(x,y)$ subject to the constraint $g(x,y)=0$ and want to find the minima or maximum, provided that there is any, then the three following conditions must be satisfied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{x}(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{y}(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda g(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda g_{x}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda g_{y}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda $ is the Lagrange multiplier. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{x}(x,y)-\lambda g_{x}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{y}(x,y)-\lambda g_{y}(x,y)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $(x,y,\lambda )=(x^{\ast },y^{\ast },\lambda ^{\ast })$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we set $F(x,y,\lambda )=f(x,y)-\lambda g(x,y)$. &lt;em&gt;Observe that&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the key point of my explanation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{x}(x,y,\lambda )=f_{x}(x,y)-\lambda g_{x}(x,y)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{y}(x,y,\lambda )=f_{y}(x,y)-\lambda g_{y}(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\lambda }(x,y,\lambda )=-g(x,y)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we must find the solution $(x^{\ast },y^{\ast },\lambda ^{\ast })$ of&#xA;the following system of 2+1 equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{x}(x,y,\lambda )=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{y}(x,y,\lambda )=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\lambda }(x,y,\lambda )=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Maximum or Minimum?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To establish whether the function $f(x,y)$ has a maximum or a minimum one must define additional conditions that take into account the 2nd derivatives of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$ &lt;em&gt;variables and&lt;/em&gt; $m$ &lt;em&gt;constraints&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a function $f$ of $n$ real variables, subject to $m$ constraints $g_{k}$, with $1\leq k\leq m$, we have $m$ Lagrange multipliers and must solve a system of $n+m$ equations. &lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-14T14:51:28.360" CommunityOwnedDate="2010-08-14T14:51:28.360" />
  <row Id="2471" PostTypeId="1" AcceptedAnswerId="2473" CreationDate="2010-08-14T15:13:23.710" Score="-9" ViewCount="265" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Problem N-69 (page 29) in &lt;a href=&quot;http://www.google.com/books?id=Snecq4MHl8YC&amp;amp;lpg=PA36&amp;amp;pg=PA29#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Group Theory: Selected Problems&lt;/a&gt; by B. Sury.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $T$  be an automorphism of order $p$(prime), of a finite group $G$, such that $T(x)=x$ iff $x=e$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that the mapping $\phi(g)=g^{-1}Tg$ is a bijection on $G$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For any $g \in G$ the product $gT(g)T^{2}(g)\cdots T^{p-1}(g)$ is the identity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove that $|G| \equiv 1 (\text{mod}\ p)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For any prime $q$ dividing $|G|$, there is a $q$ sylow subgroup such that $T(Q)=Q$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $p=3$, the group is nilpotent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a general prime $p$, the above result is proved by John Thompson, whose results i wouldn't like to use it here. I have proved the first two results and also including the fact the if $p=2$, then the group is abelian. I am in need of hints and suggestions for proving the last three of the items.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T19:39:05.800" LastActivityDate="2010-08-15T19:39:05.800" Title="Automorphism of prime order" Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="24" />
  <row Id="2473" PostTypeId="2" ParentId="2471" CreationDate="2010-08-14T16:09:42.737" Score="3" ViewCount="0" Body="&lt;p&gt;3 and 4 both follow from general facts about group actions.  If $P$ is a p-group acting on the set $S$, then # fixed points = $|S|$, mod p.  So $\langle T\rangle$ acts on $G$, and only fixes one point, so $|G|=1\pmod{p}$. Similarly, $\langle T\rangle$ acts on the set of Sylow q-groups and since the number of such groups divides $|G|$, and thus is non-zero mod p, the number of fixed points is non-zero mod p.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-14T16:09:42.737" />
  <row Id="2474" PostTypeId="1" CreationDate="2010-08-14T17:17:30.843" Score="3" ViewCount="311" Body="&lt;p&gt;In May's &quot;A Concise Course in Algebraic Topology&quot; I am supposed to calculate the fundamental group of the double torus. Can this be done using van Kampen's theorem and the fact that for (based) spaces $X, Y$: $\pi_1(X\times Y) = \pi_1(X)\times\pi_1(Y)$? Or do I need other theorems to prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this should equal $\pi_1(T)\times\pi_1(T)\times\pi_1(S^1)$ where $T$ is the torus minus a closed disc on the surface, but I do not know how to calculate $\pi_1(T)$.&lt;/p&gt;&#xA;" OwnerUserId="1013" LastActivityDate="2010-08-18T19:53:56.677" Title="Fundamental group of the double torus" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="4" />
  <row Id="2475" PostTypeId="2" ParentId="2468" CreationDate="2010-08-14T17:18:30.607" Score="2" ViewCount="0" Body="&lt;p&gt;I dunno if its been studied as a differential operator, and I kind of doubt it, but I think you could define it in a way similar to the way that you suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You consider two pairs (f,g) and (h,k) equivalent if $f/g = h/k$ or rather $fk = gh$. Then the operator L acts on equivalence classes by the following operation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ L(f,g) = (Df, Dg) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for this operator to be well defined we must enforce that $L(h,k)$ is equivalent to $L(f,g)$. This means that we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Df/Dg = Dh/Dk$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not always true. Suppose that $(f,g) = (x, x^2)$ which is equivalent to $(h,k) = (1,x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Df/Dg = \frac{1}{2x} \neq 0 = Dh/Dk$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the L'Hospital operator is well defined on the equivalence classes of pairs $(f,g)$ with $g$ non-constant with the equivalence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f,g)$ equiv $(h,k) \Leftrightarrow fk = hg $ and $ Df Dk = Dh Dg $.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-14T17:18:30.607" />
  <row Id="2476" PostTypeId="1" AcceptedAnswerId="2477" CreationDate="2010-08-14T17:21:13.397" Score="-4" ViewCount="196" Body="&lt;p&gt;Trigonometric functions like $\sin{x},\cos{x},\cdots$ all take irrational values. Considering $x \in \mathbb{R}\setminus \mathbb{Q}$ is $\sin{x}, \cos{x}$ irrational? &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-14T17:47:14.210" LastActivityDate="2010-08-14T20:36:35.633" Title="Irrationality of Trigonometric Functions" Tags="&lt;real-analysis&gt;&lt;trigonometry&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="2477" PostTypeId="2" ParentId="2476" CreationDate="2010-08-14T17:29:25.287" Score="3" ViewCount="0" Body="&lt;p&gt;There is a well known theorem (An Introduction to Theory of Numbers, by Ivan Niven and H.S Zuckerman, Theorem 6.15 in the fourth edition of the book).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta$ is a rational multiple of $\pi$, then $\sin \theta$ and $\cos \theta$ are irrational except for the cases when $\sin \theta = 0, \pm 1, \pm \frac{1}{2}$ and $\cos \theta = 0, \pm 1, \pm \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a complete characterization, though.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-14T17:42:47.260" LastActivityDate="2010-08-14T17:42:47.260" />
  <row Id="2478" PostTypeId="2" ParentId="2476" CreationDate="2010-08-14T17:32:19.057" Score="3" ViewCount="0" Body="&lt;p&gt;This paper holds the answer to your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.uni-math.gwdg.de/jahnel/Preprints/cos.pdf&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-14T17:32:19.057" CommentCount="4" />
  <row Id="2479" PostTypeId="2" ParentId="2474" CreationDate="2010-08-14T17:40:02.647" Score="0" ViewCount="0" Body="&lt;p&gt;Hi: Please see this link. This has a nice explanation as well as some more information : &lt;a href=&quot;http://math.ucr.edu/home/baez/algebraic_topology/Math205B_Mar02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/algebraic_topology/Math205B_Mar02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-14T17:40:02.647" CommentCount="4" />
  <row Id="2480" PostTypeId="2" ParentId="2476" CreationDate="2010-08-14T18:04:25.317" Score="3" ViewCount="0" Body="&lt;p&gt;Elements of $\mathbb{R} \setminus \mathbb{Q}$ are congruent mod $2 \pi$ to arbitrary elements of $\mathbb{R}$, so $\sin x$ and $\cos x$ take all of their possible values on $\mathbb{R} \setminus \mathbb{Q}$.  And of course $\sin x$ and $\cos x$ frequently take rational values on $\mathbb{R}$; just take $x = \sin^{-1} r$ for any rational $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I believe you meant to ask is when $\sin 2\pi x$ and $\cos 2\pi x$ are rational.  The answer to this question is in Moron's answer, but I see no reason not to give a proof, since this is not a hard result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It suffices to prove the result for $\cos 2 \pi x$.  Let $t = 2 \cos 2 \pi \frac{p}{q}$.  Then if $t$ is rational, it is a rational root of the $q^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev_polynomials&quot; rel=&quot;nofollow&quot;&gt;Chebyshev polynomial&lt;/a&gt; (with a modified constant term), which is a monic polynomial with integer coefficients (when correctly normalized).  By the rational root theorem, any such rational root is an integer.  It follows that $t = -2, -1, 0, 1, 2$ and all of these cases occur.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-14T20:36:35.633" LastActivityDate="2010-08-14T20:36:35.633" CommentCount="3" />
  <row Id="2481" PostTypeId="2" ParentId="107" CreationDate="2010-08-14T18:04:46.580" Score="1" ViewCount="0" Body="&lt;p&gt;As other people has said, we model the situation by making a function $F(t)$ that returns the set of balls in the box at time $t$. Within this model, we have $F(\text{11:00}) = \emptyset$. That much is unambiguous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's important to keep in mind that this is only a mathematical model. In the real world, there are many reasons to doubt the model will be accurate. For example, how can you move balls arbitrarily quickly, and how can the box have unlimited volume?.  There's no reason to think that, just because the limit is well-defined within our simplistic model, it has to be well-defined in the physical world. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-08-14T18:04:46.580" />
  <row Id="2482" PostTypeId="2" ParentId="2474" CreationDate="2010-08-14T18:09:25.603" Score="10" ViewCount="0" Body="&lt;p&gt;By van Kampen's theorem, what you get is actually&#xA;$$\pi_1(T)\ast_{\pi_i(S^1)}\pi_1(T)$$&#xA;which is an amalgamated product (a pushout in the category of groups).&#xA;Roughly speaking if you have two groups $G_1$ and $G_2$ and&#xA;embeddings $i_1$ and $i_2$ of a group $H$ in both then $G_1\ast_H\ast G_2$&#xA;is the group freely generated by the elements of $G_1$ and $G_2$&#xA;but identifying elements $i_1(h)$ and $i_2(h)$ for $h\in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\pi_1(T)$ can be computed using the fact that $T$ deformation retracts&#xA;to a bouquet of two circles. (Think about the standard torus; fix a point&#xA;and look at the circles through it going round the torus in the two natural&#xA;ways.)&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-14T18:16:03.217" LastActivityDate="2010-08-14T18:16:03.217" CommentCount="2" />
  <row Id="2483" PostTypeId="1" AcceptedAnswerId="16800" CreationDate="2010-08-14T18:58:24.240" Score="25" ViewCount="938" Body="&lt;p&gt;This is a pet idea of mine which I thought I'd share.  Fix a prime $q$ congruent to $1 \bmod 4$ and define a sequence $F_n$ by $F_0 = 0, F_1 = 1$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F_{n+2} = F_{n+1} + \frac{q-1}{4} F_n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $F_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$ where $\alpha, \beta$ are the two roots of $f(x) = x^2 - x - \frac{q-1}{4}$.  When $q = 5$ we recover the ordinary Fibonacci numbers.  The discriminant of $f(x)$ is $q$, so it splits $\bmod p$ if and only if $q$ is a quadratic residue $\bmod p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\left( \frac{q}{p} \right) = -1$, then the Frobenius morphism $x \mapsto x^p$ swaps $\alpha$ and $\beta$ (working over $\mathbb{F}_p$), hence $F_p \equiv -1 \bmod p$.  And if $\left( \frac{q}{p} \right) = 1$, then the Frobenius morphism fixes $\alpha$ and $\beta$, hence $F_p \equiv 1 \bmod p$.  In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F_p \equiv \left( \frac{q}{p} \right) \bmod p.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quadratic reciprocity in this case is equivalent to the statement that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F_p \equiv \left( \frac{p}{q} \right) \bmod p.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  Does anyone have any ideas about how to prove this directly, thereby proving quadratic reciprocity in the case that $q \equiv 1 \bmod 4$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My pet approach is to think of $F_p$ as counting the number of ways to tile a row of length $p-1$ by tiles of size $1$ and $2$, where there is one type of tile of size $1$ and $\frac{q-1}{4}$ types of tiles of size $2$.  The problem is that I don't see, say, an obvious action of the cyclic group $\mathbb{Z}/p\mathbb{Z}$ on this set.  Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-01-08T16:59:32.680" Title="Quadratic reciprocity via generalized Fibonacci numbers?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="8" />
  <row Id="2484" PostTypeId="2" ParentId="2389" CreationDate="2010-08-14T19:08:20.490" Score="3" ViewCount="0" Body="&lt;p&gt;$\sum_{i=k}^n i = \sum_{i=0}^n i - \sum_{i=0}^{k-1} i = \frac{1}{2} \left[ n(n+1)  - (k-1)k \right]$&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-14T19:08:20.490" />
  <row Id="2485" PostTypeId="1" AcceptedAnswerId="2762" CreationDate="2010-08-14T21:32:46.747" Score="1" ViewCount="285" Body="&lt;p&gt;A measurable rectangle in the cartesian product $X \times Y$ of 2 measurable spaces in $\mathbb{R}^{2}$ is of the form $A \times B$, where $A$ and $B$ are measurable subsets of $X$ and $Y$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Can a measurable function on $\mathbb{R}^{2}$ be essentially unbounded on every measurable rectangle of postive measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i did was to recognize a measurable function as limit of sequence of bounded measurable functions, but don't really know how to proceed from there.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-19T16:10:32.930" Title="Measurable Functions behaviour on $\mathbb{R}^{2}$." Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2486" PostTypeId="2" ParentId="107" CreationDate="2010-08-14T22:03:53.230" Score="0" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As others have pointed out, there is no mathematical contradiction here. We have a sequence of sets $S_i$ with increasing cardinality whose intersection is the empty set; this is perfectly consistent with the axioms of set theory. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nevertheless, there is a serious philosophical problem here. I understand that J. Thompson argued in his paper &lt;/p&gt;&#xA;&#xA;&lt;p&gt;‘Tasks and Super-Tasks’, &lt;em&gt;Analysis&lt;/em&gt;, XV, pp. 1–13, 1954-55,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that there is indeed a &lt;em&gt;logical&lt;/em&gt; contradiction in this scenario. He concluded that such supertasks (i.e. tasks which involve doing an infinite number of actions in finite time) are not merely &lt;em&gt;physically impossible&lt;/em&gt; but also &lt;em&gt;logically impossible&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In lieu of precise definitions of these terms, consider, for comparison, traveling faster than the speed of light. If the theory of relativity is correct, it may very well be &lt;em&gt;physically impossible&lt;/em&gt; but its not &lt;em&gt;logically impossible&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, consider a cake that you eat at dinner &lt;em&gt;and&lt;/em&gt; that you also don't eat at dinner. This may violate the particular laws of the universe, but its also &lt;em&gt;logically impossible&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other philosophers have disputed Thompson's argument, for example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;‘Tasks, Super-Tasks, and Modern Eleatics’, P. Benacerraf, &lt;em&gt;Journal of Philosophy&lt;/em&gt;, LIX, pp. 765–784, 1962.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that Benacerraf argued that there is no logical connection between what is inside the box at 11 pm and what is inside the box at any time before 11 pm. Our universe tends to be a &quot;continuous&quot; one, but that is not logically required. Thus Benacerraf concludes there is no obvious reason why supertasks are logically impossible. I believe &#xA;Benacerraf's refutation is accepted by most of today's philosophers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have not read the original papers. My knowledge of this comes from the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521896320&quot; rel=&quot;nofollow&quot;&gt;Paradoxes&lt;/a&gt; by R.M. Sainsbury, which was assigned reading in my freshman philosophy class. A good source for further reading is the &lt;a href=&quot;http://plato.stanford.edu/entries/spacetime-supertasks/&quot; rel=&quot;nofollow&quot;&gt;Supertask&lt;/a&gt; page of the Stanford Encyclopedia of Philosophy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-08-14T22:18:37.447" LastActivityDate="2010-08-14T22:18:37.447" />
  <row Id="2487" PostTypeId="1" AcceptedAnswerId="2495" CreationDate="2010-08-14T22:35:00.130" Score="5" ViewCount="405" Body="&lt;p&gt;There is a famous literature on $\displaystyle e = \sum\limits_{n=0}^{\infty} \frac{1}{n!}$. We know that $e$ is irrational as well as transcendental. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: For each  $x_{n}$, sequence of $\pm{1}'s$, let $$f(x_{n}) = \sum\limits_{n=0}^{\infty} \frac{x_{n}}{n!}$$ for which sequences $(x_n)$ is $f(x_{n})$ rational?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also can we write $e$ as $\displaystyle e = \sum\limits_{n=1}^{\infty} \frac{x_{n}}{n}$ where $x_{n}$ is a sequence as above.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-12-18T20:56:52.337" Title="Rationality of series $\sum \frac{1}{n!}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" />
  <row Id="2488" PostTypeId="2" ParentId="2487" CreationDate="2010-08-14T22:51:09.097" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;b&gt;Suggestion for the first question&lt;/b&gt; [I have not checked all the details myself]: Have you tried to modify the proof of the irrationality of $e$ to deal with your similar-looking series?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, referring to the standard proof given in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_e_is_irrational&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Proof_that_e_is_irrational&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which steps require modification if $\sum_{n=0}^{\infty} \frac{1}{n!}$ is replaced by &#xA;$\sum_{n=0}^{\infty} \frac{x_n}{n!}$ for $x_n \in \{ \pm 1 \}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Hint for the second question&lt;/b&gt;: are you familiar with the proof of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_rearrangement_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann Rearrangement Theorem&lt;/a&gt;?  Can you adapt it here?&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T06:23:42.993" LastActivityDate="2010-08-15T06:23:42.993" CommentCount="2" />
  <row Id="2489" PostTypeId="1" AcceptedAnswerId="2490" CreationDate="2010-08-14T23:29:46.930" Score="1" ViewCount="130" Body="&lt;p&gt;When writing software, there are often situations where I need a parameter to be a floating point number x in the range 0 &amp;lt;= x &amp;lt;= 1. I don't know of a name for that category, but I think there must be one because it's such a useful categories. Perhaps there's a name in probability theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If they don't have a name, I hereby declare them to be &quot;wombat numbers&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="1188" LastActivityDate="2010-08-14T23:37:23.640" Title="Is there a name for numbers between zero and one, inclusive?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2490" PostTypeId="2" ParentId="2489" CreationDate="2010-08-14T23:37:23.640" Score="4" ViewCount="0" Body="&lt;p&gt;Sometimes I've heard it called the &quot;unit interval&quot; as it &quot;probability measures are functions from a boolean algebra to the unit interval&quot;&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-14T23:37:23.640" CommentCount="4" />
  <row Id="2491" PostTypeId="2" ParentId="2469" CreationDate="2010-08-15T00:10:38.180" Score="5" ViewCount="0" Body="&lt;p&gt;If $K$ is any group, then $N = K \times 1$ is a normal subgroup of $G = K \times K$, with quotient $G/N$ isomorphic to $K$. But $N$ cannot be ultracharacteristic unless $K$ is trivial, since $G/(K \times 1)$ and $G/(1 \times K)$ are isomorphic. So the only condition that references $G/N$ alone and that guarantees $N$ to be ultracharacteristic is &quot;$G/N$ is trivial&quot;.&lt;/p&gt;&#xA;" OwnerUserId="631" LastActivityDate="2010-08-15T00:10:38.180" CommentCount="1" />
  <row Id="2492" PostTypeId="1" CreationDate="2010-08-15T00:14:14.180" Score="1" ViewCount="102" Body="&lt;p&gt;This question concerns  an implementation of the topmoumoute natural gradient (tonga) algorithm as described in page 5 in the paper Le Roux et al 2007 &lt;a href=&quot;http://research.microsoft.com/pubs/64644/tonga.pdf&quot; rel=&quot;nofollow&quot;&gt;http://research.microsoft.com/pubs/64644/tonga.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the basic idea is to augment stochastic gradient ascent with the covariance of the stochastic gradient estimates. Basically, the natural gradient approach multiplies a stochastic gradient with the inverse of the covariance of the gradient estimates in order to weight each component of the gradient by the variance of this component. We prefer moving into directions that show less variance during the stochastic gradient estimates: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ng \propto C^{-1} g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since updating and inverting the covariance in an online optimisation setting is costly, the authors describe a ``cheap'' approximate update algorithm as described on page 5 as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_t = \gamma \hat{C}_{t-1} + g_tg_t^T$ where  $C_{t-1}$ is the low rank approximation at time step t-1. Writing  $\hat{C}_{t} = X_tX_t^T$ with $X_t =\sqrt{\gamma} X_{t-1}\ \ g_t]$ they use an iterative update rule for the  gram matrix $X_t^T X_T = G_t = &#xA;\begin{pmatrix}\gamma G_{t-1}&amp;amp;\sqrt{\gamma} X^T_{t-1}g_t\\ \sqrt{\gamma} g^T_t X_{t-1}&amp;amp;g_t^Tg_t\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then state ``To keep a low-rank estimate of  $\hat{C}_{t} = X_tX_t^T$, we can compute its eigendecomposition and keep only the first k eigenvectors. This can be made low cost using its relation to that of the Gram matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_t= X_t^T X_T$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_t = VDV^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_t = (X_tVD^{-\frac12})D(X_tVD^{-\frac12})^T$''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it's cheaper than updating and decomposing G at every step, they then suggest that you should update X for several steps using &#xA;$C_{t+b} = X_{t+b}X_{t+b}^T$ with $X_{t+b} = \left[\gamma U_t, \ \gamma^{\frac{b-1}{2}g_{t+1}},...\ \  \gamma^{-\frac12}g_{t+b-1}, \ \gamma^{\frac{t+b}{2}g_{t+b}}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see why you can get $C_t$ from $G_t$ using the eigendecomposition. But I'm unsure about their update rule for X. The authors don't explain where U is coming from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first assumptions (by notation) was, that this is the first k eigenvectors of $C_t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this makes little sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second assumption was that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U = (X_tVD^{-\frac12})D^{\frac12}$ because then $UU^T \approx XX^T = C$, but again I'm not sure if this is reasonable, because I don't think that then&#xA;$[\gamma U, g]$ is a good approximation for $[\gamma X_{t-1}, g]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if anyone has any advice as to what U should be, that would be much appreciated. I'm not getting good results using this method and want to make sure I have implemented it as intended. &lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-08-15T00:21:54.913" LastActivityDate="2010-12-14T00:44:25.977" Title="tonga: low rank approximation of the natural gradient, question regarding Le Roux et al. 2007" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2493" PostTypeId="2" ParentId="1819" CreationDate="2010-08-15T01:48:56.800" Score="1" ViewCount="0" Body="&lt;p&gt;Consider a series of random variables $X_i$ to $X$. Let $F(x)$ be the cumulative distribution function of $X$ an $F_{n}$ be the cumulative distribution function of $X_i$. Convergence in distribution occurs when $F(x)$ is the pointwise limit of $F_{n}$, ie. $\lim_{n \to \infty} F_{n}(x)=F(x)$, whenever F is continuous at x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convergence in probability is defined as follows $\lim_{n \to \infty } Pr(|X_n-X| \ge e)=0$. It is stronger than convergence in distribution - I don't know how to prove this, but I can give the following explanation. For convergence in distribution, we only looked at the cumulative functions, so it didn't matter whether the variables were dependent or not. For convergence in probability, $X$ needs to either be a constant or $X_n$ must be dependent on X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost sure convergence is defined as follows $Pr(\lim_{n \to \infty} X_n=X)=1$. On this &lt;a href=&quot;http://mathoverflow.net/questions/35211/reducing-limits-to-a-canoncial-form&quot; rel=&quot;nofollow&quot;&gt;MathOverflow thread&lt;/a&gt; I asked if there was a way to reduce limits to some kind of canonical form. I obtained the following results where $dif_x=|X_n-X|$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Convergence in probability: $\forall e, d, n&gt;N_1(e,d): dif_x \ge e \text{ with } p &amp;lt; d $&lt;/li&gt;&#xA;&lt;li&gt;Almost sure convergence: ($\forall e, n&gt;N_2(e): dif_x &amp;lt; e) \text{ with } p=1$&lt;/li&gt;&#xA;&lt;li&gt;Alternative form: ($\forall e, n&gt;N_2(e): dif_x \ge e) \text{ with } p=0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We notice that for convergence in probability, we can reduce $p$ below any arbitrary positive value by choosing values of $n$ high enough, while in the second formula we can reduce it to 0. Additionally, the probability in the first formula looks at individual probabilities, while the second looks at probabilities over all possible values of e and n. From these forms it is clear that Almost Sure Convergence implies Convergence in probability. If convergence in probability is not occurring, then we can find some e, d, with with $p \ge d \forall n$. This alone will cause the $p$ in almost sure convergence to not be 0.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-09-11T07:20:19.580" LastActivityDate="2010-09-11T07:20:19.580" />
  <row Id="2494" PostTypeId="2" ParentId="107" CreationDate="2010-08-15T02:25:27.037" Score="2" ViewCount="0" Body="&lt;p&gt;No balls.  If $B(t)$ is the number of balls in the box at time $t$, then as you say, $B(t)$ is increasing, and $\lim_{t \to 11}B(t) = +\infty$.  But since every ball has been removed, $B(11)=0$.  All this is fine.  The &quot;paradox&quot; as I see it is that at the outset you expect the function $B$ to be continuous in some sense, but &lt;em&gt;a priori&lt;/em&gt; there's no reason why it should be, and indeed it is not.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-15T02:25:27.037" />
  <row Id="2495" PostTypeId="2" ParentId="2487" CreationDate="2010-08-15T03:22:34.900" Score="12" ViewCount="0" Body="&lt;p&gt;The numbers you describe are in fact all irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a fixed rational number $p/q$, any other rational number $a/b$ that approximates $p/q$ (but is not equal to $p/q$) can only do so to order $1/b$.  This can be seen as follows:&#xA;$$ p/q - a/b = (pb - qa)/qb,$$&#xD;&#xA;which is at least $(qb)^{-1}$ if it is not zero. So rationals cannot be approximated too well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the series $$\sum x_n/n!$$ and the partial sums $S_j$.  The partial sums $S_k$ have denominator $k!$. Their error from the infinite sum is at most $\sum_{n&amp;gt;k} 1/n!$ which (from standard upper bounds via geometric series) little oh of $1/k!$ (as in the standard proof for $e$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, the problem is that the series converges &quot;too rapidly&quot; for the sum to be rational. A rational number cannot be approximated too well (well in the denominators) by other rationals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are similar Diophantine approximation results for algebraic numbers, cf. &lt;a href=&quot;http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem&quot; rel=&quot;nofollow&quot;&gt;Roth's theorem&lt;/a&gt;. Unfortunately that's not applicable here, though, since the approximations $S_k$ are little oh of the denominator, not little oh of a power of the denominator.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-18T20:56:52.337" LastActivityDate="2010-12-18T20:56:52.337" CommentCount="5" />
  <row Id="2496" PostTypeId="1" AcceptedAnswerId="2507" CreationDate="2010-08-15T03:23:21.240" Score="1" ViewCount="173" Body="&lt;p&gt;A Banach Space is a vector space which is normed and is complete with respect to its norm (every cauchy sequence converges). Examples of Banach spaces include $\mathbb{R}^n$ and the space of bounded continuous functions with the norm $\Vert f(x) \Vert = \sup \lbrace \vert f(x) \vert \rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of all Bounded Linear Operators $\Omega$ on the space of all $C^\infty$ functions over $\mathbb{R}$ forms a Banach space:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For two operators $A, B \in \Omega$ define $(A+B)(f) = Af + Bf$, and for a constant $c \in \mathbb{R}$ define $(cA)(f) = c Af$. This makes $\Omega$ into a vector space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Define a norm by $ \Vert A \Vert_\Omega = \sup \lbrace \Vert Af \Vert : f \in C^\infty(\mathbb{R}), \Vert f \Vert \leq 1 \rbrace $, where $\Vert f \Vert$ for $f \in C^\infty(\mathbb{R})$ is some norm on that space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now my question is: Does the norm chosen for $C^\infty(\mathbb{R})$ affect the structure of $\Omega$ (i.e. its topology or whatever), and is there a natural choice for such a norm in this situation?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-15T14:38:52.453" Title="How to define a Norm on a Banach space of Operators" Tags="&lt;real-analysis&gt;&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="2497" PostTypeId="1" AcceptedAnswerId="2499" CreationDate="2010-08-15T04:36:35.660" Score="0" ViewCount="265" Body="&lt;p&gt;Probably a proof (if any exist) that calls upon Knuth's up-arrow &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation&quot; rel=&quot;nofollow&quot;&gt;notation&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Busy_beaver&quot; rel=&quot;nofollow&quot;&gt;Busy Beaver&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-08-15T22:44:06.537" CommunityOwnedDate="2010-08-15T14:44:00.880" Title="What is the biggest number ever used in a mathematical proof?" Tags="&lt;big-numbers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="2498" PostTypeId="2" ParentId="2427" CreationDate="2010-08-15T06:00:01.373" Score="6" ViewCount="0" Body="&lt;p&gt;Gerhard Michler has worked on a research program to show fairly convincingly that the possibility of infinitely many sporadic groups (with a uniform construction, but highly non-uniform properties) was quite real.  Roughly speaking the second round of sporadic groups was discovered looking for special configurations of centralizers of involutions, and he shows how this search can be continued, how it constructs almost all of the sporadic simple groups in a uniform fashion, and how it does not obviously stop there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is discussed in some detail in his books &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2266036&quot; rel=&quot;nofollow&quot;&gt;MR2266036&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2583258&quot; rel=&quot;nofollow&quot;&gt;MR2583258&lt;/a&gt;, the Theory of Finite Simple Groups, volumes I and II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at least according to him, it should not be taken for granted that there are only finitely many sporadic groups, as there is a fairly reasonable procedure for possibly producing an infinite collection of basically unrelated finite simple groups (at least, nowhere near as related as groups of a fixed Lie type and rank).&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-15T06:00:01.373" CommentCount="1" />
  <row Id="2499" PostTypeId="2" ParentId="2497" CreationDate="2010-08-15T06:04:22.780" Score="4" ViewCount="0" Body="&lt;p&gt;The mathematician Harvey Friedman observed a special finite form of Kurskal's Tree Theorem. Regarding this form, Friedman discusses the existence of a rapidly growing function he calls $TREE(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $TREE$ sequence begins $TREE(1)=1$ and $TREE(2)=3$, but $TREE(3)$ is a number so extremely large that its weak lower bound is $(A(...A(1)...))$, where the number of A's is $A(187196)$, and $A()$ is a version of Ackermann's function: $A(x) = 2↑↑...↑x$ with $x-1 ↑s$ (Knuth up-arrows). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas Graham's Number is $A^{64}(4)$, the above mentioned lower bound is $A^{A(187196)}$. As you can imagine, the $TREE$ function keeps on growing rather quickly. For a discussion on the hierarchy of fast growing functions see here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fast-growing_hierarchy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fast-growing_hierarchy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other examples of numbers greater than Graham's Number, as can be seen here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Goodstein_function#Sequence_length_as_a_function_of_the_starting_value&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Goodstein_function#Sequence_length_as_a_function_of_the_starting_value&lt;/a&gt;, although I'm not sure if this number is larger than Friedman's $TREE(3)$   &lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="647" LastEditorDisplayName="" LastEditDate="2010-08-15T22:44:06.537" LastActivityDate="2010-08-15T22:44:06.537" />
  <row Id="2500" PostTypeId="2" ParentId="2372" CreationDate="2010-08-15T06:14:50.973" Score="4" ViewCount="0" Body="&lt;p&gt;There's probably not going to be an easy way to do this...  Consider two different examples of 15-letter &quot;permutations&quot;.  Then the number of permutations with that multiset of digits depends on the proportions of the chosen digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to do this, you can sum $k!/(s(1)! s(2)! \cdots s(x)!)$ (called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;mulitnomial coefficient&lt;/a&gt;) over all partitions of $s(1)+s(2)+ \cdots +s(x)=k$ [in this partition s(i) is allowed to be zero and order is important] such that $s(i) \leq r(i)$ for all i.  The part s(i) says you have s(i) copies of the i-th letter.  The number of permutations with s(i) i-th letters is given as above, by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_action&quot; rel=&quot;nofollow&quot;&gt;Orbit Stabiliser Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, this is only one step better than the caveman's counting formula:  sum_P 1 where P is the set of permutations you want to count.  I.e. just count them one-by-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: While I'm making a few touch-ups, here's GAP code that implements the above formula.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NrPermIdent:=function(k,T)&#xA;  local PSet,x;&#xA;  x:=Size(T);&#xA;  PSet:=Filtered(OrderedPartitions(k+x,x)-1,p-&amp;gt;ForAll([1..x],i-&amp;gt;p[i]&amp;lt;=T[i]));&#xA;  return Sum(PSet,p-&amp;gt;Factorial(k)/Product(p,i-&amp;gt;Factorial(i)));&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where T is a list of bounds and k is the number of terms in the partition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; NrPermIdent(15,[4,5,8,3]);&#xA;187957770&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As another indication that finding a simple formula for these numbers is not going to be easy, observe that NrPermIdent(n,[n,k]) is equal to $\sum_{0 \leq i \leq k} {n \choose k}$ (which is considered a difficult sum to find -- see: &lt;a href=&quot;http://mathoverflow.net/questions/17202/&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17202/&lt;/a&gt;).  I remember reading somewhere (most likely in &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt;) that you can prove there is no &quot;closed-form&quot; solution for this.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-08-31T06:32:02.473" LastActivityDate="2010-08-31T06:32:02.473" CommentCount="0" />
  <row Id="2501" PostTypeId="1" AcceptedAnswerId="2515" CreationDate="2010-08-15T07:34:35.150" Score="-6" ViewCount="199" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Problem 8I (page 70–71) of &lt;a href=&quot;http://books.google.com/books?ei=YD5oTKTiNIHCceLatZAF&amp;amp;ct=result&amp;amp;id=VvTuAAAAMAAJ&amp;amp;dq=Problems+for+Mathematicians%2C+Young+and+Old&amp;amp;q=entire#search_anchor&quot; rel=&quot;nofollow&quot;&gt;Problems for Mathematicians, Young and Old&lt;/a&gt; by P. R. Halmos.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If $g$ is a polynomial that is not a translation, must the equation $g(g(z))=z$ always have a solution? Why is $g$ not a translation one may ask? The answer is, if $g$ is not a translation, then $g(g(z))-z$ cannot be a non zero constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, can we ask this for entire functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If $f$ is an entire function, that is not a translation then does $f(f(z))=z$ have a solution?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: Fundamental theorem of Algebra!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T19:25:00.763" LastActivityDate="2010-08-15T19:25:00.763" Title="Roots of Entire Functions" Tags="&lt;complex-analysis&gt;&lt;analysis&gt;&lt;complex-variables&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="2502" PostTypeId="1" AcceptedAnswerId="2516" CreationDate="2010-08-15T08:58:00.283" Score="1" ViewCount="180" Body="&lt;p&gt;I've heard large cardinals talked about, and I (think I) understand a little about how you define them, but I don't understand why you would bother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the simplest proof or whatever that requires the use of large cardinals? Is there some branch of mathematics that makes particularly heavy use of them?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-15T16:07:52.910" Title="What are large cardinals for?" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2503" PostTypeId="2" ParentId="1422" CreationDate="2010-08-15T09:34:21.720" Score="0" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eric, i had asked a similar question, in MO, which includes the condition of Differentiablility. Please check this one out. It may be of some use. &lt;a href=&quot;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-15T09:34:21.720" />
  <row Id="2504" PostTypeId="1" AcceptedAnswerId="2512" CreationDate="2010-08-15T11:03:08.890" Score="2" ViewCount="196" Body="&lt;p&gt;I just started reading some Schapira notes on &lt;a href=&quot;http://people.math.jussieu.fr/~schapira/lectnotes/AlTo.pdf&quot; rel=&quot;nofollow&quot;&gt;Algebra and Topology&lt;/a&gt;. Statement 1.7 is the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hom (Z, Ker(f, g)) ≃ Ker(Hom (Z, X) ⇉ Hom (Z, Y ))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ker(f, g) = {x ∈ X; f(x) = g(x)}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried verifying this myself by drawing a diagram, but failed. Can anyone explain why this is true?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T08:58:15.393" LastActivityDate="2010-08-16T09:04:56.390" Title="Prove Hom (Z, Ker(f, g)) ≃ Ker(Hom (Z, X) ⇉ Hom (Z, Y )) for sets" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="2505" PostTypeId="1" AcceptedAnswerId="2506" CreationDate="2010-08-15T11:04:13.943" Score="-7" ViewCount="218" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Indian National Mathematics Olympiad 2009 (INMO 2009), Day 1, Problem 2.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Define the sequence $(x_n)_{n=1}^{\infty}$ as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_n = \begin{cases}&#xA;0 &amp;amp; \text{if the number of positive divisiors of $n$ is odd}, \\&#xA;1 &amp;amp; \text{if the number of positive divisiors of $n$ is even}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Include $1$ as a positive divisor of $n$. Let $x=0.x_{1}x_{2}x_{3}.......$ be the real number whose decimal expansion contains $a_{n}$ in the $n$-th place, $n \geq 1$. Is $x$ is rational or irrational.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-15T19:31:20.577" LastActivityDate="2010-08-15T19:31:20.577" Title="Determining whether $x_{n}$ is rational or irrational." Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="2506" PostTypeId="2" ParentId="2505" CreationDate="2010-08-15T11:17:20.780" Score="4" ViewCount="0" Body="&lt;p&gt;Actually, $n$ has an odd number of positive divisors iff $n$ is a perfect square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be easily proven by pairing up the divisors $d$ with $\frac{n}{d}$, so that the number of divisors is odd iff there is a divisor $d$ such that $d=\frac{n}{d}\Leftrightarrow n=d^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, note that the number of divisors of $n$ with $p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ as it's prime factorization is $(\alpha_1+1)\ldots(\alpha_k+1)$ (because there is a bijection between the divisors $p_1^{\beta_1}\ldots p_k^{\beta_k}$ and the k-tuples $(\beta_1,\ldots,\beta_k)$, $0\leq \beta_i\leq \alpha_i$). Then the number of divisors is odd iff each $\alpha_i$ is even, or $n$ is a perfect square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This clearly implies that $x$ is irrational since it's decimal representation is not periodic (because it contains arbitrarily long sequences of ones).&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-08-15T11:17:20.780" />
  <row Id="2507" PostTypeId="2" ParentId="2496" CreationDate="2010-08-15T11:51:49.750" Score="1" ViewCount="0" Body="&lt;p&gt;First off there is no norm that can be put on $C^\infty(\mathbb{R})$, or even $C^\infty([0,1])$, that will make it into a Banach Space. I mention the last one because there are lots of obvious norms, sup norm, $L^p$ norms etc. But it will not be complete with respect to any norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly simple to see this, for the $L^2$ norm. Differentiation would then be an operator defined on the whole Banach space that has an adjoint. In fact, it is self-adjoint (this is by integration by parts). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is a theorem that any operator on a Banach space with an adjoint is automatically bounded (this is a consequence of Total Boundedness Principle). But of course differentiation is unbounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put some norm on $C^\infty(\mathbb{R})$ then yes it would greatly affect the set of bounded operators (even as a set, not just as a banach space). For example putting different $L^p$-norms would give you different duals spaces. While if you say restricted to bounded smooth functions and used the sup-norm, the dual would be measures on $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the take-away message is that the definition of bounded operators involves the norm on the domain. So if you change that norm you can greatly change the bounded operators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way most of this stuff can be found in pretty much any Functional Analysis textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like Analysis NOW by Gert Pederson.&lt;/p&gt;&#xA;" OwnerUserId="1193" LastActivityDate="2010-08-15T11:51:49.750" CommentCount="2" />
  <row Id="2508" PostTypeId="1" AcceptedAnswerId="2509" CreationDate="2010-08-15T12:23:51.247" Score="2" ViewCount="202" Body="&lt;p&gt;Hay dear!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am solving the following inequality, please look at it and tell me am i write or not because this is an example in howard anton's book, i solved it on my own as given below but in the book it is solved differently, please look at my solution and tell me am i right or not and if not then explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bOwwL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-15T20:14:52.360" LastActivityDate="2010-08-15T20:14:52.360" Title="Solving a quadratic inequality" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="2509" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T12:35:09.083" Score="4" ViewCount="0" Body="&lt;p&gt;For ab to be positive either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a and b are both positive&lt;/li&gt;&#xA;&lt;li&gt;a and b are both negative&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, a=x-5 and b=x+2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are both positive if x&gt;5. They are both negative if x&amp;lt;-2. Either of these will solve the problem&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-15T12:47:44.873" LastActivityDate="2010-08-15T12:47:44.873" CommentCount="3" />
  <row Id="2510" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T12:35:54.060" Score="0" ViewCount="0" Body="&lt;p&gt;also have to look at the case where both are negative.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2010-08-15T12:35:54.060" />
  <row Id="2511" PostTypeId="2" ParentId="2501" CreationDate="2010-08-15T12:52:57.700" Score="0" ViewCount="0" Body="&lt;p&gt;This is not a complete solution, but I'll post it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f$ is an entire function and $f\circ f$ has&#xA;no fixed points. Then neither does $f$, and so&#xA;$$f(z)=z+e^{g(z)}$$&#xA;where $g$ is also entire. Then&#xA;$$f(f(z))-z=e^{g(z)}+e^{g(f(z))}=e^{g(z)}(1+e^{g(f(z))-g(z)}).$$&#xA;For this not to vanish, it is necesdary that $g(f(z))-g(z)$&#xA;never takes any value $(2k+1)\pi i$ where $k\in\mathbb{Z}$.&#xA;By Picard's theorem, this entails that $g(f(z))-g(z)$ be constant.&#xA;Thus&#xA;$$g(f(z))-g(z)=g(z+e^{g(z)})-g(z)$$&#xA;must be constant (and not an odd multiple of $\pi i$). This&#xA;looks a bit unlikely to me, but not utterly inconceivable.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-15T12:52:57.700" />
  <row Id="2512" PostTypeId="2" ParentId="2504" CreationDate="2010-08-15T13:09:29.157" Score="2" ViewCount="0" Body="&lt;p&gt;Maybe something is missing in the original statement and is: whose kernel is that on the right hand side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's recall that $f$ and $g$ are maps from $X$ to $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then that kernel on the right is obviously&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ker (f_{*}, g_{*}) \subset \mathrm{Hom}(Z,X) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f_{*}, g_{*} : \mathrm{Hom}(Z,X) \longrightarrow \mathrm{Hom}(Z,Y)&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;are the maps defined by $\varphi \mapsto f\circ \varphi$ and $\varphi \mapsto g\circ \varphi$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the bijection you ask for simply sends every map &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : Z \longrightarrow \ker (f,g) \subset X&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to $i\circ \varphi$, where $i: \ker (f,g) \hookrightarrow X$ is the inclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi \in \ker (f_{*}, g_{*}) \ \Longleftrightarrow \ f\circ \psi = g\circ \psi \ \Longleftrightarrow \ f(\psi (z)) = g(\psi (z))  \ \text{for all} \ z \in Z&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi (z) \in \ker (f,g)   \ \text{for all} \ z \in Z \ \Longleftrightarrow \ \mathrm{im} (\psi ) \subset \ker (f, g) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\psi : Z \longrightarrow X$ factorises through the inclusion $i: \ker (f,g) \hookrightarrow X$. That is, since $i$ is injective, there exists a &lt;em&gt;unique&lt;/em&gt; $\varphi : Z \longrightarrow \ker (f,g)$ such that $\psi = i \circ \varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the map $LHS \longrightarrow RHS$ sends $\varphi $ to $i \circ \varphi$ and the map $RHS \longrightarrow LHS$ sends $\psi$ to $\varphi$ such that $\psi = i \circ \varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both compositions are the identity by definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi \mapsto \varphi \mapsto i \circ \varphi = \psi&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi \mapsto i \circ \varphi \mapsto \varphi \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: I've never seen &quot;two&quot; sets that were so much the same.   :-)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-16T01:32:45.270" LastActivityDate="2010-08-16T01:32:45.270" CommentCount="2" />
  <row Id="2513" PostTypeId="2" ParentId="2504" CreationDate="2010-08-15T13:13:59.240" Score="0" ViewCount="0" Body="&lt;p&gt;I think Hom (Z, X) ⇉ Hom (Z, Y) means some pair from this domain, rather than all possible pairs. If I am wrong, then this solution will need to be changed. If it is a single, pair, then it is pretty clear that it will be f◦, g◦&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS=$K_1$ is all k:Z→X'=Ker(f,g)$\in$X and f(x)=g(x) for x$\in$X'&#xA;RHS=$K_2$ is all k:Z→X with f◦(k)=g◦(k) or f◦(k)(z)=g◦(k)(z) for z$\in$Z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $k \in K_1$. As f(x)=g(x),f◦(k)(z)=g◦(k)(z) for all z. So $k \in K_2$&#xA;Now consider $k \in K_2$. f◦(k)=g◦(k). If k(z) not in Ker(f,g) for some z, then f◦(k)(z)$\ne$g◦(k)(z), so contradiction. So $k \in K_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there is a bijection between $k \in K_1$ and $k \in K_2$&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T09:04:56.390" LastActivityDate="2010-08-16T09:04:56.390" CommentCount="1" />
  <row Id="2514" PostTypeId="1" AcceptedAnswerId="2523" CreationDate="2010-08-15T13:23:59.043" Score="2" ViewCount="375" Body="&lt;p&gt;I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?&lt;/p&gt;&#xA;" OwnerUserId="44" LastActivityDate="2010-08-17T04:13:15.363" Title="Why can't the Polynomial Ring be a Field?" Tags="&lt;ring-theory&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="2515" PostTypeId="2" ParentId="2501" CreationDate="2010-08-15T13:28:44.290" Score="3" ViewCount="0" Body="&lt;p&gt;For entire transcendental functions (i.e. entire functions that are not polynomials), this is true but highly non trivial :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this article :&#xA;Bergweiler, Walter. Periodic points of entire functions: proof of a conjecture of Baker.  Complex Variables Theory Appl.  17  (1991),  no. 1-2, 57--72. (Reviewer: Fred Gross)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bergweiler shows the much stronger result :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem&#xA;Let $f$ be an entire transcendental function and $n \geq 2$. Then f has&#xA;infinitely many repelling periodic points of primitive period $n$.&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-15T13:28:44.290" />
  <row Id="2516" PostTypeId="2" ParentId="2502" CreationDate="2010-08-15T13:58:31.957" Score="3" ViewCount="0" Body="&lt;p&gt;I will leave the explaining of large cardinals to someone more knowleadgeable and explain one place where they are useful in tidying up things: category theory. In category theory, you are constantly facede with proper classes (the category of all sets, of all groups, etc.). To make things worse, you want to form functor categories but due to the sizes of the classes involved there is no way to do that in ZFC (and other set theories like NBG or MK soon hit a wall of their own). The Grothendieck axiom of universes is a large cardinal axiom (and I am told that it is rather mild compared to the large cardinals that set theorists routinely consider) that allows us to tidy up things for this and other constructions without paying attention to set sizes. Truth be told, this is mainly a convenience, as if one payed proper attention, and at the cost of circumlocutions (and vast tribulations) one could avoid them. But really, why go to all this trouble to settle what is a minor technicality not germane to the problem at hand, when you have this labor-savibng device at hand?&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-15T13:58:31.957" />
  <row Id="2517" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T14:34:19.203" Score="6" ViewCount="0" Body="&lt;p&gt;Because by definition, the only polynomial that can have a negative degree is 0, which is defined to have a degree of -$\infty$. Non-zero constants have degree 0.  You then have the degree equation: &lt;em&gt;deg (fg) = deg f + deg g&lt;/em&gt;  for any polynomials &lt;em&gt;f,g&lt;/em&gt;.  By inspection, any polynomial of degree &lt;em&gt;n&lt;/em&gt; $\geq$ 1 would need as an inverse a polynomial of degree -&lt;em&gt;n&lt;/em&gt;, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the &lt;em&gt;field of rational functions&lt;/em&gt;, and is precisely the set of ratios of polynomials.  It is constructed the same way that the field of rational numbers is from the ring of integers.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-15T14:34:19.203" />
  <row Id="2518" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T14:37:23.080" Score="1" ViewCount="0" Body="&lt;p&gt;Consider $\mathbb{C}[x]$ the ring of polynomials with coefficients from $\mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-15T14:37:23.080" />
  <row Id="2519" PostTypeId="2" ParentId="2496" CreationDate="2010-08-15T14:38:52.453" Score="4" ViewCount="0" Body="&lt;p&gt;I'm not aware of any natural choices for norms on $C^\infty(\mathbb{R})$ at all.  (As mentioned in comments, things like the sup or $L^p$ norms are infinite for some $C^\infty(\mathbb{R})$ functions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;unnatural&lt;/em&gt; choices, contrary to Owen Sizemore's assertion.  For example, it is known that all infinite-dimensional separable Fr&amp;eacute;chet spaces (such as $C^\infty(\mathbb{R})$ in its usual topology) have Hamel dimension $2^{\aleph_0}$, hence are isomorphic as vector spaces.  (I believe this is due to Mazur, but couldn't find the exact reference.) So if $(X, ||\cdot||_X)$ is your favorite infinite-dimensional separable normed space, there is a linear isomorphism $T : C^\infty(\mathbb{R}) \to X$ (of course it is in general horribly discontinuous).  For $f \in C^\infty(\mathbb{R})$, set $||f|| := ||Tf||_X$.  Voil&amp;agrave;, a norm on $C^\infty(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, under this norm $C^\infty(\mathbb{R})$ is isometrically isomorphic to $X$, so for all intents and purposes you are really working with $X$.  You can even make $C^\infty(\mathbb{R})$ into a Banach space or even a Hilbert space by choosing $X$ accordingly.  And the space of bounded operators on $(C^\infty(\mathbb{R}), ||\cdot||)$ is isometrically isomorphic to the space of those on $X$. Thus for any infinite-dimensional separable normed space $X$ you can realize the space of bounded operators $L(X)$ in this way, so the choice of norm on $C^\infty(\mathbb{R})$ certainly does affect the structure of $L(C^\infty(\mathbb{R}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates in some sense that the norm contains almost all of the structure of a normed space; if you drop it, you are left with just the vector space structure, which determines almost nothing.  &quot;Choosing a norm&quot; on a space is not really a sensible thing to do, since in effect it is really just choosing a normed space which could be almost totally unrelated to the original space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it should be pointed out that the space of bounded linear operators on an &lt;em&gt;incomplete&lt;/em&gt; normed space is not a Banach space.  Proof: Let $X$ be an incomplete normed space, let $\lbrace x_n\rbrace$ be a Cauchy sequence in $X$ with no limit, and let $f \in X^*$ be a nonzero bounded linear functional on $X$.  Define operators $T_n$ by $T_n x = f(x) x_n$.  Then $\lbrace T_n \rbrace$ is Cauchy in operator norm, but does not converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct theorem is that for normed spaces $X,Y$, the space $L(X,Y)$ of bounded linear operators from $X$ to $Y$ is a Banach space iff $Y$ is a Banach space.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-15T14:38:52.453" />
  <row Id="2520" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T14:52:27.693" Score="0" ViewCount="0" Body="&lt;p&gt;The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2010-08-15T20:27:53.500" LastActivityDate="2010-08-15T20:27:53.500" CommentCount="2" />
  <row Id="2521" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T15:25:54.520" Score="2" ViewCount="0" Body="&lt;p&gt;Casebash's answer is very good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a second answer. You can apply the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: If the roots $x_{1},x_{2}$ of $f(x)=ax^{2}+bx+c$ are real and $x_{1}\neq x_{2}$ (with $x_{1} &amp;lt; x_{2}$), then, the signal of $f(x)$ is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;opposite to the signal of $a$ for $x\in \left[ x_{1},x_{2}\right] $,&lt;/li&gt;&#xA;&lt;li&gt;the same of $a$ for $x\in \left] -\infty ,x_{1}\right[ \vee x\in \left]&#xA;x_{2},-\infty \right[ $.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since in your case $a=1&gt;0$, $x_{1}=-2&amp;lt;5=x_{2}$, you have $x^{2}-3x-10&gt;0$ for &#xA;$x\in \left] -\infty ,-2\right[ \vee x\in \left] 5,\infty \right[ $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Addendum:&lt;/em&gt; A possible proof of this theorem is to use the explanation of Casebash, taking into consideration that $ax^{2}+bx+c=a(x-x_1)(x-x_2)$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-15T17:05:16.697" LastActivityDate="2010-08-15T17:05:16.697" />
  <row Id="2522" PostTypeId="2" ParentId="2502" CreationDate="2010-08-15T16:07:52.910" Score="2" ViewCount="0" Body="&lt;p&gt;G. Rodrigues's specific answer gets at the general issue: large cardinals are used to examine how much more one can proof in ZFC set theory.  The first time I discovered large cardinals (in Jech's 2000 book &lt;em&gt;Set Theory&lt;/em&gt;), I was amazed.  A large cardinal is just a &quot;very big&quot; set, after all, but I did not realize that the existence of such a set changed the nature of what was mathematically provable.  For example, there is, according to Jech, the event that started it all: Ulam's work on the &lt;em&gt;problem of measure&lt;/em&gt;.  It is well-known that Lebesque measure over the reals is not defined for all sets, but it turns out to be undecidable in ZFC alone if &lt;em&gt;any&lt;/em&gt; non-trivial measure on the reals exists at all.  In order to get such a measure, one must assume the existence of a large cardinal, which is now called a &lt;em&gt;measurable cardinal&lt;/em&gt;.  So I think of large cardinals as things that change the very nature of the mathematical &quot;plumbing&quot;.  Deep stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more practical level, I think it was Dudley who said that large cardinals can be useful for seeing why a proof is failing: if a proof is not working, seeing if it fails at a large cardinal can provide insight.&lt;/p&gt;&#xA;" OwnerUserId="452" LastActivityDate="2010-08-15T16:07:52.910" />
  <row Id="2523" PostTypeId="2" ParentId="2514" CreationDate="2010-08-15T16:10:41.893" Score="19" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;\quad\rm  x \; f(x) = 1 \;$ in $\rm R[x] \;\Longrightarrow \; 0 = 1 \;$ in $\rm R \;\;$  by evaluating at $\; x = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; This has an instructive &lt;strong&gt;universal&lt;/strong&gt; interpretation: if $\rm x$ is a unit in $\rm R[x]$ then so too is every $\rm R$-algebra element $\rm r\:$, as follows by evaluating $\;\rm x \; f(x) = 1 \;$ at $\rm\; x = r \;$. Therefore to present a counterexample we need only exhibit a nonunit in some $\rm R$-algebra. A natural choice is the nonunit $\;\rm 0\in R\;$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-17T04:13:15.363" LastActivityDate="2010-08-17T04:13:15.363" CommentCount="2" />
  <row Id="2524" PostTypeId="2" ParentId="2504" CreationDate="2010-08-15T16:14:14.103" Score="4" ViewCount="0" Body="&lt;p&gt;This is a special case of the theorem that covariant hom-functors&#xA;(here $\mathrm{Hom}(Z,\underline{\ })$) preserve limits.&#xA;This is Theorem 1 in section V.4 of Mac Lane's&#xA;&lt;em&gt;Categories for the Working Mathematician&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-15T16:14:14.103" CommentCount="5" />
  <row Id="2525" PostTypeId="2" ParentId="2508" CreationDate="2010-08-15T16:41:06.920" Score="2" ViewCount="0" Body="&lt;p&gt;If you graph the function $y=x^2-3x-10$, you can see that the solution is $x&amp;lt;-2$ or $x&gt;5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UIwCt.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-08-15T16:41:06.920" />
  <row Id="2526" PostTypeId="2" ParentId="2501" CreationDate="2010-08-15T17:20:51.837" Score="4" ViewCount="0" Body="&lt;p&gt;My reputation is too low to answer Bill Dubuque's comment above (could the moderators move my answer?), but the problem is duplicated from &lt;i&gt;Problems for Mathematicians, Young and Old&lt;/i&gt; by Halmos. It's the problem 8I, p. 71. The solution is on p. 251.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2010-08-15T17:20:51.837" CommentCount="4" />
  <row Id="2527" PostTypeId="1" CreationDate="2010-08-15T18:25:46.097" Score="2" ViewCount="84" Body="&lt;p&gt;For a while I've been looking for some kind of software that will let me create and cross-reference definitions. The way I imagine this working would be, for instance: I create a file named &quot;Hopf algebra&quot;, I type in &quot;A Hopf algebra over a field k is a k-bialgebra such that...&quot; and follow with the definition of a Hopf algebra, and things like &quot;bialgebra&quot; and &quot;field&quot; would be links to those definitions. Essentially something like a DIY maths encyclopedia. Bonus things would be allowing me to put things into categories (e.g. ring theory vs. group theory) and colour-coding, and some manner of LaTeX support would be pretty essential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot be the only one who's had this idea, but so far all my Google searches have been turning up empty - I figure there's some key word I'm missing. Has anyone heard of something like this before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: What I'm looking for is a sort of better-organised and better-crossref'd software version of the giant stack of index cards I use right now, not a wiki or anything communal like that.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="1203" LastEditorDisplayName="" LastEditDate="2010-08-15T19:38:44.367" LastActivityDate="2010-08-16T08:59:16.920" Title="Software for creating a definition library" Tags="&lt;math-software&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2528" PostTypeId="2" ParentId="2527" CreationDate="2010-08-15T18:49:56.067" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://groupprops.subwiki.org/wiki/Main_Page&quot; rel=&quot;nofollow&quot;&gt;http://groupprops.subwiki.org/wiki/Main_Page&lt;/a&gt; is something like this for group theory.  There is also the &lt;a href=&quot;http://ncatlab.org/nlab/show/HomePage&quot; rel=&quot;nofollow&quot;&gt;nLab&lt;/a&gt;.  Both of them run on wiki software; do you have something else in mind?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-15T18:59:15.063" LastActivityDate="2010-08-15T18:59:15.063" CommentCount="3" />
  <row Id="2529" PostTypeId="1" AcceptedAnswerId="2546" CreationDate="2010-08-15T19:00:18.010" Score="1" ViewCount="79" Body="&lt;p&gt;This problem is taken from I.N. Herstein&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt; If $G$ is a group and $a \in G$ if of finite order and has only a finite number of conjugates in $G$, prove that these conjugates of $a$ generate a finite normal subgroup of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see a solution for this and also i would like to know whether the conjugates always generate a normal subgroup. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-15T23:34:23.133" Title="Conjugates Generating a normal subgroup." Tags="&lt;group-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="2530" PostTypeId="1" AcceptedAnswerId="2556" CreationDate="2010-08-15T19:14:02.973" Score="1" ViewCount="449" Body="&lt;p&gt;Although i am very much new to &quot;Analytic Number Theory&quot;, there are some non mathematical questions which puzzle me. First of all, why was &lt;strong&gt;G.H.Hardy&lt;/strong&gt; so much keen to have an elementary proof of the &lt;strong&gt;Prime Number Theorem&lt;/strong&gt;. He also stated that producing such a proof, will change the complexion of Mathematics, but nothing like that has happened. What was on Hardy's mind? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elementary proof although has some intricate tricks involved, i am curious to know whether the methodology can be applied for attacking more complex problems. I have seen that the analytic proof has a continuation and is not over and discuss some more interesting properties regarding the $\zeta$ function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also saw this thread &lt;em&gt;http://mathoverflow.net/questions/16735/is-a-non-analytic-proof-of-dirichlets-theorem-on-primes-known-or-possible&lt;/em&gt;. Are there any such theorem's which piques peoples curiosity in getting an elementary proof.(While writing this FLT comes to my mind!!).&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-15T19:25:24.163" LastActivityDate="2010-08-16T06:35:38.190" Title="Elementary proof of the Prime Number Theorem - Need?" Tags="&lt;soft-question&gt;&lt;intuition&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="2531" PostTypeId="2" ParentId="2355" CreationDate="2010-08-15T19:34:06.407" Score="4" ViewCount="0" Body="&lt;p&gt;Let p,p' = 2 mod 3 be prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that G is a group with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)  The 3-Sylow subgroup of G is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The number of elements of order 3 in G is 2*p*p'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: There exists a group which _in_addition_ satisfies either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii)(a); G is simple; OR (iii)(b): 3 exactly divides |G|, and G surjects onto Z/3Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Suppose that G is a group with cyclic 3-Sylow subgroups. Since all&#xA;3-Sylow subgroups are conjugate, this implies that every subgroup of&#xA;order 3 in G is also conjugate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: We may assume that G is not simple, and hence admits a proper normal&#xA;subgroup H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that H is a normal subgroup of G. If 3 divides |H|, then&#xA;H contains a subgroup of G of order 3. All such subgroups are&#xA;conjugate in G, and since H is normal, they all lie in H. Thus&#xA;G and H have the same number of elements of order 3.&#xA;Moreover, the 3-Sylow of H is clearly cyclic, and so we may&#xA;replace G by H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that (|H|,3) = 1. Then G/H still has a cyclic 3-Sylow subgroup,&#xA;and hence every element of order 3 in G/H is conjugate. This implies that&#xA;every element of order 3 in G/H lifts to an element of order 3 in G. Thus&#xA;G/H has at most&#xA;2*p*p' elements of order three. Yet the number of elements&#xA;of order 3 is (by some theorem of Frobenius?)&#xA;divisible by 2 and congruent to -1 mod 3. Thus the&#xA;number of element of G/H of order 3 is either 2 or 2*p*p'. In&#xA;the latter case, we replace G by G/H. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may now assume that G/H has exactly 2 elements of order 3.&#xA;Clearly G/H has a unique subgroup of order 3, which must be normal.&#xA;Thus G/H and G have a quotient of order |G/H|/3, and thus G contains&#xA;a normal subgroup F of order 3|H|. As above, we may replace G by F.&#xA;Note, however, that 3 exactly divides |F|, and &#xA;F surjects onto Z/3Z, and thus, by Schur-Zassenhaus&#xA;(overkill in this case, of course), F is a semi-direct product.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My understanding of Jack's argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that p' = 2, so G has 4p elements of order 3.&#xA;We still assume that the 3-Sylow of G is cyclic.&#xA;G acts by conjugation on the 2p subgroups of order p, giving&#xA;a map from G--&gt;S_2p. Let Q be one of these subgroups,&#xA;and let M be the normalizer of Q. Let P be a 3-Sylow containing Q.&#xA;Certainly [G:M] = 2p, by the orbit-stabilizer formula.&#xA; If X is a subgroup of G containing M,&#xA;then [G:X] = 1,2,p or 2p. Let N be the normalizer of P. Clearly&#xA;M contains N. Thus P is contained in M, and thus P is a 3-Sylow of M.&#xA;Similarly, P is also a 3-Sylow subgroup of X. The Sylow theorems applied&#xA;to M and X thus imply that [X:N] and [M:N] are both 1 mod 3, and&#xA;thus [X:M] is 1 mod 3. Hence, since [X:M] = 1,2,p or 2p, and, since&#xA;p = 2 mod 3, either X = M or X = G. Thus M is a maximal subgroup&#xA;of G, and hence the action of G is primitive. Now we suppose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption (*): The only primitive subgroups of S_2p are A_2p and S_2p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we deduce that some quotient of G is either A_2p or S_2p.&#xA;If G is simple, we deduce that G = A_2p, which doesn't have cyclic&#xA;3-Sylows if 2p &gt; 5, and if G is a semi-direct product of Z/3Z with&#xA;a  group of order co-prime to 3, then G cannot surject onto A_2p or&#xA;S_2p if p &gt; 2. It seems to follow that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p = 5,8 mod 9, and Assumption (*) holds, then G cannot have 4p elements&#xA;of order 3. (The congruence conditions on p ensure that the 3-Sylow of G is&#xA;cyclic).&lt;/p&gt;&#xA;" OwnerUserId="1204" LastEditorUserId="1204" LastEditorDisplayName="" LastEditDate="2010-08-17T04:17:46.750" LastActivityDate="2010-08-17T04:17:46.750" CommunityOwnedDate="2010-08-15T23:46:45.433" CommentCount="10" />
  <row Id="2532" PostTypeId="1" CreationDate="2010-08-15T19:42:56.243" Score="-3" ViewCount="74" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/2508/solving-a-quadratic-inequality&quot;&gt;Solving a quadratic inequality&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Suppose we have (x-5)(x+2) &gt; 0 &lt;br&gt;&#xA;can we solve it like this &lt;br&gt;&#xA;(x-5) &gt; 0 and (x+2) &gt;0 &lt;br&gt;&#xA;x &gt; 5     and  x &gt; -2&lt;br&gt;&#xA;please help.&lt;/p&gt;&#xA;" OwnerUserId="1050" LastActivityDate="2010-08-15T19:59:45.830" ClosedDate="2010-08-15T20:09:19.847" Title="solving inequalities" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2534" PostTypeId="2" ParentId="2532" CreationDate="2010-08-15T19:45:41.293" Score="1" ViewCount="0" Body="&lt;p&gt;If $ab&gt;0$ does it always imply that $a&gt;0$ or $b&gt;0$. You can also have both $a$ and $b$ to be less than $0$. So in your question it can also be $(x-5)&amp;lt;0$ along with $(x+2)&amp;lt;0$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-15T19:59:45.830" LastActivityDate="2010-08-15T19:59:45.830" CommentCount="5" />
  <row Id="2535" PostTypeId="2" ParentId="1829" CreationDate="2010-08-15T20:51:41.570" Score="1" ViewCount="0" Body="&lt;p&gt;On previous web sites, I've seen many people asking for more examples on Singular Value Decomposition (SVD).  Strang's LA book does this, but some folk want more.  Another thing that some undergraduates do, but then forget, but then need to do again in other courses is QR factorization (Gram-Schmidt); so, you should have multiple examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are 2 things that are pretty straight-forward for most, but examples (after the theory) would be good.&lt;/p&gt;&#xA;" OwnerUserId="1144" LastActivityDate="2010-08-15T20:51:41.570" CommunityOwnedDate="2010-08-15T20:51:41.570" />
  <row Id="2536" PostTypeId="1" AcceptedAnswerId="2541" CreationDate="2010-08-15T20:54:45.790" Score="4" ViewCount="377" Body="&lt;p&gt;Please see the problem &lt;strong&gt;5&lt;/strong&gt; of the given link: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=59&amp;amp;year=2005&amp;amp;sid=722231ab4ec5ce280584eb8f24f07656&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=59&amp;amp;year=2005&amp;amp;sid=722231ab4ec5ce280584eb8f24f07656&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It asks us to prove that $$\sum\limits_{n=1}^{\infty} \frac{1}{n^n} = \int\limits_{0}^{1} x^{x} \ dx $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Natural question which one gets by seeing this type of question is, find all continuous functions which satisfy $$\int\limits_{-\infty}^{\infty} f(x) \ dx = \sum\limits_{n= -\infty}^{\infty} f(n)$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T07:38:54.367" LastActivityDate="2010-08-16T08:13:16.903" Title="Finding $f$ such that $ \int f = \sum f$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="2537" PostTypeId="1" AcceptedAnswerId="2545" CreationDate="2010-08-15T21:06:02.550" Score="5" ViewCount="108" Body="&lt;p&gt;In a &lt;a href=&quot;http://pauli.uni-muenster.de/~munsteg/arnold.html&quot; rel=&quot;nofollow&quot;&gt;talk by V.I. Arnold&lt;/a&gt;, this is said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I was a first-year student at the Faculty of Mechanics and Mathematics of the Moscow State University, the lectures on calculus were read by the set-theoretic topologist L.A. Tumarkin, who conscientiously retold the old classical calculus course of French type in the Goursat version. He told us that integrals of rational functions along an algebraic curve can be taken if the corresponding Riemann surface is a sphere and, generally speaking, cannot be taken if its genus is higher, and that for the sphericity it is enough to have a sufficiently large number of double points on the curve of a given degree (which forces the curve to be unicursal: it is possible to draw its real points on the projective plane with one stroke of a pen).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to understand the mathematical part of this. What do I need to know to see why this makes sense? Where can I get enough of the background to understand it fully?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T02:39:15.080" Title="Toward &quot;integrals of rational functions along an algebraic curve&quot;" Tags="&lt;calculus&gt;&lt;reference-request&gt;&lt;riemann-surface&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="2538" PostTypeId="2" ParentId="2324" CreationDate="2010-08-15T21:06:29.757" Score="1" ViewCount="0" Body="&lt;p&gt;You might get &lt;em&gt;some&lt;/em&gt; results with the method you mentioned with 100 inputs to the ANN. It might be good enough for your application, but if it's not, you'll need to apply some pre-processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of pre-processing will do the trick is largely dependent on the nature of your signal and what you're trying to get. Try thinking about what should not be important (what characteristics of the signal) to end result and try to filter that out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if DC component is not important, remove it from all signals.&#xA;If amplitude is not important, scale everything to [0,1] independently.&#xA;If you have some high frequency noise, filter the signal with low-pass.&#xA;More generally, think what characteristics of the spectra of the signal might be important.&#xA;One of the tricks is to look at auto-correlation of the signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With or without pre-processing, coarser quantization or coding with Gray's code might additionally help.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-08-15T21:06:29.757" />
  <row Id="2539" PostTypeId="2" ParentId="2527" CreationDate="2010-08-15T21:19:00.933" Score="2" ViewCount="0" Body="&lt;p&gt;You could try using Wikipedia with &lt;a href=&quot;http://www.google.com/sidewiki/intl/en/index.html&quot; rel=&quot;nofollow&quot;&gt;Google SideWiki&lt;/a&gt; or any other software that lets you annotate a site.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T08:59:16.920" LastActivityDate="2010-08-16T08:59:16.920" CommentCount="1" />
  <row Id="2541" PostTypeId="2" ParentId="2536" CreationDate="2010-08-15T21:49:52.980" Score="2" ViewCount="0" Body="&lt;p&gt;Pick any sequence $f_k, -\infty &amp;lt; k &amp;lt; \infty$ such that $\sum_{n=-\infty}^{\infty} f_n$ exists, and any other two sequences $g_k, h_k, -\infty &amp;lt; k &amp;lt; \infty$ such that $\sum_{n=-\infty}^{\infty} g_n = 0$ and $h_k\rightarrow 0, k\rightarrow \pm \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $f(x) = f_k, x = k$ and construct $f(x)$ on non-integer points however you want to satisfy $f(m)=f_m$, $f(m+1)=f_{m+1}$ and $\int_{m}^{m+1} f(x) dx = f_m+g_m$ and $|\int_{m}^{y} f(x) - (y-m)(f_m-g_m) dx| &amp;lt; h_m, y\in (m,m+1)$ for all $m \in \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just as hard as finding all continuous functions on $g$ on $[0,1]$ whose integral is $1$ and $g(0)=0$, $g(1)=1$ and $|\int_0^y g(x)-y| dx &amp;lt; \varepsilon, y\in (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: When we pick $f_k$, we have fixed values for $f(x)$ on all integers, and also RHS of equality. To satisfy LHS = RHS, we have freedom to construct $f(x)$ however we want on non-integer points between fixed values on integer points since $f(x)$ should be continious.&lt;br&gt;&#xA;When we consider partial sums $\sum_{n=0}^{m} f_n$ and $\int_0^m f(x)dx$, these two can differ but they should have the same limits. This is controlled by sequence $g_n$.&lt;br&gt;&#xA;The rest is just technicality to ensure that $\int_{-\infty}^{\infty}f(x)dx$ is indeed defined. Since we choose practically values for $\int_0^m f(x)dx$ we need to ensure that $\int_0^y f(x)dx$ do not 'wiggle' too much between integer points when $m\rightarrow \infty$. This is what we use $h_n$ for.&lt;br&gt;&#xA;All given conditions should be sufficient and necessary.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="956" LastEditorDisplayName="" LastEditDate="2010-08-16T08:13:16.903" LastActivityDate="2010-08-16T08:13:16.903" CommentCount="0" />
  <row Id="2542" PostTypeId="1" AcceptedAnswerId="2564" CreationDate="2010-08-15T22:17:01.810" Score="10" ViewCount="276" Body="&lt;p&gt;Friend told me this one, I'm completely stuck but also completely fascinated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 100 boxes with apples, oranges and bananas (mixed). &lt;strike&gt;How to&lt;/strike&gt; Prove that you can pick 51 boxes and to get at least half of all apples, at least half of all oranges and at least half of all bananas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: You can take a look in the boxes.&lt;/p&gt;&#xA;" OwnerUserId="956" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T22:44:44.250" LastActivityDate="2010-08-27T22:44:44.250" Title="100 boxes of fruits - pick 51 and get at least half of each type?" Tags="&lt;combinatorics&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="3" />
  <row Id="2543" PostTypeId="1" CreationDate="2010-08-15T22:18:19.423" Score="2" ViewCount="277" Body="&lt;p&gt;I have a symmetric matrix A. How do I compute a matrix B such that B^tB=A where B^t is the transpose of B. I cannot figure out if this is at all related to the square root of A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gone through wikimedia links of square root of a matrix.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-08-16T02:20:27.840" Title="square root of symmetric matrix and transposition" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2544" PostTypeId="2" ParentId="2543" CreationDate="2010-08-15T22:22:33.383" Score="1" ViewCount="0" Body="&lt;p&gt;What you apparently want here is the Cholesky decomposition, which factors a matrix A into $BB^T$ where $B$ is a triangular matrix. However, this only works if your matrix is positive definite.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-15T22:22:33.383" CommentCount="11" />
  <row Id="2545" PostTypeId="2" ParentId="2537" CreationDate="2010-08-15T22:42:31.813" Score="7" ViewCount="0" Body="&lt;p&gt;What Arnold is describing (in a slightly oblique manner) is the theory of algebraic curves.&#xA;The idea is that if one wants to integrate some rational expression $R(x,y)dx + S(x,y) dy$&#xA;over the curve $f(x,y) = 0$, then the question of whether one can find an antiderivative&#xA;in terms of elementary functions has a positive or negative answer depending on whether&#xA;the geometric genus of the curve $f(x,y) = 0$ is zero or positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One direction is not so hard to see directly: if $f(x,y) = 0$ has geometric genus zero,&#xA;this means that we can trace out this curve in terms of a single parameter, i.e.&#xA;we can find parametric expressions $x = x(t)$ and $y = y(t)$ so that $f(x(t),y(t)) = 0$.&#xA;Then if we rewrite the integral in terms of the variable $t$, basic integral calculus (the substitution rule) lets us rewrite the integrand as a rational function of $t$, and we can&#xA;always integrate a rational function in terms of elementary functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is less obvious is that if $f(x,y) = 0$ has positive geometric genus, then it is &lt;em&gt;not&lt;/em&gt; possible to find such a parameterization of the curve (this is a non-trivial statement),&#xA;and it is &lt;em&gt;not&lt;/em&gt; possible to find an elementary antiderivative (this is related to the previous statement, but is another non-trivial deduction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first example is the curve $y^2 = (1-x^2)(1-kx^2)$ (here $k$ is some constant, neither 0 nor 1),&#xA;with the integral being $\int dx/y = \int dx/\sqrt{(1-x^2)(1 - kx^2)}$.  This is what is called an elliptic integral, and (for more or less 150 years, beginning with the invention&#xA;of calculus) people tried to find an elementary expression for it, until finally Abel and Jacobi showed that this wasn't possible, because this curve has geometric genus one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know any algebraic geometry, then a good place to start is Miles Reid's &quot;Undergaduate algebraic geometry&quot;.  The theorem you need is the one which says that&#xA;there is no rational map from a genus zero curve to a positive genus zero curve,&#xA;which I'm pretty sure is proved in that book, at least for genus one curves.&#xA;(It is not so difficult to pass from this theorem in the case of smooth curves to the &#xA;case of singular curves, but you will have more difficulty finding a treatment of the&#xA;singular curve case, which is what Arnold is talking about when he mentions double points.)&#xA;Depending on the level at which you're beginning, you might want to consult one of the algebraic geometry road-map questions on Mathoverflow; there is one asking for an undergraduate road-map for learning algebraic geometry, and a second asking for a graduate road-map.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do already know some algebraic geometry, then what you want is a historical source&#xA;that relates the geometry you know to its historical origins.  Dieudonne wrote a history of algebraic geometry, which must surely discuss this.  There will be many other historical&#xA;sources too.  For the best guide to the historical literature, you might want to ask on Mathoverflow, where there will probably be more people reading who are familiar with historical treatments of the theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should say that if you are beginning from a position of knowing no algebraic geometry, then it will take some time and effort to learn what is needed to fully understand what Arnold is discussing, especially from a standard textbook (which will likely not proceed in a straight line to where you want to go, but rather develop a more general theory, which will then be specialized to the situation you are interested in).&lt;br&gt;&#xA;So even if that is your situation, I recommend that you also do some historical reading,&#xA;to help get a better feeling for exactly what parts of an algebraic geometry text-book&#xA;you will need to read to satisfy your interest in Arnold's statement.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-16T02:39:15.080" LastActivityDate="2010-08-16T02:39:15.080" CommentCount="2" />
  <row Id="2546" PostTypeId="2" ParentId="2529" CreationDate="2010-08-15T23:34:23.133" Score="2" ViewCount="0" Body="&lt;p&gt;This called Dicman's lemma and is 14.5.7 on page 425 (1st ed) or &lt;a href=&quot;http://books.google.com/books?id=lqyCjUFY6WAC&amp;amp;lpg=PA442&amp;amp;pg=PA442&quot; rel=&quot;nofollow&quot;&gt;page 442 (2nd ed)&lt;/a&gt; of Robinson's Course in the Theory of Groups.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-15T23:34:23.133" />
  <row Id="2547" PostTypeId="2" ParentId="2492" CreationDate="2010-08-15T23:37:51.053" Score="1" ViewCount="0" Body="&lt;p&gt;So it turns out the first assumption was actually correct:  U is indeed the first k eigenvectors of C, that we calculate from G by means of the eigendecompposition $(X_tVD^{-\frac12})$.  Thanks to J. Mangaldan for his comments and especially the pointers to Sherman-Morrison-Woodbury. &lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-08-15T23:37:51.053" />
  <row Id="2548" PostTypeId="2" ParentId="2355" CreationDate="2010-08-16T00:15:12.397" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Here is my attempt, fixed by Septimus Harding:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If H is a group with 92 elements of order 3, then let G be the subgroup generated by those elements.  G has finite order by Dicman's lemma, and so G has a Sylow 3-subgroup P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P must be cyclic, lest Hall's result imply 46≡4 mod 9.  Since P is cyclic, 46≡1 mod |P|, so |P| divides 45, so |P| is 1, 3, or 9.  1 is impossible since there would be 0 elements of order 3.  3 is impossible, since then G has 46 Sylow 3-subgroups but no finite group has 46 Sylow p-subgroups.  So P is cyclic of order 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now G acts by conjugation on the 46 subgroups of order 3, and since the Sylow 3-subgroups are cyclic and (all of) their order 3-subgroups are conjugate, G acts transitively on the 46 subgroups.  Let Q = Ω(P) ≤ P be one of the subgroups of order 3, and let M = N&lt;sub&gt;G&lt;/sub&gt;(Q) be its normalizer.  Then N = N&lt;sub&gt;G&lt;/sub&gt;(P) ≤ M and [G:M] = 46.  If M ≤ X ≤ G, then both [M:N] and [X:N] are ≡1 mod 3 by Sylow, but [X:N] = [X:M][M:N], so [X:M] ≡ 1 mod 3.  However [X:M] divides 46, so [X:M] = 1 or 46, so X = M or G, and so M is a maximal subgroup of G.  So G acts primitively on 46 points (the cosets of N, aka, the subgroups of order 3), so G/Core(G,M) is a primitive group of degree 46, that is, G/Core(G,M) is the alternating or the symmetric group on 46 points.  Since the Sylow 3-subgroups of the alternating and symmetric groups (on 6 or more points) are not cyclic, we have a contradiction, and no such G exists.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This works to prove that: if n=pq for primes p,q ≡ {5,8} mod 9, and if the only primitive groups of degree pq have non-cyclic Sylow 3-subgroups (like Alt(n) and Sym(n)), then there is no group with exactly 2n elements of order 3, even though 2n ≡ 2 mod 6.  I believe primitive groups of degree pq are classified, so this could probably be made completely explicit.  140 elements is not immediately handled, since 70 = 2*5*7, but I suspect the same idea will work.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-08-16T05:43:21.307" LastActivityDate="2010-08-16T05:43:21.307" CommunityOwnedDate="2010-08-16T00:15:12.397" />
  <row Id="2549" PostTypeId="1" AcceptedAnswerId="2557" CreationDate="2010-08-16T00:39:05.017" Score="4" ViewCount="173" Body="&lt;p&gt;I recently learned that on Riemannian manifolds, one can define the &lt;em&gt;sectional curvature&lt;/em&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sectional_curvature&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sectional_curvature&lt;/a&gt;) of a (2-dimensional) plane section.  I was wondering if a similar concept exists for higher dimensional &quot;space sections.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what got me thinking about this: For 2-dimensional manifolds (surfaces), the sectional curvature is equal to $\kappa_1\kappa_2$, where $\kappa_1$ and $\kappa_2$ are the principal curvatures.  Is there a name for the quantity $\kappa_1\kappa_2\kappa_3$ for 3-manifolds, etc., and does it carry similar geometric significance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: Typesetting fixed)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-08-30T04:50:54.123" LastActivityDate="2010-08-30T04:50:54.123" Title="Are there higher-dimensional analogues of sectional curvature?" Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="2550" PostTypeId="1" AcceptedAnswerId="3152" CreationDate="2010-08-16T01:10:52.327" Score="6" ViewCount="316" Body="&lt;p&gt;Meta: Since there seems be a dearth of questions here (I am used to stackoverflow!), I think we should all try and contribute some questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part a) of the following problem appeared in one of the Putnam Exams (sorry, don't know which year exactly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to solve Part a) don't read Part b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a painting device, which given the co-ordinates of a points in the 2D plane, will colour all points on that plane black, which are at an irrational distance from the given point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially you start out with the 2D plane being white. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) You want to colour the whole plane black. What is the minimum number of points you need to feed to the painting device?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;b) Show that it is sufficient to feed $(0,0), (1,0), (\sqrt{2},0)$. &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-16T01:16:01.427" LastActivityDate="2010-08-23T20:00:21.917" Title="Irrational painting device" Tags="&lt;puzzle&gt;&lt;putnam&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="2551" PostTypeId="2" ParentId="2543" CreationDate="2010-08-16T02:20:27.840" Score="4" ViewCount="0" Body="&lt;p&gt;As J. Mangaldan says, you need your matrix $A$ to be positive definite. Since $A$, being symmetric, is always diagonalizable, this is the same as saying that it has non-negative eigenvalues. If this is the case, you can adapt alex's comment almost literally for the real case: as we've said, $A$ is diagonalizable, but, also, there exists an orthonormal base of eigenvectors of $A$. That is, there is an invertible matrix $S$ and a diagonal matrix $D$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D = SAS^t , \quad \text{with} \quad SS^t = I \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D = \mathrm{diag} (\lambda_1, \lambda_2, \dots , \lambda_n) \ ,&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a diagonal matrix and has only non-negative eigenvalues $\lambda_i$, you can take its square root&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{D} = \mathrm{diag} (\sqrt{\lambda_1}, \sqrt{\lambda_2}, \dots , \sqrt{\lambda_n} ) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then, on one hand, you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( S^t \sqrt{D} S   \right)^2 = \left( S^t \sqrt{D} S\right) \left(S^t \sqrt{D} S \right) = S^t \left( \sqrt{D}\right)^2 S = S^t D S = A \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $S^t \sqrt{D} S$ is a symmetric matrix too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( S^t \sqrt{D} S   \right)^t = S^t (\sqrt{D})^t S^{tt} = S^t \sqrt{D^t} S = S^t \sqrt{D} S  \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so you have your $B = S^t \sqrt{D} S$ such that $B^t B = A$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-16T02:20:27.840" />
  <row Id="2552" PostTypeId="1" CreationDate="2010-08-16T02:43:17.580" Score="1" ViewCount="141" Body="&lt;p&gt;i'm reading &lt;em&gt;3D math primer for graphics and game development&lt;/em&gt; By Fletcher Dunn and Ian Parberry, on page &lt;a href=&quot;http://books.google.com.hk/books?id=wCfWkc_E3GkC&amp;amp;lpg=PP1&amp;amp;ots=zIZTXKhbGJ&amp;amp;dq=3D%20Math%20Primer%20for%20Graphics%20and%20Game%20Development&amp;amp;hl=en&amp;amp;pg=PA170#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;170&lt;/a&gt;, the logarithm of quaternion is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\log \mathbf q &amp;amp;= \log \left( \begin{bmatrix} \cos \alpha &amp;amp; \mathbf n \sin \alpha \end{bmatrix} \right) \\&#xA;&amp;amp;\equiv \begin{bmatrix} 0 &amp;amp; \alpha \mathbf n \end{bmatrix}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i don't see how does $\log \left( \begin{bmatrix} \cos \alpha &amp;amp; \mathbf n \sin \alpha \end{bmatrix} \right)$ equals to $\begin{bmatrix} 0 &amp;amp; \alpha \mathbf n \end{bmatrix}$, can anyone help me out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1209" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T08:00:43.533" LastActivityDate="2010-08-16T08:34:16.463" Title="the logarithm of quaternion " Tags="&lt;quaternions&gt;" AnswerCount="2" />
  <row Id="2553" PostTypeId="2" ParentId="2112" CreationDate="2010-08-16T03:58:23.020" Score="2" ViewCount="0" Body="&lt;p&gt;If I understand you correctly, you aren't bringing up the issues raised by Gödel. You are raising the point that since it's always possible a mistake has been made, we can't be sure of the validity of a mathematical proof. There is always the possibility that even our computational examination of a proof is forgetting some facet that renders it untrue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must understand, however, that this is not a flaw in mathematics but an inherent element of reality. Maths does remain the most secure branch of knowledge, except that the way we practically do math may not be as secure as previously imagined. A fair point. &lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-08-16T03:58:23.020" CommunityOwnedDate="2010-08-16T03:58:23.020" />
  <row Id="2554" PostTypeId="2" ParentId="2552" CreationDate="2010-08-16T04:17:46.630" Score="3" ViewCount="0" Body="&lt;p&gt;I can't see the page in Google books, but what you apparently have there is the logarithm of a &lt;strong&gt;unit&lt;/strong&gt; quaternion q, which has scalar part $\cos(\theta)$ and vector part $\sin(\theta)\vec{n}$ where $\vec{n}$ is a unit vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the logarithm of an arbitrary quaternion $q=(s\;\;v)$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln q=(\ln\,|q|\;\;\frac{v}{\|v\|}\arccos\frac{s}{|q|})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|q|$ is the norm of the quaternion and $\|v\|$ is the norm of the vector part, applying that formula to a unit quaternion yields a scalar part of 0 (the logarithm of the norm of a unit quaternion is zero), and you should now be able to derive the formula for the vector part.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-16T04:17:46.630" />
  <row Id="2555" PostTypeId="2" ParentId="2536" CreationDate="2010-08-16T04:18:23.793" Score="0" ViewCount="0" Body="&lt;p&gt;I found a similar question, answered by Ramanujan. The paper is entitled &quot;Some Integrals in Ramaujan's lost notebook&quot; by Bruce C Brendt. Here is the link: &lt;a href=&quot;http://www.jstor.org/stable/4097261?cookieSet=1&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/4097261?cookieSet=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-16T04:18:23.793" />
  <row Id="2556" PostTypeId="2" ParentId="2530" CreationDate="2010-08-16T04:25:23.533" Score="5" ViewCount="0" Body="&lt;p&gt;As KCd explains in a comment, the proof of the PNT in Hardy's time seemed to be intimately connected to the complex analytic theory of the $\zeta$-function.  In fact, it was known to&#xA;be equivalent to the statement that $\zeta(s)$ has no zeroes on the half-plane $\Re s \geq  1$.&#xA;Although this equivalence may seem strange to someone unfamiliar with the subject, it is in&#xA;fact more or less straightforward.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the equivalence of PNT and the zero-freeness of $\zeta(s)$ in the region&#xA;$\Re s \geq 1$ does not lie as deep as the fact that these results are actually true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibility that one could then prove PNT in a direct way, avoiding complex analysis, seemed unlikely to Hardy, since then one would also be giving a proof, avoiding complex&#xA;analysis, of the fact that the complex analytic function $\zeta(s)$ has no zero in the region $\Re s \geq 1$, which would be a peculiar state of affairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What added to&#xA;the air of mystery surrounding the idea of an elementary proof was the possibility of accessing the Riemann hypothesis this way.  After all, if one could prove in an elementary way that&#xA;$\zeta(s)$ was zero free when $\Re s \geq 1$, perhaps the insights gained that way&#xA;might lead to a proof that $\zeta(s)$ is zero free in the region $\Re s &gt; 1/2$ (i.e.&#xA;the Riemann hypothesis), a statement which had resisted (and continues to resist)&#xA;attack from the complex analytic perspective,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, when the elementary proof of PNT was finally found, it didn't have the ramifications that Hardy anticipated (as KCd pointed out in his comment).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a modern perspective on the elementary proof, and a comparison with the complex analytic proof, I strongly recommend Terry Tao's &lt;a href=&quot;http://www.math.ucla.edu/~tao/preprints/Expository/prime.dvi&quot; rel=&quot;nofollow&quot;&gt;exposition of the prime number theorem&lt;/a&gt;.  In this exposition,&#xA;Tao is not really concerned with elementary vs. complex analytic techniques as such, but rather with&#xA;explaining what the mathematical content of the two arguments is, in a way which makes&#xA;it easy to compare them.  Studying Tao's article should help you develop a deeper sense of the mathematical content of the two arguments, and their similarities and differences.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Tao's note explains, both arguments involve a kind of &quot;harmonic analysis&quot; of the primes,&#xA;but the elementary proof works primarily in &quot;physical space&quot; (i.e. one works directly with the prime counting function and its relatives), while the complex analytic proof works much more in &quot;Fourier space&quot; (i.e. one works much more with the Fourier transforms of the prime counting function and its relatives).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding (derived in part from Tao's note) is that Bombieri's sieve is (at least in part) an outgrowth of the elementary proof, and it is in sieve methods that one can look to find modern versions of the type of arguments that appear in the elementary proof.  (As one example, see Kevin Ford's paper &lt;a href=&quot;http://www.ams.org/journals/tran/2005-357-04/S0002-9947-04-03579-2/home.html&quot; rel=&quot;nofollow&quot;&gt;On Bombieri's asymptotic sieve&lt;/a&gt;, which in its first two pages includes a&#xA;discussion of the relationship between certain sieving problems and the elementary proof.)&#xA; But I should note that modern analytic number theorists don't pursue sieve methods out of some desire for having &quot;elementary&quot; proofs.  Rather, some results can be proved by $\zeta$- or $L$-function methods, and others by sieving methods; each has their strengths and weaknesses.  They can be combined, or played off, one against the other. (The &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bombieri%E2%80%93Vinogradov_theorem&quot; rel=&quot;nofollow&quot;&gt;Bombieri--Vinogradov theorem&lt;/a&gt; is an example&#xA;of a result proved by sieve methods which, as far as I understand, is stronger than what&#xA;could be proved by current $L$-function methods; indeed, it an averaged form of the&#xA;Generalized Riemann Hypothesis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this mixing of methods is possible, I again recommend Tao's note.  Looking at it should give you a sense of how, in modern number theory, the methods of the two proofs of PNT (elementary and complex analytic) are not living in different, unrelated worlds, but are just two different, but related, methods for approaching the &quot;harmonic analysis of the primes&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-16T06:35:38.190" LastActivityDate="2010-08-16T06:35:38.190" CommentCount="3" />
  <row Id="2557" PostTypeId="2" ParentId="2549" CreationDate="2010-08-16T06:31:47.137" Score="2" ViewCount="0" Body="&lt;p&gt;Sure, there's all kinds of generalizations.  One way to think of sectional curvature is you take a 2-dimensional subspace of the tangent space, exponentiate a small neighbourhood of the origin, and take the Gauss curvature at the point of tangency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a $k$-dimensional subspace of the tangent space, exponentiate a small neighbourhood of $0$, and take the scalar curvature at the point of tangency.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another analogy you could build on is to use the comparison definition of sectional curvature -- measuring the infinitesimal deviation of the length of a circle (exponentiated from a 2-dimensional tangent space) of a given radius from that of a Euclidean circle.  You could do the same for content of spheres of arbitrary dimension in subspaces of the tangent space, etc. I imagine these give very related notions of curvature although I've never worked out the details. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-08-16T06:31:47.137" CommentCount="2" />
  <row Id="2558" PostTypeId="1" AcceptedAnswerId="2560" CreationDate="2010-08-16T06:32:03.643" Score="7" ViewCount="934" Body="&lt;p&gt;You're probably thinking, &quot;Why?&quot;  Please let me explain...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is (&lt;em&gt;very&lt;/em&gt;) well-known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall (a,b,c,x) \in \mathbb{C}^* \times \mathbb{C}^3: ax^2 + bx + c = 0 \Leftrightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some bizarre reason, I decided to try to solve $ bx + c = 0 $ using this formula by introducing a term $ \alpha x^2 $ and removing it in the limit $ \alpha \to 0 $.  Doing so with L'Hopital's rule, I find these solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle x_1 = \lim_{\alpha \to 0} {\frac{-b + \sqrt{b^2 - 4c \alpha}}{2 \alpha}} = \lim_{\alpha \to 0} {\frac{-c}{\sqrt{b^2 - 4c \alpha}}} = \frac{-c}{b}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle x_2 = \lim_{\alpha \to 0} {\frac{-b - \sqrt{b^2 - 4c \alpha}}{2 \alpha}} = \infty. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first was to be expected, but I still haven't been able to explain the second cleanly (that is, in a way other than &quot;since $ -c/b $ is gone, it couldn't be a true number&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, carrying out the analogous process one degree lower yields a root at either zero or infinity, depending on the constant.  The latter possibility (which occurs when $ c \neq 0 $) corresponds to the unsolvable case, while the former (in which $ c = 0 $) corresponds to the trivially satisfied one, so a root at zero here appears to have a vastly different meaning from $ x_1 = 0 $ above, where $ x_1 $ gives the location of the unique, genuine root of $ bx + 0 = 0 $ provided $ b \neq 0 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;why a solution at zero can have either of the two meanings just described, and&lt;/li&gt;&#xA;&lt;li&gt;whether the phantom root $ x_2 = \infty $ (obtained by treating the first-degree polynomial $ bx + c $ as a degenerate case of the second-degree one) has a meaningful interpretation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you all in advance, and sorry if my typesetting doesn't render nicely (this is my first experience).&lt;/p&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-08-16T07:46:52.787" Title="On applying the quadratic formula to a first-degree equation" Tags="&lt;polynomials&gt;&lt;roots&gt;&lt;infinity&gt;&lt;limit&gt;" AnswerCount="3" />
  <row Id="2559" PostTypeId="2" ParentId="2558" CreationDate="2010-08-16T06:38:51.787" Score="10" ViewCount="0" Body="&lt;p&gt;Just a note on your attempt to solve a degenerate quadratic: remember that the quadratic formula can be derived in two ways: solving ax²+bx+c for x, or solving a+b/x+c/x² for 1/x and then reciprocating the result. Viewing it in this manner, one equation's &quot;infinite root&quot; is the reversed equation's 0 root.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-16T06:38:51.787" CommentCount="2" />
  <row Id="2560" PostTypeId="2" ParentId="2558" CreationDate="2010-08-16T07:30:12.310" Score="10" ViewCount="0" Body="&lt;p&gt;I think the answer is to work projectively.  Rather than consider the solutions to $ax^2 + bx + c = 0$ in $\mathbb{C}$ one should think of the solutions to $aX^2 + bXY + cY^2 = 0$ in $\mathbb{P}^1(\mathbb{C})$.  Then the $a = 0$ case is easy to explain; the corresponding equation $bXY + cY^2 = 0$ has one root $(c : -b)$ which is expected and another $(1 : 0)$ which is the point at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems reasonable to me because the degeneration at $a = 0$ is something like a failure of Bezout's theorem, which is repaired precisely by working projectively.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-16T07:30:12.310" CommentCount="7" />
  <row Id="2561" PostTypeId="1" AcceptedAnswerId="2590" CreationDate="2010-08-16T07:40:51.880" Score="2" ViewCount="197" Body="&lt;p&gt;Good example of a polynomial which produces a finite number of primes is: $$x^{2}+x+41$$ which produces primes for every integer $ 0 \leq x \leq 39$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a paper H.Stark proves the following result: $X_{n}$(is the ring of &quot;algebraic integers&quot; in $Q(\sqrt{-n})$ is a prinicpal ideal domain for positive $n$ if and only if &#xA;$n = 1,2,3,7,11,19,43,67,163. $  For a reference one can see: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Harold Stark, A complete determination of the complex quadratic fields of class-number one, Michigan Math J., 14 (1967) 1-27&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider in general the polynomial $x^{2}+x + K= (x+ \alpha)(x+ \bar{\alpha})$  which we can factorize where $\alpha$ is given by $$ \alpha = \frac{{1} + \sqrt{1-4K}}{2}, \quad \bar{\alpha} = \frac{1 - \sqrt{1-4K}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can get some relationships between polynomials which produce prime in the field $Q(\sqrt{-n})$. Question is if a polynomial produces a prime, then will $X_{n}$ as defined above be a PID?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-16T16:05:10.490" Title="Behaviour of Polynomials in a PID!" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="2563" PostTypeId="2" ParentId="2558" CreationDate="2010-08-16T07:46:52.787" Score="9" ViewCount="0" Body="&lt;p&gt;Informally, the &quot;quadratic&quot; polynomial with a=0 has a second zero at the compactification point at infinity.  Graphically (working in the reals):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gOMTn.gif&quot; alt=&quot;animated graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as a goes through zero, the &quot;quadratic&quot; goes through the linear special case, where the second zero goes through infinity, crossing between the positive and negative ends of the real axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this parallels the more technical explanation given by Qiaochu Yuan.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-16T07:46:52.787" CommentCount="3" />
  <row Id="2564" PostTypeId="2" ParentId="2542" CreationDate="2010-08-16T07:50:50.187" Score="6" ViewCount="0" Body="&lt;p&gt;This is apparently a (hard) problem from the Russian Math Olympiad which no one in the exam solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for a list of questions in that exam: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&amp;amp;t=32171&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&amp;amp;t=32171&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution for this problem is here: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1367869#p1367869&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1367869#p1367869&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint that was given (by Fedor Petrov):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we have $2k$ boxes, we may&#xA;  partition them into two groups of $k$&#xA;  boxes in such a way that number of&#xA;  apples in both groups differ by at&#xA;  most the maximal number of apples in a&#xA;  single box, and the same for oranges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-16T08:02:11.877" LastActivityDate="2010-08-16T08:02:11.877" CommentCount="2" />
  <row Id="2565" PostTypeId="2" ParentId="2552" CreationDate="2010-08-16T08:34:16.463" Score="1" ViewCount="0" Body="&lt;p&gt;Just recall $\exp(\alpha i) = \cos \alpha + i \sin \alpha$ for complex numbers, the quaternion (remember a quaternion is just 3 complex numbers which all have the same real part) version is by direct analogue and take logarithm of both sides.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-16T08:34:16.463" />
  <row Id="2567" PostTypeId="1" CreationDate="2010-08-16T09:36:59.887" Score="1" ViewCount="201" Body="&lt;p&gt;Hi, I would like to know if the Lebesgue spaces $L_p$ with $ 0 &amp;lt; p &amp;lt; 1 $ are separable or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is true for $1 \leq p &amp;lt; + \infty$, but I do not find any references for the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;case  $ 0 &amp;lt; p &amp;lt; 1 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;" OwnerUserId="1222" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T11:07:59.933" LastActivityDate="2010-08-31T04:42:53.600" Title="Separability of  $ L_p $  spaces" Tags="&lt;topology&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="2568" PostTypeId="2" ParentId="2550" CreationDate="2010-08-16T09:53:01.387" Score="0" ViewCount="0" Body="&lt;p&gt;I decided to give this problem a go, but I couldn't get a complete solution. Lets look at the distances squared:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$x_1^2+y_1^2$&lt;/li&gt;&#xA;&lt;li&gt;$(x_1-1)^2+y_1^2$&lt;/li&gt;&#xA;&lt;li&gt;$(x_1 - \sqrt{2})^2+y_1^2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These must all be rational. The difference between any two rational numbers must be rational, so $-2x_1+1$ is rational and so is $x_1$. Subtracting 2. and 3., we get: $(-2 \sqrt(2)-2)x_1+3$, implying that $x_1$ must be 0 if it is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_1^2, y_1^2+1, y_1^2+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $y_1^2=a^2/b^2$ where a and b are integers. Clearly, $y_1^2+1, y_1^2+2$ are rational if and only if $a^2+b^2$ and $a^2+2*b^2$ are perfect squares. Using modular arithmetic it is pretty clear that b needs to be divisible by 2 (squares are only 0 or 1 mod 4), implying $b^2$ is divisible by 4.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-16T21:33:44.997" LastActivityDate="2010-08-16T21:33:44.997" CommentCount="4" />
  <row Id="2569" PostTypeId="2" ParentId="2550" CreationDate="2010-08-16T10:19:06.770" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose that there exists a point A that is left uncolored after feeding in the three points (0,0), (1,0), and $(\sqrt{2},0)$.  The distances from point A to each of these three points (call these b, c, and d, respectively) must be rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ryXfw.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Stewart%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Stewart's Theorem&lt;/a&gt;: $$b^2(\sqrt{2}-1)+c^2\cdot 1=\sqrt{2}(d^2+(\sqrt{2}-1)\cdot 1).$$ So $$\sqrt{2}(b^2-d^2+1)=b^2-c^2+2$$ or $$\sqrt{2}=\frac{b^2-c^2+2}{b^2-d^2+1}.$$  Since b, c, and d are rational numbers, the right side is rational, but the left side is irrational, so there is no such point A.  Thus, to color the entire plane black, it is sufficient to feed in the three points (0,0), (1,0), and $(\sqrt{2},0)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-16T10:19:06.770" CommentCount="1" />
  <row Id="2570" PostTypeId="2" ParentId="2567" CreationDate="2010-08-16T11:13:27.227" Score="0" ViewCount="0" Body="&lt;p&gt;Please refer this article. It talks about $L_{p}$ spaces for $0 &amp;lt; p \leq 1$. Link: &#xA;&lt;a href=&quot;http://www.jstor.org/stable/2041603?seq=2&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2041603?seq=2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the step functions, the ones that take rational values  and whose steps have rational endpoints there should be only countably many of those. And then you can perhaps apply the same argument, you use to prove it for $L_{p}$ spaces for $1 &amp;lt; p &amp;lt; \infty$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-16T12:09:50.020" LastActivityDate="2010-08-16T12:09:50.020" CommentCount="2" />
  <row Id="2571" PostTypeId="1" AcceptedAnswerId="2572" CreationDate="2010-08-16T11:16:39.657" Score="3" ViewCount="264" Body="&lt;p&gt;Square of an irrational number can be a rational number e.g. $\sqrt{2}$ is irrational but its square is 2 which is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there a irrational number square root of which is a rational number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it safe to assume, in general, that $n^{th}$-root of irrational will always give irrational numbers?&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-03T19:33:56.897" LastActivityDate="2011-01-03T19:33:56.897" Title="About powers of irrational numbers" Tags="&lt;number-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2572" PostTypeId="2" ParentId="2571" CreationDate="2010-08-16T11:20:43.627" Score="14" ViewCount="0" Body="&lt;p&gt;Obviously, if p is rational, then p&lt;sup&gt;2&lt;/sup&gt; must also be rational (trivial to prove).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p \in \mathbb Q \Rightarrow p^2 \in \mathbb Q. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Contraposition&quot; rel=&quot;nofollow&quot;&gt;contraposition&lt;/a&gt;, we see that if x is irrational, then &amp;radic;x must also be irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p^2 \notin \mathbb Q \Rightarrow p \notin \mathbb Q. $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By negative power I assume you mean (1/n)-th power (it is obvious that $(\sqrt2)^{-2} = \frac12\in\mathbb Q$). It is true by the statement above — just replace &lt;sup&gt;2&lt;/sup&gt; by &lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-16T11:20:43.627" CommentCount="4" />
  <row Id="2574" PostTypeId="1" AcceptedAnswerId="2577" CreationDate="2010-08-16T12:09:56.557" Score="2" ViewCount="236" Body="&lt;p&gt;In a related &lt;a href=&quot;http://math.stackexchange.com/questions/2571/about-powers-of-irrational-numbers&quot;&gt;question&lt;/a&gt; we discussed raising numbers to powers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested if anybody knows any results for raising numbers to irrational powers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, we can easily show that there exists an irrational number raised to an irrational power such that the result is a rational number. Observe ${\sqrt 2 ^ {\sqrt 2}}$. Since we do not know if ${\sqrt 2 ^ {\sqrt 2}}$ is rational or not, there are two cases.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;${\sqrt 2 ^ {\sqrt 2}}$ is rational, and we are finished.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;${\sqrt 2 ^ {\sqrt 2}}$ is irrational, but if we raise it by ${\sqrt 2}$ again, we can see that&#xA;$$\left ( \sqrt 2 ^ \sqrt 2 \right ) ^ \sqrt 2 = \sqrt 2 ^ {\sqrt 2 \cdot \sqrt 2} = \sqrt 2 ^ 2 = 2.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Either way, we have shown that there exists an irrational number raised to an irrational power such that the result is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can more be said about raising irrational numbers to irrational powers?  &lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-04T06:05:47.477" LastActivityDate="2011-01-04T06:05:47.477" Title="Real Numbers to Irrational Powers" Tags="&lt;number-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2575" PostTypeId="1" CreationDate="2010-08-16T12:17:17.150" Score="2" ViewCount="59" Body="&lt;p&gt;I have statistics of 100 questions which can be answered either &quot;yes&quot; or &quot;no&quot;:&lt;br&gt;&#xA;1) 63.3 - 36.7 (63.3% respondents answered &quot;yes&quot; and 36.7% answered &quot;no&quot;)&lt;br&gt;&#xA;2) 30.1 - 69.9&lt;br&gt;&#xA;...&lt;br&gt;&#xA;100) 88.0 - 12.0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That 100 answers would be our sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I ask the same 100 questions to 101st respondent and get new set of answers:&lt;br&gt;&#xA;1) yes&lt;br&gt;&#xA;2) yes&lt;br&gt;&#xA;...&lt;br&gt;&#xA;100) no&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need to do is some how calculate &quot;correlation&quot; value between this respondent and overall sample. Any method will be OK, I just need to get some number. Therefore there may be different &quot;right&quot; answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2010-12-22T14:44:26.220" Title="How to calculate correlation between poll statistics and one's answers?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2577" PostTypeId="2" ParentId="2574" CreationDate="2010-08-16T12:24:37.353" Score="3" ViewCount="0" Body="&lt;p&gt;Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Eulers Identity $e^{i \pi} + 1 = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gelfond-Schneider Constant: $2^\sqrt{2} = 2.66514414269022518865\cdots$ is trancendental by Gelfond-Schneider.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$i^i = 0.2078795763507619085469556198\cdots$ is also trancendental &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ramanujan constant: $e^{\pi \sqrt{163}} = 62537412640768743.999999999999250072597\cdots$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$e^\gamma$ where $\gamma$ is the Euler–Mascheroni constant. (okay nobody has proved it's irrational yet, but surely is)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-08-16T12:31:45.267" CommunityOwnedDate="2010-08-16T12:48:38.833" CommentCount="2" />
  <row Id="2578" PostTypeId="1" AcceptedAnswerId="2579" CreationDate="2010-08-16T12:36:02.527" Score="-2" ViewCount="104" Body="&lt;p&gt;A space $X$ is said to have the fixed point property if any continuous map $f: X \to X$ has a fixed point. By contraction mapping theorem we know that if $X$ is complete and $f$ is a contraction then $X$ has a fixed point which is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question, is $X$ is a space which has the fixed point property and if another space $Y$ is homemorphic to $X$, then does $Y$ have the fixed point property. Would like to see a complete proof.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-16T12:46:39.680" LastActivityDate="2010-08-16T12:46:39.680" Title="Does &quot;Homeomorphism&quot; preserve the fixed point property." Tags="&lt;analysis&gt;&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2579" PostTypeId="2" ParentId="2578" CreationDate="2010-08-16T12:42:53.213" Score="1" ViewCount="0" Body="&lt;p&gt;I second Robin Chapman's advise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;The answer is yes: Fix a homeomorphism $h\colon X\to Y$. Let $g\colon Y\to Y$ be a continuous map. Then $h^{-1}\circ f\circ h$ is a continuous map from $X$ to $X$, and hence has a fixed point $x\in X$. Define $y = h(x)$. Since $h^{-1}(f(h(x))=x$, we get $f(y)=y$.&#xA;&lt;/s&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-16T12:42:53.213" />
  <row Id="2580" PostTypeId="2" ParentId="2574" CreationDate="2010-08-16T12:44:21.417" Score="0" ViewCount="0" Body="&lt;p&gt;Forgive my lack of knowledge of how to correctly implement equations in TeX! I always particularly liked that... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{\pi} - \pi = 19.9990999791894757672664\cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would also be possible to prove that result is transcendental as well. The two cases being $e^{\pi}$ is transcendental or $e^{\pi}$ isn't. The difference of a non-transcendental number and a transcendental number is transcendental, and the only time the difference of two transcendental numbers wouldn't be transcendental is when you could extract the second from the first - That is to say, you could find some non-transcendental $x$  that would satisfy $e^{\pi} = x + \pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, my formal proof skills are not what they once were. I'm sure this is trivial, and really the result isn't even that close to 20. But I always liked it :)&lt;/p&gt;&#xA;" OwnerUserId="1200" LastEditorUserId="1200" LastEditorDisplayName="" LastEditDate="2010-08-16T12:58:36.757" LastActivityDate="2010-08-16T12:58:36.757" CommentCount="3" />
  <row Id="2581" PostTypeId="1" CreationDate="2010-08-16T12:46:51.133" Score="35" ViewCount="948" Body="&lt;p&gt;This was asked on sci.math ages ago, and never got a satisfactory answer (AFAICR)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number of sticks of integral length &gt;= $n$ whose lengths add to $n(n+1)/2$&#xA;Can these be divided into sticks of lengths $1,2,3, \ldots ,n$.  &lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-10-02T21:44:44.500" Title="Cutting sticks puzzle" Tags="&lt;combinatorics&gt;&lt;puzzle&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="14" />
  <row Id="2583" PostTypeId="2" ParentId="213" CreationDate="2010-08-16T12:53:14.677" Score="3" ViewCount="0" Body="&lt;p&gt;Graph Theory is a fantastic field. First, the fact that abstract concepts can be readily visualized makes it engaging for new students. And second, I believe it provides solid foundations into mathematical thinking like proofs, and engages the student to explore other related fields.&lt;/p&gt;&#xA;" OwnerUserId="1010" LastActivityDate="2010-08-16T12:53:14.677" CommunityOwnedDate="2010-08-16T12:53:14.677" />
  <row Id="2584" PostTypeId="1" AcceptedAnswerId="2854" CreationDate="2010-08-16T13:02:04.063" Score="2" ViewCount="281" Body="&lt;p&gt;I've got a fixed solution space defined by a minimum and maximum float which has no divisible slice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then have 0..N &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;Normal distribution&lt;/a&gt; probability sets which I need to combine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end I neeed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method to define probability set with a numeric range (not sliceable) and 0..N Gaussian functions&#xA;A function which can generate a random number in the range as defined by the calculated probabilities.&#xA;Also I know it is possible that some combinations will generate a zero solution space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm thinking the way to do it is take the normalised probability density functions and multiply them to get a new space then normalising the result. I just can't seem to break it down into algorithmic form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra Clarification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's examine the following height rules.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;STATUS Height [ 1.2 : 2.4 ] &#xA;  MODIFIER Gender[ Male ] {Height [ 1.9 ~ 0.4 ] }&#xA;  MODIFIER Nation[ Red ] { Height [ 1.7 ~ 0.2 ] }&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now assuming that we have a man from the Red nation we need to resolve these. &#xA;  We know that the original status declaration defines the entire possibility space.&#xA;  We cannot leave that space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xDRMJ.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/1ndjv.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now basically I need to find a way of combining these to get a new probability.&#xA;What I meant by slice was because its a floating point number I can't break it into a set of elements of calculate the new probability for each element. I can't slice it up into pieces.&lt;/p&gt;&#xA;" OwnerUserId="1194" LastEditorUserId="1194" LastEditorDisplayName="" LastEditDate="2010-08-17T12:54:36.897" LastActivityDate="2010-08-26T10:07:28.947" Title="Combine n Normal distribution Probability Sets in a limited float range" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="2585" PostTypeId="2" ParentId="2349" CreationDate="2010-08-16T13:05:17.240" Score="1" ViewCount="0" Body="&lt;p&gt;Heres a hint: The jacket is put on after the shirt, but is taken off before&#xA;it.&lt;/p&gt;&#xA;" OwnerUserId="1237" LastActivityDate="2010-08-16T13:05:17.240" />
  <row Id="2586" PostTypeId="1" AcceptedAnswerId="2589" CreationDate="2010-08-16T13:11:38.907" Score="2" ViewCount="73" Body="&lt;p&gt;Hi, I was going to ask this on SO but I think its more math than programming:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the sidelength, number of vertices and vertex angle in the polygon, how can I calculate the radius of its circumscribed circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polygon may have any number of sides greater than or equal to 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia entry only discusses circumscribed circle of a triangle...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; &#xA;Also, the polygon is centered around the point (0,0). So I guess I'm asking what is the distance from the origin to any of its points..&lt;/p&gt;&#xA;" OwnerUserId="1238" LastEditorUserId="1238" LastEditorDisplayName="" LastEditDate="2010-08-16T13:17:11.907" LastActivityDate="2010-08-16T13:35:37.463" Title="Radius of the circumscribed circle of a regular polygon." Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="2587" PostTypeId="2" ParentId="2586" CreationDate="2010-08-16T13:22:02.907" Score="1" ViewCount="0" Body="&lt;p&gt;Each corner of a polygon must be at equal distance from the center of the circumscribing circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, find equation for perpendicular bisectors of any to sides of the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intersection of perpendicular bisectors will give you the center of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then distance between center and any corner of the polygon is the radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT(response to edited question):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If length of a side is $L$ and number of vertices is $N$ then&#xA;Suppose points A and B defines a side and C is the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then angle ACB is $360/N$ &#xA;Suppose D is midpoint of a edge AB then angle DCA = angle DCB = 180/ N.&#xA;This implies $sin(180/N) = L/(2R) $ where R is radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $ R = \frac{L}{2sin(180/N)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pseudocode:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def get_circumradius(L,N):&#xA;    return L/(2 * math.sin(180/N))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-08-16T13:35:37.463" LastActivityDate="2010-08-16T13:35:37.463" CommentCount="1" />
  <row Id="2589" PostTypeId="2" ParentId="2586" CreationDate="2010-08-16T13:30:18.350" Score="5" ViewCount="0" Body="&lt;p&gt;Let $s$ be sidelength and $\alpha$ be vertex angle. Then using simple trigonometry you can find that&#xA;$$&#xA;\sin{\frac{\alpha}{2}} = \frac{s/2}{r}&#xA;$$&#xA;Hence&#xA;$$&#xA;r = \frac{s}{2 \sin{\frac{\alpha}{2}}}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-16T13:30:18.350" CommentCount="1" />
  <row Id="2590" PostTypeId="2" ParentId="2561" CreationDate="2010-08-16T14:04:08.643" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; The polynomial $f(x) = (x-\alpha)(x-\alpha') = x^2 + x + k$ assumes only prime values for $0 \le x \le k — 2 \iff$ $\mathbb Z[\alpha]$ is a PID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; for $(\Rightarrow)$ Show all primes $p \le \sqrt{n},\; n = 1-4k$ satisfy $(n/p) = -1$ so no primes split/ramify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For proofs, see e.g. Cohn, Advanced Number Theory, pp. 155-156, or Ribenboim, My numbers, my friends, 5.7 p.108.  Note: both proofs employ the bound $p &amp;lt; \sqrt{n}$ without explicitly mentiioning that this is a consequence of the Minkowski bound - presumably assuming that is obvious to the reader based upon earlier results. Thus you'll need to read the prior sections on the Minkowski bound. Compare Stewart and Tall, Algebraic number theory and FLT, 3ed, Theorem 10.4 p.176 where the use of the Minkowski bound is mentioned explicitly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively see the self-contained &lt;a href=&quot;http://www.jstor.org/stable/2689515&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt; which proceeds a bit more simply, employing Dirichlet approximation to obtain a generalization of the Euclidean algorithm  (the Dedekind-Rabinowitsch-Hasse criterion for a PID). If memory serves correct, this is close to the approach originally employed by Rabinowitsch when he first published this theorem in 1913.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Daniel Fendel, Prime-Producing Polynomials and Principal Ideal Domains,&lt;br&gt;&#xA;Mathematics Magazine, Vol. 58, 4, 1985, 204-210&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-16T16:05:10.490" LastActivityDate="2010-08-16T16:05:10.490" CommentCount="1" />
  <row Id="2591" PostTypeId="1" CreationDate="2010-08-16T15:11:02.367" Score="5" ViewCount="259" Body="&lt;p&gt;Quoting from &lt;a href=&quot;http://math.stackexchange.com/questions/907/correct-usage-of-the-phrase-in-the-sequel-history-alternatives/924#924&quot;&gt;this excellent answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you read enough math papers you'll find that there are certain linguistic ticks that people pick up from each other&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So here's a question (primarily for you native English speakers out there, I guess):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any particularly annoying language mistakes that you often see in mathematical writing (presumably by foreigners who are copying the mistake from other foreigners)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A list of common errors to avoid might be useful to us non-native speakers. As an example of what I have in mind, I can mention a phrase that sounds wrong to my ears, but which I have come across countless times: &quot;This allows to prove...&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-16T17:03:53.687" CommunityOwnedDate="2010-08-16T16:49:13.317" ClosedDate="2010-08-16T22:32:16.803" Title="Common English language mistakes in mathematical writing" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="2592" PostTypeId="1" AcceptedAnswerId="2750" CreationDate="2010-08-16T15:25:42.560" Score="4" ViewCount="205" Body="&lt;p&gt;A long time ago I was curious about the closed-form solutions to the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{d^{n}y}{dx^n} y = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n$ an odd number, try $y = A x^k$. Then $y^{(n)} = A k(k-1)...(k-n)x^{k-n}$. This gives the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^2 k(k-1)...(k-n) x^{2k - n - 1} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can only be true if $k = \frac{n+1}{2}$ and $n$ is odd ($k$ cannot be an integer for the formula to work, check this yourself). Furthermore one has to have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = (k(k-1)...(k-n))^{-1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is real if $n$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there are closed-form solutions to my problem for $n$ odd, and my question is if anyone can find a closed-form solution for $n=2$ or in general if $n$ is even.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-08-19T03:31:32.473" Title="Solutions to the equation $y^{(n)} y = 1$ for even $n$." Tags="&lt;real-analysis&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2593" PostTypeId="1" AcceptedAnswerId="2597" CreationDate="2010-08-16T15:26:42.000" Score="2" ViewCount="145" Body="&lt;p&gt;Given a function $f$ defined on the set of all natural numbers $\mathbb{N}$ with three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $m,n$ relatively prime, then $f(mn) = f(m)f(n)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f$ strictly increasing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(2) = 2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Find a 4th condition such that the result will be that $f(n)$ must equal $n$ for any natural number $n$. (Of course all the conditions together are needed. Your 4th condition should not make any of these three conditions redundant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is posted in my university website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: http:// mathstat.uohyd.ernet.in/noticeboard/generaldetails.php?id=12&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-16T16:09:08.773" LastActivityDate="2010-08-16T17:33:19.360" Title="Finding Extra Condition for a function to satisfy $f(n)=n$." Tags="&lt;number-theory&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2595" PostTypeId="1" CreationDate="2010-08-16T16:38:36.623" Score="6" ViewCount="277" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Image processing application in which I have a Matrix equation as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A*R=I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = 3x3 matrix (Constants)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = 3x1 matrix (Column Vector) Lets call this as actual output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = 3x1 matrix . Lets call I as Ideal output&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* denotes matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the values of matrix I, and R. I have to find what matrix A, if post multiplied by R would give me matrix I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I set his situation up in matrix algebra and solve it compute A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-AD.&lt;/p&gt;&#xA;" OwnerUserId="1249" LastActivityDate="2010-08-22T20:24:00.713" Title="Need to compute values of all the entries of a 3x3 matrix?" Tags="&lt;matrices&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="2596" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T16:50:21.757" Score="1" ViewCount="0" Body="&lt;p&gt;As Rasmus said, if you have only one (I,R) pair, then there are multiple possibilities for A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in any one such A, you can use the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R = (x,y,z)$ and assume that $x \neq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I = (a,b,c)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can choose the matrix to be&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[a/x 0 0]&#xA;[b/x 0 0]&#xA;[c/x 0 0]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-16T16:50:21.757" CommentCount="6" />
  <row Id="2597" PostTypeId="2" ParentId="2593" CreationDate="2010-08-16T16:51:44.330" Score="3" ViewCount="0" Body="&lt;p&gt;In fact, it's a theorem of Erdös that every non-decreasing multiplicative function $f:\mathbb{N}\to\mathbb{R}$ is $f(n)=n^c$ for some constant $c$. This implies that those three conditions are already enough to prove that $f(n)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a proof by Pierre Pbornsztein &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=179524#p179524&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="446" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-16T17:33:19.360" LastActivityDate="2010-08-16T17:33:19.360" CommentCount="7" />
  <row Id="2598" PostTypeId="2" ParentId="2591" CreationDate="2010-08-16T17:03:53.687" Score="2" ViewCount="0" Body="&lt;p&gt;One that confused me greatly when I first found it was &quot;We ignore if X is true.&quot;  I read it as &quot;We &lt;em&gt;disregard&lt;/em&gt; whether X is true,&quot; which made no sense in context.  Eventually I realized the authors meant &quot;We &lt;em&gt;do not know&lt;/em&gt; if X is true.&quot;  I suppose this corresponds to the fact that &quot;to ignore&quot; and &quot;to be ignorant&quot; have different meanings in English despite having the same (Latin?) root.  &lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-16T17:03:53.687" CommunityOwnedDate="2010-08-16T17:03:53.687" />
  <row Id="2600" PostTypeId="2" ParentId="1949" CreationDate="2010-08-16T17:16:07.940" Score="1" ViewCount="0" Body="&lt;p&gt;Take a look at the last big displayed equation under &quot;formal definition&quot; here.  It shows you Gauss's explicit form for a Levi-Civita connection in terms of the metric.  Since you know how the metric transforms under an isometry, and how a Lie bracket transforms under a diffeomorphism, working out how the connection transforms under an isometry amounts to putting those ingredients together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_connection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Levi-Civita_connection&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-08-16T17:16:07.940" />
  <row Id="2601" PostTypeId="1" CreationDate="2010-08-16T17:35:02.150" Score="7" ViewCount="396" Body="&lt;p&gt;Frequently, we introduce a new concept with a formal definition, then immediately say &quot;Intuitively, what this means is...&quot; What are the absolute best metaphors you've seen (for concepts of any level)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I think the comparison of modular arithmetic to arithmetic with times, on a clock face, is fantastic, because it corresponds so perfectly and because clocks are so widely known.&lt;/p&gt;&#xA;" OwnerUserId="629" LastActivityDate="2010-10-26T17:03:34.217" CommunityOwnedDate="2010-08-16T17:35:02.150" Title="Best intuitive metaphors for math concepts (of any level)" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;intuition&gt;&lt;education&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="3" />
  <row Id="2602" PostTypeId="1" AcceptedAnswerId="2624" CreationDate="2010-08-16T17:40:59.877" Score="3" ViewCount="288" Body="&lt;p&gt;I've taken an undergraduate course in ring and group theroy, but haven't studied number theory formally.   I've noticed many important results in number theory have been generalized in group/ring theory (e.g. Lagrange Thm generalizing Fermats Little Thereom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is are there any results in elementary number theory that have not been generalized using elementary group and ring theory?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-17T00:05:51.077" Title="Elementary Number theory results that are not generalized by ring or group theory" Tags="&lt;soft-question&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="2603" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T17:53:01.507" Score="3" ViewCount="0" Body="&lt;p&gt;Actually if you have 24 sets of &lt;strong&gt;I&lt;/strong&gt; and &lt;strong&gt;R&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf A \cdot \mathbf R_i = \mathbf I_i \qquad i = 1, 2, \dotsc, 24 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since there are only 9 variables for each element of A but 24&amp;times;3 = 72 linear equations, the problem is &lt;a href=&quot;http://en.wikipedia.org/wiki/Overdetermined_system&quot; rel=&quot;nofollow&quot;&gt;overdetermined&lt;/a&gt; and there may be no solutions. However, if you accept some errors &lt;strong&gt;&amp;epsilon;&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf A \cdot \mathbf R_i + \boldsymbol \epsilon_i = \mathbf I_i \qquad i = 1, 2, \dotsc, 24 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to find such an A, as this introduces 72 more variables (the problem becomes underdetermined). The best thing is of course to minimize the total error. We first write A as a column of row vectors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf A = \begin{bmatrix} \mathbf A_1 \\ \mathbf A_2 \\ \mathbf A_3 \end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf R_i \cdot \mathbf A_j + \epsilon_{ij} = I_{ij} \qquad i = 1, \dotsc, 24; j = 1, 2, 3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is exactly like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_regression_model&quot; rel=&quot;nofollow&quot;&gt;linear regression problem&lt;/a&gt; (if we minimize the sum of squares of errors)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathsf X \cdot \boldsymbol \beta_j + \boldsymbol \epsilon_j = \mathbf y_j $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\mathsf X &amp;amp;= \begin{bmatrix}&#xA;  R_{1,1} &amp;amp; R_{1,2} &amp;amp; R_{1,3} \\&#xA;  \vdots &amp;amp; \vdots &amp;amp; \vdots \\&#xA;  R_{24,1} &amp;amp; R_{24,2} &amp;amp; R_{24,3}&#xA;\end{bmatrix} \\&#xA;\boldsymbol\beta_j &amp;amp;= \begin{bmatrix}&#xA;  A_{j,1} \\&#xA;  A_{j,2} \\&#xA;  A_{j,3}&#xA;\end{bmatrix} \\&#xA;\mathbf y_j &amp;amp;= \begin{bmatrix}&#xA;  I_{1,j} \\&#xA;  \vdots \\&#xA;  I_{24,j}&#xA;\end{bmatrix}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these we can use any software supporting regression (or by hand) to find the best approximation of &lt;strong&gt;A&lt;/strong&gt;&lt;sub&gt;j&lt;/sub&gt;, and thus the whole matrix A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can be easily generalized to $N \gg 24$ pixels.)&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-16T18:46:54.267" LastActivityDate="2010-08-16T18:46:54.267" CommentCount="2" />
  <row Id="2604" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T17:53:02.143" Score="1" ViewCount="0" Body="&lt;p&gt;Let $a_{ij}$ be the number in $i$-th row and $j$-th column in the matrix $A$. Then each equation of the form $AR=I$ is basicly three linear equations with unknowns $a_{ij}$. If there are 24 sets then you have $24 \cdot 3$ linear equations in $9$ unknowns. This system might have no solution in general. But, you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_least_squares&quot; rel=&quot;nofollow&quot;&gt;Linear least squares&lt;/a&gt; to find a solution that fits best all those equations.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-08-16T17:53:02.143" />
  <row Id="2605" PostTypeId="1" CreationDate="2010-08-16T18:05:44.990" Score="1" ViewCount="70" Body="&lt;p&gt;When showing that two functors $F:A\rightarrow B$ and $G:B\rightarrow A$ are adjoint, one defines a natural bijection $Mor(X,G(Y)) \rightarrow Mor(F(X),Y)$. What if one do not require the bijection to be natural, what issues would arise? &lt;/p&gt;&#xA;" OwnerUserId="1253" LastActivityDate="2010-08-16T18:05:44.990" Title="Adjoint functors requiring a natural bijection." Tags="&lt;category-theory&gt;" CommentCount="3" />
  <row Id="2606" PostTypeId="1" AcceptedAnswerId="2609" CreationDate="2010-08-16T18:29:08.860" Score="6" ViewCount="140" Body="&lt;p&gt;I have three questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the groupification of a semigroup always exist? I believe this should be yes because for every $x$ in the semigroup one could just define an element $x'$ that should work as its inverse. But what would then happen to the product $x'y$ for $x,y$ elements of the semigroup? It feels like we get choices (or maybe not) here that messes things up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When defining the groupification, $G$, of a semigroup $S$ one require it to come with a morphism (of semigroups) $S \rightarrow G$ such that any other morphism (of semigroups) from $S$ to another group $G'$ factorizes through the previous map. Exactly which type of objects can be groupified? I guess one cannot groupify a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a broad question but is there some sense of -ification? In the example one could replace &quot;group&quot; by &quot;topological space&quot; and talk about topologyfication. Now, no such word seem to exist so I guess one could not &quot;topologyfy&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can (i think) consider the groupification functor from the category of semigroups to the category of groups and it should be adjoint to the forgetful functor from the category of groups to the category of semigroups. This would suggest that we need some sense of a forgetful functor in the first place to talk about a -ification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for this bad question, sometimes asking the right question is just as hard as answering it. &lt;/p&gt;&#xA;" OwnerUserId="1253" LastActivityDate="2010-08-16T20:12:45.263" Title="Is there some universal sense of -ification (eg, groupification) in category theory" Tags="&lt;category-theory&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="2607" PostTypeId="2" ParentId="2595" CreationDate="2010-08-16T18:31:47.980" Score="1" ViewCount="0" Body="&lt;p&gt;FIRST CASE. If you had only &lt;em&gt;one&lt;/em&gt; set of values for $I$ and $R$, as you said in your answer to Rasmus, then you'd have a system of linear equations with infinitely many solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For psychological reasons, let's write it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;XA = B \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $A$ is your $R$, $B$ your $I$ and $X$ your $A$. If we transpose we get a simultaneous system of linear equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A^t X^t = B^t \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A^t = (a_1 \ a_2 \ a_3)$ and $B^t = (b_1 \ b_2 \ b_3)$, and $(x_i \ y_i \ z_i)$ is the i-th colum $X^t$, we have the linear equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a_1 x_i + a_2 y_i + a_3 z_i = b_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $i = 1, 2, 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, assuming $a_1 \neq 0$, you can solve like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_i = \frac{b_i}{a_1} - \frac{a_2}{a_1} y_i - \frac{a_3}{a_1} z_i \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, give $y_i$ and $z_i$ the values you want and you have a solution for your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Maybe I could develop a little bit more my answer, in order to &lt;em&gt;really&lt;/em&gt; include your problem, which is one of overdeterminated simultaneous systems of linear equations, as KennyTM says, since you don't have a prefered pair of values $(R,I)$, do you? In order to handle all the $24$ pairs of values $(R,I)$ you have, maybe you should take into account this SECOND CASE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SECOND CASE. I'm sorry, but I'm changing slightly the notation again. In the end, I'll write the solution with yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;AX = B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be a (simultaneous) system of linear equations such as yours, with $A$ a $24\times 3$ matrix ($24$ rows, $3$ columns), $X$ a $3\times 3$ matrix and $B$ a $24 \times 3$ matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;Hypothesis: Let's assume that our matrix $A$ (that is, your $R$) has rang $ 3$. *&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If this is not the case, the problem is more involved.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write the first system this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_1a_1 + y_1a_2 + z_1a_3 = b_1  \ .  \qquad [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $a_i , i = 1,2,3$ are the columns of $A$, $X_1 = \begin{pmatrix} x_1 &amp;amp; y_1 &amp;amp; z_1 \end{pmatrix}^t$ the first column of $X$ and also $b_1$ is the first column of $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see in one go a geometrical interpretation of our system of equations: system [1] has a solution $\begin{pmatrix} x_1 &amp;amp; y_1 &amp;amp; z_1 \end{pmatrix}$ if and only if the vector $b_1$ belongs to the linear span generated by the columns of $A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;AX_1 = b_1 \quad \text{has a solution} \qquad \Longleftrightarrow \qquad b_1 \in [a_1, a_2, a_3]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we do if this is &lt;em&gt;not&lt;/em&gt; the case? -To look for the &lt;em&gt;nearest&lt;/em&gt; vector in $[a_1, a_2, a_3]$ to $b_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;nearest&lt;/em&gt; vector is, of course, the &lt;em&gt;orthogonal projection&lt;/em&gt; of $b_1$ onto the subspace $[a_1, a_2, a_3] $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_projection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Orthogonal_projection&lt;/a&gt; , the matrix of this orthogonal projection in the standard basis of $\mathbb{R}^{24}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_A = A (A^tA)^{-1}A^t \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the best $X = \begin{pmatrix} X_1 &amp;amp; X_2 &amp;amp; X_3\end{pmatrix}$ for you is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;X = A (A^tA)^{-1}A^tB&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B = \begin{pmatrix} b_1 &amp;amp; b_2 &amp;amp; b_3 \end{pmatrix}$. Now, let's go back to your notation: $A = X^t$, $R = A^t$ and $I = B^t$. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A =  \left( R^t (R^{tt}R^t)^{-1}R^{tt}I^t \right)^t = I R^t ((RR^t)^{-1})^t R&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ and $I$ are now the matrices with &lt;em&gt;all&lt;/em&gt; of your $R$'s and $I$'s as columns.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-17T09:16:23.623" LastActivityDate="2010-08-17T09:16:23.623" />
  <row Id="2608" PostTypeId="2" ParentId="213" CreationDate="2010-08-16T18:41:03.000" Score="3" ViewCount="0" Body="&lt;p&gt;I wish I'd learned logic much, much earlier. Obviously young students couldn't handle much depth, but at least a basic introduction to a few concepts would be nice. Just understanding the concept of axioms and deductive rules would put all of math into some perspective. When I finally understood that math was constructed with formal definitions and proofs (or, for example, that there was more than one useful way to axiomatize), I felt I'd been kept in the dark my whole life, doing something (math) that I had absolutely no understanding of.&lt;/p&gt;&#xA;" OwnerUserId="629" LastActivityDate="2010-08-16T18:41:03.000" CommunityOwnedDate="2010-08-16T18:41:03.000" />
  <row Id="2609" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T18:46:26.903" Score="6" ViewCount="0" Body="&lt;p&gt;I take the liberty of answering only the parts of your question, to which I think I can give a precise answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Yes, it is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grothendieck_group&quot; rel=&quot;nofollow&quot;&gt;Grothendieck group&lt;/a&gt; $G(N)$ of the given semigroup $S$. This construction is functorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: There is always a canonical semigroup homomorphism $S\to G(S)$, but this need not be injective in general. For example, the semigroup corresponding to $\mathbb N\cup\{\infty\}$ with the obvious addition $(n+\infty=\infty)$ is trivial!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: Concerning -ification in general, at the moment I have nothing to add to BBischof's great comments above.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-16T18:51:51.887" LastActivityDate="2010-08-16T18:51:51.887" />
  <row Id="2610" PostTypeId="2" ParentId="2602" CreationDate="2010-08-16T18:47:01.927" Score="2" ViewCount="0" Body="&lt;p&gt;I contend that every result that could reasonably be called &quot;elementary&quot; number theory also holds for $\mathbb{F}_q[x]$, the ring of polynomials in one variable over a finite field.  Quadratic reciprocity fits the bill here, for example.  This is the subject of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387953353&quot; rel=&quot;nofollow&quot;&gt;at least one book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-16T18:47:01.927" CommentCount="4" />
  <row Id="2611" PostTypeId="1" CreationDate="2010-08-16T18:50:43.073" Score="1" ViewCount="233" Body="&lt;p&gt;If given the cost to play, and the average win.  Can I calculate the edge? (probability of winning)&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2010-12-05T10:28:01.937" Title="How do I calculate gambling edge using average win?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2612" PostTypeId="1" AcceptedAnswerId="2617" CreationDate="2010-08-16T19:08:16.283" Score="2" ViewCount="150" Body="&lt;p&gt;I've recently been interested in Monte Carlo simulations and their uses, unfortunately most of the examples I find are difficult to understand for a beginner. What are some simple examples of using Monte Carlo simulations used to solve math problems? Bonus points for showing a simple example and using it to understand a more complex one.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-16T19:33:23.800" LastActivityDate="2010-08-16T22:31:18.863" Title="What are some uses for Monte Carlo simulations in mathematics?" Tags="&lt;big-list&gt;&lt;monte-carlo&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="2614" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T19:13:26.227" Score="4" ViewCount="0" Body="&lt;p&gt;To add to Rasmus's answer to 1. One can't avoid elements like&#xA;$x^{-1}y$ in the Grothendieck group. Consider say the natural numbers.&#xA;It's Grothendieck group is the integers. As it's additive we&#xA;write say four plus the inverse of 6 as $4-6$. Things aren't too&#xA;bad in the commutative case: (in additive notation) all elements&#xA;of $G(S)$ have the form $a-b$ with $a$, $b\in S$. Moreover&#xA;$a-b=c-d$ iff there is $s\in S$ with $a+d+s=b+c+s$&#xA;(and we can forget about $s$ if $S$ is cancellative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But things are nastier in the noncommutative world. Returning&#xA;to multiplicative notation, you now get elements in $G(s)$ like $ab^{-1}cd^{-1}e$&#xA;which perhaps don't simplify to shorter ones. And it's not so&#xA;easy to give a criterion for when two elements are the same&#xA;(or equivalently when one element equals the identity).&#xA;&lt;em&gt;C'est la vie!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-16T19:13:26.227" />
  <row Id="2615" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T19:20:30.440" Score="6" ViewCount="0" Body="&lt;p&gt;As BBischof says, the standard notion of -ification is to take the left adjoint of a forgetful functor.  This includes the following as special cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The groupification of a monoid or semigroup,&lt;/li&gt;&#xA;&lt;li&gt;The free group on a set, the free vector space on a set, etc.&lt;/li&gt;&#xA;&lt;li&gt;The abelianization of a group,&lt;/li&gt;&#xA;&lt;li&gt;The group ring of a group (the forgetful functor here sends a ring to its group of units),&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and many other examples.  I do not think one can reasonably talk about universal -ification without a specific choice of forgetful functor; if no good choice exists, you won't get a good notion of -ification, and on the other hand there may be more than one choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notably, I remember reading that in at least one example the natural construction is to take the &lt;em&gt;right&lt;/em&gt; adjoint, but I don't remember what this example was.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-16T19:20:30.440" CommentCount="1" />
  <row Id="2616" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T19:34:37.587" Score="0" ViewCount="0" Body="&lt;p&gt;EDIT: a previous comment posted the same answer, I just noticed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use montecarlo method to approximate pi.&#xA;You basically define a square of side length 2 and inscribe a circle of radius 1 in it. Let the center of the circle be at the origin.&#xA;Use a random number generator to pick an x-coordinate between 0 and 1 and a y-coordinate between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;number of pts in the circle / number of pts in square ~ pi/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the simplest example of a monte carlo code I know and I think it is probably the standard example.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-16T19:42:11.813" LastActivityDate="2010-08-16T19:42:11.813" />
  <row Id="2617" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T19:34:46.120" Score="7" ViewCount="0" Body="&lt;p&gt;Monte Carlo methods are very useful in numerically evaluating high-dimensional integrals. With traditional integration methods, the number of integrand evaluations required to maintain accuracy grows quickly as dimension increases.  With Monte Carlo integration, the number of integrand evaluations needed is &lt;em&gt;independent&lt;/em&gt; of dimension.  For many high-dimensional integrals, Monte Carlo methods are the only practical choice.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-16T19:34:46.120" CommentCount="3" />
  <row Id="2618" PostTypeId="2" ParentId="2455" CreationDate="2010-08-16T19:40:37.027" Score="2" ViewCount="0" Body="&lt;p&gt;Place the balls so that the corners of the pyramid have coordinates (1,1,1), (n,1,1), (1,n,1), (1,1,n). Then for the ball at (x,y,z) we see that (x,x+y,x+y+z) is an ordered triple of distinct numbers between 1 and n+2 inclusive. There are $\binom{n+2}{3}$ such triples, and they are clearly in one-to-one correspondence with the balls. Geometrically, this corresponds to drawing planes x=constant, x+y=constant, and x+y+z=constant through a given ball, and seeing where these planes intersect the x axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To compare with the case $k=2$, &quot;straighten&quot; the picture above into a right triangle. Then the diagonal lines become lines x=constant and x+y=constant, and the bottom row of balls (the blue ones) can be identified with the points where these lines intersect the x axis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also works in higher dimensions, giving $\binom{n+k-1}{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this proof really is any different from the one you suggested in the question can of course be debated...&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-16T19:40:37.027" CommentCount="0" />
  <row Id="2619" PostTypeId="1" CreationDate="2010-08-16T19:58:20.230" Score="2" ViewCount="155" Body="&lt;p&gt;Couple of days ago, i actually asked this question : &lt;a href=&quot;http://math.stackexchange.com/questions/2121/existence-of-a-normal-subgroup-with-authautg&quot;&gt;http://math.stackexchange.com/questions/2121/existence-of-a-normal-subgroup-with-authautg&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about the converse of this statement. Suppose $G$ is a finite group, with subgroups $H$ and satisfies $|Aut(G)|=|Aut(H)|$, then can anything be said about, $H$ or $G$ in terms of their structure. Will $H$ be normal or ....? Or one can even ask, that find all finite groups $G$ such that $|Aut(H)|=|Aut(G)|$ for every subgroup $H$ of $G$! I am not interested that much in this question as there seems one can have lot of Groups of this type, but i am more curious to know the behaviour of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, anyone interested can very well read this article in MO: &lt;a href=&quot;http://mathoverflow.net/questions/1075/order-of-an-automorphism-of-a-finite-group&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/1075/order-of-an-automorphism-of-a-finite-group&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-16T19:58:20.230" Title="Solving $|Aut(H)| = |Aut(G)|$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="7" />
  <row Id="2620" PostTypeId="2" ParentId="2606" CreationDate="2010-08-16T20:12:45.263" Score="4" ViewCount="0" Body="&lt;p&gt;The name of the really &lt;em&gt;really&lt;/em&gt; general construction (which is indeed in many cases the left adjoint of the forgetful functor) is called the &lt;strong&gt;completion&lt;/strong&gt; of an object. The basic idea is to find the 'most natural' or 'smallest' object having the original as a subobject, which means we can extend the idea to, say, the closure of a subset of a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/completion&quot; rel=&quot;nofollow&quot;&gt;Here's the n-lab page if you're interested...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-08-16T20:12:45.263" CommentCount="1" />
  <row Id="2621" PostTypeId="1" CreationDate="2010-08-16T20:48:51.997" Score="3" ViewCount="146" Body="&lt;p&gt;Let $A$ be a commutative Banach algebra with unit. &#xA;It is well known that if the Gelfand transform $\hat{x}$ of $x\in A$ is non-zero, then $x$ is invertible in $A$ (the so called Wiener Lemma in the case when $A$ is the Banach algebra of absolutely convergent Fourier series).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a converse of the above, let $B$ be a Banach space contained in $A$ and suppose $B$ is closed under inversion - i.e.: If $x\in B$ and $x^{-1}\in A$ then $x^{-1}\in B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Prove that $B$ is a Banach algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Must $A$ and $B$ have the same norm? If not are the norms similar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Do $A$ and $B$ have the same maximal ideal space?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-17T07:55:26.113" Title="Converse of Wiener's lemma" Tags="&lt;functional-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2622" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T21:09:11.103" Score="2" ViewCount="0" Body="&lt;p&gt;Monte Carlo methods are used extensively in &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_methods_in_finance&quot; rel=&quot;nofollow&quot;&gt;financial mathematics&lt;/a&gt; for the pricing of complex or &quot;exotic&quot; financial derivatives.  With equity options, for example, the value of the stocks in the option contract is simulated using a stochastic process and parameters that can be observed or derived from other financial instruments.  This process is computationally very intensive resulting in many investment banks having server farms of tens of thousands of machines dedicated to the pricing and risk management of derivatives.&lt;/p&gt;&#xA;" OwnerUserId="1032" LastActivityDate="2010-08-16T21:09:11.103" />
  <row Id="2624" PostTypeId="2" ParentId="2602" CreationDate="2010-08-16T22:17:26.140" Score="5" ViewCount="0" Body="&lt;p&gt;Problems for integers whose generalizations in algebraic number theory have not been developed (or at least considered) are rare.  The only examples I know of are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ancient problems without a strong relation to modern algebraic techniques.  This includes the existence of odd perfect numbers, and compositeness of Fermat numbers $2^{2^k}+1$.   Keep in mind, though, that most of the very old problems have been understood or finished off by modern techniques, such as the congruent number problem and Fermat's Last Theorem being reduced to facts about elliptic curves, or the probabilistic approach to the distribution of special types of primes (twin, Mersenne, k-term progressions, etc), and these better-understood problems are more amenable to generalization beyond the integers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Questions of combinatorial nature, such as covering congruences or arithmetic progressions (van der Waerden theorems).  Here the problems can be generalized to other rings, but most interest has focused on the integer case.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Uses of number theory in logic and computer science, such as Goedel coding, complexity of Presburger arithmetic, or other problems where iterated exponentiation appears.   Algebraic number theory is primarily about problems described by polynomial equations, and the reduction of exponential Diophantine equations to polynomial ones (as in the solution of Hilbert's 10th problem) is not direct enough to allow the use of the standard algebraic tools.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For everything else, where standard algebraic techniques or analytic ones (zeta functions, Diophantine approximation, sieves, transcendence theory, complex analysis, etc) are applicable to a problem over the integers, there has usually been an effort to search for analogues of those methods in algebraic number theory, p-adic number theory, geometry over finite fields, and complex geometry.   This applies to all the major techniques and theories developed since 1800.  So to find examples one needs to move away from &quot;central core&quot; number theory and look at questions that are not, so far, primarily studied by the methods of algebraic number theory and algebraic geometry. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-16T23:18:11.877" LastActivityDate="2010-08-16T23:18:11.877" />
  <row Id="2625" PostTypeId="2" ParentId="2612" CreationDate="2010-08-16T22:31:18.863" Score="1" ViewCount="0" Body="&lt;p&gt;Tangential to John D. Cook's reply, and also somewhat related: Monte Carlo also finds application in the solution of (partial, stochastic) differential equations, of which cubature (nobody ever uses MC in the one-dimensional case practically ;) ) is but a specialized case. As already mentioned by John, the pain dealt by the &quot;curse of dimensionality&quot; stings less with Monte Carlo.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-16T22:31:18.863" />
  <row Id="2626" PostTypeId="1" AcceptedAnswerId="2644" CreationDate="2010-08-16T22:39:14.420" Score="5" ViewCount="277" Body="&lt;p&gt;I'd like to create a basic model of the forces acting on a sail (wind sail, like a tail ship)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things I was thinking about: &#xA;1) can create a very simple model where wind is 'one' force acting on the a uniform body.&#xA;2) model wind as vectors. This is where I am a little confused on how to start. &#xA;3) adding in multiple sails. &#xA;4) I know that this would be a differential equation but after that I can't really see how it would be modeled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for someone to actually do the modelling for me, just a place to start. Like maybe some Wikipedia articles, etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2010-08-18T21:45:12.413" Title="Modelling forces acting on a sail" Tags="&lt;differential-equations&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="1" />
  <row Id="2627" PostTypeId="1" CreationDate="2010-08-16T22:54:18.983" Score="2" ViewCount="130" Body="&lt;p&gt;Let $X$ denote the blowup of $\mathbb P^2$, $E$ the exceptional divisor, and $H$ the pullback of the hyperplane class.  How can I compute $H^0(X,mE+nH)$, $H^1(X,mE+nH)$, and $H^2(X,mE+nH)$ for $m,n \in \mathbb Z$?  If I'm working analytically, how can I think of these geometrically (e.g. &quot;an element of $H^1(X,2E)$ is equivalent to a $1$-form on $\mathbb P^2$ such that...&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="1261" LastActivityDate="2010-08-17T16:11:58.447" Title="Cohomology of line bundles on the blowup of $\mathbb P^2$" Tags="&lt;algebraic-geometry&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2628" PostTypeId="1" AcceptedAnswerId="2699" CreationDate="2010-08-16T23:26:57.880" Score="2" ViewCount="256" Body="&lt;p&gt;By stressing my manipulative powers and a bit of help from &lt;em&gt;Mathematica&lt;/em&gt; I was able to show that the triple integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{8\pi}\int_0^{2\pi}\int_0^\pi\int_0^\pi\,(2\cos\,\theta\cos\,\varphi-\sin\,\theta\sin\,\varphi\cos\,\psi)^{2k}\sin\,\theta\sin\,\varphi\,\mathrm{d}\theta\mathrm{d}\varphi\mathrm{d}\psi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where k is a positive integer is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\left(\frac{2^k}{2k+1}\right)^2}\sum_{j=0}^k\frac{\binom{2j}{j}}{\binom{2k}{k}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by making use of the binomial expansion and then using a hypergeometric identity that degenerates to a ratio of gamma functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason I think there was no need for &lt;em&gt;Mathematica&lt;/em&gt; to invoke the Gauss hypergeometric function ${}_2 F_1\left(a,b;c;z\right)$ in performing the triple integration, so I'm wondering if there might be a way to show that this identity holds, using nothing more complicated than a gamma function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-21T08:29:46.533" Title="An elementary way of simplifying a trigonometric triple integral?" Tags="&lt;gamma-function&gt;&lt;closed-form&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="2629" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T00:15:17.253" Score="1" ViewCount="0" Body="&lt;p&gt;What about $F = \sum_i W \cdot S_i$, where $W$ is the wind velocity vector and $S$ is a list of all the sail unit vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see this at all being a differential equation, but if you further specify your idea of this model, I can help you a little more.&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-17T00:15:17.253" CommentCount="2" />
  <row Id="2630" PostTypeId="1" AcceptedAnswerId="2631" CreationDate="2010-08-17T01:13:38.287" Score="2" ViewCount="148" Body="&lt;p&gt;Suppose A is some algebraic structure, and x and y are two elements of the underlying set.  Is there any more concise way of stating, &quot;there exists an automorphism in A which maps x to y&quot;?.  It seems like the obvious thing to say is &quot;x and y are symmetric in A&quot;, but I've never come across this phrasing or any other in the literature, which is amazing to me since it's such an enormously useful concept.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2010-08-17T16:50:47.770" Title="Algebra terminology question" Tags="&lt;terminology&gt;&lt;universal-algebra&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2631" PostTypeId="2" ParentId="2630" CreationDate="2010-08-17T02:28:05.643" Score="2" ViewCount="0" Body="&lt;p&gt;Some authors write that $x$ and $y$ are conjugate under $G$, or $G$-conjugate, e.g. from Bourbaki, Algebra:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Definition 5.&lt;/strong&gt; Let $G$ be a group, $E$ a $G$-set and $x \in G$. An element $y \in E$ is&#xA;  &lt;em&gt;conjugate&lt;/em&gt; to $x$ under the operation of $G$ if there exists an element $\alpha \in G$ such &#xA;  that $y = \alpha x$. The set of conjugate elements of $x$ is called the &lt;em&gt;orbit&lt;/em&gt; of $x$ in $E$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-17T16:50:47.770" LastActivityDate="2010-08-17T16:50:47.770" />
  <row Id="2632" PostTypeId="1" AcceptedAnswerId="2637" CreationDate="2010-08-17T04:34:08.353" Score="0" ViewCount="270" Body="&lt;p&gt;Show that a continuous function $f:[0,1] \to \mathbb{R}$ vanishes at $0$, that is $f(0)=0$, if and only if it is the uniform limit of a sequence of odd polynomials in $[0,1]$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-17T18:26:00.063" LastActivityDate="2010-08-17T18:26:00.063" Title="Continuous functions $f$ vanishing at zero are uniform limits of odd polynomials" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;polynomials&gt;&lt;approximation-theory&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="2633" PostTypeId="1" AcceptedAnswerId="2640" CreationDate="2010-08-17T04:58:58.507" Score="0" ViewCount="236" Body="&lt;p&gt;Let $(0,\infty)$ be equipped with the Lebesgue measure, and let $1 &amp;lt; p &amp;lt; \infty$. For each $ f \in L_{p}(\lambda)$ let $$T(f)(x) = x^{-1} \int f \chi_{(0,x)} \ d\lambda \quad \text{for} \ x &gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then prove that $T$ is a bounded linear operator from $L_{p}(0,\infty)$ into itself.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-17T17:24:17.560" Title="Proving that $T$ is a bounded linear operator" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;&lt;measure-theory&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2634" PostTypeId="1" CreationDate="2010-08-17T05:09:19.613" Score="1" ViewCount="135" Body="&lt;p&gt;A Liouville number is a number which can be approximated very closely be a sequence of rational numbers (here is the rigorous definition I am working off of: &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Liouville_number&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an example of a Liouville number which cannot be approximated by a sequence of rational numbers with a denominators which are all a constant c multiplied by powers of some number a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the Louiville constant (0.110001000000000000000001...) can be approximated by the sequence 1/10, 11/10^2, 110001/10^6, etc, which is not what I am looking for because in each case the denominator is a power of 10. In this case, we would say that c=1, a=10, and the denominator is always of the form c*a^n for some positive n.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-17T05:09:19.613" Title="Certain Liouville Numbers " Tags="&lt;number-theory&gt;" CommentCount="2" FavoriteCount="0" />
  <row Id="2635" PostTypeId="2" ParentId="2632" CreationDate="2010-08-17T06:14:55.540" Score="1" ViewCount="0" Body="&lt;p&gt;This follows from the locally compact form of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Locally_compact_version&quot; rel=&quot;nofollow&quot;&gt;Stone-Weierstrass theorem&lt;/a&gt; applied to $(0, 1]$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-17T06:14:55.540" />
  <row Id="2636" PostTypeId="2" ParentId="2632" CreationDate="2010-08-17T06:27:11.617" Score="7" ViewCount="0" Body="&lt;p&gt;Problem postings will get better answers, more upvotes, and fewer downvotes, if instead of a bare problem statement they include:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Sources for the problem.  This helps readers decide if the problem is of interest to them (is it too easy or too difficult, is it likely to use techniques they know or want to learn, etc), and also whether the statement is likely to be correct.  A question from &lt;em&gt;Unsolved Problems in Real Analysis&lt;/em&gt; is quite different from a problem taken from &lt;em&gt;Beginner's Introduction to Real Analysis (chapter 1)&lt;/em&gt;.  Note well, that posting questions from olympiads, hard problem compilations, problem columns, or research articles without indicating the source can be very time-consuming for those who try to solve them, uninformed of the true difficulty.  &lt;b&gt;Repeated postings of this type will receive multiple downvotes&lt;/b&gt; every time a &quot;challenge&quot; problem is posted without information about its difficulty level.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sources for solutions, if known.   Withholding this information, when known, is a waste of other users' time and the votes will (over time) reflect that. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Proof sketches for any known special cases, or other partial progress.  This, again, indicates the techniques that might be useful and helps users understand whether the problem is of interest to them and whether they might know how to answer it.  Withholding this information, when known, is a waste of other users' time and the votes will (over time) reflect that. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;References relevant to the question, even if they do not answer it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Indication of why the result is plausible (prior to having a proof, i.e., &quot;intuition&quot;). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Isolated problem statements posed as challenges for the sake of challenge are not, by themselves, a significant contribution to a question-answer site.  As far as they are formulated in ways that waste the time of users, they become a negative.  This is useful to keep in mind when posting. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-08-17T06:27:11.617" CommentCount="6" />
  <row Id="2637" PostTypeId="2" ParentId="2632" CreationDate="2010-08-17T06:42:35.050" Score="2" ViewCount="0" Body="&lt;p&gt;Thanks to Robin Chapman, and Qiacho Yuan for giving me the hint. I tried and i think this should be the solution. Sorry for not providing the source, i got this as an email from my friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One, part is clear, that if $f$ is a uniform limit of sequence of odd polynomials then $f(0)=0$. For the converse, let $f \in C[0,1]$ and satisfies $f(0)=0$, and let $\epsilon &gt; 0$. Define $g : [-1,1] \to \mathbb{R}$ by,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1AwoZ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear $g$ is continuous and so by Weierstrass approximation theorem, we have a polynomial $p(x)$ such that $|g(x)-p(x)|&amp;lt; \epsilon$ for each $x \in [-1,1]$. Next the polynomial $p$ can be written as sum of two  polynmials $h$ and $k$ where $h$ is the one consisting of all odd terms of $p$. Then we have $h(-x)=-h(x)$ and $k(-x)=k(x)$ holds for each $x$. If $x \in [0,1]$, then we have $$|f(x)-q(x)-r(x)|=|g(x)-p(x)| &amp;lt; \epsilon$$ and $g(-x)= -f(x)$ implies $$|f(x)-q(x)+r(x)|=|p(-x)-g(-x)| &amp;lt; \epsilon$$ Using this we can deduce that $|f(x)-q(x)|&amp;lt; \epsilon$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-17T06:56:49.903" LastActivityDate="2010-08-17T06:56:49.903" CommentCount="1" />
  <row Id="2638" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T07:13:51.533" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; This is just an approach, it doesn't actually work. I don't know if it's possible to make it correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q = \frac{1}{1 - \frac{1}{p}}$ so $\frac{1}{p} + \frac{1}{q} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Holder_inequality&quot; rel=&quot;nofollow&quot;&gt;Hölder's inequality&lt;/a&gt; we have&#xA;$$&#xA;\int_{0}^{x} |f(\lambda)| d\lambda \le (\int_{0}^{x} |f(\lambda)|^{p} d\lambda)^{\frac{1}{p}} (\int_{0}^{x} |1|^{q} d\lambda)^{\frac{1}{q}} \le ||f||_{p} x^{\frac{1}{q}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;|\frac{1}{x} \int_{0}^{x} f(\lambda) d\lambda|^p \le |\frac{1}{x} \int_{0}^{x} |f(\lambda)| d\lambda|^p \le ||f||^{p}_{p} x^{\frac{1}{q}-1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now norm of $T(f)$ has the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;||T(f)||_{p} = (\int _{0}^{\infty} |T(f)(x)|^{p}dx)^{1/p} = (\int _{0}^{\infty}|\frac{1}{x} \int _{0}^{x} f(\lambda) d\lambda|^pdx)^{1/p}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $||f||^{p} = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;||T(f)||_{p} \le (\int _{0}^{\infty} c^{p} x^{\frac{1}{q}-1} dx)^{1/p} = ||f||_p (\int _{0}^{\infty} x^{-1/p})^{p}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(\int _{0}^{\infty} x^{-1/p})^{p}$ is some constant we have $||T(f)|| _{p} \le ||f|| _{p} const$, hence this operator is bounded.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-17T08:14:26.457" LastActivityDate="2010-08-17T08:14:26.457" CommentCount="3" />
  <row Id="2639" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T07:32:38.337" Score="0" ViewCount="0" Body="&lt;p&gt;Following falagar's solution one can also get the value of $ \displaystyle ||T|| = \frac{p}{p-1}$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-17T07:32:38.337" />
  <row Id="2640" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T07:33:09.847" Score="4" ViewCount="0" Body="&lt;p&gt;One (direct) way to see this is that $T$ is dominated by (a slightly modfied version of) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hardy-Littlewood_maximal_operator&quot; rel=&quot;nofollow&quot;&gt;Hardy-Littlewood maximal operator&lt;/a&gt;, and it is a theorem that this is a bounded operator on $L^p$ for $p$ greater than 1. Or, directly, one can see that $T$ is bounded on $L^\infty$ and weak-bounded on $L^1$ (because if $f$ belongs to $ L^1$ then $Tf(x) &amp;lt;= ||f||_1/x$), so the &lt;a href=&quot;http://en.wikipedia.org/wiki/Marcinkiewicz_interpolation_theorem&quot; rel=&quot;nofollow&quot;&gt;Marcinkiewicz interpolation theorem&lt;/a&gt; applies. This is really much too indirect, though, but I'd like to point out that this is a special case of something more general. A direct approach is in page 242 of &lt;a href=&quot;http://books.google.com/books?id=u5fkpAJ5Rc0C&amp;amp;pg=PA239&amp;amp;lpg=PA239&amp;amp;dq=hardy%27s+inequality+proof&amp;amp;source=bl&amp;amp;ots=MyVsFH7dqx&amp;amp;sig=JCc4As_ifrCAFHMf8Xq-MMBqufk&amp;amp;hl=en&amp;amp;ei=STlqTOiRDIGBlAfsxoC5Ag&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCQQ6AEwAjhG#v=onepage&amp;amp;q=hardy%27s%20inequality%20proof&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Inequalities&lt;/a&gt; (the page is available).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-17T08:04:15.810" LastActivityDate="2010-08-17T08:04:15.810" />
  <row Id="2641" PostTypeId="2" ParentId="2621" CreationDate="2010-08-17T07:55:26.113" Score="1" ViewCount="0" Body="&lt;p&gt;For now, I shall assume that $B$ is a closed subspace of $A$ (and is treated simply as a topological vector space for the moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Let $x \in B$; we need to prove that $x^2 \in B$. From this elementary algebra will imply that $B$ is closed under multiplication. Now, when $t$ is close to zero, we have $1-tx$ invertible in $A$, hence in $B$; so $1+tx + t^2x^2 + \dots$ lies in $B$ for $t$ close to zero. Taking the second derivative at $0$ (recall that $B$ is a &lt;em&gt;closed&lt;/em&gt; subspace) shows that $x^2 \in B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't fully understand what you are asking in (2) yet, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) No. Take $A$ to be any Banach algebra not equal to the complex numbers and $B$ the subalgebra spanned by the identity.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-17T07:55:26.113" CommentCount="4" />
  <row Id="2642" PostTypeId="2" ParentId="2632" CreationDate="2010-08-17T08:00:19.003" Score="1" ViewCount="0" Body="&lt;p&gt;There is another way that in my opinion makes it more obvious that the obtained polynomials are odd. Instead of defining an extension of $f$ to the interval $[-1,1]$ that is itself odd, we define a continuous extension $g: [-1,1] \to \mathbb{R}$ by $g(x)=0$ for $x &amp;lt; 0$. Now, for $\epsilon &gt; 0$ the Stone-Weierstrass theorem gives a polynomial $p(x)$ such that $|g(x)-p(x)| &amp;lt; \epsilon$ for all $x \in [-1,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define $q(x) = p(x) - p(-x)$ which is obviously an odd polynomium (why?) and let us compute its distance to $f$ on $[0,1]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|f(x) - q(x)| = |g(x) - p(x) + p(-x)| \leq |g(x)-p(x)| + |p(-x)|$$&#xA;$$ = |g(x)-p(x)| + |g(-x) - p(-x)| &amp;lt; 2\epsilon,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: &lt;code&gt;\begin{cases}&lt;/code&gt; and &lt;code&gt;\begin{align}&lt;/code&gt;do not work in preview. I hope this looks fine.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-08-17T08:00:19.003" />
  <row Id="2643" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T09:55:56.080" Score="2" ViewCount="0" Body="&lt;p&gt;The post &lt;em&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2009/03/23/sailing-into-the-wind-or-faster-than-the-wind/&quot; rel=&quot;nofollow&quot;&gt;The Sailing into the wind, or faster than the wind&lt;/a&gt;&lt;/em&gt;  by Professor Terence Tao and the comments to it may be of interest to you. (See my comment above and cbrulak's reply)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain &quot;the power of the wind to sail in a direction against that of the wind or to sail with a speed faster than the wind itself&quot; the author describes the one-, two- and three-dimensional sailing models.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-17T09:55:56.080" />
  <row Id="2644" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T10:21:24.860" Score="5" ViewCount="0" Body="&lt;p&gt;Assuming no edge effects/turbulence, we can calculate the force of wind on a single sail using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_equation&quot; rel=&quot;nofollow&quot;&gt;drag equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F_D} = \tfrac12 \rho \vec{u}^2 C_D A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the drag coefficient $C_D$ is going to be fairly high, for a concave sail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, however, the wind may not be hitting the sail orthogonally, but rather at some angle. We can generalise the drag equation as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F_{sail}} = \tfrac12 \rho \vec{u}^2 C_D (\hat{\vec{u}} \cdot \vec{S})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S% is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_area&quot; rel=&quot;nofollow&quot;&gt;vector area&lt;/a&gt; of the sail. We take the dot product of this vector with the unit velocity vector to resolve the wind force in the direction of the sail. This is of course an approximation, as the concave shape of the sail will come into play, but probably a good enough one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That pretty much explains the basic situation for a single sail. Now, for multiple sails, you would of course simply combine the forces on the individual sails, which will all have their own values of $C_D$ and $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F_{sails,total}} = \sum \vec{F_{sail}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you familiarise yourself with the mechanics and specific equations used here, and use that as a basic model to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some general information on the physics of sailing, you may find &lt;a href=&quot;http://www.animations.physics.unsw.edu.au/jw/sailing.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; helpful.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-08-17T10:21:24.860" CommentCount="2" />
  <row Id="2645" PostTypeId="1" AcceptedAnswerId="2646" CreationDate="2010-08-17T10:48:58.560" Score="9" ViewCount="183" Body="&lt;p&gt;Suppose there are two correlated random variable and having very small correlation coefficient (order of 10&lt;sup&gt;-1&lt;/sup&gt;). Is it valid to approximate it as  independent random variables? &lt;/p&gt;&#xA;" OwnerUserId="1273" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-17T12:14:59.223" LastActivityDate="2010-08-18T08:38:58.927" Title="probability and statistics: Does having little correlation imply independence?" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;random-variable&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="2646" PostTypeId="2" ParentId="2645" CreationDate="2010-08-17T10:52:51.410" Score="5" ViewCount="0" Body="&lt;p&gt;Independence of random variables implies that they are uncorrelated, but the reverse is not true. Hence no, such an approximation is not valid (given that information alone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A clear description is given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncorrelated&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If X and Y are independent, then they&#xA;  are uncorrelated. However, not all&#xA;  uncorrelated variables are&#xA;  independent. For example, if X is a&#xA;  continuous random variable uniformly&#xA;  distributed on [−1, 1] and Y = X²,&#xA;  then X and Y are uncorrelated even&#xA;  though X determines Y and a particular&#xA;  value of Y can be produced by only one&#xA;  or two values of X.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-08-17T13:54:46.760" LastActivityDate="2010-08-17T13:54:46.760" CommentCount="3" />
  <row Id="2647" PostTypeId="1" AcceptedAnswerId="2649" CreationDate="2010-08-17T11:57:00.337" Score="6" ViewCount="319" Body="&lt;p&gt;i) Does there exists a nonempty subset of integers $S$, such that for all $a \in S$, there exists $b,c \in S$ such that $a = b + c$, where $a, b, c$ are &lt;em&gt;distinct&lt;/em&gt; integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add:&#xA;ii) Can an $S$ with these properties be &lt;em&gt;finite&lt;/em&gt;, and also have the property that $a \in S$ implies $-a \notin S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This question is inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/2418/a-zero-sum-subset-of-a-sum-full-set&quot;&gt;another&lt;/a&gt; question, which has not been fully answered yet.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-17T12:36:30.647" LastActivityDate="2010-08-19T12:18:20.387" Title="Sets of integers in which every element is the sum of two others" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="2648" PostTypeId="2" ParentId="2647" CreationDate="2010-08-17T12:02:50.043" Score="1" ViewCount="0" Body="&lt;p&gt;The only possibilities I can think of at the moment are $ S = n \mathbb{Z} $, so for now I guess the answer is &lt;em&gt;technically&lt;/em&gt; yes.  I'll try to think of others and possibly characterize all such sets.  Great question!&lt;/p&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-08-17T12:02:50.043" CommentCount="1" />
  <row Id="2649" PostTypeId="2" ParentId="2647" CreationDate="2010-08-17T12:07:02.000" Score="10" ViewCount="0" Body="&lt;p&gt;Yes, you can take the set of integers between $-n$ and $n$ inclusive for any $n \ge 3$.&#xA;Here $0 = 1 + (-1)$, $1 = 2 + (-1)$ and $x = (x - 1) + 1$ for $x \ge 2$. Similar formulae for negative values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Answer for new version of the question&#xA;Set found by hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\{ -22, -20, -18, -16, -14, -12, -10, -2, 1, 3, 7, 8, 15, 23 \}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Answer with 8 elements. If you add -12 to this set you will get set with odd number of elements.&#xA;$$&#xA;\{ -10, -8, -6, -2, 1, 3, 4, 5 \}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-08-17T14:40:41.767" LastActivityDate="2010-08-17T14:40:41.767" CommentCount="3" />
  <row Id="2650" PostTypeId="1" AcceptedAnswerId="2660" CreationDate="2010-08-17T13:25:05.303" Score="2" ViewCount="246" Body="&lt;p&gt;In the towers of Hanoi game, how do we know that we have the optimal algorithm for solving it? I thought about this and it seemed like any deviation from the standard strategies would be putting you back a step, but I had no idea how to demonstrate this rigorously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I know is that the proof involves the Lucas correspondence between the Hanoi graph and the odd coefficients in Pascal's triangle. How is Pascal's triangle turned into a graph? I assume the coefficient are the vertices, but I don't see how you form the edges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering how to generalize the strategy to n discs and k rods because it seems like this correspondence argument wouldn't really work in the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I am wondering how the odd coefficients of the Pascal triangle are turned into a graph and whether or not there is a similar method to find and prove optimal strategy when we increase the number of rods.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-17T15:16:56.017" LastActivityDate="2010-08-17T16:37:10.110" Title="How to prove the optimal Towers of Hanoi strategy?" Tags="&lt;computer-science&gt;&lt;graph-theory&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2651" PostTypeId="2" ParentId="2645" CreationDate="2010-08-17T13:55:05.090" Score="1" ViewCount="0" Body="&lt;p&gt;If X and Y are Jointly Gaussian and X and Y are uncorrelated then X and Y are independent&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-17T13:55:05.090" CommentCount="0" />
  <row Id="2652" PostTypeId="2" ParentId="2647" CreationDate="2010-08-17T13:57:07.097" Score="2" ViewCount="0" Body="&lt;p&gt;I posted an answer to this question previously, giving a finite set. That answer was incorrect: this is a different answer, which contains &lt;em&gt;lower bounds&lt;/em&gt; on finite sets of the sort you ask for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set S such that (&amp;forall;a&amp;nbsp;&amp;isin;&amp;nbsp;S)(&amp;exist;b,c&amp;nbsp;&amp;isin;&amp;nbsp;S):(b&amp;ne;c&amp;nbsp;&amp;nbsp;&lt;em&gt;and&lt;/em&gt;&amp;nbsp;&amp;nbsp;a&amp;nbsp;=&amp;nbsp;b&amp;nbsp;+&amp;nbsp;c) we will call &lt;strong&gt;self-supporting&lt;/strong&gt;: a self-supporting set S which is disjoint from &amp;minus;S will be &lt;strong&gt;strongly&lt;/strong&gt; self-supporting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 1.&lt;/strong&gt; A self-supporting set must contain at least three positive elements, and three negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;em&gt;Proof&lt;/em&gt;]&#xA;A self-supporting set S is non-empty; because the property of self-support is preserved by negation of the elements, without loss of generality we may consider sets containing positive elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest positive element of S  can only be formed as a sum of two other positive elements; thus it has at least three positive elements 0&amp;lt;c&amp;lt;b&amp;lt;a such that a&amp;nbsp;=&amp;nbsp;b&amp;nbsp;+&amp;nbsp;c. The smallest positive element cannot be formed as a sum of smaller positive elements, so S contains negative elements. By similar arguments, it must contain at least three negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the set {&amp;minus;3,&amp;nbsp;&amp;minus;2,&amp;nbsp;&amp;minus;1,&amp;nbsp;1,&amp;nbsp;2,&amp;nbsp;3} is self-supporting, this bound is optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 2.&lt;/strong&gt; A strongly self-supporting set must contain at least four positive elements, and four negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;em&gt;Proof&lt;/em&gt;] Suppose that S is self-supporting, but only has three positive elements 0&amp;lt;c&amp;lt;b&amp;lt;a. Then a&amp;nbsp;=&amp;nbsp;b&amp;nbsp;+&amp;nbsp;c. We cannot form b as a sum of c with some other positive element of S; it then follows that b can only be formed as a sum of a and &amp;minus;c, which implies that S is not &lt;em&gt;strongly&lt;/em&gt; self-supporting. Thus, a strongly self-supporting set contains at least four positive elements; and similarly for negative elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller set in flagar's answer, {&amp;minus;10,&amp;nbsp;&amp;minus;8,&amp;nbsp;−6,&amp;nbsp;−2,&amp;nbsp;1,&amp;nbsp;3,&amp;nbsp;4,&amp;nbsp;5}, is thus a minimal-size strongly self-supporting set --- and may also be the set having elements of the smallest absolute value.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-19T12:14:05.437" LastActivityDate="2010-08-19T12:14:05.437" CommentCount="1" />
  <row Id="2653" PostTypeId="2" ParentId="201" CreationDate="2010-08-17T14:14:06.620" Score="7" ViewCount="0" Body="&lt;p&gt;Here is a sweet concise reformulation of Euclid's proof that deserves to be better known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\ \ N \: (N+1) \;$ has a larger set of prime factors than does $\:\rm N &gt; 0\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Since $\rm N+1 &gt; 1\:$ it has a prime factor $\rm P$. But $\rm P$ cannot divide $\rm N$ since $\rm N$ is coprime to $\rm N+1\ $ (viz. if $\rm\: P$ divides both $\rm N+1$ and $\rm N$ then $\rm P$ divides their difference $\rm N+1 - N = 1\:,\: $ contradiction). Thus the prime factors of $\rm\: N(N+1)$ include all those of $\rm N$ plus at least one prime $\rm P$ not dividing $\rm N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COROLLARY&lt;/strong&gt; $\ \ $ There are infinitely many primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Iterating $\rm\ N\to N\: (N+1)\ $ produces integers with an unbounded number of prime factors.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-25T22:58:45.210" LastActivityDate="2010-09-25T22:58:45.210" CommentCount="5" />
  <row Id="2654" PostTypeId="2" ParentId="2650" CreationDate="2010-08-17T15:58:50.313" Score="4" ViewCount="0" Body="&lt;p&gt;In general, with n rods and n disks, the problem is equivalent to finding a Hamiltonian path on an n-hypercube. There is no proven optimal strategy known for 4 rods. You could also read the &lt;a href=&quot;http://mathworld.wolfram.com/TowerofHanoi.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld article&lt;/a&gt; which has references for quite a few papers on the subject.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-08-17T15:58:50.313" CommentCount="1" />
  <row Id="2655" PostTypeId="1" AcceptedAnswerId="2657" CreationDate="2010-08-17T16:01:46.350" Score="4" ViewCount="282" Body="&lt;p&gt;2 is the first prime number.&#xA;3 is the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I give a prime number such as 1151024046313875220631 is there any software/website which can give the position of the prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are resources to find Nth prime. But I am having a hard time finding the reverse.&lt;/p&gt;&#xA;" OwnerUserId="1279" LastActivityDate="2010-09-08T20:28:19.610" Title="Is it possible to find the position of a prime number online?" Tags="&lt;prime-numbers&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="2656" PostTypeId="2" ParentId="2627" CreationDate="2010-08-17T16:11:58.447" Score="2" ViewCount="0" Body="&lt;p&gt;I assume you're only blowing up at one point, so here's at least a nice geometric description for $H^0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is the point blown up to $E$, then $dH-E$ is the system of plane curves of degree $d$ passing through $p$, $dH-2E$ are those that have a double point at $p$, etc.  This works with any number of blownup points, and a good exercise is using this interpretation to find all 27 lines on a cubic surface (which is the blowup at 6 points)&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-08-17T16:11:58.447" />
  <row Id="2657" PostTypeId="2" ParentId="2655" CreationDate="2010-08-17T16:12:55.640" Score="2" ViewCount="0" Body="&lt;p&gt;If you have access to Mathematica, PrimePi[x] will give you the number of primes less than x. Combined with PrimeQ, which verifies that x is indeed prime, will give you which prime number x is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have no idea how long Mathematica would take (or if it could in fact compute it) for a number that high.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-08-17T16:12:55.640" CommentCount="3" />
  <row Id="2658" PostTypeId="2" ParentId="154" CreationDate="2010-08-17T16:19:03.323" Score="14" ViewCount="0" Body="&lt;p&gt;One need only consult the history of algebra to find many informal&#xA;discussions on the existence and consistence of complex numbers. &#xA;Any &lt;em&gt;informal&lt;/em&gt; attempt to justify the existence of $\mathbb C$ &#xA;will face  the same obstacles that existed in earlier times. Namely, the lack of any rigorous (set-theoretic) foundation makes it difficult to be precise - both syntactically and semantically. Nowadays the set-theoretic foundation of algebraic structures &#xA;is so subconscious that is is easy to overlook just how much power it provides &#xA;for such purposes. But this oversight is easily remedied. One need only consult&#xA;some of the older literature where even leading mathematicians struggled&#xA;immensely to rigorously define complex numbers. For example, see the&#xA;quote below by Cauchy and Hankel's scathing critique - which is guaranteed &#xA;to make your jaw drop! (Below is an excerpt from my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zhbxkjhil.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt; on the notion &#xA;of formal polynomial rings and their quotients).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major accomplishment of the set-theoretical definition of &#xA;algebraic structures was to eliminate imprecise syntax and semantics. &#xA;By eliminating the syntactic polynomial term $a+b\cdot x+c\cdot x^2$ &#xA;in favor of its set-theoretic semantic reduction &#xA;$(a,b,c,0,0,\ldots)$, there can no longer be any doubt about the precise denotation of the symbols $\; x,\; +,\;\cdot\;,$ or about their equality, since, by &#xA;set theory definition, tuples are equal iff their components are equal. &#xA;Similarly for complex numbers  $a + b\cdot i$ &#xA;vs. their set-theoretic pair reduction $(a,b)$ discovered by Hamilton. &#xA;Before Hamilton gave this semantic reduction of $\mathbb C$ to pairs &#xA;of reals, prior syntactic constructions (e.g. by Cauchy) as &#xA;formal expressions or terms $a+b\cdot i$ were subject to heavy criticism regarding &#xA;the precise denotation of their constituent symbols, e.g. &#xA;precisely what is the meaning of the symbols $\;i, +, = \;$ ? &#xA;In modern language, Cauchy's construction of $\mathbb C$ is simply the &#xA;the quotient ring $\mathbb R[x]/(x^2+1)$, which he described essentially &#xA;as real polynomial expressions modulo $x^2+1$. However, in Cauchy's time &#xA;mathematics lacked the necessary (set-theory) foundations to &#xA;rigorously define the syntactic expressions comprising the&#xA;polynomial ring term algebra $\mathbb R[x]$, and its quotient ring of&#xA;congruence classes $\pmod{x^2+1}$. The best that Cauchy could &#xA;do was to attempt to describe the constructions in terms of &#xA;imprecise &lt;em&gt;natural&lt;/em&gt; (human) language, e.g, in 1821 Cauchy wrote: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In analysis, we call a symbolic expression any combination of &#xA;  symbols or algebraic signs which means nothing by itself but &#xA;  which one attributes a value different from the one it should &#xA;  naturally be [...] Similarly, we call symbolic equations those &#xA;  that, taken literally and interpreted according to conventions &#xA;  generally established, are inaccurate or have no meaning, but &#xA;  from which can be deduced accurate results, by changing and &#xA;  altering, according to fixed rules, the equations or symbols &#xA;  within [...] Among the symbolic expressions and equations &#xA;  whose theory is of considerable importance in analysis, one &#xA;  distinguishes especially those that have been called imaginary... -- Cauchy 1821&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While nowadays, using set theory, we can rigorously interpret such &quot;symbolic expressions&quot; &#xA;as terms of formal languages or term algebras, it was far too &#xA;imprecise in Cauchy's time to have any hope of making sense &#xA;to his colleagues, e.g. Hankel replied scathingly: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If one were to give a critique of this reasoning, we can not &#xA;  actually see where to start. There must be something &quot;which &#xA;  means nothing,&quot; or &quot;which is assigned a different value than &#xA;  it should naturally be&quot; something that has &quot;no sense&quot; or is &#xA;  &quot;incorrect&quot;, coupled with another similar kind, producing &#xA;  something real. There must be &quot;algebraic signs&quot; - are these &#xA;  signs for quantities or what? as a sign must designate something &#xA;  - combined with each other in a way that has &quot;a meaning.&quot; I do &#xA;  not think I'm exaggerating in calling this an unintelligible &#xA;  play on words, ill-becoming of mathematics, which is proud &#xA;  and rightly proud of the clarity and evidence of its concepts. -- Hankel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus it comes as no surprise that Hamilton's elimination &#xA;of such &quot;meaningless&quot; symbols - in favor of pairs of reals - &#xA;served as a major step forward in placing such numbers on a &#xA;foundation more amenable to his contemporaries. &#xA;Although there was not yet any theory of sets in which to &#xA;rigorously axiomatize the notion of pairs, they were far easier &#xA;to accept naively - esp. given the already known closely &#xA;associated geometric interpretation of complex numbers. &#xA;Hamilton introduced pairs as 'couples' in &lt;a href=&quot;http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/PureTime/PureTime.pdf&quot; rel=&quot;nofollow&quot;&gt;1837 [1]&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;p. 6: The author acknowledges with pleasure that he agrees with &#xA;  M. Cauchy, in considering every (so-called) Imaginary Equation &#xA;  as a symbolic representation of two separate Real Equations: &#xA;  but he differs from that excellent mathematician in his method &#xA;  generally, and especially in not introducing the sign  sqrt(-1) &#xA;  until he has provided for it, by his  Theory of Couples, &#xA;  a possible and real meaning, as a symbol of the couple (0,1) &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;p. 111:  But because Mr. Graves employed, in his reasoning, the &#xA;  usual principles respecting about Imaginary Quantities, and &#xA;  was content to prove the symbolical necessity without showing &#xA;  the interpretation, or inner meaning, of his formulae, the &#xA;  present Theory of Couples is published to make manifest that &#xA;  hidden meaning: and to show, by this remarkable instance, that &#xA;  expressions which seem according to common views to be merely &#xA;  symbolical, and quite incapable of being interpreted, may pass &#xA;  into the world of thoughts, and acquire reality and significance, &#xA;  if Algebra be viewed as not a mere Art or Language, but as the &#xA;  Science of Pure Time.  -- Hamilton, 1837&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not until the much later development of set-theory was it explicitly realized &#xA;that ordered pairs and n-tuples serve a fundamental foundational role.&#xA;Indeed, as Akihiro Kanamori wrote on p. 289 (17) of &#xA;his very interesting &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.95.9839&quot; rel=&quot;nofollow&quot;&gt;paper [2]&lt;/a&gt; on the history of set theory: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1897 Peano explicitly formulated the ordered pair using &#xA;  $(x; y)$ and moreover raised the two main points about the &#xA;  ordered pair: First, equation 18 of his Definitions stated &#xA;  the instrumental property which is all that is required of &#xA;  the ordered pair: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ (x,y) = (a,b)  \iff  x = a \;\;{\rm and}\;\; y = b $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Second, he broached the possibility of reducibility, writing: &#xA;  &quot;The idea of a pair is fundamental, i.e., we do not know how &#xA;  to express it using the preceding symbols.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once set-theory was fully developed one had the raw materials &#xA;(syntax and semantics) to provide rigorous constructions of &#xA;algebraic structures and precise languages for term algebras. The polynomial ring $R[x]$ is nowadays just a special case of much more general constructions of free algebras. Such equationally axiomatized algebras and their genesis via so-called 'universal mapping properties'  are topics discussed at length in any course on Universal Algebra - &#xA;e.g. see &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;Bergman [3]&lt;/a&gt; for a particularly lucid presentation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] William Rowan Hamilton. Theory of conjugate functions, or algebraic couples; with a preliminary and elementary essay on algebra as the science of pure time&lt;br&gt;&#xA;Trans. Royal Irish Academy, v.17, part 1 (1837), pp. 293-422.)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/PureTime/PureTime.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/PureTime/PureTime.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Akihiro Kanamori. The Empty Set, the Singleton, and the Ordered Pair&lt;br&gt;&#xA;The Bulletin of Symbolic Logic, Vol. 9, No. 3. (Sep., 2003), pp. 273-298.&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.95.9839&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.95.9839&lt;/a&gt;&lt;br&gt;&#xA;PS  &lt;a href=&quot;http://www.math.ucla.edu/~asl/bsl/0903/0903-001.ps&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~asl/bsl/0903/0903-001.ps&lt;/a&gt;&lt;br&gt;&#xA;PDF &lt;a href=&quot;http://ifile.it/b20c48j&quot; rel=&quot;nofollow&quot;&gt;http://ifile.it/b20c48j&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] George M. Bergman. An Invitation to General Algebra and Universal Constructions.&lt;br&gt;&#xA;PS  &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;http://math.berkeley.edu/~gbergman/245/&lt;/a&gt;&lt;br&gt;&#xA;PDF &lt;a href=&quot;http://ifile.it/yquj5w1&quot; rel=&quot;nofollow&quot;&gt;http://ifile.it/yquj5w1&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-01T19:41:44.703" LastActivityDate="2010-12-01T19:41:44.703" CommentCount="3" />
  <row Id="2659" PostTypeId="2" ParentId="2655" CreationDate="2010-08-17T16:22:14.383" Score="2" ViewCount="0" Body="&lt;p&gt;What you are looking for is the &quot;Prime Counting Function&quot;.  The closest thing to it you will find online is Wolfram Alpha:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/examples/PrimeNumbers.html&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/examples/PrimeNumbers.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1280" LastActivityDate="2010-08-17T16:22:14.383" />
  <row Id="2660" PostTypeId="2" ParentId="2650" CreationDate="2010-08-17T16:37:10.110" Score="7" ViewCount="0" Body="&lt;p&gt;I will address your first question, but not the one for larger number of rods; as far as I know, it's still generally wide open what the optimal strategy might be even for $4$ rods and a smallish number of disks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the optimal strategy for $n$ disks in $3$ rods is the &quot;usual&quot; one, you can do it by induction (which yields a recursive solution). I'm sure there are other ways of proving it, perhaps with Lucas numbers as you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the optimal strategy with $n=1$ is to simply move the disk directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you already have the optimal strategy for moving $k$ disks. To move $k+1$ disks, you need to move the largest disk from the initial rod to the terminal rod, but that is the only time it needs to move (it cannot help you with the other disks, since it must lie at the bottom at any given time, so any other moves only require further moves in the end); to move the bottom ($k+1$)st disk from the initial rod $I$ to the terminal rod $T$, you must first move the top $k$ disks out of the way; this requires moving the $k$ disks from the initial rod $I$ to the auxiliary rod $A$, and the optimal way of doing this is the optimal strategy you know for $k$ disks. Then you move the $k+1$st disk, and then you want to move the remaining $k$ disks from the auxiliary rod to the terminal one in as few moves as possible (the optimal way). So the optimal strategy for $k+1$ disks is to move the top $k$ using the optimal strategy for $k$ from $I$ to $A$, then move the largest disk from $I$ to $T$, then move the top $k$ disks using the optimal strategy for $k$ from $A$ to $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By keeping track of the actual number of moves needed at each step, you can give the number. For $n=1$, the number is $1=2^1-1$. Assume that moving $k$ disks requires $2^k-1$ moves in the optimal strategy. The optimal strategy for $k+1$ described above takes $(2^k-1) + 1 + (2^k-1) = 2^k+2^k - 1 = 2^{k+1}-1$ steps; thus, the optimal solution for $n$ disks and $3$ rods requires $2^n-1$ moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This does not generalize easily to more than $3$ rods for presumably obvious reasons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more interesting is trying to prove that the non-recursive solution gives an optimal solution; this solution only requires you to remember the last disk you moved at any given time (the recursive solution is more memory intensive, of course). Number the rods $0$, $1$, and $2$. We have three rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Never move the same disk twice in succession.&lt;/li&gt;&#xA;&lt;li&gt;You can only move a disk from the top of one rod to the top of another rod.&lt;/li&gt;&#xA;&lt;li&gt;Moving a disk from rod $i$ to rod $j$ is only valid if $i\neq j$, and either rod $j$ is empty or the top disk in rod $j$ is larger than the top disk in rod $i$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With these rules, the non-recursive algorithm has two simple steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you are moving the disk from rod $i$, and the two other rods are valid destinations, then move it to rod $i+1\mod 3$. Otherwise, move it to the only valid destination.&lt;/li&gt;&#xA;&lt;li&gt;If no move is possible, stop. Otherwise, continue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This process will solve the puzzle with $3$ rods in the minimum number of moves.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-17T16:37:10.110" CommentCount="4" />
  <row Id="2661" PostTypeId="2" ParentId="2626" CreationDate="2010-08-17T16:57:49.430" Score="1" ViewCount="0" Body="&lt;p&gt;Since you are asking for just a place to start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try Yakov Perelman, &quot;Physics can be fun&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book has two volumes. It has an article somewhere, the precise location of which I am unable to locate as I do not have the book handy, on how to explain the nautical method of &quot;tacking&quot;, ie sailing against the wind. It explains the forces on the sails using diagrams, etc.. It has also many other articles on nautical themes. I think this book is the best place to start for an amateur.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-17T16:57:49.430" />
  <row Id="2662" PostTypeId="2" ParentId="2633" CreationDate="2010-08-17T17:24:17.560" Score="1" ViewCount="0" Body="&lt;p&gt;It's a classic exercise, see for example Rudin's Real and Complex analysis (3rd ed.), exercise 14, chap.3, p. 72. There are detailed step-by-step hints, you should be fine.&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-17T17:24:17.560" CommentCount="1" />
  <row Id="2663" PostTypeId="1" AcceptedAnswerId="2667" CreationDate="2010-08-17T18:13:51.290" Score="1" ViewCount="245" Body="&lt;p&gt;I was wondering if the following is true:&#xA;In a topological space with partial order, the inf and sup for a closed subset are achievable inside the subset so that they become minimum and maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it still true if I replace &quot;a topological space with partial order&quot; with &quot;a topological space with total order&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a topological space, what other kinds of condition can make inf and sup of a subset achievable in itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an &quot;ordered&quot; (not sure what kinds of order is proper here) topological space, are inf and sup of a subset accumulation points of the subset?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More questions again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Euclidean space, are inf and sup for a closed subset inside the subset? Are they accumulation points of the subset? What if in metric space? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-25T17:06:04.900" LastActivityDate="2010-11-25T17:06:04.900" Title="when inf and sup of a subset achievable in itself?" Tags="&lt;topology&gt;&lt;general-topology&gt;&lt;order-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2664" PostTypeId="1" AcceptedAnswerId="2665" CreationDate="2010-08-17T18:21:29.653" Score="2" ViewCount="179" Body="&lt;p&gt;In the article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_space&quot; rel=&quot;nofollow&quot;&gt;metric space&lt;/a&gt; on wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If X is a complete subset of the metric space M, then X is closed in M. Indeed, a space is complete iff it is closed in any containing metric space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do the second sentences say that: &#xA;in a metric space, a subset is complete if and only if the subset is closed in the metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what is the relation between completeness and closedness for a subset in a metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-08-17T18:27:18.033" Title="completeness and closedness for a subset in a metric space" Tags="&lt;topology&gt;&lt;closed&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2665" PostTypeId="2" ParentId="2664" CreationDate="2010-08-17T18:24:18.617" Score="3" ViewCount="0" Body="&lt;p&gt;Notice: A subset is complete iff it is closed in &lt;em&gt;any&lt;/em&gt; metric space in which it is a subspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed subspace of a complete metric space is complete. On the other hand, a closed subspace of a metric space need not be complete. Take the space $(0,1)$ which is closed in itself. But it is not complete. So closedness is not enough for completeness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a compact metric space is always complete. Compactness is the right property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this and more, try out G. F. Simmons, &quot;Introduction to Topology and Modern Analysis&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-17T18:24:18.617" CommentCount="2" />
  <row Id="2666" PostTypeId="2" ParentId="2664" CreationDate="2010-08-17T18:27:18.033" Score="3" ViewCount="0" Body="&lt;p&gt;No; if that were the case, then $\mathbb{Q}$ would be complete, since it is closed in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the second sentence says is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ be a metric space. Then $X$ is complete if and only if for every metric space $M$, if $X$ is contained in $M$ then $X$ is closed in $M$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, if you have a fixed complete $X$ and a fixed metric space $M$ that contains $X$, then you can conclude that $X$ is closed in $M$ (the first sentence).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-17T18:27:18.033" CommentCount="1" />
  <row Id="2667" PostTypeId="2" ParentId="2663" CreationDate="2010-08-17T18:36:52.023" Score="4" ViewCount="0" Body="&lt;p&gt;1 and 2:  No.  Take an interval $(\alpha, \beta)$ in $\mathbb{Q}$ where $\alpha, \beta$ are irrational.  You can verify that such an interval is closed and that it contains neither its infimum nor its supremum.  The property you want, for totally ordered sets, is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_order#Completeness&quot; rel=&quot;nofollow&quot;&gt;completeness&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3:  I will assume you mean a totally ordered set with its &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_topology&quot; rel=&quot;nofollow&quot;&gt;order topology&lt;/a&gt;.  In that case, this is not always true if the infimum or supremum are contained in the subset but true otherwise; this is just a matter of working through the definitions.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-18T00:30:40.383" LastActivityDate="2010-08-18T00:30:40.383" CommentCount="12" />
  <row Id="2668" PostTypeId="2" ParentId="11" CreationDate="2010-08-17T18:58:40.717" Score="0" ViewCount="0" Body="&lt;p&gt;This should have been a comment; but I post as an answer anyway. I just want to remark that in spite of all the criticisms, all the statements in the most popular proof in the list above are true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, in the real number system, it is indeed true that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1/3 = 0.333333...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is again true that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0.33333.... * 3 = 0.99999....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would still stand by the superiority of Noah Snyder's answer, however. That's the one addressing the subtleties.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-08-17T18:58:40.717" CommunityOwnedDate="2010-08-17T18:58:40.717" />
  <row Id="2669" PostTypeId="2" ParentId="2435" CreationDate="2010-08-17T20:27:29.120" Score="-1" ViewCount="0" Body="&lt;p&gt;I am not at all sure that this would work, but it seems reasonable... (if it doesn't work, I'd like to know why)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can very simply use Monte Carlo integration to approximate pi by setting up a known area and inscribing a circle in it, and sampling the entire area with a &lt;em&gt;uniformly distributed sequence*&lt;/em&gt;, and considering the ratio of &quot;hits&quot; to the total number of &quot;shots&quot;.  (I elaborated on this slightly in &lt;a href=&quot;http://math.stackexchange.com/questions/297/simple-numerical-methods-for-calculating-the-digits-of-pi/372#372&quot;&gt;this answer&lt;/a&gt; to another question.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;* I think the uniform distribution of points is necessary in order to guarantee that there is no bias toward any particular area of the sampling space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is indeed the salient feature of uniformly distributed sequences in this context, then what would happen if we sampled with a non-uniformly distributed sequence?  I presume that we would not approximate the digits of pi very well... If this turns out to be true, then you may throw your sequence at a Monte Carlo problem with a known solution and see how it does.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a quick Matlab script approximating pi using a 2-d square.  In principle, this could be done in any dimension.  If you just have a sequence of points (1-d) you could try to approximate the length of a given segment in an interval of known length.  It also seems to me that the more points you have the better idea you will get to how far your points are from being uniformly distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Let me know if something didn't make sense here and I will try to clarify.)&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-17T20:27:29.120" CommentCount="3" />
  <row Id="2670" PostTypeId="1" AcceptedAnswerId="2678" CreationDate="2010-08-17T20:29:40.317" Score="7" ViewCount="88" Body="&lt;p&gt;Let $X$ and $Y$ be Banach spaces. A bounded operator $T\colon X\to Y$ is called Fredholm iff&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The dimension of $\ker(T)$ is finite,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The codimension of the image $\mathrm{im}(T)$ is finite,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The image $\mathrm{im}(T)$ is closed in $Y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;em&gt;Is the third condition redundant?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some lecture notes I'm working through claim that the third condition follows from the second one together with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_mapping_theorem_%28functional_analysis%29&quot; rel=&quot;nofollow&quot;&gt;open mapping theorem&lt;/a&gt;. I've checked some books on functional-analysis without finding a proof of this.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-17T21:44:21.037" LastActivityDate="2010-08-17T22:41:23.163" Title="Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" />
  <row Id="2671" PostTypeId="2" ParentId="2655" CreationDate="2010-08-17T21:01:43.203" Score="3" ViewCount="0" Body="&lt;p&gt;You can use the function prime_pi in Sage (http://sagemath.org), which is also available for free online at &lt;a href=&quot;http://sagenb.org&quot; rel=&quot;nofollow&quot;&gt;http://sagenb.org&lt;/a&gt;.   For example, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   sage: prime_pi(2011)&#xA;   305&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Like Mathematica, Sage's prime_pi function is too slow to solve your problem above.   It's also somewhat slower than Mathematica's still.  &lt;/p&gt;&#xA;" OwnerUserId="1286" LastActivityDate="2010-08-17T21:01:43.203" />
  <row Id="2672" PostTypeId="2" ParentId="1774" CreationDate="2010-08-17T21:08:43.110" Score="1" ViewCount="0" Body="&lt;p&gt;My favorite elementary number theory book is the one I published with Springer: &lt;a href=&quot;http://wstein.org/ent/&quot; rel=&quot;nofollow&quot;&gt;http://wstein.org/ent/&lt;/a&gt;   This isn't a completely shameless plug, because I was just allowed to release the PDF version legally for free (available at the above URL), which the original poster might appreciate. &lt;/p&gt;&#xA;" OwnerUserId="1286" LastActivityDate="2010-08-17T21:08:43.110" CommunityOwnedDate="2010-08-17T21:08:43.110" />
  <row Id="2673" PostTypeId="2" ParentId="1774" CreationDate="2010-08-17T21:24:49.773" Score="0" ViewCount="0" Body="&lt;p&gt;There is a list at &lt;a href=&quot;http://www.imomath.com/index.php?options=mbb|knjige&amp;amp;p=0&quot; rel=&quot;nofollow&quot;&gt;http://www.imomath.com/index.php?options=mbb|knjige&amp;amp;p=0&lt;/a&gt; . There's also the book   &lt;a href=&quot;http://www.worldscibooks.com/mathematics/7242.html&quot; rel=&quot;nofollow&quot;&gt;Problems of number theory in mathematical competitions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-17T21:24:49.773" CommunityOwnedDate="2010-08-17T21:24:49.773" />
  <row Id="2674" PostTypeId="2" ParentId="2435" CreationDate="2010-08-17T21:39:49.947" Score="3" ViewCount="0" Body="&lt;p&gt;You could use a chi square test.  Basically, this tests whether the number of draws that fall into various intervals is consistent with a uniform random distribution.   See Knuth's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201896842&quot; rel=&quot;nofollow&quot;&gt;TAOCP volume 2, Seminumerical Algorithms&lt;/a&gt; for an explanation of the chi-square test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: If the chi square test is too complicated, you could plot a histogram of your output and eyeball it to see if it looks uniform.  But if you'd like to quantify whether it's even enough, that's exactly what a chi square test is.  The test will also tell you whether the data are &lt;em&gt;too&lt;/em&gt; evenly spread out, i.e. not random enough.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-08-18T12:24:23.793" LastActivityDate="2010-08-18T12:24:23.793" CommentCount="6" />
  <row Id="2675" PostTypeId="2" ParentId="2645" CreationDate="2010-08-17T21:43:24.767" Score="2" ViewCount="0" Body="&lt;p&gt;It depends on what else you know about the relationship between the variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the correlation coefficient is the full extent of your information, then the approximation is unsafe, as Noldorin points out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, you have good external evidence that the coefficient adequately captures the level of a small linear relationship (eg, a slight dependence on some third quantity that is not germane to your analysis), then it may well be valid to approximate them as independent for some purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RVs about which you know nothing are useful abstractions -- and this is, after all, the maths site -- but real world data often exist in less of vacuum. If you're analysing in the context of a model, that may help you to work out what approximations you can get away with.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-08-17T21:43:24.767" CommentCount="1" />
  <row Id="2676" PostTypeId="2" ParentId="1774" CreationDate="2010-08-17T21:44:34.547" Score="0" ViewCount="0" Body="&lt;p&gt;Davenport's The Higher Arithmetic was my first number theory book. I think its very accessible to a high school student or beginning undergraduate student. It's quite short and very quickly readable.&#xA;If you find this treatment too informal, Niven and Zuckerman's an Introduction to the Theory of Numbers is a standard text that I think is a very well written undergraduate text, but this has already been mentioned.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-17T21:44:34.547" CommunityOwnedDate="2010-08-17T21:44:34.547" />
  <row Id="2677" PostTypeId="2" ParentId="2670" CreationDate="2010-08-17T21:54:05.153" Score="1" ViewCount="0" Body="&lt;p&gt;The third condition is redundant.  The proof I've seen is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout $X,Y$ are Banach spaces and $E$ is a closed subspace of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: if $F \subset X$ is finite dimensional then $E+F$ is closed and the image of $E$ in $X/F$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;proof: reduce to the case $\dim F = 1$, i.e. $F = \mathbb{C} x_0$ with $x_0 \not \in E$ (otherwise $E+F$ is trivially closed).  We have to show every convergent subsequence converges to a point in $E + F$.  Let $z_n = y_n + c_n x_0$ be such a sequence with limit $z_0$.  Let $\delta = dist(x_0, E) &gt; 0$.  Now $(z_n)$ is Cauchy so $|z_n - z_m| \to 0$ for $n,m$ sufficiently large.  That is, $|(c_n - c_m)x_0 - (y_n -y_m)| \ge |c_n -c_m|\delta \to 0$.  It follows that $(c_n)$ is Cauchy so converges to $c_0$.  So $y_n$ converges to $z_0 - c_0x_0 \in E$ hence $z_0 \in E + F$ as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $E+F$ is a Banach space hence so is $E+F/F = Im(E) \in X/F$ which says exactly that its closed in $X/F$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can show $TE$ is closed in $Y$.  Let $y_0 + TX, ..., y_n + TX$ be a basis for $Y/TX$ and $Y_0 = span(y_i)$.  Consider $T' \colon X \oplus Y_0 \to Y$ given by $T'(x\oplus y) = Tx + y$.  This is surjective so $T'' \colon X \oplus Y_0/(\ker T \oplus 0) \to Y$ is invertible.  By the lemma $E$ is closed in $X \oplus Y_0/(\ker T \oplus 0)$ and maps to $TE$ under the invertible map $T''$. QED.  &lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-17T21:54:05.153" />
  <row Id="2678" PostTypeId="2" ParentId="2670" CreationDate="2010-08-17T22:06:44.210" Score="5" ViewCount="0" Body="&lt;p&gt;Assume that $T$ is injective, because if it is not we know that $\textrm{ker}(T)$ is closed so we can replace $T$ by the induced map from $X/\textrm{ker}(T)$ (which is a Banach space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define $T' := X \oplus C \to Y$ by $T'(x, c) = T(x) + c$ where $C$ is a closed complement for the range of $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $T'$ is clearly bounded, linear and an isomorphism. So by the open mapping theorem $T'$ is open. Note that $\textrm{im}(T) = T'(X \oplus {0})$ which is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Offtopic: (How do you guys use LaTeX in the posts? I can't find it in the FAQ)&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-17T22:41:23.163" LastActivityDate="2010-08-17T22:41:23.163" CommentCount="2" />
  <row Id="2679" PostTypeId="2" ParentId="1413" CreationDate="2010-08-17T22:50:03.043" Score="2" ViewCount="0" Body="&lt;p&gt;The construction of the real field from the rationals using Cauchy sequences can be mimicked to construct other (characteristic zero) complete fields not isometric to the reals. Namely, instead of starting out with the metric defined by the usual (archimedean) absolute value, one can consider the $p$-adic metric ($p$ a fixed prime) and proceed in the same lines. A theorem of Ostrowski says that up to equivalence these are in fact the only ways to complete $\Bbb Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being complete, the fields ${\Bbb Q}_p$ thus obtained can be used to develop an analytic theory which is similar to the classical theory &quot;over $\Bbb R$&quot; but has some subtle differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A source of difference is that the fields ${\Bbb Q}_p$ have--unlikely $\Bbb R$--a natural subring, the radius $1$ sphere centered in $0$, a.k.a. the $p$-adic integers $\Bbb Z_p$, which is a local ring with residue field isomorphic to the field with $p$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a clue to the fact that &quot;$p$-adic analysis&quot; is deeply intertwined with the theory of finite fields.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-17T22:50:03.043" />
  <row Id="2680" PostTypeId="1" AcceptedAnswerId="2686" CreationDate="2010-08-18T00:23:22.067" Score="2" ViewCount="86" Body="&lt;p&gt;I need to create a simple formula for determining the popularity of an item based on votes and age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my current formula, which needs some work:&lt;br/&gt;&#xA;30 / (days between post date and now) * (vote count) = weighted vote&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever a vost is cast for an item it checks if its weighted vote is &gt; 300. If an item has a weighted vote more than 300 then it is promoted to the front page of my website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this formula makes it very hard for older items to be promoted.&lt;br/&gt;&#xA;30 / 1 day * 10 votes = 300 (promoted)&lt;br/&gt;&#xA;30 / 5 days * 15 votes = 90 (not promoted)&lt;br/&gt;&#xA;30 / 30 days * 30 votes = 30 (not promoted)&lt;br/&gt;&#xA;30 / 80 days * 40 votes = 15 (not promoted)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I alter the formula to make it relatively easier for older items to be promoted (IE. make the above four weighted values fairly close together)?&lt;/p&gt;&#xA;" OwnerUserId="1288" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T04:52:12.647" LastActivityDate="2010-08-18T16:36:45.980" Title="Popularity formula (using votes and age)" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2681" PostTypeId="1" CreationDate="2010-08-18T01:38:33.857" Score="0" ViewCount="85" Body="&lt;p&gt;Let $G$ be an infinite simple group. Suppose there is a subset $X$ of $G$ with&#xA;$$&#xA;X^g = gXg^{-1}=X\qquad (g \in G)&#xA;$$&#xA;which is closed under taking inverses and which even generates $G$ in finitely many steps:&#xA;$$&#xA;X^m = G&#xA;$$&#xA;for a suitable $m \ge 2.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We somehow suspect that $X$ might be equal to $G.$ &lt;i&gt;If so&lt;/i&gt;, we have to establish (in the end!) that $X$ is actually closed under the multiplication; this would make $X$ a nonidentity normal subgroup of $G,$ and hence $X=G.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose this direct approach meets serious difficulties. What are then possible &lt;i&gt;&lt;b&gt;indirect&lt;/b&gt;&lt;/i&gt; approaches here? Anyone who met this situation in their own research, or elsewhere is kindly requested to share their ideas. &lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-08-18T02:42:55.933" LastActivityDate="2010-08-18T02:42:55.933" Title="Inverse-closed subsets generating simple groups in finitely many steps" Tags="&lt;group-theory&gt;" CommentCount="7" />
  <row Id="2682" PostTypeId="1" AcceptedAnswerId="2685" CreationDate="2010-08-18T01:56:19.353" Score="1" ViewCount="167" Body="&lt;p&gt;I have several points on a graph. The graph is of a continuous curve that flows through points (0, inf), (11, 5000), and (3, 200000). Can I find the equation for that line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let's try this:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to find a function that given an input (x) becomes closer to infinity as x grows closer to 0, and closer to 0 the closer x gets to infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with &lt;code&gt;y = log(x)&lt;/code&gt; is the scale is to small (#1) and the curve it too pronounced (#2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constraints I would say &lt;code&gt;x &amp;gt;= 0 &amp;amp;&amp;amp; y &amp;gt;= 0&lt;/code&gt; at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this (for those who &lt;em&gt;must&lt;/em&gt; know or are simply curious) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a program that loops through an array of values a certain number of times. The number of times &lt;em&gt;should&lt;/em&gt; be dependent on the number of values in the array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, an array with 5000 values should be processed 11 times. An array with 250k values should be processed 3 times.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int y = // Function goes here, given x;&#xA;for(int i = 0; i &amp;lt; y; i++)&#xA;{&#xA;    // Process&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I've been playing with a graph utility&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.3}{\frac{log(x)}{24}} = y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gives me approximately what I'm looking for, though doesn't give a proper values as $x \to 0$ (or &gt; 4 actually). However &lt;em&gt;there has to be a &quot;proper&quot; way to do this&lt;/em&gt;. Rather then punch numbers in until it &quot;looks&quot; right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like the &quot;right&quot; way gives the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{C}{x^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $C$ and $n$ gives $C ~= 19.3896$ and $n ~= 0.3522$ using $(5, 11), (200, 3)$ as points for simplicity.&lt;/p&gt;&#xA;" OwnerUserId="1289" LastEditorUserId="1289" LastEditorDisplayName="" LastEditDate="2010-08-18T14:59:03.957" LastActivityDate="2010-08-18T14:59:03.957" Title="Fitting a curve to a decreasing data set undefined at zero" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="22" />
  <row Id="2683" PostTypeId="2" ParentId="1579" CreationDate="2010-08-18T02:11:55.883" Score="5" ViewCount="0" Body="&lt;p&gt;Such generalizations occur frequently as problems. Below are a couple of examples - which include references to prior literature, e.g. to an inclusion-exclusion max/min proof in Polya &amp;amp; Szego's problem book. They're problem E2229 from the Monthly 77 1970 p. 402; problem 1344 from Math. Mag. 64 1990 p. 134.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i33.tinypic.com/rlgdb9.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i38.tinypic.com/2i7mkvc.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i37.tinypic.com/5ck5ms.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-18T02:11:55.883" />
  <row Id="2685" PostTypeId="2" ParentId="2682" CreationDate="2010-08-18T03:44:50.280" Score="2" ViewCount="0" Body="&lt;p&gt;If y = 4525365/x^2.83916, that is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = \frac{4525365}{x^2.83916}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then when x = 3, y = 200000.02 and when x = 11, y = 5000.04 and when x = ∞, y = 0, and when x = 0, y = ∞.  It also satisfies y ≥ 0 whenever x &gt; 0 and x ≥ 0 whenever y &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the idea that the longer the array the less times you can afford to scan the array.  The amount of work might be equal to x*y, so you want to keep the amount of work constant, x*y = C, or y = C/x.  That does not quite fit your points, so I decided to use x^n*y = C instead.  I solved for n using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3^n*200000 = 11^n*5000, so 200000/5000 = 11^n/3^n, so 40 = (11/3)^n, so n = log(40)/log(11/3) ≈ 2.83916.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then C = 3^2.83916*200000 ≈ 4525365.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-18T03:44:50.280" CommentCount="5" />
  <row Id="2686" PostTypeId="2" ParentId="2680" CreationDate="2010-08-18T05:05:00.257" Score="2" ViewCount="0" Body="&lt;p&gt;One possibility is to take the cube root of the number of days:&#xA;$$\begin{align}&#xA;\frac{3}{\sqrt[3]{1\text{ day}}}\cdot 10\text{ votes}&amp;amp;=300&#xA;\\&#xA;\frac{3}{\sqrt[3]{5\text{ days}}}\cdot 15\text{ votes}&amp;amp;\approx 263.2&#xA;\\&#xA;\frac{3}{\sqrt[3]{30\text{ days}}}\cdot 30\text{ votes}&amp;amp;\approx 289.6&#xA;\\&#xA;\frac{3}{\sqrt[3]{80\text{ days}}}\cdot 40\text{ votes}&amp;amp;\approx 278.5&#xA;\\&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-18T05:05:00.257" />
  <row Id="2687" PostTypeId="1" AcceptedAnswerId="2689" CreationDate="2010-08-18T05:08:37.223" Score="5" ViewCount="120" Body="&lt;p&gt;I saw a statement on &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_space#Continuous_maps&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Metric_space#Continuous_maps&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;f is continuous if and only if it is continuous on every compact subset of M1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is stated for the case that f is a mapping between two metric spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the statement is true for mappings between general topological spaces?&#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the statement true for a mapping between two metric spaces? Can you give a sketch for the proof?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-08-18T06:52:51.053" LastActivityDate="2010-12-11T17:25:19.837" Title="Generalize continuity  from compact subsets to the whole domain space" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="2688" PostTypeId="2" ParentId="622" CreationDate="2010-08-18T05:25:06.697" Score="3" ViewCount="0" Body="&lt;p&gt;Representation theory is one of the keys to the Fourier transform on groups (and co-set spaces etc.). For this matter, I recommend this book:&#xA;&lt;a href=&quot;http://books.google.com/books?id=yanslHJYvAIC&amp;amp;pg=PA415&amp;amp;lpg=PA415&amp;amp;dq=sugiura+representaion+theory&amp;amp;source=bl&amp;amp;ots=j8sf2rsfnD&amp;amp;sig=Z4wjbz316q-yfVh5WO0DDQcl6Es&amp;amp;hl=en&amp;amp;ei=mWxrTP2eApCRjAfMuITiAQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBQQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Unitary representations and harmonic analysis: an introduction&#xA; By Mitsuo Sugiura&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Representation_theory&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; about representation theory, you will find several applications.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-18T05:25:06.697" />
  <row Id="2689" PostTypeId="2" ParentId="2687" CreationDate="2010-08-18T05:45:10.800" Score="8" ViewCount="0" Body="&lt;p&gt;The statement is false in general.  Let $X$ be an uncountable set with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cocountable_topology&quot; rel=&quot;nofollow&quot;&gt;cocountable topology&lt;/a&gt;.  Then all the countable subsets of $X$ have the discrete topology and are therefore closed but not compact, so the only compact subsets of $X$ are the finite subsets, which also carry the discrete topology.  It follows that every function $X \to Y$, where $Y$ is any topological space, is continuous on compact subsets of $X$.  (But, for example, if $Y$ has the same underlying set with the discrete topology, the &quot;identity&quot; function $X \to Y$ is not continuous.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  In response to your second question, the property you are looking at is equivalent to being &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactly_generated_space&quot; rel=&quot;nofollow&quot;&gt;compactly generated&lt;/a&gt;.  A topological space $X$ is compactly generated if it satisfies the following condition: a subset of $X$ is open if and only if its intersection with every compact subset is open.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are metric spaces compactly generated?  Recall that a space is &lt;a href=&quot;http://en.wikipedia.org/wiki/First-countable_space&quot; rel=&quot;nofollow&quot;&gt;first-countable&lt;/a&gt; if every point has a countable neighborhood basis.  Metric spaces are first-countable because at any point the sequence of open balls of radius $\frac{1}{n}, n \in \mathbb{N}$ forms a countable neighborhood basis.  And first-countable spaces are always compactly generated; see the proof &lt;a href=&quot;http://topospaces.subwiki.org/wiki/First-countable_implies_compactly_generated&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-18T07:04:40.013" LastActivityDate="2010-08-18T07:04:40.013" CommentCount="1" />
  <row Id="2690" PostTypeId="2" ParentId="622" CreationDate="2010-08-18T06:12:16.750" Score="2" ViewCount="0" Body="&lt;p&gt;Certainly, group actions used to model dynamical systems, for instance, Markov Chains, find extremely handy the results from representation theory.&#xA;For example, suppose you have a deck of cards, and every time you choose two cards at random and transpose them. How long you have to wait for the deck to be randomized? The (exact) answer to this question lies exclusively in the area of representation theory! See for example, &lt;a href=&quot;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521883368&quot; rel=&quot;nofollow&quot;&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521883368&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1295" LastActivityDate="2010-08-18T06:12:16.750" />
  <row Id="2691" PostTypeId="2" ParentId="2687" CreationDate="2010-08-18T06:55:27.180" Score="4" ViewCount="0" Body="&lt;p&gt;Let $X$ be a topological space.&#xA;Then $X$ has the property that a function $f:X\to Y$&#xA;(where $Y$ is an abritrary topological space) is continuous if&#xA;its restriction to each compact subspace is continuous if and only if&#xA;$X$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactly_generated_space&quot; rel=&quot;nofollow&quot;&gt;compactly generated&lt;/a&gt; topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space $X$ is compactly generated if a subset $A$ whose intersection&#xA;with any compact subspace $K$ of $X$ is open in $K$, is open in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most familiar topological spaces are compactly generated&#xA;(Euclidean spaces, manifolds, CW-complexes etc.) but not all;&#xA;see Qiaochu's answer for an example.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-18T06:55:27.180" />
  <row Id="2692" PostTypeId="2" ParentId="622" CreationDate="2010-08-18T07:02:16.650" Score="12" ViewCount="0" Body="&lt;p&gt;One comment about your sentence &quot;this seems to only trivially invoke representation theory&quot;.&#xA;It might be surprising, but such obvious representations are actually the source of much&#xA;interesting mathematics, and a lot of effort of representation theorists is devoted to studying them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: the general set up here is that we have a group (in your example $SO(n)$) acting&#xA;on a space $X$ (in your example $\mathbb R^n$), and we look at the space of functions on $X$&#xA;(let me write it $\mathcal F(X)$; in a careful treatment, one would have to think about whether&#xA;we wanted continuous, smooth, $L^2$, or some other kind of functions, but I will suppress that&#xA;kind of technical consideration here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as you observe, there is a natural representation of $G$ on $\mathcal F(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that from a certain point of view this seems trivial, because the representation is obvious.  Unlike when one first learns rep'n theory of finite groups,&#xA;where one devotes a lot of effort to constructing reps., in this context, the rep. stares you in&#xA;the face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can this be interesting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the representation $\mathcal F(X)$ will almost never be irreducible, so an important&#xA;question becomes: how does this representation decompose into irreducible representations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly we have a quite non-trivial representation theoretic problem:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we have to work out the list of irreducible representations of $G$ (which is much&#xA;like what one does in a first course on rep'n theory of finite groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, we have to figure out how $\mathcal F(X)$ decomposes, which involves representation theory (among other things, you have to develop methods for investigating this sort of question), and also often a lot of analysis (because typically $\mathcal F(X)$ will be&#xA;infinite dimensional, and may be a Hilbert space, or have some other similar sort of topological vector space structure which should be incorporated into the picture). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I should say too much more here, but I will just give some illustrative examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) If $ X = G = S^1$ (the circle group, say thought of as $\mathbb R/\mathbb Z$)&#xA;acting on itself by addition, then the solution to the problem of decomposing&#xA;$\mathcal F(S^1)$ is the theory of Fourier series.  (Note that a function on $S^1$ is the&#xA;same as a periodic function on $\mathbb R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If $ X = G = \mathbb R$, with $G$ acting on itself by addition, then&#xA;the solution to the above question (how does $\mathcal F(\mathbb R)$ decompose under the&#xA;action of $\mathbb R$) is the theory of the Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $ X = S^2$ and $G = SO(3)$ acting on $X$ via rotations, then decomposing&#xA;$\mathcal F(S^2)$ into irreducible representations gives the theory of spherical harmonics.&#xA;(This is an important example in quantum mechanics; it comes up for example in the theory&#xA;of the hydrogen atom, when one has a spherical symmetry because the electron orbits the&#xA;nucleus, which one thinks of as the centre of the sphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) If $ X = SL_2(\mathbb R)/SL_2(\mathbb Z)$ (this is the quotient of a Lie group&#xA;by a discrete subgroup, so is naturally a manifold, in this case of dimension 3),&#xA;with $G = SL_2(\mathbb R)$ acting by left multiplication, then the problem of decomposing&#xA;$\mathcal F(X)$ leads to the theory of modular forms and Maass forms, and is the first&#xA;example in the more general theory of automorphic forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Looking over the other answers, I see that this is an elaboration on  AD.'s answer.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-08-18T08:33:03.437" LastActivityDate="2010-08-18T08:33:03.437" CommentCount="0" />
  <row Id="2693" PostTypeId="2" ParentId="2655" CreationDate="2010-08-18T07:03:36.457" Score="2" ViewCount="0" Body="&lt;p&gt;If an approximation suffices, you could use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem#The_prime-counting_function_in_terms_of_the_logarithmic_integral&quot; rel=&quot;nofollow&quot;&gt;offset logarithmic integral function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-18T07:03:36.457" />
  <row Id="2694" PostTypeId="1" AcceptedAnswerId="10608" CreationDate="2010-08-18T07:47:43.093" Score="10" ViewCount="745" Body="&lt;p&gt;I have been fascinated by this problem since I first heard about it in high school.  From the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Collatz_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Collatz_problem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take any natural number n. If n is even, divide it by 2 to get n / 2, if n is odd multiply&#xA;  it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is&#xA;  that no matter what number you start with, you will always eventually reach 1. [...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Paul Erdős said about the Collatz conjecture: &quot;Mathematics is not yet ready for such&#xA;  problems.&quot; He offered $500 USD for its solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is: how important do you consider the answer to this question, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And would you speculate on what might have possessed Paul Erdős to make such an offer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Is there any reason to think that a proof of the Collatz Conjecture would be&#xA;complex (like FLT) rather than simple (like PRIMES is in P)?  And can this&#xA;characterization of FLT vs. PRIMES is in P be made more specific than a bit-length comparison?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-01T02:59:39.917" LastActivityDate="2010-11-16T20:06:52.173" CommunityOwnedDate="2010-08-18T14:51:51.837" Title="What is the importance of the Collatz conjecture?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="2" />
  <row Id="2695" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T07:53:21.800" Score="-2" ViewCount="0" Body="&lt;p&gt;I would say that it is a very interesting question, so simple and yet so hard to prove.&#xA;That is what all important questions should be like, but compared to &quot;Why are we here?&quot;, &#xA;we might actually get an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, one can spend a great deal of time trying to prove it,&#xA;maybe Erdös spent too much time on it, so he just wanted a proof of it so that he could move on.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-08-18T07:53:21.800" CommunityOwnedDate="2010-08-18T14:51:51.837" />
  <row Id="2696" PostTypeId="2" ParentId="2645" CreationDate="2010-08-18T08:38:58.927" Score="2" ViewCount="0" Body="&lt;p&gt;The correlation coefficient only measures &lt;strong&gt;linear&lt;/strong&gt; dependence of two random variables. So if they depend on each other in a non-linear way, the correlation coefficient will not catch it.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-18T08:38:58.927" CommentCount="1" />
  <row Id="2698" PostTypeId="1" CreationDate="2010-08-18T09:19:22.427" Score="3" ViewCount="74" Body="&lt;p&gt;Suppose, we have f1(n) = o(f2(n)).. (small o)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have to compare f1(f2(n)) and f2(f1(n)) asymptotically..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we say about it? Which one will be greater &amp;amp; why?&lt;/p&gt;&#xA;" OwnerUserId="933" LastActivityDate="2010-08-18T18:10:31.223" Title="Comparing composite functions asymptotically" Tags="&lt;algorithms&gt;" AnswerCount="2" />
  <row Id="2699" PostTypeId="2" ParentId="2628" CreationDate="2010-08-18T09:36:34.793" Score="5" ViewCount="0" Body="&lt;p&gt;Okay ... here we go ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To save typing, I'm going to use $A$, $B$, and $C$ for $\theta$, $\phi$, and $\psi$. So, our integral starts as this ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int\int\int \left(2 \cos A \cos B - \sin A \sin B \cos C\right)^{2k} \sin A \sin B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and expands to this ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\int\int \sum_{j=0}^{2k} {2k \choose j} \; 2^j (-1)^{2k-j} \cos^j A \; \sin^{2k-j+1} A \; \cos^j B \; \sin^{2k-j+1} B \; \cos^{2k-j} C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral over the sum will be a sum of integrals over the terms; and, since the $A$, $B$, and $C$ factors are independent, each integrated term is a product of integrals. Moreover, as mentioned in my comment on the original question, we're integrating $\cos^{2k-j}C$ over the full period of cosine, so terms with odd powers of $\cos C$ --those corresponding to odd values of $j$-- vanish. So, we take $j=2h$ and for convenience define $m := k-h$, then we write ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{h=0}^{k} 2^{2h} {2k \choose 2h} \; \left( \int \cos^{2h} A \; \sin^{2m+1} A \right) \left( \int \cos^{2h} B \; \sin^{2m+1} B \right) \left( \int \cos^{2m} C \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two integrals are identical, so we're really looking at the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{h=0}^{k} 2^{2h} {2k \choose 2h} \; \left( \int \cos^{2h} A \; \sin^{2m+1} A \right)^2 \left( \int \cos^{2m} C \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the pieces ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the reduction formula for the integral of powers of cosine, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{2\pi} \cos^{2m}C \; dC = \frac{(2m-1)(2m-3)\cdots(1)}{(2m)(2m-2)\cdots(2)}\int_{0}^{2\pi} dC = \frac{2\pi(2m-1)!!}{2^{m} m!} = \frac{2\pi(2m)!}{2^{2m} m!^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x!!$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_factorial#Double_factorial&quot; rel=&quot;nofollow&quot;&gt;double factorial&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other factor ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\pi} \cos^{2h} A \; \sin^{2m+1} A \; dA = \int_{0}^{\pi} \cos^{2h} A \; \left( 1 - \cos^{2}A\right)^m \; \sin A \; dA = \int_{-1}^{1} u^{2h} \left( 1 - u^2 \right)^m \; du$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \int_{-1}^{1} u^{2h} \sum_{n=0}^{m} {m \choose n} (-1)^n u^{2n} \; du = \sum_{n=0}^{m} (-1)^n {m \choose n} \int_{-1}^{1} u^{2n+2h} \; du = \sum_{n=0}^{m} {m \choose n} \frac{2 (-1)^n}{2n+2h+1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica evaluates the sum into a form we manipulate into something we can use: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma(h+\frac{1}{2}) m!}{\Gamma(m+h+\frac{3}{2})}  = \frac{\sqrt{\pi} (2h-1)!!}{2^h}\frac{2^{k+1}m!}{\sqrt{\pi}(2k+1)!!}= \frac{(2h)!}{2^{2h}h!}\frac{2^{k+1}m!}{(2k+1)}\frac{2^{k}k!}{(2k)!}= \frac{2^{2m+1}m!}{2k+1}\frac{(2h)!}{h!}\frac{k!}{(2k)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to a simplification bonanza in the formula for the integral ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{h=0}^{k} \;\; 2^{2h} \;\;\;\; \frac{(2k)!}{(2h)!(2m)!} \;\;\;\;  \frac{2^{4m+2}m!^2}{(2k+1)^2}\frac{(2h)!^2}{h!^2}\frac{k!^2}{(2k)!^2} \;\;\;\; \frac{2\pi(2m)!}{2^{2m} m!^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 8 \pi \sum_{h=0}^{k} \frac{2^{2(m+h)}}{(2k+1)^2}\frac{(2h)!}{h!^2}\frac{k!^2}{(2k)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 8 \pi  \frac{2^{2k}}{(2k+1)^2} \frac{k!^2}{(2k)!} \sum_{h=0}^{k}\frac{(2h)!}{h!^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 8 \pi  \left(\frac{2^{k}}{2k+1}\right)^2 {2k \choose k}^{-1} \;\; \sum_{h=0}^{k}{2h \choose h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... as expected (well, &quot;expected&quot; from your having previously forced it out of Mathematica).&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-18T09:36:34.793" CommentCount="2" />
  <row Id="2700" PostTypeId="2" ParentId="2698" CreationDate="2010-08-18T09:36:40.517" Score="0" ViewCount="0" Body="&lt;p&gt;What happens if $f(x)=x^t$ and $g(x)=e^x$&#xA;where $t$ is a positive real?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-18T09:36:40.517" />
  <row Id="2701" PostTypeId="2" ParentId="2628" CreationDate="2010-08-18T10:40:48.200" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Here is a hint&lt;/strong&gt; to an other approach: Search for the key words &quot;convolution formula&quot; &quot;Legendre polynomial&quot; &quot;addition formula&quot; and you will be en-lighted... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- A little explanation and some spherical harmonics ---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@J. Mangaldan You are totally right in particular when I was wrong in my suggestion! :/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought about was convolution on the double coset space $X=K\backslash G/K$ of $G= SO(3)$ where $K=SO(2)$. The space $X$ may be identified with $[-1,1]$ and the convolution (as induced from the translation operator on $G$) is then given by&#xA;$$f*g(x)= \frac{1}{4\pi}\int_{-1}^1\int_{0}^{2\pi}f(y)g(xy+\sqrt{1-x^2}\sqrt{1-y^2}\cos\theta) d\theta dy.$$&#xA;An interesting fact is that $f*g=g*f$ even though $G$ is not Abelian.&#xA;Choosing $f(x)=1$ and $g(x)=x^{2k}$ we get&#xA;$$\frac{1}{8\pi}\int_{-1}^1\int_{-1}^1\int_{0}^{2\pi}(xy+\sqrt{1-x^2}\sqrt{1-y^2}\cos\theta)^{2k} d\theta dydx=$$&#xA;$$\int_{-1}^1f*g(x)dx/2 = \int_{-1}^1 g*f(x)dx/2=\int_{-1}^1x^{2k}dx/2= 1/(2k+1).$$&#xA;In our case the integrand is (after changing variables) &#xA;$$(2xy + \sqrt{1-x²}\sqrt{1-y²}\cos\theta)^{2k}=$$&#xA;$$=\sum_{j=0}^{2k}{2k \choose j}(xy)^j(xy+\sqrt{1-x²}\sqrt(1-y²)\cos\theta)^{2k-j}$$&#xA;Using this we get an other convolution - but I can not see how this can be used directly. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legendre polynomials might be good since they are the characters on the space $X$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-21T08:29:46.533" LastActivityDate="2010-08-21T08:29:46.533" CommentCount="3" />
  <row Id="2702" PostTypeId="2" ParentId="2682" CreationDate="2010-08-18T10:57:09.520" Score="1" ViewCount="0" Body="&lt;p&gt;As everybody told you, there are infinitely many solutions for your problem. One particularly simple kind of them are functions like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = \frac{a}{x} + bx \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can obtain which constants $a, b$ fit with your pair of data $(11, 5.000)$ and $(3, 200.000)$ imposing them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;5.000      &amp;amp;=    \frac{1}{11}a + 11b&lt;br&gt;&#xA;\\ 200.000    &amp;amp;=    \frac{1}{3}a + 3b&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're getting a linear system of equations in which $a,b$ are the unknowns. Since $\frac{3}{11} - \frac{11}{3} \neq 0$, you have a unique solution.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-18T10:57:09.520" />
  <row Id="2703" PostTypeId="1" CreationDate="2010-08-18T10:58:26.740" Score="1" ViewCount="732" Body="&lt;p&gt;I have an examples book with a limit exercise that I can't understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit in question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x,y)=\frac{x}{x+y}$$&#xA;with $x\ne-y$;&#xA;$$\lim_{(x,y)\to(0,0)} f(x,y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then to solve it, it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{(x,y)\to(0,0)} f(x,y)&#xA;= \lim_{x\to0} f(x,mx) =\lim_{x\to 0}\frac{x}{x+mx}=\frac{1}{1+m}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me understand that?&#xA;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: Ok, just to make sure that I got it right. I have a very similar test exercise with 4 different options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following limit $$\lim_{(x,y)\to(0,0)}\frac{-x^3+3xy^2}{x^2+y^2}$$ equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. -infinite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. Doesn't exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. +infinite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My doubt is: if I consider it normally I'd say that it doesn't exist, but if I solve it using the same approach (i.e. y=mx) then the limit equals 0.&#xA;&lt;strong&gt;Which one is the right answer?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1298" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-19T21:54:07.533" LastActivityDate="2010-08-19T21:54:07.533" Title="0/0 limit question" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="2704" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T11:09:53.087" Score="8" ViewCount="0" Body="&lt;p&gt;The point is that this limit doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For it to exist, then if you take &lt;strong&gt;any&lt;/strong&gt; sequence of points $(x_n,y_n)$&#xA;converging to $(0,0)$ for which $f(x_n,y_n)$ is defined, then&#xA;$$\lim_{n\to\infty}f(x_n,y_n)$$&#xA;must exist and also be &lt;strong&gt;independent&lt;/strong&gt; of the sequence $(x_n,y_n)$.&#xA;The book is saying in essence that if you have a sequence of points of the form&#xA;$(x_n,m x_n)$ (so lying on the line with gradient $m$ though the origin)&#xA;and converging to the origin, then&#xA;$$\lim_{n\to\infty}f(x_n,y_n)=\frac1{1+m}\qquad\qquad\qquad(*)$$&#xA;which is &lt;strong&gt;not&lt;/strong&gt; independent of the sequence of points $(x_n,y_n)$,&#xA;as changing $m$ will change the limit $(*)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-18T11:09:53.087" CommentCount="7" />
  <row Id="2705" PostTypeId="2" ParentId="2474" CreationDate="2010-08-18T11:39:32.540" Score="0" ViewCount="0" Body="&lt;p&gt;This is a response to Robin Chapman's answer. (For some reason I am not able to ask this directly under his question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we get &lt;em&gt;that&lt;/em&gt; formula from van Kampen? The double torus is the union of the two open subsets that are homeomorphic to $T$ and whose intersection is $S^1$. So by van Kampen this should equal the colimit of $\pi_1(W)$ with $W \in {T,T,S^1}$. I thought the colimit in the category of groups is just the direct sum, hence the result should be $\pi_1(T) \oplus \pi_1(T) \oplus \pi_1(S^1)$.&lt;/p&gt;&#xA;" OwnerUserId="1253" LastActivityDate="2010-08-18T11:39:32.540" CommentCount="4" />
  <row Id="2706" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T12:08:30.610" Score="3" ViewCount="0" Body="&lt;p&gt;I will elaborate on Robin's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There is no&lt;/strong&gt; unique limit as (&lt;em&gt;x,y&lt;/em&gt;) goes to (0,0) --- &lt;em&gt;i.e.&lt;/em&gt; the limit does not exist. For instance, if we approach (0,0) on the &lt;em&gt;x&lt;/em&gt; axis, we obtain the limit&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim\limits_{(x,y) \to (0,0)} \dfrac{x}{x+y} = \lim\limits_{x \to 0} \dfrac{x}{x} = 1,$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and if we approach (0,0) on the &lt;em&gt;y&lt;/em&gt; axis, we obtain the limit&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim\limits_{(x,y) \to (0,0)} \dfrac{x}{x+y} = \lim\limits_{y \to 0} \dfrac{0}{y} = 0.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However,&lt;/strong&gt; we can still consider limits along different curves. The two above limits, along the &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; axes, are two examples: the limit you describe in your question is a similar limit, on the locus of the equation &lt;em&gt;y&lt;/em&gt;&amp;nbsp;=&amp;nbsp;&lt;em&gt;mx&lt;/em&gt;. &lt;em&gt;This is something different&lt;/em&gt; than whether there is &quot;a limit&quot; of f(&lt;em&gt;x,y&lt;/em&gt;) as we approach (0,0) --- in this case, because there are different limits depending on how we approach the origin, we may consider the question of what limit one has as one approaches the origin on a particular curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, a &quot;curve&quot; is a function  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\big(x(t),y(t)\big) = c(t)$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for some function $c: \mathbb R \to \mathbb R^2$. &quot;Approaching&quot; (0,0) along the curve c(&lt;em&gt;t&lt;/em&gt;) entails specifying some domain for c in which it does not cross itself (for the example of the line &lt;em&gt;y&lt;/em&gt;&amp;nbsp;=&amp;nbsp;&lt;em&gt;mx&lt;/em&gt; one could take c(&lt;em&gt;t&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;(&lt;em&gt;t&lt;/em&gt;,&amp;nbsp;&lt;em&gt;mt&lt;/em&gt;), in which case the domain can be the real numbers) and for which c(&lt;em&gt;t&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;(0,0) for some (unique) value &lt;em&gt;t&lt;/em&gt;&amp;nbsp;=&amp;nbsp;T. Then, evaluating the limit of f(&lt;em&gt;x,y&lt;/em&gt;) along the curve (&lt;em&gt;x,y&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;c(&lt;em&gt;t&lt;/em&gt;) means just taking the limit of the &lt;em&gt;composite&lt;/em&gt; function f(c(&lt;em&gt;t&lt;/em&gt;)) as &lt;em&gt;t&lt;/em&gt; approaches T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of approaching (0,0) on the curve c(&lt;em&gt;t&lt;/em&gt;)&amp;nbsp;=&amp;nbsp;(&lt;em&gt;t&lt;/em&gt;,&amp;nbsp;&lt;em&gt;mt&lt;/em&gt;), we just take T&amp;nbsp;=&amp;nbsp;0, and evaluate the limit of f(&lt;em&gt;t&lt;/em&gt;,&amp;nbsp;&lt;em&gt;mt&lt;/em&gt;) as &lt;em&gt;t&lt;/em&gt; approaches 0. Up to a change of variables, this is just what you have done above.&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="439" LastActivityDate="2010-08-18T12:08:30.610" CommentCount="3" />
  <row Id="2707" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T12:25:38.487" Score="3" ViewCount="0" Body="&lt;p&gt;Paul Erdös offered cash prizes for the solution to problems according to his assessment of their difficulty and importance. I believe his judgement of the difficulty of this problem has been shown to be correct by the fact that it remains open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this problem is very important in the sense that a large proportion of the people reading this response will have had a go at it, at one time or another. Thus its solution would be of interest to many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason for its importance is that, like Fermat's Last Theorem (Wiles's Theorem), it's easy to state and understand and thus has the potential to attract young people towards mathematics. I too learned about it in high school and could not resist its allure.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-18T22:07:05.553" LastActivityDate="2010-08-18T22:07:05.553" CommunityOwnedDate="2010-08-18T14:51:51.837" />
  <row Id="2708" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T12:36:34.347" Score="0" ViewCount="0" Body="&lt;p&gt;De l'Hôpital explains the last step.&lt;br&gt;&#xA;If both numerator and denominator approach zero, you can apply de l'Hôpital, provided the derivative of the functions in numerator and denominator exist for x=0, which is the case.&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-08-18T12:36:34.347" CommentCount="4" />
  <row Id="2709" PostTypeId="2" ParentId="2655" CreationDate="2010-08-18T12:50:38.967" Score="0" ViewCount="0" Body="&lt;p&gt;For primes smaller than your example you can use Wolfram|Alpha, as Adam S pointed out. Wolfram|Alpha has the prime Pi function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=pi(55252335667&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=pi(55252335667&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1302" LastActivityDate="2010-08-18T12:50:38.967" />
  <row Id="2710" PostTypeId="1" AcceptedAnswerId="2730" CreationDate="2010-08-18T12:55:12.593" Score="26" ViewCount="460" Body="&lt;p&gt;The &lt;em&gt;Mandelbrot set&lt;/em&gt; is the set of points of the complex plane whos orbits do not diverge. An point $c$'s &lt;em&gt;orbit&lt;/em&gt; is defined as the sequence $z_0 = c$, $z_{n+1} = z_n^2 + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of this set is well known, &lt;strong&gt;why is it that if you zoom into parts of the filaments you will find slightly deformed copies of the original shape&lt;/strong&gt;, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7HMU1.jpg&quot; alt=&quot;Detail of the Mandelbrot set&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I measured some points on the Mandelbrot, and the corresponding points from one of these smaller Mu-molecules. Comparing the orbit sequences it was possible to find points on each sequence which were very close - but this experiment did not really help me to understand anything new.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-20T20:08:03.883" Title="Why does the Mandelbrot set contain (slightly deformed) copies of itself?" Tags="&lt;fractals&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="2711" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T13:06:31.493" Score="4" ViewCount="0" Body="&lt;p&gt;For your second question: is&#xA;$$\lim_{(x,y)\to(0,0)}\frac{-x^3+3xy^2}{x^2+y^2}=0?$$&#xA;imagine a sequence of points $(x_n,y_n)$&#xA;converging to the origin, and think whether $f(x_n,y_n)\to0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be tempted to write these points in polar form:&#xA;$x_n=r_n\cos t_n$ and $y_n=r_n\sin t_n$ as then the denominator&#xA;becomes $r_n^2$ (nice) and moreover $r_n\to0$ (why?).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-18T13:13:25.613" LastActivityDate="2010-08-18T13:13:25.613" CommentCount="14" />
  <row Id="2712" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T13:13:11.853" Score="10" ViewCount="0" Body="&lt;p&gt;So many mathematicians, and famous ones among them, have tried various ways to attack this problem, and it is still as elusive as it was when first posed. So the importance of the problem is that genuinely new mathematical ideas will have to be created to solve it, and such ideas may be helpful in other domains where &quot;truly important&quot; problems are at stake. Note that Erdös himself has said something along the lines that &quot;we don't have the mathematics yet to solve this problem&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-08-18T13:13:11.853" CommunityOwnedDate="2010-08-18T14:51:51.837" CommentCount="3" />
  <row Id="2713" PostTypeId="2" ParentId="2703" CreationDate="2010-08-18T13:26:33.867" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; The limit becomes that of $x/f(x)$ along the path $y = f(x)-x$ where $f(x)\to 0$. But this limit can be whatever you desire, by choosing $f$ weaker, comparable, or stronger than $x$ at $0$. Explicitly: choosing $f$ to be one of $\; x^{1/2},\; x/c,\; x^2$ yields $\; x/f(x)\to 0,\; c,\; \infty\;$ respectively, as $x\to0^+$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-19T21:48:58.767" LastActivityDate="2010-08-19T21:48:58.767" CommentCount="3" />
  <row Id="2714" PostTypeId="2" ParentId="2698" CreationDate="2010-08-18T13:54:56.763" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $f_1 = \log x,\; f_2 = x^n \;$ for $\; n = 1/2,\; 1,\; 2\;$ shows that the answer is &quot;nothing&quot;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-18T13:54:56.763" />
  <row Id="2715" PostTypeId="1" AcceptedAnswerId="2717" CreationDate="2010-08-18T14:14:08.527" Score="4" ViewCount="223" Body="&lt;p&gt;This was asked to me by a friend. I tried this problem for a long time, but i couldn't solve it. Seems interesting though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that it is impossible to construct two different seven digit numbers, one of which is divisible by the other, out of the digits 1,2,3,4,5,6,7 (All seven digits must be in each number)&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-18T18:56:42.970" Title="Constructing a seven digit number such that one divides the other." Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2717" PostTypeId="2" ParentId="2715" CreationDate="2010-08-18T14:28:19.350" Score="10" ViewCount="0" Body="&lt;p&gt;Say the numbers are x and y and y divides x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the remainder upon dividing both numbers by 9. Both leave a remainder 1 (as sum of digits leaves a remainder 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So x = ky where 2 &amp;lt;= k &amp;lt;= 8 is not possible.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-18T14:28:19.350" CommentCount="2" />
  <row Id="2719" PostTypeId="1" AcceptedAnswerId="2724" CreationDate="2010-08-18T15:11:54.973" Score="14" ViewCount="262" Body="&lt;p&gt;consdier a cubic polynomial, p(x)=k(x-a)(x-b)(x-c) where k is some constant and a,b,c its 3 roots (not necessarily distinct, not necessarily real). It is very simple to show that if you average two roots of a cubic polynomial and compute the tangent line at their average that it will intersect the cubic polynomial at the remaining root. I tried to generalize this result to other odd degree polynomials and it did not seem to work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is so special about cubic polynomials that allows this to work that isn't true about higher degree odd polynomials?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T19:54:11.293" LastActivityDate="2010-08-18T19:54:11.293" Title="Averaging 2 roots of a cubic polynomial" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="2720" PostTypeId="2" ParentId="2715" CreationDate="2010-08-18T15:16:38.297" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;  $\; ax = b,\; (a,m)=1 \;\Rightarrow\; x = b/a \;\;{\rm mod}\; m \;$ is the &lt;strong&gt;unique&lt;/strong&gt; solution such that $\; \; 0\le x &amp;lt; m. \;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $\; m = 9, \;$ i.e. cast out nines. This is a ubiquitous technique for recreational problems of this ilk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CHALLENGE&lt;/strong&gt; Find a natural generalization to arbitrary length integers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-18T15:46:03.100" LastActivityDate="2010-08-18T15:46:03.100" />
  <row Id="2722" PostTypeId="2" ParentId="855" CreationDate="2010-08-18T17:27:14.237" Score="2" ViewCount="0" Body="&lt;p&gt;let prime number p=2k+1,&#xA;p^2-1=4k(k+1), thus: 8|p^2-1&#xA;by theorem, p^2=1(mod 3), thus 3|p^2-1&#xA;thus, 24|p^2-1&lt;/p&gt;&#xA;" OwnerUserId="1059" LastEditorUserId="1059" LastEditorDisplayName="" LastEditDate="2010-08-20T18:09:26.853" LastActivityDate="2010-08-20T18:09:26.853" CommentCount="2" />
  <row Id="2723" PostTypeId="1" AcceptedAnswerId="4569" CreationDate="2010-08-18T17:45:56.700" Score="1" ViewCount="234" Body="&lt;p&gt;I have a non-mathematician friend who is interested in re-learning algebra. I am more than happy to help, but I am in no position to judge what is a &lt;strong&gt;good&lt;/strong&gt; introductory text --- only to identify when a text is a very &lt;strong&gt;bad&lt;/strong&gt; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend is interested in starting from &quot;basics&quot; --- he's comfortable with order-of-operations, minus-times-minus-is-plus, and positive integer exponents. He's shaky with negative exponents, logarithms, and roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recommendations for good, freely-available resources?&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-18T20:02:44.677" LastActivityDate="2010-09-27T02:52:58.960" CommunityOwnedDate="2010-08-18T17:51:23.057" Title="Free introductory resources for learning algebra?" Tags="&lt;algebra-precalculus&gt;&lt;education&gt;&lt;learning&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="2724" PostTypeId="2" ParentId="2719" CreationDate="2010-08-18T17:52:48.137" Score="9" ViewCount="0" Body="&lt;p&gt;This result &lt;em&gt;does&lt;/em&gt; hold for higher-degree polynomials; you just didn't generalize it correctly.  The general statement is as follows.  Let $p(x) = (x - a) q(x)$ where $q(x)$ is &lt;em&gt;any&lt;/em&gt; differentiable function, and let $r$ be such that $q'(r) = 0$.  Then the tangent line at $(r, p(r))$ intersects the $x$-axis at $(a, 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly simple computation.  Since $p'(x) = (x - a) q'(x) + q(x)$, it follows that $p'(r) = q(r)$, so the tangent line at $(r, p(r))$ has slope $q(r)$ and hence it intersects the $x$-axis at $(r - \frac{p(r)}{q(r)}, 0) = (a, 0)$.  In particular, the above result holds for $q(x)$ a polynomial of any degree greater than or equal to $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Intuitively, when $q'(r) = 0$, the linear approximation to $p(x)$ at $r$ is $(x - a) q(r)$, so of course it has to hit the $x$-axis at $(a, 0)$.  In fact, this way of thinking about it tells you that the $r$ such that $q'(r) = 0$ are the only $r$ for which this occurs.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case that $q(x)$ is quadratic, $r$ happens to be equal to the average of the roots of $q$.  For higher-degree polynomials this is no longer the case.  Instead, all you know is that the roots of $q'(x)$ interlace between the roots of $q(x)$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-18T18:24:01.443" LastActivityDate="2010-08-18T18:24:01.443" />
  <row Id="2725" PostTypeId="2" ParentId="2723" CreationDate="2010-08-18T18:01:34.673" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.curriki.org/xwiki/bin/view/Main/WebHome&quot; rel=&quot;nofollow&quot;&gt;Curriki&lt;/a&gt; is supposed to be a good resource for this kind of stuff, but I haven't tried it out myself.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-18T18:01:34.673" CommunityOwnedDate="2010-08-18T18:01:34.673" />
  <row Id="2727" PostTypeId="2" ParentId="2723" CreationDate="2010-08-18T18:28:06.880" Score="0" ViewCount="0" Body="&lt;p&gt;There's a pretty good video course on YouTube EDU, &lt;a href=&quot;http://www.youtube.com/user/UMKC#grid/user/DE28CF08BD313B2A&quot; rel=&quot;nofollow&quot;&gt;&quot;College Algebra&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2010-08-18T18:28:06.880" CommunityOwnedDate="2010-08-18T18:28:06.880" />
  <row Id="2728" PostTypeId="2" ParentId="2715" CreationDate="2010-08-18T18:56:42.970" Score="1" ViewCount="0" Body="&lt;p&gt;7654321/1234567 = 6. something &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so x = ky where 2&amp;lt;=k&amp;lt;=6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sum of the digits are not divisible by 3 so no 3 and 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now x = ky where k = 2,4 or 5 so not possible.&lt;/p&gt;&#xA;" OwnerUserId="1093" LastActivityDate="2010-08-18T18:56:42.970" />
  <row Id="2729" PostTypeId="2" ParentId="2485" CreationDate="2010-08-18T19:28:46.130" Score="0" ViewCount="0" Body="&lt;p&gt;Maybe the following works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R_n$ be a countable basis of rectangles (for example rational coordinates) for $\mathbb{R}^2$. Define $f_0 = 0$ and given $f_n$ define $f_{n + 1}$ as $f_n$ outside $R_{n + 1}$ and $n + 1$ inside $R_{n + 1}$. $(f_n)$ is an increasing sequence of measurable functions so $f = \lim f_n$ is measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take any open rectangle $R$ and given $M &gt; 0$ take $n &gt; M$ such that $R_n \subset R$. Now $\mu(|f| &gt; M) \geq \mu(|f_n| &gt; M) \geq \textrm{area}(R_n) &gt; 0$. So it is essentially unbounded.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-08-18T19:28:46.130" />
  <row Id="2730" PostTypeId="2" ParentId="2710" CreationDate="2010-08-18T19:41:30.467" Score="9" ViewCount="0" Body="&lt;p&gt;I really like this question!  I can't yet upvote, so I'll offer an answer instead.  This is only a partial answer, as I don't fully understand this material myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f$ is a quadratic polynomial.  Suppose that there is an integer $n$ and a domain $U \subset \mathbb{C}$ so that the $n$-th iterate, $f^n$, restricted to $U$ is a &quot;quadratic-like map.&quot;  Then we'll call $f$ &lt;em&gt;renormalizable&lt;/em&gt;.   (See Chapter 7 of McMullen's book &quot;Complex dynamics and renormalization&quot; for more precise definitions.) Now renormalization preserves the property of having a connected Julia set.  Also the parameter space of &quot;quadratic-like maps&quot; is basically a copy of $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, fix a quadratic polynomial $f$ and suppose that it renormalizes.  Then, in the generic situation, all $g$ close to $f$ also renormalize using the same $n$ and almost the same $U$.  This gives a map from a small region about $f$ to the space of quadratic-like maps.  This gives a partial map from the small region to the Mandelbrot set and so explains the &quot;local&quot; self-similarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up: all of the quadratic polynomials in a baby Mandelbrot set renormalize and all renormalize in essentially the same way.  (I believe that there are issues as you approach the place where the baby is attached to the parent.)  Thus renormalization explains why the baby Mandelbrot set appears.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-08-19T22:30:51.903" LastActivityDate="2010-08-19T22:30:51.903" />
  <row Id="2731" PostTypeId="1" CreationDate="2010-08-18T19:45:10.507" Score="3" ViewCount="182" Body="&lt;p&gt;You know that a couple has two children. You go to the couple's house and one of their children, a young boy, opens the door. What is the probability that the couple's other child is a girl?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you list all possibilities for the sexes of two children, BB, BG, GB, GG, you see that 2 of the 3 pairs that have B (for boy) in them also have a girl, so the answer one could argue is 2/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, one could argue that the answer is 1/2, since the probability that any one child is a girl is 1/2, and intuitively (?) should be independent of the gender of its siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background to possibly justify posting it here: the question was asked at an interview for an actuarial/insurance type position, and the interviewer was the answer was 2/3, whereas my friend who was being interviewed (and has a masters in math) thought the answer was 1/2, even after the interviewer explained his logic. My friend felt that the interviewer wasn't taking into account the fact that it is not equally likely that a boy will open the door in the BB versus the BG combination, and one has to take into account that fact. I have no idea which is the correct answer, both sound somewhat convincing to me (I have a Ph.D. in math, but I won't mention from where in an effort to avoid embarrassing my degree granting institution!). Anyways, any help would be appreciated and I apologize if this is too simple a question for this forum.&lt;/p&gt;&#xA;" OwnerUserId="1308" LastActivityDate="2010-08-22T01:47:49.467" Title="very simple conditional probability question" Tags="&lt;probability&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="2732" PostTypeId="2" ParentId="2474" CreationDate="2010-08-18T19:53:56.677" Score="0" ViewCount="0" Body="&lt;p&gt;As a futher hint: the fundamental group of the torus $\mathbb{T}$ is generated by $a, b$ and has the relation $abAB$, where capitals denote inverses.  If you remove an open disk then you get a once-holed torus $T$.  Now the fundamental group is free (why?) and the boundary is the homotopic to the element $abAB$ (why?).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take another copy of the torus, say $\mathbb{S}$, with fundamental group generated by $c, d$ and having relation $cdCD$.  Again remove a disk to get a once holed torus $S$. Now carefully follow the answer already given, gluing $T$ and $S$, and so on. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-08-18T19:53:56.677" />
  <row Id="2733" PostTypeId="2" ParentId="2731" CreationDate="2010-08-18T20:07:25.803" Score="6" ViewCount="0" Body="&lt;p&gt;This was also posted on MathOverflow, where the answer of 1/2 was confirmed several ways.  My favorite is this: There are really &lt;em&gt;three&lt;/em&gt; independent &quot;choices&quot; are being made: the sex of the first (say, older) child, the sex of the younger child, and whether the older or younger child opens the door.  So there are &lt;em&gt;eight&lt;/em&gt; equally likely outcomes, and by considering them it's easy to get 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to mention a related game, which I believe I first saw in Martin Gardner's &lt;em&gt;Aha! Gotcha&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have three cards; one is black on both sides, one is red on both sides, and one is black on one side and red on the other.  I will let you select a card at random and look at one side, and I'll bet you even money that the other side is the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds fair.  Suppose you see black; now the card is either the black-red (which would win the game for you) or black-black (which wins for me), and at the outset, you were equally likely to have picked either one.  But in fact I win the game 2/3 of the time.  As with the puzzle, considering &lt;em&gt;all&lt;/em&gt; possible outcomes, there are 6 total &lt;em&gt;sides&lt;/em&gt; of cards.  You are equally likely to be looking at any of the 3 black sides, but only one of them (the black side of the black-red card) wins the game for you.  Another argument is to note that if you pick the black-black card, you are twice as likely to see a black side as if you pick the black-red card, so I am twice as likely to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you find this interviewer and offer to play this game with him.  You could even sweeten the odds a little to make it sound more attractive.  With high probability, you can clean him out.  Don't forget to kick back a &quot;finder's fee&quot; to your friend, or at least a nice dinner (as consolation for the job).&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-08-18T20:31:54.450" LastActivityDate="2010-08-18T20:31:54.450" CommentCount="3" />
  <row Id="2734" PostTypeId="2" ParentId="2731" CreationDate="2010-08-18T20:13:58.283" Score="3" ViewCount="0" Body="&lt;p&gt;Let A be the event &quot;Family has at least one girl&quot;.   Let B be the event &quot;Boy opens door&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pr(A and B)=1/4 *1/2 + 1/4*1/2 = 1/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AND &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pr(B)=1*1/4 + 1/4*1/2 +1/4*1/2 = 1/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Pr(A|B) = 1/4 / ( 1/2) = 1/2.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-18T20:13:58.283" />
  <row Id="2735" PostTypeId="1" AcceptedAnswerId="2748" CreationDate="2010-08-18T20:17:36.730" Score="3" ViewCount="239" Body="&lt;p&gt;Say you have a very dense matrix that is 30000x30000 elements. The very dense matrix comes from the radiosity equation, which I discussed &lt;a href=&quot;http://math.stackexchange.com/questions/2180/how-can-i-characterize-the-type-of-solution-vector-that-comes-out-of-a-matrix&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have Ax = B.  You have B, and A is 30000x30000 elements.  Solving:  You need to find x given B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving a matrix like this in C code actually introduces an interesting problem:  you cannot actually store it all in memory at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to know about ways to solve this problem &quot;in pieces&quot; -- I want to mathematically decompose the 30000x30000 matrix into smaller sub-matrices (perhaps in chunks of 30000x20?), and somehow solve that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What algorithms exist to break down / solve a matrix &quot;in pieces&quot; or steps, that can be solved regardless of memory restrictions?  It can be an iterative technique.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-20T14:51:31.840" LastActivityDate="2010-09-04T09:58:30.723" Title="Solving very large matrices in &quot;pieces&quot;" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="11" />
  <row Id="2736" PostTypeId="2" ParentId="2731" CreationDate="2010-08-18T20:18:02.937" Score="3" ViewCount="0" Body="&lt;p&gt;(Pasting my mathoverflow answer...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will assume all the obvious implicit assumptions (eg. random child being boy of girl is 50/50, boys and girls open the door uniformly, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a slightly different question, i.e. if you asked the couple if they have at least one boy, and the answer is yes, then the chance of the other one being a girl is 2/3. Intuitively, the probability is not 1/2 because in this case the answer depends on &lt;em&gt;both&lt;/em&gt; the children, i.e. it is a function of both of them considered together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you asked the couple to pick a child at random, then she/he bears no information about the other child, and consequently his/her gender does not give you any information about the sibling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case is the second case, where the child opening the door is selected at random, and she happens to be female. This does not bear any information regarding the other child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So answer is 1/2 and &lt;strong&gt;your friend is correct&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Other\ is\ B|G\ opens\ door) = P(BG|G\ opens\ door) =$&#xA;$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*1/2}{1/4+1/2*1/2} = 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be graphically understood as below -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/AcS9S.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a digression, a twist in the question can be brought about - if you take probabilities for a boy and girl to be different for opening the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg. suppose boy opens with probability $p$, girl with $q=1-p$, in a family with BG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $P(Other\ is\ B|G\ opened\ door) = P(BG)/P(G\ opened\ door) = $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*q&#xA;}{1/4+1/2*q} = \frac{2q}{2q+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This means $q = 0$ implies $P(BG|G\ opens)=0$. That makes sense, since girls don't open the door if there is a boy, so definitely the other one is girl too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Table added.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-18T21:52:37.230" LastActivityDate="2010-08-18T21:52:37.230" />
  <row Id="2737" PostTypeId="2" ParentId="2694" CreationDate="2010-08-18T20:25:47.257" Score="10" ViewCount="0" Body="&lt;p&gt;I do not think this is a conceptually important problem.  It is an example of a down-to-earth problem that can be checked numerically up to a large value and it has resisted a solution for many years.  Not all such problems are automatically important (e.g., nonexistence of odd perfect numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogy with the significance of Fermat's last theorem is apt. Before the link was made between FLT and deep conjectures of elliptic curves, there was no over-arching significance to knowing whether or not FLT was true. (I think the link between FLT and the abc-conjecture was made at around the same time.) Yes, work on FLT was responsible for useful developments in algebraic number theory, but all the same it was not clear for a long time that settling the problem, or rather finding a counterexample, would have any other repercussions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If tomorrow someone showed that the Collatz conjecture were a consequence of the abc-conjecture or some other recognizably important unsolved problem, then I would change my mind about its importance (because, as with the link between the modularity conjecture and FLT, a counterexample to Collatz would then have real implications elsewhere in mathematics).  But as long as it stays isolated, having no implications to other problems, I don't think on its own it is a mathematically profound question. The same goes for odd perfect numbers: unless someone shows the existence of an odd perfect number has effects elsewhere that we do not expect (like a counterexample to FLT having a very unexpected implication for elliptic curves), I don't think the mainstream would consider odd perfect numbers to be important either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the pedagogical side, however, I will definitely grant that this is a nice problem to show students unfamiliar with advanced mathematics that there really are unsolved math problems. People don't necessarily realize this, e.g., they may think that everything can be solved by computers or something. &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-18T20:25:47.257" CommunityOwnedDate="2010-08-18T20:25:47.257" CommentCount="1" />
  <row Id="2738" PostTypeId="2" ParentId="2626" CreationDate="2010-08-18T20:57:35.183" Score="1" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When your analysis is on a simple basis where the sail can be modeled as a plane, the force will be perpendicular to the plane.  If you need a 3-D model of sail, it's more complex.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you take what might be called the 20th century model in sailboat design, the force on the sail is resolved into a vector perpendicular to the approaching wind (lift) and a force parallel to the wind (drag). Likewise the forces on the hull are resolve to lift perpendicular to the direction of travel, and drag parallel to the direction of travel. You can find diagrams like this going back to Manfred Curry. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know about square riggers, but for the normal sloop rig, the jib and main interact too closely to consider them as independent. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One math professor who wrote about sailing was c. stanley ogilvy. I wouldn't rush out to buy his books, but if you can find them in a library, you might find them interesting. The most widely available heavy duty treatments are by Marchaj (http://en.wikipedia.org/wiki/Czes%C5%82aw_Marchaj).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="1312" LastEditorDisplayName="" LastEditDate="2010-08-18T21:45:12.413" LastActivityDate="2010-08-18T21:45:12.413" />
  <row Id="2739" PostTypeId="1" AcceptedAnswerId="2743" CreationDate="2010-08-18T21:00:44.567" Score="3" ViewCount="373" Body="&lt;p&gt;Problem 63 of the 2001 St. Petersburg Mathematical Olympiad, Second Round, 11th grade. Are there three different numbers x, y, z in [0,pi/2] such that the numbers sin x, sin y, sin z, cos x, cos y, cos z can be divided into three pairs with equal sums?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-08-19T18:58:25.337" LastActivityDate="2010-08-19T18:58:25.337" Title="sin x, sin y, sin z, cos x, cos y, cos z in [0,pi/2]" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="2740" PostTypeId="1" CreationDate="2010-08-18T21:00:46.003" Score="-2" ViewCount="108" Body="&lt;p&gt;What algorithm(s) are available for computing the result of a 3x3 medium filter in one clock cycle in hardware implementation?  Can the same algorithm(s) be applied to NxN medium filter?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1314" LastActivityDate="2010-08-19T16:58:25.670" ClosedDate="2010-08-20T07:00:09.880" Title="algorithm for computing 3x3 medium filter for hardware implementation" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2743" PostTypeId="2" ParentId="2739" CreationDate="2010-08-18T22:10:58.343" Score="4" ViewCount="0" Body="&lt;p&gt;Without loss of generality each of $x$, $y$, $z$ are $\le\pi/4$, since if for example $x\gt\pi/4$, then we can replace $x$ with $\pi/2-x$ which will not change the set $\{\cos(x),\sin(x)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now $0\lt x \lt y \lt z \le \pi/4$, and we have the following ordering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(x)\lt\sin(y)\lt\sin(z)\lt\cos(z)\lt\cos(y)\lt\cos(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the only possible pairings must be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=\sin(x)+\cos(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B=\sin(y)+\cos(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=\sin(z)+\cos(z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $A=B=C$ &#xA;(any other pairing will make one side heavier, term by term).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note $f(t)=\sin(t)+\cos(t)$ is a concave function since $\frac{d^2f(t)}{dt^2}=-\sin(t)-\cos(t)&amp;lt;0$ for $t \in (0,\pi/4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it follows that $B\gt \frac{A+C}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they can never be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Fixed glitch spotted by Tom&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-19T16:44:30.277" LastActivityDate="2010-08-19T16:44:30.277" CommentCount="13" />
  <row Id="2744" PostTypeId="1" AcceptedAnswerId="2749" CreationDate="2010-08-18T22:24:17.720" Score="2" ViewCount="156" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/2561/behaviour-of-polynomials-in-a-pid&quot;&gt;Behaviour of Polynomials in a PID!&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Prove: if $p$ is a prime, and if $n^2+n+p$ is prime for $0\leq n \leq \sqrt{p/3}$, then it is also prime for $0 \leq n \leq p-2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appeared on reddit recently, but no proof was posted. With $p=41$, it is Euler's famous prime-generating polynomial.&lt;/p&gt;&#xA;" OwnerUserId="1320" LastActivityDate="2010-08-18T22:57:56.430" ClosedDate="2010-08-19T17:06:06.377" Title="When is $n^2+n+p$ prime?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="2745" PostTypeId="2" ParentId="2592" CreationDate="2010-08-18T22:24:44.833" Score="1" ViewCount="0" Body="&lt;p&gt;Well I tried for $\frac{d^2y}{dx^2}y=1$, and I have partial solution. I will use this reply to trace the work since it supports LaTeX, I will try to complete it later (or someone else can take over).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=f(y)$, so $y=f^{-1}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $dx=f'(y)dy$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $\frac{dy}{dx}=\frac{1}{f'(y)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\frac{d^2y}{dx^2}=\frac{d}{dy}\left(\frac{1}{f'(y)}\right)\frac{dy}{dx}=-\frac{f''(y)}{(f'(y))^2}\frac{dy}{dx}=-\frac{f''(y)}{(f'(y))^3}=-\frac{g'(y)}{(g(y))^3}$ where $g(y)=f'(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving $-\frac{g'(y)}{(g(y))^3}=\frac{1}{y}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(y)=f'(y)=\frac{1}{\sqrt{\log(y^2)+c}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to solve for $f(y)$, and invert the function to get $y=f^{-1}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-18T22:24:44.833" />
  <row Id="2746" PostTypeId="1" AcceptedAnswerId="2747" CreationDate="2010-08-18T22:49:43.523" Score="13" ViewCount="596" Body="&lt;p&gt;If $n&gt;1$ is an integer, then $\sum_{k=1}^n\frac 1k$ is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know &lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand%27s_postulate&quot; rel=&quot;nofollow&quot;&gt;Bertrand's Postulate&lt;/a&gt;, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2010-09-07T15:16:57.343" Title="Is there an elementary proof that $∑_{k=1}^n 1/k$ is never an integer?" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="2747" PostTypeId="2" ParentId="2746" CreationDate="2010-08-18T22:53:47.060" Score="9" ViewCount="0" Body="&lt;p&gt;An elementary proof uses the following fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $2^s$ is the highest power of $2$ in the set $S = \{1,2,...,n\}$, then $2^s$ is not a divisor of any other integer in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider the highest power of $2$ which divides $n!$. Say that is $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the number can be rewritten as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{\sum_{k=1}^{n}{\frac{n!}{k}}}{n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest power of $2$ which divides the denominator is $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the highest power of $2$ that divides $\displaystyle \frac{n!}{k}$ is at least $t-s$. If $k \neq 2^{s}$, then this is atleast $t-s+1$ as the highest power of $2$ that divides $k$ is atmost $s-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case $k=2^s$, the highest power of $2$ that divides $\displaystyle \frac{n!}{k}$ is exactly $t-s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the highest power of $2$ that divides the numerator is atmost $t-s$. If $s &gt; 0$ (which is true if $n &gt; 1$), we are done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the above proof shows that the number is of the form $\frac{odd}{even}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-07T15:16:57.343" LastActivityDate="2010-09-07T15:16:57.343" CommentCount="2" />
  <row Id="2748" PostTypeId="2" ParentId="2735" CreationDate="2010-08-18T22:55:00.147" Score="7" ViewCount="0" Body="&lt;p&gt;The first question that should be asked of any person presenting you with a large-ish matrix: &quot;is the matrix dense or sparse?&quot; In the latter case, one definitely does not need to allocate space for the zero entries, and should thus look into special techniques  to storing them(which as I understand nowadays rely on a liberal dose of graph theory in the general case, though band matrices are still handled by storing their bands in an appropriate format).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if even after that you have the perfect excuse for having a large dense matrix (which I still believe is quite unlikely), there is a way to invert and take the determinant of a large matrix via partitioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}=\begin{pmatrix}\textbf{E}&amp;amp;\textbf{F}\\ \textbf{G}&amp;amp;\textbf{H}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\textbf{E}$ and $\textbf{H}$ are square matrices with dimensions $m\times m$ and $n\times n$ respectively, and $\textbf{F}$ and $\textbf{G}$ are dimensioned appropriately (so the dimension of $\textbf{A}$ is $(m+n)\times(m+n)$). The inverse can then be computed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{A}^{-1}=\begin{pmatrix}\textbf{E}^{-1}+\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;-\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\\-(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the parentheses emphasize common factors that you might profit from computing only once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the determinant, it is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\det\;\textbf{A}=\det\left(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}\right)\;\det\;\textbf{E}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure what seemed to be unsatisfactory with this answer, so here's a bit more of exposition: as always, the &quot;inverse&quot; here is merely notation! One would most certainly first perform LU decomposition on $\textbf{E}$ and $\textbf{H}$ first. One would also partition the right-hand side $\textbf{b}$ accordingly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{b}=\begin{pmatrix}\textbf{c}\\ \textbf{d}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $\textbf{c}$ is a length-m vector and $\textbf{d}$ is a length-n vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally multiplying the partitioned inverse with the partitioned right-hand side gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}+\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;-\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\\-(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\right)&amp;amp;(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\end{pmatrix}\cdot\begin{pmatrix}\textbf{c}\\ \textbf{d}\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which when expanded and simplified is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}\textbf{E}^{-1}\textbf{c}+\left(\textbf{E}^{-1}\textbf{F}\right)(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\textbf{c}-\textbf{d}\right)\\-(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}\left(\textbf{G}\textbf{E}^{-1}\textbf{c}-\textbf{d}\right)\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you should be able to figure out how you would use an available decomposition of $\textbf{E}$ or $\textbf{H}$, and which common subexpressions can be just computed once.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-20T13:01:28.767" LastActivityDate="2010-08-20T13:01:28.767" CommentCount="7" />
  <row Id="2749" PostTypeId="2" ParentId="2744" CreationDate="2010-08-18T22:57:56.430" Score="1" ViewCount="0" Body="&lt;p&gt;This follows by employing in &lt;a href=&quot;http://math.stackexchange.com/questions/2590&quot;&gt;Rabinowitsch's&lt;/a&gt; proof a Gauss bound, e.g. see Theorem 9.1 &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/ant-st.pdf#page=18&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-18T22:57:56.430" CommentCount="1" />
  <row Id="2750" PostTypeId="2" ParentId="2592" CreationDate="2010-08-18T23:03:53.050" Score="5" ViewCount="0" Body="&lt;p&gt;Let $y=y(x)$ and write primes for derivatives w.r.t $x$. In the equation $$y''y=1$$ the independent variable $x$ does not appear, so we can do a (somewhat magical) change of variables: in the new equation, the &lt;em&gt;independent&lt;/em&gt; variable will be $y$ and the dependent variable will be $p=y'$. If we use dots to denote derivatives with respect to $y$, a little computation using the chain rule shows that $p''=p\dot p$, so the equation becomes $$\dot ppy=1.$$ This can be rewritten as $$p\,\mathrm dp=\frac{\mathrm dy}{y},$$ which has its variables separated. It can be integrated at once, to get the relation $$\frac{p^2}{2}=\log y+c$$ with $c$ a constant. Now we recall what $p$ was, that is $y'$, and this becomes a new (first order!) ODE on $y(x)$. We have «simplified» things... (We could have obtained the same reduction od order by multiplying the original equation by $y'$, dividing it by $y$ and integrating: the trick with the $p$ substitution is, though, a general method)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same approach will similarly simplify the general case.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-18T23:11:46.453" LastActivityDate="2010-08-18T23:11:46.453" />
  <row Id="2752" PostTypeId="2" ParentId="2746" CreationDate="2010-08-18T23:28:43.727" Score="8" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; Since there is a unique denominator $2^e$ having maximal power of $2$, upon multiplying all terms through by $2^{e-1}$ one deduces the contradiction that $\; 1/2 = c/d \;$ with $\: d \:$ odd. As an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad m = 1 + \frac{1}{2} + \frac{1}{3} +\; \frac{1}{4} \;+\; \frac{1}{5} + \frac{1}{6} + \frac{1}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad\;\; 2m = 2 +\; 1 + \frac{2}{3} +\; \frac{1}{2} \;+\; \frac{2}{5} + \frac{1}{3} + \frac{2}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad -\frac{1}{2}  = \:2 +\: 1 + \frac{2}{3} - 2m  + \frac{2}{5} + \frac{1}{3} + \frac{2}{7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prior sum has all odd denominators so reduces to a fraction with odd denominator $d\:|3\cdot 5\cdot 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; I purposely avoided any use of valuation theory because Anton requested an &quot;elementary&quot; solution. The above proof can easily be made comprehensible to a high-school student.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-19T03:35:08.073" LastActivityDate="2010-08-19T03:35:08.073" />
  <row Id="2755" PostTypeId="1" AcceptedAnswerId="2761" CreationDate="2010-08-19T00:08:12.373" Score="115" ViewCount="24577" Body="&lt;p&gt;My nephew was folding laundry, and turning the occasional shirt right-side-out. I showed him a  &quot;trick&quot; where I turned it right-side-out by pulling the whole thing through a sleeve instead of the bottom or collar of the shirt. He thought it was really cool (kids are easily amused, and so am I).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he learned that you can turn a shirt or pants right-side-out by pulling the material through any hole, not just certain ones. I told him that even if there was a rip in the shirt, you could use that to turn it inside-out or right-side-out, and he was fascinated by this and asked &quot;why?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know the answer to this. Why is this the case? What if the sleeves of a long-sleeve shirt were sewn together at the cuff, creating a continuous tube from one sleeve to the other? Would you still be able to turn it right-side-out? Why? What properties must a garment have so that it can be turned inside-out and right-side-out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a lame question, but I've always wondered. I wouldn't even know what to google for, so that is why I am asking here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you know the answer to this, could you please put it into layman's terms?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Wow, I really appreciate all the participation. This is a really pleasant community and I have learned a lot here. It seems that the answer is that you need at least one puncture in the garment through which to push or pull the fabric. It appears that you can have certain handles, although it's not usually practical with clothing due to necessary stretching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accepted (a while ago actually -- sorry for not updating sooner) Dan's answer because among the answers that I understand, it is the highest ranked by this community.&lt;/p&gt;&#xA;" OwnerUserId="1319" LastEditorUserId="1319" LastEditorDisplayName="" LastEditDate="2010-09-03T17:02:47.693" LastActivityDate="2010-12-06T12:33:56.443" Title="Why can you turn clothing right-side-out?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;&lt;general-topology&gt;" AnswerCount="7" CommentCount="8" FavoriteCount="56" />
  <row Id="2756" PostTypeId="2" ParentId="2746" CreationDate="2010-08-19T00:08:20.847" Score="8" ViewCount="0" Body="&lt;p&gt;What the heck -- I'll leave my comment as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the Example on p. 13 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400intro.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400intro.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is discussed, together with (as a footnote) the strange phenomenon that this is often solved by an appeal to Bertrand's Postulate.  The discussion in the above text is intended to be &quot;didactic&quot; in that a few details are left to the reader, and I recommend it as a good exercise to flesh them out.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-19T00:08:20.847" CommentCount="6" />
  <row Id="2757" PostTypeId="2" ParentId="2601" CreationDate="2010-08-19T00:09:10.703" Score="4" ViewCount="0" Body="&lt;p&gt;The number line. For whatever elementary school grade it is, it illustrates why you need zero, and why you need negative numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The venerable cake, or probably pizza nowadays, to illustrate fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Riemann Sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coin flip as a prototypical random event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markov chain as a frog hopping from lily pad to lily pad.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="1312" LastEditorDisplayName="" LastEditDate="2010-08-19T18:24:14.407" LastActivityDate="2010-08-19T18:24:14.407" CommunityOwnedDate="2010-08-19T00:09:10.703" CommentCount="2" />
  <row Id="2758" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T00:29:24.437" Score="6" ViewCount="0" Body="&lt;p&gt;A shirt with no hole, or all holes sealed is homeomorphic to a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shirt with holes is homeomorphic to a sphere with punctures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you intuitively see that a rubber sphere with punctures can be turned inside out by continuous motion, whereas a perfect sphere cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why, let's concentrate on 2-dimentional plane, taking a circle instead of a sphere. It can be proven that any continuous deformation that inverts the circle will make it cross itself. However if the circle has 'holes', or gaps, then it topologically is equivalent to one or more line segments, which can all be inverted in place without touching each other (appropriate scaling may be necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar feature holds for a sphere too. Hence you cannot turn a shirt inside out if all the holes are sealed, but you can do it even if you have one hole (and provided that the shirt is flexible enough to go through that hole if the hole is small).&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-19T00:29:24.437" CommentCount="4" />
  <row Id="2759" PostTypeId="2" ParentId="2601" CreationDate="2010-08-19T00:41:46.580" Score="0" ViewCount="0" Body="&lt;p&gt;I can remember the following at school level -&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cartesian coordinates. Pythagoras theorem. (Metaphor - distance from the wall or from the edge of the floor.)&lt;/li&gt;&#xA;&lt;li&gt;Variables used in basic algebra. (Metaphor - common nouns, eg. &quot;an &lt;em&gt;animal&lt;/em&gt; is green and has two legs...&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Irrationals and the completeness axiom. (Metaphor - can zoom in the real line without limit.)&lt;/li&gt;&#xA;&lt;li&gt;Complex number and complex plane. (&quot;Add just solution to $x^2+1=0$ to the real line, suddenly every polynomial is solvable!&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Sets, relations. (Venn diagrams are helpful for sets.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-19T00:41:46.580" CommunityOwnedDate="2010-08-19T00:41:46.580" CommentCount="1" />
  <row Id="2760" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T00:42:01.497" Score="10" ViewCount="0" Body="&lt;p&gt;I can at least answer the question about the rip, and point you to some terms to look up.  What you are looking at is not geometry as it is usually understood but &lt;a href=&quot;http://en.wikipedia.org/wiki/Topology&quot; rel=&quot;nofollow&quot;&gt;topology&lt;/a&gt; (particularly of &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface&quot; rel=&quot;nofollow&quot;&gt;surfaces with boundary&lt;/a&gt;), which informally is the study of properties of things that are invariant under deformation.  It is often said that a topologist can't tell a coffee mug from a doughnut (it is possible to deform one into the other) and this is the attitude you should take towards questions of this kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this perspective, a rip is the same thing as the bottom of the shirt is the same thing as a sleeve: you can deform each of them into the other, so topologically they're the same thing.  (Imagine making the rip bigger, or squashing a long sleeve into a short sleeve.)  They are all just examples of what topologists call punctures.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the answer to your general question, but one property a surface must have to be able to be turned inside out is that it must be &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientability&quot; rel=&quot;nofollow&quot;&gt;orientable&lt;/a&gt;; in other words, it must have an inside and an outside to begin with!  Surfaces such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_strip&quot; rel=&quot;nofollow&quot;&gt;Mobius strip&lt;/a&gt; famously don't have this property, so it doesn't even make sense to ask about whether it's possible to turn such things inside-out or not.  (And some non-orientable surfaces can't even be realized in three dimensions...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orientable surfaces happen to have a particularly simple description: they are all topologically just spheres with some &lt;a href=&quot;http://content.answcdn.com/main/content/img/McGrawHill/Encyclopedia/images/CE701100FG0010.gif&quot; rel=&quot;nofollow&quot;&gt;handles sewn on&lt;/a&gt; and with some punctures.  To turn them inside out, at least one puncture is necessary, and if you can turn one handle inside-out then you can presumably turn all of them inside-out.  If what I said in the comments is wrong and you can't turn handles inside-out, then a necessary and sufficient condition is that there is at least one puncture and no handles.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-19T00:42:01.497" CommentCount="9" />
  <row Id="2761" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T00:42:56.970" Score="34" ViewCount="0" Body="&lt;p&gt;Everything I'm wearing is a topological sphere with holes (t-shirts have 4, pants 3, shoes and socks 1) in which cases any hole works.  Instead of a long-sleeve shirt with the arms sewn together, consider a pair of pants with legs sewn together to form a topological torus with a hole (so if you were to wear them your feet would be touching and it would be impossible to put your shoes on).  This pair of pants has two parameters which are roughly constant, the circumference of the leg and the total length of the two legs.  When it is turned inside out through the waist, these parameters swap roles, so you will have a tube about the length of a pant leg with an opening on each end, the same as if you had turned one leg inside-out and pushed the other leg through it prior to sewing.  I think this would be possible to do with real toroidal clothing as long as it is thin enough, because the process doesn't require any stretching.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-19T03:38:12.677" LastActivityDate="2010-08-19T03:38:12.677" CommentCount="16" />
  <row Id="2762" PostTypeId="2" ParentId="2485" CreationDate="2010-08-19T01:06:22.040" Score="2" ViewCount="0" Body="&lt;p&gt;The answer to this is yes! Let $g\colon\mathbb{R}\to\mathbb{R}$ be essentially unbounded on every nontrivial open interval and set $f(x,y)=g(x+y)$. Then, f is essentially unbounded on every measurable rectangle $A\times B$ of positive Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct g you can take $g_0(x)=1_{\{1&gt;x&gt;0\}}x^{-1/2}$ which is integrable but not square integrable on any neighborhood of 0. Then, letting $q_1,q_2,\ldots$ be an enumeration of the rationals, $g(x)=\sum_n2^{-n}g_0(x-q_n)$ is integrable (so, is finite almost everywhere) but not square integrable on any open interval, so is not essentially bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this construction of f works, pick any two sets $A,B\subseteq\mathbb{R}$ with positive Lebesque measure. By intersecting with a bounded interval, we can restrict to sets with finite measure. I'll write $I_A(x)$ for the indicator function of a set A and $(I_A*I_B)(x)=\int I_A(x-y)I_B(y)\,dy$ for the convolution, which is a continuous function. For any $K&gt;0$ let $C=\{x\in\mathbb{R}\colon\vert g(x)\vert\ge K\}$, which has positive measure in any nontrivial open interval. Then,&#xA;$$&#xA;\begin{align}&#xA;\mu((A\times B)\cap \{x\in\mathbb{R}^2\colon \vert f(x)\vert\ge K\})&amp;amp;=\int\int I_A(x)I_B(y)I_C(x+y)\,dx\,dy\\&#xA;&amp;amp;=\int (I_A*I_B)(x)I_C(x)\,dx.&#xA;\end{align}&#xA;$$&#xA;The convolution $I_A*I_B$ is a nonnegative function with positive integral $\int (I_A*I_B)(x)\,dx=\mu(A)\mu(B)&gt;0$, so it must be strictly positive at some point. By continuity, it is bounded below by a positive number on some open interval, so the integral displayed above is positive. Therefore, f is essentially unbounded on $A\times B$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-19T01:06:22.040" CommentCount="2" />
  <row Id="2763" PostTypeId="1" AcceptedAnswerId="2764" CreationDate="2010-08-19T01:12:52.383" Score="5" ViewCount="144" Body="&lt;p&gt;Standard deck has 52 cards, 26 Red and 26 Black. A run is a maximum contiguous block of cards, which has the same color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(R,B,R,B,...,R,B) has 52 runs.&lt;/li&gt;&#xA;&lt;li&gt;(R,R,R,...,R,B,B,B,...,B) has 2 runs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the expected number of runs in a shuffled deck of cards?&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-19T04:02:37.850" Title="What is the expected number of runs of same color in a standard deck of cards?" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;expectation&gt;" AnswerCount="1" />
  <row Id="2764" PostTypeId="2" ParentId="2763" CreationDate="2010-08-19T01:32:50.857" Score="8" ViewCount="0" Body="&lt;p&gt;The expected number of runs is 27.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_n$ be the color of the n'th card. For n&amp;lt;52, the n'th card is the end of a run if and only if $X_n\not=X_{n+1}$ and the last card in the pack is always the end of a run. So, the total number of runs is&#xA;$$&#xA;N=\sum_{n=1}^{51}1_{\{X_n\not=X_{n+1}\}}+1.&#xA;$$&#xA;The expected number of runs is&#xA;$$&#xA;\mathbb{E}[N]=\sum_{n=1}^{51}\mathbb{P}(X_n\not=X_{n+1})+1.&#xA;$$&#xA;Whatever colour the n'th card is, there are 51 remaining cards in the deck of which 26 of them are a different colour from $X_n$. So, $\mathbb{P}(X_n\not=X_{n+1})=26/51$, giving&#xA;$$&#xA;\mathbb{E}[N]=51 (26/51) + 1=27.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-19T01:32:50.857" CommentCount="2" />
  <row Id="2765" PostTypeId="1" AcceptedAnswerId="2766" CreationDate="2010-08-19T02:08:40.740" Score="-3" ViewCount="278" Body="&lt;p&gt;Prove that an integer whose decimal form consists of thirty ones, some number of zeros and some number of nines is not a perfect square.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-19T12:41:05.757" Title="Integer which is not a perfect square" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2766" PostTypeId="2" ParentId="2765" CreationDate="2010-08-19T02:14:13.330" Score="6" ViewCount="0" Body="&lt;p&gt;Consider the number modulo 9. It is 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any perfect square must be 0 or 1 or 4 or 7 modulo 9.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-19T02:14:13.330" />
  <row Id="2769" PostTypeId="2" ParentId="2746" CreationDate="2010-08-19T03:10:34.177" Score="6" ViewCount="0" Body="&lt;p&gt;I never heard of the Bertrand postulate approach before. Anton, the argument for the n-th harmonic sum not being an integer when n &gt; 1 goes back to Taeisinger in 1915. In fact, the n-th harmonic sum tends to infinity 2-adically. This naturally raises the question of the p-adic behavior of harmonic sums for odd primes p, which quickly leads into unsolved problems. I wrote a discussion of that at &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/padicharmonicsum.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/padicharmonicsum.pdf&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-19T03:10:34.177" CommentCount="1" />
  <row Id="2770" PostTypeId="2" ParentId="2765" CreationDate="2010-08-19T03:46:55.907" Score="2" ViewCount="0" Body="&lt;p&gt;Casting nines: $9|\rm N^2-3 \Rightarrow 3|\rm N \Rightarrow 9|3 \Rightarrow\Leftarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. the max power of $3$ dividing a square $&gt; 0$ must be even.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-19T12:41:05.757" LastActivityDate="2010-08-19T12:41:05.757" />
  <row Id="2771" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T04:21:20.650" Score="52" ViewCount="0" Body="&lt;p&gt;First, a warning.  I suspect this response is likely not going to be immediately comprehensibe.  There is a formal set-up for your question, there are tools available to understand what's going on.  They're not particularly light tools, but they exist and they're worthy of being mentioned.  Before I write down the main theorem, let me set-up some terminology.  The tools belong to a subject called &lt;em&gt;manifold theory&lt;/em&gt; and &lt;em&gt;algebraic topology&lt;/em&gt;.  The names of the tools I'm going to use are called things like: the isotopy extension theorem, fibre-bundles, fibrations and homotopy-groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a surface $\Sigma$, it's your shirt or whatever else you're interested in, some surface in 3-dimensional space.  Surfaces have automorphism groups, let me call it $Aut(\Sigma)$.  These are, say, all the self-homeomorphisms or diffeomorphisms of the surface.   And surfaces can sit in space.  A way of putting a surface in space is called an embedding.  Let's call all the embeddings of the surface $Emb(\Sigma, \mathbb R^3)$.  $Emb(\Sigma, \mathbb R^3)$ is a set, but in the subject of topology these sets have a natural topology as well.  We think of them as a space where &quot;nearby&quot; embeddings are almost the same, except for maybe a little wiggle here or there.  The topology on the set of embeddings is called the compact-open topology (see Wikipedia, for details on most of these definitions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so now there's some formal nonsense.  Look at the quotient space $Emb(\Sigma, \mathbb R^3)/Aut(\Sigma)$.  You can think of this as all ways $\Sigma$ can sit in space, but without any labelling -- the surface has no parametrization.  So it's the space of all subspaces of $\mathbb R^3$ that just happen to be homeomorphic to your surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Richard Palais has a really nice theorem that puts this all into a pleasant context.  The preamble is we need to think of everything as living in the world of smooth manifolds -- smooth embeddings, $Aut(\Sigma)$ is the diffeomorphism group of the surface, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two locally-trivial fibre bundles (or something more easy to prove -- Serre fibrations), this is the &quot;global&quot; isotopy-extension theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Diff(\mathbb R^3, \Sigma) \to Diff(\mathbb R^3) \to Emb(\Sigma, \mathbb R^3)/Aut(\Sigma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Diff(\mathbb R^3 fix \Sigma) \to Diff(\mathbb R^3, \Sigma) \to Aut(\Sigma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here $Diff(\mathbb R^3)$ indicates diffeomorphisms of $\mathbb R^3$ that are the identity outside of a sufficiently large ball, say.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Palais theorem, once you apply the homotopy long exact sequence of a fibration is giving you a language that allows you to translate between automorphisms of your surface, and motions of the surface in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a theorem of Jean Cerf's that $Diff(\mathbb R^3)$ is connected.  So a little diagram chase says that you that an automorphism of a surface can be realized as by a motion of that surface in 3-space if and only if that automorphism of the surface extends to an automorphism of 3-space.  For closed surfaces, the Jordan-Brouwer separation theorem gives you an obstruction to turning your surface inside-out.  But for non-closed surfaces you're out of tools.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out if you can realize an automorphism as a motion, you literally have to try to extend it &quot;by hands&quot;.  This is a very general phenomena -- you have one manifold sitting in another, but rarely does an automorphism of the submanifold extend to the ambient manifold.  You see this phenomena happening in various other branches of mathematics as well -- an automorphism of a subgroup does not always extend to the ambient group, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you try your luck and try to build the extension yourself.  In some vague sense that's a formal analogy between the visceral mystery of turning the surface inside-out and a kind of formalized mathematical problem, but of a fundamentally analogous feel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're looking for automorphisms that reverse orientation. For an arbitrary surface with boundary in 3-space, it's not clear if you &lt;em&gt;can&lt;/em&gt; turn the surface inside out.  This is because the surface might be knotted. Unknotted surfaces are examples like your t-shirt.  Let's try to cook up something that &lt;em&gt;can't&lt;/em&gt; be turned inside-out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The automorphism group of a 3-times punctured sphere has 12 path-components (12 elements up to isotopy).  There are 6 elements that preserve orientation, and 6 that reverse.  In particular the orientation-reversing automorphisms reverse the orientation of all the boundary circles.  So if you could come up with a knotted pair-of-pants (3-times punctured surface) so that its boundary circles did not admit a symmetry that reversed the orientations of all three circles simultaneously, you'd be done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this doesn't seem like a reduction to you, but it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there are things called non-invertible knots: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Invertible_knot&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Invertible_knot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we cook up a knotted pair-of-pants from that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the idea.  The non-invertible knot in the link above is sometimes called $8_{17}$.  Here is another picture of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://katlas.org/wiki/8_17&quot; rel=&quot;nofollow&quot;&gt;http://katlas.org/wiki/8_17&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a variant on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://rybu.org/photo/mathish/webstuff/ribbonsurf.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpret this image as a ribbon of paper that has three boundary circles.  One boundary circle is unknotted.  One is $8_{17}$.  The other is some other knot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that other knot isn't trivial, nor is it $8_{17}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why can't this knotted pair of pants be turned inside-out?  Well, the three knots are distinct, and $8_{17}$ can't be reversed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I know the other knot isn't $8_{17}$?  It's a hyperbolic knot and it has a different ($4.40083...$) hyperbolic volume than $8_{17}$ ($10.9859...$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI: in some sense this is one of the simplest surfaces with non-trivial boundary that can't be turned inside-out.  All discs can be turned inside-out.  Similarly, all annuli (regardless of how they're knotted) can be turned inside-out. So for genus zero surfaces, 3 boundary components is the least you can have if you're looking for a surface that can't be turned inside-out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edited to correct for Jason's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comment added later: I suggest if you purchase a garment of this form you return it to the manufacturer. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-08-19T07:50:27.730" LastActivityDate="2010-08-19T07:50:27.730" CommentCount="11" />
  <row Id="2772" PostTypeId="1" CreationDate="2010-08-19T04:25:04.937" Score="4" ViewCount="300" Body="&lt;p&gt;What exactly does blow up mean, when people say, for example, that a solution (to a pde (say)) blows up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1127" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-19T08:47:09.170" LastActivityDate="2010-08-20T02:09:13.277" Title="Blow up of a solution" Tags="&lt;soft-question&gt;&lt;terminology&gt;&lt;differential-equations&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="2773" PostTypeId="2" ParentId="2703" CreationDate="2010-08-19T04:38:13.893" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;EDITED&lt;/strong&gt;. The version I originally posted was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second problem, use the fact that squares are non-negative to show that&#xA;$$|f(x,y)| \le 3|x|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which it follows (by using the sandwich theorem) that,&#xA;$$\lim_{(x,y) \to (0,0)} f(x,y)=0$$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-08-19T05:50:34.263" LastActivityDate="2010-08-19T05:50:34.263" />
  <row Id="2774" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T05:11:47.980" Score="-2" ViewCount="0" Body="&lt;p&gt;It means that it costs more to solve the problem, relative to a smaller or more typical one.  In a practical context it has a negative connotation, but in a theoretical context it may be used in a constructive way.  For example maybe I can transform one problem into another which is known to be in P, with a polynomial &quot;blow up&quot;, thereby proving that the first problem is also in P.  On the other hand the transformation might incur a super-polynomial &quot;blow up&quot; meaning I failed to prove that.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-19T05:43:32.467" LastActivityDate="2010-08-19T05:43:32.467" CommentCount="1" />
  <row Id="2775" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T05:21:03.893" Score="1" ViewCount="0" Body="&lt;p&gt;In the context I usually hear it used, it means that a quantity is undefined or goes to infinity in the limit under consideration. For example, a function which has a singularity at that point blows up at that point (or &quot;as &lt;em&gt;x&lt;/em&gt; goes to&quot; that point).  Essentially, &quot;blows up&quot; indicates that something is being divided by zero. But I never considered it a technical term, just a bit of mathematical street lingo ;-) I do not know if this is &lt;em&gt;exactly&lt;/em&gt; what is meant when it's used in the context of PDE analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of example, I would say the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{r_0}^\infty \frac{1}{r^2}\mathrm{d}r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;blows up when $r_0 = 0$. Or that the function $\frac{1}{r^2}$ blows up at $r = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastEditorUserId="1190" LastEditorDisplayName="" LastEditDate="2010-08-20T02:09:13.277" LastActivityDate="2010-08-20T02:09:13.277" CommentCount="9" />
  <row Id="2776" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T05:28:37.000" Score="1" ViewCount="0" Body="&lt;p&gt;It denotes a singularity or discontinuity - as opposed to decaying or stable solutions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-19T05:28:37.000" />
  <row Id="2777" PostTypeId="1" CreationDate="2010-08-19T06:41:00.527" Score="2" ViewCount="272" Body="&lt;p&gt;Let $\theta_{kl}$ be an angle such that $\cos\theta_{kl}=\frac{1}{2}(\cos(\frac{2\pi k}{n})+\cos(\frac{2\pi l}{n}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that definition, if I introduce a new variable $t$ is the following a correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos(t\theta_{kl})\approx\frac{1}{2}(\cos(\frac{2\pi kt}{n})+\cos(\frac{2\pi lt}{n}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I'm actually interested in the asymptotics of $\theta_{kl}$. By a second order approximation $\theta_{kl}^2=O(\frac{k^2+l^2}{n^2})$. Is it correct? If it is, then the above holds, right? But $\theta_{kl}$ needs to be small.&lt;/p&gt;&#xA;" OwnerUserId="1330" LastEditorUserId="1330" LastEditorDisplayName="" LastEditDate="2010-08-19T07:30:48.557" LastActivityDate="2010-12-17T14:44:26.107" Title="Approximating a cosine" Tags="&lt;trigonometry&gt;&lt;approximation&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="2779" PostTypeId="2" ParentId="2777" CreationDate="2010-08-19T07:01:54.663" Score="0" ViewCount="0" Body="&lt;p&gt;If you expand the cosines in both sides of the equation as series, you get something like $1-\frac{(t\theta_{kl})^2}{2}+\frac{(t\theta_{kl})^4}{24}+\dots\approx 1-\frac{(2\pi kt/n)^2+(2\pi lt/n)^2}{4}+\dots$ ; one does not really resemble the other much...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing your update, have a look at this Wolfram Alpha &lt;a href=&quot;http://tinyurl.com/35la8ov&quot; rel=&quot;nofollow&quot;&gt;result&lt;/a&gt;, you can see that the leading term is of the form $\sqrt{k^2+l^2}\frac{t}{n}$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-19T07:39:27.040" LastActivityDate="2010-08-19T07:39:27.040" CommentCount="1" />
  <row Id="2780" PostTypeId="1" AcceptedAnswerId="2782" CreationDate="2010-08-19T07:24:15.563" Score="1" ViewCount="167" Body="&lt;p&gt;When I was studying topology I remember being able to demonstrate that the set of topological surfaces with any number of punctures (including the projective plane, Klein bottle, Moebius strip, double torus, etc.), together with their associated inverses, was &lt;em&gt;not&lt;/em&gt; a group with respect to the operations of connection and disconnection.  I could show with a sequence of drawings that it is possible to connect two Klein bottles, deform the surface, and disconnect a torus from a single remaining Klein bottle, i.e. K+K = K+T, so inverses are not well defined even though the operator is otherwise associative on that set, has an identity (the sphere), and is also Abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never made much progress in my understanding beyond this point so I am interested in any answer that can simply explain what is going on here (am I misinterpreting something else as a topological property?), else provide a reference (I'm not even sure what branch of topology addresses this).  One slightly more specific question I have is: what is an example of a collection of topologies which &lt;em&gt;does&lt;/em&gt; form a group under connection and is this ever a useful property to have?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-19T07:55:28.203" LastActivityDate="2010-08-19T08:01:23.057" Title="Basic questions about the algebra of surfaces" Tags="&lt;soft-question&gt;&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="2781" PostTypeId="2" ParentId="1870" CreationDate="2010-08-19T07:33:12.373" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.maplesoft.com/&quot; rel=&quot;nofollow&quot;&gt;Maple&lt;/a&gt; can give step-by-step solutions as well as give hints on what rule you could apply next.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-08-19T07:33:12.373" CommunityOwnedDate="2010-08-19T07:33:12.373" />
  <row Id="2782" PostTypeId="2" ParentId="2780" CreationDate="2010-08-19T07:46:32.920" Score="2" ViewCount="0" Body="&lt;p&gt;Right, surfaces under the connect-sum operation form a commutative monoid, but it's not a free commutative monoid.  It's the quotient of the free commutative monoid on two generators (projective space, torus) by the relation you mentioned (in these generators you'd express the klein bottle as the connect-sum of two projective spaces).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a monoid where the only invertible element is the sphere. The proof is that the &quot;genus&quot; is additive under connect sum, and is only zero for the sphere.  Here the 'genus' is the sum of &amp;lt; the number of projective-space summands &gt; + &amp;lt; twice the number of torus summands &gt; (for this to make sense you have to express your surface as a connect-sum of only projective spaces and tori. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure what your question is. In the sense that there's no question mark in your 1st paragraph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there's a terminology problem.  &quot;Collection of topologies&quot; literally means &quot;a collection of subsets of various sets satisfying some rules&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace surfaces with various other objects, the connect-sum operation can be a group operation.  For example, if your objects are smooth embeddings of $S^3$ in $S^6$, taken up to isotopy, then the connect-sum operation turns this collection into an infinite-cyclic group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strangely enough, embeddings of $S^j$ in $S^n$, taken up to isotopy and with operation given by connect sum form a group whenever $n &gt; j+2$.  Sometimes it's a group for silly reasons -- there is only one such embedding up to isotopy provided $2n-3j-3&gt;0$.  But when $2n-3j-3 \leq 0$ generally there are many. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-08-19T07:46:32.920" />
  <row Id="2783" PostTypeId="2" ParentId="2777" CreationDate="2010-08-19T07:57:27.480" Score="0" ViewCount="0" Body="&lt;p&gt;My reasoning is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition $\cos\theta_{kl}=\frac{1}{2}\cos\frac{2\pi k}{n}+\frac{1}{2}\cos\frac{2\pi l}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $\theta_{kl}$ is small, we take the Taylor expansion on both sides up to the second term and obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1-\theta_{kl}^2\approx\frac{1}{2}(1-\frac{4\pi^2k^2}{n^2}+1-\frac{4\pi^2l}{n^2})=1-\frac{2\pi^2k^2+2\pi^2l^2}{n^2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_{kl}=O(\frac{k+l}{n})$.&lt;/p&gt;&#xA;" OwnerUserId="1330" LastActivityDate="2010-08-19T07:57:27.480" />
  <row Id="2784" PostTypeId="2" ParentId="2780" CreationDate="2010-08-19T08:01:23.057" Score="3" ViewCount="0" Body="&lt;p&gt;Connected sum usually gives a structure of a commutative monoid but rarely of a group. Absence of inverse elements is explained (e.g. for manifolds or for knots) by Mazur swindle (&lt;a href=&quot;http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;, &lt;a href=&quot;http://terrytao.wordpress.com/2009/10/05/mazurs-swindle/&quot; rel=&quot;nofollow&quot;&gt;Terence Tao&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-19T08:01:23.057" />
  <row Id="2785" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T08:50:39.287" Score="1" ViewCount="0" Body="&lt;p&gt;It means that as time goes to infinity, the response-- be it displacement, velocity, heat, field or whatever-- will also go to infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is physically impossible and is usually due to the poor numerical/approximated methods use to solve the PDE. &lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-08-19T08:50:39.287" CommentCount="2" />
  <row Id="2786" PostTypeId="1" CreationDate="2010-08-19T09:36:07.700" Score="4" ViewCount="384" Body="&lt;p&gt;Hi,&#xA;I'm starting to read Baby Rudin (Principles of mathematical analysis) now and I wonder whether you know of any companions to it. Another supplementary book would do too. I tried Silvia's notes, but I found them a bit too &quot;logical&quot; so to say. Are they good? What else do you recommend?&lt;/p&gt;&#xA;" OwnerUserId="1333" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-19T10:57:24.307" LastActivityDate="2010-08-28T16:21:03.560" CommunityOwnedDate="2010-08-19T14:14:10.333" Title="Companions to Rudin?" Tags="&lt;soft-question&gt;&lt;analysis&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="2787" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T10:22:30.667" Score="4" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Introduction to real analysis by Bartle and Sherbert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Methods of Real Analysis by R.R. Goldberg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Mathematical Analysis by Tom Apostol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Real and Abstract Analysis by Karl Stromberg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) A radical approach to real analysis by David M Bressoud by MAA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first book is a very good book for a beginner. The next two are classics. (4) is also very good in case you want to read something advanced. The last one keeps entertaining you with some interesting examples as well as some interesting history of Real Analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Reading!!&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-19T10:22:30.667" CommunityOwnedDate="2010-08-19T14:14:10.333" />
  <row Id="2788" PostTypeId="2" ParentId="622" CreationDate="2010-08-19T10:24:54.877" Score="0" ViewCount="0" Body="&lt;p&gt;The idea of considering the representations of a group $G$ as a group of linear operators to ultimately gain informations on the group $G$ itself is an important istance of the general process of &quot;linearization&quot; that is employed systematically in many areas of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In down-to-earth (and possibly vague) terms, I would say that the only general class of problems that we are able to resolve fully are those which are linear, meaning that involve only vector spaces--best if finitely dimensional :-) --and linear maps between them. When we deal with objects that are intrinsically non-linear, a widespread strategy is to try to attach to them some linear &quot;gadgets&quot; that in some way preserve, in a different form, some of the structure of the original object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two basic examples that should be well-known to every math student:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The construction of the tensor product reduces multilinear algebra to linear algebra.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The construction of the homotopy and (co)homology groups of a topological space is an attempt to &quot;understand topology via groups and linear spaces&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The linear representations of a group $G$ respond to the same principle: they are a &quot;linear&quot; set of data attached to $G$ which hopefully can serve to characterize $G$.  &lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-19T10:24:54.877" />
  <row Id="2789" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T10:49:36.800" Score="2" ViewCount="0" Body="&lt;p&gt;There is a set of notes and additional exercises due to George Bergman.  See his web page...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.berkeley.edu/~gbergman/ug.hndts/&quot; rel=&quot;nofollow&quot;&gt;http://math.berkeley.edu/~gbergman/ug.hndts/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-08-19T10:49:36.800" CommunityOwnedDate="2010-08-19T14:14:10.333" CommentCount="3" />
  <row Id="2790" PostTypeId="2" ParentId="2647" CreationDate="2010-08-19T11:19:58.280" Score="1" ViewCount="0" Body="&lt;p&gt;A further answer, generalizing flagar's example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any N&amp;nbsp;&amp;ge;&amp;nbsp;10, there exists a set of the sort you ask for which contains at most eleven elements (with fewer than eleven in the case that N&amp;nbsp;&amp;lt;&amp;nbsp;13), and with all elements having absolute value at most N:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;{&amp;minus;N,&amp;nbsp;&amp;minus;N+2,&amp;nbsp;&amp;minus;N+4,&amp;nbsp;&amp;minus;2,&amp;nbsp;1,&amp;nbsp;3,&amp;nbsp;4,&amp;nbsp;5,&amp;nbsp;N&amp;minus;7,&amp;nbsp;N&amp;minus;6,&amp;nbsp;N&amp;minus;5}.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The shorter example presented by flagar corresponds to the case N&amp;nbsp;=&amp;nbsp;10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to this construction was to think in terms of small sequences of &lt;em&gt;almost mutually-supporting&lt;/em&gt; subsequences. For instance, the sequence &amp;minus;N, &amp;minus;N+2, &amp;minus;N+4 consist of integers which are sums of themselves, plus one of &amp;minus;2 or 4. We can include 4 in the set by including it in the sequence 3, 4, 5: we can support this by including 1 in the set (where 1 itself is supported as the sum of 3 and &amp;minus;2). We then only need a way of supporting &amp;minus;2: we do this with the other subsequence which includes N&amp;minus;6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time we have a short sequence, we have a collection of integers which is easy to &quot;support&quot; using only a small number of other integers. So one strategy to constructing such self-supporting sets of integers is to think in terms of one or more short sequences of integers, with each sequence working to support another.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-19T12:18:20.387" LastActivityDate="2010-08-19T12:18:20.387" />
  <row Id="2791" PostTypeId="1" AcceptedAnswerId="2792" CreationDate="2010-08-19T11:31:55.080" Score="4" ViewCount="222" Body="&lt;p&gt;In section 4 of the article by Afred van der Poorten's &lt;a href=&quot;http://www.ega-math.narod.ru/Apery1.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;A Proof That Euler Missed ...&lt;/em&gt;&lt;/a&gt;  the following inequality is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}\displaystyle\binom{n}{m}\leq\left\lfloor\dfrac{\ln n}{\ln p}\right\rfloor-\nu_{p}(m)\qquad(\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;In the original denoted&lt;/em&gt; $\text{ord}$ $_{p}(\cdot)$ &lt;em&gt;instead of&lt;/em&gt; $\nu_p(\cdot)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nu_{p}(k)$ is the $p$-adic valuation of $k\in\mathbb{Q}$, i. e. the exponent of the prime $p$ in the prime factorization of $k$. &#xA;I know some properties of the foor function and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}(a/b)=\nu_{p}(a)-\nu_{p}(b),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}(a\cdot b)=\nu_{p}(a)\cdot \nu_{p}(b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_{p}(n!)=\displaystyle\sum_{i\geq 1}\left\lfloor \dfrac{n}{p^{i}}\right\rfloor $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I didn't convince myself on the correct argument I should use to prove $(\ast )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: How can this inequality be proved?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-05T21:32:16.860" LastActivityDate="2010-10-05T21:32:16.860" Title="Binomial coefficients: how to prove an inequality on the $p$-adic valuation?" Tags="&lt;number-theory&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2792" PostTypeId="2" ParentId="2791" CreationDate="2010-08-19T11:58:18.307" Score="4" ViewCount="0" Body="&lt;p&gt;Since $\nu_p(n!)=\sum_i\lfloor n/p^i\rfloor$, $\nu_p\binom nm=\sum_i(\lfloor n/p^i\rfloor-\lfloor m/p^i\rfloor-\lfloor (n-m)/p^i\rfloor)$. Observe that each summand is $\le 1$, and for $i&gt;\left\lfloor\frac{\ln n}{\ln p}\right\rfloor$ it's clearly $0$. That gives $v_{p}\binom{n}{m}\le \left\lfloor\frac{\ln n}{\ln p}\right\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally observe that for $i\le\nu_p(m)$ $\lfloor m/p^i\rfloor=m/p^i$, so ($\lfloor m/p^i\rfloor+\lfloor x/p^i\rfloor=\lfloor(m+x)/p^i\rfloor$ and) the corresponding summand is also $0$. That gives the inequality in question.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-19T11:58:18.307" CommentCount="1" />
  <row Id="2793" PostTypeId="2" ParentId="622" CreationDate="2010-08-19T12:57:43.290" Score="3" ViewCount="0" Body="&lt;p&gt;Groups very, very rarely appear abstractly in (mathematical) &lt;em&gt;nature&lt;/em&gt;: when we encounter them, it is almost always in the form of a representation (a linear representation, a permutation representation, a non-linear representation in the form of automorphisms of an algebra, of a variety, of a manifold, of a group (!)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thus only natural that the study of representation theory be important! I would go further, turn your question backwards, and claim that group theory is important because it allows us to study the representations of groups.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-19T14:28:27.367" LastActivityDate="2010-08-19T14:28:27.367" />
  <row Id="2796" PostTypeId="1" AcceptedAnswerId="2797" CreationDate="2010-08-19T13:11:15.227" Score="2" ViewCount="151" Body="&lt;p&gt;I'm starting a chapter on Functions and they had the steps shown to reach the p-q equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x_{1,2} = -\frac{p}{2} \pm\sqrt{\left(\frac{p}{2}\right)^2 - q}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wanted to do the same with the Quadratic Equation. I'm using the base linear equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ax+by+c = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution I have so far is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + \frac{b}{a}x + \frac{c}{a}= 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + \frac{b}{a}x = -\frac{c}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(x + \frac{b}{2a}\right) = \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = -\frac{b}{2a} \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem comes from trying to solve the insides of the square root:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}} = \sqrt{\frac{b^2}{4a2} - \frac{c}{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sqrt{\frac{b^2}{4a^2} - \frac{c}{a} \left(\frac{4a}{4a}\right)} = \sqrt{\frac{b^2 - 4ac}{4a^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sqrt{\frac{b^2 - 4ac}{\left(2a\right)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but there is still the problem of the -(b/2a) outside of the sqrt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Also, Tex is awesome; is there a better way to do the 1,2 subscripts than _1,_2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution goes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-23T13:48:17.433" LastActivityDate="2010-08-23T13:48:17.433" Title="Help in getting the Quadratic Equation." Tags="&lt;polynomials&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="2797" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T13:29:18.430" Score="3" ViewCount="0" Body="&lt;p&gt;Sorry I don't know how to do tex on websites, but I'm trying to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just made a small mistake on the final step. In the second to last step, we actually have our full equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can collect the common factor of 1/2a and get:&#xA;$\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-19T14:13:47.840" LastActivityDate="2010-08-19T14:13:47.840" CommentCount="4" />
  <row Id="2798" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T13:41:38.353" Score="3" ViewCount="0" Body="&lt;p&gt;I find Terrence Tao's notes to be a great companion, but he deviates from the order Rudin's presented in.&#xA;&lt;a href=&quot;http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/&quot; rel=&quot;nofollow&quot;&gt;Course 1&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.math.ucla.edu/~tao/resource/general/131bh.1.03s/&quot; rel=&quot;nofollow&quot;&gt;course 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-08-19T13:41:38.353" CommunityOwnedDate="2010-08-19T14:14:10.333" />
  <row Id="2799" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T13:47:45.610" Score="0" ViewCount="0" Body="&lt;p&gt;You factored 2a out of the square root and put it in the denominator without factoring it out of -(b/2a).&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-19T13:47:45.610" />
  <row Id="2800" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T13:57:07.787" Score="0" ViewCount="0" Body="&lt;p&gt;The typical TeX way to do the 1,2 subscripts would be x_{1,2}, which yields $x_{1,2}$ as desired.  In other words, you use curly braces, like so: &quot;_{&lt;em&gt;subscripts go here&lt;/em&gt;}&quot;.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-19T13:57:07.787" CommentCount="1" />
  <row Id="2801" PostTypeId="2" ParentId="2796" CreationDate="2010-08-19T14:06:19.403" Score="1" ViewCount="0" Body="&lt;p&gt;Because I don't have edit privileges (yet?), I can't edit WWright's answer (which is correct) in order to TeX-ify it (perhaps someone with editing privileges will?), but just for the sake of making WWright's answer easier to read, here it is in TeX:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{(2a)^2}} = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As these two terms have a common denominator, we can add them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-19T14:06:19.403" CommentCount="2" />
  <row Id="2802" PostTypeId="1" AcceptedAnswerId="2836" CreationDate="2010-08-19T14:43:54.850" Score="0" ViewCount="142" Body="&lt;p&gt;In 1D case, R1 as a typical case of 1D Euclidean space is totally ordered. I was wondering if any 1D Euclidean space E1 is ordered as well? What is it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher dimensional case, are both Rn and En ordered? What are the orders?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I meant for an order for an Euclidean space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that can induce or be compatible with the topology of the Euclidean space;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or that is defined in their definition, for example, the order of R1 is the one defined in one of R1's definition as an ordered field with supreme property;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or that is used by default or most commonly. &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-08-19T17:40:40.437" LastActivityDate="2010-08-19T23:00:21.523" Title="Order on Euclidean space" Tags="&lt;set-theory&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2803" PostTypeId="2" ParentId="622" CreationDate="2010-08-19T14:44:41.260" Score="1" ViewCount="0" Body="&lt;p&gt;Representation theory plays a big role in the group-theoretic approach to special functions.&#xA;For example, Willard Miller showed that the powerful Infeld-Hill factorization / ladder method - widely exploited by physicists - is equivalent to the representation theory of four local Lie groups. This lie-theoretic approach served to powerfully unify and &quot;explain&quot; all prior similar attempts to provide a unfied theory of such classes of special functions, e.g. Truesdell's influential book &lt;a href=&quot;http://books.google.com/books?id=p1X4KKyU6kEC&quot; rel=&quot;nofollow&quot;&gt;An Essay Toward a Unified Theory of Special Functions&lt;/a&gt;. Below is the first paragraph of the introduction to Willard Miller's classic monograph &lt;a href=&quot;http://www.ima.umn.edu/~miller/lietheoryspecialfunctions.html&quot; rel=&quot;nofollow&quot;&gt;Lie theory and special functions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This monograph is the result of an attempt to understand the role &#xA;  played by special function theory in the formalism of mathematical &#xA;  physics. It demonstrates explicitly that special functions which arise in &#xA;  the study of mathematical models of physical phenomena and the &#xA;  identities which these functions obey are in many cases dictated by &#xA;  symmetry groups admitted by the models. In particular it will be shown &#xA;  that the factorization method, a powerful tool for computing eigenvalues &#xA;  and recurrence relations for solutions of second order ordinary differential &#xA;  equations (Infeld and Hull), is equivalent to the representation &#xA;  theory of four local Lie groups. A detailed study of these four groups and &#xA;  their Lie algebras leads to a unified treatment of a significant proportion &#xA;  of special function theory, especially that part of the theory which is &#xA;  most useful in mathematical physics. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also Miller's sequel &lt;a href=&quot;http://www.ima.umn.edu/~miller/separationofvariables.html&quot; rel=&quot;nofollow&quot;&gt;Symmetry and Separation of Variables&lt;/a&gt;. Again I quote from the preface:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This book is concerned with the relationship between symmetries of a &#xA;  linear second-order partial differential equation of mathematical physics, &#xA;  the coordinate systems in which the equation admits solutions via separation &#xA;  of variables, and the properties of the special functions that arise in &#xA;  this manner. It is an introduction intended for anyone with experience in &#xA;  partial differential equations, special functions, or Lie group theory, such &#xA;  as group theorists, applied mathematicians, theoretical physicists and &#xA;  chemists, and electrical engineers. We will exhibit some modern group-theoretic &#xA;  twists in the ancient method of separation of variables that can be &#xA;  used to provide a foundation for much of special function theory. In &#xA;  particular, we will show explicitly that all special functions that arise &#xA;  via separation of variables in the equations of mathematical physics can &#xA;  be studied using group theory. These include the functions of Lame, Ince, &#xA;  Mathieu, and others, as well as those of hypergeometric type. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is a very critical time in the history of group-theoretic methods in &#xA;  special function theory. The basic relations between Lie groups, special &#xA;  functions, and the method of separation of variables have recently been &#xA;  clarified. One can now construct a group-theoretic machine that, when &#xA;  applied to a given differential equation of mathematical physics, describes &#xA;  in a rational manner the possible coordinate systems in which the equation &#xA;  admits solutions via separation of variables and the various expansion &#xA;  theorems relating the separable (special function) solutions in distinct &#xA;  coordinate systems. Indeed for the most important linear equations, the &#xA;  separated solutions are characterized as common eigenfunctions of sets of &#xA;  second-order commuting elements in the universal enveloping algebra of &#xA;  the Lie symmetry algebra corresponding to the equation. The problem of &#xA;  expanding one set of separable solutions in terms of another reduces to a &#xA;  problem in the representation theory of the Lie symmetry algebra. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See Koornwinder's &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.bams/1183544920&quot; rel=&quot;nofollow&quot;&gt;review&lt;/a&gt; of this book for a very nice concise introduction to the group-theoretic approach to separation of variables.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-21T00:46:52.693" LastActivityDate="2010-08-21T00:46:52.693" />
  <row Id="2804" PostTypeId="1" AcceptedAnswerId="2807" CreationDate="2010-08-19T15:44:31.523" Score="-1" ViewCount="243" Body="&lt;p&gt;Generally given a function $f(x)$ and asked whether its Riemann Integrable or not, how does one go about solving it? Generally one takes an arbitrary partition and then tries to see whether $$U(p,f)- L(p,f) &amp;lt; \epsilon$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Comment: no, this is the definition of the Riemann (more precisely, the Darboux) integral.  It is usually very difficult to verify that a function is Riemann integrable directly from this definition and this is not what is generally done.  Rather, one proves theorems asserting that certain classes of functions are or are not Riemann integrable.  The most important such theorem appears below.  --PLC]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, we know that the &lt;em&gt;Dirichlet's function&lt;/em&gt; is not Riemann Integrable, whereas the &lt;em&gt;Thomae function&lt;/em&gt; (http://en.wikipedia.org/wiki/Thomae's_function) is Riemann Integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its useful to know (or guess) before proceeding to solve the problem, if the function is going to be Riemann Integrable or not. I would like to know, before attacking such problems, is there an intuitive idea as to whether the function can be Riemann Integrable or not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to use &lt;s&gt;Riemann - Lebesgue lemma&lt;/s&gt; &lt;a href=&quot;http://tt.lamf.uwindsor.ca/314folder/analbookfiles/RintexistLebesgue.pdf&quot; rel=&quot;nofollow&quot;&gt;Lebesgue's criterion for Riemann integrability&lt;/a&gt; here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Why not?  If not, what do you want??  -PLC]&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-22T08:26:12.953" LastActivityDate="2010-08-22T08:26:12.953" Title="Testing Riemann Integrability of a Function" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="2805" PostTypeId="2" ParentId="2735" CreationDate="2010-08-19T15:51:50.460" Score="2" ViewCount="0" Body="&lt;p&gt;This is too long for a comment, so I'll throw in this choice piece by Forman Acton in his &quot;Numerical Methods That Work&quot; on the subject of dealing with large matrices:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Whenever a person eagerly inquires if&#xA;  my computer can solve a set of 300&#xA;  equations in 300 unknowns, I must&#xA;  quickly suppress the temptation to&#xA;  retort, &quot;Yes, but why bother?&quot; There&#xA;  are, indeed, legitimate sets of&#xA;  equations that large. They arise from&#xA;  replacing a partial differential&#xA;  equation on a set of grid points, and&#xA;  the person who knows enough to tackle&#xA;  this type of problem also usually&#xA;  knows what kind of computer he needs.&#xA;  The odds are all too high that our&#xA;  inquiring friend is suffering from a&#xA;  quite different problem: he probably&#xA;  has collected a set of experimental&#xA;  data and is now attempting to fit a&#xA;  300-parameter model to it - by Least&#xA;  Squares! The sooner this guy can be&#xA;  eased out of your office, the sooner&#xA;  you will be able to get back to useful&#xA;  work - but these chaps are persistent.&#xA;  They have fitted three-parameter&#xA;  models on desk machines with no&#xA;  serious difficulties and now the&#xA;  electronic computer permits them more&#xA;  grandiose visions. They leap from the&#xA;  known to the unknown with a terrifying&#xA;  innocence and the perennial&#xA;  self-confidence that every parameter&#xA;  is totally justified. It does no good&#xA;  to point out that several parameters&#xA;  are nearly certain to be competing to&#xA;  &quot;explain&quot; the same variations in the&#xA;  data and hence the equation system&#xA;  will be nearly indeterminate. It does&#xA;  no good to point out that all large&#xA;  least-squares matrices are striving&#xA;  mightily to be proper subsets of the&#xA;  Hilbert matrix-which is virtually&#xA;  indeterminate and uninvertible - and&#xA;  so even if all 300 parameters were&#xA;  beautifully independent, the fitting &#xA;  equations would still be violently&#xA;  unstable. All of this, I repeat, does&#xA;  no good-and you end up by getting&#xA;  angry and throwing the guy out of your&#xA;  office...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...The computer center's director must&#xA;  prevent the looting of valuable&#xA;  computer time by these would-be&#xA;  fitters of many parameters. The task&#xA;  is not a pleasant one, but the&#xA;  legitimate computer users have rights,&#xA;  too. The alternative commits everybody&#xA;  to a miserable two weeks of sloshing&#xA;  around in great quantities of&#xA;  &quot;Results&quot; that are manifestly&#xA;  impossible, with no visible way of&#xA;  finding the trouble. The trouble, of&#xA;  course, arises from looking for a good&#xA;  answer to a poorly posed problem, but&#xA;  a computer director seldom knows&#xA;  enough about the subject matter to win&#xA;  any of those arguments with the&#xA;  problem's proposer, and the impasse&#xA;  finally has to be broken by&#xA;  violence-which therefore might as well&#xA;  be used in the very beginning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-19T15:51:50.460" CommunityOwnedDate="2010-08-19T15:51:50.460" CommentCount="2" />
  <row Id="2807" PostTypeId="2" ParentId="2804" CreationDate="2010-08-19T16:01:30.563" Score="6" ViewCount="0" Body="&lt;p&gt;There is a theorem which states that a bounded $f:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable if and only if $f$ is continuous almost everywhere on $[a,b]$ and is a consequence of the Monotone Convergence theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If I remember correctly, am a bit rusty)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-19T16:01:30.563" CommentCount="2" />
  <row Id="2808" PostTypeId="2" ParentId="2735" CreationDate="2010-08-19T16:02:01.610" Score="1" ViewCount="0" Body="&lt;p&gt;If all you are interested in is solving $A x = B$, I would suggest looking at iterative and/or block-iterative methods.  Noting your other question asking about positivity of solutions, I would suggest that you look at &lt;a href=&quot;http://faculty.uml.edu/cbyrne/aclafctext.pdf&quot; rel=&quot;nofollow&quot;&gt;Applied and Computational&#xA;Linear Algebra: A First Course&lt;/a&gt; by Charlie Byrne.  The MART algorithm described therein is used for finding positive $x$ solutions to $A x = B$m, and addresses many similar problems from tomography and signal processing.  If you like it get his 'Applied Iterative Methods'.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-08-19T16:02:01.610" CommentCount="4" />
  <row Id="2809" PostTypeId="2" ParentId="2485" CreationDate="2010-08-19T16:10:32.930" Score="0" ViewCount="0" Body="&lt;p&gt;Please refer Paul Halmos's book entitled &quot;Problems for Mathematicians Young and Old&quot;. Its somewhere located in the Measure theory Section. Although, i couldn't comprehend the solution fully, i would like to see a better example.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-19T16:10:32.930" CommentCount="1" />
  <row Id="2811" PostTypeId="2" ParentId="2802" CreationDate="2010-08-19T16:44:56.127" Score="0" ViewCount="0" Body="&lt;p&gt;If you are meaning a locally Euclidean whose topology is the one given by a linear order then $S^1$ is a counter-example.&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2010-08-19T16:44:56.127" />
  <row Id="2812" PostTypeId="2" ParentId="2740" CreationDate="2010-08-19T16:58:25.670" Score="1" ViewCount="0" Body="&lt;p&gt;I assume that you are looking for a Median Filter, and not a Medium filter.  Second, if you are going to do it in one clock cycle you want some sort of hardware configuration.  An algorithm, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Algorithm&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Algorithm&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematics, computer science, and related subjects, an algorithm is an effective method for solving a problem expressed as a finite sequence of instructions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're doing it in one clock cycle, your sequence of instructions is trivial and so you are looking for some sort of hardware implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching for &lt;a href=&quot;http://lmgtfy.com/?q=median+filter+one+clock+cycle&quot; rel=&quot;nofollow&quot;&gt;&quot;Median filter one clock cycle&quot;&lt;/a&gt;, brings us &lt;a href=&quot;http://authors.library.caltech.edu/747/1/PEDel96.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, giving a general construction and discussing a 5 x 5 implementation.&lt;/p&gt;&#xA;" OwnerUserId="943" LastActivityDate="2010-08-19T16:58:25.670" />
  <row Id="2814" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T17:00:40.223" Score="1" ViewCount="0" Body="&lt;p&gt;The book &lt;a href=&quot;http://community.middlebury.edu/~abbott/UA/UA.html&quot; rel=&quot;nofollow&quot;&gt;Understanding Analysis&lt;/a&gt; by Stephen Abbott is very good. So is &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=gsm-62&quot; rel=&quot;nofollow&quot;&gt;A Companion to Analysis&lt;/a&gt; by T. W. Körner.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-19T17:00:40.223" CommunityOwnedDate="2010-08-19T17:00:40.223" />
  <row Id="2815" PostTypeId="1" CreationDate="2010-08-19T17:06:53.283" Score="12" ViewCount="282" Body="&lt;p&gt;This is a final exam question in my algorithms class:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ is a taxicab number if $k = a^3+b^3=c^3+d^3$, $a,b,c,d$ are distinct positive integers. Find all taxicab number $k$ such that $a,b,c,d &amp;lt; n$ in $O(n)$ time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the problem had a typo or not, because $O(n^3)$ seems more reasonable. The best I can come up with is $O(n^2 \log n)$, and that's the best anyone I know can come up with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $O(n^2 \log n)$ algorithm: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Try all possible $a^3+b^3=k$ pairs, for each $k$, store $(k,1)$ into a binary tree(indexed by $k$) if $(k,i)$ doesn't exist, if $(k,i)$ exists, replace $(k,i)$ with $(k,i+1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Transverse the binary tree, output all $(k,i)$ where $i\geq 2$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are there any faster methods? This should be the best possible method without using any number theoretical result because the program might output $O(n^2)$ taxicab numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $O(n)$ even possible? One have to prove there are only $O(n)$ taxicab numbers lesser than $2n^3$ in order to prove there exist a $O(n)$ algorithm.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="96" LastEditorDisplayName="" LastEditDate="2010-08-19T17:22:29.907" LastActivityDate="2010-08-20T09:26:17.627" Title="Find taxicab numbers in $O(n)$ time" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="2816" PostTypeId="1" AcceptedAnswerId="2818" CreationDate="2010-08-19T17:21:15.697" Score="1" ViewCount="60" Body="&lt;p&gt;In a densely and totally ordered set, induce a order topology from the order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the dense in terms of the order and the dense in terms of the order topology equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-08-19T18:20:59.140" Title="dense in terms of order and in terms of the order topology" Tags="&lt;set-theory&gt;&lt;topology&gt;" AnswerCount="1" />
  <row Id="2817" PostTypeId="2" ParentId="2755" CreationDate="2010-08-19T17:38:30.020" Score="11" ViewCount="0" Body="&lt;p&gt;The wikipedia page for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Torus&quot; rel=&quot;nofollow&quot;&gt;torus&lt;/a&gt; has a very nice &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aInside-out_torus_%28animated,_small%29.gif&quot; rel=&quot;nofollow&quot;&gt;animation&lt;/a&gt; of a punctured torus turning inside out.  In just the same way, if you stitch together the two cuffs of a pair of pants together you can turn the result inside out.  You have to attach the cuffs in the simplest way, without knotting, to get a torus with a hole, not a Klein bottle with a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ryan says, it is not possible to turn every piece of clothing inside out,  However such &quot;bad&quot; pieces of clothing have to be knotted in some fashion.  Here is an example, perhaps a bit simpler than Ryan's, that avoids hyperbolic geometry :).  Take a pair of pants, tie the legs in an overhand knot, and stitch together the cuffs to get a knotted punctured torus.  The knot in this case is a &lt;a href=&quot;http://Trefoil&quot; rel=&quot;nofollow&quot;&gt;trefoil&lt;/a&gt; knot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll call this piece of clothing $S$.  The proof I know that $S$ cannot be turned inside out uses some ideas from low-dimensional topology. Notice that the curve of stitches &lt;a href=&quot;http://en.wikipedia.org/wiki/Incompressible_surface&quot; rel=&quot;nofollow&quot;&gt;compresses&lt;/a&gt; (bounds a disk $D$) in the inner direction.  If you could turn $S$ inside out then that motion of $S$ would also give a motion of the disk $D$.  Thus the motion sends the curve of stitches to some curve on $S$ (still not parallel to the boundary!) that compresses in the &lt;em&gt;outer direction&lt;/em&gt;.  But since the trefoil is not the &lt;a href=&quot;http://en.wikipedia.org/wiki/Unknot&quot; rel=&quot;nofollow&quot;&gt;unknot&lt;/a&gt; there is no such curve.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-08-19T19:42:34.100" LastActivityDate="2010-08-19T19:42:34.100" CommentCount="5" />
  <row Id="2818" PostTypeId="2" ParentId="2816" CreationDate="2010-08-19T17:44:26.470" Score="3" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be dense in terms of order.  Let U be an open subset.  We need to show X intersects U.  Since U is open, it contains a subset of the form (a,b) with a &amp;lt; b, since these sets form a basis for the topology (by definition of order topology).  Since X is dense in terms of order, there is an x in X such that a &amp;lt; x &amp;lt; b.  Then x is in X and (a,b), and hence x is in X and U.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose X is dense in terms of topology, meaning it intersects every open set.  Choose two elements a and b and suppose a &amp;lt; b.  Then (a,b) is an open set.  Since X is dense in terms of topology, there is an x in (a,b), and thus this x will satisfy a &amp;lt; x &amp;lt; b.  Hence, X is dense in terms of order.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-19T18:20:59.140" LastActivityDate="2010-08-19T18:20:59.140" CommentCount="2" />
  <row Id="2819" PostTypeId="2" ParentId="1880" CreationDate="2010-08-19T18:05:08.953" Score="2" ViewCount="0" Body="&lt;p&gt;It only seems odd to you because you are used to representing numbers in base 10. What if you used base 7?&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-08-19T18:05:08.953" CommentCount="3" />
  <row Id="2820" PostTypeId="2" ParentId="162" CreationDate="2010-08-19T18:19:21.250" Score="-1" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Why is &quot;the set of all sets&quot; a&#xA;  paradox? It seems like it would be&#xA;  fine, to me.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lesson of this and other set theory paradoxes is that when you define a set in a proof, you are obliged to demonstrate that the set exists and is well-defined. To me, a paradox of Russell's Paradox is that he fell into this trap since he pretty much wore himself out trying to generate a rigorous foundation for Mathematics.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-08-19T18:19:21.250" />
  <row Id="2821" PostTypeId="2" ParentId="2791" CreationDate="2010-08-19T18:25:57.900" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Theorem (Kummer):&lt;/strong&gt;  $\nu_p {n \choose m}$ is the number of carries it takes to add $m$ and $n-m$ in base $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $\lfloor \frac{\ln n}{\ln p} \rfloor$ is the maximum possible number of carries and the last $\nu_p(m)$ digits of $m$ cannot be associated to any carries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kummer's theorem itself is not hard to prove; it more or less follows from the last identity you list.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-20T21:30:40.267" LastActivityDate="2010-08-20T21:30:40.267" CommentCount="1" />
  <row Id="2822" PostTypeId="2" ParentId="2815" CreationDate="2010-08-19T18:42:21.907" Score="2" ViewCount="0" Body="&lt;p&gt;Apparently this is a solved problem and every rational solution to $x^3 + y^3 = z^3 + w^3$ is proportional to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 1 − (p − 3q)(p^2 + 3q^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = −1 + (p + 3q)(p^2 + 3q^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = (p + 3q) − (p^2 + 3q^2)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w = −(p − 3q) + (p^2 + 3q^2)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://129.81.170.14/~erowland/papers/koyama.pdf&quot; rel=&quot;nofollow&quot;&gt;http://129.81.170.14/~erowland/papers/koyama.pdf&lt;/a&gt;, Page 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see: &lt;a href=&quot;http://sites.google.com/site/tpiezas/010&quot; rel=&quot;nofollow&quot;&gt;http://sites.google.com/site/tpiezas/010&lt;/a&gt; (search for J. Binet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems like an O(n^2) algorithm might be possible. Perhaps using this more cleverly can give us an O(n) time algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. Please do update us with the solution given by your professor.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-19T18:42:21.907" />
  <row Id="2823" PostTypeId="2" ParentId="1660" CreationDate="2010-08-19T19:02:35.283" Score="0" ViewCount="0" Body="&lt;h3&gt;Schur polynomials proof&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Schur_polynomial&quot; rel=&quot;nofollow&quot;&gt;Observe&lt;/a&gt; that $D(a_1,\ldots,a_n)/D(1,\ldots,n)=s_\lambda(1,\ldots,1)$ (for $a_i=\lambda_i+i$). But by Giambelli (aka Jacobi-Trudi) &lt;a href=&quot;http://en.wikipedia.org/wiki/Giambelli_formula&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; $s_\lambda=\det h_{\lambda_i+j-1}$, so $s_\lambda(1,\ldots,1)=\det\binom{a_i}{j-1}$. Now by &lt;a href=&quot;http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/&quot; rel=&quot;nofollow&quot;&gt;Lindstrom-Gessel-Viennot lemma&lt;/a&gt; last determinant counts non-intersecting lattice paths from the set $0,1,\ldots,n$ on Y-axis to the set $a_1,\ldots,a_n$ on X-axis. Finally, observe that any such path is characterized by x-coordinates of &quot;steps down&quot; — which are subject to conditions from the definition of GZ triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/* Well, it's not that direct and it's more or less the proof from Robin Chapman's answer. Nevertheless, it explains something, I hope. */&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-19T19:02:35.283" />
  <row Id="2824" PostTypeId="2" ParentId="2786" CreationDate="2010-08-19T19:39:15.230" Score="4" ViewCount="0" Body="&lt;p&gt;Gelbaum and Olmsted, Counterexamples in Analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first real analysis/advanced calculus class is full of theorems with multiple conditions, and it can be difficult to tell which ones are necessary for what parts of the theorem. This book will provide examples for why the theorems are as they are and not otherwise.&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-19T19:39:15.230" CommunityOwnedDate="2010-08-19T19:39:15.230" />
  <row Id="2825" PostTypeId="2" ParentId="162" CreationDate="2010-08-19T19:57:08.697" Score="0" ViewCount="0" Body="&lt;p&gt;Russel's paradox arises if you consider the set $U=\left\{x:x\not\in x\right\}$. Ask yourself if $U\in U$. If you suppose so, then by the definition of unrestricted set comprehension $U\not\in U$. You have a contradiction, so it must be the opposite of what you supposed, that is, $U\not\in U$. But this is the same as saying $U$ belongs to the complement of itself, that is, $U\in U$. You now have another contradiction, but this is far worse, since you have no hypotheses. The whole theory is logically inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In set theory there are two ways for getting rid of the Russel's paradox: either you disallow the set of all sets and other similar sets (see for example the Zermelo-Fraenkel set theory), or you allow them, but you also restrict the way they are used (see for example the Morse-Kelley set theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, set comprehension says if you have a set $A$ you can have $\left\{x\in A:\phi\left(x\right)\right\}$ (notice: writing $\left\{x:\phi\left(x\right)\right\}$ is just wrong in this case, because you have to have an initial set). If you now define $U=\left\{x\in A:x\not\in x\right\}$ and you repeat the same passages as before, it only follows that $U\not\in A$. There's no contradiction and the theory is consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, you consider classes, not just sets. Sets are classes that belong to some other class, while proper classes are classes that belong to no class. Set comprehension, in this case, says you can have $\left\{x:\phi\left(x\right)\right\}$, but all its members are sets by definition. If try to reproduce Russel's paradox, you get that $U\not\in U$. If you then suppose that $U$ is a set, then you have a contradiction, so $U$ must be a proper class. This is all you get. No contradictions. The theory is consistent.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2010-08-19T19:57:08.697" />
  <row Id="2826" PostTypeId="2" ParentId="2815" CreationDate="2010-08-19T20:06:14.433" Score="5" ViewCount="0" Body="&lt;p&gt;I don't know about O(n), but I can do it in O(n^2). The main idea is to use &lt;a href=&quot;http://eli.thegreenplace.net/2008/08/23/initializing-an-array-in-constant-time/&quot; rel=&quot;nofollow&quot;&gt;initialization of an array in O(1)&lt;/a&gt; (this is the best reference that I've found, which is weird since this seems like a very important concept). Then you iterate through all the possible (a,b) pairs and do the same as step 1 in your proposed algorithm. Since a^3+b^3 &amp;lt;= 2n^3, the array needs to be of size 2n^3, but it's still initialized in O(1). Accessing an array element is O(1) like in a regular array.&lt;/p&gt;&#xA;" OwnerUserId="26" LastEditorUserId="26" LastEditorDisplayName="" LastEditDate="2010-08-19T20:20:06.613" LastActivityDate="2010-08-19T20:20:06.613" CommentCount="6" />
  <row Id="2827" PostTypeId="1" AcceptedAnswerId="2828" CreationDate="2010-08-19T20:13:39.360" Score="8" ViewCount="422" Body="&lt;p&gt;What is a good PDE book suitable for self study?   I'm looking for a book that doesn't require much prerequisite knowledge beyond undergraduate-level analysis.  My goal is to understand basic solutions techniques as well as some basic theory.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-19T20:51:50.203" LastActivityDate="2010-08-20T02:40:13.453" CommunityOwnedDate="2010-08-19T20:51:04.793" Title="Good 1st PDE book for self study" Tags="&lt;books&gt;&lt;differential-equations&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="4" />
  <row Id="2828" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:20:44.007" Score="2" ViewCount="0" Body="&lt;p&gt;The book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470054565&quot; rel=&quot;nofollow&quot;&gt;Strauss&lt;/a&gt; is pretty good for a first course. For a second one the book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821849743&quot; rel=&quot;nofollow&quot;&gt;Evans&lt;/a&gt; is nice but it requires some knowledge of measure theory and functional analysis.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-19T22:31:16.443" LastActivityDate="2010-08-19T22:31:16.443" CommunityOwnedDate="2010-08-19T20:51:04.793" CommentCount="3" />
  <row Id="2829" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:26:48.157" Score="1" ViewCount="0" Body="&lt;p&gt;I like Karl E. Gustafson's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486612716&quot; rel=&quot;nofollow&quot;&gt;Introduction to Partial Differential Equations and Hilbert Space Methods&lt;/a&gt;.  It was certainly readable after an advanced calc sequence.  You will find a few short and worthwhile conversational paragraphs throughout the book.  He also uses the technique of revisiting interesting concepts from different perspectives throughout the book.  And it's a Dover paperback, so it's cheap.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-19T20:26:48.157" CommunityOwnedDate="2010-08-19T20:26:48.157" />
  <row Id="2830" PostTypeId="1" AcceptedAnswerId="2850" CreationDate="2010-08-19T20:27:10.637" Score="2" ViewCount="67" Body="&lt;p&gt;Suppose p and q are discrete distributions over k outcomes, q is given, for which p does the following hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_1 \log q_1 + \ldots + p_k \log q_k \ge q_1 \log q_1 + \ldots + q_k \log q_k$&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-08-20T15:15:16.400" Title="When does entropy bound cross-entropy?" Tags="&lt;probability-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2831" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:33:53.087" Score="1" ViewCount="0" Body="&lt;p&gt;Logan's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387209530&quot; rel=&quot;nofollow&quot;&gt;Applied Partial Differential Equations&lt;/a&gt; might be suitable for you if you want a (relatively) quick overview of the subject, since it's not very long (about 200 pages). It's aimed at undergraduates in math, engineering and the sciences.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-19T20:33:53.087" CommunityOwnedDate="2010-08-19T20:51:04.793" />
  <row Id="2832" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T20:58:53.103" Score="0" ViewCount="0" Body="&lt;p&gt;I would recommend: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fritz John, &lt;a href=&quot;http://books.google.com/books?id=cBib_bsGGLYC&amp;amp;printsec=frontcover&amp;amp;dq=Fritz+john&amp;amp;hl=en&amp;amp;ei=Y5ptTP-MG6aHOPyC8LAL&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Partial Differential Equations&lt;/a&gt; (Applied Mathematical Sciences) ISBN: 0387906096. It is a classical Springer book that contains what you ask for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google Books might be a good start before you make your final decision. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-19T20:58:53.103" CommunityOwnedDate="2010-08-19T20:58:53.103" />
  <row Id="2833" PostTypeId="2" ParentId="2827" CreationDate="2010-08-19T21:26:46.720" Score="0" ViewCount="0" Body="&lt;p&gt;I haven't read a lot of PDE material, but I enjoyed Taylor's &lt;a href=&quot;http://books.google.com/books?id=uphqFYqYm0oC&amp;amp;printsec=frontcover&amp;amp;dq=taylor+partial+differential+equations&amp;amp;hl=en&amp;amp;ei=YaFtTNPMBYH7lwe_gZmvDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;. It's quite well-written, and also contains introductory material (like Lie derivatives), since it does things on manifolds.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-19T21:26:46.720" CommunityOwnedDate="2010-08-19T21:26:46.720" />
  <row Id="2834" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T22:17:58.437" Score="0" ViewCount="0" Body="&lt;p&gt;&quot;Blows up&quot; means it goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/x &quot;blows up&quot; at x = 0.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-08-19T22:17:58.437" />
  <row Id="2835" PostTypeId="2" ParentId="2772" CreationDate="2010-08-19T22:34:10.013" Score="6" ViewCount="0" Body="&lt;p&gt;The meaning is, of course, context-dependent...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of differential equations, that a solution to an equation with a &quot;time&quot; variable blows up usually means that the maximal domain for which it is defined is finite, so that at the endpoint of that interval something `bad' happens: either the solution goes to infinity, or it stops being smooth (in a way that makes the differential equation stop having sense, maybe), or something. This is an important phenomenon, one which causes trouble. A couple of examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Perelman's solution of the Poicaré conjecture—in a very vague sense—consists of a way to `work around' the fact that certain solutions of a (very complicated non-linear) PDE blow up; &lt;/li&gt;&#xA;&lt;li&gt;the third `Millenium' Clay &lt;a href=&quot;http://www.claymath.org/millennium/Navier-Stokes_Equations/&quot; rel=&quot;nofollow&quot;&gt;problem&lt;/a&gt; is (very roughly) the question «do the solutions of the Navier-Stokes equation blow up?».&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Consider, as a very simple example, the equation $$\frac{\mathrm dx}{\mathrm dt}=x^2.$$ This equation makes sense and satisfies the conditions for existence and uniqueness of solutions on all of the $(t,x)$-plane, but if you solve it (which is easy to do explicitely, as it has the two variabls separated) you'll see that all of its solutions have a maximal interval which is a half-line (which is bounded on the left or on the right, depending on the initial condition) and that at the finite end of that interval the solutions become unbounded. We thus say that all solutions of our equation blow up in finite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also equations which have &lt;em&gt;some&lt;/em&gt; solutions which blow up and some which live forever. One example is $$\frac{\mathrm dx}{\mathrm dt}=\begin{cases}x^2&amp;amp;\text{if $x\geq0$}\\0&amp;amp;\text{if $x\leq0$}\end{cases}$$ and you'll surely find lots of fun in trying to concoct examples where even more interesting phenomena occur.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-19T22:44:33.370" LastActivityDate="2010-08-19T22:44:33.370" />
  <row Id="2836" PostTypeId="2" ParentId="2802" CreationDate="2010-08-19T23:00:21.523" Score="6" ViewCount="0" Body="&lt;p&gt;I think the OP is asking whether, for $n &gt; 1$, the topology on $\mathbb{R}^n$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_topology&quot; rel=&quot;nofollow&quot;&gt;order topology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;no&lt;/strong&gt;.  Here is one simple way to see this: recall that a point $x$ in a topological space $X$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cut-point&quot; rel=&quot;nofollow&quot;&gt;cut-point&lt;/a&gt; if $X \setminus x$ is disconnected.  For $n &gt; 1$, $\mathbb{R}^n$ has no cut-points.  However, for any order topology, all but at most two points are cut-points, namely the points corresponding to the largest and smallest elements of the order, if any.  [To be clear, these points &lt;em&gt;may&lt;/em&gt; still be cut points, as for instance with $\mathbb{Q} \cap [0,1]$.] Otherwise $(-\infty,x)$, $(x,\infty)$ gives a separation of $X$.  Since $\mathbb{R}^n$ has more than two points (!), for $n &gt; 1$ the Euclidean topology is not an order topology.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-19T23:00:21.523" CommentCount="2" />
  <row Id="2837" PostTypeId="1" AcceptedAnswerId="5246" CreationDate="2010-08-19T23:28:10.120" Score="5" ViewCount="701" Body="&lt;p&gt;Given a line segment, denoted by it's 2 endpoints (X1, Y1) and (X2, Y2), and a circle, denoted by it's center point (Xc, Yc) and a radius R, how can I tell if the line segment is a tangent of or runs through this circle? I don't need to be able to discern between tangent or running through a circle, I just need to be able to discern between the line segment making contact with the circle in any way and no contact. If the line segment enters but does not exit the circle (if the circle contains an endpoint), that meets my specs for it making contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I need a function to find if any point of a line segment lies in or on a given circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My application is that I'm using the circle as a proximity around a point. I'm basically testing if one point is within R distance of any point in the line segment. And it must be a line segment, not a line.&lt;/p&gt;&#xA;" OwnerUserId="757" LastEditorUserId="757" LastEditorDisplayName="" LastEditDate="2010-08-19T23:54:44.083" LastActivityDate="2010-09-28T17:01:14.923" Title="How to tell if a line segment intersects with a circle?" Tags="&lt;geometry&gt;" AnswerCount="8" CommentCount="8" />
  <row Id="2838" PostTypeId="2" ParentId="2837" CreationDate="2010-08-19T23:32:48.927" Score="8" ViewCount="0" Body="&lt;p&gt;Find the intersections of the line containing the segment and the circle. This amounts to solving a quadratic equation. If there are no intersections (&lt;em&gt;i.e.&lt;/em&gt; the solutions of the corresponding equation are non-real), then your segment does not intersect the circle. Now, if there &lt;em&gt;are&lt;/em&gt; intersections, see whether they are inside the segment or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To implement this, let $P$ and $Q$ be the endpoints of your segment and $C$ and $r$ be the center and the radius of your circle. Then every point of the line through $P$ and $Q$ is given by the formula $$t P + (1-t) Q$$ for exactly one value of $t\in\mathbb R$, and the points in the segment are precisely those for which the corresponding $t$ is in the interval $[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the point corresponding to $t\in\mathbb R$ is on the circle if and only if $$\langle t P + (1-t) Q - C,t P + (1-t) Q - C\rangle = r^2,$$ where $\langle\mathord\cdot,\mathord\cdot\rangle$ is the usual inner (dot) product of vectors. If you solve this equation—this is easy, as it is a quadratic equation for $t$—, you find the $t$'s corresponding to the points of intersection of the circle and the line, and if at least one of those $t$'s belong to the interval $[0,1]$, then the &lt;em&gt;segment&lt;/em&gt; intersects the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; I tried to have &lt;em&gt;Mathematica&lt;/em&gt; do the computation for me. Assuming $C=(0,0)$ and $r=1$, as we may up to translation and rescaling, and letting $P=(p1,p2)$, $Q=(q1,q2)$, the following computes the $t$'s:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = {p1, p2};&#xA;Q = {q1, q2};&#xA;ts = t /. Solve[Norm[t P + (1 - t) Q]^2 == 1, t];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This ends up with the variable ts having the value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\frac{-p_1 q_1-p_2 q_2-\sqrt{p_1^2&#xA;   \left(-q_2^2\right)-2 p_1 q_1+2 p_2 p_1 q_1&#xA;   q_2-p_2^2 q_1^2-2 p_2&#xA;   q_2+p_1^2+p_2^2+q_1^2+q_2^2}+q_1^2+q_2^2}{-2 p_1&#xA;   q_1-2 p_2 q_2+p_1^2+p_2^2+q_1^2+q_2^2},\frac{-p_1&#xA;   q_1-p_2 q_2+\sqrt{p_1^2 \left(-q_2^2\right)-2 p_1&#xA;   q_1+2 p_2 p_1 q_1 q_2-p_2^2 q_1^2-2 p_2&#xA;   q_2+p_1^2+p_2^2+q_1^2+q_2^2}+q_1^2+q_2^2}{-2 p_1&#xA;   q_1-2 p_2 q_2+p_1^2+p_2^2+q_1^2+q_2^2}\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This was a huge formula, which will likely not fit on your browser window...)&#xA;If the expression inside the square roots is negative, there are no real $t$'s, so the line (hence, &lt;em&gt;a fortiori&lt;/em&gt; the segment) and the circle are disjoint. If not, now we need to see if at least one of the roots in in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I now tell &lt;em&gt;Mma&lt;/em&gt; to tell me when then first of the roots is in $[0,1]$, by telling&#xA;her to compute&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Reduce[0 &amp;lt;= ts[[1]] &amp;lt;= 1, Reals]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;it works for a while, and comes up with a huge answer, presumably equivalent to the original&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 &amp;lt;= ts[[1]] &amp;lt;= 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone make sense of the huge answer? (if one asks instead for the more meaningful&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Reduce[0 &amp;lt;= ts[[1]] &amp;lt;= 1 || 0 &amp;lt;= ts[[2]] &amp;lt;= 1, t, Reals]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the same thing happens)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; Please notice that I have made absolutely no attempt at being fast or particularly smart with this idea: it is just the straightforward was to set up the problem and solve it.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-20T14:57:34.153" LastActivityDate="2010-08-20T14:57:34.153" CommentCount="9" />
  <row Id="2839" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T00:35:20.227" Score="5" ViewCount="0" Body="&lt;p&gt;Maybe this is what you are getting at...  Given a line through points $(x_1,y_1)$ and $(x_2,y_2)$, and given a single point $p$ &lt;em&gt;not necessarily on the line&lt;/em&gt;, we want to find out: &lt;em&gt;If we draw a circle of radius $r$ around $p$, does that circle intersect the line?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is what you want then you are looking for the &lt;a href=&quot;http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html&quot; rel=&quot;nofollow&quot;&gt;perpendicular distance from the point to the line&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation on MathWorld may not be want you want to look at right now considering your stated familiarity with so many symbols - so let's just get right down to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that your line passes through the two points $(x_1,y_1)$ and $(x_2,y_2)$, a point $(a,b)$ not necessarily on the line, and some radius $r$ around the point $(a,b)$, in order to see if your point $(a,b)$ is within $r$ from the line, you need to check&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;if&lt;/code&gt; $r \geq \frac{|(x_1 - x_2)(a-x_1) + (y_2 - y_1)(b-y_1)|}{\sqrt{(x_1-x_2)^2 + (y_2-y_1)^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;then&lt;/code&gt; $(a,b)$ is within $r$ of the line (taken as an orthogonal distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I don't think this will work if you are working with just a line segment... This solution assumes your line is extended infinitely in both directions. &lt;strong&gt;(update: there are scenarios where this would fail)&lt;/strong&gt; I think the case of the segment itself being further away than $r$ can be taken care of in cases before you arrive at the &quot;if&quot; statement above.  It should be something along the lines of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;if&lt;/code&gt; $r \leq \sqrt{(x_1 - a)^2 + (y_1 - b)^2}$ &lt;code&gt;or&lt;/code&gt; $r \leq \sqrt{(x_2-a)^2 + (y_2 - b)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;then&lt;/code&gt; the segment is too far away to even consider checking the longer inequality above. &lt;strong&gt;(again, the previous if/then is insufficient... still thinking)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Edit:&lt;/strong&gt; This approach has been adapted and completed far more satisfactorily by &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2844#2844&quot;&gt;Isaac&lt;/a&gt;.  I will offer a &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2862#2862&quot;&gt;second approach&lt;/a&gt; that may be exactly what &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2838#2838&quot;&gt;Mariano&lt;/a&gt; did, but I'll try to keep it short sweet and simple.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-20T16:24:25.783" LastActivityDate="2010-08-20T16:24:25.783" CommentCount="4" />
  <row Id="2840" PostTypeId="2" ParentId="2827" CreationDate="2010-08-20T01:34:57.837" Score="0" ViewCount="0" Body="&lt;p&gt;I think you cannot get anything better than &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=gsm-19-r&quot; rel=&quot;nofollow&quot;&gt;Evans' book&lt;/a&gt;. Its size may be a little scaring, but it is the most clear and well written book on the subject I ever met.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-08-20T01:34:57.837" CommunityOwnedDate="2010-08-20T01:34:57.837" />
  <row Id="2841" PostTypeId="2" ParentId="2827" CreationDate="2010-08-20T02:40:13.453" Score="1" ViewCount="0" Body="&lt;p&gt;Partial Differential Equations for Scientists and Engineers by Farlow. It's Dover, so it's cheap. And it's a great first intro - very applied. If you want to follow on with a more rigorous one, you can't beat Evans (Springer - ISBN13: 978-0821207729)&lt;/p&gt;&#xA;" OwnerUserId="1116" LastActivityDate="2010-08-20T02:40:13.453" CommunityOwnedDate="2010-08-20T02:40:13.453" CommentCount="1" />
  <row Id="2843" PostTypeId="1" AcceptedAnswerId="2863" CreationDate="2010-08-20T06:51:52.177" Score="2" ViewCount="318" Body="&lt;p&gt;I am interested in image analysis and am looking for an approachable tutorial to the Wiener filter. At some point I am interested in implementing such a filter but I would like to have a deeper understanding of the algorithms I'll be writing. Does anyone have recommendations for papers, web sites, etc. that present good coverage on the topic?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-20T12:13:22.767" LastActivityDate="2010-08-20T16:32:07.287" Title="Wiener filter: A good tutorial" Tags="&lt;signal-processing&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="2844" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T07:08:49.000" Score="2" ViewCount="0" Body="&lt;p&gt;If I understand your edit correctly, you have a point $C=(x_c,y_c)$ and the line segment with endpoints $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ and you want to know whether or not $C$ is within a distance $r$ of any point on the line segment $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working from &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2839#2839&quot;&gt;Tom Stephens's answer&lt;/a&gt;, let's look at the whole line $\overleftrightarrow{PQ}$ containing the line segment $\overline{PQ}$.  The distance between $C$ and $\overleftrightarrow{PQ}$ is $d=\frac{|(x_1 - x_2)(x_c-x_1) + (y_2 - y_1)(y_c-y_1)|}{\sqrt{(x_1-x_2)^2 + (y_2-y_1)^2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If $d&gt;r$, then $C$ is too far from $\overleftrightarrow{PQ}$ to be within $r$ of any point on $\overline{PQ}$, finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $d\le r$, then $C$ is close enough to $\overleftrightarrow{PQ}$, but may not be close enough to $\overline{PQ}$.  $d$ is the perpendicular distance from $C$ to $\overleftrightarrow{PQ}$.  Consider two cases&amp;mdash;first, that $C$ and $\overline{PQ}$ are arranged such that the perpendicular distance is to a point on $\overline{PQ}$; second, that the perpendicular distance is to a point not on $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) In the first case, $\angle PQC$ and $\angle QPC$ each have measure &amp;le;90&amp;deg;, so the cosine of each angle must be nonnegative so $\frac{(x_1-x_2)^2-(x_1-x_c)^2+(x_2-x_c)^2+(y_1-y_2)^2-(y_1-y_c)^2+(y_2-y_c)^2}{2\sqrt{((x_1-x_2)^2+(y_1-y_2)^2)((x_2-x_c)^2+(y_2-y_c)^2)}}\ge 0$ and $\frac{(x_1-x_2)^2+(x_1-x_c)^2-(x_2-x_c)^2+(y_1-y_2)^2+(y_1-y_c)^2-(y_2-y_c)^2}{2\sqrt{((x_1-x_2)^2+(y_1-y_2)^2)((x_1-x_c)^2+(y_1-y_c)^2)}}\ge 0$ (from the Law of Cosines).  [ &lt;em&gt;edit 2&lt;/em&gt;: After a more careful reading of &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/2848#2848&quot;&gt;damiano's answer&lt;/a&gt;, it occurred to me that these are much more hideously complicated than necessary.  The denominators do not affect the sign at all and the numerators are somewhat simpler when expanded, so the two inequalities can be replaced by: $x_1x_2-x_1x_c+x_2x_c-x_2^2+y_1y_2-y_1y_c+y_2y_c-y_2^2\ge0$ and $x_1x_2+x_1x_c-x_2x_c-x_1^2+y_1y_2+y_1y_c-y_2y_c-y_1^2\ge0$.  These are more clearly equivalent to damiano's vector-dot-product criteria. ]  If both of these inequalities hold, then $d\le r$ is the distance from $C$ to $\overline{PQ}$, so $C$ is within $r$ of $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If those two inequalities from the Law of Cosines do not both hold, then $d$ is not the distance from $C$ to any point on $\overline{PQ}$, so the shortest distance from $C$ to any point on $\overline{PQ}$ is the distance to the closer of $P$ and $Q$: $d'=\min(\sqrt{(x_1-x_c)^2+(y_1-y_c)^2},\sqrt{(x_2-x_c)^2+(y_2-y_c)^2})$.  If $d'\le r$, then $C$ is within $r$ of $\overline{PQ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/strong&gt; (adding the numbers above, also)&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;a href=&quot;http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle/5246#5246&quot;&gt;Corey Ogburn's answer&lt;/a&gt;, I thought it might be helpful to add a diagram and discuss what parts of my original answer address which regions.  The numbers below discuss the correspondingly-numbered part of my original answer, above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rxaPF.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If $d&gt;r$, then the point $C$ is in the beige region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $d\le r$, so if $\angle PQC$ and $\angle QPC$ each have measure &amp;le;90&amp;deg;, then $C$ is in the purple region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $C$ is not in the beige or purple regions, so we only need to see whether it's in the blue or red regions (based on whether $C$ is close enough to the endpoints of the segment).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-23T15:12:20.280" LastActivityDate="2010-09-23T15:12:20.280" CommentCount="0" />
  <row Id="2845" PostTypeId="1" AcceptedAnswerId="2853" CreationDate="2010-08-20T09:18:11.370" Score="3" ViewCount="134" Body="&lt;p&gt;Suppose I have a matrix $S$ having a one-dimensional nullspace $\lbrace e \rbrace$ such that $S + ee^\top$ is a positive definite symmetric matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $b \in Range(S)$ and suppose I solve the equation $(S + ee^\top)x = b$ is there anyway I can derive the solution $x'$ of the equation $Sx' = b$?   I was trying a Sherman Morrison Woodbury type formula, but this fails since the denominator is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-08-20T18:19:34.023" Title="Solving a linear equation given the solution of another." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="2847" PostTypeId="2" ParentId="2815" CreationDate="2010-08-20T09:26:17.627" Score="1" ViewCount="0" Body="&lt;p&gt;For $O(n^2)$ (randomized) time, you can also use a hashtable of size $\Theta(n^2)$. Looking up will be constant time because the number of taxicab numbers is $O(n^2)$.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-08-20T09:26:17.627" />
  <row Id="2848" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T09:38:51.833" Score="2" ViewCount="0" Body="&lt;p&gt;If I understood what you want correctly, you need to find conditions on a line segment so that it lies entirely outside a circle; a possible reason for confusion is that mathematicians tend to call &quot;circle&quot; just the one-dimensional boundary of the &quot;disk&quot; and they call &quot;disk&quot; its interior.  In this terminology, I think that you want the segment to lie completely outside of the closed disk; this is the question that I will an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a geometric description of the answer.  Let $P$ and $Q$ be the endpoints of the segment, let $C$ be the center of the circle and let $r$ be the radius of the circle $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) We rule out the possibility that at least one among $P$ and $Q$ is contained inside the closed disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If in the triangle $PQC$ one of the two angles on the segment $PQ$ is not acute, then moving along the segment $PQ$ and away from the non-acute angle the distance from the center $C$ increases, so that there is no intersection in this case because of 1).  Otherwise, the two angles on the segment $PQ$ are acute, so that the minimum distance between the center $C$ and the points on the segment $PQ$ is achieved inside the segment.  In this case (i.e. when the two angles on $PQ$ are acute), we make sure that the distance between the center $C$ and the line joining $P$ and $Q$ is larger than $r$.  In view of 1) and the previous discussion this implies that there is no contact between the segment and the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translation into formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $dist(P,C)&gt;r$ and $dist(Q,C)&gt;r$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) if the inequalities $PC \cdot PQ &gt; 0$ and $QC \cdot QP &gt; 0$ hold, then also the inequality $||PC \times PQ|| &gt; r ||PQ||$ must hold (in the formulas, we denote by $\cdot$ the scalar product, by $\times$ the cross product, and by $||-||$ the length of a vector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicit equations, assuming that $C$ is the origin (which you can always achieve by translations):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $x_P^2+y_P^2 &gt; r^2$ and $x_Q^2+y_Q^2 &gt; r^2$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) if both $x_P^2 + y_P^2 \geq x_P x_Q + y_P y_Q$ and $x_Q^2 + y_Q^2 \geq x_P x_Q + y_P y_Q$ hold, then check that also $(x_Py_Q-x_Qy_P)^2 &gt; r^2 ((x_P-x_Q)^2 + (y_P-y_Q)^2)$ holds.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="1028" LastEditorDisplayName="" LastEditDate="2010-08-20T10:23:47.617" LastActivityDate="2010-08-20T10:23:47.617" CommentCount="1" />
  <row Id="2850" PostTypeId="2" ParentId="2830" CreationDate="2010-08-20T10:04:50.640" Score="1" ViewCount="0" Body="&lt;p&gt;p such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_i q_i^{p_i-q_i} \ge 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I arrive at this point with the following calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all I rewrite the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i p_i \log q_i \ge \sum_i q_i \log q_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i (p_i \log q_i - q_i \log q_i) = \sum_i (\log q_i^{p_i-q_i}) = \log \prod_i q_i^{p_i-q_i} \ge 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most simple deduction about p is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_i \ge q_i , \forall i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_i \ge 1$$&lt;/p&gt;&#xA;" OwnerUserId="422" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-20T15:15:16.400" LastActivityDate="2010-08-20T15:15:16.400" CommentCount="1" />
  <row Id="2851" PostTypeId="1" AcceptedAnswerId="2875" CreationDate="2010-08-20T10:06:11.717" Score="2" ViewCount="83" Body="&lt;p&gt;I am crossposting this question from Stack Overflow since it is more math then programming related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Reference to &lt;a href=&quot;http://stackoverflow.com/questions/2500499/howto-project-a-planar-polygon-on-a-plane-in-3d-space&quot;&gt;my question&lt;/a&gt; about projecting a planar Polygon to a plane, i came to another problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to project a base contour which is aligned to the x-y-plane along a projection vector from from a point P1 to a point P2. This already works, but now i also need to change the alignment of my base contour using the position of P1  and two vectors defining the new alignement base's x and y should be transformed to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use a transformation matrix M for this, which premultiplied to my projecton matrix p should give me the transformation matrix from a vertex v in the base contour to a vertex v' on the projection plane such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v' = (P x M) * v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to create a transformation matrix which transforms a vertex v relative to a coordinate system with O = (0,0,0), xdir = (1,0,0), and ydir = (0,1,0) to a vertex relative to a coordinate system with O'=(ox, oy, oz) and xdir and ydir as two arbitrary perpendicular unit vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarifications&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I use 4x4 matrices&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1357" LastActivityDate="2010-08-20T23:42:27.787" Title="Constructing an Extrusion between two non-parallel planes" Tags="&lt;matrices&gt;&lt;transformation&gt;" AnswerCount="1" />
  <row Id="2853" PostTypeId="2" ParentId="2845" CreationDate="2010-08-20T11:14:27.970" Score="5" ViewCount="0" Body="&lt;p&gt;Suppose $x$ is the solution to $(S+ee^T)x = b$. Since $b\in Range(S)$, we may write $b = Sz$ for some z. Now compute:&#xA;$$&#xA;e^T(S+ee^T)x = e^Tb = e^TSz&#xA;$$&#xA;Since S is symmetric, and e is in its nullspace, we have $e^TS = 0$. So the above equation simplifies to $e^Tx = 0$. But this implies&#xA;$$&#xA;(S+ee^T)x = Sx&#xA;$$&#xA;So x is a solution to the equation $Sx=b$ as well. As noted above, the solution to $Sx=b$ is not unique; $x + \lambda e$ is also a solution for any real $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-08-20T18:19:34.023" LastActivityDate="2010-08-20T18:19:34.023" CommentCount="6" />
  <row Id="2854" PostTypeId="2" ParentId="2584" CreationDate="2010-08-20T13:35:46.543" Score="1" ViewCount="0" Body="&lt;p&gt;First let's make sure I understood your question correctly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have a probability function that is expressed as a sum/product of N parametrized one-dimensional gaussians, each with different mean and standard deviation.&lt;/li&gt;&#xA;&lt;li&gt;You want to generate stochastic variables according to this distribution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, I reccommend you use a variation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Rejection_sampling&quot; rel=&quot;nofollow&quot;&gt;rejection sampling&lt;/a&gt;. The recipe is quite straightforward, but you might have to iterate a bit before you get an answer out of it. This is the basic outline.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You generate a uniformly distributed random number in your desired interval, x.&lt;/li&gt;&#xA;&lt;li&gt;You calculate the value of your probability distribution, p(x)&lt;/li&gt;&#xA;&lt;li&gt;You generate another uniformly distributed random number between 0 and 1, q&lt;/li&gt;&#xA;&lt;li&gt;If q &amp;lt; p(x), return x&lt;/li&gt;&#xA;&lt;li&gt;If not, start from step 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;No matter how large the temptation, do not re-use q for different iterations.&lt;/p&gt;&#xA;" OwnerUserId="2684" LastActivityDate="2010-08-20T13:35:46.543" CommentCount="5" />
  <row Id="2855" PostTypeId="1" AcceptedAnswerId="2870" CreationDate="2010-08-20T15:03:59.053" Score="1" ViewCount="187" Body="&lt;p&gt;Could someone verify this proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the following definition of divisibility $(\forall{a,b \in \mathbb{N}})[a | b \equiv (\exists{c \in \mathbb{N}})(ac=b)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement: $(\forall{a,b,c \in \mathbb{N}})[(a|b) \rightarrow (ac|bc)] $ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a,b \in \mathbb{N}$ be arbitrary.  Assuming that $a|b$, we can conclude that $\exists{d \in \mathbb{N}}$ such that $ad = b$.  By Algebra we can deduce the following sequence of statements.  If $(ad)=b$, then $(ad)c=bc$.  Therefore $(ac)d=bc$.  Hence by the definition of divisibility we know that $ac|bc$.&lt;/p&gt;&#xA;" OwnerUserId="1360" LastActivityDate="2010-08-20T22:28:45.517" Title="Divisibility Proof" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="0" />
  <row Id="2856" PostTypeId="1" AcceptedAnswerId="2860" CreationDate="2010-08-20T15:10:28.713" Score="2" ViewCount="102" Body="&lt;p&gt;I'm am trying to define a data structure to represent road networks.  The immediately obvious structure is that of a graph - a set of nodes and edges that connect pairs of nodes.  The nodes would represent things like intersections, and the edges would represent lanes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the basic concept of a graph is insufficient to describe what I want.  I also want to be able to describe how certain lanes are &quot;reachable&quot; or &quot;adjacent&quot; to other lanes.  (Imagine parallel lanes with no median between them.  You can just switch lanes.)  This seems to imply a need to have some sort of meta-edge that connects pairs of edges.  Is there any mathematical structure that sounds like this?  Is this reducible to the standard concept of a graph? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1358" LastActivityDate="2010-08-20T17:44:11.297" Title="Looking for a particular generalization of the concept of a mathematical graph" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="2857" PostTypeId="2" ParentId="2435" CreationDate="2010-08-20T15:11:14.547" Score="0" ViewCount="0" Body="&lt;p&gt;You can use the Kolmogorov Smirnov test too. It is a non parametric test, and will work on many distributions - including Uniform. The advantage it has over other tests is that it looks at the whole distribution. Also it is fairly easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to compare the CDFs of the $H_0$ distribution with the observed empirical distribution, and find the maximum difference. This difference does not depend on the distribution (hence it is non-parametric), and depends only on $n$, the number of observations you are drawing. Then you look it up against a table to see how different it is from expected range of the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Empirical CDF of your observations $\{x_0,x_1,...,x_n\}$ is $F_n(x)=\frac{1}{n}\sum_1^nI_{x_i\lt x}$. Then your test statistic will be $D_n=\sup|F(x)-F_n(x)|$. Assuming you sort your $x_n$'s in ascending order, and assuming your numbers come from Uniform[0,1] (wlg since you can scale them appropriately) it will be $D_n=\max_i(max(|x_i-\frac{i}{n}|,|x_i-\frac{i-1}{n}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you compute that, it's only a matter of looking up table or using the right software. I recommend &lt;a href=&quot;http://www.ciphersbyritter.com/JAVASCRP/NORMCHIK.HTM&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; which has an excellent tool to compute KS along with many other tests. For KS, you need to select &quot;Dn&quot;, put in the value you got, press the button, and assume interpret the result as it is truly uniform if P (computed in the form) &amp;lt; $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-20T15:11:14.547" CommentCount="2" />
  <row Id="2858" PostTypeId="2" ParentId="2856" CreationDate="2010-08-20T15:33:39.653" Score="0" ViewCount="0" Body="&lt;p&gt;May be you can use a colored graph with the convention that if there are more than edges from point A to point B, they are switchable inbetween.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit (after comment that A and B may not be same):&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if I am wrong, but geographically close points may be quite far apart topologically (i.e. when they have many intermediate closely situated points). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem boils down to given a standard graph, for each edge how to best store which edges are 'reachable'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reachability seems to be a reflexive and commutative relation (i.e. a is reachable from b implies b is reachable from a). If reachability is also a transitive relation (i.e. if i can reach b from a and c from b then i can reach c from a), then you can still use the coloring approach. All you need to do is to divide the edges into equivalence classes, and assign separate colors for each equivalence class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if it is not transitive, then the only option I see is for each edge to store a list of edges that are reachable from it.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-20T16:52:58.803" LastActivityDate="2010-08-20T16:52:58.803" CommentCount="5" />
  <row Id="2860" PostTypeId="2" ParentId="2856" CreationDate="2010-08-20T15:39:48.353" Score="3" ViewCount="0" Body="&lt;p&gt;You could make the lanes themselves nodes as well&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-08-20T15:39:48.353" CommentCount="2" />
  <row Id="2861" PostTypeId="2" ParentId="2855" CreationDate="2010-08-20T16:11:54.817" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, your proof is correct. Since it employs only the associative and commutativite axioms for multiplication it remains true in any commutative semigroup. The proof employs only the inference rules of first-order equational logic, namely that equality is an equivalence relation (reflexive, symmetric, transitive) and the rules of &lt;strong&gt;substitution&lt;/strong&gt;: $p(x) = q(x) \rightarrow p(a) = q(a)$ and &lt;strong&gt;replacement&lt;/strong&gt;: $a = b \rightarrow p(a) = p(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a famous theorem of Birkhoff, for any algebraic structure that can be defined by axioms - like comutativity, associativity, etc - that are all universally quantified &lt;strong&gt;equations&lt;/strong&gt;, said inference rules are &lt;em&gt;complete&lt;/em&gt;, i.e. an equation is provable from the axioms using said inference rules iff it is true in every structure that satisfies the axioms. However, finding such a purely equational proof can be highly difficult. For example, consider the famous Jacobson theorem that rings satisfying the axiom $x^n = x \;$ are commutative $xy = yx$.  it's notoriously difficult for $n&gt;2$ to find equational proofs of this theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the specific problem: generally scaling a polynomial $f(x)$ by a cancelable element $c$ does not change its solution set since $c f(d) = 0 \iff f(d) = 0$. In particular the  solvability of $f(x) = 0$ is not altered - which is what you are concerned with in the definition of divisibility, i.e. the solvability of $ax = b$. E.g. consider the less trivial case of quadratic $f(x)$. Note that the formula $g(a,b,c)$ for the solutions of a quadratic $ax^2+bx+c$ remains unaltered by nonzero coefficient scalings $g(ad,bd,cd) = g(a,b,c)$. As for solvability, scaling the polynomial by $d$ multiplies the discriminant $b^2-4ac$ by $d^2$ so it does not affect the solvability criterion - that the discriminant be a perfect square. But we don't need to know these explicit formulas to deduce this. Rather, we can deduce it at a conceptually higher-level, simply from the cancelability of $c$, i.e the invertibility of  scaling transformations $x\to cx$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-20T18:05:38.343" LastActivityDate="2010-08-20T18:05:38.343" />
  <row Id="2862" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T16:21:56.037" Score="1" ViewCount="0" Body="&lt;p&gt;Well, what the heck... there has been so much virtual ink spilled at this point that what's another gallon, really?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;update&lt;/strong&gt;: This is precisely the solution &lt;a href=&quot;http://math.stackexchange.com/users/274/mariano-suarez-alvarez&quot;&gt;Mariano proposed&lt;/a&gt; long before I did, it just happens that I couched this in a different notation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle behind &lt;em&gt;this&lt;/em&gt; approach is that we can find out algebraically whether the set of coordinates defining the perimeter of a circle with center $(h,k)$ and radius $r$ contains any point of the line segment connecting the points $(x_1,y_1)$ and $(x_2,y_2)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, writing the set of points $(x,y)$ lying along the line segment as&#xA;$$tx_1 + (1-t)x_2 = x$$&#xA;$$ty_1 + (1-t)y_2 = y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t$ ranges between $0$ and $1$ inclusive, and writing the equation for the circle centered at $(h,k)$ with radius $r$ as&#xA;$$(x-h)^2 + (y-k)^2 = r^2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we can simply substitute our expressions for $x$ and $y$ into the equation for the circle, and then figure out if there at least one such $t \in [0,1] \subset \mathbb{R}$.  If there is, then the perimeter of the circle and the line segment share at least one point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we agreed that we have seen enough virtual ink today, I will simply write down what that substitution looks like and then the final computation that needs to be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The substitution&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(tx_1 + (1-t)x_2 - h \right)^2 + \left(ty_1 + (1-t)y_2 - k \right)^2 = r^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a quadratic equation in $t$, note that the values of $x_1,x_2,y_1,y_2,h,k,$ and $r$ are known.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The computation&lt;/strong&gt;: &#xA;$$At^2 + Bt + C = 0$$&#xA;with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = x_{1}^{2} - 2x_{1}x_{2} + x_{2}^{2},$&lt;br&gt;&#xA;$B = 2x_{1}x_{2} - 2x_{2}^{2} - 2hx_{2} + 2hx_{2},$ and &lt;br&gt;&#xA;$C = x_{2}^{2} - 2hx_{2} + h^{2} - r^{2}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $t$ using the quadratic formula and then test whether at least one $t$ is a real number between $0$ and $1$, inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; If I have stomped on another's solution that has already been posted it is because I have not done due diligence and verified all of the solutions given previously.  Please let me know in the comments.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-20T16:46:51.480" LastActivityDate="2010-08-20T16:46:51.480" CommentCount="2" />
  <row Id="2863" PostTypeId="2" ParentId="2843" CreationDate="2010-08-20T16:32:07.287" Score="1" ViewCount="0" Body="&lt;p&gt;Detection Estimation and Modulation Theroy Volume 1 by Harry Van Trees is a good start if you have a basic understanding of Probability and Random Variables.  Any statistical signal processing book should also cover Wiener Filtering&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-20T16:32:07.287" />
  <row Id="2864" PostTypeId="1" AcceptedAnswerId="2867" CreationDate="2010-08-20T16:33:21.137" Score="7" ViewCount="332" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow&quot;&gt;hyperbolic geometry&lt;/a&gt; is a non-Euclidian geometry with constant negative curvature. It has the property that given a line and a point, many lines can be drawn containing the point that never meet the given line. The picture below (Circle Limit by M.C.Escher) is a conformal map of hyperbolic geometry to Euclidian plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rklhN.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you show that in a space with hyperbolic geometry, area of a triangle with angles $\alpha$, $\beta$, and $\gamma$ is $\Delta\propto\pi-\alpha-\beta-\gamma$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar result exists for sphere: $\Delta=R^2(\alpha+\beta+\gamma-\pi)$. I could prove this, but I suspect my proof cannot be modified for hyperbolic geometry. My proof for the sphere is fairly trivial, it makes use of symmetry and a Venn diagram.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-20T22:48:14.833" LastActivityDate="2010-08-20T23:32:20.923" Title="Area of a triangle $\propto\pi-\alpha-\beta-\gamma$" Tags="&lt;geometry&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2865" PostTypeId="1" AcceptedAnswerId="5110" CreationDate="2010-08-20T16:33:23.493" Score="7" ViewCount="285" Body="&lt;p&gt;If $(G_i,f_{ij})$ is a direct system of topological groups, is it always the case that the  topological\group-theoretical direct limit $G:=\varinjlim_iG_i$ is a topological group? (The topology on $G$ is the final topology with respect to the canonical homomorphisms $\psi_i:G_i\rightarrow G$). It is immediate from the group structure on $G$ and the definition of its topology that the inversion map is continuous. Moreover, I'm pretty certain that if the transition maps $f_{ij}$ are open, then the $\psi_i$ are open, and this gives continuity of multiplication. It's not at all clear to me that this condition is necessary though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always sort of assumed that this was true, but it's never really been an issue because I never start with a system of topological groups and take the direct limit; I always have a topological group and express it as a direct limit (e.g. the idele group of a global field, or the Cartier dual of a finite free $\mathbb{Z}_p$-module).&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2010-09-21T01:16:32.377" Title="Is a direct limit of topological groups always a topological group?" Tags="&lt;general-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="2866" PostTypeId="2" ParentId="2837" CreationDate="2010-08-20T16:37:41.633" Score="1" ViewCount="0" Body="&lt;p&gt;There are too many answers already, but since no one mentioned this, perhaps this might still be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can consider using &lt;a href=&quot;http://mathforum.org/dr.math/faq/formulas/faq.polar.html&quot; rel=&quot;nofollow&quot;&gt;Polar Coordinates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translate so that the center of the circle is the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of a line in polar coordinates is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = p \sec (\theta - \omega)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the above web page for what $\omega$ is. You can compute $p$ and $\theta$ by using the two endpoints of the segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If R is the radius of the circle, you need to find all $\theta$ in $[0, 2\pi]$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R = p \sec (\theta - \omega)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all you need to check is if this will allow the point to fall within the line segment.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-20T17:03:14.483" LastActivityDate="2010-08-20T17:03:14.483" />
  <row Id="2867" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T16:39:48.820" Score="4" ViewCount="0" Body="&lt;p&gt;I direct you to chapter 1 of &lt;em&gt;Fuchsian Groups&lt;/em&gt; by Svetlana Katok. It has the following proof of the theorem (at least where I learned it) with illustrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us define hyperbolic area to be: $$ \mu(A)=\int_{A}\frac{dxdy}{y^2}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be verified that for all $T \in$ $\text{PSL}(2,\mathbb{R})$ that $\mu(A)=\mu(T(A)).$ (This fact will be useful to us later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let us consider a triangle in this hyperbolic plane $H$. We will use the upper half plane model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt; One vertex of the triangle belongs to $\mathbb{R}\cup {\infty}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the angle at that vertex is 0. We can use transformations (recall $T$ from above?) from $\text{PSL}(2,\mathbb{R})$ to change two sides of $A$ into vertical geodesics; hence the base of the triangle is a Euclidean semicircle orthogonal to  $\mathbb{R}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can see that $$\mu(A)=\int_{A}\frac{dxdy}{y^2}=\int_{a}^{b}dx\int_{\sqrt{1-x^2}}^{\infty}\frac{dy}{y^2}=\int_{a}^{b}\frac{dx}{\sqrt{1-x^2}}.$$ When we substitute $x=cos(\theta),$ this integral becomes: $$\mu(A)=\int_{\pi - \alpha}^{\beta}\frac{-\sin(\theta)d\theta}{\sin\theta}=\pi-\alpha-\beta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt;  $A$ has no vertices in $\mathbb{R}\cup{\infty}.$ &#xA;Let the triangle have vertices $A, B,$ and $C.$ Then let the geodesic connecting $A$ and $B$ intersect the real axis at $D$. Then $\mu(A)=\mu(ACD)-\mu(BCD)$ and both are triangles as in the previous case. (You can work out the rest of the details accordingly -  draw a picture)&lt;/p&gt;&#xA;" OwnerUserId="298" LastEditorUserId="298" LastEditorDisplayName="" LastEditDate="2010-08-20T17:03:35.217" LastActivityDate="2010-08-20T17:03:35.217" />
  <row Id="2868" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T17:00:31.773" Score="5" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Observe that &lt;em&gt;defect&lt;/em&gt; of a (hyperbolic or spherical) polygon (i.e. the difference between the sum of its angles and the sum of angles of a Euclidean polygon with the same number of vertices) is an additive congruence-invariant function.&lt;/li&gt;&#xA;&lt;li&gt;Up to proportionality, there exists only one additive congruence-invariant function (for Euclidean plane it is sometimes called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bolyai%E2%80%93Gerwien_theorem&quot; rel=&quot;nofollow&quot;&gt;Bolyai–Gerwien theorem&lt;/a&gt;; one may say that it just means that area is well-defined).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence the result.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-20T17:00:31.773" CommentCount="2" />
  <row Id="2869" PostTypeId="1" AcceptedAnswerId="2935" CreationDate="2010-08-20T17:02:46.573" Score="2" ViewCount="104" Body="&lt;p&gt;The process below yields a distribution of &quot;response times&quot; (RT), and I'd like to know if there is an analytic solution to obtain the density function of this distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An RT is recorded at the earliest time ($t_1$) at which a function, $f(t)$, reaches a threshold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The threshold varies across iterations of the process according to a random uniform distribution with bounds specified by the parameters $P_1$ and $P_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(t)$ begins at $t=0$ and has a form whose derivative is defined by the function $g(t)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(t) = P_3\left (\frac{P_4}{P_4+P_5}\right )\left (1-\exp\left [-(P_4+P_5)\int_{0}^{t}h(s)ds\right ]\right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h(t)$ is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h(t) = P_6\Gamma(t,P_8,n)[1 - \Gamma(t-P_7,P_9,n)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the gamma function defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(t,\beta,n) = 1- \left ( e^{-\beta t} \right )\left ( \sum_{n-1}^{j=0} \frac{(\beta t)^{j}}{j!}  \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are in total 10 parameters in the process: $P_1$ through $P_9$, and $n$. I could of course perform a Euler simulation to generate a distribution of RT values for given a particular set of 10 parameter values, but it would be nice to avoid this because I'm hoping to use a Nelder–Mead search algorithm to find the set of 10 parameter values that maximizes the likelihood of a set of observed RT data points, and it really slows down the search if I have to compute estimates of the density function for every candidate set of parameter values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help you're able to provide!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pursuant to comments, here is a further description of the process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is an adaptation of the model presented by &lt;a href=&quot;http://dx.doi.org/10.1037/a0015156&quot; rel=&quot;nofollow&quot;&gt;Smith &amp;amp; Racliff&lt;/a&gt; (&lt;a href=&quot;http://star.psy.ohio-state.edu/coglab/People/roger/pdf/psychrev09b.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;) which models perception, attention and decision in the context of a speeded detection task. Function $h(t)$ models perception: accumulation of information about a brief visual stimulus. Function $g(t)$ models the transfer of this information to memory via attention, and function $h(t)$ is a linear ballistic decision process in the vein of &lt;a href=&quot;http://dx.doi.org/10.1016/j.cogpsych.2007.12.002&quot; rel=&quot;nofollow&quot;&gt;Brown &amp;amp; Heathcote&lt;/a&gt; (&lt;a href=&quot;http://psychology.stanford.edu/~jlm/pdfs/BrownHeathcote08.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;). I thought that supplementing an LBA as the decision mechanism in place of Smith &amp;amp; Ratcliff's diffusion process might simplify the derivation of the final density function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_1$ and $P_2$ are the upper and lower bounds of the random uniform threshold.&#xA;$P_3$ is the asymptote of $g(t)$.&#xA;$P_4$ is the intensity of the stimulus.&#xA;$P_5$ is the intensity of the background.&#xA;$P_6$ is the amplitude of the stimulus.&#xA;$P_7$ is the duration of the stimulus.&#xA;$P_8$ is the rate at which the stimulus comes on.&#xA;$P_9$ is the rate at which the stimulus turns off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smith &amp;amp; Ratcliff (2009) note that $n$ might be reasonably fixed to a value of 3, so possibly that can further simplify things. Additionally, while Smith &amp;amp; Ratcliff sought to fit their model to different stimulus intensities, I don't need this particular feature, so I'm not sure if I need parameters $P_4$ and $P_5$ (though I'm not sure how to properly remove them either).&lt;/p&gt;&#xA;" OwnerUserId="1361" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T23:05:38.743" LastActivityDate="2010-08-30T23:05:38.743" Title="Is there an analytic solution for the density function of this complex random variable?" Tags="&lt;functional-analysis&gt;&lt;gamma-function&gt;&lt;functions&gt;&lt;random-variable&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="2870" PostTypeId="2" ParentId="2855" CreationDate="2010-08-20T17:05:57.410" Score="0" ViewCount="0" Body="&lt;p&gt;The proof is valid.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-20T17:05:57.410" CommentCount="2" />
  <row Id="2871" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T17:13:55.900" Score="4" ViewCount="0" Body="&lt;p&gt;There is also a &lt;strong&gt;differential-geometric proof&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a surface with constant Gaussian curvature $K$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem&quot; rel=&quot;nofollow&quot;&gt;Gauss-Bonnet formula&lt;/a&gt; yields $\text{area}\cdot K+\text{sum of exterior angles}=2\pi$ aka $\text{area}=\text{defect}/K$. (This gives not only proportionality but also the coefficient.)&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-20T17:13:55.900" />
  <row Id="2872" PostTypeId="1" AcceptedAnswerId="2873" CreationDate="2010-08-20T17:24:57.020" Score="1" ViewCount="104" Body="&lt;p&gt;How can I use the chain rule $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ to differentiate an expression like $\frac{d}{dx}(y^3)$ ?   The solution in the book is $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$.  I'm not sure how they got it though.&lt;/p&gt;&#xA;" OwnerUserId="1360" LastEditorUserId="1360" LastEditorDisplayName="" LastEditDate="2010-08-20T17:43:03.153" LastActivityDate="2010-08-20T18:33:34.493" Title="How do you take the derivative of a variable with respect to a different variable:" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="2873" PostTypeId="2" ParentId="2872" CreationDate="2010-08-20T17:28:02.883" Score="5" ViewCount="0" Body="&lt;p&gt;In the problem, you want to differentiate the expression $\color{green}y^{\color{blue}3}$ with the variable $\color{red}x$. But there is no information on what $\color{green}y$ is. So we must assume $\color{green}y$ is an arbitrary function, i.e. $\color{green}y := \color{green}y(\color{red}x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the crucial point is to note that &lt;strong&gt;the &lt;sup&gt;3&lt;/sup&gt; itself is also a function&lt;/strong&gt;, i.e. we can define $\color{blue}z(\color{green}y) = \color{green}y^{\color{blue}3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, we could apply the chain rule &lt;sub&gt;&lt;sup&gt;(some names changed to be consistent with the problem)&lt;/sup&gt;&lt;/sub&gt;,&#xA;$$ &#xA;\frac{d \color{blue} z}{d \color{red} x} = \frac{d \color{blue} z}{d \color{green} y} \cdot \frac{d \color{green} y}{d \color{red} x}&#xA;$$&#xA;i.e. explictly,&#xA;$$&#xA;\frac{d \Bigl( \color{blue} z\bigl(\color{green}y(\color{red}x)\bigr) \Bigr)}{d \color{red} x}&#xA;= \frac{d \Bigl( \color{blue} z \bigl(\color{green}y\color{silver}{(x)}\bigr) \Bigr) }{d\bigl( \color{green}y\color{silver}{(x)}\bigr)}&#xA;\cdot \frac{d \bigl( \color{green} y (\color{red}x)\bigr) }{d \color{red} x}&#xA;$$&#xA;In $d\color{blue}z/d\color{green}y$, since no other variables (like $\color{red}x$) is directly involved, we could treat $\color{green}y$ as a variable instead of an implicit function. In $d\color{green}y/d\color{red}x$, as the function $\color{green}y(\color{red}x)$ is arbitrary, the differentiated form is the best representation already. Therefore, the expression above simplifies to&#xA;$$&#xA;\frac{d \left( \color{green}y^{\color{blue}3} \right) }{d \color{red} x}&#xA;= \underbrace{  \color{teal}{ 3 \color{green}y^2 }  }_{d(\color{green}y^{\color{blue}3})/d\color{green}y} \cdot &#xA;\frac{d \color{green} y}{d \color{red} x}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-20T18:32:45.230" LastActivityDate="2010-08-20T18:32:45.230" CommentCount="2" />
  <row Id="2874" PostTypeId="2" ParentId="2856" CreationDate="2010-08-20T17:33:48.177" Score="-1" ViewCount="0" Body="&lt;p&gt;It's not clear whether nodes for intersections are necessary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a data structure that can accomodate (state transitions resulting from) the operations &lt;em&gt;continue&lt;/em&gt; and &lt;em&gt;switch lanes&lt;/em&gt;.   The minimal set of states needed is: those pieces of road where, if you continue driving along them, the set of &quot;switch&quot; options is constant. This is the smallest set of states such that any switch operation is between different states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could impose additional nodes thought of as STATIONARY states of motion or FIXED-LOCATION geographic data, to express starting points, destinations, intersections, stopovers or geographic locations, but only the lanes on which a traveller moves are logically necessary.  At a road intersection, one can switch from lane A on one road to lane B on another road, which is expressible as a single state transition, without going through an &quot;inside the intersection&quot; state.  Having an intersection state may be troublesome, as passing through the intersection within different lane-to-lane transition paths will equate situations inside the intersection that are logically different (they support different switch options). &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-20T17:44:11.297" LastActivityDate="2010-08-20T17:44:11.297" />
  <row Id="2875" PostTypeId="2" ParentId="2851" CreationDate="2010-08-20T17:50:59.723" Score="2" ViewCount="0" Body="&lt;p&gt;You have $O\rightarrow O'$, $e_1\rightarrow e_1'$, $e_2\rightarrow e_2'$. Also using cross product to find the orthogonal, we have $e_1 \times e_2 =e_3\rightarrow e_1'\times e_2'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_1=(1,0,0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_2=(0,1,0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_3=(0,0,1,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O=(0,0,0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_1'=(xdir_x,xdir_y,xdir_z,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_2'=(ydir_x,ydir_y,ydir_z,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_3'=e_1'\times e_2'$ [here $\times$ indicates cross product - and you need to append 1 to make it a 4-vector]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O'=(o_x,o_y,o_z,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $xdir_x$, $xdir_y$ etc. are xdir and ydir's coordinates in the first system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to solve an equation in 4x4 matrices for getting the right transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{R}=\begin{pmatrix}e_1\\e_2\\e_3\\O\end{pmatrix}$, &#xA;$\textbf{S}=\begin{pmatrix}e_1'\\e_2'\\e_3'\\O'\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the required transformation $\textbf{T}$, we have $\textbf{R}\textbf{T}=\textbf{S}$ and so $\textbf{T}=\textbf{R}^{-1}\textbf{S}$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-20T23:42:27.787" LastActivityDate="2010-08-20T23:42:27.787" CommentCount="3" />
  <row Id="2877" PostTypeId="2" ParentId="2872" CreationDate="2010-08-20T18:21:23.147" Score="1" ViewCount="0" Body="&lt;p&gt;I think there is a bit of a mix up over notation - specifically, the problem is using $y$ in two different contexts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the chain rule that you need (and what you have correctly stated):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a composite function $y(u(x)) = [u(x)]^3$ (here, $y$ is the 'cubing' function), in order to take the derivative of $y$ with respect to $x$ we first take the derivative of $y$ w.r.t. $u$, then since $u$ is a function of $x$ (and should be written as $u(x)$ at this point) we use the chain rule as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{du} = 3u^{2}$, and $\frac{du}{dx} = u'(x)$ since we don't have the expression of the function $u(x)$ itself. So it looks like&#xA;$$\frac{d}{dx}([u(x)]^3) = 3[u(x)]^{2}u'(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks different because I have chosen to relabel the function $y^3$ as $u^{3}$ in order for the stated chain rule to 'look right.'  Alternatively, we could state the chain rule as $\frac{dh}{dx} = \frac{dh}{dy}\frac{dy}{dx}$ where $h = h(y(x)) = [y(x)]^3$.  Then $\frac{dh}{dy} = 3y^2$ and $\frac{dy}{dx}$ is just $y'$ again because we don't have a specific expression for that function.  Here we would have $\frac{dh}{dx} = 3y^2y'$.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-20T18:33:34.493" LastActivityDate="2010-08-20T18:33:34.493" CommentCount="4" />
  <row Id="2878" PostTypeId="2" ParentId="1880" CreationDate="2010-08-20T19:03:42.390" Score="7" ViewCount="0" Body="&lt;p&gt;Just for fun...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proof that $\displaystyle \frac{22}{7}$ is a better approximation than $\displaystyle 3.14$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we consider the amazing and well known integral formula for $\displaystyle \frac{22}{7} -\pi$ (for instance see this page: &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80&quot; rel=&quot;nofollow&quot;&gt;Proof that 22/7 exceeds pi&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \frac{22}{7} -\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 &amp;lt; \frac{22}{7} -\pi &amp;lt; \pi - 3.14$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $\displaystyle 0 &amp;lt; \frac{22}{7} - \pi$ follows trivially from the above integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx &amp;lt; \frac{1}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We split this up as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \int_{0}^{\frac{1}{2}}\frac{x^{4}(1-x)^{4}}{1+x^2} + \int_{\frac{1}{2}}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle 0$ and the second integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx &amp;lt; \int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx + \int_{\frac{1}{2}}^{1} \frac{4x^{4}(1-x)^{4}}{5}dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx = \int_{\frac{1}{2}}^{1}x^{4}(1-x)^{4}dx$$ as $\displaystyle x^4(1-x)^4$ is symmetric about $\displaystyle x = \frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also known that $$\int_{0}^{1}x^{4}(1-x)^{4}dx = \frac{1}{630}$$ (see the above page again)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx &amp;lt; \frac{1}{2*630} + \frac{4}{5*2*630} = \frac{1}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{22}{7} - \pi &amp;lt; \frac{1}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; 2(\frac{22}{7} - \frac{1}{700})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{22}{7} -  \frac{2}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{2200}{700} -  \frac{2}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{2198}{700}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi &gt; \frac{22}{7} + \frac{314}{100}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 &amp;lt; \frac{22}{7} - \pi &amp;lt; \pi - \frac{314}{100}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-04T16:11:35.890" LastActivityDate="2010-10-04T16:11:35.890" />
  <row Id="2879" PostTypeId="1" CreationDate="2010-08-20T20:05:33.780" Score="7" ViewCount="181" Body="&lt;p&gt;Take any natural number $n$, and factor it as $n=2^{e_1} 3^{e_2} 5^{e_3} ... p^{e_i}$, &#xA;where $i$ is the $i$-th prime.&#xA;Now map $n$ to the point $n \mapsto (e_1,e_2,\ldots,e_i,0,\ldots)$, where $i$ is the last prime in the factorization&#xA;of $n$.&#xA;For example, &#xA;$$n=123456789&#xA;\mapsto&#xA;(0,2,\ldots,1,\ldots,1,\ldots)$$ &#xA;because $123456789=2^0 3^2 \cdots 3607^1 \cdots 3803^1$.&#xA;So every number in $\mathbb{N}$ is mapped to a point in an arbitrarily high dimensional space.&#xA;This mapping has the property that addition of the vectors corresponds to multiplication of&#xA;the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (extremely vague!) question is: Does this viewpoint helps gain any insights into the &#xA;structure/properties of natural numbers?&#xA;Do line/planes/curves in this space mark out numerically interesting regions?&#xA;Perhaps allowing real or complex numbers?&#xA;The numbers in $\mathbb{N}$ fill this infinite-dimensional space very sparsely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is (very!) far from my research expertise, so any comments/references/links would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-22T13:59:34.660" LastActivityDate="2010-08-22T13:59:34.660" Title="Mapping natural numbers into prime-exponents space" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2880" PostTypeId="2" ParentId="2710" CreationDate="2010-08-20T20:08:03.883" Score="0" ViewCount="0" Body="&lt;p&gt;Incurring the risk of sounding too simplistic, isn't that one of the fundamental properties of a fractal [&lt;a href=&quot;http://en.wikipedia.org/wiki/Fractal&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quasi-self-similarity.&lt;/strong&gt; This is a looser form of self-similarity; the fractal appears approximately (but not exactly) identical at different scales. Quasi-self-similar fractals contain small copies of the entire fractal in distorted and degenerate forms. Fractals defined by &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation&quot; rel=&quot;nofollow&quot;&gt;recurrence relations&lt;/a&gt; are usually quasi-self-similar but not exactly self-similar. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Mandelbrot_set&quot; rel=&quot;nofollow&quot;&gt;Mandelbrot set&lt;/a&gt; is quasi-self-similar, as the satellites are approximations of the entire set, but not exact copies.&lt;/p&gt;&#xA;" OwnerUserId="1366" LastActivityDate="2010-08-20T20:08:03.883" CommentCount="1" />
  <row Id="2881" PostTypeId="2" ParentId="2879" CreationDate="2010-08-20T20:29:12.000" Score="5" ViewCount="0" Body="&lt;p&gt;This is a really broad question, so I'll just point you to two closely related notions.  (About which, I should admit, I know very little.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the function field setting, the objects you're mapping into go by the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry)#Weil_divisor&quot; rel=&quot;nofollow&quot;&gt;divisors&lt;/a&gt;.  Divisors are an important tool in algebraic geometry and the theory of Riemann surfaces, e.g. this is how people talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Roch_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann-Roch&lt;/a&gt;.  Back in the number field setting one can still define divisors and even ask for an arithmetic analogue of Riemann-Roch, which is described in the relevant chapter of Neukirch (I don't have it on hand at the moment).  This type of stuff is the domain of &lt;a href=&quot;http://en.wikipedia.org/wiki/Arakelov_theory&quot; rel=&quot;nofollow&quot;&gt;Arakelov theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, instead of just taking the exponents of the prime factorization you could embed the integers into their &lt;a href=&quot;http://en.wikipedia.org/wiki/Adele_ring&quot; rel=&quot;nofollow&quot;&gt;adele ring&lt;/a&gt;; this is, for example, the modern way to do class field theory.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-20T20:29:12.000" CommentCount="1" />
  <row Id="2882" PostTypeId="1" CreationDate="2010-08-20T20:57:03.113" Score="3" ViewCount="89" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/600/circular-permutations-with-indistinguishable-objects&quot;&gt;Circular permutations with indistinguishable objects&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;We can compute the total number of k-ary necklaces/bracelets of length n using the Euler totient function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Necklace_%28combinatorics%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Necklace_%28combinatorics%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently there is a formula for the number of k-ary necklaces/bracelets when the number of beads for each color is fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the number of necklaces/bracelets of length 5, made up of three colors (red, green, blue), with 2 red beads, 2 green beads, and 1 blue bead. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide the formula or a reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-08-20T20:57:03.113" ClosedDate="2010-08-20T21:51:05.340" Title="Number of Necklace/Bracelets With Fixed Number of Beads" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" CommentCount="2" />
  <row Id="2883" PostTypeId="1" AcceptedAnswerId="2885" CreationDate="2010-08-20T21:01:36.990" Score="1" ViewCount="269" Body="&lt;p&gt;This is a problem from &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131848682&quot; rel=&quot;nofollow&quot;&gt;Introduction to Mathematics - Algebra and Number Systems&lt;/a&gt;&quot; (specifically, exercise set 2 #9), which is one of my math texts. Please note that this isn't homework, but I would still appreciate hints rather than a complete answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If 3p&lt;sup&gt;2&lt;/sup&gt; = q&lt;sup&gt;2&lt;/sup&gt;, where $p,q \in \mathbb{Z}$, show that 3 is a common divisor of p and q.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am able to show that 3 divides q, simply by rearranging for p&lt;sup&gt;2&lt;/sup&gt; and showing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p^2 \in \mathbb{Z} \Rightarrow q^2/3 \in \mathbb{Z} \Rightarrow 3|q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure how to show that 3 divides p.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moron left a comment below in which I was prompted to apply the solution to this question as a proof of $\sqrt{3}$'s irrationality. Here's what I came up with...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;[incorrect solution...]&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct solution is provided in the comments below by Bill Dubuque.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-06T18:43:03.210" LastActivityDate="2011-01-06T18:43:03.210" Title="Show that $3p^2=q^2$ implies $3|p$ and $3|q$" Tags="&lt;number-theory&gt;&lt;proof-strategy&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="2884" PostTypeId="2" ParentId="1829" CreationDate="2010-08-20T21:06:15.857" Score="8" ViewCount="0" Body="&lt;p&gt;I am an undergrad student, and I just finished taking Application of Linear Algebra. Here is what I want from a linear algebra text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EXAMPLES WHICH ACTUALLY DEMONSTRATE THE POINT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of what I don't ever want to see: the first example in my textbook was to show how to do matrix multiplication. Simple, right? No. They used a square symmetric matrix multiplied by itself to demonstrate this, and didn't walk through the individual steps. It just gave some matrix A and said look! A.A = B. See? And of course we didn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So please! Check that your examples can only be understood one way, and that this way is the way you intend.&lt;/p&gt;&#xA;" OwnerUserId="1367" LastActivityDate="2010-08-20T21:06:15.857" CommunityOwnedDate="2010-08-20T21:06:15.857" CommentCount="1" />
  <row Id="2885" PostTypeId="2" ParentId="2883" CreationDate="2010-08-20T21:14:31.180" Score="2" ViewCount="0" Body="&lt;p&gt;Write $q$ as $3r$ and see what happens.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-20T21:14:31.180" CommentCount="3" />
  <row Id="2886" PostTypeId="2" ParentId="2879" CreationDate="2010-08-20T21:20:14.063" Score="6" ViewCount="0" Body="&lt;p&gt;In algebraic number theory such rings of &lt;strong&gt;valuation vectors&lt;/strong&gt; or &lt;strong&gt;adèles&lt;/strong&gt; or &lt;strong&gt;repartitions&lt;/strong&gt; play a key role in globally structuring local information for the global study of number fields. In particular, the topological properties of such adèle rings imply key results in algebraic number theory such as the structure of unit groups, finiteness of the class number, various results at the foundation of the Riemann-Roch theorem, etc. For example, see Tate's below review of a paper by Iwasawa (esp. the penultimate paragraph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MR0053970&lt;/strong&gt; (14,849a) 10.0X&lt;br&gt;&#xA;Iwasawa, Kenkichi. &lt;strong&gt;On the rings of valuation vectors&lt;/strong&gt;.&#xA;Ann. of Math. (2) 57, (1953). 331--356.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be a finite algebraic number field, or an algebraic function field of one variable over a finite constant field. The ring $R$ of valuation vectors over $K$ has the following properties: (1) $R$ is a semi-simple commutative ring with unit element 1; (2) $R$ is locally compact, but neither compact nor discrete; (3) $R$ has a subfield $K$ containing 1, such that $K$ is discrete in $R$ and the residue class space $R/K$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first half of his paper the author proves conversely that any topological ring having these properties is the ring of valuation vectors over a field $K$ of the type described above. The main tool used is the notion of the norm $N(\sigma,G)$ of an automorphism $\sigma$ of a locally compact group $G$, which is the factor by which the Haar measure in $G$ is stretched by $\sigma$. It is multiplicative in $\sigma$; $N(\sigma,G)=N(\sigma,H)N(\sigma,G/H)$ if $\sigma H\subset H$; and $N(\sigma,G)=1$ if $G$ is compact or discrete. If $F$ is a nondiscrete locally compact field, then the function $N(\alpha,F)$ for $\alpha\in F^\ast$ gives the &quot;normed&quot; valuation of $F$. The intersection of all closed maximal ideals $M$ of $R$ is 0, and $N(x,R)=\prod_MN(x,R/M)$ for regular $x\in R$. In particular, if $\xi\in K^\ast$, then from (3), $N(\xi,R)=1$, so that the valuations $N(\xi,R/M)$ of $K$ satisfy the ``product formula.'' From this it follows by more or less familiar methods that $K$ is an arithmetic field and $R$ its valuation vector ring. The next section contains a proof that valuation vector rings do enjoy properties (1), (2), and (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the author shows that the two central theorems of algebraic number theory can be proved directly from the topological properties of $R$. The first of these is the self-duality of $R$ with respect to $K$, on which the analog of the Riemann-Roch theorem for number fields rests. If $\chi$ is any non-trivial additive character of $R$ vanishing on $K$, then the map $x\rightarrow\chi(xy)$ is an isomorphism of $R$ onto its character group, and $\chi(x\xi)=1$ for all $\xi\in K$ if and only if $x\in K$. The essential step in the proof is to show that the kernel of the map $x\rightarrow\chi(xy)$ is a compact ideal of $R$, and is therefore 0. The second theorem states the compactness of the group of idèle classes of norm 1, which is equivalent to the finiteness of class number and unit theorem. The idèle group $J$ is the multiplicative group of regular elements of $R$. It is a locally compact group in the topology for which the convergence of $a_i$ in $J$ means the convergence of $a_i$ and $a_i{}^{-1}$ in $R$. The map $a\rightarrow N(a,R)$ is a homomorphism of $J$ with kernel $J_1$, and the compactness of $J_1/K^\ast$ is proved by a Minkowski-type argument using Haar measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout the paper, the case of function fields over non-finite constant fields is treated in parallel, using linear compactness instead of compactness.&lt;br&gt;&#xA;Reviewed by J. T. Tate&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-20T21:20:14.063" CommentCount="1" />
  <row Id="2887" PostTypeId="1" CreationDate="2010-08-20T21:43:59.903" Score="3" ViewCount="165" Body="&lt;p&gt;I solved the following problem from by book, but the answer of this problem at the end of book is x &amp;lt; = 3 Please tell me how can i get this answer.&#xA;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/n5TTa.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-20T22:30:56.840" LastActivityDate="2010-08-20T23:55:42.040" Title="Question about solving absolute values" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2888" PostTypeId="2" ParentId="2883" CreationDate="2010-08-20T21:46:12.053" Score="1" ViewCount="0" Body="&lt;p&gt;Moron's answer certainly covers your question, but as someone who's not your instructor I'd like to see a few more details in your 'proof' of the first half - can you be more specific about how $q^2/3 \in \mathbb{Z} \Rightarrow 3|q$?  While that's easy, it's not necessarily trivial, and you've elided some details there...&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-08-20T21:46:12.053" CommentCount="4" />
  <row Id="2890" PostTypeId="2" ParentId="2883" CreationDate="2010-08-20T21:50:16.997" Score="1" ViewCount="0" Body="&lt;p&gt;Think about how many times each prime factor must appear on each side of the equation, if you were to break p and q into their prime factorizations. The left side has a 3 in it, how many must the right side have, at least?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-08-20T21:50:16.997" CommentCount="1" />
  <row Id="2891" PostTypeId="2" ParentId="2887" CreationDate="2010-08-20T21:55:17.990" Score="1" ViewCount="0" Body="&lt;p&gt;By definition of absolute value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = x$ if $x &gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = -x$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given $|x-3| = 3 - x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given a real $x$, either $x&gt;3$ or $x \le 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason for splitting it this way is that we have $|x-3|$ and in order to get rid of the || we need to decide whether $x-3 &gt; 0$ or not)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we split into two cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Case 1)&lt;/code&gt;  $x &gt; 3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $x-3 &gt; 0$ and so by definition of absolute value, $|x-3| = x-3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x-3| = 3 - x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-3 = 3 -x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x = 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we assumed $x &gt; 3$, there is no solution to your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Case 2)&lt;/code&gt; $x \le 3$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $x - 3 \le 0$ and so by definition of absolute value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x-3| = -(x-3) = 3-x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore your equation is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3-x = 3-x$ which is true for any $x$ (but keep in mind that we are only considering $x \le 3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence any $x \le 3$ satisfies this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine the two solutions for both the cases and you get $x \le 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way you solved it, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 3$ or $x &amp;lt; 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you combine the two, you can say $x \le 3$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-20T22:22:45.540" LastActivityDate="2010-08-20T22:22:45.540" CommentCount="4" />
  <row Id="2893" PostTypeId="2" ParentId="2887" CreationDate="2010-08-20T22:26:49.327" Score="7" ViewCount="0" Body="&lt;p&gt;I'll answer by editing your solution slightly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the sign of $x-3$:&#xA;$$\begin{align}&#xA;x-3=3-x&amp;amp;\text{ and }x-3\ge 0&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp;-(x-3)=3-x\text{ and }&amp;amp;x-3&amp;lt;0&#xA;\\&#xA;x-3-3+x=0&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; x-3=x-3\text{ and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;2x-6=0&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; x-3-x+3=0\text{ and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;2x=6&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; 0=0\text{ and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;x=3&amp;amp;\text{ and }x\ge 3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; \text{(true) and }&amp;amp;x&amp;lt;3&#xA;\\&#xA;&amp;amp;x=3&amp;amp;\quad\text{ or }\quad&amp;amp;&amp;amp; x&amp;lt;3&amp;amp;&#xA;\\&#xA;&amp;amp;&amp;amp;x\le 3&amp;amp;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: As a further explanation of the problem as a whole, consider the graph below, where $|x-3|$ is shown in blue and $3-x$ is shown in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YQdIF.png&quot; alt=&quot;graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graphs coincide for $x\le 3$ and the blue graph is higher for $x&gt;3$, so the original equation is true for $x\le 3$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-20T22:45:56.460" LastActivityDate="2010-08-20T22:45:56.460" />
  <row Id="2894" PostTypeId="2" ParentId="2855" CreationDate="2010-08-20T22:28:45.517" Score="0" ViewCount="0" Body="&lt;p&gt;We know that $\displaystyle\frac{b}{a}$ is an integer. Therefore $\displaystyle\frac{bc}{ac}=\frac{b}{a}\in \mathbb{Z}$ i.e. $ac|bc$. Otherwise $c=0$ and then the result is trivial.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-08-20T22:28:45.517" />
  <row Id="2897" PostTypeId="2" ParentId="2887" CreationDate="2010-08-20T23:06:05.897" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; It's obvious by a shift: put $\; z = x-3 \;$ in $\; |z| = -z \iff z \le 0 \; $ Making this substitution yields $|x-3| = 3-x \iff x-3 \le 0 \iff x \le 3$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-20T23:55:42.040" LastActivityDate="2010-08-20T23:55:42.040" CommentCount="4" />
  <row Id="2898" PostTypeId="2" ParentId="2864" CreationDate="2010-08-20T23:32:20.923" Score="1" ViewCount="0" Body="&lt;p&gt;This is really a comment which complements earlier responses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For spheres the &quot;formula&quot; for the area of spherical triangle is known as Girard's Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Girard%27s_theorem#Girard.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Girard%27s_theorem#Girard.27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-20T23:32:20.923" CommentCount="1" />
  <row Id="2899" PostTypeId="1" AcceptedAnswerId="2905" CreationDate="2010-08-20T23:40:42.890" Score="7" ViewCount="533" Body="&lt;p&gt;Recently I stumbled upon someone who said he wanted to understand why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\arctan x = \int\dfrac{dx}{1+x^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was confused. This is an easy result in any integral calculus course. But then he explained that although he understood the proof, he wanted to understand it &quot;intuitively&quot;. He wanted to see why it was in terms of arclength and addition and subtraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is there an &quot;intuitive&quot; way to explain this identity?&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-21T00:04:55.253" LastActivityDate="2011-01-05T10:45:32.977" Title="Can this standard calculus result be explained &quot;intuitively&quot;" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;teaching&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="5" />
  <row Id="2901" PostTypeId="1" AcceptedAnswerId="2912" CreationDate="2010-08-21T00:00:30.363" Score="4" ViewCount="187" Body="&lt;p&gt;Let $f(x)$ be a continuous function from $\mathbb{R}\rightarrow\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's denote $k$-times repeated application of the function, $f(f(f(...f(x)...)))$ as $f^{(k)}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call any $x$ a &lt;strong&gt;periodic point&lt;/strong&gt; with period $n$ if $f^{(n)}(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that if a point with period 3 exists, then points with all possible periods exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words is it true that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exists x:f^{(3)}(x)=x\Rightarrow \forall n&gt;0 \exists y:f^{(n)}(y)=y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if so, why?&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-21T01:01:03.197" LastActivityDate="2010-08-21T20:03:24.383" Title="Period of 3 implies chaos" Tags="&lt;real-analysis&gt;&lt;dynamical-systems&gt;" AnswerCount="4" />
  <row Id="2902" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T00:05:07.673" Score="8" ViewCount="0" Body="&lt;p&gt;I believe you want &lt;a href=&quot;http://mathworld.wolfram.com/SharkovskysTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Sharkovsky's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reasonable explanation is in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kaplan, Harvey. &quot;A cartoon-assisted proof of Sarkowskiĭ's theorem.&quot;&#xA;Amer. J. Phys. 55 (1987), no. 11, 1023–1032.&#xA;MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=917121&quot; rel=&quot;nofollow&quot;&gt;917121&lt;/a&gt;&#xA;DOI:&lt;a href=&quot;http://dx.doi.org/10.1119/1.14928&quot; rel=&quot;nofollow&quot;&gt;10.1119/1.14928&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-21T00:05:07.673" CommentCount="2" />
  <row Id="2903" PostTypeId="1" AcceptedAnswerId="2904" CreationDate="2010-08-21T00:15:57.890" Score="7" ViewCount="163" Body="&lt;p&gt;I came across this 'paradox' -&#xA;$$1=e^{2\pi i}\Rightarrow 1=(e^{2\pi i})^{2\pi i}=e^{2\pi i \cdot 2\pi i}=e^{-4\pi^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized the fallacy lies in the fact that in general $(x^y)^z\ne x^{yz}$. Why doesn't it work with complex numbers even though it is valid in real case? Is it related to the fact that logarithm of complex number is not unique?&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-21T00:32:15.953" LastActivityDate="2010-08-21T20:09:52.950" Title="Why can't I combine complex powers" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="2904" PostTypeId="2" ParentId="2903" CreationDate="2010-08-21T00:28:07.273" Score="9" ViewCount="0" Body="&lt;p&gt;Even without any complex numbers: $-1=(-1)^{2\cdot\frac12}\ne((-1)^2)^{\frac12}=1^{\frac12}=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you're right, the problem is that raising to a (non-integer) power is essentially a &lt;em&gt;multivalued&lt;/em&gt; function.&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-08-21T00:28:07.273" CommunityOwnedDate="2010-08-21T00:28:07.273" CommentCount="1" />
  <row Id="2905" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T00:48:21.647" Score="18" ViewCount="0" Body="&lt;p&gt;The geometric picture is as follows.  Let $O = (0, 0), A = (1, 0), X = (1, x)$.  Then $\arctan x = \angle AOX$.  We want to understand why $\angle AOY \approx \arctan x + \frac{h}{1 + x^2}$ where $Y = (1, x + h)$ and $h$ is small; equivalently, we want to understand why $\angle XOY \approx \frac{h}{1 + x^2}$.  Since this angle is small, we equivalently want to understand why $\sin \angle XOY \approx \frac{h}{1 + x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\triangle XOY$ evidently has area $\frac{h}{2}$.  On the other hand, it has area $\frac{1}{2} |OX| |OY| \sin \angle XOY$ where $|OX| = \sqrt{1 + x^2}$ and $|OY| \approx |OX|$.  The result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The derivative follows, anyway.  The integral follows by dividing up $AX$ into little pieces and drawing a bunch of lines to $O$, then summing up all of the contributions.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-21T00:48:21.647" CommentCount="1" />
  <row Id="2906" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T01:02:48.233" Score="5" ViewCount="0" Body="&lt;p&gt;See the Monthly paper of that exact title &lt;a href=&quot;http://pb.math.univ.gda.pl/chaos/pdf/li-yorke.pdf&quot; rel=&quot;nofollow&quot;&gt;Period three implies chaos&lt;/a&gt; - which made the JSTOR &lt;a href=&quot;http://www.maa.org/news/062607jstor.html&quot; rel=&quot;nofollow&quot;&gt;All-Stars&lt;/a&gt; list as the third most frequently accessed Monthly article. See also &lt;a href=&quot;http://www.jstor.org/stable/4145161&quot; rel=&quot;nofollow&quot;&gt;A simple proof of Sharkovsky's theorem&lt;/a&gt; and this &lt;a href=&quot;http://links.jstor.org/sici?sici=0002-9890%28199711%29104%3A9%3C846%3AROST%3E2.0.CO%3B2-G&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; for some history on Sharkovsky's theorem.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-21T01:11:11.897" LastActivityDate="2010-08-21T01:11:11.897" />
  <row Id="2907" PostTypeId="1" AcceptedAnswerId="2909" CreationDate="2010-08-21T02:18:36.487" Score="-3" ViewCount="168" Body="&lt;p&gt;Let $G$ be a group. Let $G^{2}= \{ g^{2} \ | \ g \in G \}$. Can one solve the equation $G \times G \cong G^{2}$, that is finding all groups such that $G \times G$ is isomorphic to $G^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we get an idea of the structure of $G$. Generalizing this can we find groups such that $$G \times G \times \cdots \times G \cong G^{n}$$ for some $n \in \mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-21T02:26:04.990" LastActivityDate="2010-08-21T06:48:30.777" Title="Finding groups such that $G \times G \cong G^{2}$." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2908" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T02:55:32.420" Score="1" ViewCount="0" Body="&lt;p&gt;The real answer I think involves complex numbers. Factor $1+x^2 = (i-x)(-i-x)$, decompose in partial fractions, get an expression involving logarithms that ends up being $\arctan x$ as described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-21T02:55:32.420" CommentCount="3" />
  <row Id="2909" PostTypeId="2" ParentId="2907" CreationDate="2010-08-21T03:16:33.743" Score="6" ViewCount="0" Body="&lt;p&gt;It is probably worth giving an explicit example.  I believe the smallest counterexample is that in $A_4$, the only squares are the $3$-cycles, and the product of two $3$-cycles is not necessarily a $3$-cycle.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-21T03:31:47.293" LastActivityDate="2010-08-21T03:31:47.293" CommentCount="0" />
  <row Id="2910" PostTypeId="2" ParentId="1347" CreationDate="2010-08-21T03:18:02.510" Score="1" ViewCount="0" Body="&lt;p&gt;In a sense there is one model for Euclidean geometry.  However, the geometry of the sphere can be studied on spheres with different radii and, thus, different curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For critters who grew up on a hyperbolic plane there is also a parameter that measures the curvature of their world.  Some nice visuals about this and technical details can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.cornell.edu/~dwh/papers/crochet/crochet.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~dwh/papers/crochet/crochet.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-21T03:18:02.510" />
  <row Id="2911" PostTypeId="2" ParentId="2378" CreationDate="2010-08-21T05:44:06.157" Score="0" ViewCount="0" Body="&lt;p&gt;With respect to the introduction of fields, Erich Reck in his &lt;a href=&quot;http://plato.stanford.edu/entries/dedekind-foundations/#FunGroLat&quot; rel=&quot;nofollow&quot;&gt;&quot;Dedekind's Contributions to the Foundations of Mathematics&quot;&lt;/a&gt;, cites Dedekind as being the first to introduce this concept into mathematics, in the context of his work in algebraic number theory. Reck's article goes into some detail about Dedekind's other contributions to mathematics that seem like they may be of interest, for example, his work on lattices.&lt;/p&gt;&#xA;" OwnerUserId="1179" LastActivityDate="2010-08-21T05:44:06.157" CommunityOwnedDate="2010-08-21T05:44:06.157" />
  <row Id="2912" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T06:06:04.483" Score="2" ViewCount="0" Body="&lt;p&gt;I found &lt;a href=&quot;http://books.google.com/books?id=48YHnbHGZAgC&amp;amp;dq=chaos+an+introduction+to+dynamical+systems&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=wVtvTLaRE4nAsAPVt52uCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CDAQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Chaos: An Introduction to Dynamical Systems&lt;/a&gt; by Kathleen T. Alligood, Tim D. Sauer, and James A. Yorke, to be a sensible text.  A small section of the text contains a brief explanation of Sharkovsky's Theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, Sharkovsky uses this unique ordering of the natural numbers in such a way as to ensure that for each natural number n, the existence of a period-n point implies the existence of periods containing all the periods of orders higher than n. In my opinion, the text provides a reasonable simplification of this ordering and gently walks one through a decipherable step by step proof.&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-08-21T06:06:04.483" />
  <row Id="2913" PostTypeId="2" ParentId="815" CreationDate="2010-08-21T06:17:42.040" Score="4" ViewCount="0" Body="&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/David_Hilbert&quot; rel=&quot;nofollow&quot;&gt;David Hilbert&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Hilbert worked in many areas of mathematics (both pure and applied) and his work on the &quot;Hilbert Program&quot; contributed significantly to the development of modern logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find him particularly inspiring because he serves a reminder that creativity and imagination are important qualities for mathematicians to possess-according to one story, a mathematics student decided to instead become a novelist, to which Hilbert is reported to have replied &quot;He did not have enough imagination for mathematics, but he had enough for novels&quot; (see Constance Reid's book&quot; Hilbert&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1179" LastActivityDate="2010-08-21T06:17:42.040" CommunityOwnedDate="2010-08-21T06:17:42.040" />
  <row Id="2914" PostTypeId="1" AcceptedAnswerId="2920" CreationDate="2010-08-21T06:43:40.787" Score="4" ViewCount="180" Body="&lt;p&gt;Regarding &lt;a href=&quot;http://math.stackexchange.com/questions/2903/why-cant-i-combine-complex-powers&quot;&gt;this thread&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is not possible to combine complex powers in the usual way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x^y)^z = x^{yz} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was mention of multi-valued functions, is there some theory that makes this all work out correctly?&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T14:15:16.903" Title="How to combine complex powers?" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="2915" PostTypeId="2" ParentId="2907" CreationDate="2010-08-21T06:48:30.777" Score="2" ViewCount="0" Body="&lt;p&gt;Although I don't think this is a very interesting question, I'll give&#xA;an example where it holds. Just take the countable direct product&#xA;$\prod_{n=1}^\infty\mathbb{Z}$ of the additive group $\mathbb{Z}$ of integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly any nontrivial example is infinite, but this example is uncountable.&#xA;A follow-up exercise: is there a countably infinite example?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-21T06:48:30.777" CommentCount="2" />
  <row Id="2916" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T07:01:11.173" Score="2" ViewCount="0" Body="&lt;p&gt;Since $D \tan(x) = 1 + \tan(x)^2$ and $\tan(\arctan(x)) = x$ we see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\tan(\arctan(x)) &amp;amp;= x \\&#xA;(1 + \tan(\arctan(x))^2 \cdot D \arctan(x) &amp;amp;= 1 \\&#xA;D \arctan(x) &amp;amp;= \frac{1}{(1 + x)^2} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by the fundamental theorem of calculus we recover the integral.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T07:01:11.173" CommentCount="1" />
  <row Id="2917" PostTypeId="1" AcceptedAnswerId="2919" CreationDate="2010-08-21T07:12:07.843" Score="6" ViewCount="154" Body="&lt;p&gt;For two matrices $\textbf{S}$ and $\textbf{T}$, a proof of $\det(\textbf{ST})=\det(\textbf{S})\det(\textbf{T})$ is given below in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_graphical_notation&quot; rel=&quot;nofollow&quot;&gt;diagrammatic tensor notation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/75Hzw.jpg&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\det$ denotes the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can the antisymmetrizing bar be inserted in the middle because &quot;there is already antisymmetry in the index lines&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an introduction to the notation, you can refer to Figures 12.17 and 12.18 below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/UIw4m.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/a3lZi.png&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T08:17:14.950" Title="Proof of det(ST)=det(S)det(T) in Penrose graphical notation" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;tensors&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2918" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T07:13:40.883" Score="5" ViewCount="0" Body="&lt;p&gt;For complex numbers the general way of defining $x^y$ (with $x$ nonzero)&#xA;is as $\exp(ly)$ where $l$ is one of the logarithms of $x$. But $x$ does&#xA;not have a unique logarithm, it has many. If $l$ is one logarithm of $x$&#xA;then the general one is $l+2n\pi i$ with $n$ an integer. Hence the general&#xA;value of $x^y$ is&#xA;$$\exp(ly+2\pi iny)=\exp(ly)\exp(2\pi i ny).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before proceeding further note that if $y$ is an integer then $\exp(2\pi i ny)=1$&#xA;so that $x^n$ is uniquely determined. More generally if $y$ is a rational&#xA;with denominator $d$ then $\exp(2\pi i ny)$ (and so $x^y$) takes $d$ different values.&#xA;But if $y$ is irrational, then $x^y$ takes infinitely many variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the problem in hand $x^y$ takes the values $\exp(ly)\exp(2\pi i ny)$&#xA;as above, and $x^{yz}$ the values&#xA;$$\exp(lyz)\exp(2\pi inyz).$$&#xA;However in general each value of $x^y$ gives rise to infinitely many values&#xA;of $(x^y)^z$, namely&#xA;$$\exp((ly+2\pi iny)z+2\pi i mz)$$&#xA;so $(x^y)^z$ has a &quot;doubly-infinite&quot; family of values. In general, there&#xA;will be values of $(x^y)^z$ which are not values of $x^{yz}$&#xA;so the &quot;paradox&quot; in the &lt;a href=&quot;http://math.stackexchange.com/questions/2903/&quot;&gt;previous posting&lt;/a&gt; should be expected.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-21T07:13:40.883" />
  <row Id="2919" PostTypeId="2" ParentId="2917" CreationDate="2010-08-21T07:43:17.920" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KS4YK.png&quot; alt=&quot;3!&quot;&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T08:17:14.950" CommentCount="3" />
  <row Id="2920" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T08:01:09.897" Score="3" ViewCount="0" Body="&lt;p&gt;Since people keep hearing about Riemann surfaces in the three threads thus far about complex exponentiation and the logarithm being multivalued, but are probably wondering about the multivalued part, here's a visualization courtesy of Michael Trott:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tSh7x.gif&quot; alt=&quot;logarithm branch cut&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a plot of the logarithm's imaginary part over the complex plane. Note the branch cut running through the negative real axis. Analytic continuation consists of joining together copies of this cut surface above and below (corresponding to adding or subtracting integer multiples of $2\pi i$) and matching up the cuts of each piece to maintain continuity, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/goruf.gif&quot; alt=&quot;Riemann surface of logarithm&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the right hand side is the analytically continued imaginary part)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;http://functions.wolfram.com/ElementaryFunctions/Log/visualizations/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-21T08:14:57.857" LastActivityDate="2010-08-21T08:14:57.857" CommunityOwnedDate="2010-08-21T08:01:09.897" />
  <row Id="2921" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T08:05:35.317" Score="1" ViewCount="0" Body="&lt;p&gt;I guess the problem in general arises because $a^b$ can have multiple values. We have&#xA;$$\begin{eqnarray}&#xA;\log((x^y)^z) &amp;amp; = &amp;amp; z(\log(x^y)+2\pi im)=z(y(\log(x)+2 \pi in)+2\pi im)\\&#xA;&amp;amp; = &amp;amp; zy\log(x)+2\pi i(ny + m)z\\&#xA;&amp;amp; = &amp;amp; \log(x^{yz}e^{2\pi i(ny + m)z})&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems a &lt;strong&gt;correction factor&lt;/strong&gt; of $e^{2\pi i(ny + m)z}$ is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all that can be said is that $\exists m,n\in\mathbb{Z}:(x^y)^z=x^{yz}e^{2\pi i(ny + m)z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we can say that the correction term is 1 if $(ny+m)z$ is always an integer, which happens for example if we know $y,z\in\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T08:05:35.317" />
  <row Id="2922" PostTypeId="2" ParentId="2189" CreationDate="2010-08-21T08:38:23.397" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Freudenthal_magic_square&quot; rel=&quot;nofollow&quot;&gt;The Freudenthal–Tits magic square&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-08-21T08:38:23.397" CommunityOwnedDate="2010-08-21T08:38:23.397" />
  <row Id="2923" PostTypeId="1" AcceptedAnswerId="2948" CreationDate="2010-08-21T09:59:47.107" Score="1" ViewCount="234" Body="&lt;p&gt;There are at least 2 places in Wikipedia saying that X_n converges to X in mean &lt;strong&gt;in and only if&lt;/strong&gt; X_n converges to X &lt;strong&gt;in probability&lt;/strong&gt; and X_n is uniformly integrable. See the following link for example:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I found a reference in Billingsley's book saying that we only need &lt;strong&gt;convergence in distribution&lt;/strong&gt; along with uniform integrability to imply the convergence in mean. Is Wikipedia wrong, or am I missing something here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1380" LastEditorUserId="1380" LastEditorDisplayName="" LastEditDate="2010-08-21T10:12:00.060" LastActivityDate="2010-11-28T19:38:16.860" Title="Do convergence in distribution along with uniform integrability imply convergence in mean?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="4" />
  <row Id="2924" PostTypeId="1" AcceptedAnswerId="2931" CreationDate="2010-08-21T10:02:35.623" Score="2" ViewCount="81" Body="&lt;p&gt;I am looking for an element $(a_n)_{n\in\mathbb Z}\in\ell^1(\mathbb Z)$ with the property&#xA;$$&#xA;\lVert(a_n)\rVert_{\ell^1(\mathbb Z)} &gt; \lVert\sum_{n\in\mathbb Z}a_n z^n\rVert_\infty,&#xA;$$&#xA;where the norm on the right hand side denotes the sup-norm on $\mathcal C(\mathbb T)$ ($\mathbb T$ is the 1-Torus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I want to prove that the Gelfand transformation on $\ell^1(\mathbb Z)$ is not isometric.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-21T12:20:54.107" LastActivityDate="2010-08-21T13:28:03.473" Title="The Gelfand transformation on $\ell^1(\mathbb Z)$ is not isometric. Do you have an example?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2925" PostTypeId="2" ParentId="2899" CreationDate="2010-08-21T10:06:39.503" Score="4" ViewCount="0" Body="&lt;p&gt;Put $\arctan x=:\alpha$, let $n$ be large, choose points $t_k:=\tan(k\alpha/n)$  $\ (0\leq k\leq n)$ and put  $\tau_k:=\sqrt{t_k t_{k-1}}$. Then&#xA;$$\sum_{k=1}^n  {t_k - t_{k-1} \over 1 + \tau_k^2}=\sum_{k=1}^n  {t_k - t_{k-1} \over 1 + t_k t_{k-1}}= n \tan(\alpha/n).$$&#xA;Here the left side is a Riemann sum for the integral $$\int_0^{\tan\alpha}{dt \over 1+t^2}=\int_0^x{dt \over 1+t^2},$$ and the right side has limit $\alpha=\arctan x$ when $n\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-08-21T10:06:39.503" CommentCount="3" />
  <row Id="2926" PostTypeId="2" ParentId="2575" CreationDate="2010-08-21T10:09:15.117" Score="0" ViewCount="0" Body="&lt;p&gt;[In short, any of the two quantities will satisfy your need- $d$ (easy to calculate), or a monotonic function of $d$ $P(D\lt d)$ (needs bootstrapping to calculate). Both share the property that smaller they are, bigger the chance that new point comes from surveyed data. are described below. For a detailed discussion, read on.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can essentially be reformulated as a statistical hypothesis testing problem, and we need to test if the new observation comes from the surveyed population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will outline the construction of one such test. For this I will assume the questions are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_0$: New observation comes from the population. Assuming independence, the population can be characterized by a vector of probabilities $(p_1,p_2,...,p_n)$ for the $n$ questions. These $p_i$'s correspond to your calculated % of &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_1$: New observation does not come from the population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the new observation is $B=(B_1,...,B_n)$, where each $B_i$ is 1 if &quot;yes&quot; and 0 if &quot;no&quot; was entered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reasonable test statistic would be $D=\sum (B_i-p_i)^2$, which looks at the total squared deviation from the expected answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then judge if you want to label the new sample as coming from your population by evaluating $P(D\lt d)$ - if this probability is &amp;lt;5%, you can say with 95% confidence that it comes from the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing the exact distribution of $D$ under $H_0$ will be challenging, to determine $P(D\lt d)$ you can use bootstrapping - i.e. simulating observations under $H_0$ a million times and checking what proportion of times your simulated $D$ is less than observed $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a 0 to 1 valued 'correlation' like statistic, you can look at $1-P(D\lt d)$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T10:09:15.117" />
  <row Id="2927" PostTypeId="2" ParentId="2923" CreationDate="2010-08-21T10:17:06.047" Score="0" ViewCount="0" Body="&lt;p&gt;Convergence of mean (or any other feature of the distribution for that matter) does not imply weak convergence of the random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the example of an infinite sequence of iid random variables of any non degenerate distribution and see what is going wrong.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T10:17:06.047" CommentCount="2" />
  <row Id="2929" PostTypeId="2" ParentId="2923" CreationDate="2010-08-21T11:12:06.877" Score="2" ViewCount="0" Body="&lt;p&gt;Slight misunderstanding in KalEl's answer: convergence &lt;i&gt;in&lt;/i&gt; mean - which is convergence in the $L^1$ norm associated with the probability measure, and as such is a feature of the sequence of random variables itself rather than only the sequence of distributions - isn't the same thing as convergence &lt;i&gt;of&lt;/i&gt; mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, no, convergence in distribution as such isn't enough, even given uniform integrability - for instance, take two identically distributed nonzero random variables with a.s. disjoint support and alternate between them for your sequence, say (take $[0,1]^2$ with the usual Lebesgue measure for your probability space and use the indicator functions of the top and bottom half, for example), if KalEl's suggestion of an iid sequence doesn't immediately help you.&lt;/p&gt;&#xA;" OwnerUserId="1381" LastActivityDate="2010-08-21T11:12:06.877" CommentCount="1" />
  <row Id="2930" PostTypeId="2" ParentId="535" CreationDate="2010-08-21T12:03:20.307" Score="2" ViewCount="0" Body="&lt;p&gt;I presume that you are restricting $c$ to be real, so your question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which complex numbers $z$ can be roots of equations&#xA;  $$4z^4+8x^3 -3z^2 -9z+c=0$$&#xA;  as $c$ varies through the reals?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the same question as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which complex numbers is&#xA;  $$g(z)=4z^4+8x^3 -3z^2 -9z$$&#xA;  real?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly if $z$ is real then $g(z)$ is real, so we are really interested&#xA;in non-real solutions for $z$.&#xA;Now a complex number $w$ is real iff $w-\overline{w} =0$ where $\overline{w}$&#xA;is the complex conjugate of $w$. So we want to solve the equation&#xA;$$g(z)-\overline{g(z)}=0,$$&#xA;equivalently&#xA;$$4(z^4-\overline{z}^4)+6(z^3-\overline{z}^3)&#xA;-3(z^2-\overline{z}^2)-9(z-\overline{z})=0.\qquad\qquad(*)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set $z=x+yi$ with $y$ real we can simplify this using&#xA;$$z-\overline{z}=2yi,$$&#xA;$$z^2-\overline{z}^2=4xyi,$$&#xA;$$z^3-\overline{z}^3=2(3x^2-y^2)yi$$&#xA;and&#xA;$$z^4-\overline{z}^4=2(4x^3-4xy^2)yi.$$&#xA;Thus $(*)$ becomes, on cancelling $2i$,&#xA;$$y[16(x^3-xy^2)+16(3x^2-y^2)-6x-9]=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factor of $y$ in the above equation&#xA;accounts for the $x$-axis (so all real&#xA;values of $z$) and so the other solutions lie on the curve defined by&#xA;the equation&#xA;$$16(x^3-xy^2)+16(3x^2-y^2)-6x-9=0.$$&#xA;I'm not going to analyse this equation further, but it looks&#xA;a fairly typical cubic, so it probably defines an elliptic curve.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-21T12:03:20.307" />
  <row Id="2931" PostTypeId="2" ParentId="2924" CreationDate="2010-08-21T12:25:33.990" Score="4" ViewCount="0" Body="&lt;p&gt;Take for instance $(1, -1, -1)$. The $\ell^1$-norm is three. However, there is no real $x$ with $|1 - e^{ix} - e^{2ix}| $ equalling 3. Indeed, this would imply (by the equality case of the triangle inequality) that $-1 = e^{ix} = e^{2ix}$, which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-21T12:25:33.990" CommentCount="1" />
  <row Id="2932" PostTypeId="2" ParentId="2924" CreationDate="2010-08-21T12:31:00.187" Score="1" ViewCount="0" Body="&lt;p&gt;You corrected the misprint good! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f=(a_n)\in\ell^1$, the Gelfand transform $\hat{f}$ of $f$ is given by&#xA;$$\hat{f}(t)=\sum a_ne^{int}$$&#xA;hence $$\sup_t |\hat{f}(t)|=\sup_t\left|\sum a_ne^{int}\right|\leq \sum |a_n|=\|f\|_{\ell^1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also take a closer look at the Dirichlet kernel or the Féjer kernel in order to find examples where you have strict inequality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given any commutative Banach algebra we have the spectral radius formula&#xA;$$&#xA;\sup|\hat{f}|=\inf_{n\geq1} \|f^n\|^{1/n}$$&#xA;and by the Banach algebra inequality we have &#xA;$$\|f^n\|^{1/n}\leq\|f\|$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-21T13:28:03.473" LastActivityDate="2010-08-21T13:28:03.473" CommentCount="1" />
  <row Id="2933" PostTypeId="2" ParentId="535" CreationDate="2010-08-21T12:54:49.657" Score="1" ViewCount="0" Body="&lt;p&gt;We need to find all $z:\exists c\in \mathbb{R}:p_c(z)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some manipulation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}&#xA;&amp;amp;&amp;amp;p_c(z)=0\\&#xA;\Leftrightarrow&amp;amp;p_0(z)+c=0\\&#xA;\Leftrightarrow&amp;amp;Re(p_0)+i Im(p_0)+c=0\\&#xA;\Leftrightarrow&amp;amp;Re(p_0)+c=0\text{ and } Im(p_0)=0\\&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $Im(p_0(z))=0$ is a fixed set of points on the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Re(p_0)+c$ defines a surface on on $\mathbb{C}$ which varies with $c$. The surfaces that we get varying $c$ are just the translation of the surface $z=Re(p_0)$, along $z$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our required set of points are those that lie in intersection of these the sets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{z\in \mathbb{C}:Im(p_0(z))=0\}\cap\{z\in \mathbb{C}:\exists c:Re(p_0(z))+c=0\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\{z\in \mathbb{C}:Im(p_0(z))=0\}\cap\mathbb{C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\{z\in \mathbb{C}:Im(p_0(z))=0\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1:&lt;/strong&gt; As an example let's take $p_c(z)=z^2+c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot of $Im(p_0(z))=2 x y = 0$ is just the two axes, as given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/ksZ9g.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the required locus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plot of $Re(p_0(x))$ is also given below. This doesn't help in finding the locus though, and this is attached only to see why the second term in the intersection above is entire $\mathbb{C}$. Evidently, the points which intersect with the complex plane as we move the surface up and down is the whole complex plane. This will hold true for any polynomial as polynomials are holomorphic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/Iuyxg.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 2:&lt;/strong&gt; Let's take your example $p_c(z)=z^4+8z^3−3z^2−9z+c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}\&#xA; Im(p_0(z))&amp;amp;=&amp;amp;Im(p_0(x+iy))\\&#xA;&amp;amp;=&amp;amp;16 x^3 y+24 x^2 y-16 x y^3-6 x y-8 y^3-9 y\\&#xA;&amp;amp;=&amp;amp;-y \left(-16 x^3-24 x^2+16 x y^2+6 x+8 y^2+9\right)\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot of $p_0(z)=0$ (which gives your locus) is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/8Bmly.png&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-21T13:00:58.130" LastActivityDate="2010-08-21T13:00:58.130" />
  <row Id="2934" PostTypeId="2" ParentId="2914" CreationDate="2010-08-21T13:45:53.047" Score="4" ViewCount="0" Body="&lt;p&gt;The original question asked about a &lt;em&gt;theory&lt;/em&gt; that would make this all work out correctly.  The previous answers are all correct, but they did not point out the underlying, unifying factor which makes all these computations much simpler: the &lt;a href=&quot;http://www.google.ca/url?sa=t&amp;amp;source=web&amp;amp;cd=3&amp;amp;ved=0CCMQFjAC&amp;amp;url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.49.3350%26rep%3Drep1%26type%3Dpdf&amp;amp;ei=zNVvTP_9HMmRnwex0JSGCA&amp;amp;usg=AFQjCNGOK8yueIBVHnvGITOxqnxkh-2s_w&amp;amp;sig2=_Sm-9lIGcJ-evsY3BxlY0Q&quot; rel=&quot;nofollow&quot;&gt;Unwinding number&lt;/a&gt;.  The &lt;em&gt;unwinding number&lt;/em&gt; $\mathcal{K}(z)$ is defined to be&#xA;$$ \ln(\exp(z)) = z + 2\pi i\mathcal{K}(z).$$&#xA;It is an integer which tells you how many times you need to 'unwind' $z$ before you get into the range where $\ln\circ \exp$ is the identity.  This then allows a lot of identities to be 'fixed' with relative ease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also find the paper &lt;a href=&quot;http://www.apmaths.uwo.ca/~djeffrey/Offprints/AISC2000.pdf&quot; rel=&quot;nofollow&quot;&gt;Reasoning about the Elementary Functions of Complex Analysis&lt;/a&gt; by Corless, Davenport, Jeffrey, Litt and Watt quite instructive in this domain.&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="620" LastEditorDisplayName="" LastEditDate="2010-08-21T14:15:16.903" LastActivityDate="2010-08-21T14:15:16.903" CommentCount="3" />
  <row Id="2935" PostTypeId="2" ParentId="2869" CreationDate="2010-08-21T14:13:28.297" Score="1" ViewCount="0" Body="&lt;p&gt;I can't find a closed-form, but I can get partway.  Note that your $\Gamma$ can be expressed as $1-\Gamma(n,\beta t)/\Gamma(n)$ using the incomplete gamma function; it can also be expressed (in more complicated form) using the WhittakerM function (aka 1F1).  Thus so can $h(t)$.  What this really means is that $h(t)$ is holonomic and it particular satisfies a 4th order linear ordinary differential equation with polynomial coefficients.  The product-integral in $g$'s definition bumps the order by one, and then integration (to get $f$) by another order, so $f(t)$ satisfies an order $6$ LODE.  With all those parameters, it is hopeless to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a few relations - like if $P_8$ and $P_9$ are related - then you might have a chance.  Basically, unless you can knock down the complexity of what $h$ actually is, it is rather unlikely that the whole process will 'collapse' in the presence of that many free parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $n=3$ makes things much simpler.  Then $h(t)$ is a simple sum of exponentials.  It is however integrable in closed-form (but is a medium-sized mess, so I won't post it here).  But that's where things are likely to stop, since exponentials-of-exponentials do not tend to integrate in closed-form very often.  Numerically, one ought to be able to generate some very good approximations for evaluating $f(t)$ though.  I guess it depends on what you are eventually trying to do with your equations.&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="620" LastEditorDisplayName="" LastEditDate="2010-08-30T19:30:28.917" LastActivityDate="2010-08-30T19:30:28.917" />
  <row Id="2936" PostTypeId="1" CreationDate="2010-08-21T15:07:43.070" Score="2" ViewCount="136" Body="&lt;p&gt;I can understand how the Greek alphabet came to be prominent in mathematics as the Greeks had a huge influence in the math of today. Certain letters came to have certain implications about their meaning (i.e. θ is almost always an angle, never a function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why did x and y come to prominence? They seem like 2 arbitrary letters for input and output, and I can't think why we began to use them instead of a and b. Why did they become the de facto standard for Cartesian coordinates?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-21T17:19:00.267" LastActivityDate="2010-08-21T21:49:32.057" Title="Why are x and y such common variables in todays equations? How did their use originate?" Tags="&lt;notation&gt;&lt;math-history&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="2937" PostTypeId="1" AcceptedAnswerId="2940" CreationDate="2010-08-21T15:15:33.200" Score="1" ViewCount="63" Body="&lt;p&gt;In a proposal, assume $p$ is the condition and $q$ the conclusion, so the proposal is $p \to q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\neg$ be the negation on either condition or conclusion.&lt;br&gt;&#xA;what is the name for the operation from $p\to q$ to $\neg p\to\neg q$?&#xA;And what is the name for the operation from $p \to q$ to $p\to\neg q$? And from $p\to q$ to $\neg q\to\neg p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What topics in logic cover these stuffs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove $p\to q$, is it equivalent to disprove $p\to\neg q$? Is this the so-called &quot;proof by contradiction&quot;? And is it equiv to prove $\neg q\to\neg p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-08-21T15:33:59.387" LastActivityDate="2010-08-21T15:40:14.483" Title="logic operations on proposals" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="2938" PostTypeId="2" ParentId="2936" CreationDate="2010-08-21T15:20:21.217" Score="1" ViewCount="0" Body="&lt;p&gt;Classically &quot;x&quot; was always used in the sciences for denoting an unknown quantity, e.g. the &quot;X-rays&quot; of Röntgen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cajori has a nice discussion in &quot;A History of Mathematical Notations&quot;. In brief, Descartes's convention was to use letters from the earlier half (a,b,c...) for known quantities and from the latter half (x,y,z...) for unknowns. This was in the 1600s if I remember correctly.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-21T15:20:21.217" />
  <row Id="2939" PostTypeId="2" ParentId="2937" CreationDate="2010-08-21T15:24:36.503" Score="2" ViewCount="0" Body="&lt;p&gt;Briefly, $p\rightarrow q$ can be read as &quot;if $p$, then $q$&quot;; &quot;if not p then not q&quot; would be the &quot;inverse&quot;, and &quot;if not q then not p&quot; is the &quot;contrapositive&quot;, which has the same truth value as the original proposal.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-21T15:24:36.503" CommentCount="4" />
  <row Id="2940" PostTypeId="2" ParentId="2937" CreationDate="2010-08-21T15:26:58.467" Score="3" ViewCount="0" Body="&lt;p&gt;You can find a table of these &lt;a href=&quot;http://en.wikipedia.org/wiki/Contraposition#Comparisons&quot; rel=&quot;nofollow&quot;&gt;on Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{c|l} \hline&#xA;p \to q &amp;amp; \text{Implication (Conditional)} \\ &#xA;\neg p \to \neg q &amp;amp; \text{Inverse} \\ &#xA;q \to p &amp;amp; \text{Converse} \\ &#xA;\neg q \to \neg q &amp;amp; \text{Contrapositive} \\ &#xA;\phantom\neg p \to \neg q &amp;amp; \text{Contradiction} \\ \hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $p \to q$ is equivalent to $\neg p \vee q$, Therefore (Implication &amp;hArr; Contrapositive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;(p \to q) &amp;amp;\equiv (\neg p \vee q) \\&#xA;&amp;amp;\equiv (\neg\neg q \vee \neg p) \\&#xA;&amp;amp;\equiv (\neg q \to \neg p)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if we can disprove $p\to\neg q$, then we are sure both $p$ and $q$ are true, which is stronger than proving $p\to q$ that allows $p$ to be false. So they are not &quot;equivalent&quot;.&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-21T15:40:14.483" LastActivityDate="2010-08-21T15:40:14.483" CommentCount="0" />
  <row Id="2941" PostTypeId="2" ParentId="2883" CreationDate="2010-08-21T15:35:22.520" Score="2" ViewCount="0" Body="&lt;p&gt;Below is a &lt;strong&gt;conceptual&lt;/strong&gt; proof of the irrationality of square-roots. It shows that this result follows immediately from &lt;strong&gt;unique fractionization&lt;/strong&gt; -- the uniqueness of the denominator of any reduced fraction -- i.e. the least denominator divides every denominator. This in turn follows from the key fact that the set of all possible denominators of a fraction is closed under subtraction so comprises an &lt;strong&gt;ideal&lt;/strong&gt; of $\mathbb Z$, necessarily principal, since $\mathbb Z$ is a $\rm PID$. But we can easily eliminate this highbrow language to obtain the following conceptual high-school level proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ Let $\;\rm n\in\mathbb N.\;$ Then $\;\rm r = \sqrt{n}\;$ is integral if rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ Consider the set $\rm D$ of all the possible denominators $\rm d$ for $\rm r, \;$ i.e. $\;\rm D = \{ d\in\mathbb Z \;:\: dr \in \mathbb Z \;\}$.&#xA;Notice $\rm D$ is closed under subtraction: $\rm d,e \in D \Rightarrow dr,er\in\mathbb Z \Rightarrow (d-e)\:r = dr - er \in\mathbb Z.\;$&#xA;Further $\rm d\in D \Rightarrow dr\in D$ since $\rm (dr)r = dn\in\mathbb Z \;$ via $\;\rm r^2 = n\in\mathbb Z.\;$ Therefore, invoking the Lemma below, &#xA;with $\rm d $ the least positive element in $\rm D,$ we infer that $\;\rm d|dr \;$ in $\mathbb Z,\;$ i.e. $\rm\ r = (dr)/d \in\mathbb Z\:.\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ Suppose $\;\rm D\subset\mathbb Z \;$ is closed under subtraction and that $\rm D$ contains a nonzero element.&lt;br&gt;&#xA;Then $\rm D \:$ has a positive element and the least positive element of $\rm D$ divides every element of $\rm D\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ 0 \ne d\in D \Rightarrow d-d = 0\in D\Rightarrow 0-d = -d\in D\:.\ $ Hence $\rm D$ contains a positive element. Let $\rm d$ be the least positive element in $\rm D$. Since $\rm d|n \iff d\;|-n, \;$ if $\rm c\in D$ is not divisible by $\rm d$ then we &#xA;may assume that $\rm c$ is positive, and the least such element. But $\rm c-d$ is a positive element of $\rm D$ not divisible by $\rm d$ &#xA;and smaller than $\rm c$, contra leastness of $\rm c$. So $\rm d$ divides every element of $\rm D\:.\ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof exploits the fact that the denominator ideal $\rm D$ has the special property that is is closed under multiplication by $\rm\: r\:.\ $ The fundamental role that this property plays becomes clearer when one learns about Dedekind's notion of a &lt;strong&gt;conductor ideal&lt;/strong&gt;. Employing such yields a trivial one-line proof of the generalization that a Dedekind domain is integrally closed since conductor ideals are invertible so cancellable. This viewpoint serves to generalize and unify all of the ad-hoc proofs of this class of results - esp. those proofs that proceed essentially by descent on denominators. This conductor-based &lt;strong&gt;structural&lt;/strong&gt; viewpoint is not as well known as it should be - e.g. even some famous number theorists have overlooked this. See my &lt;a href=&quot;http://mathoverflow.net/questions/30313&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for further details.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-06T07:36:21.130" LastActivityDate="2011-01-06T07:36:21.130" />
  <row Id="2942" PostTypeId="2" ParentId="2936" CreationDate="2010-08-21T15:57:44.300" Score="0" ViewCount="0" Body="&lt;p&gt;I read somewhere this convention was started by &lt;a href=&quot;http://en.wikipedia.org/wiki/Ren%C3%A9_Descartes&quot; rel=&quot;nofollow&quot;&gt;Rene Descartes&lt;/a&gt;. While conceptualizing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartesian_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;coordinate system&lt;/a&gt;, he used $x$ and $y$ to denote the axes. It took root from there on and has been used ever since.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sorry I can't remember the source now, but I will cite if I do remember later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qwfnZ.jpg&quot; alt=&quot;alt Rene Descartes&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T15:57:44.300" />
  <row Id="2943" PostTypeId="1" AcceptedAnswerId="3084" CreationDate="2010-08-21T16:44:15.903" Score="6" ViewCount="141" Body="&lt;p&gt;Exercise 3B.4 on &lt;a href=&quot;http://books.google.com/books?id=pCLhYaMUg8IC&amp;amp;pg=PA84&amp;amp;lpg=PA84&quot; rel=&quot;nofollow&quot;&gt;page 84&lt;/a&gt; of Isaacs's Finite Group Theory asks one to prove a precursor of the Sylow theorems for π-subgroups, and I only know how to do it using the Sylow theorems, so I have missed the point.  I actually needed a similar (false) statement in my research, so I thought it would be wise to get the point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose G is a finite group, and N is a normal subgroup whose index and order are coprime.  Let U be a subgroup of G whose order is coprime to the order of N.  If either N or U is solvable, then show that U is contained in some subgroup H whose order is the index of N in G.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Available tools:&lt;/strong&gt; We already have Schur–Zassenhaus, and so we know that G has a subgroup whose order is the index of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if G has a normal Hall π-subgroup, then every π′-subgroup is contained in some Hall π′-subgroup.  However Hall subgroups are the next section, and containment and π-separability are even in the section after that, so appealing to Hall's generalization of Sylow's theorem will miss the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt:&lt;/strong&gt; Let M be a maximal subgroup of G containing U.  If N is not contained in M, then M∩N is a normal subgroup of M whose order is coprime to its index, and so we can find H ≤ M with U ≤ H and |H| = [M:M∩N]=[G:N], and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N is contained in every maximal subgroup of G containing U, then, …. It'd be nice if N was in the Frattini, but that's not true.  Perhaps at this point it would be wise to use solvability?&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-08-23T03:22:03.437" Title="Schur–Zassenhaus exercise: coprime subgroup contained in a complement" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="2944" PostTypeId="1" AcceptedAnswerId="2947" CreationDate="2010-08-21T17:11:29.450" Score="2" ViewCount="147" Body="&lt;p&gt;Let $f$ be a diffeomorphism, say from $\mathbb R^n$ to $\mathbb R^n$ , such as the transition map between two coordinate charts on a differentiable manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A differential $n$-form (or rather its coefficient function which is obtained by using the canonical one-chart atlas on $\mathbb R^n$) then transforms essentially by multiplication with $\mathrm{det}(Df)$, while integrals transform essentially by multiplication of the integrand with $\lvert\mathrm{det}(Df)\rvert$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the reason for the necessity to choose an orientation in order to define the integral of a top form on a differentiable manifold.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question: What is an intuitive or conceptional reason for these different transformation behaviours of forms and integrands?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-22T08:54:54.207" LastActivityDate="2010-08-22T15:54:49.553" Title="Why do differential forms and integrands have different transformation behaviours under diffeomorphisms?" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="2945" PostTypeId="1" AcceptedAnswerId="2946" CreationDate="2010-08-21T17:37:07.533" Score="3" ViewCount="104" Body="&lt;p&gt;Hi, I am reading p.22 of Dan Bump's Automorphihic forms and representations. A cusp of the congruence group acting on the upper half plane is defined to be an orbit of the action of the congruence subgroup on $\mathbb{Q} \mathbb{P}^1$. It also says that intuitively, the cusps are the places where a fundamental domain of the congruence subgroup touching the boundary of the upper half plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Why does the definition mean what the intuition says?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:11:34.137" LastActivityDate="2010-11-22T02:11:34.137" Title="Definition of cusp of a congruence group" Tags="&lt;number-theory&gt;&lt;modular-forms&gt;" AnswerCount="1" />
  <row Id="2946" PostTypeId="2" ParentId="2945" CreationDate="2010-08-21T17:41:41.080" Score="5" ViewCount="0" Body="&lt;p&gt;You should identify the upper half plane with a subspace of $\mathbb{C}$ with a subspace of the Riemann sphere.  In this identification $\mathbb{P}^1(\mathbb{R})$ is a great circle separating the upper half plane and lower half plane, and $\mathbb{P}^1(\mathbb{Q})$ is the orbit of $\infty$ under $\text{PSL}_2(\mathbb{Z})$.  This orbit breaks up into a union of orbits under any congruence subgroup and these are the points at which a fundamental domain can touch the boundary because &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;under the action of $\text{PSL}_2(\mathbb{Z})$ the fundamental domain touches the boundary only at $\infty$, and&lt;/li&gt;&#xA;&lt;li&gt;a fundamental domain for any congruence subgroup is a union of translates of fundamental domains for $\text{PSL}_2(\mathbb{Z})$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-21T17:41:41.080" CommentCount="1" />
  <row Id="2947" PostTypeId="2" ParentId="2944" CreationDate="2010-08-21T18:09:10.060" Score="2" ViewCount="0" Body="&lt;p&gt;Transforming by a diffeomorphism $f$ shouldn't make an (unsigned) integral negative.  For instance, say that the integral of $f$ over a subset $A$ of $\mathbb{R}^n$ is just the usual Lebesgue integral, no worries about differential forms of orientation.  So shifting between $f$ and $f \circ \alpha$ for $\alpha$ some map cannot possibly change the sign. This is why the absolute value signs appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can consider a &quot;volume element:&quot; this is something that transforms via $|Det f|$ for a diffeomorphism $f$.  Any differential form leads to a volume element (take its absolute value).  But the opposite is not always true. One can still define (unsigned) integration with respect to a volume element.  However, note that the cancellations in signed integration are useful. These cancellations imply that the integral of an exact form on a compact oriented manifold is zero, for instance. So restricting to volume elements loses interesting and useful information.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-22T15:54:49.553" LastActivityDate="2010-08-22T15:54:49.553" CommentCount="2" />
  <row Id="2948" PostTypeId="2" ParentId="2923" CreationDate="2010-08-21T18:16:02.743" Score="2" ViewCount="0" Body="&lt;p&gt;The Wikipedia statement is right, and the statement which you attribute to Billingsley is wrong, in some sense inherently so.  Convergence in distribution is a notion of convergence of &lt;em&gt;measures&lt;/em&gt;. A sequence of random variables can converge in distribution even if they live on totally different probability spaces, and their joint distribution never enters.  In contrast, convergence in mean (i.e. $L^1$) only makes sense for a sequence of random variables defined on the &lt;em&gt;same&lt;/em&gt; probability space (which is to say, you have to know all their joint distributions).  So it doesn't make sense for the former to imply the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if your random variables happen to be defined on the same measure space, it fails in general, as Jacob Katz pointed out.  For a very explicit counterexample, let $\lbrace X_i \rbrace_{i=1}^\infty$ be iid Bernoulli.  Trivially $X_i$ converges in distribution, and they are uniformly integrable because they are uniformly bounded.  But $E|X_i - X_j| = 1/2$ for all $i \ne j$, so the $X_i$ are not Cauchy in mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that Billingsley is in error, but I think it's more likely that you have missed some context.  Can you post the relevant text?&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-21T18:16:02.743" CommentCount="1" />
  <row Id="2949" PostTypeId="1" CreationDate="2010-08-21T18:51:34.310" Score="16" ViewCount="1850" Body="&lt;p&gt;A large part of my fascination in mathematics is because of some very surprising results that I have seen there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember one I found very hard to swallow when I first encountered it, was what is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox&quot; rel=&quot;nofollow&quot;&gt;Banach Tarski Paradox&lt;/a&gt;. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ in finite disjoint parts, rotate and translate them and rejoin (by taking &lt;em&gt;disjoint&lt;/em&gt; union), and you end up with exactly two complete balls of the same radius!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask you which are your most surprising moments in maths?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-12-24T04:51:11.767" CommunityOwnedDate="2010-08-21T19:06:41.580" Title="Which one result in maths has surprised you the most?" Tags="&lt;big-list&gt;" AnswerCount="50" CommentCount="8" FavoriteCount="10" />
  <row Id="2950" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:02:49.637" Score="28" ViewCount="0" Body="&lt;p&gt;If a function of a complex variable is once differentiable, it's infinitely differentiable.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-21T19:02:49.637" CommunityOwnedDate="2010-08-21T21:22:23.977" CommentCount="2" />
  <row Id="2951" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:05:30.913" Score="5" ViewCount="0" Body="&lt;p&gt;There exists a non-reflexive Banach space that is isomorphic to its dual.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-21T19:05:30.913" CommunityOwnedDate="2010-08-21T21:22:23.977" CommentCount="3" />
  <row Id="2952" PostTypeId="1" CreationDate="2010-08-21T19:45:15.200" Score="5" ViewCount="110" Body="&lt;p&gt;Let $\omega:\mathbb{Z}\to (0,\infty)$ and let $1\leq p&amp;lt;\infty$. Consider the space  $\ell^p(\mathbb{Z},\omega)$ of complex valued sequences $f=(a_n)_{n \in \mathbb{Z}}$ such that &#xA;$$\|f\|=\|f\|_{\ell^p(\mathbb{Z},\omega)}:=\left(\sum_{n\in\mathbb{Z}}|a_n|^p\omega(n)^p\right)^{1/p}&amp;lt;\infty.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, given two complex sequences $f=(a_n)_{n \in \mathbb{Z}}$ and $g=(b_n)_{n \in \mathbb{Z}}$ their formal convolution is defined by $f*g=(c_n)_{n\in\mathbb{Z}}$ where $c_n=\sum_{k\in\mathbb{Z}}a_kb_{n-k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to find necessary and sufficient conditions on $\omega$ such that $\ell^p(\mathbb{Z},\omega)$ is a Banach algebra. &#xA;In other words, if $f,g\in\ell^p(\mathbb{Z},\omega)$ then $f*g\in \ell^p(\mathbb{Z},\omega)$ and $\|f*g\|\leq\|f\|\cdot\|g\|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=1$ the condition is $\omega(n+k)\leq\omega(n)\omega(k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- Reformulated the problem so that $f\in\ell^p(\omega)$ is the same as $f\omega\in\ell^p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is an open problem, there are however sufficient conditions:  $\omega^{-p'}*\omega^{-p'}\leq \omega^{-p'}$ where $1/p̈́'+1/p=1$ (the history of the condition is hard to tell but it is given as Lemma 8.11 in Acta Mathematica Volume 174, Number 1, 1-84, &quot;Completeness of translates in weighted spaces on the half-line&quot; by Alexander Borichev and Håkan Hedenmalm). The proof is based on Hölder's inequality:&#xA;$$\|f*g\|^p =\sum_n \left|\sum_k a_kb_{n-k}\right|^p\omega(n)^p\leq $$&#xA;$$\leq\sum_n \left(\sum_k |a_k|^p|b_{n-k}|^p\omega(n-k)^p\omega(k)^p\right)\left(\sum_k\frac{1}{\omega(n-k)^{p'}\omega(k)^{p'}}\right)^{\frac{p}{p'}}\omega(n)^p\leq $$&#xA;$$\qquad\leq\|f\|^p\|g\|^p$$ &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-24T00:34:45.487" LastActivityDate="2010-08-25T16:41:11.567" Title="When is the weighted space $\ell^p(\mathbb{Z},\omega)$  a Banach algebra ($p&gt;1$)? " Tags="&lt;sequences-and-series&gt;&lt;functional-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="2953" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:51:39.807" Score="16" ViewCount="0" Body="&lt;p&gt;Maybe this is too obvious, but the fact that the Rationals are countable blew my mind.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-08-21T19:51:39.807" CommunityOwnedDate="2010-08-21T19:51:39.807" CommentCount="4" />
  <row Id="2954" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T19:52:53.657" Score="2" ViewCount="0" Body="&lt;p&gt;I would not rate this example as surprising, but it did provoke in me a little epiphany when I finally understood it. There is a theorem of category theory that characterizes adjunctions as a pair of functors and a pair of natural transformations satisfying a bunch of equations. Now in some sense, this is a pure formality (the proof is easy), but on the other hand, an adjunction encodes a parameterized universal property, with some implicit quantifiers (over potentially proper classes) floating around. Now think of all the adjunctions you have come across that encode huge amounts of information. The characterization theorem says that this is the same as a pair of 2-cells in a 2-category satisfying a pair of equations. Look, Ma, no quantifiers, no isomorphisms, no nothing. Just a bunch of equations in a 2-category. The single most important concept of category theory and what do we end up with? a pair of equations...&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2010-08-21T19:52:53.657" CommunityOwnedDate="2010-08-21T19:52:53.657" />
  <row Id="2955" PostTypeId="2" ParentId="2944" CreationDate="2010-08-21T19:53:33.110" Score="1" ViewCount="0" Body="&lt;p&gt;A manifold is not necessary orientable. But the &lt;strong&gt;orientation bundle&lt;/strong&gt; always exists. You can construe it as a real line bundle. Then the &lt;strong&gt;density bundle&lt;/strong&gt; is obtained by tensoring the bundle of top forms with the orientation bundle. By definition, a nonvanishing section of the orientation bundle is an &lt;strong&gt;orientation&lt;/strong&gt;, and a section of the density bundle is a &lt;strong&gt;density&lt;/strong&gt;. Compactly supported densities can be integrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locally, orientations &quot;transform&quot; by the sign of the determinant (of the jacobian matrix), top forms by the determinant, and densities by the absolute value of the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It suffices to understand the case of the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT.&lt;/strong&gt; I've just realized that, in your question, you mention manifolds only in a parenthesis, and concentrate on $\mathbb R^n$. Open subsets of $\mathbb R^n$ are of course always orientable. But the key point, again, is to understand what happens on the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the real line, it's natural to write a 1-form as $f\ dx$, and a density as $f\ |dx|$ (say with $f$ continuous). If you have a compact interval $I$, then you can integrate $f\ |dx|$ over $I$, and $f\ dx$ from one bound to the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2.&lt;/strong&gt; You can write the standard orientation of $\mathbb R$ as $|dx|/dx$, and define the integral of the form $f\ dx$ over the interval $I$ equipped with the orientation $\pm|dx|/dx$ as the integral over the nonoriented interval $I$ of the density $\pm f\ |dx|$ equal to the orientation times the form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you only need to define the integral of a density over a nonoriented (compact) interval. [See for instance de Rham's book on differentiable manifolds.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3.&lt;/strong&gt; In the same spirit, to define the $L^2$ space of a manifold, you consider &lt;strong&gt;half-densities&lt;/strong&gt;. More precisely, you complete the pre-hilbert space of compactly supported half-densities. The half-densities &quot;transform&quot; by the square root of the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general principle is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ is an $n$-manifold, then each action of $G:=GL_n(\mathbb R)$ on a manifold $F$ induces a bundle over $M$ with fiber $F$, called the &lt;strong&gt;associated bundle&lt;/strong&gt;. If the action is an $r$-dimensional $\mathbb R$-linear representation, then the associated bundle is a rank $r$ vector bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most basic example is the tangent bundle, associated to the natural representation on $\mathbb R^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\chi$ be the characters of $G$ given by the inverse of the determinant. The bundles of $n$-forms, orientations, densities, and half-densities, are the line bundles respectively associated to the following characters of $G$: $\chi$, the sign of $\chi$, its absolute value, and the square root of its absolute value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the definition of the associated bundle, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Associated_bundle#Fiber_bundle_associated_to_a_principal_bundle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Associated_bundle#Fiber_bundle_associated_to_a_principal_bundle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, the principal bundle is the &lt;strong&gt;frame bundle&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_bundle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Frame_bundle&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-22T04:58:49.807" LastActivityDate="2010-08-22T04:58:49.807" />
  <row Id="2956" PostTypeId="2" ParentId="2221" CreationDate="2010-08-21T20:01:41.377" Score="1" ViewCount="0" Body="&lt;p&gt;I think this is not possible. We need a stronger condition on the $c_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume such a measure exists, then $L:C(S^1) \to \mathbb{C}$ given by&#xA;$$L(f) = \int_{S^1} f \, d\mu$$&#xA;is a linear functional so we can extend it to $L^2(S^1)$. To see it is continuous note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|L(f)| \leq \|f\|_1 \leq \|f\|_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also write this extension as $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Riesz representation theorem we know that $L(f) = \langle f, x \rangle$ for some $x \in L^2(S^1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $L(\sum \zeta^n z^n) = \sum b_n z^n$ where $b_n = \langle \zeta^n, x \rangle$ and we know that this is equal to $\sum c_n z^n$. So that $c_n = \langle \zeta^n, x \rangle$.&#xA;But we also know that $\sum |c_n|^2 &amp;lt; \infty$. This is a stronger condition than convexity as George pointed out below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it is correct this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem looks a lot like the Hausdorff moment problem and the standard work on moment problems is probably the book by Akhiezer: The classical moment problem and some related questions in analysis. &#xA;Hafner Publishing Co., New York 1965.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-08-24T21:46:14.487" LastActivityDate="2010-08-24T21:46:14.487" CommentCount="8" />
  <row Id="2957" PostTypeId="2" ParentId="2901" CreationDate="2010-08-21T20:03:24.383" Score="0" ViewCount="0" Body="&lt;p&gt;This blog entry has a nice intro to the theorem, although no proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://divisbyzero.com/2008/12/18/sharkovskys-theorem/&quot; rel=&quot;nofollow&quot;&gt;http://divisbyzero.com/2008/12/18/sharkovskys-theorem/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T20:03:24.383" />
  <row Id="2958" PostTypeId="2" ParentId="2903" CreationDate="2010-08-21T20:09:52.950" Score="4" ViewCount="0" Body="&lt;p&gt;This is related to a note by Euler, maybe he was the first to realize that $i^i$ is real. Actually, &#xA;$$i^i = (e^{i\pi/2})^i = e^{-\pi/2}$$&#xA;on the other hand &#xA;$$i^i = (e^{i(\pi/2+2\pi n})^i = e^{-\pi/2 -2\pi n},\ n\in\mathbb{Z}.$$&#xA;So maybe it is better to say that $i^i$ is a subset of the $\mathbb{R}$ and that certain equality signs are to be understood as congruences.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-21T20:09:52.950" />
  <row Id="2959" PostTypeId="2" ParentId="2731" CreationDate="2010-08-21T20:10:04.777" Score="1" ViewCount="0" Body="&lt;p&gt;The answer is 1/2 since whether first or second child opens the door is independent of if he is a boy or a girl.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-21T20:10:04.777" />
  <row Id="2960" PostTypeId="2" ParentId="83" CreationDate="2010-08-21T20:14:41.240" Score="1" ViewCount="0" Body="&lt;p&gt;For people this age and older I know of no better way to introduce someone to mathematics than to have them look at Martin Gardner's books.  A sample of his books are listed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&quot; rel=&quot;nofollow&quot;&gt;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-21T20:14:41.240" CommunityOwnedDate="2010-08-21T20:14:41.240" />
  <row Id="2961" PostTypeId="2" ParentId="2731" CreationDate="2010-08-21T20:28:02.213" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a simple way to think about it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know one of the children is a boy (he opened the door). Therefore there are 3 possible sets of children (numbered for later reference), which you can trivially derrive from the Boy-Girl probability tree:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B&lt;sub&gt;1&lt;/sub&gt;B&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;B&lt;sub&gt;3&lt;/sub&gt;G&lt;sub&gt;4&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;5&lt;/sub&gt;B&lt;sub&gt;6&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So there are 4 possibilities for the second child:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;B&lt;sub&gt;1&lt;/sub&gt; (B&lt;sub&gt;2&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;li&gt;B&lt;sub&gt;2&lt;/sub&gt; (B&lt;sub&gt;1&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;4&lt;/sub&gt; (B&lt;sub&gt;3&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;li&gt;G&lt;sub&gt;5&lt;/sub&gt; (B&lt;sub&gt;4&lt;/sub&gt; opened the door)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the  chance is obviously 1/2 that the other child is a girl.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="181" LastEditorDisplayName="" LastEditDate="2010-08-21T20:33:39.830" LastActivityDate="2010-08-21T20:33:39.830" />
  <row Id="2962" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T21:03:36.197" Score="3" ViewCount="0" Body="&lt;p&gt;As an undergraduate, the fact that |P(x)| &gt; |X|.  I recall being surprised at how both how short and easy this was to prove, and that it implied there were infinitely many &quot;sizes&quot; of infinity.   &lt;em&gt;(The standard diagonalization of decimals proof only showed there were two sizes and took more time.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2010-08-21T21:03:36.197" CommunityOwnedDate="2010-08-21T21:03:36.197" CommentCount="4" />
  <row Id="2963" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T21:18:59.750" Score="19" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula&quot; rel=&quot;nofollow&quot;&gt;Cauchy's Integral Formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the values of an analytic function on the edge of disk (or a simple closed curve) are enough to determine all the values within the curve was very surprising to me.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-21T21:18:59.750" CommunityOwnedDate="2010-08-21T21:18:59.750" CommentCount="3" />
  <row Id="2965" PostTypeId="2" ParentId="2936" CreationDate="2010-08-21T21:49:32.057" Score="3" ViewCount="0" Body="&lt;p&gt;This question has been asked previously on MathOverflow, and answered (by Mariano Suárez-Alvarez). You can follow &lt;a href=&quot;http://mathoverflow.net/questions/30307/explanation-why-x-y-z-are-always-variables/30414#30414&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;, and I quote his response below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You'll find details on this point (and precise references) in Cajori's History of mathematical notations, ¶340. He credits Descartes in his La Géometrie for the introduction of x, y and z  (and more generally, usefully and interestingly, for the use of the first letters of the alphabet for known quantities and the last letters for the unknown quantities) He notes that Descartes used the notation considerably earlier: the book was published in 1637, yet in 1629 he was already using x as an unknown (although in the same place y is a known quantity...); also, he used the notation in manuscripts dated earlier than the book by years.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is very, very interesting to read through the description Cajori makes of the many, many other alternatives to the notation of quantities, and as one proceeds along the almost 1000 pages of the two volume book, one can very much appreciate how precious are the notations we so much take for granted!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-21T21:49:32.057" />
  <row Id="2966" PostTypeId="1" AcceptedAnswerId="2975" CreationDate="2010-08-21T21:56:15.267" Score="2" ViewCount="203" Body="&lt;p&gt;Zero is finite. If n is finite, then n+1 is finite. Hence, by induction, all numbers are finite. What is the catch?&lt;/p&gt;&#xA;" OwnerUserId="1385" LastEditorUserId="1385" LastEditorDisplayName="" LastEditDate="2010-08-21T22:46:28.147" LastActivityDate="2010-08-21T23:34:09.600" Title="Induction versus Natural Numbers" Tags="&lt;infinity&gt;&lt;axioms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2967" PostTypeId="2" ParentId="2966" CreationDate="2010-08-21T22:02:11.440" Score="2" ViewCount="0" Body="&lt;p&gt;You might call each &lt;em&gt;natural number&lt;/em&gt; finite (that's what induction applies to).  Wait until you hear about &lt;a href=&quot;http://en.wikipedia.org/wiki/Inaccessible_cardinal&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt;, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  As the comment to the question points out, the natural numbers are collectively infinite in number, if that's what you're referring to.&lt;/p&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-08-21T22:02:11.440" CommentCount="2" />
  <row Id="2968" PostTypeId="1" AcceptedAnswerId="3272" CreationDate="2010-08-21T22:09:20.263" Score="4" ViewCount="392" Body="&lt;p&gt;Each of n persons draws a card from a shuffled deck of k cards numbered from 1 to k.  There are at least as many cards as persons.  The winner is the person who is holding the largest card.  If everyone is honest, how can they mutually determine the identity of the winner, without any person learning additional information beyond that which can necessarily be inferred from n, k, the value of that person's own card, and the identity of the winner?  In the general situation with k &gt;&gt; n for any person this information includes the values of all cards except for the one held by that person, and also the relative value of any two non-winning persons' cards.  If this is not possible, in what ways can this information be minimized?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-06T18:40:04.197" LastActivityDate="2010-09-06T18:40:04.197" Title="Mental card game" Tags="&lt;puzzle&gt;&lt;cryptography&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="2969" PostTypeId="1" CreationDate="2010-08-21T22:09:55.783" Score="9" ViewCount="387" Body="&lt;p&gt;I have this wooden puzzle composed of 25 y-shaped &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentomino&quot; rel=&quot;nofollow&quot;&gt;pentominoes&lt;/a&gt;, where the objective is to assemble them into a 5x5x5 cube. After spending quite a few hours unsuccessfully trying to solve the puzzle, I finally gave up and wrote a program to try all possible combinations using backtracking. Analyzing the results revealed that for every solution found, the computer made - on average - 50 million placements and removals of pieces. This is obviously beyond my capabilities as a human, even if I can see a few steps ahead that a partial solution leads to a &quot;dead end&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is this: given that the puzzle is so symmetrical, is there some way to significantly prune the search tree (maybe even making it possible for me to solve the puzzle on my own)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XrLj1.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry for the poor quality)&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-09-13T18:05:53.573" Title="Tiling pentominoes into a 5x5x5 cube" Tags="&lt;puzzle&gt;&lt;tiling&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="4" />
  <row Id="2970" PostTypeId="2" ParentId="713" CreationDate="2010-08-21T22:11:02.430" Score="1" ViewCount="0" Body="&lt;p&gt;The other solutions posted already do a good job of explaining tensors as dyads, linear transformations, or using the ellipsoid interpretation. I thought I'd give an engineering perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In engineering applications, physical quantities (mass, velocity, force, etc.) are either &quot;scalars&quot;, &quot;vectors&quot;, or &quot;tensors&quot;. What determines which one, is how the quantity changes under a coordinate transformation. In other words: If you rotate your world view, what happens to the quantity in question? Suppose $U$ is a rotation (read: orthogonal) matrix that rotates vectors from the initial frame $F_1$ to another frame $F_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scalars, such as mass, do not change at all. An object's mass does not depend on the orientation of your frame, so $m_1$ = $m_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vectors, such as force or velocity, do change. If $v_1$ is expressed in your starting frame, then $v_2 = U^Tv_1$. Rotating the frame amounts to rotating the vector. All vector quantities tranform in exactly this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensors, such as stress or inertia, change as well. If $J_1$ is expressed in your starting frame, then $J_2 = U^TJ_1U$. Any quantity that transforms in this manner is a tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular momentum (the product of the moment of inertia and the angular velocity) is a vector, because:&#xA;$$&#xA;J_2 w_2 = U^TJ_1U U^Tw_1 = U^T(J_1 w_1)&#xA;$$&#xA;Kinetic energy is a scalar, which is intuitive, but you can verify it using the transformations above. For an object in translational and rotational motion,&#xA;$$&#xA;E = \frac{1}{2} m_1 v_1^T v_1 + \frac{1}{2} w_1^T J_1 w_1 = \frac{1}{2} m_2 v_2^T v_2 + \frac{1}{2} w_2^T J_2 w_2&#xA;$$&#xA;And so on.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-08-22T01:44:03.140" LastActivityDate="2010-08-22T01:44:03.140" />
  <row Id="2971" PostTypeId="1" AcceptedAnswerId="2972" CreationDate="2010-08-21T22:33:24.153" Score="3" ViewCount="93" Body="&lt;p&gt;Let the &lt;em&gt;class number&lt;/em&gt; $h(d)$ denote the number of distinct binary quadratic forms with discriminant $d &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better algorithm for $h$ than brute force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, by brute force I meant to generate enough forms to completely cover the space and then reducing them down to see how many equivalence classes there are.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T02:21:26.950" Title="Algorithms to compute the class number" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="2972" PostTypeId="2" ParentId="2971" CreationDate="2010-08-21T22:45:10.550" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, there are algorithms that are much better than brute force. For example, see section 5.4 in Henri Cohen's book &quot;A course in computational algebraic number theory&quot; for Shanks's baby-step giant-step algorithm $O(|D|^{1/4+\epsilon})$ - which is practical for negative discriminants $D$ up to 25 digits or more and, further, McCurley's sub-exponential algorithm (including Atkin's variant) which is $O(L(|D|)^\alpha)$ for  $\alpha = \sqrt 2 \;$ or perhaps even $\alpha = \sqrt{9/8},$ where $\; L(x) = e^{\sqrt {\ln x \ln\ln x}}$. This can handle $D$ up to 50 digits or more (nowadays, with various improvements, probably around 80 digits or more - the prior numbers are quoted from the 1993 edition of Cohen's book - currently the bible for computational algebraic number theory).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-22T02:21:26.950" LastActivityDate="2010-08-22T02:21:26.950" CommentCount="2" />
  <row Id="2973" PostTypeId="2" ParentId="2966" CreationDate="2010-08-21T22:58:52.217" Score="1" ViewCount="0" Body="&lt;p&gt;The Axiom of Infinity may be relevant to your understanding, because it concerns the existence of an infinite set, which is also an infinite &lt;em&gt;cardinal&lt;/em&gt; number.  Without it all sets are finite and equivalent to natural numbers, for the reason that you state (induction is not powerful enough).  Starting with Cantor's Theorem you can build even larger cardinals by taking the power set of the infinite set asserted to exist by the Axiom of Infinity.  There is also a technique called transfinite induction which is used to prove propositions about ordinals and cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Transfinite_induction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-21T23:10:47.983" LastActivityDate="2010-08-21T23:10:47.983" />
  <row Id="2974" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T23:18:03.930" Score="9" ViewCount="0" Body="&lt;p&gt;PRIMES is in P.  This was surprising to me both because I knew it as an open problem before it was proved, and because the proof is simple enough that I can follow the outline and understand some of the details.  The proof of FLT was not as surprising to me because comprehending it seems to require a lot of background that I don't have.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-22T00:36:51.370" LastActivityDate="2010-08-22T00:36:51.370" CommunityOwnedDate="2010-08-21T23:18:03.930" />
  <row Id="2975" PostTypeId="2" ParentId="2966" CreationDate="2010-08-21T23:34:09.600" Score="6" ViewCount="0" Body="&lt;p&gt;You have to be careful with your wording in this type of question. There is more than one type of number. Usually, what is being referred to is clear from the context, but here it is not. Natural numbers (0,1,2,3,...,etc) are all finite, and standard induction only applies to these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other types of numbers, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number&quot; rel=&quot;nofollow&quot;&gt;cardinal numbers&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_number&quot; rel=&quot;nofollow&quot;&gt;ordinal numbers&lt;/a&gt; which can be infinite. The principle of induction does not apply to these. There is an extension of induction, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;transfinite induction&lt;/a&gt;. On top of showing that the desired property holds for n+1 whenever it holds for n, transfinite induction also involves proving that the property holds in a limiting sense. The property of being finite does not survive this limit, so there is no problem with having infinite ordinal numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually when someone says &lt;em&gt;number&lt;/em&gt; without being clear about what they are referring to, they mean natural numbers, integers, real numbers or, possibly, complex numbers. In that case, there are no infinite numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the kind of thing which should cause problems in a serious mathematical discussion, but I've seen this ambiguity create confusion many times before on internet forums.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-21T23:34:09.600" />
  <row Id="2976" PostTypeId="2" ParentId="2968" CreationDate="2010-08-21T23:45:07.570" Score="7" ViewCount="0" Body="&lt;p&gt;If any person pulls the k value card, they immediately identify themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should this not happen, after a second the individual with the k-1 card identifies himself. Similarly, the k-t person identifies himself after t seconds if no one has been identified previously. On this strategy, it would take at most k-n seconds for someone to identify himself as the highest card holder, with no other information being passed besides everyone aware the value of highest card is k-n and who drew it.&lt;/p&gt;&#xA;" OwnerUserId="557" LastActivityDate="2010-08-21T23:45:07.570" CommentCount="8" />
  <row Id="2977" PostTypeId="2" ParentId="2949" CreationDate="2010-08-21T23:46:58.300" Score="8" ViewCount="0" Body="&lt;p&gt;Oh, I've been surprised a lot of times, but a particularly memorable one for me was learning the &lt;a href=&quot;http://mathworld.wolfram.com/MaximumModulusPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;maximum modulus principle&lt;/a&gt; of complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the numerics front, I still find it amazing that the humble trapezoidal rule is the best one to use for &lt;a href=&quot;http://www.jstor.org/pss/2695765&quot; rel=&quot;nofollow&quot;&gt;integrating periodic functions over a period&lt;/a&gt;, better than Simpson's rule or the other fancier quadrature methods. This can be seen by appealing to Euler-Maclaurin.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-21T23:46:58.300" CommunityOwnedDate="2010-08-21T23:46:58.300" CommentCount="7" />
  <row Id="2978" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T00:41:30.583" Score="1" ViewCount="0" Body="&lt;p&gt;The connection between syntax and model theory.  For example, you can tell that you can't define &quot;field&quot; (the algebraic structure) by equations because the category of fields doesn't have products.  In other words, a &lt;em&gt;property of the models&lt;/em&gt; controls the &lt;em&gt;logical connectives you must use to say what it is&lt;/em&gt;.  There are many results like this. &lt;/p&gt;&#xA;" OwnerUserId="318" LastActivityDate="2010-08-22T00:41:30.583" CommunityOwnedDate="2010-08-22T00:41:30.583" />
  <row Id="2979" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T00:45:50.527" Score="18" ViewCount="0" Body="&lt;p&gt;The fact that you can &lt;a href=&quot;http://new.math.uiuc.edu/optiverse/&quot; rel=&quot;nofollow&quot;&gt;turn a sphere inside out&lt;/a&gt; differentiably.&lt;/p&gt;&#xA;" OwnerUserId="318" LastActivityDate="2010-08-22T08:12:08.317" CommunityOwnedDate="2010-08-22T00:45:50.527" CommentCount="1" />
  <row Id="2980" PostTypeId="1" AcceptedAnswerId="2984" CreationDate="2010-08-22T00:45:51.510" Score="2" ViewCount="111" Body="&lt;p&gt;I'm working on a game which relies on &quot;flinging&quot; an object. That is, click and hold on the object, and then drag and release it, and it continues on the path you were dragging it. Of course, the most well-known example of flinging is with iPhone and Android devices, where you can quickly scroll down a list by quickly swiping your finger upward, giving the illusion of &quot;flinging&quot; the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tracking mouse positions (x,y) and timestamps. But I'm drawing a blank as to how I can take a list of positions and times and get out of it a velocity or curve that an object should follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are my options? Right now I am looking only for a straight-line fling action, but if it's easy to implement some sort of curve that better fits the fling, that would be good information that I might be able to integrate into the design of the game.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T17:35:23.797" LastActivityDate="2010-10-27T17:35:23.797" Title="Math behind a &quot;fling&quot;? (i.e. on a mobile touch device)" Tags="&lt;approximation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2981" PostTypeId="1" CreationDate="2010-08-22T01:02:34.860" Score="13" ViewCount="335" Body="&lt;p&gt;The elementary real-valued functions are not closed under integration.  (Elementary function has a precise definition --  see Risch algorithm in Wikipedia).  This means there are elementary functions whose integrals are not elementary.  So we can construct a larger class of functions by adjoining all the integrals of elementary functions.  You can repeat this process indefinitely.  If I understand things correctly, the set of functions that is the countable closure of this process is closed under integration.  Does any &lt;em&gt;finite&lt;/em&gt; iteration of the process achieve closure under integration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is no.  Has anyone thought about this?&lt;/p&gt;&#xA;" OwnerUserId="318" LastActivityDate="2010-11-14T20:12:40.690" Title="Closing up the elementary functions under integration." Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="2982" PostTypeId="2" ParentId="2980" CreationDate="2010-08-22T01:03:15.547" Score="0" ViewCount="0" Body="&lt;p&gt;Why not compute a cubic spline from the 5 most recent samples equidistant in time?  This will give you a smooth curve and you can use the parametric derivative of the spline as a velocity.  &lt;a href=&quot;http://mathworld.wolfram.com/CubicSpline.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/CubicSpline.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1284" LastActivityDate="2010-08-22T01:03:15.547" CommentCount="2" />
  <row Id="2983" PostTypeId="2" ParentId="2981" CreationDate="2010-08-22T01:12:48.163" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not entirely sure it can be closed; apparently more and more complicated integrals of compositions of elementary functions require further generalizations of the &quot;hypergeometric function&quot; to several variables; for instance, the integral of $\sin(\sin(x))$ requires a &lt;a href=&quot;http://mathworld.wolfram.com/KampedeFerietFunction.html&quot; rel=&quot;nofollow&quot;&gt;Kampé de Fériet&lt;/a&gt; hypergeometric function for its closed form, and the elliptic integrals (integrals containing square roots of cubic or quartic polynomials) require the &lt;a href=&quot;http://mathworld.wolfram.com/LauricellaFunctions.html&quot; rel=&quot;nofollow&quot;&gt;Lauricella&lt;/a&gt; or &lt;a href=&quot;http://mathworld.wolfram.com/AppellHypergeometricFunction.html&quot; rel=&quot;nofollow&quot;&gt;Appell&lt;/a&gt; hypergeometrics. As to whether there is &quot;the one hypergeometric to rule them all&quot;, I think that's an open question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify the little discussion Qiaochu and I had in the comments for other readers: the reason you should be interested in the hypergeometrics is that virtually all the elementary functions, and combinations thereof (sum, products, compositions, etc.) are expressible as hypergeometrics. As I have shown in the example I gave in the comments, integrating a hypergeometric function with p+q+1 arguments can require a hypergeometric function with p+q+3 arguments. If you imagine this procedure being applied by multiplying a hypergeometric function with each term of a power series, you get &lt;a href=&quot;http://functions.wolfram.com/HypergeometricFunctions/MeijerG/06/01/01/01/0002/&quot; rel=&quot;nofollow&quot;&gt;even more complicated hypergeometric functions&lt;/a&gt; (in that link, you have the multivariate generalization of the hypergeometric function known as Meijer's G function. Integrating a Meijer function can give you a &lt;a href=&quot;http://functions.wolfram.com/HypergeometricFunctions/MeijerG/21/01/01/0001/&quot; rel=&quot;nofollow&quot;&gt;Meijer function with even more arguments&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question one should be asking, I think, is that if there exists &quot;the one hypergeometric to rule them all&quot;; or to (mis)use another colloquialism: &quot;the mother of all hypergeometrics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT×2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess all I have been trying to say, after all this elaboration: unless one would be satisfied with an &quot;it's turtles all the way down&quot; sort of answer, then I think there's no way to satisfactorily resolve this question.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-22T05:08:12.987" LastActivityDate="2010-08-22T05:08:12.987" CommentCount="3" />
  <row Id="2984" PostTypeId="2" ParentId="2980" CreationDate="2010-08-22T01:30:21.920" Score="3" ViewCount="0" Body="&lt;p&gt;In the physical world, once an object is released from all external forces, it will travel in a straight line. UI design strongly suggests that interfaces work better when the user is using previous knowledge of motor control, eye movement, physics, etc. Unless you have a legitimate reason to curve after the object is released, I would personally go with linear only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, take the last few coordinates and timestamps, and average the velocity components together. Then run a quadratic or exponential function of time on each component until the change of position is negligibly small. The latter functions relates to throwing an object upward and watching it being slowed by gravity, and the former relates to shoving an object on the ground and seeing it come to a stop by friction. Since the exponential function seems like the best for the job, here's an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the last few velocities for the $x$ and $y$ components, and average them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = \frac{\Delta x}{\Delta t}; y = \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = \frac{\sum v_x}{n}; y = \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set time from release for $t$, $\beta$ as some friction constant, $v_x$ as the average velocity for each component, and $x_0$ as the initial release point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = v_x (1 - e^{-\beta t})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the 1 is to align the function to start at zero when time is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I hope this helps. I'm interested in what you're doing, so you'll have to show me when you're done. ;)&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-08-22T01:30:21.920" CommentCount="4" />
  <row Id="2985" PostTypeId="2" ParentId="2731" CreationDate="2010-08-22T01:47:49.467" Score="2" ViewCount="0" Body="&lt;p&gt;1/2. Imagine there are four houses in a row, each with two children. One has two boys; one has two girls; one has an older boy and a younger girl; one has a younger boy and an older girl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you go up to one of these houses at random and a child (chosen at random, say by coin flip) comes to the door. In four of the possible scenarios, this child is a boy. In two of &lt;em&gt;those&lt;/em&gt; scenarios, the &lt;em&gt;other&lt;/em&gt; child is also a boy. The answer is then two divided by four, or one-half.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-22T01:47:49.467" CommentCount="2" />
  <row Id="2986" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T02:09:02.170" Score="9" ViewCount="0" Body="&lt;p&gt;Euler's Polyhedral Formula: $\text{vertices} + \text{faces} - \text{edges} = 2$ for convex (more generally, sphere-like) polyhedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler discovered this about 1750 though the Greeks might well have discovered this fact. The first proof, however, was given by Legendre, using spherical geometry.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-08-22T11:24:30.980" LastActivityDate="2010-08-22T11:24:30.980" CommunityOwnedDate="2010-08-22T02:09:02.170" CommentCount="1" />
  <row Id="2987" PostTypeId="1" AcceptedAnswerId="2990" CreationDate="2010-08-22T03:40:40.760" Score="-3" ViewCount="193" Body="&lt;p&gt;I very recently solved this question: If $f$ is a continuous function on $[0,1]$ and $$\int\limits_{0}^{1} x^{n} f(x) \ dx = 0$$ then prove that $f \equiv 0$. One can use the Weierstrass approximation theorem to solve this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposing i replace $x^{n}$ by a continuous function $g(x)$, then will the result remain true. So the question becomes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Suppose $f,g$ are continuous functions on $[a,b]$ and if $f$ satisfies $$\int\limits_{a}^{b} f(x)g(x) \ dx = 0$$ for every continuous function $g$ ( Note: $g$ has the property that $g(a)=g(b)=0$), then is $f \equiv 0$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-22T07:35:35.443" LastActivityDate="2010-08-22T08:13:11.967" Title="$f$ being identically zero." Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="2988" PostTypeId="2" ParentId="2987" CreationDate="2010-08-22T05:02:17.617" Score="2" ViewCount="0" Body="&lt;p&gt;What is a nice, continuous function $ p: [a, b] \to \mathbb{R} $ that vanishes only at the endpoints (and thus does not change sign on $ (a, b) $), and how would you combine it with the continuous $ f $ to form a suitable $ g $?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not sure whether this is the right attitude here, but it's too tempting, and someone would likely have done it sooner or later.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  Since a nice proof for the question has already been given, there's no point in beating around the bush any longer:  take $ p(x) = (x-a)(b-x) $ and $ g = p \cdot f $ to conclude $ f \equiv 0 $.&lt;/p&gt;&#xA;" OwnerUserId="336" LastEditorUserId="336" LastEditorDisplayName="" LastEditDate="2010-08-22T08:13:11.967" LastActivityDate="2010-08-22T08:13:11.967" CommentCount="3" />
  <row Id="2990" PostTypeId="2" ParentId="2987" CreationDate="2010-08-22T05:40:27.583" Score="4" ViewCount="0" Body="&lt;p&gt;Here is an outline of a proof which does not use the Weierstrass Approximation Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Suppose that $f(a) = f(b) = 0$.  Then we may take $g = f$, getting $\int_a^b f(x)^2 = 0$.  Since $f^2$ is continuous and non-negative, this implies $f = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Suppose that $f$ can be $L^1$-approximated by continuous functions that vanish at the endpoints: i.e., suppose that for all $\epsilon &gt; 0$, there exists $g_n$ which is continuous on $[a,b]$, $g_n(a) = g_n(b) = 0$ and $\int_a^b |f - g_n| \leq \epsilon$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\int_a^b f^2| \leq \int_a^b |f(f-g_n)| + |\int_a^b f g_n| = \epsilon \int_a^b |f|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side approaches zero with $\epsilon$, so $\int_a^b f^2 = 0$, hence $f = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: Show that indeed every continuous function can be approximated by $g_n$'s as in Step 2.  (For instance, take $g_n$ to be equal to $f$ except for values for $x$ very close to the endpoints of the interval.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-22T05:40:27.583" CommentCount="5" />
  <row Id="2991" PostTypeId="1" AcceptedAnswerId="2995" CreationDate="2010-08-22T05:47:14.873" Score="2" ViewCount="240" Body="&lt;p&gt;Hi this is my first time posting on here... so please bear with me :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just wondering how I can solve something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25x ≡ 3 (mod 109)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone can give a break down on how to do it would be appreciated (I'm a slow learner...)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is proof that I've attempted:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Using definition of modulus we can rewrite 25x ≡ 3 (mod 109) as 25x = 3 + 109y (for some integer y). We can rearrange that to 25x - 109y = 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We use Extended Euclidean Algorithm (not sure about this part, I keep messing things up), so this is where I'm stuck at.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1397" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-22T05:53:33.413" LastActivityDate="2010-08-24T19:44:31.237" Title="Not understanding Simple Modulus Congruency" Tags="&lt;number-theory&gt;&lt;discrete-mathematics&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" />
  <row Id="2992" PostTypeId="2" ParentId="2755" CreationDate="2010-08-22T06:12:15.813" Score="35" ViewCount="0" Body="&lt;p&gt;I'm going to try to give a lighter-flavoured version of my previous answer.  I'd rather not edit the previous one anymore so here goes another response.  I want to make clear, this response is to &lt;em&gt;you&lt;/em&gt;, not your 10-year-old nephew.  How you translate this response to any person depends more on you and that person than anything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffeomorphism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page for diffeomorphism&lt;/a&gt;. In particular,the lead image &lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/5/51/Diffeomorphism_of_a_square.svg/220px-Diffeomorphism_of_a_square.svg.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look at that image I see the standard Cartesian coordinate grid, but deformed a little. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/c/c9/2D_Cartesian_Coordinates.PNG&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a &quot;big theorem&quot; in a subject called &lt;a href=&quot;http://en.wikipedia.org/wiki/Differentiable_manifold&quot; rel=&quot;nofollow&quot;&gt;Manifold Theory&lt;/a&gt; and it's name is the &quot;Isotopy Extension Theorem&quot;.  Moreover, it has a lot to do with these kinds of pictures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isotopy extension theorem is roughly this construction: say you have some rubber, and it's sitting in a medium of liquid epoxy that's near-set.  Moreover, imagine the epoxy to be multi-coloured.  So when you move the rubber bit around in the epoxy, the epoxy will &quot;track&quot; the rubber object.  If your epoxy had a happy-face coloured into it originally, after you move the rubber, you'll see a deformed happy-face.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://minimatefactory.com/blog/wp-content/uploads/2008/11/paint.jpg&quot;&gt;&#xA;&lt;img src=&quot;http://www.deloitte.com/assets/Dcom-UnitedStates/Local%20Assets/Images/Full%20Size%20Images/us_risk_paint%20swirl_200x200.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get images that look a lot like mixed paint.  Stir various blotches of paint, and the paint gets distorted.  The more you stir, the more it mixes and it gets harder and harder to see the original image.  The important thing is that the mixed paint is something of a &quot;record&quot; of how you moved your rubber object.  And if your motion of the rubber object returns it to its initial position, there is a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f : X \to X $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $X$ is all positions outside your rubber object.  Given $x \in X$ you can ask where the particle of paint at position $x$ went after the mixing, and call that position $F(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All my talk about fibre bundles and homotopy-groups in the previous response was a &quot;high level&quot; encoding of the above idea.  An intermediate step in the formalization of this idea is the solution of an ordinary differential equation, and that differential equation is essentially the &quot;paint-mixing idea&quot; above, in case you want to look at this subject in more detail later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does this mean?  A motion of an object from an initial position &lt;em&gt;back to the initial position&lt;/em&gt; gives you an idea of how to &quot;mix paint&quot; outside the object.  Or said another way, it gives you an &lt;a href=&quot;http://en.wikipedia.org/wiki/Automorphism&quot; rel=&quot;nofollow&quot;&gt;Automorphism&lt;/a&gt; of the complement, in our case that's a 1-1, continuous bijective function between 3-dimensional space remove the garment and itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find it odd but mathematicians have been studying &quot;paint mixing&quot; in all kinds of mathematical objects, including &quot;the space outside of garments&quot; and far more bizarre objects for well over 100 years.  This is the subject of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_system&quot; rel=&quot;nofollow&quot;&gt;dynamical systems&lt;/a&gt;.  &quot;Garment complements&quot; are a very special case, as these are subsets of 3-dimensional euclidean space and so they're 3-manifolds.  Over the past 40 years our understanding of 3-manifolds has changed and seriously altered our understanding of things.  To give you a sense for what this understanding is, let's start with the basics.  3-manifolds are things that on small scales look just like &quot;standard&quot; 3-dimensional Euclidean space.   So 3-manifolds are an instance of &quot;the flat earth problem&quot;.  Think about the idea that maybe the earth is like a flat sheet of paper that goes on forever.  Some people (apparently) believed this at some point. And superficially, as an idea, it's got some things going for it.  The evidence that the earth isn't flat requires some build-up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/9/97/Triangles_%28spherical_geometry%29.jpg/300px-Triangles_%28spherical_geometry%29.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, so 3-manifolds are the next step.  Maybe all &lt;em&gt;space&lt;/em&gt; isn't flat in some sense.  That's a tricky concept to make sense of as space isn't &quot;in&quot; anything -- basically by definition whatever space is in we'd call space, no?  Strangely, it's not this simple.  A guy named Gauss &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorema_Egregium&quot; rel=&quot;nofollow&quot;&gt;discovered that there is a way to make sense of space being non-flat&lt;/a&gt; &lt;em&gt;without&lt;/em&gt; space sitting in something larger. Meaning curvature is a relative thing, not something judged by some exterior absolute standard.  This idea was a revelation and spawned the idea of an &lt;em&gt;abstract manifold&lt;/em&gt;.  To summarize the notion, here is a little thought experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a rocket with a rope attached to its tail, the other end of the rope fixed to the earth.  The rocket takes off and goes &lt;em&gt;straight&lt;/em&gt; away from the earth.  Years later, the rocket returns from some other direction, and we grab both loose ends of the rope and pull.  We pull and pull, and soon the rope is tight.  And the rope doesn't move, it's taught. &lt;em&gt;as if&lt;/em&gt; it was stuck on something.  But the rope isn't touching anything except your hands.  Of course you can't see all the rope at one time as the rope is tracing out the (very long) path of the rocket. But if you climb along the rope, after years you can verify: it's finite length, it's not touching anything except where it's pinned-down on the earth.  And it can't be pulled in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what a topologist might call a &lt;em&gt;hole in the universe&lt;/em&gt;.  We have abstract conceptions of these types of objects (&quot;holes in the universe&quot;) but by their nature they're not terribly easy to visualize -- not impossible either, but it takes practice and some training.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1970's by the work of many mathematicians we started to achieve an understanding of what we expected 3-manifolds to be like.  In particular we had procedures to construct them all, and a rough idea of how many varieties of them there should be.  The conjectural description of them was called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrization_conjecture&quot; rel=&quot;nofollow&quot;&gt;geometrization conjecture&lt;/a&gt;. It was a revelation in its day, since it implied that many of our traditional notions of geometry from studying surfaces in 3-dimensional space translate to the description of all 3-dimensional manifolds.   The geometriztion conjecture was recently &lt;a href=&quot;http://en.wikipedia.org/wiki/Solution_of_the_Poincar%C3%A9_conjecture&quot; rel=&quot;nofollow&quot;&gt;proven&lt;/a&gt; in 2002. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot of this theory is that in some sense 3-dimensional manifolds &quot;crystalize&quot; and break apart in certain standard ways.  This forces any kind of dynamics on a 3-manifold (like &quot;paint mixing outside of a garment&quot;) to respect that crystalization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do I find a garment you can't turn inside-out?  I manufacture one so that its exterior crystalizes in a way I understand.  In particular I find a complement that won't allow for this kind of turning inside-out.  The fact that these things exist is rather delicate and takes work to see.  So it's not particularly easy to explain the proof.  But that's the essential idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To say a tad more, there is a certain way in which this &quot;crystalization&quot; can be extremely beautiful.  One of the simplest types of crystalizations happens when you're dealing with a finite-volume &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_3-manifold&quot; rel=&quot;nofollow&quot;&gt;hyperbolic manifold&lt;/a&gt;.  This happens more often than you might imagine -- and it's the key idea working in the example in my previous response.  The decomposition in this case is very special as there's something called the &quot;Epstein-Penner decomposition&quot; which gives a canonical way to cut the complement into convex polytopes.  Things like tetrahedra, octahedra, icosahedra, etc, very standard objects.  So understanding the dynamics of &quot;garments&quot; frequently gets turned into (ie the problem &quot;reduces to&quot;) the understanding of the geometry of convex polytopes -- the kind of things Euclid was very comfortable with.  In particular there's software called &quot;SnapPea&quot; which allows for rather easy computations of these things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.math.utk.edu/~morwen/dirichlet/gif/2vol3_dirichlet.gif&quot;&gt;&#xA;&lt;img src=&quot;http://www.math.utk.edu/~morwen/dirichlet/gif/v2678_animated.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images taken from &lt;a href=&quot;http://www.math.utk.edu/~morwen/&quot; rel=&quot;nofollow&quot;&gt;Morwen Thistlethwaite's&lt;/a&gt; webpage.  These are images of the closely-related notion of a &quot;Dirichlet domain&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image of the Dirichlet domain for the complement of $8_{17}$, the key idea in the construction of my previous post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://rybu.org/photo/mathish/chalk/MyScreenshot.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, this in the Poincare model for hyperbolic space, which gives it the jagged/curvy appearance. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-08-22T07:55:53.637" LastActivityDate="2010-08-22T07:55:53.637" CommentCount="3" />
  <row Id="2993" PostTypeId="2" ParentId="2991" CreationDate="2010-08-22T06:22:15.963" Score="3" ViewCount="0" Body="&lt;p&gt;You need to just 'divide' by 25 and get the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25x=3(mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow 25^{-1}25x=25^{-1}3 (mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow x=25^{-1}3 (mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $25^{-1}=48$, since $25*48=1200=1(mod\ 109)$. So we have -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=48*3=35(mod\ 109)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_multiplicative_inverse&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Modular_multiplicative_inverse&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-22T06:22:15.963" CommentCount="5" />
  <row Id="2994" PostTypeId="2" ParentId="2981" CreationDate="2010-08-22T06:51:34.907" Score="0" ViewCount="0" Body="&lt;p&gt;Alternate between applying the integral to each element and closing the set with respect to the elementary operations.  Let's call the smallest number of such compound iterations necessary to construct a function its &quot;rank&quot;.  So exp has rank 0, but erf, erf(exp), and erf + exp all have rank 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral of a function with rank R may have rank R or R+1.  The derivative of a function with rank R may have rank R or R-1, and the rank of the derivative of a function with rank 0 is always 0.  If the derivative of a function with rank R has rank R-1, let's call that function &quot;primitive&quot; to distinguish between the two phases of the iteration.  A non-primitive function can by definition be written in terms of primitive functions of the same rank (and possibly other functions of smaller rank).  For example, erf is primitive, but erf(erf) is not, although they are of the same rank 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose R is the largest rank.  Then the integral f of any rank-R function f' is a non-primitive rank-R function.  It is possible to express f in terms of functions that have a rank smaller than R or are primitive and have rank equal to R, and at least one function of the latter type is required.  The sum of any two primitive functions remains primitive, so in the general case we have such a function appearing in a non-additive expression.  For the same reason we can find a case where it is not the root of the expression tree.  Now we can write f as a composition of a product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = g(h * j)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where j is a primitive function of rank R and g and h are some other functions.&#xA;The derivatives of f are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f' = (h * j)' * g'(h * j)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f'' = (h * j)'' * g'(h * j) + (h * j)' * g''(h * j)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f'' contains the same terms that define f' and is therefore also of rank R, so f' is not primitive.  But there must be at least one such primitive function f', so the hypothesis of a largest rank is contradicted.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-22T09:50:54.503" LastActivityDate="2010-08-22T09:50:54.503" CommentCount="3" />
  <row Id="2995" PostTypeId="2" ParentId="2991" CreationDate="2010-08-22T07:52:27.233" Score="2" ViewCount="0" Body="&lt;p&gt;The extended euclidean algorithm is used to find x and y such that ax + by = gcd of a and b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case a = 109 and b = 25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we start as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find remainder and quotient when we divide 109 by 25 and write the remainder on the left hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 = 109 - 25*4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we get two new numbers 25 and 9. Write the remainder on the left hand side again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 = 25 - 9*2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have two new numbers, 9 and 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the extended algorithm, we use the formula for 9 in the first step &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 = 25 - (109 - 25*4)*2 = 25*9 - 109*2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 = 9 - 7*1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= (109-25*4) - (25*9 - 109*2) = 109*3 - 25*13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = 7 - 3*2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = (25*9 - 109*2) - 3*(109*3 - 25*13) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&#xA;1 = 25*48 - 109*11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus 25x - 109y = 1 for x = 48 and y = 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 25x - 109y = 3 for x = 48*3 = 144 and y = 11*3 = 33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore 144*25 = 3 (mod 109).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a number &amp;lt;= 109,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;144 = 109 + 35.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have (109+35)*25 = 3 (mod 109).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which implies 35*25 = 3 (mod 109).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus x = 35 is a solution to your equation, which we found using the extended euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-22T07:52:27.233" CommentCount="4" />
  <row Id="2996" PostTypeId="2" ParentId="2204" CreationDate="2010-08-22T08:00:11.617" Score="1" ViewCount="0" Body="&lt;p&gt;Ha, that's funny, I read &quot;plane&quot; as &quot;airplane&quot;. :-)&lt;br&gt;&#xA;(That's what happens if you talk about bugs in a math problem!)&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2010-08-22T08:00:11.617" CommentCount="1" />
  <row Id="2997" PostTypeId="2" ParentId="370" CreationDate="2010-08-22T08:07:14.007" Score="2" ViewCount="0" Body="&lt;p&gt;The first few chapters of Goldblatt's &lt;a href=&quot;http://books.google.com/books?id=VUrePAAACAAJ&quot; rel=&quot;nofollow&quot;&gt;Topoi: the categorial analysis of logic&lt;/a&gt; provide another fairly elementary introduction to the basics of category theory.&lt;/p&gt;&#xA;" OwnerUserId="1399" LastActivityDate="2010-08-22T08:07:14.007" CommunityOwnedDate="2010-08-22T08:07:14.007" />
  <row Id="2998" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T08:08:28.683" Score="3" ViewCount="0" Body="&lt;p&gt;I recall vividly the moment I learnt of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;Thomae's function&lt;/a&gt;, which is continuous at all irrational numbers and discontinuous at all rational numbers.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-22T08:08:28.683" CommunityOwnedDate="2010-08-22T08:08:28.683" CommentCount="1" />
  <row Id="2999" PostTypeId="2" ParentId="2595" CreationDate="2010-08-22T08:14:50.683" Score="2" ViewCount="0" Body="&lt;p&gt;Given some $A$, you can measure how well it performs on the ensemble of $R_1,...,R_{24}$ and $I_1,...,I_{24}$ by computing the error. For example:&#xA;$$&#xA;J = \sum_{i=1}^{24} ||A R_i - I_i ||^2&#xA;$$&#xA;Where we have used the standard Euclidean norm. The goal is to find the $A$ that makes $J$ as small as possible. We can express this more compactly by stacking the vectors $R_i$ and $I_i$ into larger 3x24 matrices. So if we let: $R = [R_1 R_2 \dots R_{24}]$ and similarly for $I$, we can rewrite our error as:&#xA;$$&#xA;J = ||AR - I||_F^2&#xA;$$&#xA;where we are now using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_norm&quot; rel=&quot;nofollow&quot;&gt;Frobenius norm&lt;/a&gt;. This optimization problem can be solved in many different ways. For example, you can &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_calculus&quot; rel=&quot;nofollow&quot;&gt;differentiate&lt;/a&gt; J with respect to A and set the result equal to zero. This yields the optimality condition:&#xA;$$&#xA;ARR^T = IR^T&#xA;$$&#xA;So any A that satisfies the above equation will be optimal in the sense that it will minimize J. If R is a full-rank matrix (likely), then the optimal A is given by:&#xA;$$&#xA;A = IR^T(RR^T)^{-1}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-22T08:14:50.683" CommentCount="1" />
  <row Id="3000" PostTypeId="2" ParentId="998" CreationDate="2010-08-22T09:36:03.033" Score="0" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://books.google.com/books?id=8zQYxx7PjqsC&quot; rel=&quot;nofollow&quot;&gt;Invitation to algebraic geometry&lt;/a&gt; by Smith et al. is also very readable. &lt;/p&gt;&#xA;" OwnerUserId="1399" LastActivityDate="2010-08-22T09:36:03.033" CommunityOwnedDate="2010-08-22T09:36:03.033" />
  <row Id="3001" PostTypeId="1" AcceptedAnswerId="3004" CreationDate="2010-08-22T09:45:04.413" Score="3" ViewCount="113" Body="&lt;p&gt;In my course of linear algebra I studied that if V is a finite dimensional vector space on the complex field, then every endomorphism of V has an eigenvector. The proof is simple: taken a polynomial f that is null on the endomorphism A (it exists because of the finite dimension of V), exploiting the algebraic closure of C we are there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gave the following geometrical interpretation: such a space can't be &lt;em&gt;fully twisted&lt;/em&gt; by any of its linear operators, we always have at least an invariant subspace of dimenson 1. This fact still appears a bit magical to me. Is there some deep or different reason for this in other parts of mathematics such as geometry or algebra? Are there other interesting geometrical consequences? Does it remain true if the space is infinite dimensional?&lt;/p&gt;&#xA;" OwnerUserId="365" LastActivityDate="2010-08-22T11:02:31.687" Title="Is there something deep in the fact that an endomorphism of a finite dimensional complex vector space has an eigenvector?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3002" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T09:54:49.523" Score="3" ViewCount="0" Body="&lt;p&gt;When I started studying elliptic curves and modular forms I was really amazed by the fact that for a normalized eigenform the Fourier coefficients are the Hecke eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-22T09:54:49.523" CommunityOwnedDate="2010-08-22T09:54:49.523" />
  <row Id="3003" PostTypeId="2" ParentId="3001" CreationDate="2010-08-22T10:03:36.643" Score="3" ViewCount="0" Body="&lt;h3&gt;Topological argument&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Suppose $A$ has no kernel. Then it induces an automorphism of the projective space $\mathbb P(V)$. Now the fact that A has an eigenvector follows from the (purely topological) &lt;a href=&quot;http://en.wikipedia.org/wiki/Lefschetz_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Lefschetz formula&lt;/a&gt; (because $\mathbb P(V)$ has non-zero Euler characteristic and any linear transformation lies in the connected component of the identity).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-22T10:40:43.977" LastActivityDate="2010-08-22T10:40:43.977" CommentCount="7" />
  <row Id="3004" PostTypeId="2" ParentId="3001" CreationDate="2010-08-22T10:18:31.477" Score="3" ViewCount="0" Body="&lt;p&gt;As to the depth and significance of the existence of eigenvectors for finite-dimensional linear operators over $\mathbb{C}$, I think you have already identified the core algebraic reason: it is that the field $\mathbb{C}$ is algebraically closed.  I think it is fair to call this a &quot;deep fact&quot; -- the task of giving a rigorous proof was the topic of Gauss's thesis work, and in fact by modern standards Gauss's (first) proof is still not completely rigorous.  (Some feel this way, anyway -- there is not universal agreement here.) Whole books have been written on various proofs of this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the depth here lies in the fact that the definition of the complex field is topological / analytic, ultimately relying on the completeness of $\mathbb{R}$.  If you start with a field $K$, the following is not very deep:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition: For a field $K$, the following are equivalent:&lt;br&gt;&#xA;(i) Every linear endomorphism of a finite dimensional $K$-vector space has an eigenvector.&lt;br&gt;&#xA;(ii) $K$ is algebraically closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see (i) &amp;rArr; (ii), use the fact that every monic polynomial of degree $n$ is the chracteristic and minimal polynomial of its &lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;companion matrix&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your second question: no, there are linear operators on infinite dimensional $\mathbb{C}$-vector spaces without eigenvalues, even &lt;em&gt;bounded&lt;/em&gt; linear operators.  A fundamental example is the Hilbert space $L^2([0,1])$ of square integrable $\mathbb{C}$-valued functions on the unit interval.  Then the &lt;strong&gt;multiplication operator&lt;/strong&gt; $M: f \mapsto xf$ has norm $1$ but is easily seen to have no eigenvectors, since the equation $xf = \lambda f$ forces $f = 0$ almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In functional analysis, there is a suitable generalization of the notion of the set of eigen*values* of a linear operator on certain infinite dimensional spaces, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrum_%28functional_analysis%29&quot; rel=&quot;nofollow&quot;&gt;spectrum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;: inspired by damiano's comment, here is what is in some sense the simplest possible example of a linear operator on an infinite dimensional space without an eigenvector: Let $K$ be any field, and let $V$ be a vector space of countably infinite dimension with basis ${e_n}_{n=1}^{\infty}$.  Then consider the &lt;strong&gt;shift operator&lt;/strong&gt; $T$ &#xA;on $V$, i.e., the unique linear operator such that for all $n \in \mathbb Z^+$, $T(e_n) = e_{n+1}$.  $T$ is very simple and easy to visualize: it just happens never to &quot;cycle back&quot; on itself.  Indeed, if $v = \sum_{n=1}^{\infty} a_n e_n \in V$ (with $a_n = 0$ &#xA;for all but finitely many $n$), then assuming that $Tv = \lambda v$ gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=1}^{\infty} \lambda a_n e_n = \sum_{n=2}^{\infty} a_{n-1} e_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular $\lambda a_1 = 0$.  It is clear that the kernel of $T$ is $0$, so if $v \neq 0$, $\lambda \neq 0$ and thus $a_1 = 0$ and then the above equation implies that $a_n = 0$ for all $n$, i.e., $v = 0$. (If $V = K[x]$ is the space of polynomials with $K$-coefficients, then with respect to the natural basis $e_n = x^{n-1}$, multiplication by $x$ &lt;i&gt;is&lt;/i&gt; the shift operator $T$.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-22T11:02:31.687" LastActivityDate="2010-08-22T11:02:31.687" CommentCount="2" />
  <row Id="3005" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T10:44:26.500" Score="3" ViewCount="0" Body="&lt;p&gt;The Thom-Pontrjagin theorem: $\Omega_n^{SO} \cong \pi_n(MSO)$. The group of equivalence classes of n-manifolds with respect to oriented cobordism is isomorphic to the n-th homotopy group of the Thom spectrum MSO. This can be generalized to include different cobordisms (unoriented, ...) and different Thom spectra. See for example &lt;a href=&quot;http://www.math.umass.edu/~weston/oldpapers/cobord.pdf&quot; rel=&quot;nofollow&quot; title=&quot;the minor thesis&quot;&gt;the minor thesis&lt;/a&gt; by T. Weston.&lt;/p&gt;&#xA;" OwnerUserId="1402" LastActivityDate="2010-08-22T10:44:26.500" CommunityOwnedDate="2010-08-22T10:44:26.500" CommentCount="0" />
  <row Id="3006" PostTypeId="1" AcceptedAnswerId="3010" CreationDate="2010-08-22T10:54:24.820" Score="10" ViewCount="700" Body="&lt;p&gt;As the title implies, It is seems that $e^x$ is the only function whoes derivative is the same as itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-22T11:00:55.777" LastActivityDate="2010-11-22T19:27:45.473" Title="Could you explain why $de^x/dx = e^x$ &quot;intuitively&quot;?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="12" CommentCount="8" FavoriteCount="6" />
  <row Id="3007" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:10:25.387" Score="0" ViewCount="0" Body="&lt;p&gt;$d/dx (e^{-x} \cdot y)=-e^{-x} \cdot y+e^{-x} \cdot dy/dx=e^{-x}(dy/dx-y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $dy/dx=y$, then $d/dx (e^{-x} \cdot y)=0$, ie $e^{-x} \cdot y=c$ or $y=c \cdot e^x$&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-22T20:59:36.120" LastActivityDate="2010-08-22T20:59:36.120" CommentCount="4" />
  <row Id="3008" PostTypeId="1" AcceptedAnswerId="3156" CreationDate="2010-08-22T11:11:38.677" Score="3" ViewCount="321" Body="&lt;p&gt;When I was first learning about the axiom of choice, a very helpful and fun exercise was to try to find a rule that maps every nonempty subset of $\mathbb R$ to one of its members.  No matter how hard you try, it was impossible to find one.  (If you try the same for the nonempty subsets of $\mathbb N$, it is pretty easy, just take the smallest member of each set.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is, has it been proved in some formal way that such a choice function cannot be completely described with a finite number of symbols? (I think this is equivalent to  the undecidability of the existence of a choice function on  $\mathcal P (\mathbb R) \setminus \{ \emptyset \}$ in ZF.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is, in ZF with the negation of AC, is it provable that there exists no choice function on $\mathcal P (\mathbb R) \setminus \{ \emptyset \}$ ?  (My guess is negative.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-08-23T12:30:14.347" LastActivityDate="2010-08-23T23:00:25.553" Title="Choice function on $\mathcal P (\mathbb R) \setminus \{ \emptyset\}$" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="3009" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:25:12.227" Score="4" ViewCount="0" Body="&lt;p&gt;Doesn't quite explain $e^x$, but looking at the series of $2^n$, i.e. $\{1,2,4,8,16,32,...,2^k,2^{k+1},...\}$ might help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see there if you take the first order difference you get back the same series, as $2^{k+1}-2^k=2^k$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-22T11:25:12.227" CommentCount="1" />
  <row Id="3010" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:25:58.480" Score="11" ViewCount="0" Body="&lt;p&gt;Well, think of &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_growth&quot; rel=&quot;nofollow&quot;&gt;exponential growth&lt;/a&gt; (like e.g. bacteria grow):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know, the more bacteria exist in a colony, the faster the colony will grow. More precisely: The growth speed of the colony $B$ is proportional to it's size ... Double size, double speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dB}{dt} \sim B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore we know the growth is exponential, since bacteria clone themselves in fixed amounts of time, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B \sim 2^{k\cdot t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it together, we can deduce that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt}2^{kt} = c \cdot 2^{kt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or with $a = 2^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt}a^t = c \cdot a^t$&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Now, how do we get the $e$? We just ask: &lt;em&gt;What base $a$ do we have to take such that $c = 1$, i.e. $\dot{B} = B$?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We simply &lt;strong&gt;call&lt;/strong&gt; that base $e$. Having such an $e$ is quite useful. We could use its special derivation traits we found above to define all exponential functions to the base $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^x = e^{x \cdot \ln a}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the factor $c$ we encountered in the above equations equals $\ln a = \log_e a$ and therefore, we can easily derive all kinds of exponential terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, $e$ turns out to be $lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \approx 2.718\ldots$&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="917" LastEditorDisplayName="" LastEditDate="2010-08-22T11:44:00.183" LastActivityDate="2010-08-22T11:44:00.183" />
  <row Id="3011" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:38:42.360" Score="4" ViewCount="0" Body="&lt;p&gt;I'll act on Casebash's proposal later, but for now... care for a movie?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nLLL7.gif&quot; alt=&quot;tangent line&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(granted, I cheated and spaced the two plots of $\exp(x)$ to clearly show the moving tangent line... but you guys should be able to understand this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(thanks to Stan Wagon!)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-22T11:51:36.440" LastActivityDate="2010-08-22T11:51:36.440" />
  <row Id="3012" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:41:27.830" Score="0" ViewCount="0" Body="&lt;p&gt;There are two components to the original question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do exponential functions generally (the class $x \mapsto ab^x$ for $b&gt;0$) have the property of being eigenfunctions of the derivative operator? That is: why do these functions f have the property $\frac{\mathrm d}{\mathrm dx} f(x) = \lambda f(x)$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are the functions $x \mapsto a \mathrm e^x$ in particular +1 eigenfunctions --- i.e. why is the scalar $\lambda$ equal to 1 for these functions in particular? (To put it another way: why is this the case when the base is equal to the particular constant e&amp;nbsp;&amp;asymp;&amp;nbsp;2.71828?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can give an intuitive answer to the first of these questions, by combining two observations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the slope at a paticular point, &lt;em&gt;e.g.&lt;/em&gt; at x&amp;nbsp;=&amp;nbsp;0. Multiplying the function f(x) by some scalar now only amplifies the function by that factor, but also amplifies the slope by that same factor. (This is just a restatement of the product rule, for constant scalar factors.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Exponential functions f(x) are self-similar: multiplying them by a positive scalar is equivalent to translating them to the left, or to the right, depending on the scalar by which you multiply: more precisely, if you multiply $\mathrm e^x$ by &lt;em&gt;c&lt;/em&gt;, why you get is $\mathrm e^{[x + \ln(c)]}$, or the same function translated  ln(&lt;em&gt;c&lt;/em&gt;) to the left. More generally, for $f(x) = ab^x$, we have &lt;em&gt;c&lt;/em&gt;f(x) = f(x + log&lt;sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;/sub&gt;(&lt;em&gt;c&lt;/em&gt;)); that is, multiplying f(x) by &lt;em&gt;c&lt;/em&gt; is equivalent to a shift to the left by log&lt;sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;/sub&gt;(&lt;em&gt;c&lt;/em&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, combine these two facts. The slope of f(x) at x&amp;nbsp;=&amp;nbsp;log&lt;sub&gt;&lt;em&gt;b&lt;/em&gt;&lt;/sub&gt;(&lt;em&gt;c&lt;/em&gt;) is &lt;em&gt;c&lt;/em&gt; times the slope of f(x) at x&amp;nbsp;=&amp;nbsp;0, because of the self-similarity of the exponential function, and because the slope of &lt;em&gt;c&lt;/em&gt;f(x) at x&amp;nbsp;=&amp;nbsp;0 is &lt;em&gt;c&lt;/em&gt; times the slope of f(x) at x&amp;nbsp;=&amp;nbsp;0. Thus the slope of f(x) is a function which is proportional to f(x), QED. &lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-22T21:12:22.567" LastActivityDate="2010-08-22T21:12:22.567" CommentCount="3" />
  <row Id="3014" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:43:44.817" Score="7" ViewCount="0" Body="&lt;p&gt;As Mangaldan says,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!}  + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + \cdots &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you accept (it's true, but it could take a while to explain why) that this series, like polynomials, can be derived &lt;em&gt;term by term&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \  \Longrightarrow \ p'(x) = a_1 + 2a_2x + \cdots + na_n x^{n-1}  \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d}{dx} e^x = \frac{1}{1!} + \frac{2x}{2!} + \cdots + \frac{nx^{n-1}}{n!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-22T11:43:44.817" CommentCount="9" />
  <row Id="3015" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T11:45:06.467" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose $\frac{d}{dx}f(x)=f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for small $h$, $f(x+h)=f(x)+hf(x)=f(x)(1+h)$. If we do this for a lot of small intervals of length $h$, we see $f(x+a)=(1+h)^{a/h}f(x)$. (Does this ring a bell already?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $x=0$ in the above, and fixing $f(0)=1$, we then have $f(1)=(1+h)^{1/h}$, which in limit as $h\rightarrow 0$ goes to $e$. And continuing $f(x)=(1+h)^{x/h}$, which goes to $e^x$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-22T11:45:06.467" CommentCount="2" />
  <row Id="3016" PostTypeId="2" ParentId="2969" CreationDate="2010-08-22T12:03:50.773" Score="1" ViewCount="0" Body="&lt;p&gt;First, a few thoughts about this specific box packing problem. It might be possible to pack this box by packing the &quot;y&quot; polycube into 25 cells in 5 identical layers. This corresponds to collapsing this 3-D shape into a polyomino in the plane, and then using it to pack a 5x5 rectangle. However, this polyomino can not tile a 5x5 rectangle. (The terminology for polyominoes is not totally standard but some use the order of a polyomino to mean the smallest number of copies of this shape to tile a rectangle with congruent copies of itself.) The order of this polyomino is 10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there is lots of material about polyominoes and to a lesser extent box packings in the books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. Golomb, Polyominoes: Puzzles, Patterns, Problems, and Packings (Revised Edition), Princeton U. Press, 1994.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G. Martin Polyominoes, MAA, 1991.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, there are a variety of papers related to polyominoes and boxes at Michael Reid's very nice site with links to others:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ucf.edu/~reid/Polyomino/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucf.edu/~reid/Polyomino/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ucf.edu/~reid/Research/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucf.edu/~reid/Research/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth, there are &quot;complexity&quot; results which govern the general problem of given a collection of tiles (either in 2-D or 3-D) when can one use them to tile some particular shape (say a rectangle or rectangular box). Here is a paper related to this of Erik Demaine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://erikdemaine.org/papers/Jigsaw_GC/&quot; rel=&quot;nofollow&quot;&gt;http://erikdemaine.org/papers/Jigsaw_GC/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://archive.ite.journal.informs.org/Vol5No3/Chlond/index.php&quot; rel=&quot;nofollow&quot;&gt;http://archive.ite.journal.informs.org/Vol5No3/Chlond/index.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-08-22T20:18:29.367" LastActivityDate="2010-08-22T20:18:29.367" CommentCount="1" />
  <row Id="3017" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T12:14:15.260" Score="6" ViewCount="0" Body="&lt;p&gt;One of the most surprising results in numerical math is &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilkinson%27s_polynomial&quot; rel=&quot;nofollow&quot;&gt;Wilkinson's polynomial&lt;/a&gt;. Wilkinson gave an example in which a very tiny change to one coefficient of a polynomial can have a drastic impact on the location of the zeros.  The change in the location of the roots is seven orders of magnitude larger than the change in the coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is an exact result.  The impact of the coefficient change is not due to numerical precision.  The point of the example, however, is that changes such as the perturbation of the coefficient are inevitable in numerical computing.)&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-22T12:14:15.260" CommunityOwnedDate="2010-08-22T12:14:15.260" CommentCount="2" />
  <row Id="3018" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T12:33:30.527" Score="0" ViewCount="0" Body="&lt;p&gt;Another example from the numerics front: it's surprising that despite the theoretical fact that Gaussian elimination can be unstable (even with pivoting!), &lt;a href=&quot;http://dx.doi.org/10.1137/0610012&quot; rel=&quot;nofollow&quot;&gt;examples&lt;/a&gt; that trigger this instability are in fact &lt;a href=&quot;http://dx.doi.org/10.1137/0611023&quot; rel=&quot;nofollow&quot;&gt;very rare in practice&lt;/a&gt;, and can be handled by a &lt;a href=&quot;http://dx.doi.org/10.1023/A:1022314201484&quot; rel=&quot;nofollow&quot;&gt;simple fix&lt;/a&gt; if they do arise.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-22T12:33:30.527" CommunityOwnedDate="2010-08-22T12:33:30.527" CommentCount="1" />
  <row Id="3019" PostTypeId="2" ParentId="2305" CreationDate="2010-08-22T12:49:47.297" Score="1" ViewCount="0" Body="&lt;p&gt;First of all, you might be interested in the work of Chas and Phillips: &quot;Self-intersection of curves on the punctured torus&quot;.  I've only skimmed their paper, but they seem to be doing something closely related to what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second I want to guess, for some reason, that the average time to compute self-intersection number is a lot slower than the average time to generate a word.  (Is that the case?  Could you tell me how you are computing minimal self-intersection numbers?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, I guess that you want to generate as few strings as possible.  I'll use $a, A, b, B$ as the generating set for $\pi_1 = \pi_1(T)$.  Looking at Lyndon words is essentially the same as applying inner automorphisms (conjugation, ie cyclic rotation) to your words.   You might also try replacing a word $w$ by its inverse $W$.  If some rotation of $W$ beats $w$ [sic], then you can throw $w$ away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other &quot;geometric automorphisms&quot; (elements of the mapping class group) &#xA;of $\pi_1$ which are very useful eg rotation of $T$ by one-quarter: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a \mapsto b \mapsto A \mapsto B \mapsto a.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also two nice reflections: either fix $b, B$ and swap $a$ with $A$, or the other way around.  Composing these gives the hyperelliptic which swaps $a$ with $A$ and swaps $b$ with $B$.  (I use python's swapcase function for this -- very simple!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any of these operations (or any compositions of these, eg the reverse of a word) produces a word $w'$ that is lexicographically before $w$, then you can throw $w$ away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if this is helpful -- I'm interested in this kind of problem. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-08-22T12:49:47.297" />
  <row Id="3020" PostTypeId="1" AcceptedAnswerId="3079" CreationDate="2010-08-22T13:51:38.763" Score="2" ViewCount="166" Body="&lt;p&gt;I am interested in the Gelfand transformation&#xA;$$&#xA;\Phi\colon\ell^1(\mathbb Z)\to\mathcal C(\mathbb T),\quad a\mapsto\sum_{n\in\mathbb Z}a_n z^n.&#xA;$$&#xA;This is an injective homomorphism of Banach algebras. It is neither isometric nor surjective. However, its image---the Wiener algebra $W$ consisting of all continuous functions on $\mathbb T$ whose Fourier series is absolutely convergent---is a subalgebra of $\mathcal C(\mathbb T)$ which is dense in the subspace topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question: Can we prove of disprove that $\Phi$ has a continuous inverse on its image $W$?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: Is $\Phi\colon\ell^1(\mathbb Z)\to W$ an isomorphism of topological algebras?&#xA;(Here $W$ carries the topology induced by the sup-norm from $\mathcal C(\mathbb T)$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-24T08:43:39.657" LastActivityDate="2010-08-24T08:43:39.657" Title="Does the Gelfand transformation on $\ell^1(\mathbb Z)$ possess a continuous inverse on its image?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3021" PostTypeId="2" ParentId="2869" CreationDate="2010-08-22T14:11:01.767" Score="0" ViewCount="0" Body="&lt;p&gt;I can't hope to do better than Jacques's answer, so I'll just offer some little numerical advice on computing your function to be fitted (but you probably already know this): make sure that whatever system you're doing your computations in has support for both the upper incomplete gamma function $\Gamma(a,x)$ and the lower incomplete gamma function $\gamma(a,x)$. With these two, we can represent your $h(t)$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h(t)=\frac1{(n-1)!^2}P_6 \gamma(n,P_8 t)\Gamma(n,P_9(t-P_7))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for doing things this way is that we mitigate the possibility of any subtractive cancellation happening when your downhill simplex optimization wanders too far out. Which brings me to my second piece of advice: you would do well to find good initial values for your $P_i$ parameters. Nelder-Mead isn't really known for being fast (at least there's no need to assume continuity!), and it also tends to wander around your parameter space unless suitably restricted.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-22T14:11:01.767" CommentCount="2" />
  <row Id="3022" PostTypeId="2" ParentId="2991" CreationDate="2010-08-22T14:30:45.563" Score="0" ViewCount="0" Body="&lt;p&gt;I meant this as a comment to the discussion after Student's answer but it seems that I don't have the option (reputation too low?) so I'll post it as an answer. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to compute quickly the inverse of 25 mod 109, note that $25=5^2$. Thus $25^{-1}=t^2$ where $t=5^{-1}$ mod 109. On the other hand, computing the inverse of 5 modulo any number $N$ ending with 9 (or 4) is immediate since it is just $(N+1)/5$.&#xA;Thus $25^{-1}=((109+1)/5)^2=22^2=48$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moral: when performing actual computations always look for easy tricks that allow shortcuts. &lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-22T14:30:45.563" />
  <row Id="3023" PostTypeId="2" ParentId="2549" CreationDate="2010-08-22T14:34:36.897" Score="1" ViewCount="0" Body="&lt;p&gt;In general the term &quot;sectional curvature&quot; is used in the n-dimensional setting. Basically one computes what is known in theory of surfaces as &quot;Gauss Curvature&quot; for the surface gotten by exponentiating every two plane inside the tangent space of the n-manifold. Then one can show that the following things to build intuition,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If two Riemann curvatures are giving the same sectional curvature for every two plane then the Riemann curvatures are equal as endomorphisms. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ricci curvature along a direction is the sum of the sectional curvatures of all possible two planes spanned by that direction and one more from a basis gotten by extending the given direction. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Scalar curvature at a point is the sum of the sectional curvatures of all possible two planes spanned by a chosen basis. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A theorem of Schur says that sectional curvature if constant on all 2-planes at a point is constant at all point of the manifold. Sectional curvature is the strongest notion of curvature and will made constant under very strong conditions like maximal symmetry or locally geodesic reflecting isometry or transitive action of the isometry group on orthonormal frames. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-22T14:34:36.897" CommentCount="0" />
  <row Id="3024" PostTypeId="1" AcceptedAnswerId="3494" CreationDate="2010-08-22T14:45:07.203" Score="3" ViewCount="97" Body="&lt;p&gt;I keep coming across calculations like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a metric on an $n+2$ dimensional manifold given as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^2 = 2dudr + 2L(u,r)du^2 -r^2d\Omega_n^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then apparently once can write down the Ricci and Einstein and other tensors as a function of n. Like for the above the Einstein tensor apparently has the following non-zero components,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_{01} = \frac{n}{r}L_r+\frac{n(n-1)(2L-1)}{2r^2}$ &#xA;$G_{22} = (n-1)[(2L-1)(\frac{2-n}{2})-2rL_r]-r^2L_{rr}$&#xA;$G_{00} = -\frac{nL_u}{r} + \frac{2nLL_r}{r} + \frac{n(n-1)L(2L-1)}{r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $G^2_2 = G^3_3 = ... = G^{n+1}_{n+1}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where the subscripts of L denote partial derivatives with respect to those variables)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fixed given n I can imagine doing the calculation either by hand or some software but I would like to know who these expressions are derived for a general n. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-28T00:26:33.527" Title="How to compute curvature tensors for general n-dimensions? " Tags="&lt;differential-geometry&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="3025" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T14:55:47.370" Score="1" ViewCount="0" Body="&lt;p&gt;Let us define a function $\exp$ as the solution to $D \exp(x) = \exp(x)$ with initial value $\exp(0) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $D \exp \circ f = D f \cdot (\exp \circ f)$ but also notice that&#xA;$$ \begin{align}&#xA;D (\exp(a(x)) \exp(b(x))) &amp;amp;= D a(x) \exp(a(x)) \exp(b(x)) + D b(x) \exp(a(x)) \exp(b(x)) \\&#xA;&amp;amp;= (D (a + b) (x)) \cdot \exp(a(x)) \exp(b(x))&#xA;\end{align} $$&#xA;so we see that $\exp(a)\exp(b)=\exp(a+b)$ which lets us write $\exp(x) = e^x$ for some as yet undetermined constant $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically we can see that the gradient of the tangent line of the curve $y = e^x$ must be between the finite differences, i.e. $\frac{e^x - e^{x-h}}{h} &amp;lt; e^x &amp;lt; \frac{e^{x+h} - e^x}{h}$. With some numerical exploration we see that $e$ is between $2$ and $3$. Specifically, the finite differences for $e=2$ and $h=0.9$ are too small and the finite differences for $e=3$ and $h=0.1$ are too big. To narrow it down more let $e=2.7$ and $h = 0.01$ to see that $2.7 &amp;lt; e &amp;lt; 3$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T15:02:02.880" />
  <row Id="3026" PostTypeId="1" CreationDate="2010-08-22T14:58:39.593" Score="1" ViewCount="154" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not very fluent in mathematical proofs. High School has, sadly, not taught me any kind of proof-theory. That's why I would like your help with my proof of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod  3).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S_m$ is the digit sum of m and $\wedge$ is a logical-AND.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to use the very little that I know:  Mathematical induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basis:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(1): S_m = 1 \wedge m = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod 3) \Rightarrow 1 = 1 (mod 3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inductive Step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n)\colon m = n \wedge S_m = \sum\limits_{k=0}^K m_k = (m_0 + m_1 + ... + m_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod 3)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n+1)\colon m = n+1 \wedge S_m = \sum\limits_{k=0}^K m_k = (m_0 + m_1 + ... + m_k + m_k+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = S_m (mod 3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble at the step of A(n+1) because I don't know how to actually prove that it's valid? Please note that $m_0, ..., m_k$ refers to the summation of the digits of the number m. I just didn't find any good documentation that showed how to depict that in mathematical terms. Feel free to correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how do curly braces work? Documentation says \{, \} should do the trick, but they are invisible to me. &lt;/p&gt;&#xA;" OwnerUserId="44" LastEditorUserId="44" LastEditorDisplayName="" LastEditDate="2010-08-23T08:48:14.290" LastActivityDate="2010-08-23T08:48:14.290" Title="Prove that $m = S_m (\mod 3)$?" Tags="&lt;induction&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="3027" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T15:04:05.933" Score="4" ViewCount="0" Body="&lt;p&gt;One conceptual way to understand the standard definition $e^x = \lim_{n\to\infty}(1+x/n)^n$ is to view it as arising from applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_method&quot; rel=&quot;nofollow&quot;&gt;Euler's approximation method&lt;/a&gt; to $y' = y$. The same method also works arbitrary higher-order constant coefficiant ODEs by converting them to linear system form and employing matrix exponentials. This is described quite nicely in Arnold's beautiful textbook &quot;Ordinary differential equations&quot;. Here's an excerpt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gK98d.jpg&quot; alt=&quot;alt text&quot;&gt; &#xA;&lt;img src=&quot;http://i.imgur.com/PV2vn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-22T15:04:05.933" CommentCount="3" />
  <row Id="3029" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:08:48.340" Score="1" ViewCount="0" Body="&lt;p&gt;I'm having a little trouble understanding your formatting/notation, but I am guessing that what you want to show is that $m = S_m$ (mod 3) where $S_m$ is the sum of the digits of the number $m$.  In the inductive, step, it looks like you are saying that $S_{n+1} = S_n + 1$.  This is not always true (e.g. it is true if $n = 23$ but not true if $n = 59$).  So you should consider the two cases where the last digit is not a 9 and where the last digit is a 9.  You'll easily see you get the same thing mod 3 and this will complete the proof.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-22T15:08:48.340" />
  <row Id="3030" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:16:56.663" Score="3" ViewCount="0" Body="&lt;p&gt;The best way to get the result is that to remember that writing $m=a_n...a_1a_0$ in base 10 is actually a shorthand for&#xA;$$&#xA;m=a_0+a_1\cdot10+...+a_n\cdot 10^n.&#xA;$$&#xA;Now it's enough to observe that $10^n\equiv1\bmod 3$ for all $n$ and thus&#xA;$$&#xA;m\equiv a_0+a_1+...+a_n \bmod 3.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-22T15:16:56.663" />
  <row Id="3031" PostTypeId="1" AcceptedAnswerId="3109" CreationDate="2010-08-22T15:22:18.313" Score="13" ViewCount="378" Body="&lt;p&gt;Groups can be defined abstractly as sets with a binary operation satisfying certain identities, or concretely as a collection of permutations of a set.  Cayley's theorem ensures that these two definitions are equivalent: any abstract group acts as a collection of permutations of its underlying set, and this action is faithful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, rings can be defined abstractly as sets with a pair of binary operations satisfying certain identities, or concretely as a collection of endomorphisms of an abelian group.  There is a &quot;Cayley's theorem&quot; here as well: any abstract ring acts as a collection of endomorphisms of its underlying abelian group, and this action is faithful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for Lie algebras seems much less clear to me.  The adjoint representation is not generally faithful, and Ado's theorem comes with qualifications and doesn't have the simplicity of the two theorems above.  For me, the problem is that I don't have a good sense of what the concrete definition of a Lie algebra is supposed to be.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that a good concrete definition of a Lie algebra is as a space of derivations on some algebra closed under commutator.  In that case, is it correct to say that a Lie algebra acts faithfully as derivations on its universal enveloping algebra?  Is this a good analogue of Cayley's theorem?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Motivation: in the books on Lie algebras I have read, the authors verify that Lie algebras which occur in nature satisfy alternativity and the Jacobi identity, but I have never seen any simple justification that these axioms are &quot;enough&quot; in the same way that Cayley's theorem tells you that the axioms for a group or a ring are &quot;enough.&quot;  There is just Ado's theorem which, again, comes with qualifications and is hard.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-23T12:05:30.773" Title="&quot;Cayley's theorem&quot; for Lie algebras?" Tags="&lt;abstract-algebra&gt;&lt;lie-algebra&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="9" />
  <row Id="3032" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:22:49.937" Score="3" ViewCount="0" Body="&lt;p&gt;The reason this works is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10 = 9 + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;modulo 3 this says that $10 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numbers like $3157$ really mean $3 \cdot 10^3 + 1 \cdot 10^2 + 5 \cdot 10 + 7$, so applying the previous congruence we get $3 \cdot 10^3 + 1 \cdot 10^2 + 5 \cdot 10 + 7 = 3 + 1 + 5 + 7$ modulo 3.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that we have a strong understanding of why the theorem is true let head toward a detailed proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us use the principle that all numbers can be &lt;em&gt;written down&lt;/em&gt; in base 10, for example the sequence of digits $(3,6,8,4)$ represents the number $4863$ (and a technicality, the empty sequence represents $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let us define the interpretation from sequences to numbers as a recursive function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align}&#xA;\text{interp}() &amp;amp;= 0 \\&#xA;\text{interp}(ds,d) &amp;amp;= d + 10 \cdot \text{interp}(ds)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align}&#xA;\text{digitalroot}() &amp;amp;= 0 \\&#xA;\text{digitalroot}(ds,d) &amp;amp;= d + \text{digitalroot}(ds)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now prove the theorem (that the interpretation of a number is equal to its digital root) by induction on the length of the number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base case $\text{interp}() = 0 = \text{digitalroot}()$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive case, assume that $\text{interp}(ds) = \text{digitalroot}(ds)$, we would like to now prove that $d + 10 \cdot \text{interp}(ds) = d + \text{digitalroot}(ds)$, cancelling d and reducing the 10 to a 1 we simply have the hypothesis, which is known to be true. That completes the proof.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-22T15:22:49.937" />
  <row Id="3033" PostTypeId="2" ParentId="3026" CreationDate="2010-08-22T15:26:45.863" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;: It suffices to work $\rm\pmod 9. \;$ Put $\rm\; S_N := $ the sum of the decimal digits of $\rm N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice $\rm\;N = 10\: n + r \;\Rightarrow\; S_N = S_n + r \;\Rightarrow\; N - S_N  = n - S_n + 9 \: n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\; N - S_N \equiv n - S_n \pmod 9$ -- the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SIMPLER&lt;/strong&gt;: For  $\rm N = a_k 10^k + \;\cdots + 10\; a_1 + a_0 \;$ put $\rm\; P(N) = a_k X^k + \;\cdots+ a_1 \; x + a_0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FACTOR THEOREM&lt;/strong&gt; $\rm\;\Rightarrow\; X-1|P(X)-P(1) \;\Rightarrow\; 9|P(10)-P(1) \;\:$ i.e. $\rm\; 9|N - S_N$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;casting nines&lt;/strong&gt; follows from the Factor Theorem purely by the &lt;strong&gt;polynomial form&lt;/strong&gt; of radix notation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another example of the &lt;strong&gt;power of polynomials!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-22T15:59:39.470" LastActivityDate="2010-08-22T15:59:39.470" />
  <row Id="3034" PostTypeId="1" AcceptedAnswerId="3050" CreationDate="2010-08-22T15:34:31.163" Score="3" ViewCount="55" Body="&lt;p&gt;One can see &lt;a href=&quot;http://books.google.com/books?id=rIyIlQwVCgkC&amp;amp;pg=PA92&amp;amp;lpg=PA92&amp;amp;dq=TBL+spacetimes&amp;amp;source=bl&amp;amp;ots=uu2rVqvCtn&amp;amp;sig=JqwYqPgtBEXK_SIwEOjZFxhNKzY&amp;amp;hl=en&amp;amp;ei=50FxTPz-Mo_mvQPI8NxB&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CB8Q6AEwAg#v=onepage&amp;amp;q=TBL%20spacetimes&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt; for TBL space-times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how the explicit expression for the function called $G$ in equations $3.108,3.108,3.110$ in the above reference is obtained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it would be nice to see some further references about TBL space-times. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-22T19:58:38.613" Title="Tolman-Bondi-Lemaitre space times " Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="3035" PostTypeId="2" ParentId="3031" CreationDate="2010-08-22T15:45:16.143" Score="3" ViewCount="0" Body="&lt;p&gt;A finite-dimensional real Lie algebra always arises as the Lie algebra of&#xA;a Lie group. Of course, the proof of this does use Ado's theorem, as well&#xA;as some Lie group theory (a subalgebra of a Lie algebra of a Lie group $G$&#xA;is also a Lie algebra of a Lie group, but not necessarily the Lie&#xA;algebra of a closed subgroup of $G$). The upshot is that over $\mathbb{R}$&#xA;the definition of (finite-dimensional) Lie algebra suffices to deal&#xA;with all Lie groups.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-22T15:45:16.143" CommentCount="2" />
  <row Id="3036" PostTypeId="2" ParentId="3031" CreationDate="2010-08-22T16:49:37.383" Score="1" ViewCount="0" Body="&lt;p&gt;Classically, the &quot;concrete&quot; definition of what a Lie algebra is is simply a subspace of $\mathrm{End}(V)$, for some vector space $V$, which is closed under commutators. It is a theorem, then, that such a thing is the same thing as a pair $(\text{vector space},\text{antisymmetric bracket})$ satisfying the Jacobi identity. This theorem is quite important when one develops the theory that way, because it means that one can describe abstractly Lie algebras through &lt;em&gt;equations&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This should be compared with the situation for &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_algebra&quot; rel=&quot;nofollow&quot;&gt;Jordan algebras&lt;/a&gt;, as explained in Jacobson's amazing book on the subject: &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_algebra#Special_Jordan_algebras&quot; rel=&quot;nofollow&quot;&gt;special Jordan algebras&lt;/a&gt;, those subspaces of $\mathrm{End}(V)$'s closed under the product $A\bullet B\dot=\tfrac12(AB+BA)$, are not characterizable through equations only; this is a consequence of the fact that the class of special Jordan algebras is not closed under quotients)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-23T04:57:37.360" LastActivityDate="2010-08-23T04:57:37.360" CommentCount="4" />
  <row Id="3037" PostTypeId="2" ParentId="3008" CreationDate="2010-08-22T17:09:14.917" Score="2" ViewCount="0" Body="&lt;p&gt;In Kunen's book on set theory, he leads one through an exercise where one creates a model of ZF (assuming ZF is consistent) in which the power set of the natural numbers is not well orderable.  Since one can prove that the power set of the naturals is in bijective correspondence with the reals, in this model, the real numbers cannot be well orderable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't have Kunen at home with me now, and it's been a year or two since I even glanced at the relevant section.  If I remember, I'll update the reference next time I have a copy of the book nearby).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there can (I think, but I'm not positive) be models where choice fails and the reals &lt;em&gt;are&lt;/em&gt; well orderable.  Note that choice refers to ALL sets, so one simply needs to rig things to fail at REALLY huge cardinal sizes but have choice work at lower levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are models where choice functions are definable: Godel's universe $L$.  For example, in $L$, EVERY set is definable.  That is, given a set $X$, there is a first order formula (with parameters) $\phi$ such that $x\in X$ iff $\phi(x)$.  Now, a function from $X$ to $Y$ (including a choice function), is nothing but a special subset of $X\times Y$, and hence will be definable in $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned about this last paragraph from assorted posting by Joel David Hamkins over at MO.  For example, see &lt;a href=&quot;http://mathoverflow.net/questions/23393/set-theory-and-vl&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/23393/set-theory-and-vl&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; I realize that this doesn't seem to address your question directly.  However, it's true that for any set $X$, the powerset of $X$ has a choice function iff $X$ can be well ordered.  See Levy's book &quot;Basic Set Theory&quot; on page 160 for details.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-08-22T17:54:24.283" LastActivityDate="2010-08-22T17:54:24.283" />
  <row Id="3038" PostTypeId="1" CreationDate="2010-08-22T17:24:14.270" Score="2" ViewCount="243" Body="&lt;p&gt;Suppose we define a topology on the real numbers as that obtained by defining open sets to be those such that the complement is finite. How does one show that this topology cannot be obtained from a distance function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  in advance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s: This is not a homework question. I am trying to self study point set topology. &lt;/p&gt;&#xA;" OwnerUserId="1127" LastActivityDate="2010-08-23T03:49:41.713" Title="A topology which cannot be obtained by a distance function" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="3039" PostTypeId="2" ParentId="3038" CreationDate="2010-08-22T17:25:26.670" Score="4" ViewCount="0" Body="&lt;p&gt;Try showing that metric topologies are Hausdorff and that this topology isn't.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-22T17:25:26.670" CommentCount="3" />
  <row Id="3040" PostTypeId="2" ParentId="2571" CreationDate="2010-08-22T17:25:51.473" Score="1" ViewCount="0" Body="&lt;p&gt;This is true precisely because the rationals $\mathbb Q$ comprise a multiplicative &lt;strong&gt;subsemigroup&lt;/strong&gt; of the reals $\mathbb R,$&lt;br&gt;&#xA;i.e. the subset of rationals is closed under the multiplication operation of $\mathbb R$. Your statement arises by taking the contrapositive of this statement - which transfers it into an equivalent statement in the complement set $\mathbb R \backslash \mathbb Q$ of irrational reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\quad\quad\quad r_1,\ldots,r_n \in \mathbb Q \;\Rightarrow\; r_1 \cdots r_n \in \mathbb Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contra+ $\rm\quad\; r_1 r_2\cdots r_n \not\in \mathbb Q \;\Rightarrow\; r_1\not\in \mathbb Q \;\:$ or $\rm\;\cdots\;$ or $\rm\;r_n\not\in\mathbb Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case $\rm\;\;\; r^n\not\in \mathbb Q \;\Rightarrow\; r\not\in \mathbb Q \;$ is the special constant case $\rm r_i = r$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the same is true if we replace $\rm\mathbb Q\subset \mathbb R$ by any subsemigroup chain $\rm G\subset H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contrapositive form is important in algebra since it characterizes &lt;strong&gt;prime&lt;/strong&gt; ideals in semigroups, rings, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-22T17:25:51.473" />
  <row Id="3041" PostTypeId="2" ParentId="3008" CreationDate="2010-08-22T18:43:23.243" Score="1" ViewCount="0" Body="&lt;p&gt;To add a bit on Jason's answer, the proof that the well ordering principle is equivalent to the axiom of choice can be transformed quite simply to a proof that if a set is well ordered then if and only if there exists a choice function on its power set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of a well ordering implies the existence of a choice function quite trivially - take the minimal of the subset.&#xA;If there exists a choice function $f$ on $P(X)$ then we can well order $X$ as following:&#xA;Pick some $x_0 \in X$, suppose for all $\alpha &amp;lt; \beta$ we have defined $x_\alpha$, define $x_{\beta} = f(X \backslash \{x_i | i &amp;lt; \beta\})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This induction covers the case of a limit ordinal as well, and the induction is bounded by some ordinal $\alpha$, otherwise $X$ is a proper class, contradiction to the assumption that it is a set. Therefore $X$ is isomorphic to some ordinal $\alpha$ and ergo - well orderable.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-08-22T19:48:31.207" LastActivityDate="2010-08-22T19:48:31.207" CommentCount="2" />
  <row Id="3042" PostTypeId="2" ParentId="3038" CreationDate="2010-08-22T18:46:07.260" Score="2" ViewCount="0" Body="&lt;p&gt;Your topology is called the &lt;em&gt;finite complement topology&lt;/em&gt; of the real line: &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_complement_topology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Finite_complement_topology&lt;/a&gt; . To show that it's not Hausdorff, try to intersect &lt;em&gt;any&lt;/em&gt; two open sets -you don't even need the points of the Hausdorff definition-, remembering the rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(X \backslash A ) \cap (X \backslash B) = X \backslash (A \cup B)  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having any Topology standard book at hand, like Munkres, might help your self study. For instance, I'm quite shure that the proof that every metric space is Hausdorff can be found there.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-23T03:49:41.713" LastActivityDate="2010-08-23T03:49:41.713" />
  <row Id="3043" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T18:54:27.917" Score="3" ViewCount="0" Body="&lt;p&gt;(Looks like Niel and &quot;J&quot; beat me to the punch while I was composing this, but I'll post my answer anyway since I like my diagram. :) ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't get too much more intuitive than the Law of Exponents ($a^x a^c = a^{x+c}$) right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce symbolic (and mental) clutter for a while, let's write $k$ for $a^c$, to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k \; a^x = a^{x+c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Multiplying the value of $a^x$ by &lt;em&gt;something&lt;/em&gt; yields the same result as adding &lt;em&gt;some other thing&lt;/em&gt; to the exponent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider what this bit of algebra tells us about the geometry of the graph of the function $y=a^x$ (shown in red in the figure below). Recall some fundamental notions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$y=k a^x$ is the result of &lt;em&gt;vertically stretching&lt;/em&gt; (scaling) the original graph by a factor of $k$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$y = a^{x+c}$ is the result of &lt;em&gt;horizontally sliding&lt;/em&gt; (translating) the original graph by a (signed) distance of $-c$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Law of Exponents tells us that (provided $k$ and $c$ are related appropriately), &lt;strong&gt;these transformed graphs are identical!&lt;/strong&gt; In the figure, the blue graph represents &lt;strong&gt;both&lt;/strong&gt; results.[*] Importantly, what points move where aren't the same under both actions; scaling moves the red point $P$ vertically onto the blue point $Q$; translating moves the red point $R$ horizontally onto $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/SHrcE.jpg&quot; alt=&quot;Diagram of Exponential Functions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as suggested by the diagram, imagine a tangent vector poking out of each point of the original graph, and consider what the transformations to do those vectors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertically stretching scales (only) the &quot;rise&quot; of the vector by factor $k$; that is, a vector with slope $m$ is pulled to achieve a slope of $m k$.&lt;/li&gt;&#xA;&lt;li&gt;Horizontally translating &lt;em&gt;has no effect&lt;/em&gt; on the vector's slope.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What can we conclude here? Why, something pretty remarkable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any point $P$ on the original graph, the point $R$ --located $c$ units to the right[**] of $P$-- is such that the slope of the tangent vector at $R$ is $k$-times the slope of the tangent vector at $P$ (where $k$ and $c$ are related appropriately).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me take this opportunity to retire $k$, since &lt;em&gt;it&lt;/em&gt; is beginning to become clutter; I'll just make the appropriate relationship explicit. Also, I'm going to retire the point name $R$, opting to describe the point instead. I'll summarize things this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the graph $y=a^x$, if the slope of tangent vector at any $P$ is $m$, then the slope of the tangent vector at the point $c$ units to the right of $P$ is $m\;a^c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Notice that there's &lt;strong&gt;nothing special&lt;/strong&gt; about the players in this game. $P$ is &lt;em&gt;any&lt;/em&gt; point on the graph, $c$ is any (horizontal) distance you care to choose; heck, even the exact value of $a$ is up for grabs. Let's use this to our advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we take $P$ to be the point where the (original, red) graph crosses the $y$-axis; that is, we take $P$ to have $x$-coordinate $0$ (and $y$-coordinate $1$, but this doesn't really matter). Then, &quot;the point $c$ units to the right of $P$&quot; can be described more simply as &quot;the point with $x$-coordinate $c$&quot;, and we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the graph $y = a^x$, if the slope of the tangent vector at the point with $x$-coordinate $0$ is $m$, then the slope of the tangent vector at the point with $x$-coordinate $c$ is $m\;a^c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Calculusian prose:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x) = a^x$, then $f^{\prime}(c) = a^c f^{\prime}(0)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Observe that we don't even need &quot;$c$&quot; in the above formula, since it's just taking the place of some $x$-coordinate. We can simply write:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x) = a^x$, then $f^{\prime}(x) = a^x f^{\prime}(0)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From here, it's pretty much a matter of definition to get to the final answer to your specific question. After all, the above holds for any (non-negative) value of $a$. Clearly, some values of $a$ correspond to graphs that cross the $y$-axis very steeply; some values correspond to graphs that cross the $y$-axis very shallowly; it's not un-reasonable to believe that there's a &lt;em&gt;convenient&lt;/em&gt; value would cause the graph to cross the $y$-axis &lt;em&gt;juuuuuuuuust right&lt;/em&gt; ... with a slope of $1$. Of course (as I mix my folklore), to name a demon is to control him, so we'll simply assign a symbol to this &quot;just right&quot; value of $a$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let &quot;e&quot; be the number such that the tangent vector to $y=e^x$ at $x=0$ has slope $1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that such a number really does exist, you don't even have to know its exact value to conclude&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f(x) = e^x$, then $f^{\prime}(x) = e^x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can then turn your attention to figuring out why $e$ happens to have the value $2.718...$ . Other answers in this thread provide insights on how those arguments proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also left as an exercise is to determine why, in the formula for the derivative of $a^x$, the &quot;$f^{\prime}(0)$&quot; factor is in fact &quot;$\log a$&quot; (the natural logarithm of $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] As Niel mentions, this illustrates that the graph of an exponential function is &quot;self-similar&quot;. I think it's important to add &quot;uni-directionally&quot; to the description, in order to distinguish it from (conventional) similarity transformations where scaling occurs &quot;omni-directionally&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[**] &quot;to the left&quot; works, too, with appropriate changes to the discussion.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-08-22T18:54:27.917" CommentCount="1" />
  <row Id="3044" PostTypeId="2" ParentId="2384" CreationDate="2010-08-22T19:01:07.340" Score="0" ViewCount="0" Body="&lt;p&gt;I'm working on www.BuzzMath.com, could fit your needs.&lt;/p&gt;&#xA;" OwnerUserId="1411" LastActivityDate="2010-08-22T19:01:07.340" CommunityOwnedDate="2010-08-22T19:01:07.340" />
  <row Id="3047" PostTypeId="1" AcceptedAnswerId="3056" CreationDate="2010-08-22T19:18:38.943" Score="5" ViewCount="151" Body="&lt;p&gt;This is from Feller's &lt;em&gt;Introduction to Probability Theory and Its Applications&lt;/em&gt;. In the context of Bernoulli trials, we define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b(k;n,p) = \binom{n}{k}p^kq^{n-k},$$&#xA;$$P\{S_n \ge r\} = \sum_{v=0}^{\infty}b(r+v;n,p).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter being the probability of having at least $r$ successes. Now, supposing $r \gt np$ and knowing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{b(k; n,p)}{b(k-1;n,p)}=\frac{(n-k+1)p}{kq}=1+\frac{(n+1)p-k}{kq},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\{S_n \ge r\} \le b(r;n,p)\frac{rq}{r-np}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Feller, it follows from the obvious fact that the terms of the series decrease faster than the terms of a geometric series with ratio $1-\frac{r-np}{rq}$. However, it's not obvious for me and I don't see how the upper bound follows.&lt;/p&gt;&#xA;" OwnerUserId="519" LastActivityDate="2010-08-22T23:11:32.360" Title="Binomial distribution and upper bound" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="3048" PostTypeId="1" AcceptedAnswerId="3052" CreationDate="2010-08-22T19:52:57.813" Score="3" ViewCount="224" Body="&lt;p&gt;What is the proper definition of a Countable Set?&lt;/p&gt;&#xA;" OwnerUserId="1414" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-25T17:00:41.440" LastActivityDate="2010-11-25T17:19:43.410" Title="Definition of a countable set" Tags="&lt;terminology&gt;&lt;discrete-mathematics&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="0" />
  <row Id="3049" PostTypeId="1" AcceptedAnswerId="3057" CreationDate="2010-08-22T19:54:55.140" Score="2" ViewCount="88" Body="&lt;p&gt;I'm quite new at measure theory and all I know I read by myself, so bear with me if this question is too trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the definition of asymptotic, can we say that for $x \rightarrow \infty$ two almost everywhere equal functions are asymptotic? I really think that this is not the case, but I haven't been able to find a counterexample. Maybe, if in general this does not hold, would it be right for real functions?&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-08-22T20:28:52.937" Title="Are almost everywhere equal functions asymptotic?" Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="3050" PostTypeId="2" ParentId="3034" CreationDate="2010-08-22T19:58:38.613" Score="3" ViewCount="0" Body="&lt;p&gt;It actually just comes from integration. Equation 3.106 from that book is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot R^2 = \left( - \frac{\partial R}{\partial t} \right)^2 = \frac FR + f $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rearranging the terms gives (note that $\dot R &amp;lt; 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dt = - \frac{dR}{\sqrt{\frac FR + f}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now perform the substitution $z = fR/F$, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dt = -\frac F{f^{3/2}} \frac{dz}{\sqrt{1+\frac1z}}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t - t_0 = \frac F{f^{3/2}} \left( \sinh^{-1}\sqrt z - \sqrt{z(z+1)}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the rest is just algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am no expert on general relativity, but the common term of that TBL spacetime seems start with &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lemaitre%E2%80%93Tolman_metric&quot; rel=&quot;nofollow&quot;&gt;Lemaitre-Tolman&lt;/a&gt;&quot;. There is a review article &lt;a href=&quot;http://arxiv.org/abs/0709.2044&quot; rel=&quot;nofollow&quot;&gt;on arXiv&lt;/a&gt;&lt;sup&gt;[1]&lt;/sup&gt; which might help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[1]: Kari Enqvist (2008). &lt;em&gt;Lemaitre–Tolman–Bondi model and accelerating expansion&lt;/em&gt;. General Relativity and Gravitation &lt;strong&gt;40&lt;/strong&gt;, 2–3, pp 451–466. DOI: 10.1007/s10714-007-0553-9.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-22T19:58:38.613" />
  <row Id="3051" PostTypeId="2" ParentId="3048" CreationDate="2010-08-22T20:00:55.863" Score="5" ViewCount="0" Body="&lt;p&gt;A non-empty set $X$ is countable if and only if there exists a surjective function $f$ from $\mathbb{N}$ onto $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Countable_set&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-22T20:00:55.863" CommentCount="6" />
  <row Id="3052" PostTypeId="2" ParentId="3048" CreationDate="2010-08-22T20:04:28.443" Score="8" ViewCount="0" Body="&lt;p&gt;A plain English definition from Kenneth Rosen's &quot;Discrete Mathematics and its Applications&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A set that is either finite or has the same cardinality as the set of positive integers is called countable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-22T20:04:28.443" CommentCount="2" />
  <row Id="3053" PostTypeId="1" AcceptedAnswerId="3058" CreationDate="2010-08-22T20:16:05.537" Score="4" ViewCount="206" Body="&lt;p&gt;Let C&lt;sup&gt;0&lt;/sup&gt; be the set of all of all (edit: bounded) continuous functions from R-&gt;R with the sup norm.  Then a &quot;path&quot; in C&lt;sup&gt;0&lt;/sup&gt; is a continuous function f:[0,1]-&gt;C&lt;sup&gt;0&lt;/sup&gt;.  The question boils down to if there is path (or continuous deformation) which transforms any one bounded continuous function into another.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-22T20:30:09.640" LastActivityDate="2010-08-23T08:32:23.463" Title="Is the set of continous functions from R-&gt;R path connected?" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="3054" PostTypeId="2" ParentId="2981" CreationDate="2010-08-22T20:19:10.803" Score="1" ViewCount="0" Body="&lt;p&gt;In a differential Galois field, integration is the antiderivative operation and not the infinite summation, but I will still call it integration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start out with simple elementary functions, say rational functions with integer coefficients, then there will be many functions that do not have integrals. If the original set of functions is finite, then only a finite number of functions need to be added.  The process can then be repeated at the next level.  The only catch is that at every stage one has to deal with constants (think of the roots of the denominators). Most functions force the addition of several integrals.  The number of functions added grows geometrically with the levels, so I would expect there is no easy way to define a countable closure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there initial sets for which the process stops? Yes, polynomials for example.  Are there initial sets for which it goes on forever?  Yes. &lt;/p&gt;&#xA;" OwnerUserId="1413" LastActivityDate="2010-08-22T20:19:10.803" CommentCount="4" />
  <row Id="3055" PostTypeId="2" ParentId="3006" CreationDate="2010-08-22T20:23:37.783" Score="0" ViewCount="0" Body="&lt;p&gt;Here's my 2 cents, just look at the def of derivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;df/dx = lim f(x+d)-f(x)/d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now the defining property of e&lt;sup&gt;x&lt;/sup&gt; is that for very small x, e&lt;sup&gt;x&lt;/sup&gt; ~ 1 + x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since e&lt;sup&gt;x + d&lt;/sup&gt; = e&lt;sup&gt;x&lt;/sup&gt;e&lt;sup&gt;d&lt;/sup&gt; ~ e&lt;sup&gt;x&lt;/sup&gt;(1+d), it's clear that by plugging everything into the definition of derivative you will get the required result.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-22T20:23:37.783" CommentCount="1" />
  <row Id="3056" PostTypeId="2" ParentId="3047" CreationDate="2010-08-22T20:27:11.830" Score="4" ViewCount="0" Body="&lt;p&gt;First notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k &gt; r$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(n-k+1)p}{kq} &amp;lt;\frac{(n-k+1)p}{rq}$ (as $n \ge k-1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\frac{(n-k+1)p}{rq} \leq \frac{(n-(r+1)+1)p}{rq} = \frac{(n-r)p}{rq}$ as the numerator is the largest when $k = r+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $1 - \frac{(n-r)p}{rq} = \frac{rq - np + rp}{rq} = \frac{r -np}{rq}$ (as $p+q=1$). &lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-22T23:11:32.360" LastActivityDate="2010-08-22T23:11:32.360" CommentCount="2" />
  <row Id="3057" PostTypeId="2" ParentId="3049" CreationDate="2010-08-22T20:28:52.937" Score="5" ViewCount="0" Body="&lt;p&gt;The short answer is no; I assume that you are working with the real numbers and the usual measure with the usual definition of asymptotic. The counterexample goes like this: any countable set has measure 0, and so choose some sequence x&lt;sub&gt;i&lt;/sub&gt; such that x&lt;sub&gt;i&lt;/sub&gt; goes to infinity (say, the positive integers). Then take any function f and define a new function g by g(x)=f(x) if x is not one of the x&lt;sub&gt;i&lt;/sub&gt; and g(x&lt;sub&gt;i&lt;/sub&gt;)=2f(x&lt;sub&gt;i&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two functions are equal almost everywhere since they differ only on the sequence x&lt;sub&gt;i&lt;/sub&gt;, which as a sequence is countable and thus have measure 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then f(x)/g(x) equals 1 when x is not one of the x&lt;sub&gt;i&lt;/sub&gt; and equals 2 on the x&lt;sub&gt;i&lt;/sub&gt;, which means that the value as you go to infinity doesn't converge, i.e. it always oscillates between 1 and 2 as you go farther and farther out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more awesomely, you can do the same trick not with the sequence x&lt;sub&gt;i&lt;/sub&gt;, but say, with the rational numbers, which are also countable and also have a measure of 0. Then the function given by h(x)=f(x)/g(x) is going to be continuous nowhere in the extended real line (i.e. no limit of h(x) as x goes to any number, or to positive or negative infinity will exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer, which I lack time to think through and write up right now, would discuss why countable sets have measures 0 and the interplay between Borel measures, topology, and the equivalence classes of functions given by a.e. equality.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-22T20:28:52.937" CommentCount="1" />
  <row Id="3058" PostTypeId="2" ParentId="3053" CreationDate="2010-08-22T20:39:20.857" Score="6" ViewCount="0" Body="&lt;p&gt;It suffices to find a path to the 0 function from any other continuous function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be any continuous function and define $F(x,t) = tf(x)$.  Assuming we can prove this is, in fact, continuous, note that $F(x,1) = f(x)$ and $F(x,0)$ is the $0$ function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is it continuous?  Let $\epsilon &gt; 0$.  Let $K = sup|f(x)|$  Choose $\delta &gt; 0 $ such that $\delta K &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we have $|t_1 - t_2| &amp;lt; \delta$ implies $|t_1f(x) - t_2 f(x)| = |(t_1 - t_2)f(x)| \leq \delta K = \epsilon$, so that this map is continuous (in fact, uniformly continuous).&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-22T20:39:20.857" />
  <row Id="3059" PostTypeId="1" AcceptedAnswerId="3078" CreationDate="2010-08-22T20:51:17.727" Score="3" ViewCount="93" Body="&lt;p&gt;It is often said that differential forms (sections of an exterior power of the cotangent bundle) are the things that you can integrate. But unless I'm being thoroughly dense differential forms are not the only things that you can integrate, c.f. the arclength form (on a 2d manifold) $ds=\sqrt{dx^2+dy^2}$, the unsigned 1-d forms $|f(x,y)dx+g(x,y)dy|$, or the unsigned area forms $|h(x,y)dx\wedge dy|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do the arclength form $ds=\sqrt{dx^2+dy^2}$, the unsigned 1-d forms |f(x,y)dx+g(x,y)dy|, and the unsigned area forms $|h(x,y)dx\wedge dy|$ live relative to the differentials $dx$ and $dy$, which I understand to live in the cotangent bundle of some 2-dimensional manifold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-08-22T22:45:55.973" Title="What is the theory of non-linear forms (as contrasted to the theory of differential forms)?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="3060" PostTypeId="1" AcceptedAnswerId="3076" CreationDate="2010-08-22T21:02:01.293" Score="6" ViewCount="175" Body="&lt;p&gt;How can we prove that it is posible to arrange numbers 1,2,3,4.........n in a row so that the average of any two of these numbers never appears between them?&lt;/p&gt;&#xA;" OwnerUserId="1414" LastActivityDate="2010-08-23T19:00:44.020" Title="Arrangement of Numbers" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="3061" PostTypeId="1" CreationDate="2010-08-22T21:26:09.130" Score="3" ViewCount="141" Body="&lt;p&gt;I haven't studied properly the theory of infinities yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let A_0 denote the set of natural numbers. Let A_i+1 denote the set whose elements are all the subsets of A_i for i=0,...,n,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand well that the cardinality of A_i+1 is always greater than the cardinality of A_i for all i in N={0,1,...,n,...}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the simplest argument which proves that there exists a set whose cardinality is greater than A_i for all i in N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1418" LastActivityDate="2010-08-22T21:36:58.887" Title="large sets. set theory." Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="3062" PostTypeId="2" ParentId="3061" CreationDate="2010-08-22T21:30:34.853" Score="5" ViewCount="0" Body="&lt;p&gt;Let $X = \bigcup_{i=0}^\infty A_i$  Then $A_i\subseteq  X$ for all i so $|A_i|\leq |X|$ for all i.  Now, consider the powerset of $X$.  Then we have $|A_i|\leq |X|&amp;lt;|P(X)|$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-22T21:30:34.853" CommentCount="1" />
  <row Id="3065" PostTypeId="2" ParentId="3061" CreationDate="2010-08-22T21:36:58.887" Score="1" ViewCount="0" Body="&lt;p&gt;Your question is closely linked to Beth numbers, their definition is (due to lack of \beth symbol in latex, I'll use plain text):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Beth null = Aleph null = the cardinality of the non-negative integers&lt;/li&gt;&#xA;&lt;li&gt;Beth (alpha+1) = the cardinality of the power-set of Beth(alpha)&lt;/li&gt;&#xA;&lt;li&gt;For a limit ordinal delta, Beth(delta) = the union of Beth(alpha) for all alpha &amp;lt; delta&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What you're asking about is Beth(omega).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading: &lt;a href=&quot;http://en.wikipedia.org/wiki/Beth_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Beth_number&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-22T21:36:58.887" />
  <row Id="3069" PostTypeId="1" CreationDate="2010-08-22T21:53:17.017" Score="5" ViewCount="219" Body="&lt;p&gt;I would like to know whether the number of infinity sets of different cardinalities is uncountable or countable. Is the proof simple?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1419" LastActivityDate="2010-08-23T12:24:06.480" Title="number of infinite sets with different cardinalities" Tags="&lt;set-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="3070" PostTypeId="2" ParentId="3069" CreationDate="2010-08-22T21:59:23.127" Score="2" ViewCount="0" Body="&lt;p&gt;If by the number of cardinalities you mean the set of all cardinalities - then there is none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The class of all cardinal numbers is a proper class, as it contains a copy of the class of ordinal numbers - which is not a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer you question, it's not countable. It's so big it's not even a set.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-22T21:59:23.127" />
  <row Id="3071" PostTypeId="2" ParentId="3053" CreationDate="2010-08-22T22:00:02.610" Score="3" ViewCount="0" Body="&lt;p&gt;Any real vector space $V$ with a &quot;reasonable&quot; topology is path connected. Given a vector $v_1\in V$ move along the line spanned by $v_1$ to $0$ and from there move to any other vector $v_2$ moving along le line spanned by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasonable means that for any $w\in V$ every open $U\ni w$ must contain an open segment around $v$ of the line spanned by $w$, which is certainly true for the case under consideration (topology under the sup norm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that $L^2({\Bbb R})$ is not reasonable, so maybe my terminology could be improved :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;Edit : *&lt;/em&gt; Of course $L^2({\Bbb R})$ IS reasonable!&#xA;$$&#xA;(\int_{\Bbb R}(f-tf)^2dx)^{1/2}=|1-t|(\int_{\Bbb R}f^2dx)^{1/2}&#xA;$$&#xA;gets as close as needed to 0 as $t\to1$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastEditorUserId="688" LastEditorDisplayName="" LastEditDate="2010-08-22T23:45:40.043" LastActivityDate="2010-08-22T23:45:40.043" CommentCount="2" />
  <row Id="3072" PostTypeId="1" AcceptedAnswerId="3116" CreationDate="2010-08-22T22:00:52.627" Score="2" ViewCount="179" Body="&lt;p&gt;I'm making a game which you can see here, if you are on Windows or Linux: &lt;a href=&quot;http://insertnamehere.org/birdsofprey/&quot; rel=&quot;nofollow&quot;&gt;http://insertnamehere.org/birdsofprey/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you click and hold your mouse on a bird, you can see I'm just swinging the bird back and forth in pendulum motion. I would like to, instead, implement a more realistic motion, where the movement of your mouse affects the swinging of the bird like a pendulum with a moving pivot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.maplesoft.com/applications/view.aspx?SID=4888&amp;amp;view=html&quot; rel=&quot;nofollow&quot;&gt;a document&lt;/a&gt; on this topic but the equations rely on knowing the pivot's acceleration (X'' and Y''), which I do not; I am only repeatedly translating the bird graphic to the current mouse position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the bird's angle (-180 to 180 degrees), angular velocity and acceleration. I will need to alter these three variables each time the mouse is moved, so I will also have the last (x,y) and the new mouse (x,y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this enough to make a good simulation of a pendulum with moving pivot?&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2010-08-23T14:49:51.430" Title="Pendulum with moving pivot" Tags="&lt;physics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3073" PostTypeId="2" ParentId="3069" CreationDate="2010-08-22T22:00:53.657" Score="4" ViewCount="0" Body="&lt;p&gt;Technically, neither one is true: the class of cardinalities is too large to form a set.  (However, having to choose between &quot;countable&quot; and &quot;uncountable&quot;, certainly &quot;uncountable&quot; seems better: there is at least &lt;em&gt;an&lt;/em&gt; uncountable set of cardinalities of sets.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is equivalent to a previous question on this site, but anyway: please see Fact 20 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and let me know if you have any further questions.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-22T22:00:53.657" />
  <row Id="3074" PostTypeId="2" ParentId="3072" CreationDate="2010-08-22T22:07:54.817" Score="2" ViewCount="0" Body="&lt;p&gt;The mouse velocity is [(current mouse position) - (previous mouse position)]/timestep (presumably you're using a constant timestep of 10ms or something?) and then the acceleration is similarly [(current mouse velocity) - (previous mouse velocity)]/timestep. (At least, approximately.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the acceleration, I guess you can use the formulas you already have.&lt;/p&gt;&#xA;" OwnerUserId="496" LastActivityDate="2010-08-22T22:07:54.817" />
  <row Id="3075" PostTypeId="2" ParentId="2949" CreationDate="2010-08-22T22:08:17.687" Score="13" ViewCount="0" Body="&lt;p&gt;Brouwer's fixed point theorem, which has several non intuitive consequences in the real-world such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that if you lay a piece of paper on your desk and trace around its outline, then crumble/wad the paper up and put it back inside the lines that there will always be a point on the paper exactly above where it started relative to the desk&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, no matter how you stir your coffee there will always be some point in the liquid that ends in the same place that it was before mixing.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-08-22T22:08:17.687" CommunityOwnedDate="2010-08-22T22:08:17.687" CommentCount="7" />
  <row Id="3076" PostTypeId="2" ParentId="3060" CreationDate="2010-08-22T22:10:48.200" Score="7" ViewCount="0" Body="&lt;p&gt;Assume the inductive hypothesis that it is true for lists smaller than n.  To reorder 1...n this way as well, split it into evens and odds, and apply the function floor((x+1)/2) to these sets to create two half-problems of the same type which by the hypothesis can be solved.  Now unapply the transformation to each half-solution using the function 2x to recover the evens and the function 2x-1 to recover the odds.  These functions are linear so the condition is still satisfied.  Now concatenate these two lists together to form the solution. This concatenation always works because the average of an even and an odd is not an integer.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-23T19:00:44.020" LastActivityDate="2010-08-23T19:00:44.020" CommentCount="3" />
  <row Id="3077" PostTypeId="2" ParentId="2384" CreationDate="2010-08-22T22:22:54.347" Score="3" ViewCount="0" Body="&lt;p&gt;If you haven't been using &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471789011&quot; rel=&quot;nofollow&quot;&gt;&quot;The Art and Craft of Problem Solving&quot; by Zeitz&lt;/a&gt;, you should. Like, right at this moment. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For neat problems... you might want to look into &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883853256&quot; rel=&quot;nofollow&quot;&gt;&quot;Which Way Did The Bicycle Go?&quot; by Konhauser, Wagon and Velleman&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="2264" LastEditorDisplayName="" LastEditDate="2010-11-16T09:32:51.477" LastActivityDate="2010-11-16T09:32:51.477" CommunityOwnedDate="2010-08-22T22:22:54.347" />
  <row Id="3078" PostTypeId="2" ParentId="3059" CreationDate="2010-08-22T22:45:55.973" Score="6" ViewCount="0" Body="&lt;p&gt;The answer to &quot;what kinds of things can you integrate&quot; depends on the context. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;b&gt;&lt;em&gt;Measurable functions&lt;/em&gt;&lt;/b&gt; are things you can integrate over measure spaces, which includes in particular measurable subsets of R^n.  &lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;&lt;em&gt;Differential forms&lt;/em&gt;&lt;/b&gt; are things you can integrate over oriented smooth manifolds -- the key thing about them is that their integrals are invariant under smooth, orientation-preserving changes of coordinates.&lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;&lt;em&gt;Densities&lt;/em&gt;&lt;/b&gt; are things that can be integrated in a coordinate-independent way on any smooth manifold, regardless of whether it has an orientation or not.  &lt;/li&gt;&#xA;&lt;li&gt;Coming full circle, every Riemannian manifold (i.e., smooth manifold endowed with a Riemannian metric) has a naturally-defined density dV, so in that context you can integrate measurable functions again: the integral of the function f is defined to be the integral of the density f dV.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All three of the expressions you asked about are examples of densities.  For details, see my book &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;, pp. 375-382.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-08-22T22:45:55.973" />
  <row Id="3079" PostTypeId="2" ParentId="3020" CreationDate="2010-08-23T00:02:57.457" Score="2" ViewCount="0" Body="&lt;p&gt;No, because if there were a continuous inverse, then $\Phi$ would be bounded below, and from this it would follow that $W$ is complete.  But it isn't, because there are continuous functions that have Fourier series that are uniformly but not absolutely convergent, and hence Cauchy sequences (the partial sums of the Fourier series) in $W$ with no limit in $W$.  There are even examples of my claim from the last sentence in the disk algebra, i.e. functions in $C(\mathbb{T})$ whose negatively indexed Fourier coefficients vanish, as was first shown by Hardy and as I discovered when trying to answer a MathOverflow question &lt;a href=&quot;http://mathoverflow.net/questions/12968/mode-of-convergence-of-a-power-series&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-08-23T00:02:57.457" CommentCount="3" />
  <row Id="3080" PostTypeId="2" ParentId="3038" CreationDate="2010-08-23T00:07:55.107" Score="2" ViewCount="0" Body="&lt;p&gt;As mentioned above, the topologies you want are those which are not metrizalbe. There are several metrication theorems that are given in terms of separation properties of the spaces. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A space is first countable if every point has a neighborhood basis&lt;/li&gt;&#xA;&lt;li&gt;A space is second countable if it has a countable basis&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All metric spaces are first countable, so if your space is not first countable then it is not metrizable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most commonly known theorem is Urysohn's metrization theorem: every second countable, regular, Hausdorff space is metrizable. Hausdorff just means that points can be separated by open sets and regular means that (nonempty) closed sets and points can be separated by open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more powerful condition is that of normality. A space is normal if disjoint closed sets can be separated by open sets. It turns out that all metrizable space are normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, these non-metrizable spaces just seem like pathological examples, but some are very important. For example, the Zariski topology is not metrizable, but is essential to the field of algebraic geometry.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2010-08-23T00:07:55.107" />
  <row Id="3081" PostTypeId="2" ParentId="83" CreationDate="2010-08-23T00:47:53.277" Score="1" ViewCount="0" Body="&lt;p&gt;Personal experiences as an elementary school math resource teacher and a 1st grade teacher have led me to believe that mathematics learning should be engaging and fun at any level (including the graduate level), but clearly it is of the utmost importance to effectively engage students at an early age, so as to hopefully instill in them a lifelong appreciation for the subject.  Most children are quite active in nature, so providing hands-on activities that keep their minds and bodies actively involved is an important task. It seems that geometry is like the fail-proof mathematical concept that can pique the interest of even the toughest audiences.  Probably because geometry is naturally hands-on and therefore easily applicable to real-life situations.  Children can see and touch geometric concepts.  It may seem trivial to adults, but numbers and number systems can in reality be quite abstract concepts to wrap some little brains around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to engage children is to give them a sense of ownership in the mathematics learning process, providing them with a keener awareness of mathematical applications, and thereby providing them with a more personal encounter with learning mathematics.  Children feel empowered when they are permitted to study things that are of interest to them.  Mathematical connections can be made to just about anything that children find curious in our present day society.  Be it the arts, dance, music, video games, computers, ipods, facebook, twitter, sports, or games, organic &quot;aha&quot; or &quot;teachable&quot; moments can easily be generated by harnessing children's intuitve curiosities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found literature (specifically picture books) a useful tool for harnessing some of these curiosities in children of all ages. Additionally, actively listening to a narrative can engage the mind and promote purposeful thinking, learning, and even critical analysis.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, another reason that I enjoy using literature to teach math is that literature tends to provide a portal for asking and generating thought provoking and meaningful questions.  One can kind of let the narrative guide the instruction.  If the kids aren't interested in it, there is no point in belaboring the issue, so move on until you find a piece of literature that does interest them.  Some books that I would recommend (based on student ratings) are &lt;a href=&quot;http://books.google.com/books?id=5BfjuWQBV6IC&amp;amp;printsec=frontcover&amp;amp;dq=the+very+greedy+triangle&amp;amp;source=bl&amp;amp;ots=VRZPvz19ql&amp;amp;sig=0-Yz0lhnPBJhwXQ_JF7Tp_w9dxU&amp;amp;hl=en&amp;amp;ei=irxxTMr-JYT6sAP96oTkDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CD0Q6AEwBg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Very Greedy Triangle&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=MzOvangyz-EC&amp;amp;dq=the+number+devil&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=87xxTNuWN5OksQO2lfm_BA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=11&amp;amp;sqi=2&amp;amp;ved=0CEIQ6AEwCg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Number Devil&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0698117530&quot; rel=&quot;nofollow&quot;&gt;Anno's Mysterious Multiplying Jar&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0590300121&quot; rel=&quot;nofollow&quot;&gt;Amanda Bean's Amazing Dream&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=SkU7BGazcusC&amp;amp;printsec=frontcover&amp;amp;dq=grapes+of+math&amp;amp;source=bl&amp;amp;ots=NL1kZAvPEg&amp;amp;sig=9P4JGvSSLVie2dHxKpJAKRl07Go&amp;amp;hl=en&amp;amp;ei=Q75xTNHVJIHWtQP8voTNCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;sqi=2&amp;amp;ved=0CCYQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Grapes of Math&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=Tncckz5fkEoC&amp;amp;dq=how+much+is+a+million&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=gb5xTN6aHIa4sAP_5YzFCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CCoQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;How Much Is a Million&lt;/a&gt;, &lt;a href=&quot;http://www.curriculumconnection.net/sircumdragonpi.htm&quot; rel=&quot;nofollow&quot;&gt;Sir Cumference and the Dragon of Pi&lt;/a&gt;, &lt;a href=&quot;http://www.gregtang.com/mathstore.asp&quot; rel=&quot;nofollow&quot;&gt;The Best of Times&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/059093998X&quot; rel=&quot;nofollow&quot;&gt;One Grain of Rice&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0590944592&quot; rel=&quot;nofollow&quot;&gt;Spaghetti and Meatballs For All&lt;/a&gt;, and we can't forget how Lewis Carroll's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0451527747&quot; rel=&quot;nofollow&quot;&gt;Alice in Wonderland&lt;/a&gt; is densely saturated with mathematical concepts...There are oodles more!  However, authors that I tend to come back to on a consistent basis are &lt;a href=&quot;http://www.gregtang.com/&quot; rel=&quot;nofollow&quot;&gt;Greg Tang&lt;/a&gt;, &lt;a href=&quot;http://www.jsworldwide.com/&quot; rel=&quot;nofollow&quot;&gt;Jon Scieszka&lt;/a&gt;, &lt;a href=&quot;http://www2.scholastic.com/browse/collection.jsp?id=492&quot; rel=&quot;nofollow&quot;&gt;Marilyn Burns&lt;/a&gt;, &lt;a href=&quot;http://cneuschwander.com/&quot; rel=&quot;nofollow&quot;&gt;Cindy Nueschwander&lt;/a&gt;, and &lt;a href=&quot;http://www.davidschwartz.com/&quot; rel=&quot;nofollow&quot;&gt;David M. Schwartz&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, just try to find ways to make mathematics learning engaging and thought provoking but also meaningful and fun!  And by all means, one doesn't always have to use a mathematics picture book to teach math concepts.  Once I read, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0590558498&quot; rel=&quot;nofollow&quot;&gt;It's Pumpkin Time&lt;/a&gt;, a book about the life cycle of pumpkins to my students and ended up creating measurement activities that corresponded with the book rather than science actitivies.  Just get creative! Children love that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More &lt;a href=&quot;http://www.childrenspicturebooks.info/articles/picture_books_for_math.htm&quot; rel=&quot;nofollow&quot;&gt;resources&lt;/a&gt; if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="219" LastEditorUserId="555" LastEditorDisplayName="" LastEditDate="2010-08-24T06:38:23.577" LastActivityDate="2010-08-24T06:38:23.577" CommunityOwnedDate="2010-08-23T00:47:53.277" CommentCount="1" />
  <row Id="3082" PostTypeId="2" ParentId="2384" CreationDate="2010-08-23T01:35:04.603" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.keypress.com/x5488.xml&quot; rel=&quot;nofollow&quot;&gt;Problem Solving Strategies: Crossing the River with Dogs and Other Mathematical Adventures&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T01:35:04.603" CommunityOwnedDate="2010-08-23T01:35:04.603" />
  <row Id="3083" PostTypeId="2" ParentId="2384" CreationDate="2010-08-23T01:36:11.713" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://arml.com/books.php&quot; rel=&quot;nofollow&quot;&gt;Books of old ARML contests&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T01:36:11.713" CommunityOwnedDate="2010-08-23T01:36:11.713" />
  <row Id="3084" PostTypeId="2" ParentId="2943" CreationDate="2010-08-23T03:22:03.437" Score="3" ViewCount="0" Body="&lt;p&gt;Let $K$ be a complement to $N$ in $G$, which exists by Schur-Zassenhaus.  Then $|K|=|G|/|N|$.  Consider the subgroup $UN$.  Since $K(UN)=G$, $|K\cap UN|=|K|\cdot |UN|/|G|=|K|\cdot|U|\cdot|N|/|G|=|U|$, since $K$ is a complement to $N$.  Thus, $K\cap UN$ is a complement to $N$ in $UN$.  Since either $U$ or $N$ is solvable, all complements to $N$ in $UN$ are conjugate, so $K\cap UN=U^g$.  Thus, $K^{g^{-1}}$ contains $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-23T03:22:03.437" CommentCount="4" />
  <row Id="3085" PostTypeId="1" CreationDate="2010-08-23T04:44:29.543" Score="5" ViewCount="450" Body="&lt;p&gt;Is there a comprehensive resource listing binomial identities? I am more interested in combinatorial proofs of such identities, but even a list without proofs will do.&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2010-10-07T19:39:11.610" Title="A comprehensive list of binomial identities?" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3086" PostTypeId="2" ParentId="3020" CreationDate="2010-08-23T04:53:18.473" Score="0" ViewCount="0" Body="&lt;p&gt;No,no,no..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wiener algebra not normed by the sup-norm. &#xA;As I see it, it might be very confusing for other readers if we just guess here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common notations are: &#xA;$W$ or $A(\mathbb{T})$ for the algebra of absolute convergent Fourier series normed by &#xA;$$\|f\|_{W}= \sum_{n=-\infty}^\infty |a_n| \quad(&amp;lt;\infty)$$ &#xA;where $f(t)=\sum_{n=-\infty}^\infty a_ne^{int}$ ($0\leq t\leq2\pi$).&#xA;That is $W=\ell¹(\mathbb{Z})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W^+$ or $A^+(\mathbb{T})$ for the algebra of absolute convergent Taylor series normed by &#xA;$$\|f\|_{W^+}= \sum_{n=0}^\infty |a_n| \quad(&amp;lt;\infty)$$ &#xA;where $f(z)=\sum_{n=0}^\infty a_nz^{n}$ ($|z|\leq1$). That is $W^+=\ell¹(\mathbb{N})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some references.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ZZk4AAAAIAAJ&amp;amp;pg=PA167&amp;amp;dq=Wiener+algebra+absolute+convergent+fourier&amp;amp;hl=en&amp;amp;ei=gfpxTL3LOovOswavkpS6Bg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CEAQ6AEwBA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=ZZk4AAAAIAAJ&amp;amp;pg=PA167&amp;amp;dq=Wiener+algebra+absolute+convergent+fourier&amp;amp;hl=en&amp;amp;ei=gfpxTL3LOovOswavkpS6Bg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CEAQ6AEwBA#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&#xA;(page 169)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wiener_algebra&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/WienerAlgebra.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/WienerAlgebra.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-23T05:05:58.843" LastActivityDate="2010-08-23T05:05:58.843" CommentCount="3" />
  <row Id="3087" PostTypeId="1" CreationDate="2010-08-23T05:09:35.357" Score="6" ViewCount="360" Body="&lt;p&gt;&quot;&lt;em&gt;Life is open book.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of widely accessible, inexpensive (or even free) computational tools and Computer Algebra Systems (TI-89, Wolfram|Alpha, etc.), much of what traditionally comprises a high school math curriculum can now easily be done by almost everyone.  Factoring polynomials, solving inequalities, graphing linear equations, differentiation and integration -- these are the types of skills high school math students spend most of their time learning, and yet all of it can be done for free by anyone with a web browser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean for the high school math curriculum?  On the one hand, we could leave it more-or-less the same, insisting that today's student learn what we learned decades ago, while banning or carefully regulating the use of these new tools.  On the other hand, we could embrace the tools and the opportunities they create to spend more math class time on different topics and skills, perhaps focusing more on analytic and synthetic problem solving and less on mechanical symbolic manipulation -- but at the risk of students never learning some basic foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how about it?  Binomial coefficients?  The angle-addition formulas for trig functions?  The conditions under which a function has an inverse?  Basic computer programming?  Keeping in mind that the vast majority of high school students do &lt;em&gt;not&lt;/em&gt; go on to become professional mathematicians, what should the high school math curriculum consist of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, I post this question (inspired by &lt;a href=&quot;http://www.theodoregray.com/BrainRot/&quot; rel=&quot;nofollow&quot;&gt;this discussion&lt;/a&gt;) here because this is a community of thoughtful mathematicians.  I recognize this discussion may belong in a different forum, but I don't what/where that forum is.  Any suggestions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-10-31T17:30:22.860" CommunityOwnedDate="2010-08-23T11:37:46.497" Title="What should the high school math curriculum consist of?" Tags="&lt;big-list&gt;&lt;education&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="3088" PostTypeId="2" ParentId="3053" CreationDate="2010-08-23T05:28:43.003" Score="2" ViewCount="0" Body="&lt;p&gt;I think it's worth to develop a little bit more Akhil's interesting remark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; If instead of ${\cal C}^0(\mathbb{R}, \mathbb{R})$, you take ${\cal C}^0([a,b], \mathbb{R})$, where $[a,b] \subset \mathbb{R}$ is a compact interval, then something interesting happens: you can give ${\cal C}^0([a,b], \mathbb{R})$ the &lt;em&gt;compact-open topology&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Compact-open_topology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Compact-open_topology&lt;/a&gt; , and this topology coincides with the one induced by the sup norm (also called the topology of &lt;em&gt;uniform convergence&lt;/em&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Uniform_convergence&lt;/a&gt; ). More generally, the same is true for ${\cal C}^0 (X , Y)$, where $X$ is any compact topological space and $Y$ any metric space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Now, for psychological reasons, let's use the notation $Y^X$ instead of ${\cal C}^0 (X , Y) $, and let $X$, $Y$ be any pair of topological spaces. Put the compact-open topology on $Y^X$ and let $Z$ be a third topological space. To every map $F: X\times Z \longrightarrow Y$ you can associate the map $\Phi (F) : Z \longrightarrow Y^X$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(\Phi (F) (z)) (x)= F(x,z) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reciprocally, to every map $\gamma :  Z \longrightarrow Y^X$ you can associate the map $\Psi (\gamma ): X \times Z \longrightarrow Y$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Psi (\gamma) (x,z) = (\gamma (z)) (x) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check easily that $\Phi$ and $\Psi$ are inverses one each other. Moreover, if $F$ is continuous, so is $\Phi (F)$. If $X$ is a locally compact Hausdorff space, the same is true for $\Psi$: $\gamma$ continuous implies $\Psi (\gamma )$ continuous. (See Munkres' &quot;Topology&quot;, theorem 46.11.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when $X$ is a locally compact Hausdorff space, you have bijections, inverses one each other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Phi : Y^{X \times Z} \longrightarrow (Y^X)^Z  \qquad \text{and} \qquad \Psi: (Y^X)^Z \longrightarrow Y^{X \times Z}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take $Z = I$, the unit interval. This bijection tells us that, when $X$ is a locally compact Hausdorff space, &lt;em&gt;paths&lt;/em&gt; $\gamma : I \longrightarrow Y^X$ and &lt;em&gt;homotopies&lt;/em&gt; $F : X \times I \longrightarrow Y$ are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; An easy topological exercise: any map $f: X \longrightarrow Y$ is homotopic to a constant map and all the constant maps are homotopic, either if&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$X$ is contractile and $Y$ is path-connected, or&lt;/li&gt;&#xA;&lt;li&gt;$Y$ is contractile.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, coming back to your situation, we could have said that ${\cal C}^0 (\mathbb{R}, \mathbb{R})$, with the compact-open topology, is always path-connected because $X = \mathbb{R}$ is locally compact, Hausdorff and contractile, and $Y= \mathbb{R}$ is path-connected (or because $X=\mathbb{R}$ is locally compact and Hausdorff and $Y= \mathbb{R}$ is contractile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notice that Jason has used the second possibility: he has constructed a &lt;em&gt;homotopy&lt;/em&gt; from the constant function $0$ to $f$. We could use the first one: $F(x,t) = f(xt)$ is a homotopy from the constant function $f(0)$ to $f$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, Jason's solution is more straightforward and doesn't need all this elementary machinery of point-set topology. But now we can say that also other function spaces, with the compact-open topology, are path-connected: every time you take as $X$ a locally compact Hausdorff space and any of the two previous possibilities, ${\cal C}^0(X,Y)$ is path-connected. For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\cal C}^0(\mathbb{R}, Y) ,\ {\cal C}^0([a,b],Y) , \ {\cal C}^0(\mathbb{R}^n,Y) ,\ {\cal C}^0(D^2,Y)  \dots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any path-connected space $Y$, are path-connected. Also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\cal C}^0(X,\mathbb{R}) , \ {\cal C}^0(X, [a,b]) , \ {\cal C}^0(X,\mathbb{R}^n) ,\ {\cal C}^0(X,D^2)  \dots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any locally compact Hausdoff space $X$, are path-connected.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-23T08:32:23.463" LastActivityDate="2010-08-23T08:32:23.463" />
  <row Id="3089" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T06:45:49.170" Score="2" ViewCount="0" Body="&lt;p&gt;Gold's theorem provides pretty convincing mathematical evidence supporting the universal grammar hypothesis in linguistics. This hypothesis is two-fold: (1) children are not presented logically with enough information to actually learn their native language; (2) hence there exists a &lt;em&gt;universal grammar&lt;/em&gt; which is encoded somehow in the human brain and which facilitates the logical gap between the positive data given to the child and the data necessary to determine the language's grammar. While the universal grammar hypothesis isn't universally accepted, it has been one of the most important ideas in linguistics so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold's theorem shows that certain classes of languages are logically not learnable. Of course, it operates in a purely formal setting. I'll provide up this setting now following the definitions and notations of &lt;a href=&quot;http://www.math.harvard.edu/theses/senior/carroll/gabriel_carroll.pdf&quot; rel=&quot;nofollow&quot;&gt;Gabriel Carroll&lt;/a&gt;, pg. 41.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a finite alphabet $\Sigma$ and let $\Sigma^*$ designate the set of finite sequences of elements of $\Sigma$. A language $L$ is a subset of $\Sigma^*$. A text of $L$ is an infinite string $w_1, w_2, \dots$ of elements of $L$ such that every element of $L$ occurs at least once. A learner for a class $\mathcal{L}$ of languages is a function $\Lambda : (\Sigma^*)^* \rightarrow \mathcal{L}$ that intuitively takes a sequence of strings of $\Sigma$ and guesses the language in $\mathcal{L}$ in which all these strings are grammatically correct. The learner $\Lambda$ learns the language $L \in \mathcal{L}$ if for every text $w_1,w_2,\dots$ of $L$ there exists a natural number $N$ such that $\Lambda(w_1,w_2,\dots,w_n) = L$ for $n \geq N$. The learner $\Lambda$ learns the class $\mathcal{L}$ if it learns each language in $\mathcal{L}$, and the class $\mathcal{L}$ is learnable if there exists a learner which learns it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Gold's theorem, first proved by Gold in his seminal &lt;a href=&quot;http://www3.isrl.illinois.edu/~junwang4/langev/localcopy/pdf/gold67limit.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (but my wording is taken from Carroll):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the class $\mathcal{L}$ contains all finite languages and at least one infinite language, then $\mathcal{L}$ is not learnable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In particular, any finite language is regular. Hence the class of regular languages is unlearnable, and it follows at once that every class of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chomsky_hierarchy&quot; rel=&quot;nofollow&quot;&gt;Chomsky Hierarchy&lt;/a&gt; is unlearnable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of Gold's theorem is, as Carroll shows, not very hard, although certainly not intuitive, and it can be reduced to a corollary of the following characterization of learnable classes of languages (Carroll, Lemma 9):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A countable class $\mathcal{L}$ of nonempty languages is learnable if and only if, for each $L \in \mathcal{L}$, there is a finite ''telltale'' subset $T \subseteq L$ such that $L$ is minimal in $\{L' \in \mathcal{L} : T \subseteq L'\}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-08-23T16:51:24.807" LastActivityDate="2010-08-23T16:51:24.807" CommunityOwnedDate="2010-08-23T06:45:49.170" />
  <row Id="3090" PostTypeId="2" ParentId="370" CreationDate="2010-08-23T06:51:29.907" Score="0" ViewCount="0" Body="&lt;p&gt;(This is really a comment to &lt;a href=&quot;http://math.stackexchange.com/questions/370/good-book-lecture-notes-about-category-theory/2997#2997&quot;&gt;mathphysicist's answer&lt;/a&gt;, but I don't have enough rep to post comments yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goldblatt's book (which is very beautifully written, by the way) is available online in its entirety &lt;a href=&quot;http://dlxs2.library.cornell.edu/m/math/browse/author/g.php&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-23T06:51:29.907" CommunityOwnedDate="2010-08-23T06:51:29.907" />
  <row Id="3091" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T07:20:57.630" Score="3" ViewCount="0" Body="&lt;p&gt;I got really struck by duality, when my professor lectured about it the first time. I think that even though the algebraic concept is easy to understand, to think that there exists a space such that all inclusions are switched always had a special place in my mind.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-08-23T07:20:57.630" CommunityOwnedDate="2010-08-23T07:20:57.630" CommentCount="2" />
  <row Id="3092" PostTypeId="2" ParentId="3069" CreationDate="2010-08-23T07:40:15.533" Score="1" ViewCount="0" Body="&lt;p&gt;Contrary to the responses thus far (and with no disrespect intended to their authors), I think we can say quite unequivocally that there are &lt;strong&gt;&lt;em&gt;strictly more&lt;/em&gt;&lt;/strong&gt; than countably-many sets with distinct infinite cardinalities, and that (ipso facto) there are uncountably many such sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers before mine fixate on the idea of cardinality itself as the concept of quantity, and correctly point out that that there isn't a &quot;set of infinite cardinalities&quot;. (There is a &lt;em&gt;class&lt;/em&gt; of infinite cardinalities, if you like that sort of thing, but perhaps not everyone likes talking about proper classes.) Because there cannot be a set of infinite cardinalities, the reasoning goes, there cannot be a cardinality (a well-defined sense of &quot;the number of&quot;) of infinite cardinalities. However, the original question isn't about &quot;cardinality&quot; &lt;em&gt;per se&lt;/em&gt;: it's about a less formal notion of &quot;the number of something&quot;, which we conventionally define by cardinality in the case of sets. As we are not talking about &quot;the set of X&quot;, however, perhaps we shouldn't be using this definition of number. What then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the original question isn't asking us &lt;em&gt;precisely&lt;/em&gt; how many infinite cardinalities there are: just whether this number is &lt;strong&gt;countable&lt;/strong&gt; or &lt;strong&gt;uncountable&lt;/strong&gt; --- a dichotomy of ordering; whether it is less-than-or-equal-to the &quot;number&quot; of natural numbers (&lt;em&gt;i.e.&lt;/em&gt; whether it can be placed in bijective correspondance with a subset of the natural numbers), or not. This is, as far as I'm aware, the definition of the terms &quot;countable&quot; and &quot;uncountable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number of infinite cardinalities cannot be put into one-to-one correspondence with any subset of the natural numbers, it follows that the number of infinities is uncountable. The fact that they cannot be put into a set is irrelevant, in my opinion: at worst, one only needs to identify a single uncountable set U of infinities, note that there is no surjection onto it from the natural numbers, and note incidentally that there are more infinities outside of the set U anyway. Any cardinal &amp;alefsym;&lt;sub&gt;&lt;em&gt;C&lt;/em&gt;&lt;/sub&gt; (identified in the usual way with an ordinal), for &lt;em&gt;C&lt;/em&gt; itself an uncountable ordinal, will suffice for U.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there are uncountably many different infinities, QED.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-08-23T07:49:51.910" LastActivityDate="2010-08-23T07:49:51.910" CommentCount="3" />
  <row Id="3093" PostTypeId="1" CreationDate="2010-08-23T07:47:44.270" Score="5" ViewCount="263" Body="&lt;p&gt;The Banach--Tarski theorem applies only in the case of three or more dimensions. In 3D, there are five regular solids, two of them being not at all obvious, and the 4D case is also interesting; but the higher-dimensional cases each yield just three &quot;solids&quot;, only one of which isn't obvious. In dynamics, particles tend to coalesce in one or two dimensions, while in four or more dimensions they tend to disperse drearily; only in 3D do they move freely but with significant local interaction. And the Poincare conjecture proved to be much harder in the 3D case than in the others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question has three parts: What other examples are there of 3D richness? Are there any underlying reasons for it? And are there fields where richness begins at a higher, but still small, number of dimensions? I would be particularly interested to learn of any theorems that hold in just a particular number (not 0, 1, or 2) of dimensions. &lt;/p&gt;&#xA;" OwnerUserId="875" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-23T10:10:47.050" LastActivityDate="2010-08-23T23:23:27.653" CommunityOwnedDate="2010-08-23T18:37:37.637" Title="Why is the 3D case so rich?" Tags="&lt;big-list&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="3" />
  <row Id="3094" PostTypeId="2" ParentId="3024" CreationDate="2010-08-23T08:02:52.710" Score="1" ViewCount="0" Body="&lt;p&gt;The full solution to this problem (with a slight difference in notation) is given in appendix A: of the following  &lt;a href=&quot;http://arxiv.org/PS_cache/gr-qc/pdf/9910/9910109v2.pdf&quot; rel=&quot;nofollow&quot;&gt;article:&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-23T08:02:52.710" />
  <row Id="3095" PostTypeId="2" ParentId="3087" CreationDate="2010-08-23T08:15:36.647" Score="2" ViewCount="0" Body="&lt;p&gt;I would include more probability, such as how to conduct a simple probabilistic experiment using statistical tests. Once you understand that, you can do a lot by simply looking up various tests. Without understanding probability, people are easily manipulated using statistics or make poor choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would include definitely include more use of technology, including Wolfram Alpha and basic mathematical programming. These tools allow people to achieve much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basic mathematical philosophy would be good as well. Game theory can be very thought provoking and is quite simple as well. We could look at other areas even though we wouldn't be able to examine them rigorously. I think students should understand that maths is based on axioms, that not all maths problems can be solved (assuming ZFC is consistent) and that many problems can't be solved efficiently (P vs NP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would focus more on exploring various mathematical puzzles or unexpected results: &lt;a href=&quot;http://math.stackexchange.com/questions/234/card-doubling-paradox&quot;&gt;Card doubling paradox&lt;/a&gt;, &lt;a href=&quot;http://www.relisoft.com/science/hats.html&quot; rel=&quot;nofollow&quot;&gt;3 hats puzzle&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty hall problem&lt;/a&gt; and other results like this.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-23T22:43:15.997" LastActivityDate="2010-08-23T22:43:15.997" CommunityOwnedDate="2010-08-23T18:46:03.097" CommentCount="3" />
  <row Id="3096" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T08:32:07.923" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Exotic_R4&quot; rel=&quot;nofollow&quot;&gt;Exotic $\mathbb R^4$&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are infinitely many non-diffemorphic smooth structures on the topological space $\mathbb R^n$ if and only if $n=4$. (Otherwise there is only one diffeomorphism class.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-23T08:32:07.923" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="1" />
  <row Id="3097" PostTypeId="1" AcceptedAnswerId="3100" CreationDate="2010-08-23T09:18:23.443" Score="5" ViewCount="172" Body="&lt;p&gt;I am aware that on a finite set the number of equivalence relations is the n-th &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_number&quot; rel=&quot;nofollow&quot;&gt;Bell's Number&lt;/a&gt;. On the other hand, the only reference I could find on the web for infinite sets was this: &lt;a href=&quot;http://www.informaworld.com/smpp/content~db=all~content=a750850336&quot; rel=&quot;nofollow&quot;&gt;On counting equivalence relations&lt;/a&gt; by D.J. Baylis, in which he proves that the set of equivalence relations on $\mathbb{N}$ is uncountable. I am interested in any sort of theorem that gives some rule about counting equivalence relations on an infinite set -something like $&amp;#x2136;_0 = \aleph_0$ and &amp;#x2136;&lt;sub&gt;n&lt;/sub&gt; = 2&lt;sup&gt;&amp;#x2136;&lt;sub&gt;n-1&lt;/sub&gt;&lt;/sup&gt; - if there are any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Mathjax does not seem to be working on the second formula with beth, feel free to edit and correct it.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-23T10:47:25.220" LastActivityDate="2010-08-23T11:33:51.977" Title="Counting equivalence relations" Tags="&lt;set-theory&gt;" AnswerCount="3" />
  <row Id="3098" PostTypeId="2" ParentId="3097" CreationDate="2010-08-23T10:06:25.573" Score="0" ViewCount="0" Body="&lt;p&gt;Let $X$ be an infinite set. Any function $f : X \to X$ determines an equivalence relation on $X$:&#xA;$$&#xA;x_1 \equiv x_2 \iff f(x_1) = f(x_2).&#xA;$$&#xA;It follows that there are at least $2^{|X|}$ equivalence relations on $X$ (see the comments below). On the other hand, any equivalence relation defines a function from $X$ to $X,$ take a complete set $S$ of representatives, send every element of a given class to its representative in $S.$ Thus there are no more than $|X|^{|X|}$ equivalence relations on $X.$ Therefore there are exactly&#xA;$|X|^{|X|}=2^{|X|}$ equivalence relations on $X.$ &lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-08-23T11:33:51.977" LastActivityDate="2010-08-23T11:33:51.977" CommentCount="7" />
  <row Id="3099" PostTypeId="2" ParentId="3097" CreationDate="2010-08-23T10:16:57.570" Score="0" ViewCount="0" Body="&lt;p&gt;I will assume that the axiom of choice holds: this may not be necessary, but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by observing that equivalence relations on a set $S$ are the &quot;same&quot; as partitions of $S$ (recall that a partition of $S$ is a set $B$ of subsets of $S$ with the properties that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\cup_{b \in B}b = S$, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) for all $b,b'\in B$ we have $b \cap b'=\emptyset$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be an infinite set, let $B(S)$ denote the set of partitions of $S$ and $P(S)$ the power set of $S$ (namely, the set of all subsets of $S$). I am going to show that $|B(S)| = |P(S)|$; note that $|P(S)| = 2^{|S|}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, use the axiom of choice to fix, for every non-empty subset $A$ of $S$, an element $c(A) \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To any function $f \colon S \to S$ we associate a partition $B_f$ of $S$ by letting $B_f = \{ f^{-1}(s) | s \in f(S) \}$; in words, $B_f$ is the partition whose elements are the non-empty fibers of the function $f$.  Any partition $B$ is obtained by this construction: we define a function $f_B \colon S \to S$ by letting $f_B(s) = c(b_s)$, where $b_s$ is the part of $B$ containing $s$. Clearly, the partition $B_{f_B}$ coincides with the partition $B$. We deduce that there is a surjection from the set of all functions $S \to S$ to the set of partitions; thus we conclude that $|B(S)| \leq |S^S| = |P(S)| = |2^S|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other inequality, let $P'(S)$ denote the set of unordered pairs of subsets of $S$ of the form $\{ A , S \setminus A \}$, where $A \subset S$ is a proper non-empty subset of $S$. Each unordered pair in $P'(S)$ determines uniquely a partition of $S$ with two parts, so that there is an injection $P'(S) \to B(S)$. We deduce that $|B(S)| \geq |P'(S)|$. To conclude it suffices to show that $|P'(S)| \geq |P(S)|$. This is easy, since there is a natural surjection $P(S) \setminus \{ S , \{\emptyset\} \} \to P'(S)$ whose fibers consist of exactly two elements, and $P(S)$ is an infinite set.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-23T10:16:57.570" />
  <row Id="3100" PostTypeId="2" ParentId="3097" CreationDate="2010-08-23T10:17:50.270" Score="3" ViewCount="0" Body="&lt;p&gt;Firstly, to address Beth numbers (as you noted, the math markup doesn't work with them. I'll use words instead)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beth numbers are cardinal numbers defined as followed:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Beth null = Aleph null&lt;/li&gt;&#xA;&lt;li&gt;Beth alpha+1 = Cardinality of the power set of Beth alpha&lt;/li&gt;&#xA;&lt;li&gt;Beth delta, for a limit ordinal delta = the union of all smaller Beth numbers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, given an infinite set $X$ and assuming the axiom of choice, we have that $|X\times X|=|X|$, therefore $|P(X\times X)| = |P(X)|$.&#xA;Define $R(X)$ as the set of all equivalence relations on $X$, then clearly $R(X) \subset P(X\times X)$ and therefore $|R(X)| \le |P(X)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand there are $2^{|X|}$ partitions of $X$ into two sets (an easy exercise in cardinal arithmetic), so for every partition as such $\{X', X\backslash X'\}$ we can define an equivalence relation with two equivalence classes, namely $X'$ and its complement in $X$. Therefore $|R(X)| \ge |P(X)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all we have that the number of equivalence relations on $X$ is $2^{|X|}$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-08-23T10:29:54.373" LastActivityDate="2010-08-23T10:29:54.373" CommentCount="3" />
  <row Id="3101" PostTypeId="1" AcceptedAnswerId="3549" CreationDate="2010-08-23T10:20:26.163" Score="4" ViewCount="402" Body="&lt;p&gt;Hi! I'm studying for math exam and one of the questions that often appears is related to derivative of a product of two functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem says that $(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof goes like this:&#xA;$f(x+h)g(x+h)-f(x)g(x)=(f(x+h)-f(x))g(x)+(g(x+h)-g(x))f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that we divide the equation by h and let h approach 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I don't understand is how they got the right side of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same problem with quotient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem says:&#xA;$\left(\frac{f(x)} {g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)} {g^2(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above comes from&#xA;$\frac {f(x+h)} {g(x+h)} - \frac {f(x)} {g(x)} = \frac{(f(x+h)-f(x))g(x)-(g(x+6h)-g(x))f(x)} {g(x+h)g(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see from where $f(x+h)g(x)$ comes, but I can't see from where $f(x)g(x)$ came.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-08-28T14:38:22.457" Title="Derivative of a product and derivative of quotient of functions theorem: I don't understand its proof." Tags="&lt;calculus&gt;" AnswerCount="9" CommentCount="4" FavoriteCount="1" />
  <row Id="3103" PostTypeId="2" ParentId="3101" CreationDate="2010-08-23T10:45:10.060" Score="2" ViewCount="0" Body="&lt;p&gt;I'll let someone else take care of proving the product formula; the quotient formula follows from applying the product formula and the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, remember that you can express $\frac{f(x)}{g(x)}$ as $f(x)\cdot\left(\frac1{g(x)}\right)$. You can now apply the product formula as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=\left(\frac{\mathrm{d}}{\mathrm{d}x} f(x)\right)\left(\frac1{g(x)}\right)+f(x)\left(\frac{\mathrm{d}}{\mathrm{d}x}\frac1{g(x)}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left addend is easily simplified:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=\frac{f^{\prime}(x)}{g(x)}+f(x)\left(\frac{\mathrm{d}}{\mathrm{d}x}\frac1{g(x)}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the right addend requires the chain rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=\frac{f^{\prime}(x)}{g(x)}+f(x)\left(-\frac1{g(x)^2}\right)\left(\frac{\mathrm{d}}{\mathrm{d}x}g(x)\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the formula follows from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you will be doing the proof using limits, just apply the steps you took in proving the product rule, with the terms being $f(x)$ and $\frac1{g(x)}$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-23T10:45:10.060" />
  <row Id="3104" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T10:57:03.550" Score="1" ViewCount="0" Body="&lt;p&gt;I think it's to do something with that in dimension 4 and above, you can have 2 commuting rotations about different axes (e.g. rotating around the xy-plane or zw-plane), so that things can be 'broken up' into separate pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related is the fact that you need 3 parameters to describe a rotation in 3-space, but 6 in 4-space.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-08-23T10:57:03.550" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="2" />
  <row Id="3105" PostTypeId="2" ParentId="3101" CreationDate="2010-08-23T11:09:04.903" Score="7" ViewCount="0" Body="&lt;p&gt;There is a more comprehensible (I hope) proof coming from thinking of derivative as a linear approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the definition of derivative&lt;sup&gt;1&lt;/sup&gt;, $f(x)=f(x_0)+f'(x_0)(x-x_0)+o(x-x_0)$ and $g(x)=g(x_0)+g'(x_0)(x-x_0)+o(x-x_0)$. Multiplying the equalities we get&#xA;\begin{multline}&#xA;f(x)g(x)=f(x_0)g(x_0)+(f'(x_0)g(x_0)+f(x_0)g'(x_0))(x-x_0)+\\&#xA;+f'(x_0)g'(x_0)(x-x_0)^2+o(x-x_0)&#xA;\end{multline}&#xA;and since $f'(x_0)g'(x_0)(x-x_0)^2=o(x-x_0)$, we can rewrite it as \[(fg)(x)=(fg)(x_0)+(f'g+fg')(x_0)\cdot(x-x_0)+o(x-x_o)\] — which exactly means that (fg)'=f'g+fg'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; $\lim_{x\to x_0}g(x)=a$ iff $g(x)=a+o(1)$, hence $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$ iff $\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)+o(1)$ which can be rewritten as $f(x)-f(x_0)=f'(x_0)(x-x_0)+o(x-x_0)$ (see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Little-o_notation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; if &quot;o(1)&quot; notation is unfamiliar to you).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-23T22:48:30.890" LastActivityDate="2010-08-23T22:48:30.890" CommentCount="5" />
  <row Id="3106" PostTypeId="2" ParentId="3031" CreationDate="2010-08-23T11:28:53.687" Score="3" ViewCount="0" Body="&lt;p&gt;Due to the Peter-Weyl theorem every finite dimensional Lie group is isomorphic to a subgroup of the&#xA;orthogonal group O(m) for some m. Therefore its Lie algebra is a subalgebra of the &#xA;Lie algebra of the orthogonal group which is the same as the Lie algebra of the spin group Spin(m).&#xA;Now, the Lie algebra of the spin group is the bivector subalgebra of the clifford algebra Cl(m).&#xA;Thus every finite dimensional Lie algebra is isomorphic to a Lie subalgebra of the bivector subalgebra of a clifford algebra.&#xA;This point of view seems analogous to the Cayley's theorem for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There exists a projective representation of the symmetric group in the Clifford algebra:&#xA;(a,b) --&gt; e_a - e_b. Thus the Clifford algebra may be seen as a kind of quantization of the symmetric group.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This construction generalizes at least to the case of affine Kac-Moody algebras which have similar&#xA;realizations in the infinite dimensional Clifford algebra.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-23T11:28:53.687" />
  <row Id="3107" PostTypeId="2" ParentId="3101" CreationDate="2010-08-23T11:49:38.350" Score="1" ViewCount="0" Body="&lt;p&gt;(Adding and subtracting the same amount, like $f(x)g(x)$, is very common in mathematics. Think of what you do when you solve a linear equation - add+subtract multiply+divide.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a geometrical proof of the product rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $T_f$ the tangent of $f(x)$ at $x_0$. It is given by $T_f(x)=f'(x_0)(x-x_0) + f(x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $T_g$ the tangent of $g(x)$ at $x_0$. It is given by $T_g(x)=g'(x_0)(x-x_0) + g(x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope of $f(x)g(x)$ at $x_0$ is $(fg)'(x_0)$ - which we want to find. That slope is the same as the slope of the product of the tangents at $x_0$: that is $(fg)'(x_0)=(T_fT_g)'(x_0)$. Multiplying through and differentiate we get&#xA;$$(T_fT_g)'(x) = ((f'(x_0)(x-x_0) + f(x_0))\cdot(g'(x_0)(x-x_0) + g(x_0)))'=$$&#xA;$$=2(x-x_0)f'(x_0)g'(x_0)+ f'(x_0)g(x_0) + f(x_0)g'(x_0)$$&#xA;so &#xA;$$(fg)'(x_0)=(T_fT_g)'(x_0) =f'(x_0)g(x_0) + f(x_0)g'(x_0).$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-23T14:18:22.180" LastActivityDate="2010-08-23T14:18:22.180" />
  <row Id="3108" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T11:59:55.523" Score="5" ViewCount="0" Body="&lt;p&gt;This is a well-known phenomenon among topologists, and although I'm not an expert, I'll give one standard answer: 3- and 4-dimensional topology are very different from topology in 5 or more dimensions because &lt;a href=&quot;http://en.wikipedia.org/wiki/Surgery_theory&quot; rel=&quot;nofollow&quot;&gt;surgery theory&lt;/a&gt; works in 5 or more dimensions.  In 3 and 4 dimensions one does not have enough &quot;wiggle room&quot; for surgery theory to be effective and this is responsible for some anomalous behavior.  This leads to some remarkable phenomena connecting 3-manifold topology to other branches of mathematics, some of which are listed in the &lt;a href=&quot;http://en.wikipedia.org/wiki/3-manifold&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;, and it makes the 4-dimensional case rather special as well.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-23T11:59:55.523" CommunityOwnedDate="2010-08-23T18:37:37.637" />
  <row Id="3109" PostTypeId="2" ParentId="3031" CreationDate="2010-08-23T12:05:30.773" Score="7" ViewCount="0" Body="&lt;p&gt;Here is a way of avoiding Ado's theorem, at the expense of using the Poincare-Birkhoff-Witt Theorem. The PBW theorem has no finite dimensionality or characteristic hypotheses, so you may like this better. Note, however, that I will realize finite dimensional Lie algebras as endomrophisms of infinite dimensional vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define a concrete Lie algebra to be a vector space $V$, and a vector subspace $\mathfrak{h}$ of $\mathrm{End}(V)$ closed under commutator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Every Lie algebra is isomorphic to a concrete Lie algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $\mathfrak{g}$ be a Lie algebra and $U$ its universal enveloping algebra. The Lie algebra $\mathfrak{g}$ acts on $U$ by left multiplication, so this gives a map $\mathfrak{g} \to U$ taking bracket to commutator. We must prove this map is injective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a basis ${ v_i }$ for $\mathfrak{g}$. Suppose that left multiplication by $\sum a_i v_i$ is $0$. Then $\left( \sum a_i v_i \right) \cdot 1 = \sum a_i v_i$ would be zero in $U$. But, by the PBW theorem, the $v_i$ are linearly independent in $U$, a contradiction. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, this special case of PBW is as hard as the whole theorem. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-08-23T12:05:30.773" CommentCount="1" />
  <row Id="3110" PostTypeId="2" ParentId="3069" CreationDate="2010-08-23T12:24:06.480" Score="0" ViewCount="0" Body="&lt;p&gt;For any set X, the set P(X) of all subsets of X has a bigger cardinality than X itself (for X is finite this is easy, for X infinite you need a clever argument from Cantor, obtaineble in any set theory text). You can do this again, obtaining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;card(X) &amp;lt; card(P(X)) &amp;lt; card(P(P(X))) &amp;lt; ...&lt;/p&gt;&#xA;" OwnerUserId="1436" LastActivityDate="2010-08-23T12:24:06.480" CommentCount="1" />
  <row Id="3111" PostTypeId="1" CreationDate="2010-08-23T12:24:28.010" Score="-1" ViewCount="251" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/2887/question-about-solving-absolute-values&quot;&gt;Question about solving absolute values&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I know the method of solving the equation like this $|2x+1|=|3x+9|$ but the problem is if the same equation would be like this $| 2x+1 | = x | 3x+9 |$, how can I solve this?&lt;/p&gt;&#xA;" OwnerUserId="1050" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T13:10:22.613" LastActivityDate="2010-08-23T14:25:39.810" ClosedDate="2010-08-23T14:34:36.357" Title="Solving equations with multiple absolute values" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="3112" PostTypeId="2" ParentId="3111" CreationDate="2010-08-23T12:37:49.043" Score="2" ViewCount="0" Body="&lt;p&gt;To solve $|f(x)|=g(x)|h(x)|$ in general, find the sets $P=\{x:f(x)g(x)\ge 0\}$ and $N=\{x:f(x)g(x)\lt 0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solve the two equations&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x)=g(x)h(x)$ - but discard solutions that are not in $P$.&lt;/li&gt;&#xA;&lt;li&gt;$f(x)=-g(x)h(x)$ - but discard solutions that are not in $N$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You will be left with the solutions of $|f(x)|=g(x)|h(x)|$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-23T12:37:49.043" />
  <row Id="3113" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T12:39:36.560" Score="1" ViewCount="0" Body="&lt;p&gt;The following is just a toy example and of course not as deep as the examples already mentioned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-dimensional space is special in the sense that $\mathbb R^1\setminus\{0\}$ is not connected---it has two connected components. In all other dimension $n\neq 1$, the space $\mathbb R^n\setminus\{0\}$ is connected.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-23T12:39:36.560" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="4" />
  <row Id="3114" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T12:51:14.180" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_%28normed_division_algebras%29&quot; rel=&quot;nofollow&quot;&gt;Hurwitz's theorem&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only normed division algebras over the real numbers appear in dimensions 1, 2, 4 and 8. &lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-23T12:51:14.180" CommunityOwnedDate="2010-08-23T18:37:37.637" />
  <row Id="3115" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T13:07:57.140" Score="2" ViewCount="0" Body="&lt;p&gt;One example of $n \geq 2$-D richness is simply the fact that $n$-by-$n$ matrices don't commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is trivial, but it means for example that every nonabelian finite group has an irreducible representation of degree $&gt;1$.  &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-23T23:23:27.653" LastActivityDate="2010-08-23T23:23:27.653" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="2" />
  <row Id="3116" PostTypeId="2" ParentId="3072" CreationDate="2010-08-23T13:37:21.200" Score="1" ViewCount="0" Body="&lt;p&gt;The actual solution may not be what you want, since it will come with perks like the birds swinging almost a full $-180^\circ$ to $180^\circ$ everytime when pulled hard and then immediately held still. Also you will have to solve a differential equation in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May I offer an alternative instead which will make it look real while being easy to implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have a simple oscillation, I assume you simply vary the angle as $\alpha(t)=r\cos(2\pi f t)+3\pi/2$ for some range $r$ and frequency $f$ in your code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add a touch of reality, you can vary the 'down' angle based on the horizontal velocity of the mouse. If the velocity is $v$, then calculate $\beta(v)=3\pi/2-\tan^{-1}(Av)$. This will give some pseudo-direction for the adjusted direction of pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also decrease the range of the oscillation as the velocity increases: $r(v)=1/(1+(Bv)^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally use the following function to get the angle: $\alpha(t)=r(v)\cos(2\pi f t)+\beta(v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to choose fractions $A$ and $B$ till you hit upon the right motion that pleases you. $A=B=0$ will give your original motion back. Larger $A$ will make the change in angle more sensitive to speed, and larger $B$ will dampen the oscillations more at a given velocity.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-23T13:37:21.200" />
  <row Id="3117" PostTypeId="2" ParentId="3111" CreationDate="2010-08-23T13:38:13.210" Score="2" ViewCount="0" Body="&lt;p&gt;Both equations you mention can be solved in the same way.  As others have said in the comments on your question here and the comments and answers on &lt;a href=&quot;http://math.stackexchange.com/questions/2887/question-about-solving-absolute-values&quot;&gt;this question&lt;/a&gt;, $|ax-b|=\begin{cases}&#xA;ax-b &amp;amp; \text{ if } x\ge \frac{b}{a} \\ &#xA;b-ax &amp;amp; \text{ if } x&amp;lt; \frac{b}{a} &#xA;\end{cases}$, where I'll call $\frac{b}{a}$ a &quot;turning point&quot; value.  So, for any equation an absolute value, there is at least one turning point value of the variable for each absolute value.  In both of your examples, the turning point values are $-\frac{1}{2}$ and $-3$.  The $n$ (2 in your examples) turning point values divide the number line into $n+1$ (3 in your examples) regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite.  Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a different example: $x|4x-1|=|3x-2|$.  The turning points are $\frac{1}{4}$ and $\frac{2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\leftarrow  &amp;amp; \frac{1}{4} &amp;amp; \text{---} &amp;amp; \frac{2}{3} &amp;amp; \rightarrow&#xA;\\&#xA;x(1-4x)=(2-3x) &amp;amp; | &amp;amp; x(4x-1)=(2-3x) &amp;amp; | &amp;amp; x(4x-1)=(3x-2)&#xA;\\&#xA;x-4x^2=2-3x &amp;amp; | &amp;amp; 4x^2-x=2-3x &amp;amp; | &amp;amp; 4x^2-x=3x-2&#xA;\\&#xA;0=4x^2-4x+2 &amp;amp; | &amp;amp; 4x^2+2x-2=0 &amp;amp; | &amp;amp; 4x^2-4x+2=0&#xA;\\&#xA;\text{no }x\in\mathbb{R} &amp;amp; | &amp;amp; x=-1\text{ or }x=\frac{1}{2} &amp;amp; | &amp;amp; \text{no }x\in\mathbb{R}&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;edit&lt;/em&gt;: Note that between any two adjacent regions, the only change in the equations at the tops of the regions is the one term with turning point at the boundary between the two regions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the solutions for each region-specific equation apply only to that region, so the solutions -1 and $\frac{1}{2}$ in the middle region only apply to the interval $[\frac{1}{4},\frac{2}{3}]$, so -1 is not a solution (it's not in the interval) but $x=\frac{1}{2}$ is a solution (and it's the only solution to $x|4x-1|=|3x-2|$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the algebraic method of solution, graphing both sides of the equation and looking for points of intersection is a good way to check your solution.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T14:25:39.810" LastActivityDate="2010-08-23T14:25:39.810" />
  <row Id="3118" PostTypeId="1" AcceptedAnswerId="3121" CreationDate="2010-08-23T13:44:55.477" Score="5" ViewCount="526" Body="&lt;p&gt;Can anyone help me out here? Can't seem to find the right rules of divisibility to show this.&lt;/p&gt;&#xA;" OwnerUserId="1397" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-23T14:00:09.563" LastActivityDate="2010-12-03T07:14:13.777" Title="If a | m and (a + 1) | m, prove a(a + 1) | m." Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="3119" PostTypeId="1" AcceptedAnswerId="3302" CreationDate="2010-08-23T13:49:57.680" Score="5" ViewCount="190" Body="&lt;p&gt;In a recent discussion, someone told me tuples in the modern meaning (in particular,  tuples are heterogeneous: that is, different elements of a tuple can belong to different sets/have different types) first appeared in Codd's &lt;a href=&quot;http://en.wikipedia.org/wiki/Tuple_relational_calculus&quot; rel=&quot;nofollow&quot;&gt;tuple calculus&lt;/a&gt;. I was surprised it would be so late, but &lt;a href=&quot;http://www.google.ru/search?tbs=bks:1,bkv%3aa,cdr:1,cd_min:03.01.0001,cd_max:31.12.1969&amp;amp;tbo=p&amp;amp;q=tuple&amp;amp;num=10&quot; rel=&quot;nofollow&quot;&gt;searching Google Books&lt;/a&gt; before 1970, I can't see any clearly heterogeneous examples, and quite a few clearly homogeneous ones (&quot;tuple of ones and zeros&quot;, &quot;tuple of natural numbers&quot;, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Сan anybody confirm that Codd introduced heterogeneous tuples or point out an earlier appearance?&lt;/p&gt;&#xA;" OwnerUserId="1437" LastActivityDate="2010-08-25T19:10:08.443" Title="When did the term &quot;tuple&quot; get its current meaning?" Tags="&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3120" PostTypeId="1" AcceptedAnswerId="3127" CreationDate="2010-08-23T13:51:32.327" Score="3" ViewCount="333" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = (0,0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B = (0,-10)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AB = AC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the angle between AB and AC, &lt;strong&gt;how are the coordinates at C calculated?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1439" LastEditorUserId="1439" LastEditorDisplayName="" LastEditDate="2010-08-23T14:36:08.180" LastActivityDate="2010-08-23T23:02:04.647" Title="Find the coordinates in an isosceles triangle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="3121" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T13:54:44.253" Score="8" ViewCount="0" Body="&lt;p&gt;It is not surprising that you are finding this difficult, because it goes beyond basic divisibility rules -- it rather requires something which is essentially equivalent to the uniqueness of prime factorization.  [&lt;b&gt;Edit&lt;/b&gt;: Actually this is comment is incorrect -- as Robin Chapman's answer shows, it &lt;em&gt;is&lt;/em&gt; possible to prove this using just divisibility rules.  In particular it is true in any integral domain.]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume $a$ and $m$ are positive integers.  The first observation is that $a$ and $a+1$ are relatively prime: i.e., there is no integer $d &gt; 1$ -- or equivalently, no prime number-- which divides both $a$ and $a+1$, for then $d$ would have to divide $(a+1) - a = 1$, so $d = 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the key step: since $a$ divides $m$, we may write $m = aM$ for some positive integer $M$.  So $a+1$ divides $aM$ and is relatively prime to $a$.  &lt;strong&gt;I claim&lt;/strong&gt; that this implies $a+1$ divides $M$.  Assuming this, we have $M = (a+1)N$, say, so altogether &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = aM = a(a+1)N$, so $a(a+1)$ divides $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is a special case of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Generalized) Euclid's Lemma: Let $a,b,c$ be positive integers.  Suppose $a$ divides $bc$ and $a$ is relatively prime to $b$.  Then $a$ divides $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A formal proof of this requires some work!  See for instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euclid's_lemma&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, proving this is essentialy as hard as proving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-23T18:06:24.263" LastActivityDate="2010-08-23T18:06:24.263" CommentCount="7" />
  <row Id="3122" PostTypeId="1" AcceptedAnswerId="3124" CreationDate="2010-08-23T14:06:32.077" Score="2" ViewCount="256" Body="&lt;p&gt;Some days ago as i had asked as to how to test the Riemann Integrability of the function. Now i was recently given this question about proving that the given function is Riemann Integrable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BqmUO.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can i show that this function is Riemann Integrable or not in the interval $[0,1]$. I tried using partitions, but it didn't work. I here don't want to use the Riemann Lebesgue lemma as i want to understand the methodology behind selecting the partitions.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-24T00:24:52.060" LastActivityDate="2010-09-12T03:10:37.317" Title="Showing $\frac{1}{x}-\left[\frac{1}{x}\right]$ is Riemann Integrable" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="3123" PostTypeId="2" ParentId="3122" CreationDate="2010-08-23T14:19:35.577" Score="0" ViewCount="0" Body="&lt;p&gt;To show that the function in Riemann-integrable you have to show that it is bounded and has at most a countable set of discontinuities in the interval you are integrating. But this is the case of an improper integral, since $1/x$ is not defined in $x = 0$, you have to follow the criterions for improper integrals, which is the existence of the limit $\lim_{x\rightarrow 0} \int_x^1 \frac{1}{x} dx$ and its finiteness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;Assuming that by $[\frac{1}{x}]$ you mean the integer part of x (the closest integer less then or equal to x), we can always assume that $[\frac{1}{x}]$ is zero, because if x less than or greater than zero, you always have a number less than one, the integer part of which is always zero. So your integral is the same as $\int_0^1 \frac{1}{x} dx - \int_0^1 [\frac{1}{x}] dx = \int_0^1 \frac{1}{x} dx$, which does not converge on that interval.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: promt comments showed that I'm way over my head...Sorry for the wrong answer.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-23T14:29:54.090" LastActivityDate="2010-08-23T14:29:54.090" CommentCount="2" />
  <row Id="3124" PostTypeId="2" ParentId="3122" CreationDate="2010-08-23T14:28:05.997" Score="5" ViewCount="0" Body="&lt;p&gt;The function is Riemann-integrable because it is bounded (it takes values in $[0,1]$) and has countably many discontinuities, namely, the points of the form $\frac{1}{n}$ and $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_integral#Properties&quot; rel=&quot;nofollow&quot;&gt;Lebesgue's criterion for Riemann integrability&lt;/a&gt; which you probably meant with &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma&quot; rel=&quot;nofollow&quot;&gt;Riemann-Lebesgue-Lemma&lt;/a&gt; and hence unfortunately didn't want to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for doing it by hand with partitions, try Akhil Mathew's hint above.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-23T19:42:43.340" LastActivityDate="2010-08-23T19:42:43.340" />
  <row Id="3125" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T14:28:49.740" Score="4" ViewCount="0" Body="&lt;p&gt;An alternate route: We show that, if $ax+by=1$ and $m$ is divisible by $a$ and $b$ then $m$ is divisible by $ab$. (Then apply this with $b=a+1$, $x=-1$ and $y=1$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $m=ak=bl$. Then $ab(xl+ky)=(ax+by)m=m$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that the hypothesis $\exists_{x,y}: ax+by=1$ is often easier to use than $GCD(a,b)=1$. The equivalence between these two is basically equivalent to unique factorization, and you can often dodge unique factorization by figuring out which of these two you really need.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-08-23T14:28:49.740" CommentCount="9" />
  <row Id="3126" PostTypeId="2" ParentId="3072" CreationDate="2010-08-23T14:32:15.317" Score="1" ViewCount="0" Body="&lt;p&gt;In case you really need to simulate accurate physics, you can use the solution given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l8GKb.png&quot; alt=&quot;alt diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture above, vector $A = g-a$ where $g$ is acceleration due to gravity, and $a$ is the acceleration of mouse which you need to compute in real time. You will have to translate $g=9.8m/s^2$ into $pixel/s^2$ and store as a constant. For $A$, you need to calculate $A_x$ and $A_y$ based on the last two frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find $\alpha(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\beta=tan^{-1}\frac{A_x}{A_y}$ and $A_1=|A|\cos(\alpha-\beta)$. Angular acceleration will be $-A_1/h$ ($h$ will vary from bird to bird).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to solve the differential equation $\frac{d^2\alpha}{dt^2}=A_1/h$ in real time to make the bird swing realistically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious way to solve the differential equation numerically is given below -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Store angular velocity $\omega=\frac{d}{dt}\alpha$. At each frame of the animation, assuming $\Delta t$ time has passed since last frame, do&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\omega\leftarrow\omega+\frac{A_1}{h}\Delta t$&lt;/li&gt;&#xA;&lt;li&gt;$\alpha\leftarrow\alpha+\omega\Delta t$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1310" LastEditorDisplayName="" LastEditDate="2010-08-23T14:49:51.430" LastActivityDate="2010-08-23T14:49:51.430" />
  <row Id="3127" PostTypeId="2" ParentId="3120" CreationDate="2010-08-23T14:36:52.330" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/strong&gt; (to match revised question)&lt;/em&gt;: Given your revised question, there is still the issue of C being on either side of the y-axis, but you have specified that AB=AC and that you are given $\mathrm{m}\angle BAC$ (the angle between AB and AC), so as in my original answer (below), the directed (trigonometric) measure of the angle from the positive x-axis to AC is $\mathrm{m}\angle BAC-90^{\circ}$ and AC=AB=10, so C has coordinates $(10\cos(\mathrm{m}\angle BAC-90^{\circ}),10\sin(\mathrm{m}\angle BAC-90^{\circ}))$.  (This matches up to one of the answers in Moron's solution; the other corresponds to the other side of the y-axis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;original answer&lt;/strong&gt; (when it was not specified that AB=AC and when the given angle was C)&lt;/em&gt;:  As suggested in the comments, there are several cases.  First, C could be on either side of the y-axis; let's assume that C has positive x-coordinate (leaving the case where it has negative x-coordinate for you to solve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, ABC could be isosceles with AB=AC, AB=BC, or AC=BC.  In the first case, $\angle B\cong \angle C$ (which cannot happen unless C is acute) and $\mathrm{m}\angle BAC=180^{\circ}-2\mathrm{m}\angle C$, so the directed (trigonometric) measure of the angle from the positive x-axis to AC is $90^{\circ}-2\mathrm{m}\angle C$ and AC=AB=10, so C has coordinates $(10\cos(90^{\circ}-2\mathrm{m}\angle C),10\sin(90^{\circ}-2\mathrm{m}\angle C))$.  The second case is similar to the first (so it's left for you to solve).  In the third case, C is equidistant from A and B, so C must lie on the perpendicular bisector of AB (as in J. Mangaldan's comment), and by symmetry this perpendicular bisector of AB also bisects $\angle ACB$; from there, you can use right triangle trigonometry to determine the coordinates of C (left for you to solve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cases where AB=AC (blue), AB=BC (red), and AC=BC (green) (lighter versions on the left side of the y-axis) are shown below for measures of angle C between 0 and 180&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JFKWo.gif&quot; alt=&quot;animated diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T16:55:07.517" LastActivityDate="2010-08-23T16:55:07.517" CommentCount="10" />
  <row Id="3128" PostTypeId="2" ParentId="3120" CreationDate="2010-08-23T14:58:45.250" Score="3" ViewCount="0" Body="&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;Polar Co-ordinates&lt;/a&gt;. A point with polar co-ordinates $(r,\theta)$ is the same point in $x,y$ co-ordinates (or as also called, rectangular co-ordinates) as $(r\cos \theta, r\sin \theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, point $C$ lies at a distance $10$ from $A$ which is the origin, so $r = 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If given angle CAB is $\alpha$, the the polar angle is either $\frac{3pi}{2}-\alpha$  or $\frac{3pi}{2}+\alpha$, i.e in polar co-ordinates $C$ is either $(10,\frac{3pi}{2}-\alpha)$ or $(10,\frac{3pi}{2}+\alpha)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It might help to draw a figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now convert back to $x,y$ co-ordinates.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-23T15:04:42.740" LastActivityDate="2010-08-23T15:04:42.740" CommentCount="7" />
  <row Id="3129" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T15:07:44.840" Score="11" ViewCount="0" Body="&lt;p&gt;The infinite-dimensional sphere is contractible.&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2010-08-23T15:07:44.840" CommunityOwnedDate="2010-08-23T15:07:44.840" CommentCount="2" />
  <row Id="3130" PostTypeId="2" ParentId="2949" CreationDate="2010-08-23T15:11:12.760" Score="1" ViewCount="0" Body="&lt;p&gt;There exists $f\colon \mathbb{N}\times\mathbb{N}\to\mathbb{N}$ which is bijective.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-08-23T15:11:12.760" CommunityOwnedDate="2010-08-23T15:11:12.760" CommentCount="1" />
  <row Id="3131" PostTypeId="2" ParentId="3120" CreationDate="2010-08-23T15:37:13.820" Score="1" ViewCount="0" Body="&lt;p&gt;Let $a,b$ and $c$ be the side lengths and $A,B$ and $C$ the angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{2}=x^{2}+\left( y+10\right) ^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^{2}=x^{2}+y^{2}=10^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b=c=10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the (Neper) theorem of tangents (corollary of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_tangents&quot; rel=&quot;nofollow&quot;&gt;Law of tangents&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C&amp;lt;\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by the theorem of sinus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c\sin A=a\sin C\iff \left( x^{2}+\left( y+10\right) ^{2}\right) \sin&#xA;C=10\sin A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compiling, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A-B}{2}=\arctan (\frac{\sqrt{x^{2}+\left( y+10\right) ^{2}}-10}{\sqrt{%&#xA;x^{2}+\left( y+10\right) ^{2}}+10}\cot \frac{C}{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C&amp;lt;\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x^{2}+(\sqrt{100-x^{2}}+10)^{2})\sin C=10\sin A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{2}=10^{2}-x^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to solve this system of four equations and four unknowns $x,y,A,B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I started this approach before the question has been updated.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-23T23:02:04.647" LastActivityDate="2010-08-23T23:02:04.647" CommentCount="7" />
  <row Id="3132" PostTypeId="2" ParentId="1706" CreationDate="2010-08-23T15:37:14.147" Score="3" ViewCount="0" Body="&lt;p&gt;I write, e.g., $\mathbb R_{&gt;0}$, $\mathbb R_{\geq0}$, $\mathbb N_{&gt;0}$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-08-23T15:37:14.147" />
  <row Id="3133" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T15:40:33.723" Score="6" ViewCount="0" Body="&lt;p&gt;There is a very specific reason why one needs 3 dimensions or more for the Banach Tarski paradox.  In dimension 3 or higher one can make rotations in independent directions,&#xA;and so the group $SO(3)$ of rotations of space contains a copy of $F_2$, the free group on&#xA;two generators.  This fact is what underlies the Banach Tarski paradox.  (The group&#xA; $F_2$ is &lt;em&gt;not amenable&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-23T15:40:33.723" CommunityOwnedDate="2010-08-23T18:37:37.637" />
  <row Id="3134" PostTypeId="2" ParentId="1706" CreationDate="2010-08-23T15:53:58.173" Score="1" ViewCount="0" Body="&lt;p&gt;I met  (in IBDP programme, UK and Poland) the followin notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\mathbb{R}^{+} = \{ x | x \in \mathbb{R} \land x &gt; 0 \} \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\mathbb{R}^{+} \cup {0} = \{ x | x \in \mathbb{R} \land x \geq 0 \} \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the explanation that $\mathbb{R}^{+}$ denotes set of positive reals and $0$ is neither positive nor negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{N}$ is possibly a sligthly different case and it usually differ from branch of mathematics to branch of mathematics. I belive that is &lt;em&gt;usually&lt;/em&gt; includes $0$ but I belive theory of numbers is easier without it. It can be easilly extended in such was to have $\mathbb{N}^+ = \mathbb{Z}^+$ denoting positive integers/naturals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course as noted before it is mainly a question of notation.  &lt;/p&gt;&#xA;" OwnerUserId="728" LastActivityDate="2010-08-23T15:53:58.173" />
  <row Id="3135" PostTypeId="1" AcceptedAnswerId="3151" CreationDate="2010-08-23T16:11:10.090" Score="3" ViewCount="138" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At 5am, a pedestrian starts walking from A to B, a distance of 30 km. At 7 am a bicyclist whose speed is 2 times the pedestrian's speed, started riding from A to B, too. After some time, the bicyclist met and passed the pedestrian. Two hours after the meeting &lt;strong&gt;the pedestrian&lt;/strong&gt; reached his destination (B). (The bicyclist reached B before the pedestrian reached it.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the speed of the pedestrian?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, first I found the distance that the pedestrian went until the bike rider reached him like that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Marked x = the pedestrian's speed.&lt;/li&gt;&#xA;&lt;li&gt;Marked y = the distance that the pedestrian went until the bike rider reached him.&lt;/li&gt;&#xA;&lt;li&gt;Time until the bike rider reached the pedestrian = y/x.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;x(y/x + 2) = 30&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;y = 30 - 2x&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the distance that the pedestrian went until the bike rider reached him was &lt;code&gt;30 - 2x&lt;/code&gt;. But how I continue from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-23T17:07:22.643" LastActivityDate="2010-08-23T19:09:57.760" Title="Traffic Homework Help" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="3136" PostTypeId="2" ParentId="3093" CreationDate="2010-08-23T16:16:02.517" Score="-2" ViewCount="0" Body="&lt;p&gt;I'm going to be a bit contrary: is this really true? Or is it just that we started studying the 3D case before we could study anything higher-dimensional, and that our brains are obviously optimized for thinking in three dimensions?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-23T16:16:02.517" CommunityOwnedDate="2010-08-23T18:37:37.637" CommentCount="1" />
  <row Id="3138" PostTypeId="1" CreationDate="2010-08-23T16:19:31.030" Score="2" ViewCount="435" Body="&lt;p&gt;I am interested in continuous functions which are not smooth, i.e. which have only finitely many derivatives (for now just taking the case f:R-&gt;R).  As far as I can tell, there are only 2 features or properties of such a function which can make it non-smooth.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The derivative of some order diverges to infinity at one or more points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The derivative of some order oscillates like sin(1/x) does at x = 0 at one or more points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Q1: Are there any other ways for a continuous function to be non-smooth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: What are some example's of non-smooth functions which satisfy a differential equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to Qiaochu comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple &quot;description of a limit failing to exist which is better than &quot;the limit fails to exist.&quot; is to specify, for example, if the limit of a sequence of natural numbers diverges to infinity or not. for example:  1,2,3,... and 1,2,1,2,1,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two examples are similar to my 1. and 2., i.e. the derivative diverges to infinity or it oscillates around without settling down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see the problem with trying to classify the ways continuous functions cannot be smooth by looking at their graphs.  One example of a non smooth function is sin(1/x), it has a certain property at x=0, another non-smooth function is |x| and it has another property at x=0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand you feel the question is not well defined because &quot;being like sin(1/x) at x=0&quot; is an intuitive concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so what are examples of non-smooth continuous functions without &quot;kinks&quot; like |x| has at 0?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two that I can think of, sin(1/x) and the Weierstrass function.  But if you think about the self similarity of the graph of the Weirstrauss function, then it's clear that the derivative of this function osscilates rapidly at every point like sin(1/x) does at x=0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of whether or not it's well-defined to say the Weierstrass function is like sin(1/x) in some sense, I am just curious to know of some other examples of non-smooth continuous functions without kinks in their graph, can someone please help?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-24T19:08:24.283" LastActivityDate="2010-09-27T22:53:45.540" Title="Ways for a continuous function to not be smooth" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="3139" PostTypeId="2" ParentId="3138" CreationDate="2010-08-23T16:28:23.623" Score="1" ViewCount="0" Body="&lt;p&gt;For Q1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depends on your definition of oscillate I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = |x|$ does not have a derivative at $0$ and would not fit your criteria for 1 or 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, your question seems to be equivalent to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In what ways can the limit of a function fail to exist at a point&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the comment of derivative of |x| being a step function and so the second derivative of |x| at 0 is infinity is nonsense, IMO. The first derivative itself does not exist, so talking about the second makes no sense. Perhaps you need to clarify exactly what you mean.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-23T16:51:29.970" LastActivityDate="2010-08-23T16:51:29.970" CommentCount="5" />
  <row Id="3140" PostTypeId="2" ParentId="3138" CreationDate="2010-08-23T16:38:54.610" Score="0" ViewCount="0" Body="&lt;p&gt;Re Q2: For smooth ODEs (i.e. something like y′=f(y) for f smooth, possibly with y taking values in Rn but depending on one parameter), existence of a solution implies regularity (i.e. smoothness) of it. This isn't particularly easy to prove, though; I know a proof is in Lang's &lt;em&gt;Differential and Riemannian Manifolds,&lt;/em&gt; and if I recall correctly the idea is to prove inductively that $y'$ is $C^1$ (or at least Lipschitz) by some sort of Gronwall inequality argument, and then get that $y'$ satisfies its own differential equation, and repeat inductively. The details are very hazy in my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of the general theory of ordinary differential equations, though, so there are probably plenty of sources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For PDEs, automatic regularity is not necessarily the case, though it is for elliptic PDEs with smooth coefficients. More generally, regularity holds for certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypoelliptic_operator&quot; rel=&quot;nofollow&quot;&gt;hypoelliptic&lt;/a&gt; operators (e.g. the heat operator). &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-23T16:38:54.610" />
  <row Id="3141" PostTypeId="2" ParentId="3138" CreationDate="2010-08-23T16:40:43.780" Score="0" ViewCount="0" Body="&lt;p&gt;To answer your second question, I offer two classical examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: The cycloid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(y^{\prime}\right)^2=\frac{2a-y}{y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: The tractrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{\prime}=-\frac{\sqrt{a^2-x^2}}{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($a$ is a parameter in both examples)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-23T16:40:43.780" />
  <row Id="3143" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T17:14:25.550" Score="17" ViewCount="0" Body="&lt;p&gt;The other answers put this in a general context, but in this example one&#xA;can be absolutely explicit. If $a\mid m$ and $(a+1)\mid m$ then there are&#xA;integers $r$ and $s$ such that&#xA;$$m=ar=(a+1)s.$$&#xA;Then&#xA;$$a(a+1)(r-s)=(a+1)[ar]-a[(a+1)s]=(a+1)m-am=m.$$&#xA;As $r-s$ is an integer, then $a(a+1)\mid m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="205" LastEditorDisplayName="" LastEditDate="2010-08-23T18:00:19.000" LastActivityDate="2010-08-23T18:00:19.000" CommentCount="1" />
  <row Id="3144" PostTypeId="2" ParentId="3135" CreationDate="2010-08-23T17:18:58.087" Score="1" ViewCount="0" Body="&lt;p&gt;In problems like this (where there are multiple moving objects that start at different times), I usually start by assigning a variable to the quantity that the question asks for: $$\text{let }x=\text{the speed of the pedestrian}$$&#xA;then assigning a variable to elapsed time from a specific instant: $$\text{let }t=\text{time (in hours) since 5am}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these variables, I then try to write expressions for the positions of each moving object in terms of $x$ and $t$, and set the position expressions equal corresponding to when the objects are in the same place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not enough to get you unstuck, please comment on this answer and let me know what you've done so far using these ideas.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T17:18:58.087" />
  <row Id="3145" PostTypeId="1" AcceptedAnswerId="3146" CreationDate="2010-08-23T17:51:01.950" Score="3" ViewCount="156" Body="&lt;p&gt;If I want to show a topological subspace is closed in an ambient space, does it suffice to know what happens on an open cover of the ambient space? More specifically,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ is a topological space with a given open cover ${ U_i }$. Suppose that $Z \subset X$ is a set such that $Z \cap U_i$ is closed in $U_i$ for all $i$. Does it follow that $Z$ is closed in X?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is clearly true if there are finitely many ${ U_i }$. At first thought, it seems unlikely to be true in the infinite case, but I'm having trouble coming up with a suitable counter-example.&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-09-06T11:02:47.323" Title="Is whether a set is closed or not a local property?" Tags="&lt;topology&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="3146" PostTypeId="2" ParentId="3145" CreationDate="2010-08-23T18:03:03.350" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, it's true: being closed is &lt;del&gt;a local property&lt;/del&gt; local on the base. Indeed, suppose $x \notin Z$. Then $x$ is in a neighborhood $U_i$ which is one of the sets in the open cover. Now $x \in U_i - Z \cap U_i$, which is an open set, so there is a neighborhood of $x$, contained in $U_i$, which does not intersect $Z \cap U_i$. This means that this neighborhood of $x$ does not intersect $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not true for closed covers (i.e. with the $U_i$ closed) in general, but it is true when they form a locally finite cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, if each point $x \in Z$ has a neighborhood $U$ such that $Z \cap U$ is closed in $U$, then $Z$ is called locally closed: this means it is the intersection of a closed subset and an open subset, but is not itself necessarily closed. &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-23T19:47:31.740" LastActivityDate="2010-08-23T19:47:31.740" CommentCount="5" />
  <row Id="3149" PostTypeId="2" ParentId="3118" CreationDate="2010-08-23T18:51:37.610" Score="8" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt; $\;\rm\quad\displaystyle \frac{m}{a},\; \frac{m}{a+1}\in\mathbb{Z} \;\;\Rightarrow\;\; \frac{m}{a} - \frac{m}{a+1} \; = \;\frac{m}{a(a+1)} \in \mathbb Z\quad\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ More generally, for $\:\rm n = bc \:-\; ad \;$ a linear combination of $\rm a, b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have $\rm\quad\quad\quad\displaystyle \frac{m}{a},\; \frac{m}{b}\in\mathbb{Z} \;\;\Rightarrow\;\; \frac{m}{a}\frac{bc}{b} - \frac{ad}{a}\frac{m}{b} = \frac{mn}{ab} \in \mathbb Z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Bezout's identity, $\rm\ n = gcd(a,b)\ $  is the least positive linear combination, so the above yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ &#xD;&#xA; a,b|m \;\Rightarrow\; ab\:|\:m\;gcd(a,b) \;\Rightarrow\; \mathfrak{m}_{a,b} |\:m\;\;\;$  for $\;\;\rm \mathfrak{m}_{a,b} := \frac{ab}{gcd(a,b)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $\:$ every common multiple $\rm m$ of $\rm a,b$ is a multiple of $\;\rm \mathfrak{m}_{a,b}. \;$ But $\rm\:\mathfrak{m}_{a,b}\:$ is also a common multiple, i.e. $\rm\ a,b|\mathfrak{m}_{a,b} \;$ viz. $\rm \frac{\mathfrak{m}_{a,b}}{a} = \;\frac{a}{a}\frac{b}{gcd(a,b)}\in\mathbb Z \;\Rightarrow\; a|\mathfrak{m}_{a,b} \;$. By symmetry, also $\rm\ b|\mathfrak{m}_{a,b} \;.\ $ Therefore $\rm \mathfrak{m}_{a,b} = lcm(a,b)$ since it's a common multiple of $\rm a,b$ that's divisibility-least, i.e. that divides every common multiple. This is the general definition of LCM in an arbitrary domain (ring without zero-divisors), namely we make the following &lt;strong&gt;universal&lt;/strong&gt; dual definitions of LCM and GCD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;LCM&lt;/strong&gt; $\quad$ If $\quad\rm a,b|c \;\iff\; [a,b]\:|\:c \quad\;$ then $\quad\rm [a,b] \;$ is an &lt;strong&gt;LCM&lt;/strong&gt; of $\;\rm a,b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;GCD&lt;/strong&gt; $\quad$ If $\quad\rm c|a,b \;\iff\; c\:|\:(a,b) \quad$ then $\quad\rm (a,b) \;$ is an &lt;strong&gt;GCD&lt;/strong&gt; of $\;\;\rm a,b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: that $\;\rm a,b\:|\:[a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. Dually $\;\rm (a,b)|\:a,b \;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such $\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\rm\;\; (a,b) = ab/[a,b] \;\;$  if $\;\rm\ [a,b] \;$  exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:  $\rm\quad\quad d|\:a,b \;\iff\; a,b\:|\:ab/d \;\iff\; [a,b]\:|\:ab/d \;\iff\; d\:|\:ab/[a,b] \quad\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conciseness of this proof arises by exploiting to the hilt the $\iff$ definition of LCM and GCD.  Compare this slick proof to many obfuscated proofs found in elementary number theory textbooks!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the theorem, GCDs exist if LCMs exist. But common multiples clearly comprise an ideal, being closed under subtraction and multiplication by any ring element. Hence in a PID the generator of an ideal of common multiples is clearly an LCM. In Euclidean domains this can be proved directly by a simple descent, e.g. in $\:\mathbb Z \;$ we have the following high-school level proof of the existence of LCMs (and, hence, of GCDs), after noting the set $\rm M$ of common multiples of $\rm a,b$ is closed under subtraction and contains $\:\rm ab \ne 0\:$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ Suppose $\;\rm M\subset\mathbb Z \;$ is closed under subtraction and that $\rm M$ contains a nonzero element.&lt;br&gt;&#xA;Then $\rm M \:$ has a positive element and the least such positive element of $\;\rm M$ divides every element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ 0 \ne m\in M \Rightarrow m-m = 0\in M\Rightarrow 0-m = -m\in M, \;$ so $\rm M$ has a positive element.&lt;br&gt;&#xA;Let $\rm m$ be the least positive element in $\rm M$. Since $\rm m|n \iff m\;|-n, \;$ if some $\rm n\in M$ isn't divisible by $\rm m$ then we &#xA;may assume this $\rm n$ is positive, and the least such element. But $\rm\: n-m \:$ is a positive element of $\rm M$ not divisible by $\rm m$ &#xA;and smaller than $\rm n$, contra leastness of $\rm n\:$. Hence $\rm m$ divides every element of $\rm M.\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T07:14:13.777" LastActivityDate="2010-12-03T07:14:13.777" />
  <row Id="3151" PostTypeId="2" ParentId="3135" CreationDate="2010-08-23T19:09:57.760" Score="1" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At 5am, a pedestrian starts walking from A to B, a distance of 30 km.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let speed of pedestrian be $v$. At time $t$, pedestrian is at distance $vt$. (Where $t$ denotes time since pedestrian started.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At 7 am a bicyclist whose speed is 2 times the pedestrian's speed, started riding from A to B, too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Speed of cyclist is $2v$. At time $t$, cyclist is at distance $(2v)(t-2)$. (Why?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After some time, the bicyclist met and passed the pedestrian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let they meet at time $t_1$. Then $vt_1=2v(t_1-2)$. (Why? Notice that $v$ cancels out, hence solve for $t_1$ from here.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two hours after the meeting the pedestrian reached his destination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pedestrian covers 30km in $t_1+2$ hours, with $t_1$ known. Hence deduce his speed.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-23T19:09:57.760" CommentCount="3" />
  <row Id="3152" PostTypeId="2" ParentId="2550" CreationDate="2010-08-23T20:00:21.917" Score="0" ViewCount="0" Body="&lt;p&gt;Just so I don't forget, here is the solution I was thinking of, for part b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x,y)$ be a point which is at a rational distance from $(0,0), (1,0)$ and $(\sqrt{2},0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $x^2 + y^2$, $(x-1)^2 + y^2$ and $(x-\sqrt{2})^2 + y^2$ are all squares of rational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference of the first and second implies $x$ is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference of the second and third implies that $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $y$ is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that for some integers $a,b,p,q$, $a \neq 0$, $b \neq 0$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $a^2 + b^2 = p^2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $a^2 + 2b^2 = q^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square (1) and multiply (2) by $a^2$ and subtract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^4 = p^4 - (aq)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that the equation $x^4 + y^2 = z^4$ has no non-trivial solutions (and in fact can be used to show that Fermat's Last theorem is true for $n=4$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for instance: &lt;a href=&quot;http://planetmath.org/encyclopedia/X4Y4z2HasNoSolutionsInPositiveIntegers.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/X4Y4z2HasNoSolutionsInPositiveIntegers.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-23T20:00:21.917" />
  <row Id="3153" PostTypeId="1" AcceptedAnswerId="3154" CreationDate="2010-08-23T22:00:32.883" Score="5" ViewCount="165" Body="&lt;p&gt;An absolute value expression such as $|ax-b|$ can be rewritten in two cases as $|ax-b|=\begin{cases}&#xA;ax-b &amp;amp; \text{ if } x\ge \frac{b}{a} \\ &#xA;b-ax &amp;amp; \text{ if } x&amp;lt; \frac{b}{a} &#xA;\end{cases}$, so an equation with $n$ separate absolute value expressions can be split up into $2^n$ cases, but is there a better way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $|2x-5|+|x-1|+|4x+3|=13$, is there a better way to handle all the possible combinations of $x\ge\frac{5}{2}$ versus $x&amp;lt;\frac{5}{2}$, $x\ge 1$ versus $x&amp;lt; 1$, and $x\ge-\frac{3}{4}$ versus $x&amp;lt;-\frac{3}{4}$?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-24T01:26:00.093" Title="What is the best way to solve an equation involving multiple absolute values?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3154" PostTypeId="2" ParentId="3153" CreationDate="2010-08-23T22:04:09.137" Score="10" ViewCount="0" Body="&lt;p&gt;For an equation with $n$ absolute values, the $n$ places where each absolute value splits into 2 cases divide the number line into $n+1$ regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite. Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the given example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\small{\begin{matrix}&#xA;\leftarrow  &amp;amp; -\frac{3}{4} &amp;amp; \text{---}   &amp;amp; 1 &amp;amp; \text{---} &amp;amp; \frac{5}{2} &amp;amp; \rightarrow&#xA;\\&#xA;\begin{matrix}-(2x-5)-(x-1)\\-(4x+3)=13\end{matrix}&#xA;&amp;amp; \begin{matrix}|\\|\end{matrix} &amp;amp;&#xA;\begin{matrix}-(2x-5)-(x-1)\\+(4x+3)=13\end{matrix} &#xA;&amp;amp; \begin{matrix}|\\|\end{matrix} &amp;amp;&#xA;\begin{matrix}-(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} &#xA;&amp;amp; \begin{matrix}|\\|\end{matrix} &amp;amp;&#xA;\begin{matrix}(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} &#xA;\\&#xA;-7x+3=13&#xA;&amp;amp; | &amp;amp;&#xA;x+9=13&#xA;&amp;amp; | &amp;amp;&#xA;3x+7=13&#xA;&amp;amp; | &amp;amp;&#xA;7x-3=13&#xA;\\&#xA;x=-\frac{10}{7}&#xA;&amp;amp; | &amp;amp;&#xA;x=4\notin[-\frac{3}{4},1]&#xA;&amp;amp; | &amp;amp;&#xA;x=2&#xA;&amp;amp; | &amp;amp;&#xA;x=\frac{16}{7}\notin[\frac{5}{2},\infty)&#xA;\end{matrix}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the solutions are $x=-\frac{10}{7}$ and $x=2$ (the values that were solutions to an equation for a particular region and were within that region).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-23T22:04:09.137" CommentCount="3" />
  <row Id="3155" PostTypeId="1" AcceptedAnswerId="3157" CreationDate="2010-08-23T22:42:04.720" Score="10" ViewCount="434" Body="&lt;p&gt;This was part of a &lt;a href=&quot;http://books.google.com/books?id=9LtfZr1snG0C&amp;amp;lpg=PP1&amp;amp;pg=PA110#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;homework problem&lt;/a&gt; from J.B. Conway's complex analysis text which I was assigned long ago but didn't get.  A few years later I was a TA for a course where the problem was assigned.  I still didn't know how to solve it, nor did any of my students.  If you can give me a simple answer, my gratefulness will far outweigh my embarrassment.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Define $f(z)=z\sin(\frac{1}{z})$ on the punctured plane.  What is the image under $f$ of a (small) punctured disk at the origin?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remarks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The corresponding problem for $\sin(\frac{1}{z})$ is straightforward to solve using the definition of $\sin$ in terms of exponentials, the fact that $\exp$ is onto the punctured plane, and periodicity.  The answer in this case is the whole plane.&lt;/li&gt;&#xA;&lt;li&gt;The corresponding problem for $z\cos(\frac{1}{z})$ is straightforward to solve using Picard's theorem, because the oddness of the function implies that 0 is the only possible excluded point, and it is easy to see that 0 is in the image by taking reciprocals of large zeros of $\cos$.  Thus the answer in this case is also the whole plane. (The same argument could be applied to the previous remark, but there no big theorems are needed.  Perhaps Picard's theorem isn't needed here either.)&lt;/li&gt;&#xA;&lt;li&gt;Another part of the problem I couldn't solve generalizes this, taking $z^n\sin(\frac{1}{z})$ when $n$ is a positive integer.  When $n$ is even, there is no problem, because the argument from the previous remark applies.  But I don't know how to solve it when $n$ is odd.  &lt;/li&gt;&#xA;&lt;li&gt;By Picard's theorem, there is at most one point missing. I would be surprised if there is a point missing, but I have no argument to back up this expectation.&lt;/li&gt;&#xA;&lt;li&gt;Although an elementary argument would be nice, I do not care whether the argument is &quot;appropriate&quot; for what is covered up to that point in the text.  This is pure curiosity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2011-01-11T09:37:14.837" Title="What is the image near the essential singularity of z sin(1/z)?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3156" PostTypeId="2" ParentId="3008" CreationDate="2010-08-23T23:00:25.553" Score="2" ViewCount="0" Body="&lt;p&gt;If there is a definable wellordering of $\mathbb{R}$, then every real number is &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_definable_set&quot; rel=&quot;nofollow&quot;&gt;ordinal definable&lt;/a&gt;. Conversely, if every real number is ordinal definable, then there is a definable wellordering of $\mathbb{R}$. Thus the assumption that there is a definable wellordering of $\mathbb{R}$ is equivalent to $\mathbb{R} \subseteq \mathrm{OD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumption is completely independent of AC in the sense that all four combinations AC + $\mathbb{R} \subseteq \mathrm{OD}$, AC + $\mathbb{R} \not\subseteq \mathrm{OD}$, &amp;not;AC + $\mathbb{R} \subseteq \mathrm{OD}$, &amp;not;AC + $\mathbb{R} \not\subseteq \mathrm{OD}$ are relatively consistent with ZF! The details of this are somewhat involved, see Thomas J. Jech's &lt;em&gt;The Axiom of Choice&lt;/em&gt; for an introduction to the various methods to prove independence from ZF.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-08-23T23:00:25.553" CommentCount="2" />
  <row Id="3157" PostTypeId="2" ParentId="3155" CreationDate="2010-08-23T23:24:32.597" Score="9" ViewCount="0" Body="&lt;p&gt;It can (apparently) be shown that the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin z = Az$ has infinitely many roots for any complex number $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For instance see this: &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;pg=PA270&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;pg=PA270&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any $\delta &gt; 0$ there must be such a root with $|z| &gt; \frac{1}{\delta}$, otherwise we can show that $\sin z - Az$ must be identically zero on some bounded disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suppose this shows that there is no value missed by $z \sin\frac{1}{z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Am a bit rusty, please pardon if incorrect).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-24T00:54:52.997" LastActivityDate="2010-08-24T00:54:52.997" CommentCount="3" />
  <row Id="3158" PostTypeId="2" ParentId="3153" CreationDate="2010-08-24T00:32:54.837" Score="4" ViewCount="0" Body="&lt;p&gt;This is merely a very special case of the powerful &lt;strong&gt;CAD (cylindrical algebraic decomposition)&lt;/strong&gt; algorithm for quantifier elimination in real-closed fields, e.g. see Jirstrand's &lt;a href=&quot;http://webber.physik.uni-freiburg.de/~hon/vorlss02/Literatur/ljung/publications/1995/1807.ps.Z&quot; rel=&quot;nofollow&quot;&gt;paper [1]&lt;/a&gt; for a nice introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] M. Jirstrand. &lt;strong&gt;Cylindrical algebraic decomposition - an introduction&lt;/strong&gt;. 1995&lt;br&gt;&#xA;Technical report S-58183, Automatic Control group, Department of Electrical Engineering&lt;br&gt;&#xA;Linkoping University, Linkoping, Sweden.&lt;br&gt;&#xA;Freely available &lt;a href=&quot;http://webber.physik.uni-freiburg.de/~hon/vorlss02/Literatur/ljung/publications/1995/1807.ps.Z&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://lib.org.by/get/_Papers/Computer_algebra/Solving%20equations/Jirstrand%20M.%20Introduction%20to%20cylindrical%20algebraic%20decomposition%20%281995%29%2840s%29.ps.gz&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T01:26:00.093" LastActivityDate="2010-08-24T01:26:00.093" CommentCount="5" />
  <row Id="3161" PostTypeId="2" ParentId="3085" CreationDate="2010-08-24T01:05:41.900" Score="4" ViewCount="0" Body="&lt;p&gt;If you do possess Concrete Mathematics, check out &lt;code&gt;5.1 Basic Identities&lt;/code&gt; and &lt;code&gt;5.2 Basic Practice&lt;/code&gt;. The list can be found at P174 &lt;code&gt;Table 174 The top ten binomial coefficient identities&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="696" LastActivityDate="2010-08-24T01:05:41.900" CommentCount="1" />
  <row Id="3162" PostTypeId="2" ParentId="3155" CreationDate="2010-08-24T01:29:45.390" Score="3" ViewCount="0" Body="&lt;p&gt;The image of $f(z)=z^n\sin(\frac1z)$ (for n odd) on any punctured disc about the origin is the whole of $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is (I think) a simplified proof along the lines of that given in Moron's reference. I'll use proof by contradiction, so suppose that there is an $a$ such that $f(z^{-1})=z^{-n}\sin(z)\not=a$ for all large $z$. By symmetry, $f(z^{-1})=g(z^2)$ for an analytic function $g$ with, for $n&gt;1$, a pole at the origin of order $(n-1)/2$. As $g(z)-a$ can have only finitely many zeros, we can decompose it as&#xA;$$&#xA;g(z)=a+z^{(1-n)/2}h(z)\prod_{i=1}^m(z-b_i)&#xA;$$&#xA;where $b_i$ are the zeros of $g(z)-a$.&#xA;Writing out $f(z^{-1})=\frac{1}{2i}z^{-n}(e^{iz}-e^{-iz})$ shows that $f(z^{-1})$ is of order $1$. So $g$ and, hence, $h$ are of order 1/2. Any entire and nonzero function of order less than 1 is constant, so $h$ is constant and we see that $f$ is a rational function, which is false.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-24T01:29:45.390" CommentCount="1" />
  <row Id="3163" PostTypeId="1" AcceptedAnswerId="3167" CreationDate="2010-08-24T01:32:41.687" Score="0" ViewCount="121" Body="&lt;p&gt;Let $p &gt;5$ be a prime number. Prove that every algebraic integer of the $p$th cyclotomic field can be represented as a sum of (finitely many) distinct units of the ring of algebraic integers of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=152&amp;amp;year=1977&amp;amp;sid=151602f87027a7ce87d3aa9421a666e9&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/resources.php?c=2&amp;amp;cid=152&amp;amp;year=1977&amp;amp;sid=151602f87027a7ce87d3aa9421a666e9&lt;/a&gt;          Question No: 4&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-24T02:05:25.180" LastActivityDate="2010-08-24T02:22:44.350" Title="Algebraic Integers in a Cyclotomic Field." Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3164" PostTypeId="2" ParentId="2110" CreationDate="2010-08-24T02:00:21.773" Score="1" ViewCount="0" Body="&lt;p&gt;If I understand the question correctly, we are given real numbers $0 \leq x_1 \leq \dots \leq x_n$, and real numbers $0 &amp;lt; t_1 &amp;lt; \dots  &amp;lt; t_n$. We also have some unknown differentiable function $x(t)$ that satisfies:&#xA;$$&#xA;x_i \leq x(t_i) &amp;lt; x_i+1  \quad for\quad i=1,2,\dots ,n&#xA;$$&#xA;Our goal is to estimate $\frac{d}{dt}x(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we are interested in the derivative at some point $t_0$. It is clear that the constraints above do not constrain the possible values of the derivative at $t_0$. Indeed, given any set of $x_i$ and $t_i$, one could find an admissible function $x(t)$ for which the derivative at $t_0$ is &lt;em&gt;anything&lt;/em&gt; we want it to be. So unless we know something more about the &quot;law of motion&quot; governing x, there isn't a reasonable way to estimate the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of more information, I'll just assume that x is a polynomial. This will enforce some sort of smoothness. Finding the lowest-degree polynomial that satisfies all the constraints can be cast as a linear program. To see how, suppose that our polynomial is of degree k. Namely, $x(t) = a_0 + a_1 t + \dots + a_{k}t^k$. Then, consider the LP:&#xA;\begin{align*}&#xA;minimize &amp;amp;\quad z \\&#xA;\text{subject to:}&amp;amp;\quad 0 \leq x(t_i) \leq z\quad i=1,\dots,n&#xA;\end{align*}&#xA;Start with $n=0$, and keep increasing until you find an optimal $z$ that is less than 1. Now that you have your polynomial approximation for $x$, you can easily evaluate its derivative anywhere.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-24T02:00:21.773" CommentCount="7" />
  <row Id="3165" PostTypeId="2" ParentId="1730" CreationDate="2010-08-24T02:13:09.173" Score="3" ViewCount="0" Body="&lt;p&gt;I would describe it as a &lt;strong&gt;verb&lt;/strong&gt;.  People are familiar with nouns and verbs so you get to shovel in a bunch of formal understanding for free that way.  Take a &lt;strong&gt;noun&lt;/strong&gt;, a real number or a vector or, even better, a food item, a TV show, or a person &amp;ndash; and think of something that operates on it.  Then the verb &lt;strong&gt;transforms&lt;/strong&gt; it and it's different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really the best thing to do is give them a positive, &lt;strong&gt;creative&lt;/strong&gt;, self-expressive assignment &amp;ndash; like an &lt;strong&gt;essay&lt;/strong&gt;.  Offer extra credit to people who find examples of functions in the real world ongoing throughout the semester.  If they bring in examples of two- or three-place operators then you get to explain why they're &lt;em&gt;right&lt;/em&gt; in front of the class and they've just brought up the lecture point for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &lt;strong&gt;the covers of magazines in the grocery store&lt;/strong&gt; and you will get the material for examples you should be using.  Take weight loss.  You could graph the number of calories taken in versus weight gain/loss.  Give them a lot of examples of functions and then the definition at the end &amp;ndash; &lt;em&gt;then&lt;/em&gt; have them look throughout the semester for more examples.  This should be in concert with or as part of a weekly one-page (or one-paragraph) essay that they submit to you with an example of something from class that related to something from life.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's another trick I've used in teaching about $(x-1)^2$ versus $x^2 - 1$, which will surely come up as well.  &lt;strong&gt;Decompose&lt;/strong&gt; it into two functions.  $x \mapsto x-1$ &lt;em&gt;relabels the abcissa&lt;/em&gt; and you've prepared the board by using a dotted or &lt;strong&gt;no ordina&lt;/strong&gt;.  Then $\cdot \mapsto \cdot^2$ allows you to plot the parabola &amp;ndash; and it's conveniently shifted to the correct place when you put the original abcissa beneath.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-22T07:13:24.917" LastActivityDate="2010-12-22T07:13:24.917" CommentCount="3" />
  <row Id="3166" PostTypeId="1" CreationDate="2010-08-24T02:20:27.613" Score="-1" ViewCount="220" Body="&lt;p&gt;Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and let $g_{k}(x) = f(x+k) - f(x)$ for each $k \in \mathbb{R}$. Suppose $g_{k} \in C^{\infty}$ on $\mathbb{R}$ for each $k \in \mathbb{R}$, then does it imply that $f \in C^{\infty}$ on $\mathbb{R}$. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-26T02:07:10.510" LastActivityDate="2010-08-26T05:59:15.793" Title="Showing $f \in C^{\infty}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3167" PostTypeId="2" ParentId="3163" CreationDate="2010-08-24T02:22:44.350" Score="2" ViewCount="0" Body="&lt;p&gt;Miklos Schweitzer is a &lt;em&gt;very hard&lt;/em&gt; contest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, solution for this (and other problems) can be found in the book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contests in Higher Mathematics, published by Springer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google books has it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=2wwXImJ2HocC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=2wwXImJ2HocC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this particular problem's solution appears here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=2wwXImJ2HocC&amp;amp;pg=PA88&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=2wwXImJ2HocC&amp;amp;pg=PA88&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T02:22:44.350" CommentCount="4" />
  <row Id="3168" PostTypeId="2" ParentId="2110" CreationDate="2010-08-24T02:24:39.633" Score="0" ViewCount="0" Body="&lt;p&gt;Since you mention that your $x_i$ is error-contaminated; the sanest solution (in the absence of other pertinent information that could be used to simplify your problem, e.g. a conjectured functional form for $x(t)$) would be probably a smoothing spline; this is a similar but probably more amenable proposal to Laurent's, with the advantage that any bad nonpolynomial behavior your data might exhibit can be confined to only a few intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what computing environment you're using, but MATLAB at least has smoothing splines in its Spline Toolbox (check the documentation for syntax and proper use); and there are FORTRAN programs developed by Carl de Boor (in fact, the MATLAB toolbox was based on these old FORTRAN programs) @ &lt;a href=&quot;http://www.netlib.org/&quot; rel=&quot;nofollow&quot;&gt;NetLib&lt;/a&gt; ; &lt;a href=&quot;http://www.netlib.org/pppack/index.html&quot; rel=&quot;nofollow&quot;&gt;pppack&lt;/a&gt; is the particular library you want.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-24T02:24:39.633" />
  <row Id="3169" PostTypeId="2" ParentId="3166" CreationDate="2010-08-24T02:39:29.277" Score="3" ViewCount="0" Body="&lt;p&gt;Your question seems incomplete, but to perhaps partly answer what I think you might be asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic Weierstrass nowhere differentiable function $w$ is periodic with period $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $g(x) = \omega(x+2\pi) - \omega(x)$, $g(x) = 0$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $g_{2\pi} \in C^{\infty}$ does not imply $f \in C^{\infty}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T02:39:29.277" CommentCount="1" />
  <row Id="3170" PostTypeId="1" AcceptedAnswerId="3180" CreationDate="2010-08-24T02:41:43.313" Score="7" ViewCount="107" Body="&lt;p&gt;Serre's criterion for affineness (Hartshorne III.3.7) states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X$ be a noetherian scheme. Suppose $H^1(X, \mathcal{F})= 0$ for every quasi-coherent sheaf on $X$. Then $X$ is affine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a more general statement (EGA II.5.2.1) that reaches the same conclusion with the hypothesis that $X$ is a separated quasi-compact scheme or one where the topological space is noetherian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't, however, understand why we need these hypotheses.  The idea of the proof is to find a set of $g_i \in \Gamma(X, O_X)$ generating the unit ideal and such that the $X_{g_i}$ are affine.  This can be done by showing the $X_{g_i}$ form a basis at each closed point using the cohomology statement (an argument which I'm pretty sure works without &lt;em&gt;any&lt;/em&gt; conditions on the scheme). Then, you take open affine sets of the form $X_f$ containing each closed point and take their union; this is an open set whose complement is closed and must contain no closed point, hence is empty. This part of the argument relies on the fact that every closed subset contains a closed point, a fact which is true under noetherian hypotheses since a scheme is a $T_0$-space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't it true that every quasi-compact $T_0$  space has a closed point? A minimal closed set must be reduced to a point. So I'm unclear why &quot;quasi-compactness&quot; alone is not enough. (It &lt;em&gt;shouldn't&lt;/em&gt; be enough. Grothendieck is not one to mince hypotheses!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Where does the standard argument break down for schemes which are only quasi-compact? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this true if $X$ is only quasi-separated and quasi-compact? (Are there counterexamples otherwise?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-08-24T06:00:11.900" Title="Why do we need noetherianness (or something like it) for Serre's criterion for affineness?" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="3171" PostTypeId="2" ParentId="154" CreationDate="2010-08-24T02:47:06.750" Score="6" ViewCount="0" Body="&lt;p&gt;Here's a way to visualize a vector &quot;pointing&quot; in the $i$ direction.  In electrical engineering, if power is traveling 100% efficiently from power generator to your home, then it's pointing in the direction $\langle 1 \rangle$ (or flip-flopping $\langle 1 \rangle$ and $\langle -1 \rangle$ for A/C).  If the power points in the direction of $\langle \sqrt{-1} \rangle$ then the wire will heat up but transmit no useful electricity.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-22T07:11:28.693" LastActivityDate="2010-12-22T07:11:28.693" CommentCount="2" />
  <row Id="3173" PostTypeId="1" CreationDate="2010-08-24T03:34:03.660" Score="0" ViewCount="289" Body="&lt;p&gt;By &quot;isomorphism&quot; I mean any structure-preserving map with a structure-preserving inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please accept my advance apology if this question is out of bounds. I sense that it's borderline, but I'm hoping it'll be considered in better taste than the typical &quot;What is your favorite X?&quot; question. I think a collection of great isomorphisms would be interesting because of what isomorphisms uniquely have the power to do: reveal deep and astonishing connections between seemingly unrelated fields of study; open a channel through which techniques and ideas can pass between disciplines; collapse two or many problems into one.)&lt;/p&gt;&#xA;" OwnerUserId="1458" LastEditorUserId="1458" LastEditorDisplayName="" LastEditDate="2010-08-24T06:10:08.730" LastActivityDate="2010-08-24T15:16:21.577" CommunityOwnedDate="2010-08-24T03:34:03.660" ClosedDate="2010-08-24T12:01:34.183" Title="What is your favorite isomorphism?" Tags="&lt;big-list&gt;&lt;math-history&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="3174" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T03:42:19.613" Score="-1" ViewCount="0" Body="&lt;p&gt;I like the idea that an $n$ dimensional vector space over the field $\mathbb{C}$ is isomorphic to $\mathbb{C}^n$&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-24T15:16:21.577" LastActivityDate="2010-08-24T15:16:21.577" CommunityOwnedDate="2010-08-24T03:42:19.613" CommentCount="2" />
  <row Id="3177" PostTypeId="1" AcceptedAnswerId="3179" CreationDate="2010-08-24T04:18:05.863" Score="5" ViewCount="155" Body="&lt;p&gt;I was having a look at &lt;a href=&quot;http://stackoverflow.com/questions/2792443/finding-the-centroid-of-a-polygon&quot;&gt;this question on SO&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, the centroid is the center of mass of an object. so, by definition its position is given by a simple mean of the positions of all the points in the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a polygon, it only has mass at the vertices. So, the centroid should be given by the arithmetic mean of the coordinates of the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;a href=&quot;http://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says centroid is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MX2km.png&quot; alt=&quot;alt text&quot;&gt;&lt;img src=&quot;http://i.imgur.com/S5n2P.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/u5kDg.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why doesn't a simple arithmetic mean work?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2010-11-11T00:13:28.900" Title="Why doesn't a simple mean give the position of a centroid in a polygon?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="3178" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T04:28:41.467" Score="4" ViewCount="0" Body="&lt;p&gt;The &lt;strong&gt;derivative product rule&lt;/strong&gt; is merely a special case of the &lt;strong&gt;congruence product rule&lt;/strong&gt; for rings, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PRODUCT RULE&lt;/strong&gt; $\rm\quad\:\:\:\: a\equiv a_1,\; b\equiv b_1 \;\Rightarrow\;\; ab \equiv a_1 b_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\quad\; ab-a_1 b_1 \equiv a(b-b_1)+(a-a_1)b_1 \equiv 0 \quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\quad\quad\quad\;\;\: f(x+t) \equiv \: f(x)\: + \: f'(x) \; t \;\pmod {t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and  $\rm\quad\quad\quad\quad g(x+t) \equiv g(x) + g'(x) \; t \;\pmod {t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\;\Rightarrow\quad\: f(x+t)g(x+t) \:\equiv\: f(x)g(x) + (f'(x)g(x) + f(x)g'(x)) \; t \:\;\pmod {t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\:\Rightarrow\quad \frac{f(x+t)g(x+t)\: - \:f(x)g(x)}{t} \equiv\: f'(x)g(x) + f(x)g'(x) \quad\:\pmod t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact this is how one universally defines derivatives in formal polynomial rings $\rm R[x]$, c.f. my post &lt;a href=&quot;http://mathoverflow.net/questions/32343&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. This yields a &lt;strong&gt;purely algebraic&lt;/strong&gt; approach to polynomial derivatives - devoid of limits or other topological notions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring $\rm R[t]/t^2 \;$ is known as the algebra of dual numbers over the ring $\rm R$. &#xA;This ring and its higher order analogs $\;\rm R[t]/t^n \;$  prove quite useful when studying (higher) derivations algebraically since they provide convenient &lt;strong&gt;algebraic models of tangent / jet spaces&lt;/strong&gt;. E.g. as above, they permit easy transfer of properties of homomorphisms to derivations -- see for example section 8.15 in Jacobson, Basic Algebra II. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dual numbers have been applied in many contexts, e.g. &lt;a href=&quot;http://people.maths.ox.ac.uk/szendroi/defth.pdf&quot; rel=&quot;nofollow&quot;&gt;deformation theory [2]&lt;/a&gt;, &#xA;&lt;a href=&quot;http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahap&quot; rel=&quot;nofollow&quot;&gt;numerical analysis [3]&lt;/a&gt; (along with &lt;a href=&quot;http://bt.pa.msu.edu/pub/papers&quot; rel=&quot;nofollow&quot;&gt;Levi-Civita fields&lt;/a&gt;), where &#xA;they're viewed simply as truncated Taylor power series, and also in &#xA;Synthetic Differential Geometry &lt;a href=&quot;http://www.jstor.org/stable/20116534&quot; rel=&quot;nofollow&quot;&gt;(SDG) [1]&lt;/a&gt;, another rigorization of &#xA;inifinitesimals based on work of Lawvere and Kock. Note that SDG employs these&#xA;nilpotent infinitesimals, unlike Abraham Robinson's nonstandard &#xA;analysis, which employs invertible infinitesimals (hence contains infinite elements). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Bell, J. L. Infinitesimals. Synthese 75 (1988) #3, 285--315.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/20116534&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/20116534&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Szendroi, B.  The unbearable lightness of deformation theory, &#xA;a tutorial introduction.&lt;br&gt;&#xA;&lt;a href=&quot;http://people.maths.ox.ac.uk/szendroi/defth.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.maths.ox.ac.uk/szendroi/defth.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] M. Berz, Differential Algebraic Techniques,&lt;br&gt;&#xA;in &quot;Handbook of Accelerator &#xA;Physics and Engineering, M. Tigner, A.Chao (Eds.)&quot; (World Scientific, 1998)&lt;br&gt;&#xA;&lt;a href=&quot;http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahape&quot; rel=&quot;nofollow&quot;&gt;http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahape&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://bt.pa.msu.edu/NA/&quot; rel=&quot;nofollow&quot;&gt;http://bt.pa.msu.edu/NA/&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://bt.pa.msu.edu/pub/papers/&quot; rel=&quot;nofollow&quot;&gt;http://bt.pa.msu.edu/pub/papers/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T14:34:48.027" LastActivityDate="2010-08-24T14:34:48.027" CommentCount="8" />
  <row Id="3179" PostTypeId="2" ParentId="3177" CreationDate="2010-08-24T04:33:04.377" Score="16" ViewCount="0" Body="&lt;p&gt;The centroid of a polygon is indeed its center of mass -- but the mass of a polygon is uniformly distributed over its surface, &lt;em&gt;not&lt;/em&gt; only at the vertices. You're right that if the mass were split evenly among the vertices only, the centroid would be the arithmetic mean of the coordinates of the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just so happens that both definitions are equivalent (mass evenly distributed over the surface vs mass at the vertices only) for simple shapes like triangles and rectangles.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-24T04:33:04.377" CommentCount="2" />
  <row Id="3180" PostTypeId="2" ParentId="3170" CreationDate="2010-08-24T05:36:18.693" Score="5" ViewCount="0" Body="&lt;p&gt;The short answer is that quasi-compactness is enough (for the statement you asked about): see lemma 3.1 in &lt;a href=&quot;http://www.math.columbia.edu/algebraic_geometry/stacks-git/coherent.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.columbia.edu/algebraic_geometry/stacks-git/coherent.pdf&lt;/a&gt; (but its essentially the argument in Hartshorne)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this business of extra hypothesis comes up because Hartshorne and Gronthendieck are proving an iff statement; that is, these hypothesis are needed to prove that if $X$ is affine then $H^1(X,F) = 0$ for every quasicoherent sheaf.  In the case of Hartshorne you need notherian hypothesis to prove &lt;/p&gt;&#xA;&#xA;&lt;p&gt;lemma II.3.3 If $I$ is an injective $A$-module and $f \in A$ then $I \to I_f$ is surjective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with this lemma you can then show if $I$ is injective then $\tilde I$ is flasque an so you can use them to calculate cohomology: $F$ is quasicoherent $\Rightarrow F = \widetilde{M}$.  If $M \to I_0 \to I_1 \to ...$ (1) is an injective resolution then $\widetilde{M} \to \widetilde{I_0} \to \widetilde{I_1} \to ...$ is flasque resolution and applying global sections just recovers (1) so all the higher cohomology vanishes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar with Gronthendieck's argument and I don't know if you can replace separated with quasi-separated.&lt;/p&gt;&#xA;" OwnerUserId="403" LastEditorUserId="403" LastEditorDisplayName="" LastEditDate="2010-08-24T06:00:11.900" LastActivityDate="2010-08-24T06:00:11.900" CommentCount="1" />
  <row Id="3181" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T05:41:25.393" Score="1" ViewCount="0" Body="&lt;p&gt;The natural isomorphism between a finite-dimensional space and its double-dual.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-08-24T05:41:25.393" CommunityOwnedDate="2010-08-24T05:41:25.393" />
  <row Id="3182" PostTypeId="2" ParentId="3177" CreationDate="2010-08-24T05:47:07.537" Score="3" ViewCount="0" Body="&lt;p&gt;Laurent is right, a polygon is not just the vertices, but the whole region. The arithmetic mean of the vertices would give the centroid if the (equal) masses were concentrated at the vertices. While that answers your question, perhaps for the future, the below might be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a mass distributed over a region (not just a polygon), &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's Theorem&lt;/a&gt; might be helpful in calculating the area and the centroid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5ZxGv.png&quot; alt=&quot;Green's Theorem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle D$ is the region of the polygon, then the x-coordinate of the center of mass is given by the area of the polygon times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \iint_{D} x dxdy$ which by Green's theorem is same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \oint_{C} \frac{x^2}{2} dy$, where the line integral is taken over the perimeter of the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For centroid we choose $\displaystyle M(x,y) = \frac{x^2}{2}$ and $\displaystyle L(x,y) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the polygon is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \iint_{D} 1 dxdy$ and can be written as a line integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For area we choose $\displaystyle M(x,y) = \frac{x}{2}$ and $\displaystyle L(x,y) = \frac{-y}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details with the computation details worked out, see the following webpage: &lt;a href=&quot;http://www.worldserver.com/turk/computergraphics/Moments.pdf&quot; rel=&quot;nofollow&quot;&gt;Computing the moments of a 2D polygon&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T00:13:28.900" LastActivityDate="2010-11-11T00:13:28.900" CommentCount="2" />
  <row Id="3183" PostTypeId="1" AcceptedAnswerId="3221" CreationDate="2010-08-24T07:34:57.887" Score="4" ViewCount="123" Body="&lt;p&gt;Here algebraic group means affine algebraic group in both instances.  Also I'm mainly interested in groups over $\mathbb{C}$.  In fact I'm taking $\pi_1(G)$ to mean the fundamental group of $G_{an}$, the analytification.  So I guess my question only applies to the base field being either $\mathbb{C}$ or $\mathbb{R}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case these groups are also Lie groups with Lie algebras.  If $\mathfrak{g}$ is a semisimple Lie algebra then there is a connection with the weight and root lattices: there is a 1-1 correspondence between connected Lie groups with Lie algebra $G$ and lattices $\Lambda$ with $\Lambda_W \supset \Lambda \supset \Lambda_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group corresponding to $\Lambda = \Lambda_R$ is always algebraic because it is the adjoint group.  A slightly more general question is then, for $\Lambda_W \supset \Lambda \supset \Lambda_R$ is the corresponding group $G_\Lambda$ affine algebraic?  &lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2010-08-24T18:20:13.960" Title="Is the universal cover of an algebraic group an algebraic group?" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-groups&gt;&lt;lie-groups&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="3184" PostTypeId="1" AcceptedAnswerId="3239" CreationDate="2010-08-24T08:21:23.537" Score="4" ViewCount="116" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; states that, for every &lt;em&gt;continuous&lt;/em&gt; symmetry of a system, there exists a conserved quantity, e.g. energy conservation for time invariance, charge conservation for U(1). Is there any similar statement for &lt;em&gt;discrete&lt;/em&gt; symmetries?&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-08-24T20:26:14.867" Title="Is there something similar to Noether's theorem for discrete symmetries?" Tags="&lt;group-theory&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="3185" PostTypeId="1" AcceptedAnswerId="3188" CreationDate="2010-08-24T08:51:19.890" Score="3" ViewCount="142" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3163/algebraic-integers-in-a-cyclotomic-field&quot;&gt;This question&lt;/a&gt; made me realize I had a misconception about the cyclotomic integers: I thought the units were exactly the roots of unity. There are only finitely many units but infinitely many integers so the question is impossible to solve unless there are more units. So what are the units of cyclotomic integers?&lt;/p&gt;&#xA;" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-24T10:27:39.677" LastActivityDate="2010-08-24T18:38:36.347" Title="What are the units of cyclotomic integers?" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3186" PostTypeId="1" AcceptedAnswerId="3187" CreationDate="2010-08-24T09:02:21.577" Score="1" ViewCount="99" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_notation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, we can use Einstein notation in&#xA;  linear algebra to distinguish easily&#xA;  between vectors and covectors: upper&#xA;  indices are used to label components &#xA;  (coordinates) of vectors, while lower&#xA;  indices are used to label components&#xA;  of covectors. However, vectors&#xA;  themselves (not their components) have&#xA;  lower indices, and covectors have&#xA;  upper indices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trying to read the Wikipedia article, but I am constantly getting confused between what represents a vector/covector and what represents a component of one of these. How can I tell?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T09:15:34.567" LastActivityDate="2010-08-24T09:15:34.567" Title="Einstein notation - difference between vectors and scalars" Tags="&lt;linear-algebra&gt;&lt;notation&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="3187" PostTypeId="2" ParentId="3186" CreationDate="2010-08-24T09:11:54.077" Score="4" ViewCount="0" Body="&lt;p&gt;A vector component is always written with 1 upper index $a^i$, while a covector component is written with 1 lower index $a_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Einstein notation, if the same index variable appear in both upper and lower positions, an implicit summation is applied, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_i b^i = a_1 b^1 + a_2 b^2 + \dotsb \qquad (*) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a vector is constructed from its component as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf a = a^1 \mathbf{\hat e}_1 + a^2 \mathbf{\hat e}_2 + \dotsb $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{\hat e}_i$ are the basis vectors. But this takes the form like (*), so if we make basis vectors to take lower indices, we will get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf a = a^i \mathbf{\hat e}_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is likely what Wikipedia means.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-24T09:11:54.077" CommentCount="2" />
  <row Id="3188" PostTypeId="2" ParentId="3185" CreationDate="2010-08-24T09:38:54.397" Score="3" ViewCount="0" Body="&lt;p&gt;I don't think there's a straightforward way to describe all the units. If $\zeta$ is a root of unity in the field, then $\frac{\zeta^k-1}{\zeta-1}$ is a unit whenever $k$ is relatively prime to the order of $\zeta$, and therefore products of such elements are units as well. This already gives you infinitely many units, and though they don't necessarily cover all units they do come close (this is a finite-index subgroup of the group of units). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look up Kummer's Lemma, for instance &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/kummer.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_unit&quot; rel=&quot;nofollow&quot;&gt;Cyclotomic unit&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2010-08-24T18:38:36.347" LastActivityDate="2010-08-24T18:38:36.347" CommentCount="2" />
  <row Id="3190" PostTypeId="2" ParentId="3185" CreationDate="2010-08-24T09:59:29.370" Score="1" ViewCount="0" Body="&lt;p&gt;Maybe this &quot;answer&quot; will help you to clarify the concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:By definition, units in a given number field $K$ are the &quot;norm one&quot; integers. That is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_K^*$={$ u\in O_K|\ |N_{K|\mathbb{Q}}(u)|=1 $}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:For the structure of the group of units, one has the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_unit_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's unit theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3:However, even in the case of cyclotomic fields, there is no explicit formula to produce the whole group of units. But we can describe its subgroup: &lt;a href=&quot;http://planetmath.org/encyclopedia/IndexOfTheGroupOfCyclotomicUnitsInTheFullUnitGroup.html&quot; rel=&quot;nofollow&quot;&gt;&quot;cyclotomic units&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="866" LastActivityDate="2010-08-24T09:59:29.370" />
  <row Id="3191" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T11:29:05.967" Score="2" ViewCount="0" Body="&lt;p&gt;Isomorphism between the &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aRiemann_surface_log.jpg&quot; rel=&quot;nofollow&quot;&gt;infinite spiral given by&lt;/a&gt; Riemann surface of $\log(z)$ and the punctured sphere.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-08-24T11:29:05.967" CommunityOwnedDate="2010-08-24T11:29:05.967" />
  <row Id="3192" PostTypeId="1" AcceptedAnswerId="3202" CreationDate="2010-08-24T11:37:08.740" Score="3" ViewCount="107" Body="&lt;p&gt;I am aware of gauge transformations and covariant derivatives as understood in Quantum Field Theory and I am also familiar with deRham derivative for vector valued differential forms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thinking of the gauge field A of the gauge group G as a Lie(G) valued 1-form on the manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't see why under a gauge transformation on A by an element $g\in G$ amounts to the following change, $A \mapsto A^g = gAg^{-1} -dgg^{-1}$ (if G is thought of as a matrix Lie Group) or in general $A_g = Ad(g)A + g^* \omega$  (where $\omega$ is the left invariant Maurer-Cartan form on G and I guess $g^*$ is pull-back of $\omega$ along left translation map by $g$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curvature is defined as $F = dA + \frac{1}{2}[A,A]$ and using this one wants to now see why does $F \mapsto F_g = gFg^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly is the expression for $A_g$ a definition or is there a derivation for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I try proving this (assuming matrix Lie groups) I am getting stuck in multiple places like what is $dA_g$ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be happy if someone can explain the explicit calculations and/or give a reference where such things are explained. Usual books which explain differential forms or connections on principal bundles don't seem to help with such calculations. &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-24T17:23:54.667" LastActivityDate="2010-08-25T08:36:17.133" Title="Gauge transformations in differential forms. " Tags="&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="3193" PostTypeId="2" ParentId="3173" CreationDate="2010-08-24T11:47:05.577" Score="0" ViewCount="0" Body="&lt;p&gt;To follow previous answers, I would say the isomorphism between a finite-dimensional vector space and its dual, even though it isn't natural (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space#CITEREFMacLaneBirkhoff1999&quot; rel=&quot;nofollow&quot;&gt;MacLane &amp;amp; Birkhoff 1999&lt;/a&gt;, §VI.4, from Wikipedia).&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-08-24T11:47:05.577" CommunityOwnedDate="2010-08-24T11:47:05.577" />
  <row Id="3194" PostTypeId="1" CreationDate="2010-08-24T12:08:06.877" Score="1" ViewCount="99" Body="&lt;p&gt;Can anyone kindly give some reference on taking trace of vector valued differential forms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like if $A$ and$B$ are two vector valued forms then I want to understand how/why this equation is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dTr(A\wedge B) = Tr(dA\wedge B) - Tr(A\wedge dB)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One particular case in which I am interested in will be when $A$ is a Lie Algebra valued one-form on some 3-manifold. Then I would like to know what is the precise meaning/definition of $Tr(A)$ or $Tr(A\wedge dA)$ or $Tr(A\wedge A \wedge A)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In how general a situation is a trace of a vector valued differential form defined? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone can give a local coordinate expression for such traces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any references to learn this would be of great help. &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-08-24T19:57:58.287" LastActivityDate="2010-08-24T19:57:58.287" Title="Taking trace of vector valued differential forms " Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="3195" PostTypeId="2" ParentId="3194" CreationDate="2010-08-24T12:23:05.983" Score="2" ViewCount="0" Body="&lt;p&gt;I do not know what you mean by vector-valued form, exactly. But your equation follows from two facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$d$ satisfies the (graded) Leinbiz equation $$d(A\wedge B) = dA\wedge B + (-1)^{\text{degree of $A$}}A\wedge dB,$$&lt;/li&gt;&#xA;&lt;li&gt;and $\mathrm{Tr}$ is linear and commutes with $d$, so that $$d\,\mathrm{Tr}(A)=\mathrm{Tr}\,dA.$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-24T12:23:05.983" CommentCount="5" />
  <row Id="3196" PostTypeId="2" ParentId="3183" CreationDate="2010-08-24T12:39:45.100" Score="3" ViewCount="0" Body="&lt;p&gt;For groups over $\mathbb{R}$ the answer is no.  The group of real points of an affine algebraic group has a faithful finite-dimensional linear representation over $\mathbb{R}$, but it is well-known that the universal cover of $\text{SL}_2(\mathbb{R})$ has no faithful finite-dimensional linear representations (unfortunately I don't know a reference for this).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-24T16:30:05.880" LastActivityDate="2010-08-24T16:30:05.880" CommentCount="4" />
  <row Id="3197" PostTypeId="1" CreationDate="2010-08-24T12:42:00.503" Score="5" ViewCount="114" Body="&lt;p&gt;If G is a simply connected Lie Group then why is every G-bundle over an orientable 3-manifold trivial? (Why is orientability important?) &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-08-24T13:40:16.203" Title="Principal bundles on 3-manifolds" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3198" PostTypeId="1" CreationDate="2010-08-24T13:39:23.017" Score="6" ViewCount="419" Body="&lt;p&gt;Is there a proof that the ratio of a circle's diameter and the circumference is the same for all circles, that doesn't involve some kind of limiting process, e.g. a direct geometrical proof?&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-24T19:33:09.483" LastActivityDate="2010-08-24T20:15:12.113" Title="Proof that Pi is constant (the same for all circles), without using limits" Tags="&lt;geometry&gt;&lt;pi&gt;" AnswerCount="3" CommentCount="16" FavoriteCount="1" />
  <row Id="3199" PostTypeId="2" ParentId="3197" CreationDate="2010-08-24T13:40:16.203" Score="1" ViewCount="0" Body="&lt;p&gt;Are you sure that orientability is necessary? The result is proved in lemma 4.1.1.  &lt;a href=&quot;http://www.math.uni-bonn.de/people/himpel/himpel_cstheory.pdf&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt; and I do not see where orientability is used.&lt;/p&gt;&#xA;" OwnerUserId="1402" LastActivityDate="2010-08-24T13:40:16.203" />
  <row Id="3200" PostTypeId="2" ParentId="3198" CreationDate="2010-08-24T13:52:18.240" Score="2" ViewCount="0" Body="&lt;p&gt;Intuitively, all circles are similar and therefore doubling the diameter also doubles the circumference.  The same applies to ratios other than 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this rigorous, we have to consider what we mean by “the length of the circumference.”  The usual rigorous definition uses integration and therefore relies on the notion of limits.  I guess that any rigorous definition of the length of a curve ultimately requires the notion of limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Rephrased a little to make the connection between the two paragraphs clearer.&lt;/p&gt;&#xA;" OwnerUserId="926" LastEditorUserId="926" LastEditorDisplayName="" LastEditDate="2010-08-24T14:07:52.803" LastActivityDate="2010-08-24T14:07:52.803" CommentCount="10" />
  <row Id="3201" PostTypeId="2" ParentId="3198" CreationDate="2010-08-24T15:22:00.390" Score="1" ViewCount="0" Body="&lt;p&gt;It seems -as far as I can understand what he was doing- that even Euclid used some sort of limiting process (the principle of exhaustion): &lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html&quot; rel=&quot;nofollow&quot;&gt;http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html&lt;/a&gt; . What Euclid is proving here is the following: let $d_1$, $d_2$ be the diameters of two circles and $A_1$, $A_2$ their areas. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{A_1} {A_2} = \frac{d_1^2}{d_2^2} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the same as saying that the proportion between the area of a circle and the square of its radius is constant: if $r_1$ and $r_2$ are the radii of our circles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2} \quad \Longleftrightarrow \quad \frac{A_1}{r_1^2} = \frac{A_2}{r_2^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, &quot;$\pi$ is constant&quot;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-24T15:22:00.390" />
  <row Id="3202" PostTypeId="2" ParentId="3192" CreationDate="2010-08-24T15:24:29.700" Score="1" ViewCount="0" Body="&lt;p&gt;The method is to use the Leibniz rule in the differentiation and and change the sign whenever the exterior derivative moves past an odd form. In addition the following identity must be used $ dgg^{-1} + g dg^{-1} = 0$, and remember that the commutator is is between odd forms thus it is with a plus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the intermediate results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2}[A_g, A_g] = \frac{1}{2}g[A, A] g^{-1} -[dgg^{-1}, g A g^{-1}] + dgg^{-1}\wedge dgg^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dA_g = g dA g^{-1} + [dgg^{-1}, g A g^{-1}] - dgg^{-1}\wedge dgg^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the required details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ d(gAg^{-1})$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Application of the Leibniz rule (Please observe the minus sign in the last term)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ d(gAg^{-1}) = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using the identities $g g^{-1} = 1$ in the first term and $ dg^{-1} = - g^{-1}dg g^{-1} $&#xA;in the last term&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ = dg g^{-1} g \wedge A g^{-1} + g dA g^{-1} +g A g^{-1} \wedge dg g^{-1} $ &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Collection of the first and last term into a commutator:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$  = g dA g^{-1} +[ dg g^{-1}, g A g^{-1} ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ d(dg g^{-1})$ &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Application of the Leibniz rule (Please observe the minus sign in the last term)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ d(dg g^{-1}) = ddg g^{-1} - dg \wedge dg^{-1}$ &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using the identities $dd = 0$ and again $ dg^{-1} = - g^{-1}dg g^{-1} $, we obtain:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ d(dg g^{-1}) = + dg g^{-1}\wedge dg g^{-1}$ &lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-08-25T08:36:17.133" LastActivityDate="2010-08-25T08:36:17.133" CommentCount="5" />
  <row Id="3203" PostTypeId="1" AcceptedAnswerId="3206" CreationDate="2010-08-24T15:38:00.727" Score="4" ViewCount="253" Body="&lt;p&gt;Does a limit that has the value of infinite exist or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've recently come across certain sources that say that if the value of a limit is infinite, then that limit does not exist. This contradicts what my calculus teachers and lecturers taught me however, that a limit doesn't exist if the right hand limit and left hand limit differ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which one is it? &lt;/p&gt;&#xA;" OwnerUserId="456" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T15:43:52.223" LastActivityDate="2010-08-25T02:53:40.783" Title="Infinite limits" Tags="&lt;calculus&gt;&lt;definition&gt;&lt;limit&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="3204" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T15:43:19.213" Score="1" ViewCount="0" Body="&lt;p&gt;The language use in my calculus class was that this is an &lt;em&gt;improper limit.&lt;/em&gt; So that strictly speaking there is no limit in the proper sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The criterion with right hand limits and left hand limits works at points really belonging to the real line (and then first of all both one-sided limits have to exist for themselves).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-24T15:43:19.213" />
  <row Id="3205" PostTypeId="1" AcceptedAnswerId="3207" CreationDate="2010-08-24T15:52:03.503" Score="1" ViewCount="54" Body="&lt;p&gt;We have a set of items $I = \{i_1, i_2, ..., i_n\}$.  Each of these items has what we call a &lt;em&gt;p&lt;/em&gt; value, which is some real number.  We want to choose a subset of $I$, call it $I'$, of size $m$ (for some $m$ with $1 \le m \le n$) such that the average of the &lt;em&gt;p&lt;/em&gt; values of the items in $I'$ falls within some specified range, $[p_l, p_u]$.  (For example, we might require an average &lt;em&gt;p&lt;/em&gt; value between 0.70 and 0.74.)  Moreover, we want to do this efficiently.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We hope to do this in O(n) time, but any polynomial time algorithm is good enough.  We certainly do not want to just try every possible subset of I of size $m$ and then check whether it satisfies the average p-value constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we will be doing this repeatedly and we want the subsets chosen to be a uniformly random distribution over all the possible such subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way of doing this?&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-24T16:07:01.457" LastActivityDate="2010-08-24T16:10:24.967" Title="Choosing subsets of a set such that the subsets satisfy a global constraint" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3206" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T15:57:43.333" Score="6" ViewCount="0" Body="&lt;p&gt;Obviously it depends on the definition of &quot;exists&quot;. Some authors explicitly work over the extended real line with $\pm\infty$ adjoined, so that such infinite limits do explicitly &quot;exist&quot; as first-class values. But there is no consensus. One needs to pay attention to the author's definitions and conventions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it is worth mention - even though this case is rather trivial - that adjoining points at infinity is a special case of various constructions that attempt to simplify matters by some type of existential closure. Below I append an excerpt from an old &lt;a href=&quot;http://google.com/groups?selm=y8zohi4mrfs.fsf_-_@martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;1996 post&lt;/a&gt; where I give a little further discussion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thread originated in a query as to whether infinity or 1/0 &#xA;could be admitted as a &quot;value&quot;, and soon drifted into discussion &#xA;of the Riemann sphere and other topological manifestations of &#xA;infinity via compactification. Below I point out a couple of &#xA;marvelous references on these topics; further I would like to &#xA;bring to your attention a much wider perspective on such topics, &#xA;namely that of existential closure as studied in model theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a beautiful exposition of points at infinity, projective &#xA;closure, compactifications, modifications, etc. in [FM][1] Chapter 7, &#xA;Points at Infinity, by H. Behnke and H. Grauert. This is volume III &#xA;in the excellent &quot;Fundamentals of Mathematics&quot; series,&#xA;which deserves to be on the bookshelf of every budding mathematician. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much deeper appreciation of the methodology behind these constructions &#xA;can be had by studying them from a model-theoretic perspective, in &#xA;particular from the standpoint of existential closure and model &#xA;completion. Kenneth Manders has written a series of thought &#xA;provoking papers [2],[3] from this perspective.&#xA;I close with an excerpt from the introduction to [2]:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The systematic adjunction of roots, or solutions to other simple &#xA;  conditions, as in formation of the complex numbers by adjoining &#xA;  imaginaries, or in adjunction of points &quot;at infinity&quot; in traditional &#xA;  geometry, may be analysed as &lt;em&gt;existential closure&lt;/em&gt; and &lt;em&gt;model &#xA;  completion&lt;/em&gt;. 'Existential closure' refers to a class of processes &#xA;  which attempt to round off a domain and simplify its theory by &#xA;  adjoining elements -- more properly, it refers to the formal &#xA;  relationship that obtains in such a process. 'Model completion' is &#xA;  one of the terms employed when this process is successful. The &#xA;  formation of the complex numbers, and the move from affine to &#xA;  projective geometry, are successes of this kind. Thus, the theory of &#xA;  existential closure gives a theoretical basis of Hilbert's &quot;method &#xA;  of ideal elements.&quot; I now sketch the theory of existential closure, &#xA;  to bring out when, how, and in what sense existential closure gives &#xA;  conceptual simplification.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[FM] Fundamentals of mathematics. Vol. III. Analysis.&lt;br&gt;&#xA;Edited by H. Behnke, F. Bachmann, K. Fladt and W. Suss.&lt;br&gt;&#xA;Translated from the second German edition by S. H. Gould.&lt;br&gt;&#xA;Reprint of the 1974 edition. MIT Press,&lt;br&gt;&#xA;Cambridge, Mass.-London, 1983. xiii+541 pp. ISBN: 0-262-52095-8 00A05 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Manders, Kenneth&lt;br&gt;&#xA;Domain extension and the philosophy of mathematics.&lt;br&gt;&#xA;J. Philos. 86 (1989), no. 10, 553--562.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2026666&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2026666&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Manders, Kenneth L.&lt;br&gt;&#xA;Logic and conceptual relationships in mathematics.&lt;br&gt;&#xA;Logic colloquium '85 (Orsay, 1985), 193--211,&lt;br&gt;&#xA;Stud. Logic Found. Math., 122,&lt;br&gt;&#xA;North-Holland, Amsterdam-New York, 1987.&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1016/S0049-237X(09)70554-3&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1016/S0049-237X(09)70554-3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T17:36:14.030" LastActivityDate="2010-08-24T17:36:14.030" />
  <row Id="3207" PostTypeId="2" ParentId="3205" CreationDate="2010-08-24T16:03:59.697" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming $m$ is an input, your problem is NP-Complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subset sum can be reduced to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $A = \{x_{1}, ..., x_{n}\}$ for which we need to find a subset of sum $S$, we can create $n$ inputs to your algorithm, with input $A$, $m$ and $p_{L} = p_{U} = \frac{S}{m}$, for $1 \leq m \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is fixed then the brute force algorithm (which you don't want) is $O(n^m)$ and so has a polynomial time algorithm.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T16:03:59.697" />
  <row Id="3208" PostTypeId="2" ParentId="3205" CreationDate="2010-08-24T16:10:24.967" Score="0" ViewCount="0" Body="&lt;p&gt;One solution I was considering was to choose items for I' successively such that the average &lt;em&gt;p&lt;/em&gt; value of the items selected so far always falls within some upper and lower bound.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, we could just choose p_l and p_u as our lower and upper bounds, respectively, and this would satisfy the average &lt;em&gt;p&lt;/em&gt; value constraint.  But this would also be too restrictive.  It could leave out a large number of possible I' where small &lt;em&gt;p&lt;/em&gt; values might balance out high &lt;em&gt;p&lt;/em&gt; values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was considering letting the upper and lower bounds start out far apart when the first items are selected and then let them narrow down to p_l and p_u when the final mth item is selected.  So we would have a cone of ranges that would become narrower as more items are picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, we would define two functions &lt;em&gt;u&lt;/em&gt; and &lt;em&gt;l&lt;/em&gt; such that the average of the first i items picked (call this p_i) would satisfy &lt;em&gt;l&lt;/em&gt;(i) &amp;lt;= p_i &amp;lt;= &lt;em&gt;u&lt;/em&gt;(i) for 0 &amp;lt;= i &amp;lt;= m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure how to define these two functions, though.  I know that &lt;em&gt;l&lt;/em&gt;(m) = p_l and &lt;em&gt;u&lt;/em&gt;(m) = p_u, but I don't know what &lt;em&gt;l&lt;/em&gt;(0) or &lt;em&gt;u&lt;/em&gt;(0) should be.  Also, should the functions be linear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comments or suggestions about this possible solution are welcome.&lt;/p&gt;&#xA;" OwnerUserId="984" LastActivityDate="2010-08-24T16:10:24.967" />
  <row Id="3209" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T16:10:57.900" Score="2" ViewCount="0" Body="&lt;p&gt;It is important to consider the context in which you are taking the limit.  When taking the limit of a sequence $\{ x_{n}\}$, we must consider what set the elements of that sequence are coming from.  For example, if $\{ x_{n}\} \subset \mathbb{R}$, then for the limit $L$ to exist (in $\mathbb{R}$) we must have that $L \in \mathbb{R}$.  If $\{ x_{n}\} \to \infty$ as $n \to \infty$, then the limit doesn't exist (in $\mathbb{R}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, there is another notion of limit called $\text{lim sup}$.  According to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0070856133&quot; rel=&quot;nofollow&quot;&gt;Baby Rudin&lt;/a&gt; (slightly modified):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;strong&gt;3.16 Definition&lt;/strong&gt; Let $\{s_{n}\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system [which includes $\pm \infty$]) such that $s_{n_{k}} \to x$ for some subsequence $\{s_{n_k}\}$.  This set $E$ contains all subsequential limits* plus possibly the numbers $+ \infty$, $-\infty$. &#xA;&lt;br&gt;&lt;br&gt;&#xA;Now define $$\begin{align}\text{lim sup} \;\; s_{n} &amp;= \sup E, \;\;\;\;\text{and} \\\ &#xA;\text{lim inf} \;\; s_{n} &amp;= \inf E.\end{align}$$&#xA;* A subsequential limit is just the limit of the subsequence, if that subsequence converges.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this definition we can discuss infinite limits (superior or inferior). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Removed my erroneous claim that finite $\text{lim sup}$'s and $\text{lim inf}$'s collapse to the regular old $\lim$.  (Counterexample: $\text{lim sup}_{x \to \infty} \sin(x) = 1$ but $\lim _{x \to \infty}\sin(x) \neq 1$.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-24T16:28:29.933" LastActivityDate="2010-08-24T16:28:29.933" />
  <row Id="3210" PostTypeId="1" CreationDate="2010-08-24T16:12:13.577" Score="0" ViewCount="186" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake&quot;&gt;-1 is not 1, so where is the mistake?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm trying to understand the exact point of failure in the following reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{\sqrt{-1}^2\sqrt{-1}^2} = \sqrt{(\sqrt{-1}\sqrt{-1})^2} = \sqrt{-1}\sqrt{-1} = \sqrt{-1}^2 = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been previously told that the problem is due to square root not being a function in C; which I found totally unhelpful. Could someone please explain the problem here in simpler terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all for your comments in trying to help me understand this. I finally do. Following is the explanation on my problems in understanding this, in case it'll be of any help to anyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem was really due to using an incorrect definition of i: $i = \sqrt{-1}$. While the correct definition would be: $i^2 = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My incorrect definition led me reasoning, such as (which superficially seemed to give expected results. I see now that this is incorrect, too):&#xA;$\sqrt{-9} = \sqrt{9 * (-1)} = \sqrt{\sqrt{9}^2 \sqrt{-1}^2} = \sqrt{(\sqrt{9} \sqrt{-1})^2} = \sqrt{9} \sqrt{-1} = 3i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, had I used the correct definition of i:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${(xi)}^2 = x^2i^2 = -x^2 = -9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 = 9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = +- 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, analyzing the equations in the original problem, I can see at least the following two errors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In the third =, I'm relying on $-1 = {\sqrt{-1}}^2$, while I should be relying on: $-1 = (+-\sqrt{-1})^2$ which would of course give two different branches. Hmm.. on the second reading, this isn't really a problem, as even with the two separate branches, both of them will lead to the result in the next step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In the fifth =, I'm relying on $\sqrt{i^4} = i^2$, which would be correct, if i was a non-negative number in R. But as i is the imaginary unit and in C: $\sqrt{i^4} = \sqrt{i} = +-(1/\sqrt{2})(1 + i)$.&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditorDisplayName="" LastEditDate="2010-08-24T20:22:18.370" LastActivityDate="2010-08-24T20:22:18.370" ClosedDate="2010-08-24T16:53:39.307" Title="Simple Complex Number Problem: 1 = -1" Tags="&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="3211" PostTypeId="2" ParentId="3210" CreationDate="2010-08-24T16:18:59.033" Score="0" ViewCount="0" Body="&lt;p&gt;You need to pay attention to branches of multivalued functions, e.g. see the Wikipedia explanation &lt;a href=&quot;http://en.wikipedia.org/wiki/Imaginary_unit&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Similar less-trivial questions often arise when symbolic mathematical sotfware systems exhibit bugs due to failure to stay on principal branches, e.g. see this &lt;a href=&quot;http://groups.google.com/group/sci.math.symbolic/browse_frm/thread/9c0201d4dda46458&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; where John McKay asks what your favorite system returns for $(-1)^{5/9} - (-1)^{2/9} - (-1)^{8/9}$. You may find such discussions instructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reader who may be interested in algorithms see for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thomas Breuer. sam@math.rwth-aachen.de&lt;br&gt;&#xA;&lt;strong&gt;Integral Bases for Subfields of Cyclotomic Fields.&lt;/strong&gt;&#xA;AAECC 8, 1997, 279-289&#xA;&lt;a href=&quot;http://www.springerlink.com/content/qnuu0knap4fl3kjt/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/qnuu0knap4fl3kjt/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abstract&lt;/strong&gt;. Integral bases of cyclotomic fields are constructed that allow to&#xA;determine easily the smallest cyclotomic field in which a given sum of roots of&#xA;unity lies.  For subfields of cyclotomic fields integral bases are constructed&#xA;that consist of orbit sums of Galois groups on roots of unity. These bases are&#xA;closely related to the bases of the enveloping cyclotomic fields mentioned above. &#xA;In both situations bases over the rationals and over cyclotomic fields are treated.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T17:04:51.540" LastActivityDate="2010-08-24T17:04:51.540" />
  <row Id="3212" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T16:32:14.977" Score="0" ViewCount="0" Body="&lt;p&gt;It depends on what you are working with. For example, one definition of the derivative of a function that the limit: $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ exists and it is finite. On the other hand, if you're talking about the limit of a sequence ${ x_n }$, then you may want to distinguish between cases when the limit is infinite, or it doesn't exist, which means, like you said, that taken two subsequences you get different values for the limit -in this case you say that the sequence oscillates-. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitions, of course, solve a lot of problems, you define in $\mathbb{R}$ $\lim_{x \rightarrow \infty} f(x) = \infty$ as follows: $\forall N &gt;0, \; \exists M &gt;0: \; x &gt; M \Rightarrow f(x) &gt; N$, then you don't really have to call in the extended number line.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-24T17:12:54.333" LastActivityDate="2010-08-24T17:12:54.333" CommentCount="2" />
  <row Id="3214" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T17:04:53.583" Score="4" ViewCount="0" Body="&lt;p&gt;An equivalent way to state the product rule is $\frac{(fg)'}{fg} = \frac{f'}{f} + \frac{g'}{g}$.  I prefer this statement because it is more intuitive: it says precisely that the &lt;em&gt;relative&lt;/em&gt; instantaneous change in $fg$ is the sum of the relative instantaneous change in $f$ and the relative instantaneous change in $g$.  In other words, this is just an expression of the approximation $(1 + a)(1 + b) \approx 1 + a + b$ when $a, b$ are both small (&quot;multiplication near the identity is addition&quot;).  In other other words, this is an expression of the familiar fact that if you increase something by 5%, then by 5% again, the total increase is just a little more than 10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for turning this into a formal proof, divide both sides of the expression you're confused about by $fg$.  (This is not strictly valid if $f$ or $g$ is equal to zero, but I'm going for clarity here.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-24T17:04:53.583" CommentCount="1" />
  <row Id="3215" PostTypeId="1" AcceptedAnswerId="3220" CreationDate="2010-08-24T17:46:48.180" Score="10" ViewCount="388" Body="&lt;p&gt;Consider the sequence defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{n+1} = \sin x_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I was able to show that the sequence $\sqrt{n} x_{n}$ converges to $\sqrt{3}$ by a tedious elementary method which I wasn't too happy about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think I did this by showing that $\sqrt{\frac{3}{n+1}} &amp;lt; x_{n} &amp;lt; \sqrt{\frac{3}{n}}$, don't remember exactly)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like it should be a standard problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a simple (and preferably elementary) proof for the fact that the sequence $\sqrt{n}x_{n}$ converges to $\sqrt{3}$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-24T17:54:50.097" LastActivityDate="2010-08-25T16:33:37.837" Title="Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="3216" PostTypeId="2" ParentId="3198" CreationDate="2010-08-24T17:58:10.050" Score="3" ViewCount="0" Body="&lt;p&gt;Limits are not involved in the problem of proving that $\pi(C)$ is independent of the circle $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In geometrical definitions of $\pi$, to a circle $C$ is associated a sequence of finite polygonal objects and thus a sequence of numbers (or lengths, or areas, or ratios of those) $\pi_k(C)$.  This sequence is thought of as a set of approximations converging to $\pi$, but that doesn't concern us here; what is important is that the sequence is &lt;em&gt;independent of the circle C&lt;/em&gt;.  Any further aspects of the sequence such as its limit or the rate of convergence will also be the same for any two circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: an example of a &quot;geometrical definition&quot; of a sequence of approximants $\pi_k(C)$ is: perimeter of a regular $k$-sided polygon inscribed in circle C, divided by the diameter of C.  Also, the use of words like &lt;em&gt;limit&lt;/em&gt; and &lt;em&gt;approximation&lt;/em&gt; above does not reflect any assumption that the sequences have limits or that an environment involving limits has been set up. We are demonstrating that if $\pi(C)$ is defined using some construction on the sequence, then whether that construction involves limits or not, it must produce the same answer for any two circles.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof that $\pi_k(C_1) = \pi_k(C_2)$  of course would just apply the similarity of polygons and the behavior of length and area with respect to changes of scale.  This argument does not assume a limit-based theory of length and area, because the theory of length and area for polygons in Euclidean geometry only requires dissections and rigid motions (&quot;cut-and-paste equivalence&quot; or &lt;em&gt;equidecomposability&lt;/em&gt;).  Any polygonal arc or region can be standardized to an interval or square by a finite number of (area and length preserving) cut-and-paste dissections.  Numerical calculations involving the $\pi_k$, such as ratios of particular lengths or areas, can be understood either as applying to equidecomposability classes of polygons, or to the standardizations.  In both interpretations, due to the similitude, the results will be the same for $C_1$ and $C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You might think that this is proving a different conclusion, that the equidecomposability version of $\pi$ for the two circles is equal, and not the numerical equality of $\pi$ within a theory that has real numbers as lengths and areas for arbitrary curved figures.  However, any real number-based theory, including elementary calculus, Jordan measure, and Lebesgue measure, is set up with a minimum requirement of compatibility with the geometric operations of dissection and rigid motion, so once equidecomposability is known, numerical equality will also follow.)   &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-24T20:15:12.113" LastActivityDate="2010-08-24T20:15:12.113" CommentCount="15" />
  <row Id="3217" PostTypeId="1" AcceptedAnswerId="3224" CreationDate="2010-08-24T17:59:50.733" Score="1" ViewCount="177" Body="&lt;p&gt;$$ \frac{4^n-1}{n-1} &gt; 4 &amp;times; 10^6 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an easy way to solve it by hand? I need to find the minimum integral value of &lt;code&gt;n&lt;/code&gt; to satisfy this equation.&lt;/p&gt;&#xA;" OwnerUserId="324" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-28T17:28:11.747" LastActivityDate="2010-08-28T17:28:11.747" Title="How to solve this equation?" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3218" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T18:05:51.363" Score="3" ViewCount="0" Body="&lt;p&gt;This answer is really very similar to the others, but I hope that&#xA;the emphasis is sufficiently different to be worth posting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a number $x$. The &lt;strong&gt;definition&lt;/strong&gt; of the derivative of $f$ is that&#xA;$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$&#xA;Let's rewrite this as&#xA;$$f'(x)=\lim_{h\to 0}F(h)$$&#xA;where we define&#xA;$$F(h)=\frac{f(x+h)-f(x)}{h}$$&#xA;for $h\ne0$. We can rewrite this definition as&#xA;$$f(x+h)=f(x)+hF(h).$$&#xA;For the same reason,&#xA;$$g(x+h)=g(x)+hG(h)$$&#xA;where&#xA;$$G(h)=\frac{g(x+h)-g(x)}{h}$$&#xA;and also&#xA;$$g'(x)=\lim_{h\to 0}G(h).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need to investigate the derivative of $f(x)g(x)$. Now by definition&#xA;this is&#xA;$$\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}.$$&#xA;Manipulating this quotient gives&#xA;$$\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=\frac{(f(x)+hF(h))(g(x)+hG(h))-f(x)g(x)}{h}$$&#xA;$$=\frac{hf(x)G(h)+hF(h)g(x)+h^2F(h)G(h)}{h}=f(x)G(h)+F(h)g(x)+hF(h)G(h).$$&#xA;Now this yields to the &quot;algebra of limits&quot;: using $\lim_{h\to 0}F(h)=f'(x)$&#xA;and $\lim_{h\to 0}G(h)=g'(x)$ we get&#xA;$$\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=f(x)g'(x)+f'(x)g(x)+0f'(x)g'(x)$$&#xA;which simplifies to $f(x)g'(x)+f'(x)g(x)$ which is exactly what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually this argument is just the same as your original but expressed&#xA;in a more pedestrian way, without any cunning sleights of hand.&#xA;As  follow-up a good exercise is to obtain the quotient rule&#xA;in a similar fashion.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-24T18:05:51.363" />
  <row Id="3219" PostTypeId="2" ParentId="3101" CreationDate="2010-08-24T18:07:47.240" Score="2" ViewCount="0" Body="&lt;p&gt;Here is an answer based on Carathéodory's formulation of differentiability. Let $a\in A\subset\mathbb R$ and $f:A\to\mathbb R$. Assume there a sequence in $A\setminus\{a\}$ converging to $a$. Then there is at most one function $\varphi:A\to\mathbb R$ which is continuous at $a$ and satisfies $$f(x)=f(a)+(x-a)\ \varphi(x)$$ for all $x$ in $A$. If such a $\varphi$ exists, we say that $f$ is &lt;strong&gt;differentiable at&lt;/strong&gt; $a$, and we put $f'(a):=\varphi(a)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that this definition is equivalent to the usual one, and, using it, it's easy to verify that the product of two functions differentiable at $a$ is differentiable at $a$ and satisfies the famous formula for the derivative of a product. (Same thing for the Chain Rule.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &quot;The Derivative à la Carathéodory&quot;, Stephen Kuhn, &lt;em&gt;The American Mathematical Monthly&lt;/em&gt;, Vol. 98, No. 1 (Jan., 1991), pp. 40-44: &lt;a href=&quot;http://www.jstor.org/pss/2324035&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2324035&lt;/a&gt; or &lt;a href=&quot;http://www.mediafire.com/?xyl23spk3ui0cga&quot; rel=&quot;nofollow&quot;&gt;http://www.mediafire.com/?xyl23spk3ui0cga&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-08-26T06:48:52.990" LastActivityDate="2010-08-26T06:48:52.990" CommentCount="2" />
  <row Id="3220" PostTypeId="2" ParentId="3215" CreationDate="2010-08-24T18:09:24.653" Score="15" ViewCount="0" Body="&lt;p&gt;Before getting into the details, let me say: The ideas I'm talking about, including this exact exapmle, can be found in chapter 8 of &lt;i&gt;Asymptotic Methods in Analaysis&lt;/i&gt; (second edition), by N. G. de Brujin. This is a really superb book, and I recommend it to anyone who wants to learn how to approximate quantities in &quot;calculus-like&quot; settings. (If you want to do approximation in combinatorial settings, I recommend Chapter 9 of &lt;i&gt;Concrete Mathematics&lt;/i&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this isn't just about $\sin$. Let $f$ be a function $0 \leq f(u) \leq u$ on an interval $[0,c]$ then the sequence $x_n:=f(f(f(\cdots f(c)\cdots)$ approaches $0$. If $f(u)=u-a u^{k+1} + O(u^{k+2})$ (with $a&gt;0$) then $x_n \approx \alpha n^{-1/k}$ and you can prove that by the same methods here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, the answer to your question. On $[0,1]$, we have&#xA;$$\sin x=x-x^3/6+O(x^5).$$&#xA;Setting $y_n=1/x_n^2$, we have&#xA;$$1/x_{n+1}^2 = x_n^{-2} \left(1-x_n^2/6+O(x_n^4) \right)^{-2} =  1/x_n^2 + 1/3 + O(x_n^2)$$&#xA;so&#xA;$$y_{n+1} = y_n + 1/3 + O(y_n^{-1}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that&#xA;$$y_n = \frac{n}{3} + O\left( \sum_{k=1}^n y_k^{-1} \right)$$&#xA;and&#xA;$$\frac{1}{n}y_n = \frac{1}{3} + \frac{1}{n} O\left( \sum_{k=1}^n y_k^{-1} \right)$$&#xA;Since we already know that $x_n \to 0$, we know that $y_n^{-1} \to 0$, so the average goes to zero and we get $\lim_{n \to \infty} y_n/n=1/3$. Transforming back to $\sqrt{n} x_n$ now follows by the continuity of $1/\sqrt{t}$.  &lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T16:33:37.837" LastActivityDate="2010-08-25T16:33:37.837" CommentCount="2" />
  <row Id="3221" PostTypeId="2" ParentId="3183" CreationDate="2010-08-24T18:12:44.720" Score="3" ViewCount="0" Body="&lt;p&gt;Over $\mathbf{C}$ I believe the answer is yes (the universal cover is algebraic), although I'm not really an expert. Here's the story as I understand it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A connected[*] semisimple linear algebraic group over a field $k$ is called simply connected if it admits no nontrivial isogeny from another connected group. (An isogeny is a surjective, flat homomorphism of algebraic $k$-groups with finite kernel.) Now just like in the Lie group story, nice algebraic groups are classified by combinatorial data. Precisely, a reductive $k$-group $G$ together with a split maximal torus $T$, is classified by a &quot;root datum&quot;, which is roughly the roots and coroots of $G$ with respect to $T$, plus the lattices of characters and cocharacters of $T$. (&quot;Roughly&quot; because when $k$ is not algebraically closed, you need to keep track of the action of the Galois group of $k$ on the lattices, too.) Correspondingly, isogenies between such groups match bijectively with appropriate &quot;morphisms&quot; between the respective root data. These theorems are a bit involved, but can certainly be found in Borel's book on linear algebraic groups, for the case you care about. (Algebraically closed field of characteristic zero, namely $\mathbf{C}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A consequence of this theory is that for any connected semisimple linear algebraic $\mathbf{C}$-group, there exists an &quot;algebraic universal cover&quot;, i.e. an isogeny $\tilde{G}\to G$ from a simply connected $\mathbf{C}$-group $\tilde{G}$. (In fact, we don't need to be working over $\mathbf{C}$ for this to be true.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the crux. I claim that if $G$ is a simply connected $\mathbf{C}$-group, then the closed points $G(\mathbf{C})$ are simply connected in the classical topology. The hypothesis that we are over $\mathbf{C}$ is crucial: $\mathrm{Sp}(2n)$ is a simply connected $\mathbf{R}$-group, but the universal cover of $\mathrm{Sp}(2n)(\mathbf{R})$ is the non-algebraic metaplectic group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me sketch the proof of the claim, which is suprisingly hard. (This was explained to me by Brian Conrad; any errors I introduce are, of course, my own.) First, it's a fact, although not a tautology, that since $G$ is connected, $G(\mathbf{C})$ is connected in the classical topology. So that's a good start. Next, by classical Lie theory, the complex Lie group $G(\mathbf{C})$ is homotopy equivalent to a maximal compact real Lie-subgroup $K$. Since $K$ is a compact manifold, this implies that $H^1(G(\mathbf{C}),\mathbf{Z})$ is finitely generated. But this group is the abelianization of $\pi_1(G(\mathbf{C}))$ (Hurewicz); and the fundamental group is already abelian because this is true for all topological groups. So $\pi_1(G(\mathbf{C}))$ is finitely generated (and abelian). In particular, it has a finite quotient. So if $G(\mathbf{C})$ were not (classically) simply connected, there would be a &lt;em&gt;finite&lt;/em&gt; covering map of complex Lie groups $G'\to G(\mathbf{C})$. Now a hard theorem of Grauert (relating $\pi_1(G(\mathbf{C}))$ to the &quot;\'etale fundamental group&quot; of $G$, which classifies the algebraic analogue of finite covering maps) implies that $G'$, as well as its analytic group structure, can uniquely be given an algebraic structure. In other words, there is an algebraic $\mathbf{C}$-group $G_0'$ with $G'=G_0'(\mathbf{C})$, and an isogeny $G_0'\to G$, such that the induced map&#xA;on $\mathbf{C}$-points is $G'\to G(\mathbf{C})$. In particular $G_0'\to G$ is a nontrivial isogeny, since on $\mathbf{C}$-points it is a nontrivial finite convering map. And this contradicts the (algebraic) simple-connectedness of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whew! So in summary, for connected semisimple linear algebraic $\mathbf{C}$-groups $G$, the universal cover of $G(\mathbf{C})$ is precisely $\tilde G(\mathbf{C})$ where $\tilde G$ is the simply connected form of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] Warning: In this answer, &quot;connected&quot; means Zariski-connected. For algebraic groups over $\mathbf{R}$, this is VERY different from the $\mathbf{R}$-points being connected in the classical topology!&lt;/p&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="1265" LastEditorDisplayName="" LastEditDate="2010-08-24T18:20:13.960" LastActivityDate="2010-08-24T18:20:13.960" />
  <row Id="3222" PostTypeId="1" AcceptedAnswerId="3229" CreationDate="2010-08-24T18:38:34.100" Score="0" ViewCount="78" Body="&lt;p&gt;Suppose $f:[0,1]-&gt;R$ is defined as f(x) = 1 if x is irrational and f(x)=0 if x is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from Real Analysis that $\int_{0}^{1} f =1$, but intuitively it seems that the &quot;area under curve&quot; is 0.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some intuitive explaination as to why $\int_{0}^{1} f =1$  (without using measure theory)?&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-08-24T18:55:33.827" LastActivityDate="2010-08-24T19:25:52.243" Title="Integral of 1 minus Dirichlet's function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3223" PostTypeId="2" ParentId="3217" CreationDate="2010-08-24T18:40:13.867" Score="1" ViewCount="0" Body="&lt;p&gt;If n has to be an integer, you can find the solution by brute force. Once you have the solution, you can show that it is correct and unique by showing that the function f(n) = (4^n - 1)/(n-1) is increasing for n &gt;= 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If n is allowed to be real, you can use your integer guess as the starting point for Newton's method and find the solution numerically to as much precision as you'd like.  Since f(n) is increasing, you can even find a good interval estimate.  if f(a) &amp;lt; 4*10^6 and f(b) &gt; 4*10^6, you know the solution is in the interval (a, b).&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-24T18:40:13.867" />
  <row Id="3224" PostTypeId="2" ParentId="3217" CreationDate="2010-08-24T18:41:16.707" Score="1" ViewCount="0" Body="&lt;p&gt;Instead of solving $\frac{4^n-1}{n-1} &gt; 4&amp;times;10^6$ try solving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{4^n}{n-1} &gt; 4&amp;times;10^6 = 4^4 &amp;times;5^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{4^{n-4}}{n-1} &gt; 5^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $n&gt; 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try a binary search method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = 20$ is too large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=15$ too large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=12$ is too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=13$ seems to give the answer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T18:41:16.707" />
  <row Id="3225" PostTypeId="2" ParentId="3217" CreationDate="2010-08-24T18:49:31.323" Score="1" ViewCount="0" Body="&lt;p&gt;Wolfram Alpha will solve it for you if you write: solve &quot;your inequality&quot;.&#xA;But as you want to solve it by hand you can rearrange it to:&#xA;$$ n &gt; (4+3\log_2{5}) + \frac{1}{2} \log_2(n-1) - \log\left(1-\frac{1}{2^{2n}}\right)/2\log(2) $$&#xA;Now the first term in brackets is about 10.96578, use that value for n in the next log term to get 1.65849... This shows you that n&gt;12.6, and so you get n=13. Note that the last term is too small to worry about.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-24T18:49:31.323" CommentCount="1" />
  <row Id="3226" PostTypeId="2" ParentId="3222" CreationDate="2010-08-24T18:53:05.153" Score="1" ViewCount="0" Body="&lt;p&gt;The rationals are a set of Lebesgue measure zero, so they don't impact the value of integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how you might see that.  Start with an epsilon &gt; 0.  You can count the rationals, so put an interval of length epsilon/2 over the first rational.  Put an interval of length epsilon/4 over the second rational, etc.  You've created a sequence of intervals of total length epsilon that cover the rationals, so they must have measure less than epsilon.  But epsilon was arbitrary, so they must have measure zero.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-24T18:53:05.153" CommentCount="4" />
  <row Id="3228" PostTypeId="2" ParentId="3203" CreationDate="2010-08-24T18:58:59.387" Score="1" ViewCount="0" Body="&lt;p&gt;There is no ambiguity about infinite limits. For instance, when we &lt;em&gt;define&lt;/em&gt; that a sequence or real numbers $(x_n)$ &quot;tends to infinity&quot;, $(x_n) \longrightarrow +\infty$, we just mean that, for each real number $N$ there is some $n_0 \in \mathbb{N}$ such that, for all $n\geq n_0$, we have $x_n &gt; N$. That is, the sequence $(x_n)$ grows indifinitely. There is no implication here that $+\infty$ &quot;exists&quot;, or doesn't, nor any problem with the fact that, certainly, $+\infty$ is not a real number. In fact, there is &lt;em&gt;no&lt;/em&gt; $\infty$ on the right hand side of the definition, but just real numbers. And symbols on the left hand side of a definition mean just you want them to mean according with what you put on the right hand side.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-25T02:37:49.383" LastActivityDate="2010-08-25T02:37:49.383" CommentCount="1" />
  <row Id="3229" PostTypeId="2" ParentId="3222" CreationDate="2010-08-24T19:25:52.243" Score="3" ViewCount="0" Body="&lt;p&gt;Some of this answer is implicit in many people's comments, but let's make it explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a question. It seems to you that the area under the curve --- of Dirichelt's function --- is zero. But Dirichelt's function is not a &quot;curve&quot;, precisely because it is discontinuous. What then do you mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider that idea of &quot;under the curve&quot;, for some possible failures of intuition.&#xA;If you take the phrase &quot;under the curve&quot; very seriously, you'll probably be thinking of something like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Darboux_integral&quot; rel=&quot;nofollow&quot;&gt;lower Darboux sum&lt;/a&gt;. This, indeed, is zero for any interval, because there is no space for any rectangles of finite width underneath &quot;the curve&quot;. The problem is that the Darboux integral is not well-defined, because the &lt;em&gt;upper&lt;/em&gt; Darboux sum for any interval is the length of the interval --- because the shortest rectangles of finite width which envelop the function are the rectangles of height 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, Dirichlet's function is not integrable by techniques such as the Darboux integral, because the upper and lower Darboux sums do not coincide. Additional techniques are required --- such as measure theory --- to define the integral at all.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-08-24T19:25:52.243" />
  <row Id="3230" PostTypeId="2" ParentId="2991" CreationDate="2010-08-24T19:38:35.290" Score="0" ViewCount="0" Body="&lt;p&gt;Here's an alternative method that is due to Gauss. Scale the congruence so to reduce the leading coefficient. We seek a multiple of $\:25\:$ that is smaller $\pmod{109}.\;$ Clearly $4\:$ works, since $\; 4*25\equiv 100 \equiv -9 \;$ has smaller absolute value than $25$. Scaling by $4$ yields $-9 x \equiv 12.\;$ Similarly, scaling this by $12$ yields $x \equiv 144 \equiv 35$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will always work if the modulus is prime. It's a special case of the Euclidean algorithm that computes inverses mod $\:\rm p\:$ prime. This is the way that Gauss proved that irreducible integers are prime (i.e. that $\rm p|ab\Rightarrow p|a$ or $\rm p|b$) - hence unique factorization; it's essentially Gauss: Disq. Arith. Art.5, 1801. It iterates  $\rm (a,p) \to (p \;mod\; a, p)\;$ i.e. $\rm a\to a' \to a'' \to \cdots,\; n' = p \;mod\; n \;$ instead of $\rm (a,p) \to (p \;mod\; a,\: a)$ as in the Euclidean algorithm. &#xA;It generates a descending chain of multiples of&lt;br&gt;&#xA;$\;\rm a$ modulo $\rm p$. See my &lt;a href=&quot;http://google.com/groups?selm=y8zu1hn1v14.fsf@nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt; for further discussion.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-24T19:44:31.237" LastActivityDate="2010-08-24T19:44:31.237" />
  <row Id="3231" PostTypeId="1" AcceptedAnswerId="3234" CreationDate="2010-08-24T19:49:51.553" Score="2" ViewCount="401" Body="&lt;p&gt;How can I convert a hexadecimal number, for example &lt;code&gt;0x1A03&lt;/code&gt; to its octal value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that one way is to convert it to decimal and then convert it to octal&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0x1A03 = 6659 = 0o15003&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a simple way to do it without the middle step (conversion to decimal or conversion to binary)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do we tend to convert it to Base10 every time?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="324" LastActivityDate="2010-08-25T19:17:48.303" Title="How to convert a hexadecimal number to an octal number?" Tags="&lt;arithmetic&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="3232" PostTypeId="1" AcceptedAnswerId="3237" CreationDate="2010-08-24T19:51:11.240" Score="1" ViewCount="50" Body="&lt;p&gt;I am having some difficulty following the concept of shrinking a set of node as given in the paper &lt;a href=&quot;http://www.zib.de/Publications/Reports/ZR-06-22.pdf&quot; rel=&quot;nofollow&quot;&gt;On the Bottleneck Shortest Path Problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shrinking a set of nodes can be done&#xA;  in linear time. More precisely, given&#xA;  a set S ⊆ V of nodes of an&#xA;  (undirected) graph G = (V,E), one can&#xA;  construct in linear time another graph&#xA;  with nodes (V \S)$\cup${vnew} (where vnew&#xA;  represents the shrunken set S), where&#xA;  v,w ∈ V \ S are adjacent if and only&#xA;  if v and w are adjacent in G (in this&#xA;  case,the edge keeps its weight), and&#xA;  vnew and w ∈ V \ S are adjacent if and&#xA;  only if there is some v ∈ S such that&#xA;  v and w are adjacent in G (in which&#xA;  case the edge receives the biggest&#xA;  weight of any edge connecting S and w&#xA;  in G).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am particularly confused about the last sentence of how vnew and w are adjacent.  Actually I am pretty much confused about it all and any clarity would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1453" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T14:03:07.433" LastActivityDate="2010-11-27T14:03:07.433" Title="Deciphering this explanation for 'shrinking a set of nodes'?" Tags="&lt;graph-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3233" PostTypeId="1" AcceptedAnswerId="3242" CreationDate="2010-08-24T19:54:08.357" Score="2" ViewCount="140" Body="&lt;p&gt;I would like help with the following problem (chapter 9, #4) from Lee's Smooth Manifolds [its not homework, I'm reading it and I got stuck on this one]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a Lie group $G$ acts smoothly and freely on a smooth manifold $M$ and the orbit space $M/G$ has a smooth manifold structure such that the quotient map $\pi: M\to M/G$ is a smooth submersion, then $G$ acts properly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its kind of a converse to the standard theorem about quotienting a manifold by a group action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints/help?&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2010-08-24T21:05:40.653" Title="Converse To Quotient Manifold Theorem [Exercise in Lee Smooth Manifolds]" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="3234" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T19:58:01.593" Score="7" ViewCount="0" Body="&lt;p&gt;A simpler way is to go through binary (base 2) instead of base 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0x1A03 = 0001 1010 0000 0011&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now group the bits in bunches of 3 starting from the right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 001 101 000 000 011&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 1 5 0 0 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is your octal representation.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T19:58:01.593" />
  <row Id="3235" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T20:00:24.187" Score="3" ViewCount="0" Body="&lt;p&gt;When you're converting among bases $2^n$, you can often do so more quickly with a comprehensive conversion table.  For example, between binary and octal, each block of 3 binary digits will convert to one octal digit:&#xA;$$\begin{matrix}&#xA;000_2=0_8 &amp;amp; 001_2=1_8 &amp;amp; 010_2=2_8 &amp;amp; 011_2=3_8&#xA;\\ &#xA;100_2=4_8 &amp;amp; 101_2=5_8 &amp;amp; 110_2=6_8 &amp;amp; 111_2=7_8&#xA;\end{matrix}$$&#xA;Similarly, when converting between binary and hexadecimal, each block of 4 binary digits converts to a single hexadecimal digit.  Because of these two facts, each block of 12 (least common multiple of 3 and 4) binary digits is 4 octal digits and 3 hexadecimal digits, so each block of 3 hexadecimal digits can be converted to a block of 4 octal digits.  The table for doing hexadecimal&amp;lt;-&gt;octal directly is quite large, though, so it's usually simpler to convert to binary as an intermediate form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to do the conversion directly by performing division in octal or hexadecimal, though this can be tricky to get used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we tend to use decimal as an intermediate form because we are most familiar and comfortable with base 10 (since we generally have 10 fingers and use it most often).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-24T20:00:24.187" />
  <row Id="3236" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T20:02:15.063" Score="3" ViewCount="0" Body="&lt;p&gt;I think the easiest way is to go through binary.  A hex digit corresponds to 4 bits, an octal digit to 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0x1A03 in binary is 0001 1010 0000 0011 (grouped into 4-bit nibbles).  If I regroup into 3-bit groups (from the right), I have 001 101 000 000 011.  That's octal 0o15003.  &lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-24T20:02:15.063" />
  <row Id="3237" PostTypeId="2" ParentId="3232" CreationDate="2010-08-24T20:08:08.513" Score="1" ViewCount="0" Body="&lt;p&gt;Draw some graph with vertex set $V$. Mark off a subset of those vertices and call it $S$. You now have two types of vertices: those in $S$ and those outside of $S$, and you have three types of edges: those connecting vertices of $S$, those connecting vertices outside of $S$, and those on the &quot;boundary&quot;, connecting vertices of $S$ to vertices outside of $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &quot;shrinking&quot; operation, we're removing all the vertices of $S$ and replacing them with a single new vertex which you're calling vnew. Visually, think of squishing all the vertices of $S$ together. So in the new graph all the vertices outside of $S$ stay put and all the vertices in $S$ merge into this new guy vnew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the edges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edges that used to connect vertices of $S$ among themselves are just gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edges that used to connect two vertices outside of $S$ stay put. Both their ends are still there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the boundary edges: each one used to connect some vertex $s$ in $S$ to some vertex $v$ outside of $S$. Quite naturally, such an edge is now replaced by a new one connecting our vnew to $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $v$ (outside of $S$) was connected to one or more vertices in $S$, it is now connected to vnew. If there are weights on the edges, the weight of this new edge between $v$ and vnew is taken to be the max of all the weights of edges connecting $v$ to anyone in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If $v$ is not connected to anything in $S$, it stays unconnected to vnew).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense now?&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-24T20:08:08.513" />
  <row Id="3238" PostTypeId="2" ParentId="3232" CreationDate="2010-08-24T20:17:12.933" Score="1" ViewCount="0" Body="&lt;p&gt;Consider the following example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you had a set of companies and for each pair of companies, you record the maximum number of emails sent by one to the other (i.e For example, when considering exchanges between companies $A$ and $B$ you pick maximum of number of emails from $A$ to $B$, and number of emails from $B$ to $A$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can model this as a graph, with the companies as nodes, join two companies by an edge if they had an email exchange. Mark the edge with the maximum number of email sent by one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose some companies decided to merge and wanted to form a new company called $vnew$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the graph model change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two companies which were not involved in the merger, the edge between them remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose companies $A,B,C$ were part of the merger, while $D$ was not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now consider the email exchange between $D$ and $vnew$, you need to consider the previous exchange between $D$ and $A$, $D$ and  $B$ and $D$ and $C$. Pick the maximum among those three, and create an edge between $D$ and the new company $vnew$ recording that exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-24T20:17:12.933" />
  <row Id="3239" PostTypeId="2" ParentId="3184" CreationDate="2010-08-24T20:26:14.867" Score="3" ViewCount="0" Body="&lt;p&gt;No, because discrete symmetries have no infinitesimal form which would give rise to the (characteristic of) conservation law. See also &lt;a href=&quot;http://dx.doi.org/10.1119/1.18109&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for a more detailed discussion.&lt;/p&gt;&#xA;" OwnerUserId="1399" LastActivityDate="2010-08-24T20:26:14.867" CommentCount="1" />
  <row Id="3241" PostTypeId="2" ParentId="3231" CreationDate="2010-08-24T20:40:39.253" Score="3" ViewCount="0" Body="&lt;p&gt;There is a trick for hex to/from octal because they relate to binary but lets ignore that and consider the general idea (which works for any bases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we want to convert to octal lets start using octal as our number system - as if were always were. So we count like this 1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20,21,22,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And our multiplication table is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|} &#xA;  \hline &#xA;     \times &amp;amp; \mathbf 1 &amp;amp; \mathbf 2 &amp;amp; \mathbf 3 &amp;amp; \mathbf 4 &amp;amp; \mathbf 5 &amp;amp; \mathbf 6 &amp;amp; \mathbf 7\\&#xA;  \hline &#xA;     \mathbf 1 &amp;amp; 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 &amp;amp; 7 \\&#xA;     \mathbf 2 &amp;amp; 2 &amp;amp; 4 &amp;amp; 6 &amp;amp; 10 &amp;amp; 12 &amp;amp; 14 &amp;amp; 16 \\&#xA;     \mathbf 3 &amp;amp; 3 &amp;amp; 6 &amp;amp; 11 &amp;amp; 14 &amp;amp; 17 &amp;amp; 22 &amp;amp; 25 \\&#xA;     \mathbf 4 &amp;amp; 4 &amp;amp; 10 &amp;amp; 14 &amp;amp; 20 &amp;amp; 24 &amp;amp; 30 &amp;amp; 34 \\&#xA;     \mathbf 5 &amp;amp; 5 &amp;amp; 12 &amp;amp; 17 &amp;amp; 24 &amp;amp; 31 &amp;amp; 36 &amp;amp; 43 \\&#xA;     \mathbf 6 &amp;amp; 6 &amp;amp; 14 &amp;amp; 22 &amp;amp; 30 &amp;amp; 36 &amp;amp; 44 &amp;amp; 52\\&#xA;     \mathbf 7 &amp;amp; 7 &amp;amp; 16 &amp;amp; 25 &amp;amp; 34 &amp;amp; 43 &amp;amp; 52 &amp;amp; 61 \\&#xA;  \hline &#xA;\end{array} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the hexadecimal digits are $0\text x1 = 1$, $0\text x2 = 2$, $0\text x3 = 3$, $0\text x4 = 4$, $0\text x5 = 5$, $0\text x6 = 6$, $0\text x7 = 7$, $0\text x8 = 10$, $0\text x9 = 11$, $0\text {xA} = 12$, $0\text xB = 13$, $0\text xC = 14$, $0\text xD = 15$, $0\text xE = 16$, $0\text xF = 17$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, we want to re-interpret the number &lt;code&gt;0x1A03&lt;/code&gt; which is equal to $ 20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3$. Now we just &quot;do the math&quot;, multiply everything out and add it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp; 20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3 \\&#xA;=&amp;amp; 10000 + 400 \cdot 12 + 3 \\&#xA;=&amp;amp; 10000 + 5000 + 3 \\&#xA;=&amp;amp; 10000 + 5000 + 3 \\&#xA;=&amp;amp; 15003&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now the fact that we have all these zeros suggests there would be a quicker method for these particular bases (normally we would not get so many zeros in our calculations).&lt;/p&gt;&#xA;" LastActivityDate="2010-08-24T21:10:39.473" CommentCount="5" />
  <row Id="3242" PostTypeId="2" ParentId="3233" CreationDate="2010-08-24T21:05:40.653" Score="1" ViewCount="0" Body="&lt;p&gt;I'm pretty sure this is true. Here's my suggested approach. There's a step I don't know how to do, so hopefully someone can fill in the gap, or you can figure it out as an exercise. (And please post a comment if you do, as I'm curious!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, show that $M$ is a principal fiber bundle with structure group $G$. That is, locally on $M/G$, $M$ looks like $(M/G)\times G$. To do this, use that $M\to  M/G$ is a submersion to find neighborhoods $U$ on $M/G$ and $V$ on $M$ where $M\to M/G$ looks like the projection onto several coordinates. The coordinates you &quot;forget&quot; in this projection give a chart on some neighborhood of $G$. (This is why the fibers of submersions are smooth manifolds; in this case, the fiber is $G$ since the action is free.)  Now use translation by the group law to &quot;spread $V$ out&quot; so that the entire preimage of $U$ in $M$ is diffeomorphic to $U\times G$, with the restriction of $M\to M/G$ to this open being the projection onto the first factor. This ensures that $M$ is a principal $G$-bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reduces the problem to showing that $G$ acts properly and freely on any principal fiber bundle with structure group $G$ (freeness being clear). To show that $G\times M\to M\times M$ is proper, it suffices to show that it is closed and that the preimage of a point in compact. The latter condition is obvious, by working locally in a trivializing neighborhood of that point. I don't immediately see how to get the closedness of the map, but I've convinced myself it's true in some examples.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2010-08-24T21:05:40.653" CommentCount="1" />
  <row Id="3243" PostTypeId="1" CreationDate="2010-08-24T21:20:15.723" Score="3" ViewCount="275" Body="&lt;p&gt;I'm trying to find out for which natural numbers n does 9^n + 1 have all of its prime factors less than 40.  If I can provide a positive answer to my title question, then I will have a proof that only n = 1 causes 9^n + 1 to have all of its prime factors less than 40.  Thank you for any help.&lt;/p&gt;&#xA;" OwnerUserId="1146" LastActivityDate="2010-11-11T00:22:17.157" Title="Does 9^(2^n) + 1 always have a prime factor larger than 40?" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="3244" PostTypeId="2" ParentId="3243" CreationDate="2010-08-24T21:44:27.093" Score="9" ViewCount="0" Body="&lt;p&gt;Let $\displaystyle a_n:=9^{2^n}+1$; reducing $\displaystyle a_n$ modulo the primes in the interval $\displaystyle [2,40]$, it is easy to find that the only times we get zero is when either the prime is 2 and $\displaystyle n$ is arbitrary, or when the prime is 17 and $\displaystyle n$ is 2.  Thus, for $\displaystyle n&gt;2$, the number $\displaystyle a_n$ has prime factors larger than 2 (&lt;strong&gt;and hence larger than 40&lt;/strong&gt;) provided it is not a power of 2 itself. Since $\displaystyle a_n \equiv 2 \pmod{4}$, we see that $\displaystyle a_n$ is never a power of two. Finally, $\displaystyle a_1=82=2 \cdot 41$ and $\displaystyle a_2=6562=2 \cdot 17 \cdot 193$, and we conclude that $\displaystyle a_n$ has a prime factor larger than 40 for $\displaystyle n \geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T00:19:12.070" LastActivityDate="2010-11-11T00:19:12.070" CommentCount="0" />
  <row Id="3245" PostTypeId="1" AcceptedAnswerId="3247" CreationDate="2010-08-24T21:59:44.630" Score="4" ViewCount="182" Body="&lt;p&gt;The $\Gamma (x)$ function has just one minimum for $x&gt;0$ . This result uses some&#xA;properties of the gamma function:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Gamma ^{\prime \prime }(x)&gt;0$ and $\Gamma (x)&gt;0$ for all $x&gt;0$&lt;/li&gt;&#xA;&lt;li&gt;$\Gamma (1)=\Gamma (2)=1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Observing the following graph (created in SWP) of $y=\Gamma (x)$ this minimum is near $x=3/2$, but likely $\min \Gamma (x)\neq \Gamma \left( 3/2\right) =\dfrac{1}{2}\Gamma \left( 1/2\right) =\dfrac{1}{2}\sqrt{\pi }$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hNj21.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that it is not possible to find analytically the exact value of $x_{\min }$, even by converting to an adequate problem in the interval $]0,1]$ and using the functional equation $\Gamma (x+1)=x\Gamma (x)$ and the reflection formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma (p)\Gamma (p-1)=\dfrac{\pi }{\sin px}\qquad $( $0\lt p\lt 1$)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Which is the best way to find $\min_{[1,2]}\Gamma (x)$ and does $x_{\min }$ lay in $[1,3/2]$ or in $[3/2,2]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Is there some useful series expansion of $\Gamma (x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Which numeric method do you suggest?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Due to the shape of $\Gamma (x)$ I thought on the one-dimensional &lt;a href=&quot;http://ntu-cg.ntu.edu.sg/ysong/conference/MemeticGradientSearch.pdf&quot; rel=&quot;nofollow&quot;&gt;Davies-Swann-Campey&lt;/a&gt; method of direct search for unconstrained optimization, which approximates a function near a minimum by successive approximating quadratic polynomials.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T10:20:35.633" LastActivityDate="2010-08-29T10:20:35.633" Title="Minimum of the Gamma Function $\Gamma (x)$ for $x&gt;0$. How to find $x_{\min}$?" Tags="&lt;calculus&gt;&lt;approximation&gt;&lt;gamma-function&gt;&lt;numerical-methods&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3246" PostTypeId="2" ParentId="3245" CreationDate="2010-08-24T22:08:24.590" Score="3" ViewCount="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://mathworld.wolfram.com/GammaFunction.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; the minimum of the Gamma function for positive $x$ is 1.46163...; in particular I guess this is enough to deduce that it is smaller than $3/2$. You can follow the links along to find some references where this is proved.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-24T22:08:24.590" CommentCount="1" />
  <row Id="3247" PostTypeId="2" ParentId="3245" CreationDate="2010-08-24T22:09:24.120" Score="4" ViewCount="0" Body="&lt;p&gt;There indeed is no closed-form for the gamma function's minimum; what you can do instead, however, is to find the positive root of the digamma function (the logarithmic derivative of the gamma function), which should be available in your computing environment.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-24T22:09:24.120" CommentCount="6" />
  <row Id="3248" PostTypeId="1" CreationDate="2010-08-24T22:43:14.593" Score="1" ViewCount="51" Body="&lt;p&gt;I have been thinking about some conjectures of mine regarding non-uniformly convergent sequences of functions.  I'm happy to discuss my work so far, if anyone wants to know, but for brevity let me just say that I am particularly interested in finding a reference which discusses some theory of non-uniformly convergent sequences of functions, specifically for functions from R&lt;sup&gt;2&lt;/sup&gt;-&gt;R.  Does anyone know of a good place to look for this?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to find a reference today for this subject and could not, is that because nobody thinks it's very interesting?  I tried google, wiki, googlebooks, ect... but there are SO many ref's for uniform convergence that come up, it's hard to find what I'm looking for.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-24T22:43:14.593" Title="Reference for theory of non-uniformly convergent sequences of functions" Tags="&lt;real-analysis&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="3249" PostTypeId="1" AcceptedAnswerId="3258" CreationDate="2010-08-24T23:04:20.163" Score="1" ViewCount="169" Body="&lt;p&gt;In a &lt;a href=&quot;http://math.stackexchange.com/questions/3138/ways-for-a-continuous-function-to-not-be-smooth&quot;&gt;recent post&lt;/a&gt; I asked what are the ways a continuous function cannot be smooth.  Apparently my question was not well posed, but since I am still thinking about these types of functions I will try one more time to ask my question in a proper way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I define a &quot;kink&quot; in the graph of a function as an abrupt, discontinuous change in the first derivative.  For example, the function f(x) = |x| has what I call a kink at x=0.  I apologize if this is not totally rigorous, but I think the idea of a &quot;kink&quot; is clear enough that someone could make it rigorous with little effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides sin(1/x) and the Weirstrauss function, what are some examples of non-smooth continuous functions without kinks, either in the graph of the function or the graph of any of it's derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i.e. I am not looking for any function for which the graph of one of the derivatives has a kink)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case anyone is interested, I edited the linked post yesterday to respond to Qiaochu's comment.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-25T01:52:29.063" Title="What are some examples of non smooth continuous functions without &quot;kinks&quot;?" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="3250" PostTypeId="1" CreationDate="2010-08-24T23:12:11.743" Score="6" ViewCount="413" Body="&lt;p&gt;I am interested in how today's professional mathematicians view the Pythagorean theorem, in terms of how the theorem fits within the axiomatic framework of mathematics.  I often come across textbooks that define length by the Pythagorean theorem, so that the theorem is in essence a definition or axiom.  In more modern mathematics such as linear algebra, is the Pythagorean theorem generally just used as the &lt;em&gt;definition&lt;/em&gt; of length?  Is it more conventional today to treat the Pythagorean theorem as a definition (or axiom) rather than a theorem?  Are there any modern proofs of the Pythagorean theorem that don't rely on Euclidean geometry (like a proof that utilizes linear algebra/the dot product, etc.)?&lt;/p&gt;&#xA;" OwnerUserId="1477" LastActivityDate="2010-08-28T17:10:05.323" Title="Where does the Pythagorean theorem &quot;fit&quot; within modern mathematics? " Tags="&lt;linear-algebra&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="3251" PostTypeId="2" ParentId="3243" CreationDate="2010-08-24T23:14:52.873" Score="12" ViewCount="0" Body="&lt;p&gt;Here is a proof with very few computations involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle x^2 + 1 = 0 (\mod p)$ for some prime $\displaystyle p$, then $\displaystyle p = 1 (\mod 4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a little less well known than the above is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle x^4 + 1 = 0 (\mod p)$ for some prime $\displaystyle p$, then $\displaystyle p = 1 (\mod 8)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle 9^{2^n}+1 = (3^{2^{n-1}})^4 + 1$ we must have that the only possible prime  &amp;lt; 40 which is a candidate is $\displaystyle 17$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we need to check the residues only for $\displaystyle 17$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $\displaystyle 3$ is a primitive root of $\displaystyle 17$, we have that $\displaystyle 3^{8} + 1 = 0 (\mod 17)$. Since $\displaystyle 3^{2^3}  = -1 ( \mod 17)$, we have that $\displaystyle 3^{2^n} + 1 = 2 (\mod 17)$ for all $\displaystyle n &gt; 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we only need to check $\displaystyle 3^{2^3} +1 = 9^{2^2} + 1 = 2&amp;times;17&amp;times;193$, which has a factor $\displaystyle &gt; 40$ and we are done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Interestingly, for any number $N$, we can show that at most a &lt;em&gt;finite&lt;/em&gt; number of numbers of the form $\displaystyle 9^{2^n}+1$ have all their prime factors  &amp;lt; $\displaystyle N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because for $\displaystyle m$ &amp;lt; $\displaystyle n$, $\displaystyle 9^{2^m}-1$ divides $\displaystyle 9^{2^n}-1$ which implies that $\displaystyle (9^{2^n} + 1, 9^{2^m} + 1) = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. any two numbers in the sequence have a gcd of $\displaystyle 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of numbers with all prime factors &amp;lt; $\displaystyle N$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another proof of the same fact is that if $\displaystyle 9^{2^m}+1 = 0 (\mod p)$ then $\displaystyle 9^{2^n}+1 = 2 (\mod p)$ for $n &gt; m$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is a different proof that the number of primes is infinite!&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T00:22:17.157" LastActivityDate="2010-11-11T00:22:17.157" CommentCount="4" />
  <row Id="3252" PostTypeId="2" ParentId="3250" CreationDate="2010-08-24T23:27:27.327" Score="4" ViewCount="0" Body="&lt;p&gt;The Pythagorean theorem follows &lt;em&gt;directly&lt;/em&gt; from the basic properties of the dot product: if $x \cdot v = 0$, then $(x + v) \cdot (x + v) = x \cdot x + v \cdot v$.  There is not much to say about this proof from a modern point of view except that it does not depend on the finite-dimensionality of the underlying space, e.g. in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space&quot; rel=&quot;nofollow&quot;&gt;Hilbert spaces&lt;/a&gt; it even generalizes to &lt;a href=&quot;http://en.wikipedia.org/wiki/Parsevals_identity&quot; rel=&quot;nofollow&quot;&gt;Parseval's identity&lt;/a&gt;.  (Of course one then has to prove that dot products induce norms, but this is the standard Cauchy-Schwarz argument.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to interpret your title question more broadly.  My impression is that mathematicians do not really think about the Pythagorean theorem directly (unless, perhaps, they are functional analysts!).  Rather, following Klein's &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlangen_program&quot; rel=&quot;nofollow&quot;&gt;Erlangen philosophy&lt;/a&gt;, they think about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;group of symmetries&lt;/a&gt; that the use of the Pythagorean theorem (really, dot products) implies.  In the plane, this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_group&quot; rel=&quot;nofollow&quot;&gt;Euclidean group&lt;/a&gt; $\text{E}(2)$, in which sits the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_group&quot; rel=&quot;nofollow&quot;&gt;orthogonal group&lt;/a&gt; $\text{O}(2)$ of linear transformations preserving the dot product (rotations and reflections).  The geometry determined by the Euclidean group is one where it makes sense to talk about lengths and angles and all the familiar tools of Euclidean geometry.  There are other &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_group&quot; rel=&quot;nofollow&quot;&gt;Lie groups&lt;/a&gt; which determine other types of geometries (the &quot;non-Euclidean&quot; ones).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-24T23:27:27.327" CommentCount="12" />
  <row Id="3253" PostTypeId="2" ParentId="3250" CreationDate="2010-08-24T23:29:45.503" Score="-1" ViewCount="0" Body="&lt;p&gt;It all has to do with non-euclidean geometry.  The basic idea is that the distance function, or to say the same thing as a professional mathematician, the &quot;metric&quot;, is super important in geometry.  You can have any kind of distance function you like, as long as it has three key properties (look them up).  This is a pretty big subject, and it all goes back to the independence of Eulcid's 5th axiom on parallel lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between two points depends on the curvature of the underlying space, that's the key idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not possible to prove that the distance between two points is given by the Pythagorean theorem; because the actual distance between two actual points is the domain of physics, not axiomatic mathematics.  The best we can do in mathematics is define what we mean by distance (with a distance function), and since it's a definition, there's nothing to prove. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story goes that Gauss realized Euclid's 5th postulate was independent because there are many geometries where it fails, e.g. on a circle or hyperboloid.  But Gauss didn't want to just tell the world he had solved the oldest problem in math, so what he did was when Riemann came to him looking for a job as a professor, he told Reimann to list 3 possible topics to give a talk on as a sort of interview for the job.  So Riemann wrote two topics down, but he couldn't think of a good third topic, and he put down &quot;foundations of geometry&quot; even though he hadn't thought very hard about it, just because he had some nice idea's which he had not fully formulated.  And since Gauss already been thinking about that, he picked the foundations of geometry; and the rest is math history.  Riemann went home and came up with a theory significantly beyond what Gauss had already known.  &lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-24T23:44:58.990" LastActivityDate="2010-08-24T23:44:58.990" CommentCount="7" />
  <row Id="3254" PostTypeId="1" CreationDate="2010-08-24T23:34:56.947" Score="4" ViewCount="580" Body="&lt;p&gt;The fact that rotation about an angle is a linear transformation is both important (for example, this is used to prove the sine/cosine angle addition formulas; see &lt;a href=&quot;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&quot;&gt;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&lt;/a&gt;) and somewhat intuitive geometrically.  However, even if this fact seems fairly obvious (at least from a diagram), how does one turn the picture proof into a formal proof?  On a related note, it seems likely that many formal proofs using a diagram will end up relying on Euclidean geometry (using angle/side congruence properties), but isn't one of the points of linear algebra to avoid using Euclidean geometry explicitly?&lt;/p&gt;&#xA;" OwnerUserId="1477" LastActivityDate="2010-08-25T23:55:01.290" Title="How do you formally prove that rotation is a linear transformation?" Tags="&lt;linear-algebra&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="4" />
  <row Id="3255" PostTypeId="2" ParentId="3254" CreationDate="2010-08-24T23:42:17.133" Score="1" ViewCount="0" Body="&lt;p&gt;First how do you non-geometrically define rotation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps in the 2D case, you can define it as multiplication by a complex number (which is in fact true!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you do that, use the definition of a Linear Transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the map T satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $T(a + b) = T(a) + T(b)$ for any vectors $a,b$ (or complex numbers if you chose to go that route) &lt;br/&gt;&#xA;2) $T(\alpha b) = \alpha T(b)$ for any scalar $\alpha$ and any vectors $a,b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, you can define complex numbers as vectors in the abstract without having to rely on any Euclidean Axioms.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T05:47:09.090" LastActivityDate="2010-08-25T05:47:09.090" CommentCount="2" />
  <row Id="3256" PostTypeId="2" ParentId="3249" CreationDate="2010-08-24T23:50:37.747" Score="2" ViewCount="0" Body="&lt;p&gt;One rigorous definition of a kink that I can think of is that the left and right derivatives exist and do not agree.  In that case, plenty of functions &quot;don't have kinks&quot; at points where they are continuous but not differentiable.  $\sin \frac{1}{x}$ and the Weierstrass function &quot;don't have kinks&quot; in the sense that neither the left nor the right derivative exist (at $0$ for the first function and everywhere for the second), and one can come up with arbitrarily badly behaved examples (so again, I do not see the point of this question) as explained for example in damiano's comment to your previous post.  If you do not understand this comment, I can elaborate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't take this the wrong way, but the fact that you ask this question and put the Weierstrass function in as an example indicates to me that you don't understand the construction of the Weierstrass function.  The point of the construction of the Weierstrass function is that although a uniformly convergent sequence of continuous functions is continuous, a uniformly convergent sequence of differentiable functions need not be differentiable.  One can imitate this construction for an extremely wide class of functions (e.g. by replacing the sines with another periodic function) and get any kind of examples you want.  So when you say &quot;the Weierstrass function,&quot; do you literally mean &quot;the Weierstrass function,&quot; or do you mean &quot;any function constructed sort of like the Weierstrass function&quot;?  If the latter, you should be more precise about this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-24T23:56:01.853" LastActivityDate="2010-08-24T23:56:01.853" CommentCount="1" />
  <row Id="3257" PostTypeId="2" ParentId="3254" CreationDate="2010-08-24T23:59:01.077" Score="0" ViewCount="0" Body="&lt;p&gt;You can prove this by looking at the appropriate diagram and using trig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what is interesting is that it is not too difficult to show that &lt;em&gt;any&lt;/em&gt; isometry of $\mathbb R^n$ (with its usual inner product) is necessarily linear.  So if you accept that rotations are isometries (which to me at least is much more self-evident than them being linear) then you get linearity immediately.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-08-24T23:59:01.077" CommentCount="1" />
  <row Id="3258" PostTypeId="2" ParentId="3249" CreationDate="2010-08-25T00:05:34.170" Score="2" ViewCount="0" Body="&lt;p&gt;This has already been said before, but let me iterate this.  The value of the derivative of a function at a point is evaluated by computing a limit, if it exists.  The limit of a sequence may not exist if at least one of the following happens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the absolute value of some of the terms grows without bounds;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you can extract two subsequences which both converge, but they converge to different values.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is important to keep in mind that both behaviors may happen for the same limit, and that for 2 you might be able to extract countably many subsequences each converging to a different value (only &quot;countably many&quot; as I am talking now of a limit of a &lt;em&gt;sequence&lt;/em&gt;). &lt;strong&gt;EDIT: in fact, you can have uncountably many different limits (see Jason's comment below).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for limits of functions is not simpler than the above, since you can &quot;embed&quot; any sequence into the limit defining the derivative of a continuous function at a point.  One of the &lt;em&gt;good&lt;/em&gt; consequences of the fact that you are dealing with a limit of a &lt;em&gt;continuous&lt;/em&gt; function is that if situation 2 occurs and the two subsequences you found have limits a&lt;code&gt;&amp;lt;&lt;/code&gt;b, then also every intermediate value $c \in [a,b]$ is the limit of a subsequence.  An elaboration of this is that the derivative of a once differentiable function satisfies the intermediate value property (&lt;a href=&quot;http://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)&quot; rel=&quot;nofollow&quot;&gt;Darboux's Theorem&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you analyze the behavior of the values of the limit that defines a derivative at a point you may find that, as you zoom in, the values you obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. also zoom in on a single real number - the derivative exists;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. are always contained in some finite interval, no matter how close to the actual limiting point you get - e.g. the function $x^2 \sin(1/x)$ near 0;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. are unbounded (possibly with alternating signs) - e.g. the function $\sqrt{x} \sin (1/x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that further pathologies include the fact that the behaviors as you approach the limit from below or from above may well be different.  Even crazier, the &quot;overall&quot; interval you find in b, or an infinite interval in the case c, may contain inside it also countably many of other intervals nested more or less arbitrarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, derivatives of continuous functions are so immensely complicated that there is very little you can say about them!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="1028" LastEditorDisplayName="" LastEditDate="2010-08-25T01:52:29.063" LastActivityDate="2010-08-25T01:52:29.063" CommentCount="3" />
  <row Id="3259" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T00:17:27.097" Score="1" ViewCount="0" Body="&lt;p&gt;If you want to be formal &lt;em&gt;and&lt;/em&gt; avoid Euclidean geometry, it seems to me that the right way to go is the other way around: first define certain linear transformations (the ones which preserve the dot product and have determinant $1$) and then prove that they have the properties you expect them to have.  More precisely, the group $\text{SO}(2)$ is abelian with Lie algebra $\mathbb{R}$, and the parameterization $\theta \mapsto \left[ \begin{array}{cc} \cos \theta &amp;amp; \sin \theta \\ - \sin \theta &amp;amp; \cos \theta \end{array} \right]$ is the exponential map from the Lie algebra to the group.  The methods for doing this should be covered in any book on Lie theory.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-25T00:23:32.170" LastActivityDate="2010-08-25T00:23:32.170" CommentCount="9" />
  <row Id="3260" PostTypeId="2" ParentId="3249" CreationDate="2010-08-25T00:20:49.260" Score="2" ViewCount="0" Body="&lt;p&gt;A real valued function is said to have a discontinuity of the first kind at a point if the left and right-hand limits at that point exist but are unequal. There are, however, only countably many such points.&#xA;These points, applied to the derivative, are kind of like &quot;kinks.&quot; A discontinuity of the second kind is not a &quot;kink&quot; since one of the  sided limits doesn't exist.  (I am attempting, possibly incorrectly, to make precise your terminology.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of $x^2 \sin(1/x^2)$ at zero is zero, so the function is differentiable everywhere, yet the derivative is unbounded in any neighborhood of the origin, which is thus a discontinuity of the second kind. More generally, &lt;a href=&quot;http://en.wikipedia.org/wiki/Volterra%27s_function&quot; rel=&quot;nofollow&quot;&gt;Volterra's function&lt;/a&gt; gives a differentiable function with a derivative with an uncountable number (even positive measure) of discontinuities of the second kind.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-08-25T00:31:45.913" LastActivityDate="2010-08-25T00:31:45.913" />
  <row Id="3261" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T00:33:31.567" Score="2" ViewCount="0" Body="&lt;p&gt;This depends on your definitions of &quot;rotation&quot; and &quot;angle&quot;, but, if you are willing to accept the sum and difference formulas of your link &lt;a href=&quot;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&quot;&gt;http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry&lt;/a&gt; , then you may follow the reasoning in Quiochu Yuan's answer there the other way around: given a vector $(x,y) \in \mathbb{R}^2$, let $\alpha$ be the angle determined by $(x,y)$ and the $x$-axis and $r= \sqrt{x^2 + y^2}$ its length. Then, of course,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;x &amp;amp;=&amp;amp; r \cos \alpha \\&#xA;y &amp;amp;=&amp;amp; r \sin \alpha&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you rotate $(x,y)$ an angle $\theta$, you'll obtain the vector $(x',y')$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;x' &amp;amp;=&amp;amp; r \cos (\alpha + \theta ) \\&#xA;y' &amp;amp;=&amp;amp; r \sin (\alpha + \theta )  \ .&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you apply those sum and difference formulas and get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;x' &amp;amp;=&amp;amp; r \cos\alpha\cos\theta - r \sin\alpha\sin\theta &amp;amp;=&amp;amp; \cos\theta \cdot x - \sin\theta \cdot y\\&#xA;y' &amp;amp;=&amp;amp; r \sin\alpha\cos\theta + r \cos\alpha\sin\theta &amp;amp;=&amp;amp; \sin\theta \cdot x + \cos\theta\cdot y \ .&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the same as saying that $(x',y')$ is obtained from $(x,y)$ multiplying with the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;\cos\theta &amp;amp; -\sin\theta   \\&#xA;\sin\theta  &amp;amp; \cos\theta&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rotation $(x,y) \mapsto (x',y')$ is the same as multiplication by a matrix, it is a linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-25T00:57:22.327" LastActivityDate="2010-08-25T00:57:22.327" />
  <row Id="3262" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T00:56:10.967" Score="6" ViewCount="0" Body="&lt;p&gt;The problem comes in defining rotation so that this whole thing doesn't wind up circular, doing so fully would undoubtedly tie us all in knots, but one can prove the linearity of rotations by observing a few choice properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rotations preserve angles&lt;/li&gt;&#xA;&lt;li&gt;They fix the origin&lt;/li&gt;&#xA;&lt;li&gt;And they preserve lengths&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now just draw a parallelogram $OACB$ (spanned by vectors $\vec{OB}, \vec{OA}$- with $\vec{OC}=\vec{OB}+\vec{OA}$), rotate and chase the angles and lengths to observe that $R(\vec{OB})+R(\vec{OA})= R(\vec{OC})= R(\vec{OB}+ \vec{OA})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sort of Euclidianish but it's the only decent way to play it (especially if we want higher dimensions) without the 'lucky guess' approach championed by Qiaochu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel pushed, you can turn the angle and length chase into norms and inner products but that would be a tragic waste of an isomorphism- here we are dealing with a geometric object in the language of vectors and to treat it non-geometrically would put us at an unnecessary disadvantage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good deal of higher mathematics depends on finding 'the best place' to solve a given problem, but the best place need not be the most abstract one: translating something about PDE into hilbert space theory may help find a solution, but it may also rob us of our geometric techniques and intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example is something called the h-cobordism theorem, which can (and does- Milnor's take is a gem) take up a small book, using formal techniques (morse theory) to prove it, but can be done in half the time with some geometric (handlebody) jazz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vector spaces are another string to our bow, a new place to do math, and at times a more rigorous place- but they are not (at least from my perspective) the only place. &lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-25T01:18:09.377" LastActivityDate="2010-08-25T01:18:09.377" CommentCount="2" />
  <row Id="3263" PostTypeId="2" ParentId="3203" CreationDate="2010-08-25T02:53:40.783" Score="0" ViewCount="0" Body="&lt;p&gt;While I'll defer to other's answers (e.g. &lt;a href=&quot;http://math.stackexchange.com/questions/3203/infinite-limits/3206#3206&quot;&gt;Bill Dubuque's&lt;/a&gt;) that in the most general sense &quot;it depends&quot; is the correct answer, it may be worth noting that in the context of AP Calculus and across numerous common AP Calculus texts (at both the AB and BC level), a limit exists if and only if it has a real value, so saying that $\underset{x\to c}\lim f(c)=\infty$ or $\underset{x\to c}\lim f(c)=-\infty$ is saying that the limit as $x$ goes to $c$ of $f(x)$ does not exist and simultaneously giving more information about how the limit does not exist.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-25T02:53:40.783" />
  <row Id="3264" PostTypeId="1" AcceptedAnswerId="3268" CreationDate="2010-08-25T04:04:14.540" Score="6" ViewCount="201" Body="&lt;p&gt;Somewhat prompted by the discussions of &lt;a href=&quot;http://math.stackexchange.com/users/232&quot;&gt;Qiaochu Yuan&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/users/1102&quot;&gt;Moron&lt;/a&gt; in this &lt;a href=&quot;http://math.stackexchange.com/questions/3254&quot;&gt;question&lt;/a&gt;, I realized that my understanding of linear/affine transformations thus far had been built on a convoluted series of circular arguments. I will now be asking a question in order to patch the gaps in my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to my innate tendency to view things geometrically, I had always taken for granted the fact that rotations, reflections, and translations, among other affine transformations, are isometric (That is to say, no matter how you move, spin, or reflect an object, all the lengths and angles do not change at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, I considered these transformations being isometries as postulates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, then, is if there is a rigorous (so probably non-geometric) way of justifying that these transformations are isometries.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-25T23:39:34.410" LastActivityDate="2010-09-01T22:42:01.117" Title="On isometric affine transformations" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="3265" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T04:43:38.243" Score="2" ViewCount="0" Body="&lt;p&gt;Maybe my first answer was not enough &quot;formal&quot;, since I didn't say explicitely what I was assuming a &quot;rotation&quot; is -nor &quot;angle&quot;. (Exercise: (1) find in which places I was using those unsaid definitions of &quot;rotation&quot; and &quot;angle&quot; and (2) which definitions are they.   :-)   )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it was a sort of &quot;high-school proof&quot;. The problem about the definition of &quot;rotation&quot;, as some answers have already pointed out, or implied, is that, if you &lt;em&gt;really&lt;/em&gt; want to be formal, you end saying that a rotation is some kind of linear transformation &lt;em&gt;by definition&lt;/em&gt;. Full stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I think I can produce an &quot;undergraduate&quot; proof of this fact, avoiding circular arguments. It has the merit that is valid in any dimension. For which I need the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Convention.&lt;/strong&gt; Whatever &quot;rotation&quot; means,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It is a map of some vector space $V$,&lt;/li&gt;&#xA;&lt;li&gt;Which has a way of measuring &quot;lengths&quot; and &quot;angles&quot; of its vectors, and&lt;/li&gt;&#xA;&lt;li&gt;&quot;Rotations&quot; preserve those &quot;lengths&quot; and &quot;angles&quot;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, a fancy way to have &quot;lengths&quot; and &quot;angles&quot; in a vector space is to have a dot product in it. So let's assume that $V$ is an euclidian vector space; for instance, a real vector space such as $\mathbb{R}^n$ with the standard dot product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, in such a $V$, &quot;length&quot; means &quot;norm&quot;, which is $\| v \| = +\sqrt{v\cdot v}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to be more careful with &quot;angles&quot; because the standard definition already involves rotations. To avoid a circular argument, we define the (non-oriented) angle determined by two vectors $v, w$ as the &lt;em&gt;unique&lt;/em&gt; real number $\theta = \widehat{vw} \in [0, \pi ]$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\cos \theta = \frac{v\cdot w}{\|v\| \|w\|} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this definition makes sense since, because of Schwartz inequality &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwartz_Inequality&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Schwartz_Inequality&lt;/a&gt; , we always have $-1 \leq \frac{v\cdot w}{\|v\| \|w\|} \leq 1$ and $\cos : [0, \pi] \longrightarrow [-1,1]$ is bijective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &quot;rotation&quot; is some kind of map $f: V \longrightarrow V$ which preserves norms and angles. Since the dot product can be expressed in terms of norms and (cosines of) angles, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v\cdot w = \|v\| \|w\| \cos\widehat{vw}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rotations preserve dot products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(v) \cdot f(w) = v\cdot w \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's show that a map that preserves the dot product is necessarily linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\| f(v+w) - f(v) -f(w) \|^2  &amp;amp;= \left( f(v+w) -f(v) -f(w) \right) \cdot \left( f(v+w) -f(v) -f(w)  \right)   \\&#xA;          &amp;amp;=  \left( f(v+w)\cdot f(v+w) - f(v+w) \cdot f(v) - f(v+w)\cdot f(w) \right) \\&#xA;          &amp;amp;{} \qquad + \left( -f(v)\cdot f(v+w) + f(v) \cdot f(v) + f(v)\cdot f(w) \right) \\&#xA;          &amp;amp;{} \qquad + \left( -f(w)\cdot f(v+w) + f(w)\cdot f(v) + f(w)\cdot f(w) \right) \\&#xA;          &amp;amp;= \left( (v+w)\cdot (v+w) - (v+w) \cdot v - (v+w)\cdot w \right) \\&#xA;          &amp;amp;{} \qquad + \left( -v\cdot (v+w) + v \cdot v + v\cdot w \right) \\&#xA;          &amp;amp;{} \qquad + \left( -w\cdot (v+w) + w\cdot v + w\cdot w \right) \\&#xA;          &amp;amp;= \| v+w - v -w \|^2 = 0  \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(v+w) = f(v) + f(w) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar computation shows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|f(\lambda v ) - \lambda f(v) \|^2 = \|\lambda v - \lambda v\|^2 = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f(\lambda v) = \lambda f(v) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, our rotation $f$ is a linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-25T23:55:01.290" LastActivityDate="2010-08-25T23:55:01.290" CommentCount="2" />
  <row Id="3266" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T04:45:15.577" Score="3" ViewCount="0" Body="&lt;p&gt;As in my comment on the question to which you refer, exactly how you define things is critical.  Suppose we consider transformations of the plane where the plane is defined to be $\mathbb{C}$ and a transformation of the plane is a 1-1 function $f:\mathbb{C}\to\mathbb{C}$.  An isometry is a transformation that preserves distances, so a transformation $f$ is an isometry if and only if $|f(a)-f(b)|=|a-b|$ for all $a,b\in\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Theorem:&lt;/em&gt; Any function of the form $f(z)=cz+d$ (direct isometry) or $f(z)=c\bar{z}+d$ (opposite isometry) where $c,d\in\mathbb{C}$ and $|c|=1$ is an isometry.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;proof:&lt;/em&gt; (direct isometry) $|f(a)-f(b)|$ $=|ca+d-(cb+d)|$ $=|ca-cb|$ $=|c(a-b)|$ $=|c|\cdot|a-b|$ $=1\cdot|a-b|$ $=|a-b|$.  (opposite isometry) $|f(a)-f(b)|$ $=|c\bar{a}+d-(c\bar{b}+d)|$ $=|c\bar{a}-c\bar{b}|$ $=|c(\bar{a}-\bar{b})|$ $=|c|\cdot|\overline{a-b}|$ $=1\cdot|a-b|$ $=|a-b|$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also note that if $f$ and $g$ are isometries, so is $f\circ g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes and for simplicity, define a translation by $a+bi$ to be the function $T_{a+bi}(z)=z+a+bi$, a rotation about 0 by $\theta$ to be the function $R_\theta(z)=(\cos\theta+i\sin\theta)z$, and reflection over the real axis to be $r_\mathbb{R}(z)=\bar{z}$.  (It should be relatively easy to show that these definitions align with any geometric definitions.)  Each of these is clearly an isometry, based on the above theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotation about some point $p$ rather than about 0, apply $T_p\circ R_\theta\circ T_{-p}$.  For a reflection over a line parallel to the real axis and passing through $bi$, apply $T_{bi}\circ r_\mathbb{R}\circ T_{-bi}$.  For a reflection over a line not parallel to the real axis, intersecting the real axis at $a$ at an angle $\theta$, apply $T_a\circ R_\theta\circ r_\mathbb{R}\circ R_{-\theta}\circ T_{-a}$.  Each of these is also an isometry because it is the composition of isometries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, any rotation, reflection, or translation of the plane is an isometry.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-25T04:45:15.577" CommentCount="3" />
  <row Id="3267" PostTypeId="2" ParentId="2968" CreationDate="2010-08-25T05:09:51.157" Score="4" ViewCount="0" Body="&lt;p&gt;You should be able to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Homomorphic_encryption&quot; rel=&quot;nofollow&quot;&gt;Fully Homomorphic Encryption Scheme&lt;/a&gt; which was recently proved to be possible by Craig Gentry from IBM (and was an important open problem in cryptography till 2009!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a fully homomorphic scheme, any circuit can be evaluated homomorphically, effectively allowing computation on encrypted input without having to decrypt it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use it in your case, each can encrypt their card value and do a homomorphic comparison (for instance see this: &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1356047&quot; rel=&quot;nofollow&quot;&gt;http://portal.acm.org/citation.cfm?id=1356047&lt;/a&gt;, caveat: I have only read the abstract). We can find out the person with the greatest value (or sort them in order) without having to divulge any non-encrypted information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is still theoretical, it might be a while before this becomes practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the wiki page for more details.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T13:13:23.530" LastActivityDate="2010-08-25T13:13:23.530" />
  <row Id="3268" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T05:13:29.397" Score="6" ViewCount="0" Body="&lt;p&gt;The first point I'd make is that translations are &lt;em&gt;not&lt;/em&gt; linear transformations because every linear transformation must take 0 to 0, and the only translation that does this is the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, here's a general way to see exactly which linear transformations are isometries (including rotations and reflections).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way is to start with the Euclidean inner product.  In the &quot;standard basis&quot;, it looks like $\langle v,w\rangle = \sum_i v_i w_i$.  Once we have this, the length of a vector $x$ is defined to be $\sqrt{\langle x,x\rangle}$.  The notation for this is $|x|$.  The angle between two vectors $x$ and $y$ is defined by $cos\theta = \frac{\langle x,y\rangle}{|x||y|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll say a function &quot;preserves the inner product&quot; if $\langle x , y\rangle = \langle f(x), f(y)\rangle$ for all $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 1.  A linear function preserves the inner product iff the linear function preserves all lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof one way is trivial.  Assuming the linear function preserves all lengths, notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x|^2 + 2\langle x,y\rangle + |y|^2 = \langle x+y,x+y\rangle = \langle f(x) + f(y), f(x) + f(y)\rangle $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= |f(x)|^2 + 2\langle f(x),f(y)\rangle + |f(y)|^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using the fact that $|x|^2 = |f(x)|^2$, we learn that $\langle x,y\rangle = \langle f(x), f(y)\rangle$.  (Note that, as an added bonus, if a linear transformation preserves length, it preserves the inner product and hence it automatically preserves angles as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing worth noting is that you can move a linear transformation around in an inner product.  That is, if $A$ is any matrix, then there is a unique matrix $B$ such that $\langle Ax, y\rangle = \langle x,By\rangle$ for all $x$ and $y$.  In fact, in an orthonormal basis, $B$ is simply given as the transpose of $A$ - that is, $B = A^t$.  The proof is simple:  let $e_i$ be an orthonormal basis.  Then $A_{ij} = \langle Ae_i, e_j\rangle = \langle e_i, Be_j\rangle = B_{ji}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we come to exactly how to recognize isometries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 2:  the matrix $A$ is an isometry iff $A^t A = Id$.  It turns out, this will automatically imply $AA^t = Id$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $A$ is an isometry.  Then $\langle x,y\rangle = \langle Ax, Ay\rangle = \langle x, A^t A y\rangle$.  Thus, we have $\langle x,y\rangle = \langle x, A^tA y\rangle$ for all $x$ and $y$.  If we set $y = x$, then we get $|x|^2 = |x||A^tA x|cos\theta$.  But since $A$ is an isometry, so is $A^t$ and hence $|A^t A x| = |x|$.  Thus, we get $|x|^2 = |x|^2 cos\theta$ which tells us $\theta = 0$, i.e, that $A^tA x = x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $A^tA = Id$, then $\langle x,y\rangle = \langle x, A^t A y\rangle = \langle Ax, Ay\rangle$, showing that $A$ is an isometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, one just verifies that every rotation matrix and reflection matrix satisfies $A^tA = Id$.  &lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-25T05:13:29.397" CommentCount="4" />
  <row Id="3269" PostTypeId="2" ParentId="2899" CreationDate="2010-08-25T06:24:28.423" Score="0" ViewCount="0" Body="&lt;p&gt;This was actually &lt;a href=&quot;http://blog.hiremebecauseimsmart.com/post/977613775/arctan#disqus_thread&quot; rel=&quot;nofollow&quot; title=&quot;my question&quot;&gt;my question&lt;/a&gt; originally.  To put the demand for intuitiveness in starker terms:  &lt;strong&gt;what does the arc of a circle have to do with the area of a square and the operation of reciprocation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone can show me a geometric figure with area $1 \over 1 + x^2$ that might be a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, they can explain how this sequence&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Times a number by itself;&lt;/li&gt;&#xA;&lt;li&gt;Add one to it;&lt;/li&gt;&#xA;&lt;li&gt;Flip that result&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;leads to &quot;a little bit of&quot; circular arc.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-08-25T06:24:28.423" CommentCount="8" />
  <row Id="3270" PostTypeId="1" AcceptedAnswerId="3288" CreationDate="2010-08-25T06:33:37.280" Score="6" ViewCount="277" Body="&lt;p&gt;I understand the technical and logical distinction between &quot;if&quot; and &quot;only if&quot; and &quot;if and only if&quot;.  But I have always been troubled by the phrase &quot;only if&quot; even though I am able to parse and interpret it.  Also in my posts on this and other sites I have frequently had to make edits to migrate the term &quot;only&quot;, sometimes across multiple structural boundaries of a sentence, which is empirical evidence to myself that I don't intuitively know the meaning of the word. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any simple rule that I can use to determine whether or not it is appropriate to use this word in a particular context in order to achieve more clarity?  In mathematical discourse, what are some other common lexical contexts, meaningful or not, in which appears the word &quot;only&quot;?  Why do I often write &quot;only&quot; in the wrong place?&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-08-25T08:52:28.513" LastActivityDate="2010-08-25T16:33:03.593" Title="What does &quot;only&quot; mean?" Tags="&lt;soft-question&gt;&lt;terminology&gt;&lt;only&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="2" />
  <row Id="3271" PostTypeId="1" AcceptedAnswerId="3274" CreationDate="2010-08-25T06:54:34.950" Score="3" ViewCount="371" Body="&lt;p&gt;How do I evaluate this function for given $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots. $&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T01:20:37.343" LastActivityDate="2010-12-03T13:53:07.570" Title="How to evaluate Riemann Zeta function" Tags="&lt;complex-analysis&gt;&lt;numerical-methods&gt;&lt;riemann-zeta&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" />
  <row Id="3272" PostTypeId="2" ParentId="2968" CreationDate="2010-08-25T07:10:43.157" Score="5" ViewCount="0" Body="&lt;p&gt;If I understand your question correctly, your problem can be seen as a multi-party extension of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Yao%27s_Millionaires%27_Problem&quot; rel=&quot;nofollow&quot;&gt;Millionaires' problem&lt;/a&gt;. In the Millionaires' problem problem, two millionaires want to know which of them is richer without revealing their wealth. There is an secure two-party protocol (due to &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.103.7844&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Andy Yao, 1982&lt;/a&gt;), and there may be even more sophisticated algorithms around these days. Maybe a simple extension of this protocol will solve your problem (every peer talks to everybody else to lean that the other guy has a higher card). However, I don't know if this simple extension is still secure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: After looking over Yao's paper I realize that he is also already addressing the multi-party problem. So you should find your solution in his paper, or follow some of the references given in &lt;a href=&quot;http://en.wikipedia.org/wiki/Secure_multi-party_computation&quot; rel=&quot;nofollow&quot;&gt;this article at Wikipedia&lt;/a&gt;. Of course, the recent homomorphic encryption scheme will also solve your problem, as has been pointed out by user Moron before. However, since the homomorphic encryption scheme is very general, it might be a little bit too complicated for your problem at hand if you are looking for a practical scheme.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-08-25T09:52:37.130" LastActivityDate="2010-08-25T09:52:37.130" CommentCount="5" />
  <row Id="3273" PostTypeId="1" CreationDate="2010-08-25T07:32:51.077" Score="6" ViewCount="175" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Weierstrass' approximation theorem for continuous functions on a compact space by using polynomials is well-known. As far as I know, there are some variants of this theorem, e.g. Stone-Weierstrass that does not refer only to polynomials as approximator functions. Where could I find these Weierstrass-like approximation theorems? On-line references are OK, but one might also point to some books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;Lucian&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2010-08-25T11:27:30.100" LastActivityDate="2010-12-24T05:29:11.530" Title="Approximation theorems" Tags="&lt;approximation&gt;&lt;approximation-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="3274" PostTypeId="2" ParentId="3271" CreationDate="2010-08-25T08:48:24.983" Score="7" ViewCount="0" Body="&lt;p&gt;Assuming you're talking about numerical evaluation, along the critical strip $0&amp;lt;\Re s&amp;lt;1$ and &quot;large&quot; $\Im s$, (which is the region of interest for many) the &lt;a href=&quot;http://www.ams.org/journals/tran/1988-309-02/S0002-9947-1988-0961614-2/S0002-9947-1988-0961614-2.pdf&quot; rel=&quot;nofollow&quot;&gt;Riemann-Siegel formula&lt;/a&gt; is standard; off the strip, what you can manage is a polyalgorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\Re s\leq0$, one can use the reflection formula for $\zeta$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(1-s)=\frac{2}{(2\pi)^s}\cos\left(\frac{s\pi}{2}\right)\Gamma(s)\zeta(s)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can consider evaluation for $\Re s&amp;gt;0$ in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $|s|$ is &quot;large enough&quot; (how large is &quot;large&quot; depends on the computing environment you're in), one can simply use the defining series, since its terms quickly diminish in magnitude. That leaves the problem of what to do for small to medium-sized $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to use the &lt;a href=&quot;http://mathworld.wolfram.com/DirichletEtaFunction.html&quot; rel=&quot;nofollow&quot;&gt;Dirichlet $\eta$ function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta(s):=-\sum_{j=1}^\infty \frac{(-1)^j}{j^s}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is related to $\zeta$ by the following identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for our interest in $\eta$ is that although this sum is slowly convergent, it is an alternating series, and a number of algorithms exist for quickly finding the sum of an alternating series numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method is the &lt;a href=&quot;http://math.stackexchange.com/questions/6625&quot;&gt;Levin transformation&lt;/a&gt;; another one, which is one of the simplest methods for numerically summing alternating series (and my personal favorite) is the Cohen-Rodriguez Villegas-Zagier algorithm. The algorithm is a bit too long to describe here, so I will just have to point you to the &lt;a href=&quot;http://projecteuclid.org/euclid.em/1046889587&quot; rel=&quot;nofollow&quot;&gt;original paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in fact identical to the approach taken by Borwein in this &lt;a href=&quot;http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P155.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T13:53:07.570" LastActivityDate="2010-12-03T13:53:07.570" CommentCount="7" />
  <row Id="3275" PostTypeId="1" AcceptedAnswerId="3282" CreationDate="2010-08-25T10:00:34.200" Score="3" ViewCount="132" Body="&lt;p&gt;I'm reading Shelah's book &quot;Proper and Improper Forcing&quot; (the first two chapters were recommended for learning the basics of forcing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a quasi-order $P$ we say that $\mathcal{I}$ is a &lt;em&gt;dense&lt;/em&gt; subset of $P$  if&#xA;$$(\forall p \in P) (\exists q \in \mathcal{I}) (p\le q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that it is open if for any $p,q \in P$ we have that $p \in \mathcal{I} \wedge p\le q$ then $q \in \mathcal{I}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is called &lt;em&gt;directed&lt;/em&gt; if every two elements in $G$ has an upper bound in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is called &lt;em&gt;downward closed&lt;/em&gt; if for every $p \in G$ and $q\in P$ if $q \le p$ then $q \in G$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset $G$ of $P$ is called &lt;em&gt;generic&lt;/em&gt; over $V$ if it is directed, downward closed and for any dense and open subset of $P$ that is in $V$, the intersection with $G$ is non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now they give proof that if $P$ has no trivial branches then a generic set cannot exist in the universe, saying that if $G \in V$ then $P\backslash G \in V$ and it is dense and open, how come? I can't figure that out.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-08-25T15:41:10.587" LastActivityDate="2010-08-25T15:46:11.657" Title="Generic sets in ZFC" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="3276" PostTypeId="2" ParentId="2899" CreationDate="2010-08-25T10:07:38.140" Score="3" ViewCount="0" Body="&lt;p&gt;Stereographic projection offers a geometric interpretation of this formula. In fact the factor&#xA;$(1+x^2)^{-1}$ is the metric of the circle in the stereographic projection coordinate. The construction is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a circle of radius 1/2, call the angle of (the radius vector of) a general &#xA;point on the circle with respect to the radius to the south pole $2\theta$.&#xA;Connect the point with the north pole and call the distance of the point of intersection of the extension&#xA;of this chord with the tangent at the south pole $x$. One can parameterize the points on the circle&#xA;by the coordinate $x$ which is called the stereographic projection coordinate. It is easy to see that $ x = tan(\theta)$.&#xA;The length of an arc of the circle between $\theta_1$ and$\theta_2$ (having the stereographic projection coordinates $x_1$ and $x_2$) is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s = 2 \int_{\theta_1}^{\theta_2} d\theta = 2\int_{x_1}^{x_2} \frac{dx}{1+x^2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction has further nice properties&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The group of plane rotations $SO(2)$ acts on $x$ by a Mobius transformation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This construction generalizes to spheres in higher dimensions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-25T10:07:38.140" />
  <row Id="3277" PostTypeId="2" ParentId="1730" CreationDate="2010-08-25T11:00:25.667" Score="5" ViewCount="0" Body="&lt;p&gt;I just wanted to add a few cents to this post to say that Isaac's statement &quot;&lt;em&gt;I've seen function machines used as low as 3rd grade&lt;/em&gt;&quot; is quite true.  I teach kindergarten and 1st grade, and I use function machines with my students, mostly when introducing the idea of complements in relation to addition and subtraction.  It's a standard part of the &quot;patterns and algebra&quot; portion of the Everyday Mathematics curriculum.  Functions are honestly NOT a challenging concept for my students to grasp when presented in this manner, hence, I am absolutely confident that your College Algebra students will be just fine! :)&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-08-25T11:00:25.667" />
  <row Id="3278" PostTypeId="1" AcceptedAnswerId="3310" CreationDate="2010-08-25T11:44:49.830" Score="3" ViewCount="222" Body="&lt;p&gt;In high school, we were often given questions of the form: &quot;What is the probability of A given B?&quot; For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the probability that two&#xA;  people were born on the same day given&#xA;  that one was born on a Tuesday?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intuitively, most people expect that knowing someone was born on a Tuesday doesn't give you any information, as it is no different from being born on a Monday or Wednesday. However, the answer given is that we are looking at pairs of people with at least one born on Tuesday (13 out of 49) and only one of these pairs has both people born on the same day, so the answer is 1 in 13. If by &quot;given one was born on Tuesday&quot;, you mean that we the birthdays of the two people are uniformly distributed between all possible pairs where at least one was born on a Tuesday&quot;, then this analysis is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the question stated just says that we are &quot;given&quot; this fact. It doesn't explain how we came to know this fact. Lets suppose someone blurted this fact out randomly. Generally it wouldn't be because they looked at groups of people and discarded them until they found a pair where at least one was born on a Tuesday. Instead, I think it would be better to model it as someone randomly getting a pair of people, then making a true statement about them. For simplicity, we will assume they statements are of the form: &quot;One of these people was born on a (INSERT DAY)&quot;. We will assume that if they were born on different days, each is equally likely. In this model, the statement that someone was born on a Tuesday actually makes no difference to the chance they were born on the same day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that part of the problem comes from giving the word &quot;given&quot; (often represented by the symbol |) a formal definition. People expect it to have the same meaning as the English word given. Anyway:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a word I can use instead of &quot;given&quot; to be clearer?&lt;/li&gt;&#xA;&lt;li&gt;Do mathematician's object to the problem being stated this way, or is this considered to be clear? If they do, why is the symbol &quot;|&quot; often pronounced as &quot;given&quot;?&lt;/li&gt;&#xA;&lt;li&gt;Have people who attended high school in other countries had this difference made clear to them?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-25T21:53:00.590" LastActivityDate="2010-08-26T11:48:14.930" Title="Probability of A given B" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="3279" PostTypeId="2" ParentId="3273" CreationDate="2010-08-25T13:40:38.083" Score="3" ViewCount="0" Body="&lt;p&gt;For approximation with Polynomials, a Weierstrass like theorem is the &lt;a href=&quot;http://mathworld.wolfram.com/MuentzsTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Muntz's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving away from polynomials, we have the classic &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series&quot; rel=&quot;nofollow&quot;&gt;Fourier Series&lt;/a&gt;. The &lt;a href=&quot;http://mathworld.wolfram.com/GeneralizedFourierSeries.html&quot; rel=&quot;nofollow&quot;&gt;Generalization&lt;/a&gt; of Fourier series gives rise to many approximation schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I wasn't able to find a single page...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-25T13:40:38.083" CommentCount="3" />
  <row Id="3280" PostTypeId="2" ParentId="3273" CreationDate="2010-08-25T13:47:12.893" Score="2" ViewCount="0" Body="&lt;p&gt;Functions belonging to reproducing kernel Hilbert spaces can be approximated by weighted discrete sums of the reproducing kernels evaluated at discrete points of the dual variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following two articles: &lt;a href=&quot;http://math.bu.edu/people/mkon/M510-1-05.pdf&quot; rel=&quot;nofollow&quot;&gt;article-1&lt;/a&gt; &lt;a href=&quot;http://www.nt.ntnu.no/users/skoge/prost/proceedings/ifac2002/data/content/00753/753.pdf&quot; rel=&quot;nofollow&quot;&gt;article-2&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-08-25T13:47:12.893" />
  <row Id="3281" PostTypeId="2" ParentId="3270" CreationDate="2010-08-25T14:40:51.300" Score="2" ViewCount="0" Body="&lt;p&gt;I think analogies in plain English are the way to internalize this... so here's one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you want to wear socks with your shoes, put your shoes on &lt;em&gt;only if&lt;/em&gt; you have already put your socks on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that &lt;em&gt;there is no other way&lt;/em&gt; to arrive at the state of having your socks and shoes on (aside from the ridiculous possibility of placing your socks over your shoes).&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-25T14:40:51.300" CommunityOwnedDate="2010-08-25T14:40:51.300" CommentCount="0" />
  <row Id="3282" PostTypeId="2" ParentId="3275" CreationDate="2010-08-25T15:35:09.833" Score="2" ViewCount="0" Body="&lt;p&gt;I'm going to use the def of &quot;directed&quot; that I found in Shelah's &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.pl/1235419814&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt;:  A set $G$ is directed if it is downward closed (if $p$ in G and if $q\leq p$ then $q\in G$) &lt;strong&gt;and&lt;/strong&gt; every two memebers of $G$ have a common upperbound in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $P-G$ is open is not too hard:  Let $p\in P-G$ and suppose $q\in G$ with $p\leq q$.  If $q$ were in $G$, then since $G$ is downwardly closed, we'd have $p\in G$, but that contradicts $p\in P-G$.  Thus, $q\notin G$ so $q\in P-G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why is $P-G$ dense?  Well, let $p\in P$.  We want to find a $q\in P-G$ with $p\leq q$.  Since there are no trivial branches, we know that above $p$ and are two points $r$ and $s$ with no common upperbound.  Since $G$ is directed, $r$ and $s$ cannot both be in $G$.  Hence, say, $r\notin G$.  Then $r\in P-G$, so we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;I'm using chapter 1 and finding the definition on page 3 of the .pdf.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-25T15:46:11.657" LastActivityDate="2010-08-25T15:46:11.657" CommentCount="4" />
  <row Id="3283" PostTypeId="1" AcceptedAnswerId="3286" CreationDate="2010-08-25T16:00:33.537" Score="11" ViewCount="542" Body="&lt;p&gt;I asked myself this question when I was a young boy playing around with the calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, I think I know the answer, but I'm not sure whether I'd be able to explain it to a child or layman playing around with a calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I'm interested in answers suitable for a person that, say, knows what the square root of a square number is, but doesn't know about sequences, functions, convergence and the like.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-08-26T00:00:03.893" Title="&quot;Why do I always get 1 when I keep hitting the square root button on my calculator?&quot;" Tags="&lt;calculus&gt;&lt;convergence&gt;&lt;calculator&gt;" AnswerCount="7" CommentCount="7" />
  <row Id="3284" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T16:11:39.617" Score="10" ViewCount="0" Body="&lt;p&gt;Make a picture of the usual spiral converging to the fixed point, in the style of &lt;img src=&quot;http://i.imgur.com/WyeV5.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this little obsession aparent on this site with 'layman' explanations? :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB.&lt;/strong&gt; The picture comes from the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Fixed_point_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;fixed points&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later.&lt;/strong&gt; To make the picture it is better to use the cosine (so you do not even have to &lt;em&gt;make&lt;/em&gt; the picture because that is what's in the wikipedia page) mainly because the iteration is seen more clearly: for the square root, the sequence converges boringly and fast to be interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS.&lt;/strong&gt; You can draw this kind of pictures, assuming you have acces to &lt;em&gt;Mathematica&lt;/em&gt;, with the following code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[t_] := Cos[t];&#xA;g = Plot[{f[t], t}, {t, -\[Pi]/2, \[Pi]/2}, &#xA;  AspectRatio -&amp;gt; 1, PlotStyle -&amp;gt; {Thick}, Epilog -&amp;gt;&#xA;   Module[{pts = NestList[f, .1, 10]},&#xA;    {&#xA;     Thin, PointSize[0.015],  &#xA;     Line@ Flatten[Map[{#, {#[[2]], #[[2]]}} &amp;amp;, Partition[pts, 2, 1], 1], 1],&#xA;     Red, Point[Partition[pts, 2, 1]]&#xA;     }&#xA;   ]&#xA;]    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-25T16:42:41.767" LastActivityDate="2010-08-25T16:42:41.767" CommentCount="5" />
  <row Id="3286" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T16:22:10.600" Score="10" ViewCount="0" Body="&lt;p&gt;Here is how I had justified it to myself when I was a kid (I was convinced :-))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x &gt; 1$ then $\sqrt x &gt; 1$ and $x &gt; \sqrt x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we keep reducing the number while still being $&gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we cannot end up at a number $&gt;1$ as then taking the square root would reduce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do end up at $1$, we stay there. Since the calculator has a limited precision, we end up at $1$, and pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the assumption here is that we do end up somewhere :-) &#xA;(Which seemed justified by the fact that the first non-zero digit after the decimal point always seemed to reduce)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T16:27:34.983" LastActivityDate="2010-08-25T16:27:34.983" CommentCount="4" />
  <row Id="3287" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T16:25:27.060" Score="6" ViewCount="0" Body="&lt;p&gt;If they know about logarithms, try this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;taking repeated square roots of a positive real number is the same as repeatedly dividing its logarithm by 2,&lt;/li&gt;&#xA;&lt;li&gt;repeatedly dividing something by 2 gets you to 0, and&lt;/li&gt;&#xA;&lt;li&gt;the exponential of 0 is 1.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-25T16:25:27.060" CommentCount="4" />
  <row Id="3288" PostTypeId="2" ParentId="3270" CreationDate="2010-08-25T16:33:03.593" Score="8" ViewCount="0" Body="&lt;p&gt;The terms &quot;only&quot; and &quot;only if&quot; mean different things.  The first is English and the second is mathematical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;only&quot; in English means &quot;just this and not others&quot;, where &quot;others&quot; could be an object or an action, depending on what &quot;only&quot; is connected with.  For example, &quot;We only proved that $p$ is prime&quot; means you showed $p$ is prime but not anything further, which I think is synonymous with &quot;We proved only that $p$ is prime&quot;; it is just a stylistic judgment as to which of those you use and I prefer the second version at the moment. However, &quot;We proved that only $p$ is prime&quot; means you proved $p$ is prime while other numbers in the argument (that may possibly have been prime) are not prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical example, which is not meant as a personal remark, is to insert &quot;only&quot; in front of each word of the sentence &quot;I love you&quot;. Each version has a completely different meaning.  This is discussed under the topic &quot;Modifiers&quot; on the page&#xA;&lt;a href=&quot;http://www.kaptest.com/GMAT/Learn-and-Discuss/Community/blogs/tag/verbal/&quot; rel=&quot;nofollow&quot;&gt;http://www.kaptest.com/GMAT/Learn-and-Discuss/Community/blogs/tag/verbal/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In math, &quot;only if&quot; has the same meaning as &quot;implies&quot;.  (Logically, it's the direction that is not &quot;if&quot;, which is how I first was able to remember its meaning.) In terms of Tom's example, &quot;shoes are on ==&gt; socks are on&quot; has the same meaning as &quot;your shoes are on only if your socks are on.&quot;&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-08-25T16:33:03.593" />
  <row Id="3289" PostTypeId="2" ParentId="2952" CreationDate="2010-08-25T16:41:11.567" Score="2" ViewCount="0" Body="&lt;p&gt;This is supposed to be a comment, but I don't seem to be able to write one :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite sure it is an open problem : it was a few years ago when I last checked up on it. It's a very difficult one. There are indeed sufficient conditions, but necessary and sufficient ones seem very hard to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you take a look at the recent work of Kuznetsova, Yu. N. on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-25T16:41:11.567" CommentCount="2" />
  <row Id="3290" PostTypeId="2" ParentId="3254" CreationDate="2010-08-25T16:50:54.517" Score="2" ViewCount="0" Body="&lt;p&gt;As an exercise in trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OWu6S.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation of rectangular co-ordinates $(x,y)$ of $P$ into the&#xA;rectangular co-ordinates $(x',y')$ of $P$ by a &lt;em&gt;rotation of the axes&lt;/em&gt; is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=x^{\prime }\cos \phi -y^{\prime }\sin \phi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=x^{\prime }\sin \phi +y^{\prime }\cos \phi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This transformation is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This way of proving the transformation of rectangular co-ordinates does not &quot;avoid using Euclidean geometry explicitly&quot; (in this case trigonometry), as writen in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the same method can be used to derive the transformation of rectangular into spherical co-ordinates, which is a non-linear transformation.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-25T20:42:21.720" LastActivityDate="2010-08-25T20:42:21.720" />
  <row Id="3291" PostTypeId="1" AcceptedAnswerId="3292" CreationDate="2010-08-25T16:52:54.200" Score="3" ViewCount="90" Body="&lt;p&gt;When calculating some probabilities, I got sums of the form&#xA;$$\sum_{j=0}^c {j+a+b \choose a} p^j,$$ for integers $a, b, c &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does someone know closed forms for these values?&lt;/p&gt;&#xA;" OwnerUserId="669" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-25T17:47:20.837" LastActivityDate="2010-08-26T04:20:07.293" Title="Estimates involving sums with binomials" Tags="&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="3292" PostTypeId="2" ParentId="3291" CreationDate="2010-08-25T16:59:16.430" Score="3" ViewCount="0" Body="&lt;p&gt;I don't have time right now to work out the exact answer, but the methods and identities in the middle third of &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/blog.php?u=14052&amp;amp;b=10590&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; will give a closed form for fixed $a$.  Use the geometric series formula to find $\sum_{j=0}^c p^{a+b+j}$, differentiate $a$ times with respect to $p$, then divide by $a! p^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a closed form for fixed $a$ isn't good enough, you should probably be more precise about which of your parameters are large and which are small.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Still don't have time to give a complete answer, but here's a fun trick.  Instead of computing the answer for fixed $a$ we can write down a generating function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle P_{b,c}(x) = \sum_{a=0}^{\infty} x^a \sum_{j=0}^c {a+b+j \choose a} p^j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then exchange the order of summation, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \begin{align}&#xA;P_{b,c}(x) &amp;amp;= \sum_{j=0}^c p^j \sum_{a=0}^{\infty} {a+b+j \choose a} x^a \\&#xA; &amp;amp;= \sum_{j=0}^c p^j \frac{1}{(1 - x)^{b+j+1}} \\&#xA; &amp;amp;= \frac{1}{(1 - x)^{b+1}} \sum_{j=0}^c \left( \frac{p}{1-x} \right)^j \\&#xA; &amp;amp;= \frac{1}{(1 - x)^{b+1}} \frac{1 - \left( \frac{p}{1-x} \right)^{c+1}}{1 - \frac{p}{1-x}} \\&#xA; &amp;amp;= \frac{(1-x)^{c+1} - p^{c+1}}{(1 - x)^{b+c+1}(1 - p - x)}. \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the coefficient of $x^a$ of this rational function is the number you want.  Not sure how useful that is for you, but you might be able to extract a useful asymptotic from it.  The identity in the second line is the binomial theorem for negative exponents.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-26T04:20:07.293" LastActivityDate="2010-08-26T04:20:07.293" CommentCount="2" />
  <row Id="3293" PostTypeId="2" ParentId="3291" CreationDate="2010-08-25T17:00:57.120" Score="0" ViewCount="0" Body="&lt;p&gt;If you don't need an exact closed form or just care about the asymptotic growth &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's Approximation&lt;/a&gt; might be useful. I've found it useful in the past for generating good asymptotic bounds on similar functions.&lt;/p&gt;&#xA;" OwnerUserId="1487" LastActivityDate="2010-08-25T17:00:57.120" CommentCount="1" />
  <row Id="3294" PostTypeId="1" AcceptedAnswerId="3983" CreationDate="2010-08-25T17:45:07.927" Score="3" ViewCount="268" Body="&lt;p&gt;I subdivide a cubic bezier curve at a given &lt;em&gt;t&lt;/em&gt; value using de Casteljau’s algorithm, which yields two cubic bezier curves. Afterwards I “scale” the second curve (proportionally).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’d like to reconnect or approximate the two curves to/with a single curve in a third step. Is that possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ujFOM.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates what I’m intending to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess reversing de Casteljau’s algorithm won’t work because I don’t have one of the intermediate points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are multiple approaches, I’d favor a simpler (faster to compute) strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;Maybe this figure makes it more clear; it shows all the points I have:&#xA;&lt;img src=&quot;http://i.imgur.com/0hh6f.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original cubic bezier curve is defined by the points $ p_{0}, p_{1}, p_{2}, p_{3} $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is divided at a given $ t $ (timing) value using De Casteljau’s algorithm, which yields the points $ q_{1}, r_{2}, i_{1}, q_{2}, r_{1}, k $ where $ k $ is the division point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two subcurves are defined by the control points $ p_{0}, q_{1}, q_{2}, k $ and $ k, r_{1}, r_{2}, p_{3} $, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scaled second subcurve is defined by the points $ k, {r}' _{1}, {r}' _{2}, {r}' _{3} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scaling is applied as follows: $ {p}' = k + (p - k) \cdot factor $ for $ r_{1}, r_{2}, p_{3} $&lt;/p&gt;&#xA;" OwnerUserId="1488" LastEditorUserId="1488" LastEditorDisplayName="" LastEditDate="2010-09-01T13:11:56.463" LastActivityDate="2010-09-03T22:36:41.800" Title="How to approximate/connect two continuous cubic bezier curves with/to a single one?" Tags="&lt;approximation&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="3295" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T18:03:26.050" Score="4" ViewCount="0" Body="&lt;p&gt;Dualizing reduces it to: repeatedly squaring $1+\epsilon$ goes to $\infty$. But that's clear by the Binomial theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely: given any positive $\;\epsilon, \;$ we can force $\;\;\; 2^n\epsilon &gt; x \;\;$ by choosing $\: n \:$ large enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we have that $\;\;\; (1+\epsilon)^{2^n} \ge 1+2^n \epsilon &gt; x \ge 1\quad$ by the Binomial Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, taking $\;2^n\:$'th roots: $\;\;\; 1+\epsilon \;\ge\; x^{1/2^n} \;\ge\; 1. \;\;$ Hence $\; x^{1/2^n}\to 1\;$ as $\;n\to\infty$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-25T20:36:45.307" LastActivityDate="2010-08-25T20:36:45.307" CommentCount="2" />
  <row Id="3296" PostTypeId="2" ParentId="1236" CreationDate="2010-08-25T18:11:37.750" Score="1" ViewCount="0" Body="&lt;p&gt;One interesting fact told to me by my father, a chem engineer, is that if you have a high pressure gas leaking into a low pressure gas through a small hole, there is a upper limit to the rate of flow. That is, no matter how high the pressure gets on the high pressure side, the rate of flow does not surpass some finite limit. I suppose this relates to the speed of sound, but I don't know. I also don't know if there a corresponding phenomenon with liquids.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-08-25T18:11:37.750" />
  <row Id="3297" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T18:22:12.597" Score="0" ViewCount="0" Body="&lt;p&gt;I think a better question is to analyze how the roots of a number X approach 1, depending on X.  I.e. x&lt;sup&gt;a&lt;/sup&gt; ~ 1 + a(lnX).  Not only does this introduce the log, it gives a natural derivation of e which is easy to remember.  I never understood what was &quot;natural&quot; about the natural log until I was an undergraduate, and I think the real value of this sequence is the way it leads so memorably and naturally to the derivation of e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see why you arrive at one from repeatedly taking square roots (as moron answered), since the sequence is strictly decreasing and bounded below by 1.  I think a motivated person of almost any age can understand that, as long as it's explained slowly and clearly.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-25T18:22:12.597" />
  <row Id="3299" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T18:57:33.837" Score="2" ViewCount="0" Body="&lt;p&gt;Ok, this one is pretty easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the theory of orthogonal matricies on wiki.  The basic idea is that the matrix corresponding to the isometric linear transformation must be orthogonal.  The proof is really pretty, but basically the idea is to show that an inner product of two vectors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(X,Y) is equal to (AX,AY) for a matrix A, if and only if the transpose of A is the same as it's inverse.  i.e. A is orthogonal.  The next step is to show that the rotation matrix is orthogonal, and that's a rigorous proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea is easily generalized to non-euclidean metrics.  Although it might be at too high a level, I absolutely love matrix lie groups by brian C hall.  The first chapter is on &lt;a href=&quot;http://books.google.com/books?id=0SQRcucskaIC&amp;amp;printsec=frontcover&amp;amp;dq=lie+groups+brian+hall&amp;amp;hl=en&amp;amp;ei=Kmd1TK2GDoH98AbApuWnBw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDEQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;google books&lt;/a&gt;, and contains some simple proofs.  This book only gets super hard in chapter 3;lie groups/algebras is one of my favorite subjects and is certainly where this line of questioning will lead you.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-08-25T18:57:33.837" />
  <row Id="3300" PostTypeId="2" ParentId="3119" CreationDate="2010-08-25T19:05:46.920" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure having really understood your question, but it seems to me that mathematicians make no difference between heterogeneous and homogeneous tuples. Consider $\left(5,b,f,8,3\right)$: one can say it is an heterogeneous tuple because the set $\left\{5,b,f,8,3\right\}$ can be partitioned into $\left\{b,f\right\}$ and $\left\{5,8,3\right\}$, letters and digits respectively. But are letters and digits of different type? What do you mean by type? In mathematics, as far as I know, there is only one type: the set type. Everything is a set. In this example letters and digits should be both defined as (particular) sets. So every tuple is homogeneous by default. Also, consider $\left(2,1,9,7\right)$: there are only digits this time, so one can say it is an homogeneous tuple. But what if I split the set of all digits into those less then $5$ (the &quot;low&quot; digit type) and those equal to or greater than $5$ (the &quot;high&quot; digit type)? It becomes heterogeneous. I would say the context is of great importance here.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2010-08-25T19:05:46.920" CommentCount="1" />
  <row Id="3301" PostTypeId="2" ParentId="3250" CreationDate="2010-08-25T19:08:45.050" Score="7" ViewCount="0" Body="&lt;p&gt;Length, in Euclidean geometry, is a relation, not a number. We say that two segments are of the same length if they are congruent to each other, and congruence is one of the undefined notions of Euclidean geometry. The Pythagorean theorem comes about as soon as you decide to map lengths to positive real numbers, and it does so because of how length and angles are related by the congruence and similarity axioms (e.g. how the Pythagorean theorem is proved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lengths, in vector spaces, can be anything that satisfies the axioms for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Normed_vector_space&quot; rel=&quot;nofollow&quot;&gt;norm&lt;/a&gt;. But to have a Pythagorean theorem, you need a notion of perpendicularity of vectors, that is, &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_product&quot; rel=&quot;nofollow&quot;&gt;an inner product&lt;/a&gt;, which exists only if the norm satisfies &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallelogram_law&quot; rel=&quot;nofollow&quot;&gt;the parallelogram law&lt;/a&gt;. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_metric&quot; rel=&quot;nofollow&quot;&gt;the taxicab norm&lt;/a&gt;, which is given by |(a,b)|=|a|+|b| does not come from an inner product and there is no Pythagorean theorem for that notion of length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural question is why does the Pythagorean theorem in Euclidean geometry correspond to the Pythagorean theorem in a finite dimensional vector space over the reals with an inner product? (there is a theorem that states that in a finite-dimensional vector space there is only one inner product up to isomorphism, so we really can talk about THE Pythagorean theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that the finite dimensional vector space over the reals with an inner product encodes the congruence and similarity axioms of Euclidean geometry. The basic exposition of (most of) this can be found in the first half of Chapter 2 of Emil Artin's wonderful book Geometric Algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a geometry of points and lines for which the following statements are true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any two points determine a line.&lt;/li&gt;&#xA;&lt;li&gt;Parallel postulate: for any point p not on line l&lt;sub&gt;1&lt;/sub&gt; there exists a unique line l&lt;sub&gt;2&lt;/sub&gt; that passes through p but doesn't intersect l&lt;sub&gt;1&lt;/sub&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Non-triviality: there exist 3 non-collinear points.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Desargues%27_theorem&quot; rel=&quot;nofollow&quot;&gt;Desargues' Theorem&lt;/a&gt; (the wiki article is horrible)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first three statements allow you to define translations (transformations of the plane that send lines to parallel lines and have no fixed points), while the fourth allows you to a construct a field over which the space of translations is a 2-dimensional vector space, and such that if you associate translations OP and OQ with (0,1) and (1,0), then any point R has a translation OR which can be written as (a,b) with a and b in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, any affine Desarguesian plane (thing that satisfies statements 1,2,3 and 4) can be identified with a 2-dimensional vector space over some field (and conversely any 2-dimensional vector space over some field corresponds to an affine Desarguesian plane). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright. Now, Euclidean geometry satisfies Desargues' Theorem, so which field does it correspond to? Well, it turns out that the fact that the geometry is ordered (i.e. we have a notion of an ordering of points on a line) means that the field has to be ordered and thus is a subfield of the real numbers (this is where we need something like the continuity or Hilbert's completeness axiom which roughly state that things that should intersect do intersect, and imply that the base field is all of the real numbers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the whole inner product shebang turns out to just be a codification of the various congruence axioms. The norm is given by choosing some vector to make a unit vector (segment) and then considering to which vector in the same direction other vectors (segments) are congruent to. The inner product is a matter of encoding the notion of angle between vectors as an inner product, i.e. as a function linear in both vectors, which essentially takes the notions of similarity, congruences and unit circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that you can do all of this in the reverse direction, and so the Euclidean plane really is a 2-d vector space over the reals with an inner product.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Addendum: a comment on symmetry groups that is too long to leave as a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every norm on a vector space has its own unit blob, that is, the set of vectors with norm less than one. Geometrically, blobs that correspond to a norm satisfy the following properties: they are convex, they are absorbing (every vector is a multiple of a vector in the blob), and they don't contain any lines through the origin. It is a theorem that any such blob corresponds to a norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, the transformations of a vector space are linear transformations, which means that the symmetry group of an object in your vector space is going to consist of the linear transformations that bijectively send an object to itself. In other words, the group of symmetries of an object is the set of invertible linear transformations that fix that object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the blob of the Euclidean norm is the unit disk, and the unit circle is the boundary of the disk. The unit circle is the set of vectors with Euclidean norm 1, and in general, the boundary of the unit blob for a norm will be the set of vectors with norm 1. The group of symmetries of the boundary of the blob then will be the set of invertible linear transformations that send vectors with norm 1 to vectors with norm 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slight caveat is that you don't want to consider just any old invertible linear transformations because when you add a norm to a vector space, you are in effect defining a topology, that is, a way to define continuity. As a result you want to work with &lt;em&gt;continuous&lt;/em&gt; linear transformations. BUT! In a finite dimensional vector space, no matter what the norm, we have that &lt;em&gt;every&lt;/em&gt; linear transformation is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that for any norm on, say the 2d vector space over the reals, the symmetry groups of the boundary of the unit blob are all invertible linear transformations of the same space. Hence, we can directly compare these groups to each other since they live in the same space (of invertible linear transformations of R&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: the boundary of the unit blob for the Euclidean norm is the unit circle, i.e. all points such that x&lt;sup&gt;2&lt;/sup&gt;+y&lt;sup&gt;2&lt;/sup&gt;=1. The boundary of the unit blob of the Taxicab norm is the unit diamond, i.e. all points such that |x|+|y|=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you think about it, any linear transformation that sends the unit diamond to itself will also send the unit circle to itself. Hence, the symmetry group for the Euclidean norm contains the symmetry group for the Taxicab norm, and in fact it contains it properly since, for example, rotation by 45 degrees fixes the circle, but doesn't fix the diamond.&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-08-26T21:19:15.677" LastActivityDate="2010-08-26T21:19:15.677" CommentCount="6" />
  <row Id="3302" PostTypeId="2" ParentId="3119" CreationDate="2010-08-25T19:10:08.443" Score="3" ViewCount="0" Body="&lt;p&gt;Tuples appear in essentially all formal treatments of set theory, and those go back to way before the 70s!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-25T19:10:08.443" CommentCount="2" />
  <row Id="3303" PostTypeId="2" ParentId="3231" CreationDate="2010-08-25T19:17:48.303" Score="1" ViewCount="0" Body="&lt;p&gt;There's a fast procedure involving no intermediate representation.  You need only four lookup tables totaling 544 entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier answers have established that the conversion can be done in blocks of 3 (working right to left).  Consider the rightmost block in your example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0xA03 = 1010 0000 0011 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to break this binary string into four groups of three bits, which I will number 1, 2, 3, 4 from the right to the left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: 011 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: 000 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: 000 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: 101 B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first depends only on the rightmost hex digit 0x3.  The second depends only on the two rightmost digits 0x03 (it gets its rightmost bit from the 0x3 and its first two bits from 0x0).  The third depends only on the second and third digits 0xA0.  The fourth depends only on the third digit 0xA.  Whence, you only need to perform the following conversions, each of which can be stored in its own static lookup table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: 0x3 --&gt; 03 [16 entries cover all possibilities]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: (0x0, 0x3) --&gt; 00 [16 * 16 entries]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: (0xA, 0x0) --&gt; 00 [16 * 16 entries]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: 0xA --&gt; 5 [16 entries].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you repeat with the next block, padding with zeros on the left as needed.  Continuing the example, the next block is 0x001:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: 0x1 --&gt; 01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: (0x0, 0x1) --&gt; 00.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can stop here because all the original input has been consumed.  The output, working backwards, is 0015003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This directly answers both parts of the original question: it provides a simple conversion without the middle step and it avoids base 10 (which is rarely used for computer conversion anyway: usually the job is done with bit shifting and masking, essentially a binary operation).  In case this procedure still looks too much like the other proposed solutions, please note that it performs absolutely no arithmetic (apart from decrementing pointers to input and output as it proceeds): it takes a string representation (hex) as its input, uses the characters to index its tables, and outputs an octal string.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T19:17:48.303" CommentCount="2" />
  <row Id="3304" PostTypeId="2" ParentId="3283" CreationDate="2010-08-25T19:33:45.380" Score="4" ViewCount="0" Body="&lt;p&gt;A nice series of solutions has already been given.  It's hard to improve on any of them.  But because this is an elementary question, it deserves an elementary answer.  (Although &quot;elementary&quot; is subjective, being able to punch buttons on a calculator requires little mathematical sophistication, so we should attempt to minimize the formality and maximize the intuition in the explanation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you begin with a value x on the calculator with x &gt; 1, note that its square root lies between 1 and the midpoint of the interval [1, x].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proofs: The midpoint is (x+1)/2.  Because ((x+1)/2)^2 - x = ((x-1)/2)^2 &gt;= 0, the midpoint is no greater than the root of x.  Non-algebraic version: draw graphs of the relations x = y^2 and y = (1+x)/2 on the same plot and note that the latter never falls below the former.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows immediately (by induction if you want to be formal) that the iterated root, although never falling below 1, decreases faster than the iterated average with 1, which obviously (on geometric or arithmetic grounds) converges to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of 0 &amp;lt; x &amp;lt; 1 is reduced to the case 1/x &gt; 1 because Sqrt(1/x) = 1/Sqrt(x).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T19:33:45.380" />
  <row Id="3305" PostTypeId="1" AcceptedAnswerId="3307" CreationDate="2010-08-25T20:05:05.883" Score="3" ViewCount="91" Body="&lt;p&gt;A function $f$ is superadditive if $f(x) + f(y) \le f(x+y)$. The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a real number $a$ exists such that for all &#xA;real numbers with $x, y\ \ge \ a $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Gamma(x) + \Gamma(y) \le \Gamma(x+y) \quad ?$$&lt;/p&gt;&#xA;" OwnerUserId="1493" LastActivityDate="2010-08-25T20:32:05.253" Title="Is the Gamma function superadditive?" Tags="&lt;real-analysis&gt;&lt;gamma-function&gt;" AnswerCount="1" />
  <row Id="3306" PostTypeId="2" ParentId="2356" CreationDate="2010-08-25T20:07:17.217" Score="4" ViewCount="0" Body="&lt;p&gt;A simple (practical, low-computation) approach to choosing among three options with equal probability exploits the fact that in a run of independent flips of an unbiased coin, the chance of encountering THT before TTH occurs is 1/3.  So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flip a coin repeatedly, keeping track of the last three outcomes.  (Save time, if you like, by assuming the first flip was T and proceeding from there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stop whenever the last three are THT or TTH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the last three were THT, select option 1.  Otherwise flip the coin one more time, choosing option 2 upon seeing T and option 3 otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T20:07:17.217" />
  <row Id="3307" PostTypeId="2" ParentId="3305" CreationDate="2010-08-25T20:16:34.340" Score="4" ViewCount="0" Body="&lt;p&gt;$a = 2$ will do, because (letting $x \ge y$ wlg),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma(x+y) \ge \Gamma(x+2) = (x+1)x \Gamma(x) \ge 6 \Gamma(x) \ge \Gamma(x) + \Gamma(x) \ge \Gamma(x) + \Gamma(y)$. &lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-25T20:32:05.253" LastActivityDate="2010-08-25T20:32:05.253" CommentCount="8" />
  <row Id="3308" PostTypeId="2" ParentId="3271" CreationDate="2010-08-25T20:32:52.090" Score="2" ViewCount="0" Body="&lt;p&gt;You should look at the work of Wadim Zudilin. In particular you should look at &#xA;&lt;a href=&quot;http://wain.mi.ras.ru/PS/zeta5-11%24.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational&quot; (Turpion link, pdf 91k, gzip ps 80k) in Russian Math. Surveys  56:4 (2001), 774--776;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-08-25T20:32:52.090" CommentCount="4" />
  <row Id="3309" PostTypeId="2" ParentId="3264" CreationDate="2010-08-25T20:35:02.703" Score="4" ViewCount="0" Body="&lt;p&gt;You implicitly define rotations, translations, and reflections as certain kinds of affine transformations.  Thus it suffices to check algebraically that they preserve distances.  (There is no need to check that they preserve angles: that follows immediately by polarization.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not terribly deep.  More insight is afforded through the usual geometric demonstration that reflections and translations are composed of a finite number of reflections.  That a reflection is an isometry is not only geometrically obvious but also easy to show algebraically.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-25T20:35:02.703" />
  <row Id="3310" PostTypeId="2" ParentId="3278" CreationDate="2010-08-25T20:58:55.200" Score="3" ViewCount="0" Body="&lt;p&gt;It is very true that statements and problems on conditional probability are often presented in an ambiguous way. The problem isn't specifically with the term &quot;given&quot;, I believe, but rather with the fact that the presentation does not make it clear what the sample space is and what are the distributions (typically some things are assumed, without comment, to be uniform).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, specifically to answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No, not that I'm aware of. If the variables are clearly defined, the use of &quot;given&quot; to indicate conditional probability is common and perfectly fine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I understand the question correctly (there seems to be a word missing), then most mathematicians do not regard such phrasings as you have mentioned to be sufficiently clear. See &lt;a href=&quot;http://blog.tanyakhovanova.com//?p=234&quot; rel=&quot;nofollow&quot;&gt;Peter Winkler's comments&lt;/a&gt; on exactly this kind of problems.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not in the country I went to high school in :-)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-25T20:58:55.200" CommentCount="1" />
  <row Id="3311" PostTypeId="1" AcceptedAnswerId="3313" CreationDate="2010-08-25T21:12:06.687" Score="1" ViewCount="79" Body="&lt;p&gt;This is a follow-up to the MathStackexchange question 3305 and MathOverflow question 23229.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bohr-Mollerup theorem states that the Gamma function is the unique function that satisfies: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\ f(x+1) = xf(x), \ (2)\ f(1) = 1, \ (3)\ \ln \circ f \ $ is convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a function $f\colon \mathbb{R^{+}} \to \mathbb{R}$  that satisfies: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\ f(x+1) = xf(x), \ (2)\ f(1) = 1, \ (3)\ $ f is superadditive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a (natural) condition which makes this function unique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: The answer is 'no' as Moron explains. And if we assume $f$ superadditive only for $x,y \ge a$ for some real $a$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I accept Moron's answer because it the correct answer to my question. Intended was the question in the sense of my first edit (and my first question). I am also curious to see an answer if condition (3) reads: (3'') $\ \ln \circ f \ $ is superadditive. Thanks to whuber for suggesting this.&lt;/p&gt;&#xA;" OwnerUserId="1493" LastEditorUserId="1493" LastEditorDisplayName="" LastEditDate="2010-08-26T21:05:24.057" LastActivityDate="2010-08-26T21:05:24.057" Title="Importance of Superadditivity of the Gamma Function" Tags="&lt;real-analysis&gt;&lt;gamma-function&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3312" PostTypeId="2" ParentId="3278" CreationDate="2010-08-25T21:13:24.437" Score="1" ViewCount="0" Body="&lt;p&gt;I'm hesitant to use an answer to respond, instead of a comment, because the answer seems to merely be a matter of semantics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a similar question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What are the odds that two people are of the same gender?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And its counterpart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What are the odds that two people of the same gender, if we knew that one of them is a boy?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are inherently different.  The answer to the first can be seen by charting all possibilities within the limitations (there are no limitations), and it can be seen that it's 2/4, or 50%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second can be seen by charting all possibilities and eliminating the ones that aren't allowed (two girls), and we'll see that the answer is 1/3, or about 33%.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a somewhat intuitive answer of &quot;how&quot; we know something, we can use a statistical model.  Let's say we wanted to find the odds of the second question.  We could conduct a survey, asking a perfect world: (where both genders are equally likely)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Anybody who has two children, with one being a boy, answer if they have two boys or not.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This survey would give you the 33% that our chart-all-the-possibilities method would lead us to expect.  There are no ambiguities because we see the answer very clearly real and evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is a maths site, and I brought up statistics, so I'm probably going to be killed in my sleep.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-08-25T21:13:24.437" CommentCount="1" />
  <row Id="3313" PostTypeId="2" ParentId="3311" CreationDate="2010-08-25T21:30:03.417" Score="4" ViewCount="0" Body="&lt;p&gt;There is no such function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(2) = f(1+1) = 1&amp;times;f(1) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By superadditivity $f(1+1) \ge f(1) + f(1)$ i.e. $ 1 = f(2) \ge 2f(1) = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not possible.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-25T21:30:03.417" />
  <row Id="3314" PostTypeId="1" CreationDate="2010-08-25T23:49:05.113" Score="2" ViewCount="167" Body="&lt;p&gt;Given &lt;code&gt;e1..e5&lt;/code&gt; as the exponents of the base units, are there any well known methods for producing a simplified form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example &lt;code&gt;m^2*kg*s^-1&lt;/code&gt; simplifies to &lt;code&gt;W*s^2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set of derived units as Name/exponent-vector pairs as input and it can be assumed all exponents are rational (and in fact most being integers and almost all not having a denominator more than about 3-4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this can be reduced to a linear algebra problem but it is VERY under defined and in fact has several possible solutions.&lt;/p&gt;&#xA;" OwnerUserId="1171" LastEditorUserId="1171" LastEditorDisplayName="" LastEditDate="2010-08-26T00:02:22.610" LastActivityDate="2010-12-24T01:44:25.940" Title="How to determine the units of a value." Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3315" PostTypeId="1" CreationDate="2010-08-25T23:50:35.547" Score="9" ViewCount="833" Body="&lt;p&gt;If i is the square root of -1, is the square root of i imaginary? Is it used or considered often in mathematics? How is it notated?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T00:24:42.643" LastActivityDate="2010-08-31T13:10:23.783" Title="what is the square root of i?" Tags="&lt;complex-numbers&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="1" />
  <row Id="3316" PostTypeId="2" ParentId="3315" CreationDate="2010-08-25T23:56:00.383" Score="10" ViewCount="0" Body="&lt;p&gt;$i^\frac1{2}=\left(e^{\pi i/2}\right)^\frac1{2}=e^{\pi i/4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{\pi i/4}=\cos\left(\frac{\pi}{4}\right)+i \sin\left(\frac{\pi}{4}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or simplified,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1+i}{\sqrt{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course the &quot;principal value&quot;; the other value (thanks Matt E!) is the &quot;negative square root&quot;, $-\frac{1+i}{\sqrt{2}}$ or in exponential form, $-e^{\pi i/4}=e^{-3\pi i/4}$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T01:36:29.243" LastActivityDate="2010-08-26T01:36:29.243" CommentCount="5" />
  <row Id="3318" PostTypeId="2" ParentId="3283" CreationDate="2010-08-26T00:00:03.893" Score="0" ViewCount="0" Body="&lt;pre&gt;&lt;code&gt;if(x == 0): sqrt(x) == 0&#xA;if(0 &amp;lt; x &amp;lt; 1): x &amp;lt; sqrt(x) &amp;lt; 1&#xA;if(x == 1): sqrt(x) == 1&#xA;if(1 &amp;lt; x ): 1 &amp;lt; sqrt(x) &amp;lt; x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Except for the cases &lt;code&gt;x == 0&lt;/code&gt; and &lt;code&gt;x == 1&lt;/code&gt;, &lt;code&gt;x != sqrt(x)&lt;/code&gt; so it must move, and it must move towards 1, so it will approach 1 to any accuracy you choose.&lt;/p&gt;&#xA;" OwnerUserId="1171" LastActivityDate="2010-08-26T00:00:03.893" CommentCount="2" />
  <row Id="3319" PostTypeId="1" CreationDate="2010-08-26T00:00:28.557" Score="5" ViewCount="343" Body="&lt;p&gt;I've heard a lot about this number e. Why is it so important? How does it fit into the 'bigger picture' of mathematics? How is it calculated and used?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T00:03:03.863" LastActivityDate="2010-08-28T23:31:39.980" Title="Why is the number e so important in mathematics?" Tags="&lt;calculus&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="3320" PostTypeId="1" CreationDate="2010-08-26T00:05:25.460" Score="4" ViewCount="115" Body="&lt;p&gt;Pi appears a LOT in trigonometry, but only because of its 'circle-significance'. Does pi ever matter in things not concerned with circles? Is its only claim to fame the fact that its irrational and an important ratio?&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2010-08-26T00:22:19.617" Title="Does the number pi have any significance besides being the ratio of a circle's diameter to its circumference?" Tags="&lt;pi&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="3321" PostTypeId="2" ParentId="3320" CreationDate="2010-08-26T00:13:19.570" Score="5" ViewCount="0" Body="&lt;p&gt;It is difficult to know if a circle is not lurking somewhere, whenever there is $\pi$, but the values of the Riemann zeta function at the positive even integers have a lot to do with powers of $\pi$: see &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for the values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, you can prove that the probability that two &quot;randomly chosen&quot; positive integers are coprime is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-26T00:13:19.570" CommentCount="1" />
  <row Id="3322" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T00:19:33.767" Score="2" ViewCount="0" Body="&lt;p&gt;Quite simply, because of the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx} (e^x) = e^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in fact the primary way of defining the constant &lt;em&gt;e&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequences of this simple fact within mathematics are myriad, and can only really be properly appreciated by studying enough calculus and other related fields.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-08-26T00:19:33.767" />
  <row Id="3323" PostTypeId="2" ParentId="3320" CreationDate="2010-08-26T00:22:19.617" Score="2" ViewCount="0" Body="&lt;p&gt;$\pi$ appears in &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirlings_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;, which is not obviously related to circles.  This means that $\pi$ appears in asymptotics related to binomial coefficients, such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle {2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the probability of flipping exactly $n$ heads and $n$ tails after flipping a coin $2n$ times is about $\frac{1}{\sqrt{\pi n}}$.  This asymptotic also suggests that on average you should flip between $n + \sqrt{\pi n}$ and $n - \sqrt{\pi n}$ heads. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T00:22:19.617" CommentCount="2" />
  <row Id="3324" PostTypeId="2" ParentId="3314" CreationDate="2010-08-26T00:28:42.330" Score="0" ViewCount="0" Body="&lt;p&gt;In general the problem of &quot;simplifying&quot; units (mathematical expressions in general) has no well defined solution - it is not even a well-defined process, in most cases. For example, there are multiple ways of reducing your given example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;m^2 kg s^-1 =&amp;gt; N m s&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;m^2 kg s^-1 =&amp;gt; J s&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;m^2 kg s^-1 =&amp;gt; W s^2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best you can really do is recursively enumerate over all such possibilities and use some heuristic to choose the &quot;best&quot; (or give the option of using any of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: intuitively I would guess that these units are representing angular momentum, in which case &lt;code&gt;N m s&lt;/code&gt; probably makes most sense, though it's quite subjective.)&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-08-26T00:28:42.330" CommentCount="1" />
  <row Id="3325" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T00:36:12.893" Score="2" ViewCount="0" Body="&lt;p&gt;One of the first exposures students have to $e$ is in the formula for simple interest compounded continuously: $$A = Pe^{rt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula can be easily derived from the formula for simple interest compounded discretely: $$A = P(1 + \frac{r}{n})^{nt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation depends on the identity $$e = \lim_{x \to \infty} (1 + \frac{1}{x})^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although this may not be the &lt;em&gt;most&lt;/em&gt; important application, it is one way in which $e$ is used and calculated.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-26T00:36:12.893" />
  <row Id="3326" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T00:45:37.513" Score="5" ViewCount="0" Body="&lt;p&gt;One could easily write entire textbooks about this subject, and I think the only way to properly appreciate the answer is just to learn a lot of mathematics.  But I will mention one perspective which suggests that $e$ ought to be fundamental to mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the perspective of &lt;a href=&quot;http://ncatlab.org/nlab/show/groupoid+cardinality&quot; rel=&quot;nofollow&quot;&gt;groupoid cardinality&lt;/a&gt;.  The definition of a groupoid is slightly technical, but for the purposes of this discussion a groupoid is a collection of groups.  In many counting problems involving groups it turns out to be natural not to count all of the objects involved, but to divide each object by the number of symmetries it has.  For example, if you draw five dots in a line on a paper and then fold the paper in half through the middle dot, two pairs of the dots have been identified and one dot has one half identified with the other half; in other words, the folding symmetry has cut the middle dot in half, so one might say that we now have $\frac{5}{2} = 2 + \frac{1}{2}$ dots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basic idea behind groupoid cardinality: the cardinality of a groupoid is the sum $\sum_x \frac{1}{|\text{Aut}(x)|}$, where the sum runs over all the isomorphism classes of objects of the groupoid (here, over all the groups in the collection, where $|\text{Aut}(x)|$ is the size of the corresponding group).  All groups here are finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is the fundamental statement.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;The groupoid cardinality of the groupoid of finite sets (e.g. the collection of symmetric groups $S_1, S_2, ...$) is $e = \sum_{n \ge 0} \frac{1}{n!}$.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, $e$ somehow embodies a fundamental property of finite sets.  It is possible to think about the whole theory of exponential generating functions in this way, in particular to think about $e^x$ and its derivative property this way.  Part of this story is described, I believe, in John Baez's &lt;a href=&quot;http://math.ucr.edu/home/baez/groupoidification/&quot; rel=&quot;nofollow&quot;&gt;the Tale of Groupoidification&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the more concrete ways to think about the relationship between the symmetric groups and $e$ is via &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;this generating function&lt;/a&gt;.  The identity described in that blog post, which sometimes goes by the name of the &quot;exponential formula,&quot; explains, among many &lt;em&gt;many&lt;/em&gt; other things, why the probability that a permutation of a large set has no fixed points is about $\frac{1}{e}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T00:45:37.513" CommentCount="4" />
  <row Id="3327" PostTypeId="2" ParentId="3264" CreationDate="2010-08-26T00:53:28.987" Score="2" ViewCount="0" Body="&lt;p&gt;Let's consider $\mathbb{R}^3$ as a vector space on $\mathbb{R}$, endowed with the standard scalar product $\langle,\rangle$ (also know as $\mathbb{E}^3$, the Euclidean space). We define a function $f: \mathbb{R}^3 \to \mathbb{R}^3$ to be an &lt;em&gt;isometry&lt;/em&gt; if $||f(u)-f(w)|| = ||u-w||$ for all $v,w \in \mathbb{R}^3$, i. e. if $f$ &lt;em&gt;preserve distances&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transaltions are isometries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define a &lt;em&gt;translation by $z$&lt;/em&gt; the map $\tau_{z}: \mathbb{R}^3 \to \mathbb{R}^3$ such that $\tau_{z}(u) = u+z$ for all $u \in \mathbb{R}^3$. Then we have $$||\tau_{z}(u)-\tau_{z}(w)|| = ||(u+z) - (w+z)|| = ||u-w||$$ for all $u,w \in \mathbb{R}^3$ so we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we consider reflections and rotations. They are both &lt;em&gt;real unitary linear maps&lt;/em&gt; (also known as &lt;em&gt;orthogonal&lt;/em&gt;, or members of $O(3)$). The following are equivalent propieties of these transformations of the space:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\langle L(u),L(w)\rangle = \langle u,w\rangle$ for all $L \in O(3)$ $u,w \in \mathbb{R}^3$ or they preserve scalar product&lt;/li&gt;&#xA;&lt;li&gt;$||L(u)|| = ||u||$ for all $L \in O(3)$ $u \in \mathbb{R}^3$ or they preserve lenghts&lt;/li&gt;&#xA;&lt;li&gt;$||L(u)|| = ||u||$ for all $L \in O(3)$ and $u \in \mathbb{R}^3$ with $||u|| = 1$ or the unit sphere is $L$-invariant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, we'll see that all real unitary transformation of the space are isometries, proving then that reflections and rotations are of this type we have finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Real unitary transformations are isometries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L \in O(3)$ then $$||L(u)-L(w)|| = \sqrt{\langle L(u)-L(w),L(u)-L(w)\rangle} = \sqrt{\langle L(u-w),L(u-w)\rangle}$$ by definition of norm and using the linearity of L, and $$\sqrt{\langle L(u-w),L(u-w)\rangle} = \sqrt{\langle u-w,u-w\rangle} = ||u-w||$$ by using one of the three proprieties above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reflections are unitary transformations (and thus isometries)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S\subseteq\mathbb{R}^3$ is a subspace, the we define reflection in respect to $S$ the linear map $\rho_{S}: S \oplus S^{\bot} \to S \oplus S^{\bot}$ such that if $u = u_{S}+u_{S^{\bot}}$ then $\rho_{S}(u) = u_{S}-u_{S^{\bot}}$. So we have $$||u_{S}±u_{S^{\bot}}|| = \sqrt{||u_{S}||^2 + ||u_{S^{\bot}}||^2}$$ and then $||\rho_{S}(u)|| = ||u||$ for all $u \in \mathbb{R}^3$, that is one of the three equivalent proprieties above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rotations are unitary transformations (and thus isometries)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a little weird. As you can read &lt;a href=&quot;http://math.stackexchange.com/questions/3254/how-do-you-formally-prove-that-rotation-is-a-linear-transformation/3259#3259&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;, perhaps the best way to &lt;em&gt;define&lt;/em&gt; rotations is to consider them as paricular members of $O(3)$, so there is nothing to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: what has been said remains true for an arbitrary space $V$ of dimension $n$ on $\mathbb{R}$ endowed with a positive product $\langle ,\rangle$ because of the isomorphism with $\mathbb{E}^n$, this is the power of linear algebra!&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-01T22:42:01.117" LastActivityDate="2010-09-01T22:42:01.117" />
  <row Id="3328" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T01:36:41.880" Score="0" ViewCount="0" Body="&lt;p&gt;The first thing to note is that e&lt;sup&gt;x&lt;/sup&gt; is defined by the well known infinite series.  But that's not actually the crucially important thing from my point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real significance comes from the exponential mapping in the the theory of continuous symmetries (Lie Groups).  The exponential mapping is a function from the &quot;infinitesimal&quot; elements of the group (Lie Algebra) to the elements of the group itself.  This is not so easy to explain unfortunately, but you may be interested to wiki the principle axis theorem for some analogies from physics that are easy to visualize intuitively (the principle axis are the lie algebra).  Clearly, the theory of continuous symmetries is of fundamental significance, and the appearance of the infinite series for e at the heart of this theory is for me quite profound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, for very general Lie Groups, we have the amazing identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = e&lt;sup&gt;t&lt;sub&gt;1&lt;/sub&gt;L&lt;sub&gt;1&lt;/sub&gt;&lt;/sup&gt;e&lt;sup&gt;t&lt;sub&gt;2&lt;/sub&gt;L&lt;sub&gt;2&lt;/sub&gt;...&lt;/sup&gt;e&lt;sup&gt;t&lt;sub&gt;N&lt;/sub&gt;L&lt;sub&gt;N&lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an element of the Lie Group A with a lie algebra L&lt;sub&gt;i&lt;/sub&gt; of size N, and some real numbers t&lt;sub&gt;i&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see the connection to the real number defined by e by representing a continuous group as a group of matrices (under matrix multiplication), and noticing that the exponential mapping is the infinite series for e&lt;sup&gt;X&lt;/sup&gt;, where X is now a matrix.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-26T01:42:17.923" LastActivityDate="2010-08-26T01:42:17.923" CommentCount="1" />
  <row Id="3329" PostTypeId="1" CreationDate="2010-08-26T02:23:20.127" Score="5" ViewCount="186" Body="&lt;p&gt;Back in the day, I was absolutely enthralled by the study of plane curves and their properties (I have &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521055857&quot; rel=&quot;nofollow&quot;&gt;Lockwood&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486442764&quot; rel=&quot;nofollow&quot;&gt;Zwikker&lt;/a&gt; to thank). I learned early on that for the purposes of generating plots on a computer (and for that matter deducing equations of &quot;derived curves&quot; and determining other special properties), one should try to find a representation in parametric equations for your plane curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I recall, in dealing with algebraic curves represented by an implicit Cartesian equation, I knew of only three tricks to derive parametric equations from an implicit equation (listed in decreasing order of effectiveness; I note that I did all these investigations even before I knew computer algebra systems existed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Convert to polar coordinates to express in the form $r=r(\theta)$; the parametric equations are then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}x&amp;amp;=r(\theta)\cos(\theta)\\y&amp;amp;=r(\theta)\sin(\theta)\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: The $y=mx$ &quot;trick&quot; (I never did get to learn the formal name for this technique); to use the implicit equation for the &lt;a href=&quot;http://mathworld.wolfram.com/FoliumofDescartes.html&quot; rel=&quot;nofollow&quot;&gt;folium of Descartes&lt;/a&gt; as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^3+y^3=3xy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^3+(mx)^3=3x(mx)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then by solving for x and using the relation $y=mx$ again,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}x&amp;amp;=\frac{3m}{1+m^3}\\y&amp;amp;=\frac{3m^2}{1+m^3}\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I remember this worked especially well for curves whose (only?) singular points are at the origin, but not very well for other curves; can anybody explain why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: Randomly replacing x or y with any of the six trigonometric functions (maybe multiplied by a convenient constant), and hope that I can easily solve for the other variable. For instance, I managed to derive the parametric equation for the &lt;a href=&quot;http://mathworld.wolfram.com/Bicorn.html&quot; rel=&quot;nofollow&quot;&gt;bicorn&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/DuerersConchoid.html&quot; rel=&quot;nofollow&quot;&gt;Dürer conchoid&lt;/a&gt; in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the only other thing I learned way after I had moved on to other things was that elliptic curves can for instance be represented as parametric equations involving the Weierstrass ℘ function or the elliptic exponential, but this is apparently limited to elliptic curves only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for my question: did I miss any other useful (general?) methods for turning an implicit Cartesian equation for an algebraic curve into parametric equations?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-28T13:11:38.130" LastActivityDate="2010-08-28T13:11:38.130" Title="Parametrizing implicit algebraic curves" Tags="&lt;geometry&gt;&lt;plane-curves&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3331" PostTypeId="2" ParentId="3250" CreationDate="2010-08-26T03:46:43.757" Score="1" ViewCount="0" Body="&lt;p&gt;Based on the previous answers, I have come to the following (possible incomplete or incorrect) conclusion: It seems that the Pythagorean theorem can be dealt with in one of two ways, either it is taken as the definition of length/distance, or it is proved by invoking the technical definitions of an inner product space.  However, proving Pythagoras via the definitions of an inner product space really just shifts the question to how one motivates these definitions (namely, why does one define the length of a vector to be the square root of its dot product with itself?).  Thus, one can say that the Pythagorean theorem is generally taken to be the &lt;em&gt;definition&lt;/em&gt; of length or distance, and the only thing left to do is &lt;em&gt;motivate&lt;/em&gt; this definition; that is, show why the Euclidean norm is such a natural system for determining length.  Qiaochu has alluded to the fact that the Euclidean norm is natural because of its large symmetry group, but I'd like someone to elaborate on this a bit if possible...&lt;/p&gt;&#xA;" OwnerUserId="1497" LastActivityDate="2010-08-26T03:46:43.757" CommentCount="6" />
  <row Id="3332" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T04:07:29.057" Score="16" ViewCount="0" Body="&lt;p&gt;Let $z=(a+bi)$ be a complex number which is a square root of $i$, that is &#xA;$$i=z^2=(a^2-b^2)+2abi.$$ &#xA;Equating real and imaginary parts we have,&#xA;$$a^2-b^2=0, 2ab=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two real solutions to this pair of equations are&#xA;$a={1 \over \sqrt{2}},b={1 \over \sqrt{2}}$ and $a=-{1 \over \sqrt{2}},b=-{1 \over \sqrt{2}}$. The two square roots of $i$ therefore are &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\pm {1 \over \sqrt{2}} (1+i)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-08-26T04:07:29.057" />
  <row Id="3333" PostTypeId="1" AcceptedAnswerId="3334" CreationDate="2010-08-26T04:09:09.350" Score="3" ViewCount="86" Body="&lt;p&gt;Ramsey number R(3,6)=18. How to construct a graph of 17 nodes which does not contain neither a clique of order 3 or an independent set of order 6. could you show me the tactics or the adjacency matrix?&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-08-26T10:02:54.803" Title="how to construct 17-vertexed graph for Ramsey number R(3,6)=18" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="3334" PostTypeId="2" ParentId="3333" CreationDate="2010-08-26T04:18:09.307" Score="3" ViewCount="0" Body="&lt;p&gt;Look at &lt;a href=&quot;http://cs.anu.edu.au/~bdm/data/ramsey.html&quot; rel=&quot;nofollow&quot;&gt;http://cs.anu.edu.au/~bdm/data/ramsey.html&lt;/a&gt; for the R(3,6)'s with &lt;em&gt;less&lt;/em&gt; than 18 vertices. The ones with 17 are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Mw2xq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apartently, an example is constructed by hand in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;J. G. Kalbfleish, &lt;em&gt;Chromatic graphs and Ramsey's theorem&lt;/em&gt;, Ph. D. thesis, University of Waterloo, January 1966.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-26T06:15:05.003" LastActivityDate="2010-08-26T06:15:05.003" CommentCount="1" />
  <row Id="3335" PostTypeId="1" AcceptedAnswerId="3496" CreationDate="2010-08-26T04:20:02.753" Score="1" ViewCount="176" Body="&lt;p&gt;I love R.Courant and F.John's &lt;em&gt;Introduction to Calculus and Analysis&lt;/em&gt; because of its wide coverage, precise description and friendly written style.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any classic textbooks like it on ODE/PDE?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-26T04:24:26.587" LastActivityDate="2010-08-28T00:46:55.493" CommunityOwnedDate="2010-08-26T04:27:38.983" Title="Could you recommend some classic textbooks on ordinary/partial differential equation?" Tags="&lt;reference-request&gt;&lt;differential-equations&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" />
  <row Id="3336" PostTypeId="1" AcceptedAnswerId="3344" CreationDate="2010-08-26T04:27:18.300" Score="1" ViewCount="248" Body="&lt;p&gt;Sometimes, there are many ways to give defintion to a mathmatical principle, for example the natural base logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about sine and cosine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-08-26T20:49:06.333" Title="How many ways to define sine and cosine?" Tags="&lt;analysis&gt;&lt;trigonometry&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="3337" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T04:36:10.410" Score="0" ViewCount="0" Body="&lt;p&gt;I am a big fan of Bentley and Cooke's &lt;a href=&quot;http://www.abebooks.com/servlet/BookDetailsPL?bi=1174162092&amp;amp;searchurl=an%3Dbentley%2Band%2Bcooke%26sts%3Dt%26x%3D82%26y%3D18&quot; rel=&quot;nofollow&quot;&gt;Linear Algebra with Differential Equations&lt;/a&gt;. It was published in the early 70's, it is black and white (i.e. not flashy), and has excellent references throughout the text (such as a footnote pointing to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/082840044X&quot; rel=&quot;nofollow&quot;&gt;Kamke&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the topics of differential equations and linear algebra are introduced side by side at the introductory level.  I think this is an ideal text for the beginning engineering student that holds him or herself to learning a bit more mathematics than their peers.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-08-26T04:36:10.410" CommunityOwnedDate="2010-08-26T04:36:10.410" CommentCount="1" />
  <row Id="3338" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T04:46:49.703" Score="0" ViewCount="0" Body="&lt;p&gt;Good books on ODE would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Differential Equations by George F. Simmons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Ordinary Differential Equations by E.Coddington&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For PDE, I request you to take a look at this link: &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=175382&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showthread.php?t=175382&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-26T04:46:49.703" CommunityOwnedDate="2010-08-26T04:46:49.703" />
  <row Id="3339" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T04:52:09.907" Score="4" ViewCount="0" Body="&lt;p&gt;I like to define them by the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y''=-y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then, choosing the initial conditions we get these two functions. This also give rise nice definition for $\pi$ being the fundamental period for the solutions of these equations.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-08-26T04:52:09.907" CommentCount="1" />
  <row Id="3340" PostTypeId="1" AcceptedAnswerId="3341" CreationDate="2010-08-26T05:07:18.977" Score="2" ViewCount="100" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/86j1p.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why do we define $f(\xi+0)$ as $\lim_{\varepsilon \rightarrow 0} f(\xi + \varepsilon^2)$ and $f(\xi-0)$ as $\lim_{\varepsilon \rightarrow 0} f(\xi - \varepsilon^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-26T06:19:26.777" LastActivityDate="2010-08-26T06:19:26.777" Title="Why we assign $1/2[f(\epsilon+0) + f(\epsilon-0)]$ as the value at the discontinuity point?" Tags="&lt;calculus&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3341" PostTypeId="2" ParentId="3340" CreationDate="2010-08-26T05:13:50.477" Score="6" ViewCount="0" Body="&lt;p&gt;It is not that we &lt;em&gt;assign&lt;/em&gt; that value: Fourier series simply converge to that independently of our desires!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-26T05:13:50.477" CommentCount="1" />
  <row Id="3342" PostTypeId="2" ParentId="3122" CreationDate="2010-08-26T05:24:08.700" Score="0" ViewCount="0" Body="&lt;p&gt;Let $\epsilon &gt; 0$ be given. There is a positive integer $n_{0}$ such that $1/n &amp;lt; \epsilon/2$ for $n &gt; n_{0}$. Choose a partition $P$ determined by $$ 0 = x_{0} &amp;lt; x_{1}= \frac{1}{n_{0}+1} &amp;lt; x_{2} &amp;lt; \cdots &amp;lt; x_{n'_{0}}= \frac{1}{n_0}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &amp;lt; x_{n'_{0}+1} &amp;lt; \cdots &amp;lt; x_{n_{1}}'= \frac{1}{n_{0}-1} &amp;lt; \cdots &amp;lt; x_{n_{n_{0}-1}'}=1$$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle x_{i} - x_{i-1} &amp;lt; \frac{\epsilon}{4n_{0}}$$   Then $$U(P,f) - L(P,f) = \frac{1}{n_{0}+1} + \sum\limits_{i=2}^{n_{0}'} (M_{i}-m_{i})(x_{i}=x_{i-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ + \sum\limits_{k=0}^{n_{0}-2} \sum\limits_{i=n_{k}'+1}^{n_{k+1}'}(M_{i}-m_{i})(x_{i}-x_{i-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &amp;lt; \frac{\epsilon}{2} + 2n_{0} \frac{\epsilon}{4n_{0}} = \epsilon$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-12T03:10:37.317" LastActivityDate="2010-09-12T03:10:37.317" />
  <row Id="3343" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T06:30:38.517" Score="10" ViewCount="0" Body="&lt;p&gt;The number $e$ is &lt;strong&gt;not&lt;/strong&gt; very important in mathematics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary it's the &lt;em&gt;exponential function&lt;/em&gt;&#xA;$$\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$$&#xA;that's important. The number $e$ is just $\exp(1)$ - just one&#xA;value of the exponential function. Compare how often one&#xA;sees the exponential function as opposed to $e$ &lt;strong&gt;by itself&lt;/strong&gt;.&#xA;In the four previous answers to this question, three focus on the&#xA;exponential function, and only one on $e$ itself.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-26T06:30:38.517" />
  <row Id="3344" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T06:38:53.573" Score="3" ViewCount="0" Body="&lt;p&gt;(This is by no means a comprehensive list.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right-triangle definition&lt;/strong&gt;: the sine (cosine) of an acute angle is the ratio of the lengths of the leg opposite (adjacent to) the given angle to the hypotenuse of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bizarre Geometric Definition&lt;/strong&gt;: the cosine of an angle in a triangle is $\frac{a^2+b^2-c^2}{2ab}$, where $a$ and $b$ are the lengths of the sides adjacent to the angle and $c$ is the length of the side opposite the angle; the sine of an acute angle is the cosine of its complement; the sine of an obtuse angle is the cosine of its supplement's complement. (&lt;em&gt;edit: this one might be even worse than I'd originally thought as a definition, so perhaps just ignore it.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Circular definition (1)&lt;/strong&gt;: the sine (cosine) of a magnitude of rotation is the vertical (horizontal) coordinate of the image of the point (1,0) under a rotation of the given magnitude centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Circular definition (2)&lt;/strong&gt;: the sine (cosine) of a directed angle with vertex at the origin and initial ray on the positive x-axis is the y-coordinate (x-coordinate) of the point of intersection of the terminal ray of the angle with the unit circle centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power series definition&lt;/strong&gt;:&#xA;$$\begin{align}&#xA;\sin x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots&#xA;\\&#xA;\cos x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exponential definition&lt;/strong&gt;:&#xA;$$\sin x=\frac{e^{ix}-e^{-ix}}{2i};\quad\quad\cos x=\frac{e^{ix}+e^{-ix}}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-26T06:47:49.147" LastActivityDate="2010-08-26T06:47:49.147" CommentCount="2" />
  <row Id="3345" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T07:19:19.593" Score="0" ViewCount="0" Body="&lt;p&gt;I like good old definition using right angled triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Sxit7.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sin\theta = b/c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cos\theta = a/c$&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-08-26T07:19:19.593" CommentCount="1" />
  <row Id="3346" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T08:09:07.367" Score="5" ViewCount="0" Body="&lt;p&gt;More generally, if you want to compute all the $n$-th roots of a complex number $z_0$, that is, all the complex numbers $z$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;z^n = z_0 \ ,  \qquad \qquad \qquad \qquad [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you should write this equation in exponential form: $z = re^{i\theta}, \ , z_0 = r_0 e^{i\theta_0}$. Then [1] becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( r e^{i \theta}\right)^n = r_0 e^{i\theta}   \qquad \Longleftrightarrow \qquad r^n e^{in\theta} = r_0 e^{i\theta_0}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have two complex numbers in polar coordinates which are equal, their moduluses must be equal clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r^n = r_0 \qquad \Longrightarrow \qquad r = +\sqrt[n]{r_0}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $r, r_0 \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the arguments, we cannot simply conclude that $n\theta = \theta_0$, but just that they differ in an integer multiple of $2\pi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n\theta = \theta_0 + 2k\pi   \qquad \Longleftrightarrow \qquad  \theta = \frac{\theta_0 + 2k \pi}{n} \quad \text{for} \quad k = 0, \pm 1 , \pm 2, \dots&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem that we have an infinite number of $n$-th roots, but we have enough with $k = 0, 1, \dots , n-1$, since for instance for $k=0$ and $k=n$ we obtain the same complex numbers. Thus, finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[n]{r_0 e^{i\theta_0}} = +\sqrt[n]{r_0} e^{i \frac{0_0 + 2k\pi}{n}} \ , \quad k = 0, 1, \dots , n-1 &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are all the complex $n$-th roots of $z_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; For $n=2$, we obtain that every complex number has exactly two square roots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\sqrt{z_0} &amp;amp;= +\sqrt{r_0}e^{i\frac{\theta_0 + 2k\pi}{2}} \ , k = 0,1  \\&#xA;           &amp;amp;= +\sqrt{r_0}e^{i\frac{\theta_0}{2}} \quad \text{and} \quad +\sqrt{r_0}e^{i\left(\frac{\theta_0}{2} + \pi \right)}  \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, since $i = e^{i\frac{\pi}{2}}$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{i} = &#xA;\begin{cases}&#xA;e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} +i \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}    \\&#xA;e^{i(\frac{\pi}{4} + \pi)} = \cos\frac{5\pi}{4} +i \sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}  \ .&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $z_0 = -1 = e^{i\pi}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{-1} = e^{i \frac{\pi}{2}} = i \quad \text{and} \quad e^{i\left( \frac{\pi}{2} + \pi\right)} = e^{i\frac{3\pi}{2}} = -i \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; For $z_0 = 1 = e^{i \cdot 0}$ and any $n$, we obtain the $n$-th &lt;em&gt;roots of unity&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[n]{1} = e^{i\frac{2k\pi}{n}} \ , \quad k= 0, 1, \dots , n-1 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if $n= 2$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{1}= e^{i \cdot 0} = 1 \quad \text{and} \quad  e^{i\pi}= -1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for $n= 4$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[4]{1} = e^{i\frac{2k\pi}{4}} \ , \quad k = 0, 1, 2, 3 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt[4]{1} = 1, i, -1 , -i \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-08-26T13:04:44.737" LastActivityDate="2010-08-26T13:04:44.737" CommentCount="3" />
  <row Id="3347" PostTypeId="2" ParentId="3329" CreationDate="2010-08-26T09:00:33.243" Score="3" ViewCount="0" Body="&lt;p&gt;This is not intended as a full answer, but as a way to try to decide if it is feasible to be able to parameterize a plane curve in an easy way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two examples you give, namely the folium and the bicorn, are examples of &quot;irreducible curves of geometric genus zero&quot;. For the purpose of this discussion you can take &quot;irreducible curve&quot; to mean the locus of zeros of an irreducible polynomial in two variables. Geometric genus zero is harder to define, though see &lt;a href=&quot;http://math.stackexchange.com/questions/1609/intuitive-explantions-for-the-concepts-of-divisor-and-genus&quot;&gt;this&lt;/a&gt; question and the various answers, especially the answer of Matt E. In the linked answer, the &quot;genus&quot; that is defined is the &quot;arithmetic genus&quot;, whereas we are interested in the &quot;geometric genus&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two genera are not unrelated, and the discrepancy between them is concentrated at the singular points of the curve. A singular point of a plane curve $C$ given as the vanishing set of a polynomial $f(x,y)$ is a point $p$ in the plane that lies on the curve (i.e. $f(p)=0$) and where both partial derivatives of $f$ vanish (i.e. $df/dx(p)=df/dy(p)=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, while the arithmetic genus of an (irreducible) plane curve $C$ of degree $d$ is simply $p_a:=(d-1)(d-2)/2$, the geometric genus of $C$ is an integer less than or equal to $p_a$. I will not dwell too much on how to compute the geometric genus, for the moment I will simply say that every singular point decreases the arithmetic genus by at least one (counting also the &quot;esoteric&quot; singular points at infinity, as you need to explain the bicorn). There are algorithmic way to decide the drop in genus that every singular point determines and thus the computation of the geometric genus can be taken as &quot;easy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Over the complex numbers.&lt;/strong&gt; It is a fact that, whenever the geometric genus of an irreducible curve $C$ is zero then you can find a parameterization of $C$ by &lt;em&gt;rational functions of a single variable $t$&lt;/em&gt;, provided that you are thinking about the complex solutions of your curve. This parameterization can be found algorithmically, and while there are various tricks to use, the substitution $y=mx$ is at the core: depending on context it might be called &quot;projection&quot; or &quot;blow up&quot;. The other trick is that you can &quot;reembed&quot; your curve by considering the vector space whose coordinates are all the monomials of some given degree $n$ and take the image of the curve under the map that sends a point in the plane in the $N$-tuple of all the evaluations of the monomials you have chosen. Instead of being more explicit, let me give an example: if the curve is the $x$-axis in the plane, and you choose the monomials $1,x,y,x^2,xy,y^2$ to reembed, then we image of the $x$-axis (i.e. the locus where $y=0$) is the set of points with coordinates $(1,x,0,x^2,0,0)$ in a 6-dimensional space. Forgetting the useless coordinates (this is an instance of projection) gives us a plane curve parameterized by $(x,x^2)$, namely the parabola $x_1^2=x_2$. In this example we started with a parameterization and ended with a parameterization. Being more refined you can also start with an implicit equation and end with (lots of) implicit equations in more variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have projections and reembeddings you can play them against one another: embed your curve in large enough space. Project away from singular points until you get back to the plane; reembed and reproject. There is a way of making sure that &quot;singular points get resolved&quot; by doing this, so that eventually you get to a curve having no singular points. Once you are there, further projections will do the trick of getting you to a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Over the real numbers.&lt;/strong&gt; Of course things are somewhat trickier. Certainly two possibilities are the folium, where the rational parameterization you had over the complex numbers &quot;just works&quot; over the reals, but there is the slightly more elaborate one of the bicorn, where you have a rational parameterization in terms of trigonometric functions. Again, by modification of the tricks discussed above, you can get every irreducible curve of genus zero to become a plane conic. Once in this form, you will find a rational parameterization if and only if you can find a point, and deciding if there is a point reduces to the computation of whether a matrix constructed out of the coefficients of the equation has eigenvalues of different signs or not. The distinction between rational parameterization in terms of a variable or in terms of trigonometric functions has to do with personal taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, this has dealt with curves of genus zero and I will not say much about higher genus. If the genus is one, there is, as you mentioned, the possibility of using the Weierstrass $\wp$. Here things become more complicated: the function used to parameterize everything is less familiar, you will probably need some transcendental numbers to get this parameterization to work, and the topological description of the space of solutions gets more in the way. In the genus zero case, the only thing that mattered was: is it empty, is it not empty. In the genus one case there is &quot;one more&quot; possibility. Of course, higher genera become harder and harder, and I will stop here, given that this is already &lt;em&gt;way&lt;/em&gt; too much for this!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-26T09:00:33.243" CommentCount="7" />
  <row Id="3348" PostTypeId="1" AcceptedAnswerId="3360" CreationDate="2010-08-26T09:11:02.423" Score="2" ViewCount="129" Body="&lt;p&gt;Construct a string by using only n characters, the string should do not contain any consecutive repeated pattern, in other words, the string must have not any substring matching /(.+)\1/, what is the length l(n) of the longest string S(n)?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n    S(n)    l(n)&#xA;1    a       1&#xA;2    aba     3&#xA;3    ?       ?&#xA;4    ?       ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;&lt;code&gt;abcabacbabcabacabcac&lt;/code&gt; has none consecutive  repeated pattern,so l(3)&gt;=20&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-26T12:36:55.710" LastActivityDate="2010-08-26T13:14:38.157" Title="The longest string of none consecutive repeated pattern" Tags="&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="3349" PostTypeId="2" ParentId="3348" CreationDate="2010-08-26T09:41:32.183" Score="0" ViewCount="0" Body="&lt;p&gt;To find a maximal string for n letters, add a new letter in every other place of a maximal string for n-1 letters.  It is still the case that no consecutive identical substrings exist.  No more letters can be added because two of them would have to appear consecutively, so this is the best that is possible.  The length of the string is 2^n-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&#xA;bab&#xA;cbcacbc&#xA;dcdbdcdadcdbdcd&lt;/p&gt;&#xA;" OwnerUserId="1284" LastActivityDate="2010-08-26T09:41:32.183" CommentCount="4" />
  <row Id="3350" PostTypeId="2" ParentId="3333" CreationDate="2010-08-26T10:02:54.803" Score="0" ViewCount="0" Body="&lt;p&gt;Here is an explicit example that is very concrete, but a near miss: a graph $G$  with 16 vertices containing no triangle and no independent set with six elements. I think that it is not possible to extend the graph $G$ by a single vertex preserving this property, but I would be interested if I could be proved wrong! For those who are interested and know about these things, the graph $G$ is the graph of configurations of the exceptional curves on a del Pezzo surface of degree four. None of this will be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Construction of&lt;/strong&gt; $G$. Start with a circle $C$ in the plane, five distinct points $p_1,\ldots,p_5$ on the circle $C$ and the 10 lines $\{\ell_{ij}\}_{1 \leq i &amp;lt; j \leq 5}$ through the five points. The graph has as vertices the circle $C$, the points $\{p_i\}$ and the lines $\{\ell_{ij}\}$, for a total of $1+5+10=16$ vertices. The edges are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the vertex $C$ is joined to each point $p_1,\ldots,p_5$, but to no line;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the vertex $p_i$ is joined to each line through itself and to the circle, but nothing else (for instance, there is no edge between different points and no edge between $p_1$ and $\ell_{23}$);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the vertex $\ell_{ij}$ is joined to the points $p_i$ and $p_j$, as well as all the lines $\ell_{kl}$ for $i,j\neq k,l$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;where, of course, $i$ and $j$ range in $\{1,\ldots,5\}$, and $i$&lt;code&gt;&amp;lt;&lt;/code&gt;$j$ when both appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now check that there are no triangles and no independent sets with six elements in $G$. The checks are easier once you realize that there is a large group of symmetries acting on $G$ (the group is larger than the obvious permutations of the indices: it has order 1920).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this graph has 16 vertices! As I mentioned above, I think that any extension of this graph will have either a triangle or an independent set of size six, but I am not sure: anyone can confirm this? One way to go about this would be to see if this graph appears as a subgraph in one of the examples that Mariano gives in his answer. My argument was not this one, and I would prefer a more conceptual argument.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-26T10:02:54.803" />
  <row Id="3351" PostTypeId="2" ParentId="2584" CreationDate="2010-08-26T10:07:28.947" Score="0" ViewCount="0" Body="&lt;p&gt;First up, you should accept Ranieri's answer, which is obviously correct. The question of &quot;slicing&quot; seems like a temporary misunderstanding that shouldn't stand in your way. Rejection sampling works with continuous distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it seems you are attempting to randomly generate individuals from a population with some number of subgroups, each of which has an expected distribution of some continuous traits. Since the traits are distributed normally, $\forall{x}: P(x) &gt; 0$, so it is notionally possible to have outlier individuals with &lt;em&gt;any&lt;/em&gt; trait combination. However, it is very likely that arbitrary distributional constraints will not be satisfiable at the population level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, imagine that you have a population divided into two genders and two colours, and specified to be 50/50 male/female and 50/50 red/blue. Then require that some trait, say foolishness, is distributed $N(1,0.1)$ for each of the subgroups men, red and blue, but $N(0.5, 0.1)$ for women. Is this possible? I'm pretty sure not. And that's without even trying to clip values into a fixed range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I suspect that you might need a somewhat better model for your probability distribution. My first inclination would be to weight the different criteria in some fashion to prioritise the distribution of some traits over others. However, I'd have to think that through properly, which I don't have time to do just now. Maybe later...&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-08-26T10:07:28.947" CommentCount="1" />
  <row Id="3352" PostTypeId="1" CreationDate="2010-08-26T10:22:08.570" Score="2" ViewCount="419" Body="&lt;p&gt;Let $f: D \to \mathbb{R}\ $ be a function on non-singular, convex domain $D \subseteq \mathbb{R}^d$ and let us assume the second-order derivatives of $f$ exist. It is well known that $f$ is convex if and only if its Hessian $\nabla^2 f(x)$ is positive semi-definite for all $x \in D$. It is also known that if $\nabla^2 f(x)$ is positive definite for all $x \in D$, we may conclude that $f$ is strictly convex (for a reference, see &lt;a href=&quot;http://www.stanford.edu/~boyd/cvxbook/&quot; rel=&quot;nofollow&quot;&gt;Boyd and Vandenberghe, 2004&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $f$ is strictly convex, we still merely know that $\nabla^2 f(x)$ is positive semi-definite for all $x \in D$. That is, there may be $x \in D$ such that $y^T \nabla^2 f(x) y = 0\ $ for some $y\not=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider $f(x)=x^4$. In this case, $f$ is strictly convex, but $f''(x)=12x^2$ and, hence, $yf''(x)y=0$ for $x=0$ and $yf''(x)y&gt;0$ for all $x\not=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, these points that ruin the complete positive definiteness seem to be very sparsely distributed within $D$. So, my question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $f$ is strictly convex, how can we characterize the set of points $X$ for which&#xA;  $\nabla^2f(x)$ is not positive definite for $x \in X$, and &#xA;  $\nabla^2f(x)$ is positive definite for $x \in D \setminus X$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, has such a set $X$ been investigated before, what properties are known, and where can I learn more about it? Any reference is welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, my guess is that on can state the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Conjecture 1:&lt;/strong&gt; The set $X$ is merely a discrete subset of $D$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since Conjecture 1 has obviously been disproved by George Lowther below, allow me to restate my guess as the following (less bold) statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Conjecture 2:&lt;/strong&gt; The set $X$ is does not contain a non-empty open ball.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or, even more cautious:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Conjecture 3:&lt;/strong&gt; The set $D \setminus X$ does contain a non-empty open ball.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-08-27T11:15:34.763" LastActivityDate="2010-08-27T11:15:34.763" Title="Positive definite Hessians from strictly convex functions" Tags="&lt;real-analysis&gt;&lt;convex-analysis&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="3353" PostTypeId="1" AcceptedAnswerId="3365" CreationDate="2010-08-26T10:40:47.077" Score="3" ViewCount="285" Body="&lt;p&gt;In the Portuguese book &lt;em&gt;Análise Matemática&lt;/em&gt; (Mathematical Analysis) by C. Sarrico, it is proved there exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\lim_{n\rightarrow +\infty}f_{n}(x)=\lim_{n\rightarrow +\infty }e^{\log f_{n}(x)},\qquad(1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{n}(x)=\dfrac{n!n^{x}}{x(x+1)(x+2)\cdots (x+n)}.\qquad(2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is stated there that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\lim_{n\rightarrow +\infty }f_{n}(x)=\Gamma (x)=\int_{0}^{+\infty}t^{x-1}e^{-t}dt\qquad x&gt;0.\qquad(3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author writes that a proof can be found in &quot;Principles of Mathematical Analysis&quot; by W. Rudin. Since I don't have it, I ask the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: How is the sketch of such a proof or other proofs of the same result?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T21:48:09.397" LastActivityDate="2010-10-04T21:48:09.397" Title="Two ways of defining the gamma function $\Gamma (x)$. How to show they are equivalent?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3354" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T11:01:49.430" Score="1" ViewCount="0" Body="&lt;p&gt;Once you define sine, you can define cosine to be the sine of the complementary angle. i.e. cos(theta) = sin(pi/2 - theta). So you could reduce your question to asking for various ways of defining sine.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-26T11:01:49.430" />
  <row Id="3355" PostTypeId="1" CreationDate="2010-08-26T11:10:46.623" Score="0" ViewCount="95" Body="&lt;p&gt;Determine the Euler characteristic of the surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M=\left\{(x,y,z); \sqrt{x^2+y^2}=1+z^{2n}, 0&amp;lt; z&amp;lt; 1\right\} $$&lt;/p&gt;&#xA;" OwnerUserId="1503" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-26T11:52:44.073" LastActivityDate="2010-08-26T11:52:44.073" Title="Euler characteristic of a compact surface" Tags="&lt;differential-geometry&gt;" CommentCount="3" />
  <row Id="3356" PostTypeId="1" AcceptedAnswerId="3358" CreationDate="2010-08-26T11:27:15.630" Score="5" ViewCount="296" Body="&lt;p&gt;I'm struggling with this former Putnam Exam problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ and $g$ are nonconstant, differentiable, real-valued functions on $R$.  Furthermore, suppose that for each pair of real numbers $x$ and $y$, $f(x + y) = f(x)f(y) - g(x)g(y)$ and $g(x + y) = f(x)g(y) + g(x)f(y)$.  If $f'(0) = 0$, prove that $(f(x))^2 + (g(x))^2 = 1$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right.  So obviously, $f(x) = \cos x$ and $g(x) = \sin x$ satisfy the conditions and also the conclusion of the problem.  But are these the unique such functions, and if so, how to prove it?  And if not, then how to prove the conclusion otherwise?&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-08-26T18:21:07.040" Title="Uniqueness of $\sin x$ and $\cos x$? (Putnam Exam problem)" Tags="&lt;calculus&gt;&lt;putnam&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" />
  <row Id="3357" PostTypeId="2" ParentId="3278" CreationDate="2010-08-26T11:48:14.930" Score="1" ViewCount="0" Body="&lt;p&gt;I don't think there is any ambiguity in the use of the word “given” here, but I do think that one needs to learn the way it is used in conditional probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If people are not familiar with how terms are used then of course there could be misunderstanding. There are some interesting comments about ambiguity in such questions in &lt;a href=&quot;http://en.wikipedia.org/wiki/Boy_or_Girl_paradox&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This sentence from the article made me smile: Many people, including professors of mathematics, argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view.) &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-26T11:48:14.930" />
  <row Id="3358" PostTypeId="2" ParentId="3356" CreationDate="2010-08-26T12:05:20.903" Score="6" ViewCount="0" Body="&lt;p&gt;Let $u(x) = f(x)^2 + g(x)^2$. It should be easy to show that $u(x+y) = u(x)u(y)$, thus $u$ must be an exponential function $u(x) = e^{Bx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as $f(2x) = f(x)^2 - g(x)^2$, we should get $0 = g(0) g'(0)$. Comparing this with $u'(0)$ we should get $B = 0$, thus $f(x)^2 + g(x)^2 = 1$.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-26T12:05:20.903" CommentCount="3" />
  <row Id="3359" PostTypeId="2" ParentId="3356" CreationDate="2010-08-26T12:23:48.873" Score="3" ViewCount="0" Body="&lt;p&gt;Since KennyTM posted a very elegant proof I will post an uncreative one that requires no thought to produce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First write out all the obvious identities that come from the information told:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x + y) = f(x)f(y) - g(x)g(y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(x) = f(x)f(0) - g(x)g(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f(0) = f(0)f(0) - g(0)g(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$g(x + y) = f(x)g(y) + g(x)f(y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$g(x) = f(x)g(0) + g(x)f(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$g(0) = f(0)g(0) + g(0)f(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Df(x + c) = Df(x)f(c) - Dg(x)g(c)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Dg(x + c) = Df(x)g(c) + Dg(x)f(c)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Df(0) = 0 = Df(0)f(0) - Dg(0)g(0) = -Dg(0)g(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Dg(0) = Df(0)g(0) + Dg(0)f(0) = Dg(0)f(0)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The $Df(0)$ equation tells us that $Dg(0) = 0 \lor g(0) = 0$ Since we don't want $Dg(0) = 0$ lets discharge that automatically now by proving it contradictory: It would imply that $Dg(0 + c) = Df(0)g(c) + Dg(0)f(c) = 0$ so $g(x)$ would be a constant but that is not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have seen that $g(0) = 0$ and now we update the identifies with this information, and that tells us is that $f(0) = 1$. We cannot determine the value of $Dg(0)$ so just call it some constant $k$. Now it is easy to recognize that we have a simple ODE problem that could be solved symbolically by a calculator or by taking eigenvalues/vectors of the matrix that defines the system. Giving the solutions $\sin(kx)$ &amp;amp; $\cos(kx)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-26T14:05:57.623" />
  <row Id="3360" PostTypeId="2" ParentId="3348" CreationDate="2010-08-26T12:28:11.080" Score="3" ViewCount="0" Body="&lt;p&gt;A theorem of Thue's (not Thue's theorem!) states that for $n = 3$ (and therefore, also for larger $n$) an infinite sequence of characters (chosen out of $n$) without repeated sub-sequences (a.k.a square-free) exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: here's a slightly related reference: J.D. Currie's paper &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.11.965&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;'There are ternary circular square-free words of length&#xA;n for n  18.'&lt;/a&gt;, which at least has a reference to Thue's paper from 1906. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: Another reference, with explicit construction of sequences of length 3, 9, 27 and so on: &lt;a href=&quot;http://arxiv.org/abs/0712.0139&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0712.0139&lt;/a&gt;. &#xA;That paper gives the following sequence of length 27: 123213231213123132312132123&lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorUserId="360" LastEditorDisplayName="" LastEditDate="2010-08-26T12:47:05.130" LastActivityDate="2010-08-26T12:47:05.130" />
  <row Id="3361" PostTypeId="2" ParentId="3348" CreationDate="2010-08-26T13:14:38.157" Score="1" ViewCount="0" Body="&lt;p&gt;Let me apologize, for it's not an answer, but merely an extended comment. I like the problem very much. It sounds very classical, and most likely the answer must be known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being reformulated, the problem is as follows: if $X$ is an $n$-letter alphabet, we have a rewriting rule&#xA;$$&#xA;w_1 s s w_2=w_1 s^2 w_2 \to w_1 w_2&#xA;$$&#xA;where $s$ is nonempty and at least one of $w_1,w_2$ must be nonempty. (In effect, any word that satisfies the condition set by a-boy must be square-free so to speak.) One can also a semigroup with $n$ generators whose relations are determined by the rewriting rule above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem as set asks whether all words over $X$ are reduced to words of certain length $l(n).$ One wonders, however, whether $n=2$ is an exceptional case, and actually there are arbitrary large reduced words. For instance, the following word of length 68 in letters $0,1,2$ is reduced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12012101201020120212012101202101210201202120121012010201202120121012&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 4-letter alphabet it is easy to find words of quite large length (I've found by means of computing such words of lenght $\ge 1000.$)&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-08-26T13:14:38.157" CommunityOwnedDate="2010-08-26T13:14:38.157" CommentCount="1" />
  <row Id="3362" PostTypeId="1" AcceptedAnswerId="3407" CreationDate="2010-08-26T13:23:08.630" Score="3" ViewCount="98" Body="&lt;p&gt;Let $f : [0, 1] \to \mathbb{R}$ be a 'well behaved' function and define $c_n$ by the Riemann sums&#xA;$$&#xA;c_n := \frac{1}{n} \sum_{k = 1}^n f(\frac{k - \theta}{n})&#xA;$$&#xA;for some given $\theta \in [0, 1]$. Then obviously $c_n$ will converge to the integral of $f$. Under which conditions on $f$ will this convergence be monotone? In particular, will the convergence be monotone if $f$ is twice differentiable with $f &gt; 0$, $f' &amp;lt; 0$ and $f'' &gt; 0$?&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-08-26T18:15:34.910" LastActivityDate="2010-08-26T19:39:44.927" Title="When do Riemann sums converge monotonically?" Tags="&lt;real-analysis&gt;&lt;convergence&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3363" PostTypeId="1" AcceptedAnswerId="3402" CreationDate="2010-08-26T13:39:19.927" Score="6" ViewCount="80" Body="&lt;p&gt;Walking on a lattice.  The number of various paths from (0,0) to (m,n)  using north and east steps is binomial coefficient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C(m+n,m)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if he needs to go back (0,0) using south and west steps, and doesn't pass by the passed points. Then what is the number of various loops walking from (0,0) to (m,n) then returning to (0,0)? Any algebraic expression for this?&#xA;&lt;img src=&quot;http://i.imgur.com/OS8Zx.png&quot; alt=&quot;alt text&quot;&gt;&#xA;btw:i asked this question before, but had not get an answer yet. Maybe I can get a good answer at here.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="1498" LastEditorDisplayName="" LastEditDate="2010-08-26T14:08:08.060" LastActivityDate="2010-08-26T18:22:44.583" Title="the number of loops on  lattice?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3364" PostTypeId="1" CreationDate="2010-08-26T13:41:22.033" Score="2" ViewCount="242" Body="&lt;p&gt;A continuous function does not always map open sets to open sets, but a continuous function will map compact sets to compact sets.  One could make list of such preservations of topological &#xA;properties by a continuous function $f$:&#xA;$$ f( \mathrm{open} ) \neq \mathrm{open} \;,$$&#xA;$$ f( \mathrm{closed} ) \neq \mathrm{closed} \;,$$&#xA;$$ f( \mathrm{compact} ) = \mathrm{compact} \;,$$&#xA;$$ f( \mathrm{convergent \; sequence} ) = \mathrm{convergent \; sequence} \;.$$&#xA;Could you please help in extending this list?&#xA;(And correct the above if I've erred!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;.  Thanks for the several comments and answers extending my list.&#xA;I was hoping that I could see some common theme among the properties preserved by a continuous&#xA;mapping, separating those that are not preserved.  But I don't see such a pattern.&#xA;If anyone does, I'd appreciate a remark.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-08-27T11:16:14.497" LastActivityDate="2010-09-28T07:03:30.743" Title="Topological properties preserved by continuous maps" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="3365" PostTypeId="2" ParentId="3353" CreationDate="2010-08-26T13:53:23.343" Score="4" ViewCount="0" Body="&lt;p&gt;I don't have a copy of Rudin, but can give you a proof of this. There's various ways of playing about with the integral form to get what you want, although I'm not sure what the cleanest method is.&#xA;One way is to use the fact that $1_{\{t\le n\}}(1-t/n)^n\to e^{-t}$ to write $\Gamma(x)$ as&#xA;$$&#xA;\begin{align}&#xA;\Gamma(x)&amp;amp;=\lim_{n\to\infty}\int_0^nt^{x-1}(1-t/n)^{n}\,dt\\&#xA;&amp;amp;=\lim_{n\to\infty}n^x\int_0^1s^{x-1}(1-s)^n\,ds.&#xA;\end{align}&#xA;$$&#xA;To prove that the first integral commutes with the limit, you could use the dominated convergence theorem. The second integral is just using the substitution $t=ns$.&#xA;It needs to be shown that this integral is equal to your function $f_n(x)$. In fact, I recognize both the function $f_n$ and the integral as &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;beta functions&lt;/a&gt;.&#xA;$$&#xA;\int_0^1s^{x-1}(1-s)^n\,ds=B(x,n+1)&#xA;$$&#xA;To find the explicit form for this, you can either use $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$ (a proof is given on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;) or repeatedly apply the identity $B(x,n+1)=B(x+1,n)n/x$ (which follows from integration by parts) and $B(x,1)=1/x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: You can also apply a very similar argument to the one above using the limit $(1+t/n)^{-n}\to e^{-t}$, which I did in my first version of this answer. Using $(1-t/n)^n\to e^{-t}$ seems cleaner though, so I edited the answer accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative method is to write&#xA;$$&#xA;\begin{align}&#xA;\Gamma(x) &amp;amp;= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{\Gamma(n+1)\Gamma(x)}{\Gamma(x+n+1)}\\&#xA;&amp;amp;= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{n!}{x(x+1)\cdots(x+n)}.&#xA;\end{align}&#xA;$$&#xA;Then, the limit you need would follow as long as it can be shown that $\Gamma(x+n+1)/\Gamma(n+1)$ approaches $n^x$ asymptotically as $n\to\infty$.&#xA;$$&#xA;\Gamma(n+x+1)=\int_0^\infty t^x t^ne^{-t}\,dt&#xA;$$&#xA;Then, the required expression depends on showing that, in the limit $n\to\infty$, to leading order, the integral only contributes for values of $t/n$ close to 1 (you can see that $t^ne^{-t}$ has its maximum at $t=n$).&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-26T18:19:13.920" LastActivityDate="2010-08-26T18:19:13.920" CommentCount="1" />
  <row Id="3366" PostTypeId="1" AcceptedAnswerId="3371" CreationDate="2010-08-26T13:53:26.640" Score="1" ViewCount="77" Body="&lt;p&gt;I was given this question (there were prior questions defining the equivelence relation ~)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;let E be the set of equivelence classes of ~&#xA; define f from E to Reals&#xA;    f:={([(x,y)],r)|x=yr,\: x,y,r in Reals}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand what it means; is the following correct?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f:[(x,y)] mapsto r&#xA;&#xA;E=(Reals X {Reals/0})/~ &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;ie the Quitient set of (Reals X {Reals/0}) by ~&#xA;((x,y) was ealier defined as an element of (Reals X {Reals/0}) ))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I understanding this right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My function takes in a equivlence class, [(x,y)]&#xA;and reaturns a real value r, furfilling the condition that x=yr?&lt;/p&gt;&#xA;" OwnerUserId="1505" LastActivityDate="2010-08-26T14:40:07.670" Title="Am I understanding this question right? (function defined with set notation)" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3367" PostTypeId="2" ParentId="3353" CreationDate="2010-08-26T13:55:04.663" Score="1" ViewCount="0" Body="&lt;p&gt;First, it should be noted that the integral representation is valid only for x with positive real part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having gotten that disclaimer away, one sketch of a proof goes as follows: the limit representation can be shown to be equivalent to the Weierstrass infinite product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{j=1}^\infty{\left(1+\frac{z}{j}\right)\exp\left(-\frac{z}{j}\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\gamma$ here is the Euler-Mascheroni constant; the representation $\lim_{k\to\infty}{\sum_{j=1}^k{\frac1{j}}-\ln(k)}$ is useful here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now show that the infinite product/limit is the same as the integral $n^z \int_0^1 (1-u)^n u^{z-1} \mathrm{d}u$, and then find a substitution for which &lt;a href=&quot;http://en.wikipedia.org/wiki/Watson%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Watson's lemma&lt;/a&gt; can be applied.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-26T13:55:04.663" CommentCount="4" />
  <row Id="3368" PostTypeId="1" AcceptedAnswerId="3369" CreationDate="2010-08-26T14:10:47.007" Score="2" ViewCount="145" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a programming problem I'm trying to solve where, in order to be efficient, I have to find the probability of some events occuring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I have 3 transactions. At each transaction there is a 5% chance that the transaction will be declined. I want to know what are the chances of 0, 1, 2 and 3 of the transactions being declined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have being trying to visualise this with a venn diagram and trying to work out the areas of each event, but unfortunately when I add up the separate events I keep on getting more than or less than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event that 0 of the transactions will be declined is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(0 transactions declined) = .95 * .95* .95 = 6859/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event that all 3 of the transactions will be declined should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(3 transactions declined) = .05 * .05 * .05 = 1/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I start running into trouble:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(2 transactions only declined) = (3 x (.05 * .05)) -3/8000 (because the A&amp;cap;B&amp;cap;C has been counted three times and we only want to know about when there are 2 transactions, and NOT when there are 2 or 3) = 57/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(1 transaction only declined) = Well p(A&amp;cap;(A'&amp;cup;B'&amp;cup;C')) x 3 would give us the answer. I make this out to be 3 x (0.05 -(2 x 20/8000) - 1/8000) = 1077/8000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I add all those events up, then I get 7994/8000. I'm not sure what the right way to go about this is. I'd like to work it out for myself, but I really need a point in the right direction. I don't just want to make this work for three transactions, but for as many as I want so I need a general formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;" OwnerUserId="1094" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-26T16:21:44.043" LastActivityDate="2010-08-26T16:21:44.043" Title="Probability of a certain number of independent events occurring" Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="3369" PostTypeId="2" ParentId="3368" CreationDate="2010-08-26T14:26:47.563" Score="3" ViewCount="0" Body="&lt;p&gt;You need to look up the binomial distribution. One online resource is Grinstead and Snell's &lt;a href=&quot;http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-08-26T14:26:47.563" CommentCount="1" />
  <row Id="3370" PostTypeId="2" ParentId="3368" CreationDate="2010-08-26T14:28:37.073" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_distribution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-26T14:28:37.073" CommentCount="1" />
  <row Id="3371" PostTypeId="2" ParentId="3366" CreationDate="2010-08-26T14:36:25.543" Score="2" ViewCount="0" Body="&lt;p&gt;Yes; the function is defined to take equivalence classes as inputs, and return reals as values. If you put in the class of $(x,y)$, which is $[(x,y)] = \{(a,b)\in \mathbb{R}\times\mathbb{R}^{\star}\,|\, (x,y)\sim (a,b)\}$, it returns a real number $r$. Which real number? The real number $r$ such that $x=yr$. (I'm using $\mathbb{R}^{\star}$ to denote the nonzero real numbers) The function takes equivalence classes as inputs, and has real numbers as outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need to make sure that this function is &lt;em&gt;well defined&lt;/em&gt;, though; that is, the definition of the function depends on the name you give the equivalence class; but the class $[(x,y)]$ has many different &quot;names&quot;; it can also be called $[(a,b)]$ for some other real numbers $a$ and $b$. So you need to make sure that if $[(x,y)]=[(a,b)]$ (that is, if $(x,y)\sim (a,b)$) then the real number $r$ such that $x=yr$ is the same as the real number $s$ such that $a=bs$. Otherwise, what you describe would not be a function.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-26T14:36:25.543" CommentCount="1" />
  <row Id="3372" PostTypeId="2" ParentId="3364" CreationDate="2010-08-26T14:37:01.227" Score="3" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Connectedness and path connectedness&lt;/li&gt;&#xA;&lt;li&gt;If $f$ is a local homeomorphism, then $f$ is an open and closed map.&lt;/li&gt;&#xA;&lt;li&gt;If $f$ is onto and the domain is normal (can separate closed sets) or the map has compact fibers, then the image will be Hausdorff if the domain is.&lt;/li&gt;&#xA;&lt;li&gt;Second-countability is preserved under open maps.&lt;/li&gt;&#xA;&lt;li&gt;The image of a simply connected space need not be simply connected. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's off the top of my head.. there are many more. Also, this should probably be community wiki.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-08-26T14:37:01.227" />
  <row Id="3373" PostTypeId="2" ParentId="3368" CreationDate="2010-08-26T14:37:42.957" Score="3" ViewCount="0" Body="&lt;p&gt;Your calculation for p(1), which I don't completely follow, is somewhat off. Another, more straightforward way of calculating this would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(\textrm{1 transaction declined}) = {3 \choose 1} \times .05 \times .95 \times .95 = \frac{1083}{8000}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this exactly accounts for the $\frac{6}{8000}$ you are missing.&lt;/p&gt;&#xA;" OwnerUserId="1507" LastActivityDate="2010-08-26T14:37:42.957" CommentCount="1" />
  <row Id="3375" PostTypeId="2" ParentId="3366" CreationDate="2010-08-26T14:40:07.670" Score="2" ViewCount="0" Body="&lt;p&gt;That looks right to me.  Recall that a function $f : X \to Y$ is formally a subset of $X \times Y$, i.e. a set of ordered pairs $(x,y), x \in X, y \in Y$, with the property that if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $X$ is the set $E$ of equivalence classes, and $Y = \mathbb{R}$.  You are told that $f$ is the set of ordered pairs $([(x,y)],r)$ for which $x = yr$.  One has to check that this is well-defined with respect to the equivalence relation ~: if $x=yr$ and $(x,y)~(x',y')$, then $x'=y'r$.  Then $f$ makes sense as a set.  To check that it is a function, one should check that if $([(x,y)], r) \in f$ and $([(x,y)], r') \in f$ (i.e. $x=yr$ and $x=yr'$) then $r=r'$ (which should follow since $y \ne 0$ by assumption).&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-08-26T14:40:07.670" />
  <row Id="3376" PostTypeId="1" AcceptedAnswerId="3377" CreationDate="2010-08-26T14:45:56.637" Score="6" ViewCount="247" Body="&lt;p&gt;Suppose $f$ and $g$ are are two polynomials with complex coefficents (i.e $f,g \in \mathbb{C}[x]$).&#xA;Let $m$ be the order of $f$ and let $n$ be the order of $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some general conditions where  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$fg= \alpha x^{n+m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some non-zero $\alpha \in \mathbb{C}$ &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-26T15:17:54.337" LastActivityDate="2010-08-28T23:33:11.693" Title="When does the product of two polynomials = $x^{k}$ ?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="3377" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T14:52:12.307" Score="11" ViewCount="0" Body="&lt;p&gt;Polynomials over $\mathbb{C}$ (in fact, over any field) are a Unique Factorization Domain (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Unique_factorization_domain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Unique_factorization_domain&lt;/a&gt;); since $x$ is an irreducible, the only way for that to happen is for $f=ax^m$ and $g=bx^n$, with $ab=\alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you don't want to bring in the sledgehammer of unique factorization, you can just do it explicitly: look at the lowest nonzero term in $f$ and the lowest nonzero term in $g$; their product will be the lowest nonzero term in $fg$, hence must be of degree $m+n$. Since the degree of the lowest nonzero term of $f$ is at most $m$ and the one of $g$ is at most $n$, you have that they must be exactly of degree $m$ and $n$, respectively, and you get the result)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-26T15:56:52.853" LastActivityDate="2010-08-26T15:56:52.853" CommentCount="1" />
  <row Id="3378" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T14:53:17.450" Score="4" ViewCount="0" Body="&lt;p&gt;The answer just occured to me.   The roots of $f$ and $g$ must be be at 0. &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-26T14:53:17.450" CommentCount="0" />
  <row Id="3379" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T14:55:32.783" Score="0" ViewCount="0" Body="&lt;p&gt;I am not sure if it would be considered a classic or not, but I absolutely love &lt;a href=&quot;http://books.google.com/books?id=MSbMZCWX2OQC&amp;amp;printsec=frontcover&amp;amp;dq=differential+equations+with+boundary-value+problems&amp;amp;source=bl&amp;amp;ots=-uLwOGzUzB&amp;amp;sig=s5kdyM3Sq0txRfdt9VeIc-XmiKI&amp;amp;hl=en&amp;amp;ei=I352TIvBNoK8lQe12IjsCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;sqi=2&amp;amp;ved=0CEMQ6AEwBQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Differential Equations with Boundary-Value Problems&lt;/a&gt; by Dennis G. Zill and Michael R. Cullen.  I feel as though the definitions are written and explained in such a way, that if one was not a mathematician, they could still grasp a basic understanding of many differential equations concepts.  I am not familiar with many other elementary differential equations textbooks, but liked that this text goes into some details regarding &lt;a href=&quot;http://planetmath.org/encyclopedia/FourierSineAndCosineSeries.html&quot; rel=&quot;nofollow&quot;&gt;Fourier series&lt;/a&gt;, &lt;a href=&quot;http://planetmath.org/encyclopedia/ApproximatingFourierIntegralsWithDiscreteFourierTransforms.html&quot; rel=&quot;nofollow&quot;&gt;integrals&lt;/a&gt;, and &lt;a href=&quot;http://planetmath.org/encyclopedia/FourierTransform.html&quot; rel=&quot;nofollow&quot;&gt;transforms&lt;/a&gt;, Legendre series, Bessel functions, and the &lt;a href=&quot;http://planetmath.org/encyclopedia/EigenvalueProblem.html&quot; rel=&quot;nofollow&quot;&gt;Sturm-Louiville problem&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still use this text as a reference for some of my graduate level ODE issues, as it is more comprehensible than most graduate level Diff-EQ texts. I also use it quite frequently when tutoring students in undergraduate level differential equations, as it contains a myriad of detailed examples and worked problems as well as a variety of real-world applications.&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-08-26T14:55:32.783" CommunityOwnedDate="2010-08-26T14:55:32.783" />
  <row Id="3380" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T14:57:11.063" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, the intuitively evident ones: all other terms in f and g must vanish.  To see this, note that the product of the constant terms of f and g equals the constant term of fg, which is zero, whence at least one of these polynomials is multiple of x.  Without any loss of generality assume it is f.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fg = x (f/x) g,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implying (f/x) g is a multiple of x^(n+m-1).  By induction this reduces us to the case n+m=0, which is trivial (because f and g then have no other terms). QED.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-26T14:57:11.063" />
  <row Id="3381" PostTypeId="2" ParentId="3376" CreationDate="2010-08-26T15:00:54.843" Score="6" ViewCount="0" Body="&lt;p&gt;It's true over any domain $\rm D$ since then $x$ is prime (via $\rm D[x]/x = D$ a domain), and products of primes always factor uniquely in domains. It fails over non-domains, e.g. $\rm x = (2x+3)(3x+2) \in \mathbb Z/6[x].$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T23:33:11.693" LastActivityDate="2010-08-28T23:33:11.693" />
  <row Id="3382" PostTypeId="1" AcceptedAnswerId="3385" CreationDate="2010-08-26T15:06:22.447" Score="6" ViewCount="244" Body="&lt;p&gt;A couple of years ago I found the following continued fraction for $\frac{1}{e-2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{e-2} = 1+\frac{1}{2 + \frac{2}{3 + \frac{3}{4 + \frac{4}{5 + \frac{5}{6 + \frac{6}{7 + \frac{7}{\cdots}}}}}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from fooling around with the well-known continued fraction for $\phi$. Can anyone here help me figure out why this equality holds? &lt;/p&gt;&#xA;" OwnerUserId="1507" LastEditorUserId="1507" LastEditorDisplayName="" LastEditDate="2010-08-26T15:23:00.980" LastActivityDate="2010-10-04T15:43:12.023" Title="Continued fraction for $\frac{1}{e-2}$" Tags="&lt;continued-fractions&gt;" AnswerCount="2" CommentCount="17" />
  <row Id="3383" PostTypeId="2" ParentId="3335" CreationDate="2010-08-26T15:21:54.960" Score="0" ViewCount="0" Body="&lt;p&gt;Though I'm not sure if it exactly meets your criteria, my favorite PDE book is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470054565&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Partial Differential Equations: An Introduction&lt;/em&gt; by Walter A. Strauss&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-26T15:21:54.960" CommunityOwnedDate="2010-08-26T15:21:54.960" />
  <row Id="3384" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T15:50:37.520" Score="1" ViewCount="0" Body="&lt;p&gt;If you understand Argand diagrams (the representation of complex numbers in the complex plane) and can envision the unit circle in it, you can easily do this in your head:&lt;br&gt;&#xA;-1 is 1 rotated over $\pi$ radians. The square root of a number on the unit circle is the number rotated over half the angle, so $i$, or $\sqrt{-1}$ is 1 rotated over $\pi/2$ radians. To find $\sqrt{i}$ you just half the angle again: $\pi/4$ radians. The corresponding real and imaginary parts are $\cos\frac{\pi}{4}$ and $\sin\frac{\pi}{4}$ resp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;LaTexified (thanks Agosti)&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-28T05:38:14.267" LastActivityDate="2010-08-28T05:38:14.267" CommentCount="8" />
  <row Id="3385" PostTypeId="2" ParentId="3382" CreationDate="2010-08-26T15:52:19.177" Score="8" ViewCount="0" Body="&lt;p&gt;Euler proved in &quot;&lt;em&gt;De Transformatione Serium in Fractiones Continuas&lt;/em&gt;&quot; &lt;strong&gt;Reference: The Euler Archive, Index number E593&lt;/strong&gt; (On the&#xA;Transformation of Infinite Series to Continued Fractions) [Theorem VI, §40 to §42] that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=\dfrac{1}{1+\dfrac{2}{2+\dfrac{3}{3+\cdots }}}=\dfrac{1}{e-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here his an explanation on how he proceeded. He stated that if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{a+\dfrac{b}{b+\dfrac{c}{c+\cdots }}}=s,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+\dfrac{a}{a+\dfrac{b}{b+\dfrac{c}{c+\cdots }}}=\dfrac{s}{1-s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, in this case, we have $a=1,b=2,c=3,\ldots $ it follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+\dfrac{1}{1+\dfrac{2}{2+\dfrac{3}{3+\cdots }}}=\dfrac{1}{e-2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Euler proves first how to transform an alternating series of a particular type into a continued fraction and then uses the expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{-1}=1-\dfrac{1}{1}+\dfrac{1}{1\cdot 2}-\dfrac{1}{1\cdot 2\cdot 3}+\ldots &#xA; .$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;REFERENCES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Euler Archive, Index number E593&lt;/strong&gt;, &lt;a href=&quot;http://www.math.dartmouth.edu/~euler/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.dartmouth.edu/~euler/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Translation of Leonhard Euler's paper by Daniel W. File&lt;/strong&gt;, The Ohio State University. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T15:43:12.023" LastActivityDate="2010-10-04T15:43:12.023" CommentCount="2" />
  <row Id="3386" PostTypeId="1" AcceptedAnswerId="3394" CreationDate="2010-08-26T16:10:28.410" Score="4" ViewCount="143" Body="&lt;p&gt;Hello guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a question in my task sheet. The question is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(43*93*47*97)/3007=X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the exact value of X. I've tried a lot, but couldn't find easier way to do it without calculator, which of course, is not allowed in exam. There are no options, they're just asking the value of X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would love if someone could help to give &lt;em&gt;Method&lt;/em&gt; to solve the problem. As I said, I know how to solve above problem with the help of calculators and I've already found the factorization with help of calc, but no luck in manual mode. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. :)&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-26T16:25:25.350" LastActivityDate="2010-08-26T20:11:45.057" Title="Division with 4 digit number in denominator" Tags="&lt;arithmetic&gt;" AnswerCount="5" CommentCount="9" />
  <row Id="3387" PostTypeId="2" ParentId="3382" CreationDate="2010-08-26T16:23:01.547" Score="7" ViewCount="0" Body="&lt;p&gt;Another possibility: remember that the numerators and denominators of successive convergents of a continued fraction can be computed using a three term recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a continued fraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\dots}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with nth convergent $\frac{C_n}{D_n}$, the recurrence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}C_n\\D_n\end{bmatrix}=b_n\begin{bmatrix}C_{n-1}\\D_{n-1}\end{bmatrix}+a_n\begin{bmatrix}C_{n-2}\\D_{n-2}\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with starting values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}C_{-1}\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$, $\begin{bmatrix}C_{0}\\D_{0}\end{bmatrix}=\begin{bmatrix}b_0\\1\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $b_j=j+1$ and $a_j=j$, you now try to find a solution for those two difference equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave to you to verify that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n=\frac{(n+3)!}{n+2}\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_n=\frac{(n+3)!}{n+2}\left(1-2\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are solutions to the two difference equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide $C_n$ by $D_n$ and take the limit as $n\to\infty$; you should get the expected result.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-26T16:23:01.547" CommentCount="6" />
  <row Id="3388" PostTypeId="1" AcceptedAnswerId="3390" CreationDate="2010-08-26T16:31:27.307" Score="1" ViewCount="130" Body="&lt;p&gt;Hello guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a problem I tried to solve, but couldn't because I don't know the method to solve it and I've never come across such problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PQ)²=RSP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where P, Q, R, S are distinct single digit natural numbers, then R=?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find out the value of R, and given option are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried solving it by option with newly created equation as Q²=RS/P and taking values of R from above options and other variables as distinct single digit numbers other than R, I've gone to a level after which I've got myself a lot confused and didn't know what to do. I think there is easier method to solve it which I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate if someone could give me a hand with this. Thanks in advance. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: The answer is c) 3&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-26T19:43:50.210" LastActivityDate="2010-08-28T01:18:31.813" Title="(PQ)²=RSP, where P,Q,R,S are distinct single digit natural numbers, then R=?" Tags="&lt;arithmetic&gt;&lt;cryptarithm&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3389" PostTypeId="1" AcceptedAnswerId="3391" CreationDate="2010-08-26T16:39:31.910" Score="1" ViewCount="211" Body="&lt;p&gt;Let $X = \mathbb{R}^n$ and let $Z$ be a closed subspace of $X$.  The inclusion map $i: Z \hookrightarrow X$ induces a map $C_0(X) \to C_0(Z)$ (via pullback).  My question is simple: what is the kernel?  The answer should either be $C_0(X - Z)$ or $C(X - Z)$ and I think I know which one is correct.  But I need a sanity check, so I thought I would post the question here.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastEditorUserId="1509" LastEditorDisplayName="" LastEditDate="2010-11-13T15:25:48.527" LastActivityDate="2010-11-13T15:25:48.527" Title="A silly question about C_0" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3390" PostTypeId="2" ParentId="3388" CreationDate="2010-08-26T16:40:51.550" Score="5" ViewCount="0" Body="&lt;p&gt;The steps are as follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A square number can only end in $1, 4, 5, 6, 9$.  To prove this one uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;.  So $P = 1, 4, 5, 6, 9$.&lt;/li&gt;&#xA;&lt;li&gt;$40^2 &gt; 999$, so $P = 1$.  &lt;/li&gt;&#xA;&lt;li&gt;Since $(PQ)^2$ ends in $1$, it follows that $Q = 1, 9$.&lt;/li&gt;&#xA;&lt;li&gt;Since $P \neq Q$, it follows that $Q = 9$.&lt;/li&gt;&#xA;&lt;li&gt;So the digits are $19^2 = 361$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T16:40:51.550" CommentCount="1" />
  <row Id="3391" PostTypeId="2" ParentId="3389" CreationDate="2010-08-26T16:42:18.213" Score="2" ViewCount="0" Body="&lt;p&gt;If $C_0$ denotes functions vanishing at infinity, then it is $C_0(X\setminus Z)$. The other candidate is much to big to be an ideal in $C_0(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, $C_0$ denotes compactly supported functions, then Robin Chapman's answer applies.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-27T13:49:08.833" LastActivityDate="2010-08-27T13:49:08.833" CommentCount="2" />
  <row Id="3392" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T16:44:37.667" Score="7" ViewCount="0" Body="&lt;p&gt;Visually, the square root of a complex number (written in polar coordinates) $(\rho,\theta)$ is the number $(\sqrt\rho,\theta/2)$.&lt;img src=&quot;http://i.imgur.com/Gxqh7.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2010-08-27T22:00:19.900" LastActivityDate="2010-08-27T22:00:19.900" CommentCount="2" />
  <row Id="3393" PostTypeId="2" ParentId="3389" CreationDate="2010-08-26T16:46:45.587" Score="2" ViewCount="0" Body="&lt;p&gt;I presume $C_0(X)$ means compactly supported continuous&#xA;functions on $X$. So you are asking which compactly supported continuous&#xA;functions on $X$ which vanish on $Z$. Now these may not be compactly supported&#xA;on $X-Z$ for there may be a point $p$ in $Z$ and a sequence of points $(p_n)$&#xA;in converging to $X$ with $f(p_n)\ne0$ but $f(p)=0$. On the other hand there&#xA;will be continuous functions on $X-Z$ which don't converge to zero on the boundary.&#xA;So neither $C(X-Z)$ nor $C_0(X-Z)$ is correct!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-26T16:46:45.587" CommentCount="1" />
  <row Id="3394" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T16:53:12.227" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\quad\rm mod \; 97: \:\quad 100 \;\equiv\; 3 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\; 3007 \;\equiv\; 30 \cdot 100 + 7$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad\;\; \;\equiv\; 30 \:\;\cdot\;\: 3 \;\: + \; 7 \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad\;\; \;\equiv\; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. &lt;strong&gt;cast out 97's&lt;/strong&gt; in analogy to &lt;strong&gt;cast out nines&lt;/strong&gt;. See also &lt;a href=&quot;http://google.com/groups?selm=y8zwsvvxdmy.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; where I discuss &lt;strong&gt;casting out 91's&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-26T17:13:50.477" LastActivityDate="2010-08-26T17:13:50.477" CommentCount="4" />
  <row Id="3395" PostTypeId="1" AcceptedAnswerId="3396" CreationDate="2010-08-26T16:57:52.420" Score="1" ViewCount="53" Body="&lt;p&gt;Hello guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one more question I need help with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Samir has some amount of rupees with him, the difference between 45.45% and 36.36% of which is Rs. 789435689. Find the amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 8683792579&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) 8673792579&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) 7573781479&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) 88667292579&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) None of these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Samir is the name of Indian male, rupess is Indian currency and Rs. is symbol for Indian currency. You can try replacing those with $ and question will be the same. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've done. I've subtracted 45.45%-36.36%=9.09. I've multiplied 789435689*9.09/100. Which should be answer, but it isn't and I don't know why. Am I doing something wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual answer is a) 8683792579.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone elaborate how its a) and if not, my method is correct or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance and really sorry for asking too many stupid questions. :(&lt;/p&gt;&#xA;" OwnerUserId="712" LastActivityDate="2010-08-26T17:19:25.260" Title="Difference between Percentage value calculation" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3396" PostTypeId="2" ParentId="3395" CreationDate="2010-08-26T17:19:25.260" Score="2" ViewCount="0" Body="&lt;p&gt;789435689 is 9.09% or 9.09/100 of the whole amount&amp;mdash;that is, $789435689=\frac{9.09}{100}x$ where $x$ is the whole amount&amp;mdash;so you want to divide by 9.09/100 rather than multiplying by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also may be worth noting that $0.\overline{36}=\frac{4}{11}$ and $0.\overline{45}=\frac{5}{11}$, so $0.\overline{09}=\frac{1}{11}$, so you are essentially multiplying by 11, which can be done by multiplying by 10 and adding the original number or by writing two copies of the number, shifted one place, and adding:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   789435689&#xA;+ 789435689&#xA;------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-26T17:19:25.260" CommentCount="1" />
  <row Id="3397" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T17:20:27.413" Score="2" ViewCount="0" Body="&lt;p&gt;This must be a multiple choice question that must be done under a time limit... so approximate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{43\cdot 93\cdot 47 \cdot 97}{3007} \approx \frac{50 \cdot 100 \cdot 50 \cdot 100}{3000}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily reduce this to $\frac{5 \cdot 100 \cdot5 \cdot 10}{3}$ which is simply $\frac{25000}{3} = 8333 \frac{1}{3}$.  &lt;em&gt;This is not all that satisfactory&lt;/em&gt; since the answer is $6063$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's be a little finer with our approximation (while being no more painful):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{43\cdot 93\cdot 47 \cdot 97}{3007} \approx \frac{40 \cdot 100 \cdot 50 \cdot 100}{3000}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we are now looking at $\frac{40 \cdot 100 \cdot 5}{3}$ which is simply $\frac{20000}{3} = 6666 \frac{2}{3}$.  That should be close enough, but we already have more information than you may think, because the true answer is less than both of these estimates.  &lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-26T18:54:54.070" LastActivityDate="2010-08-26T18:54:54.070" CommentCount="2" />
  <row Id="3398" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T17:45:15.867" Score="2" ViewCount="0" Body="&lt;p&gt;(20:30)&#xA;First we notice that 43, 47 and 97 are all prime numbers and 93=3x31. Clearly 3 is not a factor of 3007, let's see about the rest using the good ol' Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;43 is clearly not (as 43x7=280+21=301 therefore 300=43x6+42, and 427 is clearly not divisible by 43)&lt;/li&gt;&#xA;&lt;li&gt;in 47 case, 300=6x(47+3) therefore the reminder is 47-18=29, and again 297 is clearly not divisible by 47&lt;/li&gt;&#xA;&lt;li&gt;31 gives us 310=31x10 so 300=9x31+21, and 217 equals to 31x7. So we have one divisor, that is 97x31 = 3007&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore the result is 43x47x3 which should be a simple calculation.&#xA;(20:38)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 minutes, would have taken 5 if I'd began with 31 :)&#xA;I tried to describe my thoughts into words and be not very formal. I hope it's clear and it gave you some insight about how I solved it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;br&gt;&#xA;Say that it was the case where no factors are common, what's then? Then we have two options either use approximations by noticing how 97x93 is (95+2)(95-2) and same with 43x47=(45+2)(45-2) so it'd be simpler to try approximating 45x45x95x95/3000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can take the numbers you've calculated in with the Euclidean algorithm when you checked for common factors and you'd have 3007/43, 3007/47, etc. So you can just multiply their inverses for the result.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-08-26T17:50:35.273" LastActivityDate="2010-08-26T17:50:35.273" CommentCount="1" />
  <row Id="3399" PostTypeId="1" CreationDate="2010-08-26T17:49:36.227" Score="9" ViewCount="738" Body="&lt;p&gt;A function $f : \mathbb{R} \to \mathbb{R}$ is &lt;b&gt;convex&lt;/b&gt; (or &quot;concave up&quot;) provided that for all $x,y \in \mathbb{R}$ and $t \in [0,1]$,&#xA;$$f(tx + (1-t)y) \le tf(x) + (1-t)f(y).$$&#xA;Equivalently, a line segment between two points on the graph lies &lt;em&gt;above&lt;/em&gt; the graph, the region &lt;em&gt;above&lt;/em&gt; the graph is convex, etc.  I want to know why the word &quot;convex&quot; goes with the inequality in this direction, and how I can remember it.  Every reason I have heard makes just as much sense applied to the opposite inequality (&quot;concave down&quot;).&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-26T18:15:49.333" LastActivityDate="2010-12-13T05:43:05.053" CommunityOwnedDate="2010-08-28T03:33:15.160" Title="Why does &quot;convex function&quot; mean &quot;concave *up*&quot;?" Tags="&lt;soft-question&gt;&lt;real-analysis&gt;&lt;terminology&gt;&lt;functions&gt;" AnswerCount="8" CommentCount="3" />
  <row Id="3400" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T18:00:14.660" Score="6" ViewCount="0" Body="&lt;p&gt;One of my professors told me the following memorable line: &quot;A concave function looks like the roof of a cave.&quot; which helps me remember what is a concave and what is a convex function.&lt;/p&gt;&#xA;" OwnerUserId="116" LastActivityDate="2010-08-26T18:00:14.660" CommentCount="6" />
  <row Id="3401" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T18:00:58.597" Score="6" ViewCount="0" Body="&lt;p&gt;Not sure why convex is defined that way, but one way to remember is that the derivative is monotonically increasing for some convex functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe just remember that $e^x$ is conv$e^x$. (I just thought of this one!)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-26T18:00:58.597" CommentCount="3" />
  <row Id="3402" PostTypeId="2" ParentId="3363" CreationDate="2010-08-26T18:15:42.820" Score="6" ViewCount="0" Body="&lt;p&gt;The number of loops is just the number of pairs of non-intersecting paths s.t. first one goes from (0,1) to (m-1,n) and the second one goes from (1,0) to (m,n-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-intersecting paths on a lattice are counted by some &lt;a href=&quot;http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/&quot; rel=&quot;nofollow&quot;&gt;determinant formula&lt;/a&gt;. In this case it's just $\det\left(\begin{matrix}\binom{m+n-2}{m-1}&amp;amp;\binom{m+n-2}{m-2}\\\binom{m+n-2}{n-2}&amp;amp;\binom{m+n-2}{n-1}\end{matrix}\right)=\binom{m+n-2}{m-1}^2-\binom{m+n-2}{m-2}\binom{m+n-2}{n-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to prove this formula directly: a pair (path from (0,1) to (m-1,n); path from (1,0) to (m,n-1)) either forms a loop without intersection or (if paths intersect) can be (canonically) identified with a pair (path from (0,1) to (m,n-1); path from (0,1) to (m,n-1).&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-08-26T18:22:44.583" LastActivityDate="2010-08-26T18:22:44.583" CommentCount="1" />
  <row Id="3403" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T18:16:42.187" Score="1" ViewCount="0" Body="&lt;p&gt;A line is said to &quot;support&quot; the graph of a function (or indeed, any subset of the Cartesian plane) if it &quot;holds up&quot; the graph: that is, the graph lies entirely above or on the line.  (After all, gravity pulls downward!)  We might think of the union of all support lines as the &quot;ground&quot; on which the graph lies; everything else--its set-theoretic complement--is the &quot;sky&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function of the real numbers is convex if and only if its graph is the boundary between the ground and the sky.  This is a special case of the more general idea of convexity that applies to arbitrary planar regions, the same as the familiar distinction between a convex and non-convex polygon, for example.  For arbitrary regions there is no definite &quot;up&quot; and &quot;down&quot; anymore, though, so we say that a line supports a region when the region lies entirely within one of the two closed half-planes bounded by that line.  (Thus, the interior and boundary of a convex polygon form its &quot;sky&quot; and everything outside is the &quot;ground.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, calling a &quot;concave upward&quot; function &quot;convex&quot; unites two closely related familiar concepts and is justified by the universal earthbound human experience that gravity usually pulls downward.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-26T18:16:42.187" CommentCount="2" />
  <row Id="3404" PostTypeId="2" ParentId="3356" CreationDate="2010-08-26T18:21:07.040" Score="5" ViewCount="0" Body="&lt;p&gt;Here is a different way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x+y) = f(x) f(y) - g(x) g(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate wrt $y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x+y)  = f(x)f'(y) - g(x) g'(y)$ , put $y = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x) = -g(x)* g'(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g'(x) = f(x) * g'(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f(x)*f'(x) + g(x)*g'(x) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the function $f^{2}(x) + g^{2}(x)$ is constant, as it's derivative is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0) = f^2(0) - g^{2}(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(0) = 2f(0)g(0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring and adding both we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{2}(0) + g^{2}(0) = (f^{2}(0) + g^{2}(0))^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $f^{2}(0) + g^{2}(0) = 0$ then because $f^{2}(x) + g^{2}(x)$ is a constant, we get $f(x) = 0$ which implies $f$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f^{2}(0) + g^{2}(0) = 1$ and hence $f^{2}(x) + g^{2}(x) = 1$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-26T18:21:07.040" />
  <row Id="3405" PostTypeId="1" AcceptedAnswerId="3517" CreationDate="2010-08-26T19:10:14.380" Score="32" ViewCount="5619" Body="&lt;p&gt;Some years ago I was interested in the following Markov chain&#xA;whose state space is the positive integers. The chain begins at state &quot;1&quot;, &#xA;and from state &quot;n&quot; the chain next jumps to a state uniformly&#xA;selected from {n+1,n+2,...,2n}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time goes on, this chain goes to infinity, with occasional&#xA;large jumps. In any case, the chain is quite unlikely to hit any&#xA;particular large n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define p(n) to be the probability that this chain&#xA;visits state &quot;n&quot;, then p(n) goes to zero like c/n for some &#xA;constant c. In fact, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$    np(n) \to c = {1\over 2\log(2)-1} = 2.588699. \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove this convergence, I recast it as an analytic &#xA;problem. Using the Markov property, you can see that the sequence&#xA;satisfies: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$   p(1)=1\quad\mbox{ and }\quad p(n)=\sum_{\lceil n/2\rceil}^{n-1} {p(j)\over j}\mbox{ for }n&gt;1.  \qquad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some weeks, using generating functions etc. I tried and failed to find&#xA;an analytic proof of the convergence in (1). Finally, at a conference&#xA;in 2003 Tom Mountford showed me a (non-trivial) probabilistic proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the result is true, but since then I've&#xA;continued to wonder if I missed something obvious. Perhaps there is&#xA;a standard technique for showing that (2) implies (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a direct (short?, analytic?) proof of (1)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps someone who understands sequences better than I do could take a shot at this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I'm digging through my old notes on this. I now remember that I had a proof (using generating functions) that &lt;em&gt;if&lt;/em&gt; $\ np(n)$ converges, then the limit is $1\over{2\log (2)-1}$. It was the convergence that eluded me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found some curiosities like: $\sum_{n=1}^\infty {p(n)\over n(2n+1)}={1\over 2}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another update&lt;/strong&gt;: Here is the conditional result mentioned above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in Qiaochu's answer, define $Q$ to be the generating function of $p(n)/n$, that is,&#xA; $Q(t)=\sum_{n=1}^\infty {p(n)\over n} t^n$ for $0\leq t&amp;lt;1$.&#xA; Differentiating gives&#xA; $$Q^\prime(t)=1+{Q(t)-Q(t^2)\over 1-t}.$$&#xA; This is slightly different from Qiaochu's expression because&#xA; $p(n)\neq \sum_{j=\lceil n/2\rceil}^{n-1} {p(j)\over j}$ when $n=1$,&#xA; so that $p(1)$ has to be treated  separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating again and multiplying by $1-t$, we get&#xA; $$(1-t)Q^{\prime\prime}(t)=-1+2\left[Q^\prime(t)-t Q^\prime(t^2)\right],$$&#xA; that is,&#xA; $$(1-t)\sum_{j=0}^\infty (j+1) p(j+2) t^j =  -1+2\left[\sum_{j=1}^\infty (jp(j)) {t^j-t^{2j}\over j}\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\lim_n np(n)=c$ exists. Letting $t\to 1$ above the left hand side gives $c$,&#xA; while the right hand side is $-1+2c\log(2)$ and hence $c={1\over 2\log(2)-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\sum_{j=1}^\infty {t^j-t^{2j}\over j}=\log(1+t).$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New update: (Sept. 2)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an alternative proof of the conditional result that my colleague Terry Gannon&#xA; showed me in 2003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the sum $\sum_{n=2}^{2N}\ p(n)$, substitute the formula in the title,&#xA; exchange the variables $j$ and $n$, and rearrange to establish the identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1\over 2}=\sum_{j=N+1}^{2N} {j-N\over j}\ {p(j)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $jp(j)\to c$, then&#xA;$1/2=\lim_{N\to\infty} \sum_{j=N+1}^{2N} {j-N\over j^2}\  c=(\log(2)-1/2)\  c,$ so that&#xA;$c={1\over 2\log(2)-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New update: (Sept. 8)&lt;/strong&gt; Despite the nice answers and interesting discussion below, I am still holding out for an (nice?, short?) analytic proof of convergence. Basic Tauberian theory is allowed :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New update: (Sept 13)&lt;/strong&gt; I have posted a sketch of the probabilistic proof of convergence under &quot;A fun and frustrating recurrence sequence&quot; in the &quot;Publications&quot; section of my homepage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Update:&lt;/strong&gt; (Sept 15th) The deadline is approaching, so I have decided to award the bounty to T.. &#xA;Modulo the details(!), it seems that the probabilistic approach is the most likely to lead to a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sincere thanks to everyone who worked on the problem, including those who tried it but &#xA;didn't post anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, I &lt;em&gt;did&lt;/em&gt; get an answer to my question: there doesn't seem to be an easy, or standard&#xA;proof to handle this particular sequence.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-09-15T14:49:15.493" LastActivityDate="2010-11-25T23:04:30.593" Title="Convergence of $np(n)$ where $p(n)=\sum_{j=\lceil n/2\rceil}^{n-1} {p(j)\over j}$" Tags="&lt;sequences-and-series&gt;&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="6" />
  <row Id="3406" PostTypeId="2" ParentId="3399" CreationDate="2010-08-26T19:23:36.233" Score="0" ViewCount="0" Body="&lt;p&gt;I don't know if this is the original reason for the terminology, but it makes a lot of sense when you look at optimization problems. Convex functions make minimization &quot;easy&quot;, in the sense that if you have a convex function on a convex domain, it has a unique local minimum which is also the global minimum, and gradient descent works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were typically looking at &lt;em&gt;maximization&lt;/em&gt; problems instead, it would make sense to switch the meanings of convex and concave functions. But since many problems of practical and/or theoretical interest are naturally expressed as minimization (of potential energy, cost, area, etc.), we have &quot;convex&quot; defined the way it is.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-26T19:23:36.233" CommentCount="1" />
  <row Id="3407" PostTypeId="2" ParentId="3362" CreationDate="2010-08-26T19:39:44.927" Score="0" ViewCount="0" Body="&lt;p&gt;The answer to your final question is no for arbitrary $\theta$.  Specifically, let $f(x) = 1/\sqrt(x)$ for $\epsilon &amp;lt; x &amp;lt;= 1$ and define $f$ however you like for $0 \le x &amp;lt; \epsilon$ &amp;lt;&amp;lt; .01, say.  (This can be done in a way that maintains the concavity and twice-differentiability of $f$.)  With $\theta = 3/4$, the sequence of Riemann sums begins (with $n=1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2., 2.04667, 2.056, 2.0579, 2.05757, 2.05647, 2.05511, 2.05368, ...,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;peaking at $n=4$.  (The sequence is monotonic after that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem occurs because there is a section at the end of the graph of $f$, in this case near zero, that grows quickly enough to appreciably alter the very first term of the sum once $n$ grows large enough; this section is not sampled by the sum for smaller $n$.  By modifying $f$ appropriately within disjoint intervals converging to 0, you should be able to construct an example whose sequence of Riemann sums wiggles arbitrarily many times.  What this hand-waving argument does not settle is whether this process can be carried out &lt;em&gt;ad infinitum&lt;/em&gt;: the concern is that the resulting $f$ might not be continuous at 0, which conceivably is part of your concept of &quot;well-behaved&quot; (although it is not necessary for the theory of Riemann integration, which permits some points of discontinuity).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-26T19:39:44.927" CommentCount="1" />
  <row Id="3408" PostTypeId="1" CreationDate="2010-08-26T19:41:40.073" Score="11" ViewCount="287" Body="&lt;p&gt;Hi,&#xA;Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="1513" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T00:24:35.693" LastActivityDate="2010-08-27T17:57:19.797" Title="Characterizing nonconstant entire functions with modulus 1 on the unit circle" Tags="&lt;complex-analysis&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="3409" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T19:54:27.973" Score="1" ViewCount="0" Body="&lt;p&gt;The numerator is about $16000000$, so the quotient is about $5300$, say plus or minus a thousand.  Now I am going to play some tricks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is $-9 \bmod 50$ and the denominator is $7 \bmod 50$, so $X \equiv 13 \bmod 50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is $6 \bmod 9$ and the denominator is $2 \bmod 9$, so $X \equiv 3 \bmod 9$.  Hence $X \equiv 213 \bmod 450$.  So $X$ can really only be one of $4263, 4713, 5163, 5613, 6063$, and maybe $6513$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator is $8 \bmod 11$ and the denominator is $4 \bmod 11$, so $X \equiv 2 \bmod 11$.  We now know the value of $X \bmod 4950$, which is more than enough.  Looking through the above list this gives $X = 6063$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T19:54:27.973" CommentCount="3" />
  <row Id="3410" PostTypeId="2" ParentId="3386" CreationDate="2010-08-26T20:11:45.057" Score="1" ViewCount="0" Body="&lt;p&gt;You can estimate it this way very quicly:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;43*93*47*97 = (50-7).(100-7).(50-3).(100-3) = (50-7).(50-3).(100-7).(100-3)&#xA;= (100²/4 - 10.100/2 + 21) * (100² - 10.100 + 21) =&#xA;= (100.(25-5)+21) * (100.(100-10)+21) = &#xA;= 2021 * 9021 ≈ 18 000 000 (if needed for aprox: greater error around 20/2000 =&amp;gt; 1%&#xA;&#xA;using 3000 instead of 3007 (error 7/3007 around 0,2% so less than 1% previous)&#xA;&#xA;18 000 000 / 3000 = 6000 if needed, you can reduce aproximation &#xA;correcting denominator error (1%) =&amp;gt; 6000 + 60 = 6060&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1514" LastActivityDate="2010-08-26T20:11:45.057" CommentCount="3" />
  <row Id="3411" PostTypeId="2" ParentId="3405" CreationDate="2010-08-26T20:13:31.557" Score="2" ViewCount="0" Body="&lt;p&gt;Here is an approach which should give a differential equation &lt;em&gt;without&lt;/em&gt; an $f(x^2)$ term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $s_{n} = \frac{p_n}{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $s_{1} = 1$ and $s_{2} = 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily see from the recursive formula for $p_{n}$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  $s_{2n+2} = s_{2n+1}$ for $n \ge 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)  $s_{2n+1} = s_{2n} - \frac{s_n}{2n+1}$ for $n \ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these identities and putting $Q(x) = \sum {s_{i} x^{i}}$ gives us the generating function similar to what Qiaochu got (with the $Q(x^2)$ term), but we probably cannot deal with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the $Q(x^2)$ term appears is that the recurrence in (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{2n+1} = s_{2n} - \frac{s_n}{2n+1}$ has both $s_{2n}$ and $s_{n}$, in order to avoid this, we aim to eliminate $s_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{4m+1} = s_{4m} - \frac{s_{2m}}{4m+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{4m-1} = s_{4m-2} - \frac{s_{2m-1}}{4m-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $s_{2m} = s_{2m-1}$ we can eliminate that easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the first by $4m+1$, the second by $4m-1$ and subtract (also use $s_{4m} = s_{4m-1}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4m+1)s_{4m+1} = 8ms_{4m-1} - (4m-3)s_{4m-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4m+3)s_{4m+3} = (8m+4)s_{4m+1} - (4m+1)s_{4m-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This I believe would be susceptible to standard generating function methods. (I believe it will give a differential equation of the form $A(x)*f''(x) + B(x)*f'(x) + C(x)*f(x) + D(x) = 0$ where $A(x),B(x),C(x),D(x)$ are polynomials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do recheck the algebra etc if you do choose to pursue this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Incidentally, using the identities (1) and (2) above, we can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_{2} - s_{2n+1}$ =  $\sum_{j=1}^{n} {\frac{s_j}{2j+1}} = \sum_{j=1}^{n}{\frac{p_j}{j(2j+1)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $s_{2n+1}$ tends to 0 (easy to prove), we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=1}^{\infty}{\frac{p_j}{j(2j+1)}} = s_{2} = \frac{1}{2}$ which proves the curiosity you spoke of.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-27T16:40:18.890" LastActivityDate="2010-08-27T16:40:18.890" CommentCount="6" />
  <row Id="3412" PostTypeId="1" AcceptedAnswerId="3415" CreationDate="2010-08-26T20:22:09.897" Score="1" ViewCount="85" Body="&lt;p&gt;I read this fascinating article:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.newscientist.com/article/mg20627550.200-enter-the-matrix-the-deep-law-that-shapes-our-reality.html&quot; rel=&quot;nofollow&quot;&gt;http://www.newscientist.com/article/mg20627550.200-enter-the-matrix-the-deep-law-that-shapes-our-reality.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately all the other papers I find googling are just not tangible to me :-(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone please point me to some material that bridges the gap from this popular science exposition to the hard core papers that seem to pile up in the Net?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-26T20:43:16.093" Title="Accessible Intro to Random Matrix Theory (RMT)" Tags="&lt;linear-algebra&gt;&lt;random&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="3413" PostTypeId="2" ParentId="3315" CreationDate="2010-08-26T20:25:20.577" Score="1" ViewCount="0" Body="&lt;p&gt;Gives a nice concise overview:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=Sqrt(i&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=Sqrt(i&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-26T20:25:20.577" />
  <row Id="3414" PostTypeId="2" ParentId="3408" CreationDate="2010-08-26T20:34:38.407" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;b&gt;Partial answer.&lt;/b&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|f(z)|=1$ for all $|z|=1$ and $f$ is entire function, then $f:\mathbb D \to \mathbb D$, by the Maximum Modulus Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Schwarz-Lemma we can characterize, at least, the conformal ones.&lt;br&gt;&#xA;( See Functions of one Complex Variable, Conway 2ed pag. 131 )&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-08-26T20:34:38.407" />
  <row Id="3415" PostTypeId="2" ParentId="3412" CreationDate="2010-08-26T20:43:16.093" Score="2" ViewCount="0" Body="&lt;p&gt;I like very much and I think that the exposition is very good and clear in the Deift's book:&#xA;Orthogonal polynomials and random matrices: a Riemann-Hilbert approach.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-08-26T20:43:16.093" />
  <row Id="3416" PostTypeId="2" ParentId="3336" CreationDate="2010-08-26T20:49:06.333" Score="1" ViewCount="0" Body="&lt;p&gt;Try this:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=sin&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=sin&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on all the &quot;more&quot;s - works of course for &quot;cos&quot; too!&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-08-26T20:49:06.333" />
  <row Id="3417" PostTypeId="2" ParentId="3408" CreationDate="2010-08-26T20:58:04.250" Score="6" ViewCount="0" Body="&lt;p&gt;Edit: How frustrating, this is still not correct. It is not hard to make this proof work by generalizing to the functions holomorphic on unit disc and continuous on its closure, but such approach will be nothing else than proving the theorem cited by Jonas Meyer (see his answer). &lt;br&gt;&#xA;Second attempt:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that $f$ has finitely many zeros in $\mathbb{D}$ (identity principle + compactness).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Observe that if $F\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}}),$ the only zero of $F$ is in zero, and $|F|$ is constant on unit circle, then $F(z) = e^{it}z^{k}$ where $k = ord_{0}F$. &lt;br&gt;&#xA;(apply the maximum principle for $F(z)/z^{k}$ and $z^{k}/F(z)$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $z_{1},\ldots, z_{n}$ be all pairwise distinct zeros of $f$ in $\mathbb{D}$. By induction on $n$ we will show that $n = 1$ and  $z_{1} = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let $n = 1$.&lt;br&gt;&#xA;We are taking $h \in Aut(\mathbb{D})$ s.t. $h(0) = z_{1}$ and $g:= f\circ h\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$.&lt;br&gt;&#xA;Then $|g|$ is constant on unit circle, $0$ is the only zero of $g$ in $\mathbb{D}$. So $g(z) = e^{it}z^{k}$.&lt;br&gt;&#xA;Therefore $f(z) = e^{it}\cdot \left(\frac{z - z_{1}}{1 - \bar{z_{1}}z}\right)^{k}$ in $\mathbb{D}$. &lt;br&gt;&#xA;From identity principle for meromorphic functions this equality holds in $\mathbb{C}$, and gives contradiction unless $z_{1} = 0$ (otherwise $f$ is not entire).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &gt; 1$, then we're reasoning by contradiction.&lt;br&gt; &#xA;W.l.o.g. $z_{n}\neq 0$.&lt;br&gt;&#xA;We are taking $h \in Aut(\mathbb{D})$ s.t. $h(0) = z_{n}$ and $g:= f\circ h\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$, and then $h^{-1}(z_{1}),\ldots, h^{-1}(z_{n-1}), 0$ are the all zeros of $g$ in $\mathbb{D}$.&lt;br&gt;&#xA;Let $k = ord_{0}g$, and $G(z) = g(z)/z^{k}\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$.&lt;br&gt;&#xA;$G$ has $n-1$ zeros on $\mathbb{D}$, $|G(z)| = 1$ on unit circle, &lt;del&gt;thus from inductive assumption $n-1 = 1$ or $G$ is constant&lt;/del&gt; (we cannot use inductive assumption since in general $G\not\in \mathcal{O}(\mathbb{C})$!).&lt;br&gt; &#xA;If $G$ is non-constant then $n-1 = 1$ and $G(z) = e^{it}z^{l}, \ l &gt; 0$ - a contradiction with $ord_{0}g = k$. Otherwise $n - 1 = 0$ - a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally 1. and 3. proves that all entire functions with desired property are of the form $f(z) = e^{it}z^{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following reasoning is not correct (see the Malik Younsi's answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;This should be the comment to the Leandro's answer, although I cannot add comments yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W.l.o.g. $f$ is non-constant. Then by maximum modulus $f$ has zero inside $\mathbb{D}$.&#xA;Denote this zero by $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composing $f$ with an automorphism $h$ of disc s.t. $h(0) = a$ we obtain that $g = f\circ h$ satisfies $g(0)= 0$ and moreover $|g(z)|\le |z|^{k}$ inside $\mathbb{D}$, where $k$ is the order of zero of $g$ in $0$ (indeed $g(z)/z^{k}$ is entire and we may use the maximum modulus principle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Schwarz lemma there exists $\theta\in \mathbb{R}$ satisfying $g(z) = e^{i\theta}z^{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From classification of automorphisms of the disc, we have $h^{-1}(z) = e^{i\omega}\frac{z-a}{1-\bar{a}z}$ for some $\omega\in \mathbb{R}$, and from this the full classification easily follows.&lt;/del&gt;&lt;/p&gt;&#xA;" OwnerUserId="1454" LastEditorUserId="1454" LastEditorDisplayName="" LastEditDate="2010-08-27T10:09:19.460" LastActivityDate="2010-08-27T10:09:19.460" CommentCount="1" />
  <row Id="3418" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T21:11:30.640" Score="0" ViewCount="0" Body="&lt;p&gt;The ubiquity of $e$ stems from the ubiquity of $e^x$. And $e^x$ is ubiquitous due to the fact that it permits one to solve any higher-order LDE with constant coefficients - either by factoring its differential operator over $\mathbb C$ into linear factors, or by converting it into linear system form and using matrix exponentials. See the exposition by Arnold below, from his beautiful textbook &quot;Ordinary differential equations&quot;. As Arnold says &lt;strong&gt;&quot;the exponential ... gives the solution of all differential equations quite generally&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gK98d.jpg&quot; alt=&quot;alt text&quot;&gt; &#xA;&lt;img src=&quot;http://i.imgur.com/PV2vn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T23:31:39.980" LastActivityDate="2010-08-28T23:31:39.980" />
  <row Id="3419" PostTypeId="1" AcceptedAnswerId="3421" CreationDate="2010-08-26T21:14:46.133" Score="13" ViewCount="608" Body="&lt;p&gt;Consider the following expression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;16313107343153908912074032799466965289077771751767944648966669091376847859711382649033004075188224&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a 98 decimal digit number. &#xA;This can be represented as 424^37 which has just 5 digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or consider this number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16907351492333570491078177094338635851326626269508214301784177607286611534146052484795771275896190661372675631981127803129649521785142469703500691286617000071805893890889531804648801423948258750240509470456335529389117581957525380043352452775597911297901564395967893517511308057315467512494189332252686430935249121855914917866181252548011072665616976069886958296149475308550144566145651839224313331840075767830022374277939322452695654072920143693336239042875755246628767670638296599817906363150743450787176422650055877626&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 200 decimal digit number can be simply expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;loge 56 ( log 56 base e) when we discard first 6 numbers and then consider first 200 digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or log56 (discard 6) count first 200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is, is it possible to represent any and every huge random number using very few characters as possible, theoritically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.....Also&#xA;Is there any standard way to reduce it mathematically?&lt;/p&gt;&#xA;" OwnerUserId="1279" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-28T13:23:07.827" LastActivityDate="2010-10-26T10:46:37.737" CommunityOwnedDate="2010-08-26T21:14:46.133" Title="Is it possible to represent every huge number in abreviated form?" Tags="&lt;number-theory&gt;&lt;big-numbers&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="3420" PostTypeId="2" ParentId="3319" CreationDate="2010-08-26T21:23:21.227" Score="0" ViewCount="0" Body="&lt;p&gt;Read this very, very good and accessible book!&lt;br&gt;&#xA;&lt;a href=&quot;http://books.google.com/books?id=eIsyLD_bDKkC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=eIsyLD_bDKkC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-08-27T15:26:57.090" LastActivityDate="2010-08-27T15:26:57.090" CommentCount="1" />
  <row Id="3421" PostTypeId="2" ParentId="3419" CreationDate="2010-08-26T21:46:14.367" Score="41" ViewCount="0" Body="&lt;p&gt;No.  The problem is very simple: there are more huge numbers than abbreviated forms.  Suppose that I allow you to use the numerals 0-9, the English alphabet, and spaces to describe whatever numbers you want; that's still only $37$ characters. There are only $37^{140}$ expressions you can write down using these characters which is $140$ characters or shorter (the length of a tweet!), so there are only $37^{140}$ numbers that you could conceivably tweet to someone else.  In particular, since $\log_{10} 37^{140} = 219.5$, this implies that there is at least one number with $220$ digits which is impossible to tweet to someone else without using something other than numerals and English.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All right, I'll be even more generous: I'll let you use all $127$ ASCII characters. That means there are $127^{140}$ possible tweets, some of which mathematically describe some kind of number, and since $\log_{10} 127^{140} = 294.5$, there is at least one number with $295$ digits which is impossible to tweet to someone else (again, assuming that Twitter uses ASCII).  I could give you the entire Unicode alphabet and you still won't be able to tweet, for example, the kind of prime numbers that public key cryptosystems use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of questions are studied in computer science and information theory; you will find the word &lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov_complexity&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov complexity&lt;/a&gt; tossed around, as well as the word &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_(information_theory)&quot; rel=&quot;nofollow&quot;&gt;entropy&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-10-26T10:46:37.737" LastActivityDate="2010-10-26T10:46:37.737" CommunityOwnedDate="2010-08-26T21:46:14.367" CommentCount="7" />
  <row Id="3422" PostTypeId="2" ParentId="3419" CreationDate="2010-08-26T22:21:01.880" Score="10" ViewCount="0" Body="&lt;p&gt;If follows from the pigeonhole principle that any lossless compression algorithm that decreases the size of at least one number must increase the size of another - else the compression wouldn't be invertible. For deeper results consult any textbook algorithmic information theory (Kolmogorov complexity), e.g. Li; Vitanyi.  An introduction to Kolmogorov complexity and its applications.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T23:28:30.923" LastActivityDate="2010-08-28T23:28:30.923" CommunityOwnedDate="2010-08-26T22:21:01.880" />
  <row Id="3423" PostTypeId="1" AcceptedAnswerId="3427" CreationDate="2010-08-26T22:24:13.840" Score="6" ViewCount="134" Body="&lt;p&gt;From a St. Petersburg school olympiad, 11th grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove: a non constant polynomial P with non-negative integer coefficients is uniquely determined by its values P(2) and P(P(2)).&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-08-26T22:35:38.083" Title="A polynomial determined by two values." Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="3424" PostTypeId="2" ParentId="3101" CreationDate="2010-08-26T22:28:00.110" Score="3" ViewCount="0" Body="&lt;p&gt;The following is an attempt at an interpolation between (among?) the answers of Grigory M., Pierre-Yves Gaillard and Bill Dubuque (the last of which I confess that I do not completely understand, but this is one possible interpretation of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $J$ be an open interval in $\mathbb{R}$, and let $x_0 \in J$.  Following PYG's suggestion, let me begin by stating exactly what I will prove: if $f,g: J \rightarrow \mathbb{R}$ are differentiable at $x_0$, then so is their product, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(fg)'(x_0) = f'(x_0)g(x_0) + f(x_0) g'(x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will prove this &quot;congruentially&quot;, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $R$ be the ring of all functions $f: J \rightarrow \mathbb{R}$ which are continuous at &#xA;$x_0$, under the operations of pointwise addition and multiplication.  Inside $R$, consider the set $I$ of all functions $f$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x \rightarrow x_0} \frac{f(x)}{x-x_0} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that $I$ is an ideal of $R$.  This is easy to prove, but I note that it makes use &#xA;of the fact that every element of $R$ is continuous at $x_0$, hence bounded near $x_0$.  Now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For $f \in R$, $f$ lies in $I$ iff: $f(x_0) = 0$, $f$ is differentiable at $x_0$ and $f'(x_0) = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For $f \in R$, $f$ is differentiable at $x_0$ iff there exists $A \in \mathbb{R}$ such that $f \equiv f(x_0) + A(x-x_0)$.  If so, then necessarily $A = f'(x_0)$; in particular, it is uniquely determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Thus, if $f$ and $g$ are both differentiable at $x_0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$fg \equiv (f(x_0) + f'(x_0)(x-x_0))(g(x_0) + g'(x_0)(x-x_0))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\equiv f(x_0)g(x_0) + (f'(x_0)g(x_0) + f(x_0)g'(x_0))(x-x_0) + f'(x_0)g'(x_0)(x-x_0)^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\equiv f(x_0)g(x_0) + (f'(x_0)g(x_0) + f(x_0)g'(x_0))(x-x_0) \pmod I$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 2), it follows that $fg$ is differentiable at $x_0$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(fg)'(x_0) = f'(x_0)g(x_0) + f(x_0)g'(x_0)$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-26T22:28:00.110" />
  <row Id="3425" PostTypeId="1" CreationDate="2010-08-26T22:28:48.893" Score="2" ViewCount="41" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanczos_resampling&quot; rel=&quot;nofollow&quot;&gt;Lanczos 1 function&lt;/a&gt; is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{Lanczos1}(x) = \begin{cases} \frac{(\sin(\pi x))^2}{\pi^2 x^2 } &amp;amp; -1 &amp;lt; x &amp;lt; 1, x \ne 0 \\&#xA; 1 &amp;amp; x = 0 \\ &#xA;0 &amp;amp; \text{otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this function is to be implemented in a FIR filter with 4 taps, how do you calculate the filter coefficients, assuming that there are 16 phases?&#xA;Since the function is defined over the range -1 to +1, for 4 taps the coefficients will have to be shifted and spread to cover 0 to 3.&lt;/p&gt;&#xA;" OwnerUserId="1519" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T03:47:41.540" LastActivityDate="2010-08-27T03:47:41.540" Title="Lanczos 1 function for FIR filter" Tags="&lt;signal-processing&gt;" CommentCount="2" />
  <row Id="3426" PostTypeId="2" ParentId="3423" CreationDate="2010-08-26T22:32:52.670" Score="7" ViewCount="0" Body="&lt;p&gt;Look at $P(P(2))$ in base $P(2)$.  The nth place is the coefficient of $x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-08-26T22:32:52.670" CommentCount="2" />
  <row Id="3427" PostTypeId="2" ParentId="3423" CreationDate="2010-08-26T22:33:14.270" Score="7" ViewCount="0" Body="&lt;p&gt;True. If $P(x)=a_nx^n+\cdots+a_1x+a_0$ then each of the coefficients are less then $b\equiv P(2)$. Each of these coefficients can then be read off from the base-b expansion of $P(b)=P(P(2))$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-08-26T22:33:14.270" CommentCount="5" />
  <row Id="3429" PostTypeId="1" CreationDate="2010-08-26T22:54:32.400" Score="1" ViewCount="72" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lanczos_resampling&quot; rel=&quot;nofollow&quot;&gt;Lanczos-3 and Lanczos-4 functions&lt;/a&gt; are defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{Lanczos3}(x) = \begin{cases} \frac{\sin(\pi x)\sin(\pi*x/3)}{(\pi*x)*(\pi*x/3)} &amp;amp; -3 &amp;lt; x &amp;lt; 3, x \ne 0 \\&#xA; 1 &amp;amp; x = 0 \\ &#xA;0 &amp;amp; \text{otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\operatorname{Lanczos4}(x) = \begin{cases} \frac{\sin(\pi x)\sin(\pi*x/4)}{(\pi*x)*(\pi*x/4)} &amp;amp; -4 &amp;lt; x &amp;lt; 4, x \ne 0 \\&#xA; 1 &amp;amp; x = 0 \\ &#xA;0 &amp;amp; \text{otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a 4-tap FIR filter with 16 phases uses these functions, how do I calculate the coefficients for each of these functions for the filter?&lt;/p&gt;&#xA;" OwnerUserId="1520" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T03:50:51.193" LastActivityDate="2010-08-27T03:50:51.193" Title="Lanczos 3 and Lanczos 4 functions" Tags="&lt;signal-processing&gt;" CommentCount="2" />
  <row Id="3430" PostTypeId="2" ParentId="3408" CreationDate="2010-08-26T22:59:40.940" Score="1" ViewCount="0" Body="&lt;p&gt;This is meant as a comment to ifk's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how you conclude from Schwarz's lemma that $g(z)=e^{iθ}z^k$, where k is the order of zero of g in 0. Moreover, the function $f$ you obtain is not entire : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=g \circ h^{-1}$, but $h^{-1}$ is not entire : it has a pole outside $\mathbb{D}$...&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-26T22:59:40.940" CommentCount="1" />
  <row Id="3431" PostTypeId="2" ParentId="3352" CreationDate="2010-08-26T23:19:19.023" Score="1" ViewCount="0" Body="&lt;p&gt;I can't fully characterize such sets, but can say one thing - the conjecture is false.&#xA;For a counterexample on one dimension, choose a closed and nowhere dense set X (such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Cantor middle thirds set&lt;/a&gt;, which is not discrete) and let $g(x)=\min\{\vert x-y\vert\colon y\in X\}$ be the distance of a point x from X. Integrate it twice, $f(x)=\int_0^x\int_0^yg(z)\,dz\,dy$ to get a strictly convex function whose second derivative $f^{\prime\prime}=g$ vanishes precisely on X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you could take X to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Smith-Volterra-Cantor set&lt;/a&gt;, which has positive Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be used to define a strictly convex function on $\mathbb{R}^2$, $\tilde f(x,y)=f(x)+f(y)$, whose Hessian vanishes completely on a set $X\times X$ of positive Lebesgue measure.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-26T23:30:15.760" LastActivityDate="2010-08-26T23:30:15.760" CommentCount="1" />
  <row Id="3432" PostTypeId="2" ParentId="3405" CreationDate="2010-08-26T23:44:00.807" Score="4" ViewCount="0" Body="&lt;p&gt;Here is the standard generating function gruntwork.  Let $Q(x) = \sum_{n \ge 1} \frac{p(n)}{n} x^n$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q'(x) = \sum_{n \ge 1} p(n) x^{n-1} = \sum_{n \ge 1}^{\infty} x^{n-1} \sum_{j = \lceil \frac{n}{2} \rceil}^{n-1} \frac{p(j)}{j}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exchanging the order of summation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q'(x) = \sum_{j=1}^{\infty} \frac{p(j)}{j} \sum_{n=j+1}^{2j} x^{n-1} = \sum_{j=1}^{\infty} \frac{p(j)}{j} \frac{x^{2j} - x^j}{x - 1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q'(x) = \frac{Q(x^2) - Q(x)}{x - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions $Q(0) = 0, Q'(0) = 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as I recall there are theorems in Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt; that allow us to deduce asymptotic behavior about the coefficients of $Q$ from these kinds of relations; I'm going to hunt one down now, and in the meantime others can see what they can do with this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-26T23:44:00.807" CommentCount="2" />
  <row Id="3433" PostTypeId="1" AcceptedAnswerId="3435" CreationDate="2010-08-26T23:48:15.580" Score="1" ViewCount="160" Body="&lt;p&gt;I had asked this question sometime ago here: &lt;a href=&quot;http://math.stackexchange.com/questions/2536/finding-f-such-that-int-f-sum-f&quot;&gt;http://math.stackexchange.com/questions/2536/finding-f-such-that-int-f-sum-f&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i have a question which i think is more or less related to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be an increasing function (continuous of course!) with $f(1)=0$. Consider the sequence $s_{n}= ( \sum\limits_{k=1}^{n} f(k) - \int\limits_{1}^{n} f(x) dx )$. When does $s_{n}$ converge?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-27T13:11:58.400" LastActivityDate="2010-12-02T15:19:00.770" Title="Convergence of $\sum_{k=1}^{n} f(k) - \int_{1}^{n} f(x) dx$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="3434" PostTypeId="1" AcceptedAnswerId="3564" CreationDate="2010-08-27T00:56:48.797" Score="1" ViewCount="108" Body="&lt;p&gt;Suppose we have a tetrahedron defined by points (0,0,0),(1,1,0),(0,1,1),(1,0,1). Now define surface by (a,b,a + b - 2*a*b) for a,b between 0 and 1. Let E1 be the set of points inside the tetrahedron and E2 be the set of points on that surface, A the area of that surface and d be Euclidian distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that for all surfaces E3 with area not greater than A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max_{p \in E1} (\min_{q \in E2} (d(p,q))) \le \max_{p \in E1} (\min_{q \in E3} (d(p,q)))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/minimal-surface.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;g1 = Graphics3D[{Blue, Opacity[.1], &#xA;    GraphicsComplex[{{0, 0, 0}, {1, 1, 0}, {0, 1, 1}, {1, 0, 1}}, &#xA;     Polygon[{{2, 3, 4}, {3, 2, 1}, {4, 1, 2}, {1, 4, 3}}]]}];&#xA;g2 = Plot3D[a + b - 2*a*b, {a, 0, 1}, {b, 0, 1}, &#xA;   PlotStyle -&gt; Opacity[.5]];&#xA;Show[g1, g2]&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-08-28T17:47:58.447" LastActivityDate="2010-08-28T18:42:23.670" Title="Does this surface minimize maximum distance?" Tags="&lt;geometry&gt;&lt;minimal-surfaces&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3435" PostTypeId="2" ParentId="3433" CreationDate="2010-08-27T01:50:13.057" Score="5" ViewCount="0" Body="&lt;p&gt;Qiaochu was on the right track to use an integral-to-sum formula, but it sounds like you want the Abel-Plana summation formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\left(\sum_{k=m}^n f(k)-\int_m^n f(u)\mathrm{d}u\right)=\frac{f(m)}{2}-\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)\left(\frac{f(m+it)-f(m-it)}{2it}\right)\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is used for instance to evaluate the Stieltjes constants. If the expression on the right hand side is convergent, then it is equivalent to the left hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Adendum for Chandru:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitely $f(z)$ should be analytic, or at least analytic in the region where $\Re z\geq m$. Per Henrici's &quot;Applied and Computational Complex Analysis&quot;, the additional conditions are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{t\to\infty}f(u\pm it)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uniformly in $u$, and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{t\to\infty}|f(u\pm it)|\exp(\mp 2\pi t)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uniformly in $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those scratching their head on just how Abel-Plana and Euler-Maclaurin are connected, the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)|t|^{2n-2}\mathrm{d}t=\frac{|B_{2n}|}{2n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;might be of interest.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T15:19:00.770" LastActivityDate="2010-12-02T15:19:00.770" CommentCount="4" />
  <row Id="3436" PostTypeId="2" ParentId="3408" CreationDate="2010-08-27T01:51:54.877" Score="2" ViewCount="0" Body="&lt;p&gt;If a function $f$ is holomorphic in a neighborhood of the closed disk and has modulus 1 on the circle, then $f$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Blaschke_product#Finite_Blaschke_products&quot; rel=&quot;nofollow&quot;&gt;finite Blaschke product&lt;/a&gt;.  You can prove this by taking all of the zeros inside the disk counted according to multiplicity, dividing by corresponding &lt;a href=&quot;http://planetmath.org/encyclopedia/AutomorphismsOfUnitDisk.html&quot; rel=&quot;nofollow&quot;&gt;holomorphic automorphisms of the disk&lt;/a&gt; that have those zeros, and showing that the result is constant.  (This quotient and its reciprocal are analytic and bounded by 1 on the disk...)  Thus, your $f$ is a finite Blaschke product, but it also has no poles in the plane.  The only such functions are constant multiples of power functions, as ifk already said.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-08-27T17:57:19.797" LastActivityDate="2010-08-27T17:57:19.797" CommentCount="2" />
  <row Id="3437" PostTypeId="1" CreationDate="2010-08-27T02:03:18.623" Score="4" ViewCount="94" Body="&lt;p&gt;I’m interested in some sequences which have no square factor.&#xA;$$&#xA;s_i =\begin{cases}&#xA;4 &amp;amp; i=0; \\&#xA;s_{i-1}^2 - 2 &amp;amp; \text{otherwise}&#xA;\end{cases}$$&#xA;This is Lucas–Lehmer primality test sequence.  &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A003010&quot; rel=&quot;nofollow&quot;&gt;A003010 in OEIS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when ${s_i}$ has a square factor? I had checked that the sequence has no square factor for i=1,2,...,7&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-27T06:18:48.657" LastActivityDate="2010-09-03T15:52:48.587" Title="Search square factors in Lucas–Lehmer sequence" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3438" PostTypeId="1" AcceptedAnswerId="4155" CreationDate="2010-08-27T02:16:01.887" Score="0" ViewCount="251" Body="&lt;p&gt;Prove that $\{n_k\}$ is a strictly increasing sequence of positive integers then the sum of the series $$\sum\limits_{k=1}^{\infty} \frac{2^{n_k}}{(n_{k})!}$$ is an irrational number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a problem, given by a friend of mine, when we were doing some problems in Analysis. And if i remember correctly, he mentioned a theorem of &quot;Legendre&quot;, to be used for proving this result.  But this looks quite interesting. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-06T18:25:13.027" LastActivityDate="2010-09-06T18:27:17.693" Title="Nature of the series: $ \sum\limits_{k=1}^{\infty} \frac{2^{n_k}}{(n_{k})!}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" />
  <row Id="3439" PostTypeId="1" CreationDate="2010-08-27T02:32:11.643" Score="0" ViewCount="67" Body="&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt; A subgroup of a group is termed strongly potentially characteristic if there is an embedding of the bigger group in some group such that, in that embedding both the group and the subgroup become characteristic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can refer this link: &lt;a href=&quot;http://groupprops.wiki-site.com/index.php/Strongly_potentially_characteristic_subgroup&quot; rel=&quot;nofollow&quot;&gt;http://groupprops.wiki-site.com/index.php/Strongly_potentially_characteristic_subgroup&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, i have somewhat understood the definition, i would like to have some examples of such subgroups. Moreover, i would also like to know, as to whats the essence of defining such subgroups in such a manner. Where are they applicable?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-27T02:32:11.643" Title="Strongly Potentially Characteristic Subgroups." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="4" />
  <row Id="3440" PostTypeId="2" ParentId="3433" CreationDate="2010-08-27T03:09:21.327" Score="3" ViewCount="0" Body="&lt;p&gt;The question implicitly asks for a &quot;simpler&quot; or &quot;more interesting&quot; criterion for convergence.  I doubt there is one.  Intuitively, f can do almost anything between (n, f(n)) and (n+1, f(n+1)) provided it is increasing.  Thus the terms of (s(n)) can be anything.  Convergence therefore is determined by fairly arbitrary properties of f(n) as n becomes arbitrarily large.  If you don't severely restrict f--e.g., require it to be analytic or bounded and concave or something like that--you shouldn't expect to find any simpler answer than &quot;(s(n)) converges when it converges.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-27T03:09:21.327" />
  <row Id="3441" PostTypeId="1" CreationDate="2010-08-27T04:02:58.813" Score="1" ViewCount="263" Body="&lt;p&gt;If I create a 10 digit password with the following requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At least one uppercase letter A-Z - 26&lt;/li&gt;&#xA;&lt;li&gt;At least one lowercase letter a-z - 26&lt;/li&gt;&#xA;&lt;li&gt;At least one digits 0-9 - 10&lt;/li&gt;&#xA;&lt;li&gt;At least one common symbol (#,$,%,etc) - 32&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By inclusion-exclusion, I can calculate I have ~ 3.2333E+19 possible combinations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I change one of the requirements to at least TWO digits 0-9, how can I calculate the possible combinations?&lt;/p&gt;&#xA;" OwnerUserId="1524" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-08-27T04:10:02.783" LastActivityDate="2010-08-27T15:53:49.487" Title="How can I calculate the number of potential combinations in a password?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3442" PostTypeId="1" CreationDate="2010-08-27T04:19:45.673" Score="1" ViewCount="106" Body="&lt;p&gt;In particular, problems like $\lfloor\frac{10^{20000}}{3+10^{100}}\rfloor$. (From a putnam exam many years ago.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general technique for attacking such problems?&lt;/p&gt;&#xA;" OwnerUserId="632" LastEditorUserId="632" LastEditorDisplayName="" LastEditDate="2010-08-27T04:56:19.760" LastActivityDate="2010-08-27T04:56:19.760" Title="Finding the units' digit of large quotients" Tags="&lt;number-theory&gt;&lt;soft-question&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="3443" PostTypeId="2" ParentId="3442" CreationDate="2010-08-27T04:42:49.730" Score="2" ViewCount="0" Body="&lt;p&gt;In the contest-style problems I have seen which ask you to do something like this, the trick is generally that the number in question is somehow very close to or otherwise closely related to a special integer, and you should compute this special integer's digits (using, probably, modular arithmetic) and then figure out if the number you're interested in is larger or smaller, and by how much.  For example, $\phi^n$ turns out to be very close to the Lucas number $L_n$; this is because $\phi^n + \varphi^n$ (the other root of $x^2 = x + 1$) is exactly equal to $L_n$, and $|\varphi| &amp;lt; 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have in mind, which I can't quite remember the details of, is a question which asks you to compute the first six digits after the decimal point of some seemingly complicated expression; it turns out to be an integer minus a tiny amount, so the answer is $999999$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, you don't seem to be asking about a solution to the problem itself, so I will suggest for now that you think about the identity $(10^{100} + 3)(10^{100} - 3) = 10^{200} - 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best advice I can give you in general is that you should get a good sense for asymptotics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-27T04:47:56.487" LastActivityDate="2010-08-27T04:47:56.487" CommentCount="2" />
  <row Id="3444" PostTypeId="1" AcceptedAnswerId="3498" CreationDate="2010-08-27T05:01:25.447" Score="13" ViewCount="611" Body="&lt;p&gt;In &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/&quot; rel=&quot;nofollow&quot;&gt;these notes by Terence Tao&lt;/a&gt; is a proof of Stirling's formula.  I really like most of it, but at a crucial step he uses the integral identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! = \int_{0}^{\infty} t^n e^{-t}  dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;coming from the Gamma function.  I have a mathematical confession to make: I have never &quot;grokked&quot; this identity.  Why should I expect the integral on the right to give me the number of elements in the symmetric group on $n$ letters?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's not that I don't know how to prove it.  It's quite fun to prove; my favorite proof observes that it is equivalent to the integral identity $\int_{0}^{\infty} e^{(x-1)t}  dt = \frac{1}{1 - x}$.  But if someone were to ask me, &quot;yes, but why, &lt;em&gt;really&lt;/em&gt;?&quot; I would have no idea what to say.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are more intuitive ways of thinking about this identity?  Is there a probabilistic interpretation?  What kind of random variable has probability density function $\frac{t^n}{n!} e^{-t}$?  (What does this all have to do with Tate's thesis?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rough measure of what I'm looking for, your answer should make it obvious that $t^n e^{-t}$ attains its maximum at $t = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  The kind of explanation I'm looking for, as I described in the comments, is similar to &lt;a href=&quot;http://math.stackexchange.com/questions/3528/beta-function-derivation/3540#3540&quot;&gt;this explanation&lt;/a&gt; of the beta integral.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-28T14:13:15.533" LastActivityDate="2010-08-28T16:50:07.527" Title="Intuition for the definition of the Gamma function?" Tags="&lt;probability-theory&gt;&lt;intuition&gt;&lt;gamma-function&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="8" />
  <row Id="3445" PostTypeId="2" ParentId="2304" CreationDate="2010-08-27T06:14:08.463" Score="2" ViewCount="0" Body="&lt;p&gt;To solve this you can use the function theoretic description of the invariant subspaces of the shift known as Beurling's theorem.  Identify your Hilbert space with the Hardy space $H^2$ on the disk such that $U$ is identified with &quot;multiplication by $z$&quot;.  Beurling's theorem says that each invariant subspace of $U$ has the form $\phi H^2$ for a so-called inner function $\phi$, i.e., a bounded analytic function on the disk whose radial (or non-tangential) limit function has modulus 1 a.e. on the circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your problem, there are inner functions $\phi_1$ and $\phi_2$ such that $M_1=\phi_1 H^2$ and $M_2=\phi_2 H^2$.  You have $\phi_1\in M_1\subset M_2$, so $\phi_1=\phi_2 f$ for some $f\in H^2$.  The modulus of $f$ on the circle is 1 a.e. because $f=\phi_1/\phi_2$ a.e., and thus $f$ is an inner function.  &lt;em&gt;Suppose&lt;/em&gt;, for the sake of argument, that you can write $f=gh$ for some nonconstant inner functions $g$ and $h$, and let $M=\phi_2 g H^2$.  Then $M_1\subset  M\subset M_2$.  I claim that the inclusions are strict, and more specifically that $\phi_2 g$ is in $M\setminus M_1$ and $\phi_2$ is in $M_2\setminus M$.  This amounts to the same thing as saying that $1/h$ and $1/g$ (respectively) are not in $H^2$.  Note that $1/h$ and $1/g$ have modulus 1 a.e. on the circle, so if they were in $H^2$ they would in fact be in $H^\infty$ and bounded by 1 on the disk.  But $h$ and $g$ are bounded by 1 on the disk and nonconstant, so this is impossible, showing that in fact $1/h$ and $1/g$ are not in $H^2$ as claimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to be seen why $f$ has such a factorization, and this is where the hypothesis about codimension is used.  Every inner function has a factorization into a Blaschke product times a singular inner function.  If the singular part of $f$ is nontrivial, it can be factored nontrivially by scaling the corresponding singular measure by numbers between 0 and 1.  If the Blaschke part of $f$ has more than one factor, then it factors. Given that $M_1\neq M_2$, $f$ is not constant, so the only other possibility is that $f$ is a Blaschke product with a single factor, a.k.a. a holomorphic automorphism of the disk.  I claim that this would imply that $M_1$ has codimension 1 in $M_2$, and more specifically $M_2=M_1 + \mathbb{C}\phi_2$.  To see this, let $\alpha$ be the zero of $f$, and let $G=\phi_2 H$ be an element of $M_2$.  Then $G=\phi_2 f\frac{H-H(\alpha)}{f}+\phi_2 H(\alpha)\in M_1+\mathbb{C}\phi_2$.  Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details on factorizations of inner functions and more, see Chapter 2 of the book named in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As for the edit, you can't always take $M=UM_2$.  Let $0&amp;lt;|\alpha|&amp;lt;1$, and let $\phi_\alpha$ be the holomorphic automorphism of the disk that swaps 0 and $\alpha$, $\phi_\alpha(z)=\frac{\alpha-z}{1-\overline{\alpha}z}$.  Suppose that $M_2=H^2$ and $M_1=\phi_\alpha^2 H^2$.  Then your hypotheses are met, but $M_1$ is not contained in $UM_2$ because $\phi_\alpha^2$ is not in $UM_2$.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-08-27T08:03:53.797" LastActivityDate="2010-08-27T08:03:53.797" />
  <row Id="3446" PostTypeId="2" ParentId="3408" CreationDate="2010-08-27T06:20:47.470" Score="6" ViewCount="0" Body="&lt;p&gt;The only such functions are $f(z)=az^n$ with $|a|=1$.&#xA;Let $f$ be such a function and define&#xA;$$g(z)=\overline{f(\overline{z}^{-1})}^{-1}.$$&#xA;Due to the two complex conjugations, $g$ is meromorphic&#xA;on $\mathbb{C}-\{0\}$. Also $g(z)=f(z)$ when $|z|=1$.&#xA;By analytic continuation, $f(z)=g(z)$ for all nonzero $z$.&#xA;Now $g$ has a pole of order $k$ at $\infty$ where $k$&#xA;is the order of $f$ at zero. Hence $f$ is a polynomial of degree $k$.&#xA;As $k$ is also the order at $0$ then $f(z)=az^k$; clearly&#xA;$|a|=1$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-08-27T16:55:22.003" LastActivityDate="2010-08-27T16:55:22.003" CommentCount="9" />
  <row Id="3447" PostTypeId="1" AcceptedAnswerId="3466" CreationDate="2010-08-27T06:33:29.233" Score="2" ViewCount="317" Body="&lt;p&gt;Motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/3250/&quot;&gt;this&lt;/a&gt; question I would like to know whether there is an exact definition of modern mathematics. In which point in time  (century, decade) does one think, when speaking about modern mathematics. Does it refer to the Abstract Algebra?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I got -1. If this is not a correct question, please state why. And it can be closed if the question is not proper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is a genuine doubt and ignorance of mine!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; In my humble opinion instead of closing, should be better to tag it as a community wiki. Anyway I &lt;strike&gt;do not&lt;/strike&gt; have &lt;strike&gt;any&lt;/strike&gt; no objections &lt;strike&gt;against&lt;/strike&gt; in closing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 3:&lt;/strong&gt; Now is a community wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 4.&lt;/strong&gt; I have learned from the answers and comments, including the explanation for closing!&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T10:17:20.637" LastActivityDate="2010-08-29T10:17:20.637" CommunityOwnedDate="2010-08-27T12:42:06.947" ClosedDate="2010-08-27T15:41:16.057" Title="What is Modern Mathematics? Is this an exact concept with a clear meaning?" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="3448" PostTypeId="1" AcceptedAnswerId="3449" CreationDate="2010-08-27T06:34:02.243" Score="4" ViewCount="116" Body="&lt;p&gt;I'm looking for clear mathematical rules on rounding a number to n decimal places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything seems perfectly clear for positive numbers. Here is for example what I found on &lt;a href=&quot;http://math.about.com/od/arithmetic/a/Rounding.htm&quot; rel=&quot;nofollow&quot;&gt;math.about.com&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Rule One&lt;/strong&gt; Determine what your rounding digit is and look to the right side of it. If that digit is 4, 3, 2, or 1, simply drop all digits to the right of it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Rule Two&lt;/strong&gt; Determine what your rounding digit is and look to the right side of it. If that digit is 5, 6, 7, 8, or 9 add one to the rounding digit and drop all digits to the right of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what about negative numbers ? Do I apply the same rules as above ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, what is the correct result when rounding -1.24 to 1 decimal place ? -1.3 or -1.2 ?&lt;/p&gt;&#xA;" OwnerUserId="1525" LastEditorUserId="1525" LastEditorDisplayName="" LastEditDate="2010-08-27T06:41:59.453" LastActivityDate="2010-08-27T07:09:16.280" Title="rules for rounding (positive and negative numbers)" Tags="&lt;real-number-computation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3449" PostTypeId="2" ParentId="3448" CreationDate="2010-08-27T06:51:57.240" Score="3" ViewCount="0" Body="&lt;p&gt;$-1.24\approx -1.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1.21,-1.22,-1.23$ and $-1.24$ all round to $-1.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1.25,-1.26,-1.27,1.28$ and $-1.29$ all round to $-1.3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the correct result in numerical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You take the absolute value of the number, round it, and then put the $+$ or $-$ sign.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-27T07:09:16.280" LastActivityDate="2010-08-27T07:09:16.280" CommentCount="2" />
  <row Id="3450" PostTypeId="1" AcceptedAnswerId="3451" CreationDate="2010-08-27T09:43:38.900" Score="4" ViewCount="256" Body="&lt;p&gt;Factorial is defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n! = n(n-1)(n-2)...1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why mathematicians named this thing as &lt;strong&gt;FACTORIAL&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has it got something to do with factors?&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-08-27T15:02:07.757" LastActivityDate="2010-08-27T15:02:07.757" Title="Why the name 'FACTORIAL'?" Tags="&lt;soft-question&gt;&lt;combinatorics&gt;&lt;terminology&gt;&lt;functions&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="3451" PostTypeId="2" ParentId="3450" CreationDate="2010-08-27T10:08:58.140" Score="6" ViewCount="0" Body="&lt;p&gt;Well, all positive integers smaller or equal to n are factors of n!&lt;/p&gt;&#xA;" OwnerUserId="1010" LastActivityDate="2010-08-27T10:08:58.140" />
  <row Id="3452" PostTypeId="1" AcceptedAnswerId="3458" CreationDate="2010-08-27T10:09:14.437" Score="4" ViewCount="414" Body="&lt;p&gt;Which one is bigger $2^{n!}$ or $(2^{n})!$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where n∈N.&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-27T10:30:41.233" LastActivityDate="2010-08-27T20:53:34.373" Title="Which one is bigger $2^{n!}$ or $(2^{n})!$" Tags="&lt;asymptotics&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="2" />
  <row Id="3453" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T10:43:13.620" Score="1" ViewCount="0" Body="&lt;p&gt;Indefinite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n=0$: $2^{0!}=2^1=2,$ $(2^0)!=1!=1$, so $2^{0!}&gt;(2^0)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n=2$: $2^{2!}=2^2=4,$ $(2^2)!=4!=12$, so $2^{2!}&amp;lt;(2^2)!$.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-08-27T10:43:13.620" CommentCount="1" />
  <row Id="3454" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T10:44:34.617" Score="4" ViewCount="0" Body="&lt;p&gt;It's enough to take this limit: $\lim_{n \rightarrow \infty} \frac{2^{n!}}{(2^n)!}$, by the Stirling formula we get $\lim_{n \rightarrow \infty} \frac{2^{(\frac{n}{e})^n \sqrt{2 \pi n}}}{(\frac{2^n}{e})^{2^n} \sqrt{2 \pi 2^n}}$, which is asymptotic (after elevating and taking logarithms) to $\lim_{n \rightarrow \infty} \frac{\exp(\frac{n^n}{e^n})}{\exp(2^n 2^{n/2})} = \lim_{n \rightarrow \infty} \frac{\exp(\frac{n^n}{e^n})}{\exp(2^{3/2 n})}$. Now, $\frac{n^n}{e^n}$ is bigger than $2^{3/2n}$ (and it is easy to verify). So the limit is infinite, which means that $2^{n!}$ is bigger than $(2^n)!$, if $n$ is big, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Little note: I know that the asymptotic preseves constants and, even though I omitted them, it's good practice always to include them.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-08-27T10:57:23.303" LastActivityDate="2010-08-27T10:57:23.303" CommentCount="3" />
  <row Id="3455" PostTypeId="2" ParentId="3447" CreationDate="2010-08-27T10:45:04.787" Score="7" ViewCount="0" Body="&lt;p&gt;The opposite of &quot;modern&quot; is &quot;classical,&quot; and a common use of the word classical is &quot;whatever was known when I was in college.&quot;  That's a little tongue-in-cheek, but it really is how many people use the term.  See more discussion &lt;a href=&quot;http://www.johndcook.com/blog/2009/03/27/classical-science/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-08-27T10:45:04.787" CommunityOwnedDate="2010-08-27T16:05:48.460" CommentCount="5" />
  <row Id="3456" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T10:59:58.637" Score="5" ViewCount="0" Body="&lt;p&gt;Actually, $(2^n)!$ is larger if 1 &amp;lt; n &amp;lt; &lt;a href=&quot;http://www.wolframalpha.com/input/?i=log_2%28%282%5en%29!%29%3Dn!&quot; rel=&quot;nofollow&quot;&gt;4.97399743597&lt;/a&gt;. See &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+log%28log%282%5e%28n!%29%29%29+%2B+i+log%28log%28%282%5en%29!%29%29+for+n+%3D+0.1+to+10&quot; rel=&quot;nofollow&quot;&gt;this graph&lt;/a&gt; for detail.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-27T10:59:58.637" CommentCount="1" />
  <row Id="3457" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T11:05:51.433" Score="5" ViewCount="0" Body="&lt;p&gt;By proceeding straightforwardly we can answer without using Stirling's formula.&#xA;To start we show that $(n-1)!&gt;2^n$ for large n. [by induction]&#xA;First note that $5!&gt;2^6$.&#xA;Then we note that if we increase n by 1 we multiply the left hand side by n, and the right hand side by 2. Since n&gt;2 we conclude that the left hand side remains larger for bigger and bigger n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by n we get $n!&gt;n2^n$&#xA;Replacing n with $log(2^{n})$ [using base-2 logarithms] gets us: $n!&gt;2^{n}log(2^{n})$&#xA;Note that $2^{n}log(2^{n})$=&#xA;$log(2^{n})$+$log(2^{n})$+$log(2^{n})$+...$log(2^{n})$+$log(2^{n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$2^n$ terms]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&gt;log(2^{n})$+$log(2^{n}-1)$+$log(2^{n}-2)$+...log(2)+log(1) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[since this has the same amount of terms, and some of them are smaller.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=$log(2^{n}!)$ [combining logs]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting together these inequalities gives us that n!&gt;$log(2^{n!})$, and raising 2 to these powers gives $2^{n!}$&gt;$(2^{n}!)$ [for n&gt;5 since our induction only told us that $(n-1)!&gt;2^n$ for n&gt;5]&lt;/p&gt;&#xA;" OwnerUserId="1149" LastEditorUserId="1149" LastEditorDisplayName="" LastEditDate="2010-08-27T11:23:17.223" LastActivityDate="2010-08-27T11:23:17.223" CommentCount="4" />
  <row Id="3458" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T11:09:26.857" Score="18" ViewCount="0" Body="&lt;p&gt;We should expect $2^{n!}$ be larger. Note that $(2^n)! \leq (2^n)^{(2^n)} = 2^{(n 2^n)}$. &#xA;So we want to prove $n 2^n &amp;lt; n!$ eventually. (And this can be done very simply.)&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-08-27T11:14:47.690" LastActivityDate="2010-08-27T11:14:47.690" CommentCount="1" />
  <row Id="3459" PostTypeId="1" AcceptedAnswerId="3479" CreationDate="2010-08-27T11:23:38.270" Score="3" ViewCount="443" Body="&lt;p&gt;My textbook doesn't give any example of this kind of series. Could you provide some?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trigonometric series is defined in wikipedia as :&#xA;$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \cos{nx} dx\qquad (n=0,1,2,3 \dots)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \sin{nx} dx\qquad (n=1,2,3, \dots)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is fourier series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-08-27T12:29:54.243" LastActivityDate="2010-08-29T20:19:50.293" Title="Example of a trigonometric series that is not fourier series?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;fourier-series&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="3460" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T11:33:53.637" Score="-1" ViewCount="0" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Trigonometric_series&amp;amp;oldid=370829487&quot; rel=&quot;nofollow&quot;&gt;this wikipedia stub&lt;/a&gt;. A trigonometric series, which is not a Fourier series, must therefore be one that is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = A_{0}+\displaystyle\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which the relations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_{0}=\frac{1}{2\pi}\displaystyle\int^{2 \pi}_0 f(x)\,dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \cos{nx} \,dx\qquad (n=1,2,3 \dots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \sin{nx}\, dx\qquad (n=0,1,2,3, \dots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;do &lt;em&gt;not&lt;/em&gt; hold. But since {$\sin(nx),\cos(nx)$} forms a complete orthonormal basis, I cannot imagine one. &lt;strike&gt;Except for a non-convergent one perhaps...&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; As &lt;a href=&quot;http://math.stackexchange.com/users/498/j-mangaldan&quot;&gt;J. Mangaldan&lt;/a&gt; pointed out: &lt;a href=&quot;http://en.wikipedia.org/wiki/Lacunary_function#Lacunary_trigonometric_series&quot; rel=&quot;nofollow&quot;&gt;Lacunary trigonometric series&lt;/a&gt; might be the answer you search, i.e. series that cannot be analytically be continued.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-08-27T11:43:55.100" LastActivityDate="2010-08-27T11:43:55.100" CommentCount="4" />
  <row Id="3461" PostTypeId="2" ParentId="3447" CreationDate="2010-08-27T11:34:51.943" Score="5" ViewCount="0" Body="&lt;p&gt;Following in a tradition of similar purported definitions, one may say that &quot;modern mathematics is what modern mathematicians do&quot;. :-)&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-08-27T11:34:51.943" CommunityOwnedDate="2010-08-27T16:05:48.460" />
  <row Id="3462" PostTypeId="2" ParentId="3450" CreationDate="2010-08-27T12:54:41.790" Score="8" ViewCount="0" Body="&lt;p&gt;Below is the etymology, from Jeff Miller's &lt;a href=&quot;http://jeff560.tripod.com/f.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Known Uses of Some of the Words of Mathematics (F)&lt;/a&gt;. Perhaps a native French speaker can lend further insight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;FACTORIAL&lt;/strong&gt;. The earlier term faculty was introduced around 1798 by Christian Kramp (1760-1826).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Factorial&lt;/em&gt; was coined (in French as factorielle) by Louis François Antoine Arbogast (1759-1803).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Kramp withdrew his term in favor of Arbogast's term. In the Preface, pp. xi-xii, of his &quot;Éléments d'arithmétique universelle,&quot; Hansen, Cologne (1808), Kramp remarks:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...je leur avais donné le nom de facultés. Arbogast lui avait substitué la nomination plus nette et plus française de factorielles; j'ai reconnu l'avantage de cette nouvelle dénomination; et en adoptant son idée, je me suis félicité de pouvoir rendre hommage à la mémoire de mon ami. [...I've named them facultes. Arbogast has proposed the denomination factorial, clearer and more French. I've recognised the advantage of this new term, and adopting its philosophy I congratulate myself of paying homage to the memory of my friend.] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-27T12:54:41.790" CommentCount="1" />
  <row Id="3463" PostTypeId="1" AcceptedAnswerId="3470" CreationDate="2010-08-27T13:33:24.157" Score="3" ViewCount="95" Body="&lt;p&gt;I've been thinking about maps between sets. Injections, surjections and the rest. Often when thinking about some kind of map, it is interesting to say &quot;what about the maps from a set to &lt;em&gt;itself&lt;/em&gt;?&quot; Call these maps endomaps. Permutations of elements of the set are a special case of endomaps: they are bijective endomaps. Permutations have lots of interesting properties: they form groups and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about &lt;em&gt;general&lt;/em&gt; endomaps that are not necessarily bijective? Do they have any interesting properties that people study? They don't necessarily have inverses, which rules them out as forming groups, but composition of endomaps is associative, so they aren't totally devoid of interesting properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also the case that for every endomap, there exists some subset of its domain (not necessarily unique) such that it is bijective on that subset. So these maps are permutations if restricted to a particular subset. Is this enough to make endomaps interesting in their own right, or are they only studied as a part of the study of maps between sets in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[It might be obvious that this question was motivated by thinking about category theory, but there's nothing particularly categorical about the question as such...]&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-27T15:32:28.523" Title="Do endomaps of sets have interesting properties?" Tags="&lt;set-theory&gt;&lt;category-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="3464" PostTypeId="2" ParentId="3419" CreationDate="2010-08-27T14:08:27.270" Score="0" ViewCount="0" Body="&lt;p&gt;The way I understand this question, the answer is &quot;yes.&quot; Simply write the number in, say, hexadecimal notation, and it will always be shorter than in decimal (if it is &gt;=100000). If you are looking for the shortest representation, however, that is well defined only if you specify exactly what counts as a valid representation of a number. After all, every number you come up with can technically be represented in just one digit if you choose a sufficiently large base (at least one higher than the number itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the standard way to reduce it: I'd say the decimal system is the standard way of reducing numbers from just writing the appropriate number of dashes. :-) In the sense pointed out by Bill Dubuque, there is no representation that is more space efficient in principle, that is if you transfer the problem to a more abstract domain. If, instead, you are looking for the shortest representation using a given set of mathematical symbols, then no, there is no standard way to find it. I also suspect no algorithm can do better than brute force search (that is, first check all numbers representable using 1 character, then those representable in 2 characters, and so on). In any case, nobody would write a number as &quot;loge 56 (log 56 base e) when we discard first 6 numbers and then consider first 200 digits,&quot; so no conceivable set of standard symbols will account for that anyway.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-08-27T14:08:27.270" CommunityOwnedDate="2010-08-27T14:08:27.270" CommentCount="4" />
  <row Id="3465" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T14:21:41.163" Score="0" ViewCount="0" Body="&lt;p&gt;When all the coefficients are 1, the series does not converge to a function.  (It does converge in the sense of a distribution, to the Dirac Delta.).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-27T14:21:41.163" />
  <row Id="3466" PostTypeId="2" ParentId="3447" CreationDate="2010-08-27T14:22:15.513" Score="5" ViewCount="0" Body="&lt;p&gt;Further to the other answers, which are indeed correct: no technical definition exists- barks $\iff$ dog, frankly- but 'modern' &lt;em&gt;is&lt;/em&gt; a well defined concept outside of mathematics; and to a certain extent it is one to which the barkings of modern mathematics agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was once the case that mathematicians believed proofs to uncover the neccessarily true- there must be numbers working thus, and those numbers must naturally have no zero divisors- there must be a geometry built thus, and this geometry must have angles in a triangle summing to $180^o$. With the exception of the work of Euclid (whose axioms were largely seen by others as immutable anyway), the theorems of mathematics were seen as universal truths, fouded in pure logic- facts about platonic ideals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except none of this was true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the first chink in the armour of this classical mathematics came with the work of Bolyai and Gauss, constructing consistent geometries where triangles behaved unusually (turning, as we all know, into modern Hyperbolic geometry), that seeped from a change in the hitherto 'immutable' axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from here the trickle began, which rushed and swelled with time, and burst the banks of mathematics as was: axioms became plastic, changeable at will, and with them the mathematics that followed from them. New concepts were created, and concepts of concepts, enriching and enlarging the mathematical landscape in ways that generations before could not have imagined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallell to this explosion was the search for foundations for these axioms- the dying embers of platonism in the work of Frege, Russell and Whitehead; Hilbert's program, seeming at first promising, were spectacularly micturated on by Godel's incompleteness theorems. And it soon became (quite) clear, that any (provably) 'ultimate' description of mathematics was doomed to failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modernism outside of mathematics is characterised by a certain relativism- an understanding that different perspectives can lead to different (equally valid) conclusions. In modern mathematics one has the reals &lt;em&gt;and&lt;/em&gt; the p-adics, euclidean &lt;em&gt;and&lt;/em&gt;  non-euclidian geometries, topologies and metric spaces, groups, rings, algebras: sets and mereology- and we cannot claim one to be more valid than the others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In modern mathematics, our truths are absolute but crucially contingent, the children of axioms in a pluralistic universe of possible postulates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course some would say that 'modern' just means 'with categories', but that's not quite as neat- perhaps we can fit categories to 'post-modern' somehow....  &lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-08-27T14:42:07.080" LastActivityDate="2010-08-27T14:42:07.080" CommunityOwnedDate="2010-08-27T14:22:15.513" CommentCount="10" />
  <row Id="3467" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T14:34:11.527" Score="2" ViewCount="0" Body="&lt;p&gt;A standard example is &#xA;$$&#xA;f(t)= \sum_{n&gt;1} \frac{\sin(nt)}{\log(n)}$$&#xA;The conjugate of $f$ is a Fourier series but $f\not\in L^1(\mathbb{T})$ and hence is no Fourier series.&#xA;For further explanation see &lt;a href=&quot;http://books.google.com/books?id=gkpUE_m5vvsC&amp;amp;printsec=frontcover&amp;amp;dq=katznelson&amp;amp;hl=sv&amp;amp;ei=R8l3TN_tA5DLOIGhgBA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCwQ6AEwAQ#v=snippet&amp;amp;q=conjugate&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Katznelson's book page 85.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: If $f$ is not in $L^1(\mathbb{T})$ it is hard to define Fourier coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 29/8 - 2010&lt;/strong&gt;&#xA;Here is a screen dump from Zygmund's book &quot;Trigonometric Series Vol I&quot;&#xA;&lt;img src=&quot;http://i.imgur.com/BkVNH.png&quot; alt=&quot;A standard definition&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in order to be a Fourier series the coefficients are to be Fourier coefficients of some function $f$, these are well defined if $f\in L^1(\mathbb{T})$ (because these are calculated using integration against a bounded measurable function). Thus it is natural to define a trigonometric series if the coefficients are Fourier coefficients of some $L^1$ function. Hence, all answers given to this problem are in fact right! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the Riemann-Lebesgue Lemma states that the Fourier coefficients of an $L^1$-function tends to $0$ as the index approach $\pm\infty$. Thus, it is easy to construct formal trigonometric series that are not Fourier series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example $f$ given above does in fact converge everywhere and hence this is a kind of non-trivial example of a trigonometric series that is not a Fourier series.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-08-29T06:22:43.813" LastActivityDate="2010-08-29T06:22:43.813" CommentCount="2" />
  <row Id="3468" PostTypeId="2" ParentId="3463" CreationDate="2010-08-27T14:37:29.687" Score="2" ViewCount="0" Body="&lt;p&gt;Given a set $X$, the collection of all maps $s\colon X\to X$ forms a &lt;em&gt;semigroup&lt;/em&gt; under the operation of composition, called the &quot;full transformation semigroup&quot; of $X$, often denoted $\mathcal{T}_X$. Full transformation semigroups play the same role in semigroup theory as symmetric groups do in group theory (the symmetric group is the group you get by considering the bijections from a set to itself, under composition). For example, the analogue of Cayley's Theorem (&quot;every group is a subgroup of a symmetric group&quot;) holds for semigroups (&quot;every semigroup is a subsemigroup of a full transformation semigroup&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You claim that &quot;for every endomap, there exists some subset of its domain such that it is bijective on that subset&quot;. That is only true if you restrict yourself to finite sets or you allow restriction to the empty set: consider the map $f\colon\mathbb{N}\to\mathbb{N}$ given by $f(n)=n+1$. If $X$ is a nonempty subset of $\mathbb{N}$, then the minimum of $X$ is not in $f(X)$, so $f$ is not bijective on $X$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-27T14:37:29.687" CommentCount="1" />
  <row Id="3470" PostTypeId="2" ParentId="3463" CreationDate="2010-08-27T14:41:47.543" Score="3" ViewCount="0" Body="&lt;p&gt;One of the important properties of endomaps of an infinite set $X$ is that the expressive power of the logic that allows you to quantify over all endomaps of $X$ is the same as the expressive power of the full second order logic on $X.$ Roughly, this means that any relation on $X$ (a subset of $X^n$) can be modelled by means of endomaps. For instance, any equivalence relation $E$ on $X$ is modelled as follows: for every such equivalnce relation $E$ there is an endomap $f : X \to X$ such that&#xA;$$&#xA;E(x_1,x_2) \iff f(x_1)=f(x_2) \qquad (x_1,x_2 \in X). &#xA;$$&#xA;So endomaps of $X$ can tell you everything about $X,$ if you &quot;ask&quot; them properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other classes of structures, endomorphisms also carry lots and lots of information. For instance, the semigroup $\mathrm{End}(V)$ of all endomorphisms of a given vector space $V$ carries most essential information about $V,$ etc.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-08-27T14:41:47.543" />
  <row Id="3471" PostTypeId="2" ParentId="3433" CreationDate="2010-08-27T14:46:32.530" Score="0" ViewCount="0" Body="&lt;p&gt;The following theorem (which I read in Number theory: algebraic numbers and functions&#xA; By Helmut Koch)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum\limits_{n=1}^{N} g(n) = g(1) + \int\limits_{1}^{N} g(x) \mathrm{d}x + \int\limits_{1}^{N} (x - [x]) g'(x) \mathrm{d}x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tells us that the difference converges when $$\int\limits_{1}^{N} (x - [x]) g'(x) \mathrm{d}x$$ does.&lt;/p&gt;&#xA;" LastActivityDate="2010-08-27T14:46:32.530" />
  <row Id="3472" PostTypeId="2" ParentId="3444" CreationDate="2010-08-27T15:01:26.480" Score="2" ViewCount="0" Body="&lt;p&gt;You are really asking for a direct connection between some property of $n!$ and the integral.  This can be done from the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S[n] = nS[n-1], S[0] = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S[n]$ is the order of the symmetric group on $n$ elements.  The exponential generating function for this series equals $1/(1-t)$.  As in your favorite proof, replace that by $\int_{0}^{\infty} e^{(x-1)t}  dt$, expand $e^{xt}$ as a series, reverse the order of summation and integration, and you recover a power series in $t$ whose terms are precisely $\int_{0}^{\infty} t^n e^{-t}  dt/n!$.  The result follows upon term-by-term comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that's not a whole lot more insight, but it does show explicitly a connection between a defining property of $n!$ and the integral for the $\Gamma$ function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your subsequent musings go: the integrand is the probability density of a Gamma variate, of course!  One useful relation is that the Gamma distribution with parameter $n$ is the sum (i.e., convolution) of $n$ independent exponential variates (i.e., Gamma variates with $n=1$).  The expectation of an exponential variate is 1 (this is easy), whence the expectation of a Gamma variate must be $n$ (because expectations add), strongly suggesting the mode of its pdf should be near $n$ (as justified for large $n$ by the Central Limit Theorem).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-27T15:01:26.480" CommentCount="5" />
  <row Id="3473" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T15:19:16.877" Score="1" ViewCount="0" Body="&lt;p&gt;$ 2^{n!}=(2^n)^{(n-1)!} $, so we have to compare $(2^n)^{(n-1)!} $ and $ (2^n)! $. For $n\geq 6$, $(n-1)!&gt;2^n$. I may be wrong, but doesn't the result follow immediately from this? It's still true for $n=5$ though, but not for $n=4$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-08-27T15:19:16.877" CommentCount="1" />
  <row Id="3474" PostTypeId="2" ParentId="3463" CreationDate="2010-08-27T15:32:28.523" Score="3" ViewCount="0" Body="&lt;p&gt;The generalization of cycle decomposition to endomaps is quite interesting; rather than just cycles, endomaps break up into cycles in which each vertex is the root of a tree.  Counting endomaps (of which there are $n^n$) is therefore relevant to counting trees, and this is the basis of a beautiful proof due to Joyal (which I believe can be found in Bergeron, Labelle, and Leroux's &lt;a href=&quot;http://bergeron.math.uqam.ca/Species/especes.html&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Species and Tree-like Structures&lt;/a&gt;) of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayleys_formula&quot; rel=&quot;nofollow&quot;&gt;Cayley's formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set equipped with an endomap is just about the simplest kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_system&quot; rel=&quot;nofollow&quot;&gt;dynamical system&lt;/a&gt;.  The category of sets equipped with endomaps is sometimes called the category of (discrete) dynamical systems and can be a useful example in category theory, e.g. it pops up in Lawvere and Schanuel's &lt;a href=&quot;http://books.google.com/books?id=o1tHw4W5MZQC&amp;amp;printsec=frontcover&amp;amp;dq=lawvere+and+schanuel&amp;amp;source=bl&amp;amp;ots=4u3nHsy70h&amp;amp;sig=PrsIRYFkI-CxO3AtnwLKf5dHP2E&amp;amp;hl=en&amp;amp;ei=ftp3TKumIIWBlAe7tcSzCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CBsQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Conceptual Mathematics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-27T15:32:28.523" />
  <row Id="3475" PostTypeId="2" ParentId="3441" CreationDate="2010-08-27T15:45:53.410" Score="2" ViewCount="0" Body="&lt;p&gt;You have to choose 10 letters, and 2 of them must be digits. Furthermore, there must be one each of a lowercase letter, an upper case letter, and a common symbol. For the others, there are 5 choices to be made, and these are to be made from $26+26+10+32 = 94$ characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$94^5 \approx 7.339e9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;choices for the the 6 other characters that do not have to be digits.&#xA;And for the digits, there are 2 choices from 10 characters. So this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 10^2 = 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;choices. And for the one each of lower-case letters, upper-case letters, and common symbols there are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 26 * 26 * 32  = 21632$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lastly, there are $10!$ permutations of these 10 characters, so the total number of combinations of these characters is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26*26*32*100*94^5 * 10! \approx 5.76e22$&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-08-27T15:53:49.487" LastActivityDate="2010-08-27T15:53:49.487" CommentCount="1" />
  <row Id="3476" PostTypeId="1" CreationDate="2010-08-27T16:30:46.890" Score="4" ViewCount="106" Body="&lt;p&gt;In a very interesting &lt;a href=&quot;http://golem.ph.utexas.edu/category/2010/08/what_is_the_langlands_programm.html#comments&quot; rel=&quot;nofollow&quot;&gt;blog discussion&lt;/a&gt; at the $n$-category cafe, an anonymous poster made the following remark: &quot;... using the dictionary between number fields and function fields, Weil suggested that G-with-adelic-entries is analogous to the group of gauge transformations of a principal G-bundle over a Riemann surface.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how this particular piece of the number field / function field analogy is made precise. As a preliminary question, does anyone know a reference to where Weil (or someone else) explains it? I have a suspicion that it might be discussed in the notes from Weil's 1959-1960 lectures on &quot;Adeles and algebraic groups&quot;, but the copy in my local library is checked out so I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My next question is for an explanation of the analogy in what I assume is the simplest case: holomorphic vector bundles on a compact Riemann surface. There are two obstacles here. First, I don't know what a gauge transformation is, and the &quot;Mathematical Formalism&quot; section &#xA;of the Wikipedia page seems nonsensical to me. What is being transformed, and what is the transformation itself? Can anyone provide a physics-free definition, purely in the language of geometry? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given a curve $X$ over the complex numbers (or any field), we can form a topological ring $\mathbf{A}_X$ as the restricted product of the completed local rings at the closed points of $X$. The second obstacle is that when the ground field is infinite, these completions are not locally compact, so it seems unlikely that $\mathbf{A}_X$ is a good thing to consider. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, can one put the constructions of the previous two paragraphs together to produce, for any rank $n$ holomorphic vector bundle $\mathcal{E}$ on $X$, a bijection as follows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\text{gauge transformations of }\mathcal{E}}\stackrel{?}{\leftrightarrow} \mathrm{GL}_n(\mathbf{A}_X)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My final question is, what is the correct version of this when the complex numbers are replaced by  a finite field? Now we can consider algebraic vector bundles, and even better the adeles of $X$ are now a good thing to consider. So in trying to extend the putative bijection above to this case, the content seems to be in algebraizing the notion of gauge transform. Thus, I'm asking for a (second?) definition of gauge transformation which is native to algebraic geometry, if such exists.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2010-08-28T22:18:14.417" Title="How are gauge transformations of a $G$-bundle related to the adelic points of $G$?" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;&lt;riemann-surface&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3477" PostTypeId="1" CreationDate="2010-08-27T18:14:14.280" Score="2" ViewCount="45" Body="&lt;p&gt;I hope this is an appropriate forum for this question...if not, any pointers on a place to ask would be great.  If my questions is not clear, please just let me know and I'll try to add information/explanation where I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have a set of data points that I run through a &lt;a href=&quot;http://en.wikipedia.org/wiki/FLAME_clustering&quot; rel=&quot;nofollow&quot;&gt;FLAME clustering algorithm&lt;/a&gt;, and I get my set of cluster supporting objects, and the fuzzy memberships among them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, let's say I get an additional set of data, for argument's sake, let's say it doubles the amount of data.  If I want to add that to my clustering, without having to reprocess the old data (i.e. just using the cluster supporting objects, and adding these new observations to the cluster), will I be introducing some kind of bias?  To get the most &quot;accurate&quot; picture, would it be better to reprocess ALL the data, re-establish CSO's, re-compute distances, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add to the complexity, let's say I'm going to be periodically adding similar quantities of data points.  Does the answer to the above change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I guess a final question - is there a clustering algorithm (or family of algorithms) that is not subject to extreme bias in the beginning, such that added data points can be added relatively pain-free?  I'm sort of thinking that FLAME clustering, with its nice fuzziness properties is a reasonable way of going about this, but any more suggestions would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="372" LastActivityDate="2010-12-17T08:05:08.713" Title="Bias from increased information in FLAME clustering" Tags="&lt;algorithms&gt;&lt;applications&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3478" PostTypeId="3" CreationDate="2010-08-27T19:02:16.537" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Mathematics - Stack Exchange&lt;/strong&gt; is for people studying mathematics at any level and professionals in related fields.  We welcome questions about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Understanding mathematical concepts and theorems&lt;/li&gt;&#xA;&lt;li&gt;Hints on mathematical problems&lt;/li&gt;&#xA;&lt;li&gt;History and development of mathematics&lt;/li&gt;&#xA;&lt;li&gt;Solving mathematical puzzles&lt;/li&gt;&#xA;&lt;li&gt;Software that mathematicians use&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are certain subjects that, while still being on-topic here, you can get better response on our sister sites:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Research-level mathematics — &lt;a href=&quot;http://mathoverflow.net/&quot; rel=&quot;nofollow&quot;&gt;Math Overflow&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Research level Algorithm design, complexity theory, etc — &lt;a href=&quot;http://cstheory.stackexchange.com/faq&quot;&gt;Theoretical Computer Science - Stack Exchange&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Algorithm implementation/design, computer simulation and modelling, etc — &lt;a href=&quot;http://stackoverflow.com&quot;&gt;Stack Overflow&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Statistical analysis  —  &lt;a href=&quot;http://stats.stackexchange.com&quot;&gt;Statistical Analysis - Stack Exchange&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And some kinds of questions are considered off-topic:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Physics, engineering and financial questions — it is fine to ask the mathematical model and techniques, but concepts of the field would be off-topic.&lt;/li&gt;&#xA;&lt;li&gt;Typesetting equations — you can read &lt;a href=&quot;http://meta.math.stackexchange.com/questions/107/what-should-go-in-the-math-stackexchange-faq/117#117&quot;&gt;our guideline&lt;/a&gt; for writing equations on this site, or visit &lt;a href=&quot;http://tex.stackexchange.com&quot;&gt;TeX, LaTeX and Friends&lt;/a&gt; for questions about TeX&lt;/li&gt;&#xA;&lt;li&gt;Numerology&lt;/li&gt;&#xA;&lt;li&gt;Questions about the site itself — please ask on &lt;a href=&quot;http://meta.math.stackexchange.com/&quot;&gt;meta&lt;/a&gt; instead&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to get a more information of the site, you can browse our &lt;a href=&quot;http://meta.math.stackexchange.com/questions/107&quot;&gt;detailed list of FAQ&lt;/a&gt; on meta. &lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-13T18:16:35.543" LastActivityDate="2010-10-13T18:16:35.543" />
  <row Id="3479" PostTypeId="2" ParentId="3459" CreationDate="2010-08-27T19:28:30.510" Score="5" ViewCount="0" Body="&lt;p&gt;I don't know the answer (&lt;b&gt;I should have! -- see below&lt;/b&gt;), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series.  In other words, a trigonometric series is just any series of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be understood formally, i.e., this is a trigonometric series even if it doesn't converge anywhere.  A fourier series is the trigonometric series associated to an &#xA;$L^1$ function by taking $A_n$ and $B_n$ as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as a Taylor series need not need converge at any point except the central point, a Fourier series need not converge pointwise at any point.  Thus the Fourier series need not be a function and in particular the Fourier inversion theorem need not apply.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theorem of Borel asserts that given any sequence $(a_n)$ of real numbers, there exists a $C^{\infty}$-function on the real line whose Taylor series at $0$ is $\sum_{n=0}^{\infty} a_n x^n$.  In particular, if the $a_n$'s grow too rapidly, the Taylor series will diverge away from zero and the function $f$ will not be analytic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpret the question as asking whether the analogue of Borel's theorem is true for Fourier series: is every trigonometric series the Fourier series of some $L^1$ function (even if the trigonometric series does not converge to the function)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: Boo to wikipedia for asserting the answer to this question without giving a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;: As Pierre-Yves Gaillard points out, the Fourier coefficients of any $L^1$ function $f$ are uniformly bounded (by $||f||_1$), so this answers the question as I have interpreted it.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-29T20:19:50.293" LastActivityDate="2010-08-29T20:19:50.293" CommentCount="5" />
  <row Id="3480" PostTypeId="2" ParentId="3388" CreationDate="2010-08-27T19:46:08.087" Score="0" ViewCount="0" Body="&lt;p&gt;Generalizing from decimal to arbitrary radix, a quick computer search seems to indicate  that there are only a handful of radices with such a unique solution. This leads to the following &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$ If in radix $\rm M$ notation we have $\rm (PQ)^2 = RSP$ with $\rm P,Q,R,S$ distinct digits and this solution is unique in radix $\:\rm M \:$ then it is one of the following, where $\;\rm A = 10,\; B = 11, \:\ldots,\: I = 18 \;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|r|r|r|} &#xA;\hline&#xA;\rm M &amp;amp; \rm PQ &amp;amp; \rm RSP \\&#xA;\hline&#xA;6 &amp;amp; 15 &amp;amp; 321 \\&#xA;7 &amp;amp; 23 &amp;amp; 562 \\&#xA;8 &amp;amp; 17 &amp;amp; 341 \\&#xA;9 &amp;amp; 18 &amp;amp; 351 \\&#xA;10 &amp;amp; 19 &amp;amp; 361 \\&#xA;11 &amp;amp; \rm 1A &amp;amp; 371 \\&#xA;18 &amp;amp; \rm 1H &amp;amp; \rm 3E1 \\&#xA;19 &amp;amp; \rm 1\:I &amp;amp; \rm 3F1 \\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\;$ The two general solutions listed below prove that all radices $\;\rm M &gt; 26 \;$ have nonunique solution (except possibly $\rm M = 44\:$). The remaining small number of exceptional cases were verified by computer.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm  (1\: M + M-1)^2 \;=\; \;\;\: 3\: M^2 + (M\;-\;\:4) M + 1, \quad\quad M &gt; 7$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm  (4\: M + M-2)^2 \;=\; 24\: M^2 + (M-20) M + 4, \quad\quad M&gt;26,\;\; M \ne 44$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T01:18:31.813" LastActivityDate="2010-08-28T01:18:31.813" CommentCount="2" />
  <row Id="3481" PostTypeId="1" AcceptedAnswerId="3487" CreationDate="2010-08-27T20:01:15.303" Score="3" ViewCount="135" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/2395/what-are-the-recommended-textbooks-for-introductory-calculus/2446#2446&quot;&gt;an answer&lt;/a&gt; to &lt;a href=&quot;http://math.stackexchange.com/questions/2395/what-are-the-recommended-textbooks-for-introductory-calculus&quot;&gt;another question I asked&lt;/a&gt;, Isaac suggested a book that is the standard &quot;reform calculus&quot; book. In a comment, I asked what the phrase &quot;reform calculus&quot; means, and Isaac provided a link to &lt;a href=&quot;http://www.math.duke.edu/~das/essays/trends/summary.html&quot; rel=&quot;nofollow&quot;&gt;this summary page&lt;/a&gt;. It seems to be a new or revised teaching methodology. However, the page linked to doesn't provide a good explanation of what exactly the phrase &quot;reform calculus&quot; means or its history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The summary describes things like courses being &quot;leaner in terms of the number of topics in the syllabus&quot; and says that &quot;technology alters the relative importance of specific techniques, and methodology because technology offers opportunities for creating new learning environments&quot;. But what do these things actually mean for people, especially the calculus student and the professor teaching a calculus course?&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-27T21:52:09.133" Title="What is &quot;reform calculus&quot;?" Tags="&lt;calculus&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="3483" PostTypeId="1" CreationDate="2010-08-27T20:08:54.687" Score="4" ViewCount="90" Body="&lt;p&gt;I'm brushing up on my calculus, and geometry and trigonometry keep coming up. However, these are probably my weakest areas mathematically. Are there any good sites or free PDFs/ebooks that I can use to quickly (in a weekend, preferably) work through geometry and trigonometry and refresh myself in topics such as areas and volumes of various geometric shapes, trigonometric functions, relationships between angles of a given shape or structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been over 8 years since I took a geometric or trigonometric course (it was back in high school) and over 2 years since I took any course that even relied on geometry or trigonometry. I'm not looking for anything incredibly complex, either. Just something to cover the topics that I'm encountering in integral and differential calculus.&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-11T06:26:46.183" LastActivityDate="2010-09-11T06:42:13.667" CommunityOwnedDate="2010-08-27T20:08:54.687" Title="What are some good resources for brushing up on geometry and trigonometry?" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;trigonometry&gt;&lt;self-learning&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="3484" PostTypeId="1" AcceptedAnswerId="3489" CreationDate="2010-08-27T20:24:03.847" Score="2" ViewCount="212" Body="&lt;p&gt;I know how to prove $\zeta (2)=\pi ^{2}/6$ by using the &lt;em&gt;trigonometric&lt;/em&gt; Fourier series expansion of $x^{2}/4$. How can one prove the same result using the &lt;em&gt;complex&lt;/em&gt; Fourier series of $f(x)=x$ for $0\leq x\leq 1$? Any suggestion?&lt;/p&gt;&#xA;" OwnerUserId="1504" LastEditorUserId="1504" LastEditorDisplayName="" LastEditDate="2010-08-27T23:00:47.130" LastActivityDate="2010-08-28T02:41:51.660" Title="Hint on how to prove $\zeta ( 2) =\pi ^{2}/6$ using the complex Fourier  series of $f(x)=x$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;&lt;special-functions&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3485" PostTypeId="2" ParentId="3452" CreationDate="2010-08-27T20:53:34.373" Score="-2" ViewCount="0" Body="&lt;p&gt;I would say you have been extremely sloppy in formulating the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say, &quot;it depends on the particular value of n&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, for example, n=0.&lt;/p&gt;&#xA;" OwnerUserId="1536" LastActivityDate="2010-08-27T20:53:34.373" CommentCount="1" />
  <row Id="3487" PostTypeId="2" ParentId="3481" CreationDate="2010-08-27T21:22:35.727" Score="4" ViewCount="0" Body="&lt;p&gt;It is hard to describe faithfully an entire movement or &quot;ism&quot;: suppose for instance you had asked for a description of Buddhism, or Marxism, or post-modernism.  For every principle that you put forward as a &quot;plank&quot; of the movement, there is someone to say that that's a misunderstanding/oversimplification, or there is a specific submovement formed out of the violation of that principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, let me focus on a particular branch of calculus reform which is better defined (and which I have seen more of myself): the Harvard Calculus Consortium.  A very clear description, review and critique of this movement is given by Oliver Knill (who is well placed to comment on it, having been involved in the teaching and administration of calculus at Harvard for many years) here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.harvard.edu/~knill///pedagogy/harvardcalculus/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.harvard.edu/~knill///pedagogy/harvardcalculus/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I encourage you to read this first and then ask any further questions with at least one specific platform underneath your feet.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-27T21:22:35.727" CommentCount="1" />
  <row Id="3488" PostTypeId="2" ParentId="3481" CreationDate="2010-08-27T21:52:09.133" Score="2" ViewCount="0" Body="&lt;p&gt;Here's an excerpt from &lt;a href=&quot;http://ramanujan.math.trinity.edu/tumath/research/studpapers/s45.pdf&quot; rel=&quot;nofollow&quot;&gt;Reviewing Reformed Calculus&lt;/a&gt;, by Lisa Murphy, 2006.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The very beginnings of mathematics reform started in the 1960's, but the big push for Calculus&#xA;  reform started in earnest in 1989 with the publication of the National Council of Teachers of Mathematics'&#xA;  (NCTM) Principles and Standards for Mathematics Education. The NCTM published the&#xA;  Principles and Standards in response to the apathy of students towards math and the lack of academic&#xA;  success in the mathematics classroom. To combat these negative trends the NCTM outlined&#xA;  five goals for &quot;the processes of problem solving, reasoning and proof, connections, communication,&#xA;  and representation&quot; [1]. Through these goals it was hoped that students would be equipped with&#xA;  the basic skills and understanding that they would need to be successful.&#xA;  As the Principles and Standards inspired the reform of secondary mathematics education, thoughts&#xA;  of reform began to surface in the collegiate mathematics arena, especially with regard to Calculus.&#xA;  College Calculus courses were experiencing some of the same problems as secondary mathematics.&#xA;  Of the roughly 300,000 college students that are annually enrolled in an engineering-based Calculus&#xA;  course, only 140,000 earn a grade of D or higher [6]. Less than half of the students were performing&#xA;  &quot;well&quot; in their Calculus courses. Armed with statistics such as this, reform-minded professors set&#xA;  out to develop a new curriculum that would help raise the achievement level and stimulate student&#xA;  interest in mathematics.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From the reform movement, numerous curricular designs have been generated. Calculus and&#xA;  Mathematica; Calculus, Concepts, Computers and Cooperative Learning (C4L); and The Calculus&#xA;  Consortium at Harvard (CCH) are a few of the commonly used curriculums. These new curriculums&#xA;  cover the entire spectrum of reform. Some are grounded in traditional techniques but incorporate&#xA;  snippets of reform, while others differ in most aspects from the traditional approach. Despite this&#xA;  vast array, there are some basic elements that are common to all reform curriculums in varying&#xA;  degrees that separate them from the traditional Calculus curriculum.&#xA;  One of the most noticeable differences of reformed Calculus is the use of graphing calculators&#xA;  and/or computers. The graphing calculator is a critical component in the reform classroom. Many&#xA;  reform classes include a weekly lab session where students meet in a computer lab. The students&#xA;  make use of calculators and math computer programs to investigate new topics and to graphically&#xA;  see what they are working on. Most reform textbooks urge students to read through the text with a&#xA;  calculator in hand to see directly what is discussed in the text. The idea behind the incorporation of&#xA;  calculators and computers is to alleviate the heavy algebraic manipulation that students typically do&#xA;  in a traditional Calculus setting. Reform supporters argue that the removal of manipulation allows&#xA;  students to move beyond the drudgery of computation and start learning the fundamental ideas of&#xA;  Calculus. They additionally argue that topics are discussed more fully with the use of graphical&#xA;  representations.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A reformed Calculus class differs from a traditional course in methods of instruction. When&#xA;  walking into a reform classroom it is immediately clear that it is indeed a reform classroom. Most&#xA;  noticeably the teacher is no longer the central focus of the classroom experience. The lecture method&#xA;  of instruction, a standard of traditional curriculum, has a lesser place within a reform setting. The&#xA;  teacher still lectures occasionally and is available to answer questions from the students, but there is&#xA;  greater emphasis placed on cooperative learning. Reform students often work in groups to determine&#xA;  solutions or to explore concepts in a laboratory setting. This idea is rooted in the constructivist&#xA;  learning theory. Each student constructs their own meaning as they learn. Students are given the&#xA;  basic tools and from these discover how the pieces fit together to form the concept that they are&#xA;  studying. One of the primary goals of the C4L curriculum is to &quot;create situations which foster&#xA;  students to make the necessary mental constructions to learn mathematics concepts&quot; [10].&#xA;  Within the curriculum itself, the reformed method stresses the applications of Calculus. This&#xA;  emphasis hopes to justify the topics of study, which in theory raises interest in the material. In&#xA;  an effort to accomplish this, some of the mathematical rigor is removed from the curriculum. Most&#xA;  reform textbooks are void of a single proof. In the introduction to Calculus from Graphical, Numerical&#xA;  and Symbolic Point of View, the authors state that &quot;proving theorems in full generality is&#xA;  less valuable, we think, than helping students understand concretely what theorems say&quot; [9]. As&#xA;  a result of this change, a common question that arises from students new to reformed Calculus is&#xA;  &quot;Where is the math?&quot; [5].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Accompanying this application heavy curriculum is a different method of assessment. Reformed&#xA;  Calculus courses emphasize the use of writing. Projects, reports and lengthy explanations of problem&#xA;  solutions are common place within the reform classroom. In some cases, the students are graded&#xA;  more on the thoroughness and completeness of written explanations as opposed to correctness of&#xA;  answer.&#xA;  [...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The emphasis on correct explanation rather than correct answer is seen explicitly in the directions&#xA;  for the midterm. The problem also provides an example of the type of application problems that&#xA;  reformed Calculus students are accustomed to working with. This midterm question additionally&#xA;  exhibits one of the flaws that traditional professors are quick to point out. The problem asks students&#xA;  to determine when the population becomes infinite, which is a misuse of the word infinite.&#xA;  The population may become uncontrollable but it will never become infinite. Traditional professors&#xA;  argue that the misuse of mathematical terms, such as infinite, teaches students the wrong meaning&#xA;  of or concept behind the term, which results in misunderstandings in future math work.&#xA;  More generally, there is a trend in reformed Calculus moving away from individual study and&#xA;  towards a social study of Calculus. The context of learning Calculus is now placed in a more social&#xA;  setting. Students work primarily in groups to gain knowledge both from a textbook and from each&#xA;  other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-27T21:52:09.133" />
  <row Id="3489" PostTypeId="2" ParentId="3484" CreationDate="2010-08-27T23:09:53.373" Score="2" ViewCount="0" Body="&lt;p&gt;Use the definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $f$ is defined on $[-\pi, \pi]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(z) = \sum_{-\infty}^{\infty} {c_{n} e^{inz}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}{f(z)e^{-inz}} dz$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put $f(z) = z$, can you work out what $c_{n}$ turns out to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To integrate, you can try integration by parts. &lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T02:41:51.660" LastActivityDate="2010-08-28T02:41:51.660" CommentCount="2" />
  <row Id="3490" PostTypeId="1" AcceptedAnswerId="3491" CreationDate="2010-08-27T23:26:05.400" Score="2" ViewCount="179" Body="&lt;p&gt;What's the simplest formula to calculate the optimal sample size from a population size? I've been reading a little about it and they ask me stuff like confidence level or confidence interval which I have no idea about. I need to test some HTML elements in web pages to get a better idea about general properties of HTML elements, and I have no idea about he results I'll get. So I just need the most neutral possible version of the sample size calculating formula when all I know is the population size.&lt;/p&gt;&#xA;" OwnerUserId="1542" LastActivityDate="2010-09-16T22:07:20.517" Title="Optimal sample size?" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="3491" PostTypeId="2" ParentId="3490" CreationDate="2010-08-27T23:47:53.793" Score="1" ViewCount="0" Body="&lt;p&gt;It depends on your population and the confidence that you need in your results. Typically, you should know something about what you are trying to achieve. If you know your confidence level, confidence interval, and population size, you can choose a sample size that will give you the properly confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your confidence level, typically around 90% (95% and 99% are also commonly used), tells you how sure you will be that your entire population reflects your results. The confidence interval describes the range of values that you are sure about within your confidence level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sample_size&quot; rel=&quot;nofollow&quot;&gt;Wikipedia provides an overview of sample size&lt;/a&gt; methodologies. That might get you started. But unless you know how sure you want to be of your results, you can't determine a sample size. Wikipedia also provides decent definitions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Confidence_interval&quot; rel=&quot;nofollow&quot;&gt;confidence level and confidence interval&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a statistical standpoint, if you don't have clearly defined questions, you really can't analyze your data, at least using statistical methods. Perhaps you should review data and try to formulate questions, or take an analytical instead of statistical approach to solving the problem/answering your question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming a normal distribution, you can use the equation $n \geq \left(\dfrac{z_{\alpha/2} \sigma}{\delta}\right)^2$ where z is found in a table, σ is your standard deviation (which you can use either a sample standard deviation or a population standard deviation, depending on what you know), and δ is your confidence level. The value for n is your population - be sure to round up if you get a fractional n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the z-value is based on your confidence level. The value for α is a value between 1 - the decimal format of your confidence level (for a confidence level of 95%, α would be 1 - 0.95, or 0.05).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also be interested in the answers to a few questions on the Statistical Analysis Stack Exchange:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/166/how-do-you-decide-the-sample-size-when-polling-a-large-population&quot;&gt;How do you decide the sample size when polling a large population?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/276/how-large-should-a-sample-be-for-a-given-estimation-technique-and-parameters&quot;&gt;How large should a sample be for a given estimation technique and parameters?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="91" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T22:07:20.517" LastActivityDate="2010-09-16T22:07:20.517" CommentCount="2" />
  <row Id="3492" PostTypeId="2" ParentId="3484" CreationDate="2010-08-28T00:08:33.340" Score="1" ViewCount="0" Body="&lt;p&gt;This is not related but you would like to see this article: A Short Proof of ζ (2) = π2/6 T.H. Marshall American Math monthly April 2010.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-28T00:08:33.340" />
  <row Id="3493" PostTypeId="2" ParentId="3483" CreationDate="2010-08-28T00:24:46.477" Score="1" ViewCount="0" Body="&lt;p&gt;Hey there are lot of sources available on the internet. I enlist a few of them here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://www.free-ed.net/free-ed/Math/Trigonometry/trig02_SPK.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.free-ed.net/free-ed/Math/Trigonometry/trig02_SPK.asp&lt;/a&gt; This website provides an nice introduction, which is very basic in nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For geometry you can read this book by a very good Mathematician, which is available online : &lt;a href=&quot;http://math.mit.edu/~kedlaya/geometryunbound/&quot; rel=&quot;nofollow&quot;&gt;http://math.mit.edu/~kedlaya/geometryunbound/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure gigapedia.com will have lots of ebooks on this subjects. Try searching it.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-28T00:24:46.477" CommunityOwnedDate="2010-08-28T00:24:46.477" CommentCount="1" />
  <row Id="3494" PostTypeId="2" ParentId="3024" CreationDate="2010-08-28T00:26:33.527" Score="1" ViewCount="0" Body="&lt;p&gt;Your given metric is in the form of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Warped_geometry&quot; rel=&quot;nofollow&quot;&gt;warped product&lt;/a&gt;, for which there are some well established methods of computing the curvature of the total space in terms of the curvature of the warp factors. See, for example, Barrett O'Neill's &lt;em&gt;Semi-Riemannian Geometry&lt;/em&gt;, Chapter 7. For the first factor (given by $u$ and $r$), it being two dimensional, you can just use the standard results for surfaces on its intrinsic curvature. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-08-28T00:26:33.527" />
  <row Id="3495" PostTypeId="2" ParentId="3483" CreationDate="2010-08-28T00:38:53.257" Score="0" ViewCount="0" Body="&lt;p&gt;Although it's a little basic, I managed to find a &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBkQFjAA&amp;amp;url=http%3A%2F%2Fwww.videomathteacher.com%2Fuploaded%2Ffiles%2F02GeometryRefSheet.pdf&amp;amp;ei=4ll4TL7-IsP_lge_4v3rCw&amp;amp;usg=AFQjCNE44cwWueDYBTBPtf7RZcX2ZYgwQQ&amp;amp;sig2=aF6D3iZxDWn_99XSzI7lOQ&quot; rel=&quot;nofollow&quot;&gt;&quot;geometry reference sheet&quot; (direct PDF download)&lt;/a&gt;. It's got your standard two- and three-dimensional shapes, plus your law of sines, law of cosines, distance, and midpoint formulas. Although if it's been years since you had to find the volume of a sphere, this kind of thing might be useful.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2010-08-28T00:38:53.257" CommunityOwnedDate="2010-08-28T00:38:53.257" />
  <row Id="3496" PostTypeId="2" ParentId="3335" CreationDate="2010-08-28T00:46:55.493" Score="2" ViewCount="0" Body="&lt;p&gt;Since you like the writing style of Courant and John... for PDEs there are of course F. John's &lt;em&gt;Partial Differential Equations&lt;/em&gt;, and also Courant and Hilbert's &lt;em&gt;Methods of Mathematical Physics&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-08-28T00:46:55.493" CommunityOwnedDate="2010-08-28T00:46:55.493" />
  <row Id="3497" PostTypeId="1" CreationDate="2010-08-28T00:48:36.883" Score="4" ViewCount="64" Body="&lt;p&gt;Sorry if this question is too basic. It is from Fröhlich and Taylor's &quot;Algebraic Number Theory&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E/F$ be a finite Galois extension of fields, with $G=Gal(E/F)$, and let $K$ and $L$ be two subfields of $E$, containing $F$, such that $K/F$ and $L/F$ are both Galois. Let $M=Gal(E/K)$ and $N=Gal(E/L)$ be normal subgroups of $G$. Suppose ${\gamma_1,\ldots,\gamma_n}$ is a transversal for $MN$ in $G$, with $n=[G:MN]$. If $C$ is the compositum $KL$ in $E$, how can I show the map&#xA;$$ k\otimes l\mapsto (k^{\gamma_1}l,\ldots,k^{\gamma_n}l) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;induces an isomorphism between $K\otimes_F L$ and $\prod_{i=1}^n C$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear to me that this map is an $F$-algebra homomorphism, and that they both have the same dimension over $F$.  Thus surjectivity, or injectivity, would be enough.  I have not been able to figure out what the idempotents of $\prod_{i=1}^n C$ should look like in $K\otimes_F L$, so I have not been able to show surjectivity.  Meanwhile, I think injectivity should be easier to show, because if we have&#xA;$$ k_1\otimes l_1 + \cdots + k_m\otimes l_m\mapsto 0,$$&#xA;then we get a system of equations&#xA;$$ k_1l_1 + \cdots + k_ml_m=0$$&#xA;$$ \cdots$$&#xA;$$ k_1^{\gamma_n}l_1 + \cdots k_m^{\gamma_n}l_m=0.$$&#xA;Summing up the columns, I get&#xA;$$ \sum_{i=1}^m(\sum_{j=1}^n k_i^{\gamma_j})l_i=0 $$&#xA;and all this is happening in $L$.  But I can't seem to finish this argument.  Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1544" LastActivityDate="2010-08-28T16:16:55.583" Title="Algebra structure of tensor product of two Galois extensions" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3498" PostTypeId="2" ParentId="3444" CreationDate="2010-08-28T01:39:26.310" Score="7" ViewCount="0" Body="&lt;p&gt;I haven't quite got this straight yet, but I think one way to go is to think about choosing points at random from the positive reals. This answer is going to rather longer than it really needs to be, because I'm thinking about this in a few (closely related) ways, which probably aren't all necessary, and you can decide to reject the uninteresting parts and keep anything of value. Very roughly, the idea is that if you &quot;randomly&quot; choose points from the positive reals and arrange them in increasing order, then the probability that the (n+1)th point is in a small interval (t,t+dt) is a product of probabilities of independent events, n factors of t for choosing n points in the interval [0,t], one factor of $e^{-t}$ as all the other points are in $[t,\infty)$, one factor of dt for choosing the point in $(t,t+dt)$, and a denominator of $n!$ coming from the reordering. At least, as an exercise in making a simple problem much harder, here goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start with a bit of theory before trying to describe intuitively why the probability density $\frac{t^n}{n!}e^{-t}$ pops out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_process&quot; rel=&quot;nofollow&quot;&gt;homogeneous Poisson process&lt;/a&gt; (with rate parameter 1). One way to think of this is to take a sequence on independent exponentially distributed random variables with rate parameter 1, $S_1,S_2,\ldots$, and set $T_n=S_1+\cdots+S_n$. As has been commented on already, $T_{n+1}$ has the probability density function $\frac{t^n}{n!}e^{-t}$. I'm going to avoid proving this immediately though, as It would just reduce to manipulating some integrals. Then, the Poisson process $X(t)$ counts the number of times $T_i$ lying in the interval [0,t].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also look at Poisson point processes (aka, &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_random_measure&quot; rel=&quot;nofollow&quot;&gt;Poisson random measures&lt;/a&gt;, but that Wikipedia page is very poor). This is just makes rigorous the idea of randomly choosing unordered sets of points from a sigma-finite measure space $(E,\mathcal{E},\mu)$. Technically, it can be defined as a set of nonnegative integer-valued random variables $\{N(A)\colon A\in\mathcal{E}\}$ counting the number of points chosen from each subset A, such that $N(A)$ has the Poisson distribution of rate $\mu(A)$ and $N(A_1),N(A_2),\ldots$ are independent for pairwise disjoint sets $A_1,A_2,\ldots$. By definition, this satisfies&#xA;$$&#xA;\begin{array}{}\mathbb{P}(N(A)=n)=\frac{\mu(A)^n}{n!}e^{-\mu(A)}.&amp;amp;&amp;amp;(1)\end{array}&#xA;$$&#xA;The points $T_1,T_2,\ldots$ above defining the homogeneous Poisson process also define a Poisson random measure with respect to the Lebesgue measure $(\mathbb{R}_+,{\cal B},\lambda)$. Once you forget about the order in which they were defined and just regard them as a random set that is, which I think is the source of the $n!$. If you think about the probability of $T_{n+1}$ being in a small interval $(t,t+\delta t)$ then this is just the same as having $N([0,t])=n$ and $N((t,t+\delta t))=1$, which has probability $\frac{t^n}{n!}e^{-t}\delta t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can we choose points at random so that each small set $\delta A$ has probability $\mu(\delta A)$ of containing a point, and why does (1) pop out? I'm imagining a hopeless darts player randomly throwing darts about and, purely by luck, hitting the board with some of them. Consider throwing a very large number $N\gg1$ of darts, independently, so that each one only has probability $\mu(A)/N$ of hitting the set, and is distributed according to the probability distribution $\mu/\mu(A)$. This is consistent, at least, if you think about the probability of hitting a subset $B\subseteq A$. The probability of missing with all of them is $(1-\mu(A)/N)^N=e^{-\mu(A)}$. This is a multiplicative function due to independence of the number hitting disjoint sets. To get the probability of one dart hitting the set, multiply by $\mu(A)$ (one factor of $\mu(A)/N$ for each individual dart, multiplied by N because there's N of them). For n darts, we multiply by $\mu(A)$ n times, for picking n darts to hit, then divide by n! because we have over-counted the subsets of size n by this factor (due to counting all n! ways of ordering them). This gives (1). I think this argument can probably be cleaned up a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to choosing points randomly on the positive reals, this gives a probability of $\frac{t^n}{n!}e^{-t}dt$ of picking n in the interval $[0,t]$ and one in $(t,t+dt)$. If we sort them in order as $T_1\lt T_2\lt\cdots$ then $\mathbb{P}(T_1\gt t)=e^{-t}$, so it is exponentially distributed. Conditional on this, $T_2,T_3,\ldots$ are chosen randomly from $[T_1,\infty)$, so we see that the differences $T_{i+1}-T_{i}$ are independent and identically distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\frac{t^n}{n!}e^{-t}$ maximized at t=n? I'm not sure why the mode should be a simple property of a distribution. It doesn't even exist except for unimodal distributions. As $T_{n+1}$ is the sum of n+1 IID random variables of mean one, the law of large numbers suggests that it should be peaked approximately around n. The central limit theorem goes further, and gives $\frac{t^n}{n!}e^{-t}\approx\frac{1}{\sqrt{2\pi n}}e^{-(t-n)^2/{2n}}$. Stirling's formula is just this evaluated at $t=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's this to do with Tate's thesis? I don't know, and I haven't read it (but intend to), but have a vague idea of what it's about. If there is anything to do with it, maybe it is something to do with the fact that we are relating the sums of independent random variables $S_1+\cdots+S_n$ distributed with respect to the Haar measure on the multiplicative group $\mathbb{R}_+$ (edit: oops, that's not true, the multiplicative Haar measure has cumulative distribution given by log, not exp) with randomly chosen sets according to the Haar measure on the additive group $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-08-28T02:58:47.697" LastActivityDate="2010-08-28T02:58:47.697" CommentCount="2" />
  <row Id="3499" PostTypeId="1" AcceptedAnswerId="3524" CreationDate="2010-08-28T01:47:31.130" Score="0" ViewCount="82" Body="&lt;p&gt;Suppose $C \subset \mathbb{R}$ is of type $F_{\sigma}$. That is $C$ can be written as the union of $F_{n}$'s where each $F_{n}$'s are closed. Then can we prove that each point of $C$ is a point of discontinuity for some $f: \mathbb{R} \to \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I refered this link on wiki : &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thomae%27s_function&lt;/a&gt; and in the follow up subsection they given this result. I would like somebody to explain it more precisely.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-28T07:48:01.743" LastActivityDate="2010-08-31T09:57:25.443" Title="$F_{\sigma}$ subsets of $\mathbb{R}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3500" PostTypeId="2" ParentId="1890" CreationDate="2010-08-28T02:14:29.783" Score="-1" ViewCount="0" Body="&lt;p&gt;Since $f(2x-f(x))=x$ and $f$ is an injection, we get $f^{-1}(x)=2x-f(x)$. Thus we have $$f(x)-x = x - f^{-1}(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x_{0} \in [0,1]$ define the sequence $\{x_{n}\}$ by $x_{n}=f(x_{n-1})$. Then from above we have $x_{n}-x_{n-1} = x_{n-1} - x_{n-2}$. Therefore $x_{n} = x_{0} + n(x_{1}-x_{0})$. Since $|x_{n}-x_{0}| \leq 1$, we have $|x_{1}-x_{0}| \leq \frac{1}{n}$ for $n \in \mathbb{N}$. Hence $f(x_{0})=x_{1}=x_{0}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-28T02:14:29.783" CommentCount="6" />
  <row Id="3501" PostTypeId="1" AcceptedAnswerId="3584" CreationDate="2010-08-28T02:23:59.140" Score="0" ViewCount="81" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the question I've been thinking for 2 days and couldn't find an answer why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Y = √X, where both X and Y are real numbers then which of the following is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) X&gt;=0; Y&gt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) X&amp;lt;=0; Y&gt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) X&amp;lt;=0; Y&amp;lt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) X&gt;=0; Y&amp;lt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) Either a) or b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the answer is a) but I don't know why. I know a value of Y for which root of Y is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;√4 = +2, -2&#xA;√9 = +3, -3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer should be X&amp;lt;=0; Y&gt;=0, but its not. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain me why? Thanks in advance. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I think I messed up the under root symbol. (√ = Under Root / Square Root) I don't know how to get the proper symbol where I'll get a line of that symbol extended above numbers. :P&lt;/p&gt;&#xA;" OwnerUserId="712" LastEditorUserId="712" LastEditorDisplayName="" LastEditDate="2010-08-29T05:08:06.160" LastActivityDate="2010-08-29T05:11:59.420" Title="Problem related with equation consisting square root" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="3502" PostTypeId="1" AcceptedAnswerId="3505" CreationDate="2010-08-28T02:39:42.697" Score="2" ViewCount="222" Body="&lt;p&gt;How does one sum the series, $$ S = a -\frac{2}{3}a^{3} + \frac{2 \cdot 4}{3 \cdot 5} a^{5} - \frac{ 2 \cdot 4 \cdot 6}{ 3 \cdot 5 \cdot 7}a^{7} + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was asked to me by a high school student, and i am embarrassed that i couldn't solve it. Can anyone give me a hint!!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-08-28T13:58:02.783" LastActivityDate="2010-08-28T13:58:02.783" Title="Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$" Tags="&lt;sequences-and-series&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="3503" PostTypeId="2" ParentId="3502" CreationDate="2010-08-28T03:05:35.117" Score="5" ViewCount="0" Body="&lt;p&gt;You can use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx} = \frac{2 \cdot 4 \cdot 6 \cdots 2k}{3\cdot 5 \cdots (2k+1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called Wallis's product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $\displaystyle S(a) = \sum_{k=0}^{\infty} (-1)^k a^{2k+1} \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interchanging the sum and the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(a) =   \int_{0}^{\frac{\pi}{2}}{\sum_{k=0}^{\infty}{(-1)^{k}(a\sin x)^{2k+1}} dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum inside the integral is a geometric series of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x - x^3 + x^5 - \cdots = x(1 - x^2 + x^4 - \cdots)  = \frac{x}{1+x^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(a) =   \int_{0}^{\frac{\pi}{2}}{\frac{a\sin x}{1 + (a\sin x)^2}}dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now substitute $\displaystyle t = a \cos x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{0}^{a}{\frac{1}{1+a^2 - t^2}}dt = \frac{1}{2\sqrt{a^2+1}}\ln(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \ln(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a}) = \ln(\frac{(\sqrt{a^2+1}+a)^2}{(\sqrt{a^2+1}-a)(\sqrt{a^2+1}+a)}) = 2\ln(\sqrt{a^2+1}+a)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(a) = \frac{1}{\sqrt{a^2+1}}\ln(\sqrt{a^2+1}+a) = \frac{\sinh^{-1}(a)}{\sqrt{a^2+1}}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T13:42:44.377" LastActivityDate="2010-08-28T13:42:44.377" CommentCount="3" />
  <row Id="3504" PostTypeId="1" AcceptedAnswerId="3835" CreationDate="2010-08-28T03:25:23.280" Score="1" ViewCount="84" Body="&lt;p&gt;&lt;strong&gt;From a 2005 Russian olympiad. Prove that there do not exist four (pairwise) different quadratic polynomials, with leading coefficient 1, such that the sum of any two of them has a single root.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Optionally, this is what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P_i(x)=x^2+b_ix+c_i$, then $P_i+P_j$ is single-rooted iff $(b_i+b_j)^2=8(c_i+c_j)$, with root $-(b_i+b_j)/2$. If two of the six roots are equal, we're done, so suppose they're different, and that $P_1+P_2$ is single-rooted, that is, $(b_1+b_2)^2=8(c_1+c_2)$. If we want $P_1+P_3$ and $P_2+P_3$ to be single-rooted as well, we need $(b_1+b_3)^2=8(c_1+c_3)$ and $(b_2+b_3)^2=8(c_2+c_3)$, from which, taking $b_1$ and $c_1$ as known, $b_2, c_2, b_3, c_3$ could be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way looks cumbersome.)&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-02T01:55:30.690" LastActivityDate="2010-09-02T01:55:30.690" Title="Four polynomials with single-rooted sums." Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3505" PostTypeId="2" ParentId="3502" CreationDate="2010-08-28T03:31:51.030" Score="9" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad \:\;\;\rm (a^2+1) \: S' = 1 - a \: S \;\:$ by transmuting the coefficient recurrence to a differential equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\;\Rightarrow\; 1 = (a^2+1) \: S' + a \: S \; = \; f \: (f \; S)' \;\;$ for $\rm\;\; f = (a^2+1)^{1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\;\Rightarrow\; S = f^{-1} \int \; f^{-1} = \frac{\sinh^{-1}(a)}{(a^2+1)^{1/2}}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T04:36:16.937" LastActivityDate="2010-08-28T04:36:16.937" CommentCount="1" />
  <row Id="3506" PostTypeId="1" AcceptedAnswerId="3542" CreationDate="2010-08-28T04:00:36.603" Score="0" ViewCount="138" Body="&lt;p&gt;Is there a natural way to represent all the partitions of an integer set {1,2,3,...,n} as a matrix in the similar way permutations can be mapped to group of matrices?&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-08-28T12:58:51.107" Title="Matrix representation of a partition" Tags="&lt;combinatorics&gt;&lt;partitions&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="3507" PostTypeId="1" AcceptedAnswerId="3511" CreationDate="2010-08-28T04:07:44.613" Score="6" ViewCount="158" Body="&lt;p&gt;At the beginning of &lt;a href=&quot;http://math.ucr.edu/home/baez/week300.html&quot; rel=&quot;nofollow&quot;&gt;Week 300 of John Baez's blog&lt;/a&gt;,&#xA;Baez gives a proof that the &quot;number&quot; of finite sets (more specifically, the cardinality of the groupoid of all finite sets, where an object in the groupoid counts as 1/n! if it has n! symmetries) equals e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then says that this leads to a purely combinatorial proof that e^x is its own derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain the purely combinatorial proof?&lt;/p&gt;&#xA;" OwnerUserId="1546" LastActivityDate="2010-08-28T05:20:07.213" Title="Purely combinatorial proof that d/dx(e^x) = e^x" Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="3508" PostTypeId="1" AcceptedAnswerId="3518" CreationDate="2010-08-28T04:08:01.697" Score="3" ViewCount="231" Body="&lt;p&gt;A matrix determinant (naively) can be computed in O(n!) steps, or with a proper LU decomposition O(n^3) steps. This assumes that all the matrix elements are constant. If, however the matrix elements are polynomials (say univariate of max order p) at each step of the LU decomposition an element is multiplied by another element producing (on average) ever larger polynomials. Each step therefore takes longer and longer - is the cost perform the decomposition still a polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: To clarify my reasoning a bit, if polynomial (using FFT as Mangaldan suggests in the comments) takes O(m log m), and we must perform O(n^3) operations to get the LU decomposition, each step the polynomial could effectively double in degree at each multiplication*. The running time would look something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \approx O \left ( \sum_{k}^{n^3} (p \ln p)^{2k} \right ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* (it doesn't quite do that, and this is where I'm stuck)&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-08-28T05:28:02.023" LastActivityDate="2010-08-29T02:56:24.937" Title="Determinant of a polynomial matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3509" PostTypeId="2" ParentId="3502" CreationDate="2010-08-28T04:08:59.450" Score="3" ViewCount="0" Body="&lt;p&gt;Making my comments more explicit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sum of interest is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{j=0}^\infty {(-1)^j \frac{(2j)!!}{(2j+1)!!} a^{2j+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(2j)!!=2\cdot 4\cdot 6\cdots (2j)$ and $(2j+1)!!=3\cdot 5\cdot 7\cdots (2j+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify things a bit, we rearrange the series a bit to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a\sum_{j=0}^\infty {\frac{(2j)!!}{(2j+1)!!}\left(-a^2\right)^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The double factorials can be also expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2j)!!=2^j j!=2^j (1)_j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2j+1)!!=2^j \left(\frac32\right)_j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(a)_j$ is a Pochhammer symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute both expressions into the series, and then note that the series now looks like a hypergeometric series. Now you can employ the formula &lt;a href=&quot;http://functions.wolfram.com/07.23.03.3098.01&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-28T04:08:59.450" CommentCount="5" />
  <row Id="3510" PostTypeId="1" AcceptedAnswerId="3513" CreationDate="2010-08-28T04:17:02.887" Score="8" ViewCount="1316" Body="&lt;p&gt;Could you provide a proof of Euler's formula: $\exp(i t)=\cos(t)+i\sin(t)$ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T18:50:53.660" LastActivityDate="2010-11-25T09:26:35.567" Title="How to prove Euler's formula: $\exp(i t)=\cos(t)+i\sin(t)$ ?" Tags="&lt;complex-analysis&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="5" />
  <row Id="3511" PostTypeId="2" ParentId="3507" CreationDate="2010-08-28T04:39:17.760" Score="6" ViewCount="0" Body="&lt;p&gt;I am not quite sure how to translate this into groupoid cardinality language, but here is the standard proof.  Suppose $A(x) = \sum_{n \ge 0} a_n \frac{x^n}{n!}$ is an exponential generating function.  Then we should interpret $a_n$ as being the number of ways to put a certain structure on a set of size $n$.  For example, when $a_n = 1$ this is the structure of &quot;being a set.&quot;  When $a_n = n!$ this is the structure of &quot;being a totally ordered set.&quot;  And so forth.  We will call this an $A$-structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $A'(x) = \sum_{n \ge 0} a_{n+1} \frac{x^n}{n!}$ can be interpreted as having coefficients $b_n = a_{n+1}$ which count the number of ways to add an element to a set of size $n$, then put an $A$-structure on the resulting set of size $n+1$.  This is a purely combinatorial definition of differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this definition, the proof is quite obvious: there is exactly one way for a set to be a set, and there is also exactly one way to add an element to a set and then make the result a set.  So $\frac{d}{dx} e^x = e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof might seem contentless.  Try to see how it generalizes to show that $\frac{d}{dx} e^{ax} = ae^{ax}$ for any positive integer $a$, and if you're up for a challenge see if you can generalize it all the way to &lt;a href=&quot;http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/&quot; rel=&quot;nofollow&quot;&gt;this identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vaguely the proof in groupoid cardinality language goes like this.  For a finite set $X$ the groupoid of finite sets equipped with a function to $X$ has cardinality $e^{|X|}$.  (The morphisms between two objects $A \to X, B \to X$ in this category are isomorphisms $A \simeq B$ such that the obvious triangle commutes.)  One way to think about this groupoid is as the groupoid of &quot;colored&quot; sets, where $X$ is the set of colors and an isomorphism must respect color.  Then it is easy to see that an isomorphism class of colored sets where there are $|X|$ colors is the same thing as a disjoint union of isomorphism classes of $|X|$ sets, one for each color.  One gets a direct interpretation of the terms in the expansion $\left( \sum_{n \ge 0}^{\infty} \frac{1}{n!} \right)^{|X|}$ this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiation replaces $|X|^n$ with $n|X|^{n-1}$, which means that we replace functions from an $n$-element set $S$ to $X$ with functions from $S - \{ s \}$ to $X$ where $s$ ranges over all elements of $S$.  The resulting groupoid is still the groupoid of finite sets equipped with a function to $X$; in particular, it has the same cardinality.  (Note that $X$ does not really have to be a finite set of a particular size for this argument to work; it can be a &quot;formal&quot; set in the same way that $x$ is a formal variable and the resulting groupoid cardinality is a generating function instead of a number.  I think this is what the formal theory of &quot;stuff types&quot; is for, but I am not familiar with it.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-28T05:20:07.213" LastActivityDate="2010-08-28T05:20:07.213" />
  <row Id="3512" PostTypeId="2" ParentId="3510" CreationDate="2010-08-28T04:48:13.147" Score="14" ViewCount="0" Body="&lt;p&gt;Let $\mathbf{A}$ be an $n \times n$ matrix.  Recall that the system of differential equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{x}' = \mathbf{Ax}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the unique solution $\mathbf{x} = e^{\mathbf{A}t} \mathbf{x}(0)$, where $\mathbf{x}$ is a vector-valued differentiable function and $e^{\mathbf{A}t}$ denotes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_exponential&quot; rel=&quot;nofollow&quot;&gt;matrix exponential&lt;/a&gt;.  In particular, let $\mathbf{J} = \left[ \begin{array}{cc} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{array} \right]$.  Then the system of differential equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x' = y, y' = -x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions $x(0) = 1, y(0) = 0$ has the unique solution $\left[ \begin{array}{cc} x \\ y \end{array} \right] = e^{\mathbf{J}t} \left[ \begin{array}{cc} 1 \\ 0 \end{array} \right]$.  On the other hand, the above equations tell us that $x'' = -x$ and $y'' = -y$, and we know that the solutions to this differential equation are of the form $a \cos t + b \sin t$ for constants $a, b$.  By matching initial conditions we in fact find that $x = \cos t, y = \sin t$.  Now verify that on vectors multiplying by $\mathbf{J}$ has the same effect as multiplying a complex number by $i$, and you obtain Euler's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof has the following attractive physical interpretation: a particle whose $x$- and $y$-coordinates satisfy $x' = y, y' = -x$ has the property that its velocity is always perpendicular to and has proportional magnitude to its displacement.  But from physics lessons you know that this uniquely describes particles which move in a circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to interpret this proof is as a description of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_map#Lie_theory&quot; rel=&quot;nofollow&quot;&gt;exponential map&lt;/a&gt; from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_algebra&quot; rel=&quot;nofollow&quot;&gt;Lie algebra&lt;/a&gt; $\mathbb{R}$ to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_group&quot; rel=&quot;nofollow&quot;&gt;Lie group&lt;/a&gt; $\text{SO}(2)$.  Euler's formula generalizes to quaternions, and this in turn can be thought of as describing the exponential map from the Lie algebra $\mathbb{R}^3$ (with the cross product) to $\text{SU}(2)$ (which can then be sent to $\text{SO}(3)$).  This is one reason it is convenient to use quaternions to describe 3-d rotations in computer graphics; the exponential map makes it easy to interpolate between two rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  whuber's answer reminded me of the following excellent graphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/0/0e/ExpIPi.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what is happening geometrically in whuber's answer, and is essentially what happens if you apply Euler's method to the system of ODEs I described above.  Each step is a plot of the powers of $1 + \frac{i}{N}$ up to the $N^{th}$ power.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-29T13:41:05.560" LastActivityDate="2010-08-29T13:41:05.560" CommentCount="7" />
  <row Id="3513" PostTypeId="2" ParentId="3510" CreationDate="2010-08-28T04:49:16.420" Score="11" ViewCount="0" Body="&lt;p&gt;Assuming you mean $e^{ix}=\cos x+i\sin x$, one way is to use the MacLaurin series for sine and cosine, which are known to converge for all real x in a first-year calculus context, and the MacLaurin series for $e^z$, trusting that it converges for pure-imaginary z since this result requires complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MacLaurin series:&#xA;$$\begin{align}&#xA;\sin x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots&#xA;\\&#xA;\cos x&amp;amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots&#xA;\\&#xA;e^z&amp;amp;=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute $z=ix$ in the last series:&#xA;$$\begin{align}&#xA;e^{ix}&amp;amp;=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots&#xA;\\&#xA;&amp;amp;=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots&#xA;\\&#xA;&amp;amp;=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)&#xA;\\&#xA;&amp;amp;=\cos x+i\sin x&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-28T04:49:16.420" CommentCount="4" />
  <row Id="3514" PostTypeId="2" ParentId="3510" CreationDate="2010-08-28T04:56:41.357" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ Both $\:\rm e^{ix}\:$ and $\:\rm cos(x) + i \; sin(x) \:$ are solutions of $\;\rm y' = i \; y,\;\; y(0) = 1, \;$  so they are equal by the &lt;strong&gt;uniqueness theorem&lt;/strong&gt;. Alternatively, &lt;strong&gt;bisect&lt;/strong&gt; into even and odd parts the power series for $\:\rm e^{ix} \:,\;$  i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\rm f(x) \ \ &amp;amp;=\ \ \rm\frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\ &#xA;\\&#xA;\Rightarrow\quad\quad \rm e^{ix} \ \ &amp;amp;=\ \ \rm\cos(x) \ +\ i \:\sin(x)  \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARKS 1.&lt;/strong&gt;$\;$ Uniqueness theorems provide powerful tools for proving equalities for functions that satisfy certain nice differential or difference (recurrence) equations. This includes a large majority of functions encountered in theory and practice. Such ideas lie behind algorithms implemented in computer algebra systems, e.g. search the computational literature using the terms &quot;D-finite&quot; and/or &quot;holonomic&quot; functions. &#xA;For a less trivial but still easy example of this technique see my &lt;a href=&quot;http://math.stackexchange.com/questions/3505&quot;&gt;recent post&lt;/a&gt; which proves the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{sinh^{-1}(x)}{\sqrt{x^2+1}} \  = \ \ \sum_{k=0}^\infty\ (-1)^k \frac{(2k)!!}{(2k+1)!!} \: x^{2k+1}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple discrete example see my &lt;a href=&quot;http://google.com/groups?selm=y8zvdpvc099.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; where I remark that $\rm\; 13 \:|\: 3^{n+1} + 4^{2n-1} =: f_n \;$ follows from $\;\rm f_2 = f_1 = 0 \;$ and the (obvious) fact that $\;\rm f_n \;$  satisfies a monic 2nd order linear recurrence. Namely let $\;\rm S\; f_n := f_{n+1} \;$ be the shift operator. Then $\;\rm S - 3 \;$ kills $\rm\; 3^{n+1} \;$ and $\;\rm S - 16 \;$ kills $\;\rm 4^{2n-1} \;$ therefore $\;\rm (S-3)\:(S-16) = S^2 - 19\: S + 48 \;$ kills their sum $\;\rm f_n \:,\;$ i.e. $\;\rm f_{n+2} = 19\; f_{n+1} - 48\; f_n\;$. Therefore $\:\rm mod\:13:\  f_2 = f_1 = 0  \;\Rightarrow\; f_3 = 19\: f_2 - 48\: f_1 = 0 \;\Rightarrow f_4 = 0 \Rightarrow f_5 = 0 \Rightarrow \cdots\Rightarrow\; f_n = 0\: $. Hence $\:0\:$ is the unique solution of the above recurrence that satisfies the intial conditions $\;\rm f_2 = f_1 = 0. \;$ This is simply an obvious special case of the uniqueness theorem for difference equations (recurrences). Once again, by invoking a uniqueness theorem, we have greatly simplified the deduction of an equality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, above, we don't need to know the precise recurrence relation. Rather, we need only know a bound on its degree, so that we know how many initial terms are needed to determine the solution uniquely. In practice, as above, one can often easily derive simple upper bounds on the degree of the recurrence or differential equation - which makes the method even more practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; $\;$ Generalizing the above bisection into even and odd parts, one can employ n'th roots of unity to take arbitrary n-part &lt;strong&gt;multisections&lt;/strong&gt; of power series and generating functions. They often prove handy, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXERCISE&lt;/strong&gt; $\;$ Give elegant proofs of the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm\displaystyle sin(x)/e^{x} \quad\:$ has every $\rm 4k\;$'th term zero  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\rm\displaystyle cos(x)/e^{x} \quad$ has every $\rm 4k+2\;$'th term zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the posts in &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/642159e4f5db4e44&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; for various solutions and more on multisections.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-25T16:46:06.767" LastActivityDate="2010-10-25T16:46:06.767" />
  <row Id="3515" PostTypeId="2" ParentId="3490" CreationDate="2010-08-28T06:11:33.497" Score="0" ViewCount="0" Body="&lt;p&gt;what does it mean to &quot;test some HTML elements&quot; in a web page? what sort of [numerical] data do you expect to collect - and what, if anything, do you want to infer from what you get?&#xA;a particular example may help to clarify what you are driving at.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-08-28T06:11:33.497" />
  <row Id="3516" PostTypeId="2" ParentId="3476" CreationDate="2010-08-28T06:14:48.547" Score="1" ViewCount="0" Body="&lt;p&gt;I'd like a see an answer to this question, and I hope someone here or on MO can provide a good one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to address your question about gauge transformations.  Usually a gauge transformation on a principal G-bundle $P\to X$ is simply a G-equivariant map $P\to P$ which sends fibers to fibers (said another way, a gauge transformation is simply an automorphism of the principal G-bundle P).  A great place to read about the basics of gauge transformations is Atiyah and Bott's paper &quot;The Yang-Mills Equations over Riemann Surfaces&quot; (section 2 in particular).  &lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-08-28T22:18:14.417" LastActivityDate="2010-08-28T22:18:14.417" CommentCount="3" />
  <row Id="3517" PostTypeId="2" ParentId="3405" CreationDate="2010-08-28T06:33:20.990" Score="11" ViewCount="0" Body="&lt;p&gt;Update: the following probabilistic argument I had posted earlier shows only that $p(1) + p(2) + \dots + p(n) = (c + o(1)) \log(n)$ and not, as originally claimed, the convergence $np(n) \to c$.  Until a complete proof is available [edit: George has provided one in another answer] it is not clear whether $np(n)$ converges or has some oscillation, and at the moment there is evidence in both directions.  Log-periodic or other slow oscillation is known to occur in some problems where the recursion accesses many previous terms.  Actually, everything I can calculate about $np(n)$ is consistent with, and in some ways suggestive of, log-periodic fluctuations, with convergence being the special case where the bounds could somehow be strengthened and the fluctuation scale thus squeezed down to zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$p(n) \sim c/n$ is [edit: only in average] equivalent to $p(1) + p(2) + \dots + p(n)$ being asymptotic to $c \log(n)$.  The sum up to $p(n)$ is the expected time the walk spends in the interval [1,n].  For this quantity there is a simple probabilistic argument that explains (and can rigorously demonstrate) the asymptotics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Markov chain is a discrete approximation to a log-normal random walk.  If $X$ is the position of the particle, $\log X$ behaves like a simple random walk with steps $\mu \pm \sigma$ where $\mu = 2 \log 2 - 1 = 1/c$ and $\sigma^2 = (1- \mu)^2/2$.  This is true because &lt;em&gt;the Markov chain is bounded between two easily analyzed random walks with continuous steps&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Let X be the position of the particle and $n$ the number of steps; the walk starts at X=1, n=1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower bound walk $L$: at each step, multiply X by a uniform random number in [1,2] and replace n by (n+1).   $\log L$ increases, on average, by $\int_1^2 log(t) dt = 2 \log(2) - 1$ at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper bound walk $U$: at each step, jump from X to uniform random number in [X+1,2X+1] and replace n by (n+1).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L$ and $U$ have means and variances that are the same within $O(1/n)$, where the $O()$ constants can be made explicit.  Steps of $L$ are i.i.d and steps of $U$ are independent, asymptotically identical-distributed.  Thus, the Central Limit theorem shows that $\log X$ after $n$ steps is approximately a Gaussian with mean $n\mu + O(\log n)$ and variance $n\sigma^2 + O(\log n)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of steps for the particle to escape the interval $[1,t]$ is therefore $({\log t})/\mu$ with fluctuations of size $A \sqrt{\log t}$ having probability that decays rapidly in A  (bounded by $|A|^p \exp(-qA^2)$ for suitable constants).   Thus, the sum p(1) + p(2) + ... p(n) is asymptotically equivalent to $(\log n)/(2\log (2)-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is equivalent to the 2003 argument from the conference.  If the goal is to get a proof from the generating function, it suggests that dividing by $(1-x)$ may be useful for smoothing the p(n)'s.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-12T17:23:20.473" LastActivityDate="2010-11-12T17:23:20.473" CommentCount="9" />
  <row Id="3518" PostTypeId="2" ParentId="3508" CreationDate="2010-08-28T06:33:39.540" Score="4" ViewCount="0" Body="&lt;p&gt;Suppose each entry of your $n \times n$ matrix is a polynomial of degree $d$ in the variable $t$. Appealing to the cofactor expansion, we see that the determinant will be a polynomial in $t$ of degree at most $dn$; lets call it $D(t)$. So you can do the following: evaluate the determinant at $t=1,2, \ldots, dn$. You will thus know $D(1),D(2),\ldots,D(dn)$ and can find out the coefficients of $D$ by interpolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This involves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $dn$ determinant evaluations, each of which takes $O(n^3)$ operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Interpolating a polynomial of degree $dn$. This can be done by solving a $dn \times dn$ system of equations (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_interpolation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Polynomial_interpolation&lt;/a&gt; under &quot;Constructing the interpolating polynomial&quot;). There are &lt;a href=&quot;http://www.springerlink.com/content/q124843l4p368126/&quot; rel=&quot;nofollow&quot;&gt;algorithms&lt;/a&gt; which will do this in $O(dn \log^2 dn)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final number of operations will be $O(d n^4 + dn \log^2 dn)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested in knowing if its possible to do this faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I corrected an error and incorporated a suggestion of J. Mangaldan in the comments. &lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-08-29T02:56:24.937" LastActivityDate="2010-08-29T02:56:24.937" CommentCount="6" />
  <row Id="3519" PostTypeId="1" AcceptedAnswerId="3521" CreationDate="2010-08-28T06:34:51.013" Score="2" ViewCount="117" Body="&lt;p&gt;Liouville's Number is defined as $L = \sum_{n=1}^{\infty}(10^{-n!})$. Does it have other applications than just constructing a transcendental number?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Personally, I would have defined it (as &quot;Steven's Number&quot; :-)) as binary: $S = \sum_{n=1}^{\infty}(2^{-n!})$, since each digit can only be &quot;0&quot; or &quot;1&quot;: the corresponding power of 2 (instead of 10) included or not. Since according to Cantor most number are transcendental one can conjecture that this is also the case for Steven's Number. Can a proof for this be devised based on the proof for Liouville's number?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematician, so please type slowly! :-)&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-11T07:36:33.910" Title="Liouville's number revisited" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="3520" PostTypeId="2" ParentId="3483" CreationDate="2010-08-28T06:43:14.523" Score="0" ViewCount="0" Body="&lt;p&gt;If you are starting trigonometry I'd suggest you to solve Plane trigonometry book by S.L. Loney&lt;/p&gt;&#xA;" OwnerUserId="1548" LastActivityDate="2010-09-11T06:42:13.667" CommunityOwnedDate="2010-08-28T06:43:14.523" />
  <row Id="3521" PostTypeId="2" ParentId="3519" CreationDate="2010-08-28T06:43:22.343" Score="7" ViewCount="0" Body="&lt;p&gt;the number $S$ is also transcendental, for precisely the same reason as $L$: they both have rational approximations that are far too good to hold for an algebraic number. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the specific number $L$ is useful for anything else, I'm not aware of it. It's really just a simple example, one of many which can be dreamed up, of a number which fails to be algebraic since it defies Liouville's Lemma in diophantine approximation. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-28T06:43:22.343" />
  <row Id="3522" PostTypeId="1" CreationDate="2010-08-28T07:04:58.937" Score="2" ViewCount="147" Body="&lt;p&gt;I have a standard linear programming problems I want to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \min_x f^T x \text{ such that } \left\{ \begin{aligned} A\cdot x &amp;amp;\le b, \\ A_{eq}\cdot x &amp;amp;= b_{eq}, \\ lb \le x &amp;amp;\le ub. \end{aligned} \right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F^T$ is a vector that doesn't involve $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the above minimization is that, I don't know what $n$, the number of variable $x$ is — it is also a part of the minimization. &#xA;There are a few constraints governing how the vector $x$ should behave. Additionally, there are constraints linking from $x_i$ to $x_{i+1}$. This means that given $x_i$, we know how to form the constraint for $x_{i+1}$. Also, even though we don't know about $n$, but in my problem I can easily construct the corresponding $f_i$ term for each $x_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraints are complicated in the sense that it is not easy to express the constraints in the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Ax \leq a $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Bx = b $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can do, at best, is to express $a$ and $b$ involving a first order of $x$ ( i.e., no $x^2$ and above). The matrix $A$ and $B$ are known values with no involvement of $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that linear programming can be used to tackle problems such as this. But the two problems I mention above ( don't know what $n$ is, and cannot separate out the matrix/vector easily) stop me from proceeding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other techniques I can use to solve the problem?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-08-28T08:59:25.493" LastActivityDate="2010-08-28T08:59:25.493" Title="How to Minimize A Function Where The Number of Variables is Unknown" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" CommentCount="12" FavoriteCount="1" />
  <row Id="3523" PostTypeId="2" ParentId="3476" CreationDate="2010-08-28T07:08:12.473" Score="4" ViewCount="0" Body="&lt;p&gt;If $X$ is a smooth projective curve over a finite field $k$, with field of rational functions $F$, then the quotient $GL_n(F)\backslash GL_n(\mathbf A_F)/GL_n(\mathcal O_F)$ (where&#xA;$\mathcal O_F$ denotes the subring of integral adeles in $\mathbf A_F$) is in natural bijection with the set of isomorphism classes of rank $n$ vector bundles on $X$, i.e. the &#xA;set $Bun_n(k)$ of $k$-valued points of the moduli stack of rank $n$ bundles on $X$.&#xA;(And the same should be true if we replace $GL_n$ by another reductive group $G$,&#xA;and replace &quot;rank $n$ vector bundles&quot; by principal $G$-bundles; recall that principal $GL_n$-bundles are &quot;the same&quot; as rank $n$ vector bundles; one goes from the latter to the former&#xA;by passing to the associate frame bundle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The passage from a bundle to an a point in the double coset space is made as follows: given the vector bundle, one chooses an affine open subset $U = X \setminus {x_1,\ldots,x_r}$ over which&#xA;the bundle is trivialized, and also trivialiations in a n.h. $U_i$ of each of the points $x_i$.  The gluing data comparing the two trivializations on $U_i \cap U$ gives an element&#xA;$g_{x_i} \in GL_n(F_{x_i})$, where $F_{x_i}$ is the completion of $F$ at $x_i$.&#xA;If $x$ is a point distinct from the $x_i$, set $g_{x} = 1$.  We can then put all&#xA;the $g_x$ together into an adele, and its class in&#xA;the double coset space $GL_n(F)\backslash GL_n(\mathbf A_F)/GL_n(\mathcal O_F)$&#xA;is independent of all choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this map is a bijection, one first observes&#xA;that any element in the double coset space can be represented as&#xA;$(g_x)$ with $g_x = 1$ for almost all $x$, and then one interprets these $g_x$ as formal&#xA;gluing data around the points $x_i$ at which $g_{x_i} \neq 1$ and uses them to &#xA;extend the trivial bundle on $X\setminus {x_1,\ldots,x_r}$ to a bundle on all of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship with gauge transformations is  that the $g_x$ are change of &#xA;basis matrices (gauge tranformations) in formal punctured n.h.s of the points $x$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-28T07:08:12.473" CommentCount="1" />
  <row Id="3524" PostTypeId="2" ParentId="3499" CreationDate="2010-08-28T07:16:46.347" Score="3" ViewCount="0" Body="&lt;p&gt;I believe you're looking for something like the construction mentioned &lt;a href=&quot;http://mathoverflow.net/questions/165/does-the-continuous-locus-of-a-function-have-any-nice-properties/174#174&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-08-28T07:16:46.347" CommentCount="0" />
  <row Id="3525" PostTypeId="1" AcceptedAnswerId="3527" CreationDate="2010-08-28T07:48:50.807" Score="2" ViewCount="141" Body="&lt;p&gt;How to show the following identity holds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}=\pi\cot \pi z-\dfrac{1}{z}\qquad |z|&amp;lt;1$         &lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-28T08:03:38.753" Title="A series expansion of $\cot (\pi z)$" Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="3526" PostTypeId="2" ParentId="3419" CreationDate="2010-08-28T08:00:29.070" Score="-1" ViewCount="0" Body="&lt;p&gt;My guess is that for any sufficiently large n you can find integer numbers a, b and c, such that n = a^b + c and IntegerLength[N] &gt; IntegerLength[a] + IntegerLength[b] + IntegerLength[c]. Recurse for c (and a, b if they're large enough) and you'll find forms with even less digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. $n = 541^{547}  71^{73} + 29^{31} + 19^{23} + 997$ counts 1631 digits, its abbreviated form a total of 21, and probably even less is possible.&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-28T09:49:24.050" LastActivityDate="2010-08-28T09:49:24.050" CommunityOwnedDate="2010-08-28T08:00:29.070" CommentCount="0" />
  <row Id="3527" PostTypeId="2" ParentId="3525" CreationDate="2010-08-28T08:03:38.753" Score="-1" ViewCount="0" Body="&lt;p&gt;Hi, i have found a link which deals with this problem: &lt;a href=&quot;http://people.reed.edu/~jerry/311/cotan.pdf&quot; rel=&quot;nofollow&quot;&gt;people.reed.edu/~jerry/311/cotan.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-28T08:03:38.753" CommentCount="2" />
  <row Id="3528" PostTypeId="1" AcceptedAnswerId="3540" CreationDate="2010-08-28T08:34:23.640" Score="5" ViewCount="280" Body="&lt;p&gt;How do I derive the Beta function using the definition of the beta function as the normalizing constant of the Beta distribution and only common sense random experiments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure this is possible, but can't see how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{a=0}^n {n \choose a} Beta(a+1, n-a+1) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because we can imagine that we are flipping a coin $n$ times.  The $2^n$ unique sequences of flips partition the probability space.  The Beta distribution with parameters $a$ and $n-a$ can be defined as the prior over the coin's bias probability $p$ given the observation of $a$ heads and $n-a$ tails.   Since there are ${n \choose a}$ such sequences for any $n$ and $a$, that explains the scaling factor, and we know that it all sums to unity since the sequences partition the probability space, which has total measure 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't figure out is why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${n \choose a} Beta(a+1, n-a+1) = \frac{1}{n+1} \qquad \forall n \ge 0,\quad a \in {0, \dots, n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we knew that, we could easily see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Beta(a + 1,n - a + 1) = \frac{1}{(n+1){n \choose a}} = \frac{a!(n-a)!}{(n+1)!}$.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-08-28T22:17:55.707" LastActivityDate="2010-08-28T22:17:55.707" Title="Beta function derivation" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" />
  <row Id="3529" PostTypeId="2" ParentId="1747" CreationDate="2010-08-28T08:48:11.187" Score="0" ViewCount="0" Body="&lt;p&gt;This problem is a Bayesian probability problem.  You can never know whether or not the coin is fair because, as you point out yourself, even if the first n flips come up equally head and tails, you would merely have &quot;nothing to indicate [yet] that the coin is unfair&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the probability that the coin's bias is, say between 49% and 51%, by  evaluating the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_{0.51}(h+1,t+1) - I_{0.49}(h+1,t+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;incomplete beta function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the definition of the Beta distribution as the conjugate prior of the Bernoulli distribution and the definition of the incomplete beta function as its cdf.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-08-28T08:48:11.187" />
  <row Id="3530" PostTypeId="2" ParentId="3504" CreationDate="2010-08-28T08:58:40.830" Score="2" ViewCount="0" Body="&lt;p&gt;You problem statement is too vague.  I guess you mean all four polynomials and their pairwise sums are non-degenerate, otherwise we can simply pick any four straight lines whose slopes are nonzero and of different magnitudes.  There should be other restrictions as well, or else {x^2, 2x^2, 3x^2, 4x^2} will also be a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following transformation of the problem may help.  Let P_i(x) = a_ix^2 + b_ix + c_i.  If P_1+P_2 is a polynomial with a single root (actually, do you mean a double root?), it must be equal to some k_1(x-x_1)^2 with k_1 \not = 0 and x_1 \in \mathbb{R}.  Thus a_1+a_2 = k_1 and vice versa for other pairwise sums.  So we get the following matrix equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;[1 1 0 0] [a_1]   [k_1]&#xA;[1 0 1 0] [a_2]   [k_2]&#xA;[1 0 0 1] [a_3] = [k_3]&#xA;[0 1 1 0] [a_4]   [k_4]&#xA;[0 1 0 1]         [k_5]&#xA;[0 0 1 1]         [k_6].&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Call the leftmost matrix M.  Hence the vector u = (k_1, ..., k_6)^T must lie inside the the column space of M.  By inspecting the coefficients b_i and c_i, we see that v = (k_1x_1, ..., k_6x_6)^T and w = (k_1x_1^2, ..., k_6x_6^2)^T also lie inside this column space.  Now, if you can prove that this is impossible, you are done.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastEditorUserId="1551" LastEditorDisplayName="" LastEditDate="2010-08-28T09:05:02.610" LastActivityDate="2010-08-28T09:05:02.610" />
  <row Id="3531" PostTypeId="1" AcceptedAnswerId="3557" CreationDate="2010-08-28T09:48:01.023" Score="6" ViewCount="222" Body="&lt;p&gt;Find the &lt;em&gt;equation&lt;/em&gt; of the ellipse circumscribing a right triangle whose lengths of it's sides are $3,4,5$ and such that its area is the minimum possible one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may chose the origin and orientation of the $x,y$ axes as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: It can be &lt;a href=&quot;http://www.math.purdue.edu/pow/fall2008/pdf/solution8.pdf&quot; rel=&quot;nofollow&quot;&gt;proved&lt;/a&gt; [&lt;em&gt;&lt;a href=&quot;http://www.math.purdue.edu/pow/fall2008/pdf/problem8.pdf&quot; rel=&quot;nofollow&quot;&gt;Problem of the Week, Problem No. 8 (Fall 2008 Series), Department of Mathematics, Purdue University&lt;/a&gt;&lt;/em&gt;] that the area of this ellipse is $8\pi /\sqrt{3}$, without the need of using its equation, but I am also interested in finding it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: picture from &lt;a href=&quot;http://math.stackexchange.com/questions/3531/minimal-ellipse-circumscribing-a-right-triangle/3588#3588&quot;&gt;this&lt;/a&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EWwoy.png&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-08-29T22:47:24.750" LastActivityDate="2010-08-29T22:47:24.750" CommunityOwnedDate="2010-08-29T22:47:24.750" Title="Minimal Ellipse Circumscribing A Right Triangle" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;conic-sections&gt;&lt;plane-curves&gt;&lt;ellipse&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="3532" PostTypeId="1" CreationDate="2010-08-28T10:25:24.843" Score="2" ViewCount="301" Body="&lt;p&gt;Mathematicians, and esp. number theorists, are used to working with big numbers. I have noted on several occasions that lots of people don't have a clear understanding of big numbers as far as the real world is concerned. I recall a &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_thread/thread/5da7f8d21fa57d7d/29155a92aa828571?ie=utf-8&amp;amp;oe=utf-8&quot; rel=&quot;nofollow&quot;&gt;request&lt;/a&gt; for a list of all primes of less than 500 digits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is homeopathic dilutions. I understand they use dilutions like 200C, which is 1 in $10^{400}$. An absurd number in view of the fact that the total number of particles in the universe is estimated (safe margin) to be less than a googol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would you give people insight in big numbers?&lt;/strong&gt; I'm not talking about Skewes' Number or Graham's Number; for most practical purposes $10^{20}$ is equal to infinity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;To whoever voted me down: if you vote this down, please also tell me why. Thanks&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-08-28T12:28:55.447" LastActivityDate="2010-08-29T17:32:50.167" Title="How to explain Real Big Numbers?" Tags="&lt;teaching&gt;&lt;big-numbers&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="3533" PostTypeId="1" CreationDate="2010-08-28T11:01:35.350" Score="1" ViewCount="67" Body="&lt;p&gt;how could i convert a non-uniform random variable distribution to a uniform distribution ? Is there a formula ? thanks .&lt;/p&gt;&#xA;" OwnerUserId="1553" LastActivityDate="2010-08-28T11:51:53.550" Title="Converting non-uniform distribution to a uniform distribution" Tags="&lt;random-variable&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3534" PostTypeId="1" AcceptedAnswerId="3538" CreationDate="2010-08-28T11:08:04.450" Score="3" ViewCount="54" Body="&lt;p&gt;Could you please explain how the math behind the first answer to this stack overflow questions works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/3590308/testing-if-a-line-has-a-point-within-a-triangle&quot;&gt;http://stackoverflow.com/questions/3590308/testing-if-a-line-has-a-point-within-a-triangle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1554" LastActivityDate="2010-08-28T12:01:55.497" Title="Line triangle intersection." Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="3535" PostTypeId="1" AcceptedAnswerId="3536" CreationDate="2010-08-28T11:12:33.893" Score="0" ViewCount="191" Body="&lt;p&gt;Hi all, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the product of two Gaussian functions is also Gaussian function. This is stated in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, but I might need to cite a classical (text)book/paper stating this result. A book containing the proof is welcome. Could you name such a book? &lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2010-08-28T11:37:34.607" CommunityOwnedDate="2010-08-28T11:12:33.893" Title="Texbook with proof that product of two Gaussian functions is also Gaussian" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3536" PostTypeId="2" ParentId="3535" CreationDate="2010-08-28T11:24:08.583" Score="1" ViewCount="0" Body="&lt;p&gt;The product of two normal (Gaussian) random variables is &lt;em&gt;not&lt;/em&gt; normal.  That would be convenient, but it's not true.  Here are a some related theorems that are true:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;sum&lt;/em&gt; of two normal random variables is normal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The product of two &lt;em&gt;log normal&lt;/em&gt; random variables is log normal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ratio of two normal random variables is Cauchy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For more on distribution relationships, see this &lt;a href=&quot;http://www.johndcook.com/distribution_chart.html&quot; rel=&quot;nofollow&quot;&gt;chart&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you multiply two normal PDFs you get the joint PDF of a &lt;em&gt;multivariate&lt;/em&gt; normal, i.e. if f(x) is a normal PDF then g(x, y) defined by f(x)g(y) is a multivariate normal PDF, but in that case you're multiplying &lt;em&gt;densities&lt;/em&gt; and not &lt;em&gt;random variables&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I looked back at your question and I see you said Gaussian &quot;function,&quot; rather than &quot;random variable&quot;, so perhaps you had in mind the PDF result I mentioned above. If so, just write down the definition of multivariate normal and there it is.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-08-28T11:37:34.607" LastActivityDate="2010-08-28T11:37:34.607" CommunityOwnedDate="2010-08-28T11:24:08.583" />
  <row Id="3537" PostTypeId="2" ParentId="3533" CreationDate="2010-08-28T11:51:53.550" Score="1" ViewCount="0" Body="&lt;p&gt;Use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_integral_transform&quot; rel=&quot;nofollow&quot;&gt;probability integral transform&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="116" LastActivityDate="2010-08-28T11:51:53.550" />
  <row Id="3538" PostTypeId="2" ParentId="3534" CreationDate="2010-08-28T12:01:55.497" Score="4" ViewCount="0" Body="&lt;p&gt;Let's take the image from the anwer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kKod9.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of the red line as a &lt;em&gt;two dimensional plane&lt;/em&gt;, where $4x − 3y + 2 = 0$ is it's coordinate equation with some normal vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{n} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is orthogonal to the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by substituting a point into the equation, you get the distance of point and line along $\vec{n}$ (see. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hesse_normal_form&quot; rel=&quot;nofollow&quot;&gt;Hesse's normal form&lt;/a&gt;). While we don't actually care for the distance itself, we do care for it's &lt;em&gt;sign&lt;/em&gt;. Positive sign indicates the point's lying on the side $\vec{n}$ points to, a negative one indicates the opposite side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if all the values have the same sign, all triangle points lie on the same side of the line, i.e. there are no intersections. With different signs, the triangle must hit the line.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-28T12:01:55.497" />
  <row Id="3539" PostTypeId="2" ParentId="3532" CreationDate="2010-08-28T12:23:54.560" Score="2" ViewCount="0" Body="&lt;p&gt;Though I don't quite know that you actually want to hear - &lt;em&gt;what kind of numbers&lt;/em&gt; do you want to give people insight in, &lt;em&gt;whom&lt;/em&gt; and &lt;em&gt;why&lt;/em&gt;? - I'll give a few thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I) &lt;strong&gt;Real cases&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's just understanding of natural sciences - numbers of particles in the universe, number of cells in a body ... Try to first of all break down the number by using smaller parts of the example - e.g. count bacteria in a drop of water and not in a whole lake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II) &lt;strong&gt;Thought experiments (explaining probabilities, complexity etc.)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extremely big numbers arise when you try to visualize probabilities or complexities, especially when exponential growth is involved. What about getting the jackpot ten times successively or trying to solve a TSP for 100 cities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you know people aren't comfortable with that big numbers, decide:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it really important to know the number? Maybe, extremely long or extremely improbable is just the important fact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you find an easier to grasp example (special units)? Longer than the universe is old is better than &lt;em&gt;insert giant amount of milliseconds&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you describe the growth differently? &lt;em&gt;If your problem with 999 cities can be solved in a certain amount of time and you take one additional city, you'll need &lt;strong&gt;1000&lt;/strong&gt; times longer&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;III) &lt;strong&gt;Data&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially in the context of CS / cryptography, numbers can often most accurately be explained as &lt;em&gt;some data you can calculate with&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. &lt;em&gt;RSA&lt;/em&gt; (as in your link) is of course a mathematical, number-based algorithm, but in fact, you're encrypting &lt;strong&gt;data&lt;/strong&gt;, so why not say &lt;em&gt;a 500 char key&lt;/em&gt; instead of explaining the giant number involved there. &lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-28T12:23:54.560" CommentCount="1" />
  <row Id="3540" PostTypeId="2" ParentId="3528" CreationDate="2010-08-28T12:34:02.077" Score="8" ViewCount="0" Body="&lt;p&gt;For non-negative integers $a, b$ and $t \in [0, 1]$, the expression $t^a (1 - t)^b$ describes the probability of randomly selecting $a+b$ real numbers in $[0, 1]$ such that the first $a$ are in $[0, t]$ and the last $b$ are in $[t, 1]$.  The integral $\int_0^{1} t^a (1 - t)^b$ then describes the probability of randomly selecting $a+b+1$ real numbers such that the first number is $t$, the next $a$ numbers are in $[0, t]$, and the next $b$ numbers are in $[t, 1]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $ {a+b \choose b} \int_0^1 t^a (1 - t)^b$ describes the probability of randomly selecting $a+b+1$ real numbers such that the first number is $t$, some $a$ of the remaining numbers are in $[0, t]$, and some $b$ of the remaining numbers are in $[t, 1]$.  But this is the same as the probability that the first number happens to be $(a+1)^{st}$ in order, and this is just $\frac{1}{a+b+1}$.  Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 t^a (1 - t)^b = \frac{a! b!}{(a+b+1)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  I learned this proof through an exercise in a Putnam training seminar; the multidimensional generalization also works.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-28T14:10:49.590" LastActivityDate="2010-08-28T14:10:49.590" CommentCount="1" />
  <row Id="3541" PostTypeId="1" AcceptedAnswerId="3546" CreationDate="2010-08-28T12:41:18.790" Score="3" ViewCount="131" Body="&lt;p&gt;Rotations in 3 dimensions are not commutative; however they are in the plane. In classical mechanics, are we allowed to say that angular momentum is a vector because particles only rotate along a single axis? Or do we make some kind of argument that you can assign angular velocity to an object because you can approximate it locally as being along a single rotation axis and so its &quot;locally commutative&quot; and hence appropriate to call it a vector? If that is the case what happens if we have a particle rotating in a way that isn't differentiable? Is there some physical reason that particles can't rotate in that sort of way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really confused by this and haven't taken physics since high school, any help would be appreciated. :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-28T13:21:50.010" LastActivityDate="2010-08-28T14:02:21.077" Title="How can angular velocity or angular momentum be a vector?" Tags="&lt;linear-algebra&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="3542" PostTypeId="2" ParentId="3506" CreationDate="2010-08-28T12:58:51.107" Score="1" ViewCount="0" Body="&lt;p&gt;The important component of the question here seems to be: can we define a natural binary operation on the set P of partitions of $\{1,2,\ldots,n\}$ to form a group G?  &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Cayley's Theorem&lt;/a&gt; tells us that if G exists, it is isomorphic to a subgroup of the symmetric group on G.  So G will be isomorphic to a group of permutation matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, finding a &lt;em&gt;natural&lt;/em&gt; binary operation on P is going to be tricky.  Of course, P is just a set, and it would be possible to construct some highly-contrived binary operation on P, but typically it would not preserve the structure of the partitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an off-the-top-of-my-head example of why I think it should be tricky: |P| is given by the &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A000110&quot; rel=&quot;nofollow&quot;&gt;Bell Numbers&lt;/a&gt;, which can be a prime number, whence G must be a cyclic group and each non-identity element of G must somehow generate G.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-08-28T12:58:51.107" />
  <row Id="3543" PostTypeId="1" AcceptedAnswerId="3544" CreationDate="2010-08-28T13:41:26.907" Score="0" ViewCount="220" Body="&lt;p&gt;One learns evaluating indefinite integrals of the form $$ \int \frac{f(x)}{p(x)\sqrt{q(x)}} \ dx $$ where $p(x)$ and $q(x)$ are linear polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to evaluate these integrals if $p(x)$ and $q(x)$ are quadratic or cubic in nature. Is there any method for evaluating them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of such integrals are: $\displaystyle \int \frac{1}{(x^{2}+a^{2})\sqrt{x^{2}+b^{2}}} \ dx$ or $\displaystyle \int\frac{1}{(x^{3}+a^{3})\sqrt{x^{3}+b^{3}}} \ dx$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-31T17:10:32.993" Title="Evaluating Indefinite Integrals of the form $\frac{f(x)}{p(x) \sqrt{q(x)}}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="3544" PostTypeId="2" ParentId="3543" CreationDate="2010-08-28T13:46:50.493" Score="5" ViewCount="0" Body="&lt;p&gt;If $q$ is quadratic, then these can be done by trigonometric/hyperbolic&#xA;substiuitions. If $q$ is cubic then they are are elliptic integrals,&#xA;not in general expressible in elementary terms, but expressible in terms&#xA;of elementary functions and the standard examples of elliptic integrals.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-28T13:46:50.493" CommentCount="1" />
  <row Id="3545" PostTypeId="2" ParentId="3543" CreationDate="2010-08-28T13:51:37.997" Score="2" ViewCount="0" Body="&lt;p&gt;For the quadratic, first put $x = a \tan \theta$ then put $t = \sin \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(First write $p(t)$ in the form $x^2 + a^2$ where $x = t+r$, $a$ could possibly be complex)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't know about the cubic.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T13:56:38.577" LastActivityDate="2010-08-28T13:56:38.577" CommentCount="2" />
  <row Id="3546" PostTypeId="2" ParentId="3541" CreationDate="2010-08-28T14:02:21.077" Score="6" ViewCount="0" Body="&lt;p&gt;There are a couple of things at play here, which coincidentally(?) mirror the difference between mathematicians who say &quot;vector&quot; to mean anything can be added commutatively and multiplied by a scalar, and physicists who say &quot;vector&quot; to mean a 3- (or 4-)dimensional quantity that transforms properly under change of coordinates.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;While arbitrary rotations in 3 dimensions do not commute, &lt;em&gt;infinitesimal&lt;/em&gt; rotations do. (In fact any &quot;infinitesimal&quot; transformations commute, as you can see by multiplying $I + \epsilon A$ and $I + \epsilon B$ and ignoring second-order terms in $\epsilon$.) Since angular velocity can be thought of as &quot;infinitesimal rotation per infinitesimal time&quot;, it ends up being a vector (in the mathematical sense) even though rotation itself isn't. Similarly, angular momentum is the derivative of kinetic energy with respect to angular velocity, so it is a gradient, which is a (dual) vector (to angular velocity).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's an interesting mapping between these infinitesimal rotation matrices and vectors (in the physical sense) which only works in 3 dimensions. If you think of an infinitesimal rotation as $I + \epsilon A$, &lt;a href=&quot;http://mathworld.wolfram.com/InfinitesimalRotation.html&quot; rel=&quot;nofollow&quot;&gt;one can show that $A$ must be antisymmetric&lt;/a&gt;. This means its diagonal entries are 0, leaving only 3 degrees of freedom in the off-diagonal entries. Such a matrix can be associated with a vector using the cross product:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}0 &amp;amp; -\omega_z &amp;amp; \omega_y \\ \omega_z &amp;amp; 0 &amp;amp; -\omega_x \\ -\omega_y &amp;amp; \omega_x &amp;amp; 0\end{bmatrix} \mathbf{r} = \mathbf{\omega} \times \mathbf{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So angular velocity isn't &quot;naturally&quot; a vector in physical space, but rather lives in a different 3-dimensional vector space. That's why once you interpret it as a physical vector, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudovector&quot; rel=&quot;nofollow&quot;&gt;it turns out to transform slightly differently&lt;/a&gt;, as J. Mangaldan has pointed out in the comments.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-28T14:02:21.077" CommentCount="4" />
  <row Id="3547" PostTypeId="1" AcceptedAnswerId="3548" CreationDate="2010-08-28T14:16:04.017" Score="2" ViewCount="222" Body="&lt;p&gt;Wikipedia has separate pages for symmetric group and permutation group, but I don't understand what the difference between them is. A symmetric group on a set is the set of all bijections from the set to itself with composition of functions as the group action. Permutation group on a set is the set of all permutations of elements on the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't these two things the same thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one of the discussion pages, someone suggested that permutation groups don't have to include all permutations: they just have to be collections of permutations on the set, closed under composition etc. But this seems weird. First, that's not how I was taught it, and second (thanks to Cayley's theorem) it looks like it's redundant: all groups are &quot;permutation groups&quot; on this reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some subtle difference I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-28T19:47:14.960" Title="Permutation groups and symmetric groups" Tags="&lt;terminology&gt;&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="3548" PostTypeId="2" ParentId="3547" CreationDate="2010-08-28T14:27:27.347" Score="6" ViewCount="0" Body="&lt;p&gt;First, on the permutation group page, there's this line &quot;the term permutation group is usually restricted to mean a subgroup of the symmetric group.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, Cayley's theorem doesn't really make the terminology &quot;permutation group&quot; redundant. When you talk about a permutation group, I think you are implicitly giving an action of the group on a set of some objects. This is an extra data other than the group structure.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-08-28T14:27:27.347" CommentCount="1" />
  <row Id="3549" PostTypeId="2" ParentId="3101" CreationDate="2010-08-28T14:38:22.457" Score="2" ViewCount="0" Body="&lt;p&gt;As per AndrejaKo's comment, I am posting my comment as an answer. Note that there are very interesting and useful answers to this question motivating the formula for the derivative of a product: my answer is simply a technical explanation of a typo in the argument provided in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should add and subtract the quantity $f(x)g(x+h)$ from the left hand side of your equation (and the equality you wrote is indeed incorrect). Once you collect terms appropriately, everything should be clear!&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-28T14:38:22.457" />
  <row Id="3550" PostTypeId="1" AcceptedAnswerId="3571" CreationDate="2010-08-28T14:51:53.123" Score="6" ViewCount="480" Body="&lt;p&gt;In my attempts to understand just how computer algebra systems &quot;do things&quot;, I tried to dig around a bit on Gröbner bases, which are described almost everywhere as &quot;a generalization of the Euclidean algorithm and Gaussian elimination&quot;. I've tried to look for examples of Gröbner bases in action, but have been unable to find any (that can be easily understood).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go ahead and just ask for an explanation with examples from people, but I'll go one step further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General plane conics can be represented by the Cartesian equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ax^2+2bxy+cy^2+dx+fy+g=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common problem in dealing with conics is finding out if two conics intersect, and if so, where the intersection point(s) are. Usually one can do all this by eliminating variables accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I ask: how would, say, Buchberger's method, proceed on determining if two given conics intersect, and then find where they intersect?&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-13T12:24:07.137" Title="Using Gröbner bases for solving polynomial equations." Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="5" />
  <row Id="3551" PostTypeId="2" ParentId="3532" CreationDate="2010-08-28T15:06:38.940" Score="2" ViewCount="0" Body="&lt;p&gt;Very few (if any) mathematicians have significant insight regarding huge natural numbers (cf. various &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrafinitism&quot; rel=&quot;nofollow&quot;&gt;ultrafinitism&lt;/a&gt; arguments).  Perhaps the only exceptions are logicians who work with esoteric ordinal notations. This is one of the few ways one can gain any insight into arbitrarily large numbers - using various complicated inductions to show that some property holds for all naturals - thus lifting our intuition up from small naturals to arbitrarily large naturals. For example, see the &lt;a href=&quot;http://google.com/groups?selm=WGD.95Dec11023450@martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;Goodstein sequence&lt;/a&gt; (or, more graphically: the Hercules vs. Hydra game) which encodes the ordinals below $\epsilon_0 = \omega^{\omega^{\omega^{\cdot^{\cdot^\cdot}}}} \;$ into huge natural numbers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-29T17:32:50.167" LastActivityDate="2010-08-29T17:32:50.167" CommentCount="1" />
  <row Id="3552" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T16:09:52.510" Score="5" ViewCount="0" Body="&lt;p&gt;You &lt;em&gt;have&lt;/em&gt; ro read through the book[David A. Cox, John B. Little and Don O'Shea, &lt;em&gt;Ideals, Varieties, Algorithms&lt;/em&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if you start with a system of polynomial equations and compute a Groebner basis for the ideal they generate, you get a &quot;maximally triangular&quot; system of equations which is equivalent to the original one---that is why Groebner bases generalize Gaussian elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me do a simple example using &lt;a href=&quot;http://www.math.uiuc.edu/Macaulay2/&quot; rel=&quot;nofollow&quot;&gt;Macaulay 2&lt;/a&gt;. Consider the ring $\mathbb Q[x,y,z]$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i1 : R = QQ[x, y, z, MonomialOrder =&amp;gt; Lex]&#xA;&#xA;o1 = R&#xA;&#xA;o1 : PolynomialRing&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the Fermat quintic surface $x^5+y^5+z^5=1$, whose ideal is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i2 : surface = ideal (x^5 + y^5 + z^5 - 1)&#xA;&#xA;             5    5    5&#xA;o2 = ideal(x  + y  + z  - 1)&#xA;&#xA;o2 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the curve $x^2=y^2$, $y^3+1=z^3$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i3 : curve = ideal (x^2 - y^2, z^3 - y^3 - 1)&#xA;&#xA;             2    2     3    3&#xA;o3 = ideal (x  - y , - y  + z  - 1)&#xA;&#xA;o3 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The ideal of the intersection of the surface and the curve is the sum of the ideals of the intersecands:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i4 : intersection = curve + surface&#xA;&#xA;              2    2     3    3       5    5    5&#xA;o4 = ideal (x  - y , - y  + z  - 1, x  + y  + z  - 1)&#xA;&#xA;o4 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The number of points on the intersection, counting multiplicities, is the degree of the ideal:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i5 : degree intersection&#xA;&#xA;o5 = 30&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;if we now compute the degree of the radical of the ideal, we get a lower number, so not all the points are simple:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i6 : degree radical intersection&#xA;&#xA;o6 = 25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now look at the lexicographic Groebner basis of the ideal of the intersection:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i7 : ideal gens gb intersection&#xA;&#xA;              17      16      15      14      13      12       11       10      9      8       7       6      5      4      3  &#xA;o7 = ideal (9z   + 27z   + 54z   + 50z   + 15z   - 63z   - 104z   - 108z   - 35z  + 35z  + 108z  + 104z  + 63z  - 15z  - 50z  -&#xA;     ----------------------------------------------------------------------------------------------------------------------------&#xA;        2                 5            16      15      14      13      12       11      10      9       8      7       6       5&#xA;     54z  - 27z - 9, 20y*z  - 20y + 27z   + 18z   + 45z   - 75z   - 35z   - 164z   + 79z   - 10z  + 230z  - 10z  + 119z  - 164z &#xA;     ----------------------------------------------------------------------------------------------------------------------------&#xA;          4       3      2              3    3                                       16        15        14        13        12  &#xA;     - 35z  - 155z  + 45z  + 18z + 67, y  - z  + 1, 50x*z - 50x + 50y*z - 50y + 1242z   + 2718z   + 4770z   + 2220z   - 1235z   -&#xA;     ----------------------------------------------------------------------------------------------------------------------------&#xA;          11        10        9        8        7        6        5        4        3        2                 2    2&#xA;     7979z   - 7481z   - 6010z  + 1810z  + 5240z  + 9394z  + 5346z  + 1240z  - 4030z  - 4005z  - 2657z - 583, x  - y )&#xA;&#xA;o7 : Ideal of R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first generator depends only on $z$. The second one and the third, only on $z$ and $y$, and the fourth depends (linearly) on $x$ too. This is a &quot;triangular&quot; system of equations, from which you can compute the 30 points of intersection, assuming you get to solve the first equation for $z$, which is of degree 17...&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-13T12:24:07.137" LastActivityDate="2010-12-13T12:24:07.137" CommentCount="6" />
  <row Id="3554" PostTypeId="2" ParentId="3497" CreationDate="2010-08-28T16:16:55.583" Score="1" ViewCount="0" Body="&lt;p&gt;To expand on Sam's awesome comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write $K$ as $F(\alpha)$, where the minimal polynomial of $\alpha$ is $f(x)$.  Then we have a map from $L[x]$ to $K\otimes_F L$ given by&#xA;$$ a_0+a_1x+\cdots+a_nx^n\mapsto 1\otimes a_0 + \alpha\otimes a_1+\cdots + \alpha^n\otimes a_n.$$&#xA;This induces an isomorphism between $L[x]/fL[x]$ and $K\otimes_F L$. Composing with the map I've given above, we then want to show the map&#xA;$$ a(x)\mapsto (a(\alpha^{\gamma_1}),\ldots,a(\alpha^{\gamma_n}))$$&#xA;from $L[x]$ to $\prod_{i=1}^n C$, is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set &#xA;$$ f_i=\prod_{n\in N} (x-\alpha^{\gamma_in});$$&#xA;this is a polynomial in $L[x]$; now let&#xA;$$ g_i=\prod_{j\neq i} f_j; $$&#xA;then $g_i$ maps to $(0,\ldots,g_i(\alpha^{\gamma_i}),0,\ldots)$.&#xA;Since $g_i(\alpha^{\gamma_i})$ is a non-zero element of $K$, we can write $g_i(\alpha^{\gamma_i})^{-1}=h_i(\alpha^{\gamma_i})$, and so $g_ih_i$ maps to $(0,\ldots,1,\ldots)$, where the $1$ is in the $i$th place. Since $C=L(\alpha)$, this map is surjective.&lt;/p&gt;&#xA;" OwnerUserId="1544" LastActivityDate="2010-08-28T16:16:55.583" />
  <row Id="3556" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T16:21:41.727" Score="1" ViewCount="0" Body="&lt;p&gt;You might find Lecture 3 of James Carlson's CIMAT Lectures of value as well as other material on this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.utah.edu/~carlson/cimat/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.utah.edu/~carlson/cimat/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-28T16:21:41.727" CommentCount="1" />
  <row Id="3557" PostTypeId="2" ParentId="3531" CreationDate="2010-08-28T16:36:42.380" Score="4" ViewCount="0" Body="&lt;p&gt;If you look at the proof that you linked, the proof is using an Affine transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can also see that the centroid of the triangles are mapped to each other in the affine transformation, and so do the centers of the ellipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;code&gt;the ellipse you need has the same center as your triangle!&lt;/code&gt; This property uniquely defines the ellipse with the maximum area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the points $A = (-1,-4/3), B = (2, -4/3)$ and $C= (-1, 8/3)$. This is a 3-4-5 triangle whose center is origin which was gotten by starting with $(0,0), (3,0)$ and $(0,4)$ and translating so that the centroid is the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation of an ellipse whose center is the origin is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Px^2 + Qxy + Ry^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $P + 4Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(2) $4P - 8Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(3) $P -8Q/3 + 64R/9 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving these (see footnote) gives us the equation of the ellipse as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{2}/3 + xy/4 + 3y^{2}/16 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to verify this, the area of $Px^2 + Qxy + Ry^2 = 1$ is given by $\displaystyle \frac{2\pi}{\sqrt{4PR - Q^2}}$ which comes out to $\displaystyle \frac{8\pi}{\sqrt 3}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can ignore the below if you like. This is just manually solving the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $P + 4Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(2) $4P - 8Q/3 + 16R/9 = 1$ &lt;br/&gt;&#xA;(3) $P -8Q/3 + 64R/9 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting (1) and (2) gives $3P = 4Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting (2) and (3) gives $3P = 48R/9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $Q = 3P/4$ and $R = 9P/16$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus using (3) we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P - (8/3)*(3P/4) + 64/9 * (9P/16) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&#xA;$P - 2P + 4P = 1$ i.e $P = 1/3$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-28T16:57:14.703" LastActivityDate="2010-08-28T16:57:14.703" CommentCount="7" />
  <row Id="3558" PostTypeId="2" ParentId="3444" CreationDate="2010-08-28T16:42:38.230" Score="3" ViewCount="0" Body="&lt;p&gt;The geometric approach works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s compute the volume of the $2n$ dimensional ball, $D^{2n}$, in two ways.  One way is extremely clever but has been known for centuries and provides interesting insights: it’s based on Liouville’s trick.  Specifically, we will compute two integrals in polar coordinates, one of which is the volume of the ball and the other of which reduces to a product of one-dimensional integrals.  Both integrands will depend (at most) on the radial coordinate $r$, which lets us separate out  the surface area of the boundary of the ball as a common factor.  Write this surface area as $S_{2n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There’s essentially just one way to do this trick: integrate $exp(-r^2)$.  Its integral over $\mathbb{R}^{2n}$ equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{2n-1} \int_0^\infty {exp\left(- r^2 \right) r^{2n-1} dr}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, because $r^2 = x_1^2 + x_2^2 + \ldots + x_{2n}^2$, the integrand (in &lt;em&gt;Cartesian&lt;/em&gt; coordinates $\left( x_1, x_2, \ldots, x_{2n} \right)$) factors as $exp\left(-r^2 \right) = exp\left(-x_1^2 \right) \cdots exp\left(-x_{2n}^2 \right)$, each of which must be integrated from $-\infty$ to $+\infty$.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{2n-1} \int_0^\infty {exp \left(- r^2 \right) r^{2n-1} dr} = \left( \int_{- \infty}^ \infty {exp \left( -x^2 \right) dx} \right) ^{2n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will call the left hand integral $\tfrac{1}{2} \Gamma \left(n \right)$, because that is what it is (as a simple change of variables shows).  In the same notation,  $\Gamma \left(1/2 \right) =  \int_{-\infty}^\infty {exp\left(-x^2 \right) dx}$.  Algebraic re-arrangement of the foregoing yields the volume of $D^{2n}$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|D^{2n} | = S_{2n - 1} \int_0^1 {r^{2n - 1} dr}  = \frac{{S_{2n - 1} }}&#xA;{{2n}} = \frac {\Gamma \left(1/2 \right)^{2n}} { n \Gamma \left(n \right) }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was the first method: the result is a familiar one, but has been left expressed in a way that better reveals its origins in polar and Cartesian integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next way to compute the ball's volume is, I believe, new.  It is inspired by Cavalieri’s Principle: the idea that you can shift slices of a solid around without changing the volume of the solid.  The generalization is to move two-dimensional slices around &lt;em&gt;and to change their shapes while you do so,&lt;/em&gt; but in a way that does not change their areas.  It follows that the new solid has the same (hypervolume) as the original, although it might have a completely different shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will compute the volume of a region $Q_n$ in $\mathbb{R}^{2n}$.  It is conveniently described by identifying $\mathbb{R}^{2n}$ with  $\mathbb{C}^{n}$, using coordinates $z_i = \left( x_{2i - 1}, x_{2i} \right)$, in terms of which&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_n  = { \mathbf{z} \in \mathbb{C}^n :0 \leq \left| {z_1 } \right| \leq \left| {z_2 } \right| \leq  \cdots  \leq \left| {z_n } \right| \leq 1 }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these were &lt;em&gt;real&lt;/em&gt; variables, we could make the volume-preserving transformation $w_1 = z_1, w_2 = z_2 – z_1, \ldots , w_i = z_i – z_{i-1}, \ldots, w_n = z_n – z_{n-1}$, with the sole restriction that the sum of the $w_i$ (all of which are nonnegative) not exceed 1.  Because they are &lt;em&gt;complex&lt;/em&gt; variables, though, we have to consider the area of an annulus bounded by $z_{i-1}$ and $z_i$: it is proportional to $z_i^2 – z_{i – 1}^2$.  The circle of the same area has radius $w_i$ for which $w_i^2 = z_i^2 – z_{i – 1}^2$.  Therefore, &lt;em&gt;if we define new variables&lt;/em&gt; $w_i$ &lt;em&gt;according to this formula&lt;/em&gt;, we obtain a new region- - one of substantially different shape- - having the same volume.  This region is defined by $\left| {w_1 }^2 \right| + \cdots + \left| {w_n }^2 \right| \le 1$: that is, it’s our old friend $D^{2n}$.  Therefore, &lt;strong&gt;the volume of&lt;/strong&gt; $Q_n$ &lt;strong&gt;equals the volume of&lt;/strong&gt; $D^{2n}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the punch line: $Q_n$ is a fundamental domain for the action of $S[n]$, the symmetric group, on the product of $n$ disks $T^{2n} = \left( D^2 \right) ^n$; $S[n]$ acts by permuting the Complex coordinates $z_1, \ldots, z_n$.  The volume of $T^{2n}$ equals $|D^2|^n = \pi ^n$.  Writing $|S[n]|$ for the number of permutations and equating our two completely different calculations of the volume of the $2n$ ball gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi ^ n / |S[n]| = \frac {\Gamma \left(1/2 \right)^{2n}} { n \Gamma \left(n \right) }$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S[n]| = \frac{{\pi ^n n\Gamma \left( n \right)}}{{\Gamma \left( {1/2} \right)^{2n} }}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplifies: the volume formula for $n = 2$ must give the area of the unit circle, equal to $\pi$, whence $\Gamma \left( 1/2 \right)^2 = \pi$.  Finally, then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|S[n]| = n\Gamma \left( n \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will finish by remarking that Liouville's method is a perfectly natural thing to encounter when working with the multivariate Normal distribution, so it's not really an isolated trick, but is rather a pretty basic result expressing a defining property of Normal (Gaussian) variates.  There are, of course, many other ways to compute the volume of $D^{2n}$, but this one gives us the Gamma function directly.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-08-28T16:50:07.527" LastActivityDate="2010-08-28T16:50:07.527" CommentCount="3" />
  <row Id="3559" PostTypeId="2" ParentId="3531" CreationDate="2010-08-28T16:49:25.223" Score="3" ViewCount="0" Body="&lt;p&gt;Given the proof to which you linked, as Rahul Narain said in a comment, there is an affine transformation that maps the unit circle and an inscribed equilateral triangle to the desired ellipse for a particular 3-4-5 right triangle.  Represent this affine transformation by&#xA;$$\begin{pmatrix}x'\\y'\\1\end{pmatrix}=\begin{pmatrix}a_{1,1} &amp;amp; a_{1,2} &amp;amp; b_1\\a_{2,1} &amp;amp; a_{2,2} &amp;amp; b_2\\0 &amp;amp; 0 &amp;amp; 1\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}.$$&#xA;Let the 3-4-5 right triangle have vertices (0,0), (0,3), and (4,0), and let the equilateral triangle have vertices (1,0), $(-\frac{1}{2},\frac{\sqrt{3}}{2})$, and $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$.  Mapping these vertices in the order listed and solving for the parameters of the transformation gives the matrix&#xA;$$\begin{pmatrix}-\frac{4}{3} &amp;amp; -\frac{4}{\sqrt{3}} &amp;amp; \frac{4}{3}\\-1 &amp;amp; \sqrt{3} &amp;amp; 1\\0 &amp;amp; 0 &amp;amp; 1\end{pmatrix}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit circle has parametric representation $x=\cos t$, $y=\sin t$.  Applying the affine transformation to this unit circle gives the parametric representation $x=\frac{4}{3}(1-\cos t-\sqrt{3}\sin t)$, $y=1-\cos t+\sqrt{3}\sin t$ for the desired ellipse.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: Looking at &lt;a href=&quot;http://math.stackexchange.com/questions/3531/minimal-ellipse-circumscribing-a-right-triangle/3557#3557&quot;&gt;Moron's answer&lt;/a&gt;, his choice of x and y coordinates are swapped relative to mine, and as he said, his right triangle is translated to have its centroid at the origin, but accounting for these differences, my parametric equations satisfy his equation.  That is, based on his equation, an equation in x and y alone for the ellipse with parametric equations as I specified is $9x^2+12xy+16y^2-36x-48y=0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: I don't know too much of the detailed mechanics of the different representations of affine transformations.  The form that I used above is equivalent to&#xA;$$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}a_{1,1} &amp;amp; a_{1,2}\\a_{2,1} &amp;amp; a_{2,2}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}b_1\\b_2\end{pmatrix}$$&#xA;and the 1s and 0s in the third row in the earlier version are just the glue that allows an affine transformation to be represented with a single matrix multiplication.  My solution could have just as easily used this representation of the affine transformation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-28T23:04:28.197" LastActivityDate="2010-08-28T23:04:28.197" CommentCount="7" />
  <row Id="3560" PostTypeId="2" ParentId="3250" CreationDate="2010-08-28T17:10:05.323" Score="0" ViewCount="0" Body="&lt;p&gt;Perhaps these thoughts will shed some light on the discussion here. If one writes down a collection of axioms for Euclidean geometry, for example, Hilbert's axioms, the Pythagorean Theorem and discussion of a distance function do not explicitly appear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.beva.org/math323/asgn7/dec12b.htm#hilb&quot; rel=&quot;nofollow&quot;&gt;http://www.beva.org/math323/asgn7/dec12b.htm#hilb&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A model of Euclidean geometry consists of taking points as (x,y), ordered pairs of real numbers and lines as linear equations ax + by + c = 0, (a, b, c real numbers) using the Euclidean distance function as the way to find the distance between pairs of points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if one takes points as (x, y) (ordered pairs of real numbers) and lines as linear equations ax + by + c = 0 but uses |a-b| + |c- d| as the distance between the two points (a,b) and (c,d) and measures angles as one does in the Euclidean plane one gets a geometry, often called the Taxicab Plane, which obeys all of the axioms of the Euclidean Plane except for the axiom dealing with the congruence of triangles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details can be found in Eugene Krause's book Taxicab Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pyhtagorean Theorem does not hold here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For interesting subtleties, and lots about the taxicab plane, see Dawson's article available at this site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://taxicabgeometry.net/research.html&quot; rel=&quot;nofollow&quot;&gt;http://taxicabgeometry.net/research.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-28T17:10:05.323" />
  <row Id="3561" PostTypeId="1" CreationDate="2010-08-28T17:55:36.447" Score="4" ViewCount="232" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to be taking a first course in abstract algebra this fall and we are using Jacobson's Basic Algebra I for the course text. It looks like a good textbook, but it is lacking a lot of motivation (I am new to higher math), so I was wondering if there are other books to supplement my studies. Particularly historical or geometric motivations; Something to give me an idea about why we are learning about ideals, modules, etc. so I can &quot;picture&quot; what we are learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any input!&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-03T05:26:44.333" LastActivityDate="2010-09-03T05:26:44.333" CommunityOwnedDate="2010-08-28T17:55:53.997" Title="Supplement for Jacobson's Basic Algebra I" Tags="&lt;soft-question&gt;&lt;abstract-algebra&gt;&lt;math-history&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3562" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T18:19:23.343" Score="2" ViewCount="0" Body="&lt;p&gt;The example that you propose is a bit too general to serve as an introductory pedagogical example (compare e.g. the famous &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1088244&quot; rel=&quot;nofollow&quot;&gt;Kahan SIGSAM problem 9&lt;/a&gt; of determining conditions for an ellipse to lie inside the unit circle). Instead, begin with one of the earliest historical examples: Gauss's proof that every symmetric polynomial can be written uniquely as a polynomial in the elementary symmetric polynomials. This is one of the first and the simplest applications of rewriting reduction via a lexicographic order. For a nice presentation see Chapter 7 of Cox, Little, O'Shea: Ideals, Varieties and Algorithms. They also present generalizations to  the ring of invariants of a finite matrix group $\rm G \subset  GL(n,k)$. You can find online copies of said book via various ebook databases, e.g. &lt;a href=&quot;http://gigapedia.com/&quot; rel=&quot;nofollow&quot;&gt;gigapedia&lt;/a&gt; - which has links to over 20000 mathematics books.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-28T18:19:23.343" CommentCount="1" />
  <row Id="3563" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T18:21:54.240" Score="1" ViewCount="0" Body="&lt;p&gt;This is a simplistic version of the Grobner basis computation and complements Mariano's answer.  The idea behind the Buchberger algorithm is to build a basis so that you are to be able to do &quot;unique division&quot; between nonlinear polynomails.  This is done by computing (what is called the) S-polynomial between any 2 given polynomials until there are no more produced.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Decide some ordering (grlex, lex, revlex)&#xA;Initialize I = &amp;lt;f1, f2&amp;gt; &#xA;where we use the polynomials you gave above.&#xA;f_1=ax^2+2bxy+cy^2+dx+fy+g&#xA;f_2=ax^2+2bxy+cy^2+dx+fy+g&#xA;Set n=2&#xA;&#xA;Repeat the following in a loop&#xA;    Iteratively choose any two polynomials ( f_i, f_j ) from set I &#xA;    n = n + 1&#xA;    Compute f[n] = S-polynomial( f_i, f_j )&#xA;Until all S-polynomials return 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At the end of the computation, you will have a set I = {f1,f2, ..., fk} such that one of the polynomials will be in only one term (say 'x').  Solve that equation to get the one coordinate of the intersection, and then work upwards through the rest of the polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general question is related to Elimination theory, and AFAIK there are three basic methods ( I can't speak of their complexity )&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Grobner basis, as seen above&lt;/li&gt;&#xA;&lt;li&gt;Use Resultants  (example below)&lt;/li&gt;&#xA;&lt;li&gt;Wu's method of characteristic sets, which I can't say much about.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you wanted to use resultants to find the intersection of say &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E={f1(x, y,z), f2(x, y, z), f3(x, y, z)},&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;you would first compute &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E1 = {g1 = Resultant_in_x (f1, f2), g2 = Resultant_in_x(f2, f3)} &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E2 = { h1 = Resultant_in_y(g1, g2} }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The last polynomial h1 would be in one variable and amenable to solution.&lt;/p&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2010-08-28T18:21:54.240" CommentCount="4" />
  <row Id="3564" PostTypeId="2" ParentId="3434" CreationDate="2010-08-28T18:42:23.670" Score="2" ViewCount="0" Body="&lt;p&gt;No. Your defined objective is the Hausdorff distance $d_H$ between a surface and the tetrahedron -- at least, it is as long as the surface is inside the tetrahedron. For the chosen surface, the distance is $d_H(E_1,E_2) = \frac{1}{2}$, attained at $p = \left(\frac{1}{2},\frac{1}{2},0\right), q = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$. You can have $d_H(E_1,E_3) \le \frac{1}{2}$ with the area of $E_3$ arbitrarily small by letting $E_3$ be a small sphere at $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ with lots of very thin &quot;spikes&quot; reaching the boundary of the tetrahedron.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-28T18:42:23.670" />
  <row Id="3565" PostTypeId="1" AcceptedAnswerId="3569" CreationDate="2010-08-28T18:44:25.117" Score="9" ViewCount="111" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given some natural number, we can of course split it up into various sums of other naturals (e.g. $7 = 6 + 1 = 1 + 4 + 2 = \ldots$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, for $n \in \mathbb{N}$, we can a find distribution sequence (or &lt;em&gt;partition&lt;/em&gt;) $s_0,\ldots,s_u \in \mathbb{N}$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{u}s_i = n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how can I find the partition $s$ for which the overall least common multiple of all elements is maximal.&#xA;Or differently formulated the maximum product of distinct prime factors of all elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 7 \mapsto (3, 4); \Pi_{lcm} = 12 $$&#xA;$$ 7 \mapsto (1, 2, 4); \Pi_{lcm} = 4$$ &#xA;$$ \ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the first solution is the desired one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of my question is: How can I split up a sequence/string into contiguous subsequences that, when repeated and &lt;a href=&quot;http://en.wikipedia.org/wiki/Zip_%28higher-order_function%29&quot; rel=&quot;nofollow&quot;&gt;zipped together&lt;/a&gt;, will yield the longest possible resulting sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above example, I could split up a string of length $7$ in the following way:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7: abcdefg 7:  abcdefg&#xA;&#xA;    I              II&#xA;1: aaaa    3: abcabcabcabc&#xA;2: bcbc    4: defgdefgdefg&#xA;4: defg &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course, using the second distribution, the resulting sequence has a much greater period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What algorithm/approach can I use to solve this problem and maximize the product? Is this some known problem and how complex would calculating a solution be? It's not NP, I hope?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Partial solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @KennyTM pointed out in the comments, &lt;a href=&quot;http://en.wikipedia.org/wiki/Landau%27s_function&quot; rel=&quot;nofollow&quot;&gt;Landau's function $g$&lt;/a&gt; describes the maximum LCM, i.e. $g(7) = 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this actually becomes: How to actually produce the partition? Does knowledge of $g(x)$ help here, maybe for a dynamic programming solution?&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="917" LastEditorDisplayName="" LastEditDate="2010-08-28T19:47:26.520" LastActivityDate="2010-08-28T20:56:51.250" Title="Split up $n \in \mathbb{N}$ into sum of naturals with maximum LCM" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;algorithms&gt;&lt;optimization&gt;&lt;extremal-combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3566" PostTypeId="2" ParentId="3543" CreationDate="2010-08-28T18:45:00.717" Score="3" ViewCount="0" Body="&lt;p&gt;There are algorithms due to Abel, Liouville, Risch, Davenport, Trager et al. for solving the general problem of the &lt;strong&gt;algebraic&lt;/strong&gt; (vs. transcendental) case of integration in finite terms (elementary and algebraic functions). Here's an illuminating excerpt from the introduction of &lt;a href=&quot;http://hdl.handle.net/1721.1/15391&quot; rel=&quot;nofollow&quot;&gt;Barry Trager's 1984 MIT Thesis&lt;/a&gt; (coincidentally Barry was working right next door to me disproving Hardy's statement that no algorithm exists at the same time I was disproving an analogous Hardy statement while inventing algorithms for computing limits / asymptotics - while we were both student members of the MIT Mathlab Group - the research group that developed the Macsyma computer algebra system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i36.tinypic.com/2mdnoef.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i38.tinypic.com/n1feac.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i34.tinypic.com/2z9ilac.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-28T19:52:20.247" LastActivityDate="2010-08-28T19:52:20.247" CommentCount="1" />
  <row Id="3567" PostTypeId="2" ParentId="3561" CreationDate="2010-08-28T19:21:06.743" Score="2" ViewCount="0" Body="&lt;p&gt;That's a good question that - alas - does not really have a good answer - at least not a single textbook that one could point to. The best motivation for ideals and modules comes from   number theory. As such, you might find it helpful to attempt to learn some algebraic number theory in parallel, e.g. factorization theory in quadratic number fields. Unfortunately most of the algebraic number theory textbooks already assume solid algebra background, so you'll need to interpolate your own motivational examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One crucial point deserves emphasis. When first learning about new abstractions such as ideals and modules it is &lt;strong&gt;absolutely essential&lt;/strong&gt; to understand their motivating concrete instances. Thus when you are learning about ideals in algebra you should revisit your knowledge of elementary number theory and reinterpret results in terms of these new abstractions. For example, see my post &lt;a href=&quot;http://math.stackexchange.com/questions/2883/show-that-3p2q2-implies-3p-and-3q/2941#2941&quot;&gt;here&lt;/a&gt; and its links - which explains how various proofs of irrationality of sqrt's by descent on the size of a denominator are merely special cases of proofs that a denominator ideal - Dedekind's conductor ideal - is principal or invertible. From this general viewpoint the proof is a one-liner: the conductor ideal is invertible so cancelable. This shows that any Dedekind domain or PID is integrally closed - a monic generalization of the rational root test to a very wide class of rings. This is but one of many generalizations that you will meet when studying factorization theory in wider classes of rings. To better hone your intuition it is essential that you learn to easily move back-and-forth in this way between the concrete and abstract. Unfortunately most textbooks omit explicit discussion of such motivational material, so you'll need to develop it on-the-fly as need be. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-28T19:21:06.743" CommunityOwnedDate="2010-08-28T19:21:06.743" />
  <row Id="3568" PostTypeId="2" ParentId="3547" CreationDate="2010-08-28T19:47:14.960" Score="5" ViewCount="0" Body="&lt;p&gt;&quot;Permutation group&quot; usually refers to a group that is acting (faithfully) on a set; this &lt;em&gt;includes&lt;/em&gt; the symmetric groups (which are the groups of all permutations of the set), but also every subgroup of a symmetric group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although all groups can be realized as permutation groups (by acting on themselves), this kind of action does not usually help in studying the group; special kinds of actions (irreducible, faithful, transitive, doubly transitive, etc), on the other hand, can give you a lot of information about a group. For example, Jordan proved that the only finite sharply five transitive groups are $A_7$, $S_6$, $S_5$, and the Mathieu group $M_{12}$. (A &quot;sharply five transitive group&quot; is a group $G$ acting on a set $X$ with five or more elements, such that for every ten elements $a_1,\ldots,a_5,b_1,\ldots,b_5\in X$, with $a_i\neq a_j$ for $i\neq j$ and $b_i\neq b_j$ for $i\neq j$, there exists one and only one $g\in G$ such that $g\cdot a_i = b_i$). (In fact, Jordan showed that the only finite sharply $k$-trasitive gruops for $k\geq 4$ are $S_k$, $S_{k+1}$, $A_{k+2}$, $M_{11}$, and $M_{12}$; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathieu_group&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathieu_group&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You think of a permutation group as a group $G$, together with a faithful action $\sigma\colon G\times X\to X$ on a set $X$ (faithful here means that if $gx=x$ for all $x$, then $g=e$). Cayley's Theorem tells you that every group $G$ can be thought of as a permutation group, by taking $X$ to be the underlying set of $G$, and $\sigma$ to multiplication. But this gives you an embedding of $G$ into a very large symmetric group, because the set on which it is acting is large. You usually get more information if the set you are acting on is &quot;small&quot;-ish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for Cayley's Theorem is that, historically, people only considered permutation groups: collections of functions that acted on sets (the sets of roots of a polynomial, the points on the plane via symmetries, etc). Cayley was trying to abstract the notion of group; he then pointed out that his more abstract definition certainly included all the things that people were already considering, and that in fact it did not introduce any new ones in the sense that every abstract group could be considered as a permutation group. But, as he pointed out, it is sometimes more convenient or useful to consider the group abstractly, sometimes to consider it as a group of permutations. Having both viewpoints is better than having just one. I think Cayley's Theorem has more historical interest than practical interest these days, but your mileage may vary.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-08-28T19:47:14.960" />
  <row Id="3569" PostTypeId="2" ParentId="3565" CreationDate="2010-08-28T20:56:51.250" Score="3" ViewCount="0" Body="&lt;p&gt;The blog post here &lt;a href=&quot;http://amca01.wordpress.com/2010/07/10/landaus-function/&quot; rel=&quot;nofollow&quot;&gt;describes&lt;/a&gt; somebody else's efforts at coding this up. He gets an improvement over the naive approach by using some nice internal structure of the problem. &lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2010-08-28T20:56:51.250" CommentCount="1" />
  <row Id="3570" PostTypeId="2" ParentId="187" CreationDate="2010-08-28T20:57:48.987" Score="1" ViewCount="0" Body="&lt;p&gt;Let me discuss an analogous situation with regard to convex polyhedra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Archimedes, generalizing what are often called the Platonic Solids or the convex regular polyhedra (there are five of them), seems to have discovered 13 polyhedra with the property that the pattern of faces around each vertex was the same for every vertex of the polyhedron, and all of the faces of the polyhedron were regular polygons. I say seems to have done this because the manuscript that describes what he did is lost. Pappus who lived many years after Archimedes describes these polyhedra explicitly and mentions 13 convex solids. Years latter many artists and mathematicians talked about these polyhedra. Kepler explicitly mentions two infinite families of polyhedra which have this property: the prisms (consisting of two regular n-gons and n squares) and the anti-prisms (consisting of two regular n-gons and 2n equilateral triangles). Kepler purports to give a proof a proof that there 13 such solids (with at least two types of faces). I say purports because there are in fact 14 convex polyhedra which meet the local symmetry condition described above. The one Archimedes, Pappus, Kepler and others missed is often called the pseudo-rhombicuboctahedron. (Kepler created confusion in referring to 14 solids in a place other than where he gave his &quot;proof.&quot;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elongated_square_gyrobicupola&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Elongated_square_gyrobicupola&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many books to this day continue to talk about 13 Archimedean convex solids (other than the prisms and antiprisms). With the definition that Archimedes almost certainly had in mind this is wrong - there are 14 such solids. However, if one considers convex polyhedra with at least two regular polygons as faces, and where the symmetry group of the solid is transitive on the vertices (that is, the symmetry group can take any vertex to any other vertex) then there are only 13 such solids. However, almost certainly Archimedes had no knowledge of the idea of a symmetry group in the modern sense. Depending on whether one uses a local symmetry notion or a global symmetry notion there are either 14 or 13 such convex solids (aside from the prisms and antiprisms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branko Grünbaum has a nice article about this called &quot;An enduring error.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://dlib.lib.washington.edu/dspace/bitstream/handle/1773/4592/An_enduring_error.pdf?sequence=1&quot; rel=&quot;nofollow&quot;&gt;https://dlib.lib.washington.edu/dspace/bitstream/handle/1773/4592/An_enduring_error.pdf?sequence=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-28T20:57:48.987" CommentCount="1" />
  <row Id="3571" PostTypeId="2" ParentId="3550" CreationDate="2010-08-28T21:37:02.470" Score="2" ViewCount="0" Body="&lt;p&gt;In order to find the solutions of the system of conics you mention, it suffices to give a procedure to find the projection of the simultaneous vanishing set of the two conics onto some axis, for instance, the $y$-axis: the coordinates of the projection onto the $y$-axis are the $y$-coordinates of the intersection points. Knowing them allows you to substitute back into the equations the values of $y$ and solve a system of equations in a single variable $x$, thus making the problem simpler. In terms of ideals, suppose that we can find a non-zero element $r$ in the ideal generated by the two conics, that depends only on a single variable, say $y$. This means that every solution of the system has $y$-coordinate satisfying the polynomial $r$. Thus we have severely limited the choices for the $y$-coordinates of the intersection (namely, they all have to satisfy the polynomial $r$) and, provided we can actually solve the polynomial $r$ in one variable, we can then substitute the various values of $y$ we found back into the initial equations and solve for $x$, again using our algorithm for solving polynomials in a single variable. So far, so good, I hope! The question is how to produce the element $r$. This is where Gröbner bases come in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do the computation you ask, one would have to choose an appropriate monomial order. I will gloss over this, and simply &quot;do the obvious&quot;. It is an exercise for you to figure out which order to use so that the computation I am going to make is actually a Gröbner bases computation. Scattered throughout the computation there will be also a couple of special cases that I will not deal with: again, you can treat those as exercises for you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I am going to choose a generic basis, so that the first conic has the form &#xA;$$&#xA;x^2 + \alpha x + \beta ,&#xA;$$&#xA;where $\alpha$ and $\beta$ are polynomials in $y$ only (I am trying to simplify the notation as much as possible; the only &quot;assumption that I have made is that the coefficient of $x^2$ is non-zero, which can be arranged unless the &quot;conic&quot; is defined by a polynomial of degree at most one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this basis, the second equation can be assumed to have no $x^2$ term (indeed, eliminating the $x^2$ term using the first equation would be the first step in any reasonable Gröbner basis computation in which the term $x^2$ is the highest term in sight). Thus I am going to write the second conic as &#xA;$$&#xA;\gamma x + \delta&#xA;$$&#xA;where, as before, $\gamma$ and $\delta$ are polynomials in $y$ only. Again, just to fix on a definite case, I am going to assume that the polynomial $\gamma$ is non-zero. (If $\gamma$ were zero, then we would have found a polynomial in the ideal generated by the two conics which only depends on $y$: this was our goal at the start anyway! Obviously, you should worry about the case $\gamma=\delta=0$, but I won't.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute a Gröbner basis, you would now compute S-polynomials: let's do it here as well. I am going to try to eliminate the $x^2$ term from the first equation, by using the second equation. This is easy: multiply the first equation by $\gamma$ and the second one by $x$ and take the difference: we are left with the equation&#xA;$$&#xA;(\alpha \gamma - \delta) x + \beta \gamma .&#xA;$$&#xA;Now we are going to eliminate the $x$ term from this last equation using again the second equation: multiply the second equation by $(\alpha \gamma - \delta)$, the last equation by $\gamma$ and subtract to obtain&#xA;$$&#xA;\delta^2 - \alpha \gamma \delta + \beta \gamma^2 .&#xA;$$&#xA;We found an expression independent of $x$!! We are done... provided this expression is not identically zero. You can figure out what this would mean and what happens in this case. Note also that the final expression is what we would have obtained if, at the very beginning, we had &quot;solved&quot; $x=-\frac{\delta}{\gamma}$ using the second equation, substituted in the first equation and cleared the denominators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that this &quot;hybrid&quot; computation explains what is going on: Gröbner bases and Buchberger's algorithm are a systematic way of &quot;solving&quot; systems of equations. You do not have to do any thinking, once you set up the problem. But you need to set up the problem so that it computes what you want. In this case, you could have used several shortcuts to get to the answer, without following all the steps. In more complicated situations, Buchberger's algorithm might be the best way of keeping track of all the steps to be taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also comment that, except in the case in which you have a computer doing the computations for you, it is highly unlikely that Gröbner bases will help you with a specific question, unless you could have also found simple tricks to solve it right away.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-28T21:37:02.470" CommentCount="1" />
  <row Id="3572" PostTypeId="2" ParentId="3138" CreationDate="2010-08-28T22:33:49.803" Score="0" ViewCount="0" Body="&lt;p&gt;One thing you might be interested in is Darboux's Theorem, which tells you that derivatives have the intermediate value property.  There's a decent  Wikipedia page  on the topic and there was also a nice proof given by Sam B. Nadler, Jr. in the Math Monthly a little while back (Feb. 2010). &lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-08-28T22:33:49.803" />
  <row Id="3573" PostTypeId="1" AcceptedAnswerId="3615" CreationDate="2010-08-28T23:16:29.683" Score="3" ViewCount="222" Body="&lt;p&gt;I came up with this problem after discussion of taxicab geometry in math class... I thought it was a simple problem, but still pretty neat; however, I am as of yet unsure of whether my answer is correct, or logical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $[X]$ be the area of region $X$, and region $S_n$ be represented by the equation $|x-n|+|y-n|=k-n$ for all $n=0,1,2,\ldots,k-1$. Now let region $R_n$ be the region between $S_n$ and $S_{n+1}$ and $L=\displaystyle\sum_{n=0}^{k-2}{[R_n]}$. Find the smallest positive integer $k$ such that $L &gt; A$. &lt;strong&gt;($A$ is any number you can plug in)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone else verify my result of $L=\frac{5k^2-k-4}{2}$?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-08-29T20:12:38.220" LastActivityDate="2010-08-29T22:01:03.400" Title="Interesting Taxicab Problem?" Tags="&lt;geometry&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="3574" PostTypeId="1" AcceptedAnswerId="3575" CreationDate="2010-08-28T23:57:04.883" Score="2" ViewCount="95" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many &lt;strong&gt;unique&lt;/strong&gt; strings can be formed by the letters in &quot;aaamottu&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a totally incompetent math student, I simply wrote a program to calculate it. And I ended up with 3360 unique strings:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools&#xA;unique = {}&#xA;for p in itertools.permutations(list(&quot;aaamottu&quot;), 8):&#xA;    unique[p] = True&#xA;&#xA;len(unique)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how to get that number using plain ol' math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;nPr(8,8) gives 40,320. Obviously, lots of duplicates are being counted. How to I remove them?&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2010-08-29T00:17:56.540" Title="Remove duplicates from a permutation" Tags="&lt;combinatorics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3575" PostTypeId="2" ParentId="3574" CreationDate="2010-08-29T00:17:56.540" Score="2" ViewCount="0" Body="&lt;p&gt;First pretend that the letters which are the same are different, then stop pretending.  Let us suppose that the string is $a_1 a_2 a_3 m o t_1 t_2 u$; then there really would be $8! = 40320$ different words we could form.  The key insight is that we can group these words according to how they use the $a$s and the $t$s; for example, we can group together the words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1 a_2 a_3 m o t_1 t_2 u, a_1 a_2 a_3 m o t_2 t_1 u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the words are the same except for the order of the $t$s.  Since it is true that any word has a &quot;brother&quot; where the order of the $t$s is reversed, to account for the fact that we labeled the $t$s we have to divide by $2! = 2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we can group together according to how the $a$s are used; for example, we can group together the words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_1 a_2 a_3 m o t_2 t_1, a_1 a_3 a_2 m o t_2 t_1, a_2 a_1 a_3 m o t_2 t_1, a_2 a_3 a_1 m o t_2 t_1, a_3 a_1 a_2 m o t_2 t_1, a_3 a_2 a_1 m o t_2 t_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the words are the same except for the order of the $a$s.  Since it is true that any word has five &quot;partners&quot; where the order of the $a$s are different, to account for the fact that we labeled the $a$s we have to divide by $3! = 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is $\frac{8!}{3! 2!} = \frac{40320}{12} = 3360$ as desired.  This function is an example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;multinomial coefficient&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-29T00:17:56.540" />
  <row Id="3576" PostTypeId="1" AcceptedAnswerId="3577" CreationDate="2010-08-29T00:30:44.293" Score="0" ViewCount="123" Body="&lt;p&gt;In making up another problem today I came across something odd. I've been thinking it over and I can't exactly place why it's true, but after running a long Python script to check, I haven't yet found a counter example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\sum_{n=1}^{m}{n^m}\equiv 0\mod m$ true for all odd $m \ge 3$? The script showed me that each term for odd $m$ is equivalent to $n$ when taken $\mod m$ (until term $m$), and so the sum would be $\frac{m(m-1)}{2}$ which is obviously $0 \mod m$. What I am unable to understand is why $n^m\equiv n \mod m$ only for odd $m$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-08-29T00:45:46.527" Title="Why does this sum mod out to 0?" Tags="&lt;sequences-and-series&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3577" PostTypeId="2" ParentId="3576" CreationDate="2010-08-29T00:45:46.527" Score="0" ViewCount="0" Body="&lt;p&gt;If it is odd, each &lt;code&gt;n&lt;/code&gt; can be paired with &lt;code&gt;-n&lt;/code&gt;. So we get $n^m+(-n)^m=n^m-n^m=0$&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-08-29T00:45:46.527" CommentCount="2" />
  <row Id="3578" PostTypeId="1" CreationDate="2010-08-29T00:50:13.520" Score="2" ViewCount="83" Body="&lt;p&gt;Related to this question: &lt;a href=&quot;http://math.stackexchange.com/questions/2626/modelling-forces-acting-on-a-sail&quot;&gt;http://math.stackexchange.com/questions/2626/modelling-forces-acting-on-a-sail&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to create a simulation of the forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done this before with cars:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;using a free physics engine&lt;/li&gt;&#xA;&lt;li&gt;a graphics library with OpenGL, Glut and glow&lt;/li&gt;&#xA;&lt;li&gt;and a 3D model of car. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I'm wondering if there is a better/more efficient way or your way :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="608" LastActivityDate="2010-08-29T00:50:13.520" Title="physics/mathematic simulations" Tags="&lt;learning&gt;&lt;self-learning&gt;" CommentCount="2" />
  <row Id="3579" PostTypeId="1" AcceptedAnswerId="3677" CreationDate="2010-08-29T01:19:15.317" Score="7" ViewCount="314" Body="&lt;p&gt;It can be seen &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Solution1.1.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that the only numbers for which $n^{m+1}\equiv n \pmod{m}$ is true are 1, 2, 6, 42, and 1806. Through experimentation, it has been found that $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1 \bmod m$ is true for those numbers, and (as yet unproven) no others. Why is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a simple relation between $n^{m+1} \bmod{m}$ and $n^m \bmod{m}$, that would probably make this problem make more sense. It is obvious that $n^m \equiv 1 \pmod{\frac{m}{d}}$ (dividing out $n$ from both sides gives this result) for all $n$ on the interval $[1,m]$ where $d$ is a divisor of $m$. As a result of this, $n^m \bmod{m}$ takes on only values of the form $1+k \frac{m}{d} \pmod m$ where $k = -1, 0, 1$. How can it be shown that the sum of those values is equivalent to $1 \bmod{m}$?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:26:16.257" LastActivityDate="2010-09-05T04:26:16.257" Title="Why are the only numbers $m$ for which $n^{m+1}\equiv n \pmod{m}$ is true also unique for $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1 \bmod m$?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="17" FavoriteCount="1" />
  <row Id="3580" PostTypeId="1" AcceptedAnswerId="3581" CreationDate="2010-08-29T01:26:59.140" Score="0" ViewCount="42" Body="&lt;p&gt;I had a question about an instantaneous z score.  The question reads as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(z=2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think that this is 0 as z-score is the area under the curve and could be found by integrating the PDF over the range.  As the range is 0 the area also zero.  Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="1564" LastActivityDate="2010-08-29T02:06:57.483" Title="Instantaneous Z-Score Value" Tags="&lt;statistics&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="3581" PostTypeId="2" ParentId="3580" CreationDate="2010-08-29T01:36:39.840" Score="2" ViewCount="0" Body="&lt;p&gt;Since $P(z=2)$ is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_2^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then yes, the result is zero; this can be interpreted also as saying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_{-\infty}^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u-\frac1{\sqrt{2\pi}}\int_{-\infty}^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;via the Fundamental Theorem of Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, the &quot;integral&quot; covers no area on the Gaussian curve, which agrees with the result of 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(previous version)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to look at Abramowitz and Stegun just to check that you were saying what I thought you were saying; anyway, if by $P(z)$ you meant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_{-\infty}^z \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then no, $P(2)$ is not zero. The integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_{-\infty}^0 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by itself is equal to $\frac12$ (why? ;) ), and the other part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac1{\sqrt{2\pi}}\int_0^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cannot be zero as well because it is the integral of a function whose graph does not cross the x-axis.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-29T02:06:57.483" LastActivityDate="2010-08-29T02:06:57.483" CommentCount="3" />
  <row Id="3582" PostTypeId="2" ParentId="2121" CreationDate="2010-08-29T02:05:01.930" Score="2" ViewCount="0" Body="&lt;p&gt;Let $F_4 = Z/2[w]/w^2+w+1$ be the field of order 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H = (F_4^+)^n = (Z/2)^{2n}$ for $n \geq 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $Aut(H) = GL(2n,F_2)$ which has order&#xA;$(4^n-1)(4^n-2)(4^n-4)...(4^n-2*4^(n-1))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be $\langle H,a : a^3 = 1, axa^{-1} = wx \ \text{for} \ x \in H \rangle$.  So $G$ is the semi-direct product of $H$ with $Z/3$, where the action is given by multiplication by $H$.  An automorphism of $G$ must fix $H$ ($H$ consists of the set of elements of order 2).  It must also send a to one of the 2*4^n elements of order 3.  Let us consider only the automorphisms of $G$ which fix a (which is a subgroup of $Aut(G)$ of index $2*4^n$).  These are automorphisms of H which commute with multiplication by w.  Hence these are automorphisms of $F_4^n$ as an $F_4$ vector space, or elements of $GL(n,F_4)$.  Hence&#xA;$|Aut(G)| = 2*4^n* (4^n-1)(4^n-4)...(4^n-4^(n-1))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \geq 2$, it is clear that $|Aut(H)| &gt; |Aut(G)|$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-29T02:05:01.930" CommentCount="1" />
  <row Id="3583" PostTypeId="1" CreationDate="2010-08-29T04:36:09.707" Score="4" ViewCount="272" Body="&lt;p&gt;A train with infinitely many seats, one for each rational number, stops in countably many villages, one for each positive integer, in increasing order, and then finally arrives at the city. At the first village, two women board the train. At the second village, one woman leaves the train to go visit her cousin, and two other women board the train. At the third village, one woman leaves the train to go visit her cousin, and two other women board the train. At the fourth village, and in fact at every later village, the same thing keeps happening: one woman off to visit her cousin, two new women on board the train. How many women arrive at the city?&lt;/p&gt;&#xA;" OwnerUserId="1565" LastActivityDate="2010-08-30T14:12:06.803" Title="Infinite processes riddle" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="3584" PostTypeId="2" ParentId="3501" CreationDate="2010-08-29T05:11:59.420" Score="1" ViewCount="0" Body="&lt;p&gt;As has been indicated in the comments, the radical symbol is generally taken to be the principal square root function, which for nonnegative real inputs gives nonnegative real outputs.  For negative real inputs, this function gives non-real outputs, so X must be nonnegative, so Y must be nonnegative, giving (a).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-29T05:11:59.420" CommentCount="1" />
  <row Id="3586" PostTypeId="2" ParentId="3583" CreationDate="2010-08-29T06:09:19.830" Score="1" ViewCount="0" Body="&lt;p&gt;This problem isn't well defined as it relies on infinity to actually exist. There are two ways of viewing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want the limit of the number of people on the train as we approach the infinitith stop, this series is +2-1+2-1..., so a number of people with the same cardinality as the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second and seemingly more correct interpretation is: &quot;How many women stay on the train throughout the whole journey?&quot;. If we make women leave the train in the same order they board, the answer is 0, as no-one remains on the train through the whole journey. If we make the last women to board leave, then the answer is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This behaviour is not strange if you have seen the behaviour of infinite series. For example, while +2-1+2-1... approached infinity, +2-1-1+2-1-1 will always be almost 0.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-08-29T21:10:09.717" LastActivityDate="2010-08-29T21:10:09.717" CommentCount="6" />
  <row Id="3587" PostTypeId="2" ParentId="3583" CreationDate="2010-08-29T06:11:05.760" Score="7" ViewCount="0" Body="&lt;p&gt;There is not enough information to uniquely determine the answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a situation in which $0$ women get to the city: we place the women in order according to when they boarded the train, and every time a new village is reached the woman who boarded the train farthest back in time leaves.  This procedure ensures that every woman eventually leaves the train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a situation in which countably many women get to the city: every time a new village is reached one of the women who boarded at the previous village leaves.  This procedure ensures that the number of women who eventually reach the city increases by $1$ at each village.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can in fact get any finite number of women to the city.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-29T06:20:31.307" LastActivityDate="2010-08-29T06:20:31.307" CommentCount="4" />
  <row Id="3588" PostTypeId="2" ParentId="3531" CreationDate="2010-08-29T07:05:11.627" Score="2" ViewCount="0" Body="&lt;p&gt;Not much to contribute at this point except this enhanced picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OvSTr.png&quot; alt=&quot;Steiner circumellipse&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(dashed lines: axes; gray lines: medians)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the &lt;a href=&quot;http://ifile.it/xse06ua&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Mathematica&lt;/em&gt; notebook&lt;/a&gt; used to produce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks &lt;a href=&quot;http://math.stackexchange.com/users/72&quot;&gt;Isaac&lt;/a&gt;!)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-29T10:43:14.083" LastActivityDate="2010-08-29T10:43:14.083" CommentCount="9" />
  <row Id="3589" PostTypeId="2" ParentId="3579" CreationDate="2010-08-29T09:06:34.453" Score="-1" ViewCount="0" Body="&lt;p&gt;Consider all the $n$ so that $(n,m)=d$.&lt;br&gt;&#xA;The $m^{th}$ powers of these give each of the numbers with $(n',m) = (d^m,m)$ some number of times.  These sums cancel ($n'$ and $m-n'$ show up the same number of times) unless $(d^m,m) = m/2$.  Hence the sum of congruent to $0$ or $\frac{m}{2} mod m$.  This is 1 only if m is 1 or 2.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-29T09:50:01.430" LastActivityDate="2010-08-29T09:50:01.430" CommentCount="2" />
  <row Id="3590" PostTypeId="1" CreationDate="2010-08-29T12:58:49.673" Score="8" ViewCount="268" Body="&lt;p&gt;(1) The Riemann hypothesis (RH) states that all non-trivial zeros of the zeta function have real part 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The zeta function is intimately connected with the Gamma function via the functional equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second fact suggests that there is an equivalent form of RH which is expressed solely in terms of the Gamma function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the most natural form to translate RH as directly as possible (without mentioning the zeta function) into a hypothesis on the behaviour of the Gamma function?&lt;/p&gt;&#xA;" OwnerUserId="1484" LastActivityDate="2010-12-05T03:25:14.210" Title="How to express the Riemann hypothesis in terms of the Gamma function?" Tags="&lt;gamma-function&gt;&lt;riemann-hypothesis&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="3" />
  <row Id="3592" PostTypeId="1" AcceptedAnswerId="3597" CreationDate="2010-08-29T13:29:16.830" Score="1" ViewCount="393" Body="&lt;p&gt;Calculus theory problem I'm belaboring to solve... Tried it for an hour only spinning my wheels. Any hints/nudges in the right direction would be greatly appreciated. Here's the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f and g are functions that are differentiable for all real x. They have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) f'(x) = f(x) - g(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) g'(x) = g(x) - f(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) f(0) = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iv) g(0) = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that f(x) + g(x) = 6 for all x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="1568" LastActivityDate="2010-08-30T08:32:46.783" Title="Prove that f(x) + g(x) = 6" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="3593" PostTypeId="2" ParentId="3510" CreationDate="2010-08-29T13:30:24.553" Score="9" ViewCount="0" Body="&lt;p&gt;One could provide answers based on a wide range of definitions of $\exp$, $\cos$, and $\sin$ (&lt;em&gt;e.g.&lt;/em&gt;, via differential equations, power series, Lie theory, inverting integrals, infinite sums, infinite products, complex line integrals, continued fractions, circular functions, and even Euclidean geometry) as well as offering Euler's formula up as a tautology based on a definition.  But let's consider where in one's education this question arises: it's usually well before most of these other concepts are encountered.  The complex numbers have just been introduced; the Argand diagram is available; $\exp(x)$ is most likely defined as the limiting value of $(1 + x/n)^n$, and $\cos$ and $\sin$ are defined as circular functions (that is, as arclength parameters for coordinates of the unit circle).  So the question deserves a response in this context using mathematics accessible to someone at this level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, I propose that we interpret $\exp(i x)$ as the limit of $(1 + i x/n)^n$, because at least one understands how to compute the latter (as iterated multiplication of 1 by $1 + i x/n$), one at least vaguely intuits what the limit of a sequence of points in the complex plane might mean, and one has learned that $\cos(\theta) + i \sin(\theta)$ is quite literally the complex number plotted at $\left( \cos(\theta), \sin(\theta) \right)$ in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems natural to evaluate the limit by looking at its modulus and argument separately.  The modulus is relatively easy: because $|1 + i x/n| = \sqrt {  1 + (x/n)^2  }$, the modulus of its $n$th power equals $\left( 1 + (x/n)^2 \right) ^{n/2}$.  We can exploit the one limit assumed to be known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 1 + (x/n)^2 \right) ^{n/2} = \left( \left( 1 + x^2 / n^2 \right) ^ {n^2} \right) ^ {1/{2n}} \simeq \left( \exp ( x^2 ) \right) ^ {1/{2n}} \to 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There's some hand-waving in the last two steps.  This is characteristic of limiting arguments at this level.  Those of you who can see where the rigor is lacking also know exactly how to supply the missing steps.)  Whence, whatever $\exp( i x )$ might be, we deduce it should lie on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we turn our attention to the argument of the limit.  The sequence of arguments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 1 + (x/n)^2 \right) ^{1/2}, \left( 1 + (x/n)^2 \right) ^{2/2}, \ldots, \left( 1 + (x/n)^2 \right) ^{k/2}, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;obviously is non-decreasing, because it's a geometric sequence with multiplier of 1 or greater.  That is, each successive multiplication by $1 + i x / n$ is &lt;em&gt;expanding&lt;/em&gt; an original right triangle with vertices at (0,0), (1,0), and (1, $x/n$), but in the limit the amount of expansion is reduced to unity, as we have seen.  In the Argand diagram we're just laying out similar copies of this original right triangle, placing one leg of each new copy along the hypotenuse of the previous one.  In the limit, the length of the small leg (originally $x/n$) therefore remains constant.  These observations imply that near the limit, we can take all these $n$ little triangles to be essentially congruent, whence the length of the path traced out by the succession of images of the small leg must be extremely close to $n (x/n) = x$.  This pins down where on the circle $\exp( i x )$ must lie: it is the point reached by traveling (signed) distance $x$ counterclockwise along the circle beginning at (1,0).  To anyone exposed to the definition of $\sin$ and $\cos$ in terms of circular functions, it is immediate that the coordinates of this limiting location are $\left( \cos(x), \sin(x) \right)$ and we are done.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-29T13:30:24.553" />
  <row Id="3595" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T13:56:09.763" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dx} \bigl( f(x) + g(x) \bigr) = ? $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-08-29T13:56:09.763" />
  <row Id="3596" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T13:56:54.300" Score="2" ViewCount="0" Body="&lt;p&gt;Let's introduce their sum function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s(x) := f(x) + g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement to be proven thus becomes $s(x) = 6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s(0) = 5 + 1 = 6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now what derivative $s'$ do we need such that $s$ does never change it's constant value? &lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-29T13:56:54.300" />
  <row Id="3597" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T14:09:55.280" Score="4" ViewCount="0" Body="&lt;p&gt;As a follow up to Dario... If we let $s(x) := f(x) + g(x)$, $s'(x) = f'(x)+g'(x)$, which by (i) and (ii) is equal to $f(x)-g(x)+g(x)-f(x) = 0$. Since $s(0) = 6$ and $s'(x) = 0$, $s(x)$ is constant; $s(x) = 6$ for all $x$, therefore $f(x)+g(x) = 6$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-08-29T14:09:55.280" CommentCount="1" />
  <row Id="3598" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T14:21:55.807" Score="1" ViewCount="0" Body="&lt;p&gt;This problem should help you to understand and practice with two main facts about derivatives:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Derivative is a linear function&lt;/strong&gt; in fact if $f$ and $g$ are differentiable on $\mathbb{R}$ and $\alpha , \beta \in \mathbb{R}$ then $(\alpha f + \beta g)' = \alpha f' + \beta g'$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Derivative is a rate of change&lt;/strong&gt; this is more subtle, the essential fact is that if $f$ is differentiable on $\mathbb{R}$ and $f'(x) = 0$ for all $x \in \mathbb{R}$ then &lt;em&gt;the rate of change of $f$ is null&lt;/em&gt; on all $\mathbb{R}$ and then &lt;em&gt;the value of $f$ does not change&lt;/em&gt; on all $\mathbb{R}$ (pay attention, this is not always true if you break the domain of $f$ into pieces). This is a consequence of a more general and important result, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;mean value theorem&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Observe that i) and ii) in the question give you &lt;em&gt;global information&lt;/em&gt; on $f$ and $g$ (i. e. they refer to all $x \in \mathbb{R}$), while iii) and iv) give you &lt;em&gt;local&lt;/em&gt; information (in the point $x = 0$). Your question is about a &lt;em&gt;global&lt;/em&gt; propriety of the function $f+g$, so you should try to combine local and global information to reach the solution.&lt;/p&gt;&#xA;" OwnerUserId="365" LastActivityDate="2010-08-29T14:21:55.807" />
  <row Id="3600" PostTypeId="1" AcceptedAnswerId="3603" CreationDate="2010-08-29T15:37:36.130" Score="3" ViewCount="202" Body="&lt;p&gt;So, I know that every abelian (commutative) group $G$ is such that, for any subgroup $H$, the left cosets of $H$ in $G$ are the right cosets. I guess this is true even if $G$ is not abelian, but $H$ is (but I'm not sure). Is this enough to characterise the subgroup as abelian, or are there examples of non-abelian subgroups with this property?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-30T12:36:48.843" Title="Can a non-abelian subgroup be such that the right cosets equal the left cosets?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="3601" PostTypeId="2" ParentId="3600" CreationDate="2010-08-29T15:41:39.200" Score="8" ViewCount="0" Body="&lt;p&gt;The left cosets of $H$ are the same as the right cosets&#xA;of $H$ if and only if $H$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_subgroup&quot; rel=&quot;nofollow&quot;&gt;normal subgroup&lt;/a&gt; of $G$.&#xA;There are non-abelian groups all subgroups of which are normal,&#xA;for example the quaternion group of order $8$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-29T15:41:39.200" CommentCount="8" />
  <row Id="3602" PostTypeId="1" AcceptedAnswerId="3604" CreationDate="2010-08-29T15:50:12.367" Score="3" ViewCount="158" Body="&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; &lt;em&gt;Example&lt;/em&gt;. To solve a problem on evaluating the maximum of a product of $n$ real variables subject to an equality constraint on its sum $S$ ($=100$), I used the Lagrange multipliers method (which can be improved by adding the adequate conditions on the second order partial derivatives). Doing so I got a &quot;solution&quot; $x^*=(x_1,x_2,\dots ,x_n)\in \mathbb{R}^{n}$ with $x_1=x_2=\dots=x_n=100/n$. I proceeded by maximizing the single real variable function $u(t)=(100/t)^t$: the maximum is at $36\lt t^*\lt 37$. Finally I computed $u(36)\lt u(37)$. Hence, the optimum occurs at $n=37$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spite of having come to the correct solution to the problem, I was told that &lt;em&gt;such an approach does not guarantee the correct solution, in the general case&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; In order to see some limitations of treating optimization problems as illustrated by the above example but with greater generality, a few examples in which this method fails would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark&lt;/em&gt;: I hope this edited text improves the question. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; The case I have in mind is that of finding the maximum/minimum of an objective function $f(x_1,x_2,\dots,x_n)$ subject to at least one constraint $g(x_1,x_2,\dots,x_n)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a kind  of generalization of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplex_algorithm&quot; rel=&quot;nofollow&quot;&gt;linear programming simplex problem&lt;/a&gt;, two differences being &lt;em&gt;a nonlinear optimization and one integer variable&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-23T23:47:46.083" LastActivityDate="2010-12-23T23:47:46.083" Title="A naive approach for nonlinear optimization on several real variables and one natural variable. Give examples of (potential) failure" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;optimization&gt;&lt;nonlinear-optimization&gt;&lt;multivariable-calculus&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="3603" PostTypeId="2" ParentId="3600" CreationDate="2010-08-29T16:00:51.300" Score="2" ViewCount="0" Body="&lt;p&gt;For a subgroup of a group, the condition of being abelian is neither necessary nor sufficient for being &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_subgroup&quot; rel=&quot;nofollow&quot;&gt;normal&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is non-abelian, then $H=G$ is a normal but non-abelian subgroup. As another example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternating_group&quot; rel=&quot;nofollow&quot;&gt;alternating group&lt;/a&gt; is a normal but non-abelian subgroup of the symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a counterexample in the converse direction, consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_group&quot; rel=&quot;nofollow&quot;&gt;free group&lt;/a&gt; with two generators. The subgroup generated by one of the generators is abelian (isomorphic to $\mathbb Z$) but not normal. (This means that your guess above is wrong.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-08-30T12:36:48.843" LastActivityDate="2010-08-30T12:36:48.843" CommentCount="1" />
  <row Id="3604" PostTypeId="2" ParentId="3602" CreationDate="2010-08-29T17:10:36.493" Score="2" ViewCount="0" Body="&lt;p&gt;The problem, if I understand it correctly, is best appreciated geometrically by drawing a picture of the level sets of the objective function and the lattice points satisfying the constraints: the optimum (over the reals) can occur at one location, but the nearby lattice points will necessarily lie on inferior contours.  Extending those contours identifies a border of almost-optimal locations where the objective is greater than at the lattice point you chose.  All you need for a spectacular failure to occur is for there to exist a distant lattice point somewhere within this marginal area: this would correspond to a dramatically different solution satisfying all the constraints and having a better objective value than yours.  (Your solution can have arbitrarily bad objective values, too, depending on the gradient of the objective function between the Lagrange multiplier solution and your rounded approximation to it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have encountered exactly this difficulty with certain problems in statistics where the variables count things (and therefore must be integral) and the objective measures the precision of an estimate (and is to be minimized, but the change from maximization to minimization changes nothing essential).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-29T17:10:36.493" CommentCount="4" />
  <row Id="3605" PostTypeId="2" ParentId="3510" CreationDate="2010-08-29T17:14:39.880" Score="1" ViewCount="0" Body="&lt;p&gt;I'll copy my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/1133/natural-derivation-of-the-complex-exponential-function/1154#1154&quot;&gt;this question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;f(x) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;df/dx = -sin(x) + i*cos(x)&#xA;      = i*f(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;∫(1/f(x)) df = ∫i dx&#xA;ln(f(x)) = ix  + C&#xA;f(x) = e^(ix + C) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since &lt;code&gt;f(0) = 1&lt;/code&gt;, &lt;code&gt;C = 0&lt;/code&gt;, so&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;e^(ix) = cos(x) + i*sin(x)&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-08-29T17:14:39.880" CommentCount="1" />
  <row Id="3606" PostTypeId="1" CreationDate="2010-08-29T17:16:13.763" Score="3" ViewCount="172" Body="&lt;p&gt;I'm trying to express the following function more simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \begin{align} f(x) = (\sum_{n=3}^{x} \prod_{k=2}^{\lceil \sqrt n \rceil} \lceil \dfrac {n}{k} \rceil - \lfloor \dfrac{n}{k} \rfloor ) + 1 \end{align} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is kind of a vague request, but I'm trying to get away from the ceiling and floor functions and possibly the product notation, just because it feels like a hacky way of expressing what I'm trying to express.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1574" LastActivityDate="2010-08-29T20:35:58.747" Title="Function simplification" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3607" PostTypeId="2" ParentId="3606" CreationDate="2010-08-29T17:20:23.020" Score="12" ViewCount="0" Body="&lt;p&gt;$\left\lceil \frac{n}{k} \right\rceil - \left\lfloor \frac{n}{k} \right\rfloor$ is equal to $0$ or $1$ depending on whether $k$ does or does not divide $n$.  In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=2}^{ \lceil \sqrt{n} \rceil} \left( \left\lceil \frac{n}{k} \right\rceil - \left\lfloor \frac{n}{k} \right\rfloor \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equal to $0$ or $1$ depending on whether $n$ is composite or prime.  So the function you are trying to describe is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime-counting_function&quot; rel=&quot;nofollow&quot;&gt;prime counting function&lt;/a&gt;.  There really is no simpler way to describe it than &quot;the prime counting function,&quot; although a lot is known about its asymptotic behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  For functions like the prime counting function, you should replace the notion of &quot;nice formula&quot; with &quot;fast algorithm&quot; (the former being a special case of the latter).  The formula you wrote down is equivalent to a very slow algorithm: use trial division to test the primality of all the integers in your range.  There are much faster &lt;a href=&quot;http://en.wikipedia.org/wiki/Primality_test&quot; rel=&quot;nofollow&quot;&gt;primality tests&lt;/a&gt; which translate into much faster algorithms for computing $\pi(n)$ even though these tests cannot be easily translated into familiar-looking formulas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit #2:  As Shreevatsar mentions in the comments, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_theory&quot; rel=&quot;nofollow&quot;&gt;sieve theory&lt;/a&gt; is also relevant to  counting or estimating $\pi(n)$ more directly, without having to do all that primality testing.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-29T19:12:23.997" LastActivityDate="2010-08-29T19:12:23.997" CommentCount="5" />
  <row Id="3608" PostTypeId="1" CreationDate="2010-08-29T18:01:42.810" Score="1" ViewCount="126" Body="&lt;p&gt;In a recent (and to me, very beautiful) paper, entitled &quot;Relational EPR&quot;, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0604064&quot; rel=&quot;nofollow&quot;&gt;Smerlak and Rovelli&lt;/a&gt; present a way of thinking about EPR which relies upon Rovelli's previously published work on relational quantum mechanics (at see arxiv.org/abs/quant-ph/9609002 ).    In relational quantum mechanics, there is no non-locality, but the definition of when an event occurs is weakened from Einstein's strict definition and instead is localized to each observer-measurement apparatus, including subsequent observers.   There are (informal) coherence assumptions to ensure the consistency of reports from different subsequent observers.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this seems very similar to various results in modern categorical mathematics.    Is there a standard mathematical structure which well describes the structure of the space of localized measurements which Rovelli has envisioned?   I know of Isham's work on topos theory and quantum mechanics, but I think he is aiming at something a little different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking the question here because I am only an interested amateur.&lt;/p&gt;&#xA;" OwnerUserId="1575" LastEditorUserId="1575" LastEditorDisplayName="" LastEditDate="2010-11-13T10:58:12.837" LastActivityDate="2010-11-13T10:58:12.837" Title="What categorical mathematical structure(s) best describe the space of &quot;localized events&quot; in &quot;relational quantum mechanics&quot;?" Tags="&lt;category-theory&gt;&lt;physics&gt;" CommentCount="3" />
  <row Id="3609" PostTypeId="1" CreationDate="2010-08-29T18:24:10.237" Score="6" ViewCount="197" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Goodstein%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Goodstein's Theorem&lt;/a&gt; is the statement that every Goodstein sequence eventually hits 0.  It is known to be independent of Peano Arithemtic (PA), and in fact, was the first such purely number theoretic result.  It is provable in ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of phrasing this is that the theory &quot;PA + Goodstein's theorem is false&quot; is consistent (assuming PA is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Godel's completeness theorem, there must exist a model of PA in which Goodstein's theorem fails.  In fact, applying the downward Lowenheim-Skolem theorem, we may assume this model is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the interest of speaking about this result to a group of grad students (of various interests), I'd like to run this backwards.  So,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is there some known, obvious, or easy to construct countable nonstandard model of $PA$ in which Goodstein's theorem fails?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In answering this, I'm willing to accept the &quot;foundational&quot; first order logic theorems:  Godel's completeness and compactness results, the Lowenheim-Skolem theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the kind of answer I'd really like:  There is an explicit countable collection $\Sigma = \{\phi_n\}$ of first order sentences (possibly in a slightly larger language) such that $\mathbb{N}$ is a model of $PA + \Sigma_0$ for any finite $\Sigma_0\subseteq \Sigma$, and such that $PA + \Sigma$ implies Goodstein's theorem is false.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One approach I've thought of is to first enlarge the language by adding a constant symbol c.  Next, let $\phi_n$ be the statement &quot;The Goodstein sequence for $c$ takes longer than &quot;n&quot; steps to terminate&quot;.  (While I do not personally know how to encode &quot;the Goodstein sequence for $c$&quot; in first order language, I am confident it can be done, for otherwise, one could not even formulate &quot;PA proves the Goodstein sequence converges&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $\mathbb{N}$ is a model of any $PA + \Sigma_0$ by simply setting c = n+1, where n is the largest subscript of a $\phi_k$ in $\Sigma_0$ (which exists because $\Sigma_0$ is finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Godel's and Lowenheim-Skolem theorems, $PA + \Sigma$ has a countable model $M$.  Then the interpretation of $c$ in this model satisfies $\phi_n$ for all $n$, and hence the Goodstein sequence doesn't terminate for $c$ in this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the independence of Goodstein's theorem was so difficult to prove, I'm quite certain there's a mistake in this line of reasoning (though I don't know where).  I'd love for someone to patch this up into something correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, please feel free to retag as neccesary, and thank you for the reponses!&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-09-14T05:40:24.700" Title="Is there a natural model of Peano Arithmetic where Goodstein's theorem fails?" Tags="&lt;number-theory&gt;&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="3610" PostTypeId="2" ParentId="3606" CreationDate="2010-08-29T18:30:40.307" Score="0" ViewCount="0" Body="&lt;p&gt;It depends in what context you want to define that function (for computer implementations?), but following Qiaochu Yuan's post, you can e.g. use set-based notations to state the primality test more clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \left| \lbrace x \in \lbrace 3,\ldots, x \rbrace \; |\; \forall k \in \lbrace 2,\ldots,\sqrt n \rbrace . k \not | \; n \rbrace \right| + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \left| \mathbb{P} \cap \lbrace 3, \ldots, x \rbrace \right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbb{P}$ denotes the set of prime numbers or simply &lt;em&gt;let $f$ be the prime counting function&lt;/em&gt; is even simpler.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-08-29T18:30:40.307" />
  <row Id="3611" PostTypeId="2" ParentId="3592" CreationDate="2010-08-29T20:10:02.533" Score="6" ViewCount="0" Body="&lt;p&gt;Dear Bob,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you've gotten the answer, here is a little bit of a post mortem clarifying the essential points, as may come up again in future problems.  (In general, once you've solved a problem, it's a really good idea to ask for the &quot;moral(s)&quot;.  If you can't identify one, how are you better off now than before you solved it?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I think the most important single step is the recognition that the conclusion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$f(x) + g(x) = 6$ for all $x$&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be more usefully rephrased as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$f(x) + g(x)$ is a constant function, with constant value $6$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much more suggestive, because a function $f$ defined on an interval is constant if and only if its dervative is constantly equal to zero.  (Stationarity is equivalent to identically zero velocity.)  Note that half of this statement is obvious, but the other half is not: it is consequence of the Mean Value Theorem that should be emphasized both in the text and by your instructor.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you are clued in to the fact that the main thing you want to show is that $(f+g)' = f' + g' \equiv 0$ (i.e., constantly zero).  Since a constant function is determined by plugging in any point, once you know that $f+g$ is constant, seeing that it's constantly equal to SOMETHING -- in this case $6$ -- shouldn't be a problem.  So now you look and see how to get from what you're given to the conclusion that $f' + g' \equiv 0$, and you see that you're being given an expression for $f'$ and $g'$ separately.  So certainly you want to add them together and hope to get $0$; in this case, that hope is immediately fulfilled.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-30T08:32:46.783" LastActivityDate="2010-08-30T08:32:46.783" />
  <row Id="3612" PostTypeId="2" ParentId="3606" CreationDate="2010-08-29T20:22:16.507" Score="0" ViewCount="0" Body="&lt;p&gt;Presuming that you seek a concise formula (versus a fast algorithm), then one can simplify it somewhat by replacing the product by a factorial, and by employing $\rm\;\; a\perp b \;\; := \; 1 \;\;\; if \;\;\; gcd(a,b)=1 \;\;\; else \;\;\; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \pi(x) \: = \: \sum_{n\;=\;2}^x \;\: n\perp {\lfloor \sqrt{n}\rfloor}!$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-29T20:35:58.747" LastActivityDate="2010-08-29T20:35:58.747" CommentCount="4" />
  <row Id="3613" PostTypeId="1" AcceptedAnswerId="3614" CreationDate="2010-08-29T21:27:24.127" Score="3" ViewCount="244" Body="&lt;p&gt;Hi guys - I was wondering if you could help...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Math homework, I was hoping if you could check my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the equation of the line perpendicular to 3x+2y-4=0 going through point (2,-3)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y = ( -3x + 4 ) /2&#xA;y = 2&#xA;&#xA;x = - ( 2 ( y - 2 ) / 3 )&#xA;x = -3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore my equation is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for you help in advance guys.&lt;/p&gt;&#xA;" OwnerUserId="1579" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-08-29T21:31:16.617" LastActivityDate="2010-08-30T11:34:52.533" Title="Algebra - Find the equation of the line perpendicular to 3x+2y-4=0 going through point (2,-3)" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="3614" PostTypeId="2" ParentId="3613" CreationDate="2010-08-29T21:37:12.580" Score="5" ViewCount="0" Body="&lt;p&gt;If two lines are perpendicular, the product of their slopes is -1.  This is often restated as the slope of the line perpendicular to a given line is the opposite reciprocal of the slope of the given line.  For example, the line $6x-15y+3=0$ has slope $\frac{2}{5}$, so a line perpendicular to it will have slope $-\frac{5}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that fact, you should be able to determine the slope of the line for which you are finding an equation, and you know a point on the line.  Those two pieces of information should be enough for you to write an equation of the line perpendicular to $3x+2y-4=0$ going through point $(2,-3)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-29T21:37:12.580" CommentCount="1" />
  <row Id="3615" PostTypeId="2" ParentId="3573" CreationDate="2010-08-29T22:01:03.400" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vRXKB.png&quot; alt=&quot;graph for k=10&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(diagram for k=10)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each region $S_n$ is a square with sides of slope $\pm 1$, center at $(n,n)$, and side length $\sqrt{2}(k-n)$.  Each pair of successive squares is positioned such that $S_{n+1}$ mostly overlaps $S_n$, but not quite.  As the top and right vertices of every square are, respectively, on the same horizontal and vertical line and 1 unit apart (the centers of successive squares are shifted 1 unit up and 1 unit to the right), the rectangular region of $S_{n+1}$ that is not inside $S_n$ sticks out by $\frac{1}{\sqrt{2}}$ and has length equal to the side length of $S_{n+1}$, which is $\sqrt{2}(k-(n+1))$, so this rectangular region has area $k-n-1$.  Thus, &#xA;$$\begin{align}&#xA;[R_n]&amp;amp;=(\text{area of }S_n)-(\text{area of }S_{n+1})+(\text{area of rectangular region})&#xA;\\&#xA;&amp;amp;=2(k-n)^2-2(k-n-1)^2+k-n-1&#xA;\\&#xA;&amp;amp;=5k-5n-3.&#xA;\end{align}$$&#xA;Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum_{n=0}^{k-2}[R_n]&amp;amp;=\sum_{n=0}^{k-2}(5k-5n-3)&#xA;\\&#xA;&amp;amp;=5k\sum_{n=0}^{k-2}1-5\sum_{n=0}^{k-2}n-3\sum_{n=0}^{k-2}1&#xA;\\&#xA;&amp;amp;=5k(k-1)-5\left(\frac{(k-2)(k-1)}{2}\right)-3(k-1)&#xA;\\&#xA;&amp;amp;=\frac{10k^2-10k-5k^2+15k-10-6k+6}{2}&#xA;\\&#xA;&amp;amp;=\frac{5k^2-k-4}{2}.&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-29T22:01:03.400" CommentCount="3" />
  <row Id="3616" PostTypeId="2" ParentId="3602" CreationDate="2010-08-29T22:53:04.520" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a contrived example of the failure of the method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you are given the function $f$ defined for $0 \leq x \leq 50$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x) = 100 \sin\pi x, 0 \le x \le 1$ &lt;br/&gt;&#xA;$\displaystyle f(x) = x, 1 &amp;lt; x \le 50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your task is to find the integer $n$ for which $f(n)$ is maximized. Your method gives $n=0$ or $n=1$ (max of function occurs at $x = 1/2$), while the correct value we need is $n=50$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason your method works for the example you have is that the function $\displaystyle S(t) = (100/t)^t$ is strictly increasing for $0 &amp;lt; t &amp;lt; 100/e$ and strictly decreasing for $100/e &amp;lt; t$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-29T22:53:04.520" CommentCount="7" />
  <row Id="3617" PostTypeId="1" AcceptedAnswerId="3626" CreationDate="2010-08-30T00:12:15.020" Score="1" ViewCount="322" Body="&lt;p&gt;I am a statistician who wishes to learn real analysis in order to better understand the foundations of statistics. With that aim in mind I plan to go through Rudin's classic on &quot;Principles of Mathematical Analysis&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above context can I skip chapter 1? It seems to me that the material in chapter 1 is  not as important to someone with my goals. I understand that it may help in establishing the need for rigor in mathematical arguments but that is something I already appreciate. In particular, I wonder will skipping chapter 1 impede my understanding of the material in subsequent chapters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1584" LastActivityDate="2010-08-30T11:03:54.400" Title="Can I skip the first chapter in Rudin's Principles of Mathematical Analysis?" Tags="&lt;soft-question&gt;&lt;real-analysis&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="3618" PostTypeId="1" AcceptedAnswerId="3620" CreationDate="2010-08-30T00:51:48.890" Score="2" ViewCount="309" Body="&lt;p&gt;Is there any theory (analogous to Galois theory) for solving equations with irrational exponents like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x^{\sqrt{2}}+x^{\sqrt{3}}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1585" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-05T22:13:43.077" LastActivityDate="2011-01-05T22:13:43.077" Title="Solving an equation with irrational exponents" Tags="&lt;abstract-algebra&gt;&lt;irrational-numbers&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="3619" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T00:57:32.447" Score="1" ViewCount="0" Body="&lt;p&gt;I think you answered your own question. If you are willing to accept that the reals are the unique totally ordered archimedean field with the least upper bound property, then OK skip it.&#xA;But you may wish to refer back to it to know what all of these adjectives mean. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-08-30T00:57:32.447" />
  <row Id="3620" PostTypeId="2" ParentId="3618" CreationDate="2010-08-30T01:05:55.390" Score="8" ViewCount="0" Body="&lt;p&gt;The study of such equations is not &quot;abstract algebra&quot; as it is usually understood.  The reason is that to even define the function $x^{\sqrt{2}}$, for example, requires analysis; one has to prove certain properties of $\mathbb{R}$ to ensure that such a function exists.  This is in marked contrast to the case of integer or rational powers, where one has a purely algebraic definition and the background theory is equational.  To define the function $x^{\sqrt{2}}$ one has to either define $e^x$ and the logarithm or consider a limit of functions $x^{p_n}$ where $p_n$ form a sequence of rational approximations to $\sqrt{2}$, and this is irreducibly non-algebraic stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, while polynomials can be studied in an absurdly general setting, transcendental equations like those you describe are more or less restricted to $\mathbb{R}$ (or $\mathbb{C}$ if you really want to pick a branch of the logarithm).  The LHS is an increasing function of $x$, so there is at most one root, which probably one can really only compute numerically if it exists.  (Its nonexistence can be ruled out by computing local minima in $(0, 1)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another question which touches on a theme which has come up several times on math.SE, which is that exponentiation should really not be thought of as one operation.  Instead, it is a collection of related operations with various degrees of generality and applicability which happen to share the same algebraic properties, and one should not infer too much about how similar these operations are.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-08-30T01:50:09.997" LastActivityDate="2010-08-30T01:50:09.997" CommentCount="2" />
  <row Id="3621" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T02:28:20.500" Score="2" ViewCount="0" Body="&lt;p&gt;It is probably less work to read through the first chapter than to ask if reading through it is necessary! Specially in the context of you planning to read through the whole book.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-30T02:28:20.500" />
  <row Id="3622" PostTypeId="2" ParentId="3315" CreationDate="2010-08-30T03:32:35.663" Score="2" ViewCount="0" Body="&lt;p&gt;As you can see, there are several ways to find the answer to your question, from jmoy's straightforward algebra calculation, to using Euler's formula $e^{r+i\theta} = e^r(\cos(\theta)+i \sin(\theta))$, to geometric interpretations of complex numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fact that hasn't been mentioned yet is that the complex numbers are &lt;strong&gt;algebraically closed.&lt;/strong&gt;  This means that every algebraic equation using complex numbers has all of its solutions in the complex numbers.  So, the equation $x^2=i$ has both of its solutions in the complex numbers; and the equations $x^4 = -7-12i$ and $x^4+(\pi -8i)x^3+x-\sqrt{5}=0$ each have all four of their solutions in the complex numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real numbers are not algebraically closed: the equations $x^2=-1$ and $x^4+7x^2+\pi=0$ cannot be solved unless we use complex numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the main reasons complex numbers are so important; they are the &lt;strong&gt;algebraic closure&lt;/strong&gt; of the real numbers.  You will never need &quot;higher levels&quot; of imaginary numbers or new mysterious square roots; numbers of the form $a+bi$ are all you need to find any root of real or complex polynomials.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-08-30T03:32:35.663" CommentCount="4" />
  <row Id="3623" PostTypeId="1" AcceptedAnswerId="3624" CreationDate="2010-08-30T05:31:48.753" Score="1" ViewCount="147" Body="&lt;p&gt;I would like to know something like what's the first PDE etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you recommend book on the history of PDE/ODE?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T05:59:23.760" LastActivityDate="2010-09-01T23:18:15.053" CommunityOwnedDate="2010-08-30T07:43:32.373" Title="Book recommendation on the history of PDE/ODE?" Tags="&lt;reference-request&gt;&lt;math-history&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="3624" PostTypeId="2" ParentId="3623" CreationDate="2010-08-30T06:51:58.830" Score="0" ViewCount="0" Body="&lt;p&gt;You can refer to these articles: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://www.jstor.org/pss/2298771&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2298771&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) www.johnsasser.com/pdf/article13.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/index/34-XX.html#HIST&quot; rel=&quot;nofollow&quot;&gt;http://www.math.niu.edu/~rusin/known-math/index/34-XX.html#HIST&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bit of surfing yourself for things should help you out!&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-30T06:51:58.830" CommentCount="1" />
  <row Id="3625" PostTypeId="1" CreationDate="2010-08-30T09:59:38.337" Score="4" ViewCount="200" Body="&lt;p&gt;I want to fit to a dataset a power function ($y=Ax^B$). What is the best and easiest method to do this. I need the $A$ and $B$ parameters too.&#xA;I'm using in general financial data in my project, which looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;8553600 458.2&#xA;17193600    373.6&#xA;25833600    694.16&#xA;34646400    738.33&#xA;44064000    817.89&#xA;54259200    1040.67&#xA;67910400    1032.69&#xA;76291200    1222.1&#xA;84844800    1245.65&#xA;94089600    1217.44&#xA;102211200   1579.38&#xA;110592000   1859.24&#xA;118886400   1711.67&#xA;127612800   2303.62&#xA;136684800   2658.26&#xA;219196800   3669.23&#xA;225676800   3525.02&#xA;225763200   3749.27&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to implement the algorithm in a Java similar language called ActionScript.&lt;/p&gt;&#xA;" OwnerUserId="1541" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T00:55:44.507" LastActivityDate="2010-12-05T12:07:21.440" Title="easy to implement method to fit a power function (regression)" Tags="&lt;statistics&gt;&lt;algorithms&gt;&lt;regression&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="3626" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T10:05:05.417" Score="8" ViewCount="0" Body="&lt;p&gt;I'll go through the paragraphs of the third edition, motivating why you should/shouldn't consider them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;INTRODUCTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should carefully read this. This paragraph is not so important for the subsequent development, but it's fundamental for your understanding of the utility of $\mathbb{R}$, it contains an enlightening example wich shows you (with some weird algebraic trick) that &lt;em&gt;$\mathbb{Q}$ contains gaps&lt;/em&gt; and so we really need to &quot;patch&quot; it in order to do interesting things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ORDERED SETS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should read only the definitions of &lt;em&gt;bound&lt;/em&gt; and &lt;em&gt;least upper bound&lt;/em&gt; (or &lt;em&gt;supremum&lt;/em&gt;). They really matter and are used, explicitly or not, in many theorems. In particular, the latter is subtle and you should practice with it, for example with exercises 4 and 5 at the end of the chapter. The rest of the paragraphs regards ordered fields, you can skip it if you are used to work with $\mathbb{R}$ and its ordering, and this practical experience should suffice to you. Perhaps, you could find strange and interesting that $\mathbb{C}$ cannot be ordered without destroying its algebraic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FIELDS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can skip this. If your interest is real analysis, your only field will be $\mathbb{R}$ (perhaps $\mathbb{C}$ sometimes) and as said above, the practical properties of field and ordering should be enough for you (e.g. if $a,b,c \in \mathbb{R}$ and $a &amp;lt; b$ then $a + c &amp;lt; b + c$, you cannot divide by zero, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THE REAL FIELD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways to build $\mathbb{R}$. The first is &lt;em&gt;axiomatic&lt;/em&gt;: you say &quot;how I'd like to work in place that has such property&quot; and magic! You have it by axiom. The second way is &lt;em&gt;contructive&lt;/em&gt;: you take $\mathbb{Q}$, do something on it and come up with a mathematical structure that act as $\mathbb{R}$, has the properties of $\mathbb{R}$ and you eventually call it $\mathbb{R}$. It's very subtle and not practically useful, you should skip the latter method, reported in the appendix of the chapter, and know that when you follow the axiomatic method your are speaking of something that exists, in some mathematical sense. You should also consider theorem 1.20 (archimedean property and density of $\mathbb{Q}$ in $\mathbb{R}$) if you don't read the proof, at least carefully read the statement, it's very used and justify some mysterious things as: if $a \in \mathbb{R}$ and $0 \leq a &amp;lt; \epsilon$ for all $\epsilon &gt; 0$ then $a = 0$. Jump over the existence of the n-th root of a positive real, it's intuitive and you can prove it later in different (and simpler) ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THE EXTENDED REAL NUMBER SYSTEM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only it's not very useful, I think it's dangerous to introduce symbols for infinity when someone still isn't completely conscious of what infinity is and how it acts in many theorems of analysis. Skip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THE COMPLEX FIELD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the case of the real field, if you know what $\mathbb{C}$ is and how to work with it, you can safely skip this, or read it later if you need. The only thing that you probably need are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangular_inequality&quot; rel=&quot;nofollow&quot;&gt;trianglular inequality&lt;/a&gt; in theorem 1.33 and theorem 1.35, known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy-Schwartz&quot; rel=&quot;nofollow&quot;&gt;Cauchy–Schwarz inequality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EUCLIDEAN SPACES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read it, it's used in the following chapters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;APPENDIX&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said above, skip it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXERCISES&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said above, 4 and 5 are very useful. I also suggest you to work on 6 and 7, they teach you what we mean when we say things like $3^{\pi}$ or when we talk about logarithms.&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-08-30T10:26:49.283" LastActivityDate="2010-08-30T10:26:49.283" CommentCount="3" />
  <row Id="3627" PostTypeId="1" AcceptedAnswerId="3634" CreationDate="2010-08-30T10:11:14.363" Score="3" ViewCount="141" Body="&lt;p&gt;I have a matrix $F=XDX^T$, where $D$ is $m\times m$ and diagonal and $X$ is $n\times m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I compute $F^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an efficient method to update $F^{-1}$ if $D$ is updated by $D'=D+G$, where $G$ is a sparse diagonal matrix?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T16:21:33.040" LastActivityDate="2010-10-02T13:32:10.003" Title="Diagonal update of the inverse of $XDX^T$" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="3628" PostTypeId="1" AcceptedAnswerId="3660" CreationDate="2010-08-30T10:58:22.460" Score="5" ViewCount="457" Body="&lt;p&gt;Hi! I'm practicing some problems from past exams and found this one: Find nth derivative of this function:&lt;br&gt;&#xA; $f(x)=\frac {x} {x^2-1}$.&lt;br&gt;&#xA; I have no idea how to start solving this problems. Is there any theorem for finding nth derivative?&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T16:17:12.457" LastActivityDate="2010-08-30T19:34:42.577" Title="Nth derivative of a function: I don't know where to start." Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" />
  <row Id="3629" PostTypeId="2" ParentId="3617" CreationDate="2010-08-30T11:03:54.400" Score="2" ViewCount="0" Body="&lt;p&gt;From a strategic point of view: given that mathematical terminology and formalism are not &lt;em&gt;totally&lt;/em&gt; standardised, it's worth at least reading through the chapter quickly, at least to stop yourself later having to refer back to what the various terms and symbols mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This advice goes for any book, I guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of content: no one but you is in a position to really decide whether you understand the material enough to skip the chapter.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-30T11:03:54.400" CommentCount="1" />
  <row Id="3630" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T11:08:40.027" Score="8" ViewCount="0" Body="&lt;p&gt;Split it into partial fractions then differentiate.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-08-30T11:08:40.027" CommentCount="2" />
  <row Id="3631" PostTypeId="2" ParentId="3609" CreationDate="2010-08-30T11:14:49.867" Score="3" ViewCount="0" Body="&lt;p&gt;Here's the flaw in the compactness argument you sketched. In the model you construct, the interpretation of $c$ will indeed be a nonstandard number whose Goodstein sequence does not halt in a standard number of steps. But Goodstein's theorem has a universal quantifier for the number of steps, so nonstandard &quot;Goodstein sequences&quot; are also permitted. That is, when the theorem says &quot;there is a finite sequence&quot;, in a particular model that &quot;finite&quot; sequence might have the length of a nonstandard number, which cannot be excluded by the sequence of sentences in your argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to find an extension of PA where Goodstein's theorem is provable is to add enough transfinite induction to PA to formalize the usual proof of Goodstein's theorem. The details of how to do that are not so bad, although they may take too much time for an elementary class. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-08-30T16:54:07.170" LastActivityDate="2010-08-30T16:54:07.170" CommentCount="1" />
  <row Id="3632" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T11:22:22.577" Score="6" ViewCount="0" Body="&lt;p&gt;To add to Derek's hint: you will have to show the validity of the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^k}{\mathrm{d}x^k}\frac1{x}=\frac{(-1)^k k!}{x^{k+1}}$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-30T11:22:22.577" CommentCount="1" />
  <row Id="3633" PostTypeId="1" CreationDate="2010-08-30T11:24:04.010" Score="9" ViewCount="543" Body="&lt;p&gt;There are some math quizzes like:&lt;blockquote&gt;&#xA;find a function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ &#xA;such that $\phi(\phi(x)) = f(x) \equiv x^2 + 1.$&#xA;&lt;/blockquote&gt;&#xA;If such $\phi$ exists (it does in this example), $\phi$ can be viewed as a &quot;square root&quot; of $f$ in the sense of function composition because $\phi\circ\phi = f$.  Is there a general theory on the mathematical properties of this kind of square roots? (For instance, for what $f$ will a real analytic $\phi$ exist?)&lt;/p&gt;&#xA;" OwnerUserId="1551" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-04T15:13:42.063" LastActivityDate="2011-01-02T20:44:33.180" Title="Square root of a function (in the sense of composition)" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3634" PostTypeId="2" ParentId="3627" CreationDate="2010-08-30T11:24:09.673" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, there is a very efficient way to compute the updated $F^{-1}$. Begin by writing $X$ in terms of its columns: $X = [x_1 x_2 \dots x_m]$. It follows that:&#xA;$$&#xA;XGX^T = \sum_{i=1}^{m} G_{ii} x_ix_i^T = \sum_{i\in S} G_{ii} x_ix_i^T&#xA;$$&#xA;where $S \subset \{1,\dots,m\}$ is the set of indices for which $G_{ii}$ is nonzero. Suppose that $S$ contains $r$ indices. Since $G$ is sparse by assumption, $r$ is much smaller than $m$. If we assemble the columns of $X$ corresponding to indices in $S$, we obtain a new matrix $Y$, such that:&#xA;$$&#xA;XGX^T = YHY^T&#xA;$$&#xA;where $Y$ is $m \times r$, and $H$ is a diagonal matrix whose diagonal consists of all the nonzero entries along the diagonal of $G$. Now apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Woodbury_matrix_identity&quot; rel=&quot;nofollow&quot;&gt;Woodbury matrix identity&lt;/a&gt;:&#xA;$$&#xA;(F+YHY^T)^{-1} = F^{-1} - F^{-1}Y(H^{-1}+Y^TF^{-1}Y)^{-1}Y^TF^{-1}&#xA;$$&#xA;Computing the left-hand side directly is costly, as it requires inverting an $m\times m$ matrix. However, since $F^{-1}$ has already been computed, we can compute the right-hand side efficiently; the only inverse required is that of an $r\times r$ matrix, and $r$ is much smaller than $m$.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-08-30T11:24:09.673" CommentCount="1" />
  <row Id="3635" PostTypeId="2" ParentId="3613" CreationDate="2010-08-30T11:34:52.533" Score="1" ViewCount="0" Body="&lt;p&gt;Let us consider the point $P_1(x_1,y_1)$ and a straight line $r$ with equation $Ax+By+C=0$.  The straight line $s$ perpendicular to $r$ that passes through $P_1$ is given by the following two equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Bx-Ay+k=0\qquad (1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Bx_1-Ay_1+k=0\qquad (2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference (1) - (2) gives the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B(x-x_1)-A(y-y_1)=0\qquad (2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the question $A=3,B=2,x_1=2,y_1=-3$ . . . &lt;/p&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2010-08-30T11:34:52.533" />
  <row Id="3636" PostTypeId="2" ParentId="3625" CreationDate="2010-08-30T12:03:05.397" Score="5" ViewCount="0" Body="&lt;p&gt;There's the obvious approach of proceeding based on taking the logarithm of both sides of your regression formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\;y=\ln\left(ax^b\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\;y=\ln\;a+b\ln\;x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be seen to be an expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=k+bu$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v=\ln\;y$, $u=\ln\;x$, and $k=\ln\;a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a linear regression in the variables $v$ and $u$ applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, formula numbers 16, 20, 27 and 28 of &lt;a href=&quot;http://mathworld.wolfram.com/LeastSquaresFitting.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; now apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the slope $b$ and the intercept $k$ of the linear transformation, $a$ is just the exponential of the intercept ($\exp\;k$), and $b$ is the slope of the transformed line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I note here the possible caveat of the fit parameters being biased towards the data with ordinates small in magnitude. The &lt;em&gt;rigorous&lt;/em&gt; way of going about it would be to treat the parameters from the linear regression as provisional and then apply a nonlinear least-squares algorithm like Levenberg-Marquardt to the data, using the parameters from the linear regression as a starting point. This may or may not be needed though; it really depends on the data you have.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll polish off comments I gave earlier: again, the problem with using the logarithm to linearize your nonlinear function is that it tends to overemphasize the errors in the small values of y. Remember that the assumption of linear least squares is that the abscissas are accurate, but the ordinates are contaminated by error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the $y_i$ are actually of the form $\hat{y}_i\pm\sigma_i$ where the $\hat{y}_i$ are the &quot;true values&quot; (presumably unknown), and the $\sigma_i$ are inherent uncertainties. If one takes the logarithms of the $y_i$, the uncertainties are also transformed, and we have to take this into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key formula is that if the $y_i$ are transformed by a function $f(y)$, the $\sigma_i$ are transformed according to $f&amp;#39;(y_i)\sigma_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case at hand, the objective function we now have to minimize is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a,b)=\sum_i{y_i^2\;(\ln\;y_i-\ln\;a-b\ln\;x_i)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have to modify the formulae for linear regression accordingly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=\sum_i y_i^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{x}=\frac{\displaystyle \sum_i y_i^2 \ln\;x_i}{m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\sum_i y_i^2 (\ln\;x_i-\bar{x})^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereupon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=\frac{\displaystyle \sum_i y_i^2\ln\;y_i (\ln\;x_i-\bar{x})}{t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\exp\left(\frac{\displaystyle \sum_i y_i^2\ln\;y_i}{m}-b\bar{x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These should be better provisional values for later polishing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It turns out that for a separable nonlinear fit (linear in one of the parameters), the NLLS problem greatly simplifies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the &lt;em&gt;actual&lt;/em&gt; quantity that we have to minimize is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(a,b)=\sum_i{(y_i-a x_i^b)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the gradient $\nabla F(a,b)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla F(a,b)=(2a\sum_i x_i^{2b}-2\sum_i x_i^b y_i\quad 2a\sum_i \ln\left(x_i\right) x_i^{2b}-2\sum_i \ln\left(x_i\right) x_i^b y_i)^T$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;equate both components to 0, and then eliminate the linear parameter $a$, we get the univariate nonlinear equation in $b$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\sum_i x_i^b y_i\right)\left(\sum_i \ln\left(x_i\right) x_i^{2b}\right)-\left(\sum_i x_i^{2b}\right)\left(\sum_i \ln\left(x_i\right) x_i^b y_i\right)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be attacked with standard techniques, e.g. the secant method or Newton's method. But, how do we start the iteration? Why, with the provisional $b$ we got from the (weighted) linear regression earlier!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having gotten $b$, it is now a simple matter to get $a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\left(\sum_i x_i^b y_i\right) / \left(\sum_i x_i^{2b}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you now have your parameters.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T12:07:21.440" LastActivityDate="2010-12-05T12:07:21.440" CommentCount="8" />
  <row Id="3637" PostTypeId="1" AcceptedAnswerId="3638" CreationDate="2010-08-30T12:27:46.913" Score="9" ViewCount="258" Body="&lt;p&gt;I know, fundamental group of GL(n,C) is isomorphic to Z. It's written &lt;a href=&quot;http://en.wikipedia.org/wiki/General_linear_group#Complex_case&quot; rel=&quot;nofollow&quot;&gt;in wikipedia&lt;/a&gt;. Actually, I've succeed in proving this, but my proof is two pages long and very technical. I want&lt;br&gt;&#xA;1. to find some better proofs of this fact (in order to compare to mine);&lt;br&gt;&#xA;2. to find some book or article, using which I can learn, how to calculate fundamental groups and, more generally, connectedness components of maps from one space to another;&lt;br&gt;&#xA;3. to find something for the reference, which I can use in order to learn, how to write this proofs nicely, using standard terminology.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-08-30T14:00:48.943" Title="Fundamental group of GL(n,C) is isomorphic to Z. How to learn to prove facts like this?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="3638" PostTypeId="2" ParentId="3637" CreationDate="2010-08-30T13:20:45.927" Score="11" ViewCount="0" Body="&lt;p&gt;The first thing you have to do is to note that the inclusion $U(n)\to\mathrm{GL}(n,\mathbb C)$ induces an isomorphism on the fundamental groups. This can be done by noting that a loop in $\mathrm{GL}(n,\mathbb C)$ can be deformed to one in $U(n)$ by performing the Gram-Schmidt procedure at each point of the loop, and checking that this can be done continuously and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, considering the beginning of the long exact sequence for the homotopy groups of the spaces appearing in the fibration $$U(n-1)\to U(n)\to S^{2n-1}$$ which arises from the transitive linear action of $U(n)$ on $S^{2n-1}\subseteq\mathbb C^{n}$ you can prove, by induction, that the inclusion $U(1)\to U(n)$ induces an isomorphism on fundamental groups.&#xA;Then you can explicitly describe $U(1)$ as a space homeomorphic to $S^1$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-30T13:34:42.150" LastActivityDate="2010-08-30T13:34:42.150" CommentCount="0" />
  <row Id="3639" PostTypeId="2" ParentId="3637" CreationDate="2010-08-30T13:34:03.417" Score="2" ViewCount="0" Body="&lt;p&gt;For 2. and 3. I can recommend Allen Hatcher's Algebraic Topology book which can be accessed free from the author's webpage: &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/AT/ATpage.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~hatcher/AT/ATpage.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1591" LastActivityDate="2010-08-30T13:34:03.417" />
  <row Id="3640" PostTypeId="2" ParentId="3637" CreationDate="2010-08-30T14:00:48.943" Score="6" ViewCount="0" Body="&lt;p&gt;The method Mariano discusses in his answer is absolutely the way that mathematicians compute fundamental groups (and also higher homotopy groups) of Lie groups.  Here I just want to mention how his first step applies in a more general context.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Concerning $\operatorname{GL}_n(\mathbb{C})$: the unitary group $U(n)$ is a maximal compact subgroup of $\operatorname{GL}_n(\mathbb{C})$, and moreover any maximal compact subgroup is conjugate to $U(n)$.  (This can be seen by considering Hermitian forms, c.f. e.g. Section 1 of &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter9.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/8410Chapter9.pdf&lt;/a&gt;.)  Moreover, the Gram-Schmidt process gives a &lt;a href=&quot;http://en.wikipedia.org/wiki/Deformation_retraction&quot; rel=&quot;nofollow&quot;&gt;deformation retraction&lt;/a&gt; from $\operatorname{GL}_n(\mathbb{C})$ to $U(n)$, hence these two spaces are homotopy equivalent.  And in fact even more is true: there exists a finite-dimensional Euclidean space $E$ such that $\operatorname{GL}_n(\mathbb{C})$ is homeomorphic to $U(n) \times E$: this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_decomposition#Square_matrix&quot; rel=&quot;nofollow&quot;&gt;QR decomposition&lt;/a&gt;.  Moreover:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Everything in 1) goes over verbatim for $\operatorname{GL}_n(\mathbb{R})$ with the unitary group $U(n)$ replaced by the orthogonal group $O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) For any &lt;a href=&quot;http://en.wikipedia.org/wiki/Reductive_group&quot; rel=&quot;nofollow&quot;&gt;reductive group&lt;/a&gt; $G$ over $\mathbb{R}$ or $\mathbb{C}$, there exists a maximal compact subgroup $K$, any two such are conjugate in $G$, and $G$ is homeomorphic to the product of $K$ with a finite-dimensional Euclidean space.  This last fact is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iwasawa_decomposition&quot; rel=&quot;nofollow&quot;&gt;Iwasawa decomposition&lt;/a&gt;, a far-reaching generalization of the QR-decomposition.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-08-30T14:00:48.943" />
  <row Id="3641" PostTypeId="2" ParentId="3583" CreationDate="2010-08-30T14:12:06.803" Score="3" ViewCount="0" Body="&lt;p&gt;The problem is not well posed, since you don't specify which woman leaves the train at each station. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Ross%E2%80%93Littlewood_paradox&quot; rel=&quot;nofollow&quot;&gt;Ross–Littlewood paradox&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-30T14:12:06.803" />
  <row Id="3643" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T14:36:11.403" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;\;$ Upon employing partial fractions it reduces to $\;\;\;\rm D^n \:\frac{1}{x+1}. \;\;$ Now employ the Taylor series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\;  f(t+x) = \sum_{k=0}^\infty \;\; D^n \: f(x) \; \frac{t^n}{n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and note that for this problem  we've  $\rm\; f(t+x) = \frac{1}{t+x+1}$ is a geometric series with known coefficients.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such &quot;generating function&quot; approaches often work smoothly even in much more complicated problems. Indeed, there is a very powerful &lt;strong&gt;Umbral calculus&lt;/strong&gt; that frequently succeeds in computing such closed form expressions, e.g. see Steven Roman: The Umbral Calculus. 1984. For example, below is a small sample of derivative formulas for the countless number of polynomial sequences amenable to umbral calculus analysis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|r|l|} &#xA;\hline&#xA;\rm Name &amp;amp; \rm Derivative \; formula \\&#xA;\hline\hline&#xA;\rm Laguerre  &amp;amp; \rm L_n^k(x) = (D+1)^{n+k}(-x)^n \\&#xA;\rm Exponential &amp;amp; \rm\;\; e_n(x) = e^{-x}(xD)^n e^x  \\&#xA;\rm Abel &amp;amp; \rm A_n^k(x) = x \; e^{-knD} x^{n-1} \\&#xA;\rm Hermite  &amp;amp; \rm H_n^k(x) = (-1)^n e^{x^2/(2n)} (kD)^n e^{-x^2/(2n)} \\&#xA;\rm Bernoulli &amp;amp; \rm B_n^k(x) = \left(\frac{D}{e^D-1}\right)^k  x^n \\&#xA;\rm Euler  &amp;amp; \rm E_n^k(x) = \left(\frac{2}{e^D+1}\right)^k  x^n \\&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-30T16:59:11.497" LastActivityDate="2010-08-30T16:59:11.497" CommentCount="1" />
  <row Id="3644" PostTypeId="1" CreationDate="2010-08-30T15:35:10.440" Score="1" ViewCount="122" Body="&lt;p&gt;Is there a easy way to compute the coefficients of the power series which represents $\frac{x - x^k + x^{k+1}}{1-2x + x^k - x^{k+1}}$. I am currently solving this by assuming it has the form $a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots$ and solving tedious recurrences obtained from $x - x^k + x^{k+1} = (1-2x + x^k - x^{k+1}) \times (a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: If my calculations are right, the above is the generating function of the number of compositions of an integer which do not involve a given integer $k$. &lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-08-30T17:40:49.340" Title="Computing the nth coefficient of the power series representing a given rational function" Tags="&lt;combinatorics&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="3645" PostTypeId="1" AcceptedAnswerId="3689" CreationDate="2010-08-30T16:00:36.980" Score="3" ViewCount="133" Body="&lt;p&gt;Being motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/3618/solving-an-equation-with-irrational-exponents&quot;&gt;this post&lt;/a&gt;, I was wondering if there is a proof (analogous to the case of Diophantine equations) that there is no general method for solving &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcendental_equation&quot; rel=&quot;nofollow&quot;&gt;transcendental equations&lt;/a&gt;?  It seems pretty clear, intuitively, that there can be no general method; but the only reason I feel strongly about that is because I can't conceive that transcendental equations have a general method, but Diophantine equations don't.  I was never able to understand the proof for the case of Diophantine equations, so I am not in a position to even know where to begin thinking about this.  Has any work been done on this problem?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T16:16:07.277" LastActivityDate="2010-08-31T04:32:55.990" Title="Is there a proof that there is no general method to solve transcendental equations?" Tags="&lt;analysis&gt;&lt;transcendence-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3646" PostTypeId="1" AcceptedAnswerId="3666" CreationDate="2010-08-30T16:20:00.250" Score="8" ViewCount="235" Body="&lt;p&gt;Since there is a 2-1 homomorphism from SU(2)xSU(2) to SO(4) there should be a way to recover the two SU(2) matrices given an SO(4) matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I could set this up as a system of equations using the map from above and solve for the coefficients of the SU(2) matrices.  However, I wonder if anyone has already done this or can point me to the forumlas.&lt;/p&gt;&#xA;" OwnerUserId="1596" LastActivityDate="2010-08-31T06:44:13.347" Title="Recovering the two SU(2) matrices from SO(4) matrix" Tags="&lt;matrices&gt;&lt;lie-groups&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="3647" PostTypeId="2" ParentId="3644" CreationDate="2010-08-30T16:42:50.167" Score="1" ViewCount="0" Body="&lt;p&gt;There is an easier way to obtain the recurrences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First rewrite in the more manageable form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle g(x) = \frac{1-x}{1-2x+x^k-x^{k+1}} - 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle f(x) = \frac{1-x}{1-2x+x^k-x^{k+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle f(x)(1-2x+x^k-x^{k+1}) = 1-x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Leibniz's formula for the $n^{th}$ derivative of a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{r=0}^{n} {n \choose r} {f^{(n-r)'}(x)(1-2x+x^{k}-x^{k+1})^{r'}} = (1-x)^{n'}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($r'$ is the $r^{th}$ derivative)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are interested in this when $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only terms that matter are on the LHS would be for $r=0,1,k,k+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of the power series can be gotten by dividing the $n^{th}$ derivative at $0$ by $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably also see all the above if you do a partial fraction reduction of the rational function.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-08-30T17:40:49.340" LastActivityDate="2010-08-30T17:40:49.340" CommentCount="2" />
  <row Id="3648" PostTypeId="2" ParentId="3644" CreationDate="2010-08-30T17:18:43.923" Score="2" ViewCount="0" Body="&lt;p&gt;Depends on what you want to do with them.  If you want to prove things about them, there are theorems which tell you what such sequences look like in terms of the reciprocals of the roots of the denominator, but if you can't solve for the roots explicitly these are not useful for computation (although they tell you the asymptotic behavior rather precisely).  I wrote &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;some notes&lt;/a&gt; on this and other topics in generating functions if you're interested (Section 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually want to compute them for large indices, there is a way to do it faster than just running the recurrence.  The idea is that for sufficiently large indices $n$ there exists a matrix $A$ and vectors $v, w$ such that $a_n = v^T A^n w$ and one can use binary exponentiation to compute $A^n$ when $n$ is large.  In fact one can take $A$ to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;companion matrix&lt;/a&gt; of the characteristic polynomial.  A special case of this idea includes the doubling formulas for the Fibonacci numbers.  For sequences with combinatorial interpretations it is often possible to choose $A$ to be the adjacency matrix of a graph and to choose particularly nice $v$ and $w$.  Stanley's &lt;em&gt;Enumerative Combinatorics&lt;/em&gt;, I think, discusses such issues; two keywords here are &quot;regular languages&quot; and &quot;finite state machines.&quot;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-30T17:18:43.923" CommentCount="2" />
  <row Id="3649" PostTypeId="1" AcceptedAnswerId="3653" CreationDate="2010-08-30T17:26:03.807" Score="2" ViewCount="125" Body="&lt;p&gt;Let $S$ be some finite set. Let $Sym(S)$ be the symmetric group on $S$. Let $A$ be the subset of permutations $\sigma$ that are such that there exist $a,b,c \in S$ such that $\sigma(a) = b, \sigma (b) = c$ and $\sigma (c) =a$ and for every other element of $S$, the action of $\sigma$ is the identity. In short, let $A$ be the set of cycles of length 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you show that the subgroup generated by $A$ is not the whole of $Sym(S)$?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-08-30T17:54:17.950" Title="How do you prove cycles of length 3 don't generate the whole symmetric group?" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="3650" PostTypeId="2" ParentId="3649" CreationDate="2010-08-30T17:28:27.637" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the parity.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-08-30T17:28:27.637" />
  <row Id="3651" PostTypeId="2" ParentId="3645" CreationDate="2010-08-30T17:29:13.200" Score="2" ViewCount="0" Body="&lt;p&gt;Intuition may be misleading here. In fact transcendental cases are often much easier than the integer Diophantine case. For example below is a table listing the known decidability results in various rings for Hilbert's tenth problem and the full first order theory, excerpted from Bjorn Poonen's interesting paper &lt;a href=&quot;http://math.mit.edu/~poonen/papers/aws2003.pdf&quot; rel=&quot;nofollow&quot;&gt;Hilbert's tenth problem over rings of number-theoretic interest&lt;/a&gt;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/yNEos.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-30T17:29:13.200" />
  <row Id="3652" PostTypeId="2" ParentId="3649" CreationDate="2010-08-30T17:30:48.270" Score="5" ViewCount="0" Body="&lt;p&gt;Consider the action of $S_n$ on the ring of polynomials $R=\mathbb R[x_1,\dots,x_n]$ given by permutation of the variables, and consider the element $$\Delta=\prod_{1\leq i&amp;lt;j\leq n}(x_i-x_j)\in R.$$ Then every $3$-cycle fixes $\Delta$, so the whole subgroup of $S_n$ generated by $3$-cycles acts trivially on $\Delta$. Yet there are elements of $S_n$, like all transpositions, which do &lt;em&gt;not&lt;/em&gt; act trivially on $\Delta$ but map it to its opposite. Therefore $3$-cycles do not generate $S_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can alternatively use the action of $S_n$ on $\mathbb R^n$, again by permutation but now of the coordinates. Then the matrix corresponding to each $3$-cycle has determinant $1$, so every element of the subgroup generated by $3$-cycles is acts on $\mathbb R^n$ through a matrix with determinant $1$. Yet there are elements in $S_n$ whose corresponding matrix is of determinant $-1$: the same conclusion as before follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BN:&lt;/strong&gt; The point of doing this in this way is to avoid having to check first that the parity of elements of $S_n$ is well-defined; all answers so far depend on doing this (I wonder if it is at all side-steppable...). In fact, this can be massaged easily into becoming a proof of that very same fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a more or less canonical example of why we want to study representations of groups. &lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-30T17:54:17.950" LastActivityDate="2010-08-30T17:54:17.950" />
  <row Id="3653" PostTypeId="2" ParentId="3649" CreationDate="2010-08-30T17:31:30.450" Score="5" ViewCount="0" Body="&lt;p&gt;Look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_of_a_permutation&quot; rel=&quot;nofollow&quot;&gt;sign&lt;/a&gt; of a permutation. If a permutation can be decomposed into a product of even number of transpositions (aka 2-cycles), we give it a sign of 1. Otherwise give it a sign of -1. It turns out that this is a well-defined notion. The group generated by the even permutations cannot cover the odd ones, thus cannot be the whole symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The group generated by the even permutations is usually called the alternating group.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-08-30T17:31:30.450" />
  <row Id="3655" PostTypeId="1" AcceptedAnswerId="3659" CreationDate="2010-08-30T19:01:29.603" Score="1" ViewCount="145" Body="&lt;p&gt;In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1.&#xA;Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-31T00:54:59.540" LastActivityDate="2010-08-31T13:24:47.480" Title="Units of M_2(Z)" Tags="&lt;linear-algebra&gt;&lt;ring-theory&gt;&lt;elementary&gt;" AnswerCount="4" CommentCount="9" />
  <row Id="3656" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T19:03:24.810" Score="1" ViewCount="0" Body="&lt;p&gt;According to the Binomial Theorem (or using the usual formula for the sum of a geometric series with initial term 1 and common ratio $x^2$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=\frac {x} {x^2-1} = -x \frac {1} {1 - x^2} = -x \left( 1 + x^2 + x^4 + \cdots + x^{2n} + \cdots \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= -x - x^3 - x^5 - \cdots - x^{2n+1} - \cdots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the right side converges absolutely for $|x^2| &amp;lt; 1$ you can differentiate it term by term, introducing a coefficient $(2n+1)(2n) \cdots (2n+1-k+1)$ for $x^{2n+1-k}$; in other words, the coefficient of $x^j$ is $(j+1)(j+2) \cdots (j+k)$.  Dividing the entire thing through by $k!$ gives a series you can easily relate to the binomial expansion of $( 1 - x^2 ) $ to a &lt;em&gt;negative&lt;/em&gt; integral value, yielding a closed form solution.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-30T19:03:24.810" />
  <row Id="3657" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:04:27.540" Score="4" ViewCount="0" Body="&lt;p&gt;The determinant of the inverse is the inverse of the determinant, whence the determinant must be an integral unit.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-30T19:04:27.540" CommentCount="1" />
  <row Id="3658" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:04:50.833" Score="4" ViewCount="0" Body="&lt;p&gt;The map $\det:M_2(\mathbb Z)\to\mathbb Z$ is multiplicative, and the determinant of the identity matrix is $1$. It follows from this that the determinant of any inversible element of $M_2(\mathbb Z)$ must be an inversible element of $\mathbb Z$. There are only two such elements in $\mathbb Z$, &lt;em&gt;v.g.&lt;/em&gt; $1$ and $-1$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-30T19:04:50.833" />
  <row Id="3659" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:20:35.763" Score="7" ViewCount="0" Body="&lt;p&gt;As others have pointed out, the key here is that the determinant of $M^{-1}$ has to be equal to $\frac{1}{det(M)}$ on the one hand, by the properties of the determinant, and also that since $M$ and  $M^{-1}$ both have integer coefficients then their determinants must be integers. So you need an integer $e=det(M)$ such that both $e$ and $\frac{1}{e}$ are integers, and the only possibility is $e=\pm 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't consider arguments using the determinant to be &quot;elementary&quot;, perhaps the following will do: note that $ad-bc$ must divide each of $a$, $b$, $c$, and $d$. Let $D=ad-bc$; we can then write $a=Da'$, $b=Db'$, $c=Dc'$, $d=Dd'$; then $D = ad-bc=D^2(a'd'-b'c')$, so $D=D^2(a'd'-b'c')$. Therefore, since $D\neq 0$, we must have $D(a'd'-b'c')=1$, so $D|1$, hence $D=ad-bc=\pm 1$, which is what you wanted to show.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-08-31T13:24:47.480" LastActivityDate="2010-08-31T13:24:47.480" CommentCount="4" />
  <row Id="3660" PostTypeId="2" ParentId="3628" CreationDate="2010-08-30T19:34:42.577" Score="4" ViewCount="0" Body="&lt;p&gt;Maybe we can add some more help -just in case you didn't succeed to find the answer yourself yet. Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{x}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be the splitting into partial fractions. (I'm too lazy to compute the coeffitients $A$ and $B$.) Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d}{dx} \frac{x}{x^2 - 1} = -\frac{A}{(x-1)^2} - \frac{B}{(x+1)^2}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating again,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d^2}{dx^2}\frac{x}{x^2 - 1}  = \frac{2A}{(x-1)^3} + \frac{2B}{(x+1)^3} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d^3}{dx^3} \frac{x}{x^2 - 1} = - \frac{3\cdot 2 A}{(x-1)^4} - \frac{3\cdot 2 B}{(x+1)^4} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And sure enough you can find the general pattern now, can't you? Then, use induction to prove your guess.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-08-30T19:34:42.577" CommentCount="1" />
  <row Id="3661" PostTypeId="2" ParentId="3655" CreationDate="2010-08-30T19:48:18.303" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;$ Multiplicative maps preserve units: $\rm\; MN = 1 \;\;\Rightarrow\;\; d(M)\:d(N) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\rm\;\; d(1) = 1\;$ via apply $\rm d$ to $1\cdot 1 = 1\:$ then cancel $\rm d(1)\ne 0,$ valid since $\mathbb Z$ has cancelation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-30T19:56:02.413" LastActivityDate="2010-08-30T19:56:02.413" />
  <row Id="3662" PostTypeId="2" ParentId="3646" CreationDate="2010-08-30T20:27:08.237" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a start to figuring out the formulas. The first thing is to understand the map $SU(2)\times SU(2)\to SO(4)$. Let $std$ denote the standard 2-dimensional complex representation of $SU(2)$. Then the external tensor product $V=std\otimes std$ is a natural 4-dimensional complex representation of $G=SU(2)\times SU(2)$. Concretely, it's given by the homomorphism $G\to GL(V)=GL_4(\mathbf{C})$ sending $(g,h)\in G$ to the &quot;Kronecker product&quot; $g\otimes h$, which is just the matrix of $g\otimes h$ with respect to the basis for $V=\mathbf{C}^2\otimes \mathbf{C}^2$ given by $e_1\otimes e_1, e_1\otimes e_2, e_2\otimes e_1, e_2\otimes e_2$, with $e_1,e_2$ the standard basis for $\mathbf{C}^2$. Now the standard representation $std$ actually lands in $SL_2(\mathbf{C})$, since $SU(2)\subset SL_2(\mathbf{C})$. But $SL_2(\mathbf{C})=Sp_2$, which means that the standard representation $std$ is self-dual, and thus $\mathbf{C}^2$ has an $SU(2)$-invariant, anti-symmetric bilinear form. Concretely this takes a pair of column vectors $v,w\in \mathbf{C}^2$ to the determinant of the matrix with columns $v,w$. Call this bilinear form $\omega:std\otimes  std\to \mathbf{C}$. Then the external tensor product $\omega\otimes \omega:V\otimes V\to \mathbf{C}$ gives a $G$-invariant \textit{symmetric} bilinear form on $V$. In fact, this is an $SL_2(\mathbf{C})\times SL_2(\mathbf{C})$-invariant symmetric bilinear form on $V$. So we've shown that $SL_2(\mathbf{C})\times SL_2(\mathbf{C})$ actually lands in $SO(q)(\mathbf{C})$ for a certain 4-dimensional quadratic space $(V,q)$. (Here $q$ is the quadratic form associated to the sym. bilinear form $B=\omega\otimes\omega$.) By general Lie theory, this implies that the maximal compact $G$ of $SL_2(\mathbf{C})\times SL_2(\mathbf{C})$ lands in the maximal compact $SO(q)(\mathbf{R})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we really want a homomorphism $SU(2)\times SU(2)\to SO(4)$. To do this we must diagonalize $q$, over the complex number. It's not difficult to check that with respect to the tensor product basis for $V$ we were using above, the quadratic form $q$ is equal to (maybe a multiple of) $wz-xy$. That is, if $v\in V$ has coordinates $(w,x,y,z)$ with respect to the $e_i\otimes e_j$ basis, then $q(v)=wz-xy$. (Note that this is the &quot;determinant&quot; quadratic form!) You can diagonalize this form over $\mathbf{C}$ with the change of basis matrix &lt;/p&gt;&#xA;&#xA;&lt;p&gt;M= &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1   i    0   0]&#xA;&#xA;[0   0    i  -1]&#xA;&#xA;[0   0    i   1]&#xA;&#xA;[1  -i    0   0]'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;unless I screwed up the arithmetic. So conjugating the representation $V$ by this matrix (thus not changing the isomorphism type of the repn), we obtain a homomorphism $SL_2(\mathbf{C})\times SL_2(\mathbf{C})\to SO(4,\mathbf{C})$ landing in the standard 4-dimensional special orthogonal group over the complex numbers. Now the maximal compact $G$ lands in the usual group $SO(4)$, and this gives the double cover you referred to in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, I think that the 2-1 map $SU(2)\times SU(2)\to SO(4)$ is given explicitly in matrices by taking the Kronecker product and then conjugating by $M$. If I didn't make an error, one should find in particular that the condition of being in $SU(2)$ forces a pair $(g,h)$ to have real matrix entries under this map! I haven't bothered to check this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so how does that actually let you find the 2 preimages of each rotation matrix in $SO(4)$? I guess it amounts to solving a whole bunch of nonlinear equations, so that's totally disgusting and impractical. To get a nice answer to your question, you need to analyze the structure of $SU(2)$ and of $SO(4)$. The latter is described thoroughly on the Wikipedia page for $SO(4)$. The former is quite simple and is well described in M. Artin's book &lt;em&gt;Algebra&lt;/em&gt;. EDIT: I've deleted a bunch of stuff which Jason observed in a comment is totally bogus. Fortunately Robin gave another better answer which explains how to do this using quaternions.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="1265" LastEditorDisplayName="" LastEditDate="2010-08-30T22:43:44.523" LastActivityDate="2010-08-30T22:43:44.523" CommentCount="5" />
  <row Id="3663" PostTypeId="1" AcceptedAnswerId="3664" CreationDate="2010-08-30T20:49:04.227" Score="0" ViewCount="107" Body="&lt;p&gt;I can think of the set of bounded, continuous functions from R -&gt; R as a group, with composition as addition of functions.  {In other words, this group has the rule that the composition of two elements in the group, f(x) and g(x), is their point-wise sum f(x) + g(x)}  Since this set is path connected, the group is continuous, i.e. it is a lie group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the lie algebra of this group?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to use this approach to find a relationship between Lie theory and Fourier series, but I realize this might not work.  So I have decided to just ask the question in this form.  If anyone knows of any connections between Lie theory and Fourier series however, I am interested to hear about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I realize I may have made some wrong/poorly stated assertions, please point these out in comments)&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-08-30T21:51:16.580" LastActivityDate="2010-08-30T21:51:16.580" Title="Lie algebra of the bounded continuous functions." Tags="&lt;analysis&gt;&lt;fourier-series&gt;&lt;lie-groups&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="3664" PostTypeId="2" ParentId="3663" CreationDate="2010-08-30T20:58:45.247" Score="3" ViewCount="0" Body="&lt;p&gt;Since your Lie group is infinite dimensional, one has to be careful about what exactly you mean by Lie algebra, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, your «Lie group of bounded continuous functions $\mathbb R\to\mathbb R$ under composition» is not really a group... You'd have to restrict your attention to the group of homeomorphisms (or something along that line) to actually get a group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last sentence of your first paragraph is a bit misguided. You could do much worse that read Warner's book, say, on differentiable manifolds and lie groups, or Lang's, if you like infinite dimensional situations.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-08-30T21:04:21.313" LastActivityDate="2010-08-30T21:04:21.313" CommentCount="6" />
  <row Id="3666" PostTypeId="2" ParentId="3646" CreationDate="2010-08-30T21:16:25.413" Score="11" ViewCount="0" Body="&lt;p&gt;One way to think of the 2-to-1 map from $SU(2)\times SU(2)$ is via&#xA;quaternions. Regard $\mathbb{R}^4$ as $\mathbb{H}$, the set of quaternions,&#xA;and $SU(2)$ as the group of unit (that is, norm $1$) quaternions.&#xA;Then $SU(2)\times SU(2)$ acts on $\mathbf{H}$ by rotations in the following way:&#xA;$$h\mapsto q_1 hq_2^{-1}$$&#xA;is a rotation of $\mathbb{R}^4=\mathbb{H}$ for $q_1$, $q_2\in SU(2)$.&#xA;Then $(-q_1,-q_2)$ represents the same rotation as $(q_1,q_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a rotation $\phi$ of $\mathbb{H}$. If $\phi$ fixes $1$,&#xA;it's effectively a rotation of the $\mathbb{R}^3$ spanned by $i$, $j$ and $k$&#xA;and is represented by the pair $(q,q)$ where $q=\cos(t/2)+h\sin(t/2)$&#xA;where $h=ai+bj+ck$ is a unit vector along the axis of rotation, and&#xA;$t$ is the angle of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, let $q'$ be the unit quaternion defined by $\phi(1)$.&#xA;Then $\psi:h\mapsto q'^{-1}\phi(1)$ is a rotation fixing $1$ so that one&#xA;can find a quaternion $q$ such that $(q,q)$ represents $\psi$.&#xA;Then $(q'q,q)$ represents $\phi$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-30T21:16:25.413" CommentCount="1" />
  <row Id="3667" PostTypeId="1" CreationDate="2010-08-30T21:33:12.253" Score="4" ViewCount="90" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3649/how-do-you-prove-cycles-of-length-3-dont-generate-the-whole-symmetric-group&quot;&gt;Why don't 3-cycles generate the symmetric group?&lt;/a&gt; was asked earlier today. The proof is essentially that 3-cycles are even permutations, and products of even permutations are even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: do the 3-cycles generate the alternating group? Similarly, do the $k$-cycles generate the alternating group when $k$ is odd?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And do the $k$-cycles generate the symmetric group when $k$ is even? I know that transpositions ($2$-cycles) generate the symmetric group.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-08-30T21:59:42.517" Title="What do all the $k$-cycles in $S_n$ generate?" Tags="&lt;permutations&gt;&lt;symmetric-groups&gt;" AnswerCount="2" />
  <row Id="3668" PostTypeId="1" AcceptedAnswerId="3674" CreationDate="2010-08-30T21:43:22.580" Score="7" ViewCount="498" Body="&lt;p&gt;What is the value of 1^i?&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-08-31T08:48:38.780" Title="What is the value of 1^i?" Tags="&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="3669" PostTypeId="2" ParentId="3668" CreationDate="2010-08-30T21:51:55.097" Score="1" ViewCount="0" Body="&lt;p&gt;it's 1. 1 to the power of anything is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I'll elaborate. In defining $a^b$, we know intuitively what to do in certain cases. When $a$ is a positive integer and $b$ is an integer, for example. This definition is easily extended to when $a$ is real and positive, and $b$ is a real number. In the case where $a$ is negative, or complex, we run into trouble... one way to get around this is to define the power in terms of the logarithm, as $a^b = e^{b \log a}$. Then, we have the issue that the logarithm is multivalued, so we could get different answers depending on which branch we choose. This is the approach discussed in Carl's solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not believe this proposed method applies for the case where $a$ is positive and real, which is the relevant case being discussed. In the case where $a$ is positive and real, $a^b$ is unambiguously defined as $a^b = e^{b \ln a}$, where $\ln a$ is the unique real number $x$ satisfying $e^x = a$. This definition works even in the case where $b$ is complex. So we can apply this to the original problem: $1^i = e^{i\log(1)} = e^0 = 1$ (Myke, you got my +1!). In fact, $1^z = 1$ for any complex $z$. This viewpoint is shared by &lt;a href=&quot;http://mathworld.wolfram.com/Power.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;, &lt;a href=&quot;http://tinyurl.com/25xjy9z&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Power_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, for what that's worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess at some level, it's a question of semantics and preference. Why not use multiple branches and all that jazz for the case where $a&gt;0$? Because (and this is my opinion), it's unnecessary, and doesn't fit with existing definitions. The exponential function $e^z$ is well-defined via a power series and I think everyone would agree that it is single-valued, even when $z$ is complex. It makes no sense to me that $a^z$ should be any different when $a$ is real and positive.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-08-31T08:48:38.780" LastActivityDate="2010-08-31T08:48:38.780" CommentCount="8" />
  <row Id="3672" PostTypeId="2" ParentId="3667" CreationDate="2010-08-30T21:59:03.680" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, k-cycles generate the symmetric group when k is even and alternating group, when k is odd. As you've said, for k=2 you know the answer. Suppose k&gt;2.&lt;br&gt;&#xA;(1,2,...,k)(k,...,3,1,2)=(1,3,2)&lt;br&gt;&#xA;Similarly you can get any 3-cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a is even element of Sn. Then, as you know, a is a product&lt;br&gt;&#xA;$(k_1,k_2)(k_3,k_4)...(k_{4l-1},k_{4l})$&lt;br&gt;&#xA;of 2l transpositions. But product of 2 transpositions can be written using 3-cycles:&lt;br&gt;&#xA;$(k_1,k_2)(k_2,k_3)=(k_1,k_2)(k_2,k_3)*(k_2,k_3)(k_3,k_4)$.&lt;br&gt;&#xA;Any of two products, separated by * in right hand side, is either 3-cycle or unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a is odd and k is even, than a multiplied by any k-cycle is even, so we can apply the previous algorithm to it.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-08-30T21:59:03.680" />
  <row Id="3673" PostTypeId="2" ParentId="3667" CreationDate="2010-08-30T21:59:42.517" Score="9" ViewCount="0" Body="&lt;p&gt;If $n\geq5$, then the only normal subgroups of the symmetric group $S_n$ are the trivial group, the alternating group and the symmetric group itself. Since the $k$-cycles form a full conjugacy class, it follows that the subgroup they generate is normal. This determines everything if $n \geq 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically: the $k$-cycles in $S_n$ generate the alternating group if $k$ is odd and $k \ne 1$; they generate the full symmetric group if $k$ is even.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2010-08-30T21:59:42.517" />
  <row Id="3674" PostTypeId="2" ParentId="3668" CreationDate="2010-08-30T22:04:48.420" Score="22" ViewCount="0" Body="&lt;p&gt;First, a concrete example of things that can happen with complex exponentiation if you aren't careful: $1 = e^{2\pi i}$, so we can naively try to compute $1^i = (e^{2\pi i})^i = e^{(2\pi i)i} = e^{-2\pi}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal moral of that example is that the value of $1^i$ depends on the branch of the complex logarithm that you use to compute the power. You may already know that $1=e^{0+2ki\pi}$ for every integer $k$, so there are many possible choices for $\log(1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook definition of complex exponentiation states that $1^i = e^{\log(1)i}$ where &quot;log&quot; is a branch of the complex logarithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $e^{(2ki\pi)i} = e^{-2k\pi}$. If you take $k = 0$, which corresponds to using the principal branch of the logarithm, you get an answer of $1^i = e^0 = 1$. If you take $k = 2$ as in the example above, using the fact that $1=e^{2\pi i}$, you get $1^i = e^{-2\pi}$. There are infinitely many possible values for $1^i$, corresponding to different branches of the complex logarithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusing point here is that the formula $1^x = 1$ is not part of the definition of complex exponentiation, although it is an immediate consequence of the definition of natural number exponentiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second point that can be confusing is that the function $e^z$ used above is really the complex exponential function $\exp(z)$, which is defined by a power series. Otherwise, the definition of complex exponentiation would be circular. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-08-31T02:10:10.303" LastActivityDate="2010-08-31T02:10:10.303" CommentCount="13" />
  <row Id="3675" PostTypeId="1" AcceptedAnswerId="3676" CreationDate="2010-08-30T22:14:44.757" Score="1" ViewCount="133" Body="&lt;p&gt;Let there be given some conditionally convergent infinite series S.  Then let R be some real number, and Q&lt;sub&gt;k&lt;/sub&gt; a rearrangement of S such that the sum is equal to R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Q&lt;sub&gt;k&lt;/sub&gt; unique?  In other words, is there some other rearrangement W&lt;sub&gt;k&lt;/sub&gt; for which the sum of S is also R?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is that Q&lt;sub&gt;k&lt;/sub&gt; is in fact unique, because the cardinality of the set of permutations of the natural numbers (i.e. the positions of each element of the series S), is the same as the cardinality of the real numbers; and therefore the mapping from the real numbers to the rearrangements should be 1-1.  Is this line of reasoning valid, or is this result well known?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-08-30T22:44:06.913" LastActivityDate="2010-09-01T05:48:20.317" Title="Question on Riemann's rearrangement theorem" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3676" PostTypeId="2" ParentId="3675" CreationDate="2010-08-30T22:40:40.413" Score="7" ViewCount="0" Body="&lt;p&gt;The line of reasoning is not valid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find a particular $Q_k$, then I can find infinitely many other rearrangements that give the same value:  Pick any finite subset $X$ of $\mathbb{N}$ and let $\sigma$ be any permutation of $X$.  Then the re-rearrangement where we start with $Q_k$ and just mix up the subscripts in $X$ according to $\sigma$ will yield the same sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that only rearranging  a finite number of terms doesn't change the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reasoning breaks down where you say &quot;and therefore the mapping should be 1-1&quot;.  Just because two sets have the same cardinality doesn't mean any function you think of should be 1-1.  Having the same cardinality just means &quot;there is some very specific choice of function which is 1-1 and onto&quot; - it says nothing at all about a generic function you pick between two sets.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-08-30T22:40:40.413" CommentCount="3" />
  <row Id="3677" PostTypeId="2" ParentId="3579" CreationDate="2010-08-30T22:49:34.330" Score="3" ViewCount="0" Body="&lt;p&gt;Well, I've made a full proof! Part 1 was solved &lt;a href=&quot;http://mathoverflow.net/questions/37097/37155#37155&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and Part 2 was solved &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Solution1.1.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 1:&lt;/strong&gt; Any integer $m$ which satisfies the original problem also satisfies $n^{m+1} \equiv n \bmod{m}$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $p$ be a prime dividing $m$. Then $\sum_{n=1}^mn^m\equiv1\pmod p$, so $(m/p)\sum_{n=1}^{p-1}n^m\equiv1\pmod p$, so $p^2$ doesn't divide $m$. Let $g$ be a primitive root mod $p$. Then $\sum_{n=1}^{p-1}n^m\equiv\sum_{r=0}^{p-2}g^{rm}$. That's a geometric series, it sums to $(1-g^{(p-1)m})/(1-g^m)$ which is zero mod $p$ - unless $g^m=1$, in which case it sums to $-1$ mod $p$. So we must have $p-1$ dividing $m$. Looking at  $n^{m+1}\equiv n\pmod m$ and letting $n=p$, we see that $p^2$ cannot divide $m$. Now looking mod $p$, we get $n^{m+1}\equiv n\pmod p$. This is equivalent to $m+1\equiv1\pmod{p-1}$ (if $a^x \equiv a^y \bmod{n}$, then $x \equiv y \bmod{\varphi(n)}$ by Euler's theorem, and $\varphi(p) = p-1$), that is, $p-1$ divides $m$, so any integer $n^{m+1} \equiv n \bmod{m}$ as $p-1|m$ for all $m$ if $p|m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 2:&lt;/strong&gt; There are only a finite amount of integers $m$ which satisfy $n^{m+1} \equiv n \bmod{m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Since $p^2$ does not divide $m$, we may let $m = p_1 \ldots p_r$ with $p_1 &amp;lt; p_2 &amp;lt; \ldots &amp;lt; p_r$, with $p_i$ prime; as $p-1|m$ for all $p|m$, we say that $p_i-1|p_1 \ldots p_{i-1}$ for $i = 1, \ldots, r$. If we take $i = 1$, this forces $p_1-1|1$, so if $r \ge 1$, $p_1 = 2$. If $i = 2$, $p_2-1|2$, so if $r \ge 2$, $p_2 = 3$. Continuing, if $r \ge 3$, then $(p_3-1)|p_1 p_2 = 6$, so $p_3 = 7$; if $r \ge 4$, $(p_4 - 1)|p_1 p_2 p_3 = 42$, so $p_4 = 43$, as the numbers $d+1$ are not prime for other divisors $d$ of 42 larger than 6. If $r \ge 5$, then $(p_5 -1)|p_1 p_2 p_3 p_4 = 1806$, but 1, 2, 6 and 42 are the only divisors of 1806 with $d+1$ prime, so $p_5$ cannot exist. Therefore, $r \le 4$ and $m \in \{1, 2, 6, 42, 1806\}$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-08-30T22:55:58.090" LastActivityDate="2010-08-30T22:55:58.090" CommentCount="8" />
  <row Id="3678" PostTypeId="2" ParentId="3675" CreationDate="2010-08-30T23:01:52.917" Score="2" ViewCount="0" Body="&lt;p&gt;A permutation $i \to \pi(i)$ preserves convergence of sums ($\Sigma a_{\pi(i)}$ converges whenever $\Sigma a_i$ does) if and only if it maps [1,N] to a bounded number of intervals of integers, with the bound allowed to depend on the permutation but not on N.  This places strong constraints on how much &quot;motion&quot; the permutation can create. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this respect a rearrangement has to be nearly trivial in order to never change the sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is considering the problem for all convergent series simultaneously.  A single sum might be compatible with additional permutations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: permutations satisfying the condition preserve the sum of the series and not only the convergence of the partial sums.  That is, $\Sigma a_i = \Sigma a_{\pi(i)}$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-01T05:48:20.317" LastActivityDate="2010-09-01T05:48:20.317" />
  <row Id="3679" PostTypeId="1" AcceptedAnswerId="3682" CreationDate="2010-08-31T00:05:43.577" Score="5" ViewCount="184" Body="&lt;p&gt;To be more precise, I am interested in knowing &lt;strong&gt;if the intuition that &lt;em&gt;a Euclidean zero vector does not have a particular direction&lt;/em&gt; is actually correct, and if there is a rigorous formulation that would back it up.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_vector#Null_vector&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's entries&lt;/a&gt; on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Null_vector&quot; rel=&quot;nofollow&quot;&gt;zero vector&lt;/a&gt; seems to agree with that intuition, but as always, one shouldn't blindly trust Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Null_vector&quot; rel=&quot;nofollow&quot;&gt;in one place&lt;/a&gt;, it is stated that a zero vector &quot;&lt;em&gt;is orthogonal to all other vectors with the same number of components,&lt;/em&gt;&quot; while &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner-product&quot; rel=&quot;nofollow&quot;&gt;in another&lt;/a&gt;, it is stated that &quot;&lt;em&gt;two vectors can be considered orthogonal if and only if their dot product is zero, and they have non-null length.&lt;/em&gt;&quot; Correct me if I am wrong, but these two statements contradict each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question popped up in my head when I heard someone arguing that since a Euclidean vector is defined as a geometric entity that has both a magnitude and direction, and since a zero vector is a vector with length 0, then it is only fitting for a zero vector to have a &quot;&lt;em&gt;direction 0.&lt;/em&gt;&quot; I am personally inclined to say that a zero vector does not have a particular direction, but as I said, I would like to know if there is a rigorous formulation that would lead to this conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to put the question in a less &quot;discussion-inducing&quot; form, &lt;strong&gt;is there an agreement on the direction of a Euclidean zero vector?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2010-08-31T22:32:49.823" Title="What is the Direction of a Zero (Null) Vector?" Tags="&lt;intuition&gt;&lt;euclidean-geometry&gt;&lt;definition&gt;" AnswerCount="2" />
  <row Id="3680" PostTypeId="1" AcceptedAnswerId="3681" CreationDate="2010-08-31T00:07:50.123" Score="3" ViewCount="80" Body="&lt;p&gt;Consider the following data points: 1, 1, 2, 3, 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that...&#xA;the average is the total of the numbers divided by the count of numbers in the set, the median is the central value based on location in the set, the mode is the value occurring most often in the set, and the midrange is highest and lowest values divided by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for the calculation of the average based on the &lt;em&gt;unique&lt;/em&gt; values found in the set? So... (1 + 2 + 3 + 4) / 4 = 2.5? And is there a use for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pardon a possibly elementary question, I'm a programmer but I'm not exactly a math-oriented person, and this has been bugging me recently.&lt;/p&gt;&#xA;" OwnerUserId="1603" LastActivityDate="2010-08-31T14:53:40.897" Title="Is there a name (and use) for an average based on the unique values of a set of data?" Tags="&lt;average&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3681" PostTypeId="2" ParentId="3680" CreationDate="2010-08-31T00:19:20.283" Score="1" ViewCount="0" Body="&lt;p&gt;From a practical viewpoint, I wouldn't think so.  For that example $ (1, 1, 2, 3, 4) $ your procedure gives $ 2.5 $.  If the data points had instead been $ (1, 1.01, 2, 3, 4) $, however, the result would jump to about $ 2.2 $.  The most common useful properties would seem to depend continuously upon the data, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I don't know the application for which you're using that kind of average, though, it may well turn out to be helpful.  In this case, I'd recommend simply spelling it out as you did here.&lt;/p&gt;&#xA;" OwnerUserId="336" LastEditorUserId="336" LastEditorDisplayName="" LastEditDate="2010-08-31T00:24:55.170" LastActivityDate="2010-08-31T00:24:55.170" CommentCount="2" />
  <row Id="3682" PostTypeId="2" ParentId="3679" CreationDate="2010-08-31T00:20:58.113" Score="6" ViewCount="0" Body="&lt;p&gt;The zero vector has no particular direction; this is consistent with the fact that it is orthogonal to every other vector.  (It doesn't really make sense to say it has &quot;&lt;em&gt;direction 0&lt;/em&gt;&quot;, since direction is not a magnitude; &quot;direction 0&quot; makes no more sense than &quot;direction 1&quot;&#xA;or &quot;direction 5.873&quot;.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could say that it points in every direction, but with zero magnitude,&#xA;since if you take &lt;em&gt;any&lt;/em&gt; vector and multiply it by zero, you get the zero vector.  &quot;Every&#xA;direction&quot; is the same as &quot;no particular direction&quot;; it's just a different way of phrasing&#xA;things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one flaw with the traditional description of a vector as being a pair consisting of&#xA;a magnitude and a direction:  For the zero vector, the magnitude is zero, but the direction&#xA;is arbitrary.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-08-31T00:20:58.113" CommentCount="1" />
  <row Id="3683" PostTypeId="2" ParentId="3668" CreationDate="2010-08-31T01:24:40.413" Score="0" ViewCount="0" Body="&lt;p&gt;$1^{i} = e^{log(1) i} = e^{0}=1$&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-08-31T01:24:40.413" />
  <row Id="3684" PostTypeId="1" AcceptedAnswerId="3686" CreationDate="2010-08-31T01:40:20.493" Score="2" ViewCount="145" Body="&lt;p&gt;I'm relatively new to ring theory so this is probably a simple question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its easy to see that a finite ring that is commutative and has no-zero divisors (i.e. an integral domain) must have multiplicative inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if we can rearrange these properties and always get implication or produce finite rings with only 2 of the above properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly my questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does a finite ring that is commutative with multiplicative inverses always have no-zero divisors?&#xA;(EDIT: this one is pretty easy too, let xy=0 and assume x not 0. then (x^-1)xy=(x^-1)0. Hence, y=0 as desired.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is a finite ring that has multiplicative inverses and no zero divisors always commutative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these are actually simple exercises, I'd just like a hint to get started with the proof or any counterexamples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-08-31T02:40:58.157" LastActivityDate="2010-08-31T12:04:27.237" Title="When does a finite ring become a finite field?" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;&lt;elementary&gt;" AnswerCount="2" />
  <row Id="3685" PostTypeId="2" ParentId="3684" CreationDate="2010-08-31T01:55:48.003" Score="1" ViewCount="0" Body="&lt;p&gt;You might want to look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springer.com/engineering/electronics/book/978-1-4020-7039-6&quot; rel=&quot;nofollow&quot;&gt;http://www.springer.com/engineering/electronics/book/978-1-4020-7039-6&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-31T01:55:48.003" CommentCount="1" />
  <row Id="3686" PostTypeId="2" ParentId="3684" CreationDate="2010-08-31T02:23:42.300" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;  1) If $\;\rm R\;$ is finite then $\;\rm x\to r\:x\;$ is onto iff 1-1, so $\;\rm R\;$ is a field iff $\;\rm R\;$ is a domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) is &lt;a href=&quot;http://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Wedderburn's little theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-08-31T02:23:42.300" CommentCount="1" />
  <row Id="3687" PostTypeId="2" ParentId="2372" CreationDate="2010-08-31T03:43:57.950" Score="1" ViewCount="0" Body="&lt;p&gt;There are 20!/4!5!8!3! arrangements of the 20 alphabets. Consider the action of $S_5$ on the arrangements by permuting the last 5 digits. You want to compute the number of orbits. You can then try to do it by &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's formula&lt;/a&gt;. It is not difficult to do it at least for this case, since if a permutation of $S_5$ fixes an arrangement, it means that each cycle of the permutation corresponds to a color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general case, I suspect that Polya's enumeration theorem can do it, though I'm not certain.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-08-31T03:43:57.950" CommentCount="1" />
  <row Id="3688" PostTypeId="1" CreationDate="2010-08-31T04:25:13.260" Score="1" ViewCount="134" Body="&lt;p&gt;Is anything known about these integrals? Textbook suggestions are welcome&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(n,p)=\int_{x=-0.5}^p \frac{n!}{x!(n-x)!} dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n&gt;0, p\le n+0.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, as n grows $2^n-f(n,n+1/2)$ seems to be positive and monotonically decreasing...how would I find the limit?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;ListPlot[Table[2^n - NIntegrate[n!/(x! (n - x)!), {x, -1/2, n + 1/2}], {n, Range[30]}], PlotRange -&gt; All]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/gamma-integral.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-01T06:16:37.947" LastActivityDate="2010-09-01T06:16:37.947" Title="Gamma integrals" Tags="&lt;gamma-function&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="3689" PostTypeId="2" ParentId="3645" CreationDate="2010-08-31T04:32:55.990" Score="2" ViewCount="0" Body="&lt;p&gt;Yes.  The problem is that the term &quot;transcendental function&quot; is absurdly general.  For example, pick a computable bijection between the integers and the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Turing_machine&quot; rel=&quot;nofollow&quot;&gt;Turing machines&lt;/a&gt; and consider the function $f : \mathbb{R} \to \mathbb{R}$ which is $0$ on the integers corresponding to halting Turing machines, $1$ on the integers corresponding to non-halting Turing machines, and smoothly interpolated in between (for example via &lt;a href=&quot;http://en.wikipedia.org/wiki/Bump_function&quot; rel=&quot;nofollow&quot;&gt;bump functions&lt;/a&gt;) such that if $x$ is not an integer then $0 &amp;lt; f(x) &amp;lt; 1$.  Then the problem of determining the solutions to $f(x) = 1$ is equivalent to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;nofollow&quot;&gt;halting problem&lt;/a&gt;.  (And $f$ is even &lt;em&gt;smooth&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get any reasonable answer to this question one must pick a much more specific definition of transcendental function.  One must also specify what it means to find a solution to an equation; is it enough to give an algorithm which will compute it to arbitrary accuracy, or must we get the answer in &quot;closed form&quot; (whatever that means)?  There are a lot of issues here.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T04:32:55.990" CommentCount="1" />
  <row Id="3690" PostTypeId="2" ParentId="2567" CreationDate="2010-08-31T04:42:53.600" Score="2" ViewCount="0" Body="&lt;p&gt;As Chandru1 stated tentatively, the same arguments apply for $0&amp;lt; p&amp;lt;1$ as for $1\leq p&amp;lt;\infty$, so it is no surprise that a separate proof is hard to find.  For example, an introduction to $L^p$ spaces on subsets of $\mathbb{R}^n$ can be found in Chapter 8 of &lt;em&gt;Measure and integral&lt;/em&gt;  by Wheeden and Zygmund.  A proof of separability is outlined for $1\leq p&amp;lt;\infty$ in &lt;a href=&quot;http://books.google.com/books?id=YDkDmQ_hdmcC&amp;amp;lpg=PP1&amp;amp;dq=wheeden%20zygmund%20measure&amp;amp;pg=PA133#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Theorem 8.15&lt;/a&gt;.  Part of Theorem 8.16 asserts separability for $0&amp;lt; p&amp;lt;1$, but for proof they simply refer to the proof of 8.15.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-08-31T04:42:53.600" CommentCount="2" />
  <row Id="3691" PostTypeId="2" ParentId="3688" CreationDate="2010-08-31T05:27:44.920" Score="1" ViewCount="0" Body="&lt;p&gt;Not a full solution, but maybe enough for you to get a handle on the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\Gamma (x+1) = x \Gamma (x)$ and $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x}$, one obtains&#xA;$$&#xA;\Gamma(x + 1) \Gamma(n - x + 1) = \frac{(-1)^n \pi}{\sin \pi x} \prod_{k=0}^{n} (x - k) .&#xA;$$&#xA;Thus,&#xA;$$&#xA;f(n, p) = \frac{(-1)^n n!}{\pi} \int_{-0.5}^p \frac{\sin \pi x}{\prod_{k=0}^n (x-k)} d x .&#xA;$$&#xA;Using partial fractions, you can now transform $f$ into a sum of values of the sine integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: Computing the sums yields&#xA;$$&#xA;f(n, n+\frac{1}{2}) = \frac{2}{\pi} \sum_{k = 0}^n \binom{n}{k} Si (\pi k + \frac{\pi}{2})&#xA;$$&#xA;Note, that $Si (x) = \frac{\pi}{2} + O(x^{- 2})$. However, I have no idea how to leverage this to find the limit.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-08-31T07:10:29.103" LastActivityDate="2010-08-31T07:10:29.103" CommentCount="1" />
  <row Id="3692" PostTypeId="1" CreationDate="2010-08-31T06:40:55.103" Score="2" ViewCount="98" Body="&lt;p&gt;i. If it rains next Sunday, then Andy will only go to the football game if Bob does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii. If Andy will only go to the foot ball game on Sunday if it rains, then Bob will definitely go to the Sunday football game regardless of the weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut feeling is they are not equivalent. I chose the following three predicates: r: It rains next Sunday, b: Bob goes to the game, a: Andy goes to the game. And then it seems like the statements would then be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i. $r \rightarrow (b \leftrightarrow a)$ &#xA;ii. $(a \leftrightarrow r) \rightarrow b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if I choose the case where it rains on Sunday, Bob goes to the game, but Andy doesn't, statement i is false, but statement ii is true. Have I translated these correctly?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-08-31T08:14:58.773" Title="Are these statements logically equivalent?" Tags="&lt;logic&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3693" PostTypeId="2" ParentId="3646" CreationDate="2010-08-31T06:44:13.347" Score="3" ViewCount="0" Body="&lt;p&gt;In terms of SU(2) = unit quaternions, the answer can be found in the sections &quot;Isoclinic decomposition&quot; and &quot;Relation to quaternions&quot; of &lt;a href=&quot;http://en.wikipedia.org/wiki/So%284%29#Isoclinic_decomposition&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article about SO(4)&lt;/a&gt;.&#xA;(This is equivalent to Robin Chapman's answer, I presume, but the formulas are a bit more explicit.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-08-31T06:44:13.347" />
  <row Id="3694" PostTypeId="2" ParentId="3692" CreationDate="2010-08-31T07:28:23.967" Score="1" ViewCount="0" Body="&lt;p&gt;I don't think you've interpreted the end of (i) correctly: &quot;Andy will only go to the football game if Bob does.&quot;  This is not an &lt;em&gt;if-and-only-if&lt;/em&gt; statement; it is only an &lt;em&gt;only-if&lt;/em&gt; statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the same issue in your interpretation of (ii).&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-08-31T07:28:23.967" CommentCount="1" />
  <row Id="3695" PostTypeId="2" ParentId="3692" CreationDate="2010-08-31T08:14:58.773" Score="0" ViewCount="0" Body="&lt;p&gt;if we set R = &quot;Saturday it will rain&quot;, A = &quot;Andy goes to football match&quot;, B = &quot;Bob goes to football match&quot;, then your statement (i) says for triples (R,A,B) that (Y,Y,Y) and (Y,N,N) are true, while (Y,Y,N) and (Y,N,Y) are false. Nothing can be said about the other four triplets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement (ii) is indeed different: the part &quot;Andy will only go to the foot ball game on Sunday if it rains&quot; says for couples (R,A) that (Y,Y) and (N,N) are true while (Y,N) and (N,Y) are false. But the statement &quot;If S then T&quot; is always true when S is false (Latins said &quot;ex falso quodlibet&quot;, from falsehood comes everything); therefore the triple (Y,N,Y) is true.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-31T08:14:58.773" CommentCount="3" />
  <row Id="3696" PostTypeId="2" ParentId="3679" CreationDate="2010-08-31T08:17:22.677" Score="2" ViewCount="0" Body="&lt;p&gt;Since adding the zero vector to any non-zero vector does not change the direction of the latter, it cannot have a proper direction because of the parallelogram rule :-)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-08-31T08:17:22.677" />
  <row Id="3697" PostTypeId="1" AcceptedAnswerId="3714" CreationDate="2010-08-31T08:52:38.733" Score="3" ViewCount="303" Body="&lt;p&gt;I tried to find some reference on the net but couldn't find a good one. What is the advantage of pivoting in LU decomposition over regular LU decomposition? Is it something to do with matrix singular or not?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-08-31T15:04:54.693" Title="pivoting in LU decomposition" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="3698" PostTypeId="1" AcceptedAnswerId="3718" CreationDate="2010-08-31T08:59:10.580" Score="5" ViewCount="308" Body="&lt;p&gt;I know that 1 is not a prime number because $1\cdot\mathbb Z=\mathbb Z$ is, by convention, not a prime ideal in the ring $\mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since $\mathbb Z$ is a domain, $0\cdot\mathbb Z=\{0\}$ &lt;em&gt;is&lt;/em&gt; a prime ideal in $\mathbb Z$. Isn't $(p)$ being a prime ideal the very definition of $p$ being a prime element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know that this would violate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem of Arithmetic&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Apparently the answer is that a prime element in a ring is, by convention a non-zero non-unit (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_element#Divisibility.2C_prime_and_irreducible_elements&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is strange because a prime ideal of a ring is, by convention, a proper ideal but not necessarily non-zero (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_ideal&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is now: Why do we make this awkward convention?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; @Downvoter: Please explain your downvote in order to help me to improve my question.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-31T15:52:01.147" LastActivityDate="2010-08-31T15:52:01.147" Title="Why doesn't 0 being a prime ideal in Z imply that 0 is a prime number?" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;definition&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="3" />
  <row Id="3700" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T09:55:49.203" Score="2" ViewCount="0" Body="&lt;p&gt;Pivoting is necessary to reduce the effect of rounding errors. In a world with exact arithmetic you wouldn't need pivoting. With a singular matrix it would just happen that you would not be able to proceed with computing the LU decomposition beyond a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since finite computers must approximate reals with finite numbers of bits, every computation introduces a rounding error, and pivoting helps us control the accumulation of these errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of an online reference but most books on numerical analysis discuss this. An advanced source is Golub and van Loan's &lt;em&gt;Matrix Computations.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-08-31T09:55:49.203" CommentCount="1" />
  <row Id="3701" PostTypeId="2" ParentId="3499" CreationDate="2010-08-31T09:57:25.443" Score="0" ViewCount="0" Body="&lt;p&gt;One can also see this article, &quot;S.S.Kim, Amer.Math.Monthly 106 (1999), 258-259&quot;.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-31T09:57:25.443" />
  <row Id="3702" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T10:23:34.393" Score="7" ViewCount="0" Body="&lt;p&gt;The matrix $\begin{bmatrix}0 &amp;amp; 1 \\ 1 &amp;amp; 0\end{bmatrix}$ is as far from singular as you can get, yet it cannot be factorized into an LU decomposition without pivoting.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-08-31T10:23:34.393" />
  <row Id="3704" PostTypeId="1" AcceptedAnswerId="3720" CreationDate="2010-08-31T10:41:40.720" Score="3" ViewCount="319" Body="&lt;p&gt;Let's take a look at the following integrals :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\displaystyle \int\limits_{0}^{1} \frac{\log{x}}{1+x} \ dx = -\frac{\pi^{2}}{12} = -2 \sum\limits_{n=1}^{\infty} \frac{1}{n^2}= \zeta(2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For $c&amp;lt;1$ $\displaystyle \int\limits_{0}^{\frac{\pi}{2}} \arcsin(c \cos{x}) \ dx = \frac{c}{1^2} + \frac{c}{3^2} + \frac{c}{5^2} + \cdots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many infinite series which can be represented by integrals. I would like to know whether such type of representation is always possible for an infinite series?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-31T11:09:32.490" LastActivityDate="2010-08-31T20:01:37.980" ClosedDate="2010-08-31T14:51:33.513" Title="Integral Representation of  Infinite series" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="3705" PostTypeId="1" CreationDate="2010-08-31T10:49:25.463" Score="3" ViewCount="101" Body="&lt;p&gt;Say we have a set of symmetric $n \times n$ matrices $M_i$ for $1 \leq i \leq k$, elements in $\mathbb{R}$.  Suppose that for every $\boldsymbol{\lambda} = (\lambda_1, \dots , \lambda_k) \in \mathbb{R}^k$ we have that the kernel of $M_{\boldsymbol{\lambda}} = \sum_i \lambda_i M_i$ is nontrivial.  Does it follow that there exists some nonzero $n$ vector $\textbf{v}$ with $M_i \textbf{v} = 0$ for all $i$?&lt;/p&gt;&#xA;" OwnerUserId="1607" LastEditorUserId="1607" LastEditorDisplayName="" LastEditDate="2010-08-31T11:51:07.087" LastActivityDate="2010-08-31T15:26:58.350" Title="Sums of a set of symmetric matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="3706" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T11:05:11.480" Score="1" ViewCount="0" Body="&lt;p&gt;Most of the invertible square matrices admit an $LU$ decomposition with $L$ a lower triangular matrix with 1’s onthe diagonal and $U$ is an upper triangular matrix (with arbitrary nonzero diagonal entries). Such an expression is easily seen to be unique. The set of invertible matrices admitting decomposition is open and dense in the group of n by n invertible matrices $GL(n)$. A matrix belongs to this set if and only if all its leading principal minors are nonzero as mentioned in the &lt;a href=&quot;http://en.wikipedia.org/wiki/LU_decomposition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;)&#xA;If the matrix does not belong to this set it still admits an $LU$ decomposition with pivoting $A = wLU$, with $L$ and $U$ as before and $w$ is a permutaion matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric origin of these constructions is given by means of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bruhat_decomposition&quot; rel=&quot;nofollow&quot;&gt;Bruhat cell decomposition&lt;/a&gt; of $GL(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$GL(n) = \coprod_{w \in W} BwB$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B$ is the subgroup of invertibe upper triangular matrices, $W$ is the symmetric group, and $w$ is a permutation matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $w=w_0$ is the longest element of the symmetric group, given by an antidiagonal matrix with a unit antidiagonal, then: $ B w_0 = w_0 w_0^{-1} B w_0 = w_0 N $, where $N$ is the group of upper triangular matrices. Thus the set of matrices&#xA;admitting the regular LU decomposition is identified with the cell corresponding to the longest element. This cell which is dense in $GL(n)$ is called the big (Schubert) cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When, the matrix belongs to the big cell , but some of the leading principal minors are very small with respect to others, then the $LU$ decomposition will result big matrix elements, because the determinants of the principal minors appear in the denominators of the matrix elements of the $LU$ decomposition. In this case, it is customary to perform pivoting to move this matrix to another cell, to reduce the scale difference among the leading principal minors and gain numerical stability.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-08-31T15:04:54.693" LastActivityDate="2010-08-31T15:04:54.693" />
  <row Id="3707" PostTypeId="2" ParentId="3543" CreationDate="2010-08-31T11:33:10.070" Score="0" ViewCount="0" Body="&lt;p&gt;Here i have found one more method for evaluvating integrals of the form $$ I= \frac{1}{(x^{2}+a^{2})\sqrt{x^{2}+b^{2}}} \ dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = \displaystyle \sqrt{\frac{x^{2}+b^{2}}{x^{2}+a^{2}}}$, then by differentiating one can see that $$ \frac{dy}{dx} = \frac{(a^{2}-b^{2})x}{(x^{2}+a^{2})^{3/2} (x^{2}+b^{2})^{1/2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $I$ becomes $$\frac{(x^{2}+a^{2})^{1/2}}{(a^{2}-b^{2})x} \ dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also $(x^{2}+a^{2})y^{2}=x^{2}+b^{2}$ so that we have $\displaystyle x^{2}= \frac{b^{2}-a^{2}y^{2}}{y^{2}-1}$ and $$x^{2}+a^{2} = \frac{b^{2}-a^{2}}{y^{2}-1}$$ So $I$ further reduces to $$ I = \frac{1}{a^{2}-b^{2}} \int \frac{\sqrt{b^{2}-a^{2}}}{\sqrt{b^{2}-a^{2}y^{2}}} \ dy = \frac{1}{a\sqrt{b^{2}-a^{2}}} \cos^{-1}\frac{ay}{b} \ (a &amp;lt; b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $a&gt;b$ one can rearrange $$I = \frac{1}{a^{2}-b^{2}} \int \frac{\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}y^{2}-b^{2}}} \ dy$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-08-31T17:10:32.993" LastActivityDate="2010-08-31T17:10:32.993" CommentCount="2" />
  <row Id="3708" PostTypeId="1" CreationDate="2010-08-31T11:41:49.857" Score="0" ViewCount="104" Body="&lt;p&gt;Suppose $f$ is a continuous function, then can we find all $f$'s which satisfy $$\int\limits_{0}^{1} (1-(f(x))^{a})^{1/b} - (1-(f(x))^{b})^{1/a} \ dx =0 \quad (a&gt;b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation : &lt;a href=&quot;http://math.stackexchange.com/questions/316/why-is-int-01-1-x71-5-1-x51-7-dx0/345#345&quot;&gt;http://math.stackexchange.com/questions/316/why-is-int-01-1-x71-5-1-x51-7-dx0/345#345&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-31T11:41:49.857" Title="Finding all $f$'s such that $\int\limits_{0}^{1} (1-(f(x))^{a})^{1/b} - (1-(f(x))^{b})^{1/a} \ dx =0$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" />
  <row Id="3709" PostTypeId="2" ParentId="3705" CreationDate="2010-08-31T11:42:13.560" Score="3" ViewCount="0" Body="&lt;p&gt;No. As a counterexample, consider the matrices&#xA;$M_1 = \pmatrix{1 &amp;amp; 0 \cr 0 &amp;amp; 0}$, $M_2 = \pmatrix{0 &amp;amp; 1 \cr 0 &amp;amp; 0}$. Any linear combination has a row of zeroes and therefore is not invertible, but the kernels of $M_1,M_2$ are disjoint (apart from 0).&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-08-31T11:42:13.560" CommentCount="1" />
  <row Id="3710" PostTypeId="1" AcceptedAnswerId="3711" CreationDate="2010-08-31T11:54:16.687" Score="2" ViewCount="125" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to get all combinations of a six digit number where each digit is unique. Its clear that this combination will not have a 0 at the start as it will not be a valid six digit number. Help me on how to attack this problem. &lt;/p&gt;&#xA;" OwnerUserId="1611" LastEditorUserId="1611" LastEditorDisplayName="" LastEditDate="2010-08-31T13:00:13.680" LastActivityDate="2010-09-24T11:58:01.933" Title="Help me out on permutation and combination" Tags="&lt;homework&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3711" PostTypeId="2" ParentId="3710" CreationDate="2010-08-31T11:58:27.207" Score="3" ViewCount="0" Body="&lt;p&gt;First, if you're talking about a six-digit number, the order of the digits matters, so it's generally not referred to as a combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider how many possibilities there are for the first digit.  As you said, it cannot be zero, or you wouldn't have a six-digit number.  Once there is a first digit, how many possibilities are there for the second digit?  Then for the third?  And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of six-digit numbers is the product of the various numbers of possibilities for each digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As the question feels like homework to me, this is more of a hint/outline than a complete answer.  If it actually is homework, please add the homework tag.  If it's not homework, or if this doesn't get you to an answer, please comment with more information.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-08-31T11:58:27.207" CommentCount="1" />
  <row Id="3713" PostTypeId="2" ParentId="3590" CreationDate="2010-08-31T12:06:38.267" Score="2" ViewCount="0" Body="&lt;p&gt;The Wikipedia article gives a &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin_transform&quot; rel=&quot;nofollow&quot;&gt; Mellin transform &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(s)\zeta(s) =\int_0^\infty\frac{x^{s-1}}{e^x-1} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#Reciprocal&quot; rel=&quot;nofollow&quot;&gt;Dirichlet series&lt;/a&gt; over the Möbius function gives the reciprocal &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we may write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(s) = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \int_0^\infty\frac{x^{s-1}}{e^x-1} dx .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds true for every complex number s with real part &amp;gt; 1. Now let's try to enlarge the domain of validity of this representation. Riemann showed (see the book of H. M. Edwards, Riemann Zeta Function, for the details) that modifying the contour gives a formula valid for all complex s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2\sin(\pi s)\Gamma(s)\zeta(s) = i \oint_C \frac{(-x)^{s-1}}{e^x-1}dx .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin(\pi s) \Gamma(s) = \frac{i}{2} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} &#xA;\oint_C \frac{(-x)^{s-1}}{e^x-1} dx . \qquad (*)  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this formula is again only valid for s with real part &amp;gt; 1 because of the use of the Dirichlet series. Wikipedia remarks: &amp;quot;The Riemann hypothesis is equivalent to the claim that [this representation of the reciprocal of the zeta function] is valid when the real part of s is greater than 1/2.&amp;quot; Thus a possible answer to my question is: &amp;quot;The representation (*) is valid for all s with real part &amp;gt; 1/2 if and only if the RH holds.&amp;quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps someone can elaborate further to give this relation a more geometric meaning? Where are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_hypothesis&quot; rel=&quot;nofollow&quot;&gt;non-trivial zeros&lt;/a&gt; of the zeta function to be spotted in this setup? &lt;/p&gt;&#xA;" OwnerUserId="1484" LastActivityDate="2010-08-31T12:06:38.267" CommentCount="3" />
  <row Id="3714" PostTypeId="2" ParentId="3697" CreationDate="2010-08-31T13:06:19.230" Score="2" ViewCount="0" Body="&lt;p&gt;To further generalize Rahul's answer, any matrix that has a singular leading block cannot have an LU decomposition. By allowing pivoting (or in matrix factorization terms, allowing the multiplication of your original matrix by an appropriate permutation matrix), all matrices admit an LU decomposition. This is the explanation for pivoting in exact arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In inexact arithmetic, the condition &quot;singular&quot; in the explanation above is replaced by the term &quot;ill-conditioned&quot;. Every matrix has an associated condition number, which is defined as the product of the norm of the matrix and the norm of its inverse. In attempting to proceed with the LU decomposition of a matrix with an ill-conditioned leading block, you will hit a point where you have to divide by a small number (resulting in a number that may be much larger in magnitude than the matrix's original entries), which causes all sorts of trouble in the succeeding additions/subtractions that have to be done to finish the LU decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By pivoting, we avoid (or at the very least delay) the onset of encountering numbers much larger than the entries of the original matrix, which is one way precision is lost in the operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one could make the objection that the coefficients are badly scaled: for instance, if you have two equations in two unknowns, and you multiply both sides of any of the two equations by a number much smaller or much larger than the original coefficients, the solution to the system is still the same, but attempting to perform LU decomposition on the transformed system can be disastrous. This is where concepts like &quot;scaled pivoting&quot; comes in, where relative instead of absolute magnitudes are taken into account in the selection of pivots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are applications where &quot;partial pivoting&quot; (swapping of rows) is not enough; rank determination of a matrix, for instance, requires &quot;complete pivoting&quot; (swapping of both rows and columns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, LU decomposition behaves much better with pivoting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have been intentionally vague in some parts; you would do well to read Golub and Van Loan, as already recommended by jmoy, or the books &quot;Matrix Decompositions&quot; by Stewart or &quot;Applied Numerical Linear Algebra&quot; by Demmel for more rigorous versions of my explanation.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-31T13:06:19.230" />
  <row Id="3715" PostTypeId="2" ParentId="3315" CreationDate="2010-08-31T13:10:23.783" Score="-1" ViewCount="0" Body="&lt;p&gt;Fake answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/iApFD.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2010-08-31T13:10:23.783" CommentCount="2" />
  <row Id="3716" PostTypeId="2" ParentId="3698" CreationDate="2010-08-31T13:21:57.803" Score="4" ViewCount="0" Body="&lt;p&gt;The reason one permits zero as a prime ideal is because it facilitates useful reductions. For example, in many ring theoretic problems involving an ideal $\rm\; I\;$, one can reduce to the case $\rm\;I = P\;$ prime, and then reduce to $\rm\;R/P\;$, thus reducing to the case when the ring is a domain. In this case one simply says that one can factor out $\rm\; P\;$, so wlog assume $\rm\; P = 0\;$ is prime, so the ring is a domain. For example, below is an excerpt from Kaplansky's classic textbook &quot;Commutative Rings&quot;, section 1-3: G-Ideals, Hilbert Rings, and the Nullstellensatz. I've explicitly emphasized the reductions (&lt;strong&gt;reduce to...&lt;/strong&gt;) so that you can see them clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have solid evidence for the utility of the convention that the zero ideal is prime.&#xA;So why don't we adopt the same convention for the unit ideal 1, or equivalently, why don't we permit the zero ring as a domain?  There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing 1 = 0. More importantly, it implies that the  unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\;\rm R = 0\;)$ to the many arguments involving units and unit groups. For a more general perspective it's worth emphasizing that the usual rules for equational logic are not complete for &lt;strong&gt;empty structures&lt;/strong&gt; so that is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/3e8fac43081fcbf8&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; for details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the promised Kaplansky excerpt on reduction to domains by factoring out prime ideals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\rm\; I\;$ be any ideal in a ring $\rm\; R\;$. We write $\rm\; R^*\;$ for the quotient ring $\rm\; R/I\;$. &#xA;In the polynomial ring $\rm\; R[x]\;$ there is a smallest extension $\rm\; IR[x]\;$ of $\rm\; I\;$, &#xA;The quotient ring $\rm\; R[x]/IR[x]\;$ is in a natural way isomorphic to $\rm\; R^*[x].\;$ &#xA;In treating many problems, we can in this way &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; I = 0\;$, &#xA;and we shall often do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 28.&lt;/strong&gt; $\;$ Let $\rm\; M\;$ be a maximal ideal in $\rm\; R[x]\;$ and suppose that &#xA;the contraction $\rm\; M \cap R = N\;$ is maximal in $\rm\; R\;$, Then $\rm\; M\;$ can be generated &#xA;by $\rm\; N\;$ and one more element $\rm\; f\;$. We can select $\rm\; f\;$ to be a monic polynomial &#xA;which maps mod $\rm\; N\;$ into an irreducible polynomial over the field $\rm\; R/N\;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\;$ We can &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; N = 0,\;$ i. e., $\rm\; R\;$ a field, and then &#xA;the statement is immediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM 31.&lt;/strong&gt; $\;$ A commutative ring $\rm\; R\;\;$ is a Hilbert ring if and only if &#xA;the polynomial ring $\rm\; R[x] \;\;$ is a Hilbert ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\;$ If $\rm\; \;\rm R[x]\;$ is a Hilbert ring, so is its homomorphic image $\rm\; R\;$. &#xA;Conversely, assume that $\rm\; R\;$ is a Hilbert ring. Take a G-ideal $\rm\; Q\;$ in &#xA;$\rm\; R[x]\;$; we must prove that $\rm\; Q\;$ is maximal. Let $\rm\; P = Q \cap R\;$; we can &lt;strong&gt;reduce the problem to the case&lt;/strong&gt; $\rm\; P = 0\;$, which, incidentally, makes $\rm\; R\;$ a domain. &#xA;Let $\rm\; u\;$ be the image of $\rm\; x\;$ in the natural homomorphism $\rm\; R[x] \to R[x]/Q\;$. &#xA;Then $\rm\; R[u]\;$ is a G-domain. By Theorem 23, u is algebraic over R and R &#xA;is a G-domain. Since R is both a G-domain and a Hilbert ring, R is a &#xA;field. But this makes $\rm\; R[u] = R[x]/Q\;$ a field, proving $\rm\; Q\;$ to e maximal.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-31T15:08:46.357" LastActivityDate="2010-08-31T15:08:46.357" CommentCount="1" />
  <row Id="3717" PostTypeId="2" ParentId="3698" CreationDate="2010-08-31T13:41:11.140" Score="4" ViewCount="0" Body="&lt;p&gt;Generally we make nice conventions because they make the statements of theorems nice.  The theorem relevant to prime ideals is that $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.  The theorem relevant to prime elements is prime factorization (when it holds).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two concepts &lt;em&gt;almost&lt;/em&gt; coincide for principal ideals, but we must distinguish between the generic point $(0)$ and closed points, and there are good reasons for doing this.  (The zero ideal, for example, can't occur in the factorization of a nonzero ideal in a Dedekind domain.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T13:41:11.140" CommentCount="3" />
  <row Id="3718" PostTypeId="2" ParentId="3698" CreationDate="2010-08-31T14:18:57.137" Score="8" ViewCount="0" Body="&lt;p&gt;You have a point here: absolutely we want to count $(0)$ as a prime ideal in $\mathbb{Z}$ -- because $\mathbb{Z}$ is an integral domain -- whereas we do &lt;em&gt;not&lt;/em&gt; want to count $(1)$ as being a prime ideal -- because the zero ring is not an integral domain (which, to me, is much more a true fact than a convention: e.g., every integral domain has a field of fractions, and the zero ring does not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we do not want to call $0$ a prime element because, in practice, we never want to include $0$ in divisibility arguments.  Another way to say this is that we generally want to study factorization in integral domains, but once we have specified that a commutative ring $R$ is a domain, we know all there is to know about factoring $0$: $0 = x_1 \cdots x_n$ iff at least one $x_i = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one way to make this &quot;ignoring $0$&quot; convention look more natural: the notions of factorization, prime element, irreducible element, and so forth in an integral domain $R$ depend entirely on the multiplicative structure of $R$.  Thus we can think of factorization questions as taking place in the &lt;strong&gt;cancellative monoid&lt;/strong&gt; $(R \setminus 0,\cdot)$.  (Cancellative means: if $x \cdot y = x \cdot z$, then $y = z$.)  In this context it is natural to exclude zero, because otherwise the monoid would not be cancellative.  Contemporary algebraists often think about factorization as a property of monoids rather than integral domains &lt;em&gt;per se&lt;/em&gt;.  For a little more information about this, see e.g. Section 4.1 of &lt;a href=&quot;http://math.uga.edu/~pete/factorization2010.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/factorization2010.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-08-31T14:28:59.467" LastActivityDate="2010-08-31T14:28:59.467" CommentCount="6" />
  <row Id="3720" PostTypeId="2" ParentId="3704" CreationDate="2010-08-31T14:40:43.093" Score="10" ViewCount="0" Body="&lt;p&gt;There is a &lt;em&gt;very powerful&lt;/em&gt; calculus of &lt;strong&gt;multidimensional residues&lt;/strong&gt; that accomplishes what you seek, see for example the book G. P. Egorychev. Integral representation and computation of combinatorial sums. AMS, Transl. of Math. Monogr. v. 59 Providence 2nd ed. 1989 - which can be located online via gigapedia.com. Below are two illustrative examples of Egorychev's &quot;method of coefficients&quot; excerpted from the survey by Egorychev and Zima in volume 5 of Hazewinkel's Handbook of Algebra (also available online via gigapedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TiDQB.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/xTpHh.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/JMzAV.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/MAZBo.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/xOxhy.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-08-31T20:01:37.980" LastActivityDate="2010-08-31T20:01:37.980" CommentCount="0" />
  <row Id="3721" PostTypeId="2" ParentId="3680" CreationDate="2010-08-31T14:53:40.897" Score="1" ViewCount="0" Body="&lt;p&gt;You could conceive of your result as being a weighted average.  In a batch of data consisting of $n_i \ge 1$ instances of $x_i$, $1 \le i \le n$, your average is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{n} x_i / n = \sum_{i=1}^{n} \frac {1}{n_i} (n_i x_i ) / n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \sum_{i=1}^{n} \sum_{j=1}^{n_j} \left( \frac {1}{n_i} x_i \right) /  \sum_{i=1}^{n} \sum_{j=1}^{n_j} \frac {1}{n_i}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;exhibiting the weights as $\frac {1} {n_i}$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T14:53:40.897" />
  <row Id="3722" PostTypeId="2" ParentId="3705" CreationDate="2010-08-31T15:26:58.350" Score="2" ViewCount="0" Body="&lt;p&gt;Still no.  Counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = \lambda M_1 + \mu M_2 = \pmatrix{0 &amp;amp; \mu &amp;amp; \mu \cr \mu &amp;amp; \lambda &amp;amp; 0 \cr \mu &amp;amp; 0 &amp;amp; -\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $\det(M)\equiv 0$ for all $\lambda$ and $\mu$.  However, the only $(x,y,z)$ that satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M\pmatrix{x\cr y\cr z} = \pmatrix{0 &amp;amp; \mu &amp;amp; \mu \cr \mu &amp;amp; \lambda &amp;amp; 0 \cr \mu &amp;amp; 0 &amp;amp; -\lambda}\pmatrix{x\cr y\cr z} = \pmatrix{\mu(y+z)\cr \mu x + \lambda y\cr \mu x - \lambda z} = \mathbf{0}\quad \forall \lambda, \mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the zero vector.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2010-08-31T15:26:58.350" CommentCount="0" />
  <row Id="3723" PostTypeId="1" AcceptedAnswerId="3836" CreationDate="2010-08-31T15:31:40.803" Score="4" ViewCount="285" Body="&lt;p&gt;Consider the piecewise constant function $\psi:I=[-1,1] \rightarrow \mathbb{R}$ given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x) = \begin{cases} \psi_1 &amp;amp; x \leq 0, \ \psi_2 &amp;amp; x &gt; 0 \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constants $\psi_1, \psi_2 \in \mathbb{R}$.  I would like to evaluate the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_I \delta(x) \psi(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta$ is the Dirac delta distribution centered at $x=0$.  Of course, if we think about distributions in the usual way then you might say that $\delta(\psi) = \langle \delta, \psi \rangle = \psi(0) = \psi_1.$  But then the result depends on a fairly arbitrary choice when defining $\psi$: should the left- or right- half of the interval be closed?  This question doesn't seem to have a meaningful answer when dealing with a problem that arises from a physical system (say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, consider a family of distributions $\phi_\epsilon(x)$ such that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\phi_\epsilon(x)=\phi_\epsilon(-x)$,&lt;/li&gt;&#xA;&lt;li&gt;$\int_I \phi_\epsilon(x)=1$ for all $\epsilon$, and&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{\epsilon \rightarrow 0} \phi_\epsilon = \delta$,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i.e., any family of even distributions with unit mass that approaches the Dirac delta distribution as $\epsilon$ approaches zero.  (For instance, you could use the family of Gaussians $\phi_\epsilon(x) = \frac{1}{\epsilon\sqrt{\pi}}e^{-x^2/\epsilon^2}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can now think of my integral as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{\epsilon \rightarrow 0} \int_I \phi_\epsilon(x) \psi(x) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some $\epsilon &gt; 0$, the integral inside the limit can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(\epsilon) = \psi_1 \int_{-1}^0 \phi_\epsilon(x) dx + \psi_2 \int_0^1 \phi_\epsilon(x) dx = \frac{1}{2}\left( \psi_1 + \psi_2 \right) = \bar{\psi},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\bar{\psi}$ is the mean of the constant values.  Can we say, then, that $\lim_{\epsilon \rightarrow 0} u(\epsilon) = \bar{\psi}$?  It would seem so: for any $\mu &gt; 0$ there exists an $\epsilon_0$ such that $\epsilon &amp;lt; \epsilon_0$ implies $|u(\epsilon) - \bar{\psi}|&amp;lt;\mu$ (namely, $\epsilon$ is any positive constant!).  But clearly I've got a problem somewhere, because $\bar{\psi} \ne \psi_1$, i.e., this result does not agree with my earlier interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the right thing to do here?  The latter answer ($\bar{\psi}$) agrees more with my &quot;physical&quot; intuition (because it's invariant with respect to deciding which half-interval is open), but I'm concerned about rigor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Since the problem as stated is not well-posed ($\delta$ cannot be evaluated on discontinuous functions), let me give some motivation.  Imagine that I have a pair of piecewise linear functions $f,g:I^2 \rightarrow \mathbb{R}$, which are again discontinuous only at $x=0$.  I would like to integrate the &lt;em&gt;wedge product&lt;/em&gt; of $df$ and $dg$ over the domain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{I^2} df \wedge dg = \int_I \int_I \frac{\partial f}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial f}{\partial y}\frac{\partial g}{\partial x} dx dy.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider just the first term $(\partial f/\partial x)(\partial g/\partial y)$ and consider just the inner integral $\int_I \cdot dx$.  We now have (almost) the original problem: $\partial f/\partial x$ can be thought of as a $\delta$ (plus a piecewise constant), and $\partial g/\partial y$ is simply piecewise constant along the $x$-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem could be restated as: how do I integrate the wedge product $df \wedge dg$ of piecewise linear 0-forms $f$ and $g$ defined over a planar region?  Formally this problem may again be ill-posed, yet it is a real problem that comes up in the context of finite element analysis where basis functions are nonsmooth or even discontinuous.&lt;/p&gt;&#xA;" OwnerUserId="1612" LastEditorUserId="1612" LastEditorDisplayName="" LastEditDate="2010-08-31T17:59:01.973" LastActivityDate="2010-09-02T01:59:08.120" Title="Integrate product of Dirac delta and discontinuous function?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3724" PostTypeId="1" AcceptedAnswerId="3732" CreationDate="2010-08-31T16:05:51.177" Score="4" ViewCount="531" Body="&lt;p&gt;Suppose you have 100 coins. 96 of them are heavy and 4 of them are light.  Nothing is known regarding the proportion of their weights. You want to find at least one genuine (heavy) coin. You are allowed to use a weight balance twice. How do you find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavy coins all have the same weight; same for the light coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight balance compares the weight of two sides on the balance instead of giving numerical measurement of weights.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-02T01:25:52.793" LastActivityDate="2010-10-11T20:53:51.037" Title="Finding the Heavy Coin by weighing twice" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;puzzle&gt;&lt;contest-math&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="2" />
  <row Id="3725" PostTypeId="1" CreationDate="2010-08-31T17:24:32.280" Score="-1" ViewCount="145" Body="&lt;p&gt;What is the result of 2/3 + 4/5 = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I add fractions?&lt;/p&gt;&#xA;" OwnerUserId="1614" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-08-31T17:38:35.860" LastActivityDate="2010-08-31T17:38:35.860" ClosedDate="2010-08-31T20:43:49.090" Title="How do I add fractions?" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3726" PostTypeId="2" ParentId="3723" CreationDate="2010-08-31T17:25:28.610" Score="0" ViewCount="0" Body="&lt;p&gt;It seems your question contains its own answer.  $\delta$ is not defined as evaluation-at-a-point &lt;em&gt;except&lt;/em&gt; for continuous functions.  (The inner product, strictly speaking, is defined in $L^p$ spaces of &lt;em&gt;equivalence classes&lt;/em&gt; of functions and the &quot;value&quot; of such a class at a point is undefined unless you limit the evaluation to continuous functions.)  Your second method is close to one standard method of creating (and calculating with) $\delta$ as a weak limit of integrals.  (Another standard method, based on integration by parts, requires the trial function to be differentiable, so applying that method won't work here.)  Thus, you have clearly demonstrated why evaluate-at-a-point is an unsuitable definition of $\delta$ for discontinuous functions.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T17:25:28.610" />
  <row Id="3727" PostTypeId="2" ParentId="3725" CreationDate="2010-08-31T17:27:35.813" Score="2" ViewCount="0" Body="&lt;p&gt;$ \frac{2}{3} + \frac{4}{5} = \frac{2 \times 5 + 3 \times 4}{15}$ and in general if you want to add fractions of the form $\frac{a}{b} + \frac{c}{d}$ this will equal to $$\frac{ad+bc}{bd}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-08-31T17:27:35.813" />
  <row Id="3728" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:28:58.667" Score="2" ViewCount="0" Body="&lt;p&gt;If you pay me 95 of the coins in advance I'll show you how ;-).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T17:28:58.667" CommentCount="6" />
  <row Id="3729" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:45:48.853" Score="2" ViewCount="0" Body="&lt;p&gt;To avoid wasting readers' time when posting puzzles, it is useful (and will improve the upvote/downvote ratio) to provide:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;an indication of the difficulty level&lt;/em&gt;. Was this puzzle in a math-is-fun book for children, or is it a generalization of a problem from the final round of a national olympiad?  Having schoolchildren attempt a question that (they were not told) is the $n=100$ specialization of a combinatorial lemma from a research paper, is a waste of time.  Having mathematicians spend time on (what might look like) a coin weighing puzzle from the literature, but is really an easy task the high school students could solve with some patience, is a waste of resources.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;the source, if known&lt;/em&gt;.  This helps those who want to evaluate the difficulty before taking time to attempt a solution, or who would like to look up hints or answers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;indicators of whether the problem is correct or solvable as stated&lt;/em&gt;.  Problems from a book, a competition or a journal are likelier to be correct and unambiguously formulated.  Questions invented, circulated and (believed to be) solved among friends or students are more likely to have pitfalls that were missed in setting or solving the problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;clear definitions&lt;/em&gt;.  In this case, coin weighing problems involve at least two very different interpretations of &quot;use a weight balance&quot;.  Weighing can mean comparison or numerical measurement and what is possible is completely different in those settings.&#xA;Not specifying which is meant can waste the time of users who try to solve the problem in the wrong interpretation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-08-31T17:51:14.767" LastActivityDate="2010-08-31T17:51:14.767" CommentCount="4" />
  <row Id="3730" PostTypeId="2" ParentId="2968" CreationDate="2010-08-31T17:49:19.060" Score="-1" ViewCount="0" Body="&lt;p&gt;It seems to me that this puzzle as stated can be resolved by a public counter that counts down from $n$ to $n - k$ in even time steps, each of which is longer than every player's reaction time.  Each player intends to stop the counter when it displays a value equal to or lower than the player's own card; whoever first stops the counter is the winner.  This does not seem to reveal anything that is not already available from $n$, $k$, and the value of the winner's card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is impracticable when $n &gt;&gt; k$, but even then one could have the time count down in sufficiently large increments to be physically feasible.  When the number of steps is substantially larger than $k$ this would still have a good chance of identifying the winner.  Apparent ties in this procedure could be resolved by a quick run-off using the same technology (keeping the players in the runoff anonymous), giving an almost certain procedure to find a unique winner correctly.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-08-31T17:49:19.060" CommentCount="3" />
  <row Id="3731" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:49:19.420" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;My assumptions:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Your heavy coins all have the same weight; same for the light coins.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The weight balance compares the weight of two sides on the balance instead of giving numerical measurement of weights.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method:&lt;/strong&gt;&#xA;Divide the 100 coins into 2 groups A and B, each comprises 49 coins. The remaining 2 coins are labelled C and D.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Weight A and B.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1.1 A &gt; B (which also handles symmetrically B &gt; A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 24 + 1 coins, and weigh 24 vs 24.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.1 If the two sides are equal, all of them are genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.2 Otherwise the heavier side contains all genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2 A = B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 25 coins, add C to the 24 side, and weigh 25 vs 25. Call the side containing C the C-side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.1 If the C side is heavier, then C is genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.2 If the C side is equal to the other side, then C side minus C are all genuine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.3 If the C side is lighter, then the other side contains all genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation:&lt;/strong&gt;&#xA;Divide the 100 coins into 2 groups A and B, each of 49 coins, and label the remaining 2 coins as C and D.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Weight A and B.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1.1 A &gt; B, it means that B has more fake coins than A. Clearly this means that A has either 0 or 1 fake coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 24 + 1 coin, and weigh 24 coins against the other 24 coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.1 If they are equal, then they must all be genuine since there's at most one fake coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.1.2 If they aren't equal, the heavier side contains all genuine coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case where B &gt; A is symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2 A = B, it means that A and B have same number of fake coins. This number could be 1 or 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide A into 24 + 25 coins, and add C to the 24 portion, and weigh these 25 coins against the other 25.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If A and B both have 1 fake coin only, that means both C and D are fake. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If A and B both have 2 fake coins, that means both C and D are genuine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That means&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if A has 1 fake coin, then the C side is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;equal to the other side when C-side minus C are all genuine coins, and the other side     contains one fake coin.&lt;/li&gt;&#xA;&lt;li&gt;lighter than the other side when C-side contains the other fake coin as well. In this case the other side contains genuine coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if A has 2 fake coins, then the C side is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lighter than the other side when both fake coins are on the C-side. In this case the other side contains genuine coins.&lt;/li&gt;&#xA;&lt;li&gt;heavier than the other side when some fake coins are on the other side. In this case, C must be genuine anyway since A has 2 fake coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if the C side is heavier, then C is genuine.&lt;/li&gt;&#xA;&lt;li&gt;if the C side is equal to the other side, then C side minus C are all genuine coins.&lt;/li&gt;&#xA;&lt;li&gt;if the C side is lighter, then the other side contains all genuine coins.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2010-08-31T17:57:39.930" LastActivityDate="2010-08-31T17:57:39.930" />
  <row Id="3732" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T17:54:53.823" Score="6" ViewCount="0" Body="&lt;p&gt;I think this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the coins into three groups: $A$ with $33$ coins, $B$ with $33$ coins and  $C$ with $34$ coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weigh $A$ and $B$ against each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $A$ is heavier than $B$, then $A$ cannot have two or more light coins, as in that case, $A$ would be lighter (or equal to $B$). Now split $A$ into groups of $16$ plus one odd coin. Weigh the groups of $16$ against each other. If they are the same, then any of those coins is heavy. If not, then any of the heavier 16 coins is heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case when $A$ and $B$ are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibilities for $A$, $B$ and $C$ are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&#xA;+-----------+------------+-----------+&#xA;|  A        |   B        |  C        |&#xA;+-----------+------------+-----------+&#xA;|  33H      |  33H       |  30H + 4L |&#xA;|           |            |           |&#xA;|  32H + L  |  32H + L   |  32H + 2L |&#xA;|           |            |           |&#xA;|  31H + 2L |  31H + 2L  |  34H      |&#xA;+-----------+------------+-----------+&#xA;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now move one coin from $A$ to $B$ (call the resulting set $B'$) and weigh it against $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B' &gt; C$, then the coin you moved from $A$ is a heavy coin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B' = C$, then the coin you moved from $A$ is a light coin and the remaining coins in $A$ are heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B' &amp;lt; C$, then all the coins in $C$ are heavy.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T20:53:51.037" LastActivityDate="2010-10-11T20:53:51.037" CommentCount="4" />
  <row Id="3734" PostTypeId="1" AcceptedAnswerId="3739" CreationDate="2010-08-31T18:53:13.827" Score="3" ViewCount="266" Body="&lt;p&gt;This is problem taken from this link Problem 4: &lt;a href=&quot;http://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2001-eng.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.u-szeged.hu/~mmaroti/schweitzer/schweitzer-2001-eng.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find the solution anywhere and i am curious to see a solution for this problem, as i can atleast comprehend the question and it seems, that the mechanism for the solution involved will be a somewhat understandable.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:25:29.857" LastActivityDate="2010-09-05T04:25:29.857" Title="Finding the Units in the Ring $\mathbb{Z}[t][\sqrt{t^{2}-1}]$" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="3735" PostTypeId="2" ParentId="3399" CreationDate="2010-08-31T19:13:21.820" Score="1" ViewCount="0" Body="&lt;p&gt;With the caveat that it's usually more helpful to devise your own mnemonics than follow someone else's&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;here are a couple of mine, poorly drawn (the second is same as Srikant Vadali's answer):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://shreevatsa.files.wordpress.com/2009/03/convex.png&quot; alt=&quot;Convex function&quot;&gt;&#xA;&lt;img src=&quot;http://shreevatsa.files.wordpress.com/2009/03/concave1.png&quot; alt=&quot;Concave function&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;convex: smiley face&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another way of remembering them, if you recall the meanings of convex and concave outside mathematics (as in lenses, etc.), is that you &lt;strong&gt;look from below&lt;/strong&gt;: if the graph of the function viewed from below looks convex (i.e., bulging towards you) the function is convex, if it looks concave the function is concave.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yet another way is to keep in mind the definition: &quot;a function is convex if its epigraph is a convex set&quot;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Epigraph_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;epigraph&lt;/a&gt; is the set of points lying above the graph, and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_set&quot; rel=&quot;nofollow&quot;&gt;convex set&lt;/a&gt; is one in which every line segment between two points in the set lies within the set. [Actually, for me, this definition is more useful for remembering what epigraph means :-)]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-08-31T19:13:21.820" CommunityOwnedDate="2010-08-31T19:13:21.820" CommentCount="3" />
  <row Id="3736" PostTypeId="1" AcceptedAnswerId="3738" CreationDate="2010-08-31T19:37:27.057" Score="6" ViewCount="81" Body="&lt;p&gt;Hi,&#xA;Suppose $\Omega$ is a domain of the complex plane (i.e. an open and connected subset of the plane). Suppose $f$ is holomorphic on $\Omega$, and $f$ is not identically zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ has a holomorphic logarithm on $\Omega$, which means that there is a function $g$ holomorphic on $\Omega$ such that $e^g=f$. Then it is easy to show that $f$ has holomorphic $n$-th roots on $\Omega$ for each $n$, which means that for each integer $n$, there exist a function $g_n$ holomorphic on $\Omega$ such that $(g_n)^n = f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the converse true? i.e. if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then can we find a function $g$ holomorphic on $\Omega$ such that $f=e^g$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few remarks :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can prove that if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then $f$ does not vanish on $\Omega$. Therefore, we can define a holomorphic logarithm &lt;em&gt;locally&lt;/em&gt;, but is it possible to find a &lt;em&gt;global&lt;/em&gt; holomorphic logarithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, notice that $\Omega$ is not supposed simply connected, in which case the answer to my question is yes.&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-08-31T20:19:14.547" Title="A criterion for the existence of a holomorphic logarithm of a holomorphic function" Tags="&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="3737" PostTypeId="2" ParentId="1748" CreationDate="2010-08-31T20:10:19.677" Score="2" ViewCount="0" Body="&lt;p&gt;Hulek's «&lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=stml-20&quot; rel=&quot;nofollow&quot;&gt;Elementary algebraic geometry&lt;/a&gt;» is a really nice text for undergraduates.  (It's part of the AMS's student mathematical library.)&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-08-31T20:10:19.677" CommunityOwnedDate="2010-08-31T20:10:19.677" />
  <row Id="3738" PostTypeId="2" ParentId="3736" CreationDate="2010-08-31T20:19:14.547" Score="9" ViewCount="0" Body="&lt;p&gt;The condition that $f$ have a holomorphic logarithm is equivalent to&#xA;$df/f=f'(z)dz/f(z)$ being an exact differential. This is equivalent to&#xA;the integral of $df/f$ over all closed curves in $\Omega$ vanishing.&#xA;Let $C$ be a closed curve in $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f=g^n$ is an $n$-th power in $\Omega$ of a holomorphic $g$ then&#xA;$\int_C df/f=n\int_C dg/g$. But $\int_C dg/g$ is an integer multiple of $2\pi i$.&#xA;Hence $\int_C df/f$ is an integer multiple of $2\pi ni$. If this holds for&#xA;all $n$ then $\int_C df/f=0$. It follows that $f$ has a holomorphic logarithm.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-08-31T20:19:14.547" CommentCount="1" />
  <row Id="3739" PostTypeId="2" ParentId="3734" CreationDate="2010-08-31T20:21:27.620" Score="7" ViewCount="0" Body="&lt;p&gt;For any non-negative integer $n$ the element $\pm (t \pm \sqrt{t^2 - 1})^n$ has norm $1$, and there are no elements $a(t) + b(t) \sqrt{t^2 - 1}$ of norm $-1$ because $a(1)^2 - b(1)^2 \sqrt{1^2 - 1} \ge 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $a(t) + b(t) \sqrt{t^2 - 1}$ is a unit which is not one of the above units such that $\deg b$ is minimal.  Then it has norm $1$, so $a(t)^2 - b(t)^2 (t^2 - 1) = 1$.  It is not hard to see that the units $\pm 1$ are the only units for which $b = 0$, so WLOG $b$ is nonzero.  This implies that $\deg a = d+1, \deg b = d$ for some non-negative integer $d$, and moreover the leading terms of $a$ and $b$ must agree up to sign.  If the leading terms agree, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$( a(t) + b(t) \sqrt{t^2 - 1})(t - \sqrt{t^2 - 1}) = (ta(t) - b(t) (t^2 - 1)) + (tb(t) - a(t)) \sqrt{t^2 - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, which contradicts the assumption of minimality.  Similarly, if the leading terms of $a$ and $b$ are opposite in sign, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$( a(t) + b(t) \sqrt{t^2 - 1})(t + \sqrt{t^2 - 1}) = (ta(t) + b(t) (t^2 - 1)) + (tb(t) + a(t)) \sqrt{t^2 - 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, again contradicting the assumption of minimality.  So no such units exist.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-01T17:35:59.310" LastActivityDate="2010-09-01T17:35:59.310" CommentCount="3" />
  <row Id="3740" PostTypeId="2" ParentId="3561" CreationDate="2010-08-31T20:40:30.567" Score="0" ViewCount="0" Body="&lt;p&gt;I highly recommend Artin's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130047635&quot; rel=&quot;nofollow&quot;&gt;Algebra&lt;/a&gt;, which is full of great motivation and originally kindled my love for algebra.&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-08-31T20:40:30.567" CommunityOwnedDate="2010-08-31T20:40:30.567" />
  <row Id="3741" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T20:43:24.707" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; It seems that &lt;a href=&quot;http://math.stackexchange.com/users/1102/moron&quot;&gt;Moron&lt;/a&gt; has a valid objection (&lt;em&gt;see the comments&lt;/em&gt;) that renders my solution incorrect:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two sides are equal: You could have 2&#xA;  heavy and 2 light. Either all light or&#xA;  all heavy is incorrect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yaser's answer misses one case. Where&#xA;  we get H+L vs H+L giving us equal on&#xA;  the first weighing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first, I thought that even if this was the case, one could still find a heavy coin with only one more weighing. Alas, &lt;strong&gt;in every weighing scheme I could think of in the context of my solution, there was a situation where one weighing does not give the necessary information to guarantee selecting a heavy coin.&lt;/strong&gt; Perhaps there's some twisted weighing scheme I didn't think of that would guarantee a correct selection, but I really doubt that (don't take my word for it though.. it's left as an exercise to the reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3724/finding-the-heavy-coin-by-weighing-twice/3732#3732&quot;&gt;Moron's solution&lt;/a&gt; seems to be a correct one, especially that it is, according to &lt;a href=&quot;http://math.stackexchange.com/users/467/t&quot;&gt;T..&lt;/a&gt;, the same as the solution by the original proposer (Shapovalov). I still have to check it out, but it's getting late and I'm quite exhausted. For now, I think it's time for me to get some sleep and to accept the downvotes. Be gentle!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate on &lt;a href=&quot;http://math.stackexchange.com/questions/3724/finding-the-heavy-coin-by-weighing-twice/3728#3728&quot;&gt;whuber's answer&lt;/a&gt; by putting it in a slightly different form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's randomly choose 5 coins out of the 100. Let's denote this set of coins by . Since there are only 4 light coins, &lt;strong&gt;we are absolutely sure that  contains at least one heavy coin.&lt;/strong&gt; Now,  the problem is reduced to simply finding one heavy coin in  by using the weight balance twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now stop for a moment and see if you can solve the new reduced problem on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't figure it out? Well, choose any 4 coins and put 2 on one side of the balance and 2 on the other. Now, what are the possible outcomes?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;The two sides are equal.&lt;/strong&gt; This means that the 4 coins we put on the balance are &lt;em&gt;either all light or all heavy&lt;/em&gt; (&lt;strong&gt;Update: This is not true. See the update at the beginning&lt;/strong&gt;). In this case, pick any coin from the balance and compare it with the one we left out. If one is heavier, it would be trivial to select it as a heavy coin. If both are equal, then all of the coins in our initial  were actually heavy, so just select any!&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;One side is heavier.&lt;/strong&gt; This implies that &lt;em&gt;at least&lt;/em&gt; one of the 2 coins on this &quot;heavier&quot; side is a heavy coin. Take these 2 coins and compare them with each other. Now selecting a heavy coin is trivial (if they are of equal weight, then they are both heavy, so just select one!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3724/finding-the-heavy-coin-by-weighing-twice/3728#3728&quot;&gt;Whuber's answer/hint&lt;/a&gt; (&lt;em&gt;look in the comments&lt;/em&gt;) put me on the right track to this solution, so don't forget to vote his answer up!&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditorDisplayName="" LastEditDate="2010-09-01T00:00:31.437" LastActivityDate="2010-09-01T00:00:31.437" CommentCount="2" />
  <row Id="3742" PostTypeId="1" AcceptedAnswerId="3743" CreationDate="2010-08-31T21:14:37.657" Score="4" ViewCount="283" Body="&lt;p&gt;Do you have any favorite open problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me mention one of my favorites. Let $A(\mathbb{T})$ be the Wiener algebra, that is, the linear space of absolutely convergent Fourier series on the unit circle $\mathbb{T}$ $(=\mathbb{R}/2\pi\mathbb{Z}=(-\pi,\pi])$ carrying the norm &#xA;$$ f\mapsto \|f\|=\sum_{n\in\mathbb{Z}}|\hat{f}(n)|&amp;lt;\infty$$&#xA;where $\hat{f}(n)=\int_{-\pi}^\pi f(t)e^{-int}dt/2\pi$ is the $n$:th Fourier coefficient of $f$. In fact $A(\mathbb{T})$ is a unitary commutative Banach algebra. By absolute convergence it follows that each $f\in A(\mathbb{T})$ is continuous on $\mathbb{T}$. Moreover, if $f(t)\not=0$ for all $t\in\mathbb{T}$ then obviously $1/f$ is also continuous on $\mathbb{T}$ - a famous theorem of Norbert Wiener (The Wiener Lemma) states that we also have $1/f\in A(\mathbb{T})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next consider a possible quantitative refinement of the Wiener lemma: &#xA;Given $\delta&gt;0$  let $$C_\delta = \sup_{A_\delta}\|1/f\|$$&#xA;where $A_\delta={f\in A(\mathbb{T}):|f(t)|&gt;\delta,\ \|f\|\leq1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; Find &#xA;$$\delta_\inf=\inf_{\delta&gt;0} \ C_\delta&amp;lt;\infty$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark 1: It is known that $\delta_\inf\leq 1/\sqrt{2}$ and that $\delta_\inf\geq 1/2$ (see [1,2]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark 2: The problem can be treated in any commutative Banach algebra we unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] N. Nikolski, In search of the invisible spectrum, Annales de l'institut Fourier, 49 no. 6 (1999), p. 1925-1998&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] H.S. SHAPIRO, A counterexample in harmonic analysis, in Approximation Theory, Banach Center Publications, Warsaw (submitted 1975), Vol. 4 (1979), 233-236.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-01T14:31:30.020" CommunityOwnedDate="2010-08-31T21:21:43.353" Title="Favourite open problem?" Tags="&lt;big-list&gt;&lt;open-problem&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="2" />
  <row Id="3743" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T21:32:35.137" Score="2" ViewCount="0" Body="&lt;p&gt;I like this one that is simple to state but likely very difficult to prove or disprove: The irrationality of $\zeta (5)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The irrationality of $\zeta (3)$  was proved by Roger Apéry only in 1979. Despite considerable effort the picture is rather incomplete about $\zeta (s)$ for the other odd integers, $s=2t+1\gt 5$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-- Martin Aigner and Günter Ziegler, Proofs from THE BOOK.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-08-31T21:32:35.137" CommunityOwnedDate="2010-08-31T21:32:35.137" />
  <row Id="3744" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T21:46:40.883" Score="1" ViewCount="0" Body="&lt;p&gt;It is open whether or not there exists an algorithm which will determine whether an integer sequence satisfying a linear recurrence with integer coefficients (such as the Fibonacci sequence) has only nonzero terms.  See Terence Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/&quot; rel=&quot;nofollow&quot;&gt;excellent blog post&lt;/a&gt; on the subject.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T21:46:40.883" CommunityOwnedDate="2010-08-31T21:46:40.883" />
  <row Id="3746" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T22:04:28.907" Score="2" ViewCount="0" Body="&lt;p&gt;The (ir)rationality of the Euler-Mascheroni constant has always intrigued me, as it's one of those constants that is &quot;obviously&quot; irrational and yet nobody has much of a clue on how to prove that. &lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-08-31T22:04:28.907" CommunityOwnedDate="2010-08-31T22:04:28.907" />
  <row Id="3747" PostTypeId="2" ParentId="3734" CreationDate="2010-08-31T22:28:43.110" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt; $\rm\;$ It's simply a typical unit group descent. Let $\rm\: U \:$ be the group generated by the units $\rm\: (\pm t,\: 1)\:$.&lt;br&gt;&#xA; If unit $\rm\; (a,\;b) \not\in U \;$ then $\rm\;  (a,\;b)(\pm t,\; 1) = (\cdots,\;a\pm bt) \;$ is a &quot;smaller&quot; unit $\rm\:\not\in U\;$  since one of $\rm\: a\pm bt \;$ has smaller degree than $\rm\: b\:,\:$ as is easily verified using $\rm\; (a-bt)(a+bt) = 1-b^2 \:. \:\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T13:57:33.720" LastActivityDate="2010-09-01T13:57:33.720" />
  <row Id="3748" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T22:35:55.217" Score="1" ViewCount="0" Body="&lt;p&gt;Not that practical, but finding the smallest $\varepsilon$ in the statement &quot;Matrix multiplication/inversion takes $O(n^{2+\varepsilon})$ flops&quot; is still a nice open problem.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-08-31T22:35:55.217" CommunityOwnedDate="2010-08-31T22:35:55.217" />
  <row Id="3749" PostTypeId="1" AcceptedAnswerId="3755" CreationDate="2010-08-31T22:47:52.857" Score="6" ViewCount="453" Body="&lt;p&gt;When I first took linear algebra, we never learned about dual spaces. Today in lecture we discussed them and I understand what they are and everything, but I don't really understand why we want to study them within linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone knew a nice intuitive motivation for the study of dual spaces and whether or not they &quot;show up&quot; as often as other concepts in linear algebra? Is their usefulness something that just becomes more apparent as you learn more math and see them arise in different settings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I understand that dual spaces show up in functional analysis and multilinear algebra, but I still don't really understand the intuition/motivation behind their definition in the standard topics covered in a linear algebra course. (Hopefully, this clarifies my question)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-01T00:26:00.073" LastActivityDate="2010-09-06T22:13:07.150" Title="Why do we care about dual spaces?" Tags="&lt;soft-question&gt;&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="5" />
  <row Id="3750" PostTypeId="1" AcceptedAnswerId="3788" CreationDate="2010-08-31T23:05:07.650" Score="12" ViewCount="760" Body="&lt;p&gt;This problem first came to me in high school, and a couple times since, and I even assigned it for extra credit in one of my calculus classes after I became a teacher.  So I know the solution.  What I am looking for is other WAYS to obtain the solution.  I've been told there exists a solution using only arithmetic, but have never figured it out.  Other solutions using ordinary calculus, trigonometry, algebra of conic sections, and so on are also possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is usually stated in the form of a letter from an Aunt and Uncle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear niece/nephew, How are things&#xA;  going for you and your folks?  We hear&#xA;  you are doing quite well it school. &#xA;  Keep it up!  Given this success, we&#xA;  were hoping you could help us figure&#xA;  out a little dilemma.  As you know,&#xA;  our home is heated by fuel oil, and we&#xA;  have a big tank buried in the side&#xA;  yard.  The tank is a cylinder, 20 feet&#xA;  long and 10 feet in diameter, lying on&#xA;  its side five feet deep, with a narrow&#xA;  tube coming to a fill cap at ground&#xA;  level.  Your uncle has a 15 foot&#xA;  length of old pipe that we'd like to&#xA;  utilize as a dip stick in order to&#xA;  know when we are getting close to&#xA;  needing a fill-up.  We know that 0&#xA;  feet is empty, 5 feet is half full,&#xA;  and 10 feet is completely full. &#xA;  Trouble is, we don't know how to mark&#xA;  any other points. We are pretty sure&#xA;  they will not be uniformly spaced. &#xA;  What we really want is to know, within&#xA;  the nearest 0.01 foot, where to mark&#xA;  the dip stick for every multiple of&#xA;  10% from 0% to 100%.  Can you figure&#xA;  this out for us?  Of course, we will&#xA;  want to see details of your solution&#xA;  and check it ourselves, and it would&#xA;  especially help if you could draw us a&#xA;  scale model of the dip stick. Love,&#xA;  Auntie Flo and Uncle Jim&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That last sentence shows the teacher influence on the problem.&#xA;So, my challenge to this community is not to find any old solution, but to find the solution at the lowest possible grade level, so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:  To those who are focusing in on the .01 feet accuracy, I apologize.  The intent was merely to state, it is acceptable to estimate.  If the exact answer is sqrt(2)*pi/2 or some other silly thing, go ahead and just write 2.22 feet, for example.&lt;/p&gt;&#xA;" OwnerUserId="575" LastEditorUserId="575" LastEditorDisplayName="" LastEditDate="2010-09-01T23:40:29.073" LastActivityDate="2010-09-12T23:05:28.853" Title="Aunt and Uncle's fuel oil tank dip stick problem" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;arithmetic&gt;&lt;conic-sections&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="3751" PostTypeId="2" ParentId="3749" CreationDate="2010-08-31T23:13:28.700" Score="3" ViewCount="0" Body="&lt;p&gt;Absolutely yes to the last question.  There are already many applications listed in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.  I will give a few more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, together with tensor products, dual spaces can be used to talk about linear transformations; the vector space $\text{Hom}(V, W)$ of linear transformations from a vector space $V$ to a vector space $W$ is canonically isomorphic to the tensor product $V^{\ast} \otimes W$.  I assume you care about linear transformations, so you should care about tensor products and dual spaces as well.  (A really important feature of this decomposition is that it is true in considerable generality; for example, it holds for representations of groups and is a natural way to prove the orthogonality relations for characters.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dual spaces also appear in geometry as the natural setting for certain objects.  For example, a differentiable function $f : M \to \mathbb{R}$ where $M$ is a smooth manifold is an object that produces, for any point $p \in M$ and tangent vector $v \in T_p M$, a number, the directional derivative, in a linear way.  In other words, a differentiable function defines an element of the dual to the tangent space (the cotangent space) at each point of the manifold.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-08-31T23:13:28.700" CommentCount="3" />
  <row Id="3752" PostTypeId="2" ParentId="3749" CreationDate="2010-08-31T23:13:32.007" Score="1" ViewCount="0" Body="&lt;p&gt;Dual spaces are important because they show up all over the place, and you'll see them arise in all sort of settings as you learn more math.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-08-31T23:13:32.007" />
  <row Id="3753" PostTypeId="2" ParentId="3742" CreationDate="2010-08-31T23:51:18.870" Score="0" ViewCount="0" Body="&lt;p&gt;Does favorite allow me two - both about convex 3-dimensional polyhedra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Shephard's Conjecture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to cut along the edges of some spanning tree of any convex 3-dimensional polyhedron and unfold the cut open polyhedron so that the faces do not overlap (and one gets a simple polygon where the uncut edges indicate the faces of the original polyhedron in the resulting &quot;net&quot; in the plane)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Net_%28polyhedron%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Net_%28polyhedron%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Barnette's conjecture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does every 3-valent convex 3-dimensional polyhedron all of whose faces have an even number of sides admit a hamiltonian circuit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Barnette%27s_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Barnette%27s_conjecture&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-08-31T23:51:18.870" CommunityOwnedDate="2010-08-31T23:51:18.870" CommentCount="1" />
  <row Id="3754" PostTypeId="2" ParentId="3724" CreationDate="2010-08-31T23:54:55.537" Score="0" ViewCount="0" Body="&lt;p&gt;Pick any four coins and weigh two against two.  If they are imbalanced then at least one of the two coins on the heavier side is heavy and they can now be weighed against each other to identify one or both as heavy.  Otherwise replace the two coins on either side of the balance with two other coins.  If they are now imbalanced then both coins on the heavier side are heavy, and if they are still balanced then all six coins are heavy.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastActivityDate="2010-08-31T23:54:55.537" CommentCount="5" />
  <row Id="3755" PostTypeId="2" ParentId="3749" CreationDate="2010-09-01T00:44:27.447" Score="8" ViewCount="0" Body="&lt;p&gt;Let $V$ be a vector space (over any field, but we can take it to be $\mathbb R$ if you like,&#xA;and for concreteness I will take the field to be $\mathbb R$ from now on;&#xA;everything is just as interesting in that case).  Certainly one of the interesting concepts &#xA;in linear algebra is that of a &lt;em&gt;hyperplane&lt;/em&gt; in $V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $V = \mathbb R^n$, then a hyperplane is just the solution set to an equation&#xA;of the form&#xA;$$a_1 x_1 + \cdots + a_n x_n = b,$$&#xA;for some $a_i$ not all zero and some $b$.&#xA;Recall that solving such equations (or simultaneous sets of such equations) is one&#xA;of the basic motivations for developing linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now remember that when a vector space is not given to you as $\mathbb R^n$,&#xA;it doesn't normally have a canonical basis, so we don't have a canonical way&#xA;to write its elements down via coordinates, and so we can't describe hyperplanes&#xA;by explicit equations like above.  (Or better, we can, but only after choosing&#xA;coordinates, and this is not canonical.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we canonically describe hyperplanes in $V$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this we need a conceptual interpretation of the above equation.  And here linear&#xA;functionals come to the rescue.  More precisely, the map&#xA;$$\ell:(x_1,\ldots,x_n) \mapsto a_1 x_1 + \cdots a_n x_n$$&#xA;is a linear functional on $\mathbb R^n$, and so the above equation for the&#xA;hyperplane can be written as &#xA;$$\ell(v) = b.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $V$ is any vector space, and $\ell: V \to \mathbb R$ is any&#xA;non-zero linear functional (i.e. non-zero element of the dual space), then&#xA;for any $b \in \mathbb R,$ the set&#xA;$$\{v \, | \, \ell(v) = b\}$$&#xA;is a hyperplane in $V$, and all hyperplanes in $V$ arise this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this gives a reasonable justification for introducing the elements of the dual&#xA;space to $V$; they generalize the notion of linear equation in several variables&#xA;from the case of $\mathbb R^n$ to the case of an arbitrary vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you might ask: why do we make them a vector space themselves?  why do we want&#xA;to add them to one another, or multiply them by scalars?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of reasons for this; here is one: Remember how important it is,&#xA;when you solve systems of linear equations, to add equations together, or&#xA;to multiply them by scalars (here I am referring to all the steps you typically&#xA;make when performing Gaussian elimination on a collection of simultaneous linear&#xA;equations)?  Well, under the dictionary above between linear equations&#xA;and linear functionals, these processes correspond precisely to adding together&#xA;linear functionals, or multiplying them by scalars.  If you ponder this for a bit,&#xA;you can hopefully convince yourself that making the set of linear&#xA;functionals a vector space is a pretty natural thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: just as concrete vectors $(x_1,\ldots,x_n) \in \mathbb R^n$ are naturally&#xA;generalized to elements of vector spaces, concrete linear expressions&#xA;$a_1 x_1 + \ldots + a_n x_n$ in $x_1,\ldots, x_n$ are naturally generalized to linear functionals.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-01T00:44:27.447" CommentCount="5" />
  <row Id="3756" PostTypeId="2" ParentId="3749" CreationDate="2010-09-01T01:03:48.880" Score="3" ViewCount="0" Body="&lt;p&gt;There are some very beautiful and easily accessible applications of duality, adjointness, etc. in Rota's modern reformulation of the Umbral Calculus. You'll quickly gain an appreciation for the power of such duality once you see how easily this approach unifies hundreds of diverse special-function identities, and makes their derivation essentially trivial. For a nice introduction see Steven Roman's book &quot;The Umbral Calculus&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-01T01:03:48.880" CommentCount="2" />
  <row Id="3757" PostTypeId="2" ParentId="3750" CreationDate="2010-09-01T02:31:47.597" Score="6" ViewCount="0" Body="&lt;p&gt;It is sufficient to consider the circular cross-section of the tank and volumes below 50% (marks for those above 50% are the reflection image of those below 50% across the 50% mark).  Consider the radius of the tank to be 1 unit.  For some amount of oil in the tank, consider the central angle formed by the points on the circular cross-section at the top of the oil and the center of the circle--call this $\alpha$, measured in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/5pI7u.png&quot; alt=&quot;tank cross-section&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the cross-section of the oil is the area of the sector determined by $\alpha$ ($\frac{\alpha}{2\pi}\pi r^2=\frac{\alpha}{2}$) minus the area of the triangle that is part of the sector but not part of the cross-section of oil ($\frac{1}{2}ab\sin C=\frac{\sin\alpha}{2}$), $\frac{1}{2}(\alpha-\sin\alpha)$.  The portion of the circular cross-section (and hence the portion of the volume) corresponding to this angle is $\frac{\alpha-\sin\alpha}{2\pi}$.  Setting this expression equal to 0.1, 0.2, 0.3, and 0.4 and solving for $\alpha$ will give the values of $\alpha$ corresponding to 10%, 20%, 30%, and 40% full--solving here is done numerically/graphically, as there is no algebraic method to solve these equations.  For each value of $\alpha$, the distance from the center of the circle to the oil level is $\cos\frac{\alpha}{2}$, so the depth of the oil is $1-\cos\frac{\alpha}{2}$.  (Note that these are for a radius of 1 unit and need to be rescaled for the original problem's specific numbers.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-01T02:31:47.597" CommentCount="6" />
  <row Id="3758" PostTypeId="1" AcceptedAnswerId="3773" CreationDate="2010-09-01T02:55:46.223" Score="1" ViewCount="334" Body="&lt;p&gt;When implicitly finding the derivative of:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;xy^3 - xy^3*sin(x) = 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do you find the implicit derivative of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;xy^3*sin(x)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it using a &lt;em&gt;triple&lt;/em&gt; product rule of sorts?&lt;/p&gt;&#xA;" OwnerUserId="1602" LastActivityDate="2010-09-01T16:33:07.393" Title="Implicit Diferentiation... the triple product rule?" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" />
  <row Id="3759" PostTypeId="2" ParentId="3734" CreationDate="2010-09-01T03:10:32.683" Score="6" ViewCount="0" Body="&lt;p&gt;Here is another argument:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $x = t + \sqrt{t^2 - 1}$ and $y = t - \sqrt{t^2 - 1}$.    Then $x y = 1$, i.e. $y = x^{-1}$,&#xA;and $t = (x + y)/2$ , $\sqrt{t^2 - 1} = (x - y)/2$.&lt;br&gt;&#xA;Thus we see that&#xA;$\mathbb Z[t][\sqrt{t^2 - 1}] \subset \mathbb Z[1/2][x,x^{-1}].$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is easy to check that if $A$ is an integral domain, then the only units&#xA;in $A[x,x^{-1}]$ are of the form $a x^n,$ where $a \in A$ is a unit and $n$ is an integer.&#xA;Since the units in $\mathbb Z[1/2]$ are precisely the elements $\pm 2^m$ (for some $m$)&#xA;we see that the units in&#xA;$\mathbb Z[1/2][x,x^{-1}]$ are of the form $\pm 2^m x^n = \pm 2^m (t + \sqrt{t^2 - 1})^n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to check that if such an element and its inverse both actually lie in &#xA;$\mathbb Z[t][\sqrt{t^2 - 1}],$ then necessarily $m = 0$ (e.g. for norm reasons,&#xA;or just looking explicitly at their denominators), and so we get the answer that&#xA;the units are precisely the elements of the form $\pm (t + \sqrt{t^2 -1 })^n$.&#xA;(Here $n$ runs over $\mathbb Z$; this is the same set as $\pm (t \pm \sqrt{t^2 - 1})^n$,&#xA;where $n$ is now non-negative.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-01T03:10:32.683" CommentCount="2" />
  <row Id="3760" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T03:13:41.847" Score="3" ViewCount="0" Body="&lt;p&gt;It is possible to derive a 'triple product rule.' In fact, you can generalize it to an arbitrary number of products. You may ask why these rules aren't presented in these general forms. One reason for this is that writing the general formula out is a more complicated to memorize and more intimidating towards students just learning calculus. The other reason, is that having the product rule is enough to give you the general rule by a simple induction argument. Also, for practical purposes, problems can be solved with only the usual product rule being applied in multiple steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case we can apply the product rule twice:&#xA;$\frac{d}{dx}xsin(x)y^{3}=y^{3}\frac{d}{dx}(xsin(x))+3y^2\frac{dy}{dx}(xsin(x))=y^{3}(sin(x)+xcos(x))+3y^{2}\frac{dy}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also derive a 'triple product rule'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx}(f\cdot g\cdot h)=\frac{df}{dx}\cdot(h\cdot g)+\frac{dg}{dx}\cdot (f\cdot h)+\frac{dh}{dx}\cdot (f\cdot g)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you see that you still have to apply the chain rule to that to get it to fit how you need to use it in this situation and so it would be a real mess to memorize all the different special cases. This is why we just provide the product rule with 2 functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does my rambling make sense?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-01T03:20:00.040" LastActivityDate="2010-09-01T03:20:00.040" CommentCount="2" />
  <row Id="3761" PostTypeId="2" ParentId="3399" CreationDate="2010-09-01T03:23:28.117" Score="5" ViewCount="0" Body="&lt;p&gt;Lets say that you accept the definition of a convex set in higher dimensions, like a sphere in $\mathbb{R}^3$. The question I seek to provide insight into is why convex functions in one variable are defined as opening up instead of down, since this seems like an arbitrary definition. This is because, depending on how you look at the graph, you could naively view the function as bending outwards (like a convex set) or inwards (concave). However, there is a nice connection between these two things using metric spaces that I think can provide some meaning to the way it is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the metrics that you are familiar with have open balls that are convex, such as the standard metric.  But some are actually non convex. A good example of this is $ d(x,y) = \sum \sqrt{|x_i - y_i |} $. (note that $\sqrt{x}$ is not a convex function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an interesting condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a metric $d$. If for all $y,z\in E$ and $0\leq t\leq 1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d \left(x, \  t  y \; \, + \;  (1-t) z \right) \quad \leq \quad t  d(x,y) \; + \; (1-t)  d(x,z) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the open balls formed by $d$ are convex. [1] In other words, if you fix $x$ and $d(y):\mathbb{R}^n\rightarrow \mathbb{R}$ is a convex function, then the open balls are convex sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually $d(x,y) = \sum f \, (x_i,y_i)$, for some $f:\mathbb{R}^2\rightarrow\mathbb{R}$.  If we fix $x$ and $f:\mathbb{R}\rightarrow \mathbb{R}$ fits the definition of a convex function, then $d$ will also be convex, and the condition will be satisfied, giving us convex balls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So convex functions (if they can form a metric) will give you convex open balls.  A nice connection that makes the definition make more sense. Other conditions that guarantee convex open balls are discussed in the paper I reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Norfolk, T. (1991). When does a metric generate convex balls? www.math.uakron.edu/~norfolk/convex.ps&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-12T04:00:23.837" LastActivityDate="2010-12-12T04:00:23.837" CommunityOwnedDate="2010-09-01T03:23:28.117" CommentCount="7" />
  <row Id="3762" PostTypeId="2" ParentId="1748" CreationDate="2010-09-01T03:48:41.877" Score="0" ViewCount="0" Body="&lt;p&gt;Bix - Conics and Cubics is probably the most accessible. I think a high school student who did the BC calculus syllabus could read this and get a lot out of it. But I'd agree with the others about Cox, Little, O'Shea - Ideals Varieties and Algorithms being the 'best' undergraduate text.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-01T03:48:41.877" CommunityOwnedDate="2010-09-01T03:48:41.877" />
  <row Id="3763" PostTypeId="1" AcceptedAnswerId="3775" CreationDate="2010-09-01T04:15:41.187" Score="1" ViewCount="413" Body="&lt;p&gt;How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given, such a series, how does one go about solving it. Getting an &lt;em&gt;Integral Representation&lt;/em&gt; seems tough for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of going along these lines, &lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&lt;/a&gt; but couldn't succeed.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T01:18:09.083" Title="Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ " Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="3764" PostTypeId="2" ParentId="3763" CreationDate="2010-09-01T04:28:45.190" Score="1" ViewCount="0" Body="&lt;p&gt;FWIW, &lt;em&gt;Mathematica&lt;/em&gt; is able to sum this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[1]:= Sum[1/(2 n + 2 k + 1) (2 k - 1)!!/(2 k)!!, {k, 0, \[Infinity]}]&#xA;&#xA;                        3 + 2 n&#xA;        Sqrt[Pi] Gamma[-------]&#xA;                           2&#xA;Out[1]= -----------------------&#xA;        (1 + 2 n) Gamma[1 + n]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-01T04:28:45.190" />
  <row Id="3765" PostTypeId="1" CreationDate="2010-09-01T04:37:44.937" Score="1" ViewCount="70" Body="&lt;p&gt;Can someone please point me in the direction of any theory on graphs where the edge weights are not scalars but represent some relation between the nodes that is a simple function of a single variable (simple, say piecewise linear). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm interested in various basic graph properties and also thinking of the graph as representing a network. So, for example, if the graph represented a communication network over time then the edge weights would be a function represented connectivity as a function of time, how do you find a valid path between nodes? I'm looking for help both on specific algorithms but also general theory if it exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of time-extended networks where you explicitly expand out the dependence on the variable but, from what I've read, this incomplete and of limited applicability. &lt;/p&gt;&#xA;" OwnerUserId="1628" LastEditorUserId="1628" LastEditorDisplayName="" LastEditDate="2010-09-01T17:07:03.877" LastActivityDate="2010-09-01T17:07:03.877" Title="Generalized graph - edges weights are functions, not scalars" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="3766" PostTypeId="2" ParentId="3763" CreationDate="2010-09-01T04:57:00.837" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a) = \sum_{k=0}^{\infty} \frac{1\cdot3\cdots(2k-1)}{2\cdot4\cdots(2k)}\cdot a^{2n+2k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|a| \leq 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want is $$\int_{0}^{1} f(a) da$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $k^{th}$ coefficient can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}{\sin^{2k}x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(again Wallis's like product as in the other question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a) = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{k=0}^{\infty} {(a\sin x)^{2k}} dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} {\frac{1}{1-(a\sin x)^2}}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\frac{1}{1-(a\sin x)^2} = \frac{1}{\cos^2 x + (\sqrt{1-a^2}\sin x)^2} = \frac{\sec^2 x}{1+(\sqrt{1-a^2}\tan x)^2}$$ which is the derivative of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\arctan(\sqrt{1-a^2}\tan x)}{\sqrt{1-a^2}}$$ whose integral between $0$ and $\frac{\pi}{2}$ is $$\frac{\pi}{2\sqrt{1-a^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we get $$f(a) = \frac{a^{2n}}{\sqrt{1-a^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(So in fact, we have shown the identity which Qiaochu uses in his answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the required sum is $$\int_{0}^{1} \frac{a^{2n}}{\sqrt{1-a^2}}da$$ which can easily be found by the substitution $a = \sin x$ (like in Qiaochu's answer) and which gives us the result to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\frac{\pi}{2}} {\sin^{2n}x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this integral for positive integer $n$ can easily be seen to be what Mariano got. I am unsure of whether that formula holds for an arbitrary real number $n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T01:18:09.083" LastActivityDate="2010-09-02T01:18:09.083" CommentCount="3" />
  <row Id="3767" PostTypeId="2" ParentId="3623" CreationDate="2010-09-01T05:01:02.167" Score="0" ViewCount="0" Body="&lt;p&gt;If I were you I would start by looking for similar threads..  ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3335/could-you-recommend-some-classic-textbooks-on-ordinary-partial-differential-equat&quot;&gt;Could you recommend some classic textbooks on ordinary/partial differential equation?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/2827/good-1st-pde-book-for-self-study&quot;&gt;Good 1st PDE book for self study&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-01T05:01:02.167" CommunityOwnedDate="2010-09-01T05:01:02.167" />
  <row Id="3768" PostTypeId="1" AcceptedAnswerId="3770" CreationDate="2010-09-01T05:06:28.477" Score="6" ViewCount="261" Body="&lt;p&gt;From some reading, I've noticed that $\sup(\emptyset)=\min(S)$, but $\inf(\emptyset)=\max(S)$, given that $\min(S)$ and $\max(S)$ exist, where $S$ is the universe in which one is working. Is there some inherent reasoning/proof as to why this is? It seems strange to me that an upper bound of a set would be smaller than a lower bound of the same set.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-02T12:15:10.110" Title="Upper and Lower Bounds of $\emptyset$" Tags="&lt;analysis&gt;" AnswerCount="3" />
  <row Id="3769" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T05:09:35.083" Score="1" ViewCount="0" Body="&lt;p&gt;You get a formula for the derivative of a product of $n$ factors from the formula for the product of $2$ factors by doing induction. Intuitively, you do it the same way as you go from $2$ to $3$: $(fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + (fg)h' = f'gh + fg'h + fgh'$. The pattern should now be clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you already know that $(f_1\cdots f_k)' = f_1'f_2\cdots f_k + f_1f_2'f_3\cdots f_k + \cdots + f_1\cdots f_{k-1}f_k'$, then inductively you have &#xA;$$(f_1\cdots f_kf_{k+1})' = (f_1\cdots f_k)'f_{k+1} + (f_1\cdots f_k)f_{k+1}'$$&#xA;and the formula follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for $xy^3\sin(x)$, the derivative will be &#xA;$$(x)'y^3\sin(x) + x(y^3)'\sin(x) + xy^3(\sin x)'$$&#xA;with the middle summand requiring implicit differentiation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T05:09:35.083" />
  <row Id="3770" PostTypeId="2" ParentId="3768" CreationDate="2010-09-01T05:16:53.807" Score="5" ViewCount="0" Body="&lt;p&gt;The supremum is defined to be the least upper bound. An upper bound for a set $K$ is defined to be an element $b$ such that for all $k\in K$, $k\leq b$. If $K$ is empty, then the condition &quot;for all $k\in K$, $k\leq b$&quot; is true by vacuity (you have an implication of the form &quot;if $k$ is an element of the empty set, then $\lt$something happens$\gt$&quot;, and the antecedent is always false so the implication is always true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, every element of $S$ is an upper bound for $\emptyset$; since $\sup(\emptyset)$ is the least upper bound, that means that $\sup(\emptyset)=\min($upper bounds of $\emptyset) = \min(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the infimum is the greatest lower bound, and every element of $S$ is a lower bound for $\emptyset$, so $\inf(\emptyset) = \max($lower bounds of $\emptyset) = \max(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is somewhat counterintuitive that you have $\sup(\emptyset)\leq\inf(\emptyset)$, but in fact the empty set is the only one for which you can have $\sup(\emptyset)\lt \inf(\emptyset)$. The empty set often causes counterintuitive results.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T05:16:53.807" CommentCount="1" />
  <row Id="3771" PostTypeId="2" ParentId="3742" CreationDate="2010-09-01T05:17:01.043" Score="1" ViewCount="0" Body="&lt;p&gt;I am quite fond of the abc conjecture: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $q(n)$ to be the product of all distinct primes of n. The conjecture states, that for any $\epsilon &gt; 0$, there exist only finitely many triples of coprime positive integers $(a, b, c)$, satisfying $a + b = c$ for which $c &gt; q(a b c)^{1 + \epsilon}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This conjecture has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Abc_conjecture#Some_consequences&quot; rel=&quot;nofollow&quot;&gt;impressive list of consequences&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-01T05:17:01.043" CommunityOwnedDate="2010-09-01T05:17:01.043" />
  <row Id="3772" PostTypeId="2" ParentId="3768" CreationDate="2010-09-01T05:23:35.487" Score="9" ViewCount="0" Body="&lt;p&gt;A quick counterpoint to Arturo's answer: yes, the fact that $\operatorname{sup}(\emptyset) = -\infty &amp;lt; \infty = \operatorname{inf}(\emptyset)$ seems strange at first.  But it is designed, among other things, to preserve an intuitive and useful property of infima and suprema:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ and $T$ be subsets of $\mathbb{R}$ with $S \subset T$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{inf}(T) \leq \operatorname{inf}(S)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{sup}(T) \geq \operatorname{sup}(S)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, by adding elements to a set, its infimum can only get smaller and its supremum can only get larger.  Now if you take $S = \emptyset$ and let $T$ range through even all one-element subsets of $\mathbb{R}$, you see that in order for these inequalities to hold we &lt;em&gt;must&lt;/em&gt; define $\operatorname{inf}(\emptyset)$ and $\operatorname{sup}(\emptyset)$ as we did.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-01T05:29:56.220" LastActivityDate="2010-09-01T05:29:56.220" CommentCount="0" />
  <row Id="3773" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T05:33:48.730" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_differentiation&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;logarithmic differentiation&lt;/strong&gt;&lt;/a&gt; makes the n-ary generalization obvious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (abc)'= abc \;  log(abc)' = abc \;(log\; a + log\; b + log\; c)' = abc \; \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c}\bigg) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the same proof works for arbitrary length products yielding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (abc\cdots f)' = \: abc\cdots f\;\bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c} +\:\cdots\:+ \frac{f'}{f}\bigg) $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T06:09:19.707" LastActivityDate="2010-09-01T06:09:19.707" CommentCount="0" />
  <row Id="3775" PostTypeId="2" ParentId="3763" CreationDate="2010-09-01T06:09:07.147" Score="4" ViewCount="0" Body="&lt;p&gt;Observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{1}{4^n} {2n \choose n} = \frac{(2n-1)(2n-3)(2n-5)...}{2n(2n-2)(2n-4)...}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and recall that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{1}{ \sqrt{1 - x^2} } = \sum_{k \ge 0} \frac{1}{4^k} {2k \choose k} x^{2k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the desired quantity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_0^1 \frac{x^{2n}}{\sqrt{1 - x^2}} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But letting $x = \sin \theta$ this is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_0^{ \frac{\pi}{2} } \sin^{2n} \theta d \theta = \frac{\pi}{2} \frac{1}{4^n} {2n \choose n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to Mariano's closed form via the identities $\Gamma(x+1) = x \Gamma(x)$ and $\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$.  I should mention that the integral at the end of Moron's answer is quite doable; write it in terms of a cosine and use the substitution $\cos \theta = \frac{1 - t^2}{1 + t^2}$ where $t = \tan \frac{\theta}{2}$ to reduce the problem to the integral of a rational function, and then one can use one of several related methods (partial fractions, contour integration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark:&lt;/strong&gt;  The last integral identity above happens to be one of my favorite identities.  I describe a representation-theoretic and combinatorial proof of it &lt;a href=&quot;http://qchu.wordpress.com/2010/03/07/walks-on-graphs-and-tensor-products/&quot; rel=&quot;nofollow&quot;&gt;in this blog post&lt;/a&gt;.  Another approach implicitly occurs &lt;a href=&quot;http://qchu.wordpress.com/2009/10/07/extracting-the-diagonal/&quot; rel=&quot;nofollow&quot;&gt;in this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-01T06:21:16.963" LastActivityDate="2010-09-01T06:21:16.963" CommentCount="2" />
  <row Id="3776" PostTypeId="1" AcceptedAnswerId="3780" CreationDate="2010-09-01T07:22:25.440" Score="3" ViewCount="131" Body="&lt;p&gt;I was musing about a particular limit,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \prod\limits_{n &gt; 0} \bigl(1 - 2^{-n}\bigr)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we may bound 0.288&amp;nbsp;&amp;lt;&amp;nbsp;L&amp;nbsp;&amp;lt;&amp;nbsp;0.308, which we may show by taking the logarithm:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(L) = \ln \bigl( \frac{315}{1024}\bigr) + \sum\limits_{n &gt; 4} \ln\bigl(1 - 2^{-n}\bigr) &gt; \ln\bigl(\frac{315}{1024}\bigr) - \sum\limits_{n &gt; 4} 2^{-n} =\; \ln\bigl(\frac{315}{1024} \cdot \mathrm e^{-1/16}\bigr)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if this type of infinite product (or the corresponding sum of logarithms) has a name, and whether there are techniques for obtaining a closed form expression for the limit.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-09-01T19:11:27.843" LastActivityDate="2010-09-01T19:11:27.843" Title="Limit of a particular variety of infinite product/series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3777" PostTypeId="1" AcceptedAnswerId="3802" CreationDate="2010-09-01T07:22:33.540" Score="13" ViewCount="661" Body="&lt;p&gt;If memory serves, ten years ago to the week (or so), I taught first semester freshman calculus for the first time.  As many calculus instructors do, I decided I should ask some extra credit questions to get students to think more deeply about the material.  The first one I asked was this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Recall that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is said to have a &lt;strong&gt;removable discontinuity&lt;/strong&gt; at a point $x_0 \in \mathbb{R}$ if $\lim_{x \rightarrow x_0} f(x)$ exists but not does not equal $f(x_0)$.  Does there exist a function $f$ which has a removable discontinuity at $x_0$ &lt;em&gt;for every&lt;/em&gt; $x_0 \in \mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commentary: if so, we could define a new function $\tilde{f}(x_0) = \lim_{x \rightarrow x_0} f(x)$ and it seems at least that $\tilde{f}$ has a fighting chance to be continuous on $\mathbb{R}$.  Thus we have successfully &quot;removed the discontinuities&quot; of $f$, but in so doing we have changed the value at every point!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Lest you think this is too silly to even seriously contemplate, consider the function $f: \mathbb{Q} \rightarrow \mathbb{Q}$ given by $f(0) = 1$ and for a nonzero &#xA;rational number $\frac{p}{q}$, $f(\frac{p}{q}) = \frac{1}{q}$.  It is easy to see that this function has limit $0$ at every (rational) point!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I mentioned this problem to my students.  A week later, the only person who asked me about it at all was my Teaching Assistant, who was an older undergraduate, not even a math major, I think.  (I hasten to add that this was not in any sense an honors calculus class, i.e., I was pretty clueless back then.)  Thinking about it a bit, I asked him if he knew about uncountable sets, and he said that he didn't.  At that point I realized that I didn't have a solution in mind that he would understand (so still less so for the freshman calculus students) and I advised him to forget all about it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So &lt;strong&gt;my actual question is&lt;/strong&gt;: can you solve this problem using only the concepts in a non-honors freshman calculus textbook?  (In particular, without using notions of un/countability?)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[&lt;b&gt;Addendum&lt;/b&gt;: Let me say explicitly that I would welcome an answer that proceeds directly in terms of the least upper bound axiom.  Most freshman calculus books &lt;em&gt;do&lt;/em&gt; include this, albeit somewhere hidden from view of the casual readers, i.e., actual freshman calculus students.]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you can't figure out how to answer the question at all, I think the following related question helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Define a function $f: \mathbb{R} \rightarrow \mathbb{R}$ to be &lt;strong&gt;precontinuous&lt;/strong&gt; if the limit exists at every point.  For such a function, we can define $\tilde{f}$ as above.  Prove/disprove that, as suggested above, $\tilde{f}$ is indeed continuous.  [Then think about $f - \tilde{f}$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I think about it, there is an entire little area here that I don't know anything about, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The set of discontinuities of an arbitrary function is known -- any $F_{\sigma}$ set inside $\mathbb{R}$ can serve.  What can we say about the set of discontinuities of a &quot;precontinuous function&quot;?  [Edit: from the link provided in Chandru1's answer, we see that it is countable.  What &lt;em&gt;else&lt;/em&gt; can we say?  Note that taking the above example and extending by $0$ to the irrationals, we see that the set of points of discontinuity of a precontinuous function can be dense.]&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-01T08:54:48.180" LastActivityDate="2010-09-03T02:18:35.137" Title="Is there a function with a removable discontinuity at every point?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="7" />
  <row Id="3778" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T07:34:03.823" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a simple way to compute the $n$-th derivative of a product of $k$ functions. To ease the notation, stick to the case $k=3$. First follow your instinct, and then you'll see that the justification presents no difficulty. Write (with all summation indices varying from 0 to $\infty$)&#xA;$$\sum_n\frac{(fgh)^{(n)}}{n!}X^n&#xA;=\sum_i\frac{f^{(i)}}{i!}X^i\ \sum_j\frac{g^{(j)}}{j!}X^j\ \sum_k\frac{h^{(k)}}{k!}X^k$$&#xA;$$=\sum_{i,j,k}\frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}X^{i+j+k}&#xA;=\sum_n\ X^n\sum_{i+j+k=n}\frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}\quad.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In categorical language: To each $\mathbb Q$-algebra $A$ is attached the $\mathbb Q$-algebra $A':=A[[X]]$ equipped with the derivation $d/dX$ (the letter $X$ being an indeterminate). Then $A\mapsto A'$ is a &lt;strong&gt;right adjoint&lt;/strong&gt; to the forgetful functor from differential $\mathbb Q$-algebras to $\mathbb Q$-algebras. &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-09-01T07:41:21.887" LastActivityDate="2010-09-01T07:41:21.887" CommentCount="1" />
  <row Id="3779" PostTypeId="2" ParentId="3777" CreationDate="2010-09-01T07:38:06.377" Score="3" ViewCount="0" Body="&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see theorem 5.63 in this link: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=6l_E9OTFaK0C&amp;amp;pg=PA249&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=6l_E9OTFaK0C&amp;amp;pg=PA249&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-03T02:18:35.137" LastActivityDate="2010-09-03T02:18:35.137" CommentCount="6" />
  <row Id="3780" PostTypeId="2" ParentId="3776" CreationDate="2010-09-01T08:22:57.220" Score="3" ViewCount="0" Body="&lt;p&gt;The product $$\phi (x) = \prod_{n &gt; 0} (1 - x^n)$$ is called the Euler function, and is well studied. I don't know of any way to compute values at special points. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentagonal_number_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler identity&lt;/a&gt; can be used to compute numerical values.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-01T08:22:57.220" CommentCount="5" />
  <row Id="3781" PostTypeId="2" ParentId="3765" CreationDate="2010-09-01T08:25:42.667" Score="1" ViewCount="0" Body="&lt;p&gt;So, basically you have a graph G that varies over time. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    1    2    3&#xA;1  f11  f12  f13&#xA;2  f21  f22  f23&#xA;3  f31  f32  f33&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Evaluate $f_{ij}(t)$ for $t$ to get $G(t)$ graph at time t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can use graph traversal algorithms like Depth-First Search,  Breadth-First Search etc. for finding out which nodes are reachable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you are allowing traveler to  wait at nodes then problem becomes &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/%28SICI%291097-0037%28199705%2929:3%3C141%3a%3aAID-NET2%3E3.0.CO;2-H/references&quot; rel=&quot;nofollow&quot;&gt;more difficult&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can create one big graph by connecting G(t) for various time values. You can keep that graph small if you can limit time to wait at each node. &#xA;If you know period of $f_{ij}(t)$ then also you can simplify this.&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-09-01T08:39:12.193" LastActivityDate="2010-09-01T08:39:12.193" CommentCount="1" />
  <row Id="3782" PostTypeId="1" AcceptedAnswerId="3819" CreationDate="2010-09-01T08:45:09.560" Score="8" ViewCount="609" Body="&lt;p&gt;I really like mathematics, but I am not good at learning it.  I find it takes me a long time to absorb new material by reading on my own and I haven't found a formula that works for me.  I am hoping a few people out there will tell me how they go about learning math so I can try out their systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know basic things.  Should I use one book at a time or should I be reading many books on the same topic at once?  Do you stop reading when you hit on a fact that you don't understand or do you keep reading?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you read all in one go or do you do a little bit and for how long (1 hr, 2 hr or more?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you read all the chapters or do you do all the exercises before moving on from a chapter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you adjust your technique in Calculus(calculation heavy) vs. Analysis (proof heavy)? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you make notes, what do you make notes about? Do you make notes while you read or after?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think these decisions all depend, can you say what they depend on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really lost here.  I would appreciate any input.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-09-05T15:50:46.223" CommunityOwnedDate="2010-09-01T09:28:26.367" Title="How Do You Go About  Learning Mathematics?" Tags="&lt;soft-question&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="14" />
  <row Id="3783" PostTypeId="1" AcceptedAnswerId="3794" CreationDate="2010-09-01T08:51:46.267" Score="5" ViewCount="222" Body="&lt;p&gt;Can we find a polynomial $p(x) \in \mathbb{R}$ such that $\text{deg}\ p(x)&gt;1$ and which satisfies $$p^{2}(x)-1=p(x^{2}+1)$$ for all $x \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question can be very well identified with the one i have asked here: &lt;a href=&quot;http://math.stackexchange.com/questions/2346/continuous-function-satisfying-fkxfxk&quot;&gt;http://math.stackexchange.com/questions/2346/continuous-function-satisfying-fkxfxk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-01T11:15:18.370" LastActivityDate="2010-09-01T13:47:24.743" Title="Polynomials which satisfy $p^{2}(x)-1 = p(x^{2}+1)$" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="3784" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T10:12:44.070" Score="0" ViewCount="0" Body="&lt;p&gt;I fear that not only there is not a royal route to mathematics, but there is no single route which is good for everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest to start with a single book, maybe skimming at the first reading parts you are not confident with; in this way you could at least have an idea of what happens. Then you could either re-read the book more carefully, or read (carefully too) a different book whose approach maybe suits best to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for exercises, it depends on you. I find them useful to set my mind on the themes, but too many exercises are boring.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-01T10:12:44.070" CommunityOwnedDate="2010-09-01T10:12:44.070" />
  <row Id="3785" PostTypeId="1" CreationDate="2010-09-01T10:43:34.957" Score="3" ViewCount="120" Body="&lt;p&gt;For the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial^2}{{\partial}x^2}(IE\frac{{\partial^2}u}{{\partial}x^2}) = \mu(\frac{{\partial^2}u}{{\partial}t^2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $E$ a function of x, derive two ODE's by separation of variables.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I greatly appreciate your help/advice. I'm pretty unsure with how to handle derivatives like this. This is my best attempt, I've been working on this one for quite a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathbin.net/51398&quot; rel=&quot;nofollow&quot;&gt;http://mathbin.net/51398&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1642" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-01T11:07:40.710" LastActivityDate="2010-09-01T13:35:17.563" Title="Seperating a PDE into two ODE's. Made an attempt but unsure if I'm correct." Tags="&lt;differential-equations&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="1" />
  <row Id="3786" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T10:44:29.190" Score="7" ViewCount="0" Body="&lt;p&gt;Von Neumann is credited with saying that you never understand mathematics, but merely get used to it. To an extent, I find this to be true. But what is it that you're getting used to when you learn mathematics?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to the sorts of objects and properties-of-objects which are the subject of study. An excellent way to get used to them is to have examples or connections to material that you already know --- regardless of whether such examples or connections are supplied by the book itself.&lt;br&gt;&lt;br&gt;Therefore, as I read a book, I look for examples of the structures or properties which are the subject of a book. If I must, I will scan ahead in the book to hunt for them, or stop reading entirely to try and conjure one myself (or find one on the internet if I find this difficult). If I come across a theorem, I try to see how it holds for these examples --- and how attempts to construct counterexamples fail.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to certain techniques for solving problems. Getting used to them involves getting used to &lt;em&gt;seeing&lt;/em&gt; them being used, getting used to &lt;em&gt;applying&lt;/em&gt; them, and getting used to the &lt;em&gt;circumstances&lt;/em&gt; in which they may be fruitfully applied.&lt;br&gt;&lt;br&gt; Therefore, in a book, if any of the material is new to me, I do the exercises. This holds regardless of whether the exercises consist of 'calculations' or 'proofs'.&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to the notation. That is, you get used to parsing the notation and translating it mentally into an alternative, often more explicit, description of what the new notation represents. Notation is important to me, in that I find a well-chosen notation an aid to understanding --- and conversely, a poorly-chosen notation a barrier. This is especially the case when the standard notation of one field looks a lot like the standard notation of a significantly different concept that I'm already familiar with.&lt;br&gt;&lt;br&gt; Therefore, as I go through a book, I carefully decide whether to adopt the notation of the book --- or to invent my own notation for the purposes of learning the concepts, without regard to the standard usage. Forcing a change of notation as you take notes has the side-effect of forcing you to reflect often as you do it, which helps to reinforce the new concepts; consciously choosing to reject the standard notation also helps you to be more conscious of the notation you are rejecting, and therefore paradoxiacally to interpret it properly (as you are continually conscious that you must distinguish it from conventions with which you're already familiar).&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You get used to taking other people's ideas and revising them --- testing their boundaries, and perhaps even rediscovering a standard generalization --- to suit your purposes, or to arrive at a more general theory. This, of course, is something which a textbook cannot explicitly teach you. But especially as you see connections in other fields, or as you pick apart proofs in the textbook, you can ask yourself certain questions, such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Given a definition of a function or concept on the domain (0,1), is there a natural interpretation which may apply for the boundary points of 0 or 1?&lt;/li&gt;&#xA;&lt;li&gt;Given a definition of an isomorphism between two objects, and suspend the requirement of invertibility? What new relationships between objects do I obtain?&lt;/li&gt;&#xA;&lt;li&gt;Given a proof by contradiction of a result, can I obtain a direct proof of the same result --- perhaps even an efficient construction?&lt;br&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously, a lot of this process is facilitated by taking notes as one learns --- so I take notes, not just of the material in the book, but of the work I do and the reflections I make.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-09-01T10:44:29.190" CommunityOwnedDate="2010-09-01T10:44:29.190" CommentCount="3" />
  <row Id="3787" PostTypeId="2" ParentId="3777" CreationDate="2010-09-01T10:57:03.030" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;del&gt;The uncountability of $\mathbb{R}$ must be used, but perhaps we can avoid mentioning it.&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a sketch of proof.  By restriction of domain and subtraction of its limit from itself, we may consider $f:[0,1]\rightarrow\mathbb{R}$ such that $f$ has zero pointwise limit and removable discontinuity everywhere.  If we replace $f$ by $|f|$, we may further assume that $f$ is strictly positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the cardinality of a finite set $S$ by $|S|$.  Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_0 = \lbrace x: 1\le f(x)\rbrace$ and $I_n = \lbrace x: \frac{1}{n+1}\le f(x)\lt \frac{1}{n}\rbrace$ for $n\ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then each $I_n$ must be finite, or else it has an accumulation point $p\in[0,1]$ and this contradicts the assumption that the limit of $f$ at $p$ is zero.  Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_n = \lbrace x\in[0,1]: |x-y|&amp;lt;\frac{1/2}{3^{n+1}|I_n|} \textrm{ for some } y\in I_n\rbrace.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($J_n$ is empty if $I_n$ is empty.)  Hence $[0,1] = \bigcup\limits_{n\ge 0} I_n \subseteq \bigcup\limits_{n\ge 0} J_n$.  But the latter has total length $\le \frac{1}{2}$.  So we arrive at a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] Alternatively, $\lbrace J_n \rbrace_{n=1,2,...}$ is an open cover of $[0,1]$.  So there is a finite subcover.  This contradicts that the total length of all $J_n$'s is at most $1/2$.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastEditorUserId="1551" LastEditorDisplayName="" LastEditDate="2010-09-01T12:51:01.380" LastActivityDate="2010-09-01T12:51:01.380" CommentCount="3" />
  <row Id="3788" PostTypeId="2" ParentId="3750" CreationDate="2010-09-01T12:33:07.940" Score="10" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/didog.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower figure: graph of oil volume/max. volume (in %) versus oil level $l$ (in feet). The horizontal straight lines represent the area/volume ratio $A(l)/A(10)=V(l)/V(10)$ (in %)  for every multiple of 10% from 0% to 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the tank radius is $5$, the oil level with respect to the bottom of&#xA;the tank is given by $l=5-5\cos \frac{\theta }{2}$, where $\theta $ is the&#xA;central angle as shown in the figure. The area of the tank cross section&#xA;filled with oil is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(\theta )=\frac{25}{2}\theta -\frac{25}{2}\sin \theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(l)=25\arccos (\frac{5-l}{5})-\frac{25}{2}\sin (2\arccos (\frac{5-l}{5}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area ratio $A(l)/A(10)=V(l)/V(10)$ where $V(l)$ is the oil volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(l)$ denote this area ratio in percentage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(l)=\frac{100}{\pi }\arccos \left( 1-\frac{1}{5}l\right) -\frac{50}{\pi }%&#xA;\sin \left( 2\arccos \left( 1-\frac{1}{5}l\right) \right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the the sequence of $f(l)$ for $l=0,1,2,\ldots ,10$. The graph of $%&#xA;f(l)$ is shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0)=0$, $f(1)=5.2044$, $f(2)=14.238$, $f(3)=25.232$, $f(4)=37.353$, $f(5)=50$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(6)=62.647$, $f(7)=74.768$, $f(8)=85.762$, $f(9)=94.796$, $f(10)=100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: &lt;em&gt;One still needs to solve the nonlinear equation&lt;/em&gt; $f(l)-10k=0$ &lt;em&gt;for&lt;/em&gt; $k=1,2,3,4,6,7,8,9$, e. g. by the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Secant_method&quot; rel=&quot;nofollow&quot;&gt;Secant Method&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; 2: The problem of solving graphically, as shown in the secong figure, is that it would be very difficult, or impossible, to get the required accuracy of 0.01 (feet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; The oil level marks (in feet) should be placed at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,1.57,2.54,3.40,4.21,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5,5.79,6.60,7.46,8.44,10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;corresponding to the oil volume percentage of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0,10,20,30,40,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$50,60,70,80,90,100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation was based on the following $f$ function values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0)=0.0$, $f(1.5648)=10.0$, $f(2.5407)=20.0$, $f(3.40155)=30.0$, $f(4.21135)=40.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(5)=50.0$, $f(5.7887)=60.0$, $f(6.59845)=70.000$, $f(7.4593)=80.000$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(8.4352)=90.000$, $f(10)=100.0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt; Figure of marks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/j6vXn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[Rearranged to show the sequence of editions and updates.]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-12T23:05:28.853" LastActivityDate="2010-09-12T23:05:28.853" CommentCount="5" />
  <row Id="3789" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T12:39:13.593" Score="0" ViewCount="0" Body="&lt;p&gt;I think you best follow the curriculum a university publishes online. MIT has OCW for example. Your questions highly depend on yourself and what works best for you, I think you'll have to find that out on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: for me it is very hard to do the same thing for more than two hours, then I switch to something else because my mind just stops working otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-09-01T12:39:13.593" CommunityOwnedDate="2010-09-01T12:39:13.593" CommentCount="1" />
  <row Id="3790" PostTypeId="2" ParentId="783" CreationDate="2010-09-01T12:40:27.527" Score="0" ViewCount="0" Body="&lt;p&gt;The important point here, I think, is about Hom(a,a) and Hom(b,b). nm is guaranteed to be an element of the first and mn an element of the second; to be an isomorphism, both of these maps must be the identity by definition. The definition of category requires that you have at least the identity in both endomorphism sets; if neither set has non-identity elements then you get an isomorphism, since the compositions don't have anything other than the identity to be, but this needn't hold in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximally toy example: take the full subcategory of sets given by the one element set A and the two element set B. Hom(A,A) is only the identity, but Hom(B,B) has four elements, two of which are not isomorphisms.&lt;/p&gt;&#xA;" OwnerUserId="1646" LastActivityDate="2010-09-01T12:40:27.527" />
  <row Id="3793" PostTypeId="1" AcceptedAnswerId="3811" CreationDate="2010-09-01T13:35:29.013" Score="5" ViewCount="146" Body="&lt;p&gt;I have the following well-known exercise in cardinal arithmetics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\kappa, \lambda$ are cardinals such that $\lambda$ is infinite, then $\kappa^{&amp;lt;\lambda}$ equals the supremum of the $\kappa^{\theta}$, where $\theta &amp;lt; \lambda$ is a cardinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Isn't this false for $\kappa=1$? The left side is $\lambda$ and the right side $1$. Remark that $\kappa^{&amp;lt;\lambda}$ is defined to be the cardinality of the disjoint(?) union of the function sets $map(\alpha,\kappa)$, where $\alpha &amp;lt; \lambda$ is an ordinal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is this supremum understood as a cardinal supremum or an ordinal supremum? If the latter is the case, why should the supremum be a cardinal at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Anyway, I don't know how to solve this exercise. I've already seen it several times, but 1) and 2) remain obscure in the literature I know.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-02T18:43:52.797" LastActivityDate="2010-09-02T18:43:52.797" Title="cardinal exponentiation, $k^{&lt;\lambda}$" Tags="&lt;set-theory&gt;&lt;logic&gt;&lt;cardinals&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3794" PostTypeId="2" ParentId="3783" CreationDate="2010-09-01T13:42:22.430" Score="11" ViewCount="0" Body="&lt;p&gt;There is no solution. Here is the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note, that the highest coefficient of $p$ must be 1. In particular, $\lim_{x \to \infty} p(x) = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume, that there exists a real value $c$ such that $p (c) = 0$. Then&#xA;$$ 0 = (p(c)^2 - 1)^2 - 1 = p(c^2 + 1)^2 - 1 = p((c^2 + 1)^2 + 1)$$&#xA;But $(c^2 + 1)^2 + 1 &gt; c$, so any real root of $p$ would lead to another, higher root of $p$ which is impossible. So $p$ has no real root and thus $p(x) &gt; 0$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, $p(x)^2 - 1 = p(x^2 + 1) &gt; 0$ and thus $p(x) &gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a sequence $x_n$ by $x_0 = 0$ and $x_{n + 1}^2 + 1 = x_n$ with imaginary part of $x_n$ positive for $n &gt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first show, that $p(x_n) \neq 0$ for all $x_n$:&#xA;clearly, $p(x_0) \neq 0$ and $p(i)^2 - 1 = p(0) &gt; 1$ and so $p(x_1) = p(i) \neq 0$.&#xA;Assume, that $p(x_{n+1}) = 0$ for some $n &gt; 1$. Then $p(x_n) = -1$ and so $p(x_{n - 1}) = 0$. This is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the functional equation, we obtain by differentiating&#xA;$$&#xA;p(x) p'(x) = x p'(x^2 + 1)&#xA;$$&#xA;We now show by induction, that $p'(x_n) = 0$ for all $n$.&#xA;For $n = 0$, we obtain $p(0) p'(0) = 0$, and thus $p'(0) = 0$.&#xA;Assume, that $p'(x_n) = 0$, then&#xA;$$&#xA;p(x_{n+1}) p'(x_{n+1}) = x_{n + 1} p'(x_{n+1}^2+1) = x_{n+1} p'(x_n) = 0&#xA;$$&#xA;and so, $p'(x_{n+1}) = 0$ as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $x_n$ is a sequence converging to $\frac{1 + i \sqrt{3}}{2}$, and in particular consists of infinitely many distinct points, all of which are roots of $p'$ which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-09-01T13:47:24.743" LastActivityDate="2010-09-01T13:47:24.743" CommentCount="1" />
  <row Id="3795" PostTypeId="1" AcceptedAnswerId="3797" CreationDate="2010-09-01T13:55:58.633" Score="1" ViewCount="66" Body="&lt;p&gt;Let $N \subseteq M$ be a subobject in an abelian category (say, modules). A complement of $N$ in $M$ is then defined to be a subobject $Q \subseteq M$ which is maximal with respect to the condition $Q \cap N = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got just a short question what maximality means here: Does it mean that every other $Q'$ with $Q' \cap N = 0$ is contained in $Q$, or does it mean that every $Q'$ with $Q' \cap N = 0$ and $Q \subseteq Q'$ satisfies $Q = Q'$? Usually the latter is meant by maximal, right? However, then I don't understand at all the proof (sketches) in the section on essential monomorphisms and injective hulls in Gabriel's thesis.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-09-04T03:34:04.163" LastActivityDate="2010-09-04T03:34:04.163" Title="Complement of a submodule" Tags="&lt;category-theory&gt;&lt;abelian-categories&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3796" PostTypeId="2" ParentId="3742" CreationDate="2010-09-01T14:31:30.020" Score="2" ViewCount="0" Body="&lt;p&gt;Can't believe no one spoke of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Collatz_conjecture&quot; rel=&quot;nofollow&quot;&gt;Collatz conjecture&lt;/a&gt; yet.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-09-01T14:31:30.020" CommunityOwnedDate="2010-09-01T14:31:30.020" />
  <row Id="3797" PostTypeId="2" ParentId="3795" CreationDate="2010-09-01T14:50:51.753" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;Maximal&quot; means the latter, i.e. if $Q'$ contains $Q$ and $Q'\cap N = 0$, then $Q' = Q$.&#xA;Since complements are very seldom unique (as one knows already from the case of subspaces&#xA;of vector spaces), the former condition is prohibitively strong, and would almost never&#xA;be realized in practice.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-01T14:50:51.753" />
  <row Id="3798" PostTypeId="1" AcceptedAnswerId="3800" CreationDate="2010-09-01T14:53:24.840" Score="6" ViewCount="165" Body="&lt;p&gt;The presentation of the dihedral group $D_{2n}$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it incorrect to conclude that $r=s=1$ ?   In other words why doesn't this group presentation describe the trivial group.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-01T19:04:01.443" LastActivityDate="2010-09-02T20:50:36.947" Title="Presentation of  $D_{2n}$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="3799" PostTypeId="1" AcceptedAnswerId="3803" CreationDate="2010-09-01T14:57:38.597" Score="4" ViewCount="96" Body="&lt;p&gt;Let $R$ be the coordinate ring of an affine complex variety (i.e. finitely generated, commutative, reduced $\mathbb{C}$ algebra) and $M$ be an $R$ module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $s\in M$ be an element, such that $s\in \mathfrak{m}M$ for every maximal ideal $\mathfrak{m}$. Does this imply $s=0$?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-01T16:07:46.627" LastActivityDate="2010-09-01T16:07:46.627" Title="Does a section that vanishes at every point vanish?" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="3800" PostTypeId="2" ParentId="3798" CreationDate="2010-09-01T15:00:04.440" Score="5" ViewCount="0" Body="&lt;p&gt;Call $G$ the group given by that presentation. Then you can show there exists an homomorphism $\phi:G\to\mathrm{GL}_2(\mathbb R)$ which maps $s$ to $S=\left(\begin{smallmatrix}-1&amp;amp;0\\0&amp;amp;1\end{smallmatrix}\right)$ and $r$ to $R=\left(&#xA;\begin{smallmatrix}&#xA; \cos (\theta ) &amp;amp; -\sin (\theta ) \\&#xA; \sin (\theta ) &amp;amp; \cos (\theta )&#xA;\end{smallmatrix}&#xA;\right)$ with $\theta=2\pi/n$; to check this, you have to use the characteristic property of group presentations, and verify that matrices $S$ and $R$ satisfy the same relations as the generators $s$ and $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $\phi(s)\neq I$ and $\phi(r)\neq I$, it follows of course that $s\neq 1_G$ and $r\neq 1_G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can also easily find two non-identity permutations $\rho$, $\sigma\in S_n$ which satisfy the relations, and construct a map $g:G\to S_n$ such that $g(s)=\sigma$ and $g(r)=\rho$, giving another proof that $r\neq 1_G$ and $s\neq 1_G$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB&lt;/strong&gt; This is another canonical example of the things we want group representations for.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-01T15:20:47.640" LastActivityDate="2010-09-01T15:20:47.640" CommentCount="6" />
  <row Id="3801" PostTypeId="1" CreationDate="2010-09-01T15:05:52.177" Score="0" ViewCount="54" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/656/how-can-i-write-an-equation-that-matches-any-sequence&quot;&gt;How can I write an equation that matches any sequence?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm learning sequences right about now, and I'm having a really hard time finding the formula for a given sequence of numbers.  I'm worried that, at exam time I'll be spending as much time finding the formula as I am now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one problem gave me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle a_n \rangle = \{2, 8, 18, 32, 50\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, it's giving me a Math Processing error. I'm sure the code is correct; I'm also seeing this error on other people's pages. If it's me, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, one problem gave me the sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a&lt;sub&gt;n&lt;/sub&gt;) = { 2, 8, 18, 32, 50 }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a good ten to fifteen minutes tinkering around, making some diagrams, plugging in the values. I'll post some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;1&lt;/sub&gt; = 2     : +6    : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;2&lt;/sub&gt; = 8     : +10   : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;3&lt;/sub&gt; = 18    : +14   : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;4&lt;/sub&gt; = 32    : +18   : +4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&lt;sub&gt;5&lt;/sub&gt; = 50   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first checked to see by how much each values differs to the next. The +4 was the increment by how much &lt;em&gt;more&lt;/em&gt; it increased every time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I therefore came up with the following solution (among many others):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2a&lt;sub&gt;n&lt;/sub&gt; + 4a&lt;sub&gt;n-1&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This worked for the first three values in the sequence. Then it broke. Damn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually I gave up and looked at the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a&lt;sub&gt;n&lt;/sub&gt;) = (2n&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seriously? I would never have guessed this. So my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are you supposed to find a pattern? Are there any good methods for doing so?&lt;/p&gt;&#xA;" OwnerUserId="44" LastActivityDate="2010-09-01T15:05:52.177" ClosedDate="2010-09-01T15:15:28.027" Title="How to Determine a Formula for a Sequence of Numbers?" Tags="&lt;discrete-mathematics&gt;" CommentCount="3" />
  <row Id="3802" PostTypeId="2" ParentId="3777" CreationDate="2010-09-01T15:18:17.817" Score="10" ViewCount="0" Body="&lt;p&gt;I think the following works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch, I will fill in the details later if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g(x) = \lim_{t\rightarrow x} f(t)$. Then we can show that $g(x)$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h(x) = f(x) - g(x)$. Then $\lim_{t \rightarrow x} h(t)$ exists and is $0$ everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now show that $h(c) = 0$ for some $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will imply that $f(x)$ is continuous at $c$ as then we will have $f(c) = g(c) = \lim_{t-&gt;c} f(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider any point $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By limit of $h$ at $x_0$ being $0$, there is a closed interval $I_0$ (of length &gt; 0) such that $|h(x)| &amp;lt; 1$ for all $x \in I_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because, given an $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that $|h(x)| &amp;lt; \epsilon$ for all $x$ such that $0 &amp;lt; |x - x_{0}| &amp;lt; \delta$. Pick $\epsilon = 1$ and pick $I_{0}$ to be any closed interval of non-zero length in $(x_{0}, x_{0} + \delta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick any point $x_1$ in $I_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By limit of $h$ at $x_1$ being $0$, there is a closed interval $I_1 \subset I_0$ (of length &gt; 0) such that $|h(x)| &amp;lt; 1/2$ for all $x \in I_1$, by argument similar to above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing this way, we get a sequence of closed intervals $I_n$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|h(x)| &amp;lt; \frac{1}{n+1}$ for all $x \in I_n$. We also have that $I_{n+1} \subset I_n$ for each $n$, and that length $I_n$ &gt; 0. We could also arrange so that length $I_n \rightarrow 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is a point $c$ (by completeness of $\mathbb{R}$) such that $c \in \bigcap_{n=0}^{\infty}I_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that $|h(c)| &amp;lt; \frac{1}{n+1}$ for all $n$ and so $h(c) = 0$ and $f(c) = g(c)$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T16:11:30.873" LastActivityDate="2010-09-02T16:11:30.873" CommentCount="8" />
  <row Id="3803" PostTypeId="2" ParentId="3799" CreationDate="2010-09-01T15:23:18.540" Score="8" ViewCount="0" Body="&lt;p&gt;Not in general, no.  For example, if $R = \mathbb C[T]$ and $M$ is the field of fractions&#xA;of $R$, namely $\mathbb C(T)$, then (a) every maximal ideal of $R$ is principal; (b) every&#xA;element of $M$ is divisible by every non-zero element of $R$.  Putting (a) and (b) together&#xA;we find that $M = \mathfrak m M$ for every maximal ideal $\mathfrak m$ of $R$, but certainly&#xA;$M \neq 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a finitely generated example: again take $R = \mathbb C[T]$, and take $M = \mathbb C[T]/(T^2).$  Then $s = T \bmod T^2 \in \mathfrak m M$ for every $\mathfrak m$, because&#xA;$\mathfrak m M = M$ if $\mathfrak m$ is a maximal ideal other than $(T)$, and this&#xA;is clear from the choice of $s$ if $\mathfrak m = (T)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer &lt;em&gt;is&lt;/em&gt; yes if $M$ is finitely generated and torsion free.  For let $S$ be the total quotient ring of &#xA;$R$ (i.e. the product of functions fields $K(X)$ for each irreducible component $X$&#xA;of the variety attached to $R$).&lt;br&gt;&#xA;Then $M$ embeds into $S\otimes_R M$ (this is the torsion free condition), which in turn embeds into&#xA;a finite product of copies of $S$ (since it is finite type over $S$, which is just a product&#xA;of fields).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearing denominators, we find that in fact $M$ then embeds into $R^n$ for some $n$.&#xA;Thus it suffices to prove the result for $M = R^n$, and hence for $R$, in which case&#xA;it follows from the Nullstellensatz, together with the fact that $R$ is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that for any finitely generated $R$-module, if $M = \mathfrak m M$ for all $\mathfrak m$&#xA;then $M = 0$ (since Nakayama then implies that $M_{\mathfrak m} = 0$ for all&#xA;$\mathfrak m$).  Thus if $M$ is non-zero it can't be that &lt;em&gt;every&lt;/em&gt; section lies&#xA;in $\mathfrak m M$ for all $\mathfrak m$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-09-01T15:49:16.953" LastActivityDate="2010-09-01T15:49:16.953" CommentCount="1" />
  <row Id="3804" PostTypeId="2" ParentId="3798" CreationDate="2010-09-01T15:29:08.167" Score="2" ViewCount="0" Body="&lt;p&gt;A group given by presentation $G=\langle X|R\rangle$ is, by definition, the &lt;em&gt;most general&lt;/em&gt; group with a generating set $X$ and satisfying the relations $R$. Say, for simplicity, that $X=\{x_1,\ldots,x_n\}$ and $R=\{r_1(x_1,\ldots,x_n),\ldots,r_m(x_1,\ldots,x_n)\}$. Being the &quot;most general&quot; group with $n$ generators $x_1,\ldots,x_n$ that satisfy the $m$ relations $r_1,\ldots,r_m$ means that given &lt;em&gt;any&lt;/em&gt; group $H$ with elements $h_1,\ldots,h_n$ for which $r_i(h_1,\ldots,h_n)=1$ for $i=1,\ldots,m$, there will be a (unique) group homomorphism from $G$ to $H$ that maps $x_i$ to $h_i$.  It's like the &quot;free-est group&quot; with elements $x_1,\ldots,x_n$, subjected only to the relations in $R$. (The magic words are &lt;strong&gt;Van Dyck's Theorem&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are noting is that if we pick $H$ to be the trivial group, and $h_1,\ldots,h_n$ to be the trivial element, then all the relations are satisfied; but this does not tell you that the trivial group is the most general group that satisfies the given conditions, it only tells you that the trivial group is a &lt;em&gt;quotient&lt;/em&gt; of the most general group that satisfies the given conditions (as it must, since the trivial group is a quotient of everything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the dihedral group $D_{2n}$ you are looking for the most general group that has an element of order dividing $n$, an element of order dividing $2$, and such that those elements satisfy the relation $sr = r^{-1}s$ (which can be written as $srsr=1$). The trivial group certainly satisfies those relations, but it is not the most general one that does (e.g., if $n$ is even, you can look at $H=C_2\times C_2$, and identify $r$ with $(1,0)$ and $s$ with $(0,1)$). Every time you find a group and elements that satisfy the given relations, you have found a map from the group presented to the group you found, but you have not necessarily found the group presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really know that you have found the &quot;most general group&quot; that satisfies the given conditions is not always trivial. One way is to show that the group you found (call it $G$, with elements $g_1,\ldots,g_n$ playing the roles of the $x_i$) has the property ascribed to the group presented by the presentation: given any group $H$, and any elements $h_1,\ldots,h_n$ in $H$ that satisfy $r_1,\ldots,r_m$ in $H$, there exists a unique group homomorphism from your group $G$ to $H$ that maps $g_i$ to $h_i$. The trivial group does not have this property for the presentation given in $D_{2n}$, so it cannot be $D_{2n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method that works sometimes is to find a &quot;normal form&quot; for the elements of the group presented, allowing you to count exactly how many elements there are. This is not hard to do in $D_{2n}$: every element of $D_{2n}$ can be written as a product of powers of $r$ times powers of $s$. Using the rules that say that $r^n=1$, $s^2=1$, and that $sr=r^{-1}s = r^{n-1}s$, you can &quot;move&quot; all the $r$s that occur in such a product all the way to the left, and all the $s$'s to the right, so that every element of $D_{2n}$ can be written in the form $r^is^j$, with $0\leq i\lt n$ and $0\leq j\lt 2$. We do not know if each element corresponds to one and only one such expression, but certainly each element corresponds to at least one such expression. That means that $D_{2n}$ has, at most, $2n$ elements. If you can find a group $G$, generated by elements $g_1$ and $g_2$ such that $g_1^n=1$, $g_2^2=1$, and $g_2g_1=g_1^{-1}g_2$, &lt;em&gt;and&lt;/em&gt; it turns out that $G$ has exactly $2n$ elements, then you know there is a homomorphism from $D_{2n}$ to $G$ that sends $r$ to $g_1$ and $s$ to $g_2$, so that $G$ will be a quotient of $D_{2n}$; since we are assuming that $G$ has exactly $2n$ elements, and we know that $D_{2n}$ has at most $2n$ elements, then you can conclude that the map has to be injective and the kernel is trivial, and so that $D_{2n}$ is in fact isomorphic to the group $G$ you found. This is one way of &quot;realizing&quot; a presentation as a group that you can get your hands on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Note that by the property ascribed to the group given by presentation, the group must be generated by $X$; otherwise, there is a map from $G$ onto $\langle X\rangle$ that sends each $x_i$ to itself, so you get two maps from $G$ to itself sending each $x_i$ to itself (the identity, and the composite map $G\to\langle X\rangle \hookrightarrow G$); but the definition requires that there be a unique homomorphism to accomplish this.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-01T15:49:21.970" LastActivityDate="2010-09-01T15:49:21.970" CommentCount="2" />
  <row Id="3805" PostTypeId="1" AcceptedAnswerId="3807" CreationDate="2010-09-01T15:36:29.613" Score="2" ViewCount="102" Body="&lt;p&gt;Hi everybody, first question on math.stackexchange :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm studying for a Cryptography - Communication Security exam, and it involves a certain quantity of number theory - finite field theory, so be warned: this is my first encounter with these topics, and you'll have to be extra-clear with me :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I was doing pretty well with the questions and exercises in $GF(4)$… then I hit $GF(8)$ and realized I'm still missing the point :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that to represent, let's say, $x^3$, the fourth element of $GF(8)$, I just take the result of the division of my irreducible polynomial of choice, let's say $x^3+x+1$, and I'm happy. I can then build addition and multiplication tables, and switch to and from a binary representation when it seems more convenient (e.g. when XORing with something else), and this also makes me happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do I start dividing $x^0$, $x^1$, $x^2$, etc.?&lt;br&gt;&#xA;This sure seems like a sound idea when compared to “try some random polynomial“, but I can't figure why these are the candidates for being the members of the finite field? Why am I sure that this will generate all the elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question: $GF(4)$ seems to have a number of properties that are quite useful for manipulation. It seems obvious that $x+x \equiv 0$ in any $GF(2^n)$, but do $x^2 \equiv x+1$ or $x^3\equiv x^2+x\equiv 1$ always hold? Is this in any way dependent on the irreducible polynomial chosen? Are these predictable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive me for any obvious error, and thanks in advance for any help.&lt;/p&gt;&#xA;" OwnerUserId="894" LastEditorUserId="894" LastEditorDisplayName="" LastEditDate="2010-09-01T16:39:03.310" LastActivityDate="2010-09-01T21:33:50.253" Title="Powers of $x$ as members of Galois Field and their representation as remainders" Tags="&lt;number-theory&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3806" PostTypeId="2" ParentId="3758" CreationDate="2010-09-01T16:24:36.397" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ Pierre's answer is a generating function approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\rm\quad\quad\quad\;\;  f(x+t) \;=\; \sum_{n=0}^\infty \;\;  f^{(n)}(x) \; \frac{t^n}{n!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad (fgh)(x+t) \;=\; f(x+t)\: g(x+t)\: h(x+t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\;\:  \;=\; (f + f' t +\:\cdots) (g + g' t + \:\cdots) (h + h' t + \:\cdots) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\;\: \;=\; fgh + (f'gh + fg'h + fgh')\: t + \:\cdots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad (fgh)(x+t) \;=\; fgh + (fgh)' t + \:\cdots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we conclude $\:\rm(fgh)' = f'gh + fg'h + fgh'$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes to n-th order derivatives, as Pierre mentioned, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\displaystyle\rm\sum_{n=0}^\infty \: (fgh)^{(n)} \frac{t^n}{n!} \ =\ \sum_{n=0}^\infty \:\bigg(\ \sum_{i+j+k=n} \frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}\bigg) \frac{t^n}{n!} $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T16:33:07.393" LastActivityDate="2010-09-01T16:33:07.393" CommentCount="1" />
  <row Id="3807" PostTypeId="2" ParentId="3805" CreationDate="2010-09-01T16:49:02.647" Score="3" ViewCount="0" Body="&lt;p&gt;First: a Galois field is a finite field. A finite field will have $p^n$ elements for some prime $p$ and some positive integer $n$ (in fact, there is, up to isomorphism, one and only one finite field with $p^n$ elements). If a field has $p^n$ elements, then $x+x+\cdots+x=0$ ($p$ summands) for all $x$, and no smaller positive integer has that property; this is called the &lt;strong&gt;characteristic&lt;/strong&gt; of the field. For fields of order a power of $2$, you do indeed have $x+x=0$ for all $x$, but not for any other size Galois fields. However, in computer science and  cryptography there is some preference for working in fields of characteristic $2$, because they tend to be easier to represent and work with in computers (which themselves work in &quot;characteristic $2$&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by &quot;why do I start dividing $x^0$, $x^2$, $x^2$, etc.?&quot;. As you have probably seen, the field $GF(2^n)$ can always be described in terms of a  monic polynomial $P(t)$ of degree $n$ that is irreducible over $GF(2)$. This amounts to constructing the field $GF(2)[t]/(P(t))$. What this tells you is that you have the smallest ring that contains $0$, $1$, and $x$, subject to the conditions $1+1=0$, $1\cdot 1=1$, and $P(x)=0$. Any element of this ring can be written as a polynomial expression in $x$ with coefficients $0$ or $1$, $q(x) = b_mx^m+\cdots + b_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because we are assuming that $P(x)=0$, using the division algorithm we can always write any polynomial $q(t)$ as $q(t)=P(t)b(t) + r(t)$, with $r(t)=0$ or $\deg(r)\lt \deg(P)=n$. Evaluating at $x$ and using the fact that $p(x)=0$, we get that $q(x)=r(x)$, so in fact every element in this ring can be written as a polynomial expression in $x$ of degree less than $n$. The expression is unique, because if $r(x)=s(x)$ with $r$ and $s$ of degree less than $n$, then $r-s$ would be a multiple of $P(t)$ (because the latter is irreducible), and degree considerations tell you that $r=s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each element of $GF(2^n)$ can be written uniquely as $q(x)$ where $q$ has degree less than $n$. Choosing different polynomials amounts to choosing different representations for the elements, and so different rules for what $x^n$ will mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: in order for $x^2=x+1$ to hold in your Galois field $GF(p^n)$, you must have that $x^2-x-1=0$, so the polynomial $P$ you picked must divide $x^2-x-1$, and so you must be either in $GF(p)$ or in $GF(p^2)$. Similarly, for $x^3=x^2+x=1$ to hold you would need to be in $GF(p)$ or $GF(p^2)$ in order for $x^2+x-1=0$ to hold; and whether it holds depends on the polynomial $P$ that you picked to define how $x$ behaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: you divide by the polynomial so you can get a unique expression for each element in $GF(2^n)$ in terms of the distinguished element $x$ that helps you construct it (the behaviour of the element $x$ being determined by the polynomial). The condition $\alpha+\alpha=0$ will hold for every $\alpha$ in every Galois field of size $2^n$ for any $n$; but no power of $x$ smaller than $n$ will be expressible in terms of lower powers of $x$, and the expression of $x^n$ in terms of lower powers is determined by your choice of polynomial.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T16:49:02.647" CommentCount="1" />
  <row Id="3809" PostTypeId="1" AcceptedAnswerId="3814" CreationDate="2010-09-01T18:40:56.953" Score="0" ViewCount="230" Body="&lt;p&gt;Is every bounded function on an interval [a, b] is the sum of two bounded one-to-one functions on $[a, b]$?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check this easily by considering some examples, but I can't prove or disprove this fact. Can somebody help with a counterexample or a proof.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T10:26:16.887" Title="Bounded function on an interval as the sum of 2 bounded 1-1 functions" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="3810" PostTypeId="2" ParentId="3776" CreationDate="2010-09-01T18:44:23.500" Score="4" ViewCount="0" Body="&lt;p&gt;If you want to evaluate $$\prod_{n=1}^\infty (1-1/2^n)$$ numerically you will find that it's VERY close to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{1/24}\sqrt{\frac{2\pi}{\log 2}}\exp{(-\pi^2/(6\log 2))}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is about 0.2887880950866... And that's because $$\exp{(-4\pi^2/\log 2)}$$ is pretty small, approximately 1.839x10^(-25).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen from the fact that $$\Delta(-1/z) = z^{12}\Delta(z),$$ where $$\Delta(z)$$ is the cusp form of weight 12 defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta(z)=q\prod_{n=1}^\infty (1-q^n)^{24},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $$q=e^{2\pi iz}$$ for Im(z)&gt;0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just put $$z=\frac{2\pi i}{\log 2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this technique to greatly accelerate the convergence of your product for terms other than (1/2)^n.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-01T18:44:23.500" CommentCount="3" />
  <row Id="3811" PostTypeId="2" ParentId="3793" CreationDate="2010-09-01T18:52:04.667" Score="1" ViewCount="0" Body="&lt;p&gt;First, (2): if you define a cardinal to be an ordinal that is not bijectable with any strictly smaller ordinal, then suppose that $\alpha$ is the ordinal supremum of $\kappa^{\theta}$, $\theta\lt\lambda$. Let $\beta$ be a strictly smaller ordinal than $\alpha$. By the definition of supremum, there must exist a $\theta\lt\lambda$ such that $\beta\lt\kappa^{\theta}\leq\alpha$; in particular, $\kappa^{\theta}$ cannot be bijected with $\beta$ (being a cardinal), and therefore neither can $\alpha$ (that would give an embedding of $\kappa^{\theta}$ into $\beta$, and Cantor-Bernstein would give you $|\beta|=\kappa^{\theta}$, contradicting that the latter is not bijectable with any strictly smaller ordinal). Thus, $\alpha$ is not bijectable with any strictly smaller ordinal, and so must be a cardinal. So whether you define the supremum as the ordinal-sup or the cardinal-sup, you will still get a cardinal (this holds for any set of cardinals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, (1): you are correct that the definition does not match this for $\kappa=1$ (or for $\kappa = 0$); as noted by Carl in the comments, this is likely an erratum or ommission; it should hold for any $\kappa&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The definition as a sum is done running over all ordinals, rather than all cardinals, so I'm fixing this below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, (3): you are trying to show that the supremum of the $\kappa^{\theta}$ equals the sum over all ordinals $\alpha&amp;lt;\lambda$ of $|\kappa^{\alpha}$, assuming $\kappa\geq 2$ (to prevent the problems noted). This because $|\kappa^{\alpha}|=|map(\theta,\kappa)|$ by definition, and the cardinality of the disjoint union is the cardinal sum. I believe this can be shown by transfinite induction on $\lambda$ as I do below, but there probably is a simpler method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the proposition we want to show is that for any infinite &lt;em&gt;ordinal&lt;/em&gt; $\lambda$, we have&#xA;$$\sup_{\alpha\lt\lambda}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}|\kappa^{\alpha}|.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the equality holds for $\lambda=\omega$: if $2\leq \kappa\lt\aleph_0$, then $\sup\{|\kappa^n|\,|\, n=0,1,2,3,\ldots\} = \aleph_0$ and $\sum_{n=0}^{\infty}|\kappa^n| = \aleph_0$; if $\aleph_0\leq\kappa$, then $|\kappa^n|=\kappa$ for all $n$, and $\sum_{n=0}^{\infty}|\kappa^n| = \sum_{n=0}^{\infty}\kappa = \kappa\aleph_0=\kappa$, so both sides agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the result holds for $\lambda$; then the supremum of $|\kappa^{\alpha}|$ with $\alpha\lt\lambda^+$ is $|\kappa^{\lambda}|$; on the other hand, &#xA;$$\sum_{\alpha\lt\lambda^+}|\kappa^{\alpha}| = \left(\sum_{\alpha\lt\lambda}|\kappa^{\alpha}|\right) + \kappa^{\lambda} = \sup_{\alpha\lt\lambda}|\kappa^{\alpha}|+|\kappa^{\lambda}|=|\kappa^{\lambda}|,$$&#xA;where the last equality holds because $|\kappa^{\alpha}|\leq |\kappa^{\lambda}|$ for each $\alpha\lt \lambda$, so the supremum is at most $|\kappa^{\lambda}|$, and the sum of two infinite cardinals is equal to their maximum. So again the two expressions agree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we want to show that if $\lambda$ is a limit ordinal and the result holds for all $\beta\lt\lambda$, then it holds for $\lambda$. Then&#xA;$$\sup_{\alpha\lt\lambda}(|\kappa^{\alpha}|) = \sup_{\beta\lt\lambda}\left(\sup_{\alpha\lt\beta}(|\kappa^{\alpha}|\right) = \sup_{\beta\lt\lambda}\sum_{\alpha\lt\beta}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}(|\kappa^{\alpha}|).$$&#xA;So the equality holds for $\lambda$ as well. This establishes the result by transfinite induction for all infinite ordinals $\lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further Edit, 2 Sep 2010:&lt;/strong&gt; Clarify how to finish it off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the above shows that for any ordinal $\lambda$, $\sup_{\alpha\lt\lambda}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}|\kappa^{\alpha}|$. To finish the exercise, we need to show that if $\lambda$ is a cardinal (that is, an ordinal that is not bijectable with any strictly smaller ordinal), then $\sup\{|\kappa^{\alpha}|\colon\alpha$ is an ordinal and $\alpha\lt\lambda\} = \sup\{|\kappa^{\theta}|\colon \theta$ is a cardinal and $\theta\lt\lambda\}$. To see this, note that $|\kappa^{\alpha}|=|\kappa|^{|\alpha|}$, and since $\lambda$ is assumed to be a cardinal, if $\alpha\lt\lambda$, then there exists a cardinal $\theta$, $\theta\lt\lambda$, such that $|\alpha|=|\theta|$, and hence $|\kappa^{\alpha}|=|\kappa^{\theta}|$.  Thus, the two sets are equal, so their suprema are equal as well. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-02T18:43:06.160" LastActivityDate="2010-09-02T18:43:06.160" CommentCount="7" />
  <row Id="3812" PostTypeId="2" ParentId="3798" CreationDate="2010-09-01T18:55:25.270" Score="1" ViewCount="0" Body="&lt;p&gt;In the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$r^{n}=s^{2}=1, rs=sr^{-1}$ ... why is it incorrect to &lt;em&gt;conclude&lt;/em&gt; that $r=s=1$ ?  ... why doesn't this group presentation &lt;em&gt;describe&lt;/em&gt; the trivial group&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the key words are &quot;conclude&quot; and &quot;describe&quot;.  You are in effect asking what logic applies to statements about $r$ and $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that assertions about $r$ and $s$ are understood as valid in this context only if they hold in ALL groups.  The assertion &quot;if a pair of group elements $r$ and $s$ satisfies the defining relations of $D_{2n}$, then $r=s=1$&quot; is false in some groups and true in others.  It is true in groups with no subgroup of order $2n$, such as large cyclic groups of prime order.   It is false in other groups, such as the dihedral group of order 2n, or the group of order 2 where $r=1, s=-1$ is a pair satisfying the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truth in all groups is a semantic definition and can be replaced by an equivalent, syntactic definition that gives the combinatorial rules for making all valid deductions from the defining relations.  $r=s=1$ is not a valid algebraic consequence of the algebraic rules.   As in all problems of logic, showing that something does NOT follow from some rules or axioms is a matter of constructing an example (a model) where the rules apply but the assertion is false.  Here the models are groups with a pair of elements $r,s$ satisfying the relations, so for showing what is not true in the logic of generators and relations the algebraic approach doesn't provide an alternative to constructing models.  For assertions that &lt;em&gt;do&lt;/em&gt; follow from the defining relations, the algebraic definition is useful, because instead of surveying a possibly infinite collection of models to show that the statement is correct, one can perform a finite combinatorial derivation that works for all models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Added in response to comment from Matt E about normal forms: showing that any element can be placed in the normal form follows from the algebraic relations alone.  However, to use normal forms as a method of calculating in the group or understanding it, you need to show that distinct elements in the group have different normal forms. This is again the non-algebraic (or not purely algebraic) problem of showing that relations do not hold in a structure, which is done by constructing models such as actions of the group on particular spaces.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-02T20:50:36.947" LastActivityDate="2010-09-02T20:50:36.947" CommentCount="2" />
  <row Id="3813" PostTypeId="2" ParentId="3805" CreationDate="2010-09-01T19:24:26.697" Score="1" ViewCount="0" Body="&lt;p&gt;It appears that you may mistakenly believe that every element of your finite field is a power of $\rm x$. Here is a very simple counterexample: put $\rm\; f(x) = (x^5-1)/(x-1) = x^4 + x^3 + x^2 + x + 1 \;$, which is irreducible $\pmod 2$. Then $\rm\; (x-1) \: f(x) = x^5 - 1 \Rightarrow \rm x^5 \equiv 1 \pmod{f(x)}$. This implies that the powers of $\rm x$ generate only 5 elements, not all 15 nonzero elements of $\rm GF(2^4)$ constructed with $\rm f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there does exist an element whose powers generate all the nonzero elements of the finite field (said more technically: the multiplicative group is cyclic). Such generators are known as &lt;strong&gt;primitive elements&lt;/strong&gt;. However, generally there is no simple closed-form known for such generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that you didn't see this in $\rm GF(4)$ or $\rm GF(8)$ is because their multiplicative groups have prime order $3$ and $7,$ so every element $\ne 1,0 \:$ is a generator (by Lagrange's theorem the order of an element divides the order of the group which, being a prime $\rm p,$ forces the order to be $\rm p\:$ for elements $\rm \ne 1,0\:$).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-01T21:33:50.253" LastActivityDate="2010-09-01T21:33:50.253" />
  <row Id="3814" PostTypeId="2" ParentId="3809" CreationDate="2010-09-01T19:53:30.627" Score="6" ViewCount="0" Body="&lt;p&gt;This is a reply to KennyTM's comment (I can't comment yet, I don't have the reputation).&lt;br&gt;&#xA;Assume for simplicity that $[a,b] = [0,1]$.&lt;br&gt;&#xA;Then $1_{\mathbb Q}$ is the sum of the 1-1 functions $f$ and $1_{\mathbb Q} - f$, where $f(x) = x/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 1&lt;/strong&gt; A reply to Hans Lundmark's answer: If $f$ is of bounded variation on $[a,b]$, so that it can be expressed as $f = f_1 - f_2$ with $f_1, f_2$ non-decreasing, then we can express it as the sum of the bounded 1-1 functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_1(x) = f_1(x) + x$&lt;br&gt;&#xA;$g_2(x) = -f_2(x) - x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;&#xA;Conversely, if we require that the component functions $g$ and $h$ (with $f = g + h$) be continuous, then they must both be monotonic (because 1-1); so their sum is of bounded variation. Thus a continuous function on $[a,b]$ is not the sum of two bounded, continuous, 1-1 functions if it is of unbounded variation. (I know this wasn't the original question, but we've had no feedback from the original poster, so we have to make up our own conditions.)&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-09-02T09:34:11.030" LastActivityDate="2010-09-02T09:34:11.030" CommentCount="2" />
  <row Id="3815" PostTypeId="1" AcceptedAnswerId="3817" CreationDate="2010-09-01T20:28:38.880" Score="4" ViewCount="327" Body="&lt;p&gt;Proposition: For a set $X$ and its power set $P(X)$, any function $f\colon P(X)\to X$ has at least two sets $A\neq B\subseteq X$ such that $f(A)=f(B)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see how this would be true if $X$ is a finite set, since $|P(X)|\gt |X|$, so by the pidgeonhole principle, at least two of the elements in $P(X)$ would have to map to the same element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this proposition still hold for $X$ an infinite set? And if so, how does this show that the set of all singleton sets cannot exist?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-11T02:11:33.977" LastActivityDate="2010-09-11T02:11:33.977" Title="Why does the set of all singleton sets not exist?" Tags="&lt;set-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="3816" PostTypeId="2" ParentId="3809" CreationDate="2010-09-01T20:37:31.700" Score="2" ViewCount="0" Body="&lt;p&gt;A (maybe) related result, although not exactly what you're asking for: every real-valued function of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bounded_variation&quot; rel=&quot;nofollow&quot;&gt;bounded variation&lt;/a&gt; on an interval can be written as the difference of two non-decreasing functions.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-01T20:37:31.700" />
  <row Id="3817" PostTypeId="2" ParentId="3815" CreationDate="2010-09-01T20:43:50.437" Score="7" ViewCount="0" Body="&lt;p&gt;If you had a function $f\colon P(X)\to X$ such that $f(A)=f(B)\Rightarrow A=B$, then $f$ would be one-to-one, which would imply $|P(X)|\leq |X|$; since $|X|\lt |P(X)|$ holds by Cantor's Theorem, that would give you $|P(X)|\lt|P(X)|$, which is impossible. This argument does not use finiteness of $X$, so it holds for $X$ infinite as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the collection of all singletons is a set $X$. Define a map $f\colon P(X)\to X$ by $f(A) = \{A\}$; this is well defined, since $\{A\}$ is a singleton. If $A\neq B$, then $\{A\}\neq\{B\}$, so $f$ would be one-to-one. Since, by the proposition, this is impossible, we conclude that our assumption that the collection of all singletons is a set must be false. Thus, there is no &quot;set of all singletons.&quot;&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-01T20:43:50.437" CommentCount="2" />
  <row Id="3818" PostTypeId="2" ParentId="3750" CreationDate="2010-09-01T21:16:48.140" Score="2" ViewCount="0" Body="&lt;p&gt;Looking &lt;a href=&quot;http://math.stackexchange.com/questions/3750/aunt-and-uncles-fuel-oil-tank-dip-stick-problem/3757#3757&quot;&gt;@Isaac's answer&lt;/a&gt;, the appearance of &quot;$\alpha - \sin\alpha$&quot; and &quot;$1-\cos\frac{\alpha}{2}$&quot; made me think that power series would come in handy, with their respective initial terms vanishing. They didn't turn out to be as handy as I might have hoped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $w$ for the ratio of oil to the tank's capacity, and $d$ for the ratio of the oil's depth to the radius of the tank. Then, by Isaac's work ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi w = \alpha - \sin\alpha = \alpha-\sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)!}\alpha^{2m+1}= \sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m+1)!}\alpha^{2m+1}$$&#xA;$$d = 1 - \cos\frac{\alpha}{2}=1-\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $w_n$ and $d_n$ for the values associated with lopping off the power series at the $n$-th term (that is, from $m=1$ to $m=n$). In particular, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi w_1 = \frac{\alpha^3}{3!}$$&#xA;$$d_1 = \frac{\alpha^2}{2^2 2!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eliminating $\alpha$ yields the relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$32 d_1^3 = 9 w_1^2 \pi^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation, though, isn't very accurate (as shown by the $n=1$ curve in the diagram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6aIcm.png&quot; alt=&quot;Ratio of depth to radius, vs ratio of volume to capacity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take $n=2$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\pi w_2 = \frac{\alpha^3}{3!}-\frac{\alpha^5}{5!}$$&#xA;$$d_2 = \frac{\alpha^2}{2^2 2!}-\frac{a^4}{2^4 4!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... then eliminating $\alpha$ isn't quite so easy unless you do something like invoke Mathematica's &lt;code&gt;Resultant&lt;/code&gt; function, which yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4718592 d_2^5 - 13762560 d_2^4 + 10035200 d_2^3 - 768000 d_2^2 w_2^2 \pi^2 + &#xA; 3024000 d_2 w_2^2 \pi^2 - 2822400 w_2^2 \pi^2 + 1875 w_2^4 \pi^4 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also not terribly accurate at we get close to $w=0.5$ (which should have $d=1.0$). With $n=3$ we're very close, and with $n=4$ and above, we've pretty much converged on our target (at least according to the graph). I won't give the $n=4$ equation (the coefficient on $d^9$ is $2^{71} 5^4 7^2$), but here are some data points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0.0, 0.000000)$$&#xA;$$(0.1, 0.312953)$$&#xA;$$(0.2, 0.508164)$$&#xA;$$(0.3, 0.680452)$$&#xA;$$(0.4, 0.842809)$$&#xA;$$(0.5, 1.001730)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, power series and resultants aren't exactly &quot;lowest possible grade level&quot;, but this is the best I can do at this point. &lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-09-01T21:16:48.140" CommentCount="1" />
  <row Id="3819" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T23:06:25.530" Score="9" ViewCount="0" Body="&lt;p&gt;Of course everybody has their own learning style.  Here are some general suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a &lt;strong&gt;teacher&lt;/strong&gt;.  It is &lt;em&gt;hard&lt;/em&gt; to learn mathematics on your own until you have reached a certain level of mathematical sophistication; nobody is there to tell you what is important and what is unimportant.  Take courses at a university; as Agusti Roig mentioned, video lectures on MIT's &lt;a href=&quot;http://ocw.mit.edu/index.htm&quot; rel=&quot;nofollow&quot;&gt;OpenCourseWare&lt;/a&gt; are a good cheap alternative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Read&lt;/strong&gt; as much mathematics as you possibly can, from as many sources as you possibly can.  This is not limited to textbooks but extends to popular math books, blogs, expository papers, MO, math.SE... doing this will get you used to not understanding things, which is important.  It will also expose you to many fascinating ideas that will fire up your curiosity enough for you to look at the material more seriously.  As &lt;a href=&quot;http://math.stanford.edu/~vakil/potentialstudents.html&quot; rel=&quot;nofollow&quot;&gt;Ravi Vakil&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;...mathematics is so rich and infinite that it is impossible to learn it systematically, and if you wait to master one topic before moving on to the next, you'll never get anywhere. Instead, you'll have tendrils of knowledge extending far from your comfort zone. Then you can later backfill from these tendrils, and extend your comfort zone; this is much easier to do than learning &quot;forwards&quot;. &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A specific way in learning backwards is easier than learning forwards is that instead of reading the proof of a theorem in a book, you might hear about a theorem without proof, but remember that someone on a blog said something vague about a crucial step, then gradually learn enough material that suddenly you can work out the proof independently.  I have done this a handful of times, and it is quite satisfying. For example, the theorem I proved in &lt;a href=&quot;http://qchu.wordpress.com/2010/04/27/the-mckay-correspondence-i/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; is classical and extremely well-known, but I had never seen a proof of it.  I juggled around some ideas for about half a year until I figured out how to prove Lemma 6 (which I saw in a paper somewhere, again without proof), and I wrote down a proof.  Later I read a proof in an actual book, and although the second half of the proof was similar, it did not use Lemma 6.  I have yet to see a proof of Lemma 6 in print, although I am sure it is also well-known.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might sound like more work.  But guess how well I &lt;em&gt;remember&lt;/em&gt; this theorem and its proof now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do&lt;/strong&gt; as much mathematics as you possibly can.  This is not limited to textbook exercises but includes competition problems, finding alternate proofs of theorems, working out concrete examples of abstract theorems, etc.  I try to do this as much as I can on &lt;a href=&quot;http://qchu.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;my blog&lt;/a&gt;; it keeps me sharp and is also, at least for me, much more fun than reading a textbook, which I can't do for long periods of time.  This is also why I post here so often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; everything.  There are a few aspects to this.  If something is unclear or unmotivated to you, ask yourself exactly where it becomes unclear or unmotivated.  Find someone to explain it to you (for example, on math.SE!).  Read a blog post about it.  &lt;em&gt;Write&lt;/em&gt; a blog post about it!  Ask yourself how things generalize and how they connect to other things you know.  (Again, math.SE is good for this.)  The worst thing you can do is to accept what a textbook tells you as the Word of God.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;strong&gt;teach&lt;/strong&gt; as much mathematics as you possibly can.  This is the other purpose of my blog, and is an amazing test of how well you understand something.  You would be surprised how much you can learn about something by teaching it.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-01T23:14:49.710" LastActivityDate="2010-09-01T23:14:49.710" CommunityOwnedDate="2010-09-01T23:06:25.530" CommentCount="6" />
  <row Id="3820" PostTypeId="1" AcceptedAnswerId="3822" CreationDate="2010-09-01T23:13:34.313" Score="6" ViewCount="165" Body="&lt;p&gt;Must a polynomial function $f \in \mathbb{R}[x_1, \ldots, x_n]$ that's lower bounded by some $\lambda \in \mathbb{R}$ have a global minimum over $\mathbb{R}^n$?&lt;/p&gt;&#xA;" OwnerUserId="848" LastActivityDate="2010-09-02T02:29:43.950" Title="Does a polynomial that's bounded below have a global minimum?" Tags="&lt;optimization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3821" PostTypeId="2" ParentId="3623" CreationDate="2010-09-01T23:18:15.053" Score="0" ViewCount="0" Body="&lt;p&gt;SIAM (The Society for Industrial and Applied Mathematics) has collected some oral interviews and other materials that have some bearing on the history of ordinary and partial differential equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://history.siam.org/&quot; rel=&quot;nofollow&quot;&gt;http://history.siam.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a lot of material on this topic that is related to more specialized historical studies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;History of Hydraulics by Hunter Rouse and Simon Ince&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A History of the Theory of Elasticity and of the Strength of Materials ...&#xA;by Isaac Todhunter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and work on the Euler equations and the Navier-Stokes equations.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-01T23:18:15.053" CommunityOwnedDate="2010-09-01T23:18:15.053" />
  <row Id="3822" PostTypeId="2" ParentId="3820" CreationDate="2010-09-01T23:25:02.963" Score="15" ViewCount="0" Body="&lt;p&gt;No.  The polynomial $f(x, y) = (1 - xy)^2 + x^2$ is bounded below by $0$, cannot actually take the value $0$, but can take the value $\epsilon$ for any $\epsilon &gt; 0$.  I learned this example from &lt;a href=&quot;http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/30979#30979&quot; rel=&quot;nofollow&quot;&gt;Richard Dore&lt;/a&gt; on MO.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-01T23:25:02.963" />
  <row Id="3823" PostTypeId="2" ParentId="2949" CreationDate="2010-09-01T23:49:43.943" Score="11" ViewCount="0" Body="&lt;p&gt;$e^{i\pi} +1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This still blows my mind.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-01T23:49:43.943" CommunityOwnedDate="2010-09-01T23:49:43.943" />
  <row Id="3824" PostTypeId="2" ParentId="2949" CreationDate="2010-09-01T23:51:28.883" Score="7" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nine-point_circle&quot; rel=&quot;nofollow&quot;&gt;nine point circle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three sets of three points, each of which obviously determines a circle. That these three constructions &lt;em&gt;always&lt;/em&gt; give the &lt;strong&gt;same&lt;/strong&gt; circle!?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-01T23:51:28.883" CommunityOwnedDate="2010-09-01T23:51:28.883" />
  <row Id="3825" PostTypeId="2" ParentId="3782" CreationDate="2010-09-01T23:54:22.107" Score="6" ViewCount="0" Body="&lt;p&gt;I don't consider myself anywhere near an expert at this (there are a lot of things that I've tried to learn for a while but failed). However, I have spent a while trying to learn mathematics independently, so I will weigh in, but this should be taken with the usual truckload of salt.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a good book! It's easy to waste time with a poorly written textbook. I think this one is really important. Different people have different ideas about what constitutes a well-written textbook, though, so it may be desirable to look at books in a library and get a sense of them than simply following someone else's recommendation.&lt;/li&gt;&#xA;&lt;li&gt;Talk to people. I didn't realize how little I knew about algebraic geometry until I was having an email chat with someone and confused the universal properties of projective and affine space minus the origin.  That sorted out a few things in my mind.&lt;/li&gt;&#xA;&lt;li&gt;Don't worry about skipping over things that are unimportant. But try to ask someone what the important and unimportant things are.&lt;/li&gt;&#xA;&lt;li&gt;This may be just a personal preference, but I like to make copious amounts of notes when I learn something, starting from the beginning. This is one reason to keep a mathematical blog. However, creating large sets of notes is often not entirely suitable for blogging.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-09-01T23:54:22.107" CommunityOwnedDate="2010-09-01T23:54:22.107" CommentCount="3" />
  <row Id="3826" PostTypeId="2" ParentId="2949" CreationDate="2010-09-01T23:56:34.847" Score="11" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann's rearrangement theorem.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is responsible for the counter-intuitive results of, for example &lt;a href=&quot;http://math.stackexchange.com/questions/107/paradox-increasing-sequence-that-goes-to-0&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/3583/infinite-processes-riddle&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-01T23:56:34.847" CommunityOwnedDate="2010-09-01T23:56:34.847" CommentCount="1" />
  <row Id="3827" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T00:05:16.590" Score="9" ViewCount="0" Body="&lt;p&gt;Fermat's &quot;two square theorem&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G.H. Hardy's &lt;em&gt;A Mathematician's Apology&lt;/em&gt; is a book everyone should read, but for those who haven't here's something Hardy mentions that is rather surprising:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If we ignore 2) All primes fit into two classes: those that leave remainder $1$ when divided by $4$ and those that leave remainder $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This much is obvious. The surprising thing is that all of the first class, and none of the second can be expressed as the sum of two integer squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, for all prime $p$, if $p = 1 \mod 4$ then there exist $x,y$ integers such that $p = x^2 +y^2$ and if $p = 3 \mod 4$ there exists no such $x,y$&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-02T00:05:16.590" CommunityOwnedDate="2010-09-02T00:05:16.590" CommentCount="2" />
  <row Id="3828" PostTypeId="2" ParentId="3815" CreationDate="2010-09-02T00:06:16.827" Score="7" ViewCount="0" Body="&lt;p&gt;If $x$ is a set containing all singletons, $\cup x$ is a set containing all sets, which leads to Russel's paradox. Thus, in $ZF$ there is no such $x$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-02T00:06:16.827" CommentCount="3" />
  <row Id="3829" PostTypeId="2" ParentId="3815" CreationDate="2010-09-02T00:26:06.803" Score="0" ViewCount="0" Body="&lt;p&gt;If you accept that the collection of all sets (denoted by $V$ usually) is not a set (but in fact a proper class) then if you look at the function $f(x) = \{ x \}$ it is 1-1 from $V$ into the sets of all singletons, thus resulting that $V$ would be a set, in contradiction to the fact that it is in fact a proper class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course you'd have to accept that $V$ is a proper class as well, this is simpler and follows by the argument given in Arturo's answer (that the power-set of $V$ would be of the same cardinality).&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-02T00:26:06.803" />
  <row Id="3830" PostTypeId="1" AcceptedAnswerId="3885" CreationDate="2010-09-02T00:28:36.023" Score="2" ViewCount="82" Body="&lt;p&gt;Let $\kappa$ a cardinal of cofinality $\omega$; let $C \subseteq \kappa$ be a unbounded countable subset. Why is then $C$ closed (and thus a c.u.b.)? This means that if $\delta &amp;lt; \kappa$ is a limit ordinal, and $C \cap \delta$ is unbounded in $\delta$, then $\delta \in C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt that this is true; however, it's a short statement in Kunen's set theory book.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-02T12:51:04.180" LastActivityDate="2010-09-02T17:15:02.703" Title="Example of a c.u.b. set" Tags="&lt;logic&gt;&lt;cardinals&gt;" AnswerCount="2" />
  <row Id="3832" PostTypeId="2" ParentId="3768" CreationDate="2010-09-02T00:58:02.623" Score="0" ViewCount="0" Body="&lt;p&gt;A very natural explanation of the correct definition for the values of min and max on empty sets arises from their dual &lt;strong&gt;universal&lt;/strong&gt; definitions - analogous to the universal GCD, LCM definitions that I gave &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First some notation. $\;$ Write $\;\:\rm x \le S \;\iff\; x \le s,\;\: \forall\: s \in S,\;$ and dually for $\rm\; x \ge S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of $\:$ &lt;strong&gt;min&lt;/strong&gt; $\quad$ $\quad\rm x \le S \;\iff\; x \le min\ S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;max&lt;/strong&gt; $\:\quad$ $\quad\rm x \ge S \;\iff\; x \ge max\ S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For min, when $\;\rm S = \emptyset\;$ is empty, the first clause  $\;\rm x \le S\;$ in the min definition is vacuously true, so the definition reduces to $\;\rm x \le \min \emptyset\;$ is true for all $\;\rm x\;$. Hence $\;\rm \min \emptyset = \infty\;$. Dually $\;\rm \max\emptyset = -\infty\;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I remarked in said GCD, LCM post, such universal definitions often facilitate slick proofs. For some nontrivial examples of min, max flavor consider the slick proofs of the integrality of various products of binomial coefficients by employing the floor function, e.g. see Joe Roberts, Elementary Number Theory. Instead I close with an analogous slick GCD, LCM proof from my &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;mentioned post&lt;/a&gt; (see it for further details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally one has the following dual &lt;strong&gt;universal&lt;/strong&gt; definitions of LCM and GCD in arbitrary domains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;LCM&lt;/strong&gt; $\quad$ If $\quad\rm a,b|c \;\iff\; [a,b]\:|\:c \quad\;$ then $\;\quad\rm [a,b] \;\;$ is an &lt;strong&gt;LCM&lt;/strong&gt; of $\;\rm a,b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; of &lt;strong&gt;GCD&lt;/strong&gt; $\quad$ If $\quad\rm c|a,b \;\iff\; c\:|\:(a,b) \quad$ then $\quad\rm (a,b) \;$ is an &lt;strong&gt;GCD&lt;/strong&gt; of $\;\;\rm a,b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: That $\;\rm a,b\:|\:[a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. Dually $\;\rm (a,b)|\:a,b \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such $\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\rm\;\; (a,b) = ab/[a,b] \;\;$  if $\;\rm\ [a,b] \;$  exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:  $\rm\quad\quad d|\:a,b \;\iff\; a,b\:|\:ab/d \;\iff\; [a,b]\:|\:ab/d \;\iff\; d\:|\:ab/[a,b] \quad\;\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conciseness of this proof arises by exploiting to the hilt the $\iff$ definition of both LCM and GCD.  Compare to less concise, less general and less illuminating proofs often found in number theory textbooks.  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-02T12:15:10.110" LastActivityDate="2010-09-02T12:15:10.110" />
  <row Id="3833" PostTypeId="2" ParentId="3830" CreationDate="2010-09-02T00:59:35.310" Score="1" ViewCount="0" Body="&lt;p&gt;If you set $\kappa = \aleph_{\omega}$ and let $C = \{0,,1,\aleph_1, 2, \aleph_2,...\}$ (so $C$ is the union of all natural numbers together with all $\aleph_n$ for $n&gt;0$) and you let $\delta = \omega$, I think you have a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C\cap\delta = \delta$, so is automatically unbounded in $\delta$.  But $\delta$ itself isn't in $C$ by construction.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-09-02T00:59:35.310" CommentCount="1" />
  <row Id="3835" PostTypeId="2" ParentId="3504" CreationDate="2010-09-02T01:52:26.333" Score="1" ViewCount="0" Body="&lt;p&gt;So we have the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_1+b_2)^2=8(c_1+c_2)$ (i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_1+b_3)^2=8(c_1+c_3)$ (ii)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_1+b_4)^2=8(c_1+c_4)$ (iii)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2+b_3)^2=8(c_2+c_3)$ (iv)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2+b_4)^2=8(c_2+c_4)$ (v)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_3+b_4)^2=8(c_3+c_4)$ (vi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substracting (ii) from (i) we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2-b_3)(b_2+b_3+2b_1)=8(c_2-c_3)$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substracting (vi) from (v) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(b_2-b_3)(b_2+b_3+2b_4)=8(c_2-c_3)$ (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, substracting (2) from (1) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(b_2-b_3)(b_1-b_4)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar fashion we arrive at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(b_2-b_1)(b_3-b_4)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(b_1-b_3)(b_2-b_4)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last three equations show that $b_1=b_2=b_3=b_4$. Using (i) through (vi), we see that $c_1=c_2=c_3=c_4$. &lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-02T01:52:26.333" CommentCount="2" />
  <row Id="3836" PostTypeId="2" ParentId="3723" CreationDate="2010-09-02T01:59:08.120" Score="2" ViewCount="0" Body="&lt;p&gt;The solution you have suggested is a perfectly good one. The only problem is that you are venturing outside the bounds of conventional Distribution theory. Consequently, the onus is on you to prove whatever properties of your definition that you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step#Zero_argument&quot; rel=&quot;nofollow&quot;&gt;Heaviside step function&lt;/a&gt; is more usually treated as a distribution itself. Bracewell's book, &quot;The Fourier Transform and its Applications&quot; considers the problem of defining a value at the discontinuity and concludes that it mostly doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are doing seems somewhat similar to the idea of density in physics. I'd be surprised if the physicists or the electrical engineers (signal processing) people hadn't already confronted this issue. However, after looking around a little bit I am unable to find anything specific.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2010-09-02T01:59:08.120" />
  <row Id="3837" PostTypeId="2" ParentId="3820" CreationDate="2010-09-02T02:29:43.950" Score="1" ViewCount="0" Body="&lt;p&gt;@Agusti Roig. $f$ has no critical points in the interior of the disc, so you have to look for the minimum at the unit circle. One way would be to parametrize $x=\cos\theta, y=\sin\theta$, and minimize.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-02T02:29:43.950" CommentCount="2" />
  <row Id="3838" PostTypeId="1" AcceptedAnswerId="3843" CreationDate="2010-09-02T02:34:21.520" Score="2" ViewCount="95" Body="&lt;p&gt;Let $\Pi_1 : 7x-5y-2z=0, \Pi_2 : 5x-4y-z=0$ and $\mathbb{L}$ the line that passes through points $P=(-2,3,-3)$ and $Q=(-1,2,-1)$. Find all planes $\Pi$ that satisfy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Pi \cap \Pi_1 \cap \Pi_2 = \emptyset$&lt;/li&gt;&#xA;&lt;li&gt;$d(R,\Pi)=\sqrt{14} \forall R \in \mathbb{L}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far, I've found out that $\Pi_1\cap\Pi_2= { X : \lambda (1,1,1) } $ and $\mathbb{L} : X=\lambda(1,-1,2)+(-1,2,-1)$; given $\Pi : NX = NQ$, with $N$ being an orthogonal vector to the plane and $Q$ being any point in $\Pi$, it follows that $N \perp(1,1,1)$ and $N\perp(1,-1,2)$, so we can define $N=(1,1,1)\times(1,-1,2)=(3,-1,-2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{|(3,-1,-2)((-3\lambda-1,\lambda+2,-4\lambda-1)-(x,y,z))|}{||(3,-1,-2)||} = \sqrt{14}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|(3,-1,-2)(-3\lambda-1-x,\lambda+2-y,-4\lambda-1-z)| = 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-9\lambda-3-3x -\lambda-2+y +8\lambda+2+2z = \pm 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-3x+y+2z-2\lambda-3 = \pm 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I simply don't know what to do next. Any help would be really appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(R,\Pi) = \frac{|N(R-Q)|}{||N||} = \sqrt{14}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3,-1,-2)((-1,2,1)-(x,y,z)) = \pm 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-3x+y+2z-7=\pm 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can take $Q$ as either $(0,4,2)$ or $(0,2,2)$ and we get two different planes:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Pi_3: 3x-y-2z=-8$ and $\Pi_4: 3x-y-2z=-6$. Once again, thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-09-02T18:16:46.280" LastActivityDate="2010-09-02T18:16:46.280" Title="Finding all planes that satisfy two conditions." Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3839" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T02:49:53.590" Score="2" ViewCount="0" Body="&lt;p&gt;While not as surprising as, say, the countability of the rationals, and even fairly obvious to some people, the fundamental theorem of calculus joins two operations (differentiation and integration) which didn't look completely related to each other at first to me if you define them as the rate of change of a curve and the area beneath it.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-09-02T02:49:53.590" CommunityOwnedDate="2010-09-02T02:49:53.590" CommentCount="1" />
  <row Id="3840" PostTypeId="2" ParentId="3809" CreationDate="2010-09-02T03:12:10.827" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure about this (the condition is way more restrictive than being of bounded variation, anyway): if $f$ is differentiable in (0,1) and continuous in [0,1], and $m&gt;\max_{x\in[0,1]}|f'(x)|$, then $f(x)=(f(x)-mx)+mx$; the mean-value theorem guarantees that $f(x)-mx$ is injective.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-02T03:12:10.827" CommentCount="1" />
  <row Id="3841" PostTypeId="1" AcceptedAnswerId="3842" CreationDate="2010-09-02T03:26:51.857" Score="2" ViewCount="101" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; If $p &gt; 3$ is a prime and $$ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p} = \frac{a}{b}$$ then prove that $p^{4} \mid (ap-b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an exercise in Herstein which states, if $p &gt; 3$ is prime and if $$1 + \frac{1}{2} + \cdots + \frac{1}{p-1} = \frac{a}{b}$$ then $p^{2} \mid a$. (Page 116, Problem 2). But i am not sure whether this helps or not. Hints, suggestions would be of great help! &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T15:32:33.113" Title="If a prime $p$ satisfies given condition then $p^{4} \mid ap-b$" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="3842" PostTypeId="2" ParentId="3841" CreationDate="2010-09-02T03:42:39.203" Score="3" ViewCount="0" Body="&lt;p&gt;$\frac{a}{b} = \frac{p^{2}\alpha}{\beta} + \frac{1}{p}$ where $(\alpha,\beta) = (p,\beta) = 1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows immediately from this I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the exercise helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a hint for the exercise: consider $f(x) = (x-1)(x-2)\cdots(x-p+1)$. Does $p^2$ divide $f'(0)$ (derivative of f at 0)? (This is a classic proof of that exercise and should be available in textbooks)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-02T03:59:11.313" LastActivityDate="2010-09-02T03:59:11.313" CommentCount="0" />
  <row Id="3843" PostTypeId="2" ParentId="3838" CreationDate="2010-09-02T03:57:35.703" Score="2" ViewCount="0" Body="&lt;p&gt;Martín, I think you can take any value of $\lambda$, because the distance between R and $\Pi$ doesn't change if R is a point of L, since you made sure L is parallel to $\Pi$. I'm not sure, though. (Btw, I suppose the first condition is actually $\Pi\cap(\Pi_1\cap\Pi_2)$.) &lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-02T03:57:35.703" CommentCount="3" />
  <row Id="3844" PostTypeId="2" ParentId="3809" CreationDate="2010-09-02T04:04:30.253" Score="0" ViewCount="0" Body="&lt;p&gt;Let $B[a,b]$ denote the space of bounded functions on $[a,b]$. Consider $$ X = \{ f \in B[a,b] : f \ \ \text{is one - one} \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its a good exercise to show that $X$ is a residual subset ( A subset $A \subset \mathbb{R}$ is said to be residual if $\mathbb{R} \setminus A$ is of the first category) of $B[a,b]$. Then using this we can show the conclusion of the problem.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T04:04:30.253" CommentCount="3" />
  <row Id="3845" PostTypeId="2" ParentId="3841" CreationDate="2010-09-02T04:29:03.043" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\;\;\displaystyle p^4\:\big|\:ap-b = \:bp\bigg(\frac{a}{b} - \frac{1}{p}\bigg)\;\;$ via $\rm\;\; p^2|bp\;\;$ and $\rm\displaystyle\; p^2\:\big|\:\bigg(\frac{a}{b} - \frac{1}{p}\bigg)\;\;$ via &lt;a href=&quot;http://books.google.com/books?id=rey9wfSaJ9EC&amp;amp;pg=PA113&quot; rel=&quot;nofollow&quot;&gt;Wolstenholme&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-02T15:32:33.113" LastActivityDate="2010-09-02T15:32:33.113" />
  <row Id="3846" PostTypeId="2" ParentId="3815" CreationDate="2010-09-02T04:38:26.490" Score="0" ViewCount="0" Body="&lt;p&gt;The question in the title is a consequence of the &lt;a href=&quot;http://books.google.com/books?id=sxr4LrgJGeAC&amp;amp;pg=PA53&amp;amp;dq=Axiom+of+regularity&amp;amp;hl=en&amp;amp;ei=lyh_TKLtJcPGswayz53lDw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCkQ6AEwAA#v=onepage&amp;amp;q=Axiom%20of%20regularity&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;axiom of regularity&lt;/a&gt;.&#xA;In fact If there was such a set $X$, then $\{X\}\in X$. But obviously we always have $X\in\{X\}$. Hence, we would be able to construct an infinite downward chain of sets which contradicts the  axiom of regularity.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-09-02T04:58:12.737" LastActivityDate="2010-09-02T04:58:12.737" CommentCount="3" />
  <row Id="3847" PostTypeId="1" AcceptedAnswerId="3848" CreationDate="2010-09-02T06:05:37.420" Score="1" ViewCount="230" Body="&lt;p&gt;How can $n!$ be represented in terms of $2^{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean I want function $f$ such that $$n! = f(2^{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would like to know $$\frac{n!}{2^n} = g(n) = ?$$&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:25:00.143" LastActivityDate="2010-09-05T08:42:29.490" ClosedDate="2010-09-05T06:23:37.757" Title="How can $n!$ be represented in terms of $2^{n}$?" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="3848" PostTypeId="2" ParentId="3847" CreationDate="2010-09-02T06:10:39.773" Score="8" ViewCount="0" Body="Try $$f(t) = \Gamma ( 1 + \log_2 t ), $$ where $\log_2$ is the base 2 logarithm." OwnerUserId="960" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-05T08:42:29.490" LastActivityDate="2010-09-05T08:42:29.490" CommentCount="3" />
  <row Id="3849" PostTypeId="1" AcceptedAnswerId="3851" CreationDate="2010-09-02T06:22:57.290" Score="1" ViewCount="137" Body="&lt;p&gt;Let $G$ be a finite group, and let $p^{\alpha} \mid |G|$, where $p$ is a prime. Now does this imply $p \mid |Aut(G)|$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly if $|G| \leq 2$, then the Automorphism group is the trivial group, so one can see that this need not be true for $\alpha =1$. I am curious to know for higher powers of $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:24:40.170" LastActivityDate="2010-09-05T04:24:40.170" Title="Let $p$ be a prime, then does  $p^{\alpha} \mid |G| \Longrightarrow p \mid Aut(G)$?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="3850" PostTypeId="2" ParentId="3849" CreationDate="2010-09-02T06:32:32.950" Score="2" ViewCount="0" Body="&lt;p&gt;Cyclic.            &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[15char]&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-02T06:32:32.950" CommentCount="2" />
  <row Id="3851" PostTypeId="2" ParentId="3849" CreationDate="2010-09-02T06:53:04.310" Score="7" ViewCount="0" Body="&lt;p&gt;Let $P$ be a Sylow p-subgroup of $G$.  Now $G/Z(G)$ embeds in $Aut(G)$, so we can reduce to the case where $P\le Z(G)$, so $G=P\times H$, where $(|H|,p)=1$, and $P$ is abelian.  Now if $\alpha&gt;1$, then there will always be an element of order $p$ in $Aut(P)\subset Aut(G)$.  You can see this by looking at cyclic groups of order $p^m$ with $m&gt;1$, and elementary abelian p-groups of dimension greater than $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-09-02T06:53:04.310" CommentCount="4" />
  <row Id="3852" PostTypeId="1" AcceptedAnswerId="3860" CreationDate="2010-09-02T07:04:36.537" Score="13" ViewCount="1815" Body="&lt;p&gt;If A and B are square matrix such that AB = I where I is identity matrix. Show that BA = I. I do not understand anything more than the following.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Elementary row operations.&lt;/li&gt;&#xA;&lt;li&gt;Linear dependence.&lt;/li&gt;&#xA;&lt;li&gt;Row reduced forms and their relations with original matix.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If Entries of matrix are not from a mathematical structure which supports commutativity, What can we say about this prob?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : Using transform and/or inverse of a matrix MAY invite a death penalty.&lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2010-09-19T05:00:56.167" Title="if AB = I then BA = I" Tags="&lt;linear-algebra&gt;" AnswerCount="15" CommentCount="2" FavoriteCount="7" />
  <row Id="3853" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T07:18:54.513" Score="0" ViewCount="0" Body="&lt;p&gt;I was surprised to know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0! = 1$$&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-02T07:18:54.513" CommunityOwnedDate="2010-09-02T07:18:54.513" CommentCount="3" />
  <row Id="3854" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T07:20:24.273" Score="-9" ViewCount="0" Body="&lt;p&gt;Suppose,&#xA;$AB = I$ and $BA = C \ne I$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(AB) = B(I)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(BA)B = B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$CB = B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Assuming B has a right inverse}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = I $$&#xA;so our assumption $C \ne I$ is wrong.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-02T14:15:16.313" CommunityOwnedDate="2010-09-02T11:10:59.057" CommentCount="8" />
  <row Id="3855" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T07:35:03.933" Score="18" ViewCount="0" Body="&lt;p&gt;So you want to find a proof of this well-known fact, which avoids the usual &quot;indirect&quot; proofs? I've also pondered over this some time ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following general assertion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: The sequence of subspaces $... \subseteq b^{k+1} M \subseteq b^k M \subseteq ... \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no &quot;direct&quot; proof. I doubt that there is one.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-02T11:09:47.677" LastActivityDate="2010-09-02T11:09:47.677" CommentCount="6" />
  <row Id="3856" PostTypeId="2" ParentId="3849" CreationDate="2010-09-02T07:45:08.467" Score="-1" ViewCount="0" Body="&lt;p&gt;While browsing through the Internet i have got these excellent articles, which i am sure one shall thoroughly enjoy reading:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://www.jstor.org/stable/2307506?origin=JSTOR-pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2307506?origin=JSTOR-pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://www.jstor.org/stable/2033268?origin=JSTOR-pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2033268?origin=JSTOR-pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T07:45:08.467" CommentCount="5" />
  <row Id="3858" PostTypeId="2" ParentId="1344" CreationDate="2010-09-02T08:01:22.617" Score="1" ViewCount="0" Body="&lt;p&gt;Actually the result is used all the time in the basics of algebraic geometry. It provides &lt;em&gt;the&lt;/em&gt; formalization of the principle of gluing constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if want to show that fibered products $X \times_S Y$, first do it for affine schemes $X,Y,S$ using the adjunction between Spec and global sections. Now, if $S$ is arbitrary, the functor $Sch/S \to Set, Z \mapsto Hom_S(Z,X) \times Hom_S(Z,Y)$ is a sheaf and locally representable on $S$, thus representable. Thus $X \times_S Y$ exists. Now if also $X$ is arbitrary, consider the functor $Sch/X \to Set, Z \mapsto Hom_S(Z',Y)$, where $Z' = Z \to X \to S$. This is a sheaf and locally representable on $X$, thus representable, which shows that $X \times_S Y$ exists. The usual &quot;ad hoc&quot; proofs for the existence of the fibered product actually just reprove the general representabilty result in the special case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example: If $A$ is a quasi-coherent sheaf of algebras on a scheme $X$, it is possible to construct the $X$-scheme $Spec(A)$. It is very laborious to check all the details of the gluing construction, well-definedness etc. when you just want to glue the $U$-schemes $Spec(A(U))$, $U \subseteq X$ affine, together. But instead, you could just consider the functor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Sch/X \to Set, (t : Z \to X) \mapsto Hom_{\mathcal{O}&lt;em&gt;X-Alg}(A,t&lt;/em&gt;* \mathcal{O}_Z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and show that it is a sheaf (obvious) and locally on $X$ representable (usual adjunction with spectrum of a ring), so it is representable by an $X$-scheme $Spec(A)$ for which you also directly have a universal property. Again I want to emphasize: You get into a big mess when you want to construct this without using functors or universal properties. These rather abstract notions are very useful also in concrete situations, because they make you able to fix your ideas and make every construction fit together nicely. When you get more accustomed to these techniques, you stop thinking of specific functors, but you think in a &quot;functorial way&quot; and recognize, for example, why gluing constructions work.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-02T08:01:22.617" CommentCount="5" />
  <row Id="3859" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T08:09:36.297" Score="13" ViewCount="0" Body="&lt;p&gt;Since there seem to be some lingering beliefs that an approach which does not make explicit use of the finite-dimensionality could be valid, here is a familiar counterexample in the infinite dimensional case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V = \mathbb{R}[t]$ be the vector space of real polynomial functions.  Let $B: V \rightarrow V$ be differentiation: $p \mapsto p'$, and let $A: V \rightarrow V$ be anti-differentation with constant term zero: $\sum_{n=0}^{\infty} a_n t^n \mapsto \sum_{n=0}^{\infty} \frac{a_n}{n+1} t^{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are both $\mathbb{R}$-linear maps and $B \circ A$ is the identity, but $A \circ B$ is not (the constant term is lost).  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-02T08:09:36.297" CommentCount="4" />
  <row Id="3860" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T08:11:04.377" Score="10" ViewCount="0" Body="&lt;p&gt;Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2010-09-02T08:11:04.377" CommentCount="6" />
  <row Id="3861" PostTypeId="1" CreationDate="2010-09-02T08:51:28.307" Score="3" ViewCount="171" Body="&lt;p&gt;as a function of a real variable, apparently.&#xA;Part of the freedom in choosing a proof is that you get to choose what definition of $e^{ix}$ to start from -- do you use a differential equation? a power series? a definition in terms of trig functions?&#xA;Another bit of freedom is that you get to choose what definition of $\pi$ to start from.&lt;/p&gt;&#xA;" OwnerUserId="1678" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-02T09:23:07.270" LastActivityDate="2010-09-02T16:04:17.493" CommunityOwnedDate="2010-09-02T08:51:50.193" Title="What is your favorite proof that $e^{ix}$ has a period of $2\pi$?" Tags="&lt;calculus&gt;&lt;soft-question&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="3862" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T09:11:45.910" Score="2" ViewCount="0" Body="&lt;p&gt;You might have a look at the following note &quot;&lt;em&gt;Right inverse implies left inverse and vice versa&lt;/em&gt;&quot;: &lt;a href=&quot;http://www.lehigh.edu/~gi02/m242/08m242lrinv.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lehigh.edu/~gi02/m242/08m242lrinv.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="55" LastActivityDate="2010-09-02T09:11:45.910" CommentCount="3" />
  <row Id="3863" PostTypeId="2" ParentId="978" CreationDate="2010-09-02T09:28:22.613" Score="1" ViewCount="0" Body="&lt;p&gt;Once you have proved rank(AB) &amp;lt;= rank(B), you can obtain the other inequality by using transposition and the fact that it doesn't change the rank (see e.g. this &lt;a href=&quot;http://math.stackexchange.com/questions/2315/is-the-rank-of-a-matrix-the-same-of-its-transpose-if-yes-how-can-i-prove-it&quot;&gt;question&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, letting C=A^T and D=B^T, we have that rank(DC) &amp;lt;= rank(C) implies rank(C^TD^T) &amp;lt;= rank (C^T), which is rank(AB) &amp;lt;= rank(A).&lt;/p&gt;&#xA;" OwnerUserId="55" LastEditorUserId="55" LastEditorDisplayName="" LastEditDate="2010-09-02T10:03:24.290" LastActivityDate="2010-09-02T10:03:24.290" CommentCount="1" />
  <row Id="3864" PostTypeId="2" ParentId="3809" CreationDate="2010-09-02T09:45:16.660" Score="0" ViewCount="0" Body="&lt;p&gt;This is a reply to Weltschmerz's answer (I still can't comment): if $f$ is differentiable in $(0,1)$ and continuous in $[0,1]$, it can still have an unbounded derivative. For example, $f(x) = x sin (1/x)$. Even if the appropriate left- and right-derivatives exist at the end-points, the derivative can still be unbounded: for example, $f(x) = x^2 sin (1/x^2)$ on the interval $[-1,1]$; this has a well-defined derivative $0$ at $x=0$, but the derivative is unbounded in every neighbourhood of 0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, your condition is not more restrictive -- differentiability doesn't imply bounded variation.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-09-02T10:26:16.887" LastActivityDate="2010-09-02T10:26:16.887" />
  <row Id="3865" PostTypeId="2" ParentId="3861" CreationDate="2010-09-02T09:55:14.410" Score="1" ViewCount="0" Body="&lt;p&gt;$$e^{ix} = e^{i(x+T)} = e^{ix}e^{iT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to find $T$ for which $e^{iT} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rightarrow cos(T) + isin(T) = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rightarrow sin(T) = 0$$ for all $$T = 2n\pi , n = 0,1,2,3...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, period is $2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-02T09:55:14.410" CommunityOwnedDate="2010-09-02T09:55:14.410" CommentCount="1" />
  <row Id="3866" PostTypeId="2" ParentId="3809" CreationDate="2010-09-02T10:16:37.573" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=6l_E9OTFaK0C&amp;amp;pg=PA675&quot; rel=&quot;nofollow&quot;&gt;Duplicated from 13.15.11&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a hint at the end of the book.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2010-09-02T10:16:37.573" />
  <row Id="3867" PostTypeId="2" ParentId="3749" CreationDate="2010-09-02T10:59:58.580" Score="1" ViewCount="0" Body="&lt;p&gt;I still have very little mathematical experience (mainly linear algebra and a bit or real analysis) so the only answer I can give you is that dual spaces are already useful in linear algebra itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA dimension of the annihilator&lt;/strong&gt; Let $V$ be a vector space on the field $\mathbb{F}$ with $\operatorname{dim}V = n$ and $W \subseteq V$ a subspace, if $W^{0} = \{ \phi \in V^{*} | \phi_{|W} = 0 \}$ is the &lt;em&gt;annihilator&lt;/em&gt; of $W$, then $\operatorname{dim}W + \operatorname{dim}W^{0} = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(w_1, \ldots, w_k)$ be a basis of $W$ and $(w_1,\ldots,w_k,v_{k+1},\ldots,v_n)$ a completion to basis of $V$, if $(w_1^{*},\ldots,w_k^{*},v_{k+1}^{*},\ldots,v_n*)$ is the corresponding dual base of $V^{*}$ then it's easy to prove by double inclusion that $W^{0}=\langle v_{k+1}^{*},...,v_n^{*}\rangle$ and thus $\operatorname{dim}W^{0}= n-k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM dimension of the orthogonal space&lt;/strong&gt; Let $V$ be a vector space on the field $\mathbb{F}$ with $\operatorname{dim}V = n$, endowed with a &lt;em&gt;nondegenerate&lt;/em&gt; scalar product $\langle , \rangle$, then if $W \subseteq V$ is a subspace $\operatorname{dim}W + \operatorname{dim}W^{\bot} = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, notice that the statement is not obvious. In fact if a scalar product is not positive-definite $V = W \oplus W^{\bot}$ could be false (you could have $v \in W$ such that $\langle v, v\rangle$ = 0 and therefore $W\cap W^{\bot}\neq \{ 0\}$). Now by mean of the scalar product you can define a very nice isomorphism $\psi : V \to V^{*}$ such that $\psi (v) = \langle \cdot , v\rangle$ for all $v \in V$, you can think about it as the &lt;em&gt;product by $v$&lt;/em&gt;. This is an isomorphism, in fact $\operatorname{dim}V = \operatorname{dim}V^{*}$ and $\operatorname{ker}\psi = 0$ because $\langle , \rangle$ is nondegenerate. Now observe that $\psi (W^{\bot}) = W^{0}$ and thus $\operatorname{dim}W^{\bot}=\operatorname{dim}W^{0}=n-\operatorname{dim}W$ by the lemma.&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-06T22:13:07.150" LastActivityDate="2010-09-06T22:13:07.150" CommentCount="2" />
  <row Id="3868" PostTypeId="2" ParentId="3861" CreationDate="2010-09-02T11:06:24.713" Score="5" ViewCount="0" Body="&lt;p&gt;$$&#xA;e^{ix} = \cos x + i \sin x \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-02T11:06:24.713" CommunityOwnedDate="2010-09-02T11:06:24.713" CommentCount="4" />
  <row Id="3869" PostTypeId="1" AcceptedAnswerId="3871" CreationDate="2010-09-02T11:52:10.883" Score="7" ViewCount="249" Body="&lt;p&gt;Singular value decomposition (&lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_value_decomposition&quot; rel=&quot;nofollow&quot;&gt;SVD&lt;/a&gt;) and principal component analysis (PCA) are two eigenvalue methods used to reduce a high-dimensional dataset into fewer dimensions while retaining important information. Articles online say that these methods are 'related' but never specify the exact relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the intuitive relationship between PCA and SVD? As PCA uses the SVD in its calculation, clearly there is some 'extra' analysis done. What does PCA 'pay attention' to differently than the SVD? What kinds of relationships do each method utilize more in their calculations? Is one method 'blind' to a certain type of data that the other is not?&lt;/p&gt;&#xA;" OwnerUserId="1680" LastActivityDate="2010-11-21T09:48:53.223" Title="What is the intuitive relationship between SVD and PCA" Tags="&lt;analysis&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="3870" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T12:06:16.607" Score="4" ViewCount="0" Body="&lt;p&gt;I prefer to think in terms of linear operators rather than matrices. A function has a right inverse iff it is surjective and it has a left inverse iff it is injective. For a linear operator, this means that having a right inverse is equivalent to having range equal to the entire space and having a left inverse is equivalent to having trivial kernel. For a linear operator on a finite-dimensional space, the dimension of its kernel + the dimension of its range = the dimension of the entire space, so this does the job.&lt;/p&gt;&#xA;" OwnerUserId="687" LastActivityDate="2010-09-02T12:06:16.607" />
  <row Id="3871" PostTypeId="2" ParentId="3869" CreationDate="2010-09-02T12:31:53.333" Score="5" ViewCount="0" Body="&lt;p&gt;Simply put, the PCA viewpoint requires that one compute the eigenvalues and eigenvectors of the covariance matrix, which is the product $XX^T$, where X is the data matrix. Since the covariance matrix is symmetric, the matrix is diagonalizable, and the eigenvectors can be normalized such that they are orthonormal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=WDW^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, applying SVD to the data matrix $X$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=U \Sigma V^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and attempting to construct the covariance matrix from this decomposition gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=(U \Sigma V^T)(U \Sigma V^T)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=(U \Sigma V^T)(V \Sigma U^T)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since V is an orthogonal matrix ($V^T V=I$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$XX^T=U \Sigma^2 U^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the correspondence is easily seen (the square roots of the eigenvalues of $XX^T$ are the singular values of $X$, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, using the SVD to perform PCA makes much better sense numerically than forming the covariance matrix to begin with, since the formation of $XX^T$ can cause loss of precision. This is detailed in books on numerical linear algebra, but I'll leave you with an example of a matrix that can be stable SVD'd, but forming $XX^T$ can be disastrous, the Lauchli matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}1&amp;amp;1&amp;amp;1\\ \epsilon&amp;amp;0&amp;amp;0\\0&amp;amp;\epsilon&amp;amp;0\\0&amp;amp;0&amp;amp;\epsilon\end{pmatrix}^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\epsilon$ is a tiny number.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-02T14:15:11.930" LastActivityDate="2010-09-02T14:15:11.930" CommentCount="2" />
  <row Id="3872" PostTypeId="1" AcceptedAnswerId="3886" CreationDate="2010-09-02T12:49:22.513" Score="3" ViewCount="43" Body="&lt;p&gt;Let $\kappa$ ba weakly inaccessible cardinal. Why are there $\kappa$ regular cardinals $\lambda &amp;lt; \kappa$? I've tried a recursive construction, but I don't know what to do in the limit step. Supremum does not work, since then we loose regularity.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-02T17:21:05.037" Title="number of regular cardinals in a weakly inaccessible cardinal" Tags="&lt;logic&gt;&lt;cardinals&gt;" AnswerCount="1" />
  <row Id="3873" PostTypeId="1" AcceptedAnswerId="3874" CreationDate="2010-09-02T13:16:41.620" Score="3" ViewCount="68" Body="&lt;p&gt;Let us suppose to have a set of &lt;em&gt;N&lt;/em&gt; elements and to choose &lt;em&gt;m&lt;/em&gt; elements out of the set. I know I can do it in $ N \choose m $ ways; but I want to know how many of these combinations contain &lt;em&gt;k&lt;/em&gt; predefined elements (say 1,2,...,k). &lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-02T13:38:20.583" Title="Probability of having k defined elements while choosing m elements from a set of N" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="3874" PostTypeId="2" ParentId="3873" CreationDate="2010-09-02T13:19:52.267" Score="6" ViewCount="0" Body="&lt;p&gt;If you fix k elements, you are left with m-k elements to choose among N-k elements, which can be done in $N-k \choose m-k$ ways. &lt;/p&gt;&#xA;" OwnerUserId="55" LastEditorUserId="55" LastEditorDisplayName="" LastEditDate="2010-09-02T13:38:20.583" LastActivityDate="2010-09-02T13:38:20.583" />
  <row Id="3875" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T14:01:40.353" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;It follows simply by the pigeonhole principle&lt;/strong&gt;. Here is an excerpt from my &lt;a href=&quot;http://google.com/groups?selm=y8z4peo6ukm.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;old post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall (below) $\rm\; AB \:\:=\:\:\: I \:\;\Rightarrow\; BA \:\:=\:\: I\;\;\:$  easily reduces to:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$  $\rm\;\;B\;$ injective  $\rm\;\Rightarrow\:\:  B\;$ surjective, $\:$ for  linear $\rm\:B\:$ on finite dim vector space $\rm\:V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:  $\rm\;B\;$ injective $\rm\;\Rightarrow B\;$ preserves injections:  $\rm\;R &amp;lt; S \;\Rightarrow\; BR &amp;lt; BS\;$&lt;br&gt;&#xA;Hence for   $\rm\;\;\; R \;\: &amp;lt; \;\; S &amp;lt; \cdots &amp;lt; \; V\;\;$  a chain of maximum length (= dim $\rm V\:$)&lt;br&gt;&#xA;its image   $\rm\;BR &amp;lt; BS &amp;lt; \cdots &amp;lt; BV \le V\;\;\:\;$   would have length greater&lt;br&gt;&#xA;if  $\rm\;BV &amp;lt; V,\:$  therefore instead   $\rm\;\: BV  = V\:,\;\:$  i.e. $\rm\; B \;$ is surjective. $\;$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how this form of proof dramatically emphasizes the essence of the matter, namely that&lt;br&gt;&#xA;&lt;strong&gt;injective maps cannot decrease heights&lt;/strong&gt; (here = dimension = length of max subspace chain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is said standard reduction to xxxjective form. See said &lt;a href=&quot;http://google.com/groups?selm=y8z4peo6ukm.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for much more,&lt;br&gt;&#xA;including references to folklore generalizations, e.g. work of Vasconcelos in the seventies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, notice that $\rm\;\;\: AB = I \;\Rightarrow\: B\:$ injective,  since  $\rm\;A\;$  times  $\rm\;Bx = By\;$  yields  $\rm\;x = y\:,\:$  and  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm B\:$ surjective $\rm\;\Rightarrow\; BA = I \;\;$ since for all $\rm\;x\;$ exists $\rm\; y : \;\; x = By = B(AB)y = BA \: x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining them: $\rm\: AB = I \;\Rightarrow\: B\:$ injective $\rm\;\Rightarrow\; B\:$ surjective $\rm\;\Rightarrow\: BA = I$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-03T14:37:14.420" LastActivityDate="2010-09-03T14:37:14.420" CommentCount="2" />
  <row Id="3876" PostTypeId="1" AcceptedAnswerId="3879" CreationDate="2010-09-02T14:32:54.373" Score="1" ViewCount="96" Body="&lt;p&gt;For a periodic ( and not so periodic) function, it is always possible to use fourier series to find out the frequencies contained in the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about function that cannot be expressed in mathematical terms? For example, this graph ( accelerogram):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pEwli.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyway to apply a sort of fourier series, in order to find out the frequencies contained in it?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-06T11:19:52.243" Title="Find the Frequency Components of a Time Series Graph" Tags="&lt;fourier-analysis&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="3877" PostTypeId="2" ParentId="3876" CreationDate="2010-09-02T14:43:11.510" Score="2" ViewCount="0" Body="&lt;p&gt;Er... that's what FFT is for. FFT your data, and take the absolute value of the result. If you want, take their phases too.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-02T14:48:58.213" LastActivityDate="2010-09-02T14:48:58.213" CommentCount="1" />
  <row Id="3878" PostTypeId="1" AcceptedAnswerId="3896" CreationDate="2010-09-02T14:43:36.777" Score="1" ViewCount="44" Body="&lt;p&gt;This queston can be regarded as a variant of this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3799/does-a-section-that-vanihes-at-every-point-vanish&quot;&gt;http://math.stackexchange.com/questions/3799/does-a-section-that-vanihes-at-every-point-vanish&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X,Y$ be varieties, $M$ be an $\mathcal{O}_X$-Module and &#xA;$$\pi:X\times Y\rightarrow X$$&#xA;be the projection. Let $s$ be a local section of $M$, which vanishes along each &quot;horizontal line&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall y: i_y^* s=0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where denotes the map $i_y:x\mapsto (x,y)$. Does it follow, that $s=0$?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2010-09-02T17:54:16.667" LastActivityDate="2010-09-02T21:13:35.897" Title="Vanishing criterion for sections of module on a product" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3879" PostTypeId="2" ParentId="3876" CreationDate="2010-09-02T15:25:02.657" Score="4" ViewCount="0" Body="&lt;p&gt;In short - yes. You can apply a discrete FFT and that should return the complex valued spectrum of your real data. Typically you can then plot out the magnitude of the value of each point in the spectrum (a.k.a. absolute value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to take an FFT numerically (i.e. fftw routines, etc) it's also helpful to apply a window function (i.e. hamming). Your data doesn't look particularly periodic, so windowing it will help in removing the noise associated with that.&lt;/p&gt;&#xA;" OwnerUserId="1682" LastActivityDate="2010-09-02T15:25:02.657" CommentCount="1" />
  <row Id="3880" PostTypeId="2" ParentId="3861" CreationDate="2010-09-02T16:04:17.493" Score="5" ViewCount="0" Body="&lt;p&gt;My favorite has always been Walter Rudin's proof in the prologue to his &quot;Real and Complex Analysis&quot; (2nd Ed.).  Here's a sketch:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Define $\exp$ in terms of the power series.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By manipulating the series, deduce that $\exp$ is a homomorphism from the additive group to the group of complex units.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show it satisfies the usual first order ODE.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Define&lt;/em&gt; $\cos$ and $\sin$ as the real and imaginary parts of $\exp$, respectively.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Define&lt;/em&gt; $\pi$ as twice the smallest positive real root of $\cos$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Deduce that $\exp( i \pi / 2) = i$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By multiplying, conclude that $2 \pi i$ is a period of $\exp$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show, by means of the preceding properties, that no smaller period exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-02T16:04:17.493" CommunityOwnedDate="2010-09-02T16:04:17.493" />
  <row Id="3881" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T16:05:38.160" Score="15" ViewCount="0" Body="&lt;p&gt;Let $x_1, x_2, \cdots, x_n$ be a basis of the space. At first we prove that $Bx_1, Bx_2, \cdots, Bx_n$ is also a basis. To do it we need to prove that those vectors are linearly independent. Suppose it's not true. Then there exist numbers $c_1, c_2, \cdots, c_n$ not all equal to zero such that $c_1 Bx_1 + c_2 Bx_2 + \cdots + c_n B x_n = 0$. Multiplying it by $A$ from the left, we get $c_1 ABx_1 + c_2 ABx_2 + \cdots + c_n ABx_n = 0$, hence $c_1 x_1 + c_2 x_2 + \cdots + c_n x_n = 0$ and so the vectors $x_1, x_2, \cdots, x_n$ are also linearly dependent. Here we get contradiction with assumption that the vectors $x_i$ form a basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Bx_1, Bx_2, \cdots, Bx_n$ is a basis, every vector $y$ can be represented as a linear combination of those vectors. This means that for any vector $y$ there exists some vector $x$ such that $Bx = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we want to prove that $BA = I$. It is the same as to prove that for any vector $y$ we have $BAy = y$. Now given any vector $y$ we can find $x$ such that $Bx = y$. Hence $BAy = BABx = Bx = y$ by associativity of matrix multiplication.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditorDisplayName="" LastEditDate="2010-09-02T19:01:26.100" LastActivityDate="2010-09-02T19:01:26.100" CommentCount="7" />
  <row Id="3882" PostTypeId="2" ParentId="2079" CreationDate="2010-09-02T16:16:44.620" Score="0" ViewCount="0" Body="&lt;p&gt;One can even look at this article for a general reference : &lt;a href=&quot;http://www.jstor.org/stable/2160835&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2160835&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T16:16:44.620" />
  <row Id="3883" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T16:47:49.057" Score="0" ViewCount="0" Body="&lt;p&gt;Thanks guys for all the responses. Here, I was able to formulate this proof with all this material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 : &lt;strong&gt;AB&lt;/strong&gt; = &lt;strong&gt;I&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.1 :  &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; are square matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 : &lt;strong&gt;AB&lt;/strong&gt; can be thought of as we are linearly combining columns of &lt;strong&gt;A&lt;/strong&gt; using the entries in a column of &lt;strong&gt;B&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 : Since &lt;strong&gt;AB = I&lt;/strong&gt; and &lt;strong&gt;I&lt;/strong&gt; is a matrix with the property that its columns are linearly independent (We can use &lt;em&gt;Row Reduced Echelon &lt;/em&gt; (RRE) form argument here to prove this - as done in &lt;EM&gt;Hoffman &amp;amp; Kunje's Linear Algebra&lt;/EM&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.1 Every matrix have a unique RRE form. This can be proved from the fact that once you reach to a RRE form you can not modify it without disturbing its structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2 : it can be verified that row rank of a matrix is equal to its column rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2.1 : Row rank of a matrix &lt;strong&gt;A&lt;/strong&gt; = row rank of its RRE form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2.2 : Row rank of RRE from = column rank of RRE form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2.3 : Column rank of RRE form = column rank of Matrix &lt;strong&gt;A&lt;/strong&gt;. (I have a hazy proof since I have only defined elementary row operations and NOT the column operation. Anyone?). This proof, if elegant enough is different from the standard text-book proof which uses solution spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 : From 2 it follows that matrix &lt;strong&gt;AB&lt;/strong&gt; has full rank i.e. its columns are linearly independent. Else, we'll not get matrix &lt;strong&gt;I&lt;/strong&gt; which is a RRE form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.1 : From 2.2.2 it follows that its row rank is equal to its column rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 : Rank of &lt;strong&gt;A&lt;/strong&gt; = row rank of &lt;strong&gt;AB&lt;/strong&gt; = row rank of &lt;strong&gt;B&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 : From 4 its follows that there exists a matrix &lt;strong&gt;C&lt;/strong&gt; such that &lt;strong&gt;BC&lt;/strong&gt; = &lt;strong&gt;I&lt;/strong&gt; i.e. we convert &lt;strong&gt;B&lt;/strong&gt; into its RRE form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This proves the existence of such a solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 :  No we use Davidac897 steps to find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : This is the first time I am writing a proof. If there is problem in its structure, please point out. &lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2010-09-02T16:47:49.057" CommunityOwnedDate="2010-09-02T16:47:49.057" />
  <row Id="3884" PostTypeId="2" ParentId="58" CreationDate="2010-09-02T17:07:44.020" Score="1" ViewCount="0" Body="&lt;p&gt;From a physics perspective you can look at how Benford's law (or devations from it) can arise from the different stastical mechanics distributions (Boltzmann, Bose-Einstein, etc...). There is a good paper on the arVix that can get you started on this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The occurrence of the nonzero leftmost digit, i.e., 1, 2, ..., 9, of numbers from many real world sources is not uniformly distributed as one might naively expect, but instead, the nature favors smaller ones according to a logarithmic distribution, named Benford's law. We investigate three kinds of widely used physical statistics, i.e., the Boltzmann-Gibbs (BG) distribution, the Fermi-Dirac (FD) distribution, and the Bose-Einstein (BE) distribution, and find that the BG and FD distributions both fluctuate slightly in a periodic manner around the Benford distribution with respect to the temperature of the system, while the BE distribution conforms to it exactly whatever the temperature is. Thus the Benford's law seems to present a general pattern for physical statistics and might be even more fundamental and profound in nature. Furthermore, various elegant properties of Benford's law, especially the mantissa distribution of data sets, are discussed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1005.0660&quot; rel=&quot;nofollow&quot;&gt;The Significant Digit Law in Statistical Physics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-09-02T17:27:29.593" LastActivityDate="2010-09-02T17:27:29.593" />
  <row Id="3885" PostTypeId="2" ParentId="3830" CreationDate="2010-09-02T17:15:02.703" Score="3" ViewCount="0" Body="&lt;p&gt;What Kunen meant was that if $C$ is the range of a countable &lt;em&gt;increasing&lt;/em&gt; unbounded sequence in $\kappa$, that is, $C$ is unbounded and has order type $\omega$, then it will be vacuously closed, since it has no limit points below $\kappa$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation shows that the concept of club on cardinals of countable cofinality doesn't work as we want. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-02T17:15:02.703" CommentCount="3" />
  <row Id="3886" PostTypeId="2" ParentId="3872" CreationDate="2010-09-02T17:21:05.037" Score="5" ViewCount="0" Body="&lt;p&gt;Suppose that $\kappa$ is weakly inaccessible. Thus, it is a regular limit cardinal. So $\kappa=\aleph_\beta$ for some ordinal $\beta$. Since $\kappa$ is a limit cardinal, it must be that $\beta$ is a limit ordinal. Since $\kappa$ is regular, it cannot be that $\beta&amp;lt;\kappa$. So $\kappa=\aleph_\kappa$. Thus, there are $\kappa$ many cardinals below $\kappa$. All the successor cardinals $\aleph_{\beta+1}$ for $\beta&amp;lt;\kappa$ are regular, and there are $\kappa$ many of these.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-02T17:21:05.037" CommentCount="1" />
  <row Id="3887" PostTypeId="2" ParentId="2949" CreationDate="2010-09-02T17:53:56.467" Score="1" ViewCount="0" Body="&lt;p&gt;Similar to Thomae's function, I was impressed by the &lt;a href=&quot;http://mathworld.wolfram.com/DirichletFunction.html&quot; rel=&quot;nofollow&quot;&gt;Dirichlet function&lt;/a&gt;, which is not only discontinuous everywhere, but impossible to plot. The function is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=\begin{cases}&#xA;  1 \mbox{ if } x\in\mathbb{Q} \\&#xA;  0 \mbox{ if }x\notin\mathbb{Q} &#xA;\end{cases}$&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-09-02T21:43:07.253" LastActivityDate="2010-09-02T21:43:07.253" CommunityOwnedDate="2010-09-02T17:53:56.467" />
  <row Id="3888" PostTypeId="1" AcceptedAnswerId="3889" CreationDate="2010-09-02T17:57:59.947" Score="3" ViewCount="388" Body="&lt;p&gt;Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link: &lt;a href=&quot;http://web.mit.edu/rwbarton/Public/func-eq.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/rwbarton/Public/func-eq.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-07T22:03:55.617" Title="Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="3889" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T18:08:51.287" Score="10" ViewCount="0" Body="&lt;p&gt;Putting $x=1$ and $x=16$ we see that $p(2)=p(16) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p(x) = (x-2)(x-16)g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-16)(2x-2)(2x-16)g(2x) = 16(x-1)(x-2)(x-16)g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$ (x-8)g(2x) = 4(x-2)g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now see that $g(4) = g(8) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $g(x) = (x-4)(x-8)h(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$$ (x-8)(2x-4)(2x-8)h(2x) = 4(x-2)(x-4)(x-8)h(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h(2x) = h(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that $h(x)$ is a constant (as otherwise the non-identically zero polynomial $h(2x) - h(x)$ will have an infinite number of roots.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(x) = C(x-2)(x-4)(x-8)(x-16)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-02T18:08:51.287" CommentCount="1" />
  <row Id="3890" PostTypeId="2" ParentId="58" CreationDate="2010-09-02T18:38:37.677" Score="3" ViewCount="0" Body="&lt;p&gt;I recently attended a lecture on this. So I'll list a few interesting points that I remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original discovery was by Newcomb who noticed that tables of logarithms were more worn out in the front than the back. It was rediscovered by Benford in 1938.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Areas of countries, areas of rivers and sizes of populations obey this rule. A more surprising example claims the numbers that appear in newspapers also follow this rule. In spite of the many observances of Benford's law, not a lot of instances are fully explained. I believe there have been proposals to provide simple models to associate an 'exponential-like' behavior of the areas of countries, but I don't really know anything about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another note on the forensic accounting example mentioned by workmad3, I remember the lecturer pointed out that evidence pertaining to Benford's law is considered admissible in court.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are things that don't really obey Benford's law, for example: the number of pages in books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that's more or less what I can recall from the talk, hope that helps :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-02T18:38:37.677" />
  <row Id="3891" PostTypeId="1" AcceptedAnswerId="3893" CreationDate="2010-09-02T19:04:31.923" Score="-1" ViewCount="112" Body="&lt;p&gt;I have asked several questions concering the Automorphisms of groups. Some of them are here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://math.stackexchange.com/questions/2121/existence-of-a-normal-subgroup-with-authautg&quot;&gt;http://math.stackexchange.com/questions/2121/existence-of-a-normal-subgroup-with-authautg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://math.stackexchange.com/questions/2619/solving-auth-autg&quot;&gt;http://math.stackexchange.com/questions/2619/solving-auth-autg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetric group $S_{n}$ has some very interesting properties. For $n=3$ the symmetric group $S_{3}$, the automorphism of $S_{3}$ permutes with the set $\{ (1 2), (2 3) , (1 3)\}$ and no nontrivial automorphism leaves each element of the set fixed. Therefore if 2 automorphism $f,g$ permute the set in the same way, then $fg^{-1}=e \Longrightarrow f=g$. Hence $S_{3}$ has atmost $3!=6$ automorphisms. Since $Z(S_{3})=1$, therefore $\text{Aut}(S_{3}) \cong S_{3}$. One can also show that $\text{Aut}(S_{4}) \cong S_{4}$. Infact these papers prove the following result, that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group of automorphisms of the symmetric group $S_{n}$ on $n$ letters is isomorphic with $S_{n}$, except when $n = 6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Automorphisms of the symmetric group, I.E. Segal, &lt;a href=&quot;http://www.ams.org/journals/bull/1940-46-06/S0002-9904-1940-07261-1/S0002-9904-1940-07261-1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/bull/1940-46-06/S0002-9904-1940-07261-1/S0002-9904-1940-07261-1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) O. Holder, Mathematische Annalen, vol. 46 (1895), especially&#xA;pp.340-345&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other groups, other than symmetric group, which satisfy the given condition. Can we classify all such groups?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-02T20:08:54.720" Title="Finding groups such that  $|\text{Aut}(G)|=|G|$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3892" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T19:06:44.620" Score="3" ViewCount="0" Body="&lt;p&gt;For a more elementary treatment ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Fact.&lt;/b&gt; If the rows of $A$ are linearly dependent, then the rows of $A B$ are linearly dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof of fact.&lt;/em&gt; Consider the 3x3 case, where the linearly-dependent rows of $A$ are $\mathbf{a}_1$, $\mathbf{a}_2$, $\mathbf{a}_3 = h \mathbf{a}_1 + k \mathbf{a}_2$ (for some scalars $h$ and $k$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \begin{bmatrix}\mathbf{a}_1 \\ \mathbf{a}_2 \\ h\mathbf{a}_1 + k\mathbf{a}_2\end{bmatrix} = \begin{bmatrix}p &amp;amp; q &amp;amp; r \\ s &amp;amp; t &amp;amp; u \\ hp + ks &amp;amp; hq + kt &amp;amp; hr + ku \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $\mathbf{b}_1$, $\mathbf{b}_2$, and $\mathbf{b}_3$ for the rows of $B$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A B = \begin{bmatrix}p &amp;amp; q &amp;amp; r \\ s &amp;amp; t &amp;amp; u \\ hp + ks &amp;amp; hq + kt &amp;amp; hr + ku \end{bmatrix} \begin{bmatrix}\mathbf{b}_1 \\ \mathbf{b}_2 \\ \mathbf{b}_3\end{bmatrix} = \begin{bmatrix}p &#xA;\mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ (hp + ks)\mathbf{b}_1 + (hq + kt)\mathbf{b}_2 + (hr + ku)\mathbf{b}_3 \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \begin{bmatrix}p &#xA;\mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ h(p\mathbf{b_1}+q\mathbf{b}_2+r\mathbf{b}_3) + k(s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3) \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, the linear dependence of the rows of $A$ carries over to the rows of the product, proving our Fact. (This reasoning actually shows the more-precise Fact that $rank(AB)\le rank(A)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can restate the Fact this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Re-Fact.&lt;/b&gt; If the rows of $AB$ are linearly independent, then the rows of $A$ are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question: If $A B = I$, then (by the Re-Fact) the rows of $A$ must be linearly independent. This implies that $A$ can be row-reduced to a diagonal matrix with no zero entries on that diagonal: the row-reduced form of $A$ must be the Identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that row-reduction is &lt;em&gt;actually&lt;/em&gt; an application of matrix multiplication. (You can see this in the equations above, where (left-)multiplying $B$ by $A$ combined the rows of $B$ according to the entries in the rows of $A$.) This means that, if $R$ is the result of some row combinations of $A$, then there exists a matrix $C$ that &quot;performed&quot; the combinations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C A = R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If (as in the case of your problem) we have determined that $A$ can be row-reduced all the way down to the Identity matrix, then the corresponding $C$ matrix must be a (&lt;em&gt;the&lt;/em&gt;) left-inverse of $A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C A = I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's then straightforward to show that left and right inverses of $A$ must match. This has been shown in other answers, but for completeness ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A B = I \;\; \to \;\; C (A B) = C \;\; \to \;\; (C A) B = C \;\; \to \;\; I B = C \;\; \to \;\; B = C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you start thinking (&lt;em&gt;ahem&lt;/em&gt;) &quot;outside the box (of numbers)&quot; to interpret matrices as linear transformations of vectors and such, you can interpret this result in terms of mapping kernels and injectivity-vs-surjectivity and all the kinds of sophisticated things other answers are suggesting. Nevertheless, it's worth noting that this problem is solvable within the realm of &lt;em&gt;matrix multiplication&lt;/em&gt;, plain and simple.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-09-02T19:06:44.620" CommentCount="1" />
  <row Id="3893" PostTypeId="2" ParentId="3891" CreationDate="2010-09-02T19:19:49.673" Score="1" ViewCount="0" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Inner_automorphism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inner_automorphism&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The composition of two inner automorphisms is again an inner automorphism (as mentioned above:&#xA;  $(x^a)^b=x^{ab}$, and with this operation, the collection of all inner automorphisms of $G$ is itself a group, the inner automorphism group of $G$ denoted $Inn(G)$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$Inn(G)$ is a normal subgroup of the full automorphism group $Aut(G)$ of $G$.&#xA;  The quotient group&#xA;      $$Aut(G) / Inn(G)$$&#xA;  is known as the outer automorphism group $Out(G)$. The outer automorphism group measures, in a sense, how many automorphisms of G are not inner. Every non-inner automorphism yields a non-trivial element of $Out(G)$, but different non-inner automorphisms may yield the same element of $Out(G)$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By associating the element a in $G$ with the inner automorphism $f(x) = x^a$ in $Inn(G)$ as above, one obtains an isomorphism between the quotient group $G / Z(G)$ (where $Z(G)$ is the center of $G$)&#xA;  and the inner automorphism group:&#xA;      $$G / Z(G) = Inn(G)$$&#xA;  This is a consequence of the first isomorphism theorem, because $Z(G)$ is precisely the set of those elements of $G$ that give the identity mapping as corresponding inner automorphism (conjugation changes nothing).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we have that if $G$ is center-less as well all the automorphisms are inner-automorphisms (i.e. conjugation isomorphisms) then you have that $G = G / Z(G) \cong Inn(G) = Aut(G)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is a whole and full answer, but it should give you a clue I hope.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-02T20:08:54.720" LastActivityDate="2010-09-02T20:08:54.720" CommentCount="3" />
  <row Id="3894" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T19:28:29.997" Score="3" ViewCount="0" Body="&lt;p&gt;I love Moron's approach.  Because it is usually interesting to look at other solutions, consider what happens if you just write $p(x) = p_n x^n + p_{n-1}x^{n-1} + \cdots + p_0$ and write out the equation.  Comparing the coefficients of $x^k$ ($k \gt 0$) yields the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_k= \frac{2^{k-5}-1}{2^k - 1} p_{k-1}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence $p_5 = 0$, implying $p_6 = p_7 = \cdots = 0$, and the rest of the recursion yields $p_4, p_3, \ldots, p_0$ as multiples of $p_4$.  This approach, although pedestrian, also demonstrates that we need only assume $p$ is analytic near 0.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-02T19:28:29.997" />
  <row Id="3895" PostTypeId="2" ParentId="3852" CreationDate="2010-09-02T20:54:33.247" Score="13" ViewCount="0" Body="&lt;p&gt;If $\rm\:B\:$ is a linear map on a finite dimensional vector space $\rm\: V\:$ over field $\rm\:K\:,\:$ then trivially (cf. note below) by finite dimensionality there is a nonzero polynomial $\rm\:p(x)\in K[x]\;$ such that $\rm\:p(B) = 0\:$. We may assume $\rm\:p(0) \ne 0\;$ by canceling any factors of $\rm\:B\;$ from $\rm\;p(B)\;$ via left-multiplying by $\rm\:A\;$ and using $\rm\;AB = 1\;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\rm\;\;\; AB=1  \;\Rightarrow\;  (BA-1)\;\: B^n \;=\; 0\;$  for $\:\rm\;n&gt;0\;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus we have $\rm\;\; 0 \:=\: (BA-1)\: p(B) = (BA-1)\: p(0) \;\Rightarrow\;  BA=1 \quad\quad$&#xA; &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially a special case of computing inverses by the Euclidean algorithm - see my &lt;a href=&quot;http://google.com/groups?selm=y8zogksm5v0.fsf@berne.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: The existence of $\rm\;p(x)\;$ follows simply from the fact that $\rm\:V\;$ finite-dimensional implies the same for the vector space $\rm\:L(V)\:$ of linear maps on $\rm\:V\:$ (indeed if $\rm\:V\;$ has dimension $\rm n$ then a linear map is uniquely determined by its matrix of $\:\rm n^2\:$ coefficients). So $\rm\: 1,\: B,\: B^2,\: B^3,\:\cdots\;$ are $\rm\:K$-linearly dependent in $\rm\: L(V)$ which yields the sought nonzero polynomial annihilating $\rm\:B\;$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-03T13:54:04.780" LastActivityDate="2010-09-03T13:54:04.780" CommentCount="4" />
  <row Id="3896" PostTypeId="2" ParentId="3878" CreationDate="2010-09-02T21:13:35.897" Score="2" ViewCount="0" Body="&lt;p&gt;If you take $X$ to be a point, then this reduces to your previous question: you are just taking fibres at each point $y \in Y$, and so the answer is &quot;no&quot; for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to see that this isn't just being pedantic, let me explain how to bootstrap the answer&#xA;to the previous question to give examples in this case which truly have a non-trivial product&#xA;structure.  For this, suppose that $Y = \mathbb A^1$ and that $X$ is the variety attached to&#xA;the ring $A$, so that $X \times Y$ corresponds to the ring $A[t]$.  Let $M$ be the module&#xA;$A[t]/t^2,$ and let $s$ be the section $t \bmod t^2$.  Then $s$ will vanish along each &quot;horizontal line&quot;, but is non-zero.  (The case $X = $ a point gives the counterexample to&#xA;your previous question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, if $M$ is torsion free and finitely generated then you should be okay.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-02T21:13:35.897" />
  <row Id="3897" PostTypeId="1" AcceptedAnswerId="3903" CreationDate="2010-09-02T21:15:33.480" Score="2" ViewCount="239" Body="&lt;p&gt;In Exercise 5 (f) of Angus Taylor's Advanced calculus (p. 659) one is asked to find the value of the following integral if convergent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I:=\displaystyle\underset{R}{\int \int \int }\dfrac{x^{2}y^{2}z^{2}}{r^{17/2}}dV$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the unit sphere $x^{2}+y^{2}+z^{2}\leq 1$ and $r^{2}=x^{2}+y^{2}+z^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing that $\dfrac{x^{2}y^{2}z^{2}}{r^{17/2}}\leq \dfrac{r^{6}}{r^{17/2}}=r^{-5/2}$ I proved that $I$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using spherical co-ordinates $r$, $\theta $, $\phi $ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=r\sin \phi \cos \theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=r\sin \phi \sin \theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=r\cos \theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I transformed the integral $I$ into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=\displaystyle\int_{0}^{2\pi }(\int_{0}^{\pi }(\underset{\delta \rightarrow 0}{\lim }\int_{\delta }^{1}(r^{2}\sin \phi )\dfrac{x^{2}y^{2}z^{2}}{r^{17/2}}\;dr)\;d\phi )\;d\theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\displaystyle\underset{\delta \rightarrow 0}{(\lim }\int_{\delta }^{1}r^{-1/2}dr)\int_{0}^{2\pi }\cos ^{4}\theta \sin ^{2}\theta d\theta\int_{0}^{\pi }\sin ^{5}d\phi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=2\cdot \dfrac{1}{8}\pi \cdot \dfrac{16}{15}=\dfrac{4}{15}\pi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solutions the answer is $\dfrac{8}{105}\pi $. Since sometimes there&#xA;are a few book typos (in the exercises) to prevent undue copying, I ask the&#xA;following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the correct solution, $\dfrac{4}{15}\pi $ or $\dfrac{8}{105}\pi $?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE (Correction)&lt;/strong&gt;: instead of $z=r\cos \theta $ it is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=r\cos \phi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;See a comment from whuber&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral $I$ is transformed into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=\displaystyle\int_{0}^{2\pi }(\int_{0}^{\pi }(\underset{\delta \rightarrow 0}{\lim }\int_{\delta }^{1}(r^{2}\sin \phi )\dfrac{x^{2}y^{2}z^{2}}{r^{17/2}}dr)d\phi )d\theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(r^{2}\sin \phi )\dfrac{x^{2}y^{2}z^{2}}{r^{17/2}}=(r^{2}\sin \phi )\dfrac{1}{r^{17/2}}\left( r\sin \phi \cos \theta \right)&#xA;^{2}\left( r\sin \phi \sin \theta \right) ^{2}\left( r\cos \phi \right) ^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=r^{-1/2}\cos ^{2}\theta \cdot\sin ^{2}\theta \cdot\cos ^{2}\phi \cdot\sin ^{5}\phi $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the transformed integral becomes (if I am right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=\displaystyle\underset{\delta \rightarrow 0}{(\lim }\int_{\delta&#xA;}^{1}r^{-1/2}dr)\int_{0}^{2\pi }\cos ^{2}\theta\cdot\sin ^{2}\theta \;d\theta&#xA;\int_{0}^{\pi }\cos ^{2}\phi \cdot\sin ^{5}\phi \;d\phi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=2\cdot \dfrac{1}{4}\pi \cdot \dfrac{16}{105}=\dfrac{8}{105}\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct solution will be $\dfrac{8}{105}\pi $ as in the book. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-24T11:36:50.603" LastActivityDate="2010-12-24T19:33:58.383" Title="Evaluating a convergent improper triple integral over the unit sphere" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;improper-integrals&gt;&lt;multivariable-calculus&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="3898" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T21:33:02.127" Score="4" ViewCount="0" Body="&lt;p&gt;Since $\rm\;p(2x)\;$ has roots being half the roots $\rm\;r_i\;$ of $\rm\;p(x)\;$, and $\rm\;\frac{p(x)}{p(2x)} = \frac{x-16}{16(x-1)}\;$, for the roots to all cancel like that leaving only one term in the numerator and denominator, we must have simply that the map $\rm\;r\to r/2\;$ is a left shift map on the (intermediate) roots. The numerator shows the largest root is 16, and the denomimator shows half the smallest root is 1. The other intermediate roots must obey the left shift $\rm\;r_i/2 = r_{i-1},\;$ so the roots are $\rm\;2,4,8,16,\;$ i.e. $\rm\;p(x) = c(x-2)(x-4)(x-8)(x-16)\;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;$\:$ I purposely omitted a detail above since I thought it might be a homework problem. Based on the comments, this caused some confusion, so I will elaborate. Namely, notice that each linear linear factor of $\rm\;p(x)\;$ introduces a constant factor of $2\:$ in the denominator of $\rm\: p(x)/p(2x)$. But said denominator has constant factor $16$, so the $4$ factors found above are complete. In particular, $0$ cannot be a root of $\rm\:p(x).$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-03T00:20:32.310" LastActivityDate="2010-09-03T00:20:32.310" CommentCount="12" />
  <row Id="3899" PostTypeId="1" AcceptedAnswerId="3902" CreationDate="2010-09-02T21:37:05.617" Score="1" ViewCount="208" Body="&lt;p&gt;Hi All, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently prepping for uni having been a few years out of the studying loop (programming as it happens). Anyway, I've been re-reading my A-level notes/exercises and looking through OpenCourseWare stuff and I noticed this assertation in my notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $a \cos\theta \pm b \sin\theta = c$ then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \sin\theta \pm b\cos\theta = r\sin(\theta\pm\alpha)+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \cos\theta \pm b\sin\theta = r\cos(\theta\pm\alpha)+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $r \in \mathbb{N}$, $0 \leq \theta &amp;lt; 2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've heard of fourier series which have a very similar form to these equestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, is there a relation between the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please bear in mind I know I'm stepping off a cliff and into &quot;unknown unknowns&quot; territory if they are - I have absolutely no idea what harmonic analysis is and I don't (yet) understand fourier series fully, although I grasp roughly how they work. I'm just interested to know if they are related and of course any further information / direction / reading I should follow up. I ask because this was one of those &quot;odd topics&quot; at A-level that we derived in class from a geometric argument then I've never seen again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-05T04:24:00.237" LastActivityDate="2010-09-05T04:24:00.237" Title="Relation between &quot;harmonic form&quot; and fourier series?" Tags="&lt;algebra-precalculus&gt;&lt;trigonometry&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="3900" PostTypeId="2" ParentId="3888" CreationDate="2010-09-02T22:22:37.647" Score="6" ViewCount="0" Body="&lt;p&gt;Moron's solution and Dubuque's solution are essentilly the same, but I think their presentations can be made neater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $p$ is not a constant polynomial (unless it is zero).  So we may write $p(x) = C(x-z_1)(x-z_2)\ldots(x-z_n)$.  Then the given equality implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2x-z_1)(2x-z_2)\ldots(2x-z_n)(x-16) = 16(x-1)(x-z_1)(x-z_2)\ldots(x-z_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparing the leading coefficient on both sides, one sees $n=4$.  Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-\frac{z_1}{2})(x-\frac{z_2}{2})(x-\frac{z_3}{2})(x-\frac{z_4}{2})(x-16)=(x-1)(x-z_1)(x-z_2)(x-z_3)(x-z_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\lbrace \frac{z_1}{2}, \frac{z_2}{2}, \frac{z_3}{2}, \frac{z_4}{2}, 16\rbrace = \lbrace 1, z_1, z_2, z_3, z_4\rbrace$.  Now one easily sees that, up to a relabelling of indices, we must have $z_i = 2^i$.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2010-09-02T22:22:37.647" CommentCount="10" />
  <row Id="3902" PostTypeId="2" ParentId="3899" CreationDate="2010-09-02T22:53:22.233" Score="5" ViewCount="0" Body="&lt;p&gt;I can't really make sense of those equations, but it looks something like the statement that for any $ a, b \in \mathbb{R} $, then for $ r = \sqrt{a^2 + b^2 }$ and $ \phi = \arctan(b/a) $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ a \sin(\theta) + b \cos(\theta) = r \sin(\theta + \phi).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you visualize x and y axes with $ cos(\theta) $ along the x-axis and $ sin(\theta) $ along the y-axis, then this says we can identify any sinusoidal, i.e. any function like $ r \sin(\theta + \phi) $, by its cartesian coordinates $(a,b)$ in this plane, or its polar coordinate $(r, \phi )$.  $r$ is called the magnitude of the sinusodial, and $ \phi $ the phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is directly related to Fourier Analysis.  The main idea in Fourier Analysis is to decompose a function into its sinusodial components.  For instance, if $f$ is a real-valued function on the interval $ [0, 2\pi] $ that is suitably regular, it turns out you can write $f$ uniquely as an infinite sum of sine and cosines, called the Fourier Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)). $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers $ a_n $ and $ b_n $ are called the Fourier coefficients of your function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this representation of Fourier series is sometimes difficult to grasp.  But using the relationship above, we see we could have also wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = a_0 + \sum_{n=1}^{\infty} r_n \sin(nx + \phi_n) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form I think you can see more clearly what's going on: For each $n$, there is only one sinusoidal component of $f$ of frequency $n$; $r_n = \sqrt{a_n^2 +b_n^2}$ tells you its magnitude and $\phi_n$ tells you its phase.&lt;/p&gt;&#xA;" OwnerUserId="1692" LastEditorUserId="1692" LastEditorDisplayName="" LastEditDate="2010-09-03T02:51:05.743" LastActivityDate="2010-09-03T02:51:05.743" CommentCount="3" />
  <row Id="3903" PostTypeId="2" ParentId="3897" CreationDate="2010-09-02T22:55:13.307" Score="5" ViewCount="0" Body="&lt;p&gt;For this answer, I'll use $\rho$ instead of $r$, I tend to confuse things when dealing with both cylindrical and spherical coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting the original integrand into spherical coordinates gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi)\sin(\phi))^2}{\rho^{5/2}}$$&#xD;&#xA;&#xD;&#xA;which we multiply by the Jacobian $\rho^2 \sin(\phi)$ for the triple integration to give the final integrand&#xD;&#xA;&#xD;&#xA;$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}$$&#xD;&#xA;&#xD;&#xA;As for setting up the limits of the integration, we can exploit symmetry and cut up the unit sphere into octants to simplify the mathematics, thus:&#xD;&#xA;&#xD;&#xA;$$8\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$&#xD;&#xA;&#xD;&#xA;and then bring out the constant term to give&#xD;&#xA;&#xD;&#xA;$$\frac12\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$&#xD;&#xA;&#xD;&#xA;Let's separate things out:&#xD;&#xA;&#xD;&#xA;$$\frac12\int_{0}^{1}\frac{\mathrm{d}\rho}{\sqrt{\rho}}\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$&#xD;&#xA;&#xD;&#xA;The integral with respect to $\rho$ is 2, which cancels out the multiplicative factor, so we're left with&#xD;&#xA;&#xD;&#xA;$$\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$&#xD;&#xA;&#xD;&#xA;Evaluating the angular integrals gives:&#xD;&#xA;&#xD;&#xA;$$\left.\frac{\theta}{2}-\frac{\sin(4\theta)}{8}\right|_{\theta=0}^{\theta=\pi/2}=\frac{\pi}{4}$$&#xD;&#xA;&#xD;&#xA;$$\left.-\frac{5\cos(\phi)}{16}-\frac{\cos(3\phi)}{48}+\frac{3\cos(5\phi)}{80}-\frac{\cos(7\phi)}{112}\right|_{\phi=0}^{\phi=\pi/2}=\frac{32}{105}$$&#xD;&#xA;&#xD;&#xA;and we get the result $\frac{8\pi}{105}$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-11T01:52:39.690" LastActivityDate="2010-11-11T01:52:39.690" CommentCount="2" />
  <row Id="3904" PostTypeId="1" AcceptedAnswerId="3907" CreationDate="2010-09-02T23:12:07.250" Score="3" ViewCount="90" Body="&lt;p&gt;Let me first start out by saying as you might have guessed, &lt;strong&gt;this is a homework problem&lt;/strong&gt;. Therefore I am &lt;strong&gt;not looking for an answer&lt;/strong&gt; to the question. I am looking for help in how to analyze it (My textbook is horrid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a grid of 9 squares surrounded by a circle.&#xA;I can use three colors to color each region with no region touching any other region with the same color (corners don't count). **`&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I &lt;strong&gt;am asked to find the number of&#xA;  possible different colorings.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I drew a set of nodes with lines connecting neighboring regions, but I still have no idea how to figure out the number of combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also made a guess of three (which I know is wrong) since you can color the regions once with the three colors then map each color to a different one two times after the first coloring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/B7R1O.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No blue is touching any other blue regions, no white is touching any other white regions, no red is touching any other red regions.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2010-09-03T00:16:09.403" LastActivityDate="2010-09-03T03:17:31.693" Title="How can I count the number of colored combinations in a set of regions?" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="3905" PostTypeId="1" AcceptedAnswerId="3957" CreationDate="2010-09-02T23:50:54.650" Score="0" ViewCount="143" Body="&lt;p&gt;I have a set of RGB colour values detected by a camera $C_{i_{RGB}}$ which are to be described by the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{i_{RGB}} = X F_{i_{rgb}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F_{i_{rgb}}$ is the component incident light at the detector, where the components $rgb$ are fixed at narrow bands and $X$, an $(n\times 3)$ matrix, is a description of the colour band overlap (cross talk if you will) for specific frequencies.  In a simplified RGB scheme, $X$ would be described:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;\mathsf X &amp;amp;= \begin{bmatrix}&#xA;  1 &amp;amp; R_{GR} &amp;amp; R_{BR} \\&#xA;  R_{RG} &amp;amp; 1 &amp;amp; R_{BG} \\&#xA;  R_{RB} &amp;amp; R_{G} &amp;amp; 1 &#xA;\end{bmatrix}\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the diagonal entries of $X$ are $1$'s, and the off-diagonals are rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I solve for the values of $X$ only knowing $C_{i_{RGB}}$?&lt;/strong&gt; Eventually, given $C_i$ I need to find the corresponding $F_i$ using the constant $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually added colour bands (band passes) will be added which will turn the above relation into something like this (and I fear the case where I have 5, 6, and 7 band passes to deal with):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}&#xA;  C_{i_R} \\&#xA;  C_{i_G} \\&#xA;  C_{i_B} &#xA;\end{bmatrix} = \begin{align}\begin{bmatrix}&#xA;  X_{aR} &amp;amp; X_{bR} &amp;amp; X_{cR} &amp;amp; X_{dR} &amp;amp; X_{eR} &amp;amp; X_{fR} \\&#xA;  X_{aG} &amp;amp; X_{bG} &amp;amp; X_{cG} &amp;amp; X_{dG} &amp;amp; X_{eG} &amp;amp; X_{fG} \\&#xA;  X_{aB} &amp;amp; X_{bB} &amp;amp; X_{cB} &amp;amp; X_{dB} &amp;amp; X_{eB} &amp;amp; X_{fB} &#xA;\end{bmatrix}\end{align}\begin{bmatrix}&#xA;  F_{i_a} \\&#xA;  F_{i_b} \\&#xA;  F_{i_c} \\&#xA;  F_{i_d} \\&#xA;  F_{i_e} \\&#xA;  F_{i_f} &#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a graphics programming application and I lament that my first year algebra was 20 years ago.&lt;/p&gt;&#xA;" OwnerUserId="1067" LastEditorUserId="1067" LastEditorDisplayName="" LastEditDate="2010-09-03T12:34:11.863" LastActivityDate="2010-09-03T14:32:51.310" Title="How can I solve this matrix?" Tags="&lt;matrices&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="3906" PostTypeId="2" ParentId="3876" CreationDate="2010-09-03T00:18:23.320" Score="3" ViewCount="0" Body="&lt;p&gt;What you are after is known in statistics as spectral analysis. There is a whole literature on extracting frequency components from noisy data.  See &lt;a href=&quot;http://www.stat.pitt.edu/stoffer/tsa2/&quot; rel=&quot;nofollow&quot;&gt;Shumway and Stoffer (2006, chapter 4)&lt;/a&gt; for example.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2010-09-03T00:18:23.320" />
  <row Id="3907" PostTypeId="2" ParentId="3904" CreationDate="2010-09-03T00:19:07.117" Score="3" ViewCount="0" Body="&lt;p&gt;I'm not sure about this, maybe I don't understand the problem, but here it goes. If you colour the region outside the grid A, the 8 external cells of the grid must be coloured alternately with B and C, and the one in the center, with A. Thus you have two possible colourings. Same thing if you colour the region outside B, or C. Thus the total number of colourings would be 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; Indeed, there are more. If the external cells are coloured B and C, then the one in the centre may be A, and either B or C, so there are 12 different colourings now.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-03T00:54:18.047" LastActivityDate="2010-09-03T00:54:18.047" CommentCount="1" />
  <row Id="3908" PostTypeId="2" ParentId="3905" CreationDate="2010-09-03T00:37:26.793" Score="0" ViewCount="0" Body="&lt;p&gt;If you don't know &lt;em&gt;both&lt;/em&gt; $X$ and $F_i$ then you have an infinite number of solutions.&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-09-03T00:37:26.793" CommentCount="1" />
  <row Id="3909" PostTypeId="2" ParentId="3904" CreationDate="2010-09-03T00:47:09.687" Score="4" ViewCount="0" Body="&lt;p&gt;While it is valuable to attack such problems &quot;by hand,&quot; here are some generalities.  The general version of this problem is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_coloring&quot; rel=&quot;nofollow&quot;&gt;graph coloring&lt;/a&gt;.  (The graph being colored here is the graph whose vertices are the regions of the diagram where there is an edge between two regions if they are adjacent.)  Counting colorings is &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_coloring#Computational_complexity&quot; rel=&quot;nofollow&quot;&gt;known to be #P-hard&lt;/a&gt;, so it is not all that easy to do in general.  For small graphs there is a reasonable way to compute the number of colorings by any fixed number of colors $n$ by using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatic_polynomial#Deletion.E2.80.93contraction&quot; rel=&quot;nofollow&quot;&gt;deletion-contraction recurrence for the chromatic polynomial&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-03T03:17:31.693" LastActivityDate="2010-09-03T03:17:31.693" CommentCount="3" />
  <row Id="3910" PostTypeId="2" ParentId="3905" CreationDate="2010-09-03T01:11:38.343" Score="0" ViewCount="0" Body="&lt;p&gt;If I understand correctly, you're trying to correlate $n$ separate equations for $n$ different values of $C_i$ and $F_i$ into a solution for $X$; is that correct?  If so, then the easiest way to go is to turn your set of vector equations into a matrix equation!  Simply write $C = [ C_1 C_2 C_3 \ldots C_n ]$ (assuming that the $C_n$ are being written as column vectors), $F = [ F_1 F_2 F_3 \ldots F_n ]$, and then by linearity you wind up with a single matrix equation $C = X F$ which can be solved by inverting F (if F isn't invertible then you have no guaranteed solution) and writing $X = C F^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-03T01:11:38.343" CommentCount="3" />
  <row Id="3911" PostTypeId="2" ParentId="3904" CreationDate="2010-09-03T01:17:32.527" Score="1" ViewCount="0" Body="&lt;p&gt;I've found that the simplest way to attack problems like this is really by straight enumeration.  Start by coloring the outer region, then see what colors that forces or eliminates; in this case, it means that none of the eight outer cells of your grid can share that color.  Now, pick another unforced cell (e.g., the top-left corner of the grid) and see how many ways that you can color it; in this case we have two separate colors that we can use, and whichever one we use then forces the colors on all of the outer cells of the grid, leaving us with just one un-forced color (the central square of the grid can be one of two colors), etc. While this is sort-of a graph coloring problem, it's really closer to a constraint satisfaction problem in disguise, and the methods used for solving a lot of the popular pen-and-paper puzzles like Sudoku/Kakuro/Picross/etc. are pretty closely related to this.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-03T01:17:32.527" />
  <row Id="3912" PostTypeId="1" AcceptedAnswerId="3960" CreationDate="2010-09-03T02:06:57.783" Score="1" ViewCount="155" Body="&lt;p&gt;I'm trying to find out if this is a true statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming asymptotic cases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \sin(n \frac{\pi}{2}) + \log{n} = O(3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you go about solving a problem like this systematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: O = &quot;Big O&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: lg = log base 2&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-03T02:42:17.873" LastActivityDate="2010-09-03T15:11:21.847" Title="Big O comparison in asymptotic cases" Tags="&lt;asymptotics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3913" PostTypeId="1" AcceptedAnswerId="3914" CreationDate="2010-09-03T02:39:38.483" Score="10" ViewCount="297" Body="&lt;p&gt;Back in the day, before approximation methods like splines became vogue in my line of work, one way of computing the area under an empirically drawn curve was to painstakingly sketch it on a piece of graphing paper (usually with the assistance of a French curve) along with the axes, &lt;strong&gt;painstakingly&lt;/strong&gt; cut along the curve, weigh the cut pieces, cut out a square equivalent to one square unit of the graphing paper, weigh that one as well, and reckon the area from the information now available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One time, when we were faced with determining the area of a curve that crossed the horizontal axis thrice, I did the careful cutting of the paper, and made sure to separate the pieces above the horizontal axis from the pieces below the horizontal axis. Then, my boss suddenly scooped up all the pieces and weighed them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I argued that the grouped pieces should have been weighed separately, and then subtract the weights of the &quot;below&quot; group from the &quot;above&quot; group, while my boss argued that we were calculating the total area, and thus, not separating the pieces was justifiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one of us was correct?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-03T07:00:14.310" LastActivityDate="2010-09-03T08:36:49.240" Title="The right &quot;weigh&quot; to do integrals." Tags="&lt;calculus&gt;&lt;approximation&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="3914" PostTypeId="2" ParentId="3913" CreationDate="2010-09-03T03:01:06.763" Score="5" ViewCount="0" Body="&lt;p&gt;you are correct. Actually, the integral is equal to: &quot;(total above weight) minus (total below weight)&quot;, which I'm sure is what you meant. What your boss is calculating is actually&#xA;$$&#xA;\int_a^b |f(x)| dx&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-09-03T03:01:06.763" />
  <row Id="3915" PostTypeId="2" ParentId="3912" CreationDate="2010-09-03T03:10:09.737" Score="2" ViewCount="0" Body="&lt;p&gt;If $f(n)=O(g(n))$, with $g(n)&gt;0$ means that there is a a constant $M&gt;0$ that $|f(n)|\leq Mg(n)$ for every $n$, then (if you meant $O(n)$) we must show that there is a constant $M&gt;0$ that $|n\sin{(n\pi/2)}+\log{n}|\leq Mn$ for every $n$. But $|n\sin{(n\pi/2)}+\log{n}|\leq |n\sin{(n\pi/2)}| + \log{n}\leq n +\log{n}\leq 2n$ (for $n&gt; 1$), therefore $M=2$ will do if $n&gt;1$. For the case $n\to 0$, divide the inequality to prove by $n$ and note that $(\log{n})/n\to\infty$ when $n\to 0$, so it can't be true.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-03T03:48:54.393" LastActivityDate="2010-09-03T03:48:54.393" CommentCount="2" />
  <row Id="3916" PostTypeId="2" ParentId="3913" CreationDate="2010-09-03T03:21:13.250" Score="3" ViewCount="0" Body="&lt;p&gt;I guess we speak of the area &quot;under a curve&quot; when the piece of curve is above the $x$ axis. Sign does matter in the definition of the proper integral, so I would say you were right, and your boss was wrong. Doing integrals by &quot;weighing&quot; things should allow for negative weights.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-03T03:21:13.250" />
  <row Id="3917" PostTypeId="1" AcceptedAnswerId="3971" CreationDate="2010-09-03T03:43:29.130" Score="3" ViewCount="413" Body="&lt;p&gt;On my homework today, we had to find all the normal subgroups of $D_{n}$, the dihedral group of order 2n. I solved the problem by looking at how the conjugacy classes change based on whether n is even or odd and then constructed the normal subgroups as unions of the conjugacy classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 2 questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is there a better way to approach the problem than looking at the conjugacy classes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Could someone explain why we want to find all the normal subgroups of a particular group?   How does this provide us additional insight into the structure of the group we are studying? (Right now, this exercise feels more like a 'computation' to me than a way of understanding $D_{n}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-03T17:41:47.910" Title="Finding all normal subgroups of a group" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;intuition&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="3918" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T04:05:57.393" Score="1" ViewCount="0" Body="&lt;p&gt;Very vague answer to (ii) (I'm taking this course myself): if you find all normal subgroups of a given group, then you know all its quotient groups, since quotient groups are defined through normal subgroups only - if you consider all left (or right) cosets of a given group, that isn't normal, they may not conform a group. And many theorems that follow in the theory are based on quotient groups.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-03T04:05:57.393" />
  <row Id="3919" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T04:30:42.387" Score="7" ViewCount="0" Body="&lt;p&gt;(i) Conjugacy classes are certainly useful, since a normal subgroup must be a union of conjugacy classes. There are other ideas that can come into play: once you know a proper normal subgroup $N$, by taking the quotient you can obtain information about $G$ from information you may find in $G/N$, which will hopefully be easier than working directly in $G$ because $G/N$ will be smaller than $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) In fact, that is one important reason for finding all normal subgroups: being able to look at the quotient, and thus obtain information about $G$ by looking at groups that are smaller than $G$. Another is that the normal subgroups are closely associated with all possible images of $G$ under homomorphisms. Both of these facts are part of the Isomorphism Theorems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now you are probably engaged in computation and getting familiar with techniques for working with groups in general, and $D_n$ in particular; the importance of normal subgroups will likely emerge as you start using groups. For example, normal subgroups are extremely important when you are considering Galois groups (which help you study fields and field extensions). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-03T04:30:42.387" />
  <row Id="3920" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T04:32:09.093" Score="2" ViewCount="0" Body="&lt;p&gt;Finding normal subgroups of a group is a first step towards understanding how it &quot;factors&quot; into simpler groups, analogous to finding prime factors of a number.  To understand a group it then suffices to understand i) the pieces it's built out of (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_group&quot; rel=&quot;nofollow&quot;&gt;simple groups&lt;/a&gt; in its &lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_series&quot; rel=&quot;nofollow&quot;&gt;composition series&lt;/a&gt;) and ii) how those pieces fit together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case it is also just a good exercise in becoming familiar with groups and with basic constructions associated to groups.  Also, like Weltschmerz says, you should think of knowing the normal subgroups as the same thing as knowing all the quotient groups of a group, and the quotient groups of a group are like its &quot;shadows&quot; - they reveal some, but not all, of the structure of the group, and putting them together one gets a picture of the whole thing.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-03T04:32:09.093" />
  <row Id="3921" PostTypeId="1" CreationDate="2010-09-03T05:26:21.977" Score="7" ViewCount="181" Body="&lt;p&gt;Say I have a triangle with vertices &lt;code&gt;(0,0), (2,4), (4,0)&lt;/code&gt; that I want to rotate along the origin. Rotation by multiples of 90 is simple. However, I want to rotate by something a bit more complicated, such as 54 degrees. How do I figure out where the vertices would be then?&lt;/p&gt;&#xA;" OwnerUserId="1110" LastActivityDate="2010-09-03T05:38:19.880" Title="Rotations by degrees other than 90, 180, and 270." Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="3922" PostTypeId="2" ParentId="3921" CreationDate="2010-09-03T05:29:42.237" Score="12" ViewCount="0" Body="&lt;p&gt;One way is to use complex numbers.  Multiplying by $\cos\theta+i\sin\theta$ rotates by $\theta$ about 0, so you could multiply $(2+4i)(\cos 54^\circ+i\sin 54^\circ)$ to get the rotation image of (2,4).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-03T05:29:42.237" />
  <row Id="3923" PostTypeId="1" AcceptedAnswerId="3940" CreationDate="2010-09-03T05:37:58.550" Score="2" ViewCount="186" Body="&lt;p&gt;I propose the following lemma and its proof. It is related to row-null and column-null matrices - i.e. matrices whose rows and columns both sum to zero. Could you please give your opinion on the plausibility of the lemma, and the validity of the proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $Z\in\text{GL}(n,\mathbb{R})$ be a general $n\times n$ real matrix, and let $Y\in\mathcal{S}(n,\mathbb{R})$, where $\mathcal{S}(n,\mathbb{R})$ is the space of row-null column-null $n\times n$ real matrices. Then $\text{Tr}(ZY)=0$ for all $Y$ in $\mathcal{S}(n,\mathbb{R})$ if and only if $Z$ has the form $$Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&#xA;Consider the space of row-null and column-null matrices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{S}(n,\mathbb{R})= \left\{ Y_{ij}\in GL(n,\mathbb{R}):\sum_{i}Y_{ij}=0,\sum_{j}Y_{ij}=0 \right\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its dimension is&#xA;$$\text{dim}(S(n,\mathbb{R}))=N^{2}-2N+1$$&#xA;since the row-nullness and column-nullness are defined by $2N$ equations, only $2N-1$ of which are linearly independent.&#xA;Consider the following space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{G}(n,\mathbb{R})=\left\{ Z_{ij}\in GL(n,\mathbb{R}):Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its dimension is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{dim}(\mathcal{G}(n,\mathbb{R}))=2N-1$$&#xA;where $N-1$ is the contribution from the antisymmetric part and $N$ is from the symmetric part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $Y\in\mathcal{S}$ and $Z\in\mathcal{G}$, then the Frobenius inner product of two such elements is&#xA;$$&#xA;\text{Tr}(ZY) =\sum_{ij}\left[\left(p_{j}-p_{i}\right)Y_{ji}+\left(q_{j}+q_{i}\right)Y_{ji}\right]&#xA;$$&#xA;$$&#xA;=\sum_{j}(q_{j}+p_{j})\sum_{i}Y_{ji}+\sum_{i}(q_{i}-p_{i})\sum_{j}Y_{ji}=0&#xA;$$&#xA;Since $\text{dim}(\mathcal{G})+\text{dim}(\mathcal{S})=\text{dim}(GL)$ and $\mathcal{G}\perp\mathcal{S}$, then $\mathcal{G}$ and $\mathcal{S}$ must be complementary in $GL$. Therfore, if $Y$ is orthogonal to all the matrices in $\mathcal{S}$, it must lie in $\mathcal{G}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: How can I get the curly brackets {} to render in latex mode?&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="1393" LastEditorDisplayName="" LastEditDate="2010-09-03T17:59:14.190" LastActivityDate="2010-09-03T17:59:14.190" Title="The orthogonal complement of the space of row-null and column-null matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="3924" PostTypeId="2" ParentId="3921" CreationDate="2010-09-03T05:38:19.880" Score="6" ViewCount="0" Body="&lt;p&gt;In the answer to &lt;a href=&quot;http://math.stackexchange.com/questions/2429&quot;&gt;this question&lt;/a&gt;, I mentioned the formula for the rotation matrix; one merely takes the product of the rotation matrix with the coordinates (treated as 2-vectors) to get the new rotated coordinates. Note that I gave the matrix for clockwise rotation; for anticlockwise rotation, negate the angle (thus switching the sign of the two sine components).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T05:38:19.880" />
  <row Id="3925" PostTypeId="1" AcceptedAnswerId="3932" CreationDate="2010-09-03T06:12:35.230" Score="5" ViewCount="173" Body="&lt;p&gt;How does one prove that for a prime $p \geq 5$ the sum :  $$\sum\limits_{0 &amp;lt; k &amp;lt; \frac{2p}{3}} { p \choose k}$$ is divisible by $p^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each term of $\displaystyle \sum\limits_{0 &amp;lt; k &amp;lt; \frac{2p}{3}} { p \choose k}$ is divisible by $p$, only thing remains is to prove that the sum $$\sum\limits_{ 0 &amp;lt; k &amp;lt; \frac{2p}{3}} \frac{1}{p} { p \choose k}$$ is divisible by $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to evaluate this sum: $\displaystyle \frac{1}{p} { p \choose k} = \frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-03T07:47:53.743" Title="Proving that this sum $\sum\limits_{0 &lt; k &lt; \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3926" PostTypeId="1" CreationDate="2010-09-03T07:01:05.730" Score="3" ViewCount="210" Body="&lt;p&gt;A 1-1 function is called injective. What is an n-1 function called ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about homomorphisms. So perhaps homojective ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Onto is surjective. 1-1 and onto is bijective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about n-1 and onto ? Projective ? Polyjective ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think n-m and onto should be hyperjective as in hypergroups.&lt;/p&gt;&#xA;" OwnerUserId="1613" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T20:33:50.990" LastActivityDate="2010-11-24T20:33:50.990" Title="A 1-1 function is called injective. What is an n-1 function called ?" Tags="&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3927" PostTypeId="1" CreationDate="2010-09-03T07:24:55.703" Score="1" ViewCount="267" Body="&lt;p&gt;What are some interesting applications of the concept of homomorphism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: If there is a homorphism from a ring $R$ to a ring $r$ then a solution to a polynomial equation in $R$ gives rise to a solution in $r$. e.g. if $f:R \rightarrow r$ and $X^2+Y^2=0$ then $f(X^2+Y^2)=f(0), f(X^2)+f(Y^2)=0, f(X)^2+f(Y)^2=0, x^2+y^2=0$&lt;/p&gt;&#xA;" OwnerUserId="1613" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-03T07:56:42.363" LastActivityDate="2010-09-04T16:13:14.943" CommunityOwnedDate="2010-09-04T00:42:24.980" Title="Applications of the concept of homomorphism." Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="3928" PostTypeId="2" ParentId="3927" CreationDate="2010-09-03T07:32:18.657" Score="-1" ViewCount="0" Body="&lt;p&gt;Let me give you some applications of homomorphism in group theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition:&lt;/strong&gt; A group is said to be simple if it has no nontrivial homomorphic images, that is if it has no nontrivial normal subgroups. Beautiful conjectures like &quot;Non abelian simple groups of finite order has even number of elements were proved by 2 eminent group theorists, W.Feit and J.Thompson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, using the properties of homomorphism one can get some useful information about normal subgroups, which play a central role in Galois theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of applications, of Ring theory to number theory also, using the concept of homomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, here are some more of them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:R \to S$ be a homomorphism of Rings. Then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Image of an additive subgroup of $R$ is an additive subgroup of $S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Inverse image of a prime ideal is a prime ideal if both $R$ and $S$ are commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Homomorphic image of a Principal ideal ring is a Principal Ideal Ring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Inverse image of a maximal ideal in $S$ need not be a maximal ideal in $R$ unless $f$ is surjective.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-04T16:13:14.943" LastActivityDate="2010-09-04T16:13:14.943" CommunityOwnedDate="2010-09-04T00:42:24.980" />
  <row Id="3929" PostTypeId="2" ParentId="3926" CreationDate="2010-09-03T07:32:59.803" Score="6" ViewCount="0" Body="&lt;p&gt;n-1 + onto is sometimes called n-fold cover (by &lt;a href=&quot;http://en.wikipedia.org/wiki/Covering_space&quot; rel=&quot;nofollow&quot;&gt;analogy&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-09-03T07:32:59.803" />
  <row Id="3930" PostTypeId="1" AcceptedAnswerId="3933" CreationDate="2010-09-03T07:42:10.253" Score="3" ViewCount="75" Body="&lt;p&gt;I was droodling a bit and a given moment I drew the following construction:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XwnQg.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the three blue intersections are collinear (red line), no matter how I draw the construction lines. If this is always true, I assume that this a know fact [otherwise I have my first theorem! :-) erm conjecture, since I can't prove it :-(  ].&lt;br&gt;&#xA;What's the theorem called?&lt;br&gt;&#xA;TIA&lt;br&gt;&#xA;Steven  &lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-03T07:52:34.150" Title="Collinear points theorem" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="3931" PostTypeId="2" ParentId="3925" CreationDate="2010-09-03T07:42:27.643" Score="2" ViewCount="0" Body="&lt;p&gt;From your last equation one gets&#xA;$$\frac1p{p\choose k}\equiv(-1)^k\frac 1k\pmod p.$$&#xA;So the problem is equivalent to&#xA;$$\sum_{0 &amp;lt; k &amp;lt; 2p/3}(-1)^k\frac1k\equiv0\pmod{p}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $p=1979$ was set as the first problem at the &lt;a href=&quot;http://www.imo-official.org/problems.aspx&quot; rel=&quot;nofollow&quot;&gt;1979 IMO&lt;/a&gt;.&#xA;The method used for this extends to all primes, and no doubt&#xA;you can find solutions for the IMO problem out on the interweb.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-03T07:42:27.643" />
  <row Id="3932" PostTypeId="2" ParentId="3925" CreationDate="2010-09-03T07:42:33.800" Score="6" ViewCount="0" Body="&lt;p&gt;Since we are working in the field $\mathbb{F}_p$ we can write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdots k}$$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(-1)(-2) \cdots (-(k-1))}{1 \cdot2 \cdots k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $$\frac{(-1)^{k-1}}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N = [\frac{2p}{3}]$ and $M = [\frac{N}{2}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus what we need is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k}$$  $$  = \sum_{k=1}^{N} \frac{1}{k} - 2 \sum_{k=1}^{M}\frac{1}{2k}$$  $$ =\sum_{k=M+1}^{N}\frac{1}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $N+M+1 =p$ so we can rewrite as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{N} + \frac{1}{M+1} + \frac{1}{N-1} + \frac{1}{M+2} + \cdots = $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{N(M+1)} + \frac{p}{(N-1)(M+2)} + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $N-M$ terms, which is even, so each term gets paired off.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-03T07:47:53.743" LastActivityDate="2010-09-03T07:47:53.743" CommentCount="3" />
  <row Id="3933" PostTypeId="2" ParentId="3930" CreationDate="2010-09-03T07:52:34.150" Score="7" ViewCount="0" Body="&lt;p&gt;Congratulations! Looks to me like you have rediscovered &lt;a href=&quot;http://mathworld.wolfram.com/PappussHexagonTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Pappus' Hexagon Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image from the link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/SBcBj.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-03T07:52:34.150" CommentCount="7" />
  <row Id="3934" PostTypeId="2" ParentId="58" CreationDate="2010-09-03T08:08:16.830" Score="3" ViewCount="0" Body="&lt;p&gt;It may or may not be useful for detecting fraud in elections, for example the 2009 Iranian election.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sbseminar.wordpress.com/2009/06/15/benfords-law-and-the-iranian-election&quot; rel=&quot;nofollow&quot;&gt;http://sbseminar.wordpress.com/2009/06/15/benfords-law-and-the-iranian-election&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fivethirtyeight.com/2009/06/statistical-evidence-does-not-prove.html&quot; rel=&quot;nofollow&quot;&gt;http://www.fivethirtyeight.com/2009/06/statistical-evidence-does-not-prove.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-09-03T08:08:16.830" />
  <row Id="3935" PostTypeId="2" ParentId="2755" CreationDate="2010-09-03T08:18:43.480" Score="3" ViewCount="0" Body="&lt;p&gt;@Ryan-Budney, can you recommend an easy textbook that covers the background to your more complicated answer?&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-09-03T08:18:43.480" CommentCount="3" />
  <row Id="3936" PostTypeId="2" ParentId="3913" CreationDate="2010-09-03T08:36:49.240" Score="4" ViewCount="0" Body="&lt;p&gt;In my opinion, this is not really a math question.  Which procedure is correct depends what you're going to do with your calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of definition, the integral indeed measures the signed area (positive area minus negative area), as you suggest.  So your approach is computing an approximation to the definite integral $\int_a^b f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe you want the total (unsigned) area.  E.g. if you're going to lay concrete along (some real-world space corresponding to) the region bounded by the curve and the x-axis then surely you want the total area -- there's no such thing as negative concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing what you're using the calculation for, it's impossible to say.  (Essentially, you're asking us: &quot;Which of these is mathematically correct: $A-B+C$ or $A+B+C$?&quot;  Of course it depends upon what you're trying to do.) I would like to think that your boss knew what the point of it all was, so without further information I guess I would trust him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, your story arouses my curiosity.  I suppose you're not putting us on, but weighing paper cutouts is just about the last method I would ever think of to computing area (aren't you going to need a very sensitive scale or an awfully big piece of paper to get anywhere with this?).  How long ago are we talking?  What was the job?  You don't have to answer these questions, but it would be interesting to know...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-03T08:36:49.240" CommentCount="4" />
  <row Id="3937" PostTypeId="1" CreationDate="2010-09-03T09:10:48.437" Score="2" ViewCount="256" Body="&lt;p&gt;First there was arithmetic with numerical calculations (i.e. one unknown on one side of an equation). Then algebra with manipulations of variables (many unknowns anywhere in an equation). Then systems are studied that differ from ordinary arithmetic but share some of the same properties (equations where the unknowns represent all sorts of things - even functional equations) and then these properties are abstracted in abstract algebra and whole classes are studied such as groups and rings. Then category theory studies maps between structures (functorial equations), then n-category theory, then ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do we go now ? Is category theory the end of the road for the forseeable future ? &#xA;Is the only way forward to go backwards and generalize in a different direction ? (&quot;generalized equations&quot; of optimization)&lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2010-09-03T11:12:47.183" Title="What's more general than category theory?" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="3938" PostTypeId="2" ParentId="3923" CreationDate="2010-09-03T09:16:43.543" Score="2" ViewCount="0" Body="&lt;p&gt;I've checked your proof. It is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to say obvious thing: I know two methods for checking the proof is correct.&lt;br&gt;&#xA;1. Check the answer, if you can get it from something else.&lt;br&gt;&#xA;1'. If you can't get the answer independently, check some of it's properties.&lt;br&gt;&#xA;2. Check the proof line by line.&lt;br&gt;&#xA;2'. Divide your proof on parts and check each of them, using one of the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying second method is obvious. If you want, to apply first, you can try to prove the following.&lt;br&gt;&#xA;1. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.&lt;br&gt;&#xA;2. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done by using equations of the form Tr(ZY)=0 with matrices $Y\in S(n,\mathbb{R})$ of the form $Y_{ij}=a_i b_j$, where a and b are vectors of the form vectors with one coordinate equal to 1, another --- equal to -1, all other coordinates equal to 0 (like $(0,\dots,0,1,0,\dots,0,-1,0,\dots,0)$).&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-03T09:16:43.543" />
  <row Id="3940" PostTypeId="2" ParentId="3923" CreationDate="2010-09-03T10:39:13.047" Score="4" ViewCount="0" Body="&lt;p&gt;Here is an alternate way of proving your Lemma. I'm not sure if its any simpler than your proof -- but it's different, and hopefully interesting to some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be the set of $n\times n$ matrices which are row-null and column-null. We can write this set as:&#xA;$$&#xA;S = \left\{ Y\in \mathbb{R}^{n\times n} \,\mid\, Y1 = 0 \text{  and  }1^TY=0\right\}&#xA;$$&#xA;where $1$ is the $n\times 1$ vector of all-ones. The objective is the characterize the set $S^\perp$ of matrices orthogonal to every matrix in $S$, using the Frobenius inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach is to &lt;em&gt;vectorize&lt;/em&gt;. If $Y$ is any matrix in $S$, we can turn it into a vector by taking all of its columns and stacking them into one long vector, which is now in $\mathbb{R}^{n^2\times 1}$. Then $\mathop{\mathrm{vec}}(S)$ is also a subspace, satisfying:&#xA;$$&#xA;\mathop{\mathrm{vec}}(S) = \left\{ y \in \mathbb{R}^{n^2\times 1} \,\mid\, (\mathbf{1}^T\otimes I)y = 0 \text{  and  } (I \otimes \mathbf{1}^T)y = 0 \right\}&#xA;$$&#xA;where $\otimes$ denotes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_product&quot; rel=&quot;nofollow&quot;&gt;Kronecker product&lt;/a&gt;. In other words,&#xA;$$&#xA;\mathop{\mathrm{vec}}(S) = \mathop{\mathrm{Null}}(A),\qquad\text{where: }&#xA;A = \left[ \begin{array}{c} \mathbf{1}^T\otimes I \\ I \otimes \mathbf{1}^T \end{array}\right]&#xA;$$&#xA;Note that vectorization turns the Frobenius inner product into the standard Euclidean inner product. Namely: $\mathop{\mathrm{Trace}}(A^T B) = \mathop{\mathrm{vec}}(A)^T \mathop{\mathrm{vec}}(B)$. Therefore, we can apply the range-nullspace duality and obtain:&#xA;$$&#xA;\mathop{\mathrm{vec}}(S^\perp) =&#xA;\mathop{\mathrm{vec}}(S)^\perp =&#xA;\mathop{\mathrm{Null}}(A)^\perp =&#xA;\mathop{\mathrm{Range}}(A^T)&#xA;$$&#xA;So every vector in $vec(S^\perp)$ is of the form $(\mathbf{1}\otimes I)a + (I\otimes \mathbf{1})b$ for some vectors $a$ and $b$ in $R^{n\times 1}$. It follows that every matrix in $S^\perp$ is of the form $a1^T + 1b^T$. This parametrization is equivalent to the one you presented if you set $a_i = q_i-p_i$ and $b_j = q_j + p_j$.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-09-03T10:45:00.743" LastActivityDate="2010-09-03T10:45:00.743" />
  <row Id="3941" PostTypeId="1" AcceptedAnswerId="3954" CreationDate="2010-09-03T10:39:19.483" Score="7" ViewCount="373" Body="&lt;p&gt;By trignometric identities one can see that $\cos{2x} = 2 \cos^{2}(x) - 1$ can be expressed as a polynomial of $\cos{x}$, where $p(\cos{x})=2 \cos^{2}(x)-1$. Then its natural to ask the same question for the sine function. Can we express $\sin{2x}$ as a polynomial of $\sin{x}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-04T17:07:48.370" Title="Expressing $\sin(2x)$ as a polynomial of $\sin{x}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;trigonometry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="5" />
  <row Id="3942" PostTypeId="2" ParentId="3941" CreationDate="2010-09-03T10:41:04.540" Score="0" ViewCount="0" Body="&lt;p&gt;No, you can't get any simpler than $2\sin\;x\;\cos\;x$. In general, only sines of odd angle multiples ($\sin((2k+1)x)$) are expressible as polynomials in $\sin(x)$, because any powers of cosines that may appear are raised to even exponents, and thus the identity $\sin^2(x)=1-\cos^2(x)$ applies, to turn it into an expression entirely made up of powers of sines.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T10:41:04.540" CommentCount="1" />
  <row Id="3943" PostTypeId="2" ParentId="3941" CreationDate="2010-09-03T10:42:05.770" Score="0" ViewCount="0" Body="&lt;p&gt;you can express (sin2x)^2 as a polynomial of sinx&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-09-03T10:42:05.770" />
  <row Id="3944" PostTypeId="2" ParentId="3937" CreationDate="2010-09-03T11:02:23.577" Score="9" ViewCount="0" Body="&lt;p&gt;I don't really see a coherent logical progression in the branches of mathematics you're putting forward.  Mathematics isn't just about abstraction and generalizing, making things more and more general.  It's most often about solving particular problems.  Category theory was born out of algebraic topology, in many ways as a notational convenience, to make sense of the tremendous and difficult-to-follow messes algebraic topologists were producing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've ever programmed in a language like &quot;C&quot; you know the concept of a &quot;macro&quot;.  This is an idea that is re-usable in many different contexts.  You plug in different objects and the macro continues to make sense.  That's much of the point of category theory, as there are so many ideas that are duplicated over and over again in mathematics that it's confusing to give them special names.  So we call these ideas by generic names that make sense in a wide-array of contexts, like &quot;the co-product (or whatever) in the category C (name your category)&quot;, etc.  It saves time and energy.   Moreover, once you've reduced the &quot;bulk&quot; of your notation sufficiently, there is a phenomena where the concepts are lighter and easier to play with.  So by using category theory you sometimes &quot;lighten the load&quot; a little, making other discoveries perhaps a little easier (if you're lucky). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that regard what gets called &quot;category theory&quot; I think of as more of an attempt to find the natural language for certain types of ideas.  The general idea being that certain types of problems become easy when using appropriate notation. Not all, but some.  Some problems are just hard -- like the Poincare conjecture, or the Schoenflies problem, the classification of finite-simple groups, or existence and uniqueness of solutions to Navier-Stokes (and if you look at the work that's been done on these problems you will see almost no category theory at all, just a tiny little bit on the high-dimensional Poincare and Schoenflies problems). In programming category theory might be analogous to the study of data types, and how one structures memory efficiently. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-03T11:02:23.577" CommentCount="2" />
  <row Id="3945" PostTypeId="2" ParentId="3937" CreationDate="2010-09-03T11:12:47.183" Score="5" ViewCount="0" Body="&lt;p&gt;Generalization goes where problems lead it to. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-03T11:12:47.183" />
  <row Id="3946" PostTypeId="1" CreationDate="2010-09-03T11:34:07.223" Score="1" ViewCount="235" Body="&lt;p&gt;I've got a path that is made up of multiple points - i.e. 0,0  0,50  50,50  70,20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I just draw this line on the screen it looks quite harsh as it puts a sharp angle at the joining of each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I was wondering what the bezier curves algorithm/method would look like which I could call that automatically change the sharp angles to &quot;tight&quot; curves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't want the curve to be too big or generally effect the fall of main path, just soften the join out. If you have a look at the below here is a quick sample I have put together. The line on the left is what I have now, the one in the middle is the line I want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image on the right represents what I think I need the algorithm to do. Essentially, I add an additional point to each arc that makes up the join at a point which 10% away from the join, then I remove the join point and adjust the handles so that they are where the point was (not in the diagram they are slightly apart, this is just so that you can see). This is what I need to be able to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EKY7U.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1700" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-03T12:29:32.183" LastActivityDate="2010-09-08T14:24:04.060" Title="Bezier curves algorithm complex yet simple" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;bezier-curve&gt;" AnswerCount="3" />
  <row Id="3949" PostTypeId="2" ParentId="3946" CreationDate="2010-09-03T12:25:55.377" Score="0" ViewCount="0" Body="&lt;p&gt;Well, Bézier curves are always guaranteed to lie within the convex hull of the control points; it is only a matter of taste/convenience for you to adjust the control points to get the shape you wish. Try finding a program that allows you to interactively move control points around, and experiment!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T12:25:55.377" CommentCount="2" />
  <row Id="3950" PostTypeId="2" ParentId="3852" CreationDate="2010-09-03T13:32:10.677" Score="2" ViewCount="0" Body="&lt;p&gt;Coincidentally, a totally unrelated MathSciNet search turned up the following article, which gives a result along the lines of (but slightly stronger than) the one in Martin Brandenburg's answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/Hill67.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/Hill67.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem (Hill, 1967): Let $R$ be a ring satisfying the ascending chain condition on right ideals.  Then:&lt;br&gt;&#xA;a) For any $n \in \mathbb{Z}^+$, the matrix ring $M_n(R)$ also satisfies ACC on right ideals.&lt;br&gt;&#xA;b) If $R$ satisfies ACC on right ideals and $a,b \in R$ are such that $ab = 1$, then also $ba = 1$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-03T13:32:10.677" CommentCount="0" />
  <row Id="3951" PostTypeId="1" AcceptedAnswerId="3952" CreationDate="2010-09-03T13:35:25.853" Score="0" ViewCount="118" Body="&lt;p&gt;Can anyone give me an example of a vector space $V$ such that there is no norm which is complete on $V$? &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-03T14:02:55.607" Title="Vector spaces with no Complete norms " Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="3952" PostTypeId="2" ParentId="3951" CreationDate="2010-09-03T13:40:19.987" Score="2" ViewCount="0" Body="&lt;p&gt;I assume you want $V$ to be a vector space over $\mathbb{R}$ or $\mathbb{C}$.  Since the latter can be viewed as a special case of the former, let's say $V$ is an $\mathbb{R}$-vector space to fix ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A finite-dimensional subspace is always closed.&lt;br&gt;&#xA;2) A proper closed subspace is nowhere dense.&lt;br&gt;&#xA;3) There is a famous theorem about countable unions of nowhere dense subsets of a complete space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;: I decided that my comment calling for a lack of other answers after having already given an answer myself was something of a conflict of interest, so I have made this answer CW.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-03T14:02:55.607" LastActivityDate="2010-09-03T14:02:55.607" CommunityOwnedDate="2010-09-03T14:02:55.607" CommentCount="2" />
  <row Id="3953" PostTypeId="1" AcceptedAnswerId="3958" CreationDate="2010-09-03T13:43:40.270" Score="3" ViewCount="121" Body="&lt;p&gt;Define $R(0)=0=\emptyset, R(n+1)=P(R(n))$ and $R(\omega) = \cup_{n &amp;lt; \omega} R(n)$. Thus $R(\omega)$ is the set of all sets, which are build out of finitely many braces and $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following relation $E$ on $\omega$: If and only if the $n$th number in the binary representation of $m$ is $1$, then $n E m$. Now I want to construct an isomorphism $(R(\omega),\in) \cong (\omega,E)$ [This is an exercise in Kunen's set theory]. After some playing around I've come up with the following definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g : R(\omega) \to \omega, g(x) = \sum_{y \in x} 2^{g(y)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a well-defined recursion, since $rank(y) &amp;lt; rank(x)$. If $g$ is injective, then it is easy to see that $x \in y \Leftrightarrow g(x) E g(y)$. However I don't see this; neither why $g$ is surjective.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-03T14:16:23.213" LastActivityDate="2010-09-04T04:02:30.310" Title="A nice enumeration of $R(\omega)$" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3954" PostTypeId="2" ParentId="3941" CreationDate="2010-09-03T14:01:05.510" Score="9" ViewCount="0" Body="&lt;p&gt;Just for the sake of completeness, here is a proof that you cannot express it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a general proof, see the bottom of this answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we could express $\sin 2x$ as a polynomial in $\sin x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin x \cos x  = P(\sin x)$ for some polynomial $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sin 0 = 0$, $x$ is a factor of $P(x)$. Hence we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos x = Q(\sin x)$ for a polynomial $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $x = \frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ we see that $1,-1$ are roots of $Q(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we see that for 0 &amp;lt; t &amp;lt; 1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{1-t^2} = (t-1)(t+1)H(t)$ (because $Q(t) = (t-1)(t+1)H(t)$ for some polynomial $H$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By squaring we see that the degree of the left hand side polynomial is $2$ while the degree of the right hand side polynomial is $&gt;2$. This is not possible, as then their difference, which is not identically zero will have infinite roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is not possible to express $\sin 2x$ as a polynomial in $\sin x$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Case:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proof that in general $ \sin (2nx)$ cannot be written as a polynomial in $ \sin x$ based on the identity provided by J.M:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sin(2nx)}{2\cos(x)}=(-1)^{\lfloor n-1/2\rfloor}\sum_{j=0}^{\lfloor n-1/2\rfloor}{(-1)^j \sin((2j+1)x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that $\sin ((2m+1)x)$ &lt;em&gt;can&lt;/em&gt; be written as a polynomial in $\sin x$ with &lt;em&gt;integer&lt;/em&gt; coefficients (follows easily from the DeMoivre's identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $ \sin (2nx)$ could be written as a polynomial $P$. Consider the $r^{th}$ derivative of that polynomial at $0$. This must be a rational number, as the $r^{th}$ derivative of $ \sin (2nx)$ at $0$ is rational. Thus the coefficients of $P$ must be rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus using the above identity we must have that $\cos x$ is of the form $\frac{P(\sin x)}{Q(\sin x)}$ where $P$ and $Q$ are polynomials with rational coefficients. Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1-t^2} = \frac{P(t)}{Q(t)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; t &amp;lt; 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $t=\frac{1}{2}$ gives us a rational number on the RHS while an irrational number on the LHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it is not possible to represent $ \sin (2nx)$ as a polynomial in $ \sin x$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-03T15:07:14.007" LastActivityDate="2010-09-03T15:07:14.007" CommentCount="4" />
  <row Id="3955" PostTypeId="1" CreationDate="2010-09-03T14:15:19.793" Score="2" ViewCount="147" Body="&lt;p&gt;This is such a basic question I'm sure, but I've been trying to find a robust solution to it for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I have the numbers 0-100 in series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number increases, I want another number say 1000, to decrease. so that when you got to 100, it was at 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 = 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 = 500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this kind of series called?&lt;/p&gt;&#xA;" OwnerUserId="1701" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-11T04:35:16.330" LastActivityDate="2010-09-11T04:35:16.330" Title="How to reduce lower one number, while another number increases goes up increments" Tags="&lt;algebra-precalculus&gt;&lt;terminology&gt;&lt;arithmetic&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="3956" PostTypeId="2" ParentId="3926" CreationDate="2010-09-03T14:16:11.447" Score="6" ViewCount="0" Body="&lt;p&gt;IMHO, an n to 1 function should be called &quot;an n to 1 function.&quot;&#xA;Simple, decriptive, to the point.  Adding a new term in this case just muddies the waters.&lt;/p&gt;&#xA;" OwnerUserId="575" LastActivityDate="2010-09-03T14:16:11.447" CommentCount="3" />
  <row Id="3957" PostTypeId="2" ParentId="3905" CreationDate="2010-09-03T14:32:51.310" Score="1" ViewCount="0" Body="&lt;p&gt;Okay, as already mentioned the problem as posed is not tractable. However, reading between the lines, it looks to me a lot like Independent Components Analysis could work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, assuming you have a sequence of such observations from the same sources, and its reasonable to assume that the individual sources are independent of each other, then the original signals can be recovered up to scaling and order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to a nice tutorial paper by Aapo Hyvarinen:&#xA;&lt;a href=&quot;http://www.cs.helsinki.fi/u/ahyvarin/papers/NN00new.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.helsinki.fi/u/ahyvarin/papers/NN00new.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-03T14:32:51.310" CommentCount="4" />
  <row Id="3958" PostTypeId="2" ParentId="3953" CreationDate="2010-09-03T14:42:49.810" Score="1" ViewCount="0" Body="&lt;p&gt;Take the minimal $n$ such that there are $x,y \in R(\omega)$ for which $x \not= y$ and $g(x)=g(y)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g(x) = g(y)$ then we know the following: $|x| = |y|$ and $\forall u \in x \exists v \in y (g(u) = g(v))$ but since it is clear that $u \in v \Rightarrow g(u) &amp;lt; g(v)$ we have that there is some $u \in x, v\in y$ for which $g(u) = g(v) &amp;lt; g(y) = n$ which contradictions the assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;I've got no idea about the surjective part. Seems slightly tricky, but I'll get back to you when I have an answer.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $g(\emptyset) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be the minimal number that is not in the range of $g$, we can write $n = 2^{n_0}+\ldots +2^{n_k}$ where $n_0 &amp;lt; n_1 &amp;lt; \ldots &amp;lt; n_k$. Since $n_i &amp;lt; n$ we have that there is $x_i \in R(\omega)$ for which $g(x_i) = n_i$. So now we can take $x = \{x_i |0\le i\le k\}$ and it is a simple argument that $x \in R(\omega)$ and clearly $g(x) = n$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-03T14:56:33.820" LastActivityDate="2010-09-03T14:56:33.820" CommentCount="1" />
  <row Id="3959" PostTypeId="2" ParentId="3955" CreationDate="2010-09-03T14:43:38.377" Score="2" ViewCount="0" Body="&lt;p&gt;The function you want is $y=10(100-x)$. This was obtained by attempting to compute the quadratic function that passes through the three points ($n+1$ points uniquely determine an nth-degree curve), and then noting that the interpolating polynomial is actually a linear function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, computing the determinant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{vmatrix}1&amp;amp;0&amp;amp;1000\\1&amp;amp;50&amp;amp;500\\1&amp;amp;100&amp;amp;0\end{vmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the area of the &quot;triangle&quot; formed by the three points)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives a value of 0, and thus the three points are collinear. Take any two of the three points, and you can derive the equation of the line passing through them.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-03T14:43:38.377" CommentCount="1" />
  <row Id="3960" PostTypeId="2" ParentId="3912" CreationDate="2010-09-03T15:11:21.847" Score="3" ViewCount="0" Body="&lt;p&gt;Note that the notation--the use of $n$ and base-2 logs--indicates this problem comes from computer science, strongly suggesting that $n$ is an integer and we are being asked about the asymptotics as $n$ grows without bound.  When $n$ is even, the value is $\log(n)$, but when $n$ is odd, the value is either $n + \log(n)$ or $-n + \log(n)$.  Thus the growth equals $\max \left( O(\log(n)), O(-n + \log(n)), O(n + \log(n)) \right) = O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is false that $O(3) \ge O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, one wouldn't use an expression like $O(3)$, because the constant is implicit.  Conventionally, this is written $O(1)$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-03T15:11:21.847" CommentCount="2" />
  <row Id="3962" PostTypeId="2" ParentId="3437" CreationDate="2010-09-03T15:52:48.587" Score="2" ViewCount="0" Body="&lt;p&gt;It seems plausible that there may be no square factors other than $4$, or at least that there are only finitely many primes for which $p^2$ divides $s_n$. A sketchy argument follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $\tau=2 +\sqrt{3}$. Then $s_n = \tau^{2^n} + \tau^{-2^n}$. We work in the ring $R=\mathbb{Z}[\sqrt{3}]$, where $\tau$ is a unit. So $p^2$ divides $S_n$ if and only if $\tau^{2^{n+1}} \equiv 1 \mod p^2$, where our congruence is in $R$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be a prime $\geq 5$. Let $U_{p^2}$ be the unit group of $R/p^2 R$. You want to know whether the order of $\tau$ in $U_{p^2}$ is a power of $2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two cases. First, suppose there is a square root of $3$ mod $p$ (this happens when $p \equiv \pm 1 \mod 12$). This square root will lift to a solution $s$ of $s^2 \equiv 3 \mod p^2$ (we can lift by Hensel's lemma). In this case $R/p^2 \cong \mathbb{Z}/p^2 \oplus \mathbb{Z}/p^2$ and we can identify $\tau$ with the ordered pair $(2 + s, 2 - s)$. Notice that $2-s = (2+s)^{-1}$, so the two halves of the ordered pair have the same order. The unit group of $\mathbb{Z}/p^2$ is cyclic of order $p(p-1)$, and the number of elements whose orders are powers of $p$ is $2^{v_2(p-1)}$, where $v_2(k)$ is the exponent to which $2$ divides $k$. So, heuristically, I would expect the number of $p$ for which $p^2$ divides $s_n$ and there is a square root of $3$ mod $p$ to be something like&#xA;$$\sum_{p \equiv \pm 1 \mod 12} \frac{2^{v_2(p-1)}}{p(p-1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There case where where $p \equiv \pm 5 \mod 12$ is a little trickier but, if I haven't made any mistakes, the similar heuristic estimate is&#xA;$$\sum_{p \equiv \pm 5 \mod 12} \frac{2^{v_2(p+1)}}{p(p+1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is my point: those sums converge. In fact,&#xA;$$\sum_{n=1}^{\infty} \frac{2^{v_2(n)}}{n^2}$$&#xA;converges to $\pi^2/4$ (exercise!) and the above sums are less than this because they are restricted to just summing over primes. Moreover, the sum numerically looks pretty small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;expected&quot; number of such primes is a small finite quantity, and it wouldn't surprise me if $2$ were the only one.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-03T15:52:48.587" />
  <row Id="3963" PostTypeId="2" ParentId="3955" CreationDate="2010-09-03T15:54:22.283" Score="2" ViewCount="0" Body="&lt;p&gt;There is no way to uniquely answer your question without knowing more about what class of functions you permit to interpolate these points. Notice that all of your points are divisible by 50, so we may simplify a bit by scaling down by 50, yielding the points $\rm (x,y) = (0,20), (1,10), (2,0)$. Since the intermediate slopes agree $(10-20)/(1-0) = -10 = (0-10)/(2-1)\;$ it follows that a linear function interpolates these points, viz. $\rm\;y = 20-10x,\;$ or $\rm\;Y = 1000 - 10X\;$ when scaled up by 50 to $\rm (X,Y) = (50x,50y)$. But one could easily interpolate many other curves between these points. Perhaps if you tell us more about your intended application we can elaborate further.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-11T03:25:41.893" LastActivityDate="2010-09-11T03:25:41.893" />
  <row Id="3964" PostTypeId="2" ParentId="164" CreationDate="2010-09-03T16:27:40.433" Score="2" ViewCount="0" Body="&lt;p&gt;I wanted to post this as a comment to the answer by Justin L. but it is my first posting here so I couldn't. Archimedes was so proud of his proof that the volume of a sphere is two-thirds of the volume of a cylinder of the same height and diameter that he asked that a sculpture illustrating this would be placed at his tomb. The tomb was later found by the Roman orator Cicero who describes it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I managed to track down his grave. The Syracusians knew nothing about it, and indeed denied that any such thing existed. But there it was, completely surrounded and hidden by bushes of brambles and thorns. I remembered having heard of some simple lines of verse which had been inscribed on his tomb, referring to a sphere and cylinder modelled in stone on top of the grave. And so I took a good look round all the numerous tombs that stand beside the Agrigentine Gate. Finally I noted a little column just visible above the scrub: it was surmounted by a sphere and a cylinder.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The American painter Benjamin West imagined the scene in his 1797 painting &quot;Cicero Discovering the Tomb of Archimedes&quot;. &lt;a href=&quot;http://tinyurl.com/363fx3o&quot; rel=&quot;nofollow&quot;&gt;http://tinyurl.com/363fx3o&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2010-09-03T16:27:40.433" />
  <row Id="3965" PostTypeId="2" ParentId="3927" CreationDate="2010-09-03T16:42:11.653" Score="2" ViewCount="0" Body="&lt;p&gt;A professor of mine, Francisco Raggi, once said in class:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;There are two ways to study a ring. You can stare at it, and stare at it, and stare a it, until you can spot interesting features of the ring, and then you can start describing those interesting features. Or you can study its images and the way it acts on other structures, by considering homomorphic images and modules over the ring. The latter is usually easier and more fruitful.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you are studying a particular structure, you can study it in isolation: you can consider a vector space, and its subsets and subspaces, and say a lot of interesting things about that vector space; but in the end, it doesn't get you all that far. The same is true when you study a particular ring (and you can learn &lt;em&gt;a lot&lt;/em&gt; just by staring at that particular ring; think of classical Number Theory as the result of staring intently at the integers looking for interesting features), or groups, or topological spaces, etc. But it turns out that studying the &lt;em&gt;functions&lt;/em&gt; between similar objects that preserve the structure (linear transformations for vector spaces, ring and group homomorphisms, continuous maps, etc) leads to a much richer palette, with a lot more information and a lot more &lt;em&gt;structure&lt;/em&gt; to work with. In some cases, like vector spaces, the new objects have themselves the same structure that you were studying (the set of all linear transformations $T\colon\mathbf{V}\to\mathbf{W}$ can be given a natural structure of a vector space, so everything you know about vector spaces applies) or some other structure (if $A$ and $B$ are abelian groups, then the set of all group homomorphism $A\to B$ can be given a ring structure by pointwise addition and composition, so you can study &lt;em&gt;it&lt;/em&gt; as you would study a ring) which in turn gives you a wealth of information about the structures themselves. And of course, things like the Isomorphism Theorems can give you a lot of information about the structure you are looking at by looking at its images, which will often be &quot;simpler&quot; or &quot;smaller&quot; than the one you were originally concerned with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in the case were we have found a lot of good information by simply staring (as in classical number theory), just look at how big a leap forward was achieved with the introduction of congruences (homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$), or with algebraic number theory (mapping to larger rings and then trying to import the information back into $\mathbb{Z}$).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-03T16:42:11.653" CommunityOwnedDate="2010-09-04T00:42:24.980" />
  <row Id="3966" PostTypeId="1" CreationDate="2010-09-03T16:57:22.823" Score="1" ViewCount="199" Body="&lt;p&gt;Ok so I'm a programmer, I'm not a mathematician, I've got a minor in math but I didn't even do particularly well at it so please bear with some possibly really stupid thoughts.  Just please try to explain to me why what I'm asking is stupid so that I don't keep making a fool of myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently found myself thinking along the following lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the stand that it's solipsism to talk about anything that cannot be put to use in  predicting an outcome then cannot you apply the same axiom to basic mathematical operations?  For example does addition really have any meaning outside of the context of placing two objects in a box and predicting how many are inside?  Yeah, I realize that's a dangerous question to ask a board full of mathematicians but I imagine you can ask this question about any axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that in reality if you put one thing and then another into a box it does not mean that there will be two things in the box.  There will likely be two things of course but if the objects are point particles like electrons there's a chance there will be one or three, or a million electrons there.  Heck, the same is even possible with apples, just incredibly unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the mathematical wherewithal to think through what this would mean but intuitively it would seem that this might not have many implications for addition, but might for subtraction (unless you could somehow have negative amounts of particles which I won't rule out), and certainly for things like integration and derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that someone's done work along these lines before?  Has anyone actually created a system around it?  Was it useful?  Are there any accessible books or articles about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just interested&lt;/p&gt;&#xA;" OwnerUserId="1703" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-04T05:01:28.840" LastActivityDate="2010-10-01T16:24:49.200" Title="Is there a Mathematical system which is stochastic at its core?" Tags="&lt;probability-theory&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="3967" PostTypeId="2" ParentId="164" CreationDate="2010-09-03T16:59:07.623" Score="3" ViewCount="0" Body="&lt;p&gt;To give you a larger perspective, the volume of an $n-1$-sphere in $\mathbb{R}^n$ is given by $C_n r^n$, where $r$ is the radius and $C_n=\frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}$. For even $n$, this reduces to $C_n =\frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!}$, and for odd $n$, it becomes $C_n =2^{(n+1)/2}\frac{\pi^{(n-1)/2}}{n!!}$, where $n!!$ denotes the double factorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus in the case $n=3$, the constant is $2^2 \frac{\pi^1}{3!!} = \frac{4 \pi}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this is to show you that the general formula also involves factors in the denominator, and that the formula for $n=3$ is not 'random' but rather fits into a general pattern.&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2010-09-03T16:59:07.623" />
  <row Id="3968" PostTypeId="1" AcceptedAnswerId="3974" CreationDate="2010-09-03T17:06:10.590" Score="3" ViewCount="123" Body="&lt;p&gt;I hope this question is suitable for the site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently had to work with Mathematica, and the experience was, to put it kindly, unpleasing. I do not have much experience with similar programs, but I remember not liking much Matlab or Maple either. The result is that I am a mathematician who likes programming, but I never managed to learn how to work with a computer algebra system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a CAS which can be programmed using a standard language? I guess the best thing would be just an enormous library of mathematical algorithms implemented for C or Python or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know SAGE is based on Python, but as far as I understand (which is not much) it just collects preexisting open source software, so (I assume) one has to learn how to use a new tool for every different problem.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-09-04T14:58:46.557" Title="CAS with a standard language" Tags="&lt;computer-algebra-systems&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="3969" PostTypeId="2" ParentId="3927" CreationDate="2010-09-03T17:29:03.960" Score="0" ViewCount="0" Body="&lt;p&gt;Here are some striking examples from looking at the Lebesgue space $L^1(\mathbb{R})$ under Fourier transform. $L^1(\mathbb{R})$ is an &lt;em&gt;algebra&lt;/em&gt; under convolution, that is it is a vector space with a multiplication (here multiplication is the convolution). The Fourier transform $\mathcal{F}: L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an &lt;em&gt;algebra homomorphism&lt;/em&gt; from $L^1$ into a sub-algebra $\mathcal(L^1(\mathbb{R}))$ of the space $C_0$ (continuous functions that vanish at $\pm\infty$). &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In particular &#xA;$$\mathcal{F}(f*g) = \mathcal{F}(f)\cdot\mathcal{F}(g)$$ &#xA;and instead of studying convolution equations in $L^1$ we may study multiplicative equations in the Fourier image. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An other example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_tauberian_theorem&quot; rel=&quot;nofollow&quot;&gt;The Wiener Tauberian Theorem&lt;/a&gt;, that states that the translates of $f$ span a dense subspace of $L^1$ if and only if $\mathcal{F}(f)$ is non-zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-03T17:29:03.960" CommunityOwnedDate="2010-09-04T00:42:24.980" />
  <row Id="3970" PostTypeId="2" ParentId="3955" CreationDate="2010-09-03T17:34:47.683" Score="5" ViewCount="0" Body="&lt;p&gt;As you mention “series” in the question I was wondering if something as simple as the term “Arithmetic Progression” is what you're after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first arithmetic progression being $$a_n=n$$ for n=0,1,...,100 and your second being&#xA;$$b_n=1000-10n=1000-10a_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your chosen name, I was wondering if your intended application has anything to do with optics? (I ask as my wife is an optician.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not really part of the answer but on a lighter note, here's a wonderful quote involving optics from the book “Littlewood's Miscellany”:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From time to time (since 1910) there were moves to get rid of the revolting optics and astronomy set in the Mathematical Tripos. It was discovered that over a period of years no wrangler attempted a question in either subject. An equivalent form of this proposition is that every attempt to do a question in optics or astronomy resulted in a failure to get a 1st. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-03T17:34:47.683" />
  <row Id="3971" PostTypeId="2" ParentId="3917" CreationDate="2010-09-03T17:41:47.910" Score="4" ViewCount="0" Body="&lt;p&gt;I'm assuming you want to know more about group theory, so here is an answer to point you to some of the interesting things that you can learn if you are interested in computing with finite groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) If you have the conjugacy classes and there are not too many, then writing normal subgroups as unions of conjugacy classes is fast and easy.  This is often the case when you only have access to a group through its &lt;a href=&quot;http://en.wikipedia.org/wiki/Character_table&quot; rel=&quot;nofollow&quot;&gt;character table&lt;/a&gt;, such as when the Suzuki and the sporadic simple groups were being discovered, and we wanted to understand some of their (non-simple) subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a typical finite group given concretely as a permutation group, you use a special type of induction working along what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Chief_series&quot; rel=&quot;nofollow&quot;&gt;chief series&lt;/a&gt;, where you find a maximal chain of normal subgroups.  It turns out there are some fairly easy ways to find these: for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvable_group&quot; rel=&quot;nofollow&quot;&gt;solvable group&lt;/a&gt;, or any group G with an abelian quotient group, you can fairly easily and concretely find the &lt;a href=&quot;http://en.wikipedia.org/wiki/Derived_subgroup&quot; rel=&quot;nofollow&quot;&gt;derived subgroup&lt;/a&gt;, [G,G].  The quotient group is an abelian group, so every subgroup between the whole group and the derived subgroup is normal.  You then take the derived subgroup of the derived subgroup, but now you only take subgroups that are normalized by G/[G,G].  The action of G/[G,G] is relatively plain and easy to understand, so these &quot;invariant&quot; subgroups are pretty easy to find.  To finish the inductive step, you need to find the normal subgroups of G/[[G,G],[G,G]] that don't contain [G,G], and this can be done by a slightly easier version of the general subgroup lattice algorithm using what is called the first &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_cohomology&quot; rel=&quot;nofollow&quot;&gt;cohomology group&lt;/a&gt;, or &quot;derivations&quot; (basically derivatives for groups, and related in a way to the derived subgroup).  When you reach the point where the group is its own derived subgroup, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_group&quot; rel=&quot;nofollow&quot;&gt;perfect group&lt;/a&gt;, then a second algorithm begins, that identifies the simple groups involved in the top of the group, and then constructs the solvable group below them.  Sometimes you even have to repeat those last two steps, but not for groups of order less than 60&lt;sup&gt;24&lt;/sup&gt; ≈ 4.7e42 or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Often in mathematics, you want to use &quot;induction&quot;.  A normal subgroup is one of the two main ways to do induction in group theory.  Usually it is not necessary to find all normal subgroups, but rather a single (nice) chief series will do.  Some groups only have a single chief series (a fair number of dihedral groups are like this), and so finding the chief series and finding all normal subgroups is the same question.  You might try the symmetric group of degree 4 and order 24 as an example.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-03T17:41:47.910" />
  <row Id="3972" PostTypeId="2" ParentId="3897" CreationDate="2010-09-03T18:32:52.033" Score="4" ViewCount="0" Body="&lt;p&gt;As a double check, let's perform the integral in a completely different manner (so that my mistakes are unlikely to overlap yours!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is handy to use a condensed notation.  I will write $[i,j,k;a]$ for the value of $x^{2 i}y^{2 j}z^{2 k} \rho^{2 a}$ integrated over the unit sphere, where $i,j,k$ are integral and $a$ is real.  These relations are easy to establish:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$[i,j,k;a]$ is invariant under permutations of $(i,j,k)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$6 [1,1,1;a] = [0,0,0;a+3] - 3[3,0,0;a] - 18[2,1,0;a]$ is a consequence of expanding $\rho^6 = (x^2+y^2+z^2)^3$ as a multinomial, using the first property to collect equal terms, and isolating $[1,1,1;a]$ on the lhs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$[i,j,k;a+1] = [i+1,j,k;a] + [i,j+1,k;a] + [i,j,k+1;a]$ follows from $\rho^2 = x^2 + y^2 + z^2$.  Use this to compute $[2,1,0;a]$ in terms of $[3,0,0;a]$ and $[2,0,0;a+1]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$[n,0,0;a] = \frac {4 \pi} {(2 n + 1) (2 n + 3 + 2 a)}$ can be obtained via integration by parts in cylindrical coordinates or by directly performing the integration of $ z^{2 n}  \rho ^ {2 a}$ over the sphere, which is very easy to do (because the angular part only involves $\phi$ and the integrand is exact).  (The numerator is the area of the unit sphere and the factors in the denominator come from integrating a $2 n$ power (in the angular part of $z$, which separates from the $\rho$ integral) and a $2 n + 2 a + 2$ power due to the $\rho^{2 n} \rho^{2 a} \rho^2 d \rho$ term in the integrand.)  Even when $a \lt 0$, this is justified whenever $2 n + 3 + 2 a \gt 0$, because the integral still converges at the origin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From these we algebraically obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eqalign{&#xD;&#xA;[1,1,1;a] = &amp;amp;\frac{1}{6} \left( [0,0,0;a+3] - 3 [3,0,0;a] - 18 [2,1,0;a] \right) \cr&#xD;&#xA;&#xD;&#xA;= &amp;amp;\frac{4 \pi} {6} \left( \frac{1}{2 a+9} - \frac{3}{7 (2 a+9)} - \frac{18}{35(2 a+9)} \right) \cr&#xD;&#xA;&#xD;&#xA;= &amp;amp;\frac{4 \pi}{105 (2 a+9)}.&#xD;&#xA;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $2 a = -17/2$ gives the textbook answer $\frac{8 \pi}{105}$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-12-24T19:33:58.383" LastActivityDate="2010-12-24T19:33:58.383" CommentCount="6" />
  <row Id="3973" PostTypeId="1" AcceptedAnswerId="4294" CreationDate="2010-09-03T19:15:12.087" Score="1" ViewCount="307" Body="&lt;p&gt;Hi guys, I was trying to obtain this result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;integrate e^|x| dx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+e%5e|x|+dx&quot; rel=&quot;nofollow&quot;&gt;The result&lt;/a&gt; is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int e^{|x|} dx = \frac12 e^{-x} \left( (e^x-1)^2 \operatorname{sgn}(x) -2e^x +e^{2x} - 1 \right) \color{gray}{\,+ \text{constant}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is fine but I would like to see the 2 part solution for $x&gt;0$ and $x&amp;lt;0$, I know you can see that because just after before the final output I can see it but then it switches to the one line solution above.&#xA;Is there a way to see the other solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not ok to ask about a tool used in mathematics I'm sorry. You can close it.&lt;/p&gt;&#xA;" OwnerUserId="1704" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-04T15:25:30.147" LastActivityDate="2010-09-08T15:42:47.733" Title="Wolfram Alpha: How to show functions with abs() and sgn() as a piecewise function?" Tags="&lt;math-software&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3974" PostTypeId="2" ParentId="3968" CreationDate="2010-09-03T19:24:44.093" Score="1" ViewCount="0" Body="&lt;p&gt;If you're a big fan of C numerous libraries exist for doing certain tasks. MPIR/GMP are two libraries for doing arbitrary precision arithmetic. On top of these you'll find more specialised libraries like MPC which adds support for the complex field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flint is a library that sits on top of MPIR, for example, providing an array of number theory functions. PARI/GP I believe provides a similar foundation but via the GMP library.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure if you googled &quot;library for linear algebra&quot; or &quot;library for x&quot; you would find it. However, a CAS unites all of this functionality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we come to sage. From what I understand of the sage project, it is not so much a collection of open source libraries but an interface uniting them. For example, from the tutorial:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sage: a = 5   # a is an integer&#xA;sage: type(a)&#xA;&amp;lt;type 'sage.rings.integer.Integer'&amp;gt;&#xA;sage: a = 5/3  # now a is a rational number&#xA;sage: type(a)&#xA;&amp;lt;type 'sage.rings.rational.Rational'&amp;gt;&#xA;sage: a = 'hello'  # now a is a string&#xA;sage: type(a)&#xA;&amp;lt;type 'str'&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is a sage session. As you're probably aware &lt;code&gt;sage.rings.integer.Integer&lt;/code&gt; is not a built-in python type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend giving sage a go. I think it is the closest thing out there to what you're after. I personally don't find Mathematica too bad.&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-09-03T19:24:44.093" />
  <row Id="3976" PostTypeId="1" AcceptedAnswerId="3986" CreationDate="2010-09-03T19:27:16.507" Score="5" ViewCount="231" Body="&lt;p&gt;I have been thinking a little bit about the binary decimal expansion of reciprocal prime numbers; and I have a few questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://brad.chalfan.net/binq/qtob.html&quot; rel=&quot;nofollow&quot;&gt;this neat table&lt;/a&gt; which lists the binary expansion of many fractions, and I was trying to find some patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my questions, for brevity I say a natural number has a period N if the binary decimal expansion of it's reciprocal has period N: (for example, 1/7 = .001001... has period 3)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given an arbitrary natural number N, does there exist a prime number of minimum period N?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(By minimum period I mean to exclude the case that one prime has a period which is a multiple of the period of another prime.  For example, 1/3 = .0101... has period 2, and 1/5 = .00110011... has period 4; so while 1/3 has period 4, what I call it's &quot;minimum period&quot; is 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&#xA; Can two prime numbers have the same minimum period? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful result which I believe is well known, is that a natural number has a period N if and only if it is a factor of 2&lt;sup&gt;N&lt;/sup&gt; - 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a good reference that describes some theory behind the relationship between the period of the reciprocal of a natural number, and the prime factorization of that number?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-03T19:44:18.943" LastActivityDate="2010-09-04T00:24:40.890" Title="On the binary decimal expansion of the reciprocal prime's" Tags="&lt;prime-numbers&gt;&lt;decimal-expansion&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="3977" PostTypeId="1" CreationDate="2010-09-03T20:10:17.607" Score="8" ViewCount="368" Body="&lt;p&gt;Does it exist a free interactive geometry software, like GeoGebra, which works for 3D geometry? I would be able to draw spheres, great circles, and so on.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-03T20:24:28.697" LastActivityDate="2011-01-04T10:53:55.893" CommunityOwnedDate="2010-09-03T20:10:17.607" Title="3D software like GeoGebra" Tags="&lt;geometry&gt;&lt;math-software&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="2" />
  <row Id="3978" PostTypeId="1" AcceptedAnswerId="4040" CreationDate="2010-09-03T20:15:22.570" Score="7" ViewCount="657" Body="&lt;p&gt;I happened to receive this from my friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a,b \in \mathbb{N}$, such that $a^{n}+n$ divides $b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-04T18:21:39.307" LastActivityDate="2010-10-27T15:01:13.520" Title="$(a^{n}+n ) \mid (b^{n}+n) \Longrightarrow a=b$" Tags="&lt;number-theory&gt;" AnswerCount="5" CommentCount="13" FavoriteCount="3" />
  <row Id="3979" PostTypeId="2" ParentId="3977" CreationDate="2010-09-03T20:16:47.750" Score="1" ViewCount="0" Body="&lt;p&gt;You can try pgftikz &lt;a href=&quot;http://www.texample.net/tikz/examples/&quot; rel=&quot;nofollow&quot;&gt;http://www.texample.net/tikz/examples/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-03T20:24:43.840" LastActivityDate="2010-09-03T20:24:43.840" CommunityOwnedDate="2010-09-03T20:16:47.750" CommentCount="1" />
  <row Id="3982" PostTypeId="2" ParentId="3977" CreationDate="2010-09-03T22:03:31.217" Score="1" ViewCount="0" Body="&lt;p&gt;I think you can, to a limited extent, use &lt;a href=&quot;http://sketchup.google.com/&quot; rel=&quot;nofollow&quot;&gt;Google Sketchup&lt;/a&gt;, but that's not really the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cabri.com/cabri-3d.html&quot; rel=&quot;nofollow&quot;&gt;Cabri3D&lt;/a&gt; is probably the most like what you want, but it's commercial.  &lt;a href=&quot;http://www.autograph-math.com/&quot; rel=&quot;nofollow&quot;&gt;Autograph&lt;/a&gt;, also commercial, may do some of that kind of stuff, but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-04T01:28:37.913" LastActivityDate="2010-09-04T01:28:37.913" CommunityOwnedDate="2010-09-03T22:03:31.217" CommentCount="2" />
  <row Id="3983" PostTypeId="2" ParentId="3294" CreationDate="2010-09-03T22:36:41.800" Score="3" ViewCount="0" Body="&lt;p&gt;You probably already know this, but let me explicitly point out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;== exact match ==&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not possible to construct a single cubic Bezier curve that exactly matches your new curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single cubic Bezier curve has a constant &quot;curvature&quot; A along its length.&#xA;When you shrink it, the new curve has &quot;sharper&quot; curvature B (but still constant along the length of the new curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not possible to construct a single cubic Bezier curve -- since, like all single cubic curves, it has constant curvature C along its length -- such that C=A along the part of the length, and C=B along the rest of the length, where A&amp;lt;&gt;B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm mis-using the term &quot;curvature&quot; here for something that is more precisely the 3rd derivative of a curve, sometimes called the &quot;jerk&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;== approximation ==&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the simplest approximation is:&#xA;Use the initial starting point and first control knot -- p0 and p1,&#xA;and the final control knot and final ending point -- r'2 and r'3.&#xA;Use those 4 points as the start, control knots, and endpoint of a Bezier curve that approximates the one you want: a0, a1, a2, a3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approximation exactly hits the endpoints of your desired curve, and has the same initial and final slope at those endpoints, but it slightly diverges from your desired curve in the middle.&#xA;In particular, it probably doesn't go exactly through the cutpoint K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other possible approximations you could make.&#xA;The &lt;a href=&quot;http://www.tinaja.com/cubic01.asp&quot; rel=&quot;nofollow&quot;&gt;&quot;Don Lancaster's Guru's Lair Cubic Spline Library&quot;&lt;/a&gt;&#xA;may have the details I'm leaving out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is possible to nudge a1 along the line p0-p1, or to nudge a2 along the line r'2-r'3 -- or both -- such that the approximation curve not only starts and ends at the same points and slopes, but also passes through the cutpoint K.&lt;/li&gt;&#xA;&lt;li&gt;pick any 4 points along your new curve (perhaps the endpoints, the cutpoint K, and some other point) and generate a cubic Bezier curve that goes exactly through all 4 points.&lt;/li&gt;&#xA;&lt;li&gt;pick many points along your new curve -- bunched together in places where curve matching is important, perhaps near the endpoints and point K, and spaced further apart where curve matching is not so important -- and generate a cubic Bezier curve that is a least-squares best fit to those points.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-03T22:36:41.800" CommentCount="3" />
  <row Id="3984" PostTypeId="2" ParentId="297" CreationDate="2010-09-03T22:55:10.810" Score="4" ViewCount="0" Body="&lt;p&gt;Check out the&#xA;&lt;a href=&quot;http://stackoverflow.com/questions/tagged/pi&quot;&gt;questions tagged 'pi' at Stack Overflow&lt;/a&gt;.&#xA;Many of them discuss simple numerical methods for calculating any number of digits of pi.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-03T22:55:10.810" />
  <row Id="3985" PostTypeId="2" ParentId="297" CreationDate="2010-09-03T23:24:58.723" Score="1" ViewCount="0" Body="&lt;p&gt;Check out the wiki page on approximations to pi, &lt;a href=&quot;http://en.wikipedia.org/wiki/Numerical_approximations_of_%CF%80#20th_century&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Numerical_approximations_of_%CF%80#20th_century&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really love Ramanujan's method, which I recall converges extremely rapidly.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-03T23:24:58.723" />
  <row Id="3986" PostTypeId="2" ParentId="3976" CreationDate="2010-09-03T23:45:43.247" Score="2" ViewCount="0" Body="&lt;p&gt;(In the below post there are several links with the apostrophes omitted; fill them in if the links don't work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenomenon you are studying is a phenomenon in &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;.  If a prime $p$ has period $n$, this means that there is some numerator $N$ such that $\frac{N}{2^n - 1} = \frac{1}{p}$.  This is equivalent to $Np = 10^n - 1$, or $p | 2^n - 1$, or $2^n \equiv 1 \bmod p$.  The smallest $n$ for which this is true is called the &lt;strong&gt;order&lt;/strong&gt; of $2 \bmod p$, sometimes denoted $\text{ord}_p(2)$ (although this is confusingly also used to denote the greatest power of $p$ which divides $2$...).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fermats_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's little theorem&lt;/a&gt; guarantees that $\text{ord}_p(2)$ always divides $p-1$; you have already observed this yourself.  However, predicting the exact order is very difficult to do in general.  For example, knowing that the order is actually equal to $p-1$ is equivalent to knowing that $2$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive root&lt;/a&gt;, and it is not currently even known whether this is true &lt;a href=&quot;http://en.wikipedia.org/wiki/Artins_conjecture_on_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;infinitely often&lt;/a&gt;.  In any case, you should be able to find basic information about order in any good textbook on elementary number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that background out of the way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to question 1 is &lt;strong&gt;no&lt;/strong&gt;.  The only exception is $n = 6$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Zsigmondys_theorem&quot; rel=&quot;nofollow&quot;&gt;Zsigmondy's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to question 2 is &lt;strong&gt;yes&lt;/strong&gt;.  If $p$ and $n$ are relatively prime, then $p$ has period $n$ if and only if $p$ divides $\Phi_n(2)$, where $\Phi_n(x)$ is the $n^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial&quot; rel=&quot;nofollow&quot;&gt;cyclotomic polynomial&lt;/a&gt;.  (This is more or less a restatement of the condition that $p | 2^n - 1$ but $p$ doesn't divide $2^k - 1$ for $k &amp;lt; n$.)  So it suffices to show that some number of this form has more than one prime factor relatively prime to $n$.  There are two cases here which are particularly classical:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$n$ is a prime $q$.  In this case $\Phi_q(2) = 2^q - 1$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mersenne_prime&quot; rel=&quot;nofollow&quot;&gt;Mersenne number&lt;/a&gt;, and $2^{11} - 1 = 23 \cdot 89$ is the smallest composite Mersenne number, hence $23$ and $89$ both have period $11$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$n = 2^k$ for some $k$.  In this case $\Phi_{2^k}(2) = 2^{2^{k-1}} + 1$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_number#Basic_properties&quot; rel=&quot;nofollow&quot;&gt;Fermat number&lt;/a&gt;, and $\Phi_{64}(2) = 2^{32} + 1 = 641 \cdot 6700417$ is the smallest composite Fermat number, hence $641$ and $6700417$ both have period $64$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answer to question 3 is the following.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; If $n, m$ are relatively prime odd numbers, then $\text{ord}_{mn}(2) = \text{lcm}(\text{ord}_n(2), \text{ord}_m(2))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  $\text{ord}_{mn}(2)$ is the order of the element $2$ in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n&quot; rel=&quot;nofollow&quot;&gt;multiplicative group&lt;/a&gt; of $\mathbb{Z}/mn\mathbb{Z}$, which we will denote $U(mn)$.  By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese remainder theorem&lt;/a&gt;, $U(mn)$ is isomorphic to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_product_of_groups&quot; rel=&quot;nofollow&quot;&gt;direct product&lt;/a&gt; $U(m) \times U(n)$, so the order of $2$ in $U(mn)$ must be the $\text{lcm}$ of the orders of $2$ in $U(m)$ and $U(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that to compute $\text{ord}_m(2)$ for arbitrary $m$ it suffices to compute it for the odd prime power factors of $m$ and then to take the $\text{lcm}$ of the resulting numbers.  (Note that if $m$ is divisible by a power of $2$ this only contributes a leading string of zeroes to the binary expansion of $\frac{1}{m}$ and hence does not affect the computation of the period.)  Again, you can find a discussion of the Chinese remainder theorem in any good textbook on elementary number theory.  (I am particularly bad at recommending textbooks on elementary number theory because I learned mine through a summer program, not a textbook...)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-04T00:24:40.890" LastActivityDate="2010-09-04T00:24:40.890" CommentCount="1" />
  <row Id="3987" PostTypeId="1" AcceptedAnswerId="4080" CreationDate="2010-09-03T23:51:48.490" Score="3" ViewCount="188" Body="&lt;p&gt;I was going to ask this on Stack Overflow, but finally decided this was more math than programming. I may still turn out to be wrong about that, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number represented in binary, it's fairly easy to derive a decimal representation using integer division and remainders to extract digits in reverse order. You can then simply count how many digits you extracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I wondered about calculating an &lt;em&gt;exact&lt;/em&gt; number of decimal digits more efficiently by avoiding calculating the actual digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first sight this is easy. First, count the binary digits needed (ie determine the position of the highest set bit). Many processors have an operation to do this directly, and there are bit-fiddling algorithms to do it in O(log n) time where n varies with the width of a machine register. In an arbitrary position integer, you normally know how many machine words you have, so you can usually jump directly to the most significant word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the size in binary digits, you can scale by $\log 2 \over \log 10$ to get the number of decimal digits. Sort of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this scale factor is (I think) an irrational number. If you have a maximum number of digits you need to worry about, you can use a rational approximation and you only need to worry about getting the rounding right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, therefore - is there a way to determine this number of decimal digits efficiently and precisely for numbers of any size? Is there (for example) a completely different approach to the calculation that I haven't thought of?&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-09-07T01:26:38.787" LastActivityDate="2010-09-07T01:26:38.787" Title="How to calculate the number of decimal digits for a binary number?" Tags="&lt;algorithms&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3989" PostTypeId="2" ParentId="297" CreationDate="2010-09-04T01:48:27.217" Score="1" ViewCount="0" Body="&lt;p&gt;Since the AGM hasn't been mentioned yet (apart from my plug to read the Borwein and Borwein book), here's how to use the AGM to compute $\pi$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a←(1+1/√(2))/2&#xA;c←1/(8a)&#xA;s←a^2&#xA;f←1&#xA;repeat&#xA;a←(a+√((a-c)(a+c)))/2&#xA;c←(c^2)/(4a)&#xA;f←2f&#xA;s←s-fc^2&#xA;until |c| is small enough&#xA;(2a^2)/(2s-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which exploits the Legendre relation.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-04T01:48:27.217" />
  <row Id="3990" PostTypeId="1" CreationDate="2010-09-04T02:44:33.170" Score="3" ViewCount="102" Body="&lt;p&gt;Assume you have a triangle in the plane defined by its three vertices at $(x_0,y_0)$, $(x_1,y_1)$, and $(x_2,y_2)$. Is there a general expression for the moments of the triangle, where the moments are defined as averages by the following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x^n y^m \rangle = \frac{1}{A}\int_\triangle x^n y^m \, dA $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating manually to compute the first few moments, one can obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x \rangle = \frac{1}{3} \left(x_0+x_1+x_2\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x^2 \rangle = \frac{1}{6} \left(x_0^2+x_1^2+x_2^2+x_0 x_1+x_0 x_2+x_1 x_2\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x y \rangle = \frac{1}{12}\left(2 x_0 y_0+x_1 y_0+x_2 y_0+x_0 y_1+2 x_1 y_1+x_2 y_1+x_0 y_2+x_1 y_2+2 x_2 y_2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to obtain a closed form expression for general $n$ and $m$ using combinations of the coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the same be done for line segments defined by their endpoints $(x_0,y_0)$ and $(x_1,y_1)$?&#xA;Again the first few moments in this case are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x \rangle = \frac{1}{2} \left(x_0+x_1\right) $$&#xA;$$ \langle x^2 \rangle = \frac{1}{3} \left(x_0^2+x_0 x_1+x_1^2\right) $$&#xA;$$ \langle x^3 \rangle = \frac{1}{4} \left(x_0^3+x_0^2 x_1+x_0 x_1^2+x_1^3\right) $$&#xA;$$ \langle x y \rangle = \frac{1}{6} \left(2 x_0 y_0+x_1 y_0+x_0 y_1+2 x_1 y_1\right) $$&#xA;$$ \langle x^2 y \rangle  = \frac{1}{12} \left( 3 x_0^2 y_0+2 x_0 x_1 y_0+x_1^2 y_0+x_0^2 y_1+2 x_0 x_1 y_1+3 x_1^2 y_1\right)$$&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="1393" LastEditorDisplayName="" LastEditDate="2010-09-04T02:55:54.703" LastActivityDate="2010-09-04T11:02:37.463" Title="Computing the moments of a triangle" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;euclidean-geometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3991" PostTypeId="1" AcceptedAnswerId="3993" CreationDate="2010-09-04T03:02:22.437" Score="0" ViewCount="109" Body="&lt;p&gt;I'm working through a discrete maths text book and was stumped as to how to prove the closed form solution of the &lt;a href=&quot;http://mathworld.wolfram.com/HofstadterG-Sequence.html&quot; rel=&quot;nofollow&quot;&gt;Hofstadter G-Sequence&lt;/a&gt;&lt;br&gt;&#xA;$a(0) = 0$ and $a(n) = n - a(a(n-1)), n \geq 1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed form is $a(n) = \lfloor(n+1)\mu\rfloor$  where $\mu$ is the inverse of the golden ratio.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text book says to prove the following two statements on the way to proving this.&lt;br&gt;&#xA;Show that $n\mu -\lfloor n\mu \rfloor + n\mu^2 - \lfloor n\mu^2 \rfloor = 1$ for all n &gt; 0&lt;br&gt;&#xA;Show that $\lfloor(1+\mu)(1-\alpha)\rfloor + \lfloor\alpha+\mu\rfloor = 1$ for all $\alpha \in R, 0 \leq \alpha &amp;lt; 1\space and\space \alpha \ne 1-\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm relatively certain I was able to confirm those those two statements but I'm not sure where to apply them when I start on the main proof. I think it's just my weakness with algebra, particularly how to deal with the floor that's holding me up.&lt;/p&gt;&#xA;" OwnerUserId="1455" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-06T15:36:12.667" LastActivityDate="2010-09-06T15:36:12.667" Title="Proof of closed form Hofstadter G-Sequence" Tags="&lt;algebra-precalculus&gt;&lt;sequences-and-series&gt;&lt;discrete-mathematics&gt;&lt;recursive&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3992" PostTypeId="2" ParentId="3990" CreationDate="2010-09-04T03:07:59.690" Score="1" ViewCount="0" Body="&lt;p&gt;It should be possible using &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hnZe9.png&quot; alt=&quot;Green's Theorem Image from wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer my earlier answer here for more details: &lt;a href=&quot;http://math.stackexchange.com/questions/3177/why-doesnt-a-simple-mean-give-the-position-of-a-centroid-in-a-polygon/3182#3182&quot;&gt;http://math.stackexchange.com/questions/3177/why-doesnt-a-simple-mean-give-the-position-of-a-centroid-in-a-polygon/3182#3182&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And especially the link given there is more relevant here: &lt;a href=&quot;http://www.worldserver.com/turk/computergraphics/Moments.pdf&quot; rel=&quot;nofollow&quot;&gt;Computing the Moments of a 2D polygon using Green's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-04T03:17:10.493" LastActivityDate="2010-09-04T03:17:10.493" />
  <row Id="3993" PostTypeId="2" ParentId="3991" CreationDate="2010-09-04T04:00:41.780" Score="1" ViewCount="0" Body="&lt;p&gt;Our goal is to prove $$\lfloor \mu  (n+1)\rfloor  = n-\lfloor \mu  \lfloor \mu  n+1\rfloor \rfloor. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can prove  $\lfloor \mu  \lfloor \mu  n+1\rfloor \rfloor  = \lfloor (1-\mu ) (n+1)\rfloor $ then you should be able to prove the above without too much effort.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-04T04:00:41.780" CommentCount="1" />
  <row Id="3994" PostTypeId="2" ParentId="3953" CreationDate="2010-09-04T04:02:30.310" Score="2" ViewCount="0" Body="&lt;p&gt;It is easier to build the isomorphism in the other direction, and indeed we can see that the map is forced upon us. There is a unique isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, you want a map $h:\mathbb{N}\to R(\omega)$ such that $n\mathrel{E} m\iff h(n)\in h(m)$. This very equivalence tells you that you must define &lt;code&gt;$h(m)=\{ h(n) \mid n\mathrel{E} m\}$&lt;/code&gt;. This function is defined by recursion, and obviously preserves $E$ to $\in$. Thus it is injective. It is surjective because your function $g$ is the inverse. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one knows about the Mostowski collapse, then you can see immediately that $h$ is precisely the Mostowski collapse of $(\mathbb{N},E)$, which is a well-founded extensional relation. This provides another way to see that $h$ is an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-04T04:02:30.310" CommentCount="1" />
  <row Id="3995" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T04:38:29.900" Score="-1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: the proof seems incorrect as b=4, a=2, fail when n=4 (check the comments below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is a counter example when n=1, where a=6 and b=13 (as $ (6+1) | (13+1) $ ). However I shall continue for all n&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(a^n+n) | (b^n+n)$ then there is a natural number k $\geq 1$ such that $b^n+n=k(a^n+n)$. Clearly $a=b \mbox{ iff } k=1$. We shall proceed to prove that k is always 1 for all n&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing from $b^n+n=k(a^n+n)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^n+n \equiv 0 \quad (\mbox{mod } k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we allow n to vary for the same a and b, then the only value for k that will make that equation hold for all n&gt;1 is k=1, which implies that a=b.&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-04T06:47:41.943" LastActivityDate="2010-09-04T06:47:41.943" CommentCount="8" />
  <row Id="3996" PostTypeId="2" ParentId="3966" CreationDate="2010-09-04T05:00:00.780" Score="2" ViewCount="0" Body="&lt;p&gt;Your question is undoubtedly philosophical in nature: it is predicated on the premise &quot;it's solipsism to talk about anything that cannot be put to use in predicting an outcome&quot;.  But &lt;a href=&quot;http://en.wikipedia.org/wiki/Solipsism&quot; rel=&quot;nofollow&quot;&gt;solipsism&lt;/a&gt; is a philosophical term (and more fundamentally, an inherently philosophical -- as opposed to mathematical or scientific -- idea)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The philosophical view you advocate seems to be, roughly, that the truths of arithmetic are (only) empirical in nature.  As with many basic philosophical ideas, this has been propounded before and has gotten a lot of discussion.  Famously, it was the view of arithmetic espoused by &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Stuart_Mill&quot; rel=&quot;nofollow&quot;&gt;John Stuart Mill&lt;/a&gt;.  For a pretty good quick summary of his positions, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://plato.stanford.edu/entries/mill/#GeoAri&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/mill/#GeoAri&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mill's ideas on arithmetic and geometry have generally been poorly received by later philosophers.  Notably, much of the work of &lt;a href=&quot;http://en.wikipedia.org/wiki/Immanuel_Kant&quot; rel=&quot;nofollow&quot;&gt;Immanuel Kant&lt;/a&gt; is a reaction against Mill's ideas, especially his notion of the &quot;synthetic a priori&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far as I know, &lt;strike&gt;essentially no&lt;/strike&gt; very few mathematicians or philosophers of mathematics have accepted Mill's ideas.  Rather, we have followed the path of &lt;a href=&quot;http://en.wikipedia.org/wiki/Frege&quot; rel=&quot;nofollow&quot;&gt;Frege&lt;/a&gt; and viewed logic as being the foundation for mathematics.  (Later work by Russell, Godel and others have showed that there are some pitfalls and limitations of the logicist approach, but it has nevertheless been the one mathematicians have adopted for the last hundred years.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[As Qiaochu says, if you really want to talk mathematics you should remove the overtly philosophical lead to your question.  Instead, you should say more about the mathematical aspects of your question: what mathematical problem are you trying to solve?]&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-04T13:46:05.657" LastActivityDate="2010-09-04T13:46:05.657" CommentCount="2" />
  <row Id="3997" PostTypeId="2" ParentId="3973" CreationDate="2010-09-04T07:52:00.100" Score="1" ViewCount="0" Body="&lt;p&gt;You can use the following to get the two cases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Assuming[t&amp;gt;0, integrate e^|x| dx from x=0 to t]&#xA;Assuming[t&amp;lt;0, integrate e^|x| dx from x=0 to t]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or, if you don't care about choosing the constant of integration such that the two pieces fit together, simply:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Assuming[x&amp;gt;0, integrate e^|x| dx]&#xA;Assuming[x&amp;lt;0, integrate e^|x| dx]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(although then it's so easy that I don't see why you would need Wolfram Alpha).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-04T07:52:00.100" CommentCount="6" />
  <row Id="3998" PostTypeId="2" ParentId="3968" CreationDate="2010-09-04T08:04:50.717" Score="1" ViewCount="0" Body="&lt;p&gt;If you don't want to install all of Sage (which is quite big), you could perhaps try one of its components, &lt;a href=&quot;http://wiki.sympy.org/wiki/Main_Page&quot; rel=&quot;nofollow&quot;&gt;SymPy&lt;/a&gt;, on its own to be begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use C++ maybe &lt;a href=&quot;http://www.ginac.de/&quot; rel=&quot;nofollow&quot;&gt;GiNaC&lt;/a&gt; is useful (I haven't tried it).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-04T08:04:50.717" />
  <row Id="3999" PostTypeId="1" AcceptedAnswerId="4000" CreationDate="2010-09-04T08:58:14.113" Score="4" ViewCount="336" Body="&lt;p&gt;Is a prime number in the decimal system still a prime when converted to a different base?&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2010-09-04T19:54:43.680" Title="Is a prime number still a prime when in a different base?" Tags="&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="4000" PostTypeId="2" ParentId="3999" CreationDate="2010-09-04T09:10:25.080" Score="6" ViewCount="0" Body="&lt;p&gt;Here is the same question which has been asked at Mathforum: Here is the link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathforum.org/library/drmath/view/55880.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/55880.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-04T09:10:25.080" />
  <row Id="4001" PostTypeId="2" ParentId="3990" CreationDate="2010-09-04T09:42:03.227" Score="1" ViewCount="0" Body="&lt;p&gt;If you don't mind a quadruple sum as a closed form (it was the best simplification of the integral I gave in the comments that I could manage, using repeated applications of the binomial theorem), then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle x^p y^q\rangle=\displaystyle 2\sum_{k=0}^p\sum_{\ell=0}^q \frac{(-1)^{k+\ell}\binom{p}{k}\binom{q}{\ell}\nu_{k\ell}}{k+\ell+2} x_0^{p-k} y_0^{q-\ell}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \nu_{k\ell}=\sum_{i=0}^k\sum_{j=0}^\ell \frac{\binom{k}{i}\binom{\ell}{j}}{k-i+\ell-j+1}\left(x_0-x_1\right)^i \left(x_1-x_2\right)^{k-i} \left(y_0-y_1\right)^j \left(y_1-y_2\right)^{\ell-j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an explicit expression for $\langle x^p y^q\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, &lt;em&gt;Mathematica&lt;/em&gt; is unable to evaluate this (as usual with such complicated multiple sums).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the problem of finding the moments of a line segment with endpoints $\left(x_0, y_0\right)$ and $\left(x_1, y_1\right)$, the integral can be derived by noting that a set of parametric equations for the line passing through the two given points is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \begin{align*}x&amp;amp;=x_0+t(x_1-x_0)\\y&amp;amp;=y_0+t(y_1-y_0)\end{align*}\qquad 0\leq t\leq1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the two equations into the expression $x^p y^q$ and integrating over the parameter range gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_0^1\left(x_0+t\left(x_1-x_0\right)\right)^p\left(y_0+t\left(y_1-y_0\right)\right)^q\mathrm{d}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be expanded into a double sum (again by using the binomial theorem):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{j=0}^p\sum_{k=0}^q\frac{\binom{p}{j} \binom{q}{k}}{j+k+1}\left(x_1-x_0\right)^j \left(y_1-y_0\right)^k x_0^{p-j} y_0^{q-k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be &quot;simplified&quot; into the &lt;a href=&quot;http://mathworld.wolfram.com/AppellHypergeometricFunction.html&quot; rel=&quot;nofollow&quot;&gt;Appell hypergeometric function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0^p y_0^q F_1\left(1;-p,-q;2;1-\frac{x_1}{x_0},1-\frac{y_1}{y_0}\right)$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-04T11:02:37.463" LastActivityDate="2010-09-04T11:02:37.463" />
  <row Id="4002" PostTypeId="2" ParentId="2735" CreationDate="2010-09-04T09:58:30.723" Score="0" ViewCount="0" Body="&lt;p&gt;Searching for &quot;Out of core LU&quot; on google might help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, there are two simple choices:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An iterative method that merely depends on repeatedly computing A*v and perhaps A'*v for a given vector 'v'. Such a computation can be done without having the entire matrix A in RAM&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A direct method, for this the &quot;out of core LU&quot; might help.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In general, for both choices above, searching for external memory algorithms will be useful.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-09-04T09:58:30.723" CommentCount="1" />
  <row Id="4003" PostTypeId="1" AcceptedAnswerId="4006" CreationDate="2010-09-04T10:03:14.843" Score="1" ViewCount="217" Body="&lt;p&gt;We know that a group $G$ cannot be written as the set theoretic union of two of its proper subgroups. Also $G$ can be written as the union of 3 of its proper subgroups if and only if $G$ has a homomorphic image, a non-cyclic group of order 4?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this paper &lt;a href=&quot;http://www.jstor.org/stable/2695649&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2695649&lt;/a&gt; by M.Bhargava, it is shown that a group $G$ is the union of its proper normal subgroups if and only if its has a quotient that is isomorphic to $C_{p} \times C_{p}$ for some prime $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to make the condition more stringent on the subgroups. We know that Characteristic subgroups are normal. So can we have a group $G$ such that , $$G = \bigcup\limits_{i} H_{i}$$ where each $H_{i}$'s are Characteristic subgroups of $G$?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-04T13:29:24.860" LastActivityDate="2010-09-04T13:41:29.890" Title="Group as the Union of Subgroups" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="4004" PostTypeId="1" AcceptedAnswerId="4005" CreationDate="2010-09-04T11:47:03.397" Score="3" ViewCount="80" Body="&lt;h2&gt;Taken from : &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20100904034737AAfGkYF&quot; rel=&quot;nofollow&quot;&gt;answers.yahoo.com/question/index?qid=20100904034737AAfGkYF&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A mall has sweep stakes where they give you a coupon for every X dollars you spend (amount is not important). You fill the coupon with your contact details and put it in the box. Every two weeks, they draw 10 coupons, give their owners prizes, and discard the rest of the coupons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, thanks to some very big purchase we made, we have about 300 coupons (going to hurt when filling them). To get the highest probability of winning, should we put all the coupons at once, or split them evenly between the three remaining draws?&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-04T11:53:10.370" LastActivityDate="2010-09-04T13:07:17.310" Title="Probability of winning sweep stakes, put all coupons in one box or split them between boxes?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4005" PostTypeId="2" ParentId="4004" CreationDate="2010-09-04T13:07:17.310" Score="2" ViewCount="0" Body="&lt;p&gt;Let's assume that in each box there are n other tickets besides yours. So if you put them all in the first box and 10 tickets are drawn your chance of winning something is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = 1 - \prod_{k=1}^{10} \left( 1 - \frac{300}{n+301-k} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you split them, 100 each amongst three boxes (where, again, each box has n other tickets), your chance of winning something is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = 1 - \prod_{k=1}^{10} \left( 1 - \frac{100}{n+101-k} \right)^3 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now p&gt;q, but only slightly, for large n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with n=10000, $$p \approx 0.258176 \textrm{ and } q \approx 0.256004$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you're slightly better off if you split them up into three lots of 100. However, this assumes that the number of other tickets placed in the box every two weeks remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the difference between p and q decreases with increasing n.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-04T13:07:17.310" />
  <row Id="4006" PostTypeId="2" ParentId="4003" CreationDate="2010-09-04T13:16:00.683" Score="2" ViewCount="0" Body="&lt;p&gt;One way to ensure this happens is to have every maximal subgroup be characteristic.  To get every maximal subgroup normal, it is a good idea to check p-groups first.  To make sure the maximal subgroups are characteristic, it makes sense to make sure they are simply not isomorphic.  To make sure there are not too many maximal subgroups, it makes sense to take p=2 and choose a rank 2 group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasidihedral_group&quot; rel=&quot;nofollow&quot;&gt;quasi-dihedral groups&lt;/a&gt; have this property.  Their three maximal subgroups are cyclic, dihedral, and quaternion, so each must be fixed by any automorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a specific example is QD16, the Sylow 2-subgroup of GL(2,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another small example is 4×S&lt;sub&gt;3&lt;/sub&gt;.  It has three subgroups of index 2, a cyclic, a dihedral, and a 4 acting on a 3 with kernel 2.  Since these are pairwise non-isomorphic, they are characteristic too.  It also just so happens (not surprisingly, by looking in the quotient 2×2) that every element is contained in one of these maximal subgroups.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-04T13:16:00.683" CommentCount="1" />
  <row Id="4007" PostTypeId="1" AcceptedAnswerId="4013" CreationDate="2010-09-04T13:19:56.060" Score="3" ViewCount="117" Body="&lt;p&gt;One of my math professors gave me the following challenge. It isn't graded, it's just for fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1}, x \epsilon (0, 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?&lt;/p&gt;&#xA;" OwnerUserId="456" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-04T13:26:27.293" LastActivityDate="2010-09-05T19:24:10.970" Title="Finding a function that approaches another function" Tags="&lt;calculus&gt;&lt;math-software&gt;" AnswerCount="4" />
  <row Id="4008" PostTypeId="2" ParentId="4003" CreationDate="2010-09-04T13:31:06.323" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; While searching on the internet i found this paper, &quot;A remark on Hyperabelian groups&quot; by G.Baumslag where there is this remark &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Moreover so is any torsion abelian group of finite rank, since it is the union of finite characteristic subgroups.&lt;/strong&gt;  Would like to get a reference for this.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-04T13:41:29.890" LastActivityDate="2010-09-04T13:41:29.890" CommentCount="1" />
  <row Id="4009" PostTypeId="1" AcceptedAnswerId="4010" CreationDate="2010-09-04T13:32:07.540" Score="1" ViewCount="39" Body="&lt;p&gt;Hi! I'm doing some old exams and found following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Set $S={{1,2,3}}$ is given. Provide an&#xA;  example of binary operation in set S,&#xA;  binary relation in set S and a&#xA;  function $f:S\rightarrow R$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I'm thinking about $+_4$ as operation, but wouldn't it be a relation too? I can take ordered pair from $S$ and for every ordered pair associate an element form $S$ with every pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third part, I'm thinking about $y=x*1.25$ or something similar, but that part isn't so problematic for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did read the Wikipedia articles, but the difference between operations  and relations isn't clear to me.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-09-04T13:40:15.120" Title="Confusion between operation and relation: Clarification needed." Tags="&lt;relations&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4010" PostTypeId="2" ParentId="4009" CreationDate="2010-09-04T13:40:15.120" Score="3" ViewCount="0" Body="&lt;p&gt;A binary operation is a function from $S\times S \to S$ such as addition, multiplication or anything really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A binary relation is just a subset of $S^2$, that is not necessarily a function and it doesn't have to include all the elements of $S$ in one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f\colon S\to R$ is a relation, this time it's a subset of $S\times R$ however it satisfies a certain property, if you take some $s \in S$ then there is only a unique ordered pair with $s$ in it, so if you have $\langle s,r_1\rangle$ as well $\langle s,r_2\rangle$ then you can say that $r_1 = r_2$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-04T13:40:15.120" CommentCount="1" />
  <row Id="4011" PostTypeId="2" ParentId="3633" CreationDate="2010-09-04T13:42:42.280" Score="0" ViewCount="0" Body="&lt;p&gt;([Update]:  Oh sorry; I'm new here. Didn't notice the much more substantial link to the mathoverflow. But perhaps it is interesting for some newbie anyway... [/update]) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the function is a polynomial (or powerseries) with constant term =/= zero this is difficult, but sometimes a meaningful solution can be given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the function is as above, but the constant term is zero, then it is just the question of relatively simple manipulation of the formal powerseries/polynomial to construct a &quot;compositional&quot; (or &quot;iterative&quot;) root. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very good deal in L.Comtet &quot;Advanced Combinatorics&quot; at pages around 130..150 (don't have it around).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the keywords &quot;Bell-matrix&quot;, &quot;Carleman-matrix&quot; are helpful: these matrices transform the problem of funtion composition/iteration to matrix-products/matrix-powers. (The matrix-notation just implies the required formal powerseries-manipulations) . This is well established for functions f(x)  = a*x + b*x^2 + c*x^3 + ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the constant term does not vanish, f(x) = K + a*x + b*x^2 + c*x^3 + ... then things are usually much more difficult. But there is one -I think: again well established- workaround: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite f(x) as some function g(x+x0) - x0 such that now g(x) has no constant term and then apply the above mentioned procedures (solving for iterative squarerroot and the like) to g(x) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: denote the iterative root of a some function f(x) by f°0.5(x) then solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f°0.5(x) = g°0.5(x-x0)+x0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where you first must find x0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;em&gt;very&lt;/em&gt; basic introduction you may try my small essay at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;go.helms-net.de/math/tetdocs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and look for the article &quot;countinuous functional iteration&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gottfried Helms&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-09-04T15:11:09.747" LastActivityDate="2010-09-04T15:11:09.747" />
  <row Id="4012" PostTypeId="2" ParentId="4007" CreationDate="2010-09-04T14:09:57.463" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Maple, if you do&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; map(factor, sum((2*i-1)^3*x^(2*i-1),i=1..n));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result will be 2 terms, one which is shape $(x^2)^{(n+1)}*r(n,x)$ with $r(n,x)$ polynomial in $n$ and rational in $x$, and the second term is (exactly) (b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this is is actually writing&#xA;$$\sum_{i=1}^{n} (2i-1)^3x^{2i-1} = \sum_{i=1}^{\infty} (2i-1)^3x^{2i-1} - \sum_{i=N-1}^{\infty} (2i-1)^3x^{2i-1}$$&#xA;(and the term order gets flipped in Maple's output).  Both the resulting sums are 'easy' to do by hand.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2010-09-04T14:09:57.463" />
  <row Id="4013" PostTypeId="2" ParentId="4007" CreationDate="2010-09-04T14:19:01.903" Score="7" ViewCount="0" Body="&lt;p&gt;Yes looks like it must be b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one $x$ factor, so this eliminates choices c) and d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is odd, i.e $f(-x) = -f(-x)$, so this eliminates a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus b) must be the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the answer by hand you can do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = x + x^3 + x^5 + \cdots = \frac{x}{1-x^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now differentiate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f'(x)  = 1 + 3x^2 + 5x^4 + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f'(x)x = x + 3x^3 + 5x^5 + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f'(x)x)' = 1 + 3^2 x^2 + 5^2 x^4 + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by $x$, differentiate again and then multiply by $x$ to give the answer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-04T16:06:18.810" LastActivityDate="2010-09-04T16:06:18.810" CommentCount="2" />
  <row Id="4014" PostTypeId="2" ParentId="4007" CreationDate="2010-09-04T14:25:50.493" Score="0" ViewCount="0" Body="&lt;p&gt;Here's yet another way to verify that it's (b). Let Mathematica do partial fractions decomposition:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Apart[x(x^2+1)(x^4+22x^2+1)/(x^2-1)^4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives $\frac{1}{2 (x+1)}-\frac{7}{2(x+1)^2}+\frac{6}{(x+1)^3}-\frac{3}{(x+1)^4}+\frac{1}{2 (x-1)}+\frac{7}{2(x-1)^2}+\frac{6}{(x-1)^3}+\frac{3}{(x-1)^4}$, which can be expanded into power series using the geometric series and its derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is ugly, and requires you to know the answer in advance. Moron's answer is much better. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-04T14:25:50.493" />
  <row Id="4015" PostTypeId="2" ParentId="3968" CreationDate="2010-09-04T14:58:46.557" Score="-1" ViewCount="0" Body="&lt;p&gt;You can have a test with WIRIS CAS. &lt;a href=&quot;http://www.wiris.com/demo/en/&quot; rel=&quot;nofollow&quot;&gt;http://www.wiris.com/demo/en/&lt;/a&gt;&#xA;Find a free course to learn at WIRIS Campus.&#xA;The programming languaje is quite natural. Give IT a try.&lt;/p&gt;&#xA;" OwnerUserId="1716" LastActivityDate="2010-09-04T14:58:46.557" CommentCount="1" />
  <row Id="4016" PostTypeId="1" AcceptedAnswerId="4017" CreationDate="2010-09-04T15:26:52.140" Score="1" ViewCount="87" Body="&lt;p&gt;Consider the Banach space $\ell^1(\mathbb N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence $(e_n)_{n\in\mathbb N}$ feels like a kind of basis because every element $a\in\ell^1(\mathbb N)$ can be written as an absolutely convergent infinite linear combination $\sum_{n\in\mathbb N}a(n)e_n$ in a unique way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $e_n$ denotes the vector whose $n$th entry is 1 and all of whose other entries vanish.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for the Banach space $c_0(\mathbb N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above property of the sequence $(e_n)_{n\in\mathbb N}$ appropriate in order to abstractly define a basis of a Banach space? Has this been considered?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-04T15:34:21.130" LastActivityDate="2010-09-04T15:43:50.053" Title="Is there a notion of basis for Banach spaces?" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4017" PostTypeId="2" ParentId="4016" CreationDate="2010-09-04T15:43:50.053" Score="2" ViewCount="0" Body="&lt;p&gt;As Soarer points out: Yes, it is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Schauder_basis&quot; rel=&quot;nofollow&quot;&gt;Schauder basis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-04T15:43:50.053" CommunityOwnedDate="2010-09-04T15:43:50.053" />
  <row Id="4018" PostTypeId="1" AcceptedAnswerId="4019" CreationDate="2010-09-04T15:47:43.173" Score="3" ViewCount="180" Body="&lt;p&gt;Let $X = \mathcal{C}([0,1],\mathbb{R})$, be the ring of all continuous valued functions $f:[0,1] \to \mathbb{R}$. For $x \in [0,1]$, let $M_{x} = \{ f \in M \ | \ f(x)=0\}$. One can show by using compactness of $[0,1]$ that every maximal ideal is of this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extending the Question to Entire functions&lt;/strong&gt;: Let $\mathsf{C}(z)$ be the ring of complex entire functions. For $ \lambda \in \mathsf{C}$ let $M_{\lambda}$ denote the set of all entire functions which have a zero at $\lambda$. Then is $M_{\lambda}$ a maximal ideal in $\mathsf{C}(z)$, and does every ideal happen to be of this form? I don't know to prove this!&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-05T22:39:18.763" Title="Maximal Ideals in the Ring of Complex Entire Functions" Tags="&lt;analysis&gt;&lt;abstract-algebra&gt;&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="4" />
  <row Id="4019" PostTypeId="2" ParentId="4018" CreationDate="2010-09-04T16:03:00.210" Score="12" ViewCount="0" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(z)=1/\Gamma(z)$ where $\Gamma$ is the gamma function.&#xA;Then $f$ vanishes at $0$, $-1,-2,\ldots$ and nowhere else. For integers&#xA;$m$ let $f_m(z)=f(z+m)$. Then $f_m$ vanishes at $-m,-m-1,\ldots$.&#xA;Let $I$ be the ideal of the ring of the entire functions generated by the $f_m$.&#xA;Then $I$ is proper, since otherwise finitely many of the $f_m$&#xA;would generate an ideal containing the constant function $1$. But&#xA;any finite set of $f_m$ have a common zero, so this doesn't hold.&#xA;So $I\subseteq J$ for some maximal ideal $J$ (using the standard Zorn's&#xA;lemma argument). But $J\ne M_\lambda$ for any $\lambda$, as for each $\lambda$&#xA;there is some $m$ with $f_m(\lambda)\ne0$, and $f_m\in J$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-05T06:30:44.417" LastActivityDate="2010-09-05T06:30:44.417" CommentCount="2" />
  <row Id="4022" PostTypeId="1" CreationDate="2010-09-04T17:04:50.630" Score="3" ViewCount="97" Body="&lt;p&gt;I try to solve for the specific function&#xA;$f(x) = \frac{2-2a}{x-1} \int_0^{x-1} f(y) dy + af(x-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks similar to the function used to find the &lt;a href=&quot;http://mathworld.wolfram.com/RenyisParkingConstants.html&quot; rel=&quot;nofollow&quot;&gt;Renyi's parking constant&lt;/a&gt; because it came out from a simple generalization of that problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skill I have gained in my differential class can't even solve&#xA;$f(x) = f'(x-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for anyone to solve it. I just want to know the techniques for solving DE where functions and it's derivatives are evaluated at different points.(What's the terminology for this kind of DE?)&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-09-04T17:04:50.630" Title="How to solve DE that relate values of derivatives at different points?" Tags="&lt;differential-equations&gt;&lt;functional-equations&gt;" CommentCount="7" FavoriteCount="2" />
  <row Id="4023" PostTypeId="1" CreationDate="2010-09-04T17:07:04.957" Score="4" ViewCount="153" Body="&lt;p&gt;Here is a list of some applications of the exponential function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The exponential mapping in Lie theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put this first because my intuition tells me that this must be the most fundamental, or deep, way of thinking about the exponential function.  I have often been misled by my intuition however, and the main reason I feel strongly about this is because of how fundamental I consider Lie theory to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Fourier Series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Roots of Unity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Gaussian Distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Boltzmann Distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly other applications, but it always kind of bothered me that I couldn't use symmetry methods to see how (all of) these applications are related.  Is it possible that there is no way to do this, i.e. that it's just a happy accident that the exponential function has these applications and it is unrelated to any continuous symmetries?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-04T17:12:38.003" LastActivityDate="2011-01-05T16:44:36.070" Title="Is there a (deep) relationship between these various applications of the exponential function?" Tags="&lt;probability-theory&gt;&lt;fourier-series&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4024" PostTypeId="2" ParentId="3941" CreationDate="2010-09-04T17:07:48.370" Score="9" ViewCount="0" Body="&lt;p&gt;Here's a slightly different way of proving that $\sin nx$ is not&#xA;a polynomial in $\sin x$ when $n$ is even. Indeed $\sin nx$&#xA;cannot be written as $f(\sin x)$ for any function $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this all one has to do is to write down two numbers $a$&#xA;and $b$ such that $\sin a=\sin b$ but $\sin na\ne\sin nb$.&#xA;Let $a=\pi/(2n)$ and $b=\pi-a$. Then $\sin a=\sin b$. But&#xA;$na=\pi/2$ and $nb=n\pi-\pi/2$ so that $\sin na=1$ and $\sin nb=-1$&#xA;(recalling that $n$ is even).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-04T17:07:48.370" CommentCount="5" />
  <row Id="4026" PostTypeId="2" ParentId="4023" CreationDate="2010-09-04T18:09:28.090" Score="2" ViewCount="0" Body="&lt;p&gt;The most important traits of the exponential function that make $e^x$ occur in that much applications are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a)&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Euler's identity&lt;/a&gt; $e^{i\pi} + 1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its implications for the definition of complex trigonimetric functions, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{i \cdot x} = \cos x + i \sin x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(like Fourier transformation, unit roots)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b)&lt;/strong&gt; the fact that $\frac{d}{dx}e^x = e^x$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which lets $e^x$ naturally occur in many solutions of &lt;em&gt;differential equations&lt;/em&gt;, which are often the most basic formulation of natural laws (growth/decay, gaussian distribution).&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-04T18:09:28.090" CommentCount="1" />
  <row Id="4027" PostTypeId="2" ParentId="3999" CreationDate="2010-09-04T18:17:22.173" Score="1" ViewCount="0" Body="&lt;p&gt;Alas, the accepted answer is misleading (and arguably incorrect). The reason that primality (or any other purely arithmetic property) is preserved in radix representation is simply that such representation faithfully preserves all of the arithmetic operations on integers. More precisely, $\:$ if $\rm\;n\to r(n)\;$ is the map from $\rm\:n\:$ to its radix $\rm\:d\;$ representation, then it preserves addition $\rm\;r(m+n) = r(m) + r(n),\;$ and multiplication $\rm\;r(mn) = r(m)\ r(n),\;$ and $\rm\;r\;$ has an inverse $\rm\;s\;$ that similarly also preserves addition and multiplication (technically: $\rm\:r\:$ is a ring isomorphism). This readily implies that the relation  of divisibility is faithfully preserved in radix representation, because the relation of divisibility can be expressed as an equation involving only arithmetic (ring) operations (namely multiplication), and such equations are necessarily preserved by the maps $\rm\;r\;$ and $\rm\;s\;$ - indeed these maps are defined precisely so to preserve these fundamental operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's instructive to examine more closely the preservation of divisibility. First, we recall the standard notation $\rm\;a|b\; := \: a\;$ divides $\rm\:b\:,\;\;$ i.e. $\rm\:\exists \:n\in\mathbb Z : \ an = b\:,\;$ i.e. there exists an integer $\rm\:n\:$ such that $\rm\;an = b\;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\;\quad a|b \iff r(a)|r(b)\quad\quad$ &lt;strong&gt;(Divisibility Preservation by Isomorphisms)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\;(\;\Rightarrow\;)\quad a|b \;\Rightarrow\; \exists\:n\in\mathbb Z: \: an = b \;\Rightarrow\; r(a)\ r(n) = r(an) = r(b) \;\Rightarrow\; r(a)|r(b)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\;(\Leftarrow)\quad r(a)|r(b) \;\Rightarrow\;\exists\:c\in r(\mathbb Z): \: r(a)\: c = r(b)\;\Rightarrow\; r(a)\ r(n) = r(b) \;\Rightarrow\; a n = b\;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final $\;\Rightarrow\;$ above is by applying $\rm\;r\:$'s inverse $\rm\:s\:$ so to cancel the $\rm\;r\:$'s using $\rm\;\: sr = 1 =\;$ identity map.&lt;br&gt;&#xA;Note: this employs $\rm\:s\:$'s preservation of multiplication, $\:$ viz. $\rm\:s(r(a)\: r(n)) \:=\: sr(a)\: sr(n) \:=\: a\: n\;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, we conclude that primes (i.e. irreducibles) are also preserved, since they are definable purely in terms of divisibilty, viz. $\rm\;p\;$ is prime $\;\: := \:\rm\;p = ab \;\Rightarrow\; p|a\;$ or $\rm\;p|b\;$ and $\rm\;p\;$ is not a unit, i.e. not $\rm\;p|1\;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question reduces to the more fundamental  &lt;strong&gt;why is radix representation a ring isomorphism, i.e. why really does radix representation preserve the addition and multiplication operations?&lt;/strong&gt; This is a very good question that deserves a thoughtful answer. It's a serious pedagogical oversight that this topic is rarely discussed in algebra textbooks. Although most students understand this fact subconsciously, many have difficulty providing a rigorous proof (or, worse, they overlook the fact that it does require a rigorous proof). Your question would attract much more interest and receive much more interesting replies if you rephrase it in this manner. Thus I propose the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUGGESTION&lt;/strong&gt;: Edit the title of your question to say &quot;Why is radix representation a ring isomorphism, i.e. why does it preserve addition and multiplication?&quot;. Include in your question a brief description of your knowledge level so to help set the proper level for replies, e.g. do you already know any abstract algebra or elementary number theory?&#xA;Unaccept the currently accepted answer so that the software will continue to bump up the question till it gets a number of good answers. IMPORTANT: Wait at least a week before accepting any answer so that everyone has a chance to see the question. This helps serve to maximize its exposure and hence maximize your potential of receiving insightful answers. If all goes well this should result in an interesting thread that will help serve as a future reference for many similar frequently asked questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; to much more experienced readers: this problem is not as trivial as you might think at first glance (and certainly less trivial for novices). For example, the analogous problem for real numbers (or p-adics) is the subject of a famous paper [1]. For an introduction see e.g. &lt;a href=&quot;http://www.math.ou.edu/~dmccullough/teaching/slides/rotamath.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Its closing line is quite apropos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is very much in keeping with Rota’s thinking that mathematics is not just a quest to solve problems, it is&#xA;  also a quest to understand the mathematical universe as&#xA;  clearly and as deeply as possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[1] F. Faltin, N. Metropolis, B. Ross, G.-C. Rota,&lt;br&gt;&#xA;The real numbers as a wreath product.&lt;br&gt;&#xA;Advances in Math. 16 (1975), 278-304&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-04T19:54:43.680" LastActivityDate="2010-09-04T19:54:43.680" CommentCount="4" />
  <row Id="4028" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T18:23:12.220" Score="3" ViewCount="0" Body="&lt;p&gt;Chandru1 asks how we might proceed, so in that spirit let me offer an idea that provides partial results and connects this problem to a long-standing one recently characterized as a &quot;frustrating question.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's aim to show that under the conditions assumed of $a$ and $b$, $a$ necessarily divides $b$.  This additional relation suffices to show that $a = b$.  (There does not seem to be a simple demonstration of this, but it is much easier than the original problem so I'll let it go for now as an &quot;exercise.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $b &amp;lt; a^2$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \frac{b}{a} \right) ^n - \frac{b^n + n}{a^n+n} = \frac{n (b^n - a^n)}{a^n (a^n+n)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side, in the limit of large $n$, approaches zero from above.  The left hand side is the difference between $\left( \frac{b}{a} \right) ^n$ and an integer.  We conclude that eventually the fractional part of $\left( \frac{b}{a} \right) ^n$ approaches zero.  Results of Pisot, Vijayaraghavan and Andre Weil then imply that $\frac{b}{a}$ must be integral.  (See Akayama &lt;em&gt;et al.&lt;/em&gt;, &lt;em&gt;Powers of rationals modulo 1 and rational base number systems&lt;/em&gt;, &lt;a href=&quot;http://perso.telecom-paristech.fr/~jsaka/PUB/Files/RBNS-rev.pdf&quot; rel=&quot;nofollow&quot;&gt;http://perso.telecom-paristech.fr/~jsaka/PUB/Files/RBNS-rev.pdf&lt;/a&gt; .)  The intuition is that the fractional parts of powers of non-integers ought to fill the interval [0, 1) in a fairly &quot;random&quot; way.  Indeed, numerical experiments verify this for rational numbers (but not for all irrationals!): see &lt;a href=&quot;http://mathworld.wolfram.com/PowerFractionalParts.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PowerFractionalParts.html&lt;/a&gt; .  So, convergence of the fractional part to zero--which is assured by the sequence of divisibility conditions in the problem--implies the ratio $b / a$ is not behaving like a proper fraction: it must actually be an integer.  That gives us enough leverage to show the equality $a = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect a similar approach should work for $b \ge a^2$, but I haven't found it.  Indeed some of the papers in the literature note changes in the behavior of powers of $b / a$ when $b$ exceeds $a^2$, so we should be cautious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that elementary methods of number theory show that all primes dividing $a$ must also divide $b$.  That, however, doesn't seem to get us very far.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-04T18:23:12.220" CommentCount="1" />
  <row Id="4029" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T18:54:44.843" Score="4" ViewCount="0" Body="&lt;p&gt;Claim if our hypothesis holds, $a \equiv b \ (\text{mod}\ p)$ for any prime $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $n$ so that $n \equiv -a \ (\text{mod}\ p)$ and $n \equiv 1 (\text{mod}\ p-1)$ ( we can do this by &lt;strong&gt;Chinese Remainder Theorem&lt;/strong&gt;).  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{n} + n \equiv a^1 + n = a - a = 0 (\text{mod} \ p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore since $a^n + n \mid b^n + n$, $b^n + n \equiv 0 (\text{mod}\ p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^n + n \equiv b^1 + n \equiv b - a (\text{mod}\ p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore $b \equiv a \pmod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our result now follows by picking any $p &gt; b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE by BD&lt;/strong&gt; $\;$  This solution has been posted before in at least a few well-known math forums,  e.g. see my &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/b04d0459db4c2cb5&quot; rel=&quot;nofollow&quot;&gt;sci.math post on April 4,2006&lt;/a&gt;, and see Rust's post on &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&amp;amp;t=289683&quot; rel=&quot;nofollow&quot;&gt;AoPS, July 19, 2009&lt;/a&gt;. It also appeared in at least one other forum much more recently (alas, I can't recall which one). Almost surely, by now the problem and solution is listed in various problem collections, so it should be considered somewhat well-known.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-05T17:11:16.953" LastActivityDate="2010-09-05T17:11:16.953" CommentCount="17" />
  <row Id="4031" PostTypeId="1" CreationDate="2010-09-04T20:06:05.483" Score="3" ViewCount="197" Body="&lt;p&gt;Let $p_i$ denote the $i^{th}$ prime number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the smallest positive integer $k$ such that the product $n = p_1 \cdot p_2 \cdots p_k$ satisfies $\sigma(n) &gt; 3n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any positive integer $m &amp;lt; n$ satisfying $\sigma(m) &gt; 3m$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Taken From&lt;/strong&gt; (Rosen), at the end of the chapter under challenge problems&lt;/p&gt;&#xA;" OwnerUserId="1642" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-05T01:10:15.130" LastActivityDate="2010-10-05T01:10:15.130" Title="Sum of divisors and prime numbers, short proof" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="4032" PostTypeId="1" AcceptedAnswerId="4041" CreationDate="2010-09-04T20:13:20.567" Score="4" ViewCount="264" Body="&lt;p&gt;I have to prove that $||A-B||=||A+B||\Leftrightarrow AB=0$ and I was wondering if this approach is correct, or if there's a better/more elegant way to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given n-dimensional vectors A and B, we can write $||A-B||=||A+B||$ as:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{\sum\limits_{j=1}^{n}(a_j-b_j)^2}=\sqrt{\sum\limits_{j=1}^{n}(a_j+b_j)^2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring both sides and expanding the binomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits_{j=1}^{n}a^2_j-2a_jb_j+b_j^2=\sum\limits_{j=1}^{n}a^2_j+2a_jb_j+b_j^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\sum\limits_{j=1}^{n}a_jb_j=\sum\limits_{j=1}^{n}a_jb_j$, which holds true if and only if $\sum\limits_{j=1}^{n}a_jb_j=0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $AB$ is equivalent to $\sum\limits_{j=1}^{n}a_jb_j$, then $||A-B||=||A+B||\Leftrightarrow AB=0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-09-05T13:14:52.977" Title="Proving that $||A-B||=||A+B||\Leftrightarrow AB=0$" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="4034" PostTypeId="2" ParentId="4031" CreationDate="2010-09-04T20:35:12.210" Score="1" ViewCount="0" Body="&lt;p&gt;This should help you out: &lt;a href=&quot;http://www.mathhelpforum.com/math-help/f7/sum-divisors-function-question-154661.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mathhelpforum.com/math-help/f7/sum-divisors-function-question-154661.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-04T20:35:12.210" CommentCount="1" />
  <row Id="4035" PostTypeId="1" AcceptedAnswerId="4037" CreationDate="2010-09-04T21:33:30.953" Score="2" ViewCount="203" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've recently started making a small fractal app in Javascript using the famous Mandelbrot bulb (z = z^2 + c). I've been trying to find the best method of coloring the points on the complex plane, and I've come across some very interesting ideas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://linas.org/art-gallery/escape/escape.html&quot; rel=&quot;nofollow&quot;&gt;http://linas.org/art-gallery/escape/escape.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Real_Escape_Time&quot; rel=&quot;nofollow&quot;&gt;http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Real_Escape_Time&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've been focusing on this &quot;normalized iteration count algorithm&quot; and I've run into a small mental hurdle sorting out what needs to be calculated. Basically, I'm confused about the first equation that shows up in that second link. Assuming we have a complex number &lt;code&gt;Z&lt;/code&gt; which is the end result of &lt;code&gt;n&lt;/code&gt; iterations. I can't quite figure out what &quot;Zx2&quot; or &quot;zy2&quot; is and how how one might add them together or take the square root of their sum. Is that just shorthand for some basic complex operators, or is there something else going on? Basically, my problem is that I'm not particularly good at reading mathematical notation in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway...any directional guidance you can offer would be extremely helpful. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="1720" LastActivityDate="2010-09-04T22:36:03.963" Title="Continuous coloring of a Mandelbrot fractal" Tags="&lt;algorithms&gt;&lt;fractals&gt;" AnswerCount="2" />
  <row Id="4036" PostTypeId="2" ParentId="4035" CreationDate="2010-09-04T21:41:21.580" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: &quot;Zx2&quot; and &quot;Zy2&quot; are defined in the C code in the second code block &lt;a href=&quot;http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Optimisation&quot; rel=&quot;nofollow&quot;&gt;at this point in your wikibooks link&lt;/a&gt;.  They are the squares of the real and imaginary parts, respectively, of $z$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if it's exactly the same, but what I've used is:&#xA;$$n-\log\left(\frac{\log(|z|)}{\log(r)}\right)$$&#xA;where $z$ is the result of $n$ iterations and has just escaped as noted by $|z|&gt;r$ (where $r$ is the escape radius).  This generates real numbers in $[0,n)$, so you could divide by $n$ to end up with a number in $[0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-04T22:31:47.700" LastActivityDate="2010-09-04T22:31:47.700" CommentCount="1" />
  <row Id="4037" PostTypeId="2" ParentId="4035" CreationDate="2010-09-04T21:41:28.897" Score="3" ViewCount="0" Body="&lt;p&gt;Based on the C code given there, &lt;code&gt;Zx&lt;/code&gt; is just the real part $\Re z$ and &lt;code&gt;Zy&lt;/code&gt; is just the imaginary part $\Im z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the complex iteration formula for the Mandelbrot set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=z^2+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when expanded to real variables looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=x^2-y^2+u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=2xy+v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z=x+iy$ and $c=u+iv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables with a 2 appended are the squares, as can be seen from how they were defined, so for instance $|z|=\sqrt{x^2+y^2}$ is &lt;code&gt;sqrt(Zx2+Zy2)&lt;/code&gt; which is used for determining how much the iteration diverges (escapes to infinity), which is used by the coloring functions.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-04T22:36:03.963" LastActivityDate="2010-09-04T22:36:03.963" CommentCount="2" />
  <row Id="4038" PostTypeId="2" ParentId="1647" CreationDate="2010-09-04T21:58:31.473" Score="2" ViewCount="0" Body="&lt;p&gt;To expand on the suggestion to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert function&lt;/a&gt;, I'll show how it arises in your equation of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=\frac1{2\pi v^2}\exp\left(-\frac{x^2}{2v^2}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we multiply both sides by $-\pi x^2$ to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\pi x^2 y=-\frac{x^2}{2v^2}\exp\left(-\frac{x^2}{2v^2}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can now be inverted to the Lambert function (recall that the Lambert function $W(z)$ is the inverse function of $z\exp(z)$, $W(z)\exp(W(z))=z$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{x^2}{2v^2}=W(-\pi x^2 y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we can now solve for $v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v=\frac{x}{\sqrt{-2W(-\pi x^2 y)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of sign for the square root is motivated by the fact that variances are conventionally taken to be positive.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-05T23:27:59.197" LastActivityDate="2010-09-05T23:27:59.197" CommentCount="3" />
  <row Id="4039" PostTypeId="2" ParentId="3977" CreationDate="2010-09-04T23:02:13.063" Score="1" ViewCount="0" Body="&lt;p&gt;Geomview is fairly flexible, and allows you to work in Euclidean, spherical or hyperbolic geometry.  Depending on which operating system you use it might be easier or harder to get it installed.  It made it into the Ubuntu repositories recently, so if you're running Ubuntu it takes little effort to install. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it's not the easiest software to &lt;em&gt;use&lt;/em&gt; -- quite a ways away from the interface of GeoGebra. More often than not, if I need a high-quality rendered image of something 3-dimensional, I code a PovRay script and render it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blender is open-source and interactive.  It appears to be quite flexible -- it's used to produce models for 3d video games, for example.  I haven't really got comfortable using it, though. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-05T22:39:55.250" LastActivityDate="2010-09-05T22:39:55.250" CommunityOwnedDate="2010-09-04T23:02:13.063" CommentCount="3" />
  <row Id="4040" PostTypeId="2" ParentId="3978" CreationDate="2010-09-04T23:17:46.360" Score="8" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt;$\;$ from my &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/b04d0459db4c2cb5&quot; rel=&quot;nofollow&quot;&gt;sci.math post of April 4, 2006&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\begin{array}{}&#xA;p-1\:|\:n-1 \\&#xA;\quad\quad\;\rm p\:|\:a+n \\&#xA;\end{array} \;\; \Rightarrow\;\; p\;\;|\;\;a^n-a\;+\;a+n\;\;|\;\;b^n-b\;+\;b-a\;+\;a+n \;\;\Rightarrow\;\; p\:|\:b-a$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-04T23:17:46.360" CommentCount="6" />
  <row Id="4041" PostTypeId="2" ParentId="4032" CreationDate="2010-09-04T23:25:57.187" Score="7" ViewCount="0" Body="&lt;p&gt;What you've done is correct, but I think it's better to work without coordinates; just with the definition of norm in terms of the dot product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\| A \| = +\sqrt{A\cdot A} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you may observe that, since $\|A \| \geq 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|A+B\| =  \|A -B\| \  \Longleftrightarrow  \  \|A +B\|^2  =  \|A-B\|^2 .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for instance, compute the difference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \|A +B\|^2  -  \|A-B\|^2   &amp;amp;= (A+B)\cdot (A+B) - (A-B)\cdot(A-B)    \\&#xA;               &amp;amp;= A\cdot A + A\cdot B + B\cdot A + \cdots&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. I forgot to point out an obvious geometric interpretation of this result: if you draw a parallelogram with sides $A$ and $B$, then $A+B$ and $A-B$ are the diagonals of the parallelogram, right? These diagonals are equal if and only if...?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-05T13:14:52.977" LastActivityDate="2010-09-05T13:14:52.977" />
  <row Id="4042" PostTypeId="1" AcceptedAnswerId="4046" CreationDate="2010-09-05T00:56:04.653" Score="2" ViewCount="56" Body="&lt;p&gt;I am working through Theorem 6 in Chapter 8 of Hoffman and Kunze.  (The section is &lt;em&gt;Linear Functionals and Adjoints&lt;/em&gt;.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement is &lt;blockquote&gt; Let $V$ be a finite-dimensional inner product space, and $f$ a linear functional on $V$.  Then there exists a unique vector $\beta$ in $V$ such that $f(\alpha) = (\alpha | \beta)$ for all $\alpha$ in $V$.&lt;/blockquote&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;EDIT:&lt;/strong&gt;  The following is written in terms of a &lt;em&gt;real&lt;/em&gt; vector space.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fine with the fact that this essentially follows from what $f$ does to the elements of a basis for $V$, in other words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an &lt;em&gt;arbitrary&lt;/em&gt;* $\mathcal{B} = \{e_1, \ldots, e_{n} \}$ (where $n = \dim V\;\;$) we can assign $\xi_{i} = f(e_{i})$ for $i=1,\ldots, n$.  Then any $\alpha \in V$ can be written as $\alpha = x_{1}e_{1} + \cdots + x_{n}e_{n}$ so that &#xA;$$\begin{align}&#xA;f(\alpha) &amp;amp;= f(x_{1}e_{1} + \cdots + x_{n}e_{n}) \\ \\&#xA;          &amp;amp;= x_{1}f(e_{1}) + \cdots + x_{n}f(e_{n}) \\ \\&#xA;          &amp;amp;= x_{1} \xi_{1} + \cdots + x_{n}\xi_{n}.\end{align}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we can identify the $\xi_{i}$'s as being independent of $\alpha \in V$, and only having been dependent on our choice of a basis.  With this, we can simply assert that $\beta = (\xi_{1}, \ldots, \xi_{n})$ (and is in fact unique).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now&lt;/strong&gt;, this is apparently not the whole story!  *According to H. and K., the coordinates of our fixed vector $\beta$ are determined by the action of $f$ on an &lt;em&gt;orthonormal&lt;/em&gt; basis for $V$ rather than an arbitrary basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  What role does the orthogonality (or even the &lt;em&gt;'orthonormality'&lt;/em&gt;) of the basis have in this argument?  It seems to work just fine for an arbitrary basis as far as I can tell.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-09-05T01:58:55.693" LastActivityDate="2010-09-05T02:34:56.537" Title="Role of orthonormal basis for: Linear functional on a finite dimensional inner product space is 'inner product with a fixed vector in the space'" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="4043" PostTypeId="1" CreationDate="2010-09-05T01:57:01.843" Score="2" ViewCount="210" Body="&lt;p&gt;I'm trying to learn predicate logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm looking for some good resources on it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen that I learn better when I can program&#xA;So I was wondering if there was a 'predicate logic' programning language?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe an interactive tutorial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe lots of examples, something like: Predicate Logic by Example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or at least some pretty basic books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIA&lt;/p&gt;&#xA;" OwnerUserId="1724" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T02:13:54.573" LastActivityDate="2011-01-04T12:44:34.063" CommunityOwnedDate="2010-09-07T23:21:03.547" Title="What are good resources for learning predicate logic / predicate calculus? " Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4044" PostTypeId="1" AcceptedAnswerId="4053" CreationDate="2010-09-05T02:15:29.800" Score="3" ViewCount="159" Body="&lt;p&gt;This question may smell like my older question &lt;a href=&quot;http://math.stackexchange.com/questions/1778/a-probability-game&quot;&gt;http://math.stackexchange.com/questions/1778/a-probability-game&lt;/a&gt;, but this time my intentions are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above problem came from my friend. Or so I thought, it really came from me incorrectly remembering his problem. The problem he actually gave me was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you start with 1 dollar and flip a weighted coin. With probability 3/4 (call this heads) you win a dollar and with probability 1/4 (call this tails) you lose a dollar. What is the probability you eventually run out of money? (Suppose the house has as many dollar bills as needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to solve this problem in the following way:&#xA;First observe the number of tails must equal the number of heads plus one. If a string of Hs and Ts satisfy this property, call it Good. Now we want to count the number of Good strings that don't have Good proper prefixes. For a string of length 2n-1 this is just the nth Catalan number. Now we just take the appropriate summation and we arrive at the probability we eventually run out of money which turns out to be 1/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when my friend told me this problem, he said he was able to apply a clever trick and solve it almost instantly. The thing is, he doesn't remember how he did it. He says he vaguely remembers using a recurrence relation and then telescoping a sum. (Which sounds very plausible) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can you come up with a short and sweet solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, I am aware my title is poor at best. If you can think of a better one, please let me know.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-06T05:34:08.040" Title="Finding a clever solution to a game of chance." Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4045" PostTypeId="1" CreationDate="2010-09-05T02:28:12.270" Score="4" ViewCount="131" Body="&lt;p&gt;I'm currently interested in the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: X \to Y$ be a fpqc morphism of schemes. Then there is an equivalence of categories between quasi-coherent sheaves on $Y$ and &quot;descent data&quot; on $X$.  Namely, the second category consists of quasi-coherent sheaves $\mathcal{F}$ on $X$ with an isomorphism $p_{1}^*(\mathcal{F}) \simeq p_2^*(\mathcal{F})$, where $p_1, p_2: X \times_Y X \to X$ are the two projections.  &lt;strong&gt;Edit:&lt;/strong&gt; There is a further condition; a diagram involving an iterated fibered product is required to commute as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Demazure-Gabriel's &lt;em&gt;Introduction to Algebraic Geometry and Algebraic Groups,&lt;/em&gt; it is proved (under the name ffqc (sic) descent theorem) that the sequence&#xA;$$ X \times_Y X \to^{p_1, p_2} X \to Y$$ is a coequalizer in the category of locally ringed spaces under the above hypotheses. If I am not mistaken, this is the same as the theorem that says that representable functors are sheaves in the fpqc topology.  On the other hand, D-G give a fairly explicit description of the quotient space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: For a coequalizer diagram of (locally) ringed spaces, &#xA;$$A \to^{f,g} B \to C,$$&#xA;is there a descent diagram for quasi-coherent sheaves on $A,B,C$?  In particular, does the D-G form of the descent theorem directly, by itself, imply the more general one for quasi-coherent sheaves?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-05T03:41:27.047" LastActivityDate="2010-09-05T03:41:27.047" Title="Do cokernels in RingSpc automatically lead to descent?" Tags="&lt;algebraic-geometry&gt;&lt;descent&gt;" CommentCount="3" />
  <row Id="4046" PostTypeId="2" ParentId="4042" CreationDate="2010-09-05T02:34:56.537" Score="2" ViewCount="0" Body="&lt;p&gt;If the basis $\{e_1,\ldots,e_n\}$ is not orthonormal, then the inner product of the vectors $\alpha = x_1e_1 + \cdots + x_n e_n$ and $\beta = \xi_1 e_1 + \cdots + \xi_n e_n$ is not equal to $x_1\xi_1+\cdots+x_n\xi_n$.  Thus, while it is true that $f(\alpha) = x_1\xi_1+\cdots+x_n\xi_n$, it is not true that $f(\alpha)=(\alpha|\beta)$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-05T02:34:56.537" CommentCount="3" />
  <row Id="4047" PostTypeId="2" ParentId="3987" CreationDate="2010-09-05T03:58:54.607" Score="0" ViewCount="0" Body="&lt;p&gt;With no better answer, I'll just add a little to what I said in the question to get this closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are series calculations for logarithms (http://en.wikipedia.org/wiki/Logarithm) so I could in principle calculate the base 10 log directly. This would be awkward and slow, though, as for arbitrary precision integers a machine float/double isn't precise enough. I'd probably need a rational library, and the rationals involved wouldn't be simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be possible to use an imprecise base-converting factor. With an underestimate factor, you'd get an underestimate log value. Divide through by 10^x for that log value, then repeat as often as necessary to top up your underestimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that dividing through by 10^x isn't trivial unless you happen to be working in base 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably best to stick with repeated division by 10 - or powers of 10, at least. Constructing a list of powers (10^1, 10^2, 10^4, 10^8 and so on) by squaring wouldn't be too hard. Stop when you exceed the target number, then do the divisions largest to smallest. Each power should only need to be considered once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, it may be better to not build powers of 10 that won't fit in a machine word, and just do repeated division by (small powers of) 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this sounds now more like programming than math, so sorry about that.&lt;/p&gt;&#xA;" OwnerUserId="510" LastActivityDate="2010-09-05T03:58:54.607" />
  <row Id="4048" PostTypeId="1" AcceptedAnswerId="4085" CreationDate="2010-09-05T04:09:27.013" Score="3" ViewCount="168" Body="&lt;p&gt;The only place I've encountered well-definition is with proving an operation defined on an equivalence class is independent of the choice of representative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my homework, it asks us to show that the ring of formal Laurent series is well-defined, and I don't understand what exactly I need to show. However, I don't understand what I'm trying to prove. I've read the wikipedia article and don't understand how there is anything to prove in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could either point me to some better references explaining well-definition or explain what I need to do to prove that a laurent series is well-defined, I would appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-05T04:20:45.633" LastActivityDate="2010-09-05T19:35:57.807" Title="How do you show the ring of formal laurent series is well-defined?" Tags="&lt;abstract-algebra&gt;&lt;terminology&gt;&lt;homework&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4049" PostTypeId="1" AcceptedAnswerId="4051" CreationDate="2010-09-05T04:31:50.393" Score="3" ViewCount="135" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; To prove that the Ring of Complex Entire functions is neither Artinian nor noetherian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Clearly $R$ is not Artinian because it is a commutative integral domain which is not a field, and $R$ is not noetherian because its not  a factorisation domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a proof of this theorem using the Ascending Chain / Descending Chain condition for Artinian / Noetherian rings?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-05T04:52:00.290" Title="Proving that Ring of Complex Entire functions is neither Artinian nor Noetherian." Tags="&lt;analysis&gt;&lt;abstract-algebra&gt;&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4051" PostTypeId="2" ParentId="4049" CreationDate="2010-09-05T04:52:00.290" Score="4" ViewCount="0" Body="&lt;p&gt;Let $J_n$ be the ideal of entire functions vanishing on the first $n$ positive integers, and let $I_n$ be the ideal of entire functions vanishing on positive integers greater than $n$.  Then $$J_1\supset J_2\supset J_3\supset\cdots,$$ $$I_1\subset I_2\subset I_3\subset\cdots,$$ and all of these containments are proper.  One way to see that the containments are proper is to use the fact that given any sequence of complex numbers $a_1,a_2,\ldots$, there is an entire function $f$ such that $f(n)=a_n$ for each positive integer $n$.  For more on this, see &lt;a href=&quot;http://mathoverflow.net/questions/20711/does-there-exist-a-holomorphic-function-which-takes-given-values-on-the-positive&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://mathoverflow.net/questions/2944/&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt; &lt;a href=&quot;http://mathoverflow.net/questions/7328/interpolation-of-sequences-by-analytic-functions&quot; rel=&quot;nofollow&quot;&gt;questions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-05T04:52:00.290" CommentCount="1" />
  <row Id="4052" PostTypeId="2" ParentId="3946" CreationDate="2010-09-05T04:52:02.053" Score="1" ViewCount="0" Body="&lt;p&gt;It seems like you answered your own question in the last paragraph. You create the two new control points at each corner, and then replace the corner with a quadratic Bezier curve determined by the two new control points and the original corner point. So, it is not clear to me what you are asking. The &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Casteljau%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;de Casteljau algorithm&lt;/a&gt; can be used to render the curves.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-09-05T04:52:02.053" />
  <row Id="4053" PostTypeId="2" ParentId="4044" CreationDate="2010-09-05T06:11:26.577" Score="3" ViewCount="0" Body="&lt;p&gt;If $p$ is the probability that you eventually lose a dollar then we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = \frac{1}{4} + \frac{3}{4}p^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We lose a dollar, or win one and lose two eventually).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving gives $p=1$ or $p=\frac{1}{3}$. I suppose we can eliminate $p=1$ somehow and get $p=\frac{1}{3}$ as the required answer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-05T06:11:26.577" CommentCount="4" />
  <row Id="4054" PostTypeId="1" CreationDate="2010-09-05T08:39:40.427" Score="19" ViewCount="988" Body="&lt;p&gt;There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$  z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc's boundary followed in the positive direction, then for every $z$ in the disc we can write:&#xA;$$f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&#xA;&lt;strong&gt;What is the intuitive explanation of this formula?&lt;/strong&gt; (For example, but not necessary, geometrically.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.)&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-05T13:40:04.817" LastActivityDate="2010-11-02T05:45:36.757" Title="Intuitive explanation of Cauchy's Integral Formula in Complex Analysis" Tags="&lt;intuition&gt;&lt;complex-analysis&gt;" AnswerCount="10" CommentCount="4" FavoriteCount="16" />
  <row Id="4055" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T10:15:20.000" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps you can see the Cauchy's integral formula as the &quot;Winding number&quot; of a curve $C$ around any complex number $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, please see : &lt;a href=&quot;http://en.wikipedia.org/wiki/Winding_number&quot; rel=&quot;nofollow&quot;&gt;en.wikipedia.org/wiki/Winding_number&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-05T10:15:20.000" CommentCount="1" />
  <row Id="4056" PostTypeId="2" ParentId="1778" CreationDate="2010-09-05T10:17:19.317" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a completely probabilistic proof inspired by Moron in &lt;a href=&quot;http://math.stackexchange.com/questions/4044/finding-a-clever-solution-to-a-game-of-chance&quot;&gt;http://math.stackexchange.com/questions/4044/finding-a-clever-solution-to-a-game-of-chance&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let p be the probability that I end up with a net loss of 50 cents, then $p = \frac{1}{2} + \frac{1}{2} p^3$. That is to say, I either lose 50 cents, or gain a dollar and lose a dollar and 50 cents eventually. Solving for p I get 1, $(-1-\sqrt{5})/2$, and $(1-\sqrt{5})/2$ and it is not hard to see that we can dismiss the first two. Now the probability I end up with a net loss of one dollar is $p^2 = (3-\sqrt{5})/2$.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-05T10:17:19.317" />
  <row Id="4057" PostTypeId="2" ParentId="67" CreationDate="2010-09-05T10:22:57.960" Score="0" ViewCount="0" Body="&lt;p&gt;You can see this article as well: &lt;a href=&quot;http://www.cs.umd.edu/Honors/reports/ECCpaper.pdf&quot; rel=&quot;nofollow&quot;&gt;www.cs.umd.edu/Honors/reports/ECCpaper.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-05T10:22:57.960" />
  <row Id="4058" PostTypeId="1" AcceptedAnswerId="4060" CreationDate="2010-09-05T10:25:59.877" Score="1" ViewCount="117" Body="&lt;p&gt;See this SO thread: &lt;a href=&quot;http://stackoverflow.com/questions/3634106/calculating-which-item-is-next-in-a-percentage-distribution&quot;&gt;&lt;strong&gt;Calculating which item is next in a percentage distribution&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you calculate&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;max ((rand1 * 10), (rand2 * 20), (rand3 * 30), (rand4 * 40))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;for four random numbers in a range (say &lt;code&gt;1-100&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &lt;code&gt;max&lt;/code&gt; is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rand1 * 10, you choose 'A'&#xA;rand2 * 20, you choose 'B'&#xA;rand3 * 30, you choose 'C'&#xA;rand4 * 40, you choose 'D'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the expected distribution of choosing A, B, C and D?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to derive it mathematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Using &lt;a href=&quot;http://codepad.org/F6VV6J3m&quot; rel=&quot;nofollow&quot;&gt;some C code&lt;/a&gt;, the percentages come out to be 1, 10, 31 and 56 respectively)&lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2010-09-05T17:04:31.147" Title="How to calculate the expected distribution of results using random numbers?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;random-variable&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4059" PostTypeId="2" ParentId="2581" CreationDate="2010-09-05T10:53:16.123" Score="2" ViewCount="0" Body="&lt;p&gt;There are variations on the problem where the division is always possible and a proof using complete induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first variation is: [..original problem ...] where at least one of the sticks is &gt;=2n.&lt;br&gt;&#xA;The second variation is: [..original problem ...] where it is allowed to glue two of the sticks together and break once at an arbitrary position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof for the variations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case n=1 is trivial - one stick of length 1. Then we suppose the assumption holds for n and consider the case n+1. I.e. given a number of sticks of integral length &gt;= n+1 whose lengths add to (n+1)(n+2)/2 - can these be divided into sticks of lengths 1,2,3,…,n+1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break off from one of the sticks a length n+1. In the first variation we use the stick with lenght &gt;= 2n+2. This part will be the required stick of length n+1 in the solution. Because we broke off a length n+1 the remaining total length is (n+1)(n+2)/2 - (n+1) = n(n+1)/2. In the first variation the inducution step is valid, in the second variation we may not yet use the induction step because the other part of the broken stick may now be smaller then n+1. In that case the second variation allows us to glue together this part with one of the other sticks. Now the induction step is valid and we break the rest of the collection of sticks in {1,..n} and have succeeded in the division {1,.. n+1}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@downvoter: i.m.o. this try for an answer is useful, because it can show us the edges/restrictions of the original problem.&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="706" LastEditorDisplayName="" LastEditDate="2010-10-02T21:44:44.500" LastActivityDate="2010-10-02T21:44:44.500" CommentCount="6" />
  <row Id="4060" PostTypeId="2" ParentId="4058" CreationDate="2010-09-05T12:01:13.337" Score="2" ViewCount="0" Body="&lt;p&gt;The probabilities are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_i = \int_0^1 \prod_{j \in {1,2,3,4}, j\ne i} min(\frac{ix}{j},1)  dx, i = 1, 2, 3, 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;each factor in the integrand computes the probability density that the i-th random variable is larger than the j-th one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact result is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1/96, 31/288 91/288 163/288]&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-09-05T12:10:32.463" LastActivityDate="2010-09-05T12:10:32.463" CommentCount="2" />
  <row Id="4061" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T12:16:19.607" Score="8" ViewCount="0" Body="&lt;p&gt;As Chandru1 pointed out, a beautiful particular case of Cauchy's integral formula gives us a very particular &lt;em&gt;winding number&lt;/em&gt;. For the sake of simplicity, let us look only at closed paths on the boundary of the unit disc, $\gamma : [0,1] \longrightarrow S^1$. Then, the &lt;em&gt;winding number&lt;/em&gt; of $\gamma$ is the number of times $\gamma$ goes round the circumference. That is, if we write the complex number $\gamma (t)$ in its exponential form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\gamma (t) = e^{i\theta (t)} \qquad \qquad \qquad  [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you can prove that there is a &lt;em&gt;continuous&lt;/em&gt; choice for the argument $\theta (t)$ of $\gamma (t)$; that is, a continuous function $\theta : [0, 1] \longrightarrow \mathbb{R}&#xA;$ such that [1] holds for all $t$. In Algebraic Topology, $\theta$ is called a &quot;lifting&quot; of $\gamma$. Moreover, any two such continuous &quot;liftings&quot; of $\gamma$ differ necessarily by a constant integer multiple of $2\pi$: if $\widetilde{\theta}(t)$ is another (continuous) lifting of $ \gamma$, then there is a constant integer $k$, not depending on $t$, such that $\widetilde{\theta}(t) = \theta (t) + 2k\pi$ for all $t$. (This is called the &quot;lifting lemma&quot; in Algebraic Topology. For a proof in Complex Analysis, see theorem 7.1, in Stewart and Tall.) So the number of times $\gamma $ goes round the circumference (its &lt;em&gt;winding number&lt;/em&gt;) is well-defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;w (\gamma , 0) = \frac{\theta (1) - \theta (0)}{2\pi} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can easily prove (see op.cit, section 7.5), that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;w (\gamma , 0 ) = \frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Cauchy's integral formula for the constant function $f \equiv 1$, $z = 0$ (and $\gamma (t) = e^{i2\pi t}$, hence $\theta (t) =2 \pi t$) tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta = 1 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you go round the circumference just one time, well, you are going round the circumference indeed exactly once. Isn't math amazing? :-)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-05T18:04:13.617" LastActivityDate="2010-09-05T18:04:13.617" CommentCount="10" />
  <row Id="4062" PostTypeId="1" AcceptedAnswerId="4063" CreationDate="2010-09-05T13:24:02.937" Score="3" ViewCount="293" Body="&lt;p&gt;I've been struggling with figuring out how to add powers of $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the result of $i^3 + i^4 + i^5$ is $1$. But how do I get the result of $i^3 + i^4 + ... + i^{50}$? Writing it all down would be pretty mundane...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to do something with division by 4, since the &quot;power cycle&quot; of $i$ repeats every fourth power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any clues.&lt;/p&gt;&#xA;" OwnerUserId="267" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-05T13:51:29.443" LastActivityDate="2010-11-16T00:00:33.257" Title="Adding powers of i" Tags="&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="4063" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T13:31:33.700" Score="6" ViewCount="0" Body="&lt;p&gt;From $i^2=-1$ you get $i^{4n}=1$, $\quad i^{4n+1}=i$, $\quad i^{4n+2}=-1$ and $i^{4n+3}=-i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just count your positive and negative multiples of $1$ and $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $i^3+i^4+\cdots+i^{50}=0$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-05T13:42:38.977" LastActivityDate="2010-09-05T13:42:38.977" />
  <row Id="4064" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T13:35:44.510" Score="0" ViewCount="0" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $$i^{3} + i^{4} + i^{5} = 1 = i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} = i^{3} + i^{4} + \cdots +i^{4n+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $i^{50}=1 \times -1$, therefore we have the sum is $0$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-05T13:49:06.583" LastActivityDate="2010-09-05T13:49:06.583" />
  <row Id="4065" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T13:55:10.247" Score="17" ViewCount="0" Body="&lt;p&gt;Observing that $i^{3}+i^{4}+\ldots +i^{50}$ is a &lt;em&gt;geometric progression&lt;/em&gt; with ratio $i$, first term $i^3$ and $50-3+1=48$ terms, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i^{3}+i^{4}+\ldots +i^{50}=i^{3}\times \dfrac{1-i^{50-3+1}}{1-i}=i^{3}\times &#xD;&#xA;\dfrac{1-i^{48}}{1-i}=i^{2}i\times \dfrac{1-(i^{2})^{24}}{1-i}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=-i\dfrac{1-(-1)^{24}}{1-i}=-i\dfrac{1-1}{1-i}=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &quot;arithmetic&quot; corrected to &quot;geometric&quot;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-16T00:00:33.257" LastActivityDate="2010-11-16T00:00:33.257" CommentCount="4" />
  <row Id="4066" PostTypeId="1" AcceptedAnswerId="4071" CreationDate="2010-09-05T15:25:07.790" Score="2" ViewCount="124" Body="&lt;p&gt;Consider the closed interval $[0,1]$, there is $\frac{2}{3} \in [0,1]$ where $p=2$ and $q=3$. Similarly consider $[2,3]$, one can have $\frac{5}{2} \in [2,3]$ where $p=5$ and $q=2$. Does every interval of the form $[a,b]$, where $a,b \in \mathbb{R}$ contain a rational of this kind. If yes, how can we prove it?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-05T16:16:29.290" LastActivityDate="2010-09-05T16:16:29.290" Title="Rationals of the form $\frac{p}{q}$ where $p,q$ are primes in $[a,b]$." Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="4067" PostTypeId="1" AcceptedAnswerId="4079" CreationDate="2010-09-05T15:26:02.213" Score="3" ViewCount="81" Body="&lt;p&gt;In the book &lt;em&gt;Elliptic Curves - McKean &amp;amp; Moll&lt;/em&gt; we are given the outline for a proof of Eulers addition theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The (projective) quartic $\mathbf y^2 = (1-\mathbf x^2)(1-k^2 \mathbf x^2)$ has four intersections with the quadratic $\mathbf y = 1 + a \mathbf x + b \mathbf x^2$. One is located at $(\mathbf x, \mathbf y) = (0,1)$. Let the other three have covering parameters $x_1,x_2,x_3$. Check that their sum is independent of $a$ and $b$ and identify it, mod $\mathbb L$, as either $0$ or $2K$. Now compute $\mathbf x_1 + \mathbf x_2 + \mathbf x_3 = 2ab/c$ and $\mathbf x_1 \mathbf x_2 \mathbf x_3 = 2a/c$ with $c = k^2 - b^2$ and conclude that $\mathbf x_1 \mathbf y_2 + \mathbf x_2 \mathbf y_1 = \pm \mathbf x_3 (1 - k^2 \mathbf x_1^2 \mathbf x_2^2)$. What is the ambiguous sign?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is too difficult for me to solve so I have some questions that I hope someone will be able to help me with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are these covering parameters and how do we find information about their sum?&lt;/li&gt;&#xA;&lt;li&gt;Why does the information about the covering parameters matter?&lt;/li&gt;&#xA;&lt;li&gt;I have computed $\mathbf x_1 + \mathbf x_2 + \mathbf x_3$ and $\mathbf x_1 \mathbf x_2 \mathbf x_3$ but I don't see how to reach the conclusion from there. How is it done?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-05T15:56:26.190" LastActivityDate="2010-09-05T17:31:11.483" Title="Deriving Eulers Addition Theorem for Elliptic Integrals" Tags="&lt;special-functions&gt;&lt;elliptic-curves&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4068" PostTypeId="1" CreationDate="2010-09-05T15:50:23.323" Score="7" ViewCount="347" Body="&lt;p&gt;$$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots&#xA;      = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$&#xA;     $$= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-09-07T15:00:00.153" LastActivityDate="2010-09-07T20:02:28.423" Title="Explain why calculating this series could cause paradox?" Tags="&lt;sequences-and-series&gt;&lt;paradox&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="3" />
  <row Id="4069" PostTypeId="2" ParentId="3782" CreationDate="2010-09-05T15:50:46.223" Score="-1" ViewCount="0" Body="&lt;p&gt;With the Open Univeristy (http://www.open.ac.uk/) you can study at home, full or part-time, they send you reading material and regular tests to send back to be marked. It's cheaper than attending a bricks and mortar univeristy, but the fact that you are still paying for it is motivation to complete the tests the send you. After a few years you could have a degree in Mathematics from the OU. You can continue with them and study for a Masters degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They produce their own reading material which is tailored to the learning process. Much easier than reading most textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in my opinion, much better value for money than an attendance university although the latter may be suited to those who want to leave their parents' home for the first time for the social aspect and for the opportunity to rack up enormous debts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based in the UK but many course are open to anyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If after giving the OU a go you still have difficulties then you might consider that there are 3 options in life:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) be good&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) get good&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) give up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you live in denial of this and choose d) none of the above then you are falling for the sunk cost fallacy.&lt;/p&gt;&#xA;" OwnerUserId="882" LastActivityDate="2010-09-05T15:50:46.223" CommunityOwnedDate="2010-09-05T15:50:46.223" />
  <row Id="4070" PostTypeId="2" ParentId="4068" CreationDate="2010-09-05T16:07:12.710" Score="9" ViewCount="0" Body="&lt;p&gt;It's because the series for ln2 is &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_convergence&quot; rel=&quot;nofollow&quot;&gt;conditionally convergent&lt;/a&gt;.  (see also &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann's rearrangement theorem&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-05T16:07:12.710" CommentCount="14" />
  <row Id="4071" PostTypeId="2" ParentId="4066" CreationDate="2010-09-05T16:13:12.340" Score="10" ViewCount="0" Body="&lt;p&gt;Roughly speaking this asks whether the quotients of two primes&#xA;are dense in the positive reals. The answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0 &amp;lt; a &amp;lt; b$ and let $q$ be a prime.&#xA;Then there will a a prime $p$ with $a &amp;lt; p/q\le b$ if and only if&#xA;$\pi(bq) &gt; \pi(aq)$ where $\pi$ is the prime-counting function.&#xA;But by the prime number theorem, as $q\to\infty$,&#xA;$$\frac{\pi(bq)}{\pi(aq)}\sim\frac{b\log(aq)}{a\log(bq)}&#xA;=\frac{b(\log q+\log a)}{a(\log q+\log b)}\sim\frac ba&gt;1.$$&#xA;For all large enough $q$, $\pi(bq)/\pi(aq) &gt; 1$ as required.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-05T16:13:12.340" CommentCount="2" />
  <row Id="4072" PostTypeId="2" ParentId="4044" CreationDate="2010-09-05T16:14:34.773" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a more formal way to derive the answer: Define $q_n$ to be the probability that you eventually go broke when you start with $n$ dollars. Then $q_0 = 1$, $\lim_{n \to \infty} q_n = 0$, and $q_n = \frac{1}{4} q_{n-1} + \frac{3}{4} q_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equation defines a recursive sequence with constant coefficients, whose general solution is&#xA;$q_n = c_1 + c_2 3^{- n}$ (the factors $1$ and $\frac{1}{3}$ come from solving the quadratic equation of Moron's answer, which is the fundamental equation of the recursion). To satisfy the other two conditions on $q_n$, we see that $c_1 = 1$ and $c_2 = -1$ i.e.&#xA;$q_n = 1 - 3^{-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the probability to go broke starting with one dollar is $q_1 = \frac{2}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-09-05T18:18:30.160" LastActivityDate="2010-09-05T18:18:30.160" CommentCount="2" />
  <row Id="4073" PostTypeId="1" AcceptedAnswerId="4087" CreationDate="2010-09-05T16:30:53.760" Score="0" ViewCount="259" Body="&lt;p&gt;How to build the real number system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-05T16:44:56.553" LastActivityDate="2010-09-06T11:18:05.110" Title="How to build the real number system?" Tags="&lt;real-analysis&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="4074" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T16:38:00.467" Score="1" ViewCount="0" Body="&lt;p&gt;Have you learnt of dedikind cuts. If not please look at this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_cut&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dedekind_cut&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-05T16:38:00.467" />
  <row Id="4075" PostTypeId="2" ParentId="4062" CreationDate="2010-09-05T16:40:11.807" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\quad i^3 + \: i^4 \; + \:\;\cdots\;\: + \; i^k = 0\ \:\iff\: k\:\equiv\: 2 \:\pmod 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, suppose that $\rm\: z \:$ has order $\rm m&gt;1\:.$ Therefore $\rm\; z^n = 1 \iff\ m\:|n\;\;\:$ hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\quad\rm z^j + z^{j+1} + \:\cdots + z^k = 0\;\; \iff \rm\: k \:\equiv\;\; j \:-\: 1 \:\pmod m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\;\;\;\;\rm \displaystyle z^j \  (1+z+\cdots + z^{k-j}) \;=\; z^j \: \frac{1-z^{k-j+1}}{1-z} = 0 \;\iff\; \rm m\:|\:k-j+1\quad\;$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-05T21:04:47.133" LastActivityDate="2010-09-05T21:04:47.133" />
  <row Id="4076" PostTypeId="1" CreationDate="2010-09-05T16:52:45.617" Score="1" ViewCount="92" Body="&lt;p&gt;Hello i am having some problems working out how to attack this assignment, and after spending hours on it, have i resolved to ask you guys here for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been given the following differential equation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y^{4}-16y=u^{1}+u&#xA;y′′′′−16y=u′+u&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And i have been asked to determine the complete real solution. I just completely stuck on this assignment and hope that one of you guys can give me a push in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="1732" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-05T18:25:00.547" LastActivityDate="2010-09-05T18:25:00.547" Title="Determine the complete real solution to a homogeneous differential equation" Tags="&lt;homework&gt;&lt;differential-equations&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="4077" PostTypeId="2" ParentId="4058" CreationDate="2010-09-05T17:04:31.147" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose you only had three random numbers between 0 and 1. Point (x,y,z)=(rand1*10,rand2*20,rand3*30) falls uniformly inside parallelepiped with dimensions 10,20,30.&#xA;Sets x&gt;y&amp;amp;&amp;amp;x&gt;z, y&gt;x&amp;amp;&amp;amp;y&gt;z, z&gt;x&amp;amp;&amp;amp;z&gt;y partition this parallelepiped into 3 parts, as pictured below, and their relative volumes represent the fraction of the time that x, y or z are the maximum value, respectively. Working out expression for the volumes should give you the integral expression David gave above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/extreme-values.gif&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-09-05T17:04:31.147" CommentCount="2" />
  <row Id="4078" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T17:21:35.487" Score="2" ViewCount="0" Body="&lt;p&gt;Alternatively, you can look at Cauchy sequences: &lt;a href=&quot;http://en.wikipedia.org/wiki/Construction_of_the_real_numbers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Construction_of_the_real_numbers&lt;/a&gt; (and other approaches than can be found there).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-05T17:21:35.487" />
  <row Id="4079" PostTypeId="2" ParentId="4067" CreationDate="2010-09-05T17:31:11.483" Score="2" ViewCount="0" Body="&lt;p&gt;The elliptic curve is the image of $\mathbb{C}/\Lambda$ under a certain map.&#xA;Now, each point $(\mathbf{x}_0,\mathbf{y}_0)$ is the image of some point&#xA;$x$ of $\mathbb{C}/\Lambda$. This $x$ is the covering parameter. It's a complex&#xA;number unique up to translation by elements of $\Lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of $\mathbf{x}$ and $\mathbf{y}$ can be regarded as a function on&#xA;$\mathbb{C}/\Lambda$. We are interested in $f=\mathbf{y}&#xA;-1-a\mathbf{x}-b\mathbf{x}^2$. This has four zeros in each fundamental&#xA;region, and so four poles. If the zeros are $x_0,x_1,x_2,x_3$&#xA;and poles are $u_0,u_1,u_2,u_3$ then $x_0+x_1+x_2+x_3-u_0-u_1-u_2-u_3$&#xA;is an element of $\Lambda$. I suppose here, one must show that the poles&#xA;of $f$ are independent of $a$ and $b$, which leads to $x_0+x_1+x_2+x_3$&#xA;being independent of $a$ and $b$ (here $x_0$ is the covering parameter&#xA;for the point $(0,1)$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-05T17:31:11.483" />
  <row Id="4080" PostTypeId="2" ParentId="3987" CreationDate="2010-09-05T17:32:32.490" Score="4" ViewCount="0" Body="&lt;p&gt;I would post this as a comment if I could, because it seems too simple for an answer. You already said that you can find the highest bit with a single instruction (if the number fits into a register). For every result $r$ you get (that is, your number is between $2^r$ and $2^{r+1}-1$), there are only two possible numbers $k$ and $k+1$ of decimal digits. So you could just use $r$ as an index into a lookup table that stores both $k$ and $10^k$, and output $k$ if your number is less than $10^k$, or $k+1$ otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-09-05T17:32:32.490" CommentCount="3" />
  <row Id="4082" PostTypeId="1" AcceptedAnswerId="4104" CreationDate="2010-09-05T19:00:15.490" Score="0" ViewCount="73" Body="&lt;p&gt;Given 3 pairs of coordinates, x1, y1, x2, y2, x3, y3, I need a function y(x) that will return the y coordinate of any x coordinate between x1 and x3 (it can be assumed that x1 &amp;lt; x2 &amp;lt; x3).&lt;br&gt;&#xA;y(x1) must equal y1, y(x2) must equal y2, and y(x3) must equal y3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, it must never return a y coordinate that is larger than the largest of y1, y2, and y3, or smaller than the smallest of y1, y2, and y3. This is the part which I have not yet figured out how to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is not schoolwork. I'm writing an audio creation library, and these curves will be used to define volume over time.)&lt;/p&gt;&#xA;" OwnerUserId="1734" LastEditorUserId="1734" LastEditorDisplayName="" LastEditDate="2010-09-05T19:17:14.653" LastActivityDate="2010-09-06T01:54:02.867" Title="equation of a curve given 3 points and additional (constant) requirements" Tags="&lt;functions&gt;&lt;algebraic-curves&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4083" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T19:08:45.663" Score="10" ViewCount="0" Body="&lt;p&gt;Irene, if you are looking for intution then let us &lt;em&gt;assume&lt;/em&gt; that we can expand $f(\zeta)$ into a power series around $z$: $f(\zeta) = \sum_{n \geq 0} c_n(\zeta - z)^n$. Note $c_0 = f(z)$.&#xA;If you plug this into the integral and interchange the order of integration &#xA;and summation then that integral on the right side of the formula &#xA;becomes $\sum_{n \geq 0} \int c_n(\zeta - z)^{n-1}d\zeta$.  Let us also assume that an integral along a contour doesn't change if we deform the contour continuously through a region where the function is &quot;nice&quot;. So let us take as our path of integration a circle going once around the point $z$ (counterclockwise). Then you are basically reduced to &#xA;showing that $\int (\zeta - z)^{m}d\zeta$ is 0 for $m \geq 0$ and is $2\pi i$ for $m = -1$.&#xA;These can be done by direct calculations using polar coordinates with $\zeta = z + e^{it}$.  Now divide by $2\pi i$ and you have the formula.  Of course this is a hand-wavy argument in places, but the question was not asking for a rigorous proof.  Personally, this is how I first came to terms with understanding how Cauchy's integral formula could be guessed.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-09-05T19:08:45.663" />
  <row Id="4084" PostTypeId="1" AcceptedAnswerId="4086" CreationDate="2010-09-05T19:12:39.017" Score="1" ViewCount="48" Body="&lt;p&gt;Say you have a set of points and joining lines as show below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xglug.png&quot; alt=&quot;t junctions are at D, F, and E&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this diagram of triangles, T-(or Y-like) junctions are at D, F, and E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At vertex F for example, there is a triangle FDE, but no line FJ.  If there was a line FJ, then there would be no T junction at F. Two triangles would be formed instead out of triangle-ACJ as tri-AFC and tri-FCJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you find the vertices that form &quot;T&quot; (or Y-like) junctions?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-11-04T12:59:47.517" Title="How do you find T-Junctions in an array of triangles?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4085" PostTypeId="2" ParentId="4048" CreationDate="2010-09-05T19:14:08.277" Score="4" ViewCount="0" Body="&lt;p&gt;I'll make my comment into an answer so that it can be marked off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is probably not whether a formal Laurent series is well-defined by itself, but rather whether the operations you are defining on formal Laurent series (and in particular the operation of multiplication of two formal Laurent series) is well-defined, in the sense that if you take any two formal Laurent series, then the definition of &quot;product&quot; will in fact yield a formal Laurent series. When you work with formal power series (only nonnegative exponents), one usually defines the product by:&#xA;$$\left(\sum_{n=0}^{\infty}a_nx^n\right)\left(\sum_{n=0}^{\infty}b_nx^n\right) = \sum_{n=0}^{\infty}\left(\sum_{i+j=n}a_i+b_j\right)x^n.$$&#xA;In this case, since $a_k=b_k=0$ if $k&amp;lt;0$, then the definition makes sense, as each term on the right hand side is a finite sum, which makes sense. If you try doing the same thing with formal Laurent series, where the index runs from $-\infty$ to $\infty$, then it is not obvious that this definition always results in something that you can call a formal Laurent series. So one needs to check that it does in fact yield a formal Laurent series, and that the product so defined makes the set into a ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, the issue of &quot;&lt;em&gt;well-defined&lt;/em&gt;&quot;ness is not like the one when you define a function in terms of representatives (in terms of the &quot;name&quot; of an object when the object may have many different names), but rather in terms of whether the function actually makes sense for every input and yields an appropriate output. It is the same issue that arises if you try to define a map $f\colon(0,1)\to\mathbb{N}$ by taking a number in decimal expansion $0.a_1a_2a_3\ldots$ and &quot;defining&quot; $f(0.a_1a_2a_3\ldots) = \cdots a_3a_2a_1$; this map is not well-defined because for some inputs the output does not lie in the range or does not makes sense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-05T19:35:57.807" LastActivityDate="2010-09-05T19:35:57.807" />
  <row Id="4086" PostTypeId="2" ParentId="4084" CreationDate="2010-09-05T19:15:36.903" Score="0" ViewCount="0" Body="&lt;p&gt;My first stab at it is to consider each triangle in the diagram, as well as all its adjacent triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider tri-FEC first.  FEC is adjacent to ACJ, but only &lt;strong&gt;one&lt;/strong&gt; vertex is shared (C).  If there is to be no T-junction, then two vertices should have been shared between adjacent triangles.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditorDisplayName="" LastEditDate="2010-11-04T12:59:47.517" LastActivityDate="2010-11-04T12:59:47.517" />
  <row Id="4087" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T19:16:25.460" Score="2" ViewCount="0" Body="&lt;p&gt;Let's start by defining the &lt;em&gt;natural numbers&lt;/em&gt; (a version including 0) by &lt;a href=&quot;http://mathworld.wolfram.com/PeanosAxioms.html&quot; rel=&quot;nofollow&quot;&gt;Peano's Axioms&lt;/a&gt; (sometimes called the Peano Postulates):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;0 is a number.&lt;/li&gt;&#xA;&lt;li&gt;If n is a number, the successor of n is a number.&lt;/li&gt;&#xA;&lt;li&gt;0 is not the successor of a number.&lt;/li&gt;&#xA;&lt;li&gt;Two numbers of which the successors are equal are themselves equal.&lt;/li&gt;&#xA;&lt;li&gt;(induction axiom.) If a set S of numbers contains 0 and also the successor of every number in S, then every number is in S.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, the &lt;em&gt;integers&lt;/em&gt; are the natural numbers and their additive inverses and the &lt;em&gt;rational numbers&lt;/em&gt; are numbers that can be expressed as a ratio of an integer to a non-zero integer.  The &lt;strong&gt;real numbers&lt;/strong&gt; are the set of numbers that are limits of &lt;a href=&quot;http://mathworld.wolfram.com/CauchySequence.html&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequences&lt;/a&gt; of rational numbers (which is more or less a convenient way of saying that the sequence will converge in the reals).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-05T19:16:25.460" />
  <row Id="4088" PostTypeId="2" ParentId="4082" CreationDate="2010-09-05T19:20:09.303" Score="0" ViewCount="0" Body="&lt;p&gt;In general, no 2nd degree polynomial fits your requirements: If you fix its maximum/minimum at the highest/lowest of your three points, just one of the remaining two points would completely determine the curve due to its symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to use this to control volume, there are probably some additional side conditions that you need to consider anyway: If y2 is the highest point, you probably want the curve to have just one local maximum (at x2, then). I would also guess that the slope at x1 and x3 should be 0 (though that depends). In any case, you probably don't want a polynomial. The simplest solution might be to take pieces of sine curves between $0$ and $\pi$, stretch them according to your values, and attach them at x2. You could also use splines, but then you explicitly need to fix the slope at x2 to be 0.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-09-05T19:20:09.303" />
  <row Id="4089" PostTypeId="2" ParentId="4007" CreationDate="2010-09-05T19:24:10.970" Score="0" ViewCount="0" Body="&lt;p&gt;This one might be another way of approaching the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n$ is a polynomial, so it is the Taylor expansion of itself at every point; then, we can pick any point in $[0,1]$, say 0, and see what the Taylor expansions of your four possibilities are. In this case you'll see that $b)$ has a Taylor expansion at $x =0$ which is exactly the same as you $f_n$, so that's your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, since I did the calculations on my computer, I don't really know how messy it is to find the answer this way. (As Moron pointed out, also a quick check of the simmetry of the functions gives the right answer)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-05T19:24:10.970" />
  <row Id="4090" PostTypeId="1" CreationDate="2010-09-05T19:34:47.277" Score="2" ViewCount="69" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4084/how-do-you-find-t-junctions-in-an-array-of-triangles&quot;&gt;Follow up to&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have triangles as shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xh4D0.png&quot; alt=&quot;triangles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its easy to see that ABC is adjacent to ACJ:  ABC and ACJ share 2 vertices (A and C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However FEC and ACJ are &lt;strong&gt;also&lt;/strong&gt; adjacent, but only one vertex is shared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find if two triangles are adjacent when they differ in size?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-09-06T05:39:26.047" Title="How do I find if two triangles are adjacent, when they differ in size?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="4091" PostTypeId="2" ParentId="4090" CreationDate="2010-09-05T19:42:39.163" Score="0" ViewCount="0" Body="&lt;p&gt;To find all adjacent triangles to ACJ for ex, my first stab at this is to consider all triangles that share &lt;em&gt;at least one vertex&lt;/em&gt; with ACJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, ACJ shares one vertex with ADF and FEC (adjacent because an entire side is shared).  It also shares one vertex with HBA and BIC (not adjacent because they don't share a side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for two triangle t1 and t2 to be adjacent then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At least one vertex must be shared between the two triangles t1 and t2;&lt;/li&gt;&#xA;&lt;li&gt;At least one vertex &lt;em&gt;other than the shared vertex&lt;/em&gt; from t2 must fall on a line that is drawn between the &lt;em&gt;shared vertex&lt;/em&gt; &lt;strong&gt;&lt;em&gt;and an unshared vertex on t1&lt;/em&gt;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So for t1=ACJ and t2=FEC, FEC is adjacent to ACJ because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C is shared between ACJ and FEC&lt;/li&gt;&#xA;&lt;li&gt;F (from t2) falls on the line between C and A (from t1).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus ACJ and FEC are adjacent.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-09-05T19:42:39.163" CommentCount="1" />
  <row Id="4092" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T19:44:19.737" Score="3" ViewCount="0" Body="&lt;p&gt;Expanding on my comment, this result can be translated into:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A surface in R&lt;sup&gt;3&lt;/sup&gt; which satisfies the Maximum-Modulus principle is uniquely determined by specifying it's boundary&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some more intuition imagine an arbitrary fixed boundary in R&lt;sup&gt;3&lt;/sup&gt;, and now try to imagine constructing a surface whose only local maximum's or minimum's occur at this boundary.  It makes sense &quot;intuitively&quot; (at least for me) that the no local maximum or minimum condition is strong enough to force the surface to be unique.  All that is left is to show that the real and imaginary parts of a holomorphic function satisfy this condition.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-10-29T02:15:09.867" LastActivityDate="2010-10-29T02:15:09.867" CommentCount="2" />
  <row Id="4093" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T19:56:49.803" Score="3" ViewCount="0" Body="&lt;p&gt;An exercise that was a big help to me was to compute the integral of 1/z on the closed path consisting of the square from (1,-1) to (1,1) to (-1,1) to (-1,-1) to (1,-1). It illustrates how the sides with varying real differ from the sides with varying imaginary values.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-14T19:18:43.203" LastActivityDate="2010-09-14T19:18:43.203" />
  <row Id="4094" PostTypeId="1" AcceptedAnswerId="4122" CreationDate="2010-09-05T20:05:19.833" Score="2" ViewCount="136" Body="&lt;p&gt;I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-09-06T05:13:05.103" Title="How do you show that $l_p \subset l_q$ for $p \leq q$?" Tags="&lt;analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4095" PostTypeId="2" ParentId="4073" CreationDate="2010-09-05T20:07:28.650" Score="0" ViewCount="0" Body="&lt;p&gt;You have a lot of possibilities, which dependes on what you want to achieve. See, for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Construction_of_the_real_numbers&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, which also reports the following quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Few mathematical structures have&#xA;  undergone as many revisions or have&#xA;  been presented in as many guises as&#xA;  the real numbers. Every generation&#xA;  reexamines the reals in the light of&#xA;  its values and mathematical&#xA;  objectives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-09-05T20:07:28.650" />
  <row Id="4096" PostTypeId="2" ParentId="4094" CreationDate="2010-09-05T20:13:21.533" Score="4" ViewCount="0" Body="&lt;p&gt;I don't think you need to prove the inequality you have in the question; that's a bit too strong. Note that $\{x_n\}\in\ell_p$ if and only if $\left(\sum|x_n|^p\right)^{1/p}$ is finite, if and only if $\sum|x_n|^p\lt\infty$. So you really just need to show that if $\sum|x_n|^p$ is finite, then $\sum|x_n|^q$ is finite, assuming $p\leq q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to remember is two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if $p\leq q$, then for $|x|&gt;1$ you have $|x|^p\leq|x|^q$, but if $|x|&amp;lt;1$, then $|x|^p \geq |x|^q$.&lt;/li&gt;&#xA;&lt;li&gt;$\sum_{n=1}^{\infty}a_n$ converges if and only if for every $m\geq 1$, $\sum_{n=m}^{\infty}a_n$ converges.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-05T20:25:06.690" LastActivityDate="2010-09-05T20:25:06.690" CommentCount="1" />
  <row Id="4097" PostTypeId="2" ParentId="4090" CreationDate="2010-09-05T20:32:15.033" Score="0" ViewCount="0" Body="&lt;p&gt;I'm not so sure about this, but if you considered vectors FC and AC, the triangles are adjacent iff there exists some $x\in\mathbb{R}$ such that $FC = x(AC)$, and every point in AC as well as FC is contained in $L:X=\lambda(A-C)+Q$ given any $Q\in FC$.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-09-05T20:32:15.033" />
  <row Id="4098" PostTypeId="2" ParentId="4054" CreationDate="2010-09-05T21:24:34.693" Score="2" ViewCount="0" Body="&lt;p&gt;Even though I am not totally satisfied by my argument, I would say it is about average values on the curve of integration that is dedicated to analytic functions. &#xA;Lets look at a circle $C$ around $z$: If we choose points $\zeta_1,\ldots,\zeta_n$ on $C$ such that they span a regular $n$-sided polygon the average of their sum is $z$, that is&#xA;$$\sum_{k=1}^n \zeta_k\cdot\frac{1}{n}= z.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is far from Cauchy integral formula, however, since&#xA;$$1 = \frac{1}{2\pi i}\int_C \frac{1}{\zeta-z}d\zeta$$&#xA;we can write the Cauchy formula in the form &#xA;$$\frac{1}{2\pi i}\int_C\frac{f(\zeta)-f(z)}{\zeta-z}d\zeta=0.$$&#xA;Next, if we think of $f$ as a power series around $z$, we can interpret the formula in text as saying that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If we remove the constant term $f(z)$ from $f(\zeta)$ and divide out the factor $\zeta-z$, the remaining will have average 0&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-05T21:24:34.693" />
  <row Id="4099" PostTypeId="1" AcceptedAnswerId="4100" CreationDate="2010-09-05T21:46:36.033" Score="1" ViewCount="62" Body="&lt;p&gt;I'm trying to show that all graphs with 5 vertices, each of degree 2, are isomorphic to each other.  Is there a more clever way than simply listing them all out?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-09-05T22:23:29.500" Title="5-vertex graphs with vertices of degree 2" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="4100" PostTypeId="2" ParentId="4099" CreationDate="2010-09-05T21:56:28.277" Score="3" ViewCount="0" Body="&lt;p&gt;I'm assuming you do not allow multi-edges, as otherwise there is a trivial counterexample (the cycle of length $5$, versus a disconnected graph consisting of a triangle and two vertices joined by two edges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick one vertex $v_0$; it must be joined to two other distinct vertices, which we may call $v_{-1}$ and $v_1$. Each of those must be joined to another vertex; can they be joined to each other? Can they both be joined to the same vertex that is not yet listed? Consider the possibilities. Then see where each of them leads you. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-05T21:56:28.277" />
  <row Id="4101" PostTypeId="2" ParentId="4099" CreationDate="2010-09-05T22:17:54.967" Score="4" ViewCount="0" Body="&lt;p&gt;Consider this simple, powerful theorem: all vertices of a graph have even degree if and only if the set of its edges can be partitioned in cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of edges of your graph $G$ is $e(G) = \frac 1 2 \sum_{x\in V(G)} d(x) = 5$. Observe that there are not isolated vertices and that every cycle has at least three edges. Therefore the only possible partition in cycles is $E(G)$ itself, i.e. all the graphs with the required property are 5-cycles and thus they are isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="365" LastEditorUserId="365" LastEditorDisplayName="" LastEditDate="2010-09-05T22:23:29.500" LastActivityDate="2010-09-05T22:23:29.500" />
  <row Id="4102" PostTypeId="1" CreationDate="2010-09-05T22:30:02.247" Score="3" ViewCount="159" Body="&lt;p&gt;Given sample data $x_1, \ldots, x_n$ generated from a probability distribution $f(x|\theta)$ ($\theta$ being an unknown parameter), a statistic $T(x_1, \ldots, x_n)$ of the sample data is called sufficient if $f(x|\theta, t) = f(x|t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm always kinda confused by this definition, since I think of a sufficient statistic as a function that gives just as much information about $\theta$ as the original data itself (which seems a little different from the definition above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of &lt;em&gt;Bayesian&lt;/em&gt; sufficiency, on the other hand, does mesh with my intuition: $T$ is a Bayesian sufficient statistic if $f(\theta|t) = f(\theta|x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is the first definition of sufficiency important? What does it capture that Bayesian sufficiency doesn't, and how should I think about it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: I believe that every sufficient statistic is also Bayesian sufficient, but not conversely (the reverse implication doesn't hold in the infinite-dimensional case, according to Wikipedia).]&lt;/p&gt;&#xA;" OwnerUserId="1546" LastActivityDate="2010-09-14T19:18:03.517" Title="Sufficient statistics vs. Bayesian sufficient statistics" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="4103" PostTypeId="2" ParentId="4018" CreationDate="2010-09-05T22:39:18.763" Score="3" ViewCount="0" Body="&lt;p&gt;If I remember correctly from my student days, if D is a connected domain in the complex plane, then the ring &lt;em&gt;O&lt;/em&gt;(D) of holomorphic functions on D has the property that every finitely generated ideal is actually principal, but there are ideas tat cannot be finitely generated (such as in Robin's answer above)&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-05T22:39:18.763" />
  <row Id="4104" PostTypeId="2" ParentId="4082" CreationDate="2010-09-05T22:51:06.663" Score="0" ViewCount="0" Body="&lt;p&gt;What you apparently want is a monotonic interpolation scheme. As mentioned already, fitting a single polynomial to your data will result in an interpolant that does not necessarily satisfy your constraints. Using a piecewise polynomial function (a function that has a different polynomial representation at each interval) should do a better job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, the most-used class of piecewise polynomial functions is the piecewise cubic Hermite interpolant, whose representation in an interval $[x_i,x_{i+1}]$ looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_i+y_i^{\prime}\left(x-x_i\right)+c_i\left(x-x_i\right)^2+d_i\left(x-x_i\right)^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_i=\frac{3s_i-2y_i^\prime-y_{i+1}^\prime}{x_{i+1}-x_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_i=\frac{y_i^\prime+y_{i+1}^\prime-2s_i}{\left(x_{i+1}-x_i\right)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_i=\frac{y_{i+1}-y_i}{x_{i+1}-x_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and ${y_i^\prime,y_{i+1}^\prime}$ are the slopes (derivative values) of your interpolant at the corresponding points $(x_i,y_i)$, $(x_{i+1},y_{i+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you will rightly ask, &quot;how do I determine the $y_i^\prime$?&quot; There are a number of schemes to do this: one of the most popular is the cubic spline, which imposes the additional constraint that the interpolant have two continuous derivatives. The problem with this approach is that there can still be extraneous wiggles in your interpolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach to ensure the monotonicity (i.e. the curve representing your data to fit against has no spurious wiggles) of your piecewise cubic Hermite interpolant is to use the Stineman scheme. The formula for computing slopes is a bit more elaborate, so I'll just have to refer you to the &lt;a href=&quot;http://ifile.it/x1yb0qt&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;. As an aside, Stineman also proposes a &lt;em&gt;rational&lt;/em&gt; interpolant you can use with the slope determination scheme in the linked paper; you can try that one instead if for some reason using a piecewise Hermite cubic is unsatisfactory.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-06T01:54:02.867" LastActivityDate="2010-09-06T01:54:02.867" />
  <row Id="4105" PostTypeId="1" CreationDate="2010-09-05T23:37:52.187" Score="2" ViewCount="164" Body="&lt;p&gt;What's a good way of solving the Diffie–Hellman problem when those exchanging the message have chosen a low primitive root $g$ (e.g. $g=3$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you could brute force it but I'm interested in knowing whether there is a formula for solving it when you know $g^a \pmod{p}$ and $g^b \pmod{p}$ as well as $p$ and $g$ of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For those unfamiliar with the Diffie–Hellman problem the integers $g$ and $p$ (with $1 &amp;lt; g &amp;lt; p$ and $p$ being prime), $g^a \pmod{p}$ and $g^b \pmod{p}$ are public. The integers $a$ and $b$ are private integers and we want to calculate the secret key $s = g^{ab} \pmod{p}$.&lt;/p&gt;&#xA;" OwnerUserId="1737" LastEditorUserId="1737" LastEditorDisplayName="" LastEditDate="2010-09-06T09:04:13.473" LastActivityDate="2010-09-06T09:04:13.473" Title="Solving Diffie–Hellman problem for low primitive root" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;&lt;cryptography&gt;" CommentCount="7" />
  <row Id="4106" PostTypeId="1" AcceptedAnswerId="4960" CreationDate="2010-09-05T23:51:00.007" Score="4" ViewCount="303" Body="&lt;p&gt;&lt;b&gt;Bounty update&lt;/b&gt;: this can be solved by change of basis, but I'm intrigued by David's solution relying on Fourier Transform of Dirac Delta function, so the bounty is for whoever finds a way to fix his solution to give the right result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a non-negative real-valued function over $d$-dimensional real vectors as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\mathbf{x})=\exp(-\mathbf{x}' A \mathbf{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $A$ is some symmetric positive definite $d\times d$ matrix. What is the normalization factor to turn this into a valid density over the following set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_d=\{(x_1,\ldots,x_d)\in \mathbf{R}^d | \sum_i x_i=0 \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, David Bar Moshe gives a general solution to computing that integral over space orthogonal to some vector $v$, but I suspect it has a mistake because the answer depends on the norm of $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, suppose $A$ is $d$-by-$d$ identity matrix. Let $v$ be a vector of all ones. Because of symmetry, integrating over space orthogonal to $v$ should be the same as as $d-1$ dimensional Gaussian integral, ie $\pi^{(d-1)/2}$, whereas David's solution gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\pi^{(d-1)/2}}{d}$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-12T21:05:32.623" LastActivityDate="2010-09-18T23:16:59.507" Title="normalization factor for restricted density" Tags="&lt;integral&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="4107" PostTypeId="1" CreationDate="2010-09-05T23:52:36.430" Score="7" ViewCount="343" Body="&lt;p&gt;What are some examples of theories stronger than Presburger Arithmetic but weaker than Peano Arithmetic?  Are all such theories decidable?  If not, by what methods other than Gödelization can undecidability be established?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Both answers have indicated Robinson Arithmetic as an intermediate theory, but I don't understand how it can be &quot;stronger&quot; than Presburger Arithmetic in any reasonable way, because Robinson Arithmetic cannot prove commutativity of addition.  I have attempted to define &quot;stronger&quot; in the comments.  I would appreciate any suggestions for how to clarify the question and this definition.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-07T20:57:18.857" LastActivityDate="2010-09-08T04:11:15.663" Title="What are some examples of theories stronger than Presburger Arithmetic but weaker than Peano Arithmetic?" Tags="&lt;number-theory&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="4108" PostTypeId="2" ParentId="4102" CreationDate="2010-09-06T00:47:27.640" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose, we can factorize the pdf as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x|\theta) = h(x) g(t(x),\theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can achieve the above factorization then it is obvious that the estimate of $\theta$ will depend only on the value of $t(x)$. In other words, our estimate of $\theta$ will be the same for two sets of data ($x_1$, $x_2$) if $t(x_1) = t(x_2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above feature of $t(x)$ is conceptualized by calling $t(x)$ as a sufficient statistic. It is sufficient in the sense that once we know $t(x)$ we implicitly have an estimate of $\theta$ and is termed as a statistic as it is a function of the data $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a bit unsure about the relationship with bayesian sufficiency but will update my answer if I have some relevant thoughts on that issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way your definition of sufficient statistic does not seem to make sense to me as the right handside should depend on $\theta$.&lt;/p&gt;&#xA;" OwnerUserId="116" LastActivityDate="2010-09-06T00:47:27.640" CommentCount="3" />
  <row Id="4109" PostTypeId="1" CreationDate="2010-09-06T00:52:00.360" Score="0" ViewCount="124" Body="&lt;p&gt;I have been working on the following lemma for a few days. Any hints would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;Let $G$ be a finite group&lt;/em&gt;, $H \le G$ &lt;em&gt;and&lt;/em&gt; $P \in Syl_{p}(H)$. &lt;em&gt;If&lt;/em&gt; $N_{G}(P) \subseteq H$ &lt;em&gt;then&lt;/em&gt; $P$ &lt;em&gt;is a Sylow $p$-subgroup of $G$&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Partial Proof (Contrapostive).&lt;/em&gt;&lt;br&gt;&#xA;First assume that $P$ is not a Sylow $p$-subgroup of $G$. Therefore $P$ is properly contained in some Sylow $p$-subgroup of $G$, say $Q$. Next let $x \in Q \setminus P$. Therefore $P^{x} \subseteq Q^{x}=Q.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to show that infact $P^{x} = P$ so that $x \in N_{G}(P)$. Therefore $N_{G}(P) \not \subseteq H$. However I am not sure if this is true. One interesting fact is that Q contains only one Sylow $p$-subgroup of $H$. So if $P^{x} \subset H$ then the proof is complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the feeling that I may be missing something very trivial. Maybe a contradiction proof would work better. &lt;/p&gt;&#xA;" OwnerUserId="1738" LastActivityDate="2010-09-06T02:04:28.817" Title="A Nice Sylow Theory Problem" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="4110" PostTypeId="2" ParentId="4109" CreationDate="2010-09-06T01:26:45.463" Score="3" ViewCount="0" Body="&lt;p&gt;Instead of taking an &lt;em&gt;arbitrary&lt;/em&gt; $x\in Q\setminus P$, note that $N_Q(P)$ must properly contain $P$ (since $Q$ is a $p$-group, and you are assuming that $P\neq Q$). So pick $x\in N_Q(P)\setminus P$. It cannot lie in $H$, as $\langle P,x\rangle$ is a subgroup of $Q$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-06T02:04:28.817" LastActivityDate="2010-09-06T02:04:28.817" />
  <row Id="4113" PostTypeId="2" ParentId="4073" CreationDate="2010-09-06T02:46:08.337" Score="0" ViewCount="0" Body="&lt;p&gt;This is a great article that constructs the real numbers as classes of rational Cauchy sequences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.mit.edu/~tkemp/18.024/R.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.mit.edu/~tkemp/18.024/R.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1605" LastActivityDate="2010-09-06T02:46:08.337" />
  <row Id="4114" PostTypeId="1" AcceptedAnswerId="4116" CreationDate="2010-09-06T02:55:42.890" Score="6" ViewCount="328" Body="&lt;p&gt;In the definition of topological space, we require the intersection of a finite number of open sets to be open while we require the arbitrary union of open sets to be open. why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this has something to do with the following observation: $\cap_{n=1}^{\infty} (-\frac{1}{n},\frac{1}{n}) = \{0\}$ and there is some reason we don't want singletons to be considered open, I am wondering what this reason is. Am I thinking in the right direction here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-06T04:05:37.717" LastActivityDate="2010-09-06T14:50:15.403" Title="Why do we require a topological space to be closed under finite intersection?" Tags="&lt;intuition&gt;&lt;topology&gt;&lt;definition&gt;" AnswerCount="5" CommentCount="8" />
  <row Id="4115" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T03:16:07.613" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strike&gt;Minor correction: $[0,1/n]$ is not open in the usual topology of the real line.  But you could get the same thing with&lt;/strike&gt; The example $\cap_{n=1}^\infty (-1/n,1/n)=${0} is a good one.  And yes, points are not open in the usual topology on the line.  If they were, then every set, being the union of one-point sets, would be open.  This is called the discrete topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this definition can be fairly well motivated by what happens in metric spaces, of which the real line is a special case.  Call a set $U$ open if around each point in $U$ there is a ball (with respect to the metric) centered at that point and contained in $U$.  Then you can show as an exercise that intersections of finite collections of open sets are open, and that arbitrary unions of open sets are open.  However, the example above shows that you can't take arbitrary intersections.  (In the case of the real line, &quot;ball&quot; just means open interval.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does it work so well to define general topological spaces to satisfy this property of metric spaces?  I don't know.  For more on the motivation of defining topologies in terms of &quot;open sets&quot;, you may be interested in &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-06T03:42:00.263" LastActivityDate="2010-09-06T03:42:00.263" CommentCount="0" />
  <row Id="4116" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T03:25:16.147" Score="7" ViewCount="0" Body="&lt;p&gt;You need to think about what the intuition behind open sets are. One way to think about it is through neighborhoods: an open set is a set which is a neighborhood of each of its points. What is a neighborhood of a point? A neighborhood of a point $x$ is a set that contains of all points that are &quot;sufficiently close&quot; to $x$ (what does &quot;sufficiently close&quot; mean? It depends on the situation; you think of different neighborhoods perhaps specifying different degrees of closeness). In particular, any set that contains a neighborhood of $x$ is itself a neighborhood of $x$. And specifying two degrees of closeness specifies another degree of closeness that makes sense (the smaller of the two at any given place, say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: if you think about open sets as sets that are neighborhoods of all of the points they contain, then it is natural that the union of any family of pen sets will be open: each point in the union is one of the open sets, and that open set is a neighborhood, and the union contains that neighborhood and so is itself a neighborhood. So the arbitrary union of open sets should still be open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about intersection? Well, if you take two open sets $O_1$ and $O_2$, and you consider a point $x$ in $O_1\cap O_2$, then $O_1$ contains all points that are $1$-sufficiently close to $x$, and $O_2$ contains all points that are $2$-sufficiently close to $x$ (with &quot;$1$-sufficiently&quot; and &quot;$2$-sufficiently&quot; describing the two degrees of closeness required), so $O_1\cap O_2$ will contains all points that are both $1$-sufficiently and $2$-sufficiently close to $x$, and so it contains all points that are &quot;sufficiently close&quot; to $x$ for some meaning of &quot;sufficiently close&quot;, so it is also an open set. This gives you, inductively, any finite intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about arbitrary intersections? Then you run into trouble, because specfying two degrees of &quot;closeness&quot; gives you a degree of closeness (the smaller one), but an infinite number of degrees of closeness may end up excluding everything! (Just as in your example, taking the intersection of all $(-\frac{1}{n},\frac{1}{n})$, which specify all points that are $\frac{1}{n}$-close to $0$, but the intersection excludes everything). So we don't want to require that an arbitrary intersection of neighborhoods be a neighborhood, and so we don't want to require that an arbitrary intersection of open sets be an open set. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-06T03:25:16.147" />
  <row Id="4119" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T04:48:29.960" Score="3" ViewCount="0" Body="&lt;p&gt;If you think of it in the case of the topology on the real line - as soon as we leave the restriction from finite intersections we would get singletons as you have seen. &#xA;However, any set of real numbers is a union of singletons that is&#xA;$$A= \bigcup_{x\in A} \{ x \}. $$&#xA;It would follow that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;each&lt;/strong&gt; set would be open &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;each&lt;/strong&gt; set would be closed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;only finite&lt;/strong&gt; sets would be compact &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;each&lt;/strong&gt; function $f:\mathbb{R}\to X$ would be continuous, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks Jonas!)&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-09-06T07:14:50.543" LastActivityDate="2010-09-06T07:14:50.543" CommentCount="4" />
  <row Id="4120" PostTypeId="1" AcceptedAnswerId="4151" CreationDate="2010-09-06T04:56:35.983" Score="0" ViewCount="56" Body="&lt;p&gt;Given $C=XF$, and only data for $C$, solving for $X$ and $F$ has been described as &lt;a href=&quot;http://math.stackexchange.com/questions/3905/how-can-i-solve-this-matrix/3908#3908&quot;&gt;not tractable&lt;/a&gt; (which I'm content to believe, but others aren't).  If that's the case, where does the following argument fall apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $C^i=XF^i$ is written:&#xA;$\begin{bmatrix}&#xA;  C_1^i \\&#xA;  \vdots \\&#xA;  C_m^i &#xA;\end{bmatrix} = \begin{align}\begin{bmatrix}&#xA;  X_{1,1} &amp;amp; \dots &amp;amp; X_{1,n} \\&#xA;  \vdots &amp;amp;  &amp;amp; \vdots \\&#xA;  X_{m,1} &amp;amp; \dots &amp;amp; X_{m,n} &#xA;\end{bmatrix}\end{align}\begin{bmatrix}&#xA;  F_1^i \\&#xA;  \vdots \\&#xA;  F_n^i &#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One instance of data for a given $i$ of $C^i$ would produce $m$ equations and $mn+n$ unknowns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$ instances of data for $C^i$ would produce $qm$ equations and $mn+qn$ unknowns.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; the supposition that you need more equations than unknowns is sufficient, why can I contradict others statements of not tractability  by choosing whole number values to satisfy the inequality:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$qm &gt;= mn+qn$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using say $q=6,m=3,n=2$.&lt;/p&gt;&#xA;" OwnerUserId="1067" LastEditorUserId="1067" LastEditorDisplayName="" LastEditDate="2010-09-06T05:14:41.060" LastActivityDate="2010-09-06T16:23:05.307" Title="Looking for a proof of non tractability (or perhaps a definition of tractable) ..." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4121" PostTypeId="2" ParentId="4090" CreationDate="2010-09-06T05:05:24.403" Score="1" ViewCount="0" Body="&lt;p&gt;First, we need a definition of &quot;adjacent plane figures&quot;.  If we take &quot;plane figure&quot; to mean a closed plane curve (the boundary) along with its interior, then one definition might be that two plane figures are adjacent when (a) the intersection of their boundaries is a curve but not a point; (b) their interiors are disjoint.  For polygons (boundaries consist of line segments) any subtle topological issues disappear, so this will certainly do for triangles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, then I think it's probably not too hard to prove this hypothesis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Triangles ABC and DEF are adjacent iff all of these hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) At least one side of ABC (say AB) and one side of DEF (say DE) are identical as lines (not line segments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) At least one of A and B is strictly between D and E, or at least one of D and E is strictly between A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) C and F are on different sides of line AB (which is the same as line DE).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nice thing about this theorem, if true, is that it should not be hard to come up with a simple algorithm to test for adjacency in the Cartesian plane. I suspect the image processing folks have done this and much much more along the same lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the trickier cases...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://djlewis.smugmug.com/photos/996266339_m4xKQ-XL.jpg&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-09-06T05:39:26.047" LastActivityDate="2010-09-06T05:39:26.047" />
  <row Id="4122" PostTypeId="2" ParentId="4094" CreationDate="2010-09-06T05:13:05.103" Score="2" ViewCount="0" Body="&lt;p&gt;You are right @user1736.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a\leq1$ then &#xA;$$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if $p\leq q$ we have $p/q\leq1$&#xA;$$\left(\sum_n |x_n|^q\right)^{1/q}=\left(\sum_n |x_n|^q\right)^{p/qp}\leq \left(\sum_n |x_n|^{q(p/q)}\right)^{1/p}=\left(\sum|x_n|^p\right)^{1/p}$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-06T05:13:05.103" />
  <row Id="4124" PostTypeId="2" ParentId="4044" CreationDate="2010-09-06T05:34:08.040" Score="1" ViewCount="0" Body="&lt;p&gt;the assertion $q_n \to 0$ as $n \to \infty$ in michael ulm's response is perhaps not entirely obvious - and is related to the remark in moron's response: &quot;I suppose we can eliminate $p=1$ somehow..&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'll elaborate a bit more on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose $\alpha$ is the probability of winning $\$1$ and $\beta = 1-\alpha$ is the chance of losing $\$1$. two facts are clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1)\kern{15pt} q_0=1\kern{10pt} {\rm and\ for}\kern{10pt} n\ge 1,\kern{10pt}  q_n = \alpha q_{n+1} + \beta q_{n-1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern{-1.2in}(2) \kern{1.5in}  q_n = q_1^n,\qquad n\ge 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking $n = 1$ in (1) and writing $q_1 = q$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern{-1.7in}(3) \kern{1.5in}  q = \alpha q^2 + \beta,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means $q = 1$ or $q = {\beta\over\alpha}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if $\alpha \le \frac{1}{2},\ q = 1$ [because one root is $\ge 1$], while if $\alpha &gt; \frac{1}{2}$, both roots are admissible answers.&lt;br&gt;&#xA;[this is what moron also concluded.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to rule out the case $q = 1$, we can argue as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose $q = 1$ and let $S_n$ be the player's fortune at time $n$. then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = T_1 = in\kern{-1pt}f \{ n \ge 1 :  S_n = 0\}$ is finite ($wp1$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;now consider just the random walk $\{S_n : n \ge 1\}$ [forgetting about the gambling context]. we can recursively define $T_{k+1} = in\kern{-1pt}f \{n &gt; T_k : S_n = -k\} $. then $T_k$ is finite ($wp1$) for all $k\ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $\{S_{T_k} = 1-k : k \ge 1\}$ is a nonpositive subsequence of $\{S_n\}_1^\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this contradicts the SLLN which $\mathit inter\ alia$ implies that $S_n\to \infty$ when&#xA;$\alpha &gt; \frac{1}{2}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the case $q = 1$ is ruled out, and therefore $q = \frac{\beta}{\alpha} = \frac{1}{3}$ when $\alpha = \frac{3}{4}$. &lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-06T05:34:08.040" />
  <row Id="4125" PostTypeId="1" AcceptedAnswerId="4132" CreationDate="2010-09-06T06:54:04.803" Score="18" ViewCount="967" Body="&lt;p&gt;In the comments to the question: &lt;a href=&quot;http://math.stackexchange.com/questions/3978/ann-mid-bnn-longrightarrow-ab&quot;&gt;$a^n+n|b^n+n \Longrightarrow a=b$&lt;/a&gt;, there was a claim that $5^n+n$ is never prime (for integer $n&gt;0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not look obvious to prove, nor have I found a counter example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this really true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: $5^{7954} + 7954$ has been found to be prime by a computer: &lt;a href=&quot;http://www.mersenneforum.org/showpost.php?p=233370&amp;amp;postcount=46&quot; rel=&quot;nofollow&quot;&gt;http://www.mersenneforum.org/showpost.php?p=233370&amp;amp;postcount=46&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Douglas (and lavalamp)!&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-14T16:24:20.287" LastActivityDate="2010-10-14T21:29:43.933" Title="$5^n+n$ is never prime?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="4" />
  <row Id="4127" PostTypeId="2" ParentId="4106" CreationDate="2010-09-06T07:53:05.780" Score="2" ViewCount="0" Body="&lt;p&gt;First, one may assume that $\mathbf{A}$ is a symmetric matrix, since its antisymmetric part doesn't contribute to the density value. The normalization factor is equal to the reciprocal of the integral over constraint surface. This integral&#xA;can be converted into an integral over the whole of $\mathbb{R}^n$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N^{-1}=\int_{\mathbf{x} \in S_d} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x}) = \frac1{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + \sqrt{-1}k \mathbf{v}^T \mathbf{x})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mathbf{v}$ is a $d$-dimensional vector having all components equal to one. This construction can be understood by exchanging the integration order, then the integration over $k$ generates a Dirac measure concentrated on the constraint surface. The change of order is possible because the integrand is bounded. Performing a square completion of the exponent, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N^{-1}=  \frac1{2\pi}\int_{-\infty}^{\infty} \mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d} \exp(-(\mathbf{x}+\frac{\sqrt{-1}}{2}k\mathbf{A}^{-1}\mathbf{v})^T \mathbf{A}(\mathbf{x}+\frac{\sqrt{-1}}{2}k\mathbf{A}^{-1}\mathbf{v}) - \frac{k^2}{4} \mathbf{v}^T \mathbf{A}^{-2}\mathbf{v})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the Gaussian integral over $\mathbb{R}^d$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N^{-1}= \frac{(2\pi)^{\frac{d-3}{2}}}{2\det(\mathbf{A})^{\frac{1}{2}}}\int_{-\infty}^{\infty} \mathrm{d}k \exp(-\frac{k^2}{4} \mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the Gaussian integral over $k$, we obtain the final result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N^{-1} = \frac{(\pi)^{\frac{(d-1)}{2}}}{\det(\mathbf{A})^{\frac{1}{2} }(\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v})^{\frac{1}{2}}}$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-09-08T08:21:26.127" LastActivityDate="2010-09-08T08:21:26.127" CommentCount="12" />
  <row Id="4129" PostTypeId="2" ParentId="4125" CreationDate="2010-09-06T08:17:01.243" Score="1" ViewCount="0" Body="&lt;p&gt;If n is odd, then 5^n + n is always even because LSD of 5^n is always 5 for n &gt; 0. Hence, for odd n ( n &gt; 0), 5 ^n + n is composite.&lt;/p&gt;&#xA;" OwnerUserId="1744" LastActivityDate="2010-09-06T08:17:01.243" CommentCount="2" />
  <row Id="4130" PostTypeId="2" ParentId="4125" CreationDate="2010-09-06T10:18:41.600" Score="5" ViewCount="0" Body="&lt;p&gt;After reading Douglas S. Stones comment I asked mathematica to check if 5^(2*3977) + (2*3977) is prime and after about 27 seconds, found that it is indeed prime. So the claim $5^n +n$ is never prime is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It turns out the function I used in mathematica is not a deterministic algorithm. However we can still say the claim $5^n +n$ is never prime is false is most likely true.&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-09-06T11:17:45.353" LastActivityDate="2010-09-06T11:17:45.353" CommentCount="4" />
  <row Id="4132" PostTypeId="2" ParentId="4125" CreationDate="2010-09-06T10:35:33.550" Score="18" ViewCount="0" Body="&lt;p&gt;A general rule-of-thumb for &quot;is there a prime of the form f(n)?&quot; questions is, unless there exists a set of small divisors D, called a &lt;em&gt;covering set,&lt;/em&gt; that divide every number of the form f(n), then there will eventually be a prime.  See, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_number&quot; rel=&quot;nofollow&quot;&gt;Sierpinski numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running WinPFGW (it should be available from the primeform yahoo group &lt;a href=&quot;http://tech.groups.yahoo.com/group/primeform/&quot; rel=&quot;nofollow&quot;&gt;http://tech.groups.yahoo.com/group/primeform/&lt;/a&gt;), it found that $5^n+n$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Probable_prime&quot; rel=&quot;nofollow&quot;&gt;3-probable prime&lt;/a&gt; when n=7954.  Moreover, for every n less than 7954, we have $5^n+n$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To actually certify that $5^{7954}+7954$ is a prime, you could use Primo (available from &lt;a href=&quot;http://www.ellipsa.eu/public/misc/downloads.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ellipsa.eu/public/misc/downloads.html&lt;/a&gt;).  I've begun running it (so it's passed a few more pseudo-primality tests), but I doubt I will continue until it's completed -- it could take a long time (e.g. a few months).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: $5^{7954}+7954$ is officially prime.  A proof certificate was given by lavalamp at &lt;a href=&quot;http://www.mersenneforum.org/showpost.php?p=233370&amp;amp;postcount=46&quot; rel=&quot;nofollow&quot;&gt;mersenneforum.org&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-10-14T21:29:43.933" LastActivityDate="2010-10-14T21:29:43.933" CommentCount="12" />
  <row Id="4133" PostTypeId="2" ParentId="566" CreationDate="2010-09-06T10:44:28.233" Score="0" ViewCount="0" Body="&lt;p&gt;Relatively simple, but still fun...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unknown source, other than my friend telling me yesterday&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a room with 50 doors in it, each door leading to a cell that is completely sound-proof and light-proof, and each cell can hold a maximum of 1 person.  There are 50 prisoners.  There also exists a light in the middle of the main room, so that when a prisoner is pulled out of their cell, they could see the light.  There is also a light switch inside the main room which controls the light.  The 50 prisoners are the only people allowed to touch the light switch that controls the light (even the prison guard cannot touch it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before any prisoner is placed in the cells, the prison guard tells them that he will pull out, at random, any one of them at any time, and then put them back in their cell after they get a chance to turn the light on or off or do nothing at all to it. (Only one prisoner can be out of their cell at any given time)  He also guarantees them that he will pull at least one prisoner out at least once a day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is, when any one prisoner knows when all 50 prisoners have been pulled out of their cells at least one time, and he tells the prison guard this fact correctly, all prisoners will be let free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prisoners are allowed one short meeting before they all get placed in their cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also have a (reasonably) infinite amount of time to let the prison guard know when all prisoners have been let out of their cells at least one time.  (say 5 years or something to that effect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one guess allowed ever, so it better be the right answer, or else they will all be fed to the cannibals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method did they come up with to know exactly when all 50 prisoners have been out of their cells at least one time?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-09-06T10:44:28.233" CommunityOwnedDate="2010-09-06T10:44:28.233" CommentCount="1" />
  <row Id="4134" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T10:57:31.373" Score="3" ViewCount="0" Body="&lt;p&gt;I think that one should have played around with metric spaces before dealing with topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite intersection of open balls centered at $x$ is again an open ball; the radius is just the minimum of the finitely many radia. But if you take infinitely many balls, the radius is the infimum of infinitely many positive numbers, which will be $0$ in many situations (for example if the radia are $1/n$). A similar argument shows that you cannot prove in general that an infinite intersection of open subsets in a metric space is open again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since topological spaces are the natural generalization of metric spaces (replacing concrete distances by the concept of neighborhoods), the same is true for topological spaces and in the definition only finite intersections of open sets are allowed.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T10:57:31.373" />
  <row Id="4135" PostTypeId="1" CreationDate="2010-09-06T10:59:16.043" Score="2" ViewCount="80" Body="&lt;p&gt;The following is the start of basic corollary in my logic text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any set $\Gamma$ of sentences, $\Gamma\subseteq\text{Th}(\text{Mod}(\Gamma))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when $\text{Mod}(\Gamma)$ is empty? Suppose $\Gamma={p_1,\neg p_1}$, so that there is no truth value $V$ which models $\Gamma$, or otherwise $V(p_1)=\text{T}=V(\neg{p_1})$, which doesn't respect logical connectives. Then I would have something like $\Gamma\subseteq\text{Th}(\emptyset)$. If I'm working under the definition that $\text{Th}(\emptyset)$ is the set of all tautologies, then this doesn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should there be an extra condition that $\text{Mod}(\Gamma)$ be nonempty, or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-14T05:39:48.133" Title="Does this proposition hold if $\text{Mod}(\Gamma)=\emptyset$?" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" />
  <row Id="4136" PostTypeId="2" ParentId="3145" CreationDate="2010-09-06T11:02:47.323" Score="0" ViewCount="0" Body="&lt;p&gt;$U_i \setminus (Z \cap U_i)$ is open in $U_i$, thus also open in $X$. Then $X \setminus Z = \cup_ {i \in I} U_i \setminus (Z \cap U_i)$ is open in $X$, i.e. $Z$ is closed in $X$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T11:02:47.323" />
  <row Id="4137" PostTypeId="2" ParentId="4135" CreationDate="2010-09-06T11:05:42.363" Score="2" ViewCount="0" Body="&lt;p&gt;$Th(\emptyset)$ consists of every sentence: $S \models \phi$ means that $M \models \phi$ for every model $M$ in $S$, but in the case of $S=\emptyset$ this is an empty condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, the proof of $\Gamma \subseteq Th(Mod(\Gamma))$ is straight forward with the definitions and does not use $Mod(\Gamma) \neq \emptyset$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T11:05:42.363" CommentCount="2" />
  <row Id="4138" PostTypeId="2" ParentId="4073" CreationDate="2010-09-06T11:18:05.110" Score="0" ViewCount="0" Body="&lt;p&gt;The construction of $\mathbb{R}$ depends on your theory from which you want to start. For example it is reasonable for analysis to start with the axiom &quot;There is a complete ordered field&quot;, show that this field is actually unique up to unique isomorphism and call it $\mathbb{R}$. Then you may define the other number systems, too. Or we start from $\mathbb{N}$ and use Grothendieck constructions and completions to get $\mathbb{R}$ (see the other answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, we use $ZFC$ to make all these constructions work. There we have no chance to define(!) the natural numbers by means of the Peano axioms (because we have to prove the Peano axioms from $ZFC$ for $\omega$, which may be defined as the smallest inductive set). So take care of Isaac's first sentence. Also, if we start with a postulated complete ordered field, this actually follows from $ZFC$, but for the basics of analysis, this does not play any role and may be postponed to the next algebra or logic course.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-06T11:18:05.110" />
  <row Id="4139" PostTypeId="2" ParentId="3876" CreationDate="2010-09-06T11:19:52.243" Score="1" ViewCount="0" Body="&lt;p&gt;Quinn and Hannan's &lt;a href=&quot;http://books.google.com.au/books?id=mTluO2TZ8V8C&amp;amp;dq=estimation+and+tracking+of+frequency&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=4s2ETMuYIYqHcYa-pNAL&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CCoQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The estimation and tracking of frequency&lt;/a&gt; is dedicated to this topic. I can highly recommend it.&lt;/p&gt;&#xA;" OwnerUserId="567" LastActivityDate="2010-09-06T11:19:52.243" />
  <row Id="4140" PostTypeId="2" ParentId="4054" CreationDate="2010-09-06T12:16:54.660" Score="5" ViewCount="0" Body="&lt;p&gt;The trick is to remember that differentiable is really a way of saying 'locally linear', that is, close to $z$ we have $$f( \zeta )= f(z) + (\zeta -z)f'(z) +g( \zeta )$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $g(\zeta)$ is an error term that vanishes as we get close to $z$ (in fact it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Little_o#Little-o_notation&quot; rel=&quot;nofollow&quot;&gt;$o(\zeta -z)$&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, dividing through by $(\zeta -z)$ as in the integral formula gives expression under the integral sign as: $$\frac{f( \zeta )}{(\zeta -z)}= \frac{f(z)}{(\zeta -z)} + f'(z) +\frac{g( \zeta )}{(\zeta -z)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now integrate this termwise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term still vanishes as $\zeta \to z$ (by the definition of little o), which allows us to ignore it by way of a homotopy argument. You see, since we can shrink the loop $\gamma$ around $z$ arbitrarily small without changing the value of the integral (an old theorem of Cauchy's says that the integral along a closed curve is homotopy invariant $(*)$) and, as the loop gets smaller, we see that the value of the last term on $\gamma$ gets closer to vanishing- and so, morally, its integral over $\gamma$ may be bounded above in modulus by successively smaller $\epsilon$. But, because of homotopy invariance, these $\epsilon$ bound the integral over all such loops- and $\epsilon$ is arbitrary- so the integral of the last term vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term is easy- this integrates to $[\zeta f'(z)]_{\gamma (0)}^{\gamma (1)}$, with $\gamma (0)=\gamma (1)$. Plug it in to see this vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is the interesting one and others have covered it well in their posts, but I'll finish the argument anyways. Having eliminated the other terms (and observing that $f(z)$ is a constant) we have a pair of statements equivalent to the original formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\pi i}\int\frac{f(z)}{\zeta-z}d\zeta =f(z) \iff \int\frac{1}{\zeta-z}d\zeta =2\pi i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last one we can use substitution $u=\zeta-z$ and homotopy to make about the integral over a circle $\gamma (t)= e^{2\pi it}$ around the origin giving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\gamma} \frac{1}{u} du (**)= \int_0^1 \frac{\gamma '(t)}{\gamma (t)}dt= = \int_0^1 \frac{2 \pi i e^{2\pi it}}{e^{2\pi it}}dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last one is a total gift after cancellation and we all go home smiling. We could of course have morally calculated $(**)$ without resorting to technicalities by observing that the primitive of $\frac{1}{u}= log(u)$ and looking at the integrand $[log(u)]_{e^0}^{e^{(2\pi -\epsilon) i}}$ and letting $\epsilon \to 0$. But the 'right' intuition is certainly that of Augusti's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $(*)$, the intuition is another story altogether, but I may add it later if it looks worth it...&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditorDisplayName="" LastEditDate="2010-09-06T15:37:32.027" LastActivityDate="2010-09-06T15:37:32.027" />
  <row Id="4141" PostTypeId="1" AcceptedAnswerId="4143" CreationDate="2010-09-06T12:17:11.820" Score="2" ViewCount="136" Body="&lt;p&gt;According to the definitions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;algebra over a ring&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_over_a_field&quot; rel=&quot;nofollow&quot;&gt;algebra over a field&lt;/a&gt;, are &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_algebra&quot; rel=&quot;nofollow&quot;&gt;Elementary Algebra&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;Boolean Algebra&lt;/a&gt; examples of algebra over a ring or algebra over a field? How can one show that? What are the ring or field that the algebras are with respect to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-06T15:33:49.940" LastActivityDate="2010-09-06T15:33:49.940" Title="Are Elementary Algebra and Boolean Algebra Algebras over a Ring (or Field)?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="4142" PostTypeId="2" ParentId="4107" CreationDate="2010-09-06T12:32:31.550" Score="4" ViewCount="0" Body="&lt;p&gt;Peano arithmetic is relatively strong, but that standards of arithmetics, and decidable theories of arithmetic are fairly unusual.  Here's a &quot;tower&quot; of three theories between the two, of increasing strength:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Robinson_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Robinson's Q&lt;/a&gt;, which is an induction-free axiomatisation of arithmetic, that gives addition and mutliplication as its only function constant.  It cannot prove exponentiation total;&lt;/li&gt;&#xA;&lt;li&gt;Exponential function arithmetic (EFA) allows exponentiation and a very limited form of induction.  It cannot prove iterated exponetiation (a.k.a. &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetration&quot; rel=&quot;nofollow&quot;&gt;tetration&lt;/a&gt;) total;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_recursive_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Primitive recursive arithmetic&lt;/a&gt; is sometimes axiomatised as Peano arithmetic but with induction restricted to formulae without universal quantification (it is more usually given as a purely equational theory).  It cannot prove the Ackermann function total.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note how I've distinguished them by which arithmetic operators thay have, and which they cannot prove total.  This is a very natural measure of  the strength of theories of arithmetic: an essential part of proving the incompleteness theorem is providing a diagonalisation operation, which we can do only if we have multiplication.  These theories have this, and so all of these theories admit the incompleteness theorems.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-06T12:32:31.550" />
  <row Id="4143" PostTypeId="2" ParentId="4141" CreationDate="2010-09-06T14:33:09.603" Score="4" ViewCount="0" Body="&lt;p&gt;This should be a comment, I guess...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a silly remark: every algebra over a field is an algebra over a ring, because fields are rings!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A&lt;/em&gt; boolean algebra (in the technical sense of the term, more or less explained in the page &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_(structure&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Boolean_algebra_(structure&lt;/a&gt;) ) is an algebra over the field of two elemetns $\mathbb F_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;em&gt;Boolean algebra&lt;/em&gt; which is the subject of the Wikipedia page you linked to, is not an algebra in that sense. Likewise, &lt;em&gt;Elementary algebra&lt;/em&gt;, the subject of the other page you linked to, is also not an algebra in the technical sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Boolean algebra&lt;/em&gt; and &lt;em&gt;Elementary algebra&lt;/em&gt;, as the terms are used in the Wikipedia pages you linked to, are &lt;em&gt;subjects&lt;/em&gt;, not algebraic structures.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-06T14:33:09.603" CommentCount="1" />
  <row Id="4144" PostTypeId="2" ParentId="4141" CreationDate="2010-09-06T14:40:54.620" Score="1" ViewCount="0" Body="&lt;p&gt;Disclaimer: I still don't have a great grasp on a lot of abstract algebra, but I'd like to try and help when I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I think Mariano Suárez-Alvarez's comment was more to the point by addressing the issues of structures vs. subjects&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble with the boolean algebra you linked to is that there aren't inverse operations and neutral elements to union and intersection because with intersection and union each element is its own neutral element while the axioms of a field or ring state that there is one neutral element that works for your whole ring or field. Without neutral elements, you can't define inverses because the result of 'multiplying' or 'adding' an element with its inverse element is supposed to be the multiplicatively neutral or additively neutral element. So boolean algebra can't be an algebra over a ring or an algebra over a field since it is neither a ring nor field since there are no neutral elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand elementary algebra (assuming you are working with either the real or complex numbers) is a ring and also a field. We can make elementary algebra into an algebra over a field by letting the vector space be the field itself and the axioms will be satisfied; however, this is technically adding additional structure to elementary algebra (the vector space) and so its probably no technically considered 'elementary algebra.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if there are any unclear points or errors :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-06T14:50:31.693" LastActivityDate="2010-09-06T14:50:31.693" />
  <row Id="4145" PostTypeId="2" ParentId="4114" CreationDate="2010-09-06T14:50:15.403" Score="2" ViewCount="0" Body="&lt;p&gt;It is not that we &lt;em&gt;require&lt;/em&gt; open sets to be closed under finite intersection...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitions are not invented out of thin air: people had topological spaces long before anyone thought of coming up with the definition of topological spaces and in the examples they had available, open sets were closed under finite intersection and that property was in fact quite useful to get things done. In other words, the definition of topological spaces (and the requirement you mention in the title along with it) was abstracted from examples, not imposed on them.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-06T14:50:15.403" CommentCount="3" />
  <row Id="4146" PostTypeId="1" CreationDate="2010-09-06T15:25:53.557" Score="4" ViewCount="184" Body="&lt;p&gt;Hello everybody,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for somebody having a quite strong background in Mathematics, which are some good books for the domain of Operations research? I guess there are textbooks covering topics like linear and nonlinear optimization, convex optimization and quadratic programming, dynamic programming, multicriterial optimizations (did I miss something?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Lucian &lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2010-09-06T21:45:45.530" CommunityOwnedDate="2010-09-06T15:25:53.557" Title="Operations research book to start with" Tags="&lt;books&gt;&lt;optimization&gt;&lt;convex-analysis&gt;&lt;nonlinear-optimization&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="4147" PostTypeId="2" ParentId="4054" CreationDate="2010-09-06T15:28:06.283" Score="6" ViewCount="0" Body="&lt;p&gt;Cauchy's Formula has a remarkable interpretation in terms of hyperbolic geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand it, you need to know very little about hyperbolic geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you only need to know that if you define the &quot;plane&quot; as the open unit disk $D$ in $\mathbb C$, and the &quot;lines&quot; as those arcs of circle inside $D$ which are orthogonal to the unit circle $\partial D$, you get a model for the hyperbolic plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[There is a first miracle: the orientation preserving isometries of $D$ into itself are precisely the holomorphic automorphisms of $D$.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is holomorphic in a neighborhood of the closure of $D$, and $z$ is in $D$, then Cauchy's Formula says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$f(z)$ is the average of $f$ on $\partial D$ &quot;as you see it from $z$&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By this I mean that $f(z)$ is the integral of $f$ on $\partial D$ with respect to the measure which assigns to an arc in $\partial D$ the number $\theta/2\pi$ where $\theta\in[0,2\pi]$ is the angle between the (hyperbolic) half-lines going from $z$ to the end points of the arc. (That's the measure you think the arc has, as a part of your horizon, if you look at it from $z$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1 OF NOV 1, 2010.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to clarify the relationship between the Cauchy's and Poisson's Formulas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $|z|&amp;lt;1$ and $\theta$ real define the &lt;strong&gt;Cauchy kernel&lt;/strong&gt; by &#xA;$$C(\theta,z):=\frac{1}{2\pi}\ \frac{e^{i\theta}}{e^{i\theta}-z}\quad,$$&#xA;and define the &lt;strong&gt;Cauchy transform&lt;/strong&gt; of a continuous function $g$ on the unit circle $\partial D$ by &#xA;$$(Cg)(z):=\int_0^{2\pi}g(e^{i\theta})\ C(\theta,z)\ d\theta$$ &#xA;for $|z|&amp;lt;1$. In particular Cauchy's Formula says that if $g$ is the boundary value of a &lt;strong&gt;holomorphic&lt;/strong&gt; function $f$ on $D$, then $Cg=f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $|z|&amp;lt;1$ and $\theta$ real define the &lt;strong&gt;Poisson kernel&lt;/strong&gt; by &#xA;$$P(\theta,z):=C(\theta,z)+\overline{C(\theta,z)}-\frac{1}{2\pi}\quad,$$&#xA;and define the &lt;strong&gt;Poisson transform&lt;/strong&gt; of a continuous function $g$ on the unit circle $\partial D$ by &#xA;$$(Pg)(z):=\int_0^{2\pi}g(e^{i\theta})\ P(\theta,z)\ d\theta$$ &#xA;for $|z|&amp;lt;1$. In particular Poisson's Formula says that if $g$ is the boundary value of a &lt;strong&gt;harmonic&lt;/strong&gt; function $f$ on $D$, then $Pg=f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2 OF NOV 1, 2010.&lt;/strong&gt; I stole this from Bill Thurston: go to page 180 of (or search for &quot;visual&quot; in) &lt;a href=&quot;http://www.math.unl.edu/~mbrittenham2/classwk/990s08/public/thurston.notes.pdf/8a.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.unl.edu/~mbrittenham2/classwk/990s08/public/thurston.notes.pdf/8a.pdf&lt;/a&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT OF NOV 2, 2010.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stricto sensu&lt;/em&gt;, there is no geometric interpretation of the Cauchy kernel, because it is not invariant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $G$ denotes the group of biholomorphic transforms of the open unit disk $D$, then the &lt;strong&gt;Cauchy kernel&lt;/strong&gt; $C(z,\theta)d\theta$, viewed as a 1-form on $D\times\partial D$, is &lt;strong&gt;not&lt;/strong&gt; $G$-invariant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, its restriction to $\{0\}\times\partial D$ &lt;strong&gt;is&lt;/strong&gt; invariant under the stabilizer of $0\in D$ in $G$ (which is the circle group). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, this restriction extends in a unique way to a $G$-invariant 1-form on $D\times\partial D$, and this 1-form is the &lt;strong&gt;Poisson kernel&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, the (complex) vector space of $G$-invariant $(0,1)$-forms on $D\times\partial D$ is one dimensional, generated by the Poisson kernel. [This is because the action of $G$ on $D\times\partial D$ is simply transitive, and the $(0,1)$-forms on $D\times\partial D$ are the sections of a homogeneous line bundle over $D\times\partial D$.] &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-11-02T05:45:36.757" LastActivityDate="2010-11-02T05:45:36.757" CommentCount="9" />
  <row Id="4148" PostTypeId="1" AcceptedAnswerId="4149" CreationDate="2010-09-06T15:36:59.347" Score="5" ViewCount="123" Body="&lt;p&gt;I have a set containing numbers $1$ to $49$ $\{1,2,3, \cdots, 49\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to divide the set into $7$ subsets such that each subset should contain $7$ elements and sum of the elements of each subset should be $175$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to prove that such subsets exist?&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-06T15:54:19.063" LastActivityDate="2010-09-06T15:54:19.063" Title="finding seven disjoint seven element subsets of {1,2, ..., 49} with same sum" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4149" PostTypeId="2" ParentId="4148" CreationDate="2010-09-06T15:45:47.597" Score="11" ViewCount="0" Body="&lt;p&gt;Lookup &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_square&quot; rel=&quot;nofollow&quot;&gt;Magic Squares&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;22  47  16  41  10  35  4&#xA;5    23  48  17  42  11  29&#xA;30  6   24  49  18  36  12&#xA;13  31  7   25  43  19  37&#xA;38  14  32  1   26  44  20&#xA;21  39  8   33  2   27  45&#xA;46  15  40  9   34  3   28&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-06T15:45:47.597" />
  <row Id="4150" PostTypeId="2" ParentId="3888" CreationDate="2010-09-06T16:19:07.153" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ Note the equation is $\displaystyle \rm\; \frac{\sigma\:f}f \:=\; \frac{\sigma^4 g}g\ $ for $\rm\ g = x-16 \:,\:\;$ shift  $\rm\;\;\sigma\:f(x) = f(\sigma\:x) = f(2\:x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, written additively $\rm\  (\sigma-1) \; f \;=\: (\sigma^4 - 1) \; g \quad\;$ [see Note below on this additive notation] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore  $\rm\;\; f \:=\: (1+\sigma+\sigma^2+\:\sigma^3) \: g \;=\: (x-16)\:(2x-16)\:(4x-16)\:(8x-16) \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;unique upto a factor of $\rm\;h \in ker(\sigma-1) = \{h: \sigma\:h = h\:\} = \;$  constants, i.e. $\rm deg\;h = 0 \quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\;$ This reformulation of my prior answer is intended to dramatically illustrate the innate symmetry. Its striking simplicity arises precisely from the simple structure of the orbits of the shift automorphism $\rm\: \sigma\;$.  Namely, by orbit decomposition the problem reduces to one on the single orbit of an irreducible polynomial. By exploiting polynomial structure there, we reduce the problem to a trivial polynomial division by $\rm\:\sigma - 1\:$. This nicely illustrates the essence of the ideas employed to solve general difference equations (recurrences) over rational function fields - ideas at the foundation of algorithms employed in computer algebra systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully the symmetry is clearer in the above reformulation. Based on prior comments, and someone reposting my prior solution stripped of the symmetry, I fear my prior answer didn't succeed in explicitly emphasizing the beautiful innate symmetry at the heart of this problem  - a germ of the Galois theory of difference fields. Probably the problem was devised to help spur one to discover this beautiful structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;$\;\;$ The point of the additive notation is to exploit the natural polynomial structure of the action of $\rm\:\sigma\:$ on the multiplicative group generated by the elements $\rm\: \sigma^n \:f\;$ in the orbit of $\rm\: f \;$ under $\rm\:\sigma\:$. This action is best comprehended by examining it on a specific example.$\:$ Namely, recalling that $\rm\;\sigma\;f(x) \:=\; f(2\:x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\begin{align}{}&#xA;\rm \sigma\:(\:f(2^{-2}\: x)^a \; f(2^3 x)^b \;\: f(2^5 x)^c)\; =&amp;amp; \rm\;\;\: f(\:2^{-1} x)^a \;\: f(2^4 x)^b \;\: f(2^6 x)^c \\&#xA;\rm \iff \quad\;\;\: \sigma\;\:(\:(\:\sigma^{-2}\: f\;)^a \;\; (\:\sigma^3\: f\:)^b \;\; (\:\sigma^5\: f\:)^c)\; =&amp;amp; \rm\;\; \;\; (\:\sigma^{-1} \: f\:)^a \;\; (\:\sigma^4 \: f\:)^b \;\; (\:\sigma^6 \: f\;)^c \\&#xA;\rm \iff \quad\;\; \sigma\; (\:a\:\sigma^{-2} \;+\:\;\; b\:\sigma^3 \;+\;\; c\:\sigma^5)\:\; f\quad\; =&amp;amp; \rm\; (\:a\:\sigma^{-1} \: + \:\;\: b\: \sigma^4 \; +\;\;  c\: \sigma^6)\; f \\&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Galois theory literature the above action is frequently written in a highly suggestive exponential form. To illustrate this, below is the key identity in th problem at hand, expressed in this exponential notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\begin{align}{}&#xA;\rm g^{\:\sigma^4-1} \;=\;&amp;amp; \rm g^{\:(1\:+\;\sigma\:+\;\sigma^2\:+\;\:\sigma^3)\:(\sigma-1)} \\&#xA;\iff\quad\quad\rm  \frac{\sigma^4 g}g \;=\;&amp;amp; \rm (g \;\: \sigma\:g \;\:\sigma^2 g \;\:\sigma^3 g)^{\sigma - 1} \;=\; \frac{\phantom{g\;\;\:} \sigma\:g \;\;\: \sigma^2 g \;\;\:\sigma^3 g \;\;\:\sigma^4 g}{g \;\;\:\sigma\:g \;\;\:\sigma^2 g \;\;\:\sigma^3 g\phantom{\;\;\:\sigma^4 g}} \\&#xA;\end{align}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-07T22:03:55.617" LastActivityDate="2010-09-07T22:03:55.617" CommentCount="6" />
  <row Id="4151" PostTypeId="2" ParentId="4120" CreationDate="2010-09-06T16:23:05.307" Score="1" ViewCount="0" Body="&lt;p&gt;I don't think there's a general definition of tractability in the sense you probably mean, other than &quot;unsolvable under the given conditions&quot;; however &quot;given conditions&quot; are very important.&#xA;In practice conditions on a theorem aren't set beforehand but typically arise from working on the problem - you want to show X, but find out that X doesn't hold unless you also make assumption Y, and so then you ask yourself if Y is a reasonable thing to assume in the problem setting you are interested in. In the end you (hopefully) prove a theorem that has as few assumptions as you can get away with, and still is useful for the problem you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is a clear distinction between the two situations &quot;can I very likely get a practically satisfactory solution to (some impossible problem) if I add some extra constraints by applying domain knowledge&quot; and &quot;can I solve a problem that can be proved to be impossible in general&quot;. In your previous post, the problem you gave was impossible in general, but a good enough solution was tractable under mild assumptions. So, tractability is a function of how you choose to work on a problem, and specifically of which conditions or constraints you assume to hold/are happy to accept in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can recommend Imre Lakatos's &quot;Proofs and Refutations&quot;, which makes this point better than I could ever hope to and, IMHO, is one of those books that every intelligent person should read at sometime in their life. &lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-06T16:23:05.307" CommentCount="2" />
  <row Id="4152" PostTypeId="1" AcceptedAnswerId="4153" CreationDate="2010-09-06T17:58:27.650" Score="2" ViewCount="75" Body="&lt;p&gt;How can we show that the number of pairs (where $(a, b)$ is the same as $(b, a)$) with $LCM(a, b) = n$ is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(2e_1+1)(2e_2+1)...(2e_k+1)+1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=p_1^{e_1}\cdot p_2^{e_2}\cdot ...\cdot p_k^{e_k}, p_i\space prime\space \forall 1\leq i\leq k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to guess this formula for a programming problem I was working on, but I'd be interested in a deduction or proof.&lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-09-06T18:10:17.293" Title="Pairs of numbers with a given LCM" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" />
  <row Id="4153" PostTypeId="2" ParentId="4152" CreationDate="2010-09-06T18:10:17.293" Score="4" ViewCount="0" Body="&lt;p&gt;First count the pairs with $\gcd(a,b)=n$ counting $(a,b)$ and $(b,a)$ as&#xA;distinct. You have $a=\prod p_i^{r_i}$ and $b=\prod p_i^{s_i}$&#xA;and need $\max(r_i,s_i)=e_i$ for all $i$. So, given a positive integer $e$,&#xA;how many pairs $(r,s)$ of nonnegative integers are there with $\max(r,s)=e$?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-06T18:10:17.293" CommentCount="2" />
  <row Id="4154" PostTypeId="2" ParentId="3852" CreationDate="2010-09-06T18:10:36.187" Score="1" ViewCount="0" Body="&lt;p&gt;here is a try. we will use  AB = I to show that the columns of $B$ are linearly independent and then use that to show $BA$ is identity on the range of $B$ which is all of the space due to linear independence of the columns of $B$. This implies that $BA$ is identity. linear independence of the columns of  follows if we can show Bx = 0 implies x = 0. assume $Bx = 0, x = Ix = ABx = A0 = 0$. now to show that  $BA$ is an identity on range of  $B$ we have $(BA)Bx = B(AB)x = Bx $ and and we are done.&lt;/p&gt;&#xA;" OwnerUserId="1758" LastEditorUserId="1758" LastEditorDisplayName="" LastEditDate="2010-09-07T12:09:01.310" LastActivityDate="2010-09-07T12:09:01.310" />
  <row Id="4155" PostTypeId="2" ParentId="3438" CreationDate="2010-09-06T18:27:17.693" Score="0" ViewCount="0" Body="&lt;p&gt;I received this solution via email. I am TeXing it out here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can write $n!= 2^{\alpha(n)}\beta(n)$, where $\beta(n)$ is odd. From a theorem of Legendre we have $\alpha(n)=n -\gamma(n)$ where $\gamma(n)$ denotes the sum of ones in the binary representation of $n$. Moreover, have $$ \sum\limits_{k=1}^{\infty} \frac{2^{n_k}}{n_{k}!} = \sum\limits_{n=1}^{\infty} \delta_{n} \frac{2^n}{n!}$$ where $\delta_{n}=1$ if $n=n_{k}$ and $\delta_{n}=0$, otherwise. Suppose, contrary to our claim, if $$\sum\limits_{n=1}^{\infty} \delta_{n} \frac{2^n}{n!}= \frac{p}{q}$$ Write $q=2^{s}t$, where $t$ is odd. Take $N=2^{r} &gt; \max \{t,2^{s+2}\}$, it then follows that $\frac{\beta(n)}{n} \in \mathbb{N}$. Therefore $2^{s}\beta(N)\frac{p}{q}=\frac{\beta(N)}{t} p \in \mathbb{N}$. A simple calculation shows that $N!=2^{N-1}\beta(N)$. Multiplying the equality $$ \frac{p}{q} = \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} + \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^n}{n!}$$ by $2^{s}\beta(N)$, we get $$ 2^{s}\beta(N)\frac{p}{q} = 2^{s}\beta(N) \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} + 2^{s}\beta(N) \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^n}{n!} \quad (\star) $$ Note that $$ 2^{s} \beta(N) \sum\limits_{n=1}^{N} \delta_{n} \frac{2^n}{n!} = 2^{s} \sum\limits_{n=}^{N} \delta_{n} \frac{\beta(N)2^{n}}{2^{\alpha(n)}\beta(n)}$$ Since $\beta(n)$ divides $\beta(N)$ we see that the first term on the right hand side of $(\star)$ is an integer. To obtain contradiction we will show that, $$0 &amp;lt; 2^{s}\beta(N) \sum\limits_{n=N+1}^{\infty} \delta_{n} \frac{2^{n}}{n!} &amp;lt;1$$ Indeed, we have $$2^{s}\beta(N) \sum\limits_{N+1}^{\infty} \delta_{n} \frac{2^n}{n!} = 2^{s-N+1}N! \sum\limits_{n=N+1}^{\infty} \frac{2^n}{n!}$$ $$= 2^{s+2} \sum\limits_{n=N+1}^{\infty} \delta_{n}2^{n-N-1} \frac{N!}{n!} &amp;lt; \frac{2^{s+2}}{N+1} \sum\limits_{n=N+1}^{\infty} \Bigl(\frac{2}{N+2}\Bigr)^{n-N-1}$$ $$=\frac{2^{s+2}}{N+1} \frac{N+2}{N} &amp;lt; \frac{2^{s+3}}{N+1} &amp;lt; 1$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-06T18:27:17.693" />
  <row Id="4156" PostTypeId="2" ParentId="4146" CreationDate="2010-09-06T19:13:08.237" Score="2" ViewCount="0" Body="&lt;p&gt;Stephen Boyd and Lieven Vandenberghe's book is popular, and available free online:&#xA;&lt;a href=&quot;http://www.stanford.edu/~boyd/cvxbook/&quot; rel=&quot;nofollow&quot;&gt;http://www.stanford.edu/~boyd/cvxbook/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-06T19:13:08.237" CommunityOwnedDate="2010-09-06T19:13:08.237" />
  <row Id="4157" PostTypeId="2" ParentId="4107" CreationDate="2010-09-06T19:27:59.217" Score="5" ViewCount="0" Body="&lt;p&gt;There are infinitely many different theories stronger than Presburger arithmetic and weaker than Peano arithmetic. The classic examples of theories weaker than Peano arithmetic are determined by adding different amounts of induction to Robinson arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The $\Sigma^0_n$ induction axioms say that any set of numbers definable by a $\Sigma^0_n$ formula that contains 0 and is closed under successor contains every natural number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The $\Sigma^0_n$ bounding axioms say that any there is no $\Sigma^0_n$ definable function on a bounded initial segment of the natural numbers with an unbounded range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory Q of &quot;Robinson arithmetic&quot; consists of a small number of axioms about the arithmetical operations. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory I$\Sigma^0_n$ is Q plus the $\Sigma^0_n$ induction axioms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory B$\Sigma^0_n$ is Q plus the $\Sigma^0_n$ bounding axioms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is known that, for each $n$, I$\Sigma^0_{n+1}$ implies B$\Sigma^0_{n+1}$, which in turn implies I$\Sigma^0_n$, and the implications are all strict.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these theories is decidable, nor is any other consistent extension of Robinson arithmetic. The first incompleteness theorem applies to all these theories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard proof of the second incompleteness theorem applies to all the theories I$\Sigma^0_1$ and stronger.  The second incompleteness theorem can be proved for some weaker systems as well, but things get very technical for systems weaker than I$\Sigma^0_1$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-06T19:27:59.217" CommentCount="7" />
  <row Id="4158" PostTypeId="1" CreationDate="2010-09-06T19:30:50.797" Score="3" ViewCount="72" Body="&lt;p&gt;&lt;sup&gt;(This question is inspired by the &lt;a href=&quot;http://stackoverflow.com/questions/3653475/converting-prime-numbers&quot;&gt;StackOverflow question &lt;em&gt;Converting prime numbers&lt;/em&gt;&lt;/a&gt; by &lt;a href=&quot;http://stackoverflow.com/users/158163/manas&quot;&gt;@Manas&lt;/a&gt;, which turns out to be the &lt;a href=&quot;http://www.spoj.pl/problems/PPATH/&quot; rel=&quot;nofollow&quot;&gt;problem &lt;code&gt;PPATH&lt;/code&gt; of SPOJ&lt;/a&gt;.)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's construct a graph &lt;em&gt;G&lt;/em&gt; of all primes &lt;em&gt;p&lt;/em&gt; having &lt;em&gt;L&lt;/em&gt; digits in base-&lt;em&gt;n&lt;/em&gt;. Two primes &lt;em&gt;p&lt;/em&gt; and &lt;em&gt;q&lt;/em&gt; form an edge in &lt;em&gt;G&lt;/em&gt; iff they only differ in 1 digit (e.g. 1&lt;b&gt;0&lt;/b&gt;09 and 1&lt;b&gt;1&lt;/b&gt;09 form an edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some (approximated?) condition on &lt;em&gt;n&lt;/em&gt; and &lt;em&gt;L&lt;/em&gt; that &lt;em&gt;G&lt;/em&gt; is connected? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it &lt;em&gt;seems&lt;/em&gt; that when &lt;em&gt;n&lt;/em&gt; is odd &lt;em&gt;G&lt;/em&gt; is usually disconnected, while when &lt;em&gt;n&lt;/em&gt; is even (&gt;2) &lt;em&gt;G&lt;/em&gt; is usually connected.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-09-07T20:11:35.073" Title="Condition for Connectedness of Graph of Prime Numbers Differing only 1 Digit." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4160" PostTypeId="2" ParentId="4146" CreationDate="2010-09-06T19:54:16.820" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://www.math.ucla.edu/~tom/LP.pdf&quot; rel=&quot;nofollow&quot;&gt;Linear Programming&lt;/em&gt; -- &lt;em&gt;A Concise Introduction&lt;/a&gt;&lt;/em&gt; by Thomas S. Ferguson and other ebooks/lecture notes on &lt;em&gt;Optimization&lt;/em&gt; listed in &lt;a href=&quot;http://stochastix.wordpress.com/books/&quot; rel=&quot;nofollow&quot;&gt;Rod Carvalho's web notebook&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:  The classic and complete book by Hillier and Lieberman &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073017795&quot; rel=&quot;nofollow&quot;&gt;Introduction to Operations Research&lt;/a&gt;&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-06T21:45:45.530" LastActivityDate="2010-09-06T21:45:45.530" CommunityOwnedDate="2010-09-06T19:54:16.820" />
  <row Id="4161" PostTypeId="2" ParentId="29" CreationDate="2010-09-06T20:49:31.687" Score="1" ViewCount="0" Body="&lt;p&gt;In case you happen to be an Emacs user, the &lt;a href=&quot;http://www.gnu.org/software/emacs/calc.html&quot; rel=&quot;nofollow&quot;&gt;Emacs Calc&lt;/a&gt; package is a very good programmable stack-based calculator/mini-CAS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&quot;M-x calc&quot; to start, then &quot;i&quot; to open the manual in info format.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-06T20:49:31.687" CommunityOwnedDate="2010-09-06T20:49:31.687" />
  <row Id="4162" PostTypeId="1" AcceptedAnswerId="4164" CreationDate="2010-09-06T20:53:05.990" Score="5" ViewCount="83" Body="&lt;p&gt;This question isn't about the actual substance of math so much as it is about style.  I'm not going to school, so my feedback on these sorts of problems is limited.  How would you improve this proof's readability?  And if this isn't the sort of question that's suitable here, where would be a more appropriate venue?  The problem is pretty straightforward (from little Spivak):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f:\mathbb{R}^2\rightarrow \mathbb{R}$ and $D_2f=0$, show that $f$ is independent of the second variable.  If $D_1f=D_2f=0$, show that $f$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of &quot;independent of the second variable&quot; that we are using is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is independent of the second variable if for each $x\in \mathbb{R}$ we have $f(x,y_1)=f(x,y_2)$ for all $y_1,y_2\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution as I have written it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $f:\mathbb{R}^2\rightarrow \mathbb{R}$ and $D_2f=0$.  Let $x$ be arbitrary.  By integration, $\int f_2(x,y)=\int 0=C$ for some constant $C$.  It follows that $f(x,y_1)=f(x,y_2)=C$ for all $y_1,y_2\in\mathbb{R}$.  Since $x$ was arbitrary, it follows from the definition that $f$ is independent of the second variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $f$ is constant when $D_1f=D_2f=0$ we proceed by contradiction.  Assume that $f$ is not constant.  Then there exists some $x_1,y_1,x_2,y_2$ such that $f(x_1,y_1) \ne f(x_2,y_2)$.  Since $f$ is independent of the second variable, it follows that $f(x_1,y_1)=f(x_1,y_2)$ and $f(x_1,y_2) \ne f(x_2,y_2)$.  By the mean value theorem there exists $x\in(x_1,x_2)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Df_1(x,y_2)=\displaystyle\frac{f(x_2,y_2)-f(x_1,y_2)}{x_2-x_1} \ne 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the desired contradiction.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-06T21:57:34.613" Title="How can I write this proof about partial derivatives better?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4163" PostTypeId="1" AcceptedAnswerId="4166" CreationDate="2010-09-06T21:08:02.197" Score="0" ViewCount="53" Body="&lt;p&gt;If F and G are functors between two arbitrary categories C and D, does a &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_transformation&quot; rel=&quot;nofollow&quot;&gt;natural transformation&lt;/a&gt; η from F to G always exists? What is the condition for its existence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-06T21:21:40.160" LastActivityDate="2010-09-06T21:21:40.160" Title="Existence of Natural Transformation between Functors" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4164" PostTypeId="2" ParentId="4162" CreationDate="2010-09-06T21:16:51.407" Score="4" ViewCount="0" Body="&lt;p&gt;I've marked this as &quot;community wiki&quot; for myself...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument is a bit vague in the first paragraph; you have an indefinite integral, but what you actually want is the function; that is, you want $f(x,y_1)=C$. &lt;strong&gt;Edit:&lt;/strong&gt; The problem with this is that, if you recall, the indefinite integral does not produce a function, but produces a family of functions, namely all the antiderivatives of your function, and not a value of a particular function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be better to use the Fundamental Theorem of Calculus, which you can use to get specific values of a function (or to be more precise, the difference between two values of a function), and write something along these lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f\colon\mathbb{R}^2\to\mathbb{R}$ be a function such that $D_2f=0$. Fix an arbitrary $x$; then for any $y$ we have:&#xA;  $$f(x,y)-f(x,0) = \int_0^y D_2f(x,t)dt = \int_0^y 0dt = 0,$$&#xA;  so for  fixed $x$, $f(x,y)=f(x,0)$ for all $y$. Therefore, $f$ is independent of the second variable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may, of course, replace $0$ with another point above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, why not use the first part? From $D_2f=0$ you know $f$ is independent of the second variable; the same argument, by exchanging $x$ and $y$, shows that from $D_1f=0$ you get that $f$ is independent of the first variable. So you have&#xA;$$f(x_1,y_1) = f(x_1,y_2) = f(x_2,y_2)$$&#xA;for any $(x_1,y_1)$ and $(x_2,y_2)$; first equality because $f$ is independent of the second variable, second because it is independent of the first. Then you don't have to invoke something like the Mean Value Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in any case, let me point out that you seem to be assuming that $(x_1,y_1)\neq(x_2,y_2)$ implies both that $x_1\neq x_2$ and $y_1\neq y_2$, but this is not true.  You do &lt;em&gt;not&lt;/em&gt;  know a priori that $x_2\neq x_1$; all you know is that $(x_1,y_1)\neq(x_2,y_2)$, and therefore &lt;em&gt;either&lt;/em&gt; $x_1\neq x_2$ or $y_1\neq y_2$ (or both). Your argument breaks down if $x_1=x_2$, and you need to consider that case separately. I would also add at some point, in the case $x_1\neq x_2$, that &quot;we may assume without loss of generality that $x_1\lt x_2$&quot;, so that &quot;$x\in(x_1,x_2)$&quot; makes sense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-06T21:57:34.613" LastActivityDate="2010-09-06T21:57:34.613" CommunityOwnedDate="2010-09-06T21:16:51.407" CommentCount="2" />
  <row Id="4165" PostTypeId="2" ParentId="4146" CreationDate="2010-09-06T21:19:56.967" Score="1" ViewCount="0" Body="&lt;p&gt;If you want books that give a broad introduction to operations research (including optimization, queuing theory etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classics are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Operations Research - Ronald Rardin&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Operations Research - Wayne Winston&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are excellent from pedagogical point of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get into linear programming, this book is widely regarded to be one of the best:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear Programming - Vasek Chvatal.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nonlinear programming:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nonlinear programming - Dimitri Bertsekas&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1124" LastActivityDate="2010-09-06T21:19:56.967" CommunityOwnedDate="2010-09-06T21:19:56.967" />
  <row Id="4166" PostTypeId="2" ParentId="4163" CreationDate="2010-09-06T21:21:11.397" Score="3" ViewCount="0" Body="&lt;p&gt;For a natural transformation $\eta$ to exist between $F$ and $G$, you need for each object $C$ of C a morphism in $D$ $\eta(C)\colon F(C)\to G(C)$. So for an easy example in which no natural transformation exists, take D to be a category with two objects, $A$ and $B$, and in which the only arrows are $1_A\colon A\to A$ and $1_B\colon B\to B$ (the two identity arrows). Take your favorite category C with at least one object, and let $F$ be the functor that maps every object of C to $A$ and every arrow of C to $1_A$, and take $G$ to be the functor that maps every object of C to $B$ and every arrow of C to $1_B$. Then there can be no natural transformation form $F$ to $G$, since there are no morphisms from $F(C)$ to $G(C)$ for any $C$ in C.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-06T21:21:11.397" />
  <row Id="4167" PostTypeId="2" ParentId="4163" CreationDate="2010-09-06T21:21:35.730" Score="1" ViewCount="0" Body="&lt;p&gt;have you seen this:  &lt;a href=&quot;http://mathoverflow.net/questions/35731/conditions-for-natural-transformations-to-exist&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/35731/conditions-for-natural-transformations-to-exist&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-06T21:21:35.730" />
  <row Id="4168" PostTypeId="2" ParentId="3750" CreationDate="2010-09-06T21:41:55.050" Score="4" ViewCount="0" Body="&lt;p&gt;I will give the calculus-based solution myself just for the sake of argument, taking note that I am still hoping to obtain a wide variety of other solutions, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, I am going to mentally rotate the tank (or the oil within it) $90^\circ$ clockwise, cut it in half, and center it at $(0,0)$, so that the upper half of the tank is represented by $y=\sqrt{25-x^2}$ and the volume of the oil by $\int_{-5}^h \sqrt{25-x^2}\mathrm{d}x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from simple $A=\pi r^2$ that the total cross-section area of the tank is $25\pi$ and thus for the upper half from -5 to +5 is $12.5\pi$.  I will set the result of the integral to the various 10% proportions of this value, knowing that 10% of the upper half will occur at the same position as 10% of the entire circle etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral is $\frac12\left(x\sqrt{25-x^2} + 25\arcsin\left(\frac{x}{5}\right)\right)$ evaluated from -5 to h, so the equation we need to solve is:&#xA; $\frac12\left(h\sqrt{25-h^2}+25\arcsin(\frac{h}{5})\right)+\frac{25\pi}{4}=12.5\pi P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting values of .1, .2, .3, .4, and .5 successively for $P$ and using various tools to estimate $h$, my results are -3.43424, -2.45931, -1.59846, -0,788681, 0.  Adding 5 to account for the central displacement, rounding off, then reflecting these values for the right (upper) values, confirms the values given earlier by Americo.&lt;/p&gt;&#xA;" OwnerUserId="575" LastEditorUserId="575" LastEditorDisplayName="" LastEditDate="2010-09-07T11:44:57.350" LastActivityDate="2010-09-07T11:44:57.350" CommentCount="2" />
  <row Id="4169" PostTypeId="1" AcceptedAnswerId="4172" CreationDate="2010-09-06T21:50:41.957" Score="0" ViewCount="278" Body="&lt;p&gt;Let $f$ be continuous on $\mathbb{R}$. Then how to find all continuous functions satisfying $f(f(x))=f(x)+x$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-22T02:47:52.827" Title="Solving the functional Equation $f(f(x))=f(x)+x$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="4170" PostTypeId="1" CreationDate="2010-09-06T22:06:31.337" Score="2" ViewCount="284" Body="&lt;p&gt;I just started to learn mathematical logic. I'm a graduate student. I need a book with relatively more examples. Any recommendation?&lt;/p&gt;&#xA;" OwnerUserId="1451" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T01:13:12.783" LastActivityDate="2010-11-26T01:13:12.783" CommunityOwnedDate="2010-09-06T22:55:47.140" Title="Good books on mathematical logic?" Tags="&lt;logic&gt;&lt;big-list&gt;&lt;books&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="4171" PostTypeId="2" ParentId="4102" CreationDate="2010-09-06T22:07:12.740" Score="1" ViewCount="0" Body="&lt;p&gt;this response addresses part of grautur's question: how to think about the property of being &quot;sufficient&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ra fisher suggested the terminology &quot;sufficient&quot; for a statistic $t$ that satisfies the [heuristic] requirement $$\kern{-2.5in}(1)\kern{2.5in} f(x|\theta, t) = f(x|t).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[there is nothing heuristic about this way of writing the requirement if $x$ is discrete. but if $x$ is continuous, the &lt;em&gt;pdf&lt;/em&gt;  notation $f(x|t)$ is usually heuristic. an example of the latter is when $x = (x_1,\dots x_n)$, where the $\{x_i\}$ are $iid$ N($\theta$,1). here the sample mean $\bar x$ is a sufficient statistic and $(1)$ requires considering the conditional &lt;em&gt;pdf&lt;/em&gt; $$\kern{-.5in} (2)\kern{.5in} f(x_1,\dots, x_n|\bar x) \equiv f(x_1 - \bar x,\dots, x_n - \bar x | \bar x) = f(x_1 - \bar x,\dots, x_n - \bar x).$$ the last equality in $(2)$ holds since the sample deviations are independent of the sample mean in the normal case. (this fact gives an instant proof that the sample mean and variance are independent in the normal case.) here the &lt;em&gt;&quot;pdf&quot;&lt;/em&gt; on the RHS of (2) does not exist as an $n-$dimensional &lt;em&gt;pdf&lt;/em&gt; since the joint distribution of the sample deviations is singular - they sum to zero. replacing &quot;$f$&quot; in (1) by &quot;dist&quot;, removes its heuristic nature.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at any rate, the idea of (1) is that, as the conditional distribution of $x|t$ does not depend on $\theta$, one can [in principle] generate a new $x$ - call it $x^*$, say, from the known conditional distribution of $x|t$, so that $x^* \sim x$ for all $\theta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as an illustration, consider the normal example above. here, obtaining an $x^*$  is particularly easy since the deviations are independent of $\bar x$. thus one can generate $n$ &lt;em&gt;iid&lt;/em&gt; N(0,1) variables $z_1,\dots, z_n$ and let $x^* = (x_1^*,\dots. x_n^*)$, where $x_i^* = z_i - \bar z + \bar x : 1\le i\le n$.&#xA;clearly $x \sim x^*$ for all values of $\theta$, so that $x^*$ is just as &quot;good&quot; a sample from the population as the original $x$ for learning about $\theta$. clearly no one would actually want to use the sample deviations for $x^*$ in addition to its sufficient statistic $\bar x^* = \bar x$, as those deviations, the $\{z_i - \bar z\}$, were obtained by a completely extraneous random experiment having nothing to do with the actual data process. one then should agree that the original sample deviations $\{x_i - \bar x\}$ should also not be used, since they can be considered as being generated in the same way, by an extraneous random experiment, where now nature, rather than the statistician, did the generating.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-06T22:07:12.740" />
  <row Id="4172" PostTypeId="2" ParentId="4169" CreationDate="2010-09-06T23:39:25.487" Score="7" ViewCount="0" Body="&lt;p&gt;This one is a problem from a journal or from competitions at the level of the Putnam contest (see reference below).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  $g(x) = x + Af(x)$ satisfies $g(f^n(x))=A^ng(x)$ when $A^2 = A + 1$; consider the cases $n \to \pm \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source for a similar problem, with solution:  &lt;a href=&quot;http://books.google.com/books?id=-CNbGp2ZFXUC&amp;amp;pg=PA21&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=-CNbGp2ZFXUC&amp;amp;pg=PA21&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-22T02:47:52.827" LastActivityDate="2010-10-22T02:47:52.827" />
  <row Id="4173" PostTypeId="1" AcceptedAnswerId="4177" CreationDate="2010-09-06T23:55:24.823" Score="3" ViewCount="129" Body="&lt;p&gt;I asked this question on mathoverflow, but it was deemed too simple, so I'm posting here instead -- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a nice way to characterize an orthonormal basis of eigenvectors of the following $d\times d$ matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{I}-\frac{1}{d} \mathbf{v}\mathbf{v}'$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mathbf{v}$ is a $d\times 1$ vector of 1's. This is similar to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Householder_transformation&quot; rel=&quot;nofollow&quot;&gt;Householder matrix&lt;/a&gt;, except the $v's$ are not normalized. One eigenvector is $\mathbf{v}$ with corresponding eigenvalue 0, remaining eigenvalues should be 1. I'm looking for an expression in terms of unknown d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: this is covariance matrix of uniform multinomial distribution, so expression for orthonormal basis produces a linear transformation that will make variables uncorrelated for large n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: below are 5 orthonormal eigenvectors vectors I get from Gram-Schmidt for d=5...what is the expression for general d? An even bigger example -- columns of &lt;a href=&quot;http://yaroslavvb.com/upload/bigmatrix.png&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; form orthonormal basis for d=20&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{\sqrt{2}},0,0,0,\frac{1}{\sqrt{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{\sqrt{6}},0,0,\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2 \sqrt{3}},0,\frac{\sqrt{3}}{2},-\frac{1}{2 \sqrt{3}},-\frac{1}{2 \sqrt{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2 \sqrt{5}},\frac{2}{\sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update 09/08&lt;/b&gt;&#xA;I came across another interesting characterization, when d=2^k, for some k, then Walsh Functions  form orthogonal basis for this matrix. In particular, let {$\mathbf{x_i}$} represent the list of vectors of binary expansion of integers 1 to d, ie {(0,0,0),(0,0,1),(0,1,0)...}. Then, rows (and columns) of $M$ define the orthonormal basis of matrix in question, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_{ij}=(-1)^{x_i \cdot x_j}$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-13T18:02:19.070" LastActivityDate="2010-09-13T18:02:19.070" Title="Characterizing orthonormal basis of I-c v v'" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="4174" PostTypeId="1" AcceptedAnswerId="4449" CreationDate="2010-09-07T00:19:05.313" Score="1" ViewCount="112" Body="&lt;p&gt;I'm trying to construct a simple proof of the Propositional Interpolation Theorem. For the following, let $At(\phi)$ be the set of sentence symbols that occur in a sentence $\phi$. Suppose that $\psi$ is a tautological consequence of $\phi$, but neither $\neg\phi$ nor $\psi$ is a tautology, and so $\psi$ is not a tautological consequence of $\phi$ for trivial reasons. I want to show that there is some sentence $\gamma$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $\gamma$ is a tautological consequence of $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $\psi$ is a tautological consequence of $\gamma$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) $At(\gamma)\subseteq At(\phi)\cap At(\psi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From iii) $\gamma$ must be constructed from the sentence symbols found both in $\phi$ and $\psi$, so I first showed that $At(\phi)\cap At(\psi)\neq\emptyset$. Since neither $\neg\phi$ nor $\psi$ is a tautology, there must exist truth assignments $V,W$ such that $V(\neg\phi)=F$ and $W(\psi)=F$ and from this $V(\phi)=T$ and $W(\phi)=F$. If $At(\phi)\cap At(\psi)=\emptyset$ we may define a truth assignment for sentence symbols $p_i$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(p_i)=V(p_i)$ if $p_i\in At(\phi)$, $U(p_i)=W(p_i)$ if $p_i\in At(\psi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then $U(\phi)=V(\phi)=T$, but $U(\psi)=W(\psi)=F$, which contradicts the fact that $\psi$ is a tautological consequence of $\phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question now is, is there some sort of method to construct $\gamma$ such that conditions i) and ii) are satisfied?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-07T01:13:47.133" LastActivityDate="2010-09-11T22:42:03.903" Title="Constructing an example that supports the Propositional Interpolation Theorem" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4175" PostTypeId="1" AcceptedAnswerId="4187" CreationDate="2010-09-07T00:21:33.297" Score="8" ViewCount="486" Body="&lt;p&gt;Let $m,n\ge 1$ be two integers. Prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta_{mn}$ stands for the &lt;em&gt;Kronecker's delta&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I put the tag &quot;linear algebra&quot; because i think there is an elegant way to attack the problem using a certain type of matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you will enjoy. :)&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="152" LastEditorDisplayName="" LastEditDate="2010-09-07T06:10:15.277" LastActivityDate="2010-10-14T04:12:32.343" Title="Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="4176" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T00:45:11.897" Score="1" ViewCount="0" Body="&lt;p&gt;Ebbinghaus, Flum and Thomas. &lt;em&gt;Mathematical Logic&lt;/em&gt; (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387942580&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-07T00:45:11.897" CommunityOwnedDate="2010-09-07T00:45:11.897" CommentCount="1" />
  <row Id="4177" PostTypeId="2" ParentId="4173" CreationDate="2010-09-07T00:54:02.997" Score="3" ViewCount="0" Body="&lt;p&gt;If you write $P= I - \frac{1}{d}vv^t$ (I assume your $v'$ means the transpose of $v$), then you certainly have a symmetric matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P^t = \left( I- \frac{1}{d}vv^t\right)^t = I -\frac{1}{d}v^{tt}v^t = I - \frac{1}{d}vv^t = P \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it diagonalizes and has an orthonormal basis of eigenvectors. That is, there exists an orthogonal matrix $S$, ($S^{-1} = S^t$) and a diagonal matrix $D $ such that $S^tP S = D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, since $P^2 = P$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;P^2 &amp;amp;= \left( I- \frac{1}{d}vv^t\right) \left( I- \frac{1}{d}vv^t\right) \\&#xA;    &amp;amp;= I - 2\frac{1}{d}vv^t  + \frac{1}{d^2}vv^tvv^t \\&#xA;    &amp;amp;= I - 2\frac{1}{d}vv^t  + \frac{1}{d}vv^t \\&#xA;    &amp;amp;=I- \frac{1}{d}vv^t = P &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $P \neq 0,I$, its minimal polynomial is $x^2-x = x(x-1)$. So the only eigenvalues of $P$ are $0,1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for eigenvectors, as you say, $v$ is one of them, of $0$ eigenvalue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Pv = \left( I- \frac{1}{d}vv^t\right)v = v - \frac{1}{d}vv^tv = v - v = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, you can take $\frac{v}{\|v\|} = \frac{1}{\sqrt{d}} (1, \dots ,1)$ as the first vector of your orthonormal basis. The other vectors are an orthonormal basis of the orthogonal complement of $v$, $[v]^\bot$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\left( I- \frac{1}{d}vv^t\right)w = w &amp;amp;\quad\Longleftrightarrow \quad w - \frac{1}{d}vv^tw = w \\&#xA;           &amp;amp;\quad\Longleftrightarrow \quad vv^tw = 0  \\&#xA;           &amp;amp;\quad\Longleftrightarrow \quad v^tw = 0&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you only need to compute a basis for $[v]^\bot$ and orthonormalize it. In coordinates you must find the solutions of the linear equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_1 + \dots + x_d = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-1, 1, 0, \dots ,0), (-1, 0, 1, 0, \dots ,0), \dots , (-1, 0, \dots , 0,1) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now you apply the Gram-Schmidt process &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram_schmidt&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gram_schmidt&lt;/a&gt; to these vectors.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT. I think Unkz intuition is right. Before normalizing, the general pattern looks as follows. You have $d-1$ vectors of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-1, 0,\dots,0, i, -1,\dots, -1), \qquad \text{for} \quad i=1,\dots,d-1 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ is in the $d-i+1$ coordinate (so there are $i-1$ coordinates with $-1$ after it) and one more last vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(d,\dots, d) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's check that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Those first $d-1$ vectors belong to $\ker (P-I) = [(1,\dots, 1)]^\bot$ (so they are eigenvectors corresponding to the eigenvalue $1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(1,\dots, 1)\cdot (-1, 0,\dots,0, i, -1,\dots, -1) = -1 + i + (i-1)(-1)= 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Those first $d-1$ are mutually orthogonal vectors. We may assume $i&gt;j$, for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-1, 0,\dots,0, i, -1,\dots, -1) \cdot (-1, 0,\dots,0, j, -1,\dots, -1) = 1 - i +(i-1)(-1)^2 = 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you quotient out their norms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{1 + i^2 + (i-1)} = \sqrt{i^2 + i} = \sqrt{i (i+1)} \qquad \text{for} \quad i= 1, \dots, d-1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{d^2d}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you are done.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-07T03:32:17.257" LastActivityDate="2010-09-07T03:32:17.257" CommentCount="1" />
  <row Id="4179" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T01:05:03.783" Score="1" ViewCount="0" Body="&lt;p&gt;Shoenfield's &quot;Mathematical Logic&quot;.  The notation is a bit dated, but the exercises are great.&lt;/p&gt;&#xA;" OwnerUserId="851" LastActivityDate="2010-09-07T01:05:03.783" CommunityOwnedDate="2010-09-07T01:05:03.783" />
  <row Id="4180" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T01:15:14.367" Score="0" ViewCount="0" Body="&lt;p&gt;François G. Dorais and others made some &lt;a href=&quot;http://mathoverflow.net/questions/15951/propositional-logic-first-order-logic-and-higher-order-logics/15954#15954&quot; rel=&quot;nofollow&quot;&gt;great recommendations to me&lt;/a&gt; some time ago over on MathOverflow. They're fairly high-level (not exactly introductory courses) but they're good reads.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-09-07T01:15:14.367" CommunityOwnedDate="2010-09-07T01:15:14.367" />
  <row Id="4182" PostTypeId="2" ParentId="4068" CreationDate="2010-09-07T01:54:13.063" Score="2" ViewCount="0" Body="&lt;p&gt;In this chain of 4 equations,  #1, #3 and #4 are correct, and no.2 is the mistake. The equations are assertions about (limits of) some finite sums. Let $H_n = \Sigma_{j=1}^n 1/j$ and $A_n = \Sigma_{i=1}^n (-1)^{i-1} 1/i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct formula is $A_n = H_n - H_{[n/2]}$ (which is approximately $\log(n) - \log(n/2) \sim \log(2) = A_{\infty}$) , but the second equation cut off at $n$ terms is claiming $A_n = H_n - H_n$ (which is zero).&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-07T01:54:13.063" CommentCount="2" />
  <row Id="4183" PostTypeId="2" ParentId="4173" CreationDate="2010-09-07T02:25:34.227" Score="2" ViewCount="0" Body="&lt;p&gt;While I haven't taken the time to prove it, if you look at the numbers I think you'll see a very nice pattern if you multiply out all your irrational denominators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,0,0,0,1) norm=$\sqrt{1 \cdot 2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,0,0,2,-1) norm=$\sqrt{2 \cdot 3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,0,3,-1,-1) norm=$\sqrt{3 \cdot 4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1,4,-1,-1,-1) norm=$\sqrt{4 \cdot 5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5,5,5,5,5) norm=$\sqrt{5^2 \cdot 5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to work for a few other random choices of d, where you have -1 in the first column, (1..d) down the backwards diagonal, and -1 under the backwards diagonal, along with 1,1,...,1 on the bottom row.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-07T02:25:34.227" />
  <row Id="4184" PostTypeId="2" ParentId="4175" CreationDate="2010-09-07T02:35:44.010" Score="8" ViewCount="0" Body="&lt;p&gt;The vector space of polynomials in one variable has two bases $\{1, x, x^2, ... \}$ and $\{1, (x+1), (x+1)^2, ... \}$ and I believe what you've written down is equivalent to the statement that the change-of-basis matrices between these two bases multiply to the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still thinking about an inclusion-exclusion argument.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-07T02:35:44.010" CommentCount="1" />
  <row Id="4185" PostTypeId="1" AcceptedAnswerId="4190" CreationDate="2010-09-07T03:42:37.597" Score="3" ViewCount="109" Body="&lt;p&gt;consider the following game: 10 dice are tossed and those showing 3 are more are retained. [those showing 2 or less are discarded.] the remaining dice are tossed again and those showing 4 or more are retained. finally, the remaining dice are tossed once more and those showing 5 or more are retained. &#xA;if $Z$ denotes the number of dice retained after the third round of play, what is the probability that&#xA;$Z = 0$?&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-07T07:49:10.740" Title="a sequential game of dice" Tags="&lt;probability&gt;&lt;recreational-mathematics&gt;" AnswerCount="2" />
  <row Id="4186" PostTypeId="2" ParentId="29" CreationDate="2010-09-07T04:03:57.357" Score="1" ViewCount="0" Body="&lt;p&gt;If you want a quick RPN calculator with print capabilities for standard calculations, I use &lt;code&gt;free42&lt;/code&gt; which is compatible with HP42S (RPN). There is a print key to print when you need and you can configure the file used to save those prints. You can use skins to make it look the way you want, including the original HP calculator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://thomasokken.com/free42/&quot; rel=&quot;nofollow&quot;&gt;free42&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1514" LastActivityDate="2010-09-07T04:03:57.357" CommunityOwnedDate="2010-09-07T04:03:57.357" />
  <row Id="4187" PostTypeId="2" ParentId="4175" CreationDate="2010-09-07T04:05:49.773" Score="7" ViewCount="0" Body="&lt;p&gt;This follows easily from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;Multinomial Theorem&lt;/a&gt;, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 = 1^n = (1 - x + x)^n$$&#xA;$$ = \sum_{a+b+c=n} {n \choose a,b,c} 1^a \cdot (-x)^b \cdot x^c$$&#xA; $$ = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose m,k-m,n-k} 1^{m}  \cdot (-x)^{k-m} \cdot x^{n-k} $$&#xA;  $$ = \sum_{m=0}^{n} (\sum_{k=m}^{n} (-1)^{k-m} {k \choose m}{n \choose k}) x^{n-m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing coefficients now gives the result immediately.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-07T04:05:49.773" />
  <row Id="4188" PostTypeId="2" ParentId="4173" CreationDate="2010-09-07T04:24:16.157" Score="0" ViewCount="0" Body="&lt;p&gt;a slight retouching of unkz's idea is the following orthogonal basis for $v^\perp$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_1 = (1,-1,0,\dots,0),\&#xA;u_2 =  (1,1,-2,0,\dots,0),\&#xA;   \dots, \&#xA;u_{d-1} = (1,\dots,1,-d+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in this form it is pretty easy to see that the $\{u_i\}$ are in $v^\perp$ and are mutually orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i don't quite understand the remark that normalizing the $\{u_i\}$ to get an ONB is the hardest part - since $d$ is unknown.  what sort of expression &quot;in terms of [the] unknown $d$&quot; &lt;em&gt;would&lt;/em&gt; be suitable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: i would post this as a comment - if this @!*&amp;amp;*%#! site would let me.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-09-07T04:24:16.157" />
  <row Id="4189" PostTypeId="2" ParentId="4175" CreationDate="2010-09-07T05:08:29.670" Score="5" ViewCount="0" Body="&lt;p&gt;Here's another way to look at Moron's proof: the sum counts all the partitions of $[n]$ into three sets $A,B,C$ satisfying $|C|=m$, weighted according to $(-1)^{|A|}$. The identity just says that if $n \neq m$, the number of partitions with $|A|$ even is the same as those with $|A|$ odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter fact is proved by the following sign-changing involution: pick the first element which is not in $C$ (there must be one since $n \neq m$), and flip it from $A$ to $B$ or vice versa.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-07T05:08:29.670" />
  <row Id="4190" PostTypeId="2" ParentId="4185" CreationDate="2010-09-07T05:30:59.203" Score="8" ViewCount="0" Body="&lt;p&gt;The probability that a given die survives all three rounds is $(4/6)(3/6)(2/6) = 1/9$. Thus the probability that no dice survive is $(8/9)^{10} \approx 31\%$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-07T05:30:59.203" CommentCount="1" />
  <row Id="4191" PostTypeId="2" ParentId="4170" CreationDate="2010-09-07T05:39:25.800" Score="0" ViewCount="0" Body="&lt;p&gt;You should take a look at &lt;a href=&quot;http://books.google.com/books?id=4u9gQ6pctuIC&amp;amp;printsec=frontcover&amp;amp;dq=van+dalen+logic&amp;amp;source=bl&amp;amp;ots=koMWzUiwUR&amp;amp;sig=Noea7aYSfL54r63KwnunQOB1qRE&amp;amp;hl=en&amp;amp;ei=Vc-FTNL1EMjNswbKq-yaBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CCoQ6AEwBA#v=onepage&amp;amp;q=van%20dalen%20logic&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;D. Van Dalen: Logic and Structure&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-07T05:39:25.800" CommunityOwnedDate="2010-09-07T05:39:25.800" />
  <row Id="4192" PostTypeId="1" AcceptedAnswerId="4196" CreationDate="2010-09-07T06:25:10.123" Score="1" ViewCount="185" Body="&lt;p&gt;How does one find the value of the sum : $$\sum\limits_{n=1}^{p-1} [\sqrt{np} \ ]$$ where $p$ is a prime such that $p \equiv 1 (\text{mod} \ 4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i remember correctly, i got this sometime back, when i was an undergrad.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-07T07:07:19.977" Title="Finding the Value of $\sum\limits_{n=1}^{p-1} [\sqrt{np} \ ]$" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="4193" PostTypeId="1" CreationDate="2010-09-07T06:54:19.977" Score="3" ViewCount="150" Body="&lt;p&gt;In an introductory algebra course, one proves the Structure theorem for finitely generated modules over a principal ideal domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every book that I have looked in, the uniqueness part of the proof is a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a reference for a nicer proof of the uniqueness part? I am especially interested in a proof of uniqueness when the underlying ring is the ring of polynomials in one variable over a field.&lt;/p&gt;&#xA;" OwnerUserId="1725" LastEditorUserId="1725" LastEditorDisplayName="" LastEditDate="2010-09-07T08:02:57.600" LastActivityDate="2010-09-08T18:41:49.100" Title="Proving uniqueness in the structure theorem for finitely generated modules over a principal ideal domain" Tags="&lt;abstract-algebra&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4194" PostTypeId="2" ParentId="4192" CreationDate="2010-09-07T06:56:09.037" Score="4" ViewCount="0" Body="&lt;p&gt;This is the number of pairs $(a,b)$ of integers with $0 &amp;lt; a &amp;lt; p$, $b &gt; 0$&#xA;and $b^2 &amp;lt; ap$. Reversing the implied summation gives&#xA;$\sum_{b=1}^{p-1}(p-1-\lfloor b^2/p\rfloor)$. This will reduce to&#xA;summing the fractional parts of $b^2/p$, which I'll leave you to think about.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-07T06:56:09.037" />
  <row Id="4195" PostTypeId="1" CreationDate="2010-09-07T07:00:17.280" Score="2" ViewCount="72" Body="&lt;p&gt;Given complete fan $\Delta$ defining a projective toric variety (so that $\Delta$ is the normal fan of some polytope). How do one go on to find a defining ideal of the toric variety in projective space?&#xA; Or, going the other way, given binomial equations defining a projective toric variety, is there some way to recover the fan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, I am trying to recover the fan of the hypersurface $Z(x_0x_3-x_1x_2)$ in $\mathbb{P}^n$ ($n&gt;3$).&lt;/p&gt;&#xA;" OwnerUserId="1764" LastEditorUserId="1764" LastEditorDisplayName="" LastEditDate="2010-09-07T17:36:21.443" LastActivityDate="2010-09-07T17:36:21.443" Title="Equations of a projective toric variety" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="4196" PostTypeId="2" ParentId="4192" CreationDate="2010-09-07T07:07:19.977" Score="3" ViewCount="0" Body="&lt;p&gt;If you don't feel like thinking too much, here are all the details for Robin's answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's begin with the formula $\lfloor\sqrt{np}\rfloor = \sharp[C \geq 1 : np \geq C^2]$. Therefore the original sum is equal to $\sharp[n &amp;lt; p, C \geq 1 : np \geq C^2]$. Now we can change the order of summation to $\sum_{C=1}^{p-1} (p - \lceil C^2/p \rceil)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we have $\lceil C^2/p \rceil = C^2/p + (-C^2 \mod p)/p$. Substituting, the sum reduces to $\sum_{C=1}^{p-1} (p - C^2/p) - \sum_{C=1}^{p-1} (-C^2 \mod p)/p$. We can calculate the first summand explicitly using the formula for sum of squares as $p(p-1) - (p-1)(2p-1)/6 = (p-1)(4p+1)/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to calculate the second summand, notice that we are summing over all quadratic residues, twice ($-1$ is a square for our $p$). Since $p \equiv 1 \pmod{4}$, the sum of all quadratic residues equals the sum of all quadratic non-residues, and so the second summand is $-\sum_{k=1}^{p-1} k/p = -(p-1)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we get $(p-1)(4p+1)/6 - (p-1)/2 = (p-1)(2p-1)/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you were wondering, if $p$ were a prime of the other residue class, then the answer would depend on some class number.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-07T07:07:19.977" />
  <row Id="4197" PostTypeId="2" ParentId="4185" CreationDate="2010-09-07T07:49:10.740" Score="1" ViewCount="0" Body="&lt;p&gt;For a much lengthier approach, the problem can be decomposed into three smaller problems: the chance of discarding all the dice in one round, two rounds, or three rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chance of discarding all the dice in the first round is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2/6)^{10}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your chance of having $n$ dice remaining after the first round is $(2/6)^{10-n} (4/6)^n$, and your chance of discarding all the dice at this stage are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^9 (10)choose(n)(2/6)^i (4/6)^{10-i} (3/6)^{10-i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ is the number of dice eliminated in the first pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, your chance of having $m$ dice remaining after the second round is $(10)choose(n)(2/6)^{10-n} (4/6)^n (n)choose(m)(3/6)^{n-m} (3/6)^m=(2/6)^{10-n} (4/6)^n (3/6)^n$, and your chance of discarding the remaining dice at this stage is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^9 \sum_{j=1}^{9-i} (10)choose(i)(2/6)^i (4/6)^{10-i} (10-i)choose(10-i-j)(3/6)^i (4/6)^{10-i-j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i$ is the number of dice eliminated in the first pass and $j$ is the number of dice eliminated in the second pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of those three probabilities gives you the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, it works out to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in first: 0.00001694&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in second: 0.01732459&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in third: 0.29060462&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing all dice in any round: 0.30794615&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-07T07:49:10.740" CommentCount="1" />
  <row Id="4198" PostTypeId="1" AcceptedAnswerId="4210" CreationDate="2010-09-07T10:04:23.990" Score="2" ViewCount="133" Body="&lt;p&gt;I am reading Jech's Set Theory, in particular the chapter about large cardinals. After discussing measurable cardinals he moves on to weakly-compact cardinals, which have been discussed far earlier in the book.&lt;br&gt;&#xA;I went back to the chapter dealing with weakly-compact cardinals and began retracing the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, it came into this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We denote $[k]^n = \{X \subset \kappa | |X| = n\}$. If $\lambda$ is a cardinal, we denote $\kappa \to (\lambda)^2$ when for every partition of $[\kappa]^2$ into $2$ we have $H \subset \kappa$ that is of cardinality $\lambda$, and for which $[H]^2$ is strictly in one part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we say that $\kappa$ is weakly-compact if it satisfies the property $\kappa \to (\kappa)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I'm a bit lost in all those definitions, and not even sure about the $\kappa \to (\lambda)^2$ notation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are, if so, can someone help me make some sense into those definitions, and is there an equivalent definition for weakly-compact cardinals which can help me understand their properties better?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-07T19:28:18.887" Title="Weakly-compact cardinals" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" />
  <row Id="4199" PostTypeId="1" AcceptedAnswerId="4334" CreationDate="2010-09-07T10:04:56.257" Score="6" ViewCount="144" Body="&lt;p&gt;Usually, applying the conversion formulae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r^2=x^2+y^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos\;\theta=\frac{x}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin\;\theta=\frac{y}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to transform an equation in polar coordinates to an implicit Cartesian equation is quite straightforward for an equation of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=f(\cos(n\theta),\sin(n\theta))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $n$ an integer, thanks to multiple angle formulae. Polar equations of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=f\left(\cos\left(\frac{p}{q}\theta\right),\sin\left(\frac{r}{s}\theta\right)\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$, $q$, $r$ and $s$ are integers, and $\frac{p}{q}$ and $\frac{r}{s}$ are in lowest terms, are more difficult to handle, but the case where $q$, $s$ are powers of 2 is slightly eased by the existence of identities like $\tan\left(\frac{\theta}{2}\right)=\frac{\sin\;\theta}{1+\cos\;\theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really difficult ones to handle are the cases like &lt;a href=&quot;http://mathworld.wolfram.com/CayleysSextic.html&quot; rel=&quot;nofollow&quot;&gt;Cayley's sextic&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=4a\cos^3\left(\frac{\theta}{3}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in general, the cases with fractions whose denominators are not powers of 2. In particular for Cayley's sextic, I can't seem to find a quick way to exploit the appropriate multiple angle identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One &quot;cheat&quot; I have seen is to instead represent the polar equation as a pair of parametric equations, and then make the substitution $\theta=3\arctan\;t$ so that everything is represented algebraically. The problem is that apparently it takes some insight to recognize how to remove the new parameter $t$ easily. For even tougher cases like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=\cos\left(\frac{2\theta}{3}\right)-\sin\left(\frac{3\theta}{5}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the appropriate substitution after transforming to parametric equations is $\theta=\mathrm{LCM}(3,5)\arctan\;t$, but the expressions, though algebraic, look even more nightmare-ish to manipulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mathematica&lt;/em&gt; has no trouble figuring out the implicit Cartesian equation, through a judicious use of Gröbner bases:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Factor /@ GroebnerBasis[Append[Thread[{x, y} == TrigExpand[{4a Cos[t/3]^3 Cos[t], 4a Cos[t/3]^3 Sin[t]}]],&#xA;Cos[t/3]^2 + Sin[t/3]^2 == 1], {x, y}, {Cos[t/3], Sin[t/3]}, MonomialOrder -&amp;gt; EliminationOrder]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but I'm pretty sure the Cartesian equations have already been determined way before Buchberger was born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one determine the Cartesian equation of such a curve using only classical techniques?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apropos to this question, is there a quick way to determine the degree of an algebraic curve represented in polar coordinates without having to do a conversion?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-09T12:21:50.400" LastActivityDate="2010-09-09T19:08:05.050" Title="Deriving an implicit Cartesian equation from a polar equation with fractional multiples of the angle" Tags="&lt;geometry&gt;&lt;plane-curves&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4201" PostTypeId="1" AcceptedAnswerId="4209" CreationDate="2010-09-07T11:20:05.537" Score="3" ViewCount="203" Body="&lt;p&gt;Let X = complex manifold of dimension n. Thus, it's a real manifold of dimension 2n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now cohomology is a topological concept so it should not depend upon the structure given on a topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that k'th Singular cohomology of X is 0 for k &gt; 2n. We can also define a sheaf cohomology on that space using derived functor approach of Grothendieck. Then (by a result of Grothendieck) we know that k'th sheaf cohomology is 0 for k &gt; n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for constant sheaves [say &lt;strong&gt;R&lt;/strong&gt;], the sheaf cohomology agrees with singular cohomology [with coefficient &lt;strong&gt;R&lt;/strong&gt;]. Does this means that even the k'th singular cohomology of X vanishes for k &gt; n ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited] I now feel that the result which says that sheaf and singular agrees is actually this that k'th sheaf cohomology [of a complex manifold and constant sheaf] will agree with 2k'th singular cohomology [of the underlying real manifold]. Is this correct?? I would still like others to comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1767" LastEditorUserId="1767" LastEditorDisplayName="" LastEditDate="2010-09-07T11:27:52.820" LastActivityDate="2010-09-08T07:28:28.167" Title="Singular and Sheaf Cohomology" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4202" PostTypeId="1" AcceptedAnswerId="4204" CreationDate="2010-09-07T11:39:16.947" Score="14" ViewCount="738" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my Fellows asked me whether total induction is applicable to real numbers, too ( or at least all real numbers ≥ 0) . We only used that for natural numbers so far. &#xA;Of course you have to change some things in the inductive step, when you want to use it on real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that using induction on real numbers isn't really possible, since $[r,r+\epsilon]$ with $\epsilon &gt; 0$, $r \in \mathbb R$ is never empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you either give a good reason, why it isn't possible or provide an example where it is used?&lt;/p&gt;&#xA;" OwnerUserId="1768" LastEditorUserId="1768" LastEditorDisplayName="" LastEditDate="2010-09-07T18:11:44.733" LastActivityDate="2010-09-19T22:20:54.500" Title="Induction on Real Numbers" Tags="&lt;induction&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="4203" PostTypeId="2" ParentId="4202" CreationDate="2010-09-07T11:50:30.200" Score="-1" ViewCount="0" Body="&lt;p&gt;Please check this out: &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20100407075623AAB2lHH&quot; rel=&quot;nofollow&quot;&gt;http://answers.yahoo.com/question/index?qid=20100407075623AAB2lHH&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example where it cannot be applicable could be this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: All real numbers are integers. 1 is an integer. Assuming n is an integer we get $n+1$ to be an integer, which shows all real numbers are integers, a contradiction. It cant be applicable here.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-07T12:32:30.760" LastActivityDate="2010-09-07T12:32:30.760" CommentCount="1" />
  <row Id="4204" PostTypeId="2" ParentId="4202" CreationDate="2010-09-07T12:38:05.140" Score="18" ViewCount="0" Body="&lt;p&gt;Okay, I can't resist: here is a quick answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am construing the question in the following way: &quot;Is there some criterion for a subset of $[0,\infty)$ to be all of $[0,\infty)$ which is (a) analogous to the principle of mathematical induction on $\mathbb{N}$ and (b) useful for something?&quot;&#xA;The answer is yes, at least to (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me work a little more generally: let $(X,\leq)$ be a totally ordered set which has&lt;br&gt;&#xA;$\bullet $a least element, called $0$, and no greatest element.&lt;br&gt;&#xA;$\bullet$ The &lt;strong&gt;greatest lower bound property&lt;/strong&gt;: any nonempty subset $Y$ of $X$ has a greatest lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Principle of Induction on $(X,\leq)$: Let $S \subset X$ satisfy the following properties:&lt;br&gt;&#xA;(i) $0 \in S$.&lt;br&gt;&#xA;(ii) For all $x$ such that $x \in S$, there exists $y &gt; x$ such that $[x,y] \subset S$.&lt;br&gt;&#xA;(iii) If for any $y \in X$, the interval $[0,y) \subset S$, then also $y \in S$.&lt;br&gt;&#xA;Then $S = X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $S \neq X$, then the complement $S' = X \setminus S$ is nonempty, so has a least upper bound, say $y$.  By (i), we cannot have $y = 0$, since $y \in S$.  By (ii), we cannot have $y \in S$, and by (iii) we cannot have $y \in S'$.  Done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in case $(X,\leq)$ is a well-ordered set, this is equivalent to the usual statement of transfinite induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also applies to an interval in $\mathbb{R}$ of the form $[a,\infty)$.  It is not hard to adapt it to versions applicable to any interval in $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I believe that some sort of converse should be true: i.e., an ordered set with a principle of induction should have the GLB / LUB property.  [&lt;b&gt;Added&lt;/b&gt;: yes, this is true.  A totally ordered set with minimum element $0$ satisfies the principle of ordered induction as stated above iff every nonempty subset has an infimum.]  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: as for usefulness, one can use &quot;real induction&quot; to prove the three basic &lt;strong&gt;Interval Theorems&lt;/strong&gt; of honors calculus / basic real analysis.  These three theorems assert three fundamental properties of any continuous function $f: [a,b] \rightarrow \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Interval Theorem: $f$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive Proof: Let $S = \{x \in [a,b] \ | \ f|_{[a,x]} \text{is bounded} \}$.  It suffices to show that $S = [a,b]$, and we prove this by induction.&lt;br&gt;&#xA;(i) Of course $f$ is bounded on $[a,a]$.&#xA;(ii) Suppose that $f$ is bounded on $[a,x]$.  Then, since $f$ is continuous at $x$, $f$ is bounded near $x$, i.e., there exists some $\epsilon &gt; 0$ such that $f$ is bounded on &#xA;$(x-\epsilon,x+\epsilon)$, so overall $f$ is bounded on $[0,x+\epsilon)$.&lt;br&gt;&#xA;(iii) If $f$ is bounded on $[0,y)$, of course it is bounded on $[0,y]$.  Done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary: $f$ assumes its maximum and minimum values.&lt;br&gt;&#xA;Proof: Let $M$ be the least upper bound of $f$ on $[a,b]$.  If $M$ is not a value of $f$, &#xA;then $f(x)-M$ is never zero but takes values arbitrarily close to $0$, so $g(x) = \frac{1}{f(x)-M}$ is continuous and unbounded on $[a,b]$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Unfortunately in the proof I said &quot;least upper bound&quot;, and I suppose the point of proofs by induction is to remove explicit appeals to LUBs.  Perhaps someone can help me out here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second Interval Theorem (Intemediate Value Theorem): Suppose that $f(a) &amp;lt; 0$ and $f(b) &gt; 0$.  Then there exists $c \in (a,b)$ such that $f(c) = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Define $S = {x \in [a,b] \ | \ f(x) \leq 0}$.  In this case we are given that $S \neq [a,b]$, so at least one of the hypotheses of real induction must fail.  But which?&lt;br&gt;&#xA;(i) Certainly $a \in S$.&lt;br&gt;&#xA;(iii) If $f(x) \leq 0$ on $[a,y)$ and $f$ is continuous at $y$, then we must have $f(y) \leq 0$ as well: otherwise, there is a small interval about $y$ on which $f$ is positive.&lt;br&gt;&#xA;So it must be that (ii) fails: there exists some $x \in (a,b)$ such that $f \leq 0$ on &#xA;$[a,x]$ but there is no $y &gt; x$ such that $f \leq 0$ on $[a,y]$.  As above, since $f$ is continuous at $x$, we must have $f(x) = 0$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third Interval Theorem: $f$ is uniformly continuous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proof left to the interested reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, one can give even easier inductive proofs of the following basic theorems of analysis: that the interval $[a,b]$ is connected, that the interval $[a,b]$ is compact (Heine-Borel), that every infinite subset of $[a,b]$ has an accumulation point (Bolzano-Weierstrass).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Acknowledgement&lt;/b&gt;: My route to thinking about real induction was the following paper by I. Kalantari:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/Kalantari07.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/Kalantari07.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His setup is slightly different from mine -- instead of (ii) and (iii), he has the single axiom that $[0,x) \subset S$ implies there exists $y &gt; x$ such that $[0,y) \subset S$ -- and I am sorry to say that I was initially confused by it and didn't think it was correct.  But I corresponded with Prof. Kalantari this morning and he kindly set me straight on the matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, several other papers exist in the literature doing real induction (mostly in the same setup as Kalantari's, but also with some other variants such as assuming (ii) and that the subset $S$ be closed).  The result goes back at least to a 1922 paper of Khinchin, who in fact later used real induction as a basis for an axiomatic treatment of real analysis.  It is remarkable to me that this appealing concept is not more widely known -- there seems to be a serious PR problem here.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-08T08:41:43.393" LastActivityDate="2010-09-08T08:41:43.393" CommentCount="7" />
  <row Id="4205" PostTypeId="1" AcceptedAnswerId="4207" CreationDate="2010-09-07T13:35:18.387" Score="2" ViewCount="77" Body="&lt;p&gt;I'm a beginner to maths and have trouble simplyfying the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p^y \cdot (pq)^o}{p^{2y+o} \cdot q^{o-2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p^{-y} \cdot q^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-07T13:41:57.787" LastActivityDate="2010-09-07T13:52:44.417" Title="How to expand and simplify (p^y * (pq)^o) / (p^(2y+o) * q^(o-2))?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="4206" PostTypeId="2" ParentId="530" CreationDate="2010-09-07T13:44:30.737" Score="2" ViewCount="0" Body="&lt;p&gt;Note: for more on Bézout domains, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Section 8.2 of &lt;a href=&quot;http://math.uga.edu/~pete/factorization2010.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/factorization2010.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Section 12.4 of &lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-07T13:44:30.737" />
  <row Id="4207" PostTypeId="2" ParentId="4205" CreationDate="2010-09-07T13:46:39.623" Score="3" ViewCount="0" Body="&lt;p&gt;Here's the method in general, without actually working out your example. You should do that yourself to seal the concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intermediate goal is to get all the powers of p and q separated in both the numerator and denominator.  In this case, it's almost there, with the exception of (pq)^o. So expand that first, using the principle (x*y)^a = x^a * y^a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then gather the p's and q's using the properties of multiplication and exponentiation, x^a * x^b = x^(a+b). Finally, match the p's in the numerator and denominator, likewise the q's, and using the principle x^a / x^b = x^(a-b) calculate the ultimate powers of p and q. Note that there are two ways you could handle the power of p in the final answer, since it is negative.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="271" LastEditorDisplayName="" LastEditDate="2010-09-07T13:52:44.417" LastActivityDate="2010-09-07T13:52:44.417" CommentCount="4" />
  <row Id="4208" PostTypeId="2" ParentId="530" CreationDate="2010-09-07T13:54:59.220" Score="2" ViewCount="0" Body="&lt;p&gt;The easiest example I know is the ring of all algebraic integers (roots of monic polynomials with integer coefficients). As noted, it is a Bezout domain, so every finitely generated ideal is principal, and in particular for every two algebraic integers $a$ and $b$ there exist algebraic integers $\alpha$ and $\beta$ such that $\alpha a+\beta b = d$, where $d$ is a gcd for $a$ and $b$. However, the ideal $(2, 2^{1/2}, 2^{1/4}, 2^{1/8}, \ldots, 2^{1/2^{n}},\ldots)$ is not principal, so the ring is not a PID.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-07T13:54:59.220" CommentCount="6" />
  <row Id="4209" PostTypeId="2" ParentId="4201" CreationDate="2010-09-07T14:00:55.417" Score="6" ViewCount="0" Body="&lt;p&gt;The cohomological dimension of a &lt;em&gt;real&lt;/em&gt; $n$-manifold $M$ is $n$: this means that $H^i(M,\mathscr F)=0$ for each sheaf $\mathscr F$ of abelian groups on $M$ if $i&gt;n$, and that there exist sheaves $\mathcal F$ on $M$ with $H^n(M,\mathscr F)\neq0$. You'll find this proved in Bredon's book on &lt;em&gt;Sheaf theory&lt;/em&gt;, §II.16. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the cohomological dimension of a &lt;em&gt;complex&lt;/em&gt; $n$-manifold is $2n$. For example, you reach the maximum, at least for compact ones, for the constant sheaf $\mathbb R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your «Does this means that even the k'th singular cohomology of X vanishes for k &gt; n??» question is &lt;em&gt;No&lt;/em&gt; (You can answer it without determining the cohomological dimension: just consider a compact complex $n$-manifold, which is automatically oriented: what is it $2n$-th cohomology group?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question in you [Edited] paragraph also has a negative answer. Consider examples to see that it is so.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-07T16:06:24.777" LastActivityDate="2010-09-07T16:06:24.777" CommentCount="1" />
  <row Id="4210" PostTypeId="2" ParentId="4198" CreationDate="2010-09-07T14:48:47.730" Score="4" ViewCount="0" Body="&lt;p&gt;There are many ways to think of these definitions. Here's a way one can understand for example what $[\kappa]^2$ means and what it means to have a homogeneous subset, that I find intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that you have a complete undirected graph with $\kappa$ many nodes. That means that you have $\kappa$ points and you connect every pair of them with a line. Now assume that we have two colours, red and blue, and that every line between two nodes is coloured either red or blue. A subset of these $\kappa$ nodes is called homogeneous if the lines between each and every of its nodes have the same colour (that's the same as saying that it has a complete subgraph whose lines are of one colour only).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we say that $\kappa\to(\lambda)^2$ is true if no matter how we paint those lines using two colours, we can find a homogeneous set of cardinality $\lambda$. That is, for every way we colour the lines we can find $\lambda$ many nodes that every line between them is of the same colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for every function $f:[\kappa]^2\to 2$ (you can think of this as a function that sends every two elements of $\kappa$ to one of the 2 colours) we can find a subset of $\kappa$ (let's call it$H$) that has cardinality $\lambda$ such that for every $x,y,z,w\in H$ we have $f({ x,y} )=f({ z,w} )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To generalize this if for every function $f:[\kappa]^n\to\mu$ (again you can see this as a function that sends every $n$ elements of $\kappa$ to one of the $\mu$ colours, or that it partitions the subsets of $\kappa$ of cardinality $n$ into $\mu$ partitions) we can find a set $H$ such that $\left|H\right|=\lambda$ and for every $x_1,\ldots,x_n,y_1,\ldots,y_n\in H$ we have $f({ x_1,\ldots,x_n} )=f({ y_1,\ldots,y_n} )$ then we say that $\kappa\to(\lambda)^n_\mu$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrow notation, even though appears weird at first is used because the property remains true if we replace the cardinal in the left side of the arrow with a larger cardinal or if we replace any cardinal in the right side of the arrow with a smaller cardinal (if the subscript in the left side of the arrow is omitted then it is assumed that it is 2). It should be obvious that the notation only has meaning if $\lambda&amp;lt;\kappa$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many equivalent definitions of weakly compact cardinals. Jech's book mentions most of them, but here are some. If $\kappa$ is a cardinal then it is weakly compact iff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa\to(\kappa)^2$ is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa\to(\kappa)^n_\lambda$ where $\lambda&amp;lt;\kappa$ is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa$ is inaccessible and has the tree property, that is, if you have a tree with height $\kappa$ and each level of the tree has less than $\kappa$ elements then it has a branch of length $\kappa$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa$ is $\prod^1_1\textrm{-indescribable}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa$ is inaccessible and the languages $\mathcal{L}_{\kappa,\omega}$ satisfy the weak compactness theorem (hence the name). This means that in a language where we allow $\lambda$ conjunctions and disjunctions (where $\lambda&amp;lt;\kappa$), if we have a set of sentences $\Sigma$ where$\left|\Sigma\right|=\kappa$ and every subset of $\Sigma$ with less than $\kappa$ sentences is satisfiable then $\Sigma$ is satisfiable.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2010-09-07T15:59:59.743" LastActivityDate="2010-09-07T15:59:59.743" />
  <row Id="4211" PostTypeId="1" AcceptedAnswerId="4213" CreationDate="2010-09-07T15:37:21.660" Score="3" ViewCount="116" Body="&lt;p&gt;Motivation: In S.J. Patterson's &lt;em&gt;An introduction to the theory of the Riemann&#xA;Zeta-Function&lt;/em&gt; it is proved (p.132) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle -\Gamma ^{\prime }(t)/\Gamma (t)=\gamma +t^{-1}+\underset{j\geq 1}{\sum }[(j+t)^{-1}-j^{-1}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Exercise A4.1 (p.135) one is asked to show that if $|t|&amp;lt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle -\Gamma ^{\prime }(t)/\Gamma (t)=\gamma +t^{-1}+\underset{k\geq 1}{\sum }\zeta (k+1)(-t)^{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How does one prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\underset{j\geq 1}{\sum }[(j+t)^{-1}-j^{-1}]=\underset{k\geq 1}{\sum }\zeta (k+1)(-t)^{k}\qquad $for $|t|&amp;lt;1$?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-15T18:11:01.060" LastActivityDate="2010-11-15T18:11:01.060" Title="Proving identity $\displaystyle\sum_{j\geq 1}[(j+t)^{-1}-j^{-1}]=\displaystyle\sum_{k\geq 1}\zeta (k+1)(-t)^{k}$" Tags="&lt;sequences-and-series&gt;&lt;special-functions&gt;&lt;zeta-function&gt;" AnswerCount="2" />
  <row Id="4212" PostTypeId="2" ParentId="4211" CreationDate="2010-09-07T16:08:49.557" Score="5" ViewCount="0" Body="&lt;p&gt;Use&#xA;$$&#xA;\frac{1}{j + t} = \frac{1}{j} \frac{1}{1 + \frac{t}{j}} = \frac{1}{j} \sum_{k \geq 0} (-t)^k j^{- k}&#xA;$$&#xA;to obtain&#xA;$$&#xA;\sum_{j \geq 1} [ (j + t)^{-1} - j^{-1} ] =&#xA;\sum_{j \geq 1} \sum_{k \geq 1} (-t)^k j^{-k-1} = \sum_{k \geq 1} \zeta (k + 1) (-t)^k .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-07T16:08:49.557" />
  <row Id="4213" PostTypeId="2" ParentId="4211" CreationDate="2010-09-07T16:09:10.307" Score="6" ViewCount="0" Body="&lt;p&gt;Just apply $\sum_{n=0}^\infty x^n = \frac1{1-x}$. For $|t| &amp;lt; \min j = 1$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (j + t)^{-1} = j^{-1} \left( 1 + \frac tj \right)^{-1} = j^{-1} \sum_{k=0}^\infty \left( -\frac tj\right)^k, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus the LHS is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{j=1}^\infty \sum_{k=1}^\infty \frac{(-t)^k}{j^{k+1}}&#xA;= \sum_{k=1}^\infty \sum_{j=1}^\infty \frac{(-t)^k}{j^{k+1}}&#xA;= \sum_{k=1}^\infty (-t)^k \left( \sum_{j=1}^\infty \frac1{j^{k+1}} \right) &#xA;= \sum_{k=1}^\infty (-t)^k \zeta(k+1) $$&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-09-07T16:09:10.307" />
  <row Id="4214" PostTypeId="1" CreationDate="2010-09-07T16:34:27.840" Score="4" ViewCount="252" Body="&lt;p&gt;In b&lt;sup&gt;a&lt;/sup&gt; = x, b is the base, a is the exponent and x is the result of the operation. But in its logarithm counterpart, log&lt;sub&gt;b&lt;/sub&gt;(x) = a, b is still the base, and a is now the result. What is x called here? The exponent?&lt;/p&gt;&#xA;" OwnerUserId="1773" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-07T17:49:31.633" LastActivityDate="2010-09-08T05:24:03.990" Title="What is the x in $\log_b x$ called?" Tags="&lt;terminology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4215" PostTypeId="1" CreationDate="2010-09-07T17:08:21.957" Score="4" ViewCount="119" Body="&lt;p&gt;I suspect it is impossible to split a (any) 3d solid into two, such that each of the pieces is identical in shape (but not volume) to the original.  How can I prove this?&lt;/p&gt;&#xA;" OwnerUserId="1775" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-07T17:47:41.687" LastActivityDate="2010-09-07T18:09:15.317" Title="Can we slice an object into two pieces similar to the original?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="4216" PostTypeId="2" ParentId="4193" CreationDate="2010-09-07T17:18:19.717" Score="4" ViewCount="0" Body="&lt;p&gt;Let $M$ be the finitely generated module, and $M_{tor}$ its torsion submodule.&#xA;As part of the proof of the structure theorem, one shows that $M/M_{tor}$ is free&#xA;of finite rank, and that $M$ is isomorphic to $M_{tor} \oplus M/M_{tor}$,&#xA;so it suffices to show uniqueness for free modules and torsion modules separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a free module, its free rank can be computed as the maximal number of linearly independent&#xA;elements (just using the usual linear algebra definition, but applied to the free module).&#xA;So the free rank of $M/M_{tor}$ is uniquely determined.  (In fact, one can apply this&#xA;directly to $M$ --- the free part of $M$ in the decomposition given by the structure theorem&#xA;is just the maximal number of linearly independent elements of $M$ --- and so you&#xA;don't need to explicitly consider $M/M_{tor}$ if you don't want to.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now reduced to the case when $M = M_{tor}$, i.e. $M$ is torsion.  The structure theorem then gives us that $M$ is a direct sum of modules of the&#xA;form $A/\wp_j^{i_j} A$ for various non-zero prime ideals $\wp_j$ of $A$, and various indices $i_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine now that we have another such decomposition, into a direct sum of $A/\wp'_j^{i'_j}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we fix a non-zero prime ideal $\wp$ of $M$, and some $n \geq 0,$&#xA;then let $M[\wp^n]$ denote the submodule of $M$ consisting of elements annihilated by&#xA;$\wp^n$.  A direct computation with each of the decompositions in turn shows that&#xA;$M[\wp^n]/M[\wp^{n-1}]$ is an $A/\wp$-vector space of dimension equal to the number&#xA;of $j$ for which $\wp_j = \wp$ and $i_j \geq n,$ and also equal to the number of &#xA;$j$ for which $\wp'_j = \wp$ and $i'_j \geq n.$  Since $\wp$ and $n$ were arbitrary,&#xA;this shows that the two decompositions coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also make the following variant of this argument, in which we break it into two steps.&#xA;Namely, first fix a non-zero prime ideal of $A$, and consider $M[\wp^n]$.  If $n$ is larger than&#xA;any of the $i_j$ or $i'_j$, then this is the direct sum of just those $A/\wp_j^{i_j}$&#xA;for which $\wp_j = \wp$, and also is the direct sum of just those&#xA;$A/\wp'_j^{i'_j}$ for which $\wp'_j = \wp$.&#xA;Thus, replacing $M$ by $M[\wp^n]$ as $\wp$ runs over all the $\wp_j$ and $\wp'_j$, we may assume that all the $\wp_j$ and $\wp'_j$ coincide with some fixed prime ideal $\wp$.&#xA;We can then repeat the above argument, but just varying $n$, and keeping $\wp$ equal&#xA;to our given choice of prime ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the following variant, using quotients, rather than subobjects, namely,&#xA;fix a non-zero prime ideal of $A$, and consider $M/\wp^n M$.  Again, if $n$ is larger than&#xA;any of the $i_j$ or $i_j'$, then this is the direct sum of just those $A/\wp_j^{i_j}$&#xA;for which $\wp_j = \wp$, and also is the direct sum of just those&#xA;$A/\wp'_j^{i'_j}$ for which $\wp'_j = \wp$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we could consider submodules of prescribed torsion, as above,&#xA;but, since we have started by using quotients, we can finish by using quotients as well,&#xA;by considering the quotients of the form $\wp^n M/\wp^{n+1}M$ just as in&#xA;Pierre-Yves Gaillard's comment above.  As he explains, this also gives the desired uniqueness.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-09-08T07:31:41.870" LastActivityDate="2010-09-08T07:31:41.870" />
  <row Id="4217" PostTypeId="2" ParentId="4215" CreationDate="2010-09-07T17:22:52.970" Score="3" ViewCount="0" Body="&lt;p&gt;It seems that Puppe and others proved that this is impossible for &lt;em&gt;any&lt;/em&gt; strictly convex solid. See [B. L van den Waerden, Aufgabe Nr 51, &lt;em&gt;Elem. Math.&lt;/em&gt; &lt;strong&gt;4&lt;/strong&gt; (1949) &lt;strong&gt;18&lt;/strong&gt;, 140]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference comes from &lt;em&gt;Unsolved problems in geometry&lt;/em&gt; by Hallard T. Croft, K. J. Falconer and Richard K. Guy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-07T17:22:52.970" CommentCount="2" />
  <row Id="4218" PostTypeId="2" ParentId="4214" CreationDate="2010-09-07T18:00:05.650" Score="9" ViewCount="0" Body="&lt;p&gt;Argument (as you call the $x$ in any $f(x)$ argument of the function)&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-07T18:00:05.650" />
  <row Id="4219" PostTypeId="2" ParentId="4215" CreationDate="2010-09-07T18:09:15.317" Score="7" ViewCount="0" Body="&lt;p&gt;You can certainly take a rectangular box, $2^{1/3} \times 2^{2/3} \times 2$ and slice it into two boxes of size $1 \times 2^{1/3} \times 2^{2/3}$.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-07T18:09:15.317" CommentCount="5" />
  <row Id="4220" PostTypeId="1" AcceptedAnswerId="4248" CreationDate="2010-09-07T18:35:56.797" Score="6" ViewCount="318" Body="&lt;p&gt;In the course of analyzing a certain Markov chain, I once had to prove the following algebraic identity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a slick or known proof?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n$-tuples $(x_1,x_2,\dots, x_n)$ of positive real numbers define&#xA;$$\mu(x_1,x_2,\dots, x_n)=\prod_{j=1}^n {x_j\over x_j+x_{j+1}+\cdots+x_n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if $x^\ast$ is another positive real, and $1\leq k\leq n+1$, then &#xA;define $x^*_k$ to be the $(n+1)$-tuple $(x_1,x_2,\dots, x^*,\dots, x_n)$ where &#xA;$x^*$ is in the $k$th place. The identity is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n+1}\ \mu(x^\ast_k)=\mu(x_1,x_2,\dots, x_n).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&#xA;$$ {xyz\over(x+y+z)(y+z)z}&#xA; + {yxz\over(y+x+z)(x+z)z}&#xA; + {yzx\over(y+z+x)(z+x)x}={yz\over(y+z)z}.$$&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-08T05:47:50.403" LastActivityDate="2010-09-08T05:48:44.363" Title="Is this a known algebraic identity?" Tags="&lt;algebra-precalculus&gt;&lt;probability&gt;&lt;polynomials&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4221" PostTypeId="2" ParentId="4158" CreationDate="2010-09-07T18:50:18.430" Score="0" ViewCount="0" Body="&lt;p&gt;This question seems fairly advanced. You may want to consider asking it on &lt;a href=&quot;http://mathoverflow.net/&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1776" LastActivityDate="2010-09-07T18:50:18.430" />
  <row Id="4222" PostTypeId="2" ParentId="4198" CreationDate="2010-09-07T19:28:18.887" Score="2" ViewCount="0" Body="&lt;p&gt;Let me add to Ivan's answer another equivalent&#xA;characterization of weak compactness, which might appeal to&#xA;you, as it makes them resemble miniature measurable&#xA;cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, if $\kappa$ is a cardinal and&#xA;$\kappa^{&amp;lt;\kappa}=\kappa$, then $\kappa$ is weakly compact&#xA;if and only if for every transitive set $M$ of size&#xA;$\kappa$ with $\kappa\in M$, there is another transitive&#xA;set $N$ and an elementary embedding $j:M\to N$ having&#xA;critical point $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This embedding characterization admits myriad forms. For&#xA;example, one can insist that $M\models ZF^-$ or even&#xA;$M\models ZFC$, and that $M^{&amp;lt;\kappa}\subset M$, or that&#xA;every $A\subset \kappa$ can be placed into such an $M$, and so&#xA;on. One can even insist that $j\in N$, a property known as&#xA;the &lt;em&gt;Hauser&lt;/em&gt; property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These various embedding formulations of weak compactness&#xA;allow one to borrow many of the methods and techniques from&#xA;much larger large cardinals, which are most often described&#xA;in terms of embeddings, and apply them with weakly compact&#xA;cardinals. For example, using Easton support forcing&#xA;iterations, one can control the value of $2^\kappa$ while&#xA;preserving the weak compactness of $\kappa$.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-07T19:28:18.887" />
  <row Id="4223" PostTypeId="2" ParentId="4068" CreationDate="2010-09-07T20:02:28.423" Score="1" ViewCount="0" Body="&lt;p&gt;In Matt's answer I began discussing with Matt E over several comments, which I think should be written out as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Matt pointed out, this is a rearrangement of this conditionally convergent series which is why you have this sort of paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it was unclear about how this is exactly a rearrangement, as the equities seems perfectly legal - even for a conditionally convergent series.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = 1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2})$ is the first step, which is legal as you simply replace the negative terms by pairs of a positive and negative terms, but you don't change the order of summation from the original series which makes this exchange legit.&lt;/li&gt;&#xA;&lt;li&gt;$1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2}) = 1 + \frac{1}{2} + \frac{1}{3} + \ldots - 1 - \frac{1}{2} - \ldots$ &lt;strong&gt;this&lt;/strong&gt; is where things break, you've taken a conditionally convergent series and changed the order, basically we've performed infinitely many commutations in order to rearrange the series into this order, and that is what breaks the summation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The rearrangement wasn't very obvious, but it was hiding there with its big sharp pointy teeth... and when you stepped too close to its cave - it jumped out at you and bit your head off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series in the question is closely reminding me of the one my calculus teacher used when he first showed us what changing conditionally convergent series can do, although his was even less obvious.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-07T20:02:28.423" CommentCount="4" />
  <row Id="4224" PostTypeId="2" ParentId="4202" CreationDate="2010-09-07T20:09:42.483" Score="2" ViewCount="0" Body="&lt;p&gt;There's a phenomenal paper by Don Zagier which uses a sort of induction over the real numbers to give an explicit solution to some recurrence relations (which, it just so happens, calculate the Betti numbers of the moduli space of central Yang-Mills connections on bundles over Riemann surfaces).  But you don't need to understand those words to appreciate Zagier's paper.  The reference is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zagier, Don&#xA;Elementary aspects of the Verlinde formula and of the Harder-Narasimhan-Atiyah-Bott formula. (English summary) Proceedings of the Hirzebruch 65 Conference on Algebraic Geometry (Ramat Gan, 1993), 445--462,&#xA;Israel Math. Conf. Proc., 9, Bar-Ilan Univ., Ramat Gan, 1996. &lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-07T20:09:42.483" CommentCount="3" />
  <row Id="4225" PostTypeId="2" ParentId="4158" CreationDate="2010-09-07T20:11:35.073" Score="1" ViewCount="0" Body="&lt;p&gt;I think Terry Tao's paper might help here: &lt;a href=&quot;http://arxiv.org/abs/0802.3361&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0802.3361&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows that the isolated primes are of positive relative density.  I'm not sure if the method can be extended to show that there are large connected blocks.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-07T20:11:35.073" />
  <row Id="4226" PostTypeId="2" ParentId="1870" CreationDate="2010-09-07T20:35:49.157" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathxpert.com/&quot; rel=&quot;nofollow&quot;&gt;MathXpert&lt;/a&gt; supposedly fits your first 3 criteria (though I have never tried it).&lt;/p&gt;&#xA;" OwnerUserId="1386" LastActivityDate="2010-09-07T20:35:49.157" CommunityOwnedDate="2010-09-07T20:35:49.157" />
  <row Id="4227" PostTypeId="2" ParentId="262" CreationDate="2010-09-07T20:52:10.283" Score="8" ViewCount="0" Body="&lt;p&gt;Proofs from the Book, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540636986&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Proofs-BOOK-Martin-Aigner/dp/3540636986&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1780" LastActivityDate="2010-09-07T20:52:10.283" CommunityOwnedDate="2010-09-07T20:52:10.283" />
  <row Id="4228" PostTypeId="1" AcceptedAnswerId="4347" CreationDate="2010-09-07T21:40:48.790" Score="2" ViewCount="191" Body="&lt;p&gt;Say P is a convex Euclidean polytope, where the origin is not contained in any bounding hyperplane containing a facet of P, with n facets given by f ◦ x = 1 and m vertices v. (That is, each facet equation has been multiplied through as needed to obtain 1 on the right side.)&#xA;Is it true that the polytope P’ with m facets given by v ◦ x = 1 and n vertices f is combinatorially equivalent to the polar dual P* of P?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-09-08T13:23:22.503" LastActivityDate="2010-09-13T09:20:17.983" Title="“Cartesian” dual vs. polar dual of convex polytope" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="4229" PostTypeId="2" ParentId="4220" CreationDate="2010-09-07T22:23:12.113" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a proof by induction on the number of elements in the tuple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully, I got the algebra right. In any case, it would be easy to verify).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a Lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;: For $n \ge 2$ we have that $\displaystyle \mu(x_1, ..., x_n) = \frac{x_{n-1}}{x_{n-1}+x_n} \mu(x_1, .., x_{n-2}, x_{n-1} + x_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: Easy Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = x^{*}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Base Case&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ it is easy to see that $ \displaystyle \mu(y, x) + \mu(x, y) = \mu(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction Step&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $n \ge 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that (I have changed your notation slightly, I start from 0 instead of 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n-2} \mu(x_1, x_2, \dots , x_k, y, x_{k+1}, \dots,x_{n-1}, x_{n})$$&#xA;$$ = \sum_{k=0}^{n-2}\frac{x_{n-1}}{x_{n-1}+x_{n}} \mu(x_1, x_2, ..., x_k, y, \dots, x_{n-1} + x_{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is easy to see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(x_1, \dots, x_{n-1}, y, x_n) + \mu(x_1, \dots, x_{n-1}, x_n, y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  = \frac{x_{n-1}}{x_{n-1}+x_n} \mu(x_1, x_2, ..., x_{n-2}, x_{n-1} + x_n, y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{n} \mu(x_{k}^{*}) $$ &#xA;$$ = \sum_{k=0}^{n-1} \frac{x_{n-1}}{x_{n-1}+x_{n}} \mu (z_{k}^{*})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ z = (x_1, \dots, x_{n-2}, x_{n-1}+x_{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by induction hyptothesis, this is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{x_{n-1}}{x_{n-1}+x_{n}} \mu (x_1, \dots, x_{n-2}, x_{n-1}+x_{n}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $$ \mu(x_1, x_2, \dots, x_{n-1}, x_n)$$ &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-07T22:23:12.113" CommentCount="5" />
  <row Id="4230" PostTypeId="2" ParentId="4214" CreationDate="2010-09-07T22:24:53.630" Score="2" ViewCount="0" Body="&lt;p&gt;antilogarithm (noun): a compound of Greek anti-&quot;against&quot; and logarithm (qq. v.). In algebra if $log_{b}(x) = y$, y is the logarithm of x, and x is said to be the antilogarithm of y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: The Words of Mathematics: An Etymological Dictionary of Mathematical Terms Used in English, Steven Schwartzman &lt;/p&gt;&#xA;" OwnerUserId="1746" LastActivityDate="2010-09-07T22:24:53.630" CommentCount="5" />
  <row Id="4231" PostTypeId="2" ParentId="566" CreationDate="2010-09-07T22:45:17.960" Score="0" ViewCount="0" Body="&lt;p&gt;Quite simply, a monkey's mother is twice as old as the monkey will be when the monkey's father is twice as old as the monkey will be when the monkey's mother is less by the difference in ages between the monkey's mother and the monkey's father than three times as old as the monkey will be when the monkey's father is one year less than twelve times as old as the monkey is when the monkey's mother is eight times the age of the monkey, notwithstanding the fact that when the monkey is as old as the monkey's mother will be when the difference in ages between the monkey and the monkey's father is less than the age of the monkey's mother by twice the difference in ages between the monkey's mother and the monkey's father, the monkey's mother will be five times as old as the monkey will be when the monkey's father is one year more than ten times as old as the monkey is when the monkey is less by four years than one seventh of the combined ages of the monkey's mother and the monkey's father.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in a number of years equal to the number of times a monkey's mother is as old as the monkey, the monkey's father will be as many times as old as the monkey as the monkey is now, and assuming no &lt;em&gt;a priori&lt;/em&gt; knowledge of the monkeys' longevity, find their respective ages. :-D&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2010-09-07T22:45:17.960" CommunityOwnedDate="2010-09-07T22:45:17.960" CommentCount="2" />
  <row Id="4232" PostTypeId="2" ParentId="4201" CreationDate="2010-09-07T22:53:15.210" Score="3" ViewCount="0" Body="&lt;p&gt;The statement is the following: If you take a variety with the &lt;em&gt;analytic&lt;/em&gt; topology (or a complex manifold), then yes, sheaf cohomology of the constant sheaf agrees with singular cohomology. More generally, sheaf cohomology of a constant sheaf on a &lt;em&gt;locally contractible space&lt;/em&gt; agrees with singular cohomology of that space. I think there's an explanation of this somewhere in Warner's book &quot;Foundations of Differentiable Manifolds and Lie Groups&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you have an irreducible variety with the &lt;em&gt;Zariski&lt;/em&gt; topology, then sheaf cohomology $H^i$ of the constant sheaf is zero for $i &gt; 0$, because the constant sheaf is flasque. (I'm not sure -- does it make sense to talk about Zariski topology on a (non-algebraic) complex manifold?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further remarks: If what you want is to be able to deal with singular cohomology in a more &quot;algebraic&quot; way, you can do so by using the Hodge decomposition $H^n(X;\mathbb{C}) = \bigoplus_{p+q=n}H^q(X,\Omega^p)$ which expresses singular cohomology in terms of sheaf cohomologies of $\Omega^p$'s. This works if your $X$ is a compact Kaehler manifold, e.g., a smooth projective variety. If your $X$ is an algebraic variety, you can also use etale cohomology for an &quot;algebraic&quot; way of dealing with singular cohomology; see Milne's book/notes on etale cohomology for more.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-09-08T07:28:28.167" LastActivityDate="2010-09-08T07:28:28.167" CommentCount="3" />
  <row Id="4233" PostTypeId="2" ParentId="4043" CreationDate="2010-09-07T22:59:44.917" Score="0" ViewCount="0" Body="&lt;p&gt;Introduction to Logic: Predicate Logic by Prospesel is the text probably most frequently associated with the topic for introductory students from a purely academic perspective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to code in languages that are based around predicate logic (logic programming is the more formal term), there are a few options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prolog is easily the best known and without a doubt the best documented logic programming language&lt;/li&gt;&#xA;&lt;li&gt;Oz is a lesser known alternative that incorporates a number of more modern programming language features like concurrency and object orientation (modern Prolog implementations typically add some of these to the language, but it never seems a natural fit to the language)&lt;/li&gt;&#xA;&lt;li&gt;Castor is a fairly common library for C++ that builds a logic model within the syntax of C++&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1781" LastActivityDate="2010-09-07T22:59:44.917" CommunityOwnedDate="2010-09-07T23:21:03.547" />
  <row Id="4234" PostTypeId="1" AcceptedAnswerId="4235" CreationDate="2010-09-07T23:33:24.873" Score="1" ViewCount="95" Body="&lt;p&gt;Suppose {x_t} is a random walk with mean zero. (either discrete or continuous time)&#xA;Fix a time T. &#xA;What is: Probability[x_t &amp;lt; 0, for all t &amp;lt;= T]?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in words, what's the probability the random walk from 0 to T has never been above 0?&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1784" LastActivityDate="2010-09-08T00:38:04.520" Title="Random walk &lt; 0" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="4235" PostTypeId="2" ParentId="4234" CreationDate="2010-09-08T00:38:04.520" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a solution for the discrete time, simple (steps are $\pm 1$), symmetric random walk $(X_t)$&#xA;that uses the reflection principle.&#xA;You didn't specify your starting point, but presumably it is negative, say $-b$ for some $b&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to consider the equivalent problem where the random walk starts at zero, and investigate&#xA; when it hits $b&gt;0$, that is, $T_b:= \inf (t&gt;0: X_t =b )$.&#xA;Then you should be able to use equation $(*)$ to find what you are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $t,b$ have opposite parity. Then&#xA;$$P(X_t &gt; b) = P(X_t&gt;b \mid  T_b &amp;lt; t) P(T_b &amp;lt; t)={1\over 2} P(T_b &amp;lt; t).$$&#xA;The conditional probability is exactly $1/2$ because $X_t=b$ is impossible.&#xA;Given $T_b &amp;lt; t$, the sample paths satisfy $X_{T_b}=b$ at some time $T_b$ prior to $t$.&#xA;The reflection principle shows that&#xA;these paths divide equally into those with $X_t &gt; b$ and those with $X_t &amp;lt; b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us the formula  $P(T_b &amp;lt; t)=2P(X_t &gt; b)$ or $P(T_b\geq t)=P(-b\leq X_t\leq b)$.&#xA;Since the random walk is symmetric around the origin, the random variables&#xA;$T_b$ and $T_{-b}$ have the same  distribution. Thus, we can combine the two cases as follows,&#xA; taking into account that $X_t=\pm b$ is impossible,&#xA;$$P(T_b\geq t)=P(-|b|&amp;lt; X_t &amp;lt; |b|),\quad b\in {\mathbb Z}\setminus \lbrace 0\rbrace ,\ t+b\mbox{ odd}.\qquad \qquad (*)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in the special case where $b=1$, with the even time point $2t$,&#xA;we get $$P(T_1\geq 2t)=P(-1&amp;lt; X_{2t} &amp;lt; 1)=P(X_{2t}=0)={2t\choose t}\left({1\over 2}\right)^{2t}.$$   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-08T00:38:04.520" CommentCount="1" />
  <row Id="4236" PostTypeId="1" AcceptedAnswerId="4237" CreationDate="2010-09-08T00:41:52.957" Score="0" ViewCount="69" Body="&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: logs below are base 2.  (Not sure how to do subscripts here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wondering if the below equation is true when thinking asymptotically (Computer Science)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_2((n!)^n) = O(n \sin(n \frac{\pi}{2}) + \log_2{n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to compute this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing we need to take the log of both sides of the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_2((n!)^n) &amp;lt; n sin(n (pi/2)) + log_2(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;getting us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_2(log_2((n!)^n)) &amp;lt; log_2 (n  sin(n (pi/2)) + log_2(n) )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure where to go from there.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-09-08T04:17:09.247" LastActivityDate="2010-09-08T04:17:09.247" Title="Big O boundary condition truth value" Tags="&lt;asymptotics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4237" PostTypeId="2" ParentId="4236" CreationDate="2010-09-08T01:37:20.163" Score="1" ViewCount="0" Body="&lt;p&gt;If you are interested in computer science, a very useful thing to remember is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log n! = \theta(n \log n) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given this, can you tell what $f(n)$ is, if $ \log (n!^{n}) = \theta(f(n))$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would that compare to the right hand side?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-08T01:37:20.163" CommentCount="19" />
  <row Id="4238" PostTypeId="1" AcceptedAnswerId="4320" CreationDate="2010-09-08T03:12:13.693" Score="4" ViewCount="164" Body="&lt;p&gt;The two element group has a centre (of course it does, it's abelian) but no outer automorphisms. I can come up with a (convoluted) countably infinite non-abelian example. Is there a finite one? If so what is the smallest? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this sort of group? (&quot;Outerless?&quot; Since a group without a centre is &quot;centreless&quot;, or &quot;complete&quot; if it has no outermorphisms either.)&lt;/p&gt;&#xA;" OwnerUserId="1786" LastActivityDate="2010-09-09T00:19:34.470" Title="Finite non-abelian group with centre but no outer automorphism" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="4239" PostTypeId="1" AcceptedAnswerId="4240" CreationDate="2010-09-08T03:18:05.003" Score="2" ViewCount="57" Body="&lt;p&gt;Say we have a partially ordered set $(S,\preceq)$, and some subset $E\subseteq S$ such that $E$ is bounded below and $\inf E$ exists. My question is, since $S$ is not totally ordered is it possible to have two lower bounds that are not comparable, and hence be unable to say whether one lower bound is really the greatest lower bound since you cannot compare to every other lower bound? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such a thing is not possible, (which I believe is the case since it is given $\inf E$ exists) would it also be fair to say that the existence of $\inf E$ implies that the set $L(E)$ of lower bounds of $E$ is a totally ordered subset of $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-08T03:32:22.013" Title="Does the existence of an infimum imply that the set of lower bounds of a set is totally ordered?" Tags="&lt;analysis&gt;&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="4240" PostTypeId="2" ParentId="4239" CreationDate="2010-09-08T03:32:22.013" Score="1" ViewCount="0" Body="&lt;p&gt;No, because even if a partially ordered set is not totally ordered, there can be a &quot;largest&quot; element.  For a simple example, take a set $X$ with at least 2 elements, let $S$ be the power set of $X$ and let the partial order be inclusion.  Then $\inf\{X\}=X$, but the set of lower bounds of $\{X\}$ is $S$, which is not totally ordered because $X$ has more than one element.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-08T03:32:22.013" CommentCount="2" />
  <row Id="4241" PostTypeId="1" CreationDate="2010-09-08T03:43:29.700" Score="3" ViewCount="230" Body="&lt;p&gt;I was preparing for an area exam in analysis and came across a problem in the book &lt;em&gt;Real Analysis&lt;/em&gt; by Haaser &amp;amp; Sullivan. From p.34 Q 2.4.3, If the field &lt;em&gt;F&lt;/em&gt; is isomorphic to the subset &lt;em&gt;S'&lt;/em&gt; of &lt;em&gt;F'&lt;/em&gt;, show that &lt;em&gt;S'&lt;/em&gt; is a subfield of &lt;em&gt;F'&lt;/em&gt;. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so &quot;restrictive&quot;, in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2010-09-08T09:55:44.503" Title="Examples of fields of characteristic 0?" Tags="&lt;soft-question&gt;&lt;abstract-algebra&gt;&lt;reference-request&gt;" AnswerCount="4" />
  <row Id="4243" PostTypeId="2" ParentId="4202" CreationDate="2010-09-08T04:22:53.383" Score="1" ViewCount="0" Body="&lt;p&gt;If you assume a particular limit ordinal for the cardinality of the reals, you can prove new theorems about them using transfinite induction with a bounded limit case!  But I doubt these theorems have any application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers are very interesting but it is hard for me to believe in a process truly analogous to induction on a set which cannot be well-ordered.  The ability to well-order the reals depends on the Axiom of Choice, and the precise consequences of such an ordering depend on the Continuum Hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-08T04:45:36.660" LastActivityDate="2010-09-08T04:45:36.660" CommentCount="0" />
  <row Id="4244" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T04:35:23.093" Score="1" ViewCount="0" Body="&lt;p&gt;If $F$ is any field, the rational functions over it form a field $F(t)$ with the same characteristic (and cardinality, if $F$ is infinite). This field consists of all rational functions $P(t)/Q(t)$ (considered as equivalence classes, i.e. if $P_1(t) Q_2(t) = P_2(t) Q_1(t)$ then $P_1(t)/Q_1(t)$ and $P_2(t)/Q_2(t)$ are identified).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also replace polynomials with formal power series to get a different field. And you can iterate the construction or just consider rational functions in several variables.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-08T04:35:23.093" />
  <row Id="4245" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T05:04:03.750" Score="1" ViewCount="0" Body="&lt;p&gt;Yuval just posted the function field construction, which is an important example for your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned the algebraic numbers, but just to make it explicit, there are many algebraic number fields (often called just &quot;number fields&quot;).  We have $ \bar Q $, the field of &lt;em&gt;all&lt;/em&gt; algebraic numbers, but we also have $Q(\sqrt{2})$, $Q(\sqrt[5]{7})$, $Q(i,\sqrt{2},\sqrt[5]{7})$, and any other field you wish to generate over $Q$ by some tower of finite extensions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If we take $Q(\pi)$, we get a field isomorphic to the function field $Q(t)$, since $\pi$ is not algebraic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also take infinitely many different algebraic extensions of the p-adic fields, like $Q_3(i)$ or $Q_3(\sqrt[4]{2})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like big constructions, you can take the algebraic closure $\bar {Q_p}$of a p-adic field (which turns out not to be p-adically complete), then take the completion $\hat { \bar {Q_p}}$ of that to get a p-adcially complete and algebraically complete field which is isomorphic to the field of complex numbers (but not in any canonical way).&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-08T05:04:03.750" />
  <row Id="4246" PostTypeId="2" ParentId="4214" CreationDate="2010-09-08T05:24:03.990" Score="1" ViewCount="0" Body="&lt;p&gt;Another name (that I've only ever seen when someone else asked this &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBQQFjAA&amp;amp;url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FWikipedia%3AReference_desk%2FArchives%2FMathematics%2F2010_February_17&amp;amp;rct=j&amp;amp;q=%22logarithmand%22&amp;amp;ei=Xh2HTNKNKoq8cMrDxJ4I&amp;amp;usg=AFQjCNHf_ymaurzOLcW2gXQ1BM4LGDuY2Q&amp;amp;sig2=1FzceH8BR9ejZpmxUa4x8Q&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt;) is &quot;logarithmand&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-08T05:24:03.990" />
  <row Id="4247" PostTypeId="2" ParentId="4193" CreationDate="2010-09-08T05:32:50.650" Score="3" ViewCount="0" Body="&lt;p&gt;As a complement to Matt E's nice answer, let $A$ be your PID, and $p\in A$ a prime. It suffices to prove the uniqueness in the case where your finitely generated module $M$ is the product of a finite family of modules of the form $$M_i:=A/p^{i+1}A.$$ For each $j$ the quotient $p^jM/p^{j+1}M$ is an $A/pA$ vector space of finite dimension $n_j$. The multiplicity of $A/p^{i+1}A$ is then $n_i-n_{i+1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a way to see this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Form the polynomial $M(X):=\sum\ n_j\ X^j$. We have &#xA;$$M_i(X)=\frac{X^{i+1}-1}{X-1}=1+X+X^2+\cdots+X^i,$$ &#xA;and we must solve $\sum\ m_i\ M_i(X)=\sum\ n_j\ X^j$ for the $m_i$, where the $n_j$ are considered as known quantities (almost all equal to zero). Multiplying throughout by $X-1$ we get $$\sum\ m_{i-1}\ X^i-\sum\ m_i=\sum\ (n_{i-1}-n_i)\ X^i,$$ &#xA;whence the formula.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-09-08T18:41:49.100" LastActivityDate="2010-09-08T18:41:49.100" CommentCount="2" />
  <row Id="4248" PostTypeId="2" ParentId="4220" CreationDate="2010-09-08T05:48:44.363" Score="8" ViewCount="0" Body="&lt;p&gt;Consider the following experiment: there is an interval of length $x_1 + \cdots + x_n$ divided into $n$ segments of sizes $x_1,\ldots,x_n$. We sample an infinite sequence of points from the interval, and write out the segments corresponding to the points. Then, we look at the order in which the segments are &quot;discovered&quot;, i.e. which one was hit first, which was the next one, and so on. Then $\mu(x_1,\ldots,x_n)$ is the probability that the segments will be discovered in the order $x_1,\ldots,x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we add an extra segment $x^* $, and repeat the same experiment. If we just forget about this extra segment, the new experiment is just like the old one. This is the same as running the new experiment and deleting $x^* $ from the order of discovery. Your identity follows.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-08T05:48:44.363" CommentCount="4" />
  <row Id="4249" PostTypeId="2" ParentId="4238" CreationDate="2010-09-08T05:50:10.607" Score="5" ViewCount="0" Body="&lt;p&gt;$C_2\times M_{11}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steve&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-09-08T05:50:10.607" CommentCount="1" />
  <row Id="4250" PostTypeId="1" AcceptedAnswerId="4251" CreationDate="2010-09-08T06:00:57.767" Score="1" ViewCount="54" Body="&lt;p&gt;This is a follow up question to one I posted earlier. I'm trying to decide that if for $(S,\preceq)$ a partially ordered set and $E\subseteq S$, one has $L(E)=\langle s]$ for some $s\in S$ iff $\inf E$ exists, and in particular, $L(E)=\langle\inf E]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm simply curious about showing that $L(E)\subseteq\langle\inf E]$ when $\inf E$ exists. Taking some $x\in L(E)$, if $x$ is comparable to $\inf E$, then $x\preceq\inf E$, and so $x\in\langle\inf E]$. But what if $x$ and $\inf E$ are not comparable? Is it still possible to show such an inclusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $\langle s]$ is the set {$x\in S \ | \ x\preceq s$}.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-08T06:30:17.353" LastActivityDate="2010-09-08T06:31:45.597" Title="Can one show that $L(E)=\langle s]$ on a partially ordered set?" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4251" PostTypeId="2" ParentId="4250" CreationDate="2010-09-08T06:31:45.597" Score="2" ViewCount="0" Body="&lt;p&gt;If I correctly understand your question, it boils down to whether $\inf E$ is &quot;larger&quot; than or equal to every lower bound of $E$.  Isn't this the definition?  A greatest lower bound of a set must be, by definition, comparable to and &quot;larger&quot; than (or equal to) every lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the confusion arises because, as seen in your last question, having an inf does not imply that the set of lower bounds is totally ordered.  It does imply, however, that all lower bounds are comparable to the inf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is yes: If $E$ has an inf, then $L(E)=\langle \inf E]$.  Anything smaller than a lower bound is also a lower bound, which gives the right to left inclusion.  The other inclusion follows from the fact that $\inf E$ is the &quot;greatest&quot; lower bound.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-08T06:31:45.597" CommentCount="2" />
  <row Id="4252" PostTypeId="2" ParentId="3852" CreationDate="2010-09-08T06:55:08.603" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;MOTIVATION&lt;/strong&gt; $\;\:$ If vector space $\rm V\:$ has a one-one map $\rm B$ that is not onto $\rm V &gt; BV \;$ then this yields an $\infty$ descending chain of subspaces by repeatedly shifting $\rm V &gt; \: BV &gt; \;\cdots\: &gt; B^i V$ by multiplying by $\rm B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\;\;\;\: AB = 1 \;\Rightarrow\; BA=1 \;\;$ for linear maps $\rm\; A,\:B \;$ on a finite dimensional vector space $\rm\; V \;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\;\; V &gt; BV, \;$ times $\rm\; B^i \;\Rightarrow\; B^i V &gt; B^{i+1} V \;$ (else $\rm\; B^i V = B^{i+1} V, \;$ times $\rm\; A^i \Rightarrow V = BV)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\:\Rightarrow\rm\;\;\; V &gt; BV &gt; B^2 V &gt; \;\cdots \;\:$ is an $\infty$ descending chain $\rm\; \Rightarrow\; dim\: V = \infty\:,\:$ contra hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\;\; V = BV \;\Rightarrow\; (BA-1)V = (BA-1)BV = B(AB-1)V = 0 \quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\;\;$ Hence vector space $\rm\:V\;$ &lt;strong&gt;has infinite dimension&lt;/strong&gt; $\rm\;\iff V\:$ &lt;strong&gt;is Dedekind infinite&lt;/strong&gt;, i.e. $\rm\:V\:$ has an isomorphic proper subspace $\rm BV\:,\:$ viz. the theorem proves $(\Leftarrow)$ and the converse follows immediately by taking $\rm B\:$ to be a basis shift map $\rm\; (v_1,v_2,v_3,\cdots\:)\:\to\: (0,v_1,v_2,v_3,\cdots\:)\:,\;$  i.e. said more simply, a vector space is infinite iff it has a subspace chain model of &lt;strong&gt;Hilbert's infinite hotel&lt;/strong&gt;. $\:$ That is the essence.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-08T14:43:49.317" LastActivityDate="2010-09-08T14:43:49.317" CommentCount="2" />
  <row Id="4253" PostTypeId="1" AcceptedAnswerId="4292" CreationDate="2010-09-08T08:00:06.240" Score="1" ViewCount="100" Body="&lt;p&gt;Could someone give me some tips?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let e1,e2,... be iid normal mean 0 variance 1. Let Xt := e1+...+et, for t=1,2.. and X0 := 0. (So we have a discrete-time random walk whose steps are iid N(0,1))  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define first-hitting time T0 := inf{t&gt;0 : Xt &amp;lt;0}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the CDF of T0, i.e. what is Prob(T0 &amp;lt;= t) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And second (probably more difficult) question: what is ExpectedValue( X_t | T0 &gt; t )?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1784" LastEditorUserId="1784" LastEditorDisplayName="" LastEditDate="2010-09-08T08:55:23.643" LastActivityDate="2010-09-08T15:29:13.577" Title="Ruin-time in Gambler's Ruins" Tags="&lt;probability-theory&gt;&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4254" PostTypeId="2" ParentId="907" CreationDate="2010-09-08T08:36:29.900" Score="0" ViewCount="0" Body="&lt;p&gt;Mathematics is generally seen as a precise and specific language.  Surely if there is ambiguity in the phrase &quot;in the sequel&quot; it should be avoided.  There are far more accurate ways of expressing both the meanings that phrase could carry so why create needless confusion?&lt;/p&gt;&#xA;" OwnerUserId="1790" LastActivityDate="2010-09-08T08:36:29.900" />
  <row Id="4255" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T09:04:38.537" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, you are right in saying that field axioms are restrictive. Some other examples of the restriction [for finite fields] are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You can't have field of arbitrary order. Only of the order $p^n$ are possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Non-zero elements of a field form a multiplicative group. When the field is finite, this group is cyclic [it's not straightforward to prove this starting with field axioms but you should try it]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Any finite division ring is a field [Wedderburn's theorem]. This is a very surprising result as a division ring need not by commutative. But finiteness imposes it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question that you mentioned [from that book], you should note that when some set with some algebraic structure [like group, ring, field] is isomorphic [as that algebraic structure] to some other set, then that under that isomorphism, we are giving an algebraic structure to that other set. Hence in your case S' is a field. &lt;/p&gt;&#xA;" OwnerUserId="1767" LastActivityDate="2010-09-08T09:04:38.537" CommentCount="2" />
  <row Id="4256" PostTypeId="1" AcceptedAnswerId="4258" CreationDate="2010-09-08T09:17:32.093" Score="2" ViewCount="282" Body="&lt;p&gt;&lt;em&gt;This problem is taken from Vojtěch Jarník International Mathematical Competition 2010, &lt;a href=&quot;http://vjimc.osu.cz/j20/j20problems1.pdf&quot; rel=&quot;nofollow&quot;&gt;Category I, Problem 1&lt;/a&gt;. &lt;sub&gt;&lt;sup&gt;— edit by KennyTM&lt;/sup&gt;&lt;/sub&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On going through this post &lt;a href=&quot;http://math.stackexchange.com/questions/2120/does-there-exist-an-f-mathbbn-to-mathbbn-such-that-sum-fn-n2-conve&quot;&gt;http://math.stackexchange.com/questions/2120/does-there-exist-an-f-mathbbn-to-mathbbn-such-that-sum-fn-n2-conve&lt;/a&gt; i happened to get the following 2 problems into my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: \mathbb{N} \to \mathbb{N}$ be a bijection. Then does the series $$\sum\limits_{n=1}^{\infty} \frac{1}{nf(n)}$$ converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, consider the series $$\sum\limits_{n=1}^{\infty} \frac{1}{n+f(n)}$$ where $f: \mathbb{N} \to \mathbb{N}$ is a bijection. Clearly by taking $f(n)=n$ we see that the series is divergent. Does there exist a bijection such that the sum above is convergent?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-08T17:32:27.090" LastActivityDate="2010-09-09T09:23:46.643" Title="Convergence of $\sum\limits_{n=1}^{\infty} \frac{1}{nf(n)}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="4257" PostTypeId="2" ParentId="4256" CreationDate="2010-09-08T09:32:56.567" Score="1" ViewCount="0" Body="&lt;p&gt;A &lt;a href=&quot;http://www.scienzematematiche.it/forum/viewtopic.php?f=7&amp;amp;t=2322&quot; rel=&quot;nofollow&quot;&gt;couple of proofs&lt;/a&gt; in italian.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-09-08T09:32:56.567" CommentCount="1" />
  <row Id="4258" PostTypeId="2" ParentId="4256" CreationDate="2010-09-08T09:33:53.490" Score="2" ViewCount="0" Body="&lt;p&gt;Hints. For the first series, prove that&#xA;$$\sum_{n=1}^N\frac1{nf(n)}\le\sum_{n=1}^N\frac1{n^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, what would happen if $f$ were an involution,&#xA;and in each pair $(n,f(n))$ one term was much bigger than the other?&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-08T10:12:09.643" LastActivityDate="2010-09-08T10:12:09.643" />
  <row Id="4259" PostTypeId="2" ParentId="4241" CreationDate="2010-09-08T09:55:44.503" Score="1" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an analogous classification for fields of characteristic zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but it is somewhat useless and nobody would call it a classification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every field of characteristic zero has the form $Quot(\mathbb{Q}[X]/S)$, where $X$ is a set of variables and $S$ is a set of polynomials in $\mathbb{Q}[X]$ (which you may replace by the ideal generated by $S$, which must be prime). This may be improved by the existence of transcendence bases: Every field of characteristic zero has the form $Quot(\mathbb{Q}[X])[T]/S$, where $X$ and $T$ are sets of variables and $S$ consists of polynomials, which have each only one variable of $T$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-08T09:55:44.503" CommentCount="1" />
  <row Id="4260" PostTypeId="1" AcceptedAnswerId="4281" CreationDate="2010-09-08T09:58:17.100" Score="1" ViewCount="51" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have have a question which should be pretty basic commutative algebra, but I can't find a reference and I' m stuck on proving a result myself, so here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\varphi: S \to R$ be a morphism of commutative rings and let $M$ be a finitely generated $S$-module. Then, we can &quot;extend the scalars of $M$&quot; by considering $M \otimes_S R$, which is naturally an $R$-module. I would like to get a description of $\text{supp}(M \otimes_S R)$ (as an $R$-module) involving $\text{supp}(M)$ and $\varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this helps: by Eisenbud: Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a despcription of $\text{Ann}(M \otimes_S R)$ and it is not difficult to see that $\varphi(\text{Ann}(M)) \cdot R \subset \text{Ann}(M \otimes_S R)$, but I fail to prove the other inclusion.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2010-09-08T13:14:19.080" Title="Support of a module with extended scalars" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4261" PostTypeId="1" AcceptedAnswerId="4265" CreationDate="2010-09-08T10:08:15.933" Score="5" ViewCount="318" Body="&lt;p&gt;Mathematica tells me that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I have not been able to come up with a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proofs, hints, or references are all welcome.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-09-09T14:23:03.063" Title="Proving a binomial sum identity" Tags="&lt;combinatorics&gt;&lt;complex-analysis&gt;&lt;generating-functions&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="4263" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:20:08.917" Score="5" ViewCount="0" Body="&lt;p&gt;Sums of the form&#xA;$$\sum_{k=0}^n(-1)^k{n\choose k}f(k)$$&#xA;can often be attacked via the &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;Calculus of finite differences&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-08T10:20:08.917" />
  <row Id="4264" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:21:21.307" Score="3" ViewCount="0" Body="&lt;p&gt;I can't offer specific help, but I'd recommend thumbing through &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt; looking for techniques.  It has many sums that look similar, though of course the difficulty is in the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the book &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt;, but Concrete Mathematics gives an introduction to the content of A=B and in my opinion is easier to read, so I'd start with Concrete Mathematics.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-09-09T14:23:03.063" LastActivityDate="2010-09-09T14:23:03.063" />
  <row Id="4265" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:32:48.580" Score="10" ViewCount="0" Body="&lt;p&gt;You could consider the integral&#xA;$$ \int_{0}^{1} (1-x^2)^n dx .$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-08T10:32:48.580" CommentCount="1" />
  <row Id="4266" PostTypeId="1" AcceptedAnswerId="4282" CreationDate="2010-09-08T10:49:44.020" Score="1" ViewCount="49" Body="&lt;p&gt;In my previous question, &lt;a href=&quot;http://math.stackexchange.com/q/4198/622&quot;&gt;Weakly-compact cardinals&lt;/a&gt;, I was asking about weakly-compact cardinals and equivalent definitions to the basic one, which is $\kappa \to (\kappa)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of which was that $\kappa$ is inaccessible and has the &lt;em&gt;tree property&lt;/em&gt; (that is if any tree of cardinality $\kappa$ for which every level is of cardinality $&amp;lt;\kappa$ then it has a branch (i.e. a maximal chain) of cardinality $\kappa$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand the property itself and what it means. However, since $\aleph_1$ or $\aleph_\omega$ are clearly not weakly-compact cardinals, there should be a tree which contradicts this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you build this sort of tree?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-08T14:25:42.047" LastActivityDate="2010-09-08T14:25:42.047" Title="The tree property for non-weakly compact $\kappa$" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;counter-examples&gt;" AnswerCount="1" />
  <row Id="4267" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T10:54:31.803" Score="2" ViewCount="0" Body="&lt;p&gt;Your identity is a special case of the &lt;a href=&quot;http://mathworld.wolfram.com/Chu-VandermondeIdentity.html&quot; rel=&quot;nofollow&quot;&gt;Chu-Vandermonde identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}_2 F_1(-n,b;c;1)=\frac{(c-b)_n}{(c)_n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $b=\frac12$ and $c=\frac32$. More info on it is in A=B, as mentioned already by John.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-08T10:54:31.803" CommentCount="1" />
  <row Id="4268" PostTypeId="1" AcceptedAnswerId="4269" CreationDate="2010-09-08T11:17:01.167" Score="2" ViewCount="412" Body="&lt;p&gt;I know the definition of prime number when dealing with integers, but I can't understand why the following definition also works:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A prime is a quantity $p$ such that whenever $p$ is a factor of some product $a\cdot b$, then either $p$ is a factor of $a$ or $p$ is a factor of $b$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, take $4$ (which clearly is not a prime): it is a factor of $16=8\cdot 2$, so I should check that either $4\mid 8$ or $4\mid 2$. But $4\mid 8$ is true. So $4$ is a prime, which is absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that English is not my first language, so I may have easily misunderstood the above definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Let me try formalize the definition as I understood it: $p$ is prime if and only if $\exists a\exists b(p\mid a\cdot b)\rightarrow p\mid a\lor p\mid b$.&lt;/p&gt;&#xA;" OwnerUserId="1793" LastEditorUserId="1793" LastEditorDisplayName="" LastEditDate="2010-09-08T12:02:28.693" LastActivityDate="2010-09-19T22:34:53.240" Title="Alternate definition of prime number" Tags="&lt;prime-numbers&gt;&lt;definition&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="4269" PostTypeId="2" ParentId="4268" CreationDate="2010-09-08T11:23:41.937" Score="9" ViewCount="0" Body="&lt;p&gt;Note that this is &lt;strong&gt;whenever&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4|4 = 2\cdot 2$ however $4\not{|}2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the edited question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No! The word &quot;whenever&quot; comes to say that $p$ is prime if $\forall a \forall b (p | a\cdot b \Rightarrow (p | a \vee p | b))$&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-08T12:13:18.727" LastActivityDate="2010-09-08T12:13:18.727" />
  <row Id="4270" PostTypeId="1" AcceptedAnswerId="4275" CreationDate="2010-09-08T11:35:28.740" Score="1" ViewCount="160" Body="&lt;p&gt;I was wondering if a field is an algebra over itself (http://en.wikipedia.org/wiki/Algebra_over_a_field)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is a ring an algebra over itself (http://en.wikipedia.org/wiki/Algebra_(ring_theory)? If not, does the ring require to be commutative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-08T12:01:18.037" LastActivityDate="2010-09-08T18:24:30.387" Title="Is a field (ring) an algebra over itself?" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="4272" PostTypeId="1" CreationDate="2010-09-08T11:46:19.083" Score="2" ViewCount="161" Body="&lt;p&gt;I'm trying to express the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x_1, x_2) \; g(x_1, x_2) \; dx_1 dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of two real valued functions $f(x_1,x_2)$ and $g(x_1,x_2)$ in terms of their Fourier transforms $\tilde{f}(\omega_1, \omega_2)$ and $\tilde{g}(\omega_1, \omega_2)$. However, as $f(x_1,x_2)$ and $g(x_1,x_2)$ are real-valued functions and $I$ is not defined as the the product of $f(x_1,x_2)$ and $g(x_1,x_2)^\ast$ I'm not working with the complex conjugate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2 \right] \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; d\omega_1 d\omega_2 \right] dx_1 dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where it starts to get confusing. I'm not sure if I should write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_2 \tilde{g}(\omega_1, \omega_2) \; \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) } \; dx_1 \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and extract two one-dimensional dirac deltas, or write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_2 \tilde{g}(\omega_1, \omega_2) \; \;\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) + x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; dx_1 \; dx_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the hope of extracting the two-dimensional dirac delta. (I'm not even sure what the inverse Fourier transform of the two dimensional dirac delta is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate advise on how to proceed. (Thanks.)&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-09T20:53:43.363" Title="Trying to derive two dimensional version of Parseval's theorem (for real valued functions)" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4273" PostTypeId="1" AcceptedAnswerId="4377" CreationDate="2010-09-08T11:49:05.373" Score="1" ViewCount="90" Body="&lt;p&gt;Consider a situation where a person has partial knowledge, but we have a more complete picture. For example, suppose that we want to know the probability that a fish is red. Suppose that the person with partial information knows 1/3 of all fish are red, but we know that the particular species is actually red 2/3 of the time. Do these two separate probabilities have special names? If they don't have any standard names, what would you call them?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-09-10T08:55:29.000" LastActivityDate="2010-09-10T14:21:50.153" Title="Terminology for handling probabilities with partial knowledge" Tags="&lt;terminology&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="4274" PostTypeId="1" AcceptedAnswerId="4293" CreationDate="2010-09-08T12:08:34.743" Score="1" ViewCount="134" Body="&lt;p&gt;What is the area of the region bounded by $y = f_n(x)$ and the $x$-axis as $n$ gets large?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0=1-|x|$&lt;br&gt;&#xA;$f_n=1-|1-2f_{n-1}(x)|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the domain of $f$ to be the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By trying various values for $n$ and $x$  I got that $f_n(x) = 2^{n}\cdot f_0(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By integrating this I got that the area is given by $2^{n-1}\cdot x\cdot (2-|x|)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was also told that the area is in fact $1$ as $n$ approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula I came up with (guessed) seems to check out if I integrate $2^{n-1}\cdot f_0(x)$ between certain values and for a certain $n$ and compare it to manually expanding that certain $f_n$ and integrating it between the same values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is my formula wrong?&lt;/li&gt;&#xA;&lt;li&gt;How can we prove the area is 1?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="456" LastEditorUserId="456" LastEditorDisplayName="" LastEditDate="2010-09-08T13:53:37.023" LastActivityDate="2010-09-08T15:36:23.693" Title="Area of a recursive function" Tags="&lt;calculus&gt;&lt;functions&gt;&lt;recursive&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="4275" PostTypeId="2" ParentId="4270" CreationDate="2010-09-08T12:38:19.897" Score="3" ViewCount="0" Body="&lt;p&gt;A field K is an algebra over a field (the field being K). A commutative ring R is an algebra over a ring (the ring being R). &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-08T12:38:19.897" CommentCount="1" />
  <row Id="4276" PostTypeId="1" AcceptedAnswerId="4277" CreationDate="2010-09-08T12:50:07.077" Score="2" ViewCount="377" Body="&lt;p&gt;We can divide $7^{17} - 7^{15}$ by?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is $6$, but how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-08T17:49:41.067" LastActivityDate="2010-09-08T21:20:09.317" Title="We can divide $7^{17} - 7^{15}$ by?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4277" PostTypeId="2" ParentId="4276" CreationDate="2010-09-08T12:53:19.317" Score="13" ViewCount="0" Body="&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$7^{17}-7^{15}=(49-1)*7^{15}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and 48 is divisible by 6.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-08T12:53:19.317" CommentCount="23" />
  <row Id="4279" PostTypeId="1" AcceptedAnswerId="4289" CreationDate="2010-09-08T13:06:42.420" Score="2" ViewCount="162" Body="&lt;p&gt;How do I evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{\tau\rightarrow \infty} \int_{-c/\tau}^0 e^{-\beta x}(\tau x+c)^{(\lambda/\tau) -1}dx$$ to give $$1/\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I need to convert to gamma function but cannot simplify it to get the answer.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-09-10T16:13:16.397" LastActivityDate="2010-09-10T16:13:16.397" Title="evaluating $\lim_{\tau\rightarrow \infty} \int_{-c/\tau}^0 e^{-\beta x}(\tau x+c)^{(\lambda/\tau) -1}dx$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4280" PostTypeId="1" AcceptedAnswerId="4285" CreationDate="2010-09-08T13:13:17.363" Score="1" ViewCount="75" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently getting confused about indices in some spectral sequences. Assume we work in the category of modules for simplicity. Let $A^\cdot$ be a (bounded on the right) complex and let $B^\cdot$ (I don't think we have to assume anything about the boundedness of $B$). I want to compute $Ext^n(A^\cdot,B^\cdot)$, which is classically called hyperext (and sometimes denoted by $\mathbb{E}xt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, (perhaps assuming $B^\cdot$ to be bounded on the right), there exists a spectral sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^{p,q}_2 = Ext^p(A^\cdot,H^q(B^\cdot)) \Rightarrow Ext^{p+q}(A^\cdot,B^\cdot).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be an analogous by switching A and B, but I'm unsure of the indices, so my question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is $$ E^{p,q}_2 = Ext^q(A^\cdot,H^{-p}(B^\cdot)) \Rightarrow Ext^{q-p}(A^\cdot,B^\cdot) $$&#xA;  the right thing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1007" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-09-08T21:09:09.433" LastActivityDate="2010-09-08T21:09:09.433" Title="Contravariant Grothendieck Spectral Sequence" Tags="&lt;abstract-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4281" PostTypeId="2" ParentId="4260" CreationDate="2010-09-08T13:14:19.080" Score="2" ViewCount="0" Body="&lt;p&gt;The support of $M$ is the set $\{ \mathfrak q \in \text{ Spec } S \, | \, \kappa(q)\otimes_S M&#xA;\neq 0\}$, where $\kappa(\mathfrak q)$ denotes the fraction field of the domain $S/\mathfrak q$.  (Since $M$ is finitely generated, one sees by Nakayama's lemma that the stalk $M_{\mathfrak q}$ is non-zero if and only if $\kappa(\mathfrak q)\otimes_S M$ is non-zero).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The support of $R\otimes_S M$ is defined analogously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\mathfrak p$ is an element of Spec $R$, then&#xA;$$\kappa(\mathfrak p)\otimes_R (R\otimes _S M) = \kappa(\mathfrak p)\otimes_S M&#xA;= \kappa(\mathfrak p)\otimes_{\kappa(\mathfrak q)} (\kappa(\mathfrak q)\otimes_S M),$$&#xA;where $\mathfrak q$ is the preimage of $\mathfrak p$ in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\kappa(\mathfrak q) \to \kappa(\mathfrak p)$ is just an inclusion of fields,&#xA;we see that $\kappa(\mathfrak p)\otimes_R (R\otimes_S M) \neq 0$ if and only if&#xA;$\kappa(q)\otimes_S M \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the support of $R\otimes_S M$ is the preimage in Spec $R$ of the support&#xA;of $M$ in Spec $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this has a geometric interpretation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first description of the support says that, when $M$ is finitely generated, we can&#xA;check that the stalks of $M$ are non-zero by instead checking if the fibres are non-zero.&#xA;Then the computation relating the fibre of $R\otimes_S M$ at $\mathfrak p$ to&#xA;the fibre of $M$ at $\mathfrak q$ just says that the fibre of the pull-back is the pull-back of the fibre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that if you think geometrically, the result makes intuitive sense:&#xA;we are pulling back a sheaf (i.e. applying $\varphi^*$), and we can think of the sheaf as being a subset of&#xA;Spec $R$ (namely its support) with extra decoration (i.e. each point has a certain fibre&#xA;of the sheaf lying over it).  When we pull back the sheaf, we pull back the subset &#xA;(i.e. we pull back the support), and then at each point of the pull back we also pull back&#xA;the extra decoration.  Thinking about supports just involves forgetting the extra decoration.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-08T13:14:19.080" CommentCount="1" />
  <row Id="4282" PostTypeId="2" ParentId="4266" CreationDate="2010-09-08T13:22:55.863" Score="3" ViewCount="0" Body="&lt;p&gt;What you are looking for is the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Aronszajn_tree&quot; rel=&quot;nofollow&quot;&gt;Aronszajn tree&lt;/a&gt;. You can read about constructions of Aronszajn trees in any graduate level set theory text, and meanwhile, the Wikipedia page lists a summary of the basic facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;König's lemma states that $\aleph_0$-Aronszajn trees do not exist.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The existence of Aronszajn trees ($=\aleph_1$-Aronszajn trees) was proven by Nachman Aronszajn, and implies that the analogue of König's lemma does not hold for uncountable trees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The existence of $\aleph_2$-Aronszajn trees is undecidable (assuming a certain large cardinal axiom): more precisely, the continuum hypothesis implies the existence of an $\aleph_2$-Aronszajn tree, and Mitchell and Silver showed that it is consistent (relative to the existence of a weakly compact cardinal) that no $\aleph_2$-Aronszajn trees exist.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jensen proved that $V=L$ implies that there is a κ-Aronszajn tree (in fact a κ-Suslin tree) for every infinite successor cardinal κ.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cummings &amp;amp; Foreman (1998) showed (using a large cardinal axiom) that it is consistent that no $\aleph_n$-Aronszajn trees exist for any finite n other than 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If κ is weakly compact then no κ-Aronszajn trees exist. Conversely if κ is inaccessible and no κ-Aronszajn trees exist then κ is weakly compact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, let me point out a small inaccuracy in your question. The equivalence is that $\kappa$ is weakly compact iff it is inaccessible and has the tree property. It is not correct to drop the inaccessibility part, as you did, since it is consistent that $\aleph_2$ has the tree property.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-09-08T13:35:06.503" LastActivityDate="2010-09-08T13:35:06.503" />
  <row Id="4283" PostTypeId="2" ParentId="4175" CreationDate="2010-09-08T13:26:54.670" Score="2" ViewCount="0" Body="&lt;p&gt;The given quantity is the constant term in $ (-1)^m  \times { \sum_{ k \geq 0 } (-1)^{k} \binom{n}{k} (\frac{1}{x})^k } \times \sum_{k \geq 0} \binom{k}{m} x^k $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the constant term in $(-1)^m \times (1-\frac{1}{x})^n \times x^{m} \times (1-x)^{-(m+1)}$ =  $(-1)^{m+n} x^{m-n} \times (1-x)^{n-m-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m &gt;n$ clearly the constant term is 0, if $m &amp;lt; n$ then writing the above as $(-1)^{m+n} \frac{(1-x)^{n-m-1}}{x^{n-m}}$ and noting the maximum exponent of $x$ in numerator is $n-m-1$ we again see the constant term is 0. If $n=m$ then the constant term is clearly 1.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-09-08T13:26:54.670" />
  <row Id="4284" PostTypeId="1" CreationDate="2010-09-08T13:30:38.307" Score="4" ViewCount="78" Body="&lt;p&gt;On a lark, I wanted to know if one can use &lt;a href=&quot;http://mathworld.wolfram.com/PadeApproximant.html&quot; rel=&quot;nofollow&quot;&gt;Padé approximants&lt;/a&gt; to compute the exponential $\exp(z)$ of a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$. Since &lt;em&gt;Mathematica&lt;/em&gt; has a package meant for manipulating quaternions, as well as computing the exponential of a quaternion, I decided to try things out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerator $N_{pq}(z)$ and denominator $D_{pq}(z)$ of the $(p,q)$ Padé approximant for the exponential function can be computed using the following formulae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle N_{pq}(z)=\sum_{j=0}^p \frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}z^j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle D_{pq}(z)=\sum_{j=0}^q \frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}(-z)^j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after which the Padé approximant is $R_{pq}(z)=\frac{N_{pq}(z)}{D_{pq}(z)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity (and since they are the ones most used in practice), I will only consider here the &quot;diagonal&quot; case, $p=q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case where $z$ is a matrix is well-studied, and it is known that $R_{pq}(z)=N_{pq}(z)\cdot D_{pq}(z)^{-1}=D_{pq}(z)^{-1}\cdot N_{pq}(z)$. Banking on the isomorphism of quaternions to specially constructed matrices, I expected that the order of multiplication won't matter in the quaternion case, and experimentation seems to show that this holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was mystifying in my experiments is that, although in general $|R_{pp}(z)-\exp(z)|$ can be made small enough by taking $p$ large enough, one pattern I saw seems to indicate that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{pp}(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})\approx \overline{\exp(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $b&gt;0$, where the bar denotes conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see an immediate equivalent of this phenomenon in the theory of the matrix exponential, so I'm wondering why the Padé approximants to the quaternion exponential behave this way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those who want to retry my &lt;em&gt;Mathematica&lt;/em&gt; experiments:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;&amp;lt;Algebra`Quaternions`&#xA;&#xA;z=Quaternion[1.,-1.,1.,1.](*Quaternion[2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1]*)&#xA;&#xA;p=q=8;&#xA;&#xA;num=Sum[(((p+q-j)!p!)/((p+q)!j!(p-j)!)) z^j, {j, 0, p}];&#xA;&#xA;den=Sum[(((p+q-j)!q!)/((p+q)!j!(q-j)!)) (-z)^j, {j, 0, q}];&#xA;&#xA;{num**(den^-1),(den^-1)**num,Exp[z],Conjugate[Exp[z]]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-08T21:51:06.697" LastActivityDate="2010-09-08T21:51:06.697" Title="Using Padé approximants for the quaternion exponential" Tags="&lt;approximation&gt;&lt;quaternions&gt;" />
  <row Id="4285" PostTypeId="2" ParentId="4280" CreationDate="2010-09-08T13:33:18.003" Score="2" ViewCount="0" Body="&lt;p&gt;I think that you will want $B^{\bullet}$ to be bounded below (i.e. on the left), so that&#xA;after replacing either $A^{\bullet}$ by a projective resolution or $B^{\bullet}$ by an&#xA;injective resolution, the complex $Hom(\text{complex},\text{complex})$ that you compute&#xA;will be bounded below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the spectral sequence you are looking for comes about by taking this  $Hom$&#xA;complex (the one you get after replacing $A^{\bullet}$ or $B^{\bullet}$ by a projective&#xA;or injective resolution), writing it as the total complex of a double complex, and applying the standard&#xA;machine.  If you do this, you will find that the second spectral sequence that you want is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_2^{p,q} = Ext^p(H^{-q}(A^{\bullet}),B^{\bullet}) \Rightarrow Ext^{p+q}(A^{\bullet},B^{\bullet}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Your first spectral sequence comes from replacing $A^{\bullet}$ by a projective resolution.&#xA;This one comes from replacing $B^{\bullet}$ by an injective resolution.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-08T13:33:18.003" />
  <row Id="4286" PostTypeId="1" AcceptedAnswerId="4290" CreationDate="2010-09-08T13:58:32.343" Score="1" ViewCount="111" Body="&lt;p&gt;Suppose I have a 4-index, real-valued tensor $\lambda_{ij}^{ab}$ where $i$ and $j$ are from some subspace $O$ of the whole numbers, and $a$ and $b$ are from some other subspace $V,$ which is disjunct from $O.$  Suppose also that we know that $\lambda$ is antisymmetric.  If we impose the constraint that $\lambda$ is Hermitian (i.e. $\lambda_{ij}^{ab} = (\lambda_{ab}^{ij})^*=\lambda_{ab}^{ij}$ since $\lambda$ is real), and we have some function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \sum_{ij\in O, ab \in V} k_{ab}^{ij}\lambda_{ij}^{ab} + \sum_{ij\in O, ab \in V} k_{ij}^{ab}\lambda_{ab}^{ij}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $k_{ij}^{ab}$ is some antisymmetric, Hermitian tensor independent of $\lambda$) and then we take the partial derivative of that function with respect to some specific $\lambda_{a^\prime b^\prime}^{i^\prime j^\prime}$, what do we get?  What I don't understand is that if you make the substitution for the Hermitian constraint before differentiating, you can combine the two summations into one, but then differentiating gives you the same answer with a different constant, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer that the derivation I am reading gives is $k_{ij}^{ab}$, rather than $2k_{ij}^{ab}$.  Could someone explain why?&lt;/p&gt;&#xA;" OwnerUserId="1590" LastActivityDate="2010-09-08T14:56:33.960" Title="Partial Derivatives and Dependent Variables" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4287" PostTypeId="2" ParentId="3946" CreationDate="2010-09-08T14:24:04.060" Score="1" ViewCount="0" Body="&lt;p&gt;What exactly do you want?&#xA;If you want step-by-step software algorithm to draw a Bezier curve,&#xA;and the &lt;a href=&quot;http://www.tinaja.com/cubic01.asp&quot; rel=&quot;nofollow&quot;&gt;&quot;Don Lancaster's Guru's Lair Cubic Spline Library&quot;&lt;/a&gt; doesn't have the answer,&#xA;then the StackOverflow.com website is a better place to ask such questions.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-08T14:24:04.060" />
  <row Id="4288" PostTypeId="2" ParentId="4279" CreationDate="2010-09-08T14:34:57.997" Score="3" ViewCount="0" Body="&lt;p&gt;Try the subsitution $s=\beta(x+\frac{c}{\tau})$. Then use: &#xA;\begin{equation}&#xA;\int_0^a t^{b-1}e^{-t}dt=a^be^{-a}\sum_0^{\infty}\frac{a^n}{b(b+1)\ldots(b+n)}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-08T14:43:15.930" LastActivityDate="2010-09-08T14:43:15.930" CommentCount="1" />
  <row Id="4289" PostTypeId="2" ParentId="4279" CreationDate="2010-09-08T14:42:14.090" Score="3" ViewCount="0" Body="&lt;p&gt;A sketch of the proof, as I'm in a hurry - I hope I've done all the calculations correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $y = \tau x + c$ to obtain&#xA;$$&#xA;\frac{1}{\tau} \int_0^c e^{\frac{\beta}{\tau} (c - y)} y^{\lambda / \tau - 1} d y&#xA;$$&#xA;Now, partial integration yields&#xA;$$&#xA;\frac{1}{\lambda} c^{\lambda / \tau} + \frac{\beta}{\lambda \tau} \int_0^c e^{\frac{\beta}{\tau} (c - y)} y^{\lambda / \tau} dy&#xA;$$&#xA;The limit of the first summand is $\frac{1}{\lambda}$, and the integrand in the second part is bounded (maximum is obtained for $y = \frac{\lambda}{\beta}$), so the second summand will vanish when $\tau \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-08T14:42:14.090" CommentCount="1" />
  <row Id="4290" PostTypeId="2" ParentId="4286" CreationDate="2010-09-08T14:56:33.960" Score="1" ViewCount="0" Body="&lt;p&gt;It turns out that $2k_{ij}^{ab}$ is, in fact, the correct answer; it's just that when the derivation was done, the next step was to set the derivative equal to zero, which allows you to divide through by 2, and so the original deriver still got the right answer.  Sorry for the confusion!&lt;/p&gt;&#xA;" OwnerUserId="1590" LastActivityDate="2010-09-08T14:56:33.960" />
  <row Id="4291" PostTypeId="2" ParentId="4273" CreationDate="2010-09-08T15:26:49.897" Score="4" ViewCount="0" Body="&lt;p&gt;I would call the second of these a &lt;em&gt;conditional&lt;/em&gt; probability when trying to emphasise the difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of a chosen fish being red is 1/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditional probability of a fish being red, given that it is species X, is 2/3.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-09-08T15:26:49.897" CommentCount="2" />
  <row Id="4292" PostTypeId="2" ParentId="4253" CreationDate="2010-09-08T15:29:13.577" Score="2" ViewCount="0" Body="&lt;p&gt;Here is an answer for your first question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easier than dealing with simple symmetric random walk.&#xA;When the steps are diffuse,&#xA;you don't have to worry about &lt;em&gt;ties&lt;/em&gt; or &lt;em&gt;parity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking at a Brownian motion, evaluated at integer times $t$.&#xA;Then defining  $$T_0=\inf(t &gt; 0, t\in {\mathbb Z}: B_t &amp;lt; 0)$$ and integer $t &gt; 0$ &#xA;we have $$ P(T_0 &gt; t)=P(B_1 &gt; 0,B_2 &gt; 0,\dots, B_t &gt; 0)={2t\choose t}/4^t.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the answer is the same for any random walk with&#xA;symmetric, diffuse increments.&#xA;This follows from Theorem 9.11 of Kallenberg's &lt;em&gt;Foundations of Modern Probability (2nd edition)&lt;/em&gt;, for example.&#xA;(A great book, by the way!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key insight is that you should not to try to solve the&#xA;probability exactly, but rather set up a recursion that the probabilities solve.&#xA;Setting $p(t)=P(T_0&gt;t)$ for integer $t\geq 0$, you can show that&#xA;for any $t$ we have $\sum_{j=0}^t p(t-j)p(j)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probabilistic justification for the recursion formula is that&#xA;every path of a diffuse, symmetric, discrete-time random walk can be broken into two&#xA;&lt;em&gt;positive&lt;/em&gt; paths at the time $j$  where it achieves its minimum value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also find Konrad Jacob's book &lt;em&gt;Discrete Stochastics&lt;/em&gt; to be useful. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-08T15:29:13.577" />
  <row Id="4293" PostTypeId="2" ParentId="4274" CreationDate="2010-09-08T15:36:23.693" Score="4" ViewCount="0" Body="&lt;p&gt;You must be integrating between -1 and 1; otherwise, the integral diverges.  Within this region, $f_n$ is a periodic sawtooth.  This is obvious by following the steps geometrically for any one of the teeth: $-2f_{n-1}$ doubles the height of $f_{n-1}$ and flips the graph over the x-axis, which is then raised by $1$.  The absolute value creates a new &quot;tooth&quot;, and then subtracting all that from $1$ flips everything over.  The process converts each tooth into a pair of teeth of the same height but half the width, whence the new teeth have half the original area of the teeth they have replaced.  Two of them occupy the original area.  Therefore, by induction, the integrals of &lt;em&gt;all&lt;/em&gt; the $f_n$ between -1 and 1 equal the integral of $f_0$, which is 1.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-08T15:36:23.693" CommentCount="4" />
  <row Id="4294" PostTypeId="2" ParentId="3973" CreationDate="2010-09-08T15:42:47.733" Score="4" ViewCount="0" Body="&lt;p&gt;Apparently, giving the input&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Integrate[Exp[Piecewise[{{x, x &amp;gt; 0}}, -x]],x]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to WolframAlpha returns a piecewise result, which may be what you're expecting.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-08T15:42:47.733" />
  <row Id="4295" PostTypeId="2" ParentId="4261" CreationDate="2010-09-08T15:58:14.510" Score="1" ViewCount="0" Body="&lt;p&gt;There are a powerful &lt;strong&gt;algorithms generalizing telescopy&lt;/strong&gt; (Gosper, Zeilberger et al.) that easily tackle this case of the &lt;strong&gt;Chu-Vandermonde identity&lt;/strong&gt; and much more complicated sums. For example, see &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.71.3348&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which gives as an application a very interesting q-analogy - namely that L. J. Rogers' classical finite version of &lt;strong&gt;Euler's pentagonal number theorem&lt;/strong&gt; is simply the dual of a special case of a q-Chu-Vandermonde.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-08T15:58:14.510" />
  <row Id="4296" PostTypeId="2" ParentId="4270" CreationDate="2010-09-08T16:09:04.717" Score="1" ViewCount="0" Body="&lt;p&gt;[This should be a comment, I guess]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the kind of question you can answer yourself by simply checking the conditions in the definition are satisfied. You &lt;em&gt;really&lt;/em&gt; should do it, independently of what anyone answers here or elsewhere!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should know, though, that a field can be an algebra over itself in many different ways, as soon as you have a field $K$ such that there are at least two ring homomorphisms $K\to K$ (there is always one such ring homomorphism, the identity map, so to get something interesting, we need two). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to give examples of fields which do have such non-identity homomorphisms. For example, every Galois extension of $\mathbb Q$, or the field of rational functions $\mathbb C(x)$ (there is a non-identity homomorphism $\mathbb C(x)\to \mathbb C(x)$ which maps $x$ to $x^2$, and in fact many similar other ones), &amp;amp;c.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-08T16:09:04.717" CommentCount="2" />
  <row Id="4297" PostTypeId="1" CreationDate="2010-09-08T16:26:44.983" Score="0" ViewCount="150" Body="&lt;p&gt;Let $n \in \mathbb{N}$. How to find the least positive integer $\gamma$ such that $$ \sum\limits_{i=1}^{n} \cos{\theta_{n}} \leq \gamma$$ provided $$\prod\limits_{i=1}^{n} \tan{\theta_{i}} = 2^{\frac{n}{2}}$$ for any $\theta_{i} \in \Bigl(0,\frac{\pi}{2}\Bigr)$, $i=1,2,...,n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had trouble solving this problem, some time ago. If i am not mistaken, i had seen this in Loney's trigonometry book. (Although i don't remember it correctly!).&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-08T20:34:11.083" Title="Finding the least postive integer such that $\sum \cos{\theta_{i}} \leq \gamma$" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4298" PostTypeId="1" AcceptedAnswerId="4299" CreationDate="2010-09-08T16:43:58.943" Score="1" ViewCount="133" Body="&lt;p&gt;Suppose $K$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, and let $\mathcal{O}_K$ denote its ring of integers.  I want to show that for any prime number $p$, $p\mathcal{O}_K$ is not a prime ideal of $\mathcal{O}_K$. The hint I was given was to consider the factorization of $x^3-2$ over $F_p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question(s)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct was to use Kummer's criterion, which guarantees that $p\mathcal{O}_K$ is not prime as long as $x^3-2$ is reducible (has a root) in $F_p$ [and $p&gt;3$].  But how to handle the case when $p=2,3$ (I suspect it ramifies, but I can't seem to show this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second instinct was that if $p\mathcal{O}_K$ is prime in $\mathcal{O}_K$, then $\mathcal{O}_K/p\mathcal{O}_K$ would be the splitting field of $x^3-2$ over $F_p$, and so I just need to check that the degree is always less than $[K:\mathbb{Q}]=6$.  But is it always true in this case that $\mathcal{O}_K/p\mathcal{O}_K$ is the splitting field of $x^3-2$ over $F_p$?&lt;/p&gt;&#xA;" OwnerUserId="1544" LastActivityDate="2010-09-09T03:44:18.263" Title="Factoring prime ideals in a Galois extension of Q" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" />
  <row Id="4299" PostTypeId="2" ParentId="4298" CreationDate="2010-09-08T16:50:38.300" Score="3" ViewCount="0" Body="&lt;p&gt;Let's work inside $L=\mathbb{Q}(\sqrt[3]2)$.&#xA;It really should be easy to see that $2$ ramifies (there an ideal $\mathfrak{p}$&#xA;with $\mathfrak{p}^3=2\mathfrak{O}_L$). Proving that $3$ ramifies is a little trickier.&#xA;If we could find an element of $\mathfrak{O}_L$ whose minimal polynomial is Eisenstein&#xA;at $3$, that would do. An Eisenstein polynomial at $3$ would be&#xA;$$x^3+ax^2+bx+c$$&#xA;where $a$, $b$ and $c$ are divisible by $3$ and $c$ isn't divisible by $9$.&#xA;Obviously $\sqrt[3]2$ doesn't work - but is there something similar which does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;I was implicitly arguing that since the primes $2$ and $3$ aren't inert in&#xA;$L=\mathbb{Q}(\sqrt[3]2)$ then &lt;em&gt;a fortiori&lt;/em&gt; they aren't inert in its Galois&#xA;closure $K$. But no prime remains inert in a non-cyclic Galois extension&#xA;of a number field. For a prime to remain inert, its decomposition group&#xA;is the whole Galois group and its inertia group is trivial. But the&#xA;quotient of the decomposition group by the inertia group must be cyclic&#xA;(its the Galois group of the extension of residue fields)....&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-08T18:43:29.873" LastActivityDate="2010-09-08T18:43:29.873" CommentCount="5" />
  <row Id="4300" PostTypeId="2" ParentId="4270" CreationDate="2010-09-08T17:04:15.667" Score="0" ViewCount="0" Body="&lt;p&gt;The answer is clear if you know the universal view of $\rm\:R\:$-algebras, namely they are simply those rings $\rm\:A\:$ into which one can always evaluate $\rm R[x]$ for every $\rm\:a \in A\:$. More directly,  an $\rm\:R$-algebra is just a ring $\rm A\:$ containing a central subring $\rm R'$ that's a ring image of $\:\rm R\:,\:$ i.e. $\rm\: R'\:$ is either an embedding of $\rm\:R\:$ or $\rm\:R/I\:$ for some ideal $\rm\;I\in R\:.\;$ Being central is precisely the condition needed for elts in $\rm\:R'\:$ to serve as &quot;coefficients&quot; in the sense that this makes polynomial rings $\rm\: R[x]\:$ be a &lt;strong&gt;universal&lt;/strong&gt; $\rm\:R\:$-algebras. Namely, the fact that the coefficients commute with all elts of $\rm\:A\:$ is precisely what is required to make the evaluation map be a ring homomorphism $\rm\: R[x]\to A\:,\:$&#xA;viz $\rm\;\; r\; x = x\: r\;$ in $\rm\:R[x]\;\Rightarrow\;  r\: a = a\:r\;\;$ by evaluation $\rm\: x\to a\in A,\:$ i.e. by definition polynomial multiplication assumes that the coefficients commute with the indeterminates, so this property must remain true at values of indeterminates if evaluation is to be a ring homomorphism.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-08T18:24:30.387" LastActivityDate="2010-09-08T18:24:30.387" />
  <row Id="4301" PostTypeId="2" ParentId="4298" CreationDate="2010-09-08T17:07:08.013" Score="1" ViewCount="0" Body="&lt;p&gt;In general, if the ring of integers of $\mathcal{O}_K$ is of the form $\mathbb{Z}[\alpha]$, and $q(x)$ is the irreducible polynomial of $\alpha$ over $\mathbb{Q}$, then there is a natural bijection between the primes in $\mathcal{O}_K$ that lie above the rational prime $p$ and the irreducible factors of $\overline{q}(x)$ in $\mathbb{F}_p$, where $\overline{q}$ is the image of $q(x)$ under reduction modulo $p$. The bijection is that a prime $\mathfrak{p}$ of $\mathcal{O}_K$ lies above $(p)$ corresponds to the factor $\overline{Q}(x)$ if and only if $\mathfrak{p}$ is the kernel of the map $\mathbb{Z}[\alpha]\to\mathbb{F}_p[\alpha]$. So the factorization of $(p)$ in $\mathcal{O}_K$ exactly mimicks the factorization of $\overline{q}(x)$ over $\mathbb{F}_p$: same number of factors, same repetitions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to factor $x^3-2$ over $\mathbb{F}_p$; this suffices here because if $\alpha$ is a root of $x^3-2$, then $\mathcal{O}_K = \mathbb{Z}[\alpha]$ (see for example Problem 41(a)-(d) in Chapter 2 of Marcus's &lt;em&gt;Number Fields&lt;/em&gt;). &lt;strong&gt;Edit:&lt;/strong&gt; I was working in $\mathbb{Q}(\sqrt[3]{2})$ instead of the splitting field; sorry. But from the fact that they ramify in this subextension, you know they will ramify in $K$; the only question is whether they ramify as $\mathfrak{p}^6$ or as $\mathfrak{p_1}^3\mathfrak{p_2}^3$, depending on whether the prime above $p$ splits or ramifies, or is inert in $K$. Since you can go from $$\mathbb{Q}(\sqrt[3]{2})$ to $K$ by adjoining $\omega$, and the ring of integers will be, if I'm not mistaken, $\mathbb{Z}[\alpha][i]$, you can use the same theorem above (which holds for arbitrary integral extensions, not just over $\mathbb{Z}$) to get that next step, looking at $x^2+x+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $p=2$, the polynomial factors over $\mathbb{F}_p$ as $x^3$, so $(2) = \mathfrak{p}^3$ for some prime $\mathfrak{p}$ of $\mathcal{O}_K$; for $p=3$, you have $x^3 + 2 = x^3 - 1 = (x-1)^3$, so again $(3)=\mathfrak{q}^3$ for some prime $\mathfrak{q}$ of $\mathcal{O}_K$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-09T03:44:18.263" LastActivityDate="2010-09-09T03:44:18.263" CommentCount="3" />
  <row Id="4302" PostTypeId="2" ParentId="4276" CreationDate="2010-09-08T17:35:22.310" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad X^{n+2} - X^n \;=\; (X^2 - 1)\: X^n \;=\; (X-1)\: (X+1)\: X^n$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.114.548&quot; rel=&quot;nofollow&quot;&gt;Aurifeuille, Le Lasseur and Lucas&lt;/a&gt; discovered so-called &lt;a href=&quot;http://www.ams.org/journals/mcom/2006-75-253/S0025-5718-05-01766-7/S0025-5718-05-01766-7.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Aurifeuillian factorizations&lt;/strong&gt;&lt;/a&gt; of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\  D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the &lt;a href=&quot;http://homes.cerias.purdue.edu/~ssw/cun/&quot; rel=&quot;nofollow&quot;&gt;Cunningham Project&lt;/a&gt;. Below are some simple examples of such factorizations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl}&#xA;x^4 + 2^2 \quad=&amp;amp;  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\&#xA;\frac{x^6 + 3^2}{x^2 + 3} \quad=&amp;amp;  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\&#xA;\frac{x^{10} - 5^5}{x^2 - 5} \quad=&amp;amp;  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\&#xA;\frac{x^{12} + 6^6}{x^4 + 36} \quad=&amp;amp;  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-08T21:20:09.317" LastActivityDate="2010-09-08T21:20:09.317" CommentCount="3" />
  <row Id="4303" PostTypeId="2" ParentId="4256" CreationDate="2010-09-08T18:03:16.837" Score="0" ViewCount="0" Body="&lt;p&gt;For question 2, consider any f(n) which is 2^n for n which are not powers of 2. This lets us devide our sum into two parts, both of which converge.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-08T18:03:16.837" />
  <row Id="4304" PostTypeId="1" AcceptedAnswerId="4331" CreationDate="2010-09-08T18:33:42.550" Score="1" ViewCount="211" Body="&lt;p&gt;I am trying to understand &lt;a href=&quot;http://en.wikipedia.org/wiki/All-pairs_testing&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;pairwise testing&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many combinations of tests would be there for example, if&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;a&lt;/code&gt; can take values from 1 to m&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;b&lt;/code&gt; can take values from 1 to n&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;c&lt;/code&gt; can take values from 1 to p&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a&lt;/code&gt;, &lt;code&gt;b&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt; can take m, n and p distinct values respectively. &lt;strong&gt;What are the total number of pairwise combinations possible?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With a pairwise testing tool that I am testing, I am getting 40 results for m = n = p = 6. I am trying to mathematically understand how I get 40 values. &lt;/p&gt;&#xA;" OwnerUserId="324" LastEditorUserId="324" LastEditorDisplayName="" LastEditDate="2010-09-08T18:53:12.003" LastActivityDate="2010-09-09T11:51:05.527" Title="How to derive the number of pairwise combinations of a set of factors?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4305" PostTypeId="2" ParentId="4272" CreationDate="2010-09-08T18:33:58.773" Score="1" ViewCount="0" Body="&lt;p&gt;I don't see what you're doing. Are you trying to show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbb R^2} f(x) g(x) \, dx = \int_{\mathbb R^2} \widehat{f(\xi)} \overline{\widehat{g(\xi)}} \, d\xi$$?&#xA;This is usually called Parseval's identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove it as follows (even in $\mathbb R^n$), first prove by Fubini (very simple) that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbb R^2} f(x) \widehat{g(x)} \, dx = \int_{\mathbb R^2} \widehat{g(x)} f(x) \, dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $g = \overline{\hat{h}}$ and note then that $\hat{g} = \overline{h}$ (by Fourier inversion), then we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbb R^2} f(x) h(x) \, dx = \int_{\mathbb R^2} \widehat{g(x)} \overline{\widehat{h(x)}} \, dx$$&#xA;(assuming $h$ is real, otherwise we get the conjugate).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-09T19:56:42.063" LastActivityDate="2010-09-09T19:56:42.063" CommentCount="3" />
  <row Id="4306" PostTypeId="2" ParentId="4304" CreationDate="2010-09-08T18:45:30.167" Score="1" ViewCount="0" Body="&lt;p&gt;If each parameter had $10$ choices you'd be testing $300$ vs $1000$ combinations, namely hold $\rm a$ constant and vary $\rm b,c$ through $10\cdot 10 = 100$ values. Similarly hold, $\rm b$ constant; then $\rm c$. As the number of variables $\rm k$ increases you get better savings, roughly $\rm (k N)^2$ vs. $\rm N^k$, where $\rm N =$ max domain size. For QA purposes usually such rough upper bounds suffice. Do you have an intended application where you need something more precise? If so perhaps you should reveal some further details, e.g. the distribution of the sizes of the domains, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After reviewing your latest revision, it appears that the following web pages may be of interest: &lt;a href=&quot;http://www.developsense.com/pairwiseTesting.html&quot; rel=&quot;nofollow&quot;&gt;Pairwise Testing&lt;/a&gt;, which refers to various &lt;a href=&quot;http://en.wikipedia.org/wiki/Taguchi_methods&quot; rel=&quot;nofollow&quot;&gt;Taguchi methods&lt;/a&gt; such as those &lt;a href=&quot;http://www.freequality.org/html/tools.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. See also these links to introductions to &lt;a href=&quot;http://www.combinatorialtesting.com/clear-introductions-1&quot; rel=&quot;nofollow&quot;&gt;combinatorial testing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-09T02:14:11.153" LastActivityDate="2010-09-09T02:14:11.153" />
  <row Id="4308" PostTypeId="2" ParentId="4297" CreationDate="2010-09-08T19:20:14.317" Score="3" ViewCount="0" Body="&lt;p&gt;Try this. Find the maximum of $$\sum_{k=1}^n \frac{1}{\sqrt{1+{x_k}^2}},$$&#xA;where $\prod_{k=1}^n {x_k}^2=2^n,$ and $x_k = \tan\theta_k$, using Lagrange multipliers. Each $x_k$ satisfies the same equation involving the multiplier $\lambda,$&#xA;$$\lambda 2^{n+1} = \frac{{x_k}^2}{(1+{x_k}^2)^{3/2}}.$$&#xA;And so all the $x_k$ are equal and hence $x_k = \sqrt{2}.$&#xA;Therefore the maximum value of our sum is $n/\sqrt{3}$ and so the required smallest integer is&#xA;$$\lceil \frac{n}{\sqrt{3}} \rceil.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-08T20:34:11.083" LastActivityDate="2010-09-08T20:34:11.083" CommentCount="2" />
  <row Id="4309" PostTypeId="1" AcceptedAnswerId="4310" CreationDate="2010-09-08T19:21:41.827" Score="-2" ViewCount="162" Body="&lt;p&gt;Denote by $\nu(n)$ the exponent of 2 in the prime factorization of $n!$. Show that for arbitrary positive integers $a$ and $m$ there exists an integer $n&gt;1$ for which $\nu(n) \equiv \ a \ (\text{mod}\ m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While searching on the internet i found this link &lt;a href=&quot;http://web.mit.edu/yufeiz/www/imo2008/mocks-sol.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/yufeiz/www/imo2008/mocks-sol.pdf&lt;/a&gt; and problem 3 in the first page talks somewhat related, regarding the function $\nu(n)$. But i am not sure about this fact.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-08T19:33:59.877" Title="Exponent of 2 in prime factorization of $n!$" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4310" PostTypeId="2" ParentId="4309" CreationDate="2010-09-08T19:33:59.877" Score="5" ViewCount="0" Body="&lt;p&gt;You really should cite the &lt;a href=&quot;http://www.komal.hu/verseny/feladat.cgi?a=feladat&amp;amp;f=A486&amp;amp;l=en&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt; of your problem. And mention that it comes with a complete solution.&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2010-09-08T19:33:59.877" CommentCount="3" />
  <row Id="4312" PostTypeId="1" AcceptedAnswerId="5484" CreationDate="2010-09-08T20:18:51.057" Score="4" ViewCount="163" Body="&lt;p&gt;Given a sequence of row sums $r_1, \ldots, r_m$ and column sums $c_1, \ldots, c_n$, all positive, I'd like to find a matrix $A_{m\times n}$ consistent with the given row and column sums that has the fewest nonzero entries. That is, we want to minimize $|\{a_{ij} \ne 0\}|$ under the constraint that $\sum_j a_{ij} = r_i$ and $\sum_i a_{ij} = c_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also think of this as a network flow problem, with $m$ sources and $n$ sinks that must be connected with the minimum number of edges to achieve the specified flow out of each source ($r_i$) or into each sink ($c_j$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a known problem, or does it fit into a well-known class of problems? What is the computational complexity of finding the optimal solution? According to the comments on &lt;a href=&quot;http://www.reddit.com/r/math/comments/darwh/accounting_help_with_a_dash_of_linear_algebra/&quot; rel=&quot;nofollow&quot;&gt;this Reddit discussion&lt;/a&gt;, it could be related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cutting_stock_problem&quot; rel=&quot;nofollow&quot;&gt;cutting stock problem&lt;/a&gt;, but it's not exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation for this problem comes from settling debts. Suppose you have a set of people $P$ with debts between them. These debts can be settled by (1) finding the net debt $d(p)$ owed to each person $p \in P$, (2) dividing $P$ into two disjoint sets of people who are owed, $P_+ = \{p : d(p) &gt; 0\}$, and people who owe, $P_- = \{p : d(p) &amp;lt; 0\}$, and (3) having the people in $P_-$ pay the ones in $P_+$. Then the problem of resolving debts with the fewest transactions is equivalent to the above optimization problem, with $a_{ij}$ being the amount that the $i$th person in $P_-$ must give the $j$th person in $P_+$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-09-26T06:38:40.120" LastActivityDate="2010-09-26T06:38:40.120" Title="Sparsest matrix with specified row and column sums" Tags="&lt;combinatorics&gt;&lt;optimization&gt;" AnswerCount="2" />
  <row Id="4313" PostTypeId="2" ParentId="2655" CreationDate="2010-09-08T20:28:19.610" Score="2" ViewCount="0" Body="&lt;p&gt;Andrew Booker's &lt;a href=&quot;http://primes.utm.edu/nthprime/&quot; rel=&quot;nofollow&quot;&gt;Nth prime&lt;/a&gt; page is excellent... but it can't handle your example number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have custom code that can calculate values up to about 2^64, but your number is larger than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Dusart [1], we can say that its rank is somewhere between 24244547260299402427 and 24247918127257270377.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Riemann Hypothesis is true, then we know by Schoenfeld [2] that its rank is somewhere between 24245911027060346607 and 24245911157987206331.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Pierre Dusart, 'Estimates of Some Functions Over Primes without R.H.', preprint (2010), &lt;a href=&quot;http://arxiv.org/abs/1002.0442&quot; rel=&quot;nofollow&quot;&gt;arXiv:1002.0442&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Lowell Schoenfeld, 'Sharper Bounds for the Chebyshev Functions theta(x) and psi(x). II'. Mathematics of Computation, Vol 30, No 134 (Apr 1976), pp. 337-360.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-08T20:28:19.610" />
  <row Id="4314" PostTypeId="2" ParentId="1257" CreationDate="2010-09-08T20:39:47.683" Score="6" ViewCount="0" Body="&lt;p&gt;Far better than sieving in the large range ShreevatsaR suggested (which, for the 10^15-th prime, has 10^15 members and takes about 33 TB to store in compact form), take a good first guess like Riemann's R and use one of the advanced methods of computing pi(x) for that first guess.  (If this is far off for some reason -- it shouldn't be -- estimate the distance to the proper point and calculate a new guess from there.)  At this point you can sieve the small distance, perhaps just 10^8 or 10^9, to the desired number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about 100,000 times faster for numbers around the size I indicated.  Even for numbers as small as 10 to 12 digits this is faster if you don't have a precomputed table large enough to contain your answer.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-08T20:39:47.683" CommentCount="2" />
  <row Id="4315" PostTypeId="2" ParentId="4312" CreationDate="2010-09-08T21:03:18.290" Score="4" ViewCount="0" Body="&lt;p&gt;This is linear optimization in the $L_0$ (Hamming) metric which in full generality is NP-hard, so either there is a solution adapted to the specific structure of the debt settlement equations, or it is intractable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One structure the debt settlement system has is sparsity.  It can be written as $m+n$ equations in $mn$ variables, with only $2mn$ (out of $(mn)^2$) nonzero coefficients. This is a very sparse system to which $L_0$-$L_1$ transference (so-called &quot;compressed sensing&quot;) methods may apply, reducing the problem to linear programming.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-08T21:36:00.543" LastActivityDate="2010-09-08T21:36:00.543" CommentCount="2" />
  <row Id="4316" PostTypeId="1" CreationDate="2010-09-08T22:34:25.857" Score="0" ViewCount="155" Body="&lt;p&gt;Please explain your reasoning too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a lucky draw, I estimate around 240, with about 12 candies in 20 layers. The size of candies is about standard candy. I am sorry I have no more info to share.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xx3Zk.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1802" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-08T22:37:39.593" LastActivityDate="2010-09-08T22:37:39.593" ClosedDate="2010-09-08T23:06:29.800" Title="How many candies are in this picture?" Tags="&lt;geometry&gt;" CommentCount="4" />
  <row Id="4317" PostTypeId="1" AcceptedAnswerId="4318" CreationDate="2010-09-08T23:31:32.027" Score="12" ViewCount="360" Body="&lt;p&gt;For computing the present worth of an infinite sequence of equally spaced payments $(n^{2})$ I had the need to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{x^{n}}=\dfrac{x(x+1)}{(x-1)^{3}}\qquad x&gt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method I used was based on the geometric series $\displaystyle\sum_{n=1}^{\infty}x^{n}=\dfrac{x}{1-x}$ differentiating each side followed by a multiplication by $x$, differentiating a second time and multiplying again by $x$. There is at least a second (more difficult) method that is to compute the series partial sums and letting $n$ go to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a closed form for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{p}}{x^{n}}\qquad x&gt;1,p\in\mathbb{Z}^{+}\quad ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the sketch of its proof in case it exists?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-05T10:19:02.307" LastActivityDate="2010-10-05T10:19:02.307" Title="Generalizing $\sum_{n=1}^{\infty }n^{2}/x^{n}$ to $\sum_{n=1}^{\infty }n^{p}/x^{n}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="4318" PostTypeId="2" ParentId="4317" CreationDate="2010-09-08T23:48:13.087" Score="9" ViewCount="0" Body="&lt;p&gt;The general closed form is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{k=1}^{\infty} k^n x^k = \frac{1}{(1 - x)^{n+1}} \left( \sum_{m=0}^{n} A(n, m) x^{m+1} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A(n, m)$ are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_number&quot; rel=&quot;nofollow&quot;&gt;Eulerian numbers&lt;/a&gt;.  When I have time I will edit with a few more details.  If you only want the answer for a particular small value of $n$ then see Section 3 of &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;my notes on generating functions&lt;/a&gt;.  I will also mention that for a particular value of $n$ one can deduce the answer by using the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{k=0}^{\infty} {k+n \choose n} x^k = \frac{1}{(1 - x)^{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and writing $k^n$ as a linear combination of the polynomials ${k+r \choose r}$ (in $k$), for example using a finite difference table.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-08T23:48:13.087" CommentCount="6" />
  <row Id="4319" PostTypeId="2" ParentId="4317" CreationDate="2010-09-08T23:54:33.407" Score="3" ViewCount="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://mathworld.wolfram.com/GeometricDistribution.html&quot; rel=&quot;nofollow&quot;&gt;mathworld&lt;/a&gt; The $n^{th}$ moment of the geometric distribution with parameter $p$, which is $ \sum p (1-p)^n n^k $ can be expressed in terms of the polylogarithm function.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-09-08T23:54:33.407" CommentCount="1" />
  <row Id="4320" PostTypeId="2" ParentId="4238" CreationDate="2010-09-09T00:19:34.470" Score="3" ViewCount="0" Body="&lt;p&gt;The smallest example is 2×AΓL(1,8) of order 336.  A fairly standard permutation representation is generated by (+,-), (∞,0)(1,3)(2,6)(4,5), (1,2,3,4,5,6,7), and (1,2,4)(3,6,5), or you can use more GAP friendly points to get (9,10), (8,7)(1,3)(2,6)(4,5), (1,2,3,4,5,6,7), and (1,2,4)(3,6,5). AΓL(1,8) is a complete, solvable group with no normal subgroup of index 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just checked the groups ordered by size, skipping nilpotent groups since they always have outer automorphisms.  Probably one could be a bit smarter, looking for complete groups with no 2-quotients, but I wasn't sure the smallest example would be of the form 2×Complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples like 2.Sz(8) which have no outer automorphisms, have a non-trivial center, but where the quotient by the center is not complete (so it is definitely not 2×Complete).&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-09T00:19:34.470" CommentCount="1" />
  <row Id="4321" PostTypeId="2" ParentId="4304" CreationDate="2010-09-09T03:52:39.740" Score="2" ViewCount="0" Body="&lt;p&gt;After reading &lt;a href=&quot;http://www.satisfice.com/tools.shtml&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;, it seems that pairwise testing requires a set of test cases in which every pair of values from any two of the n categories occurs at least once among the test case n-tuples. In the present case, the problem is to find a minimal subset of the 6x6x6 = 216 total triples (a,b,c) such that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;each pair of values for a and b&lt;br&gt;&#xA;occurs at least once, i.e. (a,b,*),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;each pair of a and c values occurs&lt;br&gt;&#xA;at least once, i.e. (a,*,c)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;each pair of b and c values occurs&#xA;at least once, i.e. (*,b,c)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any subset satisfying these requirements must have at least 36 elements just to satisfy the (a,b,*) requirement. In the present case I think 36 test cases are also sufficient, as in the following set of triples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1, 1, 1), (1, 2, 2), (1, 3, 3), (1, 4, 4), (1, 5, 5), (1, 6, 6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2, 1, 6), (2, 2, 1), (2, 3, 2), (2, 4, 3), (2, 5, 4), (2, 6, 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3, 1, 5), (3, 2, 6), (3, 3, 1), (3, 4, 2), (3, 5, 3), (3, 6, 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4, 1, 4), (4, 2, 5), (4, 3, 6), (4, 4, 1), (4, 5, 2), (4, 6, 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5, 1, 3), (5, 2, 4), (5, 3, 5), (5, 4, 6), (5, 5, 1), (5, 6, 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6, 1, 2), (6, 2, 3), (6, 3, 4), (6, 4, 5), (6, 5, 6), (6, 6, 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example each of the three kinds of pairs occurs once and only once, i.e. there is no overlap. I don't think this will be possible in general, so it might not always be easy to come up with minimal subsets that cover all the cases.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2010-09-09T03:52:39.740" CommentCount="1" />
  <row Id="4322" PostTypeId="1" AcceptedAnswerId="4328" CreationDate="2010-09-09T06:25:46.773" Score="2" ViewCount="214" Body="&lt;p&gt;I have a 3D plane defined by three points: $P_0$ , $P_1$ and $P_2$. How to check whether a point $P$ is located right on and &lt;strong&gt;&lt;em&gt;inside&lt;/em&gt;&lt;/strong&gt; the 3D triangle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, if I have a plane defined by $({0,0,0})$, $({10,0,0})$ and $({0,10,0})$, then the point $({50,0,0})$ is considered &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; located on the plane, whereas the point  $({5,0,0})$ is. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-09T07:20:53.350" LastActivityDate="2010-09-09T08:32:03.627" Title="Check whether a point is within a 3D Triangle" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" />
  <row Id="4323" PostTypeId="1" AcceptedAnswerId="4327" CreationDate="2010-09-09T06:28:02.247" Score="11" ViewCount="470" Body="&lt;p&gt;If we have an algebraic number &amp;#945; with (complex) absolute value 1, it does not follow that &amp;#945; is a root of unity (i.e., that &amp;#945;&lt;sup&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sup&gt;=1 for some &lt;em&gt;n&lt;/em&gt;).  For example, (3/5 + 4/5 &lt;em&gt;i&lt;/em&gt;) is not a root of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we assume that &amp;#945; is an algebraic integer with absolute value 1, does it follow that &amp;#945; is a root of unity?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I know that if all conjugates of &amp;#945; have absolute value 1, then &amp;#945; is a root of unity by the argument below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimal polynomial of &amp;#945; over &lt;em&gt;Z&lt;/em&gt; is $\prod_{i=1}^d (x-\alpha_i)$, where the $\alpha_i$ are just the conjugates of &amp;#945;.  Then $\prod_{i=1}^d (x-\alpha_i^n)$ is a polynomial over &lt;em&gt;Z&lt;/em&gt; with &amp;#945;&lt;sup&gt;n&lt;/sup&gt; as a root.  It also has degree d, and all roots have absolute value 1.  But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that &amp;#945;&lt;sup&gt;n&lt;/sup&gt;=&amp;#963;(&amp;#945;) for some Galois conjugation &amp;#963;. If &amp;#963;&lt;sup&gt;m&lt;/sup&gt;(&amp;#945;)=&amp;#945;, then &amp;#945;&lt;sup&gt;n&lt;sup&gt;m&lt;/sup&gt;&lt;/sup&gt;=&amp;#945;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &amp;#945;&lt;sup&gt;n&lt;sup&gt;m&lt;/sup&gt;-1&lt;/sup&gt;=1.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-10T15:40:24.017" Title="Are all algebraic integers with absolute value 1 roots of unity?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="4" />
  <row Id="4326" PostTypeId="2" ParentId="4322" CreationDate="2010-09-09T06:44:24.010" Score="0" ViewCount="0" Body="&lt;p&gt;There may well be a more efficient algorithm, but here's a way to check that &lt;em&gt;P&lt;/em&gt; is on the triangle defined by the three points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the cross products $\vec {P_0 P_1} \times \vec {P_0 P_2} = \vec a$ and $\vec {P P_1} \times \vec {P P_2} = \vec b$.  If $\vec b = k_1 \vec a$ with $k_1 \geq 0$, then P is on the plane and is on the correct side of the line through &lt;em&gt;P&lt;sub&gt;1&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;P&lt;sub&gt;2&lt;/sub&gt;&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, compute $\vec {P P_2} \times \vec {P P_0} = \vec c$ and $\vec {P P_0} \times \vec {P P_1} = \vec d$.  If $\vec c = k_2 \vec a$ and $\vec d = k_3 \vec a$ with $k_2, k_3 \geq 0$, then &lt;em&gt;P&lt;/em&gt; lies within the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there's a lot of redundancy in this calculation.  Once we know &lt;em&gt;P&lt;/em&gt; is on the plane, we could just check one coordinate of the $\vec c$ and $\vec d$ to see that they have the correct sign.  This is just projecting the problem onto one of the coordinate planes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My old answer below explains how to check that &lt;em&gt;P&lt;/em&gt; is on the infinite plane defined by the first three points, not how to check if it lies inside the triangle defined by the first three points, which is what the question intended.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A general plane equation is &lt;em&gt;Ax + By + Cz = D&lt;/em&gt;.  The 3-dimensional vector &amp;lt;&lt;em&gt;A,B,C&lt;/em&gt;&gt; is perpendicular to the plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that there is not a unique equation of this form; you can multiply the equation by any nonzero number and get another equation of the plane.  --So you can try to find a solutions with &lt;em&gt;D=1&lt;/em&gt;, and if that doesn't work, use &lt;em&gt;D=0&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find an equation of the plane defined by your first three points, you can just plug in the fourth point to check if it satisfies the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can solve for the plane equation by hand by setting up equation from the first three points; but a more efficient method is to take the cross product of the vector $\vec {P_0 P_1}$ and the vector $\vec {P_0 P_2}$, and to use the resulting vector as &amp;lt;&lt;em&gt;A,B,C&lt;/em&gt;&gt;.  You'll find it very useful to understand the cross product if you like working in 3-D space.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-09-09T07:22:27.317" LastActivityDate="2010-09-09T07:22:27.317" CommentCount="1" />
  <row Id="4327" PostTypeId="2" ParentId="4323" CreationDate="2010-09-09T06:46:01.717" Score="12" ViewCount="0" Body="&lt;p&gt;No. There are algebraic integers on the unit circle which aren't roots of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ramanujan.math.trinity.edu/rdaileda/research/papers/p1.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; by Ryan Daileda provides some useful information and references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://en.wikipedia.org/wiki/Salem_number&quot; rel=&quot;nofollow&quot;&gt;Salem numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-09T06:46:01.717" CommentCount="2" />
  <row Id="4328" PostTypeId="2" ParentId="4322" CreationDate="2010-09-09T07:18:24.993" Score="2" ViewCount="0" Body="&lt;p&gt;Try to solve the system&#xA;$$&#xA;x * (P_1 - P_0) + y * (P_2 - P_0) = P - P_0&#xA;$$&#xA;If it is solveable, the point $P$ lies on the plane. If in addition $x \geq 0$, $y \geq 0$, and $x + y \leq 1$, then $P$ lies inside the triangle.&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="960" LastEditorDisplayName="" LastEditDate="2010-09-09T08:32:03.627" LastActivityDate="2010-09-09T08:32:03.627" CommentCount="7" />
  <row Id="4329" PostTypeId="2" ParentId="4173" CreationDate="2010-09-09T10:31:47.873" Score="1" ViewCount="0" Body="&lt;p&gt;One way is to find the householder matrix Q that maps v to a multiple of e_1 (first coordinate basis vector). Then (since Q is symmtric and orthogonal) w_2=Q*e_2 ... will be a basis of the orthogonal complement of v. &#xA;Explicitly, I get&#xA;w_k = e_k - u&#xA;where u = (v+sqrt(d)*e_1)/(d+sqrt(d))&lt;/p&gt;&#xA;" OwnerUserId="1809" LastActivityDate="2010-09-09T10:31:47.873" />
  <row Id="4330" PostTypeId="2" ParentId="4268" CreationDate="2010-09-09T11:11:21.380" Score="6" ViewCount="0" Body="&lt;p&gt;If you're willing to accept the &lt;a href=&quot;http://math.stackexchange.com/questions/4268/alternate-definition-of-prime-number&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt; (which can be proved by elementary -i.e., &quot;ideal free&quot;- methods, see the link) that states that any integer different from $\pm 1$ can be written as a unique product (up to ordering of the factors) of prime numbers, then the equivalence of both definitions of prime numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(\ m \vert p \ \Longrightarrow \ m = \pm 1 \ \text{or} \ \pm p \ ) &#xA;\qquad \Longleftrightarrow \qquad&#xA;(\ p \vert ab \ \Longrightarrow \ p \vert a \ \text{or} \ p \vert b \ )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Warning. If you feel more comfortable working only with positive integer numbers, then just drop those $\pm$ all over this answer.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\Longrightarrow)$ If $p\vert ab$, then $ab = pq$ for some integer $q$. Since both $ab$ and $pq$ can be &lt;em&gt;uniquely&lt;/em&gt; written as a product of prime numbers, $p$ must be a prime factor of $ab$. But prime factors of $ab$ are necessarily prime factors of $a$ or $b$, again by the uniqueness of the fundamental theorem of arithmetic. So, $p$ is necessarily a prime factor of $a$ or $b$; that is, $p\vert a$ or $p\vert b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\Longleftarrow)$ Again, if $m\vert p$, all prime factors of $m$ must be prime factors of $p$. But the only prime factors of $p$ are $\pm p$. So, either $m = \pm p$, or $m=\pm 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is little (big?) circular argument here, because one already uses implication $(\Longrightarrow)$ in the proof of the fundamental theorem of arithmetic. This implication is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's lemma&lt;/a&gt;. But it can be proved by elementary  methods too, that rely on &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Bézout's identity&lt;/a&gt; (whith elementary proof too: see the last link).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you can see, everything is elementary, but, perhaps, too long for a single answer and to remember in one go. While reasoning using the fundamental theorem of arithmetic may be seen as little short cut easier to remember the next time you have any doubt about the equivalence of both definitions of prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, a number satisfying the left hand side condition is called &quot;irreducible&quot;, while $p$ on the right hand side is called &quot;prime&quot;. So &quot;prime = irreducible&quot;, at least for integer numbers.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-09T11:11:21.380" CommentCount="2" />
  <row Id="4331" PostTypeId="2" ParentId="4304" CreationDate="2010-09-09T11:51:05.527" Score="3" ViewCount="0" Body="&lt;p&gt;Pairwise testing tests for all possible 2-way interactions efficiently -- I gave a quick overview here: &lt;a href=&quot;http://cstheory.stackexchange.com/questions/891/&quot;&gt;http://cstheory.stackexchange.com/questions/891/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking for strength 2 &lt;a href=&quot;http://math.nist.gov/coveringarrays/coveringarray.html&quot; rel=&quot;nofollow&quot;&gt;covering arrays&lt;/a&gt;.  In each pair of columns every pair of symbols occur -- this ensures all 2-way interactions are observed in some way.  Here's a very simple example of a covering array of strength 2 with 2 columns:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11&#xA;12&#xA;21&#xA;22&#xA;12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What castel has drawn is essentially the &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;123456&#xA;612345&#xA;561234&#xA;456123&#xA;345612&#xA;234561&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you look at each entry and write the list (r,c,s), where r is the row index, c is the column index, and s is the symbol, you will construct an orthogonal array (as depicted below) -- a covering array of strength 2 with the minimum number of rows (36).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;111&#xA;122&#xA;133&#xA;...&#xA;661&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, Latin squares exist for all orders n.  So if you have three columns (e.g. three variables) and n symbols for each variable, then you can always find a strength 2 covering array with n&lt;sup&gt;2&lt;/sup&gt; rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many combinatorial designs give rise to particularly efficient covering arrays.  Strength 2 covering arrays with more than three columns and n&lt;sup&gt;2&lt;/sup&gt; rows are equivalent to sets of &lt;a href=&quot;http://designtheory.org/library/encyc/mols/a/&quot; rel=&quot;nofollow&quot;&gt;mutually orthogonal Latin squares&lt;/a&gt; (the reference shows the construction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, if you have 40 results, then you are not using the most efficient covering array.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-09-09T11:51:05.527" />
  <row Id="4332" PostTypeId="2" ParentId="4323" CreationDate="2010-09-09T12:00:58.340" Score="10" ViewCount="0" Body="&lt;p&gt;Let x be an algebraic number with absolute value 1. Then x and its complex conjugate x* = 1/x have the same minimal polynomial. Writing f(T) for the minimal polynomial of x over Q, with degree n, the polynomials T^nf(1/T) and f(T) are irreducible over Q with root x*, so the polynomials are equal up to a scaling factor: T^nf(1/T) = cf(T). Setting T = 1, f(1) = cf(1). &#xA;Assuming x is not rational (i.e., x is not 1 or -1), f has degree greater than 1, so &#xA;f(1) is nonzero and thus c = 1. Therefore T^nf(1/T) = f(T), so f(T) has symmetric coefficients. In particular, its constant term is 1. Moreover, the roots of f(T) come in reciprocal pairs (since 1 and -1 are not roots), so n is even.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial conclusion: an algebraic number other than 1 or -1 which has absolute value 1 &#xA;has even degree over Q and its minimal polynomial has constant term 1. In particular, &#xA;if x is an algebraic integer then it must be a unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no examples of algebraic integers with degree 2 and abs. value 1 that are not roots of unity since a real quadratic field has no elements on the unit circle besides 1 and -1 and the units in an imaginary quadratic field are all roots of unity (and actually are only 1 and -1 except for Q(i) and Q(w)). Thus the smallest degree x could have over Q is 4 and there are examples with degree 4: the polynomial x^4 - 2x^3 - 2x + 1 has two roots on the unit circle and two real roots (one between 0 and 1 and the other greater than 1). &lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-09-09T12:00:58.340" CommentCount="1" />
  <row Id="4333" PostTypeId="2" ParentId="368" CreationDate="2010-09-09T12:48:34.330" Score="3" ViewCount="0" Body="&lt;p&gt;Contrast these two methods for calculating n!:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Counting permutations of a n-element set one-by-one (this is what n! counts), vs.&lt;/li&gt;&#xA;&lt;li&gt;Multiplying together the numbers 1,2,...,n.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore n! represents an amazingly efficient method for counting permutations!  So I think (and I don't think I'm the only one) that n! should probably be considered a closed form (unless you have some other clear definition of what constitutes a &quot;closed&quot; form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading, I recommend:  H. S. Wilf, What is an answer?, Amer. Math. Monthly, 89 (1982), pp. 289–292.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-09-09T12:48:34.330" CommentCount="1" />
  <row Id="4334" PostTypeId="2" ParentId="4199" CreationDate="2010-09-09T14:21:45.030" Score="3" ViewCount="0" Body="&lt;p&gt;We could do something simple, since we have:&#xA;\begin{equation}&#xA;\cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(e^{i\frac{p}{q}\theta}+e^{-i\frac{p}{q}\theta}\right)=\frac{1}{2}\left(\left(\frac{x}{r}+i\frac{y}{r}\right)^{\frac{p}{q}}+\left(\frac{x}{r}-i\frac{y}{r}\right)^{\frac{p}{q}}\right)&#xA;\end{equation}&#xA;That is &#xA;\begin{equation}&#xA;\cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}+\left(x-iy\right)^{\frac{p}{q}}\right)&#xA;\end{equation}&#xA;and similarly we have&#xA;\begin{equation}&#xA;\sin\left(\frac{p}{q}\theta\right)=\frac{1}{2i}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}-\left(x-iy\right)^{\frac{p}{q}}\right)&#xA;\end{equation}&#xA;Now we can use these to express your equations in terms of $x$ and $y$. Is this any use?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-09T18:45:05.497" LastActivityDate="2010-09-09T18:45:05.497" CommentCount="4" />
  <row Id="4335" PostTypeId="1" AcceptedAnswerId="4341" CreationDate="2010-09-09T14:51:35.530" Score="4" ViewCount="349" Body="&lt;p&gt;I'm starting a very long quest to learn about math so I can program games. I'm mostly a coorporate developer and it's somewhat boring and non exiting. When I began my career I chose it because I wanted to create games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told Linear Algebra is the best place to start - where should I go?&lt;/p&gt;&#xA;" OwnerUserId="1812" LastActivityDate="2010-09-11T07:20:06.867" CommunityOwnedDate="2010-09-09T16:15:04.960" Title="Where to start learning Linear Algebra?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" />
  <row Id="4336" PostTypeId="1" CreationDate="2010-09-09T14:52:53.860" Score="4" ViewCount="293" Body="&lt;p&gt;If the matrix is positive definite, then all its eigenvalues are strictly positive. Is the converse also true? That is, if the eigenvalues are strictly positive, then matrix is positive definite? Can you give example of 2x2 matrix with 2 positive eigenvalues but is not positive definite?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-09T16:14:39.923" LastActivityDate="2010-09-09T22:01:03.530" Title="if eigenvalues are positive, is the matrix positive definite?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="4337" PostTypeId="2" ParentId="4336" CreationDate="2010-09-09T14:57:07.710" Score="1" ViewCount="0" Body="&lt;p&gt;True. If you consider a diagonal matrix $A = diag(d_1,\cdots,d_n)$ where each diagonal entry $d_i$ is positive, then clearly it is positive definite, since $x^TAx = \sum d_ix_i^2 &gt; 0$ unless $x = 0$ ($x_i$ are the components of vector $x$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply spectral theorem for symmetric matrix to reduce to the diagonal case.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-09-09T14:57:07.710" CommentCount="0" />
  <row Id="4338" PostTypeId="2" ParentId="4335" CreationDate="2010-09-09T15:00:48.857" Score="9" ViewCount="0" Body="&lt;p&gt;Hi Sergio. MIT has a complete online course for linear algebra, complete with video lectures, lecture notes, and assignments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2005/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2005/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how much of the course is applicable to what you'll be doing in game development, but it's a solid foundation.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-09-09T15:00:48.857" CommunityOwnedDate="2010-09-09T16:15:04.960" />
  <row Id="4340" PostTypeId="2" ParentId="4336" CreationDate="2010-09-09T15:39:51.320" Score="8" ViewCount="0" Body="&lt;p&gt;I think this is false. Let $A = \begin{pmatrix} 1 &amp;amp; -3 \\ 0 &amp;amp; 1 \end{pmatrix}$ be a 2x2 matrix, in the canonical basis of $\mathbb R^2$. Then A has a double eigenvalue b=1. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\langle v, Av \rangle &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the matrix can have all its eigenvalues strictly positive, but it does not need to be symmetric. &lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-09T15:59:34.003" LastActivityDate="2010-09-09T15:59:34.003" CommentCount="3" />
  <row Id="4341" PostTypeId="2" ParentId="4335" CreationDate="2010-09-09T17:43:56.283" Score="7" ViewCount="0" Body="&lt;p&gt;You are right: Linear Algebra is not just the &quot;best&quot; place to start. It's THE place to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among all the books cited in &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_algebra&quot; rel=&quot;nofollow&quot;&gt;Wikipedia - Linear Algebra&lt;/a&gt;, I would recommend:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Strang, Gilbert, Linear Algebra and Its Applications (4th ed.) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Strang's book has at least two reasons for being recomended. First, it's extremely easy and short. Second, it's the book they use at MIT for the extremely good video Linear Algebra course you'll find in the link of Unreasonable Sin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a view towards applications (though maybe not necessarily your applications) and still elementary:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* B. Noble &amp;amp; J.W. Daniel: Applied Linear Algebra, Prentice-Hall, 1977&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Linear algebra has two sides: one more &quot;theoretical&quot;, the other one more &quot;applied&quot;. Strang's book is just elementary, but perhaps &quot;theoretical&quot;. Noble-Daniel is definitively &quot;applied&quot;. The distinction from the two points of view relies in the emphasis they put on &quot;abstract&quot; vector spaces vs specific ones such as $\mathbb{R}^n$ or $\mathbb{C}^n$, or on matrices vs linear maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe because my penchant towards &quot;pure&quot; maths, I must admit that sometimes I find matrices somewhat annoying. They are funny, specific, whereas linear maps can look more &quot;abstract&quot; and &quot;ethereal&quot;. But, for instance: I can't stand the proof that the matrix product is associative, whereas the corresponding associativity for the composition of (linear or non linear) maps is true..., well, just because it can't help to be true the first moment you write it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, at a more advanced level in the &quot;theoretical&quot; side you can use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Greub, Werner H., Linear Algebra, Graduate Texts in Mathematics (4th ed.), Springer&#xA;&#xA;* Halmos, Paul R., Finite-Dimensional Vector Spaces, Undergraduate Texts in Mathematics, Springer&#xA;&#xA;* Shilov, Georgi E., Linear algebra, Dover Publications &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the &quot;applied&quot; (?) side, a book that I love and you'll appreciate if you want to study, for instance, the exponential of a matrix is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486445542&quot; rel=&quot;nofollow&quot;&gt;Gantmacher&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, at any time, you'll need to do a lot of exercises. Lipschutz's is second to none in this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Lipschutz, Seymour, 3,000 Solved Problems in Linear Algebra, McGraw-Hill&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Enjoy!   :-)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-09T18:30:27.973" LastActivityDate="2010-09-09T18:30:27.973" CommunityOwnedDate="2010-09-09T17:43:56.283" CommentCount="2" />
  <row Id="4342" PostTypeId="2" ParentId="368" CreationDate="2010-09-09T17:57:48.713" Score="1" ViewCount="0" Body="&lt;p&gt;This is a riff on some of the comments about what might constitute an &quot;answer&quot; and what &quot;closed form&quot; might mean: although it's somewhat facetious, it's intended to prompt thoughts about these issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our base-10 number system interprets a string ${a_n}{a_{n-1}} \cdots {a_0}$ as the sum $\sum_{i=0}^{n} a_i 10^i $ (which can be, and is, computed recursively as $a_0 + 10 \left( a_1 + 10 \left( \cdots + 10 a_n \right) \cdots \right)$).  If you take the former to be an acceptable &quot;closed form&quot; representation, then why not use a slight modification of this number system?  Specifically, interpret the same string as equal to $a_0 + 2 \left( a_1 + 3 \left( \cdots + (n+1) a_n \right) \cdots \right)$ and require that $0 \le a_0 \le 1, 0 \le a_1 \le 2, \ldots, 0 \le a_n \le n$.  In this &quot;factorial&quot; number system, $n! = 10 \cdots 0$ is represented as a simple $n$-digit string: it's &quot;closed&quot;!&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-09T17:57:48.713" />
  <row Id="4343" PostTypeId="2" ParentId="4323" CreationDate="2010-09-09T18:38:21.673" Score="4" ViewCount="0" Body="&lt;p&gt;The answer to your question is no - not all algebraic integers with absolute value 1 are roots of unity - as you may have learned by now from the &lt;a href=&quot;http://ramanujan.math.trinity.edu/rdaileda/research/papers/p1.pdf&quot; rel=&quot;nofollow&quot;&gt;2005 paper of Daileda&lt;/a&gt; that was referenced in the answer supplied by Robin Chapman. However, you should be aware that &lt;strong&gt;Daileda has in fact rediscovered&lt;/strong&gt; these simple folklore results on unimodular units. These results are probably at least a half-century old - if not much older. Indeed, I recall reading similar results in papers published  by Iwasawa in the fifties or sixties. I don't have the time now to locate said Iwasawa reference, but here is another reference that suffices to prove my point. This &lt;a href=&quot;http://ci.nii.ac.jp/naid/110000010754/en&quot; rel=&quot;nofollow&quot;&gt;1979 paper of Nakahata&lt;/a&gt; includes the same results and the same simple proofs found in Daileda's paper. Moreover, Nakahata places these results naturally into a more general context. Thus I highly recommend that you consult Nakahata's paper in addition to Daileda's. I've appended its Zbl review below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lofUp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-10T15:40:24.017" LastActivityDate="2010-09-10T15:40:24.017" CommentCount="1" />
  <row Id="4344" PostTypeId="2" ParentId="4317" CreationDate="2010-09-09T18:50:11.550" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a different look:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differentiating and multiplying by $x$ gives rise to &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling Numbers of the Second Kind&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you denote the operator of differentiating and multiplying by $x$ as $D_{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} s(n,k) f^{k}(x) x^{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s(n,k)$ is the stirling number of the second kind and $f^k(x)$ is the $k^{th}$ derivative of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can easily be proven using the identity $$s(n,k) = s(n-1,k-1) + k \cdot s(n-1,k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a table for the Stirling numbers of the second kind (from the wiki page):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;n/k   0     1     2     3     4      5      6      7      8      9&#xA;0     1&#xA;1     0     1&#xA;2     0     1     1&#xA;3     0     1     3     1&#xA;4     0     1     7     6     1&#xA;5     0     1     15    25    10     1&#xA;6     0     1     31    90    65     15     1&#xA;7     0     1     63    301   350    140    21     1&#xA;8     0     1     127   966   1701   1050   266    28     1&#xA;9     0     1     255   3025  7770   6951   2646   462    36     1&#xA;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So in your case, we can start with $ f(x) = \frac{1}{1-x}$ and obtain that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{\infty} k^{n} x^k = \sum_{r=1}^{n} r! \cdot s(n,r) \frac{x^{r}}{(1-x)^{r+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in your case for $n=3$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} k^3 x^k = \frac{1! \cdot 1 \cdot x}{(1-x)^2} + \frac{2! \cdot 3 \cdot x^2}{(1-x)^3} + \frac{3! \cdot 1 \cdot x^3}{(1-x)^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{x(1-x)^2 + 6x^{2}(1-x) + 6x^3}{(1-x)^4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{x^3 + 4x^2 + x}{(1-x)^4} $$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-09T22:16:49.427" LastActivityDate="2010-09-09T22:16:49.427" CommentCount="2" />
  <row Id="4345" PostTypeId="2" ParentId="4199" CreationDate="2010-09-09T19:08:05.050" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a systematic method, but not one recommended for paper-and-pen&#xA;calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be a common denominator for all rationals $r$ such that $\sin r\theta$&#xA;or $\cos r\theta$ arises in the formula.&#xA;Let $u=\cos(\theta/n)$ and $v=\sin(\theta/n)$. There are polynomials&#xA;$\phi_n$ and $\psi_n$ such that $\cos\theta=\phi_n(u,v)$ and&#xA;$\sin\theta=\psi_n(u,v)$. The curve has a polar equation which can be written&#xA;as $f(r,u,v)=0$. Now consider the system of equations:&#xA;$$f(s^2,u,v)=0,\ \ \ x=s^2\phi_n(u,v),\ \ \ y=s^2\psi_n(u,v),\ \ \ u^2+v^2=1.$$&#xA;Yes, four equations in five variables. Using some systematic&#xA;elimination procedure, for instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis&quot; rel=&quot;nofollow&quot;&gt;Groebner bases&lt;/a&gt;, eliminate the&#xA;three variables $s$, $u$ and $v$ to get (one hopes) one equation in $x$ and $y$.&#xA;It certainly helps to have some computer package to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, for $r=4a\cos^3(\theta/3)$, the system of equations becomes&#xA;$$s^2=4au^3,\ \ \ x=s^2(u^3-3uv^2),\ \ \ y=s^2(3u^2v-v^3),\ \ \ u^2+v^2=1.$$&#xA;Well, I'm not going to do the elimination, but $s^2$ disappears readily enough....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-09T19:08:05.050" CommentCount="1" />
  <row Id="4346" PostTypeId="1" AcceptedAnswerId="4348" CreationDate="2010-09-09T19:21:49.167" Score="9" ViewCount="209" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/4169/solving-the-functional-equation-ffxfxx&quot;&gt;this&lt;/a&gt; thread, the question was to find a $f: \mathbb{R} \to \mathbb{R}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(f(x)) = f(x) + x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which was revealed in the comments to be solved by $f(x) = \varphi x$ where $\varphi$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Golden_ratio&quot; rel=&quot;nofollow&quot;&gt;golden ratio&lt;/a&gt; $\frac{1+\sqrt{5}}{2}$)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Having read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_iteration&quot; rel=&quot;nofollow&quot;&gt;iterated functions&lt;/a&gt; shortly before though, I came up with this train of thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(f(x)) = f(x) + x$$&#xA;$$\Leftrightarrow f^2 = f^1 + f^0$$&#xA;$$f^2 - f - f^0 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f^n$ denotes the $n$'th iterate of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I solved the resulting quadratic equation much as I did with plain numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 1}$$&#xA;$$f = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}\cdot f^0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally the solution&#xA;$$f(x) = \frac{1 \pm \sqrt{5}}{2} x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is: &lt;em&gt;Is it somehow allowed to work with functions in that way?&lt;/em&gt;* I know that in the above, there are denotational ambiguities as $1$ is actually treated as $f^0 = id$ ... But since the result is correct, there seems to be some correct thing in this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can I actually solve certain functional equations like this? And if true, &lt;strong&gt;how would the correct notation of the above be&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-09T23:10:28.813" LastActivityDate="2010-09-11T21:42:37.693" Title="Solving (quadratic) equations of iterated functions, such as f(f(x))=f(x)+x" Tags="&lt;real-analysis&gt;&lt;notation&gt;&lt;algebra&gt;&lt;functional-equations&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="4347" PostTypeId="2" ParentId="4228" CreationDate="2010-09-09T19:54:31.250" Score="2" ViewCount="0" Body="&lt;p&gt;Yes. Your polytope $P'$ is exactly the polar $P^*$ of $P$ with respect to the standard unit sphere. Choosing another conic instead of the unit sphere will in general give another polar. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Pole_and_polar&quot; rel=&quot;nofollow&quot;&gt;this wikpedia page&lt;/a&gt; for the two-dimensional case of polarity with respect to other conics. So your &quot;Cartesian&quot; dual corresponds to the unit sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polarity/duality of polytopes or more generally, convex bodies, is treated in many books on convexity. Unfortunately, many of them leave a lot of detail to the reader. I recommend &lt;a href=&quot;http://books.google.co.uk/books?id=SWQmykCwabwC&amp;amp;lpg=PP1&amp;amp;dq=webster%20convexity&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Webster's &lt;em&gt;Convexity&lt;/em&gt;&lt;/a&gt;, &lt;a href=&quot;http://books.google.co.uk/books?id=rvrYsRREzSYC&amp;amp;lpg=PA56&amp;amp;dq=polar%20of%20polytope&amp;amp;pg=PP1#v=onepage&amp;amp;q=polar%20of%20polytope&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Brønsted's &lt;em&gt;An Introduction to Convex Polytopes&lt;/em&gt;&lt;/a&gt; or &lt;a href=&quot;http://books.google.co.uk/books?id=dH5DK1URwQgC&amp;amp;lpg=PP1&amp;amp;dq=matousek&amp;amp;pg=PP1#v=onepage&amp;amp;q=matousek&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Matousek's &lt;em&gt;Lectures on Discrete Geometry&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1021" LastEditorUserId="1021" LastEditorDisplayName="" LastEditDate="2010-09-13T09:20:17.983" LastActivityDate="2010-09-13T09:20:17.983" CommentCount="1" />
  <row Id="4348" PostTypeId="2" ParentId="4346" CreationDate="2010-09-09T20:37:06.800" Score="5" ViewCount="0" Body="&lt;p&gt;One way to think about this is that you are assuming that $f(x) = cx$ and then solving for the value of $c$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is something much more interesting going on; you have started doing some &lt;a href=&quot;http://en.wikipedia.org/wiki/Abstract_algebra&quot; rel=&quot;nofollow&quot;&gt;abstract algebra&lt;/a&gt; without knowing it.  $f$ is not a real number, but it does live in something called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_over_a_field&quot; rel=&quot;nofollow&quot;&gt;algebra over a field&lt;/a&gt;, which is a special type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;ring&lt;/a&gt;.  In particular, $f$ lives in the algebra of continuous functions $\mathbb{R} \to \mathbb{R}$.  In this algebra there is a multiplicative identity $f(x) = x$ which plays the role of the zeroth power, an addition, a multiplication, and a scalar multiplication by real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many manipulations which are possible with ordinary real numbers are possible in rings and algebras; in particular, the first half of the proof of the quadratic formula carries through totally abstractly (the part where you complete the square).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the second half does not.  In other words, it is not true that there are only two solutions to the equation $f^2 = a$ in a general algebra.  This is because algebras are not in general &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral_domain&quot; rel=&quot;nofollow&quot;&gt;integral domains&lt;/a&gt;.  There may be none or infinitely many!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this special case $a$ is a positive real multiple of the identity, so we know it has &lt;em&gt;at least&lt;/em&gt; two square roots (even though there may be more).  These are the solutions that you found, and this method for finding them is perfectly valid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important technique.  It is often used in the case where $f$ is a differential operator as a concise way to solve linear homogeneous ODEs.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-09T20:37:06.800" CommentCount="1" />
  <row Id="4350" PostTypeId="1" AcceptedAnswerId="4352" CreationDate="2010-09-09T21:57:38.147" Score="7" ViewCount="124" Body="&lt;p&gt;I'm not sure if this a proper recurance relation &lt;em&gt;per se&lt;/em&gt; but I'd be interested in the methodology in solving a recurrence relation of the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_0 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_1 = x_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_2 = x_1Z_1 + x_2 = x_1^2 + x_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_3 = x_1Z_2 + x_2Z_1 + x_3 = x_1^3 + x_1x_2 + x_1x_2 + x_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_n = x_1 Z_{n-1} + x_2 Z_{n-2} + ... + x_n Z_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As written, each term requires the knowledge of the previous terms. Is it possible to write down a closed form for $Z_n$? For this &lt;em&gt;particular&lt;/em&gt; recurrence, I can write down the result  of $Z_n$  but I would be very interested in seeing how one can derive this from the recurrence relation itself. My gut feeling is that a generating function is lurking underneath all this.&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-09-10T00:01:28.557" Title="Solving recurrence relations that involve all previous terms" Tags="&lt;recursive&gt;&lt;generating-functions&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4351" PostTypeId="2" ParentId="4336" CreationDate="2010-09-09T22:01:03.530" Score="7" ViewCount="0" Body="&lt;p&gt;This question does a great job of illustrating the problem with thinking about these things in terms of coordinates.  The thing that is positive-definite is not a matrix $M$ but the &lt;em&gt;quadratic form&lt;/em&gt; $x \mapsto x^T M x$, which is a very different beast from the linear transformation $x \mapsto M x$.  For one thing, the quadratic form does not depend on the antisymmetric part of $M$, so using an asymmetric matrix to define a quadratic form is redundant.  And there is &lt;em&gt;no reason&lt;/em&gt; that an asymmetric matrix and its symmetrization need to be at all related; in particular, they do not need to have the same eigenvalues.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-09T22:01:03.530" CommentCount="5" />
  <row Id="4352" PostTypeId="2" ParentId="4350" CreationDate="2010-09-09T22:08:06.407" Score="7" ViewCount="0" Body="&lt;p&gt;Your gut feeling is right, but let me change your $x$s to $a$s to make the answer easier to read.  Consider the generating functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z(x) = \sum_{n \ge 0} Z_n x^n$$&#xA;$$A(x) = \sum_{n \ge 1} a_n x^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the recurrence relation you have written down is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z(x) = Z(x) A(x) + 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z(x) = \frac{1}{1 - A(x)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you know the generating function $A(x)$ in closed form, you then know the generating function $Z(x)$ in closed form.  This is one of the most basic manipulations to do with generating functions, and techniques like this are thoroughly covered in, for example, Wilf's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-09T22:08:06.407" />
  <row Id="4354" PostTypeId="2" ParentId="4350" CreationDate="2010-09-10T00:01:28.557" Score="4" ViewCount="0" Body="&lt;p&gt;To answer the general question in the title (&quot;solving recurrence relations that involve all previous terms&quot;), such relations can be thought of as &lt;em&gt;path- or history-dependent&lt;/em&gt; calculations and one tries to replace them by equivalent &lt;em&gt;state-dependent&lt;/em&gt; (Markovian) calculations that retain a finite amount of state information and keep transforming the state, but not remembering any additional information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the recurrence is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_n = S_0 + S_1 + \dots S_{n-1}$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is equivalent, for $n&gt;1$,  to $S_n = S_{n-1} + S_{n-1} = 2S_{n-1}$, and remembering the previous term is enough to reproduce the recurrence as iterated doubling (after the first two terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example in the question is also an iteration of a single transformation, but on an unbounded-dimensional state space.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-10T00:01:28.557" />
  <row Id="4355" PostTypeId="1" CreationDate="2010-09-10T01:16:08.810" Score="1" ViewCount="63" Body="&lt;p&gt;I'm having a hard time solving this problem. Would appreciate any hint! (and special thanks to  Byron Schumuland for answering my previous 2 questions. This 3rd Q is different.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let et e1,e2,... be iid normal mean 0 variance 1. Let Xt := e1+...+et, for t=1,2.. and X0 := 0. (So we have a discrete-time random walk whose steps are iid N(0,1))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define passage time T* := inf{t&gt;0 : X_t &amp;lt; -t}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the CDF of T*, i.e. what is Prob(T* &amp;lt;= t) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Equivalently: let e'_t ~ N(1,1), and T* above is simply inf{t&gt;0 : X'_t &amp;lt; 0}.)&#xA;Is there a closed form solution? (Remember we have zero correlation, just non-zero mean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reference: &#xA;&lt;a href=&quot;http://dl.dropbox.com/u/4260685/orthant.pdf&quot; rel=&quot;nofollow&quot;&gt;http://dl.dropbox.com/u/4260685/orthant.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THANK YOU!&lt;/p&gt;&#xA;" OwnerUserId="1784" LastActivityDate="2010-09-10T01:16:08.810" Title="Orthant probability / passage-time " Tags="&lt;probability&gt;&lt;statistics&gt;&lt;stochastic-processes&gt;" CommentCount="1" />
  <row Id="4356" PostTypeId="1" AcceptedAnswerId="4359" CreationDate="2010-09-10T02:32:55.873" Score="1" ViewCount="186" Body="&lt;p&gt;I have a constant area $A$   , and I can mold that area into a &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_polygon&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;regular&lt;/strong&gt;&lt;/a&gt; $n$ -side polygon, where $n&gt;=3$. The issue now is how to find the $n$  such that it can contain the most number of circles, each with a constant radius $R$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to clarify the question: the area must be a regular $n$ -side polygon. Sorry for not making this clear&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-10T07:04:47.900" LastActivityDate="2010-09-15T15:55:25.090" Title="Find the regular $n$ -side polygon of A Constant Area that can contain Most Number of Circles" Tags="&lt;geometry&gt;&lt;optimization&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="4357" PostTypeId="2" ParentId="4346" CreationDate="2010-09-10T02:41:09.587" Score="1" ViewCount="0" Body="&lt;p&gt;Assume $f$ is a linear transformation, then linear algebra lets us give it a matrix representation $f(x) = F \cdot x$. It is also the case that $f \circ g = F G$, function composition is represented by matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your derivation you do not make any distinction between $f$ and $F$, this is fine as long as you know you are doing it. Also the matrix $F$ is a 1x1 because $\mathbb{R}$ is a 1-dimensional vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{eqnarray}&#xA;f(f(x)) &amp;amp;= f(x) + x \\&#xA; \iff \\&#xA;F^2 &amp;amp;= F + I \\&#xA; \iff \\&#xA;\varphi^2 &amp;amp;= \varphi + 1&#xA;\end{eqnarray} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Where $f(x) = F \cdot x = \varphi x$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is that we cannot be sure $f$ is a linear transform just based on what was told. So it is not known whether this gives all solutions.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-10T02:41:09.587" CommentCount="1" />
  <row Id="4358" PostTypeId="1" AcceptedAnswerId="4361" CreationDate="2010-09-10T03:27:33.737" Score="1" ViewCount="114" Body="&lt;p&gt;I've been reading a bit about how the set of bounds changes for a set depending on what superset one works with. I considered the sets $S\subseteq T\subseteq\mathbb{Q}$ and worked out a few contrived examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S=T=$ {$x\in\mathbb{Q}\ | \ x^2\lt 2$}, so here $S$ is not bounded above in $T$, but it is bounded about in $\mathbb{Q}$, with $2$ being a possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $S=$ {$x\in\mathbb{Q}\ | \ x^2\lt 1$} and $T=$ {$x\in\mathbb{Q}\ | \ x\lt 2 \ \text{and}\ x\neq 1$}, then $S$ is bounded in $T$ and $\sup_\mathbb{Q} S=1$ exists, but $\sup_T S$ does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is it possible for $S$ to be bounded in $T$ where $\sup_T S$ exists, but $\sup_\mathbb{Q} S$ does not? And moreover, can both $\sup_T S$ and $\sup_\mathbb{Q} S$ exist, but not be equal? Any example of this would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-26T15:27:13.790" LastActivityDate="2010-11-26T15:27:13.790" Title="For $S\subseteq T\subseteq\mathbb{Q}$, it is possible for $\sup_T S$ to exist, but $\sup_\mathbb{Q} S$ does not?" Tags="&lt;analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="4359" PostTypeId="2" ParentId="4356" CreationDate="2010-09-10T03:35:06.723" Score="5" ViewCount="0" Body="&lt;p&gt;If $n$ is unconstrained, as it seems to be in the question, then the polygon can be taken to be an $\epsilon$-accurate approximation to the optimal figure, and we can ask directly what that figure (or figures) look like. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you don't assume the polygon is convex, the answer is trivial, in that you can get as close as desired to a collection of $k$ disjoint circles connected by very thin tubes. In this version of the problem the only constraint is that $A &gt; k \pi R^2$ from which the maximum $k$ is easily determined.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If convexity is required, finding the tightest configuration of circles -- the one whose convex hull is a minimum-area figure containing $k$ or more discs of given size -- is a hard nonlinear optimization problem which is a variant of the classical problem of &lt;em&gt;circle packing&lt;/em&gt; (finding the smallest circle enclosing $k$ unit discs).   Recent experiments indicate that the optima, even for large $k$, do not have a connectivity pattern approximating the optimal lattice packing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the polygon is convex and you are satisfied with an asymptotically optimal solution, start from a hexagonal packing of circles of radius $R$ in the whole plane, and draw a circle of area $A$ that encloses as many of these as possible, then take the convex hull of the circles inside, then approximate the convex hull closely enough by a polygon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(added: for asymmetry in high density finite packings up to n=348, see &lt;a href=&quot;http://arxiv.org/abs/1002.0604&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1002.0604&lt;/a&gt; and a long series of theory papers by the same authors.  Best known packings of small numbers of disks in circles, hexagons, squares, and other shapes are displayed at: &lt;a href=&quot;http://www2.stetson.edu/~efriedma/packing.html&quot; rel=&quot;nofollow&quot;&gt;http://www2.stetson.edu/~efriedma/packing.html&lt;/a&gt;  .)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-10T07:32:52.603" LastActivityDate="2010-09-10T07:32:52.603" CommentCount="14" />
  <row Id="4360" PostTypeId="2" ParentId="4358" CreationDate="2010-09-10T03:50:19.883" Score="3" ViewCount="0" Body="&lt;p&gt;For the first question, let $S$ be the set of all rationals less than $\pi$.  Then $S$ has no least upper bound in $\mathbb{Q}$.  On the other hand, if $T = S\cup {4}$, then 4 is the least upper bound of $S$ inside $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, let $S$ be the set of all rationals strictly less than 1, and let $T = \mathbb{Q}\cap \left([-\infty, 1)\cup [2, \infty)\right)$.  Then the least upper bound of $S$ inside $T$ is 2.  But of course the least upper bound of $S$ in $\mathbb{Q}$ is 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows the answer to your second question is &quot;Yes.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-10T03:50:19.883" CommentCount="1" />
  <row Id="4361" PostTypeId="2" ParentId="4358" CreationDate="2010-09-10T03:50:54.620" Score="5" ViewCount="0" Body="&lt;p&gt;Yes, it is possible for $\sup_T(S)$ to exist, but $\sup_\mathbb{Q}(S)$ not to exist. Yes, it's possible for both to exist and be distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the latter. Take $S=\{x\in\mathbb{Q}|0\leq x\lt 1\}$. Let $T=S\cup{{2}}$. Then $S$ is bounded above in $T$, and has a supremum, namely $2$. Indeed, for every $t\in T$, if $t\lt 2$, then there exists $s\in S$ such that $t\lt s\leq 2$, so $2$ is the supremum of $S$. And of course, the supremum of $S$ in $\mathbb{Q}$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the former, take $S=\{x\in\mathbb{Q}|0\leq x \lt \sqrt{2}\}$ (or any irrational number you please), and take $T=S\cup\{2\}$ (or any number greater than $\sqrt{2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you have partially ordered sets $P\subseteq Q$, and a subset $A$ of $P$, you can have that $A$ has a supremum in $P$ but not in $Q$; has a supremum in $Q$ but not in $P$; has suprema in neither; has the same supremum in both; or has suprema in both but they are distinct. The one thing you can say is that if both suprema exist, then $\sup_Q(A)\leq\sup_P(A)$. So in your situation, you do know that if they both exist you will have $\sup_{\mathbb{Q}}(S)\leq\sup_{T}(S)$, but you can have strict inequality.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-10T04:02:12.033" LastActivityDate="2010-09-10T04:02:12.033" CommentCount="1" />
  <row Id="4362" PostTypeId="1" AcceptedAnswerId="4364" CreationDate="2010-09-10T04:05:02.650" Score="7" ViewCount="208" Body="&lt;p&gt;Consider the following sequence $a_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = 1 + \frac{1}{2^n-2} \sum_{i=1}^{n-1} \binom{n}{i} a_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first few terms are $0,1,\frac{3}{2},\frac{13}{7},\frac{15}{7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence comes out of the analysis of a certain process, whose asymptotics we expect to be about $\log_2 n$. Numerical computations show that indeed $a_n \approx \log_2 n - C$, where $C \approx 0.318$. It should be easy to show that $a_n = \log_2 n + O(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt; Is it true that $a_n = \log_2 n - C + o(1)$ for some constant $C$? &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some sequences defined by similar recurrence relations for which the &quot;constant&quot; part actually wiggles (unfortunately the reference escapes me).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T22:48:46.400" Title="A recurrence that wiggles?" Tags="&lt;asymptotics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4363" PostTypeId="1" CreationDate="2010-09-10T04:21:49.203" Score="2" ViewCount="266" Body="&lt;p&gt;If $\varphi:U\subset \mathbb{R}^n \to \mathbb{R}^m$ is $C^1$, let $\mathrm{T}\varphi:\mathrm{T}U \to \mathrm{T}R^m$ be its tangent map. The inverse function theorem tells us that if $\ker(\mathrm{T}\varphi(x))$ is zero, $\varphi$ is injective in some neighborhood of $x$. If the kernel is non-zero, what can we say about $\varphi$ near $x$ provided we know the kernel? In particular, can we say anything about curves through $x$ whose tangents belong to this kernel? &lt;/p&gt;&#xA;" OwnerUserId="1320" LastEditorUserId="1320" LastEditorDisplayName="" LastEditDate="2010-09-10T13:07:11.803" LastActivityDate="2010-11-13T11:44:19.203" Title="Kernel of the tangent map" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="4364" PostTypeId="2" ParentId="4362" CreationDate="2010-09-10T07:14:27.373" Score="10" ViewCount="0" Body="&lt;p&gt;This is the Erdos-Renyi &quot;20 Questions&quot; problem.  You want to guess an integer in a set of N elements by selecting subsets and asking questions of the form &quot;and is it in &lt;em&gt;this&lt;/em&gt; subset?&quot;.  For the question to provide information the subset cannot be empty or the entire set of uneliminated possibilities, and if an informative question (subset) is chosen at random one gets the above recursion for the expected length of the game.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erdos and Renyi showed that the expected number of questions is $\log_2(n)$ plus a non-constant periodic function of $\log_2(n)$.  I don't remember if they allowed uninformative questions, so the problem could be slightly different, but similar log-periodic behavior should appear in both versions of the problem.  (added:  the reason it should be the same is that the dominant $\log_2(n)$ term comes from the concentration of the size of the random subset around $k/2$ where $k$ is the number of remaining possibilities, and the log-periodic part comes from the $O(\sqrt{k})$ dispersion of the subset size near $k/2$.  What happens at the very extremes of the distribution is much less influential.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-10T07:42:23.283" LastActivityDate="2010-09-10T07:42:23.283" CommentCount="1" />
  <row Id="4365" PostTypeId="1" AcceptedAnswerId="4366" CreationDate="2010-09-10T07:17:48.323" Score="5" ViewCount="103" Body="&lt;p&gt;A number $n$ is perfect if it's equal to the sum of its divisors (smaller than itself). A well known theorem by Euler states that every even perfect number is of the form $2^{p-1}(2^p-1)$ where $2^p-1$ is prime (this is what is called a &lt;strong&gt;Mersenne&lt;/strong&gt; prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this theorem proved? The converse (showing that every number of that form is perfect) is simple (the divisors of such a number are powers of 2 and powers of 2 times the Mersenne prime, and it's an easy sum to calculate) but I couldn't find a proof for the theorem itself.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-10T07:28:01.350" Title="How to show that all even perfect numbers are obtained via Mersenne primes?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="4366" PostTypeId="2" ParentId="4365" CreationDate="2010-09-10T07:28:01.350" Score="6" ViewCount="0" Body="&lt;p&gt;Many elementary number theory texts have the proof, e.g., Hardy &amp;amp; Wright's.&#xA;&lt;a href=&quot;http://primes.utm.edu/notes/proofs/EvenPerfect.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an online proof from Chris Caldwell's &lt;em&gt;Prime Pages&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-10T07:28:01.350" />
  <row Id="4367" PostTypeId="1" CreationDate="2010-09-10T07:58:40.090" Score="1" ViewCount="35" Body="&lt;p&gt;In the attached paper the writer proves the isomorphism between potential and congestion games by building a congestion game from a given potential game (theorem 1, subsection 2.7). He defines the groups M_1 and M_2 as facilities in the congestion games. If there are only 2 players in the given potential game, then M_1 and M_2 are sets of matrices. Can I look at an element from either of them as a payment matrix for a potential game? A&lt;a href=&quot;http://ie.technion.ac.il/~dov/ijcai_multi6.pdf&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;nd if so - why?&lt;/p&gt;&#xA;" OwnerUserId="1819" LastActivityDate="2010-09-10T07:58:40.090" Title="The isomorphism between congestion games and potential games" Tags="&lt;game-theory&gt;" CommentCount="1" />
  <row Id="4368" PostTypeId="1" AcceptedAnswerId="4369" CreationDate="2010-09-10T08:01:17.977" Score="5" ViewCount="189" Body="&lt;p&gt;I am trying to compute the asymptotic growth-rate in a specific combinatorial problem depending on a parameter w, using the Transfer-Matrix method. This amounts to computing the largest eigenvalue of the corresponding matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small values of w, the corresponding matrix is small and I can use the so-called power method - start with some vector, and multiply it by the matrix over and over, and under certain conditions you'll get the eigenvector corresponding to the largest eigenvalue. However, for the values of w I'm interested in, the matrix becomes to large, and so the vector becomes too large - $n&gt;10,000,000,000$ entries or so, so it can't be contained in the computer's memory anymore and I need extra programming tricks or a very powerful computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the matrix itself, I don't need to store it in memory - I can access it as a black box, i.e. given $i,j$ I can return $A_{ij}$ via a simple computation. Also, the matrix has only 0 and 1 entries, and I believe it to be sparse (i.e. only around $\log n$ of the entries are 1's, $n$ being the number of rows/columns). However, the matrix is &lt;strong&gt;not&lt;/strong&gt; symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some method more space-effective for computation of eigenvalues for a case like this?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-10T10:31:04.830" LastActivityDate="2010-09-12T10:19:35.053" Title="Computing the largest Eigenvalue of a very large sparse matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="4369" PostTypeId="2" ParentId="4368" CreationDate="2010-09-10T09:00:57.040" Score="5" ViewCount="0" Body="&lt;p&gt;You could use the Arnoldi Iteration algorithm. This algorithm only requires the matrix $A$ for matrix-vector multiplication. I'm expecting that you will be able to black-box the function $v\rightarrow Av$. What you generate is an upper Hessenberg matrix $H$ whose eigenvalues whose can be computed cheaply (by a direct method or Rayleigh quotient iteration) and which approximate the eigenvalues of $A$. Arnoldi Iteration will give the best approximation to the dominant eigenvalue so I suspect you won't have to do many iterations before you have a good estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent introduction to this is: &quot;Numerical Linear Algebra&quot; by Trefethen and Bau. (p250)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic algorithm can be found here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Arnoldi_iteration&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arnoldi_iteration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the only thing that is required to make this a fully functional algorithm is a termination condition. Since you don't seem to need the dominant eigenvalue to a high degree of accuracy I would not worry and just stop when the dominant eigenvalue estimate doesn't change too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have Matlab you can always use the built in function eigs(Afun,n,...) where Afun is the black-box function handle that computes $Av$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-10T09:29:00.427" LastActivityDate="2010-09-10T09:29:00.427" CommentCount="3" />
  <row Id="4370" PostTypeId="1" CreationDate="2010-09-10T11:02:52.957" Score="0" ViewCount="187" Body="&lt;p&gt;If $h(x_{t\wedge \tau})$ is a submartingale such that $h(0)=1$ and $\tau$ is a stopping time. Why is it that $h(x_\tau)=1_{{\tau&amp;lt;\infty}}$?. Also $h(x)$ is decreasing.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-09-10T14:26:50.317" LastActivityDate="2010-09-10T14:26:50.317" Title="stopping time and  martingale" Tags="&lt;stochastic-processes&gt;" CommentCount="11" />
  <row Id="4371" PostTypeId="1" AcceptedAnswerId="4374" CreationDate="2010-09-10T11:48:22.973" Score="3" ViewCount="332" Body="&lt;p&gt;I would like to learn Graph Theory from the beginning. It seems to me that one does not need to be familiar with many abstract type subjects to be able to understand the more basic concepts of graphs.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Which subjects should one know prior to learn Graph Theory at the introductory level?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And which book or lecture notes would you advise to learn it? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-10T18:17:40.333" LastActivityDate="2010-09-14T08:41:39.037" Title="Prerequisites for learning (basic) Graph Theory" Tags="&lt;books&gt;&lt;graph-theory&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="4372" PostTypeId="1" CreationDate="2010-09-10T11:54:39.127" Score="1" ViewCount="123" Body="&lt;p&gt;If $f$ is a positive function, the intuitive interpretation of the Riemann integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_a^b f(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the area under the curve $f$ between $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ and $g$ are smooth positive functions. Is it correct to interpret the Riemann-Stieltjes integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_a^b f(x) d g(x)$ as the volume under a &quot;ribbon&quot;, where the height of the ribbon at a point $u$ between $a$ and $b$ is determined by $f$ and the thickness is determined by $g'$?&lt;/p&gt;&#xA;" OwnerUserId="1820" LastActivityDate="2010-09-10T15:55:48.090" Title="Is this interpretation of Stieltjes integration correct?" Tags="&lt;real-analysis&gt;&lt;intuition&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="4373" PostTypeId="2" ParentId="4371" CreationDate="2010-09-10T12:11:23.777" Score="2" ViewCount="0" Body="&lt;p&gt;You don't need more than knowledge of basic notations in Mathematics to read a &lt;strong&gt;basic&lt;/strong&gt; book on Graph Theory. However, some experience in mathematics is helpful, even if the material is not used directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite books for &quot;pure&quot; graph theory is &quot;Graph Theory&quot; by Harary and &quot;Modern Graph Theory&quot; by Bollobas. They also delve into algebraic graph theory in later chapters and for that you'll need some basic linear algebra and group theory. Also, graph theory is widely used in computer science and so, for example, many chapters in the &quot;Introduction to algorithms&quot; by Corman and co. deal with algorithms on graphs (an interesting subject by itself but maybe not what you have in mind).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-10T12:11:23.777" CommentCount="2" />
  <row Id="4374" PostTypeId="2" ParentId="4371" CreationDate="2010-09-10T12:13:41.563" Score="8" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;(a)&lt;/strong&gt; Basic logic + set operations almost goes without saying (&lt;em&gt;e.g.&lt;/em&gt; logical conjunction&amp;nbsp;/&amp;nbsp;set intersections; also equivalence classes, sets and relations obtained by modding out subsets, &lt;em&gt;etc&lt;/em&gt;).&lt;br&gt;&lt;br&gt;&lt;strong&gt;(b)&lt;/strong&gt; Depending on how 'basic' you mean, you may or may not derive benefit from linear algebra (for playing with adjacency matrices), probability theory (for applying the probabilistic method, natch), elementary number theory (&lt;em&gt;e.g.&lt;/em&gt; arithmetic modulo &lt;em&gt;N&lt;/em&gt;) and experience with asymptotic analysis (knowledge of what O(&lt;em&gt;N&lt;/em&gt;), o(&lt;em&gt;N&lt;/em&gt;), &amp;omega;(&lt;em&gt;N&lt;/em&gt;), etc. refer to and how to play with them).&lt;br&gt;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I found Diestel's text quite nice. (Furthermore, it is freely available from &lt;a href=&quot;http://diestel-graph-theory.com/&quot; rel=&quot;nofollow&quot;&gt;his website&lt;/a&gt;). I also found that undoing his proofs-by-contradiction provides good exercise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-09-10T12:13:41.563" CommentCount="3" />
  <row Id="4377" PostTypeId="2" ParentId="4273" CreationDate="2010-09-10T14:21:50.153" Score="4" ViewCount="0" Body="&lt;p&gt;They are called &quot;personal probabilities&quot; in the literature on Bayesian statistics, which is the field where the possibility of different probability assessments among different observers is considered.  You could check out the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayesian_probability#Personal_probabilities_and_objective_methods_for_constructing_priors&quot; rel=&quot;nofollow&quot;&gt;Bayesian probability&lt;/a&gt;, for instance.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-10T14:21:50.153" CommentCount="4" />
  <row Id="4378" PostTypeId="2" ParentId="4362" CreationDate="2010-09-10T15:03:45.297" Score="4" ViewCount="0" Body="&lt;p&gt;Not really an answer, but too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A recurrence with a wiggle&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following was proposed by Lennart Rade&#xA;  in a 1991 issue of the &lt;em&gt;American Mathematical Monthly&lt;/em&gt;, and the&#xA;   solution appears in the January 1994 issue, on page 78.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin with $n$ fair coins all showing tails.&#xA;  Toss them all and pick up only the coins with tails.&#xA;  Repeat this procedure until all coins show heads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of coins showing tails is a Markov chain that&#xA;starts in state $n$ and ends up at 0. The transition probabilities&#xA;are binomial: $p_{n,i}={n \choose i}\left({1\over 2}\right)^n \mbox{ for }0\leq i\leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence $u_n=P_{n+1}(\mbox{chain hits }1)$ satisfies  $u_0=1$ and for $n\geq1$&#xA; $$u_n=\sum_{j=0}^n{n+1\choose j}2^{-(n+1)} u_{n-j},$$&#xA; or $$u_n=\sum_{j=1}^n{n+1\choose j}{2^{-(n+1)}\over 1-2^{-(n+1)}} u_{n-j}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this problem can be solved explicitly giving&#xA;$$u_n={n+1\over2}\sum_{k=0}^\infty 2^{-k}(1-2^{-k})^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis of this expression shows that, despite appearances, the sequence $u_n$ does&#xA; &lt;strong&gt;not&lt;/strong&gt; converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It slowly oscillates about the pseudo limit $${1\over2\ln(2)}=.72134$$&#xA;with an amplitude of about $${|\Gamma(1-2\pi i/\ln(2))|\over2\ln(2)}=.00000356.$$&lt;br&gt;&#xA;It's a fun example to show students as the oscillations are invisible graphically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For related examples see &#xA;&lt;em&gt;The asymptotic probability of a tie for first place&lt;/em&gt;,&#xA;by B. Eisenberg, G. Stengle, and G. Strang, &lt;strong&gt;Annals of Applied Probability 3&lt;/strong&gt;, 731-745 (1993). &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-10T15:03:45.297" CommentCount="1" />
  <row Id="4379" PostTypeId="2" ParentId="4372" CreationDate="2010-09-10T15:55:48.090" Score="1" ViewCount="0" Body="&lt;p&gt;You are effectively replacing $\int_a^b f(x) dg(x)$ with $\int_a^b f(x) g'(x) dx$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral#Properties_and_relation_to_the_Riemann_integral&quot; rel=&quot;nofollow&quot;&gt;A quick trip to Wikipedia&lt;/a&gt; reveals that this is perfectly fine when $g$ is absolutely continuous, or when it has a continuous derivative, but may not be valid otherwise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&lt;p&gt;If $g$ should happen to be everywhere differentiable, then the integral may still be different from the Riemann integral&#xA;&#xA;    $$\int_a^b f(x) g'(x) \, dx,$$&#xA;&#xA;for example, if the derivative is unbounded. But if the derivative is continuous, they will be the same. This condition is also satisfied if $g$ is the (Lebesgue) integral of its derivative; in this case $g$ is said to be absolutely continuous.&#xA;&#xA;&lt;p&gt;However, $g$ may have jump discontinuities, or may have derivative zero almost everywhere while still being continuous and increasing (for example, $g$ could be the Cantor function), in either of which cases the Riemann–Stieltjes integral is not captured by any expression involving derivatives of $g$.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; I didn't notice the word &quot;smooth&quot; in your question. In this light, the answer to your question is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;http://mathworld.wolfram.com/StieltjesIntegral.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld is probably a better reference&lt;/a&gt;: the two expressions are equal &quot;if $f$ is continuous and $g'$ is Riemann integrable over the specified interval&quot;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-10T15:55:48.090" />
  <row Id="4380" PostTypeId="1" AcceptedAnswerId="4382" CreationDate="2010-09-10T20:06:17.260" Score="2" ViewCount="211" Body="&lt;p&gt;If you look at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=world+population&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/x0ILQ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or this Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_countries_by_median_age&quot; rel=&quot;nofollow&quot;&gt;List of countries by median age&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QBtjT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly &lt;a href=&quot;http://en.wikipedia.org/wiki/Median&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;median&lt;/strong&gt;&lt;/a&gt; seems to be the statistic of choice when it comes to ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to explain to myself why &lt;a href=&quot;http://en.wikipedia.org/wiki/Arithmetic_mean&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;arithmetic mean&lt;/strong&gt;&lt;/a&gt; would be a worse statistic. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="324" LastEditorUserId="324" LastEditorDisplayName="" LastEditDate="2010-09-10T20:23:15.127" LastActivityDate="2010-09-12T22:17:48.857" Title="Why is median age a better statistic than mean age?" Tags="&lt;statistics&gt;&lt;average&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="4381" PostTypeId="1" AcceptedAnswerId="4383" CreationDate="2010-09-10T20:16:48.073" Score="3" ViewCount="185" Body="&lt;p&gt;Hi! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was doing a physics problem today and I needed to find when some movement is accelerating and when it is decelerating. So after some calculations, I got the equation I needed. I know that it is correct because when plotted, I can see from graph that I got the expected solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I don't know how to actually start solving the inequality. It looks like &#xA;$$k \sin(\frac {\pi x} {2})\cos(\frac {\pi x} {2})&gt;0$$ where $k$ is some huge sausage which is constant and positive.&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-11T06:21:53.403" LastActivityDate="2010-09-11T06:21:53.403" Title="How to solve this trigonometric inequality: $k \sin(\frac {\pi x} {2})\cos(\frac {\pi x} {2})&gt;0$? " Tags="&lt;algebra-precalculus&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" />
  <row Id="4382" PostTypeId="2" ParentId="4380" CreationDate="2010-09-10T20:19:13.060" Score="3" ViewCount="0" Body="&lt;p&gt;Median is what many people actually have in mind when they say &quot;mean.&quot;  It's easier to interpret the median: half the population is above this age and half are below.  Mean is a little more subtle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People look for symmetry and sometimes impose symmetry when it isn't there.  The age distribution in a population is far from symmetric, so the mean could be misleading. Age distributions are something like a pyramid.  Lots of children, not many elderly.  (Or at least that's how it is in a sort of steady state.  In the US, for example, the post-WWII baby boom generation has distorted the distribution as they age.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an asymmetrical distribution, it may be better to report the median because it is a symmetrical statistic in the sense that it splits the population in half.  Said another way, the median is symmetrical even if the distribution isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I got my logic backward when I first answered and said the mean would be lower than the median.  I meant the opposite.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-09-12T22:17:48.857" LastActivityDate="2010-09-12T22:17:48.857" CommentCount="3" />
  <row Id="4383" PostTypeId="2" ParentId="4381" CreationDate="2010-09-10T20:21:39.463" Score="7" ViewCount="0" Body="&lt;p&gt;You can use the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin 2x = 2 \sin x \cos x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you only need to determine when $$ \sin \pi x  &gt; 0$$ and this happens if and only if $$ x \in (2m, 2m+1), m \in \mathbb{Z}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-10T20:21:39.463" />
  <row Id="4384" PostTypeId="2" ParentId="4381" CreationDate="2010-09-10T20:24:25.693" Score="3" ViewCount="0" Body="&lt;p&gt;Since $k$ is positive, it is immaterial. You want to know the values of $x$ for which $\sin(\frac{\pi x}{2})$ and $\cos(\frac{\pi x}{2})$ have the same sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each function has period $4$, so we can just look at what happens on $[0,4]$. $\sin(\frac{\pi x}{2})$ is positive when $0\lt\frac{\pi x}{2}\lt\pi$ (that is, when $0\lt x\lt 2$), and negative when $2\lt x \lt 4$. $\cos(\frac{\pi x}{2})$ is positive when $0\leq \frac{\pi x}{2}\lt \frac{\pi}{2}$ and when $\frac{3\pi}{2}\lt \frac{\pi x}{2}\lt 4$ (that is, when $0\leq x\lt 1$ and when $3\lt x\lt 4$). So the two have the same sign on $(0,1)$, and on $(2,3)$. Lather, rinse, and repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this is easier to do using the identity that others have suggested; the approach of looking at the signs of the different factors is what you would use in general, or when there is no easy way of reducing the expression to a single item). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-10T20:32:54.247" LastActivityDate="2010-09-10T20:32:54.247" />
  <row Id="4385" PostTypeId="2" ParentId="4371" CreationDate="2010-09-10T20:40:17.037" Score="3" ViewCount="0" Body="&lt;p&gt;Graph Theory is indeed a very quick starting subject in the sense that one does need to have studied calculus and other mathematical subjects to get started. However, there are lots of books that are specialized towards particular purposes: applications in general, network applications, distances in graphs, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One book that I think has nice balance is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to Graph Theory (Second edition) by Douglas West, Prentice-Hall, 2001.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-09-11T00:59:56.030" LastActivityDate="2010-09-11T00:59:56.030" />
  <row Id="4386" PostTypeId="1" CreationDate="2010-09-10T20:47:45.427" Score="4" ViewCount="71" Body="&lt;p&gt;Hi, I'm a noob, and I'm not a mathematician (Although I will be a Math major next semester).  My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 68 maps I would like to use as posters on my wall at home.  They are all rectangles, and all different dimensions.  Is there an algorithm (or better yet, a website or other simple computer program I can use) to take the dimensions of all these posters as input and then compute the optimal tiling arrangement on the wall so there is little as possible &quot;wasted&quot; empty space?&lt;/p&gt;&#xA;" OwnerUserId="1825" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-10T23:23:42.210" LastActivityDate="2010-09-10T23:23:42.210" Title="Tiling Posters on a Wall" Tags="&lt;geometry&gt;&lt;tiling&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4387" PostTypeId="2" ParentId="4386" CreationDate="2010-09-10T21:00:47.747" Score="1" ViewCount="0" Body="&lt;p&gt;You are attempting to solve what is referred to as a &quot;packing problem&quot;. It's generally a hard problem but for your case I think that the existing methods will work. Try looking here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pollinimini.net/blog/rectangle-packing/&quot; rel=&quot;nofollow&quot;&gt;http://pollinimini.net/blog/rectangle-packing/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems he has already coded something. If it's not what you're after, google &quot;rectangle packing&quot; and I'm sure you'll run into something.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-10T21:00:47.747" />
  <row Id="4388" PostTypeId="2" ParentId="4386" CreationDate="2010-09-10T21:02:12.937" Score="2" ViewCount="0" Body="&lt;p&gt;This problem is known as 2-dimensional bin packing, and some algorithms for this problem can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csc.liv.ac.uk/~epa/surveyhtml.html&quot; rel=&quot;nofollow&quot;&gt;http://www.csc.liv.ac.uk/~epa/surveyhtml.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-10T21:02:12.937" />
  <row Id="4390" PostTypeId="2" ParentId="4380" CreationDate="2010-09-10T22:28:44.567" Score="1" ViewCount="0" Body="&lt;p&gt;The best statistic to summarize a distribution depends upon the distribution and what you want to use it for.  For distributions that are nicely bell-shaped, the mean, median, and mode are close together and it doesn't matter.  For skew distributions the mean is out on the skew side from the median, but it still represents the expected value of the average of a large number of samples.  The median is closer to more of the individuals than the mean.  For stranger distributions no one number can provide a useful summary.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-10T22:28:44.567" CommentCount="1" />
  <row Id="4391" PostTypeId="2" ParentId="4381" CreationDate="2010-09-10T22:31:35.077" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Slighty simpler&lt;/strong&gt;:$\;$ note $\rm\; ab &gt; 0 \iff a/b &gt; 0\;\;$ so $\;\;\sin(z)\cos(z)&gt;0 \iff \tan(z) &gt; 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is trivial using $\;\;\tan(-z) = -\tan(z)\;\;\:$ and $\:\;\;\tan(z+\pi) = \tan(z)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-10T22:42:48.923" LastActivityDate="2010-09-10T22:42:48.923" />
  <row Id="4392" PostTypeId="2" ParentId="4268" CreationDate="2010-09-11T00:05:04.600" Score="4" ViewCount="0" Body="&lt;p&gt;As far as I know, your definition&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A prime is an element p such that whenever p divides ab, then either p divides a or p divides b,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the true definition of &quot;prime&quot;. The usual one, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... an element p which cannot be expressed as a product of non-unit elements,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the definition of an &lt;em&gt;irreducible&lt;/em&gt; element. Now, in every ring all primes are also irreducible, but the converse is in general not true. In other words, there are rings where the two definitions are not equivalent. One case in which they are indeed equivalent is when the ring is a &lt;em&gt;unique factorization domain&lt;/em&gt; (the statement of the so-called &quot;fundamental theorem of arithmetic&quot; is that the integers form one such ring).&lt;/p&gt;&#xA;" OwnerUserId="1828" LastEditorUserId="1828" LastEditorDisplayName="" LastEditDate="2010-09-11T11:39:25.117" LastActivityDate="2010-09-11T11:39:25.117" CommentCount="3" />
  <row Id="4393" PostTypeId="2" ParentId="4335" CreationDate="2010-09-11T00:11:10.983" Score="2" ViewCount="0" Body="&lt;p&gt;I'll add another title (it's a bit on the theoretical side, but still at the introductory level, very readable and definitely worth):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Serge Lang, &lt;em&gt;Linear algebra&lt;/em&gt; 3rd. ed., Springer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1828" LastActivityDate="2010-09-11T00:11:10.983" CommunityOwnedDate="2010-09-11T00:11:10.983" />
  <row Id="4394" PostTypeId="2" ParentId="2395" CreationDate="2010-09-11T00:32:10.590" Score="0" ViewCount="0" Body="&lt;p&gt;If you can get a copy (which is improbable, I'm afraid),&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vladimir Smirnov, &lt;em&gt;A Course of Higher Mathematics, VOLUME 1&lt;/em&gt;, Pergamon press 1964.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a translation of a beautiful treatise on mathematical analysis which served as a textbook for generations of engineering students in the USSR. The first volume, in particular, is one of the best introductions to calculus I've ever seen.&lt;/p&gt;&#xA;" OwnerUserId="1828" LastActivityDate="2010-09-11T00:32:10.590" CommunityOwnedDate="2010-09-11T00:32:10.590" />
  <row Id="4395" PostTypeId="2" ParentId="4335" CreationDate="2010-09-11T00:48:32.547" Score="0" ViewCount="0" Body="&lt;p&gt;In my opinion, Schaum's Linear Algebra is very straight forward, has a nice balance between pure and applied, and has the advantage that it is cheap. Definitely, if you are auto-didactic, it will be easier to work through. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-09-11T00:48:32.547" CommunityOwnedDate="2010-09-11T00:48:32.547" />
  <row Id="4396" PostTypeId="2" ParentId="3955" CreationDate="2010-09-11T01:25:02.193" Score="1" ViewCount="0" Body="&lt;p&gt;Here is how you can construct a linear relation for general bounds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $x$ goes from $a$ to $b$,&lt;/li&gt;&#xA;&lt;li&gt;then $x-a$ goes from $0$ to $b-a$,&lt;/li&gt;&#xA;&lt;li&gt;then $\frac{x-a}{b-a}$ goes from $0$ to $1$,&lt;/li&gt;&#xA;&lt;li&gt;then $(x-a) \frac{d-c}{b-a}$ goes from $0$ to $d-c$,&lt;/li&gt;&#xA;&lt;li&gt;then $y := c + (x-a) \frac{d-c}{b-a}$ goes from $c$ to $d$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you plug $a=0$, $b=100$, $c=1000$, $d=0$ as in your example, you obtain $y = 1000 + (x-0) \frac{0-1000}{100-0} = 1000 - 10x$ as in the other answers.&lt;/p&gt;&#xA;" OwnerUserId="1828" LastActivityDate="2010-09-11T01:25:02.193" />
  <row Id="4397" PostTypeId="2" ParentId="250" CreationDate="2010-09-11T01:38:29.773" Score="3" ViewCount="0" Body="&lt;p&gt;&quot;The natural numbers are as many as the even natural numbers&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement is trivial and not worth to be named &quot;theorem&quot;, but it is rather counterintuitive if you don't know the meaning of &quot;as many&quot;. At least, it was for Galileo :)&lt;/p&gt;&#xA;" OwnerUserId="1828" LastEditorUserId="1828" LastEditorDisplayName="" LastEditDate="2010-09-11T01:44:31.540" LastActivityDate="2010-09-11T01:44:31.540" CommunityOwnedDate="2010-09-11T01:38:29.773" CommentCount="1" />
  <row Id="4398" PostTypeId="2" ParentId="1819" CreationDate="2010-09-11T02:22:37.813" Score="3" ViewCount="0" Body="&lt;p&gt;For $X_1$, $X_2$, $\ldots$, i.i.d., with finite variance $\sigma ^{2}$, the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem#Classical_central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;Central Limit Theorem&lt;/a&gt;&lt;/strong&gt; (convergence in distribution) says that if we wait long enough, that is, take a large enough sample, then $$ \frac{\sqrt n(\overline{X}_n-\mu)}{\sigma}$$ will have a probability distribution which is arbitrarily close to a &lt;em&gt;norm&lt;/em&gt;(0,1) distribution.  Notice the CLT doesn't say anything about the actual behavior of any particular $\overline{X}$ for large $n$, only that if we observed a whole bunch of them (all at that large $n$) and made a histogram then it would be approximately bell-shaped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Weak Law of Large Numbers&lt;/strong&gt; (convergence in probability) says that if we wait long enough (&lt;em&gt;i.e.&lt;/em&gt;, take a large enough sample) then the chance that $$ \frac{\overline{X}_n-\mu}{\sigma}$$ is nearby zero can be made arbitrarily high.  Note a couple of things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the $\sqrt{n}$ is gone, and&lt;/li&gt;&#xA;&lt;li&gt;again, nothing is said about any particular observed $\overline{X}$, only that it would be a pretty safe bet that the above is close to zero.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Strong Law of Large Numbers&lt;/strong&gt; (convergence almost surely) says that $$ \frac{\overline{X}_n-\mu}{\sigma}$$ is 100% sure to be arbitrarily close to 0, provided we wait long enough (take a large enough sample).  That is &lt;em&gt;strong&lt;/em&gt;, indeed. Again, please note a couple of things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the $\sqrt{n}$ is still gone. In fact, this is the same fellow that was in the WLLN, and&lt;/li&gt;&#xA;&lt;li&gt;this time, something is said about the &lt;em&gt;particular&lt;/em&gt; sequence of $\overline{X}_n$'s that we are watching go by.  It's still possible that it would be far away (more than $\epsilon$), but all sequences for which $\overline{X}_n$ &lt;em&gt;never&lt;/em&gt; converges to $\mu$ have probability zero.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, nothing is ever guaranteed in probability; we can do no better than 100% sure.  Also, the SLLN doesn't say anything about &lt;em&gt;how long&lt;/em&gt; we would have to wait to be within $\epsilon$, we would need the &lt;strong&gt;Law of the Iterated Logarithm&lt;/strong&gt; for something like that.  Finally, this discussion is about convergence (in distribution/probability/a.s.) to a constant, while the general definition is about convergence to another random variable (or even convergence of two sequences).  But we can regain the intuition if we think about differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see anything wrong with the other two answers given for this question, I just thought I'd offer another perspective.&lt;/p&gt;&#xA;" OwnerUserId="1830" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-09-11T06:47:32.997" LastActivityDate="2010-09-11T06:47:32.997" />
  <row Id="4399" PostTypeId="2" ParentId="4268" CreationDate="2010-09-11T03:10:58.760" Score="3" ViewCount="0" Body="&lt;p&gt;You have encountered the general ring-theoretic definition of a prime (versus irreducible) element. For general rings one distinguishes between the inequivalent properties of being &lt;strong&gt;irreducible&lt;/strong&gt;, i.e. having no nontrivial factors, and being &lt;strong&gt;prime&lt;/strong&gt;, i.e a nonunit such that if it divides a product then it divides some factor of the product. The latter property is key to unique factorization since one easily proves by induction that products of primes factor uniquely in any domain, i.e. a ring such that $\rm\; ab = 0 \;\Rightarrow\; a=0\ \  or\ \ b=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the precise definitions. Let $\rm\; a,b,p \;$ be elements of a domain $\rm\:Z\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\rm\quad\; a\;$ is a &lt;strong&gt;unit&lt;/strong&gt; $\ \ $ if $\rm\ \ a\: |\: 1\quad\quad\quad\ \ \:\; (\:a\ |\ b\ :=\ a\ \ divides\ \ b\:)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\quad$ nonunit $\rm\; p\;$ is $\quad\;$ &lt;strong&gt;prime&lt;/strong&gt; $\quad\;$ if $\;\;\;\rm p \;\;\:|\:\;\; ab \;\;\Rightarrow\;\; p|a \;\;\; or \;\;\; p|b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\quad$ nonunit $\rm\; p\;$ is $\:$ &lt;strong&gt;irreducible&lt;/strong&gt;$\;$ if $\;\;\:\rm p = ab \;\;\Rightarrow\;\; p|a \;\;\; or \;\;\; p|b\quad\;\;$ (a.k.a.$\:$ &lt;strong&gt;atom&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COROLLARY&lt;/strong&gt; $\quad\;$ prime $\;\;\Rightarrow\;\;$ irreducible, since $\;\;\;\rm p = ab \;\;\Rightarrow\;\; p|ab$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conversely&lt;/strong&gt; $\:\;$ irreducible $\;\Rightarrow\;\;$ prime $\;\;$ iff $\;$ factorizations into irreducibles are unique (upto order and units). Indeed, it is very easy to prove that a factorization into primes is unique, i.e. any other factorization into irreducibles is the same (upto order and units). &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T22:34:53.240" LastActivityDate="2010-09-19T22:34:53.240" />
  <row Id="4400" PostTypeId="1" AcceptedAnswerId="4401" CreationDate="2010-09-11T05:30:01.867" Score="5" ViewCount="438" Body="&lt;p&gt;The following probability question appeared in an &lt;a href=&quot;http://math.stackexchange.com/questions/250/a-challenge-by-r-p-feynman-give-counter-intuitive-theorems-that-can-be-transl/346#346&quot;&gt;earlier thread&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The claim was that it is not actually a mathematical problem and it is only a language problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If one wanted to restate this problem formally the obvious way would be like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: &lt;em&gt;Sex&lt;/em&gt; is defined as an element of the set $\{\text{boy},\text{girl}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: &lt;em&gt;Birthday&lt;/em&gt; is defined as an element of the set $\{\text{Monday},\text{Tuesday},\text{Wednesday},\text{Thursday},\text{Friday},\text{Saturday},\text{Sunday}\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: A &lt;em&gt;Child&lt;/em&gt; is defined to be an ordered pair: (sex $\times$ birthday).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x,y)$ be a pair of children,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define an auxiliary predicate $H(s,b) :\!\!\iff s = \text{boy} \text{ and } b = \text{Tuesday}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate $P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I don't see any other sensible way to formalize this question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To actually solve this problem now requires no thought (infact it is thinking which leads us to guess incorrect answers), we just compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;&amp;amp; P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y)) \\&#xA;=&amp;amp; \frac{P(x\text{ is a boy and }y\text{ is a boy and }(H(x)\text{ or }H(y)))}&#xA;        {P(H(x)\text{ or }H(y))} \\&#xA;=&amp;amp; \frac{P((x\text{ is a boy and }y\text{ is a boy and }H(x))\text{ or }(x\text{ is a boy and }y\text{ is a boy and }H(y)))}&#xA;        {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\&#xA;=&amp;amp; \frac{\begin{align*} &amp;amp;P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday}) \\&#xA;   + &amp;amp;P(x\text{ is a boy and }y\text{ is a boy and }y\text{ born on Tuesday}) \\&#xA;   - &amp;amp;P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday and }y\text{ born on Tuesday}) \\&#xA;   \end{align*}}&#xA;   {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\&#xA;=&amp;amp; \frac{1/2 \cdot 1/2 \cdot 1/7 + 1/2 \cdot 1/2 \cdot 1/7 - 1/2 \cdot 1/2 \cdot 1/7 \cdot 1/7}&#xA;        {1/2 \cdot 1/7 + 1/2 \cdot 1/7 - 1/2 \cdot 1/7 \cdot 1/2 \cdot 1/7} \\&#xA;=&amp;amp; 13/27&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now what I am wondering is, does this refute the claim that this puzzle is just a language problem or add to it? Was there a lot of room for misinterpreting the questions which I just missed? &lt;/p&gt;&#xA;" LastActivityDate="2010-09-13T08:58:03.033" Title="Boy Born on a Tuesday - is it just a language trick?" Tags="&lt;probability&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="5" />
  <row Id="4401" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T05:42:25.577" Score="4" ViewCount="0" Body="&lt;p&gt;There are even trickier aspects to this question. For example, what is the strategy of the guy telling you about his family? If he always mentions a boy first and not a daughter, we get one probability; if he talks about the sex of the first born child, we get a different probability. Your calculation makes a choice in this issue - you choose the version of &quot;if the father has a boy and a girl, he'll mention the boy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm aiming to is this: the question is not well-defined mathematically. It has several possible interpretations, and as such the &quot;problem&quot; here is indeed of the language; or more correctly, the fact that a simple statement in English does not convey enough information to specify the precise model for the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a simplified version without days. The probability space for the make-up of the family is {BB, GB, BG, GG} (GB means &quot;an older girl and a small boy&quot;, etc). We want to know what is $P(BB|A)$ where A is determined by the way we interpret the statement about the boys. Now let's look at different possible interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If there is a boy in the family, the statement will mention him. In this case A={BB,BG,GB} and so the probability is $1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If there is a girl in the family, the statement will mention her. In this case, since the statement talked about a boy, there are NO girls in the family. So A={BB} and so the probability is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The statement talks about the sex of the firstborn. In this case A={BB,BG} and so the probability is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line: The statement about the family looks &quot;constant&quot; to us, but it must be looked as a function from the random state of the family - and there are several different possible functions, from which you must choose one otherwise no probabilistic analysis of the situation will make sense.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-11T06:39:56.287" LastActivityDate="2010-09-11T06:39:56.287" CommentCount="24" />
  <row Id="4402" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T06:53:52.183" Score="0" ViewCount="0" Body="&lt;p&gt;The Tuesday is a red herring. It's stated as a fact, thus the probability is 1. Also, it doesn't say &quot;&lt;em&gt;only&lt;/em&gt; one boy is born on a Tuesday&quot;. But indeed, this could be a language thing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 2 children you have the following possible combinations:&lt;br&gt;&#xA;    1. two girls&lt;br&gt;&#xA;    2. a boy and a girl&lt;br&gt;&#xA;    3. a girl and a boy&lt;br&gt;&#xA;    4. two boys  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at least 1 is a boy we only have to consider the last three combinations. That gives us one in three that both are boys.&lt;br&gt;&#xA;The error which is often made is to consider 2. and 3. as a single combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;I find it completely counter-intuitive that the outcome is influenced by the day, and I simulated the problem for one million families with 2 kids. And lo and behold, the outcome is 12.99 in 27. I was wrong.&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-09-12T07:16:19.863" LastActivityDate="2010-09-12T07:16:19.863" CommentCount="4" />
  <row Id="4403" PostTypeId="1" AcceptedAnswerId="4434" CreationDate="2010-09-11T07:18:28.213" Score="3" ViewCount="51" Body="&lt;p&gt;I've been struggling with the following problem I found for a while now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $(T,\preceq)$ is a partially ordered subset of $(U,\preceq)$ and $S\subseteq T$. If $\inf_T S$ and $u=\inf_U S$ both exist, and there exists a subset $S'\subseteq T$ such that $u=\sup_U S'$, then $\inf_U S=\inf_T S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I noted that $\inf_T S\preceq\inf_U S$, since $\inf_T S\in U$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also saw that $u=\inf_U([u)\cap T)$ by the following: For $x\in S$, $u\preceq x$ and $x\in T$, so $x\in [u)\cap T$, and so $S\subseteq [u)\cap T$. So $u$ is a lower bound of $[u)\cap T$. Now suppose there exists some $y\in U$ such that $u\prec y$, but $y\preceq x$ for all $x\in [u)\cap T$. Since $S\subseteq [u)\cap T$, $y$ would be a lower bound of $S$ larger than $u$, contradicting the fact that $u=\inf_U S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping to show that $\inf_U S\preceq\inf_T S$ in order to show equality, but I can't piece it together. Can someone explain how to show this?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-11T22:32:45.570" Title="For some sets $S\subseteq T\subseteq U$, when is $\inf_T S=\inf_U S$?" Tags="&lt;analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4404" PostTypeId="2" ParentId="4335" CreationDate="2010-09-11T07:20:06.867" Score="1" ViewCount="0" Body="&lt;p&gt;Can't comment, so this will have to be a post. As well as seconding Gilbert Strang's lively lectures on the open MIT site, his book, and Artin's, which are all excellent and all already recommended, I think you could also try Paul Dawkins' online math notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tutorial.math.lamar.edu/Classes/LinAlg/LinAlg.aspx&quot; rel=&quot;nofollow&quot;&gt;http://tutorial.math.lamar.edu/Classes/LinAlg/LinAlg.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are a fair bit shorter and, given your ultimate aim, the online OpenGL tutorials at NeHe are also probably useful (because you will see lots of examples of the application of linear algebra in the area of your interest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nehe.gamedev.net/&quot; rel=&quot;nofollow&quot;&gt;http://nehe.gamedev.net/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-09-11T07:20:06.867" CommunityOwnedDate="2010-09-11T07:20:06.867" />
  <row Id="4405" PostTypeId="2" ParentId="3519" CreationDate="2010-09-11T07:36:33.910" Score="1" ViewCount="0" Body="&lt;p&gt;Liouville's Number makes it's next appearance in &lt;em&gt;Making Transcendence Transparent - Edward B. Burger, Robert Tubbs&lt;/em&gt;. The transcendence of $e + L$ is proved. (This might be the end of it's career though).&lt;/p&gt;&#xA;" LastActivityDate="2010-09-11T07:36:33.910" />
  <row Id="4406" PostTypeId="1" CreationDate="2010-09-11T07:40:51.017" Score="2" ViewCount="99" Body="&lt;p&gt;What is the precise difference or relation between these terms in logic: &#xA;Inference, Reasoning, Deduction, and Induction?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2010-10-11T17:16:40.700" Title="Difference or relation between Inference, Reasoning, Deduction, and Induction?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="4407" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T08:19:15.810" Score="2" ViewCount="0" Body="&lt;p&gt;There is always room for misinterpreting a question when one does not fully understand the language in which it is written. I think that the way mathematics and mathematicians use conditional probability is clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A|B)=P(A \cap B)/P(B).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I believe that this is the interpretation that one should take, and thus arrive at your answer of 13/27, and not search for further nuances, which are not too difficult to find.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-11T08:19:15.810" CommentCount="1" />
  <row Id="4408" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T08:31:26.883" Score="6" ViewCount="0" Body="&lt;p&gt;It is actually &lt;em&gt;impossible&lt;/em&gt; to have a unique and unambiguous answer to the puzzle without explicitly articulating a probability model for how the information on gender and birthday is generated.  The reason is that (1) for the problem to have a unique answer some random process is required, and (2) the answer is a function of which random model is used.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The problem assumes that a unique probability can be deduced as the answer. This requires that the set of children described is chosen by a random process, otherwise the number of boys is a deterministic quantity and the probability would be 0 or 1 but with no ability to determine which is the case.  More generally one can consider random processes that produce the complete set of information referenced in the problem: choose a parent, then choose what to reveal about the number, gender, and birth days of its children.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer depends on which random process is used.  If the Tuesday birth is disclosed only when there are two boys, the probability of two boys is 1.    If Tuesday birth is disclosed only when there is a sister, the probability of two boys is 0.  The answer could be any number between 0 or 1 depending on what process is assumed to produce the data. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is also a linguistic question of how to interpret &quot;one is a boy born on Tuesday&quot;.  It could mean that the number of Tuesday-born males is exactly one, or at least one child.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-13T08:58:03.033" LastActivityDate="2010-09-13T08:58:03.033" CommentCount="1" />
  <row Id="4409" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T08:35:12.623" Score="3" ViewCount="0" Body="&lt;p&gt;I guess the following two versions of the experiment provide two different answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Dave has two children. Is atleast one of them a boy who is born on Tuesday? Dave answers Yes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dave has two children. I ask him to first choose and fix one child at random, and tell me if it is a boy who was born on Tuesday. Dave answers yes he is a boy born on Tuesday.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For 1st the answer is 13/27, while the second has answer 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way in which the question is asked, it's in line with 1st, hence the answer should be 13/27.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-09-11T08:35:12.623" CommentCount="2" />
  <row Id="4410" PostTypeId="2" ParentId="154" CreationDate="2010-09-11T09:15:10.803" Score="6" ViewCount="0" Body="&lt;p&gt;Here is an example of a physical quantity which comes naturally as a complex number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_impedance&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;impedance&lt;/em&gt;&lt;/a&gt; between two nodes of a linear electrical circuit working in alternate current, which is the analogous of the resistance of a resistor and is also measured in Ohms, is a complex quantity. For instance, the impedance of an (ideal) capacitor is imaginary.&lt;/p&gt;&#xA;" OwnerUserId="1828" LastEditorUserId="1828" LastEditorDisplayName="" LastEditDate="2010-09-11T09:30:21.907" LastActivityDate="2010-09-11T09:30:21.907" />
  <row Id="4411" PostTypeId="2" ParentId="4363" CreationDate="2010-09-11T10:04:00.793" Score="1" ViewCount="0" Body="&lt;p&gt;As Wikipedia says: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with locally constant rank near a point can be put in a particular normal form near that point.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative_rule_for_inverses#Constant_rank_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Derivative_rule_for_inverses#Constant_rank_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Constant Rank Theorem is stated as Theorem (7.1) p. 47 of &lt;strong&gt;An Introduction to Differentiable Manifolds and Riemannian Geometry&lt;/strong&gt;, Revised Second Edition, William M. Boothby, Academic Press. (This is the reference given by Wikipedia.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is, for the reader's convenience, a statement of the Constant Rank Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k,n$ and $r$ be positive integers, let $a$ be in $\mathbb R^n$, let $b$ be in $\mathbb R^k$, let $f$ be a smooth map from a neighborhood of $a$ to $\mathbb R^k$ sending $a$ to $b$, and let $\ell$ be the linear map from $\mathbb R^n$ to $\mathbb R^k$ sending $x$ to $(x_1,\dots,x_r,0,\dots,0)$. Assume that the rank of the tangent map to $f$ at $x$ is equal to $r$ for all $x$ in our neighborhood of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a diffeomorphism $g$ from a neighborhood of $a$ to a neighborhood of 0 in $\mathbb R^n$, and a diffeomorphism $h$ from a neighborhood of 0 in $\mathbb R^k$ to a neighborhood of $b$, such that the equality $f=h\circ\ell\circ g$ holds in some neighborhood of $a$. &lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-09-11T10:04:00.793" CommentCount="4" />
  <row Id="4412" PostTypeId="1" AcceptedAnswerId="4413" CreationDate="2010-09-11T10:35:49.663" Score="5" ViewCount="206" Body="&lt;p&gt;We know that if $f : \mathbb{R} \to \mathbb{R}$ is a continuous function, then $f$ carries connected sets to connected sets and compact sets to compact sets. That is if $A \subset \mathbb{R}$ is connected then $f(A)$ is connected, and if $A$ is compact then $f(A)$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Suppose $f: \mathbb{R} \to \mathbb{R}$ is a function such that for every connected, compact subsets $A \subset \mathbb{R}$, $f(A)$ is connected, compact, then is $f$ continuous? If yes, i would like to see a proof.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: Does this result remain true if $f: \mathbb{R}^{2} \to \mathbb{R}$, or from any $f: \mathbb{R}^{m} \to \mathbb{R}^{n}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-11T10:48:43.390" LastActivityDate="2010-09-13T04:52:37.927" Title="Characterising Continuous functions" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="4413" PostTypeId="2" ParentId="4412" CreationDate="2010-09-11T10:42:50.813" Score="12" ViewCount="0" Body="&lt;p&gt;No. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway_base_13_function&quot; rel=&quot;nofollow&quot;&gt;Conway's base 13&lt;/a&gt; function. If we call this function $f$,&#xA;while $f$ itself is not a counterexample to your question, $g(x)=\sin f(x)$&#xA;is. The connected compact subsets of $\mathbb{R}$ are the empty set, the&#xA;one-point sets and the closed intervals $[a,b]$ with $a &amp;lt; b$. The Conway&#xA;function $f$ takes such closed intervals to $\mathbb{R}$ and so $g$&#xA;takes them to $[-1,1]$. The functions $f$ and $g$ are discontinuous everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED&lt;/strong&gt;&#xA;For functions with domain $\mathbb{R}^n$ the answer is still no. Just&#xA;take&#xA;$$(x_1,\ldots,x_n)\mapsto g(x_1)$$&#xA;where $g$ is the function above.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-11T11:54:28.047" LastActivityDate="2010-09-11T11:54:28.047" />
  <row Id="4414" PostTypeId="1" AcceptedAnswerId="4420" CreationDate="2010-09-11T11:08:06.980" Score="2" ViewCount="139" Body="&lt;p&gt;In a proof or deduction system, there are some axioms and some inference rules. To have a small proof system, axioms are defined for the primitives (for example for negation and implication in propositional logic). Now, we want to have a proof system that proves all provable formulae including the formulae contain non-primitive logical connectors (such as OR in propositional logic). The question is should we add definitions for non-primitives as axioms to the proof system, or we should do something else? In other words, do logicians accept definitions as axioms?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2010-09-21T03:46:49.333" Title="Can we define definitions as axioms in logic?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="4415" PostTypeId="2" ParentId="154" CreationDate="2010-09-11T12:34:46.353" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not a mathematician and for a long time I've struggled with the question if numbers exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics is a language. Numbers are part of its vocabulary, like &quot;apple&quot; is part of the English vocabulary. Just like you can build a sentence with the word &quot;apple&quot; you can build equations with numbers, and just like you have to follow certain rules to make a correct sentence you have to follow certain rules for your equation to hold true.&lt;br&gt;&#xA;Now, and I'm trying not go into semiotics (of which I know nothing), the word &quot;apple&quot; is not an apple: you can't eat the word &quot;apple&quot; for one thing. But it exists in the language. You can invent things in a language to represent things in the real world. Does the verb &quot;to walk&quot; exist? It represents an action in the real world, but in itself it is nothing.&lt;br&gt;&#xA;I think the same is true with numbers: numbers exist because we defined them in the language of mathematics, but they only make sense if you can connect them to something real. Natural numbers often represent the cardinality of sets: 10 may the number of real world apples in a real world basket. Rational numbers come in handy when comparing things: one apple mat be 1.2 times as large as another one. To non-mathematicians irrational numbers make less sense, for everyday non-mathematical use even pi is rational: 22/7.&lt;br&gt;&#xA;Now for complex numbers same thing: we define them in the language of mathematics, but just like irrational numbers it's much harder to make them represent something real, although especially physicists are very good at describing the real world with them.  What about other numbers like quaternions? They're part of the language, and are often required to make the grammar fit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are complex numbers real? (mind the pun! :-))&lt;br&gt;&#xA;They obviously exist in the language of mathematics, and may be used to describe real world things and events, but in themselves I don't think they exist.&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-11T12:34:46.353" CommentCount="2" />
  <row Id="4416" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T13:41:30.877" Score="0" ViewCount="0" Body="&lt;p&gt;This is discussed also on &lt;a href=&quot;http://johncarlosbaez.wordpress.com/2010/08/24/probability-puzzles-from-egan/#comment-1313&quot; rel=&quot;nofollow&quot;&gt;http://johncarlosbaez.wordpress.com/2010/08/24/probability-puzzles-from-egan/#comment-1313&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-09-11T13:41:30.877" />
  <row Id="4417" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T14:36:56.830" Score="0" ViewCount="0" Body="&lt;p&gt;This, in my opinion, is why the intuitive approach fails:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has a tendency to think that &#xA;the probability of 7*P(b AND d1) = P(b AND d1) + P(b AND d2) + ... + P(b AND d7) = P((b AND d1) OR (b AND d2) OR ... OR (b AND d7)) = P(b AND (d1 OR d2 OR ... OR d7)) = P(b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the flaw here is that, in reality, P(b AND d1) + P(b AND d2) + ... + P(b AND d7) is NOT equal to P((b AND d1) OR (b AND d2) OR ... OR (b AND d7)). This means that mentioning independent (and one might think irrelevant) information alongside with relevant information actually changes the resulting probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting consequence: if I say something like&#xA;&quot;I have two children. One of them is a boy who was born at 10:24 PM on February 10th,&quot;&#xA;The probability that I have two boys is now almost exactly the same as as the probability that I have a girl and a boy. Adding a unique or almost unique piece of information makes the stuff I want to know about the other child independent of the information I have on the first child. If I took this to the extreme and said that I have a firstborn boy, won't know anything additional about the other child.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-11T14:36:56.830" CommentCount="1" />
  <row Id="4418" PostTypeId="2" ParentId="4412" CreationDate="2010-09-11T14:55:53.767" Score="12" ViewCount="0" Body="&lt;p&gt;Much more general results are known: no statement of this form can characterize continuous functions, derivatives, Baire class 1 functions, Borel functions, measurable functions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$ There do not exist families of sets of reals $\rm\: \cal A\:,\:\cal B\;$ such that the following statement is true:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad$ for every function $\rm\; f : \mathbb R\to \mathbb R\:,\;\;\; f\:$ is continuous $\iff$ for every $\: A\in {\cal A}\:,\;\; {\rm f}\:(A) \in \cal B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the elementary proof see this &lt;a href=&quot;http://www.jstor.org/stable/2974580&quot; rel=&quot;nofollow&quot;&gt;1997 Monthly paper&lt;/a&gt; by Velleman and for more general results on classes of functions characterizable by images of sets see &lt;a href=&quot;http://at.yorku.ca/p/a/a/f/12.htm&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, whose abstract I have appended below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-empty topological spaces $X$ and $Y$ and arbitrary&#xA;families $\cal{A}\subseteq\cal{P}(X)$ and $\cal{B}\subseteq\cal{P}(Y)$ we put&#xA;$\cal{C}_{\cal{A},\cal{B}}=\{f\in Y^X\colon(\forall A\in\cal{A})(f[A]\in\cal{B})\}$.&#xA;In this paper we will examine&#xA;which classes of functions $\cal{F}\subseteq Y^X$&#xA;can be represented as $\cal{C}_{\cal{A},\cal{B}}$. We will be mainly&#xA;interested in the case&#xA;when&#xA;$\cal{F}=\cal{C}(X,Y)$ is the class of all continuous functions from $X$ into $Y$.&#xA;We prove that for non-discrete Tychonoff space $X$ the class&#xA;$\cal{F}=\cal{C}(X,\mathbb{R})$&#xA;is not equal to $\cal{C}_{\cal{A},\cal{B}}$ for any&#xA;$\cal{A}\subseteq \cal{P}(X)$ and $\cal{B}\subseteq\cal{P}(\mathbb{R})$. Thus, $\cal{C}(X,\mathbb{R})$&#xA;&lt;strong&gt;cannot be characterized by images of sets&lt;/strong&gt;.&#xA;We also show that none of the&#xA;following classes of real functions can be represented as&#xA;$\cal{C}_{\cal{A},\cal{B}}$:&#xA;upper (lower) semicontinuous functions,&#xA;derivatives, approximately  continuous functions,&#xA;Baire class 1 functions, Borel functions, and measurable functions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-13T04:52:37.927" LastActivityDate="2010-09-13T04:52:37.927" CommentCount="3" />
  <row Id="4419" PostTypeId="2" ParentId="4406" CreationDate="2010-09-11T15:28:52.267" Score="0" ViewCount="0" Body="&lt;p&gt;If your question is about philosophical matters of logic, then I think that you may find very interesting the following paper authored by a leading logician (see also other papers in the same handbook).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wilfrid Hodges, &lt;strong&gt;The scope and limits of logic&lt;/strong&gt;&lt;br&gt;&#xA;pp. 41-63 in Handbook of the Philosophy of Science: Philosophy of Logic,&lt;br&gt;&#xA;ed. Dale Jacquette, 2007, Elsevier, Amsterdam&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-11T15:28:52.267" />
  <row Id="4420" PostTypeId="2" ParentId="4414" CreationDate="2010-09-11T15:45:55.513" Score="6" ViewCount="0" Body="&lt;p&gt;Given a theory $T$, you can enlarge it with definitions in two steps. First you enlarge the alphabet $L(T)$ to a new alphabet $L'$ which will contain the new symbols of constants, functions and relations that you need. Then you enlarge $T$ itself by adding axioms of the following form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi(x_1, \dots, x_n)$ be a formula of $T$ with free variables $x_1, \dots, x_n$ and $\psi$ a new symbol of $n$-ary relation. Then you can add an axiom $$\phi(x_1, \dots, x_n) \leftrightarrow \psi(x_1, \dots, x_n).$$ This axiom &quot;defines the meaning of $\psi$&quot;. Similarly you can add new defininig axioms for functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory $T'$ with language $L'$ that you obtain in this way is called an &lt;em&gt;extension by definitions&lt;/em&gt; of $T$. It is not difficult to prove that $T'$ is a conservative extension of $T$; that is, $T$ and $T'$ prove the same formulas of $L$. In this sense one can make precise the fact that definitions do not add nothing new to a theory.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2010-09-11T15:45:55.513" />
  <row Id="4421" PostTypeId="2" ParentId="1257" CreationDate="2010-09-11T17:39:07.933" Score="1" ViewCount="0" Body="&lt;p&gt;One of India's prestigious computer scientists developed this test called as the AKS primality test. Although this doesn't answer your question, it shows as to how much progress has been made to determine the primes: Please see this article: &lt;a href=&quot;http://en.wikipedia.org/wiki/AKS_primality_test&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/AKS_primality_test&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-11T17:39:07.933" CommentCount="2" />
  <row Id="4422" PostTypeId="1" CreationDate="2010-09-11T17:57:48.697" Score="1" ViewCount="174" Body="&lt;p&gt;I am taking a fourier analysis course at the graduate level and I am unhappy with the textbook (Stein and Shakarchi).  What I am looking for is a book that is less conversational and more to the point.  Further, I am not terribly interested in applications and would rather be exposed to how Fourier Analysis fits into the broader framework of analysis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For background, I used Baby Rudin for a one-year course in advanced calculus, I am currently taking a course from Kolmogorov and Fomin's Introductory Real Analysis and I have taken complex analysis (using Conway's text, Functions of One Complex Variable) as well as topology (using Munkres as well as Engelking) at the graduate level, but I have not yet been introduced to the Lebesgue integral.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-09-12T16:58:17.337" CommunityOwnedDate="2010-09-11T17:57:48.697" Title="Fourier Analysis textbook recommendation" Tags="&lt;reference-request&gt;&lt;fourier-analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="4424" PostTypeId="2" ParentId="4202" CreationDate="2010-09-11T18:23:05.610" Score="3" ViewCount="0" Body="&lt;p&gt;Such results on continuous induction are folklore dating back over a century - much older than  the references in the Kalantari paper referenced by Pete L. Clark. I once had a large file of papers on this topic but, alas, it's long lost. But I can still recall one &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.bams/1183425067&quot; rel=&quot;nofollow&quot;&gt;old paper [1]&lt;/a&gt;  because the author was the world-renowned multi-talented Chinese linguist Yuen Ren Chao. If memory serves correct he did his thesis under Sheffer at Harvard on aspects of the continuum, so his BAMS paper was probably an offshoot of his thesis. Since Project Euclid seems to be having intermittent problems I have appended the short paper below. For some biographical information on Chao see &lt;a href=&quot;http://content.cdlib.org/view?docId=hb8779p27v&amp;amp;brand=calisphere&amp;amp;doc.view=entire_text&quot; rel=&quot;nofollow&quot;&gt;here1&lt;/a&gt; and &lt;a href=&quot;http://www.cuhk.edu.hk/lin/cbrc/interview/eng.pdf&quot; rel=&quot;nofollow&quot;&gt;here2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Yuen Ren Chao. &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.bams/1183425067&quot; rel=&quot;nofollow&quot;&gt;A note on “Continuous mathematical induction&quot;.&lt;/a&gt;&lt;br&gt;&#xA;Bull. Amer. Math. Soc. Volume 26, Number 1 (1919), 17-18. &#xA;&lt;img src=&quot;http://i.imgur.com/whVt9.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/LD8M4.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T22:20:54.500" LastActivityDate="2010-09-19T22:20:54.500" CommentCount="2" />
  <row Id="4425" PostTypeId="2" ParentId="4422" CreationDate="2010-09-11T18:39:40.483" Score="1" ViewCount="0" Body="&lt;p&gt;Here are the ones which i would recommend:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Fourier series by R.Bhatia. Link: &lt;a href=&quot;http://www.maa.org/reviews/BhatiaFourierSeries.html&quot; rel=&quot;nofollow&quot;&gt;http://www.maa.org/reviews/BhatiaFourierSeries.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Korner, T. W. Fourier Analysis. Cambridge. 1990&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second one is very good and you can comprehend it if you are familiar with the analysis which you have mentioned. I would also like to advice you to read &quot;A radical approach to Lebesgue theory of Integration&quot; by D.M.Bressoud, link &lt;a href=&quot;https://www.maa.org/EbusPPRO/DynamicSearch/ProductDetailsAdvancedSearch/tabid/176/ProductId/1498/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;https://www.maa.org/EbusPPRO/DynamicSearch/ProductDetailsAdvancedSearch/tabid/176/ProductId/1498/Default.aspx&lt;/a&gt; this is a beautiful book to learn Measure theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; You might be interested to see this link: &lt;a href=&quot;http://www.cargalmathbooks.com/#FourierAnalysis&quot; rel=&quot;nofollow&quot;&gt;http://www.cargalmathbooks.com/#FourierAnalysis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-11T19:07:12.847" LastActivityDate="2010-09-11T19:07:12.847" CommunityOwnedDate="2010-09-11T18:39:40.483" CommentCount="1" />
  <row Id="4426" PostTypeId="1" AcceptedAnswerId="4456" CreationDate="2010-09-11T18:46:04.340" Score="3" ViewCount="120" Body="&lt;p&gt;In classical real analysis I've only seen absolute continuity defined for functions on compact interval $[a,b]$, where the two equivalent definitions are: $f:[a,b]\rightarrow\mathbb{R}$ is AC if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  Given $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that $\sum_{i=1}^n |f(y_i)-f(x_i)|&amp;lt; \epsilon$ for every finite collection of nonoverlapping intervals $( (x_i,y_i) )_{i=1}^n $ each contained in $[a,b]$ with $\sum_{i=1}^n |y_i-x_i|&amp;lt; \delta$. Or,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $f'$ exists a.e. on $[a,b]$, $f'$ is integrable on $[a,b]$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \int_a^x f'(y) dy + f(a)$ for all $x \in [a,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an accepted definition for absolute continuity of a function on an open, possibly unbounded, interval $(a,b)$ where $-\infty \leq a &amp;lt; b \leq \infty$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that definition (1) extends easily to this case if we replace $[a,b]$ by $(a,b)$.  If we call this condition (1'), &lt;strike&gt;then it's easy to show (1') is equivalent to&lt;/strike&gt; (see answer by Jonas below).  A natural extension of (2) to this case would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2') $f$ is AC on an open set $U$ if, for all compact intervals $[c,d] \subset (a,b)$, $f$ is AC in the sense of (2) on $[c,d]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of these is the best extension of the definition?  I don't know enough about the notion of absolute continuity of measures to know if my extended definitions are consistent with that generalization as well.&lt;/p&gt;&#xA;" OwnerUserId="1692" LastEditorUserId="1692" LastEditorDisplayName="" LastEditDate="2010-09-13T15:59:05.423" LastActivityDate="2010-09-13T15:59:05.423" Title="Absolute continuity on an open interval of the real line?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4427" PostTypeId="2" ParentId="4422" CreationDate="2010-09-11T18:57:44.763" Score="1" ViewCount="0" Body="&lt;p&gt;I'd like to suggest &lt;em&gt;Fourier Series and Integrals&lt;/em&gt; by Dym and McKean.&#xA;It's old, but still an excellent book. Chapters 3 and 4 show how Fourier analysis fits in with some other parts of mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Preface:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The level of preparation expected is a thorough knowledge of advanced calculus.&#xA;To this must be added a willingness to believe in (or to study up on) the Lebesgue integral. &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-09-12T16:30:40.530" LastActivityDate="2010-09-12T16:30:40.530" CommunityOwnedDate="2010-09-11T18:57:44.763" />
  <row Id="4428" PostTypeId="1" AcceptedAnswerId="4429" CreationDate="2010-09-11T19:19:49.593" Score="0" ViewCount="43" Body="&lt;p&gt;What's the difference between a bar chart and a histogram?&lt;/p&gt;&#xA;" OwnerUserId="1834" LastActivityDate="2010-09-11T21:57:01.193" Title="Bar chart or histogram" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4429" PostTypeId="2" ParentId="4428" CreationDate="2010-09-11T19:23:19.257" Score="0" ViewCount="0" Body="&lt;p&gt;With bar charts, each column represents a group defined by a categorical variable; and with histograms, each column represents a group defined by a quantitative variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this link: &lt;a href=&quot;http://wiki.answers.com/Q/What_is_the_difference_between_Bar_chart_and_a_frequency_histogram&quot; rel=&quot;nofollow&quot;&gt;http://wiki.answers.com/Q/What_is_the_difference_between_Bar_chart_and_a_frequency_histogram&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-11T19:23:19.257" />
  <row Id="4430" PostTypeId="1" CreationDate="2010-09-11T19:23:50.987" Score="1" ViewCount="122" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any application or in which field this trajectories can be applied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Comment&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know any book or any site that can show me the applications, please let me know. Thanks for your time&lt;/p&gt;&#xA;" OwnerUserId="1833" LastActivityDate="2011-01-10T16:44:42.150" Title="Applications of orthogonal trajectories" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4431" PostTypeId="2" ParentId="4430" CreationDate="2010-09-11T19:26:49.500" Score="-1" ViewCount="0" Body="&lt;p&gt;If you want to see applications of orthogonal trajectories, then please see this:&#xA;&lt;a href=&quot;http://www.math.uh.edu/~etgen/Applications.pdf&quot; rel=&quot;nofollow&quot;&gt;www.math.uh.edu/~etgen/Applications.pdf &lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-11T19:36:57.657" LastActivityDate="2010-09-11T19:36:57.657" CommentCount="1" />
  <row Id="4432" PostTypeId="3" CreationDate="2010-09-11T19:36:12.317" Score="0" ViewCount="0" Body="" OwnerUserId="-1" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-11T19:36:12.317" LastActivityDate="2010-09-11T19:36:12.317" />
  <row Id="4433" PostTypeId="3" CreationDate="2010-09-11T19:36:12.317" Score="0" ViewCount="0" Body="[reference-request] are for questions that the poster wants to seek for references (books, articles, etc) in a subject." OwnerUserId="-1" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-11T19:36:12.317" LastActivityDate="2010-09-11T19:36:12.317" />
  <row Id="4434" PostTypeId="2" ParentId="4403" CreationDate="2010-09-11T19:48:40.353" Score="1" ViewCount="0" Body="&lt;p&gt;So, you have shown that $\inf_T(S)\leq \inf_U(S)=u$. You want to show that $u\leq\inf_T(S)$. Clearly, you need to use the existence of $S'$ somehow, because without the existence of $S'$ you can trivially show this is false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that $\inf_T(S)\leq\inf_U(S)$. You want to show that you do not have $\inf_T(S)\lt\inf_U(S)=u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; (added case) If $S'=\emptyset$, then $u=\inf_U(\emptyset)=\sup(U)$, so in particular $\inf_T(S)\leq u$ and we are done. So we may assume that $S'\not= \emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added later&lt;/em&gt;: Actually, this is not needed strictly speaking, but it may be a good point to mention explicitly; if $t\lt u$, then $t$ is not an upper bound for $S'$; the only way that can occur in the first place is if $S'$ is not empty, since everything is an upper bound for the empty set, so this case is covered by the assumption being made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;If $t\in T$ is such that $t\lt u$, then since $t\in U$ and $u=\sup_U(S')$, you know that there exists $s'\in S'$ such that $t\lt s'\leq u$.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; (implicitly assumed certain things were comparable, which may not be the case in general; fixed below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for every $s'\in S'$ and $s\in S$, you have $s'\leq u\leq s$ (since $u=\inf_U(S)$), and so in particular $s'\leq \inf_T(S)$, since $S'\subseteq T$. So $\inf_T(S)$ is comparable to every element of $S'$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $t\in T$ is such that $t\lt u$. Since $u=\sup_U(S')$, it follows that $t$ cannot be an upper bound for $S'$ in $U$. In particular, there exists $s'\in S'$ such that $s'\not\leq t$; so either $t\lt s'$, or else $t$ and $s'$ are incomparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, for every $t\in T$, if $t\lt u$ and $t$ is comparable to every element of $S'$, then $t\lt\inf_T(S)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\inf_T(S)\in T$, $\inf_T(S)$ is comparable to every element of $S'$ (as noted above), and $\inf_T(S)\not\lt\inf_T(S)$, you can can conclude by contrapositive that $\inf_T(S)\not\lt u$. Since you already know that $\inf_T(S)\leq u$, you are done.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-11T22:32:45.570" LastActivityDate="2010-09-11T22:32:45.570" CommentCount="6" />
  <row Id="4435" PostTypeId="1" AcceptedAnswerId="4448" CreationDate="2010-09-11T19:49:05.417" Score="0" ViewCount="113" Body="&lt;p&gt;Problem taken from the link:  &lt;a href=&quot;http://web.mit.edu/rwbarton/Public/func-eq.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/rwbarton/Public/func-eq.pdf&lt;/a&gt; I am stating the question here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which $\alpha$ does there exists a nonconstant function $f: \mathbb{R} \to \mathbb{R}$ such that $f(\alpha(x+y))=f(x)+f(y)$ for all $x,y \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly for $\alpha=1$ we see that this case is satisfied, by taking the identity function. But are there other values of $\alpha$ for which this condition is satisfied. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-11T22:42:08.930" Title="Finding $\alpha$ such that $f(\alpha(x+y))=f(x)+f(y)$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functional-equations&gt;" AnswerCount="1" />
  <row Id="4436" PostTypeId="1" CreationDate="2010-09-11T19:59:37.853" Score="1" ViewCount="58" Body="&lt;p&gt;Doing a little practice exercise, and came across the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three functions: $f(x)$, $g(x)$, and $h(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $f(x)$ and $g(x)$, such that $h(x) = f \circ g(x)$ and $g(x) = \sqrt{4} + 8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $h(x)$ is defined as $h(x) = (\sqrt{4} + 8) ^ 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure how to approach the problem. I know I'm going to have to solve this algebraically, but I just can't seem to wrap my mind around the manipulation of functions as opposed to variables and numbers.&lt;/p&gt;&#xA;" OwnerUserId="1836" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-11T21:15:09.247" LastActivityDate="2010-09-11T21:15:09.247" Title="Solving Compositions of Functions Algebraically" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="4437" PostTypeId="1" AcceptedAnswerId="4443" CreationDate="2010-09-11T20:08:18.757" Score="11" ViewCount="431" Body="&lt;p&gt;To prove the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem&quot; rel=&quot;nofollow&quot;&gt;Cayley-Hamilton&lt;/a&gt; theorem in linear algebra, my professor said that a lemma was necessary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;: Let $A \in M_n(\mathbb{K})$ be an $n\times n$ matrix over a field $\mathbb{K}$, let $b(t) \in M_n(\mathbb{K})[t]$ and $P(t) = b(t)[A-tI]$, then $P(A) = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The theorem (which says that if $f$ is an endomorphism of V, then $f$ is a solution to its characteristic polynomial), was then proven thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $B(t) = \text{adj}[A-tI]$ and $P(t) = B(t)[A-tI]$, then $P(A)=0$ but also $P(t) = \delta I$ (where $\delta = \det(A-tI)$). Since $\delta = \chi_f(t)$, so $P(A) = 0 \Rightarrow \chi_f(A) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: since we interpret the $P(t)$ of the theorem as a polynomial with matrix coefficients, isn't the whole thing kind of obvious for the properties of a polynomial ring? (Assuming we all know how to switch between matrices and endomorphisms)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-09-12T22:08:06.633" LastActivityDate="2010-09-17T23:16:20.873" Title="Is the proof of this lemma really necessary?" Tags="&lt;linear-algebra&gt;&lt;proof-strategy&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="7" />
  <row Id="4438" PostTypeId="2" ParentId="4436" CreationDate="2010-09-11T20:34:11.837" Score="1" ViewCount="0" Body="&lt;p&gt;The question, as currently posted (even after I edited it) is a mess. Are $g(x)$ and $h(x)$ meant to be &lt;em&gt;constant&lt;/em&gt; functions? Was that supposed to be $g(4) = \sqrt{4} + 8$ and $h(4)=(\sqrt{4}+8)^4$, or was it supposed to be $g(x)=\sqrt{x}+8$ and $h(x)=(\sqrt{x}+8)^4$? Something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case: a composition is a function that is obtained by first applying one function, and then taking the output of that function and plugging it into the next function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the expression $(\sqrt{4}+8)^4$. Suppose you were trying to actually compute this number. How would you go about it? First, we would take the number $4$ and plug it into the square root to obtain $\sqrt{4}$. The we would take the ouput of that, and add $8$. Finally, we would take the result we got from doing all that, and raise it to the fourth power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can think of this as the process of doing three things, one after another, where the input to each of the steps is the output of the previous step:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take the square root of what you have;&lt;/li&gt;&#xA;&lt;li&gt;Add $8$ to what you have;&lt;/li&gt;&#xA;&lt;li&gt;Raise what you have to the fourth power.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, you are told to try to express this as the result of doing &lt;em&gt;two&lt;/em&gt; things in succession: the first one is $g(x)$, and the second one will be $f(x)$. And you are told that the first thing you are doing is $g(x)=\sqrt{4}+8$. So, what else do you need to do to the output of this process in order to get to the final result? You need to raise what you are given to the fourth power. So, whatever $f(x)$ is, it should take the output of $g(x)$ and raise it to the fourth power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is $f(x)$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-11T20:34:11.837" CommentCount="6" />
  <row Id="4439" PostTypeId="2" ParentId="4346" CreationDate="2010-09-11T20:47:15.340" Score="1" ViewCount="0" Body="&lt;p&gt;Whether or not $f$ is linear, the equation can in fact be rewritten as $F^2 = F + 1$, in a suitable interpretation where to a function $f$ is associated a linear operator $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be the vector space of sums $aX + bf(X)$. Denote by $F$ the operator that takes an element of $V$ to its composition with $f$, that is,  $F (h(x)) = h(f(x))$.  Let $1$ denote the identity operator on $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $F$ is linear (from the definition), and satisfies $F^2=F+1$ as one can check on basis elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been temporarily but deliberately unclear as to whether $V$ is the 2-dimensional space of formal sums with basis vectors $x$ and $f(x)$, or the at most 2-dimensional space of functions of that form.   The question of whether $f$ is linear is the same as asking whether the second space is one-dimensional, which is no longer an algebraic question but one of analysis using regularity assumptions on $f$. (In particular, seeing $f$ as an operator or 2x2 matrix $F$, illuminates but does not quite trivialize the problem of showing that linear functions are the only continuous solutions to the functional equation. Additional arguments are needed.)   The equation $F^2 = F + 1$ holds in both interpretations.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-11T21:42:37.693" LastActivityDate="2010-09-11T21:42:37.693" />
  <row Id="4440" PostTypeId="2" ParentId="4371" CreationDate="2010-09-11T21:37:08.233" Score="1" ViewCount="0" Body="&lt;p&gt;I am learning some graph theory myself as an independent study in college.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with a very simple, but informative text, Introductory Graph Theory by Chatrand. It is a Dover book, and can be bought for very cheap on Amazon.&#xA;I then went to my university library and took out Modern Graph Theory by Bollobas. It is not an easy book, but it is incredibly clear and informative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also great introductions to be found online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rather lengthly sample of Bollobas's &lt;I&gt;Modern Graph Theory&lt;/I&gt;:&#xA;&lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;printsec=frontcover&amp;amp;source=gbs_slider_thumb#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;printsec=frontcover&amp;amp;source=gbs_slider_thumb#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1838" LastEditorUserId="1838" LastEditorDisplayName="" LastEditDate="2010-09-12T03:32:20.460" LastActivityDate="2010-09-12T03:32:20.460" CommentCount="1" />
  <row Id="4441" PostTypeId="2" ParentId="4437" CreationDate="2010-09-11T21:41:03.873" Score="5" ViewCount="0" Body="&lt;p&gt;It's not clear to me what the exact problem is here, but there is a common pitfall&#xA;concerning polynomials with matrix coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a polynomial $P(t)=\sum_i B_i t^i$ we can certainly substitute a matrix&#xA;$A$ for $t$ to get $P(A)=\sum_i B_i A^i$. But there is a catch when we multiply&#xA;polynomials. If $Q(t)$ is another polynomial with matrix coefficients, then we may&#xA;have&#xA;$$(PQ)(A)\ne P(A)Q(A).$$&#xA;A sufficient condition for $(PQ)(A)=P(A)Q(A)$ to hold is that $A$ commutes&#xA;with the coefficients of $Q$. But when it doesn't then in general&#xA;$(PQ)(A)\ne P(A)Q(A)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-11T21:41:03.873" CommentCount="3" />
  <row Id="4442" PostTypeId="1" AcceptedAnswerId="4454" CreationDate="2010-09-11T21:46:34.420" Score="3" ViewCount="276" Body="&lt;p&gt;I've been reading Ravi Vakil's notes on algebraic geometry &lt;a href=&quot;http://math.stanford.edu/~vakil/216blog/FOAGsept1010public.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; and am having a hard time connecting the standard definition of the localization of a module to the definition in terms of universal properties (page 24). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody either explain the connection or provide some kind of 'roadmap' of the material necessary to understand this connection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize if this question is vague or unclear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is an attempt to reformulate my original question to make it more precise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we did not know the definition of the localization of a module and we just saw the diagram presented on page 24, how do we recover the usual definition and how do we determine that anything else satisfying those properties is isomorphic to the usual definition?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-21T19:29:16.827" LastActivityDate="2010-09-21T19:29:16.827" Title="Learning to think categorically (localization of rings and modules)" Tags="&lt;category-theory&gt;&lt;modules&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="4443" PostTypeId="2" ParentId="4437" CreationDate="2010-09-11T21:56:20.730" Score="12" ViewCount="0" Body="&lt;p&gt;The reason we need the lemma is that from $P(t)=b(t)(A-tI)$ one cannot directly conclude that $P(A)=b(A)(A-AI)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is a &lt;em&gt;commutative&lt;/em&gt; ring, then there is a natural map $R[t]\to R^R$ which is a ring homomorphism (we endow $R^R$ with the pointwise ring structure: $(f+g)(r) = f(r)+g(r)$, and $fg(r) = f(r)g(r)$ for every $r\in R$). If $p(t)=q(t)s(t)$, then for every $r\in R$ you have that $p(r)=q(r)s(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't work if $R$ is not commutative. For example, taking $p(t) = at$, $q(t) = t$ and $s(t)=a$, you have $p(t)=q(t)s(t)$ in $R[t]$ (since $t$ is central in $R[t]$ even when $R$ is not commutative), but $p(r) = ar$ while $q(r)s(r) = ra$. So you get $p(r)=q(r)s(r)$ if and only if $a$ and $r$ commute. Thus, while you can certainly define a map $\psi\colon R[t]\to R^R$ by&#xA;$$\psi(a_0+a_1t+\cdots+a_nt^n)(r) =  a_0 + a_1r + \cdots + a_nr^n,$$&#xA;this map is not a ring homomorphism when the ring is not commutative. This is the situation we have here, where the ring $R$ is the ring $n\times n$ matrices over $\mathbb{K}$, which is not commutative when $n\gt 1$. In particular, from $P(t) = B(t)(A-tI)$ one cannot simply conclude that $P(A)=B(A)(A-AI)$. This implicitly assumes that your map $M_n(\mathbb{K})[t]\to M_n(\mathbb{K})^{M_n(\mathbb{K})}$ is multiplicative, which it is not in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your $A$ happens to be central in $M_n(\mathbb{K})$, then it is true that the induced map $M_n(\mathbb{K})[t]\to M_n(\mathbb{K})$ is a homomorphism. But then you would be assuming that your $A$ is a scalar multiple of the identity. It would also be true if the coefficients of the polynomial $b(t)$ centralize $A$, but you are not assuming that.  So you do need to prove that in this case you have $P(A)=b(A)(A-AI)$, since it does not follow from the general set-up (the way it would in a commutative setting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. In fact, this is the subtle point where the proof that a polynomial over a field of degree $n$ has at most $n$ roots breaks down for skew fields/division rings. If $K$ is a division ring, then the division algorithm holds for polynomials with coefficients over $K$, so one can show that for every $p(t)\in K[t]$ and $a(t)\in K[t]$, $a(t)\neq 0$, there exist unique $q(t)$ and $r(t)$ such that $p(t)=q(t)a(t) + r(t)$ and $r(t)=0$ or $\deg(r)\lt \deg(a)$. From this, we can deduce that for every polynomial $p(t)$ and for every $a\in K$, we can write $p(t) = q(t)(t-a) + r$, where $r\in K$. But the proof of the Remainder and Factor Theorems no longer goes through, because we cannot go from $p(t)=q(t)(t-a)+r$ to $p(a)=q(a)(a-a)+r$; and you cannot get the recursion argument to work, because from $p(t)=q(t)(t-a)$, and $p(b)=0$ with $b\neq a$, you cannot deduce that $q(b)=0$. For instance, over the real quaternions, we have $p(t)=t^2+1=(t+i)(t-i)$, but $p(j)=j^2+1\neq 2k = ij-ji = (j+i)(j-i)$.  I remember when I first learned the corresponding theorems for polynomial rings, the professor challenging us to identify all the field axioms used in the proofs of the Remainder and Factor Theorem; none of us spotted the use of commutativity in the evaluation map.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-12T17:51:10.830" LastActivityDate="2010-09-12T17:51:10.830" CommentCount="2" />
  <row Id="4444" PostTypeId="2" ParentId="4428" CreationDate="2010-09-11T21:57:01.193" Score="1" ViewCount="0" Body="&lt;p&gt;I disagree with wiki answers, which although it's partially correct, completely misses the main point.  The &lt;em&gt;heights&lt;/em&gt; in a bar chart represent counts.  You can use a bar chart for this purpose with categorical data or with binned &quot;continuous&quot; variates; either is fine.  The &lt;em&gt;areas&lt;/em&gt; in a histogram represent &lt;em&gt;relative frequencies&lt;/em&gt; or &lt;em&gt;proportions per unit value&lt;/em&gt; of a continuous variate.  A histogram thus is a discrete approximation to a frequency or probability density function, whereas a bar chart has no such meaning (except accidentally when it's really serving as a histogram!).  The distinction comes to the fore when the bars in a histogram have varying widths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Freedman, Pisani, Purves, &lt;em&gt;Statistics&lt;/em&gt;.  (Any of the first through fifth editions should be fine.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-11T21:57:01.193" />
  <row Id="4445" PostTypeId="1" AcceptedAnswerId="4446" CreationDate="2010-09-11T22:02:05.623" Score="2" ViewCount="100" Body="&lt;p&gt;From any textbook on fourier analysis: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is easily shown that for $f$ and $g$, both $2 \pi$-periodic functions on $[-\pi,\pi]$, we have $$(f \ast g)(x) = \int_{-\pi}^{\pi}f(x-y)g(y)\;dy = \int_{-\pi}^{\pi}f(z)g(x-z)\;dz = (g \ast f)(x),$$ by using the substitution $z = x-y.$&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;I don't doubt that this is true, but I cannot figure out what happened to the negative sign coming from $dy = -dz\;$ after the change of variable $z = x - y$.  In particular, after the change of variable $z = x-y,\;$ I am coming up with $ -\int_{-\pi}^{\pi}f(z)g(x-z)\;dz$. What am I missing here?&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-09-11T22:17:44.833" Title="Proving commutativity of convolution $(f \ast g)(x) = (g \ast f)(x)$" Tags="&lt;fourier-analysis&gt;&lt;convolution&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4446" PostTypeId="2" ParentId="4445" CreationDate="2010-09-11T22:14:20.233" Score="6" ViewCount="0" Body="&lt;p&gt;You need to check the bounds on your integral. since $y$ ranges from $-\pi$ to $\pi$, you'll have $z = x-y$ ranging from $x+\pi$ to $x-\pi$. Therefore:&#xA;$$&#xA;\int_{-\pi}^{\pi}f(x-y)g(y)dy = -\int_{x+\pi}^{x-\pi}f(z)g(x-z)dz =&#xA;\int_{x-\pi}^{x+\pi}f(z)g(x-z)dz = \int_{-\pi}^{\pi}f(z)g(x-z)dz&#xA;$$&#xA;in the second-last step, I swapped the two bounds on the integral (this changes the sign). In the final step, I shifted both bounds on the integral by $-x$, which does not change the value because we are integrating over an interval of length $2\pi$ and the function is $2\pi$-periodic.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-09-11T22:14:20.233" CommentCount="1" />
  <row Id="4447" PostTypeId="2" ParentId="4445" CreationDate="2010-09-11T22:17:44.833" Score="1" ViewCount="0" Body="&lt;p&gt;Here is something I've sometimes wondered about. If $f,g$ are both nonnegative proving &#xA;commutativity of convolution can be done without a tedious change of variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, let $X$ be a random variable with density $f$ and let $Y$ be a random variable with density $g$. Its easy to see that $f$ convolved with $g$ is the density of $X+Y$ (or in your case $X+Y ~{\rm mod} ~2 \pi$). By commutativity of addition, the density of $X+Y$ is the same as the density of $Y+X$ and we are done! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if an argument of this sort can work in the general case. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-09-11T22:17:44.833" />
  <row Id="4448" PostTypeId="2" ParentId="4435" CreationDate="2010-09-11T22:19:08.367" Score="5" ViewCount="0" Body="&lt;p&gt;$f(0+0)=f(0) + f(0)$ implies $f(0)=0$. Set $y=0$ to get $f(\alpha x)=f(x)$ so that $f(\alpha (x+y))=f(x+y)$ and the function is additive.  If $\alpha \neq 1$ this means $f((\alpha - 1)x)=0$ and hence $f=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is too easy to be on rwbarton's list, and in fact I cannot find this problem in the file that you linked.  Which problem number is it in the list? &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-11T22:42:08.930" LastActivityDate="2010-09-11T22:42:08.930" CommentCount="2" />
  <row Id="4449" PostTypeId="2" ParentId="4174" CreationDate="2010-09-11T22:36:56.227" Score="4" ViewCount="0" Body="&lt;p&gt;Induction on the number of sentence symbols is the kind of proof Craig gave originally, and is what logicians call a syntactic proof. It's also possible to give a semantic or model-theoretic proof. Bell and Slomson give one for first-order logic in Chapter 7, section 4 of their text &lt;a href=&quot;http://amzn.com/0486449793&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Models and Ultraproducts&lt;/em&gt;.&lt;/a&gt; I haven't seen a model-theoretic proof for propositional logic.  (Submitted as an answer because I don't have enough rep yet to comment.)&lt;/p&gt;&#xA;" OwnerUserId="1813" LastEditorUserId="1813" LastEditorDisplayName="" LastEditDate="2010-09-11T22:42:03.903" LastActivityDate="2010-09-11T22:42:03.903" CommentCount="1" />
  <row Id="4450" PostTypeId="2" ParentId="4442" CreationDate="2010-09-11T22:44:58.487" Score="3" ViewCount="0" Body="&lt;p&gt;In my opinion, the most enlightening (and the simplest) way to present the universal construction of localizations (and fractions) is to use instead of the pair construction the natural presentation in terms of generators and relations. This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (and to avoid the many tedious verifications required in the pair approach). Moreover, this approach is much more conceptual. Indeed, the pairs in the pair construction are nothing but normal forms for the polynomial terms in the presentation based approach.  For details of this approach see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, and Voloch's: &lt;a href=&quot;http://www.ma.utexas.edu/users/voloch/Homework/fractions.pdf&quot; rel=&quot;nofollow&quot;&gt;Rings of fractions the hard way&lt;/a&gt;. Note: presumably Voloch's title is a joke - since the presentation based approach is actually the easiest way - in fact both Rotman's and Voloch's expositions can be simplified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're just beginning to understand universal constructions then I highly recommend that you peruse the beautiful exposition in George Bergman's &lt;a href=&quot;http://www.math.berkeley.edu/~gbergman/245&quot; rel=&quot;nofollow&quot;&gt;An Invitation to General Algebra and Universal Constructions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also find illuminating Paul Cohn's historical article &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf#page=15&quot; rel=&quot;nofollow&quot;&gt;Localization in general rings, a historical survey&lt;/a&gt; - as well as other papers in that &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf&quot; rel=&quot;nofollow&quot;&gt;volume [1]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Ranicki, A.(ed). &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf&quot; rel=&quot;nofollow&quot;&gt;Noncommutative localization in algebra and topology&lt;/a&gt;. ICMS 2002&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T22:14:03.513" LastActivityDate="2010-09-19T22:14:03.513" CommentCount="1" />
  <row Id="4451" PostTypeId="2" ParentId="4437" CreationDate="2010-09-11T23:21:58.943" Score="1" ViewCount="0" Body="&lt;p&gt;The problem is quite simple: the definition of polynomial multiplication assumes that the coefficients commute with the indeterminates since e.g. $\rm\; r\ x = x\ r\:.\:$ So for evaluation to preserve polynomial arithmetic, i.e. for it to be a ring homomorphism, one can only evaluate polynomials in rings where the coefficient ring is central, i.e. where coefficients commute with all other ring elements. This concept is so important that it has a name, viz. $\rm\:R\:$-algebras are those rings $\rm\:A\:$ in which polynomials in $\rm R[x]$ can be evaluated. More precisely, this means that $\rm\:A\:$ contains an image $\rm\:R'$ of $\rm\:R\:$ and that for all $\rm\;a\in A\:$ the evaluation map $\rm\:x\to a\:$ is a ring homomorphism $\rm\:R[x]\to A\:$. In particular, evaluating $\rm\;\; r\: x = x\: r\;\;$ at $\rm\:x = a\in A\:$ implies that $\rm\;R'$ is central, i.e. it commutes with all elements of $\rm\:A\:$. That's precisely all that's required for the existence of a unique evaluation homomorphism $\rm\:\epsilon_a : x\to a\:$ for every $\rm\:a\in A\:.\:$ This permits one to view the polynomial ring $\rm R[x]$ as the universal $\rm\:R$-algebra generated by $\rm\:x\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T23:16:20.873" LastActivityDate="2010-09-17T23:16:20.873" CommentCount="2" />
  <row Id="4452" PostTypeId="2" ParentId="4400" CreationDate="2010-09-11T23:58:43.577" Score="0" ViewCount="0" Body="&lt;p&gt;Well, given the unstated assumption that the writer is a mathematician and therefore not using regular english, then I agree with the 13/27 answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in everyday english, from &quot;there are two fleems, one is a glarp&quot; we all infer that the other is not a glarp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &quot;there are two fleems, one is a glarp, which is snibble&quot; we would still infer that the other is not a glarp. Whereas from &quot;there are two fleems, one is a glarp which is snibble&quot; (absence of comma, or when spoken, difference in intonation) we would infer that the other is not a snibble glarp, but it could still be an unsnibble glarp.&lt;/p&gt;&#xA;" OwnerUserId="1840" LastActivityDate="2010-09-11T23:58:43.577" CommentCount="2" />
  <row Id="4453" PostTypeId="1" CreationDate="2010-09-12T00:20:11.347" Score="3" ViewCount="72" Body="&lt;p&gt;This is a simple terminology question.  Let $S$ be a (let's say smooth)&#xA;surface in $\mathbb{R}^3$, and $p$ a point&#xA;on $S$.  Suppose the principle curvatures $\kappa_1$ and $\kappa_2$ at $p$ are both negative.&#xA;I am imagining $p$ sitting at the bottom of a dent in the surface.&#xA;Is there an accepted term to describe such a point?&#xA;The difficulty is that the Gaussian curvature $\kappa_1 \kappa_2$ is positive, so intrinsically&#xA;$p$ is no different than if it were on a bump rather than a dent. &#xA;I could make up my own term of course, e.g., &lt;em&gt;valley point&lt;/em&gt; or &lt;em&gt;cup point&lt;/em&gt;, but I'd rather follow convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-12T01:53:46.720" LastActivityDate="2010-09-12T01:53:46.720" Title="Terminology for point in dent in surface?" Tags="&lt;terminology&gt;&lt;differential-geometry&gt;" CommentCount="3" />
  <row Id="4454" PostTypeId="2" ParentId="4442" CreationDate="2010-09-12T01:15:06.697" Score="1" ViewCount="0" Body="&lt;p&gt;As the Rotman's, or Voloch's, constructions of Dubuque's answers show, you &lt;em&gt;can't&lt;/em&gt; recover &lt;em&gt;exactly&lt;/em&gt; &quot;the usual definition&quot; of the localization of a ring from its universal property. That is, a universal property won't tell you, in general, how to construct an object that verifies it (as you can see, there are at least, two constructions for the localization, both verifying the universal property: how are you going to choose among them?). What a universal property will tell you is that all the guys verifying it are necessarily isomorphic: for instance, the construction in Vakil's notes, and Rotman's are necessarily isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely, let's stay with rings and let $\gamma' : A \longrightarrow B$ be some guy that satisfies the universal property of the localization. Let $\gamma: A \longrightarrow S^{-1}A$ denote the construction of the localization explained in Vakil's notes, 2.3.3. Then, since both $S^{-1}A$ and $B$ satisfy the universal property, you'll have morphisms $\widetilde{\gamma} : S^{-1}A \longrightarrow B$ and $\widetilde{\gamma'}: B \longrightarrow S^{-1}A$ such that $\widetilde{\gamma'}\gamma = \gamma'$ and $\widetilde{\gamma}\gamma' = \gamma$. Let's show that both $\widetilde{\gamma}$ and $\widetilde{\gamma'}$ are inverses of each other. For instance,  $\widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B$ because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(\widetilde{\gamma'} \widetilde{\gamma})\gamma' = \widetilde{\gamma'}\gamma = \gamma' \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\mathrm{id}_B$ verifies the same identity clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{id}_B \gamma'= \gamma' \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, because of the uniqueness of the universal property, you must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\gamma}\widetilde{\gamma'} = \mathrm{id}_{S^{-1}A} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can (and must) do is to verify that the &lt;em&gt;usual&lt;/em&gt; definition (particular construction) of the localization verifies the universal property. That is, given some morphism of rings $\varphi : A \longrightarrow B$ such that $\varphi s \in B$ is a unit for every $s \in S$, then there is a unique morphism of rings $\widetilde{\varphi} : S^{-1}A \longrightarrow B$ such that $\widetilde{\varphi}\gamma = \varphi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: you can easily check that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} \left( \frac{a}{s}  \right) = \frac{\varphi (a)}{\varphi (s)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is well-defined (that is, if $\frac{a}{s} = \frac{b}{t}$, then $\widetilde{\varphi}\left( \frac{a}{s}  \right) = \widetilde{\varphi}\left( \frac{b}{t}  \right) $), is a morphism of rings, verifies $\widetilde{\varphi} \gamma = \varphi$ and is unique (that is, if you had another $\psi : S^{-1}A \longrightarrow B$ such that $\psi\gamma = \varphi$, then $\psi = \widetilde{\varphi}$).&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-12T06:55:16.627" LastActivityDate="2010-09-12T06:55:16.627" CommentCount="1" />
  <row Id="4455" PostTypeId="1" AcceptedAnswerId="4457" CreationDate="2010-09-12T02:13:38.003" Score="3" ViewCount="62" Body="&lt;p&gt;I fear I over simplified the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any partially ordered set $(A,\leq)$, let $A^* = A- ${$ maxA,minA $} if $maxA$ and $minA$ exist. Show the inclusion $(A^*,\leq)\hookrightarrow (A,\leq)$ is sup-continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I took any set $B\subseteq A^*$ such that $supB$ exists. Then $\iota(B)=B$, and $\iota(supB)=supB$. Then $\iota(supB)=supB=sup(\iota(B))$, and so $\iota$ is sup-continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems too simple so I'm sure I've misinterpreted something. Can someone point out the source of error?&lt;/p&gt;&#xA;" OwnerUserId="1841" LastActivityDate="2010-09-12T03:44:46.230" Title="Showing the Inclusion is sup-continuous." Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="4456" PostTypeId="2" ParentId="4426" CreationDate="2010-09-12T02:58:27.513" Score="1" ViewCount="0" Body="&lt;p&gt;(1') is not equivalent to (2').  For example, $f(x)=x^2$ satisfies (2') on $(-\infty,\infty)$ but not (1'). It is not even uniformly continuous.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition (2'), being AC on all compact subintervals, is a condition I have at least seen used, and it is the right one if you want to extend the equivalence to being an indefinite integral.  Namely, it is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If $a\lt c \lt b$, then $f(x)=f(c)+ \int_c^x f'$ for all $x\in(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only a slight modification of (2), which must be made because $f$ may be unbounded or otherwise undefined at $a$ and $f'$ may not be integrable on $(a,x)$, even if $f$ satisfies the stronger condition (1') (e.g., $f(x)=x$ on $(-\infty,\infty)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you asked if this is the &quot;best&quot; definition or if there is an &quot;accepted&quot; definition, and of that I am not sure.  I have not seen anyone write &quot;$f$ is AC on $(a,b)$&quot; when they mean (2') holds.  For the case $(-\infty,\infty)$, I have seen simply &quot;f is AC on bounded intervals&quot;.  On the Wikipedia page, they use the phrase &quot;locally absolutely continuous&quot; in the section on the relationship to measures.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-12T03:15:06.287" LastActivityDate="2010-09-12T03:15:06.287" CommentCount="2" />
  <row Id="4457" PostTypeId="2" ParentId="4455" CreationDate="2010-09-12T03:38:49.213" Score="2" ViewCount="0" Body="&lt;p&gt;The set of upper bounds depends on whether you are in A or in A*, so technically you have to check that sup(B) actually is the least upper bound of B in A.  It would still be an upper bound of B, but there are 2 new contenders, max(A) and min(A).  Of course sup(B) &amp;lt; max(A), so max(A) is a larger upper bound and presents no problem.  On the other hand, min(A) &amp;lt; sup(B), so if min(A) is an upper bound for B in A, then it must in fact be the sup of B in A.  This could only happen if B is empty, and it could happen.  For example, let A={0,1,2} with the usual order and let B be the empty set.  Then &quot;sup(B)&quot; in A*={1} is 1, but the sup of B in A is 0.  This problem occurs only because A* has a least element, thus making the empty set have a sup in A*.  Everything is fine if B is nonempty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I've correctly hit on the point you were stumbling over, perhaps it would help to adjust notation to keep in mind that sups are conditional, e.g., $\sup_A (B)$ for the sup of $B$ in $A$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-12T03:44:46.230" LastActivityDate="2010-09-12T03:44:46.230" CommentCount="4" />
  <row Id="4458" PostTypeId="1" CreationDate="2010-09-12T03:56:11.690" Score="6" ViewCount="146" Body="&lt;p&gt;My goal here is to make my own custom &lt;a href=&quot;http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/projPoly.html&quot; rel=&quot;nofollow&quot;&gt;&quot;polyhedral map&quot;&lt;/a&gt; of Earth. If you print out something from the &lt;a href=&quot;http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/Foldout/foldout.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Map Fold-outs&quot;&lt;/a&gt; page, you will have something almost exactly like what I'm trying to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 2 triangles. One is a spherical triangle drawn on a 3D globe. By definition, each edge of a spherical triangle is part of a great circle. When you look at that 3D globe, there are a bunch of cities, coastlines, etc. that are (hopefully) accurately plotted on that 3D globe, inside that spherical triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other triangle is a flat, plane (2D), straight-edged, Euclidean triangle. On paper. At the moment the interior of that triangle is blank white paper, but eventually I want to draw a copy of all those cities, coastlines, etc. into that area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every map projection will map that 3D spherical triangle to a 2D image. Then it's easy to (in 2D) rotate and slide and shrink, and perhaps skew, that image until the 3 corners exactly line up with the 3 corners of my plane triangle. If I stack the results of a bunch of different map projections on top of each other, even though I've forced the 3 corners to exactly line up, each projection will put the cities in a slightly different location. Unfortunately, many projections take cities that are slightly inside the spherical triangle and draw them slightly outside the flat 2D triangle. (One symptom of this problem is that the sides of the spherical triangle are mapped to plane (2D) curves whose endpoints match the corners properly, but they bulge outward slightly from perfectly straight lines drawn between the corners). That leads to the city being completely missing from my polyhedral map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd rather not run the risk of some Polish mathematician getting upset that Warsaw is inexplicably missing from the flat, plane (2d), paper map that I've made :-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm looking for map projections that &quot;keep it inside the lines&quot;. I need 3 specific great circles (the edges of the spherical triangle) to be mapped to straight lines on paper. I don't care about other great circles -- straight, non-straight, whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What map projections meet that criteria? And can you give me a link to the (x,y) = f(lat, long) equations for that projection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hear that the map projection used for the Dymaxion map meets that criteria; and someone told me that he thought it used the Chamberlin trimetric projection.&#xA;Alas, when I use my (extremely rough and probably buggy) implementation of Chamberlin trimetric projection to map the spherical triangle formed by its 3 base points to the plane, I seem to be getting a shape that is almost a triangle, but the three &quot;lines&quot; curve and bulge out. Is that a bug in my code, or is it supposed to do that?&#xA;So apparently either (a) I'm using the wrong equations -- so where can I find the right equations? Or (b) Dymaxion actually uses some other projection -- so where can I find the equations for that projection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for answers of the form &quot;The gnomonic projection meets your criteria. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gnomonic_projection&quot; rel=&quot;nofollow&quot;&gt;gnomonic projection equations&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get a list of several map projections that meet that critera.&#xA;I know that the gnomonic projection is not the only one, because&#xA;the Collignon projection and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Peirce_quincuncial_projection&quot; rel=&quot;nofollow&quot;&gt;Peirce quincuncial projection&lt;/a&gt; can map an octant (a spherical triangle with 90 degree corners) to a straight-edged plane triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Forgive me for reposting &lt;a href=&quot;http://stackoverflow.com/questions/3305056/how-do-i-map-a-spherical-triangle-to-a-plane-triangle&quot;&gt;this question from StackOverflow&lt;/a&gt;. I hope that a few people here will know a few answers to this that no one at StackOverflow seems to know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several responses have mentioned that the &lt;em&gt;only&lt;/em&gt; way to make every straight line on the plane map correspond to a great circle on the globe is to use the gnomonic projection (or some linear affine transformation of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's technically correct, but irrelevant to my question.&#xA;I'm not asking for a projection that maps &lt;em&gt;every&lt;/em&gt; straight line on the plane map correspond to a great circle on the globe.&#xA;I'm asking for projections that map 3 particular straight lines -- the boundaries of one facet of my map -- to 3 particular great circles on the globe -- the great circles connecting a few particular carefully-chosen locations on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While certainly the gnomonic projection is &lt;em&gt;one&lt;/em&gt; way to map the cities, coastlines, etc. in a spherical triangle to a plane triangle, it is not the only way. As an existence proof, as I mentioned before, the Collignon projection and the Peirce quincuncial projection also map the octant -- a spherical triangle -- to a straight-edged plane triangle.&#xA;(What are the equations for the &lt;a href=&quot;http://www.grunch.net/synergetics/map/dymax.html&quot; rel=&quot;nofollow&quot;&gt;Dymaxion map projection&lt;/a&gt;?&#xA;I hear that it is yet another way to map a spherical triangle to a plane triangle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas, some people do not like the area distortion, length distortion, and the angle distortion of the gnomonic projection.&#xA;I'm hoping that by allowing a little distortion in the paths of &lt;em&gt;some&lt;/em&gt; great circles (mapping them to slightly curved paths on the plane), I will end up with less area distortion, length distortion, and angle distortion.&#xA;(I see that the Peirce quincuncial projection has zero angle distortion at most points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;there are infinitely many smooth mappings from the sphere to the plane that satisfy that criterion. ... you can always take any such mapping and &quot;distort&quot; it a little in the interior.&quot;&#xA;Yes, but most of them make area distortion, length distortion, and angle distortion &lt;em&gt;worse&lt;/em&gt;, and have horrifically complicated equations describing them.&#xA;I'm looking for projections that are &lt;em&gt;better&lt;/em&gt; than the gnomonic projection, preferably with relatively simple equations describing the projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hesitate to say that I'm looking for the projection that &lt;em&gt;minimizes&lt;/em&gt; distortion, because I don't want several pages of horrifically complicated equations that happen to minimize some specific arbitrarily-weighted average of area distortion, length distortion, and angle distortion.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastEditorUserId="1705" LastEditorDisplayName="" LastEditDate="2010-09-13T15:52:36.827" LastActivityDate="2010-09-13T15:52:36.827" Title="How do I map a spherical triangle to a plane triangle?" Tags="&lt;spherical-geometry&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="4459" PostTypeId="2" ParentId="4437" CreationDate="2010-09-12T05:00:26.510" Score="0" ViewCount="0" Body="&lt;p&gt;I'd just like to make the following three observations as a minor complement to the other answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OBSERVATION 1.&lt;/strong&gt; What the Cayley-Hamilton Theorem says is $$\det&#xA;\begin{pmatrix}&#xA;a_{11}-A &amp;amp; a_{12}&amp;amp;\cdots&amp;amp;a_{1n}\\ &#xA;a_{21}&amp;amp;a_{22}-A&amp;amp;\cdots&amp;amp;a_{2n}\\ &#xA;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\vdots\\ &#xA;a_{n1}&amp;amp;a_{n2}&amp;amp;\cdots&amp;amp;a_{nn}-A&#xA;\end{pmatrix}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OBSERVATION 2.&lt;/strong&gt; The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on &lt;a href=&quot;http://www.iecn.u-nancy.fr/~gaillard/atiyah_macdonald_p_21.pdf&quot; rel=&quot;nofollow&quot;&gt;page 21&lt;/a&gt; (proof of Proposition 2.4) of &lt;strong&gt;Introduction to Commutative Algebra&lt;/strong&gt; by Atiyah and MacDonald. The argument can be phrased as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be a commutative ring; let $n$ be a positive integer; let $A=(a_{ij})\in M_n(K)$ be an $n$ by $n$ matrix with entries in $K$; let $\chi$ be its characteristic polynomial; define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}\,A-a_{ij}$; observe $$\sum_i\ \ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Applying (a trivial case of) Fubini's Theorem to the double sum $\sum_{i,j}\ c_{jk}\ b_{ij}\ e_i$, we get $\chi(A)=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OBSERVATION 3.&lt;/strong&gt; It's easy to define the ring $R[X]$ of polynomials in the indeterminate $X$ with coefficients in the &lt;strong&gt;noncommutative&lt;/strong&gt; $R$. But when you think in terms of universal properties, you see that this construction is not very natural. So, it's better, I think, not to introduce it just to prove the Cayley-Hamilton Theorem. &lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-09-12T05:00:26.510" />
  <row Id="4460" PostTypeId="2" ParentId="2969" CreationDate="2010-09-12T07:23:21.410" Score="0" ViewCount="0" Body="&lt;p&gt;I realized about 80% through that this answer may have too many parts to implement, so I'm also going to leave a shorter one.&lt;br&gt;&lt;br&gt;&#xA;If you prioritize filling some cells over others, you can at least prune the computer's search tree. I suggest filling the center cube, followed by the six face centers, followed by the eight corners, finishing with everything else.&lt;br&gt;&lt;br&gt;&#xA;Here, let me number the squares of a Y pentomino:&lt;br&gt;1&lt;br&gt;23&lt;br&gt;4&lt;br&gt;5&lt;br&gt;By starting with the Y filling the center cube, the search space can be reduced by a factor of 12 by holding this piece's orientation constant. Note that only cells 2, 3, and 4 can occupy the center of the cube. If 2 or 4 is the center cell, then five face centers will remain to be filled; if it's 3, then all six will remain.&lt;br&gt;&lt;br&gt;&#xA;For the five or six centers, the Y occupying each center can be placed in one of 36 ways, disregarding overlap, as follows:&lt;br&gt;&lt;br&gt;&#xA;-The &quot;spine&quot; (the cubes 1245) can point in any of four directions&lt;br&gt;-Again, only 2, 3, or 4 can fill these cubes&lt;br&gt;-The full Y can lie on the face in three ways for each cube: on either Y face for all three, plus either along the spine for 2 and 4 or alongthe furthest opposing face for 3&lt;br&gt;&lt;br&gt;&#xA;Now for the eight corners. Only 1 and 5 can occupy these cubes, so for each of these corners we can work on partial placements for just the spines along the edges. Now, each corner is adjacent to three three-square edges, although some of the twelve edges are certainly part-occupied by Y's placed in the previous phase. The corners to be filled need to be prioritized by fewest available edges.&lt;br&gt;&lt;br&gt;&#xA;-If, at any point, an unfilled corner has 0 free edges, then either the last &quot;temporary edge placement&quot; or the entire face-center configuration needs to be advanced to the next one&lt;br&gt;-Whenever a corner piece is laid along an edge and the other corner sharing that edge is not yet filled, that other corner should be re-prioritized for having one fewer free edge&lt;br&gt;-If some unfilled corner has only one free edge, a spine is placed along that corner and edge&lt;br&gt;-And if no unfilled corner has fewer than two free edges, pick and note a &quot;temporary edge placement&quot; to fix if it leads to a contradiction. (I guarantee at least two corners will have no more than two edges. If the first placement is wrong, the second placement should be used but not noted as temporary.)&lt;br&gt;&lt;br&gt;&#xA;This gives eight partial placements and six or seven total placements. This leaves eleven or ten Y's to place within a volume of 63 or 58, respectively, which includes at least 26 cubes that are absolutely not filled regardless of how exactly the corner Y's are placed (each one still has up to four possibilities for the placement of the 3 cube), a much smaller search space.&lt;/p&gt;&#xA;" OwnerUserId="1845" LastActivityDate="2010-09-12T07:23:21.410" />
  <row Id="4461" PostTypeId="2" ParentId="2969" CreationDate="2010-09-12T07:32:13.777" Score="0" ViewCount="0" Body="&lt;p&gt;For a simpler solution path, I again suggest filling the center cube first to reduce symmetries, but make sure cube 1 or 5 is on the top layer (or bottom, depending on which you want to fill first). Then you can fill the first layer (possibly prioritizing the corners), followed by the subsequent layers.&lt;/p&gt;&#xA;" OwnerUserId="1845" LastActivityDate="2010-09-12T07:32:13.777" />
  <row Id="4462" PostTypeId="1" AcceptedAnswerId="6225" CreationDate="2010-09-12T09:08:12.540" Score="5" ViewCount="150" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathreference.com/ag,fgaf.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; page contains a result which it refers to as the UFD field lemma. I was wondering if anybody knew of any other references which discuss this result--this page is the only place I've seen it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UFD field lemma appears to assert that if $R$ is a unique factorization domain containing infinitely many prime elements, if $F$ is the field of fractions of $R$, and if $A$ is a finitely generated $R$-algebra which is a field and is algebraic over $F$, then $A$ does not contain $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for other sources because I find the exposition on that page a little hard to follow.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-10-06T20:40:03.507" LastActivityDate="2010-10-06T21:12:40.120" Title="The UFD field lemma" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4463" PostTypeId="2" ParentId="4368" CreationDate="2010-09-12T09:49:10.053" Score="0" ViewCount="0" Body="&lt;p&gt;Please search for &lt;strong&gt;external memory linear algebra&lt;/strong&gt; or &lt;strong&gt;out of core&lt;/strong&gt; computation to see if you find implementations / ideas for how to manage these huge 'v' vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, since computing the largest eigenvalue + corresponding eigenvector are in a way core parts of Google's pagerank algorithm, you might find useful leads on how to perform this huge computation by using ideas similar to those used for pagerank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you are content with approximate answers, other options like random projections open up.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-09-12T09:49:10.053" />
  <row Id="4465" PostTypeId="2" ParentId="4368" CreationDate="2010-09-12T10:19:35.053" Score="0" ViewCount="0" Body="&lt;p&gt;For a computer package that solves the eigenvalue for large sparse matrix problem. Use &lt;a href=&quot;http://www.caam.rice.edu/software/ARPACK/&quot; rel=&quot;nofollow&quot;&gt;ARPACK&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://en.wikipedia.org/wiki/ARPACK&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The package is designed to compute a&#xA;  few eigenvalues and corresponding&#xA;  eigenvectors of large sparse or&#xA;  structured matrices, using the&#xA;  Implicitly Restarted Arnoldi Method&#xA;  (IRAM) or, in the case of symmetric&#xA;  matrices, the corresponding variant of&#xA;  the Lanczos algorithm.the Lanczos algorithm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-12T10:19:35.053" />
  <row Id="4466" PostTypeId="1" AcceptedAnswerId="4493" CreationDate="2010-09-12T10:57:24.120" Score="4" ViewCount="139" Body="&lt;p&gt;I'm working my way through Niven's Introduction to Number Theory, and the wording of the following problem is making me unsure of my answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that there is a one-to-one correspondence between twin primes and numbers $n$ such that $n^2-1$ has just four positive divisors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I felt an obvious bijection would be $f\colon A\rightarrow B\colon (p,p+2)\mapsto p+1$, where $A$ is the set of all pairs of twin primes, and $B$ is the set of all positive $n$ such that $n^2-1$ has only four positive divisors. $f$ is injective, and for any such $n$, if $n^2-1=(n-1)(n+1)$, has only four positive divisors, they must be $1,(n-1),(n+1),n^2-1$, implying that $n-1$ and $n+1$ are twin primes, and thus $(n-1,n+1)$ would be a suitable preimage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I assumed the wording of the problem meant I should show a bijection from unordered pairs of twin primes to positive integers, but I'm not sure if the problem intended for me to find a bijection that takes a single prime that happens to be a twin prime, to any such $n$, not necessarily positive which is problematic since both $n$ and $-n$ give the same $n^2-1$. Have I interpreted this problem correctly? If not, do other bijections exist with the differing domains and ranges I mentioned above?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-15T10:42:42.653" Title="Bijection between twin primes and numbers $n$ such that $n^2-1$ has exactly four positive divisors" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4467" PostTypeId="1" AcceptedAnswerId="4478" CreationDate="2010-09-12T11:24:41.593" Score="1" ViewCount="984" Body="&lt;p&gt;I got this interesting question in my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we prove that if $a \in  \mathbb{N}$ then $\sqrt{a}$ is an integer or an irrational number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we extend this result?  That is, can it be shown that if $a,b \in \mathbb{N}$, then $a^{1/b}$ is an integer or an irrational?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-05T00:37:02.193" LastActivityDate="2011-01-05T00:37:02.193" Title="$a^{1/2}$ is either an integer or an irrational number." Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;irrational-numbers&gt;" AnswerCount="9" CommentCount="11" FavoriteCount="2" />
  <row Id="4468" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T11:32:07.030" Score="11" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: If $a$ and $b$ are positive integers, then $a^{1/b}$ is either irrational or an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a^{1/b}=x/y$ where $y$ does not divide $x$, then $a=(a^{1/b})^b=x^b/y^b$ is not an integer (since $y^b$ does not divide $x^b$), giving a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I subsequently found a variant of this proof on Wikipedia, under &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality&quot; rel=&quot;nofollow&quot;&gt;Proof by unique factorization&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bracketed claim is proved below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt;:  If $y$ does not divide $x$, then $y^b$ does not divide $x^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Unique prime factorisation&lt;/a&gt; implies that there exists a prime $p$ and positive integer $t$ such that $p^t$ divides $y$ while $p^t$ does not divide $x$.  Therefore $p^{bt}$ divides $y^b$ while $p^{bt}$ does not divide $x^t$.  Hence $y^b$ does not divide $x^b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[OOC: This answer has been through several revisions (some of the comments below might not relate to this version)]&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T10:40:59.017" LastActivityDate="2010-09-13T10:40:59.017" CommentCount="10" />
  <row Id="4469" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T11:47:59.613" Score="5" ViewCount="0" Body="&lt;p&gt;The general theorem is that a natural number $a$ has a rational square root if and only if the multiplicity of every prime factor of $a$ is even.  For example $2^43^611^2$ has a rational square root but $5^411^3$ does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if a natural number has a rational square root, that square root is always obtained by halving the multiplicity of each prime factor, and so the square root is also a natural number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same principle works for $n$th roots, $n &gt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-12T11:47:59.613" CommentCount="1" />
  <row Id="4470" PostTypeId="1" AcceptedAnswerId="4473" CreationDate="2010-09-12T12:08:16.930" Score="2" ViewCount="43" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallelepiped&quot; rel=&quot;nofollow&quot;&gt;Wikipedia it is stated that the volume of the parallelepiped&lt;/a&gt; given its edge  lengths $a,b,c$, and the internal angles between the edges $\alpha ,\beta ,\gamma $ is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=abc\sqrt{1+2\cos \alpha \cos \beta \cos \gamma -\cos ^{2}\alpha&#xA;- \cos^{2}\beta - \cos ^{2}\gamma }\qquad$(*).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was not able to derive it by using the determinant formula and expressing $\cos \alpha ,\cos \beta ,\cos \gamma $ in terms of $a,b,c$ and $\alpha&#xA;,\beta ,\gamma $. For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}=ab\cos \alpha $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Could you give a hint on how can the formula (*) be proved?&lt;/p&gt;&#xA;" OwnerUserId="1504" LastEditorUserId="1504" LastEditorDisplayName="" LastEditDate="2010-09-12T13:35:24.507" LastActivityDate="2010-09-12T13:35:24.507" Title="Deriving volume of parallelepiped as a function of edge lengths and angles between the edges" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4471" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T12:10:41.543" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: An &lt;em&gt;algebraic integer&lt;/em&gt; is a solution of a monic polynomial with integer coefficients. This set is closed by sums and products and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: If an algebraic integer is rational, then it is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: Apply rational roots theorem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This theorem proves that $\sqrt[b]{a}$ is irrational unless $a$ is a $b$-th power.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-12T12:10:41.543" />
  <row Id="4472" PostTypeId="2" ParentId="4470" CreationDate="2010-09-12T12:55:29.160" Score="1" ViewCount="0" Body="&lt;p&gt;Please see this link, under trigonometric approach: &lt;a href=&quot;http://www.ehow.com/how_6545076_prove-formula-volume-cuboids.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ehow.com/how_6545076_prove-formula-volume-cuboids.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-12T12:55:29.160" />
  <row Id="4473" PostTypeId="2" ParentId="4470" CreationDate="2010-09-12T13:01:33.887" Score="4" ViewCount="0" Body="&lt;p&gt;The volume satisfies $V=|\det D|$ where $D$ is the matrix&#xA;$$\pmatrix{a_1&amp;amp;a_2&amp;amp;a_3\\&#xA;b_1&amp;amp;b_2&amp;amp;b_3\\&#xA;c_1&amp;amp;b_2&amp;amp;c_3}$$&#xA;where $\mathbf{a}=(a_1,a_2,a_3)$ is the vector corresponding to $a$ etc.&#xA;Thus&#xA;$$V^2=\det(DD^t)=\det&#xA;\pmatrix{\mathbf{a}\cdot\mathbf{a}&#xA;&amp;amp;\mathbf{a}\cdot\mathbf{b}&#xA;&amp;amp;\mathbf{a}\cdot\mathbf{c}\\&#xA;\mathbf{b}\cdot\mathbf{a}&#xA;&amp;amp;\mathbf{b}\cdot\mathbf{b}&#xA;&amp;amp;\mathbf{b}\cdot\mathbf{c}\\&#xA;\mathbf{c}\cdot\mathbf{a}&#xA;&amp;amp;\mathbf{c}\cdot\mathbf{b}&#xA;&amp;amp;\mathbf{c}\cdot\mathbf{c}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can express $V^2$ in terms of $a$, $b$, $c$ and the various&#xA;cosines. &lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-12T13:01:33.887" />
  <row Id="4474" PostTypeId="2" ParentId="4430" CreationDate="2010-09-12T13:34:24.417" Score="0" ViewCount="0" Body="&lt;p&gt;Perhaps what you mean is how differential equations can be applied to find othogonal trajectories?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html&quot; rel=&quot;nofollow&quot;&gt;http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-12T13:34:24.417" />
  <row Id="4475" PostTypeId="2" ParentId="4458" CreationDate="2010-09-12T13:47:51.423" Score="1" ViewCount="0" Body="&lt;p&gt;You might want to look at the relatively recent book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portrait of the Earth: A Mathematician Looks at Maps by Timothy Freeman&lt;/p&gt;&#xA;&#xA;&lt;p&gt;American Mathematical Society, Providence, 2002&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-12T13:47:51.423" />
  <row Id="4476" PostTypeId="1" AcceptedAnswerId="4507" CreationDate="2010-09-12T13:48:47.790" Score="1" ViewCount="138" Body="&lt;p&gt;The Fourier transform can be defined on $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, and we can extend this to $X:=L^2(\mathbb{R}^n)$ by a density argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by Plancherel we know that $\|\widehat{f}\|_2 = \|f\|_2$, so the Fourier transform is an isometry on this space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question now is, what is a theorem that guarantees that the Fourier transform has a fixed point on $L^2$? I know the Gaussian is a fixed point, but I'm also interested in other integral transforms, but I just take the Fourier transform as an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Banach Fixed Point Theorem does not work here since we don't have a contraction (operator norm $&amp;lt; 1$). Can we apply the Tychonoff fixed point theorem? Then we would need to show that there exists a non-empty compact convex set $C \subset X$ such that the Fourier transform restricted to $C$ is a mapping from $C$ to $C$. Is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a fixed point, what would be a way to show it is unique? By linearity we obviously have infinitely many fixed points of we have at least two of them.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-26T16:14:35.587" Title="Fixed point Fourier transform (and similar transforms)." Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4477" PostTypeId="2" ParentId="2418" CreationDate="2010-09-12T14:13:50.823" Score="1" ViewCount="0" Body="&lt;p&gt;This is not intended as an answer, but clearly is too long for comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some approaches I have been playing around with; I would appreciate any comments, advice, or suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the condition s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt; is the key thing about this problem.  It seems like in order to state this as a problem (instead of as a conjecture), it must have been solved by someone, otherwise where does |T| $\leq$ 7 come from?  Since we don't know for sure if the problem has been solved, I am assuming that |T| $\leq$ 7 is possibly wrong and just trying to work out the consequences of the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition is very constrictive, and it's consequences are not at all readily apparent.  One question I have is, &lt;strong&gt;how many independent elements can a set which satisfies the condition have?&lt;/strong&gt;  The idea of an independent element is not rigorous for a set like this, but the idea is to figure out how to formulate the idea of independent element so it is rigorous.  Given a set of integers H, with |H| less than 15, is it possible to add numbers (by combining elements of H) so that H has 15 elements and satisfies the condition; if so, can we say something about the original size of H?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this direction I have made the following (slightly trivial) observations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set which satisfies the condition contains:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at least two (distinct) positive numbers&lt;/li&gt;&#xA;&lt;li&gt;at least two (distinct) negative numbers&lt;/li&gt;&#xA;&lt;li&gt;at least two (distinct) numbers which are the sum of a positive and a negative number&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This follows from the fact that each such set will contain a largest positive number which must be the sum of only positive numbers (similarly for negative); and the smallest positive number must be the sum of a positive and negative (similar for smallest negative number).  I'm not sure how this relates to the (poorly defined) notion of &quot;independent element&quot;, but it feels like a start in that direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is that the condition s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt; can be &quot;unpacked&quot; by applying it recursively to s&lt;sub&gt;j&lt;/sub&gt; and s&lt;sub&gt;k&lt;/sub&gt;.  My hope is that I can show that there will always be one element which can be &quot;unpacked&quot; so that it is equal to 8 elements in S, one of which is itself.  This would of course show that the sum of the other 7 is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;= s&lt;sub&gt;l&lt;/sub&gt; + s&lt;sub&gt;m&lt;/sub&gt; + s&lt;sub&gt;n&lt;/sub&gt; + s&lt;sub&gt;p&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= s&lt;sub&gt;q&lt;/sub&gt; + s&lt;sub&gt;r&lt;/sub&gt; + s&lt;sub&gt;s&lt;/sub&gt; + s&lt;sub&gt;t&lt;/sub&gt; + s&lt;sub&gt;u&lt;/sub&gt; + s&lt;sub&gt;v&lt;/sub&gt; + s&lt;sub&gt;w&lt;/sub&gt; + s&lt;sub&gt;i&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next idea is to show that any set satisfying the condition will contain what I call a &quot;7 cycle&quot;, which is inspired from the totally anti-symmetric unit tensor.  A &quot;7 cycle&quot; is a kind of symmetry for the index's of s&lt;sub&gt;i&lt;/sub&gt; = s&lt;sub&gt;j&lt;/sub&gt; + s&lt;sub&gt;k&lt;/sub&gt;, I will just write down what a 7 cycle looks like, because that is more clear than explaining it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt; = s&lt;sub&gt;2&lt;/sub&gt; + s&lt;sub&gt;3&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;2&lt;/sub&gt; = s&lt;sub&gt;3&lt;/sub&gt; + s&lt;sub&gt;4&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;3&lt;/sub&gt; = s&lt;sub&gt;4&lt;/sub&gt; + s&lt;sub&gt;5&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;4&lt;/sub&gt; = s&lt;sub&gt;5&lt;/sub&gt; + s&lt;sub&gt;6&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;5&lt;/sub&gt; = s&lt;sub&gt;6&lt;/sub&gt; + s&lt;sub&gt;7&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;6&lt;/sub&gt; = s&lt;sub&gt;7&lt;/sub&gt; + s&lt;sub&gt;1&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;7&lt;/sub&gt; = s&lt;sub&gt;1&lt;/sub&gt; + s&lt;sub&gt;2&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's possible to show that any set satisfying the condition contains a 7 cycle, then it's clear that the sum  of s&lt;sub&gt;1&lt;/sub&gt;...s&lt;sub&gt;7&lt;/sub&gt; will be zero (since it is equal to twice itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is a combinatorial graph theory approach.  Clearly, a set satisfying the condition generates a directed graph (as pointed out in the MO link).  For clarity, each number in S corresponds to a vertex, and if for example, s&lt;sub&gt;1&lt;/sub&gt; = s&lt;sub&gt;2&lt;/sub&gt; + s&lt;sub&gt;3&lt;/sub&gt;, then there will be edges directed from s&lt;sub&gt;2&lt;/sub&gt; to s&lt;sub&gt;1&lt;/sub&gt; and from s&lt;sub&gt;3&lt;/sub&gt; to s&lt;sub&gt;1&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A question I have which I can't seem to make progress on, is: &lt;strong&gt;what is the minimum possible number of vertex's which have outgoing edges?&lt;/strong&gt; (a vertex has an outgoing edge if it can be added to an element of S to obtain another element of S, also the assumptions used in the previously given &quot;weaker statement&quot; led to the conclusion that each vertex has an outgoing edge, and this was used to show |T| $\leq$ 7).  My hope is that once I know the minimum number of vertex's with outgoing edges, I can use some sort of combinatorial reasoning to show that there will be a closed cycle with a maximum length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling &quot;zero sum subset&quot; quickly leads to references for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-sum_problem&quot; rel=&quot;nofollow&quot;&gt;EGZ theorem&lt;/a&gt;, and by thinking about the problem backwards, it makes sense that since this is one of the best known results for zero sum problems; that if this problem has already been solved this result might be involved in the solution somehow.  I can't quite wrap my head around exactly how it will be involved, but since it is a result dealing with the integers mod N, the idea would be to somehow show that the condition forces some (possibly elaborate) modular arithmetic structure on the set S which satisfies it.  (my understanding is that EGZ says that given an arbitrary set of 2n - 1 integers in $\mathbb{Z}$ mod n, there will be a subset of n elements whose sum is congruent to 0 mod n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another idea is to reformulate the problem in terms of polynomials.  Given a set S which satisfies the condition, consider the 15th degree polynomial whose roots are the elements of S.  Then these roots possess the symmetry that every root is the sum of two other roots; these types of polynomials might have some property which forces the sum of at most 7 elements to be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't feel strongly that any of these approaches will ultimately prove successful, but hopefully by making this post I can help keep this question alive and inspire others to work on it.  &lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-12T20:33:13.723" LastActivityDate="2010-09-12T20:33:13.723" CommentCount="2" />
  <row Id="4478" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T14:47:42.780" Score="5" ViewCount="0" Body="&lt;p&gt;These (standard) results are discussed in detail in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/4400irrationals.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/4400irrationals.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the second handout for a first course in number theory at the advanced undergraduate level.  Three different proofs are discussed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A generalization of the proof of irrationality of $\sqrt{2}$, using the decomposition of any positive integer into a perfect $k$th power times a $k$th power-free integer, followed by Euclid's Lemma.  (For some reason, I don't give all the details of this proof.  Maybe I should...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A proof using the functions $\operatorname{ord}_p$, very much along the lines of the one Carl Mummert mentions in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) A proof by establishing that the ring of integers is integrally closed.  This is done directly from unique factorization, but afterwards I mention that it is a special case of the Rational Roots Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also remark that every proof I have ever seen of this fact uses the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) in some form.  [&lt;b&gt;Edit&lt;/b&gt;: I have now seen Robin Chapman's answer to the question, so this is no longer quite true.]  However, if you want to prove any particular case of the result, you can use a brute force case-by-case analysis that avoids FTA.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-13T07:30:29.880" LastActivityDate="2010-09-13T07:30:29.880" CommentCount="4" />
  <row Id="4479" PostTypeId="1" AcceptedAnswerId="4487" CreationDate="2010-09-12T15:11:52.683" Score="3" ViewCount="131" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{M}$ be an n-dimensional manifold endowed with an affine connection $\nabla$. Let $\gamma_1:[a,b]\rightarrow M$ and $\gamma_2:[c,d]\rightarrow \mathcal{M}$ be two curves with the same initial and final points, that is,&#xA;$p=\gamma_1(a)=\gamma_2(c), q=\gamma_1(b)=\gamma_2(d)$. Take  $X\in T_p\mathcal{M}$. Parallelly propagating $X$ along $\gamma_1$ and $\gamma_2$ we obtain two vectors $X_1, X_2\in T_q\mathcal{M}$, respectively. Let $R$ be the curvature tensor of the connection, $R(X,Y)Z=\nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z -\nabla_{[X,Y]}Z$, and $\tau$ its torsion, $\tau(X,Y)=\nabla_X Y-\nabla_Y X -[X,Y]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: How can I compare the two vectors $X_1$ and $X_2$? Can I write the difference $(X_2-X_1)$ in terms of $R,\tau$ and the curves?&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-09-12T17:52:45.653" Title="Parallel transport of a vector along two distinct curves" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4480" PostTypeId="1" AcceptedAnswerId="4499" CreationDate="2010-09-12T15:21:08.620" Score="7" ViewCount="299" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vzhVr.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the radius of the circles if the size of the larger square is 1x1.&lt;br&gt;&#xA;Enjoy!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(read about the &lt;a href=&quot;http://mathworld.wolfram.com/SangakuProblem.html&quot; rel=&quot;nofollow&quot;&gt;origin of sangaku&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="923" LastEditorDisplayName="" LastEditDate="2010-09-13T14:44:59.953" LastActivityDate="2010-09-13T14:44:59.953" Title="sangaku - a geometrical puzzle" Tags="&lt;geometry&gt;&lt;puzzle&gt;&lt;puzzle&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="4481" PostTypeId="2" ParentId="4422" CreationDate="2010-09-12T16:58:17.337" Score="0" ViewCount="0" Body="&lt;p&gt;Oops I misread the question, but I actually like Shakarchi and Stein volume 1!  In my opinion the second volume is a lot better than the first one however.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-12T16:58:17.337" CommunityOwnedDate="2010-09-12T16:58:17.337" />
  <row Id="4482" PostTypeId="2" ParentId="4479" CreationDate="2010-09-12T17:05:46.370" Score="1" ViewCount="0" Body="&lt;p&gt;The difference depends on the homotopy classes of the paths, not only local data like the curvature and torsion.  I don't think there is a simpler answer than &quot;the difference in the parallel transport along the two paths&quot;.  But see the (superb) Wikipedia page on holonomy, especially the sections on the Ambrose-Singer theorem and affine connections.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Holonomy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Holonomy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-12T17:05:46.370" />
  <row Id="4483" PostTypeId="1" AcceptedAnswerId="4486" CreationDate="2010-09-12T17:06:05.153" Score="4" ViewCount="226" Body="&lt;p&gt;Any linear map between two finite-dimensional vector spaces can be represented as a matrix under the bases of the two spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if one or all of the vector spaces is infinite dimensional, is the linear map still represented as a matrix under their bases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is matrix of infinite dimension, what is it used for if not used as a representation of a linear map between vector spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-09-12T20:20:40.280" Title="Matrix of Infinite Dimension" Tags="&lt;abstract-algebra&gt;&lt;matrices&gt;&lt;vector-space&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4485" PostTypeId="2" ParentId="4480" CreationDate="2010-09-12T17:42:13.457" Score="5" ViewCount="0" Body="&lt;p&gt;Let $r$ be the length the radius of the circles, and let $\theta$ be the measure of the (smaller) angle made at the corner of the big square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9Gurs.png&quot; alt=&quot;Labeled image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The width of the square is equal to two radii and the projection of a double diameter (a quadruple-radius), so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)\hspace{1.0in}4r\cos\theta=1-2r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the four right triangles, we see that the center circle's diameter is equal to the difference in the lengths of the legs; since the hypotenuse has length $1$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2)\hspace{1.0in}2r = \cos\theta - \sin\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, we simply need to eliminate $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying (2) through by $4r$ and substituting in from (1) ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8 r^2 = 4r\cos\theta - 4r \sin\theta = 1 - 2r - 4r \sin\theta$$&#xA;$$4r \sin\theta = 1 - 2r - 8 r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}16r^2 &amp;amp;=&amp;amp; (4r \cos\theta)^2 + (4 r \sin\theta)^2 \\ &#xA;&amp;amp;=&amp;amp; ( 1 - 2r )^2 + ( 1 - 2r - 8 r^2 )^2 \\&#xA;&amp;amp;=&amp;amp; 2 - 8 r - 8 r^2 + 32r^3 + 64 r^4 \end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = 32 r^4 + 16 r^3 - 12 r^2 - 4 r + 1 = (2r+1)(2r-1)(8 r^2 + 4 r - 1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots of the polynomial are $\pm1/2$ and $(-1\pm\sqrt{3})/4$. We can eliminate three of them from consideration to conclude that $r = (-1+\sqrt{3})/4$.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-09-12T17:42:13.457" CommentCount="3" />
  <row Id="4486" PostTypeId="2" ParentId="4483" CreationDate="2010-09-12T17:46:45.500" Score="5" ViewCount="0" Body="&lt;p&gt;Sure, if $T:V\to W$ is a linear transformation between vector spaces $V$ and $W$ with bases $B$ and $C$, respectively, then $T$ can be described in terms of the coordinates with respect to these bases, thus yielding a &quot;matrix&quot;.  How closely this relates to the usual notion of matrix depends on the nature of $B$ and $C$.  In the usual notion, you take bases that are not only finite, but ordered, so that it makes sense to talk about the 1st row, etc., of the matrix; that is, you make all bases indexed by sets of the form $\{1,2,\ldots,n\}$.  The closest to this in the infinite dimensional setting would be to have bases indexed by the positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, suppose $B=\{v_j\}_{j\in J}$ and $C=\{w_i\}_{i\in I}$, where $I$ and $J$ are sets.  Then the matrix of $T$ can be described as a function $M:I\times J\to F$, where $F$ is the base field, by taking $M(i,j)$ to be the coefficient of $w_i$ in the $C$-expansion of $Tv_j$.  Such matrices are &lt;em&gt;column-finite&lt;/em&gt;, in the sense that for each $j\in J$, the set of $i\in I$ such that $M(i,j)\neq0$ is finite.  Conversely, each column-finite matrix, in this sense, corresponds uniquely to a linear transformation between $V$ and $W$.  Coordinate-wise addition of such matrices corresponds to addition of the linear transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also extend multiplication.  Suppose that $S:W\to X$ is a linear transformation and that $X$ has basis $D=\{x_k\}_{k\in K}$.  Let $N:K\times I\to F$ denote the $C$-$D$ matrix of $S$.  Then $ST:V\to X$ has $B$-$D$ matrix $NM:K\times J\to F$ defined by &#xA;$$(NM)(k,j)=\sum_{i\in I}N(k,i)M(i,j).$$  In particular, note that this sum is always finite because $M$ is column-finite.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-12T18:02:09.120" LastActivityDate="2010-09-12T18:02:09.120" />
  <row Id="4487" PostTypeId="2" ParentId="4479" CreationDate="2010-09-12T17:52:45.653" Score="3" ViewCount="0" Body="&lt;p&gt;If your two paths are homotopic you can make a comparison between your two vectors. And yes the comparison involves an integral over the homotopy of a function of curvature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example Theorem 13.6.4 in Pressley's &quot;Elementary Differential Geometry&quot; (Google books will bring up the statement of the theorem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your paths are not homotopic you're out of luck, as T describes. There are things you can say of course but it's not clear what you're looking for.  You should think of an example of a Riemann manifold you're interested in, to get a sense for how bad it can get. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-12T17:52:45.653" CommentCount="3" />
  <row Id="4488" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T18:03:03.590" Score="3" ViewCount="0" Body="&lt;p&gt;As muad points out, you can also obtain this as an easy consequence of the Rational Root Theorem: if $a_nx^n+\cdots+a_0$ is a polynomial with integer coefficients, and $\frac{p}{q}$ is a rational root with $\gcd(p,q)=1$, then $p|a_0$ and $q|a_n$ (plug in, clear denominators, factor out). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you look at the polynomial $x^b-a$, with $b$ and $a$ positive integers, then a rational root must be of the form $\frac{p}{q}$, with $\gcd(p,q)=1$, and $q|1$. Thus, it must be an integer. So if it has a rational root, then the root is integral. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-12T18:03:03.590" />
  <row Id="4489" PostTypeId="1" AcceptedAnswerId="4494" CreationDate="2010-09-12T18:48:45.160" Score="-3" ViewCount="245" Body="&lt;p&gt;I have got an interesting problem from one of my friends:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x,y,z$ are positive and satisfy $xy = z^{2}+z+3$, then prove that one of the Diophantine equations: $$a^{2}+11b^{2}=4x, \quad a^{2}+11b^{2}=4y$$ has a solution, $(a_{1},b_{1})$ with $a_{1},b_{1}$ being odd numbers.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T03:35:03.410" LastActivityDate="2010-09-21T03:35:03.410" Title="Proving that the given Diophantine equation has a solution" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="4490" PostTypeId="2" ParentId="4483" CreationDate="2010-09-12T20:10:39.423" Score="3" ViewCount="0" Body="&lt;p&gt;What, however, must be understood that the role of matrices when one works with linear operators of infinite-dimensional vector spaces is a (very) marginal one. Familiar techniques such as the use of determinants, traces etc. no longer work. For instance, any determinant-like function on $\mathrm{GL}(V)$ where $V$ is an infinite-dimensional vector space over a field is necessarily the trivial one, since any element of $\mathrm{GL}(V)$ is a product of commutators (the group $\mathrm{GL}(V)$ is perfect, in other words; a result by Alex Rosenberg of 1958.)&lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-09-12T20:20:40.280" LastActivityDate="2010-09-12T20:20:40.280" />
  <row Id="4491" PostTypeId="2" ParentId="4467" CreationDate="2010-09-12T20:25:03.167" Score="7" ViewCount="0" Body="&lt;p&gt;Here is a simple &lt;em&gt;conceptual&lt;/em&gt; $\:$ proof of the irrationality of sqrts - completely from first principles.&lt;br&gt;&#xA;The key idea is that the least positive denominator of a fraction $\rm\:r\:$ divides &lt;em&gt;every&lt;/em&gt;  denominator of $\rm\:r\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ \ \rm r = \sqrt{a}\ $ is integral if rational,$\:$ for $\:\rm a\in\mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\quad$  Put $\ \ \displaystyle\rm r = \frac{c}d ,\;$ least $\rm\; d&amp;gt;0\:.\;$ $\  \displaystyle\rm\sqrt{a}\; = \frac{a}{\sqrt{a}} \ \Rightarrow\ \frac{c}d = \frac{ad}c \ \Rightarrow\ d\:$ divides $\rm\: c\ $ via this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\;$ The least positive denominator of a fraction $\:\rm r\:$ divides every denominator of $\rm\:r\;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ n,m\ $ denominators $\: \Rightarrow\: $ so is $\rm\ n-m\ $ via $\;\rm n\:r,\: m\:r\in \mathbb Z \ \Rightarrow\: (n-m)\:r\in \mathbb Z\:,\:$ now apply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; $\ \ $ If a set of positive integers $\rm\: M\:$ satisfies $\rm n &amp;gt; m \in M \ \Rightarrow\ \: n-m \in M$ &lt;br/&gt; then every element of $\rm\:M\:$ is a multiple of the least element $\rm\:m \in M$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\:$ If not there'd be a least nonmultiple $\rm\:n\in M\:,$ contra  $\rm\:n-m \in M\:$ is a nonmultiple of $\rm\:m\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\:$ The above LEMMA is quite fundamental to factorization. I frequently  refer to it by the suggestive moniker &lt;strong&gt;unique fractionization&lt;/strong&gt;, since it's equivalent to  uniqueness of factorizations into irreducibles (indeed one easily sees that it is equivalent to Euclid's Lemma, which implies that irreducibles are prime).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T19:31:11.163" LastActivityDate="2010-12-02T19:31:11.163" CommentCount="4" />
  <row Id="4492" PostTypeId="2" ParentId="4480" CreationDate="2010-09-12T20:39:04.157" Score="2" ViewCount="0" Body="&lt;p&gt;Somewhat related to Don's solution: From the figure, we see that the four triangles are 1: congruent, and 2: right triangles. The hypotenuse of one triangle has length 1, and if we let $\theta$ be the smaller of the two angles of the right triangle, and use $r$ to denote the radius of one circle, then the Pythagorean relation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos^2\;\theta+(\cos\;\theta-2r)^2=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can now be solved as a simultaneous equation with any of the other two equations Don obtained, or we can use another equation, the expression for the &lt;a href=&quot;http://mathworld.wolfram.com/Incircle.html&quot; rel=&quot;nofollow&quot;&gt;inradius&lt;/a&gt; $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r^2=\frac{(s-1)(s-\cos\;\theta)(s+2r-\cos\;\theta)}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s=\frac{1+\cos\;\theta+(\cos\;\theta-2r)}{2}$ is the semiperimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we eliminate $\cos\;\theta$ and solve the two equations here for $r$, we find that the roots of the resulting quartic equation are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\frac{\pm 1\pm\sqrt{3}}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we carry out Don's approach as well, we find that only one &lt;strong&gt;positive&lt;/strong&gt; value of $r$ is consistent with both systems, and thus has to be the correct answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\frac{-1+\sqrt{3}}{4}$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-12T20:46:24.930" LastActivityDate="2010-09-12T20:46:24.930" CommentCount="1" />
  <row Id="4493" PostTypeId="2" ParentId="4466" CreationDate="2010-09-12T20:59:35.247" Score="0" ViewCount="0" Body="&lt;p&gt;As Yuval points out, (F(n-1=p, n+1=q) := n)^2-1 = pq is the obvious interpretation.  If you want to interpret the problem as specifying as a domain for the bijection a single prime p for which either p+2 or p-2 is also prime, or two primes which are part of the same twin pair but given in either order, or if n is allowed to be negative while p and q are conventionally positive, then (in any of those cases) a bijection exists if and only if twin primes are infinite in number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you extend &lt;em&gt;both&lt;/em&gt; the domain and range in this way then it still works: F(-n+1=p, -n-1=q) := n.  Also if you extend the range and extend the domain by allowing negative primes then it still works without adding to the definition of F.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastEditorUserId="1284" LastEditorDisplayName="" LastEditDate="2010-09-12T21:48:36.947" LastActivityDate="2010-09-12T21:48:36.947" />
  <row Id="4494" PostTypeId="2" ParentId="4489" CreationDate="2010-09-12T20:59:40.950" Score="-2" ViewCount="0" Body="&lt;p&gt;As baudrillard points out the answer is on this link: Mathematical Olympiad in China: problems and solutions by Bin Xiong, Peng Yee Lee - Page 115&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-12T20:59:40.950" CommunityOwnedDate="2010-09-12T20:59:40.950" />
  <row Id="4495" PostTypeId="1" AcceptedAnswerId="4567" CreationDate="2010-09-12T21:00:51.220" Score="6" ViewCount="283" Body="&lt;p&gt;An ordinary eigenvector can be viewed as a vector on which the operator acts by only stretching (without rotating) it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar intuition behind generalized eigenvectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: By generalized eigenvectors I'm referring to vectors in the kernel of $(T-\lambda I)^k$ for some $k$.&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2010-09-13T06:23:00.797" LastActivityDate="2010-09-13T18:47:28.640" Title="The intuition behind generalized eigenvectors" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="4496" PostTypeId="2" ParentId="4495" CreationDate="2010-09-12T21:09:56.267" Score="6" ViewCount="0" Body="&lt;p&gt;Consider first eigenvectors associated to $0$; these are elements of the nullspace. Now, it is possible for a vector to not lie in the nullspace, but for its image to lie in the nullspace (for example, consider the linear transformation $T\colon \mathbb{R}^2\to\mathbb{R}^2$ given by $T(x,y) = (y,0)$; then $T(0,1) = (1,0)$, and $T(T(0,1))=(0,0)$. Or for the image of the image to lie in the nullspace... etc. If you want to think about all the vectors that will &lt;em&gt;eventually&lt;/em&gt; map to zero if you keep applying $T$ (which seems like a sensible thing to think about in many circumstances), then you are looking for the vectors that lie in union of the nullspaces of $T^n$, $n=1,2,\ldots$. Notice that $\mathbf{N}(T^n)\subseteq\mathbf{N}(T^{n+1})$, so this is an increasing union. Once a vector maps to $0$ under a power of $T$, it stays there. These vectors are &lt;em&gt;precisely&lt;/em&gt; the generalized eigenvectors of $T$ associated to $\lambda=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what about arbitrary eigenvalues? Instead of thinking of an eigenvector associated to $\lambda$ as a vector on which $T$ acts by stretching/compressing, think of an eigenvector associated to $\lambda$ as an element of the nullspace of $T-\lambda I$. The generalized eigenvector associated to $\lambda$ are then the vectors that lie in $\mathbf{N}((T-\lambda I)^n)$ for some positive integer $n$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-12T21:09:56.267" CommentCount="3" />
  <row Id="4497" PostTypeId="1" CreationDate="2010-09-12T23:01:39.370" Score="2" ViewCount="140" Body="&lt;p&gt;What is a conditional double limit and how to compute it for $\lim_{(x,y)\rightarrow\infty,x\leq y}\frac{2x-1}{x-1}+\frac{x}{y}$?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-09-12T23:17:13.473" LastActivityDate="2010-09-13T22:22:43.270" CommunityOwnedDate="2010-09-13T22:26:13.870" Title="What is a conditional double limit and how to compute it?" Tags="&lt;classical-analysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="4498" PostTypeId="1" CreationDate="2010-09-12T23:11:20.007" Score="4" ViewCount="222" Body="&lt;p&gt;I take it there isn't a classification of finite complete groups yet. Has someone put together a classification of small complete groups? I.e. $S_4$, $\text{Aut}(G)$ for $G$ simple, $\text{Hol}(C_p)$ for $p$ odd prime are complete. The smallest complete group not of one of these forms is $H$ of order $\left|H\right|$, which generalises to a class $P$. The smallest complete group not of one of these forms is $J$, and so on, until we run out of ideas.&lt;/p&gt;&#xA;" OwnerUserId="1857" LastActivityDate="2010-10-01T12:43:32.170" Title="classification of small complete groups" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="4499" PostTypeId="2" ParentId="4480" CreationDate="2010-09-13T01:14:31.620" Score="8" ViewCount="0" Body="&lt;p&gt;Here's a nicer approach ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/fBUgX.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two segments emanating from a point and ending at points of tangency with a circle, we know that those segments are congruent. In the figure, the three vertices of the right triangle give rise to three fundamental lengths; note that, since the &quot;green&quot; angle is a right angle, the lengths of the tangent segments are equal to the radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pythagoras tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(b+r)^2 + ( r + a )^2 = ( a + b )^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we also have that $a+b=1$ (the side of the square), and that $b=a+2r$ (via tangent circles on the &quot;outside&quot; of the right triangle). From these relations, we find that $a=1/2-r$ and $b=1/2+r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1/2+2r)^2 + ( 1/2 )^2 = ( 1 )^2$$&#xA;$$1/4+2r+4r^2+1/4=1$$&#xA;$$8r^2+4r-1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots are $(-1\pm\sqrt{3})/4$, and we select the positive value: $r = (-1+\sqrt{3})/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Américo noted: The sides of the triangle have lengths $r+a=1/2$, $3r+a=\sqrt{3}/2$, and $1$, so that we have a 30-60-90 triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I like that there's but the one extraneous value this time, rather than the three in my first attempt. Is there yet a better approach that yields the answer directly, with &lt;em&gt;no&lt;/em&gt; extraneous values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit.&lt;/b&gt; There is. Immediately after we have deduced that $a=1/2-r$, we know that the short leg of the triangle has length $1/2$, so that its longer leg is $\sqrt{3}/2$. Since that longer leg is also $a+3r=1/2+2r$, we have that $r=(-1+\sqrt{3})/4$.)&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-09-13T01:25:43.373" LastActivityDate="2010-09-13T01:25:43.373" CommentCount="3" />
  <row Id="4500" PostTypeId="2" ParentId="4497" CreationDate="2010-09-13T01:42:47.590" Score="1" ViewCount="0" Body="&lt;p&gt;The limit depends on the exact way $x,y$ approach infinity. The first summand $(2x-1)/(x-1)$ actually tends to $2$ (always). The second summand can tend to any limit in $[0,1]$, or to no limit at all (in which case the limit of the entire sum does not exist). To get a limit of $0$, use $y = x^2$. To get a limit of $0 &amp;lt; a\leq 1$, use $x = ay$. To get no limit, you can use $x = y(sin(y)+2)/3$ [thanks AD!], as well as many other options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you could define $\lim \sup$ as the supremum of all limits attainable for some sequence $(x_n,y_n)$, and then $\lim \sup_{(x,y)\rightarrow\infty, x\leq y} x/y = 1$ (and similarly $\lim \inf_{(x,y)\rightarrow\infty, x\leq y} x/y = 0$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-09-13T07:27:34.767" LastActivityDate="2010-09-13T07:27:34.767" CommentCount="3" />
  <row Id="4501" PostTypeId="1" AcceptedAnswerId="4502" CreationDate="2010-09-13T04:09:54.390" Score="5" ViewCount="121" Body="&lt;p&gt;In a inner product space with inner product $\langle\ ,\ \rangle$ and real or complex line as its base field, for each point $x$ in the space, is $\langle x,-\rangle$ continuous function on the second argument, and is $\langle - ,x\rangle$ continous function on the first argument? &quot;Continuous&quot; is defined respect to the topology induced by the inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-13T04:46:29.767" LastActivityDate="2010-09-13T14:11:13.623" Title="Is Inner product continous when one arg is fixed?" Tags="&lt;inner-product-space&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="4502" PostTypeId="2" ParentId="4501" CreationDate="2010-09-13T04:50:42.973" Score="7" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix x in the inner product space, and let $f(y) = \langle y, x \rangle$ denote the inner product function.  Note that this is a linear functional -- that is, it is linear in y, and maps vectors to scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a well-known theorem that linear functionals are continuous (on the entire space) if and only if they are bounded.  Here, &quot;bounded&quot; means that there exists a constant M such that $|f(y)| \leq M|y|$ for all y in the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the inner product functional is bounded now follows from the Cauchy-Schwarz Inequality: $|f(y)| \leq |x||y|.$&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-09-13T04:50:42.973" CommentCount="1" />
  <row Id="4503" PostTypeId="2" ParentId="4501" CreationDate="2010-09-13T04:51:20.257" Score="5" ViewCount="0" Body="&lt;p&gt;If we define $T_x v = \langle v,x\rangle$ then $T_x$ is a linear functional. Since an inner product induces a norm, and thus it is continuous if and only if it is bounded on the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $|T_x v | = |\langle v,x\rangle | \le ||v||\cdot ||x||$ and for $v$ on the unit circle, i.e. $||v|| = 1$ we have that $T_x$ is indeed bounded and therefore continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is similar if you fix the left argument, however it may not be a linear functional if it is a complex vector space, but rather antilinear (that is linear up to conjugation). The norm is unaffected by that, though, so it's not a big step to overcome.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-13T05:12:15.050" LastActivityDate="2010-09-13T05:12:15.050" />
  <row Id="4504" PostTypeId="2" ParentId="4501" CreationDate="2010-09-13T04:51:59.490" Score="4" ViewCount="0" Body="&lt;p&gt;Fix $x$. We have $|\langle x,y\rangle - \langle x,z\rangle| = |\langle x,y-z\rangle|\leq ||x||||y-z||$ by the Cauchy-Schwartz inequality. This gives you easily that $\langle x,-\rangle\colon \mathbf{V}\to\mathbb{F}$ is not only continuous, but uniformly continuous. Similarly for $\langle -,x\rangle$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-13T14:11:13.623" LastActivityDate="2010-09-13T14:11:13.623" />
  <row Id="4505" PostTypeId="2" ParentId="4458" CreationDate="2010-09-13T04:53:16.880" Score="0" ViewCount="0" Body="&lt;p&gt;The uniqueness of gnomonic projections is a consequence of the uniqueness of line-preserving transformations of the plane.  If you have map projections $P_1$ and $P_2$ that take all arcs of great circles to segments of lines, then $P_1 \circ P_2^{-1}$ is a transformation of the plane (or of one part of the plane onto another) that sends line segments to line segments. [In case the $\circ$ notation is not clear, I mean the transformation that results from taking the target of $P_2$, un-projecting it back to the sphere, and projecting that pre-target from the sphere back to the plane using $P_1$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that, within the class of topology-preserving (continuous and one-to-one) transformations between connected regions in the plane, any collinearity-preserving transformation is the restriction of a single linear transformation.   If you allow points at infinity (for the gnomonic projection these are the projection of points on the equator of the sphere) then it is also true that any collinearity-preserving transformation is the restriction of a single projective transformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence for gnomonic projections is that if you consider the projection plane to be a rigid object, with an x-y coordinate system drawn on it, but that coordinatized plane can be moved around in space, then any map projection from a sphere (or parts of the sphere) to the plane that takes arcs of great circles to segments or lines, is a gnomonic projection onto &lt;em&gt;some&lt;/em&gt; position of the plane.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-13T04:53:16.880" CommentCount="2" />
  <row Id="4506" PostTypeId="1" AcceptedAnswerId="4509" CreationDate="2010-09-13T06:47:12.453" Score="1" ViewCount="166" Body="&lt;p&gt;We, know that there are infinitely many primes of the form $4n-1,4n+1,5n-1,\cdots, \text{etc}$. I saw these things in Apostol's &lt;em&gt;Introduction to Analytic Number theory&lt;/em&gt; textbook. I would like to have an argument working for $n^{2}$. The first expression which came to my mind was $4n^{2}+3$, which gives a prime for $n=1,2,3,4,5$. For $n=6$ it gives $147$ which is divisible by 3. For $n=7$ it gives $199$ which is again a prime. Then for $n=11$, it gives $487$ which is again a prime. I would like to know whether there are &lt;em&gt;infinitely many primes of the form $4n^{2}+3$&lt;/em&gt;? If yes, then a proof!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-13T06:52:13.287" LastActivityDate="2010-09-13T07:11:29.300" Title="Are there infinitely many primes of the form $4n^{2}+3$?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4507" PostTypeId="2" ParentId="4476" CreationDate="2010-09-13T06:48:26.437" Score="0" ViewCount="0" Body="&lt;p&gt;My Functional Analysis Fu has gotten bit weak lately, but I think the following should work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schauder fixed point theorem says, that a continuous function on a compact convex set in a topological vector space has a fixed point. Because of isometry, the Fourier transform maps the unit ball in $L^2$ to itself. Owing to the Banach Alaoglu theorem, the unit ball in $L^2$ is compact with respect to the weak topology. The Fourier transform is continuous in the weak topology, because if $( f_n, \phi ) \to (f, \phi)$ for all $\phi \in L^2$, then&#xA;$$&#xA;(\hat{f}_n, \phi) = (f_n, \hat{\phi}) \to (f, \hat{\phi}) = (\hat{f}, \phi).&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-09-13T06:48:26.437" CommentCount="5" />
  <row Id="4508" PostTypeId="2" ParentId="4467" CreationDate="2010-09-13T06:48:26.500" Score="3" ViewCount="0" Body="&lt;p&gt;To answer Pete's comment as to how to prove integral&#xA;closure of $\mathbb{Z}$ without using the UFD property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a/b$ be a rational ($a$, $b\in\mathbb{Z}$) which is&#xA;integral over $\mathbb{Z}$. Let $R=\mathbb{Z}[a/b]$. Then&#xA;$R$ is a finitely-generated $\mathbb{Z}$-module. It follows that&#xA;$b^n R\subseteq\mathbb{Z}$ for some $n$. We reduce to proving&#xA;the lemma that if $R$ is a ring with &#xA;$\mathbb{Z}\subseteq R\subseteq N^{-1}\mathbb{Z}$ for some&#xA;nonzero integer $N$ then $R=\mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various ways of proving this. For instance if&#xA;$R\ne\mathbb{Z}$ there is an element of $R$ strictly between two&#xA;consecutive integers (this is the division algorithm) and so&#xA;an element $x$ of $R$ strictly between $0$ and $1$. If $y$ is the least&#xA;such number then considering $y^2$ gives a contradiction.&#xA;Alternatively, $M=xN$ is an integer and&#xA;$R\subseteq M^{-1}\mathbb{Z}$ so we can always replace $N$ by a smaller integer etc.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-13T06:48:26.500" CommentCount="7" />
  <row Id="4509" PostTypeId="2" ParentId="4506" CreationDate="2010-09-13T07:11:29.300" Score="5" ViewCount="0" Body="&lt;p&gt;Many people &quot;would like to have an argument working for $n^2$&quot;, but what is available at the moment (and for the last two centuries) are conjectures.  For any list of integer polynomials there is a conjecture on how often all polynomials on the list are prime:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is extremely hard to prove that any natural set of integers of density 0 contains infinitely many primes.  It is known for the set of values of $x^2 + y^4$ but not for the values of any single-variable polynomial of degree higher than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymptotic formula in the Bateman-Horn conjecture isn't necessarily the most general expression of what people in the field believe to be true (and it is probably a lot older than Bateman and Horn's article that formally codified it), but it does subsume many earlier conjectures on primes of the form $n^2+1$, prime twins and k-tuplets, Schinzel's Hypothesis and Buniakowsy's conjecture.  You can calculate from the formula the predicted frequency of $n$ such that $4n^2 + 3$ is prime.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-13T07:11:29.300" CommentCount="2" />
  <row Id="4510" PostTypeId="1" AcceptedAnswerId="4566" CreationDate="2010-09-13T09:02:07.990" Score="6" ViewCount="271" Body="&lt;p&gt;In many combinatorial enumeration problems it is possible to find a rational generating function (i.e. the quotient of two polynomials) for the sequence in question. The question is - given the generating function, how can we find (algorithmically) the values of the sequence, i.e. the coefficients of the corresponding power series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for a rational generating function, the sequence satisfies a recurrence relation given by the coefficients of the polynomial in the denominator, so it's really just the question of finding the finite initial values.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T18:21:33.887" Title="Finding the power series of a rational function" Tags="&lt;real-analysis&gt;&lt;combinatorics&gt;&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="4511" PostTypeId="1" AcceptedAnswerId="4519" CreationDate="2010-09-13T10:23:05.620" Score="1" ViewCount="90" Body="&lt;p&gt;I found the following in the wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_class#Squares_to_top_Pontryagin_class&quot; rel=&quot;nofollow&quot;&gt;Euler class&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If the rank r is even, then this cohomology class $e(E) \cup e(E)$ equals the top Pontryagin class $p_{r/2}(E)$.&#xA;Under the splitting principle, this corresponds to the square of the Vandermonde polynomial equaling the discriminant: the Euler class corresponds to the Vandermonde polynomial, the basic alternating polynomial, while the top Pontryagin class corresponds to the discriminant, a symmetric polynomial.&#xA;More formally, the Euler class of a direct sum of line bundles is the Vandermonde polynomial (orientation determines the order of the line bundles up to sign), while top Pontryagin class is the discriminant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody clarify me this? In particular I don't understand the sense of &quot;the Euler class of a direct sum of line bundles is the Vandermonde polynomial&quot;.  I checked the references but there is nothing about that.&lt;/p&gt;&#xA;" OwnerUserId="1860" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-13T10:27:50.697" LastActivityDate="2010-09-13T11:26:56.987" Title="euler class and vandermonde polynomial" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4512" PostTypeId="1" AcceptedAnswerId="4549" CreationDate="2010-09-13T10:23:45.590" Score="3" ViewCount="69" Body="&lt;p&gt;I came across the following theorem that for any class $K$ of truth assignments, $K\subseteq Mod(Th(K))$, and if $K$ is axiomatizable, then $K=Mod(Th(K))$. Together these seem to me to imply that there is some class $K$ such that $K\neq Mod(Th(K))$, however, I can't find any class $K$ that would verify my assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know that $K$ would have to contain more than one truth assignment, since if $K$ were to have only one assignment $V$, $Th(V)$ has a unique model $V$, so then $K=V=Mod(Th(K))$. Does anyone know of any example where $K\neq Mod(Th(K))$, or at least how to construct one?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-13T17:03:52.947" Title="Is it possible for $Mod(Th(K))\neq K$ for some class $K$ of truth assignments?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4513" PostTypeId="1" CreationDate="2010-09-13T10:44:29.463" Score="0" ViewCount="177" Body="&lt;p&gt;I would like to find the center of arc from the starting point, the endpoint, and i,j values&lt;/p&gt;&#xA;" OwnerUserId="1861" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-13T11:08:07.073" LastActivityDate="2010-09-13T11:08:07.073" Title="How to calculate center of arc from startpoint, endpoint and i,j values?" Tags="&lt;geometry&gt;&lt;algorithms&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="4514" PostTypeId="2" ParentId="4497" CreationDate="2010-09-13T10:48:59.640" Score="1" ViewCount="0" Body="&lt;p&gt;Just adding some extra comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First $$\frac{2x-1}{x-1}=\frac{2-1/x}{1-1/x}\to\frac{2-0}{1-0}=2$$&#xA;as $x\to\infty$ (this is independent of how the limit is taken).&#xA;Secondly, we note that&#xA;$$0\leq\frac{x}{y}\leq1$$&#xA;the first inequality comes from that both $x,y$ are $&gt;0$ (since they should tend to $\infty$), the second comes from $x\leq y$ (divide by $y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we have &#xA;$$2\leq\liminf_{(x,y)\to\infty,x\leq y} \frac{2x-1}{x-1} + \frac{x}{y}\leq\limsup_{(x,y)\to\infty,x\leq y} \frac{2x-1}{x-1} + \frac{x}{y}\leq3$$&#xA;At the boundary of the limit, that is when $x=y$ we would get 3. To reach the liminf we only have to find a way to press $x/y$ as low as we can (recall that we already know it is $x/y\ge0$) - at this point you have do some testing, you want $x$ to tend slower to infinity than $y$, $x=\log y$ or $x=\sqrt{y}$ works fine! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the linear choice $x=ay$, earlier suggested by Yuval Filmus above works fine to see that all limits between 2 and 3 are possible. The example in the the comment of Yuval Filmus shows that there are choices of $x$'s such that no limit is attained.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-13T10:48:59.640" CommentCount="0" />
  <row Id="4515" PostTypeId="1" AcceptedAnswerId="4516" CreationDate="2010-09-13T10:51:46.270" Score="4" ViewCount="523" Body="&lt;p&gt;I've been using the sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a series converges then the limit of the sequence is zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as a criterion to prove that a series diverges (when lim != 0) and I can understand the rationale behind it, but I can't find a &lt;strong&gt;formal proof&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2010-09-13T12:24:41.803" Title="Proof: Series converges =&gt; the limit of the sequence is zero" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4516" PostTypeId="2" ParentId="4515" CreationDate="2010-09-13T10:57:18.580" Score="5" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left ( \sum_{k = 1}^{n + 1} a_k - \sum_{k = 1}^{n} a_k \right ) = \lim_{n \to \infty} a_{n + 1} $$&#xA;And both sums will converge to the same number so the limit is zero. This is by far the easiest proof I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the Cauchy criterion in disguise by the way, so you could use that too.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-13T12:24:41.803" LastActivityDate="2010-09-13T12:24:41.803" CommentCount="6" />
  <row Id="4518" PostTypeId="2" ParentId="4515" CreationDate="2010-09-13T11:16:13.927" Score="2" ViewCount="0" Body="&lt;p&gt;If we know that the sequence converges and merely wish to show it converges to zero, then a proof by contradiction gives a little more intuition here (although the direct proofs are simple and beautiful). Assume $a_n\to a$ with $a&gt;0$, then for all $n&gt;N$ for some large enough $N$ we have $a_n &gt; a/2$ (take $\varepsilon = a/2$ in the definition of the limit). Now the sum diverges: $\sum_{n&gt;N}a_n &gt; \sum_{n&gt;N}a/2 = \infty$. A similar argument works when $a&amp;lt;0$.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-13T11:53:11.037" LastActivityDate="2010-09-13T11:53:11.037" CommentCount="2" />
  <row Id="4519" PostTypeId="2" ParentId="4511" CreationDate="2010-09-13T11:26:56.987" Score="1" ViewCount="0" Body="&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chern%E2%80%93Weil_homomorphism&quot; rel=&quot;nofollow&quot;&gt;Chern-Weil theory&lt;/a&gt; (see for example these &lt;a href=&quot;http://data.imf.au.dk/publications/ln/2003/imf-ln-2003-69.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; by Johan Dupont) , characteristic classes such as the Euler and Pontryagin classes correspond to invariant polynomials (under the bundle structure group) of the curvature form of the vector bundle. Basically, what the &lt;a href=&quot;http://en.wikipedia.org/wiki/Splitting_principle&quot; rel=&quot;nofollow&quot;&gt;splitting principle&lt;/a&gt; tells us is that these polynomials can be &quot;formally&quot; written in terms of pullbacks of curvature forms of line bundles which correspond to the &quot;eigenvalues&quot; of the curvature form (viewed as an antisymmetric endomorphism).The invariant polynomial corresponding to the Euler class is the Vandermonede polynomial. For a given rank of the vector bundle, one can always write these polynomials in terms of the vector bundle curvature form and the use of the line bundle curvature forms is for simplifying the algebraic work.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-09-13T11:26:56.987" CommentCount="2" />
  <row Id="4520" PostTypeId="1" AcceptedAnswerId="4522" CreationDate="2010-09-13T12:11:29.667" Score="4" ViewCount="62" Body="&lt;p&gt;Let $X$ be a contractible space. If $x_0 \in X$, it is not necessarily true that the pointed space $(X,x_0)$ is contractible (i.e., it is possible that any contracting homotopy will move $x_0$).  An example is given in 1.4 of Spanier: the comb space.  However, this space is contractible as a pointed space if the basepoint is in the bottom line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a contractible space which is not contractible as a pointed space for &lt;em&gt;any&lt;/em&gt; choice of basepoint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this will have to be some kind of pathological space, because for CW complexes, we have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Whitehead_theorem&quot; rel=&quot;nofollow&quot;&gt;Whitehead theorem&lt;/a&gt;. (So I'm not completely sure that the Whitehead theorem is actually a statement about the &lt;em&gt;pointed&lt;/em&gt; homotopy category, but hopefully I'm right.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-13T12:18:37.577" LastActivityDate="2010-09-13T12:21:19.947" Title="Does contractibility imply contractibility with a basepoint?" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4521" PostTypeId="1" AcceptedAnswerId="4523" CreationDate="2010-09-13T12:16:08.380" Score="2" ViewCount="205" Body="&lt;p&gt;$$\arccos\left(\frac{1-x^2}{1+x^2}\right) = 2\arctan{x}$$ for $x \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not even sure what kind of math to try? :(&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-13T12:52:59.240" LastActivityDate="2010-10-04T18:26:36.017" Title="How can I prove this trigonometric equality?" Tags="&lt;trigonometry&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="4522" PostTypeId="2" ParentId="4520" CreationDate="2010-09-13T12:21:19.947" Score="3" ViewCount="0" Body="&lt;p&gt;Yes.  See exercise 7 &lt;a href=&quot;http://books.google.com/books?id=tru-TWQFDAIC&amp;amp;lpg=PP1&amp;amp;dq=using%20the%20borsuk%20ulam%20theorem&amp;amp;pg=PA7#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-09-13T12:21:19.947" CommentCount="1" />
  <row Id="4523" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T12:23:52.773" Score="5" ViewCount="0" Body="&lt;p&gt;Put $t=\arctan x$. Then $x=\tan t$.&#xA;You have to prove that $2t=\arccos(1-x^2)/(1+x^2)$.&#xA;This is more-or-less the same as $(1-x^2)/(1+x^2)=\cos2t$. If you put&#xA;$x=\tan t$ into $(1-x^2)/(1+x^2)$ what do you get?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-13T12:23:52.773" />
  <row Id="4524" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T12:24:55.407" Score="1" ViewCount="0" Body="&lt;p&gt;One way would be to use the complex logarithm representations of the inverse trigonometric functions. For this answer, I'll be more pedestrian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the cosines of both sides, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\left(\arccos\left(\frac{1-x^2}{1+x^2}\right)\right)=\cos\left(2\arctan\left(x\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\cos\left(2\arctan\left(x\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use the double angle formula,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=2\cos^2\left(\arctan\left(x\right)\right)-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;switch to secants&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\frac{2}{\mathrm{sec}^2\left(\arctan(x)\right)}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use a Pythagorean identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\frac{2}{1+\tan^2\left(\arctan(x)\right)}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now you have a function and its inverse composed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-x^2}{1+x^2}=\frac{2}{1+x^2}-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this last one is a definite identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as already noted, the left side (of the original identity) is even; the right side odd. We check that the two expressions agree at $x=0$, and both have the same limiting behavior as $x\to\infty$. The identity is thus only valid for nonnegative $x$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-13T21:04:30.483" LastActivityDate="2010-09-13T21:04:30.483" CommentCount="3" />
  <row Id="4525" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T12:50:21.960" Score="5" ViewCount="0" Body="&lt;p&gt;Setting $x=\tan a$ your equality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arccos \dfrac{1-x^{2}}{1+x^{2}}=2\arctan x\qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos 2a=\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}.\qquad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove (2) which is &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_trigonometric_identities&quot; rel=&quot;nofollow&quot;&gt;listed in the wikipedia&lt;/a&gt; we can take the duplication formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos 2a=\cos ^{2}a-\sin ^{2}a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $\cos ^{2}a+\sin ^{2}a=1$, we establish&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos 2a=\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a+\sin ^{2}a}=\dfrac{1-\tan&#xA;^{2}a}{1+\tan ^{2}a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; One must note that $\arctan x$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_and_odd_functions&quot; rel=&quot;nofollow&quot;&gt;odd function&lt;/a&gt;, while $\arccos \frac{1-x^{2}}{1+x^{2}}$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_and_odd_functions&quot; rel=&quot;nofollow&quot;&gt;even&lt;/a&gt; one. Only for $x\geq 0$ are both sides of $(1)$ equal. Identity $(2)$ is valid for both positive and negative values of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UZHEW.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blue: $y=\arccos((1-x^2)/(1+x^2))$ for $x\ge 0$;  Red: $y=2\arctan x$ &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T18:26:36.017" LastActivityDate="2010-10-04T18:26:36.017" CommentCount="3" />
  <row Id="4526" PostTypeId="1" AcceptedAnswerId="4538" CreationDate="2010-09-13T13:07:23.073" Score="9" ViewCount="276" Body="&lt;p&gt;I would like to collect a list of mathematical concepts that have been named after mathematicians, which are now used in lowercase form (such as &quot;abelian&quot;).  This question is partly motivated by my former supervisor, who joked (something like):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You know you've made it as a mathematician when they start using your name in lowercase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;em&gt;community wiki&lt;/em&gt;, please:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One mathematician per answer.  Please update already-existing answers if that mathematician already exists.&lt;/li&gt;&#xA;&lt;li&gt;Please give a brief description of the concept and the person (ideally with links to wikipedia).&lt;/li&gt;&#xA;&lt;li&gt;This is not about whether or not you agree with the capitalisation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Expect your answer to be edited otherwise.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-13T14:22:07.607" LastActivityDate="2010-09-13T20:45:24.727" CommunityOwnedDate="2010-09-13T13:07:23.073" Title="Mathematical concepts named after mathematicians that have become acceptable to spell in lowercase form (e.g. abelian)?" Tags="&lt;big-list&gt;&lt;terminology&gt;&lt;mathematicians&gt;" AnswerCount="12" CommentCount="5" />
  <row Id="4527" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:13:48.567" Score="6" ViewCount="0" Body="&lt;p&gt;I've seen &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;jacobian&lt;/a&gt;&quot; in lower case in a few books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the uninitiated: it refers to either a matrix of partial derivatives, or its determinant. This is named after &lt;a href=&quot;http://en.wikipedia.org/wiki/Carl_Gustav_Jacob_Jacobi&quot; rel=&quot;nofollow&quot;&gt;Carl Gustav Jacob Jacobi&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources using lowercase &quot;jacobi&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Claus Bendtsen, Preliminary Experiences with Extrapolation Methods for Parallel Solution of Differential, in Parallel scientific computing, 1994.&lt;/li&gt;&#xA;&lt;li&gt;Foundations of Modern Analysis, By J. Dieudonne.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="498" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:36:20.377" LastActivityDate="2010-09-13T14:36:20.377" CommunityOwnedDate="2010-09-13T13:13:48.567" />
  <row Id="4528" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:18:54.437" Score="9" ViewCount="0" Body="&lt;p&gt;For something even more common, &quot;boolean&quot; (e.g. &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_data_type&quot; rel=&quot;nofollow&quot;&gt;boolean type&lt;/a&gt;&quot;, &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28structure%29&quot; rel=&quot;nofollow&quot;&gt;boolean algebra&lt;/a&gt;&quot;) is increasingly not being capitalized as well. &lt;a href=&quot;http://en.wikipedia.org/wiki/George_Boole&quot; rel=&quot;nofollow&quot;&gt;George Boole&lt;/a&gt; formalized the manipulation of logical statements, and thus his name crops up frequently in connection to logic.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T13:39:33.170" LastActivityDate="2010-09-13T13:39:33.170" CommunityOwnedDate="2010-09-13T13:18:54.437" />
  <row Id="4529" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:48:53.747" Score="1" ViewCount="0" Body="&lt;p&gt;Noetherian rings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noetherian_ring&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Noetherian_ring&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in honor of Emmy Noether:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Emmy_Noether&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Emmy_Noether&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lectures on modules and rings, By Tsit-Yuen Lam&lt;/li&gt;&#xA;&lt;li&gt;Commutative algebra, By Oscar Zariski, Pierre Samuel, Irvin Sol Cohen&lt;/li&gt;&#xA;&lt;li&gt;Model theory and modules, By Mike Prest&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(plus many others)&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:41:29.983" LastActivityDate="2010-09-13T14:41:29.983" CommunityOwnedDate="2010-09-13T13:48:53.747" CommentCount="1" />
  <row Id="4530" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:50:56.303" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid&quot; rel=&quot;nofollow&quot;&gt;Euclid&lt;/a&gt;'s name has been attached to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_space&quot; rel=&quot;nofollow&quot;&gt;euclidean space&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;euclidean algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:08:33.927" LastActivityDate="2010-09-13T14:08:33.927" CommunityOwnedDate="2010-09-13T13:50:56.303" CommentCount="1" />
  <row Id="4531" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:51:39.810" Score="8" ViewCount="0" Body="&lt;p&gt;&lt;code&gt;cartesian&lt;/code&gt;. Many things are described as cartesian, after &lt;a href=&quot;http://en.wikipedia.org/wiki/Ren%C3%A9_Descartes&quot; rel=&quot;nofollow&quot;&gt;René Descartes&lt;/a&gt;. In philosophy things attributed to Descartes tend to keep the capital when described as &quot;Cartesian&quot;, but in mathematics, I've seen &quot;cartesian product&quot; lower case a few times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's weird is that this isn't more prevalent, given that Descartes is older than Abel by almost 200 years, and I conjecture that cartesian products, cartesian planes and the like come up at least as much as abelian groups... And since the adjectival form of &quot;Descartes&quot; isn't just the name+ending, it's strange how common the capitalised form has remained...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google books reveals that lowercase &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cartesian_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;cartesian plane&lt;/a&gt;&quot; was used in the following books (for example):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Algebra Through Practice: Volume 2, Matrices and Vector Spaces,  By T. S. Blyth, E. F. Robertson&lt;/li&gt;&#xA;&lt;li&gt;Cracking the MCAT with CD-ROM, By James L. Flowers&lt;/li&gt;&#xA;&lt;li&gt;Symmetry And Spectroscopy Of Molecules, By K Veera Reddy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="172" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:26:12.987" LastActivityDate="2010-09-13T14:26:12.987" CommunityOwnedDate="2010-09-13T13:51:39.810" />
  <row Id="4532" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:52:31.240" Score="6" ViewCount="0" Body="&lt;p&gt;Many things have been named gaussian after &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss&quot; rel=&quot;nofollow&quot;&gt;Carl Friedrick Gauss&lt;/a&gt;: &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;gaussian distribution&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_integer&quot; rel=&quot;nofollow&quot;&gt;gaussian integers&lt;/a&gt;, etc.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:28:37.853" LastActivityDate="2010-09-13T14:28:37.853" CommunityOwnedDate="2010-09-13T13:52:31.240" />
  <row Id="4533" PostTypeId="1" AcceptedAnswerId="4535" CreationDate="2010-09-13T13:56:32.420" Score="1" ViewCount="125" Body="&lt;p&gt;I've got two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1). I would like to know if by definition the Sobolev space norm is defined only by differentials with respect to one variable i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel \; = \; \left[ \int_{\mathbb{R}} \left( \left| f \right|^p \; + \; \left| \frac{\partial f}{\partial x} \right|^p \; + \ldots + \; \left| \frac{\partial^m f}{\partial x^m} \right|^p \right) dx \right]^{\frac{1}{p}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the corresponding Sobolev space norm be if $f$ were a function of two variables e.g. $f(x,y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2). If the Sobolev space norm exists for multivariate functions, is it correct to say that the Beppo-Levi space norm a special case of the Sobolev space norm? From the little I've read, it appears that the Beppo-Levi space norm is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel \; = \; \int_{\mathbb{R}^2} \sum_{i = 0}^m {m \choose i} \left( \frac{\partial^m f}{\partial x^i \partial y^{m - i}} \right)^2 dx dy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the partial derivative is raised to the power of 3. Would this still be a Beppo-Levi space norm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: based on Robin's answer and the paper &quot;Spline functions and stochastic filtering&quot; by Christine Thomas-Agnan, here's my understanding of the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sobolev norm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel^p \; = \; \sum_{ 0 \: \le \: ( \: \mid \alpha \mid \: = \: \alpha_1 \: + \: \cdots \: + \: \alpha_n \: ) \: \le \: p}\int_{\mathbb{R}^n} \; \left| \frac{\partial^{\mid \alpha \mid} f}{\partial x_1^{\alpha_1} \: \cdots \: \partial x_n^{\alpha_n}} \right|^p \: dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beppo-Levi norm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\parallel f \parallel^m \; = \; \int_{\mathbb{R}^n} \; \sum_{\alpha_i \: + \: \cdots \: + \: \alpha_n \: = \: m } {m \choose {\alpha_i! \; \cdots \; \alpha_n!}} \left| \frac{\partial^m f(x)}{\partial x_1^{\alpha_1} \cdots \partial x_n^{\alpha_n} } \right|^2 dx$&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-09-15T20:35:01.523" LastActivityDate="2010-09-15T20:35:01.523" Title="Sobolev Space norm and Beppo-Levi Space norm" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4534" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T13:58:53.970" Score="2" ViewCount="0" Body="&lt;p&gt;The algebraic structure formed by a set with a commutative binary operation is described as &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_group&quot; rel=&quot;nofollow&quot;&gt;abelian&lt;/a&gt;.  If G is a group, then the quotient group G/[G,G] (where [G,G] is the commutator subgroup) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelianization#Abelianization&quot; rel=&quot;nofollow&quot;&gt;abelianization&lt;/a&gt; of G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Niels_Henrik_Abel&quot; rel=&quot;nofollow&quot;&gt;Niels Henrik Abel&lt;/a&gt; wikipedia page suggests that he pretty much invented group theory.  [[citation needed]]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-09-13T13:58:53.970" CommunityOwnedDate="2010-09-13T13:58:53.970" />
  <row Id="4535" PostTypeId="2" ParentId="4533" CreationDate="2010-09-13T14:06:04.017" Score="2" ViewCount="0" Body="&lt;p&gt;The Sobolev norm is also defined for regions in $\mathbb{R}^n$. You would typically&#xA;have&#xA;$$\|f\|^p=\sum_{(i_1,\ldots,i_n)}\int\left|&#xA;\frac{\partial^I f}{\partial x_1^{i_1}\cdots\partial x_n^{i_n}}&#xA;\right|^pd\mathbf{x}$$&#xA;where $I=i_1+\cdots+i_n$ and the sum is over all tuples with $0\le I\le m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-13T14:06:04.017" CommentCount="7" />
  <row Id="4536" PostTypeId="1" AcceptedAnswerId="4564" CreationDate="2010-09-13T14:37:53.260" Score="9" ViewCount="209" Body="&lt;p&gt;I'm looking at the random sequence $x_n$ with $x_0=x_1=1$ and &#xA;\begin{equation}&#xA;x_{n+1}=2x_n\pm x_{n-1}&#xA;\end{equation}&#xA;where we choose the $\pm$ sign independently with equal probability. Now considering the recurrence relations $x_{n+1}=2x_n+x_{n-1}$ and $x_{n+1}=2x_n-x_{n-1}$ separately, it is clear that $x_n\rightarrow\infty$ as $n\rightarrow\infty$. However, the sequence $x_n^{1/n}$ seems to tend to a limit near 1.91 (got this from numerical computation and some brute force by a Monte Carlo simulating the $\pm$ sign). Thus the sequence $x_n^{1/n}$ seems to convenge almost surely. I was wondering if anyone could show that the sequence was indeed almost surely converging and/or work out the limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;br&gt;&#xA;This comment shows that $\lim_{n\rightarrow\infty} |x_n|^{1/n}$ converges almost surely. Let $y_n=\frac{x_n}{2^n}$ then $2^{n+1}y_{n+1}=2^{n+1}y_n\pm 2^{n-1}y_{n-1}$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_{n+1}=y_n\pm \frac{1}{4}y_{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Embree-Trefethen showed that $\lim_{n\rightarrow\infty} |y_n|^{1/n}$ converges almost surely. See Embree, M.; Trefethen, L. N. (1999), &quot;Growth and decay of random Fibonacci sequences&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, finding the almost sure limit accurately or analytically is proving difficult at the moment. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-17T05:45:13.823" LastActivityDate="2010-09-21T15:30:09.140" Title="Solving randomized recurrence relation" Tags="&lt;probability&gt;&lt;approximation&gt;&lt;recursive&gt;" AnswerCount="3" />
  <row Id="4537" PostTypeId="1" AcceptedAnswerId="4547" CreationDate="2010-09-13T14:39:25.920" Score="1" ViewCount="210" Body="&lt;p&gt;In India, generally during the graduate years, we follow a course work pattern, unlike many places in the U.S where students are exposed to research during their undergraduate years itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As, a graduate student, we have basic courses like, Algebra, Analysis, Rings and Modules, Measure theory, Topology, functional analysis etc.. Generally topology is one subject, which i don't find that much of interest. But in some universities, students are forced to take Topology, and Algebraic topology, during their graduate years, and i have seen many students facing trouble, as they have to study a subject which is not of their interests. My question, would be, for a student whose research are in Analytic and Algebraic Number theory, does he needs to know Algebraic Topology?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-13T16:13:09.910" CommunityOwnedDate="2010-09-13T14:39:25.920" Title="Algebraic Topology Need!" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="4538" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T14:40:26.240" Score="5" ViewCount="0" Body="&lt;p&gt;Like &quot;cartesian&quot;, &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Algorithm&quot; rel=&quot;nofollow&quot;&gt;algorithm&lt;/a&gt;&quot; is not only a lowercased but a derivative form of the name of a mathematician (&lt;a href=&quot;http://en.wikipedia.org/wiki/Al-Khw%C4%81rizm%C4%AB&quot; rel=&quot;nofollow&quot;&gt;al-Khwarismi&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1867" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-13T14:44:11.460" LastActivityDate="2010-09-13T14:44:11.460" CommunityOwnedDate="2010-09-13T14:40:26.240" CommentCount="2" />
  <row Id="4539" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T14:41:14.677" Score="1" ViewCount="0" Body="&lt;p&gt;I have found the term &quot;karoubianization&quot; in the literature (although google only finds for me instances where it is capitalized...). It is a synonym for &quot;pseudo-abelianization&quot; or &quot;idempotent-completion&quot;, and refers to the process whereby one enlarges a category universally so that all its idempotents split. The word refers to Max Karoubi.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-13T14:48:46.953" LastActivityDate="2010-09-13T14:48:46.953" CommunityOwnedDate="2010-09-13T14:41:14.677" CommentCount="1" />
  <row Id="4541" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T15:29:58.107" Score="3" ViewCount="0" Body="&lt;p&gt;David Rusin compiled &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/98/MSC.names&quot; rel=&quot;nofollow&quot;&gt;a list of mathematicians whose names appear in the 1991 MSC&lt;/a&gt;. (That's the &quot;mathematics subject classification&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 357. He claims Abel's name is the only one among these which appears noncapitalized in the MSC. The other people who have been named in this question (Noether, Descartes, Euclid, Karoubi, Gauss, Boole, Jacobi) all appear in the MSC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the 1991 MSC is nearly twenty years old, and is likely to reflect more-formal-than-average usage.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-09-13T15:29:58.107" CommunityOwnedDate="2010-09-13T15:29:58.107" CommentCount="1" />
  <row Id="4542" PostTypeId="2" ParentId="4521" CreationDate="2010-09-13T15:32:46.893" Score="2" ViewCount="0" Body="&lt;p&gt;Differentiate the difference $$\arccos\left(\frac{1-x^2}{1+x^2}\right) - 2\arctan{x},$$ simplify the resulting expression and find $0$ when $x&gt;0$, concluding that the difference is constant. Now evaluate at $1$, to see that the difference is in fact constantly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N.B.:&lt;/strong&gt; This is of course quite unenlightenling, but you always know that if two things are equal, their difference is constant so you never lose much by trying to show that it is constant and zero. Specially if you have a computer algebra system at hand that can do the derivatives and simplifications for you!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-13T15:46:10.360" LastActivityDate="2010-09-13T15:46:10.360" />
  <row Id="4544" PostTypeId="1" AcceptedAnswerId="4548" CreationDate="2010-09-13T15:53:18.607" Score="10" ViewCount="446" Body="&lt;p&gt;Hi,&lt;br&gt;&#xA;the fact that Ramanujan's Constant $e^{\pi \sqrt{163}}$ is almost integer ($262 537 412 640 768 743.99999999999925...$) doesn't seem to be a coincidence, but has to do with the $163$ appearing in it. Can you explain why it's almost-but-not-quite integer in layman's terms (I'm not a mathematician)?&lt;br&gt;&#xA;TIA&lt;br&gt;&#xA;Steven&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-13T16:23:36.007" Title="Why is $e^{\pi \sqrt{163}}$ almost integer?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="4545" PostTypeId="2" ParentId="4544" CreationDate="2010-09-13T16:05:10.260" Score="0" ViewCount="0" Body="&lt;p&gt;As QY pointed out this cannot be explained in layman terms: Here are the papers which you would like to see: &lt;a href=&quot;http://www.isibang.ac.in/~sury/ramanujanday.pdf&quot; rel=&quot;nofollow&quot;&gt;www.isibang.ac.in/~sury/ramanujanday.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.isibang.ac.in/~sury/episq163.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.isibang.ac.in/~sury/episq163.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-13T16:05:10.260" />
  <row Id="4546" PostTypeId="2" ParentId="4544" CreationDate="2010-09-13T16:05:34.727" Score="15" ViewCount="0" Body="&lt;p&gt;This is quite a challenge to express in &quot;layman's terms&quot;, but the&#xA;reason is that&#xA;$$j\left(\frac{1+\sqrt{-163}}{2}\right)$$&#xA;is an integer where $j$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/J-invariant&quot; rel=&quot;nofollow&quot;&gt;$j$-function&lt;/a&gt;. When you substitute&#xA;$(1+\sqrt{-163})/2$ into the $q$-expansion (see the wikipedia page)&#xA;of $j$, all terms save the first two are small, and the first two equal&#xA;$$-\exp(\pi\sqrt{163})+744.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that this $j$-value is an integer is due to the quadratic&#xA;field $\mathbb{Q}(\sqrt{-163})$ having&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Class_number_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;class number&lt;/a&gt; one, or equivalently that all positive-definite&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_quadratic_form&quot; rel=&quot;nofollow&quot;&gt;integer binary quadratic forms&lt;/a&gt; of discriminant $-163$ are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;I'll try to explain the connection with binary quadratic forms. Consider&#xA;a quadratic form&#xA;$$Q(x,y)=ax^2+bxy+cy^2$$&#xA;with $a$, $b$ and $c$ integers. I'll only consider forms $Q$ which are&#xA;&lt;em&gt;primitive&lt;/em&gt;, so that $a$, $b$ and $c$ have no common factor $ &gt; 1$,&#xA;and &lt;em&gt;positive-definite&lt;/em&gt;, that is $a &gt; 0$ and the &lt;em&gt;discriminant&lt;/em&gt;&#xA;$D=b^2-4ac &amp;lt; 0$. There is a notion of equivalence of quadratic forms,&#xA;and two primitive positive-definite forms $Q$ and $Q'(x,y)=a'x^2+b'xy+c'y^2$&#xA;(necessarily also of discriminant $D$) are equivalent if and only if&#xA;$$j\left(\frac{b+\sqrt{-D}}{2a}\right)&#xA;=j\left(\frac{b'+\sqrt{-D}}{2a'}\right).$$&#xA;For each possible discriminant there are only finitely many equivalence&#xA;classes. Thus we get a finite set of $j$-values for each discriminant, and&#xA;the big theorem is that they are the solutions of a monic algebraic equation&#xA;with integer coefficients. When there is only one class the equation has the&#xA;form $x-k=0$ where $x$ is an integer, and the $j$-value must be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommended reference for this is David Cox's book&#xA;&lt;a href=&quot;http://books.google.co.uk/books?id=pSMlAQAAIAAJ&amp;amp;q=cox+primes&amp;amp;dq=cox+primes&amp;amp;hl=en&amp;amp;ei=0E2OTIKIHZOCswb3ysmAAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;Primes of the form $x^2+ny^2$&lt;/a&gt;. But these results appear towards&#xA;the end of this 350-page book.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-13T16:23:36.007" LastActivityDate="2010-09-13T16:23:36.007" CommentCount="1" />
  <row Id="4547" PostTypeId="2" ParentId="4537" CreationDate="2010-09-13T16:13:09.910" Score="3" ViewCount="0" Body="&lt;p&gt;Well, I'm not an expert in Number Theory and I don't know if your interests may include in the future things like &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_conjectures&quot; rel=&quot;nofollow&quot;&gt;Weil conjectures&lt;/a&gt; and $\zeta$-functions for algebraic varieties, or on the contrary, you'll try to avoid any contact with Algebraic Geometry and Homological Algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, just in case, your interests lead you towards the first issues, then you'll have a nasty encounter with something called $\ell$-adic and étale cohomologies. I wouldn't like to be in your shoes at that moment, without having seen before any other simpler cohomology (as the singular cohomology you're going to learn in Algebraic Topology) and the classical Lefchstez fixed-point theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, except you're not going to use Algebraic Geometry at all in your research, nor Homological Algebra, or your use of the first is limited to the most classical aspects, you'll have to be familiar with sheaf cohomology and derived functors such as $\mathrm{Ext}$ and $\mathrm{Tor}$. It isn't impossible to learn sheaf cohomology without knowing a word of singular cohomology, and the same applies for derived functors, but you'll clearly have a tremendous gap at that point.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-13T16:13:09.910" CommunityOwnedDate="2010-09-13T16:13:09.910" />
  <row Id="4548" PostTypeId="2" ParentId="4544" CreationDate="2010-09-13T16:14:10.200" Score="12" ViewCount="0" Body="&lt;p&gt;I do not think &quot;why&quot; has a reasonable layman's-terms answer, but let me at least explain &quot;how&quot; with a simpler example of such a numerical coincidence.  If one takes powers of the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ it is not hard to see that they are close to integers.  For example, $\phi^{20} = 15126.999934...$.  One might ask an analogous question about why these numbers are close to integers.  The answer is that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi^n + \varphi^n = L_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L_n$ is the $n^{th}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas_number&quot; rel=&quot;nofollow&quot;&gt;Lucas number&lt;/a&gt; (an integer), and where $\varphi = \frac{1 - \sqrt{5}}{2}$ has absolute value less than $1$.  As $n$ gets larger, $\varphi^n$ gets smaller, so $\phi^n$ becomes a better and better approximation to the integer $L_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar, but much more complicated, phenomenon is happening here.  The reason $e^{\pi \sqrt{163}}$ is so close to an integer is that it, plus a small error term, gives a special formula for that integer.  But &quot;why&quot; this formula exists is a rather &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_multiplication&quot; rel=&quot;nofollow&quot;&gt;long and complicated story&lt;/a&gt; (as Robin Chapman hints at) and I do not think there is any reasonable way to talk about it in layman's terms.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-13T16:14:10.200" CommentCount="2" />
  <row Id="4549" PostTypeId="2" ParentId="4512" CreationDate="2010-09-13T17:03:52.947" Score="3" ViewCount="0" Body="&lt;p&gt;Your use of the term &quot;truth assignments&quot; indicates that you are asking this question in the context of propositional logic, rather than the context of model theory and first order logic, where it often arises. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is Yes, there are $K$ for which $K\neq Mod(Th(K))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the propositional logic context, we have a collection $K$ of truth assignments of a fixed set of propositional variables to {true,false}. That is, $K$ is a set of rows in the truth table. In this case, $Th(K)$ is the set of propositional assertions true in all those rows, and $Mod(Th(K))$ is the set of all models of all those assertions. Since the rows appearing in $K$ do model all those assertions, we see that immediately that $K\subset Mod(Th(K))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is a counterexample showing that $K\neq Mod(Th(K))$ is possible. Suppose that there are infinitely many propositional variables $p_0,p_1,\ldots$, and let $K$ be the rows for which at most finitely many $p_n$ are true. In this case, I claim $K\neq Mod(Th(K))$. The reason is that any statement $\varphi$ involves only finitely many variables, and so if it is not tautological, that is, if it fails anywhere, then it will fail in a model for which only finitely many variables are true (just ignore all variables not in $\varphi$). Thus, $Th(K)$ includes only the tautologies, and so $Mod(Th(K))$ includes all truth assignments. So this is a counterexample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not difficult to see that any counterexample will involve infinitely many propositional variables, since if there are only finitely many variables, then the assingments in $K$ can be explicitly described and the theory axiomatized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of first order logic, we take $K$ to be a set of first order structures, and there are numerous examples of $K$ not forming an elementary class, as in the link provided by Asaf. One example is the collection $K$ of all countably infinite structures (of a fixed consistent theory in a countable language). This will not be $Mod(Th(K))$, since by Lowenheim-Skolem, there are arbitrary large structures of any consistent theory.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-13T17:03:52.947" CommentCount="2" />
  <row Id="4550" PostTypeId="2" ParentId="4536" CreationDate="2010-09-13T17:08:04.447" Score="0" ViewCount="0" Body="&lt;p&gt;To start with: the sequence can be written as &#xA;\begin{equation}&#xA;    x_{n+1} = 2 x_{n} + e_{n+1} x_{n-1}   \; [1]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e_n$ is a sequence of iid Bernoully variables taking the values $(1,-1)$ with equal probabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, taking expectations, and because $e_{n+1}$ and $x_{n-1}$ are independent, it results that $E(x_n) = 2^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, I'd say that, if the sequence $x_n^{1/n}$ converges, it should converge to 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;== continued ==&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see if we can compute the variance.&#xA;Multiplying by $x_{n+1}$ and taking expectations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  E(x^2_{n+1})=2E(x_n x_{n+1}) + E(e_{n+1} x_{n+1} x_{n-1})  \; [2]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the first term we multiply [1]  by  $x_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  E(x_n x_{n+1})= 2 E(x_n^2) + E(e_{n+1} x_{n} x_{n-1}) = 2 E(x_n^2)  \; [3]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the second term we multiply [1] by  $e_{n+1} x_{m-1}$ and recall that $e_{n+1}^2 =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;   E(e_{n+1} x_{n+1} x_{n-1}) =  2 E(e_{n+1} x_{n-1} x_{n}) + E(e_{n+1}^2) = E(e_{n+1}^2)  \; [4]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;   E(x_{n+1}^2) = 4 E(x^2_n) + E(x^2_{n-1}) \; [5]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;====&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated: this doesnt lead where I expected to ... I thought that the linearly-normalized sequence (say, $x_n / 2^n$ ) would have an asymtotically vanishing variance, but this does not appear to be so.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-09-13T18:33:08.490" LastActivityDate="2010-09-13T18:33:08.490" CommentCount="5" />
  <row Id="4551" PostTypeId="1" AcceptedAnswerId="4552" CreationDate="2010-09-13T17:21:45.143" Score="1" ViewCount="506" Body="&lt;p&gt;If A,B non empty, upper bounded sets and &lt;strong&gt;A+B={a+b}&lt;/strong&gt;, how can I prove that &lt;strong&gt;sup(A+B)=supA+supB&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T23:08:43.327" LastActivityDate="2010-11-21T23:08:43.327" Title="How can I prove sup(A+B)=supA+supB if A+B={a+b}" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4552" PostTypeId="2" ParentId="4551" CreationDate="2010-09-13T17:25:11.050" Score="4" ViewCount="0" Body="&lt;p&gt;Show that $\sup(A+B)$ is less than or equal to $\sup(A)+\sup(B)$ by showing that the latter is an upper bound for $A+B$. Then show that $\sup(A)+\sup(B)$ is less than or equal to $\sup(A+B)$ by showing that $\sup(A+B)$ is an upper bound for $A+\sup(B)$ and that $\sup(A+\sup(B)) = \sup(A)+\sup(B)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-13T17:25:11.050" />
  <row Id="4553" PostTypeId="2" ParentId="4551" CreationDate="2010-09-13T17:27:31.970" Score="0" ViewCount="0" Body="&lt;p&gt;Set x=supA, y=supB. Given $a\in A, b\in B$ we have $a\le x, b\le y$ so $a+b\le x+y$ so it's an upper bound. Now take $\varepsilon &gt; 0$ and find $a,b$ such that $a&gt;x-\varepsilon/2, b&gt;y-\varepsilon/2$ and you have $a+b &gt; x+y - \varepsilon$. That suffices (since it means that every potential &quot;smaller upper bound&quot; $x+y-\varepsilon$ is not really an upper bound).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-13T17:27:31.970" />
  <row Id="4554" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T17:33:27.877" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Wronskian&quot; rel=&quot;nofollow&quot;&gt;wronskian&lt;/a&gt;, named after Józef Hoene-Wroński, sometimes appears uncapitalised (though it doesn't on Wikipedia, it does in several of the first hits on Google Scholar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wronskian of a set of n functions, as a function of x, is defined as the determinant of the matrix whose (i,j)'th entry is the (i-1)'th derivative of the j'th function, evaluated at x.&lt;/p&gt;&#xA;" OwnerUserId="1835" LastActivityDate="2010-09-13T17:33:27.877" CommunityOwnedDate="2010-09-13T17:33:27.877" CommentCount="1" />
  <row Id="4556" PostTypeId="1" AcceptedAnswerId="4559" CreationDate="2010-09-13T17:42:43.110" Score="2" ViewCount="138" Body="&lt;p&gt;Let $c$ be a postive integer that is not prime. Show that there is some positive integer $b$ such that $b|c$ and $b \leq \sqrt{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this can be proved by contradiction, but I'm not sure how to approach it. Usually I write the proof in the form $P \to Q$, and then if we can prove $P \land \neg Q$ is false, $P \to Q$ must be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I wrote it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c$ is a composite, positive integer, &lt;b&gt;then&lt;/b&gt; $b|c$ and $b \leq \sqrt{c}$, for some positive integer $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that as long as I assume that $b \not| c$ or $b &gt; \sqrt{c}$, then this is still valid as $\neg Q$- that is, I don't have to assume the converse of both parts of Q?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on, if $b &gt; \sqrt{c}$, and $b|c$, then $br=c$ for some integer $r$, which means $r &amp;lt; \sqrt{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where I get stuck.&lt;/p&gt;&#xA;" OwnerUserId="1868" LastActivityDate="2010-09-13T17:57:54.647" Title="Understanding proof by contradiction" Tags="&lt;number-theory&gt;&lt;proof-strategy&gt;&lt;contradiction&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="4557" PostTypeId="2" ParentId="4556" CreationDate="2010-09-13T17:52:06.207" Score="1" ViewCount="0" Body="&lt;p&gt;What you want is to assume that &lt;strong&gt;every&lt;/strong&gt; b that divides c is &quot;too large&quot; and derive a contradiction; however, you don't need proof by contradiction here at all. If b divides c and is too large, then it's easy to show directly that $c/b$ also divides c and is small enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also can be phrased with contradiction: assume $c/b &gt; \sqrt{c}$ and $b &gt; \sqrt{c}$ then $c=(c/b)\cdot b &gt; \sqrt{c}\cdot\sqrt{c}=c$ - contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-13T17:52:06.207" />
  <row Id="4558" PostTypeId="2" ParentId="4556" CreationDate="2010-09-13T17:55:28.150" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;I:&lt;/strong&gt; That there exists some $b$ with $b|c$ is true by the definition (c is not prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we assume that - as &lt;strong&gt;II&lt;/strong&gt; - there is no $b \leq \sqrt c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there &lt;strong&gt;has&lt;/strong&gt; to be &lt;em&gt;some&lt;/em&gt; $b$ due to &lt;strong&gt;I&lt;/strong&gt;, this $b$ now has to meet $b &gt; \sqrt c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $z = \frac{c}{b}$ surely is an integer that divides $c$ too, and, furthermore, $z \leq \sqrt c$. Therefore, by assuming &lt;strong&gt;II&lt;/strong&gt;, we disprove &lt;strong&gt;II&lt;/strong&gt; - Contradiction, q.e.d.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-13T17:55:28.150" />
  <row Id="4559" PostTypeId="2" ParentId="4556" CreationDate="2010-09-13T17:57:54.647" Score="5" ViewCount="0" Body="&lt;p&gt;When you negate statements with quantifiers, you need to flip the quantifiers.  For example, the opposite of &quot;every dog is black&quot; is not &quot;every dog is not black,&quot; but &quot;there exists some dog which is not black.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the correct negation of &quot;there exists some $b$ such that $b | c$ and $b \le \sqrt{c}$,&quot; if you want to keep the condition that $b$ is a divisor, is &quot;for all $b$ such that $b | c$, $b &gt; \sqrt{c}$.&quot;  This is the statement from which you can prove a contradiction: for every $b$ that satisfies this condition, $\frac{c}{b}$ must also satisfy this condition, giving the contradiction $c &gt; c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And everywhere you want the condition that $b &gt; 1$ or else the problem is trivial as KennyTM says in the comments.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-13T17:57:54.647" CommentCount="1" />
  <row Id="4560" PostTypeId="2" ParentId="2969" CreationDate="2010-09-13T18:05:53.573" Score="0" ViewCount="0" Body="&lt;p&gt;While you talk about pruning the search tree through symmetry, generally the best way I've seen to approach these problems is by breaking cells down into classes or layers; for instance, many of the classic polyomino non-tiling problems can be solved with suitable colorings, and constraints provided by colorings can help reduce the search space immensely (e.g., the Soma cube result that the 'T' piece must have its spine on an edge of the cube).  The thing that jumps out at first glance is the 'intermediate' layer of cells, the 26 cells that are neither on the outermost face of the cube nor the center cell.  Whatever piece occupies the center must also occupy at least (and in fact, exactly) three of these cells, and any piece that's not wholly on an outer layer occupies at least one, with many placements forcing several cells to be occupied; it seems intuitively like they might be at a premium and that that's a constraint worth investigating.  The section on the Soma cube in the last volume of &lt;em&gt;Winning Ways For Your Mathematical Plays&lt;/em&gt; has some discussion of these sorts of colorings, and that might be a good place to start.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-13T18:05:53.573" />
  <row Id="4562" PostTypeId="1" AcceptedAnswerId="4563" CreationDate="2010-09-13T18:18:45.433" Score="0" ViewCount="285" Body="&lt;p&gt;If A, B two non empty sets of real numbers and for every a from A and b from B&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;a &amp;lt; b&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;how can I prove that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;supA ≤ infB&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-25T16:59:19.413" LastActivityDate="2010-11-25T16:59:19.413" Title="If a &lt; b =&gt; sup A ≤ inf B" Tags="&lt;real-analysis&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="4563" PostTypeId="2" ParentId="4562" CreationDate="2010-09-13T18:23:33.673" Score="4" ViewCount="0" Body="&lt;p&gt;I don't think this is true. Let $A=(0,1)$ and $B=(1,2)$ then $supA=1$ and $infB=1$ but $A$ and $B$ satisfy your conditions. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-13T18:23:33.673" CommentCount="3" />
  <row Id="4564" PostTypeId="2" ParentId="4536" CreationDate="2010-09-13T18:29:37.443" Score="3" ViewCount="0" Body="&lt;p&gt;If you replace 2 with 1 in your recurrence, you get the &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_Fibonacci_sequence&quot; rel=&quot;nofollow&quot;&gt;random Fibonacci sequence&lt;/a&gt;. &lt;a href=&quot;http://www.ams.org/journals/mcom/2000-69-231/S0025-5718-99-01145-X/home.html&quot; rel=&quot;nofollow&quot;&gt;Viswanath has proven&lt;/a&gt; that the random Fibonacci sequence almost surely has exponential growth (and gives the growth constant).  I haven't carefully read the proof, though, so I can't comment on how it could be adapted to the case you're interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/stable/2237962&quot; rel=&quot;nofollow&quot;&gt;Kesten and Furstenberg&lt;/a&gt; showed quite a while ago that the limit you say is around 1.91 at least exists, by restating the problem as one about multiplication of random matrices.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-09-13T18:29:37.443" CommentCount="1" />
  <row Id="4565" PostTypeId="2" ParentId="350" CreationDate="2010-09-13T18:42:45.433" Score="0" ViewCount="0" Body="&lt;p&gt;For quick review Schaum's Discrete Mathematics is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have time, read &lt;strong&gt;Foundations of Computer Science&lt;/strong&gt; By Aho-Ullman. It's free and available online. Around 800 pages.&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-13T18:42:45.433" CommunityOwnedDate="2010-09-13T20:09:02.853" />
  <row Id="4566" PostTypeId="2" ParentId="4510" CreationDate="2010-09-13T18:43:31.310" Score="3" ViewCount="0" Body="&lt;p&gt;If $$f(x) = \frac{P(x)}{Q(x)}$$ where $P,Q$ are polynomials, then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)Q(x) = P(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Leibniz's formula for differentiating a product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{r=0}^{n} {n \choose r} f^{r}(x) Q^{n-r}(x) = P^{n}(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you can evaluate at $0$ for $n=1,2,\dots$ and you can obtain $(n+1)^{th}$ derivative of $f$ at $0$ from the first $n$ derivatives using the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient you need would be $\frac{f^{n}(0)}{n!}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be easily doable by a computer. No integration etc needed.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-13T18:43:31.310" />
  <row Id="4567" PostTypeId="2" ParentId="4495" CreationDate="2010-09-13T18:47:28.640" Score="2" ViewCount="0" Body="&lt;p&gt;Some intuition is afforded by the Jordan Normal Form, and even more is obtained by understanding a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jordan form of an endomorphism decomposes it as the direct sum of &quot;Jordan blocks&quot;; if you have an intuition for what a block is doing, you can understand the entire direct sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Jordan block geometrically is the sum of two operations.  One rescales everything by a constant, the eigenvalue.  The other essentially collapses the module onto a codimension one submodule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collapsing is the nilpotent part.  A geometric intuition is obtained by considering the simplest nontrivial example, the linear map defined (in coordinates) by $(x,y) \to (y,0)$: two dimensions are collapsed onto one essentially by forgetting the first dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Jordan block, the nilpotent operation is generally $(x,y, \ldots, z) \to (y, \ldots, z, 0)$.  This establishes a hierarchy in the module: the last dimension generates the kernel of $T$, the last two dimensions are killed by $T^2$, and so on.  Iterating $T$ enough times eventually produces the zero map--that's what it means to be nilpotent.  Thus the kernel of $(T-\lambda I)^k$ picks up all the Jordan blocks associated with eigenvalue $\lambda$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $\lambda$, &lt;em&gt;up to some &quot;error&quot; term&lt;/em&gt; generated by certain of the other generalized eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two dimensions, then, a Jordan block effects a transformation that can be written in suitable coordinates (given by two of the generalized eigenvectors) as $(x,y) \to (\lambda x + y, \lambda y)$: an isothety and skew transformation together.  The geometric picture isn't really any different for higher-dimensional blocks.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-13T18:47:28.640" />
  <row Id="4568" PostTypeId="2" ParentId="153" CreationDate="2010-09-13T18:56:55.120" Score="1" ViewCount="0" Body="&lt;p&gt;It seems you want to give it a try and play around a while before you jump into, if at all, seriously. I recommend you learn Python. Then use Sage. You'll have lot of fun, I bet. It's awesome. Learning python not only will lead you to sage, it'll lead you to several other python based computing environments. See python's website for further reference.&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-13T18:56:55.120" />
  <row Id="4569" PostTypeId="2" ParentId="2723" CreationDate="2010-09-13T19:00:08.430" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;Khan Academy&lt;/a&gt; offers dozens of videos on particular Algebra topics, among thousands of videos on other academic topics.&lt;/p&gt;&#xA;" OwnerUserId="1869" LastEditorUserId="1995" LastEditorDisplayName="" LastEditDate="2010-09-27T02:52:58.960" LastActivityDate="2010-09-27T02:52:58.960" CommunityOwnedDate="2010-09-13T19:00:08.430" />
  <row Id="4571" PostTypeId="2" ParentId="4497" CreationDate="2010-09-13T19:40:22.727" Score="1" ViewCount="0" Body="&lt;p&gt;It seems to me that the limit of the function does not exist, since the function approaches to different limits along different ways. However, we can compute iterated limits:&#xA;\begin{equation}&#xA;\lim_{x\rightarrow\infty}\lim_{y\rightarrow\infty}\frac{2x−1}{x−1}+\frac{x}{y}=2&#xA;\end{equation}&#xA;and&#xA;\begin{equation}&#xA;\lim_{y\rightarrow\infty}\lim_{x\rightarrow y}\frac{2x−1}{x−1}+\frac{x}{y}=3.&#xA;\end{equation}&#xA;The non-existence of the double limit is also implicated by the non-equality of the iterated limits.&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-09-13T22:22:43.270" LastActivityDate="2010-09-13T22:22:43.270" CommentCount="4" />
  <row Id="4572" PostTypeId="1" AcceptedAnswerId="4574" CreationDate="2010-09-13T19:47:37.310" Score="7" ViewCount="575" Body="&lt;p&gt;In the question here: &lt;a href=&quot;http://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number&quot;&gt;http://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was an answer by Douglas S Stones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&#xA;If $\sqrt{a}=x/y$ where $y$ does not divide $x$, then $a=(\sqrt{a})^2=x^2/y^2$ is not an integer (since $y^2$ does not divide $x^2$), giving a contradiction.&#xA;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two people whose opinions I respect claimed that this proof approach was &quot;totally bogus/circular&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really see how this is circular, or bogus for that matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't the result follow immediately from unique factorization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: What is wrong with that answer?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-15T20:37:15.570" LastActivityDate="2010-09-15T20:37:15.570" ClosedDate="2010-09-15T23:04:46.297" Title="Is this proof of $a^{1/2}$ being either integer or irrational circular/incorrect?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="38" FavoriteCount="2" />
  <row Id="4573" PostTypeId="2" ParentId="4572" CreationDate="2010-09-13T19:56:58.343" Score="2" ViewCount="0" Body="&lt;p&gt;The result does follow from unique factorization, but the point being made here is that the stated claim - that if $y$ doesn't divide $x$, it's also true that $y^2$ doesn't divide $x^2$ - is &lt;strong&gt;equivalent&lt;/strong&gt; to the problem statement, so it's circular to use it to prove the problem statement (or in any case it doesn't address the meat of the problem).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Let me try to make it clearer why I think this proof is not a proof in the sense that no mathematical work has been done.  We want to show that if $a$ is a positive integer, $\sqrt{a}$ is either an integer or irrational.  What does that mean?  That means if $\sqrt{a} = \frac{p}{q}$ where $p, q$ are positive integers, then $q | p$.  Equivalently, if $a = \frac{p^2}{q^2}$ where $p, q$ are positive integers, then $q | p$.  Equivalently, if $q^2 | p^2$, then $q | p$.  Equivalently, if $q$ does not divide $p$, then $q^2$ does not divide $p^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done no mathematical work so far.  All I have done is unpack definitions.  The statement that I have ended up with is 1) exactly as hard to prove as the statement I started with, and 2) true in all of the same rings as the statement I started with.  A crucial part of the problem - that we are working in $\mathbb{Z}$ - has not yet been used.  To claim that the statement is &quot;obvious&quot; from here is to ignore an essential nontrivial property of $\mathbb{Z}$, namely that it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Integrally_closed&quot; rel=&quot;nofollow&quot;&gt;integrally closed&lt;/a&gt; (which follows from unique factorization).  There is a reason this property has a name.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-13T20:49:07.357" LastActivityDate="2010-09-13T20:49:07.357" CommentCount="24" />
  <row Id="4574" PostTypeId="2" ParentId="4572" CreationDate="2010-09-13T20:12:27.690" Score="4" ViewCount="0" Body="&lt;p&gt;It sounds like we only need to be clear about what is being asserted here, because there shouldn't be any dispute.  Evidently you want to say that when $a$ is an integer, $\sqrt{a}$ cannot be a non-integral rational number.  Assume the contrary in order to achieve a contradiction; that is, suppose there exists a rational number $x/y$ whose square equals $a$ but which itself is non-integral.  Because $x/y$ is not an integer, $y$ does not divide $x$.  Then $y^2$ does not divide $x^2$, whence $x^2/y^2 = (x/y)^2 = a^2$ is nonintegral, the contradiction.  &lt;em&gt;Ergo,&lt;/em&gt; this is a valid argument.  (Like all logical arguments, it's a complete tautology.  But that's not a circularity!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to a request in the comments below the main question, I am happy to point out that one possible justification for the final step in this demonstration is the unique factorization property of the integers.  (UF is not strictly necessary, though: the assertion that for all $x, y$ in a ring $x^2 \mid y^2 \Rightarrow x \mid y$ in and of itself does not guarantee unique factorization.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-15T18:22:49.877" LastActivityDate="2010-09-15T18:22:49.877" CommentCount="19" />
  <row Id="4575" PostTypeId="2" ParentId="4562" CreationDate="2010-09-13T20:24:59.137" Score="1" ViewCount="0" Body="&lt;p&gt;With the edited problem: if $a\lt b$ for all $a\in A$ and $b\in B$, why is it that $\sup(A)\leq\inf(B)$? Is every element of $A$ a lower bound for $B$? If so, what does that tell you about the elements of $A$ relative to $\inf(B)$? And given that, what does &lt;em&gt;that&lt;/em&gt; tell you about $\sup(A)$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-13T20:24:59.137" />
  <row Id="4576" PostTypeId="1" CreationDate="2010-09-13T20:29:02.237" Score="2" ViewCount="117" Body="&lt;p&gt;&lt;strong&gt;The problem:&lt;/strong&gt; Let X be a product of equidimensional spheres of arbitrary dimension, say k, and G a finite group acting freely on X. Assume that the induced G-action on the Z_2-cohomology ring of X is trivial. We then have a Serre spectral sequence with the E_2-page given by the expression H*(G; H*(X; Z_2)) and converging to H*(X/G; Z_2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first non-zero differential is certainly d^{k+1}. Let M be the image of the transgression (hence M is a subgroup of H^{k+1}(G; Z_2)). Now suppose that M = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I deduce that E_{\infty} = E_{k+2} = E_2? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; I've been trying to learn some spectral sequences these past few days. After a while I had decided I kind of have a grasp of what is going on and have been trying to understand some research-level examples. The one described above comes from the paper &lt;em&gt;Free compact group actions on products of spheres&lt;/em&gt; by R. Oliver. In fact, he takes G to be the alternating group on four letters, but I don't think this has anything to do with my question. (My guess is that it &lt;em&gt;somehow&lt;/em&gt; follows from the multiplicative structure of the spectral sequence.)&lt;/p&gt;&#xA;" OwnerUserId="1875" LastActivityDate="2010-10-28T16:44:08.287" Title="Why does this particular Serre spectral sequence collapse at the E_{k+2}-page?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4577" PostTypeId="1" CreationDate="2010-09-13T20:34:38.340" Score="0" ViewCount="88" Body="&lt;p&gt;The following is an example given by my professor, but there is an equality that I don't understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a partially ordered set $(S,\preceq)$ is the union of three sets such that $S=X\cup Y\cup Z$ such that $x\preceq y$ and $x\preceq z$ for all $x\in X, y\in Y, z\in Z$, but no $y\in Y$ and no $z\in Z$ are comparable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $T=S\cup$ {$\nu,\eta$} and $T'=S\cup$ {$\nu$} where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\prec\nu\prec y$ for all $x\in X$ and all $y\in Y$,&#xA;$x\prec\eta\prec z$ for all $x\in X$ and all $z\in Z$,&#xA;but $\nu$ and $\eta$ are not comparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It continues to say that $L(Y)=(\nu]\cap S=X=L(Z)$. However, I don't see how the first equality holds. Since $\nu=\inf_{T'}Y$, it would follow that $\nu\in L(Y)$, but isn't $\nu\notin (\nu]\cap S$, since $\nu\notin S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I have mistook $L(Y)$ as $L_{T'}(Y)$, and so it is actually not the case that $\nu\in L(Y)$? Thanks for any answer to my doubts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, is there a different way to TeX { and } on this site? { and \left{ don't seem to translate.)&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-01-12T04:44:44.183" Title="Set Equality between Lower Bounds and and an Intersection of a Subset and Interval" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4578" PostTypeId="2" ParentId="4526" CreationDate="2010-09-13T20:45:24.727" Score="2" ViewCount="0" Body="&lt;p&gt;Here they talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_and_tauberian_theorems&quot; rel=&quot;nofollow&quot;&gt;abelian and tauberian theorems&lt;/a&gt; after (Niels-Henrik Abel and Alfred Tauber), however it is also common to use &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/1084/Abelian-theorem&quot; rel=&quot;nofollow&quot;&gt;Abelian theorem&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=_G4k2onFlQwC&amp;amp;pg=PA65&amp;amp;lpg=PA65&amp;amp;dq=Adjective+tauberian+Lennart+carleson&amp;amp;source=bl&amp;amp;ots=n7eubjLKk7&amp;amp;sig=xIDZ82wz-eugYyDGox7ipnDHILs&amp;amp;hl=en&amp;amp;ei=U4qOTOqrK4nHswafh6WVAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q=Adjective%20tauberian%20Lennart%20carleson&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Tauberian theorem&lt;/a&gt;. Also, in the last paper, by Lennart Carleson, we can read about &quot;this thread&quot; -- that is about people that becomes adjectives.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-13T20:45:24.727" CommunityOwnedDate="2010-09-13T20:45:24.727" />
  <row Id="4579" PostTypeId="2" ParentId="4577" CreationDate="2010-09-13T20:48:38.043" Score="1" ViewCount="0" Body="&lt;p&gt;I do not think you know that $\nu=\inf_{T'}(Y)$. Since $S\subset T'$, you know that if $\inf_{T'}(Y)$ and $\inf_S(Y)$ both exist, then $\inf_S(Y)\preceq\inf_{T'}(Y)$; but you could have equality (e.g., if $Y$ has a minimum). If you have equality, then $\nu$ is not the infimum of $Y$ in $T'$. You do know that $\nu\preceq\inf_{T'}(Y)$, because $\nu$ is a lower bound for $Y$ in $T'$, but the  inequality could be strict. I would argue that you do not know that $L(Y)=(\nu]\cap S$, because it excludes the possibility that there is a $y\in Y$ that is a lower bound for $Y$. For the same reason, $L_{T'}(Y)$ cannot equal $(\nu]\cap S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, presumably you are also assuming that $Y$ has no minimum. If that's the case, then I think that $L(Y)$ is meant to be $L_S(Y)$, rather than $L_{T'}(Y)$ or $L_T(Y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm assuming that $L(A)$ means &quot;lower bounds of $A$&quot;; it's possible it is supposed to mean &quot;strict lower bounds of $A$&quot; (that is, $L(A) = \{x|x\prec y$ for all $y\in Y\}$), in which case my objection above vanishes. In any case, I suspect that $L(Y)$ is meant to be $L_S(Y)$, $S$ being the set that we started with so the one that it makes sense to take as understood from context.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-14T04:19:34.840" LastActivityDate="2010-09-14T04:19:34.840" />
  <row Id="4580" PostTypeId="1" CreationDate="2010-09-13T20:53:44.400" Score="0" ViewCount="64" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/4561/show-process-of-solution&quot;&gt;Show process of solution? &lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Show process of solution? [reposted]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: no python program or programming needed/necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cryptogram - two variations - most likely by substitution - please help me solve:&#xA;Explain the process I need to take. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cfcfcf&lt;br&gt;&#xA;adgddlb&lt;br&gt;&#xA;cfcicmc&lt;br&gt;&#xA;gddo&lt;br&gt;&#xA;bcfcfc&lt;br&gt;&#xA;dgddg&lt;br&gt;&#xA;cfcjcfc&lt;br&gt;&#xA;dgdhgo&lt;br&gt;&#xA;cfcfb&lt;br&gt;&#xA;ehk&lt;br&gt;&#xA;cb  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ab&lt;br&gt;&#xA;cdccdcdce&lt;br&gt;&#xA;fgfgfgfgf&lt;br&gt;&#xA;cdcdcdcdchc&lt;br&gt;&#xA;fifjfb&lt;br&gt;&#xA;cdcdcdchk&lt;br&gt;&#xA;flmgfgn&lt;br&gt;&#xA;bcoco  &lt;/p&gt;&#xA;" OwnerUserId="1871" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-13T21:13:51.697" LastActivityDate="2010-09-13T21:13:51.697" ClosedDate="2010-09-13T21:39:22.320" Title="Show process of solution?  [reposted]" Tags="&lt;cryptography&gt;" CommentCount="7" />
  <row Id="4581" PostTypeId="1" AcceptedAnswerId="4595" CreationDate="2010-09-13T21:05:57.657" Score="9" ViewCount="228" Body="&lt;p&gt;The following is a problem in an algebra textbook, probably a well-known fact, but I just don't know how to google it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K/k$ be a finite field extension. Then $K(X)/k(X)$ is also finite with the same degree as $K/k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously if $v_1,...,v_n$ is a $k$-Basis of $K$, any polynomial in $K(X)$ can be written as a $k(X)$-linear combination of $v_1,...,v_n$, but I have no idea what to do with a nontrivial denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there perhaps a more elegant way of proving this?  &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2010-09-16T11:14:54.187" Title="Why does K-&gt;K(X) preserve the degree of field extensions?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" />
  <row Id="4582" PostTypeId="1" AcceptedAnswerId="4583" CreationDate="2010-09-13T22:45:16.783" Score="4" ViewCount="260" Body="&lt;p&gt;I am reviewing Calc 2 material and I came across a problem which asked me to explain why $x^\pi$  does not have a Taylor Series expansion around x=0. To me it seems that it would have an expansion but it would just be 0, so maybe it's not a suitable expansion. It doesn't have any holes and it is infinitely differentiable so I don't know why it couldn't have an expansion. &lt;/p&gt;&#xA;" OwnerUserId="1880" LastActivityDate="2010-09-14T01:19:00.630" Title="Explanation of Maclaurin Series of x^pi" Tags="&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;&lt;analytic&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="4583" PostTypeId="2" ParentId="4582" CreationDate="2010-09-13T22:58:35.483" Score="5" ViewCount="0" Body="&lt;p&gt;Consider what happens for higher powered derivatives. $\frac{d}{d^{4}x}x^{\pi}=(\pi)(\pi -1)(\pi -2)(\pi -3)x^{\pi -4}$ You can think of this as $c \frac{1}{x^{k}}$ for some positive real number k which is not defined at 0.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-13T22:58:35.483" CommentCount="3" />
  <row Id="4584" PostTypeId="1" AcceptedAnswerId="4608" CreationDate="2010-09-13T23:25:16.033" Score="1" ViewCount="66" Body="&lt;p&gt;The [infinite] Ramsey theorem states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $n$ and $k$ be natural numbers. Every partition $\{X_1,\ldots ,X_k\}$ of $[\omega]^n$ into $k$ pieces has an infinite homogeneous set. Equivalently, for every $F\colon [\omega]^n\to \{1, . . . , k\}$ there exists an infinite $H \subseteq \omega$ such that $F$ is constant on $[H]^n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $[X]^k := \{ Y \subset X | |Y| = k\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when $k=2$ we can interpret this as coloring edges of a complete graph. But what happens when $k&gt;2$, is there some geometrical or graphic, in a similar sense, to which we can turn this state into?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-14T07:02:56.020" Title="A &quot;geometrical&quot; representation for Ramsey's theorem" Tags="&lt;combinatorics&gt;&lt;set-theory&gt;&lt;intuition&gt;" AnswerCount="1" />
  <row Id="4585" PostTypeId="2" ParentId="4467" CreationDate="2010-09-13T23:31:58.853" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezouts_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout identity&lt;/a&gt; yields a strikingly simple proof - as I discovered as a teenager (inspired by Dedekind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;\;\rm r = \sqrt{n}\;\;$ is integral if rational, $\:$ for $\:\rm n\in\mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\;\;\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1  \;\Rightarrow\; ad-bc \;=\; 1\;$  for some $\:\rm c,d \in \mathbb{Z}\;\;$ by Bezout   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus $\;\;\rm 0 = (a-br) (c+dr) \:=\: ac-bdn + r  \;\Rightarrow\;  r \in \mathbb{Z}\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays my favorite proof is the 1-line gem using Dedekind's &lt;strong&gt;conductor ideal&lt;/strong&gt;. As I explained at length &lt;a href=&quot;http://mathoverflow.net/questions/30313&quot; rel=&quot;nofollow&quot;&gt;elsewhere&lt;/a&gt;, it beautifully encapsulates the denominator descent in ad-hoc &quot;elementary&quot; irrationality proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also my &lt;a href=&quot;http://math.stackexchange.com/4491&quot;&gt;other post&lt;/a&gt; here. That post concisely proves the theorem after first proving from scratch the fundamental lemma that the least positive denominator of a fraction  must divide every other denominator. The proof of the theorem follows immediately from this &quot;principal&quot; denominator theorem, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;$\;\;$  Put $\displaystyle\;\rm \sqrt{n} \:= \frac{a}b\:,\;\;$ least $\rm b&gt;0\:.\;\:$ $\displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \;\Rightarrow\; \frac{a}b = \frac{nb}a \;\Rightarrow\; b\:|\:a\;\Rightarrow\; \sqrt{n}\in\mathbb Z$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T21:53:46.307" LastActivityDate="2010-09-19T21:53:46.307" />
  <row Id="4586" PostTypeId="2" ParentId="4582" CreationDate="2010-09-13T23:44:32.417" Score="6" ViewCount="0" Body="&lt;p&gt;I decided to elaborate on my comment above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $x^\pi$ if $x$ is negative?  Trying to approximate with $x^r$ for rational $r$ is problematic, depending on which rational numbers you use to approximate; should the answer be positive, negative, imaginary?  Perhaps more reasonable would be to take $x^\pi=e^{\pi \log(x)}$ for some suitably chosen branch of the logarithm on $(-\infty,0)$, the most common choice being $\log x=\ln(-x)+\pi i$.  Your function then becomes &#xA; $$&#xA;   f(x) = \left\{&#xA;     \begin{array}{lr}&#xA;       e^{\pi \ln x} &amp;amp; : x&gt;0 \\&#xA;       0 &amp;amp; : x=0 \\&#xA;       e^{i\pi^2}e^{\pi \ln(-x)} &amp;amp; :x&amp;lt;0&#xA;     \end{array}&#xA;   \right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it is not real-valued.  You can still go ahead and try to take its derivatives, but this piecewise representation may make it less surprising that there is going to be a singularity at zero.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-13T23:44:32.417" CommentCount="2" />
  <row Id="4587" PostTypeId="1" AcceptedAnswerId="4588" CreationDate="2010-09-13T23:50:36.443" Score="7" ViewCount="127" Body="&lt;p&gt;If I want to take the derivative of $ax^n$, I will get $anx^{n-1}$. If I were to take the derivative again, I get $an(n-1)x^{n-2}$.&lt;br&gt;&#xA;We can generalize this for integer k easily to get the kth derivative $a\frac{n!}{(n-k)!} x ^{n-k}$. But what about for a more general k?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have some name? Has it been widely studied? If so, can you show how to generalize this formula for kth derivative of $ax^n$, and explain how it works? If not, is there a good reason it is impossible?&lt;/p&gt;&#xA;" OwnerUserId="1881" LastActivityDate="2010-09-14T00:36:10.343" Title="Is it meaningful to take the derivative of a function a non-integer number of times?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4588" PostTypeId="2" ParentId="4587" CreationDate="2010-09-14T00:36:10.343" Score="5" ViewCount="0" Body="&lt;p&gt;To expand on Jonas's comment: Yes, it makes sense. For the case of the power function, one can consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma(n+1)}{\Gamma(n-\alpha+1)}x^{n-\alpha}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the $\alpha$-th derivative of the power function $x^n$, where $\Gamma(z)$ is the gamma function, the generalization of the factorial to the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, one has a number of definitions for so-called &quot;fractional derivatives&quot;, or, as &lt;a href=&quot;http://www.sciencedirect.com/science/bookseries/00765392&quot; rel=&quot;nofollow&quot;&gt;Spanier and Oldham&lt;/a&gt; prefer to call it, the &quot;differintegral&quot;. Negative values of $\alpha$ in expressions like the one given above correspond to integration, positive values correspond to differentiation, and in general $\alpha$ can be complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot of things to look at (Caputo derivatives, Riemann-Lioville integrals, Grunwald-Lednikov series), and I suggest you look at the book I linked to first, and then search around the web. Have fun!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-14T00:36:10.343" CommentCount="1" />
  <row Id="4589" PostTypeId="1" CreationDate="2010-09-14T00:43:13.930" Score="1" ViewCount="115" Body="&lt;p&gt;I am seeking methods to see if a polynomial $f \in O(\mathbb{C^2},0)$ is irreducible. The subject is really new to my and I am studying for myself, for which I don't see about this subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would thank them a lot they could provide me some bibliography where to read about this topic&lt;/p&gt;&#xA;" OwnerUserId="1882" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-14T00:47:32.017" LastActivityDate="2010-09-14T02:48:01.320" Title="Irreducible Polynomials in two complex variables" Tags="&lt;reference-request&gt;&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="4590" PostTypeId="2" ParentId="4582" CreationDate="2010-09-14T01:19:00.630" Score="3" ViewCount="0" Body="&lt;p&gt;As a graphical supplement to Jonas's and WWright's answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gK1wy.png&quot; alt=&quot;real and imaginary parts of x^\pi&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a plot of the real and imaginary parts of $(x+iy)^\pi$ in the complex plane. Note the cut running across the negative real axis. This cut is precisely the reason why you cannot have a Maclaurin expansion; polynomials cannot exhibit cuts, and a Maclaurin expansion amounts to approximating your function with a sequence of polynomials.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-14T01:19:00.630" CommentCount="5" />
  <row Id="4591" PostTypeId="2" ParentId="4467" CreationDate="2010-09-14T01:34:51.157" Score="2" ViewCount="0" Body="&lt;p&gt;So... we can prove something pretty general here: the only rational roots of polynomials $x^n + \cdots + a_0$ with integer coefficients are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, suppose $p/q$ is a rational root, in lowest form, then $p^n = -q(a_0q^{n-1} + a_1pq^{n-2} + \cdots + a_{n-1}p^{n-1})$. Now, if $q&gt;1$, then any prime divisor of $q$ also divides $p^n$, and hence $p$. But this contradicts our assumption that $p/q$ is in lowest form, so we conclude that $q=1$, so the root is integral.&lt;/p&gt;&#xA;" OwnerUserId="423" LastEditorUserId="423" LastEditorDisplayName="" LastEditDate="2010-09-14T02:07:13.703" LastActivityDate="2010-09-14T02:07:13.703" CommentCount="5" />
  <row Id="4592" PostTypeId="2" ParentId="4589" CreationDate="2010-09-14T02:48:01.320" Score="2" ViewCount="0" Body="&lt;p&gt;Here's one idea: if $f(x, y)$ is reducible, then the projective closure of $f(x, y) = 0$ has at least two components, which intersect by Bezout's theorem.  These intersections are singularities and therefore can be found by setting all partial derivatives to zero.  If no singularities exist then $f$ is irreducible.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-14T02:48:01.320" CommentCount="1" />
  <row Id="4593" PostTypeId="1" CreationDate="2010-09-14T03:01:06.940" Score="1" ViewCount="64" Body="&lt;p&gt;I want to know the tools and some source codes for generating these amazing pictures.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Portal%3aMathematics/Featured_picture_archive&quot; rel=&quot;nofollow&quot;&gt;Featured_picture_archive in wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-09-14T03:01:06.940" ClosedDate="2010-09-14T05:40:43.403" Title="How to generate these mathematical (animated) pictures" Tags="&lt;math-software&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="4594" PostTypeId="1" CreationDate="2010-09-14T03:11:31.470" Score="2" ViewCount="144" Body="&lt;p&gt;It seems to be an accepted belief based on decades of experience that naive algorithms are not &#xA;adequate to solve NP-complete problems in a reasonable amount of time. Even those who believe P = NP seem to be looking for an algorithm with a very clever representation, so far without success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the observations above be formalized? Let's use the full, unrestricted Clique problem as an example. Suppose it could be shown that no clever representation exists for the Clique problem. That is, suppose it could be proven for every algorithm that either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the algorithm is correct and uses an internal representation such that no two distinct input cliques    lead to the same internal representation,&#xA;or &lt;/li&gt;&#xA;&lt;li&gt;the algorithm is incorrect. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Would this be a worthwhile result? How important would it be? Or is it wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there already a proof that CLIQUE or SAT, for example, cannot be solved in polynomial time by performing operations on cliques or Boolean expressions respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References with your answer would be helpful. Thanks&lt;/p&gt;&#xA;" OwnerUserId="1884" LastActivityDate="2010-09-14T09:27:23.250" Title="Question about the P versus NP Problem" Tags="&lt;computer-science&gt;&lt;np-complete&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4595" PostTypeId="2" ParentId="4581" CreationDate="2010-09-14T03:38:20.003" Score="6" ViewCount="0" Body="&lt;p&gt;The missing step is to show that $v_1,\ldots,v_n$ is linearly independent over $k(x)$.  So suppose that there exist rational functions $r_1(x),\ldots,r_n(x)$, not all the zero function, such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_1(x) v_1 + \ldots + r_n(x) v_n = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may write $r_i(x) = n_i(x)/d(x)$, i.e., let $d(x)$ be a common denominator of the rational functions.  Multiplying through by $d(x)$, we get a polynomial dependence relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_1(x) v_1 + \ldots + n_n(x) v_n = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $N(x)$ be the greatest common divisor of $n_1(x),\ldots,n_r(x)$.  (Such things exist since the polynomial ring $k[x]$ has a division algorithm, hence is a Unique Factorization Domain.)  Dividing through by $N(x)$, we get, say,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_1(x) v_1 + \ldots + N_n(x) v_n = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\operatorname{gcd}(N_1,\ldots,N_n) = 1$.  In particular, not all of the $N_i$'s are divisible by $x$, so plugging in $x = 0$ gives a nontrivial linear dependence relation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_1(0) v_1 + \ldots + N_n(0) v_n = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;over $k$, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more sophisticated language, we are showing that the extensions $K/k$ and $k(x)/k$ are &lt;strong&gt;linearly disjoint&lt;/strong&gt;.  This can be rephrased in terms of tensor products, for  instance...but in the end the above proof is the simplest I can think of.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: Robin Chapman and Steve D are right: I misread the question and thought that the OP had already worked out that a basis $v_1,\ldots,v_n$ of $K/k$ also spans $K(x)$ over $k(x)$, but in fact this is the hardest part of the argument.  As the OP says, it is easy to see that the set of all $k(x)$-linear combinations of the $v_1,\ldots,v_n$ contains all elements of $K[x]$.  As Robin says, it is enough to show that this span also contains all elements $\frac{1}{Q(x)}$ with $Q(x) \in K[x] \setminus {0}$, and a nice way to see this is to show that every $Q \in K[x]$ divides some nonzero polynomial $q \in k[x]$, for if $Q(x) g(x) = q(x)$, $\frac{1}{Q} = = \frac{1}{q} \cdot g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robin gives a nice argument for this: essentially he extends the norm map from a finite dimensional field extension to polynomials.  I might as well complete my answer, and I might as well do it in a different way, so here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to assume that $Q$ is irreducible in the UFD $K[x]$, i.e., that $\mathcal{P} = (Q)$ is a prime ideal.  Put $\mathfrak{p} = \mathcal{P} \cap k[x]$.  In great generality, the restriction of a prime ideal to a subring is again a prime ideal (this is even true for the preimage of a prime ideal under an arbitrary ring homomorphism, and is very easy to show).  What we want to show is that $\mathfrak{p} \neq 0$.  But $K[x]$ is a free $k[x]$-module of dimension $n = [K:k]$.  In particular, the extension $K[x] / k[x]$ is finitely generated as a module, hence an &lt;strong&gt;integral&lt;/strong&gt; extension, and in such an extension any maximal ideal pulls back to a maximal ideal.  So $\mathfrak{p}$ is maximal, hence nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These facts follow immediately from the definition of integral extensions: see e.g. Proposition 160 and Corollary 168 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a possible virtue of this argument is that one does not need to treat the separable and inseparable cases differently.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-14T08:01:11.157" LastActivityDate="2010-09-14T08:01:11.157" CommentCount="1" />
  <row Id="4597" PostTypeId="1" AcceptedAnswerId="4598" CreationDate="2010-09-14T04:31:26.257" Score="5" ViewCount="441" Body="&lt;p&gt;Coming from an understanding of Fermat's primality test, I'm looking for a clear explanation of the Miller-Rabin primality test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically: I understand that for some reason, having non-trivial square roots of 1 mod &lt;code&gt;p&lt;/code&gt; means that &lt;code&gt;p&lt;/code&gt; is definitely composite; and I gather that you can find these non-trivial square roots by squaring &lt;code&gt;x&lt;/code&gt;, but I don't really understand what these reasons are. Specific examples of non-trivial roots of a composite number would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="1885" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-14T06:00:27.883" LastActivityDate="2010-12-07T17:04:02.510" Title="Simple explanation and examples of the Miller-Rabin Primality Test" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="4598" PostTypeId="2" ParentId="4597" CreationDate="2010-09-14T05:14:20.197" Score="5" ViewCount="0" Body="&lt;p&gt;Suppose $p$ was prime, and $y$ was a non-trivial square root of $1$ mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we must have that $y^2 = 1 \mod p$ and so $(y-1)(y+1) = 0 \mod p$. This implies that either $y = 1 \mod p$ or $y = -1 \mod p$, which implies that $y$ is a trivial square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if there is a non-trivial square root of $1$ mod $p$, then $p$ has to be composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a non trivial square root of a composite, consider $p = 15$. We have that $4^2 = 16 = 1 \mod 15$. Thus $15$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the witness in the primality test is not necessarily a non-trivial square root of $1$ mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact about non-trivial square roots can be used to prove that if $p$ is prime, then for any $a$ relatively prime to $p$, some power of $a$ from a given set of powers (the powers are based on the even factors of $p-1$) must be $-1$ or a specific odd power of $a$ (again based on factor of $p-1$) must be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for some $a$ none of the above set of powers is $-1$ and the specific odd power is not $1$, then it must be the fact that $p$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be shown that for composite $p$, the chances of finding such $a$ is atleast $3/4$. This $a$ is the witness in the primality test and is not necessarily a non-trivial square root of $1$ mod $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The squaring that is done is to get the powers described above which are based on the factors of $p-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki page has really got a lot of good information (including the exact powers of $a$ that need to be taken): &lt;a href=&quot;http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test&quot; rel=&quot;nofollow&quot;&gt;Miller Rabin Primality Test&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-14T05:57:43.760" LastActivityDate="2010-09-14T05:57:43.760" CommentCount="2" />
  <row Id="4599" PostTypeId="2" ParentId="4597" CreationDate="2010-09-14T05:19:37.370" Score="2" ViewCount="0" Body="&lt;p&gt;If $p$ is prime then $\mathbb{Z}_p$ (integers modulo $p$) is a &lt;strong&gt;field&lt;/strong&gt;. It is a basic result in algebra that in a field, a polynomial of degree $n$ has at most $n$ roots, and so the polynomial $x^2-1$ has exactly two roots: $1$ and $-1$ (which exist in every field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is composite, then $\mathbb{Z}_n$ is never a field because not all elements have an inverse; it it well known that $a\in \mathbb{Z}_n$ has an inverse if and only if $a$ is relatively prime to $n$. Let's look at the case $n$ is the product of two primes, $n=pq$. In this case we can do arithmetic in $\mathbb{Z}_n$ by doing arithmetic in $\mathbb{Z}_p$ and $\mathbb{Z}_q$ and combining the results using the Chinese remainder theorem (which basically states that $\mathbb{Z}_n\cong\mathbb{Z}_p\times\mathbb{Z}_q$. Since $\mathbb{Z}_p$ and $\mathbb{Z}_q$ are both fields, $1$ has two roots in each of them. For every combination of a root from $\mathbb{Z}_p$ and a root from $\mathbb{Z}_q$ we'll get a root of 1 in $\mathbb{Z}_n$, meaning we'll get 4 roots of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major challenge of the Miller-Rabin test is to show that there is a &quot;large&quot; chance to stumble upon a non-trivial root while squaring random elements of $\mathbb{Z}_n$, and the proof, although not difficult, is not immediate either.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T05:19:37.370" CommentCount="6" />
  <row Id="4600" PostTypeId="1" AcceptedAnswerId="5210" CreationDate="2010-09-14T05:22:47.953" Score="1" ViewCount="61" Body="&lt;p&gt;I'm just starting to introduce myself to first-order logic, and I find it very confusing. As such, I'm not entirely sure how well I constructed the following. For $L$ a first-order language, let $L$ be equipped with the following nonlogical symbols: a unary function $\dot{+}$, a binary function $\dot{1}$, and a 3-ary relation $\dot{&amp;lt;}$, give an example of an $L-structure$ with an infinite universe. (Here the functions have nothing to do with $+$, $1$ and $\lt$ as they typically are.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to let $\mathbb{R}$ be the universe, and for $r,s,t\in\mathbb{R}$, I defined the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{+}\colon\mathbb{R}\to\mathbb{R}\colon \dot{+}(r)\mapsto r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{1}\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}\colon \dot{1}(r,s)\mapsto rs$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{&amp;lt;}\subseteq\mathbb{R}\times\mathbb{R}\times\mathbb{R}\colon (r,s,t)\in \dot{&amp;lt;}$ iff $r\lt s\lt t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $(\mathbb{R},+,-,\times)$ then a valid $L-structure$ with an infinite universe? $\mathbb{R}$ is infinite and both functions are closed, and $\dot{&amp;lt;}$ is in the correct cartesian product, so I can't see any problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is correct, I was surprised by the amount of freedom in defining the nonlogical symbols. This led me to ask if there's a particular limit to $L-structures$ depending on the universe and the functions in the language. For example, suppose I had a finite universe $U=\{a,b,c,d\}$, with a unary function $-$, and two binary functions $+,\times$. In this case, would there be $16^4$ possible $L-structures$? My thinking was that for the binary functions, there are $16$ elements in the domain, and $4$ elements in the codomain, and for $-$, there are 4 in each for a total of $(16)(4)(16)(4)(4)(4)$ ways to define all the functions, which accounts for all possible functions, relations, and constants. Or am I totally off base? Thanks for any input.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-22T13:21:54.113" Title="Constructing an L-structure with an Infinite Universe, and Counting L-structures on a Finite Universe." Tags="&lt;logic&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4601" PostTypeId="1" AcceptedAnswerId="4602" CreationDate="2010-09-14T05:30:07.387" Score="6" ViewCount="147" Body="&lt;p&gt;The word &quot;quad&quot; generally means 4. Quadratics don't have 4 of anything. Can anyone explain where the name comes from?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T05:36:07.857" LastActivityDate="2010-09-14T05:36:07.857" Title="Why are quadratic equations called quadratics?" Tags="&lt;algebra-precalculus&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="4602" PostTypeId="2" ParentId="4601" CreationDate="2010-09-14T05:35:21.427" Score="8" ViewCount="0" Body="&lt;p&gt;From &lt;a href=&quot;http://mathworld.wolfram.com/Quadratic.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Latin prefix quadri- is used to indicate the number 4, for example, quadrilateral, quadrant, etc. However, it also very commonly used to denote objects involving the number 2. This is the case because quadratum is the Latin word for square, and since the area of a square of side length $x$ is given by $x^2$, a polynomial equation having exponent two is known as a quadratic (&quot;square-like&quot;) equation. By extension, a quadratic surface is a second-order algebraic surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-14T05:35:21.427" CommentCount="1" />
  <row Id="4603" PostTypeId="1" AcceptedAnswerId="4605" CreationDate="2010-09-14T05:45:55.633" Score="6" ViewCount="345" Body="&lt;p&gt;I'm studying for qualifying exams and ran into this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if ${a_n}$ is a nonincreasing sequence of positive real&#xA;numbers such that $\sum_n{a_n}$ converges, then $\lim_{n \rightarrow \infty}{n a_n} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of the limit, this is equivalent to showing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\forall \epsilon &gt; 0 \; \exists n_0 \; \mbox{such that} \;&#xA;|n a_n| &amp;lt; \epsilon \; \forall n &gt; n_0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\forall \epsilon &gt; 0 \; \exists n_0 \; \mbox{such that} \;&#xA;a_n &amp;lt; \frac{\epsilon}{n} \; \forall n &gt; n_0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!&lt;/p&gt;&#xA;" OwnerUserId="1761" LastEditorUserId="1761" LastEditorDisplayName="" LastEditDate="2010-09-14T06:03:22.813" LastActivityDate="2010-09-14T14:03:56.783" Title="Series converges implies $\lim{n a_n} = 0$" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="4604" PostTypeId="2" ParentId="4597" CreationDate="2010-09-14T06:00:58.723" Score="2" ViewCount="0" Body="&lt;p&gt;Below is little-known general yet simple answer to your question about nontrivial sqrts $\rm (mod\ m)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\;$ One may quickly factor $\rm m&amp;gt;1\:$ given a polynomial with more roots mod $\rm\: m\:$ than its degree, viz. suppose that,  modulo $\rm m,\;$ the polynomial $\rm\: f(x)\ne 0\:$ has degree $\rm\:n\:$ but has $\rm\:n+1 \:$ distinct roots $\rm\:r_{\:i}\:$.  Then one of $\rm\;gcd(m\;,\:r_{\:i} - r_{\:j}\:),\; i\ne j \:$ must yield a proper factor of $\rm\:m\;$. For if that failed, then all of the gcds must be improper so $1\:,\:$ not $\rm\;m \;$ since $\rm\; i\ne j\;\Rightarrow\; r_{\:i}\: \not\equiv r_{\:j}\ (mod\ m) \:$. Now an induction using the Factor Theorem yields $\rm\;f(x) = (x-r_1)\cdots(x-r_{n+1})\; g(x),\;\; g(x) \ne 0 \;\;$ contra $\rm\;\:\deg\: f = n \;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE&lt;/strong&gt; $\rm\;(deg\ f = 1)\;\:$ Suppose, modulo $\rm\:m\:$, that $\rm\; 0 \:\ne\: f \:=\: a\:x\;$ has a &quot;nonntrivial&quot; root $\rm\: b\ne 0\:$. Then $\:\rm gcd(m,b)\:$ is a proper factor of $\rm\: m\:$. More directly: $\rm\; m\:|\:ab\:,\;$ not $\rm\; m|a\;\:$ nor $\rm\: m|b \;\Rightarrow\; gcd(m,b)\;$ is a proper factor of $\rm\:m\:,\:$ i.e. if $\rm\: m\:$ fails the Prime Divisor Property then $\rm\: m\:$ is &quot;constructively&quot; composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE&lt;/strong&gt; $\rm\;(deg\ f = 2)\;\:$ Suppose, modulo $\rm\:m\:$, that $\rm\; x^2 = 1\;$ has a &quot;nontrivial&quot; square root $\rm\: b\ne \pm1\:$. Then $\;\rm gcd(m,b\pm 1)\;$ yields a proper factor of $\rm\: m\:$ when $\rm\: m\:$ is odd. This idea lies at the heart of many integer factorization algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above proof easily extends to yield a proof of the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ Ring $\rm\: R\:$ is a domain $\iff\:$ every $\rm 0 \ne f(x)\in R[x]\:$ has at most $\rm\: deg\ f(x)\:$ roots in $\rm R$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ $(\Rightarrow)\;$ Employ induction on the polynomial degree, as in the proof of the above Theorem. $(\Leftarrow)\ \ $ If $\rm\; a \ne 0\;$ then  the polynomial $\rm\: a\:x\:$ has only the root $\rm\; x = 0\:,\:$ hence $\rm\ ab = 0 \ \Rightarrow\ b = 0\:,\:$ therefore $\rm\: R\:$ has no zero divisors.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T17:04:02.510" LastActivityDate="2010-12-07T17:04:02.510" CommentCount="3" />
  <row Id="4605" PostTypeId="2" ParentId="4603" CreationDate="2010-09-14T06:08:26.923" Score="7" ViewCount="0" Body="&lt;p&gt;Some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $S_{n} = \sum_{k=1}^{n} a_{k}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then what is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \to \infty} S_{2n} - S_{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now can you use the fact that $a_{n}$ is non-increasing to upper bound a certain term of the sequence $na_{n}$ with a multiple of $S_{2n} - S_{n}$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-14T14:01:52.847" LastActivityDate="2010-09-14T14:01:52.847" CommentCount="2" />
  <row Id="4607" PostTypeId="2" ParentId="4581" CreationDate="2010-09-14T06:24:46.793" Score="7" ViewCount="0" Body="&lt;p&gt;This is quite a subtle problem, and there are important differences between&#xA;the finite extension and infinite extension cases. The key is to show&#xA;that $v_1,\ldots,v_n$ form a $k(X)$-basis of $K(X)$. Pete has dealt with&#xA;their linear indepdendence but it remains to show that their $k(X)$-span is&#xA;$K(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f\in K(X)$. Then $f=g/h$ where $g$ and $h$ are polynomials over $K$. The key&#xA;is that there is a non-zero polynomial $h_1$  with coefficients in $k$&#xA;such that $h\mid h_1$. In the separable case one can take $h_1$ be&#xA;the product of Galois conjugates of $h$; in general one may need to take&#xA;a positive power of that. The upshot is that $f=g_1/h_1$ where $h_1$&#xA;has coefficients in $k$. This reduces the problem to showing that $g_1$&#xA;is in the $k(X)$-span of the $v_i$, which you have already done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subtlety is that this breaks down for infnite non-algebraic extensions.&#xA;If $a\in K$ is transcendental over $k$ then $1/(X+a)$ is not in the $k(X)$-linear&#xA;span of the elements of $K$. In short, even though $K$ and $k(X)$ are&#xA;linearly disjoint over $k$, $K(X)\ne K\cdot k(X)$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (15/9/2010)&#xA;The real distinction isn't between finite and infinite extensions&#xA;(as I said above)&#xA;but rather between algebraic and transcendental extensions. To summarize:&#xA;the natural map&#xA;$$K\otimes_k k(X)\to K(X)$$&#xA;is always injective, and is surjective if and only if $K/k$ is&#xA;an algebraic extension.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-15T17:31:28.907" LastActivityDate="2010-09-15T17:31:28.907" />
  <row Id="4608" PostTypeId="2" ParentId="4584" CreationDate="2010-09-14T06:56:55.233" Score="1" ViewCount="0" Body="&lt;p&gt;Partitioning $[\omega]^{(n)}$ into $k$ pieces can be interpreted as colouring the complete $n$-uniform hypergraph, with vertex set $\omega$, in $k$ colours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=2$ the complete $2$-uniform hypergraph is the complete graph with vertex set $\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a definition of hypergraph see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergraph&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hypergraph&lt;/a&gt;. Briefly it is the generalisation of a graph where edges can join more than two vertices. And by complete $n$-uniform I mean all the edges contain exactly $n$ vertices and every possible edge of size $n$ is in the hypergraph. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-14T07:02:56.020" LastActivityDate="2010-09-14T07:02:56.020" CommentCount="1" />
  <row Id="4609" PostTypeId="1" AcceptedAnswerId="4664" CreationDate="2010-09-14T07:06:11.427" Score="1" ViewCount="54" Body="&lt;p&gt;I have data modelled to be Gaussian-distributed, and then displayed on a logarithmic scale and like to know some properties of the displayed data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effektively, I ignore that a Gaussian $X$ (with positive mean) may be negative and need mean, standard deviation and full width at half maximum of $\log X$. Unfortunately, the probability density is sufficiently ugly to prevent me from calculating this myself. Has this distribution a special name that I might look for?&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-09-14T07:30:05.237" LastActivityDate="2010-09-14T16:29:19.703" Title="Characteristics of a Gaussian on a logarithmic scale?" Tags="&lt;probability-theory&gt;&lt;normal-distribution&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="4611" PostTypeId="1" AcceptedAnswerId="4613" CreationDate="2010-09-14T07:19:19.770" Score="2" ViewCount="58" Body="&lt;p&gt;Given a quadratic form with discriminant $D$, what does the class number of $\mathbb{Q}(\sqrt{D})$ tell us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question is inspired by a comment on the question &lt;a href=&quot;http://math.stackexchange.com/questions/4489/proving-that-the-given-diophantine-equation-has-a-solution&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" LastActivityDate="2010-09-14T07:34:27.720" Title="What does the class number tell us about a quadratic form?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="4612" PostTypeId="2" ParentId="4510" CreationDate="2010-09-14T07:23:21.070" Score="2" ViewCount="0" Body="&lt;p&gt;If one is computing by hand then the following small observation is useful: If $f(x) = P(x)/Q(x)$ and $Q(x) = x^d(1 - g(x))$ for some $d \geq 0$ and some polynomial $g(x)$ (which we may assume after rescaling&#xA;$P(x)$ and $Q(x)$), then using the formula for a geometric series we find that&#xA;$$f(x) = x^{-d}P(x)(1 + g(x) + g(x)^2 + \cdots ).$$&#xA;For explicit pen and paper computations, I normally find this is the simplest way to&#xA;compute the Taylor series for $f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-14T07:23:21.070" />
  <row Id="4613" PostTypeId="2" ParentId="4611" CreationDate="2010-09-14T07:34:27.720" Score="4" ViewCount="0" Body="&lt;p&gt;I guess you mean a binary quadratic form, i.e. of the form $a x^2 + b x y + c y^2$,&#xA;with $a,b,c$ integers and of discrmimant $D$ (i.e. $b^2 - 4 a c = D$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbb Q(\sqrt{D})$ has class number one, then the conditions for solving&#xA;$a x^2 + b x y + c y^2 = p$ ($p$ a prime not dividing $D$) depend only on the residue class&#xA;of $p$ mod $D$, in fact only on the value of the Kronecker symbol of $p$ mod $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the class group is just a product of cyclic groups of order 2 (note that this can't&#xA;be detected by the class number alone, which e.g. can't distinguish between $C_2 \times C_2$&#xA;and $C_4$) then the condition for solving $a x^2 + b x y + c y^2 = p$ depends only on the&#xA;residue class of $p$ mod $D$, but one has to consider not just the Kronecker symbol mod $D$,&#xA;but  other Kronecker symbols modulo various divisors of $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the class group is not a product of cyclic groups of order 2 (e.g. if the class number&#xA;is not a power of two) then there is no congruence condition on $p$ which guarantees being able to solve $a x^2 + b x y + c y^2$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-14T07:34:27.720" />
  <row Id="4614" PostTypeId="1" CreationDate="2010-09-14T07:51:46.150" Score="0" ViewCount="53" Body="&lt;p&gt;is it practicable to render cartoon game map by using Latex. Imagine a game role jumping on Latex.&#xA;&lt;img src=&quot;http://i.imgur.com/2plyu.png&quot; alt=&quot;alt text&quot;&gt;&#xA;[{ \to &lt;em&gt;{\infty \infty \infty \infty \infty \infty \infty {\infty ^{ \in  \in { \in ^{aaaaa{a&lt;/em&gt;{_{&lt;em&gt;{&lt;/em&gt;{_{&lt;em&gt;{fffffffffffff}}}}}}}}}}}}} \bot  \bot  \bot {\angle ^{ -  -  -  -  -  - }}&lt;/em&gt;{_{oooo} \wedge  \wedge }{\hbox{\ \hbox{$\mid$}\kern -1em&#xA;\lower .5em \hbox{$\leftarrow$}}} ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw,why does not Latex work here? I copied it from Mathtype.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-09-14T07:51:46.150" ClosedDate="2010-09-14T08:19:27.127" Title="game map rendering by Latex " Tags="&lt;recreational-mathematics&gt;&lt;latex&gt;" CommentCount="1" />
  <row Id="4615" PostTypeId="1" CreationDate="2010-09-14T08:14:06.063" Score="1" ViewCount="57" Body="&lt;p&gt;Is there any compilation of graph encoding algorithms? I know about Prufer and Huffman encoding. But papers say, prufer is not good enough to represent Minimum Spanning Trees in the sense it &lt;em&gt;may&lt;/em&gt; represent different MSTs with the same string. Any clue?&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-14T08:14:06.063" Title="How many graph encoding algorithms are there?" Tags="&lt;graph-theory&gt;" CommentCount="2" />
  <row Id="4617" PostTypeId="2" ParentId="4371" CreationDate="2010-09-14T08:41:39.037" Score="1" ViewCount="0" Body="&lt;p&gt;For easy read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Combinatorics and Graph Theory by Harris&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Introduction to Graph Theory by Wilson&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-14T08:41:39.037" />
  <row Id="4618" PostTypeId="2" ParentId="4603" CreationDate="2010-09-14T08:56:07.027" Score="4" ViewCount="0" Body="&lt;p&gt;Try to stay away from quantifier-laden formulas, which make the problem harder to understand, and draw a picture.   There is an equivalent problem for decreasing functions (as in the Integral Test for convergence) and the picture makes it obvious what is true in that case.  Having seen the continuous proof, run the same argument for the sequence, or specialize the function to a sequence by using step functions or approximation thereof. I won't spoil the &quot;aha!&quot; proof-by-picture experience by posting more details, but it is quite easy once you draw the graph.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-14T08:56:07.027" />
  <row Id="4619" PostTypeId="1" AcceptedAnswerId="4645" CreationDate="2010-09-14T09:20:53.783" Score="4" ViewCount="382" Body="&lt;p&gt;Please mention ONE problem that is NP-Hard but not NP-Complete? And, explain why. I see some papers assert &lt;strong&gt;Degree Constrained Minimum Spanning Tree&lt;/strong&gt; is an NP-Hard problem and some say it's NP-Complete. Why so? Is it something that we don't have clear idea about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-11-24T12:01:14.003" Title="Mention 1 problem that is NP Hard but not NP Complete and why" Tags="&lt;np-complete&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="4620" PostTypeId="2" ParentId="4594" CreationDate="2010-09-14T09:27:23.250" Score="2" ViewCount="0" Body="&lt;p&gt;There are formal notions of Natural Proofs (Razborov-Rudich) and of &lt;em&gt;proofs&lt;/em&gt; that relativize (Baker-Gill-Solovay) and both are known to not work for showing $P \neq NP$ .  You are asking about relativizable &lt;em&gt;algorithms&lt;/em&gt;, because any algorithm that works only on the &quot;external shell&quot; or &quot;superficial interface&quot; of SAT or Max-Clique will probably work equally in environments where Turing machines are connected to an oracle, and P=NP can be true or false depending on the oracle.  At least for oracles with P &amp;lt; NP the algorithm will not run in polynomial time, and I think this is true for random oracles, so possibilities for SAT solvers that use only externally visible features to work in polynomial time are fairly limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also problems where algorithms that use only the external interface are outperformed by algorithms that take apart the internal representation of the data. For example, sorting by comparisons requires $n \log n$ operations compared to the linear-time radix sorting.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-14T09:27:23.250" />
  <row Id="4621" PostTypeId="1" AcceptedAnswerId="4622" CreationDate="2010-09-14T09:30:34.807" Score="1" ViewCount="70" Body="&lt;p&gt;Below are the common set of curves that are used in a DJ mixer on the crossfader. I have a software equivalent and am using the &quot;Transition&quot; type curve but would like to include some different curve types. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On both the x and y axis' my range is from 0-1. What I usually do is have a function that gives me one side of the curve. (i.e. just the red from the diagrams below) Then as I need two levels (the red and the blue) for a given x value I invert the x value and feed it to the same formula (something like invertedX = x * -1 + 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the formula for the Dipped curve in the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra credit goes to those who can give me the formulas to the other curves &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Intermediate&lt;/li&gt;&#xA;&lt;li&gt;Constant Power&lt;/li&gt;&#xA;&lt;li&gt;I'm pretty sure Slow Fade, Slow Cut and Fast Cut are all the same formula with just a parameter difference or two.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have Transition (the easiest of them all)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IO7KF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="676" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T12:43:07.867" LastActivityDate="2010-09-14T12:43:07.867" Title="Simple Formula for Curve of DJ Crossfader volume : Dipped" Tags="&lt;algebra-precalculus&gt;&lt;functions&gt;&lt;algebraic-curves&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4622" PostTypeId="2" ParentId="4621" CreationDate="2010-09-14T09:49:23.603" Score="1" ViewCount="0" Body="&lt;p&gt;Intermediate is clearly  $y = 1 - x$ and $y = x$; should be similarly related to transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;want&lt;/em&gt; to say that Dipped is a parabola (not sure it is; hard to tell); in which case it would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ y = (x-1)^2 $ and $ y = x^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are many parabolas that fit to the points (0,0) and (1,1), or (0,1) and (1,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name &quot;Power&quot; seems to imply &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_power&quot; rel=&quot;nofollow&quot;&gt;Power in Sound&lt;/a&gt;, which would mean logarithms.  I could imagine fitting logarithms into that, but perhaps it would be easier to just use 1-dipped.  That is, $ y = 1 - (x-1)^2 $ and $ y = 1 - x^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fade/cut/cut could be cubic formulas, shifted upwards +0.5.  So they'd be variations of $x^3 + 0.5$ stretched vertically/horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll get back to this when I'm less sleepy, heh.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-09-14T09:49:23.603" CommentCount="1" />
  <row Id="4624" PostTypeId="2" ParentId="4621" CreationDate="2010-09-14T09:59:48.427" Score="1" ViewCount="0" Body="&lt;p&gt;For the slow fade may be you can take the Gaussian curve $y = e^{-x^{2}}$ and here is the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gdbEu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the transition you could try $y=-|x|$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-14T09:59:48.427" CommentCount="2" />
  <row Id="4625" PostTypeId="2" ParentId="4619" CreationDate="2010-09-14T10:01:39.670" Score="5" ViewCount="0" Body="&lt;p&gt;The difference is that NP-complete means both NP-hard and in NP. Sometimes it's not important to mention that something is in NP even if it is, so NP-hard is said instead. I don't think there's some sophisticated motive behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The halting problem is a classical example of NP-hard but not in NP problem; it can't be in NP since it's not even decidable, and it's NP-hard since given any NP-language L and an NP machine M for it, then the reduction from L to HP goes like this: reduce the input x to the input (M',x) where M' is a machine that runs M on x in all the possible ways (M is nondeterministic so there are several such ways, but a finite number since M is polynomial) and halts if M accepts in at least one of these computations. Of course, M' is not polynomial, but the reduction (producing (M',x) given x) is polynomial, so this is a polynomial-time reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes people are not satisfied with HP since it's undecidable. It's much more difficult to give a concrete example of a decidable problem, since most of the everyday problems are in PSPACE and it is yet unknown if PSPACE is different than P (if it is, than every PSPACE-complete language, such as TQBF (The language of all true quantified boolean formulas) will do. For now we need to go as far as NEXP-complete problems (since the hierarchy theorems show that NP is different than NEXP). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete problems for NEXP can be constructed from complete graph problems for NP (such as Hamiltonian cycle) if we change the representation of the graph - instead of being given directly, it is given via a black-box algorithm that takes two vertices as input and outputs whether there is an edge between them or not (so the size of the representation of the graph is exponentially smaller than it would be for if the edges were given directly).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-14T11:58:48.733" LastActivityDate="2010-09-14T11:58:48.733" CommentCount="2" />
  <row Id="4626" PostTypeId="1" AcceptedAnswerId="5374" CreationDate="2010-09-14T10:13:18.280" Score="9" ViewCount="583" Body="&lt;p&gt;&lt;em&gt;Notation&lt;/em&gt;: $b_{0}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( a_{n}/b_{n}\right) $ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction&quot; rel=&quot;nofollow&quot;&gt;Gauss Notation for generalized continued fractions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Description of the &lt;a href=&quot;http://mathworld.wolfram.com/Bauer-MuirTransformation.html&quot; rel=&quot;nofollow&quot;&gt;Bauer-Muir transformation&lt;/a&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Based on &lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;pp. 76-77&lt;/a&gt; of Lisa Lorentzen and Haakon Waadeland's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444892656&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; (A), chapter II, theorems 11, and Theorem 7;  also section 5 of J. Mc Laughlin, and Nancy J. Wyshinski's online &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0402/0402462v2.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (C), and section 5 of Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son's online &lt;a href=&quot;http://www.math.uiuc.edu/~berndt/articles/bhss.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (B) ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a convergent c.f. $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=\lim_{n\rightarrow \infty }A_{n}/B_{n}$ and a sequence ${w_{n}}$ we can construct a new c.f $\underset{n=1}{\overset{\infty }{\mathbb{K}}}({c_{n}/d_{n}})=\lim_{n\rightarrow \infty }C_{n}/D_{n}$ (if convergent) which is its &lt;em&gt;Bauer-Muir transform with respect to&lt;/em&gt; ${w_{n}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By theorem 11, chapter II, of Lisa Lorentzen and Haakon Waadeland's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444892656&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; (A), &lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;pp.76-77&lt;/a&gt;, the relations between $A_{n},B_{n}$ and $C_{n},D_{n}$ are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{n}=A_{n}+A_{n-1}w_{n}$, $D_{n}=B_{n}+B_{n-1}w_{n}$, with the initial conditions $C_{-1}=1,D_{-1}=0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for $n\geq 1$,  $\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})\neq 0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{n}=a_{n-1}\lambda_{n}/\lambda_{n-1},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda _{n-1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n\geq 2,$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=w_{0}+\dfrac{\lambda_{1}}{b_{1}+w_{1}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}(c_{n}/d_{n})}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements of $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})$ are computed by an application of Theorem 7, chapter II, of Lisa Lorentzen and Haakon Waadeland's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444892656&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; (A), which transforms a sequence into a continued fraction. &lt;em&gt;I was able to derive&lt;/em&gt; $c_{n}$ &lt;em&gt;but not&lt;/em&gt; $d_{n}$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Application to the $\log (1+t)$ expansion. By choosing $w_{0}=w_{1}=0,w_{n}=(n-1)t$, for $n\geq 2$, one can derive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left(&#xA;n+1\right) -kt\right) \right) =\dfrac{t}{2+\underset{n=3}{\overset{\infty }{%&#xA;\mathbb{K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right)&#xA;t\right) \right) }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_logarithm&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;) the expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1+t)=\displaystyle\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}t^{n}}{n}=\dfrac{t}{1+%&#xA;\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left(&#xA;n+1\right) -nt\right) \right) },$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be improved with respect to the convergence speed to this one &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1+t)=\dfrac{t}{1+\dfrac{t}{2+\underset{n=3}{\overset{\infty }{\mathbb{%&#xA;K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right) t\right)&#xA;\right) }}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation of $c_n$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how I got $c_n$. In page 77 of reference A it is proved that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{n-1}D_n-D_{n-1}C_n=(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{n-2}D_{n-1}-D_{n-2}C_{n-1}=(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n\ge 2$ Theorem 7 of reference A (derived from the fundamental c.f. recurrence) gives  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n=\dfrac{C_{n-1}D_n-D_{n-1}C_n}{-(C_{n-2}D_{n-1}-D_{n-2}C_{n-1})}=\dfrac{(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n}{-(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the determinant formula we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n-1}B_{n-2}-A_{n-2}B_{n-1}=-a_{n-1}(A_{n-2}B_{n-3}-A_{n-3}B_{n-2}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n=a_{n-1}\dfrac{\lambda_n}{\lambda_{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;QUESTION 1: How does one prove&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda_{n-1}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;from&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$d_{n}=\dfrac{C_{n}D_{n-2}-D_{n}C_{n-2}}{C_{n}D_{n-1}-D_{n}C_{n-1}}\qquad\text{for}\quad n\ge 2$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;and&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})$,&#xA;  $C_{n}=A_{n}+A_{n-1}w_{n}$,&#xA;  $D_{n}=B_{n}+B_{n-1}w_{n}$ &lt;strong&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A - Lisa Lorentzen and Haakon Waadeland, &lt;em&gt;&lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;Continued Fractions and Applications&lt;/a&gt;&lt;/em&gt;, North-Holland, Amsterdam, 1992. (&lt;a href=&quot;http://calcauxprobteor.files.wordpress.com/2010/09/cfwa-bmtpdf.pdf&quot; rel=&quot;nofollow&quot;&gt;pdf file of pp. 76-77&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B - Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son, &lt;em&gt;&lt;a href=&quot;http://www.math.uiuc.edu/~berndt/articles/bhss.pdf&quot; rel=&quot;nofollow&quot;&gt;A Transformation Formula in Rogers--Ramanujan Continued Fraction&lt;/a&gt;&lt;/em&gt;. (section 5) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C - J. Mc Laughlin, and Nancy J. Wyshinski, &lt;em&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0402/0402462v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Real numbers with polynomial continued fraction expansions&lt;/a&gt;&lt;/em&gt;, arXiv, 2004. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-05T09:31:16.683" LastActivityDate="2010-10-05T09:31:16.683" CommunityOwnedDate="2010-09-18T10:50:23.373" Title="How do I prove the partial denominators formula of the Bauer-Muir transformation of a generalized continued fraction ? " Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;continued-fractions&gt;&lt;computational-mathematics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="4627" PostTypeId="1" AcceptedAnswerId="5203" CreationDate="2010-09-14T10:28:27.603" Score="2" ViewCount="194" Body="&lt;p&gt;Hey, I'm looking for a guide on how I find Q given the following, where $a$ and $b$ are constants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{dQ}{dt} = \frac{a + Q}{b}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the answer and working for specific case I'm trying to solve but do not understand the steps involved. A guide on how I can solve this, with an explanation of each step would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-21T19:27:43.467" LastActivityDate="2010-09-22T07:25:38.223" Title="How do I find a function from a differential equation? " Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="4628" PostTypeId="2" ParentId="4627" CreationDate="2010-09-14T10:58:10.750" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Your particular differential equation can be solved without any calculation if you know the solutions of $$\frac{dQ}{dt} = Q$$, i.e. $C\exp(t), C\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, if $f$ is a solution to $\frac{dQ}{dt} = \frac 1b Q$, then $f-a$ is a solution to your equation, therefore is suffices to only look at $a=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain rule can be used to see, that if $f$ solves $\frac{dQ}{dt} = Q$, then $f(t/b)$ solves $\frac{dQ}{dt} = \frac 1b Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore all your solutions are of the form $C\exp(t/b) - a, C\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old Post:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these differential equations can be solved by &quot;Separation of Variables&quot;. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia&lt;/a&gt; for a guide. It's got examples! =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-09-14T13:23:52.040" LastActivityDate="2010-09-14T13:23:52.040" CommentCount="5" />
  <row Id="4630" PostTypeId="2" ParentId="4363" CreationDate="2010-09-14T11:33:59.377" Score="0" ViewCount="0" Body="&lt;p&gt;Here, I think is what the OP is after- we can say:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The image of the curve under the map will be '$C^1$' (that is as a given parametrisation from an interval to $\mathbb{R}^m$) if the original curve is (under similar caveats of what that means): to see this, choose a $C^1$ parameterisation $\gamma :I \to U$ and observe that $\varphi \circ \gamma$ must also be $C^1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\varphi \circ \gamma$ need not be injective: let $m=n=2$ and let $\psi$ be a bump function of height one centred around the x- axis, then the map $(x,y) \mapsto (x, (1- \psi )y )$ makes any parametrisation of the y-axis non-injective&lt;/li&gt;&#xA;&lt;li&gt;The preimage of a curve in $\mathbb{R}^m$ in $U$, even if it exists and is unique (that is, the preimage gives you something that looks like a bona fide curve rather than, say, several curves meeting at a point, or worse), need not be smooth: take $\varphi(x,y)= (x, y^2)$ and $\gamma=${$(t,t^3)| t \in [0, \infty)$}&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Indeed, I am fairly sure that the first bullet point is *the only interesting thing we can say about such curves. Fun excercise/ research practice: try to come up with more statements about such curves and see if you can find a counter example to each...&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2010-09-14T11:33:59.377" CommentCount="2" />
  <row Id="4631" PostTypeId="1" AcceptedAnswerId="4633" CreationDate="2010-09-14T11:42:43.140" Score="1" ViewCount="129" Body="&lt;p&gt;Why are differential equations called differential equations?&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T14:36:57.943" LastActivityDate="2010-09-14T14:36:57.943" Title="Why are differential equations called differential equations?" Tags="&lt;calculus&gt;&lt;terminology&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="4632" PostTypeId="1" AcceptedAnswerId="4652" CreationDate="2010-09-14T11:44:50.143" Score="2" ViewCount="156" Body="&lt;p&gt;I am trying to identify the general case algorithm for counting the different ways dice can add to a given number.  For instance, there are &lt;a href=&quot;http://wizardofodds.com/gambling/dice.html&quot; rel=&quot;nofollow&quot;&gt;six ways to roll a seven&lt;/a&gt; with two 6-dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've spent quite a bit of time working on this (for a while my friend and I were using Figurate numbers, as the early items in the series match up) but at this point I'm tired and stumped, and would love some assistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we've got something to this effect (apologies for the feeble attempt at mathematical notation - I usually reside on StackOverflow):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;count(x):&#xA;   x = min(x,n*m-x+n)&#xA;   if x = n&#xA;      1&#xA;   else&#xA;      some sort of (recursive?) operation&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first line simplifies the problem to just the lower numbers - where the count is increasing.  Then, if we're looking for the count of the minimum possible (which is also now the max because of the previous line) there is only one way to do that so it is 1, no matter the n or m.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2010-09-14T15:06:16.907" Title="How can I (algorithmically) count the number of ways n m-sided dice can add up to a given number?" Tags="&lt;algorithms&gt;&lt;probability&gt;&lt;probability-theory&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4633" PostTypeId="2" ParentId="4631" CreationDate="2010-09-14T11:49:20.940" Score="4" ViewCount="0" Body="&lt;p&gt;Because they are equations (with the variable being a function, not a number) that involve a function and its derivatives (the functions obtained by differentiating it).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T11:49:20.940" />
  <row Id="4634" PostTypeId="1" AcceptedAnswerId="4659" CreationDate="2010-09-14T11:52:09.570" Score="3" ViewCount="199" Body="&lt;p&gt;I know the &lt;a href=&quot;http://en.wikipedia.org/wiki/Four_color_theorem&quot; rel=&quot;nofollow&quot;&gt;four colour theorem&lt;/a&gt; was solved by a computer checking a large number of cases. What I don't understand is why there are only a large but finite number of cases. It seems like there should be infinitely many planar graphs... What is the intuition behind cutting this down to a finite number of configurations?&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-09-15T10:45:34.653" LastActivityDate="2010-09-15T10:45:34.653" Title="Intuition behind the &quot;large but finite search space&quot; of the proof of the four colour theorem?" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="4635" PostTypeId="2" ParentId="4619" CreationDate="2010-09-14T11:53:02.537" Score="1" ViewCount="0" Body="&lt;p&gt;As noted in the earlier answers, &lt;strong&gt;NP&lt;/strong&gt;-hard means that any problem in &lt;strong&gt;NP&lt;/strong&gt; can be reduced to it. This means that any complete problem for a class (e.g. &lt;strong&gt;PSPACE&lt;/strong&gt;) which contains &lt;strong&gt;NP&lt;/strong&gt; is also &lt;strong&gt;NP&lt;/strong&gt;-hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get a problem which is &lt;strong&gt;NP&lt;/strong&gt;-hard but not &lt;strong&gt;NP&lt;/strong&gt;-complete, it suffices to find a computational class which (&lt;em&gt;a&lt;/em&gt;)&amp;nbsp;has complete problems, (&lt;em&gt;b&lt;/em&gt;)&amp;nbsp;provably contains &lt;strong&gt;NP&lt;/strong&gt;, and (&lt;em&gt;c&lt;/em&gt;)&amp;nbsp;is provably different from &lt;strong&gt;NP&lt;/strong&gt;. So:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;While &lt;strong&gt;PSPACE&lt;/strong&gt; contains &lt;strong&gt;NP&lt;/strong&gt;, and has complete problems, the containment is not yet known to be strict. Therefore,&lt;strong&gt;PSPACE&lt;/strong&gt;-complete problems do not suffice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The same goes for any complete problem for any class in the polynomial hierarchy. If the hierarchy &lt;em&gt;does not&lt;/em&gt; collapse, any complete problem for any class in the hierarchy (aside from &lt;strong&gt;P&lt;/strong&gt;, &lt;strong&gt;NP&lt;/strong&gt;, or &lt;strong&gt;coNP&lt;/strong&gt;) would suffice; but without a proof that the PH doesn't collapse, we can't use such problems&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We &lt;em&gt;can&lt;/em&gt; prove that &lt;strong&gt;NEXP&lt;/strong&gt; (the class of problems solvable in &lt;em&gt;exponential time&lt;/em&gt; on a nondeterministic Turing Machine) strictly contains &lt;strong&gt;NP&lt;/strong&gt;, and has complete problems. So any &lt;strong&gt;NEXP&lt;/strong&gt;-complete problem is &lt;strong&gt;NP&lt;/strong&gt;-hard but not &lt;strong&gt;NP&lt;/strong&gt;-complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-09-14T14:46:24.167" LastActivityDate="2010-09-14T14:46:24.167" />
  <row Id="4636" PostTypeId="1" AcceptedAnswerId="4650" CreationDate="2010-09-14T12:06:35.343" Score="1" ViewCount="85" Body="&lt;p&gt;A polyomino is a connected subset of $\mathbb{Z}^2$ - a set of squares joined along their edges such that the resulting form is connected (or, more shortly, a generalized form of Tetris cube). A polyomino can be easily thought of as graph with vertices being elements of $\mathbb{Z}^2$ and edges between two vertices of hamming distance 1 (i.e. with one coordinate equal and the other different by 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian path/cycle problem is the problem of determining for a given graph whether it contains a path/cycle that visits every vertex exactly once. It is known to be NP-complete for general graphs and even for planar graphs. My question is whether it's still NP-complete when restricting the graphs to be polyominoes (and if it is NP-complete, how it is shown).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T13:46:50.203" Title="The Hamiltonian problem on Polyominoes" Tags="&lt;computer-science&gt;&lt;graph-theory&gt;&lt;np-complete&gt;&lt;polyominoes&gt;" AnswerCount="1" />
  <row Id="4637" PostTypeId="2" ParentId="4627" CreationDate="2010-09-14T12:09:52.217" Score="3" ViewCount="0" Body="&lt;p&gt;Yet, another method for solving this differential equation is to look at it as a &lt;em&gt;linear differential equation&lt;/em&gt;, whose general form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y'(x) = a(x) y(x) + b(x) \ , \qquad\qquad\qquad [1]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a(x), b(x)$ are arbitrary functions depending on the variable $x$. In your case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = t \ , y(x) = Q(t) \ , b(x) = \frac{a}{b} \quad \text{and}\quad a(x) = \frac{1}{b} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general procedure for solving [1] is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; First, try to solve the associated &lt;em&gt;homogeneous&lt;/em&gt; linear differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y' = a(x)y \ .  \qquad\qquad\qquad    [2]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy: the general solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = K e^{A(x)} \ ,   \qquad\qquad\qquad   [3]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K\in \mathbb{R}$ is an arbitrary constant and $A(x) = \int a(x)dx$ is a primitive function of $a(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Once you have the general solution [3] of [2], you apply &lt;em&gt;variation of constants&lt;/em&gt;; that is, you look for solutions of [1] of the following kind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = K(x) e^{A(x)} \ .   \qquad \qquad \qquad  [4]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we have replaced the arbitrary constant $K$ by an arbitrary unkown function $K(x)$ (hence the name &quot;variation of constants&quot;) to be determined. How? Imposing that we want [4] to be a solution of our first differential equation [1]. It goes like this: if you replace $y$ in [4] into [1], you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K'(x) e^{A(x)} + K(x) A'(x) e^{A(x)} = a(x) K(x) e^{A(x)} + b(x) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A'(x) = a(x) $, this is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K'(x) e^{A(x)} = b(x) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K(x) = \int b(x)e^{-A(x)}dx + C \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C \in \mathbb{R}$ is an arbitrary constant. Now you put this $K(x)$ into [4] and get the general solution of your differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y(x) = Ce^{A(x)} + e^{A(x)}\int b(x) e^{-A(x)}dx \ . \qquad\qquad\qquad   [5]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I've never could remember formula [5], I use to repeat the whole process for each particular linear differential equation, which is not hard and you can do it for yours.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-14T12:09:52.217" CommentCount="1" />
  <row Id="4638" PostTypeId="1" CreationDate="2010-09-14T12:11:31.153" Score="2" ViewCount="119" Body="&lt;p&gt;A at 45,10 ,  B at 10,20,  AB=AC and angle C=20 degree find the coordinates of C .suggest the formula so i can write code in Perl.&lt;/p&gt;&#xA;" OwnerUserId="1890" LastActivityDate="2010-09-14T19:00:53.433" Title="Find the coordinates in an isosceles triangle if the triangle it self is in positive axis" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="4639" PostTypeId="2" ParentId="4638" CreationDate="2010-09-14T12:27:25.813" Score="1" ViewCount="0" Body="&lt;p&gt;You have the coordinates of A and B, so you can compute the distance AB.  AB=AC, so you then know the distance AC.  Let the coordinates of C be (x,y).  Apply the distance formula to A and C and set the result equal to the distance you already computed.  This equation guarantees that AB=AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the angle at C is determined by the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$.  These vectors can be found by subtracting the coordinates of C from A and B (respectively).  $(CA)(CB)\cos C = \overrightarrow{CA}\cdot\overrightarrow{CB}$, and you know the distances CA = AC = AB and the measure of C, and you can compute CB and the dot product in terms of (x,y).  This equation guarantees that the measure of C is 20&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve the system resulting from the two equations above to find the coordinates of C.  There are almost certainly 2 solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See also &lt;a href=&quot;http://math.stackexchange.com/questions/3120/find-the-coordinates-in-an-isosceles-triangle&quot;&gt;this question&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T16:19:14.913" LastActivityDate="2010-09-14T16:19:14.913" />
  <row Id="4640" PostTypeId="1" AcceptedAnswerId="4642" CreationDate="2010-09-14T12:33:21.410" Score="1" ViewCount="73" Body="&lt;p&gt;When given an algrebraic expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ 6A + 7{A^2}\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and asked to simplify, would the correct answer be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)&#xA;$$ \ A(6 + 7A)\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)&#xA;That this cannot be simplified further&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-14T12:39:44.953" LastActivityDate="2010-09-14T15:55:58.170" Title="Simplification or Alternate form" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4642" PostTypeId="2" ParentId="4640" CreationDate="2010-09-14T12:39:33.503" Score="3" ViewCount="0" Body="&lt;p&gt;&quot;Simplify&quot; is not particularly well-defined.  For the expression by itself, I think (b) might be the more common textbook answer; if the expression were the denominator of a rational expression, (a) is more likely to be useful.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-14T12:39:33.503" />
  <row Id="4643" PostTypeId="1" AcceptedAnswerId="4654" CreationDate="2010-09-14T12:58:51.490" Score="1" ViewCount="101" Body="&lt;p&gt;There are a number of ways to describe this problem. I shall name a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Submultisets&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(M, f)$ be a multiset where $M = {x_1, ... x_k}, |M| = k$ and $f(x_i) = m_i$, i ranging from 0 to k inclusive. Find the number of submultisets $(S,g)$ of $(M,f)$ of size n, $|(S, g)| = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Integer Partitions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n, x_1, ... x_k, m_1, ... m_k \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the number of solutions of the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1 + x_2 + ... x_k = n$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that $0 \le x_i \le m_i$, i of course ranging from 0 to k inclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unordered combinations of size n&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a collection of objects M have k types of objects, where there are $m_i$ identical objects for each type, i from 0 to k inclusive. How many ways can I pick n elements from M, given the restriction that I cannot pick more than $m_i$ from each type?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I've made a mistake somewhere, each of these problems should be identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ordinary undergrad combinatorics literature, there is a method for picking the number of ways to solve this, without the $m_i$ restrictions I mentioned (or at least $m_i \ge n$. That is call the multiset coefficient. There is also the number of ways of ordering all of the elements of M, provided you use all of them. That is called the multinomial coefficient. This, of course, uses only n elements from M, and does not include ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution I've come across, given the 5 measly hours I've researched the problem, was in Combinatorics and Graph Theory by Harris et. al, which mentions &lt;a href=&quot;http://books.google.com/books?id=C-0y7LI2FFMC&amp;amp;printsec=frontcover&amp;amp;dq=harris+combinatorics&amp;amp;hl=en&amp;amp;ei=6nCPTOGyLcGB8gazx_WhDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q=almaalabama&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;a crude method&lt;/a&gt; for counting the number of strings of length 4 of a given set of letters &quot;Alma, Alabama&quot;. They enumerate 4 different patterns by hand, and use well known equations on those patterns. It works when n and k are small, but is inefficient otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I should have made my question a bit more explicit. I'm looking for an efficient method to do it by hand. The recurrence equations are helpful, but it's not quite what I was looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there is a simple equation that provides the answers. &lt;em&gt;A recurrence equation is likely to catch all of the cases, but may not be efficient for computing by hand. I don't think there is an efficient way to do this, if k is large and n is small. Or is there?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgive me for bringing up the concept of a multiset. I thought it was in the standard literature. The size of a multiset $|(M, f)|$ is just the summation of the multiplicities $|(M, f)| = \sum_{i} f(x_i)$. It's not the same as the size of the support $|M|$. They are only equal when $f(x_i) = 1$ for all of the elements, just a regular set.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2010-09-14T20:03:33.847" LastActivityDate="2010-09-14T21:20:49.373" Title="An efficient method for computing the number of submultisets of size n, of a given multiset" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4645" PostTypeId="2" ParentId="4619" CreationDate="2010-09-14T13:31:16.480" Score="4" ViewCount="0" Body="&lt;p&gt;I will answer the following part of the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And, explain why. I see some papers assert Degree Constrained Minimum Spanning Tree is an NP-Hard problem and some say it's NP-Complete. Why so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some people say “the Degree Constrained Minimum Spanning Tree problem (DCMST) is NP-hard” for a reason, other people say “DCMST is NP-complete” for a different reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is explained in the other answers, the word “NP-complete” means that a problem belongs to NP and is NP-hard.  Note that NP is a class of decision problems (that is, problems whose answers are yes/no).  Some people avoid saying “DCMST is NP-complete” and choose to say “DCMST is NP-hard” because DCMST is a function problem (that is, not a decision problem) and therefore cannot belong to NP by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not the end of the story.  Other people &lt;em&gt;do&lt;/em&gt; say “DCMST is NP-complete.”  When they say this, they consider the &lt;em&gt;decision version&lt;/em&gt; of the problem implicitly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The function version of DCMST: Given a graph G with edge weights and a positive integer d, find a minimum-weight spanning tree of G whose maximum degree is at most d.&lt;/li&gt;&#xA;&lt;li&gt;The decision version of DCMST: Given a graph G with edge weights and positive integers d and k, decide whether G has a spanning tree whose maximum degree is at most d and whose total weight is at most k.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, if you accept the convention to refer to the decision version as DCMST, it is perfectly fine to say that DCMST is NP-complete.  In computational complexity theory, this convention is often used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the decision version of DCMST can be solved in polynomial time if and only if the function version of DCMST can be (“if” is easy, “only if” is slightly more difficult).  This is why calling the function version and the decision version by the same name causes little confusion.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-14T13:31:16.480" CommentCount="1" />
  <row Id="4646" PostTypeId="2" ParentId="4609" CreationDate="2010-09-14T13:38:49.803" Score="0" ViewCount="0" Body="&lt;p&gt;You can work out the distribution using the transformation theorem.  If you have a normal density truncated to positive values, the log of that distribution has PDF&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2}{\sqrt{2\pi}} \exp(x -\exp(2x)/2)$&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-14T13:38:49.803" CommentCount="1" />
  <row Id="4648" PostTypeId="1" AcceptedAnswerId="4661" CreationDate="2010-09-14T13:40:32.997" Score="3" ViewCount="105" Body="&lt;p&gt;I made an assertion in &lt;a href=&quot;http://math.stackexchange.com/questions/4107/what-are-some-examples-of-theories-stronger-than-presburger-arithmetic-but-weaker&quot;&gt;What are some examples of theories stronger than Presburger Arithmetic but weaker than Peano Arithmetic?&lt;/a&gt; that Q has higher consistency strength than Pres, Presburger arithmetic; i.e., Q proves the consistency sentence for Pres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in fact, I only know something weaker, that Q can formalise the provability predicate for Hilbert systems, and so prove, say, that Peano arithmetic proves Pres consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a direct proof of consistenct of Pres in Q?&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-14T15:45:12.443" Title="Can Robinson's Q prove Presburger arithmetic consistent?" Tags="&lt;logic&gt;&lt;proof-theory&gt;&lt;consistency-proofs&gt;&lt;incompleteness&gt;" AnswerCount="1" />
  <row Id="4649" PostTypeId="2" ParentId="4634" CreationDate="2010-09-14T13:45:56.010" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps this expository account of an &quot;improved&quot; proof of the 4-color theorem will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.math.gatech.edu/~thomas/FC/fourcolor.html&quot; rel=&quot;nofollow&quot;&gt;http://people.math.gatech.edu/~thomas/FC/fourcolor.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-14T13:45:56.010" />
  <row Id="4650" PostTypeId="2" ParentId="4636" CreationDate="2010-09-14T13:46:50.203" Score="4" ViewCount="0" Body="&lt;p&gt;This problem (both the path version and the circuit version) was shown to be NP-complete by Itai, Papadimitriou and Szwarcfiter [IPS82] by a reduction from a special case of the Hamiltonian circuit problem for planar graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[IPS82] Alon Itai, Christos H. Papadimitriou and Jayme Luiz Szwarcfiter.  Hamilton paths in grid graphs.  &lt;em&gt;SIAM Journal on Computing&lt;/em&gt;, 11(4):676–686, 1982.  &lt;a href=&quot;http://dx.doi.org/10.1137/0211056&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1137/0211056&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-14T13:46:50.203" />
  <row Id="4651" PostTypeId="2" ParentId="4603" CreationDate="2010-09-14T14:03:56.783" Score="1" ViewCount="0" Body="&lt;p&gt;You might also do this the other way around. What if &#xA;$$\lim_{n\to\infty}na_n \not=0,$$&#xA;that is if the limit does not exist or if it is positive, what does this tell you about $a_n$? What about sub-sequences of $(a_n)$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-14T14:03:56.783" />
  <row Id="4652" PostTypeId="2" ParentId="4632" CreationDate="2010-09-14T14:12:25.033" Score="2" ViewCount="0" Body="&lt;p&gt;Since the order matters i.e 2+3+4 is different from 3+4+2, you can use generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways to get a sum $S$ by rolling $n$ dice (with numbers $1,2,\dots,m$) is the coefficient of $x^S$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x+x^2+\dots+x^m)^n = x^n(\frac{1-x^{m}}{1-x})^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= $$ x^n(1-x^{m})^{n}(\sum_{k=0}^{\infty} {n+k-1 \choose k} x^k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of ways is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{rm+k=S-n} {n \choose r} {n+k-1 \choose k} (-1)^{r}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-14T15:06:16.907" LastActivityDate="2010-09-14T15:06:16.907" CommentCount="8" />
  <row Id="4653" PostTypeId="1" CreationDate="2010-09-14T14:15:18.113" Score="-1" ViewCount="179" Body="&lt;p&gt;HI All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want a funny maths story that must be logically high to understand and funny enough with the use of maths concept .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The targeted audience are students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;nothing particular any topic from cloud of maths is acceptable provided that it must be understood by children :P&lt;/p&gt;&#xA;" OwnerUserId="1894" LastEditorUserId="1894" LastEditorDisplayName="" LastEditDate="2010-09-14T14:25:55.443" LastActivityDate="2010-09-14T14:53:04.023" ClosedDate="2010-09-14T15:24:23.227" Title="Maths funny Story needed for a presentation" Tags="&lt;math-history&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4654" PostTypeId="2" ParentId="4643" CreationDate="2010-09-14T14:23:31.043" Score="1" ViewCount="0" Body="&lt;p&gt;I don't really understand the description of the SubMultiset problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I do understand the description of the integer partition problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be solved by generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of solutions to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1 + x_2 + \dots + x_k = n$ with $0 \le x_i \le m_i$ is the coefficient of $x^n$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (1+x+\dots+x^{m_1})(1+x+\dots+x^{m_2})\dots(1+x+\dots+x^{m_k}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{\prod_{i}^{k}(1-x^{m_i+1})}{(1-x)^{k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = (\sum_{j=0}^{\infty}{k+j-1 \choose j} x^j) \prod_{i}^{k}(1-x^{m_i+1})$$ &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-14T14:23:31.043" CommentCount="2" />
  <row Id="4655" PostTypeId="2" ParentId="4632" CreationDate="2010-09-14T14:23:42.873" Score="0" ViewCount="0" Body="&lt;p&gt;Continuing Moron's answer, you can break the final sum as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{r=0}^{\lfloor S/(m+1) \rfloor} (-1)^r \binom{n-1+S-r(m+1)}{n-1} \binom{n}{r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for your example, $m=6$, $n=2$, $S=7$, and the formula gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{8}{1} \binom{2}{0} - \binom{1}{1} \binom{2}{1} = 8 - 2 = 6 $$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-14T14:23:42.873" />
  <row Id="4656" PostTypeId="2" ParentId="4581" CreationDate="2010-09-14T14:40:38.417" Score="2" ViewCount="0" Body="&lt;p&gt;There is also a more elementary argument explaining why $v_{1},\ldots, v_{n}$ spans $K(X)/k(X)$. To understand it one doesn't need to know separable field extensions nor integral ring extensions, only a little bit of linear algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E = span_{k(X)} \lbrace v_{1},\ldots, v_{n}\rbrace$. Note that $K[X]\subset E$ implies $K[X]E\subset E,\ 1\in E$, as $v_{1},\ldots, v_{n}$ is a basis of $K/k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an arbitrary $h\in K[X]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cdot h:K(X)\ni a\mapsto a\cdot h\in K(X)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an isomorphism of linear spaces over $k(X)$ ($h^{-1}$ defines inverse linear mapping $\cdot h^{-1}$). Furthermore $\cdot h(E)\subset E$ as $K[X]\subset E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cdot h|_{E}: E\to E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a monomorphism of finite dimensional linear spaces over $k(X)$, so in fact an isomorphism.&#xA;Therefore $\cdot h^{-1}|_{E}:E\to E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular $(\cdot h^{-1})(1) = h^{-1}\in E$, and $gh^{-1}\in E$ for every $g\in K[X]$.&lt;/p&gt;&#xA;" OwnerUserId="1454" LastActivityDate="2010-09-14T14:40:38.417" CommentCount="0" />
  <row Id="4657" PostTypeId="2" ParentId="4653" CreationDate="2010-09-14T14:53:04.023" Score="3" ViewCount="0" Body="&lt;p&gt;You can borrow from the idea Futurama presented in one of their latest episodes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://theinfosphere.org/The_Prisoner_of_Benda&quot; rel=&quot;nofollow&quot;&gt;http://theinfosphere.org/The_Prisoner_of_Benda&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the idea can be understood by children and its a nice problem which can be presented in a funny way.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-14T14:53:04.023" CommentCount="1" />
  <row Id="4658" PostTypeId="1" CreationDate="2010-09-14T14:53:33.890" Score="7" ViewCount="549" Body="&lt;p&gt;If you were to flip a coin 150 times, what is the probability that it would land tails 7 times in a row?  How about 6 times in a row?  Is there some forumula that can calculate this probability?&lt;/p&gt;&#xA;" OwnerUserId="1895" LastActivityDate="2010-09-30T19:00:50.207" Title="What is the probability of a coin landing tails 7 times in a row in a series of 150 coin flips?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" />
  <row Id="4659" PostTypeId="2" ParentId="4634" CreationDate="2010-09-14T15:19:51.157" Score="13" ViewCount="0" Body="&lt;p&gt;I'm not sure whether or not you understand how the proof of the $4$ color map theorem worked. I'll sketch that, and then make some philosophical comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by proving the $6$ color map theorem. Without loss of generality, we may assume that all the faces of our graph are triangular, because adding edges just makes a graph harder to color. (We are coloring vertices.) Let $V$, $E$ and $F$ be the number of vertices, edges and faces. Every edge is in two faces, and every face has three edges, so $F=(2/3)E$. We have $V-E+F=2$ by Euler's relation, so $V=E/3 +2$. Since the sum of all vertex degrees is $2E$, the average degree of a vertex is $2E/V = (2E)/(E/3 + 2) &amp;lt; 6$. So the average degree is less than $6$ and there must be a vertex $v$ of degree $\leq 5$. Delete $v$, $6$-color the rest of the graph (by induction), and add $v$ back. Since the vertex has degree $\leq 5$, there is some color which does not border $v$, color $v$ that color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the $5$ color map theorem requires one new idea. Find the vertex $v$ as before, color the rest of the graph and add it back. But we have a problem if the $v$ has degree $5$ and its neighbors all have different colors. The idea to get around this is to consider the subgraphs $G_{ij}$ consisting of vertices of colors $i$ and $j$, for different pairs $(i,j)$. Because $G_{ij}$ and $G_{k \ell}$ can not cross, for $i$, $j$, $k$ and $\ell$ distinct, some of these graphs must be disconnected. By considering the finitely possible many arrangements of these graphs, one sees that, in each case, one can take some component of $G_{ij}$ and switch the colors $i$ and $j$ on that component in order to make $v$ have only $4$ distinct colors of neighbor. See, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Five_color_theorem#Outline_of_the_proof_by_contradiction&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $4$ color map theorem requires the recoloring idea of the previous paragraph, plus one more idea. Instead of showing that there is always a vertex of degree $3$, $4$ or $5$, one shows that one of a finite list of larger configurations is unavoidable. This list tends to have 600-2000 members, depending on exactly which copy of the proof you read. The idea to showing that this list is unavoidable is an averaging argument like the one I used; the technical term to search for is &lt;a href=&quot;http://en.wikipedia.org/wiki/Discharging_method_%28discrete_mathematics%29&quot; rel=&quot;nofollow&quot;&gt;&quot;discharging&quot;&lt;/a&gt;. For each of these hundreds of configurations, one removes it from the graph, colors the rest, and analyzes the possible positions of the graphs $G_{ij}$. In each case one can recolor some of the components of the $G_{ij}$ in such a way that, when the configuration is put back in, the whole graph will be colorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, computers are used in two ways. The less interesting one is that there is a routine which verifies that the discharging argument is correct, finds the possible positions of the $G_{ij}$ and checks that they can be recolored. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more interesting one is that, when the routine in the previous paragraph fails, the computer perturbs the discharging argument, finds a new resulting list of configurations and tries again.&#xA;It is basically a genetic process, where the computer tweaks the discharging algorithm until it finds a list of configurations which works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leap of faith which Appel and Haken took was to believe that there was some sufficiently complex discharging algorithm, and some list of configurations for which this procedure would work, and they just had to search long enough. Since then, this search has been implemented many times, with the details of the search algorithm different in each case, and many different solutions have been found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my mind, the whole thing reminds me of &lt;a href=&quot;http://en.wikipedia.org/wiki/Parallel_evolution&quot; rel=&quot;nofollow&quot;&gt;parallel evolution&lt;/a&gt;. There are lots of ways to make a wing, and if you leave an organism for long enough in an environment that rewards flight then it will develop one. But all of those ways are so complex that a human intellegence couldn't engineer them. Similarly, there are lots of discharging algorithms that prove the $4$ color map theorem, but they are all too bizarre for a human mathematician to find directly.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-14T15:19:51.157" CommentCount="6" />
  <row Id="4661" PostTypeId="2" ParentId="4648" CreationDate="2010-09-14T15:45:12.443" Score="2" ViewCount="0" Body="&lt;p&gt;I believe that Theorem 1 of Bezboruah and Shepherdson 1976 [1] covers your question, at least in spirit. Their theory $T_0$ is a finite theory extending $Q$. Quoting their paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem 1.&lt;/strong&gt; Let $L$ be any formal system with a recursive set of axioms, a finite number of finitary and recursive rules of inference including modus ponens and having $A \to A$ as a theorem for all sentences $A$. Let&#xA;  $$Con_L =_{df}\quad \lnot(\exists y,z)(Th_L(y) \land Th_L(z) \land neg(z,y))$$&#xA;  where $\text{Th}_L$, $\text{neg}$ are given in Definition 3 below. Then $\text{Con}_L$ is not provable in $T_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The authors, however, express the common doubt that consistency proofs in Q are philosophically meaningful. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;We must agree with Kreisel that this is devoid of any philosophical interest and that in such a weak system this formula cannot be said to express consistency but only an algebraic property which in a stronger system (e.g. Peano arithmetic P) could reasonably be said to express the consistency of Q.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The (well known?) difficulty here is that Q can formalize the provability predicate but cannot verify the Hilbert-Bernays derivability conditions for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: A. Bezboruah and J. C. Shepherdson, &quot;Gödel's Second Incompleteness Theorem for Q&quot;, &lt;em&gt;The Journal of Symbolic Logic&lt;/em&gt; Vol. 41, No. 2 (Jun., 1976), pp. 503-512, &lt;a href=&quot;http://www.jstor.org/stable/info/2272251&quot; rel=&quot;nofollow&quot;&gt;JStor&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-14T15:45:12.443" CommentCount="3" />
  <row Id="4662" PostTypeId="2" ParentId="4640" CreationDate="2010-09-14T15:55:58.170" Score="1" ViewCount="0" Body="&lt;p&gt;There can be no meaningful answer without a prior definition of &quot;simpler&quot;. What is simplest or most useful is highly context dependent. For example, in some particular contexts it might prove &quot;simpler&quot; to employ $\rm\;A^2 + 6A(A+1)\;$ instead of either of the above forms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-14T15:55:58.170" />
  <row Id="4663" PostTypeId="2" ParentId="4638" CreationDate="2010-09-14T16:10:26.273" Score="2" ViewCount="0" Body="&lt;p&gt;We have $A(45,10),B(10,20),C(x_c,y_c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=B=20^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}=\pi /9$ rad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=180^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}-40^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}=140^{%&#xA;%TCIMACRO{\U{ba}}%&#xA;%BeginExpansion&#xA;{{}^o}%&#xA;%EndExpansion&#xA;}=7\pi /9$ rad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ziKPQ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's make the following change of variables: $x=X+45,y=Y+10$ (translation of axes). Then $A$ becomes the origin of the $XY$ referential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector $\overrightarrow{AB}$ can be written in this $XY$ referential as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow{AB}=(5\sqrt{53}\cos \left( \pi -\arctan \frac{2}{7}\right)&#xA;,5\sqrt{53}\sin \left( \pi -\arctan \frac{2}{7}\right) )=(-35,10)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the vector $\overrightarrow{AC}$ as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow{AC}=(5\sqrt{53}\cos \left( \pi -\frac{7\pi }{9}-\arctan &#xA;\frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\frac{7\pi }{9}-\arctan \frac{%&#xA;2}{7}\right) )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore in the original referencial $xy$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{C}=5\sqrt{53}\cos \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi&#xA;\right) +45\approx 78.239$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_{C}=5\sqrt{53}\sin \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi&#xA;\right) +10\approx 24.837$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-14T19:00:53.433" LastActivityDate="2010-09-14T19:00:53.433" />
  <row Id="4664" PostTypeId="2" ParentId="4609" CreationDate="2010-09-14T16:29:19.703" Score="3" ViewCount="0" Body="&lt;p&gt;No, it does not have a standard name.  By analogy to the lognormal distribution it might be dubbed the &quot;exponential-normal&quot; distribution.  But the analogy is imperfect, because you have to truncate the normal part at zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrals to compute the moments (or, equivalently, the characteristic function) do not have closed forms.  You will need to compute them numerically.  They are obtained by integrating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{\sqrt{2 \pi } \sigma } x^k e^{x-\frac{\left(e^x-\mu \right)^2}{2 \sigma ^2}} dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;over the entire real line.  (This comes from the pdf for a Normal distribution of $y$ with mean $\mu$ and standard deviation $\sigma$ and substituting $y = exp(x)$.)  $k = 1$ yields the mean; $k = 2$ gives the second moment; subtracting the square of the mean from that is the variance; and the square root of the variance is the standard deviation.  Finding the width at half maximum also requires numerical methods (but convergence should be rapid).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-14T16:29:19.703" CommentCount="1" />
  <row Id="4665" PostTypeId="1" AcceptedAnswerId="4673" CreationDate="2010-09-14T16:49:26.483" Score="3" ViewCount="109" Body="&lt;p&gt;Hi,&#xA;It is a well known theorem that any doubly connected domain can be conformally mapped onto an annulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the simpler version :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $D$ is a bounded domain whose boundary is two non-intersecting circles. Then $D$ can be conformally mapped onto an annulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the proof of this should be easier, that the conformal map would be just a linear fractional transformation. However, I'm having trouble constructing it. Could someone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="1897" LastActivityDate="2010-09-14T18:44:31.647" Title="Conformal mapping of a doubly connected domain onto an annulus" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4666" PostTypeId="1" AcceptedAnswerId="4677" CreationDate="2010-09-14T17:05:08.740" Score="3" ViewCount="150" Body="&lt;p&gt;In measure theory, we have &quot;lambda systems&quot; and &quot;pi systems&quot;.  Pearl's message passing algorithm has &quot;labmda messages&quot; and &quot;pi messages&quot;.  Is there a reason that lambda and pi go together?&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-09-14T20:01:25.120" Title="Why do lambda and pi go together?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4667" PostTypeId="1" CreationDate="2010-09-14T17:30:49.437" Score="2" ViewCount="103" Body="&lt;p&gt;Is it possible to quantify the number of dimensions in combinatorial spaces.  The space I am particularly interested is all partitions of a set, bounded by the Bell number, where objects in this space are particular partitions.&lt;/p&gt;&#xA;" OwnerUserId="1453" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T02:31:34.547" LastActivityDate="2010-11-24T02:31:34.547" Title="Quantifying dimensionality of combinatorial space" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4668" PostTypeId="2" ParentId="4643" CreationDate="2010-09-14T18:15:02.590" Score="0" ViewCount="0" Body="&lt;p&gt;Counting submultisets is the same as asking for the number of divisors of $\prod p_i^{m_i}$ (where the $p_i$ are distinct but otherwise arbitrary), so it's just $\prod(m_i+1)$.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-14T18:15:02.590" CommentCount="4" />
  <row Id="4669" PostTypeId="1" CreationDate="2010-09-14T18:18:32.560" Score="5" ViewCount="122" Body="&lt;p&gt;Could anyone give an example of a sequence of &lt;em&gt;differentiable&lt;/em&gt; (real) functions that uniformly converges to a &lt;em&gt;differentiable&lt;/em&gt; function, but the derivatives of which don't converge to the derivative of the limit function?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-14T19:17:06.103" Title="Example: Function sequence uniformly converges, its derivatives don't. " Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;intuition&gt;" AnswerCount="2" />
  <row Id="4670" PostTypeId="2" ParentId="4510" CreationDate="2010-09-14T18:21:33.887" Score="2" ViewCount="0" Body="&lt;p&gt;Sometimes partial fractions decomposition is useful. Especially if there are only simple poles, since then each partial fraction can be expanded in a geometric series.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-14T18:21:33.887" />
  <row Id="4671" PostTypeId="2" ParentId="4669" CreationDate="2010-09-14T18:25:45.673" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the function $f_{n}: [0,2\pi] \to \mathbb{R}$ defined by $f_{n}(x)=n^{-1/2}\sin{nx}$ and let $f:[0,2\pi] \to \mathbb{R}$ be the zero function, that is $f(x)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here i work out the details. Since $\sin{x}$ oscillates between -1 and 1, we have $d_{\infty}(f_{n},f) \leq n^{-1/2}$, where $d_{\infty}$ is the uniform metric defined as $d_{\infty}(f,g)= \sup_{x \in [0, 2\pi]} |f(x)-g(x)|$. Since $\frac{1}{\sqrt{n}} \to 0$, using squeeze test we see that $f_{n} \to f$ uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas $f_{n}'(x)= \sqrt{n}\cos{nx}$ so we have $|f_{n}'(0-f'(0)|=\sqrt{n}$, which says that $f_{n}'$ does not converge pointwise to $f'$. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-14T19:17:06.103" LastActivityDate="2010-09-14T19:17:06.103" CommentCount="0" />
  <row Id="4672" PostTypeId="2" ParentId="4669" CreationDate="2010-09-14T18:26:07.623" Score="8" ViewCount="0" Body="&lt;p&gt;This is from the book, Counterexamples in Analysis by Gelbaum and Olmstead. (Google books link: &lt;a href=&quot;http://books.google.com/books?id=cDAMh5n4lkkC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=cDAMh5n4lkkC&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is under:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A sequence of infinitely differentiable functions converging uniformly to zero, the sequence of whose derivatives diverges everywhere.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{n}(x) = \frac{\sin nx}{\sqrt{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is on page 76, Chapter 7.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-14T18:26:07.623" />
  <row Id="4673" PostTypeId="2" ParentId="4665" CreationDate="2010-09-14T18:44:31.647" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose your outer circle is the unit circle $C_1=\{z:|z|=1\}$.&#xA;All you have to do is to get a linear fractional transformation&#xA;taking $C_1$ to itself and the inner circle $C_2$ to a circle centred&#xA;at the origin. You might as well rotate $C_2$ so that its centre&#xA;is on the real axis. Then $C_2$ meets the real axis in points $a$ and $b$&#xA;with $-1 &amp;lt; a &amp;lt; b &amp;lt; 1$. The required linear fractional transformation&#xA;will have real coefficients and take $-1$, $a$, $b$, $1$ to $-1$,&#xA;$-c$, $c$, $1$ where $0 &amp;lt; c &amp;lt; 1$. The typical linear fractional transformation&#xA;fixing $-1$ and $1$ has the form&#xA;$$f_u:z\mapsto\frac{z+u}{uz+1}.$$&#xA;We are considering $u$ real and for $u$ to take the unit disc to itself,&#xA;$|u| &amp;lt; 1$. We need $f_u(a)=-f_u(b)$. This is a quadratic equation in $u$ with two&#xA;real roots, one of which satsifies $|u| &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-14T18:44:31.647" CommentCount="1" />
  <row Id="4674" PostTypeId="2" ParentId="4658" CreationDate="2010-09-14T19:06:24.917" Score="8" ViewCount="0" Body="&lt;p&gt;I'll sketch a solution; details are left to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you flip your coin, think about what data you would want to keep track of to see whether $7$ heads have come up yet. You'd want to know: Whether you have already won and what the number of heads at the end of your current sequence was. In other words, there are $8$ states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$: We have not flipped $7$ heads in a row yet, and the last flip was $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B$: We have not flipped $7$ heads in a row yet, and the last two flips was $TH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$: We have not flipped $7$ heads in a row yet, and the last three flips were $THH$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$: We have not flipped $7$ heads in a row yet, and the last seven flips were $THHHHHH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H$: We've flipped $7$ heads in a row!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are in state $A$ then, with probability $1/2$ we move to state $B$ and with probability $1/2$ we stay in state $A$. If we are in state $B$ then, with probability $1/2$ we move to state $C$ and with probability $1/2$ we move back to state $A$. $\ldots$ If we are in state $G$, with probability $1/2$ we move forward to state $H$ and with probability $1/2$ we move back to state $A$. Once we are in state $H$ we stay there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, define $M$ to be the matrix&#xA;$$\begin{pmatrix} &#xA;1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 1/2 &amp;amp; 0 \\&#xA;1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 0   &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 0   &amp;amp; 1/2 &amp;amp; 1 &#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the entries of $M^n$ give the probability of transitioning from one given state to another in $n$ coin flips. (Please, please, please, do not go on until you understand why this works! This is one of the most standard uses of matrix multiplication.) You are interested in the lower left entry of $M^{150}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a computer algebra system can compute this number for you quite rapidly. Rather than do this, I will discuss some interesting math which comes out of this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt; &#xA;&#xA;&lt;p&gt;(1) The &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem&quot; rel=&quot;nofollow&quot;&gt;Perron-Frobenius theorem&lt;/a&gt; tells us that $1$ is an eigenvalue of $M$ (with corresponding eigenvector $(0,0,0,0,0,0,0,1)^T$, in this case) and all the other eigenvalues are less then $1$. Let $\lambda$ be the largest eigenvalue less than $1$, then probability of getting $7$ heads in a row, when we flip $n$ times, is approximately $1-c \lambda^n$ for some constant $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) You might wonder what sort of configurations of coin flips can be answered by this method. For example, could we understand the case where we flip $3$ heads in a row before we flip $2$ tails in a row? (Answer: Yes.) Could we understand the question of whether the first $2k$ flips are a palindrome for some $k$? (Answer: No, not by this method.) In general, the question is which properties can be recognized by &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite-state_machine&quot; rel=&quot;nofollow&quot;&gt;finite state automata&lt;/a&gt;, also called the regular languages. There is a lot of study of this subject. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) See chapter $8.4$ of &lt;em&gt;Concrete Mathematics&lt;/em&gt;, by Graham, Knuth and Patashnik, for many more coin flipping problems.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-14T19:06:24.917" />
  <row Id="4675" PostTypeId="2" ParentId="4658" CreationDate="2010-09-14T19:11:06.607" Score="8" ViewCount="0" Body="&lt;p&gt;Here are some details; I will only work out the case where you want $7$ tails in a row, and the general case is similar.  I am interpreting your question to mean &quot;what is the probability that, at least once, you flip at least 7 tails in a row?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_n$ denote the number of ways to flip $n$ coins such that at no point do you flip more than $6$ consecutive tails.  Then the number you want to compute is $1 - \frac{a_{150}}{2^{150}}$.  The last few coin flips in such a sequence of $n$ coin flips must be one of $H, HT, HTT, HTTT, HTTTT, HTTTTT$, or $HTTTTTT$.  After deleting this last bit, what remains is another sequence of coin flips with no more than $6$ consecutive tails.  So it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n+7} = a_{n+6} + a_{n+5} + a_{n+4} + a_{n+3} + a_{n+2} + a_{n+1} + a_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions $a_k = 2^k, 0 \le k \le 6$.  Using a computer it would not be very hard to compute $a_{150}$ from here, especially if you use the matrix method that David Speyer suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, let's see what we can say approximately.  The asymptotic growth of $a_n$ is controlled by the largest positive root of the characteristic polynomial $x^7 = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, which is a little less than $2$.  Rearranging this identity gives $2 - x = \frac{1}{x^7}$, so to a first approximation the largest root is $r \approx 2 - \frac{1}{128}$.  This means that $a_n$ is approximately $\lambda \left( 2 - \frac{1}{128} \right)^n$ for some constant $\lambda$, which means that $\frac{a_{150}}{2^{150}}$ is roughly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda \left( 1 - \frac{1}{256} \right)^{150} \approx \lambda e^{ - \frac{150}{256} } \approx 0.56 \lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although $\lambda$ still needs to be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  So let's approximate $\lambda$.  I claim that the generating function for $a_n$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) = 1 + \sum_{n \ge 1} a_{n-1} x^n = \frac{1}{1 - x - x^2 - x^3 - x^4 - x^5 - x^6 - x^7}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because, by iterating the argument in the second paragraph, we can decompose any valid sequence of coin flips into a sequence of one of seven blocks $H, HT, ...$ uniquely, except that the initial segment does not necessarily start with $H$.  To simplify the above expression, write $A(x) = \frac{1 - x}{1 - 2x + x^8}$.  Now, the partial fraction decomposition of $A(x)$ has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(x) = \frac{\lambda}{r(1 - rx)} + \text{other terms}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda, r$ are as above, and it is this first term which determines the asymptotic behavior of $a_n$ as above.  To compute $\lambda$ we can use l'Hopital's rule; we find that $\lambda$ is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to \frac{1}{r}} \frac{r(1 - rx)(1 - x)}{1 - 2x + x^8} = \lim_{x \to \frac{1}{r}} \frac{-r(r+1) + 2r^2x}{-2 + 8x^7} = \frac{r^2-r}{2 - \frac{8}{r^7}} \approx 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my official guess at the actual value of the answer is $1 - 0.56 = 0.44$.  Anyone care to validate it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sequences like $a_n$ count the number of words in objects called &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_language&quot; rel=&quot;nofollow&quot;&gt;regular languages&lt;/a&gt;, whose enumerative behavior is described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation#Linear_homogeneous_recurrence_relations_with_constant_coefficients&quot; rel=&quot;nofollow&quot;&gt;linear recurrences&lt;/a&gt; and which can also be analyzed using &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite-state_machine&quot; rel=&quot;nofollow&quot;&gt;finite state machines&lt;/a&gt;.  Those are all good keywords to look up if you are interested in generalizations of this method.  I discuss some of these issues in my &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on generating functions&lt;/a&gt;, but you can find a more thorough introduction in the relevant section of Stanley's &lt;a href=&quot;http://www-math.mit.edu/~rstan/ec/&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-14T19:51:28.350" LastActivityDate="2010-09-14T19:51:28.350" CommentCount="10" />
  <row Id="4676" PostTypeId="2" ParentId="4102" CreationDate="2010-09-14T19:12:04.467" Score="0" ViewCount="0" Body="&lt;p&gt;I think I see where your confusion is rooted.  The notation f(x|θ) is chosen in the first place because there's a implication of a causal relationship between theta (the inferred cause) and x (the resulting observations.)  And also because we usually have simple pdf representing the density on x given θ. This function is necessarily invertible, so nothing forces us to write things this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the problem you presented, we are given x and inferring θ.  So, f(x|θ) is read backwards.  It could have been written L(θ|x) (the likelihood of θ given x).  And then your intuition is almost prefect, except for one flaw discussed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write that a statistic t is sufficient iff L(θ|x) = L(θ|t,x), or equivalently, L(θ|t,x) = L(θ|t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flaw with your equation is that you need to condition on both t and x. Because even if L(θ|x) = L(θ|t), t could be based on information other than (just) x.  For example, if x comprises n examples, θ could be the result of a different n examples.  In that case, the combination of t and x would be like seeing 2n examples, and would almost always yield a different result.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-09-14T19:18:03.517" LastActivityDate="2010-09-14T19:18:03.517" />
  <row Id="4677" PostTypeId="2" ParentId="4666" CreationDate="2010-09-14T20:01:25.120" Score="3" ViewCount="0" Body="&lt;p&gt;The measure theory ideas of $\pi$-system and $\lambda$-system were introduced by Dynkin in his book &lt;em&gt;Die Grundlagen der Theorie der Markoffschen Prozesse&lt;/em&gt; (1961 German translation of 1959 Russian original); a note at the end of the book mentions they are new, but doesn't explain why they are so called.  My guess has been that $\pi$ is for &quot;product&quot; and $\lambda$ is for &quot;limit,&quot; or some Russian cognates thereof; I'm not sure there's any connection between the letters themselves.  Although Professor Dynkin has recently retired, he still has an office here at Cornell; if I see him and I think of it, I may ask him about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I don't know anything about message passing, so I can't say whether Pearl's terminology is related or a coincidence.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-14T20:01:25.120" />
  <row Id="4678" PostTypeId="1" CreationDate="2010-09-14T20:08:52.257" Score="1" ViewCount="152" Body="&lt;p&gt;Let $G$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Directed_graph&quot; rel=&quot;nofollow&quot;&gt;directed graph&lt;/a&gt; which contains N vertex's, and which satisfies the following condition: each vertex of $G$ has at least one incoming edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For clarity, a vertex $V_1$ has an &quot;incoming edge&quot; if there is an edge $E$ which connects $V_1$ to some other vertex on the graph, $V_2$, for which $E$ is directed from $V_2$ to $V_1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I believe it follows that $G$ contains a closed cycle.  The reason I am posting this is because the result seems straightforward/obvious, and yet it would apparently solve an &lt;a href=&quot;http://mathoverflow.net/questions/16857/existence-of-a-zero-sum-subset&quot; rel=&quot;nofollow&quot;&gt;open problem on MO regarding zero sum subsets&lt;/a&gt;, so I am concerned that I may have a mistake.  (To be frank, I am not comfortable posting on MO)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is to show that starting with a set of N vertex's, and no edges.  That it is impossible to construct a directed graph with no closed cycles by adding one (directed) edge at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a set of N vertex's be given, and let us attempt to construct a directed graph with no closed cycle by adding one directed edge at a time.  Choose some arbitrary vertex $V_1 \in G$, then by the defining condition of $G$, $V_1$ must have an incoming edge, say from $V_2$.  It's clear that $V_2$ has an incoming edge, if this edge originates from $V_1$, then the graph would have a closed cycle, so we choose some other vertex which connects to $V_2$: and in this way we are forced to create a chain of N-1 vertex's, $V_1$, has an edge coming from $V_2$, which has an edge coming from $V_3$, ...$V_{N-2} $, which has an edge coming from $V_{N-1}$.  It's clear that if any of the vertex's in this chain has an incoming edge originating from another vertex on the chain, that $G$ will contain a closed cycle.  So once we construct a chain of N-1 distinct vertex's, there is one vertex left, and since this vertex must have an incoming edge originating from another vertex which is already on the chain, $G$ must contain a closed cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let me explain why I think this fairly trivial result solves the &lt;a href=&quot;http://mathoverflow.net/questions/16857/existence-of-a-zero-sum-subset&quot; rel=&quot;nofollow&quot;&gt;open zero sum subset problem&lt;/a&gt; on MO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question asks: given a finite set of real numbers $S$ with the property that every number in the set can be written as the sum of two numbers in the set (not necessarily different), if there exists a subset $S$ which sums to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, clearly a set $S$ which satisfies the condition of the problem stated above corresponds to a directed graph for which each vertex has at least one incoming edge (each real number in the set $S$ is associated with a vertex, and if given three numbers $a,b,c \in S$, with $a = b + c$, then the vertex associated with $a$ will have incoming edges from the vertex's associated with $b$ and $c$).   Therefore the graph corresponding to an arbitrary S which satisfies the condition contains a closed cycle.  By adding the numbers in the closed cycle, we obtain a sum of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a + b + c + .... = a$, hence it follows that $b + c + ... = 0$, i.e. a set $S$ which satisfies the condition must contain a zero sum subset.  Have I made a mistake somewhere, or does this solve the problem? &lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-09-14T20:29:47.717" LastActivityDate="2010-09-14T20:42:03.677" Title="Does this solve the zero sum subset problem?  " Tags="&lt;graph-theory&gt;" AnswerCount="0" CommentCount="4" />
  <row Id="4680" PostTypeId="1" AcceptedAnswerId="4681" CreationDate="2010-09-14T21:00:16.327" Score="4" ViewCount="129" Body="&lt;p&gt;As an example, consider the polynomial f(x) = x^3 + x - 2 = (x - 1)(x^2 + x + 2) which clearly has a root x = 1.&#xA;But we can also find the roots using Cardano's method, which leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = ((28/27)^(1/2) + 1)^(1/3) - ((28/27)^(1/2) - 1)^(1/3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and 2 other roots.&#xA;It's easy to check numerically that this expression is really equal to 1, but is there a way to derive it algebraically which isn't equivalent to showing that this expression satisfies f(x) = 0?&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-09-15T03:55:47.930" Title="Simplification of expressions containing radicals" Tags="&lt;algebra&gt;&lt;roots&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4681" PostTypeId="2" ParentId="4680" CreationDate="2010-09-14T21:12:14.693" Score="6" ViewCount="0" Body="&lt;p&gt;Yes.  The first thing to try is to guess that $\sqrt[3]{ \left( \sqrt{ \frac{28}{27} } \pm 1 \right) } = \pm \frac{1}{2} + \sqrt{a}$ for some $a$.  Cubing both sides then gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{9} \sqrt{21} \pm 1 = \pm \frac{1}{8} + \frac{3}{4} \sqrt{a} \pm \frac{3}{2} a + a \sqrt{a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $1 = \frac{1}{8} + \frac{3a}{2}$ gives $a = \frac{7}{12}$, and we can verify that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3}{4} \sqrt{a} + a \sqrt{a} = \frac{1}{8} \sqrt{21} + \frac{7}{72} \sqrt{21} = \frac{2}{9} \sqrt{21}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  If this method doesn't work then the problem becomes harder.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-14T21:12:14.693" CommentCount="4" />
  <row Id="4682" PostTypeId="1" AcceptedAnswerId="4688" CreationDate="2010-09-14T21:12:57.730" Score="2" ViewCount="67" Body="&lt;p&gt;I'm a software engineer and have been asked to research a Voronoi implementation in four dimensions. I'm not asking for &quot;teh codez&quot; but &lt;em&gt;am&lt;/em&gt; interested in approachable tutorials on Voronoi decomposition and/or implementations in any dimension. Any help here would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-14T22:08:10.453" LastActivityDate="2010-09-15T05:38:59.310" Title="Voronoi decomposition implementation in four dimensions?" Tags="&lt;algorithms&gt;&lt;reference-request&gt;&lt;discrete-geometry&gt;" AnswerCount="2" />
  <row Id="4683" PostTypeId="1" AcceptedAnswerId="4684" CreationDate="2010-09-14T21:46:05.027" Score="4" ViewCount="232" Body="&lt;p&gt;I know that a continuous function which is a BV may not be absolutely continuous.  Is there an example of such a function?   I was looking for a BV whose derivative is not Lebesgue integrable but I couldn't find one.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-09-14T23:01:42.567" LastActivityDate="2010-09-14T23:01:42.567" Title="Continuous and  bounded variation does not imply absolutely continuous." Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4684" PostTypeId="2" ParentId="4683" CreationDate="2010-09-14T22:01:46.720" Score="4" ViewCount="0" Body="&lt;p&gt;The Devil's staircase function does the trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its derivative is almost surely zero with respect to Lebesgue measure,&#xA;so the function is not absolutely continuous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/DevilsStaircase.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/DevilsStaircase.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-14T22:01:46.720" CommentCount="1" />
  <row Id="4685" PostTypeId="2" ParentId="4683" CreationDate="2010-09-14T22:10:53.583" Score="3" ViewCount="0" Body="&lt;p&gt;Byron already answered your main question, but your last sentence is another matter.  You want a BV function whose derivative is not integrable, but such things don't exist.  In particular, if $f$ is monotone on $[a,b]$, then $f'$ exists a.e., is Lebesgue integrable, and $\int_a^b f' \leq f(b)-f(a)$.  Thus half of the fundamental theorem of calculus holds, so to speak.  General BV functions are differences of monotone functions, so their derivatives are also Lebesgue integrable.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-14T22:10:53.583" CommentCount="2" />
  <row Id="4687" PostTypeId="1" CreationDate="2010-09-14T23:15:33.803" Score="5" ViewCount="102" Body="&lt;p&gt;My frustration started after hours of searching failed to turn up a formula for the vertex of a parabola in the general form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is already well known, the discriminant $\Delta=B^2-4AC$ can be used to diagnose the type of conic one has if it it given in the general form. Other useful formulae I found in the midst of searching include the formula for the slope of the principal axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\;\theta=\frac{B}{A-C+(\mathrm{sgn}\;B)\sqrt{B^2+(A-C)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the eccentricity (intended only for elliptic or hyperbolic cases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\varepsilon=\left(\sqrt{\frac12-\frac{(\mathrm{sgn}\;\Delta)(A+C)}{2\sqrt{B^2+(A-C)^2}}}\right)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the coordinates of the center for a central conic (I won't list it here so you can figure it out or find it yourself ;) ), but there were no formulae given for finding the coordinates of the vertex of a parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, for the parabola case, I could rotate axes, find the vertex through completing the square, and then rotate back, but I was hoping somebody already went through the trouble of deriving a formula so I don't have to reinvent the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to my query: is there a comprehensive compilation (book, article) somewhere of formulae related to dealing with conic sections? The implicit Cartesian form is what I'm currently dealing with, but lists relating to other forms (polar, parametric) are welcome as well.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-14T23:15:33.803" Title="Comprehensive compilation of conic section formulae" Tags="&lt;geometry&gt;&lt;reference-request&gt;&lt;conic-sections&gt;&lt;plane-curves&gt;" CommentCount="7" />
  <row Id="4688" PostTypeId="2" ParentId="4682" CreationDate="2010-09-14T23:24:44.577" Score="4" ViewCount="0" Body="&lt;p&gt;The standard survey is &lt;a href=&quot;http://www.pi6.fernuni-hagen.de/publ/tr198.pdf&quot; rel=&quot;nofollow&quot;&gt;Aurenhammer and Klein&lt;/a&gt;.  Have you looked at that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This would be a comment, but I don't have enough rep on this site yet]&lt;/p&gt;&#xA;" OwnerUserId="826" LastActivityDate="2010-09-14T23:24:44.577" />
  <row Id="4689" PostTypeId="1" CreationDate="2010-09-14T23:47:54.910" Score="1" ViewCount="40" Body="&lt;p&gt;In signal detection, an observer is assigned the task of discerning the presence (or absence) of some signal with accompanying noise. There are four possible outcomes: a hit, a miss, a false alarm, or a correct rejection. An iso-sensitivity function is the function that describes the relation between correct and false reports of a difference, with p(H) on the Y axis and p(FA) on the X axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p(H) = p + (1-p)FA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and p(FA) = g  (guess)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and p, g both lie between 0 and 1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can someone walk me through how to derive the prediction for the iso-sensitivity function in ROC space? (Or point me to some source that can clearly explain this)?&lt;/p&gt;&#xA;" OwnerUserId="1902" LastActivityDate="2010-09-14T23:47:54.910" Title="Deriving Isosensitivity Functions in ROC space from elementary signal detection parameters" Tags="&lt;calculus&gt;&lt;functions&gt;" CommentCount="1" />
  <row Id="4690" PostTypeId="2" ParentId="3405" CreationDate="2010-09-15T00:02:09.473" Score="0" ViewCount="0" Body="&lt;p&gt;Not a proof, but a conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$s_{n} = \frac{p(n)}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like the sequence $$\{(n+2)(n+3)s_{2n}\}$$ is monotonically decreasing! I verified this for the first 5000 terms or so, using a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement that the above sequence is monotonically decreasing is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture 1.1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (n+4) s_{2n+1} &amp;lt; (n+2) s_{2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to prove that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone else can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others, feel free to edit this to add your observations if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some python code&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from fractions import Fraction&#xA;&#xA;s = []&#xA;s.append(0)&#xA;s.append(Fraction(1,1))&#xA;s.append(Fraction(1,2))&#xA;&#xA;for i in xrange(3,1000):&#xA;    if (i %2 == 0):&#xA;        s.append(s[i-1])&#xA;    else:&#xA;        s.append(s[i-1] - s[(i-1)/2]/i )&#xA;&#xA;&#xA;for i in xrange(2,499):&#xA;    if (s[2*i]*(i+3) &amp;gt;= s[2*i-2]*(i+1)):&#xA;        print &quot;Failed&quot;&#xA;        print i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-15T04:16:10.193" LastActivityDate="2010-09-15T04:16:10.193" CommunityOwnedDate="2010-09-15T00:02:09.473" CommentCount="1" />
  <row Id="4691" PostTypeId="1" CreationDate="2010-09-15T02:04:14.107" Score="2" ViewCount="408" Body="&lt;p&gt;Is there a way to derive the Maclaurin series for $\frac{1}{(1-x)}$ after finding the Maclaurin series for $(1+x)^n$ which is $\displaystyle\sum\limits_{k=0}^\infty \frac{f^k(0)}{k!}*x^k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the original equation I could substitute -1 for n and -x for x but I don't see how I could do that in the summation. The intention of this is getting the next series without having to do the expansion from scratch.&lt;/p&gt;&#xA;" OwnerUserId="1880" LastActivityDate="2010-09-15T11:20:35.807" Title="Maclaurin Series of 1/(1-x) derived from maclaurin series of (1+x)^n" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="4692" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T02:08:44.560" Score="1" ViewCount="0" Body="&lt;p&gt;Try division? 1/(1-x) = 1 + x + x^2 + x^3 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: I'm not sure why you would try to derive it from the series for (1+x)^n ) &lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-09-15T02:08:44.560" CommentCount="1" />
  <row Id="4693" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T02:15:09.933" Score="6" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/BinomialSeries.html&quot; rel=&quot;nofollow&quot;&gt;binomial series&lt;/a&gt; works for any real $n$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-15T02:15:09.933" CommentCount="2" />
  <row Id="4694" PostTypeId="1" AcceptedAnswerId="4709" CreationDate="2010-09-15T02:20:53.777" Score="1" ViewCount="207" Body="&lt;p&gt;While teaching the concept of vector spaces, my professor mentioned that addition and multiplication aren't necessarily what we &lt;em&gt;normally&lt;/em&gt; call addition and multiplication, but any other function that complies with the eight axioms needed by the definition of a vector space (for instance, associativity, commutativity of addition, etc.). Is there any widely used vector space in which alternative functions are used as addition/multiplication?&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-09-15T11:33:54.470" Title="What are some alternative definitions of vector addition and scalar multiplication?" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="4695" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T02:49:29.250" Score="1" ViewCount="0" Body="&lt;p&gt;Many of the operations commonly encountered in vector spaces, such as modular arithmetic, functional convolution, or even p-adic arithmetic, ultimately are related to the addition and multiplication with which you are familiar.  One interesting exception is Conway's field of &lt;em&gt;nimbers&lt;/em&gt; (which is effectively a one-dimensional vector space).  I wouldn't say it's &quot;widely used,&quot; but it's well enough known to qualify as a genuine example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References: &lt;a href=&quot;http://mathoverflow.net/questions/6455/nimber-multiplication&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6455/nimber-multiplication&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Surreal_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Surreal_number&lt;/a&gt; (nimbers form a subfield of the surreal numbers).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-15T02:49:29.250" />
  <row Id="4696" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T02:50:19.483" Score="3" ViewCount="0" Body="&lt;p&gt;This example is certainly not &quot;widely used&quot;, but I think it's worth thinking about anyway.  This answer comes from an MO post by John Goodrick: &lt;a href=&quot;http://mathoverflow.net/questions/9402/pedagogical-question-about-linear-algebra&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/9402/pedagogical-question-about-linear-algebra&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll quote the entirety of his post here in case you don't feel like clicking on the link (and since I've done no work on my own, I'll make this post CW)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You could try giving the following example: the set of all positive  real numbers, considered as a vector space over the field R, with vector addition given by multiplication and scalar multiplication given by taking exponents.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As a first step, you could verify that this satisfies a few of the vector-space axioms, and then let students check the rest of them (say, as homework). Then, you could ask questions like, &quot;what is the dimension of this vector space?&quot; or, &quot;give an example of a (nontrivial) linear transformation from this space into R^3.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-09-15T02:50:19.483" CommunityOwnedDate="2010-09-15T02:51:22.770" />
  <row Id="4697" PostTypeId="2" ParentId="4680" CreationDate="2010-09-15T03:03:43.943" Score="7" ViewCount="0" Body="&lt;p&gt;There are very general algorithms known for radical denesting. Below is the structure theorem which lies at the foundation of these algorithms. It widely generalizes the heuristic employed by Qiaochu in his answer. It may be employed heuristically - in a similar manner as Qiaochu - to perform complicated denestings, without requiring much comprehension of the underlying theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Bloemer's FOCS '91, '92 papers you'll find his polynomial-time algorithms for radical denesting. Informally, the Denesting Structure Theorem says that if a radical $\rm\; r^{1/d} \;$ denests in any radical extension $\rm\; F' \;$ of its base field $\rm\; F \;$, then a suitable multiple $\rm\; q b\: r \;$ of &#xA;the radicand $\rm\; r \;$ must already denest in the field $\rm\; F' \;$ defined by the &#xA;radicand. More precisely &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DENESTING STRUCTURE THEOREM&lt;/strong&gt;$\;\; \;$  Let $\rm\; F \;$ be a real field and &#xA;$\rm\; F' = F(q_1^{1/d1},\ldots,q_k^{1/dk}) \;$ be a real radical extension of $\rm\; F \;$ &#xA;of degree $\rm\; n \;$. By $\rm\; B = \{b_0,\ldots, b_{n-1}\}$ denote the standard &#xA;basis of $\rm\; F' \;$ over $\rm\; F \;$. If $\rm\; r \;$ is in $\rm\; F' \;$ and $\rm\; d \;$ is a positive integer such &#xA;that $\rm\; r^{1/d} \;$ denests over $\rm\; F \;$ using only real radicals, that is, &#xA;$\rm\; r^{1/d} \;$ is in $\rm\; F(a_1^{1/t_1},\ldots,a_m^{1/t_m}) \;$ for some positive integers &#xA;$\rm\; t_i \;$ and positive $\rm\; a_i \in F \;$, then there exists a nonzero $\rm\; q \in F \;$ and a &#xA;$\rm\; b \in B \;$ such that $\rm\; (q b r)^{1/d} \in F' \;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. multiplying the radicand by a $\rm\; q \;$ in the base field $\rm\; F \;$ &#xA;and a power product $\rm\; b \;=\; q_1^{e_1/d_1}\cdots q_k^{e_k/d_k} \;$ we can &#xA;normalize any denesting so that it denests in the field defined &#xA;by the radicand. E.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \;\;=\; \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$&#xA;normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example with nontrivial $\rm\:b$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;normalises to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\rm\; F=\mathbb Q,\ F' = \mathbb Q(\sqrt 6),\ n=2,\ B = \{1,\sqrt 6\},\ d=2,\ q=1/3,\ b= \sqrt 6\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure theorem also hold for complex fields except that &#xA;in this case one has to assume that $\rm\; F \;$ contains enough roots of &#xA;unity (which may be computationally expensive in practice, to &#xA;wit doubly-exponential complexity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the complexity of even simpler problems involving &#xA;radicals is currently unknown. For example, no polynomial time &#xA;algorithm is known for determining the sign of a sum of real &#xA;radicals $\rm\; \sum{c_i\: q_i^{1/r_i}} \;$ where $\rm\; c_i,\: q_i \;$ are rational numbers and &#xA;$\rm\; r_i \;$ is a positive integer. Such sums play an important role in &#xA;various geometric problems (e.g. Euclidean shortest paths and &#xA;traveling salesman tours). Even though testing whether such a sum &#xA;of radicals is zero can be decided in polynomial time, this is &#xA;of no help in determining the sign, it only shows that if sign &#xA;testing is in $\rm\; NP \;$ then it is already in $\rm\; NP \cap \text{co-NP} \;$. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-15T03:55:47.930" LastActivityDate="2010-09-15T03:55:47.930" CommentCount="2" />
  <row Id="4698" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T03:07:04.753" Score="4" ViewCount="0" Body="&lt;p&gt;My favorite example is the set of subsets of a set under the operation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_difference&quot; rel=&quot;nofollow&quot;&gt;symmetric difference&lt;/a&gt; (otherwise known as bitwise XOR).  This forms a vector space over the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_field&quot; rel=&quot;nofollow&quot;&gt;finite field&lt;/a&gt; $\mathbb{F}_2$.  This example is important in computer science, coding theory, combinatorics, ...&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-15T03:07:04.753" />
  <row Id="4699" PostTypeId="1" CreationDate="2010-09-15T03:53:04.560" Score="1" ViewCount="85" Body="&lt;p&gt;Among the smooth 1-manifolds (with or without boundary) which embed into $\mathbb{R}^2$, which ones can be represented by a single parametrization $z = (x,y) = f(t)$, for $t \in I$, where $I$ is an interval (not necessarily open or closed), and $f$ is smooth (i.e. infinitely differentiable)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask is that in my reading, I've come across two definitions for &quot;line integral&quot; (which I'm sure turn out to be equivalent):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is the standard definition given in most multivariable calculus and complex analysis classes, which relies on defining a &quot;curve&quot; as a (sufficiently nice) function (or image set) $z = \gamma(t)$.  That is: $\int_\gamma f(z) dz = \int_a^b f(g(t))g'(t) dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is the more high-powered definition involving 1-manifolds and differential 1-forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is not really about whether or not these two definitions are equivalent per se, but rather about how much generality is lost by looking at only the &quot;special&quot; 1-manifolds which admit representations as $z = \gamma(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My own thoughts: (1) Do these 1-manifolds end up being the 1-manifolds with an atlas consisting of one chart?  I'm thinking not, because the circle cannot be given an atlas with one chart, yet can still be parametrized by a single function. (2) Does the Implicit Function Theorem have any role to play?)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-09-15T06:46:28.080" LastActivityDate="2010-09-15T06:46:28.080" Title="Which smooth 1-manifolds can be represented by a single smooth parametrization?" Tags="&lt;complex-analysis&gt;&lt;differential-topology&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="4700" PostTypeId="1" AcceptedAnswerId="4712" CreationDate="2010-09-15T04:26:54.693" Score="2" ViewCount="182" Body="&lt;p&gt;Here's the problem (from little Spivak):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\begin{cases}&#xA;e^{-x^{-2}} &amp;amp; x\ne 0 \\&#xA;0 &amp;amp; x=0 &#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f$ is a $C^\infty$ function, and $f^{(i)}(0)=0$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, here's my proof.  It gets pretty ugly at the end and I do a little handwaving (every term is of the form ... pretty much &quot;because I say so&quot;, and the last step should be a better form of induction instead of &quot;successive applications of the rule...&quot; but the exact way to say it correctly escapes me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For arbitrary $p,q:\mathbb{R}\rightarrow\mathbb{R}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(e^{q(x)}p(x))'=e^{q(x)} q'(x)p(x) + e^{q(x)} p'(x)=e^{q(x)}(q'(x)p(x)+p'(x))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q(x)=-x^{-2}$, $p_0(x)=1$, and $p_{n+1}(x)=q'(x)p_{n}(x)+p_n'(x)=\frac{2}{x^3} p_n(x) + p_n'(x)$.  Every term of $p_n(x)$ is therefore of the form $a x^{-n}$ for some $a,n\in\mathbb{R}$ and $n&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f^{(n)}(x)=e^{q(x)} p_n(x)=e^{-x^{-2}}p_n(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the definition of derivative,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;f^{(n+1)}(0) &amp;amp;= \lim_{h\rightarrow 0}\frac{f^{(n)}(h)}{h}\&#xA;&amp;amp;= \lim_{h\rightarrow 0}\frac{e^{-x^{-2}} p_n(h)}{h}\&#xA;&amp;amp;= \lim_{h\rightarrow 0}\frac{h^{-1} p_n(h)}{e^{h^{-2}}}\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since every term of $p_n(h)$ is of the form $a h^{-m}$ for some $a,m\in\mathbb{R},n&gt;0$, it follows that every term of $h^{-1} p_n(h)$ is also of the same form.  Thus if we can show that $\lim_{h\rightarrow 0}\frac{ah^{-m}}{e^{h^{-2}}}=0$, the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying L'Hopital's rule results in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{h\rightarrow 0}\frac{-m h^{-m-1}}{e^{h^{-2}} (-\frac{2}{h^3})}=\lim_{h\rightarrow 0}\frac{2m h^{-m+2}}{e^{h^{-2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Successive applications of L'Hopital's rule result in multiplying numerator by some constant and increasing the exponent of $h$ by 2.  Thus apply the rule $m/2+1$ times will give a positive exponent, and the limit will be clearly 0.&lt;/p&gt;&#xA;" OwnerUserId="326" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-15T17:30:07.710" LastActivityDate="2010-09-15T17:30:07.710" Title="Improve my proof about this $C^\infty$ function." Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4701" PostTypeId="2" ParentId="4682" CreationDate="2010-09-15T05:38:59.310" Score="2" ViewCount="0" Body="&lt;p&gt;Even though you didn't ask for 'teh codez' :), &lt;a href=&quot;http://www.qhull.org/html/qvoronoi.htm&quot; rel=&quot;nofollow&quot;&gt;qvoronoi&lt;/a&gt; is a pretty good implementation of Voronoi diagrams for low dimensions (i.e 4). It actually computes the Delaunay triangulation, which in turn is constructed by a convex hull algorithm (qhull) one dimension higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convex hull algorithm itself &lt;a href=&quot;http://www.qhull.org/html/index.htm#algorithm&quot; rel=&quot;nofollow&quot;&gt;is a combination of&lt;/a&gt; &quot;the 2-d Quickhull algorithm with the n-d beneath-beyond algorithm [c.f., Preparata &amp;amp; Shamos '85]&quot;&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-09-15T05:38:59.310" CommentCount="1" />
  <row Id="4702" PostTypeId="2" ParentId="4581" CreationDate="2010-09-15T08:48:59.140" Score="1" ViewCount="0" Body="&lt;p&gt;Here is what I believe I understand of the other answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K/k$ be an extension and $X$ an indeterminate. We identify $k(X)\otimes_kK$ with $k(X)\otimes_{k[X]}K[X]$, and denote this domain by $A$. Everything will follow from the &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;OBSERVATION.&lt;/strong&gt; Any element of $A$ can be written as &#xA;  $$\frac{1}{p(X)}\otimes P(X)$$ with $p(X)$ in $k[X]$, $p(X)\not=0$, $P(X)$ in $K[X]$. Its image in $K(X)$ is $P(X)/p(X)$. In particular, the natural map from $A$ to $K(X)$ is injective. Thus we can (and will) view $A$ as the subdomain of $K(X)$ formed by the elements which can be written, in the above notation, as $P(X)/p(X)$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fraction field of $A$, regarded as a subfield of $K(X)$, is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product_of_fields#Compositum_of_fields&quot; rel=&quot;nofollow&quot;&gt;compositum&lt;/a&gt; of $k(X)$ and $K$. It is denoted $k(X)K$, and coincides with $K(X)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a$ is in $K$ and $1/(X-a)$ is in $A$, then $a$ is algebraic over $k$. So, $A$ is &lt;strong&gt;not&lt;/strong&gt; a field if $K/k$ is transcendental. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K/k$ has finite degree, the domain $A$, being finite dimensional over $k(X)$, is a field, and is thus equal to $K(X)$. The same conclusion holds if $K$ is a union of finite degree extensions of $k$, that is, if $K/k$ is algebraic. &lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-09-16T08:59:44.957" LastActivityDate="2010-09-16T08:59:44.957" />
  <row Id="4703" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T09:15:34.800" Score="2" ViewCount="0" Body="&lt;p&gt;The expression $\displaystyle \sum_{k=0}^{\infty} \frac{f^k(0)}{k!} x^k$ is not just the Maclaurin series for functions of the form $f(x)=(1+x)^n$; it is a formula for the Maclaurin series of any function infinitely differentiable at $0$.  It gives us the power series in $x$ with all derivatives at $x=0$ matching the derivatives of $f(x)$ at $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use this formula to find the Maclaurin series for $f(x)=\frac{1}{1-x}$, you need to compute all of the derivatives of $f(x)$.  --Fortunately, there is a simple pattern you can observe and prove (by induction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the derivatives, you can plug in $x=0$, simplify your summation formula, and you'll have your answer.  &lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-15T09:15:34.800" />
  <row Id="4704" PostTypeId="2" ParentId="4466" CreationDate="2010-09-15T10:42:42.653" Score="0" ViewCount="0" Body="&lt;p&gt;I did this same problem for a class on number theory, the book omits a constraint that n&gt;3 since 3^2 -1 = 8 which is divisible by 1,2,4,8 and 4 is not prime it does however hold for all n&gt;3. Intuitively this is a consequence of the fact that 2 is the only even prime and that the product of three consecutive integers is divisible by 6 (n-1 being divisible by two would typically eliminate its candidacy for being prime EXCEPT for the n-1 = 2 case). Proving that a one-to-one correspondence exists is done by proving a that there is an iff statement relating (n-1),(n+1) being prime and (n-1)(n+1) having 4 dividers. Since this would imply that the set of all n such that n^2-1 has 4 dividers is equal to the set of n s.t. (n-1),(n+1) are prime and the function mapping a set back to itself is a trivial bijective map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One direction is pretty easy, (n-1),(n+1) prime =&gt; s = (n-1)(n+1) = p1*p2 therefore s has 4 dividers, 1,s,p1, and p2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other direction is left for you to figure out,&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-09-15T10:42:42.653" />
  <row Id="4705" PostTypeId="1" AcceptedAnswerId="4747" CreationDate="2010-09-15T11:00:54.413" Score="2" ViewCount="348" Body="&lt;p&gt;I am looking for an intuitive explanation of $\liminf$ and $\limsup$ for sequence of sets and how it corresponds to $\liminf$ and $\limsup$ for sets of real numbers. I researched online but cannot find a good comparison. Any link, reference or answer very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, what is $\liminf$ and $\limsup$ of real number sequences $a_n=(-1)^n$ and $b_n=1/n$. Corresponding to this, what is the $\liminf$ and $\limsup$ of sequence of sets $A_n=\{(−1)^n\}$ and $B_n=\{1/n\}$?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T20:34:11.740" LastActivityDate="2010-11-24T20:34:11.740" Title="limit inferior and superior for sets vs real numbers" Tags="&lt;measure-theory&gt;&lt;elementary-set-theory&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="4706" PostTypeId="1" CreationDate="2010-09-15T11:11:32.410" Score="1" ViewCount="86" Body="&lt;p&gt;Under the condition that the spaces (or maybe just the total) are connected and locally path connected, is then the a covering the same as a homeomorphism? &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-09-15T11:28:49.493" Title="Covering for connected and locally path connected spaces" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4707" PostTypeId="2" ParentId="4691" CreationDate="2010-09-15T11:20:35.807" Score="5" ViewCount="0" Body="&lt;p&gt;Elaborating on J.M. answer, you can procede this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1+x)^n = 1+ n x + \frac{1}{2}(n-1) n x^2 + \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$, more generally,&#xA;$$(1+x)^\alpha = 1 +  \sum_{k = 1}^{n} \binom{\alpha}{k} x^k + o(x^n)$$ for any real $\alpha$, where $\binom{\alpha}{k} = \frac{(\alpha -1)(\alpha-2)\ldots (\alpha - k +1)}{k!}$ is the binomial coefficient extended to any real number. Then, you get that $$(1-x)^n = 1 - nx + \frac{1}{2}(n-1)n x^2 - \frac{1}{6}(n-2)(n-1)n x^3 + \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, you only have to plug in $n = -1$ to get $$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots$$&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-15T11:20:35.807" />
  <row Id="4708" PostTypeId="2" ParentId="4706" CreationDate="2010-09-15T11:28:49.493" Score="-3" ViewCount="0" Body="&lt;p&gt;I would suggest you to take a look at this article: &lt;a href=&quot;http://arxiv.org/pdf/0801.4967v3&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/0801.4967v3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-15T11:28:49.493" CommentCount="4" />
  <row Id="4709" PostTypeId="2" ParentId="4694" CreationDate="2010-09-15T11:33:54.470" Score="3" ViewCount="0" Body="&lt;p&gt;What do you &quot;normally&quot; call addition and multiplication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Just&lt;/em&gt; those operations with real numbers, or &lt;em&gt;all&lt;/em&gt; kind of addition and multiplication &quot;derived&quot; from the well-known operations with real numbers, or that &quot;look like&quot; these operations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, in the first case, you have plenty of elementary and widely used vector spaces with operations which are &lt;em&gt;not&lt;/em&gt; those of real numbers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\mathbb{R}^2$, the set of ordered pairs of real numbers $(x,y)$, is a real vector space, with addition and multiplication defined as $(x,y) + (u,v) = (x+u, y+v)$ and $\lambda (x,y) = (\lambda x , \lambda y)$. These operations are defined using the &quot;normal&quot; addition and multiplication of real numbers, but are &lt;em&gt;not&lt;/em&gt; the &quot;normal&quot; addition and multiplication of real numbers just because $(x,y)$ is &lt;em&gt;not&lt;/em&gt; a real number.&lt;/li&gt;&#xA;&lt;li&gt;${\cal C}^0 (\mathbb{R}, \mathbb{R})$, the set of continuous functions $f:\mathbb{R} \longrightarrow \mathbb{R}$, is a real vector space, with addition and multiplication defined point-wise; that is $(f+g)(x) = f(x) + g(x)$ and $(\lambda f)(x) = \lambda f(x)$. Again, these addition and multiplication are defined using the &quot;normal&quot; addition and multiplication of real numbers, but are not the &quot;normal&quot; addition and multiplication of real numbers for the same reason.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Z}/2\mathbb{Z}$, the set of integers mod 2, is a $\mathbb{Z}/2\mathbb{Z}$-vector space, with addition and multiplication $\widetilde{m} + \widetilde{n} = \widetilde{n+m}$ and $\widetilde{\lambda}\widetilde{m} = \widetilde{\lambda m}$, where $\widetilde{m}$ denotes the class of $m$ mod 2. Ditto.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{R}(x)$, the field of rational functions $\frac{p(x)}{q(x)}$, where $p(x), q(x) \in \mathbb{R}[x]$ are polynomials, $q(x) \neq 0$, is a $\mathbb{R}(x)$-vector space, with addition and multiplication $\frac{p(x)}{q(x)} + \frac{r(x)}{s(x)} = \frac{p(x) s(x) + r(x) q(x)}{q(x)s(x)} $ and $\frac{p(x)}{q(x)}  \frac{r(x)}{s(x)} = \frac{p(x)r(x)}{q(x)s(x)}$. Ditto.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{C}$, the set of complex numbers, is a $\mathbb{C}$-vector space, whit the addition and multiplication of complex numbers. Ditto.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{K}^n$, the set of ordered families $(x_1, \dots , x_n)$ of elements of any field $\mathbb{K}$, is a $\mathbb{K}$-vector space, with addition and multiplication defined as in example 1. Examples 3, 4 and 5 are particular cases of this one with $n=1$ and $\mathbb{K} =$ $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{R}(x)$ and $\mathbb{C}$, respectively. Example 1 is also a particular case, with $n=2$ and $\mathbb{K} = \mathbb{R}$. Addition and multiplication in $\mathbb{K}$ may have nothing in common with the operations with real numbers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-15T11:33:54.470" />
  <row Id="4710" PostTypeId="2" ParentId="4705" CreationDate="2010-09-15T11:43:40.493" Score="0" ViewCount="0" Body="&lt;p&gt;As I know the limits of sequences of sets, the limit superior consists of all points contained in infinitely many of these sets, the limit inferior are all points contained in almost all sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\limsup A_n = \{x | x\in A_n \text{ for infinitely many }n\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\liminf A_n = \{x | x\in A_n \text{ for all but finitely many }n\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, they do not correspond with the limits of sequences of real numbers in an obvious way. I've come to the conclusion that they are poorly named, but I may lack insight here. =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-09-15T12:05:23.763" LastActivityDate="2010-09-15T12:05:23.763" CommentCount="4" />
  <row Id="4711" PostTypeId="1" AcceptedAnswerId="4777" CreationDate="2010-09-15T12:08:49.927" Score="3" ViewCount="212" Body="&lt;p&gt;If I have an arbitrary shape, I would like to fill it only with 45-45-90&#xA;triangles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aim is to get a Tangram look, so it's related to &lt;a href=&quot;http://stackoverflow.com/questions/3717178/how-can-i-fill-an-outline-with-predefined-tangram-shapes&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with 45-45-90 triangles would be an amazing start. After&#xA;the shape if filled I imagine I could pick adjacent triangles, either&#xA;2 or 4 and draw squares and parallelograms instead, but just getting&#xA;the outline estimated with packed 45-45-90 triangles would be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I get started ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;@J.M.'s comment makes perfect sense, which means I should first make sure my shape is suited for this. Here's a sketch I did to illustrate this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ROZGP.gif&quot; alt=&quot;hand&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black shape is the 'arbitrary shape', the blue path is the path I should be filling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I see the first step is to 'estimate' arbitrary paths with lines at right or 45 degree angles. The second step would be the initial question, packing 45-45-90 triangles into the shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hints for estimating random angles lines with 45/90 degrees angled lines or 45-45-90 triangle packing ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gone ahead with a naive approach to estimate an arbitrary path(outline) using only straight or diagonal(45 degrees) lines.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function estimate45(points:Vector.&amp;lt;Point&amp;gt;):Vector.&amp;lt;Point&amp;gt; {&#xA;    var result:Vector.&amp;lt;Point&amp;gt; = new Vector.&amp;lt;Point&amp;gt;();&#xA;    var pNum:int = points.length,angle:Number,pi:Number = 3.141592653589793,qpi:Number = pi*.25,d:Number = 0;&#xA;    for(var i:int = 0 ; i &amp;lt; pNum ; i++){&#xA;        if(i == 0) angle = Math.atan2(points[i].y,points[i].x);&#xA;        else {&#xA;            angle = Math.atan2(points[i].y-points[i-1].y,points[i].x-points[i-1].x);&#xA;            d = Math.sqrt((points[i].x-points[i-1].x)*(points[i].x-points[i-1].x)+(points[i].y-points[i-1].y)*(points[i].y-points[i-1].y))&#xA;        }&#xA;        //constraint to 45 (1. scale to values between 0,1 (/qpi) 2. round, 3. do whatever(convert to degrees/radians as needed) &#xA;        angle = (Math.round(angle/qpi)) * 45 / 57.2957795;&#xA;        if(i == 0) result.push(new Point(Math.cos(angle)*d,Math.s(angle)*d));&#xA;        else       result.push(new Point(result[i-1].x+Math.cos(angle)*d,result[i-1].y+Math.s(angle)*d));&#xA;    }&#xA;    return result;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I loop trough the the path (an ordered list of points) and I calculate the angle and radius of each line(cartesian to polar I think). Then I 'round' the angle to 45 degrees,&#xA;and draw a line to a 45 degrees constrained version of the original angle, and keep the same length for the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't very good way to do it, especially for consecutive lines with similar angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some tests:&#xA;&lt;img src=&quot;http://i.imgur.com/gFFAK.gif&quot; alt=&quot;crescent&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The faded red is the original, the green is the estimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Vgiec.gif&quot; alt=&quot;hand 45&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Américo Tavares's suggestion is great though. I could use this approach for bitmap graphics too, not just vector graphics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I would go with this approach, I imagine I would do something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;get a mosaic(create a grid of boxes to cover the size of the shape)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;boxes_xnum = floor(w/box_size)&#xA;boxes_ynum = floor(h/box_size)&#xA;for(y to boxes_ynum):&#xA;    for(x to boxes_xnum):&#xA;        grid.addBitmap(copyPixels(source,x*box_size,y*box_size,box_size,box_size));//copypixels(source,x,y,width,height)&#xA;&#xA;for box in grid:&#xA;   if(box.nonAlphaPixels/box.totalPixels &amp;gt; .75): fullBox&#xA;   else:&#xA;        checkDiagonalType()//is it like this / or like this \&#xA;        checkFillSide()//which of the two sides should be filled&#xA;        //which means I should check for something constrained to 45 degrees angles \| or _\ or |/ or /_&#xA;        //in the case of halfs go for random diagonal ? &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I think about this better,&lt;br&gt;&#xA;when I loop though the pixels of a box, keep a pixel count per box 'quadrant'(top-left).&#xA;If the non transparent pixels in one quadrant in larger than .5 or .75 it's marked as used.&#xA;Based on how many and which of the 4 quadrants are used in a box, a diagonal with direction is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2CLf7.jpg&quot; alt=&quot;solve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense, or am I over complicating this ?&lt;/p&gt;&#xA;" OwnerUserId="304" LastEditorUserId="304" LastEditorDisplayName="" LastEditDate="2010-09-16T14:46:59.817" LastActivityDate="2010-09-16T16:29:51.100" Title="How can I pack 45-45-90 triangles inside an arbitrary shape ?" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="4712" PostTypeId="2" ParentId="4700" CreationDate="2010-09-15T13:16:58.637" Score="2" ViewCount="0" Body="&lt;p&gt;You seem to have two main observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;statement $A_{n}$ (for $n \ge 1$):&#xA;$$f^{n}(x) = e^{q(x)}p_{n}(x), x \neq 0$$&#xA;where $$p_{n}(x) = \sum \frac{a_{r}}{x^r}, x \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;statement $B_{n}$ (for $n \ge -1$):&#xA;$$\lim_{h \to 0} \frac{1}{h^{n} e^{h^{-2}}}  = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result you have is that $A_{n-1}$ implies $A_{n}$ and that $B_{n-2}$ implies $B_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you rewrite those formally using the right form of induction, with proper base cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have those, you finally can prove the third statement $C_{n}$  (for $n \ge 1$) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f^{n}(0) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't explicitly spelled this out in your proof, but you are showing that $C_{n}$ (combined with the $A_{i}$ and $B_{j}$) implies $C_{n+1}$ which can again be formalized by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance the statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{(n+1)}(0) = \lim_{h\rightarrow 0}\frac{f^{(n)}(h)}{h}$$ is implicitly using the fact that $f^{n}(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-15T13:16:58.637" CommentCount="1" />
  <row Id="4713" PostTypeId="1" AcceptedAnswerId="4715" CreationDate="2010-09-15T13:29:47.237" Score="1" ViewCount="210" Body="&lt;p&gt;I've just come up with this question as I'm studying for a number theory midterm. If $p$ and $q$ are different prime numbers, and it's known that $2^p \equiv 1 \bmod{q}$, then $q\equiv 1 \bmod{p}$. I've tried a few cases and it seems to be true. How can it be proved?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T15:26:13.887" LastActivityDate="2010-09-17T15:26:13.887" Title="If the order divides a prime P then the order is P (or 1)" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="4715" PostTypeId="2" ParentId="4713" CreationDate="2010-09-15T13:57:24.333" Score="8" ViewCount="0" Body="&lt;p&gt;$\rm\displaystyle\;\ 2^p,\;  2^{q-1} \equiv 1 \;\Rightarrow\; 2^{gcd(p,q-1)} \equiv 1\ (mod\  q)\;\Rightarrow\; gcd(p,q-1) = p\; $ (not $1$ else $\rm \:q|2^1-1\:$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said in group theory language: if an element $\ne 1$ has order dividing a prime $\:\rm p\;$ then it must have order $\rm p\:$. In ring theory language: if a principal ideal$\;\ne 1$ contains an irreducible element then that element generates the ideal (which is why the minimal polynomial is sometimes called the &quot;irreducible polynomial&quot;). Here the group / ideal is simply the so-called &lt;strong&gt;order ideal&lt;/strong&gt; $\rm\; \{n : x^n = 1 \}$ which, being nonempty and closed under subtraction, comprises a subgroup / ideal of $\:\mathbb Z\:$. The Euclidean algorithm implies that ideals in $\mathbb Z$ are principal, so every element of a nonzero ideal is a multiple of the least positive element. For an order ideal this simply says that every &quot;possible&quot; order is a multiple of the least possible order $\rm m\:$, $\:$ i.e. $\rm\; x^n = 1 \;\Rightarrow\; m\:|\:n\;$. Compare this ring-theoretic proof to the ubiquitous group-theoretic proof using Lagrange's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple additive example: the order ideal of a fraction is the &lt;strong&gt;denominator ideal&lt;/strong&gt; $\rm\; \{n : n\: x \in \mathbb Z \:\}\:$. Here the analogous result is that if a proper fraction can be written with a prime denominator then it is the least possible denominator (in the multiplicative sense), i.e. it divides ever other possible denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denominator ideals and fractional ideals can yield significant number-theoretical simplifications - in much the same way that fractional arithmetic often serves to simplify integer arithmetic in elementary number theory. For example, see my post on &lt;a href=&quot;http://math.stackexchange.com/questions/4491&quot;&gt;irrationality proofs using the denominator ideal&lt;/a&gt; or, more generally, see my proof employing Dedekind's &lt;a href=&quot;http://mathoverflow.net/questions/30313&quot; rel=&quot;nofollow&quot;&gt;conductor ideal&lt;/a&gt; - a sort of universal denominator for an entire extension ring. Follow the links there to my sci.math posts for much further discussion on this and related topics.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T15:19:44.253" LastActivityDate="2010-09-17T15:19:44.253" />
  <row Id="4717" PostTypeId="1" CreationDate="2010-09-15T14:05:37.410" Score="2" ViewCount="53" Body="&lt;p&gt;Is there a reliable method for testing how invertible a Hermitian Toeplitz matrix is without going through the work of actually inverting it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A determinant is obviously easy to compute, but I'm not sure what threshold to compare it against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm I am using is Levinson recursion.  The function can detect the divide-by-zero condition, but I am interested in also detecting/scoring the matrices that are &lt;em&gt;almost&lt;/em&gt; singular.&lt;/p&gt;&#xA;" OwnerUserId="1908" LastActivityDate="2010-09-15T14:20:13.547" Title="fast and reliable method for testing invertibility of Hermitian Toeplitz matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4718" PostTypeId="2" ParentId="4717" CreationDate="2010-09-15T14:20:13.547" Score="2" ViewCount="0" Body="&lt;p&gt;It's the condition number $\kappa$ you should be looking at, not the determinant. Since Levinson is an O(n²) method, you can hook it up to the &lt;a href=&quot;http://doi.acm.org/10.1145/50063.214386&quot; rel=&quot;nofollow&quot;&gt;Hager-Higham condition estimator&lt;/a&gt; for determining how near to singularity your matrix is. The classical heuristic is that one loses $\log_{b}\kappa$ base-$b$ significant figures in solving a linear system.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-15T14:20:13.547" CommentCount="1" />
  <row Id="4719" PostTypeId="1" AcceptedAnswerId="4724" CreationDate="2010-09-15T14:50:44.543" Score="4" ViewCount="86" Body="&lt;p&gt;Apologies for the simple question. Is it possible to analytically minimize a function such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = a^x + c x (x-1) \qquad x &gt; 0, \quad 0 &amp;lt; a,c &amp;lt; 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing else can be said about the constants $a$ or $c$. &lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1910" LastEditorDisplayName="" LastEditDate="2010-09-15T15:22:08.600" LastActivityDate="2010-09-15T15:45:25.863" Title="Minimization of a function including quadratic and exponential terms." Tags="&lt;optimization&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="4720" PostTypeId="1" AcceptedAnswerId="4723" CreationDate="2010-09-15T14:53:26.083" Score="7" ViewCount="368" Body="&lt;p&gt;I'm taking a course in differential geometry, and have here been introduced to the wedge product of to vectors defined (in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132125897&quot; rel=&quot;nofollow&quot;&gt;Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo&lt;/a&gt;) by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbf{u}$, $\mathbf{v}$ be in $\mathbb{R}^3$. $\mathbf{u}\wedge\mathbf{v}$ in $\mathbb{R}^3$ is the unique vector that satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mathbf{u}\wedge\mathbf{v})\cdot\mathbf{w} = \det\;(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ for all $\mathbf{w}$ in $\mathbb{R}^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to clarify, $(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ is the 3×3 matrix with $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ as its columns, in that order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: is there &lt;em&gt;any&lt;/em&gt; difference between this and the regular cross product or vector product of two vectors, as long as we stay in $\mathbb{R}^3$? And if there is no difference, then why introduce the wedge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="1911" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T01:35:13.070" LastActivityDate="2010-09-17T01:35:13.070" Title="Wedge product and cross product - any difference?" Tags="&lt;linear-algebra&gt;&lt;terminology&gt;&lt;differential-geometry&gt;&lt;cross-product&gt;&lt;exterior-algebra&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="3" />
  <row Id="4721" PostTypeId="2" ParentId="4719" CreationDate="2010-09-15T14:58:20.430" Score="2" ViewCount="0" Body="&lt;p&gt;I think there is no analytical solution to $f^{\prime }(x)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{\prime }(x)=2cx-c+\left( \ln a\right) a^{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I mean no solution in elementary functions.&lt;/p&gt;&#xA;" OwnerUserId="1504" LastEditorUserId="1504" LastEditorDisplayName="" LastEditDate="2010-09-15T15:18:25.763" LastActivityDate="2010-09-15T15:18:25.763" />
  <row Id="4722" PostTypeId="2" ParentId="3405" CreationDate="2010-09-15T15:27:50.337" Score="2" ViewCount="0" Body="&lt;p&gt;Here is an observation: It looks like $(2n) p(2n) - c$ converges to $0$ much faster than $(2n+1) p(2n+1)-c$ does. Here is a plot of the points $(\log n, n p(n) -c)$, with blue points for$n$ even and red points for $n$ odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/R6asV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be easiest to first prove the result for $n$ even, then use the fact that $(2n) p(2n) - (2n-1) p(2n-1) = p(2n) + p(2n-1)$ to get that the odd and even points must have the same limit. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-15T15:27:50.337" CommentCount="2" />
  <row Id="4723" PostTypeId="2" ParentId="4720" CreationDate="2010-09-15T15:39:08.383" Score="7" ViewCount="0" Body="&lt;p&gt;No difference at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to write a little proof, but the software on this page seems to have forgotten how to write maths.  :-(   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway: I assume that by &quot;regular cross/vector product&quot; you mean the definition with coordinates as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to compute both sides of your equation $(u\wedge v ) \cdot w = \det (u, v, w)$ with your definition using coordinates for the cross / vector product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this equation defines a &lt;em&gt;unique&lt;/em&gt; vector, has just &lt;em&gt;one&lt;/em&gt; solution, once you've checked your definition with coordinates verifies it, that means that your definition with coordinates agrees with the present one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. I forgot. The adventage of this &quot;new&quot; definition is that it is &quot;coordinate-free&quot;. &lt;em&gt;So what?&lt;/em&gt; -I can hear someone whispering at the end of the classroom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So, if you ever wonder if your &quot;former&quot; definition could be affected just in case you change the (positive orthonormal) basis in which you're writing your vectors, the answer, you betcha, is &quot;no&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-15T17:57:13.470" LastActivityDate="2010-09-15T17:57:13.470" CommentCount="1" />
  <row Id="4724" PostTypeId="2" ParentId="4719" CreationDate="2010-09-15T15:45:25.863" Score="5" ViewCount="0" Body="&lt;p&gt;Alright, here's my expansion of my comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's rearrange the equation in VFG's answer like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ln\;a}{2c}=\left(\frac12-x\right)a^{-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and multiply both sides by $\sqrt{a}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)a^{\frac12-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're close to a form where we can now use the Lambert function, but let's change bases first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're much closer now, so we multiply both sides by a factor of $\ln\;a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{a}(\ln\;a)^2}{2c}=\left(\frac{\ln\;a}{2}-x\ln\;a\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then perform the inversion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)=\frac{\ln\;a}{2}-x\ln\;a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is now easily solved for $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac12-\frac1{\ln\;a}W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Lambert function has two real branches $W(z)$ (principal branch) and $W_{-1}(z)$; one is real for $z\geq-e^{-1}$, and the other is real only in the interval $[-e^{-1},0)$. If both $a$ and $c$ are in the interval $(0,1)$, the argument of the Lambert function would be much greater than 1, and thus we can dispose of the $W_{-1}$ branch, as well as consider $W(z)$ only for $z\geq 0$, which is true for $a$ and $c$ in the interval $(0,1)$. Thus, the Lambert solution using the principal branch is valid.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-15T15:45:25.863" CommentCount="1" />
  <row Id="4725" PostTypeId="1" AcceptedAnswerId="4727" CreationDate="2010-09-15T16:01:02.187" Score="3" ViewCount="77" Body="&lt;p&gt;I have a stocastic event (say a weighted coin toss) that produces a positive outcome (heads) according to some unknown probability P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given N total events (coin tosses) and n positive outcomes (heads), how can I measure the likelihood that n/N is a good approximation of P?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, as N grows, it becomes more likely than n/N is close to the value P, but how can this convergence be described mathematically?&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-09-17T19:29:53.720" Title="Deriving the probability of an event from limited trial data" Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="4726" PostTypeId="2" ParentId="4725" CreationDate="2010-09-15T16:13:34.967" Score="1" ViewCount="0" Body="&lt;p&gt;The wiki on &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;convergence of random variables&lt;/a&gt; is what you probably want to read. It discusses various forms of convergence and in particular &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_probability&quot; rel=&quot;nofollow&quot;&gt;convergence in probability&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables#Almost_sure_convergence&quot; rel=&quot;nofollow&quot;&gt;almost sure convergence&lt;/a&gt; are what you need to take a look at.&lt;/p&gt;&#xA;" OwnerUserId="116" LastActivityDate="2010-09-15T16:13:34.967" />
  <row Id="4727" PostTypeId="2" ParentId="4725" CreationDate="2010-09-15T16:15:42.640" Score="3" ViewCount="0" Body="&lt;p&gt;While I realise this isn't a full answer, the convergence you discuss is guaranteed by the weak and strong laws of large numbers. You might also be interested in Chebychev's inequality, which would say $$\mathbb{P}(|n/N - p| &gt; \alpha) \leq \frac{\sigma^2}{\alpha^2}$$ where $\sigma^2$ is the variance of your estimator, $n/N$. Finally the Central Limit Theorem (CLT) proves that as $N \to \infty$ the estimator becomes normally distributed about $p$ with variance $\frac{\sigma^2}{N}$. &lt;/p&gt;&#xA;" OwnerUserId="1910" LastActivityDate="2010-09-15T16:15:42.640" />
  <row Id="4728" PostTypeId="2" ParentId="4705" CreationDate="2010-09-15T16:32:11.570" Score="2" ViewCount="0" Body="&lt;p&gt;Here are two facts that may help with the analogy.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $x_n$ be a real sequence; let $A_n = (-\infty, x_n)$.  Then $\limsup A_n = (-\infty, \limsup x_n)$ or $(-\infty, \limsup x_n]$.  Similarly, $\liminf A_n = (-\infty, \liminf x_n)$ or $(-\infty, \liminf x_n]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $A_n$ be subsets of some set $\Omega$, and let $1_{A_n} : \Omega \to \mathbb{R}$ be the function which is 1 on $A_n$ and 0 on $A_n^c$.  Then $\limsup 1_{A_n}(\omega) = 1_{\limsup A_n}(\omega)$ for all $\omega \in \Omega$, and similarly $\liminf 1_{A_n}(\omega) = 1_{\liminf A_n}(\omega)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-15T16:32:11.570" CommentCount="2" />
  <row Id="4730" PostTypeId="2" ParentId="4720" CreationDate="2010-09-15T17:07:16.000" Score="6" ViewCount="0" Body="&lt;p&gt;I would have left this as a comment, but apparently I can't comment yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your final question, why introduce the wedge, the point is that the wedge product (as explained in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra#Formal_definitions_and_algebraic_properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;) is a notion that generalizes to R^n and indeed any vector space---in general the output is what is called a &quot;bivector&quot;. Now, it just so happens that in R^3 there is a natural identification between bivectors and vectors, and the vector that corresponds to the (bivector) wedge product is indeed the cross product, but this does not work in other vector spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up, the notion of cross product is specific to R^3, whereas wedge product makes sense in any vector space.&lt;/p&gt;&#xA;" OwnerUserId="1914" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-15T17:24:17.300" LastActivityDate="2010-09-15T17:24:17.300" CommentCount="1" />
  <row Id="4731" PostTypeId="1" CreationDate="2010-09-15T18:03:22.457" Score="2" ViewCount="214" Body="&lt;p&gt;For functions on $\mathbb{R}^2$, it lets you classify extreme points. When function is log-likelihood, it gives volume of the region where maximum likelihood estimates tend to fall. Are there other places it comes up?&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-16T20:50:26.483" LastActivityDate="2010-09-16T21:46:51.393" Title="Determinant of the Hessian, what is it good for?" Tags="&lt;calculus&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4732" PostTypeId="1" AcceptedAnswerId="4740" CreationDate="2010-09-15T18:34:15.633" Score="4" ViewCount="145" Body="&lt;p&gt;I am  trying to solve a ODE that arises from a Hamilton-Jacobi-Bellman (HJB) equation. The equation is $$\frac{1}{2}b^2(1-\rho_s^2)\psi''-\frac{1}{2}\left(\frac{\mu-r}{\sigma}\right)^2\frac{(\psi')^2}{\psi''}+[ru+\theta a+b\rho_s(\mu-r)(1-\frac{2}{\sigma})]\psi'=0,$$ where $\mu, r, \sigma, \theta, a, \rho_s, b$ are constant. I want to determine $\psi'(u)$ (so that I get an integral form for $\psi(u)$). I have tried guessing (trial and error method) forms of the solution but didn't get far. I also tried the Legendre transform, but could not get the linear form. These are the method that I have seen being used in with these problems. &lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-09-15T21:17:12.803" LastActivityDate="2010-09-15T22:03:08.357" Title="Solution of a Hamilton-Jacobi-Bellman (HJB) equation" Tags="&lt;differential-equations&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4733" PostTypeId="1" AcceptedAnswerId="4748" CreationDate="2010-09-15T18:48:49.570" Score="5" ViewCount="288" Body="&lt;p&gt;Let $f$ be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\begin{cases} x \sin(\frac{1}{x}) &amp;amp; x\ne 0 \\ 0 &amp;amp; x=0 \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)$  is not absolutely continuous so it &lt;strike&gt;cannot&lt;/strike&gt; &lt;em&gt;might not&lt;/em&gt; satisfy the Lusin N condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a direct proof of that it does not?  i.e. I wanted to know how to construct a set of zero measure which does not satisfy Lusin's Condition for this function.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-15T20:54:00.430" LastActivityDate="2010-09-16T02:24:28.570" Title="Direct proof that f(x)=x sin(1/x)  does not satisfy Lusin N condition." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="4734" PostTypeId="2" ParentId="4572" CreationDate="2010-09-15T18:55:10.053" Score="6" ViewCount="0" Body="&lt;p&gt;It seems to me that maybe the best way to describe the situation is this: Douglas Stone's original answer to the original question consisted of rephrasing the question in such a way as to make it accessible to a proof using basic properties on the integers (specifically, unique factorization).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, one thing which is being lost in all this discussion is just how important it can be to rephrase a question! Sure, the process of rephrasing contains &quot;no math&quot; as Qiaochu has pointed out.  But that doesn't make it useless (and I wouldn't use the word circular here either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding ways to rephrase questions so that they become accessible to the methods available is a basic skill beginning students of mathematics need to learn.  For example, much of the material in the early chapters of modern linear algebra books consists of teaching students how to rephrase questions in linear algebra so that they can be solved by row reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't accept Douglas Stone's answer as a complete solution to the problem if it were turned in by a student in an elementary number theory class, just as in my linear algebra class, reducing a problem to a question of row reduction isn't a complete solution.  But if a student came to me and said he or she was stuck on the problem, the first thing I'd try to do is get them to rephrase the question in precisely the way Douglas Stone did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pleasantly, the community has pointed out (both here and at the original question) exactly how to finish the proof after rephrasing it in this useful way.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-15T18:55:10.053" CommentCount="5" />
  <row Id="4735" PostTypeId="1" CreationDate="2010-09-15T19:17:00.060" Score="1" ViewCount="157" Body="&lt;p&gt;Cut three slices from a cheese with the form of a cube. Find the way of cutting which is most &quot;efficient&quot; - I think cutting it by the edge.(i.e the volume of the cube to be the most biggest possible). What will be that volume?&lt;br&gt;&#xA;Full surface area: 600 cm^2&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-15T20:24:52.893" LastActivityDate="2010-09-15T20:24:52.893" Title="Volume of a Cube" Tags="&lt;geometry&gt;" CommentCount="21" FavoriteCount="1" />
  <row Id="4736" PostTypeId="2" ParentId="4731" CreationDate="2010-09-15T19:41:41.533" Score="1" ViewCount="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132125897&quot; rel=&quot;nofollow&quot;&gt;Do Carmo&lt;/a&gt;, exercise 22 in page 173, the Hessian gives you the second fundamental form of a surface, as John pointed out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely, let $S$ be a surface, $p\in S$ a point and $h: S \longrightarrow \mathbb{R}$ be the height function of $S$ relative to $T_p S$; that is, $h(q) = (q-p)\cdot N(p)$, where $N(p)$ is the normal vector to the surface $S$ at the point $p$. Then the Hessian of the function $h$ at $p$ is the second fundamental form of $S$ at $p$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-15T19:41:41.533" CommentCount="1" />
  <row Id="4737" PostTypeId="1" CreationDate="2010-09-15T19:51:58.437" Score="11" ViewCount="255" Body="&lt;p&gt;Does there exists a family of discs $\lbrace D_{n}\rbrace_{n=1}^{\infty}$ in euclidean plane such that:&lt;br&gt;&#xA;-the radius of $D_{n}$ is $\frac{1}{n+1}$&lt;br&gt;&#xA;-each $D_{n}$ is contained in the unit disc&lt;br&gt;&#xA;and&lt;br&gt;&#xA;-$D_{n}\cap D_{m} = \emptyset$ for each $n\neq m$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not sure what tags are appropriate for this kind question, so if You have any suggestions, You're welcome to inform me about it via comments)&lt;/p&gt;&#xA;" OwnerUserId="1454" LastActivityDate="2010-10-03T15:46:37.793" Title="Packing disjoint family of discs with radii $\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4},\ldots$ inside the unit disc" Tags="&lt;geometry&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="4738" PostTypeId="1" AcceptedAnswerId="4790" CreationDate="2010-09-15T20:18:03.397" Score="0" ViewCount="163" Body="&lt;p&gt;Suppose $X$ is a metric space. A real valued function $f$ defined on a metric space $X$ is said to be of first baire class, if $f$ is a pointwise, limit of a sequence of continuous functions on $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A famous well know theorem asserts that if &lt;em&gt;$f$ is in the first baire class, then the set of points of discontinuity is of the first category&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Using this definition, can we characterize for what &lt;em&gt;Metric Spaces&lt;/em&gt;, will the points of continuity of a real valued function, be dense in itself.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, since we are talking about metric spaces, lets consider the 2 famous ones, complete and Compact. A compact metric space, is &quot;Complete and Totally Bounded&quot; by definition, so lets not look into it. So lets look at complete metric spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question basically, is if, $X$ is a complete metric space, and $f$ is as above(1st Baire class), then is the points of continuity of $f$ dense in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some exercises related, but i didn't take it from here. So dont misunderstand me. &lt;a href=&quot;http://books.google.com/books?id=CoSIe7h5mTsC&amp;amp;pg=PA166&amp;amp;dq=f+is+of+first+baire+class,+points+of+continuity&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=CoSIe7h5mTsC&amp;amp;pg=PA166&amp;amp;dq=f+is+of+first+baire+class,+points+of+continuity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-16T15:29:25.780" LastActivityDate="2010-09-16T15:29:25.780" Title=" Points of continuity of a function being dense" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4740" PostTypeId="2" ParentId="4732" CreationDate="2010-09-15T22:03:08.357" Score="1" ViewCount="0" Body="&lt;p&gt;Assume $\psi' = f$, then the equation is of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Af' + B\frac{f^2}{f'} + (cu+d)f = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By putting $$g = \frac{f'}{f} = (\log f)'$$ we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ag + \frac{B}{g} + (cu+d) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a quadratic in $g$ and can be easily solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get $$ \psi' = f = e^{\int g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-15T22:03:08.357" CommentCount="1" />
  <row Id="4741" PostTypeId="1" AcceptedAnswerId="4792" CreationDate="2010-09-15T22:07:56.313" Score="6" ViewCount="166" Body="&lt;p&gt;Suppose $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ is a $C^\infty$&#xA;curve, parameterized by arc length ($\left\|\alpha'(t)\right\| = 1$),&#xA;and with $\alpha(0) = \alpha(\ell)$.  Show that&#xA;there exists a $t_0 \in [0,\ell]$ such that&#xA;$&#xA;\left\| \alpha''(t_0)\right\| \geq \frac{2\pi}{\ell}.&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/WD65E.png&quot; alt=&quot;loop&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had that $\alpha'(0) = \alpha'(\ell)$, then the result would&#xA;follow from Fenchel's theorem, but we don't. Actually there is a nice&#xA;extension to Fenchel's theorem described in Toponogov's book&#xA;``Differential Geometry of Curves and Surfaces: A Concise Guide'',&#xA;under the name Fenchel-Reshetnyak, which gives lower bounds on the&#xA;integral curvature in an open curve. This can be adapted to our&#xA;situation to show that $\left\| \alpha''(t_0)\right\| \geq&#xA;\frac{\pi}{\ell}$, but that bound is only half as good as it&#xA;presumably could be. It makes sense that the integral curvature bound&#xA;will be only half as good if we don't constrain $\alpha'(0)$&#xA;w.r.t. $\alpha'(\ell)$ (in particular, one may be the negative of the&#xA;other), but nonetheless it seems that the original bound on the&#xA;&lt;i&gt;maximum&lt;/i&gt; curvature in the loop should still stand. How can we&#xA;recover it?&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2010-09-15T22:13:02.567" LastActivityDate="2010-09-16T23:47:45.373" Title="A good lower bound on the maximum curvature in a loop" Tags="&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="4742" PostTypeId="1" CreationDate="2010-09-15T22:36:08.850" Score="2" ViewCount="258" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor wants us to do this problem to refresh ourselves with substitution. We have to solve:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Integral of sqrt(1 + sqrt(x))&#xA;Integral of sqrt(1 + sqrt(1 + sqrt(x)))&#xA;...etc...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If someone could point me in the right direction with the first one, I think I can handle the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cheers,&#xA;Mike&lt;/p&gt;&#xA;" OwnerUserId="1917" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-09-16T00:26:25.417" LastActivityDate="2010-09-16T09:21:16.290" Title="Integral of sqrt(1 + sqrt(x))" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="4744" PostTypeId="1" AcceptedAnswerId="4750" CreationDate="2010-09-16T02:11:30.943" Score="5" ViewCount="230" Body="&lt;p&gt;Given that I have two formal power series:&#xA;$$ A(x) = \sum_{k \ge 0} a_k x^k $$&#xA;$$ B(x) = \sum_{k \ge 0} b_k x^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cauchy Product gives a series&#xA;$$ C(x) = \sum_{k \ge 0} c_k x^k $$&#xA;$$ c_k  = \sum_{n=0}^k a_n b_{k-n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which comes from taking the product of the two series $C(x)=A(x)B(x)$. What then, in terms of $A(x)$ and $B(x)$, is this series?&#xA;$$Y(x) = \sum_{k \ge 0} a_k b_k x^k $$&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-09-16T02:27:41.210" LastActivityDate="2010-09-18T17:08:01.813" Title="Formal power series coefficient multiplication" Tags="&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="4745" PostTypeId="2" ParentId="4742" CreationDate="2010-09-16T02:11:45.247" Score="2" ViewCount="0" Body="&lt;p&gt;One technique usually worth trying in integration is to be extremely optimistic: ask yourself what is &quot;hard&quot; about the integrand and then make a substitution that simplifies it, even if the substitution looks awful.  In your case, the &quot;hard&quot; part about $\sqrt{1 + \sqrt{x}}$ is that the argument of the outer root is not evidently a perfect square, so it doesn't simplify.  So substituting $u^2$ for $1 + \sqrt{x}$ would be worth considering.  (You will obtain an integrand that is a polynomial in $u$.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T02:11:45.247" CommentCount="0" />
  <row Id="4746" PostTypeId="2" ParentId="4744" CreationDate="2010-09-16T02:21:25.970" Score="-1" ViewCount="0" Body="&lt;p&gt;Stupid answer: $Y\left(x\right)=\delta_{ik}A\left(x\right)B\left(1\right)$, for $\delta_{ik}$ the Kronecker Delta, and $i$ being the index on $B\left(x\right)$.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-09-16T02:21:25.970" CommentCount="3" />
  <row Id="4747" PostTypeId="2" ParentId="4705" CreationDate="2010-09-16T02:21:31.497" Score="1" ViewCount="0" Body="&lt;p&gt;Let's deal with the examples first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\liminf$ of the real sequence $\{a_n\}$ with $a_n = (-1)^n$ is $-1$; the $\limsup$ of the same sequence is $1$. For the sequence $\{b_n\}$, with $b_n=\frac{1}{n}$, since the sequence converges to $0$, every subsequence converges to $0$ so both $\liminf$ and $\limsup$ are equal to $0$. You can think of the $\liminf$ as the infimum of the all the limits of all converging subsequences of the sequence, and the $\limsup$ is the supremum of all the limits of all converging subsequences of the sequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definitions given by Jens, for $A_n=\{(-1)^n\}$ the $\limsup$ is $\{-1, 1\}$, because both elements show up in infinitely many of the $A_n$, and the $\liminf$ is empty, because no element occurs in all but finitely many of the $A_n$. For $B_n$, they are both empty because no element occurs in infinitely many of the sets, nor in all but finitely many (each element occurs in only one $B_n$). Again, $\liminf$ is the collection of all points that are in all but finitely many of the sets, while $\limsup$ is the collection of all points that are in infinitely many of the sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you are looking at the &lt;em&gt;wrong&lt;/em&gt; sets if you want your sets to be related to your sequences. As Nate Eldredge points out, what you should be looking at is the set $A_n = (-\infty,a_n)$, or $A_n = (-\infty,(-1)^n)$. Using &lt;em&gt;that&lt;/em&gt; definition, you have that $\limsup A_n = (-\infty,1)$ (as expected, since $\limsup a_n = 1$; each of these numbers occurs in infinitely many of the $A_n$), and $\liminf A_n=(-\infty,-1)$ because those are the only ones that occur in all but finitely many of the sets (in fact, in all; every other number that occurs in any $A_n$ occurs only in the $A_n$ with even $n$, so is missing from infinitely many $A_n$); while if you let $B_n = (-\infty,\frac{1}{n})$, then $\liminf B_n=\limsup B_n = (-\infty,0]$ (again, as expected, since the limit inferior and limit superior of $b_n$ are both equal to $0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the reason you seem to be getting hung up is that there seems to be little relation between the limits inferior and superior of a $a_n$, and the limits inferior and superior of the sequence of sets $\{a_n\}$. But the point that Nate Eldredge made is that these are not the sets you want to associate with the sequence $a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may remember that a sequence $\{a_n\}$ converges to $L$ if and only if every subsequence $\{a_{n_k}\}$ converges to $L$. Also, every sequence contains a monotone sequence, so if we allow $\infty$ and $-\infty$ as &quot;limits&quot;, it follows that every sequence will necessarily have a converging subsequence. So one can ask: &quot;what are all the points $M$ for which there is a subsequence of $\{a_n\}$ that converges to $M$?&quot; One can view the limits inferior and superior in terms of this set: the limit inferior of the sequence is the smallest number $\ell$ (including possibly $\infty$ or $-\infty$) for which there is a subsequence of $\{a_n\}$ converging to $\ell$. The limit superior is the largest number $L$ for which there is a subsequence of $\{a_n\}$ that converges to $L$. As it happens, the limit exists if and only if $\ell=L$. The limits inferior and superior can also be defined by&#xA;$$\liminf a_n = \lim_{n\to\infty}(\inf\{a_m|m\geq n\}) = \sup_n\left(\inf\{a_m|m\geq n\}\right)$$&#xA;and&#xA;$$\limsup a_n = \lim_{n\to\infty}(\sup\{a_m|m\geq n\}) = \inf_n\left(\sup\{a_m|m\geq n\}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viewed like this, you can perhaps see a bit more of a connection with the limits inferior and superior of a sequence of sets. If $\{A_n\}$ is a sequence of sets, then the limits inferior and superior are defined to be:&#xA;$$\liminf A_n = \cup_{n=1}^{\infty}\left(\cap_{m=n}^{\infty} A_m\right)$$&#xA;and&#xA;$$\limsup A_n = \cap_{n=1}^{\infty}\left(\cup_{m=n}^{\infty} A_m\right).$$&#xA;Think of an intersection as taking &quot;smallest&quot; thing in common (so like an infimum), and think of union as taking &quot;largest&quot; thing in common (so like a supremum). The limit inferior is the supremum of the infima, while the limit superior is the infimum of the suprema. It is now a nice exercise to verify that $\liminf A_n$ is the collection of all things which are in all but finitely many of the $A_i$, while $\limsup A_n$ is the collection of all things which are in infinitely many of the $A_i$ (as described by Jens). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you connect a sequence $\{a_n\}$ to sets so that the limits inferior and superior correspond in some way? You can't just take $A_n = \{a_n\}$, because then each $A_n$ knows nothing about what came before or after; you lose all information that could tell you something about subsequences. You could try letting $A_n =\{a_m|m\geq n\}$, and that will even work in some instances, but the problem here is that the information you are losing is that in the real numbers, a sequence may converge to a number even if &lt;em&gt;no term&lt;/em&gt; in the sequence equals the limit; then the limit is never going to show up in any of the sets, and it's not going to show up in the limits inferior nor superior of the sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the solution? The limit inferior of a sequence is going to be a lower bound for all but finitely many of the terms of the sequence (if there were infinitely many terms of the sequence strictly smaller than $\liminf a_n$, then you would be able to get a subsequence from among them that converges to something strictly smaller than $\liminf a_n$, a contradiction). This suggests that what you want to do is let $A_n$ be the collection of all lower bounds to $a_n$; then the limit inferior of the $A_n$ will be the collection of all things that are lower bounds to all but finitely many terms of the sequence, exactly the set you want to consider to find $\liminf a_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the limit superior of $a_n$? You can define it dually, as the smallest of all numbers that upper bounds for all but finitely many terms of the sequence (this will lead to the formula that says that $\limsup a_n = -\liminf(-a_n)$). Or you can try to define it in terms of the lower bounds again: any number $k$ strictly smaller than the limit superior must have infinitely many terms of the sequence larger than $k$ (otherwise, no subsequence could converge to something larger, so no subsequence could converge to the limit superior). That is: look at the collection of all things that are lower bounds for &lt;em&gt;infinitely many&lt;/em&gt; of the $a_n$, and the supremum of that will be the limit superior.  So we again look at $A_n = (-\infty,a_n)$ (the set of all lower bounds to $a_n$), and consider the limit superior of the $A_n$; this is the collection of all numbers that are lower bounds to infinitely many of the $a_n$, and so its supremum will be $\limsup a_n$. So that's why you consider the set $A_n(-\infty,a_n)$ instead of the set $\{a_n\}$, and where they come from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways of associating to each $a_n$ an appropriate set; in this situation, note that $\sup A_n = a_n$ for each $n$, that $\sup(\liminf A_n) = \liminf (\sup A_n) = \liminf a_n$, and $\sup(\limsup A_n) = \limsup(\sup A_n) = \limsup a_n$, which makes this association pretty nice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps clarify it further.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-16T05:02:01.010" LastActivityDate="2010-09-16T05:02:01.010" CommentCount="5" />
  <row Id="4748" PostTypeId="2" ParentId="4733" CreationDate="2010-09-16T02:24:28.570" Score="3" ViewCount="0" Body="&lt;p&gt;The function sends each set of measure zero to a set of measure zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\subset\mathbb R$ be a null set (i.e. $A$ has measure 0).  Then $f(A)=f(A\cap \{0\})\cup f(A\cap(\mathbb{R}\setminus \{0\}))$.  The first set in the union has at most one point, so we need only worry about the second.  The set $A\cap(\mathbb{R}\setminus \{0\})$ can be expressed as a countable union of sets of the form $A\cap [a,b]$ with $0\lt a$ or $b\lt 0$, and therefore, since the image of a union is the union of the images, $f(A\cap(\mathbb{R}\setminus \{0\}))$ can be expressed as a countable union of sets of the form $f(A\cap [a,b])$ with $0\lt a$ or $b\lt 0$.  For each such $a$ and $b$, $f$ is continuously differentiable in a neighborhood of $[a,b]$, and hence the restriction of $f$ to $[a,b]$ is absolutely continuous (e.g. by the fundamental theorem of calculus for $C^1$ functions).  Therefore the image of the null set $A\cap[a,b]$ under $f$ is null.  Countable unions of null sets are null, so this shows that $f(A)$ is null.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-16T02:24:28.570" CommentCount="3" />
  <row Id="4749" PostTypeId="1" AcceptedAnswerId="4752" CreationDate="2010-09-16T02:28:19.010" Score="1" ViewCount="105" Body="&lt;p&gt;For plane curves, twice the quadratic term of the Taylor expansion at a point on that curve is precisely the curvature. I know of one proof as featured in &lt;a href=&quot;http://www.math.cornell.edu/~hubbard/vectorcalculus.html&quot; rel=&quot;nofollow&quot;&gt;Hubbard and Hubbard&lt;/a&gt;, but I was wondering if there was another more elegant solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-16T02:45:13.877" LastActivityDate="2010-09-16T04:56:46.857" Title="A nice proof of the fact that the Quadratic Term of the Taylor polynomial is Curvature" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;&lt;curvature&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4750" PostTypeId="2" ParentId="4744" CreationDate="2010-09-16T02:33:44.670" Score="8" ViewCount="0" Body="&lt;p&gt;It is known as the &lt;strong&gt;Hadamard product&lt;/strong&gt; $\rm\; f \star g\;$. There is no closed form for general hadamard products, but some classes of functions are known to be closed under such products, e.g. rational power series. However, algebraic series are not generally closed under Hadamard products, the standard example being $\rm\quad f = g = (1-4x)^{-1/2} \;=\; \sum \binom{2n}{n}\ x^n\:.\quad$ However, $\rm\: f\:$ rational, $\rm\: g\:$ algebraic $\rm\;\Rightarrow\; f\star g\;$ algebraic. Also &lt;a href=&quot;http://www-math.mit.edu/~rstan/pubs/pubfiles/45.pdf&quot; rel=&quot;nofollow&quot;&gt;D-finite power series&lt;/a&gt; are closed under Hadamard products, i.e. power series satisfying a linear differential equation with polynomial coefficients or, equivalently, series whose coefficients satisfy a linear recursive equation with polynomial coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hadamard products can also be defined in terms of diagonals of multivariate series (and vice versa), e.g. see this &lt;a href=&quot;http://pictor.math.uqam.ca/~plouffe/OEIS/archive_in_pdf/005_a084.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;, where you'll find some interesting connections with finite automata which, e.g. help one to easily observe that algebraic series over $\rm\mathbb{F}_p$ &lt;em&gt;are&lt;/em&gt; closed under Hadamard product (which fails in characteristic 0)&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-16T03:40:03.010" LastActivityDate="2010-09-16T03:40:03.010" />
  <row Id="4751" PostTypeId="2" ParentId="4744" CreationDate="2010-09-16T02:33:49.990" Score="6" ViewCount="0" Body="&lt;p&gt;$Y$ is known as the Hadamard product of $A$ and $B$, and there is no simple way to find it from $A$ and $B$ in general.  (If you don't believe me, set $A = B = e^x$.)  The closest thing I know to a formula is (with all the analytic caveats that makes everything converge)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y(r) = \frac{1}{2\pi } \int_0^{2\pi} A(e^{i \theta}) B(re^{-i \theta}) d \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which follows by &lt;a href=&quot;http://en.wikipedia.org/wiki/Parsevals_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's theorem&lt;/a&gt;.  If $A$ and $B$ are sufficiently simple (for example if they are rational functions) then it is possible to evaluate this integral, but in general there's not much you can do; $Y$ can be much more complicated than $A$ or $B$.  (Another idea is to write the above as a contour integral, and if the integrand ends up being meromorphic you can try to use the residue theorem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing Hadamard products is a special case of computing the diagonal of a two-variable generating function, a problem which I describe with a few examples &lt;a href=&quot;http://qchu.wordpress.com/2009/10/07/extracting-the-diagonal/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; using the residue theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course for very special $A$ it is possible to say more, e.g. when $a_k$ is a polynomial in $k$.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-16T02:51:24.297" LastActivityDate="2010-09-16T02:51:24.297" CommentCount="1" />
  <row Id="4752" PostTypeId="2" ParentId="4749" CreationDate="2010-09-16T02:43:52.113" Score="2" ViewCount="0" Body="&lt;p&gt;I'm unfamiliar with your reference, so please forgive me if this might be reminiscent of its solution.  I'll use the Serret-Frenet formulae.  For a twice differentiable curve $\gamma$ parameterized by arclength, one of them asserts that the normal component of the derivative of the (unit) tangent vector is the curvature $\kappa$.  (This is often taken as the definition of curvature.)  To simplify things, but wlg, let's use coordinates in which the point in question is the origin, so that the Taylor expansion is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma(t) = \left(a_1 t + \frac{a_2}{2}t^2 + \cdots, b_1 t + \frac{b_2}{2}t^2 + \cdots \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whence the unit tangent vector at the origin is $T = (a_1, b_1)$ and its derivative is $(a_2, b_2)$.  To obtain the component normal to $T$, it suffices to compute the inner product of this derivative with the rotated tangent vector $(-b_1, a_1)$ (because the tangent vector has unit length), yielding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa = -a_2 b_1 + b_2 a_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This general result, specialized to the case where $\gamma$ is of the form $(t, f(t))$ for a function $f(t) = 0 + b_1 t + \frac{b_2}{2}t^2 + \cdots$, reduces (because $a_1 = 1$ and $a_2 = 0$) to $b_2$, &quot;twice the quadratic term in the Taylor expansion&quot; at this point.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T02:43:52.113" CommentCount="3" />
  <row Id="4753" PostTypeId="2" ParentId="4742" CreationDate="2010-09-16T03:13:22.113" Score="1" ViewCount="0" Body="&lt;p&gt;This is the answer from &lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+of+sqrt%281+%2B+sqrt%28x%29%29&quot; rel=&quot;nofollow&quot;&gt;Wolframalpha&lt;/a&gt;: note that you have to use the substitution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u=\sqrt{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first. Note that Wolframalpha able to give you the full step by step solution.&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-16T09:21:16.290" LastActivityDate="2010-09-16T09:21:16.290" />
  <row Id="4754" PostTypeId="2" ParentId="4749" CreationDate="2010-09-16T04:56:46.857" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a completely different answer inspired by J.M.'s description of the text.  It might give a bit more geometric insight.  It essentially shows that (a) twice the second coefficient in the Taylor series describes a parabola closely approximating the curve at the origin, (b) the curvature is determined by a circle closely approximating the curve at the origin, and (c) the curvature of the Taylor series parabola is twice that coefficient.  In other words, when we're talking about curvature we're really talking about the second-order behavior of the curve and that's exactly what the quadratic term in the Taylor series describes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant definition of the curvature here is the reciprocal of the radius of the osculating circle (taken to be negative if the circle lies below the curve, with 1/infinity = 0).  An osculating (&quot;kissing&quot;) circle is found by taking three points close to the point under consideration (which again might as well be the origin), fitting a circle to them, letting the points independently converge to the origin, and seeing what the circle settles down to.  Assuming the circle exists, we can find it by considering some special configurations of points.  Choose the three consisting of the origin itself, with coordinates (0,0), and the points at nonzero distance $x$ to both sides of the origin.  Their coordinates are given by the Taylor series.  One is is at $(x, \frac{b}{2}x^2 + O(x^3))$ and the second is at $(-x, \frac{b}{2}(-x)^2 + O((-x)^3))$.  The center of the circle is equidistant from all three.  If we choose $x$ small enough (witness some geometric hand-waving here that can be made rigorous at the expense of making the following algebra a little harder), the situation is essentially symmetric around the origin and we expect the center to be located directly above (or below) the origin at a point $(0,a)$.  Consider the distances from $(0,a)$ to the origin and to the point corresponding to $x$: their squares must be equal.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^2 = (x-0)^2 + (\frac{b}{2}x^2 - a)^2 + O(x^3)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at least to an excellent approximation when $x$ is small.  Some terms cancel and an easy algebraic simplification gives (remember, we're solving for $a$ in terms of $x$ and $b$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = 1 - a b + \frac{b^2}{4}x^4 + O(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implying $1/a = b$ in the limit as $x \to 0$.  Since $1/a$ is the curvature, our task is done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, as I have warned, this lacks rigor, but the rigor can be supplied (I recall Michael Spivak does this early in Volume II of his Differential Geometry series) and what it lacks in sophistication may be compensated by the elementary geometric intuition it affords of both curvature and the Taylor series.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T04:56:46.857" CommentCount="1" />
  <row Id="4755" PostTypeId="2" ParentId="4742" CreationDate="2010-09-16T05:25:19.160" Score="-1" ViewCount="0" Body="&lt;p&gt;Suprisingly useful hint: $x = 1+(x-1)$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-16T05:25:19.160" />
  <row Id="4757" PostTypeId="1" CreationDate="2010-09-16T06:17:34.643" Score="1" ViewCount="177" Body="&lt;p&gt;In a sense, I did just ask this question.  However, since the question is about how to write a beautiful looking proof and my proof is entirely rewritten, it seems like it should be a new question.  What I'm looking for here are (aside from any actual mathematical problems in the proof of course) is advice on how to better structure what I'm writing or to better communicate the ideas.  Anything from overall structure, word choice, what letters I give my variables, anything at all.  Anyhow, the question from Spivak:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\begin{cases}&#xA;e^{-x^{-2}}&amp;amp; x\ne 0\\&#xA;0 &amp;amp; x=0\\&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f$ is a $C^\infty$ function, and $f^{(i)}(0)=0$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is my proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we show that $f\in C^\infty$ for $x\ne 0$ by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $-g(x)=\frac{1}{x^2}$.  Then we can write $f$ as $f(x)=e^{g(x)}$.  Let $P={p:\mathbb{R}\rightarrow\mathbb{R} : p(x)=\sum \frac{a_n}{x^n},a_n\in\mathbb{R}}$.  Now we can show by induction that $f^{(n)}(x)=e^{g(x)}p_n(x)$ for some $p_n \in P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_0=\frac{1}{x^0}=1$.  Then $p\in P$ and $f(x)=e^{g(x)}p_0(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $p_n\in P$ such that $p_n=\sum_{i=0}^\infty\frac{a_i}{x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;(e^{g(x)}p_n(x)) &amp;amp;= e^{g(x)}g'(x) p_n(x) + e^{g(x)} p_n'(x)\\&#xA;&amp;amp;= e^{g(x)}(g'(x) p_n(x) + p_n'(x))\\&#xA;&amp;amp;= e^{g(x)}(\frac{2}{x^3} p_n(x) + p_n'(x))\\&#xA;&amp;amp;= e^{g(x)}(\frac{2}{x^3} \sum \frac{a_i}{x^i} + \sum \frac{-ia_i}{x^{i+1}})\\&#xA;&amp;amp;= e^{g(x)}(\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}})\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}}\in P$, it follows that $p_n\in P \Rightarrow p_{n+1}\in P$.  Thus by induction, $f^{(n)}=e^{g(x)}p_n(x)$ where $p_n\in P$ and $f\in C^\infty$ for $x\ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we show by induction that $f^{(n)}(0)=0$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we prove that for all $n \ge -1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(n)=\lim_{h\rightarrow 0} \frac{1}{h^n e^{\frac{1}{h^2}}}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=-1$ we have $\lim_{h\rightarrow 0}\frac{1}{h^{-1} e^{\frac{1}{h^2}}}=\lim_{h\rightarrow 0}\frac{h}{e^{\frac{1}{h^2}}}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=0$ we have $\lim_{h\rightarrow 0}\frac{1}{h^{0} e^{\frac{1}{h^2}}}=\lim_{h\rightarrow 0}\frac{1}{e^{\frac{1}{h^2}}}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that $m(n-1)=0$ and $m(n-2)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;m(n) &amp;amp;= \lim_{h\rightarrow 0} \frac{1}{h^n e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \lim_{h\rightarrow 0} \frac{h^{-n}}{e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \lim_{h\rightarrow 0} \frac{-nh^{-n-1}}{e^{\frac{1}{h^2}} (-2 h^{-3})}\tag{L'H\^{o}pital's rule}\\&#xA;&amp;amp;= \frac{n}{2} \lim_{h\rightarrow 0} \frac{h^{-n-1+3}}{e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \frac{n}{2} \lim_{h\rightarrow 0} \frac{1}{h^{n-2}e^{\frac{1}{h^2}}}\\&#xA;&amp;amp;= \frac{n}{2} m(n-2)=0\\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by induction, $m(n)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{(n)}(x)=e^{g(x)}(\sum \frac{2a_i}{x^{i+3}} + \sum \frac{-ia_i}{x^{i+1}})=\sum 2a_i m(i+3)+\sum-im(i+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that $f^{(n)}=0$ for all $n$.&lt;/p&gt;&#xA;" OwnerUserId="326" LastEditorUserId="326" LastEditorDisplayName="" LastEditDate="2010-09-17T01:46:40.937" LastActivityDate="2010-10-17T02:45:36.573" Title="Improve my proof about this $C^\infty$ function even more!" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;proof-strategy&gt;" AnswerCount="1" />
  <row Id="4758" PostTypeId="1" AcceptedAnswerId="4760" CreationDate="2010-09-16T06:59:57.377" Score="1" ViewCount="184" Body="&lt;p&gt;I first noticed this is true for the integers of the sequence $9, 99, 999, 9999,\dots$, since for some term $a_n=10^n-1$ in the sequence and $p$ a prime other than $2$ or $5$, we have $a_n\equiv 0 (mod\ p)\Leftrightarrow 10^n\equiv 1 (mod\ p)$ and we can find infinitely many $n=k(p-1)$ so that $10^n=(10^{p-1})^k\equiv 1^k\equiv 1 (mod\ p)$ by Fermat's little theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious, is there a similar way to show this is true for repunits $1, 11, 111,\dots$?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-19T21:30:05.707" Title="Why do primes other than 2 and 5 divide infinitely many repunits?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="4759" PostTypeId="1" CreationDate="2010-09-16T07:03:46.947" Score="1" ViewCount="225" Body="&lt;p&gt;Hi guys, I wasnt sure where to ask this but since this is an algorithmic question here it goes. I've come face to face with a math problem and can't seem to get over it for the last couple of days. It goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given an adding machine that sums a set of N+1 digits&#xA;  consisting of positive integers 1 to N&#xA;  as it's given the numbers (e.g. the&#xA;  machine is given 3 as the first number&#xA;  and outputs 3. It's then given 6 as&#xA;  the second number and outputs 9. It's&#xA;  given 11 as the third number and&#xA;  outputs 20. Etcetera until it has&#xA;  processed N+1 numbers). One (and only&#xA;  one) of the digits is repeated. How do&#xA;  you determine which number is&#xA;  repeated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems like a trick question and I'd be really pissed off it is just that a question to which the answer is 'not possible' - any ideas here?&lt;/p&gt;&#xA;" OwnerUserId="1922" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-16T13:50:43.397" LastActivityDate="2010-09-16T13:50:43.397" Title="Need help with a math trick question" Tags="&lt;algebra-precalculus&gt;&lt;average&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4760" PostTypeId="2" ParentId="4758" CreationDate="2010-09-16T07:08:00.653" Score="7" ViewCount="0" Body="&lt;p&gt;Note that the $n$-th repunit is $r_n=(10^n-1)/9$.&#xA;If $p$ is a prime and $p\ne 3$ then $p\mid r_n$ iff $p\mid(10^n-1)$.&#xA;(This is due to Euclid's lemma that $p\mid ab$ implies $p\mid a$&#xA;or $p\mid b$.) So when $p\notin\{2,3,5\}$ the problem reduces to the one&#xA;you've done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the case $p=3$ remains. Now remember that $3\mid n$ iff $3$ divides&#xA;the &lt;em&gt;digital sum&lt;/em&gt; of $n$ (the sum of the decimal digits of $n$)....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-16T07:08:00.653" CommentCount="1" />
  <row Id="4761" PostTypeId="2" ParentId="4759" CreationDate="2010-09-16T07:14:47.423" Score="6" ViewCount="0" Body="&lt;p&gt;We know that $1+2+\ldots +N = \frac{1}{2}N(N+1)$. Suppose the end amount is $T$ then the repeated number is $T-\frac{1}{2}N(N+1)$.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-16T07:14:47.423" CommentCount="3" />
  <row Id="4762" PostTypeId="2" ParentId="4720" CreationDate="2010-09-16T08:03:57.763" Score="-1" ViewCount="0" Body="&lt;p&gt;There is a difference. Both products take two vectors in $\mathbb{R}^3$. The cross product gives a vector in the same $\mathbb{R}^3$ and the wedge product gives a vector in a different $\mathbb{R}^3$. The two output vector spaces are indeed isomorphic and if you choose an isomorphism you can identify the two products. However this isomorphism is a choice, or to put it another way depends on fixing a convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In higher dimensions the wedge product gives a vector in a vector space of higher dimension and so no identification is possible.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-09-16T08:03:57.763" CommentCount="1" />
  <row Id="4763" PostTypeId="2" ParentId="4720" CreationDate="2010-09-16T08:05:43.583" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, this definition is chosen for a reason, as the unique solution to a pedagogical problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do Carmo's definition is awkward and redundant in 3 dimensions, but it is the only one among the usual definitions for the cross product that when generalized to $n$ dimensions (there is a cross-product of $n-1$ vectors in $R^n$) is rigorous, visibly basis independent, and produces a vector in the same space --- all without a long digression on multilinear algebra, duality, and other sophisticated theory.  Consider the other definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The right hand rule does work in $n$ dimensions, and is coordinate-independent, but it is hard to describe without a formula which hand is the &quot;right&quot; one, and verifying that the result is multilinear can be awkward if one does not have a reliable formalism of $n$-dimensional Euclidean geometry.   I doubt most students would consider this approach rigorous even if, in principle, it can be made so.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The exterior algebra relies on theory not known by all audiences of Do Carmo's book (e.g, engineers).  It does not produce a vector in $R^n$ without even more theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The definition as a determinant of the vector coordinates can be confusing, with vectors and scalars mixed together as entries of one matrix.  It is also geometrically meaningful only when proven independent of the choice of (orthonormal) basis for the coordinates. In 3-dimensions this is known by the equivalence with the right hand rule and by drawing pictures, techniques that require a long development to perform in $n$ dimensions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-16T08:12:24.537" LastActivityDate="2010-09-16T08:12:24.537" />
  <row Id="4764" PostTypeId="1" AcceptedAnswerId="4767" CreationDate="2010-09-16T08:53:14.743" Score="3" ViewCount="272" Body="&lt;p&gt;We know that the sine function takes it values between $[-1,1]$. So is the set $$A = \{ \sin{n} \ : \ n \in \mathbb{N}\}$$ dense in $[-1,1]$. Generally, for showing the set is dense, one proceeds, by finding out what is $\overline{A}$ of this given set. And if $\overline{A} = [-1,1]$, we are through with the proof, but i having trouble here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly can one do this with cosine function also, that is proving $B= \{ \cos{n} \ : \ n \in \mathbb{N}\}$ being dense in $[-1,1]$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-16T11:37:04.880" Title="Sine function dense in $[-1,1]$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" />
  <row Id="4765" PostTypeId="1" AcceptedAnswerId="4766" CreationDate="2010-09-16T08:56:02.620" Score="1" ViewCount="113" Body="&lt;p&gt;I have a grid made up of overlapping 3x3 squares like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BaY9s.png&quot; alt=&quot;Grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers on the grid indicate the number of overlapping squares. Given that we know the maximum number of overlapping squares (9 at the middle), and the size of the squares (3x3), is there a simple way to calculate the rest of the number of overlaps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. I know the maximum number of overlaps is 9 at point (2,2) and the square size is 3x3. So given point (3,2) how can I calculate that there are 6 overlaps at that point?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2010-09-16T11:05:58.773" Title="Grid of overlapping squares" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="4766" PostTypeId="2" ParentId="4765" CreationDate="2010-09-16T09:15:06.143" Score="1" ViewCount="0" Body="&lt;p&gt;The maximum number of overlaps is always $k^{2}$ (where $k$ is the small square's side length) (assuming the larger square is always of side length $2k-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of overlaps at any other point $(i,j)$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(i,2k-i) \cdot \min(j,2k-j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a pointer towards a proof consider the first row and the first column of the larger square which must always be the sequence $(1,2,\ldots,k,\ldots,2,1)$.  Then look at why the other rows and columns are the products of their intersection with the edge row and column.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-09-16T09:15:06.143" CommentCount="1" />
  <row Id="4767" PostTypeId="2" ParentId="4764" CreationDate="2010-09-16T09:18:13.813" Score="5" ViewCount="0" Body="&lt;p&gt;The hard part is to show that for any $x$ such that $0 \le x \le 2\pi$, and any $\epsilon&gt;0$ there exists a real number $y$ and two integers $m$ and $n$ such that $|y-x|&amp;lt;\epsilon$ and $n=2\pi m+y$. Hint: break up $[0,2\pi]$ into small subintervals, remember that $\pi$ is irrational and apply the pigeonhole principle.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-16T09:18:13.813" />
  <row Id="4768" PostTypeId="1" CreationDate="2010-09-16T10:09:04.857" Score="4" ViewCount="359" Body="&lt;p&gt;How we can show every permutation is either even or odd,but not both......I can't arrive at a proof for this ..... Can anybody give me the proof...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance...&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2010-09-16T15:35:07.063" Title="every permutation is either even or odd,but not both......" Tags="&lt;permutations&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="4769" PostTypeId="1" AcceptedAnswerId="4771" CreationDate="2010-09-16T10:13:05.400" Score="0" ViewCount="91" Body="&lt;p&gt;I have a suspicion that the following expression is true, however my algebra skills aren't brilliant, so any help would be appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it possible to infer $x = \frac{y + z}{2}$ from $a^{\frac{1}{x}} = b^{\frac{1}{y}} + c^{\frac{1}{z}}$ and $a = b + c$&lt;/p&gt;&#xA;" OwnerUserId="1923" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-16T13:49:34.157" LastActivityDate="2010-09-16T13:49:34.157" Title="algebraic puzzle" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="4770" PostTypeId="2" ParentId="4764" CreationDate="2010-09-16T10:13:43.853" Score="2" ViewCount="0" Body="&lt;p&gt;Jmoy's is a good hint, but I think an even nicer approach is to&#xA;prove that the points $(\cos n,\sin n)$ are dense on the unit circle,&#xA;and then deduce the results for the individual terms.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-16T10:13:43.853" CommentCount="2" />
  <row Id="4771" PostTypeId="2" ParentId="4769" CreationDate="2010-09-16T10:20:44.077" Score="3" ViewCount="0" Body="&lt;p&gt;Let $a=2$, $b=c=1$. Also let $x=1$ then $a=b+c$ and $a^{\frac{1}{x}}=2$. Regardless of the values of $y$ and $z$ then $1^{\frac{1}{y}}+1^{\frac{1}{z}}=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we cannot infer that $x=\frac{y+z}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counterexample:&#xA;Let $a=2$, $b=c=1$, $x=1$ and $y=z=2$ then $b+c=2=a$ and $b^{\frac{1}{y}}+c^{\frac{1}{z}}=2=a^{\frac{1}{x}}$. But $x\neq \frac{y+z}{2}=2$. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-16T10:31:21.043" LastActivityDate="2010-09-16T10:31:21.043" CommentCount="2" />
  <row Id="4772" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T10:20:47.133" Score="3" ViewCount="0" Body="&lt;p&gt;One way is to define the sign of a permutation $\sigma$ using the polynomial $\Delta = \Pi (x_i-x_j)$ with $1\le i &amp;lt; j \le n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that $\sigma(\Delta) = \Pi (x_{\sigma(i)}-x_{\sigma(j)})$ satisfies $\sigma(\Delta)=\pm\Delta$. Now define the sign by $sign(\sigma)=\frac{\Delta}{\sigma(\Delta)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little more work you can show that this function is a homomorphism of groups, and that on transpositions it return -1. Therefore, if $\sigma=\tau_1\cdots\tau_k$ is a way to write $\sigma$ as a product of transpositions, we have $sign(\sigma)=(-1)^k$ and so for even permutations (permutations whose sign is 1) k must always be even, and for odd permutations (whose sign is -1) it must always be odd.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-16T10:20:47.133" CommentCount="3" />
  <row Id="4773" PostTypeId="1" CreationDate="2010-09-16T10:35:48.917" Score="0" ViewCount="348" Body="&lt;p&gt;A sequence of sets is defined as $A_n=\{x \in [0,1] : |\sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})}| \geq p\}$ for some positive $p\geq0$. What is $\limsup_{n\to\infty}A_n(p)$ and $\liminf_{n\to\infty}A_n(p)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $x$, the function $f_n(x) = \sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})}$ is $1$ or $-1$ based on number of paritions $n$. So, $|f_n(x)|$ is always $1$ for any given $x$. Hence, if $p &gt; 1$, $A_n$ is empty set for all $n$. Hence both $\limsup$ and $\liminf$ is empty. If $p \leq 1$, then $A_n = [0,1]$ for all $n$. Therefore, both $\limsup$ and $\liminf$ is $[0,1]$. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I experimented further by changing $f_n(x)$ definition as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x) = \sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})} ~~~~\mbox{if } 0 \leq x &amp;lt; 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x) = x^n ~~~~\mbox{if } 1/2 \leq x \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $\limsup_{n\to\infty}A_n(p)$ and $\liminf_{n\to\infty}A_n(p)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still trying to solve this.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T23:07:22.160" LastActivityDate="2010-12-22T00:44:27.000" Title="limit superior and limit inferior of the given sequence of sets" Tags="&lt;measure-theory&gt;&lt;limit&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="4774" PostTypeId="2" ParentId="4765" CreationDate="2010-09-16T11:05:58.773" Score="1" ViewCount="0" Body="&lt;p&gt;If you are just considering $3\times 3$ squares then the number of overlapping squares at the $(i,j)$ is the number of $1\times 1$ squares (including itself) which are internal neighbours. i.e. neighbouring squares which are not on the edge.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-16T11:05:58.773" />
  <row Id="4775" PostTypeId="2" ParentId="4581" CreationDate="2010-09-16T11:08:27.053" Score="2" ViewCount="0" Body="&lt;p&gt;Let $K/k$ be a field extension. Then $K \otimes_k k(X)$ is a localization of the integral domain $K \otimes_k k[X] = K[X]$, thus the natural map $K \otimes_k k(X) \to K(X)$ is injective. Now we have the following equivalence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $K/k$ is algebraic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Every polynomial over $K$ divides some polynomial over $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $K \otimes_k k(X) = K(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \Rightarrow b$: Because of uniqueness of the division algorithm, it suffices to prove divisibilty in an algebraic extension. So take a splitting field of the polynomial and reduce to the case of linear factors $x-a$. But they divide the minimal polynomial over $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b \Rightarrow c$: $K \otimes_k k(X)$ is the localization of $K[X]$ at all nontrivial polynomials over $k$. Since we have $b$, we localize at all nontrivial polynomials  over $K$, i.e. we get $K(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c \Rightarrow a$: Take $u \in K$. Then $1/(X-u) = p/q$ for $p \in K[X], q \in k[X]$, i.e. $p (X-u)$ is a polynomial over $k$ with root $u$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if $K/k$ is algebraic, we have $[K(X):k(X)] = [K:k]$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-16T11:14:54.187" LastActivityDate="2010-09-16T11:14:54.187" CommentCount="2" />
  <row Id="4776" PostTypeId="2" ParentId="4764" CreationDate="2010-09-16T11:37:04.880" Score="0" ViewCount="0" Body="&lt;p&gt;While searching on google, i got this JSTOR link: &lt;a href=&quot;http://www.jstor.org/stable/2688681&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2688681&lt;/a&gt; which answers the question in an intricate way.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-16T11:37:04.880" />
  <row Id="4777" PostTypeId="2" ParentId="4711" CreationDate="2010-09-16T11:44:40.270" Score="3" ViewCount="0" Body="&lt;p&gt;I show here my interpretation of your question and comments until now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Y6yoF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to define the triangle sides lengths (one is of course enough) depending on the resolution one needs. In the figure I approximated the red shape by the green right triangles with two equal angles. The shape boundary determines the triangles over or closed to it, the criteria being that at least 50% of the triangle area lies inside the shape. The inner triangles can be chosen more freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: there are a few mistakes in the sketch.  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-16T16:29:51.100" LastActivityDate="2010-09-16T16:29:51.100" CommentCount="1" />
  <row Id="4778" PostTypeId="1" AcceptedAnswerId="4786" CreationDate="2010-09-16T11:51:08.257" Score="4" ViewCount="116" Body="&lt;p&gt;Assume $M$ is a set, in which all axioms of $ZF - P + (V=L)$ hold. Does then $M$ believe that there exists an uncountable ordinal? I mean, why should the class of all countable ordinal numbers be a set?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-16T12:55:42.130" LastActivityDate="2010-09-16T13:35:24.383" Title="Uncountable ordinals without power set axiom" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4779" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T12:15:49.947" Score="2" ViewCount="0" Body="&lt;p&gt;I claim all three &quot;proofs&quot; of this result (currently) on &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_of_a_permutation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; are incomplete.  Let's begin with the observation that, assuming that identity permutation is not an odd permutation, then the result follows fairly easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: Assuming the identity permutation is not an odd permutation, then all permutations are either even xor odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;:  Let $\sigma$ be both an even and an odd permutation.  Then there exists transpositions $t_i$ and $s_j$ such that \[\sigma=t_1 \circ t_2 \circ \cdots \circ t_k=s_1 \circ s_2 \circ \cdots \circ s_m\] where $k$ is even and $m$ is odd.  Note that \[\sigma^{-1}=s_m \circ s_{m-1} \circ \cdots \circ s_1.\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the identity permutation $\sigma \circ \sigma^{-1}$ is the odd permutation \[t_1 \circ t_2 \circ \cdots \circ t_k \circ s_m \circ s_{m-1} \circ \cdots \circ s_1,\] giving a contradiction.  Thus completing the proof of the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that we cannot just assume that the identity permutation not an odd permutation (yes, it's an even permutation, but what's to stop it being an odd permutation also?), it does not follow from the definition and it is the base case of the &quot;induction&quot;.  To prove it, we need to show that the identity permutation cannot be decomposed into an odd number of transpositions (without using the theorem itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can deduce from the above theorem that either (a) all permutations are both even and odd or (b) all permutations are either even xor odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On wikipedia: Proof 1 states that the identity permutation is an even permutation (which it is) then assumes that the identity permutation is not also an odd permutation (this is analogous to assuming that a closed set is not an open set).  Proof 2 essentially says that we can switch signs by multiplying by a transposition (which is fine, if you know &lt;em&gt;a priori&lt;/em&gt; that there exists some permutation that is not even or not odd).  Proof 3 neglects the possibility that an even-length word might be equal to an odd-length word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same problem arises in:  &lt;a href=&quot;http://www.jstor.org/stable/2316272&quot; rel=&quot;nofollow&quot;&gt;An Historical Note on the Parity of Permutations&lt;/a&gt;, T. L. Barlow, American Mathematical Monthly Vol. 79, No. 7 (Aug. - Sep., 1972), pp. 766-769.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;:  Keith Conrad gave a proof that the identity permutation is not an odd permutation &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sign.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  It is almost a page long (and is the majority of the proof of the result in question).&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-09-16T13:23:27.850" LastActivityDate="2010-09-16T13:23:27.850" CommentCount="6" />
  <row Id="4780" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T12:16:37.730" Score="1" ViewCount="0" Body="&lt;p&gt;This is overkill, but it follows from general facts about Coxeter groups as outlined &lt;a href=&quot;http://qchu.wordpress.com/2010/06/26/coxeter-groups/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  In particular, it follows from the fact that $S_n$ has presentation $s_i^2 = 1, (s_i s_{i+1})^3 = 1, s_i s_j = s_j s_i, |i - j| \ge 2$ (which follows from the faithfulness of the geometric representation) that the homomorphism $s_i \to -1$ is well-defined.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T12:16:37.730" CommentCount="1" />
  <row Id="4781" PostTypeId="1" CreationDate="2010-09-16T12:58:18.613" Score="3" ViewCount="89" Body="&lt;p&gt;The complexity class RL is described at the complexity zoo as: Has the same relation to L as RP does to P. The randomized machine must halt with probability 1 on any input. It must also run in polynomial time (since otherwise we would just get NL).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is - how can we get NL when omitting the demand for polynomial time? I have a solution of my own but it seems strange to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution: If suffices to solve ST-CON. We can use the same NL algorithm for ST-CON (guessing a path) with one major difference - we count the number of steps we make, and if it suppresses the number of vertices in the graph, we restart the computation, without remembering ANYTHING.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means we can play this game indefinitely. If the graph is not ST-connected, then we'll never halt, but if it's ST-connected we'll halt in probability 1 (this is the same as saying that a geometric random variable obtains a finite value in probability 1). However, since we do not halt for NO-instances, this solution &quot;feels wrong&quot; to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another solution? And is my solution correct?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-16T13:14:08.263" LastActivityDate="2010-12-06T22:55:46.200" Title="Complexity - why is RL=NL when omitting the demand for polynomial run-time?" Tags="&lt;algorithms&gt;&lt;probability&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="4782" PostTypeId="2" ParentId="4773" CreationDate="2010-09-16T12:59:37.720" Score="1" ViewCount="0" Body="&lt;p&gt;I believe you have got the first example correct now. For the second example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p&gt;1$ then since $f_n(x)\leq 1$ for all $x\in(\frac{1}{2},1]$ we have $A_n=\emptyset$. If $p\leq 1$ then $f_n(x)\geq p$ if and only if $x\geq p^{\frac{1}{n}}$. Thus $A_n=[0,\frac{1}{2})\cup [p^{\frac{1}{n}},1]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity for $p=1$: I take the set $[1,1]=\{1\}$ so this agrees with what you suspected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p\geq 1$, $A_n$ doesn't depend on $n$ so the $\liminf$ and $\limsup$ are easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p&amp;lt;1$ we have $A_{n+1}\subset A_n$. So $\liminf=\limsup=\lim=[0,\frac{1}{2})\cup\{1\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always get this from $\liminf A_n=\bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty} A_m$ and similarly $\limsup A_n=\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty} A_m$&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-16T18:52:05.473" LastActivityDate="2010-09-16T18:52:05.473" CommentCount="8" />
  <row Id="4783" PostTypeId="1" AcceptedAnswerId="4784" CreationDate="2010-09-16T13:20:23.217" Score="4" ViewCount="331" Body="&lt;p&gt;We know that $$ 2^{n}= (1+1)^{n} = \sum\limits_{k=0}^{n} {n \choose k}$$ I was asked to solve this limit, $$\lim_{n \to \infty} \ \sum\limits_{k=0}^{n} {n \choose k}^{-1}=? \quad \text{for} \ n \geq 1$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-12T16:13:56.180" Title="Finding $\displaystyle \lim_{n \to \infty} \sum\limits_{k=0}^{n} { n \choose k}^{-1}$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4784" PostTypeId="2" ParentId="4783" CreationDate="2010-09-16T13:26:55.173" Score="9" ViewCount="0" Body="&lt;p&gt;The terms from $k = 2$ to $k = n-2$ are all at most $\frac{1}{ {n \choose 2} }$ by unimodality, hence their sum is at most $\frac{2}{n-1}$.  So the entire sum is at most $2 + \frac{2}{n} + \frac{2}{n-1}$.  The limit is $2$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T13:26:55.173" CommentCount="3" />
  <row Id="4785" PostTypeId="1" CreationDate="2010-09-16T13:27:46.853" Score="2" ViewCount="70" Body="&lt;p&gt;For a sequence of positive integers $a_1, \ldots, a_n$ and a base ring $R$ there is a graded ring $R[x_1,\ldots, x_n]$ where $x_i$ is in degree $a_i$.  We can then apply Proj and get a scheme, and this is usually called a weighted projective space; if all of the $a_i$ are 1, then the resulting scheme really is projective space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the way that this arises is as the quotient of $\mathbb{A}^n \setminus 0$ by an action of the multiplicative group, given by $(x_1,\ldots,x_n) \simeq (\lambda^{a_1} x_1, \ldots, \lambda^{a_n} x_n)$ for all $\lambda$.  This is a &quot;coarse&quot; group quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an alternative version where one instead takes the associated quotient stack/orbifold, and this has a number of nice properties (including possession of a line bundle $\mathcal{O}(1)$); this is true more generally of a graded ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the appropriate terminology for the stack-theoretic version of this construction if &quot;weighted projective space&quot; is taken?&lt;/p&gt;&#xA;" OwnerUserId="1925" LastActivityDate="2010-09-16T13:27:46.853" Title="Terminology for weighted projective spaces" Tags="&lt;terminology&gt;&lt;algebraic-geometry&gt;" CommentCount="2" />
  <row Id="4786" PostTypeId="2" ParentId="4778" CreationDate="2010-09-16T13:35:24.383" Score="4" ViewCount="0" Body="&lt;p&gt;No. One cannot prove that $\omega_1$ exists in ZF - P, with or without V = L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $HC$ of hereditarily countable sets always satisfies ZF - P. (This is straightforward to check axiom per axiom.) Of course, $\omega_1 \notin HC$ but moreover every set in $HC$ is countable or finite as witnessed by a function in $HC$. Therefore, $HC$ knows that every set in $HC$ is countable and hence $HC$ is a model of ZF - P + &quot;every set is at most countable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throwing in V = L into the mix doesn't help. Indeed, $HC^L = L_{\omega_1^L}$ (see my answer to your &lt;a href=&quot;http://mathoverflow.net/questions/38570/inclusions-of-constructible-sets-are-not-definable/38576#38576&quot; rel=&quot;nofollow&quot;&gt;recent MO question&lt;/a&gt; for a proof) is a model of ZF - P + V = L + &quot;every set is at most countable.&quot;&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-09-16T13:35:24.383" CommentCount="4" />
  <row Id="4787" PostTypeId="1" CreationDate="2010-09-16T13:42:39.973" Score="3" ViewCount="177" Body="&lt;p&gt;Let's take the numbers 0-10.  Their mean is 5, and the individual deviations from 5 are&lt;br&gt;&#xA;&lt;code&gt;-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5&lt;/code&gt;&lt;br&gt;&#xA;And so the average (magnitude of) deviation from the mean is $30/11 \approx 2.72$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not the standard deviation.  The standard deviation is $\sqrt{10} \approx 3.16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first mean-deviation is a simpler and by far more intuitive definition of the &lt;em&gt;&quot;standard-deviation&quot;&lt;/em&gt;, so I'm sure it's the first definition statisticians worked with.  However, for some reason they decided to adopt the second definition instead.  What is the reasoning behind that decision?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-09-16T13:48:19.807" LastActivityDate="2010-09-16T16:15:24.087" Title="Motivation behind standard deviation?" Tags="&lt;intuition&gt;&lt;statistics&gt;&lt;standard-deviation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="4788" PostTypeId="1" AcceptedAnswerId="4794" CreationDate="2010-09-16T14:20:22.483" Score="10" ViewCount="490" Body="&lt;p&gt;Despite its name, its often claimed that the fundamental theorem of algebra (which shows that the Complex numbers are algebraically closed - this is not to be confused with the claim that a polynomial of degree n has &lt;strong&gt;at most&lt;/strong&gt; n roots) is not considered fundamental by algebraists as it's not needed for the development of modern algebra. My question is - what &lt;strong&gt;are&lt;/strong&gt; the major uses of the theorem, and to which extent can they justify the claim that the theorem is fundamental for &lt;strong&gt;something&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example I think of is the Jordan canonical form for matrices, but I don't think it suffices.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-17T16:58:23.690" LastActivityDate="2010-09-17T16:58:23.690" Title="How fundamental is the fundamental theorem of algebra?" Tags="&lt;terminology&gt;&lt;algebra&gt;&lt;complex-numbers&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" />
  <row Id="4789" PostTypeId="2" ParentId="4787" CreationDate="2010-09-16T14:28:19.333" Score="2" ViewCount="0" Body="&lt;p&gt;Squaring is nicer than taking the absolute value, e.g. it is smooth.  It also leads to a definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Variance&quot; rel=&quot;nofollow&quot;&gt;variance&lt;/a&gt; which has nice mathematical properties, e.g. it is additive.  But for me the theorem that really justifies using standard deviation over the mean absolute error is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;central limit theorem&lt;/a&gt;.  The central limit theorem is at work whenever we measure the mean and standard deviation of a distribution we assume to be normal (e.g. heights in a population) and use that to make predictions about the entire distribution, since a normal distribution is completely specified by its mean and standard deviation.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-16T14:48:34.323" LastActivityDate="2010-09-16T14:48:34.323" CommentCount="6" />
  <row Id="4790" PostTypeId="2" ParentId="4738" CreationDate="2010-09-16T14:33:34.057" Score="1" ViewCount="0" Body="&lt;p&gt;Yes.  The Baire category theorem asserts that in a complete metric space, a set of first category not only is not the whole space, but actually has dense complement.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-16T14:33:34.057" />
  <row Id="4791" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T15:03:33.453" Score="4" ViewCount="0" Body="&lt;p&gt;Well, there is something called &quot;Algebraic Geometry&quot;. You may check the first pages of most of the books on that subject and you'll see something like: &quot;Let $k$ be an algebraically closed field...&quot; [for instance, $\mathbb{C}$]. Indeed, this is the beginning of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387902449&quot; rel=&quot;nofollow&quot;&gt;Hartshorne's book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason? It's much harder to do geometry over the real numbers, where all your theorems about the set of solutions of some system of polynomial equations (the primary object of Algebraic Geometry) can miserably fail, just because there are equations like $x^2 + 1 = 0$ with no solution at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the existence of polynomial equations like $x^2 + 1 = 0$ with no solution over the reals, ruins the basic bijection between prime  ideals in $\mathbb{R}[x_1, \dots , x_n]$ and irreducible algebraic sets  in the affine space $\mathbb{R}^n$, which is central for Algebraic Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean you cannot do Algebraic Geometry over the reals, only it's another business.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-16T17:44:59.567" LastActivityDate="2010-09-16T17:44:59.567" CommentCount="5" />
  <row Id="4792" PostTypeId="2" ParentId="4741" CreationDate="2010-09-16T15:04:32.990" Score="3" ViewCount="0" Body="&lt;p&gt;This smells like a problem in the Calculus of Variations and can be solved as such.  We therefore make a slight reformulation: among all $C^3$ curves of length $l$ originating and terminating at the same point, find the one(s) for which the maximum curvature is smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's begin by developing some intuition.  In general, a minimax problem like this has a solution where the objective (the curvature in this case) is as constant as possible.  A circle is an obvious candidate for a solution and indeed a circle of length $l$ has constant radius $l / (2 \pi)$ and constant curvature of $2 \pi / l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a tiny simplification, by choosing our linear units of measurement let's henceforth assume $l=1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique of the Calculus of Variations is to assume we have a solution and perturb it a little, showing that any perturbation, no matter how small, cannot decrease the value of the objective.  To motivate this in the present case, though, I find it more appealing to consider how one might go about reducing the maximum curvature of any unit-length loop.  A point of maximum curvature is part of a relatively sharp &quot;bump&quot; on the curve.  If we push that bump inwards a little, we should be able to decrease its sharpness and at the same time we slightly decrease the total length of the curve.  To make up for the latter, uniformly dilate the entire curve relative to its point of origin until the dilated curve again has unit length.  This uniformly &lt;em&gt;decreases&lt;/em&gt; all curvatures along the loop.  In this fashion the maximum curvature has strictly decreased.  The only way this operation can fail is when there is no &quot;bump&quot;: the curvature everywhere is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this go through rigorously, use an adapted orthonormal frame for the curve: $T(t) = \alpha'(t)$ is the tangent, $N(t)$ is the inward-pointing unit normal with $T'(t) = \kappa(t)N(t)$, and $B(t)$ is the unit binormal, with $N'(t) = -\kappa(t)T(t) + \tau(t)B(t)$ and $B'(t) = -\tau(t)N(t)$.  ($\tau$ is the &lt;em&gt;torsion&lt;/em&gt;.  These Serret-Frenet formulae generalize readily to higher dimensions and in two dimensions just forget about $B$.)  Notice that $\kappa \ge 0$.  Let $\epsilon &gt; 0$ be small and let $\delta(t)$ be a smooth non-negative function in an open neighborhood of $[0,1]$, vanishing at $0$ and $1$ (in order to keep the endpoints of the new curve fixed).  Set the &quot;pushed&quot; curve to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(t) = \alpha(t) + \epsilon \delta(t) N(t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute the curvature of $\psi$ to first order in $\epsilon$ using the Serret-Frenet formulae.  I find that its square becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa^2 + 2 \kappa \left( \delta'' - \delta \tau^2 \right) \epsilon + O(\epsilon^2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the curvature is not constant.  Then there is a closed interval of maximum curvatures (perhaps reducing to a point) and within some neighborhood of that interval all curvatures are strictly less than the maximum.  We can make $\delta$ zero outside this larger neighborhood, let it increase slowly enough (and choose $\epsilon$ sufficiently small) so that $\delta''$ keeps the new curvature less than the maximum within the outer neighborhood, and make $\delta''$ strictly negative at all points within the interval of maximum curvature.  This procedure strictly decreases the maximum curvature within the outer interval.  You can also check, again to second order in $\epsilon$, that the length of $\psi'$ decreases by $\epsilon \delta \kappa \ge 0$, so we will be able to apply our dilation trick in order to keep the total loop length constant after this perturbation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little more formally (and using less work, actually), the preceding equation shows that when $\kappa$ is constant and $\tau = 0$, we would need $\delta'' \lt \delta \tau^2 = 0$ everywhere in order to decrease the curvature, which implies $\delta$ is identically zero in the unit interval.  Otherwise (namely, if $\kappa$ is not constant or $\tau \gt 0$), it will be possible to decrease the curvature subject to maintaining the unit length of the curve.  Curves of constant curvature and zero torsion are circular arcs, and the only circular arc that returns to its starting point is a full circle.  Therefore the maximum curvature in &lt;em&gt;any&lt;/em&gt; sufficiently smooth unit-length loop can be no less than $2 \pi$ and is strictly greater than that if the loop is not a circle, QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the same technique works in higher dimensions, too, although I have not done the calculations.  (There are additional vectors in the adapted frame, and therefore additional torsion terms, but this doesn't seem to present any obstacle to carrying out the same program.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-16T23:47:45.373" LastActivityDate="2010-09-16T23:47:45.373" CommentCount="7" />
  <row Id="4793" PostTypeId="2" ParentId="4768" CreationDate="2010-09-16T15:35:07.063" Score="0" ViewCount="0" Body="&lt;p&gt;It is amazing. It seems no one (including the Wikipedia contributors) is aware of the link between the parity of a permutation and the determinant of the associated permutation matrix :)&lt;/p&gt;&#xA;" OwnerUserId="1927" LastActivityDate="2010-09-16T15:35:07.063" CommentCount="1" />
  <row Id="4794" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T15:59:11.423" Score="8" ViewCount="0" Body="&lt;p&gt;I can immediately think of four important areas for which the FTA is actually quite fundamental. Maybe other people may come up with more contributions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Algebraic Geometry (as already touched in Agusti Roig's answer). In particular, we should couple the FTA with Lefschetz's Principle which basically says that the complex projective space is sort of &quot;universal&quot; environment for algebraic geometry in characteristic 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Classification of algebraic structures over local or global fields, where one uses constantly that $\Bbb C$ has no non trivial finite extensions (a trivial consequence of the FTA)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory of representations of groups, where the TFA plays a big role making the theory of complex representations simpler.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Galois theory and algebraic number theory, where the FTA allows to realize $\Bbb C$ as a good environment for studying finite extensions of $\Bbb Q$ (such in the basic resut that a number field of degree $n$ always admits $n$ independent embedings into $\Bbb C$) and allowing the use of the &quot;geometry of numbers&quot; (Minkowski's theorem) to prove important arithmetic results such as the finiteness of the class number and the structure of the units (Dirichlet's theorem)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-16T15:59:11.423" />
  <row Id="4795" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T16:08:30.833" Score="1" ViewCount="0" Body="&lt;p&gt;Perhaps it is worth expanding my answer.  (Forgive me if I get any details wrong.)  If you are interested in Lie groups, you are interested in their classification.  A connected Lie group is the quotient of its universal cover by a discrete subgroup of its center, so to classify connected Lie groups it suffices to classify simply connected Lie groups, then compute their centers.  This classification is essentially equivalent to the classification of finite-dimensional real Lie algebras.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi_decomposition&quot; rel=&quot;nofollow&quot;&gt;Levi decomposition&lt;/a&gt; says that to classify such Lie algebras it is more or less enough to classify solvable and semisimple Lie algebras.  The former classification problem, as I understand it, is wild; however the latter is tractable.  The basic idea is first to study &lt;em&gt;complex&lt;/em&gt; semisimple Lie algebras and then to study their real forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big reason this is possible is that the complex numbers are algebraically closed.  This means that eigenvalues always exist, and from there the entire classification is possible.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T16:08:30.833" />
  <row Id="4796" PostTypeId="2" ParentId="4787" CreationDate="2010-09-16T16:15:24.087" Score="6" ViewCount="0" Body="&lt;p&gt;Your guess is correct: least absolute deviations was the method tried first historically.  The first to use it were astronomers who were attempting to combine observations subject to error.  Boscovitch in 1755 published this method and a geometric solution.  It was used later by Laplace in a 1789 work on geodesy.  Laplace formulated the problem more mathematically and described an analytical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legendre appears to be the first to use least squares, doing so as early as 1798 for work in celestial mechanics.  However, he supplied no probabilistic justification.  A decade later, Gauss (in an 1809 treatise on celestial motion and conic sections) asserted axiomatically that the arithmetic mean was the best way to combine observations, invoked the maximum likelihood principle, and then showed that a probability distribution for which the likelihood is maximized at the mean must be proportional to $\exp(-x^2 / (2 \sigma^2))$ (now called a &quot;Gaussian&quot;) where $\sigma$ quantifies the precision of the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The likelihood (when the observations are statistically independent) is the product of these Gaussian terms which, due to the presence of the exponential, is most easily maximized by minimizing the negative of its logarithm.  Up to an additive constant, the negative log of the product is the sum of the squares (all divided by a constant $2 \sigma^2$, which will not affect the minimization).  Thus, even historically, the method of least squares is intimately tied up with likelihood calculations and averaging.  There are plenty of other modern justifications for least squares, of course, but this derivation by Gauss--with the almost magical appearance of the Gaussian, which had first appeared some 70 years early in De Moivre's work on sums of Bernoulli variables (the Central Limit Theorem)--is memorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This story was researched, and is ably recounted, by Steven Stigler in his &lt;em&gt;The History of Statistics - The Measurement of Uncertainty before 1900&lt;/em&gt; (1986).  Here I have merely given the highlights of parts of chapters 1 and 4.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-16T16:15:24.087" CommentCount="2" />
  <row Id="4797" PostTypeId="1" AcceptedAnswerId="4798" CreationDate="2010-09-16T17:24:57.733" Score="1" ViewCount="105" Body="&lt;p&gt;Suppose A is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadamard_matrix&quot; rel=&quot;nofollow&quot;&gt;Hadamard matrix&lt;/a&gt; of size $d$. Let A be in normalized in a sense that first row and first column are all ones. What is the sum of rows? I tried random Hadamard matrices and seem to get $d,0,0,0,0,\ldots$&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-09-23T17:18:23.430" Title="What is sum of rows of Hadamard matrix" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4798" PostTypeId="2" ParentId="4797" CreationDate="2010-09-16T17:29:39.057" Score="5" ViewCount="0" Body="&lt;p&gt;Follows immediately from the property of Hadamard Matrices: Any two distinct rows are orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the other rows are orthogonal to the all ones row, the sum of the elements in each of those rows must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To answer Qiaochu's query:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MUltiplying a row or column of a Hadamard matrix by -1 gives yet another Hadamard matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this operation, we can easily normalize a Hadamard matrix to make the elements of the first row to be all ones.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-23T17:18:23.430" LastActivityDate="2010-09-23T17:18:23.430" CommentCount="3" />
  <row Id="4799" PostTypeId="2" ParentId="4788" CreationDate="2010-09-16T17:49:36.060" Score="10" ViewCount="0" Body="&lt;p&gt;One cannot over-emphasize the point that one of the primary benefits of working over an algebraically closed field such as $\mathbb C$ is the immense simplification gained by &lt;strong&gt;linearizing&lt;/strong&gt; what would otherwise be much more complicated &lt;strong&gt;nonlinear&lt;/strong&gt; phenomena. The ability to factor polynomials completely into linear factors over $\mathbb C$ enables widespread linearization simplifications of diverse problems. An example familiar to any calculus student is the fact that integration of rational functions is much simpler over $\mathbb C$ (vs. $\mathbb R$) since partial fraction decompositions involve at most linear (vs. quadratic) polynomials in the denominator. Analogously, one may reduce higher-order constant coefficient differential and difference equations (i.e. recurrences) to linear (first-order) equations by factoring them as linear operators over $\mathbb C$ (i.e. &quot;operator algebra&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, such &lt;strong&gt;simplification by linearization&lt;/strong&gt; was at the heart of the development of abstract algebra. Namely, Dedekind, by abstracting out the essential linear structures (ideals and modules) in number theory, greatly simplified the prior nonlinear Gaussian theory (based on quadratic forms). This enabled him to exploit to the hilt the power of linear algebra. Examples abound of the revolutionary breakthroughs that this brought to number theory and algebra - e.g. it provided the methods needed to generalize the law of quadratic reciprocity to higher-order reciprocity laws - a longstanding problem that motivated much of the early work in number theory and algebra. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T00:09:12.653" LastActivityDate="2010-09-17T00:09:12.653" CommentCount="10" />
  <row Id="4800" PostTypeId="1" CreationDate="2010-09-16T18:22:11.177" Score="2" ViewCount="129" Body="&lt;p&gt;Suppose I have an number of distances from an unknown location to a known location. I can use these distances and the known locations to draw a number of circles. The point where all the circles intersect is my unknown location. This is easy to solve. However, if my distance measurements have some error, I will get a cluster of points around my unknown location. The question is, for each pair of intersection points, how can I tell which solution is closer to my unknown point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every solution I have come up with involves many special cases, such as when all the known locations are inside one circle, or all the known locations are co-linear, and so on, buy I am hoping to come up with a more elegant solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets make this more simple. I have 2 circles that have at most 2 intersection points. If they don't intersect, I can easily choose the point between them. If they do intersect at 2 points, how can I use a third circle to choose which one of these 2 intersections is the correct one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current plan is to do a comparison between the distance between the radius of the third circle and the intersection points and choose the smaller one. The problem with this is that I don't know how much error this approach can tolerate before you choose the wrong side. If I knew that, I could put a goal on my measurement method to try to reduce the error by X%&lt;/p&gt;&#xA;" OwnerUserId="1932" LastEditorUserId="1932" LastEditorDisplayName="" LastEditDate="2010-09-17T12:56:14.577" LastActivityDate="2010-09-17T12:56:14.577" Title="Choosing solutions for the intersections of n number of circles" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4802" PostTypeId="2" ParentId="4800" CreationDate="2010-09-16T19:07:58.613" Score="4" ViewCount="0" Body="&lt;p&gt;One handles error with a probability model.  In this case, the typical error in a distance measurement is likely proportional to the distance itself.  The errors themselves are often thought of as accumulated small nearly independent errors, allowing one to invoke the Central Limit Theorem and suppose, at least as a reasonable hypothesis, that the errors are normally distributed.  We usually assume there's no systematic bias in the errors: they should average out to zero in principle.  A simple model is obtained by supposing the errors are independent.  (Other suppositions can be treated, depending on how the measurements were made, but it quickly gets complicated.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with just three unknowns to estimate: the true coordinates $(x,y)$ of your location and the precision of the relative errors, usually expressed as their (common) standard deviation $\sigma$.  Your data consist of $n$ measured distances to the known locations $(x_i, y_i)$, say $d_i, i = 1 \ldots n$.  Mathematically, these assumptions translate to the following.  The probability of observing $d_i$ equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{2 \pi} \sigma \delta_i} \exp \left( -\frac{(d_i - \delta_i)^2}{2 \sigma^2 \delta_i^2)} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta_i = \delta_i(x,y) = \sqrt{(x - x_i)^2 + (y - y_i)^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of your data $(d_1, d_2, \ldots, d_n)$ is the product of these probabilities.  This is the &lt;em&gt;likelihood&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reasonably tractable estimator maximizes the likelihood (as a function of the unknown parameters $x$, $y$, and $\sigma$).  This is usually done by maximizing the negative logarithm of the likelihood, which will be a sum of terms, one for each data value.  It's nonlinear and a little nasty, but we know geometrically that there will be at least one global optimum and, quite likely, exactly one.  Many methods exist to solve this, available in statistical packages, numerical optimization routines, and general-purpose math software like &lt;em&gt;Mathematica&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the optimal arguments $(x,y)$ tell you where your location likely is.  The optimal value of $\sigma$ estimates the typical relative error in the distance measurement: you can check that it's a reasonable value.  There are ways to extract confidence intervals for all three parameters (from the Hessian of the log likelihood).  A joint confidence interval for $(x,y)$ gives you an ellipse in which the true point is likely to be.  Statistical packages will give you this information.  If you are doing this by hand, you can analyze any other candidate solution $(x',y')$, such as the intersection of a few circles, by evaluating the likelihood there and comparing it to the optimum likelihood.  If the logarithms differ by less than $2$, the candidate solution is consistent with the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, as you've seen, you're practically forced to take a statistical approach.  With real data you find that some circles don't even intersect and that overall the set of data has little or no internal consistency.  You &lt;em&gt;have&lt;/em&gt; to model the error somehow.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-09-16T19:19:00.463" LastActivityDate="2010-09-16T19:19:00.463" />
  <row Id="4803" PostTypeId="1" CreationDate="2010-09-16T19:13:24.393" Score="3" ViewCount="153" Body="&lt;p&gt;If $n$ is an integer, is there a nice way to write the partial fraction expansion of $\frac{1}{x^n - 1}$? I figure that if $\zeta$ is the $n$-th root of unity, then for some coefficients $a_0, a_1, \ldots, a_{n-1}$ we may write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{x^n - 1} = \frac{a_0}{x - 1} + \frac{a_1}{x - \zeta} + \frac{a_2}{x - \zeta^2} + \ldots + \frac{a_{n-1}}{x - \zeta^{n-1}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for $0 \leq i \leq n -1$,&#xA;$$&#xA;a_i = \lim_{x \to \zeta^{i}} \frac{x - \zeta^{i}}{x^n - 1} = \frac{1}{(\zeta^{i} - 1) \ldots (\zeta^{i} - \zeta^{i-1}) (\zeta^{i} - \zeta^{i + 1}) \ldots (\zeta^{i} - \zeta^{n - 1})}.&#xA;$$&#xA;Is there a simpler expression for this and if so, how could I see it easily?&lt;/p&gt;&#xA;" OwnerUserId="1933" LastEditorUserId="1933" LastEditorDisplayName="" LastEditDate="2010-09-16T19:21:05.420" LastActivityDate="2010-09-16T19:34:55.040" Title="The partial fraction expansion of $\frac{1}{x^n - 1}$. " Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4804" PostTypeId="2" ParentId="4803" CreationDate="2010-09-16T19:29:30.823" Score="6" ViewCount="0" Body="&lt;p&gt;Use l'Hopital's rule.  $a_i = \lim_{x \to \zeta^i} \frac{1}{nx^{n-1}} = \frac{\zeta^i}{n}$.  Note that the identity I am claiming is true is equivalent to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=0}^{n-1} (\zeta^i)^k = \begin{cases} n \text{ if } n | k \\ 0 \text{ otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by taking the generating function of both sides.  More generally, the coefficient of $\frac{1}{x - a}$ in the partial fraction decomposition of $\frac{P(x)}{Q(x)}$ is (as long as $a$ is a simple root of $Q$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \to a} \frac{(x - a) P(x)}{Q(x)} = \frac{P(a)}{Q'(a)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula is tremendously useful (for example I recently used it on &lt;a href=&quot;http://math.stackexchange.com/questions/4658/what-is-the-probability-of-a-coin-landing-tails-7-times-in-a-row-in-a-series-of-1/4675#4675&quot;&gt;this math.SE question&lt;/a&gt;) but does not seem to be widely known.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-16T19:34:55.040" LastActivityDate="2010-09-16T19:34:55.040" CommentCount="2" />
  <row Id="4805" PostTypeId="1" AcceptedAnswerId="4809" CreationDate="2010-09-16T19:51:20.197" Score="0" ViewCount="68" Body="&lt;p&gt;What do you call this function?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = x * (1.0 - x)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does it have a name?  It’s cropped up in at least two statistical formulas I've been asked to program so far.  (In both cases, &lt;code&gt;x&lt;/code&gt; was something called a &lt;em&gt;p-value&lt;/em&gt; (for what it's worth).)&lt;/p&gt;&#xA;" OwnerUserId="984" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-16T21:51:11.370" LastActivityDate="2010-09-16T21:51:11.370" Title="If x is a p-value, what do we call x*(1-x)?" Tags="&lt;terminology&gt;&lt;statistics&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="4806" PostTypeId="1" CreationDate="2010-09-16T19:53:36.750" Score="1" ViewCount="104" Body="&lt;p&gt;While learning about limits and derivatives, I came accross the following problem on one of Stewart's exercises book. I've been trying to wrap my head around it but I haven't got anywhere useful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$ \forall x f''(x) $ exists&lt;/li&gt;&#xA;&lt;li&gt;$ \exists c \in \mathbb{R} $ / $\forall x \neq c, f'(x) &gt; 0 \wedge f'(c) = 0 $&lt;/li&gt;&#xA;&lt;li&gt;Then, $(c, f(c))$ is an &lt;strong&gt;inflection point&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From this I can gather that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ is continuous&lt;/li&gt;&#xA;&lt;li&gt;$f$ is increasing $\forall x \neq c$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My intuition tells me that $(3)$ is false since I might be able to come up with a function defined by parts that contradicts the statement, but I haven't found a way to prove this. Any pointers would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1935" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-16T21:38:48.490" LastActivityDate="2010-09-16T21:38:48.490" Title="Finding the inflection point for a function" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="4807" PostTypeId="2" ParentId="4806" CreationDate="2010-09-16T20:01:57.357" Score="3" ViewCount="0" Body="&lt;p&gt;Because the second derivative exists everywhere, you also know that $f'$ is continuous everywhere. Now think about the function $f'(x)$; it is positive to the left and to the right of $c$, and is $0$ at $c$, and it is continuous. Does that tell you something about $f''$, and hence about concavity?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-16T20:01:57.357" CommentCount="3" />
  <row Id="4808" PostTypeId="1" AcceptedAnswerId="4816" CreationDate="2010-09-16T20:02:03.207" Score="5" ViewCount="322" Body="&lt;p&gt;In this wikipedia, article &lt;a href=&quot;http://en.wikipedia.org/wiki/Circle#Area_enclosed&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Circle#Area_enclosed&lt;/a&gt; its stated that the circle is the closed curve which has the maximum area for a given arc length. First, of all i would like to see different proofs, for this result. (If there are any elementary ones!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One, interesting observation, which one can think while seeing this problem, is: How does one propose such type of problem? Does, anyone take all closed curves, and calculate their area to come this conclusion. I don't think thats the right intuition.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-17T15:46:08.957" Title="Why Circle encloses largest Area?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;analysis&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" />
  <row Id="4809" PostTypeId="2" ParentId="4805" CreationDate="2010-09-16T20:23:21.037" Score="4" ViewCount="0" Body="&lt;p&gt;If &lt;code&gt;Z&lt;/code&gt; can take values 1 or 0 with probability &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;1-x&lt;/code&gt;, respectively, we usually call &lt;code&gt;x*(1-x)&lt;/code&gt; the &lt;strong&gt;variance&lt;/strong&gt; of Z.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the more general &lt;a href=&quot;http://en.wikipedia.org/wiki/Variance#Definition&quot; rel=&quot;nofollow&quot;&gt;variance&lt;/a&gt; formula, and works because Z is what we call an &lt;a href=&quot;http://en.wikipedia.org/wiki/Indicator_function#Mean.2C_Variance_and_covariance&quot; rel=&quot;nofollow&quot;&gt;indicator variable&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2010-09-16T20:23:21.037" />
  <row Id="4810" PostTypeId="2" ParentId="4758" CreationDate="2010-09-16T20:43:35.940" Score="4" ViewCount="0" Body="&lt;p&gt;The essence of the matter is both strikingly simple and extremely general. Namely, there are infinitely many solutions to your congruence simply because the solutions lie in the orbit of an invertible map on a finite set. But such permutations decompose into finite &lt;strong&gt;cycles&lt;/strong&gt; - so they repeat infinitely often upon iteration.  Below I discuss your simple linear example and then I show how the technique generalizes to nonlinear recursions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we convert $\rm\;f_n = 10^n-1 \;$ to shift form. Note $\rm\;f_{n+2}-f_{n+1} = 9\cdot 10^{n+1} = 10\ (f_{n+1}-f_n) \;$ which yields the recurrence relation $\rm\;\: f_{n+2} \;=\; 11\ f_{n+1}-10\ f_n \:$.  This is simpler as a shift map on pairs: $\rm\;S : (f_n,f_{n+1})\to (f_{n+1},f_{n+2}) =\: (f_{n+1},11\ f_{n+1} - 10\ f_n) \;$. Extended to all pairs it yields the  map $\rm\;S : (a,b)\to (b,\:11\:b-10\:a = c)\:$  with the obvious inverse $\rm\;(b,c)\to (a = (11\:b-c)/10,\:b)\:$ which clearly exists in any ring $\rm\:R\:$ where $10\:$ has an inverse. In particular, let $\rm\;R =\mathbb Z\pmod m \;$ be the ring of integers modulo $\rm\:m&gt;1\:$, for any $\rm\:m\:$ coprime to $10$. Since $\rm\:R \:$ is finite, so too is its set of pairs $\rm  R^2$. But on such pairs the shift map $\rm\;S \;$ is an invertible map on a finite set, so it is a permutation. Therefore its orbit decomposition consists of finite cycles. Thus the values $\rm f_n$ all occur in the orbit of the initial condition pair $\rm (f_0,f_1) = (0,9)$, and each value necessarily occurs infinitely many times since the shift map successively steps along the finite circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method also works for nonlinear recurrences with invertible shift map, e.g. from one of my old posts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXERCISE&lt;/strong&gt; $\;\;$ A sequence $\rm f_n$ satisfies the recurrence $\rm\; f_{n+2} = f_{n+1}^{\:2} - f_n\:,\:$ with $\rm\; f_1 = 39,\; f_2 = 45 \;$.&lt;br&gt;&#xA;Prove that $\: 1986 \:$ divides infinitely many terms of the sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;\;$ The shift map is $\rm\; (a,b)\to (b,\;b^2-a = c)\;$ with inverse $\rm\; (b,c)\to (a = b^2-c,\; b)\;$&lt;br&gt;&#xA;and maps $\;(39,45)\to (45,0) \;\pmod {1986}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\;\;$ When the recurrence is linear the shift map has a matrix representation, which may be used to quickly compute the terms of the recurrence by fast exponentiation algorithms (e.g. repeatedly squaring). However, neither the matrix nor any other consequences of linearity are required to deduce the cyclicity properties above. Often many students (and even some professors) don't realize this, which can lead to obfuscated solutions (compared to the simple approach above). I've even encountered some solutions that completely overlook the innate symmetry and, as a result, end up completely reinventing the wheel, i.e. discovering the innate cycle structure completely from scratch - not realizing that it is simply a special case of the cycle structure of permutations, i.e. orbit decomposition. The moral of the story is: always look for any innate symmetries of a problem before diving in head-first to apply more brute-force techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE 2&lt;/strong&gt; $\;\;$ For another very nice application of orbit decompositions see my &lt;a href=&quot;http://math.stackexchange.com/questions/4150&quot;&gt;recent post&lt;/a&gt; which illustrates how they prove key for finding polynomial and rational function solutions of recurrences.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-19T21:30:05.707" LastActivityDate="2010-09-19T21:30:05.707" />
  <row Id="4811" PostTypeId="2" ParentId="4731" CreationDate="2010-09-16T20:44:14.473" Score="5" ViewCount="0" Body="&lt;p&gt;Yep. The condition that the Hessian is nonsingular at every critical point means that the function in question is suitably &quot;nice,&quot; and is called a Morse function. Morse functions are useful because their inverse images $f^{-1}((-\infty, a])$ behave very nicely.  Basically, as $a$ varies, you get things that are (up to homotopy equivalence) obtained by attaching cells when $a$ passes through a critical value (the type of cell being determined by the type of the critical point).  So since every compact manifold admits a Morse function (the &quot;typical&quot; function is, in fact, Morse--I believe this is a standard transversality idea in differential topology), it follows that every compact manifold is homotopy equivalent to a CW complex. &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-16T21:46:51.393" LastActivityDate="2010-09-16T21:46:51.393" />
  <row Id="4812" PostTypeId="2" ParentId="4808" CreationDate="2010-09-16T20:55:39.897" Score="3" ViewCount="0" Body="&lt;p&gt;As for the second question, the result is quite intuitive.  First of all one can readily see that we can suppose WLOG that the curve is convex, so already it cannot be too far from being circle-like.  Second, it's not hard to see that stretching the curve out so as to make it non-uniform causes it to enclose less area, e.g. consider the analogous problem for rectangles (or even try it for $n$-gons).  The intuition here is that a kink in the boundary of an area does not create much extra area but creates extra arc length.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the hard part is to justify these intuitions.  Probably the proof that comes closest proceeds via Steiner symmetrization; there is a link at the Wikipedia article.  There is also a neat Fourier-analytic proof.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-16T20:55:39.897" />
  <row Id="4813" PostTypeId="2" ParentId="3405" CreationDate="2010-09-16T21:04:49.547" Score="11" ViewCount="0" Body="&lt;p&gt;After getting some insight by looking at some numerical calculations to see what np(n)-c looks like for large n, I can now describe the convergence in some detail. First, the results of the simulations strongly suggested the following.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for n odd, np(n) converges to c from below at rate O(1/n).&lt;/li&gt;&#xA;&lt;li&gt;for n a multiple of 4, np(n) converges to c from above at rate O(1/n&lt;sup&gt;2&lt;/sup&gt;).&lt;/li&gt;&#xA;&lt;li&gt;for n a multiple of 2, but not 4, np(n) converges to c from below at rate O(1/n&lt;sup&gt;2&lt;/sup&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This suggests the following asymptotic form for p(n).&#xA;$$&#xD;&#xA;p(n)=\frac{c}{n}\left(1+\frac{a_1(n)}{n}+\frac{a_2(n)}{n^2}\right)+o\left(\frac{1}{n^3}\right)&#xD;&#xA;$$&#xD;&#xA;where a&lt;sub&gt;1&lt;/sub&gt;(n) has period 2, with a&lt;sub&gt;1&lt;/sub&gt;(0)&amp;nbsp;=&amp;nbsp;0, a&lt;sub&gt;1&lt;/sub&gt;(1)&amp;nbsp;&amp;lt;&amp;nbsp;0 and a&lt;sub&gt;2&lt;/sub&gt;(n) has period 4 with a&lt;sub&gt;2&lt;/sub&gt;(0)&amp;nbsp;&amp;gt;&amp;nbsp;0 and a&lt;sub&gt;2&lt;/sub&gt;(2)&amp;nbsp;&amp;lt;&amp;nbsp;0. In fact, we can expand p(n) to arbitrary order [Edit: Actually, not quite true. See below]&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;{}p(n)=c\sum_{r=0}^s\frac{a_r(n)}{n^{r+1}}+O(n^{-s-2})&amp;amp;&amp;amp;(1)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;and the term a&lt;sub&gt;r&lt;/sub&gt;(n) is periodic in n, with period 2&lt;sup&gt;r&lt;/sup&gt;. Here, I have normalized a&lt;sub&gt;0&lt;/sub&gt;(n) to be equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute all of the coefficients in (1). As p satisfies the recurrence relation&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle p(n+1)=(1+1/n)p(n)-1_{\lbrace n\textrm{ even}\rbrace}\frac2np(n/2) -1_{\lbrace n=1\rbrace}.&amp;amp;&amp;amp;(2)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;we can simply plug (1) into this, expand out the terms $(n+1)^{-r}=\sum_s\binom{-r}{s}n^{-r-s}$ on the left hand side, and compare coefficients of 1/n.&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle a_r(k+1)-a_r(k)=a_{r-1}(k)-\sum_{s=0}^{r-1}\binom{-s-1}{r-s}a_{s}(k+1)-1_{\lbrace k\textrm{ even}\rbrace}2^{r+1}a_{r-1}(k/2).&amp;amp;&amp;amp;(3)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;Letting &amp;#257;&lt;sub&gt;r&lt;/sub&gt; be the average of a&lt;sub&gt;r&lt;/sub&gt;(k) as k varies, we can average (3) over k to get a recurrence relation for &amp;#257;&lt;sub&gt;r&lt;/sub&gt;. Alternatively, the function f(n)&amp;nbsp;=&amp;nbsp;&amp;Sigma;&lt;sub&gt;r&lt;/sub&gt;&amp;#257;&lt;sub&gt;r&lt;/sub&gt;n&lt;sup&gt;-r-1&lt;/sup&gt; must satisfy f(n+1)&amp;nbsp;=&amp;nbsp;(1+1/n)f(n)&amp;nbsp;-&amp;nbsp;f(n/2)/n which is solved by f(n)&amp;nbsp;=&amp;nbsp;1/(n+1)&amp;nbsp;=&amp;nbsp;&amp;Sigma;&lt;sub&gt;r&amp;ge;0&lt;/sub&gt;(-1)&lt;sup&gt;r&lt;/sup&gt;n&lt;sup&gt;-r-1&lt;/sup&gt;, so we get &amp;#257;&lt;sub&gt;r&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;(-1)&lt;sup&gt;r&lt;/sup&gt;. Then, (3) can be applied iteratively to obtain a&lt;sub&gt;r&lt;/sub&gt;(k+1)-a&lt;sub&gt;r&lt;/sub&gt;(k) in terms of a&lt;sub&gt;s&lt;/sub&gt;(k) for s&amp;nbsp;&amp;lt;&amp;nbsp;r. Together with &amp;#257;&lt;sub&gt;r&lt;/sub&gt;, this gives a&lt;sub&gt;r&lt;/sub&gt;(k) and it can be seen that the period of a&lt;sub&gt;r&lt;/sub&gt;(k) doubles at each step. Doing this gives a&lt;sub&gt;r&lt;/sub&gt;&amp;nbsp;&amp;equiv;&amp;nbsp;(a&lt;sub&gt;r&lt;/sub&gt;(0),...,a&lt;sub&gt;r&lt;/sub&gt;(2&lt;sup&gt;r&lt;/sup&gt;-1)) as follows&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp; a_0=(1),\\\\&#xD;&#xA;&amp;amp; a_1=(0,-2),\\\\&#xD;&#xA;&amp;amp; a_2=(7,-3,-9,13)/2&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;These agree with the numerical simulation mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: I tried another numerical simulation to check these asymptotics, by successively subtracting out the leading order terms. You can see it converges beautifully to the levels a&lt;sub&gt;0&lt;/sub&gt;, a&lt;sub&gt;1&lt;/sub&gt;, a&lt;sub&gt;2&lt;/sub&gt; but, then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZDNzb.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... it seems that after the a&lt;sub&gt;2&lt;/sub&gt;n&lt;sup&gt;-3&lt;/sup&gt; part, there is an oscillating term! I wasn't expecting that, but it there is an asymptotic of the form cn&lt;sup&gt;-r&lt;/sup&gt;sin(alogn+&amp;theta;), where you can solve to leading order to obtain r&amp;nbsp;&amp;asymp;&amp;nbsp;3.54536, a&amp;nbsp;&amp;asymp;&amp;nbsp;10.7539.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update 2: I was re-thinking this question a few days ago, and it suddenly occured how you can not only prove it using analytic methods, but give a full asymptotic expansion to arbitrary order. The idea involves some very cool maths! (if I may say so myself). Apologies that this answer has grown and grown, but I think it's worth it. It is a very interesting question and I've certainly learned a lot by thinking about it. The idea is that, instead of using a power series generating function as in Qiaochu's answer, you use a Dirichlet series which can be inverted with &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Perron's formula&lt;/a&gt;. First, the expansion is as follows,&#xA;$$&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;\displaystyle&#xD;&#xA;p(n)=\sum_{\Re[r]+k\le \alpha}a_{r,k}(n)n^{-r-k}+o(n^{-\alpha}),&amp;amp;&amp;amp;(4)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;for any real &amp;alpha;. The sum is over nonnegative integers k and complex numbers r with real part at least 1 and satisfying r+1&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;r&lt;/sup&gt; (the leading term being r=1), and a&lt;sub&gt;r,k&lt;/sub&gt;(n) is a periodic function of n, with period 2&lt;sup&gt;k&lt;/sup&gt;. The reason for such exponents is that the difference equation (2) has the continuous-time limit p'(x)&amp;nbsp;=&amp;nbsp;p(x)/x&amp;nbsp;-&amp;nbsp;p(x/2)/x which has solutions p(x)&amp;nbsp;=&amp;nbsp;x&lt;sup&gt;-r&lt;/sup&gt; for precisely such exponents. Splitting into real and imaginary parts r&amp;nbsp;=&amp;nbsp;u+iv, all solutions to r+1&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;r&lt;/sup&gt; lie on the line (1+u)&lt;sup&gt;2&lt;/sup&gt;+v&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;4&lt;sup&gt;u&lt;/sup&gt; and, other than the leading term u=1, v=0, there is precisely one complex solution with imaginary part 2n&amp;pi;&amp;nbsp;&amp;le;&amp;nbsp;vlog2&amp;nbsp;&amp;le;2n&amp;pi;+&amp;pi;/2 (positive integer n) and, together with the complex conjugates, this lists all possible exponents r.&#xA;Then, a&lt;sub&gt;r,k&lt;/sub&gt;(n) is determined (as a multiple of a&lt;sub&gt;r,0&lt;/sub&gt;) for k&amp;nbsp;&gt;&amp;nbsp;0 by substituting this expansion back into the difference equation as I did above. I arrived at this expansion after running the simulations plotted above (and T..'s mention of complex exponents of n helped). Then, the Dirichlet series idea nailed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the Dirichlet series with coefficients p(n)/n&#xA;$$&#xD;&#xA;L(s)=\sum_{n=1}^\infty p(n)n^{-1-s},&#xD;&#xA;$$&#xD;&#xA;which converges absolutely for the real part of s large enough (greater than 0, since p is bounded by 1). It can be seen that L(s)&amp;nbsp;-&amp;nbsp;1 is of order 2&lt;sup&gt;-1-s&lt;/sup&gt; in the limit as the real part of s goes to infinity. Multiplying (2) by n&lt;sup&gt;-s&lt;/sup&gt;, summing and expanding n&lt;sup&gt;-s&lt;/sup&gt; in terms of (n+1)&lt;sup&gt;-s&lt;/sup&gt; on the LHS gives the functional equation&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle&#xD;&#xA;(s-1+2^{-s})L(s)=s+\sum_{k=1}^\infty(-1)^k\binom{-s}{k+1}(L(s+k)-1).&amp;amp;&amp;amp;(5)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;This extends L to a meromorphic function on the whole complex plane with simple poles precisely at the points -r with real part at least one and r+1&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;r&lt;/sup&gt;, and then at -r-k for nonnegative integers k. The pole with largest real component is at s&amp;nbsp;=&amp;nbsp;-1 and has residue&#xA;$$&#xD;&#xA;{\rm Res}(L,-1)={\rm Res}(s/(s-1+2^{-s}),-1)=\frac{1}{2\log2-1}.&#xD;&#xA;$$&#xD;&#xA;If we &lt;em&gt;define&lt;/em&gt; p'(n) by the truncated expansion (4) for some suitably large &amp;alpha;, then it will satisfy the recurrence relation (2) up to an error term of size O(n&lt;sup&gt;-&amp;alpha;-1&lt;/sup&gt;) and its Dirichlet series will satisfy the functional equation (5) up to an error term which will be an analytic function over R[s]&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; (being a Dirichlet series with coefficients o(n&lt;sup&gt;-&amp;alpha;-1&lt;/sup&gt;)). It follows that p' has simple poles in the same places as p on the domain R[s]&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; and, by choosing a&lt;sub&gt;r,0&lt;/sub&gt; appropriately, it will have the same residues. Then, the Dirichlet series associated with q(n)&amp;nbsp;=&amp;nbsp;p'(n)&amp;nbsp;-&amp;nbsp;p(n) will be analytic on R[s]&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; We can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Perron's formula&lt;/a&gt; to show that q(n) is of size O(n&lt;sup&gt;&amp;beta;&lt;/sup&gt;) for any &amp;beta;&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha; and, by making &amp;alpha; as large as we like, this will prove the asymptotic expansion (4). Differentiated, Perron's formula gives&#xA;$$&#xD;&#xA;dQ(x)/dx = \frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}x^sL(s)\,ds&#xD;&#xA;$$&#xD;&#xA;where Q(x)&amp;nbsp;=&amp;nbsp;&amp;Sigma;&lt;sub&gt;n&amp;le;x&lt;/sub&gt;q(n) and &amp;beta;&amp;nbsp;&amp;gt;&amp;nbsp;-&amp;alpha;. This expression is intended to be taken in the sense of distributions (i.e., multiply both sides by a smooth function with compact support in (0,&amp;infin;) and integrate). If &amp;theta; is a smooth function of compact support in (0,&amp;infin;) then&#xA;$$&#xD;&#xA;\begin{array}&#xD;&#xA;\displaystyle\sum_{n=1}^\infty q(n)\theta(n/x)/x&amp;amp;\displaystyle=\int_0^\infty\theta(y)\dot Q(xy)\,dy\\\\&#xD;&#xA;&amp;amp;\displaystyle=\frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}x^sL(s)\int\theta(y)y^s\,dy\,ds=O(x^\beta)\ \ (6)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;We obtain the final bound, because by integration by parts, the integral of &amp;theta;(y)y&lt;sup&gt;s&lt;/sup&gt; can be shown to go to zero faster than any power of 1/s, so the integrand is indeed integrable and the x&lt;sup&gt;&amp;beta;&lt;/sup&gt; term can be pulled outside.&#xA;This is enough to show that q(n) is itself of O(n&lt;sup&gt;&amp;beta;&lt;/sup&gt;). Trying to finish this answer off without too much further detail, the argument is as follows. If q(n)n&lt;sup&gt;-&amp;beta;&lt;/sup&gt; was unbounded then would keep exceeding its previous maximum and, by the recurrence relation (2), it would take time at least &lt;a href=&quot;http://en.wikipedia.org/wiki/Big-omega_notation&quot; rel=&quot;nofollow&quot;&gt;&amp;Omega;(n)&lt;/a&gt; to get back close to zero. So, if &amp;theta; has support in [1,1+&amp;epsilon;] for small enough &amp;epsilon;, the integral $\int\theta(y)\dot Q(ny)\,dy$ will be of order q(n) at such times and, as this happens infinitely often, it would contradict (6). Phew! I knew that this could be done, but it took some work! Possibly not as simple or direct as you were asking for, but Dirichlet series are quite standard (more commonly in analytic number theory, in my experience). However, maybe not really more difficult than the probabilistic method and you do get a whole lot more. This approach should also work for other types of recurrence relations and differential equations too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I added a much more detailed writeup on my blog, fleshing out some of the details which I skimmed over here. See &lt;a href=&quot;http://almostsure.wordpress.com/2010/10/06/asymptotic-expansions-of-a-recurrence-relation/&quot; rel=&quot;nofollow&quot;&gt;Asymptotic Expansions of a Recurrence Relation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-19T19:04:11.270" LastActivityDate="2010-11-19T19:04:11.270" CommunityOwnedDate="2010-11-19T18:58:39.777" CommentCount="10" />
  <row Id="4814" PostTypeId="1" AcceptedAnswerId="4824" CreationDate="2010-09-16T21:14:45.120" Score="2" ViewCount="116" Body="&lt;p&gt;I know that many types graphs have unique names, ie. a directed graph where each node has exactly one outbound edge is known as a &lt;code&gt;functional&lt;/code&gt; graph. Do the graphs with exactly 2 inbound/outbound edges have a name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus&lt;/strong&gt;: If the 2 inbound/outbound graph is named, is the nth inbound/outbound graph named?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T01:28:31.090" LastActivityDate="2010-09-17T19:56:26.913" Title="What are the names of graphs with 2 in/outbound edges?" Tags="&lt;terminology&gt;&lt;graph-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4815" PostTypeId="1" AcceptedAnswerId="4817" CreationDate="2010-09-16T21:16:38.457" Score="4" ViewCount="130" Body="&lt;p&gt;Given an infinite dimensional normed linear space, how would one show that it is possible to fit an infinite collection of non-overlapping balls of radius 1/4 in the unit ball?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess one can immediately reduce the problem to a normed linear space of countably infinite dimensions. The solution seems clear if the concept of orthogonality exists, but not every normed linear space has an inner product so it's not possible to apply something like Gram-Schmidt to produce an orthogonal basis. Is there any way around this, or there another approach that can be used?&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-09-16T21:33:39.653" LastActivityDate="2010-09-16T23:23:22.190" Title="Fitting an infinite collection of balls in an infinite dimensional unit ball " Tags="&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="4816" PostTypeId="2" ParentId="4808" CreationDate="2010-09-16T22:27:00.530" Score="4" ViewCount="0" Body="&lt;p&gt;There are relatively simple proofs in textbooks on calculus of variations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more elementary approaches a convex figure is deformed, in discrete steps or through a continuous unbending process, toward a circle, and two things need to be proved: convergence to the circle, and increase of the isoperimetric ratio throughout the flow. Usually one step is easy and the other is difficult, requiring non-elementary methods to make rigorous.   It is also necessary to make explicit what class of curves is considered: rectifiable, piecewise smooth, or something else.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest argument I know that is elementary and rigorous is to prove the finite-dimensional approximation, that for fixed edge lengths of a polygon, there is a maximum area (by compactness) and (by elementary geometry or Lagrange multipliers) it is the one where all vertices are on a circle. Then, use this to prove that any smooth curve, if it beats the circle, has a finite polygonal approximation that beats the inscribed polygon. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-16T22:27:00.530" />
  <row Id="4817" PostTypeId="2" ParentId="4815" CreationDate="2010-09-16T23:03:28.723" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a standard lemma you could use. If $M$ is a closed proper subspace of a normed linear space $E$, then for all $\epsilon\gt0$ there is an $x\in E$ of norm 1 whose distance to $M$ is greater than $1-\epsilon$.  (E.g., &lt;a href=&quot;http://books.google.com/books?id=maRLP-UBquYC&amp;amp;lpg=PP1&amp;amp;pg=PA56#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here's a proof&lt;/a&gt; in Lemma 3-6.10 of Tsoy-Wo Ma's &lt;em&gt;Classical analysis on normed spaces&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how you could use it.  Let $B$ denote the unit ball of an infinite dimensional normed space $X$.  Let $x_1\in X$ have norm 1, and let $M_1$ be the span of $\{x_1\}$.  By the lemma there is an $x_2\in X$ of norm 1 whose distance to $M_1$ is greater than $\frac{2}{3}$.  Let $M_2$ be the span of $\{x_1,x_2\}$, and let $x_3\in X$ have norm 1 and distance greater than $\frac{2}{3}$ to $M_2$.  Repeat countably infinitely many times to obtain a sequence $x_1,x_2,\ldots$ of elements of $X$ of norm 1 with pairwise distances greater than $\frac{2}{3}$.  Note that the lemma will always apply, because each $M_k=\operatorname{span}\{x_1,x_2,\ldots,x_k\}$ is finite dimensional, hence closed and proper.  Then the balls of radius $\frac{1}{4}$ centered at the points $\frac{3}{4}x_1,\frac{3}{4}x_2,\ldots$ are disjoint and contained in $B$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-16T23:23:22.190" LastActivityDate="2010-09-16T23:23:22.190" CommentCount="3" />
  <row Id="4818" PostTypeId="2" ParentId="4808" CreationDate="2010-09-16T23:34:06.463" Score="3" ViewCount="0" Body="&lt;p&gt;As Qiaochu Yuan pointed out, this is a consequence of the isoperimetric inequality that relates the length $L$ and the area $A$ for any closed curve $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;4\pi A \leq L^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a circumference of radius $r$ such that $2\pi r = L$, you obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A \leq \frac{L^2}{4\pi} = \frac{4 \pi^2 r^2}{4\pi} = \pi r^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the area $A$ enclosed by the curve $C$ is smaller than the area enclosed by the circumference of the same length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As for the proof of the isoperimetric inequality&lt;/strong&gt;, here is the one I've learnt as undergraduate, which is elementary and beautiful, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go round your curve $C$ counterclockwise. For a plane vector field $(P,Q)$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's theorem&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\oint_{\partial D}(Pdx + Qdy) = \int_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)  dxdy\ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply it for the vector field $(P,Q) = (-y,x)$ and when $D$ is the region enclosed by your curve $C = \partial D$. You obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \frac{1}{2} \oint_{\partial D} (-ydx + xdy) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, parametrize $C= \partial D$ with arc length: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\gamma : [0,L] \longrightarrow \mathbb{R}^2 \ ,\qquad \gamma (s) = (x(s), y(s)) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0= xy \vert_0^L = \int_0^L x'yds + \int_0^L xy'ds \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \int_0^L xy'ds = -\int_0^L x'yds \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So enough for now with our curve $C$. Let's look for a nice circumference to compare with!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, $[0,L]$ being compact, the function $x: [0,L] \longrightarrow \mathbb{R}$ will have a global maximum and a global minimum. Changing the origin of our parametrization if necessary, me may assume the minimum is attained at $s=0$. Let the maximum be attained at $s=s_0 \in [0,L]$. Let $q = \gamma (0)$ and $p = \gamma (s_0)$. (If there are more than one minimum and more than one maximum, we choose one of each: the ones you prefer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x'(0) = x'(s_0) = 0$, we have vertical tangent lines at both points $p,q$ of our curve $C$. Draw a circumference between these parallel lines, tangent to both of them (a little far away of $C$ to avoid making a mess). So the radius of this circumference will be $r = \frac{\| pq \|}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the origin of coordinates at the center of this circumference. We parametrize it with the same $s$, the arc length of $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma (s) = (\overline{x}(s), \overline{y}(s)) \ , \quad s \in [0, L] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, $\overline{x}(s)^2 + \overline{y}(s)^2 = r^2$ for all $s$. If we choose $\overline{x}(s) = x(s)$, this forces us to take $ \overline{y}(s) = \pm \sqrt{r^2 - \overline{x}(s)^2}$. In order that $\sigma (s)$ goes round all over our circumference counterclockwise too, we choose the minus sign if $0\leq s \leq s_0$ and the plus sign if $s_0 \leq s \leq L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We are almost done, just a few computations left.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\overline{A}$ denote the area of our circumference. So, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \int_0^L xy'ds = \int_0^L \overline{x}y'ds \qquad \text{and} \qquad \overline{A}= \pi r^2 = -\int_0^L\overline{y}\overline{x}'ds = -\int_0^L\overline{y} x'ds \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;A + \pi r^2  &amp;amp;= A + \overline{A} = \int_0^L (\overline{x}y' - \overline{y}x')ds \\&#xA;            &amp;amp;\leq \int_0^L \vert \overline{x}y' - \overline{y}x'\vert ds     \\&#xA;            &amp;amp;= \int_0^L \vert (\overline{x}, \overline{y})\cdot (y', -x')\vert ds   \\&#xA;            &amp;amp;\leq \int_0^L \sqrt{\overline{x}^2 + \overline{y}^2} \cdot \sqrt{(y')^2+ (-x')^2}ds \\&#xA;            &amp;amp;= \int_0^L rds = rL \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last inequality is Cauchy-Schwarz's one and the last but one equality is due to the fact that $s$ is the arc-length of $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A + \pi r^2 \leq rL \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the geometric mean is always smaller than the arithmetic one,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sqrt{A\pi r^2} \leq \frac{A + \pi r^2}{2} \leq \frac{rL}{2} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A \pi r^2 \leq \frac{r^2L^2}{4}  \qquad \Longrightarrow \qquad 4\pi A \leq L^2 \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-17T02:33:49.903" LastActivityDate="2010-09-17T02:33:49.903" />
  <row Id="4819" PostTypeId="2" ParentId="4757" CreationDate="2010-09-16T23:52:12.517" Score="1" ViewCount="0" Body="&lt;p&gt;I think the first part of your proof is a little clumsy. For $x\neq 0$ you don't need to use induction: note that what you have for $x\neq 0$ is a composite of $C^\infty$ functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you write  $(e^{g(x)}p_n(x))'=e^{g(x)}g′(x)p_n(x)+e^{g(x)}p_n′(x)$ you are utilizing the chain rule. Thus, you are assuming that the composite $(e^{g(x)}p_n(x))$ is differentiable. Of course this function is differentiable, but this is what you want to prove: that $f^{(n)}$ is differentiable for $x\neq 0$. The point is that you are proving even more: you're giving the general form of the derivatives.  &lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-09-16T23:52:12.517" CommentCount="3" />
  <row Id="4820" PostTypeId="1" AcceptedAnswerId="4831" CreationDate="2010-09-17T00:29:17.893" Score="3" ViewCount="73" Body="&lt;p&gt;If we know the rational homology of X is 0, can we get some information about the rational homology of X/G, where G is a finite group? Thank you very much for the answers!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-09-17T16:13:10.693" Title="Is there any relation about rational homology of X and X/G" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="4821" PostTypeId="1" CreationDate="2010-09-17T00:39:55.013" Score="2" ViewCount="129" Body="&lt;p&gt;I've somewhat got this question down but I'm only half way. &#xA;Q: How many 7-digit even numbers less than 3 000 000 can be formed using all the digits 1,2,2,3,5,5,6?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I figured that there's about 4 possible approaches&#xA;1 _ _ _ _ _ 6&#xA;1 _ _ _ _ _ 2&#xA;2 _ _ _ _ _ 6&#xA;2 _ _ _ _ _ 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I fill in the middle? I tried 5! factorial and dividing out the similar factorials but I didn't get the right answer &lt;/p&gt;&#xA;" OwnerUserId="1945" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T08:03:07.690" LastActivityDate="2010-09-18T00:32:16.700" Title="Permutations with identical items" Tags="&lt;combinatorics&gt;" AnswerCount="3" />
  <row Id="4822" PostTypeId="2" ParentId="4808" CreationDate="2010-09-17T00:43:03.727" Score="0" ViewCount="0" Body="&lt;p&gt;First you can propose this problem as: The area $A$ encompassed by any simple closed rectifiable curve $C$, of length $L$, satisfacts the inequality $A\geq \frac{L^2}{4\pi}$, and equality occurs, if and only if, $C$ is a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only proof I have done for this was using Parseval's identity (and therefore Fourier series), so it's not elementary (but it's rather simple if you know the aforementioned identity). Thought if you want I can post that proof.&lt;/p&gt;&#xA;" OwnerUserId="1605" LastActivityDate="2010-09-17T00:43:03.727" />
  <row Id="4823" PostTypeId="2" ParentId="4821" CreationDate="2010-09-17T00:56:34.037" Score="1" ViewCount="0" Body="&lt;p&gt;Try the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;multinomial coefficient.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1892" LastActivityDate="2010-09-17T00:56:34.037" />
  <row Id="4824" PostTypeId="2" ParentId="4814" CreationDate="2010-09-17T01:10:29.217" Score="3" ViewCount="0" Body="&lt;p&gt;This is known as a $d$-regular graph (at least if the graph is undirected). If your graph is a sub-graph of a bigger graph, then it is a $d$-factor. Finally, for $d=2$ it's also called a &lt;i&gt;cycle cover&lt;/i&gt;, since it's a collection of cycles (if it's finite). For $d=1$ it's called a &lt;i&gt;matching&lt;/i&gt;. For $d=3$ it is known as a &lt;i&gt;cubic graph&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For directed graphs it's usually not clear what is supposed to be regular, the in-degree, the out-degree or both.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-09-17T19:56:26.913" LastActivityDate="2010-09-17T19:56:26.913" CommentCount="0" />
  <row Id="4825" PostTypeId="2" ParentId="4821" CreationDate="2010-09-17T01:19:16.293" Score="3" ViewCount="0" Body="&lt;p&gt;We actually have only two cases: 1... and 2... .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, we have to fit the numbers 2,2,3,5,5,6 somehow. Imagine that we tag each of the repeated elements somehow, e.g. 2',2&quot;,3,5',5&quot;,6. If we care about the tags, then there are $6!$ possible ways to arrange the numbers. Now each untagged arrangement corresponds to $2!2!$ different tagged arrangements (consider the order in which the tags on the $2$s and $5$s appear). In total, this gives $6!/2!2!= 180$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, we have only one repeated element $5$ so there are $6!/2! = 360$ possibilities, for a total of $540$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if we have $k$ unique elements repeated $t_1,\ldots,t_k$ times (respectively), then the answer is going to be $(t_1+\cdots+t_k)!/t_1!\cdots t_k!$, using the same reasoning. This is known as a multinomial coefficient since it appears in the multinomial theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x_1 + \cdots + x_k)^n = \sum_{t_1 + \cdots + t_k = n} n!/t_1!\cdots t_k! x_1^{t_1} \cdots x_k^{t_k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the sum is over all non-negative integers summing to $n$. The multinomial coefficient is sometimes denoted $\binom{n}{t_1\ldots t_k}$, although one of these numbers is really redundant. When $k=2$ we get the binomial coefficient, and we usually omit $t_2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-09-17T01:19:16.293" CommentCount="1" />
  <row Id="4826" PostTypeId="1" AcceptedAnswerId="5009" CreationDate="2010-09-17T01:37:56.247" Score="6" ViewCount="163" Body="&lt;p&gt;Here's the setup: Let $U,V$ open $\subset \mathbb{R}^n$, and let $u:V\rightarrow \mathbb{R}$ be harmonic, and $v:U\rightarrow V$ be conformal, i.e. $v$ is $C^1$ and the Jacobian $J_v(x)$ is a scalar multiple of an orthogonal transformation for all [; x\in U ;].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove $u\circ v$ is harmonic. [I've seen this stated as a fact in a few places without reference, namely here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_map#Uses&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Conformal_map#Uses&lt;/a&gt; , but maybe my hypotheses are slightly different and this is not true at all]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen a proof that if $u$ is $C^2$ and $T$ is an orthogonal transformation then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta (u \circ T) = \Delta(u) \circ T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm thinking that to show $u\circ v$ is harmonic, we can use the fact that $v$ acts locally as its Jacobian, which is an orthogonal transformation, and move the Laplacian onto $u$ and conclude $u\circ v$ is harmonic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm having trouble making this idea precise.  After glancing at my copy of baby Rudin, my hunch is to use the inverse function theorem or constant rank theorem, but I'm unsure how to apply those.  Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="1692" LastEditorUserId="1692" LastEditorDisplayName="" LastEditDate="2010-09-17T19:12:12.670" LastActivityDate="2010-09-19T19:02:10.397" Title="Harmonic function composed with conformal map is harmonic (in R^n)" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4827" PostTypeId="1" AcceptedAnswerId="4829" CreationDate="2010-09-17T01:41:36.120" Score="2" ViewCount="56" Body="&lt;p&gt;I came across the two congruences that for $p$ an odd prime,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1^2\cdot 3^2\cdot 5^2\cdots (p-2)^2\equiv (-1)^{(p+1)/(2)}(mod\ p)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^2\cdot 4^2\cdot 6^2\cdots (p-1)^2\equiv (-1)^{(p+1)/(2)}(mod\ p)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this? It seems to me to be closely related to Wilson's Theorem, and possibly Fermat's little theorem since $p-1=2k$ for some $k$, and then $j^{2k}\equiv 1(mod\ p)$ for all the evens and odds $j$ less than $p$, but I wasn't able to complete this train of thought.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-17T03:21:42.447" Title="Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}(mod\ p)$?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="4829" PostTypeId="2" ParentId="4827" CreationDate="2010-09-17T01:56:02.783" Score="2" ViewCount="0" Body="&lt;p&gt;Modulo $p$, you have $k=-(p-k)$ and so both products are equal to $(p-1)!(-1)^{(p-1)/2} \equiv (-1)^{(p+1)/2} \pmod{p}$ by Wilson's theorem.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-09-17T03:21:42.447" LastActivityDate="2010-09-17T03:21:42.447" CommentCount="3" />
  <row Id="4830" PostTypeId="2" ParentId="4788" CreationDate="2010-09-17T02:40:25.903" Score="3" ViewCount="0" Body="&lt;p&gt;It's easy to be jaded about FTA after mathematical history has run for another two centuries.  Sure, it is not the number one most important result any more, or the center of any research program (though understanding the algebraic closure of Q could be considered as half of number theory).  But consider the situation around 1800.  In addition to solution of algebraic equations one had new methods of constructing numbers, using power series, integrals and other limits.  Algebra and number theory dealt with the first situation, to a limited extent, and analysis showed that the second type of construction could be iterated but still stay within the same realm of numbers.  There was still the possibility that solving equations with $\pi$ and $e$ as coefficients could require an entirely new type of super-transcendental analysis.  Fundamental Theorem of Algebra is self-defeating in this sense: it shows that nothing more was needed than complex numbers.  But this is not clear in a world where you don't know that FTA is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea what algebraic geometry looked like without complex numbers, look up Newton's classification of degree 3 algebraic curves in the plane, $P_3(x,y)=0$, using real coordinates.  The reason this work is obscure today is that there are many dozens of cases compared to the complex projective version.   As in Lie groups and topology, looking from the universal covering (C) downward, modulo some Galois-theoretic details (R), is usually easier than working from the bottom up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to evaluate the integral, from $- \infty$ to $+\infty$, of a rational function (one with integer coefficients would illustrate the point).  The answer will involve $\pi$ and the usual method for finding it will use the specific location of the roots in the complex plane, so is more specific to complex numbers than the existence of roots in an algebraic closure.  There are some non-usual methods that stay entirely within the real numbers, but they are nonstandard because they are more complicated, and harder to understand and adapt to other problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebraic geometry in general has a transcendental part -- periods, Hodge theory, uniformization, etc -- that, in the present state of knowledge, cannot be fully substituted by algebraic methods over fields of characteristic 0 (or p).  Sometimes Lefschetz principle or reduction to positive characteristic can be used, sometimes not, or the theory is unknown.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-17T02:40:25.903" CommentCount="1" />
  <row Id="4831" PostTypeId="2" ParentId="4820" CreationDate="2010-09-17T02:49:24.020" Score="3" ViewCount="0" Body="&lt;p&gt;When $G$ is finite, the rational cohomology of $X/G$ is the fixed point set $H^*(X;\mathbb{Q})^G$.  This is proven in Grothendieck's Tohoku paper (Theorem 5.3.1 and the Corollary to Proposition 5.3.2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the rational cohomology of $X$ is trivial, the same is true for $X/G$. And rationally the cohomology and homology are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For paracompact Hausdorff spaces, these cohomology groups can be taken to be the Cech cohomology groups.  Note that if $X$ is homotopy equivalent to a CW complex, then Cech cohomology agrees with singular cohomology.  You might also want to look at Oscar Randall-Williams comments here: &lt;a href=&quot;http://mathoverflow.net/questions/18898/grothendiecks-tohoku-paper-and-combinatorial-topology/30015#30015&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/18898/grothendiecks-tohoku-paper-and-combinatorial-topology/30015#30015&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-09-17T16:13:10.693" LastActivityDate="2010-09-17T16:13:10.693" CommentCount="5" />
  <row Id="4832" PostTypeId="2" ParentId="4737" CreationDate="2010-09-17T03:04:53.763" Score="2" ViewCount="0" Body="&lt;p&gt;I think Apollonian circle packing is relevant here.  It's certainly not identical with the question's topic, but it's enough to convince me that the answer is yes-- you can fit disks of radii 1/2, 1/3, 1/4,... inside a disk of radius 1.  Hopefully someone more knowledgeable can make a rigorous argument, and solve the question for a disk of radius 5/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;http://arxiv.org/abs/0811.2236&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; on arXiv by Alex Kontorovich and Hee Oh demonstrating that for a bounded Apollonian circle packing, the number of circles with curvature at most N is asymptotic to cN&lt;sup&gt;a&lt;/sup&gt; where a=1.30...  This is good news for our problem!  Apollonian packing greedily fits the largest possible circle in the gap created by any 3 mutually tangent circles, and the number of circles with radius at least 1/N grows as N&lt;sup&gt;a&lt;/sup&gt;, where we just need it to grow as N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture from Wikipedia of an Apollonian packing, labeled by curvature (1/r):&#xA;&lt;img src=&quot;http://i.imgur.com/pGCwR.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer circle has radius 1, and we want to find space for disks with curvature 2,3,4,5,...  Compare the list of curvatures in the picture with the list we want to find space for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,2,3,3,6,6,6,6,11,11,11,11,14,14,14,14,15,15,18,18,18,18,23,23,23,23,26,26,26,26,27,27,&#xA;2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;27,27,30,30,30,30,35,35,35,35,35,35,38,38,38,38,38,38,38,38,...&#xA;34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everywhere the curvature in the lower list is $\geq$ the curvature in the upper list, we can just fit our disk inside the corresponding circle in the Apollonian packing.  The only case in the list that doesn't fit is 10-- but since we've left unused space several times already, it's clear we can make room for it.  Of course, shifting things to make room breaks the Apollonian packing (though we can leave three-quarters of it intact); but it sure looks like we have plenty of space left, and the asymptotic formula also suggests we are safe.  I haven't looked closely at the paper for any explicit bounds that would make this rigorous. I realize it is just a sketch, but hopefully a convincing one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second Apollonian packing picture is analogous to the radius=5/6 case, scaled up 5x.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/SRTtF.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we want to fit disks with curvature 10,15,20,25,30,35,40,45,50,...  We run into trouble already with 30 (the disk of radius 1/6), and we'll have to shift things around quite a bit to fit our disks in.  The Apollonian packing does not lead to a solution as easily as it seems to in the radius 1 case.  Nonetheless, the asymptotic formula suggests that if we can pack the first N disks (for N perhaps not very large), that there will be plenty of room for the rest.  One would have to work out some details to find the value of N.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-09-17T03:47:30.980" LastActivityDate="2010-09-17T03:47:30.980" CommentCount="1" />
  <row Id="4833" PostTypeId="1" CreationDate="2010-09-17T03:22:11.793" Score="3" ViewCount="63" Body="&lt;p&gt;I forget for a while, we don't need the compactness condition here right?&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-09-17T07:10:04.400" Title="Does noncompact manifold or orbifold have the homotopy type,of CW complex?" Tags="&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="4834" PostTypeId="1" AcceptedAnswerId="4857" CreationDate="2010-09-17T03:40:37.583" Score="2" ViewCount="140" Body="&lt;p&gt;If I want to find the number of all possible groupings of a set of numbers, including all groupings with only a single number and groupings will all numbers is there a better method than this. For this use I don't want to count empty groups and order is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the summation of the number of permutations with repetition of size i where i goes from 1 to the number of items in my collection?&lt;br&gt;&#xA;s = number of elements in the collection&lt;br&gt;&#xA;i = number I'm summing over from one to s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{s} s^i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;basically I want to know this:&#xA;1. will this count all possible groupings except for the empty group?&#xA;2. how do you get math equations to show up in questions for future use?&#xA;3. is there a better way to express this (if you can understand what I'm trying to say)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok small example. if I have a collection (1,2,3) I want to count the possible collections that could be made from is such as: &#xA;(1),(2),(3)&#xA;(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),&#xA;(1,1,1),(1,1,2),(1,1,3),(1,2,1),(1,2,2),(1,2,3),(1,3,1),(1,3,2),(1,3,3),(2,1,1),(2,1,2),(2,1,3),(2,2,1),(2,2,2),(2,2,3),(2,3,1),(2,3,2),(2,3,3),(3,1,1),(3,1,2),(3,1,3),(3,2,1),(3,2,2),(3,2,3),(3,3,1),(3,3,2),(3,3,3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for the collection (1,2,3) I can make 39 collections from it not exceeding the length of the original collection.&lt;/p&gt;&#xA;" OwnerUserId="1582" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-17T17:24:59.643" LastActivityDate="2010-09-17T17:24:59.643" Title="Given a collection of items, how many ordered subsets can I make?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="4836" PostTypeId="1" AcceptedAnswerId="4837" CreationDate="2010-09-17T05:37:22.933" Score="2" ViewCount="74" Body="&lt;p&gt;Let $M$ be a 2n dimensional manifold, and let $A$ be a n-dimensional cycle on M. I want to compute the self-intersection $(A.A)$ of A with itself. Let $\eta_A$ be the form in $H^n(M, \mathbb{R})$ given by the Poincare duality isomorphism, i.e. $\int_B \eta_A=(A.B)$ for an n-cycle B. Then by the rule that intersection of cycles corresponds to taking wedge product, we have $\int_M \eta_A \wedge \eta_B=(A.B)$. Taking A=B, since $\eta_A \wedge \eta_A =0$, I get $(A.A)=0$, which I don't think is the case. Where am I going wrong? &lt;/p&gt;&#xA;" OwnerUserId="1947" LastActivityDate="2010-09-17T05:48:51.757" Title="computing intersection number via differential forms" Tags="&lt;differential-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4837" PostTypeId="2" ParentId="4836" CreationDate="2010-09-17T05:48:51.757" Score="6" ViewCount="0" Body="&lt;p&gt;$\omega \wedge \eta = (-1)^{ab}\eta \wedge \omega$ where $a$ is the dimension of $\omega$ and $b$ is the dimension of $\eta$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your &quot;since $\eta_A \wedge \eta_A=0$&quot; comment is only true when $n$ is odd.&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-17T05:48:51.757" CommentCount="4" />
  <row Id="4839" PostTypeId="2" ParentId="4833" CreationDate="2010-09-17T07:10:04.400" Score="4" ViewCount="0" Body="&lt;p&gt;According to &lt;em&gt;The Topology of CW-Complexes&lt;/em&gt; by Lundell and Weingram&#xA;(Van Nostrand Reinhold, 1969) the answer is yes for (separable) manifolds.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-17T07:10:04.400" CommentCount="1" />
  <row Id="4840" PostTypeId="1" CreationDate="2010-09-17T07:23:18.387" Score="6" ViewCount="95" Body="&lt;p&gt;I have a terrain, which is represented by  &lt;a href=&quot;http://en.wikipedia.org/wiki/Polygon_mesh&quot; rel=&quot;nofollow&quot;&gt;one mesh with a lot of polygons&lt;/a&gt; as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jZRSC.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This terrain will be cut by a plane at a certain level. So there are volumes of the terrains that are located above the plane ( cut volume), and volumes that are located below the plane ( fill volume). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, how do I obtain the cut/ fill volume? My current approach is simply take one mesh at a time, and then form a tetrahedron with the plane, and compute the volume. &lt;strong&gt;But this is slow&lt;/strong&gt;. Is there other better approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach that I have in mind, is to try to form Bezier surface for the terrain, and then try to use integration to compute the volume. But I don't know how to proceed with this. Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Terminology updated&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-19T08:32:53.333" LastActivityDate="2010-09-19T08:32:53.333" Title="Find the Volume Enclosed by Terrain and a Certain Sea Level" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="4841" PostTypeId="1" CreationDate="2010-09-17T08:05:46.290" Score="5" ViewCount="821" Body="&lt;p&gt;How can we prove that inverse of upper(lower) triangular matrix is upper(lower) triangular...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody answer to this question.......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.........&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-17T08:10:34.963" LastActivityDate="2010-09-18T17:06:21.803" Title="Inverse of a triangular matrix( both upper &amp; lower ) is triangular" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="4842" PostTypeId="2" ParentId="4841" CreationDate="2010-09-17T08:16:49.233" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose that $U$ is upper. The $i$th column $x_i$ of the inverse is given by $Ux_i=e_i$ where $e_i$ is the $i$th unit vector. By backward subsitution you can see that $(x_i)_j=0$ for $i+1\leq j\leq n$. I.e all the entries in the $i$th column of the inverse below the diagonal are zero. This is true for all $i$ and hence the inverse $U^{-1}=[x_1|\ldots|x_n]$ is upper triangular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing works for lower triangular using forward subsitution.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-17T08:16:49.233" />
  <row Id="4843" PostTypeId="2" ParentId="4841" CreationDate="2010-09-17T09:07:27.647" Score="2" ViewCount="0" Body="&lt;p&gt;I'll add nothing to alext87 answer, or J.M. comments. Just &quot;display&quot; them.   :-)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remeber that you can compute the inverse of a matrix by reducing it to &lt;a href=&quot;http://math.stackexchange.com/questions/1822/is-reducing-a-matrix-to-row-echelon-form-useful-at-all/1831#1831&quot;&gt;row echelon form&lt;/a&gt; and solving the simultaneous systems of linear equations $ (A \vert I)$, where $A$ is the matrix you want to invert and $I$ the unit matrix. When you have finished the process, you'll get a matrix like $(I\vert A^{-1})$ and the matrix on the right, yes!, is the inverse of $A$. (Why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, half of the work is already done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;a^1_1 &amp;amp; a^1_2 &amp;amp; \cdots &amp;amp; a^1_{n-1}     &amp;amp; a^1_n     &amp;amp; 1 &amp;amp; 0      &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0 \\&#xA;      &amp;amp; a^2_2 &amp;amp; \cdots &amp;amp; a^2_{n-1}     &amp;amp; a^2_n     &amp;amp;   &amp;amp; 1      &amp;amp; \cdots &amp;amp; 0 &amp;amp; 0  \\&#xA;      &amp;amp;       &amp;amp; \ddots &amp;amp; \vdots        &amp;amp; \vdots    &amp;amp;   &amp;amp;        &amp;amp; \ddots &amp;amp; \vdots  &amp;amp; \vdots \\&#xA;      &amp;amp;       &amp;amp;        &amp;amp; a^{n-1}_{n-1} &amp;amp; a^{n-1}_n &amp;amp;   &amp;amp;        &amp;amp;        &amp;amp; 1 &amp;amp; 0 \\&#xA;      &amp;amp;       &amp;amp;        &amp;amp;               &amp;amp; a^n_n     &amp;amp;   &amp;amp;        &amp;amp;        &amp;amp;   &amp;amp; 1&lt;br&gt;&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens when you do back substitution starting with $a^n_n$ and then continuing with $a^{n-1}_{n-1}$...?&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-17T09:07:27.647" />
  <row Id="4844" PostTypeId="1" CreationDate="2010-09-17T09:24:22.037" Score="10" ViewCount="206" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N$ point charges are distributed in the unit ball in $\mathbb{R}^k$, $k=2,3$. Given locations of the particles $x_1,\ldots,x_N$ the potential energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=\sum_{j=1}^{N-1}\sum_{k=j+1}^N |x_j-x_k|^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|x_j-x_k|$ is Euclidean distance between $x_j$ and $x_k$. I'm interested in both the minimal value of $E$ over all possible locations of the particles in the unit ball and what this configuration looks like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the Unit Interval&lt;/strong&gt;&#xA;For $k=1$ the $N$ charges are distributed on the interval $[-1,1]$ according to the roots of the $N+1$th Chebshev polynomial. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev_polynomials#Roots_and_extrema&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chebyshev_polynomials#Roots_and_extrema&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-28T15:33:49.910" LastActivityDate="2010-10-18T02:40:32.613" Title="Distributions of point charges" Tags="&lt;numerical-methods&gt;&lt;physics&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="4845" PostTypeId="1" AcceptedAnswerId="4847" CreationDate="2010-09-17T10:46:40.253" Score="1" ViewCount="136" Body="&lt;p&gt;This paper &lt;a href=&quot;http://arxiv.org/pdf/0911.1933&quot; rel=&quot;nofollow&quot;&gt;arxiv.org/pdf/0911.1933&lt;/a&gt; discusses, regarding the irrationality of certain trigonometric functions. Recently, i encountered this problem which says states the given function, $$ f(n)=\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$$ is irrational for every odd $n \geq 3$. But i couldn't find the proof anywhere. Can anyone provide me with the proof.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-17T11:05:12.800" Title="Irrationality of $ \frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4846" PostTypeId="1" AcceptedAnswerId="4856" CreationDate="2010-09-17T11:02:26.057" Score="3" ViewCount="464" Body="&lt;p&gt;Everyone in this community i think would be familiar with International Mathematical Olympiad, which is an International Mathematics Competition held for high school students, with many countries participating from around the world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats interesting to note is many of the IMO participants have gone to win the Fields Medal. Notable personalities would be &quot;Terence Tao (2006)&quot;, &quot;Ngo Bau Chau (2010)&quot;, Grigori Perelman..etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know, what advantage does an IMO student possess over a normal student in terms of Mathematical Research. Does, the IMO competition help the student in becoming a good research Mathematician. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-17T17:46:17.557" CommunityOwnedDate="2010-09-17T11:02:26.057" Title="Advantages of IMO students in Mathematical Research" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="4847" PostTypeId="2" ParentId="4845" CreationDate="2010-09-17T11:05:12.800" Score="7" ViewCount="0" Body="&lt;p&gt;You could find the proof in &lt;a href=&quot;http://books.google.com/books?id=CuTpgZaLahkC&amp;amp;pg=PA40&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt;: &lt;i&gt;Proofs from the Book&lt;/i&gt; by Martin Aigner, Günter M. Ziegler, Karl H. Hofmann, pages 40-41&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2010-09-17T11:05:12.800" CommunityOwnedDate="2010-09-17T11:05:12.800" CommentCount="4" />
  <row Id="4848" PostTypeId="2" ParentId="213" CreationDate="2010-09-17T11:37:12.163" Score="0" ViewCount="0" Body="&lt;p&gt;Not unpopular, but I wish I had studied the theory of &lt;strong&gt;Rings and Fields&lt;/strong&gt;, and basic &lt;strong&gt;Topology&lt;/strong&gt; earlier, because in my opinion both these branches appear in many interesting subjects studied relatively early during one's undergraduate studies.&#xA;For example, the whole concept of a minimal polynomial of a linear transformation $T$ is more intuitive (at least for me) when viewed as the generator of the ideal of polynomials such that $P(t)=0$.&#xA;Metric Spaces (studied in the introductory Topology course at my university) appear as early as in calculus, and are generally a basis to many definitions there. Also, many algebraic structures can be viewed as topologies, which can sometimes give a new insight or assist in proofs (a favourite of mine is a topological proof of the infinity of the primes, which can be found in &lt;em&gt;Proofs from the Book&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-17T11:37:12.163" CommunityOwnedDate="2010-09-17T11:37:12.163" />
  <row Id="4849" PostTypeId="2" ParentId="4846" CreationDate="2010-09-17T11:46:22.357" Score="17" ViewCount="0" Body="&lt;p&gt;I would say that olympiads build some, but far from all, of the skills needed to excel at mathematical research. I'd compare it to running 100 meters versus playing soccer. Usain Bolt is probably a better soccer player than the vast majority of the population, because he could outsprint anyone and because he's generally in fantastic shape. But that doesn't mean he's going to be able to play on a top team. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a successful researcher requires the ability to learn new fields of mathematics, and develop ways of thinking about them that others haven't. It requires the discipline to spend months or years returning to a problem and trying new angles on it. It requires, or at least is strongly aided by, the ability to communicate and &quot;sell&quot; one's results, in writing and in talks. It requires the ability to write good definitions, that will be useful and cover the boundary cases correctly. It requires the ability to form an intelligent guess as to which unproven statements are true and which are false. It requires the ability to hold a complex argument in one's head and play with it. It requires, or at least is strongly aided by, the ability to find clever technical arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that olympiads are very helpful in developing the last skill, somewhat helpful in developing the fifth and sixth, and not at all in developing the first four. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I definitely, at some points in my research, find myself needing lemmas which would be fair to put on an IMO or a Putnam exam. And when that I happens I feel myself relaxing, because I know I can do that. But I also spend a lot of my time trying to learn how to think about a subject, or figuring out what to prove, or trying to figure out how broadly a phenomenon holds. And those are not skills which I found olympiad training helpful in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, I was very competitive in my High School years, but not a world level competitor; I was the first alternate to the US team in 1998 and, during my senior year, I regularly came in somewhere in the top 10 spots in national contests. I've heard similar views expressed by Kiran Kedlaya and Ravi Vakil, who were much more impressive competitors.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-17T11:46:22.357" CommunityOwnedDate="2010-09-17T11:46:22.357" />
  <row Id="4850" PostTypeId="2" ParentId="4846" CreationDate="2010-09-17T12:02:29.173" Score="9" ViewCount="0" Body="&lt;p&gt;Keep in mind that what you have here is a correlation, not a causation. While doubtlessly Olimpiad Training would help develop some skills necessary for research, I think it is likely that many of the strongest mathematicians participate in these competitions when they are in high school and go on to do research later.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-17T12:02:29.173" CommunityOwnedDate="2010-09-17T12:02:29.173" CommentCount="9" />
  <row Id="4851" PostTypeId="1" AcceptedAnswerId="5208" CreationDate="2010-09-17T12:25:36.453" Score="2" ViewCount="80" Body="&lt;p&gt;Let $p : E \rightarrow B$ be a cover s.t. Aut(E) acts transitively on $p^{-1}(b)$ for some fix $b \in B$. Then $E/Aut(E)$ is homeomorphic to $B$ where $Aut(E) \subset Cov(E,E)$. The problem is that I don't even understand how the map E/Aut(E) -&gt; B is defined, would someone please give me this definition? Apparently it is &quot;clear&quot; but I fail to see this. &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-09-22T15:07:40.383" Title="Cover E -&gt; B gives a homeomorphism E/Aut(E) -&gt; B" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4853" PostTypeId="2" ParentId="250" CreationDate="2010-09-17T12:59:33.820" Score="4" ViewCount="0" Body="&lt;p&gt;One that Feynman would have rejected: There are always two antipodal points on the Earth that are the same temperature. (This follows from the fact that a continuous scalar field on a circle has a diameter whose endpoints have the same value)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One that Feynman might have preferred:&#xA;Draw a triangle. Draw a circle through the midpoints of the circle's sides. This circle also passes through the the foot of each altitude and the midpoint of the line segment from the orthcentre to each vertex. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nine-point_circle&quot; rel=&quot;nofollow&quot;&gt;The nine point circle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-17T12:59:33.820" CommunityOwnedDate="2010-09-17T12:59:33.820" />
  <row Id="4854" PostTypeId="1" AcceptedAnswerId="4877" CreationDate="2010-09-17T13:03:03.767" Score="6" ViewCount="104" Body="&lt;p&gt;I am obsessed with &lt;a href=&quot;http://en.wikipedia.org/wiki/Nine-point_circle&quot; rel=&quot;nofollow&quot;&gt;the nine point circle&lt;/a&gt;. I was thinking, is there a generalisation to aribtrary tetrahedra and spheres? What about higher dimensions? For each face of the tetrahedron, there is a nine point circle. Do these circles all lie on a sphere?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-17T19:00:29.907" Title="Does the nine point circle generalise to some theorem about n-spheres and n-simplices?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4855" PostTypeId="2" ParentId="4854" CreationDate="2010-09-17T13:11:08.130" Score="4" ViewCount="0" Body="&lt;p&gt;Take a look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/mj55u41710w625j8/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/mj55u41710w625j8/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedron&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tetrahedron&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also this paper that deals with sphere analogues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/3617834&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/3617834&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-09-17T13:28:58.697" LastActivityDate="2010-09-17T13:28:58.697" />
  <row Id="4856" PostTypeId="2" ParentId="4846" CreationDate="2010-09-17T14:04:43.390" Score="14" ViewCount="0" Body="&lt;p&gt;Training for competitions will help you solve competition problems - that's all. These are not the sort of problems that one typically struggles with later as a professional mathematician - for many different reasons. First, and foremost, the problems that one typically faces at research level are not problems carefully crafted so that they may be solved in certain time limits. Indeed, for problems encountered &quot;in the wild&quot;, one often does not have any inkling whether or not they are true. So often one works simultaneously looking for counterexamples and proofs. Often solutions require discovering fundamentally new techniques - as opposed to competition problems - which typically may be solved by employing variations of methods from a standard toolbox of &quot;tricks&quot;. Moreover, there is no artificial time limit constraint on solving problems in the wild. Some research level problems require years of work and immense persistence (e.g. Wiles proof of FLT). Those are not skills that can be measured by competitions. While competitions might be used to encourage students, they should never be used to discourage them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a great diversity among mathematicians. Some are prolific problem solvers (e.g. Erdos) and others are grand theory builders (e.g. Grothendieck). Most are somewhere between these extremes. All can make significant, surprising contributions to mathematics. History is a good teacher here. One can learn from the masters not only from their mathematics, but also from the way that they learned their mathematics. You will find much interesting advice in the (auto-)biographies of eminent mathematicians. Time spent perusing such may prove much more rewarding later in your career than time spent learning yet another competition trick. Strive to aim for a proper balance of specialization and generalization in your studies.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T17:46:17.557" LastActivityDate="2010-09-17T17:46:17.557" CommunityOwnedDate="2010-09-17T14:04:43.390" CommentCount="2" />
  <row Id="4857" PostTypeId="2" ParentId="4834" CreationDate="2010-09-17T14:20:50.137" Score="3" ViewCount="0" Body="&lt;p&gt;Seems like you need&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{s} s^i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a geometric progression with the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{s(s^{s} -1)}{s-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $s=3$ this gives $\frac{3(3^3 -1)}{3-1} = 39$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-17T14:20:50.137" CommentCount="1" />
  <row Id="4858" PostTypeId="1" CreationDate="2010-09-17T14:22:03.873" Score="1" ViewCount="106" Body="&lt;p&gt;Let $S$ be a (n-1)-submanifold of a n-manifold $M$ and that be a submanifold of (n+1)-manifold V. (All of the manifolds are assumed orientable) Let $V$ carry a metric of signature $(1,n)$. Using terminology as in Einstein's Theory let $n$ be a future directed unit normal to $M$ and $m$ be an unit normal to $S$ pointing outward which is tangent to $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $h$ be a metric on $S$ , $g'$ be a metric on $M$ and $g$ be a metric on $V$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{\mu \nu}' = g_{\mu \nu} + n_\mu n_\nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_{\mu \nu} = g_{\mu \nu} + n_\mu n_\nu - m_\mu m _\nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In each of the cases $\mu$ and $\nu$ run over the dimensions of the lowest dimensional manifold whose metric is involved) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know as to what is the motivation for the above choice of metrics that is often made. What makes the above &quot;natural&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To pull-back a metric on a submanifold, one needs to fix an embedding first. &#xA;Here what is the implicit embedding? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Now the LaTeX is compiling} &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-09-18T11:29:27.783" LastActivityDate="2010-09-18T11:29:27.783" Title="A particular method of pulling back a metric on a submanifold" Tags="&lt;differential-geometry&gt;" CommentCount="6" />
  <row Id="4859" PostTypeId="1" AcceptedAnswerId="4864" CreationDate="2010-09-17T15:21:25.000" Score="5" ViewCount="108" Body="&lt;p&gt;Let $C$ be a small pre-additive category. Let $R(C)$ denote its category ring, that is,&#xA;$$&#xA;R(C)=\bigoplus_{a,b\in \mathrm{Ob}(C)} C(a,b)&#xA;$$&#xA;as Abelian group, where the direct sum runs over all object $a$, $b$ of $C$. The multiplication in $R(C)$ is given by composition of composable morphisms and 0 for uncomposable morphisms (extended by bilinearity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This constuction is functorial: An additive functor $C\to D$ between small pre-additive categories induces a ring homomorphism $R(C)\to R(D)$ between the corresponding category rings in a canocical way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we have a functor $R$ from the category of small pre-additive categories (with additive functors as morhisms) to the category of rings (with ring homomorphisms as morphisms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the category of small pre-additive categories has a 2-categorical structure given by natural transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence my question: Does this 2-categorical structure have a counterpart in the category of rings? More precisely, &lt;strong&gt;is there a natural notion of 2-morphisms between ring homomorphisms turning $R$ into a 2-functor?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-17T20:25:02.957" LastActivityDate="2010-09-19T09:02:35.137" Title="Can we turn the functor &quot;category ring&quot; into a 2-functor in a natural way?" Tags="&lt;abstract-algebra&gt;&lt;category-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="4860" PostTypeId="2" ParentId="4841" CreationDate="2010-09-17T15:44:04.247" Score="2" ViewCount="0" Body="&lt;p&gt;Another method is as follows. An invertible upper triangular matrix has the form&#xA;$A=D(I+N)$ where $D$ is diagonal (with the same diagonal entries as $A$)&#xA;and $N$ is upper triangular with zero diagonal. Then $N^n=0$ where $A$&#xA;is $n$ by $n$. Both $D$ and $I+N$ have upper triangular inverses:&#xA;$D^{-1}$ is diagonal, and $(I+N)^{-1}=I-N+N^2-\cdots +(-1)^{n-1}N^{n-1}$.&#xA;So $A^{-1}=(I+N)^{-1}D^{-1}$ is upper triangular.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-17T15:44:04.247" CommentCount="1" />
  <row Id="4861" PostTypeId="2" ParentId="4808" CreationDate="2010-09-17T15:46:08.957" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Assuming that such a closed curve exists&lt;/strong&gt;, it is fairly straightforward to show that it must be a circle. For example, check out this &lt;a href=&quot;http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; for a proof.  It is interesting that the existence hypothesis is the &lt;em&gt;hard&lt;/em&gt; part of the proof.  &lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2010-09-17T15:46:08.957" />
  <row Id="4862" PostTypeId="1" CreationDate="2010-09-17T16:19:58.933" Score="3" ViewCount="514" Body="&lt;p&gt;Ok this is a bit of an unanswerable question, but hopefully someone will answer. As I have been going through college &amp;amp; high school there has been a kind of &quot;path&quot; through which you learn math. For example I went through basic math, prealgebra, algebra, precalc, calculus 1-3. My question is where to go after calculus three?(which is where I am at now btw). It looks to me like math really starts branching after calculus. I worried that I'll end up in classes that I can't fully understand because I didn't have the prerequisite knowledge. Do I go to linear algebra, real analysis, number theory, another branch ...? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working towards a degree in mathematics and would like to get the most out of every class, So if anybody would tell me what worked/didn't work for them I would be grateful.&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2010-09-22T20:21:51.227" CommunityOwnedDate="2010-09-17T17:15:13.250" Title="Where to go after calculus?" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;learning&gt;" AnswerCount="8" CommentCount="18" FavoriteCount="2" />
  <row Id="4863" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T16:31:33.200" Score="1" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One of the disappointments experienced by most mathematics students is that they never get a course in mathematics. They get courses in calculus, algebra, topology, and so on, but the divison of labor in teaching seems to prevent these different topics from being combined into a whole. In fact, some of the most important and natural questions are stifled because they wall on the wrong side of topic boundary lines. ... Thus, if students are to feel they really know mathematics by the time they graduate, there is a need to unify the subject.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from Stillwell - Mathematics and Its History&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So you could take this time to study some of the history of mathematics. Another thing which is useful is to study some formal logic (the stuff that makes proofs) - since (at least in my experience) this is not taught very much in classes.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-17T16:31:33.200" CommunityOwnedDate="2010-09-17T17:15:13.250" CommentCount="1" />
  <row Id="4864" PostTypeId="2" ParentId="4859" CreationDate="2010-09-17T16:33:00.127" Score="2" ViewCount="0" Body="&lt;p&gt;Rings are precisely those pre-additive categories with exactly one object. Thus the category of rings (noncommutative here) is a $2$-category. If $f,g : R \to S$ are $1$-morphisms, then a $2$-morphism $f \to g$ is an element $s \in S$ such that for all $r \in R$, we have $s f(r) = g(r) s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take a natural transformation $\eta : F \to G$ between additive functors $F,G : C \to D$. Then we may take $s = \sum_{x \in C} \eta(x)$ as an element of $R(D)$. But wait, this sum does not have to be finite. Then let's define $R(-)$ simply as the product and not the direct sum. This causes other problems, see the answer of Agusti Roig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the desired equation $s F(r) = G(r) s$ for some morphism $r : x \to y$ in $C$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{u : F(u)=F(x)} \eta(u) F(r) = \sum_{v : G(v)=G(x)} G(r) \eta(v)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be true only if $F$ and $G$ are injective on objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my answer would be: No, unfortunately $R$ cannot be made into a $2$-functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But somehow, $R$ &lt;em&gt;should&lt;/em&gt; be a $2$-functor, and perhaps we can modify the whole setting a but, so that it works.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-19T09:02:35.137" LastActivityDate="2010-09-19T09:02:35.137" />
  <row Id="4865" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T16:44:57.550" Score="3" ViewCount="0" Body="&lt;p&gt;A good History of Math course would probably be enjoyable, and give you a good idea of what things are and where they lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the mathematical point of view, Linear algebra is often a good &quot;first advanced mathematics course&quot;, a good jumping-off point: it is still concrete enough that you won't get lost in a sea of abstraction (a possible issue with abstract algebra depending on how it is taught), cover entirely new ideas (&quot;advanced calculus&quot; can feel like you are just re-treading the same ground you already know, and depending on the specific topics analysis might also feel like it), but it should make you work through proofs and concepts in a way with which you probably have not done so far. In addition, linear methods will show up all over the place later on, so it would prove useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that same vein, Number Theory can be a really good &quot;first abstract course in mathematics&quot;, while sticking close to things you are very familiar with (the integers and rationals) while also probably delivering some exciting surprises. It often surprises a lot of people just how &lt;em&gt;much&lt;/em&gt; of mathematics arises out of number theory (complex analysis and abstract algebra, to name just two). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to stick to the applied side, differential equations is a good place to go as well. Linear algebra would be useful there, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I would suggest linear algebra or number theory first (if you can also get a good history of math course, do that as well), then decide if you want to go towards abstraction (in which case, head to abstract algebra, mathematical analysis, or whichever of linear algebra or number theory you did not take) or more towards applications (differential equations, a good advanced probability/statistics course, or a discrete mathematics course). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is indeed the case that mathematics starts branching out, but some of the most interesting things happen where the branches meet; it would be ideal to be able to take a good one semester or one year sequence in the major areas (analysis, algebra, differential equations, topology, logic/set theory, number theory), then go on to more advanced courses in whichever area(s) you find interesting. But the truth is that this is very hard to do: not only would such a wide choice not be available except in the largest universities, but it would also mean a &lt;em&gt;lot&lt;/em&gt; of your time. I did my undergraduate in Mexico, where all I did in college was mathematics courses, and it basically took six semesters before that had been covered (in addition to the calculus sequence, a linear algebra sequence, an abstract algebra sequence, a mathematical analysis sequence, differential equations, discrete mathematics, complex analysis, probability and statistics, plus some other stuff to &quot;fill in the corners&quot;; it would be barely possible to do it in two years if you are not taking the calculus sequence, but not if you are also taking other coursework as you would be in most institutions in the United States).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-17T16:44:57.550" CommunityOwnedDate="2010-09-17T17:15:13.250" CommentCount="5" />
  <row Id="4866" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T16:47:11.060" Score="2" ViewCount="0" Body="&lt;p&gt;Large parts of mathematics as it has evolved in the 20th century are quite independent of calculus.  Game theory, graph theory, combinatorics, discrete mathematics, etc. are often courses which are taught without a specific Calculus prerequisite. The order in which one studies mathematics is to some extent related to what one's goals and &quot;natural talents&quot; are. For example, there are parts of geometry which don't require calculus (directly) while there are other parts of geometry where using Calculus methods are very central. If one majors in mathematics typically there is a &quot;recommended route&quot; of courses at the school one is studying at. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-17T16:47:11.060" CommunityOwnedDate="2010-09-17T17:15:13.250" CommentCount="2" />
  <row Id="4867" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T17:13:43.583" Score="2" ViewCount="0" Body="&lt;p&gt;If you're interested in building on your calculus knowledge in a way that builds links to many other areas of mathematics, a natural direction to go would be to take some linear algebra and a differential geometry course.  From there you could take some analysis and then a manifolds course.  At that point the subject of Lie groups acts as something of a unifying picture for all of the above said courses -- it fits everything together in a very pleasant way and gives you many pleasant connections between group theory, linear algebra and analysis.  Stillwell's book &quot;Naive Lie Theory&quot; is something you could pick up right now and start reading. When reading the book you'll likely be able to spot what you're comfortable with and what you aren't, and go from there. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-17T17:13:43.583" CommunityOwnedDate="2010-09-17T17:15:13.250" />
  <row Id="4868" PostTypeId="1" AcceptedAnswerId="4892" CreationDate="2010-09-17T17:15:27.010" Score="4" ViewCount="220" Body="&lt;p&gt;Let $A \subset \mathbb{C} $ be an open set containing the closed unit disc.&#xA;Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it follow that $f(z) = a z^{n} \frac{cz^{m}-b}{1-cz^{m}\bar{b}} $ for some $a,b,c \in \mathbb{C}$ s.t. $|c|=1$ $|a|=1$, $|b|&amp;lt;1$ and some $n,m\ge 0 $?&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-17T17:48:47.530" LastActivityDate="2010-09-19T05:47:47.120" Title="Mapping that takes unit circle to unit circle" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="4869" PostTypeId="1" AcceptedAnswerId="4938" CreationDate="2010-09-17T17:16:59.183" Score="6" ViewCount="335" Body="&lt;p&gt;In real analysis, we can define a line integral, but we also define (earlier) the regular definite integral.&#xA;Why is it that in complex analysis we are interested only in a line integral?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-10-20T23:53:13.067" Title="Why is an integral of a complex function defined as a line integral?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="4870" PostTypeId="1" AcceptedAnswerId="4873" CreationDate="2010-09-17T17:22:34.210" Score="1" ViewCount="112" Body="&lt;p&gt;Let's assume X,Y are arbitrary sets, and f is a function between them. What must those sets, and f, satisfy, in order for us to be able to define an integral of f in a way that &quot;makes sense&quot;?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-17T21:36:19.200" Title="$f:X \rightarrow Y$. What are the conditions on $f$,$X$,$Y$ in order for an integral of $f$ to be defined?" Tags="&lt;analysis&gt;&lt;functions&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4871" PostTypeId="2" ParentId="4862" CreationDate="2010-09-17T17:26:11.257" Score="0" ViewCount="0" Body="&lt;p&gt;I would strongly recommend you to read the book, &quot;Differential Calculus For Normed Linear Banach Spaces&quot; by Kalyan Mukherjee. Its published by the Hindustan Book Agency. I think it is one of those rare wonderful books which bridges the gap between calculus and manifold theory. It has beautiful sections on differentiating and Taylor expanding and integrating functions between arbitrary finite dimensional euclidean planes. He masterfully uses all this to motivate Lie Groups and calculates tangents to curves in Lie Groups. He further explains subtle things like differentiating matrix multiplication and determinant maps. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-09-17T17:26:11.257" CommunityOwnedDate="2010-09-17T17:26:11.257" />
  <row Id="4872" PostTypeId="1" AcceptedAnswerId="4879" CreationDate="2010-09-17T18:26:14.560" Score="3" ViewCount="195" Body="&lt;p&gt;I've read through the elementary proof of why there exists a solution $x$ to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$ for $p$ an odd prime. Is there a group theory generalization for this fact as well? &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-18T00:14:11.493" Title="Group theory proof of existence of a solution to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4873" PostTypeId="2" ParentId="4870" CreationDate="2010-09-17T18:39:34.167" Score="5" ViewCount="0" Body="&lt;p&gt;First of all, the question is not &quot;what conditions must $X$ and $Y$ satisfy&quot; but &quot;what &lt;em&gt;structures&lt;/em&gt; must $X$ and $Y$ have?&quot;  Second of all, there is not one most general definition of the integral; depending on what you consider an &quot;integral,&quot; there are several definitions which generalize in different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most basic requirement is that the integral is a linear operator.  For this to make sense, as Akhil says, $Y$ needs some kind of linear structure.  If $Y$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_group&quot; rel=&quot;nofollow&quot;&gt;abelian group&lt;/a&gt; and $X$ a finite set, then a perfectly sensible definition of the integral of $f$ is $ \sum_{x \in X} f(x)$, which is in fact a linear operator from the abelian group of functions $X \to Y$ to $Y$ (in fact the unique one, up to scaling, which is invariant under the symmetry group of $X$).  But of course the finite case is not very interesting.  (Note that in order for the sum to make sense above without an ordering on $X$ it is necessary that $Y$ be abelian.)  More generally one can stipulate that the only &quot;integrable functions&quot; are the finitely supported ones, in which case this definition alwaysm makes sense.  But again, a pretty boring definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the infinite case, for example if $X = \mathbb{N}$, we would like to define the integral as some kind of limit, for example maybe $\lim_{n \to \infty}  \sum_{k=1}^{n} f(k)$, because we would intuitively like to &quot;sum over $X$&quot; (but can't, at least not directly).  For this to make sense $Y$ needs a notion of convergence and $f$ needs to satisfy a &quot;smallness&quot; condition.  At a bare minimum, then, it needs to be a topological abelian group (say complete with respect to a norm), but there are some additional technicalities and I think they are usually resolved by requiring that $Y$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_space&quot; rel=&quot;nofollow&quot;&gt;Banach space&lt;/a&gt; over a &lt;a href=&quot;http://en.wikipedia.org/wiki/Complete_field&quot; rel=&quot;nofollow&quot;&gt;complete field&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another feature of the infinite case is that, if indicator functions of &quot;small&quot; subsets of $X$ are to be integrable, then a notion of integral on functions out of $X$ implies a notion of &quot;size&quot; or measure on $X$.  The usual way people handle this is to require that $X$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;measure space&lt;/a&gt;, in which case (together with the structures on $Y$ described above) the usual construction of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_integration&quot; rel=&quot;nofollow&quot;&gt;Lebesgue integral&lt;/a&gt; goes through.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-17T21:36:19.200" LastActivityDate="2010-09-17T21:36:19.200" />
  <row Id="4877" PostTypeId="2" ParentId="4854" CreationDate="2010-09-17T19:00:29.907" Score="5" ViewCount="0" Body="&lt;p&gt;An &lt;em&gt;orthocentric&lt;/em&gt; tetrahedron is one for which the altitudes from the vertices to the opposite faces are concurrent (this is not true for all tetrahedrons).  For an orthocentric tetrahedron, there exists a sphere (the 24-point sphere) that intersects each face of the tetrahedron in its 9-point circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also the summary of a talk by Steve West (&quot;Discovering Theorems Using Cabri 3-D&quot;) in &lt;a href=&quot;http://mmcchicago.org/dinners08-09/November08.pdf&quot; rel=&quot;nofollow&quot;&gt;the November 2008 issue of &lt;em&gt;Points &amp;amp; Angles&lt;/em&gt; (PDF)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-17T19:00:29.907" />
  <row Id="4878" PostTypeId="2" ParentId="4872" CreationDate="2010-09-17T19:11:11.813" Score="4" ViewCount="0" Body="&lt;p&gt;Presumably you refer to the following proof, which contradicts Lagrange's or Fermat's little Theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\rm\; p = 4n+3\;$ is prime. Then $\rm\ x^2 = -1 \;\Rightarrow\; x^{p-1} = (x^2)^{2n+1} = -1 \pmod{p}\;\Rightarrow\Leftarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A finite field of order $\rm\:4n+3\:$ has subgroup of squares of order $\rm\:2n+1\:$. But one easily shows that in any group of odd order $\rm\; 2n+1\:$ the equation $\rm\; x^2 = a\;$ has the solution $\rm\:  a^{n+1}\;$ (due to Lagrange in 1769). Specializing $\rm\; a = -1\:$ we conclude that $\rm\; x^2 = -1\:$ has no roots. On the other hand, the equation always has a root in a finite field of order $\rm 4n+1$ by the case $\rm k=4$ of &lt;a href=&quot;http://www.math.wisc.edu/~jensen/frobtalk.pdf&quot; rel=&quot;nofollow&quot;&gt;Frobenius's  theorem (1907)&lt;/a&gt; - which says that if $\rm k$ divides the order of a finite group $\rm G$ then $\rm k$ also divides the number of $\rm k$'th roots of $1$ in $\rm G\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-17T21:04:49.020" LastActivityDate="2010-09-17T21:04:49.020" />
  <row Id="4879" PostTypeId="2" ParentId="4872" CreationDate="2010-09-17T19:18:32.203" Score="6" ViewCount="0" Body="&lt;p&gt;I would sketch the group theoretic fact as follows: Let $G$ be a finite abelian group. Then the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) For every $d$, there are at most $d$ solutions to $g^d=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $G$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) For $n$ dividing $|G|$, there are precisely $\phi(n)$ elements of order $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only place that I know of where hypothesis (1) shows up naturally is considering the multiplicative groups of finite fields. From this we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an element of order $d$ in $\mathbb{F}_{p^r}$ if and only if $d| p^r-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure there is any more to say.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-17T19:18:32.203" />
  <row Id="4880" PostTypeId="2" ParentId="4725" CreationDate="2010-09-17T19:29:53.720" Score="1" ViewCount="0" Body="&lt;p&gt;This is the problem of &lt;em&gt;confidence intervals&lt;/em&gt; in statistics.  Your question is about the original problem of that type, the  &quot;confidence interval for a proportion in binomial sampling&quot; (wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval&lt;/a&gt; ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymptotic theory, when $Np(1-p)$ is very large, is described by the Central Limit Theorem, which says that $n = pN + A\sqrt{Np(1-p)}$ where the probability distribution of $A$ is the famous Gaussian (normal, &quot;bell shaped&quot;) curve.  Qualitatively, fluctuations of $n$ around the expected value are usually within a small factor of $\sqrt{Np(1-p)}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For limited data, the story is more complicated, with no perfect solution, many proposals, and a large literature.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-17T19:29:53.720" />
  <row Id="4881" PostTypeId="1" AcceptedAnswerId="4887" CreationDate="2010-09-17T19:42:51.613" Score="3" ViewCount="80" Body="&lt;p&gt;Let $T$ be the operator from $L^2(\mathbb R^n)$ to $L^2(\mathbb R^n)$ that is given by $Tf := f * g$ where $g$ is in $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I now find that the spectrum of $T$ is equal to the essential range of $\hat{g}$? How is the spectrum of $T$ related to the invertibility of the operator $G\hat{f} = \hat{f}\hat{g}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hat denotes the Fourier transform.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-09-17T21:06:31.267" Title="Spectrum of a convolution operator." Tags="&lt;functional-analysis&gt;&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="4882" PostTypeId="1" AcceptedAnswerId="4883" CreationDate="2010-09-17T19:42:58.150" Score="1" ViewCount="79" Body="&lt;p&gt;Define a binary step-order set as a finite set S of positive integers s&lt;sub&gt;i&lt;/sub&gt;, i = 0 to n-1, where s&lt;sub&gt;i&lt;/sub&gt; ≡ 2&lt;sup&gt;i&lt;/sup&gt; (mod 2&lt;sup&gt;i+1&lt;/sup&gt;) for each i from 0 to n-1. So, i is the power of 2 appearing in the prime factorization of s&lt;sub&gt;i&lt;/sub&gt;. It’s elementary that binary step-order sets have distinct subset sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the least upper bound on the number of pairs of disjoint subsets of an n-element binary step-order set S whose sums differ by 1?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2010-09-17T20:43:14.167" Title="What is the least upper bound on the number of pairs of disjoint subsets of a binary step-order set S whose sums differ by 1?" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="4883" PostTypeId="2" ParentId="4882" CreationDate="2010-09-17T20:26:01.663" Score="1" ViewCount="0" Body="&lt;p&gt;The answer is F&lt;sub&gt;n+1&lt;/sub&gt;, where {F&lt;sub&gt;i&lt;/sub&gt;} is the Fibonacci sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma – Let S be a binary step-order set with n elements. For any positive integer k, F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound for the number of pairs of disjoint subsets of S whose sums differ by k, where {F&lt;sub&gt;i&lt;/sub&gt;} is the Fibonacci sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceeding by math induction, first prove the cases n = 1 and n = 2. For n = 1, the only possible pair of disjoint subsets is (S, φ); thus, F&lt;sub&gt;2&lt;/sub&gt; = 1 is an upper bound for the number of pairs of disjoint subsets with sums differing by k. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n ≥ 2, observe that finding a pair of disjoint subsets whose sums differ by k is equivalent to finding an n-tuple of coefficients &lt;strong&gt;d&lt;/strong&gt; ∊ {-1, 0, 1}&lt;sup&gt;n&lt;/sup&gt; such that ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; = k – the elements of the higher-sum subset have coefficients 1, the elements of the lower-sum subset (if any – this may be φ) have coefficients -1, and the elements in neither subset have coefficients 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 2, we have s&lt;sub&gt;0&lt;/sub&gt; odd and s&lt;sub&gt;1&lt;/sub&gt; even. So, there are two cases; s&lt;sub&gt;0&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt; and s&lt;sub&gt;0&lt;/sub&gt; &amp;lt; s&lt;sub&gt;1&lt;/sub&gt;.&#xA;For s&lt;sub&gt;0&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt;, the only possible coefficient ordered pairs d are (1, 0), (0, 1), (1, -1) and (1, 1); otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is not positive. Furthermore, these all yield distinct values of ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt;, because s&lt;sub&gt;0&lt;/sub&gt; + s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;0&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt;, and because s&lt;sub&gt;1&lt;/sub&gt; ≠ s&lt;sub&gt;0&lt;/sub&gt; – s&lt;sub&gt;1&lt;/sub&gt; because the former is even and the latter is odd. So, in this case, 1 is an upper bound for the number of disjoint subset pairs whose sums differ by k.&#xA;For s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;0&lt;/sub&gt;, the only possible coefficient ordered pairs d are (1, 0), (0, 1), (-1, 1) and (1, 1); otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is not positive. We have s&lt;sub&gt;0&lt;/sub&gt; + s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;1&lt;/sub&gt; &gt; s&lt;sub&gt;0&lt;/sub&gt;, so the only possible way to have ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; = k for two distinct &lt;strong&gt;d&lt;/strong&gt; is if s&lt;sub&gt;0&lt;/sub&gt; = s&lt;sub&gt;1&lt;/sub&gt; – s&lt;sub&gt;0&lt;/sub&gt;; i.e., s&lt;sub&gt;1&lt;/sub&gt; = 2s&lt;sub&gt;0&lt;/sub&gt;.  So, in this case, F&lt;sub&gt;3&lt;/sub&gt; = 2 is an upper bound for this case and the general n = 2 case of the number of disjoint subset pairs whose sums differ by k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume that the lemma is true for all positive integers from 1 through n ≥ 2. Consider an n+1 element binary step-order set S&lt;sub&gt;n+1&lt;/sub&gt;. Let S&lt;sub&gt;n&lt;/sub&gt; be the n-element set obtained by omitting the odd element of S&lt;sub&gt;n+1&lt;/sub&gt; and dividing all the other elements by 2 – clearly S&lt;sub&gt;n&lt;/sub&gt; is a binary step-order set. Let S&lt;sub&gt;n-1&lt;/sub&gt; be the n-1 element set obtained by omitting the odd element of S&lt;sub&gt;n&lt;/sub&gt; and dividing all the other elements by 2 – clearly S&lt;sub&gt;n-1&lt;/sub&gt; is also a binary step-order set. So s&lt;sub&gt;n,i&lt;/sub&gt; = s&lt;sub&gt;n+1,i+1&lt;/sub&gt; / 2 and s&lt;sub&gt;n-1,i&lt;/sub&gt; = s&lt;sub&gt;n,i+1&lt;/sub&gt; / 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case where k is even. The coefficient of s&lt;sub&gt;n+1,0&lt;/sub&gt; must be 0 for any &lt;strong&gt;d&lt;/strong&gt; in the solution set; otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is odd and ≠ k.  Each solution &lt;strong&gt;d&lt;/strong&gt; = (0, d&lt;sub&gt;n+1,1&lt;/sub&gt;, …, d&lt;sub&gt;n+1,n&lt;/sub&gt;) corresponds to a solution d&lt;sub&gt;n&lt;/sub&gt; for k/2 in S&lt;sub&gt;n&lt;/sub&gt; via bijection, where d&lt;sub&gt;n,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i&lt;/sub&gt; for i = 0 through n – 1. So, F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound for the number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by an even k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case where k is odd. The coefficient d&lt;sub&gt;n+1,0&lt;/sub&gt; of s&lt;sub&gt;n+1,0&lt;/sub&gt; must be either 1 or -1 for any &lt;strong&gt;d&lt;/strong&gt; in the solution set; otherwise ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; is even and ≠ k.  Now, for both of k and s&lt;sub&gt;n+1,0&lt;/sub&gt;, we have cases ≡ 1 (mod 4) and ≡ 3 (mod 4); as a consequence, exactly one of k + s&lt;sub&gt;n+1,0&lt;/sub&gt; and k - s&lt;sub&gt;n+1,0&lt;/sub&gt; is ≡ 2 (mod 4), and the other is ≡ 0 (mod 4).  Say k + s&lt;sub&gt;n+1,0&lt;/sub&gt; ≡ 0 (mod 4), and k - s&lt;sub&gt;n+1,0&lt;/sub&gt; ≡ 2 (mod 4); the other case is similar &amp;amp; will be omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each solution &lt;strong&gt;d&lt;/strong&gt; = (1, d&lt;sub&gt;n+1,1&lt;/sub&gt;, …, d&lt;sub&gt;n+1,n&lt;/sub&gt;) corresponds to a solution &lt;strong&gt;d&lt;sub&gt;n&lt;/sub&gt;&lt;/strong&gt; for |( k - s&lt;sub&gt;n+1,0&lt;/sub&gt;)/2| in S&lt;sub&gt;n&lt;/sub&gt; via bijection, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If k &gt; s&lt;sub&gt;n+1,0&lt;/sub&gt;, d&lt;sub&gt;n,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i+1&lt;/sub&gt; for i = 0 through n – 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If k &amp;lt; s&lt;sub&gt;n+1,0&lt;/sub&gt;, d&lt;sub&gt;n,i&lt;/sub&gt; = -d&lt;sub&gt;n+1,i+1&lt;/sub&gt; for i = 0 through n – 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound for the number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by k with d&lt;sub&gt;n+1,0&lt;/sub&gt; = 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the solutions &lt;strong&gt;d&lt;/strong&gt; where d&lt;sub&gt;n+1,0&lt;/sub&gt; = -1.  We have ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; = k, so ∑ d&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; + s&lt;sub&gt;n+1,0&lt;/sub&gt; ≡ 0 (mod 4). Hence, for the ordered n+1 tuple &lt;strong&gt;d’&lt;/strong&gt; where d’&lt;sub&gt;n+1,0&lt;/sub&gt; = 0 and d’&lt;sub&gt;n+1,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i&lt;/sub&gt; for i = 1 through n, we have   ∑ d’&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; ≡ 0 (mod 4). Therefore, d’&lt;sub&gt;n+1,1&lt;/sub&gt; = d&lt;sub&gt;n+1,1&lt;/sub&gt; = 0; otherwise ∑ d’&lt;sub&gt;i&lt;/sub&gt;s&lt;sub&gt;i&lt;/sub&gt; ≡ 2 (mod 4).  Each solution &lt;strong&gt;d&lt;/strong&gt; = (-1, 0, d&lt;sub&gt;n+1,2&lt;/sub&gt;, …, d&lt;sub&gt;n+1,n&lt;/sub&gt;) corresponds to a solution d&lt;sub&gt;n-1&lt;/sub&gt; for( k + s&lt;sub&gt;n+1,0&lt;/sub&gt;)/4 in S&lt;sub&gt;n-1&lt;/sub&gt; via bijection, where d&lt;sub&gt;n-1,i&lt;/sub&gt; = d&lt;sub&gt;n+1,i+2&lt;/sub&gt; for i = 0 through n – 2. So, F&lt;sub&gt;n&lt;/sub&gt; is an upper bound for the number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by k with d&lt;sub&gt;n+1,0&lt;/sub&gt; = -1. As we have exhausted the two cases for d&lt;sub&gt;n+1,0&lt;/sub&gt;, F&lt;sub&gt;n+2&lt;/sub&gt; = F&lt;sub&gt;n+1&lt;/sub&gt; + F&lt;sub&gt;n&lt;/sub&gt; is an upper bound for the total number of disjoint subset pairs of S&lt;sub&gt;n+1&lt;/sub&gt; whose sums differ by an odd k, or any positive integer k./&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the lemma to k = 1, conclude that F&lt;sub&gt;n+1&lt;/sub&gt; is an upper bound on the number of pairs of disjoint subsets of an n-element binary step-order set S whose sums differ by 1. To show that F&lt;sub&gt;n+1&lt;/sub&gt; is the least upper bound, we’ll construct a sequence of binary step-order sets where this upper bound is realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let S&lt;sub&gt;1&lt;/sub&gt; = {1}, S&lt;sub&gt;2&lt;/sub&gt; = {1,2}, and S&lt;sub&gt;3&lt;/sub&gt; = {3,2,4}. Given S&lt;sub&gt;n&lt;/sub&gt; for n ≥ 2, S&lt;sub&gt;n+1&lt;/sub&gt; is constructed by setting s&lt;sub&gt;n+1,0&lt;/sub&gt; = 3, and s&lt;sub&gt;n+1,i&lt;/sub&gt; = 2s&lt;sub&gt;n,i-1&lt;/sub&gt;for i = 1 to n. (The set elements are listed in the order of increasing power of 2 in the prime factorization of the element.) There are 2 solutions for &lt;strong&gt;d&lt;/strong&gt; for S&lt;sub&gt;2&lt;/sub&gt;; namely (1,0) and (-1,1). There are 3 solutions for &lt;strong&gt;d&lt;/strong&gt; for S&lt;sub&gt;3&lt;/sub&gt;; namely (1,-1,0), (1,1,-1) and (-1,0,1). (Because the 0 term of the sum is either 3 or -3, the subtotal of the other terms must be either -2 or 4 to have the sum of 1.) The F&lt;sub&gt;n+2&lt;/sub&gt; solutions for S&lt;sub&gt;n+1&lt;/sub&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The F&lt;sub&gt;n+1&lt;/sub&gt; n+1 tuples d defined by setting d&lt;sub&gt;n+1,0&lt;/sub&gt; = 1 and  d&lt;sub&gt;n+1,i&lt;/sub&gt; = -d&lt;sub&gt;n,i-1&lt;/sub&gt; for i = 1 to n from all the solutions d&lt;sub&gt;n&lt;/sub&gt; for S&lt;sub&gt;n&lt;/sub&gt;; plus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The F&lt;sub&gt;n&lt;/sub&gt; n+1 tuples d defined by setting d&lt;sub&gt;n+1,0&lt;/sub&gt; = -1, d&lt;sub&gt;n+1,1&lt;/sub&gt; = 0 and  d&lt;sub&gt;n+1,i&lt;/sub&gt; = d&lt;sub&gt;n-1,i-1&lt;/sub&gt; for i = 1 to n from all the solutions d&lt;sub&gt;n-1&lt;/sub&gt; for S&lt;sub&gt;n-1&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I couldn't work out fully how to do sigma notation; hopefully the sum limits are clearly implied.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-09-17T20:43:14.167" LastActivityDate="2010-09-17T20:43:14.167" CommentCount="2" />
  <row Id="4884" PostTypeId="1" AcceptedAnswerId="4885" CreationDate="2010-09-17T20:49:45.077" Score="1" ViewCount="127" Body="&lt;p&gt;In the book Introduction to Linear Optimization by Bertsimas Dimitri, a polyhedron is defined as a set $ \lbrace x \in \mathbb{R^n} | Ax \geq b \rbrace $, where A is an m x n matrix and b is a vector in $\mathbb{R^m}$. What it means is that a polyhedron is the intersection of several halfspaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball can also be viewed as the intersection of infinitely many halfspaces. So I was wondering if a ball is also a polyhedron by that definition or by any other definition that you might use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-17T23:02:37.393" LastActivityDate="2010-09-17T23:02:37.393" Title="Is a ball a polyhedron?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4885" PostTypeId="2" ParentId="4884" CreationDate="2010-09-17T20:59:39.307" Score="3" ViewCount="0" Body="&lt;p&gt;No a ball is not a polyhedron, even by this definition. In your definition the matrix $A$ is of size $m\times n$, where $m\in\mathbb{N}$ thus the matrix is finite. The integer $m$ is an upper bound on the number of halfspaces which intersect to form the polyhedron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason $m$ is an upper bound is because suppose $A$ has two rows identical. Then there are two hyperspaces which are parallel so at least one of them does not form any part of the polyhedron.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-17T20:59:39.307" />
  <row Id="4886" PostTypeId="2" ParentId="4884" CreationDate="2010-09-17T21:00:19.213" Score="2" ViewCount="0" Body="&lt;p&gt;The usual definition of a polyhedron requires that either one intersects a finite number of half-spaces, or one takes the convex hull of a finite set of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the book Convex Polytopes by Branko Grünbaum (either the first or second edition).&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-17T21:00:19.213" CommentCount="2" />
  <row Id="4887" PostTypeId="2" ParentId="4881" CreationDate="2010-09-17T21:06:31.267" Score="5" ViewCount="0" Body="&lt;p&gt;The point is that the Fourier transform $\mathcal{F}$&#xA;conjugates $T$ to the multiplication&#xA;operator $S:h\mapsto\hat g h$ (i.e. $\mathcal{F}(f\ast g)=\hat g\mathcal{F}(f)$).&#xA;Hence $T$ and $S$ have the same spectrum.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-17T21:06:31.267" CommentCount="1" />
  <row Id="4889" PostTypeId="1" CreationDate="2010-09-17T21:19:10.037" Score="1" ViewCount="293" Body="&lt;p&gt;In many elementary (and not-so-elementary) Euclidean geometry texts, a (simple) polygon is said to be &lt;em&gt;tangential&lt;/em&gt;&amp;nbsp; if it is convex and has an inscribed circle (i.e., a circle that intersects and is tangent to each side of the polygon).  The assumption of convexity is not needed: I've come up with a rather laborious proof that every polygon with an inscribed circle is convex.  But I'd like to find either a simple elementary proof or a reference to a proof in the literature.  (By &quot;elementary,&quot; I mean using only standard facts of axiomatic Euclidean geometry.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a reference for a proof of this fact (elementary or not)?  Or can anyone think of a straightforward elementary proof?  You can use any definition of &quot;convex polygon&quot; that you like, but the easiest one to work with is that for each edge, the vertices not on that edge lie on one side of the line through that edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Interestingly, the corresponding fact for circumscribed circles--i.e., that every polygon with a circumscribed circle is convex--is quite easy to prove: If P has a circumscribed circle, any two nonadjacent sides of P are non-intersecting chords of the circle; it is easy to show that both endpoints of each chord lie on the same side of the line through the other, and from there it is an easy matter to prove that P is convex.)&lt;/p&gt;&#xA;" OwnerUserId="1421" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-17T23:54:25.870" LastActivityDate="2010-09-18T01:42:20.337" Title="Proof that every polygon with an inscribed circle is convex?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4890" PostTypeId="2" ParentId="4870" CreationDate="2010-09-17T21:28:41.303" Score="2" ViewCount="0" Body="&lt;p&gt;In Stefan Rolewicz book &lt;a href=&quot;http://books.google.com/books?id=qVmHL2M5I9oC&amp;amp;dq=rolewicz+Metric+linear+spaces&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=1teTTMnnIouPswaolsX3CQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CCcQ6AEwAw#v=onepage&amp;amp;q=riemann&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Metric Linear Spaces&lt;/a&gt; there are different kinds of general and useful results regarding integration on &quot;more abstract spaces&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example  $f:[0,1]\to X$ where $X$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/F-space&quot; rel=&quot;nofollow&quot;&gt;$F$-space&lt;/a&gt;, he discuss the Riemann-integral and if, in addition, $X$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Locally_convex&quot; rel=&quot;nofollow&quot;&gt;locally convex&lt;/a&gt; then the Bochner-Lebesgue integral is more fruitful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, though I like your question I believe that in the end I think it is better to have some application in mind and then, whenever needed, try to look for some integration methods for the situation. Sometimes it is good to be open to alternative methods in other areas of mathematics that might be more handy, for example &quot;abstract maximum principles&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-17T21:28:41.303" />
  <row Id="4891" PostTypeId="2" ParentId="4889" CreationDate="2010-09-17T22:05:04.180" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;: Suppose that a simple polygon has an inscribed circle.  Without loss of generality, pick a &quot;first&quot; edge and let the circle be on the &quot;right&quot; side of that edge.  The angle that is on the same side as the circle--that is, to the right--between the first edge and the second edge must have measure less than 180&amp;deg;.  Similarly, the angle between the second and third edges that is also on the right must also have measure less than 180&amp;deg;, and so on, so that all of the angles on the right side of the polygon's perimeter (whether it is the inside or the outside) must have measure less than 180&amp;deg; and all of the angles on the left side must have measure greater than 180&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sum of the interior angles of a simple polygon with &lt;em&gt;n&lt;/em&gt; sides is 180&amp;deg;(&lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;ndash;&amp;nbsp;2), the average measure of an interior angle of a simple polygon with &lt;em&gt;n&lt;/em&gt; sides is 180&amp;deg;&amp;nbsp;&amp;ndash;&amp;nbsp;360&amp;deg;/&lt;em&gt;n&lt;/em&gt;, which is strictly less than 180&amp;deg;.  So, since the angles on the left side of the perimeter all have measure greater than 180&amp;deg;, their average is greater than 180&amp;deg;, so the left cannot be the interior of the polygon and the right side must be the interior, so the inscribed circle must be in the interior of the polygon and the internal angles all have measure less than 180&amp;deg;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;original answer&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the polygon is non-self-intersecting, then two consecutive sides of the polygon correspond to two consecutive points on tangency on the circle, with the angle formed by the two sides subtending the minor arc between the points of tangency.  The measure of the angle of the polygon is half the difference between the measure of the major and minor arcs between the points of tangency.  The greatest possible difference would be the degenerate case where the minor arc has measure 0&amp;deg; and the major arc has measure 360&amp;deg;, giving the angle measure 180&amp;deg;; for non-degenerate cases, the difference in the arc measures must be less than 360&amp;deg;, so the angle measure must be less than 180&amp;deg;.  This applies to every pair of consecutive sides, so every interior angle of the polygon has measure less than 180&amp;deg;, so the polygon is convex.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: Alternately, and blatantly assuming that the inscribed circle must be in the interior of the polygon, suppose that a polygon is not convex, so that there is a vertex at which the interior angle has measure greater than 180&amp;deg;.  For a circle to be tangent to the two sides that meet at that vertex, the circle must be exterior to the polygon at that vertex (so as to be in the non-reflex side of that angle) and thus cannot be the inscribed circle.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-18T01:31:56.330" LastActivityDate="2010-09-18T01:31:56.330" CommentCount="8" />
  <row Id="4892" PostTypeId="2" ParentId="4868" CreationDate="2010-09-17T22:37:12.357" Score="2" ViewCount="0" Body="&lt;p&gt;Quoting part of my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/3408/characterizing-nonconstant-entire-functions-with-modulus-1-on-the-unit-circle&quot;&gt;this other question&lt;/a&gt; is my answer to yours (and Robin basically already said it in the comments):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a function $f$ is holomorphic in a neighborhood of the closed disk and has modulus 1 on the circle, then $f$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Blaschke_product#Finite_Blaschke_products&quot; rel=&quot;nofollow&quot;&gt;finite Blaschke product&lt;/a&gt;.  You can prove this by taking all of the zeros inside the disk counted according to multiplicity, dividing by corresponding &lt;a href=&quot;http://planetmath.org/encyclopedia/AutomorphismsOfUnitDisk.html&quot; rel=&quot;nofollow&quot;&gt;holomorphic automorphisms of the disk&lt;/a&gt; that have those zeros, and showing that the result is constant.  (This quotient and its reciprocal are analytic and bounded by 1 on the disk...)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course I'm assuming $A$ is connected.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-17T22:37:12.357" CommunityOwnedDate="2010-09-17T22:38:35.357" CommentCount="1" />
  <row Id="4893" PostTypeId="1" AcceptedAnswerId="4895" CreationDate="2010-09-17T23:27:17.700" Score="3" ViewCount="170" Body="&lt;p&gt;Why is saying $\frac{dy}{dx}\frac{dy}{dx}=y\frac{d^2y}{(dx)^2 }$ not valid? Does Leibniz notation (and thinking of it as an infinitesimal quotient) not work for higher-order derivatives?&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-09-18T11:43:30.850" Title="Can Leibniz Notation Be Treated As a Quotient?" Tags="&lt;calculus&gt;&lt;notation&gt;" AnswerCount="3" />
  <row Id="4894" PostTypeId="1" CreationDate="2010-09-17T23:39:21.200" Score="2" ViewCount="266" Body="&lt;p&gt;It is not quite intuitive , at least from its origin. Could any one can give me an intuitive explanation?Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-12-07T21:51:50.023" LastActivityDate="2010-12-07T23:29:51.383" Title="Why does symplectic geometry have many applications in mathematics" Tags="&lt;geometry&gt;&lt;soft-question&gt;" AnswerCount="4" />
  <row Id="4895" PostTypeId="2" ParentId="4893" CreationDate="2010-09-17T23:48:21.867" Score="4" ViewCount="0" Body="&lt;p&gt;Consider $y=3x$ then $\frac{dy}{dx} =3$. So the left hand side of your equation would be 9. However, $\frac{d^{2}y}{dx^{2}} =0$, so the right hand side of your equation would be 0 for this example, but $0\neq 9$. Thus, the function $y=3x$ provides a counterexample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leibniz notation does work for higher derivatives, the nth derivative of y is denoted by $\frac{d^{n}y}{dx^{n}}$. However, treating the derivative as a quotient is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Abuse_of_notation#Derivative&quot; rel=&quot;nofollow&quot;&gt;abuse of notation&lt;/a&gt; and is considered more as a convenient tool to help students remember certain rules like the chain rule. There is actually a way of treating infinitesimals rigorously is treated in what is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-standard_analysis&quot; rel=&quot;nofollow&quot;&gt;non-standard analysis&lt;/a&gt;, but this isn't really the most common approach.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-17T23:48:21.867" />
  <row Id="4896" PostTypeId="1" AcceptedAnswerId="4911" CreationDate="2010-09-17T23:54:03.710" Score="2" ViewCount="97" Body="&lt;p&gt;Say we have a undirected, connected graph that represents a network of streets.  How would you prove that there always exists a tour of the network, where one ``drives'' in both directions on every street exactly once?&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2010-09-18T10:52:04.190" Title="Network of streets" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4897" PostTypeId="2" ParentId="4896" CreationDate="2010-09-18T00:09:43.980" Score="3" ViewCount="0" Body="&lt;p&gt;What you are looking to prove is that the graph is &lt;a href=&quot;http://en.wikipedia.org/wiki/Eulerian_path&quot; rel=&quot;nofollow&quot;&gt;Eulerian&lt;/a&gt;, which is true as each vertex has an even degree.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-18T00:09:43.980" CommentCount="4" />
  <row Id="4898" PostTypeId="2" ParentId="4896" CreationDate="2010-09-18T00:12:02.480" Score="0" ViewCount="0" Body="&lt;p&gt;How do you represent the fact that you want to traverse each edge twice?  How do you know whether you can trace a graph with a single continuous path?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-18T00:12:02.480" CommentCount="2" />
  <row Id="4899" PostTypeId="2" ParentId="4872" CreationDate="2010-09-18T00:14:11.493" Score="2" ViewCount="0" Body="&lt;p&gt;Probably the simplest way to state a corresponding group theory result is the generalization of Frobenius's theorem mentioned by Bill Dubuque. The generalization says that if $G$ is a finite group, then the number of solutions in $G$ to $x^n=1$ is a multiple of $\gcd(|G|,n)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the congruence $x^2\equiv -1\pmod{p}$, the solutions to this congruence are exactly the solutions to $x^4\equiv 1 \pmod{p}$ that are not solutions to $x^2\equiv 1\pmod{p}$. The order of the group here is $p-1$, so the number of solutions to the first is a multiple of $\gcd(4,p-1)$; the number of solutions of the second is a multiple of $2$, and we know the two: $1$ and $-1$. Since every solution to $x^4\equiv 1\pmod{p}$ is a solution to $x^2\equiv 1 \pmod{p}$, there are solutions to $x^2\equiv -1\pmod{p}$ if and only if $\gcd(4,p-1) = 4$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-18T00:14:11.493" CommentCount="1" />
  <row Id="4900" PostTypeId="1" AcceptedAnswerId="4906" CreationDate="2010-09-18T00:21:52.723" Score="3" ViewCount="134" Body="&lt;p&gt;From an olympic book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $m$ be an odd number and $a_1, ... , a_m$ integers. $x=(x_1,...,x_m)$ is a permutation of the integers $1, ..., m$ and $f$ at $x$ is defined by $f(x)=x_1a_1 + ... + x_m a_m$. Prove that there exist two permutations $x$ and $y$ such that $f(x)-f(y)$ is divisible by $m!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of a few facts. For instance, if all $a_i$ are odd (or even), any permutations of $x$ results in values of $f$ of the same parity, but there are only $m!/2$ (for $m&gt;1$) residues module $m!$ of the same parity, and an application of the Pigeon Hole Principle yields the conclusion. So in the general case I'd like to say something like there's a greater number of permutations that give a value of $f$ of one parity than the other. I don't know how to show this.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-18T07:14:57.937" Title="pigeon-hole and parity" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4901" PostTypeId="2" ParentId="4821" CreationDate="2010-09-18T00:32:16.700" Score="2" ViewCount="0" Body="&lt;p&gt;It appears that both previous answers missed that the number was to be even.  So the real count is 5!/2! + 5!/2!/2! + 5!/2! + 5!/2! = 210&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-18T00:32:16.700" CommentCount="1" />
  <row Id="4902" PostTypeId="2" ParentId="4889" CreationDate="2010-09-18T01:42:20.337" Score="2" ViewCount="0" Body="&lt;p&gt;The definition of &lt;em&gt;simple&lt;/em&gt; establishes that the polygon itself equals the intersection of the half-planes tangent to the circle at the points where the polygon's sides contact the circle.  Any nonempty intersection of halfplanes is convex.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-18T01:42:20.337" CommentCount="6" />
  <row Id="4903" PostTypeId="2" ParentId="4894" CreationDate="2010-09-18T02:47:28.480" Score="1" ViewCount="0" Body="&lt;p&gt;Symplectic geometry is applicable in lot of areas such as Classical mechanics, Hamiltonian mechanics. While browsing through the web i have found a reason as to why SG is good for classical mechanics. You may read it here: &lt;a href=&quot;http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html&quot; rel=&quot;nofollow&quot;&gt;http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-18T02:47:28.480" CommentCount="2" />
  <row Id="4904" PostTypeId="2" ParentId="4841" CreationDate="2010-09-18T06:04:04.877" Score="2" ViewCount="0" Body="&lt;p&gt;Personally, I prefer arguments which are more geometric to arguments rooted in matrix algebra. With that in mind, here is a proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, two observations on the geometric meaning of an upper triangular invertible linear map. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Define $S_k = {\rm span} (e_1, \ldots, e_k)$, where $e_i$ the standard basis vectors. &#xA;Clearly, the linear map $T$ is upper triangular if and only if $T S_k \subset S_k$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $T$ is in addition invertible, we must have the stronger relation $T S_k = S_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if $T S_k$ was a strict subset of $S_k$, then $Te_1, \ldots, Te_k$ are $k$ vectors in a space of dimension strictly less than $k$, so they must be dependent: $\sum_i \alpha_i Te_i=0$ for some $\alpha_i$ not all zero. This implies that $T$ sends the &lt;b&gt;nonzero&lt;/b&gt; vector $\sum_i \alpha_i e_i$ to zero, so $T$ is not invertible. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With these two observations in place, the proof proceeds as follows. Take any $s \in S_k$. Since $TS_k=S_k$ there exists some $s' \in S_k$ with $Ts'=s$ or $T^{-1}s = s'$. In other words, $T^{-1} s$ lies in $S_k$, so $T^{-1}$ is upper triangular. &lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-09-18T17:06:21.803" LastActivityDate="2010-09-18T17:06:21.803" CommentCount="3" />
  <row Id="4905" PostTypeId="1" CreationDate="2010-09-18T06:41:39.513" Score="6" ViewCount="248" Body="&lt;p&gt;The following seems to hold in numerical simulations, is it true?&#xA;$$\lim_{n\to \infty} \int_0^1 dx \frac{n! 2^{-n} n}{(n x)!(n-n x)!\sqrt{x(1-x)}}=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a combination of two known integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to \infty} \int_0^1 dx \frac{n! 2^{-n} n}{(n x)!(n-n x)!}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 dx \frac{1}{\sqrt{x(1-x)}}=\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First one follows from asymptotic &lt;a href=&quot;http://www.physicsforums.com/attachment.php?attachmentid=27909&amp;amp;d=1283240321&quot; rel=&quot;nofollow&quot;&gt;expansion&lt;/a&gt; of Gamma and the second is done using trig substitution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: $x!$ is short for $\Gamma(x+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limit convergence is apparent for $n$ between 10 and $10^6$, $\log_{10} n$ gives an almost perfect fit to the log of deviation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/error-convergence.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation -- the last integral is an instance of &quot;Information Volume&quot;, an estimate of number of statistically distinguishable distributions in the Bernoulli model, used as a measure of model complexity . A major annoyance is that this integral fails to exist for some popular models, like the geometric distribution family. Perhaps we can fix this by instead estimating the number of observation sequences that are well fit by some distribution in the model? That's the first integral.&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-09-18T20:50:47.637" LastActivityDate="2010-09-19T08:47:07.610" Title="Limit of Gamma integrals" Tags="&lt;gamma-function&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="4906" PostTypeId="2" ParentId="4900" CreationDate="2010-09-18T07:14:57.937" Score="3" ViewCount="0" Body="&lt;p&gt;Let&#xA;$$S=\{(x_1,\ldots,x_m):x_1,\ldots,x_m\textrm{ is a permutation of }1,\ldots m\}.$$&#xA;So for a vector $a=(a_1,\ldots,a_m)\in\mathbb{Z}^n$ we need to show there are&#xA;distinct $x$, $y\in S$ with $x\cdot a\equiv y\cdot a$ (mod $m!$). If this&#xA;wasn't true, $x\cdot a$ runs through all congruence classes modulo $m!$&#xA;as $x$ runs through $S$. Therefore&#xA;$$\left(\sum_{x\in S}x\right)\cdot a\equiv\sum_{k=0}^{m!-1}k\pmod{m!}.$$&#xA;Each coordinate of $X=\sum_{x\in S}x$ equals $(m-1)!\sum_{j=1}^m j$.&#xA;But since $m$ is odd, this is divisible by $m!$. Hence $X\cdot a\equiv0$&#xA;(mod $m!$) but $\sum_{k=0}^{m!-1}k\not\equiv0$ (mod $m!$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T07:14:57.937" CommentCount="2" />
  <row Id="4907" PostTypeId="2" ParentId="4893" CreationDate="2010-09-18T08:23:06.837" Score="2" ViewCount="0" Body="&lt;p&gt;Your intuition is correct that the derivative OPERATOR can be multiplied together in such a manner i.e. $\frac{d}{dx}\frac{d}{dx} = \frac{d^2}{dx^2}$. However, the expression $\frac{dy}{dx} = \frac{d}{dx}(y)$ is an evaluated derivative on the function $y$ we write the $y$ in the numerator with $d$ as a notational convenience. When you evaluate a derivative, it no longer behaves like the derivative operator and is now simply a function. Another way to look at it is if you consider $\frac{d}{dx}$ as a &quot;function&quot; then the product $\frac{d}{dx}\frac{d}{dx}$ is simply the composition of two derivative functions, conversely $\frac{dy}{dx} \frac{dy}{dx}$ is the product of two evaluated functions. So we can restructure your question of follows : does $f(x)*f(x) = f(f(x))$ ? the answer is obviously no. However, $(f\circ f)(x) = f(f(x))$ is true by the definition of composition of functions. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-18T08:48:54.713" LastActivityDate="2010-09-18T08:48:54.713" />
  <row Id="4908" PostTypeId="1" AcceptedAnswerId="4909" CreationDate="2010-09-18T09:01:11.907" Score="4" ViewCount="283" Body="&lt;p&gt;Sometime back i watched the documentary of Andrew Wiles proving the Fermat's Last theorem. A truly inspiring video and i still watch it whenever i am in a depressed mood. There are certain things(infact many) which i couldn't follow and i would like it to be explained here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Taniyama-Shimura conjecture. In the video it's said that that an elliptic curve is a modular form in disguise. I would want someone to explain this statement. I have seen the definition of a Modular form in Wikipedia, but i can't correlate this with an elliptic curve. The definition of an elliptic curve is simply a cubic equation of the form $y^{2}=x^{3}+ax+b$. How can it be a modular form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Next, there was a mention of this Mathematician named &quot;Gerhard Frey&quot; who seems to considered this question of what could happen if there was a solution to the equation, $x^{n}+y^{n}=z^{n}$, and by considering this he constructed a curve which is not modular, contradicting Taniyama-Shimura. If he had constructed such a curve then what was the need for Prof. Ribet to actually prove the Epsilon conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, here i would like to know this answer: How many of you agree with Wiles, that possibly Fermat could have fooled himself by saying that he had a proof of this result? I certainly disagree with his statement. Well, the reason, is just instinct!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:12:57.537" LastActivityDate="2010-11-22T02:12:57.537" Title="A question on FLT and Taniyama Shimura" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;algebraic-number-theory&gt;&lt;elliptic-curves&gt;&lt;modular-forms&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4909" PostTypeId="2" ParentId="4908" CreationDate="2010-09-18T09:18:34.297" Score="6" ViewCount="0" Body="&lt;p&gt;The elliptic curve is not a modular form. The idea that there is&#xA;a modular form $f$ associated to $E$. It satisfies&#xA;$$f(z)=\sum_{n=1}^\infty c_n q^n$$&#xA;where $q=\exp(2\pi i z)$, $c_1=1$ and (with finitely many exceptions)&#xA;for prime $p$, the equation $y^2=x^3+ax=b$ has $p-c_p$ solutions $(x,y)$&#xA;considered modulo $p$. It also satisfies various other conditions&#xA;that I won't spell out (that it's a &quot;newform&quot; for a modular group $\Gamma_0(N)$&#xA;etc.)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T09:18:34.297" CommentCount="2" />
  <row Id="4911" PostTypeId="2" ParentId="4896" CreationDate="2010-09-18T10:52:04.190" Score="2" ViewCount="0" Body="&lt;p&gt;This is essentially the same as Isaac’s comment on &lt;a href=&quot;http://math.stackexchange.com/questions/4896/network-of-streets/4897#4897&quot;&gt;Moron’s answer&lt;/a&gt;, but I’ll post it as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that by “one ‘drives’ in both directions on every street exactly once,” you mean “one ‘drives’ in &lt;em&gt;each&lt;/em&gt; direction on every street exactly once.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A directed graph has an Eulerian circuit (i.e. a circuit which uses every edge exactly once) if and only if it is strongly connected and each vertex has equal in-degree and out-degree.  This fact can be proved in the same way as the well-known fact that an undirected graph has an Eulerian circuit if and only if it is connected and each vertex has even degree.  See a textbook on graph theory for a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your question, let &lt;em&gt;G&lt;/em&gt; be the connected undirected graph representing the network of streets.  You are looking for an Eulerian circuit (or an Eulerian path if you do not require the tour to end at the starting point) in the directed graph &lt;em&gt;G&lt;/em&gt;′ obtained by replacing each edge in &lt;em&gt;G&lt;/em&gt; by a pair of directed edges in both directions.  It is easy to see that the directed graph &lt;em&gt;G&lt;/em&gt;′ satisfies the two conditions above, and therefore &lt;em&gt;G&lt;/em&gt;′ has an Eulerian circuit.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-18T10:52:04.190" />
  <row Id="4912" PostTypeId="2" ParentId="4859" CreationDate="2010-09-18T11:38:49.393" Score="0" ViewCount="0" Body="&lt;p&gt;This is not an answer, but an elementay question about this category ring $R({\cal C})$. Because of Martin's remark, I'll use the product notation: elements of $R({\cal C})$ are tuples &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f= (f_{ab}) \in \prod_{a, b \in {\mathrm ob} {\cal C}} {\cal C}(a,b) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which every $f_{ab} : a \longrightarrow b$ is a morphism of ${\cal C}$. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way: are we still in our universe when we make such a product? I mean: this won't be a &quot;ring&quot; at all, that is: a small ring.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem that Martin has pointed out comes from the fact that, when you multiply two of these tuples $f=(f_{ab})$ and $g= (g_{ab})$ a sum appears in the $ab$-component of $g\cdot f$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_x g_{xb}\circ f_{ax} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it so?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-18T20:05:55.780" LastActivityDate="2010-09-18T20:05:55.780" CommentCount="1" />
  <row Id="4913" PostTypeId="2" ParentId="4893" CreationDate="2010-09-18T11:43:30.850" Score="2" ViewCount="0" Body="&lt;p&gt;Since&#xA;$$f''(a)=\lim_{h\to0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$&#xA;under suitable conditions, I suppose one could conceptualize&#xA;the $d^2 f$ in $d^2 f/dx^2$ as a &quot;second-order&quot; infinitesimal difference:&#xA;i.e., as $f(x+2dx)-2f(x+dx)+f(x)$. Mind you, I've never liked thinking&#xA;in terms of &quot;infinitesimals&quot; :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T11:43:30.850" />
  <row Id="4914" PostTypeId="1" AcceptedAnswerId="4924" CreationDate="2010-09-18T11:54:21.823" Score="1" ViewCount="261" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the Cartesian equation for the perpendicular bisector of the line joining A(2,3) and B(0,6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do I do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-18T20:30:28.980" LastActivityDate="2010-09-18T20:30:28.980" Title="Cartesian Equation for the perpendicular bisector of a line." Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="5" />
  <row Id="4915" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T11:57:45.883" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a hint. The mid point of the line joining $A$ and $B$ is $(1, \frac{9}{2})$. Use this to find the equation of the line passing through this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As, J.M says the product of the slopes of the perpendiculars is $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slope of the line passing through $A$ and $B$ is $\displaystyle m_{1}= \frac{6-3}{0-2} = -\frac{3}{2}$. Therefore the slope of the line perpendicular to it is $m_{2}= \frac{2}{3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the required equation of the line is $(y-\frac{9}{2})=\frac{2}{3}(x-1)$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-18T12:17:31.120" LastActivityDate="2010-09-18T12:17:31.120" />
  <row Id="4916" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T11:59:00.220" Score="3" ViewCount="0" Body="&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Get the slope and the midpoint of the segment joining your two given points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Recall the relationship of the slopes of two perpendicular lines&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the point-slope form of the equation of a line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-18T11:59:00.220" />
  <row Id="4917" PostTypeId="2" ParentId="4908" CreationDate="2010-09-18T13:04:03.790" Score="2" ViewCount="0" Body="&lt;p&gt;The starting point of the link between elliptic curves and modular forms is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a topological point of view, elliptic curves are just (2-dimensional) tori, i.e. products $S^1\times S^1$, where $S^1$ is a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A torus has always an invariant never vanishing tangent field. Dually, one can find a non-vanishing invariant differential form $\omega$ on every elliptic curve $E$. It is a useful exercise to write $\omega$ in terms of the coordinates $x$ and $y$ when $E$ is given as a Weierstrass cubic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a modular parametrization $\pi:X_0(N)\rightarrow E$ you can pull-back the form $\omega$ to $X_0(N)$ and in terms of the coordinate $z$ in the complex upper halfplane $\pi^*(\omega)=f(z)dz$ for some holomorphic function $f(z)$. It is basically immediate that $f(z)$ is a modular form of weight $2$.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-18T13:04:03.790" CommentCount="12" />
  <row Id="4919" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T13:24:42.747" Score="2" ViewCount="0" Body="&lt;p&gt;The perpendicular bisector of the segment $AB$ is the locus of points&#xA;$P$ equidistant from $A$ and $B$, that is $|AP|=|BP|$. It's&#xA;easier to consider the equation $|AP|^2=|BP|^2$ which, when $A=(a,b)$,&#xA;$B=(c,d)$ and $P=(x,y)$ becomes&#xA;$$(x-a)^2+(y-b)^2=(x-c)^2+(y-d)^2$$&#xA;and can be simplified further....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-18T13:24:42.747" CommentCount="2" />
  <row Id="4920" PostTypeId="1" AcceptedAnswerId="4933" CreationDate="2010-09-18T14:52:03.653" Score="2" ViewCount="167" Body="&lt;p&gt;According to my textbook, the second derivative of $y^{2}+xy-x^{2}=9$ is $\frac{90}{(2y+x)^{3}}$. The problem states &quot;Express $\frac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$.&quot; I've tried for two days straight now, and I can't get that answer. I am convinced the book has a typo.&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-09-18T18:15:41.697" Title="Second Derivatives Using Implicit Differentiation" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4921" PostTypeId="2" ParentId="4840" CreationDate="2010-09-18T15:02:21.430" Score="1" ViewCount="0" Body="&lt;p&gt;How close do you have to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you just find the center of the each triangle in space then divide the cutting plane into a 2d grid and do a series of rectangular volume calculations using the length and width of the grid section and the average height of the triangle midpoint above/below that grid section?&lt;/p&gt;&#xA;" OwnerUserId="1582" LastEditorUserId="1582" LastEditorDisplayName="" LastEditDate="2010-09-18T15:22:57.477" LastActivityDate="2010-09-18T15:22:57.477" CommentCount="5" />
  <row Id="4922" PostTypeId="2" ParentId="4920" CreationDate="2010-09-18T15:07:27.587" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: treat $y$ as a function $y(x)$, so differentiating $xy(x)$ should give something like $x y^{\prime}(x)+y(x)$. Differentiate expressions twice, and solve for $y^{\prime\prime}(x)$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-18T15:07:27.587" CommentCount="4" />
  <row Id="4923" PostTypeId="2" ParentId="4869" CreationDate="2010-09-18T16:13:12.483" Score="3" ViewCount="0" Body="&lt;p&gt;Let f:C to C be a complex-valued function. If you want to do ''regular'' integration, by which I assume you mean integrating over 2-dimensional regions in the complex plane, then you would have that the integral of f(x+iy)=u(x,y)+iv(x,y) over some region A in R&lt;sup&gt;2&lt;/sup&gt; equals the integral of the real component over A plus i*integral of the imaginary component of f (by sensible definition). I suspect it turns out that the theory of ''regular'' integration of complex-valued functions of the plane reduces quite literally to the theory of ''regular'' integration of real-valued functions functions of the plane, and hence is of not much interest and not much discussed as a theory of complex integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now line integrals are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can associate with your complex-valued function f a vector-field on the plane Pdx+Qdy given by f(x+iy)=P(x,y)+iQ(x,y) where P and Q are the real-valued real and imaginary parts of f as function of the real plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition that f be differentiable (i.e. the Cauchy-Riemann equations) translates to requiring that the associated vector field have vanishing curl, which means that the vector field Pdx+Qdy is conservative, which in turn means that it has path-invariant line integrals, which are precisely the line integrals of the complex-valued function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This though only works on simply connected domains, but it gives enough reason to consider line integrals of (differentiable) complex-valued functions as something special and worthy of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: if we can think of complex-valued functions as vector-fields, why do we bother with having a theory of complex-valued functions with its own terminology? The answer is that the language of vector fields is only applicable to the simplest bootstrapping case of the complex plane; as soon as you leave that haven (for, say, meromorphic functions that have poles, multiple variables, etc) it all becomes much different and then the vector-fields simply do not reflect what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED:&#xA;For a rigorous exposition of the above sketch, check out the papers on Polya vector fields on Paul Fishback's web-site: &lt;a href=&quot;http://faculty.gvsu.edu/fishbacp/complex/complex.htm&quot; rel=&quot;nofollow&quot;&gt;http://faculty.gvsu.edu/fishbacp/complex/complex.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-09-18T18:33:03.293" LastActivityDate="2010-09-18T18:33:03.293" />
  <row Id="4924" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T16:30:50.033" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ The equation is  $\rm\;\: 2\ (A-B)\cdot (x,y) \;=\; |A|-|B|\;\;\;$ where $\rm\;\;\; |(a,b)| \ =\  a^2 + b^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, if worked out, yields $\rm\;\: (-4,6)\cdot (x,y) \;=\; \;36 \;- 13\;\;\:$ for $\rm\; A = (0,6),\;\; B = (2,3)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, after simplifying, yields the equation $\rm\; 6y =\; 4x+23$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-18T17:55:35.957" LastActivityDate="2010-09-18T17:55:35.957" CommentCount="3" />
  <row Id="4925" PostTypeId="2" ParentId="250" CreationDate="2010-09-18T16:50:06.530" Score="2" ViewCount="0" Body="&lt;p&gt;Morely's trisection theorem -- not discovered until 1899.  &lt;a href=&quot;http://www.mathpages.com/home/kmath376/kmath376.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mathpages.com/home/kmath376/kmath376.htm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trisect the three angles of a triangle. Take the intersection of each trisector with its nearest trisector of the nearest other vertex. These three points form an equilateral triangle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With bisectors, they all meet in a point.  That trisectors should give an equilateral triangle is pretty surprising.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-09-18T16:50:06.530" CommunityOwnedDate="2010-09-18T16:50:06.530" CommentCount="1" />
  <row Id="4926" PostTypeId="2" ParentId="4744" CreationDate="2010-09-18T17:08:01.813" Score="2" ViewCount="0" Body="&lt;p&gt;For more on Hadamard Products or Hadamard Multiplication Theorem look for papers by Louis R. Bragg in American Mathematical Monthly, Jan. 1999, pp 36-42 or SIAM J. Math. Anal., Vol. 17, 1986, pp 220-230 for starters.&lt;/p&gt;&#xA;" OwnerUserId="1974" LastActivityDate="2010-09-18T17:08:01.813" />
  <row Id="4927" PostTypeId="1" AcceptedAnswerId="4929" CreationDate="2010-09-18T17:24:27.460" Score="1" ViewCount="228" Body="&lt;p&gt;How to &quot;show&quot;, that if x=1, then its powers will end with 1, if x=5, it's powers end with 5.&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-18T20:47:05.410" LastActivityDate="2010-09-22T13:48:47.127" Title="If x=1, then its powers end with 1" Tags="&lt;number-theory&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="4928" PostTypeId="2" ParentId="4914" CreationDate="2010-09-18T17:24:44.900" Score="1" ViewCount="0" Body="&lt;p&gt;Generally, &lt;a href=&quot;http://math.stackexchange.com/questions/4914/cartesian-equation-for-the-perpendicular-bisector-of-a-line/4916#4916&quot;&gt;J.M.'s answer&lt;/a&gt; is what I'd suggest at the high school level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is a somewhat common problem on timed math contests (or part of a problem), and in that settings, I'd take advantage of the fact that in the form $ax+by=c$, $\langle a,b\rangle$ is a vector perpendicular to the line:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find the vector $\overrightarrow{AB}=\langle x_a,y_a\rangle$, which is perpendicular to the line you want.&lt;/li&gt;&#xA;&lt;li&gt;The midpoint of $\overline{AB}$ is $A+\frac{1}{2}\overrightarrow{AB}=\langle x_m,y_m\rangle$, which is a point on the line.&lt;/li&gt;&#xA;&lt;li&gt;An equation for the line is $\langle x_a,y_a\rangle\cdot\langle x,y\rangle=\langle x_a,y_a\rangle\cdot\langle x_m,y_m\rangle$ or $x_ax+y_ay=x_ax_m+y_ay_m$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T17:24:44.900" />
  <row Id="4929" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T17:28:35.923" Score="5" ViewCount="0" Body="&lt;p&gt;I am presuming you are talking about the last digit of a number in base 10 being 1 and 5 implying the powers of that number will also end with 1 and 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will give you a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give two numbers whose last digits are 1, what does their product end with?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-18T17:28:35.923" CommentCount="11" />
  <row Id="4930" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T17:29:36.980" Score="11" ViewCount="0" Body="&lt;p&gt;Use mathematical induction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show that $x^1$ ends in 1 (or 5).&lt;/li&gt;&#xA;&lt;li&gt;Show that if $x^n$ ends in 1 (or 5), then $x^{n+1}$ also ends in 1 (or 5).  (Think about multiplying a number than ends in 1 by 1 (or ends in 5 by 5).)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T17:29:36.980" />
  <row Id="4931" PostTypeId="1" AcceptedAnswerId="4936" CreationDate="2010-09-18T18:14:20.677" Score="0" ViewCount="202" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qBfas.png&quot; alt=&quot;http://img683.imageshack.us/img683/7001/mathw.png&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Show that BE is the perpendicular bisector to AC&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle, and therefore said it was not the perpendicular bisector, I'm probably making a silly mistake, so any advice would be helpfull!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot :)!&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-19T03:59:12.470" LastActivityDate="2010-09-19T04:37:25.237" Title="Perpendicular bisector." Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="4932" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T18:14:41.033" Score="1" ViewCount="0" Body="&lt;p&gt;If you know that addition and multiplication are well-defined modulo $n$, then just work (mod 10).&lt;br&gt;&#xA;Let $x_0 \in$ {1,5} be the ones digit of $x$.  Then for any $k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^k \equiv x_0^k \equiv x_0$ (mod 10).&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2010-09-18T18:14:41.033" />
  <row Id="4933" PostTypeId="2" ParentId="4920" CreationDate="2010-09-18T18:15:41.697" Score="2" ViewCount="0" Body="&lt;p&gt;Based on &lt;a href=&quot;http://imgur.com/4DDa7.png&quot; rel=&quot;nofollow&quot;&gt;your work&lt;/a&gt; as linked in the comments on &lt;a href=&quot;http://math.stackexchange.com/questions/4920/second-derivatives-using-implicit-differentiation/4922#4922&quot;&gt;J.M.'s answer&lt;/a&gt;, you've very nearly got it (except for a typo in differentiating *: there's a dy/dx where there should be a d/dx, but the mathematics that follows is correct as if it were d/dx).  The numerator you have is &#xA;$$\begin{align}&#xA;-2 ((2 x - y)^2 &amp;amp;+ (2 x - y) (2 y + x) - (2 y + x)^2)&#xA;\\&#xA;&amp;amp;=-10x^2+10xy+10y^2&#xA;\\&#xA;&amp;amp;=10(y^2+xy-x^2)&#xA;\\&#xA;&amp;amp;=10\cdot9&#xA;\\&#xA;&amp;amp;=90.&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T18:15:41.697" CommentCount="3" />
  <row Id="4934" PostTypeId="2" ParentId="4927" CreationDate="2010-09-18T18:19:05.497" Score="5" ViewCount="0" Body="&lt;p&gt;If you know modular arithmetic: $\rm\; e\equiv 1,5\;\Rightarrow\; e^2\equiv e \;\Rightarrow\; e^k\equiv e\pmod{10}\;$ for all $\rm\; k&gt;0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, use induction: suppose $\rm\;\; e^2 = e + 10j\;$ and  $\rm\; (e + 10 m)^k = e + 10n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $\rm\; (e + 10 m)^{k+1} = (e + 10n) (e + 10m) = e + 10(j+em+en+10mn)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; Note how the use of modular arithmetic has allowed us to rephrase the problem in a completely arithmetical form that makes the induction step utterly trivial - namely that $\rm 1^k \equiv 1$ or, more generally that $\rm\; e^2 \equiv e\;\Rightarrow e^k \equiv e\;$ for $\rm\: k&gt;0$. This is but one very simple illustration of how one may often reduce problems about integers to simpler problems about integers modulo $\rm\:m\:$. Because the integers modulo $\rm\:m\:$ satisfy many of  the same arithmetical laws as the integers (associative, commutative, distributive, etc) - i.e. they form a ring - we can reuse our strong intuition about integers to help discover proofs in this simpler structure integers modulo $\rm\:m\:$. And, if we're lucky, we can sometimes lift these simple solutions from the integers modulo $\rm\:m\:$ up to the integers. If you study abstract algebra you'll see many more examples of problems solved by passing to such simpler structures (technically they are known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Quotient_ring&quot; rel=&quot;nofollow&quot;&gt;quotient objects&lt;/a&gt; - structures modulo a congruence, i.e. a structure-compatible equivalence relation). Said in an informal way, such &lt;strong&gt;reduction to quotient objects&lt;/strong&gt; is a method to &lt;strong&gt;divide and conquer&lt;/strong&gt; mathematically.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-22T13:48:47.127" LastActivityDate="2010-09-22T13:48:47.127" CommentCount="2" />
  <row Id="4935" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:19:44.653" Score="1" ViewCount="0" Body="&lt;p&gt;To show that the segments are perpendicular, consider their slopes.  The product of the slopes of perpendicular lines is -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that BE bisects AC, show that the distance AE is equal to the distance EC.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T18:19:44.653" />
  <row Id="4936" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:21:49.947" Score="1" ViewCount="0" Body="&lt;p&gt;Consider the slopes of AE and EB. AE has a slope of -1, whereas EB has a slope of 1. Two lines on a plane are perpendicular if their slopes multiply to -1, so since -1*1=-1, the lines are in fact perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-18T18:21:49.947" CommentCount="1" />
  <row Id="4937" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:39:39.277" Score="1" ViewCount="0" Body="&lt;p&gt;BE is perpendicular to AC, that is correct. You don't need any analytic geometry to see that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AE = EC, so E is the midpoint of AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AB = CB, so ABC is isosceles triangle with B being the the vertex opposite the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus BE is perpendicular to AC, as the line joining the vertex opposite the base (B) to the midpoint of base (E) is perpendicular to the base (AC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, analytic geometry is probably more useful in similar situations.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-18T19:00:09.837" LastActivityDate="2010-09-18T19:00:09.837" />
  <row Id="4938" PostTypeId="2" ParentId="4869" CreationDate="2010-09-18T18:41:43.833" Score="6" ViewCount="0" Body="&lt;p&gt;Let's suppose, imitating the real case, the we want to integrate the expression&#xA;$f(z) dz$, where $f$ is a function of the complex variable $z$.  What would this mean?&#xA;Well, imitating the real case, we find that we have to form Riemann sums&#xA;$\sum_{i = 1}^{n-1} f(z_i) (z_{i+1} - z_i)$, and then let the absolute value of the differences$| z_{i+1} - z_i|$ tend to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sum is easily seen to be the line integral along the polygonal arc joining $z_1$ to&#xA;$z_n$ via $z_2, z_3, \ldots ,$ of the piecewise constant function whose value is $f(z_i)$ on&#xA;the arc joining $z_i$ to $z_{i+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can form the limit in any reasonable sense, we will get a line integral of $f(z) dz$&#xA;along the curve which is the limit of these polygonal arcs as&#xA;$n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the limit we get will depend not just on the end points $a$ and $b$,&#xA;but (at least &lt;em&gt;a priori&lt;/em&gt;)  on the particular path that arises as the limit of these polygonal arcs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing this, we now see that to define the integral $\int_a^b f(z) dz$, we should&#xA;choose a path joining $a$ to $b$, and choose the $z_i$ to lie along this path.&#xA;The resulting integral will depend on this choice of path &lt;em&gt;a priori&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we are seeing is that although passing from the real to the complex numbers involves&#xA;passing from a one-dimensional to a two-dimensional situation, taking an integral&#xA;of $f(z) dz$ still involves choosing a finite sequence of points and then passing to the&#xA;limit.  In the real case, these points fill out an interval (in the limit), while in the complex case they will fill out a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to write an integral over an area in terms of complex variables, you would&#xA;need to do integrals like $\int f(z) dzd\overline{z}$; this now involves adding up Riemann sums in which the terms involve products $\Delta z\Delta\overline{z}$ of a small change in $z$ and an independent small change in $\overline{z}$.  One computes that this&#xA;is the same as the integral $-2i\int f(z) dxdy$, which is a usual double integral in the plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: If we try to generalize the one real variable case to the one complex variable case,&#xA;we find that we are forced to choose a path along which to take the integral, because&#xA;the complex numbers lie in a two-dimensional plane, while the idea of forming Riemann sums to compute the integral requires us to choose some arc along which the points $z_i$ lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamentally, adding up small changes $\Delta z$ (weighted by values of the function $f$)&#xA;requires looking at changes in $z$ along some one-dimensional object.  Thus $\int f(z) dz$ has to be a line integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding up products of indepenent small&#xA;changes $\Delta z$ and $ \Delta \overline{z}$ (weighted by vaues of the function $f$)  requires moving in two independent directions, and so takes place over a two-dimensional&#xA;region.  Thus $\int f(z) dz d \overline{z}$ has to be a double integral.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-20T23:53:13.067" LastActivityDate="2010-10-20T23:53:13.067" />
  <row Id="4939" PostTypeId="2" ParentId="4931" CreationDate="2010-09-18T18:57:22.877" Score="0" ViewCount="0" Body="&lt;p&gt;You can also use the simple general formula that I mentioned in my &lt;a href=&quot;http://math.stackexchange.com/questions/4914/cartesian-equation-for-the-perpendicular-bisector-of-a-line/4924#4924&quot;&gt;prior post&lt;/a&gt;, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ The equation is  $\rm\;\: 2\ (A-B)\cdot (x,y) \;=\; |A|-|B|\;\;\;$ where $\rm\;\;\; |(a,b)| \ =\  a^2 + b^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, if worked out, yields $\rm\;\: (-8,8)\cdot (x,y) \;=\; \;17 \;- 9\;\;\:$ for $\rm\; A = (4,-1),\;\; B = (0,3)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, after simplifying, yields the equation $\rm\;\;\: y \;=\; x-1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-18T18:57:22.877" />
  <row Id="4941" PostTypeId="1" AcceptedAnswerId="4943" CreationDate="2010-09-18T19:08:10.340" Score="2" ViewCount="68" Body="&lt;p&gt;I was reviewing my class notes and found the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The name 'torsion' comes from topology and refers to spaces that are twisted, ex. Mobius band&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our notes we used the following definition for torsion element and torsion module:&#xA;An element m of an R-module M is called a torsion element if $rm=0$ for some $r\in R$. &#xA;A torsion module is a module which consists solely of torsion elements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between torsion modules and twisted spaces? Was the definition of torsion module somehow motivated from topological considerations of twisted spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really see any obvious connection. I'm taking my first topology class this semester, so I apologize if this is something you learn about later in courses like algebraic topology, but I haven't been able to find any explanation of this.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-09-18T19:17:51.037" LastActivityDate="2010-09-18T19:17:51.037" Title="Relationship between torsion modules and topology" Tags="&lt;soft-question&gt;&lt;terminology&gt;&lt;intuition&gt;&lt;topology&gt;&lt;modules&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4942" PostTypeId="1" CreationDate="2010-09-18T19:08:15.440" Score="0" ViewCount="129" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The perpendicular bisector of the line&#xA;  joining A(0,1) and C(-4,7) intersects&#xA;  the x-axis at B and the y-axis at D.&#xA;  Find the area of the quadrilateral&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="1976" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-19T03:48:57.737" LastActivityDate="2010-09-19T06:33:39.527" Title="Area of a quadrilateral" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="4943" PostTypeId="2" ParentId="4941" CreationDate="2010-09-18T19:14:21.010" Score="6" ViewCount="0" Body="&lt;p&gt;When you compute the homology groups of &quot;twisted&quot; spaces (which are abelian groups), you (sometimes) find that they contain non-zero torsion elements; furthermore, the presence of these particular elements in the homology is due to the twisting (in that, when you compute the homology groups, you see that is the twisting in the space that causes the calculation to give rise to torsion elements).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you haven't studied algebraic topology yet, I won't say more here.  Hopefully, while necessarily vague, the above description gives you some feeling for the meaning of the remark in your notes.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-18T19:14:21.010" />
  <row Id="4944" PostTypeId="2" ParentId="4941" CreationDate="2010-09-18T19:15:31.693" Score="5" ViewCount="0" Body="&lt;p&gt;The definition of torsion in modules is a generalization of the definition of torsion in $\mathbb{Z}$-modules, e.g. abelian groups.  Torsion in abelian groups refers to elements of finite order, and this in turn relates to topology because to any topological space we can associate abelian groups called (integral) &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;homology groups&lt;/a&gt;, and torsion in these groups is suggestive of a kind of &quot;twistedness&quot; in the space.  The simplest example of this is in the first integral homology of closed surfaces; the group has torsion if and only if the surface is non-orientable, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Klein_bottle&quot; rel=&quot;nofollow&quot;&gt;Klein bottle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-18T19:15:31.693" />
  <row Id="4945" PostTypeId="2" ParentId="4942" CreationDate="2010-09-18T19:17:41.793" Score="1" ViewCount="0" Body="&lt;p&gt;Tag the question as homework and show what you've already tried, otherwise it will be (rightfully) closed soon.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Now, we just had &lt;a href=&quot;http://math.stackexchange.com/questions/4931/perpendicular-bisector&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;find middle of $\overrightarrow{AC}$&lt;/li&gt;&#xA;&lt;li&gt;create a line's equation $b: y = m\cdot x + c$ for the bisector&lt;/li&gt;&#xA;&lt;li&gt;compute the intersections of $b$ and the coordinate axis&lt;/li&gt;&#xA;&lt;li&gt;compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadrilateral#Area_of_a_convex_quadrilateral&quot; rel=&quot;nofollow&quot;&gt;area&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-18T19:17:41.793" CommentCount="1" />
  <row Id="4946" PostTypeId="2" ParentId="4942" CreationDate="2010-09-18T19:19:26.537" Score="3" ViewCount="0" Body="&lt;p&gt;The perpendicular bisector of the segment $AC$, by definition, passes through the midpoint of the segment. Look at your drawing. The segment $BD$ is a base to the triangles $BCD$ and $BAD$. Note also that the heights of the triangles are equal.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-18T19:19:26.537" />
  <row Id="4947" PostTypeId="1" AcceptedAnswerId="4954" CreationDate="2010-09-18T19:34:37.793" Score="0" ViewCount="106" Body="&lt;p&gt;I have to show that any line parallel to Ax+ By + c =0 is of the form Ax + By + k =0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I show this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-18T20:29:49.010" LastActivityDate="2010-09-19T11:34:27.870" Title="Equation of a line" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;analytic-geometry&gt;" AnswerCount="4" />
  <row Id="4948" PostTypeId="1" AcceptedAnswerId="4951" CreationDate="2010-09-18T19:35:35.967" Score="1" ViewCount="122" Body="&lt;p&gt;&lt;strong&gt;The Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;70% of all vehicles pass inspection. Assuming vehicles pass or fail independently. What is the probability:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) exactly one of the next 3 vehicles  passes&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) at most 1 of the next 3 vehicles  passes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to a) is .189. The way I calculated it was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(success) * P(fail) * P(fail) + &#xA;P(fail) * P(success) * P(fail) + &#xA;P(fail) * P(fail) * P(success) = .7*.3*.3 + .3*.7*.3 + .3*.3*.7 = .189&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I summed the 3 possible permutations of 1 success and 2 failures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For b) the answer is .216. To get that answer you take your answer to a) and add the probability of exactly 0 successes which is P(fail) * P(fail) * P(fail) =&gt; &lt;code&gt;.189 + .3*.3*.3 = .216&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why the probability of exactly 0 successes doesn't follow the pattern of exactly 1 success. Why doesn't the &quot;formula&quot; work:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P(fail) * P(fail) * P(fail) + &#xA;P(fail) * P(fail) * P(fail) + &#xA;P(fail) * P(fail) * P(fail) = .3*.3*.3+.3*.3*.3+.3*.3*.3 = .081&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=&amp;gt; .189 + .081 = .27 (not .216)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm wondering if I calculated the answer to a) the wrong way, and it was merely a coincidence that I got the right answer!&lt;/p&gt;&#xA;" OwnerUserId="1977" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-19T01:57:52.823" LastActivityDate="2010-09-19T01:57:52.823" Title="What is wrong with my reasoning?" Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="4949" PostTypeId="2" ParentId="4947" CreationDate="2010-09-18T19:38:46.163" Score="3" ViewCount="0" Body="&lt;p&gt;Examples don't show it in any case. You cannot prove that something holds for infinitely many things by doing nothing except checking finitely many examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do is remember that two lines are parallel if and only if they have the same slope or they are both vertical (of the form $x=k$ for some constant $k$). Then figure out what the slope of an arbitrary line $\alpha x+\beta y+\gamma=0$ would be in terms of $\alpha$, $\beta$, and $\gamma$ (careful when dividing by constants to consider what happens if the constant you are dividing by is equal to zero). Then see what this means if you want to find all lines $\alpha x+\beta y+\gamma = 0$ that are parallel to a given line $Ax+By+c=0$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-18T19:46:31.370" LastActivityDate="2010-09-18T19:46:31.370" CommentCount="2" />
  <row Id="4950" PostTypeId="2" ParentId="4947" CreationDate="2010-09-18T19:44:09.873" Score="0" ViewCount="0" Body="&lt;p&gt;You can solve the following system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{cases}&#xA;Ax+By+c=0,\\&#xA;Ax+By+k=0&lt;br&gt;&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you subtract the second line from the firs, you obtain $c-k=0$. Then, if $c\ne k$, you have no solution: the lines are parallel.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-18T19:47:31.967" LastActivityDate="2010-09-18T19:47:31.967" CommunityOwnedDate="2010-09-18T19:44:43.477" CommentCount="2" />
  <row Id="4951" PostTypeId="2" ParentId="4948" CreationDate="2010-09-18T19:44:42.060" Score="6" ViewCount="0" Body="&lt;p&gt;The case that the &lt;em&gt;first&lt;/em&gt; car passes and the next two fail is &lt;em&gt;different&lt;/em&gt; from the case that the first car fails, the second passes, and the third fails. But there is only one case in which all cars fail: namely, &lt;em&gt;all cars fail&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it with coin tosses: if you flip a penny, a nickel, and a dime (different coins, in case you are not in the U.S. or Canada), then there is only &lt;em&gt;one&lt;/em&gt; way for all three coins to come out tails,  namely, the penny comes out tails, the nickel comes out tails, and the dime comes out tails. But there are three ways in which exactly one coin comes out heads: you can have the penny come out heads and the nickel and dime come out tails; you can have the nickel come out heads and the penny and dime come out tails; or you can have the dime come out heads and the penny and nickel come out tails. These are three different outcomes, but you do not have three different ways in which the three can come out tails.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-18T19:54:25.133" LastActivityDate="2010-09-18T19:54:25.133" />
  <row Id="4952" PostTypeId="2" ParentId="4948" CreationDate="2010-09-18T19:46:02.347" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not too confident with my probability, but I'll try my hand at an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first part, you are considering the next 3 trials as successive events so you can have a success in either the first spot, second spot, or third spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part b, you require all 3 events to be failures so there is only 1 way that the 3 failures can occur in the next 3 events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make any sense? I never felt probability was my strong point :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-09-18T19:46:02.347" />
  <row Id="4953" PostTypeId="2" ParentId="4948" CreationDate="2010-09-18T19:48:54.787" Score="3" ViewCount="0" Body="&lt;p&gt;In part a, there are exactly three ways for one out of three cars to pass, they are the three possibilities that you added. But there is only one way for all the cars to fail! The first car must fail, the second car must fail, and the third car must fail. Since there is only one way for this to happen you only consider this one probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw there are also three ways for exactly two cars to pass and only one way for all three of them to pass. Hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-09-18T19:48:54.787" />
  <row Id="4954" PostTypeId="2" ParentId="4947" CreationDate="2010-09-18T20:16:31.697" Score="1" ViewCount="0" Body="&lt;p&gt;(Based on &lt;a href=&quot;http://math.stackexchange.com/questions/4947/equation-of-a-line/4950#4950&quot;&gt;zar's answer&lt;/a&gt; and Moron's comment there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two lines $Ax+By+C=0$ and $Dx+Ey+F=0$ are parallel, then they have no points of intersection (or infinitely many if you allow a line to be parallel to itself).  In either case, the system of equations &#xA;$$\begin{cases}&#xA;Ax+By+C=0,\\&#xA;Dx+Ey+F=0 &#xA;\end{cases}$$&#xA;will not have a single, unique solution.  The system of equations is equivalent to&#xA;$$\begin{bmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{bmatrix}\cdot\begin{bmatrix}&#xA;x\\&#xA;y&#xA;\end{bmatrix}+\begin{bmatrix}&#xA;C\\&#xA;F&#xA;\end{bmatrix}=0$$&#xA;or&#xA;$$\begin{bmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{bmatrix}\cdot\begin{bmatrix}&#xA;x\\&#xA;y&#xA;\end{bmatrix}=\begin{bmatrix}&#xA;-C\\&#xA;-F&#xA;\end{bmatrix}.$$&#xA;This matrix equation has a single unique solution if and only if the coefficient matrix $\begin{bmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{bmatrix}$ is invertible, which is true if and only if its determinant is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the lines are parallel, then &#xA;$$\begin{vmatrix}&#xA;A &amp;amp; B\\ &#xA;D &amp;amp; E&#xA;\end{vmatrix}=AE-BD=0$$&#xA;or&#xA;$$\frac{A}{D}=\frac{B}{E}$$&#xA;which means that $\frac{A}{D}(Dx+Ey+F)=Ax+By+\frac{FA}{D}=0$, so $Dx+Ey+F=0$ can be written in the form $Ax+By+k=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: from Arturo Magidin's comment below, to handle the cases where $D=0$ or $E=0$ (in which $A/D = B/E$ makes no sense):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $D=0$ then $A=0$ or $E=0$, but you cannot have both $D$ and $E$ zero. So $D=0$ implies $A=0$, and $B\neq 0\neq E$. Similarly, if $E=0$, then $B=0$ and $A\neq 0 \neq D$. And you get the similar conclusion without forgetting about horizontal and vertical lines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="72" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-19T03:49:55.520" LastActivityDate="2010-09-19T03:49:55.520" CommentCount="2" />
  <row Id="4955" PostTypeId="2" ParentId="4942" CreationDate="2010-09-18T20:21:36.047" Score="1" ViewCount="0" Body="&lt;p&gt;If the quadrilateral is ABCD--that is, if AC and BD are its diagonals--then the quadrilateral has perpendicular diagonals, so its area is half the product of the lengths of the diagonals (which you can find from the coordinates of A, B, C, and D).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-18T20:21:36.047" />
  <row Id="4956" PostTypeId="1" AcceptedAnswerId="4958" CreationDate="2010-09-18T20:36:46.120" Score="-6" ViewCount="244" Body="&lt;p&gt;Suppose in the plane, $\mathbb{R}^{2}$ we are allowed to move from $(x, y)$ to one of the four points $(x, y \pm 2x)$, $(x \pm 2y, y)$. Prove that if we start at the point $(1, \sqrt{2})$, then we cannot return there after finitely many moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a problem asked in the Kurschak competition, in the year 1989. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; As Bill says,  One is not allowed to reverse the prior move.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-18T21:08:29.097" LastActivityDate="2010-09-19T16:04:45.893" Title="Not returning to the same point after finitely many moves in $\mathbb{R}^{2}$." Tags="&lt;geometry&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="18" />
  <row Id="4957" PostTypeId="2" ParentId="4905" CreationDate="2010-09-18T20:52:55.890" Score="2" ViewCount="0" Body="&lt;p&gt;Limit (and even asymptotic series) of your integral can be calculated using &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%27s_method&quot; rel=&quot;nofollow&quot;&gt;Laplace's method&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;. The solution can be done in 5 steps:&lt;br&gt;&#xA;1. Prove, that integral over $[0,1/n]$ can be neglected. In order to do this, use Stirling's approximation for all factorials except $(nx)!$, which can be approximated by a constant.&lt;br&gt;&#xA;2. Prove, that integral over $[1/n,\delta]$ can be neglected (for $\delta&amp;lt;1/2$). In order to do this, use Stirling's approximation for all factorials (note, that $z!/(\sqrt{2 \pi z} (z/e)^z))$ is bounded from both sides by positive constants if $z&gt;=1$).&lt;br&gt;&#xA;3. Using symmetry you know, that your limit is equal to $\lim_{n\to\infty}\int_{\delta}^{1-\delta}dx(\dots)$.&lt;br&gt;&#xA;4. Use Stirling's approximation to approximate factorials. You will get the following:&lt;br&gt;&#xA;$$\int_{\delta}^{1-\delta}dx\frac{n!2^{-n}n}{(xn)!(n-xn)!\sqrt{x(1-x)}}=(1+O(n^{-1}))\sqrt{\frac{n}{2\pi}}\cdot$$&#xA;$$\int_{\delta}^{1-\delta} x^{-1}(1-x)^{-1}exp\bigl(-n(x\ln x+(1-x)\ln(1-x)+\ln 2)\bigr)dx$$&lt;br&gt;&#xA;5. Apply Laplace's method to the last integral and get it's asymptotic.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-18T20:52:55.890" CommentCount="2" />
  <row Id="4958" PostTypeId="2" ParentId="4956" CreationDate="2010-09-18T21:22:52.613" Score="4" ViewCount="0" Body="&lt;p&gt;It is not hard to see that no nontrivial element of $\text{GL}_2(\mathbb{Z})$ can fix $(1, \sqrt{2})$.  The problem is therefore equivalent to showing that the group generated by $\left[ \begin{array}{cc} 1 &amp;amp; 2 \\ 0 &amp;amp; 1 \end{array} \right]$ and $\left[ \begin{array}{cc} 1 &amp;amp; 0 \\ 2 &amp;amp; 1 \end{array} \right]$ is free on these two generators.  This group injects into $\text{PSL}_2(\mathbb{Z})$ and its image there is (if you prefer, also the image of) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_subgroup&quot; rel=&quot;nofollow&quot;&gt;congruence subgroup&lt;/a&gt; $\Gamma(2)$ and there is a topological proof that it is free based on its action on the upper half plane which I sketch very briefly &lt;a href=&quot;http://qchu.wordpress.com/2010/03/12/fractional-linear-transformations-and-elliptic-curves/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  (The basic idea is that a point in $\mathbb{H}/\Gamma(2)$ is the same thing as an elliptic curve equipped with a choice of Legendre normal form $y^2 = x(x - 1)(x - \lambda)$.  The moduli space of all such curves is $\mathbb{C} - \{0, 1 \}$, which has universal cover $\mathbb{H}$ and fundamental group the free group on two generators.)  Presumably more elementary proofs are known.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-18T21:22:52.613" CommentCount="1" />
  <row Id="4959" PostTypeId="1" AcceptedAnswerId="4991" CreationDate="2010-09-18T21:26:36.597" Score="1" ViewCount="72" Body="&lt;p&gt;In the area of convex analysis and the area of optimization in their general sense, are convex subsets assumed to be in vector spaces or topological vector spaces? Are convex functions defined to be from convex subsets in vector spaces or topological vector spaces to general field or some special field such as $\mathbb{R}$ or $\mathbb{C}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now instead of considering the general sense, what are the special cases that have been extensively studied and have significance in application?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-09-19T13:27:55.917" Title="study objects of convex analysis and optimization" Tags="&lt;optimization&gt;&lt;convex-analysis&gt;&lt;convex-optimization&gt;" AnswerCount="1" />
  <row Id="4960" PostTypeId="2" ParentId="4106" CreationDate="2010-09-18T22:31:21.533" Score="1" ViewCount="0" Body="&lt;p&gt;This is a correction of the answer of David Bar Moshe &lt;strong&gt;edit:&lt;/strong&gt; and generalizing it for the case, when $\mathbf{A}$ is degenerate. Using formulas for Fourier and inverse Fourier transforms we can write&#xA;$$f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y f(y)e^{iky}. (1)$$&#xA;Suppose $S_d(y)=$&quot;$S_d$ shifted by $(y,0,\dots,0)$&quot; and&#xA;$f(y)=\int_{\mathbf{x} \in S_d(y)} \mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x}).$&#xA;Then using the formula (1) we will get&#xA;$$N^{-1}=f(0) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y f(y)e^{iky}  =$$&#xA;$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k\int_{-\infty}^{\infty} \mathrm{d}y \int_{\mathbf{x} \in S_d(y)}\mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + iky) =$$&#xA;$$\frac{\|\mathbf{v}\|}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d}\mathrm{d}\mathbf{x} \exp(-\mathbf{x}^T \mathbf{A}\mathbf{x} + ik \mathbf{v}^T \mathbf{x}),$$&#xA;where $\mathbf{v}$ is vector, orthogonal to the plane we are interested in. For example, we can take $\mathbf{v}=(1,\dots,1)$. Note $\|\mathbf{v}\|$ in the last integral, appearing from change of coordinates from $y$ and coordinates in $S_d$ to coordinates in $\mathbb{R}^d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $N^{-1}=$&#xA;$$\frac{\|\mathbf{v}\|}{2\pi}\int_{-\infty}^{\infty} \mathrm{d}k \int_{\mathbf{x} \in \mathbb{R}^d}\mathrm{d}\mathbf{x} \exp(-(\mathbf{x}+\frac{i}{2}k\mathbf{A}^{-1}\mathbf{v})^T \mathbf{A}(\mathbf{x}+\frac{i}{2}k\mathbf{A}^{-1}\mathbf{v}) - \frac{k^2}{4} \mathbf{v}^T \mathbf{A}^{-1}\mathbf{v})=$$&#xA;$$\frac{\|\mathbf{v}\|}{2\pi} \frac{\pi^{d/2}}{\sqrt{\det{A}}} \frac{2\sqrt{\pi}}{\sqrt{\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}}}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\det{A}}\sqrt{\mathbf{v}^T \mathbf{A}^{-1}\mathbf{v}}}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\mathbf{v}^T \mathbf{C}\mathbf{v}}},$$&#xA;where $\mathbf{C}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjugate_matrix&quot; rel=&quot;nofollow&quot;&gt;adjugate matrix&lt;/a&gt; to $\mathbf{A}$. Initial integral and the answer both are continuous in $\mathbf{A}$ when $A$ restricted to $S_d$ is not degenerate. From this we can conclude, that equation&#xA;$N^{-1}=\frac{\|\mathbf{v}\| \pi^{(d-1)/2}}{\sqrt{\mathbf{v}^T \mathbf{C}\mathbf{v}}}$&#xA;is true whenever $\mathbf{A}$ restricted to $S_d$ is not degenerate (i.e. even if $\mathbf{A}$ is degenerate).&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-09-18T23:16:59.507" LastActivityDate="2010-09-18T23:16:59.507" />
  <row Id="4961" PostTypeId="1" AcceptedAnswerId="4971" CreationDate="2010-09-18T23:54:21.547" Score="16" ViewCount="877" Body="&lt;p&gt;I was just looking at a calculus textbook preparing my class for next week on complex numbers. I found it interesting to see as an exercise a way to calculate the usual freshman calculus integrals $\int e^{ax}\cos{bx}\ dx$ and $\int e^{ax}\sin{bx}\ dx$ by taking the real and imaginary parts of the &quot;complex&quot; integral $\int e^{(a + bi)x} \ dx$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is if you know of other &quot;relatively interesting&quot; results that can be obtained easily by using complex numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be something that one can present to engineering students taking the usual calculus sequence, but I'm also interested in somewhat more advanced examples (if they're available under the condition that the process to get them is somewhat easy, or not too long). Thank you all.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-27T04:38:14.523" CommunityOwnedDate="2010-09-19T00:13:12.210" Title="Interesting results easily achieved using complex numbers" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="17" CommentCount="5" FavoriteCount="13" />
  <row Id="4962" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T00:16:20.763" Score="2" ViewCount="0" Body="&lt;p&gt;If you feel like picking up a good book on this topic, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691118221&quot; rel=&quot;nofollow&quot;&gt;&quot;Dr. Euler's Fabulous Formula&quot; by Nahin&lt;/a&gt; provides many beautiful applications of complex numbers to other parts of mathematics (he is an electrical engineer).&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-20T23:56:06.797" LastActivityDate="2010-09-20T23:56:06.797" CommunityOwnedDate="2010-09-19T00:16:20.763" CommentCount="2" />
  <row Id="4963" PostTypeId="1" AcceptedAnswerId="5582" CreationDate="2010-09-19T00:51:02.560" Score="6" ViewCount="222" Body="&lt;p&gt;Suppose $A$ is a commutative algebra over $\mathbb{R}$ with unity. $\mathbb{R}$-linear map $\xi\colon A\to A$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivation_%28abstract_algebra%29&quot; rel=&quot;nofollow&quot;&gt;derivation&lt;/a&gt; of $A$ iff $\xi(ab)=a\xi(b)+\xi(a)b$ for any $a,b\in A$.&lt;br&gt;&#xA;If $\gamma\colon \mathbb{R}\to A$, $a\in A$, then we say, that $a=\frac{\partial}{\partial t}| _ {t=\tau} \gamma(t)$ iff $h(a) =\frac{\partial}{\partial t}|_{t=\tau} h(\gamma(t))$ for any $\mathbb{R}$-linear map $h\colon A\to\mathbb{R}$.&lt;br&gt;&#xA;Suppose $\xi$ is a derivation of $A$. Then $\Phi\colon A\times\mathbb{R}\to A$ is it's flow iff $\Phi(a,0)=a$ for any $a\in A$ and&lt;br&gt;&#xA;$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt; (with sketches of the proofs)&lt;br&gt;&#xA;1. Algebra $C^\infty(M)$ of smooth functions on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed_manifold&quot; rel=&quot;nofollow&quot;&gt;closed manifold&lt;/a&gt; $M$ --- yes (if I haven't made a mistake), any derivation possesses a flow. This, I believe, can be checked using &lt;a href=&quot;http://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem&quot; rel=&quot;nofollow&quot;&gt;Picard-Lindelof theorem&lt;/a&gt;.&lt;br&gt;&#xA;2. Algebra $C^\infty((0,1))$ of smooth functions on an interval --- no, $\frac{\partial}{\partial x}$ does not possess a flow.&lt;br&gt;&#xA;3. Algebra $C^\infty([0,1])$ of smooth functions on a segment --- yes, any derivation possesses a flow, but it's not always unique (for example, for $\frac{\partial}{\partial x}$ it is not). In order to prove this, one can consider an embedding of $[0,1]$ to some closed manifold $N$ and prolong any function from $[0,1]$ to $N$. Then use example 1.&lt;br&gt;&#xA;4. Algebra $C^\infty(\mathbb{R})$ --- no, because it is isomorphic to the algebra from example 2.&lt;br&gt;&#xA;5. Algebra $\mathbb{R}[x]$ --- no. In order to prove this one can consider derivation $\xi=x^2\frac{\partial}{\partial x}$ and manually solve equation (1) for $a=x$. Any solution locally should be of the form $\frac{x}{1+xt}$. It is not in $\mathbb{R}[x]$.&lt;br&gt;&#xA;6. Algebra $\mathbb{R}[x,y]/(x^2+y^2-1)$ --- no. In order to prove this take $\xi = y (x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x})=\sin(\varphi)\frac{\partial}{\partial \varphi}$ and solve equation (1) manually (locally) in polar coordinates (take, for example, $a=y$). Check that the answer is not a polynomial.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-09-19T19:54:08.407" LastActivityDate="2010-09-27T18:24:31.163" Title="In which commutative algebras does any derivation possess a flow?" Tags="&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;&lt;differential-equations&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="4964" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:01:09.213" Score="4" ViewCount="0" Body="&lt;p&gt;You might get some ideas from this MathOverflow thread:&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/30156/demystifying-complex-numbers/30185&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/30156/demystifying-complex-numbers/30185&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T03:59:19.373" LastActivityDate="2010-09-21T03:59:19.373" CommunityOwnedDate="2010-09-19T01:01:09.213" />
  <row Id="4965" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:01:49.913" Score="21" ViewCount="0" Body="&lt;p&gt;Without complex numbers, it's a mystery why the power series for $1/(1+ x^2)$ centered at the origin has radius of convergence 1.  The function is infinitely differentiable, no strange behavior at 1 or -1, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the complex domain, there are singularities at $\pm i$ which explains why the radius of convergence is 1.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-19T01:01:49.913" CommunityOwnedDate="2010-09-19T01:01:49.913" CommentCount="3" />
  <row Id="4966" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:54:08.453" Score="3" ViewCount="0" Body="&lt;p&gt;The prime number theorem can be proved without complex analysis, but the elementary proof is generally considered much more difficult than the standard one (which deduces it from analytic properties of the zeta-function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if Dirichlet's theorem on primes in arithmetic progressions can be proved without complex analysis (one special case can using cyclotomic polynomials), but the analytic proof is the only one I've seen.  &lt;strong&gt;Edit:&lt;/strong&gt; Adrian below says that you can avoid complex analysis, but it won't look as nice. KCd has given references to specific papers. In either case, note that the analytic proof chronologically preceded the elementary one.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-09-19T13:54:06.360" LastActivityDate="2010-09-19T13:54:06.360" CommunityOwnedDate="2010-09-19T01:54:08.453" CommentCount="4" />
  <row Id="4967" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T01:58:33.013" Score="1" ViewCount="0" Body="&lt;p&gt;Since one of my interests is the geometry of plane curves, the examples I'll be giving reflect this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, a lot of manipulative advantages occur when using the Argand form of a plane curve $z=f(t)+ig(t)$ instead of the parametric form $(x\;\;y)=(f(t)\;\;g(t))$: translation is the addition of an appropriate complex quantity, rotation corresponds to a multiplication by a factor $\exp(i\theta)$, and compare the formula for the curvature of a parametric curve (formula 13 &lt;a href=&quot;http://mathworld.wolfram.com/Curvature.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) with the formula for the curvature for a curve in Argand form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa=\frac{\Im(\bar{z}^{\prime}z^{\prime\prime})}{|z^{\prime}|^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll stop here, because at this point I should refer you to Zwikker's &quot;Advanced Plane Geometry&quot;, it's an oldie but goodie.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-19T01:58:33.013" CommunityOwnedDate="2010-09-19T01:58:33.013" CommentCount="1" />
  <row Id="4968" PostTypeId="1" AcceptedAnswerId="4969" CreationDate="2010-09-19T02:02:47.307" Score="4" ViewCount="129" Body="&lt;p&gt;We know this is true for commutative ring, but if $S\subset R$ is a left and right Ore set, and $S^{-1}R$ its localization by this Ore set, is this always a flat $R$-module?&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-26T05:54:41.077" LastActivityDate="2010-09-27T15:35:52.190" Title="Are localized rings always flat as R-modules?" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4969" PostTypeId="2" ParentId="4968" CreationDate="2010-09-19T02:08:13.267" Score="5" ViewCount="0" Body="&lt;p&gt;This is Proposition 2.1.16 in McConnel+Robson's book on &lt;em&gt;Noncommutative Noetherian Rings&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-19T02:08:13.267" CommentCount="2" />
  <row Id="4970" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T02:13:45.607" Score="2" ViewCount="0" Body="&lt;p&gt;Yet another example close to my heart: one of the &lt;a href=&quot;http://www.siam.org/books/100digitchallenge/&quot; rel=&quot;nofollow&quot;&gt;hundred-dollar hundred-digit problems&lt;/a&gt; asked for the value of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\lim_{\varepsilon\to 0} \int_{\varepsilon}^1 \frac1{t}\cos\left(\frac{\ln\;t}{t}\right) \mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution was to transform it into an oscillatory integral with infinite limit, but the slickest solution to this problem was to use a contour in the complex plane that side-stepped the numerical difficulties with having an oscillatory integrand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\Re\left(\int_0^1 t^{i/t-1} \mathrm{d}t\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by using an appropriate contour, even the humble trapezoidal rule suffices to handle the integral.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-19T02:13:45.607" CommunityOwnedDate="2010-09-19T02:13:45.607" CommentCount="2" />
  <row Id="4971" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T02:34:26.107" Score="20" ViewCount="0" Body="&lt;p&gt;There are too many examples to count.  Let me just mention one that is particularly concrete: how many subsets of an $n$-element set have a cardinality divisible by $3$ (or any positive integer $k$)?  In other words, how do we evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\lfloor \frac{n}{3} \rfloor} {n \choose 3k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in closed form?  Although the statement of this problem does not involve complex numbers, the answer does: the key is what is known in high school competition circles as the roots of unity filter and what is known among real mathematicians as the discrete Fourier transform.  Starting with the generating function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1 + x)^n = \sum_{k=0}^{n} {n \choose k} x^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we observe that the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \omega^k + \omega^{2k} = \begin{cases} 3 \text{ if } 3 | k \\ 0 \text{ otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega = e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{ \lfloor \frac{n}{3} \rfloor} {n \choose 3k} = \frac{(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $1 + \omega = -\omega^2$ and $1 + \omega^2 = - \omega$, this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{ \lfloor \frac{n}{3} \rfloor} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega^2)^n}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula can be stated without complex numbers (either by using cosines or listing out cases) but both the statement and the proof are much cleaner with it.  More generally, complex numbers make their presence known in combinatorics in countless ways; for example, they are pivotal to the theory of asymptotics of combinatorial sequences.  See, for example, Flajolet and Sedgewick's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-19T02:34:26.107" CommunityOwnedDate="2010-09-19T02:34:26.107" CommentCount="2" />
  <row Id="4973" PostTypeId="2" ParentId="4931" CreationDate="2010-09-19T04:12:27.713" Score="1" ViewCount="0" Body="&lt;p&gt;I don't understand why do you say &quot;I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle&quot;, because, to me at least, Pythagoras tells me that it &lt;em&gt;is&lt;/em&gt; a right angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, you must know that Pythagoras' theorem actually is an &quot;if and only if&quot;. That is, if you have a triangle with two sides named $u$ and $v$, then the third one is $u-v$ and Pythagoras says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u\cdot v = 0 \qquad \Longleftrightarrow \qquad \| u \|^2 + \|v\|^2 = \|u-v\|^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if sides (cathetuses) $u$ and $v$ are perpendicular to each other, &lt;em&gt;then&lt;/em&gt; the sum of their squares is equal to the square of the third side (hypotenuse) $u-v$. But the reciprocal is also true, if the equality on the right hand side of the displayed formula above holds, &lt;em&gt;then&lt;/em&gt; $u$ and $v$ are perpendicular to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is easy if you have already seen some Linear Algebra (if not, you may skip what follows, and go for the actual computations for your problem down below). So, the proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|u\|^2+  \|v\|^2 - \|u-v\|^2 = u\cdot u + v\cdot v - (u-v)\cdot (u-v) = -2u\cdot v \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your problem, let's compute the three sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|be\| = \sqrt{3^2 + 3^2} = \sqrt{18}     \ , \quad \|ce\|= \sqrt{2^2 + 2^2} = \sqrt{8}   \quad \text{and} \quad \|bc\|= \sqrt{1^2 + 5^2} = \sqrt{26} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the quantities appearing in the right hand side of the &quot;extended&quot; Pythagoras' theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\|be\|^2 + \|ce\|^2 = 18 + 8 = 26 \qquad \text{and} \qquad \|bc\|^2 = 26 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $be$ and $ce$ are perpendicular to each other.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-19T04:12:27.713" />
  <row Id="4974" PostTypeId="2" ParentId="4931" CreationDate="2010-09-19T04:20:30.987" Score="2" ViewCount="0" Body="&lt;p&gt;A meta-analysis: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation with Pythagoras' theorem should in fact work for the points indicated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If done correctly it has to be consistent with any correct method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the result of the Q &amp;amp; A is not to treat the posted solutions as correct (maybe because there are several of them that all support the problem statement, sound authoritative, are posted by high-ranked math.SE users, or you checked the reasoning in detail, or whatever) but simultaneously disregard the original computational evidence that another equally valid principle seems to give a different answer.  Tracing back to locate the source of the inconsistency should be worthwhile.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-19T04:20:30.987" />
  <row Id="4975" PostTypeId="1" CreationDate="2010-09-19T04:28:05.337" Score="2" ViewCount="149" Body="&lt;p&gt;For context I am developing a ray-tracer for a computer science class, and want to implement some more advanced shapes than just spheres.  So while this is related to schoolwork, I'm not asking you to do my work for me, the work is implementing the programming, and it's the math I don't understand, so I'm just looking for help understanding how the math works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand how to calculate the intersection point, and the normal vector from that point, of several &lt;a href=&quot;http://homepage.univie.ac.at/herwig.hauser/bildergalerie/gallery.html&quot; rel=&quot;nofollow&quot;&gt;algebraic surfaces&lt;/a&gt;.  I am at the very frustrating point of knowing what I need to do, and how it is theoretically done, but not really grasping how to actually do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I need to take the equation for the line and substitute the x, y, and z variables in the surface equation for the equivalent portions of the line equation, but as soon as I sit down to do that, I immediately hit a mental brick wall.  As for the normal calculations, I'm really lost, I'm not even sure there is a general case way to calculate the normals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd love some help on how to calculate the intersection and normal of some of these shapes, and any sort of general case rules for these calculations would be fantastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;While real general case solutions would be super awesome, it's ok to assume the shapes are in their standard orientation, not rotated or transformed at all - just positioned and (maybe) scaled. This make the problem much simpler, I believe. If there are other limitations you can use to make the problem even simpler, that's likely fine.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditorDisplayName="" LastEditDate="2010-09-19T10:36:52.527" LastActivityDate="2010-09-20T00:03:34.130" Title="Calculating Intersections of Lines and Algebraic Surfaces" Tags="&lt;geometry&gt;&lt;algebra-precalculus&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4976" PostTypeId="2" ParentId="4931" CreationDate="2010-09-19T04:37:25.237" Score="2" ViewCount="0" Body="&lt;p&gt;To expand on Agusti's point further, here's a strategy that only makes use of Pythagoras (remember that the distance formula for two points &lt;em&gt;is&lt;/em&gt; a result of the Pythagorean relation!): You have two adjacent triangles $\triangle AEB$ and $\triangle BEC$; compute the distances of all relevant segments, show that the two triangles are right triangles (thus proving the &quot;perpendicular&quot; part) and then show that $\overline{AE}\cong\overline{EC}$ (thus proving the &quot;bisector&quot; part).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-19T04:37:25.237" />
  <row Id="4977" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T04:37:48.623" Score="16" ViewCount="0" Body="&lt;p&gt;As Dylan Wilson comments in Byron's link, you can derive most of &lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonometric_identity&quot; rel=&quot;nofollow&quot;&gt;trigonometric identities&lt;/a&gt; thanks to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i\theta} = \cos \theta + i \sin \theta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I never could learn by heart the formula for the cosine or the sine of the sum of two angles, but on one hand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i(\alpha + \beta)} = e^{i\alpha} \cdot e^{i\beta} = (\cos\alpha \cos\beta -\sin\alpha\sin\beta) + i (\sin\alpha\cos\beta + \cos\alpha \sin\beta) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And on the other hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i(\alpha + \beta)} = \cos(\alpha + \beta) + i\sin(\alpha + \beta) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\cos(\alpha + \beta) = \cos\alpha \cos\beta -\sin\alpha\sin\beta&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha \sin\beta \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-19T07:48:42.670" LastActivityDate="2010-09-19T07:48:42.670" CommunityOwnedDate="2010-09-19T04:37:48.623" CommentCount="3" />
  <row Id="4978" PostTypeId="2" ParentId="4975" CreationDate="2010-09-19T04:45:44.783" Score="1" ViewCount="0" Body="&lt;p&gt;The &quot;normal vector&quot; to a surface is easy enough: compute the partial derivatives with respect to both dependent variables (with respect to $x$ and $y$ if you have $z=f(x,y)$; with respect to parameters $u$ and $v$ if you have a parametric representation $(x\;\;y\;\;z)=(f(u,v)\;\;g(u,v)\;\;h(u,v))$), and then take the cross product of the vectors obtained, (with the option of normalizing it to have unit length).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, I think you're better off trying to do surfaces with &quot;easy&quot; parametric representations: the manipulations are much easier than if you have them in implicit Cartesian form.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For polyhedra, finding normals to the faces is an even easier task: take the barycenter/centroid of the face under consideration, subtract the components of that from the components of any two selected vertices in the face, and take the cross product of the two resulting vectors (with again the option to normalize to unit length afterward).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-20T00:03:34.130" LastActivityDate="2010-09-20T00:03:34.130" CommentCount="6" />
  <row Id="4979" PostTypeId="2" ParentId="3852" CreationDate="2010-09-19T05:00:56.167" Score="0" ViewCount="0" Body="&lt;p&gt;i agree to &quot; TheMachineCharmer &quot; good concern.&lt;/p&gt;&#xA;" OwnerUserId="1981" LastActivityDate="2010-09-19T05:00:56.167" />
  <row Id="4980" PostTypeId="2" ParentId="4868" CreationDate="2010-09-19T05:47:47.120" Score="0" ViewCount="0" Body="&lt;p&gt;The answer is No. Recall the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\subset\mathbb{C}$ be an open set containing the closed unit disc. Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it follow that &#xA;$$f(z)=az^{n}\,\frac{cz^{m}-b}{1-\bar{b}cz^{m}}$$ &#xA;for some $a,b,c\in\mathbb{C}$ s.t. $|c|=1$, $|a|=1$, $|b|\le1$ and some $n,m\ge0$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B$ be the set of analytic functions from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$, and let $C$ the set of rational fractions of the indicated form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is an element of $C$ with $0&amp;lt;|b|&amp;lt;1$ and $m\ge1$, then $z\mapsto f(z)^2$ is in $B$, but, having a multiple zero $z_0\not=0$, is not in $C$. &lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2010-09-19T05:47:47.120" />
  <row Id="4981" PostTypeId="2" ParentId="4942" CreationDate="2010-09-19T06:13:29.510" Score="1" ViewCount="0" Body="&lt;p&gt;Making Weltschmerz's answer explicit, and very much related to Isaac's answer: note that the quadrilateral can be split into two triangles. The area of one triangle is half the product of the length of $\overline{BD}$ (the base) and half the length of $\overline{AC}$ (the height). Doubling that gives an area expression that is exactly what Isaac stated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for computing the coordinates  of $B$ and $D$, here's a twofer method: once you can compute the equation of the line from the point-slope form, transform the equation you have into the &quot;two-intercept&quot; form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x}{a}+\frac{y}{b}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence your x- and y-intercepts are (a,0) and (0,b).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-19T06:33:39.527" LastActivityDate="2010-09-19T06:33:39.527" />
  <row Id="4982" PostTypeId="2" ParentId="4905" CreationDate="2010-09-19T08:37:21.843" Score="6" ViewCount="0" Body="&lt;p&gt;This is nothing mysterious, it is the continuous analogue of $\Sigma 2^{-n} \binom{n}{k} =  1$ with factorials interpolated by Gamma functions that vary quite smoothly in the &lt;em&gt;middle range&lt;/em&gt; where the sum is concentrated and has approximately Gaussian distribution, $k = n/2 + O(\sqrt{n})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio between the discrete step function and continuous interpolation is $1+O(1/n)$ in the middle range (using the first few terms of Stirling asymptotic expansion for $\Gamma$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $1/\sqrt{x(1-x)}$ is irrelevant because it is effectively a constant factor $2 + O(1/n)$ in the middle range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle range has width less than $n^{1/2 + \epsilon}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, writing the integral in terms of $u = nx$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \int_0^n \binom{n}{u} 2^{-n} du \frac {1} {\sqrt{x(1-x)}} \sim 2 \int_0^n \binom{n}{u} 2^{-n} du \sim 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with an additive error of $O(n^{\epsilon - 1/2})$. The $\epsilon$ can be a small positive constant or an infinitesimal depending on how you dissect the integral into middle and tails (we have to choose a cut for each $n$, such that the entire middle range is covered as $n \to \infty$; any set of choices will prove convergence, but the rate of convergence depends on the choices).&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-19T08:47:07.610" LastActivityDate="2010-09-19T08:47:07.610" CommentCount="6" />
  <row Id="4983" PostTypeId="1" CreationDate="2010-09-19T10:31:50.380" Score="4" ViewCount="374" Body="&lt;p&gt;My (maybe wrong) impression is that while probability is widely used in science (for example, in statistical mechanics), it is rarely seen in pure mathematics. Which leads me to the question - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there some interesting application of Probability Theory in pure mathematics, outside Probability Theory itself?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-09-19T16:55:36.493" LastActivityDate="2010-09-27T17:19:39.553" CommunityOwnedDate="2010-09-19T18:23:53.840" Title="Applications of Probability Theory in pure mathematics" Tags="&lt;big-list&gt;&lt;probability&gt;&lt;probability-theory&gt;&lt;applications&gt;" AnswerCount="8" CommentCount="1" />
  <row Id="4984" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T10:41:09.070" Score="1" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see this link: &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_number_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Probabilistic_number_theory&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-19T10:41:09.070" CommentCount="0" />
  <row Id="4985" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T10:58:49.080" Score="8" ViewCount="0" Body="&lt;p&gt;The topic of Probabilistic Combinatorics. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_method&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Probabilistic_method&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a powerful way of giving non-constuctive existence proofs for lots of different (finite) mathematical structures and determining their properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this the highlight of all my undergraduate probability courses. If you are interested in learning more I would recommend: &quot;The Probabilistic Method&quot; by Alon and Spencer.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T10:58:49.080" CommentCount="1" />
  <row Id="4986" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T10:59:23.567" Score="2" ViewCount="0" Body="&lt;p&gt;One of my favorite examples is a proof of the Weierstrass approximation theorem, the fact that polynomials are dense in the space of continuous functions over an interval.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like the proof because it's elementary and the conclusion comes as a surprise.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-19T10:59:23.567" CommentCount="3" />
  <row Id="4987" PostTypeId="2" ParentId="4947" CreationDate="2010-09-19T11:34:27.870" Score="1" ViewCount="0" Body="&lt;p&gt;It is a preliminary and useful exercise to show that the line $Ax+By+C=0$ and the vector $v=(A,B)$ are perpendicular (one says that $v$ is the normal vector of the line). With these, all amounts to observe that two lines are parallel if and only if their normal vectors are parallel.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-19T11:34:27.870" />
  <row Id="4988" PostTypeId="1" AcceptedAnswerId="5123" CreationDate="2010-09-19T12:24:45.797" Score="2" ViewCount="126" Body="&lt;p&gt;I have a problem, which has $n$ variables. Let's call these variables as $a_1$, $a_2$, $a_3$ ... $a_n$. Each of these variables have their own range (i.e, $a_{imin} \leq a_i \leq a_{imax}$ for all $i$).  And most importantly, each $a_n$ is ranked in terms of importance. Without a loss of generality one can assume that that $a_1$ is more important than $a_2$, which in turn is more important than $a_3$, and so on. This means that as long as the constraint is satisfied, it is better to have a smaller $a_1$ compare to have a smaller $a_2$, and a smaller $a_3$ is better compare to a smaller $a_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These variables are required to fulfill one ( yes, just one!!) constraint function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{min} \leq f(a_1,a_2,a_3, .... a_n) \leq F_{max}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now is, how to construct a minimizing function that satisfy the above condition? If I have two variables ($a_1$ and $a_2$), and both of them are between or equal to 0 and 9, then I can easily write the minimizing function as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=10a_1 + a_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given that $a_1$ is more important than $a_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for complicated case, I am a bit confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Maybe I should clarify further. My point is, for the above case, if $a_1=3, a_2=4$ and $a_1=4, a_2=3$ both satisfy the constraint equation, $a_1=3, a_2=4$ should be selected. &lt;strong&gt;So how to construct $L$ in this case to reflect the situation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2 : Referring to &lt;a href=&quot;http://math.stackexchange.com/questions/4988/formulate-a-minimizing-function-that-prefers-a-minimal-variable-over-another/5044#5044&quot;&gt;the answer&lt;/a&gt; by Rahul, it seems that it is not possible to formulate such a minimizing function. Is there other methods of attack I can use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 3: &lt;a href=&quot;http://math.stackexchange.com/questions/4988/formulate-a-minimizing-function-that-prefers-a-minimal-variable-over-another/4992#4992&quot;&gt;alext87&lt;/a&gt; answer is a good start, it contains general guidelines on how to proceed. But I want a hard, quantitative rule that tells me how to select for $lambda_i$ for all $i$. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-21T01:57:19.580" LastActivityDate="2010-09-21T05:06:34.000" Title="Formulate A Minimizing Function that Prefers A Minimal Variable Over Another" Tags="&lt;optimization&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="4989" PostTypeId="1" AcceptedAnswerId="4994" CreationDate="2010-09-19T12:44:31.903" Score="1" ViewCount="71" Body="&lt;p&gt;Hello! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question may seem simple, but I have no idea where to start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in design phase of a hobby electronics project and need your help. Basically, I'm going to have two platforms with electronics on both of them. I'm planning to have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stepper_motor&quot; rel=&quot;nofollow&quot;&gt;stepper motor&lt;/a&gt; on the bottom platform which will be connected to top platform by an axle. I plan to use a microcontroller to precisely control the intensity of angular velocity of the motor and the angle by which it will rotate. I will also need to use wires to connect the bottom platform to the top platform. My idea is to have wires run on the outside of the axle and be long enough that the axle will be able to rotate by some angle of $±θ$. In normal operation the angle shouldn't be grater than 2π, so I want to avoid constructing some complex mechanism which will transfer current irrelevant of the axle's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image (I apologize for my bad drawing skills) shows the two states which are interesting (at least in my opinion) here. The top is when the axle is in neutral position and angle by which it is rotated is 0. While the image does not show it, the wire will be loose. The bottom image shows what I expect will happen after the axle has been rotated by a certain angle. &#xA;The two black rectangles are soldering joints by which the cable will be connected to the bottom and the top platforms. The red line is wire.&lt;br&gt;&#xA;&lt;a href=&quot;http://i.imgur.com/OSeFP.png&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s direct link to the image. It is more clear there.&#xA;&lt;img src=&quot;http://i.imgur.com/OSeFP.png&quot; alt=&quot;Image of the problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my question: Is I have an axle with length L and radius R, how would I calculate length l of a cable running form the top of the axle to the bottom after axle has rotated by a certain angle θ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have no idea which tags would be appropriate, so do feel free to edit them.&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-09-19T13:20:41.003" LastActivityDate="2010-09-20T01:14:37.283" Title="Calulating length of cable running along exterior of an axle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4990" PostTypeId="1" AcceptedAnswerId="4996" CreationDate="2010-09-19T12:57:24.710" Score="3" ViewCount="877" Body="&lt;p&gt;The birational change of variables $(u,v) = (\frac{36+y}{6x},\frac{36-y}{6x})$ maps $u^3+v^3=1$ to $y^2 = x^3 - 432$ which has discriminant $-2^{12}\cdot 3^9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using pari/gp we can compute the torsion subgroup:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;? elltors(ellinit([0,0,0,0,-432]))&#xA;%1 = [3, [3], [[12, 36]]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This says the torsion subgroup has order 3, is $\mathbf{Z}/3\mathbf{Z}$ and is generated by $(12,36)$ (which corresponds to $1^3+0^3=1^3$). The reason it has order 3 is because this also includes the projective solution $[0:0:1]$ of $X^3+Y^3=Z^3$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: By Nagell-Lutz one only needs to solve $y^2 = x^3 - 432$ in integers for $y=0$ and $y^2|2^{12}\cdot 3^9$ (which is a simple generate and test) to compute the elements of the torsion subgroup 'on paper'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The group of rational points for this curve is then (by Mordell's Theorem) of the form $\mathbf{Z}^r \times \mathbf{Z}/3\mathbf{Z}$ where $r$ is the rank of the curve. If we can show the rank is 0 then this would prove fermats last theorem for $n = 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be shown directly the rank of this curve is 0?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-18T00:01:51.407" Title="How could I calculate the rank of this elliptic curve?" Tags="&lt;number-theory&gt;&lt;elliptic-curves&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4991" PostTypeId="2" ParentId="4959" CreationDate="2010-09-19T13:16:58.720" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Q1:&lt;/strong&gt; Are convex subsets assumed to be in vector spaces or topological vector spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convex sets are easily defined in topological vector spaces over a subfield of $\mathbb{C}$, following the usual definition: $A$ is a convex set if $\forall x,y\in A$ we have $tx+(1-t)y\in A$ for $t\in[0,1]$. Basically means we can draw a line between any two points and the line segment is contained in $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-euclidean spaces convexity generalises to geodesically convex set see &quot;Convex functions and optimization methods on Riemannian manifolds&quot; by Constantin.  There are a few other generalisations. For the most abstract there are Convexity spaces which is an axiomatic description but I can't remember much about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q2:&lt;/strong&gt; Are convex functions defined to be from convex subsets in vector spaces or topological vector spaces to general field or some special field such as R or C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly I have seen a convex function defined without the need for its domain to be a convex set at all! It required the function to only be convex on any convex subset of its domain. (I don't think is standard though.)&#xA;Secondly, let $f:I\rightarrow K$ be a convex function where $I$ is a convex subset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K=\mathbb{R}$ then $f$ is convex iff $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$, $t\in[0,1]$. Is suspect this could be generalised to any ordered field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K=\mathbb{C}$ then we can't use this definition as $\mathbb{C}$ is not a totally ordered field. Instead we have $\mathbb{C}\sim\mathbb{R}^2$ and use the hessian: $f$ is convex iff $\nabla^2f$ is positive semi-definite. However this requires $f\in\mathcal{C}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q3:&lt;/strong&gt; What are the special cases that have been extensively studied and have significance in application?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of the applications in convex optimisation are with a real vector space and a convex real-valued function. There are so many applications for this the mind boggles. A short memory dump is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Least squares problems&lt;/li&gt;&#xA;&lt;li&gt;Linear programming: network flow problems, mircoeconomics.&lt;/li&gt;&#xA;&lt;li&gt;Quadratic programming: This forms the basis of so many optimization algorithms&lt;/li&gt;&#xA;&lt;li&gt;Entropy maximisation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-19T13:27:55.917" LastActivityDate="2010-09-19T13:27:55.917" />
  <row Id="4992" PostTypeId="2" ParentId="4988" CreationDate="2010-09-19T13:25:05.347" Score="1" ViewCount="0" Body="&lt;p&gt;You can turn this into an optimization problem. For each $a_i$ set up a parameter $\lambda_i$, where $\lambda_i$ you determine to be the importance of the variable being small. For example as $a_i$ is more important to be small than $a_{i+1}$ you will select $\lambda_i&gt;\lambda_{i+1}$. Then solve the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min \lambda_1a_1+\ldots+\lambda_na_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\text{min}}\leq f(a_1,a_2,\ldots,a_n)\leq F_{\text{max}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{imin}\leq a_i\leq a_{imax}$ for all i&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T13:25:05.347" CommentCount="5" />
  <row Id="4993" PostTypeId="2" ParentId="4989" CreationDate="2010-09-19T13:28:48.473" Score="2" ViewCount="0" Body="&lt;p&gt;Looks like you can approximate the cable's equation with a helix, $(x\quad y\quad z)=(r\cos(m t)\quad r\sin(m t)\quad \frac{h t}{2\pi})$ where $m$ is the number of turns, $h$ is the height of the cylinder, $r$ is the radius of the cylinder, and the parameter $t$ runs from $0$ to $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compute the arclength of this one, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{4\pi^2 r^2 m^2+h^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be a nice starting point, unless there's a better way to approximate the behavior of the cable.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-20T01:14:37.283" LastActivityDate="2010-09-20T01:14:37.283" CommentCount="1" />
  <row Id="4994" PostTypeId="2" ParentId="4989" CreationDate="2010-09-19T13:36:41.717" Score="4" ViewCount="0" Body="&lt;p&gt;This is a total and utter guess, but suppose you fix one endpoint of the wire and unroll the axle to make a rectangle? This is the same trick used to justify the rectangle in the surface area of a cylinder. Then the wire is the diagonal of a rectangle of width $L$ and height $r\theta$. So by the Pythagorean theorem, $l = \sqrt{L^2+r^2\theta^2}$.&lt;/p&gt;&#xA;" OwnerUserId="854" LastActivityDate="2010-09-19T13:36:41.717" CommentCount="3" />
  <row Id="4996" PostTypeId="2" ParentId="4990" CreationDate="2010-09-19T14:00:39.650" Score="2" ViewCount="0" Body="&lt;p&gt;I don't know much about this, but there is a theorem due to combined work of Zagier, Kolyvagin, Gross, etc, that says that if the analytic rank of the elliptic curve $E$ is 0 then it's rank is in fact 0. So you can compute the analytic rank for instance using SAGE and then by this theorem you will have shown that the rank is 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is like cheating since I think there ought to be some simpler way to do this, but you'll have to wait for the experts to respond you. For the time being, you can look at Theorem 5.8 and the definition preceding it in the following paper &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/bull/2002-39-04/S0273-0979-02-00952-7/S0273-0979-02-00952-7.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/bull/2002-39-04/S0273-0979-02-00952-7/S0273-0979-02-00952-7.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-09-19T14:00:39.650" CommentCount="3" />
  <row Id="4997" PostTypeId="1" AcceptedAnswerId="5000" CreationDate="2010-09-19T14:19:15.433" Score="0" ViewCount="67" Body="&lt;p&gt;Let $I$ be a finite nonempty set, and $D$ a nonvoid compact subset of an N-dimensional Euclidean space. The functions $V:D\rightarrow\mathbb{R}$ and $u_i:D\rightarrow\mathbb{R}$ are given. They are all continuous, and, moreover, each $u_i$ satisfies $u_i(\lambda p + (1-\lambda) q) = \lambda u_i( p) + (1-\lambda) u_i(q)$ when $\lambda \in (0,1)$ (I don't believe this last piece of information is important, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ and $u_i$ also satisfy: (P) if $u_i(p) = u_i(q)$ for all $i\in I$, then $V(p) = V(q)$. This condition allows us to construct a mapping $W:L\rightarrow\mathbb{R}$, where $L =$ {$ (u_1(p),...,u_I(p)): p \in D$}, and $W$ is defined as $W(u) = V(p)$ for any $p \in D$ such that $u = (u_1(p),...,u_I(p))$. Condition (P) ensures W is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is $W$ continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: I am sketching a &quot;proof&quot; from a (published) paper in &quot;applied math&quot; I found online, but that &lt;em&gt;I cannot verify&lt;/em&gt;. Let $(u^n)$ be a sequence in $L$ converging to $u \in L$. Let $(p_n)$ be a sequence in $D$ such that $u^n = (u_1(p_n),...,u_I(p_n))$ for all $n$, and $p \in D$ satisfy $u = (u_1(p),...,u_I(p))$. Because $D$ is compact, there exists a convergent subsequence $p_{n_{k}}\rightarrow p^\prime$. Because $V$ is continuous, we obtain $V(p_{n_{k}})\rightarrow V(p^\prime)$. Moreover, since each $u_i$ is also continuous, we have $\lim u^n = \lim u^{n_{k}} $, and hence $(u_1(p),...,u_I(p)) = (u_1(p^\prime),...,u_I(p^\prime))$. By property (P), we have $V(p) = V(p^\prime)$. Finally, &lt;em&gt;the article goes on and concludes&lt;/em&gt; that $W(u^n) = V(p_n) \rightarrow V(p^\prime) = V(p) = W(u)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think the conclusion follows from the previous arguments. In particular, nothing guarantees that $V(p_n)$ converges to $p$ in the first place, right? Nevertheless, I cannot think of an easy counter-example...&lt;/p&gt;&#xA;" OwnerUserId="1987" LastActivityDate="2010-09-19T15:23:43.323" Title="Continuity of an induced function" Tags="&lt;real-analysis&gt;&lt;counter-examples&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4998" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T14:44:40.490" Score="1" ViewCount="0" Body="&lt;p&gt;You may find some interesting examples on this MathOverFlow thread:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-19T14:44:40.490" CommunityOwnedDate="2010-09-19T16:58:00.910" />
  <row Id="4999" PostTypeId="2" ParentId="4968" CreationDate="2010-09-19T14:51:20.470" Score="2" ViewCount="0" Body="&lt;p&gt;As I mentioned in a prior &lt;a href=&quot;http://math.stackexchange.com/questions/4442/learning-to-think-categorically-localization-of-rings-and-modules/4450#4450&quot;&gt;post here&lt;/a&gt; there is a wealth of information on noncommutative localizations in Ranicki, A.(ed). &lt;a href=&quot;http://www.maths.ed.ac.uk/~aar/books/nlat.pdf&quot; rel=&quot;nofollow&quot;&gt;Noncommutative localization in algebra and topology. ICMS 2002&lt;/a&gt;. In particular, there you will find an interesting paper on this very topic by Beachy: &quot;On flatness and the Ore condition&quot;. Below is general reference information for flatness in the commutative case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very nice treatment of flatness in Bourbaki's &quot;Commutative Algebra&quot; - which begins with an excellent chapter on flat modules before turning to localizations in Chapter 2 (see Theorem 2.41. p. 68 for the result you seek). Also perhaps of interest is the following motivational remark from the introduction&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The study of the passage from a ring $\rm A$ to a local ring $\rm A$, or to a completion &#xA;  $\rm \hat A$ brings to light a feature common to these two operations, the property of &#xA;  flatness of the $\rm A$-modules $\rm A$, and $\rm \hat A$, which allows amongst other things &#xA;  the use of tensor products of such $\rm A$-modules with arbitrary $\rm A$-modules somewhat &#xA;  similar to that of tensor products of vector spaces, that is, without all the precautions &#xA;  surrounding their use in the general case. The properties associated with this notion, which &#xA;  are also applicable to modules over non-commutative rings, are  the object of study in Chapter I. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also Atiyah and Macdonald, Corollary 3.6 and Proposition 3.10 pp. 40-41.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-27T15:35:52.190" LastActivityDate="2010-09-27T15:35:52.190" CommentCount="4" />
  <row Id="5000" PostTypeId="2" ParentId="4997" CreationDate="2010-09-19T15:23:43.323" Score="0" ViewCount="0" Body="&lt;p&gt;You can think of your $u_i$ as the components of a continuous&#xA;map $U:D\to\mathbb{R}^n$ for some $n$. So you have the condition&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $U(p)=U(q)$ then $V(p)=V(q)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so $V=W\circ U$ for some map from $L=U(D)$ to $\mathbb{R}$.&#xA;To show $W$ is continuous, it suffices to prove that $U:D\to L$&#xA;is a quotient map, that is $X\subseteq L$ is closed in $L$ iff $U^{-1}(X)$&#xA;is closed in $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By continuity if $X$ is closed in $L$ then $U^{-1}(X)$ is closed&#xA;in $D$, so suppose that $Y=U^{-1}(X)$ is closed in $D$. As $D$ is compact,&#xA;then $Y$ is compact (Heine-Borel), and as $U$ is continuous, $U(Y)$ is a compact&#xA;subspace of $\mathbb{R}^n$. Thus $U(Y)$ is a closed subset of $\mathbb{R}^n$&#xA;and &lt;em&gt;a fortiori&lt;/em&gt; of $L$. But $U(Y)=X$, as $U$ maps $D$ onto $L$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-19T15:23:43.323" CommentCount="1" />
  <row Id="5002" PostTypeId="2" ParentId="4961" CreationDate="2010-09-19T16:16:07.843" Score="1" ViewCount="0" Body="&lt;p&gt;The Cauchy integral formula's are pretty easy to prove from the Cauchy-Riemann equation, especially if your class already knows Green's theorem.  I think these are my favorite &quot;easily achieved results&quot;; because these formula's are central to complex analysis, the proof's are rather easy, and the applications are extremely powerful.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are discussing complex numbers in a calculus class, it seems natural to discuss applications to integration theory in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-09-19T16:16:07.843" CommunityOwnedDate="2010-09-19T16:16:07.843" />
  <row Id="5003" PostTypeId="1" CreationDate="2010-09-19T16:21:10.127" Score="0" ViewCount="166" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/3633/square-root-of-a-function-in-the-sense-of-composition&quot;&gt;Square root of a function (in the sense of composition)&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm interested in solving equations of the form $f(f(x))=g(x)$ for $x\in\mathbb{R}$ where $g(x)$ is a known function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: if $g(x)=x$ then we $f(f(x))=x$ so one solution is $f(x)=x$, $\forall x\in\mathbb{R}$. However, another is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=x+1$ if $x\in(0,1]$,  $f(x)=x-1$ if $x\in(1,2]$ and $f(x)=x$ otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence there are infinitely many solutions to this equation when $g(x)=x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; Find an $f$ where $f(f(x))=e^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra:&lt;/strong&gt; Are there extra constraints that could be placed on $f$ so the solution is unique?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T16:21:10.127" ClosedDate="2010-09-19T22:15:29.917" Title="Solving $f(f(x))=g(x)$ equations." Tags="&lt;functional-equations&gt;" CommentCount="6" />
  <row Id="5004" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T16:54:50.140" Score="2" ViewCount="0" Body="&lt;p&gt;There is a strong connection between probability theory and areas of real analysis such as linear PDE and potential theory.  If $X_t$ is a continuous time stochastic process on a state space $S$, the function $u(x,t) = E_x[f(X_t)]$ tends to satisfy an equation like $\partial_t u + Lu = 0$, where $L$ is some interesting operator.  ($E_x$ denotes that the process should start at the point $x \in S$, and $f$ is a real-valued function on $S$.)  The classical example is when $X_t$ is Brownian motion on $\mathbb{R}^n$; then $L$ is the Laplace operator.  So one can study heat equations, harmonic functions, spectral theory, functional inequalities, and lots of other &quot;analysis&quot; topics by thinking about a corresponding stochastic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of Dirichlet forms makes this correspondence quite precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the state space $S$ is a manifold, then the study of the operator $L$ tends to be closely related to the geometry of $S$, so one can also study differential geometry through probability theory.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-19T16:54:50.140" CommunityOwnedDate="2010-09-19T16:55:05.467" />
  <row Id="5005" PostTypeId="1" AcceptedAnswerId="5006" CreationDate="2010-09-19T18:31:11.157" Score="3" ViewCount="143" Body="&lt;p&gt;In my daughter's class of 23, three students and the teacher all share the same birthday. Of course, there are 365 days in the year, and the first case of the shared birthday is not counted in the probability. But is the likelihood of this just (1/365)*3 = .0082? For my curiosity, leap years can be omitted.&lt;/p&gt;&#xA;" OwnerUserId="1990" LastActivityDate="2010-09-19T19:06:34.193" Title="Birthday Probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="5006" PostTypeId="2" ParentId="5005" CreationDate="2010-09-19T18:42:23.453" Score="1" ViewCount="0" Body="&lt;p&gt;No, 1/365^3 is (roughly) the probability that, given a date and 3 people, those 3 people have a birthday on that date.  This is much smaller than the probability that, given a date and 23 people, 3 of those people have their birthday on that date.   It seems now that you want, given 24 people (and no date fixed beforehand), the probability that 4 of those people have their birthday on the same date.  A generalization of this question can be found at &lt;a href=&quot;http://stats.stackexchange.com/questions/1308/extending-the-birthday-paradox-to-more-than-2-people&quot;&gt;http://stats.stackexchange.com/questions/1308/extending-the-birthday-paradox-to-more-than-2-people&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-19T19:06:34.193" LastActivityDate="2010-09-19T19:06:34.193" CommunityOwnedDate="2010-09-19T18:55:55.683" />
  <row Id="5007" PostTypeId="1" AcceptedAnswerId="5047" CreationDate="2010-09-19T18:51:01.750" Score="3" ViewCount="130" Body="&lt;p&gt;Seems straight-forward but I've been unable to get it right. Here are my steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$&#xA;$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x$$&#xA;$$-\sqrt{28-2y} = x + c$$&#xA;$$(28-2y) = (x+c)^2$$&#xA;$$y = -1/2x^2 - cx - c^2/2 + 14$$&#xA;$$c = {-2, 10}$$&#xA;$$\Rightarrow y = -1/2x^2 - 10x - 36$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've checked it over many countless times but for the life of me I can't figure out why it won't work. I've tried plugging the result back into the original equation and it seems to me like it checks out if you take the negative of the square root..&lt;/p&gt;&#xA;" OwnerUserId="1989" LastEditorUserId="1989" LastEditorDisplayName="" LastEditDate="2010-09-19T19:41:12.673" LastActivityDate="2010-09-20T08:26:11.113" Title="Solving separable differential equation" Tags="&lt;differential-equations&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="5008" PostTypeId="1" AcceptedAnswerId="5011" CreationDate="2010-09-19T18:58:11.280" Score="1" ViewCount="114" Body="&lt;p&gt;What does the following term mean?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;category of geometrically reduced schemes of finite type over some field&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I know what a category and a field is but I cannot translate any of the middle bits)&lt;/p&gt;&#xA;" LastActivityDate="2010-11-06T08:31:51.333" Title="What is the meaning of &quot;reduced schemes of finite type&quot;?" Tags="&lt;terminology&gt;" AnswerCount="2" />
  <row Id="5009" PostTypeId="2" ParentId="4826" CreationDate="2010-09-19T19:02:10.397" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an attempt at a solution: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_0\in U$, since $U$ is open there is a $\delta&gt;0$ such that $B(x_0,\delta)\subset U$ and by Talyor's theorem (using $v\in\mathcal{C}^1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v(x) = v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0)$  for some $x_{\epsilon}\in B(x_0,\delta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that $\Delta u(v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0))=0$ for $\forall x\in B(x_0,\delta)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the fact that $\nabla v(x_{\epsilon})^T$ is orthogonal and the theorem stated in the question we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta u(\nabla v(x_{\epsilon})^T(x-x_0))=\Delta (u)\nabla v(x_{\epsilon})^T(x-x_0)=0$  $\forall x\in B(x_0,\delta)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $v(x_0)$ is the zero vector then we are done. Else $v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0)$ is a translation of $\nabla v(x_{\epsilon})^T(x-x_0)$ by a constant vector $v(x_0)$ which is an orthogonal transformation. Let this translation be denoted as the linear orthogonal operator $T_{v(x_0)}$ then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta u(T_{v(x_0)}[\nabla v(x_{\epsilon})^T(x-x_0)]) = \Delta (u\circ T_{v(x_0)})\nabla v(x_{\epsilon})^T(x-x_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $v(x_{\epsilon})^T$ is orthogonal and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta (u\circ T_{v(x_0)})\nabla v(x_{\epsilon})^T(x-x_0)= \Delta (u)\circ T_{v(x_0)}(\nabla v(x_{\epsilon})^T(x-x_0))=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $T_{v(x_0)}$ is orthogonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_0$ was any point in $U$ we have showed that $\Delta(u\circ v)=0$ in $U$ and thus is a harmonic function. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-19T19:02:10.397" CommentCount="2" />
  <row Id="5010" PostTypeId="2" ParentId="5007" CreationDate="2010-09-19T19:03:01.003" Score="2" ViewCount="0" Body="&lt;p&gt;Your second line should be&#xA;$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-19T19:03:01.003" CommentCount="1" />
  <row Id="5011" PostTypeId="2" ParentId="5008" CreationDate="2010-09-19T19:12:26.003" Score="6" ViewCount="0" Body="&lt;p&gt;You need to consult a text on scheme theory, for instance&#xA;Hartshorne's &lt;em&gt;Algebraic Geometry&lt;/em&gt;. A scheme of finite type&#xA;over a field $k$ is one with a finite cover of spectra of&#xA;rings of the form $k[x_1,\ldots,x_m]/I$ where $I$ is an ideal.&#xA;A reduced scheme is a scheme where for each open set $U$ the ring&#xA;of functions defined over $U$ has no nonzero nilpotent elements.&#xA;It's geometrically reduced if it remains reduced when the base&#xA;field is changed to its algebraic closure.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-19T19:12:26.003" CommentCount="1" />
  <row Id="5012" PostTypeId="2" ParentId="4990" CreationDate="2010-09-19T20:22:32.287" Score="2" ViewCount="0" Body="&lt;p&gt;For some of the programs that output a rank, such as MAGMA, when they do so it is a rigorous certificate that the rank is the one displayed, and not only an apparent rank calculated from the L-function assuming the Birch and Swinnerton-Dyer conjecture.  So although for some cases such as $L(E,1) \neq 0$ or using parity etc, the state of knowledge on BSD is enough to conclude the rank from accurate approximate calculation on the L-function and other invariants, these tricks are not needed given the program output.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-19T20:22:32.287" />
  <row Id="5013" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T20:50:28.710" Score="1" ViewCount="0" Body="&lt;p&gt;You might find interesting various probabilistic plausibility arguments for the $\rm\:3n+1$ conjecture. For an introduction see &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/lagarias/&quot; rel=&quot;nofollow&quot;&gt;Lagarias's survey&lt;/a&gt; excerpted below, and for more sophisticated arguments see his paper &lt;a href=&quot;http://g4gardner.pbworks.com/f/mm-lagarias3XP1.pdf&quot; rel=&quot;nofollow&quot;&gt;How random are the 3x+1 function iterates?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Introduction.&lt;/strong&gt;$\;$ The $\rm\:3x + 1\:$ problem, also known as the Collatz problem, the Syracuse problem,&#xA;Kakutani's problem, Hasse's algorithm, and Ulam's problem, concerns the behavior of the &#xA;iterates of the function which takes odd integers $\rm\:n\:$ to $\rm\:3n + 1\:$ and even integers $\rm\:n\:$ to $\rm\:n/2\:$. The &#xA;$\rm\: 3x + 1\:$ Conjecture asserts that, starting from any positive integer $\rm\:n\:$, repeated iteration of this &#xA;function eventually produces the value $\rm\:1\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\rm\:3x + 1\:$ Conjecture is simple to state and apparently intractably hard to solve. It shares &#xA;these properties with other iteration problems, for example that of aliquot sequences (see Guy &#xA;[36], Problem B6) and with celebrated Diophantine equations such as Fermat's last theorem. Paul &#xA;Erdos commented concerning the intractability of the $\rm\:3x + 1\:$ problem: &quot;Mathematics is not yet &#xA;ready for such problems.&quot; Despite this doleful pronouncement, study of the $\rm\:3x + 1\:$ problem has &#xA;not been without reward. It has interesting connections with the Diophantine approximation of &#xA;$\log_2 3\:$ and the distribution (mod $\rm\:1\:$) of the sequence $\rm\:\{(3/2)^k : k = 1,2,\ldots\}\:$, with questions of ergodic theory on the $\rm\:2\:$-adic integers $\rm\:\mathbb Z_2\:$, and with computability theory--a generalization of the $\rm\:3x + 1\:$ problem has been shown to be a computationally unsolvable problem. In this paper I &#xA;describe the history of the $\rm\:3x + 1\:$ problem and survey all the literature I am aware of about this &#xA;problem and its generalizations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.1. A heuristic argument.&lt;/strong&gt; $\;$ The following heuristic probabilistic argument supports the $\rm\:3x + 1\:$ &#xA;Conjecture (see [28]). Pick an odd integer $\rm\:n_0\:$ at random and iterate the function $\rm\:T(n) = n/2\ \text{if n is even}\:$ $\rm\:\text{else}\ (3n+1)/2\:$ until another  odd integer $\rm\:n_1\:$ occurs. Then $\rm\:1/2\:$ of the time $\rm\:n_1 = (3 n_o + 1)/2,\ 1/4\:$ of the time, $\rm\:n_1 = (3 n_o + 1)/4,\ 1/8\:$ of the time $\rm\:n_1 = (3 n_o + 1)/8\:$ and so on. If one supposes that the function $\rm\:T\:$ is sufficiently &quot;mixing&quot; that successive odd integers in the trajectory of $\rm\:n\:$ behave as though they were drawn at &#xA;random (mod $\rm\:2^k\:$) from the set of odd integers (mod $\rm\:2^k\:$) for all $\rm\:k\:$, then the expected growth in size &#xA;between two consecutive odd integers in such a trajectory is the multiplicative factor &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3}{2}^{1/2}\frac{3}{4}^{1/2}\frac{3}{8}^{1/2}\cdots \ = \ \frac{3}{4} &amp;lt; 1\:.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently this heuristic argument suggests that on average the iterates in a trajectory tend to &#xA;shrink in size, so that divergent trajectories should not exist. Furthermore it suggests that the total &#xA;stopping time $\rm\:\sigma_x(n)\:$ is (in some average sense) a constant multiple of $\rm\:\log n\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the viewpoint of this heuristic argument, the central difficulty of the $\rm\:3x + 1\:$ problem lies &#xA;in understanding in detail the &quot;mixing&quot; properties of iterates of the function $\rm\:T(n)\ (mod\ 2^k)\:$ for &#xA;all powers of $\rm\:2\:$. The function $\rm\:T(n)\:$ does indeed have some &quot;mixing&quot; properties given by &#xA;Theorems B and K below; these are much weaker than what one needs to settle the $\rm\:3x + 1\:$ Conjecture. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-19T20:50:28.710" CommunityOwnedDate="2010-09-19T20:50:28.710" />
  <row Id="5014" PostTypeId="2" ParentId="4983" CreationDate="2010-09-19T21:10:38.803" Score="1" ViewCount="0" Body="&lt;p&gt;In addition to examples, the overall &quot;gestalt&quot; answer to the question is: probability is used everywhere in mathematics.  It is a basic idea on the level of algorithm, algebraic structure, geometry, calculus, or other very ubiquitous things.  It has become a very popular source of questions and intuitions in research, in all fields.   Knowing that this is true, it is not surprising that many examples of theoretical uses of probability can be posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a basic language, it is also true that many of the uses of probability are basic, and do not go far beyond the idea of a probability distribution, frequencies of events, expectations, related combinatorics and so on.  But in some fields, advanced results in probability theory are constantly being used.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-19T21:18:28.140" LastActivityDate="2010-09-19T21:18:28.140" CommunityOwnedDate="2010-09-19T21:10:38.803" />
  <row Id="5015" PostTypeId="1" CreationDate="2010-09-19T21:52:30.600" Score="1" ViewCount="120" Body="&lt;p&gt;I'm sure this is a solved problem but I don't know what to search for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set of positive numbers (256 at the moment but it could get bigger) and I need to find another set of positive numbers such that for each number in the original set, there is a sub set of the new set that sums to within some error bound of the number:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Give $S\subset\mathbb{R}^+$&lt;/li&gt;&#xA;&lt;li&gt;Find a small set $P\subset\mathbb{R}^+$&lt;/li&gt;&#xA;&lt;li&gt;Sutch that $\forall i\in S, (\exists P_i\subset P,|i-\sum P_i|\le\epsilon)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The best solution I have found so far is itterative:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;find $M = max(S_n)$.&lt;/li&gt;&#xA;&lt;li&gt;find $m = min(x | x \in S_n \wedge x &gt; M/2)$&lt;/li&gt;&#xA;&lt;li&gt;$P_{n+1} = P_n \cup m$&lt;/li&gt;&#xA;&lt;li&gt;$S_{n+1} = (x | x \in S_n \wedge x &amp;lt; m) \cup (x - m | x \in S_n \wedge x \ge m)$&lt;/li&gt;&#xA;&lt;li&gt;Stop when the approximation is good enough (i.e. $\forall x\in P_n,x\le\epsilon $)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1171" LastEditorUserId="1171" LastEditorDisplayName="" LastEditDate="2010-09-23T14:56:29.940" LastActivityDate="2010-09-23T14:56:29.940" Title="Minimal set with subsets that sum to given values" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5016" PostTypeId="1" AcceptedAnswerId="5054" CreationDate="2010-09-19T21:54:32.000" Score="2" ViewCount="93" Body="&lt;p&gt;Suppose $\mathcal{F}$ is a famliy of analytic functions of the unit disc.   Suppose also that&#xA;$( Re(f(z)) )^2 \ne ( Im(f(z)) ) $for all $|z|&amp;lt;1$ and all $f \in \mathcal{F}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_normality_test&quot; rel=&quot;nofollow&quot;&gt;Fundamental Normality test&lt;/a&gt;   that $\mathcal{F}$ is a normal family. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a for elementary way of showing $\mathcal{F}$ is a normal family without invoking the Fundamental Normality Test?&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-19T23:05:05.237" LastActivityDate="2010-09-20T10:31:29.050" Title="Normal Families" Tags="&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5018" PostTypeId="1" AcceptedAnswerId="5024" CreationDate="2010-09-19T22:26:24.410" Score="4" ViewCount="586" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fell in love with mathematics a bit too late when I've already taken decisions regarding my future, career-wise. Now I would like to learn math on my own but I'm a bit confused as where to start. My knowledge of mathematics is comparable to that of a 15-16 year old highschool freshman. I would like to know how would you (if you were in my position) start your learning adventure on your own. There are a lot of resources online, of that I'm sure but I would like to follow a path. I'm pretty sure I can learn mathematics on my own and been thinking about this decision for almost a year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1991" LastActivityDate="2010-09-25T11:37:29.643" CommunityOwnedDate="2010-09-19T23:19:34.053" Title="How to start with mathematics?" Tags="&lt;learning&gt;&lt;self-learning&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="4" />
  <row Id="5019" PostTypeId="1" AcceptedAnswerId="5041" CreationDate="2010-09-19T22:30:09.340" Score="3" ViewCount="203" Body="&lt;p&gt;Why are row reductions so useful in linear algebra? It is easy to get lost in the mechanical solving of equations. I know I can get a matrix into reduced row echelon form. But what are the outcomes of this? What can this mean?&lt;/p&gt;&#xA;" OwnerUserId="1992" LastActivityDate="2010-09-20T13:33:19.623" Title="What are we solving with row reduction?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="5020" PostTypeId="2" ParentId="4975" CreationDate="2010-09-19T22:44:09.240" Score="1" ViewCount="0" Body="&lt;p&gt;Perhaps this more elementary description could help.&#xA;Let $e$ be the eye/camera, and $v$ a line-of-sight vector.&#xA;You want to solve simultaneously $e + t v$ with the surface you want to view, solving for $t$.&#xA;If you have two or more surfaces, don't try to intersect them with one another,&#xA;which can be algebraically complex, but rather let the ray tracing (effectively) do it for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a surface patch $S$ (perhaps a Bezier surface) parametrized by $a$ and $b$.&#xA;So now you want to solve simultaneously for $(t, a, b)$.  If $S$ is a sphere or cylinder,&#xA;this amounts to quadratic equations.  If $S$ is a cubic patch, it will reduce to solving&#xA;cubic equations.  If $S$ is a torus, degree-4 equations. Once you have $(a,b)$, you can get the normal vector at that point from your parametric equations, as&#xA;J.M. describes.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-09-19T22:44:09.240" />
  <row Id="5021" PostTypeId="2" ParentId="5019" CreationDate="2010-09-19T22:46:29.707" Score="4" ViewCount="0" Body="&lt;p&gt;So you have a matrix $A$ and vectors $x,b$, to solve $Ax=b$ you apply row reductions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So considering a concrete example, if we wanted to solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc} 1 &amp;amp; 1 \\ 1 &amp;amp; 2\end{array}\right) x = \left(\begin{array}{c} 5 \\ 8\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the row that changes the bottom row into bottom row minus top:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc} 1 &amp;amp; 1 \\ 0 &amp;amp; 1\end{array}\right) x = \left(\begin{array}{c} 5 \\ 3\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which tells us that $x = \left(\begin{array}{c} 2 \\ 3\end{array}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The secret is that the second line is actually&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cc} 1 &amp;amp; 0 \\ -1 &amp;amp; 1\end{array}\right) \left(\begin{array}{cc} 1 &amp;amp; 1 \\ 1 &amp;amp; 2\end{array}\right) x = \left(\begin{array}{cc} 1 &amp;amp; 0 \\ -1 &amp;amp; 1\end{array}\right) \left(\begin{array}{c} 5 \\ 8\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a row operation corresponds to multiplying both sides by a matrix - Why is this useful for solving a system? Well it's just the same as when you solve any equation, for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2x+3=7x-4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can add 4 to both sides and then subtract 2x from both to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7=5x$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That is the meaning of row operations, and why they help to solve a system - it's also interesting to note that if we start out with $AX=I$ and use row operations until it's in the form $IX=B$ then the inverse of $A$ has been found. Viewing the row operations as matrices, lets say it took 3 operations $R_1, R_2, R_3$ then we have $R_3 R_2 R_1 A X = I X = R_3 R_2 R_1 I$ so $A^{-1} = X = R_3 R_2 R_1$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T22:46:29.707" />
  <row Id="5023" PostTypeId="2" ParentId="5018" CreationDate="2010-09-19T23:11:56.337" Score="0" ViewCount="0" Body="&lt;p&gt;Pardon me if I'm understating your current knowledge, but are you familiar with geometry, elementary algebra and pre-calculus? Those are pretty standard courses for high school students, and I'm sure most standard textbooks is fine to get the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are already familiar with that, computational calculus and computational linear algebra is usually the next step, and books like Tom Apostol's Calculus Vol. 1-2 are a nice introduction to single and multivariable calculus with some basic linear algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're comfortable with computational mathematics, it might be nice to move on to more theoretical mathematics, like linear algebra, abstract algebra, and real analysis. Some good texts in my opinion are Friedberg, Insel and Spence for Linear algebra, Dummit and Foote for Abstract Algebra, and Rudin for Real Analysis. Those are just some suggestions that I would pick and choose parts which I would want to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently self-teaching myself as well, so I don't have as much experience as many others here, so they can probably be more helpful.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-19T23:17:33.603" LastActivityDate="2010-09-19T23:17:33.603" />
  <row Id="5024" PostTypeId="2" ParentId="5018" CreationDate="2010-09-19T23:13:24.627" Score="3" ViewCount="0" Body="&lt;p&gt;I hope this is useful, I am just describing what worked well for me in the past. I recommend using real physical books (for example, I used &lt;em&gt;Mathematics for the Million - Lancelot Hogben&lt;/em&gt; at this stage and I would recommend this particular book also because I am very fond of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can a very good grip of the basic mathematical concepts by programming them (e.g. algorithms to convert between bases, factoring quadratics, finding square roots by newtons method when you get to calculus) - if you are just programming for programmings sake you will probably not learn very much mathematics (or at least, most of that which you do learn is by accident), but if you do it as an act of &lt;em&gt;detailed explanation&lt;/em&gt; it can be an excellent process to find much clearer understanding of whatever it is you're learning about (and as a bonus you can run them to automatically solve problems that you already know how to do!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In finding a path it would be useful for you to make a list of the more advanced theorems or applications which fascinate you (you will have to choose your own but I mean things like why are the orbits of planets in newtons universe ellipses, why you cannot square the circle or how raytracing works etc.) - then find out which setting they live in, then study the building blocks needed to get started working on the background - it doesn't even matter if you give up after a couple months extremely hard work by discovering that you cannot possibly study this stuff yet because you will still have got a lot out of it.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-19T23:13:24.627" />
  <row Id="5025" PostTypeId="1" AcceptedAnswerId="5038" CreationDate="2010-09-19T23:20:32.243" Score="1" ViewCount="113" Body="&lt;p&gt;Let $e(n)$ be the number of partitions of $n$ with even number of even parts and let $o(n)$ denote the number of partitions with odd number of even parts. In Enumerative Combinatorics 1, it is claimed that it is easy to see that $\sum_{n \geq 0} (e(n) - o(n)) x^n = \frac{1}{(1-x) \times (1+x^2) \times (1 - x^3 ) \times (1+x^4) \times ... }$. I have been racking my head over this for the past few hours, and I can't see any light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed, that $e(n) - o(n) = 2e(n) - p(n)$ where $p(n)$ is the number of partitions of $n$, so the above claim is equivalent to showing $\sum_{ n \geq 0} e(n)x^n = \frac{1}{2} \frac{1}{(1-x)(1-x^3)(1-x^5)...}( \frac{1}{(1-x^2)(1-x^4)....} + \frac{1}{(1+x^2)(1+x^4)........})$, and similarly, it is equivalent to  $\sum_{ n \geq 0} o(n)x^n = \frac{1}{2} \frac{1}{(1-x)(1-x^3)(1-x^5)...}( \frac{1}{(1-x^2)(1-x^4)....} - \frac{1}{(1+x^2)(1+x^4)........})$, but these identities appear more difficult than the original one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints and suggestions appreciated.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-09-20T03:28:20.987" Title="partitions with even number of even parts - partitions with odd number of even parts" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="5026" PostTypeId="2" ParentId="5018" CreationDate="2010-09-19T23:23:10.687" Score="1" ViewCount="0" Body="&lt;p&gt;I am a work in progress myself, but I think I can give some advice.  I think tackling mathematics as an older person has some advantages because you have a lot more life experience than a 15-16 year old person.  You know yourself better.  This can be helpful in terms of keeping a consistent work ethic.  I base these observations on tutoring a wide number of age groups (at a local community college).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, unfortunately, your question is a bit broad and I can't really tell from your description what a 15-16 year old high school freshman might know in your country.  In North America, preparation can vary greatly between school districts.  So my advice has to be general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would set myself a goal of learning the basic topics that are needed at the high school level in the local school system: read the books and do the exercises.  Ensuring that you don't have major gaps in the topics covered in your school system has the advantage of making it easier to enter the formal system if at some point later you find that you want to take a class perhaps at a local community college.  Such classes are likely to be tuned to the educational background of the typical high school student.  As an independent learner myself, I occasionally have the problem of weird gaps in my background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics at the high school level is straightforward I think.  If you read the book and do all the exercises, you can get through most of the curriculum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's important to have social feedback.  I have a few ideas in this area:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can participate in online mathematics community such as this site.  In addition to this site, I like &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/index.php&quot; rel=&quot;nofollow&quot;&gt;Art of Problem Solving&lt;/a&gt;.  I think it's worth it to interact with others and keep your enthusiasm high.  I also like the Mathematics Category on Yahoo Answers.  There are a large variety of questions and you can compare your answers to others.  The questions are quite often in the high school to early college range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you get to a level where you feel confident enough, you might tutor either as a volunteer or as a way to make some extra money.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can start your own blog.  I haven't tried it, but a few people have reported that it's a great way to learn new things.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-09-19T23:23:10.687" CommunityOwnedDate="2010-09-19T23:23:10.687" />
  <row Id="5027" PostTypeId="2" ParentId="5019" CreationDate="2010-09-19T23:44:10.017" Score="6" ViewCount="0" Body="&lt;p&gt;The main point of row operations is that they do not change the solution set of the underlying linear system.  So when you take a system of linear equations, write down its (augmented) coefficient matrix, and row reduce that matrix, you get a new system of equations that has the same solutions as the original system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the coefficient matrix is in reduced row echelon form, it is particularly easy to understand the solution set of the linear system.  In particular, you can readily tell whether the system is consistent or inconsistent, and when the system is consistent, you can readily see how a choice for the values of the free (non-pivotal) variables leads to a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend David Lay's book as a textbook for this material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that sometimes gets lost in the process of Gaussian elimination (the usual algorithm for putting a matrix into reduced row echelon form) is just how similar this process is to the approach, often taught in high school algebra, of solving for one variable in a particular equation and then eliminating its occurrences from all the other equations.  When you produce a pivot in a row of the coefficient matrix, this corresponds to solving the equation for that variable.  When you add or subtract (multiples of) this pivotal row from the other rows to produce zeros in the pivotal column, the new rows represent the equations you get by substituting in for the pivotal variable.  In a sense, Gaussian elimination is just an excellent way to keep track of the process of solving for variables and then eliminating them from the other equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example.  We start with the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3x + 6y + 3z = 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x + y  + 7z  = 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding matrix is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left[\begin{array}{rrr|r} 3 &amp;amp; 6 &amp;amp; 3 &amp;amp; 3\\&#xA; 2 &amp;amp; 1 &amp;amp; 7 &amp;amp; 6\end{array}\right]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing the first row/equation by 2 gives you the equation $x + 2y + z = 1$, which you can think of as $x = 1 - 2y - z$.  You can plug that expression into the second equation to get&#xA;$2(1-2y -z) + y + 7z = 6$, which simpifies to $-3y + 5z = 4$.  Notice that when you row reduce by dividing the first row by 2 and then subtracting twice row 1 from row 2, you get the same thing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left[\begin{array}{rrr|r} 3 &amp;amp; 6 &amp;amp; 3 &amp;amp; 3\\&#xA; 2 &amp;amp; 1 &amp;amp; 7 &amp;amp; 6\end{array}\right] \to  \left[\begin{array}{rrr|r} 1 &amp;amp; 2 &amp;amp; 1 &amp;amp; 1\\&#xA; 2 &amp;amp; 1 &amp;amp; 7 &amp;amp; 6 \end{array}\right] \to  \left[\begin{array}{rrr|r} 1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 1\\&#xA; 0 &amp;amp; -3 &amp;amp; 5 &amp;amp; 4\end{array}\right].$&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-09-20T03:07:27.073" LastActivityDate="2010-09-20T03:07:27.073" CommentCount="2" />
  <row Id="5029" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T00:12:16.573" Score="3" ViewCount="0" Body="&lt;p&gt;A good place to sample quick starting parts of mathematics is through the books of Martin Gardner, who sadly, died recently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&quot; rel=&quot;nofollow&quot;&gt;http://www.york.cuny.edu/~malk/biblio/martin-gardner-biblio.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggestion is to sample the essays in the wonderful book (1034 pages):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Princeton Companion to Mathematics (2008)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited by Timothy Gowers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Princeton U. Press, Princeton, NJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These essays cover a wide swath of mathematics and there are suggestions about learning more about areas that may catch your interest.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-20T00:12:16.573" CommunityOwnedDate="2010-09-20T00:12:16.573" />
  <row Id="5030" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T00:27:17.597" Score="3" ViewCount="0" Body="&lt;p&gt;I can only speculate on your background, so I would recommend only that in addition to the stuff you picked up from school, learn how to program (insert language/environment of choice here). It can greatly help to have another tool apart from pen and paper to try out whatever piece of mathematics you're trying to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that little prerequisite... it's open season for problems! Math problems don't have to be hard, advanced, or a &quot;hot topic&quot;; all they have to be is to be captivating and fascinating to you! Look around encyclopedias like MathWorld or Wikipedia so you can get a feel of what's out there (sometimes the &quot;Random Entry&quot; function is helpful in this regard if you don't know where to start), and if you find something you like, you can start from there! In addition to that, the bibliographies should give you ideas on what to type into the search engine should the encyclopedia entry prove inadequate in giving you what you want. If you find your background is less than adequate in understanding fully what you saw, that's okay too; this is one good thing about being in the Internet age, the ability to search for information without having to leave your seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a further supplement, go to the nearest public library and take a look at the math section. Since not all the great books are already online, so to speak, a look at the printed pages can be helpful.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-20T00:27:17.597" CommunityOwnedDate="2010-09-20T00:27:17.597" CommentCount="2" />
  <row Id="5031" PostTypeId="1" AcceptedAnswerId="5032" CreationDate="2010-09-20T00:53:07.813" Score="5" ViewCount="153" Body="&lt;p&gt;I was reading about isomorphisms and homomorphisms on general structures, and first came across the definition of an injective homomorphism, or an embedding. This made me curious, is it possible for two structures $A$ and $B$ to be embeddable in each other, yet no isomorphism exists between them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some looking around, I let the structures be $A=\mathbb{R}$ and $B=[-1,1]$ with $f\colon [-1,1]\to\mathbb{R}\colon r\mapsto r$ and $g\colon\mathbb{R}\to [-1,1]\colon r\mapsto \frac{2}{\pi}\arctan(r)$. If I refrain from defining any relations, functions, or distinguished elements in the universes of $A$ and $B$, then it is vacuously true that $f$ and $g$ are homomorphisms. Also, $[-1,1]$ and $\mathbb{R}$ would not be isomorphic since $[-1,1]$ has a maximum and minimum element. (Or would this require me to define $\lt$ on $[-1,1]$?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some other structures, even contrived ones, where such embeddings $f$ and $g$ exist, but $A$ and $B$ are still not isomorphic?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-09-20T02:19:09.570" LastActivityDate="2010-09-20T03:38:37.437" Title="Can two structures be embeddable in each other, but not isomorphic?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5032" PostTypeId="2" ParentId="5031" CreationDate="2010-09-20T00:54:00.837" Score="7" ViewCount="0" Body="&lt;p&gt;Yes. This was discussed on MO and many examples were given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold&lt;/a&gt; and the links thereon.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-20T00:54:00.837" CommentCount="2" />
  <row Id="5033" PostTypeId="2" ParentId="5015" CreationDate="2010-09-20T01:50:35.353" Score="1" ViewCount="0" Body="&lt;p&gt;First, from your description, it appears that P=S is a valid, but likely non-optimal[1], solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin with this. Sort the set from smallest to largest. For every member of the set, if there exists a subset that sums to that member, remove that member. Do not repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] For S=[1,2,4,8] then optimal P=[1,2,4,8]&lt;/p&gt;&#xA;" OwnerUserId="1995" LastEditorUserId="1995" LastEditorDisplayName="" LastEditDate="2010-09-21T04:36:17.883" LastActivityDate="2010-09-21T04:36:17.883" CommentCount="5" />
  <row Id="5034" PostTypeId="2" ParentId="5019" CreationDate="2010-09-20T01:52:26.740" Score="2" ViewCount="0" Body="&lt;p&gt;Well, Dan and muad have already told you everything, but let me stress the two main points (in my opinion):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Row reduction is just the same as solving one equation for one unkown in terms of the others and then plugging the obtained expression into the remaining equations. (But without having to write down the unkowns all the time.) In a nutshell, &lt;strong&gt;row reduction = solving your system&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Row reduction doesn't change the solution of the system of linear equations. (So the target in performing row reduction is to obtain a simpler system, with the same solutions as the original one, but such that solutions can be read directly from it.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The second statement can be proved formally, but I think that, if I can convince you that &lt;strong&gt;(1)&lt;/strong&gt; is true, you won't need this formal proof, will you? (Because, when you solve your system, you're sure that, if you don't make mistakes, the solution you arrive at is, well, &lt;em&gt;the solution&lt;/em&gt;. You don't change the solution of a system when you perform your operations in order to solve it, do you? So, if row reduction = solving your system and you don't change the solutions of your system while solving it, then row reduction doesn't change the solutions of your system.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's see how row reduction = solving your system. Say you're trying to solve the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;5x - 3y &amp;amp;= 3 \\&#xA;2x + 4y &amp;amp;= 2&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You realize that the second equation looks nicer than the first one, so you decide to start with it and &lt;strong&gt;interchange&lt;/strong&gt; both equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;2x + 4y &amp;amp;= 2    \\&#xA;5x - 3y &amp;amp;= 3 &#xA;\end{align}&#xA;\qquad \qquad \qquad \qquad \textbf{Step one}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you solve the first equation for $x$. You &lt;strong&gt;divide by two&lt;/strong&gt; your, now, first equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x + 2y = 1&#xA;\qquad \qquad \qquad \qquad \textbf{Step two}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solve it for $x$, $x = 1 -2y$, and &lt;strong&gt;plug it into the second one&lt;/strong&gt;, that is $5(1-2y) - 3y = 3$. You obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-13y = -2&#xA;\qquad \qquad \qquad \qquad \textbf{Step three}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you solve the second equation for $y$, &lt;strong&gt;dividing by&lt;/strong&gt; $-13$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = \frac{2}{13}&#xA;\qquad \qquad \qquad \qquad \textbf{Step four}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you perform &lt;strong&gt;back subtitution&lt;/strong&gt;, that is you plug this $y$ into the first equation, $x = 1- 2\frac{2}{13}$, getting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = \frac{9}{13}&#xA;\qquad \qquad \qquad \qquad \textbf{Step five}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you know &lt;strong&gt;the solution&lt;/strong&gt; of your system is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x &amp;amp;= \frac{9}{13} \\&#xA;y &amp;amp;= \frac{2}{13}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we are going to do &lt;strong&gt;exactly the same&lt;/strong&gt;, but with &lt;strong&gt;row reduction&lt;/strong&gt;. Our system of equations is the same as its augmented matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;5 &amp;amp; -3 &amp;amp; 3 \\&#xA;2 &amp;amp; 4  &amp;amp; 2&#xA;\end{array}&#xA;\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step one&lt;/strong&gt;, we've interchanged both equations. Now, we interchange the two rows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;2 &amp;amp; 4  &amp;amp; 2 \\&#xA;5 &amp;amp; -3 &amp;amp; 3&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step one}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step two&lt;/strong&gt;, we divided by two the first equation. Now, we divide by two the first row:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 2  &amp;amp; 1 \\&#xA;5 &amp;amp; -3 &amp;amp; 3&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step two}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step three&lt;/strong&gt;, we plugged $x = 1 -2y$ into the second equation. Now, we substract five times the first equation from the second:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 2  &amp;amp; 1 \\&#xA;0 &amp;amp; -13 &amp;amp; -2&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step three}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step four&lt;/strong&gt;, we divided the second equation by $-13$. Now, we divide the second row by $-13$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 2  &amp;amp; 1 \\&#xA;0 &amp;amp; 1 &amp;amp; \frac{2}{13}&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step four}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Step five&lt;/strong&gt;, we performed back substitution. Now, we substract $2$ times the second row from the first one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(&#xA;\begin{array}{rr|r}&#xA;1 &amp;amp; 0  &amp;amp; \frac{9}{13} \\&#xA;0 &amp;amp; 1  &amp;amp; \frac{2}{13}&#xA;\end{array}&#xA;\right)&#xA;\qquad \qquad \qquad \qquad \textbf{Step five}&lt;br&gt;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;strong&gt;the solution&lt;/strong&gt; of your system is in the third column, because the system that corresponds to this matrix is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x &amp;amp;= \frac{9}{13} \\&#xA;y &amp;amp;= \frac{2}{13}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-20T07:20:56.887" LastActivityDate="2010-09-20T07:20:56.887" />
  <row Id="5035" PostTypeId="1" CreationDate="2010-09-20T02:17:59.813" Score="5" ViewCount="665" Body="&lt;p&gt;Is there any formula for this series?&lt;/p&gt;&#xA;" OwnerUserId="1997" LastActivityDate="2010-09-20T14:11:32.997" Title="1 + 1/2 + 1/3 + ... + 1/n = ?" Tags="&lt;sequences-and-series&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5036" PostTypeId="1" CreationDate="2010-09-20T02:22:51.703" Score="3" ViewCount="131" Body="&lt;p&gt;For the definition of moduli space of Riemann surface, it seems that it is not based on the definiton in terms of functor(representable...) but rather put a topology on the set of isormophism classes. Is there any difference between these? Or if I miss something? Is there any reference ? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-09-20T04:17:20.900" Title="moduli space of Riemann surface" Tags="&lt;topology&gt;&lt;riemann-surface&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5037" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T02:43:30.307" Score="12" ViewCount="0" Body="&lt;p&gt;There is no formula for the nth partial sum of the harmonic series, only approximations. A well known approximation by Euler is that the nth partial sum is approximately $$ln(n) + \gamma $$ where $\gamma$ is the Euler–Mascheroni constant and is close to 0.5772. The amount of error in this approximation gets arbitrarily small for sufficiently large values of n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well known fact in mathematics is that the harmonic series is divergent which means that if you add up enough terms in the series you can make their sum as large as you wish.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-09-20T02:43:30.307" CommentCount="2" />
  <row Id="5038" PostTypeId="2" ParentId="5025" CreationDate="2010-09-20T03:28:20.987" Score="3" ViewCount="0" Body="&lt;p&gt;Now,&#xA;$$ \frac{1}{(1-x)(1+x^2)(1-x^3)\dots} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = (1+x+x^2+\dots)(1-x^2+x^4-\dots)(1+x^3+x^6+\dots)(1-x^4+x^8-\dots)\dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a term which provides a negative coefficient to $x^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we pick $x^{2n_2}$ from the second term, $x^{4n_4}$ from the fourth and so on,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $-1$ is $$(-1)^{n_2 + n_4 + \dots}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this will be negative if and only if $n_2 + n_4 + \dots$ is odd which comes from $o(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positive ones come from $e(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum (e(n) - o(n))x^n = \frac{1}{(1-x)(1+x^2)(1-x^3)\dots}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-20T03:28:20.987" CommentCount="1" />
  <row Id="5039" PostTypeId="2" ParentId="5036" CreationDate="2010-09-20T03:29:02.857" Score="2" ViewCount="0" Body="&lt;p&gt;From the mention of representability I assume you mean Riemann surfaces as algebraic curves, not the complex-analytic or conformal moduli problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Representing the moduli functor would mean you have object in a &quot;reasonable&quot; category (e.g., a variety, scheme, or algebraic space) that is a universal family of curves of whatever kind you are considering.  But this doesn't exist for algebraic curves without further restrictions (eg., stability) and rigidification data (such as marked points) to get a &quot;fine moduli space&quot;.  However, if by topology you mean a Grothendieck topology then you can define such a thing, i.e., a suitable category of coverings associated to the problem, as a virtual moduli object that you can consider as a space in the sense that cohomology and other geometrical invariants can be associated to it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical and much admired exposition that explained this to the world is &quot;Picard Groups of Moduli Problems&quot; by David Mumford.   Among other achievements it shows how to work with moduli stacks, without using the word &quot;stack&quot; (which may have been coined later, I don't know).&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-20T03:29:02.857" CommentCount="4" />
  <row Id="5040" PostTypeId="2" ParentId="5031" CreationDate="2010-09-20T03:38:37.437" Score="4" ViewCount="0" Body="&lt;p&gt;There are unenlightening examples if you don't ask for the image of the embedding to be large.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let A be a square and B an annulus, considered as topological spaces.  A small copy of each one can be placed inside the other but they are not isomorphic in the topological category. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bidirectional embedding is an equivalence relation, as is bidirectional dense (epimorphic) embedding, so at least linguistically the more natural problem in a given category is to ask what is the same about A and B if they are equivalent in this sense.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-20T03:38:37.437" />
  <row Id="5041" PostTypeId="2" ParentId="5019" CreationDate="2010-09-20T03:51:09.947" Score="2" ViewCount="0" Body="&lt;p&gt;The point of row reduction is to transform the system into an equivalent form that makes it simpler to deduce key information. In particular the reduced row echelon form immediately reveals fundamental information such as the  &lt;strong&gt;column space&lt;/strong&gt;, spanned by the pivot columns, the &lt;strong&gt;rank&lt;/strong&gt;, the &lt;strong&gt;nullspace&lt;/strong&gt;, etc. Generally such information can not be so simply deduced for a matrix in general form. Here is a worked example from Strang's Linear Algebra, 4ed p. 84:&#xA;&lt;img src=&quot;http://i.imgur.com/R8yqm.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/6LjBp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-20T03:51:09.947" />
  <row Id="5042" PostTypeId="2" ParentId="5036" CreationDate="2010-09-20T04:17:20.900" Score="3" ViewCount="0" Body="&lt;p&gt;In the complex analytic setting one can also work in terms of representing functors and so on; unfortunately, just as in the algebraic setting, it will turn out that the natural functor is not representable without either imposing level structure or passing to a larger category of stack-like objects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the complex analytic study of Riemann surfaces and their moduli has its own traditions, coming from Teichmuller theory, the theory of the mapping class group, the analytic theory of Ahlfors and Bers, and so on, and these traditions come with their own view-points, in which emphasizing the representability of the functor by a certain stack-like object may not be of paramount importance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to see why this might be, consider that if $f: X \to S$ is a smooth proper map of complex analytic manifolds (e.g. a family of Riemann surfaces over a smooth base), then by Ehresmann's theorem, $f$ is a fibre-bundle in the $\mathcal C^{\infty}$-category.  So one sees that&#xA;the fibres of $f$ don't vary topologically, and one can try to consider directly how the complex structure varies.  (This is the Kodaira--Spencer view-point on deformation theory.)  The fact that one has these vivid geometric view-points available means that there is less pressure to use categorical language to describe the situation.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-20T04:17:20.900" CommentCount="3" />
  <row Id="5043" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T04:57:39.863" Score="6" ViewCount="0" Body="&lt;p&gt;Let $H_n=1+1/2+\cdots+1/n$, as usual. Then, for $n&gt;1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n =\log(n+1/2) + \gamma + \varepsilon(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 &amp;lt; \varepsilon(n) &amp;lt; 1/(24n^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if you can use a high-precision estimate rather than an exact answer this may be good enough.  For example, $1+1/2+1/3+\cdots+1/10^{100}$ is between&lt;/p&gt;&#xA;&#xA;&lt;p&gt;230.83572496430610126240565755851882319115230819881722120213855733164212869451291269453666757225157658376140985147843194582191305052276721850285291090752309248454422116629840542211766342541591511108644544&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;230.83572496430610126240565755851882319115230819881722120213855733164212869451291269453666757225157658376140985147843194582191305052276721850285291090752309248454422116629840542211766342541591511108644546&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the numbers are identical up to their last digit.  If you need higher precision, a series expansion will yield formulas that are increasingly precise for large values of $n$.  (For small values of $n$, calculate directly...)&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-20T05:04:19.720" LastActivityDate="2010-09-20T05:04:19.720" CommentCount="1" />
  <row Id="5044" PostTypeId="2" ParentId="4988" CreationDate="2010-09-20T04:58:59.010" Score="2" ViewCount="0" Body="&lt;p&gt;I think my point in the comments was not taken, so I'll elaborate here and make this a proper answer. As far as I understand, you want to find the feasible point with the smallest $a_1$. If there is more than one such point, you want to find the one among them with the smallest $a_2$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such objective function that will work for all constraints $f$. Taking your example $L = 10x + y$ (I've renamed the variables as it'll make things easier later), I can choose the constraint to be $f(x,y) = 100x + y \in [1,2]$, making the feasible set to be a convex quadrilateral with vertices at $(0.01,0)$, $(0.02,0)$, $(0,1)$, and $(0,2)$. Now by your criteria you would like to choose the point $(0,1)$ as it minimizes $x$, but if you work out the values of $L$, you'll find that it is instead minimized at $(0.01,0)$. So the objective function you could &quot;easily write&quot; does not give you the desired solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, suppose $L$ is a function that gives you the desired solution for all constraints. Pick two values $x_1 &amp;lt; x_2$ and consider two corresponding &quot;slices&quot; of the domain $S_1 = \{(x_1,y) : y_{\min} \le y \le y_{\max}\}$ and $S_2 = \{(x_2,y) : y_{\min} \le y \le y_{\max}\}$. Then $L$ attains strictly lower values on $S_1$ than on $S_2$, because all points in $S_1$ are preferred over all points in $S_2$. On the other hand, $L$ must increase by a nonzero amount over $S_1$ itself, because $L(x_1,y_{\max}) &gt; L(x_1,y_{\min})$; call the difference $\Delta L(x_1)$. So, $L(x_2,y_{\min}) &gt; L(x_1,y_{\max}) = L(x_1,y_{\min}) + \Delta L(x_1)$. This is the key point — if $x_2 &gt; x_1$, $L(x_2,y_{\min}) &gt; L(x_1,y_{\min}) + \Delta L(x_1)$ — that is, keeping $y$ fixed, moving along $x$ by &lt;em&gt;any&lt;/em&gt; arbitrarily small amount must cause a finite increase in $L$ independent of how little you move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, the proof is easy to wrap up. There are uncountably many $x$s in $[x_{\min},x_{\max}]$, yet $L(x,y_{\min})$ must increase by a positive amount $\Delta L(x)$ at each $x$. So $L(x_{\max},y_{\min}) \ge L(x_{\min},y_{\min}) + \sum_x \Delta L(x)$, but this is impossible because &lt;a href=&quot;http://mathforum.org/library/drmath/view/66534.html&quot; rel=&quot;nofollow&quot;&gt;the sum of uncountably many positive numbers must diverge&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-20T04:58:59.010" CommentCount="9" />
  <row Id="5046" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T06:28:24.687" Score="6" ViewCount="0" Body="&lt;p&gt;I'm also an older person trying to teach myself math. I took math till high school and since then I have taught myself to beginning-graduate level in one area (analysis) and bits and pieces of other areas at an undergraduate level. My suggestions,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Time is limited, don't waste it on popular books.&lt;/strong&gt; Study high-school and undergraduate textbooks and do the problems.  It may be a good idea to     go to an older generation of textbooks since they are less chatty than the current ones.      The &lt;a href=&quot;http://www.ocf.berkeley.edu/~abhishek/chicmath.htm&quot; rel=&quot;nofollow&quot;&gt;Chicago Undergradute Mathematics Bibliography&lt;/a&gt; is useful. But don't obsess too much on which book to read. At this level it is hard to get things very wrong. Pick one book and persevere with it.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Work in sprints.&lt;/strong&gt; It takes    time to get into a&#xA;productive frame    of mind. So trying to do&#xA;math for a little    while each day&#xA;does not lead to any    progress and&#xA;results in frustration.    Instead,   dedicate&#xA;a week of your vacation or at least a &#xA;whole weekend for learning something &#xA;new. Between these sprints revise what you have learned by doing more problems.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Get it right the first time.&lt;/strong&gt; While studying a text if you feel that you don't understand something fully, stop and work on your difficulty. Look at other books. Ask for help on forums like this. Go and brush up on the prerequisites if need be. But never leave a gap. Don't skip more than one exercise in ten.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Be    patient.&lt;/strong&gt; A good undergraduate     training takes 4 years of     work with teachers and peers. My      guess is it will take at least a      dozen years to achieve the same level   of mathematical maturity working      part-time on your own. But at the end your understanding will be better than an undergraduate's since you will have had the benefit of greater intellectual maturity and more time to reflect.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Is it worth it?&lt;/strong&gt; Math is no doubt beautiful. But the time you give to it comes at the cost of your career, personal life and other interests. Are you willing to make the necessary sacrifices? If you are then it might be a good idea to make the sacrifices upfront and go back to school to learn math. The chances of fulfillment would be much higher.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-20T06:28:24.687" CommunityOwnedDate="2010-09-20T06:28:24.687" CommentCount="3" />
  <row Id="5047" PostTypeId="2" ParentId="5007" CreationDate="2010-09-20T06:31:51.513" Score="2" ViewCount="0" Body="&lt;p&gt;As Robin Chapman has pointed out, the correct choice is $c=-2$, not $c=10$. This gives&#xA;$$y(x)=\frac{28-(x-2)^2}{2} = -\frac{x^2}{2} + 2x + 12.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this formula describes a parabola $y(x)$ which hits its highest point $y=14$ at $x=-2$, and this is a decreasing function for $x&gt;-2$, which disagrees with the ODE. (Since the square root in the right-hand side of the ODE can't be negative, the solution $y(x)$ can't be decreasing.) Moreover, the right-hand side of the ODE is undefined for $y&gt;14$, so the solution can't continue to increase beyond 14. The conclusion is that the only possible continuation is $y(x)=14$ for all $x&gt;-2$. Hence, the solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(x)=\frac{28-(x-2)^2}{2} = -\frac{x^2}{2} + 2x + 12, \quad -4 \le x \le -2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(x)=14, x&gt;-2.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-09-20T08:26:11.113" LastActivityDate="2010-09-20T08:26:11.113" CommentCount="1" />
  <row Id="5048" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T06:33:58.260" Score="0" ViewCount="0" Body="&lt;p&gt;Well there are some formulae for the harmonic numbers. For example, one of the most well known is&#xA;$ H_n = \int_0^1 {1 - x^n \over 1 - x} dx$, which has certain useful properties. However, it is not in any way simpler than the original definition. And if you're looking for numerical approximations rather than exact identities, you certainly are better off with the approximation $H_n= \ln (n+1/2) +\gamma + \epsilon (n)$&lt;/p&gt;&#xA;" OwnerUserId="1999" LastActivityDate="2010-09-20T06:33:58.260" />
  <row Id="5049" PostTypeId="1" AcceptedAnswerId="5051" CreationDate="2010-09-20T07:59:14.853" Score="4" ViewCount="151" Body="&lt;p&gt;According this problem/solution set from an MIT class (http://ocw.mit.edu/courses/mathematics/18-100c-analysis-i-spring-2006/exams/exam1_sol.pdf), the assertion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Every collection of disjoint intervals in R is countable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is True, because &quot;every interval contains a rational number&quot;, and the rationals are countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me this should be False, with possible counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{ [x,x] | x is an element of R}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie the set of all singelton intervals on R. Why isn't this a valid counterexample? &lt;/p&gt;&#xA;" OwnerUserId="2003" LastEditorUserId="2003" LastEditorDisplayName="" LastEditDate="2010-09-20T11:00:42.273" LastActivityDate="2010-09-20T11:00:42.273" Title="Countability of disjoint intervals" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5050" PostTypeId="2" ParentId="5049" CreationDate="2010-09-20T08:28:51.477" Score="1" ViewCount="0" Body="&lt;p&gt;Because &quot;singleton interval&quot; is usually not considered to be an interval.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-20T08:28:51.477" CommentCount="1" />
  <row Id="5051" PostTypeId="2" ParentId="5049" CreationDate="2010-09-20T08:29:37.840" Score="5" ViewCount="0" Body="&lt;p&gt;Your thinking is correct; the set of all singleton sets of &lt;strong&gt;R&lt;/strong&gt; is certainly uncountable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the question meant something like &quot;Every collection of disjoint &lt;em&gt;open&lt;/em&gt; intervals in &lt;strong&gt;R&lt;/strong&gt; is countable.&quot;  (In this case, the claim that each interval contains a rational number is valid.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there was some convention in the course that &quot;interval&quot; meant open interval, or excluded singleton sets; perhaps it's simply a mistake.  Either way, it's good that you noticed this detail!&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-09-20T08:29:37.840" CommentCount="1" />
  <row Id="5052" PostTypeId="1" AcceptedAnswerId="5339" CreationDate="2010-09-20T09:39:39.077" Score="1" ViewCount="203" Body="&lt;p&gt;I have heard that Legendre functions are important in number theory. Can any one tell me how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Legendre function of the first kind $P_s$ is defined by&#xA;\begin{eqnarray*}P_s(x) =&amp;amp; \frac{1}{2\pi}\int_{-\pi}^{\pi}\left(x+\sqrt{x^2-1}\cos\theta\right)^s d\theta&lt;br&gt;&#xA;\newline&#xA;=&amp;amp;\frac{1}{\pi}\int_0^1\left(x+\sqrt{x^2-1}(2t-1)\right)^s\frac{dt}{\sqrt{t(1-t)}},&amp;amp; s\in\mathbb{C}, x\ge 1.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-15T21:16:22.587" Title="Legendre functions in number theory" Tags="&lt;number-theory&gt;&lt;special-functions&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="5053" PostTypeId="1" AcceptedAnswerId="5055" CreationDate="2010-09-20T10:21:45.337" Score="5" ViewCount="137" Body="&lt;p&gt;The German mathematician &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin_Eichler&quot; rel=&quot;nofollow&quot;&gt;Martin Eichler&lt;/a&gt; once stated that there were five fundamental operations of mathematics: addition, subtraction, multiplication, division, and modular forms. This was also mentioned in Simon Singh's book on Fermat's last theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about modular forms beyond the &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_forms&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry&lt;/a&gt;, and am quite confused by this remark. The usual operations seem to be different entities (two numbers input, one number output) than a modular form (one number input, one number output).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have some deeper insight into Martin Eichler's assertion?&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:10:57.960" LastActivityDate="2010-11-22T02:10:57.960" Title="How are modular forms fundamental operations?" Tags="&lt;modular-forms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5054" PostTypeId="2" ParentId="5016" CreationDate="2010-09-20T10:31:29.050" Score="2" ViewCount="0" Body="&lt;p&gt;Your condition means that  if $\gamma$ is the parabola $x^2-y=0$ then for any $f\in\mathcal{F}$ we have $f(\mathbb{D})\cap\gamma=\emptyset$. Thus &#xA;$\mathcal{F}=\mathcal{G}\cup \mathcal{H}$ where each $g$ in $\mathcal{G}$ maps the disk in the open set $x^2-y&gt;$0 and each $h$ in $\mathcal{H}$ maps the disk in the open set $x^2-y&amp;lt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just note that these two open sets are biholomorphic to the unit disk by uniformization theorem, thus $\mathcal{G}$ and $\mathcal{H}$ are normal families.  If you don't want to use uniformization theorem try to prove by hand that the open sets are biholomorphic to the disk.&lt;/p&gt;&#xA;" OwnerUserId="1860" LastActivityDate="2010-09-20T10:31:29.050" CommentCount="3" />
  <row Id="5055" PostTypeId="2" ParentId="5053" CreationDate="2010-09-20T10:44:14.377" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/6955/modular-forms-eichler-quote&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6955/modular-forms-eichler-quote&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-20T10:44:14.377" CommunityOwnedDate="2010-09-20T14:34:30.470" CommentCount="1" />
  <row Id="5056" PostTypeId="2" ParentId="5052" CreationDate="2010-09-20T10:50:48.317" Score="-2" ViewCount="0" Body="&lt;p&gt;Did you refer wiki, the function is useful in harmonic analysis: &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_function#Legendre_function_as_characters&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Legendre_function#Legendre_function_as_characters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-20T13:19:06.453" LastActivityDate="2010-09-20T13:19:06.453" CommunityOwnedDate="2010-09-20T13:19:06.453" CommentCount="2" />
  <row Id="5057" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T13:24:30.727" Score="2" ViewCount="0" Body="&lt;p&gt;This question made me recall this wonderful formula for $H_n$ due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Gregory Fontana&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \gamma + \log n + {1 \over 2n} - \sum_{k=2}^\infty { (k-1)! C_k \over n(n+1)\ldots(n+k-1)},&#xA;\qquad \textrm{ for } n=1,2,3,\ldots,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficients $C_k$ are the Gregory coefficients given by&#xA;$${ z \over \log(1-z)} = \sum_{k=0}^\infty C_k z^k \qquad \textrm{ for } |z|&amp;lt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I remember proving this as a student, about 30 years ago, I've just spent a frustrating 90 minutes trying to recover the proof without success, so unfortunately I cannot sketch it here. PS: I'll post a question shortly, and perhaps someone can beat me to it and put me out of my misery :-)&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-20T13:24:30.727" />
  <row Id="5058" PostTypeId="1" AcceptedAnswerId="5234" CreationDate="2010-09-20T13:33:17.043" Score="14" ViewCount="420" Body="&lt;p&gt;My question was inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/5035/1-1-2-1-3-1-n&quot;&gt;this&lt;/a&gt; stackexchange question. For the last 90 minutes I have been trying to prove this formula due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Gregorio Fontana&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \gamma + \log n + {1 \over 2n} - \sum_{k=2}^\infty { (k-1)! C_k \over n(n+1)\ldots(n+k-1)},&#xA;\qquad \textrm{ for } n=1,2,3,\ldots,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_n = \sum_{k=1}^n 1/k$ and the coefficients $C_k$ are the Gregory coefficients given by&#xA;$${ z \over \log(1-z)} = \sum_{n=0}^\infty C_k z^k \qquad \textrm{ for } |z|&amp;lt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit frustrating as it's something I recall proving as a student many years ago. I have a vague recollection that I began with something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \int_0^1 {1-(1-x)^n \over x } \textrm{d}x,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but my attempts to follow on from there have failed. Can you help?&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-22T22:16:34.700" LastActivityDate="2010-09-22T22:16:34.700" Title="Formula for the harmonic series $H_n = \sum_{k=1}^n 1/k$ due to Gregorio Fontana." Tags="&lt;sequences-and-series&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="5059" PostTypeId="2" ParentId="5019" CreationDate="2010-09-20T13:33:19.623" Score="1" ViewCount="0" Body="&lt;p&gt;i think of gaussian elimination or row reduction of a matrix A as giving you a simpler basis for the row space. it is only a small step from there by looking at the reduction of augmented matrix to find the compatibility condition on the right hand side b and solve ax = b&#xA;in that case.&lt;/p&gt;&#xA;" OwnerUserId="1758" LastActivityDate="2010-09-20T13:33:19.623" />
  <row Id="5060" PostTypeId="2" ParentId="5035" CreationDate="2010-09-20T14:11:32.997" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a way to interpret the harmonic numbers combinatorially:&#xA;$$H_n=\dfrac {\genfrac{[}{]}{0pt}{}{n+1}{2}}{n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\genfrac{[}{]}{0pt}{}{n+1}{2}$ is the absolute value of the Stirling number of the first kind, namely, the number of permutations of ${1,2,\dots,n,n+1}$ that have exactly $2$ cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These satisfy the following recurrence:&#xA;$$\genfrac{[}{]}{0pt}{}{n}{k}=\genfrac{[}{]}{0pt}{}{n-1}{k-1}+(n-1)\genfrac{[}{]}{0pt}{}{n-1}{k}$$ which makes it algebraically obvious that they are related to the Harmonic numbers the way they are, though there is also a purely combinatorial proof. The virtue of this interpretation is that you can prove a whole host of crazy identities involving the Harmonic numbers by just translating the natural identities for the Stirling numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stirling numbers of the first kind are also the coefficients of $x(x-1)(x-2)\cdots(x-(n-1))$, so their absolute values are the coefficients of $x(x+1)(x+2)\cdots (x+n-1)$, and so in particular&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The n&lt;sup&gt;th&lt;/sup&gt; Harmonic number is the coefficient of $x^2$ in $\frac1{n!}x(x+1)(x+2)\dots(x+n)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of this can be found in Chapter 7 of Benjamin &amp;amp; Quinn's wonderful book Proofs That Really Count.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-09-20T14:11:32.997" />
  <row Id="5061" PostTypeId="1" AcceptedAnswerId="5067" CreationDate="2010-09-20T15:34:56.363" Score="2" ViewCount="137" Body="&lt;p&gt;I managed to prove that if $\phi(0)$ and $\forall n(n\in\mathbb{N},\phi(n)\rightarrow\phi(n+1))$ then $\forall n(n\in\mathbb{N}\rightarrow\phi(n))$. The proof is based on the fact that $\{m\in\mathbb{N}:\phi(m)\}$ is an inductive set. Now I'd like to show (but I can't) that if you take a $k\in\mathbb{N}$, if $\phi(k)$ and $\forall n(n\in\mathbb{N},k\leq n,\phi(n)\rightarrow\phi(n+1))$ then $\forall n(n\in\mathbb{N},k\leq n\rightarrow\phi(n))$. The problem is I cannot show that $\{m\in\mathbb{N}:k\leq m,\phi(m)\}$ is an inductive set, since it is obvious, for example, that it doesn't contain zero as an element (except if $k=0$). What can I do instead? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2013" LastActivityDate="2010-09-20T16:06:38.437" Title="Induction principle" Tags="&lt;induction&gt;" AnswerCount="2" />
  <row Id="5062" PostTypeId="2" ParentId="5061" CreationDate="2010-09-20T15:39:34.700" Score="0" ViewCount="0" Body="&lt;p&gt;You could define $\psi$ on $\mathbb{N}$ by $\psi(n)=\phi(n+k)$, apply your first result to $\psi$, and then reinterpret for $\phi(n)=\psi(n-k)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-20T15:39:34.700" CommentCount="5" />
  <row Id="5063" PostTypeId="1" AcceptedAnswerId="5066" CreationDate="2010-09-20T15:46:37.340" Score="2" ViewCount="174" Body="&lt;p&gt;Ok this is for a HW but I'm not looking for a handout...just a hint to get me on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea where to begin proving this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if A is a symmetric matrix, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^TAy = y^TAx$$&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-20T17:07:09.550" LastActivityDate="2010-09-20T18:43:00.450" Title="How do I prove that $x^TAy = y^TAx$ if A is symmetric?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="4" />
  <row Id="5064" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T15:48:12.963" Score="3" ViewCount="0" Body="&lt;p&gt;Prove that whenever $A$ and $B$ are matrices for which you can compute the product $AB$, then $$(AB)^t=B^tA^t$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next apply $(\mathord-)^t$ to the left hand side of your equation, and compare the result to the right hand side, keeping in mind that both sides are actually numbers (well, $1$-by-$1$ matrices)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-20T15:48:12.963" />
  <row Id="5065" PostTypeId="1" CreationDate="2010-09-20T16:02:23.313" Score="3" ViewCount="188" Body="&lt;p&gt;For a continuous scalar field on a circle, there is a diameter of the circle such that the endpoints of the diameter have the same value. If you think of the scalar field as &quot;temperature&quot;, then what this says is that there are points on opposite sides of the circle that are the same temperature: isothermic antipodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a continuous scalar field on a sphere, the same is true: there are isothermic antipodes. (Just consider any great circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is there more you can say? Can you say for example that there is a closed loop on the surface of the sphere such that every point on the loop has the same value as the other endpoint of its diameter?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-09-24T15:44:58.387" Title="Does a continuous scalar field on a sphere have continuous loop of &quot;isothermic antipodes&quot;" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5066" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T16:05:48.257" Score="10" ViewCount="0" Body="&lt;p&gt;$x^tAy$ is a scalar. So $x^tAy=(x^tAy)^t$. Can you continue from here?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-20T16:05:48.257" CommentCount="0" />
  <row Id="5067" PostTypeId="2" ParentId="5061" CreationDate="2010-09-20T16:06:38.437" Score="2" ViewCount="0" Body="&lt;p&gt;Prove instead that the set $\{m\in\mathbb{N}\colon m\lt k\}\cup\{m\in\mathbb{N}\colon\phi(m)\}$ is inductive, and thus equals $\mathbb{N}$. Then use this to deduce what you want. (That is, take every element for which $\phi(m)$ holds, and then &quot;throw in for free&quot; every number strictly less than $k$, since you do not care what happens to them, and show that this is the set of all natural numbers). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-20T16:06:38.437" />
  <row Id="5068" PostTypeId="2" ParentId="5065" CreationDate="2010-09-20T16:12:09.417" Score="0" ViewCount="0" Body="&lt;p&gt;As for what more you can say in higher dimensions, take a look at the Borsuk-Ulam Theorem: &lt;a href=&quot;http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-09-20T16:12:09.417" CommentCount="5" />
  <row Id="5069" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T16:14:06.070" Score="3" ViewCount="0" Body="&lt;p&gt;A result that might be particularly suitable to students in a calculus sequence is complex partial fraction decomposition.  Over the real numbers, you get denominators with both linear and quadratic factors, but over the complex numbers, algebraic closure means you only have to think about linear factors.  For example, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 + x^2} = \frac{1}{2} \left( \frac{1}{1 + ix} + \frac{1}{1 - ix} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{1 + x^2}  dx = \frac{1}{2} \left( i \log(1 - ix) - i \log(1 + ix) \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It turns out that this function is just the arctangent in disguise.)  So complex numbers give you a simple algorithm which you can use, in principle, to integrate any rational function.  For example, in &lt;a href=&quot;http://math.stackexchange.com/questions/4803/the-partial-fraction-expansion-of-frac1xn-1&quot;&gt;this recent math.SE thread&lt;/a&gt; it is shown that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 - x^n} = \frac{1}{n} \left( \sum_{k=0}^{n-1} \frac{1}{1 - \zeta^k x} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta = e^{ \frac{2 \pi i}{n} }$ is a primitive $n^{th}$ root of unity.  This implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{1 - x^n} dx = \frac{1}{n} \left( \sum_{k=0}^{n-1} -\zeta^{-k} \log (1 - \zeta^k x) \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not an integral most students would suspect is expressible in closed form!  The analogous computation of $\int \frac{1}{1 + x^n} dx$ would even allow you to express the sum $\sum_{k=0}^{\infty} \frac{(-1)^k}{kn+1}$ in closed form (which is just the definite integral from $0$ to $1$).  These examples illustrate another application of the discrete Fourier transform, but the method I'm describing here is totally general.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-20T21:46:49.423" LastActivityDate="2010-09-20T21:46:49.423" CommunityOwnedDate="2010-09-20T16:14:06.070" CommentCount="1" />
  <row Id="5070" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T16:43:04.623" Score="6" ViewCount="0" Body="&lt;p&gt;I'd like to mention another combinatorial result, this time about asymptotics.  Let $a_n$ denote the number of ways that $n$ horses can finish in a race (with ties); in other words, $a_n$ is the number of &quot;lists of (nonempty) sets&quot; with $n$ total members.  Generating function methods allow you to deduce that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(z) = \sum_{n=0}^{\infty} \frac{a_n}{n!} z^n = \frac{1}{1 - (1 - e^z)} = \frac{1}{2 - e^z}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is meromorphic, so the asymptotic behavior of $a_n$ is controlled by its poles.  The dominant pole occurs at $z = \ln 2$ with residue $-\frac{1}{2}$, which means that the asymptotic behavior of $a_n$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_n}{n!} \approx \frac{1}{2 (\ln 2)^{n+1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so real-variable.  Where do complex numbers come in?  The &lt;strong&gt;error&lt;/strong&gt; in this approximation is controlled by the remaining poles of $A(z)$, all of which are complex.  The next two most dominant poles are at $\ln 2 \pm 2 \pi i$ with the same residue, which means that the asymptotic behavior of the error is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_n}{n!} - \frac{1}{2 (\ln 2)^{n+1}} \approx \frac{1}{2 (\ln 2 + 2 \pi i)^{n+1}} + \frac{1}{2 (\ln 2 - 2 \pi i)^{n+1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $\frac{1}{\ln 2 + 2 \pi i} = r e^{i \theta}$ where $r = \frac{1}{\sqrt{(\ln 2)^2 + 4 \pi^2}}$ and $\theta = -\arctan \frac{2 \pi}{\ln 2}$, it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_n}{n!} - \frac{1}{2(\ln 2)^{n+1}} \approx r^{n+1} \cos (n+1)\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the error in the above approximation is quasi-periodic.  (Of course there are infinitely many poles, each pair of which also contributes quasi-periodic terms, but as $n$ becomes large these terms become less and less important, most of the time.)  This is a phenomenon you can easily see for yourself by actually computing the error for several consecutive values of $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think about this: even if you correctly guessed the asymptotic behavior of $a_n$ (for instance by making a table of the values $\frac{a_n}{n!}$ (or its inverse, if you are interested in the probability that there are no ties in the race) and noticing that the number of digits grows linearly), and even if you computed experimentally that the error in the approximation is quasi-periodic, how on earth could you possibly have deduced the value of either $r$ or $\theta$ without complex numbers?  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-20T19:44:14.303" LastActivityDate="2010-09-20T19:44:14.303" CommunityOwnedDate="2010-09-20T16:43:04.623" />
  <row Id="5071" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T16:59:42.500" Score="3" ViewCount="0" Body="&lt;p&gt;I know of several ways to prove the Fundamental Theorem of Algebra using only basic tools from complex analysis. Suppose $P(z)$ is a non-constant polynomial. $|P(z)|$ is large outside of a large disk, and inside that disk since $1/P(z)$ is analytic, $1/|P(z)|$ is bounded by the maximum it has on the boundary. Thus $1/P(z)$ is a bounded analytic function on the whole plane so by Liouville's Theorem it is constant, a contradiction. Thus $P(z)$ has a zero somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really only uses the fact that polynomials are analytic, and analytic functions satisfy the maximum principle. Liouville's Theorem takes a little more work, but should be covered in any first complex analysis course.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-09-20T16:59:42.500" CommunityOwnedDate="2010-09-20T16:59:42.500" />
  <row Id="5072" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T17:04:00.743" Score="3" ViewCount="0" Body="&lt;p&gt;This condition can actually be used as an equivalent definition of $A$ being symmetric.  The key point is that both sides are bilinear in $x$ and $y$, so it suffices to prove the result when $x$ and $y$ are basis vectors, say $e_i$ and $e_j$.  What does the condition say then?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-20T17:04:00.743" />
  <row Id="5074" PostTypeId="2" ParentId="5063" CreationDate="2010-09-20T17:56:36.643" Score="0" ViewCount="0" Body="&lt;p&gt;compute $[x_1, x_2]A[y_1 y_2]^T$ for a 2 x 2 symmetric matrix $A = [[a, b], [b, c]]$.&#xA;you get $ax_1y_1 + b(x_1y_2 + x_2y_1) + cx_2y_2$. this expression is clearly symmetric&#xA;in $x$ and $y$ i.e., does not change when $x$ and $y$ are exchanged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for an n x n matrix you may want to use sigma notation.  &lt;/p&gt;&#xA;" OwnerUserId="1758" LastEditorUserId="1758" LastEditorDisplayName="" LastEditDate="2010-09-20T18:43:00.450" LastActivityDate="2010-09-20T18:43:00.450" />
  <row Id="5076" PostTypeId="1" AcceptedAnswerId="5082" CreationDate="2010-09-20T18:17:06.570" Score="2" ViewCount="528" Body="&lt;p&gt;Given $f(x) = \frac{1}{2}x^TAx + b^Tx + \alpha $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is an nxn symmetric matrix, b is an n-dimensional vector, and alpha a scalar. Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigtriangledown _{x}f(x) = Ax + b$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = \bigtriangledown ^{2}_{x}f(x) = A$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this simply a matter of taking a derivative with respect to X, how would you attack this one?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2014" LastEditorDisplayName="" LastEditDate="2010-09-20T18:30:52.767" LastActivityDate="2010-09-20T18:56:13.823" Title="what does ∇ (upside down triangle) symbol mean in this problem" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5077" PostTypeId="1" AcceptedAnswerId="5078" CreationDate="2010-09-20T18:24:33.607" Score="0" ViewCount="50" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any rectangle, that if you divide it into another rectangle (or any other quadrilateral), the &lt;strong&gt;relation&lt;/strong&gt; between the two different edges is the same in both rectangles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example - &lt;strong&gt;Orginal rectangle:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/WF3f0.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Divided Rectangle:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/raRto.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these illustrations aren't really exact, it's just to clear the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-20T18:26:08.280" LastActivityDate="2010-09-20T18:31:46.527" Title="Relation of different edges in rectangles" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5078" PostTypeId="2" ParentId="5077" CreationDate="2010-09-20T18:31:41.593" Score="1" ViewCount="0" Body="&lt;p&gt;This amounts to solving an equation. For the case you illustrate (where we divide the longer side in half), you fall into one of two cases: if $b\gt 2a$, so that in the second rectangle the horizontal sides are still the longer sides, you want $\frac{b/2}{a}=\frac{b}{a}$. For this to occur, you would need $ba = 2ba$, which is impossible since both $a$ and $b$ are assumed positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b\leq 2a$, then in the second rectangle it is the vertical side that are longest, so you want $\frac{a}{b/2} = \frac{b}{a}$. This gives you $2a^2 = b^2$, or $\sqrt{2}a=b$ (since $a$ and $b$ are lengths, they will be positive). So, for example, if $a$ is $1$ in the first figure, you need $b=\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it will all depend on what proportions you want. For example, one can consider the case in which you &quot;cut off&quot; a square of side length $a$ from your original rectangle, leaving you with a rectangle with sides $a$ and $b-a$. Then you would want this latter rectangle to have the same proportions as the original one, which means you want $\frac{b}{a}=\frac{a}{b-a}$ (turns out you cannot do it if $b-a\gt a$). This leads to an equation that you can solve for $b$ in terms of $a$ (or solve for $a$ in terms of $b$), giving you the proportions that the original rectangle must have. Etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-20T18:31:41.593" />
  <row Id="5079" PostTypeId="2" ParentId="5077" CreationDate="2010-09-20T18:31:46.527" Score="3" ViewCount="0" Body="&lt;p&gt;I presume you want $b/a=a/(b/2)$, that is $(b/a)^2=2$ so the ratio of sides&#xA;is $\sqrt 2$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/A4_paper_size#The_A_series&quot; rel=&quot;nofollow&quot;&gt;A series&lt;/a&gt; of paper sizes is based on this ratio.&#xA;To the nearest millimetre the ratio of the long to the short size&#xA;is $\sqrt2$, so tearing a sheet of A4 along a line through its centre&#xA;parallel to its short sides gives two sheets of A5 etc.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-20T18:31:46.527" />
  <row Id="5080" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T18:32:32.130" Score="3" ViewCount="0" Body="&lt;p&gt;There are tons of applications of complex numbers in places you would not expect. Number theory is one, for example. This example is from &lt;em&gt;The Art and Craft of Problem Solving&lt;/em&gt; by Zeitz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition: If $m$ and $n$ are integers that can be written as the sum of two squares, then $mn$ can also be written as the sum of two squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $m=a^2+b^2$ and $n=c^2+d^2$ and define $z=(a+bi)(c+di)$. Then $|z|^2=|a+bi|^2 |c+di|^2=mn$. Since $\Re(z)=ac-bd$ and $\Im(z)=bc+ad$ are both integers and $|z|^2$ equals the sum of the squares of its real and imaginary components, then $|z|^2=mn$ is the sum of two squares.&lt;/p&gt;&#xA;" OwnerUserId="1999" LastEditorUserId="1999" LastEditorDisplayName="" LastEditDate="2010-09-20T19:13:23.457" LastActivityDate="2010-09-20T19:13:23.457" CommunityOwnedDate="2010-09-20T18:32:32.130" />
  <row Id="5081" PostTypeId="2" ParentId="5076" CreationDate="2010-09-20T18:47:49.380" Score="0" ViewCount="0" Body="&lt;p&gt;$\bigtriangledown f$ finds the direction of maximal change in f.&lt;/p&gt;&#xA;" OwnerUserId="1758" LastActivityDate="2010-09-20T18:47:49.380" />
  <row Id="5082" PostTypeId="2" ParentId="5076" CreationDate="2010-09-20T18:56:13.823" Score="5" ViewCount="0" Body="&lt;p&gt;$\nabla f = (\partial f/\partial x_1, \ldots, \partial f/\partial x_n)^t$ denotes the vector of partial derivatives of $f$ and is a completely standard notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\nabla^2 f$ seems to be used here in an unusual way, namely to denote the Hessian (the matrix of all second order partial derivatives), $(\partial^2 f/\partial x_i \partial x_j)_{i,j=1}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The usual meaning of $\nabla^2 f$ is the Laplacian, $\partial^2 f/\partial x_1^2 + \ldots + \partial^2 f/\partial x_n^2$.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-20T18:56:13.823" CommentCount="3" />
  <row Id="5083" PostTypeId="1" AcceptedAnswerId="5086" CreationDate="2010-09-20T19:10:45.387" Score="1" ViewCount="57" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to partition given $p(x)$ using a generating function as my textbook on discrete mathematics explains it in detail. However, I want to know if it is possible to restrict the source elements of the partition? So, given I want to find the partition of $x$ is it possible to do it using only elements from $S={y_1, y_2, y_3, ..., y_p : y_i \in \mathbb{Z} \forall i}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example might be partition any number using only 4 and 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a homework question of sorts but this is &lt;strong&gt;not&lt;/strong&gt; the actual homework question (because I can solve that - I just can't find a general answer for it yet and this is a way I think I can generalise (possibly)).&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-09-20T19:54:15.903" Title="Partitioning Integers using only elements from a specific set" Tags="&lt;number-theory&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5084" PostTypeId="1" AcceptedAnswerId="5085" CreationDate="2010-09-20T19:22:14.977" Score="3" ViewCount="77" Body="&lt;p&gt;Given a set of vectors $\{x_1,\ldots,x_n\}$, how can I find a set of orthonormal vectors $\{y_1,\ldots,y_m\colon m\leq n\}$ spanning the same space spanned by $\{x_i\}$?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-20T19:27:13.807" LastActivityDate="2010-09-20T19:28:01.827" Title="Orthogonal Vectors" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="5085" PostTypeId="2" ParentId="5084" CreationDate="2010-09-20T19:28:01.827" Score="5" ViewCount="0" Body="&lt;p&gt;Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram-Schmidt_Process&quot; rel=&quot;nofollow&quot;&gt;the Gram-Schmidt process&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-20T19:28:01.827" />
  <row Id="5086" PostTypeId="2" ParentId="5083" CreationDate="2010-09-20T19:48:26.263" Score="2" ViewCount="0" Body="&lt;p&gt;It's the same generating function method.  If $p_S(n)$ denotes the number of partitions of $n$ using only positive integers in some set $S \subset \mathbb{N}$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} p_S(n) x^n = \prod_{s \in S} \frac{1}{1 - x^s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A popular choice is $S = \{ 1, 5, 10, 25 \}$ (the problem of making change).  When $S$ is finite the above function is rational and it is possible to give a closed form for $p_S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is about whether such a partition exists at all, for sufficiently large $n$ this is possible if and only if the greatest common divisor of the elements of $S$ divides $n$.  For small $n$ there are greater difficulties; see the Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;Frobenius problem&lt;/a&gt;.  When $|S| = 2$ (say $S = \{ a, b \}$) and the elements of $S$ are relatively prime, the largest $n$ for which no such partition exists is known to be $ab - a - b$.  When $a = 4, b = 9$ this gives $23$, so for any $n \ge 24$ such a partition always exists.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-20T19:54:15.903" LastActivityDate="2010-09-20T19:54:15.903" CommentCount="1" />
  <row Id="5087" PostTypeId="2" ParentId="5018" CreationDate="2010-09-20T19:57:12.203" Score="3" ViewCount="0" Body="&lt;p&gt;Khan Academy has excellent short videos you can take at your own pace, from basic arithmetic to advanced calculus. He has a very conversational style as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2022" LastActivityDate="2010-09-20T19:57:12.203" CommunityOwnedDate="2010-09-20T19:57:12.203" />
  <row Id="5088" PostTypeId="1" AcceptedAnswerId="5090" CreationDate="2010-09-20T20:06:43.977" Score="3" ViewCount="108" Body="&lt;p&gt;&lt;em&gt;Just out of pure curiosity ...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I want to connect the two points $(0|0)$ and $(1|1)$ with the graph of some continuous and differentiable function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f : [0; 1] \to [0; 1]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and let $s$ be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arc_length#Finding_arc_lengths_by_integrating&quot; rel=&quot;nofollow&quot;&gt;arc length&lt;/a&gt; of that function in $[0; 1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the function with minimum $s$ that satisfies the above conditions is $f(x) = x$ with $s = \sqrt 2$. So for $s = \sqrt 2$, exactly one matching function can be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;what happens to the number of these functions if $s$ increases?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, more functions can be found to match the given arc length - uncountably many more I suppose due to the nature of the real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But intuitively, I'd think that the &lt;em&gt;number of such functions grows even more the greater $s$ gets&lt;/em&gt;, since there is more &quot;space&quot; the graph can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, despite continuum cardinality, &lt;strong&gt;are there any means of measuring the number of such functions&lt;/strong&gt; against $s$ or is it all the same once that minimal way of $f(x) = x$ as been taken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And would this change if we limited the ways of constructing such functions to e.g. some elementary ones?&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-09-20T20:48:46.797" Title="Number of continuous $[0; 1] \to [0; 1]$ functions for given arc length" Tags="&lt;functional-analysis&gt;&lt;functions&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5089" PostTypeId="2" ParentId="5065" CreationDate="2010-09-20T20:25:00.070" Score="1" ViewCount="0" Body="&lt;p&gt;For any continuous scalar field $f$ on the sphere, you can define a continuous scalar field $g$ as $g(p) = f(p) - f(\bar{p})$, where $\bar{p}$ is the antipodal point of $p$. Unless $g$ is zero everywhere, there is a point $q$ for which $g(p) &gt; 0$ and $g(\bar{q}) = -g(q) &amp;lt; 0$. As $g$ is continuous, a closed contour separating $q$ from $\bar{q}$ exists on which $g$ is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; The last statement &quot;seems obvious&quot; to me, but I don't have enough knowledge of compact topological spaces or manifolds to make it rigorous. Perhaps a counterexample exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit 2:&lt;/em&gt; See &lt;a href=&quot;http://math.stackexchange.com/questions/5065/does-a-continuous-scalar-field-on-a-sphere-have-continuous-loop-of-isothermic-an/5127#5127&quot;&gt;T..'s answer&lt;/a&gt; for a sketch of a counterexample. Apparently I was thinking about the existence of a &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Continuum_%28topology%29#Definitions&quot; rel=&quot;nofollow&quot;&gt;curve&lt;/a&gt;&lt;/em&gt; while the question asked for a &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;loop&lt;/a&gt;&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-09-21T06:39:25.573" LastActivityDate="2010-09-21T06:39:25.573" CommentCount="4" />
  <row Id="5090" PostTypeId="2" ParentId="5088" CreationDate="2010-09-20T20:29:28.300" Score="3" ViewCount="0" Body="&lt;p&gt;There is probably a sophisticated answer to this; let me just mention an unsophisticated one.  Several possible discrete analogues of the question you ask exhibit the desired behavior.  Consider, for example, the problem of finding a path from $(0, 0)$ to $(n, n)$ in the lattice $\mathbb{Z}^2$ where the only allowable steps are between diagonally adjacent lattice points.  Let the length of such a path be the total number of steps.  Then, of course, there is only one path of length $n$, which is the obvious diagonal one.  However, there are ${n+2 \choose 2} + n$ paths of length $n+2$, already a big leap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general answer can be computed as follows.  In a path of length $n+2k$ we need to choose $2j$ of the steps to be up-left or down-right (for some $j$ between $0$ and $k$) and we need to choose $n+2k-2j$ of the steps to be up-right or down-left.  Of the first set of steps, half need to be of one type and half need to be the other, and of the second set of steps, $n+k-j$ need to be of one type and $k-j$ need to be the other.  This gives the total number of paths of length $n+2k$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{k} {n+2k \choose 2j} {2j \choose j} {n+2k-2j \choose k-j}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dominant term when $n$ is large compared to $k$ is the term associated to $j = k$, which is ${n+2k \choose 2k}$ and hence grows asymptotically like $\frac{n^{2k}}{(2k)!}$.  So, again when $n$ is large compared to $k$, it is indeed true that the number of allowable paths grows with the length of the path.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-20T20:29:28.300" />
  <row Id="5091" PostTypeId="2" ParentId="5088" CreationDate="2010-09-20T20:48:46.797" Score="1" ViewCount="0" Body="&lt;p&gt;Here's my take on an intuitive answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are $2^{\aleph_0}$ continuous functions from the unit interval to itself, it is clear that there can be at most that many functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given $s &gt; \sqrt{2}$ (which as you well observed the minimal distance), we define $t = s - \sqrt{2}$ and for any $x\in (0,1)$ we can construct a function that has the wanted arc length, simply by &quot;pulling&quot; the value of $f(x)$ smoothly in a way that extends the length of the arc by $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The construction is very clear when you don't request $f$ to be differentiable, as you can just stick a triangle in the proper length, however the idea should be clear)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have at least $|(0,1)| = 2^{\aleph_0}$ functions as such, which means that you have exactly that many.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-20T20:48:46.797" />
  <row Id="5092" PostTypeId="1" AcceptedAnswerId="5093" CreationDate="2010-09-20T21:47:22.200" Score="4" ViewCount="270" Body="&lt;p&gt;I studied applied math, so each course (except abstract algebra) was dedicated to solution of a similar problems. After those courses it seems that every branch of mathematics has a developed theory with a few unsolved problems. I believe that is wrong. I found out that there is no general approach to Diophantine equations (the 10th Hilbert problem), to problems like &quot;Collatz conjecture&quot; and to nonlinear differential equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the question is: &lt;strong&gt;what is the other branches of mathematics (collections of similar problems) without a general method to solve them?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2016" LastEditorUserId="2016" LastEditorDisplayName="" LastEditDate="2010-09-21T12:14:37.597" LastActivityDate="2010-09-21T13:44:45.603" CommunityOwnedDate="2010-09-20T21:59:25.957" Title="Branches of mathematics not having a general method to solve" Tags="&lt;diophantine-equations&gt;&lt;open-problem&gt;&lt;conjectures&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5093" PostTypeId="2" ParentId="5092" CreationDate="2010-09-20T21:58:43.243" Score="8" ViewCount="0" Body="&lt;p&gt;Diophantine equations are not special; any branch of mathematics in which it is possible to ask sufficiently strong questions will suffer a similar fate due to the existence of problems like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;nofollow&quot;&gt;halting problem&lt;/a&gt; and the possibility of asking a question equivalent to the halting problem (or something similar).  For example, group theory has such problems: it is generally impossible to decide from a presentation of a group whether it is trivial or not.  See &lt;a href=&quot;http://mathoverflow.net/questions/11540/what-are-the-most-attractive-turing-undecidable-problems-in-mathematics&quot; rel=&quot;nofollow&quot;&gt;this list of problems&lt;/a&gt; on MathOverflow.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-20T21:58:43.243" CommunityOwnedDate="2010-09-20T21:59:25.957" CommentCount="1" />
  <row Id="5094" PostTypeId="1" AcceptedAnswerId="5096" CreationDate="2010-09-20T22:45:53.223" Score="0" ViewCount="111" Body="&lt;p&gt;I think this is obvious and I am just missing something but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given some finitely presentable algebra, can we always reduce the relations to commutator relations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And moreover, if yes, then is there a reasonable algorithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, I feel like I should know this, so I apologize if it is obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-20T23:04:10.703" LastActivityDate="2010-09-21T03:27:25.343" Title="Reducing Relations to Commutator Relations" Tags="&lt;abstract-algebra&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5095" PostTypeId="1" AcceptedAnswerId="5103" CreationDate="2010-09-20T22:59:42.003" Score="5" ViewCount="128" Body="&lt;p&gt;I know the limits in the categories of groups, abelian groups and topological spaces and was wondering about the same thing.&lt;/p&gt;&#xA;" OwnerUserId="2026" LastActivityDate="2010-09-21T00:53:31.440" Title="what are the product and coproduct in the category of topological groups" Tags="&lt;category-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5096" PostTypeId="2" ParentId="5094" CreationDate="2010-09-20T23:01:23.320" Score="4" ViewCount="0" Body="&lt;p&gt;(Apologies for putting this in the form of an answer - I haven't got enough rep yet to make it a comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about elements of finite order?  It seems like even a simple algebra such as the one generated by $a^m = b^n = e$ couldn't possibly have a commutator presentation, unless I'm missing something immensely obvious...&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-20T23:01:23.320" CommentCount="1" />
  <row Id="5097" PostTypeId="2" ParentId="4961" CreationDate="2010-09-20T23:31:32.633" Score="3" ViewCount="0" Body="&lt;p&gt;I just remembered something that at the moment seemed pretty nice to me. Usually in an elementary number theory course one is taught that the integer solutions to the diophantine equation $x^2 + y^2 = z^2$ are given by the formulas $x = a^2 - b^2$, $y = 2ab$ and $z = a^2 + b^2$. By working in a naive way in the ring of Gaussian integers $\mathbb{Z}[i] = \{ a + bi \mid a, b \in \mathbb{Z} \}$ one can get these formulas without much effort. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We factor the equation as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x + iy)(x - iy) = z^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then since the right hand side is a square, the two factors on the left must be squares of numbers in $\mathbb{Z}[i]$ so for instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x + iy = (a + bi)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and squaring this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x + iy = a^2 - b^2 + 2abi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now equating real and imaginary parts shows that $x = a^2 - b^2$ and $y = 2ab$. From this one easily gets $z = a^2 + b^2$. Of course all of these manipulations need some more details but it would make for a nice example after the formulas have been derived in the usual way (without complex numbers).&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-09-20T23:38:37.350" LastActivityDate="2010-09-20T23:38:37.350" CommunityOwnedDate="2010-09-20T23:31:32.633" />
  <row Id="5098" PostTypeId="1" CreationDate="2010-09-20T23:57:08.490" Score="0" ViewCount="143" Body="&lt;p&gt;For the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) = x^2/y$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;given that $x$ and $y$ are integers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the easiest/fastest way to find at least one tuple $(x_0,y_0)$ solution to $f(x,y)=4$ or any other number really. I know that $(2,1)$ would work, but is there a process for finding the solution to any integer that is equal to $f(x,y)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any easy way to just enter this in and calculate it in Mathematica, MATLAB, Sage, etc.? Is there any way to work this out by hand besides just guessing which values would work?&lt;/p&gt;&#xA;" OwnerUserId="1815" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T03:29:58.507" LastActivityDate="2010-09-21T03:29:58.507" Title="Finding positive integer solutions to functions with only positive integer input" Tags="&lt;math-software&gt;&lt;mathematica&gt;&lt;matlab&gt;&lt;diophantine-equations&gt;" AnswerCount="4" />
  <row Id="5099" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T00:03:12.930" Score="0" ViewCount="0" Body="&lt;p&gt;In &lt;em&gt;Mathematica&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;FindInstance[x^2/y == 4, {x, y}, Integers, n]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;n&lt;/code&gt; is an integer returns &lt;code&gt;n&lt;/code&gt; possible instances.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-21T00:03:12.930" CommentCount="4" />
  <row Id="5100" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T00:11:16.983" Score="1" ViewCount="0" Body="&lt;p&gt;This could help: $x^2/y=a$, where $x,y,a$ are integers. If $a$ is a perfect square, $y=1$ and $x=\sqrt{a}$. If it isn't a perfect square, suppose $a=p_1^{\alpha_1}...p_k^{\alpha_k}$. Then any $y$ of the form $p_1^{\beta_1}...p_k^{\beta_k}$ such that $\alpha_i +\beta_i$ is even will do, because then $ay$ will be a perfect square. Moreover, $ay$ times any other perfect square is good too.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-21T00:11:16.983" />
  <row Id="5101" PostTypeId="1" AcceptedAnswerId="5102" CreationDate="2010-09-21T00:23:59.827" Score="1" ViewCount="336" Body="&lt;p&gt;I'm learning about proof by contrapositive and by mathematical induction in a computer science class. I'm banging my head trying to solve this problem and would like some help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the sum of the first n even numbers is n^2 + n&#xA;(a) indirectly by assuming that the sum of the first n odd numbers is n^2&#xA;(b) directly by mathematical induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no problem doing (b). But I can't figure out how to do (a) using indirect proof. I can only come up with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of first n numbers is n^2 + n =&gt; numbers are even&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrapositive: numbers are odd =&gt; sum of first n numbers is not n^2 + n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This just doesn't seem logical to me. Any hints would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2028" LastActivityDate="2010-09-21T00:46:04.013" Title="Indirect proof that sum of first n even numbers is n^2 + n" Tags="&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5102" PostTypeId="2" ParentId="5101" CreationDate="2010-09-21T00:32:42.227" Score="3" ViewCount="0" Body="&lt;p&gt;Using the contrapositive isn't the only indirect way of proving a theorem, and I highly doubt it's what your instructors had in mind here.  As a hint for part (a): what can you say about the first $n$ even numbers (individually) compared to the first $n$ odd numbers?  Can you find some convenient correspondence between, say, the 5th even number and the 5th odd number?&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-21T00:32:42.227" CommentCount="4" />
  <row Id="5103" PostTypeId="2" ParentId="5095" CreationDate="2010-09-21T00:44:40.223" Score="10" ViewCount="0" Body="&lt;p&gt;The product of topological groups is simply the product of the underlying groups with the product topology. The universal property is easily verified. The coproduct topology is more complicated. Again, the underlying group is just the coproduct of the underlying groups, which is also called the free product. Here is a description of the topology:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(G_i) (i \in I)$ our family of topological groups. Then consider the class of all topological groups $H$, such that there is a surjective group homomorphism $G:=\coprod_{i \in I} G_i \to H$, such that the composites $G_i \to H$ are continuous. Notice that this class is essentially a set. Namely, consider the set of all quotient groups $G / N$ endowed with some topology. In particular, we may define the topology on $G$ to be the initial one with respect to all maps $G \to H$ as above. Then $G$ is a topological group. For example, the inversion map $G \to G$ is continuous since the composite with every $G \to H$ as above is the composite with $G \to H$ and the inversion map $H \to H$. The universal property is also easy to verify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works in general: If $\tau$ is some type of algebraic structures, then the category of topological $\tau$-algebras is complete and cocomplete, and the forgetful functor to $\tau$-algebras preserves and creates limits and colimits. You can also prove this by applying Freyd's representabilty theorem. In the above example, I have written down a solution set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is hard to describe this topology explicitely (and thus, for instance, proving separation axioms). See for example &lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S0166864101000852&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S0166864103000890&quot; rel=&quot;nofollow&quot;&gt;that&lt;/a&gt; article.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-21T00:53:31.440" LastActivityDate="2010-09-21T00:53:31.440" CommentCount="2" />
  <row Id="5104" PostTypeId="1" AcceptedAnswerId="5113" CreationDate="2010-09-21T00:45:06.603" Score="1" ViewCount="69" Body="&lt;p&gt;When studying group theory, I was interested to learn that a the quotient set $G/N$ of a group $G$ is again a group with $\pi\colon G\to G/N$ a homomorphism when $N$ is a normal subgroup of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more general set of criteria for this situation outside of group theory? As in, for any set $U$ and an equivalence relation $\sim$ on $U$, is there a similar set of criteria for $\sim$ such that $\pi\colon U\to U/\sim$ is a homomorphism in that it respects all functions and relations on $U$? Where by this I mean that $R^{U}(a_1,\dots,a_{\nu_L(R)})\Leftrightarrow R^{U/\sim}(\pi(a_1),\dots,\pi(a_{\nu_L(R)}))$ and $\pi(f^U(a_1,\dots,a_{\nu_L(f)}))=f^{U/\sim}(\pi(a_1),\dots,\pi(a_{\nu_L(f)}))$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-21T04:03:16.997" Title="Quotienting a Set by an Equivalence Relation such that the Natural Projection is a Homomorphism." Tags="&lt;abstract-algebra&gt;&lt;logic&gt;" AnswerCount="2" />
  <row Id="5105" PostTypeId="2" ParentId="5101" CreationDate="2010-09-21T00:46:04.013" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8UmRp.png&quot; alt=&quot;grid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram should help you see the proper approach to proving this.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-21T00:46:04.013" />
  <row Id="5106" PostTypeId="2" ParentId="5104" CreationDate="2010-09-21T00:47:27.913" Score="1" ViewCount="0" Body="&lt;p&gt;Any homomorphism $A \to B$ of structured sets which is surjective has this property.  This is because any function $f : A \to B$ defines an equivalence relation on $A$ whose equivalence classes are the subsets $f^{-1}(b), b \in B$.  So write down all the conditions necessary for $f$ to be a morphism and you'll get all the conditions you need on the corresponding equivalence relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for a surjective function $f : G \to H$ between groups we require that $f(ab) = f(a) f(b)$.  This implies that the preimage of the identity in $H$ is a subgroup $F$ of $G$, and from there it follows that the preimage of $h \in H$ is the coset $g F = F g$ for any $g \in f^{-1}(h)$, hence that $F$ is normal.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-21T02:34:09.403" LastActivityDate="2010-09-21T02:34:09.403" CommentCount="8" />
  <row Id="5107" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T00:59:30.733" Score="0" ViewCount="0" Body="&lt;p&gt;One obvious solution to f(x,y)=a is x=y=a, since a^2/a = a.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-21T00:59:30.733" />
  <row Id="5108" PostTypeId="1" AcceptedAnswerId="5111" CreationDate="2010-09-21T01:01:49.050" Score="12" ViewCount="437" Body="&lt;p&gt;I'm taking a course in Complex Analysis,&#xA;and the teacher mentioned that if we do not&#xA;restrict our attention to analytic functions,&#xA;we would just be looking at functions from $\mathbb{R}^2$ to $\mathbb{R}^2$.&lt;br&gt;&#xA;What I don't understand is why this is not true&#xA;when we do restrict our attention to analytic functions.&#xA;I understand that complex analytic functions&#xA;have different properties than real functions on $\mathbb{R}^2$,&#xA;but I don't understand why this is so.&lt;br&gt;&#xA;If I look at a complex number $z$ as a vector in $\mathbb{R}^2$,&#xA;then isn't differentiability of $w=f(z)$ in $\mathbb{C}$ defined the same way&#xA;as differentiability of $(u,v)=F(x,y)$ in $\mathbb{R}^2$ ?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-09-21T10:50:49.930" Title="How is $\mathbb{C}$ different than $\mathbb{R}^2$?" Tags="&lt;complex-analysis&gt;&lt;complex-variables&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="7" />
  <row Id="5109" PostTypeId="2" ParentId="5108" CreationDate="2010-09-21T01:10:20.613" Score="3" ViewCount="0" Body="&lt;p&gt;The essential point of complex analysis is that the answer to this question is a resounding no.  Differentiability of a function $f$ in the complex sense requires taking a limit in which the limiting variable is a complex number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(z) = \lim_{w\to z} \frac{f(w) - f(z)}{w-z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this limit to exist, one needs convergence uniformly in all directions (think through the $\epsilon$-$\delta$ definition of this limit).  On the other hand, a function from $\mathbb{R}^2\to \mathbb{R}^2$ (or more generally $\mathbb{R}^n\to \mathbb{R}^m$) is differentiable (in the sense explained in Ryan's answer) at all points in an open neighborhood if and only if all the partial derivatives exist at every point and are continuous on the given neighborhood.  Of course that forces all the directional derivatives to exist, but doesn't require the same sort of uniformity from different directions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Wikipedia page is a good place to learn some more: &lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Holomorphic_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cauchy-Riemann equations, as Ryan's answer explains, relate the directional derivatives of the real and complex parts of a holomorphic function and give rise to all sorts of interesting properties.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-09-21T03:37:31.913" LastActivityDate="2010-09-21T03:37:31.913" CommentCount="7" />
  <row Id="5110" PostTypeId="2" ParentId="2865" CreationDate="2010-09-21T01:16:32.377" Score="2" ViewCount="0" Body="&lt;p&gt;You can construct colimits by a rather general procedure. See my answer &lt;a href=&quot;http://math.stackexchange.com/questions/5095/what-are-the-product-and-coproduct-in-the-category-of-topological-groups&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topologies mentioned by Agusti are not correct. The continuity of the group laws does not hold in general. The problem is that quotient topologies do not commute with product topologies.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-21T01:16:32.377" CommentCount="2" />
  <row Id="5111" PostTypeId="2" ParentId="5108" CreationDate="2010-09-21T01:27:57.953" Score="17" ViewCount="0" Body="&lt;p&gt;For a function $f : \mathbb R^2 \to \mathbb R^2$ &quot;differentiable&quot; at a point $x \in \mathbb R^2$ means you have a linear approximation $f'_x$ which satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{y \to x} \frac{f(x)-f(y)-f'_x(x-y)}{|x-y|} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying that $f$ is complex analytic is the constraint that $f'_x$ is a complex linear function for all $x$.  &quot;Complex linear&quot; means that not only is it true that $f'_x(av+bw)=af'_x(v)+bf'_x(w)$ for $a, b \in \mathbb R$, but it also holds for $a,b \in \mathbb C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to say $L$ is &quot;Complex linear&quot; is that $L$ is a regular (real linear) function plus $L(iv)=iL(v)$ for all vectors $v$.   Stated in terms of the derivative, the directional derivative of $f$ in the direction $(0,1)$ is $i$ times the directional derivative in the direction $(1,0)$.  In component notation these are the &quot;Cauchy-Riemann&quot; equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where if you write $f(z) = f(x+iy) = u(x,y) + iv(x,y)$ translates to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}, &#xA;   \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives the key picture.  A complex linear map means that the map looks like the composite of a rotation together with a re-scaling map $v \longmapsto av$ where $a \in \mathbb R$.  These maps are called &quot;conformal&quot;.  The nice things about conformal linear maps is they preserve all angles -- they do not always preserve length.  So the &quot;nice&quot; thing about complex differentiable maps is that if you have any collection of curves in the plane, and you apply your complex differentiable function to them, it preserves the angles of intersection of your curves.  That's a very special property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: An instructive example would be to think through two different functions from $\mathbb R^2$ to $\mathbb R^2$.  The first function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y) \longmapsto x(\cos(y),\sin(y))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the second function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y) \longmapsto e^x(\cos(y),\sin(y))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first function preserves the angles between the coordinate grid lines -- curves like $x=a$ and $y=b$ in the coordinate plane.  The first function is not complex differentiable but the second is!  So this means that the second function preserves all angles (not just the coordinate lines $x=a, y=b$).  Can you spot curves in the domain which intersect in some angle $\theta$, but when after you compose them with the 1st function, their angle of intersection is not $\theta$? &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-21T01:53:54.667" LastActivityDate="2010-09-21T01:53:54.667" CommentCount="1" />
  <row Id="5112" PostTypeId="2" ParentId="5098" CreationDate="2010-09-21T02:57:03.617" Score="0" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematica is great. Mathematica is a crutch. If you depend on a calculator, you won't get the meaning of your answer and you won't know what to do with it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you know nothing about the RHS, to get a 'minimal' answer by generate and test, go in rings around the origin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{0,0}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1,0},{0,1},{-1,0},{0,-1}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{2,0},{1,1},{0,2},{-1,1},{-2,0},{-1,-1},{0,-2},{1,-1}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&#xA;(is that how Mathematics's FindInstance works?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the RHS is $x^2/y$, to understand the problem, what is the thing  that causes immediate problems with integers (getting integer solutions)? It's the division. If $x$ divides $y^2$ (in order to make f an integer) $x$ has to divide $y$. That is, beyond yrudoy's answer, you can make $x$ be any divisor of $y$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2010-09-21T02:57:03.617" />
  <row Id="5113" PostTypeId="2" ParentId="5104" CreationDate="2010-09-21T03:07:41.240" Score="2" ViewCount="0" Body="&lt;p&gt;Assuming I've understood your question correctly and you are only interested in equivalence relations that respect &lt;em&gt;some&lt;/em&gt; given operations and/or relations (as opposed to &quot;all&quot; operations and relations at the same time, which I suspect would lead you to only the identity or the total relation), let me give you the punchline first: an equivalence relation will &quot;respect the operations and the relations&quot; if and only if it is a substructure when considered as a subset of $U\times U$, where $U\times U$ is given the &quot;coordinatewise&quot; structure in the obvious way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept you are looking for is that of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Congruence_relation#Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;congruence&lt;/strong&gt;&lt;/a&gt;, and is studied in universal algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you have a set $U$ with operators $\{f_i\}_{i\in I}$, with $f_i$ of arity $n_i$; and relations $\{R_j\}_{j\in J}$ with $R_j$ and $m_j$-ary relation. Let $\sim$ be an equivalence relation which &quot;respects the operations&quot;, so that if $a_1,\ldots,a_{n_i},b_1,\ldots,b_{n_i}\in U$, $a_k\sim b_k$, then $f_i(a_1,\ldots,a_{n_i})\sim f_i(b_1,\ldots,b_{n_i})$, and and if $c_1,\ldots,c_{m_j},d_1,\ldots,d_{m_j}\in U$ with $c_k\sim d_k$ and $(c_1,\ldots,c_{m_j})\in R_j$, then $(d_1,\ldots,d_{m_j})\in R_j$. Then $\sim$, considered as subset of $U\times U$ is closed under each of $f_i\times f_i$, and the restrictions of $R_j\times R_j$ work just as needed to make $\sim$ into a substructure of $U\times U$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $\sim$ is an equivalence relation that is also a substructure of $U\times U$, then if $a_k\sim b_k$ you have $(a_1,b_1),\ldots,(a_{n_i},b_{n_i})\in \sim$, so $$f_i\times f_i\bigl((a_1,b_1),\ldots,(a_{n_i},b_{n_i})\bigr) = \bigl(f_i(a_1,\ldots,a_{n_i}),f_i(b_1,\ldots,b_{n_i})\bigr)\in\sim,$$&#xA;hence $f_i(a_1,\ldots,a_{n_i})\sim f_i(b_1,\ldots,b_{n_i})$; similarly with the relations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an equivalence relation &quot;respects the operations and relations&quot; if and only if it is a substructure of $U\times U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a theorem that is completely analogous to the usual Isomorphism Theorems for groups and rings, for congruences (in fact, the theorems for groups and rings are special cases). If you are interested in this, you can see Gratzer's book (titled &quot;Universal Algebra&quot;), or George Bergman's &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;An Invitation to General Algebra and Universal Constructions&lt;/a&gt;. (Though the material on universal algebra is in Chapter 8; Gratzer will get to this much quicker). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You need to be careful with relational algebras; in general, defining a substructure is a bit tricky; if you think of functions as special kinds of relations, you can see that you need some kind of &quot;relational closure&quot; for your set to be a substructure; but if you define substructures adequately, then the condition on $\sim$ that it be a substructure of $U\times U$ will take care of it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-21T04:03:16.997" LastActivityDate="2010-09-21T04:03:16.997" CommentCount="2" />
  <row Id="5114" PostTypeId="2" ParentId="5094" CreationDate="2010-09-21T03:27:25.343" Score="4" ViewCount="0" Body="&lt;p&gt;No, you cannot even do it in groups. For if you could, consider the identity $x^n=e$ with $n&gt;1$. If this were equivalent to a set of commutator words, then every abelian group would satisfy the relation, but of course this is false: only those abelian groups that are of exponent $n$ satisfy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, in groups every identity is equivalent to a pair of identities, one a commutator word and one a power word; equivalently, if $w$ is a word in the free group $F$ on countably many generators $x_1,\ldots,x_n,\ldots$, then there exist words $w_1$ and $w_2$, with $w_1=x_1^a$ for some positive integer $a$, and $w_2\in [F,F]$, such that for all groups $G$, $G$ satisfies $w$ as an identity (every evaluation of $w$ in $G$ is trivial) if and only if $G$ satisfies $w_1$ and $w_2$ as identities. This is due to Bernhard Neumann; write $w=x_1^{n_1}\cdots x_m^{n_m}c$, with $c$ a commutator word, by using the collection process. Then let $a=\gcd(n_1,\ldots,n_m)$, and $w_2=c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least in groups you also need &quot;power identities&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-21T03:27:25.343" />
  <row Id="5115" PostTypeId="1" CreationDate="2010-09-21T03:45:23.580" Score="3" ViewCount="723" Body="&lt;p&gt;So for the formula 1/x, If you were to add up all y values from x=1 to x=∞, wouldn't the sum approach a number because even though you are always adding, aren't you just adding smaller and smaller numbers? Wouldn't this mean that it approached a certain number?&lt;/p&gt;&#xA;" OwnerUserId="2033" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-21T14:39:27.730" LastActivityDate="2010-10-02T20:03:09.010" Title="Why does 1/x diverge?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;integral&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5116" PostTypeId="2" ParentId="4414" CreationDate="2010-09-21T03:46:49.333" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, absolutely you can introduce definitions as axioms.  That's what Metamath does:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://us.metamath.org/mpegif/mmset.html#definitions&quot; rel=&quot;nofollow&quot;&gt;http://us.metamath.org/mpegif/mmset.html#definitions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hss.cmu.edu/philosophy/techreports/181_Avigad.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.hss.cmu.edu/philosophy/techreports/181_Avigad.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which defines a formal system with definitions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We extend the foundational framework in two ways. First, we allow for explicit deﬁnitions of new predicates and functions on the universe of sets. And,&#xA;  second, we allow function symbols to denote functions that are only partially de-&#xA;  ﬁned, using a logic of partial terms. We call the resulting formal system DZFC .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-21T03:46:49.333" />
  <row Id="5117" PostTypeId="2" ParentId="5115" CreationDate="2010-09-21T03:54:19.810" Score="14" ViewCount="0" Body="&lt;p&gt;Ignoring the 1, if you group together the first term, then the next two terms, then the next four terms, and so on, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is greater than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where now each group is exactly equal to 1/2.  This shows that the sum of the first $2^n$ terms is at least $1 + 1/2 \cdot n$, and so the sum of all the terms is unbounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know a little bit of calculus, $\int dx/x=\log x$, so $1+1/2+1/3+\cdots+1/n\ge\int_1^ndx/x=\log n-\log1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/5035&quot;&gt;http://math.stackexchange.com/questions/5035&lt;/a&gt; which discusses precisely how fast the series diverges.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-21T03:54:19.810" CommentCount="1" />
  <row Id="5118" PostTypeId="2" ParentId="5115" CreationDate="2010-09-21T03:59:58.100" Score="4" ViewCount="0" Body="&lt;p&gt;Yes it is true that the numbers you are adding are getting smaller and smaller. The key is that they do not get small quick enough. There are many proofs that can be found easily online (search for proof that the harmonic series diverges) that show that you can add up enough terms of the harmonic series to make its sum as large as you wish. The key is that the rate of growth of the (partial sums of the) harmonic series is logarithmic. Even though $\ln(x)$ grows &lt;strong&gt;extremely&lt;/strong&gt; slowly, it can still be made larger than any fixed value given sufficiently large $x$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-09-21T03:59:58.100" />
  <row Id="5119" PostTypeId="1" AcceptedAnswerId="5121" CreationDate="2010-09-21T04:26:32.180" Score="1" ViewCount="354" Body="&lt;p&gt;If $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \dotsb$ are sigma algebras, what is wrong with claiming that $\cup_i\mathcal{F}_i$ is a sigma algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems closed under complement since for all $x$ in the union, $x$ has to belong to some F_i, and so must its complement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems closed under countable union, since for any countable unions of $x_i$ within it, each of the $x_i$ must be in some $F_j$, and so we can stop the sequence at any point and take the highest $j$ and we know that all the $x_i$'s up to that point are in $F_j$, and thus so must be their union.  There must be some counterexample, but I don't see it.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-10-07T03:38:17.343" Title="Sigma algebra question" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="5120" PostTypeId="2" ParentId="5119" CreationDate="2010-09-21T04:37:12.093" Score="0" ViewCount="0" Body="&lt;p&gt;If $x_1\in F_1$, $x_2\in F_2$, and so forth, then the infinite union $x_1 \cup x_2 \cup \cdots$ does not lie in any one $F_j$, and therefore does not lie in the union of the $F_j$'s.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-21T04:37:12.093" CommentCount="1" />
  <row Id="5121" PostTypeId="2" ParentId="5119" CreationDate="2010-09-21T04:42:48.320" Score="5" ViewCount="0" Body="&lt;p&gt;The problem arises in the countable union; your argument is correct as far as it goes, but from the fact that $\cup_{i=1}^n x_i\in \cup_{i=1}^{\infty}F_i$ for each $n$ you cannot conclude that $\cup_{i=1}^{\infty} x_i$  lies in $\cup_{i=1}^{\infty} F_i$: the full union must be in one of the $F_j$ in order to be in $\cup_{i=1}^{\infty}F_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explicit example, take $X=\mathbb{N}$; let $F_n$ be the sigma algebra that consists of all subsets of $\{1,\ldots,n\}$ and their complements in $X$. Now let $x_i=\{2i\}$. Then each $x_i$ is in $\cup F_i$, but the union does not lie in any of the $F_k$, hence does not lie in $\cup F_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; In this example, $\cup_{i=1}^{\infty}F_n$ is the algebra of subsets of $X$ consisting of all subsets that are either finite or cofinite, so any infinite subset with infinite complement will not lie in the union, and such a set can always be expressed as a countable union of elements of $\cup F_i$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-21T19:46:24.373" LastActivityDate="2010-09-21T19:46:24.373" />
  <row Id="5122" PostTypeId="1" AcceptedAnswerId="5133" CreationDate="2010-09-21T05:03:41.010" Score="0" ViewCount="89" Body="&lt;p&gt;I was considering the group of automorphisms on a vector space $\mathbb{Q}^3$ with the binary operation of addition. Is there a general description of the elements of this group? At first, it seemed to be that it would consist of bijective linear transformations, but I'm not sure how informative that is. Is there a specific category that this automorphism group would fall under?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-09-21T08:28:51.533" Title="Is there a particular characterization of $Aut(\mathbb{Q}^3)$?" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5123" PostTypeId="2" ParentId="4988" CreationDate="2010-09-21T05:06:34.000" Score="4" ViewCount="0" Body="&lt;p&gt;The kind of ordering you specify is known to economists as 'lexicographic preferences'. There is no objective function which can represent these preferences. Here is an outline of the proof for two variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume on the contrary that there is such a function $f(x,y)$. It must be the case for any $x$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,0) &amp;lt; f(x,1),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can find a rational number, say $q(x)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,0) &amp;lt; q(x) &amp;lt; f(x,1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $x &amp;lt; y$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q(x) &amp;lt; f(x,1) &amp;lt; f(y,0) &amp;lt; q(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $q(\cdot)$ is a one-to-one mapping from the set of real numbers into the set of rational numbers which is impossible since the set of real numbers is uncountable while the set of rational numbers is countable.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-21T05:06:34.000" CommentCount="4" />
  <row Id="5124" PostTypeId="1" AcceptedAnswerId="5125" CreationDate="2010-09-21T05:12:27.760" Score="1" ViewCount="233" Body="&lt;p&gt;I'm actually looking for more intuition behind what's going on than a simple answer.  I understand that the Borel sets are the smallest sigma algebra that contain all of the open sets.  I can see how open intervals and their unions are in this set, and how it makes sense to measure them.  It's hard to see why you'd want to measure rational numbers (the probability that a normally distributed variable takes on a rational number?), and more importantly, why the sigma algebra should have to contain rational numbers (why can't we get away with a sigma algebra that doesn't contain the set of rational numbers?)&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-09-29T06:54:30.900" Title="Why are the sets of rational and irrational numbers Borel sets (over the reals)?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5125" PostTypeId="2" ParentId="5124" CreationDate="2010-09-21T05:17:06.120" Score="1" ViewCount="0" Body="&lt;p&gt;If you want your sigma algebra to contain all of the one-point sets, so in particular if it's the Borel sigma-algebra generated by a $T_1$ topology, then taking countable unions yields all of the countable sets, including the set of rational numbers in your case.  There are sigma-algebras of subsets of the real numbers that don't contain the rational numbers, but if you're OK with open intervals, then you have to be OK with points.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-21T05:17:06.120" CommentCount="1" />
  <row Id="5126" PostTypeId="1" CreationDate="2010-09-21T05:31:10.720" Score="1" ViewCount="135" Body="&lt;p&gt;Can anyone recommend good, free online articles or books to learn Discrete mathematics? When I google'd for them, I came across few resources..but don't know whether they are good to start learning with. It's better to know a tutorial or a book that get you started in Discrete Mathematics. Can readers of SE Mathematics can suggest a good starting point?&lt;/p&gt;&#xA;" OwnerUserId="2036" LastActivityDate="2010-09-21T11:14:46.630" CommunityOwnedDate="2010-09-21T08:16:45.550" Title="Free resources to start learning Discrete Mathematics " Tags="&lt;books&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5127" PostTypeId="2" ParentId="5065" CreationDate="2010-09-21T05:51:03.533" Score="2" ViewCount="0" Body="&lt;p&gt;[edit: as predicted, there is a counterexample with no continuous loop. See addition below.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you may not quite get a loop, only a topological continuum (a compact connected subset) on the sphere whose complement contains multiple components.  The continuum can be gotten by elaborating Rahul's answer to choose a suitable component of the $g=0$ locus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of topologically wild continua such as the &quot;Warsaw circle&quot; suggests that you can draw such a creature on the sphere or projective plane and then extend to a continuous function that would give a counterexample.  Or you could take a field that has the equator as the locus of isothermal antipodes ($g=0$) and try to perform a (antisymmetric) bending construction that modifies parts of the equator, turning it into a wild curve that cannot be traced by a continuous loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[added:  the extension construction would work as follows.  Take two opposite points on the equator.  Join them with a wild continuum in one hemisphere, and the antipode of that continuum in the opposite hemisphere. Define $f(x)$ to be the distance to the wild thing, in one hemisphere, and the negative of distance to the wild thing, in the opposite hemisphere.   Hence $f(x) = -f(-x)$ on the whole sphere, and $f=0$ only on the wild construction that cannot be traversed continuously by a path.] &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-24T15:44:58.387" LastActivityDate="2010-09-24T15:44:58.387" CommentCount="2" />
  <row Id="5128" PostTypeId="2" ParentId="5122" CreationDate="2010-09-21T06:21:56.903" Score="2" ViewCount="0" Body="&lt;p&gt;The distinction between preserving addition and also preserving scalar multiplication is illusory in this case, because any vector can be divided by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction resurfaces, however, for the same problem with $Q$ replaced by (almost) any other field, such as the real numbers.  There preserving scalar multiplication is an additional property that does not follow from addition being conserved.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-21T06:21:56.903" CommentCount="5" />
  <row Id="5129" PostTypeId="1" AcceptedAnswerId="5130" CreationDate="2010-09-21T06:26:58.610" Score="3" ViewCount="169" Body="&lt;p&gt;Hi - I have the Cartesian coordinates of the hypotenuse 'corners' in a right angle triangle. I also have the length of the sides of the triangle. What is the method of determining the coordinates of the third vertex where the opposite &amp;amp; adjacent sides meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Kevin.&lt;/p&gt;&#xA;" OwnerUserId="1529" LastEditorUserId="1529" LastEditorDisplayName="" LastEditDate="2010-09-21T06:35:40.807" LastActivityDate="2010-09-21T12:48:56.120" Title="Given coordinates of hypotenuse, how can I calculate coordinates of other vertex?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5130" PostTypeId="2" ParentId="5129" CreationDate="2010-09-21T06:37:13.393" Score="5" ViewCount="0" Body="&lt;p&gt;You have two points $A=(a,b)$ and $B=(c,d)$ and want a point $P$&#xA;at given distances from $A$ and $B$, say $l$ and $m$. Then $|PA|^2=l^2$&#xA;and $|PB|^2=m^2$ that is&#xA;$$(x-a)^2+(y-b)^2=l^2\qquad\qquad(1)$$&#xA;and&#xA;$$(x-c)^2+(y-d)^2=m^2.\qquad\qquad(2)$$&#xA;Subtracting (2) from (1) gives a linear equation. Use this to eliminate&#xA;one variable from (1). This yields a quadratic equation in the other variable.&#xA;Solving this will give the two possible positions for $P$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-21T06:37:13.393" CommentCount="1" />
  <row Id="5131" PostTypeId="1" AcceptedAnswerId="5135" CreationDate="2010-09-21T07:30:17.017" Score="1" ViewCount="137" Body="&lt;p&gt;Here is a question that I can't find the answer to in my notes or my textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any relation between $\text{Sym}_n$ and $\text{Sym}_k$ where $k &amp;lt; n$? Is $\text{Sym}_k \subset \text{Sym}_n$? Since $k! |  n!\text{ } \forall  k &amp;lt; n $ they satisfy Lagrange's theorem, but of course that doesnt guarantee the existence of a subgroup&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use this result to show that that there is a transitive action of a finite group $G$ on a smaller finite set $S$ by showing that $G$ or a subgroup of $G$ is isomorphic to $\Sigma (S)$ (The permutation group of $S$) which has a natural transitive action. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-21T08:30:45.137" LastActivityDate="2010-09-21T11:50:45.277" Title="Relationships between Symmetric Groups" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;homework&gt;&lt;symmetric-groups&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5132" PostTypeId="2" ParentId="5131" CreationDate="2010-09-21T07:37:57.587" Score="2" ViewCount="0" Body="&lt;p&gt;Yes. $Sym_k &amp;lt; Sym_n$ for $n &amp;lt; k$ and it corresponds to a subgroup of all permutations with the same $k-n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is, fix $k-n$ elements first, then look at all the transpositions that keep them in place. This subgroup of $S_k$ is isomorphic to $S_n$.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-21T07:37:57.587" />
  <row Id="5133" PostTypeId="2" ParentId="5122" CreationDate="2010-09-21T08:28:51.533" Score="2" ViewCount="0" Body="&lt;p&gt;A vector space over $\mathbb{Q}$ is the same as a torsionfree divisible abelian group. The product $p/q * v$ is the unique vector so that $q * p/q v = p v$. In particular, every group homomorphism between vector spaces over $\mathbb{Q}$ is automatically $\mathbb{Q}$-linear. Thus, for example, the group automorphisms of $\mathbb{Q}^n$ may be described by $GL_n(\mathbb{Q})$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-21T08:28:51.533" />
  <row Id="5134" PostTypeId="1" CreationDate="2010-09-21T09:23:27.450" Score="-1" ViewCount="78" Body="&lt;p&gt;Let $F$ be a Banach function space and $(f_n) \subseteq F$. Then is it true that $\|\liminf f_n \|_F \leq \liminf \|f_n\|_F$? If so, how to prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't hold in general, what about the special case when $f_n \geq 0$ for all $n$?&lt;/p&gt;&#xA;" OwnerUserId="2041" LastActivityDate="2010-09-21T10:40:42.597" Title="Interchanging lim inf and norm" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="5135" PostTypeId="2" ParentId="5131" CreationDate="2010-09-21T10:30:18.830" Score="5" ViewCount="0" Body="&lt;p&gt;$S_n$ is not contained in $S_m$ when $n&amp;lt;m$ if you define $S_k$ to be the set of all bijections $\{1,\dots,k\}\to\{1,\dots,k\}$, simply because an element of $S_n$ has domain $\{1,\dots,n\}$ and elements of $S_m$ have domain $\{1,\dots,m\}$, and the two sets are &lt;em&gt;different&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is a silly problem, so surely there is some way out...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose again that $n&amp;lt;m$. There is a map $\phi:S_n\to S_m$ such that whenever $\pi\in S_n$, so that $\pi:\{1,\dots,n\}\to\{1,\dots,n\}$ is a bijection, then $\phi(\pi):\{1,\dots,m\}\to\{1,\dots,m\}$ is the map given by $$\phi(\pi)(i)=\begin{cases}\pi(i),&amp;amp;\text{if $i\leq n$;}\\i,&amp;amp;\text{if $i&amp;gt;n$.}\end{cases}$$ You can easily check that $phi$ is well-defined (that is, that $\phi(\pi)$ is a bijection for all $\pi\in S_n$) and that moreover $\phi$ is a group homomorphism which is &lt;em&gt;injective&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from this that the image of $\phi$ is a subgroup of $S_m$ which is isomorphic to $S_n$. It is usual to &lt;em&gt;identify&lt;/em&gt; $S_n$ with its image $\phi(S_n)\subseteq S_m$, and then we can say that «$S_n$ is a subgroup of $S_m$», but this is nothing but a &lt;em&gt;façon de parler&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted, though, that there are &lt;em&gt;many&lt;/em&gt; injective homomorphisms $S_n\to S_m$, so there are many ways to identify a subgroup of $S_m$ with $S_n$. The one I described above is nice because it looks very natural.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-21T11:50:45.277" LastActivityDate="2010-09-21T11:50:45.277" CommentCount="1" />
  <row Id="5136" PostTypeId="2" ParentId="5134" CreationDate="2010-09-21T10:40:42.597" Score="1" ViewCount="0" Body="&lt;p&gt;In the special case of $F = L^1(\Omega)$ for a measure space $\Omega$, your result is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fatou%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Fatou's lemma&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that the answer to your question in general would have to depend on the particular norm.  For the Lebesgue $L^1$ norm the answer is &quot;yes,&quot; but I wouldn't think you could say any more in general.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a particular Banach space in mind? Your question implies you're looking at some sort of space of real-valued functions. Otherwise lim inf wouldn't make sense.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-21T10:40:42.597" />
  <row Id="5137" PostTypeId="2" ParentId="5108" CreationDate="2010-09-21T10:50:49.930" Score="5" ViewCount="0" Body="&lt;p&gt;Analytic functions map tiny disks to tiny disks.  (Of course that's not rigorous, but you could make it rigorous by putting in the right limit language.)  Analytic functions can shift, stretch, and rotate disks, but they can't flip disks over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smooth functions of two real variables can map disks to ellipses.  That is, they can stretch a disk more in one direction than in another.  Complex analytic functions can't do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex conjugation is not analytic because it flips disks over.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-21T10:50:49.930" />
  <row Id="5138" PostTypeId="2" ParentId="5126" CreationDate="2010-09-21T11:14:46.630" Score="1" ViewCount="0" Body="&lt;p&gt;Try MIT Open Courseware -- Course 6.042J (Mathematics for Computer Science) on &lt;a href=&quot;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/&lt;/a&gt; would be a good place to start. &lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-09-21T11:14:46.630" CommunityOwnedDate="2010-09-21T11:14:46.630" />
  <row Id="5139" PostTypeId="1" AcceptedAnswerId="5181" CreationDate="2010-09-21T11:56:36.070" Score="2" ViewCount="247" Body="&lt;p&gt;This an Exercise from Apostol's analytic number theory. I have been struggling with this problem for quite some time. Looks elementary though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to prove that this quantity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Biggl[\frac{8n+13}{25}\Biggr] - \Biggl[\frac{n-12 - \Bigl[\frac{n-17}{25}\Bigr]}{3}\Biggr]$$ is independent of $n$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-21T21:59:50.207" Title="Proving $\Bigl[\frac{8n+13}{25}\Bigr] - \Bigl[\frac{n-12 - \Bigl[\frac{n-17}{25}\Bigr]}{3}\Bigr]$ is independent of $n$." Tags="&lt;analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5140" PostTypeId="2" ParentId="5139" CreationDate="2010-09-21T12:09:28.623" Score="8" ViewCount="0" Body="&lt;p&gt;Write $n=25m+r$ with $0\leq r&amp;lt;25$, and replace in the equation.You'll see that the result only depends on $r$. Now make the $25$ computations corresponding to each of the possible values of $r$, and check that the result is $4$ in all those finitely many cases.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-21T12:09:28.623" CommentCount="4" />
  <row Id="5141" PostTypeId="2" ParentId="5129" CreationDate="2010-09-21T12:48:56.120" Score="2" ViewCount="0" Body="&lt;p&gt;While I'd use the same algebra as &lt;a href=&quot;http://math.stackexchange.com/questions/5129/given-coordinates-of-hypotenuse-how-can-i-calculate-coordinates-of-other-vertex/5130#5130&quot;&gt;Robin Chapman's solution&lt;/a&gt;, my first thought on this problem yields a third circle equation.  The circumcenter of a right triangle is at the midpoint of its hypotenuse.  Given the endpoints of the hypotenuse, $A=(a,b)$ and $B=(c,d)$, and letting $h$ be the length of the hypotenuse, the circumcircle has equation&#xA;$$\left(x-\frac{a+c}{2}\right)^2+\left(y-\frac{b+d}{2}\right)^2=\left(\frac{h}{2}\right)^2.$$&#xA;This may look a bit intimidating in symbols, but isn't really any different to work with for solving than the other two circles.  I don't know that this offers any advantages, though.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-21T12:48:56.120" CommentCount="2" />
  <row Id="5143" PostTypeId="1" AcceptedAnswerId="5189" CreationDate="2010-09-21T13:36:22.633" Score="1" ViewCount="103" Body="&lt;p&gt;I have a series of inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{1min} \leq y_{1}(x) \leq y_{1max}$$&#xA;$$y_{2min} \leq y_{2}(x) \leq y_{2max}$$&#xA;$$..$$&#xA;$$y_{nmin} \leq y_{n}(x) \leq y_{nmax}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x\in\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, is there a &lt;strong&gt;general method&lt;/strong&gt; that allows me to find the range of $x$ that satisfies the above inequalities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are easy if there is just one inequality, and $y_1(x)$ is at most a quadratic function. But I am looking for a general solution here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: My $y$ function can roughly be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\Biggl[\frac{(x^2+bx+c)^f}{(dx+e)^g}\Biggr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ and $e$ are small comparatively.$(dx+e)^g$ must be positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one of the $y_i(x)$, $f=5/3, g=2/3$, but in another, $f=1, g=1$. It is safe to assume that both $f,g\in\mathbb{Q}$, that is they are &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_number&quot; rel=&quot;nofollow&quot;&gt;rational numbers&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b,c,d,e\in\mathbb{R}$ , and they vary from one $y_i(x)$ to another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Edit:Let's make this question a little general; instead of letting $y(x)$ depends on one variable, what if $y$ depending on multiple variables? That is, how to find the range for $(x_1,x_2,x_3,..., x_i)$, given that  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{1min} \leq y_{1}(x_1,x_2,x_3,..., x_i) \leq y_{1max}$$&#xA;$$y_{2min} \leq y_{2}(x_1,x_2,x_3,..., x_i) \leq y_{2max}$$&#xA;$$..$$&#xA;$$y_{nmin} \leq y_{n}(x_1,x_2,x_3,..., x_i) \leq y_{nmax}$$&#xA;&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-22T00:47:47.083" LastActivityDate="2010-09-22T00:47:47.083" Title="Find the range of $x$, given $y_{min} \leq y(x) \leq y_{max}$, where $y(x) $ can be any function ( Updated)" Tags="&lt;optimization&gt;&lt;numerical-methods&gt;&lt;nonlinear-optimization&gt;" AnswerCount="2" CommentCount="28" />
  <row Id="5144" PostTypeId="2" ParentId="5092" CreationDate="2010-09-21T13:44:45.603" Score="3" ViewCount="0" Body="&lt;p&gt;There is no general method for solving differential equations.  There are general methods for proving existence and uniqueness of solutions, but not for computing solutions (if that's even possible) or for studying detailed properties of solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make a tiny change to a differential equation and it can go from being well understood to being an area of research.  Or it can go from something one group of mathematicians study to something another group studies.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-21T13:44:45.603" CommunityOwnedDate="2010-09-21T13:44:45.603" CommentCount="1" />
  <row Id="5145" PostTypeId="1" AcceptedAnswerId="5155" CreationDate="2010-09-21T13:52:11.043" Score="4" ViewCount="277" Body="&lt;p&gt;The CASIO fx-7400G calculator is very quick at doing $(1+1/1000000)^{1000000}$, giving an output of 2.718280469.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica (v7) takes over a minute, running on an Intel Core2 Duo processor, and its output is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dP0ac.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With n=10000000, my patience ran out. The calculator is quick again with an output of 2.718281693.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know why this happens?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-09-21T16:12:17.713" LastActivityDate="2010-09-21T16:30:21.757" Title="Casio hand calculator vs Mathematica at (1+1/n)^n" Tags="&lt;math-software&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="5146" PostTypeId="2" ParentId="5143" CreationDate="2010-09-21T13:54:22.603" Score="1" ViewCount="0" Body="&lt;p&gt;For a start, suppose we can find the interval $I_i\subseteq\mathbb{R}$ such that $y_{imin}\leq y_i(x)\leq y_{imax}$. Then the interval $I$ which satisfies all the constraints is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=\bigcap_{i=1}^n I_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can concentrate on finding $I_i$ separately for each $1\leq i\leq n$. Here I'm sure we will need some conditions on $y_i(x)$. Do your functions have compact support? Are they continuous? Do they oscillate wildly in any region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the general case - finding a feasible point&lt;/strong&gt;&#xA;You mentioned that $y_i(x)$ are continuous. Lets suppose that in addition they are continuously differentiable. I'll do the problem with $1$ constraint for easy of notion but the general case is exactly the same though more difficult to write down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite the constraints as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_1(x)\geq y_{1min}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-y_1(x)\geq -y_{1max}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $A(x) = \left(\nabla y_1(x),\nabla(-y_1(x))\right)^T$ (the jacobian matrix of the constraints). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also let $b=\left(y_{1min},-y_{1max}\right)^T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to find a $x_0$ such that $Ax_0\geq b$. We do this by solving the following (phase-1) problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $x_{guess}\in\mathbb{R}^n$. Let $r=\min(b-Ax_{guess},0)$ and solve the problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min_{x\in\mathbb{R}^n,\mu\in\mathbb{R}}$  $\mu$ subject to $Ax + \mu r\geq b$ and $\mu\geq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you may not think that we have come very far but because $(x,\mu)=(x_{guess},1)$ is an initial feasible point for this minimisation problem it can be solved by standard software. If the optimal point gives an optimal value $&gt;0$ then the orignal problem is not feasible. If the optimal value is $0$ then the orignal problem is feasible and the optimal point is a feasible point for the orignal problem. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-21T14:40:11.597" LastActivityDate="2010-09-21T14:40:11.597" CommentCount="4" />
  <row Id="5147" PostTypeId="2" ParentId="5145" CreationDate="2010-09-21T14:11:08.300" Score="1" ViewCount="0" Body="&lt;p&gt;I think your calculator might use $$ a^b = \exp(b\ln a)$$ together with precalculated $\exp$ and $\ln$ tables to speed things up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I would =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-09-21T14:11:08.300" />
  <row Id="5148" PostTypeId="1" AcceptedAnswerId="5151" CreationDate="2010-09-21T14:19:19.650" Score="1" ViewCount="66" Body="&lt;p&gt;An affine transformation is a linear transformation followed by a translation. They are &lt;a href=&quot;http://en.wikipedia.org/wiki/Morphism&quot; rel=&quot;nofollow&quot;&gt;morphism&lt;/a&gt; between affine spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rigid transformation consists of a rotation and a translation. I was wondering what are the spaces between which the rigid transformations can become morphisms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what are the spaces between which the rotation transformations can become morphisms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same question for similarity transformation and projective transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-21T15:16:47.767" LastActivityDate="2010-09-21T15:36:04.607" Title="What are the most special spaces between which rigid transformations preserve the structures of the spaces" Tags="&lt;rigid-transformation&gt;&lt;morphism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5149" PostTypeId="2" ParentId="5145" CreationDate="2010-09-21T14:19:43.443" Score="6" ViewCount="0" Body="&lt;p&gt;Try&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N[1 + 1/100000000000000, 2000]^100000000000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which makes &lt;em&gt;Mma&lt;/em&gt; do the computation with floating point numbers (but using 2000 exact digits, waaaaay more that your calculator!) In my computer,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[41]:= Timing[N[1 + 1/100000000000000, 2000]^100000000000000]&#xA;&#xA;Out[41]= {0.001999, &#xA;2.71828182845903164395114517625107361345759538109318401368108700081263&#xA;8880450911203511655259489457817904923933478503365902065203002443715271&#xA;2825230612837327582882359132201752486932090411441742152459981748434424&#xA;6796125955942676709724311384945123182525138185474706841748184641523835&#xA;6399319640050121555981826495805200410019220302352025402844917823115708&#xA;4890277526840712036588689004850449217720498524694546241483265385120337&#xA;3036215908323704165098770345249059641531345794163630084215403881301114&#xA;0222776646139077637586730556318554208747513123989912904288599990729397&#xA;4105912301684777926066523131379164170616992617301710309080427339172694&#xA;8652209704120914116266089477162936814414049552884109177338585344418934&#xA;4664761359706737137907335308204793693031274512475102742160130609557028&#xA;0606249997104370287581131479088448270923696582800846475772654584603126&#xA;9574917252623375481387197279590503661010907784866738170370137382154502&#xA;5239391867955746777742097465819430570361920416259681558416618161699424&#xA;5237267230224583604177051168028000345047412788448526660107314695031108&#xA;5918243291907671190858585383905903649083888525646705126721235172922992&#xA;7474539583108858881111281802649615453982269311194004892438663488977893&#xA;2481678450112257387961011092938028584251024526367562391996230685117325&#xA;0436239706780584781622006285082275199325910403142256578320132681839729&#xA;1241283875634170613592405031777334496908146321259758762276254422018659&#xA;2501711558155518458153097538288499261588998555672920722850605105301158&#xA;3203321297849315979775985916474141976627081937753054953188751346134290&#xA;6581791976723248211158139510858187401773833843382750617470712252506403&#xA;3501718283248275223376526908879174360214195321187111170739858908536480&#xA;0084276450965142494861141929264441784847261049017482036450996057648280&#xA;0994695573343998546256052750841029336112669690421548936514156835398676&#xA;3390097789579643086512481978170502348966487726085752931492687574582562&#xA;2555386293362518511636564659786083245494967887013590756572099854766124&#xA;625831647757374695456737571}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;takes 0.002 seconds.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-21T14:19:43.443" CommentCount="3" />
  <row Id="5151" PostTypeId="2" ParentId="5148" CreationDate="2010-09-21T14:29:46.070" Score="1" ViewCount="0" Body="&lt;p&gt;If by &quot;rotation&quot; you mean &quot;orientation-preserving isometry of $\mathbb{R}^n$ fixing the origin,&quot; then &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the spaces between which rigid transformations are morphisms are the &quot;affine oriented (finite-dimensional) inner product spaces&quot; over $\mathbb{R}$, by which I mean &lt;a href=&quot;http://math.ucr.edu/home/baez/torsors.html&quot; rel=&quot;nofollow&quot;&gt;torsors&lt;/a&gt; over a finite-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientation_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;oriented&lt;/a&gt; inner product space over $\mathbb{R}$,&lt;/li&gt;&#xA;&lt;li&gt;the spaces between which rotations are morphisms are the oriented finite-dimensional inner product spaces over $\mathbb{R}$, and&lt;/li&gt;&#xA;&lt;li&gt;the spaces between which projective transformations are morphisms are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_space&quot; rel=&quot;nofollow&quot;&gt;projective spaces&lt;/a&gt; over $\mathbb{R}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I cannot off the top of my head think of a good name for the spaces between which similarities are morphisms.  I &lt;em&gt;think&lt;/em&gt; you are looking for &quot;conformal affine spaces,&quot; e.g. torsors over $\mathbb{R}^n$ equipped with a notion of oriented angle (but not the full inner product).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-21T15:36:04.607" LastActivityDate="2010-09-21T15:36:04.607" CommentCount="1" />
  <row Id="5152" PostTypeId="1" CreationDate="2010-09-21T14:43:15.710" Score="0" ViewCount="120" Body="&lt;p&gt;I would like to know what is an eulerian approach and an Lagragian approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone can recommend some references?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2045" LastActivityDate="2010-09-21T15:39:40.733" Title="Eulerian vs Lagrangian approach" Tags="&lt;calculus&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="5153" PostTypeId="1" AcceptedAnswerId="5171" CreationDate="2010-09-21T14:46:18.590" Score="2" ViewCount="163" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Goursat%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Goursat's Lemma&lt;/a&gt; characterizes the subgroups of direct products. Is there a similar characterization for the subgroups of semidirect products? What about if I'm only interested in the normal subgroups?&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-09-21T19:47:50.883" Title="What are the subgroups of a semidirect product?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5154" PostTypeId="2" ParentId="5115" CreationDate="2010-09-21T14:55:23.287" Score="3" ViewCount="0" Body="&lt;p&gt;Your logic fails much more obviously for the series&lt;br&gt;&#xA;$1.1 + 1.01 + 1.001 + 1.0001 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though each term is smaller than the last, it is clearly divergent (think of $1 + 1 + 1 + \cdots$)&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-09-21T14:55:23.287" CommentCount="1" />
  <row Id="5155" PostTypeId="2" ParentId="5145" CreationDate="2010-09-21T15:05:55.810" Score="5" ViewCount="0" Body="&lt;p&gt;General purpose computer algebra systems, unlike most calculators, are able to perform both exact/symbolic computations (e.g. exact integer arithmetic) in addition to approximate numerical computations (e.g. floating point). Thus when you present the problem in exact terms - as you do above - they will preserve that form - so as not to lose any information. Thus your problem will be interpreted as a computation in exact rational arithmetic, i.e. raising a rational number to an integer power. If instead you desire a real approximation of the result then you must explicitly specify such, e.g. using N[...] in Mma. This is but one of the many complexities that arise when one passes from special-purpose computation systems (e.g. calculators) to general purpose computer algebra systems (speaking as one who has often tried to tackle less trivial such complexities while working as a developer of the Macsyma computer algebra system). &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-21T16:30:21.757" LastActivityDate="2010-09-21T16:30:21.757" />
  <row Id="5157" PostTypeId="2" ParentId="4536" CreationDate="2010-09-21T15:30:09.140" Score="0" ViewCount="0" Body="&lt;p&gt;I've been testing a new math search engine and found a similar discussion on the mathoverflow, there is link to search results (http://uniquation.com/en/solutions.aspx?query=F_{n%2B1}%3D2F_n%2BF_{n-1})&lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2010-09-21T15:30:09.140" />
  <row Id="5158" PostTypeId="2" ParentId="5160" CreationDate="2010-09-21T15:39:40.733" Score="1" ViewCount="0" Body="&lt;p&gt;There is a Eulerian vs. Lagrangian distinction in fluid mechanics:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-21T15:39:40.733" />
  <row Id="5159" PostTypeId="1" CreationDate="2010-09-21T16:16:21.723" Score="3" ViewCount="126" Body="&lt;p&gt;Let $C$ be the curve in $\mathbb{C}^2$ given by $f(x,y)=x^3+y^3-y=0$ and let g:$C \to \mathbb{C}$ be the projection $g(x,y)=x$. I want to find where this map $g$ is ramified, and do this by computing the sheaf of relative differentials. I have $(3y^2-1)dy=-3x^2dx$. Now, by the definition of relative differentials I thought I set $dx=0$, and since ramified points are where this sheaf is nonzero, I get that dy is not killed when $3y^2-1=0$ but I don't think that gives the right answer. On the other hand, the derivative $dy/dx=\frac{-3x^2}{3y^2-1}$ vanishes when x=0, and that gives the right answer. Apparently I should be setting $dy=0$ but that doesn't make sense since the target of $g$ is $Spec\ \mathbb{C}[x]$. I'm confused why my sheaf of differentials approach isn't working - clearly I just have a basic misunderstand about it and I was hoping someone could clear if up for me.&lt;/p&gt;&#xA;" OwnerUserId="2046" LastActivityDate="2010-12-24T21:44:25.883" Title="stuck on computing points of ramification of a map of curves via sheaf of relative differentials" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5160" PostTypeId="1" CreationDate="2010-09-21T16:24:49.670" Score="1" ViewCount="68" Body="&lt;p&gt;I am reading about active contours models. It is said that this can be broadly classified as either parametric active contour models and geometric active contour models. The first one are represented explicitly as parameterized curves in a Lagragian framework, while the geometric active contours are represented implicitly as level sets of a 2-dimensional function that evolves in a Eulerian framework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an Eulerian approach and an Lagragian approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2045" LastActivityDate="2010-10-21T17:01:54.407" Title="Active Contour Models - Lagrangian/Eulerian approches" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="5161" PostTypeId="1" AcceptedAnswerId="5162" CreationDate="2010-09-21T16:43:06.410" Score="3" ViewCount="296" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know the name of this kind of polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(x)=x^n+a_{1}x^{n-1}+\ldots+a_{n-1}x+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $a_{i}\in\lbrace0,1\rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-30T01:32:14.937" LastActivityDate="2010-11-30T01:32:14.937" Title="What is the name of this kind of polynomial?" Tags="&lt;terminology&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="5162" PostTypeId="2" ParentId="5161" CreationDate="2010-09-21T17:21:29.123" Score="10" ViewCount="0" Body="&lt;p&gt;They're known as &lt;strong&gt;Newman polynomials&lt;/strong&gt;. They are often studied in contexts where one is interested in learning what interesting consequences result from placing such restrictions on the coefficients, for example &lt;a href=&quot;http://www.ams.org/journals/mcom/2009-78-265/S0025-5718-08-02138-8/S0025-5718-08-02138-8.pdf&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;.  Erdos and Littlewoood posed several questions about the effects this has on the minimum modulus of the polynomial on the unit circle, e.g. see &lt;a href=&quot;http://www.ams.org/journals/proc/1992-116-02/S0002-9939-1992-1101984-4/S0002-9939-1992-1101984-4.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-21T17:35:28.707" LastActivityDate="2010-09-21T17:35:28.707" CommentCount="2" />
  <row Id="5163" PostTypeId="1" AcceptedAnswerId="5169" CreationDate="2010-09-21T17:34:00.923" Score="1" ViewCount="173" Body="&lt;p&gt;So first off all, this WAS homework. Submitted it about 3 hours ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be an $m \times m$ matrix. then show that if rank A = m, then Ax = 0 has a unique solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My roommate said this is wrong but here is how I solved it. &#xA;very simply, i stated that if the rank of A is m that means the number of leading ones are the same as the columns. Hence there exists a determinant. According to theorms, if a matrix has a determinant then it has an inverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So keeping that in mind, we do this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A * A^(-1) x = 0 * A^(-1)  &amp;lt;--- multipling both sides by the inverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I x = 0 where I is the identity matrix and hence showing that if rank A = m then Ax = 0 has a unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is this completely wrong? is there an alternative?&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-09-23T02:56:15.583" Title="Finding a unique solution given rank A = m" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5164" PostTypeId="2" ParentId="5163" CreationDate="2010-09-21T17:42:42.327" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, the rank is the number of pivots in Gaussian elimination. But you must explicitly state how you utilize this to conclude that the determinant is nonzero - viz. by pigeonhole, all diagonal elements must be 1, not 0. For further insight see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Invertible_matrix&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; for at least 17 properties that are equivalent to $\rm A$ being invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one alternative solution: since the matrix has full column rank, the columns are linearly independent. Therefore any solution, being a linear combination of the columns, is necessarily unique. Said slightly more generally: $\rm\ A\:$ has full column rank $\iff$ $\rm\:A$ is injective (1-1) as a linear map $\rm\iff\ ker\ A = \:0$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-21T19:20:23.707" LastActivityDate="2010-09-21T19:20:23.707" />
  <row Id="5165" PostTypeId="2" ParentId="5163" CreationDate="2010-09-21T17:46:12.803" Score="4" ViewCount="0" Body="&lt;p&gt;Your reasoning is not quite right. Every square matrix has a determinant regardless if it has an inverse or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct statement is: A square matrix has an inverse iff its determinant is non-zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the easiest way to solve your problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A=[a_1|\ldots|a_m]$    i.e. $a_i$ is the $i^{\text{th}}$ column of $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Recall: $\text{rank}(A) = \text{dim}(\text{Span}(a_1,\ldots,a_m))$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rank($A$)=m iff Span$(a_1,\ldots,a_m)=\mathbb{R}^m$ This comes from the definition of rank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $m$ vectors $a_1,\ldots,a_m$ that span an $m$ dimension space $\mathbb{R}^m$ and thus $a_1,\ldots,a_m$ are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0=Ax=x_1a_1+\ldots + x_ma_m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by definition of linearly independence, $x_i=0$ for $1\leq i\leq m$. Hence the zero vector is the unique solution.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-21T18:43:29.647" LastActivityDate="2010-09-21T18:43:29.647" />
  <row Id="5166" PostTypeId="1" CreationDate="2010-09-21T19:02:42.603" Score="1" ViewCount="265" Body="&lt;p&gt;Hi all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have all points where my curve pass through, but I need to get the coordinates of the control points to be able to draw the curve. How can I do to calculate this points?&lt;/p&gt;&#xA;" OwnerUserId="2050" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-22T01:48:57.900" LastActivityDate="2010-11-02T18:37:10.093" Title="algorithm to calculate the control points of a cubic Bezier curve" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="5167" PostTypeId="1" AcceptedAnswerId="5176" CreationDate="2010-09-21T19:06:15.070" Score="0" ViewCount="241" Body="&lt;p&gt;I was reading this book &quot;Rational points on Elliptic curves&quot; by J.Silverman, and J.Tate, 2 prominent figures in Number theory and was very intrigued after reading the first couple of pages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connection between Algebra and Geometry is displayed in a professional manner in that book. One can generally ask this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's clear that the Mathematicians had to develop some laws in order to make the points on an Elliptic curve to a group. The first axioms of adding 2 points on an elliptic curve is clear. But i would like to know how does one come up with the mechanism of the remaining axioms(like associativity) which are actually intricate in nature. What is the motivation behind it. Can we define an axiom in an other way so that the point on the elliptic curve still forms a group? Also why to define the group axioms only on for this type of curves. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A very much related question was asked in MO link: &lt;a href=&quot;http://mathoverflow.net/questions/6870/why-is-an-elliptic-curve-a-group&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/6870/why-is-an-elliptic-curve-a-group&lt;/a&gt; but i actually expect more about the mechanism of the intricate axioms. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-21T19:15:08.160" LastActivityDate="2010-09-21T20:41:51.347" Title="Group Law for an Elliptic curve" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;elliptic-curves&gt;" AnswerCount="3" CommentCount="15" FavoriteCount="1" />
  <row Id="5168" PostTypeId="2" ParentId="5153" CreationDate="2010-09-21T19:10:19.390" Score="0" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While searching on the net i found this article. See whether it helps or not: &lt;a href=&quot;http://www.springerlink.com/content/l272110h87u05667/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/l272110h87u05667/fulltext.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one can't access this then try: &lt;a href=&quot;http://dx.doi.org/10.1007/BF01058705&quot; rel=&quot;nofollow&quot;&gt;dx.doi.org/10.1007/BF01058705&lt;/a&gt; or &lt;a href=&quot;http://ams.org/mathscinet-getitem?mr=1148867&quot; rel=&quot;nofollow&quot;&gt;ams.org/mathscinet-getitem?mr=1148867&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Jack.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-21T19:47:50.883" LastActivityDate="2010-09-21T19:47:50.883" CommentCount="4" />
  <row Id="5169" PostTypeId="2" ParentId="5163" CreationDate="2010-09-21T19:12:05.420" Score="2" ViewCount="0" Body="&lt;p&gt;There are several misstatements and one error in your solution, though the &lt;em&gt;idea&lt;/em&gt; is essentially correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it is false that if the matrix has rank $m$ then &quot;the number of leading ones are he same as the columns.&quot; The matrix $A$ may have no &quot;leading ones&quot; whatsoever. What you meant, presumably, is that the *&lt;em&gt;row-echelon form&lt;/em&gt; (or the row-reduced form) of $A$ will have as many leading $1$s as there are columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: determinants exist for square matrices regardless of the form of the row-echelon form; the determinant of a square matrix is nonzero if and only if the matrix is invertible, if and only if the number of leading $1$s in the row-echelon form of $A$ equals the number of columns. This has to be proven in some way; for example: the elementary row operations will either multiply the determinant by $-1$ in the case of row exchanges, will multiply it by a nonzero constant $\alpha$ in the case of multiplying a row by $\alpha\neq 0$, or will leave the determinant without change in the case that you add the multiple of one row to another row. Thus, performing elementary row operations will not change whether the determinant is or is not equal to $0$, in the sense that if you obtain $B$ from $A$ by a finite sequence of elementary row operations, then $\det(A)=0$ if and only if $\det(B)=0$. Since the determinant of an upper triangular matrix is just the product of the diagonal entries, the determinant of a square matrix in row echelon form will be $1$ if the leading $1$s are all in the diagonal, and $0$  otherwise. So in your case, the row echelon form of $A$ has nonzero determinant, and therefore (by what we saw before) the determinant of $A$ is nonzero, so $A$ has an inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, your next line is incorrect. You go from&#xA;\[ A\mathbf{x}=\mathbf{0}\]&#xA;to&#xA;\[ AA^{-1}\mathbf{x} = \mathbf{0}A^{-1}.\]&#xA;This not only does not follow, it does not even make sense! First, because if $A$ is an $m\times m$ matrix, then $\mathbf{x}$ and $\mathbf{0}$ are $m\times 1$ matrices; but for $\mathbf{0}A^{-1}$ to make sense, the number of columns of $\mathbf{0}$ (namely, 1) has to equal the number of rows in $A^{-1}$ (namely $m$); and this is not true unless you are dealing with $1\times 1$ matrices. Second, even if you &lt;em&gt;could&lt;/em&gt; multiply both sides by $A^{-1}$ on the right, then you would not get $AA^{-1}\mathbf{x}$ on the left hand side of the equation, you would get $A\mathbf{x}A^{-1}$, and then you need to remember that matrix multiplication is not commutative, so in general you cannot go from $\mathbf{x}A^{-1}$ to $A^{-1}\mathbf{x}$ (as noted above, it may not even make sense to write one of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you multiplied by $A^{-1}$ on the wrong side, and you did it incorrectly. Rather, you want to multiply by $A^{-1}$ on the &lt;em&gt;left&lt;/em&gt; of each side of the equality, not on the right. That is, you want to go from&#xA;\[ A\mathbf{x}=\mathbf{0}\]&#xA;to&#xA;\[ A^{-1}(A\mathbf{x}) = A^{-1}\mathbf{0}.\]&#xA;And from here you can proceed as you did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... partial, but definitely not full credit.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T02:56:15.583" LastActivityDate="2010-09-23T02:56:15.583" CommentCount="2" />
  <row Id="5170" PostTypeId="1" AcceptedAnswerId="5172" CreationDate="2010-09-21T19:20:33.360" Score="1" ViewCount="85" Body="&lt;p&gt;So I have some circles that look kinda like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img80.imageshack.us/img80/792/mmmcircles.png&quot; width=&quot;800&quot; height=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm given the radius of the circle with center point A which is also the distance AB, the distance AB between the two center points on the x axis (they share same y values for the centers), and the distance CD which is the height of the shape created by the intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also assume that despite my drawing the circle ARE perfect circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to find the distance BD which is also the radius of the circle centered on point B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help.&#xA;&lt;br&gt;--AvatarOfChronos&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2010-09-21T19:27:20.497" Title="Find the radius of a circle based off of its intersection with another." Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5171" PostTypeId="2" ParentId="5153" CreationDate="2010-09-21T19:26:29.230" Score="3" ViewCount="0" Body="&lt;p&gt;The short answer is that there is nothing nearly so nice as Goursat's Lemma. You can certainly reduce easily to the case where $\pi_K(H)=K$, much like you can reduce Goursat's Lemma to the case of a subdirect product, but after that it gets complicated. To give you an idea, here are three references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theorem of Rosenbaum (&lt;em&gt;Die Untergruppen von halbdirekten Produkten&lt;/em&gt;, Rostock. Math. Kolloq. No. 35 (1988), 21-30) gives (from the MathScieNet Review MR991728 (90c:20032)):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; A set $U$ of elements of the semidirect product $G=NK$ with $N\triangleleft G$ is a subgroup of $G$ if and only if&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$UN\cap K$ and $U\cap K$ are subgroups of $G$;&lt;/li&gt;&#xA;  &lt;li&gt;$U\cap N$ is a subgroup and $UK\cap N$ is a collection of $U\cap N$-cosets in $N$; and&lt;/li&gt;&#xA;  &lt;li&gt;There is a mapping $\varphi$ defined for all $g\in UK\cap N$ mapping $(U\cap K)g$ onto some coset $n(U\cap N)$, with $n\in N$, satisfying $\varphi(g_1g_2)=g_2^{-1}\varphi(g_1)g_2\varphi(g_2)$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The criterion was then used by Gutiérrez-Barrios to develop a criterion for a set of elements to be a normal subgroup of the semidirect product (&lt;em&gt;Die Normalteiler von halbdirekten Produkten.&lt;/em&gt; Wiss. Z. Pädagog. Hochsch. Erfurt/Mühlhausen Math.-Natur. Reihe &lt;strong&gt;25&lt;/strong&gt; (1989), no. 2, 108-114. MR1044548 (91b:20029))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usenko (&lt;em&gt;Subgroups of semidirect products&lt;/em&gt;, english translation in &#xA;Ukrainian Math. J. &lt;strong&gt;43&lt;/strong&gt; (1991), no. 7-8, 982-988 (1992), MR1148867 (92k:20045)) uses crossed homomorphisms to study subgroups of semidirect products. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-21T19:26:29.230" />
  <row Id="5172" PostTypeId="2" ParentId="5170" CreationDate="2010-09-21T19:27:20.497" Score="2" ViewCount="0" Body="&lt;p&gt;First, note that segment AB bisects segment CD.  Call their point of intersection E, which is also the midpoint of CD.  Since you know the length of CD, you know the length of CE.  The measure of angle BAC (which could also be called EAC) is $\sin^{-1}\left(\frac{CE}{AC}\right)$.  Apply the Law of Cosines to triangle ABC to find BC=BD: &#xA;$$BC^2=AC^2+AB^2-2\cdot AC\cdot AB\cdot\cos\left(\sin^{-1}\left(\frac{CE}{AC}\right)\right).$$&#xA;Fill in the known lengths and solve.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-21T19:27:20.497" CommentCount="2" />
  <row Id="5173" PostTypeId="2" ParentId="5167" CreationDate="2010-09-21T19:52:23.680" Score="6" ViewCount="0" Body="&lt;p&gt;I agree that from the chord-and-tangent definition of the group law it is far from clear why associativity should hold.  This definition is the one presented in Silverman and Tate because all the other ways you might define the group law require &lt;strong&gt;much&lt;/strong&gt; more background; however, these are the definitions from which associativity makes more sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one approach an elliptic curve is a group of the form $\mathbb{C}/\Lambda$ where $\Lambda$ is a &lt;strong&gt;lattice&lt;/strong&gt;, e.g. a discrete subgroup of $\mathbb{C}$ isomorphic to $\mathbb{Z}^2$.  (The motivation for looking at such quotients comes basically from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniformization_theorem&quot; rel=&quot;nofollow&quot;&gt;uniformization theorem&lt;/a&gt;).  Such a quotient forms a compact Riemann surface, topologically a torus, hence it admits no nonconstant holomorphic functions; therefore we would like to describe the meromorphic functions.  A natural way to do this is to write down a function and then try to average over $\Lambda$, which (if you do it right) will lead you to the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_elliptic_function&quot; rel=&quot;nofollow&quot;&gt;Weierstrass elliptic function&lt;/a&gt; $\wp(z)$.  By examining its poles and the poles of its derivative $\wp'(z)$ we can deduce the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g_2, g_3$ are certain constants associated to $\Lambda$.  In fact the map from $\mathbb{C}/\Lambda$ to the elliptic curve $y^2 = 4x^3 - g_2 x - g_3$ given by $z \mapsto (\wp(z), \wp'(z))$ is an isomorphism.  This forms the connection between the complex-analytic picture and the algebraic picture in terms of Weierstrass normal forms.  The group law here is just regular addition, e.g. the sum of the points $(\wp(a), wp'(a))$ and $(\wp(b), wp'(b))$ is just $(\wp(a+b), wp'(a+b))$, and so it is obvious that associativity holds.  The fact that if $a + b + c = 0$ then the corresponding points are collinear is then a relatively simple computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Historically, this is not how the first group laws on elliptic curves were discovered.  For the elliptic-function perspective I recommend Stevenhagen's &lt;a href=&quot;http://websites.math.leidenuniv.nl/algebra/ellipticfunctions.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on the subject&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another approach an elliptic curve (over $\mathbb{C}$, for simplicity) is a smooth projective curve of genus $1$.  The category of smooth projective curves over $\mathbb{C}$ is known to be equivalent to the category of compact Riemann surfaces, which is part of the connection between this definition and the above definition.  To any smooth projective curve $C$ one can associate its &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_variety&quot; rel=&quot;nofollow&quot;&gt;Jacobian variety&lt;/a&gt;, which is a group of the form $\mathbb{C}^g/\Lambda$ where $g$ is the genus which naturally occurs when one thinks about integration over such a curve.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel-Jacobi_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Jacobi theorem&lt;/a&gt; asserts that the Jacobian variety of a curve is isomorphic to its &lt;a href=&quot;http://en.wikipedia.org/wiki/Picard_group&quot; rel=&quot;nofollow&quot;&gt;Picard group&lt;/a&gt; (the group of &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry)&quot; rel=&quot;nofollow&quot;&gt;divisors&lt;/a&gt; modulo principal divisors), and the points of the Picard group of a curve of genus $1$ can be put into natural bijection with the curve itself (once one chooses an identity), so the group law on the Jacobian variety (which is perfectly natural) then gives a group law on the original curve (again once one chooses an identity).  The connection to collinear points comes from the fact that lines determine principal divisors, which are zero in the Picard group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your last question, elliptic curves are, as it turns out, the only projective curves which can be given the structure of algebraic groups; see the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Abelian_variety&quot; rel=&quot;nofollow&quot;&gt;abelian varieties&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-21T19:59:55.597" LastActivityDate="2010-09-21T19:59:55.597" />
  <row Id="5174" PostTypeId="1" AcceptedAnswerId="5187" CreationDate="2010-09-21T19:55:51.983" Score="0" ViewCount="118" Body="&lt;p&gt;I hope you will excuse my vague title. Let's define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\varsigma \ \colon\ \mathbb N\to \mathbb Z\ \colon\ k\mapsto (-1)^{\lfloor \frac{k-1}{2}\rfloor}$$&#xA;and&#xA;$$S(m,n)=\displaystyle\sum_{k=m}^n k\cdot \varsigma(k)$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find every integer $n\ge 1$ such that $S(1,\lfloor n/2\rfloor)=S(\lfloor n/2\rfloor+1,n)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $-(n+1)\le \delta(n)\le 2(n+1)$, when $\delta(n)=S(1,\lfloor n/2\rfloor)-S(\lfloor n/2\rfloor+1,n)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-09-21T22:35:47.367" LastActivityDate="2010-09-21T23:59:49.583" Title="About a certain signed sum of consecutive integers ($\sum_{k=1}^n k\cdot \varsigma(k)=\sum_{k=n}^{2n} k\cdot \varsigma(k)$)" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5175" PostTypeId="2" ParentId="5167" CreationDate="2010-09-21T20:41:17.853" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;This is a failed attempt at constructing the group law with associativity built in&lt;/strong&gt;, It is still interesting to consider though - I will explain why at the end.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Every line intersects 3 points on an elliptic curve, call them $P,Q,R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can pick some designated point $\mathcal O$ as the identity on the curve and then set the ternary relation $P+Q+R = \mathcal O$ to hold for all triples which lie on a line (N.B. I could just write this $C(P,Q,R)$ but I use the suggestive notation of $+$ and $=$ instead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by the group axioms we have $P + \mathcal O + -P = \mathcal O$, so we know how to compute the inverse of a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any two points $P,Q$ we could also find a third $Q$ easily, so that $P + Q + R = \mathcal O$. We can now define the binary operation of addition as $P + Q = -R$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now it is possible to reduce the proof of associativity of the group law to a purely combinatoric statement.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$(P + Q) + R$ means that $P + Q + -PQ = \mathcal O$ and $PQ + R + -PQR = \mathcal O$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$P + (Q + R)$ means that $Q + R + -QR = \mathcal O$ and $P + QR + -PQR' = \mathcal O$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(and also the extra relations between positive and negative terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These relations define an &lt;a href=&quot;http://en.wikipedia.org/wiki/Incidence_structure&quot; rel=&quot;nofollow&quot;&gt;incidence structure&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{(P,Q,-PQ),(PQ,O,-PQ),(PQ,R,-PQR),(PQR,O,-PQR),(Q,R,-QR),(QR,O,-QR),(P,QR,-PQR'),(PQR',O,-PQR')}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the interesting bit! We would hope that a Fano plane style situation would force PQR to be equal to PQR' but it doesn't happen. So we didn't get any proof out of this, but we do get a sense of the difficulty of this associativity problem: There is something special about the &lt;em&gt;shape&lt;/em&gt; of the elliptic curve which forces these points to be equal. That implies that any proof or construction of this group law &lt;strong&gt;must&lt;/strong&gt; use the geometric properties of the curve in an essential way.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-21T20:41:17.853" CommentCount="3" />
  <row Id="5176" PostTypeId="2" ParentId="5167" CreationDate="2010-09-21T20:41:51.347" Score="6" ViewCount="0" Body="&lt;p&gt;Dear Chandru, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group law on an elliptic curve was not discovered in a vacuum.  It came up in the context&#xA;of abelian integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y^2 = f(x)$, where $f(x)$ is a cubic in $x$ be an elliptic curve; call it $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elliptic integrals are integrals of the form $$\int_{a}^x dx/y = \int_a^x \frac{d x}{\sqrt{&#xA;f(x)}}.$$  (Here $a$ is some fixed base-point.) They come up (in a slightly transformed manner) when computing the arclength of an ellipse (whence their name).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If was realized at some point (in the 1600s or 1700s) (at least in special cases) that if &#xA;you apply certain substitutions to $x$, you can double the value of the integral, or that&#xA;if you apply certain substitusions of the form $x = \phi(x_1,x_2)$, the integral you compute is the sum of the individual integrals for $x_1$ and $x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real understanding came from the work of Abel and Jacobi.  (Unfortunately I don't know&#xA;the precise history or attributions here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they realized (in modern terms) is that, if we fix the base-point $a$ and let $b$&#xA;vary, then the elliptic integral is giving a multivalued map from the elliptic curve $E$&#xA;to $\mathbb C$,&#xA;and that the formula $\phi(x_1,x_2)$ mentioned above shows us a way to add points on the&#xA;elliptic curve, so that this map is a (multi-valued) group homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the inverse of this multi-valued map gives a single-valued map (which is how we are&#xA;more used to thinking about it) &#xA;$$\mathbb C \to E,$$&#xA;which is a homomorphism when we give $\mathbb C$ its addivite structure and $E$ the&#xA;group law coming from $(x_1,x_2)\mapsto \phi(x_1,x_2)$.   The kernel of this map turns out&#xA;to be a lattice $\Lambda$, so that we get an isomorphism $\mathbb C/\Lambda \cong E.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula $\phi(x_1,x_2)$ turns out to be precisely the formula describing addition&#xA;on the elliptic curve via chords and tangents, and there are lots of theoretical explanations for it, as you can find on the linked MO page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a higher genus curve $C$, it turns out that there is not just the one holomorphic differential $dx/y$, but $g$ linearly independent such (if the curve has genus $g$),&#xA;say $\omega_1, \ldots,\omega_g$.&#xA;Furthermore, if we fix a particular differential $\omega_i$, then there is &lt;em&gt;no&lt;/em&gt; formula&#xA;$\phi(x_1,x_2)$ such that the sum of the integrals of $\omega$ for $x_1$ and $x_2$ is&#xA;equal to the integral of $\omega$ for $\phi(x_1,x_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what Abel and Jacobi found is that, if we consider the map &#xA;$$(x_1,\ldots,x_g) \mapsto (\sum_{i = 1}^g \int_a^{x_i} \omega_1,&#xA;\ldots,\sum_{i = 1}^g \int_a^{x_i} \omega_g),$$&#xA;which gives a multi-valued map $$Sym^g C \to \mathbb C$$&#xA;(here $Sym^g C$ denotes the $g$th symmetric power of $C$, so it is the product of&#xA;$g$ copies of $C$, modulo the action of the symmetric group on the $g$ factors),&#xA;then we can find a formula $\phi(x_{1,1},\ldots,x_{1,g},x_{2,1},\ldots,x_{2,g})$&#xA;such that $\phi$ defines a group law on $Sym^g C$ (at least generically) and&#xA;such that this map is a homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this map&#xA; becomes well-defined and (generically) single valued if we quotient out the target by an appropriate lattice $\Lambda$ (the period lattice), to get a birational map&#xA;$$Sym^g C \to \mathbb C^g /\Lambda.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The target here is called the Jacobian of $C$, and can be identified with $Pic^0(C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, to generalize the addition law on an elliptic curve to higher genus curves,&#xA;you have to consider unordererd $g$-tuples of points (where $g$ is the genus), and add those (not just individual&#xA;points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The relationship with $Pic^0(C)$ is that if $x_1,\ldots,x_g$ are $g$ points on $C$,&#xA;then $x_1 + \ldots + x_g - g a$ is a degree zero divisor on $C$, and every degree zero&#xA;divisor is linearly equivalent to a divisor of this form, and, generically, to a unique&#xA;such divisor.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-21T20:41:51.347" />
  <row Id="5177" PostTypeId="1" CreationDate="2010-09-21T21:11:53.667" Score="2" ViewCount="93" Body="&lt;p&gt;Do generalized hypergeometric functions &#xA;$${}_p F_q(a_1,\ldots,a_p; b_1, \ldots,b_q; z) $$&#xD;&#xA;with $p = q+1$ always possess a singularity at $z=1$, independent of the their parameters $a_1,\ldots,a_p$ and $b_1,\ldots,b_q$ under the provision that all cases of finite polynomials are excluded? Can someone mention a reliable source for such a statement or provide a counterexample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2052" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T23:50:41.000" LastActivityDate="2010-12-20T23:50:41.000" Title="A singularity of hypergeometric functions" Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5178" PostTypeId="2" ParentId="250" CreationDate="2010-09-21T21:18:16.207" Score="3" ViewCount="0" Body="&lt;p&gt;It is possible for a group of people to hold a secret ballot election in which all communication is done publicly. This is one of the many surprising consequences of the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Secure_multi-party_computation&quot; rel=&quot;nofollow&quot;&gt;secure multiparty computation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, the ballots are only &quot;secret&quot; under some reasonable cryptographic assumptions. I guess Feynman may have objected to this.)&lt;/p&gt;&#xA;" OwnerUserId="1776" LastActivityDate="2010-09-21T21:18:16.207" CommunityOwnedDate="2010-09-21T21:18:16.207" />
  <row Id="5179" PostTypeId="1" AcceptedAnswerId="5183" CreationDate="2010-09-21T21:34:59.040" Score="3" ViewCount="143" Body="&lt;p&gt;What are the benefits of using a conjugate linear inner product in a complex vector space vs a simple linear inner product?  That is, why do we demand that (y,x) = (x,y)* as opposed to (y,x)=(x,y)?  Of course, this ensures that (x,x) is real and thus makes an easy definition of norm, but is that necessary?  &lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-09-21T22:14:27.290" Title="Is complex conjugation needed for valid inner product?" Tags="&lt;inner-product-space&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="5180" PostTypeId="2" ParentId="5159" CreationDate="2010-09-21T21:39:34.387" Score="0" ViewCount="0" Body="&lt;p&gt;This should not under any circumstances be considered rigorous, but intuitively here is what is happening in my mind. If you project onto the &quot;$x$-axis&quot;, then think of the tangent space of the curve $C$ at some point $(a,b)$. If the tangent direction is &quot;vertical&quot;, then any vector in that space will project to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after projecting, there is no &quot;$y$-component&quot; to any of the tangent vectors, and hence the $dy$ always evaluates to $0$. So in some sense $g$ is transverse to the projection precisely when there is some &quot;$x$-direction&quot;. This is precisely when $-3x^2dx\neq 0$.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-09-21T21:39:34.387" />
  <row Id="5181" PostTypeId="2" ParentId="5139" CreationDate="2010-09-21T21:59:50.207" Score="6" ViewCount="0" Body="&lt;p&gt;There is a simpler way (in terms of manual computations) than the other answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the following facts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is an integer then $$ A -[x] = [A-x + \delta] $$ where $\delta = 0$ if $x$ is an integer, and $\delta = 1$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &gt; 1$ is a positive integer then $$ [\frac{[x]}{n}] = [\frac{x}{n}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{n-12 - [\frac{n-17}{25}]}{3}] = [\frac{n-12-\frac{(n-17)}{25}+\delta}{3}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta = 0$ iff $n=17 \mod 25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = [\frac{24n-283+25\delta}{75}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $24n+39 = r \mod 75$ where $0 \leq r &amp;lt; 75$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $r$ must be divisible by $3$. Thus $0 \leq r \leq 72$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $24n+39 = 75m+r$, then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{8n+13}{25}] = [\frac{24n+39}{75}] = m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $$[\frac{24n-283+25\delta}{75}] = [\frac{75m+r-375+53+25\delta}{75}]$$&#xA;= $$m-5 + [\frac{r + 53 + 25\delta}{75}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\delta = 0$ iff $n=17 \mod 25$ iff $r = 72$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $r &amp;lt; 72$, $\delta = 1$, for which we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{r + 53 + 25\delta}{75}] = [\frac{r + 53 + 25}{75}]  = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $r = 72$, we have $\delta=0$ for which we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\frac{r + 53 + 25\delta}{75}] = [\frac{r + 53}{75}] = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the whole expression is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m - ((m-5)+1) = 4 $$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-21T21:59:50.207" />
  <row Id="5182" PostTypeId="2" ParentId="5179" CreationDate="2010-09-21T22:01:00.577" Score="3" ViewCount="0" Body="&lt;p&gt;You need to define the inner product this way (in a complex vector space) to have a good definition of the norm of a vector. Let $V$ be this vector space, and take $X\in V$. Then, the norm of X is defined as $|X|=\sqrt{(X,X)}$. This definition is good ($|X|$ is well defined for all $X$) if $(y,x)=(x,y)^*$. If this is not the case, for example if $(y,x)=(x,y)$, then $ (x,x) $ need not be positive. More: $(x,x)$ need not be real. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aother thing, related to the above one: in metric spaces, given an inner product we define a norm (the same way we did above) and, from the norm, a distance function. This distance function determines the topology of the space (the topology induced by the inner product). To do this, we need to be able to define in a good way the norm of a vector, and the problem appears again.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-09-21T22:01:00.577" />
  <row Id="5183" PostTypeId="2" ParentId="5179" CreationDate="2010-09-21T22:14:27.290" Score="6" ViewCount="0" Body="&lt;p&gt;It is in fact necessary. The inner product axioms without the conjugation are inconsistent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $u$, $v$, $w$ are vectors and $c$ is a scalar)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\langle cu, v\rangle = c\langle u, v\rangle$&lt;/li&gt;&#xA;&lt;li&gt;$\langle u,v\rangle = \langle v,u\rangle$&lt;/li&gt;&#xA;&lt;li&gt;If $u \neq 0$, then $\langle u,u\rangle$ is a positive real number&lt;/li&gt;&#xA;&lt;li&gt;$\langle u+v,w\rangle = \langle u,w\rangle + \langle v,w\rangle$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In fact, 1-3 alone are inconsistent. Indeed, let $u$ be any nonzero vector, so $\langle u,u\rangle &gt; 0$ by condition 3. But if $i = \sqrt{-1}$, then $\langle iu, iu\rangle = i\langle u, iu\rangle$ (by 1) $ = i\langle iu, u\rangle$ (by 2) $ = i^2\langle u, u\rangle$ (by 1) $ = -\langle u, u\rangle &amp;lt; 0$, contradicting condition 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot is that you can choose: either conjugate one side of condition 2, giving you the axioms for an inner product, or get rid of condition 3, giving you the axioms for a symmetric bilinear form. You could also consider a weaker version of 3, like requiring that if $u\neq 0$, then $\langle u, v\rangle \neq 0$ for some $v$. That gives you nondegenerate symmetric bilinear forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there's nothing wrong with bilinear forms on complex vector spaces; they're just not inner products. They're disjoint concepts, unlike in real vector spaces, where inner products are just special symmetric bilinear forms. In some ways, bilinear forms are nicer than inner products, since you don't have to worry about complex conjugation. However, bilinear forms over the complex numbers do not give rise to norms, which means they don't endow vector spaces with good geometry. Inner products do, hence their ubiquity. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-21T22:14:27.290" />
  <row Id="5184" PostTypeId="2" ParentId="5177" CreationDate="2010-09-21T22:31:36.307" Score="1" ViewCount="0" Body="&lt;p&gt;For the $p=q+1$ case, taking the branch cut from 1 to ∞ is conventional so that you can speak of the &lt;em&gt;principal branch&lt;/em&gt; of ${}_{q+1} F_q$ ; the function is multivalued outside the unit disk, and 1 and ∞ are branch points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For unit argument, the key word &quot;Saalschützian&quot; is what you might take into account; see &lt;a href=&quot;http://www.ams.org/journals/proc/1992-114-01/S0002-9939-1992-1068116-2/S0002-9939-1992-1068116-2.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; by Bühring for more information.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-21T22:31:36.307" />
  <row Id="5185" PostTypeId="1" AcceptedAnswerId="5186" CreationDate="2010-09-21T22:36:10.747" Score="6" ViewCount="174" Body="&lt;p&gt;Prompted by today's Minute Math question on the MAA site (&lt;a href=&quot;http://amc.maa.org/mathclub/5-0,problems/T-problems/T-web,ia/2005web/tb05-12-ia.shtml&quot; rel=&quot;nofollow&quot;&gt;http://amc.maa.org/mathclub/5-0,problems/T-problems/T-web,ia/2005web/tb05-12-ia.shtml&lt;/a&gt;), I started thinking about the probability that the sum of the numbers rolled on a set of $n$ dice is prime, particularly the asymptotics as $n\rightarrow\infty$.  Heuristics strongly suggest that this is proportional to $1/\mathrm{ln}\ n$, and in fact, that it's $1/\mathrm{ln}\ n - O(1/\mathrm{ln}^2n)$, but I'm wondering how I would go about getting better asymptotics on the second term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, the heuristic argument goes something like this: assume for concreteness' sake that we're rolling 6-sided dice.  Then the sum of the dice is closely approximated by a normal variable with mean $\mu=7n/2$ and variance $\sigma^2=35n/12$, and since the PNT says that the 'probability' of an integer $n$ being prime is roughly $1/\mathrm{ln}\ n$, we should be able to integrate that probability with respect to the normal distribution:&#xA;$$p = {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} {1\over\mathrm{ln}\ t} dt$$&#xA;And since $1/\mathrm{ln}\ t$ is monotonic, the value of the integral is bounded by the values we get by replacing its term in the integral with its maximum and minimum values on the integration interval:&#xA;$${1\over\mathrm{ln}\ 6n} {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} dt &amp;lt; p &amp;lt; {1\over\mathrm{ln}\ n} {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} dt$$&#xA;Both of the integrals in the latter formula are essentially 1 (by definition), so we get ${1\over\mathrm{ln}\ 6n} &amp;lt; p &amp;lt; {1\over\mathrm{ln}\ n}$; replacing $\mathrm{ln}\ 6n$ by $(\mathrm{ln}\ 6 + \mathrm{ln}\ n)$ and using the binomial formula gives the heuristic approximation I alluded to above.  This leads me to a couple of questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&lt;li&gt;How safe is the heuristic argument above?  I know that the PNT gives good bounds on the number of primes in an interval (on the order of $n^{1/2}$ here, which in particular means that the error from the prime-counting would be $O(n^{-1/2})$ and so much smaller than the inverse-log terms above), but my analytic number theory isn't good enough to know whether 'weighting' by the normal distribution would throw off the classical proofs.&#xA;&lt;li&gt;How would I go about evaluating the integral above?  Obviously the bounds I use bring it in to a fairly small range, but it seems as though to get a second term in my asymptotics I'd need to be able to at least approximate the integral, and there aren't any obvious tricks that look like they'd handle it well...&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-21T23:05:33.727" Title="What's the probability that a sum of dice is prime?" Tags="&lt;number-theory&gt;&lt;probability&gt;&lt;prime-numbers&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5186" PostTypeId="2" ParentId="5185" CreationDate="2010-09-21T22:57:45.520" Score="2" ViewCount="0" Body="&lt;p&gt;For (1.), Rosser and Schoenfeld (sp?) non-asymptotic estimates bound $\pi(n)$ between functions of the form $x/(\log x + C)$ and this should be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2) your integrals are rapid decaying and incredibly close to the integrals on the whole real line.  O($n$) standard deviations from the average is quite an unlikely event.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-21T23:05:33.727" LastActivityDate="2010-09-21T23:05:33.727" CommentCount="3" />
  <row Id="5187" PostTypeId="2" ParentId="5174" CreationDate="2010-09-21T23:59:49.583" Score="1" ViewCount="0" Body="&lt;p&gt;The easiest way to attack this is elementary; if we define $S_0(n) = S(1,n)$, then $S(m,n) = S_0(n)-S_0(m)$, so all that's needed is to analyze $S_0$.  That most easily happens by cases; a bit of quick calculation and some simple algebra shows that $S_0(4k+1) = 1$, $S_0(4k+3) = 0$, $S_0(4k+2) = 4k+3$, and $S_0(4k) = -4k$.  From there it's straightforward to find the possible cases for $\delta(n)$ based on $n$ mod $8$; in particular, it seems like $\delta(n)$ will be zero whenever $n = 8k$ (where $S_0(n/2) = S_0(n)/2$) or $n=8k+7$ (where $S_0(n) = S_0(\lfloor n/2 \rfloor) = 0$).&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-09-21T23:59:49.583" />
  <row Id="5188" PostTypeId="2" ParentId="5018" CreationDate="2010-09-22T00:01:24.820" Score="1" ViewCount="0" Body="&lt;p&gt;I want to point out that there are, topicwise, different paths from the most suggested standard one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by &quot;jmoy&quot;, time is limited - the comments telling you to learn some physics or  programming first or on the way seem to ignore this somewhat. Sure it is good and will pay off to have these things as background, but it is not indispensable. Apparently you already have discovered your fascination for mathematics, so you don't have to back it up with these things that surround mathematics. Physics and programming can give you good intuition, but so can doing mathematics itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, going through the whole standard curriculum with lots of calculus and coordinate geometry is by no means the only way to come into mathematics!! Certainly a basic knowledge of these things will be indispensable &lt;em&gt;at some point&lt;/em&gt;. If you actually enjoy studying calculus, then do it, it's an excellent start! But don't just do it because it is the standard curriculum in US colleges. Often people who went through this can't imagine having learned things in a different way, but that is perfectly possible, and normal in many countries (I went through a German school and university and learned much less calculus on the way than an average US student - it didn't do me any harm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all you are wanting to study mathematics for having a rewarding experience, and while there will be some frustration on the way, it should feel somewhat rewarding right from the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are three alternative routes - in case they pique your curiosity more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Number Theory&lt;/em&gt; is a field which seems perfect for self study: It is concerned with objects that you know very well already - the natural numbers - yet it is a deep and difficult field and connects to virtually every other part of mathematics. This is mathematics of an entirely different flavour than you would see following the calculus path, and it can bring you just as far. In particular it will bring you to a point where you want to study calculus because it tells you things about number theory.&#xA;For this road, pick up a book on Elementary Number Theory (e.g. &lt;a href=&quot;http://modular.math.washington.edu/ent/&quot; rel=&quot;nofollow&quot;&gt;(1)&lt;/a&gt;, free for download -  but if it's too hard going, there are more basic ones also) and work your way through that. Accompany it with the book &quot;Fearless Symmetry&quot; &lt;a href=&quot;http://books.google.com/books?id=h7oUr88S7JQC&amp;amp;printsec=frontcover&amp;amp;dq=fearless+symmetry&amp;amp;source=bl&amp;amp;ots=MzULIGz4SJ&amp;amp;sig=pI76wlCbxjZok0-1psJhclj4Hd8&amp;amp;hl=de&amp;amp;ei=7xyZTKHhNcH7OZOirNMP&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCEQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;(2)&lt;/a&gt; an excellent popular math book, and not a waste of time! It leads you very far into things which connect to current research (the so-called &quot;Langlands program&quot;) but is very friendly written. It gives you a path to follow; you get intuitive ideas for example about &quot;group theory&quot;, and whenever you get to such a point where a new concept (like &quot;group&quot;) is introduced, you should accompany it with a more formal treatment, where you can see the basic theorems about that structure and get some exercises to solve. I suggest Michael Artin's &quot;Algebra&quot; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130047635&quot; rel=&quot;nofollow&quot;&gt;(3)&lt;/a&gt; which contains all you will need to know for a long time on this way. It starts at a low level, which either is already ok for you, or will take only little intermediate reading to bridge the gap.&#xA;(If you start wondering when the calculus comes in, pick up Serre's &quot;Course in Arithmetic&quot; &lt;a href=&quot;http://books.google.com/books?id=Ohj9NNk6btUC&amp;amp;printsec=frontcover&amp;amp;dq=course+in+arithmetic&amp;amp;hl=en&amp;amp;ei=Y0OZTJ3rFY-MOOPN9KsP&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;(4)&lt;/a&gt; and jump to part II, but only after you got some feeling for what's in the first half of Artin's &quot;Algebra&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Combinatorics&lt;/em&gt; (e.g. with the book &quot;Proofs that really count&quot; &lt;a href=&quot;http://books.google.com/books?id=FXLGzXwbwIAC&amp;amp;dq=proofs+that+really+count&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=d0CZTJaSKcvKOMPd1JgP&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CCQQ6AEwAw#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;(5)&lt;/a&gt;) - this is also a topic which needs no motivation from physics or any other area about which you would have to learn things first. As in number theory you can see how one develops mathematical reasoning to answer questions that make sense immediately. And again it can lead you the point where you want to study other mathematics, e.g. calculus, because it makes sense to pursue those questions in the context of what you already know...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Logic&lt;/em&gt;, e.g. starting with Carnielli, Epstein: &quot;Computability&quot; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534546447&quot; rel=&quot;nofollow&quot;&gt;(6)&lt;/a&gt; - this is a basic introduction into logic and computabiliy theory, written for philosophers, and therefore starting very gently. It also paves a way into mathematics via a motivating question - When should we call something &quot;computable&quot;? - and leads you all the way to a proof of Gödel's famous undecidability theorems. Again this is connected to a lot of mathematics. Maybe you would after that like to go on studying non-classical logics, which will lead you to algebra, lattice theory and topology...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever route you choose (and I have to emphasize again that the calculus route is very good as well, if you feel like it), talk to people about the things you study, ask and answer questions, e.g. work as a tutor, or spend time here in the forum! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enjoy your journey to mathematics!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="2048" LastEditorDisplayName="" LastEditDate="2010-09-25T11:37:29.643" LastActivityDate="2010-09-25T11:37:29.643" CommunityOwnedDate="2010-09-22T00:01:24.820" CommentCount="2" />
  <row Id="5189" PostTypeId="2" ParentId="5143" CreationDate="2010-09-22T00:43:30.657" Score="2" ViewCount="0" Body="&lt;p&gt;If $f$ and $g$ are rational the problem is very easy.  Multiply out by the denominator, take the $d^{th}$ power of both sides where $d$ is the least common denominator of $f$ and $g$, and you've reduced to the polynomial case, which is easy (check the extrema and so forth).  This is why it is important to provide details.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-22T00:43:30.657" CommentCount="1" />
  <row Id="5190" PostTypeId="2" ParentId="5115" CreationDate="2010-09-22T00:51:17.320" Score="1" ViewCount="0" Body="&lt;p&gt;Let's try a series that diverges more quickly.  The sequence $b_n = \sqrt{n}$ clearly grows to infinity, but slowly.  The sequence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_n = b_{n+1} - b_n = \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;goes to zero.  Nevertheless, it is clear that the series $a_1 + a_2 + a_3 + ... $ diverges.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-22T00:51:17.320" />
  <row Id="5191" PostTypeId="1" CreationDate="2010-09-22T01:41:11.400" Score="13" ViewCount="472" Body="&lt;p&gt;A common way to define a group is as the group of structure-preserving transformations on some structured set.  For example, the symmetric group on a set $X$ preserves no structure: or, in other words, it preserves only the structure of being a set.  When $X$ is finite, what structure can the &lt;em&gt;alternating&lt;/em&gt; group be said to preserve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a way of making the question precise, is there a natural definition of a category $C$ equipped with a faithful functor to $\text{FinSet}$ such that the skeleton of the underlying groupoid of $C$ is the groupoid with objects $X_n$ such that $\text{Aut}(X_n) \simeq A_n$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I've been looking for a purely combinatorial answer, but upon reflection a geometric answer might be more appropriate.  If someone can provide a convincing argument why a geometric answer is more natural than a combinatorial answer I will be happy to accept that answer (or Omar's answer).&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-28T15:37:36.763" LastActivityDate="2010-09-30T10:00:06.137" Title="What structure does the alternating group preserve?" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;&lt;category-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="5192" PostTypeId="2" ParentId="5166" CreationDate="2010-09-22T02:28:15.560" Score="1" ViewCount="0" Body="&lt;p&gt;When what you already have is a set of points where your curve must pass through, Bézier is not what you want; you should be using a parametric (cubic) spline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that your points $\mathbf{P}_i=(x_i,y_i)$, $i=1\dots n$ represent a general curve (as opposed to a function $f(x)$, for which simply applying the usual cubic spline algorithm on your points suffices), Eugene Lee proposed &lt;a href=&quot;http://www.cs.bgu.ac.il/~leonid/na105/Splines/Lee.pdf&quot; rel=&quot;nofollow&quot;&gt;centripetal parametrization&lt;/a&gt; to generate suitable parameter values $t_i$ associated with your $\mathbf{P}_i$. The prescription for generating the $t_i$ (in its most general form) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ffWgx.gif&quot; alt=&quot;centripetal&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\left\| \cdot \right\|$ is the (Euclidean) length, $e$ is an adjustable exponent in the interval $[0,1]$, and $t_1=0$. (A usual value for $e$ is 0.5, but $e=1$ is sometimes used as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, one applies the usual cubic spline algorithm to the sets $(t_i,x_i)$ and $(t_i,y_i)$, from which you now have your parametric spline. (The periodic spline is recommended for closed curves, and the &quot;not-a-knot&quot; spline for all other cases.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MATLAB Spline Toolbox has support for this parametrization scheme, though it shouldn't be too hard to write your own implementation.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-22T02:28:15.560" />
  <row Id="5193" PostTypeId="1" CreationDate="2010-09-22T02:59:52.830" Score="6" ViewCount="107" Body="&lt;p&gt;The torsion of a curve in $\mathbb{R}^3$ indicates how much it twists around. The curl of a vector field indicates how much the vector field twists around. Is there a relation between the curl of a vector field and the torsion of a curve through that vector field at a given point?&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2010-09-22T03:59:19.987" Title="Relation between torsion of a curve and the curl of a vector field" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5194" PostTypeId="1" AcceptedAnswerId="5195" CreationDate="2010-09-22T03:34:19.700" Score="1" ViewCount="39" Body="&lt;p&gt;Fix a finite group $G$, and look at all its irreducible representations/$\mathbb{C}$. It is said in Serre's book that &quot;there cannot be any $\mathbb{C}$-linear relation between the matrix coefficients of these irreducible representations (of course we fix a set of bases first), because of the orthogonality formulas&quot;. I don't understand the quoted part. Can anyone explain that? I think I must be missing something obvious here.&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2010-09-22T03:43:23.977" Title="No linear relation between matrix coefficients of all the irreducible repn of finite group" Tags="&lt;group-theory&gt;&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="5195" PostTypeId="2" ParentId="5194" CreationDate="2010-09-22T03:43:23.977" Score="3" ViewCount="0" Body="&lt;p&gt;Essentially because a set of pairwise orthgonal nonzero vectors must be&#xA;linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here orthogonality is with respect to a complex (sesquilinear)&#xA;positive-definite inner product, and we can regard each matrix entry&#xA;as a vector of length $|G|$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-22T03:43:23.977" CommentCount="1" />
  <row Id="5196" PostTypeId="2" ParentId="5193" CreationDate="2010-09-22T03:59:19.987" Score="5" ViewCount="0" Body="&lt;p&gt;The paper &quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.13.9738&quot; rel=&quot;nofollow&quot;&gt;Curvature Measures of 3D Vector Fields and Their Applications&lt;/a&gt;&quot; by Weinkauf et al. describes how to compute the curvature and torsion of the tangent curves in terms of the spatial derivatives of the underlying vector field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going the other way and recovering any information about the vector field from its tangent curves is not necessarily possible. As a simple counterexample, consider that all vector fields of the form $\mathbf{v}(r,\theta,z) = f(r)\mathbf{e}_\theta$ in cylindrical coordinates have exactly the same tangent curves (circles around the $z$-axis) but completely different curls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the difficulty is that the properties of the tangent curve can only tell you about the derivatives of the vector field along the parallel direction; you get no information about how the vector field varies in the plane normal to the curve.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-22T03:59:19.987" />
  <row Id="5197" PostTypeId="2" ParentId="5191" CreationDate="2010-09-22T04:22:08.603" Score="4" ViewCount="0" Body="&lt;p&gt;The alternating group preserves orientation, more or less by definition. I guess you can take C to be the category of simplices together with an orientation. I.e., the objects of C are  affinely independent sets of points in some R^n together with an orientation and the morphisms are affine transformations taking the vertices of one simplex to the vertices of another (we don't require anything special for the orientation: some morphisms preserve it and some invert it). Of course this is cheating since if you actually try to define orientation you'll probably wind up with something like &quot;coset of the alternating group&quot; as the definition. On the other hand, some people find orientations of simplices to be a geometric concept, so this might conceivably be reasonable to you.&lt;/p&gt;&#xA;" OwnerUserId="1070" LastActivityDate="2010-09-22T04:22:08.603" CommentCount="2" />
  <row Id="5198" PostTypeId="1" AcceptedAnswerId="5200" CreationDate="2010-09-22T05:02:19.277" Score="2" ViewCount="168" Body="&lt;p&gt;How does one find a simple expression of the one below applying the binomial theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n k \cdot 2^k{ n \choose k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx}(x^k)=kx^{k-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1 + x)^n = \sum_{k=0}^n { n \choose k}x^k = \sum_{k=0}^n { n \choose k}\frac{d}{dx}(x^k) = \sum_{k=0}^n { n \choose k}kx^{k-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(1+x)^{n-1} = \sum_{k=1}^n k{ n \choose k}x^{k-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$nx(1+x)^{n-1} = \sum_{k=1}^n k{ n \choose k}2^{k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if substitute &lt;strong&gt;x = 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n(1+2)^{n-1} = \sum_{k=1}^n k{ n \choose k}2^{k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="2060" LastEditorUserId="2060" LastEditorDisplayName="" LastEditDate="2010-09-22T06:33:21.900" LastActivityDate="2010-09-22T15:17:46.350" Title="Finding a simple expression - Binomial Theorem" Tags="&lt;homework&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="5200" PostTypeId="2" ParentId="5198" CreationDate="2010-09-22T05:09:42.863" Score="2" ViewCount="0" Body="&lt;p&gt;Since this looks like homework, I won't tell you how to do it with the binomial theorem and will tell you how to do it bijectively instead.  How many ways are there to choose a committee of $k$ people out of $n$ people, split the committee into Democrats and Republicans, then elect a President from that committee, for some $k$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, the answer is $\displaystyle \sum_{k=1}^{n} k \cdot 2^k {n \choose k}$.  On the other hand, one can first choose the President, then split the non-Presidents into Democrats, Republicans, and people not on the committee, then choose a party for the President, which can be done in $2n \cdot 3^{n-1}$ ways.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for how to do it through the binomial theorem, here is a different but related application.  Recall the geometric series formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 - x} = \sum_{k=0}^{\infty} x^k.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if we take the derivative of both sides?  We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1 - x)^2} = \sum_{k=1}^{\infty} kx^{k-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $x = \frac{1}{2}$, we then conclude that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{\infty} \frac{k}{2^{k-1}} = 4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem requires essentially the same trick.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-22T05:49:01.817" LastActivityDate="2010-09-22T05:49:01.817" CommentCount="1" />
  <row Id="5201" PostTypeId="1" AcceptedAnswerId="5202" CreationDate="2010-09-22T05:19:17.853" Score="1" ViewCount="104" Body="&lt;p&gt;Before I begin, note that this is significantly outside what I've studied, so if it's a load of crap just let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up till now, everything I've covered assumes the same domain for all variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to define an arbitrary one-to-one relationship between elements in two different sets, and want each of the variables x and y in the formula below to be in the domain of its particular set:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;X = {X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;2&lt;/sub&gt;, X&lt;sub&gt;3&lt;/sub&gt;}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Y = {Y&lt;sub&gt;1&lt;/sub&gt;, Y&lt;sub&gt;2&lt;/sub&gt;, Y&lt;sub&gt;3&lt;/sub&gt;}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, I want:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x ∈ X, y ∈ Y&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If possible, how would I go about specifying this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I've phrased the formula&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;∀x ∃y [ A(x, y) ∧ ¬A(¬x, y) ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where A(x, y) indicates an relationship between x and y. Is this statement well formed? Particularly the ¬x in ¬A(¬x, y)? And does it mean what I think it means - That A(x, y) for all x ∈ X except the x in question, and for the y in question, must be false in order for the statement ot be true? That is, does it define a one-to-one relationship?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot,&#xA;Wyatt&lt;/p&gt;&#xA;" OwnerUserId="2058" LastActivityDate="2010-09-22T20:28:58.890" Title="different domains for each variable in predicate logic" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5202" PostTypeId="2" ParentId="5201" CreationDate="2010-09-22T05:53:00.740" Score="4" ViewCount="0" Body="&lt;p&gt;No, I don't think it makes sense to write $\neg x$ in this context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to specify that $x\in X$ and $y\in Y$ is to make them clauses in your statement. For example, to write that for every $x\in X$ there should exist a $y\in Y$ such that $A(x,y)$ holds, you would usually write:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\forall x \left(x\in X\rightarrow \exists y(y\in Y\wedge A(x,y)\right).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the second part: to say that no other $z\in X$ will also be associated to $y$, the easiest why is to say that anyone who is associated to $y$ is going to be equal to $x$; so you would want a clause $A(z,y)\rightarrow x=z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it all together:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\forall x\left( x\in X\rightarrow \exists y(y\in Y\wedge A(x,y)\wedge \forall z(z\in X\wedge A(z,y)\rightarrow x=z))\right).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is also possible to use common (but perhaps slightly abusive) notation and write&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\forall x\in X\Bigl(\exists y\in Y\Bigl(\forall z\in X\Bigl(A(x,y)\wedge (A(z,y)\rightarrow x=z)\Bigr)\Bigr).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; As noted in several comments, the above does not completely describe a one-to-one &lt;em&gt;function&lt;/em&gt;; it only describes a relation with domain $X$, codomain $Y$, such that for each $y\in Y$ there is at most one $x\in X$ such that $A(x,y)$ holds; this does not fully describe a function (unless you happen to know some other things about $X$ and $Y$, e.g., that they are both finite and of equal size, in which case the other condition will follow), nor a bijective function. In general, the clause you want for &quot;function-ness&quot; is the dual of the condition you have: for all $x\in X$, and for all $y\in Y$, $A(x,y)\wedge A(x,y')\rightarrow y=y'$ (so each input has at most one output). For bijectivity (so you have a true correspondence) you also need that for all $y\in Y$ there exists $x\in X$ such that $A(x,y)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-22T15:46:56.413" LastActivityDate="2010-09-22T15:46:56.413" CommentCount="1" />
  <row Id="5203" PostTypeId="2" ParentId="4627" CreationDate="2010-09-22T07:25:38.223" Score="1" ViewCount="0" Body="&lt;p&gt;To deal well with substitution of variables in differential equations you should know when there is a differential operator and when there is a derivative. In the first case a variable is linked with differential operator and changing it you should also change a differential operator, in the second a derivative is just a function so you can simultaneously change the same argument in all equation's functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see on your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dQ}{dt}=\frac{a+Q}{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And rewrite it using a differential operator notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[Q(t)]=\frac{a+Q(t)}{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define a functional map $F_1(f)(x)=f(x)-a$, obvious this map is bijection and let $g$ be a solution of $F_1(g)=Q$. So $Q(t)=F_1(g)(t)=g(t)-a$ and we rewrite the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[g(t)-a]=\frac{a+g(t)-a}{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduce it and calculate a derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g'(t)=\frac{1}{b}g(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And back to a differential operator notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[g(t)]=\frac{1}{b}g(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the trick with a map $F_2(f)(x)=f(\frac{1}{b}x)$, let $w$ be a solution of $F_2(w)=g$, so $g(t)=F_2(w)(t)=w(\frac{1}{b}t)$ and substitute it into the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}[w(\frac{1}{b}t)]=\frac{1}{b}w(\frac{1}{b}t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate a derivative and reduce:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w'(\frac{1}{b}t)=w(\frac{1}{b}t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is no a differential operator, so we change an argument $\frac{1}{b}t$ to $t$ in whole equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w'(t)=w(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's known that a solution is $w(t)=C_1e^t$ and $Q=F_1(F_2(w))$ so $Q(t)=C_1e^{\frac{1}{b}t}-a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very verbose solution, but then understand it you can just use rules of substitution of variables and solve an equation in a few lines.&lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2010-09-22T07:25:38.223" CommunityOwnedDate="2010-09-22T07:25:38.223" />
  <row Id="5205" PostTypeId="1" CreationDate="2010-09-22T09:36:14.377" Score="2" ViewCount="256" Body="&lt;p&gt;I apologise for the really basic question. This didn't really fit on any other StackExchange website so the Maths one was the closest one where I could ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really Basic Question- What are 4 sided shapes called again? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like how triangles are 3 sided shapes, octagons are 8 sided shapes, ... What are the 4 sided ones called then?&lt;/p&gt;&#xA;" OwnerUserId="2062" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-22T14:39:18.223" LastActivityDate="2010-09-22T14:39:18.223" Title="What Are 4 Sided Shapes Called Again?" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="5206" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T09:36:33.927" Score="1" ViewCount="0" Body="&lt;p&gt;Wait, never mind. It's quadrilateral. :)&lt;/p&gt;&#xA;" OwnerUserId="2062" LastActivityDate="2010-09-22T09:36:33.927" />
  <row Id="5207" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T10:24:15.247" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quadrilateral&quot; rel=&quot;nofollow&quot;&gt;The word &lt;strong&gt;quadrilateral&lt;/strong&gt; is made of the words &lt;strong&gt;quad (meaning &quot;four&quot;)&lt;/strong&gt; and &lt;strong&gt;lateral (meaning &quot;of sides&quot;)&lt;/strong&gt;.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-22T10:24:15.247" />
  <row Id="5208" PostTypeId="2" ParentId="4851" CreationDate="2010-09-22T10:44:53.467" Score="2" ViewCount="0" Body="&lt;p&gt;Let's remove this question from the list of unanswered questions by up-voting this community wiki answer stating that Agusti Roig's comment above contains the right answer.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T10:44:53.467" CommunityOwnedDate="2010-09-22T10:44:53.467" />
  <row Id="5209" PostTypeId="2" ParentId="5201" CreationDate="2010-09-22T11:06:05.337" Score="0" ViewCount="0" Body="&lt;p&gt;There are lots of ways to write a formula A(x,y) that defines a bijection between two sets. If you're trying to keep the depth of quantifiers to a minimum, you could use&#xA;$$&#xA;\begin{align}&#xA;&amp;amp;  (\forall x)(\exists y) [ x \in X \to y \in Y \land A(x,y)] \\&#xA;&amp;amp; {} \land (\forall y)(\exists x) [ y \in Y \to x \in X \land A(x,y)]\\&#xA;&amp;amp; {} \land (\forall x,x',y,y') ([x \in X \land y \in Y) \\&#xA;&amp;amp; \qquad {} \to [(A(x,y) \land A(x',y) \to x=x'] &#xA; \land [A(x,y) \land A(x,y') \to y = y'])]\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be written more elegantly with the abbreviations $(\forall x \in X) $ and $(\exists x \in X)$:&#xA;$$&#xA;\begin{align}&#xA;(\forall x \in X)\Phi &amp;amp;\equiv (\forall x) [ x \in X \to \Phi] \\&#xA;(\exists x \in X)\Phi &amp;amp;\equiv (\exists x) [ x \in X \land \Phi] \\&#xA;\end{align}&#xA;$$&#xA;like so: &#xA;$$&#xA;\begin{align}&#xA;&amp;amp;  (\forall x \in X)(\exists y\in Y) A(x,y) \land (\forall y \in Y )(\exists x \in X) A(x,y)\\&#xA;&amp;amp; {} \land (\forall x \in X)(\forall x' \in X)(\forall y \in Y)(\forall y' \in Y)([(A(x,y) \land A(x',y) \to x=x'] \\&#xA;&amp;amp; \qquad \qquad {} \land [A(x,y) \land A(x,y') \to y = y'])\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-09-22T12:14:16.893" LastActivityDate="2010-09-22T12:14:16.893" />
  <row Id="5210" PostTypeId="2" ParentId="4600" CreationDate="2010-09-22T13:08:04.003" Score="2" ViewCount="0" Body="&lt;p&gt;You are absolutely right with your understanding. The notion of L-structure is very loose, it is just a way of attaching (any) meaning to the symbols. It is only after this that it makes sense to ask whether a given formal statement is true in a given L-structure - i.e. whether the L-structure is a &lt;em&gt;model&lt;/em&gt; of your statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very roughly, the area of model theory called Classification theory looks how many isomorphism classes of models for a given set of statements there are - of course you have to fix a cardinality, otherwise two models can't be isomorphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could continue your exercise of classifying all L-structures of cardinality 4 (which you did correctly) by classifying all isomorphism classes of such L-structures. An isomorphism of L-structures would be a bijection which preserves the relation and is compatible with the functions (see any book for a formal definition). You will see that lots of them are non-isomorphic - but maybe you won't want to make a complete list, it seems quite a lot of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was then the classification of models of the empty set of sentences. If you now require your L-structures to render a certain set of sentences true, you get a subset of your L-structures (usually easier to see) and can ask the same about this subset - your classification from before is of course still valid. This gives you a small taste of Classification theory (which of course in reality proceeds by very advanced and more effective methods, which don't feel like this exercise at all).&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-22T13:08:04.003" />
  <row Id="5211" PostTypeId="2" ParentId="4862" CreationDate="2010-09-22T13:18:34.913" Score="0" ViewCount="0" Body="&lt;p&gt;I have suggested three alternative paths to the usual calculus one in &lt;a href=&quot;http://math.stackexchange.com/questions/5018/how-to-start-with-mathematics/5188#5188&quot;&gt;this&lt;/a&gt; answer!&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-22T13:18:34.913" CommunityOwnedDate="2010-09-22T13:18:34.913" />
  <row Id="5212" PostTypeId="2" ParentId="4600" CreationDate="2010-09-22T13:21:54.113" Score="1" ViewCount="0" Body="&lt;p&gt;The reason that you have so much freedom is that you have not required that the structure satisfy any axioms at all. This is somewhat like saying &quot;Let $f$ be an arbitrary function&quot; - of course there are lots of examples.  If you added a list of axioms that the structure has to satisfy, it would reduce the freedom you have in constructing the structure. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-09-22T13:21:54.113" />
  <row Id="5213" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T13:57:14.117" Score="1" ViewCount="0" Body="&lt;p&gt;Shorter version is &quot;quadrangle&quot;.&lt;/p&gt;&#xA;" OwnerUserId="817" LastActivityDate="2010-09-22T13:57:14.117" CommentCount="3" />
  <row Id="5214" PostTypeId="2" ParentId="5205" CreationDate="2010-09-22T14:11:12.273" Score="0" ViewCount="0" Body="&lt;p&gt;I have a draft article where I 'should' use 'quadrilateral' dozens of times. I refuse, stating up front that I will use 'quad' instead.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2010-09-22T14:11:12.273" CommunityOwnedDate="2010-09-22T14:11:12.273" CommentCount="2" />
  <row Id="5215" PostTypeId="1" CreationDate="2010-09-22T14:22:38.240" Score="1" ViewCount="37" Body="&lt;p&gt;Given $X_i, Y_i \in SO_3, i=1..n, n&gt;2$, find $A,B$ that satisfies $AX_iB=Y_i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$, there are an infinite number of solutions, a trivial one is:&lt;br&gt;&#xA;$B=I$&lt;br&gt;&#xA;$A=Y_1X_1^T$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=2$, there should be exactly one solution. One could do the following:&lt;br&gt;&#xA;Compute $A$ from the equation where $i=1$: $A=Y_1B^{-1}X_1^{-1}$,&lt;br&gt;&#xA;Substituting this into equation where $i=2$: $Y_1B^{-1}X_1^{-1}X_2B=Y_2$,&lt;br&gt;&#xA;Now, let $C=X_1^{-1}X_2$ and $D=Y_1^{-1}Y_2$,&lt;br&gt;&#xA;We get: $B^{-1}CB=D$,&lt;br&gt;&#xA;Which can be rewritten as: $CB=BD$&lt;br&gt;&#xA;I'm already lost at this point. How do I solve for $B$, given $D,C$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now for $n&gt;2$, how can I find a solution that minimizes the least squares error? Is it possible to rearrange the $n$ equations into a linear system at all? Or do I need to look into using some kind of iterative approach or linear programming instead?&lt;/p&gt;&#xA;" OwnerUserId="2063" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-22T14:30:13.770" LastActivityDate="2010-09-22T14:48:06.623" Title="Looking for closed-form least squares solution of a simplified inverse kinematics problem" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="5216" PostTypeId="1" AcceptedAnswerId="5242" CreationDate="2010-09-22T14:32:45.077" Score="2" ViewCount="117" Body="&lt;p&gt;I have an array of &quot;lines&quot; each defined by 2 points. I am working with only the line segments lying between those points. I need to search lines that could continue one another (relative to some angle) and lie on the same line (with some offset)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean I had something like 3 lines&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4uzcj.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved some mathematical problem (formulation of which is my question) and got understanding that there are lines that could be called relatively one line (with some angle K and offset J)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9YEFd.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course by math formulation I meant some kind of math formula like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NsKKl.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I know the algorithm I used such simple approach like one presented by Victor Liu &lt;a href=&quot;http://stackoverflow.com/questions/3769885/how-to-formulate-such-problem-mathematicaly-line-continuation-search/3770275#3770275&quot;&gt;http://stackoverflow.com/questions/3769885/how-to-formulate-such-problem-mathematicaly-line-continuation-search/3770275#3770275&lt;/a&gt; but its Pusedo solution code and what I need is problem formulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is its formalisation into math using some Matrix and other math words (to write into some paper=) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;guys - I am a programmer - sorry=) &lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-22T17:17:36.690" LastActivityDate="2010-09-22T17:42:06.953" Title="How to formulate such problem mathematicaly? (line continuation search)" Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="0" />
  <row Id="5217" PostTypeId="1" AcceptedAnswerId="5227" CreationDate="2010-09-22T14:47:45.047" Score="4" ViewCount="96" Body="&lt;p&gt;The problem asks us to calculate:&lt;br&gt;&#xA;$ \sum_{i = 0}^{n}(-1)^i \binom{n}{i} \binom{n}{n-i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I tried solving is:&lt;br&gt;&#xA;The given sum is the coefficient of $x^n$ in $ (1+x)^n(1-x)^n $, which is $ (1 - x^2)^n $.&#xA;The coefficient of $x^n$ in $(1 -x^2)^n$ is $ (-1)^{n/2} \binom{n}{n/2} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I doing it right?&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="813" LastEditorDisplayName="" LastEditDate="2010-09-22T15:57:08.537" LastActivityDate="2010-09-22T15:57:08.537" Title="a combinatorial exercise" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5218" PostTypeId="2" ParentId="5215" CreationDate="2010-09-22T14:48:06.623" Score="1" ViewCount="0" Body="&lt;p&gt;$\mathbf{C}\mathbf{B}-\mathbf{B}\mathbf{D}=\mathbf{0}$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester_equation&quot; rel=&quot;nofollow&quot;&gt;Sylvester equation&lt;/a&gt;, for which algorithms like &lt;a href=&quot;http://doi.acm.org/10.1145/361573.361582&quot; rel=&quot;nofollow&quot;&gt;Bartels-Stewart&lt;/a&gt; or &lt;a href=&quot;http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=1102170&quot; rel=&quot;nofollow&quot;&gt;Hessenberg-Schur&lt;/a&gt; apply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't seen your more general problem in my notes, but I'll update this answer when I find something.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-22T14:48:06.623" CommentCount="1" />
  <row Id="5219" PostTypeId="1" AcceptedAnswerId="5228" CreationDate="2010-09-22T14:57:18.170" Score="4" ViewCount="298" Body="&lt;p&gt;Assume $k,n$ are positive integers. The sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+\ldots +\dfrac{1}{k+n-1}+\dfrac{1}{k+n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is not an integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $k=0$ is proved in this question &quot;&lt;a href=&quot;http://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-k1n-1-k-is-never-an-integer&quot;&gt;Is there an elementary proof that $\sum_{k=1}^{n}1/k$ is never an integer?&lt;/a&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="2044" LastActivityDate="2010-09-22T16:43:28.570" Title="How do I prove this sum is not an integer" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="5220" PostTypeId="1" AcceptedAnswerId="5222" CreationDate="2010-09-22T15:05:14.607" Score="2" ViewCount="109" Body="&lt;h3&gt;Background:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I was studying Theorem 2.3.3 from Introduction to Graph Theory by W. B. West. The main idea of his proof is as follows:&#xA;&lt;br&gt;&#xA;&lt;b&gt;T&lt;/b&gt;, resulting tree.&#xA;&lt;br&gt;&#xA;&lt;b&gt;T&lt;sup&gt; *&lt;/sup&gt;&lt;/b&gt;, spanning tree of minimum weight.&#xA;&lt;br&gt;&#xA;Let, T $\neq$ T&lt;sup&gt; *&lt;/sup&gt;, then there are edges in T&lt;sup&gt; *&lt;/sup&gt; that are not in T.&#xA;We first consider only 1 edge. Let's name it e. Hence, T&lt;sup&gt; *&lt;/sup&gt; has all the edges that are in T, except e. Now, if we add e to T&lt;sup&gt; *&lt;/sup&gt;, we should get a cycle, say 'C'. This implies, C has an edge that is not in T. Let's name it e'. Now we get spanning tree T&lt;sup&gt; *&lt;/sup&gt;+ e - e'.&#xA;Now, let us think what happened when Kruskal ran and produced T. It examined all edges in G. Including e and e'. But, it included only e. So, w(e)$\leq$w(e').&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus T&lt;sup&gt; *&lt;/sup&gt;+ e - e' is a spanning tree with weight at most T&lt;sup&gt; *&lt;/sup&gt; that agrees with T for a &lt;em&gt;longer initial list&lt;/em&gt; of edges than T&lt;sup&gt; *&lt;/sup&gt; does.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What does it mean actually?&lt;/li&gt;&#xA;&lt;li&gt;Anybody knows any other proof?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-09-22T15:16:29.410" Title="Prove : In a connected weighted Graph G, Kruskal's Algorithm constructs a minimum spanning tree" Tags="&lt;graph-theory&gt;&lt;proof-strategy&gt;" AnswerCount="1" />
  <row Id="5221" PostTypeId="2" ParentId="4851" CreationDate="2010-09-22T15:07:40.383" Score="-1" ViewCount="0" Body="&lt;p&gt;Let's honor Rasmus' creative handling of the software restrictions by downvoting &lt;a href=&quot;http://math.stackexchange.com/questions/4851/cover-e-b-gives-a-homeomorphism-e-aute-b/5221#5221&quot;&gt;this&lt;/a&gt; self-referential community wiki answer.&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-22T15:07:40.383" CommunityOwnedDate="2010-09-22T15:07:40.383" CommentCount="1" />
  <row Id="5222" PostTypeId="2" ParentId="5220" CreationDate="2010-09-22T15:16:29.410" Score="2" ViewCount="0" Body="&lt;p&gt;If T is a tree and one joins two vertices of this tree that are not already joined by an edge with a new edge, one gets a unique circuit C. Now, what happens if one removes an edge from C different from the edge just added to T? One gets a new tree.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-22T15:16:29.410" CommentCount="2" />
  <row Id="5224" PostTypeId="2" ParentId="4862" CreationDate="2010-09-22T15:19:11.307" Score="1" ViewCount="0" Body="&lt;p&gt;I do not know what &quot;Calculus 1-3&quot; entails, but if the terms &quot;complete ordered field&quot; and &quot;least upper bound property&quot; don't mean anything to you, I'd start with a first course in real analysis. The usual stuff: Cauchy sequences, absolutely and conditionally convergent series, the $\varepsilon$-$\delta$ definition of continuity, the intermediate value theorem, a rigorous construction of the Riemann integral using upper and lower sums$\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All (at least non-nonstandard) analysis flows from this, up to and including metric spaces, topology (to a lesser degree admittedly), differential geometry, complex analysis, functional analysis and meassure theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paraphrasing the late Randy Pausch: Fundamentals. Fundamentals. Fundamentals. You need to have the basics straight or the fancy stuff ain't gonna work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there I'd look at linear algebra. As has been mentioned earlier, it's nice and concrete but still very useful and shows up in a lot of places. For example in abstract algebra when studying field extensions(a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of abstract algebra, it takes up an interesting place in undergraduate mathematics. I find it hard to explain, but it's like it's sitting in a corner off by itself doing its own thing - at least until you get to for example algebraic topology. But algebraic topology isn't usually taught to undergraduates (at least that I know of), so algebra feels rather disconnected from the rest of an undergraduate mathematics education.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what it boils down to is that I'd stay away from abstract algebra beyond a first course (because all undergraduate math students need to at least know groups, rings and fields) unless you either 1) loved the first course, 2) find out that algebraic tools are needed for something else you like, or 3) the rules say you must do more algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Though probably not in a first course in abstract algebra since Galois theory can't really be taught until after you've done the basics of groups, rings and fields which take up at least a semester.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-09-22T15:19:11.307" CommunityOwnedDate="2010-09-22T15:19:11.307" />
  <row Id="5225" PostTypeId="2" ParentId="4498" CreationDate="2010-09-22T15:19:57.477" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not sure you've asked a question with an answer, but I'll try to address what I can.  Surely people have looked at small complete groups, but people do not tend to publish such things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest complete group that is not S4, Aut(simple), nor Hol(p) is S3 x Hol(5).  Direct products of complete groups are often but certainly not always complete.  You might try to find reasonable conditions on when the direct product of completes is complete.  Other people have written down some conditions, but I can't recall where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite family omitted are the AΓL(1,q) of semi-linear transformations of a one-dimensional vector space.  If q is prime, this is just Hol(q).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next two groups not covered might be interesting to understand.  They are of the form G⋉V where G is complete and V is an absolutely irreducible G-module (G being AΓL(1,9) or Hol(5), V being smallest non-trivial dimension and rational).  I don't think all such groups are complete, but it would be interesting to know conditions on G,V to make them complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think one will get any good sort of classification as there are just weird complete groups.  &lt;s&gt;One group V of order 128 has an automorphism G of order 7 (and several other outer automorphisms), but the semi-direct product G⋉V of order 128*7 is complete.&lt;/s&gt; (SmallGroup(432,520) and SG(1344,816) are almost as weird).  I at least find it a little disturbing to have small subgroups of Hol(V) be complete when neither V nor Hol(V) is complete.  The other G⋉V examples are like this too, but perhaps I find it less disturbing when V is abelian.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-09-30T15:23:13.437" LastActivityDate="2010-09-30T15:23:13.437" CommentCount="3" />
  <row Id="5226" PostTypeId="1" AcceptedAnswerId="13394" CreationDate="2010-09-22T15:35:46.090" Score="3" ViewCount="78" Body="&lt;p&gt;Call a finite group A &lt;strong&gt;affine&lt;/strong&gt; if it has a normal, self-centralizing, complemented, elementary abelian subgroup V.  Such a group A is a semi-direct product G⋉V where V is a vector space of dimension n over GF(p) and G is a group of matrices in GL(n,p).  The elements of A can be written as matrices [ g, v ; 0, 1 ] where g in G, v in V and 0,1 are row vectors of the appropriate length.  Conversely, given G ≤ GL(V), V a vector space over GF(p), A=G⋉V is an affine group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the full affine group, AGL(n,p) where G is GL(n,p) and V is GF(p)^n, that is A is all (n+1)×(n+1) matrices of the form [ g, v ; 0, 1 ] where g is n×n invertible, v is anything, 0 is a zero vector, and 1 is just a 1×1 identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V becomes a G-module and its G-module structure has a large influence on the group theoretic structure of A.  In particular, V contains no &quot;trivial&quot; (central) summand as a G-module iff A is centerless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Supposing A is centerless, what conditions on V ensure A is a complete group, that is, so that A is also &quot;outerless&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See the previous questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4238/finite-non-abelian-group-with-centre-but-no-outer-automorphism&quot;&gt;4238: Finite non-abelian group with centre but no outer automorphism&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4498/classification-of-small-complete-groups&quot;&gt;4498: Classification of small complete groups&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-07T16:53:38.637" Title="When is an affine group complete?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5227" PostTypeId="2" ParentId="5217" CreationDate="2010-09-22T15:36:32.957" Score="3" ViewCount="0" Body="&lt;p&gt;Your solution is correct for even $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is odd then your last sentence should read &quot;The coefficient of $x^n$ in $(1-x^2)^n$ is $0$.&quot; This is because only even powers of $x$ occur when expanding $(1-x^2)^n$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-22T15:42:07.033" LastActivityDate="2010-09-22T15:42:07.033" CommentCount="1" />
  <row Id="5228" PostTypeId="2" ParentId="5219" CreationDate="2010-09-22T15:38:13.810" Score="5" ViewCount="0" Body="&lt;p&gt;The proof that &lt;a href=&quot;http://math.stackexchange.com/questions/2752&quot;&gt;I gave in that thread&lt;/a&gt; works just as well here. It depends only on the fact that in any contiguous sequence of integers (here denominators) the maximal power $\rm 2^k$ that divides any element occurs in precisely one element. Indeed, after the first (necessarily odd) multiple of $\rm 2^k$, the next multiple would, by contiguity, be an even multiple, so a multiple of $\rm 2^{k+1}$, contra maximality. Here is said proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\;$ Since there is a unique denominator $\rm 2^k$ having maximal power of $2$, upon multiplying all terms through by $\rm 2^{k-1}$ one deduces the contradiction that $\rm\; a/2 = b/c \;$ with $\rm \; a,\ c \:$ odd. As an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad m = \frac{1}{2} + \frac{1}{3} +\; \frac{1}{4} \;+\; \frac{1}{5} + \frac{1}{6} + \frac{1}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad\;\; 2m = \; 1 + \frac{2}{3} +\; \frac{1}{2} \;+\; \frac{2}{5} + \frac{1}{3} + \frac{2}{7} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\Rightarrow\quad -\frac{1}{2}  = \; 1 + \frac{2}{3} - 2m  + \frac{2}{5} + \frac{1}{3} + \frac{2}{7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prior sum has all odd denominators so reduces to a fraction with odd denominator $\rm d\:|3\cdot 5\cdot 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; $\:$ I purposely avoided any use of valuation theory because Anton requested an &quot;elementary&quot; solution. The above proof can easily be made comprehensible to a high-school student. The proof is trivial to anyone who knows valuation theory: namely the sum has a lone dominant term with 2-adic value $\rm v_2&amp;lt;0\:$ so, by the &lt;strong&gt;domination principle&lt;/strong&gt;, the sum has 2-adic value $\rm v_2&amp;lt;0\:,\:$ i.e. the sum has even denominator in lowest terms, so it is nonintegral.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-22T16:43:28.570" LastActivityDate="2010-09-22T16:43:28.570" CommentCount="1" />
  <row Id="5229" PostTypeId="2" ParentId="5219" CreationDate="2010-09-22T15:38:55.540" Score="2" ViewCount="0" Body="&lt;p&gt;{Differences of harmonic numbers are not integers}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\sum_{k=m}^n {1\over k}$. Find the&#xA;largest power of $2^r$ that divides one of the denominators between $m$ and $n$.&#xA;There can be no even multiples of $2^r$ between $m$ and $n$, hence there&#xA; is only &lt;strong&gt;one&lt;/strong&gt; odd multiple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $2^{r-1}(1/m+\cdots+1/n)$ is equal to $1/2o_1$ plus a bunch of&#xA;fractions with odd denominator. Adding them gives one fraction with&#xA;an odd denominator, write it as $x/o_2$.&#xA;Here $o_1$ and $o_2$ are odd integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $2^{r-1}(1/m+\cdots+1/n)=1/2o_1+x/o_2$ so that&#xA;$${1\over m}+\cdots+{1\over n}={2xo_1+o_2\over 2^r o_1 o_2}.$$&#xA;Therefore $2^r$ divides the denominator.&#xA;If $n\not=m$, then $r\geq1$ and the sum is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cf. Exercise 6.21 (page 311) of Graham, Knuth, and Patashnik.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-22T15:38:55.540" />
  <row Id="5230" PostTypeId="1" AcceptedAnswerId="5237" CreationDate="2010-09-22T15:55:06.450" Score="2" ViewCount="113" Body="&lt;p&gt;Suppose $\mu$ is a measure. Is there any difference in meaning between the notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f(x)d\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f(x) \mu(dx)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastEditorUserId="1820" LastEditorDisplayName="" LastEditDate="2010-09-22T16:27:14.247" LastActivityDate="2010-09-22T16:55:27.000" Title="Notation question: Integrating against a measure" Tags="&lt;notation&gt;&lt;integral&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5231" PostTypeId="1" AcceptedAnswerId="5961" CreationDate="2010-09-22T16:05:06.420" Score="1" ViewCount="182" Body="&lt;p&gt;I have coordinates for 4 vertices/points that define a plane and the normal/perpendicular.&#xA;The plane has an arbitrary rotation applied to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I 'un-rotate'/translate the points so that the plane has rotation 0 on x,y,z ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to get the plane rotation from the plane's normal:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rotationX = atan2(normal.z,normal.y);&#xA;rotationY = atan2(normal.z,normal.x);&#xA;rotationZ = atan2(normal.y,normal.x);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this correct ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I apply the inverse rotation to the position vectors ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to create a matrix with those rotations and multiply it with the vertices,&#xA;but it doesn't look right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, I've wrote a simple test using &lt;a href=&quot;http://processing.org/&quot; rel=&quot;nofollow&quot;&gt;Processing&lt;/a&gt; and can be seen &lt;a href=&quot;http://lifesine.eu/so/vertex_rotation/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float s = 50.0f;//scale/unit&#xA;PVector[] face = {new PVector(1.08335042,0.351914703846,0.839020013809),&#xA;new PVector(-0.886264681816,0.69921118021,0.839020371437),&#xA;new PVector(-1.05991327763,-0.285596489906,-0.893030643463),&#xA;new PVector(0.909702301025,-0.63289296627,-0.893030762672)};&#xA;PVector n = new PVector(0.150384, -0.500000, 0.852869);&#xA;PVector[] clone;&#xA;&#xA;void setup(){&#xA;  size(400,400,P3D);&#xA;  smooth();&#xA;  clone = unRotate(face,n,true);&#xA;}&#xA;&#xA;void draw(){&#xA;  background(255);&#xA;  translate(width*.5,height*.5);&#xA;  if(mousePressed){&#xA;    rotateX(map(mouseY,0,height,0,TWO_PI));&#xA;    rotateY(map(mouseX,0,width,0,TWO_PI));&#xA;  }&#xA;  stroke(128,0,0);&#xA;  beginShape(QUADS);&#xA;    for(int i = 0 ; i &amp;lt; 4; i++) vertex(face[i].x*s,face[i].y*s,face[i].z*s);&#xA;  endShape();&#xA;  stroke(0,128,0);&#xA;  beginShape(QUADS);&#xA;    for(int i = 0 ; i &amp;lt; 4; i++) vertex(clone[i].x*s,clone[i].y*s,clone[i].z*s);&#xA;  endShape(); &#xA;}&#xA;//get rotation from normal&#xA;PVector getRot(PVector loc,Boolean asRadians){&#xA;  loc.normalize();&#xA;  float rz = asRadians ? atan2(loc.y,loc.x) : degrees(atan2(loc.y,loc.x));&#xA;  float ry = asRadians ? atan2(loc.z,loc.x) : degrees(atan2(loc.z,loc.x));&#xA;  float rx = asRadians ? atan2(loc.z,loc.y) : degrees(atan2(loc.z,loc.y));&#xA;  return new PVector(rx,ry,rz);&#xA;}&#xA;//translate vertices&#xA;PVector[] unRotate(PVector[] verts,PVector no,Boolean doClone){&#xA;  int vl = verts.length;&#xA;  PVector[] clone;&#xA;  if(doClone) {&#xA;    clone = new PVector[vl];&#xA;    for(int i = 0; i&amp;lt;vl;i++) clone[i] = PVector.add(verts[i],new PVector());&#xA;  }else clone = verts;&#xA;  PVector rot = getRot(no,false);&#xA;&#xA;  PMatrix3D rMat = new PMatrix3D();&#xA;  rMat.rotateX(-rot.x);rMat.rotateY(-rot.y);rMat.rotateZ(-rot.z);&#xA;  for(int i = 0; i&amp;lt;vl;i++) rMat.mult(clone[i],clone[i]);&#xA;  return clone;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Any syntax/pseudo code or explanation is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What trying to achieve is this:&#xA;If I have a rotated plane:&#xA;&lt;img src=&quot;http://i.imgur.com/bZ1fn.png&quot; alt=&quot;rotated plane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can move the vertices to have something that would have no rotation:&#xA;&lt;img src=&quot;http://i.imgur.com/ogFR0.png&quot; alt=&quot;plane with no rotations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@muad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand. I thought I was using matrices for rotations.&#xA;PMatrix3D's rotateX,rotateY,rotateZ calls should done the rotations for me.&#xA;Doing it manually would be declaring 3d matrices and multiplying them.&#xA;Here's a little snippet to illustrate this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; PMatrix3D rx = new PMatrix3D(1,          0,          0,  0,&#xA;                               0, cos(rot.x),-sin(rot.x),  0,&#xA;                               0, sin(rot.x),cos(rot.x) ,  0,&#xA;                               0,          0,          0,  1);&#xA;  PMatrix3D ry = new PMatrix3D(cos(rot.y), 0,sin(rot.y),  0,&#xA;                                        0, 1,0         ,  0,&#xA;                              -sin(rot.y), 0,cos(rot.y),  0,&#xA;                                        0, 0,0         ,  1);&#xA;  PMatrix3D rz = new PMatrix3D(cos(rot.z),-sin(rot.z), 0, 0,&#xA;                               sin(rot.z), cos(rot.z), 0, 0,&#xA;                               0         ,          0, 1, 0,&#xA;                               0         ,          0, 0, 1);&#xA;  PMatrix3D r = new PMatrix3D();&#xA;  r.apply(rx);r.apply(ry);r.apply(rz);&#xA;&#xA;  //test&#xA;  PMatrix rmat = new PMatrix3D();rmat.rotateX(rot.x);rmat.rotateY(rot.y);rmat.rotateZ(rot.z);&#xA;  float[] frmat = new float[16];rmat.get(frmat);&#xA;  float[] fr    = new float[16];r.get(fr);&#xA;  println(frmat);println(fr);&#xA;&#xA;/*&#xA;Outputs:&#xA;[0] 0.059300933&#xA;[1] 0.09312407&#xA;[2] -0.99388695&#xA;[3] 0.0&#xA;[4] 0.90466285&#xA;[5] 0.41586864&#xA;[6] 0.09294289&#xA;[7] 0.0&#xA;[8] 0.42198166&#xA;[9] -0.9046442&#xA;[10] -0.059584484&#xA;[11] 0.0&#xA;[12] 0.0&#xA;[13] 0.0&#xA;[14] 0.0&#xA;[15] 1.0&#xA;[0] 0.059300933&#xA;[1] 0.09312407&#xA;[2] -0.99388695&#xA;[3] 0.0&#xA;[4] 0.90466285&#xA;[5] 0.41586864&#xA;[6] 0.09294289&#xA;[7] 0.0&#xA;[8] 0.42198166&#xA;[9] -0.9046442&#xA;[10] -0.059584484&#xA;[11] 0.0&#xA;[12] 0.0&#xA;[13] 0.0&#xA;[14] 0.0&#xA;[15] 1.0&#xA;&#xA;*/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="304" LastEditorUserId="304" LastEditorDisplayName="" LastEditDate="2010-09-22T22:28:31.207" LastActivityDate="2010-11-24T23:35:50.863" Title="How can I remove rotations from points defining a plane ?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" />
  <row Id="5232" PostTypeId="1" CreationDate="2010-09-22T16:17:23.533" Score="3" ViewCount="79" Body="&lt;p&gt;I spent a little time looking around the Wikipedia and Wolfram articles on Penrose Tiling, the Domino Problem, Wang Tiles, etc., but I'm having a little trouble getting into them. A lot of these resources are fairly non-linear in that they often have circular dependencies of concepts that you need to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I can tell that aperiodicity is important somehow, but I don't know why and I'm not quite sure about the actual definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep seeing tiling linked in unexpectedly (e.g. L-systems, the Entscheidungsproblem) or in non-mathematical places (games, geometric art... even Neal Stephenson's &lt;em&gt;Anathem&lt;/em&gt;) that it seems important and I'd really like to know more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with that said: does anyone have any resources they'd recommend? I'm looking for something fairly complete or at least introductory, not just an article about a certain facet. Books are fine, visual or interactive resources are great. Would love descriptions of practical implications and relationships with other fields.&lt;/p&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2010-09-22T17:46:27.240" Title="Getting started with aperiodic tiling" Tags="&lt;tiling&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5233" PostTypeId="1" CreationDate="2010-09-22T16:34:50.537" Score="6" ViewCount="250" Body="&lt;p&gt;When teaching abstract vector spaces for the first time, it is handy to have some really weird examples at hand, or even some really weird non-examples that may illustrate the concept. For example, a physicist friend of mine uses &quot;color space&quot; as a (non) example, with two different bases given essentially {red, green, blue} and {hue, saturation and brightness} (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Color_space&lt;/a&gt;). I say this is a non-example for a number of reasons, the most obvious being the absence of &quot;negative color&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, what are some bizarre and vivid examples of vector spaces you've come across that would be suitable for a first introduction?&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-09-23T00:58:21.423" Title="Vivid examples of vector spaces?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="5234" PostTypeId="2" ParentId="5058" CreationDate="2010-09-22T16:43:40.970" Score="10" ViewCount="0" Body="&lt;p&gt;Let's rewrite the formula, using $C_1=1/2$ as&#xA;$$S_n=\gamma+\log n-H_{n-1}$$&#xA;where&#xA;$$S_n=\sum_{k=1}^\infty\frac{(k-1)!C_k}{n(n+1)\cdots(n+k-1)}.$$&#xA;Use the formula&#xA;$$\frac{(k-1)!}{n(n+1)\cdots(n+k-1)}dx=\int_0^1 x^{n-1}(1-x)^{k-1}$$&#xA;which is a special case of the beta integral.&#xA;Therefore&#xA;$$S_n=\int_0^1x^{n-1}\sum_{k=1}^\infty C_k(1-x)^k&#xA;=\int_0^1\left(\frac{1}{1-x}+\frac{1}{\log x}\right)x^{n-1}dx.$$&#xA;When $n=1$ we get&#xA;$$S_1=\int_0^1\left(\frac{1}{1-x}+\frac{1}{\log x}\right)dx=\gamma$$&#xA;by a well-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; which can be derived from the aymptotic&#xA;$$\int_t^\infty\frac{e^{-x}}{x}dx=-\log t-\gamma+O(t)$$&#xA;as $t\to0$, for the exponential integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction it suffices to consider the difference&#xA;$$S_n-S_{n+1}=\int_0^1\left(x^{n-1}+\frac{x^{n-1}-x^n}{\log x}\right)dx&#xA;=\frac1n-\int_0^\infty\frac{e^{-ny}-e^{-(n+1)y}}{y}dy.$$&#xA;Using the identity&#xA;$$\int_0^\infty\frac{e^{-ay}-e^{-by}}{y}=\log\frac ba$$&#xA;for $0 &amp;lt; a &amp;lt; b$ &#xA;which follows by integrating $e^{-xy}$ over the region $[a,b]\times[0,\infty)$&#xA;we get&#xA;$$S_n-S_{n+1}=\frac1n-\log\frac{n+1}{n}.$$&#xA;The desired formula for $S_n$ now follows by induction.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-22T22:09:09.710" LastActivityDate="2010-09-22T22:09:09.710" CommentCount="2" />
  <row Id="5235" PostTypeId="2" ParentId="5232" CreationDate="2010-09-22T16:44:32.807" Score="1" ViewCount="0" Body="&lt;p&gt;Try &lt;a href=&quot;http://www.ams.org/bookstore-getitem/item=STML-1&quot; rel=&quot;nofollow&quot;&gt;Miles of Tiles&lt;/a&gt; by Charles Radin.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-09-22T16:44:32.807" />
  <row Id="5236" PostTypeId="2" ParentId="5233" CreationDate="2010-09-22T16:47:32.023" Score="3" ViewCount="0" Body="&lt;p&gt;I like the color example.  It shows how the idea of a basis is useful, even though it's not a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barycentric coordinates are another example of something like a vector space but not a vector space.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-22T16:47:32.023" CommentCount="2" />
  <row Id="5237" PostTypeId="2" ParentId="5230" CreationDate="2010-09-22T16:55:27.000" Score="7" ViewCount="0" Body="&lt;p&gt;There is no difference in meaning. I grabbed five books at random from my shelf and four out of five use the form $\int f(x) \mu(dx)$, while one uses the form  $\int f(x) d\mu(x)$. The form suppressing the variable of integration $\int f d\mu$ is also very common. I have never seen $\int f(x) d\mu$, which seems to me a weird hybrid. I work in probability so there may be some bias.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-09-22T16:55:27.000" CommunityOwnedDate="2010-09-22T16:55:27.000" CommentCount="3" />
  <row Id="5238" PostTypeId="1" AcceptedAnswerId="5245" CreationDate="2010-09-22T17:27:40.383" Score="1" ViewCount="123" Body="&lt;p&gt;For my homework, I have been asked to rationalise and simplify this surd;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{11}{3\sqrt{3}+7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time I do this I get the wrong answer. The method I am using is;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{11}{3\sqrt3-7} \times \frac{3\sqrt3-7}{3\sqrt3-7} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ended up with $\frac{33+11\sqrt3-77}{9+3+21+7\sqrt3-21-7\sqrt3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ends up no where near the right answer, even once it is simplified, can someone tell me where i'm going wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-09-22T17:34:29.293" LastActivityDate="2010-09-23T03:01:01.600" Title="Rationalising the denominator" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="5239" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:36:44.000" Score="1" ViewCount="0" Body="&lt;p&gt;After multiplying the numerator and denominator by $3\sqrt3-7$&#xA;the new denominator is &#xA;$$(3\sqrt 3+7)(3\sqrt3-7)=(3\sqrt3)^2-7^2=27-49=-22$$&#xA;a nice integer to divide by.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-22T17:36:44.000" />
  <row Id="5240" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:37:09.383" Score="1" ViewCount="0" Body="&lt;p&gt;Do you mean rationalise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{11}{3\sqrt{3}-7} \qquad \text{?}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And are you sure you're trying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{11}{3\sqrt{3}-7} \cdot \frac{3\sqrt{3} + 7}{3\sqrt{3} + 7}\qquad \text{?}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, &lt;a href=&quot;http://en.wikipedia.org/wiki/Rationalisation_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; is almost always of great help too.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-23T03:01:01.600" LastActivityDate="2010-09-23T03:01:01.600" />
  <row Id="5241" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:38:19.957" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; The general trick is&#xA;$$&#xA;\frac{1}{\sqrt{a}+b}=\frac{\sqrt{a}-b}{(\sqrt{a}+b)(\sqrt{a}-b)}=\frac{\sqrt{a}-b}{a-b^2}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T17:38:19.957" />
  <row Id="5242" PostTypeId="2" ParentId="5216" CreationDate="2010-09-22T17:42:06.953" Score="2" ViewCount="0" Body="&lt;p&gt;Write one procedure that takes seg1 and seg2 as input, and returns how close their closest endpoints are.  Write another procedure that computes the minimum angle between the lines containing seg1 and seg2.&#xA;Finally, mix the two with some weights to determine a &lt;em&gt;continuation score&lt;/em&gt;: how good seg2 serves as a continuation of seg1.  Then select the seg2 that has the best continuation score for seg1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you have, say, $10^5$ segments, then you will need to make it more efficient, which is an entirely different topic.)&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-09-22T17:42:06.953" />
  <row Id="5243" PostTypeId="2" ParentId="5233" CreationDate="2010-09-22T17:43:45.657" Score="1" ViewCount="0" Body="&lt;p&gt;I like the example $C([0,1])$ of continuous functions on the interval (or something similar). It is familiar-looking but shows that there is not always a natural choice of basis.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T17:43:45.657" CommentCount="5" />
  <row Id="5244" PostTypeId="2" ParentId="5232" CreationDate="2010-09-22T17:46:27.240" Score="2" ViewCount="0" Body="&lt;p&gt;Chaim Goodman-Strauss has quite a few papers on his web page about this topic, including this specific one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://comp.uark.edu/~strauss/papers/newsmall.pdf&quot; rel=&quot;nofollow&quot;&gt;http://comp.uark.edu/~strauss/papers/newsmall.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goodman-Strass's publications page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://comp.uark.edu/~strauss/papers/index.html&quot; rel=&quot;nofollow&quot;&gt;http://comp.uark.edu/~strauss/papers/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-22T17:46:27.240" />
  <row Id="5245" PostTypeId="2" ParentId="5238" CreationDate="2010-09-22T17:47:58.703" Score="2" ViewCount="0" Body="&lt;p&gt;You're mistakenly multiplying $\rm\; a * b\sqrt{3} \ =\ ab + a\sqrt{3}\:\,\;$ but $\rm\; ab\:\sqrt{3}\;$ is correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words $\rm\; b\:\sqrt{3}\;$ means $\rm b * \sqrt{3}\:,\;$ not $\rm\; b + \sqrt{3}\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to rationalize the denominator use $\rm\; (a+b\sqrt 3)\:(a-b\sqrt 3)\ =\ a^2 - 3 b^2$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-22T17:47:58.703" />
  <row Id="5246" PostTypeId="2" ParentId="2837" CreationDate="2010-09-22T18:06:33.650" Score="-2" ViewCount="0" Body="&lt;p&gt;Dear God you guys over complicated things...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being a math expert along the lines of you guys, I set the problem down and came back to it later. Upon looking it over again, I realized I could really simplify this by using the given data to look at it as different geometry. Instead of a line segment and a circle and testing for intersection, I used the radius of the circle to convert the line to a rectangle and then checked if the center of the circle (as just a point, no longer as a circle) is inside the new rectangle.&#xA;&lt;img src=&quot;http://i.imgur.com/NRO80.png&quot; alt=&quot;Line Circle Intersect Using New Geometry&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse my rectangle not having perfect dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although my answer meets my needs for my specific application, to better answer my own question, about the semi circles at the end of the rectangle (to test if an endpoint of the line segment is contained in the circle), I can compare the distance between an endpoint and the center of the circle, if that distance is less than the radius of the circle, the circle contains the end point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My application was for a Tower Defense game. The line represented some sort of laser stretched across the trail that enemies (the circle) would walk past, I needed to know when the laser contacted the enemy. The laser endpoints are attached to towers that are not positioned on the path the enemy walks down, so the enemy's circle could not contain an endpoint. I realize I was unclear during the original question about some of these details.&lt;/p&gt;&#xA;" OwnerUserId="757" LastEditorUserId="757" LastEditorDisplayName="" LastEditDate="2010-09-25T20:12:43.303" LastActivityDate="2010-09-25T20:12:43.303" CommentCount="2" />
  <row Id="5247" PostTypeId="2" ParentId="5231" CreationDate="2010-09-22T18:10:55.087" Score="1" ViewCount="0" Body="&lt;p&gt;Try to represent rotations using matrices instead of angles - then finding the inverse is easy.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-22T18:10:55.087" CommentCount="5" />
  <row Id="5248" PostTypeId="1" AcceptedAnswerId="5322" CreationDate="2010-09-22T19:09:03.090" Score="5" ViewCount="878" Body="&lt;p&gt;A famous exercise which one encounters while doing Complex Analysis(Residue theory): is to prove that the given integral $$\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, can anyone prove this without using Residue theory. I actually thought of doint this: $$\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \lim_{t \to \infty} \int_{0}^{t} \frac{1}{t} \Bigl( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \Bigr) \ dt$$ but i don't see how $\pi$ comes here, since we need the answer to be equal to $\frac{\pi}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-31T02:06:54.593" LastActivityDate="2010-10-31T02:06:54.593" Title="Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="5" />
  <row Id="5249" PostTypeId="2" ParentId="5248" CreationDate="2010-09-22T19:19:06.853" Score="6" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_integral&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dirichlet_integral&lt;/a&gt; for a proof using differentiation under the integral sign.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-22T19:19:06.853" CommunityOwnedDate="2010-09-22T19:19:06.853" CommentCount="1" />
  <row Id="5250" PostTypeId="1" CreationDate="2010-09-22T20:01:31.330" Score="3" ViewCount="146" Body="&lt;p&gt;I am trying to find a formula that will enable me to calculate the illumination of the moon down to one thousandth of a percent, given the gregorian year, month, day, and hour is known. I am unsure if this is the place to ask this sort of a question, but can anyone help me on this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2070" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-31T20:40:36.337" LastActivityDate="2010-10-31T20:40:36.337" Title="Illumination calculations?" Tags="&lt;physics&gt;&lt;applications&gt;" CommentCount="7" />
  <row Id="5251" PostTypeId="2" ParentId="5248" CreationDate="2010-09-22T20:07:13.260" Score="5" ViewCount="0" Body="&lt;p&gt;In my &lt;a href=&quot;http://problemasteoremas.files.wordpress.com/2008/06/tayloradvancedcalculus1.jpg?w=351&amp;amp;h=490&quot; rel=&quot;nofollow&quot;&gt;Advanced Calculus book (Angus Taylor)&lt;/a&gt; it is shown that, if $a\gt 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt=\arctan\dfrac{x}{a}\qquad (1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&gt;0$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\int_0^{\infty}\dfrac{\sin xt}{t}dt=\dfrac{\pi}{2}\qquad (2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;follows from $(1)$, observing that the integrand is $G(0)$ for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(a)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt\qquad (3)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is uniformly convergent when $a\ge 0$, and $G(a)$ approaches $G(0)$ as $a$ tends to $0^+$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Answer to Qiaochu: $(1)$ is proved as an application of the following theorem to $F(x)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let $F(x)=\displaystyle\int_c^{\infty}f(t,x)dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be convergent when $a\le x\le b$. Let $\dfrac{\partial f}{\partial x}$ be continuous in $t,x$ for $c\lt t$,$a\lt x\lt b$, and let $\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$ converge uniformly on $[a,b]$. Then $F'(x)=\displaystyle\int_c^{\infty}\dfrac{\partial f}{\partial x}dt$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PS. It was a surprise to me to see the evaluation of $F(x)=\displaystyle\int_0^{\infty}\dfrac{e^{-at}\sin xt}{t}dt$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=\dfrac{\pi}{2}\qquad$ if $x&gt;0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=-\dfrac{\pi}{2}\qquad$ if $x&amp;lt;0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=0\qquad\qquad$ if $x=0$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-09-22T22:11:15.680" LastActivityDate="2010-09-22T22:11:15.680" CommentCount="5" />
  <row Id="5252" PostTypeId="2" ParentId="4862" CreationDate="2010-09-22T20:21:51.227" Score="0" ViewCount="0" Body="&lt;p&gt;Many of the suggestions here are good ones, if perhaps a little advanced. A course on multivarible calculus and a course on ordinary differential equations seem like the logical progression to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an undergrad maths course, this would typically be accompanied by courses on linear algebra and abstract algebra.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastActivityDate="2010-09-22T20:21:51.227" CommunityOwnedDate="2010-09-22T20:21:51.227" />
  <row Id="5254" PostTypeId="2" ParentId="5201" CreationDate="2010-09-22T20:28:58.890" Score="0" ViewCount="0" Body="&lt;p&gt;It seems to me what you're trying to do is formulate a first-order theory. This has non-logical axioms. Your theory uses two one-place predicates, say &quot;F&quot; for the Xs and &quot;G&quot; for the Ys. It has one non-logical axiom, perhaps, which says there are Fs and there are Gs. Then it has another which says nothing is both F and G. Then you have a two-place predicate &quot;R&quot; to represent the relationship, and another non-logical axiom for that, one that uses &quot;F&quot; and &quot;G&quot; as well as &quot;R&quot;. From the point of view of logic, there's still only one domain; it's your non-logical axioms that divide it up. -- It's the predicates and axioms that are important, not the variables &quot;x&quot; and &quot;y&quot;. Shoenfield's text &lt;em&gt;Mathematical Logic&lt;/em&gt; is good for examples of first-order theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: by non-logical axiom I mean an axiom that's not itself already provable in first-order logic.&lt;/p&gt;&#xA;" OwnerUserId="1813" LastActivityDate="2010-09-22T20:28:58.890" />
  <row Id="5255" PostTypeId="1" AcceptedAnswerId="5256" CreationDate="2010-09-22T20:36:34.023" Score="3" ViewCount="85" Body="&lt;p&gt;Simplicial sequences generalize the familiar &quot;linear&quot;, &quot;triangular&quot;, and &quot;tetrahedral&quot; number sequences. (A line segment is a 1-simplex, a triangle is a 2-simplex, a tetrahedron is a 3-simplex, and so on, so I'm calling these simplicial sequences.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the n-th d-simplicial number is the binomial coefficient C(n+d-1,d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These pictures below graphically show the n-th element (modulo 512) of 512 simplicial sequences side by side. The white dots in a given frame have coordinates (d,C(n+d-1,d) mod 512), where d runs from 0 through to 511, and the value of n is displayed at the bottom of the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you are using a &quot;modern browser&quot; (i.e. not IE6,7,8), you can see an animation running through all values of n &lt;a href=&quot;http://www.brainjam.ca/stackoverflow/simplicialNumbers.htm&quot; rel=&quot;nofollow&quot;&gt;by following this link&lt;/a&gt;.  The animation allows you to start and stop, and even save individual frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch the animation, you'll see that binary visual patterns emerge, in ways that are threaded through the image and increase and decrease in strength.  The images below for n=5 and n=256 indicate the extremes .. for n=5 the points look randomly distributed, for n=256 they are strongly patterned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether there are any explanations for the emergence and disappearance of visual binary patterns as we iterate through n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/eur9v.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/WIZ0x.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2010-09-22T21:37:41.203" LastActivityDate="2010-09-22T21:37:41.203" Title="Is there an explanation for the patterns formed by the binomial coefficients C(n+d-1,d) mod 512?" Tags="&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;&lt;experimental-mathematics&gt;" AnswerCount="1" />
  <row Id="5256" PostTypeId="2" ParentId="5255" CreationDate="2010-09-22T20:48:42.823" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Theorem (Kummer):&lt;/strong&gt;  The greatest power of $p$ dividing ${m+n \choose n}$ is the number of carries it takes to add $m$ and $n$ in base $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular the greatest power of $2$ dividing ${n+d-1 \choose d}$ is the number of carries it takes to add $n-1$ and $d$ in base $2$, and so depends entirely on the binary digits of both $n-1$ and $d$.  (When $n = 256$, for example, there are likely to be a lot of carries, which is why the position of the dots is so restricted in that case.  More generally the same happens when $n-1$ has lots of $1$s in its binary representation.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above theorem places a strong restriction on where the white dots can be, which I think will be clearer visually if you plot things in the shape of &lt;strong&gt;Pascal's triangle&lt;/strong&gt; instead of whatever it is you're doing now (for example $\bmod 2$ you will get the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_triangle&quot; rel=&quot;nofollow&quot;&gt;Sierpinski triangle&lt;/a&gt;.  More generally I think you should try plotting the entries of Pascal's triangle $\bmod 512$ where the modulus determines the color of a block.  This won't be an animation but will allow you to consolidate all of the patterns you're seeing into one image which I think will ultimately be clearer).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas_theorem&quot; rel=&quot;nofollow&quot;&gt;Lucas' theorem&lt;/a&gt; for a partial explanation of patterns in the exact remainder $\bmod 512$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-22T20:57:08.937" LastActivityDate="2010-09-22T20:57:08.937" CommentCount="2" />
  <row Id="5257" PostTypeId="2" ParentId="5248" CreationDate="2010-09-22T21:36:29.747" Score="19" ViewCount="0" Body="&lt;p&gt;I believe this can also be solved using double integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible (if I remember correctly) to justify switching the order of integration to give the equality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}(\int_{0}^{\infty} e^{-xy} \sin x \ dy) \ dx = \int_{0}^{\infty}(\int_{0}^{\infty} e^{-xy} \sin x \ dx)\ dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty} e^{-xy} \sin x \ dy = \frac{\sin x}{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads us to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}(\frac{\sin x}{x}) \ dx = \int_{0}^{\infty}(\int_{0}^{\infty} e^{-xy} \sin x \ dx)\ dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the right hand side can be found easily, using integration by parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int e^{-xy} \sin x \ dx = -e^{-xy}{\cos x} + y \int e^{-xy} \cos x \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = -e^{-xy}{\cos x} + y (e^{-xy}\sin x - y \int e^{-xy} \sin x \ dx)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \frac{ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\int_{0}^{\infty} e^{-xy} \sin x \ dx = \frac{1}{1+y^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\int_{0}^{\infty}(\frac{\sin x}{x}) \ dx = \int_{0}^{\infty}\frac{1}{1+y^2}\ dy = \frac{\pi}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-22T23:05:16.203" LastActivityDate="2010-09-22T23:05:16.203" CommentCount="3" />
  <row Id="5258" PostTypeId="1" CreationDate="2010-09-22T21:36:39.433" Score="-1" ViewCount="257" Body="&lt;p&gt;Evaluate these Integrals: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $\int_{C} \ f(z) \ dz$ where $f(z) = \frac{z+2}{z}$ and $C$ is the semicircle $\gamma(t) = 2e^{it}$ for $0 \leq t \leq \pi$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) $f(z) = \pi e^{\pi \bar{z}}$ where $C$ is the boundary if the square with lowest left vertex at the origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you use Residue Theorem?&lt;/p&gt;&#xA;" OwnerUserId="2074" LastActivityDate="2010-12-21T23:44:26.060" Title="Complex Integral" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="5259" PostTypeId="1" AcceptedAnswerId="5287" CreationDate="2010-09-22T22:36:01.287" Score="2" ViewCount="182" Body="&lt;p&gt;I'm trying to approximate $\pi$ using Monte Carlo integration; I am approximating the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\limits_0^1\!\frac{4}{1+x^2}\;\mathrm{d}x=\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is working fine, and so is estimating the error (variance), $\sigma$. However, when I then try to use importance sampling with a Cauchy(0,1) distribution, things start to go wrong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n}\sum\limits_{i=0}^n\frac{f(x_i)}{p(x_i)}=\frac{1}{n}\sum\limits_{i=0}^n\frac{\frac{4}{1+x^2}}{\frac{1}{\pi(1+x^2)}}=\frac{1}{n}\sum\limits_{i=0}^n\frac{4\pi(1+x^2)}{1+x^2}=\frac{1}{n}\sum\limits_{i=0}^n4\pi=4\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously something's wrong, since the mean is computed independently of the random variables I generate. Where is this going wrong? Is the distribution too close to $f$? &lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-23T14:16:00.327" LastActivityDate="2010-09-23T14:16:00.327" Title="Approximating $\pi$ using Monte Carlo integration" Tags="&lt;integral&gt;&lt;approximation&gt;&lt;numerical-methods&gt;&lt;monte-carlo&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5260" PostTypeId="2" ParentId="5258" CreationDate="2010-09-22T22:42:50.010" Score="1" ViewCount="0" Body="&lt;p&gt;For (i) you can write $f(z)=1+\frac{2}{z}$. Integral for the first term can be done directly, for the second you can use, that $\frac{2}{z} dz = \frac{2}{-z} d(-z)$ and, therefore, integral over semicircle is equal to $\frac12$ of the integral over the whole circle, which can be done by residue theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii) I would calculate integrals for 4 edges of the square one by one directly, writing $z=x+iy$. I would not use residue theorem. I would use &quot;integral of the exponent&quot; instead.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-22T22:42:50.010" CommentCount="1" />
  <row Id="5261" PostTypeId="2" ParentId="5191" CreationDate="2010-09-22T22:46:22.590" Score="0" ViewCount="0" Body="&lt;p&gt;Here is one idea, although I do not find it very satisfying.  An object of $C$ is a finite set $X$ equipped with $\frac{|X|!}{2}$ (or $1$ if $|X| = 1$) total orders, all of which are &lt;strong&gt;even&lt;/strong&gt; with respect to each other (in other words, basically a coset of $A_n$ in $S_n$).  A morphism between two objects in $X$ is a map of sets preserving these orders (in other words, take one of the orderings on $X$ and apply a function $f : X \to Y$ to its elements.  The result, after throwing out repeats, must be compatible with an ordering on $Y$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more or less a discretization of Omar's answer.  Again, I would like to do better than this, or at least see the data described above packaged in a more satisfying way.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-22T22:46:22.590" />
  <row Id="5262" PostTypeId="1" AcceptedAnswerId="5264" CreationDate="2010-09-22T23:18:55.690" Score="2" ViewCount="116" Body="&lt;p&gt;I have 2 numbers (a &amp;amp; b) but I need a formula (or a how to) to find which 2 numbers (c &amp;amp; d) will add together to give a and times together to give b. &#xA;So&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c + d = a&#xA;c * d = b &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2075" LastActivityDate="2010-09-23T02:55:40.663" Title="Formula(/How) to find 2 numbers that add together to give one number and times to give another" Tags="&lt;recreational-mathematics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5263" PostTypeId="2" ParentId="5233" CreationDate="2010-09-22T23:21:58.090" Score="1" ViewCount="0" Body="&lt;p&gt;The real vector space of all fibonacci sequences (the first two values are arbitrary) is quite instructive. Or the subspace of all smooth functions satisfying the differential equation $f'=f$. It is quite illuminating that elementary linear algebra has a interplay with analysis.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-22T23:21:58.090" />
  <row Id="5264" PostTypeId="2" ParentId="5262" CreationDate="2010-09-22T23:31:58.390" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the quadratic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-c)(x-d) = x^2 -(c+d)x + cd = x^2 - ax +b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the roots of $$x^2 -ax + b = 0$$ are your numbers $c$ and $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ a \pm \sqrt{a^2 -4b}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this page out for more information: &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation&quot; rel=&quot;nofollow&quot;&gt;Quadratic Equation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-22T23:31:58.390" CommentCount="1" />
  <row Id="5265" PostTypeId="2" ParentId="5262" CreationDate="2010-09-22T23:32:58.130" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad (X - c)\: (X - d) \; =\; X^2 - a\ X + b $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-22T23:32:58.130" />
  <row Id="5266" PostTypeId="1" AcceptedAnswerId="5284" CreationDate="2010-09-22T23:36:08.967" Score="1" ViewCount="116" Body="&lt;p&gt;In my statistical signal modelling class we were given the following task:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, let's assume E(y) = E(x) = 0. Show that if the real variables x and y are normally distributed, then the equation E(y|x) = corr(y,x)*x/var(x) holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were also given a hint: write the assisting variable s = y - corr(x,y)*x/var(x) and show that it's uncorrelated with x, meaning that E(sx) = E(s)E(x) = 0. Because s and x together are normally distributed, it follows that s and x are uncorrelated, ie. the conditional probability of s doesn't depend on x. Now you can easily write y using s, and calculate E(y|x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been hacking at this for 3 hours without much progress so any help is appreciated&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2010-09-23T06:05:59.897" Title="Proving the conditional probability E(y|x) = corr(y,x)*x/(stdev(x)^2)" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5267" PostTypeId="1" AcceptedAnswerId="5268" CreationDate="2010-09-22T23:46:35.733" Score="2" ViewCount="232" Body="&lt;p&gt;Hi there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a programming exercise I need to do, but I just can't figure out the math part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to check if 3 of 6 lines intersect in the same point. I am given the equation ax+by=c, and I input random numbers in a,b,c. How would I go about solving this mathematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="2078" LastEditorDisplayName="" LastEditDate="2010-09-22T23:58:53.790" LastActivityDate="2010-09-23T01:48:20.000" Title="How to find where 3 lines intersect." Tags="&lt;geometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5268" PostTypeId="2" ParentId="5267" CreationDate="2010-09-22T23:52:29.170" Score="3" ViewCount="0" Body="&lt;p&gt;You're trying to check for the condition of &lt;a href=&quot;http://mathworld.wolfram.com/Concurrent.html&quot; rel=&quot;nofollow&quot;&gt;concurrency&lt;/a&gt; of three lines; the last determinant formula involving the coefficients of your three lines in that link may be of help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've shown that your three lines are concurrent, take any two of the three and solve as simultaneous equations (two equations in two unknowns) to get the coordinates of the intersection point.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-22T23:52:29.170" CommentCount="3" />
  <row Id="5269" PostTypeId="2" ParentId="5267" CreationDate="2010-09-23T00:46:18.733" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;$\;$ &lt;strong&gt;Divide and conquer&lt;/strong&gt;. $\:$ Split the lines into two sets of three lines. Compute the intersection points of each set (at most three). If either intersection set has cardinality one then you're done. Else check if any intersection point from one set lies on any line in the other set. Total operation count: 6 line intersections and 6 line evaluations.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if the lines need not be distinct then you need to make slight modifications to the above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-23T01:12:26.247" LastActivityDate="2010-09-23T01:12:26.247" CommentCount="3" />
  <row Id="5270" PostTypeId="2" ParentId="5233" CreationDate="2010-09-23T00:58:21.423" Score="2" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The positive real numbers, where 1 is the &quot;zero vector,&quot; &quot;scalar multiplication&quot; is really numerical exponentiation, and &quot;addition&quot; is really numerical multiplication. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Simplicial complexes, as in algebraic topology, are another good example, but perhaps this is even &lt;em&gt;too&lt;/em&gt; weird. Still, it might be fun to throw out the idea that mathematicians like to add triangles to each other to get quadrilaterals and negate them to reverse orientation, but they'll probably have to take it on faith that this is actually useful. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-23T00:58:21.423" CommentCount="4" />
  <row Id="5271" PostTypeId="2" ParentId="5267" CreationDate="2010-09-23T01:48:20.000" Score="1" ViewCount="0" Body="&lt;p&gt;The first step to write a program is to break down a problem into small parts.  I would do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Write a function to decide whether given three lines intersect at one point.  &lt;a href=&quot;http://math.stackexchange.com/questions/5267/how-to-find-where-3-lines-intersect/5268#5268&quot;&gt;J. M.’s answer&lt;/a&gt; gives some material to do this.&lt;/li&gt;&#xA;&lt;li&gt;Iterate over all combinations of three lines out of six.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the program does not work, try to test the two parts separately.  Smaller, simpler programs are easier to verify.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-23T01:48:20.000" CommentCount="2" />
  <row Id="5272" PostTypeId="2" ParentId="5262" CreationDate="2010-09-23T02:42:14.620" Score="2" ViewCount="0" Body="&lt;p&gt;Since I'm not so smart to imagine that the solution has anything to do with Vieta formulae or roots of polynomials,  ;-)  I'd try the following. First, let me call your unknowns $c,d$ with letters that look like unkowns (I've tried to do the problem with $c,d$ and at a certain point I forgot which among $a,b,c,d$ were the unkowns and which the data). So let $x = c$ and $y =d$. Your system becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x + y &amp;amp;= a \\&#xA;xy    &amp;amp;= b   \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, do what you would do in those cases. For instance, solve the first equation for $x$ in terms of $y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = a -y  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substitute this expression for $x$ into the second equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(a-y)y = b \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, surprise!, you've got a polynomial equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y^2 -ay + b = 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that looks like the ones I've been told. (So this Vieta must have been a really smart guy, but you've rediscovered his trick.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, once you've got the solutions $y$ of this second degree equation, you should use $x = a-y$ to find $x$... (Just to find out that if $y$ is one of the roots of the last polynomial equation, $x$ is the other one.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-23T02:55:40.663" LastActivityDate="2010-09-23T02:55:40.663" CommentCount="7" />
  <row Id="5273" PostTypeId="2" ParentId="227" CreationDate="2010-09-23T03:09:14.133" Score="2" ViewCount="0" Body="&lt;p&gt;Some advice from the eminent macroeconomist Thomas Sargent&#xA;&lt;a href=&quot;http://homepages.nyu.edu/~ts43/math_courses.html&quot; rel=&quot;nofollow&quot;&gt;http://homepages.nyu.edu/~ts43/math_courses.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-23T03:09:14.133" CommunityOwnedDate="2010-09-23T03:09:14.133" CommentCount="1" />
  <row Id="5274" PostTypeId="1" AcceptedAnswerId="5276" CreationDate="2010-09-23T03:19:56.110" Score="2" ViewCount="277" Body="&lt;p&gt;While working on a proof showing that all functions limited to the domain of real numbers can be expressed as a sum of their odd and even components, I stumbled into a troublesome roadblock; namely, I had no clue how one divides the function into these even and odd parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking up a solution for the proof, I found these general formulas for the even and odd parts of a function f(n):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Sl6Hk.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I understand that in an even function f(n) = f(-n) and that in an odd function f(-n) = -f(n), I still don't get how these general formulas for the even and odd parts were obtained. Can someone guide me through the logic?&lt;/p&gt;&#xA;" OwnerUserId="1956" LastActivityDate="2010-09-23T04:06:16.363" Title="How do I divide a function into even and odd sections? " Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5275" PostTypeId="2" ParentId="5274" CreationDate="2010-09-23T03:31:05.070" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose $f(x)=g(x)+h(x)$ with $g$ even and $h$ odd.  Then $f(-x)=g(-x)+h(-x)=g(x)-h(x)$.  Think of $f(x)=g(x)+h(x)$ and $f(-x)=g(x)-h(x)$ as a system of two equations in two unknowns, $g(x)$ and $h(x)$, to solve for $g$ and $h$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-23T03:31:05.070" CommentCount="1" />
  <row Id="5276" PostTypeId="2" ParentId="5274" CreationDate="2010-09-23T03:32:02.367" Score="9" ViewCount="0" Body="&lt;p&gt;Suppose you could write a function $f(x)$ as the sum of an even and an odd function; call them $E(x)$ and $O(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, you would have&#xA;\[f(x) = E(x)+O(x)\]&#xA;and you would also have&#xA;\[f(-x) = E(-x) + O(-x) = E(x) - O(x)\]&#xA;with the latter equation because we are assuming $E$ is even and $O$ is odd, so $E(x)=E(-x)$ and $O(-x) = -O(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding both equations you get $f(x)+f(-x) = 2E(x)$. Subtracting the second equation from the first gives you $f(x)-f(-x)=2O(x)$. Now solve for $E(x)$ and $O(x)$, and you get the formulas you see in the solution. Then you check that the answer does indeed work (that is, you check that the formulas you found do give you an even and an odd function in all cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: pretend you already know the answer, and try to deduce conditions that the answer must satisfy (these will be &lt;em&gt;necessary&lt;/em&gt; conditions); if things go well, you'll get enough information about what they &lt;em&gt;must&lt;/em&gt; be like to figure out &lt;em&gt;what&lt;/em&gt; they are.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T04:06:16.363" LastActivityDate="2010-09-23T04:06:16.363" CommentCount="1" />
  <row Id="5277" PostTypeId="1" AcceptedAnswerId="5279" CreationDate="2010-09-23T05:10:47.343" Score="4" ViewCount="384" Body="&lt;p&gt;How do I mathematically determine if a number is prime?  I'm writing a program that the user inputs any integer and from that the program determines whether the number is prime, but how do I go about that?&lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-27T19:26:26.430" LastActivityDate="2010-09-27T19:26:26.430" Title="Determine whether a number is prime" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="1" />
  <row Id="5278" PostTypeId="1" AcceptedAnswerId="5285" CreationDate="2010-09-23T05:16:47.097" Score="6" ViewCount="223" Body="&lt;p&gt;This question is taken from Dummit and Foote(14.9 #6). Any help will be appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Show that if $t$ is transcendental over $\mathbb{Q}$, then $\mathbb{Q}(t,\sqrt{t^3-t})$ is not a purely transcendental extension of $\mathbb{Q}$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've got so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abbreviate $\sqrt{t^3-t}$ as $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've shown that the transcendence degree is 1, so the problem boils down to showing that $\mathbb{Q}(t,u) \supset \mathbb{Q}(f(t,u)/g(t,u))$ strictly, for all polynomials $f,g$ in two variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for contradiction that $\mathbb{Q}(t,u) = \mathbb{Q}(f(t,u)/g(t,u))$. Look at this field as $\mathbb{Q}(t)[x]/(x^2-(t^3-t))$, with $\bar{x}=u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then since $t$ and $u$ are generated by $f/g$, we have that for some polynomials $a,b,c,d$ in 1 variable, $a(\frac{f(t,x)}{g(t,x)})/b(\frac{f(t,x)}{g(t,x)})-t \in (x^2-(t^3-t))$, and $c(\frac{f(t,x)}{g(t,x)})/d(\frac{f(t,x)}{g(t,x)}) - x \in (x^2-(t^3-t))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then tried playing with degrees, but I haven't found a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2080" LastActivityDate="2010-09-23T18:00:35.540" Title="Why is $\mathbb{Q}(t,\sqrt{t^3-t})$ not a purely transcendental extension of $\mathbb{Q}$?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="5279" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:19:22.370" Score="5" ViewCount="0" Body="&lt;p&gt;Have the program find the remainder when dividing the input (say n) by 2, 3, 4, ..., $\sqrt n$  (or the following integer if $\sqrt n$ is not an integer.) If this value ever leaves a remainder of zero then your number is composite and you can stop checking divisors. If the remainder is non-zero for all of these values then your number is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more efficient ways to check primeness but this is probably the easiest way to program it and I imagine will suffice for your purposes. Unless you are checking very very large numbers? &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-09-23T05:36:26.947" LastActivityDate="2010-09-23T05:36:26.947" CommentCount="2" />
  <row Id="5280" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:23:11.573" Score="6" ViewCount="0" Body="&lt;p&gt;There are many different algorithms for primality testing. See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Primality_test&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; for an introduction and see Henri Cohen's book &quot;A course in computational algebraic number theory&quot; for further details. See also Caldwell's &lt;a href=&quot;http://primes.utm.edu/prove/index.html&quot; rel=&quot;nofollow&quot;&gt;Prime Pages&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-23T05:36:27.897" LastActivityDate="2010-09-23T05:36:27.897" />
  <row Id="5281" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:32:48.977" Score="14" ViewCount="0" Body="&lt;p&gt;For very small numbers (less than a million), trial division is the best way: divide by 2, 3, 5, and so on until the square root of the number.  If you find a factor, the number is composite; otherwise, the number is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers there are better methods, but choosing which one depends on how much work you're willing to put into the program.  It is now known that there are no BPSW-pseudoprimes below $2^{64}$, so if you can write that test (see &lt;a href=&quot;http://www.trnicely.net/misc/bpsw.html&quot; rel=&quot;nofollow&quot;&gt;here for details&lt;/a&gt;) then you have a very quick test for primality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only need to test up to $2^{32}$, you can simply check if the number is a 2-strong pseudoprime.  If so, test if it's one of 2314 exceptions (this can be done in 12 or 13 steps with a binary search); if the test fails or it's an exception, the number is composite, otherwise prime.  (You can go higher than $2^{32}$ if you're willing to build an appropriate table of exceptions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers, the work is usually split into two parts: determining with high probability (say, 99.99999999%) that the number is prime, then actually proving that it is.  What type of proof depends on the form and size of the number.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-23T05:32:48.977" CommentCount="2" />
  <row Id="5282" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T05:47:36.980" Score="1" ViewCount="0" Body="&lt;p&gt;Algorithm posted by &lt;a href=&quot;http://math.stackexchange.com/questions/5277/determine-whether-a-number-is-prime/5279#5279&quot;&gt;jericson&lt;/a&gt; is the best for basic purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO, for &lt;em&gt;programming competitions&lt;/em&gt; and for &lt;em&gt;practical purposes&lt;/em&gt; &lt;strong&gt;randomized algorithms&lt;/strong&gt; are best. Rabin-Miller is my favorite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;http://www.topcoder.com/tc?module=Static&amp;amp;d1=tutorials&amp;amp;d2=primalityTesting&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Rabin-Miller primality testing algorithm code @ TopCoder&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As primes are in P, there is deterministic, polynomial time algorithm called &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/AKS_primality_test&quot; rel=&quot;nofollow&quot;&gt;AKS primality test.&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-09-23T12:52:46.370" LastActivityDate="2010-09-23T12:52:46.370" CommentCount="3" />
  <row Id="5283" PostTypeId="2" ParentId="5278" CreationDate="2010-09-23T06:01:30.980" Score="6" ViewCount="0" Body="&lt;p&gt;By definition a purely transcendental extension of $\mathbf{Q}$ is the function field of a rational variety over $\mathbf{Q}$, namely the fraction field of a polynomial ring $\mathbf{Q}[x_1,\ldots, x_n]$ for some $n$. Call your field $K$. Obviously the transcendence degree of $K$ over $\mathbf{Q}$ is 1, so it's enough to show that $K$ is not isomorphic to a rational function field in 1 variable, i.e. to $\mathbf{Q}(t)$. Now by inspection $K$ is the function field of the elliptic curve $E:y^2 = x^3-x$ over $\mathbf{Q}$. It's a standard fact, shown in various books concerned with algebraic curves, that $K\cong \mathbf{Q}(t)$ if and only if $E$ is birational to the projective line $\mathbf{P}^1_{\mathbf{Q}}$. This is not the case, because (for example) the genus of $E$ is 1 (say by the degree-genus formula for plane curves), the genus of $\mathbf{P}^1$ is zero, and the genus is a birational invariant of a curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to solve this problem with &quot;elementary&quot; (non-geometric) methods off the top of my head; hopefully someone else can explain how to do this.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2010-09-23T06:01:30.980" />
  <row Id="5284" PostTypeId="2" ParentId="5266" CreationDate="2010-09-23T06:05:59.897" Score="2" ViewCount="0" Body="&lt;p&gt;Note that $corr(x,y) = E(xy)-E(x)E(y) = E(xy)$ and $var(x) = E(x^2)-E(x)^2=E(x^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(sx) = E(yx) - \frac{corr(y,x)}{var(x)}E(x^2) = E(yx) -\frac{E(xy)}{E(x^2)}E(x^2) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then also $y=s+\frac{corr(x,y)x}{var(x)}$ so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(y|x) = E(s+\frac{corr(x,y)x}{var(x)}|x) = E(s|x) + E(\frac{corr(x,y)x}{var(x)}|x) = E(s) + \frac{corr(x,y)x}{var(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $E(s)=0$ giving the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to justify all these formulas and steps.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T06:05:59.897" CommentCount="2" />
  <row Id="5285" PostTypeId="2" ParentId="5278" CreationDate="2010-09-23T06:45:32.530" Score="12" ViewCount="0" Body="&lt;p&gt;For convenience change the base field to $k=\mathbb{C}$&#xA;(if the extension were purely transcendental before&#xA;then it would be purely transcendental after).&#xA;As the extension has transcendance degree 1, then&#xA;if the extension were purely transcendental it would equal $k(x)$&#xA;for some $x$. Then there are nonconstant rational functions&#xA;$f(x)$ and $g(x)$ such that&#xA;$$g(x)^2=f(x)^3-f(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see we can write $f(x)=u(x)/w(x)^2$&#xA;and $g(x)=v(x)/w(x)^3$ in lowest terms&#xA;where $u$, $v$ and $w$ are polynomials.&#xA;We find that&#xA;$$\phi(x)=\frac{g'(x)}{3f(x)^2-1}=\frac{f'(x)}{2g(x)}.$$&#xA;If fact $\phi(x)$ is a polynomial. Otherwise the denominator&#xA;of $\phi$ has a factor $x-a$. Expressing $\phi$ in terms of $u$, $v$&#xA;and $w$ (details?) we see that this implies that&#xA;$$2g(a)=3f(a)^2-1=0$$&#xA;as well as&#xA;$$g(a)^2=f(a)^3-f(a).$$&#xA;This is impossible. As $f$ and $g$ are nonconstant, $\phi$ is a nonzero&#xA;polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now replace $f(x)$ and $g(x)$ by $f(1/x)$ and $g(1/x)$. Then $\phi(x)$&#xA;is replaced by $-x^{-2}\phi(1/x)$ so that is also a polynomial, which&#xA;it isn't. So we get the required contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; The above was composed in a rush, and I cut a few corners,&#xA;so some extra details. We have $f'(x)=\textrm{polynomial}/w(x)^3$ and so&#xA;$$\phi(x)=\frac{f'(x)}{g(x)}=\frac{\textrm{polynomial}}{v(x)}.$$&#xA;Similarly&#xA;$$\phi(x)=\frac{\textrm{polynomial}}{3u(x)^2-w(x)^4}.$$&#xA;If $x-a$ divides the denominator of $\phi(x)$ then $v(a)=0=u(a)^2-w(a)^4$.&#xA;We can't have $w(a)=0$ as then $x-a$ would divide $u(x)$ and $w(x)$&#xA;contrary to $u(x)/v(x)^2$ being in lowest terms. Hence $f(a)$ and $g(a)$&#xA;make sense and we get $g(a)=3f(a)^2-1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really a geometric argument. On the elliptic curve&#xA;$$E:\qquad\qquad z^2=t^3-t$$&#xA;the &quot;invariant differential&quot;&#xA;$$\omega=\frac{dt}{2z}=\frac{dz}{3t^2-1}$$&#xA;would have no poles on the (projective curve) $E$. But were $E$ rational&#xA;then $\omega=\phi(x)dx$ has no poles on the projective line; but every&#xA;nonzero differential on the projective line has a pole. The above is just&#xA;a naive version of this argument, which works for all elliptic curves.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-23T18:00:35.540" LastActivityDate="2010-09-23T18:00:35.540" CommentCount="2" />
  <row Id="5286" PostTypeId="1" CreationDate="2010-09-23T07:24:51.937" Score="5" ViewCount="240" Body="&lt;p&gt;I'm developing some instructional material for a Calculus 1 class and I wanted to know from experience for yourself, tutoring others, and/or helping people on this site where is the most difficulty in Calculus? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had any good methods of helping people that would be very helpful. &lt;/p&gt;&#xA;" OwnerUserId="2083" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T14:38:11.700" LastActivityDate="2010-09-23T14:44:28.830" CommunityOwnedDate="2010-09-23T11:15:58.493" Title="What concepts were most difficult for you to understand in Calculus?" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;big-list&gt;&lt;teaching&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5287" PostTypeId="2" ParentId="5259" CreationDate="2010-09-23T07:33:27.643" Score="2" ViewCount="0" Body="&lt;p&gt;This is quite a common error when doing Monte Carlo integration. The support of the random variable you choose must be equal to the range of integration. Though the Cauchy distribution has support on $\mathbb{R}$ we can adapt it slightly to make it work here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that: $\int_0^1 \frac{1}{\pi(1+x^2)} dx = \frac{1}{\pi}\tan^{-1}(1)=\frac{1}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $g(x) = \frac{4}{\pi(1+x^2)}$ for $x\in(0,1)$ is a probability density with support on $(0,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;using this we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{n}\sum_{i=0}^n \frac{f(x_i)}{g(x_i)} = \pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a problem! Since $g(x) = \frac{1}{\pi}f(x)$ you have found exactly the right probability distribution to use to evaluate $\int_0^1f(x)dx$! The error is zero for any sample, regardless of the size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, I needed to know how to integrate $\int_0^1\frac{1}{\pi(1+x^2)}$ in the first place to form the probability distribution. Thus for arbitrary functions it is impossible to get this situation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note the Monte Carlo is not a very good way of approximating integrals in general. Far better deterministic methods are quadrature rules. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T07:33:27.643" CommentCount="6" />
  <row Id="5288" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T07:35:51.060" Score="0" ViewCount="0" Body="&lt;p&gt;The area formula, namely, how could&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{\infty} y_i \delta x_i=\int y dx$$&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-23T07:35:51.060" CommunityOwnedDate="2010-09-23T11:15:58.493" />
  <row Id="5289" PostTypeId="1" CreationDate="2010-09-23T07:48:12.340" Score="1" ViewCount="99" Body="&lt;p&gt;This is a similar question with &lt;a href=&quot;http://math.stackexchange.com/questions/5143/find-the-range-of-x-given-y-min-leq-yx-leq-y-max-where-yx-can&quot;&gt;this one&lt;/a&gt;. But with a bit of twist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{1min} \leq y_{1}(x_1,x_2) \leq y_{1max}$$&#xA;$$y_{2min} \leq y_{2}(x_1,x_2) \leq y_{2max}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_i(x_1,x_2)=\Biggl[\frac{(x_{1}^2x_{2}+b_{i}x_{1}x_{2}+c_{i})^{f_i}}{(x_{1}(2+x_{2})+e_{i})^{g_i}}\Biggr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_1=g_1=1, f_2=5, g_2=2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_{i},c_{i},e_{i},x_1,x_2\in\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: How to find the range of $x_{1},x_{2}$ that satisfy the above inequalities?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;range&lt;/em&gt; of $x_{1},x_{2}$ is defined as all the $x_1,x_2$ pairs that satisfy the above inequalities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Maybe I'll relax the question a bit: let's assume that $b_{1}=b_{2}, c_{1}=c_{2}, e_{1}=e_{2}$.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-24T06:47:51.483" LastActivityDate="2010-09-24T06:47:51.483" Title="Find Range of $x_1,x_2$, where $y_{min} \leq y(x_1, x_2) \leq y_{max}$" Tags="&lt;optimization&gt;" CommentCount="7" />
  <row Id="5290" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T07:53:51.670" Score="8" ViewCount="0" Body="&lt;p&gt;The hardest thing for me was to understand what is meant when someone writes $\mathrm d x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still don't know...&lt;/p&gt;&#xA;" LastActivityDate="2010-09-23T07:53:51.670" CommunityOwnedDate="2010-09-23T11:15:58.493" CommentCount="2" />
  <row Id="5291" PostTypeId="1" AcceptedAnswerId="5292" CreationDate="2010-09-23T08:32:35.283" Score="4" ViewCount="177" Body="&lt;p&gt;Hi &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble answering the following question:&lt;br&gt;&#xA;&lt;br&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By integrating the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = \frac{R + z}{z(R - z)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;round a suitable contour, prove that, for $0 \leq r &amp;lt; R$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\theta) + r^2}\mathrm{d}\theta = 1   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I &lt;em&gt;think&lt;/em&gt; they want me to do is to use the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_C f(z) \mathrm{d}z  =  \int_{z_1}^{z_2} f(z) \mathrm{d}z = \int_a^bf(z(t))\cdot z'(t)\mathrm{d}t  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where C is a contour from $z_1$ to $z_2$ and $z(a) = z_1$ and $z(b) = z_2$. Then I should find a contour $z(t)$ in terms of $r$ and $t$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z(t))\cdot z'(t) = \frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then with this contour I could find values for $z_1$ and $z_2$ meaning I could evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{z_1}^{z_2} f(z) \mathrm{d}z  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which should equal 2$\pi$ or something. However so far I have not been able to find a suitable z(t). The closest I have come is $z(t) = re^{i\theta}$ which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z(t))\cdot z'(t) = \frac{R^2\cdot i - 2Rr\sin(t) + r - r^2\cdot i}{R^2 - 2Rr\cos(t) + r^2}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this entire approach could be wrong. I've got to the stage when you start to question your whole approach and feel like resorting to violence...According to Einstein, only insane people do the same thing over and over again and expect different results......Oh wait that's me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First of all a big thank you to everyone who has helped me here - your hints are like a life raft to a drowning man! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To Mariano: I have not done anything on residues at this stage in the course - so the question &lt;em&gt;should&lt;/em&gt; be answerable without using them.  That technique does look very powerful though and I hope I will get a good understanding of it in the future...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To Hans Lundmark: Your link helped me to see the bigger picture put this equation in context (wow its more that a figment of my prof's imagination!)...Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To J. M: Your note helped a lot to remove the log jam in my head but there are still some issues that remain with my answer. I would very much appreciate your comments on my 2nd attempt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 1)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $f(z)$ can be decomposed using partial fractions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = \frac{R + z}{z(R - z)}  = \frac{1}{z} + \frac{2}{R - z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then integrated with respect to $z$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int f(z)dz = log(z) - 2log(R - z) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information we can evaluate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_C f(z)dz $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where C is the positively oriented circle $z = re^{i\theta}$. Because log(z) is not defined at its branch cut we cannot integrate around the whole circle but must split the circle into halves $C_1$ and $C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_1$ denote the right half of the circle where $ -\frac{\pi}{2} &amp;lt;= \theta &amp;lt;= \frac{\pi}{2}$ and consider the branch: $log(z) = ln(r) + i\theta$ where $-\pi &amp;lt;= \theta &amp;lt;= \pi$. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_1} f(z) = [log(z) - 2log(R - z)]_{-ri}^{ri} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_1} f(z)dz = \pi{i} - 2log(R - ri) + 2log(R + ri) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C_2$ denote the left half of the circle where $\frac{\pi}{2} &amp;lt;= \theta &amp;lt;= \frac{3\pi}{2}$ and consider the branch: $log(z) = ln(r) + i\theta$ where $0 &amp;lt;= \theta &amp;lt;= 2\pi$. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_2} f(z) = [log(z) - 2log(R - z)]_{ri}^{-ri} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C_2} f(z)dz = \pi{i} - 2log(R + ri) + 2log(R - ri) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C} f(z)dz = \int_{C_1} f(z)dz + \int_{C_2} f(z)dz  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{C} f(z)dz = 2\pi{i}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the part I am currently having difficulty with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_C f(z) \mathrm{d}z  =  \int_a^bf(z(t))\cdot f'(t) dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;imply that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im\int_C f(z) \mathrm{d}z  =  \int_a^b\Im[{f(z(t))}\cdot f'(t) dt] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because if it did I could say that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Im(2\pi{i}) = \int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\theta) + r^2}\mathrm{d}\theta   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the proof would be complete...But I dont know if I can do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can some one help please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to J.M step 3 should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the integral operator is linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int z(t) \mathrm{d}t=\int \Re z(t) \mathrm{d}t+i\int \Im z(t) \mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therfore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_C f(z) \mathrm{d}z  =  \int_a^b\Re{f(z(t))}\cdot f'(t) dt + i\int_a^b\Im{f(z(t))}\cdot f'(t) dt $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{0}^{2\pi}\Re[{f(z(t))}\cdot f'(t) dt] = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_C f(z) \mathrm{d}z = 2{\pi}i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Im[f(z(t))\cdot z'(t)] = \frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_C f(z) \mathrm{d}z  =  i\int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2{\pi}i  =  i\int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2{\pi}  =  \int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2{\pi}  =  \int_0^{2\pi}\frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2}dt  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the proof is complete. Awesome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps some other dude out there who is trying to understand this stuff...&lt;/p&gt;&#xA;" OwnerUserId="2081" LastActivityDate="2010-09-25T17:40:36.797" Title="How to find a suitable contour to integrate round?" Tags="&lt;complex-analysis&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="5292" PostTypeId="2" ParentId="5291" CreationDate="2010-09-23T08:58:53.410" Score="3" ViewCount="0" Body="&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im\left(\frac{R+r\exp(i\theta)}{r\exp(i\theta)(R-r\exp(i\theta))}(i r\exp(i\theta))\right)=\frac{R^2-r^2}{R^2+r^2-2Rr\cos\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for your second attempt, remember that the integral is a &lt;em&gt;linear&lt;/em&gt; operator, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int z(t) \mathrm{d}t=\int \Re z(t) \mathrm{d}t+i\int \Im z(t) \mathrm{d}t$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-24T07:37:25.970" LastActivityDate="2010-09-24T07:37:25.970" CommentCount="3" />
  <row Id="5293" PostTypeId="1" AcceptedAnswerId="5299" CreationDate="2010-09-23T09:09:58.437" Score="1" ViewCount="172" Body="&lt;p&gt;An orthogonal matrix is a matrix $A$ over the reals such that $A^t=A^{-1}$ (its transpose is its inverse). The Frobenius norm over $n\times n$ real matrices is given by $\|A\| = \sqrt{trace(A^tA)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come across the following claim: The distance (induced by the Frobenius norm) between any two (non equal) orthogonal matrices is $\sqrt{n}$. I can't find a proof for this claim, but no refutation either (of course, if the difference between two orthogonal matrices is itself an orthogonal matrix the claim is clear, but I don't know if that's true either).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-23T12:37:42.073" Title="The distance between orthogonal matrices induced by the Frobenius norm" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5295" PostTypeId="2" ParentId="5293" CreationDate="2010-09-23T09:23:55.657" Score="5" ViewCount="0" Body="&lt;p&gt;I don't think this is true. Let $A=I_2$ and $B$ be 2 by 2 matrix with $b_{11}=b_{22}=0$ and $b_{12}=b_{21}=1$. Then both these matrices are orthogonal but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $||A-B||_F = 2\neq \sqrt{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even easier counterexample: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A=1$ and $B=-1$ the 1 by 1 orthogonal matrices. Then $A-B=2$ so $||A-B||_F=2\neq 1$. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-23T09:29:08.910" LastActivityDate="2010-09-23T09:29:08.910" />
  <row Id="5297" PostTypeId="1" CreationDate="2010-09-23T09:55:21.410" Score="9" ViewCount="440" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/2865/is-a-direct-limit-of-topological-groups-always-a-topological-group&quot;&gt;Direct limit&lt;/a&gt;, Martin rightly pointed out that my naive construction (now deleted) of the colimit (direct limit) of topological abelian groups was wrong. He shows how to do it properly (at least the coproduct) &lt;a href=&quot;http://math.stackexchange.com/questions/5095/what-are-the-product-and-coproduct-in-the-category-of-topological-groups&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, I've been lurking some of the literature about the subject and this problem of the colimit of topological groups (about which I had previously no idea) seems at the same time classical and topical. For instance, &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.kjm/1250518067&quot; rel=&quot;nofollow&quot;&gt;this 1998's paper&lt;/a&gt; points out that the Encyclopedic Dictionary of Mathematics, second edition, MIT (1987), article 210, has made the same mistake that I did, stating that the direct limit of topological groups, with the &lt;em&gt;inductive limit of topologies&lt;/em&gt; (my naive attempt) has always a continuous multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors of this paper show a counter-example (example 1.2, page 553) and here is my question: I must be absolutely dumb, but don't understand it. Could anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who don't have access to the paper, here is the example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G_n = \mathbb{Q} \times \mathbb{R}^n$ with the usual topology. Imbed $G_n \hookrightarrow G_{n+1}$ as $x \mapsto (x,0)$. Then, as a plain abelian group, $G = \varinjlim_{n} G_n= \mathbb{Q} \times \prod'\mathbb{R}$, where $\prod'\mathbb{R}$ denotes the &lt;em&gt;weak&lt;/em&gt; or &lt;em&gt;restricted&lt;/em&gt; product, which is the way guys in this area call the &lt;em&gt;direct sum&lt;/em&gt;; that is, elements of $\prod'\mathbb{R}$ are infinite tuples $( x_1, \dots , x_n, \dots )$, in which all of its components $x_n \in \mathbb{R}$ are zero, except a finite number of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;inductive limit topology&lt;/em&gt; is the finest one that makes all the inclusions $G_n \hookrightarrow G$ continuous. That is, $U \subset G$ is open if and only if $U\cap G_n$ is open for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see how they show that, with this inductive topology, the &quot;product&quot; (in fact, addition) $\mu : G \times G \longrightarrow G$ is &lt;strong&gt;not&lt;/strong&gt; continuous ($\mu$ is the operation induced as an honest colimit of groups -no topologies- by the operations $\mu_n : G_n \times G_n \longrightarrow G_n$, which I assume are the usual additions of those linear spaces). In plain English:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(x_0, x_1, \dots , x_n , \dots ) + (y_0, y_1, \dots , y_n , \dots ) = (x_0+y_0, x_1 + y_1 , \dots , x_n + y_n , \dots ) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this situation it's enough to produce an open neighbourhood $U \subset G$ of the neutral element $e \in G$ such that $V^2$ is not in $U$, for any open neighbourhood $V$ of $e$ -where, I assume, $V^2$ means $V + V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here is the guy that is supposed to ruin (and sure it does) my naive attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U = \left\{ &#xA;x = (x_0, x_1, \dots , x_n , \dots ) \quad \vert \quad \vert x_j\vert &amp;lt; \vert \cos (jx_0) \vert \ , 1 \leq j&#xA;\right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This guy is an open set of $G$ because $x_0$ being a rational number guarantees $\cos (jx_0) \neq 0$ for all $j$. Assume there is an open neighbourhood $V$ of $e$ such that $V^2 \subset U$. Then, $V \cap G_j$ contains an open interval $(-\varepsilon_j , \varepsilon_j)$ in $\mathbb{R}$ with $\epsilon_j &gt; 0$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-\varepsilon_0 , \varepsilon_0) \times (-\varepsilon_j , \varepsilon_j) \subset &#xA;\left\{&#xA;(x_0 , x_j) \in \mathbb{Q} \times \mathbb{R} \quad \vert \quad \vert x_j\vert &amp;lt; \vert \cos (jx_0) \vert&#xA;\right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here come the two final sentences of the example I don't understand: &lt;strong&gt;This is impossible if $2j\varepsilon_0 &gt; \pi$. A contradiction.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints or remarks (even humiliating ones) will be welcome.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-24T05:29:07.693" LastActivityDate="2010-10-03T05:53:24.397" Title="Colimit of topological groups (again)" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="3" />
  <row Id="5298" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T10:16:45.010" Score="1" ViewCount="0" Body="&lt;p&gt;I think the entire concept behind integration is hard to grasp for students who are not familiar with analysis. They tend to think of it only as the &quot;inverse operation of derivatives&quot;, which is quite restrisctive, in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-23T10:16:45.010" CommunityOwnedDate="2010-09-23T11:15:58.493" />
  <row Id="5299" PostTypeId="2" ParentId="5293" CreationDate="2010-09-23T10:40:17.970" Score="4" ViewCount="0" Body="&lt;p&gt;I don't know where you found your claim, but it seems that the distance induced by the Frobenius norm between any two orthogonal matrices can be any real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\|A - B\|^2 &amp;amp;= \mathrm{tr} \left( (A-B)^t (A-B)    \right) \\&#xA;            &amp;amp;= \mathrm{tr} \left( (A^t -B^t) (A-B)  \right)  \\&#xA;            &amp;amp;= \mathrm{tr} (A^tA -A^tB - B^tA + B^tB)    \\&#xA;            &amp;amp;= \mathrm{tr} (2I) - \mathrm{tr}(2A^tB)  \\&#xA;            &amp;amp;= 2n - 2\mathrm{tr}(A^tB)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last but one equality is due to the fact that $A^tA = B^tB = I$ and $(B^tA)^t = A^tB$ and a matrix and its transpose have the same trace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take $A = I$ and we've got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\| I - B\|^2 = 2n - 2\mathrm{tr}(B)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any orthogonal matrix $B$. So, if you take as $B$ the family of orthogonal matrices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;\cos\theta  &amp;amp;  -\sin\theta \\&#xA;\sin\theta  &amp;amp;  \cos\theta&#xA;\end{pmatrix} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta \in [0, 2\pi]$, their traces $\mathrm{tr} (B) = 2\cos\theta$ can be any real number between $-2$ and $2$. So, their Frobenius distances to the unit matrix $I$ can be any real number from $0$ to $\sqrt{8}$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-23T12:37:42.073" LastActivityDate="2010-09-23T12:37:42.073" />
  <row Id="5300" PostTypeId="1" AcceptedAnswerId="5301" CreationDate="2010-09-23T10:42:06.343" Score="0" ViewCount="191" Body="&lt;p&gt;a. For every real number c, neither f nor g is continuous at c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. For every real number c, the sum f + g is continuous at c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. For every real number c, the product fg is continous at c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: consider functions defined one way on rationals and another on irrationals&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-09-23T17:30:14.290" Title="Give examples of functions f and g so that:" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5301" PostTypeId="2" ParentId="5300" CreationDate="2010-09-23T10:49:42.680" Score="3" ViewCount="0" Body="&lt;p&gt;Take $f=1_\mathbb Q$ (that is, $f(x)=1$ if $x$ is rational and $f(x)=0$ otherwise) and $g=1-f$. Then neither $f$ nor $g$ is continuous at any point but $f+g$ and $fg$ are very simple.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-23T10:49:42.680" />
  <row Id="5302" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T11:13:59.747" Score="4" ViewCount="0" Body="&lt;p&gt;I really struggled with the $\epsilon-\delta$ definition of limits, especially for non-linear functions.  This was also my first exposure to proof, as in: Prove that $$\lim_{x \to 2} (x^2 + 3) = 7$$ and I had a hard time with it at first.  To be clear, computing these limits was no problem, but using the definition to prove they were correct really confused me.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-09-23T11:13:59.747" CommunityOwnedDate="2010-09-23T11:15:58.493" CommentCount="5" />
  <row Id="5303" PostTypeId="1" AcceptedAnswerId="5304" CreationDate="2010-09-23T12:48:32.990" Score="3" ViewCount="561" Body="&lt;p&gt;The motivation for this question is : &lt;a href=&quot;http://math.stackexchange.com/questions/4066/rationals-of-the-form-fracpq-where-p-q-are-primes-in-a-b&quot;&gt;http://math.stackexchange.com/questions/4066/rationals-of-the-form-fracpq-where-p-q-are-primes-in-a-b&lt;/a&gt; and some other problems in Mathematics which looks as if they are elementary but their proofs are very much sophisticated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to consider two famous questions: First the &quot;Fermat's Last Theorem&quot; and next the unproven &quot;Goldbach conjecture&quot;. These questions appear elementary in nature, but require a lot of Mathematics for even comprehending the solution. Even the problem, which i posed in the link is so elementary but i don't see anyone even giving a proof without using the prime number theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is: Why is this happening? If i able to understand the question, then i should be able to comprehend the solution as well. A Mathematician once quoted: Mathematics is the understanding of how nature works. Is nature so complicated that a common person can't understand as to how it works, or is it we are making it complicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, i appreciate the beauty of Mathematics also: Paul Erdos proof of Bertrands postulate is something which i admire so much because, of its elementary nature. But at the same time i have my scepticism about FLT and other theorems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have stated 2 examples of questions which appear elementary, but the proofs are intricate. I know some other problems, in number theory which are of this type. Are there any other problems of this type, which are not Number Theoritical? If yes, i would like to see some of them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-28T00:03:51.660" Title="Why do statements which appear elementary have complicated proofs?" Tags="&lt;soft-question&gt;&lt;philosophy&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="0" />
  <row Id="5304" PostTypeId="2" ParentId="5303" CreationDate="2010-09-23T13:57:51.133" Score="11" ViewCount="0" Body="&lt;p&gt;Simple questions often have complex answers, or no answers at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to ask me why the sky was blue, to give a complete answer I would have to describe the heliocentric model, the earth's atmosphere, and the electromagnetic spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you asked me how my computer connected to the internet, the answer would take a considerable amount of time to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you asked my whether there was life on other planets, I wouldn't be able to give an answer; we just don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, it is impossible to give a simple answer to a simple question. This is true in any field you might encounter, ranging from the sciences to the humanities. And it is true in mathematics,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics allows us to formalize our questions within an axiomatic structure. It lets us ask our questions more precisely. But it in no way guarantees that the simplicity of the question would translate into simplicity of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other simple problems which cannot be answered simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Four color theorem states that any map made up of continuous regions can be colored with 4 colors such that each region gets 1 color and no two adjacent regions get the same color. The proof is quite complex, requiring the use of a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scheinerman's conjecture is the conjecture that &quot;every planar graph is the intersection graph of a set of line segments in the plane&quot; (sourced from &lt;a href=&quot;http://en.wikipedia.org/wiki/Scheinerman%27s_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Scheinerman%27s_conjecture&lt;/a&gt;). However, the proof was only completed in 2009, and is fairly difficult.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-23T13:57:51.133" CommentCount="1" />
  <row Id="5305" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T13:58:33.647" Score="-2" ViewCount="0" Body="&lt;p&gt;Not sure that it's possible. If you look &lt;a href=&quot;http://www.claymath.org/millennium/Riemann_Hypothesis/&quot; rel=&quot;nofollow&quot;&gt;http://www.claymath.org/millennium/Riemann_Hypothesis/&lt;/a&gt; you'll see that one of the major unsolved problems in maths is that currently, finding the distrubution of prime numbers is impossible(hence it's a Millenium Prize Problem). However, this only applies for very large numbers. For all practical purposes, you could just use what TheMachineCharmer&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2010-09-23T13:58:33.647" CommentCount="2" />
  <row Id="5306" PostTypeId="2" ParentId="5286" CreationDate="2010-09-23T14:44:28.830" Score="2" ViewCount="0" Body="&lt;p&gt;At the school I was taught to look at the derivative as the instantaneous rate of change and that fit well with applications in physics. But later, when I was learning Economics in college, I had to learn to look at the derivative as the best linear (affine) approximation, and a differentiable function as a function which had 'good' linear approximations. That is also the intuition that generalizes to many variables. I wish it had been discussed in my early calculus classes.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-23T14:44:28.830" CommunityOwnedDate="2010-09-23T14:44:28.830" CommentCount="1" />
  <row Id="5307" PostTypeId="1" CreationDate="2010-09-23T15:01:34.520" Score="2" ViewCount="68" Body="&lt;p&gt;I'm trying to find a percentage to describe how close a generated frequency distribution is to an ideal distribution, but am having a total blank. Basically I need to rate distributions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say I have the below frequency distributions for a sample size of say 100:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   0&#xA;2   0&#xA;3   0&#xA;4   100&#xA;5   0&#xA;6   0&#xA;7   0&#xA;8   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The above is my ideal distribution and would receive a rating of 100%&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   1&#xA;2   4&#xA;3   15&#xA;4   50&#xA;5   20&#xA;6   6&#xA;7   3&#xA;8   1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The above would also be considered ideal as well with the spike falling on the same frequency and the count is sufficiently  larger then the surrounding values so a rating of 100% (or close to)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   15&#xA;2   50&#xA;3   20&#xA;4   8&#xA;5   4&#xA;6   2&#xA;7   1&#xA;8   0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This while having a sufficient sized spike the spiking is happening on the wrong frequency so I want it to be weighted more towards spike happening on the right frequency then the size of the spike. So this might get rated at 50%&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A   B&#xA;1   5&#xA;2   30&#xA;3   10&#xA;4   4&#xA;5   6&#xA;6   10&#xA;7   30&#xA;8   5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then something like this with two large spikes, but one being close to the required frequency might receive a rating of 20%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I'm after a formula to rate a distribution to how close it is to ideal with more weighting towards getting the frequency right (ie the closer the spike is to the ideal frequency the better the score, but the score will also be effected by the the size of the spike and other spikes etc etc.)&lt;/p&gt;&#xA;" OwnerUserId="2089" LastActivityDate="2010-09-23T15:13:12.867" Title="Quantifying difference between two frequency distributions" Tags="&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="5308" PostTypeId="2" ParentId="5307" CreationDate="2010-09-23T15:06:09.973" Score="2" ViewCount="0" Body="&lt;p&gt;Maybe the statistics stack exchange site will give you better answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/4/assessing-the-significance-of-differences-in-distributions&quot;&gt;http://stats.stackexchange.com/questions/4/assessing-the-significance-of-differences-in-distributions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link seems to give the answers you seek. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-23T15:13:12.867" LastActivityDate="2010-09-23T15:13:12.867" CommentCount="2" />
  <row Id="5309" PostTypeId="1" AcceptedAnswerId="5310" CreationDate="2010-09-23T15:18:21.037" Score="3" ViewCount="61" Body="&lt;p&gt;Let G be a topological group and let $s_1$ and $s_2$ be loops in G (both loops are based at the identity e of G). Is it true that the loop $s_1s_2$ (where the multiplication is the one of the group structure of G) is equal, in $\pi_1(G,e)$, to the loop $s_1*s_2$ where this product is given by first going around $s_1$ and then $s_2$ (i.e., do we have $[s_1s_2] = [s_1*s_2]$)? If yes, what is the proof for this? &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-09-23T15:27:18.733" Title="Topological group: Multiplying two loops is homotopic to linking these paths?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="5310" PostTypeId="2" ParentId="5309" CreationDate="2010-09-23T15:27:18.733" Score="2" ViewCount="0" Body="&lt;p&gt;Yes and this is known as the Eckmann-Hilton argument. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-23T15:27:18.733" CommentCount="1" />
  <row Id="5311" PostTypeId="2" ParentId="227" CreationDate="2010-09-23T15:32:47.960" Score="0" ViewCount="0" Body="&lt;p&gt;I think game theory is becoming increasingly important for economists. In addition to classical models such as matrix games (zero-sum and examples like Prisoner's Dilemma and Chicken) and the work of John Nash on equilibria and bargaining, there is the Gale-Shapley model for two-markets, mechanism design (a recent Nobel Memorial Prize was awarded for this work), auctions, fair division methods, group decision making, etc.. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-23T15:32:47.960" CommunityOwnedDate="2010-09-23T15:32:47.960" />
  <row Id="5312" PostTypeId="1" AcceptedAnswerId="5313" CreationDate="2010-09-23T16:27:49.820" Score="3" ViewCount="183" Body="&lt;p&gt;Suppose we have a 3x3x3 cube of cubes (i. e. 27 cubes) and consider the graph whose vertices are the cubes, and let an edge connect two iff they have touching faces.  I'm quite sure there is no Hamilton path that starts at the middle cube and ends in a corner cube, though I'm not sure how to prove it without tedious casework.&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2010-09-24T20:40:01.887" Title="3-dimensional graph of cubes" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5313" PostTypeId="2" ParentId="5312" CreationDate="2010-09-23T16:30:39.640" Score="4" ViewCount="0" Body="&lt;p&gt;If you give 3D coordinates to the cube nodes $(x,y,z)$. Two nodes are adjacent iff they differ by exactly $1$ in exactly one of the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $x+y+z$. The parity(whether it is even or odd) of this flips each time you move to an adjacent node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parity of the center cube and the corner cube are different. In 26 moves, you will end up at a node with the same parity as the initial node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no Hamiltonian Path can exist which starts at the center cube and ends at a corner cube.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T16:30:39.640" CommentCount="2" />
  <row Id="5314" PostTypeId="1" CreationDate="2010-09-23T16:58:23.503" Score="3" ViewCount="89" Body="&lt;p&gt;Consider the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given a multiset (a set with repetitions allowed) of $2n+1$ real numbers, say $S = \{r_1, \dots, r_{2n+1}\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers are such that for every $k$, the multiset $S - \{r_k\}$ can be split into two multisets of size $n$ each, such that the sum of numbers in one multiset is same as the sum of numbers in the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that all the numbers must be equal.( i.e. $r_{i} = r_{j}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please stop reading further if you want to try and solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this problem can easily be solved using Linear Algebra. We have a set of $2n+1$ linear equations, which corresponds to a matrix equation $Ar = 0$. It can be shown that $A$ has rank at least $2n$ which implies the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, is there any solution to this problem which does not involve any linear algebra? &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T18:11:29.450" Title="Odd number of reals with equal partitions" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="5315" PostTypeId="1" AcceptedAnswerId="5320" CreationDate="2010-09-23T16:59:56.350" Score="4" ViewCount="179" Body="&lt;p&gt;Let $X$ be a set, and $d,d'$ two metrics on $X$.  Consider the identity map $i : (X,d) \to (X,d')$ as a map of metric spaces.  There are (at least) three reasonable notions of equivalence for $d$ and $d'$, in increasing order of strength:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$i$ is a homeomorphism, i.e. $d$ and $d'$ induce the same topology on $X$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$i$ and $i^{-1}$ are uniformly continuous.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$i$ is bilipschitz, i.e. $C_1 d' \le d \le C_2 d'$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would like to know what terms are used for these notions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, Mathworld says that the term &quot;equivalent&quot; refers to sense 1.  This seems counterintuitive since, for instance, sense 1 does not preserve completeness.  Munkres's &lt;em&gt;General Topology&lt;/em&gt; uses &quot;metrically equivalent&quot; for sense 2.  Does this agree with people's experience of standard usage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; I will point out that 3 implies 2 implies 1 (since Lipschitz implies uniformly continuous implies continuous) but converses are false.  Let $X = \mathbb{R}$, let $d_1(x,y) = |x-y|$, $d_2(x,y) = |x-y| \wedge 1$, $d_3(x,y) = |\phi(x)-\phi(y)|$, where $\phi : \mathbb{R}\to (0,1)$ is your favorite homeomorphism.  In sense 1 all three are equivalent, in sense 2 $d_1 \sim d_2 \not\sim d_3$, and in sense 3 all are inequivalent.  In particular note that $d_1, d_2$ are complete but $d_3$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; Further confusing the matter is the fact that if $X$ is a vector space and $d,d'$ are induced by norms $||\cdot||, ||\cdot||'$, then all three senses coincide, and sense 3 is usually taken as the definition of &quot;equivalent norms.&quot;&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-09-23T18:40:30.957" LastActivityDate="2010-09-23T18:40:30.957" Title="Notions of equivalent metrics" Tags="&lt;terminology&gt;&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5316" PostTypeId="2" ParentId="5277" CreationDate="2010-09-23T17:08:19.857" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;How do I mathematically determine if a number is prime?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number is n, then dividing it by every prime number less than or equal to sqrt(n) and showing that there is a remainder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of different &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_primes&quot; rel=&quot;nofollow&quot;&gt;sieve&lt;/a&gt; solutions for finding prime numbers, the oldest and most famous of which is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Sieve of Eratosthenes&lt;/a&gt;.  These are generally easy to programme and can, for example, find all of the primes below 100,000 in a few milliseconds on a modern processor.&lt;/p&gt;&#xA;" OwnerUserId="1032" LastEditorUserId="1032" LastEditorDisplayName="" LastEditDate="2010-09-27T09:42:35.480" LastActivityDate="2010-09-27T09:42:35.480" />
  <row Id="5317" PostTypeId="2" ParentId="5300" CreationDate="2010-09-23T17:30:14.290" Score="2" ViewCount="0" Body="&lt;p&gt;This isn't specific to the real numbers.  The domain, range, sum and product could be specified (almost) arbitrarily.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S= f+g$ and $P=fg$.  These are continuous functions.  $f$ and $g$ are roots of a quadratic equation with continuous coefficients, $z^2 - Sz + P$.  There is a formula expressing the solution continuously in terms of $S$ and $P$, using one extraction of a square root.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible source of discontinuity is in the selection of which sign of the square root is used to define $f$ and which one for $g$.   To attain discontinuity everywhere, one needs to have $S^2 \neq 4P$ everywhere (so that $f,g$ are always distinct) and for the set where the plus sign is used for determining $f$ to be dense with dense complement.  Hence, a splitting of the domain into two dense sets (such as rational and irrational) is not an artificial feature of the answer posted earlier, it is necessary and sufficient for the solution. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-09-23T17:30:14.290" />
  <row Id="5318" PostTypeId="2" ParentId="5314" CreationDate="2010-09-23T17:57:23.190" Score="3" ViewCount="0" Body="&lt;p&gt;You can't avoid some sort of algebra, because the statement is false in a commutative group where $nx = 0$ has nontrivial solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow use of the linear algebra fact that rank is the same over any field containing the coefficients of the equations, it is enough to consider rational (and thus integer) solutions and extra structure is available.  One can then avoid use of determinants or matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Sigma$ is the sum of all elements, $\Sigma - r_i$ is even and thus all $r_i$ have the same parity.  We can replace each $r_i$ by $(r_i-r_k)/2$ and get a smaller solution, where $r_k$ is the smallest of the numbers.  This process ends at the zero solution, and is reversible, so the original solution has all numbers equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added:  you can consider this a use of either the real or 2-adic metric on integers, so this must correspond to a linear algebra argument using inequalities or reduction mod 2^(2n+1) on the system of equations or its determinant.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-23T18:11:29.450" LastActivityDate="2010-09-23T18:11:29.450" CommentCount="4" />
  <row Id="5319" PostTypeId="2" ParentId="5248" CreationDate="2010-09-23T18:01:22.667" Score="4" ViewCount="0" Body="&lt;p&gt;Here is a sketch of another elementary solution based on a proof in Bromwich's Theory of Infinite Series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\sin(2k+1)x-\sin(2k-1)x = 2\cos2kx\sin x$ and summing from k=1 to k=n we have&#xA;$$\sin(2n+1)x = \sin x \left( 1+ 2 \sum_{k=1}^n \cos 2kx \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence&#xA;$$ \int_0^{\pi/2} {\sin(2n+1)x \over \sin x} dx = \pi/2. \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y=(2n+1)x$ and this becomes&#xA;$$ \int_0^{(2n+1)\pi/2} {\sin y \over (2n+1) \sin (y/(2n+1))} dy = \pi/2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $\lim_{n \to \infty} (2n+1) \sin { y \over 2n+1} = y$ it suggests that there is a proof lurking in there somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's put&#xA;$$\begin{align}&#xA;I_n &amp;amp;=  \int_0^{n\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx&#xA;\ &amp;amp;= \sum_{k=0}^{n-1} \int_{k\pi/(2n+1)}^{(k+1)\pi/(2n+1)} {\sin(2n+1)x \over \sin x} dx.&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we have $I_n = u_0 – u_1 + u_2 \cdots + (-1)^{n-1}u_{n-1},$ where $u_k$ is a decreasing sequence of positive terms. We can see this from the shape of the curve&#xA;$y = \sin(2n+1)x / \sin x,$ which crosses the x-axis at $\pi/(2n+1), 2\pi/(2n+1),\ldots,n\pi/(2n+1).$ (I said that this is just a sketch, you have to check the details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the sequence $I_n$ converges, and by (1) it converges to $\pi/2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we make the substitution $y=(2n+1)x$ we see that&#xA;$$u_k = \int_{k\pi}^{(k+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $I_n$ can be written as an alternating sequence of decreasing positive terms we can truncate the sequence wherever we like and the value of $I_n$ lies between two successive partial summations. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{0}^{2m\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy &amp;lt; I_n &amp;lt;&#xA; \int_{0}^{(2m+1)\pi} {\sin y \over (2n+1) \sin (y/(2n+1))} dy. \qquad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any m such that $2m+1 \le n.$ (Take $m=[\sqrt{n}],$ say, $n \ge 6.$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\left| { \sin y \over y} - {\sin y \over (2n+1) \sin(y/(2n+1))} \right| &#xA;&amp;lt; { \pi^2(2m+1)^2 \over 3(2n+1)^2}$$&#xA;and so this difference tends to zero uniformly in the interval $0 \le y \le (2m+1)\pi$&#xA;and so by taking the $\lim_{n \to \infty}$ in (2) we obtain&#xA;$$\int_0^{\infty} { \sin x \over x } dx = { \pi \over 2}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-23T18:01:22.667" />
  <row Id="5320" PostTypeId="2" ParentId="5315" CreationDate="2010-09-23T18:08:45.897" Score="1" ViewCount="0" Body="&lt;p&gt;I don't know if there is a standard, but I can provide two sample points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burago, Burago, and Ivanov in &quot;A course in metric geometry&quot;, p. 9&#xA;call definition (3) Lipschitz equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dugundji in &quot;Topology&quot; calls those metrics satisfying (1) equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burago et al. is available online: www.math.psu.edu/petrunin/papers/alexandrov/bbi.pdf&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2010-09-23T18:16:53.353" LastActivityDate="2010-09-23T18:16:53.353" CommentCount="1" />
  <row Id="5321" PostTypeId="2" ParentId="5315" CreationDate="2010-09-23T18:14:56.790" Score="0" ViewCount="0" Body="&lt;p&gt;I would call (1) topologically equivalent or though as yasmar mentions some just use equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would call (3) strongly equivalent but in numerical analysis we often use just equivalent (as in your norm equivalent example) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly confident that (1) iff (2) and certain that (3) implies (1). &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T18:14:56.790" CommentCount="2" />
  <row Id="5322" PostTypeId="2" ParentId="5248" CreationDate="2010-09-23T18:18:28.747" Score="10" ViewCount="0" Body="&lt;p&gt;Here's another way of finishing off Derek's argument. He proves&#xA;$$\int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2.$$&#xA;Let&#xA;$$I_n=\int_0^{\pi/2}\frac{\sin(2n+1)x}{x}dx=&#xA;\int_0^{(2n+1)\pi/2}\frac{\sin x}{x}dx.$$&#xA;Let&#xA;$$D_n=\frac\pi2-I_n=\int_0^{\pi/2}f(x)\sin(2n+1)x\ dx$$&#xA;where&#xA;$$f(x)=\frac1{\sin x}-\frac1x.$$&#xA;We need the fact that if we define $f(0)=0$ then $f$ has a continuous&#xA;derivative on the interval $[0,\pi/2]$. Integration by parts yields&#xA;$$D_n=\frac1{2n+1}\int_0^{\pi/2}f'(x)\cos(2n+1)x\ dx=O(1/n).$$&#xA;Hence $I_n\to\pi/2$ and we conclude that&#xA;$$\int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-23T18:18:28.747" CommentCount="1" />
  <row Id="5323" PostTypeId="2" ParentId="5315" CreationDate="2010-09-23T18:28:51.380" Score="1" ViewCount="0" Body="&lt;p&gt;The first one is simply the definition of topological equivalence - it verbatim extends to general topological spaces (not necessarily metric spaces). I don't have a name for the second one. The third, which is used often, is called &quot;Lipschitz equivalence&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-23T18:28:51.380" />
  <row Id="5324" PostTypeId="2" ParentId="5297" CreationDate="2010-09-23T18:53:41.657" Score="2" ViewCount="0" Body="&lt;p&gt;This means that $j\epsilon_0 &gt; \frac{\pi}{2}$. Hence $-j\epsilon_0 &amp;lt; -\frac{\pi}{2}$. By the density of rationals, this tells you that there exists a sequence $q_1, q_2,\dots,q_n$ of rational numbers in $(-\epsilon_0,\epsilon_0)$ such that $jq_n\rightarrow \frac{\pi}{2}$. Hence $|\cos(jq_n)|\rightarrow 0$. Combining this with $|x_j| &amp;lt; |\cos(jq_n)|$ for all $n$ gives the contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-23T18:53:41.657" CommentCount="2" />
  <row Id="5325" PostTypeId="1" AcceptedAnswerId="5327" CreationDate="2010-09-23T19:07:55.263" Score="0" ViewCount="85" Body="&lt;p&gt;I have a series of terms $\{t_n : t_n = a_n x_n\}$, and I want to talk about the product of each term except $t_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would any of these be an appropriate way to say that?  I like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i \ne j} t_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or maybe this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=1,i \ne j}^n t_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a normal thing to do, to put qualifiers on an index like that, or is there a better way to do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would it be better to make a subset, $\{b_n : a_n, n \ne j \}$ and use that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=1}^n b_i$$&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-09-23T20:03:17.657" Title="What is the correct way to write product or sum in capital pi or capital sigma notation when you wish to exclude an index?" Tags="&lt;soft-question&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5326" PostTypeId="2" ParentId="4844" CreationDate="2010-09-23T19:08:16.377" Score="0" ViewCount="0" Body="&lt;p&gt;This problem seems related to evenly distributing points across the surface of a sphere (specifically, that is very similar to your k=2 case). That problem is addressed in a programming competition challenge named &lt;a href=&quot;http://www.spoj.pl/problems/PSPHERE/&quot; rel=&quot;nofollow&quot;&gt;PSPHERE&lt;/a&gt; at &lt;a href=&quot;http://www.spoj.pl/&quot; rel=&quot;nofollow&quot;&gt;SPOJ&lt;/a&gt;. The solutions there are not public, but perhaps approaching the leading contestants on that particular challenge could prove helpful.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-23T19:08:16.377" />
  <row Id="5327" PostTypeId="2" ParentId="5325" CreationDate="2010-09-23T19:21:40.023" Score="4" ViewCount="0" Body="&lt;p&gt;It is normal to write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{\begin{matrix}i=1 \\ i \ne j \end{matrix}}^n t_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but another good notation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{t \in T \setminus \{t_j\}} t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $T = \{t_n : t_n = a_n x_n\}$, also note that it is only defined up to permutation)&lt;/p&gt;&#xA;" LastActivityDate="2010-09-23T20:03:17.657" CommentCount="4" />
  <row Id="5328" PostTypeId="1" CreationDate="2010-09-23T19:38:23.503" Score="2" ViewCount="195" Body="&lt;p&gt;I have the standard logical equivalence:&#xA;$(p\rightarrow q)\wedge(q\rightarrow r)\Leftrightarrow p\rightarrow (q\wedge r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using several distributive laws I was able to get it down to:&#xA;$(\neg p\wedge\neg q) \vee (\neg p\wedge r) \vee  (q\wedge r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must be missing some manipulation I can do to reduce this.&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T19:46:45.363" LastActivityDate="2010-09-23T21:47:09.907" Title="Proof of logical equivalence" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="5329" PostTypeId="1" AcceptedAnswerId="5345" CreationDate="2010-09-23T20:00:24.023" Score="1" ViewCount="79" Body="&lt;p&gt;I have to prove that if $P$ is a $R$-module , $P$ is projective $\Leftrightarrow$ there is a family $\{x_i\}$ in $P$ and morphisms $f_i\colon P\to R$ such that for all $x\in P$&#xA;$$ x= \sum_{i\in I} f_i(x)x_i$$&#xA;where for each $x\in P$, $f_i(x)=0$ for almost all $i\in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-23T20:03:22.763" LastActivityDate="2010-09-23T21:58:44.937" Title="Projective modules" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5330" PostTypeId="1" AcceptedAnswerId="5341" CreationDate="2010-09-23T20:15:24.113" Score="1" ViewCount="36" Body="&lt;p&gt;I am reading the book Stochastic Simulation, Algorithms and Analysis by Asmussen. On page 39, when talking about inversion method for sampling a distribution, he gave an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an r.v. distributed as the overshoot X − a given X &gt; a can be generated as $F^{-1} (U (1-F (a)) − a$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $U$ is a r.v. uniform in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think the distribution is $F^{-1} (U (1-F (a)) + F(a)) - a$. I was wondering which is correct and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-09-23T20:58:54.077" LastActivityDate="2010-09-23T21:28:32.360" Title="A problem of inversion method for sampling" Tags="&lt;probability&gt;&lt;monte-carlo&gt;&lt;simulation&gt;" AnswerCount="1" />
  <row Id="5331" PostTypeId="1" CreationDate="2010-09-23T20:30:33.460" Score="6" ViewCount="153" Body="&lt;p&gt;A Blaschke product is a function of the form&#xA;$$B(z):=z^k\prod_{n=1}^{\infty}\frac{a_n-z}{1-\overline{a_n}z}\frac{|a_n|}{a_n}$$&#xA;where the $a_n$ are the non-zero zeros of $B$, and satisfie $\sum_{n=1}^{\infty}(1-|a_n|) &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blashke products are holomorphic and bounded by 1 on the unit disk. A well known theorem asserts that $B$ has radial limits almost everywhere on the unit circle, i.e. that the limit&#xA;$$\lim_{r \rightarrow 1} B(re^{i \theta})$$ exist for almost every $\theta$. I'm looking for an example of Blashke product such that the radial limit does not exist at a certain point, say $1$ for example. In particular, a Blaschke product with zeros in $(0,1)$ such that &#xA;$$\limsup_{r \rightarrow 1}|B(r)| =1$$ would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a construction or reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-12-25T20:44:26.253" Title="On radial limits of Blaschke Products" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5333" PostTypeId="1" AcceptedAnswerId="5336" CreationDate="2010-09-23T20:38:49.070" Score="2" ViewCount="68" Body="&lt;p&gt;The modular group is the group $G$ consisting of all linear fractional transformations $\phi$ of the form&#xA;$$\phi(z)=\frac{az+b}{cz+d}$$&#xA;where $a,b,c,d$ are integers and $ad-bc=1$. I have read that $G$ is generated by the transformations $\tau(z)=z+1$ and $\sigma(z)=-1/z$. Is there an easy way to prove this? In particular, is there a proof that uses the relation between linear fractional transformations and matrices? Any good reference would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik &lt;/p&gt;&#xA;" OwnerUserId="1197" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-23T20:50:22.003" LastActivityDate="2010-09-23T20:55:25.240" Title="On the generators of the Modular Group" Tags="&lt;linear-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5334" PostTypeId="1" AcceptedAnswerId="5353" CreationDate="2010-09-23T20:39:53.030" Score="4" ViewCount="132" Body="&lt;p&gt;So, I &lt;a href=&quot;http://math.stackexchange.com/questions/5267/how-to-find-where-3-lines-intersect&quot;&gt;asked a question&lt;/a&gt; about how to find if 3 lines are concurrent. I built the algorithm I needed,and it was working well, until I started doubting my power of judgement. So my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are 2 equal lines also intersecting?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, 9x+9y=9;7x+7y+7 ? Because now, if I would draw these lines and add &lt;em&gt;any&lt;/em&gt; line to the comparison, these 3 lines would be concurrent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;help?&lt;/p&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-24T05:52:25.167" LastActivityDate="2010-09-24T05:52:25.167" Title="Do 2 equal lines also count as intersecting?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="5335" PostTypeId="2" ParentId="5331" CreationDate="2010-09-23T20:41:19.550" Score="0" ViewCount="0" Body="&lt;p&gt;If you have the zeros on (0,1), as you suggest, then certainly &#xA;$$\liminf_{r\to 1}|B(r)|=0$$&#xA;(as long as you have infinite numbers of zeros of course). &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-09-23T20:41:19.550" CommentCount="1" />
  <row Id="5336" PostTypeId="2" ParentId="5333" CreationDate="2010-09-23T20:49:28.917" Score="8" ViewCount="0" Body="&lt;p&gt;Yes; this statement is essentially equivalent to the Euclidean algorithm.  I discuss these issues in &lt;a href=&quot;http://www.artofproblemsolving.com/blog/10588&quot; rel=&quot;nofollow&quot;&gt;this old blog post&lt;/a&gt;.  (A very brief sketch: by applying the generators and the inverses to an arbitrary element of the modular group it is possible to perform the Euclidean algorithm on $a$ and $c$ (or maybe it's $a$ and $b$).  The rest is casework.)  You can think of this as a form of row reduction, which is generalized by the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;Smith normal form&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a geometric proof using the action on the upper half plane which is given, for example, in the relevant section of Serre's &lt;em&gt;Course in Arithmetic.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-23T20:55:25.240" LastActivityDate="2010-09-23T20:55:25.240" CommentCount="2" />
  <row Id="5338" PostTypeId="1" AcceptedAnswerId="5342" CreationDate="2010-09-23T21:19:23.557" Score="2" ViewCount="102" Body="&lt;p&gt;Let φ(w)ψ(x) be a solution to the heat problems. Given that ψ(−1) = ψ(1) = 0 &#xA;prove that it is not possible for ψ(x) and ψ′′ (x) to be strictly positive for all x such that &#xA;−1 &amp;lt; x &amp;lt; 1. You may use any facts, such as the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, this is homework. But we have been at it for hours now,and could use a little nudge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $ψ(x), ψ''(x)$ are positive.&#xA;then the mean value theorem states $ψ'(c) = 0$ according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ψ'(c) = \frac{ψ(1)-ψ(−1)}{1+(-1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ψ(c) = 0$, then we know there exists a tangent parallel to x-axis. This is the extrema which is strictly positive. &lt;em&gt;Here is where we get stuck&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we assumed $ψ''(c)$ is positive then we know that there exists a minimum due to the the derivative test. But how does this help us prove that the range from -1 to 1 is NOT strictly positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little background info:&#xA;This is for real analysis. Both me and roommate have not taken a prior proofs course (which is acutally the pre-req). So we are not experts are proving so if any of you take the time to answer, please dumb it down as much as you can. &lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-23T21:26:36.077" LastActivityDate="2010-09-23T21:37:36.713" Title="Proving a function is below zero using the mean value theorm." Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="5339" PostTypeId="2" ParentId="5052" CreationDate="2010-09-23T21:19:26.577" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps this has something for you: &lt;a href=&quot;http://wis.kuleuven.be/analyse/walter/coimbra.pdf&quot; rel=&quot;nofollow&quot;&gt;http://wis.kuleuven.be/analyse/walter/coimbra.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uses the Legendre polynomials in proving irrationality of certain numbers.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T21:19:26.577" />
  <row Id="5340" PostTypeId="1" CreationDate="2010-09-23T21:23:02.227" Score="1" ViewCount="277" Body="&lt;p&gt;If the probability of an event is &lt;code&gt;1 / ∞&lt;/code&gt; and &lt;code&gt;∞&lt;/code&gt; trials are conducted, how many times will the event occur — &lt;code&gt;0&lt;/code&gt;, &lt;code&gt;1&lt;/code&gt;, or &lt;code&gt;∞&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2010-09-24T02:48:05.317" Title="Probability and Infinity" Tags="&lt;probability&gt;&lt;infinity&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="5341" PostTypeId="2" ParentId="5330" CreationDate="2010-09-23T21:28:32.360" Score="1" ViewCount="0" Body="&lt;p&gt;Let $Y=X$ with support on $[a,\infty)$. If $f_X$ is the p.d.f of $X$ and $f_Y$ is the pd.f. of $Y$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_Y(y) = \frac{f_X(y)}{1-F(a)}$  for $y&gt;a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_Y(y) = \frac{F_X(y)-F_X(a)}{1-F_X(a)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $U=F_Y(Y)$ then as usual we have $U\sim Uni[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U=\frac{F_X(Y)-F_X(a)}{1-F_X(a)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging gives us &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y=F_X^{-1}(U(1-F_X(a))+F_X(a))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the over shoot r.v. $X-a$ for $X&gt;a$ can be generated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_X^{-1}(U(1-F_X(a))+F_X(a))-a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with you. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-23T21:28:32.360" />
  <row Id="5342" PostTypeId="2" ParentId="5338" CreationDate="2010-09-23T21:37:36.713" Score="1" ViewCount="0" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\psi(x)$ is strictly positive, what does it say about the trend of $\psi'(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\psi''(x)$ is strictly positive, what does it say about the trend of $\psi'(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be useful to draw a graph.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T21:37:36.713" CommentCount="10" />
  <row Id="5343" PostTypeId="1" CreationDate="2010-09-23T21:37:45.247" Score="2" ViewCount="98" Body="&lt;p&gt;I'm looking at the quantum group $SU_q(2)$ (over ${\mathbb C}$) and can't see why it has no zero divisors. It's clear that $M_q(2)$, the quantum $2 \times 2$ matrices have no zero divisors, but I can't seem to see why/if this extends to the quotient over $det_q - 1$. I think one can argue it from the grading on $M_q(2)$, but that's just a guess.&lt;/p&gt;&#xA;" OwnerUserId="2099" LastActivityDate="2010-11-23T03:44:19.833" Title="Zero divisors in $SU_q(2)$" Tags="&lt;abstract-algebra&gt;&lt;zero-divisor&gt;&lt;quantum-groups&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5344" PostTypeId="2" ParentId="5328" CreationDate="2010-09-23T21:47:09.907" Score="6" ViewCount="0" Body="&lt;p&gt;Umm... maybe I am missing something, but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $p$ is false, $q$ is true and $r$ is false, then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(p\rightarrow q)\wedge(q\rightarrow r)$ is false&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p\rightarrow (q\wedge r)$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't see how you can prove the equivalence.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-23T21:47:09.907" />
  <row Id="5345" PostTypeId="2" ParentId="5329" CreationDate="2010-09-23T21:58:44.937" Score="3" ViewCount="0" Body="&lt;p&gt;Put the f&lt;sub&gt;i&lt;/sub&gt; together to form one giant f from P to R&lt;sup&gt;(I)&lt;/sup&gt;, the direct sum of I copies of the ring R.  The condition that x = Sum f_i(x) x_i just means that there is some g:R&lt;sup&gt;(I)&lt;/sup&gt;→P such that g(f(x)) = x, namely g((r1,r2,...)) = r1*x1 + r2*x2 + .... In other words, P is a direct summand of the free module R&lt;sup&gt;(I)&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-09-23T21:58:44.937" CommentCount="4" />
  <row Id="5347" PostTypeId="1" AcceptedAnswerId="5360" CreationDate="2010-09-23T23:58:16.693" Score="1" ViewCount="143" Body="&lt;p&gt;In Stochastic Simulation: Algorithms and Analysis by Søren Asmussen, on Page 38&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Poisson r.v. $N$ with rate $\lambda$ ($P(X = n) = e^{−\lambda} \frac{\lambda^n}{n!}$) can be constructed using the relation between the exponential distribution and the Poisson process: generate $X_1 , X_2 , ...$ as i.i.d. exponentials with rate $\lambda$ one at a time until the random time at which the sum exceeds 1, say $X_1 + · · ·+ X_N &amp;lt; 1 &amp;lt; X_1 + · · ·+ X_{N +1}$ . Then N is the desired Poisson r.v. To reduce the number of evaluations of special functions, use&#xA;                      $N = max \lbrace n \geq 0 : \prod_{i=1}^n U_i &gt; e^{-\lambda} \rbrace$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $U_i$ is r.v. uniform in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering why to &quot;reduce the number of evaluations of special functions&quot;, the $N$ in the last equation has Poisson distribution with rate λ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-24T05:49:11.787" LastActivityDate="2010-09-24T05:49:11.787" Title="Sample Poisson Distribution" Tags="&lt;probability&gt;&lt;monte-carlo&gt;&lt;simulation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5348" PostTypeId="2" ParentId="5343" CreationDate="2010-09-24T00:13:14.040" Score="1" ViewCount="0" Body="&lt;p&gt;The algebra can be constructed as an iterated Ore extension starting from the base field. By standard properties of such extensions, it is then a domain.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-24T00:49:11.097" LastActivityDate="2010-09-24T00:49:11.097" CommentCount="2" />
  <row Id="5349" PostTypeId="1" CreationDate="2010-09-24T00:35:45.427" Score="3" ViewCount="106" Body="&lt;p&gt;I'm looking for different proofs (references or sketch of main ideas) of the structure theorem for finitely generated modules over a PID. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, a comparison in terms of clarity, elegance or usefulness would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-09-24T03:00:09.187" LastActivityDate="2010-09-24T03:00:09.187" Title="Proofs of the structure theorem for finitely generated modules over  a PID" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" CommentCount="2" />
  <row Id="5350" PostTypeId="2" ParentId="5340" CreationDate="2010-09-24T00:35:57.423" Score="1" ViewCount="0" Body="&lt;p&gt;Sorry byoogle, I misinterpreted the point of your question. So now I feel compelled to try to answer. I don't know much about probability, but by chance I stumbled across something today that might be interesting to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to frame a question that I think is close to what you are asking, and is also answered by this nice one page paper I found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm looking for $x \in [0,1]$, and each &quot;instant&quot; I &quot;look&quot;, I select a set $A_\alpha \subset [0,1]$. Unfortunately, the probability that $x \in A_\alpha$ is always $0$. But, I have at least this: my set from instant $\alpha$ includes all the previous sets I selected. It seems little consolation though, because at any given $\alpha$ the probability that I have $x \in A_\alpha$ is still 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I am allowed to go on looking indefinitely, is the probability that $x$ is &quot;found&quot; zero? Well, it seems that the probability can in fact be 1, if we choose well. At least that is how I interpret this:&#xA;&lt;a href=&quot;http://www.math.harvard.edu/~elkies/Misc/quickie.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.harvard.edu/~elkies/Misc/quickie.pdf&lt;/a&gt; &lt;br&gt;&#xA;When you read that, you can read &quot;measure zero&quot; or &quot;negligible&quot; as &quot;probability zero&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a fair chance that my analogy is broken, and hopefully that will be pointed out if that's the case. Our &quot;instants&quot; can't be countable, in other words we need to be continously looking, rather than just looking every microsecond or so. Also, my analogy seems to be using the ordering on the real numbers, and maybe that is a logical stumble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps, or is at least amusing.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2010-09-24T00:58:23.260" LastActivityDate="2010-09-24T00:58:23.260" CommentCount="3" />
  <row Id="5352" PostTypeId="2" ParentId="5340" CreationDate="2010-09-24T01:37:56.520" Score="2" ViewCount="0" Body="&lt;p&gt;Here's one possible answer.  Say each event occurs with probability $\varepsilon&gt;0$ and there are $1/\varepsilon$ total events.  Taking the limit as $\varepsilon\to0^+$, the event occurs 0 times with probability $1/e\approx36.8\%$, once with probability $1/e\approx36.8\%$, twice with probability $1/2e\approx18.4\%$, three times with probability $1/6e\approx6.13\%$, and so on: the Poisson distribution with $\lambda=1$.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-24T01:37:56.520" CommentCount="2" />
  <row Id="5353" PostTypeId="2" ParentId="5334" CreationDate="2010-09-24T01:39:51.163" Score="3" ViewCount="0" Body="&lt;p&gt;In the analytical geometry version of Euclidean Geometry we need a convenient way of thinking of parallelism and there is also the interest in constructing the real projective plane (where there are no parallel lines) from  Euclidean Geometry. (The key idea is that we think of all the lines which have the same slope as having a special additional point on them at &quot;infinity&quot; (thereby creating new lines in the projective plane) and all these new points, one for each slope, lie on a new line, the line at &quot;infinity.&quot;) In the analytical geometry framework of Euclidean Geometry lines are parallel when they have the same slope but there is the problem that a nonzero multiple of an equation looks like a  different equation but consists of all of the same points. Furthermore, one would like parallelism to be an equivalence relation, which means it obeys being reflexive, symmetric and transitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equivalence_relation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is CONVENIENT to think of a line as being parallel to itself. Now all lines with the same slope can be thought of as parallel. (All the lines with undefined slope are also parallel to each other; they are the vertical lines.) Furthermore, we can think of the Euclidean plane's lines as being partitioned in classes by this parallelism relation.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-09-24T01:39:51.163" />
  <row Id="5354" PostTypeId="2" ParentId="5340" CreationDate="2010-09-24T02:18:49.220" Score="6" ViewCount="0" Body="&lt;p&gt;There are meaningful ways to work out probabilities on infinite spaces, but your question is not well-defined.  We could define situations matching your question with the answer 0, 1, $\infty$, or any other finite answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability distributions on finite sets are pretty easy to write down and work with.  For rolling a single die, the probability distribution is {p(1)=1/6, p(2)=1/6, p(3)=1/6, p(4)=1/6, p(5)=1/6, p(6)=1/6}.  From this, we can ask and answer many different questions-- for example, the probability of rolling a 3 three times in a row is 1/216.  However, the probability of rolling a 7 is 0; and it doesn't matter how many times you roll, you'll never get a 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability distributions on infinite sets require a lot more background to define and work with.  You need a &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_measure&quot; rel=&quot;nofollow&quot;&gt;probability measure&lt;/a&gt; and you can compute integrals to find probabilities.  (Measure theory is usually a graduate-level topic, and it requires very careful, abstract thought to get the details straight.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An instructive and fun problem to read about is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;Random Walk&lt;/a&gt;.  Suppose I start at (0,0) on an infinite 2-dimensional grid; and each second, I go up, down, left, or right one unit.  (Each direction has probability 1/4 every second.)  The probability space is the set of &lt;em&gt;all&lt;/em&gt; possible walks, which is an infinite set.  The probability measure on this space is determined by the fact that each direction has equal probability every second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though there are many walks in the probability space that never return to (0,0), the subset of walks that &lt;em&gt;do&lt;/em&gt; return to (0,0) has measure 1.  This implies that on a 2-dimensional random walk, the expected number of returns to (0,0) is infinite.  However, the expected number of times I take a diagonal step is 0 (because, by my definition of the problem, it never happens).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also consider the 3-dimensional random walk.  Interestingly, on a 3-dimensional random walk, the expected number of returns to (0,0,0) is about 0.516385.  (You can think of this as the average number of returns if you sample many 3-dimensional random walks.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results requires some work to obtain, and you always have to start with a precisely defined problem.  If you think about them and read other examples on your own, it should help you gain some understanding of infinite probability spaces, probability 0 events, and probability 1 events.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-09-24T02:48:05.317" LastActivityDate="2010-09-24T02:48:05.317" />
  <row Id="5356" PostTypeId="2" ParentId="5297" CreationDate="2010-09-24T04:03:48.703" Score="0" ViewCount="0" Body="&lt;p&gt;I would have added a comment, but I fear the answer would overlap the allowed character limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, maybe I'm missing something, but I think I see a clear contradiction. Here we go...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_j = \{(x_0, x_j)\in\mathbb{Q}\times\mathbb{R}: |x_j| &amp;lt; |\cos(jx_0)|\}$ Let $V_j = (-\epsilon_0,\epsilon_0)\times(-\epsilon_j,\epsilon_j)$. The claim is that for all $j$, we have $V_j\subset S_j$. Fix $j$ and assume this inclusion holds. On the other hand, if $2j\epsilon_0 &gt; \pi$, there is a sequence $q_1,q_2,\dots$ of rationals in $(-\epsilon_0,\epsilon_0)$, such that $jq_n\rightarrow \frac{\pi}{2}$. Let $x_j\in (-\epsilon_j, \epsilon_j)$ not equal to zero. Since the inclusion $V_j\subset S_j$ holds, we have $|x_j| &amp;lt; |\cos(jq_n)|$ for all $n$. Clearly impossible, since $x_j &gt; 0$. &lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-24T04:03:48.703" CommentCount="1" />
  <row Id="5357" PostTypeId="1" AcceptedAnswerId="5369" CreationDate="2010-09-24T04:55:03.190" Score="2" ViewCount="130" Body="&lt;p&gt;I know that the product rule is generalised by Leibniz's general rule and the chain rule by Faà di Bruno's formula, but what about the quotient rule? Is there a generalisation for it analogous to these? Wikipedia mentions both  Leibniz's general rule and Faà di Bruno's formula for the product and the chain rule, but rather nothing for the quotient rule. &lt;/p&gt;&#xA;" OwnerUserId="2104" LastEditorUserId="2104" LastEditorDisplayName="" LastEditDate="2010-09-24T05:01:11.980" LastActivityDate="2010-09-24T11:45:43.443" Title="What's the generalisation of the quotient rule for higher derivatives?" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="5358" PostTypeId="2" ParentId="5357" CreationDate="2010-09-24T05:00:12.950" Score="1" ViewCount="0" Body="&lt;p&gt;Quotient Rule is actually Product Rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(u/v) = D(uw)$ where $w = 1/v$.&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-24T05:00:12.950" CommentCount="1" />
  <row Id="5359" PostTypeId="2" ParentId="4961" CreationDate="2010-09-24T05:17:34.757" Score="1" ViewCount="0" Body="&lt;p&gt;The useful identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{2} + \cos x + \cos 2x + \cdots + \cos nx = \frac{\sin (n+x/2)}{2 \sin (x/2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which appears in the study of Fourier series is most easily proven by replacing every $\cos(kx)$ on the left-hand side with $$\cos(kx)=\frac{e^{ikx}+e^{-ikx}}{2}$$ and applying the formula for the sum of the geometric series. &lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-09-24T05:17:34.757" CommunityOwnedDate="2010-09-24T05:17:34.757" />
  <row Id="5360" PostTypeId="2" ParentId="5347" CreationDate="2010-09-24T05:36:13.623" Score="3" ViewCount="0" Body="&lt;p&gt;Note that by inverse sampling we can generate exponential r.v. with rate $\lambda$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=F^{-1}(U)$  where $F(x) = 1-e^{-\lambda x}$, is the c.d.f of exponential r.v. with rate $\lambda$ and $U\sim Uni[0,1]$. Thus $F^{-1}(U)=\frac{-\log(1-U)}{\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $1-U\sim Uni[0,1]$ so if $U_i\sim Uni[0,1]$ then  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_i=\frac{-\log(U_i)}{\lambda}$ is exponential with rate $\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_1+\ldots +X_n= -\frac{1}{\lambda}(\log(U_1)+\ldots +\log(U_n)) = -\frac{1}{\lambda}\log(U_1U_2\ldots U_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is $X_1+\ldots + X_n&amp;lt;1$ iff $\prod_{i=1}^nU_i&gt;e^{\lambda}$. Clearly in both cases $N$ is actually the same random variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why reduce the evaulation of special functions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\lambda=10^{-3}$ (small). In the first method, where you generate each $X_i$ in turn add them up and check if totals less than one, you would expect to generate $10^3$ exponentials.  Each one requires you to evaluate the logarithm (with argument in $(0,1)$) and simulate a uniform r.v.. So you have 1 thousand evaluations of a special function, in the region where it is hard to approximate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second method, you expect to generate $10^3$ uniform r.v. and check if the product is greater than $e^{-10^3}$. This method requires you to calculate $e^{-10^3}$ but only once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both methods generate what you desired but the second method will be much more efficient on a computer. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-24T05:36:13.623" />
  <row Id="5361" PostTypeId="2" ParentId="4961" CreationDate="2010-09-24T06:42:40.913" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2AaSE.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem:&#xA;&lt;strong&gt;The line segments joining the centers of opposite squares are perpendicular and of equal length.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $2a,2b,2c,2d$ be complex numbers running along the edges of the quadrilateral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition for quadrilateral to close up is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a + b + c + d = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let origin be at vertex where $2a$ begins &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p = a + ia$ i.e. $a$ + ($a$ rotated though 90 degrees) &lt;br /&gt;&#xA;$ q = 2a + (1+i)b$&lt;br /&gt;&#xA; $r = 2a + 2b + (1+i)c$&lt;br /&gt;&#xA;$ s = 2a + 2b + 2c + (1+i)d$&lt;br /&gt;&lt;br /&gt;&#xA; $A = s-q$  and $B = r - p$ &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we want to show that A and B are perpendicular and equal i.e. $B = iA$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A + iB = (a+b+c+d) + i(a+b+c+d) = 0 $ since $a + b + c + d = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image, theorem, proof and everything is taken from &lt;strong&gt;&lt;a href=&quot;http://usf.usfca.edu/vca//&quot; rel=&quot;nofollow&quot;&gt;Visual Complex Analysis by Tristan Needham.&lt;/a&gt;&lt;/strong&gt;[One of the Best Books. Highly Recommended.]&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-09-24T06:42:40.913" CommunityOwnedDate="2010-09-24T06:42:40.913" CommentCount="2" />
  <row Id="5362" PostTypeId="1" CreationDate="2010-09-24T09:07:45.970" Score="3" ViewCount="97" Body="&lt;p&gt;Let $A$ be a set of infinite cardinals. Assume that for every regular $\lambda$, the subset $A \cap \lambda$ of $\lambda$ is not stationary. Then I want to prove that there is an injective function $g$ on $A$ which satisfies $g(\alpha) &amp;lt; \alpha$ for all $\alpha \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an exercise in Kunen's set theory. I dont' understand if $g$ should be a map from $A \to A$. But then the claim would be false if $A = \{\kappa\}$. But if $g$ may have arbitrary ordinal values, then the claim is somehow trivial, also without the assumptions. So I don't know what this exercise about.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-25T05:10:16.097" Title="set of infinite cardinals admits an injective regressive function" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5363" PostTypeId="1" AcceptedAnswerId="5387" CreationDate="2010-09-24T09:52:58.310" Score="1" ViewCount="228" Body="&lt;p&gt;There is something in the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_product&quot; rel=&quot;nofollow&quot;&gt;free product&lt;/a&gt; of two groups that annoys me, and it's this &quot;word&quot; thing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If G and H are groups, a word in G and H is a product of the form&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;     s_1 s_2 \dots s_m,&#xA;  $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where each $s_i$ is either an element of G or an element of H. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what is this &quot;word&quot; guy? Does it come out of the blue? Does it come from some sort of new operation that I can perform with the two sets $G$ and $H$ -in addition to the well-known ones of union, intersection, Cartesian product...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunatelly, I think there is nothing new under the sun of set operations: it's easy to realise that words can be identified with elements of some Cartesian product (see below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_m )  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Cartesian product is a well-established set-theoretical operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried to translate the rest of Wikipedia's definition &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Such a word may be &lt;em&gt;reduced&lt;/em&gt; using the following operations:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Remove an instance of the identity element (of either G or H).&#xA;  Replace a pair of the form $g_1g_2$ by its product in $G$, or a pair $h_1h_2$ by its product in $H$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Every reduced word is an alternating product of elements of G and elements of H, e.g.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;  g_1 h_1 g_2 h_2 \dots g_r h_r.&#xA;  $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The free product $G ∗ H$ is the group whose elements are the reduced words in G and H, under the operation of concatenation followed by reduction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in an elementary set setting. First, consider the set of &quot;unreduced&quot; tuples of elements of $G$ and $H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U =  G \sqcup H \sqcup (G\times G) \times (G\times H) \sqcup (H\times G) \sqcup (H\times H) \sqcup (G\times G \times G) \sqcup \dots &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concisely:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the following formula may be less messier than the one I wrote previously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U = \bigsqcup_{r \geq 1} (S_1 \times \cdots \times S_r)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S_i = G$ or $S_i = H$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, elements of $U$ are ordered tuples (&lt;em&gt;unreduced&lt;/em&gt; ones)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_m)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each $s_i$ is either an element of $G$ or an element of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product of two unreduced tuples is defined by concatenation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, \dots , s_m) \cdot (t_1,  \dots , t_n) = (s_1, \dots , s_m, t_1 , \dots , t_n) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the following equivalence relation in the set of unreduced tuples $U$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_{i-1}, 1, s_{i+1}, \dots , s_n) \sim (s_1, s_2, \dots, s_{i-1}, s_i, \dots , s_n) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $1$ is either the unit element of $G$ or the one of $H$. And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(s_1, s_2, \dots , s_i,s_{i+1}, \dots , s_r) \sim (s_1, s_2, \dots , s_is_{i+1}, \dots , s_r )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whenever two adjacent $s_i, s_{i+1} \in G$ or $s_i, s_{i+1} \in H$ at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want, you may call the equivalence class of a tuple under this equivalence relation a &lt;em&gt;reduced&lt;/em&gt; tuple. So every reduced tuple is an alternating one, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(g_1, h_1, \dots , g_r , h_r) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $g_i \in G$ and $h_i \in H$ for all $i = 1, \dots , r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the free product of $G$ and $H$ as the quotient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G*H = U/\sim \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, one verifies that concatenation is well-defined on unreduced tuples and gives $G*H$ a group structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After performing this elementary exercise I understand perfectly well why nobody defines the free product in this way, but I still wanted to ask:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this correct?&lt;/li&gt;&#xA;&lt;li&gt;Is it written somewhere?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-26T16:28:17.277" LastActivityDate="2010-09-26T16:28:17.277" Title="Set-theoretical description of the free product?" Tags="&lt;abstract-algebra&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5364" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T10:09:52.663" Score="1" ViewCount="0" Body="&lt;p&gt;Well, I think that the usual definition of the free product is just a slightly less formal version of what you've written, which, by the way, is certainly correct. I haven't seen it completely written out like that before but maybe others have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, one can avoid the whole messy construction and prove (using, for example, a version of the adjoint functor theorem) that the category Grp has small colimits. The free product is just the coproduct in Grp, so that would suffice for all practical purposes, and it is a rigorous formulation, if that's what you're worried about.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-09-24T10:09:52.663" CommentCount="5" />
  <row Id="5365" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T11:19:25.500" Score="1" ViewCount="0" Body="&lt;p&gt;This is not an answer to your questions, but I guess I would point out that the “words” in the definition in Wikipedia can be given a structure: consider the free monoid freely generated by the set &lt;i&gt;G&lt;/i&gt;⊔&lt;i&gt;H&lt;/i&gt;, and just call its elements &lt;em&gt;words in G and H&lt;/em&gt;.  Now you can introduce an equivalence relation and define the free product as the quotient (similarly to what you did), verifying that the operation on the quotient is well-defined and makes it a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is not at all shorter than your written-down definition if we write down the definition of the free monoid and the definition of the equivalence relation.  I just wanted to point out that the “words” are a natural concept and that this two-step construction may be slightly tidier than constructing the free product of groups in one big step (based on sets).&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-24T11:19:25.500" CommentCount="1" />
  <row Id="5366" PostTypeId="1" CreationDate="2010-09-24T11:24:50.330" Score="4" ViewCount="214" Body="&lt;p&gt;I apologise for this question, as it is no doubt very simple, but I've never been very confident with proofs. Our lecturer today (in a course related to maths but not mathematical itself) was playing with doing the modulus of powers, and used the above fact - $(x.y)\; mod\; m\; ==\; ((x\; mod\; m).y)\; mod\; m$ - to show a quick way to do it. She mentioned offhand that the proof was short and easy, and that we should try it. Well, I did. And I'm not really sure that what I've done is mathematically sound. So I'd very much appreciate it if anyone could give me some help with how to do this proof correctly - googling for it turned up many pages which just state it as simple fact. I would not call my proof short or easy, so I reckon I've just got completely the wrong end of the stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When considering any number, denoted n, in regards to another number m, we can write it as $n_b$ + $n_r$, where the former is some multiple of m, and the latter is the remainder when you take n mod m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we expand the original formula, $(x.y)\; mod\; m$ like this, we get $((x_b + x_r)(y_b + y_r))\; mod\; m$. This then expands to $(x_b y_b + x_b y_r + x_r y_b + x_r y_r)\; mod\; m$. The first three terms in the brackets are all some multiple of m, and so when they are taken modulus m they are 0. The final term we do not know, so we are left with: $(x_r y_r)\; mod\; m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can consider the left hand side. $((x\; mod\; m).y)\; mod\; m$ is equivalent to $((x_r).y)\; mod\; m$. Expanding y gives us: $((x_r)(y_b + y_r))\; mod\; m$. Multiplying out then gives $(x_r y_b + x_r y_r)\; mod\; m$. As before, the first term in the brackets are multiples of m and so are 0 when taken mod m, and the last one we do not know, so we are left with: $(x_r y_r)\; mod\; m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both the LHS and RHS boil down to the same thing, the equality holds.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="2108" LastEditorDisplayName="" LastEditDate="2010-09-24T12:17:59.507" LastActivityDate="2011-01-02T23:00:02.660" Title="Very simple question, but what is the proof that x.y mod m == ((x mod m).y) mod m?" Tags="&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="5368" PostTypeId="2" ParentId="5366" CreationDate="2010-09-24T11:43:57.320" Score="2" ViewCount="0" Body="&lt;p&gt;I guess you already know how to prove that (&lt;i&gt;s&lt;/i&gt;+&lt;i&gt;t&lt;/i&gt;) mod &lt;i&gt;m&lt;/i&gt; = ((&lt;i&gt;s&lt;/i&gt; mod &lt;i&gt;m&lt;/i&gt;)+(&lt;i&gt;t&lt;/i&gt; mod &lt;i&gt;m&lt;/i&gt;)) mod &lt;i&gt;m&lt;/i&gt;.  If not, you should first prove it, because you are using it in your proof.  Other than this, your proof is correct except for a tiny point that you have not stated what &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;b&lt;/sub&gt; and &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;r&lt;/sub&gt; are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I call your proof easy, but slightly verbose.  Most importantly, you do not have to break &lt;i&gt;y&lt;/i&gt; into &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;b&lt;/sub&gt; and &lt;i&gt;y&lt;/i&gt;&lt;sub&gt;r&lt;/sub&gt;.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-09-24T11:43:57.320" CommentCount="4" />
  <row Id="5369" PostTypeId="2" ParentId="5357" CreationDate="2010-09-24T11:45:43.443" Score="1" ViewCount="0" Body="&lt;p&gt;As others have already said, you just apply the product rule to $f.g^{-1}.$ However, the is an&#xA;&lt;a href=&quot;http://www.jstor.org/pss/2324425&quot; rel=&quot;nofollow&quot;&gt;American Mathematical Monthly article&lt;/a&gt; on how NOT to do it, which you may find instructive.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-24T11:45:43.443" />
  <row Id="5370" PostTypeId="1" AcceptedAnswerId="5372" CreationDate="2010-09-24T11:48:43.340" Score="4" ViewCount="217" Body="&lt;p&gt;Please suggest a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:13:55.503" LastActivityDate="2010-11-11T15:13:55.503" Title="How many number of three digit numbers lying between 100 and 999 (inclusive) and having only two consecutive digits identical ?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5371" PostTypeId="2" ParentId="3710" CreationDate="2010-09-24T11:58:01.933" Score="2" ViewCount="0" Body="&lt;p&gt;This seems simple : The digits will be from the set {0,1,2,3,4,5,6,7,8,9} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of possibility for first digit (Most significant digit) = &lt;strong&gt;9&lt;/strong&gt; (since we can't use 0)&#xA;Number of possibility for second digit = &lt;strong&gt;9&lt;/strong&gt; &#xA;Number of possibility for third digit  = &lt;strong&gt;8&lt;/strong&gt;&#xA;...&#xA;Number of possibility for sixth digit (Least significant digit)  = &lt;strong&gt;5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence total possibility = &lt;strong&gt;9*9*8*7*6*5&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-09-24T11:58:01.933" />
  <row Id="5372" PostTypeId="2" ParentId="5370" CreationDate="2010-09-24T12:08:32.873" Score="4" ViewCount="0" Body="&lt;p&gt;Ok. Ask yourself first how many contain 00. Next, how many contain 11 (it's 17, you work it out). Then consider the number that contain 22, etc. You will get 9+17*9=162. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-24T12:08:32.873" CommentCount="1" />
  <row Id="5373" PostTypeId="2" ParentId="2293" CreationDate="2010-09-24T12:10:49.987" Score="1" ViewCount="0" Body="&lt;p&gt;So I just stumbled across a very cool, easy to read paper by Eckmann called &quot;Social Choice and Topology A Case of Pure and Applied Mathematics.&quot; It's a survey of a result that Eckmann proved a long time ago about spaces that have &quot;n-means,&quot; and the application of this result to social theory. Basically, if you have some function that assigns a single preference when given the preferences of a population, and this function is &quot;continuous&quot; and fair (you can switch the order of the population and it won't change the output, and if everyone has the same preference then that's what you get out), then this says a LOT about the space of preferences. Given a few assumptions about the space of preferences (I dunno how reasonable they are or realistic, since I have no idea what it would mean for a space of preferences to be a cell complex...), one can show right off the bat that the fundamental group is abelian (we have an H-space structure), and later on he basically classifies all possible spaces that admit social choice functions. Anyway, I thought this was pretty cool as an example of beautiful mathematics and a quirky &quot;application.&quot;&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-09-24T12:10:49.987" CommunityOwnedDate="2010-09-24T12:10:49.987" CommentCount="1" />
  <row Id="5374" PostTypeId="2" ParentId="4626" CreationDate="2010-09-24T12:53:21.430" Score="1" ViewCount="0" Body="&lt;p&gt;Answer to Question 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &quot;Die Transformation von Bauer und Muir&quot;, §7 of Oskar Perron, &lt;em&gt;Die Lehre von den Kettenbrüchen&lt;/em&gt;, Band II, 1957.&lt;/p&gt;&#xA;" OwnerUserId="2044" LastActivityDate="2010-09-24T12:53:21.430" CommunityOwnedDate="2010-09-24T12:53:21.430" CommentCount="1" />
  <row Id="5375" PostTypeId="2" ParentId="2293" CreationDate="2010-09-24T13:29:51.577" Score="4" ViewCount="0" Body="&lt;p&gt;This isn't a specific application but an area of application.  Robert Ghrist and others use algebraic topology as a way of integrating local data about sensor networks into global information.  For example, you may want to determine whether there are any holes in your sensor coverage.  Here's an &lt;a href=&quot;http://www.johndcook.com/blog/2010/09/13/applied-topology-and-dante-an-interview-with-robert-ghrist/&quot; rel=&quot;nofollow&quot;&gt;interview&lt;/a&gt; with Ghrist where he elaborates on this idea.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-24T13:29:51.577" CommunityOwnedDate="2010-09-24T13:29:51.577" />
  <row Id="5376" PostTypeId="2" ParentId="5366" CreationDate="2010-09-24T14:07:04.347" Score="4" ViewCount="0" Body="&lt;p&gt;Here's the &lt;strong&gt;product rule&lt;/strong&gt; proof expressed in both divisibility form and congruence form, using the standard notation: $\rm\ \ \ a\ |\ b \ :=\ a\:$ divides $\rm\: b\:,\;$ and $\rm\ \; a\equiv b\ \ (mod\ m)\: \iff\: m\:|\:a-b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\: m\ |\ X-x\ $ and $\rm\ m\:|\:Y-y \ \Rightarrow\ m\:|\:XY - \: xy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\quad\ \ \ \  m\:|\:(X-x)\:y\ +\ X\:(Y-y) \ \ \ =\ \ \ XY - \: xy \quad$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\quad\quad\ \: X\equiv x\quad\:$ and $\rm\quad\: Y\equiv y \ \ \ \ \Rightarrow\:\ \ XY \equiv xy\ \ \ (mod\ m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: $\rm\ \ \ 0 \ \equiv\ (X-x)\:y\ +\ X\:(Y-y) \ \ \: \equiv\ \ \ XY - \: xy \ \ \ (mod\ m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the congruence notation eliminates the cumbersome manipulation of relations (divisibility). Indeed, the relations are replaced by a generalized equality (congruence) which, being compatible with multiplication (as above) and addition (similar proof), enables us to exploit our well-honed intuition manipulating integer equations - which immediately generalizes to manipulating congruences (mod m). When you study abstract algebra you'll learn that this is a very special case of a quotient or residue ring. This product rule arises in many analogous contexts, e.g. see my post on the &lt;a href=&quot;http://math.stackexchange.com/questions/3178&quot;&gt;product rule for derivatives&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-02T23:00:02.660" LastActivityDate="2011-01-02T23:00:02.660" CommentCount="3" />
  <row Id="5377" PostTypeId="1" AcceptedAnswerId="5405" CreationDate="2010-09-24T14:24:55.983" Score="6" ViewCount="556" Body="&lt;p&gt;I came across the following &quot;proof&quot; for the inconsistency of ZFC and can't find the flaw in it (if there is one...):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct a Turing machine A which sequentially runs on all proofs in ZFC and checks if the claim &quot;A does not halt or ZFC is inconsistent&quot; is proved. If such a proof is found, A halts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that A can know its encoding - see Kleene's recursion theorem - and even if it can't this &quot;obstacle&quot; can be overcome. So it seems to me such A is constructible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if A halts then it is because it has found a proof for &quot;A does not halt or ZFC is inconsistent&quot;. Since A halted, it is implied that ZFC is inconsistent. So either A does not halt, or ZFC is inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (and my guess is that this is the part where this proof &quot;cheats&quot;) we have just &lt;strong&gt;proved&lt;/strong&gt; the claim &quot;A does not halt or ZFC is inconsistent&quot;, so such a proof exists in ZFC. So A must halt on this proof - and we have seen that this implies that ZFC is inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the mistake?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1818" LastEditorDisplayName="" LastEditDate="2010-09-24T14:37:39.790" LastActivityDate="2010-12-24T03:53:16.367" Title="&quot;Proof&quot; that ZFC is inconsistent using Turing machines" Tags="&lt;logic&gt;&lt;fake-proofs&gt;&lt;computability&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="1" />
  <row Id="5378" PostTypeId="1" AcceptedAnswerId="5379" CreationDate="2010-09-24T14:30:15.703" Score="2" ViewCount="248" Body="&lt;p&gt;Hi,&lt;br&gt;&#xA;I understand that there are different types of infinity: one can (even intuitively) understand that the infinity of the reals is different from the infinity of the natural numbers. Or that the infinity of the even numbers is the same as that of the natural numbers. How many types of infinity are there? Or is this infinite itself?&lt;br&gt;&#xA;TIA&lt;br&gt;&#xA;Steven&lt;/p&gt;&#xA;" OwnerUserId="923" LastActivityDate="2010-09-25T22:14:30.547" Title="Types of infinity" Tags="&lt;infinity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="5379" PostTypeId="2" ParentId="5378" CreationDate="2010-09-24T14:33:04.053" Score="3" ViewCount="0" Body="&lt;p&gt;Not only infinite - it's &quot;so big&quot; that there is no infinite set so large as the collection of all types of infinity (in Set Theoretic terms, the collection of all types of infinity is a class, not a set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily see that there are infinite types of infinity via Cantor's theorem which shows that given a set A, its power set P(A) is strictly larger in terms of infinite size (the technical term is &quot;cardinality&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cantor%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-24T14:33:04.053" CommentCount="1" />
  <row Id="5380" PostTypeId="2" ParentId="5378" CreationDate="2010-09-24T14:40:01.110" Score="1" ViewCount="0" Body="&lt;p&gt;well, there is no definite answer, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some mathematicians contend that the concept of (actual) infinite itself is ill-defined, and you may have only a potential infinite, the one you have when you say that a straight line segment may be extended at will, but you do not ave a whole straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you accept Cantor theory, the cardinal numbers (the ones which count how many objects there are in a set) are infinite; you start from &amp;alefsym;&lt;sub&gt;0&lt;/sub&gt; which corresponds to the integers, and go on. With the usual axioms, you cannot decide whether there are or not infinite cardinal numbers strictly greater than &amp;alefsym;&lt;sub&gt;0&lt;/sub&gt; and strictly less than the cardinality of the real numbers, but anyway the infinite numbers are infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also consider ordinal numbers (the ones which consider the ordered elements of a set); they are many more than the cardinal numbers ;-)&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-24T14:40:01.110" />
  <row Id="5381" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T14:43:55.670" Score="1" ViewCount="0" Body="&lt;p&gt;In the question above, it seems that the description is far too complicated as it actually is. I will elaborate on the construction of coproducts of groups or monoids a bit. This will go beyond the scope of the specific question, but I hope it's helpful in other situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First assume $M,N$ are sets. What is their coproduct? Obviously it's the disjoint union, which I will denote by $M + N$. Now assume $M,N$ are monoids. What is their coproduct? Well we have to introduce products of elements in $M + N$. Thus consider $(M + N)^{&amp;lt;\omega}$, the set of finite sequences with entries in $M + N$. This is a monoid by concatenation, but the inclusions $M,N \to (M + N)^{&amp;lt;\omega}$ are no homomorphisms. Well, then let's force it! Mod out the smallest congruence relation $\sim$ on our monoid, which satisfies $1 \sim (1_M), (m,m') \sim (mm')$ for all $m,m' \in M$ and similarily for elements of $N$. Then the quotient $(M + N)^{&amp;lt;\omega} / \sim$ is obviously a monoid (since $\sim$ is a congruence relation), and the universal property of the coproduct is also verified easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what about groups? The correct definition of a group involves the operations of the underlying monoid and the inversion. But in practice, this inversion can be reconstructed from the rest of the data ($x=y^{-1}$ iff $xy=1$), and actually the category of groups is a full subcategory of the category of monoids. If the monoids $M,N$ above happen to be groups, their coproduct turns out to be a group, and clearly the universal property then also holds with respect to groups. Thus we have constructed the coproduct in the category of groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's turn back to monoids. The above construction of $M \coprod N$ is rather general, it shows the existence of colimits in every finitary algebraic category, but what can be said about the &lt;em&gt;elements&lt;/em&gt;? Do they have a canonical representation? Now there are two ways of doing it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) The elegant, short, &quot;geometric&quot; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The long, tedious one. This one is preferred in textbooks ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, b) means that you write down the set of reduced words, endow it with a terribly complicated monoid structure, show that every monoid axiom is satisfied, and finally check the universal property. Good luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a) you just use that $M \coprod N$ exists. We have constructed it, but the construction does not answer such simple questions as: Do $M$ and $N$ intersect in $M \coprod N$ trivially? Therefore we just use the existence of the coproduct together with the structure maps $i : M \to M \coprod N, j : N \to M \coprod N$. The idea is now to define an action of $M \coprod N$ on another object. This may be a geometric one, but in our case it's our desired set of reduced words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observe that the elements in $M \coprod N$, which are products of elements, which are in the image of $i$ or $j$, constitute a submonoid, which verifies the same universal property. In other words, every element in $M \coprod N$ is such a product. In such a product we may replace $i(m) i(m')$ by $i(mm')$ and cancel $i(1)$, similarily for $j$. Thus every element is in the image of the canonical map $X \to M \coprod N$, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X := \{(... ,m_1,n_1,m_2,n_2, ...)  : m_i \in M - \{1\}, n_i \in N - \{1\}\}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we prove that this map $X \to M \coprod N$ is a bijection, i.e. every element of the coproduct has a &lt;em&gt;unique&lt;/em&gt; representation as $... i(m_1) j(n_1) i(m_2) ...$. To do this, we define an action of the monoid $M \coprod N$ on $X$, i.e. a monoid homomorphism $M \coprod N \to Sym(X), (m \mapsto (x \mapsto mx))$, which should imitate the usual multiplication. By the universal property, it is enough to construct this homomorphism on $M$ and on $N$. If $m \in M$ and $x \in X$, define $mx$ as follows: If $x$ starts with an element in $N$, just concatenate with $(m)$. If $x$ starts with an element of $M$, which is not inverse to $m$, then multiply $m$ in the first entry. Otherwise delete the first entry. Similarily the homomorphism $N \to Sym(X)$ is defined. The resultung homomorphism $M \coprod N \to Sym(X)$ can be composed with the evaluation at the empty sequence to get a map $M \coprod N \to X$, which turns out to be a left inverse to $X \to M \coprod N$. Thus, $X \to M \coprod N$ is a bijection.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-09-24T14:51:02.093" LastActivityDate="2010-09-24T14:51:02.093" />
  <row Id="5382" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T14:49:06.107" Score="3" ViewCount="0" Body="&lt;p&gt;You've just proved that &quot;either A does not halt or ZFC is inconsistent&quot; must be true, not that this statement is provable in ZFC. There exist unprovable true statements in ZFC (if it is consistent). See Gödel's first incompleteness theorem.&lt;/p&gt;&#xA;" OwnerUserId="519" LastActivityDate="2010-09-24T14:49:06.107" CommentCount="3" />
  <row Id="5383" PostTypeId="1" AcceptedAnswerId="5385" CreationDate="2010-09-24T14:54:40.403" Score="1" ViewCount="84" Body="&lt;p&gt;Consider the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XsLEN.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got $g'(1) = f'(1) = 2$, which is correct according to the expected answer. However, I get that $a$ can be any value, as it doesn't affect anything and $g$ is differentiable no matter what $a$ is (am I missing something here?). The expected answer is that $a$ must be $2$ however. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain why $a$ needs to be $2$?&lt;/p&gt;&#xA;" OwnerUserId="456" LastActivityDate="2010-09-24T14:57:45.340" Title="Conditions for differentiability" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="5384" PostTypeId="2" ParentId="5383" CreationDate="2010-09-24T14:56:42.663" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;: if $g$ is supposed to be differentiable at $x=1$, the left and right derivatives of $g$ should be identical. Now, find the value of $a$ that would make it so.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-24T14:56:42.663" />
  <row Id="5385" PostTypeId="2" ParentId="5383" CreationDate="2010-09-24T14:57:45.340" Score="2" ViewCount="0" Body="&lt;p&gt;The derivative of g from the right at $x=1$ is a, and from the left it's $f^\prime(1) = 2$. They must be equal.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-24T14:57:45.340" CommentCount="2" />
  <row Id="5386" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T15:23:19.603" Score="2" ViewCount="0" Body="&lt;p&gt;For such questions, it is hard to get an intuitive understanding. I think the best way to intuitively understand it is by formalizing it using model-theory. So lets look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you define your A as a set in ZFC, using an $\cal{L}_1$-Formula, and denote the property of &quot;halting&quot; as $\cal{H}$, which can also be expressed in ZFC. So by your assumption, you have $ZFC\models{\cal H}(A)\leftrightarrow(ZFC\models\bot \vee ZFC\models\lnot{\cal H}(A))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little more theory, you can imply therefore $ZFC+{\cal H}(A)\models (ZFC\models\bot\vee ZFC\models\lnot{\cal H}(A))$, and therefore, $ZFC+{\cal H}(A)\models (ZFC\models\bot)$. That is your one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other direction goes with $ZFC+\lnot{\cal H}(A)\models \lnot (ZFC\models\bot \vee ZFC\models\lnot{\cal H}(A))$ (which follows from the first assumption), i.e. $ZFC+\lnot{\cal H}(A)\models (ZFC\not\models\bot) \wedge (ZFC\not\models\lnot{\cal H}(A))$. That is, in every model of ZFC, in which $\lnot{\cal H}(A)$ holds, every &quot;sub-model&quot; which can be created doint meta-theory inside that one, does neither model $\bot$ nor model $\lnot{\cal H}(A)$.&lt;/p&gt;&#xA;" OwnerUserId="1858" LastActivityDate="2010-09-24T15:23:19.603" />
  <row Id="5387" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T15:35:56.497" Score="4" ViewCount="0" Body="&lt;p&gt;In answer to your question about whether this is written somewhere: the construction of free products in my book &lt;em&gt;Introduction to Topological Manifolds&lt;/em&gt; proceeds very much along these lines. (I'm not the only one who's written it down, but I don't have other references handy.) I don't go through the set-theoretic description of the set of ordered tuples in as much detail as you do, but it can be described more simply as just the disjoint union of the sets $(G\amalg H)^n$ over all n.  I like the concreteness of this construction as a way of giving one a handle on what the free product looks like.  Once the group is constructed, it's a relatively easy matter to prove after the fact that it is uniquely determined up to isomorphism as the coproduct in the category of groups.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-24T15:35:56.497" CommentCount="7" />
  <row Id="5388" PostTypeId="2" ParentId="5370" CreationDate="2010-09-24T15:42:54.137" Score="6" ViewCount="0" Body="&lt;p&gt;In your comment, you said that the given solution was 9 &amp;bull; 9 + 1 &amp;bull; 9 + 8 &amp;bull; 9 = 162.  I'll attempt to explain a logic that yields that calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the 3-digit numbers that start with two identical digits.  There are 9 choices of the first digit (and inherently the second digit): 11, 22, 33, 44, 55, 66, 77, 88, and 99 (not 00 because the number is in the range 100-999).  For each of these, there are 9 choices of the final digit (0-9, except whatever digit was already chosen for the first two).  So, there are 9 &amp;bull; 9 such numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that the number ends with two identical digits.  There are 10 choices for the last digit (and inherently the second-to-last digit): 00, 11, 22, 33, 44, 55, 66, 77, 88, and 99, but we need to treat 00 separately from the rest.  If the number ends with 00, then the first digit can be 1-9, so 9 choices, so 1 &amp;bull; 9.  If the number ends with 11-99, there are 8 choices of first digit (1-9 except the digit already chosen), so 9 &amp;bull; 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I have the 8 and 9 transposed in the final term, this is term-by-term the same expression as in the solution you gave.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-24T15:42:54.137" CommentCount="1" />
  <row Id="5389" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T15:52:58.030" Score="3" ViewCount="0" Body="&lt;p&gt;You might find &lt;a href=&quot;http://www.scottaaronson.com/democritus/lec3.html&quot; rel=&quot;nofollow&quot;&gt;Scott Aaronson's lecture&lt;/a&gt; on this useful. Search for the phrase &quot;Talk to the axioms!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, suppose ZFC is consistent. I'll work with Alex's simplification, where $A$ searches for a ZFC proof of &quot;A does not halt.&quot; Then there is a model $M$ of ZFC+&quot;$A$ halts&quot;. In $M$, there is a set $S$ which satisfies the whatever sentence of ZFC you have cooked up to encode &quot;a halting sequence of states for $A$&quot;. There is a set $P$ which satisfies the first order sentence you have cooked up to encode &quot;a ZFC proof that $A$ does not halt.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of the model $M$, there is no contradiction. The model sees that $P$ encodes a ZFC proof that $A$ doesn't halt, the model sees that $A$ does halt, so the model believes that ZFC is inconsistent. Since Con(ZFC) is not an axiom of ZFC, the model sees nothing wrong with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an outside perspective, $S$ does not actually look like a sequence of states for the Turing machine $A$, and $P$ does not look like a proof. They are just objects of the model which the model thinks have those features, because it has some strange interpretation of set theory. Even if you could construct $M$ (and I think there are obstacles to $M$ being computable in any reasonable way), you could not convert $P$ into an actual ZFC proof that $A$ does not halt, nor could you convert $S$ into a halting sequence for $A$. Any attempt to convert them into an actual proof or actual halting sequence will run into the same difficulty Scott describes in trying to extract a proof of NOT(Con(PA)) from a model of PA+NOT(Con(PA))&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-09-24T15:52:58.030" CommentCount="0" />
  <row Id="5390" PostTypeId="2" ParentId="5378" CreationDate="2010-09-24T15:58:03.450" Score="3" ViewCount="0" Body="&lt;p&gt;A very nice introduction to the many different notions of infinity in mathematics is Rudy Rucker's book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691001723&quot; rel=&quot;nofollow&quot;&gt;Infinity and the Mind&lt;/a&gt;. Unlike many other popularizations, this is written by someone who did a Ph.D. on the topic. Moreover, Rucker has gone to great lengths to make the presentation faithful to the mathematics but still accessible to an educated layperson. Below is an excerpt on the Alephs.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;stevenh: your recent comment &lt;a href=&quot;http://math.stackexchange.com/questions/201/is-there-possibly-a-largest-prime-number/2653#2653&quot;&gt;to this answer&lt;/a&gt; is based upon a misunderstanding. It is not only correct, but also beautiful, to reformulate Euclid's proof in the form: $\rm N(N+1)$ has more prime factors than does $\rm N$. For further details see my comments there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/z7owE.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/kcYlt.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/zfeyJ.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/kH2we.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-25T22:14:30.547" LastActivityDate="2010-09-25T22:14:30.547" />
  <row Id="5391" PostTypeId="1" AcceptedAnswerId="5414" CreationDate="2010-09-24T16:06:53.790" Score="3" ViewCount="112" Body="&lt;p&gt;If we have a covering $p:H\rightarrow G$ where $G$ is a topological group, then $H$ is also a topological group. The multiplication function can be defined as follows. Consider the map $f:H\times H \rightarrow G$ which is a composition of the map $p\times p$ and the multiplication function on $G$. Choose $h\in p^{-1}(e)$ where $e$ is the identity element of $G$. If $$f_* (\pi_1(H\times H,(h,h))) \subset p_*(\pi_1(H,h)),$$ then $f$ can be lifted to a map $g:H\times H \rightarrow H$  such that $p\circ g = f$ and $g(h,h) = h$. Suppose we have shown the &quot;if&quot; part, then $g$ should function as our multiplication map on $H$. But given any $x\in H$, why do we know that $g(x,h) = x$ and that $g(x,h)$ does &lt;strong&gt;not&lt;/strong&gt; equal any other element of $p^{-1}(p(x))$?&lt;/p&gt;&#xA;" OwnerUserId="1907" LastEditorUserId="1907" LastEditorDisplayName="" LastEditDate="2010-09-24T19:01:34.190" LastActivityDate="2010-09-25T01:43:46.533" Title="Covering of a topological group is a topological group" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="5392" PostTypeId="1" CreationDate="2010-09-24T17:14:00.647" Score="5" ViewCount="205" Body="&lt;p&gt;While I understand the intuition behind expectation, I don't really understand the meaning of variance and moment. &#xA;What is a good way to think of those two terms?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-09-25T23:22:57.000" Title="Intuitive explanation of variance and moment in Probability." Tags="&lt;probability&gt;&lt;intuition&gt;&lt;probability-theory&gt;" AnswerCount="4" FavoriteCount="4" />
  <row Id="5393" PostTypeId="2" ParentId="5392" CreationDate="2010-09-24T17:17:39.800" Score="2" ViewCount="0" Body="&lt;p&gt;The variance is the expected squared deviation between the random variable and its expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It measures how much the random variable scatters or spreads around its expectation.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-24T17:17:39.800" CommentCount="3" />
  <row Id="5394" PostTypeId="2" ParentId="5363" CreationDate="2010-09-24T17:21:45.837" Score="3" ViewCount="0" Body="&lt;p&gt;You can see essentially the same construction in two different ways in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;George Bergman's &lt;em&gt;An Invitation to General Algebra and Universal Constructions&lt;/em&gt;&lt;/a&gt; in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/Ch.2.ps&quot; rel=&quot;nofollow&quot;&gt;Chapter 2&lt;/a&gt; (link is to a postscript file) for the free group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you define &lt;em&gt;&quot;the set of all terms in the elements of the set $X$ under the formal group operations $\mu$, $i$, $e$&quot;&lt;/em&gt; to mean a set which is given with functions symb${}_T\colon X\to T$, $\mu_T\colon T^2\to T$, $i_T\colon T\to T$, and $e_T\colon T^0\to T$, such that each of these maps is one-to-one, their images are disjoint, and $T$ is the union of the images, and $T$ is generated by symb${}_T(X)$ under the operations $\mu_T$, $i_T$, and $e_T$. Such a set exists (it can be constructed inductively with enough care; given in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/Ch.1.ps&quot; rel=&quot;nofollow&quot;&gt;Chapter 1&lt;/a&gt; of the same notes). Then one defines an apropriate equivalence relation $\sim$ on $T$; the set $T/\sim$ gives the underlying set of the free group, and one defines the operations in the free group via representatives in the natural way. Bergman labels this &quot;the logician's approach&quot; (section 2.2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative construction (&quot;the classical construction&quot;, section 2.4) gives &quot;free groups as groups of words&quot;. Again, you start with a set $X$, and let $T$ be the set of all group-theoretic terms of $X$; identify $X$ with its image under symb, and one defines a subset $T_{red}$ of &quot;reduced terms&quot; (defining what this means appropriately) and then defining operations $\otimes$, ${}^{(-)}$, and $e_T$ on this set to make it into a group. Proving it is a group can be done either in the straightforward but tedious way, or by using &quot;van der Waerden's trick&quot; (embed the set $T_{red}$ into a group of permutations, and check that the operations you defined correspond to the operations in the image, so that &quot;group&quot;-ness gets inherited). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the free product, you let $X$ be the disjoint union of the underlying sets of $G$ and $H$, and either adds to the equivalence relation (in the &quot;logician's approach&quot;), or restricts the definition of &quot;reduced words&quot; (in the &quot;classical approach&quot;), in essentially the way you did. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-24T17:21:45.837" CommentCount="1" />
  <row Id="5396" PostTypeId="1" CreationDate="2010-09-24T17:36:48.840" Score="1" ViewCount="79" Body="&lt;p&gt;$v^i$ be a vector on a Riemannian 3-manifold with metric $g_{ij}$ embedded inside a 3+1 space-time such that for some constant $N_M$ it satisfies the inequality $g_{ij}v^iv^j \leq  N_M ^2$. Let $K$ be a symmetric rank-2 tensor on the 3-manifold. Then apparently the following holds,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vert K_{ij} v^i v^j \vert \leq \vert K \vert _g N_M ^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like some sort of a Cauchy-Schwartz inequality but given that $K$ is a tensor as described, I don't understand what the notation on the RHS means. For a rank-2 symmetric tensor $K$ what does $\vert K \vert _g$ mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one knows that for some function $N$, $g_{ij}v^iv^j \leq  N^2$ where the function $N$ is itself bounded between constants as, $N_m \leq N \leq N_M$ then using inequalities like the above one can apparently show the following bound,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int _{t_1} ^t \frac{1}{N} (-v^i \partial &lt;em&gt;i N - \frac{dN}{dt} + K&lt;/em&gt;{ij}v^iv^j) dt \leq -2log N_m + \frac{1}{N_m} \int _{t_1} ^t (\vert \nabla N \vert _g N_M + \vert K \vert _g N_M ^2 )dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(for some fixed $t_1$ and $t$ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand that first &quot;log&quot; term in the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also once the above bound is shown does it follow that the integral can be unbounded above or below depending solely on the property of the function $N$ ? If yes then what would be needed of $N$ to make the integral unbounded above or below? &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-09-24T17:59:27.037" LastActivityDate="2010-09-24T17:59:27.037" Title="Modulus of a tensor? " Tags="&lt;differential-geometry&gt;" CommentCount="4" />
  <row Id="5397" PostTypeId="1" AcceptedAnswerId="5401" CreationDate="2010-09-24T17:46:26.540" Score="0" ViewCount="146" Body="&lt;p&gt;How do I find a (longitude, latitude) point (any will do) on a circle where the only info I have is a (longitude, latitude) center point, and a radius measured in Feet (ft.)?&lt;/p&gt;&#xA;" OwnerUserId="2113" LastEditorUserId="2113" LastEditorDisplayName="" LastEditDate="2010-09-24T18:33:15.820" LastActivityDate="2010-09-24T18:33:15.820" Title="How to find a (longitude, latitude) point on a circle when given only the center (longitude, latitude) point and radius measured in Feet" Tags="&lt;circle&gt;&lt;geodesy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="5398" PostTypeId="2" ParentId="5392" CreationDate="2010-09-24T17:47:12.987" Score="4" ViewCount="0" Body="&lt;p&gt;Variance of a r.v. $X$ is $Var(X) = \mathbb{E}[(X-\mathbb{E}(X))^2]$.  Variance is used as a measure for the dispersion of a random variable. If you are interested in how far a random variable is from what you expect; perhaps you would consider $\mathbb{E}(|X-\mathbb{E}(X)|)$ but see: &lt;a href=&quot;http://stats.stackexchange.com/questions/118/standard-deviation-why-square-the-difference-instead-of-taking-the-absolute-val&quot;&gt;http://stats.stackexchange.com/questions/118/standard-deviation-why-square-the-difference-instead-of-taking-the-absolute-val&lt;/a&gt; for why this we use the square instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all except the first moment (i.e. the mean) the central moments (centred about the mean) are far more interesting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the $r$th (central) moment of $X$, that is $\mathbb{E}[(X-\mathbb{E}(X))^r]$.  The lower moments ($r&amp;lt;5$) are related to properties of the distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=0$ is $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=1$ is $0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=2$ is the variance which is a measure of the spread from the mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=3$ is related to skewness which is a measure of the asymmetry of the distribution. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Skewness&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Skewness&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r=4$ is related to kurtois which is a measure of the 'peakedness' of a distribution. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Kurtosis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kurtosis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $r\geq5$ they become slightly harder to compute. The $5th$ central moment is a measure for the asymmetry of the distributions tails, but I would use skewness instead. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-24T19:08:56.037" LastActivityDate="2010-09-24T19:08:56.037" CommentCount="5" />
  <row Id="5399" PostTypeId="1" AcceptedAnswerId="5403" CreationDate="2010-09-24T17:55:20.583" Score="2" ViewCount="118" Body="&lt;p&gt;I'm trying to understand how the performance of financial investments are measuring when the asset has multiple points of investment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say I invest &lt;code&gt;$1000&lt;/code&gt; in a stock. After 6 months, it's worth &lt;code&gt;$1200&lt;/code&gt;. At this point, I invest an additional &lt;code&gt;$2000&lt;/code&gt;, for a total value of &lt;code&gt;$3200&lt;/code&gt;. If after an additional 6 months, the entire investment is worth &lt;code&gt;$4000&lt;/code&gt;, how would I calculate the overall yield or effective interest rate? Normally, you'd do (final-initial)/initial, but in this example there are multiple initial values, but only one final. How would I reconcile this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Note, some values are supposed to be dollar amounts, but for some reason the site's markup is interpreting the dollar sign as the start of italics...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Escaped dollar amounts (thanks KennyTM). &lt;/p&gt;&#xA;" OwnerUserId="2114" LastEditorUserId="2114" LastEditorDisplayName="" LastEditDate="2010-09-24T19:06:41.017" LastActivityDate="2010-09-25T15:09:11.527" Title="Measuring Financial Investment Performance" Tags="&lt;finance&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="5400" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T17:57:12.420" Score="0" ViewCount="0" Body="&lt;p&gt;The problem is that if the machine manages to prove it - it halts. and therefore ZFC inconsistent.&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2010-09-24T17:57:12.420" CommentCount="1" />
  <row Id="5401" PostTypeId="2" ParentId="5397" CreationDate="2010-09-24T18:24:25.353" Score="4" ViewCount="0" Body="&lt;p&gt;Assuming you are using a spherical model of the earth, a simple way is to proceed due north or due south, for then each degree of latitude has a constant length (with value extremely close to 10000000/90 meters, because the French revolutionaries originally defined the meter as 10^-7 times the distance from the equator to the North Pole along the Paris meridian!).  For large radii or points close to poles you will have to adjust the latitude appropriately to account for passing over the pole (and add 180 degrees to the longitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, starting at (lon, lat) = (-90, 40) with a radius of 10,000 feet, you would convert that to 10,000 / (39.37/12) = 3048.006 meters, then convert that to 3048.006 * 90 / 10000000 = 0.0274321 degrees, landing you at (-90, 40.0274321).  (Notice I write the longitude first, because it's the horizontal coordinate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more accuracy, &lt;em&gt;don't&lt;/em&gt; look up the approximate spherical radius of the earth: use the geoid appropriate for the datum of your coordinates.  Appropriate formulae appear in books on geodesy and earth projections.  A good resource is the USGS Professional Paper 1395, &lt;em&gt;Map Projections--A Working Manual&lt;/em&gt;, by John Snyder.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-24T18:24:25.353" />
  <row Id="5402" PostTypeId="2" ParentId="5331" CreationDate="2010-09-24T18:26:14.623" Score="1" ViewCount="0" Body="&lt;p&gt;Let $c_n$ be a sequence dense in $S^1$ and define $a_n = \left ( 1 + \frac{1}{n^2} \right ) c_n$. Then $1 - |z_n| = \frac{1}{n^2}$ so this are the zeros of a Blaschke product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $c_n = r e^{i \phi_n}$ and fill this in in the Blaschke product. Then we if $B_n(z,r)$ is the term inside the Blaschke product, then we can try to evaluate if $\sum 1 - B_n(z,r)$ converges. We can evaluate the convergence of this with the integral test and then we see the limit does not exist as $r \to 1^{-}$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-26T20:16:59.990" LastActivityDate="2010-09-26T20:16:59.990" />
  <row Id="5403" PostTypeId="2" ParentId="5399" CreationDate="2010-09-24T18:41:24.773" Score="2" ViewCount="0" Body="&lt;p&gt;Excel has a function for this called IRR.  To roll your own you just need to find the rate of return that produces the final value.  You have investments i1, i2, ... that have been in the account for m1, m2, ... months.  For a given return per month you should have i1*(1+r)^m1 + i2*(1+r)^m2 + ...  Now just search for r that makes this equal to the final value.  For some patterns of cash flow, there may be multiple values of r that satisfy the equation.  You can use your favorite root finder, but bisection is quite robust.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-24T18:41:24.773" CommentCount="4" />
  <row Id="5404" PostTypeId="2" ParentId="5392" CreationDate="2010-09-24T18:43:48.710" Score="1" ViewCount="0" Body="&lt;p&gt;There are many good answers possible, depending on how one's intuition works and one's experience with related things like averages, $L^p$ norms, etc.  One answer that builds on your intuition of an expectation is to think of the higher central moments as &lt;em&gt;weighted&lt;/em&gt; expectations.  The central third moment, for example, is defined as the expectation of the third powers of deviations from the mean.  Split a third power into two parts: one is the deviation from the mean (let's call it, after common statistical parlance, the &quot;residual&quot;) and the other is the squared deviation from the mean (&lt;em&gt;ergo&lt;/em&gt;, the squared residual).  Treat the latter as a weight.  As the residual gets larger in size, these weights get larger (in a quadratic fashion).  The third central moment is the weighted &quot;average&quot; of the residuals.  (I have to put &quot;average&quot; in quotes because, strictly speaking, to deserve being called an average one would require the weights themselves to average out to 1, but they don't usually do that.)  Thus, the third central moment gives disproportionately greater emphasis to larger residuals compared to smaller ones.  (It is possible for that emphasis to be so great that the third central moment doesn't even exist: if the probability doesn't decay rapidly enough with the size of the residual, the weights can swamp the probability, with the net effect of creating an infinitely great third central moment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can understand, and even analyze, all other central moments in the same way, even fractional moments and absolute moments.  It should be immediately clear, for instance, that the higher the moment, the more the weights are emphasized at extreme values: higher central moments measure (average) behavior of the probability distribution at greater distances from the mean.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-09-24T18:43:48.710" />
  <row Id="5405" PostTypeId="2" ParentId="5377" CreationDate="2010-09-24T19:13:44.730" Score="7" ViewCount="0" Body="&lt;p&gt;Here's my take on it. Use $(A \uparrow)$ to mean $A$ does not halt, and $(A \downarrow)$ to mean $A$ does halt. Let $\Phi$ be any sentence; the question uses ~Con(ZFC) but this is not material. Let $\mathrm{Pvbl}(b)$ be the standard ZFC-formalized provability predicate, which says that there is a coded proof of the formula with number $b$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is right that, by the recursion theorem, we can create a specific machine $A$ such that&#xA;$$&#xA;(A \downarrow) \Leftrightarrow \mathrm{Pvbl}((A\uparrow) \lor \Phi)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, because of the specific form of the formula $(A \downarrow)$, ZFC is able to prove&#xA;$$&#xA;(A \downarrow) \to \mathrm{Pbvl}(A\downarrow)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working in ZFC, assume $(A \downarrow)$. Then we know $\mathrm{Pvbl}(A \downarrow)$ and $\mathrm{Pvbl}((A \uparrow) \lor \Phi)$.  ZFC is able to prove enough about the Pvbl predicate to ensure that&#xA;$$\mathrm{Pvbl}(\psi) \land \mathrm{Pvbl}(\lnot \psi \lor \theta) \rightarrow \mathrm{Pvbl}(\theta)&#xA;$$&#xA;for all $\psi$ and $\theta$. So we can obtain $\mathrm{Pvbl}(\Phi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: what we have obtained is: &quot;A does not halt or Pvbl($\Phi$)&quot;. In the fourth paragraph of the question, it instead claims we can obtain &quot;A does not halt or $\Phi$&quot;. That is a stronger statement, and this is the first error I see in the proof. In the fourth paragraph, it is silently assumed that provability implies truth, but this is not correct for formalized provability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ZFC does not prove $\mathrm{Pvbl}(\Phi) \to \Phi$ in general. In particular, ZFC does not prove $\mathrm{Pvbl}(0=1) \to (0=1)$ because there are models of ZFC in which both $\mathrm{Pvbl}(0=1)$ and $0 \not = 1$ are satisfied.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: Löb's theorem addresses this exact question. Applied to ZFC, it states that if ZFC proves $\mathrm{Pvbl}(\Phi) \to \Phi$ then ZFC already proves $\Phi$. So, in particular, ZFC does not prove that Pvbl(~Con(ZFC)) implies ~Con(ZFC). &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-09-25T02:19:39.400" LastActivityDate="2010-09-25T02:19:39.400" CommentCount="4" />
  <row Id="5407" PostTypeId="2" ParentId="5312" CreationDate="2010-09-24T19:26:58.917" Score="2" ViewCount="0" Body="&lt;p&gt;Here is a more general framework for problems of this kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bipartite graph is one for which the vertices can be divided into two sets so that any edges only join vertices which are not in the same same one of these two sets. Bipartite graphs can be vertex colored with two colors and any circuits they have must have even length. If the two color sets of the bipartite graph differ in size then the graph can not have a hamiltonian circuit and if the two color sets differ by two the graph can not have a hamiltonian path. If the two color sets differ by 1 any hamiltonian path must start and end at vertices of the same color. If a bipartite graph has color sets of equal size it may or may not have a hamiltonian circuit. For example the 1x1x1 cube has a hamiltonian circuit but the rhombic dodecahedron does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A famous conjecture of David Barnette suggests that every planar 3-connected 3-valent bipartite graph has a hamiltonian circuit. Another way of putting this is that the graphs of 3-valent convex polyhedra all of whose faces have an even number sides have hamiltonian circuits. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-09-24T20:40:01.887" LastActivityDate="2010-09-24T20:40:01.887" />
  <row Id="5408" PostTypeId="1" AcceptedAnswerId="5409" CreationDate="2010-09-24T19:55:58.140" Score="2" ViewCount="217" Body="&lt;p&gt;Suppose $x_{1}&gt;0$, and consider the sequence, $\{x_{n}\}$ defined as follows: $$x_{n+1}=\log(1+x_{n}) \quad n\geq 1 $$ Find the value of $\displaystyle \lim_{n \to \infty} nx_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble solving it. One thing is clear, that since $x_{n}&gt;0$ and $x_{n+1} &amp;lt; x_{n}$, we can have a sequence which converges an $f$ which satisfies $f=\log(1+f)$, so that $f=0$. Any way as to how we can proceed from here.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-24T22:31:40.427" LastActivityDate="2010-09-24T22:31:40.427" Title="Limit of the sequence $nx_{n}$ where $x_{n+1} = \log (1 +x_{n})$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="5409" PostTypeId="2" ParentId="5408" CreationDate="2010-09-24T20:19:48.617" Score="8" ViewCount="0" Body="&lt;p&gt;Using the methods similar to the answer by David Speyer here: &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&quot;&gt;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I suggest you read that first)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe we get the answer to be $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the fact that $$\log (1+x) = x - \frac{x^2}{2} + O(x^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $$y_{n}  = \frac{1}{x_n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{n+1} = \frac{1}{x_n(1 - x_{n}/2 + O((x_{n})^2))} = y_{n} +  \frac{1}{2} + O(x_{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives rise to (again similar to that answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{y_{n}}{n} = \frac{1}{2} + \frac{1}{n}O(\sum_{k=1}^{n-1} x_{k})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_{n} \rightarrow 0$ we have that $$\frac{y_{n}}{n} \rightarrow \frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus $$nx_{n} \rightarrow 2$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-24T20:19:48.617" />
  <row Id="5410" PostTypeId="2" ParentId="4844" CreationDate="2010-09-24T21:18:12.523" Score="1" ViewCount="0" Body="&lt;p&gt;You can look at one particular configuration at&#xA;&quot;Animated (Java) Illustrations {of} &lt;a href=&quot;http://mathpages.com/home/polyjava.htm&quot; rel=&quot;nofollow&quot;&gt;24 Electrons on a Sphere&lt;/a&gt;&quot;&#xA;and a few more at&#xA;&lt;a href=&quot;http://mathpages.com/home/kmath005/kmath005.htm&quot; rel=&quot;nofollow&quot;&gt;&quot;Min-Energy Configurations of Electrons On A Sphere&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1705" LastEditorUserId="1705" LastEditorDisplayName="" LastEditDate="2010-09-27T22:03:00.287" LastActivityDate="2010-09-27T22:03:00.287" CommentCount="2" />
  <row Id="5411" PostTypeId="1" CreationDate="2010-09-24T21:58:21.090" Score="2" ViewCount="234" Body="&lt;p&gt;I would like to recover the probability mass function (pmf) from the characteristic function (CF) of a discrete probability distribution using &lt;i&gt;Mathematica&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt; Ideally, I'd like to do calculations like this example (not necessarily this simple). To compute the pmf f&lt;sub&gt;X+X&lt;/sub&gt;[x] of the sum of two iid discrete uniform distributions X with support S={1,...,6}, it seems reasonable to try something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Subscript[\[CurlyPhi], X + X][t_] := &#xA; CharacteristicFunction[DiscreteUniformDistribution[{1, 6}], t]^2&#xA;&#xA;Subscript[f, X + X][x_] := &#xA; InverseFourierTransform[Subscript[\[CurlyPhi], X + X][t], t, x]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but &lt;b&gt;InverseFourierTransform&lt;/b&gt; doesn't recover the pmf from the CF of discrete distributions (it does recover pdfs from the CFs of continuous distributions). Am I using the wrong function? (There are other candidates, but &lt;b&gt;InverseFourier&lt;/b&gt; only works on lists of numbers, and &lt;b&gt;InverseZTransform&lt;/b&gt; doesn't seem to work here either.) Am I forgetting to set some necessary options to the inverse function? Or is there just no built-in to recover the pmf from the CF of a discrete probability distribution?&lt;/p&gt;&#xA;" LastActivityDate="2010-09-27T20:50:46.520" Title="Recovering the probability mass function from the characteristic function of a discrete probability distribution using Mathematica" Tags="&lt;probability&gt;&lt;fourier-analysis&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="5412" PostTypeId="2" ParentId="5399" CreationDate="2010-09-24T22:09:13.207" Score="4" ViewCount="0" Body="&lt;p&gt;My interpretation of your question, my hand calculations and explanations are as follows. For periods of six months you have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of period 1 you invest $C_{0}=1000$ (dollars). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the end of period 1 $C_{0}$ worths $F_{1}=1200$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of period 2 you invest the additional capital of $%&#xA;C_{1}=2000$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the end of period 2 your investement worths $F_{2}=4000$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gk7Gh.jpg&quot; alt=&quot;alt text&quot;&gt; &lt;strong&gt;Figure: Cash Flows&lt;/strong&gt; of -1000, -2000 and +4000 dollars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote by $r$ the anual nominal interest rate of your investement. In &#xA;$n$ periods of six months the investement worths $(1+r/2)^{n}$ per currency unit of invested capital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial investement $C_{0}$ worths $C_{0}(1+r/2)^{2}$ at the end of&#xA;period 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional capital $C_{1}$ worths $C_{1}(1+r/2)$ at the end of period 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{0}(1+r/2)^{2}+C_{1}(1+r/2)=F_{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1000(1+r/2)^{2}+2000(1+r/2)=4000$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+r/2)^{2}+2(1+r/2)=4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2r+\dfrac{1}{4}r^{2}+3=4.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is: $r=2\sqrt{5}-4\approx 0.47214\approx 47.214\%$ anual&#xA;nominal rate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the effective interest rate, you could find how much would you&#xA;need to invest so that at a nominal rate of $r$ you yould have $4000$ in 1&#xA;year (2 periods):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P\cdot 1.2361^{2}=4000$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=4000/1.2361^{2}=2617.9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anual effective interest rate is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i_{eff}=\dfrac{4000}{2617.9}-1=0.52794\approx 52,794\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: The future value $F_{n}$ at the end of period $n$ of a present value &#xA;$P$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{n}=P\left( 1+\dfrac{r}{m}\right) ^{n}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r=i_N$ is the anual nominal interest rate compounded $m$ periods per year. In your case $m=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Rigg, Bedworth and Randhawa, &lt;em&gt;Engineering Economics&lt;/em&gt;, Mc Graw Hill, 4th ed., 1996.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All errors and omissions are of mine, of course.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-25T15:09:11.527" LastActivityDate="2010-09-25T15:09:11.527" />
  <row Id="5413" PostTypeId="2" ParentId="5362" CreationDate="2010-09-25T00:18:37.227" Score="0" ViewCount="0" Body="&lt;p&gt;I'm no specialist, but see how this goes.  When you say &quot;the subset A∩λ of λ is not stationary&quot;, stationary in what?  Presumably not stationary in λ.  So there is a closed and unbounded subset of λ that is disjoint from A.  Pick any element of this subset greater than sup{g(α):α&amp;lt;λ}&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-25T00:18:37.227" CommentCount="1" />
  <row Id="5414" PostTypeId="2" ParentId="5391" CreationDate="2010-09-25T01:43:46.533" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the map $k: H\to H$ given by $k(x) = g(x,h)$.  Then for any $x\in H$, we have $$p\circ k(x)  = p\circ g(x,h) = m\circ (p\times p)(x,h) = m(p(x),e) = p(x),$$ &#xA;which implies that $k$ is an automorphism of the covering $p$ (which you can also think of as a lift of $p$).  Note also that $k(h)  = g(h,h) = h$.  Thus $k$ and the identity are both automorphisms of $p$ that agree at a point, so they are equal. This implies $g(x,h)=x$ for all $x$.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-25T01:43:46.533" CommentCount="3" />
  <row Id="5415" PostTypeId="2" ParentId="201" CreationDate="2010-09-25T03:16:39.083" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a really nice proof that there are an infinite number of primes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\prod(1-p_i^{-2})^{-1} = \zeta(2) = \pi^2/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\pi$ is transcendental, the RHS is irrational. Therefore there must be an infinite number of terms in the product on the left.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastEditorUserId="1820" LastEditorDisplayName="" LastEditDate="2010-09-25T04:34:12.230" LastActivityDate="2010-09-25T04:34:12.230" />
  <row Id="5416" PostTypeId="1" AcceptedAnswerId="5518" CreationDate="2010-09-25T04:58:05.933" Score="4" ViewCount="412" Body="&lt;p&gt;The derivative of a function $f: C \rightarrow C$ with respect to $\overline{z}$ is defined as&#xA;$$ \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) $$&#xA;where $\frac{\partial}{\partial x}$ is defined as the operator which maps $u+iv$ to $u_x + iv_x$, and $\frac{\partial}{\partial y}$ is defined likewise. I am wondering whether there is a geometric way to think about this quantity. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-09-27T08:46:24.137" Title="geometric meaning of differentiation with respect to the complex conjugate of $z$" Tags="&lt;complex-analysis&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="5417" PostTypeId="2" ParentId="5362" CreationDate="2010-09-25T05:10:16.097" Score="2" ViewCount="0" Body="&lt;p&gt;Hi Martin. The exercise is asking for a map from $A$ to the ordinals, not to itself. Why do you think the exercise is trivial? If $\lambda$ is regular and $A\cap\lambda$ is non-stationary (in $\lambda$) then there is a club subset of $\lambda$ disjoint from $A$. You can then pick $g(\lambda)$ in this club. You need to ensure, however, that this element is not $g(\tau)$ for some prior $\tau$, and this requires some argument: If, as suggested in another answer, you are to pick something larger than $\sup(g(\alpha)\mid\alpha\in A\cap\lambda)$, you need to argue that this supremum is strictly below $\lambda$. But this is not automatic (or true in general), as $A$ may contain all cardinals up to and including some inaccessible. Also, $A$ may contain singular cardinals, and you need to define $g$ on them as well. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-09-25T05:10:16.097" CommentCount="4" />
  <row Id="5418" PostTypeId="1" CreationDate="2010-09-25T05:17:25.233" Score="1" ViewCount="86" Body="&lt;p&gt;in constrained Lagrangian optimization what is a general way to figure out how the optimal point varies with respect to parameters of the constraints? For example maximize x*y when x + y &amp;lt; k, and I wanted to find dx*/dk and dy*/dk where (x*,y*) is the optimal point. How would I do that?&lt;/p&gt;&#xA;" OwnerUserId="2121" LastActivityDate="2010-09-25T05:49:29.387" Title="Lagrangian (constrained) optimization - rate of change of optimal values." Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5419" PostTypeId="2" ParentId="5418" CreationDate="2010-09-25T05:49:29.387" Score="2" ViewCount="0" Body="&lt;p&gt;The optimal point may not be differentiable at a value of $k$ where a constraint goes from binding to non-binding. At any other value of $k$ just differentiate with respect to $k$ the (equality) first order conditions for the constraints which bind and solve for $\partial x^*/\partial k$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at any standard optimization book, for eg. Luenberger &lt;em&gt;Linear and Non-linear Programming&lt;/em&gt; or Sundaram's &lt;em&gt;A First Course in Optimization Theory&lt;/em&gt;,&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-25T05:49:29.387" CommentCount="1" />
  <row Id="5420" PostTypeId="1" CreationDate="2010-09-25T05:54:40.267" Score="8" ViewCount="220" Body="&lt;p&gt;I'm working on a multiple-precision library. I'd like to make it possible for users to ask for higher precision answers for results already computed at a fixed precision. My $\mathrm{sqrt}(x)$ can pick up where it left off, even if $x$ changes a bit, because it uses Newton-Raphson. But $\exp(x)$ computed using a Maclaurin series or continued fraction has to be computed from scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an iterative refinement (i.e. Newton-Raphson, gradient descent) method for computing $\exp(x)$ that uses only arithmetic and integer roots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know Newton-Raphson can solve $\log(y)-x=0$ to compute $\exp(x)$. I am specifically not asking for that. Newton-Raphson can also solve $\exp(y)-x=0$ to compute $\log(x)$. Note that each requires the other. I have neither right now as an arbitrary-precision function. I have arithmetic, integer roots, and equality/inequality tests.)&lt;/p&gt;&#xA;" OwnerUserId="2122" LastEditorUserId="2122" LastEditorDisplayName="" LastEditDate="2010-09-25T16:30:28.770" LastActivityDate="2010-11-17T20:57:10.527" Title="Iterative refinement algorithm for computing exp(x) with arbitrary precision" Tags="&lt;algorithms&gt;&lt;numerical-methods&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="5422" PostTypeId="2" ParentId="5416" CreationDate="2010-09-25T09:33:58.763" Score="1" ViewCount="0" Body="&lt;p&gt;I think the hypothesis that f is a totally differentiable function is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic first result is that, f is holomorphic if and only if $\frac{\partial f}{\partial \overline{z}}=0$, and in that case $f' = \frac{\partial f}{\partial z}$, i.e &quot;Holomorphic functions are independent of  $\overline{z}$ and depend only on z&quot;, by Cauchy-Riemann.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpreting f as a real function of two independent real variables x, y, where z = x + i y. Hence $x = \frac{z + \overline{z}}{2}$, and $y = \frac{z - \overline{z}}{2i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimizing f with respect to $\overline{z}$ and applying the chain rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial f}{\partial \overline{z}} = \frac{\partial x}{\partial \overline{z}}_z\frac{\partial f}{\partial x}_y +  \frac{\partial y}{\partial \overline{z}}_z\frac{\partial f}{\partial y}_x$&#xA;$= \frac{1}{2} ( \frac{\partial f}{\partial x}_x + i \frac{\partial f}{\partial y}_y)$&#xA;and to minimize over all z = x + iy (since minimizing w.r.t x (resp. y) is to have the partial derivative = 0), then $\frac{\partial f}{\partial \overline{z}} = 0$.&#xA;see &lt;a href=&quot;http://people.ccmr.cornell.edu/~muchomas/P480/Notes/dft/node10.html&quot; rel=&quot;nofollow&quot;&gt;http://people.ccmr.cornell.edu/~muchomas/P480/Notes/dft/node10.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be mentioned that both $\frac{\partial f}{\partial \overline{z}}$ along with $\frac{\partial f}{\partial z}$ are known as Wirtinger derivatives(operators). I'm not sure if geometric intuition is necessary, but these derivatives give the Wirtinger calculus. One builds a theory (a &quot;symbolical tool&quot;?!!) that's analogous to the real case that is applied even to nonholomorphic functions (see &lt;a href=&quot;http://books.google.com.eg/books?id=CC0dQxtYb6kC&amp;amp;pg=PA67&amp;amp;lpg=PA67&amp;amp;dq=Wirtinger+calculus&amp;amp;source=bl&amp;amp;ots=70YeqX9-k6&amp;amp;sig=asCVvTFAvG_y3GrPXSnHnNphl9Q&amp;amp;hl=en&amp;amp;ei=i5mdTJ6yNIjH4AakrrjIDg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwADge#v=onepage&amp;amp;q=Wirtinger%20calculus&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&quot;Theory of complex functions&quot;, Remmert, page 67-69&lt;/a&gt; for a quick development), or the beginnings of &quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.6515&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;The Complex Gradient Operator and the CR-Calculus, ECE275A – Lecture Supplement&lt;/a&gt;&quot;, which is more elementary but could be useful. &lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-09-25T09:47:14.573" LastActivityDate="2010-09-25T09:47:14.573" />
  <row Id="5424" PostTypeId="1" CreationDate="2010-09-25T11:08:13.333" Score="2" ViewCount="147" Body="&lt;p&gt;Without strenuous arithmetic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a programme I can download to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the quadratic residues modulo 5^4 or 5^5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2127" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-25T11:21:35.063" LastActivityDate="2010-09-29T02:19:23.543" Title="How can I find out whether a number is a quadratic residue in a large modulo?" Tags="&lt;number-theory&gt;&lt;math-software&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5425" PostTypeId="2" ParentId="5416" CreationDate="2010-09-25T11:30:46.813" Score="3" ViewCount="0" Body="&lt;p&gt;For getting geometric intuition about complex analysis Tristram Needham's book &lt;a href=&quot;http://books.google.com/books?id=ogz5FjmiqlQC&amp;amp;dq=visual+complex+analysis&amp;amp;printsec=frontcover&amp;amp;source=bn&amp;amp;hl=en&amp;amp;ei=j9qdTJ2kDMrKOLDomZ8M&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CC8Q6AEwBQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&quot;Visual Complex Analysis&quot;&lt;/a&gt; seems to be the best place you can turn to! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, reading from page 216 on (and you can safely just jump there, if you already have arrived at questions like the one you are asking here), you will learn that holomorphic functions are the ones which look locally like compositions of rotations, translations and rescalings (I mean if you really draw the complex plane as a plane), i.e. they are the ones which carry disks to disks (and I mean the real round disks, with boundary a circle :-). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since the derivation of a function $f$ with respect to $\bar{z}$ is zero iff $f$ is holomorphic, you could say that it measures how far $f$ is from being holomorphic or, in view of the above characterization, how much a disk can get deformed (e.g. into an ellipsoid) when you apply $f$.&lt;/p&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="2048" LastEditorDisplayName="" LastEditDate="2010-09-25T11:49:50.690" LastActivityDate="2010-09-25T11:49:50.690" />
  <row Id="5426" PostTypeId="2" ParentId="5424" CreationDate="2010-09-25T11:31:51.597" Score="4" ViewCount="0" Body="&lt;p&gt;If you want to know the squares modulo $p^m$ where $p$ is an odd prime,&#xA;they are congruent to $0$ or to numbers $a p^{2k}$ where $0\le k &amp;lt; m/2$&#xA;and the Legendre symbol $\left(\frac ap\right)=1$. For large primes $p$,&#xA;Legendre symbols can be calculated by regarding them as&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_symbol&quot; rel=&quot;nofollow&quot;&gt;Jacobi symbols&lt;/a&gt; and using the law of quadratic reciprocity&#xA;for Jacobi symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't regard $5$ as a large prime. :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-25T11:31:51.597" CommentCount="2" />
  <row Id="5427" PostTypeId="1" CreationDate="2010-09-25T13:42:11.600" Score="1" ViewCount="138" Body="&lt;p&gt;I'm studying on K. Hrbacek and T. Jech, Introduction to Set Theory. In the third chapter, they prove the usual properties of the strict ordering on natural numbers in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They prove transitivity using induction.&lt;/li&gt;&#xA;&lt;li&gt;They prove irreflexivity using induction and transitivity.&lt;/li&gt;&#xA;&lt;li&gt;They prove asymmetry by contradiction and using transitivity and irreflexivity.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is nothing wrong with this, but I'd like, if possible, to prove the previous properties (in particular irreflexivity and asymmetry) independently and using induction only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $\phi(x)$ be $x\not\in x$. Of course $\phi(0)$ is true ($0\not\in 0$ since $0$ is the empty set). Now I should suppose that $\phi(x)$ is true and show that $\phi(x+1)$ is true, too. $\phi(x+1)$ stands for $x+1\not\in x+1$, that is $x+1\not\in x$ and $x+1\not=x$. But here I cannot go on, mainly because $x+1$ is at the left of the relations symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hint? Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $x+1$ is defined as $x\cup\{x\}$.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastEditorUserId="2128" LastEditorDisplayName="" LastEditDate="2010-09-25T17:48:53.300" LastActivityDate="2010-09-25T20:57:35.340" Title="Strict ordering on natural numbers" Tags="&lt;set-theory&gt;&lt;induction&gt;&lt;natural-number&gt;" AnswerCount="2" />
  <row Id="5428" PostTypeId="2" ParentId="5297" CreationDate="2010-09-25T13:43:33.377" Score="1" ViewCount="0" Body="&lt;p&gt;This is a new version of the answer. The comments refer to previous versions. Martin Brandenburg's comment was especially helpful. A comment of Dan Ramras, somewhere else in this thread, drew my attention to Lemma 5.5 &lt;a href=&quot;http://www.iecn.u-nancy.fr/~gaillard/milnor_64.pdf&quot; rel=&quot;nofollow&quot;&gt;page 64&lt;/a&gt; in Milnor and Stasheff's &lt;strong&gt;Characteristic Classes&lt;/strong&gt;. Thank you also to Agusti Roig for his wonderful question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me state the Lemma of Milnor and Stasheff mentioned above: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Let $A_1\subset A_2\subset\cdots$ and $B_1\subset B_2\subset\cdots$ be sequences of locally compact spaces with inductive limits $A$ and $B$ respectively. Then the product topology on $A\times B$ coincides with the inductive limit topology which is associated with the sequence $A_1\times B_1\subset A_2\times B_1\subset\cdots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof (which can be found &lt;a href=&quot;http://www.iecn.u-nancy.fr/~gaillard/milnor_64.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) shows in fact that the following more technical statement also holds: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Let $A_1\subset A_2\subset\cdots$ and $B_1\subset B_2\subset\cdots$ be as above. For each $i$ let $C_i\subset A_i$ and $D_i\subset B_i$ be subspaces. Assume $C_i\subset C_{i+1}$ and $D_i\subset D_{i+1}$ for all $i$, and call $C$ and $D$ the respective inductive limits. Then the product topology on $C\times D$ coincides with the inductive limit topology which is associated with the sequence $C_1\times D_1\subset C_2\times D_2\subset\cdots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if the $C_i$ are topological groups, then so is $C$ (assuming, of course, that the topological group structure on $C_i$ is induced by that of $C_{i+1}$). This seems to indicate that the alleged counterexample of Tatsuuma et al. is not really a counterexample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know no example of an inductive limit of topological groups which is not a topological group. (I think that such examples do exist. It would be interesting to know if $(\mathbb R^\infty)^\infty$ is a topological group.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's prove (2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $W$ be a subspace of $C\times D$ which is open in the inductive limit topology. For each $i$ let $W_i$ be an open subset of $A_i\times B_i$ which has the same intersection with $C_i\times D_i$ as $W$. Let $(c,d)$ be in $W\cap(C_i\times D_i)$. There is a compact neighborhood $K_i$ of $c$ in $A_i$, and a compact neighborhood $L_i$ of $d$ in $B_i$, such that $K_i\times L_i\subset W_i$. There is also a compact neighborhood $K_{i+1}$ of $K_i$ in $A_{i+1}$, and a compact neighborhood $L_{i+1}$ of $L_i$ in $B_{i+1}$, such that $K_{i+1}\times L_{i+1}\subset W_{i+1}$. Then $K_{i+1}\cap C_{i+1}$ is a neighborhood of $K_i\cap C_i$ in $C_{i+1}$, and $L_{i+1}\cap D_{i+1}$ is a neighborhood of $L_i\cap D_i$ in $D_{i+1}$. The union $U$ of the $K_i\cap C_i$ is open in $C$, and the union $V$ of the $L_i\cap D_i$ is open in $D$, the spaces $C$ and $D$ being equipped with the inductive limit topology. Moreover we have $(c,d)\in U\times V\subset W$.&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditorDisplayName="" LastEditDate="2010-10-03T05:53:24.397" LastActivityDate="2010-10-03T05:53:24.397" CommunityOwnedDate="2010-10-03T05:29:23.220" CommentCount="2" />
  <row Id="5429" PostTypeId="2" ParentId="213" CreationDate="2010-09-25T14:33:17.487" Score="0" ViewCount="0" Body="&lt;p&gt;My answer is: fundamental concepts and methods of both first order logic and set theory. I really wish I learned them much earlier, since all mathematics is based on them.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-09-25T14:33:17.487" CommunityOwnedDate="2010-09-25T14:33:17.487" />
  <row Id="5430" PostTypeId="1" AcceptedAnswerId="5435" CreationDate="2010-09-25T16:23:39.410" Score="2" ViewCount="65" Body="&lt;p&gt;This question stumped me on a recent calculus test... I would really like an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The region $\mathbf{R}$ is bounded by $h(x) = -0.8(x-2)^2 + 8$ and $p(x) = x\cos(x) - \ln(2x+2)$. Find the lateral surface area of the solid formed when $\mathbf{R}$ is revolved around the line $y=11$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to convert both functions into terms of y so that I could get the arc length, but that didn't get me very far. Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2131" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-25T16:26:43.407" LastActivityDate="2010-09-25T17:16:43.820" Title="Lateral surface area of a region" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5431" PostTypeId="1" AcceptedAnswerId="5434" CreationDate="2010-09-25T16:38:45.680" Score="3" ViewCount="144" Body="&lt;p&gt;Say I have a function $f(x) = ax^3 + bx^2 + cx + d$ where $a &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that for a high enough value of $x$, the $x^3$ term will dominate and I can say $f(x) \in O(x^3)$, but this doesn't seem very formal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal definition is $f(x) \in O(g(x))$ if constants $k, x_0 &gt; 0$ exist, such that $0 \le f(x) \le kg(x)$ for all $x &gt; x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what are appropriate values for $k$ and $x_0$? It's easy enough to find ones that apply (say $k = a + b + c + d$). By the formal definition, all I have to do is show that these numbers exist, so does it actually matter which numbers I use? For some value of $x$, $k$ could be anywhere from $1$ to $a + b + c + d + ... $. From my understanding, it doesn't matter what numbers I pick as long as they 'work', but is this right? It seems too easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1957" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-25T22:21:09.597" LastActivityDate="2010-09-25T22:21:09.597" Title="Formally proving that a function is $O(x^n)$" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="5432" PostTypeId="1" AcceptedAnswerId="5515" CreationDate="2010-09-25T16:53:58.793" Score="1" ViewCount="242" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The well known formula for perfect numbers is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_n=2^{n-1}(2^{n}-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$496=1+2+4+8+16+31+62+124+248$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;one can see that the first pattern is a sequence of powers of $2$ that stops at $16$, the second pattern starts with a prime number, in this case $31$, the rest of them are multiples of $31$, e.g. $2\cdot31, 4\cdot31$ and $8\cdot31$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I found that the pattern proceeds after $31$, for example $31=2^5-2^0$, $62=2^6-2^1$, $124=2^7-2^2$ and finally $248=2^8-2^3$, so the perfect number can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$496=1+2+4+8+16+(2^5-2^0)+(2^6-2^1)+(2^7-2^2)+(2^8-2^3)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$496=(1+2+4+8+16+32+64+128+256) -(1+2+4+8)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the formula follows very naturally from this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched but didn't find this formulation anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, is this something new? Have anyone seen this somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-09-26T23:41:41.400" Title="Perfect numbers, the pattern continues..." Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5433" PostTypeId="2" ParentId="5424" CreationDate="2010-09-25T17:03:07.043" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Euler criterion&lt;/a&gt; is easily generalized to yield the following test for squareness (mod $\rm n\:$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $ Let $\rm\ a,\:n\:$ be integers, with $\rm\:a\:$ coprime to $\rm\:n\ =\ 2^e \:p_1^{e_1}\cdots p_k^{e_k}\:,\ \ p_i\:$ primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad \ x^2\ =\ a\ \ (mod\ n)\ $ is solvable for $\rm\:x\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad \iff\ \ \: a^{(p_i-1)/2} \ \ \equiv\ \ 1\ \ (mod\ p_i)\quad\quad\ \ $  for all $\rm\ i\le k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\ $ and $\rm\quad\ \ e&gt;1 \:\Rightarrow\: a\equiv 1\ \ (mod\ 2^{2+\delta}\:),\ \ \delta = 1\ \ if\ \ e\ge 3\ \ else\ \ \delta = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;: See Ireland and Rosen,  A Classical Introduction to Modern Number Theory, Proposition 5.1.1 p.50.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\:$ The criterion is very practical if one knows a full factorization of $\rm\:n\:$, since the exponentiations may be quickly computed by repeated squaring.&#xA;Beware that the criterion cannot be expressed equivalently as a simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_symbol&quot; rel=&quot;nofollow&quot;&gt;Jacobi symbol&lt;/a&gt; calculation, e.g. $\rm(8|15) = 1\ $ but $8$ is not a square (mod $15$).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T02:19:23.543" LastActivityDate="2010-09-29T02:19:23.543" />
  <row Id="5434" PostTypeId="2" ParentId="5431" CreationDate="2010-09-25T17:16:38.247" Score="1" ViewCount="0" Body="&lt;p&gt;The argument you are getting at as I understand is is, roughly: $x^n \in O(x^n)$ and thus $kx^n \in O(x^n)$, so $f(x)$ acts like $O(x^3) + O(x^2) + O(x) + O(1)$ which can be reduced to $O(x^3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the theorem we would like to prove now is that for $n\geq m$: $f \in O(x^n)$ and $g \in O(x^m)$ implies $f + g \in O(x^n)$. Once we have this you just add up the monomials of the polynomial and that proves the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at what we have, from the definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f \in O(x^n) \Rightarrow \exists x_0, k,\;\; \forall x&gt;x_0,\;\; f(x) \leq kx^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g \in O(x^m) \Rightarrow \exists x_1, k',\;\; \forall x&gt;x_1,\;\; g(x) \leq kx^m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_2$ be the maximum of $x_0$ and $x_1$, $k''$ be the maximum of $k$ and $k'$ and add these inequalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall x&gt;x_2,\;\; (f+g)(x) \leq kx^n + k'x^m \leq k''(x^n + x^m) \leq 2 k'' x^n $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the pair of values $(x_2,2k'')$ prove that $f+g \in O(x^n)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To consider abstract functions like this makes the proof very easy, but it is clear that the values we exhibit to prove the existential may not be the &lt;em&gt;best&lt;/em&gt;, although we still prove the theorem in an effective way. In particular, you could do a very detailed analysis of the functions in specific cases to get very tight bounds - in this lucky case it's not needed which is why the theorem is easier to prove in the abstract case.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T17:16:38.247" />
  <row Id="5435" PostTypeId="2" ParentId="5430" CreationDate="2010-09-25T17:16:43.820" Score="3" ViewCount="0" Body="&lt;p&gt;First, let's shift the functions down so that instead of revolving about $y=11$, we're revolving about the x-axis, $y=0$: $h_1(x)=h(x)-11$, $p_1(x)=p(x)-11$.  Let $a&amp;lt; b$ be the x-coordinates of the points of intersection of $h_1(x)$ and $p_1(x)$ (in the vicinity of -2 and 5)--that is, the least and greatest x-coordinates of any points in $\mathbf{R}$.  There are two lateral surface areas to compute: the &quot;inner&quot; part bounded by $h_1$, and the &quot;outer&quot; part bounded by $p_1$.  For each, the surface area is given by $\int_{a}^{b}2\pi y(x)\sqrt{1+(y'(x))^2}dx$, replacing $y$ with $h_1$ or $p_1$ (and $y'$ with its derivative) as appropriate.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-25T17:16:43.820" CommentCount="1" />
  <row Id="5436" PostTypeId="2" ParentId="5432" CreationDate="2010-09-25T17:29:28.050" Score="1" ViewCount="0" Body="&lt;p&gt;You've observed that $P_5 = 2^4(2^5-1) = 496$ can also be written as the sum of the first 9 powers of two minus the sum of the first four powers of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sums of powers of two&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powers of two written in binary look like $1, 10, 100, 1000, \cdots$ but you can also write them like this $1, 1+1, 11+1, 111+1, \cdots$. This explains why (now in decimal notation) $1 + 2 + 4 + 8 = 15$ and the general sums of this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nine and four&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well obvious $4 = 5-1$, and $9 = 2\cdot 5 - 1$, but where do these come from? Well just multiply out the formula! $2^{n-1}(2^n-1) = 2^{2n-1}-2^{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains why the form $(1 + 2 + 4 + 8 + \cdots) - (1 + 2 + \cdots)$ works.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T17:29:28.050" CommentCount="3" />
  <row Id="5437" PostTypeId="2" ParentId="5427" CreationDate="2010-09-25T17:37:15.960" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming that $0 = \{\}$ and $x + 1 = \{x,\{x\}\}$ are the definitions you use, the recursive induction step would be to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x \notin x \Rightarrow \{x,\{x\}\} \notin \{x,\{x\}\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By extensionality what we really need to show is the two cases&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\{x,\{x\}\} \neq \{x\}$&lt;/li&gt;&#xA;&lt;li&gt;$\{x,\{x\}\} \neq x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first is immediate since one is a singleton and the other is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part we notice that if it were true, we would have $x \in x$ but that contradicts the induction hypothesis.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T17:37:15.960" CommentCount="5" />
  <row Id="5438" PostTypeId="2" ParentId="5432" CreationDate="2010-09-25T17:46:51.900" Score="2" ViewCount="0" Body="&lt;p&gt;The patterns you're after tell us nothing about the perfect(ness) of a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they hold even if the number is not perfect, for example, take n = 6. $2^5 (2^6-1) = 2016 = 2^{10} + 2^9 + ... + 2^5$. Which is valid for any n, since in binary, 2^6-1 is 6-1=5 1's from left, and multiplication by 2^5 is adding 5 zeros from right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the following also hold for any n,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^5 (2^6-1) = 2^0 + ... + 2^{6-1} + (2^6-2^0) + (2^{6+1} + 2^1) + ... + (2^{2*6-2} - 2^{6-2})$.&#xA;$= 1 + ... + 32 + (2^6 - 2^0) + ... + (2^10 - 2^4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but 2016 is not perfect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $2^n-1$ must be prime so that $2^{n-1}(2^n-1)$ is the unique prime factorization of the number, and in which case, the terms in  $\sum_{k=0}^{n-1} 2^k + \sum_{k=0}^{n-2} 2^k(2^n - 1)$ are precisely the divisors of $2^{n-1}(2^n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: &lt;strong&gt;If P is an even perfect number, then P = $2^{k-1} (2^k - 1)$ for some k&gt;1 with (2^k - 1) prime.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose P is even perfect. Then we can find k&gt;1 such that $P = 2^{k-1}m$, for m odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $2^{k-1} - 1 = 1 + ... + 2^{k-2}$ (as explained earlier, 2^{k-1} - 1 has k-2 ones). &#xA;i.e explaining why 31 = 1 + 2 + ... + 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $P = (2^{k-1}-1 + 1)m$. Then $P = (1 + ... + 2^{k-2})m + m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The case m is prime:&lt;/strong&gt; The proper divisors of P are then $1,2, ..., 2^{k-1}, m, 2m, ..., 2^{k-2}m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since P is assumed perfect, $P = 1 + 2 + ... + 2^{k-1} + m + 2m + ... + 2^{k-2}m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $P = (1 + ... + 2^{k-2})m + m$ as shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $m = 1 + 2 + ... + 2^{k-1} = 2^{k}-1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e $P = 2^{k-1} (2^k - 1)$ (note 2^{k}-1 prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The case for m is not prime:&lt;/strong&gt; Assume without loss of generality $m = p_1 p_2$, neither a unit or even. Then the divisors (hopefully I didn't miss a divisor) of P are:&#xA;$1,2, ..., 2^{k-1}, p_1, 2p_1, ... 2^{k-1}p_1, p_2, 2p_2, ... 2^{k-1}p_2, 2m, ..., 2^{k-2}m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the sum of the divisors and equating with $P = (1 + ... + 2^{k-2})m + m$. Then &#xA;$m = 1 + 2 + ... + 2^{k-1} + (1 + ... + 2^{k-1})p_1 + (1 + ... + 2^{k-1})p_2 = (2^{k} - 1) (1 + p_1 + p_2)$, but then $p_1 = (1 + p_1 + p_2)$, i.e p_2=-1 or $p_2 = (1 + p_1 + p_2)$, in either case a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-09-25T23:58:50.547" LastActivityDate="2010-09-25T23:58:50.547" CommentCount="3" />
  <row Id="5439" PostTypeId="1" CreationDate="2010-09-25T18:07:44.677" Score="1" ViewCount="90" Body="&lt;p&gt;I'm trying to learn about the Selberg zeta function, but it seems like introductory texts assume more knowledge of Riemannian geometry than I'm comfortable with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some basic questions that someone might be able to help with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the composition of two closed geodesics itself a closed geodesic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is composition of geodesics a commutative operation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can all geodesics be decomposed into compositions of primitive closed geodesics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has comments or references, I would appreciate them.&lt;/p&gt;&#xA;" OwnerUserId="1820" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-01T16:28:10.980" LastActivityDate="2010-10-01T16:28:10.980" Title="Some basic questions about the Selberg zeta function" Tags="&lt;geometry&gt;&lt;zeta-function&gt;" CommentCount="4" />
  <row Id="5440" PostTypeId="2" ParentId="5431" CreationDate="2010-09-25T18:18:25.783" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose you have a function $f(x)$ and $g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really simple method (that works when $f(x)$ and $g(x)$ are polynomials) of determining a constant that works is the following. Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the limit exists and if a constant $0\leq C&amp;lt; \infty$. Then $C+\epsilon$ for any $\epsilon&gt;0$ is a constant that works. To see this just apply the definition of the limit. $\forall \epsilon&gt;0$ there exists a $x_0(\epsilon)$ such that $\forall x&gt;x_0(\epsilon)$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left|\frac{f(x)}{g(x)}-C\right|&amp;lt;\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{f(x)}{g(x)} &amp;lt; C+\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know the constant from calculating the limit and know the existence of $x_0$. Since you are always considering asymptotics when using this definition you are never concerned with the value of $x_0$ (only that it exists)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not matter what constants you use and someone could easily use different constants to get $f(x)=O(g(x))$. This notation is to compare the growth rate of two functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}$ does not exist or is hard to calculate then as long as you can bound it above you still have $f(x)=O(g(x))$&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-25T18:35:41.037" LastActivityDate="2010-09-25T18:35:41.037" />
  <row Id="5441" PostTypeId="1" CreationDate="2010-09-25T18:37:31.767" Score="1" ViewCount="149" Body="&lt;p&gt;Can you advise some good books covering areas which are required for serious probability theory studies (e.g. measure theory, functional analysis)? Preferably this book should have some problem sets to work on. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1463" LastActivityDate="2010-09-30T15:22:57.440" CommunityOwnedDate="2010-09-26T11:14:44.290" Title="Supplementary reading for probability theory studies" Tags="&lt;soft-question&gt;&lt;probability&gt;&lt;books&gt;&lt;self-learning&gt;" AnswerCount="7" CommentCount="3" />
  <row Id="5442" PostTypeId="2" ParentId="5431" CreationDate="2010-09-25T18:43:02.207" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad\rm  ax^3 + bx^2 + cx + d\  \le \ (|a|+|b|+|c|+|d|)\ x^3 \ $  for $\rm\ x &gt; 1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-25T18:43:02.207" />
  <row Id="5443" PostTypeId="2" ParentId="5441" CreationDate="2010-09-25T18:55:21.170" Score="1" ViewCount="0" Body="&lt;p&gt;Take a look at &quot;Measure Theory and Probability&quot; by A.K. Basu. This book has lots of problem sets in them and contains some serious probabilty theory. It developed measure theory with probability as the main application. Also through-out the book there are examples which were really helpful.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-25T18:55:21.170" CommunityOwnedDate="2010-09-28T07:16:29.983" CommentCount="1" />
  <row Id="5444" PostTypeId="2" ParentId="5441" CreationDate="2010-09-25T18:56:04.320" Score="2" ViewCount="0" Body="&lt;p&gt;For a broad overview in one book at a graduate level, it is hard to beat &lt;em&gt;Foundations of Modern Probability (Second edition)&lt;/em&gt; by Olav Kallenberg. His presentation is classroom tested, and many standard complicated proofs have been made very concise. I'm a big fan of this book!    &lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-09-30T15:22:57.440" LastActivityDate="2010-09-30T15:22:57.440" CommunityOwnedDate="2010-09-25T18:56:04.320" />
  <row Id="5445" PostTypeId="2" ParentId="201" CreationDate="2010-09-25T19:05:21.913" Score="4" ViewCount="0" Body="&lt;p&gt;$$\sum_p \frac{1}{p}=\infty$$&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges&quot; rel=&quot;nofollow&quot;&gt;Here is a link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If there was finitely many primes, then the sum would be finite.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-15T19:40:24.540" LastActivityDate="2010-10-15T19:40:24.540" />
  <row Id="5446" PostTypeId="2" ParentId="5303" CreationDate="2010-09-25T20:02:58.737" Score="10" ViewCount="0" Body="&lt;p&gt;I suppose there is a way to formalize your question into a relatively precise one.   Let $P$ be the set of all propositions in the language of Zermelo-Frankel set theory for which there exist proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a natural number $k$ we want to have an upper bound $l(k)$ on the minimal-length of the proof of any proposition $p$ from $P$ such that the length of the proposition $p$ is at most $k$. Here I'm using &quot;length&quot; in the sense of the number of ASCII characters it would take to write the proposition (or proof of the proposition, respectively). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably $l(k)$ is a non-computable function. Since if $l(k)$ were computable, given a proposition $p$ of length $k$ you'd have a &quot;simple&quot; procedure to find a proof of any statement in Zermelo-Frankel set theory, provided one exists.  The idea would be to iterate through all propositions of length at most $l(k)$ and verify they're proofs of $p$.  :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine such a statement is known to logicians.  If it is known, I would call it the &quot;rationalization is hard theorem&quot;.&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-25T20:08:08.670" LastActivityDate="2010-09-25T20:08:08.670" CommentCount="4" />
  <row Id="5447" PostTypeId="1" AcceptedAnswerId="5448" CreationDate="2010-09-25T20:46:04.593" Score="1" ViewCount="184" Body="&lt;p&gt;A problem that is an example of this is $x^2+16$. I have to know how to factor this over the set of complex numbers. How do you do that? I used to know it's just been a long time.&lt;/p&gt;&#xA;" OwnerUserId="2135" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-26T16:52:18.107" LastActivityDate="2010-09-27T10:00:27.323" Title="How do you factor an equation over the set of complex numbers?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="5448" PostTypeId="2" ParentId="5447" CreationDate="2010-09-25T20:50:07.613" Score="3" ViewCount="0" Body="&lt;p&gt;Notice that $16 = -(4i)^2$, so $x^2 + 16$ is a difference of two squares and can be written as $(x - 4i)(x + 4i)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-09-25T20:50:07.613" CommentCount="2" />
  <row Id="5449" PostTypeId="2" ParentId="5427" CreationDate="2010-09-25T20:57:35.340" Score="1" ViewCount="0" Body="&lt;p&gt;As the induction hypothesis is clearly not strong enough, how about strengthening your proof obligation? Just adding $x+1\not\in x$ will mean you'll also need $x+2\not\in x$, so I'd suggest $y\not\in x$ for all $y$ such that $x\le y$.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2010-09-25T20:57:35.340" CommentCount="2" />
  <row Id="5450" PostTypeId="1" AcceptedAnswerId="5453" CreationDate="2010-09-25T21:00:50.127" Score="2" ViewCount="210" Body="&lt;p&gt;I'm stuck trying to solve an inequality of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$; specifically, $\left|\frac{2x + 5}{x + 1}\right| \geq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried the approach that was taught for solving inequalities in general, and solved first for x when the part enclosed in absolute value signs is  $\geq 1$, and then solved for when it is $\leq -1$, but plugging in numbers for the resulting domain $[-4, -2]$ back into the original inequality resulted in some answers that were not $\geq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where have I gone wrong in trying to solve this inequality? How is this different from solving an inequality like $\left|\frac{3}{x + 1}\right| \geq 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Okay, since I'm clearly messing this up horribly(and now realize that I was absent-mindedly taking the intersection and not the union of my results in spite of having not mixed up the two many times before), I'll edit with the specifics of the solution I was trying and which had me confused before I asked the question. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left|\frac{2x + 5}{x + 1}\right| \geq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meaning 1: $\frac{2x + 5}{x + 1} \geq 1$ or 2: $\frac{2x + 5}{x + 1} \leq -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the other inequality: multiplied -1 by $x + 1$, again forgetting to consider $x + 1$ being negative: $2x + 5 \leq -x - 1$ , rearrange and divide by 3 for $x \leq -2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Domain is the union, so $[-4, \infty] \cup [-\infty, -2]$, so $R - { 1 }$. This is obviously not right, as plugging in -3 shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in inequalities where the denominator is some function g(x), if I split it and choose to multiply, will I have to solve each part of the inequality twice(once assuming g(x) is positive, a second time assuming it's negative) and then check which solution makes sense by plugging in numbers?&lt;/p&gt;&#xA;" OwnerUserId="1956" LastEditorUserId="1956" LastEditorDisplayName="" LastEditDate="2010-09-25T23:05:00.250" LastActivityDate="2010-09-26T00:34:45.693" Title="How do I solve inequalities of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5451" PostTypeId="2" ParentId="5450" CreationDate="2010-09-25T21:07:40.203" Score="4" ViewCount="0" Body="&lt;p&gt;The solution set of the inequality $$|f(x)/g(x)|\geq1$$ is equal to the union of the solutions sets of the inequalities $$f(x)/g(x)\leq-1$$ and $$f(x)/g(x)\geq1,$$ so solve these two and compute the union.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-25T21:07:40.203" CommentCount="11" />
  <row Id="5452" PostTypeId="2" ParentId="5447" CreationDate="2010-09-25T21:25:47.457" Score="2" ViewCount="0" Body="&lt;p&gt;In general, it is true that every polynomial splits (i.e., factors into linear factors) over the complex number field. However, finding this factorization for polynomials of degree up to four is manageable, and factoring something of degree higher than 4 is impossible (in the sense that there does not exist, for a general polynomial of degree 5 or higher, an algorithm for finding its roots). Well, actually, this isn't entirely true. I should say that there is no formula for the roots of a general quintic expressed by radicals. There are other methods: for instance, the so called ultraradicals, approximations, etc.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-25T21:25:47.457" />
  <row Id="5453" PostTypeId="2" ParentId="5450" CreationDate="2010-09-25T21:36:27.947" Score="4" ViewCount="0" Body="&lt;p&gt;Your answer of $[-4,-2]$ appears to be the solution set to $\left|\frac{2x + 5}{x + 1}\right| \leq 1$.  Without seeing all of your work, I can't be sure what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for inequalities of this type (actually, pretty much any inequalities involving a quotient with variables in the denominator), the method I'd suggest is the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Split it into two inequalities to solve separately: $\frac{f(x)}{g(x)} \geq 1$ or $-1\geq \frac{f(x)}{g(x)}$.  Once these are solved, take the union of their solution sets (only one of these inequalities must be true for your original inequality to hold).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each of these two inequalities, manipulate them so that the comparison is to 0 instead of 1 or -1.  For example, $\frac{f(x)}{g(x)} \geq 1$ can be rewritten as $\frac{f(x)}{g(x)}-1\geq 0$ or $\frac{f(x)-g(x)}{g(x)} \geq 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When solving an inequality that compares an expression of the form $\frac{a(x)}{b(x)}$ to zero, the solution set will be the union of intervals whose endpoints are at the zeros of $a(x)$ or $b(x)$.  That is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;determine the zeros of $a(x)$ and $b(x)$&lt;/li&gt;&#xA;&lt;li&gt;plot the zeros on a number line&#xA;&lt;ul&gt;&#xA;&lt;li&gt;use open circles (excluded values) for the zeros of $b(x)$&lt;/li&gt;&#xA;&lt;li&gt;if the inequality is $\leq$ of $\geq$, use closed circles (included values) for the zeros of $a(x)$&lt;/li&gt;&#xA;&lt;li&gt;if the inequality is $&amp;lt;$ or $&amp;gt;$, use open circles for the zeros of $a(x)$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;for each region between two zeros, test a value in the inequality you are solving&lt;/li&gt;&#xA;&lt;li&gt;if that value makes the inequality true, then the region is part of the solution set&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is sometimes called the &quot;boundary algorithm&quot; or the &quot;test-point method&quot; for solving inequalities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  Regarding &quot;if I split it and choose to multiply, will I have to solve each part of the inequality twice,&quot; well...  sort of.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tackling the first: multiplied 1 by $x + 1$, forgetting to account for the fact that $x + 1$ could potentially be negative: $2x + 5 \geq x + 1$, then rearrange for $x \geq -4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, you should have $(2x+5\geq x+1\text{ and }x\geq -1)\text{ or }(2x+5\leq x+1\text{ and }x&amp;lt;-1)$.  In solving, remember that the &quot;and&quot; becomes an intersection of solution sets and the &quot;or&quot; becomes a union of solution sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'd indicated in the comments, I use the method I originally described in order to avoid this sort of casework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit 2&lt;/em&gt;:  As an example of the method that I describe above, let's solve $\frac{2x+5}{x+1}\geq 1$.  Subtract 1 from both sides, get a common denominator on the left side, and combine the terms to get $\frac{x+4}{x+1}\geq 0$.  The only zero of the numerator is -4; the only zero of the denominator is -1.  Graphing these on a number line gives:&#xA;$$\leftarrow\!\!\underset{-4}{-\!\!\bullet\!\!-}\!\!-\!\!\underset{-1}{-\!\!\circ\!\!-}\!\!\rightarrow$$&#xA;(open circle at -1 because $x=-1$ makes the quotient undefined, closed circle at -4 because $x=-4$ makes the numerator 0, so the quotient is 0, and the inequality is $\geq 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are 3 intervals to test: $(-\infty,-4]$, $[-4,-1)$, and $(-1,\infty)$.  Relatively simple-to-test values in the intervals are -5, -2, and 0, respectively. $\frac{-5+4}{-5+1}=\frac{-}{-}=+\geq 0$; $\frac{-2+4}{-2+1}=\frac{+}{-}=-\not\geq 0$; $\frac{0+4}{0+1}=\frac{+}{+}=+\geq 0$.  So, the solution is $(-\infty,-4]\cup(-1,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To complete the original problem, you still need to solve $\frac{2x+5}{x+1}\leq -1$ and take the union of that solution with the set we just found.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-26T00:34:45.693" LastActivityDate="2010-09-26T00:34:45.693" />
  <row Id="5454" PostTypeId="1" AcceptedAnswerId="5456" CreationDate="2010-09-25T21:46:33.617" Score="1" ViewCount="96" Body="&lt;p&gt;How do we go from $$\frac{1}{1+(x/2)+(x^2/3)+(x^3/4)+\dots}$$ to $$1-(x/2)-(x^2/12)-\dots$$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could consolidate the series in denominator in some form, then I could use binomial expansion in numerator by raising it to power -1, but I don't see the one in denominator in any known form other than converting it to log.&lt;/p&gt;&#xA;" OwnerUserId="754" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-25T22:48:19.407" LastActivityDate="2010-09-25T22:48:19.407" Title="How is this series in denominator converted to a series in numerator?" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="5455" PostTypeId="2" ParentId="5450" CreationDate="2010-09-25T21:56:12.230" Score="3" ViewCount="0" Body="&lt;p&gt;You can also rewrite in equivalent form as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^2(x) \ge g^2(x) \wedge g(x) \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (f(x)+g(x))(f(x)-g(x)) \ge 0 \wedge g(x) \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular case, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (3x+6)(x+4) \ge 0 \wedge x+1 \ne 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x \ge -2 \vee x \le -4) \wedge x \ne -1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\wedge$ stand for AND and $\vee$ stands for OR, so the solution set is $x$ is not -1 AND either $x \ge -2$ OR $x \le -4$, i.e $$\mathbb{R} - ((-4,-2) \cup \{1\})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you seem to have the answer reversed, my guess it that you multiplied one of you inequalities by a negative number at some stage and forgot to reverse the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance $$ x &amp;lt; 1$$&#xA;Mulitplying by $-1$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -x &gt; -1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you forget to reverse the inequality it gives $$ -x &amp;lt; -1$$ which is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-25T21:56:12.230" CommentCount="2" />
  <row Id="5456" PostTypeId="2" ParentId="5454" CreationDate="2010-09-25T22:14:02.963" Score="2" ViewCount="0" Body="&lt;p&gt;Using $$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - \frac{x}{\log(1-x)} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the series for $$\frac{x}{\log(1-x)}$$ is well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the question asked on this very site here: &lt;a href=&quot;http://math.stackexchange.com/questions/5058/formula-for-the-harmonic-series-h-n-sum-k1n-1-k-due-to-gregorio-fontana&quot;&gt;http://math.stackexchange.com/questions/5058/formula-for-the-harmonic-series-h-n-sum-k1n-1-k-due-to-gregorio-fontana&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the page here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&lt;/a&gt;. (search the page for Gregory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series expansion is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{\log(1-x)} = \sum_{k=0}^{\infty} C_{k} x^{k} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $C_{k}$ are called as Gregory coefficients. The wiki page I linked above tells you how they can be calculated using a recursive formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots} = - \sum_{k=0}^{\infty} C_{k} x^k = 1 - \frac{x}{2} - \frac{x^2}{12} - \frac{x^3}{24} - \dots$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-25T22:14:02.963" CommentCount="2" />
  <row Id="5457" PostTypeId="2" ParentId="5454" CreationDate="2010-09-25T22:20:25.113" Score="1" ViewCount="0" Body="&lt;p&gt;Let $f(x) = \sum f_n x^n$ be a formal power series.  To compute the inverse of $1 - x f(x)$, write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1 - x f(x)} = \sum_{n \ge 0} x^n f^n(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the coefficient of $x^k$ it suffices to compute the contributions from the first $k$ terms on the RHS, so this results in a finite algorithm which will let you compute any particular coefficient of the inverse.  If $f$ has special properties (for example it is a polynomial or more generally meromorphic) then often one can even find a nice closed form.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-25T22:20:25.113" />
  <row Id="5458" PostTypeId="2" ParentId="5416" CreationDate="2010-09-25T23:00:31.773" Score="1" ViewCount="0" Body="&lt;p&gt;This goes along with the question of how to interpret&#xA;$$&#xA;\frac{\partial}{\partial z} \;=\; \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)&#xA;$$&#xA;for non-holomorphic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible answer is that $\partial f / \partial z$ is the average value of &#xA;$$&#xA;\frac{f(w) - f(z)}{w - z}&#xA;$$&#xA;as $w$ takes values in a small circle centered at $z$.  To be precise:&#xA;$$&#xA;\frac{\partial f}{\partial z}(z) \;=\; \lim_{r\to 0}\frac{1}{2\pi}\int_0^{2\pi} \frac{f(z+re^{it}) - f(z)}{re^{it}} dt.&#xA;$$&#xA;Similarly, $\partial f / \partial \overline{z}$ is the average value of&#xA;$$&#xA;\frac{f(w) - f(z)}{\overline{w} - \overline{z}}&#xA;$$&#xA;as $w$ takes values in a small circle centered at $z$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-25T23:00:31.773" />
  <row Id="5459" PostTypeId="2" ParentId="201" CreationDate="2010-09-25T23:18:23.443" Score="2" ViewCount="0" Body="&lt;p&gt;Another proof is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the numbers $$9^{2^n} + 1, \ \  n = 1,2,\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $$9^{2^n} + 1 = 0 \mod p$$ then we have that, for $ m &gt; n$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9^{2^m} + 1 = (9^{2^n})^{2^{m-n}} + 1 = (-1)^{2^{m-n}} + 1 = 1+1 = 2 \mod p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if one term of the sequence is divisible by a prime, none of the next terms are divisible by that prime, i.e. if you write out the factors of the terms of the sequence, each term of this sequence gives rise to a prime not seen before!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a curiosity, it can be shown that each number in the sequence has at least one prime factor &gt; 40. See this question on this very site: &lt;a href=&quot;http://math.stackexchange.com/questions/3243/does-92n-1-always-have-a-prime-factor-larger-than-40/&quot;&gt;http://math.stackexchange.com/questions/3243/does-92n-1-always-have-a-prime-factor-larger-than-40/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-25T23:18:23.443" CommentCount="2" />
  <row Id="5460" PostTypeId="2" ParentId="5392" CreationDate="2010-09-25T23:22:57.000" Score="1" ViewCount="0" Body="&lt;p&gt;The localization and the dispersion probabilistic measures can be &quot;seen&quot; as the corresponding momentums of mechanical systems of &quot;probabilistic masses&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expectation has the following mechanical interpretation. Given that $F(x)$ is the &quot;probalilistic mass&quot; contained in the interval $0\le X\lt x$ (in one dimention), the mathematical expectation of the random variable $X$ is the static momentum with respect to the origin of the &quot;probabilistic masses&quot; system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance is the mechanical analog of the inertia momentum of the &quot;probabilistic masses&quot; system with respect to the center of masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance of the random variable $X$ is the 2nd order momentum of $X-m$ where $m$ is the expectation of $X$, i.e. $m=E(X)=\displaystyle\int x dF(x)$ ($F(x)$ is the repartion function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $k$-order momentum is the expectation of $(X-m)^k$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-09-25T23:22:57.000" />
  <row Id="5461" PostTypeId="2" ParentId="213" CreationDate="2010-09-26T00:02:24.270" Score="0" ViewCount="0" Body="&lt;p&gt;I wish I'd learned about special functions earlier.  The subject is a treasure trove of results that were commonly known a century ago but now few people know.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-26T00:02:24.270" CommunityOwnedDate="2010-09-26T00:02:24.270" />
  <row Id="5462" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T01:05:01.477" Score="1" ViewCount="0" Body="&lt;p&gt;Pollard, &lt;em&gt;A User's Guide to Measure Theoretic Probability&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-26T01:05:01.477" CommunityOwnedDate="2010-09-28T07:16:29.983" />
  <row Id="5463" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T01:06:01.020" Score="1" ViewCount="0" Body="&lt;p&gt;Dudley, &lt;em&gt;Real Analysis and Probability&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-26T01:06:01.020" CommunityOwnedDate="2010-09-28T07:16:29.983" />
  <row Id="5464" PostTypeId="2" ParentId="5377" CreationDate="2010-09-26T02:18:55.970" Score="1" ViewCount="0" Body="&lt;p&gt;Here's a simplified form of your original argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct a Turing machine A which sequentially runs on all proofs in ZFC and checks if the claim &quot;A does not halt&quot; is proved. If such a proof is found, A halts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that A can know its encoding - see Kleene's recursion theorem - and even if it can't this &quot;obstacle&quot; can be overcome. So it seems to me such A is constructible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if A halts then it is because it has found a proof for &quot;A does not halt&quot;. This is impossible, since A halted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have just proved the claim &quot;A does not halt&quot;, so such a proof exists in ZFC. So A must halt on this proof - a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully this version makes the error more clear.  The proof that &quot;A does not halt&quot; assumes the consistency of ZFC, and is therefore a proof in ZFC + Con(ZFC), not ZFC.  In fact, the above argument is essentially a proof that Con(ZFC) is not provable in ZFC.)&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-26T02:18:55.970" CommentCount="1" />
  <row Id="5465" PostTypeId="1" CreationDate="2010-09-26T03:09:51.650" Score="5" ViewCount="172" Body="&lt;p&gt;I'm talking about the shape made up of a rectangle with a semi-circle at each end.  Does it have a particular name?&lt;/p&gt;&#xA;" OwnerUserId="2139" LastActivityDate="2010-09-26T12:07:00.697" Title="What is the proper geometric description of a the oval used for a horse racetrack?" Tags="&lt;geometry&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="5466" PostTypeId="1" AcceptedAnswerId="5475" CreationDate="2010-09-26T03:15:00.497" Score="0" ViewCount="167" Body="&lt;p&gt;I was told to write the 3 cube roots of 27 in rectangular form how is this to be done? I know how to do cube roots on something like 5+0i in polar form, but how do you do it with just a number?&lt;/p&gt;&#xA;" OwnerUserId="2135" LastActivityDate="2010-09-26T09:25:09.030" Title="How do you find the cube roots of  just a number?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5468" PostTypeId="2" ParentId="5465" CreationDate="2010-09-26T03:34:07.780" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wiktionary.org/wiki/obround&quot; rel=&quot;nofollow&quot;&gt;Obround&lt;/a&gt;, apparently.  I don't know Wiktionary's source.  This definition of obround does not appear in OED, for example.  Googling indicates that this definition is commonly used for machine parts having this shape.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-09-26T04:27:43.150" LastActivityDate="2010-09-26T04:27:43.150" CommentCount="1" />
  <row Id="5469" PostTypeId="2" ParentId="3724" CreationDate="2010-09-26T03:34:23.287" Score="0" ViewCount="0" Body="&lt;p&gt;The answer from Soarer has a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is not correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.2.2 If the C side is equal to the other side, then C side minus C are all genuine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that group A has 2 fake coins. In that case we could have 1 fake at the C size (within the 24 coins) and another fake coin on the other size (within 25 coins). C is no fake when A has two fakes. This will make C side equal to the other side also and we could not claim that C side minus C are all genuine.&lt;/p&gt;&#xA;" OwnerUserId="2140" LastActivityDate="2010-09-26T03:34:23.287" />
  <row Id="5470" PostTypeId="2" ParentId="5465" CreationDate="2010-09-26T03:42:31.557" Score="5" ViewCount="0" Body="&lt;p&gt;I'm pretty sure they actually use four &lt;a href=&quot;http://mathworld.wolfram.com/CornuSpiral.html&quot; rel=&quot;nofollow&quot;&gt;clothoid&lt;/a&gt; arcs joined together in practice, e.g. &lt;a href=&quot;http://dx.doi.org/10.1111/j.2042-3306.1975.tb03240.x&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. This has a lot to do with the fact that the clothoid is the curve whose curvature is directly proportional to its arclength; an abrupt variation in curvature would equate to an abrupt variation in centripetal force, which can be bad for the racehorses (or even racecars, for that matter).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a simulated clothoid track drawn in &lt;em&gt;Mathematica&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XfWQ8.png&quot; alt=&quot;clothoid track&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to show that the bends are honest-to-goodness clothoids, I drew the clothoid corresponding to the lower right portion of the track in full (the dashed gray one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parametrization used is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x\qquad y)=\left(\sqrt{\frac{\pi}{2}}C\left(\sqrt{\frac{2}{\pi}}s\right)\qquad \sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}s\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C(x)$ and $S(x)$ are the &lt;a href=&quot;http://dlmf.nist.gov/7.2.iii&quot; rel=&quot;nofollow&quot;&gt;Fresnel integrals&lt;/a&gt;; I leave you to verify using those expressions that the curvature of the clothoid is indeed directly proportional to the arclength.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-26T05:31:53.197" LastActivityDate="2010-09-26T05:31:53.197" CommentCount="7" />
  <row Id="5471" PostTypeId="2" ParentId="5466" CreationDate="2010-09-26T03:44:11.583" Score="2" ViewCount="0" Body="&lt;p&gt;$z^\frac{1}{n} = +\sqrt[n]{\vert z\vert} e^{ i \frac{\arg (z) + 2k\pi }{n} } $, where $\arg(z)$ is the principal argument of the complex number $z$, and $k = {0, 1, ... n-1}$. In this case, since the complex number $z$ is purely real, $\arg(z) = 0$. Therefore we are left with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27^\frac{1}{3} = \sqrt[3]{27} e^{i\frac{2k\pi}{3} }$, which, iterating over $k$, gives us the three roots $3, 3e^{i\frac{2\pi}{3}}$, and $3e^{i\frac{4\pi}{3}} $. Now, to express them in rectangular form, simply use Euler's formula; $e^{ix} =  \cos(x) + i \sin(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final answers should then be $3, \frac{3}{2}(-1+i\sqrt{3})$, and $\frac{3}{2}(-1-i\sqrt{3})$.&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-26T09:25:09.030" LastActivityDate="2010-09-26T09:25:09.030" CommentCount="2" />
  <row Id="5472" PostTypeId="2" ParentId="5466" CreationDate="2010-09-26T03:44:40.397" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, a number has three cube roots. Think about how you would factor the polynomial $x^3-27$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-26T03:44:40.397" CommentCount="1" />
  <row Id="5473" PostTypeId="1" AcceptedAnswerId="5474" CreationDate="2010-09-26T03:46:19.667" Score="1" ViewCount="111" Body="&lt;p&gt;First of all, I had no idea what to title this question, so feel free to change it to something more appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set defined as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a\frown b\mid a \in A \wedge b \in \{c\frown d\mid c \in C \wedge d \in D\}\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that logically equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a\frown (c\frown d)\mid a \in A \wedge c \in C \wedge d \in D\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $\frown$ is an associative relation.&lt;/p&gt;&#xA;" OwnerUserId="2141" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T20:35:16.637" LastActivityDate="2010-11-24T20:35:16.637" Title="Expansion of terms in set definition?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="5474" PostTypeId="2" ParentId="5473" CreationDate="2010-09-26T03:55:27.147" Score="1" ViewCount="0" Body="&lt;p&gt;The sets are equal, regardless of whether $\frown$ is associative.  For instance, a straightforward &quot;element chase&quot; shows this. More directly, note that $x=a\frown b$ for some $b$ of the form $c\frown d$ means precisely the same thing as $x=a\frown(c\frown d)$ for some $c$ and $d$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-09-26T03:55:27.147" CommentCount="2" />
  <row Id="5475" PostTypeId="2" ParentId="5466" CreationDate="2010-09-26T03:59:49.687" Score="2" ViewCount="0" Body="&lt;p&gt;If you know how to find the cube roots of $5+0i$ in polar form, you're most of the way there already.  $27=27+0i$, so you can use the same techniques on it as on $5+0i$ to get the polar form of the cube roots.  To get to rectangular form from polar form, evaluate the polar form&amp;mdash;that is, $$re^{i\theta}=r(\cos\theta+i\sin\theta)=r\cos\theta+(r\sin\theta)i.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-26T03:59:49.687" />
  <row Id="5476" PostTypeId="2" ParentId="5416" CreationDate="2010-09-26T04:11:06.040" Score="4" ViewCount="0" Body="&lt;p&gt;The dee zee bar operator in the secret blogging seminar: &lt;a href=&quot;http://sbseminar.wordpress.com/2009/01/08/what%C2%B4s-up-with-dee-zee-bar/&quot; rel=&quot;nofollow&quot;&gt;What´s up with dee zee bar&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="776" LastActivityDate="2010-09-26T04:11:06.040" />
  <row Id="5477" PostTypeId="1" CreationDate="2010-09-26T04:49:16.450" Score="2" ViewCount="76" Body="&lt;p&gt;I have attempted to google for this, but the searching is marred by the close relationship of Jordan Algebras and Quantum Mechanics. I have been passively thinking of this question for ages, tonight I finally post it. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;My question is fairly simple. Similar to the universal enveloping algebra for Lie algebras, we should be able to form a universal enveloping algebra for Jordan algebras. So question one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we indeed construct this &quot;Universal enveloping algebra&quot;?&lt;/strike&gt;(Thanks Matt)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can form a Universal enveloping algebra for Jordan algebras. So;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there some sort of Serre relations for Jordan algebras?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then as we do for Lie algebras,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we find &quot;quantum Jordan 'Serre' relations&quot; for quantum Jordan algebras?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or just;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we deform the Jordan Universal Enveloping algebra in a similar way that we deform Lie Enveloping algebras to their quantized versions? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would be happy with an explanation or a reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-01T06:04:02.560" LastActivityDate="2010-10-01T06:04:02.560" Title="Quantum Jordan Algebras(Still wondering!)" Tags="&lt;abstract-algebra&gt;&lt;reference-request&gt;&lt;jordan-algebras&gt;" AnswerCount="1" />
  <row Id="5478" PostTypeId="2" ParentId="5477" CreationDate="2010-09-26T05:31:48.860" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, there is a notion of universal enveloping algebra for Jordan algbera; it is left adjoint&#xA;to the functor that associates a Jordan algebra to an associative algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no analogue of the Poincare--Birkhoff--Witt theorem for Jordan algebras,&#xA;and so the map from the Jordan algebra to its universal enveloping algebra need not be injective.  (I forget where I learned this, but it should be discussed in most texts that discuss Jordan algebras; indeed, there is an adjective --- which I also forget --- used to separate those Jordan algebras that do embed into their enveloping algebra from those that&#xA;don't.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for deformation quantization, I don't know.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-26T05:31:48.860" />
  <row Id="5480" PostTypeId="1" AcceptedAnswerId="5492" CreationDate="2010-09-26T05:40:06.160" Score="3" ViewCount="301" Body="&lt;p&gt;Or, in other words, why are all functions not surjective? Isn't any subset of the codomain which isn't part of the image rather arbitrary?&lt;/p&gt;&#xA;" OwnerUserId="2058" LastActivityDate="2010-09-26T14:04:16.200" Title="How is the codomain for a function defined?" Tags="&lt;functions&gt;" AnswerCount="6" CommentCount="1" />
  <row Id="5481" PostTypeId="1" AcceptedAnswerId="5491" CreationDate="2010-09-26T05:46:10.890" Score="1" ViewCount="105" Body="&lt;p&gt;I am looking at Perturbation Theory by E.J. Hinch. The author introduces Van Dyke's matching rule: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(m term inner)(n term outer) = (n term outer)(m term inner)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be used to match the terms in an inner solution to the the terms in the outer solution for some perturbed differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hinch when comparing with introducing intermediate variables says: 'Van Dyke's matching rule usually works and is more convenient.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know an example where this matching rule doesn't work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;In section $5.2.6$ of the book above, the author gives an example of where the matching rule is meant to fail. Is this an example where the matching rule actually fails or just shows some care in comparing terms is required? In particular, would the intermediate variables approach have any problems?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-26T08:38:38.417" LastActivityDate="2010-09-26T09:11:16.160" Title="Van Dyke's matching rule" Tags="&lt;differential-equations&gt;&lt;perturbation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5482" PostTypeId="1" CreationDate="2010-09-26T06:01:38.737" Score="2" ViewCount="140" Body="&lt;p&gt;Could someone please help me understand this proof given in an article by William Miller &lt;/p&gt;&#xA;&#xA;&lt;p&gt;its supposed to follow from the prime number theorem that given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x)$ which is the sum of all primes less than or equal to $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\theta(x)$ which is the sum of the logarithm of all primes less than or equal to x &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x)\sim \frac{x^2}{2\log x}$ and $\theta(x) \sim x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following identity is used: $\theta(x) = \int_1^x \log(t)\mathrm{d}(\pi(t))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\pi(t)$ is the prime counting function. I don't understand why this is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\sim$ means asymptotic to i.e. $\lim_{n\to\infty} \frac{f(x)}{g(x)}=1$&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-26T06:14:03.830" LastActivityDate="2010-09-26T06:45:00.243" Title="Prime Number theorem and the prime counting function" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5483" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T06:28:31.547" Score="4" ViewCount="0" Body="&lt;p&gt;This is something that trips people up the more they dwell on it.  And depending on how you like to think about mathematical objects, you have different preferred answers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More often than not, I tend to think of a function as consisting of three things: a domain $A$, a range $B$ and a subset $S$ of $A \times B$ which satisfies this condition, that for any $a \in A$ there is a unique $b \in B$ such that $(a,b) \in S$.  To be precise, a &quot;function&quot; $f$ is a triple $(A,B,S)$ where $S \subset A\times B$ satisfies the above criterion. We say $f : A \to B$ and $f(a) = b$ provided $(a,b) \in S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above formalism, you can't describe a function &lt;em&gt;without&lt;/em&gt; specifying the domain and range before hand. To be specific, here are two different ways of specifying the function that in a calculus class would just be called $x^2$:  $f=(\mathbb R, \mathbb R, \{(x,x^2) : x \in \mathbb R\})$.   $g=(\mathbb R, [0,\infty), \{(x,x^2) : x \in \mathbb R\})$.  So in this formalism, $f$ is not an onto function, while $g$ is an onto function.  Although $f(x)=g(x)$ for all $x$ in the domain of $f$ and $g$, $f \neq g$ since they have different ranges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't use a formalism like the above, yes, the range (or co-domain as you say) is just any arbitrary set containing the image.  Frequently this is taken care of by considering the range to be a typical &quot;universal&quot; set. In a standard calculus course what happens is people never really mention the range, they only talk about the image. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-26T06:28:31.547" CommentCount="7" />
  <row Id="5484" PostTypeId="2" ParentId="4312" CreationDate="2010-09-26T06:36:47.443" Score="1" ViewCount="0" Body="&lt;p&gt;It turns out that, as T.. had guessed, the problem is indeed NP-hard. I realised this after participating in some discussions in the comments of &lt;a href=&quot;http://www.toastandtesla.com/engineering/maths/good-friendship-requires-good-accounting/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; where the same problem turned up again, and &lt;a href=&quot;http://www.reddit.com/r/math/comments/dioxr/resolving_debts_to_your_friends_with_minimal/c10huhd&quot; rel=&quot;nofollow&quot;&gt;the corresponding Reddit discussion&lt;/a&gt;. The solution I am posting here is based on the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a lemma: The optimal solution for $m+n$ people cannot have more than $m+n-1$ nonzero entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is by inductive construction of a solution with $m+n-1$ entries. For $m = n = 1$, the solution trivially has 1 entry. When $m+n &gt; 2$, let $a_{11} = \min(r_1,c_1)$. This satisfies one row or column (or both, if $r_1 = c_1$), leaving a subproblem of size $m+n-1$ whose solution has at most $m+n-2$ entries. This completes the proof of the lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a solution exists with fewer than $m+n-1$ entries. Consider the network flow view of the problem, with $m+n$ nodes connected by edges. If there are fewer than $m+n-1$ edges, the underlying undirected graph is disconnected. Then each connected component is &quot;self-sufficient&quot; in that it has no net in- or out-flow. That means that the sum of the corresponding subset of $\{r_i\} \cup \{-c_j\}$ is zero. So, this problem includes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;subset sum problem&lt;/a&gt;, which is NP-complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, suppose an instance of the subset sum problem is given as a set of integers $X = \{x_1,\ldots,x_k\}$, and the problem is to determine whether there is a nonempty subset of $X$ whose sum is zero. Let these be the debts $d(p)$ among a set of people $P$ as in the question, introducing an additional person with $d(p) = -\sum X$ to balance the books. Then there is a solution with fewer than $|P|-1$ entries if and only if there is a subset of $X$ with sum zero. (The above lemma is useful for proving the &quot;if&quot; part.) Therefore the subset sum problem can be reduced to this problem.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-26T06:36:47.443" />
  <row Id="5485" PostTypeId="2" ParentId="5482" CreationDate="2010-09-26T06:45:00.243" Score="2" ViewCount="0" Body="&lt;p&gt;The notation&#xA;$$\int \log(t)\ d(\pi(t))$$&#xA;is an example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stieltjes_integral&quot; rel=&quot;nofollow&quot;&gt;Stieltjes integral&lt;/a&gt;. Some like this (I don't)&#xA;when discussing summation, as integrating with a step function $f(t)$&#xA;inside the $d$ notation is essentially a summation. Here $\pi(t)$&#xA;is constant, save at prime values of $t$ where it jumps by $1$. This means&#xA;that&#xA;$$\int_a^b g(t)\ d(\pi(t))=\sum_{p\ \mathrm{prime}}g(p)\times 1.$$&#xA;Here the summation is over all primes in the interval from $a$ to $b$&#xA;(I can't remember whay you do about the endpoints; one reason I don't&#xA;like the notation). If you like you can think that&#xA;$$\int_a^b g(t)\ d(\pi(t))=\int_a^b g(t)\pi'(t)\ dt$$&#xA;where $\pi'$ is seen as a generalized function, so as a bunch of delta-functions&#xA;at the primes. Again one has to do the right thing (or is it the left thing? :-))&#xA;at the endpoints.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-26T06:45:00.243" />
  <row Id="5486" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T06:50:11.617" Score="1" ViewCount="0" Body="&lt;p&gt;Yes it is arbitrary, and in a sense this is exactly the problem. When we have a function we would like to be able to talk about the inverse of the function (assuming it is 1-1.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following: I have some function $f$ that maps $\mathbb{N} \rightarrow \mathbb{R}$ such that $$f(x) = (e^\sqrt{3.7x}+137.2)^3$$ The point here is that it is much easier to say that my codomain is $\mathbb{R}$ and my function is not surjective than to explicitly list the image of my function and say that $f$ is surjective on this subset of $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that every output will be a real number, but it is hard to say exactly which numbers I will get out of $f$ and which real numbers I can never get. To get to the point, if my function was surjective I know that I could plug any real number into $f^{-1}$. But I know it's not and I therefore have to be very careful with my inputs to $f^{-1}$, restricting my attention only to $f$'s image (otherwise the inverse is undefined.) In this case, we say that $f$ does not have the 'nice' property of surjectivity.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-09-26T07:07:46.933" LastActivityDate="2010-09-26T07:07:46.933" CommentCount="0" />
  <row Id="5487" PostTypeId="1" AcceptedAnswerId="5488" CreationDate="2010-09-26T07:49:39.227" Score="-1" ViewCount="86" Body="&lt;p&gt;What is the k(e) and r in Coulomb's law (scalar form)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coulomb%27s_law&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2145" LastActivityDate="2010-09-26T08:13:13.377" ClosedDate="2010-09-26T11:24:53.393" Title="What is the k(e) and r in Coulomb's law?" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5488" PostTypeId="2" ParentId="5487" CreationDate="2010-09-26T08:13:13.377" Score="0" ViewCount="0" Body="&lt;p&gt;$k$ is the columb constant, and $r$ is the distance between two charges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the thing that you really don't understand?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-09-26T08:13:13.377" />
  <row Id="5489" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T08:30:59.407" Score="1" ViewCount="0" Body="&lt;p&gt;In the same sense you could ask why all functions aren't total functions. The real question is perhaps &quot;what is the use of writing $f:\ S\rightarrow T$&quot; or even more general &quot;what is the use of writing &quot;$v \in S$&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd expect that in most cases where it is mentioned explicitly it's a way to say &quot;this object belongs to this class of objects&quot; and is done because it belonging to that class is somehow considered relevant or useful at that point. This might sound a bit obvious, but I don't think there is much more to it.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2010-09-26T08:30:59.407" CommentCount="6" />
  <row Id="5490" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T08:53:49.270" Score="0" ViewCount="0" Body="&lt;p&gt;As far as I know, relations can be defined in two ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given two sets $A$ and $B$, any subset $R$ of $A\times B$ is a relation. The domain is $\{x\in A:\exists y(y\in B,(x,y)\in R)\}$. The codomain is $\{y\in B:\exists x(x\in A,(x,y)\in R)\}$. Obviously the domain is a subset of $A$ and the codomain is a subset of $B$. Not every relation is surjective.&lt;/li&gt;&#xA;&lt;li&gt;Every set that only contains ordered pairs is a relation. The domain is $\{x\in\bigcup\bigcup R:\exists y((x,y)\in R)\}$. The codomain is $\{y\in\bigcup\bigcup R:\exists x((x,y)\in R)\}$. Every relation is surjective.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My preferred definition is the first one, but you may like the second one, instead.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-09-26T08:53:49.270" />
  <row Id="5491" PostTypeId="2" ParentId="5481" CreationDate="2010-09-26T09:11:16.160" Score="0" ViewCount="0" Body="&lt;p&gt;Okay I have found an example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{xx} + \frac{1}{x}f_x+f_x^2 +\epsilon ff_x=0$ for $x&gt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=0$ on $x=1$ and $f\rightarrow 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is something close to $\log(1+\log(x)/\log(1/\epsilon))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Van Dyke's rule fails because when expanding an infinite number terms are of the same order (in size). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with the intermediate variable approach since we are scaling by $\epsilon^{-\alpha}$ with $\alpha \in[0,1]$ we are able to choose $\alpha$ so that there are only a finite number of terms the same order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example is completely contrived though, I would feel much better with a more natural example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your effort J.Meyer and J.M.&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-09-26T09:11:16.160" />
  <row Id="5492" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T09:46:45.360" Score="7" ViewCount="0" Body="&lt;p&gt;Often what is important is not functions but &lt;em&gt;collections&lt;/em&gt; of functions, especially subsets of the set of functions between two sets $X$ and $Y$, some of which will be surjective and some of which won't.  What also doesn't get emphasized a lot at the level of elementary set theory is the compositional structure of functions, e.g. a function $f : X \to Y$ can be composed with a function $g : Y \to Z$ to give a function $fg : X \to Z$.  For the purposes of studying this compositional structure (for example if $X = Y = Z$) it is generally important not to require that $f$ and $g$ be surjective or you will miss out on structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a simple way to define a dynamical system is just as a function $f : X \to X$.  Whether $f$ is surjective or not is an important aspect of classifying the dynamics of $f$, or in other words of classifying the behavior of the sequences $\{ x, f(x), f^2(x), f^3(x), ... \}$ for various $x$.  This sequence is not well-defined if you pretend that the domain and codomain of $f$ are different just because the range and the codomain aren't equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another way of saying that a function really consists of three pieces of data (a domain, a codomain, and the mapping from one to the other), but the reason these three pieces of data are all important is really the compositional structure.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-26T09:46:45.360" />
  <row Id="5493" PostTypeId="2" ParentId="5465" CreationDate="2010-09-26T12:07:00.697" Score="0" ViewCount="0" Body="&lt;p&gt;It's called a &lt;em&gt;stadium&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-09-26T12:07:00.697" CommentCount="1" />
  <row Id="5494" PostTypeId="1" AcceptedAnswerId="5517" CreationDate="2010-09-26T12:10:38.490" Score="1" ViewCount="211" Body="&lt;p&gt;How can one build a 3x3 linear equation system (3 equations with 3 variables) if the variables are known, for example, a=1, b=2 and c=3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="2145" LastActivityDate="2010-12-10T18:06:59.877" Title="How to build a linear equation system?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="5495" PostTypeId="2" ParentId="5494" CreationDate="2010-09-26T12:12:55.787" Score="2" ViewCount="0" Body="&lt;p&gt;It's ill-posed; you can construct any &lt;strong&gt;nonsingular&lt;/strong&gt; matrix $\mathbf A$ to be multiplied with $(1\;2\;3)^T$ for the corresponding right hand side.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T18:06:59.877" LastActivityDate="2010-12-10T18:06:59.877" CommentCount="0" />
  <row Id="5496" PostTypeId="2" ParentId="329" CreationDate="2010-09-26T12:24:39.960" Score="1" ViewCount="0" Body="&lt;p&gt;Manin and Panchishkin's &lt;a href=&quot;http://lib.org.by/info/M_Mathematics/MT_Number%20theory/Manin%20I.,%20Panchishkin%20A.%20Introduction%20to%20modern%20number%20theory..%20fundamental%20problems,%20ideas%20and%20theories%20%28ISBN%203540203648%29%282ed.,%20Springer,%202005%29%28520s%29_MT_.pdf&quot; rel=&quot;nofollow&quot;&gt;Introduction to Modern Number Theory&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2010-09-26T12:24:39.960" CommunityOwnedDate="2010-09-26T12:24:39.960" CommentCount="2" />
  <row Id="5499" PostTypeId="2" ParentId="5480" CreationDate="2010-09-26T14:04:16.200" Score="3" ViewCount="0" Body="&lt;p&gt;Here is an answer which directly addresses the question in the title:  the codomain has to be given as part of the information telling you what the functions is; it can't be deduced (or in the language of question, &quot;defined&quot;) if all you know is the domain and values of&#xA;the function.  It is an extra piece of data.  (This is why it seems arbitrary to you; you are thinking about how to determine the codomain from the other data, which can't be done!&#xA;You have to be told what it is as part of the initial description of the function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that this is &lt;em&gt;incompatible&lt;/em&gt; with one traditional definition of a function as being a set of ordered pairs.  The set of ordered pairs definition determines the domain and values of the function, but not the codomain.   (I guess that some people do use this definition of function; for them, a function doesn't have a codomain separate from its image.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define a function which has a domain and a codomain, one should instead use the scheme&#xA;described by Ryan: a function is a triple (domain $A$, codomain $B$, set of elements of&#xA;$A\times B$ determining its values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why we introduce the concept of codomain, Qiaochu's answer describes this.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-26T14:04:16.200" />
  <row Id="5500" PostTypeId="1" CreationDate="2010-09-26T14:26:12.770" Score="1" ViewCount="140" Body="&lt;p&gt;Can we represent all algebraic structures in First-Order logic?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2010-09-26T21:52:53.773" Title="Can we represent all algebraic structures in First-Order logic?" Tags="&lt;logic&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5501" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T14:48:51.350" Score="1" ViewCount="0" Body="&lt;p&gt;Billingsley, &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471007102&quot; rel=&quot;nofollow&quot;&gt;Probability and Measure&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-26T14:48:51.350" CommunityOwnedDate="2010-09-26T14:48:51.350" CommentCount="1" />
  <row Id="5502" PostTypeId="1" AcceptedAnswerId="5509" CreationDate="2010-09-26T14:50:12.323" Score="0" ViewCount="45" Body="&lt;p&gt;In relation to this geometric construction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AoQDP.png&quot; alt=&quot;construction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where D is the center of both circles, if the inner radius (&lt;em&gt;x&lt;/em&gt; = length of line segments DA and DE), the angle &lt;em&gt;φ&lt;/em&gt; = ∠CAB, and the length Δ&lt;i&gt;g&lt;/i&gt; of line segment AB are given, what is the length Δ&lt;i&gt;x&lt;/i&gt; of line segment AC?&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-09-26T15:28:01.870" LastActivityDate="2010-09-26T15:42:36.620" Title="What is the difference in radii of two concentric circles given an angle and length of a triangle that is inscribed in the annulus?" Tags="&lt;geometry&gt;&lt;geometric-construction&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5503" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T14:50:41.923" Score="1" ViewCount="0" Body="&lt;p&gt;A quite new and german one: Klenke, &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540255451&quot; rel=&quot;nofollow&quot;&gt;Wahrscheinlichkeitstheorie&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-26T14:50:41.923" CommunityOwnedDate="2010-09-26T14:50:41.923" CommentCount="1" />
  <row Id="5504" PostTypeId="2" ParentId="5500" CreationDate="2010-09-26T15:17:53.973" Score="1" ViewCount="0" Body="&lt;p&gt;I am not sure I understand your question. If, by definition, an algebraic structure consists of a set $A\ne\emptyset$ (or a family of such sets) together with a collection of finitary functions on $A$ (functions $f:A^n\to A$ or even ${\mathbf f}:A^n\to A^m$), finitary relations (subsets of $A^k$), and constants (elements of $A$), then 'algebraic structures' are precisely the structures that first-order concerns itself with (you have to use a multi-sorted version of first-order if you want to appropriately work in the setting of &quot;families of sets&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any such structure can be axiomatized in a first-order language as long as you are willing to allow enough function, relation, and constant symbols into your language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the Lowenheim-Skolem theorems will prevent you from uniquely characterizing infinite structures unless you impose additional conditions that are not first-order (like &quot;completeness&quot;, for the reals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps you mean something else. If you are interested in &lt;em&gt;families&lt;/em&gt; of such structures, which is more natural, then the answer is no in general (simple groups, for example). And many interesting &quot;structures&quot; that appear in algebra are second (or higher) order in nature, and typically first-order falls short to capture them explicitly.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-09-26T15:17:53.973" />
  <row Id="5505" PostTypeId="1" CreationDate="2010-09-26T15:24:22.467" Score="6" ViewCount="195" Body="&lt;p&gt;If you look at &lt;a href=&quot;http://math.stackexchange.com/questions/3573/interesting-taxicab-problem&quot;&gt;this previous problem&lt;/a&gt;, I asked how to find the sum of all the areas between two taxicab geometry circles. However, upon learning about &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space#Motivation&quot; rel=&quot;nofollow&quot;&gt;$\ell^p$ norms&lt;/a&gt;, I thought it would be pretty interesting to extend the problem to all $\ell^p$ norm circles, not just $\ell^1$ (taxicab).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p=1$, then the result has already been found (the total area is $\frac{5k^2-k-4}{2}$). If $p = \infty$, then each &quot;circle&quot; is just a square, and the area is also easily found (I'm too tired to think about it, but I think it would just be $4(k^2-1)$). Is there, however, a general formula for the area of each circle and the total area of the regions between circles in terms of $k$ and $p$; that is, what is the equation for the area of each overlapping region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of an individual circle, if I did it correctly, is the area of a Lamé curve with $r = p$ and a radius of $k-n$ (see the &lt;a href=&quot;http://math.stackexchange.com/questions/3573/interesting-taxicab-problem&quot;&gt;linked problem&lt;/a&gt;), which equals $\displaystyle 4(k-n)^2\frac{(\Gamma(1+\frac{1}{p})^2)}{\Gamma(1+\frac{2}{p})}$. This can be reduced to $\displaystyle 2(k-n)^2 \frac{\Gamma(\frac{1}{p})^2}{p \Gamma(\frac{2}{p})}$ (see equations 41 and 42 &lt;a href=&quot;http://mathworld.wolfram.com/GammaFunction.html#eqn41&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some explanatory pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=5, p=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/lic2e.png&quot; alt=&quot;k=5, p=1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=5, p=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/KYOs7.png&quot; alt=&quot;k=5, p=2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=5, p=3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/0mFHv.png&quot; alt=&quot;k=5, p=3&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-02T18:57:36.037" LastActivityDate="2010-10-05T04:03:42.103" Title="Extension of previous problem, involving $\ell^p$ norm circles" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;functional-analysis&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="5506" PostTypeId="2" ParentId="5505" CreationDate="2010-09-26T15:29:17.017" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/Superellipse.html&quot; rel=&quot;nofollow&quot;&gt;superellipses&lt;/a&gt; would seem to fit your bill, as long as $p &amp;lt; \infty$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-26T15:29:17.017" CommentCount="5" />
  <row Id="5507" PostTypeId="2" ParentId="329" CreationDate="2010-09-26T15:31:55.163" Score="4" ViewCount="0" Body="&lt;p&gt;It depends on the level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an undergraduate interested in &lt;strong&gt;algebraic&lt;/strong&gt; number theory, I would strongly suggest (parts of) Serre's &lt;em&gt;Cours d'arithmetique&lt;/em&gt; and also Samuel's &lt;em&gt;Théorie algébriques des nombres&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a graduate student aiming at a future of research work in number theory, Cassels &amp;amp; Fröhlich is a must.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-26T15:31:55.163" CommunityOwnedDate="2010-09-26T15:31:55.163" />
  <row Id="5509" PostTypeId="2" ParentId="5502" CreationDate="2010-09-26T15:42:36.620" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the triangle DAB. You know two sides and an angle. The length of the third side is $x + \Delta x$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-26T15:42:36.620" CommentCount="1" />
  <row Id="5510" PostTypeId="1" AcceptedAnswerId="5608" CreationDate="2010-09-26T17:44:57.837" Score="2" ViewCount="571" Body="&lt;p&gt;I was maths student but now I'm a software engineer and almost all my knowledge about maths formulas is perished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my client wants to calculate optimal price for perishable products. He has come across an article named &lt;strong&gt;Optimal Pricing Policies for Perishable Products&lt;/strong&gt; by Miguel F. Anjos¤ Russell C.H. Cheng Christine S.M. Currie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a formula which is called Karush-Kuhn-Tucker condition. I'm completely baffled by looking at it, may be my basics are completely vanished that's why I became completely blank. I understood the basic requirement and has come up with a proper solution for that but I'm not sure whether my solution is completely addressing the condition given in this article. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody explain me in plain english about this condition so that I can convert it to algorithm written in a programming language. Or if there is any programming library available for this condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry to ask without giving full information about this condition. I'm assuming someone will know about this condition, if not I'll have to post that entire formula which is of around 2-3 pages. :(&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-27T16:18:38.750" LastActivityDate="2010-12-17T21:19:40.990" Title="Karush-Kuhn-Tucker condition - Lagrange multiplier" Tags="&lt;differential-equations&gt;&lt;integral&gt;&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="5511" PostTypeId="2" ParentId="5441" CreationDate="2010-09-26T17:46:53.417" Score="1" ViewCount="0" Body="&lt;p&gt;Jacod and protter is great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability Theory: A Comprehensive Course by Klenke is worth looking at too.&lt;/p&gt;&#xA;" OwnerUserId="2152" LastActivityDate="2010-09-26T17:46:53.417" CommunityOwnedDate="2010-09-26T17:46:53.417" />
  <row Id="5512" PostTypeId="2" ParentId="5510" CreationDate="2010-09-26T18:19:38.540" Score="3" ViewCount="0" Body="&lt;p&gt;The KKT conditions are necessary conditions for a constrainted optimization problem to have a local minimum. For complex problems these conditions normally motivate the development of minimization algorithms rather than trying to find the values that satisfy the KKT conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is black-boxed optimization software and it is advised to take advantages of these. Perhaps a SQP (Sequential quadratic programming) would be ideal for this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optimization toolbox in Matlab is a really easy to use. The function fmincon is likely to solve your problem. See: &lt;a href=&quot;http://www.mathworks.com/products/optimization/&quot; rel=&quot;nofollow&quot;&gt;http://www.mathworks.com/products/optimization/&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are alternatives though I personally have never used them.  See: &lt;a href=&quot;http://wwwasdoc.web.cern.ch/wwwasdoc/minuit/minmain.html&quot; rel=&quot;nofollow&quot;&gt;http://wwwasdoc.web.cern.ch/wwwasdoc/minuit/minmain.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-26T18:35:13.687" LastActivityDate="2010-09-26T18:35:13.687" CommentCount="2" />
  <row Id="5513" PostTypeId="1" AcceptedAnswerId="5514" CreationDate="2010-09-26T19:19:26.007" Score="0" ViewCount="107" Body="&lt;p&gt;I have a series of problems that involve events that happen one after another, and where the usual formulas like P(A/B)=P(AB)/P(A) don't seem to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a horse race there are 3 horses. E12&#xA;  means horse 1 finishes before horse 2.&#xA;  E123 means horse 1 finishes before 2,&#xA;  which finishes before 3, etc.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If &lt;strong&gt;P(E13)=2/3&lt;/strong&gt; and &lt;strong&gt;P(E23)=1/2&lt;/strong&gt; then&#xA;  &lt;strong&gt;P(E123)=?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I go about solving this kind of problems?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2010-09-26T19:46:00.677" Title="How to solve this kind of Conditional Probability problems?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="5514" PostTypeId="2" ParentId="5513" CreationDate="2010-09-26T19:46:00.677" Score="2" ViewCount="0" Body="&lt;p&gt;This is &lt;strong&gt;not&lt;/strong&gt; a conditional probability question.&#xA;Your particular example has insufficient information to solve it.&#xA;Here $E123$ is the event of the horses finishing in the order 1, 2, 3.&#xA;So (assuming the horses can't tie for a position)&#xA;$$P(E13)=P(E132)+P(E123)+P(E213),$$&#xA;$$P(E23)=P(E123)+P(E213)+P(E231)$$&#xA;and of course&#xA;$$1=P(E123)+P(E132)+P(E213)+P(E231)+P(E312)+P(E321).$$&#xA;Knowing $P(E13)$ and $P(E23)$ isn't enough to determine any one&#xA;of the $P(Eijk)$s.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-26T19:46:00.677" />
  <row Id="5515" PostTypeId="2" ParentId="5432" CreationDate="2010-09-26T19:56:43.197" Score="3" ViewCount="0" Body="&lt;p&gt;What you have discovered is the special case $\rm\ a = 2,\ m = n-1\ $ of the following simple identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\ a^{m}\:\frac{a^n-1}{a-1}\: =\: \frac{a^{m+n}-1}{a-1} - \frac{a^m-1}{a-1}\: =\: 1+a+\cdots+a^{m+n-1} - (1+a+\cdots+a^{m-1})$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it yields $\ 11111000 = 11111111 - 111\ $ for $\rm\ a = 10,\ m = 3,\ n = 5\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, except in the special case when the left hand side is an even perfect number, the right hand side is not the (rearranged) sum of the divisors of the left hand side, so it bears little relation to perfect numbers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-26T23:41:41.400" LastActivityDate="2010-09-26T23:41:41.400" CommentCount="2" />
  <row Id="5516" PostTypeId="2" ParentId="5494" CreationDate="2010-09-26T20:06:49.030" Score="3" ViewCount="0" Body="&lt;p&gt;If this is homework, it's a pretty strange one.   :-?  Anyway, do you like the following system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;x  &amp;amp;=  1   \\\&#xD;&#xA; y &amp;amp;=  2  \\\&#xD;&#xA; z &amp;amp;=  3&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Too much easy, I see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, try this: your system must fulfil the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;A &#xD;&#xA;\begin{pmatrix}&#xD;&#xA;1 \\\&#xD;&#xA;2 \\\&#xD;&#xA;3&#xD;&#xA;\end{pmatrix}  = b&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is a $3\times 3$ matrix and $b$ is a column vector with $3$ coordinates (a $3\times 1$ matrix). Write $a_1, a_2, a_3$ for the columns of $A$. Then this equation is the same as the following equality between vectors in $\mathbb{R}^3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;a_1 + 2a_2 + 3a_3 = b \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick your three (linearly independent) favourite vectors in $\mathbb{R}^3$, put in the places of $a_1, a_2, a_3$, compute the resulting $b$ and you have another system of linear equations with solution $x=1, y=2$, and $z=3$. For instance, if you take $a_1 = (1\ 0\ 0)^t$, $a_2 = (0\ 1 \ 0)^t$ and $a_3 = (0\ 0\ 1)^t$, you'll obtain my first example.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT. Maybe I should add something to explain the necessity of my misterious condition that $a_1, a_2, a_3$ must be &quot;linearly independent&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is the following: if you pick your three vectors randomly (that is to say, you put no matter what coefficients into the matrix $A$), you could obtain systems like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;x + y + z = 6 &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(take $a_1 = a_2 = a_3 = (1\ 0\ 0)^t$). Of course this system has the desired solution $x=1, y=2,$ and $ z=3$, but also an infinite number of them. In fact, the solution set is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(x, y, z) = (1, 2, 3) + \lambda (1, 0, -1) + \mu (0, 1, -1)&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all values of $\lambda, \mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you fill in the matrix $A$ randomly, the probability of &lt;em&gt;not&lt;/em&gt; obtaining three linearly independent vectors $a_1, a_2, a_3$ is zero, but, it won't take you too much time and effort to verify that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\det (A) \neq 0&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;before submitting your homework.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-10T18:05:15.563" LastActivityDate="2010-12-10T18:05:15.563" />
  <row Id="5517" PostTypeId="2" ParentId="5494" CreationDate="2010-09-26T20:12:48.813" Score="2" ViewCount="0" Body="&lt;p&gt;If I understand correctly, you have the desired solution for a system of three equations in 3 unknowns and want to know how to construct a system with those solutions.  This is a fairly common task for people who teach high-school algebra.  The way that I typically do it is to make up the coefficients of the variables first, so something like:&#xA;$$\begin{align}&#xA;3a+4b+5c&amp;amp;=&#xA;\\&#xA;8a-5b+7c&amp;amp;=&#xA;\\&#xA;-5a+b+c&amp;amp;=&#xA;\end{align}$$&#xA;then substitute in the desired solutions to get the constant terms on the right side of each equation (using your $a=1$, $b=1$, and $c=3$):&#xA;$$\begin{align}&#xA;3\cdot 1+4\cdot 2+5\cdot 3&amp;amp;=26&#xA;\\&#xA;8\cdot 1-5\cdot 2+7\cdot 3&amp;amp;=19&#xA;\\&#xA;-5\cdot 1+2+3&amp;amp;=0&#xA;\end{align}$$&#xA;so that one system with your given solutions would be:&#xA;$$\begin{align}&#xA;3a+4b+5c&amp;amp;=26&#xA;\\&#xA;8a-5b+7c&amp;amp;=19&#xA;\\&#xA;-5a+b+c&amp;amp;=0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: I should probably note (similar to the edited note in &lt;a href=&quot;http://math.stackexchange.com/questions/5494/how-to-build-a-linear-equation-system/5516#5516&quot;&gt;Agusti Roig's answer&lt;/a&gt;) that this method only guarantees that the given values are a solution to the system, not that they are the only solution to the system.  As with nearly any problem, it is worthwhile to check that the resulting system has only the solution(s) you want by solving it.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-27T14:07:35.747" LastActivityDate="2010-09-27T14:07:35.747" />
  <row Id="5518" PostTypeId="2" ParentId="5416" CreationDate="2010-09-26T20:35:48.577" Score="3" ViewCount="0" Body="&lt;p&gt;Geometrically one can see what's going on by analyzing $f$'s induced map on tangent vectors.  You'll see that the operator $\frac{\partial f}{\partial \overline{z}}$ is measuring the failure of a function to respect multiplication by $i$ between domain tangent vectors and range tangent vectors.  It is helpful to think of $f$ mapping from a domain $C$ ~ $R^2$ to a distinct range $C$ ~ $R^2$.  Any differentiable function $f$ will induce a map from tangent vectors in the domain to tangent vectors in the range, denoted here $v \rightarrow f'(v)$.  This is the derivative of $f$ along the vector $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both our domain and range are complex and thus have a concept of multiplication by $i$ on tangent vectors.&#xA;Multiplying by $i$ is like rotating a vector counter-clockwise 90 degrees, e.g. at a point $z=(x,y)$, the vector $\frac{\partial}{\partial x}$ gets sent to $\frac{\partial}{\partial y}$ under multiplication by $i$.  The beauty of holomorphic functions is that they respect this 90 degree rotation between the domain and range.  That is, if $f$ is holomorphic, then $f'(iv) = i f'(v)$ where the first $i$ multiplication is in the domain and the second represents $i$ multiplication in the range.  So, from the image of a vector in the domain (the derivative in one direction at some point), you can infer the image of $f'$ in every direction at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We write this as, for all $v$, $f'(iv) = if'(v)$.  The failure to achieve holomorphicity is thus $f'(iv)-if'(v)$ being different from $0$ for some $v$.  The operator $\frac{\partial f}{\partial \overline{z}}$ is essentially this measurement for a specific v.  Let v be the unit vector $\frac{-\partial}{\partial y}$.  Then $i v$ is $\frac{\partial}{\partial x}$.  The above formula $f'(iv)-if'(v)$ becomes $f'(\frac{\partial}{\partial x}) - i f'(-\frac{\partial}{\partial y})$ = $\frac{\partial f}{\partial x} + i (\frac{\partial f}{\partial y})$ = $2 \frac{\partial f}{\partial \overline{z}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there's a certain symmetry here.  While functions where $\frac{\partial f}{\partial \overline{z}}=0$ are holomorphic, functions where $\frac{\partial f}{\partial z}=0$ are anti-holomorphic; $f'$ sends the vector $i v$ to $- i f'(v)$ for all $v$.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-09-26T20:35:48.577" />
  <row Id="5520" PostTypeId="2" ParentId="5500" CreationDate="2010-09-26T21:52:53.773" Score="2" ViewCount="0" Body="&lt;p&gt;The subject of &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;universal algebra&lt;/a&gt; is the study of algebraic structures as classes of first-order structures in languages having only function symbols as opposed to relation symbols. This subject is regarded by many as much closer to algebra than the rest of model theory, allowing one to combine methods from model theory with a stronger category-theoretic perspective. A central concern are the classes of algebras, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Variety_%28universal_algebra%29&quot; rel=&quot;nofollow&quot;&gt;varieties&lt;/a&gt;, defined by a system of algebraic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Variety_(universal_algebra)#Birkhoff.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Birkoff's theorem&lt;/a&gt; states that a class of algebraic structures is a variety of algebras if and only if it is closed under homomorphisms, subalgebras and products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar corresponding theorem in first order logic is due to Kiesler and Shelah: two structures are elementary equivalent if and only if they have a common ultrapower, and one can use this to characterize which collections of algebras form an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_class&quot; rel=&quot;nofollow&quot;&gt;elementary class&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-26T21:52:53.773" />
  <row Id="5521" PostTypeId="2" ParentId="5416" CreationDate="2010-09-26T22:30:25.000" Score="3" ViewCount="0" Body="&lt;p&gt;This is a rephrasing of aspects of what several people above have said. In particular Weaam's 3rd paragraph shows you how to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\mathbb R$-linear function $f : \mathbb R^2 \to \mathbb R^2$ can be decomposed uniquely into a sum $f = Cf+Af$ of two $\mathbb R$-linear functions $Cf, Af : \mathbb R^2 \to \mathbb R^2$, where $Cf$ is &quot;complex linear&quot; meaning that $Cf(av+bw)=aCf(v)+bCf(w)$ for all complex numbers $a,b$ and vectors $v,w\in \mathbb R^2$.   $Af$ is complex anti-linear $Af(av+bw)=\overline{a}Af(v)+\overline{b}Af(w)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given an arbitrary Frechet ($\mathbb R$-differentiable) function $f$ defined on an open subset of $\mathbb R^2$,&#xA;$\partial f/\partial z$ is the complex-linear part of the derivative of $f$.  $\partial f/\partial \overline{z}$ is the complex anti-linear part of the derivative of $f$.   So this explains why $f$ is complex differentiable if and only if $\partial f/\partial \overline{z} = 0$.  In the notation of the previous paragraph, $\partial f/\partial \overline{z} = Af'$, and $\partial f/\partial z = Cf'$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-26T22:30:25.000" />
  <row Id="5522" PostTypeId="1" CreationDate="2010-09-26T22:34:19.713" Score="1" ViewCount="225" Body="&lt;p&gt;What are some examples of cutting-edge research involving fractals or self-similar structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who's actively contributing high-quality research in this field?&lt;/p&gt;&#xA;" OwnerUserId="2154" LastActivityDate="2010-09-27T03:38:32.643" Title="Where have fractals gone since Mandelbrot?" Tags="&lt;fractals&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="5523" PostTypeId="1" AcceptedAnswerId="5612" CreationDate="2010-09-27T00:20:49.510" Score="2" ViewCount="74" Body="&lt;p&gt;It's said on Wikipedia page about &lt;a href=&quot;http://en.wikipedia.org/wiki/Post%27s_lattice&quot; rel=&quot;nofollow&quot;&gt;Post's lattice&lt;/a&gt; that DM clone(all monotone self-dual functions) has majority function(xy + xz + yz) as one of its bases. What is the proof of this fact? &lt;/p&gt;&#xA;" OwnerUserId="2156" LastActivityDate="2010-09-28T01:50:16.483" Title="Monotone self-dual boolean functions clone" Tags="&lt;logic&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5524" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T00:54:25.027" Score="0" ViewCount="0" Body="&lt;p&gt;Fractal &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractal_antenna&quot; rel=&quot;nofollow&quot;&gt;antennae&lt;/a&gt; lend theory to the design of fractal solar panels.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-27T00:54:25.027" />
  <row Id="5525" PostTypeId="2" ParentId="2581" CreationDate="2010-09-27T01:17:23.687" Score="1" ViewCount="0" Body="&lt;p&gt;It is possible to brute force a list of possible solutions for every possible combination of sticks. Ignoring permutations... For n=1 there is a single solution. Ditto for n=2. For n=3 there are two solutions for one combination of sticks. For n=4 there are three solutions for two combinations of sticks. For n=5 I think there are ten solutions for four combinations of sticks. For n=6 I think there are 24 solutions. A completely baseless and uniformed search of the OEIS suggests this sequence might continue with 130 solutions for n=7 and 504 solutions for n=8. This trend &lt;em&gt;suggests&lt;/em&gt; that there are increasingly more solutions for any such n, and a non-decreasing number of solutions for the &quot;hardest&quot; combination of sticks for a given n, and the pattern by which such solutions can be enumerated may suggest a proof of the original question.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-27T01:17:23.687" CommentCount="3" />
  <row Id="5526" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T01:28:24.273" Score="-1" ViewCount="0" Body="&lt;p&gt;Behold the mandelbulb &lt;a href=&quot;http://www.skytopia.com/project/fractal/mandelbulb.html&quot; rel=&quot;nofollow&quot;&gt;http://www.skytopia.com/project/fractal/mandelbulb.html&lt;/a&gt;  !!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oddly enough the mandelbulb looks almost organic.&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2010-09-27T01:28:24.273" />
  <row Id="5527" PostTypeId="1" AcceptedAnswerId="5530" CreationDate="2010-09-27T02:32:41.370" Score="1" ViewCount="98" Body="&lt;p&gt;This is mostly related to doing large modular exponentiation by hand. For example, a problem I was doing was to find the last 3 digits of $7^{9729}$; that is, find $7^{9729}\bmod{1000}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the simplest concept for Euler's theorem, I found that $7^{400}\equiv 1\pmod{1000}$, since $\varphi(1000)=400$. Using Carmichael's theorem, I found a smaller number, $7^{100}\equiv 1\pmod{1000}$, as $\lambda(1000)=100$. Now, by manually multiplying it out, I found that $7^{20}\equiv 1 \pmod{1000}$, and that is the first $n$ for which that is true, meaning I just need to find $7^9 \bmod{1000}$, making the answer 607.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to arrive at this answer without multiplying it out each time for every number I get? For example, could I do something like $13^{12937}\bmod{1000}$ without sitting around modding out multiples of $13^4$? (I know that the first $n$ for 13 would be 100, so no less than using Carmichael's theorem, but I want to know if there are other ways to find numbers lower than those given by Carmichael's theorem)&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-09-27T04:13:32.837" Title="How do I find the lowest $n$ for which $a^n \equiv 1 \pmod{b}$?" Tags="&lt;algebra&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="5528" PostTypeId="2" ParentId="5527" CreationDate="2010-09-27T03:05:04.210" Score="1" ViewCount="0" Body="&lt;p&gt;As I discuss in &lt;a href=&quot;http://math.stackexchange.com/questions/3976/on-the-binary-decimal-expansion-of-the-reciprocal-primes&quot;&gt;my answer to this question&lt;/a&gt;, this is a hard quantity to compute in general.  For practical purposes you should use &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;exponentiation by squaring&lt;/a&gt; to test the possibilities (and you should definitely use exponentiation by squaring for computing remainders in general; this is how computer algebra systems do it).  Note that using Carmichael's theorem is not trivial, as it requires that you know the prime factorization of $b$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-27T03:05:04.210" CommentCount="5" />
  <row Id="5529" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T03:17:03.313" Score="0" ViewCount="0" Body="&lt;p&gt;render 3D fractals based on the original 2D mandelbrot formula, such as mandelbox and mandelbulb + many more - using free open source software called &quot;&lt;a href=&quot;http://sourceforge.net/projects/mandelbulber/&quot; rel=&quot;nofollow&quot;&gt;mandelbulber&lt;/a&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="2158" LastEditorUserId="2158" LastEditorDisplayName="" LastEditDate="2010-09-27T03:29:42.487" LastActivityDate="2010-09-27T03:29:42.487" />
  <row Id="5530" PostTypeId="2" ParentId="5527" CreationDate="2010-09-27T03:28:58.353" Score="4" ViewCount="0" Body="&lt;p&gt;If $\rm\: gcd(a,10)=1\:$ then order of $\rm\: a\:$ in $\:\mathbb Z/1000 \:$ must be a divisor of $100 = \lambda(1000)$. You can compute the order simply and quickly by computing in order  $\rm a^2, a^4, a^5, a^{10}, a^{20}, a^{25}, a^{50}\:$ by squaring or multiplying previous entries. This requires at most 5 squaring and 2 multiplication operations $\rm (mod\ 1000)$. Obviously the same sort of optimized divisor lattice searching works for any modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively one can use the following well-known simple order algorithm for groups&#xA;&lt;img src=&quot;http://i.imgur.com/77uZH.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://groups.csail.mit.edu/cis/theses/sutherland-phd.pdf&quot; rel=&quot;nofollow&quot;&gt;This thesis&lt;/a&gt; is a good reference on order algorithms in generic groups. Here's the abstract:&#xA;&lt;img src=&quot;http://i.imgur.com/VAw3T.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-27T04:13:32.837" LastActivityDate="2010-09-27T04:13:32.837" CommentCount="3" />
  <row Id="5531" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T03:29:46.897" Score="2" ViewCount="0" Body="&lt;p&gt;As pointed out in the comments, this is a tremendously broad area and there are many people doing a lot of very good research, so I can only say a little bit about the small part of it that I'm familiar with.  When dealing with &quot;chaotic&quot; dynamical systems (a notion which can be made precise in a few different ways), one often finds that the chaotic behaviour of the dynamics is intimately connected to the presence of various geometric structures in the phase space that are best characterised as fractals -- the Julia sets found in complex dynamics are examples of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To study these structures, one uses various dimensional quantities, such as Hausdorff dimension.  It turns out that some fundamental dynamical quantities such as topological entropy can also be defined as &quot;dimensions&quot; of a sort, so that there are deep connections between fractal geometry and chaotic dynamics.  One of the standard (advanced) references on this is &quot;Dimension Theory in Dynamical Systems&quot;, by Yakov Pesin.  A more introductory exposition (with apologies for self-advertising) is &quot;Lectures on Fractal Geometry and Dynamical Systems&quot;, by Pesin and Climenhaga.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of the sort of thing that occurs, suppose you have a dynamical system with phase space $X$ and an observable function $\phi\colon X\to \mathbb{R}$.  You take measurements of $\phi$ as time goes along, and calculate its average value as time goes to infinity.  This average value is a function of your starting position $x\in X$.  Let $K_\alpha$ denote the set of points for which that average value is equal to $\alpha$; then typically speaking, there is a range of values of $\alpha$ for which $K_\alpha$ is a fractal with quite an intricate structure.  This is called a &lt;em&gt;multifractal decomposition&lt;/em&gt;.  One can define a &lt;em&gt;multifractal spectrum&lt;/em&gt; by $B(\alpha) = \dim_H(K_\alpha)$, and it turns out (somewhat miraculously) that the function $\alpha\mapsto B(\alpha)$ is actually concave and analytic in a great many (important) examples!  This multifractal analysis has deep connections to thermodynamic formalism and statistical properties of chaotic dynamical systems; a good reference is &quot;Thermodynamics of Chaotic Systems&quot;, by Beck and Schl&amp;ouml;gl.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2010-09-27T03:29:46.897" />
  <row Id="5532" PostTypeId="2" ParentId="5522" CreationDate="2010-09-27T03:38:32.643" Score="-2" ViewCount="0" Body="&lt;p&gt;Do you know why more complex viruses (e.g., HIV, AIDS-virus and SARS corona virus) tend to be more fatal? this is due to their un-smooth surfaces having fractal dimension.&#xA;Using fractal technology there is a way to stop the virus&#xA;The paper is written by &quot;Ji-Huan He&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His website &lt;a href=&quot;http://works.bepress.com/ji_huan_he/&quot; rel=&quot;nofollow&quot;&gt;http://works.bepress.com/ji_huan_he/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can download the original paper there.&lt;/p&gt;&#xA;" OwnerUserId="2158" LastActivityDate="2010-09-27T03:38:32.643" CommentCount="2" />
  <row Id="5533" PostTypeId="2" ParentId="2581" CreationDate="2010-09-27T03:55:25.067" Score="0" ViewCount="0" Body="&lt;p&gt;This is not a solution, but a potential restatement of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a single stick of length n(n+1)/2 that can be broken in any single place, leaving two pieces of length &gt;=n. The resulting two sticks can be broken into 1...n easily, possibly in multiple ways. Can you show that for any position of that break that leaves a stick of length &gt;=2n, there exists a subdivision of the stick such that the longer stick can be broken into any possible two pieces of length &gt;=n? If so, I think that you may have the desired proof.&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-09-27T03:55:25.067" />
  <row Id="5534" PostTypeId="1" AcceptedAnswerId="5548" CreationDate="2010-09-27T04:23:19.910" Score="5" ViewCount="262" Body="&lt;p&gt;Let random variable $X\sim\text{Binomial}\left(a+b,\frac{a}{a+b}\right)$, where $a$ and $b$ are positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove that $\mathbb{P}[X\geq\mathbb{E}[X]]\leq\frac{3}{4}$, which appears to be true numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a suggestion on how to proceed?&#xA;The complication in this particular problem is that the condition is not a strict inequality &quot;$&gt;$&quot;.&#xA;I've tried the Chernoff bound but it's not tight enough.&lt;/p&gt;&#xA;" OwnerUserId="1023" LastEditorUserId="1023" LastEditorDisplayName="" LastEditDate="2010-10-24T17:54:31.377" LastActivityDate="2010-10-24T17:54:31.377" Title="Proving an upper bound for Prob[X&gt;=E[X]]" Tags="&lt;probability&gt;&lt;random-variable&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="5535" PostTypeId="2" ParentId="227" CreationDate="2010-09-27T05:29:23.870" Score="1" ViewCount="0" Body="&lt;p&gt;Looks like you're set on linear algebra and real analysis, both of which are great choices. You can never know enough linear algebra, and if you haven't taken an analysis class yet, it'll definitely get you thinking more &quot;mathematically&quot; than in many of your previous classes, which like Katie said, is definitely one of the main goals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to take a look into probability theory. You mentioned that you talked to a couple of finance professors, so I'm guessing that finance might be a possibility in your potential field of research. In that case, probability will be a pretty key component. It's also useful to know it in a lot of other areas of economics as well. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-09-27T05:29:23.870" CommunityOwnedDate="2010-09-27T05:29:23.870" />
  <row Id="5537" PostTypeId="1" AcceptedAnswerId="5576" CreationDate="2010-09-27T06:15:09.690" Score="8" ViewCount="145" Body="&lt;p&gt;This question is motivated by the following homework problem.  I'm trying to explicitly compute the homeomorphism $f:S^2 \rightarrow \mathbb{CP}^1$ by using stereographic projection and considering $\mathbb{CP}^1 = \mathbb{C}\cup {\infty}$.  I'll want to prove that this is an isometry, where $S^2$ has the standard angle metric and $\mathbb{CP}^1$ has the Fubini-Study metric given by $d(\overline{x},\overline{y})=2\cos^{-1}|(x,y)|$, where $x,y\in \mathbb{C}^2$ are unit vectors (and presumably $(-,-)$ is the usual Hermitian inner product).  Later, I'll use this to explicitly compute the Lie group homomorphism $U(2)\rightarrow SO(3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My stereographic projection is from the north pole, takes the equator to the unit circle, and puts the south pole at the origin.  What I've gotten so far is that for $z\not= 1$, \begin{equation*} f(x,y,z)=\left( \frac{x}{1-z} , \frac{y}{1-z} \right) = \frac{x+iy}{1-z} = [x+iy : 1-z ], \end{equation*} where these are coordinates in $\mathbb{R}^2$, $\mathbb{C}$, and $\mathbb{C}\subseteq \mathbb{CP}^1$ respectively.  This is troublesome, because philosophically I'd expect that I should be able to define this for $(x,y,z)\not= (0,0,1)$ and then end up with a function to projective space that extends continuously over the north pole; that's sort of the point of projective space, to make $\infty$ into just another point.  However, it is not immediately obvious that this works, although luckily \begin{equation*} \left| \frac{x+iy}{1-z} \right| = \sqrt{ \frac{|x+iy|^2}{(1-z)^2} } = \sqrt{ \frac{1-z^2}{(1-z)^2}}, \end{equation*} and the limit of this expression as $z\rightarrow 1^-$ is indeed $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, fair enough.  This ends up extending to a continuous function after all.  But: &lt;strong&gt;Am I wrong in my philosophical understanding of projective space?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For what it's worth, I tried using my calculations to verify that $f$ is an isometry, and it didn't look like it was going to work out.  So maybe I really am just doing something wrong.)&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-09-27T15:40:24.537" LastActivityDate="2010-09-28T07:31:38.220" Title="Does projectivizing always fix problems at infinity?  (Or, am I making a mistake somewhere?)" Tags="&lt;differential-geometry&gt;&lt;lie-groups&gt;&lt;projective-space&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="5538" PostTypeId="1" CreationDate="2010-09-27T07:05:40.413" Score="2" ViewCount="147" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the question asked in a written test conducted by a company. The question sounded stupid to me. May be its not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Given the area of the coin to be 'A'. If the probability of getting a tail, head and the edge are same, what is the thickness of the coin?&lt;/p&gt;&#xA;" OwnerUserId="1611" LastActivityDate="2010-09-27T10:05:11.777" Title="Find thickness of a coin" Tags="&lt;probability&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5539" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T07:38:21.937" Score="1" ViewCount="0" Body="&lt;p&gt;If you assume that the probability of getting tail, head or edge only depends on the surface area that those regions of the coin have, then you see that the area of the edge must also equal $A$. If you denote by $T$ the thickness of the coin, then the area of the edge is given by $T \cdot C$, where $C$ is the circumference of the coin. Now, as you are given $A$, you can determine $C = 2 \sqrt{A \pi}$ and thus you get $T = \frac{A}{C} = \frac{A}{2 \sqrt{A \pi}} = \frac{\sqrt{A\pi}}{2\pi}$&lt;/p&gt;&#xA;" OwnerUserId="1766" LastEditorUserId="1766" LastEditorDisplayName="" LastEditDate="2010-09-27T09:59:23.747" LastActivityDate="2010-09-27T09:59:23.747" CommentCount="2" />
  <row Id="5540" PostTypeId="1" AcceptedAnswerId="5546" CreationDate="2010-09-27T08:14:13.247" Score="3" ViewCount="237" Body="&lt;p&gt;Generally one can see that there are infinite number of solutions for this equation $$a^{2}+b^{2}=c^{2}$$ by taking multiples of the solution $3,4$ and $5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can i use this as a fact to prove, that $k^{2} + (k+1)^{2}$ is a perfect square for infinitely many $k \in \mathbb{N}$? Any hints, suggestions would be helpful. If not, then how do i prove this fact!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-09-27T15:51:39.387" LastActivityDate="2010-09-27T15:56:20.440" Title="$k^{2}+(k+1)^{2}$ being a perfect square for infinitely many $k$." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5541" PostTypeId="2" ParentId="4576" CreationDate="2010-09-27T08:14:17.583" Score="1" ViewCount="0" Body="&lt;p&gt;To see that $E_{\infty}=E_{k+2}$: your $E_2$ page is supported in the 0th and kth rows, so any differentials beyond $d^{k+1}$ must be zero because they start and/or end at entries that are 0 (and have been since the $E_2$ page).  Remember that the differential $d^j$ goes $j$ to the right and $j-1$ downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that $E_{k+2}=E_2$ follows from the fact that the transgression is 0.  In fact, since the coefficients are trivial (and the cohomology rings of $G$ and $X$ are both finitely generated in each degree), then $E_2 = H^&lt;em&gt;(G;H^&lt;/em&gt;(X)) \cong H^&lt;em&gt;(G) \otimes H^&lt;/em&gt;(X)$, and under this identification all the differentials must satisfy a signed Leibniz rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to make sense of your &quot;Serre SS&quot;, since it's not the one I'm familiar with.  The one I'm used to has $E_2^{p,q} = H^p(B;H^q(F;\mathbb{Z})) \Rightarrow H^{p+q}(E;\mathbb{Z})$; here, $F \hookrightarrow E \rightarrow B$ is a fibration, and we assume either that $\pi_1(B)=0$ or at least that $\pi_1(B)$ acts trivially on $H^q(F)$.  (Otherwise you get twisted coefficients, which are awesome but harder to deal with.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the setup you've got, it sounds like you have a fiber bundle with discrete fiber (i.e., a covering space) $G \hookrightarrow X \rightarrow X/G$, where by $G$ I actually mean the underlying set of the group $G$.  So $H^0(G;\mathbb{Z}) = \mathbb{Z}^{\oplus |G|}$ and $H^q(G;\mathbb{Z})=0$ for $q&gt;0$.  I'd imagine that the assumption of triviality of the induced $G$-action on $H^*(X)$ would guarantee that the coefficients are untwisted, although I'm not sure this is right.  If it is, the $E_2$ term of my Serre SS for this would only have nonzero entries along the bottom row; this reflects the close connection between the (co)homology of a base space and a covering space.  So the SS has &lt;em&gt;no&lt;/em&gt; nonzero differentials, again because all possible differentials $d^i$ ($i\geq 2$) either start at a trivial group or end at a trivial group or both.  So $E_2 = E_\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess probably that paper is using some fancier SS for group actions that I've never heard of.  Since usually the condition for untwisted coefficients is that $\pi_1(B)$ acts trivially on $H^q(F)$, I'm inclined to think that perhaps your cohomology is actually group cohomology or something.  I don't know much about this, but I'm pretty sure that $H^q_{group}(G)=H^q_{sing}(BG)$, where $BG=K(G,1)$ is the classifying space of $G$.  This would mesh with what I've said so far, because $\pi_1(K(G,1))=G$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-09-28T15:44:43.150" LastActivityDate="2010-09-28T15:44:43.150" CommentCount="1" />
  <row Id="5542" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T08:22:17.747" Score="-1" ViewCount="0" Body="&lt;p&gt;I think that the question is stupid, because it does not take into account the fact that even with a really thick coin it's easier to tilt it when it lands on the edge than when it lands on a face.&lt;br&gt;&#xA;This said, as Sebastian wrote down in formulas you should think at the edge of the coin as a surface, and not at the single line where the coin ends up.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-09-27T08:22:17.747" CommentCount="2" />
  <row Id="5543" PostTypeId="1" AcceptedAnswerId="5556" CreationDate="2010-09-27T08:44:07.763" Score="5" ViewCount="82" Body="&lt;p&gt;Again I have trouble with some exercises in Kunen's set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following, let $\kappa &gt; \omega$ a cardinal. Then I want to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $|H(\kappa)| = 2^{&amp;lt;\kappa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $H(\kappa)=R(\kappa)$ if and only if $\kappa = \beth_{\kappa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1) the inequality $\geq$ is easy, but I cannot prove $\leq$. In 2) I have no idea. It is always true that $H(\kappa) \subseteq R(\kappa)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: It seems that LaTeX here does not have the Beth symbol.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-27T19:03:13.797" Title="Cardinality of H(k)" Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="5544" PostTypeId="2" ParentId="5416" CreationDate="2010-09-27T08:46:24.137" Score="0" ViewCount="0" Body="&lt;p&gt;The formulas&#xA;$$\frac{\partial f}{\partial z} \;=\; \frac{1}{2}\left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y}\right)$$ $$\frac{\partial f}{\partial \bar{z}} \;=\; \frac{1}{2}\left(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}\right)$$&#xA;are just the trigonometric-to-exponential Fourier coefficient transformations&#xA;$$c_1=\frac{1}{2}(a_1-ib_1)$$ $$c_{-1}=\frac{1}{2}(a_1+ib_1)$$&lt;/p&gt;&#xA;" OwnerUserId="2161" LastActivityDate="2010-09-27T08:46:24.137" />
  <row Id="5545" PostTypeId="2" ParentId="1519" CreationDate="2010-09-27T08:47:56.927" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Note that you're only attaching the &lt;em&gt;boundary&lt;/em&gt; of the 4-cell.&lt;/strong&gt;  This is a map $S^3 \rightarrow S^2$.  The resulting topological space has a completely intact copy of the interior of the 4-cell.  This is similar to making $\mathbb{RP}^2$ by gluing a disk to a circle by the degree-2 map $\partial D^2 = S^1 \rightarrow S^1$ (given, say, in complex coordinates by $z\mapsto z^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the issue of having high-dimensional spheres &quot;wrap around&quot; low-dimensional cells: up to homotopy, this can be phrased as a question about &lt;em&gt;homotopy groups&lt;/em&gt;, that is, maps from spheres into some space $X$.  (This doesn't address the question of singularities, but I think it is still helpful for trying to think about these gluing constructions.)  These are groups $\pi_k(X)$ for all $k\geq 1$; when $k=1$, this just the fundamental group.  In general, they are quite poorly understood.  Even for the simplest possible CW complexes $X=S^n$ (except $n=1$), we don't know all the homotopy groups!  This might sound like an easy problem, but it's really....REALLY...nontrivial.  Now, when you make CW complexes, you're gluing on cells $e^k$ one by one; these are really just maps from $\partial e^k = S^{k-1}$ to the stuff you already had (the &quot;$(k-1)$-skeleton&quot;).  People often only care about the homotopy types of these maps.  In summary, while it's certainly a good idea to try and visualize these things, it's also good to be able to deal abstractly with gluing constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres&quot; rel=&quot;nofollow&quot;&gt;homotopy groups of spheres&lt;/a&gt; is really interesting, and contains a sort of picture of the Hopf fibration.  Basically, you want to think of it as a bunch of circles that somehow fill out $S^3$, which is just $\mathbb{R}^3$ with one extra point at $\infty$.  Interestingly, any two of these circles are linked.  So you can kind of think of it locally as the unit circle in the xy-plane, and then you can move that circle in 2 dimension's worth of directions, and in every one of those directions that you push the circle off itself you're going to get a new, nearby circle that's linked to the original one.  (Of course, there is precisely an $S^2$'s worth of circles!)  You can take the circle going through $\infty$ to be the z-axis.  So by what I just said, all of these other circles need to wind once around the z-axis too.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-09-27T08:47:56.927" />
  <row Id="5546" PostTypeId="2" ParentId="5540" CreationDate="2010-09-27T08:54:13.960" Score="7" ViewCount="0" Body="&lt;p&gt;You can find infinitely many by considering solutions of Pell's equation $ m^2 - 2a^2=-1.$&#xA;Your (3,4,5) solution comes from$$(7-5\sqrt{2})(7+5\sqrt{2})=-1,$$&#xA;where $m=7$ and $a=5$.&#xA;For example $$(7-5\sqrt{2})^3$$ yields&#xA;$$696^2+697^2 = 985^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to make it a bit more clear note that&#xA;$$\left( {m-1 \over 2} \right)^2 + \left( {m+1 \over 2} \right)^2 = a^2.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-27T09:01:39.200" LastActivityDate="2010-09-27T09:01:39.200" />
  <row Id="5547" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T09:10:46.327" Score="4" ViewCount="0" Body="&lt;p&gt;This problem is also considered in the book &lt;a href=&quot;http://books.google.co.uk/books?id=QiuqPejnweEC&amp;amp;pg=PA10&amp;amp;lpg=PA10&amp;amp;dq=%22how+thick+should+a+coin+be%22&amp;amp;source=bl&amp;amp;ots=uI9vTy3Egy&amp;amp;sig=S7SjnQTJ-SSwygsV2fTfm4lEOTo&amp;amp;hl=en&amp;amp;ei=516gTP_rHM-TjAfyvtWBDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBkQ6AEwAA#v=onepage&amp;amp;q=%22how%20thick%20should%20a%20coin%20be%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Fifty challenging problems in probability with solutions&lt;/a&gt; by Frederick Mosteller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more in depth study on bias in coin tossing you could read:&#xA;&lt;a href=&quot;http://comptop.stanford.edu/preprints/heads.pdf&quot; rel=&quot;nofollow&quot;&gt;http://comptop.stanford.edu/preprints/heads.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-27T09:10:46.327" />
  <row Id="5548" PostTypeId="2" ParentId="5534" CreationDate="2010-09-27T09:22:31.547" Score="4" ViewCount="0" Body="&lt;p&gt;Here is an answer giving the outline of a proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Bound is tight&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bound is true then it is tight: If $a=1$, $b=1$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{P}[X\geq \mathbb{E}[x]] = \mathbb{P}[X\geq 1] =1-\mathbb{P}[X=0] = 1-\frac{1}{2}\frac{1}{2}=\frac{3}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Getting a feel for the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb (though can be made precise) if $a&gt;5$ and $b&gt;5$ then we can approximate the Binomial by normal random variable by the central limit theorem. That is $X\sim N(a,Var(X))$ approximately.  The error in this approximation becomes increasing small as $a$ and $b$ increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\mathbb{P}[X\geq \mathbb{E}[x]] \approx \frac{1}{2}&amp;lt;\frac{3}{4}$ for $a&gt;5$ and $b&gt;5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make precise, first read this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then: Box, Hunter and Hunter (1978). Statistics for experimenters. Wiley. p. 130&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are left with three cases: a and b both small, a large and b small, b large and a small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$b$ large and $a\leq 5$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b+a&gt;100$ then a poisson approximation to the Binomial r.v. becomes appealling. That is $X\sim Po(a)$ approximately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\mathbb{P}(X\geq a)=1-\sum_{i=0}^{a-1}\frac{e^{-a}a^i}{i!}\leq 0.63&amp;lt;\frac{3}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For explicit error bounds look in A bound on the Poisson-binomial relative error by Teerapabolarn (2007). Theorem 2.1 contains a good error bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$b\leq 5$ and $a$ large&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y=a+b-X\sim  \text{Bin}(a+b,\frac{b}{a+b})$ now if $b+a&gt;100$ $Y$ is approximately Poisson and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{P}(X\geq a)\approx\mathbb{P}(Y\leq b) &amp;lt;0.45 \leq \frac{3}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remaining cases&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the cases above are not quite mutually exclusive, but there a finite number of cases left. The cases left are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $a\leq 5$ and $b\leq 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $b\leq 5$ and $a\leq 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this can be checked on a computer. To give the bound. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-09-28T12:52:11.903" LastActivityDate="2010-09-28T12:52:11.903" CommentCount="3" />
  <row Id="5549" PostTypeId="1" AcceptedAnswerId="5554" CreationDate="2010-09-27T09:48:02.227" Score="0" ViewCount="260" Body="&lt;p&gt;If $\displaystyle \frac{a^n + b^n}{a^{n-1} + b^{n-1}}$ , ( a != b and a,b,n are real numbers ) is the Arithmetic mean &lt;strong&gt;or&lt;/strong&gt; Geometric mean between a and b then find the values of n for both means respectively.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-09-27T10:10:05.670" LastActivityDate="2010-10-06T15:45:41.487" Title="A simple algebra problem" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="5550" PostTypeId="1" AcceptedAnswerId="5553" CreationDate="2010-09-27T09:57:45.970" Score="2" ViewCount="117" Body="&lt;p&gt;I want to show that if $F:\mathbb{R^n}\to\mathbb{R^m}$ has a derivative of rank = r at a point $p$, then there is a neighbourhood of that point where the rank of the function doesn't decrease, i.e. $\exists N=B(p, \delta)$ such that $rk(DF_x) \geq rk(DF_p) ~ \forall x\in N$  &lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditorDisplayName="" LastEditDate="2010-09-27T10:03:23.590" LastActivityDate="2010-09-27T12:13:18.800" Title="Why doesn't small perturbations of a matrix decrease its rank?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5551" PostTypeId="2" ParentId="5447" CreationDate="2010-09-27T10:00:27.323" Score="0" ViewCount="0" Body="&lt;p&gt;The easiest way to factor polynomials is by finding roots: if $p(a)=0$ then $p(x)=(x-a)q(x)$ with $q(x)$ being a polynomial of smaller degree than $p(x)$ and you can repeat the process by finding a root of $q(x)$. So if you have a root $a$, you simply compute $q(x)=p(x)/(x-a)$ and continue recursively. This works over any field - the power of the Complex numbers stems from the fact that this process &lt;strong&gt;always&lt;/strong&gt; works in the complex number - every polynomial has a root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, finding such a root is not always an easy task. For polynomials of degree 2 it is trivial using the well known formula $z_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ which gives the roots of the polynomial $az^2+bz+c$. In simpler cases, such as $x^2+16$ you gave above, one can solve directly ($x^2=-16$ and then taking roots you find $\pm 4i$). However, for higher degrees the task becomes more difficult. There is a formula for polynomials of degree 3 and 4 but it is more complex; and for degree 5 you don't have a simple formula anymore (with &quot;simple&quot; meaning &quot;involving only the four operations of arithmetic and taking roots&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those cases, it is simpler to find an approximate solution and use the division method described above. A typical method for finding approximate roots of polynomials is the Newton-Rapson method.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-27T10:00:27.323" />
  <row Id="5552" PostTypeId="2" ParentId="5538" CreationDate="2010-09-27T10:05:11.777" Score="1" ViewCount="0" Body="&lt;p&gt;I assumed that the probability of getting a head, tail or edge depended on the angle from the centre of the coin that the side lies in.  So the head, tail and edge must each occupy 120 degrees when viewed along the axis of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dw3GQ.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram above the angles at the centre are all (meant to be) 60 degrees and the radius of each face is $\sqrt{A/\pi}$.  A small amount of trigonometry later and I found the edge length to be $\sqrt{\frac{4A}{3\pi}}$.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-09-27T10:05:11.777" />
  <row Id="5553" PostTypeId="2" ParentId="5550" CreationDate="2010-09-27T10:11:25.013" Score="9" ViewCount="0" Body="&lt;p&gt;Use the continuity of the determinant. If the rank is $r$ the Jacobian matrix&#xA;has an $r$-by-$r$ submatrix $A$ with nonzero determinants. Peturbing the point&#xA;will perturb the Jacobian matrix and so $A$. As the determinant is a continuous&#xA;function of its entries a small enough perturbation will leave the new $A$&#xA;having nonzero determinant, and so the new Jacobian matrix will have rank $\ge r$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-27T10:11:25.013" CommentCount="1" />
  <row Id="5554" PostTypeId="2" ParentId="5549" CreationDate="2010-09-27T10:16:17.500" Score="3" ViewCount="0" Body="&lt;p&gt;For the current version of the problem the answer is $n=1$ (for arithmetic mean) and $n=1/2$ (for geometric mean).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first the case&#xA;$$&#xA;\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a + b}{2}&#xA;$$&#xA;This equality implies $2a^n + 2b^n = a^{n} + ab^{n-1} + b^{n} + ba^{n-1}$. Hence $a^n + b^n = ab^{n-1} + ba^{n-1}$ and so $(a^{n-1} - b^{n-1})(a-b) = 0$. Hence (since $a \ne b$) we have $n=1$. It's easy to check that $n=1$ indeed satisfies the conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the case&#xA;$$&#xA;\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \sqrt{ab}&#xA;$$&#xA;This equality implies $a^n + b^n = a^{n-1/2}\sqrt{b} + \sqrt{a}b^{n-1/2}$. Hence $(a^{n-1/2} - b^{n-1/2})(\sqrt{a}-\sqrt{b}) = 0$. So (since $a \ne b$) we have $n=1/2$. It's easy to check that $n=1/2$ indeed satisfies the conditions.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-09-27T10:16:17.500" CommentCount="2" />
  <row Id="5555" PostTypeId="2" ParentId="5537" CreationDate="2010-09-27T10:45:25.247" Score="0" ViewCount="0" Body="&lt;p&gt;&quot;Am I wrong in my philosophical understanding of projective space?&quot;&lt;br&gt;&#xA;I think your understanding is incomplete in some ways.&lt;br&gt;&#xA;E.g. your remark &quot;resolve singularities by replacing a point with a copy of projective space&quot; doesn't make much sense to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think I didn't understand the question correctly. But I couldn't ask any clarification in comments to the question like others do, because of lack of reputation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may help in your understanding of &quot;projective spaces&quot; to realize that their resolvement of &lt;code&gt;\infty&lt;/code&gt; singularities uses homogene coordinates, e.g. in &lt;code&gt;\mathbb{CP}^1 = \mathbb{C}\cup {\infty}&lt;/code&gt; one uses coordiantes \rho\cdot (z, 1) for elements in &lt;code&gt;\mathbb{C}&lt;/code&gt; and &lt;code&gt;\rho (1, 0)&lt;/code&gt; for &lt;code&gt;{\infty}&lt;/code&gt;. Then the mapping should involve both coordinates and account for the free factor &lt;code&gt;$\rho != 0$&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="706" LastEditorDisplayName="" LastEditDate="2010-09-28T07:31:38.220" LastActivityDate="2010-09-28T07:31:38.220" CommentCount="4" />
  <row Id="5556" PostTypeId="2" ParentId="5543" CreationDate="2010-09-27T11:20:38.220" Score="2" ViewCount="0" Body="&lt;p&gt;1) To show that $|H(\kappa)| \leq 2^{&amp;lt;\kappa}$, we need to code elements of $H(\kappa)$ using subsets of ordinals $&amp;lt; \kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in H(\kappa)$ and let $a$ be the transitive closure of $\{x\}$. Fix a bijection $f:|a|\to a$ with $f(0) = x$ and let $E \subseteq |a|^2$ be defined by $\xi \mathrel{E} \zeta$ iff $f(\xi) \in f(\zeta)$. Then $x$ is recovered as the value of $0$ in the transitive collapse of $(|a|,E)$. The pair $(|a|,E)$ can be encoded as a subset of $|a|^2 &amp;lt; \kappa$. This encoding process shows that $|H(\kappa)| \leq 2^{&amp;lt;\kappa}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) To see that $H(\kappa) \subseteq R(\kappa)$, note that the rank of an element $x \in H(\kappa)$ is smaller than $|a|^+$, where $a$ is once again the transitive closure of $\{x\}$. Since $|R(\kappa)| = \beth_\kappa$, we see that $H(\kappa) = R(\kappa)$ can only hold when $2^{&amp;lt;\kappa} = \beth_\kappa$. Note that $2^{\lambda} &amp;lt; \beth_\kappa$ for every $\lambda &amp;lt; \beth_\kappa$. Therefore, $H(\kappa) = R(\kappa)$ entails that $\kappa = \beth_\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\kappa = \beth_\kappa$ entails that $H(\kappa) = R(\kappa)$ it suffices to check that every element of $R(\kappa)$ has transitive closure of size less than $\beth_\kappa$. In fact, this holds for every limit ordinal $\kappa$: if $x \in R(\kappa)$ then $x \in R(\alpha)$ for some $\alpha &amp;lt; \kappa$ and hence the transitive closure of $x$ has size at most $|R(\alpha)| \leq \beth_\alpha &amp;lt; \beth_\kappa$.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditorDisplayName="" LastEditDate="2010-09-27T19:03:13.797" LastActivityDate="2010-09-27T19:03:13.797" CommentCount="3" />
  <row Id="5557" PostTypeId="2" ParentId="5550" CreationDate="2010-09-27T11:41:01.770" Score="2" ViewCount="0" Body="&lt;p&gt;The &quot;why doesn't ... decrease its rank&quot; is because it is hard to decrease rank.  The rank is almost everywhere at its highest possible value, with special exceptions that are destroyed by perturbation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be a maximum rank $r(f) \quad$ that holds at almost all points, and possibly some lower-rank points contained in lower-dimensional pieces of the domain of $f$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower rank is an unstable property. Perturbation away from a low-rank point destroys low rank and replaces it by maximum rank, unless you perturb in special directions that remain inside the lower-dimensional locus where the rank is below maximum.   At the &quot;generic point&quot;, the rank is the highest one possible, $r(f)$, and for any point, all or (in the case of low-rank points, almost all) of the nearby points to which one can perturb, will be generic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in Robin Chapman's answer this is because low rank requires some functions (determinants) computed from $Df$ to equal zero, while maximum rank requires them to be nonzero.  The first property is unstable and the second is stable with respect to perturbation.  It is also true that the rank cannot decrease through a small perturbation, because the set of determinants that are nonzero can only expand as one moves to a close enough nearby point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sard%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sard%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-27T12:13:18.800" LastActivityDate="2010-09-27T12:13:18.800" />
  <row Id="5558" PostTypeId="1" AcceptedAnswerId="5570" CreationDate="2010-09-27T11:44:31.777" Score="6" ViewCount="303" Body="&lt;p&gt;Please suggest an approach for this task.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:13:18.873" LastActivityDate="2010-11-11T15:13:18.873" Title="Find the  sum to n terms of the series $\displaystyle \frac{1} {1.2.3.4} + \frac{1} {2.3.4.5} + \frac{1} {3.4.5.6} $..." Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="5559" PostTypeId="1" CreationDate="2010-09-27T11:45:57.837" Score="0" ViewCount="80" Body="&lt;p&gt;Hi. I'm looking for some FOSS/GPL programs (or Python libraries) for dealing with sparse matrices. I haven't found much online about these. Can someone please point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2163" LastActivityDate="2010-09-27T14:12:25.760" Title="Good computer programs for dealing with sparse matrices" Tags="&lt;matrices&gt;&lt;math-software&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5560" PostTypeId="2" ParentId="5559" CreationDate="2010-09-27T11:59:15.413" Score="2" ViewCount="0" Body="&lt;p&gt;I haven't really seen any serious numerical computing being done in Python; FORTRAN is still standard, and &lt;a href=&quot;http://www.netlib.org/&quot; rel=&quot;nofollow&quot;&gt;Netlib&lt;/a&gt; has a number of (free!) libraries for solving sparse linear systems, sparse eigenproblems, ... etc. You are of course welcome to construct an interface from Python to these, or probably even better, translate these into Python. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-27T11:59:15.413" />
  <row Id="5561" PostTypeId="2" ParentId="5559" CreationDate="2010-09-27T12:13:18.613" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.caam.rice.edu/software/ARPACK/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;ARPACK&lt;/strong&gt;&lt;/a&gt; is the most common and effective solution to problems involving large matrices (both sparse and dense). It's a FORTRAN library (with C++ bindings) and a very stable/mature one at that. I know personally that it is used extensively in academia, so I strongly recommend you give it a go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of Python, I believe &lt;a href=&quot;http://www.scipy.org/&quot; rel=&quot;nofollow&quot;&gt;SciPy&lt;/a&gt;/NumPY actually includes wrappers for ARPACK, though you'll need to investigate this a bit more. The actual implementation is written in C, so efficiency is not at all a problem.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-09-27T14:12:25.760" LastActivityDate="2010-09-27T14:12:25.760" CommentCount="5" />
  <row Id="5563" PostTypeId="1" CreationDate="2010-09-27T12:34:55.677" Score="0" ViewCount="101" Body="&lt;p&gt;hi everybody could you people give some of shortcut methods to do algebra calculation&lt;/p&gt;&#xA;" OwnerUserId="1328" LastActivityDate="2010-09-27T12:34:55.677" ClosedDate="2010-09-27T15:14:29.307" Title="easy way to do algebra calculations?" Tags="&lt;algebra&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="5564" PostTypeId="1" AcceptedAnswerId="5692" CreationDate="2010-09-27T12:37:06.477" Score="7" ViewCount="279" Body="&lt;p&gt;I am reading a Set Theory book by Kunen. He presents first-order logic and claims that if a set of sentences in inconsistent, then it proves every possible sentence. Since he does not explicitly specify the inference rules, I became curious as to how fundamental this property of inconsistent systems is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is what is the simplest proof, with the least use of assumptions, of the vague claim that &quot;inconsistent systems can prove anything&quot; - in particular I'm interested in the assumptions about the system needed to prove this - is it true only for first order logic? Only for first order logic with the &quot;standard&quot; rules of inference (Modus ponens and GEN)? Or is it such a basic truth that it can be proved for every &quot;reasonable&quot; proof system (and what is &quot;reasonable&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-09-29T07:40:45.423" Title="Why an inconsistent formal system can prove everything?" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" />
  <row Id="5565" PostTypeId="2" ParentId="5564" CreationDate="2010-09-27T12:51:36.213" Score="1" ViewCount="0" Body="&lt;p&gt;From my recollection, it goes something like this: if $K$ is a first-order inconsistent theory, there exists, by definition, a formula $C$ such that $\vdash_{K} C$ and $\vdash_{K}\neg C$. If $D$ is an arbitrary formula in $K$ then we have the following chain (proof):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$C$&lt;/li&gt;&#xA;&lt;li&gt;$\neg C$&lt;/li&gt;&#xA;&lt;li&gt;$\neg C \Rightarrow (C\Rightarrow D)$ (Tautology)&lt;/li&gt;&#xA;&lt;li&gt;$C \Rightarrow D$ (2,3, Modus Ponens)&lt;/li&gt;&#xA;&lt;li&gt;$D$ (1,4, Modus Ponens).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, if inconsistency is defined the same way for logics of any other order, I think this proof wouldn't change. I don't see anything explicit about first-orderness in it, but correct me if I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-27T12:51:36.213" CommentCount="2" />
  <row Id="5566" PostTypeId="2" ParentId="5564" CreationDate="2010-09-27T13:18:40.877" Score="4" ViewCount="0" Body="&lt;p&gt;If $T$ is an inconsistent set of first order theorems (or axioms for the proof), then it is possible to prove from $T$ for some $\alpha$ both $\alpha$ and $\neg \alpha$. So without the loss of generality we can assume that $T$ includes $\alpha$ and $\neg\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $\beta$ is whatever first order sentence that you want to prove.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\alpha$ (in $T$)&lt;/li&gt;&#xA;&lt;li&gt;$\beta \to \alpha$ (easily verified to be true since $\alpha$ is an axiom of $T$)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \alpha \to \neg \beta$ &lt;/li&gt;&#xA;&lt;li&gt;$\neg \alpha$ (axiom of $T$)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \beta$ (inferred from 3 &amp;amp; 4)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \beta \to \alpha$ (holds for the same reason in 2, further more we have $\neg\beta$)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \alpha \to \neg\neg\beta$&lt;/li&gt;&#xA;&lt;li&gt;$\neg \neg \beta$ (inferred from 4,7)&lt;/li&gt;&#xA;&lt;li&gt;$\neg \neg \beta \to \beta$ (tautology)&lt;/li&gt;&#xA;&lt;li&gt;$\beta$ (inferred from 8,9)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So you see, you can prove pretty much anything you want from $\{\alpha ,\neg\alpha\}$ for some first order sentence $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-09-27T14:17:08.203" LastActivityDate="2010-09-27T14:17:08.203" CommentCount="5" />
  <row Id="5567" PostTypeId="2" ParentId="5540" CreationDate="2010-09-27T13:44:48.540" Score="6" ViewCount="0" Body="&lt;p&gt;This is obvious by ascent in the &lt;a href=&quot;http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726&quot; rel=&quot;nofollow&quot;&gt;ternary tree of Pythagorean triples&lt;/a&gt;. Explicitly, ascent  yields this formula:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm (x,x+1,z) \to (X,X+1,Z),\ \ X = 3x+2z+1,\ \ Z = 4x+3z+2\:.\ \ $  For example  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm (3,4,5)\to (20, 21, 29)\to (119, 120, 169)\to (696, 697, 985)\to (4059, 4060, 5741)\to\cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This corresponds to always taking the middle branch in the tree, see below&#xA;&lt;img src=&quot;http://i27.tinypic.com/ohimfs.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;The reflection that yields the descent in the triples tree is the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (x,y,z)\; \mapsto (x,y,z) - 2 \dfrac{(x,y,z)\cdot(1,1,1)}{(1,1,1)\cdot(1,1,1)} (1,1,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad = (x,y,z) - 2 \; (x+y-z) \; (1,1,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad\quad = (-x-2y+2z, \; -2x-y+2z, \; -2x-2y+3z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ascend the tree by inverting this reflection, combined with trivial sign-changing reflections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing in this way enables one to reflectively generate the entire tree of primitive Pythagorean triples. This has a beautiful geometric interpretation in terms of refelections - see my said &lt;a href=&quot;http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726&quot; rel=&quot;nofollow&quot;&gt;MathOverflow post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-27T15:56:20.440" LastActivityDate="2010-09-27T15:56:20.440" CommentCount="0" />
  <row Id="5568" PostTypeId="1" AcceptedAnswerId="5575" CreationDate="2010-09-27T14:02:17.930" Score="4" ViewCount="261" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/CatalansConstant.html&quot; rel=&quot;nofollow&quot;&gt;Catalan's Constant&lt;/a&gt; and quite a few other mathematical constants are known to have an infinite continued fraction (see the bottom of that webpage). On  wikipedia (I'm sorry, I can't post anymore hyperlinks because of my low rep.), a condition for irrationality of that continued fraction is given (see 'generalized continued fractions'). It is said that a given continued fraction converges to an irrational limit if $ b_n &gt; a_n $ in the continued fraction $b_0 + a_0/(b_1 + a_1/(b_2 + a_2/(...b_n+a_n)))$ for some sufficiently large $n$. In the webpage I provided you with, however, the degree of the polynomial $a_n$ of the continued fractions is bigger than the one of $b_n$. Therefore, all values of $b_n$ will never exceed all values of $a_n$, even after some large $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Why does the degree of $a_n$ need to be smaller than the degree of $b_n$ in order for a continued fraction representation of a constant to be irrational? I think I read somewhere it had to do with something like 'Tietschzes criterion' (but I'm not sure). (bonus question: Does anyone know where a proof of this 'criterion for irrationality' can be found?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T18:55:06.253" LastActivityDate="2011-01-06T18:55:06.253" Title="Why are some mathematical constants irrational by their continued fraction while others aren't ?" Tags="&lt;number-theory&gt;&lt;reference-request&gt;&lt;continued-fractions&gt;&lt;irrational-numbers&gt;&lt;approximation-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5569" PostTypeId="2" ParentId="5540" CreationDate="2010-09-27T14:02:39.833" Score="0" ViewCount="0" Body="&lt;p&gt;Just to add to Derek's excellent answer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pell's equation becomes essential in getting &lt;em&gt;all&lt;/em&gt; the solutions to $$k^2 + (k+1)^2 = j^2$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $gcd(k,k+1) =1 $ we may assume that (by the formula for all primitive pythagorean triples) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m^2 - n^2 = 2mn \pm 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a quadratic in $m$ and solving for $m$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m = n \pm \sqrt{2n^2 \pm 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus any solution of $$k^2 + (k+1)^2 = j^2$$ can be used to get a solution to the equation $$2n^2 \pm 1 = b^2$$ and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-27T14:02:39.833" />
  <row Id="5570" PostTypeId="2" ParentId="5558" CreationDate="2010-09-27T14:03:43.960" Score="8" ViewCount="0" Body="&lt;p&gt;$\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{k+3}{k(k+1)(k+2)(k+3)} - \dfrac{k}{k(k+1)(k+2)(k+3)})$&#xA;$ = \dfrac{1}{3}(\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} \dfrac{1}{2*3} = \dfrac{1}{18}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@moron Yes, you are right. For the first n terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit] more details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \sum_{k=1}^{n} \dfrac{1}{3} (\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$&#xA;$= \dfrac{1}{3} [\sum_{k=1}^{n} \dfrac{1}{k(k+1)(k+2)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$&#xA;$= \dfrac{1}{3} [\sum_{k=0}^{n-1} \dfrac{1}{(k+1)(k+2)(k+3)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$&#xA;$= \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$&#xA;&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="2164" LastEditorDisplayName="" LastEditDate="2010-09-27T16:29:40.460" LastActivityDate="2010-09-27T16:29:40.460" CommentCount="8" />
  <row Id="5571" PostTypeId="2" ParentId="5549" CreationDate="2010-09-27T14:18:46.397" Score="1" ViewCount="0" Body="&lt;p&gt;Put $\rm\ A = a^{n-1},\ B = b^{n-1}\:$ below and the results follow immediately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ =\ \frac{a+b}2\ \iff\ (a-b)\ (A-B)\ =\ 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ = \ \ \sqrt{ab}\ \ \ \:\iff\ (\sqrt a - \sqrt b)\ (A\sqrt{a}-B\sqrt{b})\ =\ 0$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T15:45:41.487" LastActivityDate="2010-10-06T15:45:41.487" />
  <row Id="5572" PostTypeId="2" ParentId="5558" CreationDate="2010-09-27T14:49:50.117" Score="6" ViewCount="0" Body="&lt;p&gt;This is similar to what has been said by Branimir, but shows how we can extend the result to&#xA;$$\sum_{k=1}^n {1 \over k(k+1) \cdots (k+m)}, \qquad m \in \mathbb{N}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can build up the result from the identities&#xA;$${1 \over k(k+1)} = {1 \over k} - { 1 \over k+1}, \qquad (1)$$&#xA;$${1 \over k(k+1)(k+2)} = {1 \over 2} \left( {1 \over k(k+1)} - { 1 \over (k+1)(k+2)} \right),$$&#xA;$${1 \over k(k+1)(k+2)(k+3)} = &#xA;{1 \over 3} \left( {1 \over k(k+1)(k+2)} - { 1 \over (k+1)(k+2)(k+3)} \right), \quad \textrm{ etc...}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $S_1 = \sum_{k=1}^n {1 \over k(k+1)},$&#xA;$S_2 = \sum_{k=1}^n {1 \over k(k+1)(k+2)},$ etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing for $S_1$ using (1) all terms on RHS cancel to get the classic&#xA;$$S_1 = \sum_{k=1}^n {1 \over k(k+1)} = 1 – {1 \over n+1} = {n \over n+1}.$$&#xA;We then sum the series for $S_2$ using this result obtained for $S_1,$ and so on.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-27T14:49:50.117" CommentCount="1" />
  <row Id="5573" PostTypeId="2" ParentId="5558" CreationDate="2010-09-27T14:54:16.097" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\ \frac{1}{(k+1)(k+2)(k+3)(k+4)} = \frac{1}{6(k+1)} - \frac{1}{2(k+2)}+\frac{1}{2(k+3)}-\frac{1}{6(k+4)}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ f(k+1)-f(k)\: = $ above $\rm\displaystyle\  \Rightarrow\  f(k) \:=\: c_0 + \frac{c_1}{k+1}\ \:+\:\ \frac{c_2}{k+2}\ \:+\:\ \frac{c_3}{k+3}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating yields $\rm\ c_0,c_1,c_2,c_3 \ =\ 1/18,\ -1/6,\ 1/3,\ -1/6$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For remarks on the group theory behind rational indefinite summation see &lt;a href=&quot;http://math.stackexchange.com/questions/4150&quot;&gt;my post here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-27T14:54:16.097" />
  <row Id="5574" PostTypeId="1" AcceptedAnswerId="5580" CreationDate="2010-09-27T15:19:08.110" Score="3" ViewCount="114" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d(n)$ represent the divisor function as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(n)=\displaystyle\sum\limits_{k|n}1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the divisor summatory function as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(x)=\displaystyle\sum\limits_{n \leq x}d(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the following triangular representation for the values of $D(n)$&#xA;&lt;img src=&quot;http://i.imgur.com/J0ZFW.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values on the right are the sum of all elements in a row.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above picture is the result of the following observation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $v_{m}(n)$ be the greatest power of $m$ that divides $n$ with $ m,n \in  \mathbb{N}$  , so we get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(n)=\displaystyle\sum\limits_{m=2}^{\infty}v_{m}(p^{n}), p \in  \mathbb{P}$ where $p$ is a fixed prime number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't try to prove this. I don't know how to do it, but hopefully some one will have some idea on how to prove or disprove this conjecture.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if this is a known fact. I don't have a proof but I've tested lots of values and woks all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-09-27T20:37:39.393" LastActivityDate="2010-09-27T20:37:39.393" Title="A triangular representation for the divisor summatory function, $D(x)$" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="5575" PostTypeId="2" ParentId="5568" CreationDate="2010-09-27T15:22:45.500" Score="5" ViewCount="0" Body="&lt;p&gt;You can find the &lt;a href=&quot;http://www.maths.unsw.edu.au/~angell/5535/chapter7.ps&quot; rel=&quot;nofollow&quot;&gt;proofs here&lt;/a&gt;, taken from Chrystal's Algebra - which is one of the best references on continued fractions. I suspect that if you read Chrystal then almost all of your questions will be answered. See also chapter 9 of Fowler's &quot;The mathematics of Plato's Academy&quot; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-27T16:13:29.007" LastActivityDate="2010-09-27T16:13:29.007" CommentCount="4" />
  <row Id="5576" PostTypeId="2" ParentId="5537" CreationDate="2010-09-27T15:28:44.587" Score="3" ViewCount="0" Body="&lt;p&gt;One complication in your situation is that you are mixing real and complex coordinates.&lt;br&gt;&#xA;If you were considering a map from a complex curve to $\mathbb{CP}^1$, then the kind of computation you are trying to make would work out more straightforwardly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you are looking at a map of real analytic manifolds, not complex analytic ones&#xA;(concretely, you are working with the variables $x,y,z$, which are real coordinates), the&#xA;point of view you have adopted is perhaps not quite as natural.  Nevertheless, it can be&#xA;made to work, as follows:&#xA;$$[x+iy:1-z] \text{ (which is where you finished) }&#xA;= [x^2 + y^2: (1-z)(x - i y)]$$&#xA;$$ = [ 1 - z^2: (1-z)(x-iy)] = [1+z:(x-iy)].$$&#xA;This rewriting of your map to $\mathbb{CP}^1$ is now well-defined in a neighbourhood of $(0,0,1)$ on the sphere.  (The fact that I introduced a complex conjugate of $x + i y$ to&#xA;facilitate the computation is related to the real vs. complex issue mentioned above.  This is also essentially the same computation you made to check that your map tends to $\infty$ as $z \to 1$, just rewritten in homogeneous coordinates.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-27T15:28:44.587" CommentCount="2" />
  <row Id="5577" PostTypeId="1" AcceptedAnswerId="5585" CreationDate="2010-09-27T15:57:58.893" Score="2" ViewCount="164" Body="&lt;p&gt;According &lt;a href=&quot;http://oeis.org/A180926&quot; rel=&quot;nofollow&quot;&gt;A180926&lt;/a&gt;, the elements of the set&#xA;{$a:\exists m,n|60a=5n^2+5n=3m^2+3m$} satisfy the following recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=\frac{62a_{n-1}+1+\sqrt{(48a_{n-1}+1)(80a_{n-1}+1)}}{2}$$ &#xA;$$a_1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how this can be derived?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-27T16:42:25.387" LastActivityDate="2010-09-28T15:50:39.207" Title="Recurrence relation for $a_{n}$ where $6a_{n}$ and $10a_{n}$ are both triangular." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5578" PostTypeId="2" ParentId="1519" CreationDate="2010-09-27T16:30:14.270" Score="2" ViewCount="0" Body="&lt;p&gt;Tom, I'm not sure I see how it's getting any harder in passing from a torus to a projective space.  In your $\mathbb CP^2$ case, you have $\mathbb CP^1$ sitting inside of it, and the boundary of a regular (tubular) neighbourhood of the $\mathbb CP^1$ is $S^3$.  And $D^4$ has $S^3$ as its boundary, so the attaching map is tautological.  The normal bundle is the missing data and that's what your CW-decomposition is ignoring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially what always happens.  Perhaps the conceptual hump you're dealing with is that you're asking for CW-decompositions of manifolds.  Morse functions generically only build homotopy-equivalences to CW-complexes, they do not put CW-structures on the manifold without some significant work. Moreover, CW-decompositions ignore some of the most essential properties of the manifold, like smooth structures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead you work with handle decompositions, what I state in my first paragraph is basically a generality -- critical points amount to handle attachments and the gluing instructions are always given in a direct way from the flow lines of the Morse function's (suitably normalized) gradient. So the handle decomposition is &lt;em&gt;on&lt;/em&gt; the given manifold -- unlike the CW-case where you only have a homotopy-equivalence to a CW-complex. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-09-27T16:30:14.270" />
  <row Id="5579" PostTypeId="1" CreationDate="2010-09-27T16:30:23.987" Score="1" ViewCount="64" Body="&lt;p&gt;In blind experiments subjects are randomly assigned to one of groups.  The most commonly used solution is to use (equivalent of) a coin toss, with the same probability to be assigned to each group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read about two modifications of a procedure of random assignment in blind tests:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;fair / just coin&lt;/strong&gt;, where probability of assignment to group A is proportional to the number of subjects assigned to other group (group B) divided by number of all assigned subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution is meant to have more equal distribution of subjects to groups (as truly random coin can result in one group to be over-represented).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known as &lt;strong&gt;covariate-minimization randomization&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;result-based probability&lt;/strong&gt;, where probability of assignment to given group is proportional to how well this group performs in experiment (e.g. in drug test).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known as &lt;strong&gt;response-adaptive randomization&lt;/strong&gt;, or &lt;strong&gt;outcome-adaptive randomization&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How based on results of such not fully random assignment method can one recover (calculate) the same estimates and error bounds as for random sampling?&lt;/p&gt;&#xA;" OwnerUserId="2167" LastActivityDate="2010-09-27T17:12:50.193" Title="Random assignment in blind experiments and &quot;fair / just coin&quot;" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;random&gt;" AnswerCount="1" />
  <row Id="5580" PostTypeId="2" ParentId="5574" CreationDate="2010-09-27T16:42:24.107" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, this is true.  Write $D(x) = \sum_{n \le x} d(n) = \sum_{n \le x} \sum_{d | n} 1 = \sum_{d \le x} \lfloor \frac{x}{d} \rfloor$; this is equivalent to the pattern you observe.  The last step is exchanging the order of summation together with the observation that the number of times a number $d$ appears in the double sum is the number of numbers less than or equal to $x$ it divides.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-27T16:42:24.107" CommentCount="1" />
  <row Id="5581" PostTypeId="1" CreationDate="2010-09-27T16:46:08.483" Score="0" ViewCount="203" Body="&lt;p&gt;(Wade's rewording)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a phrase that contains 4 words and 22 total letters (not counting the three spaces between the words). One letter appears four times, no letter appears three times, three letters appear twice, and 12 letters appear once each. What is this phrase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;======================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@@$$##&amp;amp;&amp;amp;&amp;amp;%%%%????????????&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 words and the &amp;amp; represents a whitespace.&lt;br&gt;&#xA;@,#,$,% represent one letter each (4 total letters).&lt;br&gt;&#xA;Each ? represents one of the other 22 letter of the alphabet (12 total letters).&lt;br&gt;&#xA;16 different letters used.&lt;br&gt;&#xA;ciphertext also needs to be rearranged - it is 'scrambled.'  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I approach this?&lt;br&gt;&#xA;Not looking for the answer really.&lt;br&gt;&#xA;I just need a tool to unscramble the phrase and decrypt the ciphertext simultaneously.&lt;br&gt;&#xA;Decrypto only does the later. Thank you.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Noldorin: This is not homework. From elsewhere. &#xA;@Kenny: How are encryption/decryption methodologies not math? This is an applied math problem, whereas your preferences tend to be theoretical/ pure math?&lt;/p&gt;&#xA;" OwnerUserId="2168" LastEditorUserId="2168" LastEditorDisplayName="" LastEditDate="2010-09-27T18:10:59.440" LastActivityDate="2010-09-27T21:12:34.590" Title="Solve cryptogram - ciphertext given" Tags="&lt;cryptography&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" />
  <row Id="5582" PostTypeId="2" ParentId="4963" CreationDate="2010-09-27T16:57:56.233" Score="1" ViewCount="0" Body="&lt;p&gt;@ fiktor: due to the low amount of reputation points I have at the moment, I can't comment yet. I just wanted to point out that, since I suspect this is a research-level question, there's a website called www.mathoverflow.com which might be an (even) better platform to launch this question from.  &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-09-27T18:24:31.163" LastActivityDate="2010-09-27T18:24:31.163" CommentCount="1" />
  <row Id="5583" PostTypeId="2" ParentId="5579" CreationDate="2010-09-27T17:12:50.193" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a paper that may be helpful regarding your second point: &lt;a href=&quot;http://www.johndcook.com/power_bias.pdf&quot; rel=&quot;nofollow&quot;&gt;Power and bias in adaptively randomized clinical trials&lt;/a&gt;.  When looked at from the Bayesian standpoint, you simply compute posterior probabilities.  The randomization mechanism doesn't matter.  If you want to get frequentist operating characteristics, you can run simulations.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-27T17:12:50.193" />
  <row Id="5584" PostTypeId="2" ParentId="4983" CreationDate="2010-09-27T17:19:39.553" Score="2" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem&quot; rel=&quot;nofollow&quot;&gt;Erdős–Kac theorem&lt;/a&gt; shows that the (log of the log of the) prime factors of a number are Poisson/normally distributed.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-09-27T17:19:39.553" CommunityOwnedDate="2010-09-27T17:19:39.553" />
  <row Id="5585" PostTypeId="2" ParentId="5577" CreationDate="2010-09-27T17:22:00.967" Score="5" ViewCount="0" Body="&lt;p&gt;Since we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 5(n^2 +n) = 3(m^2 +m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying by $4$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5(4n^2 + 4n + 1) - 5 = 3(4m^2 + 4m + 1) -3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $x = 2n+1$ and $y = 2m+1$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5x^2 - 3y^2 = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalized Pell's equation of the form $ax^2 - by^2 = c$ and I believe this is a solved problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, see this: &lt;a href=&quot;http://books.google.com/books?id=YD6UlfcRlGQC&amp;amp;pg=PA81&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=YD6UlfcRlGQC&amp;amp;pg=PA81&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should give us all the solutions, and so make the recurrence easy to verify.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-27T17:22:00.967" CommentCount="8" />
  <row Id="5586" PostTypeId="1" AcceptedAnswerId="5589" CreationDate="2010-09-27T18:22:43.087" Score="5" ViewCount="100" Body="&lt;p&gt;Suppose I embed a manifold-with-boundary $M$ in some $\mathbb{R}^n$.  Are there conditions (necessary, sufficient, or both) that can help determine when the topological boundary of $M$ is equal to the manifold boundary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;topological boundary,&quot; I'm referring to $\text{Bd } M$, which is the closure minus the interior (relative to $\mathbb{R}^n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;manifold boundary,&quot; I mean the boundary $\partial M$ that is specified in the definition of &quot;manifold-with-boundary.&quot;&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2011-01-02T06:27:46.397" LastActivityDate="2011-01-02T06:27:46.397" Title="When does the topological boundary of an embedded manifold equal its manifold boundary?" Tags="&lt;topology&gt;&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5587" PostTypeId="1" AcceptedAnswerId="5590" CreationDate="2010-09-27T18:25:08.720" Score="1" ViewCount="53" Body="&lt;p&gt;The title and tags are a little goofy since I'm not quite sure what to call this.  This may be more of a computer science question, but I thought I'd ask here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two integers n and m that are arbitrarily (logically) related with integers x and y, respectively, where all values are finite and &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if there's a way to combine n and x, and m and y, into two numbers (integer, decimal, whatever) such that neither will overlap.  Additionally, I need to be able to perform some kind of operation to reverse the previous process to once again obtain n and x, and m and y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this isn't clear.  Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2010-09-27T18:39:56.247" Title="Related Integer Combination" Tags="&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5588" PostTypeId="1" AcceptedAnswerId="5592" CreationDate="2010-09-27T18:25:52.523" Score="3" ViewCount="173" Body="&lt;p&gt;I recently learned about the Classification Theorem for compact 2-manifolds.  Is there a similar classification theorem for ALL 2-manifolds, not just the compact ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, is there a theorem which classifies the 2-manifolds with boundary?&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-09-28T05:00:44.507" Title="Classification Theorem for Non-Compact 2-Manifolds?  2-Manifolds With Boundary?" Tags="&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5589" PostTypeId="2" ParentId="5586" CreationDate="2010-09-27T18:31:50.703" Score="5" ViewCount="0" Body="&lt;p&gt;If you embed a smooth $m$-manifold $M$ smoothly in $\mathbb{R}^n$&#xA;then the &quot;topological boundary&quot; of $M$ is the closure of $M$. As&#xA;locally each point of $M$ has a neighbourhood in $\mathbb{R}^n$&#xA;where $M$ looks like $\mathbb{R}^m$, then no point of $M$ is interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $m=n$ and $M$ is compact then, yes (at least in the smooth case)&#xA;the topological and manifold boundaries coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the above hold for topological embeddings but won't swear&#xA;to it; they need not be locally flat (see nasties like the Alexander&#xA;horned sphere),&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-27T18:31:50.703" CommentCount="2" />
  <row Id="5590" PostTypeId="2" ParentId="5587" CreationDate="2010-09-27T18:39:23.933" Score="3" ViewCount="0" Body="&lt;p&gt;You could employ a pairing function, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor's&lt;/a&gt; or &lt;a href=&quot;http://szudzik.com/ElegantPairing.pdf&quot; rel=&quot;nofollow&quot;&gt;others&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-27T18:39:23.933" />
  <row Id="5592" PostTypeId="2" ParentId="5588" CreationDate="2010-09-27T18:56:03.307" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, there's a classification theorem for non-compact 2-manifolds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper gives the classification for triangulable 2-manifolds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/stable/1993768&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/1993768&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That an arbitrary (2nd countable, Hausdorff) topological 2-manifold admits a triangulation is fairly classical.  Ahlfors book &quot;Riemann Surfaces&quot; has a proof.  There are others available, see for example this list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/17578/triangulating-surfaces&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17578/triangulating-surfaces&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all you're interested in is compact manifolds with boundary, you get that classification immediately from the closed manifold case.  Because if you have a compact manifold with boundary, its boundary is a disjoint union of circles.  So cap those circles off with discs to produce a closed manifold.  So compact manifolds with boundary are classified by the closed manifold you get by &quot;capping off&quot; and the number of boundary circles you started with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-compact manifolds have a more delicate classification -- think for example about the complement of a Cantor set in a compact surface. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-09-28T05:00:44.507" LastActivityDate="2010-09-28T05:00:44.507" CommentCount="2" />
  <row Id="5593" PostTypeId="2" ParentId="5581" CreationDate="2010-09-27T19:10:17.413" Score="6" ViewCount="0" Body="&lt;p&gt;Captain Kirk was wondering why his journal was not more excited to hear about his exploits. He said&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LOG, YOU SEEM FAINTHEARTED!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More seriously, I dumped the letters EEEETTAAOOINSHRDLUGFMY into a multiword anagrammer, and it immediately found more than 24000 answers, of which this was one of the more amusing ones. What I deduce from this experience is that, without the knowledge that you are looking for a common English phrase, this puzzle is dramatically under constrained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you look for a file of common English phrases which you can test for this property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't math, but there is a bit of a mathematical lesson here -- it is worth doing a quick test to see whether you have enough data to answer a question before working too hard on it.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-09-27T19:36:28.547" LastActivityDate="2010-09-27T19:36:28.547" CommunityOwnedDate="2010-09-27T19:10:17.413" CommentCount="2" />
  <row Id="5594" PostTypeId="2" ParentId="5581" CreationDate="2010-09-27T19:59:23.460" Score="3" ViewCount="0" Body="&lt;p&gt;As David Speyer notes, the problem is too unconstrained for a solution as it stands. You can make some guesses using frequency tables for letters in English, but that's about the only place where some math will come in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were a &quot;straight cryptogram&quot; (the letters are placed in the correct order, whether with the correct word spacing or in blocks), then you are a little closer to a problem in which mathematics can come in. You would have frequency tables, plus also tables on digraphs or trigraphs (blocks of 2 or 3 letters, when certain letters tend to occur next to each other); the two together can give you a lot of information (though the shorter the text, the less useful information they give: you need larger cryptograms to avoid the information getting lost in the &quot;noise&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see some of the general procedures to attack &quot;classic&quot; cryptographic systems like monoalphabetic ciphers (where each letter in the original is replaced by another letter, the same one throughout, and the order of the letters is not modified), and polyalphabetic ciphers (where the substitutions depend on the location of the letter; these range from simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;nofollow&quot;&gt;Vigenère substitution&lt;/a&gt; all the way to substitutions like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Enigma_machine&quot; rel=&quot;nofollow&quot;&gt;Engima cipher&lt;/a&gt;), or transposition systems (where the order of the letters is changed, but the letters themselves are not), then I can recommend &lt;em&gt;Elementary Cryptanalysis: A mathematical approach&lt;/em&gt; by Abraham Sinkov (Mathematical Association of America, New Mathematical Library, ISBN 0883856220). Pretty much all of them are obsolete as far as modern cryptography goes, but it can give you a taste of how (and when) mathematics comes in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your problem as stated, however, the role of mathematics is so limited as to become nearly useless. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-27T21:12:34.590" LastActivityDate="2010-09-27T21:12:34.590" CommentCount="6" />
  <row Id="5595" PostTypeId="1" AcceptedAnswerId="5596" CreationDate="2010-09-27T20:15:47.413" Score="11" ViewCount="390" Body="&lt;p&gt;This is a neat little problem that I was discussing today with my lab group out at lunch. Not particularly difficult but interesting implications nonetheless&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine there are a 100 people in line to board a plane that seats 100. The first person in line realizes he lost his boarding pass so when he boards he decides to take a random seat instead. Every person that boards the plane after him will either take their &quot;proper&quot; seat, or if that seat is taken, a random seat instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the probability that the last person that boards will end up in his/her proper seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, and this is the part I'm still pondering about. Can you think of a physical system that would follow this combinatorial statistics? Maybe a spin wave function in a crystal etc...&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-11-19T21:26:27.500" Title="Taking Seats on a Plane" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="5596" PostTypeId="2" ParentId="5595" CreationDate="2010-09-27T20:30:13.373" Score="12" ViewCount="0" Body="&lt;p&gt;This is a classic puzzle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that the probability that the last person ends in up in his proper seat is exactly $\frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that the fate of the last person is determined the moment either the first or the last seat is selected! This is because the last person will either get the first seat or the last seat. Any other seat will necessarily be taken by the time the last guy gets to 'choose'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since at each choice step, the first or last is equally probable to be taken, the last person will get either the first or last with equal probability: $\frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, no clue about a physical system.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-27T20:30:13.373" CommentCount="1" />
  <row Id="5597" PostTypeId="2" ParentId="5411" CreationDate="2010-09-27T20:50:46.520" Score="1" ViewCount="0" Body="&lt;p&gt;Here's one way to solve the problem. The trick is to substitute &lt;code&gt;-I*Log[x]&lt;/code&gt; for t in the characteristic function $\varphi[t]$. This turns it into a generating function. Then extract the coefficient of $x^n$ from the resulting series using the &lt;b&gt;SeriesCoefficient&lt;/b&gt; built-in. The coefficient of $x^n$ is the value of the pmf f at n. Here's a simple example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;        (* Characteristic function (CF) of the discrete probability&#xA;           distribution for summing two 6-sided dice *)&#xA;In[1]:= \[CurlyPhi][t_] :=&#xA;          CharacteristicFunction[DiscreteUniformDistribution[{1, 6}], t]^2&#xA;&#xA;        (* Recover the probability mass function f from this CF *)&#xA;In[2]:= f[n_] := SeriesCoefficient[\[CurlyPhi][-I*Log[x]], {x, 0, n}]&#xA;&#xA;        (* Show the range of the pmf f over its domain, {2,..,12} *)&#xA;In[3]:= f[#] &amp; /@ Range[2, 12]&#xA;Out[3]= {1/36, 1/18, 1/12, 1/9, 5/36, 1/6, 5/36, 1/9, 1/12, 1/18, 1/36}&#xA;&#xA;        (* Note that applying f to a number outside its domain (like 1)&#xA;           produces 0 *)&#xA;In[4]:= f[1]&#xA;Out[4]= 0&#xA;&#xA;        (* Now use this technique to show the range of the pmf of summing three&#xA;           6-sided dice *)&#xA;In[5]:= Map[&#xA;          Function[n, &#xA;            SeriesCoefficient[&#xA;              CharacteristicFunction[&#xA;                DiscreteUniformDistribution[{1, 6}], -I*Log[x]]^3, {x, 0, n}]], &#xA;                  Range[3, 18]]&#xA;Out[5]= {1/216, 1/72, 1/36, 5/108, 5/72, 7/72, 25/216, 1/8, 1/8, 25/216, 7/72,&#xA;          5/72, 5/108, 1/36, 1/72, 1/216}&#xA;&#xA;        (* And verify that the values in the range of this pmf sum to 1 *)&#xA;In[6]:= Fold[Plus, 0, %]&#xA;Out[6]= 1&#xA;&lt;/pre&gt;&#xA;" LastActivityDate="2010-09-27T20:50:46.520" />
  <row Id="5601" PostTypeId="2" ParentId="5568" CreationDate="2010-09-27T22:42:04.467" Score="2" ViewCount="0" Body="&lt;p&gt;The continued fraction expansion of any possible real number falls in one of 2 possible categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the continued fraction expansion has a finite length if and only if the number is rational.&lt;/li&gt;&#xA;&lt;li&gt;the continued fraction expansion never ends if and only if the number is irrational.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;quite a few other mathematical constants are known to have an infinite continued fraction&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, and all such constants are therefore all irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Catalan's Constant ... known to have an infinite continued fraction&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a proof? My understanding is that we don't yet know whether or not Catalan's Constant has an finite continued fraction -- therefore we don't yet know whether or not it is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Why does the degree of an need to be smaller than the degree of bn in order for a continued fraction representation of a constant to be irrational?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is required for the sequence to converge on some specific value.&#xA;If the degree of an is &lt;em&gt;larger&lt;/em&gt;, that doesn't mean the result is rational -- it indicates that it doesn't converge to any real number, and therefore isn't a &quot;constant&quot; at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to make the &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction#Conditions_for_irrationality&quot; rel=&quot;nofollow&quot;&gt;Wikipedia &quot;generalized continued fraction&quot; article&lt;/a&gt; a little less confusing?&lt;/p&gt;&#xA;" OwnerUserId="1705" LastActivityDate="2010-09-27T22:42:04.467" CommentCount="2" />
  <row Id="5602" PostTypeId="1" AcceptedAnswerId="5605" CreationDate="2010-09-27T23:13:35.370" Score="2" ViewCount="65" Body="&lt;p&gt;I'm reading Jech's Set Theory, and in the chapter about measurable cardinals there is a theorem that if $\kappa$ is real-measurable but not measurable then it is $\le 2^{\aleph_0}$ and so and so. (Corollary 10.10)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a cardinal number be real-measurable without being measurable? Can't a measure be &quot;destructed&quot; (as in opposite of constructed) into a trivial $0,1$ measure?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-09-28T00:00:14.313" Title="Real-measurable cardinals that are not measurable ones" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" />
  <row Id="5603" PostTypeId="2" ParentId="5577" CreationDate="2010-09-27T23:13:57.980" Score="8" ViewCount="0" Body="&lt;p&gt;Here's a complete solution, including a derivation of the recurrence relation in &lt;a href=&quot;http://oeis.org/A180926&quot; rel=&quot;nofollow&quot;&gt;A180926&lt;/a&gt;. First, notice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ 60\:a \:=\: 5\:(n^2+n)\ =\ 3\:(m^2+m)\ \ $  is easily transformed into the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Pell%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Pell equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \: 5\:x^2-3\:y^2 \:= \ 2\ \ \ $  for $\rm\ \ \ (x,\:y)\: =\: (2\:n+1,\:2\:m+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ \ \ \: x^2 - 1\ =\ 4\: (n^2+n)\ =\ 48\:a,\quad\ y^2 - 1\ =\ 4\: (m^2+m)\ =\ 80\:a\quad\quad\quad\ \ (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://www.alpertron.com.ar/QUAD.HTM&quot; rel=&quot;nofollow&quot;&gt;standard Pell theory&lt;/a&gt; $\rm\ \: (x,y) \:\to\: (X,Y)\:=\: (4\:x+3\:y\ \ ,\ \ 5\:x+4\:y)\ $ is a solution too, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\quad\Rightarrow\quad}\ X+Y\omega \: =\: (4+5\:\omega)\:(x+y\:\omega) \: = \: (4\:x+3\:y + (5\:x+4\:y)\:\omega) \: ,\ \ \ \omega \:=\: \sqrt{3/5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\Rightarrow\quad  X- Y\omega \: =\: (4-5\:\omega)\:(x-y\:\omega)\quad\quad\:$ via conjugate prior equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\Rightarrow\quad  5\: X^2 - 3\: Y^2 \: =\  5\: x^2 - 3\: y^2\: =\: 2\ \ $ via multiply the prior two equations, then scale by 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This yields a recurrence to generate a new solution $\rm (X,Y) = (x_{n+1},\:y_{n+1})$ from a known solution $\rm\: (x,\:y) = (x_n,\:y_n)\:.\:$ We square this recurrence to obtain a recurrence for $\rm (A,\:a) = (a_{n+1},\:a_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely $\rm\ \: (4\:x+3\:y)^2 =\ X^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad \Rightarrow\quad\ 24\ x\:y\ =\ X^2-1 - 16\:(x^2-1) - 9\:(y^2-1) - 24$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 48\:A - 16\:(48\:a) - 9\:(80\:a) - 24\quad$ by $\ (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 24\ (2\:A - 62\: a - 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad \Rightarrow\quad\ x^2\ y^2\ =\ (2\:A - 62\: a - 1)^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also $\rm\quad\quad\quad\: x^2\ y^2\ =\ (48\: a + 1)\ (80\: a + 1)\quad$ by $\ (*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ \ (2\ a_{n+1} - 62\ a_n - 1)^2 \ = \ (48\ a_n+1)\ (80\ a_n+1)\ \ $  as was to be proved.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-28T15:50:39.207" LastActivityDate="2010-09-28T15:50:39.207" CommentCount="2" />
  <row Id="5604" PostTypeId="2" ParentId="5602" CreationDate="2010-09-27T23:36:29.937" Score="2" ViewCount="0" Body="&lt;p&gt;It is a good question. The answer is that &lt;a href=&quot;http://en.wikipedia.org/wiki/Measurable_cardinal#Real-valued_measurable&quot; rel=&quot;nofollow&quot;&gt;real-valued measurable cardinal&lt;/a&gt; need not be strongly inaccessible, whilst every measurable cardinal is strongly inaccessible. Indeed, it is consistent that the continuum itself is a real-valued measurable cardinal, but the continuum can never be a measurable cardinal, since every measurable cardinal is strongly inaccessible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, part of what you claim is true: Solovay proved that every real-valued measurable cardinal $\kappa$ is fully measurable (with a two-valued measure) in an inner model of the universe. That is, if $\kappa$ is a real-valued measurable cardinal, then there is a definable transitive class $W$ satisfying ZFC in which $\kappa$ is an actual measurable cardinal. The class $W$ is defined directly from the real-valued measure on $\kappa$, and this provides a sense in which the measure is deconstructed to form a 2-valued measure.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-27T23:36:29.937" />
  <row Id="5605" PostTypeId="2" ParentId="5602" CreationDate="2010-09-28T00:00:14.313" Score="2" ViewCount="0" Body="&lt;p&gt;Two-valued measures behave very differently from real-valued measures. For example, suppose $\mathcal{U}$ is a countably complete ultrafilter on a set $X$ and suppose that $f:X\to2^\omega$ is an injection. There is a $b \in 2^\omega$ such that&#xA;$$B_n = \{ a \in X : f(a)(n) = b(n) \} \in \mathcal{U}$$&#xA;for every $n &amp;lt; \omega$. By countable completeness, $B = \bigcap_{n&amp;lt;\omega} B_n \in \mathcal{U}$. But $B$ contains exactly one element (namely $f^{-1}(b)$) since $f$ is an injection. Therefore, $\mathcal{U}$ is a principal ultrafilter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument shows that the first measurable cardinal is larger than $2^{\aleph_0}$. Indeed, a slightly more general argument can be used to show that a measurable cardinal must be inaccessible. However, this argument cannot be carried out with a real-valued measure. In fact, it is possible (assuming the consistency of a measurable cardinal) for Lebesgue measure to be extended to a measure defined on all subsets of $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-09-28T00:00:14.313" CommentCount="1" />
  <row Id="5606" PostTypeId="1" AcceptedAnswerId="5610" CreationDate="2010-09-28T00:09:26.330" Score="1" ViewCount="83" Body="&lt;p&gt;Hi people. Can you help me realize why this is true? I can tell you that $P_i$ and $P_j$ are probabilities, i.e. $0 \leq P_i, P_j \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty ijP_iP_j \leq \sum_{i=1}^\infty \sum_{j=1}^\infty j^2P_jP_i$.&lt;/p&gt;&#xA;" OwnerUserId="2174" LastActivityDate="2010-09-28T01:38:05.997" Title="Help me realize this inequality of infinite sums." Tags="&lt;sequences-and-series&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="5607" PostTypeId="1" AcceptedAnswerId="5620" CreationDate="2010-09-28T00:33:48.630" Score="3" ViewCount="76" Body="&lt;p&gt;Suppose one wants to prove that a 1-dimensional integral proper scheme over an algebraically closed field is projective. This is a step in how Hartshorne has you prove that any 1-dim proper scheme (over an algebraically closed field) is projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method is that first note the normalization is non-singular and hence projective. Consider $f: \tilde{X}\to X$, and let $\mathcal{L}$ be a very ample sheaf on $\tilde{X}$. The goal is to prove that there is an effective divisor $D=\sum P_i$ such that $\mathcal{L}(D)\simeq \mathcal{L}$ and such that each $f(P_i)$ are non-singular points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the proof follows from the fact that there merely exists some very ample sheaf with that property, but the exercise seems to imply that any very ample sheaf satisfies this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm missing something really obvious, but take a proper curve over $k$, say $C$, that has a single singularity. Blow-up that singularity, $\pi: \tilde{C}\to C$, this is the normalization. Take a point $P\in \tilde{C}$ such that $\pi(P)$ is the singularity. It seems to me that $D=P$ is an effective divisor, so $\mathcal{L}(D)$ is an invertible sheaf. If the above is correct, then $\mathcal{L}(D)$ cannot be very ample. Does anyone have a simple reason for this?&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-09-28T08:14:21.607" Title="Very ample sheaf on a blowup" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="5608" PostTypeId="2" ParentId="5510" CreationDate="2010-09-28T00:59:06.213" Score="5" ViewCount="0" Body="&lt;p&gt;I've glanced at the paper you linked to. I have a few remarks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It looks the authors of the paper managed to derive explicit closed-form expressions for the stationarity conditions for a special class of functions. This is sometimes possible for very specific problems. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you remember your calculus, for a univariate function $f(x)$, the minimum or maximum is found when $f'(x) = 0$. You can think of the KKT as a generalization of this idea for an optimization problem with constraints. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The reason I (and the authors) say &quot;stationarity&quot; conditions instead of &quot;optimality&quot; conditions is because KKT conditions are necessary but insufficient conditions for optimality. For a true guarantee of optimality, you need to test your solution against SSOCs (Sufficient Second Order Conditions). Again, the second order conditions are analogous to the Calculus-101 idea of determining whether $f''(x)$ is positive or negative, which tells you whether the solution of $f'(x) = 0$ is a minimum or maximum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nevertheless in this case, under the assumptions in the paper, in the absence of saddle points, we may presume the solution to be optimal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This means you &lt;strong&gt;do not have to use any optimization algorithm/library in this particular case&lt;/strong&gt;. For a given pricing function, all you have to do is evaluate a function based on the results given in the paper. This is a very nice property -- it means the optimization is implicitly done for you. See Example 1 in the paper for a worked example. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general, the KKT does not provide an algorithm for finding an optimum; it's only a test for stationarity. However, in very specific cases (i.e. with regularity conditions assumed, convex, no inequalities, etc.), the KKT conditions can provide closed-form expressions for the optimal solution of a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact... this is the idea behind the Method of Lagrange Multipliers. The Lagrange Multiplier method is simply a special case of the KKT conditions with no inequality constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Side Note&lt;/em&gt;: one of the reasons behind the difficulty in using the KKT as a practical algorithm to find stationary/optimal points is due to the &quot;complementarity conditions&quot; in the KKT system (see Wikipedia article).  when you have inequality constraints, you have to satisfy a set of constraints known as &quot;complementarity constraints&quot;. There can be nasty, nonconvex, degenerate bilinear terms in these constraints, and special techniques are required to handle them -- enumerative methods are required for obtaining global optimality, and therefore the problem is NP-hard. There is a whole new field of research in burgeoning area of complementarity problems (and its related problem class, MPECs = mathematical programs with equilibrium constraints)... currently there's a lot of work being done on determining different types of stationarity conditions for this class of problems. It is outside the scope of the question, but I thought I'd just let you know what's out there.&lt;/p&gt;&#xA;" OwnerUserId="1124" LastEditorUserId="1124" LastEditorDisplayName="" LastEditDate="2010-09-28T01:22:31.307" LastActivityDate="2010-09-28T01:22:31.307" CommentCount="3" />
  <row Id="5609" PostTypeId="1" AcceptedAnswerId="5611" CreationDate="2010-09-28T01:32:41.710" Score="0" ViewCount="206" Body="&lt;p&gt;I'm trying to figure out the limit of this sequence. I feel like I'm close to understanding it, I just need a little push.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lbrace\sum_{n=1}^{k}\frac{1}{\sqrt{k^2 + n}}\rbrace_{k=1}^\infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that I need to show that this sequence lies between $\lbrace \frac{k}{\sqrt{k^2 + k}}\rbrace_{k=1}^\infty$ and $\lbrace \frac{k}{\sqrt{k^2 + 1}}\rbrace_{k=1}^\infty$, and that both of those sequences converge to 1. However, I'm not quite sure how to show that both of them converge to 1, or why the first sequence would lie between them. The summation and the square roots are confusing me, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Okay, so I have the notion that I should divide everything in the convergent sequences by k, but I don't think I'm doing the algebra right. For the first sequence, dividing by k seems to get me $\frac{1}{\sqrt{k + 1}}$, which seems to converge to 0 as k approaches $\infty$ since $\frac{1}{\sqrt{k + 1}} \leq \frac{1}{k}$. Similarly, dividing everything in the latter sequence seems to give $\frac{1}{\sqrt{k + \frac{1}{k}}}$, which would also converge to zero as k approaches $\infty$. I don't feel like I'm distributing the k properly over the square root.&lt;/p&gt;&#xA;" OwnerUserId="2175" LastEditorUserId="2175" LastEditorDisplayName="" LastEditDate="2010-09-28T04:29:15.193" LastActivityDate="2010-09-28T04:29:15.193" Title="Finding the limit of $\lbrace\sum_{n=1}^{k}\frac{1}{\sqrt{k^2 + n}}\rbrace_{k=1}^\infty$" Tags="&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5610" PostTypeId="2" ParentId="5606" CreationDate="2010-09-28T01:38:05.997" Score="5" ViewCount="0" Body="&lt;p&gt;For a given pair $(m,n)$ with $m\ne n$, consider the sum of the two terms corresponding to $(i,j) = (m,n)$ and $(i,j)=(n,m)$.  For the summation on the left, this will be&#xA;$$&#xA;mnP_mP_n + nmP_nP_m = 2mnP_mP_n,&#xA;$$&#xA;while for the sum on the right this will be&#xA;$$&#xA;n^2P_n P_m + m^2 P_m P_n = (n^2+m^2)P_mP_n.&#xA;$$&#xA;Since $2mn \le m^2+n^2$ for all $m$ and $n$, it follows that&#xA;$$&#xA;\sum_{i \ne j} ijP_iP_j \le \sum_{i\ne j} j^2 P_j P_i.&#xA;$$&#xA;But&#xA;$$&#xA;\sum_{i = j} ijP_iP_j = \sum_{i = j} j^2 P_j P_i,&#xA;$$&#xA;so this proves the given inequality.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-28T01:38:05.997" />
  <row Id="5611" PostTypeId="2" ParentId="5609" CreationDate="2010-09-28T01:42:18.963" Score="2" ViewCount="0" Body="&lt;p&gt;Since this looks like homework, I will give a hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the largest among $\sqrt{k^2+1}, \sqrt{k^2+2},\dots, \sqrt{k^2+k}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your next step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the mistake you are doing is substituting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sqrt{k^2+k}}{k}$ with $\sqrt{\frac{k^2+k}{k}}$ instead of $\sqrt{\frac{k^2+k}{k^2}}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T03:49:19.593" LastActivityDate="2010-09-28T03:49:19.593" CommentCount="6" />
  <row Id="5612" PostTypeId="2" ParentId="5523" CreationDate="2010-09-28T01:50:16.483" Score="2" ViewCount="0" Body="&lt;p&gt;I proved this a few years ago as a homework assignment. Even if it's not &lt;em&gt;your&lt;/em&gt; homework assignment, that makes me reluctant to give a complete solution; however, the hints below should be helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 1:&lt;/strong&gt; The proof I submitted is along the following lines: given a monotone self-dual function $f$, show by induction that for all sufficiently small nonnegative integers $l$, there is another function $g_l$ (of the same arity as $f$) such that $||f-g_l||_1 = 2l$ and $g_l$ is in the clone. (Here $||f-g||_1$ is the &quot;$l_1$-norm,&quot; which in this case is just the number of $x$ for which $f(x)\neq g(x)$.) The basis case is obvious: pick an arbitrary function you know is in the clone. For the inductive step, you have to figure out how to take $g_l$ and produce $g_{l-1}$, which I'll leave out, but a hint is that there is a $g_{l-1}$ with $||g_l - g_{l-1}||_1 = 2$, i. e., $g_{l-1}$ is almost the same as $g_l$. The conclusion (by induction) is the case $l=0$, but $g_0$ has to be $f$, so $f$ is in the clone. (Informally: you can start at some function in the clone and &quot;walk&quot; by simple steps to any other monotone self-dual function of the same arity, and adjacent functions on the walk can be obtained from one another using the majority function.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 2:&lt;/strong&gt; However, there's a much neater argument by induction on the arity of $f$. To simplify future notation, let $\bar{x} = (x_4,\ldots,x_n)$, so $f(x_1,\ldots,x_n) = f(x_1,x_2,x_3,\bar{x})$. Observe that if&#xA;$f:\{0,1\}^n\to \{0,1\}$ is monotone and self-dual and $n&gt; 3$, then the&#xA;three functions of $n-1$ arguments $$f_1(x,y,\bar{x}) =&#xA;f(x,y,y,\bar{x}),$$ $$f_2(x,y,\bar{x}) = f(y,x,y,\bar{x}),$$&#xA;$$f_3(x,y,\bar{x}) = f(y,y,x,\bar{x}),$$ are monotone and self-dual&#xA;as well, and moreover, they determine $f$. (Indeed, out of $x_1,x_2,x_3$,&#xA;at least 2 must be equal, so for example, if $x_1 = x_2$, then&#xA;$f(x_1,\ldots,x_n) = f_3(x_3,x_2,\bar{x})$.) All you have to do now is reconstruct $f$ from $f_1,f_2,f_3$ using only the majority function. Thankfully,  the construction is very simple, although it may not be obvious at first why it works: $$f(x_1,\ldots,x_n) = M(f_1(x_1,x_3,\bar{x}), f_2(x_2,x_1,\bar{x}), f_3(x_3,x_2,\bar{x}))$$ (where $M$ is the majority function). (Hint: prove the case $x_1=1,x_2=x_3=0$, and the rest should become obvious. You need monotonicity here, but surprisingly not self-duality.) The basis case is $n\leq 3$ (where self-duality is needed). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, well; I guess that's almost a complete solution anyway, but I don't feel like going back and removing steps from the argument at the risk of making it cryptic and useless. :-)&lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-28T01:50:16.483" />
  <row Id="5613" PostTypeId="2" ParentId="5609" CreationDate="2010-09-28T02:13:47.860" Score="0" ViewCount="0" Body="&lt;p&gt;You want to take the limits of your two sequences as they approach infinity. if you are in a calculus class, you can use L'Hôpital's rule (http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule). Otherwize, you should devide both the numerator and denominator by k, and then bring the k in the denominator into the radical. Then, as you go to infinity the stuff inside the radical goes to 1, and since 1/sqrt(1)=1, that becomes your answer.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-28T02:13:47.860" CommentCount="1" />
  <row Id="5614" PostTypeId="1" AcceptedAnswerId="5638" CreationDate="2010-09-28T04:15:14.757" Score="1" ViewCount="162" Body="&lt;p&gt;Let $N$ be a normal subgroup of a finite group $G$. Let $S$ belonging to $G$ be a conjugacy class of elements in $G$, and assume that $S$ belongs to $N$. Prove that $S$ is a union of $n$ conjugacy classes in $N$, all having the same cardinality, where $n$ equals the index $[G : NCent(x)]$ of the group generated by $N$ and the centralizer $Cent(x)$ in $G$ of and element $x$ belonging to $S$.&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-28T14:45:56.877" LastActivityDate="2010-09-28T14:51:48.860" Title="a question about group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5615" PostTypeId="2" ParentId="5588" CreationDate="2010-09-28T04:33:17.677" Score="4" ViewCount="0" Body="&lt;p&gt;Amazingly, the complete classification of noncompact 2-manifolds with boundary was not completed until 2007.  Here's the reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. O. Prishlyak and K. I. Mischenko, &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2308580&quot; rel=&quot;nofollow&quot;&gt;Classification of noncompact surfaces with&#xA;boundary&lt;/a&gt;, Methods Funct. Anal. Topology 13 (2007), no. 1, 62–66.  &lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-28T04:33:17.677" />
  <row Id="5616" PostTypeId="1" AcceptedAnswerId="5623" CreationDate="2010-09-28T05:35:49.283" Score="1" ViewCount="136" Body="&lt;p&gt;If  S1, S2, and S are the sums of n terms, 2n terms and to infinity of a G.P. Then, find the value of S1(S1-S).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:Nothing is given about the common ratio.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-09-28T22:41:46.897" Title="Geometric Progression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="2" />
  <row Id="5617" PostTypeId="1" CreationDate="2010-09-28T05:50:49.190" Score="4" ViewCount="339" Body="&lt;p&gt;Recently, I came across this problem in my textbook. Any hints on how to solve this using dynamic programming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Professor Stewart is consulting for the president of a corporation that is planning a company party. The company has a hierarchical structure; that is, the supervisor relation forms a tree rooted at the president. The personnel office has ranked each employee with a conviviality rating, which is a real number. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend. Professor Stewart is given the tree that describes the structure of the corporation, using the left-child, right-sibling representation. Each node of the tree holds, in addition to the pointers, the name of an employee and that employee’s conviviality ranking. Describe an algorithm to make up a guest list that maximizes the sum of the conviviality ratings of the guests.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2178" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-28T08:21:45.810" LastActivityDate="2010-10-28T14:44:08.577" Title="Party planning problem" Tags="&lt;algorithms&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="3" />
  <row Id="5618" PostTypeId="2" ParentId="5616" CreationDate="2010-09-28T06:05:23.293" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad\:$  In $\rm\  \ (1-X)\ (1-(1-X))\ =\ 1-X^2-(1-X)\ \ \ $ put $\rm\ \ \ X = x^n\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then multiply both sides by $\rm\  1/(1-x)^2\ =\ S/(1-x)\:.\ \ $  More generally one has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ (1-x^a)\:(1-x^b)\ =\ (1-x^a) + (1-x^b) - (1-x^{a+b})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \Rightarrow\quad\quad S_a\ S_b\ =\ S\ (S_a + S_b - S_{a+b})\:,\quad S_n = \displaystyle\frac{1-x^n}{1-x},\quad S = S_\infty = \frac{1}{1-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes to arbitrary products $\rm\: S_{a}\: S_b\: S_c\cdots S_k\:$ using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;Inclusion–exclusion principle.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-28T22:41:46.897" LastActivityDate="2010-09-28T22:41:46.897" />
  <row Id="5619" PostTypeId="2" ParentId="3364" CreationDate="2010-09-28T06:18:20.320" Score="0" ViewCount="0" Body="&lt;p&gt;These examples may be silly, but just to add to your list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Sequential compactness (i.e. every sequence has a convergent subsequence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Countable compactness (i.e. every &lt;em&gt;countable&lt;/em&gt; open cover has a finite subcover)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $\sigma$-compactness (i.e. the space is a countable union of compact sets)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less trivially, as mentioned in the Wikipedia article, the Lindelof property and separability are both preserved.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-09-28T06:23:21.680" LastActivityDate="2010-09-28T06:23:21.680" />
  <row Id="5620" PostTypeId="2" ParentId="5607" CreationDate="2010-09-28T06:48:58.417" Score="2" ViewCount="0" Body="&lt;p&gt;A very ample invertible sheaf $\mathcal L(D)$ is characterized by there being an embedding $\tilde{C} \hookrightarrow \mathbb P^n$ so that $D$ is the intersection of $\tilde{C}$&#xA;with a hyperplane $H$ in $\mathbb P^n$.  If a hyperplane intersects $\tilde{C}$ in the point&#xA;$P$ with multiplicity one, then $\tilde{C}$ has to be a degree one curve in $\mathbb P^n$, which is to say, a line.  Thus $\tilde{C}$ must have genus zero.  Thus all points on $\tilde{C}$ must be linearly equivalent, and so in this case $\mathcal L(P) = \mathcal L(P')$ for any&#xA;point $P'$ on $\tilde{C}$ (just to say the same thing a different way, we have&#xA;that $\tilde{C} = \mathbb P^1$, and that $\mathcal L(D) = \mathcal O(1)$ on this&#xA;$\mathbb P^1$).&#xA;In particular, we can take $P'$ to be a point whose image in $C$ is non-singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to phrase the argument, which avoids this precise intersection theoretic analysis, is to say that if $\mathcal L$ is very ample on $\tilde{C}$, then it comes from an embedding&#xA;$\tilde{C} \hookrightarrow \mathbb P^n$ by pulling back $\mathcal O(1)$, and so is of the&#xA;form $\mathcal L(D)$ for any intersection of $\tilde{C}$ with a hyperplane in $\mathbb P^n$.  But we can always choose this hyperplane so that its intersection with $\tilde{C}$ misses any given finite set of points.  So we can always choose this hyperplane so that &#xA;$D$ consists only of points whose images in $C$ are non-singular, which is what we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, when thinking about what very ample means, you want to relate it to the corresponding projective embedding, and then use geometric facts about hyperplane intersections.  (Primarily: that there are a lot of hyperplanes (!), so you can hope to get a lot of control over the $D := \tilde{C} \cap H$ for which your very ample invertible&#xA;sheaf equals $\mathcal L(D)$.  Although it isn't relevant here, because you are just working with curves, a typical tool that you might apply to understand the possible such $D$ is Bertini's theorem.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-09-28T08:14:21.607" LastActivityDate="2010-09-28T08:14:21.607" CommentCount="1" />
  <row Id="5621" PostTypeId="2" ParentId="3364" CreationDate="2010-09-28T07:03:30.743" Score="0" ViewCount="0" Body="&lt;p&gt;Following from muad's comment one could say that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $A \subseteq X_2$ is open, then $f^{-1}(A) \subseteq X_1$ is open &lt;br /&gt;&#xA;2) same thing for closed sets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I think are more simple conditions for characterizing a continuous function as opposed to those in your list, since the concept of open and closed sets are &quot;simpler&quot; than compact of connected sets.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-28T07:03:30.743" CommentCount="2" />
  <row Id="5622" PostTypeId="2" ParentId="5607" CreationDate="2010-09-28T07:42:50.917" Score="2" ViewCount="0" Body="&lt;p&gt;Notice that in your 2nd paragraph you say the goal is to prove that «there is an effective divisor $D=\sum P_i$ such that $\mathcal{L}(D)\simeq \mathcal{L}$ and such that each $f(P_i)$ are non-singular points», but in your 4th paragraph you are picking a non singular point and asking if there is an effective divisor with a specific support. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two things have quantifiers in different places. In particular, the claim in your 2nd paragraph does not say that you can pick the support of the divisor as you want!&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-28T07:50:47.420" LastActivityDate="2010-09-28T07:50:47.420" />
  <row Id="5623" PostTypeId="2" ParentId="5616" CreationDate="2010-09-28T08:57:42.210" Score="4" ViewCount="0" Body="&lt;p&gt;I change your notation from S1, S2 and S to $S_{n},S_{2n}$ and $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of $n$ terms of a geometric progression of ratio $r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{1},u_{2},\ldots ,u_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{n}=u_{1}\times \dfrac{1-r^{n}}{1-r}\qquad (1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the sum of $2n$ terms of the same progression is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{2n}=u_{1}\times \dfrac{1-r^{2n}}{1-r}\qquad (2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the sum $S$ exists, it is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S=\lim S_{n}=u_{1}\times \dfrac{1}{1-r}\qquad (3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &quot;answer is S(S1-S2)&quot;, we have to prove this identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{n}(S_{n}-S)=S(S_{n}-S_{2n})\qquad (4).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging $(1)$, $(2)$ and $(3)$ into $(4)$ we have to prove the following equivalent algebraic identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{1}\times \dfrac{1-r^{n}}{1-r}\left( u_{1}\times \dfrac{1-r^{n}}{1-r}%&#xA;-u_{1}\times \dfrac{1}{1-r}\right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=u_{1}\times \dfrac{1}{1-r}\left( u_{1}\times \dfrac{1-r^{n}}{1-r}-u_{1}\times \dfrac{1-r^{2n}}{1-r}\right)&#xA;\qquad (5)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, after simplifying $u_1$ and the denominator $1-r$, becomes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{1-r^{n}}{1}\left( \dfrac{1-r^{n}}{1}-\dfrac{1}{1}\right) =\left( \dfrac{%&#xA;1-r^{n}}{1}-\dfrac{1-r^{2n}}{1}\right) \qquad (6)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 1-r^{n}\right) \left( -r^{n}\right) =-r^{n}+r^{2n}\iff 0=0\qquad (7)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $(7)$ is true, $(5)$ and $(4)$ are also true.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-28T10:31:24.270" LastActivityDate="2010-09-28T10:31:24.270" CommentCount="2" />
  <row Id="5624" PostTypeId="1" AcceptedAnswerId="5629" CreationDate="2010-09-28T09:23:32.537" Score="3" ViewCount="46" Body="&lt;p&gt;for a finite group G and a trivial abelian G module A, there is the short exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \rightarrow Ext^1 (G_{ab},A) \rightarrow H^2(G,A) \rightarrow Hom (H_2 (G,Z), A) \rightarrow 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a description for the connecting maps. Specially, I want to use the representation of $H_2(G,Z)$ as $M(G)=[F,F]\cap R/ [F,R]$ where $G\cong F/R$ and F is free.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-28T12:25:33.993" Title="connecting maps in the universal coefficients theorem" Tags="&lt;cohomology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5625" PostTypeId="1" CreationDate="2010-09-28T10:00:42.893" Score="1" ViewCount="137" Body="&lt;p&gt;Greetings, dear Mathematicians!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to implement base-9 to base-3 conversion in O(log(n)) on 4-tuple ((stateFrom, encounterWhat, Command, StateTo), where Command can be &amp;lt; (turn left), &gt; (turn right), # (stop execution), or symbol from an alphabet) Turing machine, finite from the left side and infinite from the right side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your answers!&lt;/p&gt;&#xA;" OwnerUserId="2181" LastActivityDate="2010-10-15T03:58:15.300" Title="Turing machine to convert base-9 number to base-3 in O(log(n))" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5626" PostTypeId="2" ParentId="5614" CreationDate="2010-09-28T10:16:29.237" Score="0" ViewCount="0" Body="&lt;p&gt;for $x \in S$ define a function from $G/ NCent(x)$ to conjugacy classes in N by $gNCent(x) \mapsto [gxg^{-1}]$. This is well defined because changing by an element of Cent(x) does nothing to x, and changing by an element of N changes $gxg^{-1}$ to another element in the same conjugacy class of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that this is onto the conjugacy classes in S. also this is injective, because if $[hxh^{-1}]=[gxg^{-1}]$ then there is an $m\in N$ such that $(g^{-1}mh)x=x(g^{-1}mh)$, so you have that $g^{-1}h (h^{-1}mh) \in Cent(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $m_i\in N$ be the elements such that ${ m_i x m_i ^{-1} }$ are all the elements in the conjugacy class of x in N. From this we get that $(g_i m_i g_i^{-1}) (g_i x g_i^{-1}) (g_i m_i g_i^{-1})^{-1} $ are different elements in the conjugacy class of $(g_i x g_i^{-1})$ so $ | [x] | \leq | [gxg^{-1}] | $, but there is nothing special about taking x from S so we have the opposite inequality as well.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-28T10:16:29.237" />
  <row Id="5627" PostTypeId="1" AcceptedAnswerId="5628" CreationDate="2010-09-28T11:37:43.690" Score="2" ViewCount="174" Body="&lt;p&gt;Helping a buddy with his &quot;intro to math&quot; course for CompSci. I'm afraid my 'leet math skills are already giving out, so any help would be appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;f&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt; are both surjective functions, then so is the composition &lt;em&gt;f&lt;/em&gt; o &lt;em&gt;g&lt;/em&gt;. (so far so good).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect the converse to be true too: If &lt;em&gt;f&lt;/em&gt; o &lt;em&gt;g&lt;/em&gt; is a surjective function, does that mean &lt;em&gt;f&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt; are surjective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my research so far it seems that in this case &lt;em&gt;f&lt;/em&gt; is indeed surjective but &lt;em&gt;g&lt;/em&gt; is not necessarily so. Can anybody explain why, please? An example would rock!&lt;/p&gt;&#xA;" OwnerUserId="2183" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T02:30:44.593" LastActivityDate="2010-11-24T07:19:41.617" Title="Surjection on composed function?" Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;" AnswerCount="4" />
  <row Id="5628" PostTypeId="2" ParentId="5627" CreationDate="2010-09-28T12:04:17.360" Score="4" ViewCount="0" Body="&lt;p&gt;It is important to consider the codomain when talking about the `surjectivity' of a function. Let us give a real-valued example of functions for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g(x)$ be any function that maps the real number line to the interval $(-1, 1)$ (eg. sin) and let $f(x) = \frac{x}{x^2-1}$ be considered only over the interval $(-1, 1)$. Notice that $f(x)$ is continuous over this interval, and the range of f is the real line (You can graph x/(x^2-1) at wolframalpha.com to see this as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is not necessary that $g(x)$ be surjective. In fact, the real number line is isomorphic to any open interval.&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditorDisplayName="" LastEditDate="2010-09-28T12:11:42.830" LastActivityDate="2010-09-28T12:11:42.830" CommentCount="1" />
  <row Id="5629" PostTypeId="2" ParentId="5624" CreationDate="2010-09-28T12:25:33.993" Score="3" ViewCount="0" Body="&lt;p&gt;By «connecting maps» you mean the maps that appear in the short exact sequence you wrote?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, you will find the details of the construction of the sequence in Hilton-Stammbach's book &lt;em&gt;A Course on homological algebra&lt;/em&gt;, GTM4. The result is theorem VI.15.1, but the meat of the construction is in section §V.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the expressions in terms of the isomorphisms between $H_2(G,\mathbb Z)$ and $M(G)$, well, then you have to write down both and compose them :) I do not have time now to do the computation, but you should &lt;em&gt;really&lt;/em&gt; do it yourself!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-09-28T12:25:33.993" />
  <row Id="5630" PostTypeId="2" ParentId="5627" CreationDate="2010-09-28T12:28:14.370" Score="1" ViewCount="0" Body="&lt;p&gt;Consider $f: X \rightarrow Y$, $g: Y \rightarrow Z$, then $g \circ f: X \rightarrow Z$. If it is surjective, it means that for any $z \in Z$ there exists $x \in X$ such that $(g \circ f)(x) = g(f(x)) = z$; thus, if $y = f(x)$, then $g(y) = z$, which shows that $g$ is surjective&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-09-28T12:28:14.370" />
  <row Id="5631" PostTypeId="2" ParentId="5564" CreationDate="2010-09-28T12:34:21.673" Score="0" ViewCount="0" Body="&lt;p&gt;If you assume natural deduction, this is easy to explain. Let $\Sigma$ be an inconsistent set of axioms. This means you can prove both $\phi$ and $\lnot\phi$ for some statement $\phi$: in symbols $\Sigma\vdash\phi$ and $\Sigma\vdash\lnot\phi$. Now you suppose the negation of whatever statement $\psi$ you like to prove, but since, by monotonicity, $\Sigma\cup\{\lnot\psi\}\vdash\phi$ and $\Sigma\cup\{\lnot\psi\}\vdash\lnot\phi$, you can apply negation elimination obtaining $\Sigma\vdash\psi$.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-09-28T12:34:21.673" />
  <row Id="5632" PostTypeId="1" AcceptedAnswerId="5633" CreationDate="2010-09-28T12:54:57.133" Score="0" ViewCount="137" Body="&lt;p&gt;I think this is really weird. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to solve the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2x^2+3x+5)^3+(2x^2+3x+5)^2=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a polynomial equation of 6 degrees. It should have 6 roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;wolframalpha&lt;/a&gt; only returns 4 roots! Anything I miss?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the command I use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;solve ((2x^2+3x+5)^3)+((2x^2+3x+5)^2)=0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Ah, Wolframalpha doesn't return repeated roots. So now case solved.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-09-28T13:06:09.000" LastActivityDate="2010-09-28T13:15:43.557" Title="WolframAlpha Returns only 2 Roots for a Polynomial Equation of 6 Degree" Tags="&lt;wolframalpha&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="5633" PostTypeId="2" ParentId="5632" CreationDate="2010-09-28T13:15:43.557" Score="3" ViewCount="0" Body="&lt;p&gt;Trying&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Solve[(2x^2 + 3x + 5)^3 + (2x^2 + 3x + 5)^2 == 0, x]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;on WolframAlpha returns four roots; the truth of the matter is that the equation has two roots of multiplicity two, so that you have six roots in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure why the multiplicities aren't indicated...&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-28T13:15:43.557" />
  <row Id="5634" PostTypeId="1" CreationDate="2010-09-28T13:24:34.177" Score="2" ViewCount="95" Body="&lt;p&gt;I'm trying to convey to some folks some logical and bitwise operators such as AND, OR, XOR, &amp;amp;c in programming languages such as C &amp;amp;| Java.  Though these I guess could be considered &quot;boolean&quot; operators there is no &quot;Boolean Algebra&quot; in here, just the results of various operation, yes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never heard of Boolean &lt;strong&gt;Arithmetic&lt;/strong&gt;, but would one consider these simple questions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;5 AND 8 = ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;boolean arithmetic (operating bitwise) or rather &lt;em&gt;logical&lt;/em&gt; or &lt;em&gt;bitwise&lt;/em&gt; arithmetic?&lt;/p&gt;&#xA;" OwnerUserId="2184" LastActivityDate="2010-09-28T15:02:02.490" Title="Arithmetic?  Logical/Boolean/Bitwise operations?" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5635" PostTypeId="2" ParentId="5634" CreationDate="2010-09-28T13:40:49.277" Score="4" ViewCount="0" Body="&lt;p&gt;The terms &lt;em&gt;boolean&lt;/em&gt;, &lt;em&gt;binary&lt;/em&gt;, and &lt;em&gt;bitwise&lt;/em&gt; arithmetic are all equivalent in my mind. However, it is most commonly referred to as &lt;em&gt;binary arithmetic&lt;/em&gt; in my experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Boolean algebra&lt;/em&gt;&lt;/a&gt; however is a more general field (encompassing more than just arithmetic), and is really synonymous for &lt;em&gt;classical propositional calculus&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-09-28T13:40:49.277" />
  <row Id="5636" PostTypeId="2" ParentId="5627" CreationDate="2010-09-28T13:49:36.963" Score="3" ViewCount="0" Body="&lt;p&gt;The example to keep in mind is the following: take $X=\{x\}$ be a one element set, $Y=\{1,2\}$, a two elements set; and $Z=\{a\}$ a one-element set. Let $f\colon X\to Y$ be given by $f(x)=1$, and $g\colon Y\to Z$ by $g(1)=g(2)=a$. Then $g\circ f\colon X\to Y$ is bijective (one-to-one and onto), $f$ is one-to-one but not onto, and $g$ is onto but not one-to-one. So even if you know &lt;em&gt;more&lt;/em&gt; about the composition, still the best you can conclude is that the second function is onto, but not necessarily the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: if a composition $g\circ f$ is one-to-one, the best you can say is that $f$ is one-to-one; if $g\circ f$ is onto, then the best you can say is that $g$ is onto. Intuitively, in the one-to-one case, the second function can be ill-behaved outside the image of $f$, but the composition &quot;will never know&quot;. And in the onto-case, $g$ can have a lot of redundancy, so that $f$ does not have to get to &lt;em&gt;every&lt;/em&gt; point in the domain of $g$, just to &quot;enough&quot; of them.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-09-28T13:49:36.963" />
  <row Id="5637" PostTypeId="2" ParentId="5617" CreationDate="2010-09-28T14:02:42.817" Score="3" ViewCount="0" Body="&lt;p&gt;I'll give you a hint without referring to the representation of the graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a tree structure of a corporation, you can either have the root (president) attend the party or not. Say the root has n children. If the root is not in the party, you can have a party which is the &quot;sum&quot; of n different parties, each representing the subtree at each of the n children (this is the recursive step). If the root is in the party, then you again have the sum of n different parties, but this time each &quot;subparty&quot; &lt;em&gt;cannot&lt;/em&gt; have its root included (this is another recursive step).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can reformat this recursion into a dynamic programming solution, or solve it recursively with memoization.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-09-28T14:02:42.817" CommentCount="0" />
  <row Id="5638" PostTypeId="2" ParentId="5614" CreationDate="2010-09-28T14:35:18.560" Score="3" ViewCount="0" Body="&lt;p&gt;The fact that $S$ is a union of conjugacy classes of $N$ should be immediate; also, the sizes are the same: let $x\in S$, and let $U$ be the $N$-conjugacy class of $x$. If $T$ is any other $N$-conjugacy class contained in $S$, then let $y\in T$. There exists $g\in G$ such that $gxg^{-1}=y$. Then $|T|=|gUg^{-1}|=|U|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real meat of the problem is showing that the &lt;em&gt;number&lt;/em&gt; of conjugacy classes is the index of $NC_G(x)$ in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you see a problem in which you need to show that the number of something or other in a group equals the index of a subgroup $H$ in $G$, this should suggest finding a map from $G$ to the set you want, and then showing that the images of two elements coincide if and only if they lie in the same coset of $H$ in $G$. This will then give you that the number of distinct images is the index of $H$ in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here what you want to do is find a map from $G$ to the conjugacy classes of $N$ contained in $S$, and arrange matters so that two elements of $G$ map to the same thing if and only if they lie in the same coset of $NC_G(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... how to map it? The obvious thing to do is to take a conjugacy class of $N$ contains in $S$, represented by some $x\in S$, and map it to the conjugacy class of $gxg^{-1}$ (which lies in $N$ because $N$ is assumed to be normal). If $x\sim_N y$ (that is, $x$ is conjugate to $y$ in $N$) then $y=nxn^{-1}$, so $gyg^{-1} = gnxn^{-1}g^{-1} = (gng^{-1})x(gng^{-1})$, and since $gng^{-1}\in N$, then $gyg^{-1}\sim_N gxg^{-1}$. So this map is well-defined on $N$-conjugacy classes. Since $S$ is a conjugacy class, $gxg^{-1}\in S$ as well, so the image of the $N$-conjugacy class of $x$ is also an $N$-conjugacy class contained in $S$. That is: conjugation maps $N$-conjugacy classes contained in $S$ to $N$-conjugacy classes contained in $S$. And note that this map is onto: if you pick &lt;em&gt;any&lt;/em&gt; $x\in S$, then every element of $S$ is of the form $gxg^{-1}$ for some $g\in G$, so you can pick your favorite $x\in S$, and use it as a basis: you will &quot;hit&quot; every $N$-conjugacy class that makes up $S$ this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when will $g$ and $h$ do the exact same thing to every $N$-conjugacy class contained in $N$? Suppose that $gxg^{-1}\sim_N hxh^{-1}$ for some $x\in S$. Then there exists $n\in N$ such that $ngxg^{-1}n^{-1} = hxh^{-1}$, hence $h^{-1}ngxg^{-1}n^{-1}h = x$; that is, there exists $n\in N$ such that $h^{-1}ng\in C_G(x)$ (we're getting somewhere... we've gotten to the centralizer of $x$ in $G$). This means that $ng C_G(x)=hC_G(x)$. Not quite what we want: that $n$ is  &quot;on the wrong side&quot; so to speak. Ah, but $N$ is normal. So $ng = (gg^{-1})ng = g(g^{-1}ng) = gn'$ for some $n'\in N$, so we have $gn'C_G(x)=hC_G(x)$. Thus, of $gxg^{-1}\sim_N hxh^{-1}$, then there exists $n'\in N$ such that $gn'C_G(x)=hC_G(x)$. And this is equivalent to $gNC_G(x) = hNC_g(x)$. (Almost there!) Conversely, suppose that $gNC_G(x) = hNC_g(x)$; then there exists $n\in N$ and $c\in C_G(x)$ such that $h = gnc$. Then $hxh^{-1} = (gnc)x(gnc)^{-1} = gn(cxc^{-1})n^{-1}g^{-1} = gnxn^{-1}g^{-1}$ (since $cx=xc$); since $n$ is normal, $gn = gn(g^{-1}g) = (gng^{-1})g = n'g$ for some $n'\in N$, so $hxh^{-1} = gnxn^{-1}g^{-1} = n'gxg^{-1}n'^{-1}$, hence $hxh^{-1}\sim_N gxg^{-1}$. That is: $g$ and $h$ map the $N$-conjugacy class of $x$ to the same $N$-conjugacy class if and only if $g$ and $h$ are in the same left coset of $NC_G(x)$ in $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since every element of $S$ is $G$-conjugate to $x$, this is enough: we can get to &lt;em&gt;any&lt;/em&gt; $N$-conjugacy class contained in $N$ by conjugating $x$ by some element of $G$. And we have just seen that two elements of $G$ have the same image if and only if they are in the same left cosets of $NC_G(x)$. So the number of distinct images, that is, the number of $N$-conjugacy classes that make up $S$ (and here we can use &quot;number&quot; to mean cardinality, even in the infinte case!), is exactly the same as the number of left cosets of $NC_G(x)$ in $G$. That is, the number of conjugacy classes is $[G:NC_G(x)]$, as claimed. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-28T14:51:48.860" LastActivityDate="2010-09-28T14:51:48.860" />
  <row Id="5639" PostTypeId="1" AcceptedAnswerId="5642" CreationDate="2010-09-28T14:47:08.623" Score="4" ViewCount="314" Body="&lt;p&gt;A simple question here, which likely demands a somewhat complex answer... Or rather, a set of related questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the advantages of typed lambda calculus over untyped lambda calculus in terms of proof theory?&lt;/li&gt;&#xA;&lt;li&gt;Specifically, Church's original Lambda Calculus was untyped and allows arbitrarily high-order functions. What are the limitations with respect to constructing a proof calculus from it?&lt;/li&gt;&#xA;&lt;li&gt;Are not untyped and typed lambda calculi inherently higher-order formal systems?&lt;/li&gt;&#xA;&lt;li&gt;What are the reasons for using complex (e.g. polymorphic/dependent) type theories over the simple type theory in lambda calculus? Are they more 'powerful' in some sense; if so, how exactly?&lt;/li&gt;&#xA;&lt;li&gt;Do semantics (interpretation) have anything to say here, with respect to typed and untyped theories, especially in terms of soundness and completeness?&lt;/li&gt;&#xA;&lt;li&gt;The well-known proof verifier &lt;em&gt;Coq&lt;/em&gt;, which (I believe) uses a language of higher-order complex-typed lambda calculus to represent proofs in constructive (intuistionistic) mathematics. I have read that the theory behind it (the &lt;em&gt;Calculus of Constructions&lt;/em&gt;) is essentially an extension of the Curry-Howard isomorphism to higher-order logic. Are there any elaborations/clarifications I should be aware of here?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-09-28T15:52:16.553" Title="The Power of Lambda Calculi" Tags="&lt;logic&gt;&lt;proof-theory&gt;&lt;lambda-calculus&gt;&lt;formal-systems&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="5640" PostTypeId="2" ParentId="5634" CreationDate="2010-09-28T15:02:02.490" Score="6" ViewCount="0" Body="&lt;p&gt;But there is boolean algebra here. Namely, you can consider an $\rm n$-bit integer as a characteristic function for a subset of $\rm\: n = \{0,1,\ldots,n-1\}$, viz. bit $\rm i$ of $\rm k$ is $1$ iff $\rm i$ lies in the subset represented by $\rm k$. The bitwise boolean operations suffice to define all of the boolean algebra operations in this powerset boolean algebra.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-28T15:02:02.490" />
  <row Id="5641" PostTypeId="1" AcceptedAnswerId="5643" CreationDate="2010-09-28T15:38:11.287" Score="8" ViewCount="273" Body="&lt;p&gt;I'm going to ask a very silly question, so I'm begging you to be understanding if it is absolutely trivial, or if it's an exercise in some Bourbaki. I'm afraid of asking you, because the question entails this particular case: is the addition of (a finite, but variable quantity of) real numbers a continuous function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, let me try to explain what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm insisting in doing exercices about topological groups, see if I can understand something about those previous &lt;a href=&quot;http://math.stackexchange.com/questions/2865/is-a-direct-limit-of-topological-groups-always-a-topological-group&quot;&gt;question 1&lt;/a&gt;, &lt;a href=&quot;http://math.stackexchange.com/questions/5095/what-are-the-product-and-coproduct-in-the-category-of-topological-groups&quot;&gt;question 2&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/5297/colimit-of-topological-groups-again&quot;&gt;question 3&lt;/a&gt;. This one should be more elementary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a topological Abelian group, written additively, and let $I$ be an arbitrary set. I don't mind if you take $G = \mathbb{R}$, with the usual Euclidian topology and the usual addition of real numbers, but you &lt;strong&gt;cannot&lt;/strong&gt; assume $I$ is finite, nor countable. Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G^I = \prod G = \prod_\alpha G_\alpha , \qquad \text{where}\ G_\alpha = G \qquad \text{for all}\ \alpha \in I \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote  $(x_\alpha)$ the elements of $\prod G$. This is an Abelian topological group with the usual &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_topology&quot; rel=&quot;nofollow&quot;&gt;product topology&lt;/a&gt; and operations defined component-wise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;(x_\alpha) + (y_\alpha) &amp;amp;= (x_\alpha + y_\alpha)  \\&#xA;-(x_\alpha)              &amp;amp;= (-x_\alpha)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also the &lt;em&gt;weak product&lt;/em&gt; $\prod' G$ (aka, &lt;em&gt;direct sum&lt;/em&gt;). That is, elements of $\prod' G$ are those tuples $(x_\alpha) \in \prod  G$ such that $x_\alpha = 0$, except for a finite number of indexes $\alpha \in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\prod'G $ is of course a subgroup of $\prod G$, $\prod' G \subset \prod G$. Let's consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Subspace_topology&quot; rel=&quot;nofollow&quot;&gt;subspace topology&lt;/a&gt; on $\prod' G$, induced from the product topology on $\prod G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have a well-defined addition map that we hadn't in $\prod G$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum : \prod ' G \longrightarrow G \ , \qquad  (x_\alpha ) \mapsto \sum_\alpha x_\alpha \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense, despite the fact $I$ is not necessarily an ordered set because, given an element $(x_\alpha) \in \prod'G$, if $x_{\alpha_1}, \dots , x_{\alpha_n}$ are its non-zero components, we don't have a canonical choice in order to perform the addition $x_{\alpha_1} + \cdots + x_{\alpha_n}$. Nevertheless, since the sum is associative and commutative, we may do it as we please: the result will always be the same. Hence $\sum_\alpha x_\alpha$ is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; Is this map $\sum$ continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you don't like too much abstraction in your life, but prefer to be very specific: is the sum of real numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum: \prod' \mathbb{R} \longrightarrow \mathbb{R} \ , \qquad (x_\alpha) \mapsto \sum_\alpha x_\alpha&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;continuous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark 1.&lt;/strong&gt; The question seems silly, doesn't it? Well, maybe it is, but the first answer that came to my mind is wrong, as far as I can see: you cannot say &quot;this $\sum$ is continuous because it is a composition of continuous maps&quot;; namely, the iteration of the addition of $G$. Yeah, but: &lt;em&gt;which&lt;/em&gt; composition? Notice that there is no canonical choice for doing the addition in an specific order -there could be no order at all in $I$. So every time you compute $\sum_\alpha x_\alpha$ you can change the order in which you perform the (continuous) sums $x_\alpha + x_\beta$. So, $\sum$ is &lt;em&gt;not&lt;/em&gt; a composition of a finite number of specific maps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit. Previous Remark 2 was wrong.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-09-30T07:30:10.030" LastActivityDate="2010-09-30T07:30:10.030" Title="Is addition continuous?" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="5642" PostTypeId="2" ParentId="5639" CreationDate="2010-09-28T15:52:16.553" Score="3" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;What are the advantages of typed lambda calculus over untyped lambda calculus in terms of proof theory? &lt;em&gt;and also&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Specifically, Church's original Lambda Calculus was untyped and allows arbitrarily high-order functions. What are the limitations with respect to constructing a proof calculus from it?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simply, typed lambda calculus has a proof theory, and untyped lambda calculus doesn't because it lacks a normal-form theorem.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Are not untyped and typed lambda calculi inherently higher-order formal systems?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is tricky, because there are two ways of looking at higher-orderness.  They are higher-order when you look at the terms, since the definition of higher order is abstraction over higher-order entities like functions, and lambda abstraction is abstraction.  But note, under the formulae-as-types correspondence, the proposaitions in logic are associated with the types of the lambda terms, and in, say, the simply-typed lambda calculus, there is no abstraction over anything.  This point confused me when I first studied Church's simple theory of types, because it is a higher-order calculus based on simply-typed lambda calculus, where the propositions are formed using the lambda terms, rather than, as with the formulae-as-types corespondence, being the types.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;What are the reasons for using complex (e.g. polymorphic/dependent) type theories over the simple type theory in lambda calculus? Are they more 'powerful' in some sense; if so, how exactly?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They do add power.  Under the formulae-as-types correspondence, the simply-typed lambda calculus has as its matching logic propositional calculus, adding dependent types adds universal and existential quantification at higher finite types to its logic.  Polymorphic types allow types allowing all usual mathematical entities to be constructed (although without the usual theory of provability) and has high proof-theoretic consistency: Strong normalisation of system F has (over a base theory such as RCL0) the same strength as consistency second order arithmetic, even though it lacks an induction principle.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Do semantics (interpretation) have anything to say here, with respect to typed and untyped theories, especially in terms of soundness and completeness?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They are much less useful with these sorts of theories than model theory is with classical logic, although there are good applications of category theory to the semantics of type theory.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The well-known proof verifier Coq, which (I believe) uses a language of higher-order complex-typed lambda calculus to represent proofs in constructive (intuistionistic) mathematics. I have read that the theory behind it (the Calculus of Constructions) is essentially an extension of the Curry-Howard isomorphism to higher-order logic. Are there any elaborations/clarifications I should be aware of here?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yes, that polymorphic types are tricky, and non-conservative over the base theory.  I'd recommend sarting an exploration of formulae-as-types with Martin-Löf's dependent type theory.  If you want to work with a proof assistant, there is &lt;a href=&quot;http://wiki.portal.chalmers.se/agda/pmwiki.php&quot; rel=&quot;nofollow&quot;&gt;Agda&lt;/a&gt;, which is a functional programming language whose type system is Martin-Löf's type theory.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-28T15:52:16.553" CommentCount="9" />
  <row Id="5643" PostTypeId="2" ParentId="5641" CreationDate="2010-09-28T16:47:07.040" Score="6" ViewCount="0" Body="&lt;p&gt;In the case where $I$ is infinite and $G$ has a non trivial topology, then the answer is no.&#xA;let $(x_\alpha) \in \prod 'G_\alpha$ and let $U$ be an open set containing $( x_\alpha)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;U comes from an open set in the regular product topology, so we can assume that it is $\prod U_\alpha \cap \prod'G_\alpha$ where only for finite number $U_\alpha \neq G$. the assumption that $I$ is infinite gives at least one $U_\beta=G$ so for all $(y_\alpha)$ such that $y_\alpha = 0 \forall \alpha\neq \beta$  we have $(x_\alpha)+(y_\alpha) \in U$ so $\sum (U) = G$, because we can choose $y_\beta$ to be any element of G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if $V\neq G$ is an open neighborhood of $\sum (x_\alpha)$ then there is no open neighborhood U of $(x_\alpha)$ such that $\sum (U) \subseteq V$. &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-28T16:47:07.040" CommentCount="1" />
  <row Id="5644" PostTypeId="2" ParentId="2837" CreationDate="2010-09-28T17:01:14.923" Score="0" ViewCount="0" Body="&lt;p&gt;This dude have some nice info about this topic!&#xA;Click his name to see more math stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/geometry/sphereline/&quot; rel=&quot;nofollow&quot;&gt;http://local.wasp.uwa.edu.au/~pbourke/geometry/sphereline/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2189" LastActivityDate="2010-09-28T17:01:14.923" />
  <row Id="5645" PostTypeId="1" AcceptedAnswerId="7905" CreationDate="2010-09-28T17:05:24.147" Score="4" ViewCount="217" Body="&lt;p&gt;I want to show that there is a finite conjunction $\phi$ of axioms of $ZF$, such that every transitive proper class $M$, which satisfies $\phi$, is already a model of $ZF$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an exercise in Kunen's set theory. There is a hint, it seems to be useful to apply the Reflection principle to the union $M = \cup_{\alpha} M \cap R(\alpha)$. But I don't know with which axioms we can do that (we can only use finitely many!), and why this yields an ordinal which is independent from $M$. Please give me only a hint, because basically I want to solve this on my own, but I don't know how to start with the hint above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what is the &quot;philosophical&quot; reason that we cannot deduce from this, that $ZF$ is finitely axiomatizable (which is wrong)? I mean I cannot prove that this $\phi$ above proves every axiom, but is there also a deeper reason for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There was an answer with some hints, but it was deleted...&#xA;I still don't know how to produce this strange sentence $\phi$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-10-26T11:52:22.077" LastActivityDate="2010-10-26T12:26:52.290" Title="ZF is almost finitely axiomatizable" Tags="&lt;set-theory&gt;&lt;logic&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5646" PostTypeId="1" AcceptedAnswerId="5654" CreationDate="2010-09-28T19:48:05.417" Score="3" ViewCount="222" Body="&lt;p&gt;I've been trying to understand how  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${x^3-12x+9}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;factors to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-3) (x^2+3 x-3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What factoring rule does this follow? The net result seems to be similar to what is attained through the sum/difference of cubes factoring pattern, but the signs are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, what type of problem is this, so I can make better and more relevant searches for help on future questions. Is it a cubic trinomial?&lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2010-09-28T23:27:48.137" Title="Factoring a Cubic Polynomial" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="5647" PostTypeId="1" AcceptedAnswerId="5651" CreationDate="2010-09-28T19:53:21.637" Score="2" ViewCount="101" Body="&lt;p&gt;How would you find the image of ${ x+yi : -\pi &amp;lt; x &amp;lt; \pi, y &gt; 0 }$ under $\sin(z)$?  I see that $\text{Re } \sin(x+yi) = \sin x \cosh y$ and $\text{Im } \sin(x+yi) = \cos x \sinh y$, but I'm not sure what to do next since $\sinh$ and $\cosh$ are unbounded.&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2010-09-28T20:24:30.960" Title="Image of sin(z) over a complex set" Tags="&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="5648" PostTypeId="1" AcceptedAnswerId="5722" CreationDate="2010-09-28T19:57:36.457" Score="6" ViewCount="165" Body="&lt;p&gt;If $h(x,t)$ is measurable and the measures involved are $\sigma$-finite, does there exist a sequences of functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_n(x,t) = \sum_{j=1}^{N_n} f_{j,n}(x) 1_{F_{j,n}}(t)$$ &#xA;where the sets are pairwise disjoint (in $j$) such that $h_n \to h$ pointwise almost everywhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if we fix $x$ we can get a sequence like that but we cannot just make the coefficients depending on $x$ since the sets $F_{j,n}$ will be different. Can I somehow combine them? Do I need the $\sigma$-finiteness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is positive, we can use this to prove Minkowski's inequality for integrals quite easily.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-09-29T15:28:12.087" LastActivityDate="2010-09-29T18:17:02.123" Title="Sequence of measurable functions." Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="5649" PostTypeId="2" ParentId="5646" CreationDate="2010-09-28T20:00:03.967" Score="2" ViewCount="0" Body="&lt;p&gt;One way to see that is that $x=3$ is a root of $x^3 - 12x  + 9$. So $(x-3)$ will be a factor (by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Factor_theorem&quot; rel=&quot;nofollow&quot;&gt;Factor Theorem&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can try to get $x-3$ somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way you can do that is to rewrite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^3-12x + 9 = x^3 - (3^3 - 3^3) - 12(x - 3+3)  + 9$$&#xA;$$ = x^3 - 27 + 27 - 12(x - 3) - 36 + 9 = x^3 - 27 - 12(x-3) + (9 -36 + 27)$$&#xA;$$ = (x-3)(x^2+3x+9) - 12(x-3) = (x-3)(x^2+3x-3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we used the fact that $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $r$ is a root of $f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0}$, then $f(x) - f(r) = f(x)$ gives us a way to factorize $f(x)$ as $(x-r)g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0} - (a_{n}r^{n} + a_{n-1}r^{n-1} + \dots +a_{0})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = a_{n}(x^n - r^n) + a_{n-1}(x^{n-1} - r^{n-1}) + \dots + a_{1}(x-r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like $x^3 - a^2 = (x-a)(x^2 + ax + a^2)$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^n - r^n = (x-r)(x^{n-1} + rx^{n-1} + \dots + r^{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = (x-r) (a_{n}(x^{n-1} + rx^{n-2} + \dots + r^{n-1}) + a_{n-1}(x^{n-2} + \dots +r^{n-1}) + \dots + a_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we know a root, we can also try using &lt;a href=&quot;http://www.sosmath.com/algebra/factor/fac01/fac01.html&quot; rel=&quot;nofollow&quot;&gt;Polynomial Long Division&lt;/a&gt; to get the other factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cubics, the roots can be found without the need to guess. Check this out: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method&quot; rel=&quot;nofollow&quot;&gt;Cardano's Method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T22:17:37.967" LastActivityDate="2010-09-28T22:17:37.967" CommentCount="4" />
  <row Id="5650" PostTypeId="1" AcceptedAnswerId="5667" CreationDate="2010-09-28T20:14:32.190" Score="2" ViewCount="156" Body="&lt;p&gt;Right, this is an exercise in Apostol, which i am not being able to solve. I was able to prove this result for a small case, that is the case when $n=2$, $[x] + \Bigl[x + \frac{1}{2}\Bigr]=[2x]$, but i am struggling with the generalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$\sum\limits_{k=0}^{n-1} \Biggl[x + \frac{k}{n}\Biggr] = [nx]$$, where $[ \ ]$ denotes the greatest integer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be proved via induction, i ask this because i have shown it to bealready true for $n=2$. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-29T21:12:36.650" Title="Proving $\sum\limits_{k=0}^{n-1} \Bigl[x + \frac{k}{n}\Bigr] = [nx]$" Tags="&lt;functions&gt;" AnswerCount="2" />
  <row Id="5651" PostTypeId="2" ParentId="5647" CreationDate="2010-09-28T20:17:25.490" Score="3" ViewCount="0" Body="&lt;p&gt;What shape do you get when you fix y and consider all values of x? Then see how the shape depends on y. Of course, you could try the reverse as well (fix x and vary y first), but my way, you don't have to deal with the &quot;y &gt; 0&quot; condition until the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, draw a picture showing the images of the lines x=constant and y=constant. You get two families of curves that are orthogonal to one another (since sin(z) is holomorphic and thus generally preserves angles). I promise that this will be very enlightening. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-28T20:17:25.490" />
  <row Id="5652" PostTypeId="2" ParentId="5650" CreationDate="2010-09-28T20:24:09.220" Score="5" ViewCount="0" Body="&lt;p&gt;It is enough to prove it for $0 &amp;lt; x &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $M$ be an integer such that, $\frac{M}{n} \le x &amp;lt; \frac{M+1}{n}$ where $0 \le M &amp;lt; n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $[nx] = M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $0 \le k \le n-M-1$, we have that $[x+\frac{k}{n}] = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n-1 \ge k &gt; n-M-1$ we have that $[x + \frac{k}{n}] = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think induction can be used here.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T20:31:00.237" LastActivityDate="2010-09-28T20:31:00.237" />
  <row Id="5653" PostTypeId="2" ParentId="5647" CreationDate="2010-09-28T20:24:30.960" Score="2" ViewCount="0" Body="&lt;p&gt;The image is the whole complex plane minus the interval $[-1,1]$ (on the real axis) and the negative imaginary axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mai.liu.se/~halun/complex/domain_coloring-unicode.html#fig%3asin&quot; rel=&quot;nofollow&quot;&gt;Here is a figure&lt;/a&gt; illustrating the mapping. (The color-coding is explained at the top of that page.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-28T20:24:30.960" CommentCount="2" />
  <row Id="5654" PostTypeId="2" ParentId="5646" CreationDate="2010-09-28T20:29:57.350" Score="4" ViewCount="0" Body="&lt;p&gt;By the &lt;a href=&quot;http://mathworld.wolfram.com/RationalZeroTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Rational Zero Theorem&lt;/a&gt; all the rational roots of $x^{3}-12x+9$ must have a numerator which is a factor of $9$ and a denominator which is a factor of $1$. Therefore they have to be of the form $\frac{9}{1}=9$ or $\frac{3}{1}=3$. Let $f(x)=x^{3}-12x+9$. Since $f(9)=630$ and $f(3)=0$, $3$ is a root of $f(x)$. So it can be factored as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{3}-12x+9=(x-3)\left( ax^{2}+bx+c\right) =ax^{3}+\left( b-3a\right)&#xA;x^{2}+\left( c-3b\right) x-3c$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing coefficients we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=1,b-3=0\iff b=3,-3c=9\iff c=-3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{3}-12x+9=(x-3)\left( x^{2}+3x-3\right) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.  Or we could apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Ruffini%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Ruffini's rule&lt;/a&gt; to find the coefficients of $ax^{2}+bx+c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS. As commented by user1827 &quot;The rational zero theorem&quot; also permits &quot;$1/1=1$ and $-9,-3,-1$. But they are not zeros.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, since $f(3)=0$ we can factor $f(x)$ rightaway, without taking into consideration all the remaining possibilities.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-28T23:27:48.137" LastActivityDate="2010-09-28T23:27:48.137" CommentCount="4" />
  <row Id="5655" PostTypeId="2" ParentId="5646" CreationDate="2010-09-28T20:38:04.710" Score="3" ViewCount="0" Body="&lt;p&gt;It's not a special case of any general factorization rule (except perhaps the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt;).&lt;br&gt;&#xA;However, one easy ad-hoc way to derive the factorization is to notice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\  f(x) + 3\:(x-3)\ =\ x\: (x^2 - 9)  $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ \ f(x)\ =\ (x-3)\ (x\: (x+3) - 3)$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-28T20:38:04.710" />
  <row Id="5656" PostTypeId="1" AcceptedAnswerId="5674" CreationDate="2010-09-28T20:46:44.590" Score="1" ViewCount="95" Body="&lt;p&gt;Looking at a 2&lt;sup&gt;nd&lt;/sup&gt;-order Taylor series approximation of the function $f$, I have this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t_1) = f(t_0) + hf'(t_0) + {h^2\over 2}f''(t_0) + O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say I approximate $f''(t0)$ with a 2&lt;sup&gt;nd&lt;/sup&gt;-order central difference method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f''(t) = {1\over 2}{f'(t+h) - f'(t-h)\over h} + O(h^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the resulting error of this method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive approach would be to substitute the central difference equation into the Taylor series, giving something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t_1) = f(t_0) + hf'(t_0) + {h\over 4}(f'(t_0+h)-f'(t_0-h)) + {1\over 2}O(h^4) + O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that plausible? Would error actually &lt;em&gt;decrease&lt;/em&gt; (go from 2&lt;sup&gt;nd&lt;/sup&gt;-order to 4&lt;sup&gt;th&lt;/sup&gt;-order)?&lt;/p&gt;&#xA;" OwnerUserId="2192" LastEditorUserId="2192" LastEditorDisplayName="" LastEditDate="2010-09-28T21:01:43.040" LastActivityDate="2010-09-29T15:35:36.130" Title="How do I combine &quot;error of order&quot; terms in numerical analysis?" Tags="&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5657" PostTypeId="1" AcceptedAnswerId="5673" CreationDate="2010-09-28T20:54:48.890" Score="3" ViewCount="127" Body="&lt;p&gt;I made a picture of the problem here:&#xA;&lt;img src=&quot;http://i.imgur.com/phRsj.png&quot; alt=&quot;(statement of the problem, rendered as an image)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the link does not work, read this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ be a non-zero vector in $\mathbb{R}^n$. Let S be the set of all orthogonal vectors to $a$ in $\mathbb{R}^n$. I.e., for all $x \in \mathbb{R}^n$, $a\cdot x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that the interior of S is empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I show that for every point in S, all &quot;close&quot; points are either in the complement of S or in S itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I attempted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u\in B(r,x) = \{ v \in \mathbb{R}^n : |v - x| &amp;lt; r \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $|u - x| &amp;lt; r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $|a||u - x| &amp;lt; |a|r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Cauchy-Schwarz, $|a\cdot(u-x)| \leq |a||u - x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $|a\cdot u - a\cdot x| &amp;lt; |a|r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u\in S$, then either $u\in S^{\text{int}}$ or $u\in \delta S$.  ($\delta$ denotes boundary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u\in S^{\text{int}}$, then $B(r,x) \subset S$, and $a\cdot u = 0$. But then the inequality becomes $|a|r &gt; 0$ which implies $B(r,x) \subset S \forall r &gt; 0$, but this is impossible since it would also imply that $S = \mathbb{R}^n$ and $S^c$ is empty, which is false. Therefore, if $u\in S$, then $u\in\delta S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\forall u\in B(r, x)$ such that $u\in S$, $u\in\delta S$. Thus $S^{\text{int}}$ is empty.&lt;/p&gt;&#xA;" OwnerUserId="2193" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-28T23:28:03.907" LastActivityDate="2010-09-29T04:13:52.067" Title="Prove that the interior of the set of all orthogonal vectors to &quot;a&quot; is empty." Tags="&lt;topology&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="5659" PostTypeId="2" ParentId="5191" CreationDate="2010-09-28T21:24:13.810" Score="2" ViewCount="0" Body="&lt;p&gt;You can do with something very slightly weaker than an ordering: an identification of each $n$-element set with a single &quot;universal&quot; (unordered) $n$-set.  This data canonically identifies any two $n$-element sets and thus associates a permutation to any bijection of such sets.  Even bijections can then be defined in terms of the cycle structure of the permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not much of an improvement, but you are in effect asking for a lifting of the alternating group to a groupoid of &lt;em&gt;maps&lt;/em&gt; between finite sets.  It is hard to see how to determine whether a bijection of finite sets is even (reducing to the usual notion when the sets are the same, and also &quot;transporting structure&quot; along the whole category) without having a coordinatization of the sets.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-28T22:21:54.793" LastActivityDate="2010-09-28T22:21:54.793" />
  <row Id="5660" PostTypeId="2" ParentId="5657" CreationDate="2010-09-28T21:29:46.407" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the function $\phi:x\in\mathbb R^n\mapsto x\cdot a\in\mathbb R$, which is clearly linear. If $x_0$ is a point in the interior of your set, compute the derivatives of $\phi$ at $x_0$. Now look at the Taylor development of $\phi$ at $x_0$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more geometric way: suppose that $x$ is orthogonal to $a$ (and that $a\neq0$) and consider the sequence $(x_n)_{n\geq1}$ with $x_n=x+\tfrac1na$ for each $n$. Show that $x_n\to x$ when $n\to\infty$ and then see why this is useful for you.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-09-28T21:54:57.497" LastActivityDate="2010-09-28T21:54:57.497" CommentCount="0" />
  <row Id="5661" PostTypeId="1" AcceptedAnswerId="5663" CreationDate="2010-09-28T22:11:03.640" Score="2" ViewCount="162" Body="&lt;p&gt;I was reading about Euler's totient function on wikipedia, and it eventually led me to this book on google:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=1MTcYrbTdsUC&amp;amp;pg=PA74&quot; rel=&quot;nofollow&quot;&gt;Page 74 of the book, Prime numbers: the most mysterious figures in math&#xA; By David G. Wells&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the books lists many assertions without proof, only references which I can't find. One I could not solve for myself said that not all even numbers are values of $\phi(n)$. The sequence of even non-values of $\phi(n)$ starts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$14, 26, 34, 38, 50, 62,\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about it for a while, I've made little headway. Looking at 14 specifically, I suppose if such a solution $a$ to $\phi(x)=14$ were to exist, $(a,m_i)=1$ for $1\leq i\leq 14$, for some $m_i\lt a$ and so there must also exist inverses $\overline{a_i}$ such that $\overline{a_i}a\equiv 1 \pmod {m_i}$ for each $i$. I'm looking for some contradiction, possibly of the Chinese Remainder Theorem, to show such $a$ cannot exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some way to generalize which values are not taken by $\phi$, or at least explain why this is the case? I was hoping to see why $14$ is the least such integer such that this is true, but values $1$ to $13$ must be indeed taken. I suppose this would also explain why 26 is the next such value that is not taken, while $15$ to $25$ are.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-28T22:20:29.707" LastActivityDate="2010-09-28T23:09:58.773" Title="Generalizing values which  Euler's-totient function does not take." Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5663" PostTypeId="2" ParentId="5661" CreationDate="2010-09-28T22:47:38.607" Score="3" ViewCount="0" Body="&lt;p&gt;It's just casework on the prime factorization of $n$.  If $\phi(n) = 14$ then $n$ can't be divisible by any primes larger than $7$ (because $p-1$ cannot divide $14$ for such primes), and it can't be divisible by any of $5, 7$ because $4, 6$ don't divide $14$.  So it can only be divisible by $2$ or $3$.  Since $14$ is not divisible by $3$, $3$ can only divide $n$ at most once, so $n$ is of the form $2^k 3$.  But then $\phi(n) = 2^k$; contradiction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $\phi(n) = 26$ then $n$ can't be divisible by any primes larger than $13$, and it can't be divisible by any of $5, 7, 11, 13$ because $4, 6, 10, 12$ don't divide $26$.  So it can only be divisible by $2$ or $3$, and then the same argument as above works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any particular potential value of $\phi(n)$ a similar, but longer, argument should work; in particular it should be possible to test in finite time whether a particular candidate works.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-09-28T22:47:38.607" CommentCount="1" />
  <row Id="5664" PostTypeId="2" ParentId="5661" CreationDate="2010-09-28T22:50:54.370" Score="3" ViewCount="0" Body="&lt;p&gt;From the comments to &lt;a href=&quot;http://www.research.att.com/~njas/sequences/?q=A005277&amp;amp;sort=0&amp;amp;fmt=0&amp;amp;language=english&amp;amp;go=Search/%22A005277%22&quot; rel=&quot;nofollow&quot;&gt;A005277&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p is prime then the following two statements are true. I. 2p is in the sequence iff 2p+1 is composite (p is not a Sophie Germain prime). II. 4p is in the sequence iff 2p+1 and 4p+1 are composite. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Dec 30 2005 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This covers most of your cases, 50 is covered by the next comment.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-09-28T22:50:54.370" CommentCount="2" />
  <row Id="5666" PostTypeId="1" AcceptedAnswerId="5669" CreationDate="2010-09-28T23:09:53.210" Score="3" ViewCount="133" Body="&lt;p&gt;I was in the middle of writing the same old geographic distance calculation using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Haversine_formula&quot; rel=&quot;nofollow&quot;&gt;Haversine formula&lt;/a&gt; when it occurred to me: shouldn't there be simpler way to do this? Haversine is of course derived from the Law of Cosines. But in thinking about this problem, I ran across &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_trigonometry&quot; rel=&quot;nofollow&quot;&gt;Napier's Rules&lt;/a&gt; for right-angled spherical triangles. It seems like Napier's Rules should apply, after all Latitude small circles and Longitude great circles always intersect at right angles, so you should be able to draw a right-angled spherical triangle where the hypotenuse connects any two points (unless the case devolves into a line or single point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I'm applying Napier's Rules correctly, if our delta latitude and delta longitude in radians are $a$ and $b$ respectively, the angle $c$ should trace the arc between the two points. So Napier says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(\pi-c) = \cos(a) \cos(b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which should then simplify to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = \arcsin(\cos(a) \cos(b))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I try to verify this with a couple test points, the result doesn't match the Haversine formula. Is there a mistake in my (admittedly rusty) algebra or is my mistake in assuming I can apply Napier's Rules to this problem?&lt;/p&gt;&#xA;" OwnerUserId="2195" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-28T23:22:48.820" LastActivityDate="2010-09-28T23:22:48.820" Title="Napier's Rules applied to spherical distance calculations" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;spherical-geometry&gt;" AnswerCount="1" />
  <row Id="5667" PostTypeId="2" ParentId="5650" CreationDate="2010-09-28T23:16:08.783" Score="6" ViewCount="0" Body="&lt;p&gt;Both sides are equal since they count the same set: the RHS counts naturals $\rm\:\le n\:x\:$. The LHS counts them in a unique mod $\rm\ n\ $ representation, $\:$ viz. $\rm\ \: j \:\le\: x+k/n\: \iff \ j\:n-k \:\le\: n\:x\:,\ \ j&gt;0 \le k &amp;lt; n\:$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\:$ That every natural has a unique representation of form $\rm \: j\:n-k \ \ \:$ where $\rm\ \ \: j&gt;0 \le k &amp;lt; n\ \ \ $ is simply a slight variant of the &lt;strong&gt;Division Algorithm&lt;/strong&gt; where one utilizes negative (vs. positive) remainders.$\ \ $ To derive this negative form, simply  perform the following transformation on the positive remainder form $\rm\ q\: n + r\ \to\ (q+1)\:n + r-n\ $ if $\rm\ r\ne 0\:$, i.e. inc the quotient, dec the remainder by the dividend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the result is equivalent to the Division Algorithm, whose normal proof is indeed by &lt;em&gt;induction&lt;/em&gt;. One could give a direct inductive proof of the result if, instead of invoking the Division Algorithm by name, one unwinds or inlines this inductive proof directly into the proof of the result - much as the same way that the classic &lt;a href=&quot;http://www.math.uiuc.edu/~dan/ShortProofs/uf.html&quot; rel=&quot;nofollow&quot;&gt;Lindenmann - Zermelo direct proof of unique factorization of naturals&lt;/a&gt; inlines a division / Euclidean algorithm based descent proof of the fundamental Prime Divisor property $\rm\ p|ab\ \Rightarrow\ p|a\ \ or\ \ p|b\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T21:12:36.650" LastActivityDate="2010-09-29T21:12:36.650" />
  <row Id="5668" PostTypeId="2" ParentId="5191" CreationDate="2010-09-28T23:20:45.783" Score="2" ViewCount="0" Body="&lt;p&gt;$A_n$ is the symmetry group of the chamber of the Tits building of $\mathbb{P}GL_n$. The shape of this chamber is independent of what coefficients you insert into the group scheme $\mathbb{P}GL_n$, just the number and configuration of chambers changes. If you insert the finite fields $\mathbb{F}_p$ then you get finite simplicial complexes as buildings, and the smaller $p$ gets, the fewer chambers you have. You can analyse and even reconstruct the group in terms of its action on this building. The natural limit case would be just having one chamber and the symmetry group of this chamber - the Weyl group - is $A_n$. This is how Tits first thought that there is a limit case to the sequence of finite fields - which he called the field with one element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe somewhat more algebraically you can think in terms of Lie algebras - as I said the shape of the chamber does not change with different coefficients. The reason is that it is determined just by the Lie algebra of the group and thus describable by a Dynkin diagram or by a root system (ok, geometry creeps in again). The Wikipedia page about &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_group&quot; rel=&quot;nofollow&quot;&gt;Weyl groups&lt;/a&gt; tells you that the Weyl group of the Lie algebra $sl_n$ is $S_n$. If have no experience with Lie algebras, but maybe you can get $A_n$ the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can get hold of it, you can read Tits' original account, it's nice to read (but geometric) see the reference on this &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_with_one_element&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Aha, I found a link now: Lieven Le Bruyn's &lt;a href=&quot;http://matrix.cmi.ua.ac.be/fun/index.php/kapranov-smirnov-on-f_un.html&quot; rel=&quot;nofollow&quot;&gt;F_un page&lt;/a&gt; is back online. You can look there under &quot;papers&quot; and find Tits' article. And, since you are picking up the determinant ideas, you should definitely take a look at Kapranov/Smirnov!&lt;/p&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="2048" LastEditorDisplayName="" LastEditDate="2010-09-29T17:43:34.493" LastActivityDate="2010-09-29T17:43:34.493" CommentCount="5" />
  <row Id="5669" PostTypeId="2" ParentId="5666" CreationDate="2010-09-28T23:22:29.867" Score="4" ViewCount="0" Body="&lt;p&gt;Latitude &quot;lines&quot; are not spherical-geometry lines, as they are not great circles, so the &quot;triangles&quot; you are getting are not spherical triangles.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-09-28T23:22:29.867" CommentCount="7" />
  <row Id="5670" PostTypeId="1" CreationDate="2010-09-29T00:27:28.717" Score="4" ViewCount="184" Body="&lt;p&gt;There is a lemma in my introduction to analysis book for which the proof eludes me. I was hoping to get some clarification. Here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: If $\lbrace b_n \rbrace_{n=1}^\infty$ converges to B and B $\not=$ 0, then there is a positive real number $M$ and a positive integer $N$ such that if $n \geq N$, then $\mid b_n \mid \geq M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Since $B \not= 0, \frac{\mid B \mid}{2} = \epsilon &gt; 0.$ There is $N$ such that if $n \geq N, \mid b_n - B \mid &amp;lt; \epsilon.$ Let $M = \frac{\mid B \mid}{2}.$ Thus for $n \geq N$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mid b_n \mid = \mid b_n - B + B \mid \geq \mid B \mid - \mid b_n - B \mid \geq \mid B \mid - \frac{\mid B \mid}{2} = \frac{\mid B \mid}{2} = M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lemma makes intuitive sense to me because of course if a sequence converges to a number other than 0, its absolute value will always be above some number $M$ for $n \geq N$. However, I get lost in the inequalities of the proof. I don't understand how the author came up with $\mid b_n - B + B \mid \geq \mid B \mid - \mid b_n - B \mid$. I understand that he added and subtracted $B$ to $b_n$ for the left side, but I'm not so clear on the how he came up with $\mid B \mid - \mid b_n - B \mid$. It looks like an iteration of the triangle inequality, but I don't quite see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, the book had an exercise for a variation on this lemma that involved proving $b_n \geq M$ for all $n$. The proof for that involved setting $M = min \lbrace b_1, b_2, ..., b_n \rbrace$, which made a lot more sense to me than the inequalities for this proof.&lt;/p&gt;&#xA;" OwnerUserId="2175" LastEditorUserId="2175" LastEditorDisplayName="" LastEditDate="2010-09-30T03:53:22.440" LastActivityDate="2010-09-30T03:53:22.440" Title="Clarification of a sequence convergence proof." Tags="&lt;sequences-and-series&gt;&lt;proof-strategy&gt;" AnswerCount="2" />
  <row Id="5671" PostTypeId="2" ParentId="5657" CreationDate="2010-09-29T00:34:24.367" Score="0" ViewCount="0" Body="&lt;p&gt;I would do it this way. First of all, by choosing $\frac{a}{\|a\|}$ as the last vector of an orthonormal basis of $\mathbb{R}^n$, you may assume that $a = (0,\dots , 0,\alpha), \alpha \in \mathbb{R}, \alpha \neq 0$. Hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S =\left\{ (x_1 , \dots , x_n) \in \mathbb{R}^n\ \vert \ x_n = 0    \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick any $x = (x_1 , \dots , x_{n-1} , 0) \in S$ and let's show that, for any $r&gt;0$, $B(r,x) $ is not included in $S$, so the interior of $S$ will be empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To this end it's enough to produce a single vector in $B(r,x)$ which is not in $S$, right? Ok, take a look at $v = (x_1, \dots , x_{n-1}, \frac{r}{2})$ and compute (draw a picture in $\mathbb{R}^3$ too to convince yourself).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-09-29T00:34:24.367" />
  <row Id="5672" PostTypeId="2" ParentId="5670" CreationDate="2010-09-29T00:51:25.863" Score="3" ViewCount="0" Body="&lt;p&gt;You are right, it's just the triangle inequality. $B=b_n+(B-b_n)$. So $|B|\leq |b_n| + |B-b_n|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, shouldn't it be $n \geq N$ in your second para?&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-29T00:51:25.863" CommentCount="2" />
  <row Id="5673" PostTypeId="2" ParentId="5657" CreationDate="2010-09-29T01:00:08.310" Score="1" ViewCount="0" Body="&lt;p&gt;It is a tautology to say &quot;every point in S, all 'close' points are either in the complement of S or in S itself&quot;. Also I am not sure how you use $a|r|&gt;0$ to conclude that  $B(r,x) \subset S \, \forall r&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway you don't need to think of all &quot;close&quot; points. You just need to prove that arbitrarily close to any $x \in S$ there is at least one point which is not in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: think of points of the form $x+\lambda a$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-09-29T03:25:21.597" LastActivityDate="2010-09-29T03:25:21.597" CommentCount="1" />
  <row Id="5674" PostTypeId="2" ParentId="5656" CreationDate="2010-09-29T01:03:50.563" Score="2" ViewCount="0" Body="&lt;p&gt;If we start with the Taylor expansion (I'll change variables here, too many subscripts confuse me):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}f^{\prime\prime}(x)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the derivative of this w.r.t. $h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x+h)=f^{\prime}(x)+h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the version of this with $h$ replaced by negative $h$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x-h)=f^{\prime}(x)-h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subtracting the third expression from the second expression gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x+h)-f^{\prime}(x-h)=2h f^{\prime\prime}(x)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we see that the even powers drop out of this error expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we solve for $f^{\prime\prime}(x)$ like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime\prime}(x)=\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substitute in the first expression,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can take the $O(h^2)$ within the parentheses out as an $O(h^4)$ term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the leading term after the replaced portion is $O(h^3)$, thus simplifying to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+O(h^3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we see that the formula has $O(h^3)$ error: cutting $h$ in half decreases the error by a factor of $2^3=8$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-29T15:35:36.130" LastActivityDate="2010-09-29T15:35:36.130" CommentCount="7" />
  <row Id="5675" PostTypeId="1" CreationDate="2010-09-29T02:29:54.573" Score="3" ViewCount="104" Body="&lt;p&gt;Me and a team of programmers are programming a robot to play a game in a competition. We did not create the hardware, and all teams will use the same type of robot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;&lt;strong&gt;GAME DESCRIPTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game has two players opposing each other and are both trying to achieve an objective. This objective is to move to a position to pick up a payload then return to another position to return the payload. Each team has their own payload and their own target to bring them to, and it is impossible to pick up the other team's payload. The team that brings their payload to the target first wins and the game is ended. There is a time limit of 210 seconds, and if the games times out, the team who held the payload the longest wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is a little more complicated than that. The robots also have the ability to face the opponent and &quot;push them back&quot; from any range (the closer they are the more&#xA;forcefully they push). Also, if a robot is pushed out of the bounds of the playing field, they drop their payload and it is moved back inside the playing field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;&lt;strong&gt;GAME THEORY QUESTIONS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, is this a zero-sum game or not? I am new to game theory math and I am not completely sure if it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how does the minimax theorem apply to this specific game? I understand how it works, but I do not know what the values to account for would be in this game (would it be the difference in how long each team has held the payload somehow combined with how close they are to bringing to the target?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really am not sure at all how to calculate any of this, but if anyone can at least point me in the right direction for coming up with an effective strategy system for this I would be very appreciative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for your time, this project is very important to me and your help means a lot. If I need to clarify anything please ask, I am not sure if I included all the information needed for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2196" LastActivityDate="2010-09-29T03:43:02.943" Title="Zero-Sum Game Theory" Tags="&lt;game-theory&gt;" AnswerCount="2" />
  <row Id="5676" PostTypeId="1" CreationDate="2010-09-29T02:34:58.383" Score="8" ViewCount="409" Body="&lt;p&gt;By looking at an integral and bounding the error?&lt;/p&gt;&#xA;" OwnerUserId="1838" LastActivityDate="2010-09-30T17:44:02.330" Title="How closely can we estimate $\sum_{i=0}^n \sqrt{i}$" Tags="&lt;sequences-and-series&gt;&lt;integral&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="1" />
  <row Id="5677" PostTypeId="2" ParentId="5675" CreationDate="2010-09-29T02:40:56.580" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, this is a zero sum game, since both players are fighting over a single &quot;win&quot;. What that means in terms of strategy is that hurting your opponent is as effective as helping yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how you would use minmax theory in this situation. As far as I know, game theory tools are geared towards games with a relatively small number of possible situations and moves, whereas in your game, you have a very large number of possible situations (because the position of the robots is variable).&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-29T02:40:56.580" />
  <row Id="5678" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T02:49:06.793" Score="2" ViewCount="0" Body="&lt;p&gt;Since square root is a monotonically increasing function, that summation will be between the integral of sqrt(i) from 0 to n and the integral of sqrt(i) from 1 to n-1.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-09-29T02:49:06.793" />
  <row Id="5679" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T03:01:32.453" Score="3" ViewCount="0" Body="&lt;p&gt;The standard approach should be along the lines of $\sum_{i\le n}\sqrt i=\sqrt n\sum_{i\le n}\sqrt{\frac{i}{n}}$, so $\frac1{n\sqrt n}\sum_{i\le n}\sqrt i=\sum_{i\le n}\sqrt{\frac{i}{n}}\frac1n\to\int_0^1\sqrt x{} dx=\frac23$, or $\sum_{i\le n}\sqrt i\sim\frac23 n\sqrt n$. What techniques do you know to estimate the error between integrals and their Riemann sums?&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-09-29T03:01:32.453" CommentCount="1" />
  <row Id="5680" PostTypeId="2" ParentId="5675" CreationDate="2010-09-29T03:43:02.943" Score="0" ViewCount="0" Body="&lt;p&gt;This is a zero sum game &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_game&quot; rel=&quot;nofollow&quot;&gt;differential game&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your strategy is a &lt;em&gt;function&lt;/em&gt; from the state of the board (everyone's position and velocity and how long they have held their payload) to your control inputs. Your payoff is 1 if you win and 0 if you lose. [Which is why this is a zero sum game, since the sum of payoffs over all players equals 1]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, the actual solved examples of differential games I have seen have been for simple two player problems. You may be better off using some kind of heuristic for the contest.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-09-29T03:43:02.943" />
  <row Id="5681" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T03:59:57.267" Score="5" ViewCount="0" Body="&lt;p&gt;The &quot;sophisticated&quot; way to do this is using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin formula&lt;/a&gt; - look under the heading &quot;Asymptotic expansion of sums&quot;. You'll want to start your sum at 1, not at 0, to avoid dividing by zero.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-09-29T03:59:57.267" />
  <row Id="5682" PostTypeId="2" ParentId="5657" CreationDate="2010-09-29T04:13:52.067" Score="0" ViewCount="0" Body="&lt;p&gt;The condition that $x$ be orthogonal to $a$, i.e. that $x$ lies in $S$, is that $x \cdot a = 0$.  Imagine perturbing $x$ by a small amount, say to $x'$.  If $x$ were in the interior,&#xA;than one would have $x' \cdot a = 0$ as well, provided that $x'$ is very close to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about whether this is possible for every $x'$.  (Hint: $x'$ has to be close to $x$,&#xA;i.e. $x - x'$ has to be small.  But it can point in any direction!)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-29T04:13:52.067" />
  <row Id="5683" PostTypeId="2" ParentId="5670" CreationDate="2010-09-29T04:24:56.963" Score="2" ViewCount="0" Body="&lt;p&gt;Just a remark (that is too long to put in a comment box): The idea of the proof is just this, that if $x_n$ approaches $B$,&#xA;then eventually (i.e. for large enough $n$) it has to be within distance $|B/2|$ of $B$, &#xA;and once this happens we can be sure that $x_n \geq B/2$ (if $B$ is positive) or that&#xA;$x_n \leq B/2$ (if $B$ is negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't clear just from how I've written it, draw  picture: label $B$,&#xA;draw the interval of points within distance $|B/2|$ of $B$, which is the interval&#xA;$[B/2, 3B/2]$ (or switch the endpoints if $B$ is negative), and then just randomly&#xA;draw in a sequence of points getting closer and closer to $B$.  You'll see that eventually&#xA;you have to put them all inside this interval, so they will all be at least $B/2$&#xA;(or at most $B/2$ if $B$ is negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manipulations with absolute values are just a slightly tortuous verification that&#xA;if $x$ is within distance $|B/2|$ of $B$, then $x$ is at least $B/2$ (or at most $B/2$&#xA;is $B$ is negative).  It is easy to write down statements that are obvious when you say&#xA;them out like this, or when you draw a picture, but which become a bit of a battle when&#xA;you try to derive them formally from the triangle inequality.  The fact that the author&#xA;is treating the case when $B$ is negative at the same time adds another small but distracting layer of complexity, because you have to write $|B/2|$ rather than $B/2$ for the distance under consideration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you become more familiar with these sorts of manipulations, when you come to a point&#xA;like this in a text where you recognize some kind of triangle inequality manipulations, it is often easier (and good training!) just to draw a picture and figure out the manipulations for yourself, rather than working through somebody else's slightly painful manipulations on the page.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-29T04:24:56.963" CommentCount="3" />
  <row Id="5684" PostTypeId="1" AcceptedAnswerId="5687" CreationDate="2010-09-29T05:21:05.217" Score="2" ViewCount="146" Body="&lt;p&gt;An exercise from Herstein asks to prove that the number of elements of order $p$, $p$ a prime in $S_{p}$ is $(p-1)!+1$. I would like somebody to help me out on this and also i would like to know whether can we prove Wilson's theorem which says $(p-1)! \equiv -1 \ (\text{mod} \ p)$ using this result.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-09-29T23:17:03.257" LastActivityDate="2010-09-29T23:17:03.257" Title="Number of Elements of order $p$ in $S_{p}$" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="5686" PostTypeId="2" ParentId="5648" CreationDate="2010-09-29T06:20:37.350" Score="1" ViewCount="0" Body="&lt;p&gt;The answer here is no. I am assuming you're trying to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\sigma$-finite measure spaces $\Omega_1, \Omega_2$, and $\Omega = \Omega_1\times\Omega_2$ carries the product measure, then if $h: \Omega\rightarrow \mathbb{R}$ is a measurable function, then $h$ is point-wise approximated almost everywhere by the sequence $h_n$ given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_n(x,t) = \sum_{j=1}^{N_n}f_{j,n}(x)1_{F_{j,n}}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, in general this is not the case since the sets $F_{j,n}$ will generally depend on $x$. If $\Omega_1$ is countable, the situation is not too bad: the family of sets $\{F_{j,n}(x): x\in \Omega_1\}$ is countable. In general, however, the situation is hopeless. I think using $\Omega_1 = \Omega_2 = \mathbb{R}_{\geq 0}$ and $h(x,t) = x^2 + e^t$, or something to that effect, should do as a counter example. Play with it.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-29T06:20:37.350" CommentCount="3" />
  <row Id="5687" PostTypeId="2" ParentId="5684" CreationDate="2010-09-29T06:35:25.243" Score="3" ViewCount="0" Body="&lt;p&gt;Maybe you mean the number of elements of order dividing $p$ (so that you are including the identity)?  (Think about the case $p = 3$ --- there are two three cycles, not three of them.)  For the general question, think about the possible cycle structure of an element of order $p$&#xA;in $S_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can go from the formula in your question to Wilson's theorem by counting the number of $p$-Sylow subgroups (each contains $p-1$ elements of order $p$), and then appealing to Sylow's theorem.  (You will find that there are $(p-2)!$ $p$-Sylow subgroups, and by Sylow's theorem this number is congruent to $1$ mod $p$.  Multiplying by $p-1$, we find that&#xA;$(p-1)!$ is congruent to $-1$ mod $p$.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-29T06:35:25.243" />
  <row Id="5688" PostTypeId="2" ParentId="5124" CreationDate="2010-09-29T06:54:30.900" Score="0" ViewCount="0" Body="&lt;p&gt;It isn't that we &quot;Want them to be,&quot; but that they must be simply from definition: Single point sets are necessarily closed in the topology of $\mathbb{R}$, hence are in $\mathcal{B}$ - the Borel $\sigma$-algebra. Since $\mathbb{Q}$ is a countable union of single point sets, it must be in $\mathcal{B}$. Hence so is $\mathbb{Q}^c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing else, this at least allows some nice examples. For instance, define $f(x)$ to be identically $1$ on $\mathbb{Q}$ and identically $0$ on $\mathbb{Q}^c$. $f$ is not Riemann integrable, however, $f$ is Lebesgue measurable, and integrates to zero in the Lebesgue sense (which is something one would expect).&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2010-09-29T06:54:30.900" />
  <row Id="5689" PostTypeId="1" CreationDate="2010-09-29T07:19:26.897" Score="1" ViewCount="59" Body="&lt;p&gt;What is the general solution to the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little stuck because all the techniques I know are unable to solve it.&lt;/p&gt;&#xA;" OwnerUserId="2148" LastActivityDate="2010-09-29T07:23:53.117" Title="Solution to 2nd order PDE" Tags="&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5690" PostTypeId="2" ParentId="5689" CreationDate="2010-09-29T07:23:53.117" Score="4" ViewCount="0" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_equation&quot; rel=&quot;nofollow&quot;&gt;heat equation&lt;/a&gt;. Don't expect a single formula for the general solution, because the problem is too complicated for that. There are, however, fairly general formulas which apply under special circumstances (Fourier series solution for the initial–boundary value problem, etc.). &lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-09-29T07:23:53.117" />
  <row Id="5691" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T07:23:58.463" Score="1" ViewCount="0" Body="&lt;p&gt;You will find some really nice approximations at &lt;a href=&quot;http://www.mathkb.com/Uwe/Forum.aspx/undergrad/3288/sequence-sum&quot; rel=&quot;nofollow&quot;&gt;MathKB&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: You'll also find a couple of exact formulas there, though not written with elementary functions of course.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-29T07:40:17.620" LastActivityDate="2010-09-29T07:40:17.620" />
  <row Id="5692" PostTypeId="2" ParentId="5564" CreationDate="2010-09-29T07:40:45.423" Score="5" ViewCount="0" Body="&lt;p&gt;It doesn't have to: logics which don't are called paraconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important paraconsistent logic is relevance logic, which repudiates the K axiom:&#xA;$$\alpha \rightarrow \beta \rightarrow \alpha$$&#xA;and replaces it by axioms that do not allow there to be unused assumptions.  This is equivalent to saying weakening, the principle that if $\Gamma \vdash \alpha$ then $\Gamma'\vdash \alpha$ for $\Gamma\subset\Gamma'$.  This blocks derivations such as Weltschmertz's, which appeals to the K axiom once, Asaf's which uses it twice; Francesco appeals to monotonicity in his proof, which is another name for weakening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not difficult to see that this also blocks proofs of everything from a contradictory pair of propositions in a logic satisfying compactness, since one can prove inductively about such proof systems that if $\alpha\rightarrow\beta$, then all positive atoms in $\beta$ must occur either negatively in $\beta$ or positively in $\alpha$.  So if our contradictory pair (over an assumption) takes the form $\alpha\rightarrow\beta$ and $\alpha\rightarrow\neg\beta$, we need to prove for any $\gamma$ that $\alpha\rightarrow\gamma$.  But if we choose $\gamma$ to be any positive atom not occuring in $\alpha$, our inductive proof tells us this cannot be done.  We need compactness here, to be ensure that the basis for all contradictory pairs can be expressed by a finitary formula.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-09-29T07:40:45.423" />
  <row Id="5693" PostTypeId="1" CreationDate="2010-09-29T08:03:47.577" Score="1" ViewCount="196" Body="&lt;p&gt;first show you only need to consider squares of functions as f.g = 1/4 [(f+g)sqr - (f-g)sqr]. show then that you only need to consider only positive valued functions becuase f(x).g(x)=|f(x)|sqr. then , if 0 &amp;lt;=f(x) &amp;lt;= M on [a,b] show that f sqr(x) - f sqr(y) &amp;lt;= 2M (f(x)-f(y)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does anyone know how i would answer this ??&lt;/p&gt;&#xA;" OwnerUserId="2201" LastActivityDate="2010-09-29T14:37:29.703" Title="proof : product of two riemann integrals are integrable." Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="5694" PostTypeId="2" ParentId="5693" CreationDate="2010-09-29T08:14:09.137" Score="5" ViewCount="0" Body="&lt;p&gt;This follows from Lebesgue's characterization of Riemann integrable functions&#xA;as bounded functions continuous outside a set of Lebesgue measure zero.&#xA;This characterization is usually the swiftest way of deciding on the&#xA;Riemannn integrability of a function.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-29T08:14:09.137" CommentCount="4" />
  <row Id="5695" PostTypeId="2" ParentId="5684" CreationDate="2010-09-29T08:21:17.313" Score="2" ViewCount="0" Body="&lt;p&gt;Every element of order $p$ in $S_p$ is a $p$-cycle. The symmetric group $S_{p-1}$ acts transitively on these $p$ cycles.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-09-29T08:21:17.313" CommentCount="2" />
  <row Id="5696" PostTypeId="1" AcceptedAnswerId="5704" CreationDate="2010-09-29T08:25:08.957" Score="1" ViewCount="128" Body="&lt;p&gt;I think Cauchy's integral formula and the Hilbert transform can be used to prove one direction, but is this an equivalence or only an implication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; for clarification:&#xA;Is a function $f : \mathbb C \to \mathbb C, z\mapsto f(z)$ analytical $\Leftrightarrow$ The Fourier-Transform $\mathcal F\{f\}(\omega) = N \int_{\mathbb R} f(z) e^{i\omega z} dz$ (choose whatever normalization $N$ you like, I prefer symmetry $N=\sqrt{2\pi}$) is zero for all $\omega&amp;lt;0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or shorter: Is the following true? $f$ analytical $\Leftrightarrow$ $\mathrm{supp}_{\mathcal F\{f\}}=\mathbb R^+$&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-29T10:54:29.867" LastActivityDate="2010-09-29T18:19:17.247" Title="Is a function analytical on C iff its Fourier-transform vanishes for negative frequencies?" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5697" PostTypeId="1" AcceptedAnswerId="5732" CreationDate="2010-09-29T08:48:42.770" Score="6" ViewCount="266" Body="&lt;p&gt;I will restate the 3-D version of the problem. In how many ways can you color a regular cube with 2 colors up to a rotational isometry. The answer is of course a special case of Burnsides Lemma which can be used to show that the number of distinct face permutations is $\frac{1}{24}(N^6 + 3N^4 + 12N^3 + 8N^2)$ where $N$ is the number of colors used, $2$ in this case, which gives us an answer of $10$ distinct permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how can you expand this to a tesseract, and then more generally, to any hypercube. The rotational isometries of a cube are somewhat simple to comprehend, but the rotational isometries o hypercube are difficult to grasp (even after an hour of playing with the 4d rubiks cube app)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thought was to consider the expansion of a tesseract as 8 interconnected cubelets. For the two color case each one of these cubelets has 10 distinct states. Which gives us $10^8$ non-distinct permutations of the hypercube. Or more generally, this reduces to how many distinct ways one can color a 8 faced 3-dimensional figure using 10 colors. So the complicated question of 4-dimensional isometries reduces to 3 dimensional rotational isometries of a octahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm a physicist so what the hell do I know :D&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-29T09:04:51.830" LastActivityDate="2010-09-30T18:04:50.063" Title="Coloring the faces of a hypercube" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="5698" PostTypeId="1" AcceptedAnswerId="5701" CreationDate="2010-09-29T09:04:24.670" Score="1" ViewCount="142" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/548/explanation-of-method-for-showing-that-0-0-is-undefined&quot;&gt;Explanation of method for showing that 0 / 0 is undefined&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;always i wonder this question why we cant calculate 0/0 in real numbers. where is my mind. i cant think anything logically&lt;/p&gt;&#xA;" OwnerUserId="2202" LastActivityDate="2010-09-29T10:02:56.333" ClosedDate="2010-09-29T14:06:04.673" Title="why we cant calculate 0/0 why is it undefinable?" Tags="&lt;zeros&gt;&lt;zero-divisor&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5699" PostTypeId="2" ParentId="5696" CreationDate="2010-09-29T09:10:31.603" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not sure if I got your question right, but as I understand it the answer is no. Let's stick with the fourier transform as an operator $L^2(\mathbb{R}) \to L^2(\mathbb{R})$.&#xA;One of the basic facts about the fourier transform is that compactly supported functions transform into analytic functions. So as a counterexample just pick the characteristic function of $[-1,1]$. Its fourier transform is analytical but it certainly has &quot;negative frequencies&quot;, namely, it has frequencies almost everywhere in the interval $[-1,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To make this more explicit: Let $f := \hat \chi_{[-1,1]}$. Then $f$ is analytic (in fact, we use the definition $\hat u(\xi) = \int e^{i\xi x} f(x) d x$ then $f(\xi)=2 \sin \xi / \xi$ (and $f(0)=2$), which is analytic). By the Fourier inversion theorem, $\hat f(x)=\hat{\hat \chi}_{[-1,1]} (x)=2\pi \chi _{[-1,1]}(-x)$ , whose support is $[-1,1]$. So $f$ is a counterexample to your statement.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="1609" LastEditorDisplayName="" LastEditDate="2010-09-29T11:38:04.110" LastActivityDate="2010-09-29T11:38:04.110" CommentCount="7" />
  <row Id="5700" PostTypeId="2" ParentId="5697" CreationDate="2010-09-29T09:12:33.813" Score="1" ViewCount="0" Body="&lt;p&gt;I presume you want to colour the &lt;em&gt;faces&lt;/em&gt; of the cube?&#xA;If so there are $2^6$ colourings as there are six faces.&#xA;But do you want two colourings which are equivalent under&#xA;rotations (or rotations and reflections) to count as as the same?&#xA;If so then this is a classic application of&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's lemma&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;If you want to do the same in higher dimensions then you need a good&#xA;handle on the symmetries of an $n$-cube. Its convenient to centre the&#xA;cube at the origin and take its vertices to be all points $(\pm1,\ldots,\pm1)$.&#xA;Then all symmetries of the cube fix the origin and their matrices&#xA;are &quot;signed&quot; permutation matrices: each row and column is all zero save for&#xA;one entry which is $\pm1$. The rotations correspond to signed permutation&#xA;matrices of determinant 1. The faces are the sets defined by $x_i=1$ and&#xA;$x_i=-1$. Again use Burnside's lemma but the book-keeping gets fiddlier.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-29T12:06:39.490" LastActivityDate="2010-09-29T12:06:39.490" CommentCount="1" />
  <row Id="5701" PostTypeId="2" ParentId="5698" CreationDate="2010-09-29T09:17:13.343" Score="1" ViewCount="0" Body="&lt;p&gt;The simple answer is that division by zero is not well defined. As Andrea pointed out the fraction $\frac{0}{0}$ can be restated as $x$ s.t. $x*0 = 0$ which is true for all x in the real numbers. However, we still are motivated to find some kind of resolution to this issue, what exactly $ does $ $\frac{0}{0}$ equal to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this we have to turn to the world of calculus, $\frac{0}{0}$ is a standard limit case that we can examine. The answer in that case depends on the function you are taking the limit of. And somewhat surprisingly (or not), it can be equal to any real number. Just to illustrate this, I want to give you a simple example for the three cases, where the limit is equal to $0$, $N$, or $\pm\infty$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_1(x) = \frac{Nx}{x}$$&#xA;$$f_2(x) = \frac{x^2}{x}$$&#xA;$$f_3(x) = \frac{x}{x^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of these three cases, you see that plugging in zero gives you the undefined expression of $\frac{0}{0}$. However, if you take the limit of $f_n(x)$ as $x\rightarrow0$ you see that the three functions tend to completely different values. One tends to $N$ the second tends to $0$ and the third tends to $\pm\infty$ depending on the direction you are coming from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we resolve the issue in some sense, because now we can understand how this beast behaves, but on the other hand we still have the ambiguity that we had in the beginning.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-09-29T10:02:56.333" LastActivityDate="2010-09-29T10:02:56.333" CommentCount="4" />
  <row Id="5702" PostTypeId="2" ParentId="1730" CreationDate="2010-09-29T09:33:16.630" Score="1" ViewCount="0" Body="&lt;p&gt;A function is something that takes a number , twists it around and spits out another number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once they grasp that, you can talk about how you can have functions that work with things besides numbers. And how generally, a functions can twist around any object, tangible or mathematical and spit out another object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can move on to explaining the concept of domain and range. Domain is type of objects that your function can accept, and range is the type of object that your function can conceivably spit out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good concrete example. A soda machine is a function that maps (don't use that word though, it scares students at their first encounter with it) the domain of coins to the range of soda's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once they have the image in their head and some intuition THEN you go back and discuss the formal definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least that what I do with my algebra students&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-09-29T09:33:16.630" />
  <row Id="5703" PostTypeId="2" ParentId="5698" CreationDate="2010-09-29T09:47:00.633" Score="1" ViewCount="0" Body="&lt;p&gt;Perhaprs the simplest way to understand this is to use &quot;words from the elementary school&quot;. In that time the teacher asked me how much is 12/4 and why. So my answer was 3, bacause $3\cdot 4 = 12$. Imagine now that the teacher asks you how much is 0/0? You can say for example 1 and by the previous argument you are right, because $1\cdot 0 = 0$. But on the other side you could say 0 or -1 or any real number. In this case many answers seem to be correct, so in such situations in mathematics the expression cannot be well-defined.&lt;/p&gt;&#xA;" OwnerUserId="2035" LastActivityDate="2010-09-29T09:47:00.633" />
  <row Id="5704" PostTypeId="2" ParentId="5696" CreationDate="2010-09-29T11:44:28.857" Score="1" ViewCount="0" Body="&lt;p&gt;So, you are only interested in the Fourier transform of $f$&#xA;restricted to the real line. Typically the restriction&#xA;of such a function won't be in $L^2(\mathbb{R})$ (or any $L^p(\mathbb{R})$&#xA;for that matter) and won't have polynomial growth, so won't be a tempered&#xA;distribution. I don't know a way in which such a function (e.g. $f(z)=\exp(z)$)&#xA;could be said to have a Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But any compactly supported distribution on $\mathbb{R}$ will have&#xA;a Fourier transform that is an entire analytic function.&#xA;Some other tempered distributions also have this property, notably&#xA;the Gaussian $g(z)=\exp(-z^2)$. As Florian points out, that even when&#xA;an entire function has a Fourier transform, it need not be supported&#xA;on the positive reals; there is absolutely no bias towards positivity&#xA;or negativity in its support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in Fourier transforms of analytic functions, you should&#xA;look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem&quot; rel=&quot;nofollow&quot;&gt;Paley-Wiener theorem&lt;/a&gt; which translates between properties&#xA;of the one and of the other.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-29T18:19:17.247" LastActivityDate="2010-09-29T18:19:17.247" CommentCount="1" />
  <row Id="5705" PostTypeId="2" ParentId="1099" CreationDate="2010-09-29T11:54:18.223" Score="4" ViewCount="0" Body="&lt;p&gt;Starting with the very well known iteration formula that creates the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mandelbrot_set&quot; rel=&quot;nofollow&quot;&gt;Mandelbrot set&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_n^2+c$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c=z_0=x_0+iy_0=\text{Re}(c)+i\text{Im}(c)$ is a complex constant which is the starting point of the trajectory generated in the complex plane by the iteration, in &lt;em&gt;&lt;a href=&quot;http://fraktalwelten.wordpress.com/2010/01/11/doppelpot/&quot; rel=&quot;nofollow&quot;&gt;Doppelpot&lt;/a&gt;&lt;/em&gt; , from the German blog &lt;em&gt;&lt;a href=&quot;http://fraktalwelten.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Fraktale Welten&lt;/a&gt; &lt;/em&gt;by &lt;em&gt; &lt;a href=&quot;http://tamagothi.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Nachtwaechter&lt;/a&gt;&lt;/em&gt;, it is explained that, when we have two complex exponentiations the fractals generated by the recursive formula are normally beautiful, particularly those of  Julia sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;a href=&quot;http://fraktalwelten.wordpress.com/2010/01/11/doppelpot/#comment-3186&quot; rel=&quot;nofollow&quot;&gt;suggested&lt;/a&gt;  to the author to try the following formulae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+2}=z_{n+1}^{2}+z_{n}+c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_{n}^{z_{n}c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_{n}^{z_{n}+c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+2}=z_{n+1}^{3}+c^{z_{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_{n+1}=z_{n}^{c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://fraktalwelten.wordpress.com/2010/02/22/funf-formeln/&quot; rel=&quot;nofollow&quot;&gt;Fünf Formeln&lt;/a&gt; the fractals are shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reproduced some with the author's permission in this &lt;a href=&quot;http://problemasteoremas.wordpress.com/2010/02/25/alguns-dos-fractais-mostrados-no-fraktalen-welten-gerados-por-formulas-de-iteracao-que-sugeri/&quot; rel=&quot;nofollow&quot;&gt;entry&lt;/a&gt; of my blog. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This one is generated by  $z_{n+2}=z_{n+1}^{2}+z_{n}+c$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LNuSp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this by $z_{n+2}=z_{n+1}^{3}+c^{z_{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/P3Ysd.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-30T12:18:43.050" LastActivityDate="2010-09-30T12:18:43.050" />
  <row Id="5706" PostTypeId="1" AcceptedAnswerId="5708" CreationDate="2010-09-29T12:48:00.587" Score="0" ViewCount="108" Body="&lt;p&gt;Let $G$ be a (not necessarily finite) group with the property that for each subgroup $H$ of $G$, there exists a `retraction' of $G$ to $H$ (that is, a group homomorphism from $G$ to $H$ which is identity on $H$). Then, we claim :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$G$ is abelian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each element of $G$ has finite order.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each element of $G$ has square-free order.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let $g$ be a nontrivial element of $G$ and consider a retraction $T : G \to \langle{g\rangle}$ which is identity on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic and so, it is abelian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than this i don't know how to prove the other claims of the problem. Moreover, a similar problem was asked in Berkeley Ph.D exam, in the year 2006, which actually asks us to prove that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $G$ is finite and there is a retraction for each subgroups $H$ of $G$, then $G$ is the products of groups of prime order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-29T14:48:53.463" Title="Showing $G$ is the product of groups of prime order." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5707" PostTypeId="1" CreationDate="2010-09-29T12:55:33.773" Score="2" ViewCount="71" Body="&lt;p&gt;If I have an equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(x)\frac{\partial^2u}{\partial x^2} + r(x)\frac{\partial^2u}{\partial x\partial y} + q(x)\frac{\partial^2 u}{\partial y^2}=f(x,y,u)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $f$ maybe contains first partial derivatives for $u$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a worked example of how to solve this using the method of characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="2148" LastActivityDate="2010-09-30T19:51:11.527" Title="Characteristics for 2nd order differential equations" Tags="&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5708" PostTypeId="2" ParentId="5706" CreationDate="2010-09-29T13:20:03.193" Score="4" ViewCount="0" Body="&lt;p&gt;Let $H$ be a subgroup of $K$ which is a subgroup of $G$.&#xA;If there's a retraction of $G$ onto $H$, it restricts to&#xA;a retraction of $K$ onto $H$. So if a group $G$ has this&#xA;property (let's say it's &quot;retractible&quot;) then each subgroup of $G$&#xA;is retractible. Which cyclic groups are retractible?&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-29T14:45:18.367" LastActivityDate="2010-09-29T14:45:18.367" CommentCount="3" />
  <row Id="5709" PostTypeId="2" ParentId="5693" CreationDate="2010-09-29T14:37:29.703" Score="4" ViewCount="0" Body="&lt;p&gt;This is not a problem I would assign as homework (at least, not without substantial guidance).  Rather, it is one of the fundamental results of the subject -- the subject being advanced calculus / elementary real analysis -- and as such I would expect any instructor / textbook to supply a proof.  For instance, Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; covers this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those readers who do not have access to any basic analysis textbooks, see Corollary 8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/243integrals2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/243integrals2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Robin says, the result also follows from Lebesgue's criterion of Riemann integrability: now &lt;em&gt;that&lt;/em&gt;'s something I would leave as an exercise, since finding this short argument on one's own helps to drive home the power of the Lebesgue criterion.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-09-29T14:37:29.703" />
  <row Id="5710" PostTypeId="2" ParentId="5706" CreationDate="2010-09-29T14:48:53.463" Score="0" ViewCount="0" Body="&lt;p&gt;Let $g$ be a nontrivial element of $G$ and consider a retraction $T : G \to \langle{g\rangle}$ which is identity on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic and so, it is abelian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $[G,G]$ is contained in $Ker(T)$. Since $g \notin Ker(T)$, $g \notin [G,G]$. As $g$ is an arbitrary nontrivial element of $G$, this means that $[G,G] = {e}$; that is, $G$ is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at any element $g \in G$ and consider a retraction $T:G \to \langle{g^2 \rangle}$.&#xA;$T(g)$ is in $\langle{g^2 \rangle}$ means $T(g) = g^{2r}$ for some $r$. Also, $T(g^2)=g^2$ means then that&#xA;$g^{4r}=g^2$; that is, $g^{4r-2} = e$. As $4r-2$ is not zero, we get that $g$ has finite order.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-29T14:48:53.463" />
  <row Id="5711" PostTypeId="2" ParentId="5707" CreationDate="2010-09-29T14:54:35.983" Score="1" ViewCount="0" Body="&lt;p&gt;See these &lt;a href=&quot;http://research.nianet.org/~grossman/lectnotes3_09-ch5.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; on the method of characteristics for second-order PDEs, where the method is applied to steady isentropic flow (gasdynamics)&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T15:00:48.453" LastActivityDate="2010-09-29T15:00:48.453" />
  <row Id="5712" PostTypeId="1" AcceptedAnswerId="5714" CreationDate="2010-09-29T16:24:58.593" Score="4" ViewCount="113" Body="&lt;p&gt;I've recently spent some time learning (the very basics of) K-theory for $C^*$-algebras and topological K-theory. Actually, my main fields of interest are PDEs and related topics, in particular functional calculus for unbounded operators, Sobolev/Bessel potential/Besov spaces, interpolation theory, semigroups and so on. You get the idea.&#xA;Now I would like to deepen (and broaden) my knowledge about the machinery of K-theory, ideally by learning some interesting connection to said PDE-related topics such as for example some PDE-relevant result admitting a proof with a k-theoretic flavour. Or some general idea how K-theory might provide insight into (or an interesting point of view on) some PDE-related results or concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd be very thankful, and I hope this request is not too broad, if you could provide me with some examples of interesting relations, if they exist, between K-theory and mentioned PDE-related topics.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastEditorUserId="1403" LastEditorDisplayName="" LastEditDate="2010-09-30T06:30:25.283" LastActivityDate="2010-09-30T06:30:25.283" Title="Connections between K-Theory and PDEs?" Tags="&lt;topology&gt;&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="5713" PostTypeId="1" AcceptedAnswerId="5715" CreationDate="2010-09-29T16:31:41.140" Score="3" ViewCount="159" Body="&lt;p&gt;This is (a slight paraphrase) of question 1.3.14 in Chang and Keisler's Model Theory book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Show that for each natural number $n$, there is a language $L_n$ and finite model $M_n$ of $L$ such that $M_n$ has precisely $n$ undefinable elements.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, an element $x\in M$ is definable if there is a (first order) formula in $L$, called $\phi$, such that $x$ is the unique element of $M$ satisfying $\phi$.  Of course, &quot;undefinable&quot; means &quot;not definable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is starred, indicating that it is more difficult than a standard problem in that book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang and Keisler remark that $n=1$ is the only difficult case.  In that spirit, here is the proof for all $n\neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L_n$ have a single 2 place predicate symbol (I'm thinking of $L_n = \{ &amp;lt; \}$).  Let $M_n$ be the partial order with minimum a and with elements $b_1,...,b_{n}$ with $a &amp;lt; b_i$ for all $i$ and the $b_i$ pairwise incomparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that a is definable:  it uniquely satisfies $\phi(x) =$ for all y, $x\leq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $n =0$, there are no $b_i$, and hence in this model, we have 0 undefinable elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &gt; 1$, then I claim that all the $b_i$ are undefinable.  The short answer is that any permutation of the $b_i$ can be extended to a unique isomorphism of $M_n$.  Hence, for any formula $\phi$, we have $\phi(b_i)$ iff $\phi(b_j)$ for all $b_j$.  Thus, no $\phi$ can single out any particular $b_i$, so each $b_i$ is undefinable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof fails completely for $n=1$, for then $b_1$ is the unique element that satisfies &quot;b_1 is not a&quot;.  Or more in the spirit of first order logic, $b_1$ is the unique element satisfying $\phi(x) =$ there is a $y$ such that for all $z$, $y&amp;lt; z$ and $x$ is not equal to $y$.&quot;  Incidentally, this proves that any such model that works for $n=1$ must have at least 3 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is an example of a language with finite model having precisely one undefinable element?  Is the smallest cardinality of such a model known?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-09-29T16:38:47.560" LastActivityDate="2010-11-23T12:00:56.430" Title="What is an example of a finite model in first order logic having a unique undefinable element?" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="3" />
  <row Id="5714" PostTypeId="2" ParentId="5712" CreationDate="2010-09-29T16:35:54.333" Score="5" ViewCount="0" Body="&lt;p&gt;I would suggest the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atiyah%E2%80%93Singer_index_theorem&quot; rel=&quot;nofollow&quot;&gt;Atiyah–Singer index theorem&lt;/a&gt; if this counts as a PDE-related topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a theorem about elliptic differential operators on compact manifolds. The original proof uses K-theory.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-29T16:35:54.333" CommentCount="1" />
  <row Id="5715" PostTypeId="2" ParentId="5713" CreationDate="2010-09-29T16:54:21.850" Score="6" ViewCount="0" Body="&lt;p&gt;If your languages necessarily have the = relation, which is a common assumption, then the case n=1 is impossible in a finite model. The reason is that if a model $M$ has $k$ elements $x_1$, $x_2$, ... $x_k$ for finite $k\gt 1$ and each $x_i$ is defined by $\varphi_i(x)$ for $i\lt k$, then the remaining element $x_k$ is defined by the formula $\neg\varphi_1(x)\wedge\cdots\wedge \neg\varphi_{k-1}(x)$. If $M$ has only one element, then it is defined by the formula $x=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we do not insist that $=$ is in the language, then let $L$ be the empty language, having no relations at all. In this case, there are no atomic formulas and hence no well-formed formulas to define elements, so a one-point model has exactly one undefinable element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we allow infinite models, then we can easily arrange to have exactly one undefinable element, even in a language with $=$. For example, consider the lanuage with infinitely many constant symbols, and have a model where all these constants are interpreted by different elements, and there is one extra un-named object. That extra object will be the only non-definable element, even when $=$ is in the language.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-09-29T16:54:21.850" CommentCount="1" />
  <row Id="5716" PostTypeId="2" ParentId="5713" CreationDate="2010-09-29T16:58:13.757" Score="5" ViewCount="0" Body="&lt;p&gt;You can't have equality as a relation in the language that is interpreted as equality in the model, otherwise you can define the undefinable element as that (one) which is not equal to any of the definable ones.  Most languages in common use have equality so it is hard to think of a natural example that describes, e.g., algebraic structures or combinatorial objects. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-09-29T17:03:41.610" LastActivityDate="2010-09-29T17:03:41.610" CommentCount="1" />
  <row Id="5717" PostTypeId="1" AcceptedAnswerId="5719" CreationDate="2010-09-29T17:10:19.397" Score="0" ViewCount="204" Body="&lt;p&gt;Looking for a hint on show to show convexity in a set....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f\colon \mathbb{R}^n \rightarrow \mathbb{R}$ be a convex function and let $c$ be some constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the set $s=\{x \in \mathbb{R}^n \mid f(x) \le c \}$ is convex.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-30T18:06:54.723" LastActivityDate="2010-09-30T18:06:54.723" Title="How to show a set is convex" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5718" PostTypeId="1" AcceptedAnswerId="5751" CreationDate="2010-09-29T17:12:34.920" Score="3" ViewCount="82" Body="&lt;p&gt;Let $K$ be a number field and $L$ an infinite algebraic extension of $K$. Fix a non-trivial absolute value $v$ on $K$ (so $v$ is induced either by an embedding into the complex numbers or by a prime ideal in the ring of integers of $K$). If $K_v$ is the corresponding completion and $\overline{K}_v$ a choice of algebraic closure of $K_v$, then the extensions of $v$ to an absolute value on $K$ are in bijection with the $Gal(\overline{K}_v/K_v)$-orbits of $Hom_{K-alg}(L,\overline{K}_v)$ (this is described in detail in, e.g., Neukirch's &lt;em&gt;Algebraic Number Theory&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question regards the case when $v$ is non-Archimedean, arising from a prime ideal $\mathfrak{p}$ with residue characteristic $p$. In this case, &lt;strong&gt;is there a bijection between prime ideals of $\mathscr{O}_L$ lying above $\mathfrak{p}$ and places of $L$ above $v$?&lt;/strong&gt; Since $\mathscr{O}_L$ is not generally going to be Dedekind (though it is a one-dimensional, integrally closed domain), we don't get an (additive) discrete valuation in the usual way from a prime ideal of $\mathscr{O}_L$, but if $w$ is a non-Archimedean absolute value on $L$ extending $v$, then by restricting to finite sub-extensions $L_i$ of $L/K$ we get sort of a coherent sequence of prime ideals $\mathfrak{p}_i$ in the integer rings of the $L_i$. Perhaps there is a way to convert this sequence of primes into a single prime ideal of $\mathscr{O}_L$ lying above each of the $\mathfrak{p}_i$? I thought maybe some kind of compactness\inverse limit type argument might work, but I'm not sure anymore...even so, if such an argument were to work, it would probably just show the existence of such a prime, as opposed to determining it uniquely. Alternatively, if I look at the maximal ideal of the valuation ring of $w$ in $L$ and intersect it with $\mathscr{O}_L$, I should get a (hopefully) non-zero prime ideal of $\mathscr{O}_L$ that maybe fits the bill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be the wrong approach entirely (and maybe the answer to my question is just &quot;no&quot;). The reason I'm interested is because, for example, in Washington's book on cyclotomic fields, he (in an appendix) defines the decomposition group of a prime ideal in an infinite Galois extension of number fields, but makes no mention of the decomposition group of a &lt;em&gt;place&lt;/em&gt; of an infinite algebraic extension of a number field. When one starts considering objects like the maximal unramified abelian $p$-extension of a $\mathbb{Z}_p$-extension of a number field, surely this (possibly more general) notion becomes relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would greatly appreciate if anyone could set me straight on this issue or point me in the direction of a reference where it's discussed. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2010-09-30T07:53:57.700" Title="What is the relationship between non-Archimedean places of infinite extensions of number fields and primes in the ring of integers?" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="5719" PostTypeId="2" ParentId="5717" CreationDate="2010-09-29T17:13:00.920" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Well, just write down a convex combination of elements in s and verify that it belong to s. You will find the convexity of f useful for this.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-09-29T17:13:00.920" CommentCount="5" />
  <row Id="5720" PostTypeId="1" AcceptedAnswerId="5726" CreationDate="2010-09-29T18:12:31.190" Score="0" ViewCount="78" Body="&lt;p&gt;What are linear combinations of a column space?   What equation to they satisfy? Why are they useful in leaner algebra? &lt;/p&gt;&#xA;" OwnerUserId="1992" LastActivityDate="2010-09-29T19:16:18.000" Title="What are linear combinations of a column space? " Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5721" PostTypeId="1" AcceptedAnswerId="5724" CreationDate="2010-09-29T18:13:53.370" Score="0" ViewCount="106" Body="&lt;p&gt;What is the homogeneous problem?   What is the purpose of null space of a vector in this context? &lt;/p&gt;&#xA;" OwnerUserId="1992" LastEditorUserId="1992" LastEditorDisplayName="" LastEditDate="2010-09-29T18:29:01.080" LastActivityDate="2010-09-29T19:03:36.190" Title="What is the homogeneous problem? " Tags="&lt;linear-algebra&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5722" PostTypeId="2" ParentId="5648" CreationDate="2010-09-29T18:17:02.123" Score="2" ViewCount="0" Body="&lt;p&gt;In contrast to William, I think the answer is yes.  I had to prove something similar once; here is a suitably adapted argument.  It might be more complicated than necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, reduce to the case that $(\Omega_1, \mu_1), (\Omega_2, \mu_2)$ are finite measure spaces and $h$ is bounded.  Let $\mathcal{P}$ be the set of all functions on $\Omega_1 \times \Omega_2$ of the form $F(x,y) = f(x) g(y)$ where $f$ is bounded and measurable and $g$ is simple; such functions can be written in the form you seek.  Let $\mathcal{Q}$ be the linear span of $\mathcal{P}$; these functions can also be written in the desired form.  Let $\mathcal{L}$ be the closure of $\mathcal{Q}$ in $L^1(\mu_1 \times \mu_2)$ and let $\mathcal{L}_b$ be the bounded functions from $\mathcal{L}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\mathcal{L}_b$ is a vector space which is closed under bounded convergence (by the dominated convergence theorem), contains the constants and contains $\mathcal{P}$.  $\mathcal{P}$ is closed under multiplication and contains all functions of the form $1_{A \times B}$ with $A \subset \Omega_1$, $B \subset \Omega_2$ measurable; the collection of such sets $A \times B$ generates the product $\sigma$-algebra.  By the functional version of the Dynkin $\pi$-$\lambda$ theorem (references below), $\mathcal{L}_b$ contains all bounded measurable functions; in particular it contains $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ is in $\mathcal{L}$ which is the $L^1$ closure of $\mathcal{Q}$, there is a sequence $\mathcal{Q} \ni h_n \to h$ in $L^1$.  Then some subsequence converges almost everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the functional $\pi$-$\lambda$ theorem, see Theorem 8.2 of &lt;a href=&quot;http://www.math.ucsd.edu/~bdriver/Cornell%20Summer%20Notes%202010/Lecture_Notes/Probability%20Tools%20with%20Examples.pdf&quot; rel=&quot;nofollow&quot;&gt;Bruce Driver's probability notes&lt;/a&gt;. A reference is also given to C. Dellacherie, &lt;i&gt;Capacit&amp;eacute;s et processus stochastiques&lt;/i&gt;, page 14.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-29T18:17:02.123" CommentCount="2" />
  <row Id="5723" PostTypeId="2" ParentId="155" CreationDate="2010-09-29T18:29:31.810" Score="4" ViewCount="0" Body="&lt;p&gt;The techniques used for indefinite integration of elementary functions are actually quite simple in the transcendental (vs. algebraic) case, i.e. the case where the integrand lies in a purely transcendental extension of the field of rational functions $\rm\mathbb C(x)$. Informally this means that the integrand lies in some tower of fields $\rm\mathbb C(x) = F_0 &amp;lt; \cdots &amp;lt; F_n = \mathbb C(x,t_1,\ldots,t_n)$ which is built by adjoining an exponential or logarithm of an element from the prior field, i.e $\rm\  t_{i+1} =\: exp(f_i)\ $ or $\rm\ t_{i+1} =\: log(f_i)\ $ for $\rm\ f_i \in F_i$ where $\rm t_{i+1}$ is transcendental over $\rm F_i\:.\ $ For example $\rm\ exp(x),\ log(x)\ $ are transcendental over $\rm\mathbb C(x)$ but $\rm\ exp(2\ log(x)) = x^2\ $ is not. Now, because $\rm\ F_{i} = F_{i-1}(t_{i})$ is transcendental it has particularly simple structure, viz. it is isomorphic to the field of rational functions in one indeterminate $\rm\:t_i\:$ over $\rm\ F_{i-1}\ $. In particular, this means that one may employ well-known rational function integration, techniques such as expansions into partial fractions. This, combined with a simple analysis of the effect of differentiation on the degree of polynomials $\rm\  p(t_i)$, quickly leads to the fundamental result of Liouville on the structure of antiderivatives, namely they must lie in the same field $\rm F$ as the integrand except possibly for the addition of constant multiples of log's over $\rm F$. With this structure theorem in hand, the transcendental case reduces to elementary computations in rational function fields. This transcendental case of the algorithm is so simple that it may be  easily comprehended by anyone who has mastered a first course in abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the full-blown algebraic case of the algorithm requires nontrivial results from the theory of algebraic functions. Although there are some simple special case algorithms for sqrt's and cube-roots (Davenport, Trager) the general algorithm requires deep results about points of finite order on abelian varieties over finitely generated ground fields. This algebraic case of the integration algorithm was discovered by Robert Risch in 1969 - who did his Berkeley Ph.D. on this topic (under Max Rosenlicht).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a very nice introduction to the theory see Max Rosenlicht's Monthly paper, available &lt;a href=&quot;http://www.jstor.org/stable/2318066&quot; rel=&quot;nofollow&quot;&gt;from JSTOR&lt;/a&gt; and &lt;a href=&quot;http://vigo.ime.unicamp.br/~fismat/Integration.pdf&quot; rel=&quot;nofollow&quot;&gt;also here&lt;/a&gt;. This exposition includes a complete proof of the Liouville structure theorem along with a derivation of Liouville's classic criterion for $\rm\int f(z)\: e^{g(z)}\: dz\ $ to be elementary, for $\rm\: f(z),\: g(z)\in \mathbb C(x)$. For algorithms see &lt;a href=&quot;http://dspace.mit.edu/handle/1721.1/15391&quot; rel=&quot;nofollow&quot;&gt;Barry Trager's 1984 MIT thesis&lt;/a&gt; and Manual Bronstein: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540214933&quot; rel=&quot;nofollow&quot;&gt;Symbolic Integration I: Transcendental Functions.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I implemented the integration algorithm in Macsyma (not the much older Maxima) so, perhaps due to this experience,  my judgment of simplicity might be slightly biased. However, the fact that the basic results are derived in a handful of pages in Rosenlicht's paper yields independent evidence for such claims.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T18:36:49.347" LastActivityDate="2010-09-29T18:36:49.347" />
  <row Id="5724" PostTypeId="2" ParentId="5721" CreationDate="2010-09-29T18:47:51.987" Score="5" ViewCount="0" Body="&lt;p&gt;The answer to what I think you're asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The homogeneous problem is: &quot;Given a matrix $A$, find all solutions to the problem $Ax = 0$, where $x$ is a vector of appropriate dimension, and '0' is the vector of all zeros&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nullspace of $A$ is precisely the set of all such solutions. There is always at least one solution to the homogeneous problem. Indeed, $x=0$ (the vector of all zeros) is always a solution. In the case where $A$ is a square and invertible matrix, $x=0$ is the only solution. In general, there can be other solutions, and the set of all solutions (the Nullspace) is actually a subspace. To see why, notice that if $x_1$ and $x_2$ both satisfy $Ax=0$, then so does $a_1 x_1 + a_2 x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question. Best of luck!&lt;/p&gt;&#xA;" OwnerUserId="1356" LastActivityDate="2010-09-29T18:47:51.987" CommentCount="2" />
  <row Id="5725" PostTypeId="2" ParentId="1099" CreationDate="2010-09-29T18:50:41.613" Score="4" ViewCount="0" Body="&lt;p&gt;Isaac, the auxiliary question that we should ask ourselves before answering your question is: &lt;strong&gt;what is the &quot;right&quot; (i. e., most mathematically relevant/interesting) generalization of the Mandelbrot set to other parametrized families of functions?&lt;/strong&gt; Of course, you can take some arbitrary holomorphic family of functions $f_c(z)$ that depends holomorphically on the parameter $c$, and for each $c$, pick some starting value $z_0$ (which depends on $c$ in an essentially arbitrary way) and color the plane somehow based on the long-term behavior of the iterates. However, as some of the other posts show, you typically end up with objects that don't really resemble the Mandelbrot set, so the status of this construction as the &quot;right&quot; generalization of the M-set is very dubious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I think is the right answer to my auxiliary question: the boundary of the standard Mandelbrot set is the &lt;em&gt;bifurcation locus&lt;/em&gt; for the holomorphic family $f_c(z) = z^2 + c$: it's the set of $c$ at which the long-term dynamical behavior of $f_c$ is not continuous. (For example, if you cross the boundary and move from the inside of the M-set to the outside, you go from having a finite-valued attracting orbit to not having one (with only the fixed point at infinity attracting)). This concept applies equally well to other holomorphic families, and this is what I think is the right generalization of the M-set: the bifurcation locus of a family $f_c(z)$ of holomorphic functions that depends holomorphically on $c$. More generally, we can think of not just the bifurcation locus, but a more complete &quot;catalog&quot; of the dynamics of the family: a partition of the complex plane into the bifurcation set plus the union of a bunch of open sets in which the long-term dynamical behavior varies continuously (which for the M-set, would be a partition of the plane into the outside, the boundary, and all of the cardiods and bulbs (including those in smaller copies of the M-set found near the boundary)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer &lt;em&gt;your&lt;/em&gt; question: yes, these objects have been studied in some generality, but I'm not an expert in the field and thus can't tell you everything about them. However, for example, it is known that every nonempty bifurcation locus contains copies of the boundary of the standard M-set (or the bifurcation locus of the family $z\mapsto = z^d + c$ for some $d$, the &quot;degree-$d$ M-set&quot;), and moreover, that these copies are dense in every bifurcation locus. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-09-29T18:50:41.613" />
  <row Id="5726" PostTypeId="2" ParentId="5720" CreationDate="2010-09-29T18:59:11.673" Score="3" ViewCount="0" Body="&lt;p&gt;A linear combination of vectors, say ${ x_1, \ldots , x_n}$, of a vector space on some field $\mathbb{K}$ is $$\sum_{i = 1}^{n} \lambda_i x_i$$, where $\lambda_i \in \mathbb{K}$ and not all of them are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kind of combinations are important in linear algebra because you can define &quot;linear independence&quot; and &quot;generators&quot;, which, if combined, give you a &quot;basis&quot; for the vector space (you can find all the definitions here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, ${x_1, \ldots , x_n }$ vectors are said to be independent, if $$\sum_{i = 1}^{n} \lambda_i x_i = 0 \Rightarrow \lambda_i = 0 \quad \forall i$$. This definition, as you can see, uses the concept of linear combination. On the same note, you can say that hose vectors generate the space $V$ if $\forall x \in V \quad x = \sum_{i = 1}^{n} \lambda_i x_i$ for appropriate $\lambda_i \in \mathbb{K}$. As you can see, also this definition uses the concept of linear combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the rest of your question, I can't really grasp its meaning.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-09-29T19:16:18.000" LastActivityDate="2010-09-29T19:16:18.000" CommentCount="2" />
  <row Id="5727" PostTypeId="2" ParentId="5721" CreationDate="2010-09-29T19:03:36.190" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ For $\rm A\:$ linear, $\rm\ A\:X_1 = B = A\:X_2 \ \iff\ 0 \:=\: A\:X_1 - A\:X_2 = A\:(X_1-X_2)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the general solution of $\rm\ \ \ \:A\:X = B\ $ is the sum of any particular solution plus a solution of the associated &quot;homogeneous&quot; equation $\rm\ A\:X = 0\:$. This property holds true for every linear operator, e.g. for matrices, linear differential equations, linear recurrences, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-29T19:03:36.190" />
  <row Id="5728" PostTypeId="2" ParentId="5676" CreationDate="2010-09-29T19:07:48.493" Score="14" ViewCount="0" Body="&lt;p&gt;This was an interesting challenge, to try and come up with an estimate for this sum in an elementary way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimate I got was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $C$ which appears to be close to $-0.207$ (which I guess will be $\zeta(-1/2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By $a_{n} \sim b_{n}$ I mean $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe here is a completely elementary proof of that fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the inequality for $x &gt; 0$ and $k &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{x} \le \frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows easily by the arithmetic mean $\ge$ geometric mean inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{k}^{k+1} \sqrt{x} \ dx \le \int_{k}^{k+1} (\frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}) \ dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{(k+1)^2 - k^2}{4\sqrt{k}} + \frac{\sqrt{k}}{2} = \sqrt{k} + \frac{1}{4\sqrt{k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\sum_{k=1}^{n-1} \int_{k}^{k+1} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{n} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\sqrt{k}} &amp;lt; \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sqrt{k} -\sqrt{k-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=1}^{n-1} \frac{1}{4\sqrt{k}} &amp;lt; \frac{\sqrt{n-1}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \frac{\sqrt{n-1}}{2} + \sum_{k=1}^{n-1} \sqrt{k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $$S_{n} = \sum_{k=1}^{n} \sqrt{k}$$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{n} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \sqrt{n} - \frac{\sqrt{n-1}}{2} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \frac{\sqrt{n}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $$G_{n} = S_{n} - \frac{2}{3} n^{3/2} - \frac{\sqrt{n}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $$G_{n} \ge -\frac{2}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily show that (using tedious but not too complicated algebra&lt;sup&gt;*&lt;/sup&gt;) $$G_{n+1} &amp;lt; G_{n}$$ and so $G_{n}$ is a convergent sequence, as it is bounded below and monotonically decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there exists a constant $C$ (the limit of $G_{n}$) such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt; For the sake of completeness, we show that $G_{n+1} &amp;lt; G_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $$6(G_{n}-G_{n+1}) = 4(n+1)\sqrt{n+1} + 3\sqrt{n+1} - 6\sqrt{n+1} - 4n\sqrt{n} - 3\sqrt{n}$$&#xA;$$ = \sqrt{n+1} + 4n(\sqrt{n+1} -\sqrt{n}) - 3\sqrt{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $\sqrt{n+1} + \sqrt{n}$ does not change the sign, so we look at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{n+1}(\sqrt{n+1} + \sqrt{n}) + 4n - 3\sqrt{n}(\sqrt{n+1} + \sqrt{n})$$&#xA;$$ = 2n+1 - 2\sqrt{n^2 + n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$ (2n+1)^2 = 4n^2 + 4n + 1 &gt; 4n^2 + 4n = (2\sqrt{n^2+n}) ^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2n+1 &gt; 2\sqrt{n^2+n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_{n} &gt; G_{n+1}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-29T23:04:12.800" LastActivityDate="2010-09-29T23:04:12.800" CommentCount="2" />
  <row Id="5729" PostTypeId="2" ParentId="5720" CreationDate="2010-09-29T19:12:42.697" Score="2" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Column_space&quot; rel=&quot;nofollow&quot;&gt;column space&lt;/a&gt; of a matrix is simply the space spanned by all linear combinations of the columns of the matrix. If $\rm \:A_i\:$ denotes the $\rm i$'th column of $\rm \:A\:$ then $\rm \ b = Ax = A_i\: x_1 + \cdots + A_n\: x_n$ has a solution iff $\rm\:b\:$ lies in the column space. Said in geometric language, the column space spans the image (range) of the linear map corresponding to the matrix $\rm\: A\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-09-29T19:12:42.697" />
  <row Id="5731" PostTypeId="2" ParentId="5697" CreationDate="2010-09-29T19:53:45.247" Score="2" ViewCount="0" Body="&lt;p&gt;First let's state the special case of Burnside's lemma that is relevant here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  Let $G$ be a finite group acting on a finite set $X$.  The number of ways to color the elements of $X$ with $z$ different colors, up to the action of $G$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{|G|} \sum_{g \in G} z^{c(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c(g)$ is the number of cycles in the cycle decomposition of $g$ acting on $X$.  (&lt;a href=&quot;http://qchu.wordpress.com/2009/06/16/gila-iii-the-orbit-counting-lemma-and-baby-polya/&quot; rel=&quot;nofollow&quot;&gt;Proof&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $X$ is the set of faces of a hypercube.  In $n$ dimensions there are $2n$ such faces.  $G$ is the subgroup of index $2$ in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperoctahedral_group#Subgroups&quot; rel=&quot;nofollow&quot;&gt;hyperoctahedral group&lt;/a&gt; with determinant $1$, also known as the Coxeter group $D_n$.  So our job now is to count, for each $k$, the number of elements of $D_n$ with $k$ cycles in the action on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that to analyze the action of $D_n$ on the faces it suffices to analyze the action of $D_n$ on the midpoints of the faces.  But these are precisely the $2n$ points $(0, 0, ... \pm 1, ..., 0, 0)$, so writing the elements of $D_n$ as signed permutation matrices is very well-suited to analyzing their action on these points; in particular, it suffices to figure out the answer for a single signed cycle.  But this turns out to be very simple: there are either one or two cycles depending on whether the product of the signs is $-1$ or $+1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It might be helpful here to play with a specific example.  Consider $\left[ \begin{array}{cc} 0 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; -1 \\ -1 &amp;amp; 0 &amp;amp; 0 \end{array} \right]$ acting on the six points $(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)$ to get a feel for what's going on in the general case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I think it's easiest to work with generating functions because the combinatorics get a little messy.  Begin with the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{m \ge 0} Z(S_n) t^n = \exp \left( z_1 t + \frac{z_2 t^2}{2} + \frac{z_3 t^3}{3} + ... \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Z(S_n)$ is the &lt;a href=&quot;http://qchu.wordpress.com/2009/06/21/gila-v-the-polya-enumeration-theorem-and-applications/&quot; rel=&quot;nofollow&quot;&gt;cycle index polynomial&lt;/a&gt; for the action of $S_n$ on $\{ 1, 2, ... n \}$.  Each $z_i$ is the term that controls cycles of length $i$.  We want to modify this generating function so that it tells us how the cycles in $D_n$ work.  There are $2^i$ signed cycles of length $i$ which come in two flavors: half of them have positive sign product (two unsigned cycles) and half of them have negative sign product (one unsigned cycle), so to keep track of the total number of unsigned cycles we should replace $z_i$ with $2^{i-1} z^2 + 2^{i-1} z$.  We also have to keep in mind that the determinant of a signed cycle is its sign product multiplied by $(-1)^{i+1}$, and we only want permutations with determinant $1$.  So the generating function we want is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} f_n(z) \frac{t^n}{n!} = \frac{1}{2} \left( \exp \left( \sum_{i \ge 1} \frac{2^{i-1} z^2 + 2^{i-1} z) t^i}{i} \right) + \exp \left( \sum_{i \ge 1} (-1)^{i+1} \frac{2^{i-1} z^2 - 2^{i-1} z) t^i}{i} \right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_n(z) = \sum_{g \in D_n} z^{c(g)}$.  After some simplification the above becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} \frac{1}{|D_n|} f_n(z) t^n = \frac{1}{(1 - t)^{(z^2+z)/2}} + (1+t)^{(z^2-z)/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $z = 2$ gives, at last, the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} \frac{1}{|D_n|} f_n(2) t^n = \frac{1}{(1 - t)^3} + 1 + t.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for $n \ge 2$ we simply have $\frac{1}{|D_n|} f_n(2) = {n+2 \choose 2}$.  This is such a simple answer that there should be a direct proof of it.  I'll keep working on it.  (There is a rather straightforward proof of the corresponding result with &quot;hypercube&quot; replaced by &quot;simplex,&quot; so my guess is something along those lines is possible here.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-09-29T20:00:18.270" LastActivityDate="2010-09-29T20:00:18.270" CommentCount="1" />
  <row Id="5732" PostTypeId="2" ParentId="5697" CreationDate="2010-09-29T20:30:38.770" Score="3" ViewCount="0" Body="&lt;p&gt;Once someone else has done the hard work and come up with the answer,&#xA;as Qiaochu has done, then it's easy to prove it on an &lt;em&gt;ad hoc&lt;/em&gt;&#xA;basis as this interloper (me) will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an $n$-cube with faces coloured red and blue. Consider a set&#xA;of $n$ balls. Colour the $j$-th ball red, blue or green according to the&#xA;colours of the $j$-th pair of faces in the cube, that is if they both have&#xA;the same colour give the ball that colour, else if they have different&#xA;colours colour the ball green. Now a symmetry operation on the cube&#xA;will shuffle the balls so the number of each colour of balls will be the same.&#xA;Conversely given the $n$ coloured balls there will be various coloured&#xA;cubes giving rise to the colour distribution of the balls, but these&#xA;will all be equivalent under the full symmetry group of the cube.&#xA;However for $n\ge 2$ each $2$-coloured cube has a non-rotational symmetry,&#xA;so any two coloured cubes giving rise to the same colour distribution&#xA;of balls are related by a rotation. Hence the number of $2$-coloured $n$-cubes&#xA;up rotation is the same as the number of triples $(r,b,g)$ of nonnegative&#xA;integers adding to $n$ and there are ${n+2\choose 2}$ of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can repeat this with $k$-colours for the cube. This time one needs&#xA;${k+1\choose 2}$ colours for the balls, and the argument ensuring that&#xA;rotations and the full symmetry group give the same answer requires&#xA;$n&gt;{k\choose 2}$ (why?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; (30/9/2010)&#xA;One can get the general result using these considerations. For&#xA;$n&gt;{k\choose 2}$ one gets&#xA;$${n+(k^2+k)/2-1\choose (k^2+k)/2-1}$$&#xA;colourings up to rotations. For $n\le {k\choose 2}$ there are $k$-colourings&#xA;having no improper (determinant $-1$) symmetries. But any such colouring&#xA;has no nontrivial symmetries at all. If two opposite faces have&#xA;the same colour, one can reflect through a hyperplane parallel to them.&#xA;So assume that no colour is opposite itself. Then if two pairs of opposite&#xA;faces have the same two colours between them, one can reflect in a plane&#xA;$x_i=x_j$or $x_i=-x_j$. Hence in these &quot;special&quot; colourings each pair&#xA;of opposite faces has a distinct pair of distinct colours. Up to symmetry&#xA;there are&#xA;$${(k^2-k)/2\choose n}$$&#xA;of these special colourings. So for $n\le{k\choose 2}$ there are&#xA;$${(k^2-k)/2\choose n}+{n+(k^2+k)/2-1\choose (k^2+k)/2-1}$$&#xA;up to rotational symmetries.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-09-30T18:04:50.063" LastActivityDate="2010-09-30T18:04:50.063" CommentCount="1" />
  <row Id="5733" PostTypeId="2" ParentId="120" CreationDate="2010-09-29T20:47:39.027" Score="-1" ViewCount="0" Body="&lt;p&gt;1 is not a prime. Is 1 a prime ,so it is the only prime. Every number have primefactor 1 ... so &quot;the primes&quot; are prime and not prime to the same time.&lt;/p&gt;&#xA;" OwnerUserId="2208" LastActivityDate="2010-09-29T20:47:39.027" CommentCount="2" />
  <row Id="5734" PostTypeId="1" AcceptedAnswerId="5750" CreationDate="2010-09-29T21:31:58.827" Score="1" ViewCount="67" Body="&lt;p&gt;Suppose $G$ is a compact topological group.  We can construct the profinite completion of $G$; let's call this $\Gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Assuming that we know nothing about the (original) topology of $G$ other than that it is compact, is there anything we can say linking the topology of $G$ to the topology of $\Gamma$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the answer to this question is &quot;no&quot; since it seems to me that we usually regard $G$ as an abstract group when thinking about constructing its profinite completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If not (and, like I said, I assume the answer to (1) is &quot;no&quot;) is there a way to construct a profinite completion (or something like this) of $G$ that takes into account the topology we already have?&lt;/p&gt;&#xA;" OwnerUserId="2209" LastActivityDate="2010-09-30T05:31:12.057" Title="Relationship between topology of a compact group and the topology of its profinite completion." Tags="&lt;group-theory&gt;&lt;topology&gt;" AnswerCount="1" />
  <row Id="5735" PostTypeId="2" ParentId="120" CreationDate="2010-09-29T21:34:28.633" Score="3" ViewCount="0" Body="&lt;p&gt;It's worth emphasizing that, in addition to unique factorization, there are also somewhat deeper structural reasons underlying the convention that $\:1\:$ is neither prime nor a prime ideal, but $\:0\:$ is. Below I discuss the motivations for these differing conventions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important motivation for including the zero ideal as prime is that this facilitates powerful reductions. For example, in many ring theoretic problems involving an ideal $\rm\; I\;$, one can reduce to the case $\rm\;I = P\;$ prime, and then reduce to $\rm\;R/P\;$, therefore reducing to the case when the ring is a domain. In this case one simply says that one can factor out $\rm\; P\;$, so wlog assume $\rm\; P = 0\;$ is prime, hence the ring is a domain. For example, at the end of this post is an excerpt from Kaplansky's classic textbook &quot;Commutative Rings&quot;, section 1-3: G-Ideals, Hilbert Rings, and the Nullstellensatz, where I've explicitly highlighted a few prototypical examples of such reductions - cf. &lt;strong&gt;reduce to...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have solid evidence for the utility of the convention that the zero ideal is prime.&#xA;So why don't we adopt the same convention for the unit ideal $1$ or, equivalently, why don't we permit the zero ring as a domain?  There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $1 = 0\:$. More importantly, it implies that the  unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\;\rm R = 0)$ to the ubiquitous arguments involving units and unit groups. More generally it's worth emphasizing that the usual rules for equational logic are not complete for &lt;em&gt;empty structures&lt;/em&gt;. That is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/3e8fac43081fcbf8&quot; rel=&quot;nofollow&quot;&gt;thread&lt;/a&gt; for further details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the mentioned Kaplansky excerpt on reduction to domains by factoring out prime ideals. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\rm\; I\;$ be any ideal in a ring $\rm\; R\;$. We write $\rm\: R^{*}\:$ for the quotient ring $\rm\; R/I\;$. &#xA;  In the polynomial ring $\rm\; R[x]\;$ there is a smallest extension $\rm\; IR[x]\;$ of $\rm\; I\;$. The quotient ring $\rm\; R[x]/IR[x]\;$ is in a natural way isomorphic to $\rm\; R^*[x].\;$ In treating many problems, we can in this way &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; I = 0\;$, and we shall often do so.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;THEOREM 28.&lt;/strong&gt; $\;$ Let $\rm\; M\;$ be a maximal ideal in $\rm\; R[x]\;$ and suppose that &#xA;  the contraction $\rm\; M \cap R\ =\ N\;$ is maximal in $\rm\; R\;.\ $ Then $\rm\; M\;$ can be generated by $\rm\; N\;$ and one more element $\rm\; f\:.\ $ We can select $\rm\; f\;$ to be a monic polynomial which maps mod $\rm\; N\;$ into an irreducible polynomial over the field $\rm\; R/N\;$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\;$ We can &lt;strong&gt;reduce to the case&lt;/strong&gt; $\rm\; N = 0,\;$ i. e., $\rm\; R\;$ a field, and then &#xA;  the statement is immediate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;THEOREM 31.&lt;/strong&gt; $\;$ A commutative ring $\rm\; R\;\;$ is a Hilbert ring if and only if &#xA;  the polynomial ring $\rm\; R[x] \;\;$ is a Hilbert ring. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; $\;$ If $\rm\; \;\rm R[x]\;$ is a Hilbert ring, so is its homomorphic image $\rm\; R\;$. Conversely, assume that $\rm\; R\;$ is a Hilbert ring. Take a G-ideal $\rm\: Q\:$ in $\rm\; R[x]\;$; we must prove that $\rm\: Q\:$ is maximal. Let $\rm\: P = Q \cap R\:$; we can &lt;strong&gt;reduce the problem to the case&lt;/strong&gt; $\rm\ P = 0\:,\ $ which, incidentally, makes $\rm\ R\ $ a domain. Let $\rm\; u\;$ be the image of $\rm\; x\;$ in the natural homomorphism $\rm\ R[x] \ \to\  R[x]/Q\ .\ \ $ Then $\rm\; R[u]\;$ is a G-domain. $\ $ By Theorem 23, $\rm\ u\ $ is algebraic over $\rm\ R\ $ and $\rm\ R\ $ is a G-domain. Since $\rm\:R\:$ is both a G-domain and a Hilbert ring, $\rm\ R\ $ is a field. $\ $ But this makes $\rm\ R[u] = R[x]/Q\ $ a field, $\ $ proving $\rm\ Q\ $ to be maximal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T05:25:50.567" LastActivityDate="2010-12-07T05:25:50.567" />
  <row Id="5736" PostTypeId="1" AcceptedAnswerId="5737" CreationDate="2010-09-29T21:58:52.043" Score="1" ViewCount="21" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone shed a light on how to solve something like this?&#xA;I've been looking for permutations but so far I found it very confusing.&#xA;As I'm not at home atm, I can't reach for my math books either for an answer.&#xA;Maybe you can help? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a 5x5 grid&lt;/li&gt;&#xA;&lt;li&gt;each cell can have 2 possible values: 0 or 1&lt;/li&gt;&#xA;&lt;li&gt;each resulting grid &quot;layout&quot; must be bilateral-symmetric ( on the vertical axis )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;What's the number of possible variations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="1969" LastEditorUserId="1969" LastEditorDisplayName="" LastEditDate="2010-09-29T22:06:30.000" LastActivityDate="2010-09-29T22:06:30.000" Title="Possible variations in a bilateral symmetric system" Tags="&lt;permutations&gt;" AnswerCount="1" />
  <row Id="5737" PostTypeId="2" ParentId="5736" CreationDate="2010-09-29T22:05:27.367" Score="0" ViewCount="0" Body="&lt;p&gt;You only need to consider all possible $3\times 5$ grids, from which you can get a symmetrical $5\times 5$ grid by mirroring the first two columns about the third. Each cell in the $3\times 5$ grid has 2 possibilities, independent of all the others, so the total number of possibilities is $2^{3\times 5} = 32768$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-09-29T22:05:27.367" CommentCount="1" />
  <row Id="5738" PostTypeId="1" AcceptedAnswerId="5741" CreationDate="2010-09-29T22:18:31.353" Score="1" ViewCount="185" Body="&lt;p&gt;This question is from the book How to Prove It and I'm having trouble getting started with it.  The book provides the hint &quot;first prove a &amp;lt; 0&quot; however I can't figure out how to get that far with the provided givens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a≠0 and b≠0 and a &amp;lt; 1/a &amp;lt; b &amp;lt; 1/b prove a &amp;lt; -1&lt;/p&gt;&#xA;" OwnerUserId="2211" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-09-30T10:19:41.543" LastActivityDate="2010-10-01T03:08:07.393" Title="help with a proof:  Given a≠0 and b≠0 and a &lt; 1/a &lt; b &lt; 1/b prove a &lt; -1" Tags="&lt;algebra-precalculus&gt;&lt;proof-strategy&gt;" AnswerCount="6" CommentCount="0" />
  <row Id="5739" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:25:16.040" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: Try to prove $a &amp;lt; 0$ by contradiction, i.e. assume $a &gt; 0$ and try to derive a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-29T22:25:16.040" />
  <row Id="5740" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:30:08.330" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: What does a&amp;lt;1/a say about a? What does b&amp;lt;1/b say about b?&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-09-29T22:30:08.330" />
  <row Id="5741" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:32:23.673" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose that $a$ were positive. Then, multiplying the string of ineqs. (doesn't change their direction) by $a$ gives $a^2 &amp;lt;1 &amp;lt; ba &amp;lt; a/b$. From here we see that $b&gt;1$ (why?), but then, from the last ineq., $b^2 a &amp;lt; a$ which is impossible (why?). Therefore $a&amp;lt;0$. Now multiply again by $a$ and conclude.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-29T22:32:23.673" CommentCount="2" />
  <row Id="5742" PostTypeId="2" ParentId="5738" CreationDate="2010-09-29T22:58:42.047" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ \ $ Multiplying $\rm\ a &amp;lt; 1/a\ \:$ by $\rm\ a^2\ $ yields $\rm\:\ a^3 &amp;lt; a\ $. By symmetry $\rm\ b^3 &amp;lt; b\ $. So both $\rm\: a\:$ and $\rm \:b\:$ must lie in the intervals where the graph of $\ x^3 - x &amp;lt; 0\:,\ $ i.e. either in  $(-\infty,-1)$ or in $(0,1)\:$. But $\rm\ a \in (0,1)\ \Rightarrow\ 1/a &gt; 1\ $ contra $\rm\ 1/a &amp;lt; b &amp;lt; 1\ $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CXxGs.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-29T23:05:34.013" LastActivityDate="2010-09-29T23:05:34.013" CommentCount="2" />
  <row Id="5743" PostTypeId="1" AcceptedAnswerId="5744" CreationDate="2010-09-29T23:16:52.600" Score="1" ViewCount="68" Body="&lt;p&gt;In the literature it is stated that to each quadratic irrational $\gamma=\frac{P+\sqrt{D}}{Q}$ there is a corresponding ideal $I=[|Q|/\sigma , (P+\sqrt{D})/\sigma]$, where $\sigma=1$, if $\Delta \equiv0$ mod $4$ and $\sigma=2$, otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the case of $\frac{2+\sqrt{13}}{3}$ the associated ideal must be $I=[3/2, (2+\sqrt{13})/2]$ which makes no sense, as $N(I)=3/2$ is supposed to be a rational integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong here? &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-09-30T02:19:15.367" Title="Ideals in a Quadratic Number Fields" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="5744" PostTypeId="2" ParentId="5743" CreationDate="2010-09-29T23:41:35.470" Score="2" ViewCount="0" Body="&lt;p&gt;Below is a proof of the standard equivalences between forms, ideals and numbers, excerpted from section 5.2, p. 225 of Henri Cohen's book &quot;A course in computational algebraic number theory&quot;. Note that your quadratic number is not of the form specified in this equivalence, viz. $\rm\ \tau = (-b+\sqrt{D})/(2a)\:,\:$ and $\rm\: 4\:a\:|\:(D-b^2)\:,\:\:$ i.e. $\rm\ a\:|\:N(a\tau)\:,\:$ a condition equivalent to the $\rm\mathbb Z$-module $\rm\ a\:\mathbb Z + a\tau\ \mathbb Z\ $ being an ideal when $\rm\:D\:$ and $\rm\:b\:$ have the same parity, e.g. see &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/ant/ant02.pdf&quot; rel=&quot;nofollow&quot;&gt;Proposition 2.8 p.18 in Franz Lemmermeyer's notes.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/eIFwF.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/bs8E6.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-30T02:19:15.367" LastActivityDate="2010-09-30T02:19:15.367" />
  <row Id="5745" PostTypeId="1" AcceptedAnswerId="5771" CreationDate="2010-09-30T00:27:56.233" Score="-1" ViewCount="217" Body="&lt;p&gt;In Quaternion mathematics, normally $$i ^ {2} = j ^ {2} = k ^ {2} = i*j*k = {-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am looking at a different type of Formula where $$i ^ {j ^ k} = {-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit again, since somehow this remains elusive to explain...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With functions, $f$ and all permutations of the subset variables $N$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $i$ is used in cases of non-real $f(x)=N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $i ^ j$ is used in cases of non-real $f(x,y)=N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $i ^ {j ^ k}$ is used in cases of non-real $f(x,y,z) = N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course must obey other permutations of $x$, $y$, and $z$ such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $k ^ i$ is used in cases of non-real $f(z,x) = N$; etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;EDIT NOT A RING&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have given a bit more thought about this, and since $x$, $y$ and $z$ are possibly inter-changable depending on the orientation in real space, consider the possibility:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i ^ {j ^ k} = j ^ {k ^ i} = k ^ {i ^ j} = -1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would be an &quot;exponential-type ring&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;EDIT MAYBE IT IS A RING?&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some important fundamental differences (or formulae) I should be looking at in how i, j and k compare? (more importantly is the second formula even possible?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT TO FURTHER ELABORATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider in the second example a hypothetical anti-DeSitter space, where the real spatial directions &lt;code&gt;X, Y, and Z&lt;/code&gt; are real components, and &lt;code&gt;i, j and k&lt;/code&gt; are imaginary spatial directions that associate to each one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i^2 = -1$; for $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j^2 = -1$; for $y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k^2 = -1$; for $z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, I am looking at a &lt;em&gt;different&lt;/em&gt; approach to &lt;code&gt;Riemann Tensors&lt;/code&gt;, my approach is to use exponentiation, rather than multiply. I am hoping someone that has experience with this type of thing can point out glaring flaws in this type of approach, and apologies in advance if I have errors in the reasoning here, its been some time since I've worked out anything this complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have further details &lt;a href=&quot;http://physicsmathforums.com/showthread.php?t=2343&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; &lt;a href=&quot;http://physicsmathforums.com/showthread.php?t=2416&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://physicsmathforums.com/showthread.php?t=2417&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; regarding my problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT FOR SIMPLE CASE EXAMPLE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What values are possible or valid for $a$, $b$, and $c$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(ai)^{(bj)^{(ck)}} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $i^2 = j^2 = k^2 = -1$&lt;/p&gt;&#xA;" OwnerUserId="2213" LastEditorUserId="2213" LastEditorDisplayName="" LastEditDate="2010-09-30T10:54:05.580" LastActivityDate="2010-10-01T03:33:23.723" Title="Differences in Quaternion properties?" Tags="&lt;complex-numbers&gt;&lt;quaternions&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="5746" PostTypeId="1" CreationDate="2010-09-30T03:29:43.493" Score="1" ViewCount="207" Body="&lt;p&gt;How do I use second order Taylor method to solve a system of non-linear equations? Is there a good reference that gives details? I found mentions of it in a dozen of numerical analysis books, but no examples&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, $f:\mathbb{R}^n \to \mathbb{R}^m$, solve $f(\mathbf{x})=\mathbf{0}$ using second order Taylor expansion of $f$ around initial guess $\mathbf{x_0}$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-30T04:00:18.637" LastActivityDate="2010-09-30T16:18:11.627" Title="Second order Taylor method to solve system of equations" Tags="&lt;sequences-and-series&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5747" PostTypeId="1" AcceptedAnswerId="5748" CreationDate="2010-09-30T04:28:16.190" Score="1" ViewCount="149" Body="&lt;p&gt;A variation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Roots_of_unity&quot; rel=&quot;nofollow&quot;&gt;Root of Unity&lt;/a&gt; problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find all possible answers to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^n = i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $$i^2 = -1$$&lt;/p&gt;&#xA;" OwnerUserId="2213" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-09-30T05:26:20.187" LastActivityDate="2010-09-30T09:45:10.913" Title="How can you find the complex roots of i?" Tags="&lt;complex-numbers&gt;&lt;roots&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="5748" PostTypeId="2" ParentId="5747" CreationDate="2010-09-30T04:38:09.700" Score="4" ViewCount="0" Body="&lt;p&gt;If the polar form of $z$ is $z=r(\cos\theta + i\sin\theta)$, there are $n$ distinct solutions to the equation $w^n = z$: $w=\sqrt[n]{r}(\cos\frac{\theta +2\pi k}{n}+ i \sin\frac{\theta +2\pi k}{n})$, where $k=0,1,...,n-1$. In your case, $z=i$, whose polar form is given by $r=1$, $\theta = \pi /2$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-30T04:38:09.700" />
  <row Id="5749" PostTypeId="2" ParentId="5747" CreationDate="2010-09-30T04:41:44.857" Score="2" ViewCount="0" Body="&lt;p&gt;Generally, the answers would be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[n]{i}\omega_n^j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega_n=\exp\left(\frac{2i\pi}{n}\right)$ is a root of unity, and $j=0\dots n-1$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-09-30T04:41:44.857" />
  <row Id="5750" PostTypeId="2" ParentId="5734" CreationDate="2010-09-30T05:31:12.057" Score="3" ViewCount="0" Body="&lt;p&gt;In principle, the answer to question (1) should be &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for question (2), it seems that a basic construction would be to form a &quot;profinite completion&quot; of the topological group $G$ by taking the inverse limit of all quotients $G/N$, where $N$ ranges over all finite-index &lt;em&gt;open&lt;/em&gt; normal subgroups of $G$.  The resulting completion $\hat{G}$ will have the property that the canonical function $G\to\hat{G}$ is continuous, and it should be somehow universal with respect to this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $\mathbb{Z}_2^\infty$ be the direct sum of infinitely many copies of $\mathbb{Z}_2$, which is a subspace of the infinite product $\mathbb{Z}_2^\omega$.  Then $\mathbb{Z}_2^\infty$ has many finite-index subgroups that are not open (e.g. the subgroup of elements with an even number of $1$'s), but it seems to me that every finite-index open subgroup should be the intersection of an open subgroup of $\mathbb{Z}_2^\omega$ with $\mathbb{Z}_2^\infty$.  Then the resulting topological profinite completion of $\mathbb{Z}_2^\infty$ ought to be $\mathbb{Z}_2^\omega$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-09-30T05:31:12.057" CommentCount="2" />
  <row Id="5751" PostTypeId="2" ParentId="5718" CreationDate="2010-09-30T07:53:57.700" Score="3" ViewCount="0" Body="&lt;p&gt;Since $\mathcal O_L$ is the union of $\mathcal O_{L_i}$, where $L_i$ runs over all the finite subextensions, giving a prime ideal $\mathfrak p$ in $\mathcal O_L$ is the same as giving a compatible collection of primes ideals $\mathfrak p_i$ in each $\mathcal O_{L_i}$.  (We set&#xA;$\mathfrak p_i := \mathfrak p \cap \mathcal O_{L_i}$, and $\mathfrak p = \cup_i \mathfrak p_i$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $L$ is the union of the $L_i$, giving an absolute value $v$ on $L$ is the same as giving compatible absolute values $v_i$ on the various $L_i$.  (Take $v_i$ to be the restriction to $L_i$ of $v$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the preceding two remarks, we see that the bijection between primes ideals and non-Archimedean valuations in the case of finite extensions extends to a corresponding bijection in the case of infinite extensions.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-09-30T07:53:57.700" CommentCount="1" />
  <row Id="5752" PostTypeId="2" ParentId="5747" CreationDate="2010-09-30T09:45:10.913" Score="1" ViewCount="0" Body="&lt;p&gt;Also, observe that if $z^n=i$ then $z^{4n}=1$. Thus, the complex numbers you're looking for are particular $4n$-th roots of $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know that the $m$-th roots of 1 (any $m$) can be written as powers of a single well-chosen one (a primitive root), it shouldn't be too hardto find exactly which $4n$-th roots have the desired property.&lt;/p&gt;&#xA;" OwnerUserId="688" LastActivityDate="2010-09-30T09:45:10.913" />
  <row Id="5753" PostTypeId="2" ParentId="5738" CreationDate="2010-09-30T09:48:53.903" Score="0" ViewCount="0" Body="&lt;p&gt;Robin-presumably that is ' not equal to' as opposed to factorial!&lt;/p&gt;&#xA;" OwnerUserId="2216" LastActivityDate="2010-09-30T09:48:53.903" CommunityOwnedDate="2010-09-30T09:48:53.903" />
  <row Id="5754" PostTypeId="2" ParentId="5191" CreationDate="2010-09-30T10:00:06.137" Score="0" ViewCount="0" Body="&lt;p&gt;Let &lt;em&gt;Vect&lt;/em&gt; be the category of vector spaces over some field and let &lt;em&gt;Fin&lt;/em&gt; be the category of finite sets and bijections. There is a functor &lt;em&gt;f&lt;/em&gt; from &lt;em&gt;Fin&lt;/em&gt; to &lt;em&gt;Vect&lt;/em&gt; that maps a set &lt;em&gt;S&lt;/em&gt; to the top exterior power of the free vector space spanned by &lt;em&gt;S&lt;/em&gt;. Let &lt;em&gt;g&lt;/em&gt; be the functor from the one-object, one-morphism category to &lt;em&gt;Vect&lt;/em&gt; that is the inclusion of a one-dimensional vector space. Then the category you are looking for is the comma category &lt;em&gt;(f/g)&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-09-30T10:00:06.137" CommentCount="1" />
  <row Id="5755" PostTypeId="1" AcceptedAnswerId="5843" CreationDate="2010-09-30T10:32:50.683" Score="6" ViewCount="104" Body="&lt;p&gt;I'm looking for a convenient upper bound on the integral $\int_y^\infty x^k \exp(-(x-\mu)^2/2) dx$ for (possibly large) positive integer k.  This is equivalent to finding higher moments of a truncated normal distribution. A bound that works for non-integer k as well would be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course &quot;convenient&quot; is in the eye of the beholder, but I'd like some sort of fairly simple expression that I can use in further calculations.  For example, an upper bound of the form $f(x) \exp( -g(x))$ where where $f$ and $g$ are low-degree polynomials would be great. I'm more interested in simplicity of form than in obtaining the tightest possible bound.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-09-30T14:48:56.203" LastActivityDate="2010-10-01T20:09:45.113" Title="Bounding higher moments of truncated normal" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="5756" PostTypeId="1" AcceptedAnswerId="5776" CreationDate="2010-09-30T10:35:01.987" Score="5" ViewCount="65" Body="&lt;p&gt;What is the largest (acyclic, connected) transitively reduced acyclic graph on $n$ vertices, for every $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many edges $e$ does it have?  How does $e$ grow as a function of $n$?  Faster than $n^2/4$?&lt;/p&gt;&#xA;" OwnerUserId="2217" LastEditorUserId="2217" LastEditorDisplayName="" LastEditDate="2010-09-30T11:20:40.093" LastActivityDate="2010-09-30T18:52:55.323" Title="How big are transitively reduced graphs?" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="5757" PostTypeId="2" ParentId="5755" CreationDate="2010-09-30T11:09:43.610" Score="1" ViewCount="0" Body="&lt;p&gt;Let $I_k=\int_y^\infty x^k\mathrm{exp}(-(x-\mu)^2/2)dx$ and $a_k=y^{k}\mathrm{exp}(-(y-\mu)^2/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integration by parts gives $I_k=a_{k-1}+(k-1)I_{k-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for even $n$,&#xA;\begin{multline}&#xA;I_n=a_{n-1}+(n-1)a_{n-3}+(n-1)(n-3)a_{n-5}+\cdots\newline+(n-1)(n-3)\cdots 1 a_1+(n-1)(n-3)\cdots 1 I_0.&#xA;\end{multline}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to expand this answer and deduce an estimate from the above.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-09-30T11:15:56.733" LastActivityDate="2010-09-30T11:15:56.733" CommunityOwnedDate="2010-09-30T11:19:34.993" CommentCount="3" />
  <row Id="5758" PostTypeId="1" AcceptedAnswerId="5759" CreationDate="2010-09-30T12:14:30.857" Score="2" ViewCount="104" Body="&lt;p&gt;If you start with an infinite set, you can have a sequence of nested sets which converge to a single point. (ie Intersection of (-1/n, 1/n) as n-&gt;infinity))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at no time during the sequence is there a first element with only one point. In fact, for any finite (but unbounded) n, the number of points in the interval is uncountably large. So, how do you understand the idea that this infinite intersection contains just one point? How do you make sense of it?&lt;/p&gt;&#xA;" OwnerUserId="2003" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-24T02:31:12.190" LastActivityDate="2010-11-24T02:31:12.190" Title="Understanding of convergence of intersections of sets" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;elementary-set-theory&gt;&lt;infinity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5759" PostTypeId="2" ParentId="5758" CreationDate="2010-09-30T12:28:52.460" Score="3" ViewCount="0" Body="&lt;p&gt;A point is in the intersection if it's in &lt;em&gt;every&lt;/em&gt; set of the sequence. Focus on that. No matter how close to $0$ a point is (excluding $0$ itself), there will be a sufficiently large $n$ such that the set $(-1/n,1/n)$ does not contain it, and therefore that point will not be in the (infinite) intersection, because, again, a point is in the intersection iff and it is in every set specified by the intersection.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-09-30T12:28:52.460" />
  <row Id="5760" PostTypeId="1" AcceptedAnswerId="5764" CreationDate="2010-09-30T13:02:10.917" Score="1" ViewCount="61" Body="&lt;p&gt;Suppose $x_1, x_2, y_1, y_2$ are known integers which satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_1 \equiv \gamma x_1 \pmod c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_2 \equiv \gamma x_2 \pmod c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ and $c$ are unknown integers.   Is there a way to determine the values of $\gamma$ and $c$? &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-09-30T13:08:59.750" LastActivityDate="2010-09-30T13:56:39.927" Title="System of Congruences" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5761" PostTypeId="2" ParentId="5758" CreationDate="2010-09-30T13:02:17.637" Score="4" ViewCount="0" Body="&lt;p&gt;I think the source of the confusion is this: taking the cardinality of a set seems like such a simple operation that we subconsciously expect it to be continuous.  We might think that simple limiting operations like countable nested intersections should give convergent cardinalities.  But the simple fact is that they don't; our subconscious guess is just wrong, and this example proves it.  Cardinality is not continuous in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to the standard &quot;paradox&quot; of the supertask where at time $1-2^{-n}$ you add to a box two balls numbered $2n-1$ and $2n$, and remove the one numbered $n$.  Despite the fact that the cardinality of the balls in the box is always increasing, in the limit at time 1 the box is empty.  Again, the paradox arises only if you feel that cardinality should be continuous.  If you realize that it isn't, then you no longer see such examples as paradoxical.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-30T13:02:17.637" CommentCount="2" />
  <row Id="5762" PostTypeId="2" ParentId="5756" CreationDate="2010-09-30T13:25:23.663" Score="2" ViewCount="0" Body="&lt;p&gt;Bipartite graph is the largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Suppose G is a graph, v is it's vertex. Let $L(v)$ is equal to the maximal length of a directed path, which starts in $v$. Let $L(G)=max_{v\in G} L(v)$. If $L(G)=1$, the graph is bipartite. In order to prove, that bipartite graph is the largest, I will prove, that for any graph $G$ such that $L(G)&gt;1$ there exists graph $H$ with the same number of vertices and edges, such that $L(G)&gt;L(H)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that any vertex with $L(v)=0$ has only inbound connections. This connections are from vertices w with $L(w)=1$. We will do graph $H$ in two steps.&lt;br&gt;&#xA;Step 1. Remove all vertices with $L(v)=0$, and edges to them. Note that $L(G)$ has been decreased.&#xA;Step 2. Add all the vertices, removed on step 1. Add edges, removed on step 1, but reverse their direction. Call the new graph $H$. Note, that it has the same number of vertices and edges and $L(H) &amp;lt; L(G)$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-09-30T13:25:23.663" CommentCount="1" />
  <row Id="5763" PostTypeId="1" AcceptedAnswerId="5780" CreationDate="2010-09-30T13:29:45.010" Score="3" ViewCount="203" Body="&lt;p&gt;I had asked this question: &lt;a href=&quot;http://math.stackexchange.com/questions/4412/characterising-continuous-functions&quot;&gt;http://math.stackexchange.com/questions/4412/characterising-continuous-functions&lt;/a&gt; some time back, and this question is more or less related to that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a function $f: \mathbb{R} \to \mathbb{R}$ and suppose the set  $G = \{ (x,f(x) : x \in \mathbb{R}\}$ is connected and closed in $\mathbb{R}^{2}$, then does it imply $f$ is continuous?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-09-30T21:35:51.343" Title="Characterizing continuous functions based on the graph of the function." Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5764" PostTypeId="2" ParentId="5760" CreationDate="2010-09-30T13:50:45.917" Score="2" ViewCount="0" Body="&lt;p&gt;No, because if c=pq, for primes p and q, then both congruences hold mod p and mod q as well. So c is not uniquely determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $y_1=y_2=11$ and $x_1=x_2=1$ and $\gamma=5$, c could be 1,2,3 or 6.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-09-30T13:56:39.927" LastActivityDate="2010-09-30T13:56:39.927" />
  <row Id="5766" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T15:40:17.687" Score="0" ViewCount="0" Body="&lt;p&gt;suppose you have a function $f:\mathbb{R} \rightarrow \mathbb{R}$ which is not continuous and its graph is closed and connected. wlog assume f is not continuous at zero and f(0)=0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;first, I want to show that when x approaches zero then either f(x) approaches 0 or goes to infinity. &#xA;for every b&gt;a&gt;0 let $ X_{a,b} = \{ x&gt;0 \mid a &amp;lt; f(x) &amp;lt; b \}  $ . if this set is not empty, then its infimum cannot be 0, because then the closure of the graph will give us that $ a \leq f(0) \leq b $ (you can find a series $a_n \rightarrow 0 \subseteq X_{a,b}$ and a sub-series such that $f(a_{n_i})$ is monotone and thus converges in $R^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now I want to use the connectedness to show that f(x) cannot go to infinity when x goes to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ isn't continuous at zero, so wlog there is an $\varepsilon &gt; 0$ such that $ X_{2\varepsilon, \infty}$ is not empty and its infimum is 0. $X_{\varepsilon, 2\varepsilon}$ has infimum &gt; 0 (if not empty) - denote it by $\delta$. if you have $0 &amp;lt; x_0 &amp;lt; \delta$ such that $ x_0 \notin X_{2 \varepsilon, \infty} $ (and so $f(x_0) &amp;lt; \varepsilon $ ) then you have the open sets $ A = (0, x_0) \times (1.5 \varepsilon, \infty) $ and $ R^2 - \bar A$ which separates the garph of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;otherwise, if $ (0,\delta) \subseteq X_{2\varepsilon, \infty} $ then you can take the sets $A = \{x\leq 0 \} \cup (0, \delta/2) \times (-\infty, 1.5 \varepsilon)$ and $R^2 - \bar A$.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-09-30T15:40:17.687" />
  <row Id="5767" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T15:57:21.707" Score="0" ViewCount="0" Body="&lt;p&gt;The answer is yes.  Here's one way to prove it.  (There might be a slicker way, but this seems to work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $G$ is connected and closed. Let $a\in\mathbb R$ be arbitrary, and let $\epsilon&gt;0$ be given.  Because $(a,f(a)-\epsilon)\notin G$, the fact that the complement of $G$ is open implies that there is a product neighborhood of the form $(a-\delta,a+\delta)\times (f(a)-\epsilon-c, f(a)-\epsilon+c)$ contained in the complement of $G$.  This means that  $|x-a|&amp;lt;\delta$ implies that one of the following two inequalities holds:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x)\ge f(a)-\epsilon+c&gt;f(a)-\epsilon$, or&lt;/li&gt;&#xA;&lt;li&gt;$f(x)\le f(a)-\epsilon-c$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If there is any $x\in (a-\delta,a+\delta)$ so that the second inequality holds, say $f(b)\le f(a)-\epsilon-c$ (without loss of generality, we may assume $b\lt a$), then the graph of $f$ does not intersect the following set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{(b,y): y\ge f(a)-\epsilon \}\cup \{(x,f(a)-\epsilon): b\le x \le a\} \cup \{(a,y): y\le f(a)-\epsilon\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See the diagram below.)  This broken line disconnects $G$, contradicting the assumption that $G$ is connected.  Therefore inequality (1) holds when $|x-a|&amp;lt;\delta$.&lt;br&gt;&#xA;A similar argument shows that $f(x)\lt f(a)+\epsilon$ when $|x-a|&amp;lt;\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting these together, we conclude that $|x-a|&amp;lt;\delta$ implies $|f(x)-f(a)|&amp;lt;\epsilon$, so $f$ is continuous at $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OrP64.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-09-30T15:57:21.707" />
  <row Id="5768" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T16:00:40.263" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, I think so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, observe that such $f$ must have the intermediate value property.  For suppose not; then there exist $a &amp;lt; b$ with (say) $f(a) &amp;lt; f(b)$ and $y \in (f(a),f(b))$ such that $f(x) \ne y$ for all $x \in (a,b)$.  Then $A = (-\infty,a) \times \mathbb{R} \cup (-\infty,b) \times (-\infty,y)$ and $B = (b, +\infty) \times \mathbb{R} \cup (a,+\infty) \times (y,+\infty)$ are disjoint nonempty open subsets of $\mathbb{R}^2$ whose union contains $G$, contradicting connectedness.  (Draw  a picture.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take some $x \in \mathbb{R}$, and suppose $f(x) &amp;lt; y &amp;lt; \limsup_{t \uparrow x} f(t) \le +\infty$. Then there is a  sequence $t_n \uparrow x$ with $f(t_n) &gt; y$ for each $n$.  By the intermediate value property, for each $n$ there is $s_n \in (t_n, x)$ with $f(s_n) = y$.  So $(s_n, y) \in G$ and $(s_n,y) \to (x,y)$, so since $G$ is closed $(x,y) \in G$ and $y = f(x)$, a contradiction.  So $\limsup_{t \uparrow x} f(t) \le f(x)$.  Similarly, $\limsup_{t \downarrow x} f(t) \le f(x)$, so $\limsup_{t \to x} f(t) \le f(x)$.  Similarly, $\liminf_{t \to x} f(t) \ge f(x)$, so that $\lim_{t \to x} f(t) = f(x)$, and $f$ is continuous at $x$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-30T16:00:40.263" CommentCount="2" />
  <row Id="5769" PostTypeId="2" ParentId="5746" CreationDate="2010-09-30T16:18:11.627" Score="1" ViewCount="0" Body="&lt;p&gt;One way to solve f(x) = y is to minimize g(x) = (f(x) - y)^2.  You could do this by taking the quadratic approximation to g(x) that comes from the second order Taylor series centered at your initial guess, finding its minimum, and letting that minimum be the starting guess for the next iteration. This isn't necessarily the best algorithm, but it's easy to understand and implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For high-dimensional problems, the work is in solving the system of equations to minimize the quadratic approximation.  A great deal of research has gone into doing that step cleverly, taking advantage of sparse matrix structure etc.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-09-30T16:18:11.627" CommentCount="4" />
  <row Id="5770" PostTypeId="2" ParentId="5738" CreationDate="2010-09-30T16:42:18.513" Score="0" ViewCount="0" Body="&lt;p&gt;$\ $ If $\rm\ a &gt; 0\ $ then $\rm\ a &amp;lt; b\ $ times $\rm\ 1/a &amp;lt; 1/b\ \Rightarrow\ 1 &amp;lt; 1$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ a &amp;lt; 0\ \:$ and $\rm\:\ a &amp;lt; 1/a\ \:\Rightarrow\:\ a^2 &gt; 1 \ \:\Rightarrow\:\ a &amp;lt; -1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T03:08:07.393" LastActivityDate="2010-10-01T03:08:07.393" />
  <row Id="5771" PostTypeId="2" ParentId="5745" CreationDate="2010-09-30T16:48:32.307" Score="2" ViewCount="0" Body="&lt;p&gt;This may be a complete stupid thing, for which I'll lose some hundred points of reputation, but here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand anything at all about your functionss $f$ &quot;and all permutations of the subset variables N...&quot;, nor the rest of your question. But I think that expressions like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^{j^k} \ , \qquad j^{k^i} \ , \qquad \text{and}\qquad   k^{i^j} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could make sense. However, they don't seem to be equal to $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's remember how we define $x^y$ for real numbers $x, y$, $x&gt;0$. We put&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x^y = e^{y\log x} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;don't we? After choosing a determination for the complex logarithm, we define also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^z = e^{z \log i }\ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to compute $\log i$. For this we can rely on Euler's formula, which, for a real number $\theta$, says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e^{i\theta} = \cos\theta + i \sin\theta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ask a complex number $z = x + iy$ to be a logarithm for $i$, we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i = e^z = e^{x + iy} = e^x e^{iy} = e^x (\cos y + i \sin y) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0 = e^x \cos y \qquad \text{and} \qquad 1 = e^x\sin y \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may take $y = \frac{\pi}{2}$ and then $x= 0$. Hence a logarithm for $i$ is $\log i = i\frac{\pi}{2}$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^z = e^{zi\frac{\pi}{2}} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good: what about quaternions? Well, we don't need to define exponentials and logarithms for them, but just notice that Euler's formula only needs two things to be true: (a) the Taylor series for $\exp z$, $\cos z$ and $\sin z$ and (b) &quot;something&quot;, called $i$, that can be added with real numbers and itself, multyplied by real numbers and such that $i^2 = -1$. Right? Okay. So, for the same reasons as for $i$, we may take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log i = i\frac{\pi}{2} \ , \qquad &#xA;\log j = j\frac{\pi}{2} \ , \qquad \text{and} \qquad&#xA;\log k = k\frac{\pi}{2} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can compute $j^k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;j^k = e^{k\log j} = e^{k j\frac{\pi}{2}} = e^{-i \frac{\pi}{2}} = \cos \left( \frac{-\pi}{2} \right) +i \sin \left( \frac{-\pi}{2} \right) = -i \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;i^{j^k} = i^{-i} = e^{-i \log i} = e^{-i i \frac{\pi}{2}} = e^{\frac{\pi}{2}} \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-01T03:33:23.723" LastActivityDate="2010-10-01T03:33:23.723" CommentCount="5" />
  <row Id="5772" PostTypeId="1" AcceptedAnswerId="5822" CreationDate="2010-09-30T17:09:45.773" Score="3" ViewCount="198" Body="&lt;p&gt;I've proved $\neg \neg \beta \rightarrow \beta$ is a theorem but I can't figure out a way to do the same for $\beta \rightarrow \neg \neg \beta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems the proof would use Axiom 2 and the deduction theorem (which allows $\beta$ to be an axiom) - but i've endlessy tried values to no avail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 1: $A \rightarrow ( B \rightarrow A )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 2: $( A \rightarrow ( B \rightarrow C ) ) \rightarrow ( ( A \rightarrow B ) \rightarrow (A \rightarrow C) ) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom 3: $( \neg B \rightarrow \neg A) \rightarrow ( ( \neg B \rightarrow A) \rightarrow B )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to clarify A, B, C $\alpha$ and $\beta$ are propositions (i.e. assigned True or False). $\rightarrow$ and $\neg$ have the standard logical meanings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\TeX$ification does not work in the IE9 beta.&lt;/p&gt;&#xA;" OwnerUserId="2222" LastEditorUserId="2222" LastEditorDisplayName="" LastEditDate="2010-10-01T16:05:59.743" LastActivityDate="2010-10-01T16:05:59.743" Title="Prove that $\beta \rightarrow \neg \neg \beta$ is a theorem using standard axioms 1,2,3 and MP." Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" />
  <row Id="5773" PostTypeId="2" ParentId="5676" CreationDate="2010-09-30T17:44:02.330" Score="4" ViewCount="0" Body="&lt;p&gt;In case you were wondering, like me, Moron's excellent proof adapts easily to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \sqrt[3]{2} + \dots + \sqrt[3]{n} \sim \frac{3}{4}n^{4/3} + \frac{\sqrt[3]{n}}{2} + C,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $C.$ In this case $C = \zeta(-1/3) \approx -0.277343.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, as before, $a_n \sim b_n$ means $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to the previous proof, we use the AM-GM inequality to show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{x} \le \frac{x}{3k^{2/3}} + \frac{2k^{1/3}}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing from $k=1$ to $n-1$ and integrating we arrive at&#xA;$$ \frac{3}{4}n^{4/3} - \frac{3}{4} \le \sum_{k=1}^{n-1}\sqrt[3]{n} +&#xA;\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n&gt;1.$$&#xA;And so,&#xA;$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + n^{1/3} - \frac{3}{4} - \frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n&gt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le 1+ \int_1^n x^{-2/3} dx,&#xA;\qquad n&gt;1,$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le \frac{1}{2}n^{1/3} - \frac{1}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And hence&#xA;$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + \frac{1}{2}n^{1/3} - \frac{5}{12}, \qquad n&gt;1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the previous argument, set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_n = \sum_{k=1}^{n}\sqrt[3]{n} - \frac{3}{4}n^{4/3} - \frac{1}{2}n^{1/3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $G_n \ge -5/12,$ and we can show $G_n – G_{n+1} &gt; 0$ by showing that&#xA;$$\frac{(n+1)^{4/3} – n^{4/3}}{(n+1)^{1/3} + n^{1/3}} &gt; \frac{2}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as before, $G_n$ is convergent, since it is bounded below and monotonically decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this argument also adapts easily to the more general case&#xA;$\sum_{k=1}^{n}\sqrt[r]{n},$ for $r \in \mathbb{N},$ where I'm guessing, we'll find&#xA;$$1 + \sqrt[r]{2} + \dots + \sqrt[r]{n} \sim \frac{r}{r+1}n^{(r+1)/r} + \frac{\sqrt[r]{n}}{2} + C,$$ &#xA;where $C = \zeta(-1/r).$ However, I confess to not having checked the details of this further generalisation.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-09-30T17:44:02.330" />
  <row Id="5774" PostTypeId="1" AcceptedAnswerId="6019" CreationDate="2010-09-30T17:54:20.380" Score="2" ViewCount="98" Body="&lt;p&gt;I have recently been working out the full game tree for tic-tac-toe, just for fun.  I am using the well known equivalence relation of rotations/reflections to simplify this tree in the standard way (which begins by noting that there are only 3 opening moves: edge, corner, and center).  I actually could not find an image of the full game tree for tic-tac-toe, so if anyone can provide a link I'd appreciate that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This work motivates the following;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let there be given an N x N grid; and let $m$ be a natural number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I$ be the set of all possible ways to place $m$ copies of the letter $X$ in the grid, and $m$ copies of the letter $O$ in the grid (with the restriction that we can only place one letter per space of the grid; in other words, just imagine playing a game of tic-tac-toe on an N x N board for an even number of moves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many ways are there to completely break the rotational/reflection symmetry of an N x N grid by placing $m$ copies of $X$ and $m$ copies of $O$?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-10-01T01:36:26.670" LastActivityDate="2010-10-04T13:26:21.830" Title="Counting the number of ways to break the symmetry of an N x N grid by placing one X and one O at a time" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5775" PostTypeId="1" AcceptedAnswerId="5782" CreationDate="2010-09-30T18:37:53.607" Score="4" ViewCount="153" Body="&lt;p&gt;Given is a sequence $\langle a_1,a_2,\ldots,a_n\rangle$ over the alphabet $\{1,2,\ldots,m\}$ chosen uniformly at random among the $m^n$ possibilities. What is the expected size of the &lt;em&gt;set&lt;/em&gt; $\{a_1,a_2,\ldots,a_n\}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m=n$ it &lt;a href=&quot;http://rgrig.blogspot.com/2010/09/expected-size-of-set.html&quot; rel=&quot;nofollow&quot;&gt;seems&lt;/a&gt; the answer tends to $(1-1/e)n$ as $n\to\infty$, but I don't know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bumped into this while benchmarking some code for hashtables, so I wouldn't be surprised if it is a standard result in the hash world.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2010-09-30T22:30:51.203" Title="How many bins do random numbers fill?" Tags="&lt;combinatorics&gt;&lt;probability&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" />
  <row Id="5776" PostTypeId="2" ParentId="5756" CreationDate="2010-09-30T18:52:55.323" Score="3" ViewCount="0" Body="&lt;p&gt;Since the transitively reduced graph $G$ is acyclic, there is a corresponding undirected simple graph $G'$, which can be formed by ignoring the direction of the edge. The number of edges in $G$ is same as the number of edges in $G'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $G$ to be transitively reduced, a necessary condition is the $G'$ must not contain any triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known (&lt;a href=&quot;http://en.wikipedia.org/wiki/Tur%C3%A1n%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Turan's/Mantel's theorem&lt;/a&gt;) that any simple undirected graph on $n$ vertices and more than $n^2/4$ edges has a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also known that the undirected graph on $n$ vertices with maximum edges and no triangles is the complete bipartite graph $K_{[n/2],[n/2]}$ (for instance check out: &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA28&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA28&lt;/a&gt;, exercise 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your comments, it looks like you found a $G$ whose corresponding $G'$ is $K_{[n/2],[n/2]}$ and so that proves it.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-09-30T18:52:55.323" CommentCount="1" />
  <row Id="5777" PostTypeId="1" AcceptedAnswerId="5781" CreationDate="2010-09-30T18:55:15.780" Score="2" ViewCount="437" Body="&lt;p&gt;I know i'm supposed to do sin(3x) - sin x = 0 but beyond that im stuck.. i tryed expanding sin(3x) but that didn't help  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;i want the value of x in the interval [0, 2π)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2227" LastEditorUserId="2227" LastEditorDisplayName="" LastEditDate="2010-09-30T19:09:28.747" LastActivityDate="2010-09-30T20:04:32.573" Title="sin(3x) = sin(x)" Tags="&lt;trigonometry&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5778" PostTypeId="1" CreationDate="2010-09-30T18:58:20.143" Score="1" ViewCount="169" Body="&lt;p&gt;This is probably a very simple question to all of you, but I have never taken a physics class, so I am lost for what I need to do for a programming problem. I will need to specify a force by inputting the newtons x, newtons y, and newtons z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate these 3 forces based on the desired angle and desired force for that direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time, this project is important to me.&lt;/p&gt;&#xA;" OwnerUserId="2228" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:46:29.377" LastActivityDate="2010-11-19T16:46:29.377" Title="Physics: Calculating the Forces Needed from an Angle and a Force" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="5779" PostTypeId="2" ParentId="4658" CreationDate="2010-09-30T19:00:50.207" Score="4" ViewCount="0" Body="&lt;p&gt;With respect to the exact answer, the recursion is related to the  &lt;a href=&quot;http://mathworld.wolfram.com/Fibonaccin-StepNumber.html&quot; rel=&quot;nofollow&quot;&gt;Fibonacci n-Step Numbers&lt;/a&gt;  (look at the phrase below the table).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to an approximate/asympotic solution: it can also be obtained by probabilistic reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of throwing N (150) coins, lets consider the (thought) experiment of throwing M alternate &lt;em&gt;runs&lt;/em&gt; (a ‘run’ is a sequence of consecutive tails/heads). That is, instead of throwing N iid Bernoulli random variables (two values with prob=1/2), we throw M iid geometric random variables ( p(1)=1/2 p(2)=1/4 p(3)=1/8 ...) which we interpret as the length of each alternate run of consecutive tails/head in the coins sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected value of the the runs (in both experiments) is 2. If we choose M=N/2, then the expected total number of coins (in the seconde experiment) will be N, and so we can expect (informally) that the two experiments are asympotically equivalent (in the original experiment the number of coins is fixed, the number of runs is a random variable; in the second, the reverse; this could be related to the use of different ensembles in statistical physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the modified experiment it’s easy to compute the probability that no run exceeds a length L: it’s just an  (1-(1/2)^(L-1))^(N/2)&#xA;If we consider just tails, the number of &quot;trials&quot; would be N/4 instead. This, in our case, (L=150, N=6) gives P=0.44160&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-09-30T19:00:50.207" />
  <row Id="5780" PostTypeId="2" ParentId="5763" CreationDate="2010-09-30T19:03:12.147" Score="9" ViewCount="0" Body="&lt;p&gt;This &lt;a href=&quot;http://www.jstor.org/stable/2324521&quot; rel=&quot;nofollow&quot;&gt;Monthly paper&lt;/a&gt; has short simple proofs of the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;$\ $ TFAE if $\rm\ f: \mathbb R\to \mathbb R\ $ has a closed graph in $\:\mathbb R^2$&lt;br&gt;&#xA;(a)$\rm\ \ f\ $ is continuous.&lt;br&gt;&#xA;(b)$\rm\ \  f\ $ is locally bounded.&lt;br&gt;&#xA;(c)$\rm\ \ f\ $ has the intermediate value property.&lt;br&gt;&#xA;(d)$\rm\ \ f\ $ has a connected graph in $\rm\mathbb R^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally the result is merely a special case of R. L. Moore's 1920 characterization of a topological line as a locally compact metric space that is separated into two connected sets by each of its points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per request, I've appended the proof of the theorem below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vnoai.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/UyIrJ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-09-30T21:35:51.343" LastActivityDate="2010-09-30T21:35:51.343" CommentCount="8" />
  <row Id="5781" PostTypeId="2" ParentId="5777" CreationDate="2010-09-30T19:03:21.460" Score="4" ViewCount="0" Body="&lt;p&gt;We have $\sin{3x} = 3\sin{x} - 4\sin^{3}{x}$ which says that we have to solve the equation $$3\sin{x} - 4\sin^{3}{x} - \sin{x}=0$$, that is $2 \sin{x} - 4\sin^{3}{x}=0$. Take $y = \sin{x}$ and so you have $$2y-4y^{3}=0 \Longrightarrow 2y(1-2y^{2})=0$$ and then see what happens. I hope this helps you out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can even try this $$\sin{3x} - \sin{x} = 2 \cos\Bigl(\frac{3x+x}{2}\Bigr) \cdot \sin\Bigl(\frac{3x-x}{2}\Bigr) = 2\cos{2x} \cdot \sin{x}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-09-30T20:04:32.573" LastActivityDate="2010-09-30T20:04:32.573" CommentCount="9" />
  <row Id="5782" PostTypeId="2" ParentId="5775" CreationDate="2010-09-30T19:10:24.803" Score="11" ViewCount="0" Body="&lt;p&gt;Define the indicator random variable $I_i$ for $1 \leq i \leq m$ as $1$ if alphabet $i$ is present in the set ${a_1,\dots,a_n}$. Then the size of the set is simply $\sum_{i=1}^m I_i$. The expectation of this can be easily computed by linearity of expectation. The probability that $I_i$ equals $1$ is given by $1-\left( \frac{m-1}{m} \right)^n$ and therefore the expected size of the set is $m \left[ 1- \left( 1 - \frac{1}{m} \right)^n \right]$. For $m=n$, the limiting value is indeed as you mentioned in the question.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-09-30T19:10:24.803" />
  <row Id="5783" PostTypeId="1" CreationDate="2010-09-30T19:37:11.613" Score="1" ViewCount="87" Body="&lt;p&gt;please help.. In my notes, it is the V itself. I just want to confirm this. Can you also explain and give an example if that is so...&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2010-09-30T19:40:39.770" Title="What is the center of V, the Klein 4 group?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5784" PostTypeId="2" ParentId="5783" CreationDate="2010-09-30T19:40:39.770" Score="3" ViewCount="0" Body="&lt;p&gt;Every abelian group is its own center.  If you look at the definitions you will see this.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-09-30T19:40:39.770" CommentCount="1" />
  <row Id="5785" PostTypeId="2" ParentId="5777" CreationDate="2010-09-30T19:47:23.353" Score="8" ViewCount="0" Body="&lt;p&gt;You could use the fact that $\sin x=\sin y$ if and only if either&#xA;$x-y$ is an even integer times $\pi$ or $x+y$ is an odd integer times $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-09-30T19:47:23.353" />
  <row Id="5787" PostTypeId="2" ParentId="3966" CreationDate="2010-09-30T20:56:21.630" Score="0" ViewCount="0" Body="&lt;p&gt;The first thing that comes to mind is Quantum Electrodynamics. I don't really know much more than the name, but the first chapters of Richard Feynmans's Six Easy Pieces (I think that was the one) are pretty accessible. They raise the problems of trying to keep track of single photons, and how they behave completely differently from regular solid objects. With a bunch of photons you can only keep track of the distribution of photons (more or less). Look for Feynmann vidoes on the web. He was entertaining and you won't regret the time.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-09-30T20:56:21.630" />
  <row Id="5788" PostTypeId="1" AcceptedAnswerId="5790" CreationDate="2010-09-30T21:03:57.143" Score="2" ViewCount="102" Body="&lt;p&gt;it has been years since I have done logs, I remember something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\log_z(y)} = y^{\log_z(x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $z$ is the base)&#xA;Is that correct?  It doesn't seem so, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^{\log_2(4)} \neq 4^{\log_2(3)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right in that assumption?&#xA;If I am, does&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\log_z(y)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;convert to anything easily?&lt;/p&gt;&#xA;" OwnerUserId="2232" LastActivityDate="2010-10-01T01:31:55.180" Title="simple/dumb logarithmic conversion question" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5789" PostTypeId="2" ParentId="5778" CreationDate="2010-09-30T22:01:50.540" Score="1" ViewCount="0" Body="&lt;p&gt;Isaac's comment asking how the angles are given is a good one.  If you specify that and the coordinate system the problem will be well posed.  Guessing that it is in azimuth $(\phi )$ and elevation $(\theta )$, with $\phi =\theta =0$ along the positive x axis and positive z up,&#xA;$F_x=F\cos(\theta )*\cos(\phi ), F_y=F\cos(\theta )*\sin(\phi ), F_z=F\sin(\theta )$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-19T01:50:41.683" LastActivityDate="2010-11-19T01:50:41.683" />
  <row Id="5790" PostTypeId="2" ParentId="5788" CreationDate="2010-09-30T22:27:08.903" Score="3" ViewCount="0" Body="&lt;p&gt;We do have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{\log _{z}y}=y^{\log _{z}x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\log_{z}y)\log_{z}x=(\log_{z}x)\log _{z}y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical relation is an equality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3^{\log _{2}4}=4^{\log _{2}3}=9$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-09-30T22:35:03.687" LastActivityDate="2010-09-30T22:35:03.687" />
  <row Id="5791" PostTypeId="1" CreationDate="2010-09-30T22:27:30.120" Score="0" ViewCount="224" Body="&lt;p&gt;Here is the dot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aQBA2.jpg&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is how I arrived at it »&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4LBJi.jpg&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty simple IFS using a 2 x 2 grid as the base for the iterations. Is there a way to describe this point as well as its siblings, all of which lie on the dotted red line(pic 2), using an equation?&lt;/p&gt;&#xA;" OwnerUserId="2234" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-09-30T23:24:03.020" LastActivityDate="2010-11-02T06:44:16.823" Title="How do I plot this red dot?" Tags="&lt;sequences-and-series&gt;&lt;fractals&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="5792" PostTypeId="2" ParentId="5775" CreationDate="2010-09-30T22:30:51.203" Score="0" ViewCount="0" Body="&lt;p&gt;This is dealt with in depth at &lt;a href=&quot;http://www.ds.unifi.it/VL/VL_EN/urn/urn8.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ds.unifi.it/VL/VL_EN/urn/urn8.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1789" LastActivityDate="2010-09-30T22:30:51.203" />
  <row Id="5793" PostTypeId="1" CreationDate="2010-09-30T23:44:46.070" Score="-2" ViewCount="92" Body="&lt;p&gt;In matlab, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i have 12 variables, i want to calculate the each columns mean and subtract from each data dimension (every variable in column) and get a new matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help&lt;/p&gt;&#xA;" OwnerUserId="2235" LastActivityDate="2010-10-01T10:47:35.380" Title="How can i subtract the mean from each of the data dimension in matlab?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5794" PostTypeId="2" ParentId="5788" CreationDate="2010-10-01T00:42:53.970" Score="2" ViewCount="0" Body="&lt;p&gt;If $\rm\:\ \ell\:X\ =\ \log_Z{X}\ \ $ then $\rm\ \ell(X^{\:\ell\:{Y}})\ =\ \ell X\ \ell Y\ =\ \ell Y\ \ell X\ =\ \ell(Y^{\:\ell\: X})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, without taking logarithms $\rm\displaystyle\ \ \: X^{\:\ell\:{Y}}\ \ =\ Z^{\:\ell\: X\ \ell\: Y}\ =\ Z^{\:\ell\: Y\ \ell\: X}\ =\ Y^{\:\ell\: X}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your final example is a special case $\rm\ \ Y\ =\ Z^n\:$, $\rm \ $ but it also has a simple direct proof,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;namely $\rm\ \ \: X^{\:\ell\: Z^{\:n}}\ =\ X^n\ =\ Z^{\:\ell\: X^n}\ =\ Z^{n\:\ell\: X}\:$.$\:$ Yours is $\rm\: X = 3,\ Z = 2 = n$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T01:31:55.180" LastActivityDate="2010-10-01T01:31:55.180" />
  <row Id="5795" PostTypeId="1" AcceptedAnswerId="5798" CreationDate="2010-10-01T01:18:33.547" Score="4" ViewCount="76" Body="&lt;p&gt;Suppose I have a weakly sequentially continuous linear operator T between two normed linear spaces X and Y (i.e. $x_n \stackrel {w}{\rightharpoonup} x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel {w}{\rightharpoonup} T(x)$ in $Y$). Does this imply that my operator T must be bounded?   &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-10-01T02:59:39.523" LastActivityDate="2010-10-01T03:26:16.743" Title="weak sequential continuity of linear operators" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="5796" PostTypeId="2" ParentId="4498" CreationDate="2010-10-01T02:03:53.727" Score="0" ViewCount="0" Body="&lt;p&gt;This is not a classification, but it is interesting to notice that every finite complete group can be obtained by the following process: start with any finite centerless group $G$ and iterate the automorphism group operations to build the automorphism tower:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G\to\mathop{Aut}(G)\to \mathop{Aut}(\mathop{Aut}(G))\to\cdots$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each group maps into the next by the inner automorphisms. If $G$ is centerless, then so is $\mathop{Aut}(G)$ and all the groups in the tower are centerless. The tower terminates at a complete group, a centerless group that is isomorphic to $\mathop{Aut}(G)$ by the canonical map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a beautiful theorem of Wielandt (1939) that every finite centerless group has an automorphism tower that terminates in finitely many steps. Thus, every finite complete group arises in this fashion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is also considered at &lt;a href=&quot;http://mathoverflow.net/questions/5635/does-autaut-autg-stabilize&quot; rel=&quot;nofollow&quot;&gt;this classic MO question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truly interesting things occur if one is willing to push things harder by continuing the iteration transfinitely. That is, having built the finite part of the automorphism tower, which has an associated system of mappings, one may simply take the direct limit to produce the limit group $G_\omega$ at stage $\omega$ and continue the iteration transfinitely with &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G_0\to G_1\to G_2\to\cdots G_\omega\to G_{\omega+1}\to\cdots\to G_\alpha\to\cdots$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where one takes the automorphism group at successor ordinal stages and direct limits of the system at limit ordinal stages. Simon Thomas proved the wonderful theorem that every centerless group leads in this transfinite manner to a complete group, where the process stops (see &lt;a href=&quot;http://www.jstor.org/pss/2044505&quot; rel=&quot;nofollow&quot;&gt;Proceedings AMS article&lt;/a&gt;). I extended this result by proving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; Every group $G$ has a terminating transfinite automorphism tower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof proceeds by showing that every $G$ leads eventually to a centerless group in the transfinite tower, whose subsequent tower then terminates by Thomas' theorem.  You can see &lt;a href=&quot;http://arxiv.org/abs/math/9808014&quot; rel=&quot;nofollow&quot;&gt;my paper at the ArXiv&lt;/a&gt;, published in the &lt;a href=&quot;http://www.jstor.org/stable/119135&quot; rel=&quot;nofollow&quot;&gt;Proceedings of the AMS&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an open question how tall the automorphism tower of a finite group can be. Examples are known with nontrivial centers that do not terminate at any finite stage and make it out to $\omega+n$ for any desired $n$, but the smallest upper bound is on the order of the least inaccessible cardinal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simon Thomas is writing a book on the subject of the automorphism tower problem, and the preliminary versions that I have seen are excellent.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-10-01T12:43:32.170" LastActivityDate="2010-10-01T12:43:32.170" />
  <row Id="5798" PostTypeId="2" ParentId="5795" CreationDate="2010-10-01T02:21:31.977" Score="2" ViewCount="0" Body="&lt;p&gt;In my original answer I only mentioned that it works for $Y$ complete, but as Nate pointed out in a comment, I never actually used completeness of $Y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes. Weakly convergent sequences in a normed space are bounded, as a consequence of the uniform boundedness principle applied to the dual space (which is a Banach space) and the fact that a convergent sequence of real (or complex) numbers is bounded. If $T$ is unbounded, then there is a sequence $x_1,x_2,\ldots$ in $X$ converging in norm (and hence weakly) to 0 such that $\|T(x_n)\|\to\infty$, so by the previous sentence this implies that $T(x_1),T(x_2),\ldots$ does not converge weakly. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-01T03:26:16.743" LastActivityDate="2010-10-01T03:26:16.743" CommentCount="5" />
  <row Id="5799" PostTypeId="1" AcceptedAnswerId="5811" CreationDate="2010-10-01T02:56:04.147" Score="2" ViewCount="97" Body="&lt;p&gt;Consider a canonical eliptical cone C with its vertex at the origin, with height h, and with a base given by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;(x/a)&lt;sup&gt;2&lt;/sup&gt;+(y/b)&lt;sup&gt;2&lt;/sup&gt; = 1; z = h&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;a, b not equal to 0.  Given a point p = (p&lt;sub&gt;x&lt;/sub&gt;, p&lt;sub&gt;y&lt;/sub&gt;, p&lt;sub&gt;z&lt;/sub&gt;) how would you determine if p is interior to C?  I realize I can first check to see if p is outside of the elliptical right cylinder where one base the base of the elliptical cone and the other is on the XY plane, so for grins, assume that check has already been made, and p is in fact inside that cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious solution I see is to linearly scale a and b by p&lt;sub&gt;z&lt;/sub&gt;/h and see if p is inside the ellipse:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;(h/p&lt;sub&gt;z&lt;/sub&gt;)&lt;sup&gt;2&lt;/sup&gt;[(x/a)&lt;sup&gt;2&lt;/sup&gt;+(y/b)&lt;sup&gt;2&lt;/sup&gt;]=1; z = p&lt;sub&gt;z&lt;/sub&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;Are there other approaches?  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2010-10-01T04:39:12.180" Title="How do I determine if a point is interior to an elliptical cone?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5801" PostTypeId="1" AcceptedAnswerId="5813" CreationDate="2010-10-01T03:22:36.873" Score="4" ViewCount="266" Body="&lt;p&gt;This is an old exam problem I found online. For any positive integers $a,b$, one has $a^4|b^3$ implies $a|b$. Clearly if $a^4|b^3$, then $a|b^3$. It seemed simple on first reading, but I can't figure out how to show that $a|b$ follows, since $a$ is not necessarily prime. Is there some observation I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-08T16:27:11.213" Title="For any positive integers $a,b$, one has $a^4|b^3$ implies $a|b$?" Tags="&lt;number-theory&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5802" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T03:37:36.380" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose $p$ is prime and $p^k$ divides $a$ (and no higher power of $p$ divides $a$).  Then $p^{4k}$ divides $a^4$ and hence $p^{4k}$ divides $b^3$.  It follows that $p^{4k/3}$ divides $b$.  In particular, $p^k$ must divide $b$ since $4k/3 \geq k$.  Doing this for all $p$ shows that $a|b$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-01T03:37:36.380" CommentCount="3" />
  <row Id="5804" PostTypeId="1" AcceptedAnswerId="13355" CreationDate="2010-10-01T03:48:41.257" Score="7" ViewCount="181" Body="&lt;p&gt;I think this is asked as a standard exercise in books about wavelets (e.g. exercise 7.2 in &lt;a href=&quot;http://books.google.com/books?id=hbVOfWQNtB8C&amp;amp;pg=370&quot; rel=&quot;nofollow&quot;&gt;Mallat's book&lt;/a&gt;), but I couldn't find a proof.  Let $\phi$ be a scaling function (see definition below).  I would like to learn why &#xA;$$\sum_{k\in\mathbb Z} \phi(x-k) = 1 $$&lt;br&gt;&#xA;almost everywhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; A sequence of subspaces $\{V_j: j\in \mathbb{Z}\}$ of $L^2(\mathbb R)$ is&#xA;called a &lt;em&gt;multiresolution analysis&lt;/em&gt;  if it satisfies the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$V_j \subset V_{j+1}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\bigcap_{j}V_j = \{0\}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\overline{\bigcup_jV_j} = L^2(\mathbb R)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$f(x)\in V_j$ if and only if $f(2x) \in V_{j+1}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;There exists a function $\phi \in V_0$ such that&#xA;$\{\phi(x-k)\}_{k\in\mathbb Z}$ is an orthogonal basis for $V_0$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The function $\phi$ here is called as a &lt;em&gt;scaling function&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T10:27:48.813" LastActivityDate="2010-12-09T10:27:48.813" Title="Integer translates of a scaling function" Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5805" PostTypeId="1" AcceptedAnswerId="5808" CreationDate="2010-10-01T04:07:22.047" Score="2" ViewCount="147" Body="&lt;p&gt;Suppose I have a 8x6 rectangle, with its lower left corner at the origin (0, 0). I want to scale this rectangle by 0.5 at an anchor point (3, 3). So the resulting rectangle is 4x3, but I cannot figure out how to compute the distance from the origin to the lower left corner of the new rectangle. Help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2239" LastActivityDate="2010-10-01T04:36:09.453" Title="Scaling at an arbitrary point and figuring out the distance from origin" Tags="&lt;geometry&gt;&lt;matrices&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5806" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T04:08:41.013" Score="10" ViewCount="0" Body="&lt;p&gt;It's simple: $\rm\ a^3|b^3\ \Rightarrow\ (b/a)^3\in\mathbb Z\ \Rightarrow\ b/a\in \mathbb Z\ \ $  by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate, the &lt;strong&gt;Rational Root Test&lt;/strong&gt; implies, as a special case, that if $\rm\ r\ $ is a rational root of an integer coefficient polynomial that is &lt;strong&gt;monic&lt;/strong&gt; (i.e. has leading coefficient equal to 1), then $\rm\:r\:$ is necessarily an integer. Hence, in the case above, since $\rm\ r = b/a\ $ is a root of the monic polynomial $\rm\ x^3 - n\ $ for some $\rm n\in \mathbb Z\:$, we deduce that $\rm\ r = b/a\ $ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is merely a special case of results equivalent to the irrationality of $\rm\:n\:$'th roots. Namely, any domain $\rm Z$ satisfying the following equivalent conditions is called &lt;strong&gt;root-closed&lt;/strong&gt;. This is a weaker condition than being integrally-closed, i.e. satisfying the monic Rational Root Test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\:$ TFAE for $\rm\: a,b\: $ in domain $\rm\:Z\:,\ \: r \in Q \:=\:$ fraction field of $\rm\: Z\:,\ n\in \mathbb N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\rm\ \ r = \sqrt[n]a  \ \Rightarrow\ r \in Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\rm\ \ r^n \in \:Z \:\ \Rightarrow\  r \in Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\rm\ \ \ a^n\:|\:b^n \:\ \Rightarrow\:\  a\:|\:b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) $\rm\ \ (a^n) =  (b^n,\: c^n) \ \Rightarrow\  (a) = (b,c)\ $  as ideals in $\rm\:Z$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T15:30:56.830" LastActivityDate="2010-10-01T15:30:56.830" CommentCount="1" />
  <row Id="5808" PostTypeId="2" ParentId="5805" CreationDate="2010-10-01T04:20:49.670" Score="3" ViewCount="0" Body="&lt;p&gt;Scaling by 0.5 about the center (3,3) is the same as translating by (-3,-3), scaling by 0.5 about (0,0), then translating by (3,3).  So, to find the coordinates of the lower left corner of the new rectangle, take the coordinates of the lower left corner of the original rectangle, subtract 3 from each coordinate, multiply each coordinate by 0.5, and add 3 to each coordinate.  Once you have the point's coordinates, you can find its distance from the origin.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-01T04:20:49.670" CommentCount="8" />
  <row Id="5809" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T04:21:37.420" Score="0" ViewCount="0" Body="&lt;p&gt;As you said, $a\vert b$. Hence every prime factor of $a$ is a prime factor of $b$. So we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a = p_1^{m_1} \dots p_r^{m_r} \qquad \text{and} \qquad b = p_1^{n_1} \dots p_r^{n_r} q_1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q_1$ is a positive integer number such that none of the $p_i$ divides it. So in order to see that $a\vert b$ it's enough to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n_i \geq m_i \qquad \text{for all} \ i = 1, \dots , r \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $n_i &amp;lt; m_i$ for some $i$. Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a^4 \vert b^3 \qquad \Longleftrightarrow \qquad b^3 = a^4q_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some positive integer $q_2$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p_1^{3n_1} \dots p_r^{3n_r} q_1^3 = p_1^{4m_1} \dots p_r^{4m_r}q_2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $q_2$ may have, or may have not, some of the $p_i$ as prime factors (but $q_1$ hasn't). In any case, we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;3n_i = 4m_i + t_i \qquad \text{for all} \ i= 1, \dots , r \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_i \geq 0$ for all $i$. But we have assumed that there is some $i$ such that $n_i &amp;lt; m_i$. So, for this $i$, we would have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;4m_i + t_i = 3n_i &amp;lt; 3m_i \qquad \Longleftrightarrow \qquad m_i + t_i &amp;lt; 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is impossible, since both $m_i, t_i \geq 0$. A contradiction. Hence $n_i \geq m_i$ for all $i$ and so $a\vert b$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-01T04:21:37.420" />
  <row Id="5810" PostTypeId="2" ParentId="5805" CreationDate="2010-10-01T04:36:09.453" Score="1" ViewCount="0" Body="&lt;p&gt;If you want to scale by .5 with anchor (a,b), then (x,y) will be sent to ((a+x)/2,((b+y)/2), the midpoint of the segment connecting the two points.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-01T04:36:09.453" />
  <row Id="5811" PostTypeId="2" ParentId="5799" CreationDate="2010-10-01T04:39:12.180" Score="5" ViewCount="0" Body="&lt;p&gt;Write down the equation for the cone:  it is &#xA;$$(x/a)^2 + (y/b)^2 - (z/h)^2 = 0.$$&#xA;Now the interior of the cone will be defined by the inequalities&#xA;$$(x/a)^2 + (y/b)^2 - (z/h)^2 &amp;lt; 0 \text{ and } 0 &amp;lt; z &amp;lt; h.$$&#xA;So given a point $(x,y,z)$, assuming that $0 &amp;lt; z &amp;lt; h$ (so that at least the $z$-coordinate&#xA;is compatible with lying in the interior of the cone), you evaluate $(x/a)^2 + (y/b)^2 - (z/h)^2$ and consider whether or not it is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course just a rephrasing of the description you give in your post,&#xA;but it is the most practical test that I can think of, and should be easy to program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: I am writing $(x,y,z)$ rather than $(p_x,p_y,p_z)$ just to make the notation simpler.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-01T04:39:12.180" />
  <row Id="5812" PostTypeId="1" AcceptedAnswerId="5814" CreationDate="2010-10-01T05:25:00.387" Score="2" ViewCount="87" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I'm at right place, as I asked this question at stackoverflow.com where the question was closed and sugested I go ask here. I'll try to be more comprehensive also, looking at couple of replies I got there&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is simple for those who know how algorhitms work, and as I'm physics student I'm not sure about the answer. I had exam in Statistics where professor suggested that formula I used (red outlined) in determining &quot;Least-Squares Regression Line&quot; was computationally inferior to the one derived below (green outline). My simple thinking was that first one uses n squares while derived (n+1) squares, but supposedly that's not enough. Can someone point why could derived formula be computationally more effective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hm, as new user I can't post image. Please look at the image here: &lt;a href=&quot;http://i.imgur.com/cd9ex.gif&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/cd9ex.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2240" LastActivityDate="2010-10-01T06:09:13.433" Title="Simple algorithm effectiveness (Least-Squares Regression Line in Terms of Sample Variances)" Tags="&lt;algorithms&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="5813" PostTypeId="2" ParentId="5801" CreationDate="2010-10-01T05:57:43.487" Score="6" ViewCount="0" Body="&lt;p&gt;This may be a situation where it is enlightening to ask (and answer) a more general question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: let $m$ and $n$ be positive integers.  Show that the following are equivalent:&lt;br&gt;&#xA;(i) For all positive integers $a,b,$ $a^m \ | \ b^n$ implies  $a \ | \ b$.&lt;br&gt;&#xA;(ii) $m \geq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the other answers given should generalize to this situation.  As people who are familiar with my number theory notes probably know, I like arguments using the functions $\operatorname{ord}_p(n)$, defined to be the largest number $a$ such that $p^a \ | \ n$.  The &quot;local to global principle&quot; for divisibility in the integers is: for $x,y \in \mathbb{Z}^+$, $x \ | \ y$ iff for all primes $p$, $\operatorname{ord}_p(x) \leq \operatorname{ord}_p(y)$.  [More generally, this is true in any unique factorization domain.  &lt;b&gt;Added&lt;/b&gt;: in fact, with appropriate modifications it is true in any &lt;a href=&quot;http://en.wikipedia.org/wiki/Krull_ring&quot; rel=&quot;nofollow&quot;&gt;Krull domain&lt;/a&gt;.  This actually came up in my own research recently...]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the present situation, $a^m \ | \ b^n$ implies that for all prime numbers $p$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \cdot \operatorname{ord}_p(a) = \operatorname{ord}_p(a^m) \leq \operatorname{ord}_p(b^n) = n \cdot \operatorname{ord}_p(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $m \geq n$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \cdot  \operatorname{ord}_p(a) \leq m \cdot \operatorname{ord}_p(a) \leq n \cdot \operatorname{ord}_p(b)$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\operatorname{ord}_p(a) \leq \operatorname{ord}_p(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that (ii) implies (i).  I'll leave the other direction to the reader as a simple exercise in understanding what's going on here, with the hint that one should prove the contrapositive: assume $m &amp;lt; n$ and take $a$ and $b$ to be powers of the same prime number. &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-08T16:27:11.213" LastActivityDate="2010-10-08T16:27:11.213" CommentCount="1" />
  <row Id="5814" PostTypeId="2" ParentId="5812" CreationDate="2010-10-01T06:09:13.433" Score="3" ViewCount="0" Body="&lt;p&gt;You're comparing the two variance formulae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_x^2=\frac1{n}\sum_{i=1}^n \left(x_i-\bar{x}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_x^2=\frac1{n}\left(\sum_{i=1}^n x_i^2\right)-\bar{x}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the computational effort needed, the second formula looks better: it only requires one loop to pass through your data, as opposed to the two needed for the first one (first for computing the mean, and then for adding up the squared deviations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth of the matter, however, is that the first formula makes much more computational sense than the second one. In practice, the two terms being subtracted in the second form can be very large; if the variance is small, you lose a lot of significant figures in the subtraction of two nearly-equal quantities (&lt;em&gt;subtractive cancellation&lt;/em&gt;). Worst comes to worst, you may even end up with a negative variance with the second formula!&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-01T06:09:13.433" CommentCount="5" />
  <row Id="5815" PostTypeId="2" ParentId="370" CreationDate="2010-10-01T07:15:09.930" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;MATH 4135/5135: Introduction to Category Theory&lt;/em&gt; by Peter Selinger&#xA;(17pp). Concise course outline. Only wish it covered more topics. Available in PS or PDF format.&#xA;&lt;br&gt; &lt;a href=&quot;http://www.mscs.dal.ca/~selinger/4135/&quot; rel=&quot;nofollow&quot;&gt;http://www.mscs.dal.ca/~selinger/4135/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Handbook of Categorical Algebra&lt;/em&gt; (Encyclopedia of Mathematics and its Applications) by Francis Borceux. Rigorous. Comprehensive. This is NOT free, but you can see the contents/index/excerpts at the publisher's web site, listed below. This is a three volume set: &lt;br&gt;&#xA;(v. 1) Basic Category Theory, 364pp. (ISBN-13: 9780521441780)&#xA;&lt;br&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521441780&#xA;&lt;br&gt;&#xA;(v. 2) Categories and Structures, 464pp. (ISBN-13: 9780521441797)&#xA;&lt;br&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521441797&#xA;&lt;br&gt;&#xA;(v. 3) Sheaf Theory, 544pp. (ISBN-13: 9780521441803)&#xA;&lt;br&gt;http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521441803&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reprints in Theory and Applications of Categories (TAC)&lt;/em&gt;. This site has 18 books and articles on category theory in PDF, including several by F.W. Lawvere.&#xA;&lt;br&gt;http://www.tac.mta.ca/tac/reprints/index.html&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abstract and Concrete Categories-The Joy of Cats&lt;/em&gt; by Jirı Adamek, Horst Herrlich, and George E. Strecker (524pp). Free PDF. Published under the GNU Free Documentation License. Mentioned already by Seamus in reference to Wikipedia's external links for Category Theory, but worth repeating, because it's very readable.&#xA;&lt;br&gt;http://katmat.math.uni-bremen.de/acc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A Gentle Introduction to Category Theory (the calculational approach)&lt;/em&gt; by Maarten M. Fokkinga (80pp).&#xA;&lt;br&gt;http://wwwhome.cs.utwente.nl/~fokkinga/mmf92b.html&lt;/p&gt;&#xA;" LastActivityDate="2010-10-01T07:15:09.930" CommunityOwnedDate="2010-10-01T07:15:09.930" />
  <row Id="5816" PostTypeId="2" ParentId="5793" CreationDate="2010-10-01T10:47:35.380" Score="0" ViewCount="0" Body="&lt;p&gt;Something like this should do the trick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;myNewArray  =  myOldArray - mean(myOldArray,1);&lt;/p&gt;&#xA;" OwnerUserId="2244" LastActivityDate="2010-10-01T10:47:35.380" CommentCount="1" />
  <row Id="5817" PostTypeId="1" AcceptedAnswerId="5873" CreationDate="2010-10-01T10:50:34.227" Score="3" ViewCount="146" Body="&lt;p&gt;Given a differential equation of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon \frac{d^ny}{dx^n} + \sum_{k=0}^{n-1} a_k(x)\frac{d^ky}{dx^k}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the WKB method says to choose the ansatz $y\sim exp({\frac{i\phi(x)}{\epsilon}})A(x,\epsilon)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A(x,\epsilon) = \sum_{k=0}^{\infty} A_k(x)\epsilon^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered what is the motivation for this educated guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-10-02T09:47:07.130" Title="The WKB method: motivation" Tags="&lt;differential-equations&gt;&lt;perturbation-theory&gt;" AnswerCount="1" />
  <row Id="5818" PostTypeId="1" AcceptedAnswerId="5829" CreationDate="2010-10-01T11:37:35.233" Score="2" ViewCount="385" Body="&lt;p&gt;This is an exercise from Apostol, which i have been struggling for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a prime $p$, how does one show that $${n \choose p} \equiv \biggl[\frac{n}{p}\biggr] \ (\text{mod} \ p)$$ Note that $\Bigl[\frac{n}{p}\Bigr]$ denotes the integral part of $\frac{n}{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to know as to how does one try to solve this problem. Well, what we need is to show is whenever one divides ${n \choose p}$ by a prime $p$ the remainder is the integral part of $\frac{n}{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $${ n \choose p} = \frac{n!}{p! \cdot (n-p)!}$$ Now $n!$ can be written as $$n!= n \cdot (n-1) \cdot (n-2) \cdots (n-p) \cdots 2 \cdot 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i am really struggling in getting the integral part.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-04T02:41:07.190" Title="Proving ${n \choose p} \equiv \Bigl[\frac{n}{p}\Bigr] \ (\text{mod} \ p)$" Tags="&lt;number-theory&gt;" AnswerCount="6" />
  <row Id="5819" PostTypeId="1" AcceptedAnswerId="5821" CreationDate="2010-10-01T11:50:04.010" Score="6" ViewCount="217" Body="&lt;p&gt;Let's say we have some topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiom $T_1$ states that for any two points $y \neq x$, there is an open neighborhood $U_y$ of $y$ such that $x \notin U_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we say that a topological space is $T_4$ if it is $T_1$ and also satisfies that for any two closed, non-intersecting sets $A,B$, there are open neighborhoods $U_A,U_B$ respectively, such that $U_A\cap U_B = \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone give an example of a topological space which satisfies the second condition of $T_4$, but which is not $T_1$?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-12-01T19:35:26.107" Title="Why is $T_1$ required for a topological space to be $T_4$?" Tags="&lt;topology&gt;&lt;general-topology&gt;&lt;counter-examples&gt;&lt;examples&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="5821" PostTypeId="2" ParentId="5819" CreationDate="2010-10-01T12:03:06.167" Score="14" ViewCount="0" Body="&lt;p&gt;Consider a topological space $X$ with two (different) points and the trivial topology (that is, only $\emptyset$ and $X$ are open sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This space satisfies the second condition of $T_4$: the only two closed non-intersecting sets are $\emptyset$ and $X$ and you can take as open neighborhoods $U_\emptyset = \emptyset$ and $U_X = X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $X$ is not $T_1$, as you can easily check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1$ (also called &lt;em&gt;Fréchet&lt;/em&gt; space) is equivalent to the fact that every point is a closed set (exercise: prove this!). So, if you had a &lt;em&gt;regular&lt;/em&gt; space ($T_3$) $X$ that was not $T_1$, you could not say that $X$ is $T_2$ (or &lt;em&gt;Hausdorff&lt;/em&gt;); that is, you couldn't say that $T_3 \Rightarrow T_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is necessary to add the $T_1$ condition to $T_3$ and $T_4$ in order to have the beautiful sequence of implications that makes all of us happy :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_4 \ \text{(normal)} \ \Longrightarrow \&#xA;T_3 \ \text{(regular)} \ \Longrightarrow \&#xA;T_2 \ \text{(Hausdorff)} \ \Longrightarrow \&#xA;T_1 \ \text{(Fréchet)}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-01T17:46:03.360" LastActivityDate="2010-10-01T17:46:03.360" />
  <row Id="5822" PostTypeId="2" ParentId="5772" CreationDate="2010-10-01T12:08:14.550" Score="5" ViewCount="0" Body="&lt;p&gt;I'll use the deduction theorem, so I'll assume $\beta$ and need to prove $\neg\neg\beta$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\beta$ (assumption)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\beta\to (\neg\neg\neg\beta\to\beta)$ (axiom 1)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\neg\neg\neg\beta\to\beta$ (modus ponens using 1 and 2)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\neg\neg\neg\beta\to\neg\beta$ (you have proved that $\neg\neg\beta\to\beta$ is a theorem)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\neg\neg\neg\beta\to\neg\beta)\to((\neg\neg\neg\beta\to\beta)\to\neg\neg\beta)$ (axiom 3)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\neg\neg\neg\beta\to\beta)\to\neg\neg\beta$ (modus ponens using 4 and 5)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\neg\neg\beta$ (modus ponens using 3 and 6)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2010-10-01T12:08:14.550" CommentCount="2" />
  <row Id="5823" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T12:09:31.810" Score="2" ViewCount="0" Body="&lt;p&gt;Consider ${n \choose p}$ as $(n)(n-1)...(p+2)(p+1)/(n-p)!$. That can be written as $(n)/(n-p) \cdot (n-1)/(n-p-1) \cdot \cdots \cdot (p+1)/(1)$. Notice that these are all of the form $(x+p)/x$. Thus, terms where $x$ is  not divisible by $p$ can be reduced modulo $p$ to $x/x=1$. The rest of the terms are going to be &#xA;$(p [n/p])/((p [n/p]-1))) \cdot \cdots \cdot (3p)/(2p) \cdot (2p)/(p)$. This product telescopes to $(p [n/p])/p = [n/p]$.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-01T16:04:02.720" CommentCount="5" />
  <row Id="5824" PostTypeId="1" CreationDate="2010-10-01T12:30:39.897" Score="8" ViewCount="196" Body="&lt;p&gt;Let us say that a Rubik's cube in a particular configuration is in a particular &quot;state&quot;. All other configurations of this cube (other &quot;states&quot;), which can be achieved by rotations of the cube can be thought to be connected to each other... rotations are like walking the edges of the graph where the different states are the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if our Rubik's cube is of the type where we can peel off the stickers easily and put them back too, then we can swap 2 of the stickers in a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the cube cannot be solved any more. Our cube has moved to a new state which is disconnected from the previous graph built above. Also, all the states reachable from this new state are also disconnected from the other graph. (otherwise the cube would be solvable). So now we have 2 graphs which are disconnected from each other. A third set of states might be disconnected from both the above states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all possible reasonable color assignments ( by reasonable, I mean to preserve the number of tiles for each color=9), how many such disconnected graphs exist for a 3x3x3 Rubik's cube?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditorDisplayName="" LastEditDate="2010-10-02T02:43:29.877" LastActivityDate="2010-10-02T02:43:29.877" Title="How many disconnected graphs of the Rubik's cube exist?" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5825" PostTypeId="2" ParentId="5824" CreationDate="2010-10-01T12:55:23.830" Score="2" ViewCount="0" Body="&lt;p&gt;You can calculate this using Burnside's Lemma, &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Burnside%27s_lemma&lt;/a&gt;, once you know the group of transformations for Rubik's cube.&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-10-01T12:55:23.830" />
  <row Id="5826" PostTypeId="1" CreationDate="2010-10-01T13:54:16.557" Score="1" ViewCount="135" Body="&lt;p&gt;How to prove connect sum of two manifolds doesn't depend on the choices of balls(which would be cutted) and different gluing of boundary spheres? Is it still true in differential category? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2248" LastEditorUserId="2248" LastEditorDisplayName="" LastEditDate="2010-10-01T14:12:24.690" LastActivityDate="2010-10-04T16:37:27.990" Title="a basic question about connect sum of two manifolds" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="5827" PostTypeId="1" AcceptedAnswerId="5841" CreationDate="2010-10-01T15:37:55.637" Score="1" ViewCount="234" Body="&lt;p&gt;This is a problem which appeared in one of my tests, which i wasn't able to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be a uncountable set. Let $S$ be the collection of subsets of $\Omega$ given by: $A \in S$ if and only if $A$ is countable or $A^{c}$ is countable. Suppose $f: \Omega \to \mathbb{R}$ is a real measurable function. Prove that there exists a $y \in \mathbb{R}$ and a countable set $B$ such that the $f(x)=y$ is on $B^{c}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-01T18:06:28.700" LastActivityDate="2010-10-01T23:15:45.617" Title="Measurable function remaining constant" Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="5828" PostTypeId="1" AcceptedAnswerId="5836" CreationDate="2010-10-01T15:51:24.840" Score="2" ViewCount="122" Body="&lt;p&gt;(Apologies for the vague way the question is stated.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the function like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $f(x) = x$ for $0\leq x \leq a$ (for some $a \geq 15 \in \mathbb{R}$) $\land$ $f(x) = a$ if $ x &gt; a$ . For $x&amp;lt;0$, it doesn't really matter what $f(x)$ looks like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has someone studied this type of function? Does it have a 'nicer' definition than the one stated above? Any references on the subject? It is not absolutely necessary for the function to have &lt;em&gt;all&lt;/em&gt; of the above characteristics. If you know a function that is similar to it, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation:&#xA;I'm looking for a function that assigns a constant to 'infinite values' of x and does not 'deform' finite values (too much). So the function should satisfy $f(\zeta(1))=a$ for some finite value of $a \in \mathbb{R}$ and $f(3)=3$, for example. It doesn't really matter whether $f(b)=b$ or any multiple of b or that $f(x)$ is some finite polynomial, as long as it's 'easy' to find the value of $f(b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller &lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2010-10-01T18:12:46.290" Title="Does the following function have a name and what properties does it have?" Tags="&lt;reference-request&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5829" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T16:18:03.913" Score="5" ViewCount="0" Body="&lt;p&gt;It's easy: $\:$ if $\rm\ n\equiv j\ \: (mod\ p)\:\: $ and $\rm\: \bigl[\frac{n}{p}\bigr] = k\;, \: $ then pairing factors top $\equiv$ bottom $\rm\:(mod\ p)\:$ we have  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad {n \choose p}\ =\ \frac{n\:(n-1)\:\cdots\:(n-p+1)}{p\:(p-1)\:\cdots\: 1} $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad \phantom{{n \choose p}}\ =\ \frac{n}{j}\ \frac{n-1}{j-1}\cdots\frac{kp+1}{1}\ \frac{kp}p\ \frac{kp-1}{p-1}\:\cdots\frac{n-p+2}{j+2}\ \frac{n-p+1}{j+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\rm\:n\equiv j\:,\: $ all of the above fractions are $\rm\: \equiv 1\:$ except the middle-shown one $\rm\: (kp)/p = k$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very special case of much more general arithmetical results on binomial coefficients. For example, see Andrew Granville's very interesting survey &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/&quot; rel=&quot;nofollow&quot;&gt;The Arithmetic Properties of Binomial Coefficients&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; The above proof is a special case of a very simple &lt;a href=&quot;http://math.stackexchange.com/questions/2192&quot;&gt;purely arithmetical proof&lt;/a&gt; that I devised to show that binomial coefficients are integral. Namely, the same idea of exploiting the innate symmetry by aligning the numerators and denominators $\rm\:(mod\ p)\:$ extends to yield a simple algorithm that, given a binomial coefficient and a prime $\rm p$, rewrites the binomial coefficient as a product of fractions whose denominators are all coprime to $\rm p$. This implies that no prime divides the lowest-terms denominator, so that we may therefore conclude that the binomial coefficient is integral. For an example and further discussion see my &lt;a href=&quot;http://math.stackexchange.com/questions/2192&quot;&gt;prior post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T19:50:52.380" LastActivityDate="2010-10-01T19:50:52.380" />
  <row Id="5830" PostTypeId="2" ParentId="3966" CreationDate="2010-10-01T16:24:49.200" Score="1" ViewCount="0" Body="&lt;p&gt;You are looking at the problem of mathematics without &lt;em&gt;knowing&lt;/em&gt; mathematics. Most of what you write applies to Chinese, provided you do not understand Chinese. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing: A question like «does addition really have any meaning outside of the context of placing two objects in a box and predicting how many are inside?» is meaningless to anyone who understands arithmetic and, better, &quot;higher arithmetic&quot;: it is obvious to him/her that addition is meaningful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, your &quot;axiom&quot; regarding solipsism is quite against observation.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-01T16:24:49.200" />
  <row Id="5831" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T16:56:34.653" Score="6" ViewCount="0" Body="&lt;p&gt;A useful result in such problems is &lt;a href=&quot;http://planetmath.org/encyclopedia/LucassTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Lucas's Theorem&lt;/a&gt; which states that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is a prime and if $a = \sum_{i=0}^{k} a_{i} p^{i},  \  0 \le a_i &amp;lt; p \ $ i.e. the $a_i$ are digits of $a$ in base $p$ and similarly $b = \sum_{i=0}^{k} b_{i}p^{i}$ (pad with zeroes if required) then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${a \choose b} = \prod_{i=0}^{k} {a_i \choose b_i} \mod p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case $n=a$ and $b=p$. Since $b = p$ we have $b_1 = 1$ and rest of the $b_i = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${a \choose p} = {a_1 \choose 1} = a_1 \mod p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that $a_1 = [\frac{a}{p}] \mod p$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-01T16:56:34.653" />
  <row Id="5832" PostTypeId="2" ParentId="5595" CreationDate="2010-10-01T17:03:20.077" Score="4" ViewCount="0" Body="&lt;p&gt;This analysis is correct, but not complete enough to convince me.  For example, why is the fate of the last person settled as soon as the first person's seat chosen?  Why will any other seat but the first person's or the last person's be taken by the time the last person boards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to fill in the holes for myself this way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last person's fate is decided as soon as anybody chooses the first person's seat (nobody is now in a wrong seat, so everybody else gets their assigned seat, including the last person) or the last person's seat (the last person now won't get their correct seat).  Any other choice at any stage doesn't change the probabilities at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rephrasing... at each stage, either the matter gets settled and there is a 50/50 chance it gets settled each way for the last person's seat, or the agony is just postponed.  The matter can thus be settled at any stage, and the probabilities at that stage are the only ones that matter -- and they are 50/50 no matter what stage. Thus, the overall probability is 50/50.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2010-10-01T17:03:20.077" CommentCount="0" />
  <row Id="5833" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T17:28:48.993" Score="5" ViewCount="0" Body="&lt;p&gt;Here's another way to look at it. Suppose $n=kp+j,$ for $0 \le j \le p-1.$ Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (p-1)! {n \choose p} = \frac{n(n-1) \cdots (n-p+1)}{p} = \left( {n-j \over p} \right)&#xA;\prod_{i=0,i\ne j }^{p-1} (n-i)$$&#xA;$$ \equiv k(p-1)! (\textrm{ mod } p). $$&#xA;Since the product runs through a complete set of non-zero residues mod p.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-01T17:39:26.450" LastActivityDate="2010-10-01T17:39:26.450" CommentCount="10" />
  <row Id="5834" PostTypeId="1" CreationDate="2010-10-01T17:49:19.413" Score="1" ViewCount="47" Body="&lt;p&gt;I'm looking for good online resources for 2nd order system dynamics. Any recommendations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for stuff that ideally includes discussion of Q, damping ratio, overshoot, bode plots, for systems with transfer functions  of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_1(s) = \frac{1}{\tau_2^2 s^2 + \tau_1 s + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_2(s) = \frac{\tau_1 s + 1}{\tau_2^2 s^2 + \tau_1 s + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_3(s) = \frac{s}{\tau_2^2 s^2 + \tau_1 s + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt; I've done change of variables to rewrite as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_1(\sigma) = \frac{1}{\sigma^2 + 2\zeta\sigma + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_2(\sigma) = \frac{2\zeta\sigma + 1}{\sigma^2 + 2\zeta\sigma + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_3(\sigma) / \omega_0 = \frac{\sigma}{\sigma^2 + 2\zeta\sigma + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma = \tau_2 s = s / \omega_0$, $\omega_0 = 1 / \tau_2$, and $\zeta = \tau_1 / 2\tau_2$ to normalize out the time scaling factor and end up with a transfer function of the single variable $\zeta$ (damping factor) which I think is the conventional treatment.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-10-01T18:18:57.637" LastActivityDate="2010-12-30T18:44:30.240" CommunityOwnedDate="2010-10-01T18:19:29.117" Title="good online resources for 2nd-order system dynamics" Tags="&lt;control-theory&gt;&lt;online-resources&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5835" PostTypeId="2" ParentId="5834" CreationDate="2010-10-01T18:03:16.463" Score="0" ViewCount="0" Body="&lt;p&gt;Searching in Google itself gives you a great deal of information. I got this link which explains has the content which you need.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/SysDyn/SysDyn2.html&quot; rel=&quot;nofollow&quot;&gt;http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/SysDyn/SysDyn2.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next, gigapedia.com is an excellent source for those who are in need of books online. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-01T18:03:16.463" CommunityOwnedDate="2010-10-01T18:03:29.587" CommentCount="1" />
  <row Id="5836" PostTypeId="2" ParentId="5828" CreationDate="2010-10-01T18:12:46.290" Score="2" ViewCount="0" Body="&lt;p&gt;f(x)=min(x,a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rewrite this in terms of the absolute value function as (x+a)/2-|x-a|/2, the average minus half the difference. If you prefer this can be seen as getting the graph of the absolute value function, and manipulating it to fit the function you want. This is possible since they are both formed of two straight lines meeting at a point.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-10-01T18:12:46.290" CommentCount="0" />
  <row Id="5837" PostTypeId="1" CreationDate="2010-10-01T18:14:29.650" Score="3" ViewCount="220" Body="&lt;p&gt;A student of mine has cooked up a new graphical notation for computing with knots on surfaces. The trouble is, writing up his results is difficult due to his new notation. Is there a good &quot;drawing tool&quot; for mathematics that anyone can suggest?&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-10-03T00:01:40.123" Title="A good, free, graphics package for mathematics?" Tags="&lt;math-software&gt;" AnswerCount="5" />
  <row Id="5838" PostTypeId="2" ParentId="5837" CreationDate="2010-10-01T18:24:42.500" Score="0" ViewCount="0" Body="&lt;p&gt;LaTeX is a good way to write all mathematical formulas and graphical notations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a lot more than that, but personnaly, I use it frequently only for that.&lt;/p&gt;&#xA;" OwnerUserId="2252" LastActivityDate="2010-10-01T18:24:42.500" />
  <row Id="5839" PostTypeId="1" AcceptedAnswerId="5850" CreationDate="2010-10-01T18:45:17.327" Score="2" ViewCount="317" Body="&lt;p&gt;Given that I know the point coordinates of A and B on segment AB and the expected length of a perpendicular segment CD crossing the middle of AB, how do I calculate the point coordinates of segment CD? &lt;/p&gt;&#xA;" OwnerUserId="2250" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-02T18:40:52.153" LastActivityDate="2010-10-05T19:48:17.120" Title="Finding point coordinates of a perpendicular bisector" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;analytic-geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="5840" PostTypeId="1" AcceptedAnswerId="5845" CreationDate="2010-10-01T18:47:34.920" Score="2" ViewCount="132" Body="&lt;p&gt;I have read a few weeks ago that some mathematical researchers have discover that the minimum number of movements to solve any initial configuration of a Rubik's cube has been downsized to 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do they calculate such a number ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2252" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-02T01:43:52.543" LastActivityDate="2010-10-02T05:52:22.320" Title="How is done the calculation of the minimum number of movement to solve any configuration of Rubik's Cube ?" Tags="&lt;combinatorics&gt;&lt;recreational-mathematics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5841" PostTypeId="2" ParentId="5827" CreationDate="2010-10-01T18:48:20.660" Score="3" ViewCount="0" Body="&lt;p&gt;Note, that the countable intersection of co-countable sets (i.e., sets who complement is countable) is co-countable ( since its complement is a countable union of countable sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the union of the inverse image of the sets $[n,n+1]$ under as $n$ varies over all integers is all of $\Omega$. Since the inverse image of each $[n,n+1]$ is either countable or co-countable, so at least one of them must be co-countable (since $\Omega$ is uncountable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $[n_1,n_1+1] = [a_1,b_1] $ is a set with co-countable (and hence uncountable) inverse image. Clearly, one of $[n_1, n_1 + 1/2]$ and $[n_1+1/2,n_1+1]$ must have a co-countable inverse image, call it $[a_2,b_2]$, ... proceeding in this manner we get a sequence of nested intervals $[a_n,b_n]$ each of whose inverse image is co-countable and $\lim_{n\to\infty} b_n - a_n = 0$, their intersection consists of a single point, say $y$, and  $f^{-1}({y}) = \cap f^{-1}( [a_n,b_n] )$ being an intersection of co-countable sets is co-countable. Call $B^{c} = f^{-1}({y})$, we are done.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-01T23:15:45.617" LastActivityDate="2010-10-01T23:15:45.617" CommentCount="1" />
  <row Id="5842" PostTypeId="1" AcceptedAnswerId="5844" CreationDate="2010-10-01T18:49:18.760" Score="2" ViewCount="189" Body="&lt;p&gt;I'd like to know if the function $ y = f(x) : [0,1] \rightarrow [0,1]$ defined implicitly by the trascendental equation $$\displaystyle y = e^{(y-1)/x}$$ is &quot;well known&quot; (name, properties) or is related to some other well known function.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-10-01T19:26:17.413" LastActivityDate="2010-10-02T12:00:13.993" Title="Implicit function  y = e^((y-1)/x)" Tags="&lt;functions&gt;&lt;implicit&gt;" AnswerCount="2" />
  <row Id="5843" PostTypeId="2" ParentId="5755" CreationDate="2010-10-01T18:54:48.807" Score="2" ViewCount="0" Body="&lt;p&gt;I assume $y \gt 0$ and $y \gg \mu$.  If you replace $x^k$ by $y^k \exp {\left( (k/y)x - k \right) }$ you will overestimate the $x^k$ term (because this is the exponential of the first two terms of the MacLaurin series of $\log(x^k)$ expanded around $x=y$, the series is alternating, and the remainder term is negative).  Completing the square yields a closed-form formula for an upper bound, one of whose factors is a Gaussian integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{2\pi }y^k \exp \left( {\frac{k (k+2 y (-y+\mu ))}{2 y^2}} \right) \Phi \left(\frac{k}{y}-y+\mu \right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will work extremely well when $k$ is large compared to $y$ and $\mu$ because then most of the mass of the integral is concentrated at its lower limit where the exponential upper bound to $x^k$ is a good approximation.  To avoid exponential overflow, use logarithms to compute the product.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-01T20:09:45.113" LastActivityDate="2010-10-01T20:09:45.113" CommentCount="1" />
  <row Id="5844" PostTypeId="2" ParentId="5842" CreationDate="2010-10-01T19:02:00.540" Score="2" ViewCount="0" Body="&lt;p&gt;It is solvable employing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W function&lt;/a&gt; as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation is $\rm\quad\quad\ Y \ \: =\ e^{\:Z-YZ}\ \: $ for $\rm\: \ Z\ =\: -\frac{1}X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\ \ Y\:Z\ e^{Y\:Z}\ \:=\:\ Z\ e^Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\ \  Y\:Z\ \ \:=\:\ \ W(Z\ e^Z) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\iff\ \ Y\: =\: -X\ W(-\frac{1}X e^{-1/X}) $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-02T02:46:31.823" LastActivityDate="2010-10-02T02:46:31.823" />
  <row Id="5845" PostTypeId="2" ParentId="5840" CreationDate="2010-10-01T19:07:56.737" Score="2" ViewCount="0" Body="&lt;p&gt;See the discussion and references on the associated Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Optimal_solutions_for_Rubik%27s_Cube&quot; rel=&quot;nofollow&quot;&gt;Optimal solutions for Rubik's Cube&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-01T19:07:56.737" CommentCount="1" />
  <row Id="5846" PostTypeId="2" ParentId="5839" CreationDate="2010-10-01T19:13:05.830" Score="3" ViewCount="0" Body="&lt;p&gt;Knowing the coordinates of A and B, you can find the slope of the line AB and the coordinates of the midpoint M of segment AB.  The slopes of perpendicular lines have product -1, so you can use the slope of AB to find the slope of CD.  With this slope and the point M, you can write an equation for the line CD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that the points C and D could be anywhere on that line, so long as they are the given length apart, so I suspect there is a bit of information missing.  If M is the midpoint of CD as well (so that AB and CD are the perpendicular bisectors of each other), then C and D are each half the given length from M, so you can write an equation for the circle with radius half the length and center M, and find where the line CD intersects the circle (solve the system of the two equations).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-01T19:13:05.830" />
  <row Id="5847" PostTypeId="2" ParentId="5826" CreationDate="2010-10-01T19:16:56.157" Score="1" ViewCount="0" Body="&lt;p&gt;First of all, you need your surfaces to be connected, otherwise the result is not true: let $X = \mathbb{T}^2 \sqcup \mathbb{S}^2$ and $Y = \mathbb{T}^2$. ($\mathbb{T}^2$ a torus, $\mathbb{S}^2$ a sphere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the connected sum $X \oplus Y$ &lt;em&gt;does&lt;/em&gt; depend on the choice of the disks. It doesn't matter which disk you choose on $Y$, of course, but if you choose your disk on $X$ belonging to $\mathbb{T}^2$, then you get $2\mathbb{T}^2 \sqcup \mathbb{S}^2$ and if you choose your disk belonging to $\mathbb{S}^2$, you obtain $\mathbb{T}^2 \sqcup \mathbb{T}^2$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-04T16:37:27.990" LastActivityDate="2010-10-04T16:37:27.990" />
  <row Id="5848" PostTypeId="1" AcceptedAnswerId="5852" CreationDate="2010-10-01T20:13:52.973" Score="0" ViewCount="282" Body="&lt;p&gt;This is the question which i am attempting to solve, and it seems to difficult to get rid of the exponents.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that a the two cubic curves $Y^3 = X^2 + X^3$ and $X^3 = Y^2 + Y^3$ intersect in nine points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated. By the way i would also like to know whether is there any general method for solving such problems. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-04T18:59:27.707" LastActivityDate="2010-10-06T18:16:28.757" Title="Intersection of Cubic curves." Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;&lt;plane-curves&gt;&lt;algebraic-curves&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="5849" PostTypeId="2" ParentId="5848" CreationDate="2010-10-01T20:20:22.343" Score="1" ViewCount="0" Body="&lt;p&gt;Like your earlier question about how to prove the associativity of the group law on elliptic curves, this too is a consequence of general results on intersection theory, esp. (variants of) &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezouts_theorem&quot; rel=&quot;nofollow&quot;&gt;Bezout's theorem&lt;/a&gt;. For computational purposes you can employ various constructive elimination techniques, e.g Grobner bases, various triangularization methods, etc, which are available in most computer algebra systems&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-01T20:26:29.420" LastActivityDate="2010-10-01T20:26:29.420" />
  <row Id="5850" PostTypeId="2" ParentId="5839" CreationDate="2010-10-01T20:29:45.717" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt;: Let $M$ be the midpoint of $[CD]$. I assume that $M$ belongs to $[AB]$, as shown above. If it is not the case, then this is only a particular solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KxkHD.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\left( x_{A},y_{A}\right) ,B(x_{B},y_{B}),C(x_{C},y_{C}),D(x_{D},y_{D})&#xA;$ be the four points. By translation and rotation of the $x,y$ axes as shown&#xA;in the figure, we can center $[A,B,C,D]$. The translation corresponds to the&#xA;change of variables&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\overline{x}+X$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\overline{y}+Y,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{x}=\dfrac{x_{A}+x_{B}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{y}=\dfrac{y_{A}+y_{B}}{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the rotation corresponds to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=x^{\prime }\cos \theta -y^{\prime }\sin \theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=x^{\prime }\sin \theta +y^{\prime }\cos \theta ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta =\arctan \dfrac{y_{B}-y_{A}}{x_{B}-x_{A}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining both transformations, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\overline{x}+x^{\prime }\cos \theta -y^{\prime }\sin \theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\overline{y}+x^{\prime }\sin \theta +y^{\prime }\cos \theta .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denoting the distance between the points $C$ and $D$ by $\overline{CD}=d_{CD}&#xA;$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{C}^{\prime }=x_{D}^{\prime }=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{C}^{\prime }=\dfrac{d_{CD}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{D}^{\prime }=-\dfrac{d_{CD}}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above transformations we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=\left( x_{C},y_{C}\right) =\left( \overline{x}-\dfrac{d_{CD}}{2}\sin&#xA;\theta ,\overline{y}+\dfrac{d_{CD}}{2}\cos \theta \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D=\left( x_{D},y_{D}\right) =\left( \overline{x}+\dfrac{d_{CD}}{2}\sin&#xA;\theta ,\overline{y}-\dfrac{d_{CD}}{2}\cos \theta \right) ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\overline{x}$ and $\theta $ are given above. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-05T19:48:17.120" LastActivityDate="2010-10-05T19:48:17.120" CommentCount="3" />
  <row Id="5851" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T20:32:00.997" Score="3" ViewCount="0" Body="&lt;p&gt;The binomial theorem together with the identity $(x+y)^p = x^p + y^p \mod p$ is useful for this type of problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we want the the $x^p$ coefficient of $(1+x)^{Ap+B} = (1+x^p)^A (1+x)^B = (1 + Ax^p + \dots)(1+X)^B$ where $0 \leq B &amp;lt; p$. This coefficient comes only from multiplication of $(Ax^p)$ by $(1)$.  So the result mod p is $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution is to use combinatorics.  Partition the set of integers from 1 to $n$ into $A$ blocks of size $p$ and a (perhaps empty) remainder of size $B$, as above.  Any subset is either a complete block or it contains a proper nonempty subset of at least one block.  Rotating the elements in the first partially filled block organizes the $\binom{n}{p}$ p-subsets into collections of size $p$, plus a &quot;remainder&quot; consisting of complete $p$-blocks, and there are $A$ of the those.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-01T20:32:00.997" CommentCount="1" />
  <row Id="5852" PostTypeId="2" ParentId="5848" CreationDate="2010-10-01T20:43:15.100" Score="4" ViewCount="0" Body="&lt;p&gt;In addition to the usual generalities on Bezout's theorem and resultants, notice that for this system, $X^2 + Y^2 = 0$, so that there are 6 finite solutions (3 on each line $X = \pm iY$, organized as 1 transverse intersection point per line, plus one double point at (0,0) on each line, leading to a quadruple intersection of the curves at 0)  and accounting for the other 3 solutions, the solutions &quot;at infinity&quot; with $X^3=Y^3$, i.e., the asymptotic directions $(X/Y) \to \omega, \omega^3=1, |X| \to \infty$.  In other words, this example is built to show the need for complex projective geometry, and the counting of points according to their multiplicities, as the environment where the Bezout 3x3=9 calculation works.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the generalities are: if f(x,y)=0 and g(x,y)=0 are the polynomials, you can write down a basis for k[x,y]/(f=g=0), and check that as a vector space it is of dimension 9. If you meant that there are 9 distinct intersection points, you can check that by writing down the action of the multiply-by-x operator on this vector space and seeing whether it has repeated roots. This can all be done by hand but is not as efficient as what is done in computer algebra systems. )&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-06T18:16:28.757" LastActivityDate="2010-10-06T18:16:28.757" CommentCount="3" />
  <row Id="5853" PostTypeId="2" ParentId="123" CreationDate="2010-10-01T21:03:01.563" Score="1" ViewCount="0" Body="&lt;p&gt;Most curves that look parabolic are actually Catenaries, which is based in the hyperbolic cosine function. A good example of a Catenary would be the Gateway Arch in Saint Louis, Missouri. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tractrix, which is based in the hyperbolic secant, is also known as the pursuit curve, which models objects like cargo trucks turning corners, or a dog on a porch starting to chase a car in the street.&lt;/p&gt;&#xA;" OwnerUserId="2254" LastActivityDate="2010-10-01T21:03:01.563" />
  <row Id="5854" PostTypeId="2" ParentId="5837" CreationDate="2010-10-01T21:15:23.110" Score="0" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also something called as TeXmacs which you might be interested in. As Xavier quote LaTeX is the best way to start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you are in need of a drawing package i list the following 3 packages which are commonly used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;pstricks&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;PgfTikZ&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gnuplot&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please google the names and you shall find lot of interesting things.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-01T21:15:23.110" />
  <row Id="5856" PostTypeId="2" ParentId="5827" CreationDate="2010-10-01T21:30:36.700" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a fun way to write the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define on $(\Omega, \mathcal{S})$ the probability measure $P(A) = 0$ if $A$ is countable, $P(A)=1$ if $A^c$ is countable.  Then $f$ can be seen as a random variable $X$.  Since all events have probability $0$ or $1$, all events, and hence all random variables, are independent.  Now there must be some $N$ with $P(|X| \le N) &gt; 0$ (since $\Omega = \bigcup_{N=1}^\infty \{|X| \le N\}$ and $P$ is countably additive), hence $P(|X| \le N) = 1$.  So $X$ is a.s. bounded and in particular is $L^2$.  But $Var(X) = Cov(X,X) = 0$ since $X$ is independent of itself!  So $X = EX$ a.s., i.e. except on a countable set.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-01T21:30:36.700" />
  <row Id="5857" PostTypeId="2" ParentId="5848" CreationDate="2010-10-01T21:31:15.850" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose $(x,y)$ satisfies both equations. Substituting one equation into the other, we obtain $x^2+y^2=0$. The only real solution to this equation are $x=y=0$, as pointed out by Moron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $x\neq 0$, and so $x$ is a complex number. Then $y = \pm i x$. Eliminating $y$ from the other equation yields two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$y = ix$, and $-ix^3 = x^2 + x^3$. We assume $x\neq 0$, so we have: $-ix = 1 + x$. This yields the solution: $x = (-1+i)/2, y=(-1-i)/2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$y = -ix$, and $ix^3 = x^2 + x^3$. Similarly to before, we have: $ix = 1+x$. This yields the solution: $x = (-1-i)/2, y=(-1+i)/2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, these are the only possible solutions... There are &lt;strong&gt;three&lt;/strong&gt; solutions, not nine.&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-10-02T05:13:03.113" LastActivityDate="2010-10-02T05:13:03.113" CommentCount="1" />
  <row Id="5858" PostTypeId="1" AcceptedAnswerId="5865" CreationDate="2010-10-01T21:39:33.073" Score="1" ViewCount="135" Body="&lt;p&gt;I have a two state dynamical system. The two state variables are $P$ and $Z$ and $a,b,c,d$ are parameters. The system equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{dt}=a\cdot P-b\cdot PZ=P\left(a-bZ\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dZ}{dt}=c\cdot PZ-d\cdot Z=Z\left(cP-d\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying both sides by $\frac{1}{PZ}$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{dt}\frac{1}{PZ}=a\frac{1}{Z}-b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dZ}{dt}\frac{1}{PZ}=c-d\frac{1}{P}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying with the original system equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dZ}{dt}\left(a\frac{1}{Z}-b\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dP}{dt}\left(c-d\frac{1}{P}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the difference of the last two equations gives:&#xA;$0=\frac{dZ}{dt}\left(a\frac{1}{Z}-b\right)+\frac{dP}{dt}\left(d\frac{1}{P}-c\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the chain rule, the total time derivative of my conserved quantity is:&#xA;$\frac{d}{dt}E\left(P,Z\right)=\frac{dZ}{dt}\frac{\partial}{\partial Z}E\left(P,Z\right)+\frac{dP}{dt}\frac{\partial}{\partial P}E\left(P,Z\right)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so then the following should hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial}{\partial Z}E\left(P,Z\right)=\left(a\frac{1}{Z}-b\right)$ and&#xA;$\frac{\partial}{\partial P}E\left(P,Z\right)=\left(d\frac{1}{P}-c\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;E\left(P,Z\right)=\int\left(a\frac{1}{Z}-b\right)\partial Z=\int\left(d\frac{1}{P}-c\right)\partial P&#xA;$&#xA;$=a\ln{Z}-bZ+C_{Z}\left(P\right)=d\ln{P}-cP+C_{P}\left(Z\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally: &#xA;$E\left(P,Z\right)=a\ln{Z}+dln{P}-bZ-cP$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; and this quantity is conserved! I had a bad calculus mistake.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="2256" LastEditorDisplayName="" LastEditDate="2010-10-02T00:03:43.957" LastActivityDate="2010-10-02T00:04:48.137" Title="Determine a conserved quantity in a dynamical system Lotka-Volterra" Tags="&lt;calculus&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5859" PostTypeId="2" ParentId="3977" CreationDate="2010-10-01T22:01:46.423" Score="3" ViewCount="0" Body="&lt;p&gt;There will be a public preview of GeoGebra 3D soon. Keep an eye on the GeoGebra forums :)&lt;/p&gt;&#xA;" OwnerUserId="2257" LastActivityDate="2010-10-01T22:01:46.423" CommunityOwnedDate="2010-10-01T22:01:46.423" CommentCount="1" />
  <row Id="5860" PostTypeId="2" ParentId="2158" CreationDate="2010-10-01T22:07:06.190" Score="5" ViewCount="0" Body="&lt;p&gt;Here is another proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use Legendre's formula for the exact power of a prime $p$ which divides $n!$ which is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=1}^{\infty} [\frac{n}{p^k}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[x]$ is the integer part of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coupled with $$ \sum_{i=1}^{n}  [x_i] \le [\sum_{i=1}^{n} x_i]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get that if $\sum_{i=1}^{n} a_i = N$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{n} [\frac{a_i}{p^k}] \le [\frac{N}{p^k}] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so any prime power which divides $(a_1)! \dots (a_n)!$ divides $N!$ and so $\frac{N!}{(a_1)! \dots (a_n)!}$ is an integer.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-01T22:07:06.190" />
  <row Id="5861" PostTypeId="1" CreationDate="2010-10-01T22:24:58.737" Score="7" ViewCount="171" Body="&lt;p&gt;Let $(\Omega, \mathcal{F}, P)$ be a probability space, $X$ an&#xA;integrable random variable, $\mathcal{G} \subset \mathcal{F}$ a&#xA;$\sigma$-field.  The &lt;b&gt;conditional expectation&lt;/b&gt; of $X$ given&#xA;$\mathcal{G}$ is by definition the unique random variable $Y$ which is&#xA;$\mathcal{G}$-measurable and satisfies $E[Y;A] = E[X;A]$ for all $A&#xA;\in \mathcal{G}$.  Proving the uniqueness of $Y$ is easy, but&#xA;existence is harder.  I am looking for a nice existence proof with&#xA;minimal prerequisites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional proof is to invoke the Radon-Nikodym theorem: the&#xA;signed measure $\nu(A) = E[X;A]$ on $(\Omega, \mathcal{G})$ is&#xA;absolutely continuous to $\mu = P|_\mathcal{G}$, so take $Y$ to be the&#xA;Radon-Nikodym derivative, and it clearly has the desired properties.&#xA;But the proofs I know of the Radon-Nikodym theorem, while elementary,&#xA;are somewhat involved (at least 2 pages, even if you only do the&#xA;absolutely continuous case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another proof is to first take $X$ with finite variance, and note that&#xA;$K = L^2(\Omega, \mathcal{G}, P)$ is a closed subspace of the Hilbert&#xA;space $H = L^2(\Omega, \mathcal{F}, P)$; then take $Y$ to be the&#xA;orthogonal projection of $X$ onto $K$.  Again, it is then easy to see&#xA;that $Y$ has the desired properties.  But this is not as suitable&#xA;for students with no functional analysis background.  You can develop&#xA;the necessary facts from scratch but it's a little tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am wondering if anyone knows of a simple proof, preferably using&#xA;only basic measure theory and probability facts.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-06T15:55:10.333" Title="Slick constructions of conditional expectation" Tags="&lt;probability-theory&gt;&lt;reference-request&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="5862" PostTypeId="2" ParentId="5818" CreationDate="2010-10-01T22:26:41.750" Score="1" ViewCount="0" Body="&lt;p&gt;Below is a sketch of the generating-function-based proof of Lucas's theorem - which is mentioned in passing in a number of other posts here. Below we will see that the original problem is just a special linear case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ (1 + X)^{\:a + b P + c P^2} \ \ \ (mod\ P)\quad\quad$ with $\rm\quad\quad 0 \le a,\: b,\: c &amp;lt; P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad  =\ \ \: (1 + X)^a \ (1 + X^P)^b \ (1 + X^{P^2})^c \ \ \ (mod\ P)\quad\quad$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad  =\ \ \:  (1 + \:\binom{\:a\:}1 \ X\ \ \ \:+ \ \:\binom{\:a\:}2\ X^2\ \ \  +\  \binom{a}3\ X^3 \ \ \ + \:\cdots\: + \ X^{a}\ )$&lt;br&gt;&#xA;$\rm\quad\quad\ \ *\   (1 + \binom{b}1 \ X^P \ + \ \binom{b}2\: X^{2P} \ \: +\ \binom{b}3\ X^{3P} \ \:+\: \cdots\: + \ X^{bP}) $&#xA;$\rm\quad\quad\ \ * \ (1 +\: \binom{\:c\:}1 \ X^{P^2} + \ \binom{\:c\:}2\ X^{2P^2} +\ \binom{c}3\ X^{3P^2} +\: \cdots\: + \ X^{cP^2}) \quad\ \ (mod\ P)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ \ \sum\ \binom{a}A \binom{b}B \binom{c}C\  X^{\: A + BP + CP^2}\quad\quad\quad\: (mod\ P)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\rm\quad \ \ \binom{a}A \binom{b}B \binom{c}C = \binom{a + bP + cP^2}{A + BP + CP^2}\ \ \ (mod\ P)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \Rightarrow\quad\ b\ \ =\ \ \binom{b}1 \quad\ \: =\quad\ \: \binom{a+bp}p\ \ \ $ for $\rm\ \  B = 1,\ A = C = c = 0,\ P = p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the result sought with $\rm\ n = a+bp\ \Rightarrow\ [n/p] = b\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other proofs see Granville's delightful survey &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/&quot; rel=&quot;nofollow&quot;&gt;The Arithmetic Properties of Binomial Coefficients&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T02:41:07.190" LastActivityDate="2010-10-04T02:41:07.190" />
  <row Id="5863" PostTypeId="1" CreationDate="2010-10-01T22:43:35.017" Score="1" ViewCount="245" Body="&lt;p&gt;If you are given of a ring, how do you find its ideals? &lt;/p&gt;&#xA;" OwnerUserId="2258" LastActivityDate="2010-10-02T06:15:29.217" Title="If you are given of a ring, how do you find its ideals?" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="5864" PostTypeId="2" ParentId="5837" CreationDate="2010-10-01T23:51:08.280" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://asymptote.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Asymptote&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-01T23:51:08.280" />
  <row Id="5865" PostTypeId="2" ParentId="5858" CreationDate="2010-10-02T00:04:48.137" Score="1" ViewCount="0" Body="&lt;p&gt;I have edited the original post to include the correct derivation. My mathematical weakness showed when I confused integration with differentiation!&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-10-02T00:04:48.137" />
  <row Id="5866" PostTypeId="2" ParentId="5837" CreationDate="2010-10-02T00:13:15.560" Score="1" ViewCount="0" Body="&lt;p&gt;On unix or Mac environments try xfig. There are windows simulators. Picture environments can be used with LaTeX to draw, but they are a pain (we used this in &quot;The Classical and Quantum 6j symbols&quot;). We used xfig for both &quot;Knotted Surfaces and Their Diagrams&quot; and &quot;Surfaces in 4-Space.&quot; Otherwise, get a good set of Rapidograph pens, some blue pencil, and lots of white-out. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-10-02T00:13:15.560" CommentCount="1" />
  <row Id="5867" PostTypeId="2" ParentId="5842" CreationDate="2010-10-02T01:10:59.630" Score="4" ViewCount="0" Body="&lt;p&gt;To give a more detailed solution, we start by rearranging the original equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y\exp\left(\frac{1-y}{x}\right)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y\exp\left(\frac1{x}\right)\exp\left(-\frac{y}{x}\right)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-y\exp\left(-\frac{y}{x}\right)=-\exp\left(-\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{y}{x}\exp\left(-\frac{y}{x}\right)=-\frac1{x}\exp\left(-\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we can now invert to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{y}{x}=W\left(-\frac1{x}\exp\left(-\frac1{x}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=-xW\left(-\frac1{x}\exp\left(-\frac1{x}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the Lambert function has &lt;em&gt;two&lt;/em&gt; real branches, only the &quot;principal branch&quot; gives a nontrivial solution; the solution corresponding to the $W_{-1}(\cdot)$ branch is $y=1$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-02T12:00:13.993" LastActivityDate="2010-10-02T12:00:13.993" />
  <row Id="5868" PostTypeId="1" AcceptedAnswerId="5872" CreationDate="2010-10-02T04:03:12.450" Score="1" ViewCount="135" Body="&lt;p&gt;Given lines $\mathbb{L}_1 : \lambda(1,3,2)+(-1,3,1)$, $\mathbb{L}_2 : \lambda(-1,2,3)+(0,0,-1)$ and $\mathbb{L}_3 : \lambda(1,1,-2)+(2,0,1)$, find a line $\mathbb{L}$ such that $\mathbb{L}$ is parallel to $\mathbb{L}_1$, $\mathbb{L}\cap\mathbb{L}_2 \neq \emptyset$ and $\mathbb{L}\cap\mathbb{L}_3 \neq \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb{L}$ must be parallel to $\mathbb{L}_1$, then $\mathbb{L}:\lambda(1,3,2)+(x,y,z)$ but I can't figure out how to get that (x,y,z) point. I'd like to be given just a slight nod because I'm sure the problem is really easy. Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-02T04:20:04.417" LastActivityDate="2010-10-03T13:12:35.357" Title="Finding a line that satisfies three conditions." Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="5869" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T04:20:51.297" Score="1" ViewCount="0" Body="&lt;p&gt;$\mathbb L \cap \mathbb L_2 \ne \emptyset$ implies that for some value of $\lambda$, your equation for $\mathbb L$ is on $\mathbb L_2$.  Since substituting $\lambda' = \lambda - c$ doesn't change the line $\mathbb L$, you can take $\lambda = 0$ without loss of generality.  That is, $(x,y,z)$ is on $\mathbb L_2$, or $(x,y,z) = \lambda_2(-1,2,3) + (0,0,-1)$. ...&lt;/p&gt;&#xA;" OwnerUserId="2236" LastActivityDate="2010-10-02T04:20:51.297" CommentCount="3" />
  <row Id="5870" PostTypeId="2" ParentId="5840" CreationDate="2010-10-02T05:52:22.320" Score="2" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://www.cube20.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.cube20.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2010-10-02T05:52:22.320" />
  <row Id="5871" PostTypeId="2" ParentId="5863" CreationDate="2010-10-02T06:15:29.217" Score="3" ViewCount="0" Body="&lt;p&gt;What Bill say in the comments is that there is no method to determine all ideals of a ring. Even if you specialize and look just for the maximal ideals in a given ring it might imply hard work! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great example is to look at $H^\infty(\mathbb{D})$ which is the collection of all power series $$f(z)=\sum_{n\ge0}a_nz^n$$ such that &#xA;$$\sup_{|z|&amp;lt;1}|f(z)|&amp;lt;\infty.$$&#xA;In other words, the space of all bounded analytic functions on the unit disc $\mathbb{D}$.&#xA;It is fairly easy to see that $R=H^\infty(\mathbb{D})$ is a ring under point-wise operations. &#xA;However, for long it was unknown how the maximal ideals ideals looks like. That problem was solved in 1962 by L. Carleson and is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Corona_theorem&quot; rel=&quot;nofollow&quot;&gt;Corona Theorem&lt;/a&gt;. I recommend reading this popular text regarding the &lt;a href=&quot;http://www.abelprisen.no/en/prisvinnere/2006/lennartcarleson/koronateoremet.html&quot; rel=&quot;nofollow&quot;&gt;Corona theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this day no one knows for sure how the the maximals looks like in $H^\infty(\mathbb{D}\times\mathbb{D})$, that is the space of bounded analytic functions in the bi-disc in $\mathbb{C}^2$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-02T06:15:29.217" CommentCount="7" />
  <row Id="5872" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T07:32:12.837" Score="5" ViewCount="0" Body="&lt;p&gt;Your first step is correct. Let's write the vector equations of your line $\mathbb{L}$ and the remaining ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\mathbb{L}_{\ \ } \qquad (x,y,z) &amp;amp;= (x_0, y_0, z_0) + \lambda (1, 3,2) \\&#xA;\mathbb{L}_2 \qquad (x,y,z) &amp;amp;= (0,0,-1) + \lambda_2 (-1, 2,3) \\&#xA;\mathbb{L}_3 \qquad (x,y,z) &amp;amp;= (2,0,1) + \lambda_3 (1,1,-2)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb{L}\cap\mathbb{L}_2 \neq \emptyset$, you may take the point $(x_0, y_0, z_0) \in \mathbb{L}_2$. So the equation for $\mathbb{L}$ must look like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathbb{L} \qquad (x,y,z) = (-\lambda_2 , 2\lambda_2, -1 + 3\lambda_2 ) + \lambda (1, 3,2)    \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $\mathbb{L}\cap\mathbb{L}_3 \neq \emptyset$, you should find the common point thanks to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(-\lambda_2 , 2\lambda_2, -1 + 3\lambda_2 ) + \lambda (1, 3,2) = (2,0,1) + \lambda_3 (1,1,-2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat in disguise, this is a $3\times 3$ system of linear equations, with unkowns $\lambda, \lambda_2$ and $\lambda_3$. Solve it. In particular, find $\lambda_2$ and you have your line $\mathbb{L}$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-02T08:15:36.193" LastActivityDate="2010-10-02T08:15:36.193" CommentCount="3" />
  <row Id="5873" PostTypeId="2" ParentId="5817" CreationDate="2010-10-02T09:47:07.130" Score="2" ViewCount="0" Body="&lt;p&gt;Look at the introduction to Chapter 10 in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387989315&quot; rel=&quot;nofollow&quot;&gt;Bender &amp;amp; Orszag&lt;/a&gt;. (&quot;Click to look inside&quot;, then search for &quot;WKB&quot; and go to p.484 in the search results.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-02T09:47:07.130" CommentCount="2" />
  <row Id="5874" PostTypeId="1" AcceptedAnswerId="5894" CreationDate="2010-10-02T11:39:46.653" Score="2" ViewCount="127" Body="&lt;p&gt;Given a set $E$ which is bounded, measurable and $m^{\ast}(E)=x &gt;0$, then for each $y \in (0,x)$ we can find a measurable set $A \subset E$, such that $m^{\ast}(A)=y$. To see this one considers measure as a continuous function and applies the Intermediate Value property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we extend the question in the following manner: Suppose $E$ is a set which has finite measure, then for each positive $x &amp;lt; m^{\ast}(E)$ prove that there exists a perfect set $A \subset E$, such that $m^{\ast}(A)=x$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-02T15:11:20.557" LastActivityDate="2010-10-02T21:25:18.457" Title="Existence of a perfect measurable set" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="5875" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T11:58:59.940" Score="3" ViewCount="0" Body="&lt;p&gt;I'd like to add the following to Agustí Roig's splendid answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of the general line passing through both $\mathbb{L}_2$ and $\mathbb{L}_3$ is given by $\underline{r} = \underline{a}+ \mu(\mathbb{L}_2 - \mathbb{L}_3),$ where $\underline{a}$ is some point on this line and its direction is $\mathbb{L}_2 - \mathbb{L}_3.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that the direction is (1,3,2), so set&#xA;$\lambda (1,3,2) = \mathbb{L}_2 - \mathbb{L}_3.$&#xA;That is&#xA;$$\lambda (1,3,2) = (0,0,-1)+\lambda_2(-1,2,3) – (2,0,1) - \lambda_3(1,1,-2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we obtain $\lambda_2 = -2$, $\lambda_3 = 2$ and $\lambda = -2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $\lambda_2 = -2$ in the equation of $\mathbb{L}_2$ gives the point (2,-4,-7), which is equivalent to w=0 in my comment to Agustí's answer. I hope that this adds some value for you.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-02T12:08:54.683" LastActivityDate="2010-10-02T12:08:54.683" />
  <row Id="5876" PostTypeId="2" ParentId="3627" CreationDate="2010-10-02T13:32:10.003" Score="0" ViewCount="0" Body="&lt;p&gt;What if rank(G) is m? Is there any efficient way?&lt;/p&gt;&#xA;" OwnerUserId="2265" LastActivityDate="2010-10-02T13:32:10.003" CommentCount="1" />
  <row Id="5877" PostTypeId="1" AcceptedAnswerId="5883" CreationDate="2010-10-02T14:07:17.043" Score="6" ViewCount="439" Body="&lt;p&gt;The web is littered with any number of pages (&lt;a href=&quot;http://planetmath.org/encyclopedia/ProofOfThuesLemma.html&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;) giving an existence and uniqueness proof that a pair of squares can be found summing to primes congruent to 1 mod 4 (and also that there are no such pairs for primes congruent to 3 mod 4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, none of the stuff I've read on the topic offers any help with actually efficiently finding (ie other than a straight search up to sqrt(p)) the concrete values of such squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to actually find them ?&lt;/p&gt;&#xA;" OwnerUserId="2266" LastActivityDate="2010-10-06T13:33:13.530" Title="Efficiently finding two squares which sum to a prime" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="5878" PostTypeId="2" ParentId="531" CreationDate="2010-10-02T14:19:56.223" Score="2" ViewCount="0" Body="&lt;p&gt;Greg Egan has some fun with this idea in one of his best short stories, &quot;Luminous&quot; (published in the collection of the same name).  A pair of researchers are exploring an apparent &quot;defect&quot; in mathematics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;You still don't get it, do you,&#xA;  Bruno? You're still thinking like a&#xA;  Platonist. The universe has only been&#xA;  around for fifteen billion years. It&#xA;  hasn't had time to create infinities.&#xA;  The far side can't go on&#xA;  forever-because somewhere beyond the&#xA;  defect, there are theorems that don't&#xA;  belong to any system. Theorems that&#xA;  have never been touched, never been&#xA;  tested, never been rendered true or&#xA;  false.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Terrific stuff!&lt;/p&gt;&#xA;" OwnerUserId="2266" LastEditorUserId="2266" LastEditorDisplayName="" LastEditDate="2010-10-03T22:58:53.233" LastActivityDate="2010-10-03T22:58:53.233" />
  <row Id="5879" PostTypeId="1" AcceptedAnswerId="5897" CreationDate="2010-10-02T14:27:39.830" Score="0" ViewCount="83" Body="&lt;p&gt;I am having a little trouble understanding some results from a program i wrote. I think i have the correct results but i need to understand the theory behind it a little more (i am a programmer, not mathematician).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My program generates symmetric k-regular graphs (single undirected edges). It generates an adjacency matrix and i take that adjacency matrix and find it's canonical isomorph. I compare the canonical isomorph to the list of all previous ones i have found to determine it if's a new k-regular graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is an example and i think it is correct but i just need to understand the result a little more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 6&lt;br /&gt;&#xA;k = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adjacency Matrix 1: &lt;br /&gt;&#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 1 0 0 0 &lt;br /&gt;&#xA;1 1 0 0 0 0 &lt;br /&gt;&#xA;0 0 0 0 1 1 &lt;br /&gt;&#xA;0 0 0 1 0 1 &lt;br /&gt;&#xA;0 0 0 1 1 0 &lt;br /&gt;&#xA;&lt;/code&gt;&#xA;&lt;br /&gt;&lt;br /&gt;&#xA;Canonical Isomorph:&#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 1 0 0 0 &lt;br /&gt;&#xA;1 1 0 0 0 0 &lt;br /&gt;&#xA;0 0 0 0 1 1 &lt;br /&gt;&#xA;0 0 0 1 0 1 &lt;br /&gt;&#xA;0 0 0 1 1 0 &lt;br /&gt;&#xA;&lt;/code&gt;&#xA;&lt;br /&gt;&#xA;The relabeling Label:&lt;br /&gt; &#xA;0 1 2 3 4 5&lt;br /&gt;&#xA;(which is the identity)&#xA;&lt;br /&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adjacency Matrix 2: &#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 0 1 0 0 &lt;br /&gt;&#xA;1 0 0 0 1 0 &lt;br /&gt;&#xA;0 1 0 0 0 1 &lt;br /&gt;&#xA;0 0 1 0 0 1 &lt;br /&gt;&#xA;0 0 0 1 1 0 &lt;br /&gt;&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canonical Isomorph: &#xA;&lt;code&gt;&lt;br /&gt;&#xA;0 1 1 0 0 0 &lt;br /&gt;&#xA;1 0 0 0 1 0 &lt;br /&gt;&#xA;1 0 0 0 0 1 &lt;br /&gt;&#xA;0 0 0 0 1 1 &lt;br /&gt;&#xA;0 1 0 1 0 0 &lt;br /&gt;&#xA;0 0 1 1 0 0 &lt;br /&gt;&#xA;&lt;/code&gt;&#xA;Relabeling: &lt;br /&gt;&#xA;0 1 2 5 3 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this look correct? Labeling means each vertex is labeled from 0-5.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-12-30T21:26:24.213" Title="Canonical Isomorphic Graphs" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5880" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T14:33:27.897" Score="5" ViewCount="0" Body="&lt;p&gt;You can try using the &lt;a href=&quot;http://www.numbertheory.org/courses/MP313/lectures/lecture13/page4.html&quot; rel=&quot;nofollow&quot;&gt;Hermite-Serret&lt;/a&gt; algorithm, which is briefly as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a $z$ such that $z^2 = -1 \mod p$. Apply the Euclidean algorithm to $p$ and $z$ stopping when the first pair of remainders $(x,y)$ go below $\sqrt{p}$. Hermite proved that $x^2 + y^2 = p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $z^2 = -1 \mod p$, you can probably guess a $z$ and start with $z^{p-1} = 1 \mod p$ and start taking square roots or as Robin noted (in a now deleted comment/his answer) guess an $a$ and see if $z = a^{\frac{p-1}{4}}$ is a candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been improvements to this by BrillHart. Derek was kind enough to hunt down the reference to that here: &lt;a href=&quot;http://jstor.org/pss/2005889&quot; rel=&quot;nofollow&quot;&gt;http://jstor.org/pss/2005889&lt;/a&gt; and a free link (thanks to J.M) &lt;a href=&quot;http://www.ams.org/journals/mcom/1972-26-120/S0025-5718-1972-0314745-6/S0025-5718-1972-0314745-6.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/mcom/1972-26-120/S0025-5718-1972-0314745-6/S0025-5718-1972-0314745-6.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-05T13:54:35.157" LastActivityDate="2010-10-05T13:54:35.157" CommentCount="6" />
  <row Id="5881" PostTypeId="1" AcceptedAnswerId="5882" CreationDate="2010-10-02T15:28:16.143" Score="2" ViewCount="91" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following assertion in my notes from last year that I'm trying hard to digest, but I think it isn't true:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $p$ is prime $\Leftrightarrow$ if $p | ab$ then either $p | a$ or $p | b$ or both.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A valid proof must prove both directions, so:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if $p$ is prime, let $a_p$ be the prime factorisation of $a$ and likewise $b_p$ be the prime factorisation of $b$. By the fundamental theorem of arithmetic, if $p|ab$ then $p|a$ or $p|b$ or both.&lt;/li&gt;&#xA;&lt;li&gt;However, assuming $p|ab$ and that $p|a$ or $p|b$ or both, I can find a case that doesn't work. Let $p=4$, $a=7$, $b=8$. Then $4|56$ does imply $4|8$ ($p|b$) but $4$ is not prime.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? The rest of my notes use this result quite a lot (albeit mostly from the point of view of $p$ prime which is clearly true). Unfortunately I'm not sure I'll be happy to progress unless I'm happy this statement is true...&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-10-02T17:28:29.477" Title="Prime divisibility" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;divisibility&gt;" AnswerCount="2" />
  <row Id="5882" PostTypeId="2" ParentId="5881" CreationDate="2010-10-02T15:36:01.467" Score="5" ViewCount="0" Body="&lt;p&gt;What you are missing is the following: The criterion &quot;$p|ab$ implies that $p|a$ or $p|b$&quot; has to hold &lt;strong&gt;for all natural numbers $a$ and $b$&lt;/strong&gt; for $p$ to be prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, $p=4$ is not prime, because $4|36$ but not $4|6$.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-02T15:36:01.467" CommentCount="1" />
  <row Id="5883" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T15:46:07.847" Score="10" ViewCount="0" Body="&lt;p&gt;In practice this comes to the same algorithm as Moron's, but I'll give&#xA;my two penn'orth anyway. There are two stages: (i) find $z$ with $z^2\equiv -1$&#xA;(mod $p$), and (ii) use $z$ to find $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage (i). If $a$ is a quadratic nonresidue modulo $p$ then $a^{(p-1)/2}&#xA;\equiv-1$ (mod $p$) so we can take $z\equiv a^{(p-1)/4}$ (mod $p$)&#xA;(and that is easily computed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_exponentiation&quot; rel=&quot;nofollow&quot;&gt;modular exponentiation&lt;/a&gt;).&#xA;How does one find quadratic nonresidues? Well just over half of all&#xA;numbers $a$ such that $1 &amp;lt; a &amp;lt; p-1$ &#xA;are quadratic nonresidues, so taking $a$ at random&#xA;will need at most two goes on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stage (ii). Compute the greatest common divisor of $p$ and $z+i$ using the&#xA;Euclidean algorithm for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_integer&quot; rel=&quot;nofollow&quot;&gt;Gaussian integers&lt;/a&gt;. The answer will&#xA;be $x+yi$ where $x^2+y^2=p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p\equiv 1$ (mod $2^k$) where $k\ge3$ one can speed up stage (i) a bit&#xA;on average. Compute $w=a^{(p-1)/2^k}$ modulo $p$. If $w\equiv\pm1$ we lose&#xA;but otherwise keep squaring $w$. Then we eventually get to $-1$ and before&#xA;that we have the required $z$. This sometimes wins even if $a$ is a quadratic&#xA;residue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect the Gaussian gcd above and Moron's Hermite-Serret algorithm amount&#xA;to the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is some rough-and-ready Python code to compute $a$ and $b$ such that&#xA;$p=a^2+b^2$. The function that does this is 2sq. It will behave erratically&#xA;if fed p not a prime congruent to 1 modulo 4.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def mods(a, n):&#xA;    if n &amp;lt;= 0:&#xA;        return &quot;negative modulus&quot;&#xA;    a = a % n&#xA;    if (2 * a &amp;gt; n):&#xA;        a -= n&#xA;    return a&#xA;&#xA;def powmods(a, r, n):&#xA;    out = 1&#xA;    while r &amp;gt; 0:&#xA;        if (r % 2) == 1:&#xA;            r -= 1&#xA;            out = mods(out * a, n)&#xA;        r /= 2&#xA;        a = mods(a * a, n)&#xA;    return out&#xA;&#xA;def quos(a, n):&#xA;    if n &amp;lt;= 0:&#xA;        return &quot;negative modulus&quot;&#xA;    return (a - mods(a, n))/n&#xA;&#xA;def grem(w, z):&#xA;    # remainder in Gaussian integers when dividing w by z&#xA;    (w0, w1) = w&#xA;    (z0, z1) = z&#xA;    n = z0 * z0 + z1 * z1&#xA;    if n == 0:&#xA;        return &quot;division by zero&quot;&#xA;    u0 = quos(w0 * z0 + w1 * z1, n)&#xA;    u1 = quos(w1 * z0 - w0 * z1, n)&#xA;    return(w0 - z0 * u0 + z1 * u1,&#xA;           w1 - z0 * u1 - z1 * u0)&#xA;&#xA;def ggcd(w, z):&#xA;    while z != (0,0):&#xA;        w, z = z, grem(w, z)&#xA;    return w&#xA;&#xA;def root4(p):&#xA;    # 4th root of 1 modulo p&#xA;    if p &amp;lt;= 1:&#xA;        return &quot;too small&quot;&#xA;    if (p % 4) != 1:&#xA;        return &quot;not congruent to 1&quot;&#xA;    k = p/4&#xA;    j = 2&#xA;    while True:&#xA;        a = powmods(j, k, p)&#xA;        b = mods(a * a, p)&#xA;        if b == -1:&#xA;            return a&#xA;        if b != 1:&#xA;            return &quot;not prime&quot;&#xA;        j += 1&#xA;&#xA;def sq2(p):&#xA;    a = root4(p)&#xA;    return ggcd((p,0),(a,1))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-06T08:21:18.800" LastActivityDate="2010-10-06T08:21:18.800" CommentCount="3" />
  <row Id="5884" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T16:22:30.517" Score="5" ViewCount="0" Body="&lt;p&gt;Just for amusement, here is an exact formula for finding $a$ so that $a^2 + b^2 = p$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a  = \dfrac{1}{2}\sum_{x = 0}^{p-1} \bigl( \dfrac{x^3 - x}{p} \bigr),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\bigl( \dfrac{a}{p} \bigr)$ denotes the Legendre symbol of $a$ mod $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: At Bill Dubuque's suggestion, I am noting here that this exact formula is not&#xA;an &lt;em&gt;efficient&lt;/em&gt; means to compute the value of $a$ (hence my prefatory &quot;Just for amusement ...&quot;).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-04T05:32:38.230" LastActivityDate="2010-10-04T05:32:38.230" CommentCount="6" />
  <row Id="5885" PostTypeId="2" ParentId="5881" CreationDate="2010-10-02T17:28:29.477" Score="1" ViewCount="0" Body="&lt;p&gt;You could also take a look at this &lt;a href=&quot;http://math.stackexchange.com/questions/4268/alternate-definition-of-prime-number&quot;&gt;previous question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-02T17:28:29.477" CommentCount="1" />
  <row Id="5886" PostTypeId="2" ParentId="5868" CreationDate="2010-10-02T17:49:49.377" Score="1" ViewCount="0" Body="&lt;p&gt;Notation can help us see what's going on.  Let's systematically parametrize the lines: let $\mathbb{L_2}$ start at $\mathbf{a} = (0,0,-1)$ and have a direction vector $\mathbf{u} = (-1,2,3)$ and similarly let $\mathbf{b} = (2,0,1)$, $\mathbf{v} = (1,1,-2)$ determine $\mathbb{L_3}$.  All we need to know about $\mathbb{L_1}$ is its direction vector, $\mathbf{w} = (1,3,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there exists a solution then we can start at the origin $\mathbf{a}$ of $\mathbb{L_2}$ and reach the origin $\mathbf{b}$ of $\mathbb{L_3}$ via a path that takes us along $\mathbb{L_2}$, thence in the direction $\mathbf{w}$ of $\mathbb{L_1}$ to a point on $\mathbb{L_3}$, and finally back along $\mathbb{L_3}$ to $\mathbf{b}$.  Let $\lambda$, $\rho$, and $\mu$ be the multiples of the direction vectors $\mathbf{u}$,  $\mathbf{w}$, and  $\mathbf{v}$ that are traveled, respectively.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{b - a} = \lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solution line $\mathbb{L}$ passes through $\mathbf{a} +  \lambda \mathbf{u}$ in the $\mathbf{w}$ direction.  All we need to do, then, is find $\lambda$.  A simple way is to find a linear form that kills both $\mathbf{w}$ and $\mathbf{v}$ but not $\mathbf{u}$ and apply it to both sides.  The obvious one is to take a dot product with some vector that is mutually perpendicular to $\mathbf{w}$ and $\mathbf{v}$.  All vectors are multiples of the cross product $\mathbf{v} \times \mathbf{w}$, so let's use it.  Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w}) =(\lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}) \cdot (\mathbf{v} \times \mathbf{w}) = \lambda \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing gives the solution (if it exists!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} = -2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A parametrization of the line $\mathbb{L}$ therefore is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{a} + \mathbf{u} \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} + t \mathbf{w}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a real parameter $t$.  To check the solution, reverse the roles of the lines $\mathbb{L_2}$ and $\mathbb{L_3}$, thereby finding $\mu$ and the point along $\mathbb{L_3}$.  (We don't have to recompute the triple product in the denominator; only its sign changes.)  The corresponding point (on $\mathbb{L_3}$ now) indeed lies on the solution line (with $t = 2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach immediately shows a solution does exist and is unique if and only if the triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is nonzero; i.e., the three direction vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are not coplanar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this all is just another way of reproducing the previous solutions.  Its possible advantages are its geometric clarity and the straightforward formulas it provides for the answer.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-03T13:12:35.357" LastActivityDate="2010-10-03T13:12:35.357" CommentCount="1" />
  <row Id="5887" PostTypeId="2" ParentId="5839" CreationDate="2010-10-02T18:09:03.357" Score="0" ViewCount="0" Body="&lt;p&gt;Well, I'm not too good at proving things, but here goes my attempt. This is a general result for when we don't know one of the endpoints of $\overline{CD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have points $A(a_x,a_y)$ and $B(b_x,b_y)$, and length of the perpendicular bisector $d$. Let the midpoint of $\overline{AB}$ be $\displaystyle M\left(\frac{a_x+b_x}{2},\frac{a_y+b_y}{2}\right)$, and let the point $Q$ be the point $d$ units away from $M$ along the perpendicular bisector $\overline{QM}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope of $\overline{AB}$ is $\displaystyle\frac{b_y-a_y}{b_x-a_x}$, so the slope of the perpendicular bisector is $\displaystyle -\frac{b_x-a_x}{b_y-a_y}$. The equation of the perpendicular bisector in vector form is the collection of all points $P$ where $P(t)=M+\langle a_y-b_y,b_x-a_x \rangle t$, $t \in \mathbb{R}$ ($M$ being the midpoint of $\overline{AB}$). Now, making the vector into a unit vector is much easier, and since the magnitude of the vector ($\sqrt{(a_y-b_y)^2+(b_x-a_x)^2}$) is simply the length $AB$, we can say that the point equation is $P(t)=M+\frac{1}{AB}\langle a_y-b_y,b_x-a_x \rangle t$, and thus the point $Q$ would be the point returned by the equation when $t=d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All possibilities for the points $C$ and $D$ can be found by finding the right number for $t$; namely, the first point $P(t)$ on the range of $t \in [-d,0]$ ($t = 0$ making it the midpoint of $\overline{AB}$), and the second point then being $P(t+d)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct me in any way necessary! I feel like that $P(t)$ equation will give you all the points possible if you use the correct range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I think an advantage of this method is that if the question were posed something like finding the endpoints of the perpendicular bisector with a certain length, where $C$ was, say, $\frac{2}{7}$ths of the distance along the line away, all it would take is plugging in $t=\pm\frac{2d}{7},\mp\frac{5d}{7}$.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-02T18:28:31.227" LastActivityDate="2010-10-02T18:28:31.227" CommentCount="2" />
  <row Id="5888" PostTypeId="2" ParentId="5861" CreationDate="2010-10-02T18:19:29.450" Score="1" ViewCount="0" Body="&lt;p&gt;For the basic case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $X$ and $Y$ are random variables on a probability space $(\Omega, \mathcal{F}, P)$ where $E[|Y|] &amp;lt; \infty$. Further assume that $X$ and $Y$ have joint probability distribution $f_{X, Y}(x,y)$. Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = \int_{\mathbb{R}} \frac{f_{X,Y}(x,y)}{f_X(x)} \, dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_X$ is the marginal density of $X$. now $g$ is the conditional expectation $E[Y|X = x]$ from elementary probability theory. Now we can see that $E[Y|X] = g(X)$. Now $g$ is $\sigma(X)$-measurable so now we need to check:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_A g(X) \, dP = \int_A Y \, dP \textrm{ for $A$ in $\sigma(X)$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the partial-averaging-property, so we get the conditional expectation. Well, this is just some syntax-manipulation so I'll skip that. I can add it if you want.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-06T15:55:10.333" LastActivityDate="2010-10-06T15:55:10.333" CommentCount="2" />
  <row Id="5889" PostTypeId="1" CreationDate="2010-10-02T19:00:20.743" Score="6" ViewCount="168" Body="&lt;p&gt;Let $G = A *_C B$ be an amalgamated free product of groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: suppose $C$ and $G$ are finitely generated, can we prove that so is $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to prove it by contradiction. Any suggestion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks...&lt;/p&gt;&#xA;" OwnerUserId="2270" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-02T19:27:43.247" LastActivityDate="2010-10-06T19:24:47.760" Title="Finite generation in amalgamated free products" Tags="&lt;group-theory&gt;&lt;category-theory&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="5891" PostTypeId="2" ParentId="5115" CreationDate="2010-10-02T20:03:09.010" Score="0" ViewCount="0" Body="&lt;p&gt;Improving Charles answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $$\sum\limits_{n=0}^{\infty} x^{n} = \frac{1}{1-x} \quad \text{for} \ |x|&amp;lt;1 \quad \cdots \ (1)$$ Differentiating term by term, we obtain, $$\sum\limits_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^{2}}, \quad  \text{for} \ |x|&amp;lt;1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating (1) term by term we get, $$\sum\limits_{n=0}^{\infty} \frac{1}{n+1}x^{n+1} = \int\limits_{0}^{x} \frac{1}{1-t} \ dt = -\log{1-x}$$ or we have $$\log{1-x} = -\sum\limits_{n=1}^{\infty}\frac{1}{n} x^{n} \quad \text{for} \ |x|&amp;lt;1 \quad \cdots \ (2)$$ Next putting $x = \frac{m-1}{m}$ we have $$\sum\limits_{n=1}^{\infty} \frac{1}{n} \Bigl(\frac{m-1}{m}\Bigr)^{n}  = -\log\Bigl(1 - \frac{m-1}{m}\Bigr) = -\log{\frac{1}{m}}=\log{m}$$ Hence we have $$\sum\limits_{n=1}^{\infty} \frac{1}{n} \geq \sum\limits_{n=1}^{\infty} \frac{1}{n} \Bigl(\frac{m-1}{m}\Bigr)^{n} = \log{m}$$ for all $m$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-02T20:03:09.010" />
  <row Id="5892" PostTypeId="1" CreationDate="2010-10-02T20:12:27.387" Score="8" ViewCount="418" Body="&lt;p&gt;As it is required for most students who wish to do a Ph.D in maths in the US to sit the GRE subject specific mathematics exam, I hope this question will be of interest to the mathematical community and will not be closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the exam was &quot;rescaled&quot; (made more difficult) in 2001 and I have only been able to find 2 past &quot;rescaled&quot; papers, one of which is available on the official website, the other available &lt;a href=&quot;http://www.math.ucsb.edu/mathclub/GRE/GRE9768.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Are other past papers available elsewhere? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In actual fact, the exam was not made more difficult (see comments), so any past paper would be of interest.&lt;/p&gt;&#xA;" OwnerUserId="2272" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-10-05T09:06:28.970" LastActivityDate="2010-10-31T11:49:06.070" CommunityOwnedDate="2010-10-03T16:28:14.450" Title="GRE past papers" Tags="&lt;reference-request&gt;&lt;education&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="6" />
  <row Id="5894" PostTypeId="2" ParentId="5874" CreationDate="2010-10-02T21:25:18.457" Score="4" ViewCount="0" Body="&lt;p&gt;Suppose first that $E$ is contained in the closed interval $[a,b]$ and let $E' = [a,b] \setminus E$. By definition of Lebesgue measure, we can find an open set $G$ containing $E'$ such that $m(G) = m(E') + \varepsilon = (b - a) - m(E) + \varepsilon$, where $\varepsilon$ is small enough. Then $F = [a,b] \setminus G$ is a compact set contained in $E$ with measure $m(F) = m(E) - \varepsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have chosen $\varepsilon$ small enough that $x \leq m(E)-\varepsilon = m(F)$. Your intermediate value argument proves the existence of some point $c \in [a,b]$ such that $m(F \cap [a,c]) = x$. By the Cantor-Bendixson Theorem, we have a decomposition $F \cap [a,c] = K \cup H$ where $K$ is perfect and $H$ is countable. Since $m(H) = 0$, it follows that $m(K) = m(F \cap [a,c]) = x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $E$ is unbounded, then pick $[a,b]$ such that $m(E \cap [a,b]) = m(E) - \delta$, where $\delta$ is small enough, and apply the above argument to $E \cap [a,b]$.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2010-10-02T21:25:18.457" />
  <row Id="5895" PostTypeId="2" ParentId="5877" CreationDate="2010-10-02T21:28:50.797" Score="3" ViewCount="0" Body="&lt;p&gt;One may quickly compute a representation of a prime $\rm\:p\equiv 1\ (mod\ 4)\:$ as a sum of two squares by employing the Euclidean GCD algorithm in $\rm\mathbb Z[\:i\:]\:$ and an algorithm for computing square roots $\rm\:(mod\ p)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\ $  Let $\rm\ \: c = \sqrt{-1}\ \:(mod\ p)\ $ and $\rm\ gcd(p,\:i-c)\: = \: a+b\:i\ $.$\ $ Then $\rm\ p = a^2 + b^2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ We shall show:  $\:$(1)$\ $ Representing $\rm\: p\:$ as a sum of squares is equivalent to finding a proper splitting $\rm\: p = \alpha\:\beta\ $ in $\rm\: \mathbb Z[\:i\:]\:$. $\ $(2)$\ $ Since $\rm\: \mathbb Z[\:i\:]\:$ has a Euclidean algorithm, a proper splitting of $\rm\: p\:$ can be computed by GCD with a suitable splitting of some multiple of $\rm\: p\ $. (3)$\ \:$ A suitable splitting of some multiple of $\rm\: p\:$ arises by factoring $\rm\ \: x^2+1\ \ (mod\ p\:)\:$,$\ $ i.e.$\ $ by computing $\rm\ \ \sqrt{-1}\ \ (mod\ p\:)\:$. $\:$  Below are the proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\:$ If $\rm\:\alpha|p\ $ properly then, conjugating, $\rm\:\alpha'|p\ $ properly. Multiplying them yields $\rm\:\alpha\:\alpha'|p^2\:$ properly in $\:\mathbb Z\:$.  But  in $\rm\mathbb{Z}\:$ the only proper factor$&gt;0$ of $\rm\:p^2\:$ is $\rm\: p\:$, thus $\rm\: p = \alpha\:\alpha' = (a+b\:i)(a-b\:i) = a^2 + b^2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\ $ If $\rm\ p\ \gamma \:=\: \alpha\:\beta\ $  and $\rm\:p\:$ does not divide $\rm\:\alpha\:$ nor $\rm\:\beta\:$,$\ $ then $\rm\: gcd(\alpha,p)\:$ is a $\:$ &lt;em&gt;proper&lt;/em&gt;$\ $ factor of $\rm\:p\ $.&lt;br&gt;&#xA;Else $\rm\ gcd(\alpha,p) = p\:$ or $1$. If the gcd is $\rm p$ then $\rm\: p\:|\:\alpha\:$ contra hypothesis.  Otherwise, if $\rm\ gcd(\alpha,p)=1\ $ then by Euclid's Lemma, $\rm\ \alpha\:|\:p\:\gamma \ \Rightarrow\ \alpha\:|\:\gamma\:$,$\ $ so $\rm\  \gamma/\alpha = \beta/p\ \Rightarrow\ p\:|\:\beta\:$,$\ $ again contra hypothesis. Note: generally, in rings, GCDs are unique only up to unit multiples. Here the units are $\rm\ i^n = \pm 1,\:\pm i\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\rm\ \ x^2 + 1 \ =\ (x-c)\:(x+c) + p\  f(x)\ $ in $\rm\mathbb\ Z[x]\ \: \Rightarrow\:  -p\ f(i)\ =\ (i-c)\:(i+c)\ $ in $\rm\mathbb\ Z[\:i\:]\ \ $ by evaluation at $\rm\: x = i\:$. This splitting is suitable to split $\rm\:p\:$ by (2) since $\rm\:p\:$ doesn't divide $\rm\: i\pm c\ \:$ in $\rm\ \: Z[\:i\:]\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ There are many variations on the Euclidean algorithm in $\rm\mathbb \ Z[\:i\:]\ $ that are employed in practice, e.g. employing continued fractions, binary quadratic forms, etc. Also, there are also at least a few algorithms for computing sqrts $\rm\:(mod\ p)\:$, e.g. by factoring polynomials $\rm\:(mod\ p)\:$, by elliptic curves (Schoof), and and algorithm of Tonelli and Shanks. For much further information see Henri Cohen's book &quot;A course in computational algebraic number theory&quot;. Here is an excerpt on the beautiful algorithm of Cornacchia:&#xA;&lt;img src=&quot;http://i.imgur.com/fG9OE.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/rdIY5.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/FjAUY.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T05:46:51.930" LastActivityDate="2010-10-04T05:46:51.930" />
  <row Id="5896" PostTypeId="1" AcceptedAnswerId="5903" CreationDate="2010-10-02T22:17:43.243" Score="6" ViewCount="236" Body="&lt;p&gt;$\sum_{n=1}^{\infty} \frac{\varphi(n)}{n}$ where $\varphi(n)$ is 1 if the variable $\text n$ has the number $\text 7$ in its typical base-$\text10$ representation, and $\text0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposed to find out if this series converges or diverges. I think it diverges, and here is why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that there is a series whose partial sums are always below our series, but which diverges. Compare some of the terms of each sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{7} &gt; \frac{1}{8}$&lt;br/&gt;&#xA;$\frac{1}{70} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\frac{1}{71} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\frac{1}{72} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\text ... $&lt;br/&gt;&#xA;$\frac{1}{79} &gt; \frac{1}{80}$&lt;br/&gt;&#xA;$\text ... $&lt;br/&gt;&#xA;$\frac{1}{700} &gt; \frac{1}{800}$&lt;br/&gt;&#xA;$\text ... $&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And continue in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously some terms are left out of the sequence on the left, which is fine since our sequence of terms on the left is already greater than the right side. Notice the right side can be grouped into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{8} + \frac{1}{8} + ... $ because we will have $10$ $\frac{1}{80}$s, $100$ $\frac{1}{800}$s, etc etc. Thus we are adding up infinitely many 1/8s. This is similar to the idea of the divergence of the harmonic series. So, my conclusion is that it diverges. A bunch of other students in my real analysis class have come to the conclusion that is, in fact, convergent, and launched into a detailed verbal explanation about comparison with a geometric series that I couldn't follow without seeing their work. Is my reasoning, like they suspect, flawed? I can't see how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry about the poor format, I'm new to TeX and couldn't figure out how to format a piecewise function (it was telling me a my \left delimiter wasn't recognized).&lt;/p&gt;&#xA;" OwnerUserId="2276" LastEditorUserId="2276" LastEditorDisplayName="" LastEditDate="2010-10-02T22:51:24.530" LastActivityDate="2010-10-03T23:13:03.293" Title="An interesting series." Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="5897" PostTypeId="2" ParentId="5879" CreationDate="2010-10-02T22:43:13.530" Score="1" ViewCount="0" Body="&lt;p&gt;Looks okay to me.  The first graph is two disjoint copies of C_3 whereas the second graph is C_6.  They are not isomorphic and they receive a different canonical label (which is correct).  And there's no other 2-regular simple graphs on 6 vertices.  The permutation listed is applied to the rows and columns of the matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything more specific you want to clarify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also compare your output with that of &lt;a href=&quot;http://cs.anu.edu.au/~bdm/nauty/&quot; rel=&quot;nofollow&quot;&gt;nauty&lt;/a&gt;.  It also has a function &lt;em&gt;geng&lt;/em&gt; that can be used to generate graphs with various properties.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-02T22:43:13.530" CommentCount="2" />
  <row Id="5898" PostTypeId="2" ParentId="5896" CreationDate="2010-10-02T23:11:36.427" Score="3" ViewCount="0" Body="&lt;p&gt;Your argument seems fine to me...  so I'll give the argument that popped into my head when I read the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum phi(n)/n for n congruent to 7 (mod 10) and multiply by 10 (which does not affect divergence/convergence).  Note that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10/7  &amp;gt; 1/7 +1/8 +...+1/16&#xA;10/17 &amp;gt; 1/17+1/18+...+1/26&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so on.  The result is &quot;larger&quot; than the harmonic series minus the first six terms (which diverges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, I actually prefer the OP's proof since it's self-contained (i.e. doesn't require the harmonic series).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-02T23:11:36.427" CommentCount="1" />
  <row Id="5899" PostTypeId="1" CreationDate="2010-10-02T23:26:08.467" Score="0" ViewCount="79" Body="&lt;p&gt;If the order of a group is pqr with p q r primes, then there exist three elements A B C with order p q r. Is it possible that the order of the subgroup generated by A and B has the order of pqr ?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-10-03T03:32:53.533" Title="a question about finite group" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5900" PostTypeId="2" ParentId="5837" CreationDate="2010-10-03T00:01:40.123" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://inkscape.org/&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt; is a good open source drawing package, and it &lt;a href=&quot;http://www.johndcook.com/blog/2009/12/22/including-latex-in-inkscape/&quot; rel=&quot;nofollow&quot;&gt;integrates with LaTeX&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-03T00:01:40.123" />
  <row Id="5901" PostTypeId="1" AcceptedAnswerId="5920" CreationDate="2010-10-03T00:09:03.130" Score="7" ViewCount="197" Body="&lt;p&gt;Suppose $V, V'$ are subspaces of dimension $d$ of a vector space $X$. Then there is a subspace $W$ of $X$ of codimension $d$ such that $W \cap V = W \cap V' = { 0 }$. &#xA;This can be proved by choosing an explicit basis for $X$ which contains a basis for $V$ and a basis for $V'$ and a basis for $V \cap V'$.  On the other hand, there should be a nice way to do this without choosing a basis. Can anyone explain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclosure: This came up when I was doing a homework problem. However, I'm just going to use the non-basis-free approach when I write up my answer.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-03T06:34:25.707" Title="How to show that you can find two subspaces that don't intersect" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="5902" PostTypeId="1" AcceptedAnswerId="5912" CreationDate="2010-10-03T01:42:35.153" Score="2" ViewCount="145" Body="&lt;p&gt;I was wondering what are the differences and relations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;between geometry and topology; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;between differential geometry and differential topology;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;between algebraic geometry and algebraic topology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they studying different objects? Such as different mathematical structures/spaces? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if they study the same object, but study different aspects/properties of the same object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading their wikipedia pages really confuses me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-10-03T05:12:38.343" Title="geometry and topology" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5903" PostTypeId="2" ParentId="5896" CreationDate="2010-10-03T02:23:50.617" Score="4" ViewCount="0" Body="&lt;p&gt;Yes your argument seems fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another argument is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the integers into blocks of 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[1 \dots 10] [11 \dots 20] [21 \dots 30] \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each block will have an number with digit $7$ in it and so the series has sum at least&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{10} + \frac{1}{20} + \dots + \frac{1}{10n} + \dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{1}{10}(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \dots)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is divergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another argument which uses the following useful result (the main reason for posting my answer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A = \{a_i\}$ is a sequence of natural numbers such that $\sum_{i=1}^{\infty} \frac{1}{a_i}$ converges then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; of $A$ is zero. (This is an exercise in Ivan Niven's book on Number theory and for a proof see here: &lt;a href=&quot;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/5967#5967&quot;&gt;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/5967#5967&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case the density of the numbers under consideration is at least $\frac{1}{10}$ and thus the sum cannot be convergent.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-03T23:13:03.293" LastActivityDate="2010-10-03T23:13:03.293" CommentCount="0" />
  <row Id="5904" PostTypeId="1" CreationDate="2010-10-03T02:35:58.657" Score="2" ViewCount="136" Body="&lt;p&gt;Does the basis of an Euclidean space have to be ordered by definition? Or can be left unordered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering about what is the morphism (i.e. the mapping that can preserve all the structures) on Euclidean spaces? Is it &lt;a href=&quot;http://planetmath.org/encyclopedia/Rotate.html&quot; rel=&quot;nofollow&quot;&gt;Euclidean transformation&lt;/a&gt; (rigid transformation), consisting of rotation, translation and reflection? Or the reflection is not part of it, because the basis of a Euclidean space must be ordered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-03T04:19:27.420" LastActivityDate="2010-10-03T12:06:21.923" Title="Must Basis of an Euclidean Space Be Ordered" Tags="&lt;euclidean-geometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="5905" PostTypeId="2" ParentId="5902" CreationDate="2010-10-03T02:55:27.563" Score="3" ViewCount="0" Body="&lt;p&gt;Your question is quite vague.  There are two parallel dynamics that go on when you learn mathematics.  On one level, mathematics is extremely specific so as you learn one subject in detail it appears as if all you know is that subject and there appears to be no relation to any other subject. But once you get to understand something you start noticing patterns -- standard examples that appear in many different fields, but in different guises, standard constructions many of which fit into categorical or other natural frameworks that are beyond the specifics of one field.  So to some extent there are broad unifying themes between subjects in mathematics.  In that regard there's many connections between subjects labelled by names where you combine two of the words from the set {geometry(ic), topology, algebra(ic)}.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at its most coarse, primitive level, there are some big differences.  Algebraic geometry is about the study of algebraic varieties -- solutions to things like polynomial equations. Geometric topology is largely about the study of manifolds -- which are like varieties but with no singularities, i.e. homogeneous objects.  Algebraic topology you could say is more about the study of homotopy-type or &quot;holes in spaces&quot;.  These are all inaccurate descriptions as in some sense subjects definitions are shaped by their histories.  I'd say for example that Algebraic topology is more defined by the nature of the tools it employs. While geometric topology is more motivated by objects it wants to prove theorems about.  That can seem like an artificial distinction, too, since isn't a &quot;tool&quot; an &quot;object&quot;?  Geometric topology is very much motivated by low-dimensional phenomena -- and the very notion of low-dimensional phenomena being special is due to the existence of a big tool called the Whitney Trick, which allows one to readily convert certain problems in manifold theory into (sometimes quite complicated) algebraic problems.  The thing is the Whitney trick fails in dimensions $4$ and lower.  In that regard, geometric topology has some characteristics of a grumpy old man who is really set on figuring out something specific. And algebraic topology in some sense has more of the air of the person that follows the natural lay-of-the-land from some formal perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it's much more common in the geometric topology community to go to a talk that's an illustration of an idea, or a problem illustrated entirely by examples: generating tables of knots or a census of manifolds, or testing a hypothesis by computer experiement, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after all that is said, there are many links between these fields and others, so it frequently difficult to disambiguate them except in rather pat, artificial ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if that helps at all.  But that's a generic 1st response. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-03T02:55:27.563" CommentCount="1" />
  <row Id="5906" PostTypeId="1" AcceptedAnswerId="5940" CreationDate="2010-10-03T02:58:34.453" Score="4" ViewCount="119" Body="&lt;p&gt;Is a triangle (its sides and the region enclosed by its sides) in a 2D Euclidean space $\mathbb{E}^2$ a manifold? I was thinking to use the identity mapping as its charts, but for each point on the sides of the triangle, there is no neighborhood of it can be mapped to an open subset in $\mathbb{E}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-10-03T19:16:11.357" Title="is triangle a manifold ?" Tags="&lt;manifold&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="5907" PostTypeId="2" ParentId="5899" CreationDate="2010-10-03T03:32:53.533" Score="2" ViewCount="0" Body="&lt;p&gt;Yes. Consider the group of 2x2 matrices whose entries come from the field of 7 elements.  Take A = [ -1, 0 ; 0, 1 ], B = [ 2, 1 ; 0, 1 ], C = [ 1, 1 ; 0, 1 ].  Then A,B,C generate a group of order 42=2*3*7 called Hol(7) = AGL(1,7).  A has order 2, B has order 3, C has order 7, but the subgroup generated by A and B also has order 42.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically the same trick used to generate the the non-abelian group of order 6 using two elements of order 2.  You can't take them from the same Sylow 2-subgroup, but if you choose from two Sylow 2-subgroups things work.  For p=2, q=3, r=7, we cannot choose two from the same Hall {p,q}-subgroup, lest we get a subgroup of order pq, so we choose from two different Hall 6-subgroups, and generate the whole group.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-10-03T03:32:53.533" />
  <row Id="5909" PostTypeId="1" AcceptedAnswerId="5917" CreationDate="2010-10-03T04:01:19.087" Score="2" ViewCount="151" Body="&lt;p&gt;Consider a simple random walk on a undirected, connected graph. This is the random walk which, at every time step, moves to a random neighbor, with all neighbors being equally likely. Lets assume every node has a self-loop to avoid issues associated with periodicity. Then, the following (surprising to me) fact is true: &lt;b&gt;under the stationary distribution, every edge is traversed with the same probability&lt;/b&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get some intuition for why this holds, and I am wondering whether someone &#xA;can provide an explanation for this phenomenon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, I know the standard proof which&#xA;observes that $\pi_i = d_{i}/\sum_k d_k$ satisfies the equations for the stationary distribution, and so $\pi_i p_{ij} = 1/\sum_k d_k$ whenever the edge $(i,j)$ is present in the graph. This proof, however, feels more like a lucky calculation to me than an explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the fact in boldface is false for directed graphs. Its also false for the corresponding two-step chain, i.e. the Markov chain with probability transition matrix $P^2$, where $P$ is the transition matrix of the simple random walk. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastEditorUserId="2120" LastEditorDisplayName="" LastEditDate="2010-10-03T04:15:35.890" LastActivityDate="2010-10-03T06:06:45.550" Title="simple random walks on undirected graphs" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5910" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T04:02:30.953" Score="4" ViewCount="0" Body="&lt;p&gt;Bases must be ordered, otherwise you couldn't speak of coordinates as &lt;em&gt;ordered&lt;/em&gt; tuples. And this could make things pretty messy. For instance, if the standard basis for $\mathbb{R}^2$ could be either $e_1, e_2$ or $e_2, e_1$ simultaneously, then it would be difficult to speak of &quot;the point of coordinates $(2,1)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflexions send bases to bases, because they are isomorphisms. But this doesn't mean they necessarily send a particular base to the &lt;em&gt;same&lt;/em&gt; particular base. So they can change the order of the vectors of the basis. They don't preserve &lt;a href=&quot;http://en.wikipedia.org/wiki/Orientation_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;orientation&lt;/a&gt;, but this is an extra piece of structure, not included in the definition of Euclidean space.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-03T10:34:30.693" LastActivityDate="2010-10-03T10:34:30.693" CommentCount="2" />
  <row Id="5911" PostTypeId="2" ParentId="5896" CreationDate="2010-10-03T04:11:03.563" Score="3" ViewCount="0" Body="&lt;p&gt;Here's another proof that your sum diverges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the sum $\sum_{n-1}^\infty (1-\phi(n))/n$.  This is the sum of the reciprocals of integers which &lt;i&gt;don't&lt;/i&gt; have a 7 in their decimal expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of integers $n$ with $1-\phi(n)=1$ and $1 \le n &amp;lt; 10^k$ is $9^k - 1$. (We can choose each of $k$ decimal digits to be anything but 7, except we don't want to choose all zeroes.) Thus the numbers of integers $n$ with $1-\phi(n) = 1$ and $10^{k-1} \le n &amp;lt; 10^k$ is $(9^k - 1) - (9^{k-1}-1) = 8 \times 9^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&#xA;$$ S_k = \sum_{n=10^{k-1}}^{10^k - 1} {1-\phi(n) \over n} $$&#xA;has $8 \times 9^{k-1}$ nonzero terms; they are each at most $10^{k-1}$. So $S_k \le 8 \times (9/10)^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinite sum is then&#xA;$$ \sum_{k=1}^\infty S_k \le 8 \times \sum_{k=1}^\infty (9/10)^{k-1} = 8 \times 10 = 80 $$&#xA;and in particular it's finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the harmonic series $\sum_{n=1}^\infty 1/n$ diverges; removing terms whose sum converges (like the sum that I just showed to converge) won't change that. So your series diverges.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2010-10-03T04:57:16.833" LastActivityDate="2010-10-03T04:57:16.833" CommentCount="1" />
  <row Id="5912" PostTypeId="2" ParentId="5902" CreationDate="2010-10-03T04:24:49.217" Score="5" ViewCount="0" Body="&lt;p&gt;As a supplement to Ryan's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential geometry typically studies Riemannian metrics on manifolds, and properties of them.  A typical differential geometry result is the &lt;em&gt;sphere theorem&lt;/em&gt;, stating that if $M$ is a closed manifold equipped with a Riemannian metric for which the sectional curvatures lie in the half-open interval $(1/4, 1]$, then $M$ is a sphere.  Note that the basic object is a manifold equipped with a Riemannian metric (a &lt;em&gt;Riemannian manifold&lt;/em&gt;), and the &lt;em&gt;curvature&lt;/em&gt; of the metric plays a key role in the statement of the theorem.  These are typical features of problems/theorems in differential geometry.  Note though that the conclusion of the theorem involves a statement about the topology of $M$; so there is certainly overlap between differential geometry and the concerns of topology.  (One might say that the sphere theorem is a global result, using geometric hypotheses to draw topological conclusions.  One can also have local results, in which topology plays no role in the hypothesis or conclusions: e.g. that a Riemanninan manifold with everywhere zero curvature is locally isometric to Euclidean space; one can also have global results that begin with topology and conclude with geometry: e.g. that any compact orientable surface of genus 2 or higher admits a Riemannian metric with constant curvature $-1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential topology refers to results about manifolds that are more directly topological, and don't refer to metric structures.  The generalized Poincare conjecture (that a closed manifold that is homotopic to a sphere is homeomorphic to one) is an example; another simpler example is Ehresmann's theorem, stating that a submersion between closed manifolds (or more generally, any &lt;em&gt;proper&lt;/em&gt; submersion) is a fibre bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these distinctions can be subtle, and may not always be well-defined, but a typical distinction bewteen geometry and topology in general (and which is borne out in the&#xA;preceding discussion) is that geometry studies metric properties of spaces, while topology studies questions which don't involve metric notions (it is the study of &lt;em&gt;pure shape&lt;/em&gt;, if you like; the old name &lt;em&gt;analysis situs&lt;/em&gt; also sheds some light on the meaning of topology).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One caveat is that classically, Euclidean geometry branched not only into other geometries&#xA;such as hyperbolic geometry (this branching being a precursor to Riemann's introduction of the general notions of Riemannian manifold and curvature), but gave rise to another branch known as projective geometry.  In projective geometry, metric notions of distance and angle aren't studied (because they are not preserved by projective transformations), but notions such as being a straight line, or being a conic section, are.  &lt;em&gt;Algebraic geometry&lt;/em&gt; is the modern subject which developed out of projective geometry (among other sources; see &lt;a href=&quot;http://math.stackexchange.com/questions/5167/group-law-for-an-elliptic-curve/5176#5176&quot;&gt;this answer&lt;/a&gt; for a discussion of a quite different problem --- computing elliptic integrals --- which was another historical precursor to algebraic geometry).   In algebraic geometry one studies &lt;em&gt;varieties&lt;/em&gt;, which are solution sets to polynomial equations; thus in its elementary form it feels a lot like what is called &lt;em&gt;analytic geomery&lt;/em&gt; in high-school,&#xA;namely studying figure in the plane, or in space, cut out by equations in the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this &lt;em&gt;geometry&lt;/em&gt; (as opposed to &lt;em&gt;topology&lt;/em&gt;, say)?  Because it turns out that when the functions one is using to cut out figures, or describe maps between figures, are restricted to be polynomial, the objects one obtains are quite rigid, in a way very similar to the way more traditional Euclidean geometry figures are rigid.  So one has the sensation of doing geometry, rather than topology.  (In topology, by contrast, things feel rather fluid, since one is allowed to deform objects in fairly extreme ways without changing their essential topological nature.)  And in fact it turns out that there are deeper connections between algebraic and metric geometry: for example, for a compact orientable surface of genus at least 2, it turns out that the possible ways of realizing this surface as an algebraic variety over the complex numbers are in a natural bijection with the possible choices of a constant curvature -1 metric on the surface.  A more recent example is given by the Calabi conjecture proved by Yau.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-03T05:12:38.343" LastActivityDate="2010-10-03T05:12:38.343" CommentCount="2" />
  <row Id="5913" PostTypeId="2" ParentId="5901" CreationDate="2010-10-03T04:51:48.407" Score="1" ViewCount="0" Body="&lt;p&gt;So let $G$ be the space of all co-dimension d subspaces. ($G$ is&#xA;naturally a manifold). I claim that if $W$ is any dimension $d$ subspace,&#xA;the set of $X$ s.t. $X \cap W = 0$ is an open, dense subset of $G$.&#xA;Open is easy. If you have $X_i$ each with non-trivial intersection with&#xA;W and $X_i \to X$ then you can pick lines $L_i$ in $W$ intersect $X_i$ and find&#xA;some limit (perhaps of a subsequence) so that $L_i \to L$. Then $L$ is in&#xA;$W \cap X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show dense is also not hard. You need to show that if $X$ intersects&#xA;$W$, then you can change $X$ by a little so that it no longer does. This&#xA;really isn't hard but it is annoying to do without ever picking a&#xA;basis of anything. How about this. You can think of $X$ as the image&#xA;of a map $Y \to Z$ ($Y$ is a space of dim $n-d$, $Z$ is your big space). $X$&#xA;intersects $W$ trivially iff the map $Y \to Z/W$ is an injection. Pick&#xA;some $U$ so that $Z = U + V$. Then we can think of our map as $Y \to U+V$,&#xA;and we want the map $Y \to U$ to be injective. But $Y$ and  $U$ have the same&#xA;dimension and it is easy to modify a map between such spaces by&#xA;epsilon to make it a bijection (for example add a small multiple of&#xA;some particular bijection).&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-03T04:51:48.407" CommentCount="1" />
  <row Id="5914" PostTypeId="1" AcceptedAnswerId="5988" CreationDate="2010-10-03T05:18:39.663" Score="3" ViewCount="196" Body="&lt;p&gt;How to prove that the only integral solutions to the equation $$y^{2}=x^{3}-1$$ is $x=1, y=0$. &#xA;I rewrote the equation as $y^{2}+1=x^{3}$ and then we can factorize $y^{2}+1$ as $$y^{2}+1 = (y+i) \cdot (y-i)$$ in $\mathbb{Z}[i]$. Next i claim that the factor's $y+i$ and $y-i$ are co-prime. But i am not able to show this. Any help would be useful. Moreover, i would also like to see different proofs of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extending&lt;/strong&gt; Consider the equation $$y^{a}=x^{b}-1$$ where $a,b \in \mathbb{Z}$ and $(a,b)=1$ and $a &amp;lt; b$. Then is there any result which states about the nature of the solution to this equation.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-03T05:30:58.677" LastActivityDate="2010-10-04T07:07:14.327" Title="Integral solutions to $y^{2}=x^{3}-1$." Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5915" PostTypeId="2" ParentId="5914" CreationDate="2010-10-03T05:38:24.663" Score="2" ViewCount="0" Body="&lt;p&gt;If $x$ is even then $x^3$ is divisible by 4, so $x^3-1 \equiv 3 \pmod 4$ and this cannot be a square. Thus $x$ is odd and $y$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if a prime $p$ divides both $y+i$ and $y-i$ then it divides their difference $2i$. Thus $p=2$ (up to units), but then $p$ divides $y$. That's impossible since it divides $y+i$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I was too hasty in writing this, as pointed out by Moron. The prime 2 ramifies in $\mathbb{Z}[i]$, and is (up to units) the square of $1+i$. Sorry for the confusion.&lt;/p&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2010-10-04T07:07:14.327" LastActivityDate="2010-10-04T07:07:14.327" />
  <row Id="5916" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T05:38:36.380" Score="2" ViewCount="0" Body="&lt;p&gt;Orientation does not play any role in Euclidean geometry. So it is most natural to define a Euclidean space as a finite-dimensional vector space with an inner product. Lengths, distances and angles are then defined in terms of this inner product. We are interested in transformations which preserve lengths and angles, which are precisely the orthogonal transformations &lt;em&gt;including reflections&lt;/em&gt;. There is no need to explicitly choose a basis at any point.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-03T05:38:36.380" />
  <row Id="5917" PostTypeId="2" ParentId="5909" CreationDate="2010-10-03T06:06:45.550" Score="2" ViewCount="0" Body="&lt;p&gt;First, think of each edge as a unit-length path connecting two vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine that instead of a random walk on the vertices, we consider a continuous random walk that moves along edges at unit speed.  Every time we arrive at a vertex, we randomly decide which edge coming out of the vertex to traverse next.  Note that the sequence of vertices and edges that we traverse will be the same as in the discrete-time random walk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that we fill the graph with a very large number of particles, each of which is performing a continuous walk.  We start the particles at random times in the interval $[0,1]$, so that each vertex continuously has particles going through it, and then wait until we are in the steady state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for me, it seems obvious at this point that the graph will be uniformly filled with particles, with each edge having the same density.  Every vertex will have particles coming in from all of its edges at the same rate, and these particles will randomly distribute themselves among the outgoing edges.  This situation is clearly stable, so it must be the steady state.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-03T06:06:45.550" CommentCount="3" />
  <row Id="5918" PostTypeId="2" ParentId="5791" CreationDate="2010-10-03T06:17:21.190" Score="1" ViewCount="0" Body="&lt;p&gt;It's not very clear what you're trying to ask, but what you're describing looks like a region &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadtree&quot; rel=&quot;nofollow&quot;&gt;quadtree&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each step in your IFS corresponds to a choice of child node (or subquadrant), and the tree path of any given point is the same as the list of numeric input steps in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not really matter whether you store the point as an arbitrary-precision coordinate pair or tree traversal: the binary digits required to store its coordinates correspond exactly to its quadtree traversal path.&lt;/p&gt;&#xA;" OwnerUserId="2273" LastEditorUserId="2273" LastEditorDisplayName="" LastEditDate="2010-10-03T06:37:14.780" LastActivityDate="2010-10-03T06:37:14.780" />
  <row Id="5919" PostTypeId="1" AcceptedAnswerId="5976" CreationDate="2010-10-03T06:30:08.717" Score="7" ViewCount="119" Body="&lt;p&gt;So if I'm given $H$, when I can conclude there is a group $G$ such that $H\cong G/Z(G)$? It's easy to show that non-trivial cyclic groups are &lt;em&gt;not&lt;/em&gt; of this form. More generally, any group with the property that all finitely generated subgroups are cyclic (e.g. $\mathbb{Q},\mathbb{Q}/\mathbb{Z}$) cannot be of this form. Are there are any others?&lt;/p&gt;&#xA;" OwnerUserId="546" LastActivityDate="2010-10-04T01:43:30.420" Title="When can a (finite) group be written as the quotient of some other group by its center?" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5920" PostTypeId="2" ParentId="5901" CreationDate="2010-10-03T06:34:25.707" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose first that $V+ V' = X$.  Then the natural map $X/(V\cap V') \to X/V \times X/V'$&#xA;is an isomorphism.  Choose a subspace of the target of this isomorphism&#xA;that projects isomorphically onto each factor (i.e. the graph of an isomorphism between the two factors; such an isomorphism exists since the two factors have the same dimension).&#xA;Its preimage under the natural map is a subspace of $X/(V\cap V')$ which&#xA;meets each of $V$ and $V'$ trivially.  Now choose any  subspace $W$&#xA;of $X$ that projects isomorphically onto this preimage; this is then a subspace of&#xA;$X$ that maps isomorphically onto each of $X/V$ and $X/V'$, and hence is a codimension $d$&#xA;subspace with the desired property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general (i.e. if $V + V' \neq X$) then the above gives a codimension $d$ subspace $W'$ of &#xA;$V + V'$ meeting $V$ and $V'$ trivially.   Choose $W''$ to be any subspace of $X$ which&#xA;maps isomorphically onto $X/(V + V')$.  The sum $W' + W''$ is then a codimension $d$ subspace&#xA;of $X$ meeting $V$ and $V'$ trivially.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-03T06:34:25.707" CommentCount="1" />
  <row Id="5921" PostTypeId="1" CreationDate="2010-10-03T06:58:10.770" Score="3" ViewCount="128" Body="&lt;p&gt;F &lt;strong&gt;y&lt;/strong&gt;(w) &gt;= F &lt;strong&gt;x&lt;/strong&gt;(w), for all w.  Prove P{&lt;strong&gt;y&lt;/strong&gt;&amp;lt;&lt;strong&gt;x&lt;/strong&gt;} &gt;= 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had this question on the exam and boy did it stump me.  It was simple to understand why it was true, but I tried several different paths and got nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2010-10-03T17:36:25.637" LastActivityDate="2010-10-23T22:42:55.797" Title="For all w, CDF of y at w is greater than the CDF of x at w.  Prove that the probability ( y &lt; x ) &gt;= 0.5" Tags="&lt;probability-theory&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5922" PostTypeId="1" CreationDate="2010-10-03T07:16:08.217" Score="1" ViewCount="115" Body="&lt;p&gt;So I wanted to figure out a meaningful probability for the card deck order problem.  I chose arbitrary values in an attempt to show the magnitude of permutations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you and everyone on the Earth right&#xA;  now shuffled a million randomized&#xA;  decks every millisecond that universe&#xA;  has existed, the probability that any&#xA;  two decks at one moment were the same would be:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the probability?&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-10-04T05:40:19.807" Title="Probability of card decks being in the same order for N shufflers over X amount of time?" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="5923" PostTypeId="2" ParentId="5922" CreationDate="2010-10-03T07:18:02.393" Score="0" ViewCount="0" Body="&lt;p&gt;Here is what I came up with and my assumptions.  Please point out any errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I attempted to find the probability of 2 people having the same deck order of N people shuffling decks in one trial.  I could not figure out how to model it in a more standard way, so my leap of faith was that for N people, there were (N+N-1+N-2+...)=Nt.  For 6.7e9 people on Earth, that's 2.2445*^19 pairs of people.  The probability that one pair had the same deck order was (1/52!) = 1/8e67 = p1.  Since the events are not independent, I need to multiply by (52!)!/(52!-N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I wanted the probability that any of them had a match.  So that would be 1 minus the probability that none of them had any matches:&lt;br&gt;&#xA;&lt;code&gt;(1 - (52!)!/(52!-N)*(p1)^0*(1 - p1)^(Nt)) = (1 - (52!)!/(52!-N)*(1 - p1)^(Nt)) = p2.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I wanted to do a million randomized trials every millisecond that the universe existed.  So that's 1.4e10 years * 365 days/year * 24 hours/day * 3600e3 ms / hour * 1e9 trials/ms = T = 4.5e29.  Again, I wanted at least one match, that would bring it to:&lt;br&gt;&#xA;&lt;code&gt;(1 - (1 - p2)^(T))&lt;/code&gt; = &#xA;&lt;code&gt;(1 - (1 - (1 - (52!)!/(52!-N)*(1 - p1)^(Nt)))^(T))&lt;/code&gt; =&lt;br&gt;&#xA;&lt;code&gt;(1-(1-(1-(52!)!/(52!-6.7e9)*(1-1/8e67)^(2.2445*^19)))^(4.5e29))&lt;/code&gt; = &lt;code&gt;6.31 × 10^-20&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2010-10-03T17:57:43.763" LastActivityDate="2010-10-03T17:57:43.763" CommentCount="5" />
  <row Id="5924" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T08:19:32.470" Score="3" ViewCount="0" Body="&lt;h3&gt;On bases&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Bases do not have to be ordered. I respectfully present the following as a counterpoint to Agustí Roig's response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, a basis is simply a &lt;em&gt;set&lt;/em&gt; of vectors, by definition; and sets are unordered. If you wish to enumerate the vectors in a basis, you will of course do so in &lt;em&gt;some&lt;/em&gt; order, but this order is arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This arbitrary order of enumeration is the reason why we describe vectors as tuples. The order of the coefficients in a tuple matters &lt;em&gt;only in as much as it must be in agreement with the arbitrary order which was selected for the basis vectors&lt;/em&gt;. To wit: let &lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;y&lt;/strong&gt; be two arbitrary linearly independent vectors. The tuple [&amp;thinsp;7&amp;nbsp;&amp;nbsp;5&amp;thinsp;] with respect to the (enumeration of the) basis &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt; represents the same vector as the tuple [&amp;thinsp;5&amp;nbsp;&amp;nbsp;7&amp;thinsp;] with respect to the (&lt;em&gt;different&lt;/em&gt; enumeration of the &lt;em&gt;same&lt;/em&gt;) basis &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt;. The tuple is just a &lt;em&gt;representation&lt;/em&gt; of a vector, relative to an &lt;em&gt;arbitrarily chosen&lt;/em&gt; order for the basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more foundationally: &quot;tuples&quot; can be regarded as functions from the integers (&lt;em&gt;e.g.&lt;/em&gt; the indices '1', '2', etc.) to the reals, complex numbers, or whichever set you draw your coefficients from. So [&amp;thinsp;7&amp;nbsp;&amp;nbsp;5&amp;thinsp;] can be thought of as 'really being' the function mapping&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'1'&amp;nbsp;$\mapsto$&amp;nbsp;7, &lt;br&gt;&#xA;  '2'&amp;nbsp;$\mapsto$&amp;nbsp;5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our enumeration of the basis has a similar role: choosing the enumeration &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;strong&gt;v&lt;/strong&gt;&lt;sub&gt;2&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt; is equivalent to defining a mapping&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'1'&amp;nbsp;$\mapsto$&amp;nbsp;&lt;strong&gt;x&lt;/strong&gt;, &lt;br&gt;&#xA;  '2'&amp;nbsp;$\mapsto$&amp;nbsp;&lt;strong&gt;y&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do I have '1' and '2' in quotes? Because they're just labels; any other labels would do just as well. For instance, I could replace '1' and '2' with the vectors &lt;strong&gt;x&lt;/strong&gt; and &lt;strong&gt;y&lt;/strong&gt; themselves. Then we could define vectors by coefficient functions such as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;x&lt;/strong&gt;&amp;nbsp;$\mapsto$&amp;nbsp;7, &lt;br&gt;&#xA;  &lt;strong&gt;y&lt;/strong&gt;&amp;nbsp;$\mapsto$&amp;nbsp;5;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that is, the coefficient 7 is associated to the vector &lt;strong&gt;x&lt;/strong&gt;, and the coefficient 5 is associated to &lt;strong&gt;y&lt;/strong&gt;, to represent the vector 7&lt;strong&gt;x&lt;/strong&gt;&amp;nbsp;+&amp;nbsp;5&lt;strong&gt;y&lt;/strong&gt;. This is what we really mean anyway; never mind any ordering of the vectors in your basis. With this, the arbitrary ordering of the basis disappears, leaving nothing but what it is we really mean by it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: when we order our bases, its only to make it easier to present things &amp;mdash; it is not part of the definition, or part of the structure of the objects we really care about.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;On morphisms&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A morphism on a Euclidean space ought to preserve all of the structures of Euclidean geometry. In particular, it should map circles to other circles, to preserve the structures guaranteed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_geometry#Axioms&quot; rel=&quot;nofollow&quot;&gt;third postulate&lt;/a&gt;; so it must be an &lt;em&gt;isothety&lt;/em&gt; (a rigid transformation up to scaling). This includes, but is not restricted to, the rigid transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you make any argument why the morphisms should consist &lt;em&gt;only&lt;/em&gt; of rigid transformations (excluding isothetic maps such as &lt;strong&gt;v&lt;/strong&gt;&amp;nbsp;$\mapsto$&amp;nbsp;2&lt;strong&gt;v&lt;/strong&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-10-03T12:06:21.923" LastActivityDate="2010-10-03T12:06:21.923" />
  <row Id="5926" PostTypeId="2" ParentId="5904" CreationDate="2010-10-03T10:23:33.440" Score="2" ViewCount="0" Body="&lt;p&gt;This began as a reply to Jyotirmoy's comment on Agustí's answer, but it became way too long. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I stated in my comment on Agustí's answer, some textbooks define a basis as a set, and I think this is wrong. Obviously, some people disagree (just read some of the other answers and comments). I'll try to explain why a basis should never be thought of as a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the columns of the following matrix form a basis for $\mathbb{R}^2$? $$\begin{bmatrix}1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 1\end{bmatrix}$$ The answer is &quot;yes&quot; if a basis is a set, but obviously the right answer is &quot;no.&quot; This makes it clear that whatever sort of collection to which we apply concepts like &quot;linear [in]dependence&quot; or &quot;basis&quot;, it must allow repeated vectors; otherwise, a lot of basic theorems from linear algebra would require extra clauses to deal with matrices with repeated rows or columns. This rules out sets as our collections of choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out what our collection of choice should be, let's revisit the definition of a basic concept like linear dependence (in somewhat imprecise &quot;collection-agnostic&quot; language (whose meaning should nevertheless be clear)): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The vectors $v_1,\ldots,v_n$ are linearly dependent if there are scalars $c_1,\ldots,c_n$, not all zero, such that $c_1v_1+\cdots+c_nv_n=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's obvious here that the natural object to apply such a concept to is an indexed tuple of vectors (indexed by $\{1,\ldots,n\}$ above, but we could formulate the concept equally well for an arbitrary finite index set): we begin with an indexed tuple of vectors and then consider a tuple of scalars indexed by the same set, and then we apply a function to these two tuples that is now completely specified (taking the linear combination). We could have started with a completely unindexed, unordered collection of vectors with repetition allowed (i. e., a &quot;multiset&quot;), but then to consider a linear combination, we would have had to choose some indexing of the multiset first and then to consider some scalars indexed by our arbitrary index set in order to form a linear combination. The core point is that the natural input to the &quot;linear combination&quot; operation is a tuple of vectors and a tuple of scalars indexed by the same finite set. This is the fundamental operation on a vector space, so to me, it makes sense to formulate everything in terms of indexed tuples of vectors whenever possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I talked about indexing lists of vectors in the previous paragraph, but indexing isn't the same as ordering. When I say &quot;order,&quot; though, I'm actually being a bit imprecise. It's usually not the order relation itself that matters, but the choice of index set does matter. For example, if you want to write down the coordinate matrix of a vector or the matrix of a linear transformation relative to a basis, then your index set pretty much has to be $\{1,...,n\}$ (or $\{0,...,n-1\}$ if you've ever programmed in a language other than Matlab or Fortran) because whatever you're using to index the elements of the basis, you're also using to index the rows or columns of your matrices. Since one of the main uses of bases is to reduce things to matrices, the sets $\{1,...,n\}$ are very convenient for indexing. In this very common context, all the basic concepts like &quot;basis,&quot; &quot;linear independence,&quot; &quot;spanning,&quot; etc. would apply to finite &lt;em&gt;sequences&lt;/em&gt; of vectors. In no context, however, do they naturally apply to finite &lt;em&gt;sets&lt;/em&gt; of vectors. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-10-03T10:23:33.440" CommentCount="4" />
  <row Id="5927" PostTypeId="1" AcceptedAnswerId="5930" CreationDate="2010-10-03T10:31:41.490" Score="1" ViewCount="195" Body="&lt;p&gt;Hi there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a problem with the binomial coefficient $\binom{5}{7}$. I know that the solution is zero, but I have problems to reproduce that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\displaystyle \binom{5}{7}=\frac{5!}{7!\times(5-7)!}=\frac{5!}{7!\times(-2)!}=\frac{120}{5040\times-2}=\frac{120}{-10080}=-\frac{1}{84}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is my mistake?&lt;/p&gt;&#xA;" OwnerUserId="2280" LastActivityDate="2010-10-08T04:08:42.413" Title="Problem with binomial coefficients." Tags="&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="5928" PostTypeId="1" AcceptedAnswerId="5950" CreationDate="2010-10-03T10:33:17.680" Score="4" ViewCount="111" Body="&lt;p&gt;The definition of the Mandelbrot set that most enthusiasts first encounter is that of the set of all complex numbers $c$ for which the sequence $z_{n+1} = z_n^2 + c$ starting from $z_0 = 0$ does not diverge. For convenience, let us name this familiar quadratic map $P_c(z) = z^2 + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that an equivalent definition of the Mandelbrot set is as the &quot;connectedness locus&quot; of the Julia sets of $P_c$. That is, $c$ is in the Mandelbrot set if and only if its corresponding filled Julia set $K_c = \{z : P_c^n(z) \not\rightarrow \infty\}$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are these definitions equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the equivalence boils down to the statement that $K_c$ is connected if and only it contains $0$, but I don't know how this is proved.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-03T17:37:41.680" Title="Why are these two definitions of the Mandelbrot set equivalent?" Tags="&lt;complex-analysis&gt;&lt;fractals&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5929" PostTypeId="1" CreationDate="2010-10-03T11:17:55.953" Score="1" ViewCount="109" Body="&lt;p&gt;what is the useness in math, I know it can provide topological invariant, any other things?&lt;/p&gt;&#xA;" OwnerUserId="2248" LastActivityDate="2010-10-03T19:01:43.920" Title="Why do we need to study mathematical aspects of quantum field theory, especially axiomatic pincture" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5930" PostTypeId="2" ParentId="5927" CreationDate="2010-10-03T11:30:20.510" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, now I see the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, (-2)! really isn't defined. And I can't use the factorial method if $n\notin\mathbb{N}$. So I have to go these way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\displaystyle \binom{5}{7}=\frac{5\times4\times3\times2\times1\times0\times-1}{7!}=\frac{0}{7!}=0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $k&gt;n$ the solution will always be zero, because the numerator has always the factor zero.&lt;/p&gt;&#xA;" OwnerUserId="2280" LastActivityDate="2010-10-03T11:30:20.510" />
  <row Id="5931" PostTypeId="2" ParentId="5921" CreationDate="2010-10-03T11:51:57.280" Score="1" ViewCount="0" Body="&lt;p&gt;For definiteness, let us assume that $X$ and $Y$ are positive random variables taking values &#xA;in $[0, \infty)$,and $F_Y(w)\le F_X(w)$, where $F_X(x), F_Y(y)$ are the CDFs and the equalities&#xA;take place only at zero and in the limit $w \rightarrow \infty$. For simplicity, let us assume further that &#xA;both CDFs are strictly increasing. The latter condition implies that we can invert the x-CDF and substitute in the y-CDF&#xA;to obtain the function $F_Y(F_X)$. In this function, both independent ($F_X$) and dependent ($F_Y$)&#xA;variables range in $[0, 1]$ and the value of F_Y is always less than F_X (except at 0 and 1).&#xA;(It is easy to see that, if one plots the x-CDF and y-CDF on the same axes, and see that at every value of the x-CDF, the y-CDF lies to the left of the x-CDF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $P(Y\lt X)$ is equal to the area under the line of $F_Y(F_X)$ , which is less than the area of a right isosceles rectangle of unit sides which is equal to 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Here is the proof (as asked by Eruditass) that the required probability can be computed from the area of under the $F_Y$ graph as a function of $F_X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the convolution formula to compute the probability density function of $Y-X$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{Y-X}(w) = \int_{0}^{\infty}f_X(x) f_Y(w+x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of which, the required probability is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X) = P(Y -X&amp;lt;0) = \int_{-\infty}^{0} f_{Y-X}(w)  dw$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the convolution formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{-\infty}^{0} dw  \int_{0}^{\infty} dx f_X(x) f_Y(w+x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the order of integration (the integration is over bounded functions) and performing a change of variables $ u = w +x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{0}^{\infty} dx f_X(x)  \int_{-\infty}^{x} du   f_Y(u) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of the cumulative distribution function, the second integral is just the Y-CDF&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{0}^{\infty} dx f_X(x)  F_Y(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again using the definition of the CDF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_X(x) dx = dF_X(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the integration variable from $x$ to $F_X$, we obtain the final result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(Y &amp;lt; X)  = \int_{0}^{1} dF_X  F_Y$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-10-04T16:26:46.673" LastActivityDate="2010-10-04T16:26:46.673" CommentCount="3" />
  <row Id="5932" PostTypeId="1" AcceptedAnswerId="5967" CreationDate="2010-10-03T12:04:43.417" Score="8" ViewCount="189" Body="&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/5896/an-interesting-series/5903#5903&quot;&gt;answer&lt;/a&gt; to a question about a series, a theorem was stated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $A= \{a_i \}$ is a set such that $\sum_{i = 1}^{\infty} \frac{1}{a_i}$ converges, then $d(A) = 0$, where $d(A)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; of the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it? &lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-10-04T19:58:36.130" Title="Theorem on natural density" Tags="&lt;number-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="5933" PostTypeId="2" ParentId="5932" CreationDate="2010-10-03T12:30:31.807" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT nr 2: Please forget about this answer; as pointed out in the comments, it wasn't correct! (I've removed the text it here; see the edit history for the embarrasing details. Next time, I'll try to think properly before I post...)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-04T03:31:11.477" LastActivityDate="2010-10-04T03:31:11.477" CommentCount="5" />
  <row Id="5934" PostTypeId="1" CreationDate="2010-10-03T13:10:45.330" Score="0" ViewCount="123" Body="&lt;p&gt;Suppose&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X) = A&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;P(Z|Y) = B&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where X and Y are independent. How does one determine P(Z|X,Y)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't studied maths in years, so what I post below may be entirely off base, but I don't like to ask a question without some attempt, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I read, Baye's theorem is used to calculate update a probability based on new information. If the prior probability is P(Z|X), and the posterior P(Z|X,Y):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = (P(Y|Z,X)P(Z|X))/P(Y)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Some of the above was worked out on intuition, and I'd love for someone to give me an explanation of how to properly apply Baye's theorem to update an already dependent result, every discussion I found dealt with the prior probability as a simple probability, not a conditional). Since Y is independent of X, this is equal to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = (P(Y|Z)P(Z|X))/P(Y)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To digress:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Y|Z) = P(Z|Y)P(Y)/P(Z)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(By Baye's theorem, again), so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = ((P(Z|Y)P(Y)/P(Z))P(Z|X))/P(Y)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;P(Z|X,Y) = (P(Z|Y)P(Y)P(Z/X))/P(Y)P(Z)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ignoring the degenerate case P(Y) = 0:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P(Z|X,Y) = (P(Z|X)P(Z|Y))/P(Z)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone first confirm whether or not this is correct? Secondly, can this be extended to more than two dependencies? It seems to work, but honestly my brain started melting when I tried to figure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if possible, explain simply why it works? I always like to know intuitively to reason for a result. I assume it's dependent on P(Z) because its necessary to find the overlap between P(X) and P(Y), but aside from that it's a bit hazy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if my guess is correct and the term P(Z) is necessary to determine the overlap between P(X) and P(Y), what extra information would I need and what equation would I have to solve in order to calculate P(Z|X,Y) without knowledge of P(Z)? This one I haven't been able to make any progress on, because I can't see where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise this is an awfully large question, and thank you all very much in advance for any contributions. If any parts of the question are too large (the last part, particularly, seems a candidate), please feel free to ignore them. This is all out of curiousity, so I don't specifically need an answer to any individual part.&lt;/p&gt;&#xA;" OwnerUserId="2058" LastActivityDate="2010-10-03T15:26:16.140" Title="Probability of consequent if dependent on two independent antecedents" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="5935" PostTypeId="1" CreationDate="2010-10-03T13:29:24.257" Score="1" ViewCount="58" Body="&lt;p&gt;I have some discrete points, with time $t_n$ and value $u(t_n)$. I perform discrete fourier transormation using values $u(t_n)$. Now I have some complex values, as I understood absolute value is amplitude and argument is frequency, right? well my teacher said i need to compute &lt;strong&gt;something&lt;/strong&gt; using $t_n$'s, $u(t_n)$'s, $A_n$'s and $\omega_n$'s (assuming $A_n$ is absolute value, $\omega_n$ is argument). It's somehow related to statistics variance and uncertainity principle. As you see, I don't know about Fourier transformations :D. Now I've made a program that performs DFT and builds a plot using $\omega_n$'s as $Ox$ axis and $A_n$'s as $Oy$ axis. It looks like this: &lt;a href=&quot;http://yfrog.com/muscreenshot2otp&quot; rel=&quot;nofollow&quot;&gt;screenshot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Value i need to compute is $\Delta T \Delta \omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1=\frac{\sum_{n=1}^N u(t_n)*t_n}  {\sum_{n=1}^N u(t_n)}$&#xA;$T_2=\frac{\sum_{n=1}^N u(t_n)*t_n^2}  {\sum_{n=1}^N u(t_n)}$&#xA;$\Delta T = \sigma^2 = 2\sqrt{T_2-T_1^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and same for $\Delta \omega$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_1=\frac{\sum_{n=1}^N A_n*\omega_n}  {\sum_{n=1}^N A_n}$&#xA;$W_2=\frac{\sum_{n=1}^N A_n*\omega_n^2}  {\sum_{n=1}^N A_n}$&#xA;$\Delta \omega = \sigma^2 = 2\sqrt{W_2-W_1^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What my teacher wants me to compute and which values should I expect as corresponding to theory? Can you give me step by step tutorial from source values to result? May be there's something wrong with the task.&lt;/p&gt;&#xA;" OwnerUserId="2281" LastEditorUserId="2281" LastEditorDisplayName="" LastEditDate="2010-10-03T15:33:02.183" LastActivityDate="2010-10-03T15:33:02.183" Title="Link between time and complex amplitude argument, DFT." Tags="&lt;fourier-analysis&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="5937" PostTypeId="1" AcceptedAnswerId="5939" CreationDate="2010-10-03T15:17:00.147" Score="1" ViewCount="293" Body="&lt;p&gt;&lt;code&gt;(4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove it is right equality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2284" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T03:53:52.017" LastActivityDate="2010-10-04T11:13:36.483" Title="How to prove an identity in radicals?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" />
  <row Id="5938" PostTypeId="2" ParentId="5934" CreationDate="2010-10-03T15:26:16.140" Score="1" ViewCount="0" Body="&lt;p&gt;Unfortunately your result is not correct. Consider two coin tosses (with a fair coin). Let X be: &quot;You get heads in the first toss&quot; and Y be &quot;You get heads in the second toss&quot;. Then X and Y are independent events. Let Z be &quot;You get at least one head&quot;. Then P(Z)=3/4, but P(Z|X,Y)=P(Z|X)=P(Z|Y)=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem in your argument is that P(Y|Z,X) is not necessary equal to P(Y|Z), or in other words, when constrained to Z, X and Y need no longer be independent.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-03T15:26:16.140" CommentCount="1" />
  <row Id="5939" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T15:31:00.583" Score="2" ViewCount="0" Body="&lt;p&gt;Simplify both sides to $8+6\sqrt{5}$, as they are both equal to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just square the RHS to see this and rationalise the denominators on the LHS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the rationalisation of the denominators on the LHS (which is very quick) you obtain $(3+\sqrt{5})^2-6,$ and hence the $8+6\sqrt{5}.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-03T16:06:49.753" LastActivityDate="2010-10-03T16:06:49.753" />
  <row Id="5940" PostTypeId="2" ParentId="5906" CreationDate="2010-10-03T15:37:08.807" Score="6" ViewCount="0" Body="&lt;p&gt;Here's a variant of my comment: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A triangle is a topological manifold with boundary (and so can be made abstractly into a smooth manifold with boundary). But as a subspace of Euclidean space it is not a smooth manifold with boundary. On the other hand, a triangle is a &quot;smooth manifold with corners&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of stratified enhancements on the manifold concept. Manifold with no boundary is the &quot;base&quot; manifold concept. You can add boundary or various other stratifications and at some point you can let your space degenerate to the point that anything is more or less a &quot;manifold with enough degeneracies...&quot;  Common terms for highly-degenerate manifold types are things like &quot;stratified spaces&quot; and &quot;manifolds with singularities&quot; or &quot;pseudomanifolds&quot;.  Orbifolds are another variation on this thread of ideas. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-03T15:37:08.807" />
  <row Id="5941" PostTypeId="2" ParentId="4737" CreationDate="2010-10-03T15:40:19.380" Score="1" ViewCount="0" Body="&lt;p&gt;The following is the rigorous construction of desired packing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following picture:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/vm68m.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;On this picture discs with curvatures 2,2,3,3,6,6,6,6,11,11,11,11 are packed inside a unit disc. One can prove correctness of this image by solving quadratic equations. I want to cut circles with curvatures 2,3,4,5,6,... form these: use already obtained circles with curvatures 2 and 3. Use the second circle with curvature 2 for cutting 4,6,8,10,16,... (all even curvatures, starting with 4) using repeating the same procedure scaled to the circle with curvature 2. Use the second 3 for cutting 5 and 9. Use 6 for cutting 7. Use 11 for 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have circles with radii 6, 6 and 6. Use the same procedure to obtain circles with curvatures 6*2, 6*3, 6*4, ... = 12, 18, 24, ... from them, each repeating 3 times. Use 12, 12 and 12 for 13, 15, 17; 18 for 19, 21, 23; 24 for 25, 27, 29 and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check, that words &quot;use the same procedure&quot; are ok, because if we will use the scheme, described above to cut circles one by one (2, then 3, then 4, then 5, then 6, then 7 in this order) immediately repeating the steps in described smaller circles, we will never use result of the step before the step itself.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-03T15:46:37.793" LastActivityDate="2010-10-03T15:46:37.793" />
  <row Id="5942" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T15:40:51.907" Score="1" ViewCount="0" Body="&lt;p&gt;You want to show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First multiply out the squares&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{16}{14-6\sqrt{5}} - \frac{186-60\sqrt{6}}{31-10\sqrt{6}} = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then simplify it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{16}{14-6\sqrt{5}} - 6 = 2\sqrt{61+24\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now we can start getting rid of the square roots on the right hand side by squaring both sides&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{256}{376-168\sqrt{5}} - \frac{192}{14-6\sqrt{5}} + 36 = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at this point it makes sense to rationalize the denominators of the fractions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{256}{376-168\sqrt{5}}\right)\left(\frac{376+168\sqrt{5}}{376+168\sqrt{5}}\right) - \left(\frac{192}{14-6\sqrt{5}}\right)\left(\frac{14+6\sqrt{5}}{14+6\sqrt{5}}\right) + 36 = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$376 + 168\sqrt{5} - 168 - 72\sqrt{5} + 36 = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;collecting like terms now gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(376 - 168 + 36) + (168 - 72)\sqrt{5} = 244 + 96 \sqrt{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is easily seem to be equal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another approach is, using the fact that algebraic numbers cannot be too close together - compute the first few digits of both sides of the original identity and compare them.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-03T15:40:51.907" CommentCount="4" />
  <row Id="5943" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T15:55:48.177" Score="2" ViewCount="0" Body="&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{4^{2}}{\left( 3-\sqrt{5}\right) ^{2}}-\dfrac{\left( 6-5\sqrt{6}%&#xA;\right) ^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{4^{2}\left( 5-\sqrt{6}%&#xA;\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}}{%&#xA;\left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{\left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{%&#xA;\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}}{5776},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \dfrac{4}{3-\sqrt{5}}\right) ^{2}-\dfrac{\left( 6-5\sqrt{6}\right)&#xA;^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=2\sqrt{61+24\sqrt{5}}\quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is enough to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left(&#xA;3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}%&#xA;\right) ^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=11552\sqrt{61+24\sqrt{5}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But (see &lt;em&gt;addendum&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left(&#xA;3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}%&#xA;\right) ^{2}=3656\sqrt{5}+46208.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 34656\sqrt{5}+46208\right) ^{2}=11552^{2}\left( 61+24\sqrt{5}%&#xA;\right) \quad (2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 3656\sqrt{5}+46208\right) ^{2}=3202768896\sqrt{5}+8140370944$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the right hand side is equal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$11552^{2}\left( 61+24\sqrt{5}\right) =3202768896\sqrt{5}+8140370944,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which proves your equality.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left(&#xA;3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=16\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}\left( 5-\sqrt{%&#xA;6}\right) ^{2}-\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right)&#xA;^{2}\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=5776\left( 3+\sqrt{5}\right) ^{2}-34656=34656\sqrt{5}+46208$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt; 2: The second equality stated in the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}\quad (3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{8}{14-6\sqrt{5}}-3=\sqrt{61+24\sqrt{5}}=\sqrt{61+\sqrt{2880}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LHS can be transformed into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{8}{14-6\sqrt{5}}-3=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}\times \dfrac{14+6\sqrt{5}}{14+6\sqrt{5}%&#xA;}=\dfrac{64+48\sqrt{5}}{16}=4+3\sqrt{5}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{61+\sqrt{2880}}=4+3\sqrt{5}.\quad (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply to the LHS the following &lt;strong&gt;general transformation involving radicals&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{A+\sqrt{B}}=\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}+\sqrt{\dfrac{A-\sqrt{%&#xA;A^{2}-B}}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A=61,B=2880$, then $\sqrt{A^{2}-B}=\sqrt{61^{2}-2880}=29$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}=3\sqrt{5},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\dfrac{A-\sqrt{A^{2}-B}}{2}}=4,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T11:13:36.483" LastActivityDate="2010-10-04T11:13:36.483" CommentCount="6" />
  <row Id="5944" PostTypeId="1" CreationDate="2010-10-03T16:11:05.990" Score="1" ViewCount="152" Body="&lt;p&gt;I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come up with my own solution, but I'm not sure if it's right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before substituting values back in, I got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac\pi{16} \left( (1+4x^2)^{3/2} 2x - \frac32 \ln\sqrt{1+4x^2+2x} + x\sqrt{1+4x^2} \right)  \text{ from }0\text{ to }\sqrt2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please verify this? For my integration by parts, I set $u = \sec^3\theta - \sec\theta$ and $dv = \sec^2\theta d\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much in advance.&lt;/p&gt;&#xA;" OwnerUserId="2131" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-03T17:15:22.853" LastActivityDate="2010-12-08T06:31:59.413" Title="Lateral surface area by hand" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5945" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T16:24:53.097" Score="-1" ViewCount="0" Body="&lt;p&gt;Trivial. Lets simplify the left part:&#xA;multiply first fraction to the sum of expressions in its denominator, and get whole part in second fraction and do with second fraction the same that we did with first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\frac{4(3+\sqrt{5})}{9-5})^2 - (\frac{25-19 - \sqrt{6}}{5-\sqrt{6}}) = (3+\sqrt{5})^2 - (5-\frac{19(5+\sqrt{6})}{25-6})^2 = (3+\sqrt{5})^2 - 6 = 8 + 6\sqrt{5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets prove this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$8+6\sqrt{5}=2\sqrt{61+24\sqrt{5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;divide by two and square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4+3\sqrt{5} = \sqrt{61+24\sqrt{5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16+45+24\sqrt{5}=61+24\sqrt{5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats all.&lt;/p&gt;&#xA;" OwnerUserId="2281" LastActivityDate="2010-10-03T16:24:53.097" />
  <row Id="5946" PostTypeId="1" AcceptedAnswerId="5948" CreationDate="2010-10-03T16:41:48.443" Score="-4" ViewCount="112" Body="&lt;p&gt;sqrt(3 - sqrt(5)) * (3 + sqrt(5)) * (sqrt(10) - sqrt(2))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please solve this? I've been trying to do that two hours! The result is 8.&lt;/p&gt;&#xA;" OwnerUserId="2284" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T11:01:42.187" LastActivityDate="2010-10-04T11:01:42.187" Title="Simplifying a radical expression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5947" PostTypeId="2" ParentId="5889" CreationDate="2010-10-03T16:54:27.373" Score="3" ViewCount="0" Body="&lt;p&gt;Let &lt;code&gt;$\{g_i\}$&lt;/code&gt; be a set of generators for $G$ and &lt;code&gt;$\{c_j\}$&lt;/code&gt; a set of generators for $C$.  For each $i$, write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_i= a_{i,1}b_{i,1}\ldots a_{i,n}b_{i,n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as usual, where $a_k\in A$, $b_k\in B$.  Clearly, the finite set &lt;code&gt;$\{a_{i,k},b_{i,k}\}$&lt;/code&gt; generates $G$.  Now, we shall see that &lt;code&gt;$\{a_{i,k}\}\cup\{c_j\}$&lt;/code&gt; generates $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any $a\in A$ can be written as a product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=\alpha_1\beta_1\ldots\alpha_m\beta_m$   (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha_l$ is a product of $a_{i,k}^{\pm1}$'s, hence is in $A$, and $\beta_l$ is a product of $b_{i,k}^{\pm1}$'s, hence is in $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An expression like the right hand side of (1) is called &lt;em&gt;reduced&lt;/em&gt; if every $\alpha_i\notin C$ for $i&gt;1$ and every $\beta_i\notin C$ for $i&amp;lt; m$.  In other words, there's no 'obvious' way to reduce $m$ by combining one of the terms with its neighbouring terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Normal Form Theorem for amalgamated products (as Lilly says below, this is Theorem 1 is Serre's &lt;em&gt;Trees&lt;/em&gt;---this is where we use that the maps $C\to A$ and $C\to B$ are injective) asserts that every element has a unique reduced expression (up to the obvious ambiguity that we can insert elements of $C$ in the $A$-terms and cancel them in neighbouring $B$-terms).  But (1) gives us two expressions for $a$, and the left hand side is clearly reduced.  It follows that the right hand side is not reduced, unless $m=1$ and $\beta_1\in C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $m&gt;1$ we know that the right hand side is not reduced, which means that some $\alpha_l\in C$ for $l&gt;1$ or some $\beta_l\in C$ for $l&amp;lt; m$.  Let's assume that $\beta_l\in C$---the case of $\alpha_l\in C$ is identical.  Now we can write $\beta_l$ as a product of $c_j^{\pm1}$'s. We now know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha'_l=\alpha_l\beta_l\alpha_{l+1}$ (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a product of $a_{i,k}^{\pm1}$'s and $c_j^{\pm1}$'s.  Use (2) to simplify the right hand side of (1), so it becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$a=\alpha_1\beta_1\ldots\beta_{l-1}\alpha'_l \beta_{l+1} \ldots \beta_m$&lt;/code&gt; . (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product on the right hand side of (3) has fewer terms than the product on the right hand side of (1), so we can continue by induction until the right hand side is reduced, at which point we have an expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=\alpha_1\beta_1$ (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\beta_1\in C$ and $\alpha_1$ a product of $a_{i,k}^{\pm1}$'s and $c_j$'s.  So we are done.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-10-06T19:24:47.760" LastActivityDate="2010-10-06T19:24:47.760" CommentCount="7" />
  <row Id="5948" PostTypeId="2" ParentId="5946" CreationDate="2010-10-03T17:04:48.507" Score="3" ViewCount="0" Body="&lt;p&gt;You want to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\sqrt{3 - \sqrt{5}}) (3 + \sqrt{5}) (\sqrt{10} - \sqrt{2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{3 - \sqrt{5}} \sqrt{(3 + \sqrt{5})^2} \sqrt{(\sqrt{10} - \sqrt{2})^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;merging the top level square roots gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{(3 - \sqrt{5}) (3 + \sqrt{5})^2 (\sqrt{10} - \sqrt{2})^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying this out gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{144 - 24 \sqrt{20} + 48 \sqrt{5} - 8 \sqrt{100}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pulling a factor of 48 into the radical to get like terms gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{64 - 24 \sqrt{20} + 24 \sqrt{20}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is $8$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-03T17:04:48.507" CommentCount="2" />
  <row Id="5949" PostTypeId="1" CreationDate="2010-10-03T17:28:13.697" Score="2" ViewCount="47" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that if we are given two modules $M,N$ over a ring $R$, and a pairing between them&#xA;$M \otimes_R N \to R$, we can construct a perfect pairing $M'\otimes_R N' \to R$ by taking kernels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this construction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1007" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T00:17:22.407" LastActivityDate="2010-10-04T00:17:22.407" Title="is there a name for this construction? (taking a pairing and obtaining a perfect one)" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;terminology&gt;" CommentCount="1" />
  <row Id="5950" PostTypeId="2" ParentId="5928" CreationDate="2010-10-03T17:37:41.680" Score="4" ViewCount="0" Body="&lt;p&gt;Let $S_1$ be a circle of radius $100$ in the complex plane centered at the origin.  Clearly everything outside of $S_1$ diverges to $\infty$ under iteration of $P_c$, so the filled Julia set lies entirely inside of $S_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the preimage $S_2$ of $S_1$ under $P_c$.  This will be a smaller curve (close to a circle, with a radius of approximately $\sqrt{100} = 10$), which again must contain the entire filled Julia set.  Iterating this process, we obtain a sequence $S_1,S_2,\ldots$ of closed curves, each of which contains the filled Julia set in its interior.  In fact, since any point outside of the filled Julia set goes to $\infty$ under iteration of $P_c$, the intersection of the interiors of the curves $S_n$ is precisely the filled Julia set.  (See &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/8/8e/Time_escape_Julia_set_from_coordinate_%28phi-2%2C_0%29.jpg&quot; rel=&quot;nofollow&quot;&gt;this picture&lt;/a&gt; for an example.  The curves separating the different shades of orange are the iterated preimages of some large circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this reasoning is not quite correct, because the preimage of a closed curve under $P_c$ is not always a single closed curve.  Sometimes it is one closed curve, and sometimes it is two closed curves with disjoint interiors.  If we repeatedly take the preimages of $S_1$, we may find that $S_n$ is a union of a very large number of closed curves!  Note, however, that these curves still contain the filled Julia set entirely inside of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is the key bit: &lt;strong&gt;the preimage of a curve will have one component if and only if $0$ lies in the the interior of the preimage&lt;/strong&gt;.  This is because $0$ is the critical point of the map $P_c$.  Thus the preimage of a curve is either a single curve that surrounds $0$, or two curves, neither of which surrounds $0$.  (In the latter case, the two curves are actually negatives of one another, i.e. symmetric across the origin.)  Therefore, there are exactly two cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The point $0$ lies in the filled Julia set.  In this case, each preimage $S_n$ must be a single curve, so the intersection of the interiors is connected.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The point $0$ lies outside the filled Julia set.  In this case, some preimage $S_n$ does not encircle $0$, so it must have two components.  Then each successive preimage will have twice as many curves as the last, and the resulting filled Julia set is homeomorphic to the Cantor set. (See &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/en/6/65/Julia_0.45_0.1428.png&quot; rel=&quot;nofollow&quot;&gt;this picture&lt;/a&gt; for an example.  The curves separating the different shades of blue are the iterated preimages of some large circle.  You ought to be able to see the first step at which the curve separates into two components.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-03T17:37:41.680" />
  <row Id="5951" PostTypeId="2" ParentId="5922" CreationDate="2010-10-03T17:43:48.393" Score="2" ViewCount="0" Body="&lt;p&gt;Let's make some assumptions for the purposes of calculation. Suppose 6.7 billion people have been alive for 13.5 billion years each shuffling one deck a thousand times a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of rounds of shuffles $N=1000 \times 3600 \times 24 \times 365 \times1.35 \times 10^{10}.$&#xA;So $N = 4.25736 \times 10^{20}.$&#xA;Let $n=52! \approx 8.0658 \times 10^{67}$ and let $m=6.7 \times 10^9.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When everyone shuffles the cards once in the first millisecond the probability that they are all different is&#xA;$$ \prod_{i=1}^{m-1} \left( 1- \frac{i}{n} \right) \approx 1 - \frac{m(m-1)}{2n},$$&#xA;ignoring the cross products since $n$ is much bigger than $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each shuffle is independent, so after $N$ steps the probabilty that we have had no match is&#xA;$$\left( 1 - \frac{m(m-1)}{2n} \right)^N \approx 1 - \frac{m(m-1)N}{2n}, \qquad (1)$$&#xA;since $m(m-1)/2n$ is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that there has been a match is 1 minus RHS of (1), that is&#xA;$$\frac{m(m-1)N}{2n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make this about $1.18 \times 10^{-28}$ with the figures I've used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;REMARK: Note that if you drop the &quot;at one moment&quot; requirement. That is, if you now ask if two shuffles have ever been the same, the problem is slightly quicker. The total number of shuffles that have ever taken place is $mN$ and so the probability that they were all different (as in the birthday problem) is $\prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ where as before $n=52!,$ So the probability that we've had a repeat is  $1- \prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ which, similar to before, is approximately $mN(mN-1)/2n,$ or $5 \times 10^{-8}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that as the index $i$ in the product gets larger, better approximations are obtained by passing to logarithms (as on the wikipedia page).&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-04T05:40:19.807" LastActivityDate="2010-10-04T05:40:19.807" CommentCount="3" />
  <row Id="5952" PostTypeId="1" AcceptedAnswerId="5958" CreationDate="2010-10-03T18:37:00.610" Score="3" ViewCount="74" Body="&lt;p&gt;Are there any positive integer solutions (n,x) to the equation (x)(x+1)=n! except (2,1) and (3,2)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not (as I suspect is the case), how do you prove that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, is there a way to approach the Diophantine equation n!=P(x) where P is a polynomial?&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T01:03:22.963" LastActivityDate="2010-11-15T01:03:22.963" Title="Factorial equaling a polynomial" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="5953" PostTypeId="2" ParentId="5944" CreationDate="2010-10-03T18:44:43.120" Score="2" ViewCount="0" Body="&lt;p&gt;Not a complete answer, but maybe a bit of help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface area is $\int_{x=0}^{\sqrt{2}} (2\pi y) ds = \int_{0}^{\sqrt{2}} 2 \pi x^2 \sqrt{1+(2x)^2} dx$, and the antiderivate is, according to Mathematica,&#xA;$$\frac{\pi}{32} \left(2 \sqrt{4 x^2+1} \left(8 x^3+x\right)-\sinh^{-1}(2x)\right).$$&#xA;The inverse of $\sinh$ can be expressed using $\ln$, so you should be able to compare this to your expression.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-03T18:44:43.120" CommentCount="0" />
  <row Id="5954" PostTypeId="2" ParentId="5937" CreationDate="2010-10-03T18:54:39.820" Score="0" ViewCount="0" Body="&lt;p&gt;Specialize $\rm\ \ \ b=3,\ \ c = 5,\ \ d = 6 \ \Rightarrow\ a = 4\ \ $ in this simple derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\displaystyle \bigg(\frac{b^2-c}{b-\sqrt{c}}\bigg)^2 - \bigg(\frac{d -5\sqrt{d}}{5-\sqrt d}\bigg)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\quad \:(\: b \ \: + \:   \sqrt{c}\ )^{\:2} \ \ \:-\ \ \ (\:-\:\sqrt{d}\:)^{2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ \  2\ (\:a + b \sqrt{c}\ )\:, \quad 2\ a\ =\ b^2+c-d $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ \ 2\:\sqrt{a^2+b^2\:c+2\:a\:b\sqrt{c} } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\:$ Replacing numbers by functions makes  the proof both simpler and more general - similar to your &lt;a href=&quot;http://math.stackexchange.com/questions/5972&quot;&gt;recently asked question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T04:54:34.390" LastActivityDate="2010-10-04T04:54:34.390" />
  <row Id="5955" PostTypeId="1" AcceptedAnswerId="5957" CreationDate="2010-10-03T19:01:25.853" Score="1" ViewCount="36" Body="&lt;p&gt;If $\phi:X\rightarrow Y$ is a rational map of varieties, why is the graph of $\phi$ well-defined? That is, if $\langle \rho,U\rangle$ and $\langle \psi,V\rangle$ are two representatives of $\phi$ (so that $U,V\subseteq X$ open, and $\rho\vert_{U\cap V}=\psi\vert_{U\cap V}$), why is the closure of the image of $\rho$ equal to the closure of the image of $\psi$?&lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2010-10-03T19:12:39.090" Title="Why is the graph of a rational map independent of the choice of representative?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5956" PostTypeId="2" ParentId="5929" CreationDate="2010-10-03T19:01:43.920" Score="1" ViewCount="0" Body="&lt;p&gt;One can prove many things (enumerative results in algebraic geometry, results in the geometric Langlands program, mirror symmetry) by making physics-style computations in various quantum field theories.  Currently, one is not able to make such arguments rigorous in general, and hence they become a method for producing conjectures in mathematics, not theorems.  One would like to upgrade their status to being proved theorems (!), and one approach would be to find&#xA;a framework in which the field-theoretic computations become valid mathematical deductions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One current framework in which one can do something like this is that supplied by the &lt;em&gt;bordism hypothesis&lt;/em&gt; of Baez and Dolan, recently proved by Jacob Lurie.    This provides a rigorous means of constructing certain topological field theories, which is already beginning to find applications.  But (for example) this framework is not encompassing enough to explain mathematically the results of Kapustin and Witten in the direction of deducing geometric Langlands from mirror symmetry.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-03T19:01:43.920" />
  <row Id="5957" PostTypeId="2" ParentId="5955" CreationDate="2010-10-03T19:12:39.090" Score="3" ViewCount="0" Body="&lt;p&gt;Because $U\cap V$ is Zariski dense in both $U$ and $V$ (by irreducibility of $X$), and so both graphs&#xA;are obtained by closing up the graph of $\rho_{| U\cap V}$ (which of course coincides with the graph of $\phi_{| U \cap V}$).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-03T19:12:39.090" CommentCount="1" />
  <row Id="5958" PostTypeId="2" ParentId="5952" CreationDate="2010-10-03T19:13:54.860" Score="5" ViewCount="0" Body="&lt;p&gt;This question was discussed on &lt;a href=&quot;http://mathoverflow.net/questions/39210/solve-in-positive-integers-nmm1&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt;, and the question is open.&lt;/p&gt;&#xA;" OwnerUserId="425" LastActivityDate="2010-10-03T19:13:54.860" />
  <row Id="5959" PostTypeId="1" AcceptedAnswerId="6001" CreationDate="2010-10-03T19:15:53.807" Score="0" ViewCount="86" Body="&lt;p&gt;I suspect this is easy, but that I'm missing something obvious: If a ring homomorphism $f:R\rightarrow S$ is such that a maximal ideal $m$ in $S$ does not have $f^{-1}(m)$ maximal in $R$, must there be some non-unit element $x\in R$ whose image under $f$ is a unit in $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is just that some &quot;localization&quot; must have occurred. In the classic example of $\mathbb{Z}\hookrightarrow\mathbb{Q}$, the zero ideal becomes maximal because 2, 3, 5, ... all become units.&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-10-03T19:39:06.897" LastActivityDate="2010-10-04T05:28:28.583" Title="Is the failure of MaxSpec to be functorial due to homomorphisms which take non-units to units?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="5961" PostTypeId="2" ParentId="5231" CreationDate="2010-10-03T19:16:56.350" Score="2" ViewCount="0" Body="&lt;p&gt;From your comments, what I understand of your problem is that you have the coordinates of an arbitrarily oriented rectangle centred on the origin, and you want to find the rotation that will bring it to an axis-aligned rectangle on the $xy$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u$ and $v$ be the unit vectors that should be mapped to the axis-oriented unit vectors $e_x$ and $e_y$ respectively. You can get these by subtracting adjacent points of the rectangle and normalizing. Then you want a rotation $R$ which satisfies $Ru = e_x$, $Rv = e_y$, and $Rn = e_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can express this as $R[u\;v\;n] = [e_x\;e_y\;e_z] = I$. Then $R$ equals $[u\;v\;n]^{-1}$, which is simply $[u\;v\;n]^T$ since $u$, $v$, and $n$ are an orthonormal set. To be more explicit, the rotation matrix you want is:&#xA;$$R = \begin{bmatrix}u_x &amp;amp; u_y &amp;amp; u_z \\ v_x &amp;amp; v_y &amp;amp; v_z \\ n_x &amp;amp; n_y &amp;amp; n_z\end{bmatrix}.$$&#xA;If you really like Euler angles (&lt;code&gt;rotateX&lt;/code&gt;, &lt;code&gt;rotateY&lt;/code&gt;, &lt;code&gt;rotateZ&lt;/code&gt;), there are ways to convert a rotation matrix like above to Euler angles, but they're ugly. You're best off using the rotation matrix explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if your rectangle is &lt;em&gt;not&lt;/em&gt; centred on the origin, and you want to perform the rotation keeping its centre (say $c$) fixed, you'll have to get the rotated coordinates of a point $p$ not simply as $Rp$ but as $R(p-c) + c$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-03T19:16:56.350" CommentCount="6" />
  <row Id="5962" PostTypeId="1" CreationDate="2010-10-03T19:39:03.560" Score="3" ViewCount="269" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $A$ is a differential one form then $A\wedge A .. (more\text{ }than\text{ }2\text{ }times) = 0$ Then how does the $A\wedge A \wedge A$ make sense in the Chern-Simon's form, $Tr(A\wedge dA + \frac{2}{3} A \wedge A \wedge A)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this anti-commutative nature of wedge product does not work for Lie algebra valued one-forms since tensoring two vectors is not commutative. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the vector space in which the vector valued differential form is taking values is $V$ with a chosen basis ${v_i}$ then books use the notation of $A = \Sigma _{i}  A_i v_i$ where the sum is over the dimension of the vector space and $A_i$ are ordinary forms of the same rank. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know whether this $A_i v_i$ is just a notation for $A_i \otimes v_i$ ?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly say $B$ is a vector valued differential form taking values in $W$ with a chosen basis ${w_i}$ and in the same notation, $B = \Sigma _j B_j w_j$. Then the notation used is that, $A \wedge B = \Sigma _{i,j} A_i \wedge B_j v_i \otimes w_j$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if in the above $A_i \wedge B_j v_i \otimes w_j$ is just a notation for $A_i \wedge B_j \otimes v_i \otimes w_j$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ and $B$ are vector bundle valued differential forms (like say the connection-1-form $\omega$ or the curvature-2-form $\omega$) then how is $Tr(A)$ defined and why is $d(Tr A) = Tr( d A)$ and $Tr(A \wedge B) = Tr(B \wedge A)$ ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is $A \wedge A \wedge A \wedge A = 0$ ? for $A$ being a vector bundle valued $1$-form or is only $Tr(A \wedge A \wedge A \wedge A) = 0$ ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If A and B are two vector bundle valued $k$ and $l$ form respectively then one defines $[A,B]$ as , $[A,B] (X_1,..,X_{k+l}) = \frac{1}{(k+l)!} \Sigma _{\sigma \in S_n} (sgn \sigma) [A (X_{\sigma(1)},X_{\sigma(2)},..,X_{\sigma(k)}) , B (X_{\sigma(k+1)},X_{\sigma(k+2)},..,X_{\sigma(k+l)})]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if say $k=1$ then $[A,A] (X,Y) = [A(X),A(Y)]$ and $[A,A] = 2A \wedge A$.&#xA;The Cartan structure equation states that, $d\Omega = \Omega \wedge \omega - \omega \wedge \Omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some people write this as, $d\Omega = [\Omega,\omega]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not clear to me. Because if the above were to be taken as a definition of the $[,]$ then clearly $[A,A]=0$ contradicting what was earlier derived. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-10-03T20:19:48.943" LastActivityDate="2010-10-19T21:12:29.033" Title="Some questions about differential forms " Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="4" />
  <row Id="5963" PostTypeId="2" ParentId="5959" CreationDate="2010-10-03T19:43:38.580" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; Consider the image of a maximal ideal containing $f^{-1}(m)$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T19:43:38.580" CommentCount="2" />
  <row Id="5964" PostTypeId="1" CreationDate="2010-10-03T19:43:41.963" Score="2" ViewCount="71" Body="&lt;p&gt;I play a game known as Bejeweled (I'm sure some of you have heard of it) which has the following properties:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a planar graph with every vertex having degree 4, any path of 3 (or more) contiguous colors is considered &quot;complete.&quot;  Additionally, any two adjacent vertices may be &quot;swapped&quot; to produce said path.  If such a swap exists the board is also said to be &quot;complete.&quot;  You cannot swap if it does not produce a path of 3 or more contiguous colors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AALIM.jpg&quot; alt=&quot;bejeweled&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment I'm ignoring the property that this path must be in a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to do is to limit the number of colors I have so that the board will always be complete (having a path, or a single swap produces the path).  Now obviously this is trivially true in the 2 color case, but I want the maximal number of colors.  Or more precisely,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the minimum number of colors that I need to color a planar graph, with every vertex having degree 4, such that there is no contiguous path of 3 identical colors or such a path cannot be produced by a single swapping of two nodes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T00:46:59.717" LastActivityDate="2010-11-26T00:46:59.717" Title="What is the minimal coloring of a planar graph with swaps?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5965" PostTypeId="1" CreationDate="2010-10-03T19:48:40.820" Score="2" ViewCount="79" Body="&lt;p&gt;Can anyone explain how to do these types of questions for me? I am having trouble wrapping my head around it. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spherical soap bubble is seen to undergo oscillations when slightly deformed. Assuming&#xA;that the frequency of these oscillations is a function of the surface tension S of the soap bubble,&#xA;the size of the soap bubble as given by the radius r of the original spherical bubble and the density&#xA;of soap, say p. Use dimensional analysis to construct a possible dependence for f in terms of S, r&#xA;and p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spherical soap bubble of radius 1cm oscillates at 1 Hz. What would be the anticipated frequency*&#xA;of oscillation of a soap bubble of radius 2cm?&lt;/p&gt;&#xA;" OwnerUserId="2292" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T00:08:11.667" LastActivityDate="2010-10-04T00:08:11.667" Title="Dimensional Analysis" Tags="&lt;physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5966" PostTypeId="2" ParentId="5927" CreationDate="2010-10-03T19:51:42.583" Score="1" ViewCount="0" Body="&lt;p&gt;By definition $\rm\binom{n}k$ is the coefficient of $\rm x^k$ in $\rm (1+x)^n$ so it is $0$ for $\rm k &gt; n\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T19:51:42.583" />
  <row Id="5967" PostTypeId="2" ParentId="5932" CreationDate="2010-10-03T19:55:46.123" Score="3" ViewCount="0" Body="&lt;p&gt;If the natural density of $A = \{a_i\}$ exists, then we can show that it must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle S_{n} = \frac{|A \cup [1,n]|}{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \{\frac{n}{a_n}\}$ is a subsequence of $S_{n}$ and so if the limit is $\displaystyle 2\delta &gt; 0 $ then we have that for all $\displaystyle n &gt; N_0$, $\displaystyle \frac{n}{a_n} &gt; \delta$ and so $\displaystyle \frac{1}{a_n} &gt; \frac{\delta}{n}$ for all $\displaystyle n &gt; N_0$ and so $\displaystyle \sum \frac{1}{a_n}$ diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem is actually showing that the limit exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that $\liminf$ is zero: If the limit was $\displaystyle 2\delta &gt; 0$ then for all $n &gt; N_{0}$, $S_{n} &gt; \delta$ and an argument similar to above works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $\displaystyle  \limsup S_n = 2\delta &gt; 0$. Then there is a subsequence $\displaystyle S_{N_1}, S_{N_2}, ..., S_{N_k}, \dots $ which converges to $\displaystyle 2\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can choose the subsequence so that $\displaystyle S_{N_i} &gt; \delta$ and $\displaystyle  N_{k+1} &gt; \frac{2N_{k}}{\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the number of elements of $\displaystyle  A$ in the interval $\displaystyle (N_{k}, N_{k+1}]$ is atleast $\displaystyle  \delta N_{k+1} - N_k \ge \delta N_{k+1} - \frac{\delta N_{k+1}}{2} \ge \frac{\delta N_{k+1}}{2}$ and so the sum of reciprocals in that interval is atleast $\displaystyle \frac{\delta N_{k+1}}{2} \frac{1}{N_{k+1}} =  \displaystyle \frac{\delta}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the sum of reciprocals must diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\displaystyle \limsup S_n = 0 = \liminf S_n$ and thus $\displaystyle \lim S_n = 0$ and thus the natural density is zero.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-03T20:04:07.450" LastActivityDate="2010-10-03T20:04:07.450" />
  <row Id="5968" PostTypeId="1" CreationDate="2010-10-03T20:23:29.270" Score="2" ViewCount="113" Body="&lt;p&gt;My friend is taking a course on probabilities and come across a problem that has a solution we both have trouble understanding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single printer prints, on average, 22 jobs an hour. Let's use a simplification where no job takes more than 2 minutes to complete. Using these assumptions, calculate the upper limit for the probability that, when a job arrives to the printer, the printer is processing another job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution was to use the Poisson distribution: 1 - ((22/30)^0)*e^(-22/30)/1! = 1 - e^(-22/30) = ~0.52&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble believing this: wouldn't simply 22/30 make more sense? Even though I can see how enqueueing the printing jobs is Poisson distributed, how come that goes for the waiting times, too? Or have they messed up with the solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: after sending feedback the assistant said the problem was, indeed, ambiguous. Thanks for the help.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastEditorUserId="2076" LastEditorDisplayName="" LastEditDate="2010-10-15T12:15:20.937" LastActivityDate="2010-10-15T12:15:20.937" Title="Using Poisson distribution to estimate if a printer is available? (I have the solution but don't understand it) [SOLVED]" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="5969" PostTypeId="1" AcceptedAnswerId="5980" CreationDate="2010-10-03T20:55:47.123" Score="3" ViewCount="177" Body="&lt;p&gt;Say you have two groups $G = \langle g \rangle$ with order $n$ and $H = \langle h \rangle$ with order $m$. Then the product $G \times H$ is a cyclic group if and only if $gcd(n,m)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to figure out how to start proving this. I have tried with some examples, where I pick $(g,h)$ as a candidate generator of $G \times H$. I see that what we want is for the cycles of $g$ and $h$, as we take powers of $(g,h)$, to interleave such that we do not get $(1,1)$ until the $(mn)$-th power. However, I am having a hard time formalizing this and relating it to the greatest common divisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints are much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2174" LastActivityDate="2010-10-03T22:44:35.837" Title="Product of two cyclic groups is cyclic iff their orders are co-prime" Tags="&lt;group-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="5970" PostTypeId="2" ParentId="5965" CreationDate="2010-10-03T21:00:05.173" Score="3" ViewCount="0" Body="&lt;p&gt;Dimensional Analysis says ignore the variables you have and manipulate what you've got in terms of dimensions. For example, let any measure of $mass=M$, let any measure of $length=L$, let any measure of $time=T$. $Force=F=MLT^-2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then surface tension is $\frac{MT^2}{L}$. Radius is simple, it is L whilst density is mass divided by area. Area in this case is $\pi*L^2$ and mass $M$ so density is $\frac{M}{\pi L^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency is period, or $\frac{1}{T}$, correct? Then, you simply need to attempt to equate both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your case, $Frequency = f(S, r, p)$. Well, $\frac{1}{T} = \sqrt{\frac{L}{MT^2}}*\sqrt{\frac{M}{\pi L^2}} * \sqrt{L} = \sqrt{\frac{L^2M}{\pi L^2T^2M}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully you can see these expressions are equivalent except for a constant $\pi$. Constants don't have dimension, so they don't matter. However, you might also see we have three expressions we've already defined, as such, $F = \sqrt{S*r*p}$. We know this to be true because the units work as expected and you can check it based on the values being what you expect.&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-10-03T21:00:05.173" />
  <row Id="5971" PostTypeId="2" ParentId="5927" CreationDate="2010-10-03T21:05:11.313" Score="2" ViewCount="0" Body="&lt;p&gt;$(-2)!$ is actually infinite. A more palatable way to phrase that, perhaps, is in terms of the reciprocal factorial: $1/(-2)! = 0$. We only need the recurrence relation $n! = n(n-1)!$, or in terms of reciprocal factorials: $$\frac{1}{(n-1)!} = n\cdot\frac{1}{n!}.$$ That means $\frac{1}{(-2)!} = \frac{0\cdot (-1)}{0!} = 0$. Then $\binom{5}{7} = \frac{5!}{7!}\cdot \frac{1}{(-2)!} = 0$, QED. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-10-03T21:05:11.313" />
  <row Id="5972" PostTypeId="2" ParentId="5946" CreationDate="2010-10-03T21:15:00.697" Score="2" ViewCount="0" Body="&lt;p&gt;Square it, then put $\rm\ \: a = 3,\ b = 2\:a-1 = 5,\ c = 2\ \ $ in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad (a-\sqrt b)\ (a+\sqrt b)^2\ ((\sqrt b - 1)\sqrt c)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ (a^2 \: -\: b)\ \:(a+\sqrt b)\ (b+1-2\sqrt b)\ c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad =\ (a^2\:-\: b)^2\ 2\:c\ \ \ \: $ via $\rm\:\ \ b+1 \ =\ 2\:a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\:$ Replacing numbers by functions makes it both simpler and more general, as in your &lt;a href=&quot;http://math.stackexchange.com/questions/5954&quot;&gt;prior question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T21:15:00.697" CommentCount="1" />
  <row Id="5973" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T21:18:01.600" Score="0" ViewCount="0" Body="&lt;p&gt;Use these facts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Cyclic groups of the same order are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://mathforum.org/library/drmath/view/73205.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/73205.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-03T21:18:01.600" />
  <row Id="5975" PostTypeId="1" CreationDate="2010-10-03T21:18:29.243" Score="0" ViewCount="74" Body="&lt;p&gt;Here is the question for my homework:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mother wants to invest 7000 USD for her son's future education. She invests a portion of the money in a bank certificate of deposit (CD account) which earns 4% and the remainder in a savings bond that earns 7%. If the total interest earned after one year is $420.00, how much money was invested in the CD account?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is my steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amount invested in the savings bond can be represented as 7000 - x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interest earned in one year on the CD account is 0.04x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the interest earned on the savings bond in one year can be represented as: 0.07(7000-x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total interested earned after one year is 420.00 USD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is now 0.04x+0.07(7000-x) = 420.00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to SIMPLFY it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n - 420.00 = 0.03x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I get the number N?&lt;/p&gt;&#xA;" OwnerUserId="2294" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T01:48:18.633" LastActivityDate="2010-10-04T01:48:18.633" Title="Getting a number from interest equation" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5976" PostTypeId="2" ParentId="5919" CreationDate="2010-10-03T21:24:33.677" Score="10" ViewCount="0" Body="&lt;p&gt;The groups you are looking for are called &lt;em&gt;capable groups&lt;/em&gt;. The determination of all capable group (even all finite capable groups) is very far from done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earliest result is due to Baer, who characterized exactly which groups that are direct sums of cyclic groups are capable (Baer, Reinhold. &lt;em&gt;Groups with preassigned central and central quotient groups&lt;/em&gt;, Trans. Amer. Math. Soc. &lt;strong&gt;44&lt;/strong&gt; (1938), 387-412; in fact Baer considered the question of when a group has a specified center and central quotient, each of which is a direct sum of cyclic groups. You obtain the characterization of capable groups as a corollary). For finitely generated abelian groups, this becomes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $G$ be a finitely generated abelian group, written as a direct sum of cyclic groups&#xA;  $$G = C_{a_1}\oplus\cdots\oplus C_{a_n}$$&#xA;  with $a_1|a_2|\cdots|a_n$ ($C_0$ denotes the infinite cyclic group). Then $G$ is capable if and only if $n\gt 1$ and $a_{n-1}=a_n$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a generalization of this result, replacing abelian groups with $p$-groups and replacing the direct sum with the nilpotent product (&lt;em&gt;Capability of nilpotent products of cyclic groups&lt;/em&gt; and &lt;em&gt;Capability of nilpotent products of cyclic groups II&lt;/em&gt;, J. Group Theory &lt;strong&gt;8&lt;/strong&gt;, no. 4 (2005), 431-452; and J. Group Theory &lt;strong&gt;10&lt;/strong&gt; no. 4 (2007), 441-451):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $p$ be a prime and let $c$ be a positive integer, $c\leq p$. If $G$ is the $c$-nilpotent product of cyclic groups&#xA;  $$G = C_{p^{a_1}}\coprod^{c}\cdots\coprod^{c} C_{p^{a_n}}$$&#xA;  where $\coprod^{c}$ represents the $c$-nilpotent product, and $1\leq a_1\leq\cdots\leq a_n$. Then $G$ is capable if and only if $n\gt 1$ and $a_n\leq a_{n-1}+\lfloor\frac{c-1}{p-1}\rfloor$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last condition is in fact necessary in general for $p$-groups:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt;(2005) Let $G$ be a nilpotent $p$-group of class $c\gt 0$, and let $x_1,\ldots,x_n$ be a generating set for $G$ with $x_i$ of order $p^{a_i}$, $a_1\leq a_2\leq\cdots\leq a_n$. If $G$ is capable, then $n\gt 1$ and $a_n\leq a_{n-1}+\lfloor\frac{c-1}{p-1}\rfloor$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The extra-special capable $p$-groups were not classified until 40 years after Baer (F.R. Beyl, U. Felgner, and P. Schmid. &lt;em&gt;On groups ocurring as central factor groups&lt;/em&gt;, J Algebra &lt;strong&gt;61&lt;/strong&gt; (1979), 161-177). The only extra-special capable $p$-groups are the nonabelian groups of order $p^3$ and exponent $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from these classes, there is a full classification of the capable $2$-generated $p$-groups of class two. You can find it the paper (with Robert F. Morse) &lt;em&gt;Certain homological functors for $2$-generated $p$-groups of class 2&lt;/em&gt;, in Computational Group Theory and the Theory of Groups II, Contemporary Mathematics &lt;strong&gt;511&lt;/strong&gt; (2010), pp 127-166, American Mathematical Society. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of known necessary conditions, but in general sufficient conditions are harder to find. Phillip Hall commented 70 years ago (&lt;em&gt;The classification of prime-power groups&lt;/em&gt;, J. Reine Angew. Math. &lt;strong&gt;182&lt;/strong&gt; (1940) 130-141) that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question of what conditions a group $G$ must fulfil in order that it may be the central quotient of another group $H$, $G\cong H/Z(H)$, is an interesting one. But while it is easy to write down a number of necessary conditions, it is not so easy to be sure that they are sufficient.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T01:43:30.420" LastActivityDate="2010-10-04T01:43:30.420" CommentCount="1" />
  <row Id="5977" PostTypeId="2" ParentId="5975" CreationDate="2010-10-03T21:28:45.323" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure what n is supposed to represent but instead of it you should still have that 7000*.07. Now you are left with just a number on the left and you can proceed to solve for x. $$.07(7000)-420.00 = 0.03x$$&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-03T21:28:45.323" CommentCount="6" />
  <row Id="5978" PostTypeId="2" ParentId="5944" CreationDate="2010-10-03T21:31:53.327" Score="2" ViewCount="0" Body="&lt;p&gt;I assume you have in your earlier steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ \sigma = 2\pi \int_0^\sqrt{2} x^2 \sqrt{1+4x^2} dx \]&lt;br&gt;&#xA;\[ \tan \theta = 2x \; \mathrm{and} \; \sec \theta = \sqrt{1+4x^2} \]&lt;br&gt;&#xA;\[ \sigma = \frac{\pi}2 \int_{x=0}^{x=\sqrt{2}} \sec^3\theta \tan^2\theta d\theta. \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have a messy antiderivative you worked out, and want to check whether it's right?  Why not see what happens if you take the derivative of your expression above?  If everything is fine, you should get back to $2\pi x^2 \sqrt{1+4x^2}$.  I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ d \sigma = \frac{\pi}{16} \left[ 24x^2\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - \frac{3(8x+2)}{4(1+4x^2+2x)} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx. \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there are a few problems here.  More things should be cancelling out.  The middle term (from $\ln$) is conspicuous for involving $\sqrt{1+4x^2+2x}$ instead of $\sqrt{1+4x^2}$.  I'm pretty sure the $2x$ snuck under the radical by accident, and the term should really be $\ln | \sqrt{1+4x^2}+2x |$.  If I fix that in your expression and derive again, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ d \sigma = \frac{\pi}{16} \left[ 24x\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - 3(1+4x^2)^{-\frac 12} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx. \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are some like terms and related terms, but they won't cancel out far enough, because some of your coefficients are wrong.  Maybe from here you can find your mistakes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The correct antiderivative in terms of $x$ never involves a factor of 3, unless you count the term $(1+4x^2)^{\frac 32}$.&lt;/li&gt;&#xA;&lt;li&gt;If you have a good antiderivative in terms of $\theta$, you might want to plug in to that equation instead of converting back to $x$, by finding values for $\tan \theta$ and $\sec \theta$ at $x=0$ and $x=\sqrt{2}$.  (At $x=\sqrt{2}$, $\sec \theta$ is rational!)&lt;/li&gt;&#xA;&lt;li&gt;When you have a final answer, compare to my result $\frac{\pi}{32}\left[102\sqrt{2} - \ln\left(3+2\sqrt{2}\right)\right]$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2236" LastActivityDate="2010-10-03T21:31:53.327" />
  <row Id="5979" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T21:35:14.430" Score="1" ViewCount="0" Body="&lt;p&gt;The order of $G\times H$ is $n.m$. Thus, $G\times H$ is cyclic iff it has an element with order $n.m$. Suppose $gcd(n.m)=1$. This implies that $g^m$ has order $n$, and analogously $h^n$ has order $m$. That is, $g\times h$ has order $n.m$, and therefore $G\times H$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now that $gcd(n.m) &gt;1$. Let $g^k$ be an element of $G$ and $h^j$ be an element of $H$. Since the lowest common multiple of $n$ and $m$ is lower than the product $n.m$, that is, $lcm(n,m) &amp;lt; n.m$, and since $(g^k)^{lcm(n,m)} = e_{G}$, $(h^j)^{lcm(n,m)} = e_{H}$, we have $(g^k\times h^j)^{lcm(n,m)} = e_{G\times H}$. It follows that every element of $G\times H$ has order lower than $n.m$, and therefore $G\times H$ is not cyclic.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-10-03T21:35:14.430" />
  <row Id="5980" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T21:39:48.537" Score="1" ViewCount="0" Body="&lt;p&gt;Note that $|G\times H|=|G||H|=nm$; so $G\times H$ is cyclic if and only if there is an element of order $nm$ in $G\times H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any group $A$, if $a,b\in A$ commute with one another, $a$ has order $k$, and $b$ has order $\ell$ then the order of $ab$ will divide lcm$(k,\ell)$ (prove it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take an element of $G\times H$, written as $(g^a,h^b)$, where $G=\langle g\rangle$, $H=\langle h\rangle$, $0\leq a\lt n$, $0\leq b\lt m$. Then $(g^a,h^b)=(g^a,1)(1,h^b)$. In this case, what is the order? Under what conditions can you get an element of order exactly $nm$, which is what you need?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-03T21:39:48.537" CommentCount="7" />
  <row Id="5981" PostTypeId="2" ParentId="5968" CreationDate="2010-10-03T21:48:55.033" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;This was my first answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your answer 22/30 ignores the fact that the job preceding yours may itself have been held up by a previous job.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Charles and tsiki point out, this is wrong: it's the Poisson-derived answer that ignores this. So now I'll try to answer the question as posed:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry, the probability of a job having been started less than 2 minutes prior to yours is the same that a job will be started less than 2 minutes after yours. This is a Poisson distribution.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-10-04T08:46:07.137" LastActivityDate="2010-10-04T08:46:07.137" CommentCount="1" />
  <row Id="5982" PostTypeId="1" AcceptedAnswerId="5994" CreationDate="2010-10-03T22:24:35.550" Score="0" ViewCount="206" Body="&lt;p&gt;Say I have the square root of 2/9. How would I convert this to a mixed radical where the radicand is a &lt;strong&gt;whole number&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can convert the square root of 8 to a mixed radical easily (2 sqrt 2) but this fraction is tripping me up.&lt;/p&gt;&#xA;" OwnerUserId="2296" LastEditorUserId="2296" LastEditorDisplayName="" LastEditDate="2010-10-03T23:00:26.617" LastActivityDate="2010-10-04T01:45:39.430" Title="Converting a radical to a mixed radical " Tags="&lt;roots&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5983" PostTypeId="1" AcceptedAnswerId="6011" CreationDate="2010-10-03T22:41:26.817" Score="4" ViewCount="142" Body="&lt;p&gt;Let n = p^k be a power of a prime. Let G be a transitive permutation group on n letters. Show that there exists a ﬁxed-point free element of order p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that there must be a fixed point:&#xA;The transitivity of the group implies that there is only 1 orbit. Burnside's lemma then implies that the average number of fixed points for an element in the group is 1. Since the identity fixes everything, there is a fixed point free element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that there exists an element of order p:&#xA;Again we use that the group is transitive and that there is therefore only one orbit. Since there is only one orbit, the order of the orbit of x is p^k. By the orbit-stabilizer theorem, the order of the stabilizer of some element x is equal to the order of the group divided by p^k. Since the order of a stabilizer is an integer, p^k divides the order of the group. By Cauchy's theorem, there is an element of order p in the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot, however, show that there is an element of order p with no fixed points. Does anyone have any hints?&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-04T09:22:46.490" Title="fixed-point free elements in transitive permutation groups" Tags="&lt;group-theory&gt;" AnswerCount="1" />
  <row Id="5984" PostTypeId="2" ParentId="5969" CreationDate="2010-10-03T22:44:35.837" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ \ gcd(m,n) &gt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\ lcm(m,n) &amp;lt; mn$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\ $ has all elts of order $\rm &amp;lt; mn$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\ $ is noncyclic&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-03T22:44:35.837" />
  <row Id="5985" PostTypeId="2" ParentId="5982" CreationDate="2010-10-03T22:48:16.673" Score="0" ViewCount="0" Body="&lt;p&gt;What do you mean by the radicand is a &quot;whole number&quot;? In your example $\sqrt{8}=2\sqrt{2}$ in both sides of the equation the radicand is a whole number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, it doesn't really matter because the problem as you stated it doesn't have a radical at all when you simplify: $\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-03T22:48:16.673" CommentCount="1" />
  <row Id="5986" PostTypeId="1" CreationDate="2010-10-03T23:26:39.807" Score="3" ViewCount="177" Body="&lt;p&gt;My question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cashier has a total of 114 bills, made up of tens and hundreds. The total value of the money is 6450 USD. How much of each kind does he have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has [BLANK] tens and [BLANK] hundreds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure there is an algebra equation for this. There is the hard way of doing it on paper (that takes long) or the short way of putting it equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help?&lt;/p&gt;&#xA;" OwnerUserId="2294" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-15T03:22:53.767" LastActivityDate="2010-11-15T03:22:53.767" Title="$\$10$ and $\$100$ bill equation" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="5987" PostTypeId="2" ParentId="5986" CreationDate="2010-10-03T23:29:49.070" Score="4" ViewCount="0" Body="&lt;p&gt;Let $T$ be the number of tens, and let $H$ be the number of hundreds. You want to find &lt;em&gt;integer&lt;/em&gt; solutions to the equation&#xA;$$ 10T + 100H = 6450.$$&#xA;This is what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations&quot; rel=&quot;nofollow&quot;&gt;linear Diophantine equation&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: And in fact, you have another piece of information which I missed on my first read-through, as noted by jericson:&#xA;$$H+T = 114.$$&#xA;Once you have the two equations, this is no longer a Diophantine equation problem, but a simple system of two linear equations in two unknowns.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-03T23:41:17.263" LastActivityDate="2010-10-03T23:41:17.263" CommentCount="5" />
  <row Id="5988" PostTypeId="2" ParentId="5914" CreationDate="2010-10-03T23:51:06.067" Score="4" ViewCount="0" Body="&lt;p&gt;Alon Amit's answer is not right, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2$ is not prime in the Gaussian integers! $(1+i)$ divides $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument upto the fact that the prime $p$ divides $y+i$, $y-i$ and $2i$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $p|2i$ we consider $p = 1+i$. Now any multiple of $1+i$ is of the form $2x + i2y$ or $2x+1 + i(2y+1)$. Since $y$ is even, $y+i$ cannot be divisible by $1+i$ and thus $y+i$ and $y-i$ are co-prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $y+i$ to be a perfect cube (upto units) easily gives us $y=0$ and so $x=1, y=0$ is the only solution.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-03T23:51:06.067" />
  <row Id="5989" PostTypeId="2" ParentId="5986" CreationDate="2010-10-04T00:18:18.083" Score="6" ViewCount="0" Body="&lt;p&gt;If all 114 were 10's then their sum is 1140, which is 5310 too low. So we need to swap some 10's for 100's. Each swap adds 90, so we need 5310/90 = 59 swaps, yielding 59 100's, $\:$ hence 114 - 59 = 55 10's.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-04T00:18:18.083" />
  <row Id="5991" PostTypeId="2" ParentId="5986" CreationDate="2010-10-04T00:55:57.307" Score="2" ViewCount="0" Body="&lt;p&gt;Dan,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have $$H+T=114$$ and $$10T+100H=6450$$ as above, write $$H=114-T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substitute this into the second equation above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10T+100(114-T)=6450$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand, and simplify. &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-10-04T00:55:57.307" CommentCount="3" />
  <row Id="5992" PostTypeId="2" ParentId="5932" CreationDate="2010-10-04T01:31:46.113" Score="2" ViewCount="0" Body="&lt;p&gt;Given a set $A = (\cup a_i) \subset N$ with positive lim-sup density, it is possible to partition $N$ into intervals on which $A$ has density $\geq d$.  If the intervals grow in size at a suitable rate (for this problem, exponential growth is enough), there will be a positive number $C&gt;0$ such that on each interval, $\Sigma_{a \in A} 1/a \geq C$, and this makes the sum over all intervals infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a partition of the positive integers into intervals $[1,n_1], [n_1 + 1, n_2], [n_2 + 1,n_3], \dots$ such that the density of $A$ is at least $d &gt; 0$ on each interval.  This can be done for any $d &amp;lt; L$ where $L$ is the lim sup density of $A$.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By joining together adjacent intervals, one can also require that the sequence $n_k$ grows fast enough that density in $[1,n_i]$ is not strongly affected by what happens in preceding intervals $[1,n_j]$ with $j &amp;lt; i$.  In particular, one can keep or remove the subinterval $[1,n_{i-1}]$ with small effect on density questions in $[1,n_i]$.    For this problem, fast enough growth will mean strictly exponential growth, i.e., $n_{i+1} / n_i &gt; p $ for constant $p &gt; 1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The most efficient packing of a given number of integers into an interval $[m,n]$ (where efficiency means minimim sum of $1/a$ with $a$ ranging over the subset) is to place the whole subset as close as possible to $n$.  If the density is at least $d$, the number of integers to be placed is at least $d(n-m+1)$, and then $\Sigma 1/a$ is at least $d(n-m)/n = d(1-m/n)$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the exponentially growing intervals with $m/n &amp;lt; 1/p$ the sum in each interval of the partition of $N$ is bounded below by $C=d(1-1/p) &gt; 0$, so the sum of all $1/a_i$ is infinite.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-04T19:58:36.130" LastActivityDate="2010-10-04T19:58:36.130" />
  <row Id="5993" PostTypeId="1" CreationDate="2010-10-04T01:44:39.080" Score="0" ViewCount="155" Body="&lt;p&gt;I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a 6% sales tax, a shirt cost 15.37 USD. Find the original cost of the shirt before sales tax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly how do I approach this? Thanks&lt;/p&gt;&#xA;" OwnerUserId="2294" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T01:48:56.840" LastActivityDate="2010-10-04T03:02:03.823" Title="Sales Tax equation" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5994" PostTypeId="2" ParentId="5982" CreationDate="2010-10-04T01:45:39.430" Score="1" ViewCount="0" Body="&lt;p&gt;$\sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{\sqrt{9}} = \frac{\sqrt{2}}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this what you wanted?&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-10-04T01:45:39.430" CommentCount="3" />
  <row Id="5995" PostTypeId="2" ParentId="5993" CreationDate="2010-10-04T01:47:44.233" Score="0" ViewCount="0" Body="&lt;p&gt;US\$15.37 is 106% of the original cost. So solve for the cost.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-04T01:47:44.233" />
  <row Id="5996" PostTypeId="2" ParentId="5993" CreationDate="2010-10-04T01:48:03.780" Score="1" ViewCount="0" Body="&lt;p&gt;Let x equal the original cost of the shirt. Then x plus the sales tax (.06 of x) equals 15.37. Solve for x.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-04T01:48:03.780" />
  <row Id="5997" PostTypeId="1" CreationDate="2010-10-04T02:06:48.470" Score="0" ViewCount="166" Body="&lt;p&gt;I have been tasked with creating a C++ program (with GDI+ for graphics) that takes a set of user defined points and creates polynomial curve through them. For extra credit, I have to support a parametric polynomial for any number of points. The thing is.. he didn't explain what a parametric polynomial is! So I need some help understanding how to figure one out based on a set of points.. in layman's terms, please. I've taken Calc 1, so I'm not that up-to-par on my calculus skills, just so you know what level I'm at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2301" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T10:44:28.620" LastActivityDate="2010-10-04T10:44:28.620" Title="Graphing a Parametric Polynomial based on a given set of points" Tags="&lt;geometry&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="5998" PostTypeId="2" ParentId="5993" CreationDate="2010-10-04T02:16:44.237" Score="0" ViewCount="0" Body="&lt;pre&gt;&lt;code&gt;15.37 = 1.06x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where &lt;em&gt;x&lt;/em&gt; is the original cost of the t-shirt.&#xA;The 6% sales tax would make the t-shirt cost 106% of the original price, so now just do some simple algebra to isolate &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2296" LastEditorUserId="2296" LastEditorDisplayName="" LastEditDate="2010-10-04T03:02:03.823" LastActivityDate="2010-10-04T03:02:03.823" CommentCount="4" />
  <row Id="5999" PostTypeId="2" ParentId="5997" CreationDate="2010-10-04T02:27:32.550" Score="0" ViewCount="0" Body="&lt;p&gt;I suspect that by 'parameter' your professor means that your code should accept a number supplied by the user specifying how many data points they would like to input.  Then your code should write a polynomial to fit that data.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is fairly simple to write a unique $(n-1)$-degree polynomial given $n$ data points.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Curve_fitting&quot; rel=&quot;nofollow&quot;&gt;Start here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-10-04T02:27:32.550" CommentCount="2" />
  <row Id="6000" PostTypeId="1" AcceptedAnswerId="6012" CreationDate="2010-10-04T04:21:23.283" Score="5" ViewCount="137" Body="&lt;p&gt;The essence of my question is, if I have a Hermitian matrix that is linearly dependent on a set of parameters and I have an estimate of its eigenvalues, is there a &quot;simple&quot; way to determine the values of the parameters?  Ideally, I would also like to have some measure of the goodness of the fit and the degree of variation within the parameters.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a materials physicist, I often have to create a simple quantum mechanical model from either experimental data or a more complex calculation.  For the smaller problems (8x8, with 10 params), the parameters can be found by painstakingly working through the various relationships among the parameters, due to symmetry, etc.  But, this method is specific to each problem and does not scale well to larger problems.  For instance, one system I'm looking at would require a 20x20 matrix with 21 parameters, and that is without including spin!  Alternatively, there is the brute force method of simulated annealing, which involves taking a random walk through the parameter space, and slowly decreasing the step size in the hopes that the calculation will get stuck in the global minimum.  Neither of these methods is particularly appealing, so I'd like some ideas on how to approach this in a consistent manner. &lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2010-10-04T09:44:32.807" Title="Fitting a parameter dependent matrix to its eigenvalues" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6001" PostTypeId="2" ParentId="5959" CreationDate="2010-10-04T05:28:28.583" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not sure that I properly understand the question, especially in light of your acceptance of Bill Dubuque's answer and your accompanying comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, maybe these remarks will be useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $R/f^{-1}(m) \to S/m$ is a map from an integral domain to a field, and if the&#xA;source is not also a field (i.e. if $f^{-1}(m)$ is not itself maximal), then there are certainly&#xA;non-units in $R/f^{-1}(m)$ that map to units in $S/m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interpretation of your question is whether, in this context, there must necessarily be a non-unit of $R$ that maps to a unit in $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &quot;no&quot; in general.  Here is a counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a domain which is not a field, and let $I$ by any subset of $R\setminus \{0\}$ which generates $R\setminus \{0\}$ as a monoid under multiplication.  Let $S$ be the polynomial ring $R[\{x_i\}_{i \in I}].$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Q$ denotes the fraction field of $R$, then there is a natural surjection&#xA;$S \to Q$ given by mapping $x_i$ to $1/i$.  (This is where we use the assumption that&#xA;$I$ generates $R\setminus \{0\}$, so as to get a surjection.)&#xA;The kernel of this surjection is a maximal ideal $m$ of $S$, whose preimage in $R$ is the&#xA;zero ideal (hence not maximal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, since $S$ is a polynomial ring over $R$, any non-unit in $R$ remains a non-unit in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional comments: if $R$ is Jacoboson (e.g. $\mathbb Z$, or a polynomial ring over a field), then any map between finite type $R$-algebras preserves the corresponding MaxSpecs.&#xA;On the other hand, for such domains, the set $I$ is necessarily infinite, and so $S$ is necessarily infintely generated as an $R$-algebra (and hence no contradiction ensues!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can get finite type examples by taking $R$ to be e.g. a DVR.  Then $I$ can be taken&#xA;to consist of a single element (the uniformizer), and so $S = R[x]$ is a polynomial ring&#xA;in a single variable.  For example, if $R = \mathbb Z_p$ (the $p$-adic integers) then&#xA;we have &#xA;$$\mathbb Z_p \hookrightarrow \mathbb Z_p[x] \mapsto \mathbb Z_p[x]/(p x - 1) = \mathbb Q_p,$$&#xA;and so $p x - 1$ generates a maximal ideal in $\mathbb Z_p[x]$ whose preimage is not maximal, but every non-unit in $\mathbb Z_p$ remains a non-unit in $\mathbb Z_p[x]$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-04T05:28:28.583" CommentCount="1" />
  <row Id="6002" PostTypeId="1" AcceptedAnswerId="6004" CreationDate="2010-10-04T06:20:03.537" Score="0" ViewCount="49" Body="&lt;p&gt;I was reading through some notes on closure operations, and the example was given that for a poset $(S,\preceq)$, the operation on $(\mathcal{P}(S),\subseteq)$ given by $E\mapsto LU(E)$ is a closure operation. Here $LU(E)$ is the set of lower bounds of the upper bounds of $E$ for some $E\subseteq S$. I verified that indeed $E\subseteq LU(E)$, and if $E\subseteq E'$, then $LU(E)\subseteq LU(E')$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if I take a family of sets $\xi\subseteq \mathcal{P}(S)$, such that $\xi$ consists only of $LU$-closed subsets of $S$, is it then true that the union is again $LU$-closed? That is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bigcup_{E\in\xi}E=\bigcup_{E\in\xi}LU(E)=LU(\bigcup_{E\in\xi}E)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the first equality comes from the fact that $E=LU(E)$. I see that for $e\in\bigcup_{E\in\xi}E$, then $e\in LU(\bigcup_{E\in\xi}E)$, however, the other containment has eluded me. I know that the Kuratowski closure axioms require a closure operator to preserve unions in topology, but is such an equality even the case for partially ordered sets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; The reason I ask is that there is the following exercise in the notes, and here $\mathcal{L}(S)\subseteq\mathcal{P}(S)$ is the set of all $LU$-closed subsets of $S$. I want to show for any family $\xi\subseteq \mathcal{L}(S)$ of $LU$-closed subsets of $S$, then $\sup\xi$ exists in $\mathcal{L}(S)$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sup\xi=LU(\bigcup_{E\in\xi}E).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\sup\xi$ in $\mathcal{P}(S)$ is the union of all sets $E\in\xi$. My strategy was if one could show that $\cup_{E\in\xi}E=LU(\cup_{E\in\xi}E)$, then it follows that $\cup_{E\in\xi}E$ does indeed exist in $\mathcal{L}(S)$, and thus $\sup\xi=\cup_{E\in\xi}E=LU(\cup_{E\in\xi}E)$. Perhaps I have misinterpreted the question?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-04T07:52:35.483" LastActivityDate="2010-10-04T08:18:16.697" Title="Is the union of a family of $LU$-closed sets also $LU$-closed?" Tags="&lt;set-theory&gt;&lt;topology&gt;" AnswerCount="1" />
  <row Id="6003" PostTypeId="2" ParentId="5962" CreationDate="2010-10-04T06:40:28.743" Score="1" ViewCount="0" Body="&lt;p&gt;I've seen some of this stuff before, but I'm definitely not an expert.  Here's my take (although I'm likely wrong on at least some of it, hopefully not too much):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can think of this as a matrix of 1-forms.  Then, $A \wedge A$ means &quot;matrix multiplication, where multiplication is wedge product&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Yes, I think $A_i v_i$ really means $A_i \otimes v_i$.&lt;/li&gt;&#xA;&lt;li&gt;I think also $A_i \wedge B_j v_i \otimes w_j$ means $A_i \wedge B_j \otimes v_i \otimes w_j$, although there might be some issue of skew-symmetrization.&lt;/li&gt;&#xA;&lt;li&gt;A vector-bundle-valued differential k-form locally looks like $\omega \otimes s$, where $\omega$ is a usual k-form and $s$ is a section.  Then $d$ should act by the Leibniz rule (once you've fixed a connection).  I'd imagine that the trace is just trace in the usual sense, and then it's obvious that this commutes with $d$ and that $Tr$ admits cyclic permutations of the factors in its argument.&lt;/li&gt;&#xA;&lt;li&gt;I think that whether $A\wedge A\wedge A\wedge A$ must be zero depends on the dimension of the vector bundle where it takes its values.  If for example it's the constant 1-dimensional bundle, then this is no different from usual 1-forms.&lt;/li&gt;&#xA;&lt;li&gt;Your last question is a little unclear, sorry.  Could you clarify exactly what you're referring to?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll respond to your comments in order.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $A$ takes a vector field and spits out a matrix, we can consider each of the entries of the matrix (in a fixed basis, of course) as an ordinary 1-form.  If you want to wedge two matrix-valued forms, you just write out the matrices, act like you're multiplying matrices like you learned a long time ago (i.e. take the dot product of rows with columns), but then when you want to simplify the expression, you consider the product to be the wedge.  So, e.g. if&#xA;$$ A = (a,b;c,d) $$&#xA;is a 2x2 matrix-valued 1-form &lt;em&gt;(sorry, for whatever reason my matrices aren't working out)&lt;/em&gt;, this means that $a$, $b$, $c$, and $d$ are just ordinary 1-forms.  So evaluating on a vector $v$ gives the 2x2 matrix (of numbers)&#xA;$$ A(v) = ( a(v) , b(v) ;c(v) , d(v) ). $$&#xA;If you want to calculate $A\wedge A$, this is the 2x2 matrix-valued 2-form&#xA;$$ A\wedge A = ( a, b ;c,d) \wedge ( a , b ;c ,d ) $$&#xA;$$ = ( a\wedge a + b\wedge c , a\wedge b + b\wedge d ; c\wedge a + d\wedge c , c\wedge b+d\wedge d) $$&#xA;$$ = ( b\wedge c , b\wedge (d-a) ;c\wedge (a-d) , c\wedge b ). $$&lt;/li&gt;&#xA;&lt;li&gt;This should be more or less answered by (1).  The decomposition of the form $\omega \otimes s$ is just saying that a vector-bundle-valued form can be thought of as an ordinary form (that's $\omega$), except that you need to decide which direction in the fiber it takes you (that's $s$).  This specializes to the case where the bundle is the trivial $\mathbb{R}$-bundle.  Then $s$ is actually just a function, and we're recovering the fact that $n$-forms are a module over $C^\infty(M,\mathbb{R})$.&lt;/li&gt;&#xA;&lt;li&gt;Like you said in your first comment, a connection takes in tangent vectors to the total space.  From here, this is exactly what I said in (2).&lt;/li&gt;&#xA;&lt;li&gt;Well, any form is nilpotent -- if you wedge it with itself enough times, you eventually get 0 for dimensional reasons.  (This assumes that the (co)tangent space is finite-dimensional, of course.  I have no idea what happens in the case that it isn't.)  The only obvious reason (to me) that $A\wedge A\wedge A\wedge A$ would be traceless is if it's 0.  But then again, I don't know anything about Chern-Simons theory.  Or maybe it's a trick along the lines of the proof that the commutator of two square matrices is always traceless (? I'm pretty sure that's true, and follows from the fact that $Tr(AB)=Tr(BA)$, which you should note does &lt;em&gt;not&lt;/em&gt; imply that $Tr(ABC)=Tr(ACB)$, only that you can &lt;em&gt;cyclically&lt;/em&gt; permute the matrices so e.g. $Tr(ABC)=Tr(CAB)=Tr(BCA)$).&lt;/li&gt;&#xA;&lt;li&gt;All I know is that back in linear algebra, you learn that the trace of a matrix is invariant under basis change.  In other words, we can have a well-defined notion of the trace of a linear map.  Of course, this requires that the dimensions of the domain and codomain agree, which doesn't seem like it'll be true in the example you gave (of a vector-valued differential form).  However, if they do agree, then (in a basis) you can consider a form to be matrix-valued, and then the trace is in all likelihood just the sum of the diagonal entries, which are themselves ordinary differential forms.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-06T08:01:23.200" LastActivityDate="2010-10-06T08:01:23.200" CommentCount="7" />
  <row Id="6004" PostTypeId="2" ParentId="6002" CreationDate="2010-10-04T07:35:51.490" Score="2" ViewCount="0" Body="&lt;p&gt;This isn't even true in finite posets.&#xA;Let $S=\{a,b,c\}$ where the partial order is&#xA;defined by $a &amp;lt; c$, $ b &amp;lt; c $ and $a$ and $b$ are incomparable.&#xA;Then $\{a\}$ and $\{b\}$ are LU-closed but&#xA;$\{a,b\}$ isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;The supremum you seek is not&#xA;$\bigcup_{E\in\xi}E$ but rather $LU(\bigcup_{E\in\xi}E)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-04T08:18:16.697" LastActivityDate="2010-10-04T08:18:16.697" CommentCount="1" />
  <row Id="6005" PostTypeId="1" AcceptedAnswerId="6006" CreationDate="2010-10-04T07:43:46.317" Score="1" ViewCount="83" Body="&lt;p&gt;Anybody knows whether there is a list, say first 1000 sphenic numbers? It would be much helpful to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="1869" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-04T08:25:53.513" LastActivityDate="2010-10-04T08:25:53.513" Title="Where can I find a list of sphenic numbers?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6006" PostTypeId="2" ParentId="6005" CreationDate="2010-10-04T07:52:11.817" Score="5" ViewCount="0" Body="&lt;p&gt;The first 10000 are listed &lt;a href=&quot;http://oeis.org/classic/b007304.txt&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. (Found via &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphenic_number&quot; rel=&quot;nofollow&quot;&gt;Wikipedia: Sphenic number&lt;/a&gt; and &lt;a href=&quot;http://oeis.org/classic/A007304&quot; rel=&quot;nofollow&quot;&gt;OEIS&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T07:52:11.817" />
  <row Id="6007" PostTypeId="2" ParentId="5997" CreationDate="2010-10-04T08:03:35.490" Score="1" ViewCount="0" Body="&lt;p&gt;It's not clear to me whether this is just an exercise for its own sake, or meant to serve some real purpose. If is is meant to be like a drawing program where you want a nice curve that &quot;follows&quot; the points that the user selects, the interpolating polynomial curve is probably not a very good idea, since it will oscillate wildly when you have many points (and most likely go way outside of the screen). In such a situation, &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zier_curve&quot; rel=&quot;nofollow&quot;&gt;Bézier curves&lt;/a&gt; should be more useful.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T08:03:35.490" CommentCount="2" />
  <row Id="6008" PostTypeId="1" CreationDate="2010-10-04T08:49:46.183" Score="4" ViewCount="134" Body="&lt;p&gt;Hi, folks!&lt;br&gt;&#xA;I have the following problem:&lt;br&gt;&#xA;Suppose you have a coin that has chance p of landing heads. Suppose you flip the coin N times and let&lt;br&gt;&#xA;X denote the number of &quot;head runs&quot; in N flips. A &quot;head run&quot; is defined as any sequence of heads.&lt;br&gt;&#xA;For example the sequence HHTHHHHHTTTTHHTHT contains 4 head runs. I want to compute E(X) and Var(X).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a difficulty especially with variance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="2304" LastActivityDate="2010-10-05T10:24:50.807" Title="The Expectation and the Variance of the runs" Tags="&lt;probability-theory&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" />
  <row Id="6009" PostTypeId="1" AcceptedAnswerId="6016" CreationDate="2010-10-04T08:52:16.870" Score="1" ViewCount="149" Body="&lt;p&gt;1.Find the number of ways in which a mixed double tennis game can be arranged between 10 players consisting of 6 men and 4 women.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach : Since,if every pair of selection of men and women gives rise to 4 matches so &lt;strong&gt;C(6,2) * C(4,2) * 4.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.Find the number of ways in which a mixed double game can be arranged from amongst nine married couples, if no husband and wife play in the same game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach : &lt;strong&gt;C(9,2) * C(7,2) * 4.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this two questions comes from my test paper I would be grateful if somebody check whether this is correct or not since my answers doesn't matched to the given solution in the paper.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-19T07:45:59.363" Title="Two question on Permutation and Combination" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6010" PostTypeId="2" ParentId="6008" CreationDate="2010-10-04T08:59:26.910" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Monte Carlo method&lt;/strong&gt; (actually flipping the coin)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $x_i$ is the number of &quot;head runs&quot; in trial $i$, you can compute the sample expectation (or rather, the mean) after $n$ trials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X]=\sum\limits_{i=0}^n x_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance is calculated in a similar manner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V[X]=E[X^2]-E[X]^2=\sum\limits_{i=0}^n x_i^2 - \left(\sum\limits_{i=0}^n x_i\right)^2$$&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditorDisplayName="" LastEditDate="2010-10-04T14:17:26.210" LastActivityDate="2010-10-04T14:17:26.210" CommentCount="3" />
  <row Id="6011" PostTypeId="2" ParentId="5983" CreationDate="2010-10-04T09:22:46.490" Score="2" ViewCount="0" Body="&lt;p&gt;As you ask only for hints, here are two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Show that some/any $p$-Sylow subgroup $P$ of $G$ acts transitively on the $n$ letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Then consider the action of a central element of $P$ of order $p$.&lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2010-10-04T09:22:46.490" CommentCount="5" />
  <row Id="6012" PostTypeId="2" ParentId="6000" CreationDate="2010-10-04T09:44:32.807" Score="2" ViewCount="0" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ M = \Sigma_{i=1}^{n} t_i M_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_i$ are the unknown parameters and $M_i$ are known matrices and $n$ is the number of parameters, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N$ be the size of the matrix $M$ and $\lambda_j$, $j = 1, . . . , N$ its eigenvalues, then one can use the identities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ tr(M^k) = \Sigma_{j=1}^{N} \lambda_j^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain N polynomial equations of the parameters, for example for $k=2$, the equation has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Sigma_{i=1}^{n} \Sigma_{j=1}^{n} t_i t_j tr(M_i M_j) = \Sigma_{j=1}^{N} \lambda_j^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the righthand side is known and the polynomial coefficients of the left hand side are traces of products of the known submatrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the parameters can be found in principle from a numerical solution of the polynomial equation system.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-04T09:44:32.807" CommentCount="3" />
  <row Id="6013" PostTypeId="1" CreationDate="2010-10-04T10:05:55.423" Score="1" ViewCount="152" Body="&lt;p&gt;I have used logic for specification of security policies in a security model. One of my reviewers asked a question that &quot;why did you used logic and not used algebra for this purpose, and what is your justification?&quot;.&#xA;The first answer receives in my mind is that algebra is more suitable for using in modeling the behaviors (dynamic aspects). For example for specification of security protocols. However, logic is more suitable for specification of properties (static aspects). For example for specification of the security properties we require to satisfy in a computing environment, or the security properties we desire to be satisfied in a security protocol. In other words, algebra is suitable to say HOW, but logic is suitable to say WHAT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions regarding this issue are:&#xA;1- Is my initial answer to this question correct or not?&#xA;2- What are the other answers for this question?&#xA;3- What are the main references describing where we should use logic, and where we should use algebra in their applications in computer science?&lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2010-10-04T17:32:40.470" Title="Applications of Logic and Algebra in Computer Science" Tags="&lt;logic&gt;&lt;algebra&gt;" AnswerCount="3" />
  <row Id="6014" PostTypeId="2" ParentId="5997" CreationDate="2010-10-04T10:33:39.387" Score="0" ViewCount="0" Body="&lt;p&gt;As I mentioned in &lt;a href=&quot;http://math.stackexchange.com/questions/5166&quot;&gt;the answer to this question&lt;/a&gt;, &lt;em&gt;parametric (cubic) splines&lt;/em&gt; which are represented as &lt;em&gt;piecewise polynomials&lt;/em&gt; in each coordinate $(x,y)$ are the most useful choice, when one wants to draw a curve passing through points that do not necessarily represent a function. (The problem with Bézier is that it treats the input points as a convex hull; i.e. the curve is within the set of points but does not pass through them.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hS0XJ.png&quot; alt=&quot;spline vs. Bezier&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparison of a parametric spline (red) and a Bézier curve (blue).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An interpolating polynomial is bad form as well for arbitrary point sets; the Runge phenomenon is one vivid example.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-04T10:40:22.963" LastActivityDate="2010-10-04T10:40:22.963" CommentCount="5" />
  <row Id="6015" PostTypeId="2" ParentId="6013" CreationDate="2010-10-04T11:00:50.930" Score="2" ViewCount="0" Body="&lt;p&gt;At some level, algebra and logic are equivalent.  See &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883853272&quot; rel=&quot;nofollow&quot;&gt;Logic as Algebra&lt;/a&gt; by Paul Halmos and Steven Givant.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-04T11:00:50.930" CommentCount="1" />
  <row Id="6016" PostTypeId="2" ParentId="6009" CreationDate="2010-10-04T11:09:56.230" Score="2" ViewCount="0" Body="&lt;p&gt;The answer to (1) is 2*C(6,2)*C(4,2) (and not with a 4), since M1, W1 playing M2, W2 is the same as M2, W2 playing M1, W1. I don't think the question wants you to take account of which end they're playing from. You've made a similar error in part(2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it more clear you've included a factor 2 for swapping the women over, say, in our set of four people, and then included another factor of 2 for swapping the men over, but have of course ended up with the same match when both swaps have taken place.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-04T11:21:47.560" LastActivityDate="2010-10-04T11:21:47.560" CommentCount="2" />
  <row Id="6017" PostTypeId="1" AcceptedAnswerId="6021" CreationDate="2010-10-04T12:02:46.100" Score="2" ViewCount="153" Body="&lt;p&gt;How does one find the solution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{dy}{dx}\left( 1-\left( 1-t\right) x-x^{2}\right) -\left( 1+h\left( 1+t\right) +x\right) y=0\quad ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is an integer constant and $t$ is a real constant between $0$ and $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$($ In Roger Apéry, &lt;em&gt;Interpolations de Fractions Continues et Irrationalité de certaines Constantes&lt;/em&gt;, Bull. section des sciences du C.T.H.S., n.º3, p.37-53, the solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=(1-x)^{-1-h}(1+tx)^{h}.)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The sequence $(v_{h,n})$ in $y=f_{h}(x)=\displaystyle\sum_{n\ge 0}v_{h,n}x^n$  satisfies a recurrence related to $\log (1+t)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Copy of the original with the equation and solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DGW2b.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt; 2: I transcribe the comment in the 1st answer: &quot;the corrected differential equation above agrees with the recurrence in your excerpt so there is clearly a typo in the printed differential equation.&quot;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-04T17:48:26.850" LastActivityDate="2010-10-04T17:48:26.850" Title="Solving a differential equation related to  $\log (1+t)$" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="6018" PostTypeId="2" ParentId="6013" CreationDate="2010-10-04T12:49:32.743" Score="3" ViewCount="0" Body="&lt;p&gt;I don't like this kind of question &amp;mdash;not yours, your referee's&amp;mdash; it's enough for authors to solve problems, the existence of other routes that other people might prefer is besides the point, unless there is some relative disadvantage that can the referee can point out.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Is my initial answer to this question correct or not?&lt;/em&gt; &amp;mdash; It's not what I would say.  There are plenty of approaches to handle mutability in logic, of which dynamic logic is probably the most widely applied.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;What are the other answers for this question?&lt;/em&gt; &amp;mdash; I should think that ion your shoes I would say something like: &quot;The choice of logic vs. algebra is much less important than the design decisions made in formalising the specification.  This is what seemed to me to be most well suited at the time.&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;What are the main references describing where we should use logic, and where we should use algebra in their applications in computer science?&lt;/em&gt; &amp;mdash; I don't know of anything along the lines of what you are asking.  Blackburn, de Rijke &amp;amp; Venema (2001) &lt;em&gt;Modal Logic&lt;/em&gt; talks about the similarities and differences between algebraic and model-theoretic techniques - I left with the impression that the biggest issue was whether you preferred to work with completeness theorems or representation theorems.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To say something in addition to John's answer, here was a thread on Math Overflow, &lt;a href=&quot;http://mathoverflow.net/questions/11974/is-there-a-relationship-between-model-theory-and-category-theory&quot; rel=&quot;nofollow&quot;&gt;Is there a relationship between model theory and category theory&lt;/a&gt;, that I think gives some impression on how little distance there is seen to be between algebra and logic.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-10-04T12:49:32.743" CommentCount="3" />
  <row Id="6019" PostTypeId="2" ParentId="5774" CreationDate="2010-10-04T13:26:21.830" Score="2" ViewCount="0" Body="&lt;p&gt;I think you're trying to ask:  how many such grids admit a non-trivial symmetry (in the dihedral group)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For fixed m&lt;/em&gt;&amp;ge;1, asymptotically almost all (0,1,-1)-matrices with exactly m 1's and m -1's do not admit a non-trivial symmetry (rotation/reflection).  For these matrices, we can consider the X's as the 1's and the (letter) O's as the -1's.  The (number) 0's represent the empty cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since m is fixed, there are ${{N^2} \choose {m,m,N^2-2m}}=N^{4m}/m!^2+o(N^{4m})$ (0,1,-1)-matrices in total with exactly m 1's and m -1's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of such matrices that are preserved under transposition is \[\sum_{i,j} {N \choose {i,j,n-i-j}} {{N \choose 2} \choose (m-i)/2} {{{N \choose 2}-(m-i)/2} \choose (m-j)/2}\] where i is the number of 1's on the main diagonal, j is the number of -1's on the main diagonal.  Therefore we require 0&amp;le;i+j&amp;le;m and m-i and m-j even.  A crude upper bound to the above summand is $\mathrm{const} \cdot N^{i+j+2(m-i)/2+2(m-j)/2} \leq N^{3m}=o(N^{4m})$.  Since m is fixed, there is only a finite number of pairs i,j which is summed over, so the overall result is $o(N^{4m})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of such matrices that admit any of the other non-trivial symmetries is bounded above by the above formula also.  So asymptotically almost all (0,1,-1)-matrices with exactly m 1's and m -1's do not admit a non-trivial symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I suspect that this result would be true without the fixed m condition, but I can't think of how to prove it off-hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in answer to the question, unless you're dealing with a very small case, and unless you're deliberately trying to create a symmetry, you're unlikely to create a matrix that has a non-trivial symmetry.  For the search tree, most of the time you will be able to identify 8 nodes corresponding to equivalent games of naughts and crosses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[PS: it would be messy (although possible) to find an exact formula along these lines for the number of such matrices that admit a non-trivial symmetry.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-04T13:26:21.830" CommentCount="1" />
  <row Id="6020" PostTypeId="1" AcceptedAnswerId="6022" CreationDate="2010-10-04T13:58:04.677" Score="0" ViewCount="105" Body="&lt;p&gt;Let $\mathcal{F}$ be an open cover of a metric space $(X,d)$. I have a question concerning some algebraic aspects of $\mathcal{F}$ and $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, is it always possible to consider $(X,d)$ a subspace of some bigger space $(X^{\prime},d)$, where $d$ is the inherited metric from $X^{\prime}$? If this is possible, can we say that $\mathcal{F}$ is a subset of the power set of $X^{\prime}$? What are the conditions that we must state for this to be true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if all of that is true, can we say that $\mathcal{F}$ is an algebra of sets? Of course, for it to be true, $\mathcal{F}$ should contain both $X^{\prime}$ and $\emptyset$, but, whatever the cover is, we can always add those two sets and not change anything about the cover, right? So, finally, whatever $\mathcal{F}$ is, can we say that $\mathcal{F} \cup X^{\prime} \cup \emptyset$ is an algebra of sets over $X^{\prime}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some motivation: in one of my exams I could prove something just saying &quot;It is enough to notice that the cover is an algebra of sets&quot;, but I was unsure if that was always true.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-10-04T15:59:49.390" Title="A question about open covers" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6021" PostTypeId="2" ParentId="6017" CreationDate="2010-10-04T14:31:45.970" Score="6" ViewCount="0" Body="&lt;p&gt;If the given solution is correct then the posted differential equation is wrong. Instead it should be as follows, with the corrected terms underlined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^{\prime }\ \left( 1-\left( 1-t\right) x - \underline{tx^{2}}\right) -\left( 1+h\left( 1+t\right) + \underline{tx}\right)\ y\ =\ 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which of course is trivially integrable since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{y'}y\ =\ \frac{1+h}{1-x}\ +\: \frac{ht}{1+tx} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: the corrected differential equation above agrees with the recurrence in your excerpt so there is clearly a typo in the printed differential equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T17:24:12.233" LastActivityDate="2010-10-04T17:24:12.233" CommentCount="4" />
  <row Id="6022" PostTypeId="2" ParentId="6020" CreationDate="2010-10-04T14:42:05.470" Score="3" ViewCount="0" Body="&lt;p&gt;Your first question is rather vague: you can always take $X'=X$, $d'=d$, and then $(X,d)$ is a subspace of $(X',d')$ with the inherited metric. Presumably you first of all want $X$ &lt;em&gt;properly contained&lt;/em&gt; in $X'$ (&quot;bigger&quot; is a too vague, sorry!). But - do you also want some other property for $X'$, besides merely containing $X$? Perhaps &quot;complete&quot;, or something along those lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all you want is a metric space that properly contains $X$ and with induced metric, then the answer is yes. For example, if $X$ has only one element, then you can let $X'=X\cup\{p\}$, where $p\notin X$, and take the discrete metric. If $X$ has more than one element, you can take $X'=X\times X$, and define a metric by $d'((a,b),(x,y)) = d(a,x) + d(b,y)$ (the &quot;taxicab norm&quot;). To verify that this is indeed a metric, note that since $d(r,s)\geq 0$ for all $r,s\in X$, we have $d'((a,b),(x,y))\geq 0$; and $d'((a,b),(x,y))=0$ if and only if $d(a,x)=d(b,y)=0$, if and only if $a=x$ and $b=y$, if and only if $(a,b)=(x,y)$. Also, $d'((a,b),(x,y)) = d(a,x)+d(b,y) = d(x,a)+d(y,b) = d'((x,y),(a,b))$. Finally, for the triangle inequality, we have that &#xA;$$\begin{array}{rcl}&#xA;d'((a,b),(r,s))+d'((r,s),(x,y)) &amp;amp;=&amp;amp; d(a,r)+d(b,s) + d(r,x) + d(s,y)\\&#xA;&amp;amp; \geq &amp;amp; d(a,x) + d(b,y) = d'((a,b),(x,y)).&#xA;\end{array}$$&#xA;So $(X',d')$ is a metric space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now fix $x_0\in X$, and embed $X$ into $X'$ by mapping $a\in X$ to $(a,x_0)$. This is an embedding, and for all $a,b\in X$, $d'((a,x_0),(b,x_0)) = d(a,b)+d(x_0,x_0)=d(a,b)$, so if we identify $X$ with its image in $X'$ you get that $d$ is the restriction of $d'$ to $X$. Since we are assuming that $X$ has more than one element, (the image of) $X$ is properly contained in $X'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;$\mathcal{F}$ is an open cover for a metric space $(X,d)$&quot;, I interpret this to mean that each element of $\mathcal{F}$ is an open subset of $X$ and that $\cup\mathcal{F}=X$. You seem to be thinking that the sets may be strictly larger than $X$, but in the absence of a given overset, this makes no sense. Thus, we can talk about open covers of $[0,1]$ &lt;em&gt;in $\mathbb{R}$&lt;/em&gt; (open subset of $\mathbb{R}$ whose union contains $[0,1]$); but if we just say &quot;an open cover of $\mathbb{R}$&quot;, we assume the sets are subsets  of $\mathbb{R}$ absent any other context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $X$ is a set, $X'$ contains $X$, and $\mathcal{F}$ is a family of subsets of $X$, then $\mathcal{F}$ is also a family of subsets of $X'$, hence a subset of the power set of $X'$; so the answer to your second question (&quot;can we say $\mathcal{F}$ is a subset of the power set of $X'$?&quot;) is an absolute yes. The fact that we are dealing with a family of open sets or any such thing is completely immaterial; this is simply a consequence of the meaning of &quot;subset&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for being an algebra: an algebra of sets must be closed under unions and complements. If you have an open cover, it is not necessarily closed under unions, and it will &lt;em&gt;usually&lt;/em&gt; not be closed under complements (the complements will be closed; if they are also part of the open cover, then each element of the open cover except for $X$ and $\emptyset$ (if they are there) would disconnect the space). So &quot;most of the time&quot; an open cover will not be closed under complementation and hence not be an algebra. (&quot;Closing&quot; the cover under unions is fairly easy; just throw in the unions of any two elements until you are closed under unions; all of them will be open, and enlarging an open cover by adding open sets will not change the fact that it is an open cover).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-04T15:59:49.390" LastActivityDate="2010-10-04T15:59:49.390" CommentCount="3" />
  <row Id="6023" PostTypeId="2" ParentId="5826" CreationDate="2010-10-04T15:22:16.763" Score="0" ViewCount="0" Body="&lt;p&gt;Now, I think I've got a possible real answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once connection assumed, I'm going to develop my previous comment in order to prove that, at least for compact connected surfaces, connected sums do NOT depend on anything at all. Feel free to point out if I'm missing something or any mistakes in my reasonings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of the independence of the disks and &quot;gluing&quot; homeomorphisms can be stated as follows: Let $X$ and $Y$ be two topological surfaces, $D$ a disc in $X$ and $D'$ a disc in $Y$. Let $h: \partial D \longrightarrow  \partial D'$ be a homeomorphism. Is it necessarily true that $X \oplus_h Y $ does not depend on $h$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, you could proceed as follows. First of all, you need the theorem of classification of surfaces, which says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classification theorem.&lt;/strong&gt; Let $X$ be a connected compact surface. Then $X$ is homeomorphic to one, and only one, of the following &lt;em&gt;standard surfaces&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The sphere $\mathbb{S}^2$.&lt;/li&gt;&#xA;&lt;li&gt;A connected sum of $g$ toruses $g\mathbb{T}^2 = \mathbb{T}^2 \oplus   \dots \oplus \mathbb{T}^2 $.&lt;/li&gt;&#xA;&lt;li&gt;A connected sum of $g$ projective planes $g\mathbb{RP}^2 = \mathbb{RP}^2 \oplus   \dots \oplus \mathbb{RP}^2  $.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warning.&lt;/strong&gt; There is no circular argument here when talking about &quot;the&quot; connected sum of toruses or projective planes, provided you make the following convention. For the time being, this connected sum just means that we are representing spheres, connected sums of toruses, or projective planes by means of the standard polygonal presentations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(a)&lt;/strong&gt; $\mathbb{S}^2 = aa^{-1}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b)&lt;/strong&gt; $g \mathbb{T}^2 =a_1b_1a_1^{-1}b_1^{-1} \dots a_gb_ga_g^{-1}b_g^{-1}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(c)&lt;/strong&gt; $g \mathbb{RP}^2 = a_1a_1 \dots a_ga_g$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;em&gt;define&lt;/em&gt; the connected sum of the standard surfaces by choosing the same &quot;standard&quot; small (such that its boundary doesn't touch the polygon) $D = D'$ disk centered at the barycenters of the polygons and with $h = \mathrm{id} : \partial D \longrightarrow \partial D'$ as homeormorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there could be still a little problem here with orientations: you could &quot;glue&quot;, for instance, two spheres with this convention in two apparently different ways: &quot;side by side&quot;, or one inside the other. Fortunately, for surfaces at least, this doesn't matter (you can &quot;extract&quot; the inner sphere without changing the homeomorphism type).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with this definition, you can prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 1.&lt;/strong&gt; For the standard surfaces, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; $\mathbb{S}^2 \oplus X \cong X$, for any standard surface $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; $g\mathbb{T}^2 \oplus g'\mathbb{T}^2 \cong (g+g')\mathbb{T}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; $g\mathbb{RP}^2 \oplus g'\mathbb{RP}^2 \cong (g+g')\mathbb{RP}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(4)&lt;/strong&gt; $g\mathbb{T}^2 \oplus g'\mathbb{RP}^2 \cong (2g +g')\mathbb{RP}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we already know what the connected sum of the standard surfaces (with the aforementioned convention) is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In fact, we can say more.&lt;/strong&gt; For standard surfaces, keeping the &quot;standard&quot; disk fixed once for all we can choose &lt;em&gt;any&lt;/em&gt; homeomorphism $h: \partial D \longrightarrow \partial D'$ we please. To see this we must rely on the aforementioned classification theorem and the computation of the homology of the standard surfaces, which says, among other things, that the knowledge of $H_1(X)$ suffices in order to classify our surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when we try to compute, for instance, the homology of $X = g\mathbb{T}^2 \oplus g'\mathbb{T}^2$, we may use Mayer-Vietoris, choosing $g\mathbb{T}^2 \backslash V$  and $g'\mathbb{T}^2\backslash V'$ as a cover for $g\mathbb{T}^2 \oplus g'\mathbb{T}^2 $, where $V$ and $V'$ are the interior of $D$ and $D'$, respectively. We have $\mathbb{S}^1 = (g\mathbb{T}^2 \backslash V) \cap (g'\mathbb{T}^2 \backslash V')$ and the interesting part of the Mayer-Vietoris sequence is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0 \longrightarrow H_2(X) \longrightarrow H_1(S^1) \stackrel{f}{\longrightarrow} H_1( g\mathbb{T}^2 \backslash V  )\oplus H_1 ( g'\mathbb{T}^2\backslash V' ) \longrightarrow H_1(X) \longrightarrow 0 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $f$ is the morphism induced by the inclusions $\mathbb{S}^1 \hookrightarrow g\mathbb{T}^2 \backslash V$ and $\mathbb{S}^1 \hookrightarrow g'\mathbb{T}^2 \backslash V' $. Now, the first morphism is just the identity on $\mathbb{S}^1 \longrightarrow \mathbb{S}^1 \subset  g\mathbb{T}^2 \backslash V$ and the second one is $h:\mathbb{S}^1 \longrightarrow \mathbb{S}^1 \subset  g'\mathbb{T}^2 \backslash V' $. But any homeomorphism $\mathbb{S}^1 \longrightarrow \mathbb{S}^1$ is homotopic to $\pm \mathrm{id} : \mathbb{S}^1 \longrightarrow \mathbb{S}^1$. So, an easy computation shows that $f = 0$ and the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;VERY IMPORTANT REMARK.&lt;/strong&gt; Here is the point where, for standard surfaces at least, the &quot;gluing&quot; homeomorphism $h: \partial D = \mathbb{S}^1 \longrightarrow \mathbb{S}^1 = \partial D'$ &quot;disappears&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next result we need is that we can &quot;move&quot; our disks wherever we please on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 2.&lt;/strong&gt; Let $X$ be a connected surface and $D, D'$ two disks in $X$. Then there exists a homeomorphism $H: X \longrightarrow X$ such that $H(D) = D'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of this result is more involved than the previous ones and needs &lt;a href=&quot;http://en.wikipedia.org/wiki/Sch%C3%B6nflies_problem&quot; rel=&quot;nofollow&quot;&gt;Schönflies theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we are prepared to prove the independence of the connected sum. Let $X$ and $Y$ be two connected compact surfaces, $D\subset X$ and $D'\subset Y$ two disks inside of them, $h: \partial D \longrightarrow \partial D'$ a homeomorphism. Let's show that $X \oplus_h Y$ doesn't depend on $h$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Because of the classification theorem $X$ and $Y$ are homeomorphic to two standard surfaces: $S_X$ and $S_Y$, respectively. Let $\phi : X \longrightarrow S_X$ and $\psi : Y \longrightarrow S_Y$ denote the homeomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; These homeomorphisms $X \cong S_X$ and $Y \cong S_Y$ don't need to identify $D$ and $D'$ with our standard disks on the standard surfaces $S_X$ and $S_Y$, but for this we have proposition 2, which tells that we can &quot;move&quot; $D$, $D'$ to those standard disks using two homeomorphisms $H_X : X \longrightarrow X$ and $H_Y: Y \longrightarrow Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; Now we glue our standard surfaces along their standard disks using the homeomorphism induced by $h$ on them: $h' = H_Y\psi h  \phi^{-1} H_X^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(4)&lt;/strong&gt; Because our &quot;we can say more&quot;, the connected sum $S_X \oplus_{h'} S_Y$ does not depend on $h'$. Hence, neither $X\oplus_h Y \cong S_X \oplus_{h'} S_Y$ depends on $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V. Navarro, P. Pascual, &quot;Topologia Algebraica&quot; (Algebraic Topology), Edicions UB (1999).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J.M. Lee, &quot;Introduction to topological manifolds&quot;, Springer (2000).&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-04T16:24:59.357" LastActivityDate="2010-10-04T16:24:59.357" />
  <row Id="6024" PostTypeId="1" AcceptedAnswerId="6025" CreationDate="2010-10-04T15:55:01.473" Score="0" ViewCount="66" Body="&lt;p&gt;I recently started with 2d transformations in my class. I was just working with a program when I realized the formula I am using rotates the object spirally inwards. I have no idea whats over about this. &#xA;I expected my figures to be rotated about a point but rather they all deform with every degree of rotation. I used a single point &amp;amp; realized that the point was moving spirally inwards with every degree I moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me &amp;amp; let me know why it falls in rather than following circle or if that is what is expected of 2d rotation. I somehow felt its something related to how we view in 2d plane (or the imaginary z index here) but simply can't get around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am following &lt;a href=&quot;http://en.wikipedia.org/wiki/Transformation_matrix&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; &amp;amp; talking about this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ldY8x.png&quot; alt=&quot;alt text&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;x' = xcosθ − ysinθ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;y' = xsinθ + ycosθ&lt;/p&gt;&#xA;" OwnerUserId="2305" LastActivityDate="2010-10-04T16:16:08.507" Title="2d rotation falls spirally inwards?" Tags="&lt;transformation&gt;&lt;computational-geometry&gt;&lt;rigid-transformation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6025" PostTypeId="2" ParentId="6024" CreationDate="2010-10-04T16:16:08.507" Score="6" ViewCount="0" Body="&lt;p&gt;My guess is that this is a &lt;em&gt;very&lt;/em&gt; classic error in computer graphics implementation; you have to make sure that you don't accidentally use the new value of X when computing the new value of y!  For instance, this code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;x = x*cos(theta) - y*sin(theta);&#xA;&lt;br/&gt;&#xA;y = x*sin(theta) + y*cos(theta);&#xA;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;will actually perform the following operation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;x'  =x cos &amp;theta; - y sin &amp;theta;&#xA;&lt;br/&gt;&#xA;y' = x cos &amp;theta; sin &amp;theta; + y (cos &amp;theta; - sin&lt;sup&gt;2&lt;/sup&gt;&amp;theta;)&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and for small values of &amp;theta; (i.e., incremental rotation) will generally cause the point &amp;lt;x,y&amp;gt; to spiral in towards &amp;lt;0,0&amp;gt;.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-04T16:16:08.507" CommentCount="3" />
  <row Id="6026" PostTypeId="1" AcceptedAnswerId="6027" CreationDate="2010-10-04T16:16:14.843" Score="1" ViewCount="117" Body="&lt;p&gt;In the $d$ variable polynomial ring $R=k[x_{1},\cdots,x_{d}]$ show that&#xA;$0, x_{1}R, (x_{1},x_{2})R, \cdots , (x_{1},x_{2},...,x_{d})R$ is a strictly increasing sequence of prime ideals and there is no longer such chain.  how do i prove this claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, i am thinking that for this problem this result may be helpful: for a ring, $R$, to be Noetherian, one formulation dictates that any ascending chain of ideals in R terminates. But i just can't get going with it.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-04T16:21:33.497" LastActivityDate="2010-10-04T18:09:11.197" Title="Sequence of Prime ideals in a Polynomial Ring" Tags="&lt;abstract-algebra&gt;&lt;algebraic-geometry&gt;&lt;homework&gt;&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="6027" PostTypeId="2" ParentId="6026" CreationDate="2010-10-04T16:43:31.457" Score="4" ViewCount="0" Body="&lt;p&gt;A standard proof uses Noether normalization. For a superb introduction to (Krull) dimension of rings see Chapter 8, &quot;Introduction to dimension theory&quot; in Eisenbud's &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387942696&quot; rel=&quot;nofollow&quot;&gt;Commutative algebra, with a view toward algebraic geometry&lt;/a&gt;&quot;, esp. Section 8.2.1, and for the proof see Section 13.1.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-04T16:43:31.457" CommentCount="2" />
  <row Id="6028" PostTypeId="1" AcceptedAnswerId="6073" CreationDate="2010-10-04T17:15:35.760" Score="-1" ViewCount="190" Body="&lt;p&gt;I know how to count derivatives, but I actually don't understand what are those? What do they show? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia explanation: &lt;code&gt;Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity&lt;/code&gt;. I understand nothing what is written here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, if you can, explain me as simple as you can it. Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="2284" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-04T18:33:05.843" LastActivityDate="2010-10-05T01:48:17.147" Title="Simple way to understand what derivative is" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="6029" PostTypeId="2" ParentId="6008" CreationDate="2010-10-04T17:19:07.890" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/2763/what-is-the-expected-number-of-runs-of-same-color-in-a-standard-deck-of-cards&quot;&gt;This question&lt;/a&gt; is very similar and should give some pointers.&lt;/p&gt;&#xA;" OwnerUserId="1910" LastActivityDate="2010-10-04T17:19:07.890" />
  <row Id="6030" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T17:28:00.150" Score="2" ViewCount="0" Body="&lt;p&gt;If one thing, $A$, changes when you change something else, $B$, then the &lt;em&gt;derivative of $A$ with respect to $B$&lt;/em&gt; is the rate of change of $A$ when you make very small changes in $B$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-04T17:28:00.150" CommentCount="1" />
  <row Id="6031" PostTypeId="2" ParentId="6013" CreationDate="2010-10-04T17:32:40.470" Score="0" ViewCount="0" Body="&lt;p&gt;You may want to read about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Curry%E2%80%93Howard_correspondence&quot; rel=&quot;nofollow&quot;&gt;Curry-Howard isomorphism&lt;/a&gt; which, for example, tells us that types in typed lambda calculus (thus in programming languages like OCaml, SML, Haskell, among others) actually are propositions in some logic formalism, and that values of these types are proofs of the corresponding propositions. It is e.g used to develop proof assistants that actually also are functional programming languages (Agda, Coq, Epigram, ...). Regarding algebra, you may want to read &quot;Elements of Programming&quot;, by Alexander Stepanov. &lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-10-04T17:32:40.470" />
  <row Id="6032" PostTypeId="2" ParentId="6008" CreationDate="2010-10-04T17:38:12.360" Score="2" ViewCount="0" Body="&lt;p&gt;You can consider the random variable $Y$=number of alternations (from head to tail or viceversa) in the sequence of $N$ trials. It's clear that $Y$ follows a Binomial distribution, with values in $[0,N-1]$ and $p=1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(Y) = (N - 1) /2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Var(Y) = (N - 1) /4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation with $X$ takes slightly different forms depending on whether the first ocurrence was a head, and whether $N$ is even or odd.&#xA;This makes the problem a little cumbersone, but solvable nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a quick approximate  (presumably quite good with N moderately large) solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Y$ is odd, then: $X=(Y+1)/2$. We assume that this holds always, and then compute expectation and variance by linearity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(X) \approx (N-1)/4 + 1/2 \approx N/4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Var(X) \approx (N-1)/16  \approx N/16$&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-04T17:38:12.360" CommentCount="1" />
  <row Id="6033" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T17:45:10.163" Score="1" ViewCount="0" Body="&lt;p&gt;Graphically, when you have the x axis, and the y axis with the graph of a (derivable) function f, then the derivative (the &lt;strong&gt;number&lt;/strong&gt;) of f at x0 basically tells you how much the value of the function changes if you move to x0+1. However, this won't be very accurate, since the graph may be very different at x0+1. This is why we have the derivative of f (the &lt;strong&gt;function&lt;/strong&gt;) which gives the above number for any &quot;x0&quot; you give it. &#xA;So it gives the coefficient of the tangent line at each point. It's only about how your function grows for an infinitesimal &quot;move&quot; around some x0.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-10-04T17:45:10.163" />
  <row Id="6034" PostTypeId="2" ParentId="6026" CreationDate="2010-10-04T18:09:11.197" Score="1" ViewCount="0" Body="&lt;p&gt;Bill is right that the most standard proof uses Noether normalization.  Another proof exploiting the fact that $R$ is a Hilbert-Jacobson ring can be found in Section 8.2 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exposition here follows Kaplansky's &lt;em&gt;Commutative Rings&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-04T18:09:11.197" CommentCount="1" />
  <row Id="6035" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T18:12:37.163" Score="0" ViewCount="0" Body="&lt;p&gt;You can see this post as well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.math.pitt.edu/~sparling/061/23561/23561notes/node5.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.pitt.edu/~sparling/061/23561/23561notes/node5.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a geometric visualization please see &lt;a href=&quot;http://www.splung.com/content/sid/1/page/differentiation&quot; rel=&quot;nofollow&quot;&gt;http://www.splung.com/content/sid/1/page/differentiation&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-04T18:12:37.163" CommentCount="0" />
  <row Id="6036" PostTypeId="1" AcceptedAnswerId="6037" CreationDate="2010-10-04T18:18:16.423" Score="1" ViewCount="79" Body="&lt;p&gt;Let $S$ be a string of length $n$. Each character of $S$ has probability $p$ of being 'A' and probability $1-p=q$ of being 'B'. $R$ is the number of occurrences of the substring 'AB' in $S$. I'd like to determine $\mathbf{E}[R]$ and $\mathbf{var}[R]$ . I can determine the expectation via a recursive definition and total expectation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{E}[R] = pq\left(1+\mathbf{E}\left[R_{n-1}\right]\right)+\left(1-pq\right)\mathbf{E}\left[R_{n-1}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not sure how I can approach the problem for determining the variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried defining a Probability Mass Function but it looks like it will get messy very soon.&#xA;$\Pr(R=x) = \binom{n-x}{x}\cdot\left(pq\right)^{x} \cdot \text{probability of no other ABs}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that &quot;no other&quot; probability ends up being, for example for $x=1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{u}p^i q^{u-1} \cdot \sum_{i=0}^{v}p^i q^{u-1}$ summed over all $u+v+2=n$&#xA;but I think there might be a better way.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-10-04T18:33:45.897" Title="Moments on number of occurrences of substring" Tags="&lt;probability&gt;&lt;recursive&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6037" PostTypeId="2" ParentId="6036" CreationDate="2010-10-04T18:33:45.897" Score="4" ViewCount="0" Body="&lt;p&gt;I presume all the characters are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_1,\ldots,X_{n-1}$ be the random variables defined as follows:&#xA;$X_i=1$ if there is $AB$ starting at the $i$-th position, otherwise $X_i=0$.&#xA;Then $R=X_1+\cdots+X_{n-1}$. By linearity of expectation $E(R)=\sum_i E(X_i)$&#xA;and $E(R^2)=\sum_{i,j} E(X_i X_j)$. Now $E(X_i)=pq$ so $E(R)=(n-1)pq$.&#xA;IF $i=j$ then $X_i X_j=X_i^2=X_i$; if $|i-j|=1$ then $X_i X_j=0$ and&#xA;if $|i-j|\ge2$ then $X_i$ and $X_j$ are independent so that $E(X_i X_j)=E(X_i)^2&#xA;=p^2 q^2$. Now it's just book-keeping to find $E(R^2)$ and so also $Var(R)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-04T18:33:45.897" CommentCount="2" />
  <row Id="6038" PostTypeId="1" CreationDate="2010-10-04T18:52:45.167" Score="3" ViewCount="106" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_billiards&quot; rel=&quot;nofollow&quot;&gt;this article in Wikipedia&lt;/a&gt;: A billiard is a dynamical system in which a particle alternates between motion in a straight line and specular reflections from a boundary. When the particle hits the boundary it reflects from it without loss of speed. Billiard dynamical systems are Hamiltonian idealizations of the game of billiards, but where the region contained by the boundary can have shapes other than rectangular and even be multidimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is motivated by a videogame I've been playing lately, which can be seen at &lt;a href=&quot;http://www.youtube.com/watch?v=LLLmfwxNJYU&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=LLLmfwxNJYU&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the &quot;physics&quot; of the game involves several billiards in a polygon, and the player has to slash off pieces of the polygon while avoiding the billiards.  Also, the removed piece has to be void of billiards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been assuming that the distribution of billiards is in the long run uniform, in some hand waving sense.  That's assuming that initial distributions and velocities are random.  Is that true, or can the polygon be shaped in various ways to make that distribution non-uniform?  In other words, are certain regions of a polygon are more likely to be void than other regions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I believe that a term like &quot;ergodic&quot; applies to this, but I'm not confident using it). &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2010-10-04T18:52:45.167" Title="Polygonal billiards and uniform distribution" Tags="&lt;random&gt;&lt;dynamical-systems&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="6039" PostTypeId="1" AcceptedAnswerId="6205" CreationDate="2010-10-04T19:34:47.573" Score="4" ViewCount="578" Body="&lt;p&gt;Let $p(x)$ be a polynomial with integer coefficients, which is not constant. Then is this condition possible: $$p(x)=3^{x}$$ whenever $x \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: &lt;a href=&quot;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-05T01:52:21.237" LastActivityDate="2010-10-06T15:47:50.797" Title="Polynomial satisfying $p(x)=3^{x}$ for $ x \in \mathbb{N}$" Tags="&lt;calculus&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;diophantine-equations&gt;" AnswerCount="10" CommentCount="13" />
  <row Id="6040" PostTypeId="1" AcceptedAnswerId="6048" CreationDate="2010-10-04T19:36:27.490" Score="0" ViewCount="62" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The strength of glue is known to follow a normal distribution with a mean of 59 and a standard deviation of 3.4.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Find the mean strength needed such that the probability  that the glue is weaker than 55 is less than 5%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To solve this, I just guessed and checked in Excel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=NORMDIST(55, 50.1, 3.4, 1)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=NORMDIST(55, 50.2, 3.4, 1)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;=NORMDIST(55, 50.3, 3.4, 1)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;until the cell was close to .05. The answer I got was  60.6, which appears to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That probably isn't the &lt;em&gt;proper&lt;/em&gt; way to solve the problem. I searched my book for a problem like it but didn't find any examples. I did find this &quot;Percentiles of an Arbitrary Normal Distribution&quot; formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(100p)th percentile for normal(mu, signma) = mu + [(100p)th for standard normal] * sigma&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to apply it (or maybe that formula doesn't help at all). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the smart way to solve the problem?&lt;/p&gt;&#xA;" OwnerUserId="1977" LastActivityDate="2010-10-05T06:02:25.053" Title="Find mean of normal distrubution such that it fits a condition" Tags="&lt;probability&gt;&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="6041" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T19:45:24.350" Score="12" ViewCount="0" Body="&lt;p&gt;No, a polynomial can't grow exponentially as $x\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T19:45:24.350" CommentCount="3" />
  <row Id="6042" PostTypeId="2" ParentId="6028" CreationDate="2010-10-04T19:52:51.887" Score="0" ViewCount="0" Body="&lt;p&gt;You can find some nice introductory lectures on derivatives in various &lt;a href=&quot;http://www.khanacademy.org/search?page_search_query=derivative&quot; rel=&quot;nofollow&quot;&gt;videos  from Khan's Academy&lt;/a&gt;. The site has a free online collection of a couple thousand videos on mathematics, science, history, and economics. It's worth exploring when you're looking for introductory lectures on such topics.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-05T00:31:12.803" LastActivityDate="2010-10-05T00:31:12.803" />
  <row Id="6043" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T19:59:23.297" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 3^x$ for $x \in \mathbb N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $a_n (x+1)^n + a_{n-1} (x+1)^{n-1} + \cdots + a_0 = 3 a_n x^n + 3 a_{n-1} x^{n-1} + \cdots + 3 a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the coefficient of $x^n$ on the LHS is $a_n$ and on the RHS it is $3 a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contradicts the assumption.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-04T19:59:23.297" CommentCount="4" />
  <row Id="6044" PostTypeId="1" AcceptedAnswerId="6047" CreationDate="2010-10-04T20:06:51.853" Score="3" ViewCount="268" Body="&lt;p&gt;As a part of my studies in ring theory, I've encountered the concept of an idempotent element, i.e., an element $e$ such that $e^2=e$.&#xA;Are there some interesting examples of rings with idempotent elements?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-10-05T07:35:22.180" Title="Examples of rings with idempotent elements" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;&lt;examples&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="1" />
  <row Id="6045" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:09:31.880" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose that $p(x)$ exists and has degree $n$.  $\lim_{x\to\infty}\frac{p(x)}{x^{n+1}}=0$ and $\lim_{x\to\infty}\frac{3^x}{x^{n+1}}\to\infty$, but $\frac{p(x)}{x^{n+1}}=\frac{3^x}{x^{n+1}}$ for all $x\in\mathbb{N}$, so it is impossible for the first limit to be finite and the second limit to be unbounded.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-04T20:09:31.880" />
  <row Id="6047" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T20:18:06.507" Score="6" ViewCount="0" Body="&lt;p&gt;As you realize, all rings have the idempotents $0$ and $1$, so the&#xA;question is whether they have any others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a commutative ring has a non-trivial idempotent, then it is&#xA;isomorphic to a product of two non-trivial rings. The same&#xA;is true for a non-commutative ring, as long as the idempotent&#xA;lies in its centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is a positive integer which is not a prime power then&#xA;$\mathbb{Z}/n\mathbb{Z}$ has nontrivial idempotents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrix rings tend to have lots of idempotents, but not usually in their centres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A group algebra $KG$ for a finite group $G$ over a characteristic zero&#xA;field $K$ has the&#xA;central idempotent $|G|^{-1}\sum_{g\in G}g$ and usually others.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-04T20:18:06.507" CommentCount="0" />
  <row Id="6048" PostTypeId="2" ParentId="6040" CreationDate="2010-10-04T20:18:28.393" Score="1" ViewCount="0" Body="&lt;p&gt;Ok. I will show you an example.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x$ is distributed normally with mean $a=34$ and standard deviation $\sigma=12$. Find $x_0$ such that $x&gt;x_0$ with probability $0.95$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose $x_0=a-\delta$.  We need to find $\delta$, such that $P(x&gt;a-\delta)=0.95$ or, equivalently, $P(x-a\leq-\delta)=0.05$. First note, that normal distribution is symmetric. Therefore $P(x-a\geq\delta)=0.05$ and $P(|x-a|\geq\delta)=0.1$ or $P(|x-a|&amp;lt;\delta)=0.9$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution#Standard_deviation_and_confidence_intervals&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you can find, that $\delta=n\sigma$, where $n=1.644853626951$ (just search for this number on &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution#Standard_deviation_and_confidence_intervals&quot; rel=&quot;nofollow&quot;&gt;the page&lt;/a&gt; to see the table, I've used to find it). Therefore the answer is&#xA;$$x_0=a-\delta=a-n\sigma=14.3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Your task is actually the same. The difference is that in my task we know $a$ and we are asked to find $a-\delta$ and in your case we know $a-\delta$ and we are asked to find $a$. That's simple. Just write $a= (a-\delta)+\delta$. I don't know Excel, but if your use of it is right, you can find the answer of your task by &lt;code&gt;= NORMINV(0.95, 55, 12)&lt;/code&gt;. Actually, the answers in the following tasks are th same (here $a,\sigma,p$ are given; you can replace $a$, $\sigma$ by any positive numbers and $p$ by any number in $[0,1]$):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$x$ is distributed normally with mean $a$ and standard deviation $\sigma$. Find $b$ such that $P(x\leq b)=p$.  &lt;/li&gt;&#xA;  &lt;li&gt;$y$ is distributed normally with standard deviation $\sigma$. It is known, that $P(y\geq a)=p$. Find $b$ --- the mean of the distribution of $y$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's because if $x$ is distributed normally with parameters $a$, $\sigma$, then distribution of $z=x-a$ depends only on $\sigma$ (it is independent of $a$) and symmetric (for each $\delta$ we have $p(z&amp;lt;\delta)=p(z&gt;-\delta)$). You can see, that tasks are the same by denoting $x-a$ in the first task by $z$ (and replacing $x$ by $z+a$) and denoting $y-b$ in the second by $z$ (and replacing $y$ by $z+b$). Then both tasks have the following form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$z$ is distributed normally with mean 0 and standard deviation $\sigma$. Find $b$ such that $P(z\geq b-a)=p$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-05T06:02:25.053" LastActivityDate="2010-10-05T06:02:25.053" CommentCount="3" />
  <row Id="6049" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:23:02.060" Score="3" ViewCount="0" Body="&lt;p&gt;If so then then the values of $\rm\: q(x) = p(x^2)\:$ are all powers of 3, contradicting the fact that infinitely many primes must occur among the values of a nonconstant $\rm\: p(x) \in \mathbb Z[x]\:$, by way of a simple Euclid-like proof. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-05T00:16:24.960" LastActivityDate="2010-10-05T00:16:24.960" />
  <row Id="6050" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:25:00.047" Score="2" ViewCount="0" Body="&lt;p&gt;Here's another argument (at the other end of the real line!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a polynomial would have to be non-constant, which means that it tends to $\infty$ or $-\infty$ as $x\to-\infty$. But this is impossible, since $3^x \to 0$ as $x\to-\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-04T20:25:00.047" CommentCount="1" />
  <row Id="6051" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:25:57.300" Score="10" ViewCount="0" Body="&lt;p&gt;Yet another proof: if $p$ is a nonzero polynomial with real coefficients&#xA;then&#xA;$$\lim_{n\to\infty}\frac{p(n+1)}{p(n)}=1$$&#xA;which certainly doesn't hold for $p(x)=3^x$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-04T20:25:57.300" />
  <row Id="6052" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T20:46:40.070" Score="5" ViewCount="0" Body="&lt;p&gt;Nontrivial idempotents are intimately connected to direct product decompositions. Generally, any idempotent  $\rm e^2 = e$ yields a decomposition $\rm\  R  =  e R + (1-e) R\ $  known as the &lt;a href=&quot;http://myyn.org/m/article/pierce-decomposition/&quot; rel=&quot;nofollow&quot;&gt;Pierce Decomposition&lt;/a&gt;, and vice versa. This extends to any finite set of idempotents with sum 1 which are mutually orthogonal: $\rm\ e_j\: e_k = 0\ $ if $\rm\ j \ne k\ $. For example, the Chinese Remainder Theorem (&lt;strong&gt;CRT&lt;/strong&gt;) for rings has this form.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-04T20:57:43.867" LastActivityDate="2010-10-04T20:57:43.867" />
  <row Id="6053" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T20:54:01.407" Score="10" ViewCount="0" Body="&lt;p&gt;A polynomial of degree $n$ has the property that $p(x+1) - p(x)$ is a polynomial of degree $n-1$.  After taking $n$ finite differences you reach a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-04T20:54:01.407" CommentCount="2" />
  <row Id="6054" PostTypeId="2" ParentId="6039" CreationDate="2010-10-04T21:00:41.690" Score="7" ViewCount="0" Body="&lt;p&gt;The answer depends on the ring (which the question did not specify) where $x$ and $p(x)$ take their values. Polynomials with integer coefficients, and non-negative powers of integers such as $3^n$, can be evaluated in any ring, so the question makes sense in this generality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a ring where 2=0 (such as the integers modulo 2), $3^n = 1^n$, and any polynomial with $p(0)=p(1)=1$ is an example.  It should be an interesting problem to determine all rings where there is such a $p(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED: Let's consider the situation over an arbitrary commutative ring $R$ with unit.  $Q(x) = p(x+1)- 3p(x)$ is zero (as a function) for all $x$ in the image of $\mathbb{Z}$ in $R$.  If $R$ is of characteristic different from $2$, $\deg Q = \deg P$ (as polynomials).  In characteristic 0 having zeros at all integers is enough to force $Q = 0$ as a polynomial, and there are no examples.  In finite characteristic $n &gt; 1$, we also have $p(n) = 3^n = p(0) = 1$ so that $n | (3^n - 1)$. If $n$ is prime then $n | (3^n - 3)$ so that $n | 2$ and thus $n=2$.  If $n$ is not prime, and $q$ is a prime dividing $n$, then an example in $R$ is an example in $R/q$ which has prime characteristic $q$ and therefore $q=2$.  So the only remaining possibility is a ring where $2^k = 0$, for $k &gt; 1$.  For such rings there are examples because in the binomial expansion of $(1+2)^n = 1 + 2n + 2n(n-1) + 4n(n-1)(n-2)/3 + 2n(n-1)(n-2)(n-3)/3 + \dots$ all terms are congruent modulo $2^k$ to polynomials with integer coefficients, and all terms of degree higher than $2^k - 1$ vanish on the integers modulo $2^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED-2:  one can see that $2^k=0$ more directly by writing $p(n+1)-p(n) = 2p(n)$ (for images of integer $n$ in our ring).  This means that $\Delta^k p(n) = 2^k p(n)$ and for $k &gt; deg(P)$ this will be zero, then set $n=0$ to get $0 = 2^k p(0)= 2^k$.  [Note that this argument applies to $p(n)$ for integer $n$ only, not $p(x)$ as an abstract polynomial or a polynomial evaluated at non-integer values of $x$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: there are polynomials $p(x)$ such that $p(n)=3^n$ for all $n \geq 0$ (with the polynomial considered as a function on the commutative ring-with-unit $R$) if and only if in this ring, $2^k = 0$ for some positive integer $k$.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-05T19:01:11.673" LastActivityDate="2010-10-05T19:01:11.673" CommentCount="0" />
  <row Id="6055" PostTypeId="1" AcceptedAnswerId="6076" CreationDate="2010-10-04T21:03:11.163" Score="2" ViewCount="115" Body="&lt;p&gt;I'm learning category theory by self-study. I have a couple of texts, and they both talk about how we ought to try not to think so much about the equality between objects in categories. Rather, the important relation is isomorphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they go on to talk about &quot;Cat&quot;, the category of (small) categories. One of them even segues into it by considering &quot;categories themselves as structured objects. The `morphisms' between them that preserve their structure are called functors.&quot; But then it becomes clear that equality of functors depends on equality of objects, which isn't part of the structure of categories! (Or at least, a part we're not supposed to think about.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are natural transformations, which inherit equality from the underlying equality of morphisms in the codomain category. They induce a natural isomorphism on functors. Treating functors, modulo natural isomorphism, as morphisms in a category of categories, seems to me to be the obvious &quot;Cat&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, not only is this not &quot;Cat&quot;, but I couldn't find it among the loads of exotic examples of categories I'm given. Apparently it wasn't even deemed worthy of mention. This is a bit disappointing. Doesn't it at least have a standard name and snappy acronym? Can someone point me somewhere where I can learn about its basic properties somewhere? Or is there something defective about it that I've missed? &lt;/p&gt;&#xA;" OwnerUserId="2311" LastActivityDate="2010-10-05T03:19:58.107" Title="&quot;Cat&quot; modulo natural isomorphism?" Tags="&lt;category-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="6056" PostTypeId="1" AcceptedAnswerId="6069" CreationDate="2010-10-04T21:33:24.197" Score="1" ViewCount="141" Body="&lt;p&gt;Are there any good places for me to sell off my mathematics books online especially Springer and Dover books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I thought perhaps this was off topic, but then I thought everybody in math probably has the problem of having bought pricey books that they would prefer to get some money back for, at one point or another.)&lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1004" LastEditorDisplayName="" LastEditDate="2010-10-04T21:39:12.933" LastActivityDate="2010-10-05T00:03:34.907" Title="Reselling Books" Tags="&lt;books&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6057" PostTypeId="1" CreationDate="2010-10-04T22:02:40.107" Score="2" ViewCount="210" Body="&lt;p&gt;I have an image encoded in the form of a list of points, like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;335.0 2743.0 &#xA;335.0 2754.0 &#xA;325.0 2754.0 &#xA;304.0 2743.0 &#xA;304.0 2733.0 &#xA;283.0 2723.0 &#xA;273.0 2702.0 &#xA;273.0 2691.0 &#xA;273.0 2681.0 &#xA;273.0 2670.0 &#xA;283.0 2670.0 &#xA;294.0 2670.0 &#xA;304.0 2670.0 &#xA;325.0 2670.0 &#xA;335.0 2681.0 &#xA;346.0 2702.0 &#xA;346.0 2712.0 &#xA;356.0 2723.0 &#xA;346.0 2733.0 &#xA;346.0 2743.0 &#xA;346.0 2733.0 &#xA;346.0 2723.0 &#xA;356.0 2702.0 &#xA;356.0 2670.0 &#xA;356.0 2660.0 &#xA;367.0 2660.0 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is a line drawn between each point to make the image - if you sketch the points above (I'd advise doing it programatically) you'll see a lower case 'a'. This is part of a project regarding online character recognition. I am currently investigating the effects of several pre-processing techniques that can be applied to such a recognition system, and one technique that many of the papers that I have read apply is to 'smooth' the image, usually via a Gaussian blur. Unfortunately, they all proclaim it as &quot;easy&quot;, and therefore they neglect to mention &lt;em&gt;how&lt;/em&gt; one goes about doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking for quite some time now, but I find myself still unable to understand how I would take the idea and apply it to an image made up of a set of points like the one above. From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_blur&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; article I have the function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(X) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have no idea how I actually apply that to my list of points to get a smoothed image. All of the examples online are both incredibly complicated and also for pixel-images, where every pixel in your image matters. Here, I only actually have the 20 or so points. So, what I am looking for is some explanation and advice regarding how I would apply a Gaussian blur in my way of representing the image - what do I apply $G(X)$ to (each co-ordinate?), how I calculate $\sigma^2$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The disclaimer&lt;/strong&gt;: As stated, this is part of an assignment. However, this is not breaking any rules or 'cheating' in any form. The assignment is to investigate many features of online character recognition, not to magically know how to do one of the possible techniques for pre-processing the characters. Should I gain any knowledge of how to do the smoothing from this site, the site (and the individual, should they wish) will of course be credited in full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thank you for any help. I hope this question is placed correctly (I felt it was more mathematical than program-based, since it's more about the concept of Gaussian blurs rather than the implementation - I'm fairly sure I'll be able to handle that part) and that it is clear. If it is not, please just ask and I will clarify any points I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I also apologise for the tag. I'm fairly sure it's inaccurate. But there's no 'Gaussian' tag. Tis odd.)&lt;/p&gt;&#xA;" OwnerUserId="693" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-10-05T08:57:34.567" LastActivityDate="2010-10-05T08:57:34.567" Title="How do I apply a Gaussian Blur (low-pass filter) to an image made up from a set of points?" Tags="&lt;signal-processing&gt;&lt;image-processing&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="6058" PostTypeId="2" ParentId="6056" CreationDate="2010-10-04T22:24:49.457" Score="0" ViewCount="0" Body="&lt;p&gt;Amazon will buy back a lot of textbooks at pretty fair rates. Check &lt;a href=&quot;http://www.amazon.com/Textbooks-Trade-In/b/ref=bhp_9p2_A_txbq410_05?ie=UTF8&amp;amp;node=2205237011&amp;amp;pf_rd_m=ATVPDKIKX0DER&amp;amp;pf_rd_s=center-6&amp;amp;pf_rd_r=0RH8PF985VSEMWFKG1SV&amp;amp;pf_rd_t=101&amp;amp;pf_rd_p=1275995182&amp;amp;pf_rd_i=283155&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; out.&lt;/p&gt;&#xA;" OwnerUserId="2302" LastActivityDate="2010-10-04T22:24:49.457" />
  <row Id="6059" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:04:37.253" Score="5" ViewCount="0" Body="&lt;p&gt;Let $X$ be a disconnected space. Then the ring $C(X)$ of complex continuous functions has an idempotent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, this is a motivating example, because if $R$ has a nontrivial idempotent, then $Spec(R)$ is disconnected. Since elements of $R$ can be viewed (loosely) as functions, we can construct an element of $R$ which is &quot;one&quot; on one half of $Spec(R)$ and &quot;zero&quot; on the other. (To make this precise, one thinks in terms of the sheaf of regular functions on the scheme $Spec(R)$.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-04T23:04:37.253" />
  <row Id="6060" PostTypeId="1" AcceptedAnswerId="6115" CreationDate="2010-10-04T23:11:29.707" Score="5" ViewCount="142" Body="&lt;p&gt;Suppose there are a finite number of disjoint unit-radii disks in the &#xA;plane, each spinning clockwise or counterclockwise at the same&#xA;angular velocity.&#xA;The plane is filled with a thin fluid layer,&#xA;and the disks can be viewed as spinning fan blades&#xA;determining vectors of fluid motion tangent to the disks.&#xA;Is the resulting flow and vector field throughout the plane known?&#xA;My initial intuition is that there should be something like a&#xA;Voronoi diagram demarcating boundaries of regions of influence.&#xA;But in exploring a bit I find it may even be nontrivial to&#xA;determine the flow between just two counter-rotating vortices.&#xA;For example, the following image was&#xA;computed by &lt;a href=&quot;http://www.bugman123.com/&quot; rel=&quot;nofollow&quot;&gt;Paul Nylander&lt;/a&gt;&#xA;based on a paper by&#xA;O.S. Kerr and J.W. Dold,&#xA;&quot;Periodic Steady Vortices in a Stagnation Point Flow,&quot;&#xA;&lt;a href=&quot;http://www.city.ac.uk/sems/mathematics/research/fluids/vortices.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;J. Fluid Mech.&lt;/em&gt;, 276, 307-325 (1994)&lt;/a&gt;.&#xA;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/QK72h.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am quite unschooled in this topic, pointers to&#xA;relevant literature might suffice.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit1&lt;/b&gt;. I've now asked a revised version of &lt;a href=&quot;http://mathoverflow.net/questions/41132/vortex-voronoi-diagram&quot; rel=&quot;nofollow&quot;&gt;this question on Math Overflow&lt;/a&gt;,&#xA;incorporating the clarifying suggestions of Rahul.  I might hit a fluid dynamics expert there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit2&lt;/b&gt;. Thanks to Rahul and David Bar Moshe here, and Willie Wong and Bob Terrell&#xA;on MO, I have a much broader understanding of the problem, and could likely compute a numerical &#xA;solution if needed.  I appreciate the help!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-06T01:04:05.323" LastActivityDate="2010-10-06T04:08:22.573" Title="Vortex Voronoi diagram?" Tags="&lt;geometry&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6061" PostTypeId="2" ParentId="6055" CreationDate="2010-10-04T23:26:52.950" Score="0" ViewCount="0" Body="&lt;p&gt;I'm not sure to understand what you mean, but maybe the operation for which you are asking for a name is &lt;em&gt;skeleton&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting Mac Lane's &quot;Categories for the working mathematician&quot; (chapter IV, section 4), &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In any category ${\cal C}$ [not just $\mathbf{Cat}$], a &lt;em&gt;skeleton&lt;/em&gt; of ${\cal C}$ is any full subcategory ${\cal A}$ such that any object of ${\cal C}$ is isomorphic to exactly one object of ${\cal A}$. Then ${\cal A}$ is equivalent to ${\cal C}$ and the inclusion $K: {\cal A} \longrightarrow {\cal C}$ is an equivalence of categories. (...) &lt;em&gt;A category is equivalent to any of its skeletons.&lt;/em&gt; For instance, the category of all finite sets has as skeleton the full subcategory with objects all finite ordinal numbers $0, 1, 2, \dots , n , \dots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, Mac Lane adds&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A category is called &lt;em&gt;skeletal&lt;/em&gt; when any two isomorphic objects are identical; i.e., when the category is its own skeleton.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance, the category with objects all finite ordinal numbers is skeletal: there you can forget about the picky distinction between &quot;equal&quot; and &quot;isomorphic&quot; objects. In real life, there is another circumstance in which isomorphic objects are treated as being &quot;equal&quot; for all purposes, namely, when you have a &lt;em&gt;canonical&lt;/em&gt; isomorphism, for instance one which is derived from a universal property such as a the one verified by coproducts.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-04T23:26:52.950" CommentCount="3" />
  <row Id="6062" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:30:02.463" Score="4" ViewCount="0" Body="&lt;p&gt;The endomorphism ring $\mathrm{End}(V)$ of a given vector space $V$ is riddled with idempotents. For instance, all projections $\pi=\pi_{L_1,L_2}$ of $V$ onto $L_1$ along $L_2$ where $V=L_1 \oplus L_2$ is a direct decompositon of $V$ are idempotents ($\pi(x)=\pi(x_1+x_2)=x_1$ where $x \in V$ and $x_k \in L_k,$ $k=1,2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are many hard/nice questions concering idempotents in group rings.  &lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2010-10-04T23:30:02.463" />
  <row Id="6063" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:37:04.083" Score="3" ViewCount="0" Body="&lt;p&gt;Actually there are rings in which every element is idempotent. They are called boolean rings. Stone duality tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X \mapsto C_0(X,\mathbb{F}_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a duality between locally compact totally disconnected hausdorff spaces and boolean rings. Compactness here corresponds to unitality. For example, every finite boolean ring is isomorphic to $(\mathbb{F}_2)^n$ for some $n$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-04T23:37:04.083" />
  <row Id="6064" PostTypeId="2" ParentId="6055" CreationDate="2010-10-04T23:43:09.907" Score="0" ViewCount="0" Body="&lt;p&gt;If $C$ is a $2$-category, every hom-set is a category and isomorphy yields a congruence relation on them. The quotient $\tilde{C}$ is then a $1$-category. It's objects are the objects of $C$, but the morphisms are the $1$-morphisms of $C$ modulo $2$-isomorphy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to compare $C$ with $2$-categories, it does not make any sense to consider $\tilde{C}$. However, this construction becomes useful when you want to compare with $1$-categories, that is, usual categories. For example, the functor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Ring \to \tilde{Cat}, R \mapsto Mod(R), \phi \mapsto \phi^*$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\phi^* \psi^* \cong (\phi \psi)^*$ holds in $Cat$, thus these morphisms are equal in $\tilde{Cat}$ and we get, indeed, a functor.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-04T23:43:09.907" CommentCount="2" />
  <row Id="6065" PostTypeId="2" ParentId="6057" CreationDate="2010-10-04T23:53:34.983" Score="2" ViewCount="0" Body="&lt;p&gt;The operation is better described in the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_filter&quot; rel=&quot;nofollow&quot;&gt;Gaussian Filter&lt;/a&gt;.  First you need to convert your list of points to an image, in essence filling in the line between the points.  If you use 8 bits, those cells will have darkness 255 while all others will be 0.  Or perhaps you give the lines a thickness and calculate how much of each pixel is covered.  Now the image is a large array of numbers.  The value of sigma is yours to choose.  From sigma you choose will calculate a small array that represents G(X).  If sigma is one pixel, you can represent G(X) only over +-3 pixels, so it is 7x7.  To calculate the blurred darkness of a pixel, you take the original darknesses of the 7x7 neighborhood of the pixel, multiply them by the G(X) array and add.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="73" LastEditorDisplayName="" LastEditDate="2010-10-05T08:54:56.597" LastActivityDate="2010-10-05T08:54:56.597" CommentCount="1" />
  <row Id="6066" PostTypeId="2" ParentId="6044" CreationDate="2010-10-04T23:54:44.730" Score="1" ViewCount="0" Body="&lt;p&gt;You can compute idempotents in any square matrix ring $M_n(K)$, $K$ a field, as follows:  $A \in M_n(K)$ is idempotent if and only if $A^2 = A$. But this means that the polynomial $p(t) = t(t-1)$ annihilates $A$. So the &lt;a href=&quot;http://en.wikipedia.org/wiki/Minimal_polynomial_%28linear_algebra%29&quot; rel=&quot;nofollow&quot;&gt;minimal polynomial&lt;/a&gt; $m_A(t)$ of $A$ must divide $p(t)$. This leaves only four possibilities for $m_A(t)$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$m_A(t) = 1$, in which case the unit matrix equals the zero matrix $I=0$ and $K = 0$.&lt;/li&gt;&#xA;&lt;li&gt;$m_A(t) = t$, in which case $A = 0$.&lt;/li&gt;&#xA;&lt;li&gt;$m_A(t) = t-1$, in which case $A=I$.&lt;/li&gt;&#xA;&lt;li&gt;$m_A(t) = t(t-1)$ is the only non trivial one. It means that your matrix $A$ has only two eigenvalues: $0$ and $1$. For instance, &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$&#xA;A=&#xA;\begin{pmatrix}&#xA;0 &amp;amp; 0 \\&#xA;0 &amp;amp; 1&#xA;\end{pmatrix} \ .&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all of prime factors in the minimal polynomial have multiplicity $1$, all diagonalize, so it's fairly easy to get a picture of them: all are of the form $A = S^{-1}D S$, with $D$ a diagonal matrix with just $0$ and $1$ on the main diagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrices of Olod's non-trivial projections belong to this case.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-05T07:35:22.180" LastActivityDate="2010-10-05T07:35:22.180" />
  <row Id="6067" PostTypeId="2" ParentId="6056" CreationDate="2010-10-04T23:54:44.730" Score="2" ViewCount="0" Body="&lt;p&gt;I find it's pretty easy to sell books on e-bay.  Local used bookstores rarely go for good math books, unfortunately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to try the big technical book resellers -- like Powell's Books, they're based in Portland Oregon.  Most on-line searches for books (like Google Books) will search Powell's stock.  So that might be one of the best options. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-04T23:54:44.730" />
  <row Id="6068" PostTypeId="1" AcceptedAnswerId="6074" CreationDate="2010-10-04T23:56:15.170" Score="2" ViewCount="78" Body="&lt;p&gt;I know that a function of bounded variation on a closed interval can be written as a difference of increasing functions. The same is true is the function is of bounded variation on all of R, but I can't come up with a proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-05T02:07:06.240" Title="Bounded variation: Jordan's theorem on all of R?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6069" PostTypeId="2" ParentId="6056" CreationDate="2010-10-05T00:03:34.907" Score="2" ViewCount="0" Body="&lt;p&gt;You can also list your books on amazon to sell rather than have amazon buy them from you&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-05T00:03:34.907" />
  <row Id="6070" PostTypeId="2" ParentId="6039" CreationDate="2010-10-05T00:16:36.270" Score="7" ViewCount="0" Body="&lt;p&gt;$\rm\ p(x) = 3^x\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(2x) = p(x)^2\:$  on $\rm\:\mathbb N \ \Rightarrow\ p(2x) = p(x)^2\:$ contra degree comparison.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-05T00:16:36.270" CommentCount="4" />
  <row Id="6071" PostTypeId="1" AcceptedAnswerId="6072" CreationDate="2010-10-05T00:44:22.240" Score="0" ViewCount="92" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A school play ran for two nights, with&#xA;  audiences totaling 1390 adults and&#xA;  students. They paid 4285 dollars for&#xA;  admission. One adult ticket cost 4 dollars&#xA;  and one student ticket cost 2.50 dollars. The&#xA;  ratio of adults to students was 3:5&#xA;  for the first night and 2:3 on the&#xA;  second night. How many students&#xA;  attended each night?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have determined that there was 850 students and 540 adults on &lt;em&gt;both&lt;/em&gt; nights, but for the life of me couldn't figure out how to incoporate those ratios into finding how much students were &lt;em&gt;per day&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is marked as a homework question so please don't give me the full answer but how to go about it.&#xA;I also have the answers if that is any help: for the first night, there were 400 students on the first night, and 450 on the second.&lt;/p&gt;&#xA;" OwnerUserId="2315" LastEditorUserId="2315" LastEditorDisplayName="" LastEditDate="2010-10-05T01:12:37.087" LastActivityDate="2010-10-05T01:19:12.573" Title="Linear Algebra Word Question" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="6072" PostTypeId="2" ParentId="6071" CreationDate="2010-10-05T01:19:12.573" Score="0" ViewCount="0" Body="&lt;p&gt;I have verified that the adult/student totals you got is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a total of $x$ people (i.e. including students and adults) attended the first day and total $y$ people attended the second day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the total number of students in terms of $x$ and $y$?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-05T01:19:12.573" CommentCount="24" />
  <row Id="6073" PostTypeId="2" ParentId="6028" CreationDate="2010-10-05T01:48:17.147" Score="1" ViewCount="0" Body="&lt;p&gt;I find the visual way of thinking about it to be the easiest: if you look at the graph of $f$ and zoom in to the point $(x,f(x))$, the graph will eventually start looking very much like a line.  That line is the &quot;tangent line&quot; to $f$ at $x$, and its slope is the derivative of $f$ at $x$.  (Some functions won't ever start looking like a line, no matter how far you zoom in.  One example is $f(x)=\left\vert x\right\vert$, at $(0,0)$.  We say that this function isn't differentiable there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal definition of the derivative, as&#xA;$$ f^\prime(x)=\lim_{a\rightarrow x}\frac{f(x)-f(a)}{x-a},$$&#xA;is really just another, more mathematical, way to describe &quot;zooming in&quot; and the construction of a tangent line.  If you think about it, the expression inside the limit is just the slope formula for a line going through $(x,f(x))$ and $(a,f(a))$.  This line is called a &quot;secant line.&quot;  If we let $a=x$, then we only have one point and so we don't have a unique line anymore.  But if we instead ensure that $a\ne x$ but that $a$ gets closer and closer to $x$, the secant lines approach the tangent line that we saw above.  This is just the same &quot;zooming in&quot; I was talking about above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're less of a visual person, it's often helpful to think of a physical quantity, like velocity.  Imagine driving a car or riding a bike in a straight line.  At any instant, you have a pretty good idea of how fast you're going &quot;right now,&quot; even if your speed is in the middle of changing.  Ryan Budney mentioned the example of a car with a speedometer above.  The speedometer can tell you your speed at any specific time.  This is just the derivative of your position: if you let the line be the $y$-axis and time be the $x$-axis, and graph your journey, the slope of a tangent line at a point will be exactly the speedometer reading at the that point.  On the other hand, you can also measure how much time it takes for you to get from $a$ to $b$: this is giving you the slope of a secant line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;i&gt;instantaneous&lt;/i&gt; velocity = slope of tangent line or derivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while &lt;i&gt;average&lt;/i&gt; velocity = slope of secant line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all derivatives &quot;with respect to time,&quot; but you can easily take the derivative with respect to other things, as long as you have a function relating them.  I'm not sure what your science background is, but this is the kind of thing that pops up often in school science experiments: the rate of change of the volume of a gas with respect to pressure, etc. &lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-05T01:48:17.147" />
  <row Id="6074" PostTypeId="2" ParentId="6068" CreationDate="2010-10-05T02:07:06.240" Score="1" ViewCount="0" Body="&lt;p&gt;You can use the positive and negative variation functions as usual.  The only difference is that you don't have a left endpoint to use.  Assuming BV on all of $\mathbb{R}$, a finite limit exists for each of these functions as you approach $-\infty$, and you could use this in place of the left endpoint.  More generally, it is enough to assume that the function is BV on all bounded intervals, but for this you will want to fix a finite base point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f$ is BV on all bounded intervals.  We want to find nondecreasing functions $g$ and $h$ such that $f=g-h$.  For nonnegative $x$, define $g(x)$ to be $f(0)$ plus the positive variation of $f$ on $[0,x]$, and define $h(x)$ to be the negative variation of $f$ on $[0,x]$.  For negative $x$, define $g(x)$ to be $f(0)$ minus the positive variation of $f$ on $[x,0]$, and define $h(x)$ to be minus the negative variation of $f$ on $[x,0]$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-05T02:07:06.240" CommentCount="1" />
  <row Id="6075" PostTypeId="1" CreationDate="2010-10-05T03:02:30.793" Score="1" ViewCount="60" Body="&lt;p&gt;I have the following situation: let $m,n$ be integers such that $m|n$ and let $\zeta_m$, $\zeta_n$ denote  primitive $m$ and $n$th roots of unity. Then we have the inclusion of  fields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{Q}\subset \mathbb{Q}(\zeta_m) \subset \mathbb{Q}(\zeta_n)$$&#xA;Now suppose we also have primes (where $(p,n)=1$)&#xA;$$(p)\subset \mathbb{Z}$$&#xA; and then &#xA;$$\mathfrak{p}\subset \mathbb{Z}(\zeta_m)$$&#xA; lying over $(p)$ and &#xA;$$\mathfrak{P}\subset \mathbb{Z}(\zeta_n)$$&#xA; lying over $\mathfrak{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a congruence in $\mathbb{Q}(\zeta_n)$ of the form $a\equiv b \pmod{\mathfrak{P}}$, where $a,b$ are actually elements of $\mathbb{Q}(\zeta_m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I say about the congruence properties of $a,b$ in $\mathbb{Q}(\zeta_m)$? Also, if I take the trace or the  norm down to $\mathbb{Q}$, can I say anything about their congruence properties there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2316" LastEditorUserId="2316" LastEditorDisplayName="" LastEditDate="2010-10-05T03:12:28.023" LastActivityDate="2010-11-04T07:44:18.083" Title="Congruences mod primes in Galois extensions" Tags="&lt;algebraic-number-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" />
  <row Id="6076" PostTypeId="2" ParentId="6055" CreationDate="2010-10-05T03:19:58.107" Score="2" ViewCount="0" Body="&lt;p&gt;You are correct that in practice one does not care so much whether two functors are equal, but rather whether they are naturally isomorphic.  The difficulty is that typically one does not want to completely forget about the natural isomorphism either!  (Which is what happens if one quotients out by natural isomorphisms as you suggest.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical example (which I hope will make sense to you) is to consider for each topological space the category of vector bundles on $X$. So we have a category $Vect_X$. If $f: X \to Y$ is a continuous map of spaces,&#xA;and $\mathcal V$ is a vector bundle on $Y$, then one can pull-back $\mathcal V$ to form&#xA;a vector-bundle $f^* \mathcal V$ on $Y$.  So we get a functor $f^*: Vect_Y \to Vect_X$.&#xA;If now $g: Y \to Z$ as well, then one sees that $(g f)^* $ is naturally isomorphic to&#xA;$f^* g^*$,  say by some natural isomorphism $c_{f,g}: (g f)^* \cong f^* g^*.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morally, one would like to say that $X \mapsto Vect_X$ and $f \mapsto f^* $ gives a contravariant functor from the category of topological spaces to the category of categories Cat.  In practice, because we don't have equality between $(f g)^* $ and $f^* g^* $, we don't get such a functor, although we do get a functor into your suggested category &quot;Cat&quot;.&#xA;The problem is that in practice, one wants to remember the natural isomorphisms $c_{f , g}$,&#xA;which satisfy some important properties: for example, if $h: Z \to W$ is a third map,&#xA;then $f^* c_{g,h} \circ c_{f, h g} = c_{f,g}h^* \circ c_{gf, h}.$  (If you write this out, it is a commutative square that relates the various ways to pull back $\mathcal V$ by&#xA;$h$, $g$, and $f$, taking into account the associative law $(hg)f = h(g f)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really, one wants to work in a more sophisticated structure then either Cat or &quot;Cat&quot;,&#xA;namely a structure in which the objects are categories, the morphisms are functors, &#xA;and in which we add an explicit extra layer of structure, so-called 2-morphisms, which are natural isomorphisms between functors.  One then develops a theory in which one morally&#xA;regards functors as equal if they coincide up to natural isomorphism (i.e. up to a diagram involving a 2-morphism), but one also keeps track of the 2-morphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting structure is called a 2-category, and is part of the study of higher category theory.  (Everything written above should provide some background for understanding Martin Brandenburg's answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory has a lot in common with homotopy theory: in topology, one can pass to a category in which objects are spaces and maps are continuos maps modulo homotopy, but in lots of applications one wants to remember not just that maps are homotopic, but one actually wants to remember the homotopy; often then there are homotopies between homotopies,&#xA;homotopies between homotopies between homotopies, and so on.  Similarly in category theory one can introduce not just the notion of 2-category, but notions of n-categories, in which&#xA;there are 3-morphisms beweeen the 2-morphisms, etc., up to n-morphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your structure &quot;Cat&quot; is analogous to passing to the homotopy category in topology; it is interesting, but forgets information one often wants to remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you search for &quot;higher category theory&quot;, you will find an enormous amount of material.&#xA;One good reference is the n-category cafe and the n-Lab.  Another place to look is at the&#xA;various manuscripts on Jacob Lurie's web-page (at Harvard).  What you will find is a rather intricate, and rapidbly evolving theory, blending category theory and homotopy theory in a fascinating (although sometimes daunting!) way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, your idea and your question are far from misguided, but in fact are pointing at one of the most active areas of modern research in category theory and related areas!&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-05T03:19:58.107" CommentCount="3" />
  <row Id="6077" PostTypeId="1" AcceptedAnswerId="6086" CreationDate="2010-10-05T03:53:12.637" Score="1" ViewCount="101" Body="&lt;p&gt;So I have to find the maximum possible error $dR$ in calculating equivalent resistance for three resistors, $\displaystyle\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ if the maximum error in each is 0.5%; $R_1=25\ \Omega$, $R_2=45\ \Omega$, $R_3=50\ \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, originally I did $dR_1=dR_2=dR_3=0.005$, and then did $\displaystyle\frac{dR}{R^2}=\frac{dR_1}{R_1^2}+\frac{dR_2}{R_2^2}+\frac{dR_3}{R_3^2}$ and solved for $dR$... now I realize now that that doesn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall when doing an example problem we took the derivative like normal except when doing the chain rule, replacing it with the partial; for example, when $A=\ell w$, $dA = \frac{\partial A}{\partial \ell} d\ell+\frac{\partial A}{\partial w} dw$ (finding error in area of rectangle) and plugged in what I know. How would that work here? Was I close in my original attempt? I feel like I'm not sure where to put all the partials now that there's a bunch of reciprocals everywhere.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-10-05T13:45:06.927" Title="Calculating total error based on error of variables" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6078" PostTypeId="1" CreationDate="2010-10-05T04:46:56.493" Score="0" ViewCount="90" Body="&lt;p&gt;Let $G$ be a group of order $pqr$, where $p$, $q$ and $r$ be three distinct primes. By Cauchy's theorem there exist three elements, $a$, $b$ and $c$, whose orders are $p$, $q$ and $r$, respectively. If the subgroup generated by $a$ and $b$ is the whole group, then I wonder if it is possible that there exists a proper normal subgroup of $G$.&lt;/p&gt;&#xA;" OwnerUserId="2318" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-05T04:55:03.043" LastActivityDate="2010-10-05T08:05:10.820" Title="A question about finite group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6079" PostTypeId="2" ParentId="6078" CreationDate="2010-10-05T05:10:53.010" Score="4" ViewCount="0" Body="&lt;p&gt;No, a group of order $pqr$ cannot be simple, regardless of whether you can find elements $a$ and $b$ of orders $p$ and $q$ that generate $G$ or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p\lt q\lt r$, and let $\ell$, $m$, and $n$ be the number of $p$-Sylow, $q$-Sylow, and $r$-Sylow subgroups of $G$, respectively. Any two distinct Sylow subgroups will intersect trivially, because they are all cyclic of prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ were simple, then $\ell,m,n\gt 1$. Also, $n|pq$, $m|pr$, and $\ell|qr$. Since $n$ must be congruent to $1$ modulo $r$ and $p,q\lt r$, then we must have $n=pq$. Since $m$ must be congruent to $1$ modulo $q$ and $p\lt q$, the smallest value possible for $m$ is $r$. So the smallest values they can have are $n=pq$, $m=r$, and $\ell=q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the distinct $p$-Sylow subgroups give at least $q(p-1)$ nonidentity elements. The distinct $q$-Sylow subgroups give at least $r(q-1)$ nonidentity elements; and the distinct $r$-Sylow subgroups give at least $pq(r-1)$ nonidentity elements. So $G$ will have at least&#xA;$$\begin{array}{rcl} q(p-1) + r(q-1) + pq(r-1) + 1 &amp;amp;=&amp;amp; &#xA;pqr + pq + rq - q - r - pq + 1\\&#xA;&amp;amp; = &amp;amp;pqr + rq +1 - q-r&#xA;\end{array}$$&#xA;distinct elements. But this is strictly larger than $pqr$, since $r$ and $q$ are greater than 1, so $qr\geq q+r$. This is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contradiction arises from the assumption that $G$ is simple; hence $G$ is not simple.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-05T05:22:00.113" LastActivityDate="2010-10-05T05:22:00.113" />
  <row Id="6080" PostTypeId="1" AcceptedAnswerId="6081" CreationDate="2010-10-05T05:12:18.323" Score="7" ViewCount="199" Body="&lt;p&gt;I was wondering if there is a nice way to see that $\mathbb{RP}^{3}$ is orientable without using  tools of algebraic topology, like homology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only think I could think of was to argue that $\mathbb{RP}^{3}=\mathbb{R}^3 \cup \mathbb{RP}^{2}$ and perhaps you could argue that to get back to any starting position you have to cross the $\mathbb{RP}^{2}$ boundary but I'm pretty sure that what I'm thinking is nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a question on the homework for one of my topics courses and I plan on asking the professor about it tomorrow, but I was curious to see if anyone had any interesting ways of thinking about or picturing this space.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-06T02:23:09.410" Title="Orientability of $\mathbb{RP}^3$" Tags="&lt;geometry&gt;&lt;intuition&gt;&lt;homework&gt;&lt;projective-space&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="6081" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T05:43:07.597" Score="9" ViewCount="0" Body="&lt;p&gt;The orientation on the universal cover, $S^3$, descends. It takes some checking of details, but you can do it for $\mathbb{R}P^n$ in exactly the same way for $n$ odd. Actually, much more generally if you have a covering $\pi: X\to B$, and $Aut_\pi (X)$ is cyclic and generated by some orientation preserving diffeo $f: X\to X$, then if $X$ is orientable, the orientation will descend to $B$. In this case the diffeo is the antipodal map which is orientation preserving on $S^n$ if and only if $n$ is odd.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastEditorUserId="2000" LastEditorDisplayName="" LastEditDate="2010-10-06T02:23:09.410" LastActivityDate="2010-10-06T02:23:09.410" />
  <row Id="6082" PostTypeId="1" AcceptedAnswerId="6123" CreationDate="2010-10-05T05:53:04.420" Score="3" ViewCount="83" Body="&lt;p&gt;I thought I would post (as a puzzle) one of my favourite results in combinatorics.  I actually use variants of this result in research quite often.  It's not impossible that someone will post an unexpected proof that could be used in future research.  But even if that does not occur, I hope it will be a fun exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;em&gt;orthomorphism&lt;/em&gt; of the cyclic group $\mathbb{Z}_n$ is a permutation $\sigma:\mathbb{Z}_n \rightarrow \mathbb{Z}_n$ such that $i \mapsto \sigma(i)-i$ is also a permutation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that orthomorphisms of $\mathbb{Z}_n$ exist if and only if $n$ is odd.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, one direction is easy:  $\sigma(i)=2i$ is an orthomorphism of $\mathbb{Z}_n$ when $n$ is odd.  So what's left is to prove the non-existence of orthomorphisms of $\mathbb{Z}_n$ when $n$ is even.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-05T14:35:44.220" Title="When does an orthomorphism of the cyclic group exist?" Tags="&lt;combinatorics&gt;&lt;group-theory&gt;&lt;puzzle&gt;&lt;permutations&gt;" AnswerCount="1" />
  <row Id="6083" PostTypeId="1" AcceptedAnswerId="6085" CreationDate="2010-10-05T05:54:06.960" Score="1" ViewCount="237" Body="&lt;p&gt;If $(r, \theta)$ are polar coordinates on $\mathbb{R}^2\backslash \left\{ (0,0)\right\}$, then how do I show/prove that &#xA;$d\theta =\dfrac{x dy - y dx}{x^2 + y^2}$?&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-05T09:11:55.267" LastActivityDate="2010-10-05T16:23:06.303" Title="How do you show that $d\theta = \frac{x dy - y dx }{x^2 + y^2}$?" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="6084" PostTypeId="2" ParentId="6083" CreationDate="2010-10-05T06:15:59.670" Score="5" ViewCount="0" Body="&lt;p&gt;On $R^2\setminus\{0\}$ we have $\theta=\Im(\ln(x+iy))$ (here $\Im$ stands for &quot;the imaginary part of&quot;). Therefore $$d\theta = \Im\left(\frac1{x+iy}\right)dx+\Im\left(\frac{i}{x+iy}\right)dy=\Im\left(\frac{x-iy}{x^2+y^2}\right)dx+\Im\left(\frac{i(x-iy)}{x^2+y^2}\right)dy=$$$$\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy=\frac{xdy-ydx}{x^2+y^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-05T07:51:14.650" LastActivityDate="2010-10-05T07:51:14.650" />
  <row Id="6085" PostTypeId="2" ParentId="6083" CreationDate="2010-10-05T06:24:31.193" Score="12" ViewCount="0" Body="&lt;p&gt;$$x=r\cos\theta$$&#xA;$$y=r\sin\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;$$dx=\cos\theta\thinspace dr -r\sin\theta\thinspace d\theta$$&#xA;$$dy=\sin\theta\thinspace dr +r\cos\theta\thinspace d\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $d\theta$ and using $\sin^2\theta+\cos^2\theta=1$&#xA;$$d\theta={{dy\thinspace\cos\theta-dx\thinspace\sin\theta}\over {r}}$$&#xA;Multiplying and dividing by $r$&#xA;$$={{dy\thinspace (r\cos\theta)-dx\thinspace (r\sin\theta)}\over {r^2}}$$&#xA;Using the definition of $x$ and $y$ in terms of polar coordinates&#xA;$$={{x\thinspace dy-y\thinspace dx}\over {x^2+y^2}}$$&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-05T06:24:31.193" CommentCount="1" />
  <row Id="6086" PostTypeId="2" ParentId="6077" CreationDate="2010-10-05T06:28:26.800" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;em&gt;dR&lt;sub&gt;i&lt;/sub&gt;&lt;/em&gt; denotes the absolute error, but that 0.5% is a relative error. Therefore, you should use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;dR_1 &amp;amp;= 0.5\% \times R_1 = 0.125\, \Omega \\&#xA;dR_2 &amp;amp;= 0.5\% \times R_2 = 0.225\, \Omega \\&#xA;dR_2 &amp;amp;= 0.5\% \times R_3 = 0.25\, \Omega&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is correct for obtaining a conservative value of &lt;em&gt;dR&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(However, if you can ensure the quantities are distributed normally and uncorrelated, the errors are added together quadratically:&#xA;$$ \Delta(A+B) = \sqrt{\Delta A^2 + \Delta B^2}, $$&#xA;i.e. &#xA;$$ \left(\frac{\Delta R}{R^2}\right)^2 =&#xA;\left(\frac{\Delta R_1}{R_1^2}\right)^2 &#xA;+ \left(\frac{\Delta R_2}{R_2^2}\right)^2&#xA;+ \left(\frac{\Delta R_3}{R_3^2}\right)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be too advanced for your course.)&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-05T13:45:06.927" LastActivityDate="2010-10-05T13:45:06.927" />
  <row Id="6087" PostTypeId="1" AcceptedAnswerId="6091" CreationDate="2010-10-05T06:50:02.733" Score="1" ViewCount="200" Body="&lt;p&gt;This is an Exercise from Apostol's book. Many of them seem tough and here is one of them which i am struggling with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-07T01:44:05.467" Title="For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$" Tags="&lt;number-theory&gt;&lt;functions&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="6088" PostTypeId="2" ParentId="6057" CreationDate="2010-10-05T07:15:06.643" Score="2" ViewCount="0" Body="&lt;p&gt;The industry standard way of doing this is is to calculate a &quot;mask&quot; of sorts, and apply it to each one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;mask&quot;, I mean, say, a function for the current point based on the points before and after.  For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{1}{9}[x-2] + \frac{2}{9}[x-1] + \frac{3}{9}[x] + \frac{2}{9}[x+1] + \frac{1}{9}[x+2] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[p]$ is the value of the pixel at $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to find the new pixel value at, say, pixel $4$, you'd use $ \frac{1}{9}[2] + \frac{2}{9}[3] + \frac{3}{9}[4] + \frac{2}{9}[5] + \frac{1}{9}[6] $ (remembering that, say, $[3]$ is the value of pixel 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that's left to do is then apply your mask to every pixel.  (What should you use for pixels that are close to the edge?  That's up to you.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that to be a true blurring mask, your coefficients must add up to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;Gaussian Blur&quot; is just applying a special mask based off of the Gaussian Curve to your pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, you make the &quot;coefficient&quot; of each term a number based off of the Gaussian Curve at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_Blur#Sample_Gaussian_matrix&quot; rel=&quot;nofollow&quot;&gt;wikipedia lists&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = 0.00038771[x-3] + 0.01330373[x-2] +&#xA;0.11098164[x-1] + 0.22508352[x] + 0.11098164[x+1] +&#xA;0.01330373[x+2] + 0.00038771[x+2]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as a mask with σ = 0.84089642 and a &quot;cut-off&quot; radius of 3, although you can pick your own standard deviation and cut-off radius to your liking.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-10-05T07:15:06.643" />
  <row Id="6089" PostTypeId="2" ParentId="6075" CreationDate="2010-10-05T07:18:08.693" Score="1" ViewCount="0" Body="&lt;p&gt;If $\mathcal{A}_m$ is the ring of integers of $\mathbb{Q}(\zeta_m)$, then $\mathfrak{B} \cap \mathcal{A}_m = \mathfrak{p}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say $a \equiv b \pmod{\mathfrak{B}}$ where $a,b \in \mathbb{Q}(\zeta_m)$.  Does this mean $a,b$ are algebraic integers (in $\mathcal{A}_m$)?  If not, it must mean that there is some denominator $d \in \mathcal{A}_m$, $d \notin \mathfrak{B}$ such that $d(a-b) \in \mathfrak{B}$.  (Just like we could say $\frac{2}{3} \equiv \frac{7}{3} \pmod{5}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, the above fact gives us $a \equiv b \pmod{\mathfrak{p}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can relate Norm(a) to Norm(b) or Trace(a) to Trace(b) modulo (p).  Consider $1 \equiv 2i \pmod{(1-2i)}$ in the Gaussian integers $\mathbb{Z}[i]$.  Taking norm or trace gives us nothing modulo (5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can, however, use $\mathfrak{p} \cap \mathbb{Z} = (p)$ to say that ${\rm Norm}(a-b) \in (p)$, if $a,b$ are algebraic integers.  If, however, we have some denominator $d \in \mathfrak{p}'$ where $\mathfrak{p}'$ is a conjugate ideal to $\mathfrak{p}$, then the norm will give us powers of $p$ in the numerator and denominator, so we have to be more careful.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-05T07:18:08.693" />
  <row Id="6090" PostTypeId="1" AcceptedAnswerId="6160" CreationDate="2010-10-05T07:26:11.997" Score="2" ViewCount="255" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been reading about Riemann Zeta function and have been thinking about it for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anything been published regarding upper bound for the real part of zeta function zeros as the imaginary part of the zeros tend to infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-05T08:31:23.300" LastActivityDate="2010-12-24T23:30:36.527" Title="On Zeta function zeros in the critical strip" Tags="&lt;number-theory&gt;&lt;riemann-zeta&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="6091" PostTypeId="2" ParentId="6087" CreationDate="2010-10-05T07:43:15.927" Score="6" ViewCount="0" Body="&lt;p&gt;$\begin{align*}(\sqrt{n} + \sqrt{n+1})^2 &amp;amp;= 4n+2 - 2\left(n+1/2 - \sqrt{n(n+1)}\right) \\ &amp;amp;= 4n+2 - 2(AM(n,n+1) - GM(n,n+1)) \\ &amp;amp;\in (4n+1, 4n+2).\end{align*}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The first line is just algebra. In the second line, $AM$ and $GM$ are respectively the arithmetic and geometric means. To get the third line: $n &amp;lt; GM(n,n+1) &amp;lt; AM(n,n+1)=n+1/2$ by the AM-GM inequality, so $0 &amp;lt; AM(n,n+1)-GM(n,n+1) &amp;lt; 1/2$, and the third line follows.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are no perfect squares between $4n+1$ and $4n+2$, and thus no integers between $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$, QED.&lt;/p&gt;&#xA;" OwnerUserId="2032" LastEditorUserId="2032" LastEditorDisplayName="" LastEditDate="2010-10-05T08:11:42.177" LastActivityDate="2010-10-05T08:11:42.177" CommentCount="3" />
  <row Id="6092" PostTypeId="2" ParentId="6077" CreationDate="2010-10-05T08:02:29.957" Score="3" ViewCount="0" Body="&lt;p&gt;This is easy, since $R$ is monotonic as a function of $R_1,R_2,R_3$ (which is obvious from physical considerations: increasing one of the resistances can only increase the total resistance). Thus, to compute error bounds for $R$, it suffices to consider the minimum and maximum possible values for your three resistors: the minimum corrected value of $R$ is $f(0.995R_1, 0.995R_2, 0.995R_3)$, and the maximum corrected value is $f(1.005R_1, 1.005R_2, 1.005R_3)$, where $f(x,y,z) = (x^{-1} + y^{-1} + z^{-1})^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a further simplification is that $f$ is homogeneous of degree 1, i. e., $f(cx, cy, cz) = cf(x,y,z)$ (again, this is obvious intuitively: choosing different units for your resistances multiplies both the individual and total resistances by the same proportionality constant). Thus, $f(0.995R_1, 0.995R_2, 0.995R_3) = 0.995R$, and $f(1.005R_1, 1.005R_2, 1.005R_3) = 1.005R$. Thus, the maximum relative error in the total resistance is also 0.5%. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-10-05T08:02:29.957" CommentCount="2" />
  <row Id="6093" PostTypeId="2" ParentId="6078" CreationDate="2010-10-05T08:05:10.820" Score="1" ViewCount="0" Body="&lt;p&gt;More generally, let $G$ be a finite group with order $p_1p_2\cdots p_n$, where $p_1&amp;lt; p_2&amp;lt; \cdots &amp;lt; p_n$ are distinct primes.  Then by the normalizer-centralizer theorem, the Sylow $p_1$-subgroup is central in its normalizer, so there is a normal $p_1$-complement in $G$ by Burnside's transfer theorem.  Inductively, we can then show the Sylow $p_n$-subgroup is normal in $G$.&lt;/p&gt;&#xA;" OwnerUserId="641" LastActivityDate="2010-10-05T08:05:10.820" />
  <row Id="6094" PostTypeId="1" AcceptedAnswerId="6095" CreationDate="2010-10-05T08:13:21.650" Score="1" ViewCount="65" Body="&lt;p&gt;Given an affine variety, we can forget it into a variety. The other way round, given a variety $X$ we can look at &#xA;$$spec(\mathcal O_X(X)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this necessarily an affine variety (a priori its just an affien scheme)?&#xA;If so, am I right that this implys that taking spec of the global functions is left adjoint to the forgetful functor on the level of varieties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what does this functor intuitively? If you feed in an affine variety, it spits out the same variety again. If you apply it to a complete irreducible variety (for example $\mathbb P^n$) it gets transformed into a point. &lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-10-05T08:43:19.683" Title="Turning varieties into affine varieties?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="6095" PostTypeId="2" ParentId="6094" CreationDate="2010-10-05T08:43:19.683" Score="4" ViewCount="0" Body="&lt;p&gt;The functor $Sch \to Ring^{op}, X \mapsto \mathcal{O}_X(X)$ is left adjoint to the functor $\text{Spec} : Ring^{op}\to Sch$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a variety you mean a separated integral scheme $X$ of finite type over a field $k$? Then $O_X(X)$ does not have to be of finite type over $k$. Namely, there are integral $k$-subalgebras $A,B$ of a $k$-algebra $C$ such that $A,B$ are of finite type, but $A \cap B$ is not. Then glue $Spec(A)$ and $Spec(B)$ along $Spec(C)$ to get a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-05T08:43:19.683" CommentCount="3" />
  <row Id="6096" PostTypeId="2" ParentId="6087" CreationDate="2010-10-05T08:54:25.330" Score="5" ViewCount="0" Body="&lt;p&gt;Note that $$\frac{1}{2} &gt; \sqrt{n+\frac{1}{2}}-\sqrt{n} &gt; \sqrt{n+1}- \sqrt{n+\frac{1}{2}} &gt; 0,$$&#xA;since $f(x)=\sqrt{x+1/2}-\sqrt{x}$ is a decreasing function. The inequalities and the fact that $f(x)$ is decreasing follow from noting that $\sqrt{x+1/2} - \sqrt{x} = 1/2(\sqrt{x+1/2} + \sqrt{x}).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can write&#xA;$$ 1&gt; \left( \sqrt{n+\frac{1}{2}}-\sqrt{n} \right) - \left( \sqrt{n+1}- \sqrt{n+\frac{1}{2}} \right) &gt; 0.$$&#xA;Hence $$1 &gt; 2\sqrt{n+\frac{1}{2}} - \left( \sqrt{n+1}+\sqrt{n} \right) &gt; 0.$$&#xA;From which the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity we've shown that we can write&#xA;$$\sqrt{n+1}+\sqrt{n} + r = 2\sqrt{n+\frac{1}{2}} \quad \textrm{ for } 0 &amp;lt; r &amp;lt; 1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we note that we do not straddle an integer since $\sqrt{4n+1} &amp;lt; \sqrt{n} + \sqrt{n+1},$ and there are no integers between $\sqrt{4n+1}$ and $\sqrt{4n+2}.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-05T10:53:51.137" LastActivityDate="2010-10-05T10:53:51.137" CommentCount="1" />
  <row Id="6097" PostTypeId="2" ParentId="5964" CreationDate="2010-10-05T09:12:19.540" Score="2" ViewCount="0" Body="&lt;p&gt;The answer is three.  As for a proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above two colours is not enough.  Every square grid colouring must have either alternating colours uniformly across the grid, or somewhere there are two adjacent vertices of the same colour.  In the former case then any swap of vertices will create a three-in-a-row.  If there are two in a row of the same colour and we assume it is not possible to get to a three in a row configuration, then all the vertices marked X below must not be colour 1.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.X..X.&#xA;XX11XX&#xA;.X..X.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With only two colours this clearly creates a problem as all the X vertices must be colour 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that 3 colours suffices consider the colouring below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;123123&#xA;231231&#xA;312312&#xA;123123&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The pattern can be extended to cover an infinitely large grid and it's not possible to swap to get three-in-a-row or even to get a coloured path of length three.&lt;/p&gt;&#xA;" OwnerUserId="1750" LastActivityDate="2010-10-05T09:12:19.540" />
  <row Id="6098" PostTypeId="2" ParentId="6008" CreationDate="2010-10-05T10:24:50.807" Score="0" ViewCount="0" Body="&lt;p&gt;Ok, i managed with &lt;a href=&quot;http://math.stackexchange.com/questions/6036/moments-on-number-of-occurrences-of-substring&quot;&gt;Robin Chapman's help&lt;/a&gt;  compute the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X1,X2,...,XN be the random variables defined as follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Xi=1 if there is a run of heads starting at the position i, otherwise Xi=0.&lt;br&gt;&#xA;Then X=X1+X2+...+XN.&lt;br&gt;&#xA;By linearity of expectation E(X)=E(X1)+...+E(XN).&lt;br&gt;&#xA;Now, a run of heads can start at the position i only by the condition &quot;a tail at the position i-1&quot;. That means P(Xi=1)=(1-p)*p=q*p.&lt;br&gt;&#xA;But at i=1 there is no i=0.&lt;br&gt;&#xA;Hence P(X1=1)=p.&#xA;Last step: E(Xi)=1*P(Xi=1)+0*P(Xi=0)=P(Xi=1)&lt;br&gt;&#xA;Finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E(X)=p+(N-1)*p*q&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of p=1/2 E(X)=(N+1)/4&#xA;and if N=&gt;oo then E(X)=&gt;N/4  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Congrats leonbloy! Your approach was very fruitful. Now, to get exact variance seems to be rather tricky. But in any case, many thanks anyway.&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-10-05T10:24:50.807" />
  <row Id="6099" PostTypeId="1" CreationDate="2010-10-05T10:49:44.183" Score="5" ViewCount="137" Body="&lt;p&gt;If $M$ is a differentiable manifold, De Rham's theorem gives for each positive integer $k$ an isomorphism&#xA;$Rh^k : H^k_{DR}(M,\mathbb R) \to H^k_{singular}(M,\mathbb R)$. On the other hand, we have a canonical map $H^k_{sing}(M,\mathbb Z) \to H^k_{singular}(M,\mathbb R)$ . Allow me to denote (this is not standard)  its image by $\tilde {H}^k_{singular }(M,\mathbb Z)$. My question is : how do you recognize if, given a closed differental $k$- form $\omega$ on $M$, its image &#xA;$Rh^k([\omega]) \in H^k_{singular}(M,\mathbb R) $ is actually in &#xA;$\tilde {H}^k_{singular}(M,\mathbb Z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would very much appreciate a concrete answer, ideally backed up by one or more explicit calculations. Thank you for your attention.&lt;/p&gt;&#xA;" OwnerUserId="2324" LastActivityDate="2010-10-06T09:55:55.863" Title="Integral classes in de Rham cohomology" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="6100" PostTypeId="1" CreationDate="2010-10-05T11:11:43.373" Score="1" ViewCount="147" Body="&lt;p&gt;Please list both the problem/area and justify why it is important philosophically. This question doesn't cover questions that are only important within the philosophy of mathematics itself.&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-10-13T18:03:02.257" CommunityOwnedDate="2010-10-05T11:11:43.373" Title="What mathematical questions or areas have philosophical implications outside of mathematics?" Tags="&lt;philosophy&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="1" />
  <row Id="6101" PostTypeId="1" CreationDate="2010-10-05T11:35:36.583" Score="2" ViewCount="267" Body="&lt;p&gt;This question is intended to complement &lt;a href=&quot;http://math.stackexchange.com/questions/6100/what-mathematical-questions-or-areas-have-philosophical-implications-outside-of-m&quot;&gt;What mathematical questions or areas have philosophical implications outside of mathematics?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-10-13T18:02:36.500" CommunityOwnedDate="2010-10-05T11:35:36.583" Title="What are the most important questions or areas of study in the philosophy of mathematics?" Tags="&lt;philosophy&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="6102" PostTypeId="2" ParentId="6099" CreationDate="2010-10-05T11:49:37.413" Score="6" ViewCount="0" Body="&lt;p&gt;By Poincare duality a $k$-form will be integral iff its integral over all&#xA;(smooth) singular $k$-cycles is an integer.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-05T11:49:37.413" />
  <row Id="6103" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T11:57:19.307" Score="6" ViewCount="0" Body="&lt;p&gt;One can also see that $\mathbb{R}P^3$ is a Lie group, and that all Lie groups are orientable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\mathbb{R}P^3$ is a Lie group, first think of $S^3$ as the collection of unit quaternions.  The subset $\{1,-1\}$ is not only a subgroup, but it is a normal subgroup of $S^3$.  Hence, $S^3/\{1,-1\}$ is a Lie group.  In fact, the elmement -1 acts as the antipodal map.  (It turns out, that as a Lie group, $\mathbb{R}P^3$ is isomorphic to $SO(3)$, the 3x3 real matrices satisfying $AA^t = A^tA = Id$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is every (connected component of a) Lie group orientable?  Well, suppose $G$ is a (connected) Lie group with identity $e$.  At the identity, pick any ordered basis $e_i$ for $T_e G$ (i.e., pick an orientation on the vector space $T_e G$.  To figure out what what orientation at $g\in G$ should be, notice the map $L_g:G\rightarrow G$ given by $L_g(h) = gh$ maps $e$ to $g$.  Hence its differential $d_e L_g$ maps $T_e G$ to $T_g G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $L_g$ is actually a diffeomorphism (with inverse $L_{g^{-1}}$), so the map $d_e L_g$ must also be an isomorphism.  Now, simply take as an ordered basis $d_e L_g e_i$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-05T11:57:19.307" />
  <row Id="6104" PostTypeId="1" CreationDate="2010-10-05T12:15:21.090" Score="0" ViewCount="68" Body="&lt;p&gt;Is it possible to make up a &lt;em&gt;distinct&lt;/em&gt; number that is made up of four tuples. One of which indicates the direction of the arrangement of the other three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, 9*4*7 = 252. 252 &lt;em&gt;is distinct&lt;/em&gt; in the sense that you cannot get it as a multiplication of 3 distinct single digit numbers other than 9, 4, 7. Now, the problem is 252 could mean, any of the three orders. Say, 947, 497, 974. Is there any way to add a sense of direction/arrangement to it?&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-10-05T13:01:08.910" Title="Numbers that represent four tuples" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6105" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T12:16:47.673" Score="4" ViewCount="0" Body="&lt;p&gt;Here is a rather brief summary of the major positions on a couple of important philosophy of mathematics questions. First a list of questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Are we creating or discovering when we do mathematics?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What are mathematical entities?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;How do you explain &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences&quot; rel=&quot;nofollow&quot;&gt;the unreasonable effectiveness of mathematics in natural science&lt;/a&gt;?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An early answer to these question was Plato's suggestion that mathematical entities (triangles, squares, numbers...) exist in the world of forms. So numbers really do exist, just in some kind of rarified universe &lt;a href=&quot;http://en.wikipedia.org/wiki/Allegory_of_the_Cave&quot; rel=&quot;nofollow&quot;&gt;that we only have partial access to&lt;/a&gt;. The idea of a &quot;Platonic heaven&quot; of numbers is not a particularly popular position any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immanuel Kant thought that geometry and arithmetic were somehow innate (indeed, they were preconditions to experience of space and the passage of time). That is, without having an &lt;em&gt;a priori&lt;/em&gt; understanding of space and number, we wouldn't be able to perceive the world around us as we do. Mathematical entities, for Kant, were real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logicism is a position that grew out of Russell and Whitehead's work on mathematical logic. The idea was that logic is somehow universal, and that all of mathematics is built up from these basic axioms, which are uncontroversially, universally, absolutely True. (With a capital &quot;T&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formalism started at roughly the same time with Hilbert. Here the idea is that mathematics is a game we play, with certain rules (axioms) laid down somewhat arbitrarily. This shares a lot with logicisim, but is importantly different in its attitude to the status of the axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structuralism is a much more recent position (although there are echoes of the ideas in Hilbert, Poincare and others). The idea here is that mathematics is about &lt;em&gt;structures.&lt;/em&gt; We learn about structures through abstracting away incidental details to recognise patterns. What exactly structures are depends on what kind of structuralist you ask. Some (like Stewart Shapiro) think that structures really are out there in the world, somehow. That is, structures exist over and above the things that satisfy the structural relations. Others (like Michael Resnik) think they are not quite this real. Resnik's brand of structuralism looks a lot like &lt;em&gt;nominalism:&lt;/em&gt; Nominalism about mathematical entities grows out of a more &lt;a href=&quot;http://en.wikipedia.org/wiki/Nominalism&quot; rel=&quot;nofollow&quot;&gt;general position&lt;/a&gt; towards abstract entities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a commenter said above, philosophy isn't about results. I think the most important contribution of philosophy is to ask the big questions, like the ones above, and to map out the logical space of possible responses to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have restricted myself to philosophy of &lt;em&gt;mathematics&lt;/em&gt; and deliberately avoided philosophy of logic, which is its own cottage industry. Also, this summary reflects my own interests and biases and should not be taken as a representative summary of all the debates and positions taken in philosophy of maths. A more balanced summary can be found at the excellent &lt;em&gt;Stanford Encyclopedia of Philosophy&lt;/em&gt; &lt;a href=&quot;http://plato.stanford.edu/entries/philosophy-mathematics/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by Leon Horsten.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-10-05T12:16:47.673" CommunityOwnedDate="2010-10-05T12:16:47.673" CommentCount="2" />
  <row Id="6106" PostTypeId="2" ParentId="6100" CreationDate="2010-10-05T12:24:26.470" Score="3" ViewCount="0" Body="&lt;p&gt;Goedel's incompleteness theorem is used by some (e.g. Roger Penrose) as part of a justification for why computers will never achieve consciousness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that all infinite dimensional separable Hilbert spaces are isomorphic has philosophical implications for the metaphysics of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various results in dynamical systems theory related to chaotic systems limit what can be said about predictability and about what it means for a system to be deterministic. For example &lt;a href=&quot;http://www.ams.org/journals/bull/1991-24-01/S0273-0979-1991-15953-7/S0273-0979-1991-15953-7.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Ornstein and Weiss (warning: it's huge and will take a long time to download on slow connections) has been used to suggest that the distinction between deterministic and stochastic systems is flawed.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-10-05T12:24:26.470" CommunityOwnedDate="2010-10-05T12:24:26.470" CommentCount="6" />
  <row Id="6107" PostTypeId="1" CreationDate="2010-10-05T12:51:02.377" Score="1" ViewCount="52" Body="&lt;p&gt;i have lineA with points are(1,1) and  (8,8) and another lineB with points(2,2) and (4,4) i like to prove lineB lies on LineA ,how can do this?&lt;/p&gt;&#xA;" OwnerUserId="2325" LastActivityDate="2010-10-05T13:13:18.930" Title="how do find the line lies on another line" Tags="&lt;geometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6109" PostTypeId="2" ParentId="6107" CreationDate="2010-10-05T12:55:11.633" Score="2" ViewCount="0" Body="&lt;p&gt;The points are all on the line $y=x$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-05T12:55:11.633" />
  <row Id="6110" PostTypeId="2" ParentId="6107" CreationDate="2010-10-05T12:57:35.057" Score="0" ViewCount="0" Body="&lt;p&gt;Hint: &lt;strike&gt; Take the perpendicular distance and prove that the distance is 0.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the 2 point slope form.Equation of the line passing through $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given bye $$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}$$ and observe that they are same.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-05T13:13:18.930" LastActivityDate="2010-10-05T13:13:18.930" CommentCount="0" />
  <row Id="6111" PostTypeId="2" ParentId="6104" CreationDate="2010-10-05T13:01:08.910" Score="1" ViewCount="0" Body="&lt;p&gt;Yes. Try creating your number like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose your 3-tuple has values (a,b,c). Then generate your distinct number as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^a * 3^b * 5^c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these are prime bases, this number will be unique for any 3 values and will be recoverable, in that you can factor your number and get back all 3 numbers and their orders. To extend this to any n-tuple just use more prime number bases. &lt;/p&gt;&#xA;" OwnerUserId="2003" LastActivityDate="2010-10-05T13:01:08.910" />
  <row Id="6112" PostTypeId="2" ParentId="6107" CreationDate="2010-10-05T13:01:23.460" Score="0" ViewCount="0" Body="&lt;p&gt;In general you can use a determinant expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{vmatrix}x_1&amp;amp;y_1&amp;amp;1\\x_2&amp;amp;y_2&amp;amp;1\\x_3&amp;amp;y_3&amp;amp;1\end{vmatrix}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to verify that three points $(x_i,y_i)$, $i=1\dots 3$ are collinear. As Hans points out though, that the coordinates of all four points that you have have their abscissas and ordinates identical makes them obviously collinear.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-05T13:01:23.460" />
  <row Id="6113" PostTypeId="1" AcceptedAnswerId="6114" CreationDate="2010-10-05T13:03:01.973" Score="1" ViewCount="69" Body="&lt;p&gt;A quick question-Something that I learned that I forgot throughout the years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a question asked you for a multiple of 9 and 12, which would be less than 40, would &quot;36&quot; be an answer, as both 9 and 12 can multiply to get 36, yet it fits under the criteria for being under 40?&lt;/p&gt;&#xA;" OwnerUserId="2062" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-05T14:13:02.427" LastActivityDate="2010-10-05T14:13:02.427" Title="What's a Multiple?" Tags="&lt;arithmetic&gt;" AnswerCount="1" />
  <row Id="6114" PostTypeId="2" ParentId="6113" CreationDate="2010-10-05T13:06:41.687" Score="4" ViewCount="0" Body="&lt;p&gt;Yes you are correct with your reasoning. Multiples of $9$ are of the form $9m$ where $m \in \mathbb{Z}$, the set of integers.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-05T13:06:41.687" />
  <row Id="6115" PostTypeId="2" ParentId="6060" CreationDate="2010-10-05T13:10:23.830" Score="1" ViewCount="0" Body="&lt;p&gt;Within the framework of &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_flow&quot; rel=&quot;nofollow&quot;&gt;Potential flow&lt;/a&gt;,the velocity field of a two dimensional potential flow is the gradient of the real part of a complex function $w(z)$ which is analytical outside regions of singularity. The imaginary part of the complex potential is the stream function, whose constant loci are the streamlines, i.e., the integral curves of the velocity field.&#xA;For example for a single vortex loacted at $z_0$, the potential function is given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w(z) = -\frac{i \Gamma}{2 \pi} log(z-z_0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\Gamma$ is (the constant) vorticity). It is easy to verify that this poential function&#xA;gives rize to an angular flow around $z_0$ of velocity inversly proportional to the distance from $z_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real part of the complex potential saisfies the two dimensional Laplace equation and by linearity, new solutions can be obtained by superposition. In the case of the multiple vortices, the complex potential function takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w(z) = \sum_n -\frac{(-1)^n i\Gamma}{2 \pi} log(z-z_n)$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z_n$ are the vortex centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I can think of a possible candidate of a Voronoi diagram for the vortices by defining the cell walls as the loci of stagnation points in which the fluid velocity is zero. These points are characterized by the condition: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dw(z)}{dz}=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggestion will result a simple solution in the case of vortices rotating in the same direction, but not in the case of counter-rotating vortices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: From a superficial reading of the referred article, the vortex solution is a perturbation to a stagnation point flow. This is a flow towards a wall or a corner.&#xA;To take into account the existence of this wall, one must add vortices from the other side of the wall according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Method_of_images&quot; rel=&quot;nofollow&quot;&gt;method of images&lt;/a&gt;. May be the stagnation point loci of the complete solution (including the images) will have the required Voronoi cell structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the correct comment by Rahul, the stagnation points are certainly discrete&#xA;and do not have one dimensional loci. I think that that the suggestion can be salvaged if we define&#xA;the Voronoi cell walls as passing through the stagnation points and having a normal in the direction of the line&#xA;connecting the two vortices.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-10-05T16:29:29.913" LastActivityDate="2010-10-05T16:29:29.913" CommentCount="2" />
  <row Id="6116" PostTypeId="1" AcceptedAnswerId="6119" CreationDate="2010-10-05T13:37:55.930" Score="1" ViewCount="48" Body="&lt;p&gt;I've been doing some maths work using the rate of flow of liquids. I've used various models for the flow and various methods to integrate these models. The one thing that is confusing me is the difference between constants of proportionality, and constants of integration?&lt;/p&gt;&#xA;" OwnerUserId="1699" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-05T14:08:37.167" LastActivityDate="2010-10-05T14:08:37.167" Title="What is the difference between constants of proportionality and constants of integration?" Tags="&lt;terminology&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="6117" PostTypeId="1" AcceptedAnswerId="6127" CreationDate="2010-10-05T13:40:56.157" Score="1" ViewCount="66" Body="&lt;p&gt;According to Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let G be a covering group of H. The&#xA;  kernel K of the covering homomorphism&#xA;  is just the fiber over the identity in&#xA;  H and is a discrete normal subgroup of&#xA;  G.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is easy to show that the kernel is a normal subgroup, but why is it discrete? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this would be true if the identity of H was open, but I cannot show this (and I don't even know if it is true/the right way to prove that K is discrete). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;If we assume that the definition of &quot;cover space&quot; does not require the fibres to be discrete and we assume that H is connected and locally path-connected, does it still follow that the kernel is discrete? &lt;/p&gt;&#xA;" OwnerUserId="1907" LastEditorUserId="1907" LastEditorDisplayName="" LastEditDate="2010-10-05T14:55:33.127" LastActivityDate="2010-10-05T15:25:47.943" Title="Why is the kernel of a covering group discrete? " Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="6118" PostTypeId="1" CreationDate="2010-10-05T13:46:12.883" Score="-1" ViewCount="98" Body="&lt;p&gt;I was just wondering if there is any single English word for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call things quantifiable, measurable etc. but none of these exactly fit the the question posted in subject.&lt;/p&gt;&#xA;" OwnerUserId="2327" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-05T14:00:27.930" LastActivityDate="2010-10-05T14:00:27.930" ClosedDate="2010-10-05T14:23:19.657" Title="What is the single word to describe a situation that can not be expressed as a mathematical formula" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6119" PostTypeId="2" ParentId="6116" CreationDate="2010-10-05T13:54:26.843" Score="4" ViewCount="0" Body="&lt;p&gt;They are &lt;em&gt;completely different animals&lt;/em&gt;. About the only thing they have in common is the words &quot;constant of&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two quantities, $p$ and $q$, are &quot;proportional&quot;, that means that growth in one will be &quot;mirrored&quot; by growth in the other, and vice-versa. The growth need not be exactly the same (for example, your weight is [roughly] proportional to your height, but an increase in weight of 10 pounds does not correspond to an increase in height of 10 inches). There is a certain &quot;scaling&quot; that goes on. The &quot;constant of proportionality&quot; is what represents that scaling. We will have something of the form $p=kq$, with $k$ a constant; that means that for every unit that $q$ increases, $p$ will increase by $k$ units. That's the constant of proportionality. It is a constant, that depends on the quantities being considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do an indefinite integral, on the other hand, you are trying to find &lt;em&gt;all&lt;/em&gt; antiderivatives of a function. That is, $\int f(x)dx$ represents a &lt;em&gt;family&lt;/em&gt; of functions, namely, all functions $F(x)$ such that $F'(x)=f(x)$. Functions will generally have a &lt;em&gt;lot&lt;/em&gt; of antiderivatives. But one can prove that if $F(x)$ and $G(x)$ are two functions such that $F'(x)=G'(x)=f(x)$, then $F$ and $G$ will just be vertical translates of each other; that is, there will exist a constant $c_0$ such that $G(x)=F(x)+c_0$. In order to represent &lt;em&gt;all&lt;/em&gt; antiderivatives of a function $f(x)$, then, it is enough to find a single antiderivative $F(x)$, because then every other antiderivative will look like &quot;$F(x)+c$&quot; for some constant $c$. So we write &#xA;$$\int f(x)dx = F(x)+C$$&#xA;to represent the entire family of antiderivatives; this means: &quot;the antiderivatives of $f(x)$ are all the functions of the form $F(x)+C$, where $C$ is a constant.&quot; Pick a constant, you get an antiderivative. Pick a different constant, you get a different antiderivative. If $G$ is any antiderivative, then there will be some constant $C$ such that $G(x)=F(x)+C$. We call $C$ the &quot;constant of integration.&quot; So $C$ is not a specific constant, but rather corresponds to the fact that antiderivatives are families, not specific functions. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-05T13:54:26.843" />
  <row Id="6120" PostTypeId="2" ParentId="6118" CreationDate="2010-10-05T13:55:12.747" Score="0" ViewCount="0" Body="&lt;p&gt;For a mathematical function which cannot be expressed as a formula we say that it has no closed-form expression. &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-05T13:55:12.747" CommentCount="3" />
  <row Id="6121" PostTypeId="2" ParentId="6118" CreationDate="2010-10-05T13:59:44.290" Score="0" ViewCount="0" Body="&lt;p&gt;I can think of a few that might fit the bill: unpredictable, incalculable, incomputable. But maybe you'd be better off asking on: &lt;a href=&quot;http://english.stackexchange.com/questions&quot;&gt;http://english.stackexchange.com/questions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-05T13:59:44.290" />
  <row Id="6122" PostTypeId="2" ParentId="6099" CreationDate="2010-10-05T14:06:05.467" Score="1" ViewCount="0" Body="&lt;p&gt;I think Robin's result is also clear if we look at the definition of the isomorphism between De Rham and (${\cal C}^\infty$) singular cohomologies. (Ok, we are both saying the same thing, but I like to say it this way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember (see Dupont, Curvature and characteristic classes, Springer LNM 640) the definition of the isomorphism at the cochain level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I : \Omega^n (M) \longrightarrow C^n (M) \ , &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\Omega^n(M)$ are the degree $n$-forms on $M$, $C^n(M)$ the ${\cal C}^\infty$ singular $n$-cochains. Then, for $\omega \in \Omega^n (M)$, the value of $I(\omega )$ on a ${\cal C}^\infty $ singular $n$-simplex $\sigma$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I (w)_{\sigma}  = \int \sigma^*  \omega \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the integral is taken over the standard $n$-simplex $\Delta^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for $[I(w)]$ to be an integral class it suffices that all values of $I(w)$ on cycles $\tau = \sum_i \lambda_i\sigma_i$ are integer numbers; that is, if and only if $I(w)_\tau = \sum_i \lambda_i\int \sigma_i^*\omega \in \mathbb{Z}$ for all cycles $\tau$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T00:29:14.807" LastActivityDate="2010-10-06T00:29:14.807" CommentCount="1" />
  <row Id="6123" PostTypeId="2" ParentId="6082" CreationDate="2010-10-05T14:35:44.220" Score="4" ViewCount="0" Body="&lt;p&gt;I suppose you are talking about the additive group $\mathbb{Z}_n$ (i.e. the numbers modulo $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I am missing something, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is even then $\sum_{i=1}^{n} i \ne 0 \mod n$. So if $\sigma(i) -i$ were a permutation we would have that $\sum_{i=1}^{n} (\sigma(i) - i) \ne 0 \mod n$, but it is $0$ as $\sigma(i)$ is a permutation too.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-05T14:35:44.220" CommentCount="1" />
  <row Id="6124" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T14:40:58.833" Score="4" ViewCount="0" Body="&lt;p&gt;$RP3$ can be represented as a three dimensional ball $D3$ with the antipodal points of the bounding sphere identified. One can draw a small loop on the surface of the bounding sphere and compute the direction of the normal according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Right-hand_rule&quot; rel=&quot;nofollow&quot;&gt;right hand rule&lt;/a&gt;. Now, performing the same computation for the antipodal loop, the normal will have the same direction. Thus there is a consistent choice of the normal. Notice that the same argument will fail in the case of the nonorientable $RP^2$.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-05T14:40:58.833" CommentCount="0" />
  <row Id="6125" PostTypeId="1" AcceptedAnswerId="6126" CreationDate="2010-10-05T14:54:51.797" Score="1" ViewCount="105" Body="&lt;p&gt;Let's say you have a set of numbers S and you want to obtain subsets of S, s&lt;sub&gt;1&lt;/sub&gt;, s&lt;sub&gt;2&lt;/sub&gt;,...s&lt;sub&gt;i&lt;/sub&gt;, where i&amp;lt;N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a particular operation that will group the numbers that are &quot;close to each other&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will give an example to clarify the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a set S which contains {1.4, 2, 2.5, 2.7, 14, 16, 49, 57, 58}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's say that you can have maximum of N=4 subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you might end up with a result that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s&lt;sub&gt;1&lt;/sub&gt; = {1.4, 2, 2.5, 2.7}&lt;br/&gt;&#xA;s&lt;sub&gt;2&lt;/sub&gt; = {14, 16}&lt;br/&gt;&#xA;s&lt;sub&gt;3&lt;/sub&gt; = {49}&lt;br/&gt;&#xA;s&lt;sub&gt;4&lt;/sub&gt; = {57, 58}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for either the name what such a problem would be called (which I can then use to research and write an algorithm).  If you can provide a simple solution, even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help with this.&lt;/p&gt;&#xA;" OwnerUserId="2329" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-05T14:59:05.887" LastActivityDate="2010-10-06T06:22:51.217" Title="Grouping numbers that are &quot;close to each other&quot;" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="6126" PostTypeId="2" ParentId="6125" CreationDate="2010-10-05T15:19:13.367" Score="4" ViewCount="0" Body="&lt;p&gt;There are many solutions to this problem extant, each rediscovered many times.  Search &quot;statistical clustering methods&quot; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cluster_analysis&quot; rel=&quot;nofollow&quot;&gt;&quot;cluster analysis.&quot;&lt;/a&gt;  A particularly nice one for your problem is called &lt;a href=&quot;http://en.wikipedia.org/wiki/K-means_clustering&quot; rel=&quot;nofollow&quot;&gt;&quot;K-means&quot;&lt;/a&gt;.  This was rediscovered by cartographers (!) where it is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Jenks_Natural_Breaks_Optimization&quot; rel=&quot;nofollow&quot;&gt;&quot;Jenks' method&quot;&lt;/a&gt; (and is worth mentioning because it's used by some popular mapping software, so perhaps more people have heard of this than of any other method).  The idea behind K-means is to minimize the population-weighted sum of variances of the subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the challenge in most clustering algorithms lies in determining the number of clusters; usually that needs to be specified (as $N = 4$ in this example) or at least hinted at by the user.  For instance, there always exists a minimum K-means solution: partition $S$ into singletons.  If you require $i &amp;lt; n$, merge the two nearest neighbors but leave all the other singletons as they are: the sum of variances is then just one-quarter the square of the smallest difference among the $x_i$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-05T15:58:08.887" LastActivityDate="2010-10-05T15:58:08.887" CommentCount="2" />
  <row Id="6127" PostTypeId="2" ParentId="6117" CreationDate="2010-10-05T15:25:47.943" Score="3" ViewCount="0" Body="&lt;p&gt;By definition, if $f:Y \to X$ is a covering space and $x \in X$,&#xA;then there is some neighbourhood $U$ of $x$ such that $f^{-1}(U)$ is a union of open sets&#xA;$V_i$ such that $f$ restricted to each $V_i$ is a homeomorphism.  In particular, &#xA;$f^{-1}(x) \cap V_i$ consists of a single point, and so each point of $f^{-1}(x)$ is open&#xA;in the induced topology on $f^{-1}(x)$.  Thus, as has already been pointed out, the fibres&#xA;of a covering map are discrete.  (This is not part of the standard definition of covering&#xA;space, but is a consequence of it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, you probably should explain in more detail what you mean by &quot;If we assume ...&quot;.&#xA;What kind of map do you actually want to consider?&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-05T15:25:47.943" />
  <row Id="6128" PostTypeId="1" CreationDate="2010-10-05T15:32:56.283" Score="2" ViewCount="98" Body="&lt;p&gt;I am trying to understand, in as simple terms as possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to define integration for non-orientable manifolds, and&lt;/li&gt;&#xA;&lt;li&gt;why it is impossible to do so using only differential forms.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In particular, I've seen some discussion of using &quot;densities&quot; instead of $n$-forms for integration, but am not really clear on why densities are required.  In other words, is it really impossible to define integration on nonorientable manifolds using forms alone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am of course aware that any $n$-form must vanish somewhere on a nonorientable manifold, so we cannot find a volume form, hence cannot use the standard definition of integration.  I think the reason I'm not finding this answer satisfying is that it is a bit tautological: we can't define integration with respect to volume forms because there are no volume forms. But why must we define integration with respect to a (global) volume form in the first place? Is there really no other way to do it using locally-defined forms? Thinking of a manifold as a collection of local charts is common in geometry, and I'm having trouble understanding why this approach doesn't work in the case of integration.&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="2330" LastEditorDisplayName="" LastEditDate="2010-10-06T15:58:43.900" LastActivityDate="2010-10-06T15:58:43.900" Title="Is the notion of density really needed to define integration on nonorientable manifolds?" Tags="&lt;geometry&gt;&lt;calculus&gt;&lt;differential&gt;" AnswerCount="1" CommentCount="17" FavoriteCount="1" />
  <row Id="6129" PostTypeId="2" ParentId="6080" CreationDate="2010-10-05T15:45:10.037" Score="4" ViewCount="0" Body="&lt;p&gt;$RP^{2k-1}$ is a quotient of a codimension 1 submanifold with induced orientation ($S^{2k-1}$)in $R^{2k}$, by an orientation-preserving transformation of $R^{2k}$.  Because there is a consistent notion of normal vector (i.e., of &quot;inside&quot; and &quot;outside&quot;) on the submanifold, preserving orientation of the surrounding manifold also preserves orientation on the submanifold.  Transporting the orientation along a path between two points of the submanifold that are identified in the quotient can be viewed as transporting the orientation in the surrounding manifold, and this has to be consistent (have positive determinant on the local frames) because of the orientability of the surrounding manifold.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-05T15:45:10.037" />
  <row Id="6132" PostTypeId="2" ParentId="6083" CreationDate="2010-10-05T16:17:52.473" Score="1" ViewCount="0" Body="&lt;p&gt;First you need a definition of $\theta$ (and/or $d \theta$) to compare with the form involving $dx$ and $dy$.    In coordinates any definition is only &quot;local&quot; and limited to regions of $(x,y)$ that exclude loops around $0$ (which would modify $\theta$ by $2\pi$).  In practice any local definition will be as a function of $x/y$ or of $y/x$ according to which line or half-line from $0$ is excluded as a &quot;branch cut&quot;.  e.g., $\theta = \arctan(y/x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta$ is defined using trigonometric functions one then needs a definition of those functions not using angles, or the whole exercise is logically vacuous, defining $\theta$ in terms of $\theta$ (using polar coordinates is an example of this).  An angle-free definition is possible but it is ad hoc if there is no clear relation to geometry, such as defining inverse sine or tangent functions as integrals.  A more &quot;correct&quot; definition is to use arc length on the projection of $(x,y)$ to the unit circle.  Here if $u(x,y)=(x,y)/\sqrt{x^2+y^2}$ is the projection one would have $d\theta = |u(x+dx,y+dy) - u(x,y)|$ up to second order terms in $dx$ and $dy$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-05T16:23:06.303" LastActivityDate="2010-10-05T16:23:06.303" />
  <row Id="6133" PostTypeId="2" ParentId="566" CreationDate="2010-10-05T16:35:27.453" Score="0" ViewCount="0" Body="&lt;p&gt;From New Scientist some years ago:  20 teams play a round robin tournament, each gets 1 point for a win, 0 for a loss, and there are no ties.  Each team's score is a square number.  How many upsets occurred? An upset is defined as team A defeating team B where B scored more total points than A.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-05T16:35:27.453" CommunityOwnedDate="2010-10-05T16:35:27.453" />
  <row Id="6134" PostTypeId="2" ParentId="2273" CreationDate="2010-10-05T17:25:21.780" Score="0" ViewCount="0" Body="&lt;p&gt;I recommend the book &quot;How to prove it&quot; by Daniel Velleman. The associated (free) package for learning formal aspects of proof, called Proof Designer, gives the user some practice in learning how complicated finding such paths can be if one goes completely formal! &lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-10-05T17:25:21.780" />
  <row Id="6135" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T17:45:48.213" Score="3" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Formal Undecidability and related topics (Tarski truth definability, Godel incompleteness, The hydra problem, Matiyasevich's negative solution of the 10th problem)&lt;/li&gt;&#xA;&lt;li&gt;Proof Theory (Gentzen's consistency proof, Girard's linear logic, Barendregt's Lambda Cube)&lt;/li&gt;&#xA;&lt;li&gt;The Continuum (Cantor's diagonalization, Cohen and Godel's independence proof, Robinson's analysis)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-05T17:45:48.213" CommunityOwnedDate="2010-10-05T17:45:48.213" CommentCount="6" />
  <row Id="6136" PostTypeId="1" CreationDate="2010-10-05T17:56:27.483" Score="3" ViewCount="203" Body="&lt;p&gt;Any curve can be considered to be made of sufficiently small straight lines. What is the name of the theorem which states this fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked the same question at &lt;a href=&quot;http://news.ycombinator.com/item?id=1760209&quot; rel=&quot;nofollow&quot;&gt;Hacker News&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: in the Hacker News question, the OP included the following link, which I think clarifies his/her intent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://musr.physics.ubc.ca/~jess/hr/skept/Math/node10.html&quot; rel=&quot;nofollow&quot;&gt;http://musr.physics.ubc.ca/~jess/hr/skept/Math/node10.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--PLC.]&lt;/p&gt;&#xA;" OwnerUserId="2332" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-05T21:52:42.087" LastActivityDate="2010-10-06T01:35:18.803" Title="Is there a mathematical theorem that states the equivalence of curved and straight lines" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="6137" PostTypeId="2" ParentId="6136" CreationDate="2010-10-05T18:37:04.533" Score="1" ViewCount="0" Body="&lt;p&gt;Maybe what you need is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Stone-Weierstrass Theorem&lt;/a&gt;, which you can use to show (roughly) that a continuous function on a compact set can be approximated arbitrarily well by continuous piecewise-linear functions. The boolean ring version of the theorem is directly applicable here.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-05T18:37:04.533" CommentCount="10" />
  <row Id="6138" PostTypeId="2" ParentId="566" CreationDate="2010-10-05T18:40:30.243" Score="0" ViewCount="0" Body="&lt;p&gt;A personal favorite (albeit not ideally-worded): Jamie has a windowbox where he plants a row of iris flowers.  He plants in just two colors - blue and yellow - but never plants two yellow irises next to each other (the result is just too garish).  Assuming that he wants to keep the same number of flowers in his box every day, how many flowers will he need if he wants a different arrangement of blue and yellow every day for a year?&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-05T18:40:30.243" CommunityOwnedDate="2010-10-05T18:40:30.243" />
  <row Id="6139" PostTypeId="2" ParentId="6136" CreationDate="2010-10-05T19:03:37.177" Score="5" ViewCount="0" Body="&lt;p&gt;For many calculations that one might want to perform on a &quot;reasonable&quot; curve (e.g., one coordinatized by functions that are differentiable, or piecewise differentiable/smooth/analytic/nice), taking a finely spaced mesh of points on the curve and computing the result on the polygon joining those points, instead of the curve, will produce an answer close to the one for the curve.  The smaller the spacing, the closer the result will be to the result for the curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantities approximable in this way include area enclosed by a curve, arc length, integrals of given functions along the curve, winding number around a point, splines, parametrizations, and others.  Quantities not approximable in this way include curvature, which will be zero on the sides of any polygon used as a substitute for the curve, and integer-valued &quot;global&quot; quantities such as number of tangent lines.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-05T19:03:37.177" />
  <row Id="6140" PostTypeId="1" AcceptedAnswerId="6144" CreationDate="2010-10-05T19:05:12.547" Score="2" ViewCount="312" Body="&lt;p&gt;Allow me to preface this by saying this is not a homework problem. If I had had this thought four years ago when I was taking calculus, I probably could do it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to calculate the limit as  $n \to \infty$ of $1-\left(\frac{n-1}{n}\right)^n$ -&#xA;- it's a constant I'm inclined to call &quot;natural chance of success&quot;. I have estimated this value to be ~0.632121 but would very much like to see how it could be calculated aside from the brute-force method i employed earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of this would be... consider n = 2 (a coin). You are given two flips of the coin to get what you pick - what is the chance you'll get your chosen outcome, assuming of course it's a fair coin. The best way to go about this would be to say there's a 1/2 chance of failing, and you have 2 flips. This means you have $(1/2)^2$ chance of failure, being 1/4. 1-1/4 is 3/4, so your chance of success here is 3/4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider n = 6 (standard die). You get six rolls to get the number of your choice (assuming a fair die again). Again, you have a 5/6 chance to not get your choice, and 6 rolls at 5/6 chance would be $(5/6)^6$, or ~.334, giving you a ~.665 chance of success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm curious as n increases to infinity, what is your chance of success? Now again, I've estimated this with a double precision float (in Java) to be 0.63212 (actually, this was the point at which it could simply gain no more precision on the value, n = 296536) but this doesn't really give insight to the derivation of the number, merely its value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm hoping someone a little fresher on their integrals than I can help me out here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2333" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T00:18:41.290" LastActivityDate="2010-10-06T02:06:35.027" Title="Help with a specific limit  $\left(  \dfrac{n-1}{n}  \right)^n$ as $n \rightarrow  \infty$" Tags="&lt;probability&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="6141" PostTypeId="1" AcceptedAnswerId="6147" CreationDate="2010-10-05T19:05:23.197" Score="0" ViewCount="36" Body="&lt;p&gt;Hmmm... this question might be a bit low-brow. I'm no mathematician. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that I have an unordered sequence of fuzzy values $T = (t_1 \ldots t_n)$, $T \in (0,1]^n$. Each value is a positive indication towards some observation (hence, the values are not probabilities in that a value of 0.1 does not indicate a value of 0.9 against, just that we have a weak positive affirmation of the observation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take an example, let's say that the observation is &quot;whether or not there has been a hurricane in the US today&quot;: we have evidence from a weather forecast two days ago saying that there would be one today (0.1), there's been a high level of traffic on twitter with the keyword hurricane (0.2) and there's been breaking news reports about a hurricane in Florida (0.7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to formalise a means of aggregating these readings into a final score (the absolute value of the score makes no difference, but the relative value compared to other aggregations does). For what I'm doing, I want each fuzzy value to reduce the cumulative &quot;uncertainty&quot; about the observation. I also need to set an external $max\in(0,1]$ value on the aggregation result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_0(T,max) = 0$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_k(T,max) = (max - F_{k-1})t_k + F_{k-1}$; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(T,max) = F_n(T,max)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the hurricane example, let $max = 1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_0 = 0$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_1 = (1 - 0) * 0.1 + 0$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_2 = (1 - 0.1) * 0.2 + 0.1 = 0.28$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_3 = (1 - 0.28) * 0.7 + 0.28 = 0.784$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This aggregation is nice (for my scenario) because it rewards high fuzzy values better than &#xA;multiple medium or low values, it's order indepedent, and the answer is in $(0,max]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a lot of trouble formalising this fairly simple idea... don't have the necessary know-how or lingo (probably clear by now). So I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Has anyone seen a similar such aggregation somewhere?&lt;/li&gt;&#xA;&lt;li&gt;How should I call this sequence of values, where the order is not important, and duplicates should be maintained? An unordered sequence? A set with duplicates? An unordered n-tuple?&lt;/li&gt;&#xA;&lt;li&gt;How should I formulate this function? Should I formulate it for an unordered pair first, and then for an unordered tuple? Is there a neat way of demonstrating commutativity?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any help or nudges in the right direction would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2331" LastActivityDate="2010-10-05T20:33:41.057" Title="Aggregating multiple fuzzy values for a given observation" Tags="&lt;set-theory&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="6142" PostTypeId="1" AcceptedAnswerId="6184" CreationDate="2010-10-05T19:16:53.577" Score="1" ViewCount="53" Body="&lt;p&gt;Let $P$ be some partial order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that $x$ and $y$ are &lt;em&gt;compatible&lt;/em&gt; if $\exists r\in P (r\le x \wedge r\le y)$, we denote this by $x \perp y$. Otherwise, we say that $x$ and $y$ are &lt;em&gt;incompatible&lt;/em&gt;.&lt;br&gt;&#xA;An order $P$ is called &lt;em&gt;separative&lt;/em&gt; if whenever $\neg (x \le y)$ then there exists some $r\le x$ which incompatible with $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Jech's &quot;Set Theory&quot; he proves a lemma (14.11) which construct a separative quotient, $Q$, and a mapping $h$ for a (general) partially ordered set, $P$, that satisfies the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x\le y \Rightarrow h(x) \preceq h(y)$&lt;/li&gt;&#xA;&lt;li&gt;$x \perp y \iff h(x) \perp h(y)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And $h$ is onto $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction is by taking a quotient of $P$ over the equivalence relation $x \sim y \iff \forall z(z \perp x \leftrightarrow z \perp y)$, and $[x] \preceq [y] \iff \forall z \le x(z \perp y)$. (And obviously enough, $h(x) = [x]$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today I was trying to prove the following lemma (which does not appear on Jech's book):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be some partial order, and $Q$ its separative quotient. If $[x] \preceq [y]$ then there exist $x' \in [x], y' \in [y]$ such that $x' \le y'$.&lt;br&gt;&#xA;(That is to say that the separative mapping is &quot;almost&quot; order-preserving.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively it seems right, but it might very well be a not-common misbelief. Any hints, partial proofs or complete proofs (granted that I haven't proved it myself yet - in which case I'll rush to the nearest computer and update) will be most welcomed.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-10-05T19:46:33.620" LastActivityDate="2010-10-06T07:53:22.927" Title="Separative Quotients, and the Induced Order" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6143" PostTypeId="2" ParentId="6140" CreationDate="2010-10-05T19:26:30.593" Score="3" ViewCount="0" Body="&lt;p&gt;Your question is motivated by probability, but in itself it's pure and elemental calculus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\to \infty} (1 - \frac{1}{x})^x  = e ^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that, your number is 0.63212...&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-05T19:26:30.593" />
  <row Id="6144" PostTypeId="2" ParentId="6140" CreationDate="2010-10-05T19:29:15.100" Score="9" ViewCount="0" Body="&lt;p&gt;You are essentially trying to find the limit of $\left(1-\frac{1}{n}\right)^n$ as $n\to\infty$. Rewrite this as $e^{n\ln(1 - \frac{1}{n})}$. Since the exponential function is continuous, you have&#xA;$$\lim_{n\to\infty} e^{n\ln(1-\frac{1}{n})} = e^{\lim_{n\to\infty}(n\ln(1-\frac{1}{n})}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute that limit, rewrite and use L'Hopital's:&#xA;$$\begin{array}{rcl}&#xA;\lim_{n\to\infty}n\ln\left(1 - \frac{1}{n}\right) &amp;amp; = &amp;amp; \lim_{n\to\infty}\frac{\ln(1-\frac{1}{n})}{\frac{1}{n}}\\&#xA;&amp;amp; = &amp;amp; \lim_{n\to\infty} \frac{\left(\frac{1}{1-\frac{1}{n}}\right)\left(1-\frac{1}{n}\right)'}{(n^{-1})'}\\&#xA;&amp;amp; = &amp;amp; \lim_{n\to\infty}\frac{\quad\frac{n^{-2}}{1-\frac{1}{n}}\quad}{-n^{-2}}\\&#xA;&amp;amp; = &amp;amp; \lim_{n\to\infty}\frac{-1}{1-\frac{1}{n}} = -1.&#xA;\end{array}$$&#xA;So, since you wanted $1$ minus this limit, your limit equals $1-e^{-1}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-05T19:29:15.100" CommentCount="4" />
  <row Id="6145" PostTypeId="1" AcceptedAnswerId="6154" CreationDate="2010-10-05T20:10:40.073" Score="1" ViewCount="98" Body="&lt;p&gt;Suppose we have an atom at every point with integer coordinates in $\mathbb{R}^d$. Take a ($d-1$)-dimensional hyperplane going through $\mathbf{0}$ and orthogonal to $(1,1,1,\ldots)$. What is the name of the lattice formed by atoms in that plane?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-10-05T21:57:46.430" Title="What is the name of this lattice?" Tags="&lt;geometry&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6146" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T20:31:42.037" Score="2" ViewCount="0" Body="&lt;p&gt;Paul &lt;strong&gt;Benacerraf's identification problem&lt;/strong&gt; is an argument against Platonism: Many different models satisfy Peano's arithmetic axioms. Which of these are really the Natural numbers? I.e., the uniquely intended set? We are unable to pick one account to the exclusion of all the others. Adding axioms won't help. Any such axiomatic system will be satisfied by multiple models. For example, is the number Three given by $3 = \{ \emptyset, \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}$ or by $3 = \{ \{ \{ \emptyset \} \} \}$? Benacerraf concludes that numbers are not sets, as most mathematicians believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent trend for dealing with Benacerraf's identification problem is &lt;strong&gt;fictionalism&lt;/strong&gt;, a form of structuralism: &quot;Fictionalism holds that mathematical theories are like fiction stories such as fairy tales and novels. Mathematical theories describe fictional entities, in the same way that literary fiction describes fictional characters.&quot; See &lt;a href=&quot;http://plato.stanford.edu/entries/philosophy-mathematics/#Fic&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/philosophy-mathematics/#Fic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fictionalists include Hartry Fields, John Burgess, and Charles Chihara.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-05T20:31:42.037" CommunityOwnedDate="2010-10-05T20:31:42.037" CommentCount="2" />
  <row Id="6147" PostTypeId="2" ParentId="6141" CreationDate="2010-10-05T20:33:41.057" Score="0" ViewCount="0" Body="&lt;p&gt;I'm at least acquainted with the idea of fuzzy logic, as distinct from probabilities. Having said that: while you are not looking for a probabilistic sort of formula, you have stumbled upon one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking &lt;em&gt;max&lt;/em&gt;&amp;nbsp;=&amp;nbsp;1, your formula can be finessed to the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;F&lt;sub&gt;k+1&lt;/sub&gt;&lt;br&gt;&amp;nbsp;&amp;nbsp;=&amp;nbsp;(1&amp;nbsp;&amp;minus;&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;)&amp;thinsp;t&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;+&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;&lt;br&gt;&amp;nbsp;&amp;nbsp;=&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;+&amp;nbsp;t&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;&amp;minus;&amp;nbsp;F&lt;sub&gt;k&lt;/sub&gt;t&lt;sub&gt;k&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is the same thing as the &quot;probability&quot; of at least one of event A or event B happening, where A and B are independent random events, occurring with probability F&lt;sub&gt;k&lt;/sub&gt; and t&lt;sub&gt;k&lt;/sub&gt; respectively. &amp;mdash; This is why your construction is order independent (and in fact can be proven so using this observation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say the same thing in different terms: if you take the &quot;fuzzy AND&quot; (logical conjunction) of two qualities with fuzzy strengths &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; to be the product &lt;em&gt;xy&lt;/em&gt;, then the formula &lt;em&gt;x&lt;/em&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;y&lt;/em&gt;&amp;nbsp;&amp;minus;&amp;nbsp;&lt;em&gt;xy&lt;/em&gt; is the corresponding &quot;fuzzy OR&quot; which will satisfy de Morgan's laws, &lt;em&gt;i.e.&lt;/em&gt; NOT(&lt;em&gt;x&lt;/em&gt;&amp;nbsp;AND&amp;nbsp;&lt;em&gt;y&lt;/em&gt;) = (NOT&amp;nbsp;&lt;em&gt;x&lt;/em&gt;)&amp;nbsp;OR&amp;nbsp;(NOT&amp;nbsp;&lt;em&gt;y&lt;/em&gt;), if you take NOT&amp;nbsp;&lt;em&gt;x&lt;/em&gt;&amp;nbsp;=&amp;nbsp;(1&amp;nbsp;&amp;minus;&amp;nbsp;&lt;em&gt;x&lt;/em&gt;). Of course, you may not want this if you don't want a small positive fuzzy value to represent a large fuzzy value for its negation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that whatever you choose &lt;em&gt;max&lt;/em&gt; to be, you can re-scale it to get identical behaviour to the case &lt;em&gt;max&lt;/em&gt;&amp;nbsp;=&amp;nbsp;1. So the utility of this interpretation is unlikely to hinge on that particular parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your other questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A collection in which number is important, but order is not, is a &lt;em&gt;bag&lt;/em&gt; or &lt;em&gt;multi-set&lt;/em&gt; (these are synonyms). Furthermore, that's all that it is; they're basically &quot;set-like collections&quot; which are equivalent to a histogram.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I would call the function &quot;fuzzy OR&quot; or &quot;fuzzy disjunction&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to make the fuzzy OR of a collection (a &quot;bag&quot;) well-defined, it suffices to show commutativity (which makes it well-defined for bags of size at most two), and then show that it is associative &amp;mdash; that is, that OR(&lt;em&gt;x&lt;/em&gt;,&amp;nbsp;OR(&lt;em&gt;y&lt;/em&gt;,&lt;em&gt;z&lt;/em&gt;))&amp;nbsp;=&amp;nbsp;OR(OR(&lt;em&gt;x&lt;/em&gt;,&lt;em&gt;y&lt;/em&gt;),&amp;nbsp;&lt;em&gt;z&lt;/em&gt;) &amp;mdash; which ensures that if it is well-defined for bags of size &lt;em&gt;k&lt;/em&gt;, it is also well-defined for bags of size &lt;em&gt;k&lt;/em&gt;&amp;nbsp;+&amp;nbsp;1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-05T20:33:41.057" CommentCount="3" />
  <row Id="6148" PostTypeId="1" AcceptedAnswerId="6149" CreationDate="2010-10-05T20:44:21.713" Score="3" ViewCount="102" Body="&lt;p&gt;There's a theorem that every planar graph can be colored with 4 colors in such a way that no 2 adjacent vertices have the same color. Is the opposite true as well?&lt;/p&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2010-10-05T20:46:37.347" Title="If a graph can be colored with max 4 colors, is it planar?" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="6149" PostTypeId="2" ParentId="6148" CreationDate="2010-10-05T20:46:37.347" Score="13" ViewCount="0" Body="&lt;p&gt;No. Consider $K_{3,3}$, the graph with two sets of 3 vertices each such that every vertex in one set is connected to every vertex in the other. It's not planar but can be colored with just 2 colors. More generally, take any dense bipartite graph - it's still 2-colorable, but far from planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A picture of $K_{3,3}$ (along with $K_5$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/fowUR.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-10-05T20:46:37.347" CommentCount="2" />
  <row Id="6150" PostTypeId="2" ParentId="6140" CreationDate="2010-10-05T20:47:37.300" Score="6" ViewCount="0" Body="&lt;p&gt;If you know one of the definitions for the number $e$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e=\lim_{n\to\pm\infty}\Big(1+\frac{1}{n}\Big)^n,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you could calculate your limit in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{n\to\infty}\Big(1-\frac{1}{n}\Big)^n =&#xA;\lim_{n\to\infty}\left[\Big(1+\frac{1}{-n}\Big)^{-n}\right]^{-1} =&#xA;\frac{1}{e}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-10-05T20:47:37.300" />
  <row Id="6151" PostTypeId="2" ParentId="6101" CreationDate="2010-10-05T21:07:35.083" Score="1" ViewCount="0" Body="&lt;p&gt;George Lakoff and Rafael Núñez started studying &lt;i&gt;embodied mathematics&lt;/i&gt; pretty recently.  The idea is that our ideas of mathematics are inextricable from our humanity as opposed to being platonic truths, and can all be understood in terms of metaphors for real-world concepts and our learning/acquisition of these metaphors.  To take a simple example, whenever we add numbers, our brains will always essentially be adding things to a pile, because that's what addition is to us.  If I've understood it correctly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their theories really haven't caught on among mathematicians, in part because they're not mathematicians themselves, which is honestly a pretty big weakness among philosophers of math.  In my opinion, a lot of what they say is really silly (my brain has nothing to do with $\mathbb{R}$ being the unique completion of $\mathbb{Q}$ as well as the largest Archimedean field), but there is a kernel of truth there: we don't study math by taking logical step after logical step, we study it by thinking about things we have ways of comprehending.  For instance, we live in three-dimensional space, and so it's hard to study higher-dimensional things that we therefore can't visualize.  I think it would be a useful philosophical project to understand how the mathematicians who study such objects conceptualize them, and how &lt;i&gt;that&lt;/i&gt; relates to our human nature, but that would obviously require a larger mathematical background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important question is the origin of mathematical taste and beauty.  This is probably more important to mathematicians than to outsiders who don't have as much of a sense of it.  (Yet I'm convinced that a cool proof can be appreciated by anyone, if it's explained properly!)  I don't know who, if anyone, has written about this, but it definitely exists and is worthy of study.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-05T21:07:35.083" CommunityOwnedDate="2010-10-05T21:07:35.083" />
  <row Id="6152" PostTypeId="1" CreationDate="2010-10-05T21:46:15.077" Score="0" ViewCount="86" Body="&lt;p&gt;Suppose we have a linear ODE of the form $\dfrac{dy}{dt}= Ay$. We know that $y= e^{tA}  y(0)$ is a solution to the ODE, where $e^{tA}$ represents the matrix exponential. Suppose we also knew that this is stable in the sense of Lyapunov (i.e for any $\varepsilon &gt;0$, there exists a  $\delta &gt; 0$  such that if $\vert y(0) \vert &amp;lt; \delta$, then $\vert y(t) \vert &amp;lt; \varepsilon$ for all time $t &gt; 0$). How does that affect the norm of the matrix $e^{tA}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so the actual problem is as follows. If a linear ODE $y'= Ay$ is Lyapunov stable and if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ also Lyapunov stable? &lt;/p&gt;&#xA;" OwnerUserId="1127" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T00:15:41.610" LastActivityDate="2010-10-06T00:15:41.610" Title="Matrix exponential and Stability of Linear ODE" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="6153" PostTypeId="1" CreationDate="2010-10-05T21:55:34.610" Score="1" ViewCount="135" Body="&lt;p&gt;Hey, sorry for the non so mathematical question but that's the only place where I'd thought someone would answer this so here we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing a chemistry homework and I have to answer this very question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If 50 grams of a particular mineral containing copper nitrate (II) gave, after appropriate treatment 1.6933 grams of pure copper. For this ore, determine the percentage content by weight of copper and the percentage content by weight of copper nitrate (II) in ore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone help with this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, any help will be very much appreciated :)&lt;/p&gt;&#xA;" OwnerUserId="2338" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-05T22:48:48.817" LastActivityDate="2010-10-05T23:58:20.343" Title="Chemistry: Gravimetric determination of copper in ore" Tags="&lt;homework&gt;&lt;chemistry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6154" PostTypeId="2" ParentId="6145" CreationDate="2010-10-05T21:57:46.430" Score="4" ViewCount="0" Body="&lt;p&gt;It's called $A_{d-1}$.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Root_system#An&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Root_system#An&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-05T21:57:46.430" CommentCount="4" />
  <row Id="6155" PostTypeId="1" AcceptedAnswerId="6156" CreationDate="2010-10-05T22:25:59.347" Score="2" ViewCount="229" Body="&lt;p&gt;Hi. I'm having a little trouble with the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate $F'(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says we have to use substitution but I don't see why the answer can't just be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x-\sin^{2}x$&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-05T22:43:07.457" LastActivityDate="2010-12-01T21:17:49.570" Title="Derivative of Integral" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="6156" PostTypeId="2" ParentId="6155" CreationDate="2010-10-05T22:42:17.653" Score="10" ViewCount="0" Body="&lt;p&gt;If you had $G(x)=\int_{1}^{x}(t-\sin^{2}t) dt$ (note that it's just $x$ at the upper limit of the integral), then $G'(x)=x-\sin^2x$.  But, you have $F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt=G(x^2)$, so $F'(x)=G'(x^2)\cdot2x=(x^2-\sin^2x^2)2x$ (using the chain rule).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-05T22:42:17.653" CommentCount="1" />
  <row Id="6157" PostTypeId="2" ParentId="6155" CreationDate="2010-10-05T22:44:27.940" Score="1" ViewCount="0" Body="&lt;p&gt;Because the upper limit of the integral is x^2, not x.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-05T22:44:27.940" />
  <row Id="6158" PostTypeId="2" ParentId="6152" CreationDate="2010-10-05T22:45:36.090" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;OK, so the actual problem is as follows. If a linear ODE $y'= Ay$ is Lyapunov stable and if $B= G^{-1}AG$ where $G$ is an invertible matrix, is $B$ also Lyapunov stable?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, is the system $z'=Bz$ Lyapunov stable around the origin? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the two systems are related by a change of basis. If $y=Gz$, then the system $y'= Ay$ can be written as $(Gz)' = A(Gz)$, that is, $z'=(G^{-1}AG)z = Bz$. And we know that $|y| = |Gz| \leq ||G||. |z|$, where $||G|| := sup_{|x|=1}|G.x|$ is the norm of the matrix $G$ (prove it!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $G$ is invertible, $z=G^{-1}y$, an analogous argument leads to $|z| \leq ||G^{-1}||.|y|$. I think these inequalities can help you solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-10-05T22:45:36.090" />
  <row Id="6159" PostTypeId="2" ParentId="5877" CreationDate="2010-10-05T22:55:01.363" Score="2" ViewCount="0" Body="&lt;p&gt;Below is an algorithm to compute a representation $\rm\ p = a^2 + b^2\ $ for an integer $\rm\ p = 1\ \: (mod\ 4)\:.\ \ $ The algorithm is coded in $\ $ &lt;a href=&quot;http://en.wikipedia.org/wiki/Macsyma&quot; rel=&quot;nofollow&quot;&gt;$\rm Macsyma$&lt;/a&gt; $\ $ but it can easily be translated into almost any another language. The only Macsyma specific subroutines that you might need to implement in another language are exponentiation $\rm (mod\ p)$ (by repeated squaring) and the floor function in $\rm \mathbb Q[\:i\:]\:$. While this algorithm is reasonably efficient, the point is not to optimize efficiency but, rather, to emphasize the simplicity of the solution employing the Euclidean algorithm in $\rm\ \mathbb Z[\:i\:]\:.\ $ For the underlying theory see my &lt;a href=&quot;http://math.stackexchange.com/questions/5895&quot;&gt;companion post&lt;/a&gt;. For a faster algorithm see e.g. Collins, &lt;a href=&quot;http://www.csie.nuk.edu.tw/~cychen/gcd/A%20Fast%20Euclidean%20Algorithm%20for%20Gaussian%20Integers.pdf&quot; rel=&quot;nofollow&quot;&gt;A fast Euclidean algorithm for Gaussian integers&lt;/a&gt;, and see also Agarwal and Frandsen, &lt;a href=&quot;http://www.springerlink.com/content/j4up5k6v8n7ujh6d/&quot; rel=&quot;nofollow&quot;&gt;Binary GCD Like Algorithms for Some Complex Quadratic Rings.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/* $\:$ For $\rm\ p = 1\ \: (mod\ 4)\ $ compute $\rm\ \alpha = a + b\:i\ $ with $\rm \ p = \alpha\:\alpha' = a^2 + b^2\ \ $*/  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\text {psplit(p) := block([}\ \:\text{j : 0,}\ \ \text{e : floor(p/4),}\ \ \text{algebraic : true ], }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\text{block([ modulus : p ],}\quad\quad\quad\quad\ $ /* search for $\rm\ j = \sqrt{-1}\ \ (mod\ p)\ $ */&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\ \text{for}\ \: \text{ a : 2}\ \ \text{while}\ \ \text{j^2 # -1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\ \ \text{do}\ \: \text{ j : rat(a)^e ),}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \text{x : p,}\ \ \: \text{y : j - %i,} \quad\quad\quad\quad\quad\quad\quad $ /* find $\rm\  gcd(p,\:j - i)\ \in\ \mathbb Z[\:i\:]\ $ by Euclidean algorithm */&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \text{do ( if}\ \text{  y = 0}\ \ \text{then}\ \text{ return(x), }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \ \text{j : rat(x - y * floor(rat(x/y))), }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \ \text{x : y,}\ \text{ y : j )) }$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T13:33:13.530" LastActivityDate="2010-10-06T13:33:13.530" />
  <row Id="6160" PostTypeId="2" ParentId="6090" CreationDate="2010-10-05T22:57:30.667" Score="7" ViewCount="0" Body="&lt;p&gt;The term in analytic number theory is &quot;zero-free regions&quot;.  Any proof of the prime number theorem will produce such a region, and the region is equivalent to the error term in bounds for $\pi(x)$ and to the lower bounds in nonvanishing theorems $|\zeta (1+it)|&gt; 0$. At present, the known zero-free regions are asymptotic to the line $Re(s)=1$: at height $h$ all zeros are at least at distance $d(h)$ from the line with $\lim_{|h| \to \infty} d(h) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Added: on the subject of later improvements, if any region not asymptotic to $Re(s)=1$ were demonstrated it would be a giant advance in number theory, comparable to Wiles' proof of the modularity conjecture and Fermat's Last Theorem. In the analogous function field case there is an algebraic technique for boosting Beta &amp;lt; 1 to Beta &amp;lt; (1/2)+epsilon, and the latter is the Riemann hypothesis.  You can be sure that if a zero free region were proven that had finite distance from the boundary of the critical strip, we would all have heard about it.  Some known bounds are listed at &lt;a href=&quot;http://www.math.uiuc.edu/~ford/wwwpapers/zeros.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/~ford/wwwpapers/zeros.pdf&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-08T19:09:29.930" LastActivityDate="2010-10-08T19:09:29.930" CommentCount="1" />
  <row Id="6161" PostTypeId="1" CreationDate="2010-10-05T23:19:19.427" Score="1" ViewCount="126" Body="&lt;p&gt;&lt;B&gt;Ahem. I can only now laugh at how blind I am to not see such a simple answer, LMAO. Thanks to Isaac ♦, Paul VanKoughnett, and Bill Dubuque! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case closed!&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you factor  $6x^2 + 18x - 60$  over a set of complex numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I did it myself, I had  $6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not 100% sure about my answer, so I am asking anyone who knows how to do this to help me confirm the answer! (:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;- EDIT -&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, first thing that I did was to factor out 6.&#xA;$6(x^2 + 3x - 10)$ , then I changed the expression into a perfect square, so : &#xA;$6(x^2 - 3x + 2.25) - 12.25$ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that:  $6(x - 1.5)^2 + (-12.25)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I took the square roots of that, so I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6(1 - 1.5)^2 + 3.5i)$  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus my answer:  $6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I see that it is wrongggggg~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher did not go over it much; she was rushing throughout the whole lesson and her accent does not make it any easier. The textbook does not have anything on it. I looked around the index and there is nothing. -&lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;em&gt;_&lt;/em&gt;_&lt;/strong&gt;-&lt;/p&gt;&#xA;" OwnerUserId="2339" LastEditorUserId="2339" LastEditorDisplayName="" LastEditDate="2010-10-05T23:49:11.580" LastActivityDate="2010-10-06T00:11:51.377" Title="Factoring over a set of complex numbers . " Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="6162" PostTypeId="2" ParentId="6155" CreationDate="2010-10-05T23:23:20.867" Score="3" ViewCount="0" Body="&lt;p&gt;You could apply the following more general rule (&lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_integral_rule&quot; rel=&quot;nofollow&quot;&gt;Leibnitz rule&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule&quot; rel=&quot;nofollow&quot;&gt;chain rule&lt;/a&gt;) -- although Isaac's answer is more straightforward for this derivative: If &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(x)=J(u,v,x)=\displaystyle\int_{u(x)}^{v(x)}f(t,x)dt,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, under suitable conditions, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I^{\prime }(x)=\displaystyle\int_{u(x)}^{v(x)}\dfrac{\partial f(t,x)}{\partial x}dt+f(v(x),x)v^{\prime }(x)-f(u(x),x)u^{\prime }(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x)=I(x)=\displaystyle\int_{1}^{x^{2}}(t-\sin ^{2}t)dt,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(x)=1,v(x)=x^{2},f(t,x)=(t-\sin ^{2}t),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial f(t,x)}{\partial x}=0,v^{\prime }(x)=2x,u^{\prime }(x)=0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F^{\prime }(x)=I^{\prime }(x)=(x^{2}-\sin ^{2}x^{2})2x=2x^{3}-2x\sin&#xD;&#xA;^{2}x^{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The conditions are: $u(x),v(x)$ are differential functions, $f(t,x)$ is a&#xA;continuous function and $\partial f/\partial x$ exists and is continuous.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T21:17:49.570" LastActivityDate="2010-12-01T21:17:49.570" />
  <row Id="6163" PostTypeId="2" ParentId="6161" CreationDate="2010-10-05T23:26:38.523" Score="0" ViewCount="0" Body="&lt;p&gt;$$\begin{align}&#xA;6x^2+18x-60&amp;amp;=6(x^2+3x-10)&#xA;\\&#xA;&amp;amp;=6(x+5)(x-2)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I expand yours:&#xA;$$6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)=6x^2-18x+76.875$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm not entirely sure what went wrong.  How did you arrive at your answer?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: Given your work, it's worth having another look.  You factored out the 6, as I did.  Then, you went to complete the square, which is reasonable, but you somehow went from $+3x$ to $-3x$ and there are some missing/misplaced parentheses:&#xA;$$\begin{align}&#xA;6(x^2+3x-10)&amp;amp;=6(x^2+3x+2.25-12.25)&#xA;\\&#xA;&amp;amp;=6((x+1.5)^2-12.25)&#xA;\\&#xA;&amp;amp;=6((x+1.5)^2-(3.5)^2)&#xA;\\&#xA;&amp;amp;=6(x+1.5+3.5)(x+1.5-3.5)&#xA;\\&#xA;&amp;amp;=6(x+5)(x-2)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize, the ideas you tried to use would have worked, but there were one or more instances where something went wrong with +/-, and you should be careful with parentheses and the factored-out leading coefficient (the 6).&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-06T00:11:51.377" LastActivityDate="2010-10-06T00:11:51.377" CommentCount="3" />
  <row Id="6164" PostTypeId="2" ParentId="6161" CreationDate="2010-10-05T23:27:32.063" Score="0" ViewCount="0" Body="&lt;p&gt;Polynomials over the complex numbers still have the unique factorization property, so if a polynomial has a real factorization, that will also be its complex factorization.  This also means that you can go about factorizing in the same way you would over the reals.  The ways you already know for quadratics (quadratic formula, completing the square, etc.) still work fine, except now you keep complex answers instead of throwing them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check your answer, just multiply the factors back together, using complex multiplication.  In this case, your answer isn't right, since $(1.5-3.25i)(1.5+3.25i)=1.25^2+3.25^2=12.8125$.  In fact, this polynomial has real roots, which I'm sure you can find if you look at $6(x^2+3x-10)$ for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a reminder, a quadratic has real roots if and only if its discriminant is nonnegative.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: appears somebody else got to the answer first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt; in light of your most recent edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you really understand completing the square.  That's actually okay, since you can do everything with the quadratic formula, but at your level in school you'll probably get tested on that specific technique, so you might as well learn it.  Also, it's how you derive the quadratic formula in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, you made three mistakes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you had $x^2-3x-10$ instead of $x^2+3x-10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you forgot some parentheses when you completed the square.&#xA;$$6(x^2-3x-10)=6((x^2-3x+2.25)-12.25)=6(x-1.5)^2-73.5.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, &quot;taking the square root&quot; didn't really work out.  Here, it's better to work with the equation $f(x)=0$ than just the expression $f(x)$.  Then, you can move the $12.25$ to the other side of the equation and take the square root, giving $\pm 3.5$, not $\pm 3.5i$.  (Also, you changed $3.5i$ to $3.25i$ on the way back.)  When you add it back in, this gives you the expected answer, $6(x-5)(x+2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That isn't the answer to the equation the first way you wrote it, but I can't tell which is the right equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the moral is that you can check your work at nearly any step of these types of problems.  If you do this frequently, you can avoid little mistakes like these.  Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-06T00:08:09.243" LastActivityDate="2010-10-06T00:08:09.243" CommentCount="2" />
  <row Id="6165" PostTypeId="2" ParentId="6161" CreationDate="2010-10-05T23:29:09.157" Score="1" ViewCount="0" Body="&lt;p&gt;The answer is wrong since $\rm\ 3.25^2 \ne 10\:\ $ It appears that you made the error of computing the discriminant as $\rm\ 9^2 - 24\cdot 60\ = 39^2\ $ instead of $\rm\ 18^2 -24\cdot 60 = 42^2\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simpler: apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation&quot; rel=&quot;nofollow&quot;&gt;quadratic formula&lt;/a&gt; to $\rm\ p(x)/6 = x^2 + 3x -10\:$, or apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root test&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-05T23:51:59.420" LastActivityDate="2010-10-05T23:51:59.420" />
  <row Id="6167" PostTypeId="2" ParentId="6153" CreationDate="2010-10-05T23:58:20.343" Score="1" ViewCount="0" Body="&lt;p&gt;To determine the percentage of copper (II) nitrate in the ore, you need to use a so-called &quot;gravimetric factor&quot;, which converts &quot;% Cu&quot; into &quot;% Cu(NO&lt;sub&gt;3&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;&quot;. Since the two have the same amount of copper on a molar basis, multiplying the percentage Américo got with an appropriate gravimetric factor will give you the percent by weight of copper nitrate in the ore.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-05T23:58:20.343" CommentCount="1" />
  <row Id="6168" PostTypeId="1" AcceptedAnswerId="6169" CreationDate="2010-10-06T00:56:01.987" Score="3" ViewCount="141" Body="&lt;p&gt;I was wondering what are the general spaces that the concepts limsup and liminf can apply to?&#xA;Is complete lattice one of them? Also How about metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are limsup and liminf specified with respect to? A subset? A sequence/net/filter base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many kinds of definitions for limsup and liminf in these various cases? Are they equivalent? If not, what are the conditions for them to be equivalent?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-06T05:55:26.863" LastActivityDate="2010-10-06T05:55:26.863" Title="Definitions for limsup and liminf " Tags="&lt;definition&gt;" AnswerCount="1" />
  <row Id="6169" PostTypeId="2" ParentId="6168" CreationDate="2010-10-06T01:01:23.167" Score="1" ViewCount="0" Body="&lt;p&gt;The notion of a $\limsup$ of a filtered directed set makes sense. Namely, let $A$ be a filtered directed set and $x_\alpha, \alpha \in A$ be an $A$-indexed family in $\mathbb{R}$. Then one can define the $\limsup$ as the infimum of $\sup_{\beta &gt; \alpha} x_{\beta}$ over all $\alpha$. This makes sense for the $\liminf$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One needs the ordering of the range set to define the limsup, though.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-06T01:01:23.167" />
  <row Id="6170" PostTypeId="2" ParentId="6136" CreationDate="2010-10-06T01:35:18.803" Score="3" ViewCount="0" Body="&lt;p&gt;&quot;Calculus is the acknowledgment that anything but a linear function is far too complicated for us to handle.  Differentiation provides the techniques to tame functions by making them locally linear, and &lt;strong&gt;integration comprises the rules for sticking the local pieces together again&lt;/strong&gt;.&quot; -- Alf van der Poorten, &lt;em&gt;Notes on Fermat's Last Theorem&lt;/em&gt; (Lecture IX).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point of view, one might nominate the &lt;em&gt;Fundamental Theorem of Calculus&lt;/em&gt; (to which van der Poorten is alluding) as an answer to this question.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-06T01:35:18.803" CommentCount="2" />
  <row Id="6171" PostTypeId="1" CreationDate="2010-10-06T01:41:23.887" Score="2" ViewCount="102" Body="&lt;p&gt;I was under the impression that all 2-forms are the wedge (^) of two 1-forms. Is it possible to have a 2-form that you can't write as A^B with A,B 1-forms?&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-06T02:45:09.253" LastActivityDate="2010-10-06T02:45:09.253" Title="Can you find a 2-form not written as the wedge of two 1-forms?" Tags="&lt;exterior-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6172" PostTypeId="2" ParentId="6171" CreationDate="2010-10-06T01:44:26.640" Score="9" ViewCount="0" Body="&lt;p&gt;Yes, it is possible. (And you should find an example yourself: I will not deprive you of the joy of finding it :) )&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-06T01:44:26.640" CommentCount="11" />
  <row Id="6173" PostTypeId="2" ParentId="6140" CreationDate="2010-10-06T01:49:52.157" Score="4" ViewCount="0" Body="&lt;p&gt;If you expand $\lim_{n\to\infty}\Big(1-\frac{1}{n}\Big)^n,$ by using the binomial formula, you get $1 - \frac{n}{n} +  \frac{n*(n-1)}{2*n^2} -  \frac{n*(n-1)*(n-2)}{6*n^3} + ...$&#xA;As n goes to infinity, this approaches $1 - \frac{n}{n} +  \frac{n*n}{2*n^2} -  \frac{n*n*n}{6*n^3} + ... $, which equals $1 - 1 +  \frac{1}{2} -  \frac{1}{6} + ...$ This is exactly the Taylor expansion of $e^x$ when $x=-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, wolfram alpha can actually match constants to approximations, like so:&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=0.632121&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=0.632121&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That may be a resource that you could find use for in the future.&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="796" LastEditorDisplayName="" LastEditDate="2010-10-06T02:06:35.027" LastActivityDate="2010-10-06T02:06:35.027" />
  <row Id="6174" PostTypeId="1" AcceptedAnswerId="6175" CreationDate="2010-10-06T01:52:40.920" Score="1" ViewCount="60" Body="&lt;p&gt;I'm studying for an exam an I came across a problem that I am having difficultly solving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{F}$ is a family of analytic functions on the closed unit disc, $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{D} |f|^{2} dA \le 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $f \in \mathcal{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I conclude that $\mathcal{F}$ is locally bounded?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-06T02:50:04.013" Title="Locally bounded Family" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6175" PostTypeId="2" ParentId="6174" CreationDate="2010-10-06T02:50:04.013" Score="3" ViewCount="0" Body="&lt;p&gt;Yes. Cauchy's formula states that $$f(z) = \frac{1}{2\pi} \int_{\theta} f(z + re^{i \theta}) d\theta$$&#xA;which gives an expression for $f(z)$ in terms of the average around a circle. &#xA;By integration it follows that &#xA;$$ f(z) = \frac{1}{r_0^2 \pi} \int_{r \leq r_0, \theta} f(z+re^{i \theta}) r dr d\theta$$&#xA;which implies that if the square integral of $f$ is bounded, then (by Cauchy-Schwarz) $f$ is locally bounded by a bound depending on the square integral of $f$ and the circle in question. From this your claim follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the point is that $f$ is the average of its values in a neighborhood. Note that this implies that the space of holomorphic functions in an open set $U$ such that $\int_{U} |f|^2$ is finite is actually a Hilbert space (i.e., complete) under the usual inner product. &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-06T02:50:04.013" CommentCount="2" />
  <row Id="6176" PostTypeId="2" ParentId="6090" CreationDate="2010-10-06T03:53:13.073" Score="2" ViewCount="0" Body="&lt;p&gt;I would suggest you read the book &quot;Riemann's Zeta Function&quot; by H.M.Edwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question, this is one theorem (from the above book)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;De La Vallee Poussin's Theorem (1899)&lt;/strong&gt; : There exist constants $c &gt; 0, K &gt; 1$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \beta &amp;lt; 1 - \frac{c}{\log \gamma}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all roots $\rho = \beta + i \gamma$, in the range $\gamma &gt; K$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure there will be more such theorems (the book itself mentions that the above theorem has been improved upon). The book also mentions that (as of 1974) that the bound $\beta &amp;lt; 1$ (real part of the root) has not been improved!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-06T03:53:13.073" CommentCount="2" />
  <row Id="6177" PostTypeId="2" ParentId="6060" CreationDate="2010-10-06T03:55:20.780" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Edit: It appears an identical idea has, with far greater detail, &lt;a href=&quot;http://mathoverflow.net/questions/41132/vortex-voronoi-diagram/41225#41225&quot; rel=&quot;nofollow&quot;&gt; already been given to you by jvkersch&lt;/a&gt;. I am humbled. I should also point out that my example below, which was only meant as an illustration, would not be a steady-state solution in a physical fluid, because the interaction between the vortices themselves would cause them to move.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Bar Moshe's idea reminded me of some work in vector field visualization which does indeed use stagnation points to divide the fluid domain into something like &quot;regions of influence&quot;. I believe the initial paper which introduced the idea was Helman and Hesselink's &quot;Visualizing Vector Field Topology in Fluid Flows&quot; (&lt;a href=&quot;http://graphics.ethz.ch/teaching/scivis_common/Literature/HelmanHesselink91.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF copy&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, because we assumed that the flow is incompressible and irrotational, in a generic configuration the velocity can only be zero at saddle points, where the flow points inward along two directions and outward along two directions. Streamlines along these directions are called the &lt;em&gt;separatrices&lt;/em&gt; of the saddle point. If you place two particles close together on different sides of a separatrix and let them following the flow, they will diverge at the saddle point and follow disparate long-term trajectories. So these separatrices divide space into regions where the global topology of the streamlines is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example I cooked up in Matlab because I thought it would look pretty. There are four point vortices in a square, whose circulations are -1, -1, -2 and 1 going clockwise from top left. Here's what the direction of the velocity field looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kuHf3.png&quot; alt=&quot;quiver plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram below, the separatrices divide space into regions around vortices (and clusters thereof). In each distinct region, the streamlines wind around a particular set of vortices. You can see the saddle points as the points where four arcs come together. I believe this sort of diagram is what you were hoping for when you asked for something like a Voronoi diagram around the vortices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qg78W.png&quot; alt=&quot;separatrices&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a complete picture, there ought to be arrows on the separatrices to indicate the direction of flow, but I couldn't figure out how to do that in Matlab.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-06T04:08:22.573" LastActivityDate="2010-10-06T04:08:22.573" />
  <row Id="6178" PostTypeId="2" ParentId="5889" CreationDate="2010-10-06T03:56:13.773" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose f_1:C-&gt;A, f_2:C-&gt;B and C is generated by a finite set Q. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = &amp;lt; ger(A) , ger(B) | rel(A) , rel(B) , {f_1(q)=f_2(q), q in Q} &gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this presentation we can eliminate at most finitely many generators of A, namely those coming from C. Therefore if A is not finitely generated then G is not finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="2342" LastActivityDate="2010-10-06T03:56:13.773" CommentCount="4" />
  <row Id="6179" PostTypeId="1" CreationDate="2010-10-06T04:09:42.187" Score="11" ViewCount="369" Body="&lt;p&gt;I'm a private tutor of High school &amp;amp; pre-engineering level mathematics. Obviously, only those people who have some trouble with maths take tuitions. Most of my students are actually good. I mean they are not dumb. Its just that they are &quot;don't like&quot;, &quot;hate&quot;, &quot;scared of&quot;, &quot;bored&quot; (these are their own words) of mathematics. They feel maths is very &lt;em&gt;abstract&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't blame them. Its their teachers and the books they follow. Most of the mathematics books just tell the mathematical concepts either bluntly or in an abstract way. I'm looking for some books which teach mathematics in &quot;fun&quot;, &quot;interesting&quot;, &quot;&lt;strong&gt;intutive&lt;/strong&gt;&quot;, &quot;&lt;strong&gt;thought provoking&lt;/strong&gt;&quot; way. Of course, as a tutor I'm doing my best to make maths simple &amp;amp; intuitive but books also help me in getting new ideas. I really like &lt;a href=&quot;http://betterexplained.com&quot; rel=&quot;nofollow&quot;&gt;http://betterexplained.com&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the topics covered in my tutions&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Set theory (Sets, Relations &amp;amp; Functions)&#xA;Trigonometry&#xA;Complex Numbers&#xA;Linear Algebra&#xA;Number theory&#xA;Geometry (st.lines, triangles, circles, parabola, hyperbola, 3D: spheres, cubes, etc.)&#xA;Calculus (Limits, Differential &amp;amp; Integral calculus)&#xA;Probability (permutations &amp;amp; combinations)&#xA;Statistics&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm looking for any kind of resources books, papers, websites, videos, softwares, simulations or even suggestions. I need to cultivate interest at the same time make them good at mathematics.&lt;/p&gt;&#xA;" OwnerUserId="2264" LastActivityDate="2010-10-07T00:54:04.920" CommunityOwnedDate="2010-10-06T04:31:09.617" Title="Teaching mathematics in intuitive and thought provoking way" Tags="&lt;books&gt;&lt;learning&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="6180" PostTypeId="2" ParentId="6179" CreationDate="2010-10-06T05:05:59.923" Score="3" ViewCount="0" Body="&lt;p&gt;It depends upon the student's motivation.  When you are hired for high school, is it by the parents when the student has a problem?  And in pre-engineering is it the student?  This is a big difference.  If the student isn't motivated, I have no ideas.  If s/he is motivated, showing what math is good for, as problems in the area of interest is a good approach.  I've done a bit of this and what worked there was just to take their text and work exercises.  The one-on-one work seemed to help understanding.  For the general high school situation, again assuming motivation, there are many problem books that require creativity.  Martin Gardner and Peter Winkler are the names I would first think of, but there are many more.  Also the first problems at projecteuler.org give you a lot to work with.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-06T05:05:59.923" CommunityOwnedDate="2010-10-06T05:05:59.923" />
  <row Id="6181" PostTypeId="1" CreationDate="2010-10-06T05:57:19.060" Score="0" ViewCount="35" Body="&lt;p&gt;Has anyone had experience diagramming extensive form games in ditaa? I haven't quite figured out how to make diagonal lines in ditaa that scale well for a tree diagram (well, at least top-down oriented trees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping this application is what gives me an excuse to play with ditaa, but I wanted to consult others before diving in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might belong on another stack site, but I wasn't sure.&lt;/p&gt;&#xA;" OwnerUserId="1325" LastActivityDate="2010-10-06T05:57:19.060" Title="Is it feasible to diagram extensive form games using ditaa?" Tags="&lt;game-theory&gt;" CommentCount="3" />
  <row Id="6182" PostTypeId="2" ParentId="6125" CreationDate="2010-10-06T06:22:51.217" Score="0" ViewCount="0" Body="&lt;p&gt;There is a somewhat related operation, &lt;i&gt;condensation&lt;/i&gt;, in the theory of linear orderings. For example, you can take a linear ordering and identify any two elements $x,y$ which are connected by some finite chain, i.e. there are elements $x = v_1,\ldots,v_n = y$ such that $v_1 &amp;lt; \cdots &amp;lt; v_n$ and there are no elements 'in between', i.e. no $z$ such that $v_i &amp;lt; z &amp;lt; v_{i+1}$. This will compress all copies of $\mathbb{Z}$ and $\mathbb{N}$ (or its opposite) to a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular condensation can be used to (de)construct all countable linear orders, if I remember correctly. There are also some other condensations, take a look at Rosenstein's &lt;i&gt;Linear orderings&lt;/i&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-06T06:22:51.217" />
  <row Id="6183" PostTypeId="2" ParentId="6155" CreationDate="2010-10-06T07:37:55.533" Score="0" ViewCount="0" Body="&lt;p&gt;Well according to me the answer is $$F'(x) = \frac{d}{dx}(x^{2}) \cdot [x^{2}-\sin^{2}(x^{2})] - \frac{d}{dx}(1) \times \text{something} = 2x \cdot \Bigl[x^{2} -\sin^{2}(x^{2})\Bigr] - 0$$ would be the answer.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-06T07:37:55.533" />
  <row Id="6184" PostTypeId="2" ParentId="6142" CreationDate="2010-10-06T07:40:16.590" Score="1" ViewCount="0" Body="&lt;p&gt;My friend came up with a good counter-example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZSjRv.png&quot; alt=&quot;Counter example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it is easy to verify that $[x] \prec [y]$ but $x \not{&amp;lt;} y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if anyone has any further insights into the lemma or this whole idea, I'd very like to hear.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-10-06T07:53:22.927" LastActivityDate="2010-10-06T07:53:22.927" />
  <row Id="6185" PostTypeId="2" ParentId="6179" CreationDate="2010-10-06T09:36:37.373" Score="4" ViewCount="0" Body="&lt;p&gt;There is no silver bullet - the hardest part of the project is getting the student curious about maths again. Once you do that, teaching them the skills to pass the exams is comparatively a breeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, one is &lt;em&gt;usually&lt;/em&gt; hired by parents when the student has a problem, and not before, so motivating the student is a key part of initial sessions. It helps to remember that good students who have gone off maths were once good students who enjoyed maths; you need to try and find out what they liked about maths when they enjoyed it, and when they stopped enjoying it and why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find out why the student is there - often students need to pass maths to do something in the future that they are interested in (to get in to college perhaps); focussing on the goal can motivate some students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with J.M. and Ross that puzzles can be an effective way to engage good students whose interest in maths has been diverted.&#xA;For example, a favourite of mine for kindling interest in trig is &lt;a href=&quot;http://rec-puzzles.org/index.php/Table%20in%20Corner&quot; rel=&quot;nofollow&quot;&gt;Table in Corner&lt;/a&gt; where elementary trig is used to solve a problem that appears to have too little information given (I drop the quarter circle condition when I state the problem - it provides an opportunity to discuss the reasonableness of the two possible solutions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elementary number theory is another rich source of good puzzles, and even tricks like divisibility tests can come in handy. I remember getting one student interested in practising the long division algorithm by coming up with exact multiples of horrible small primes like 13 and 17 for her to divide. She knew I couldn't possibly know the 17 times table to several thousand places, so where was I getting these numbers from? We discussed the plausibility of her suggestions and I kept her guessing for most of the session before I explained. She got the algorithm, and she got to understand essentially how it works in one go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that abstraction means different things to different people. Students often say maths is abstract because they only see it employed in textbooks, where every problem is clearly contrived, absolutely divorced from the real world, and one of a collection of standard types (puzzles usually avoid the same problem by being contrived but non-standard).  Since textbooks never give plausible real-world applications you can talk about applications often, in fact it is helpful to have a bank of non-contrived applications for most of the topics you want to teach. Books by the likes of Ian Stewart and Keith Devlin can help you here. Usually I prefer to say stuff like &quot;who cares, it's fun&quot; (and then give an application a bit later on), but then that's me - YMMV.&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2010-10-06T09:36:37.373" CommunityOwnedDate="2010-10-06T09:36:37.373" />
  <row Id="6186" PostTypeId="2" ParentId="6099" CreationDate="2010-10-06T09:55:55.863" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an explicit procedure based on the isomorphism between the de-Rham and Cech cohomologies &#xA;for smooth manifolds based on R. Bott and L.W. Tu's &lt;a href=&quot;http://books.google.com/books?id=S6Ve0KXyDj8C&amp;amp;printsec=frontcover&amp;amp;dq=Differential+forms+in+algebraic+topology+By+Raoul+Bott,+Loring+W.+Tu&amp;amp;source=bl&amp;amp;ots=_kbe5djU5m&amp;amp;sig=DEVBdAp8SnxkgKWJkrWMHq-BurE&amp;amp;hl=en&amp;amp;ei=qT-sTK2uIcvHswbTwPzEBA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;sqi=2&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;: Differential forms in algebraic topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description will be given for a three form but it can be generalized &#xA;along the same lines to forms of any degree. We suppose that the manifold has a finite good cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data needed is the transition functions between the coordinate charts (which will be denoted by: $U_\alpha$, $U_\beta$, etc.) &#xA;and of course, the coordinate expression of the given form on each chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a three form F, then by the Poincare lemma, on  $U_\alpha,  F = dB_\alpha$, where $B_\alpha$ are two forms on $U_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by the Poincare lemma on $U_\alpha \cap U_\beta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_\alpha-B_\beta = dA_{\alpha\beta}$ , where $A_{\alpha\beta}$ are one forms on $U_\alpha \cap U_\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since on $U_\alpha \cap U_\beta \cap U_\gamma$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha})=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by the Poincare lemma&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha} = d\phi_{\alpha\beta\gamma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: $\phi_{\alpha\beta\gamma}$ are zero forms on $U_\alpha \cap U_\beta \cap U_\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, Since on: $U_\alpha \cap U_\beta \cap U_\gamma \cap U_\delta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta})=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta} = C_{\alpha\beta\gamma\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: $C_{\alpha\beta\gamma\delta}$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential form F is integral iff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{\alpha\beta\gamma\delta}= 2 \pi n_{\alpha\beta\gamma\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n_{\alpha\beta\gamma\delta}$ are integers on all quadruple intersections.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-06T09:55:55.863" CommentCount="1" />
  <row Id="6187" PostTypeId="1" AcceptedAnswerId="6194" CreationDate="2010-10-06T10:08:59.800" Score="3" ViewCount="84" Body="&lt;p&gt;Hi everybody!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a usual category $\mathcal{C}$ one can pull the colim out of the Hom like this: $Hom_\mathcal{C}(colim A_i,B)=lim Hom_\mathcal{C}(A_i,B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a corresponding statement for hocolims - lets say in simplicial sets, but if there are more general statements, that's even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. I could imagine&#xA;$$Hom_{\mathcal{C}}(hocolim A_i,B)=lim Hom_{Ho(\mathcal{C})}(A_i,B)$$&#xA;- maybe one needs to have $B$ fibrant and the A_i cofibrant here, i.e. that the Homs on the right are $\mathbb{R}Homs$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the internal Hom in simplicial sets I could also imagine versions like this:&#xA;$$Hom(colim A_i,B)=holim Hom(A_i,B)$$&#xA;$$Hom(colim A_i,B)=holim \mathbb{R}Hom(A_i,B)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the right statement and what is the place to learn this hocolim-yoga?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;N.B.&lt;/p&gt;&#xA;" OwnerUserId="2345" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-07T07:19:31.883" LastActivityDate="2010-10-07T07:19:31.883" Title="How does hocolim relate to Hom?" Tags="&lt;category-theory&gt;&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="6188" PostTypeId="1" CreationDate="2010-10-06T10:33:50.197" Score="1" ViewCount="98" Body="&lt;p&gt;i have arc A ,i know startangle,endangle,startPoint,Endpoint,centre,radius of arc and i have point B, i like to find point B lies or not in arc A , i need formula or algorithum for this&lt;/p&gt;&#xA;" OwnerUserId="1861" LastActivityDate="2010-11-05T13:44:17.937" Title="how to find point lie on the arc" Tags="&lt;geometry&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6189" PostTypeId="2" ParentId="6188" CreationDate="2010-10-06T10:36:29.527" Score="1" ViewCount="0" Body="&lt;p&gt;Did you take a look here: &lt;a href=&quot;http://mathforum.org/library/drmath/view/54365.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/library/drmath/view/54365.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-06T12:58:25.220" LastActivityDate="2010-10-06T12:58:25.220" CommunityOwnedDate="2010-10-06T12:58:25.220" CommentCount="2" />
  <row Id="6190" PostTypeId="1" AcceptedAnswerId="6219" CreationDate="2010-10-06T11:51:32.463" Score="6" ViewCount="157" Body="&lt;p&gt;The card game &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%2528game%2529&quot; rel=&quot;nofollow&quot;&gt;Set&lt;/a&gt; lead to the following question. Lets call a subset $A$ of $(\mathbb{Z}/3)^n$ dependent, if there is $\{x,y,z\}\subset A$ with $x+y+z=0$. (So unlike the case of linear dependence we are not allowing any coefficients here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(n)$ denote the maximal size of a independent subset of $(\mathbb{Z}/3)^n$. &#xA;Is there an explicit expression / a recursion for $f(n)$? Can anything be said about its asymptotic behaviour (like $f\in O(c^n)$ for some minimal $c$)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\{0;1\}^n$ is independent, we know that $f(n)\ge 2^n$. And so $c\ge 2$.&#xA;The card game set deals with the case $n=4$ and $f(4)=20$ if i remember correctly.&lt;/p&gt;&#xA;" OwnerUserId="707" LastEditorUserId="707" LastEditorDisplayName="" LastEditDate="2010-10-06T14:33:48.323" LastActivityDate="2010-10-07T00:12:18.137" Title="A question related to the card game &quot;Set&quot;" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6191" PostTypeId="2" ParentId="6188" CreationDate="2010-10-06T11:59:51.823" Score="0" ViewCount="0" Body="&lt;p&gt;Assuming you're talking about an arc of a circle, the first question to ask is whether the point lies on the circle.  This just means that the distance from the center of the circle to your point is equal to the radius.  Then find out the angle between the radius to your point and, say, a horizontal line, and figure out whether the point is within the bounds of the arc.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-06T11:59:51.823" CommentCount="1" />
  <row Id="6194" PostTypeId="2" ParentId="6187" CreationDate="2010-10-06T12:41:32.687" Score="1" ViewCount="0" Body="&lt;p&gt;A formula like the one you are asking for could be the following (Bousfield-Kan, &quot;Homotopy limits, completions and localizations&quot;, chapter XII, proposition 4.1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{hom}_* (\mathrm{hocolim}\ \mathbf{A}, B) \cong \mathrm{holim}\ \mathrm{hom}_* (\mathbf{A}, B) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $B$ is a pointed simplicial set, $\mathbf{A} : I \longrightarrow \Delta^{\mathrm{o}}\mathbf{Set}_*$ a functor from a small category $I$ to the category of pointed simplicial sets, and for pointed simplicial sets $A, B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{hom}_* (A,B) \in \Delta^{\mathrm{o}}\mathbf{Set}_*&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the &lt;em&gt;pointed simplicial function space&lt;/em&gt; which $n$-simplices are maps in $\Delta^{\mathrm{o}}\mathbf{Set}_* $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( \Delta [n] \times A \right) / \left( \Delta [n] \times * \right) \longrightarrow B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(op.cit., chapter VIII, 4.8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't go through the details, but, if it annoys you, it seems to me that you can drop the &quot;pointed&quot; thing everywhere just deleting &quot;pointed&quot;, $*$, and $\Delta [n] \times * $ in what Bousfield-Kan say.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-06T13:01:08.490" LastActivityDate="2010-10-06T13:01:08.490" CommentCount="1" />
  <row Id="6195" PostTypeId="1" CreationDate="2010-10-06T12:43:37.593" Score="15" ViewCount="503" Body="&lt;p&gt;You play a game flipping a fair coin.  You may stop after any trial, at which point you are paid in dollars &#xA;the percentage of heads flipped.  So if on the first trial you flip a head, you should stop and earn \$100&#xA;because you have 100% heads.  If you flip a tail then a head, you could either stop and earn \$50,&#xA;or continue on, hoping the ratio will exceed 1/2.  This second strategy is superior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A paper by Medina and Zeilberger (&lt;a href=&quot;http://arxiv.org/abs/0907.0032&quot; rel=&quot;nofollow&quot;&gt;arXiv:0907.0032v2 [math.PR]&lt;/a&gt;) says that it is an unsolved&#xA;problem to determine if it is better to continue or stop after you have flipped 5 heads in 8 trials: accept \$62.50 or hope for more.  It is easy to simulate this problem and it is clear&#xA;from even limited experimental data that it is better to continue (perhaps more than 70% chance you'll improve over \$62.50):&#xA;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/ZVHVw.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;br /&gt;&#xA;My question is basically: Why is this difficult to prove?  Presumably it is not that difficult&#xA;to write out an expression for the expectation of exceeding 5/8 in terms of the cumulative binomial distribution.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-07T17:58:58.357" Title="Why is this coin-flipping probability problem unsolved?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="5" />
  <row Id="6196" PostTypeId="1" AcceptedAnswerId="6198" CreationDate="2010-10-06T12:59:01.127" Score="4" ViewCount="166" Body="&lt;p&gt;Is there a reasonable statistic test one can do to standard random number generators (say, one of those that come built in in Python libs) which shows they are not really random?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By reasonable I mean to exclude silly things like «generate a looong list and see that it is periodic»)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; Notice I am looking for a test that &lt;em&gt;fails&lt;/em&gt; on standard generators.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-06T17:29:45.927" LastActivityDate="2010-10-06T17:29:45.927" Title="The pseudoness of pseudorandom number generators" Tags="&lt;statistics&gt;&lt;random&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6197" PostTypeId="2" ParentId="6196" CreationDate="2010-10-06T13:06:36.777" Score="2" ViewCount="0" Body="&lt;p&gt;Searches for Marsaglia (George), works by him or about his PRNG tests, would uncover a lot of information.  e.g.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diehard_tests&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Diehard_tests&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;was found by looking for &quot;marsaglia test&quot;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-06T13:18:05.897" LastActivityDate="2010-10-06T13:18:05.897" CommentCount="2" />
  <row Id="6198" PostTypeId="2" ParentId="6196" CreationDate="2010-10-06T14:09:03.790" Score="5" ViewCount="0" Body="&lt;p&gt;The standard reference is Knuth, The Art of Computer Programming, Volume 2, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201896842&quot; rel=&quot;nofollow&quot;&gt;Seminumerical&#xA;Algorithms&lt;/a&gt;. Anyone with serious interest in this topic should read Knuth's extensive exposition here. In addition to theoretical discussion, he presents tests including frequency, serial, gap, poker, coupon collector's, permutation, run, maximum-of-t,&#xA;collision, birthday spacings, and serial correlation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marsaglia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Diehard_tests&quot; rel=&quot;nofollow&quot;&gt;DIEHARD&lt;/a&gt; suite of statistical test includes birthday spacings, overlapping permutations, ranks of 31x31 and 32x32 matrices, ranks of 6x8 matrices, monkey tests on 20-bit Words,monkey tests OPSO, OQSO, DNA, count the 1's in a stream of bytes, count the 1's in specific bytes, parking lot, minimum distance, random spheres, squeeze, overlapping&#xA;sums, runs, and craps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.itl.nist.gov/div898/pubs/ar/ar2000/node9.html&quot; rel=&quot;nofollow&quot;&gt;NIST Statistical Test Suite&lt;/a&gt; includes frequency, block frequency, cumulative sums, runs, long runs,Marsaglia's rank, spectral (based on the Discrete Fourier Transform), nonoverlapping template matchings, overlapping template matchings, Maurer's universal statistical, approximate entropy (based on the work of Pincus, Singer and Kalman), random&#xA;excursions (due to Baron and Rukhin), Lempel-Ziv complexity, linear complexity, and serial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for tests that fail, as Knuth mentions when discussing a typical generator in section 3.6 p. 188 &quot;Caution: The numbers generated by ran_array fail the birthday spacings test of Section 3.3.2J, and they have other deficiencies that sometimes show up in high-resolution simulations (see exercises 3.3.2-31 and 3.3.2-35)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T14:46:04.913" LastActivityDate="2010-10-06T14:46:04.913" />
  <row Id="6199" PostTypeId="1" AcceptedAnswerId="6592" CreationDate="2010-10-06T14:17:51.730" Score="4" ViewCount="284" Body="&lt;p&gt;I have been stumped for long by this exercise (3.12(d)) from Stokey and Lucas's &lt;em&gt;Recursive Methods in Economic Dynamics&lt;/em&gt;. Would greatly appreciate any hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi: X \to Y$ and $\psi: X \to Y$  be lower hemicontinuous correspondences (set-valued functions), and suppose that for all $x \in X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(x)=\{y \in Y: y \in \phi(x) \cap \psi(x)\}\neq \emptyset$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if $\phi$ and $\psi$ are both convex valued, and if $\mathrm{int} \phi(x) \cap \mathrm{int} \psi(x) \neq \emptyset$, then $\Gamma(x)$ is lower hemicontinuous at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[A correspondence $\Gamma: X \to Y$ is said to be &lt;em&gt;lower hemicontinuous&lt;/em&gt; at $x \in X$ if $\Gamma(x)$ is nonempty and if, for every $y \in \Gamma(x)$ and every sequence $x_n \to x$, there exists $N \geq 1$ and a sequence $\{y_n\}_{n=N}^\infty$ such that $y_n \to y$ and $y_n \in \Gamma(x_n)$, all $n \geq N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively this means that the graph of $\Gamma(x)$ cannot suddenly broaden out.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: We can assume that $X$ and $Y$ are subsets of $\mathbf{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-08T15:12:02.357" LastActivityDate="2010-10-13T13:29:04.093" Title="Lower hemicontinuity of the intersection of lower hemicontinuous correspondences" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;convex-analysis&gt;&lt;economics&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="6200" PostTypeId="1" CreationDate="2010-10-06T14:56:09.460" Score="2" ViewCount="80" Body="&lt;p&gt;I've been trying to solve this problem from Elements of Applied&#xA;Bifurcation Theory, but even with the hints given I haven't found out&#xA;how to proceed yet. I would appreciate any further hints or insights&#xA;you could give me in order to make some progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's exercise 2.6.(2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following system of partial differential equations is the FitzHugh-&#xA;Nagumo caricature of the Hodgkin-Huxley equations modeling the nerve&#xA;impulse propagation along an axon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}-f_a(u)-v$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial v}{\partial t} = bu $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u(x,t)$ represents the membrane potential, $v=v(x,t)$ is a&#xA;&quot;recovery&quot; variable, $f_a(u)=u(u-a)(u-1)$, $1&gt;a&gt;0,b&gt;0,-\infty &amp;lt; x &amp;lt;&#xA;\infty$, and $t&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traveling waves are solutions to these equations of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(x,t)=U(\xi)$, $v(x,t)=V(\xi)$, $\xi = x+ct$, where $c$ is an a&#xA;priori unknown wave propagation speed. The functions $U(\xi)$,&#xA;$V(\xi)$ are the wave profiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Derive a system of three ordinary differential equations for the&#xA;profiles with &quot;time&quot; $\xi$. (Hint: introduce an extra variable:&#xA;$W=\dot{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Check that for all $c&gt;0$ the system for the profiles (the wave&#xA;system) has a unique equilibrium with one positive eigenvalue and two&#xA;eigenvalues with negative real parts. (Hint: First, verify this&#xA;assuming that eigenvalues are real. The, show that the characteristic&#xA;equation cannot have roots on the imaginary axis, and finally, use the&#xA;continuous dependence of the eigenvalues on the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Conclude that the equilibrium can be either a saddle or a saddle-&#xA;focus with a one-dimensional unstable and a two-dimensional stable in-&#xA;variant manifold, and find a condition on the system parameters that de-&#xA;fines a boundary between these two cases. Plot several boundaries in the&#xA;$(a,c)$-plane for different values of $b$ and specify the region&#xA;corresponding to saddle-foci. (Hint: at the boundary the&#xA;characteristic polynomial $h(\lambda)$ has a double root&#xA;$\lambda_0:h(\lambda_0)=h'(\lambda_0)=0.$).&lt;/p&gt;&#xA;" OwnerUserId="2348" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-06T17:36:58.000" LastActivityDate="2010-10-06T17:59:17.500" Title="FitzHugh- Nagumo caricature of the Hodgkin-Huxley equations" Tags="&lt;differential-equations&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6203" PostTypeId="1" AcceptedAnswerId="6257" CreationDate="2010-10-06T15:38:12.927" Score="1" ViewCount="197" Body="&lt;p&gt;If I give you the following definition of the set $A$, how could you prove it is equal the set of the natural numbers without an explicit definiton for the latter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $A$ is inductively defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $0 \in A$; and &lt;br/&gt;&#xA;ii) $\forall n$, a natural number, if $n \in A$, then $n+1 \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily prove that $A$ is contained in the natural numbers, but I'm failing to see how to prove the converse without a similar definition for the natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking the time to read me.&lt;/p&gt;&#xA;" OwnerUserId="2349" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-27T20:20:35.383" LastActivityDate="2010-12-27T20:20:35.383" Title="Prove that the natural numbers are present on an inductive definition of another set" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;&lt;induction&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6204" PostTypeId="2" ParentId="6128" CreationDate="2010-10-06T15:43:21.860" Score="2" ViewCount="0" Body="&lt;p&gt;On an orientable manifold, we define &lt;em&gt;integration of functions with respect to a volume form&lt;/em&gt;. On a non orientable manifold, there are no volume forms, so we have to do something else!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-06T15:43:21.860" CommentCount="2" />
  <row Id="6205" PostTypeId="2" ParentId="6039" CreationDate="2010-10-06T15:47:50.797" Score="1" ViewCount="0" Body="&lt;p&gt;$\rm\ p(x+1) = 3\:p(x)\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(x+1) = 3\:p(x)\ \Rightarrow\ p(-1) = 1/3\ $ contra $\rm\ p(x)\in \mathbb Z[x]\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, continuing,  $\rm\ p(-n) = 3^{-n}\: \Rightarrow \ p(x)\:p(-x) = 1\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(-x)\:p(x) = 1\ $ contra degree comparison.  This proof works over much more general coefficient rings, e.g. $\mathbb Q\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-06T15:47:50.797" CommentCount="0" />
  <row Id="6206" PostTypeId="2" ParentId="6203" CreationDate="2010-10-06T15:55:05.287" Score="2" ViewCount="0" Body="&lt;p&gt;It seems to me that your 'proof' that the set $A \subseteq \mathbb{N}$ must have a mistake because basically you're just defining what Tom Apostol calls an inductive set in his Calculus book. I mean, $A$ can just be a set of real numbers and could perfectly well satisfy your two propierties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance you can take $A = [0, \infty[$ and then $A$ satisfies both properties, but nonetheless $A$ is not contained in $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact Apostol's book defines $\mathbb{N}$ as the set of all numbers that belong to every inductive set.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-06T15:55:05.287" CommentCount="11" />
  <row Id="6207" PostTypeId="2" ParentId="6203" CreationDate="2010-10-06T16:11:10.790" Score="0" ViewCount="0" Body="&lt;p&gt;That's not an inductive definition of $A\:$. Rather, it's an inductive proof that $\ A \supseteq\mathbb N\:$. If you know additionally that $\: A \subseteq \mathbb N\:$ then you can conclude that $\:A = \mathbb N\:$. That's standard mathematical induction.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-06T16:11:10.790" CommentCount="2" />
  <row Id="6208" PostTypeId="1" AcceptedAnswerId="6210" CreationDate="2010-10-06T16:22:25.843" Score="4" ViewCount="190" Body="&lt;p&gt;Let $f$ be an analytic function on the open unit disk domain $D$.  Suppose also that $f$ is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is bounded I beleive that $f$ can be continuously extended to the closed unit disk.&#xA;I know that the zeros of $f$ in the open disk $D$ are isolated.   Are the zeros of $f$ in the closed unit disk also necessarily isolated?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-07T22:37:58.370" Title="Isolated zeros on closure of a domain" Tags="&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6209" PostTypeId="2" ParentId="566" CreationDate="2010-10-06T16:23:21.533" Score="1" ViewCount="0" Body="&lt;p&gt;If you have 32 2x1 dominoes then you can cover an 8x8 board easily enough; if you throw away a domino and cut a 1x1 square from each end of a diagonal of the board, can you cover the remaining shape with the remaining dominoes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard this decades ago at university.&lt;/p&gt;&#xA;" OwnerUserId="1809" LastEditorUserId="1995" LastEditorDisplayName="" LastEditDate="2010-10-07T00:11:03.083" LastActivityDate="2010-10-07T00:11:03.083" CommunityOwnedDate="2010-10-06T16:23:21.533" CommentCount="1" />
  <row Id="6210" PostTypeId="2" ParentId="6208" CreationDate="2010-10-06T17:02:19.013" Score="9" ViewCount="0" Body="&lt;p&gt;It is not true that $f$ can be extended continuously to the closed disk.  The set of bounded analytic functions on the open disk is called the Hardy algebra, denoted $H^\infty$, and the set of analytic functions on the disk that have continuous extensions to the closed disk is called the disk algebra, often denoted $A(\mathbb{D})$.  Elements of the Hardy algebra do have radial limits (or even nontangential limits) a.e. on the boundary, allowing the Hardy algebra to be embedded in $L^\infty$ of the circle as the subspace of functions whose negatively indexed Fourier coefficients vanish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements of the Hardy algebra must be nonzero a.e. on the circle (unless they are identically zero).  In fact, this is even true for elements of $H^1$, i.e., for integrable functions on the circle whose negatively indexed Fourier coefficients vanish.  It is a theorem of Rudin that given any closed subset $E$ of the unit circle with (one-dimensional) Lebesgue measure 0, and given any complex valued continuous function $g$ on $E$, there is an element of the disk algebra whose restriction to $E$ is $g$.  In particular, when $g$ is the zero function, this answers your question about what the zero sets can be, and shows that the zeros need not be isolated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://projecteuclid.org/DPubS?verb=Display&amp;amp;version=1.0&amp;amp;service=UI&amp;amp;handle=euclid.nmj/1118795360&amp;amp;page=record&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s the quickest reference I could find to Rudin's theorem, in an open access article (see page 5), although I'm fairly certain that all of the claims I've made can be found in Hoffman's &lt;em&gt;Banach spaces of analytic functions&lt;/em&gt; or Duren's &lt;em&gt;$H^p$ spaces&lt;/em&gt;, both available in Dover editions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some explicit examples of bounded analytic functions on the open disk with no continuous extension to the closed disk (in other words, elements of $H^\infty\setminus A(\mathbb{D})$), consider &lt;a href=&quot;http://eom.springer.de/b/b016630.htm&quot; rel=&quot;nofollow&quot;&gt;Blaschke products&lt;/a&gt; with infinitely many zeros.  If $f$ is such a function, then $f$ has boundary values of modulus 1 a.e., and since $f$ has infinitely many zeros, the set of zeros has a limit point $x$ on the circle.  If $f$ had a continuous extension to the closed disk, this would imply respectively that $f$ has modulus 1 everywhere on the circle, while at the same time $f(x)=0$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-07T22:37:58.370" LastActivityDate="2010-10-07T22:37:58.370" CommentCount="1" />
  <row Id="6211" PostTypeId="2" ParentId="6179" CreationDate="2010-10-06T17:15:03.520" Score="11" ViewCount="0" Body="&lt;p&gt;Provided your student has a strength, you can play to that.  In some sense that's what finding &quot;applications&quot; is about -- appealing to someone's immediate values.  But there's also a danger in that, if all you do is appeal to someone's immediate values you may be forced to build up a rather contrived and limited notion of what mathematics is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in teaching it's also very useful to take students to foreign territory where they're weak, and build up their appreciation for what they may have found useless or &quot;abstract&quot;.  Finding applications isn't the only way to do this -- simply providing re-interpretations of things they know that seem complicated can be useful.  If something they found hard becomes easier in a different setting, this will frequently be perceived as useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm teaching an introductory calculus class this semester.  In the last few classes we've been differentiating trig functions. Recently students have been asking questions like &quot;can we have a formula sheet on the exams?&quot;  The bulk of the number of things to remember is starting to appear heavy.  Their concern is that things like the double-angle formula or the formula for the derivative of $\csc(x)$ are difficult to remember.   So I spent a little bit of time on how you can get by, not remembering much at all about trig functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the pencil-sketch of the idea that you can flesh-out: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take it as an axiom people know what the graphs of $\sin(x)$ and $\cos(x)$ look like -- and students should be able to deduce these things from &quot;$(\cos(x),\sin(x))$ is the point on the unit circle that you get to if you lay-out a string of length $x$ along the unit circle, starting from $(1,0)$ and going in a counter-clockwise direction&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you remember what $\sin'(x)$ is?  You sketch the graph of $\sin(x)$, lay down several tangent lines, and plot the slopes of the tangent lines.  Observe that the slopes vary continuously, and fill in the gaps appropriately with your knowledge of the local maxima and minimal of $\sin(x)$.  Then you observe the graph looks like $\cos(x)$, so you conclude $\sin'(x)$ must be $\cos(x)$.  This isn't a formal conclusion, more of a seat-of-the-pants &quot;well, how complicated can $\sin'(x)$ be?&quot; type argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the double-angle formula for $\cos(2x)$.  How do you remember that?  The idea is that we &lt;em&gt;know&lt;/em&gt; there is a formula that relates $\cos(2x)$ to either $\cos(x)$ or $\sin(x)$ but people tend not to have the exact form on immediate recall.  I sketch the graph of $\cos(x)$, then sketch the graph of $\cos(2x)$ as the previous graph with double the frequency.  $\cos(2x)$ has twice the frequency -- you can also get twice the frequency by looking at $\cos^2(x)$ or $\sin^2(x)$ -- so spend some time sketching the graph of those functions.  This amounts to knowing the graph of $\sin(x)$ and $\cos(x)$, plus the graph of $x^2$ -- to know that $\cos^2(x)$ for example not only has twice the frequency but that when $\cos^2(x)=0$ the function looks quadratic -- like $x^2$ around those points.  From there it's just a scaling argument to say $2\cos^2(x)$ looks like $\cos(2x)+1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the point of all this is that, with enough patience and geometric insight, all these formulas that students encounter you can cast them as things that, &lt;em&gt;by their very nature&lt;/em&gt; they must be easily re-discoverable with very little probing.  In particular, provided you're comfortable enough graphing functions it's less energy (and really far more informative) to remember the forms of these identities by intuiting them from your rough expectations.  In that way you can re-cast identities like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in a different light.  The fact that this identity has such a simple form is somewhat remarkable -- but once you know it has a simple form, re-discovering it via non-rigorous arguments and heuristics is rather immediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, this line of reasoning isn't great for all students as it assumes a certain familiarity and an intuition with graphing that many students may not have.  But once students have the knowledge that this is something they can think about, and that other people think this way -- that the discovery process is full of intuition and guesses before anything rigorous -- this sets them on a path. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-10-06T17:21:13.743" LastActivityDate="2010-10-06T17:21:13.743" CommunityOwnedDate="2010-10-06T17:15:03.520" CommentCount="3" />
  <row Id="6212" PostTypeId="2" ParentId="6190" CreationDate="2010-10-06T17:41:30.893" Score="1" ViewCount="0" Body="&lt;p&gt;Based on &lt;a href=&quot;http://www2.truman.edu/~erickson/presentation7.ppt&quot; rel=&quot;nofollow&quot;&gt;this presentation&lt;/a&gt; (PPT; &lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache%3a-7EZGJG-DWEJ%3awww2.truman.edu/~erickson/presentation7.ppt+proof+of+minimum+number+of+%22set+cards%22+to+guarantee+a+set&amp;amp;cd=1&amp;amp;hl=en&amp;amp;ct=clnk&amp;amp;gl=us&quot; rel=&quot;nofollow&quot;&gt;Google-based HTML conversion&lt;/a&gt;), I think a formula for your $f(n)$ is:&#xA;$$f(n)=\begin{cases}&#xA;\frac{2^{n+2}-4}{3} &amp;amp; \text{ if }n\text{ is even}&#xA;\\&#xA;\frac{2^{n+2}-5}{3} &amp;amp; \text{ if }n&gt;1\text{ is odd}&#xA;\end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-06T17:41:30.893" />
  <row Id="6213" PostTypeId="1" AcceptedAnswerId="6215" CreationDate="2010-10-06T17:46:03.613" Score="2" ViewCount="77" Body="&lt;p&gt;So if you have a distribution &lt;code&gt;fx(x)&lt;/code&gt; and you observed x=a, how do you calculate &lt;code&gt;E{x | x=a} = integral (x*fx(x|x=a) from -inf to inf&lt;/code&gt;, or more specifically, &lt;code&gt;fx(x|x=a)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble when using Bayes or finding &lt;code&gt;Fx(x|x=a)&lt;/code&gt; because I end up with &lt;code&gt;P(x=a)&lt;/code&gt; which is just a constant.  Is there a simple example somewhere? Searching only leads me to bivariate distributions.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2010-10-06T17:51:16.300" LastActivityDate="2010-10-06T18:25:41.597" Title="How do you calculate conditional expectation for a single continuous r.v.?" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6214" PostTypeId="2" ParentId="6200" CreationDate="2010-10-06T17:59:17.500" Score="2" ViewCount="0" Body="&lt;p&gt;Here's something to get you started: $u_t=c U'(\xi)$, $u_{xx}=U''(\xi)$, and $v_t=c V'(\xi)$. Plugging this into the system of PDEs gives you a system of two ODEs for $U(\xi)$ and $V(\xi)$. The first of these ODEs is second order in $U$, but you can turn it into a first order equation by replacing $U'(\xi)$ by an auxiliary function $W(\xi)$ and replacing $U''(\xi)$ by $W'(\xi)$, and then augmenting the system with the extra equation $U'=W$. Now you have a system of three first order ODEs (i.e., a dynamical system), and can proceed to part (b).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-06T17:59:17.500" />
  <row Id="6215" PostTypeId="2" ParentId="6213" CreationDate="2010-10-06T18:04:58.960" Score="2" ViewCount="0" Body="&lt;p&gt;Unless I misunderstood your problem: your conditional density is just a Dirac delta function in $x=a$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_x(x|x=a) =\delta(x-a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $E(x | x=a) = a$ &#xA;... as intuition says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to apply the defintion, you can use a limiting procedure (because, for a continuous variable the event $X=a$ has actually zero probability), considering instead the event that $X$ takes a value around $a$, in a neighborhood of length $dx$.&#xA;But that's not necessary. You already know that the join probability -the one that goes in the numerator- is zero if $x\ne a$. Hence, as the conditional density must integrate to one, it must be a dirac delta function.  &lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-10-06T18:25:41.597" LastActivityDate="2010-10-06T18:25:41.597" CommentCount="2" />
  <row Id="6216" PostTypeId="1" AcceptedAnswerId="6217" CreationDate="2010-10-06T18:10:30.797" Score="6" ViewCount="91" Body="&lt;p&gt;Hi All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first post here...not really a math expert, but certainly enjoy the challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I working writing a random string generator and would like to know how to calculate how many possible combinations there are for a particular combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am generating a string of 2numbers followed by 2 letters (lowercase) e.g. 12ab&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the calculation would be (breaking it down) &#xA;number combinations 10*10=100&#xA;letter combinations 26*26=676&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of possible combinations is 100*676=67600, but this seems a lot to me so I'm thinking I am off on my calculations!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please point me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thx&lt;/p&gt;&#xA;" OwnerUserId="2351" LastActivityDate="2010-10-06T18:36:17.827" Title="Calculate combinations of characters." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6217" PostTypeId="2" ParentId="6216" CreationDate="2010-10-06T18:29:25.730" Score="2" ViewCount="0" Body="&lt;p&gt;You are right. That is the most basic/fundamental procedure for counting in combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sometimes called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_product&quot; rel=&quot;nofollow&quot;&gt;Rule of product&lt;/a&gt;, or &lt;strong&gt;multiplication principle&lt;/strong&gt; or &lt;a href=&quot;http://people.richland.edu/james/lecture/m170/ch04-not.html&quot; rel=&quot;nofollow&quot;&gt;fundamental counting principle&lt;/a&gt;, and it can be visualized as a &lt;a href=&quot;http://www.ltcconline.net/greenl/courses/201/probdist/trees.htm&quot; rel=&quot;nofollow&quot;&gt;tree&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-10-06T18:36:17.827" LastActivityDate="2010-10-06T18:36:17.827" CommentCount="1" />
  <row Id="6218" PostTypeId="2" ParentId="6216" CreationDate="2010-10-06T18:29:48.393" Score="0" ViewCount="0" Body="&lt;p&gt;Some nomenclature: when you say &quot;2 number&quot;, you really mean &quot;2 &lt;strong&gt;digits&lt;/strong&gt;&quot;. Also, you need to specify if the digits can be anything or not (for example, do you allow leading zeroes?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each of the two digits can be anything, 0 through 9, and each of the letters can be anything, a through z, then your computation is correct. If you think about it, you can see why the number is not off: any particular pair of letters have 100 numbers that can go before them to make the string. Each particular letter going first has 26 possible &quot;second letters&quot;, and each of them has 100 possible pairs of digits to go. So there are already 2600 possible strings of the form xxxa. Another batch for xxxb, etc. They add up very quickly. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-06T18:29:48.393" />
  <row Id="6219" PostTypeId="2" ParentId="6190" CreationDate="2010-10-06T18:50:06.283" Score="2" ViewCount="0" Body="&lt;p&gt;Mathematicians call these &quot;independent sets&quot; &lt;em&gt;capsets&lt;/em&gt;, and they're fairly widely studied. But surprisingly, we don't know much about their asymptotic behavior! We have a lower bound of order c^n for some c &gt; 2; and we have an upper bound of order about 3^n/n (although this is not in the least trivial to prove!) Most mathematicians seem to suspect that the upper bound can't be improved to c^n for c strictly less than 3, but no one's sure how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information (perhaps a little out-of-date, but I don't think much has changed) see Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2007/02/23/open-question-best-bounds-for-cap-sets/&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2353" LastActivityDate="2010-10-06T18:50:06.283" />
  <row Id="6220" PostTypeId="1" AcceptedAnswerId="6222" CreationDate="2010-10-06T18:55:11.880" Score="7" ViewCount="209" Body="&lt;p&gt;Can this expression be further simplified : $ \sum_{k=0}^{n-1} { n -1 \choose k } (-2)^{k} (2n - k)! $? This is the coefficient of $x^{2n}$ in the formal power series expansion of $(1-2x)^{n-1} \times \sum_{ k \geq 0} k! x^k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I came across this when trying to solve a problem using inclusion-exclusion principle, I am not mentioning the original problem because I am interested in this sum as an independent problem.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-06T19:00:36.970" LastActivityDate="2010-10-07T02:02:29.680" Title="Can this sum be simplified?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="6221" PostTypeId="1" CreationDate="2010-10-06T19:18:09.600" Score="3" ViewCount="168" Body="&lt;p&gt;I'm currently teaching myself calculus and I'm probably trying to run before I can walk, but I've been working on this problem..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to find the correct result for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty }(2e^{-3x}+4e^{-7x})^2dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by expanding it to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}4e^{-6x}+16e^{-10x}+16e^{-14x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then working from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better/more general approach I could have taken? I've attempted to solve it using substitution but haven't had any success...&lt;/p&gt;&#xA;" OwnerUserId="2352" LastActivityDate="2010-11-14T04:44:19.153" Title="Evaluating $\int_{0}^{\infty }(2e^{-3x}+4e^{-7x})^2dx$" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6222" PostTypeId="2" ParentId="6220" CreationDate="2010-10-06T19:18:59.850" Score="5" ViewCount="0" Body="&lt;p&gt;It's a confluent hypergeometric function $\rm\ (2n)! \phantom{f}_1 F_1(1-n;\: -2\:n;\: -2)\ $ which can be expressed in closed form in terms of modified Bessel functions of the first kind, namely&#xA;&lt;img src=&quot;http://i.imgur.com/jb2KZ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may well have a simpler closed form, since the various hypergeometric simplifiers in CAS are not always optimal (above is via Mma). It is not known by the OEIS Superseeker so you may wish to submit it there.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T19:27:20.237" LastActivityDate="2010-10-06T19:27:20.237" />
  <row Id="6223" PostTypeId="2" ParentId="6221" CreationDate="2010-10-06T19:37:44.793" Score="1" ViewCount="0" Body="&lt;p&gt;Using substitution you can actually solve this, but generally it's more or less the same thing. Put $t=e^{-x}$ therefore you have $dx = -\frac{1}{t} \ dt$. Then your integral reduces to $$-\int\limits_{1}^{0} \Bigl[2t^{3}+4t^{7}\Bigr]^{2} \cdot \frac{dt}{t}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-06T19:37:44.793" CommentCount="2" />
  <row Id="6224" PostTypeId="1" AcceptedAnswerId="6233" CreationDate="2010-10-06T20:31:01.510" Score="3" ViewCount="52" Body="&lt;p&gt;Theorem 2 in a &lt;a href=&quot;http://www.ams.org/journals/jams/2000-13-03/S0894-0347-00-00337-4/&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Borodin, Okounkov, and Olshanski  states that the discrete Bessel kernel $J(x,y,\theta)$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{\theta} \frac{J_x J_{y+1} - J_{x+1} J_y}{x-y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $J_x = J_x(2\sqrt{\theta})$ is the Bessel function of order $x$ and argument $2\sqrt{\theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have any ideas on how to determine the asymptotics of this kernel in the case that $x = \alpha \sqrt{n}, y = \beta \sqrt{n}$ where $\alpha,\beta$ are distinct real numbers that differ from $\pm 2$?&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-10-06T22:44:47.977" Title="The discrete Bessel kernel" Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="6225" PostTypeId="2" ParentId="4462" CreationDate="2010-10-06T21:12:40.120" Score="2" ViewCount="0" Body="&lt;p&gt;I think you might find it helpful to first comprehend the essence of the matter in a slightly simpler context, e.g. see the &lt;a href=&quot;http://groups.google.com/groups?selm=y8zfxg12ckl.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;proof that I presented here&lt;/a&gt;. The key ideas are already there. Generally, I recommend Kaplansky, Commutative Rings, for the circle of ideas  around the generalized Nullstellensatz (Goldman, Krull, Zariski).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-06T21:12:40.120" />
  <row Id="6226" PostTypeId="1" AcceptedAnswerId="6229" CreationDate="2010-10-06T21:15:28.747" Score="3" ViewCount="86" Body="&lt;p&gt;Why does the following integral converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{0\leq |x_1|\leq|x_2|\leq\dots\leq |x_n|\leq 1}\frac{(1-\sum_{i=1}^n\log |x_i|)^r}{|x_1x_2\dots x_n|^{1/2}}dx_1dx_2\dots dx_n&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $r$ is any positive real number.&lt;/p&gt;&#xA;" OwnerUserId="1101" LastActivityDate="2010-10-06T21:57:08.857" Title="Convergence of integral" Tags="&lt;calculus&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6227" PostTypeId="1" CreationDate="2010-10-06T21:24:50.520" Score="1" ViewCount="171" Body="&lt;p&gt;$&#xA;3= \sum_{n=1}^{t} \frac{1}{1.08^n}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that it is $3 = 1.08^{-t}(12.5 \times 1.08^t{-12.5})$ (from Wolfram Alpha, but I'm not sure how to get it. I tried solving as a geometric series, I had problems and didn't get the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that $ t\approx3.56592$, which seems like it's correct, but I have no idea where the 12.5 and all that stuff came from. Unfortunately my calculus book doesn't help much, as it is mainly focus on infinite series.&lt;/p&gt;&#xA;" OwnerUserId="2355" LastActivityDate="2010-10-07T12:00:29.380" Title="Sum of Series with negative exponents" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="6228" PostTypeId="1" AcceptedAnswerId="8227" CreationDate="2010-10-06T21:42:57.857" Score="0" ViewCount="108" Body="&lt;p&gt;I'd like to understand why $ \int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i $ (the second equality), where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j^i = \rho \frac{dx^i}{dt} $ is the current density 4-vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{j} = \rho \mathbf{v}$ is the current density 3-vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ j^i = (c\rho, \mathbf{j}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho$ is the charge density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dS_i$ is the element $-dx-dy-dz+cdt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you able to explain me this equality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="2354" LastActivityDate="2010-10-29T01:14:36.893" Title="Why $ \int{\rho}dV =  \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i $ ???" Tags="&lt;integral&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6229" PostTypeId="2" ParentId="6226" CreationDate="2010-10-06T21:47:35.627" Score="4" ViewCount="0" Body="&lt;p&gt;You can show (for example, using l'Hôpital's rule) that $\frac{(1-\log(t))^r}{t^{1/2}}$ is dominated by $\frac{1}{t^{2/3}}$ for sufficiently small $t$.  If you use this along with the fact that that $\sum_{i=1}^n\log|x_i|=\log|x_1\cdots x_n|$, you have your integrand dominated by $|x_1\cdots x_n|^{-2/3}$ in a neighborhood of the set where it is undefined, and bounded on the complement of this neighborhood.  Does that get you where you want to be?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to keep things simple, you may as well just consider $0\leq x_1\leq x_2\cdots\leq x_n\leq1$, because the whole integral can be broken up into $2^n$ equal pieces depending on the various choices of sign.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-06T21:57:08.857" LastActivityDate="2010-10-06T21:57:08.857" CommentCount="2" />
  <row Id="6231" PostTypeId="2" ParentId="6227" CreationDate="2010-10-06T21:52:05.617" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\ $ Put $\rm\ a = 1.08\:$. Multiply both sides by $\rm\ a^t\ $ to get$\:$ RHS $\rm\: = \ \sum_{k=0}^{t-1}\ a^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;where the $12.5$ came from&quot;, $\rm\ 1/(a-1) = 1/0.08 = 100/8 = 12.5$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-06T22:35:34.510" LastActivityDate="2010-10-06T22:35:34.510" />
  <row Id="6232" PostTypeId="2" ParentId="4844" CreationDate="2010-10-06T22:02:46.347" Score="1" ViewCount="0" Body="&lt;p&gt;The most comprehensive webpage I have found is &lt;a href=&quot;http://www.ogre.nu/sphere.htm&quot; rel=&quot;nofollow&quot; title=&quot;Evenly distributing N points on a sphere&quot;&gt;about evenly distributing N points on a sphere&lt;/a&gt;. To be more general, this is known as the seventh of Stephen Smale's problems: the &quot;optimal&quot; distribution of points on the 2-sphere. It is still unsolved.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-10-06T22:02:46.347" />
  <row Id="6233" PostTypeId="2" ParentId="6224" CreationDate="2010-10-06T22:44:47.977" Score="2" ViewCount="0" Body="&lt;p&gt;You could start with the asymptotics for $J_x$ and apply trig identities to your kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8tWuG.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-06T22:44:47.977" CommentCount="1" />
  <row Id="6234" PostTypeId="1" CreationDate="2010-10-06T22:47:03.750" Score="20" ViewCount="584" Body="&lt;p&gt;One way of looking at the Vieta product&#xA;$${2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$&#xA;is as the infinite product of a series of successive 'approximations' to 2, defined by $a_0 = \sqrt{2}$, $a_{i+1} = \sqrt{2+a_i}$ (or more accurately, their ratio to their limit 2).  This allows one to see that the product converges; if $|a_i-2|=\epsilon$, then $|a_{i+1}-2|\approx\epsilon/2$ and so the terms of the product go as roughly $(1+2^{-i})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the sequence of infinite radicals $a_0=1$, $a_{i+1} = \sqrt{1+a_i}$ converges exponentially to the golden ratio $\phi$, and so the same sort of infinite product can be formed:&#xA;$$\Phi = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$&#xA;and an equivalent proof of convergence goes through.  The question is, what's the value of $\Phi$?  The usual proof of Vieta's product by way of the double-angle formula for sin doesn't translate over, and from what I know of the logistic map it seems immensely unlikely that there's any function conjugate to the iteration map here in the same way that the trig functions are suitably conjugate to the version in the Vieta product.  Is there any other approach that's likely to work, or is $\Phi$ just unlikely to have any formula more explicit than its infinite product?&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-12-31T00:44:29.980" Title="What's the value of this Vieta-style product involving the golden ratio?" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;elementary&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="11" />
  <row Id="6235" PostTypeId="1" AcceptedAnswerId="6236" CreationDate="2010-10-06T22:51:21.207" Score="1" ViewCount="83" Body="&lt;p&gt;and in the other way, orthogonal nullspace and range require symmetry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R&lt;/p&gt;&#xA;" OwnerUserId="2357" LastActivityDate="2010-10-06T23:24:27.297" Title="What is the best way to show an idempotent matrix whose nullspace and range are orthogonal must be symmetric?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="6236" PostTypeId="2" ParentId="6235" CreationDate="2010-10-06T23:13:14.167" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: any $v\in V$ can be written &lt;em&gt;uniquely&lt;/em&gt; as $v_1+v_2$, where $v_1$ is in the nullspace of $A$ and $v_2$ in the range space. Thus written, $Av=v_2$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-06T23:13:14.167" CommentCount="3" />
  <row Id="6237" PostTypeId="2" ParentId="6235" CreationDate="2010-10-06T23:19:02.847" Score="2" ViewCount="0" Body="&lt;p&gt;Let $f^2=f$. Then $f$ is self-adjoint iff $f = f^* f$ iff $\langle fx,y \rangle=\langle f^* f,y \rangle $ iff $\langle fx,y \rangle =\langle fx,fy \rangle$ iff $\langle fx,fy-y \rangle =0$ for all $x,y$ iff $ran(f)$ and $ker(f)$ are orthogonal.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditorDisplayName="" LastEditDate="2010-10-06T23:24:27.297" LastActivityDate="2010-10-06T23:24:27.297" CommentCount="2" />
  <row Id="6238" PostTypeId="2" ParentId="6195" CreationDate="2010-10-06T23:27:29.217" Score="9" ViewCount="0" Body="&lt;p&gt;I accept Qiaochu's answer &quot;Have you tried actually doing that?&quot; I did try now, and now I can appreciate the challenge. :-) The paper I cited refers to another by Chow and Robbins from 1965&#xA;that has a beautiful formulation, much clearer than the cummulative binomials with which I struggled.  Let me explain it, because it is really cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the natural strategy I mentioned in the comments (and echoed by Raynos),&#xA;let $f(n,h,t)$ be the expected payoff&#xA;if you start with $h$ heads and $t$ tails, and let the game continue no more than $n$ further trials.  Then there is an easy recursive formulation for $f$:&#xA;$$f(n,h,t) = \max \left( \frac{1}{2} f(n,h+1,t) + \frac{1}{2} f(n,h,t+1) , h/(h+t) \right) $$&#xA;because you have an equal chance of increasing to $h+1$ heads or to $t+1$ tails on the next&#xA;flip if you continue, and you get the current ratio if you stop.&#xA;Now, when $h+t=n$, you need to make a &quot;boundary&quot; assumption.  Assuming the law of&#xA;large numbers for large $n$ leads to the reasonable value $\max ( 1/2, h/n )$ in this case.&#xA;So now all you need to do is fill out the table for all $h$ and $t$ values up to $n=h+t$.&#xA;The real answer is the limit when &#xA;$n \rightarrow \infty$, but using large $n$ approximates this limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the Medina and Zeilberger paper was released, in fact just about three weeks ago,&#xA;a very careful calculation using the above recursive formulation was made by Julian Wiseman and reported on &lt;a href=&quot;http://www.jdawiseman.com/papers/easymath/chow_robbins.html&quot; rel=&quot;nofollow&quot;&gt;this web page&lt;/a&gt;.  The conclusion is to me remarkable: &quot;Choosing to continue lowers the expected payoff [from 0.625] to 0.62361957757.&quot;&#xA;This is still not a proof, but the &quot;answer&quot; is now known. &#xA;So my &quot;it is clear from even limited experimental data that&quot; was completely wrong! &#xA;I am delighted to learn from my mistake.  &lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-06T23:33:39.617" LastActivityDate="2010-10-06T23:33:39.617" />
  <row Id="6239" PostTypeId="2" ParentId="6227" CreationDate="2010-10-06T23:29:26.880" Score="4" ViewCount="0" Body="&lt;p&gt;I would solve for $t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(t)=\displaystyle\sum_{n=1}^{t}\dfrac{1}{1.08^{n}}=3.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of a geometric progression with first term $u_{1}$ and ratio $r$ is&#xA;given $(\ast)$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(t)=u_{1}\times \dfrac{1-r^{t}}{1-r}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{1.08}\times \dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{1-\dfrac{1}{%&#xA;1.08}}=3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{0.08}=3\iff \left( \dfrac{1}{1.08}%&#xA;\right) ^{t}=0.76.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying logarithms, we obtain the &quot;time&quot; $t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t\log \left( \dfrac{1}{1.08}\right) =\log 0.76,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\dfrac{\log 0.76}{-\log 1.08}\approx 3.5659,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that we need more than $3$ periods and less than $4$, at an interest rate of $8\%$, coumpounded per period, to get a total of $3$ currency units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\ast)$ Derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{t}\right) -\left(&#xA;ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-r)S=u_{1}-ru_{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\dfrac{u_{1}-ru_{t}}{1-r}=\dfrac{u_{1}-u_{1}r^{t}}{1-r}$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-07T12:00:29.380" LastActivityDate="2010-10-07T12:00:29.380" />
  <row Id="6241" PostTypeId="2" ParentId="6179" CreationDate="2010-10-07T00:54:04.920" Score="2" ViewCount="0" Body="&lt;p&gt;This is not a broad answer like those informative ones already posted, but just a single resource that may not be well known: the book &lt;a href=&quot;http://www.howround.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;How Round is Your Circle&lt;/em&gt;&lt;/a&gt;. It emphasizes the connections between the physical and mathematics, as its subtitle indicates: &lt;em&gt;Where Engineering and Mathematics Meet&lt;/em&gt;.  It is likely too advanced in many spots for your clientele, but it has a wealth of material that can be adapted.  I've had 8th and 9th-grade children build pantograph linkages (covered in Chapter 3) and learn that vector addition is commutative in a very physical way.  You won't get that directly from this book, but this book can give you such ideas.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-07T00:54:04.920" CommunityOwnedDate="2010-10-07T00:54:04.920" />
  <row Id="6242" PostTypeId="2" ParentId="6087" CreationDate="2010-10-07T01:05:22.640" Score="3" ViewCount="0" Body="&lt;p&gt;The proof follows immediately from the  following variant of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&quot; rel=&quot;nofollow&quot;&gt;AM-GM inequality&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\ \ 0 &amp;lt; n &amp;lt; m \ \Rightarrow\ 3\:n+m\ &amp;lt;\ (\sqrt{n}\ \ +\: \ \sqrt{m}\ )^2\ &amp;lt;\  2\:n+2\:m  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\:\ m = n+1 \ \:\Rightarrow\  4\:n+1\ \ &amp;lt;\ (\sqrt{n} + \sqrt{n+1})^2\ &amp;lt;\ 4\:n+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first inequality has an easy proof: expand the middle term, then subtract $\rm\ n+m\ $ from all the terms. Then it reduces on the left to  $\rm\ n\ &amp;lt; \sqrt{nm}\ $ via $\rm\ n^2 &amp;lt; nm\ $, and to the AM-GM inequality on the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such minor variations on the $\:$ AM-GM $\:$ arise not too infrequently in practice (e.g. in competition problems).$\:$  Thus it is worthwhile to point them out in their full generality to help aid in recognizing them &quot;in the wild&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-07T01:44:05.467" LastActivityDate="2010-10-07T01:44:05.467" CommentCount="1" />
  <row Id="6243" PostTypeId="2" ParentId="6220" CreationDate="2010-10-07T02:02:29.680" Score="6" ViewCount="0" Body="&lt;p&gt;As a way of restating Bill's answer, what you have can in fact be expressed in terms of the so-called &lt;a href=&quot;http://dlmf.nist.gov/18.34.E1&quot; rel=&quot;nofollow&quot;&gt;(generalized) Bessel polynomial&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_n(x;a)=(n+a-1)_n \left(\frac{x}{2}\right)^n {}_1 F_1 \left(-n;-2n-a+2;\frac{2}{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the Kummer hypergeometric series degenerates to a polynomial here because both the numerator and denominator parameters are negative integers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your original expression, then, in terms of the (generalized) Bessel polynomial, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-2)^{n-1}(n+1)!y_{n-1}(-1;4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The references in the DLMF can point you to papers where the (generalized) Bessel polynomials have been studied; for an elementary treatment, see Chihara's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0677041500&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;An Introduction to Orthogonal Polynomials&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-07T02:02:29.680" />
  <row Id="6244" PostTypeId="1" AcceptedAnswerId="6247" CreationDate="2010-10-07T02:10:17.703" Score="8" ViewCount="650" Body="&lt;p&gt;It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-10-11T14:09:27.680" Title="Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?  " Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="4" />
  <row Id="6245" PostTypeId="1" AcceptedAnswerId="6246" CreationDate="2010-10-07T02:23:42.610" Score="3" ViewCount="118" Body="&lt;p&gt;I'm currently studying basic Combinatorics for a college course and my professor is awful (and that is being generous).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I'm looking for good resources to learn basic Combinatorics so that I can prepare for his exams and ace the course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2010-10-07T02:42:36.493" Title="Good resources (book or otherwise) to learn/study basic Combinatorics" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6246" PostTypeId="2" ParentId="6245" CreationDate="2010-10-07T02:42:36.493" Score="4" ViewCount="0" Body="&lt;p&gt;IMO there are two great references at two different levels. The first is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131001191&quot; rel=&quot;nofollow&quot;&gt;Brualdi&lt;/a&gt;. This book is awesome. It is super clear, super straightfoward, and very readable by yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is much more advanced, but even better! It is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642072534&quot; rel=&quot;nofollow&quot;&gt;Aigner&lt;/a&gt;'s GTM on the subject. This book is fairly advanced, but rarely have I seen such a beautiful exposition of any mathematical subject. His proofs are impressively elegant, and his exercises are very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-07T02:42:36.493" />
  <row Id="6247" PostTypeId="2" ParentId="6244" CreationDate="2010-10-07T03:14:48.747" Score="13" ViewCount="0" Body="&lt;p&gt;As Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\alpha_1 v_1+\cdots+\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, $\mathbb{R}$ cannot be finite dimensional over $\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your further question in the comments, whether a vector space over $\mathbb{Q}$ is finite dimensional &lt;em&gt;if and only if&lt;/em&gt; the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb{Q}$ that are countable as sets. The simplest example is $\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\mathbb{Q}$, which is a countable set, and has dimension $\aleph_0$, with basis $\{1,x,x^2,\ldots,x^n,\ldots\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Of course, if $V$ is a vector space over $\mathbb{Q}$, then it has &lt;em&gt;countable dimension&lt;/em&gt; (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\mathbb{R}$ infinite dimensional over $\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is &lt;em&gt;uncountably&lt;/em&gt;-dimensional over $\mathbb{Q}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-07T04:09:02.470" LastActivityDate="2010-10-07T04:09:02.470" CommentCount="4" />
  <row Id="6248" PostTypeId="2" ParentId="5119" CreationDate="2010-10-07T03:38:17.343" Score="0" ViewCount="0" Body="&lt;p&gt;By continuity, the set of $\mathcal{F}_{i}$'s converge to the set $\cup_{i=1}^{\infty}\mathcal{F}_i$. The limiting set is a $\sigma$-algebra by definition of the problem. What is invalid about this argument?&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-07T03:38:17.343" CommentCount="6" />
  <row Id="6250" PostTypeId="2" ParentId="6244" CreationDate="2010-10-07T04:01:41.873" Score="34" ViewCount="0" Body="&lt;p&gt;The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $\rm\: \mathbb Q$-independent set of reals. Consider the set consisting of the logs of all primes $\rm\: p_i\:$. Then if $\rm\ \: c_1\ log\ p_1 +\:\cdots\: + c_n\ log \ p_n \: =\ 0\:,\ \: c_i\in\mathbb Q\:,\: $ multipyling by a common denominator we can assume that all $\rm\ c_i \in \mathbb Z\:,\:$ and, exponentiating, we obtain $\rm\: p_1^{\:c_1}\cdots p_n^{\:c_n} = 1\ \Rightarrow\ c_i = 0,\:$ for all $\rm\:i\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-07T04:31:18.470" LastActivityDate="2010-10-07T04:31:18.470" CommentCount="4" />
  <row Id="6251" PostTypeId="1" AcceptedAnswerId="6252" CreationDate="2010-10-07T04:13:49.453" Score="4" ViewCount="67" Body="&lt;p&gt;Suppose $f$ is an analytic function defined on the unit disk, $D$.   I want to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{D} f(\omega) dA(\omega)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using a change of variable.   Suppose $\phi$ is a conformal map of the $D$ onto itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{D} f(\omega) dA(\omega) = \int_{D} f(\phi(z)) |\phi^{'}(z)|^{2}  dA(z) $?,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\phi^{'}$ is the derivative of $\phi$. &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-07T04:30:05.703" Title="Change of Variable (conformal map)" Tags="&lt;complex-analysis&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6252" PostTypeId="2" ParentId="6251" CreationDate="2010-10-07T04:30:05.703" Score="2" ViewCount="0" Body="&lt;p&gt;Because $|\phi'(z)|^2$ is the determinant of the Jacobian of $\phi$, this follows from the substitution formula for open sets in $\mathbb{R}^n$, as seen for example in &lt;a href=&quot;http://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;, which contains further references for the more general result.  Technically, to apply the result directly you would break up each side into real and imaginary parts.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-07T04:30:05.703" />
  <row Id="6253" PostTypeId="2" ParentId="2949" CreationDate="2010-10-07T04:47:31.343" Score="1" ViewCount="0" Body="&lt;p&gt;For me it would be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem&quot; rel=&quot;nofollow&quot;&gt;Green-Tao&lt;/a&gt; theorem, which states: For any natural number $k$, there exist $k$-term arithmetic progressions of primes.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-07T04:47:31.343" CommunityOwnedDate="2010-10-07T04:47:31.343" CommentCount="0" />
  <row Id="6254" PostTypeId="1" CreationDate="2010-10-07T04:50:55.197" Score="3" ViewCount="116" Body="&lt;p&gt;To give you some background, I have a grasp on the basics of statistics and probability theory and even remember touching Bayes theorem at the university data mining course. But being a few years away from the university made my math got &lt;em&gt;extremely&lt;/em&gt; rusty (so much for last-minute pre-exam cramming). While I remember various random basic concepts, there are a lot of gaps in my understanding of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a good material (a book, a site, or otherwise equally accessible medium) to revise the fundamentals and go beyond basics? I'd like a book that can be actually read &lt;em&gt;as a book&lt;/em&gt; (most statistics books are really dry and are close to being reference material, rather than a book).&lt;/p&gt;&#xA;" OwnerUserId="2360" LastActivityDate="2010-10-07T06:01:44.723" CommunityOwnedDate="2010-10-07T04:53:02.180" Title="Recommend a statistics fundamentals book" Tags="&lt;statistics&gt;&lt;books&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6255" PostTypeId="2" ParentId="6254" CreationDate="2010-10-07T04:55:41.140" Score="1" ViewCount="0" Body="&lt;p&gt;Alex i recommend you these two books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to Probability by Sheldon Ross&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An Introduction to Probability Theory and Its Applications: William Feller&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both are very good books for an introductory level and the second one is a classic. It's referred by many people.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-07T04:55:41.140" CommunityOwnedDate="2010-10-07T04:55:41.140" />
  <row Id="6256" PostTypeId="2" ParentId="6254" CreationDate="2010-10-07T06:01:44.723" Score="3" ViewCount="0" Body="&lt;p&gt;Freedman, Pisani and Purves, &lt;em&gt;Statistics&lt;/em&gt;, followed by Freedman, &lt;em&gt;Statistical Models&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-07T06:01:44.723" CommunityOwnedDate="2010-10-07T06:01:44.723" CommentCount="2" />
  <row Id="6257" PostTypeId="2" ParentId="6203" CreationDate="2010-10-07T08:00:48.810" Score="0" ViewCount="0" Body="&lt;p&gt;An inductive set $A$ is any set with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\left\{\right\}\in A$&lt;/li&gt;&#xA;&lt;li&gt;$\forall x\left(x\in A\rightarrow S\left(x\right)\in A\right)$, where $S\left(x\right)$ is the successor of $x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can prove that $\mathbb{N}$ is an inductive set, and that if $A$ is any inductive set, then $\mathbb{N}\subseteq A$, but the other way around is false, that is, you can't prove that $A\subseteq\mathbb{N}$. Take for example $A=\mathbb{N}\cup\left\{a\right\}$, where $a$ is anything except a natural number. $A$ is inductive, but it is obvious that $A\subseteq\mathbb{N}$ is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, as far as I know, the best definition of the set of natural numbers is the following one: $\forall x\left(x\in\mathbb{N}\leftrightarrow\forall A\left(x\in A\right)\right)$, where $A$ is any inductive set.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-10-07T08:00:48.810" CommentCount="2" />
  <row Id="6258" PostTypeId="1" AcceptedAnswerId="6260" CreationDate="2010-10-07T08:16:07.097" Score="4" ViewCount="393" Body="&lt;p&gt;I've come across a paper that mentions this as a fact...where can I find the proof?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-10-07T19:25:11.063" Title="matrices commute=common set of eigenvectors?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6260" PostTypeId="2" ParentId="6258" CreationDate="2010-10-07T08:27:04.560" Score="11" ViewCount="0" Body="&lt;p&gt;Suppose that $A$ and $B$ are $n\times n$ matrices, with complex entries say, that commute.&lt;br&gt;&#xA;Then we decompose $\mathbb C^n$ as a direct sum of eigenspaces of $A$, say&#xA;$\mathbb C^n = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_m}$, where $\lambda_1,\ldots,&#xA;\lambda_m$ are the eigenvalues of $A$, and $E_{\lambda_i}$ is the eigenspace for $\lambda_i$.&#xA;(Here $m \leq n$, but some eigenspaces could be of dimension bigger than one, so we needn't&#xA; have $m = n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one sees that since $B$ commutes with $A$, $B$ preserves each of the $E_{\lambda_i}$:&#xA;if $A v = \lambda_i v, $ then $A (B v) = (AB)v = (BA)v = B(Av) = B(\lambda_i v) = \lambda_i Bv.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we considered $B$ restricted to each $E_{\lambda_i}$ separately, and decompose&#xA;each $E_{\lambda_i}$ into a sum of eigenspaces for $B$.   Putting all these decompositions together, we get a decomposition of $\mathbb C^n$ into a direct sum of spaces, each&#xA;of which is a simultaneous eigenspace for $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: I am cheating here, in that $A$ and $B$ may not be diagonalizable (and then the statement of your question is not literally true), but in this case, if you replace &quot;eigenspace&quot; by &quot;generalized eigenspace&quot;, the above argument goes through just as well.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-07T19:25:11.063" LastActivityDate="2010-10-07T19:25:11.063" CommentCount="4" />
  <row Id="6261" PostTypeId="2" ParentId="6258" CreationDate="2010-10-07T08:30:05.263" Score="8" ViewCount="0" Body="&lt;p&gt;This is false in a sort of trivial way.  The identity matrix $I$ commutes with every matrix and has eigenvector set all of the underlying vector space $V$, but no non-identity matrix has this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that two matrices which commute and are also &lt;em&gt;diagonalizable&lt;/em&gt; are &lt;a href=&quot;http://en.wikipedia.org/wiki/Diagonalizable_matrix#Simultaneous_diagonalization&quot; rel=&quot;nofollow&quot;&gt;simultaneously diagonalizable&lt;/a&gt;.  The proof is particularly simple if at least one of the two matrices has distinct eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-07T08:43:11.853" LastActivityDate="2010-10-07T08:43:11.853" CommentCount="3" />
  <row Id="6262" PostTypeId="1" AcceptedAnswerId="6264" CreationDate="2010-10-07T09:43:08.133" Score="0" ViewCount="111" Body="&lt;p&gt;I'm trying to figure out how many ways are there to stack N cubes on top of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get a particular stack of 4 cubes let's say, the order of the cubes doesn't matter, just which sides are being shown does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm generally finding 2 numbers: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) 6 * 24^(n-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) 3 * 24^(n-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct answer, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand the question better, you can refer to the Instant Insanity puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Instant_Insanity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Instant_Insanity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-10-07T10:01:23.413" LastActivityDate="2010-10-08T00:45:45.987" Title="Number of ways to stack N cubes" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6263" PostTypeId="1" AcceptedAnswerId="6271" CreationDate="2010-10-07T10:18:55.953" Score="4" ViewCount="231" Body="&lt;p&gt;This question is mainly to understand the meaning of my professor's correction to a proof of a theorem I gave during an oral examination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was to show that $\text{diam}A = \text{diam}\bar{A}$ where $\text{diam}$ is the diameter of a subspace of a metric space and $\bar{A}$ is the closure of the subset $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proceded the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\bar{A} \supseteq A$, then we have $\text{diam}\bar{A}\geq \text{diam}A$. To prove the other inequality, we procede this way: $\forall x,y \in \bar{A}$ there exist $a,b \in A$ such that $d(x,a) &amp;lt; \epsilon$ and $d(y,b) &amp;lt; \epsilon$ with $\epsilon &gt; 0$. Then we know that $$d(x,y) \leq d(x,a) + d(y,b) + d(a,b) &amp;lt; 2 \epsilon + d(a,b) \leq 2 \epsilon + \text{diam}A$$ and by the arbitrarity of $\epsilon$ the inequality follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor said that I coudn't do it this way and that, instead, I should do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x,y \in \bar{A}$ there exist $a_n,b_n \in A$ such that $d(x,a_n) &amp;lt; \frac{1}{n}$ and $d(y,b_n) &amp;lt; \frac{1}{n}$, then we would have $$d(x,y) \leq d(x,a_n) + d(y,b_n) + d(a_n,b_n) &amp;lt; \frac{2}{n} + d(a_n,b_n)$$, the next step would be to take the sup of it all this way: $$\sup_{x,y \in \bar{A}} d(x,y) \leq \frac{2}{n}+ \sup_{n \in \mathbb{N}} d(a_n,b_n)$$ from which the inequality follows because of the archimedean property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how are these two proofs different? Why is mine wrong?&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-10-23T17:55:17.097" Title="On two different proofs" Tags="&lt;proof-strategy&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="6264" PostTypeId="2" ParentId="6262" CreationDate="2010-10-07T11:00:36.487" Score="1" ViewCount="0" Body="&lt;p&gt;Your solutions presuppose that the cubes are identical; this is not the&#xA;case in Instant Insanity. If they are all distinguishable you are&#xA;missing a factor of $N!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller possible solution is valid under the assumption&#xA;that a stack is regarded as identical with that stack &#xA;produced by turning the original stack upside down. The larger&#xA;solution regards these two stacks as different.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-07T11:00:36.487" CommentCount="1" />
  <row Id="6265" PostTypeId="1" AcceptedAnswerId="6266" CreationDate="2010-10-07T13:27:16.890" Score="4" ViewCount="131" Body="&lt;p&gt;If $\sum a_n$ and $\sum b_n$ both converge and one of them absolutely then the Cauchy product $\sum c_n$ converges to $\sum a_n \sum b_n$. ($c_n = \sum_{k = 0}^n a_k b_{n - k}$), by Mertens Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if both converge conditionally then the product does not have to converge as $a_n = b_n = (-1)^n/n$ shows.&#xA;My question now is: What if $\sum a_n$ and $\sum b_n$ both converge conditionally and $\sum c_n$ converges, then is it always true that $\sum c_n$ converges to the product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this is not homework, I'm already past the real analysis part.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-07T13:42:58.830" Title="Conditional convergence, Mertens theorem." Tags="&lt;real-analysis&gt;&lt;elementary&gt;" AnswerCount="1" />
  <row Id="6266" PostTypeId="2" ParentId="6265" CreationDate="2010-10-07T13:42:58.830" Score="6" ViewCount="0" Body="&lt;p&gt;This follows readily from Abel's convergence theorem: if $\sum_0^\infty a_n$&#xA;converges then&#xA;$$\sum_0^\infty a_n=\lim_{x\to1^-}\sum_0^\infty a_n x^n.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-07T13:42:58.830" CommentCount="1" />
  <row Id="6267" PostTypeId="2" ParentId="6195" CreationDate="2010-10-07T14:13:16.657" Score="1" ViewCount="0" Body="&lt;p&gt;@joseph: on the rhs of your recursion relation for $f(n,h,t)$, shouldn't the $n$ be $n-1$?&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-07T14:13:16.657" CommentCount="1" />
  <row Id="6268" PostTypeId="1" AcceptedAnswerId="6269" CreationDate="2010-10-07T14:23:08.193" Score="1" ViewCount="206" Body="&lt;p&gt;Suppose $f: \mathbb{R} \to \mathbb{R}$ is a continuous function which is 1-1, then does $f$ map Borel sets onto Borel sets?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T07:33:58.040" LastActivityDate="2010-12-08T07:33:58.040" Title="Continuous and 1-1 $\Longrightarrow$ Borel Sets $\longrightarrow$ Borel sets?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6269" PostTypeId="2" ParentId="6268" CreationDate="2010-10-07T14:56:17.320" Score="7" ViewCount="0" Body="&lt;p&gt;The continuous image of a Borel set in a Polish space is analytic. So it is Lebesgue measurable. If the complement of that analytic set is analytic then it is Borel (a theorem by Suslin). Now we know that for injective functions we have $f(A \setminus B) = f(A) \setminus f(B)$ so the complement is analytic too, hence the continuous one-to-one image of a Borel set is Borel in a Polish space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is unnecessarly cluttered but it is the thing that pops up in my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f$ maps open sets to open sets. This is because injective functions $f:\mathbb{R} \to \mathbb{R}$ are strictly monotonic, so it maps open intervals to open intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $D := \{Q \in P(\mathbb{R}) : f(Q) \textrm{ is Borel}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\mathbb{R}$ is in $D$, it is closed under complements and unions (because it is injective), thus $D$ is a $\sigma$-algebra. So let $B$ be the Borel $\sigma$-algebra and $O$ the open sets of $\mathbb{R}$. Then $B \subset D$ if $O \subset D$. So we get that for all Borel sets the image is Borel.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-07T15:31:14.710" LastActivityDate="2010-10-07T15:31:14.710" CommentCount="5" />
  <row Id="6270" PostTypeId="2" ParentId="2949" CreationDate="2010-10-07T15:16:21.143" Score="1" ViewCount="0" Body="&lt;p&gt;Aside from some results I found amazing that have already been mentionned, Lagrange's Theorem in group theory is one that amazed me for some time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who don't know about it, it tells us that the order of any subgroup of a group $G$ divides the order of $G$.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastEditorUserId="1865" LastEditorDisplayName="" LastEditDate="2010-10-07T15:21:45.950" LastActivityDate="2010-10-07T15:21:45.950" CommunityOwnedDate="2010-10-07T15:16:21.143" />
  <row Id="6271" PostTypeId="2" ParentId="6263" CreationDate="2010-10-07T15:36:50.567" Score="7" ViewCount="0" Body="&lt;p&gt;Just to elaborate on some of the comments: looking at your argument, you can see that&#xA;there is an implicit step which it might be good to make explicit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your last displayed equation reads&#xA;$$d(x,y) \leq \cdots \leq 2\epsilon + diam(A).$$&#xA;Now the implict step is that you want to take a sup over all pairs $x,y$ of elements&#xA;of $\overline{A}$ to conclude that&#xA;$$diam(\overline{A}) \leq 2\epsilon + diam(A).$$&#xA;Writing out this step, we see that $x$ and $y$ have disappeared, but that the quantity&#xA;$\epsilon$ persists.  And now, for this to be valid, we see that it is important to note&#xA;that $\epsilon$ can be fixed independenty of $x$ and $y$.  Luckily, this is possible: if we&#xA;fix $\epsilon &gt; 0$ in advance, then for any pair $x,y$ of points in $\overline{A}$, we &lt;em&gt;can&lt;/em&gt; find $a$ and $b$ in $A$ within distance $\epsilon$ of them, as you note.  But it is important to state that $\epsilon$ is fixed in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for being pedantic about this is that in other contexts, one may try to make similar arguments in which it turns out that $\epsilon$ can't be fixed in advance.  This kind of issue (whether a certain quantity (in your case, the quantity $\epsilon$) can be chosen &lt;em&gt;uniformly&lt;/em&gt;  in terms of some other quantities (in your case, $x$ and $y$) &#xA;comes up constantly in analysis, which is why people make a big deal about it in introductory courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final remark: the last step of your professor's argument is misstated (as was also noted).&#xA;One should add one more step to the second-to-last line, to conclude that&#xA;$$d(x,y) \leq \ldots \leq \frac{2}{n} + d(a_n,b_n) \leq \frac{2}{n} + diam(A)$$&#xA;(this is the same as the implicit step in your argument which I mentioned above),&#xA;and &lt;em&gt;then&lt;/em&gt; pass to sups, to find that&#xA;$$\sup_{x,y} d(x,y) \leq \frac{2}{n} + diam(A).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, the only differences between this (corrected) argument and yours are:&#xA;(a) using $1/n$ instead of $\epsilon$, which is just a matter of taste and style;&#xA;(b) making the implicit step explicit; and&#xA;(c) making it explicit that $n$ is fixed in advance of considering any particular&#xA;pair $x$ and $y$ (which is pedantic, but as I already remarked, important).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-23T17:55:17.097" LastActivityDate="2010-10-23T17:55:17.097" />
  <row Id="6272" PostTypeId="1" CreationDate="2010-10-07T16:08:23.040" Score="1" ViewCount="59" Body="&lt;p&gt;I am building an android application that records accelerometer data during sleep, so as to analyze sleep trends and optionally wake the user near a desired time during light sleep. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already built the component that collects and stores data, as well as the alarm. I still need to tackle the beast of displaying sleep data in a really meaningful way, one that preferably also lends itself to analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of pictures say two thousand words: (I can only post one link due to low rep)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://code.google.com/p/electricsleep/downloads/list&quot; rel=&quot;nofollow&quot;&gt;Unfiltered accelerometer activity summed @ 30 second intervals with n data points AND The same data, filtered by averaging n/100 parts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The translucent white line indicates movements that would set off the alarm, if enabled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, neither of these are optimal solutions- the first is a little hard to understand for the average user, and the second, which is easier to read, hides a lot of what is really going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there is a better way to filter the data- I just don't have the mathematical background to know where to start looking (just look at my tags, lolz.) I'm looking for suggestions, so I can at least have some reference point to start researching a better solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2365" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-07T16:30:42.143" LastActivityDate="2010-10-07T16:30:42.143" Title="Meaningful way to reinterpret a sleep-movement-graph/set" Tags="&lt;statistics&gt;" CommentCount="6" FavoriteCount="0" />
  <row Id="6273" PostTypeId="1" AcceptedAnswerId="6275" CreationDate="2010-10-07T16:18:04.363" Score="3" ViewCount="167" Body="&lt;p&gt;I am not a mathematician(IANAM), however I wish I could be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: I often find this at the bottom of pages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page generated in 0.00013899803161621&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, I come across &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page generated in 9.0122222900391E-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that time mean. &#xA;I tried searching wikipedia for E and maths but found the &lt;a href=&quot;http://en.wikipedia.org/wiki/E_%28mathematical_constant%29&quot; rel=&quot;nofollow&quot;&gt;e mathematical constant&lt;/a&gt; with a scary looking graph. My guess is E stand for Exponential or something and -5 is the power it is raised to. And the time is a really small number. But that doesn't make sense when compared to the other time in the question is 0.00013899803161621 bigger than 9.0122222900391E-5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it means x10^-5, then 9.0122222900391E-5 will be 0.000090122222900391 which is smaller than 0.00013899803161621 &lt;/p&gt;&#xA;" OwnerUserId="2366" LastEditorUserId="2366" LastEditorDisplayName="" LastEditDate="2010-10-07T16:32:56.190" LastActivityDate="2010-10-07T16:32:56.190" Title="What doe E mean in 9.0122222900391E-5?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="6274" PostTypeId="1" CreationDate="2010-10-07T16:22:25.387" Score="3" ViewCount="276" Body="&lt;p&gt;In a book about radar signal processing, they are estimating the squared distance between two objects as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d^2 = \frac{(R_p-R_o)^2}{\sigma^2_R} + \frac{(\theta_p-\theta_o)^2}{\sigma^2_\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where R is a radius, $\theta$ is an angle and $\sigma^2$ is the variance of the radius and angle respectivly. p is the predicted target and o is an observation. The book is Blackman, S. - Design and Analysis of Modern Tracking Systems&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just me or is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimensional_analysis&quot; rel=&quot;nofollow&quot;&gt;dimentional analysis&lt;/a&gt; for this equation a bit off?&lt;/p&gt;&#xA;" OwnerUserId="2367" LastEditorUserId="2367" LastEditorDisplayName="" LastEditDate="2010-10-07T20:43:39.217" LastActivityDate="2011-01-07T03:44:39.250" Title="Weird distance estimate?" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="6275" PostTypeId="2" ParentId="6273" CreationDate="2010-10-07T16:22:36.450" Score="7" ViewCount="0" Body="&lt;p&gt;It means 9.0122222900391 times $10^{-5}$. It's known as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Scientific_notation&quot; rel=&quot;nofollow&quot;&gt;scientific notation&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-07T16:22:36.450" CommentCount="8" />
  <row Id="6276" PostTypeId="2" ParentId="6273" CreationDate="2010-10-07T16:23:01.863" Score="4" ViewCount="0" Body="&lt;p&gt;E-5 just means x 10^(-5) i.e. divided by 100000&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-10-07T16:23:01.863" />
  <row Id="6277" PostTypeId="1" CreationDate="2010-10-07T16:53:35.450" Score="1" ViewCount="129" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; question simplified to remove confusion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a sorted list of items with indexes from 1 to &lt;code&gt;N&lt;/code&gt;, and given &lt;em&gt;only&lt;/em&gt; an index number &lt;code&gt;i&lt;/code&gt; and the maximum index &lt;code&gt;N&lt;/code&gt;, is there a simple function which will return the two possible indexes that would be reached next in a binary search of the list?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i = 6&#xA;N = 8&#xA;LeftSearchIndex(i, N) = 5&#xA;RightSearchIndex(i, N) = 7&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-10-07T19:07:32.607" LastActivityDate="2011-01-05T19:44:39.360" Title="Is there a simple function which can be used to determine the two &quot;next&quot; search indexes in a binary search?" Tags="&lt;algorithms&gt;&lt;discrete-mathematics&gt;&lt;searching&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6278" PostTypeId="2" ParentId="6277" CreationDate="2010-10-07T17:05:24.320" Score="1" ViewCount="0" Body="&lt;p&gt;No longer relevant, keeping answer around for comments.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-07T19:25:46.210" LastActivityDate="2010-10-07T19:25:46.210" CommentCount="16" />
  <row Id="6279" PostTypeId="2" ParentId="6195" CreationDate="2010-10-07T17:58:58.357" Score="3" ViewCount="0" Body="&lt;p&gt;This seems to be related to Gittins Indices. Gittins Indices are a way of solving these kind of optimal stopping problems for some classes of problems, and basically give you a way of balancing how much you are expected to gain given your current knowledge and how much more you &lt;em&gt;could&lt;/em&gt; gain by risking obtaining more information about the process (or probability of flipping heads, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruno&lt;/p&gt;&#xA;" OwnerUserId="2368" LastActivityDate="2010-10-07T17:58:58.357" CommentCount="1" />
  <row Id="6280" PostTypeId="1" AcceptedAnswerId="6283" CreationDate="2010-10-07T18:05:01.580" Score="13" ViewCount="361" Body="&lt;p&gt;Let $\pi(x)$ be the number of primes less than or equal to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One observes that, $\pi(6) \mid 6$, $\pi(8) \mid 8$. So does $\pi(x) \mid x$ for only finitely many $x$, or is this fact true for infinitely many $x$?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-08T02:06:26.897" LastActivityDate="2010-10-08T02:06:26.897" Title="$\pi(x) \mid x$ for how many $x$?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" />
  <row Id="6282" PostTypeId="1" AcceptedAnswerId="6284" CreationDate="2010-10-07T18:45:19.383" Score="1" ViewCount="167" Body="&lt;p&gt;Let us consider that there are &lt;strong&gt;G girl students&lt;/strong&gt; and &lt;strong&gt;B boy students&lt;/strong&gt; in a class, we need to arrange them in a single row, but arrangement of students should be in order to minimize the &lt;strong&gt;gender regularity*&lt;/strong&gt;. Find the minimum gender regularity among all possible arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The gender regularity of an arrangement is the maximum number of students of the same gender (all girls or all boys) that appear consecutively.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:49:12.317" LastActivityDate="2010-10-16T17:49:12.317" Title="Minimizing Gender Regularity in a linear arrangement of boys and girls." Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="6283" PostTypeId="2" ParentId="6280" CreationDate="2010-10-07T18:47:19.847" Score="18" ViewCount="0" Body="&lt;p&gt;Well, it's not hard to show that for every natural $k&gt;2$ equation $x=k\pi(x)$ has a positive solution.&#xA;Proof by contradiction: Imagine, that $x\ne k\pi(x)$ for every natural $x$. But then - for $x=2$:$x-k\pi(x)=2-k&amp;lt;0$ and for very large $x$: $x-k\pi(x)\sim x(1-\frac{k}{\ln x})&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there should be such $t$, that $t-k\pi(t)&amp;lt;0$ and $(t+1)-k\pi(t+1)&gt;0$. But then from one hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;t+1-k\pi(t+1)-(t-k\pi(t))\ge 2&#xA;$$&#xA;as a difference of positive integer and negative integer, but from the other hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;t+1-k\pi(t+1)-(t-k\pi(t))=1-k(\pi(t+1)-\pi(t))\le 1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contradiction.&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="425" LastEditorDisplayName="" LastEditDate="2010-10-07T19:00:19.800" LastActivityDate="2010-10-07T19:00:19.800" CommentCount="4" />
  <row Id="6284" PostTypeId="2" ParentId="6282" CreationDate="2010-10-07T19:36:38.393" Score="3" ViewCount="0" Body="&lt;p&gt;I will try to show the solution is ceil(M / N+1) where M is the larger of B and G, N is the lesser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best case is M = N, where we need merely arrange the genders alternately, and regularity = ceil(M / M+1) = 1.  Worst case is a single-gender class, i.e N=0 in which regularity = ceil(M/1) = M, i.e. the whole class size.  In all other cases, clearly 1 &amp;lt; regularity &amp;lt; M+N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now work quasi-inductively.  If N = 1, then we place the singleton in the center of the line, with half of the others to each side.  If the larger group is even, there are M/2 to each side (the ceil of a whole number is itself.)  If the larger group is odd, there will be one more on one side of the single, and it is this group that gives regularity = ceil(M/2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If N = 2, do a similar process, divide the M into three groups, placing the two in between them (picture ---&lt;em&gt;----&lt;/em&gt;--- where the * are the small group, - the larger).  If the larger group is divisible by 3, there are M/3 in each division.  If not, distribute the 1 or 2 remainder over distinct divisions, and the largest will contain ceil(M/3) students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As N increases toward M, repeat the concept, dividing the larger group into M/N+1 divisions, place the N between them, and distribute the remainder.&lt;/p&gt;&#xA;" OwnerUserId="575" LastActivityDate="2010-10-07T19:36:38.393" />
  <row Id="6285" PostTypeId="2" ParentId="3085" CreationDate="2010-10-07T19:39:11.610" Score="1" ViewCount="0" Body="&lt;p&gt;The most comprehensive list I know of is H.W. Gould's &lt;em&gt;Combinatorial Identities.&lt;/em&gt;  It is available directly from him if you contact him.  He also has some pdf documents available for download from his &lt;a href=&quot;http://www.math.wvu.edu/~gould/&quot; rel=&quot;nofollow&quot;&gt;web site&lt;/a&gt;.  Although he says they do &quot;NOT replace [&lt;em&gt;Combinatorial Identities&lt;/em&gt;] which remains in print with supplements,&quot; they still contain many more binomial identities even than in &lt;em&gt;Concrete Mathematics&lt;/em&gt;.  In general, Gould's work is a great resource for this sort of thing; he has spent much of his career collecting and proving combinatorial identities. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-07T19:39:11.610" CommentCount="1" />
  <row Id="6286" PostTypeId="2" ParentId="6280" CreationDate="2010-10-07T20:28:34.083" Score="3" ViewCount="0" Body="&lt;p&gt;This is Sloane's &lt;a href=&quot;http://oeis.org/classic/A057809&quot; rel=&quot;nofollow&quot;&gt;A057809&lt;/a&gt;, but unfortunately there's not much information there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heuristically, the chance that $\pi(x)|x$ is $\log x/x$ which suggests that there are $0.5\log^2 x$ such numbers up to x, that is, infinitely many.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-07T20:28:34.083" CommentCount="1" />
  <row Id="6287" PostTypeId="1" AcceptedAnswerId="6290" CreationDate="2010-10-07T20:29:39.683" Score="2" ViewCount="72" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to simplify the following booleans:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = ¬A¬B¬C¬D + ¬A¬B¬CD + ¬A¬BC¬D + ¬A¬BCD + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = ¬A¬B + ¬B¬D + AC¬D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My solution is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = ¬A¬B (¬C¬D + ¬CD + C¬D + CD) + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;  = ¬A¬B (¬C (¬D + D) + C (¬D + D)) + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;  = ¬A¬B (¬C + C) + A¬B¬C¬D + A¬BC¬D + ABC¬D&#xA;  = ¬A¬B + A¬D (¬B¬C + ¬BC + BC) &#xA;  = ¬A¬B + A¬D (¬B¬C + (¬BC + ¬BC) + BC) // law of idempotency&#xA;  = ¬A¬B + A¬D (¬B (¬C + C) + C (¬B + B))&#xA;  = ¬A¬B + A¬D (¬B + C)&#xA;  = ¬A¬B + A¬B¬D + AC¬D&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see.. my answer varies with the solution by the 2nd term. Any ideas? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2371" LastActivityDate="2010-10-07T21:22:40.763" Title="Boolean Simplification" Tags="&lt;logic&gt;&lt;algebra&gt;" AnswerCount="2" />
  <row Id="6289" PostTypeId="2" ParentId="6287" CreationDate="2010-10-07T21:07:38.527" Score="0" ViewCount="0" Body="&lt;p&gt;I agree with you.  The last three terms all have A required, so anything that comes out of them (your last two terms) need an A.  Are you sure the given solution is right?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-07T21:07:38.527" CommentCount="1" />
  <row Id="6290" PostTypeId="2" ParentId="6287" CreationDate="2010-10-07T21:20:35.123" Score="2" ViewCount="0" Body="&lt;p&gt;All your work is right.  The first two terms of your answer simplify to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Y = ¬A¬B + A¬B¬D&#xA;  = ¬B(¬A + A¬D)&#xA;  = ¬B(¬AD + ¬A¬D + A¬D) // split the first term&#xA;  = ¬B(¬AD + ¬A¬D + ¬A¬D + A¬D) // idempotence&#xA;  = ¬B(¬A + ¬D) // combine the first two and last two terms&#xA;  = ¬A¬B + ¬B¬D.&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I wouldn't have believed it myself at first, but idempotence lets the &lt;code&gt;¬A&lt;/code&gt; &quot;spit out&quot; a &lt;code&gt;¬A¬D&lt;/code&gt;, which is, after all, just a subset.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-07T21:20:35.123" CommentCount="3" />
  <row Id="6292" PostTypeId="1" AcceptedAnswerId="6298" CreationDate="2010-10-07T23:15:20.493" Score="1" ViewCount="77" Body="&lt;p&gt;I am trying to calculate the average distance between all vertices of a complete $k$-ary tree. A complete $k$-ary tree is a tree such that all vertices have $k$ children except for the leafs of the tree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the tree has $r$ levels, where level 0 is the root. So the height of the tree is $r-1$. First i have to know how many vertices there are in the tree. For this i would sum up the vertices on each level of the tree which would give me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{r}k^i = k^1 + k^2 + ...+ k^r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(is there an easier way of representing this?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how many ways are there to choose paths in this tree that are of length $1, 2, 3, ..., r-1, ..., 2(r-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2(r-1)$ because we can start from on leaf and go all the way up to the root and all the way down again to a vertex that is on a different branch that the initial vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to count all these. For length 1 i can easily count those (i think):&#xA;$r-1 * \sum_{i=i}^{r}k^i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But level two is a little trickier because you can go up two parents or go up one parent and down to the child on the same level as the initial vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One i figure out the total distance between all vertices i have to divide it by the total number of vertices to get the average, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-10-08T21:27:08.200" Title="complete $k$-ary tree: average distance between all vertices" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6293" PostTypeId="2" ParentId="6292" CreationDate="2010-10-08T00:38:38.123" Score="1" ViewCount="0" Body="&lt;p&gt;Your sum $\sum_{i=1}^{r}k^i = k^1 + k^2 + ...+ k^r$ should run from $0$ to $r-1$.  This is a geometric series which can be summed in the usual way to $(k^r-1)/(k-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried counting the average distance from a node on each level but it became a mess.  A slicker way is to count how many times each edge gets used when you count all the paths.  It will get used if one of the vertices is above it and the other is not.  For example, an edge attached to the root has a tree of height r-1 above it, so will get used $(k^{(r-1)}-1)/(k-1) * ((k^r-1)-(k^{(r-1)}-1)/(k-1)$ times.  An edge at the top of the tree will get used if and only if its leaf is one of the vertices so will get counted $(k^r-1)/(k-1)-1$ times.  An edge at level $i$ will get used $(k^{(r-i-1)}-1)/(k-1) * ((k^r-1)-(k^{(r-i-1)}-1)/(k-1)$ times and there are $k^{(i+1)}$ of them where i goes from 0 to r-2.  Then divide by the number of pairs of vertices.  So you want $\sum_{i=0}^{r-2} k^{(i+1)}*(k^{(r-i-1)}-1)/(k-1) * ((k^r-1)-(k^{(r-i-1)}-1)/(k-1)$ divided by $((k^r-1)/(k-1)-1)*(k^r-1)/(k-1)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a mess, but I'm sure Mathematica can do it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-08T21:27:08.200" LastActivityDate="2010-10-08T21:27:08.200" CommentCount="1" />
  <row Id="6294" PostTypeId="2" ParentId="6262" CreationDate="2010-10-08T00:45:45.987" Score="0" ViewCount="0" Body="&lt;p&gt;I would say 6*24(n-1).  The first cube can be placed in 6 ways, just choosing the top face.  Then you select one side as the front.  The second cube can then be placed in 24 ways as you choose the top and then which one will be front.  Each successive cube has 24 ways to be placed.  So you multiply all these numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it is that each cube has 24 possibilities, but they fall into groups of four by rotating the whole stack.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-08T00:45:45.987" />
  <row Id="6295" PostTypeId="1" AcceptedAnswerId="6296" CreationDate="2010-10-08T00:47:13.110" Score="0" ViewCount="189" Body="&lt;p&gt;Suppose you have two distinct large primes $p$ and $q$.&#xA;Explain how you can find an integer $x$ such that $x^2 \equiv 49$&#xA;(mod $pq$), $x \not\equiv\pm 7$ (mod $pq$).&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-09T16:01:24.587" LastActivityDate="2010-10-09T16:03:14.070" Title="Find x such that $x^2 \equiv 49$ (mod $pq$), $x \not\equiv\pm 7$ (mod $pq$)" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;prime-numbers&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6296" PostTypeId="2" ParentId="6295" CreationDate="2010-10-08T01:10:11.323" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ pq\:|\:(x-7)(x+7)\:,\ $ not $\rm\ pq\:|\:x\pm7\ \Rightarrow\ p\:|\:x- 7,\ q\:|\:x+7\ $ or vice versa.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you need $\rm\: -7\not\equiv 7\ $ so $\rm\ p,\ q\ $ coprime to $\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-08T02:03:36.410" LastActivityDate="2010-10-08T02:03:36.410" />
  <row Id="6297" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T01:26:42.237" Score="7" ViewCount="0" Body="&lt;p&gt;I  found the simplicity of &lt;A HREF=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Pick's Theorem&lt;/A&gt; pretty surprising when I first stumbled across it.&lt;/p&gt;&#xA;" OwnerUserId="2271" LastActivityDate="2010-10-08T01:26:42.237" CommunityOwnedDate="2010-10-08T01:26:42.237" CommentCount="1" />
  <row Id="6298" PostTypeId="2" ParentId="6292" CreationDate="2010-10-08T01:27:31.840" Score="1" ViewCount="0" Body="&lt;p&gt;Here's another approach to trying to find the total distance: assume that we have some 'initial' node on level $m$, and we're trying to find the distance to the nodes at equal or lower levels.  Start with the equal level: there are $(k-1)$ siblings, at distance 2; then $k(k-1)$ 'first cousins' (as each of our parent's $k-1$ siblings has $k$ children, at distance 4; $k^2(k-1)$ 'second cousins', at distance , ..., up to $k^{m-1}(k-1)$ 'root cousins' (i.e., children going through the root node).  Now, consider e.g. our first cousins; each of those will bring $k$ children at distance 4+1=5, $k^2$ grandchildren at distance 4+2=6, ..., $k^{n-m}$ great-great-...grandchildren, each of which contributes $4+(n-m)$ distance.  A bit of careful working through the algebra should be enough to sum these correctly (distribute out that factor of 4 first!), and that'll be enough to get the total distance from a given node on level $m$ to all nodes on deeper levels; then just sum that over all depths.  The one little 'gotcha' is that you want to divide the distance-sum from a node on level $m$ to all nodes on the same level by 2, otherwise you'll double-count it, but that's easily accounted for.  I can try and work out the algebra later this evening, but I think it'd make a pretty good exercise.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-08T01:27:31.840" CommentCount="1" />
  <row Id="6299" PostTypeId="1" CreationDate="2010-10-08T01:35:30.443" Score="0" ViewCount="103" Body="&lt;p&gt;Suppose I have two machines, A and B. A encrypts a message m and outputs the ciphertext m^e (mod n). B outputs c such that c = m^e (mod p) and c = m^e + 1 (mod q). How can I use A and B to find p and q? I am allowed to choose m and n.&lt;/p&gt;&#xA;" OwnerUserId="2375" LastActivityDate="2010-10-08T01:53:48.823" Title="Break RSA given a correct and faulty implementation" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;prime-numbers&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" />
  <row Id="6300" PostTypeId="2" ParentId="6299" CreationDate="2010-10-08T01:48:05.870" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Consider $\rm\ gcd(A(m)-B(m),\:pq)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-08T01:53:48.823" LastActivityDate="2010-10-08T01:53:48.823" CommentCount="2" />
  <row Id="6302" PostTypeId="2" ParentId="6295" CreationDate="2010-10-08T02:01:48.180" Score="4" ViewCount="0" Body="&lt;p&gt;It's helpful to know the multiplicative structure of $(\mathbb{Z}/n\mathbb{Z})^*$.  It is always a product of cyclic subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For powers of 2, we have $(\mathbb{Z}/2^k\mathbb{Z})^* \equiv C_2 \times C_{2^{k-2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd primes &lt;em&gt;p&lt;/em&gt;, we have $(\mathbb{Z}/p^k\mathbb{Z})^* \equiv C_{(p-1)p^{k-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the structure for the prime powers is enough; we can use the Chinese Remainder Theorem to see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mathbb{Z}/p_1^{i_1}\cdots p_r^{i_r}\mathbb{Z})^* = (\mathbb{Z}/p_1^{i_1}\mathbb{Z})^* \times \cdots \times (\mathbb{Z}/p_r^{i_r}\mathbb{Z})^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $(\mathbb{Z}/pq\mathbb{Z})^* = C_{p-1} \times C_{q-1}$ has four solutions to $x^2=7^2$, and Bill's answer shows very nicely how to find them.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-08T02:01:48.180" />
  <row Id="6303" PostTypeId="1" AcceptedAnswerId="6304" CreationDate="2010-10-08T02:33:05.510" Score="2" ViewCount="189" Body="&lt;p&gt;Geometric multiplicity of an eigenvalue of a matrix is the dimension of the corresponding eigenspace.  The algebraic multiplicity is its multiplicity as a root of the characteristic polynomial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that &lt;em&gt;the geometric multiplicity of an eigenvalue cannot be greater than the algebraic multiplicity&lt;/em&gt;.  This fact can be shown easily using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_normal_form&quot; rel=&quot;nofollow&quot;&gt;Jordan normal form&lt;/a&gt; of a matrix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a more elementary way to prove this fact, possibly longer but without using the Jordan normal form?  (This is an exercise in &lt;a href=&quot;http://books.google.com/books?id=nZmpQgAACAAJ&amp;amp;dq=inauthor%3a%22Erwin+Kreyszig%22&amp;amp;hl=en&amp;amp;ei=JYKuTMe1G4KC8gbmzcTABA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CDsQ6AEwAw&quot; rel=&quot;nofollow&quot;&gt;Kreyszig's book on functional analysis&lt;/a&gt;, and given the author's style, I suspect that he did not intend the solution to use Jordan form, because otherwise I guess he would have given a hint about that. But I might be wrong.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-10-08T02:41:52.527" Title="Geometric multiplicity of an eigenvalue" Tags="&lt;linear-algebra&gt;" AnswerCount="1" />
  <row Id="6304" PostTypeId="2" ParentId="6303" CreationDate="2010-10-08T02:41:52.527" Score="7" ViewCount="0" Body="&lt;p&gt;You don't need the Jordan form: suppose the geometric multiplicity of $\lambda$ is $k$, and let $\gamma=\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$ be a basis for the corresponding eigenspace. Extend the basis $\gamma$ to a basis $\beta$ for $F^n$, and let $Q$ be the change-of-basis matrix. Then the characteristic polynomials of $A$ and $Q^{-1}AQ$ are the same. The upper left $k\times k$ block of $Q^{-1}AQ$ is simply $\lambda I_k$, and, the $(n-k)\times k$ block under it is all zeroes. So the characteristic polynomial of $Q^{-1}AQ$ is a multiple of $(\lambda - t)^k$ hence the algebraic multiplicity of $\lambda$ is at least $k$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-08T02:41:52.527" CommentCount="1" />
  <row Id="6305" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T02:42:16.150" Score="11" ViewCount="0" Body="&lt;p&gt;I think one of my favorites would be Gödel's incompleteness theorem, which tells us that a formal system cannot be demonstrated to be consistent.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-08T02:42:16.150" CommunityOwnedDate="2010-10-08T02:42:16.150" />
  <row Id="6306" PostTypeId="2" ParentId="29" CreationDate="2010-10-08T02:42:51.233" Score="2" ViewCount="0" Body="&lt;p&gt;I just discovered a software called &lt;a href=&quot;http://www.spacetime.us/&quot; rel=&quot;nofollow&quot;&gt;Spacetime&lt;/a&gt;. Its free, available for windows/mac/iphone and very powerful. You can:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Graph 2D, 3D and 4D function,&#xA;  parametric, polar, vector fields,&#xA;  implicit, contour, and fractal plots&#xA;  with the tap of a stylus or click of a&#xA;  mouse.  Move, zoom, rotate and trace&#xA;  graphs in real-time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it supports writing scripts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UI is good, and the software comes with a great tutorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qI8Jr.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-10-08T02:42:51.233" CommunityOwnedDate="2010-10-08T02:42:51.233" />
  <row Id="6307" PostTypeId="2" ParentId="6282" CreationDate="2010-10-08T02:48:29.493" Score="1" ViewCount="0" Body="&lt;p&gt;Assume that there are B boys and G girls. Let us say (Without loss of generality) that B is no smaller than G. Let us make the arrangement by lining up the girls and then lining up the boys among them. In this arrangement, there are G+1 places the boys can be (before girl 1, between girl 1 and 2, ... , between girl G-1 and G, and after girl G). By the pigeonhole principal, you will have at least ceil(B / G+1) boys in one of these locations, and since there are no other conditions, this minimum is achievable.&lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-08T02:48:29.493" />
  <row Id="6308" PostTypeId="2" ParentId="5927" CreationDate="2010-10-08T04:08:42.413" Score="0" ViewCount="0" Body="&lt;p&gt;As Bill's answer implies, this really is a problem of definition.  The definition of $\binom{n}{k}$ is not $\frac{n!}{k!(n-k)!}$.  Depending on your point of view, it's either &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Bill's answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) the number of distinct sets of $k$ elements that can be formed from a set containing $n$ elements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or even (a bit more obscurely) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) the solution to the recurrence relation $R^n_k = R^{n-1}_k + R^{n-1} _{k-1}$ with initial conditions $R^0_0 = 1$ and $R^0_k = 0$ for $k \neq 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any of cases (1), (2), and (3) it is not hard to prove that you get $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ when $n \geq k$ and $\binom{n}{k} = 0$ when $n &amp;lt; k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point is: The factorial formula for $\binom{n}{k}$ does not apply when $n &amp;lt; k$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-08T04:08:42.413" />
  <row Id="6309" PostTypeId="1" AcceptedAnswerId="6319" CreationDate="2010-10-08T04:59:45.483" Score="2" ViewCount="82" Body="&lt;p&gt;I am trying to derive the following equation for an analytic function, $f$, defined on the unit disc, $D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z)=\frac{1}{\pi}\int_{D} \frac{f(w)}{(1- \bar{w}z)^{2}}dA(w)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{\bf Question}: Anyone have any ideas on how to derive this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been able to verify that the above is true for $z =0$.   In attempting to derive for the general case, I used conformal maps to arrive at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z)=\frac{1}{\pi}\int_{D} f(w) \left| \frac{-1+|z|^{2}}{(-1+w\bar{z})^{2}} \right|^{2} dA(w)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunatly this is not the formula I was trying to derive,&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-10-08T18:48:40.690" LastActivityDate="2010-10-08T18:48:40.690" Title="Derivation of a &quot;Weighted&quot; Mean Value Thm" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6310" PostTypeId="1" AcceptedAnswerId="6312" CreationDate="2010-10-08T05:08:12.630" Score="1" ViewCount="201" Body="&lt;p&gt;How does one prove that if $n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $ \displaystyle n \mid \frac{a^{n}-b^{n}}{a-b}$ where $a,b, n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i thought of is to consider $$(a-b)^{n} \equiv a^{n} + (-1)^{n}b^{n} \ (\text{mod} \ n)$$ and if we suppose that $n$ is odd then we have, $$(a-b)^{n} \equiv a^{n} -b^{n} \ (\text{mod} \ n)$$ and since $n \mid (a^{n} - b^{n})$ we have $$(a-b)^{n} \equiv 0 \ (\text{mod} \ n) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think i am far away from the conclusion of the problem, but this is what i could work on regarding the problem. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-08T05:39:23.340" LastActivityDate="2010-10-10T18:56:32.583" Title="$n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $n \mid \frac{a^{n}-b^{n}}{a-b}$" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6311" PostTypeId="2" ParentId="6274" CreationDate="2010-10-08T05:26:09.893" Score="0" ViewCount="0" Body="&lt;p&gt;The formula is not even close.  You are right that d should have dimensions of length and the right hand side is unitless.  Worse, the formula doesn't reduce properly if the errors are zero.  Are you sure it isn't $d^2/\sigma(d)^2$ on the left?  That would fix the units.  The cosine formula gives the distance estimate you want: $d^2=R_p^2+R_o^2-2*R_p*R_o*cos(\theta_p-\theta_o)$.  When you mix in sigmas you are trying to estimate the error in d based on the errors in the measurements.  So you should use the formulas you have for combining measurement errors to give the error in the measured quantity.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-09T02:45:47.217" LastActivityDate="2010-10-09T02:45:47.217" CommentCount="2" />
  <row Id="6312" PostTypeId="2" ParentId="6310" CreationDate="2010-10-08T05:44:42.973" Score="6" ViewCount="0" Body="&lt;p&gt;Use the fact that $x \equiv y \pmod{p^{\ell}}$ (with $\ell &gt; 0$) implies $x^p \equiv y^p \pmod{p^{\ell+1}}$ to treat the case where $n$ is a prime power.  But note that to prove the statement for $n$, it suffices to prove the statement separately for each prime power dividing $n$.  [I am of course leaving out details, but perhaps the OP might enjoy trying to fill in the sketch.]&lt;/p&gt;&#xA;" OwnerUserId="631" LastEditorUserId="631" LastEditorDisplayName="" LastEditDate="2010-10-08T05:52:25.810" LastActivityDate="2010-10-08T05:52:25.810" />
  <row Id="6313" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T06:17:04.913" Score="5" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss-Bonnet_Theorem&quot; rel=&quot;nofollow&quot;&gt;Gauss-Bonnet theorem&lt;/a&gt;.  The integral of the curvature of a manifold, a totally geometric concept and one that looks dependent on the embedding, is equal to $2\pi$ times its Euler characteristic, an algebraic homotopy invariant.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-08T06:17:04.913" CommunityOwnedDate="2010-10-08T06:17:04.913" />
  <row Id="6314" PostTypeId="1" AcceptedAnswerId="6338" CreationDate="2010-10-08T06:21:42.363" Score="16" ViewCount="819" Body="&lt;p&gt;Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?&lt;/p&gt;&#xA;" OwnerUserId="546" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-08T07:49:05.047" LastActivityDate="2010-11-15T17:24:41.013" Title="Is $[0,1]$ a countable disjoint union of closed sets?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" />
  <row Id="6315" PostTypeId="2" ParentId="6314" CreationDate="2010-10-08T06:31:06.180" Score="0" ViewCount="0" Body="&lt;p&gt;No. $[0,1]$ cannot be written as the union of countable disjoint closed intervals . Try to use the Baire Category theorem. You may also refer this post &lt;a href=&quot;http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-08T06:31:06.180" CommentCount="8" />
  <row Id="6316" PostTypeId="2" ParentId="2949" CreationDate="2010-10-08T09:01:43.973" Score="1" ViewCount="0" Body="&lt;p&gt;The fact that the axiom of foundations (also known as regularity in some places) is independent of the rest of the axioms, and not only you can have an infinitely descending chain $x_0 \ni x_1 \ni \ldots$ but you can have $x = \{ x \}$ and even more than that! $P(a) \in a$ is possible as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crazy set theoretic voodoo, that's what this is!&#xA;(And I'm loving every single bit of it :))&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-08T09:01:43.973" CommunityOwnedDate="2010-10-08T09:01:43.973" CommentCount="1" />
  <row Id="6317" PostTypeId="1" AcceptedAnswerId="6318" CreationDate="2010-10-08T10:36:52.747" Score="1" ViewCount="59" Body="&lt;p&gt;It seems to me not, since this would seem to imply that for all functors &lt;em&gt;F&lt;/em&gt; and all objects &lt;em&gt;A&lt;/em&gt; in &lt;strong&gt;C&lt;/strong&gt; there exists a morphism &lt;em&gt;F(A) -&gt; A&lt;/em&gt; (making all functors co-pointed?). However, intuitively it seems like the identity functor acts like a terminal object; a monad &lt;em&gt;M&lt;/em&gt; on &lt;strong&gt;C&lt;/strong&gt; is a monoid on &lt;strong&gt;[C, C]&lt;/strong&gt; where the &quot;unit&quot; is a natural transformation &lt;em&gt;η : I -&gt; M&lt;/em&gt;, while for a monoidal set &lt;em&gt;S&lt;/em&gt; in &lt;strong&gt;Set&lt;/strong&gt; the unit is a function &lt;em&gt;e : 1 -&gt; S&lt;/em&gt;. So am I misunderstanding something, or are my intuitions leading me astray?&lt;/p&gt;&#xA;" OwnerUserId="2378" LastActivityDate="2010-10-08T10:54:33.210" Title="Is the identity functor the terminal object of the category of endofunctors on C?" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="6318" PostTypeId="2" ParentId="6317" CreationDate="2010-10-08T10:48:01.960" Score="1" ViewCount="0" Body="&lt;p&gt;The difference between those two examples is that in $\textbf{Set}$ the monoidal operation is the categorical product (so the identity object is the terminal object), whereas this is not true in the category of endofunctors on $\mathbf{C}$.  (I believe the latter has a product if and only if $\mathbf{C}$ does, and then it is the pointwise product.  It follows that the terminal object, if it exists, is the functor which sends all objects to $\mathbf{1}$ and all morphisms to the unique morphism $\mathbf{1} \to \mathbf{1}$.  In particular, it's not the identity functor.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-08T10:54:33.210" LastActivityDate="2010-10-08T10:54:33.210" CommentCount="4" />
  <row Id="6319" PostTypeId="2" ParentId="6309" CreationDate="2010-10-08T12:06:10.470" Score="5" ViewCount="0" Body="&lt;p&gt;First, you're missing a factor of $1/\pi$ on the right-hand side, as can be seen by taking $f=1$ and $z=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the identity, you can use Green's formula, which translated into complex variables takes the form $$\int_C F(w,\bar{w}) dw = 2i \iint \frac{\partial F}{\partial \bar{w}} dA(w).$$&#xA;This transforms your area integral into a line integral around the unit circle $C$. To evaluate this line integral, use that $\bar{w}=1/w$ when $|w|=1$; this gives you an integral in $w$ alone (no $\bar{w}$) so that you can use Cauchy's formula to finish the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've checked that it works, but I omit the details so that I don't spoil your pleasure too much. :-)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-08T12:06:10.470" CommentCount="1" />
  <row Id="6321" PostTypeId="1" AcceptedAnswerId="6565" CreationDate="2010-10-08T13:56:07.787" Score="8" ViewCount="267" Body="&lt;p&gt;Any number less than 1 can be expressed in base g as $\sum _{k=1}^\infty {\frac {D_k}{g^k}}$, where $D_k$ is the value of the $k^{th}$ digit. If we were interested in only the non-zero digits of this number, we could equivilantly express it as $\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}}$, where $Z(k)$ is the position of the $k^{th}$ non-zero digit base $g$ and $C_k$ is the value of that digit (i.e. $C_k = D_{Z(k)}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider all the numbers of this form $(\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}})$ where the function $Z(k)$ eventually dominates any polynomial. Is there a proof that any number of this form is transcendental?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have found a paper demostrating this result for the case $g=2$; it can be found here: &lt;a href=&quot;http://www.escholarship.org/uc/item/44t5s388?display=all&quot; rel=&quot;nofollow&quot;&gt;http://www.escholarship.org/uc/item/44t5s388?display=all&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="796" LastEditorDisplayName="" LastEditDate="2010-10-09T06:29:16.160" LastActivityDate="2010-10-11T21:26:28.743" Title="Have all numbers with &quot;sufficiently many zeros&quot; been proven transcendental?" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;number&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6322" PostTypeId="1" CreationDate="2010-10-08T14:13:40.920" Score="2" ViewCount="98" Body="&lt;p&gt;I've been self studying abstract algebra from these &lt;a href=&quot;http://jmilne.org/math/CourseNotes/gt.html&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt;. I've ran across the following Lemma:&#xA;&lt;img src=&quot;http://i.imgur.com/HZM3A.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusing part is that $X$ is a &quot;any subset&quot; not any &quot;subgroup&quot;, and that the term &quot;normal set&quot; appeared in place where I expected &quot;normal subgroup&quot;. The definition of a &quot;normal set&quot; has not been provided and I couldn't find it with a quick Googling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is a &quot;normal set&quot; and why is the proof of that Lemma obvious ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-10-08T14:13:40.920" Title="The smallest normal set containing a subset $X$, what is a &quot;normal set&quot; ?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="4" />
  <row Id="6323" PostTypeId="1" CreationDate="2010-10-08T15:29:39.190" Score="2" ViewCount="111" Body="&lt;p&gt;I understand in mathematics there are many &quot;quotienting &quot; proceduce, is this the only reason that we consider equivariant theory for different &quot;unequivariant&quot; theory? Are there any more applications for equivariant theory?Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-12-08T15:45:21.453" Title="How equivariant theory (like equivariant cohomology) arise" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="6324" PostTypeId="1" AcceptedAnswerId="6326" CreationDate="2010-10-08T16:11:25.480" Score="10" ViewCount="488" Body="&lt;p&gt;This came up in proving non-regularity of a certain language (powers of 2 over the ternary alphabet). Any clue to the above equation could help me move forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Of course, $x = 1, y = 1$ is a solution. I am looking for non-trivial solutions.&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="813" LastEditorDisplayName="" LastEditDate="2010-10-08T16:20:33.547" LastActivityDate="2010-10-09T04:37:24.277" Title="Are there integer solutions to $9^x - 8^y = 1$?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="6325" PostTypeId="2" ParentId="6324" CreationDate="2010-10-08T16:17:48.403" Score="13" ViewCount="0" Body="&lt;p&gt;The only solution is $x=y=1$. This is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan%27s_conjecture&quot; rel=&quot;nofollow&quot;&gt;Catalan's conjecture&lt;/a&gt; which was proven by Mihailescu in 2002: the only solution in natural numbers of $x^a - y^b = 1$ with $a,b\gt 1$ is $x=3$, $a=2$, $y=2$, and $b=3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as Moron points out, it can be deduced much more elementarily.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-08T16:33:11.307" LastActivityDate="2010-10-08T16:33:11.307" />
  <row Id="6326" PostTypeId="2" ParentId="6324" CreationDate="2010-10-08T16:26:15.627" Score="29" ViewCount="0" Body="&lt;p&gt;Except $x=1$ and $y=1$ there aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^{2x} - 1 = 8^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (3^x + 1)(3^x - 1)  = 8^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^x + 1 = 2^m, 3^x - 1 = 2^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$2^m - 2^n = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&#xA;$$ 2^n(2^{m-n} - 1) = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $n=1$ and $m=2$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-08T17:08:03.653" LastActivityDate="2010-10-08T17:08:03.653" CommentCount="6" />
  <row Id="6327" PostTypeId="2" ParentId="6323" CreationDate="2010-10-08T16:33:31.103" Score="3" ViewCount="0" Body="&lt;p&gt;When you pass to the quotient by an action you may disastrously break some (or all!) interesting structure, so you look for ways of doing things in the quotient without actually constructing it. Doing things equivariantly upstairs is one of the ways to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the quotient, when it is a sensible object, may not contain all the information you want: for example, there are situations in which you get the same quotient by dividing a space by the action of two different groups, yet you &lt;em&gt;want&lt;/em&gt; to have different results. So you work equivariantly and are happy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-08T16:39:04.413" LastActivityDate="2010-10-08T16:39:04.413" />
  <row Id="6328" PostTypeId="1" AcceptedAnswerId="6334" CreationDate="2010-10-08T16:53:01.353" Score="2" ViewCount="109" Body="&lt;p&gt;Suppose $P(x,y)$ gives transition probabilities of a random walk. I've seen $Pf=f$ being called the discrete version of Laplace's equation. In what sense are they analogous?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-10-09T01:44:36.650" Title="discrete version of Laplacian" Tags="&lt;probability-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6329" PostTypeId="1" CreationDate="2010-10-08T17:00:37.243" Score="1" ViewCount="103" Body="&lt;p&gt;Convex polyhedron $P$ is a subset of $\mathbb{R}^n$ that satisfies system of linear inequalities&#xA;\begin{align}&#xA;a_{11}x_1 + \cdots + a_{1n}x_n &amp;amp; \sim_1\, c_1 \\&#xA;&amp;amp; \vdots \\&#xA;a_{p1}x_1 + \cdots + a_{pn}x_n &amp;amp; \sim_p\, c_p,&#xA;\end{align}&#xA;where $\sim_i \in \{\leq,\geq\}$. It can be alternatively represented by two finite sets of generators $V, W \subseteq \mathbb{R}^n$: $$P = \text{conv}(V) + \text{cone}(W),$$ where conv(V) denotes all convex combinations of points in V and cone(W) all nonnegative linear combinations of points in W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if we allow $\sim_i$ to be from $\{\geq,&gt;,\leq,&amp;lt;\}$. Is there some similar representation in terms of generating points for such sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I possess no knowledge of this area of mathematics, so I apologize if I got the terminology wrong or if this question is just plain stupid.)&lt;/p&gt;&#xA;" OwnerUserId="2383" LastEditorUserId="2383" LastEditorDisplayName="" LastEditDate="2010-10-08T17:42:36.013" LastActivityDate="2010-10-13T08:21:28.233" Title="Convex polyhedron with open faces" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6330" PostTypeId="2" ParentId="6329" CreationDate="2010-10-08T17:23:52.060" Score="2" ViewCount="0" Body="&lt;p&gt;No.  Unfortunately, it can't even be done in the 1D case.  For instance, consider the inequalities $x &gt; 0$ and $x &amp;lt; 1$.  Then $P$ is the open interval $(0,1)$.  You can't take $V = \{0,1\}$ because then conv$(V)$ is the closed interval $[0,1]$.  Taking $V$ to be any other two points between 0 and 1 would generate a conv$(V)$ that is a strict subset of $P$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-08T17:23:52.060" CommentCount="2" />
  <row Id="6331" PostTypeId="2" ParentId="6328" CreationDate="2010-10-08T17:35:00.133" Score="3" ViewCount="0" Body="&lt;p&gt;The infinitesimal generator for Brownian motion in $\mathbb R^n$ (and, in general, in a Riemannian manifold) is $\tfrac12\Delta$. Your matrix $P$ plays exactly the same rôle in the discrete version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the equation: the equation $\tfrac12\Delta f=0$ is, in the case of Brownian motion in $\mathbb R^n$, the equation for stationary states. Your equation is the same thing in the discrete version.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-08T17:46:53.200" LastActivityDate="2010-10-08T17:46:53.200" />
  <row Id="6332" PostTypeId="1" CreationDate="2010-10-08T17:44:49.213" Score="2" ViewCount="82" Body="&lt;p&gt;&quot;Gromov Witten potential&quot;, when does &quot;potential&quot; mean here? What does the whole thing mean in physics? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-08T17:44:49.213" Title="What does &quot;Gromov Witten potential&quot; the &quot;potential&quot; mean" Tags="&lt;algebraic-geometry&gt;&lt;physics&gt;" CommentCount="0" />
  <row Id="6333" PostTypeId="2" ParentId="6328" CreationDate="2010-10-08T18:05:33.403" Score="4" ViewCount="0" Body="&lt;p&gt;Here is a low-brow answer.  If you accept that the discrete version of the second derivative of a function $f$ is $f(x+1) - 2f(x) + f(x-1)$, then the discrete Laplacian, say in two dimensions, is $\Delta f(x, y) = \frac{f(x+1, y) + f(x-1, y) + f(x,y+1) + f(x,y-1)}{4} - f(x,y)$.  But the first term is just the transition probabilities of a random walk on $\mathbb{Z}^2$ where one moves to each of the four horizontally or vertically adjacent neighbors with equal probability.  A similar statement is true in $n$ dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally one can define a &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Laplace_operator#On_Graphs&quot; rel=&quot;nofollow&quot;&gt;discrete Laplacian on any graph&lt;/a&gt; which mimics the usual Laplacian.  In fact there is &lt;a href=&quot;http://front.math.ucdavis.edu/math.PR/0001057&quot; rel=&quot;nofollow&quot;&gt;an entire textbook&lt;/a&gt; by Doyle and Snell dedicated to working out potential theory on graphs.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-08T18:05:33.403" />
  <row Id="6334" PostTypeId="2" ParentId="6328" CreationDate="2010-10-08T18:20:21.707" Score="4" ViewCount="0" Body="&lt;p&gt;Some analogies between discrete and continuous Laplacians were discussed in MO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you search for &quot;discrete harmonic&quot; or &quot;discrete Laplacian&quot; there may be more there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added: the analogy is not as simple as the fact that the lattice Laplace operator converges to the continuous Laplacian as the spacing shrinks to zero.  There is a rotational symmetry of the continuous Laplacian which is an essential part of the geometric theory. For the analogies to be meaningful, at least some of the geometry of the continuum story should be be visible in the lattice or graph Laplacians, and for this eigenvalue inequalities, Hodge theory, vector bundles, zeta functions and other constructs are considered for graphs, in addition to the more easily perceived analogies with harmonic theory and Brownian motion.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-09T01:44:36.650" LastActivityDate="2010-10-09T01:44:36.650" />
  <row Id="6335" PostTypeId="1" AcceptedAnswerId="6348" CreationDate="2010-10-08T18:23:03.220" Score="6" ViewCount="141" Body="&lt;p&gt;Using generating functions one can see that the $n^{th}$ Bell number, i.e., the number of all possible partitions of a set of $n$ elements, is equal to $E(X^n)$ where $X$ is a Poisson random variable with mean 1. Is there a way to explain this connection intuitively?&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-08T22:57:39.597" LastActivityDate="2010-10-08T22:57:39.597" Title="Bell numbers and moments of the Poisson distribution" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="6336" PostTypeId="1" AcceptedAnswerId="6342" CreationDate="2010-10-08T18:32:36.663" Score="3" ViewCount="221" Body="&lt;p&gt;If not, is there only a limited rangle of polynomials for which the root can be found?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $u=x^{\frac{3}{2}}+x$, is $x$ expressible in terms of $u$?&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2010-10-08T21:32:55.553" Title="Are all polynomials solvable?" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6337" PostTypeId="2" ParentId="6336" CreationDate="2010-10-08T19:06:36.403" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, there is just a limited amount of polynomials for which we can find the exact, i.e. algebraic roots by some general formula (like it's possible for quadratic polynomials). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel-Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Ruffini-theorem&lt;/a&gt; states that there is no such general solution for polynomials of degree five or higher in terms of radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean they don't have roots (in fact they &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra&quot; rel=&quot;nofollow&quot;&gt;always&lt;/a&gt; have), but one has to write them down or approximate them in other ways.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As to your equation (note that this is no longer a polynomial equation, which would require it's exponents to be integers, but a general power sum)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u = x ^ {\frac{3}{2}} + x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We substitute $k = x^{\frac{1}{2}}$ and now have to solve a cubic (polynomial) equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k^3 + k^2 - u = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For degree three, we can use a well-known but kinda unwieldy &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardano_formulae#Roots_of_a_cubic_function&quot; rel=&quot;nofollow&quot;&gt;general solution formula&lt;/a&gt; and finally obtain zero to three results.&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-08T20:26:59.870" LastActivityDate="2010-10-08T20:26:59.870" CommentCount="9" />
  <row Id="6338" PostTypeId="2" ParentId="6314" CreationDate="2010-10-08T19:07:30.290" Score="15" ViewCount="0" Body="&lt;p&gt;The answer is no. In fact, as Steve D said, we have a theorem that holds for a wide class of spaces, which includes closed intervals, circles, balls and cubes. It was proved by Sierpinski in $1918$ $[1]$. You can find the proof in the book &quot;General Topology&quot; by Ryszard Engelking, but I'll post here since it's not easy to find it online. First a definition: a topological space is called a continuum if it is a compact connected Hausdorff space. The precise statement is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem (Sierpinski).&lt;/strong&gt; If a continuum $X$ has a countable cover $\{X_i\}_{i=1}^{\infty}$ by pairwise disjoint closed subsets, then at most one of the sets $X_i$ is non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove this we'll need the following lemmas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma $1$.&lt;/strong&gt; Let $X$ be a continuum. If $F$ is a non-trivial closed subset of $X$, then for every component $C$ of $F$ we have that $\text{Bd}(F) \cap C$ is non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. Let $x_0$ be in $C$. Since $X$ is Hausdorff compact, quasicomponents coincide with components, so $C$ is the intersection of all open-closed sets in $F$ which contain $x_0$. Suppose that $C$ is disjoint from $\text{Bd}(F)$. Then, by compactness of $\text{Bd}(F)$, there is one open-closed set $A$ in $F$ containing $x_0$ and disjoint from $\text{Bd}(F)$. Take an open set $U$ such that $A = U \cap F$. Thus the equality $A \cap \text{Bd}(F) = \emptyset$ implies that $A = U \cap \text{Int}(F)$, so $A$ is open in $X$. But $A$ is also closed in $X$, and contains $x_0$, so $A=X$. But then $\text{Bd}(F) = \emptyset$, which is not possible since $F$ would be non-trivial open-closed in $X$. $\bullet$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma $2$.&lt;/strong&gt; If a continuum $X$ is covered by pairwise disjoint closed sets $X_1, X_2, \ldots$ of which at least two are non-empty, then for every $i$ there exists a continuum $C \subseteq X$ such that $ C \cap X_i = \emptyset$ and at least two sets in the sequence $C \cap X_1, C \cap X_2, \ldots$ are non-empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. If $X_i$ is empty then we can take $C = X$; thus we can assume that $X_i$ is non-empty. Take $j \ne i$ such that $X_j \ne \emptyset$. Since $X$ is Hausdorff compact, there are disjoint open sets $U,V \subseteq X$ satisfying $X_i \subseteq U$ and $X_j \subseteq V$. Let $x$ be a point of $X_j$ and $C$ the component of $x$ in the subspace $\overline{V}$. Clearly, $C$ is a continuum, $ C \cap X_i = \emptyset$ and $ C \cap X_j \ne \emptyset$. By the previous lemma, $C \cap \text{Bd}( \overline{V}) \ne \emptyset$ and since $X_j \subseteq \text{Int}(\overline{V})$, there exist a $k \ne j$ such that $C \cap X_k \ne \emptyset$. $\bullet$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can prove the theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. Assume that at least two of the sets $X_i$ are non-empty. From lemma $2$ it follows that there exists a decreasing sequence $C_1 \supseteq C_2 \ \supseteq \ldots$ of continua contained in $X$ such that $C_i \cap X_i = \emptyset$ and $C_i \ne \emptyset$ for $i=1,2, \ldots$ The first part implies that $\bigcap_{i=1}^{\infty} C_i = \emptyset$ and from the second part and compactness of $X$ it follows that $\bigcap_{i=1}^{\infty} C_i \ne \emptyset$. $\bullet$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hausdorff hypothesis is fundamental. For example, consider $X$ a countable infinite set with the cofinite topology. Then $X$ is compact, connected and a $T_1$-space. However, we can write $X$ as a disjoint union of countable singletons, which are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[1]$ Sierpinski, W: Un théorème sur les continus, Tohoku Math. J. $13$ ($1918$), $300-305$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-15T17:07:32.963" LastActivityDate="2010-11-15T17:07:32.963" />
  <row Id="6339" PostTypeId="1" AcceptedAnswerId="6343" CreationDate="2010-10-08T19:24:14.447" Score="2" ViewCount="102" Body="&lt;p&gt;Suppose that $A \in {\cal S}_+^n$ is a symmetric positive semidefinite matrix. Let $B = {\rm sign}(A)$, where the sign is taken elementwise. Is the resulting matrix $B$ always positive semidefinite? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, under what conditions can we say that $B \in {\cal S}_+^n$ ?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-03T21:48:52.257" Title="Is a &quot;sign matrix&quot; obtained from a symmetric positive-semidefinite matrix itself symmetic positive-semidefinite?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" />
  <row Id="6340" PostTypeId="1" CreationDate="2010-10-08T19:38:41.347" Score="1" ViewCount="119" Body="&lt;p&gt;Just to clarify, I want to show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If f is entire and $\int_{\mathbb{C}} |f|^p dxdy &amp;lt;\infty$, then f=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I can show that this is the case for p=2, but I'm not sure about other values of p...&lt;/p&gt;&#xA;" OwnerUserId="2386" LastActivityDate="2010-10-14T07:13:28.553" Title="How do you show that an L^p entire (holomorphic on the complex plane) function is 0?" Tags="&lt;complex-analysis&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="6341" PostTypeId="2" ParentId="6340" CreationDate="2010-10-08T19:46:34.637" Score="3" ViewCount="0" Body="&lt;p&gt;Use Hölder's inequality and Cauchy's integral formula to show the function and its derivatives all vanishes at zero.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-08T19:46:34.637" CommentCount="1" />
  <row Id="6342" PostTypeId="2" ParentId="6336" CreationDate="2010-10-08T20:03:04.637" Score="9" ViewCount="0" Body="&lt;p&gt;The answer to your question depends entirely on what &quot;can be found&quot; means. I will note, before we start, that the expression you have with fractional exponents is not generally considered a polynomial: a polynomial in the variable $x$, with real coefficients, is an expression of the form&#xA;$$a_0 + a_1x + a_2x^2+ \cdots + a_nx^n$$&#xA;where $a_0,a_1,\ldots,a_n$ are real numbers, and $n$ is a nonnegative integer. Note that all the powers of the variable are &lt;em&gt;integral&lt;/em&gt; powers. If $a_n\neq 0$, we say the polynomial has degree $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n\leq 4$, then there are formulas that express all the roots of the polynomial in terms of the coefficients ($a_0$, $a_1,\ldots,a_n$). For $n=1$, that is a polynomial $a_0+a_1x$ with $a_1\neq 0$, the solution is simply $x=-\frac{a_0}{a_1}$. For $n=2$, that is a polynomial $a_0 + a_1x + a_2x^2$ with $a_2\neq 0$, you get the well-known quadratic formula that gives the two roots:&#xA;$$ r_1 = \frac{-a_1 + \sqrt{a_1^2 - 4a_0a_2}}{2a_2}\quad\text{and}\quad r_2 = \frac{-a_1-\sqrt{a_1^2-4a_0a_2}}{2a_2}.$$&#xA;When $n=3$ and when $n=4$ there are also formulas to express all the roots of the polynomial in terms of the coefficients and using only the operations of addition, subtractions, multiplication, division, and root extraction. For $n=3$, these are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardano_formulae&quot; rel=&quot;nofollow&quot;&gt;Cardano formulae&lt;/a&gt;, and for $n=4$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_equation&quot; rel=&quot;nofollow&quot;&gt;the solution&lt;/a&gt; is due to Ferrari.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no similar formulas when $n\geq 5$; this is the celebrated &lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel-Ruffini&lt;/a&gt; theorem. This does not mean we have no way of finding the roots, just that there is no formula that applies to all polynomials that gives the roots in terms of the coefficients, using only certain kinds of operations. For example, quintic equations (degree 5 polynomials) can be solved using other more complicated kinds of functions and operations (theta functions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are plenty of methods for finding &lt;em&gt;approximate&lt;/em&gt; values for roots of polynomials, or in fact values that are &quot;as close as you want&quot; to the roots of the polynomial. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sturm%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Sturm's theorem&lt;/a&gt; can help you locate the roots approximately, and then you can combine it with &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;Newton's method&lt;/a&gt; to find very good approximations to the roots of the polynomial. See for example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Root-finding_algorithm&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on root-finding algorithms&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, this turns on whether &quot;finding an approximation&quot; qualifies as &quot;can be found&quot; (or even if &quot;can be found, theoretically, given enough time to work&quot; qualifies as &quot;can be found&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your &quot;also&quot; question, you want to express $x$ in terms of $u$ if $u=x^{3/2}+x$. This is slightly different from &quot;solving a polynomial&quot;; you are really trying to find a formula for the inverse of $x^{3/2}+x$ (assuming it has one, which it does because the function is increasing). You can turn it into a problem of solving a polynomial because if you let $z=x^{1/2}$, then you have $u=z^3+z^2$, which is equivalent to $z^3+z^2-u=0$; this is a cubic polynomial, so the roots of the polynomial can be expressed in terms of the coefficients (in this case, $1$, $1$, $0$, and $-u$) using Cardano's formulae. Then you would replace $x^{1/2}$ for $z$ in the case in which $z$ is nonnegative and real, and squaring gives you the answer. So, yes, it can be done.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-08T20:03:04.637" />
  <row Id="6343" PostTypeId="2" ParentId="6339" CreationDate="2010-10-08T20:20:42.663" Score="2" ViewCount="0" Body="&lt;p&gt;A small symmetric perturbation of the identity matrix is positive definite, but the corresponding sign matrix need not be positive semidefinite.  For example, let $A$ be the 3-by-3 matrix with 1 on the diagonal, and, say, -1/100 in the other 6 entries.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have nothing to say about your second question.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-08T20:20:42.663" />
  <row Id="6344" PostTypeId="1" AcceptedAnswerId="6347" CreationDate="2010-10-08T20:37:39.853" Score="1" ViewCount="94" Body="&lt;p&gt;I am looking for a good reference that could clearly explain finite field arithemtic.  Specifically I want to understand the importance and utility of the field generator polynomial and why it is important in multiplication.&lt;/p&gt;&#xA;" OwnerUserId="2390" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-08T22:55:58.883" LastActivityDate="2010-10-08T22:55:58.883" Title="Good Resources for Understanding Finite Field Arithemtic" Tags="&lt;number-theory&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6345" PostTypeId="1" CreationDate="2010-10-08T20:59:52.720" Score="2" ViewCount="73" Body="&lt;p&gt;I know this question is easy, but for the life of me, I cannot remember what we call this thing. Googling for this has offered no help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an object $A$ and a second object $B$(let them be groups if you so choose). We wish to consider and action of $A$ on $B$. Moreover there is a subobject $C \hookrightarrow B$(subgroup) which is annihilated by the action of $A$, i.e. the restriction of the action of $A$ on $B$ to $C$ sends $C$ to the zero object(the zero in $B$ which corresponds to the trivial group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it would be the kernel of the action, but this term is reserved for something else(in particular those objects which fix everything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this should be referred to as Torsion, and in particular, in the back of my mind, I keep thinking it is called the $A$-Torsion of $B$. But I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what this has been called in the past?&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-08T21:07:27.253" LastActivityDate="2010-11-07T21:44:17.970" Title="The name for a subobject(subgroup) which is annihilated by action." Tags="&lt;group-theory&gt;&lt;terminology&gt;&lt;category-theory&gt;&lt;action&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6346" PostTypeId="2" ParentId="6344" CreationDate="2010-10-08T21:02:54.527" Score="0" ViewCount="0" Body="&lt;p&gt;When I was an undergrad, I was trying to learn this material for the purpose of finite geometry. I found that even with little background in Abstract Algebra, that &lt;a href=&quot;http://www-ma4.upc.es/~simeon/IFG.pdf&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; by Simeon Ball and Zsuzsa Weiner, was very readable. The beginning in particular focuses on what you asked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to mention, it's free!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-08T21:02:54.527" />
  <row Id="6347" PostTypeId="2" ParentId="6344" CreationDate="2010-10-08T21:09:44.537" Score="0" ViewCount="0" Body="&lt;p&gt;One area where finite fields is widely applied is in the theory of error-correcting codes. A very nice book that treats these two subjects jointly is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Error-Correcting Codes and Finite Fields by Oliver Pretzel; Clarendon Press, Oxford, 1992.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-08T21:09:44.537" />
  <row Id="6348" PostTypeId="2" ParentId="6335" CreationDate="2010-10-08T21:14:34.140" Score="5" ViewCount="0" Body="&lt;p&gt;One way may be to use these facts.  You can decide if this is intuitive enough or not. :)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$B_n = \sum_{k=0}^n S(n,k)$, where $S(n,k)$ is a Stirling number of the second kind. ($S(n,k)$ counts the number of ways to partition a set of $n$ elements into sets of size $k$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stirling numbers of the second kind are used to convert ordinary powers to falling powers via $x^n = \sum_{k=0}^n x^{\underline{k}} S(n,k)$, where $x^{\underline n} = x(x-1)(x-2) \cdots (x-n+1)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The factorial moments of a Poisson(1) distribution are all 1; i.e., $E[X^{\underline{n}}] = 1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Putting them together yields &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X^n] = \sum_{k=0}^n E[X^{\underline{k}}] S(n,k) = \sum_{k=0}^n  S(n,k) = B_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Facts 1 and 2 are well-known properties of the Bell and Stirling numbers.  Here is a quick proof of #3.  The second step is the definition of expected value, using the Poisson probability mass function.  The second-to-last step is the Maclaurin series expansion for $e^x$ evaluated at 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X^{\underline{n}}] = E[X(X-1)(X-2) \cdots (X-n+1)] = \sum_{x=0}^{\infty} (x)(x-1) \cdots (x-n+1) \frac{e^{-1}}{x!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_{x=n}^{\infty} (x)(x-1) \cdots (x-n+1) \frac{e^{-1}}{x!} = \sum_{x=n}^{\infty} \frac{x!}{(x-n)!} \frac{e^{-1}}{x!} = \sum_{y=0}^{\infty} \frac{e^{-1}}{y!} = e/e = 1.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-08T21:14:34.140" CommentCount="1" />
  <row Id="6349" PostTypeId="2" ParentId="6323" CreationDate="2010-10-08T21:15:03.657" Score="0" ViewCount="0" Body="&lt;p&gt;i'm not sure what the context is for your first question, but in reply to the second one, i can mention applications in statistics that come with the word 'equivariant' attached. [they involve 'homogeneous spaces', topological spaces on which act a group of isomorphisms of interest.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a simple example might suffice to illustrate the idea. [the ensuing is motivated by changing scales of measurement, such as celsius to kelvin or celsius to fahrenheit.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose ${\mathbf X} = (X_1,\cdots,X_n)$, where the coordinates are independent observations from the same normal distribution $N(\mu, \sigma^2)$ and we want to estimate $\sigma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the {$X_j$} are transformed in an affine way, it does not change the model of normality for the data, but it could change $\sigma$ [or $\mu$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider first a location shift $X\to X+a$, where $a$ is some constant. [the notation is meant to suggest that the shift transformation on $\mathbb R$ is extended to $\mathbb R^n$ by applying it to each coordinate of $\mathbf X$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is a corresponding transformation induced on the parameters, taking ($\mu, \sigma)\to (\mu+a, \sigma)$. as this transformation of the data does not change $\sigma$, it is reasonable to require of an estimator $\hat\sigma (X_1,\cdots,X_n)$ of $\sigma$ that it be invariant under shifts; that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-57pt (1)\kern 57pt\hat\sigma (X_1+a,\cdots,X_n+a) = \hat\sigma (X_1,\cdots,X_n)\ \forall a\in\mathbb R.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since (1) holds for all $a$, we can set $a = -\bar X$ in (1), where $\bar X$ is the sample mean. then (1) implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-70pt (2)\kern 70pt\hat\sigma (X_1,\cdots,X_n) = \hat\sigma (X_1-\bar X,\cdots,X_n-\bar X).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so any estimator of $\sigma$ satisfying (1) can depend on the data only thru ${\mathbf D} := (X_1-\bar X,\cdots,X_n-\bar X)$, the [vector of] deviations from the sample mean. [${\mathbf D}$ is a &lt;em&gt;maximal&lt;/em&gt;$\kern2pt$ invariant for shift transformations in that it indexes the orbits in $\mathbb R^n$ for the shifts $(x_1,\cdots,x_n)\to (x_1+a,\cdots,x_n+a)$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;another group of affine transformations acting on the data is scale changes: $X\to X+b,\ b &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in turn, we get the following induced transformations: $(\mu,\sigma)\to (b\mu,b\sigma)$ and&#xA;${\mathbf D}\to b{\mathbf D} = (b(X_1-\bar X),\cdots,b(X_n-\bar X))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since applying $b$ to the data rescales $\sigma$, this suggests [using (2)] that $\hat\sigma$ should be required to satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-20pt (3)\kern 20pt\hat\sigma(b(X_1-\bar X),\cdots,b(X_n-\bar X)) = b\hat\sigma (X_1-\bar X,\cdots,X_n-\bar X), \forall b&gt;0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) requires that $\hat\sigma$ be equivariant for scale changes of the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;letting $D^2 = \sum_1^n (X_j - \bar X)^2 = |{\mathbf D}|^2$ denote the sum of the squared deviations from $\bar X$, putting $b = 1/D$ in (3) gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern-55pt (4)\kern 55pt \hat\sigma (X_1,\cdots,X_n) = D\hat\sigma \big( \frac{X_1-\bar X}{D},\cdots,\frac{X_n-\bar X}{D}\big).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the vector ${\mathbf S} := \frac{\mathbf D}{|\mathbf D|} = \big( \frac{X_1-\bar X}{D},\cdots,\frac{X_n-\bar X}{D}\big)$ in (4) is a maximal invariant for the affine group on $\mathbb R$ extended [coordinatewise] to $\mathbb R^n$. note that its distribution does not depend on $(\mu,\sigma)$ [and it is thus an example of what is called an ancillary statistic]. in fact, $\mathbf S$ is also independent of $(\bar X, D)$. the multiplier of $D$ [$\hat\sigma ({\mathbf S})$] in (4) is usually chosen to be a constant [else $\hat\sigma$ is a randomized estimator].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[it is also well-known that $\bar X$ and $D$ are independent in the normal case, so that&#xA;$\bar X, D$ and $\mathbf S$ are mutually independent. this is due to the fact that $\bar X$ and ${\mathbf D}$ are obtained from projections of $\mathbf X$ into two orthogonal subspaces,  and $\bf 1^\perp$, where  ${\bf 1}$ = (1,$\cdots$,1) $\in \mathbb R^n$. this implies that $\bar X$ and $\mathbf D$ are independent - and that $\mathbf D$ is spherically distributed in $\mathbf 1^\perp$, so that its length $D$ and orientation $\mathbf S$ are independent as well.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this argument extends to jointly estimating $(\mu,\sigma)$ and shows that the nonrandomized equivariant estimators are of the form $(\hat\mu, \hat\sigma) = (\bar X +bD), cD)$, where $b$ and $c$ are constants. the common choices $b=0$ and $c = \frac{1}{\sqrt{n-1}}$ give $\hat\mu = \bar X$ and &#xA;$\hat\sigma^2 := s^2 = \frac{\sum (X_j-\bar X)^2}{n-1}$, which are unbiased for $\mu$ and $\sigma^2$, respectively, [$s^2$ being the sample variance].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the transformation group involved here is perhaps not so exciting. the situation becomes more interesting in the multivariate normal case. see, &lt;em&gt;e.g.&lt;/em&gt; &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.aos/1176346733&quot; rel=&quot;nofollow&quot;&gt;eaton (1983) or muirhead (1982)&lt;/a&gt;  or &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.lnms/1196285102&quot; rel=&quot;nofollow&quot;&gt;eaton (2007)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-08T21:15:03.657" CommentCount="0" />
  <row Id="6350" PostTypeId="2" ParentId="6345" CreationDate="2010-10-08T21:20:12.277" Score="1" ViewCount="0" Body="&lt;p&gt;in linear algebra, the subspace annihilated by a linear mapping $A$ is the nullity of $A$.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-08T21:20:12.277" CommentCount="3" />
  <row Id="6351" PostTypeId="2" ParentId="6344" CreationDate="2010-10-08T21:29:15.933" Score="0" ViewCount="0" Body="&lt;p&gt;Try the book &lt;a href=&quot;http://www.ams.org/bookstore-getitem/isbn=0-8218-4418-0&quot; rel=&quot;nofollow&quot;&gt;Finite Fields and Applications&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-08T21:29:15.933" />
  <row Id="6352" PostTypeId="2" ParentId="6336" CreationDate="2010-10-08T21:32:55.553" Score="2" ViewCount="0" Body="&lt;p&gt;Your equation can be solved: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+u%3Dx%5E%283%2F2%29%2Bx+for+x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=solve+u%3Dx%5E%283%2F2%29%2Bx+for+x&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics that decides when a polynomial is solvable by radicals is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_theory&quot; rel=&quot;nofollow&quot;&gt;Galois Theory&lt;/a&gt; and is one of the hallmarks of modern algebra.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-08T21:32:55.553" />
  <row Id="6353" PostTypeId="1" CreationDate="2010-10-08T23:42:19.380" Score="2" ViewCount="124" Body="&lt;p&gt;Suppose $U$ and $V$ are 2 dimensional subspaces of $\mathbb{R}^4$. How do you determine the dimension of $U \cap V$? I know that $\text{dim}(U + V) = \text{dim}(U)+ \text{dim}(V) - \text{dim}(U \cap V)$. So it seems that $\text{dim}(U \cap V) = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2392" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-09T03:10:39.053" LastActivityDate="2010-10-09T03:10:39.053" Title="How do you determine dimension" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6354" PostTypeId="2" ParentId="6353" CreationDate="2010-10-08T23:48:36.053" Score="1" ViewCount="0" Body="&lt;p&gt;If you have B and B', basis of U and V respectively, you can check if the vectors of B are linearly independent with those of B'. If so, then $\dim(U\cap V) = 0$. If not, you may be able to write a new basis of $U+V$ eliminating the linearly dependent vectors. $\dim(U\cap V)$ is the amount of vectors you have eliminated from $U+V$'s basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-09T02:36:33.083" LastActivityDate="2010-10-09T02:36:33.083" CommentCount="3" />
  <row Id="6355" PostTypeId="1" CreationDate="2010-10-08T23:54:21.337" Score="0" ViewCount="46" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an algorithm, that can process an amount of (GPS-) positions and determine, if the actual position gets closer to a way point. &#xA;What is the &quot;best&quot; way to filter variability in the positions? So far, I've implemented a moving average over the last positions and compare that to the new one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make sense to use a low-pass filter? Any other ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2393" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-09T00:53:52.733" LastActivityDate="2010-10-09T00:53:52.733" Title="Algorithm for approaching a position" Tags="&lt;geometry&gt;&lt;statistics&gt;" CommentCount="2" />
  <row Id="6357" PostTypeId="1" AcceptedAnswerId="6362" CreationDate="2010-10-09T00:32:48.293" Score="4" ViewCount="225" Body="&lt;p&gt;Hello,&#xA;I'm currently studying the notion of a quotient topology. The one thing I'm having trouble with understanding is what we're actually doing to the points as we're identifying them. Say we have a [0,1] x [0,1] in E^2 (euclidean 2-space) with the subspace topology and we partition X into:&#xA;1) The set {(0,0),(1,0),(0,1),(1,1)} of four corner points&#xA;2) sets of pairs of points (x,0),(x,1) where 0&#xA;&#xA;&lt;p&gt;I wonder whether there's a any natural way to see how to &quot;place&quot; the points? This example is of course kinda easy, but just given this set, is there any way to figure out how it looks geometrically? If we'd have say the set consisting of all points from a fixed distance from a fixed point, we'd identify it as a circle, is there anything similar here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now, for the last question here:&#xA;Let B^n denote the unit ball in n-dimensiona euclidean space, and let S^(n-1) denote its boundary. Consider the partition of B^n which has as members:&#xA;1) the set S^(n-1);&#xA;b) the individual points of B^n-S^(n-1). &#xA;OK, when we now identify S^(n-1) it collapses into one point, right? Exactly what does it collapse into and where do we &quot;put it&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope you can shed some light on this.&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2010-10-10T02:35:15.503" Title="Quotient topologies and equivalence classes." Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="6358" PostTypeId="2" ParentId="6353" CreationDate="2010-10-09T02:26:58.037" Score="3" ViewCount="0" Body="&lt;p&gt;First: it is incorrect, in general, to even write $\dim(U\cup V)$, because $U\cup V$ is &lt;em&gt;almost never&lt;/em&gt; a subspace: it is a subspace if and only if $U\subseteq V$ or $V\subseteq U$. Not being a subspace, it doesn't even make sense to talk about its dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, what you probably meant is the correct equation&#xA;$$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$&#xA;(though I prefer to put the intersection on the left hand side, because in that form it is valid even in the infinite dimensional case). Here, $U+V$ is the smallest subspace that contains $U$ and $V$, and it happens to equal the set of all vectors of the form $u+v$ with $u\in U$ and $v\in V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation does not give you the full answer, because there are several situations that can occur: you could have $U$ and $V$ intersect trivially: this is what happens when $U+V=\mathbb{R}^4$. For an explicit example, you could have $U=\{(a,b,0,0):a,b\in\mathbb{R}\}$, and $V=\{(0,0,c,d) : c,d\in\mathbb{R}\}$. Here, $\dim(U\cap V)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could have that $U$ and $V$ intersect in a one-dimensional subspace (for example, take $U$ as above, but take $V=\{(0,b,c,0) : b,c\in\mathbb{R}\}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could have $U=V$, in which case the intersection has dimension $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can say is: if the spaces are distinct, then the intersection will either have dimension 1 or dimension 0. But that is all you can say with the given information.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-09T02:38:30.113" LastActivityDate="2010-10-09T02:38:30.113" />
  <row Id="6359" PostTypeId="2" ParentId="6357" CreationDate="2010-10-09T02:38:20.660" Score="10" ViewCount="0" Body="&lt;p&gt;There are two things you can do with Topology: (1) try to imagine what's going on intuitively, and (2) write it down rigorously. Both are useful, necessary steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if I understand your first example, you have the unit square $[0,1]^2 \subset \mathbb{R}^2$ with the subspace topology. Let me write the coordinates of its points $(\theta, z) \in [0,1]^2$, for psychological reasons that will become apparent in a minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the intuitive side, what you are doing is bending your square $[0,1]^2$ in such a way that  sides $\theta = 0$ and $\theta = 1$ become closer and closer... You put some glue on them, stick them... Ok: and what do you &quot;see&quot; now? -A cylinder, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Except for the fact that in real life you cannot stick just two segments -or it's really difficult-, you can try to do this experiment with a piece of paper and some glue.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the rigorous side, you may do the following: you recover that parametrization of the cylinder you have probably seen in some geometry or integral calculus course&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi: [0,1]^2 \longrightarrow \mathbb{R}^3 \ , \qquad \varphi (\theta , z) = (\cos (2\pi\theta ), \sin (2\pi\theta ), z) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $\varphi$ is obviously continuous (&quot;obviously&quot; means: &quot;someone taught you this in some calculus course a year ago&quot;) and surjective, if we restrict the codomain to the cylinder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 = 1 \ , \ z \in [0,1]      \right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we are looking to $\varphi$ as a map $\varphi : [0,1]^2 \longrightarrow C$. Notice that also this way it is continuous with the subspace topologies on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\varphi $ is &lt;em&gt;almost&lt;/em&gt; injective too, except for those points $(0,z)$ and $(1,z)$ that have the same image &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi (0,z) = (1,0,z) = \varphi (1,z) \ , \qquad \text{for all}\qquad z \in [0,1] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to fix this lack of injectivity, you say: &quot;Ok, let's identify (&quot;identify&quot; is the picky, correct, way to say &quot;glue&quot; in Topology) each $(0,z)$ with its corresponding $(1,z)$ for all $z$.&quot; So, you are defining an equivalence relation $\sim$ among the points of the square $[0,1]^2$, the one generated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(0,z) \sim (1,z) \qquad \text{for all} \qquad z \in [0,1] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You call $X = [0,1]^2/\sim$ the space obtained from the square $[0,1]^2$ after gluing (sorry, &quot;identifying&quot;) each $(0,z)$ with its corresponding $(1,z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write $\widetilde{(\theta ,z)} \in X$ the equivalence classes produced by this equivalence relation. Notice that, if $\theta \neq 0, 1$, you just have $\widetilde{(\theta , z)} = \left\{ (\theta , z) \right\}$, but $\widetilde{(0,z)} = \widetilde{(1,z)} = \left\{ (0,z), (1,z)   \right\}$. That is, points of $X$ are &quot;the same&quot; as the points of $[0,1]^2$, except for those $(0,z)$ and $(1,z)$ that are now glued togehter (&quot;identified&quot;, yes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In doing this, we also get a natural map (&quot;projection&quot;, &quot;identification&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\pi : [0,1]^2 \longrightarrow X \ , \qquad \pi (\theta , z) = \widetilde{(\theta , z)} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is continuous by definition of the quotient topology on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to our $\varphi$: since it identifies the same points as $\sim$ does, $\varphi$ induces a well-defined map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : X \longrightarrow C \ , \qquad \widetilde{\varphi}\widetilde{(\theta , z)} = \varphi (\theta ,z) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since we have glued the points with the same image by $\varphi$,  this new $\widetilde{\varphi}$ is bijective, but it's also continuous, because of the universal property of the quotient topology and the fact that $\widetilde{\varphi} \circ \pi = \varphi$. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last step. A continuous bijective map doesn't need to be a homeomorphism, as you probably know: the inverse $\widetilde{\varphi}^{-1}$ is not necessarily continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could we do? Well, you can try to write down an explicit formula for $\widetilde{\varphi}^{-1}$ and check its continuity directly, but this is hard and painful, so nobody does it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, everybody resorts to the following marvelous, great, second to none trick (the best one you can buy in elementary Topology; btw, &quot;trick&quot; means &quot;proposition&quot;: you can prove it, of course): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;$X$ is compact, because it is a quotient of a compact space (namely, $[0,1]^2$). $C$ is Hausdorff, because it is a subspace of a Hausdorff space (namely, $\mathbb{R}^3$). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The GTET (Greatest Trick in Elementary Topology)&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, I have a continuous, bijective map $\widetilde{\varphi} : X \longrightarrow C$ between a compact space and a Hausdorff one. &lt;strong&gt;Hence&lt;/strong&gt; $\widetilde{\varphi}$ is a homeomorphism.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Isn't that fantastic?   :-)  (Don't forget it the next time you need to prove that some map is a homeomorphism: it will save your life.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can also prove rigorously your first intuition: your space is a cylinder, $X \cong C$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-09T23:12:28.000" LastActivityDate="2010-10-09T23:12:28.000" CommentCount="2" />
  <row Id="6360" PostTypeId="2" ParentId="6357" CreationDate="2010-10-09T02:38:51.353" Score="1" ViewCount="0" Body="&lt;p&gt;Trying to place points in topology is risky.  The geometric intuition it gives can be very useful, but can also be very misleading.  Formally, a topology is defined by the open sets regardless of any capability to imagine them.  See &quot;Counterexamples in Topology&quot; for examples of the problems you face here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collapsing points can change the shape of a space considerably.  I think your specific example of collapsing the boundary of $B^n$ gives you $S^n$ as you have a compact n-dimensional manifold without boundary in which every loop is contractible to a point, but I am willing to be corrected. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-09T02:38:51.353" CommentCount="4" />
  <row Id="6361" PostTypeId="2" ParentId="6324" CreationDate="2010-10-09T04:37:24.277" Score="3" ViewCount="0" Body="&lt;p&gt;Equation $\rm\ 3^{2x}-2^{3y}=1\ $ is an instance of various special cases of &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalans_conjecture&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First,$\ $ making the specialization $\rm\ \ \: z,\:p^n = 3^x,2^{3y}\ $ below yields $\rm\ x = 1 = y\ $ as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt;$\ \ $ $\rm z^2 - p^n = 1\ \ \Rightarrow\ \ z,\:p^n = \:3\ ,\:2^3\ $  or $\ 2,\:3\ $ for $\rm\ \ z,\:p\:,n\in \mathbb N,\ \ p\: $ prime  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ \ (z+1)\:(z-1)\: =\: p^n\ \  \Rightarrow\ \ z+1 = p^{\:j},\ \ z-1 = p^k\ $ for some $\rm\ j,\:k\in \mathbb N$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \:\Rightarrow\ \ \ \ 2\ =\ p^{\:j} - p^k\ =\ p^k\: (p^{\:j-k}-1) \ \Rightarrow\ p^k=2\ $ or $\rm\ p^k = 1 \ \Rightarrow\ \ldots$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it's simply the special case $\rm\: X = 3^x,\ Y = 2^y\: $ of $\rm\ X^2 - Y^3 = 1\:,\: $ solved by Euler in 1738. Nowadays one can present this solution quite easily using elementary properties of $\rm\ \mathbb Z[\sqrt[3]{2}]\:$, e.g see p.44 of Metsankyla: &lt;a href=&quot;http://www.ams.org/journals/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture: another old diophantine problem solved&lt;/a&gt;. See also &lt;a href=&quot;http://mathoverflow.net/questions/39561/is-there-an-elementary-way-to-find-the-integer-solutions-to-x2-y31&quot; rel=&quot;nofollow&quot;&gt;this MO thread&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/25674/integer-points-of-an-elliptic-curve&quot; rel=&quot;nofollow&quot;&gt;this MO thread&lt;/a&gt; and Schoof: &lt;a href=&quot;http://www.maths.mq.edu.au/~alf/Humid%20Summer/catalan.pdf&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture&lt;/a&gt;. Note also that Catalan equations are a special case of the theory of generalized Fermat (FLT) equations, e.g. see &lt;a href=&quot;http://www.math.mcgill.ca/~darmon/pub/Articles/Expository/09.Ribenboim/paper.pdf&quot; rel=&quot;nofollow&quot;&gt;Darmon's exposition&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-09T04:37:24.277" />
  <row Id="6362" PostTypeId="2" ParentId="6357" CreationDate="2010-10-09T08:30:59.550" Score="5" ViewCount="0" Body="&lt;p&gt;As for your second example, Ross Milikan's intuition is right. Let's see how we can arrive to the conclusion that $B^n/S^{n-1} \cong S^n$ examining what happens in low dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For $n=1$&lt;/strong&gt;, the (closed) unit ball is the interval $B^1 = [-1,1]$ and its boundary the &quot;sphere&quot; $S^0 = \left\{ -1, 1\right\}$. Now, if we try to imagine what happens to $B^1$ when we glue (&quot;identify&quot;) the points $-1$ and $1$, we see the segment $[-1,1]$ bending, points $-1$ and $1$ approaching one to each other... Our glue provider has already put some of it on them, so when they meet, they stick together and we get... a circumference! So $B^1 / S^0 \cong S^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, before going to higher dimensions: how can we prove this? Well, it turns out that, with some adaptations, we already know how to do it. First, notice that $B^1 \cong [0,1]$ and here $S^0 = \left\{ 0,1 \right\}$. So we resort again to a parametrization; this time, of the circumference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : [0,1] \longrightarrow \mathbb{R}^2 \ ,\qquad \varphi (\theta ) = (\cos (2\pi\theta) , \sin (2\pi \theta) )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and proceed along the lines of the former cylinder example: $\varphi$ identifies the points of the &quot;sphere&quot; $S^0$, $\varphi (0) = (1,0) = \varphi (1)$, so $\varphi$ induces a well-defined map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : B^1/S^0 \longrightarrow  S^1 \ , \qquad \widetilde{\varphi}(\widetilde{\theta}) = \varphi (\theta) \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is continuous and bijective. Now use the GTET (Greatest Trick of Elementary Topology) you already know to conclude that $\widetilde{\varphi} $ is a homeomorphism  $B^1/S^0 \cong S^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so good. What's next? &lt;strong&gt;Let's take a look at $n=2$.&lt;/strong&gt; The unit ball $B^2$ is now a disk and its boundary the circumference $S^1$. We put some glue all over $S^1$ and we bend the disk and at the same time shrink the circumference making it become smaller and smaller till it's just one point. What do we get? A sphere!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may suspect that $B^2 / S^1 \cong S^2$. In order to prove it, we may try to reproduce our &quot;visual&quot; homeomorphism with formulae. Let's put our disk $B^2$ in $\mathbb{R}^3$ in the $xy$-plane, centered at $(0,0,0)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 \leq 1 \ , \ z= 0  \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we put our sphere centered at $(0,0, 1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 + (z-1)^2 = 1 \right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the South Pole is $(0,0,0)$ -the center of the disk $B^2$-, the North Pole is $(0,0,2)$ and the equator lies in the plane $z= 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our &quot;visual&quot; homeomorphism bends the disk over the sphere and in doing this shrinks its boundary till it is identified with the North Pole of the sphere. So, let's try to do the same with a map like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : B^2 \longrightarrow S^2 \ , \qquad \varphi (x,y) = (f(r)x , f(r)y, 2r) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $r= + \sqrt{x^2+y^2}$. Let's analyse what this kind of map is doing:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When going from $r = 0$ to $r=1$, we want the points $\varphi (x,y)$ go up the sphere, from height $z=0$ (South Pole) to $z=2$ (North Pole). So $z$ has to climb up at double speed as $r$ goes from the center of the disk to its boundary. Hence we put that $2$.&lt;/li&gt;&#xA;&lt;li&gt;To make things easier, we want the projections of the points $\varphi (x,y)$ onto the $xy$-plane to lie in the same direction as the point of the disk $(x,y)$ they come from. So this projection should be something like $(\lambda x, \lambda y)$. At the same time this lambda will vary with the distance $r$ to the center of the disk in order that $\varphi (x,y)$ remains on the sphere. So we try this $(f(r)x , f(r)y)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To find which $f(r)$ we need, we impose the condition that $\varphi (x,y) $ lies on the sphere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1 = x^2 + y^2 + (z-1)^2 = f(r)^2 r^2 + (2r-1)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we get $f(r) = 2 \sqrt{\frac{1-r}{r}}$. So our map is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi (x,y) = &#xA;\begin{cases}&#xA;\left( 2 \sqrt{\frac{1-r}{r}}x , 2 \sqrt{\frac{1-r}{r}}y, 2r  \right) &amp;amp; \text{if}\ (x,y) \neq (0,0) \\&#xA;(0,0,0) &amp;amp; \text{if}\ (x,y) = (0,0) \ .&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise.&lt;/strong&gt; Prove that $\varphi$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we proceed much in the same way as in the cylinder example. By construction, $\varphi$ is clearly surjective. But not injective: all the points in the boundary $r=1$ of the disk  go to the North Pole. So we quotient out this boundary and, as before, $\varphi$ induces a bijective continuous map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : B^2/S^1 \longrightarrow S^2 \ , \qquad \widetilde{\varphi}\widetilde{(x,y)} = \varphi (x,y) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We apply again our nuclear weapon GTET and we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All right, &lt;strong&gt;what about higher dimensions?&lt;/strong&gt; -We've got no geometric or visual intuition there (at least, I've got none; if you think you can &quot;see&quot; what happens in, say, $\mathbb{R}^4$, congratulations, but just in case do a visit to your psychiatrist and tell him/her). Anyway, formulae allow us to go beyond our visual intuition. We put our ball and sphere inside $\mathbb{R}^{n+1}$ as before:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2 \leq 1 \ , \ x_{n+1} = 0 \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2   + (x_{n+1}-1)^2 = 1 \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we try some map of the same kind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : B^n \longrightarrow S^n \ , \qquad \varphi (x_1, \dots , x_n ) = (f(r)x_1, \dots , f(r)x_n, 2r) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercises.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Complete the details.&lt;/li&gt;&#xA;&lt;li&gt;This is not the same homeomorphism we previously found for $n=1$. Write this general formula for $n=1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-10T02:35:15.503" LastActivityDate="2010-10-10T02:35:15.503" />
  <row Id="6363" PostTypeId="2" ParentId="6321" CreationDate="2010-10-09T08:40:23.810" Score="1" ViewCount="0" Body="&lt;p&gt;Unless I am confused (and I hope someone will correct me!) you are describing a subset of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville numbers&lt;/a&gt; and they are all transcendental.  The point is - the numbers you describe are not rational, but are very very well approximated by rational numbers.  I think, also, that it suffices for $Z(k)$ to be super-quadratic (so its difference $Z(k+1) - Z(k)$ is super-linear).  The canonical example is discussed on the same webpage.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-10-09T09:04:51.113" LastActivityDate="2010-10-09T09:04:51.113" CommentCount="5" />
  <row Id="6364" PostTypeId="1" CreationDate="2010-10-09T08:58:46.173" Score="7" ViewCount="244" Body="&lt;p&gt;In particular for two real numbers $a$ and $b$, I'd like to know if there are formulas for $\zeta (a+b)$ and $\zeta (a-b)$ as a function of $\zeta (a)$ and $\zeta (b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I could find online is a paper by Harry Yosh &quot;General Addition Formula for Meromorphic Functions Derived from Residue Theorem&quot; in some little known journal, but unfortunately I have no access to it and don't know if it would answer my question. Maybe this is well-known and I didn't search correctly...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated, thanks!&lt;/p&gt;&#xA;" OwnerUserId="2398" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-09T11:33:57.857" LastActivityDate="2010-12-01T18:55:59.120" Title="Are there addition formulas for the Riemann Zeta function?" Tags="&lt;number-theory&gt;&lt;special-functions&gt;&lt;riemann-zeta&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="6365" PostTypeId="1" AcceptedAnswerId="6438" CreationDate="2010-10-09T09:08:01.233" Score="0" ViewCount="267" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to solve this one. I've tried Dirichlet's test but to no avail.&#xA;Any help is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2399" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-09T11:36:27.880" LastActivityDate="2010-10-10T17:55:56.297" Title="Convergence of $\sum_{n=1}^{\infty} \frac{\sin (n x)}{2^n}$" Tags="&lt;sequences-and-series&gt;&lt;fourier-series&gt;" AnswerCount="2" />
  <row Id="6366" PostTypeId="2" ParentId="6365" CreationDate="2010-10-09T09:15:24.220" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; To show absolute convergence use that $\sin$ is bounded by 1 (I assume that you want $x$ to take real values).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-09T09:15:24.220" CommentCount="1" />
  <row Id="6367" PostTypeId="2" ParentId="6364" CreationDate="2010-10-09T09:35:57.320" Score="2" ViewCount="0" Body="&lt;p&gt;I've never seen any such formulas. If there were any, the would most likely be listed in &lt;a href=&quot;http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/&quot; rel=&quot;nofollow&quot;&gt;the Zeta section at functions.wolfram.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-09T09:35:57.320" />
  <row Id="6368" PostTypeId="2" ParentId="6364" CreationDate="2010-10-09T09:46:50.070" Score="9" ViewCount="0" Body="&lt;p&gt;Almost certainly, the answer is no. Consider $a$ with real part in the interval (1/2,1). The Riemann hypothesis states that $\zeta(a)\not=0$. If $\zeta(a)$ were a function of $\zeta(2+a)$ and $\zeta(2)$ then this would convert the RH into a statement about the values of $\zeta(s)$ on $5/2&amp;lt;\Re[s]&amp;lt;3$, which would be just too easy.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-10-09T09:54:25.557" LastActivityDate="2010-10-09T09:54:25.557" />
  <row Id="6369" PostTypeId="1" AcceptedAnswerId="6373" CreationDate="2010-10-09T11:57:44.643" Score="1" ViewCount="110" Body="&lt;p&gt;How does one &lt;strike&gt;find the no of positive integers such that&lt;/strike&gt; find all &lt;em&gt;possible&lt;/em&gt; numbers such that $$\lfloor{\sqrt{n}\rfloor} \mid n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i did was to subsitute $n=t^{2}$ so that the equation becomes $\lfloor{t\rfloor} \mid t^{2}$ But this means that we want $t^{2} = k \lfloor{t\rfloor}$, where $k \in \mathbb{N}$. I don't really know what to do from here. By the way, this problem is in Apostol. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-09T15:36:34.200" LastActivityDate="2010-10-09T15:36:34.200" Title="Finding the Number of Positive integers such that $\lfloor{\sqrt{n}\rfloor} \mid n$" Tags="&lt;number-theory&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6370" PostTypeId="1" AcceptedAnswerId="6374" CreationDate="2010-10-09T13:35:19.900" Score="5" ViewCount="277" Body="&lt;p&gt;A map $s : \mathbb{N} \to X$ is a &lt;em&gt;computable sequence&lt;/em&gt; in $(X,\nu_X)$ when there exists a computable map $f : \mathbb{N} \to \mathbb{N}$ such that $s(n) = \nu_X(f(n))$ for all $n \in \mathrm{dom}(\nu_X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess would be, &quot;A map s taking N onto X is a computable sequence in the ??? (X, nu??) when there exists a computable map f taking N onto N such that s at n equals ??? of f ??? for all n elements of the domain of nu ???.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am searching for a way to read it aloud that encodes all the elements of the sentence into speech.&lt;/p&gt;&#xA;" OwnerUserId="2163" LastActivityDate="2010-10-10T12:15:35.987" Title="How can I read this mathematical sentence aloud in English?" Tags="&lt;terminology&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="6371" PostTypeId="1" AcceptedAnswerId="6375" CreationDate="2010-10-09T13:51:42.060" Score="2" ViewCount="147" Body="&lt;p&gt;In the UK the lottery uses numbers 1 to 49 and a total of six numbers are picked. It has been said may times that there is as much chance of numbers 1 2 3 4 5 6 to be picked as any other random combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the first 3 numbers to come out are 1,2 and 3. What are the chances of a number between 1 and 10 coming out next Vs a number between 11 and 20? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are obviously less numbers between 1 and 10 now that we already lost 1 to 3, so surely the probability is that a number between 11 and 20 is more likely? In which case, the chances of a lottery selection of 1,2,3,4,5,6 is less likely than 2,12,21,28,32,47 for example...&lt;/p&gt;&#xA;" OwnerUserId="2400" LastActivityDate="2010-10-09T15:43:46.663" Title="Lottery probability" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6372" PostTypeId="2" ParentId="6370" CreationDate="2010-10-09T14:05:08.647" Score="4" ViewCount="0" Body="&lt;p&gt;I would say &quot;A map s is a computable sequence when there exists a computable map f satisfying certain properties&quot; while writing down the property.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-09T14:05:08.647" CommentCount="2" />
  <row Id="6373" PostTypeId="2" ParentId="6369" CreationDate="2010-10-09T14:18:30.247" Score="8" ViewCount="0" Body="&lt;p&gt;I suppose you want to find all &lt;em&gt;possible&lt;/em&gt; numbers such that $\displaystyle [\sqrt{n}] \mid n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $\displaystyle n$ is not a perfect square, then there must be some $\displaystyle k$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  k^2 &amp;lt; n &amp;lt; (k+1)^2$. We have that $\displaystyle k = [\sqrt{n}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only numbers in the range $\displaystyle  k^2 &amp;lt; n &amp;lt; (k+1)^2$ which are divisible by $k$ are $\displaystyle k^2+k$ and $\displaystyle k^2+2k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the numbers $\displaystyle n$ such that $\displaystyle [\sqrt{n}] \mid n$ are of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle k^2, k^2+k, k^2+2k$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-09T14:18:30.247" CommentCount="1" />
  <row Id="6374" PostTypeId="2" ParentId="6370" CreationDate="2010-10-09T15:05:08.833" Score="11" ViewCount="0" Body="&lt;p&gt;&quot;A map ess from en to ex is a computable sequence in ex nu-ex when there exists a computable map eff from en to en such that ess of en equals nu-ex of eff of en for all en in the domain of nu-ex.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyphens are meant to indicate that the pause between these syllables is shorter than between two separated words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I am not verbalizing that the X is subscripted, the ordered pair, or the difference between $n$ and $\mathbb{N}$. I assume you are talking about a situation where you are writing this statement on the board as you speak. Communicating any significant amount of math in a purely oral manner is incredibly hard because mathematical notation is much denser than ordinary spoken language; I find that I have to leave out some detail in order for the sentence to fit in my listener's head.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-09T15:05:08.833" CommentCount="4" />
  <row Id="6375" PostTypeId="2" ParentId="6371" CreationDate="2010-10-09T15:16:21.393" Score="2" ViewCount="0" Body="&lt;p&gt;The chances of the next number being between 1 and 10 is $\frac{7}{49}$, as opposed to the probability of it being between 11 and 20 being $\frac{10}{49}$. So, among other things, it is less likely that a lottery ticket will have only numbers between 1 and 10, as opposed to numbers between 1 and 20. However, that does not mean that a given ticket with numbers between 1 and 10 is less likely then a given ticket with numbers between 1 and 20. The fact that there are more tickets with numbers between 1 and 20 exactly cancels with the higher chance of getting such a ticket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;You have a bag of pens; a red pen, and 99 blue pens numbered 1-99. The chance of you getting a red pen is only .01 whereas the chance of oyu getting a blue pen is .99, but the chance of getting any given blue pen (say, #42) is .99 / 99 = .01, which is the same as the chance of getting the red pen.  &lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2010-10-09T15:16:21.393" />
  <row Id="6376" PostTypeId="1" CreationDate="2010-10-09T15:30:28.190" Score="2" ViewCount="108" Body="&lt;p&gt;For instance, say $G = \langle x , y \ | \ x^{12}y=yx^{18} \rangle$. I want to know what is the &lt;a href=&quot;http://groupprops.subwiki.org/wiki/Normal_subgroup_generated_by_a_subset&quot; rel=&quot;nofollow&quot;&gt;normal closure&lt;/a&gt; of $y$ in $G$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, what are the standard approaches to compute the normal closure of a subset of a finitely presented group? Are there algorithms?  &lt;/p&gt;&#xA;" OwnerUserId="2342" LastActivityDate="2010-10-13T13:23:04.123" Title="Normal closure in groups" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6377" PostTypeId="2" ParentId="6376" CreationDate="2010-10-09T15:42:48.880" Score="6" ViewCount="0" Body="&lt;p&gt;You can compute the normal closure by computing the quotient, and then considering the kernel of the quotient homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example you gave, let $N$ be the normal closure of $y$ in $G$.  Then $G/N$ has presentation&#xA;$$&#xA;\langle x,y \mid x^{12}y = yx^{18},y=1\rangle&#xA;$$&#xA;This presentation reduces to $\langle x \mid x^{12} = x^{18}\rangle$, which is the same as $\langle x \mid x^6 = 1\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $G/N$ is a cyclic group of order 6, and $N$ is the kernel of the homomorphism $G\to G/N$. In particular, the normal closure of $y$ consists of all words for which the total power of $x$ is a multiple of 6.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-09T15:42:48.880" CommentCount="1" />
  <row Id="6378" PostTypeId="2" ParentId="6371" CreationDate="2010-10-09T15:43:46.663" Score="2" ViewCount="0" Body="&lt;p&gt;Having chosen 1,2,3 all possibilities (there are 46*45*44/6) that include 1,2,3 are equally likely and all possibilities that do not include 1,2,3 have probability zero.  The psychological error comes from grouping &quot;numbers less than 10&quot; together.  As Moron commented you could also group &quot;numbers containing a digit 2&quot;.  Each of these are legal groupings and the probability of selecting all numbers from within them can be calculated.  But it is still true that any particular combination is equally likely at the outset.  A draw of 13,18,29,33,36,45 looks &quot;random&quot; to our eye and we group it with all other &quot;random looking&quot; combinations.  From this point of view, a &quot;random looking&quot; combination is highly likely.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-09T15:43:46.663" />
  <row Id="6379" PostTypeId="2" ParentId="6295" CreationDate="2010-10-09T16:03:14.070" Score="0" ViewCount="0" Body="&lt;p&gt;You could solve $y^2\equiv1$ (mod $pq$) with $y\not\equiv\pm1$ (mod $pq$)&#xA;and then take $x=7y$. :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-09T16:03:14.070" CommunityOwnedDate="2010-10-09T16:03:14.070" />
  <row Id="6380" PostTypeId="1" AcceptedAnswerId="6439" CreationDate="2010-10-09T16:05:28.807" Score="7" ViewCount="89" Body="&lt;p&gt;Let $X$ be a Hausdorff space and $A$ a closed subspace. Suppose the inclusion $A \hookrightarrow X$ is a cofibration. Let $f, g: X \to Y$ be maps that agree on $A$ and which are homotopic. Are they homotopic relative to $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motivation for asking this question comes from the following result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $i: A \to X, j: A \to Y$ be cofibrations. Suppose $f: X \to Y$ is a map which makes the natural triangle commutative. Suppose $f$ is a homotopy equivalence. Then $f$ is a cofiber homotopy equivalence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I'm having trouble adapting the proof in Peter May's book of this to the question I asked. Nonetheless, the standard examples of pairs of maps which are homotopic but not with respect to which some subset on which they agree (say, the identity map of a comb space and its collapsing to a suitably chosen point), don't seem to involve NDR pairs. &lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-10T17:25:49.837" Title="Are homotopic maps over a cofibration homotopic relative to the cofibration?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6381" PostTypeId="1" CreationDate="2010-10-09T16:30:03.147" Score="2" ViewCount="90" Body="&lt;p&gt;Is there any relation between homology of a space with local coefficient (in $Q$ vector space) and the homology with $Q$ coefficient?Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-12-07T05:35:16.680" Title="homology with local coefficient" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6382" PostTypeId="1" CreationDate="2010-10-09T17:00:17.160" Score="0" ViewCount="163" Body="&lt;p&gt;(Hint: notice that (n+1)(n!+1) = (n+1)!+n+1)&lt;/p&gt;&#xA;" OwnerUserId="2401" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-09T17:37:52.347" LastActivityDate="2010-10-09T21:17:49.833" Title="Let n be a positive integer, n! denotes the factorial of n. Let d = gcd(n! + 1, (n + 1)! + 1). Show that d divides n." Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="6383" PostTypeId="1" AcceptedAnswerId="6386" CreationDate="2010-10-09T17:38:15.293" Score="1" ViewCount="95" Body="&lt;p&gt;Let's say I am distributing a bunch of encryption keys, where if two users have the same key, they can communicate.  I have 20 different keys, and I am giving each user 2.  Since encryption keys don't disappear when I hand them out, it can be considered 'with replacement', but I'm not giving the same user the same key twice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what the probability that two users can communicate under these circumstances.  Doing a bit of the math from what I know, each user has 20 * 19 = 380 permutations of keys.  The order that they get the keys in doesn't matter, though, so it's actually 190 different combinations of keys.  So, two users have 190 possible combinations of four keys, and what is the probability that one of their two keys is in common.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That reasoning is about as far as I get.  How does one move forward from there?  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2402" LastEditorUserId="2402" LastEditorDisplayName="" LastEditDate="2010-10-09T18:19:18.467" LastActivityDate="2010-10-10T00:53:05.037" Title="Basic probability... probability that two users share a key out of a keypool of 20?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6384" PostTypeId="2" ParentId="6382" CreationDate="2010-10-09T17:43:17.953" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the definition of GCD and the fact that $$(n+1)!+1 = n! \times (n+1) +1 = n! \cdot n + (n!+1)$$ We know that $d \mid (n!+1)$ since $d$ is the &lt;strong&gt;GCD&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-09T17:43:17.953" CommentCount="3" />
  <row Id="6385" PostTypeId="2" ParentId="6382" CreationDate="2010-10-09T17:59:07.643" Score="3" ViewCount="0" Body="&lt;p&gt;The given hint shows that $\rm\ n\ $ is an integral linear combination of $\rm\ n!+1\ $ and $\rm\ (n+1)! + 1\:,\: $ so $\rm\ n\ $ is divisible by all common divisors, including the GCD. In fact we can go further and show that the GCD = 1. Namely, since the GCD divides the coprime numbers $\rm\:n\:$ and $\rm\ n!+1\ $ it must be 1. Below is an alternate derivation using explicit gcd laws, along with an explicit Bezout linear representation of the GCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $\rm\ k = n!+1\ $ below shows that the GCD equals $\rm\ gcd(n!+1,n) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\ \  gcd(k,(n+1)k-n)\ =\ gcd(k,n)\ \ $ via $\rm\ \ n = (n+1)k - ((n+1)k-n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad$ recalling $\rm\quad\quad\ \  gcd(k,m)\ =\ gcd(k,\:jk\pm m)\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad$ since if $\rm\ \ d|k\ $ then $\rm\ \: d|m\ \ \iff\ \: d\ |\ jk\pm m\quad\ \ $ Above $\rm \ j = n+1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact we can unwind the above to obtain an explicit Bezout linear representation of the GCD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\quad \rm n\ =\ n!\ ((n+1)!+1) + (n-(n+1)!)\ (n!+1) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing the above through by $\rm\:n\:$ shows that the gcd is 1! $\ $ But, alas, I fear I've exclaimed too much ...&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-09T21:17:49.833" LastActivityDate="2010-10-09T21:17:49.833" CommentCount="5" />
  <row Id="6386" PostTypeId="2" ParentId="6383" CreationDate="2010-10-09T17:59:09.550" Score="1" ViewCount="0" Body="&lt;p&gt;In this case, the probability is just $1$ minus the probability that they &lt;em&gt;can't&lt;/em&gt; communicate, i. e. that they've been given $4$ different keys.  The number of possible arrangements of keys among the two users is $190^2=36100$, since, as you noted, they each have $190$ possible pairs.  The number of ways they can be given four different keys in total is $\frac{20\times 19\times 18\times 17}{4}=29070$.  The numerator should be obvious, and we are dividing by $4$ to account for swaps in either Alice's or Bob's keys.  So the probability is $1-(29070/36100)=37/190$.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-09T17:59:09.550" CommentCount="5" />
  <row Id="6387" PostTypeId="1" AcceptedAnswerId="6388" CreationDate="2010-10-09T19:24:40.357" Score="2" ViewCount="256" Body="&lt;p&gt;So I was bored and decided to figure out the indefinite integral of the absolute value function, $|x|$. Using integration by parts ($u=|x|, dv=dx$, $dx = \text{sgn}(x)=\frac{|x|}{x}$), it can be shown that $\displaystyle\int |x| dx = \frac{x |x|}{2}+C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I decided to take the integral again, finding that $\displaystyle\int\left(\int |x| dx \right) dx=\frac{x^2 |x|}{3}+C$. Continuing, I found the pattern in the title, that the $n$th indefinite integral of $|x|$ is $\displaystyle\frac{x^n |x|}{n+1}+C$. Is there a way to prove this general result?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-09T19:30:07.987" LastActivityDate="2010-11-11T20:08:36.847" Title="How can I prove $\displaystyle\underbrace{\int \ldots \int}_{n} |x| dx = \frac{x^n |x|}{n+1}+C$?" Tags="&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="6388" PostTypeId="2" ParentId="6387" CreationDate="2010-10-09T19:33:44.530" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, there is a way: Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_induction&quot; rel=&quot;nofollow&quot;&gt;Mathematical induction&lt;/a&gt;. I don't add more details because I think it's elementary, isn't it?&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-09T19:33:44.530" CommentCount="7" />
  <row Id="6389" PostTypeId="2" ParentId="6387" CreationDate="2010-10-09T19:57:06.080" Score="3" ViewCount="0" Body="&lt;p&gt;The $C$ in the title of this question should be replaced by an arbitrary polynomial $p(x)$ of degree $\le n-1$. After all, we are talking of an $n$-fold integration here.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-09T19:57:06.080" CommentCount="1" />
  <row Id="6390" PostTypeId="2" ParentId="6364" CreationDate="2010-10-09T20:01:49.957" Score="6" ViewCount="0" Body="&lt;p&gt;There is a functional equation relating values of $\zeta$ at $s$ and $(1-s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were formulas relating values along a continuous family of 1-dimensional curves, such as $x + y = C$, one would get a differential equation for $\zeta$, or some comparably strong constraint.  It is known that $\zeta$ does not satisfy any ODE's with algebraic functions as coefficients. Of course there could be gamma functions or other more complicated coefficients but prospects for this kind of additional structure in $\zeta$ seem dim.  There are no extra functional equations for finite field zetas, for example.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-01T18:55:59.120" LastActivityDate="2010-12-01T18:55:59.120" CommentCount="2" />
  <row Id="6391" PostTypeId="1" CreationDate="2010-10-09T20:09:05.127" Score="1" ViewCount="116" Body="&lt;p&gt;I'm looking for a fast algorithm to perform division of large numbers (by hand). Traditional long division just isn't fast enough for my needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of these cases, I'm only looking for the modulus/remainder of the division, so if there is a fast algorithm specifically for this purpose, it would suffice. (But, if there is a way to get the quotient as well, it would also help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Now that I think about it, the algorithm for synthetic division/synthetic substitution involves no division (and addition/multiplication is typically easier and faster than division). Is there perhaps a clever way to adapt this algorithm for fast numerical division?&lt;/p&gt;&#xA;" OwnerUserId="2403" LastEditorUserId="2403" LastEditorDisplayName="" LastEditDate="2010-10-10T01:46:25.713" LastActivityDate="2010-10-10T01:46:25.713" Title="Fast algorithm for modular division (residue)" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6392" PostTypeId="2" ParentId="6391" CreationDate="2010-10-09T20:22:37.470" Score="2" ViewCount="0" Body="&lt;p&gt;The key idea is to make a good estimate of the quotient based on the most significant digits of the dividend and divisor. This is described in detail in the division algorithm presented in section 4.3.1 of Knuth, The art of computer programming, Volume 2, Seminumerical algorithms - the standard reference. Similar ideas (Lehmer at al.) are used in optimizing the computation of the remainder sequence in the Euclidean algorithm for the GCD, which is also described in Knuth's book, viz. p. 345, section 4.5.2.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-09T20:22:37.470" />
  <row Id="6393" PostTypeId="1" AcceptedAnswerId="6396" CreationDate="2010-10-09T20:52:44.067" Score="3" ViewCount="292" Body="&lt;p&gt;I am trying to understand left/right cosets in group theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the example in my text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G = \lbrace 1, a, b, c, d ,e \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets define the group operation $.$ by the following table, where the entry at row $x$ and column $y$ gives $x.y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ex. $d.e = b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&amp;nbsp;&lt;b&gt;1&lt;/b&gt;&amp;nbsp;&lt;b&gt;a&lt;/b&gt;&amp;nbsp;&lt;b&gt;b&lt;/b&gt;&amp;nbsp;&lt;b&gt;c&lt;/b&gt;&amp;nbsp;&lt;b&gt;d&lt;/b&gt;&amp;nbsp;&lt;b&gt;e&lt;/b&gt;&lt;br /&gt;&#xA;&lt;b&gt;1&lt;/b&gt;&amp;nbsp;1&amp;nbsp;a&amp;nbsp;b&amp;nbsp;c&amp;nbsp;d&amp;nbsp;e&lt;br /&gt;&#xA;&lt;b&gt;a&lt;/b&gt;&amp;nbsp;a&amp;nbsp;b&amp;nbsp;1&amp;nbsp;d&amp;nbsp;e&amp;nbsp;c&lt;br /&gt;&#xA;&lt;b&gt;b&lt;/b&gt;&amp;nbsp;b&amp;nbsp;1&amp;nbsp;a&amp;nbsp;e&amp;nbsp;c&amp;nbsp;d&lt;br /&gt;&#xA;&lt;b&gt;c&lt;/b&gt;&amp;nbsp;c&amp;nbsp;e&amp;nbsp;d&amp;nbsp;1&amp;nbsp;b&amp;nbsp;a&lt;br /&gt;&#xA;&lt;b&gt;d&lt;/b&gt;&amp;nbsp;d&amp;nbsp;c&amp;nbsp;e&amp;nbsp;a&amp;nbsp;1&amp;nbsp;b&lt;br /&gt;&#xA;&lt;b&gt;e&lt;/b&gt;&amp;nbsp;e&amp;nbsp;d&amp;nbsp;c&amp;nbsp;b&amp;nbsp;a&amp;nbsp;1&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is no problem i understand this. But then we get the left and right cosets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a group and let $H \leq G$. A left coset of $H$ in $G$  ($G / H$) is a set of the form $gH = \lbrace gh : h \in H \rbrace$ for some $g \in G$. A right coset of $H$ in $G$ ($H$ \ $G$) is a set of the form $Hg = \lbrace hg : h \in H \rbrace$ for some $g \in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples that i am trying to figure out how they are generated. I guess i just do not understand the theory fully. I would like a little help explaining and possible a few more examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, a, b \rbrace = \lbrace \lbrace1, a, b\rbrace , \lbrace c, d, e \rbrace\rbrace$&lt;br /&gt;&#xA;$\lbrace 1, a, b \rbrace$ \ $\lbrace 1, a, b, c, d, e\rbrace = \lbrace \lbrace 1, a, b \rbrace, \lbrace c, d, e \rbrace \rbrace$&lt;br /&gt;&#xA;$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, c \rbrace = \lbrace \lbrace 1, c \rbrace, \lbrace a, d \rbrace, \lbrace b, e \rbrace \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few more examples which i want to figure out are:&lt;br /&gt;&#xA;$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, e \rbrace = ?$&lt;br /&gt;&#xA;$\lbrace 1, d \rbrace$ \ $\lbrace 1, a, b, c, d, e \rbrace = ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="43" LastEditorDisplayName="" LastEditDate="2010-10-09T21:39:27.093" LastActivityDate="2010-10-10T00:32:59.547" Title="Left/Right Cosets" Tags="&lt;group-theory&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="6394" PostTypeId="2" ParentId="6370" CreationDate="2010-10-09T20:57:34.077" Score="0" ViewCount="0" Body="&lt;p&gt;I would try to keep it really simple. Rather than verbalizing all the function and set symbols and subscripts, I would just say, &lt;strong&gt;&quot;A &lt;em&gt;computable sequence&lt;/em&gt; is the result of composing a &lt;em&gt;notation&lt;/em&gt; after a &lt;em&gt;computable function&lt;/em&gt;.&quot;&lt;/strong&gt; I assume you're talking Computable Analysis theory and by $\nu_X$ you intend a &lt;em&gt;notation&lt;/em&gt; $\nu_X:\subseteq \Sigma^* \to X$. Of course, this reading assumes your audience already knows what you mean by a &lt;em&gt;notation&lt;/em&gt; and a &lt;em&gt;computable function&lt;/em&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-09T20:57:34.077" />
  <row Id="6395" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:30:19.507" Score="1" ViewCount="0" Body="&lt;p&gt;gprime, thinking about cosets in the following context may help you out. Leg $G$ be $\mathbb{R}^3$, the group operation will be vector addition. Suppose you want to know the left cosets of $H$, where $H$ is a straight line through the origin, which as you know, is a subspace. You look at the sets &quot;$x+H$&quot;. Fix such an $x$, and see what happens when you add $x+h_1$, $x+h_2$, etc. where $h_1, h_2, ...$ are elements of $H$. You get a straight line that is parallel to $H$, but no longer passes through the origin. Which point does it surely pass through? For one, $x$, because, being $H$ a subspace of $\mathbb{R}^3$, $x+0$ is one of the elements of $x+H$. If you change the $x$ for $y$, another point not in the same straight line that is $x+H$, you'll obtain another straight line parallel to $H$. The left cosets (which in this case are the same as the right cosets, because vector addition is commutative) are thus all straight lines parallel to $H$. Since in this case the left and right cosets are the same, all these parallel straight lines form a group themselves, with the group operation defined simply as this: to add two straight lines, just take one point from on one of them, another point on the other, and add them, and all points that you obtain doing this will be a new straight line parallel to $H$. This new group is called the quotient group.&#xA;As your textbook says, however not all left cosets will be right cosets, in which case there's no defining a quotient group.&#xA;Well, I hope I didn't make it worse, but that particular example was helpful for me.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-09T21:30:19.507" />
  <row Id="6396" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:30:36.543" Score="7" ViewCount="0" Body="&lt;p&gt;If your table is correct, it should be the symmetric group in three letters, with &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; the 3-cycles, and &lt;strong&gt;c&lt;/strong&gt;, &lt;strong&gt;d&lt;/strong&gt;, and &lt;strong&gt;e&lt;/strong&gt; the transpositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Now. Your first equations is, alas, nonsense. The set that contains $1$, $a$, $b$, $c$, $d$, and $e$ is certainly not equal to the set whose elements are the sets $\{1,a,b\}$ and $\{c,d,e\}$. You are writing nonsense.  The rest are a bit better because you are looking at cosets.&lt;/strike&gt; Fixed in problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ has six different subgroups: the trivial subgroup $\{1\}$; the whole subgroup $G$; one subgroup of order three, $H=\{1,a,b\}$, and three subgroups of order two: $K_1=\{1,c\}$, $K_2=\{1,d\}$, and $K_3=\{1,e\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the left cosets of $H$ in $G$? They are the sets $1H$, $aH$, $bH$, $cH$, $dH$, and $eH$. As it happens, $1H=aH=bH = H$, and $cH=dH=eH=\{c,d,e\}$. You can verify it exlicitly; for instance, &#xA;$$aH = \{ ah : h \in H\} = \{a1, aa, ab\} = \{a, b, 1\} = H$$&#xA;and&#xA;$$cH = \{ ch: h \in H\} = \{c1, ca, cb\} = \{c, e, d\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right costs of $H$ are the same as the left cosets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what are the left cosets of $K_3=\{1,e\}$ in $G$? They are the sets $1K_3$, $aK_3$, $bK_3$, $cK_3$, $dK_3$, and $eK_3$. They are:&#xA;$$\begin{array}{rcl}&#xA;1K_3 &amp;amp; = &amp;amp; \{1k : k \in K_3\} = \{11, 1e\} = \{1,e\} = K_3\\&#xA;aK_3 &amp;amp; = &amp;amp; \{ak : k \in K_3\} = \{a1, ae\} = \{a, c\}.\\&#xA;bK_3 &amp;amp; = &amp;amp; \{bk : k \in K_3\} = \{b1, be\} = \{b, d\}.\\&#xA;cK_3 &amp;amp; = &amp;amp; \{ck : k \in K_3\} = \{c1, ce\} = \{c, a\} = aK_3.\\&#xA;dK_3 &amp;amp; = &amp;amp; \{dk : k \in K_3\} = \{d1, de\} = \{d, b\} = bK_3.\\&#xA;eK_3 &amp;amp; = &amp;amp; \{ek : k \in K_3\} = \{ei, ee\} = \{e, 1\} = K_3.&#xA;\end{array}$$&#xA;So there are three distinct left cosets, and they are $\{1,e\}$, $\{a,c\}$, and $\{b,d\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the right cosets of $\{1,d\}$ in $G$? They are $K_21$, $K_2a$, $K_2b$, $K_2c$, $K_2d$, and $K_2e$. Let us be a bit smarter this time, rather than computing them directly: we know that any two &lt;em&gt;distinct&lt;/em&gt; cosets are disjoint, and that the right coset of $x$ contains $x$. So $K_21$ contains both $1$ and $d$, hence must be equal to the right coset of $d$, $K_2d$; that is, $K_21=K_2d = \{1,d\}$. The right coset of $a$ contains $a$ and contains $da = c$, hence must be equal to the right coset of $c$, $K_2c$; indeed it is, as $dc = a$; so $K_2a=K_2c=\{a,c\}$. And the right coset of $b$, $K_2b$, contains $b$ and $db=e$, hence must equal the right coset of $e$, $K_2e$, which it does (since $de=b$ and $1e=1$). So $K_2b=K_2e=\{b,e\}$. So the three right cosets of $K_2$ in $G$ are $\{1,d\}$, $\{a,c\}$, and $\{b,e\}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-09T22:06:13.987" LastActivityDate="2010-10-09T22:06:13.987" CommentCount="2" />
  <row Id="6397" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:40:12.660" Score="3" ViewCount="0" Body="&lt;p&gt;The fundamental idea of a coset is that of taking a subgroup and &quot;translating&quot; it to fill up $G$.  A coset is just one such &quot;translate.&quot;  No coset but $H$ itself is a subgroup of $G$, since they don't contain $1$, but there is a lot of ground that can be covered by treating them like elements of a group or set instead.  Since the cardinality of the cosets is constant, they represent ways of dividing up $G$, and you get nice formulas like $|G|=|H|[H:G]$, where $[H:G]$ is called the &lt;strong&gt;index&lt;/strong&gt; of $H$ in $G$ and is equal to the number of cosets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the biggest use of cosets is to set up the idea of a &lt;strong&gt;quotient group&lt;/strong&gt;.  You may have seen the general idea -- formally dividing one mathematical structure by a substructure -- in other fields, like topology.  What you do in almost every case is add, to the laws defining your mathematical structure, another law that sets everything in the substructure equal to a point, or the identity, or zero.  Then you try to make everything else work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In group theory, in particular, the set $G/H$ (or $H\backslash G$) always has an induced group structure when $H$ is a normal subgroup.  And in fact, it's easy most of the time to just think of its elements as abstract and indivisible group elements.  But sometimes, especially when you're proving that this works, it's useful to think of them as equivalence classes instead, where the equivalence relation is of the form $a\sim b$ when $a=bh$ for some $h\in H$.  Of course, such an equivalence class is just the coset $bH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give one nice example with abelian groups before moving on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quotient group $\mathbb{R}/\mathbb{Z}$ is the set of cosets of the form $x+\mathbb{Z}$, with $(x+\mathbb{Z})+(y+\mathbb{Z})=(x+y)+\mathbb{Z}$, which is just the induced addition.  But you can also think of the elements of the group as &lt;em&gt;numbers&lt;/em&gt; of the form $0\le x&amp;lt;1$, with addition being done &quot;modulo $\mathbb{Z}$&quot;: that is, $x+y$ in $\mathbb{R}/\mathbb{Z}$ is equal to the &lt;em&gt;fractional part&lt;/em&gt; of $x+y$ in $\mathbb{R}$.  So all we really have is a circle with an addition operation.  One place where this pops up is the group formed by numbers $e^{i\theta}$, the unit circle in $\mathbb{C}$, under multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another, weirder example is $\mathbb{R}/\mathbb{Q}$.  I don't know much about it in terms of group theory, but it gives rise to a standard example of a non-measurable set in analysis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another use of cosets will pop up when you study group actions.  I think this is just called the &lt;strong&gt;coset action&lt;/strong&gt;.  Basically, given a subgroup $H$, $G$ acts on $G/H$ by multiplication: $(g,kH)\mapsto gkH$.  Surprisingly, this action works differently than $G$'s normal multiplication on itself.  I don't know any good examples of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the coset questions you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the line $\lbrace 1,a,b,c,d,e\rbrace=\lbrace\lbrace 1,a,b\rbrace,\lbrace c,d,e\rbrace\rbrace$ looks like a typo or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the &lt;em&gt;right&lt;/em&gt; cosets of $\lbrace 1,a,b\rbrace$ in $G$.  If you look at the top left corner of the multiplication table, you see that right multiplication by $a$ or $b$ just gives you the same set $\lbrace 1,a,b\rbrace$, though its elements are permuted.  Multiplying by $c,d,$ or $e$ gives something different.  We don't have to check that these all give the same coset, since the sizes of the cosets are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second example, you can look at the two columns labeled $1$ and $c$.  Then just scan down the rows: $a\mapsto\lbrace a,d\rbrace,b\mapsto\lbrace b,e\rbrace,c\mapsto\lbrace c,1\rbrace$, and so on.  These are all cosets, and identifying three distinct ones is easy.  They also show you something important: if, for example, $aH=\lbrace a,d\rbrace$, you know for a fact that $dH$ is the same set, simply because $H$ contains $1$.  Thus, the elements of a coset always give the same coset when multiplied by the subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's do the same thing for the left cosets of $\lbrace 1,e\rbrace$.  Again, ignore everything but the $1$ and $e$ columns, and scan down.  We get $aH=\lbrace a,c\rbrace =cH,bH=\lbrace b,d\rbrace =dH$.  And, of course, $eH=H$ since $e\in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right cosets of $\lbrace 1,d\rbrace$ will be similar, but you'll be using the rows $1$ and $d$ instead of columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck, and I hope this was at least somewhat helpful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what book are you using?  I learned all this from Artin, but it's possible that you might be getting a different perspective if you have a different book.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-09T21:40:12.660" CommentCount="1" />
  <row Id="6398" PostTypeId="2" ParentId="6393" CreationDate="2010-10-09T21:45:19.707" Score="0" ViewCount="0" Body="&lt;p&gt;Some general notes: the notation G/H is used to denote the set of left cosets of H in G. Note that this is a &lt;em&gt;set&lt;/em&gt; whose elements are also sets (in particular, they are the left cosets of H). Similarly, the notation H\G is used to denote the set of right cosets of H in G; again, this is a &lt;em&gt;set&lt;/em&gt; whose elements are sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's look at the first three equations. Unfortunately, the first one makes no sense (OP has since edited the first equation), so let's start with the second one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notationally, as per what I wrote above, {1,a,b} \ {1,a,b,c,d,e} denotes the set of right cosets of {1,a,b}. To actually compute this, you take each element of H={1,a,b} and multiply on the right by each element of G={1,a,b,c,d,e}. So for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H*1 = {1,a,b}*1 = {1*1, a*1, b*1} = {1,a,b} = H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's too easy. Try the element a:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H*a = {1,a,b}*a = {1*a, a*a, b*a} = {a,b,1} = H, where we're using the multiplication table given to figure out each of these products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceed similarly for each of the remaining elements of G (which are b,c,d,e). That is, compute the cosets H*b, H*c, H*d, H*e in exactly the same way we did above. You should find that every coset is either H or {c,d,e}. Thus, H\G = { H , {c,d,e} }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third equation, you want to compute the &lt;em&gt;left&lt;/em&gt; cosets of the subset {1,c}. You do this exactly as I've done above, except remember to multiply on the &lt;em&gt;left&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the two examples you'd like to see worked out, we again proceed exactly as above. The only difference is that the first example is &lt;em&gt;left&lt;/em&gt; cosets of {1,e} while the second example is &lt;em&gt;right&lt;/em&gt; cosets of {1,d}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend working through these examples on your own. Working with cosets can be a little goofy at first, but doing the legwork on your own is really invaluable. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2010-10-09T21:53:36.893" LastActivityDate="2010-10-09T21:53:36.893" CommentCount="1" />
  <row Id="6401" PostTypeId="1" CreationDate="2010-10-09T23:52:00.067" Score="0" ViewCount="42" Body="&lt;p&gt;The fist &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_algebra&quot; rel=&quot;nofollow&quot;&gt;Weyl algebra&lt;/a&gt; $A_{1}$ can be represented by the following matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=\left(\begin{array}{ccccc}&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 3 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q=\left(\begin{array}{ccccc}&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots\\&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where P corresponds to the differential operator and Q corresponds to multiplication by x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in how this correspondence generalizes for $A_{n}$ because it seems like the language of matrices is in adequate for higher n but I'm afraid I don't know enough algebra to know what the appropriate tool to use would be.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-09T23:52:00.067" Title="Generalizing the matrix formulation of the Weyl algebra" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" CommentCount="2" />
  <row Id="6402" PostTypeId="2" ParentId="6370" CreationDate="2010-10-10T00:28:35.237" Score="4" ViewCount="0" Body="&lt;p&gt;David Speyer wrote how I would say it in practise, in a context where I was writing it on a black/whiteboard. Here's how I would say it in a pub or walking down the street:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Let's define a 'representation map for X' [&lt;em&gt;or your own preferred jargon&lt;/em&gt;] to be just some partial function nu, from the natural numbers to X. Then we can define &lt;em&gt;a computable sequence for that representation map nu&lt;/em&gt; to be any function s, from the natural numbers to X, which [is consistent with&amp;nbsp;/&amp;nbsp;agrees with&amp;nbsp;/&amp;nbsp;extends] the composition of nu with a computable function f on the natural numbers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using natural language, choose your nouns wisely and characterize them. Do you care about the ordered pair $(X,\nu_X)$, or really just the map $\nu_X$ (for which $X$ is just the background against which the idea is presented)? What is the &lt;em&gt;role&lt;/em&gt; of the partial map $\nu_X$ in the idea you are communicating? Do you care about the integers $f(n) \in \mathop{\mathrm{dom}}(\nu_X)$ over which you quantify, or really just the domain of the composite function $\nu_X \circ f$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identify the main characters in the synopsis of your play, and their roles: you will have a better chance of transporting the objects and morphisms of your idea faithfully to your interlocutors.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-10-10T12:15:35.987" LastActivityDate="2010-10-10T12:15:35.987" CommentCount="2" />
  <row Id="6403" PostTypeId="2" ParentId="6393" CreationDate="2010-10-10T00:32:59.547" Score="1" ViewCount="0" Body="&lt;p&gt;Cosets arise when you want to model the idea that certain elements of a group are &lt;em&gt;effectively equal&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, rather that looking at one coset at a time it is best to think of all possible cosets of a subgroup. Then you will find that cosets partition a group $G$ into equivalence classes such that two elements of a class differ by an element of the subgroup $H$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example sometimes in number theory we do not want to distinguish two numbers that differ by a multiple of a given number $n$. Say $n=4$. Then let $G$ be the set of integers under addition and $H$ be the set of multiples of $4$. The cosets of $H$ are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H\cdot 0=\{\ldots,-4,0,4,8,\ldots\}$$&#xA;$$H\cdot 1=\{\ldots,-3,1,5,9,\ldots\}$$&#xA;$$H\cdot 2=\{\ldots,-2,2,6,10,\ldots\}$$&#xA;$$H\cdot 3=\{\ldots,-1,3,7,11,\ldots\}$$&#xA;$$H\cdot 4=\{\ldots,0,4,8,12,\ldots\}$$&#xA;$$H\cdot 5=\{\ldots,1,5,9,13,\ldots\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $H\cdot 0 =H\cdot 4$ and $H\cdot 1 =H\cdot 5$. In fact there are only $4$ distinct cosets, each corresponding to one congruence class of integers modulo $4$. If we care only about the remainder of an integer after division by $4$ then all the elements in a coset are equivalent and we can think of them as a single entity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take another example, let $G$ be the group possible rotations of a point on a unit circle and we care only about where a point ends up on the circle after the rotation. Then let $H$ be rotations that are a multiple of $2\pi$. The cosets of $H$ will now be rotations which differ by multiples of $2\pi$ and which therefore have the same effect on the final position of the point rotated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, two elements $g_1$ and $g_2$ of $G$ are defined to be equivalent ($g_1 \equiv g_2$) if $g_1 \cdot g_2^{-1} \in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove the following&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For all $g$ in $G$, $g \equiv g$ since the identity element belongs to $H$ as it is a subgroup.&lt;/li&gt;&#xA;&lt;li&gt;If $g_1 \equiv g_2$ the $g_2 \equiv g_1$ since if an element belongs to $H$ then so does its inverse.&lt;/li&gt;&#xA;&lt;li&gt;If $g_1 \equiv g_2$ and $g_2 \equiv g_3$ then $g_1 \equiv g_3$ since $H$ is closed under multiplication.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This show that $\equiv$ is a genuine equivalence relation. We define the equivalence class of an element $x$ of $G$ as&#xA;$$[x]=\{y \in G\mid x \equiv y\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$x \in [x]$&lt;/li&gt;&#xA;&lt;li&gt;For any $x,y \in G$, either $[x] \cap [y] =\emptyset$ or $[x]=[y]$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So the equivalence classes are either equal or disjoint and they cover $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, back to cosets. The equivalence classes we have defined above are the same as the right cosets. If $y \equiv x$ then $yx^{-1}=h$ for some $h \in H$ or $y=hx$. So $[x]=Hx$. If we had instead defined our equivalence relation by the condition $x^{-1}y \in H$ then we would have got the left cosets.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-10T00:32:59.547" />
  <row Id="6404" PostTypeId="2" ParentId="6383" CreationDate="2010-10-10T00:53:05.037" Score="3" ViewCount="0" Body="&lt;p&gt;A solution which doesn't involve quite so large numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;without loss of generality, say Alice gets keys 1 and 2. (Renumber the keys to make this the case.) Then there are ${20 \choose 2} = (20\times 19)/2 = 190$ possible pairs of keys that Bob can get. Of these, ${18 \choose 2} = (18 \times 17)/2 = 153$ include just keys chosen from ${ 3, \ldots, 20 }$. So the probability that Alice and Bob can't communicate is $153/190$, and the probability that they can is one minus this, or $37/190$.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-10T00:53:05.037" />
  <row Id="6405" PostTypeId="1" CreationDate="2010-10-10T01:39:45.623" Score="-1" ViewCount="183" Body="&lt;pre&gt;&lt;code&gt;2x-3y = 6&#xA;&#xA;y = -3&#xA;&#xA;Find X.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;OK so -3 * -3 = 9 so upon completing ... I get: x = 7.5&#xA;So then 2(7.5)-3(-3) = 24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am WAY off.. what am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have: 2x - 3(-3) = 6&#xA;First I do -3 * -3 and I get +9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Now I have 2x + 9 = 6&#xA;Then I minus 9 from 9 and 6&#xA;So now I have 2x = -3&#xA;Then I do 2 divided by 2 and -3 and i get:&#xA;x = - 0.67&lt;/p&gt;&#xA;" OwnerUserId="2294" LastEditorUserId="2294" LastEditorDisplayName="" LastEditDate="2010-10-10T02:11:11.057" LastActivityDate="2010-10-10T11:34:18.407" Title="Find X in equation.. Simple enough right?" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="6406" PostTypeId="1" CreationDate="2010-10-10T02:08:18.003" Score="4" ViewCount="100" Body="&lt;p&gt;Suppose we have two schemes $X, Y$ and a map $f: X\to Y$. Then we know that $Hom_X(f^*\mathcal{G}, \mathcal{F})\simeq Hom_Y(\mathcal{G}, f_*\mathcal{F})$, where $\mathcal{F}$ is an $\mathcal{O}_X$-module and $\mathcal{G}$ an $\mathcal{O}_Y$-module (and the Homs are in the category of $\mathcal{O}_X$-modules etc). This gives a natural map $f^* f_* \mathcal{F}\to \mathcal{F}$, just by setting $\mathcal{G}=f_* \mathcal{F}$ and looking at where the identity map goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any well-known conditions on the map or sheaves that give this is an isomorphism? For instance, I was looking through a book and saw that the map is surjective if $\mathcal{F}$ is a very ample invertible sheaf (and maybe some more hypothesis on the map and $X$ and $Y$ were assumed as well).&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-10T02:08:18.003" Title="Pullback and Pushforward Isomorphism of Sheaves" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="6407" PostTypeId="1" AcceptedAnswerId="6408" CreationDate="2010-10-10T02:18:25.580" Score="3" ViewCount="167" Body="&lt;p&gt;Suppose you would like to find the two tangent lines that support two given disks in the plane&#xA;to the same side.  Parameterizing the circles using $( \cos \theta, \sin \theta )$,&#xA;I find that ultimately I am computing the roots of a 4th-degree polynomial in $x = \cos \theta$&#xA;(two of whose roots are imaginary, and two real).&#xA;But I may not have formalized this in the most perspicacious manner.&#xA;Rather than detail my approach, let me just pose the general question of whether&#xA;there is a way to compute the tangent lines without effectively finding the roots of&#xA;a 4th-degree polynomial. With two solutions one might hope for a 2nd-degree polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum&lt;/b&gt;.&#xA;I should have made clear that what I need are the coordinates of the two points of tangency for&#xA;two circles arbitrarily placed in the plane.  Nevertheless, both Ross's and Moron's solutions&#xA;can answer this version of the question as well.  With appropriate translation, scaling, and rotation, I can place the two circles so that one is a unit-radius circle centered on the origin,&#xA;and the other has $r \ge 1$ with its center on the $+x$ axis.  Then I am in the situation Moron drew.&#xA;Knowing the side lengths of the triangles yields $\sin \theta$ where $\theta$ is the&#xA;angle of the triangle on the $x$-axis, and from that I can compute the coordinates&#xA;of the points of tangencies&#xA;in terms of $\sin \theta$ and $\cos \theta$.  Much simpler than my brute-force calculation.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-10T18:41:14.210" LastActivityDate="2010-10-11T19:53:49.743" Title="Tangent to two disks: Roots of a 4th-degree polynomial?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="6408" PostTypeId="2" ParentId="6407" CreationDate="2010-10-10T02:59:52.143" Score="2" ViewCount="0" Body="&lt;p&gt;I hope I understood the question correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following (sorry for the crude image):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XtwA4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using similar triangles we get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{D'+D}{D'} = \frac{R}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily solve for $D'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r,R$ are the radii of the circles and and $D$ is the distance between the centres of the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D'$ is the distance between the common point of the tangents to the center of the smaller circle.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-10T02:59:52.143" CommentCount="2" />
  <row Id="6409" PostTypeId="2" ParentId="6407" CreationDate="2010-10-10T03:00:09.543" Score="2" ViewCount="0" Body="&lt;p&gt;In the wikipedia article&lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_lines_to_circles&quot; rel=&quot;nofollow&quot;&gt; tangent lines to circles&lt;/a&gt; there is a construction that appears to be second order.  You construct the tangents from the center of the smaller circle to a circle, centered on the center of the larger, with radius the difference of the radii.  Your tangents are parallel to these, offset by the smaller radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the broken link-an extra / got in the way&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-11T19:53:49.743" LastActivityDate="2010-10-11T19:53:49.743" CommentCount="1" />
  <row Id="6410" PostTypeId="1" CreationDate="2010-10-10T03:27:52.747" Score="1" ViewCount="165" Body="&lt;p&gt;I know that a universal quantifier can be distributed over conjunction and not disjunction, but I'm having a hard time wrap my head around it. Why is this the case? Is there an example of a statement that would demonstrate this principle?&lt;/p&gt;&#xA;" OwnerUserId="2403" LastActivityDate="2010-10-11T01:16:04.503" Title="Distribution of Universal Quantifiers" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6411" PostTypeId="1" AcceptedAnswerId="6455" CreationDate="2010-10-10T04:06:38.390" Score="1" ViewCount="126" Body="&lt;p&gt;Let $(X,||\cdot||_X)$, $(Y,||\cdot||_Y)$ be a pair of normed linear spaces, and $(X \times Y, ||\cdot||_{X \times Y})$ the induced product space and norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(x,y)$ is an element in $X \times Y$, is it true that $||(x,y)||_{X \times Y} \lt \delta$ implies that $||x||_{X \times Y} \lt \delta$, where $(||x||_{X\times Y}=||(x,0)||_{X\times Y}$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been true with every product norm that I have encountered, but I am not sure if it is true in general and/or if there is an easy way to show it. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-10T19:50:45.393" LastActivityDate="2010-10-10T20:05:49.500" Title="Relation between the norms on X and Y and the induced product norm on X x Y" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6412" PostTypeId="1" AcceptedAnswerId="6425" CreationDate="2010-10-10T04:25:00.083" Score="2" ViewCount="98" Body="&lt;p&gt;I'm guessing that this is a standard enough problem, but can't seem to find an intuitive solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an undirected weighted graph $G = (V,E,f)$ with vertices $V$, edges $E$ and $f$ the weighting function for an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set $\mathbf{I} = \{ I_1, ..., I_n \}$ such that for $I \in \mathbf{I}$, $I \subset V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to perform a minimum-cut type operation on $G$ to produce $G'$ such that no pair of vertices in a set $I \in \mathbf{I}$ are reachable from each other, and such that the cut minimises the summation of edge weights removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can tell, I'm not a maths wizard, but I need to implement this for small graphs. (Preferably in the next two hours :/). Should I be looking at something like linear programming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers would be greatly appreciated. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Not a homework problem (unfortunately)... if I had had the type of education where this was a homework problem, I wouldn't need to ask now. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background is that I have a set of equivalence relations which hold between a set of things. I can perform the transitive closure of these equivalence relations to determine equivalence classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these relations are fallible. Afterwards, I can look at an equivalence class $\{A, B, C, D, E\}$ and say hey, crap!: thing $A$ and thing $B$ are different, and thing $B$ and $C$ are different. I need to break up this equivalence class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I look at the original non-transitive graph of equivalences. I want to do a minimum cut, but generally the number of vertices in the graph are small, and ties frequently occur. I want to avoid trivial decisions on what to cut. Turns out I can weight the original equivalence relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the original non-transitive weighted graph of equivalences, and the set of sets of things I know to be different, I want to perform some minimal-cut of the graph, and then reapply transitive closure over the sub-graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated for a sleep-deprived Java hacker (even if you hate Java hackers).&lt;/p&gt;&#xA;" OwnerUserId="2331" LastEditorUserId="2331" LastEditorDisplayName="" LastEditDate="2010-10-10T04:51:32.270" LastActivityDate="2010-10-12T14:59:13.333" Title="Minimum cuts for weighted graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6413" PostTypeId="2" ParentId="6411" CreationDate="2010-10-10T04:29:32.173" Score="0" ViewCount="0" Body="&lt;p&gt;Your question doesn't show up well, but the wikipedia entry &lt;a href=&quot;http://en.wikipedia.org/wiki/Normed_vector_space&quot; rel=&quot;nofollow&quot;&gt;Normed vector space&lt;/a&gt; claims that if you have the product of a finite number of normed vector spaces, the sum of the norms in the individual spaces is a norm in the product.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-10T04:29:32.173" CommentCount="5" />
  <row Id="6414" PostTypeId="1" CreationDate="2010-10-10T04:47:06.387" Score="5" ViewCount="245" Body="&lt;p&gt;...subject to the conditions that (i) the projection be smooth and that (ii) smooth sections correspond to smooth vector fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This homework problem is really bugging the hell out of me.  Of course it can be checked locally, so we'll look at an open neighborhood $U\in \mathbb{R}^n$ and check that $TU\cong U\times \mathbb{R}^n$ (where the isomorphism is as topological spaces with sheaves of functions that we declare to be smooth).  On the one hand, it's easy to see that $U \times \mathbb{R}^n$ satisfies these two conditions, since the projection is literally just a projection (in coordinates) and the correspondence is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X$ a vector field --&gt; $s = (x_1,\ldots, x_n, X(\partial/\partial x_1), \ldots, X(\partial/\partial x_n))$&lt;/li&gt;&#xA;&lt;li&gt;$s=(s_1,\ldots, s_n)$ a smooth section --&gt; $X = \sum_i s_i\cdot\partial/\partial x_i $.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I'm having a lot of trouble showing that the obvious sheaf (which I'll denote $O_{TU}$) is the unique one with properties (i) and (ii).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose $\mathcal{F}$ is another sheaf of functions on $U\times \mathbb{R}^n$.  Suppose that $f \in \mathcal{F}(V) \backslash \mathcal{O}_{TU}$ for some open set $V\subseteq U$.  There really doesn't seem to be any way to show that this violates property (i), because what it really means is that for any $g\in O_U$, $g \circ \pi \in O_{TU}$, and the hypothesis of this statement begins with a function on $U$, not $TU$.  On the other hand, I don't see any way to make a vector field on $U$ out of $f$, and (according to &lt;a href=&quot;http://mathoverflow.net/questions/37423/is-there-an-easy-way-to-describe-the-sheaf-of-smooth-functions-on-a-product-manif&quot; rel=&quot;nofollow&quot;&gt;this question&lt;/a&gt;) it feels likely that I can't necessarily get a section that detects the failure of $f$ to be $O_{TU}$-smooth.  Nor am I having much luck deriving a contradiction from $f \in O_{TU}(V) \backslash \mathcal{F}(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd welcome any suggestions as to how I should proceed.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-12-18T22:44:25.993" Title="Why is the manifold structure on the tangent bundle unique?" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" />
  <row Id="6415" PostTypeId="1" CreationDate="2010-10-10T04:58:32.847" Score="0" ViewCount="162" Body="&lt;p&gt;first question: Let a, b, c be distinct primes larger than 3. x = abc. Show that if p is a positive integer and p^2 ≡ 9 (mod x), then p ≡ +- 3 (mod a), p ≡ +- 3 (mod b) and p ≡ +- 3 (mod c).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;second question: Using the fact that 1001 = (7)(11)(13) find the eight solutions to y^2 ≡ 9 (mod 1001) in the set { 1, 2, 3 ... 1000 }&lt;/p&gt;&#xA;" OwnerUserId="2406" LastActivityDate="2010-10-10T21:12:41.350" Title="Two little number theory questions" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6416" PostTypeId="2" ParentId="6410" CreationDate="2010-10-10T05:18:57.067" Score="-1" ViewCount="0" Body="&lt;p&gt;Matt E, in his comment, gives two statements. He shows that a universal quantifier cannot be distributed over disjunction, because if it could, a false statement would follow logically from a true one. The statement &quot;A universal quantifier can be distributed over disjunction&quot; is itself a universal statement, and requires only one counterexample to &lt;strong&gt;disprove&lt;/strong&gt; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A universal quantifier can be distributed over conjunction&quot; is also a universal statement. It cannot be &lt;strong&gt;proved&lt;/strong&gt; by giving examples. There might be some counterexample we didn't think of. So how do we know it's true? -- Good question!&lt;/p&gt;&#xA;" OwnerUserId="1813" LastActivityDate="2010-10-10T05:18:57.067" />
  <row Id="6417" PostTypeId="2" ParentId="6405" CreationDate="2010-10-10T05:24:56.893" Score="1" ViewCount="0" Body="&lt;p&gt;In your first version, getting $x=7.5$, I suspect that when you substituted -3 for $y$, you wrote $2x-9=6$ (which gives $2x=15$ so that $x=7.5$) instead of the $2x+9=6$ (which is correct) that you have in the second version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said in the second version, $2x+9=6$ does give $2x=-3$; at this point, you need to undo the multiplication-by-2 on the left side in order to have just x, so you are dividing both sides of the equation by 2.  This should give that $x$ is the result of -3 divided by 2 (whereas the $x=-0.67$ is an approximation of 2 divided by -3).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-10T05:24:56.893" />
  <row Id="6418" PostTypeId="2" ParentId="6364" CreationDate="2010-10-10T07:04:15.037" Score="1" ViewCount="0" Body="&lt;p&gt;Many thanks for the answers, all three are interesting in their own right (I'm not registered so can't upvote nor comment, apologies).&lt;/p&gt;&#xA;" OwnerUserId="2407" LastActivityDate="2010-10-10T07:04:15.037" />
  <row Id="6419" PostTypeId="2" ParentId="6415" CreationDate="2010-10-10T07:12:34.167" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\rm\ \ a|x|(p-3)(p+3)\ \Rightarrow \ a\:|\:p-3\ \: or\ \: a\:|\:p+3\:,\:$ since $\rm\:a\:$ is prime. Combine this with the Chinese remainder theorem to solve the second question. See also my &lt;a href=&quot;http://math.stackexchange.com/questions/6296&quot;&gt;answer here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-10T07:12:34.167" />
  <row Id="6421" PostTypeId="1" AcceptedAnswerId="6422" CreationDate="2010-10-10T09:06:47.943" Score="3" ViewCount="122" Body="&lt;p&gt;As I'm trying to improve my maths I've now reached the subject of differentiation of more general exponential functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already studied the so called &quot;chain rule&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my book tells me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=a^x \rightarrow y^{\prime}=a^x\ln(a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there's the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = 10^{-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the above rule, this would lead me to believe that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x) = 10^{-x} \ln(10)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the book tells me it's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{\prime}(x) = -10^{-x} \ln(10)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why's that? They say they also used the chain rule, but if I use the chain rule I get something totally different:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = 10^{-x}$$&#xA;$$f^{\prime}(x) = -x(10)^{-x-1}\cdot(-1*1)$$&#xA;$$f^{\prime}(x) = x(10)^{-x-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong in all these cases?&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-10T17:49:04.870" LastActivityDate="2010-10-10T22:15:16.057" Title="How does differentiation of &quot;more general exponential functions&quot;  work" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6422" PostTypeId="2" ParentId="6421" CreationDate="2010-10-10T09:28:26.897" Score="5" ViewCount="0" Body="&lt;p&gt;Let $N: \mathbb{R} \to \mathbb{R}$ be the negation function, i.e. $N(x) = -x$ and let $g(x) = 10^x$. Then it should be clear that your $f = g \circ N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x) = (g \circ N)'(x) = g'(N(x))N'(x) = \log 10 \cdot 10^{-x} \cdot (-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Learn to use $\LaTeX$ instead of code. It takes up less space and looks prettier.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-10T09:28:26.897" CommentCount="4" />
  <row Id="6423" PostTypeId="2" ParentId="6405" CreationDate="2010-10-10T11:34:18.407" Score="1" ViewCount="0" Body="&lt;p&gt;2x-3y = 6; y = -3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x - 3(-3) = 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x + 9 = 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x = -3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = -(3/2)&lt;/p&gt;&#xA;" OwnerUserId="2409" LastActivityDate="2010-10-10T11:34:18.407" />
  <row Id="6424" PostTypeId="1" CreationDate="2010-10-10T14:10:30.063" Score="2" ViewCount="86" Body="&lt;p&gt;The space of connection has no measure right?&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-10T14:10:30.063" Title="How does physicist compute path integral in Chern-Simons theory" Tags="&lt;topology&gt;&lt;physics&gt;" FavoriteCount="1" />
  <row Id="6425" PostTypeId="2" ParentId="6412" CreationDate="2010-10-10T14:32:05.887" Score="1" ViewCount="0" Body="&lt;p&gt;So you want to find a minimum cost cut so that for each $I \in \mathbf{I}$, $I$ is 'cut' by the set of edges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general this is NP-Hard, even in the case when each $|I| = 2$ (i.e each set has exactly two elements). (I believe this is called as the multi-cut problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\mathbf{I} = \{ \{s, t\}\}$, this becomes the max-flow min-cut problem which is solvable in polynomial time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-10T14:32:05.887" CommentCount="1" />
  <row Id="6426" PostTypeId="2" ParentId="6407" CreationDate="2010-10-10T15:11:31.620" Score="2" ViewCount="0" Body="&lt;p&gt;The degree 4 polynomial appears because the circles have 4, not 2, tangent lines algebraically: two internal and two external.  In the question, the phrase &quot;tangent lines that &lt;em&gt;support&lt;/em&gt;&quot; singles out the external tangents, which are lines of support, but this is not a distinction that can be made algebraically given a coordinatization of the problem.  The subtlety is that the equations of the circles contain only $r^2$ and $R^2$ while a choice of which pair of tangents lines counts as &quot;internal&quot; and which pair is &quot;external&quot; requires a choice of sign for $r/R$ to determine the relative orientation of the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal and external pairs of tangent lines to two circles can be constructed by ruler and compass, using one intersection of circles in each case, which in coordinates is solution of a quadratic equation.  The diagram posted in the other answer, with $(R-r)$, is the same as the one for the Euclidean construction.  The quartic with roots corresponding to the tangent lines is a product of two quadratics $Q_{int}(x)Q_{ext}(x)$ for the internal and external construction problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the 4 tangent lines the difference between the internal and external pairs can be seen algebraically, as choices of sign for different square roots.  This naturally partitions the set of 4 roots into two pairs.  However, a further labelling of the pairs as &quot;internal&quot; and &quot;external&quot; requires a further choice beyond a coordinatization of the plane.   This is because the coefficient field of the Cartesian equations defining the tangent lines is $k(d_x,d_y,R^2,r^2)$ where $d_x$ and $d_y$ are the $x,y$ distance between centers and $k$ is the underlying field of coordinates, but the selection of which pair of tangents is external relies on a value of $r/R$ or $(R-r)^2$ and this requires some sign choices for the radii, or equivalently, an extension of the coefficient field by some square roots. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-10T19:27:13.790" LastActivityDate="2010-10-10T19:27:13.790" />
  <row Id="6427" PostTypeId="1" AcceptedAnswerId="6429" CreationDate="2010-10-10T15:22:08.143" Score="2" ViewCount="88" Body="&lt;p&gt;Let's say I have a bag of coins, which contains 1 quarter, 2 dimes, 3 nickles and 4 pennies.  If I were to randomly pull out 3 coins, on average, how much money would I get?&lt;/p&gt;&#xA;" OwnerUserId="2410" LastActivityDate="2010-10-10T17:51:32.187" Title="Average amount of money received from pulling three random coins" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;average&gt;" AnswerCount="3" CommentCount="20" />
  <row Id="6428" PostTypeId="1" CreationDate="2010-10-10T15:34:36.010" Score="3" ViewCount="96" Body="&lt;p&gt;What I currently meet are all fiber bundles.&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-11-09T17:59:53.627" Title="Is there anywhere we use a fibration which is not a fiber bundle" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6429" PostTypeId="2" ParentId="6427" CreationDate="2010-10-10T15:53:53.970" Score="5" ViewCount="0" Body="&lt;p&gt;Each coin is pulled 30% of the time.  So on average you will get 30% of 64 cents=19.2 cents&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-10T15:53:53.970" CommentCount="2" />
  <row Id="6430" PostTypeId="1" AcceptedAnswerId="6434" CreationDate="2010-10-10T16:01:22.007" Score="0" ViewCount="87" Body="&lt;p&gt;The vector space $V(n,d)$ has dimension $n$ and the vector coordinates are from set &#xA;$\Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d:=|\Sigma|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many vectors can the vector space contain? Why?&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="2411" LastEditorDisplayName="" LastEditDate="2010-10-10T16:14:12.500" LastActivityDate="2010-10-10T17:01:28.767" Title="Maximal number of vectors in vector space" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6431" PostTypeId="1" AcceptedAnswerId="6432" CreationDate="2010-10-10T16:11:14.810" Score="2" ViewCount="132" Body="&lt;p&gt;While searching a question about fibre bundles, which was asked here, i got directed to Vector bundles. I noticed this word &quot;Hairy Ball&quot; which sounded eccentric and made a search at Wikipedia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the hairy ball theorem related to this statement: You can't comb the hair on a coconut.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-10T17:02:03.897" Title="Hairy Ball theorem and its applications." Tags="&lt;soft-question&gt;&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="6432" PostTypeId="2" ParentId="6431" CreationDate="2010-10-10T16:30:25.430" Score="3" ViewCount="0" Body="&lt;p&gt;I think the picture at the top right of the wikipedia page does a pretty good job at visualization of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that if you think of each hair on a coconut as a vector with base at a point on the sphere, then we have a notion of what it means for these to change continuously. So combing the hairs is essentially the analogy for making all the hairs lie tangent to the sphere and for them to vary continuously. This cannot be done, since if you have a continuous vector field on the sphere, it must be zero somewhere. The way the picture shows this is that you get hairs that point outward or &quot;normal&quot; to the sphere (which I think is not a really good way to think about the theorem) and hence have not been combed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that you can't just comb that one part down, or else it won't be continuous. There will be some &quot;overlapping&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-10T16:30:25.430" />
  <row Id="6433" PostTypeId="2" ParentId="6430" CreationDate="2010-10-10T16:39:50.107" Score="1" ViewCount="0" Body="&lt;p&gt;V(n,d) contains d&lt;sup&gt;n&lt;/sup&gt; vectors. As you know, each vector in V(n,d) can be described as an n-tuple (v&lt;sub&gt;1&lt;/sub&gt;,&amp;nbsp;...,&amp;nbsp;v&lt;sub&gt;n&lt;/sub&gt;) where each coefficient v&lt;sub&gt;j&lt;/sub&gt; is drawn from the coefficient-set &amp;Sigma;. It is straightforward to show that the number of such tuples is |&amp;Sigma;|&lt;sup&gt;n&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;d&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular: if &amp;Sigma; is a finite field (such as the integers modulo 2), then the vector space has finitely many elements. Otherwise, it has infinitely many elements.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-10T16:39:50.107" />
  <row Id="6434" PostTypeId="2" ParentId="6430" CreationDate="2010-10-10T16:55:04.357" Score="2" ViewCount="0" Body="&lt;p&gt;If $\rm V$ is an $\rm n$-dimensional vector space over a field $\rm F$ then it has cardinality $\rm |V| = |F|^n$ if both $\rm |F|$ and $\rm n$ are finite; otherwise $\rm |V| = n\:|F| = max(n,|F|)\:$, as follows from basic properties of cardinal arithmetic. For example, this implies that the dimension of $\mathbb R$ over $\mathbb Q\:$ is $\rm |\mathbb R| &gt; |\mathbb Q| = |\mathbb N|\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-10T17:01:28.767" LastActivityDate="2010-10-10T17:01:28.767" />
  <row Id="6435" PostTypeId="2" ParentId="6431" CreationDate="2010-10-10T17:02:03.897" Score="3" ViewCount="0" Body="&lt;p&gt;More of a comment than an answer, but it didn't fit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the hairy ball theorem works on any even-dimensional sphere.  However, odd-dimensional spheres &lt;em&gt;do&lt;/em&gt; admit nonvanishing vector fields: if you use the usual embedding of $S^{2k-1}$ in $\mathbb{R}^{2k}$, then at the point $(x_1,y_1,\ldots,x_k,y_k)$ you can put the (nonzero) tangent vector $(y_1,-x_1,\ldots,y_k,-x_k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural generalization is: How many linearly independent tangent vector fields can exist on $S^{n-1}$?  As it turns out, the answer only depends on how many factors of 2 there are in $n$.  So e.g. $S^3$ and $S^{35}$ both admit exactly 3 linearly independent tangent vector fields, because $3+1=2^2$ and $35+1=2^2\cdot 9$.  This is crazy!&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-10-10T17:02:03.897" CommentCount="2" />
  <row Id="6436" PostTypeId="2" ParentId="6428" CreationDate="2010-10-10T17:05:05.237" Score="0" ViewCount="0" Body="&lt;p&gt;I think that most examples of fibrations that you can think of off the top of your head are going to be fiber bundles (although this is probably only because of my limited experience).  On the other hand, it's much easier to just prove that things are fibrations, because the definition is much weaker.  As far as homotopy theory is concerned, this is going to be the same thing.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-10-10T17:05:05.237" />
  <row Id="6437" PostTypeId="1" CreationDate="2010-10-10T17:22:56.677" Score="1" ViewCount="225" Body="&lt;p&gt;Five microcomputers are to be connected to three printers.How many connections are&#xA;necessary between computers and printers in order to ensure that whenever any three&#xA;computers require a printer the printers are available?&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-09T21:14:25.380" LastActivityDate="2010-11-09T21:14:25.380" Title="Pigeon Hole Principle" Tags="&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6438" PostTypeId="2" ParentId="6365" CreationDate="2010-10-10T17:24:51.910" Score="8" ViewCount="0" Body="&lt;p&gt;If you want, you can actually calculate what it converges to by appealing to complex numbers.  The sum is the imaginary part of &#xA;$$&#xA;\sum \frac{e^{nix}}{2^n} = \sum \left(\frac{e^{ix}}{2}\right)^n = \frac{2}{2-e^{ix}},&#xA;$$&#xA;so your sum converges to $\frac{2 \sin x}{5 - 4\cos x}$.&lt;/p&gt;&#xA;" OwnerUserId="848" LastEditorUserId="848" LastEditorDisplayName="" LastEditDate="2010-10-10T17:55:56.297" LastActivityDate="2010-10-10T17:55:56.297" />
  <row Id="6439" PostTypeId="2" ParentId="6380" CreationDate="2010-10-10T17:25:49.837" Score="2" ViewCount="0" Body="&lt;p&gt;I think the machinery of obstruction theory deals with the special case where the spaces are  skeleta of a CW-complex.  Here's the setup (I'm basically just copying from pp. 6-7 of Mosher &amp;amp; Tangora here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ be simply-connected for simplicity.  First, let $B$ be a complex and $A$ be a subcomplex.  Let $f:A\cup B^n\rightarrow Y$.  Then we get an &lt;em&gt;obstruction cochain&lt;/em&gt; $c(f)\in C^{n+1}(B,A;\pi_n(Y))$ (i.e. a function on relative $(n+1)$-cells with values in $\pi_n(Y)$).  Similarly, let $K$ be a complex; then for any two maps $f,g:K\rightarrow Y$ that agree on $K^{n-1}$, we similarly get a &lt;em&gt;difference cochain&lt;/em&gt; $d(f,g)\in C^n(K;\pi_n(Y))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the two results.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Theorem: There is a map $g:A\cup B^{n+1}\rightarrow Y$ agreeing with $f$ on $A\cup B^{n-1}$ iff $[c(f)]=0 \in H^{n+1}(B,A;\pi_n(Y))$.&lt;/li&gt;&#xA;&lt;li&gt;Theorem: The restrictions of $f$ and $g$ to $X^n$ are homotopic rel $K^{n-1}$ iff $d(f,g)=0 \in C^n(K;\pi_n(Y))$.  They are homotopic rel $K^{n-2}$ iff $[d(f,g)]=0 \in H^n(K;\pi_n(Y))$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-10-10T17:25:49.837" CommentCount="2" />
  <row Id="6440" PostTypeId="2" ParentId="6427" CreationDate="2010-10-10T17:26:09.710" Score="5" ViewCount="0" Body="&lt;p&gt;Ross Millikan's answer is entirely correct, but perhaps it deserves slightly more elaboration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected amount of money you receive is the sum of the expected amounts of money that each coin gives you, which is the value of the coin times the probability that you draw it.  (Linearity of expectation!)  The probability that you draw each coin is the same: it is one minus the probability that you do not draw it, or $1 - \frac{ {9 \choose 3} }{ {10 \choose 3} } = \frac{3}{10}$.  So the expected amount of money you receive is $\frac{3}{10}$ of the total value of the coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I am basically treating coins with the same value as distinguishable (pretend they have the same value but look different; it can't affect the answer to the question).  The question of distinguishability is a red herring.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-10T17:26:09.710" />
  <row Id="6441" PostTypeId="1" AcceptedAnswerId="6443" CreationDate="2010-10-10T17:51:22.747" Score="1" ViewCount="102" Body="&lt;p&gt;&lt;strong&gt;What number of positive integers satisfying the given inequality :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              C(n + 1,n – 2) – C(n + 1,n – 1) &amp;lt;= 100 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What I did so far :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C(n + 1,n – 2) – C(n + 1,n – 1) &amp;lt;= 100 &#xA;=&amp;gt; C(n + 1,3)  – C(n + 1,2) &amp;lt;= 100 &#xA;=&amp;gt; n(n+1)(n-4) &amp;lt;= 600&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now to derive after this ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This is not my homework, It originated my test paper without proper explanation of answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:The time constraint is only a mint so please suggest shortest possible method.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-10T18:58:03.017" LastActivityDate="2010-10-10T19:49:30.447" Title="What number of positive integers satisfying the given inequality ?" Tags="&lt;combinatorics&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6442" PostTypeId="2" ParentId="6427" CreationDate="2010-10-10T17:51:32.187" Score="4" ViewCount="0" Body="&lt;p&gt;Ross Millikan has already given a very elegant solution, but if you're still in doubt, this problem is small enough that you can easily verify it by letting a computer enumerate all the possibilities. Here's a Python program to do it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;coins = [25, 10, 10, 5, 5, 5, 1, 1, 1, 1]&#xA;total = count = 0&#xA;for k in range(10):&#xA;    for l in range(10):&#xA;        for m in range(10):&#xA;            if k!=l and k!=m and l!=m:&#xA;                total += coins[k] + coins[l] + coins[m]&#xA;                count += 1&#xA;print &quot;Got&quot;, total, &quot;cents in&quot;, count, &quot;draws; on average&quot;, total*1.0/count, &quot;cents.&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The answer comes instantly: &quot;Got 13824 cents in 720 draws; on average 19.2 cents.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-10T17:51:32.187" CommentCount="3" />
  <row Id="6443" PostTypeId="2" ParentId="6441" CreationDate="2010-10-10T17:56:12.567" Score="2" ViewCount="0" Body="&lt;p&gt;To solve $n(n+1)(n-4)\le 600$ algebraically, expand the left side and subtract 600 from both sides, so that you're comparing a polynomial to 0, find the zeros of the polynomial&amp;mdash;which divide the number line into regions&amp;mdash;and test a value in each region to see which regions satisfy the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simpler, especially since you're only looking for positive integer solutions, would be to graph $y=x(x+1)(x-4)-600$ and look for where it is below the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;: Under very tight time constraints and without a calculator/computer, I'd have approached the original problem differently.  The original statement  ${n+1\choose n–2}–{n+1\choose n–1}\le 100$  (as well as the restatement ${n+1\choose 3}–{n+1\choose 2}\le 100$) can be interpreted as a statement about the (n+1)st row of Pascal's triangle: that the difference between the numbers 2 and 3 positions in from the edge is less than or equal to 100.  It should be clear that the difference between adjacent entries in a row of Pascal's triangle is increasing, so it is sufficient to write successive rows until the statement fails:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/DTQS9.png&quot; alt=&quot;Pascal's triangle, through the 11th row&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bold/darkest numbers are the relevant ones:&#xA;n=1: $0-1\le 100$;&#xA;n=2: $1-3\le 100$;&#xA;n=3: $4-6\le 100$;&#xA;n=4: $10-10\le 100$;&#xA;n=5: $20-15\le 100$;&#xA;n=6: $35-21\le 100$;&#xA;n=7: $56-28\le 100$;&#xA;n=8: $84-36\le 100$;&#xA;n=9: $120-45\le 100$;&#xA;n=10: $165-55&gt;100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that the lightest numbers, below and to the left of the relevant numbers, are not used at all, so you don't actually need to compute them.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-10T19:49:30.447" LastActivityDate="2010-10-10T19:49:30.447" CommentCount="6" />
  <row Id="6444" PostTypeId="2" ParentId="6441" CreationDate="2010-10-10T18:11:20.713" Score="2" ViewCount="0" Body="&lt;p&gt;$n(n+1)(n-4) \le 600$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is just a matter of trial and error, as $n(n+1)(n-4)$ is increasing for $n \ge 4$. For $ 0 &amp;lt; n &amp;lt; 4$ it is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=10$, the value is $660 &gt; 600$ and for $n=9$ the value is $450$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number is $9$ got by $1 \le n \le 9$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-10T18:11:20.713" CommentCount="1" />
  <row Id="6445" PostTypeId="1" AcceptedAnswerId="6449" CreationDate="2010-10-10T18:20:44.087" Score="1" ViewCount="110" Body="&lt;p&gt;Given finite-dimensional vector spaces $V,W$, there is an isomorphism $\text{Hom}(V,W) \rightarrow V^* \otimes W$. In particular, any linear map $\phi : V \rightarrow W$ has a tensor expansion $\sum v^*_i \otimes w_i$ where $v^*_i \in V^*, w_i \in W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if one chooses dual bases $\{x_i\}, \{e_i\}$ of $V^*$ and $V$, then $\sum x_i \otimes e_i \in V^* \otimes V$ is a tensor expansion for the identity map on $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to intuitively understand tensor expansions of linear maps?&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-10T19:40:11.677" LastActivityDate="2010-10-10T19:40:11.677" Title="Intuition behind tensor expansions of linear maps" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;intuition&gt;&lt;linear&gt;" AnswerCount="3" />
  <row Id="6446" PostTypeId="2" ParentId="6437" CreationDate="2010-10-10T18:22:37.657" Score="1" ViewCount="0" Body="&lt;p&gt;mcAfee, here's a hint: call P1, P2, P3 the printers and C1, ..., C5 the computers. Suppose P1 is connected to C1, C4, C5; P2 is connected to C2, C4, C5, and P3 is connected to C3, C4, C5. That is 9 connections (and this is not the only way to connect them; you should try others). Check out whether this works, and if it does, try and see whether it's possible to do it with 8 connections. &lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-10T18:22:37.657" />
  <row Id="6447" PostTypeId="1" AcceptedAnswerId="6450" CreationDate="2010-10-10T18:34:23.107" Score="2" ViewCount="90" Body="&lt;p&gt;Let $T_f g = f \cdot g$ where $f, g, f \cdot g$ are in $H^2(\mathbb{D})$ (where $H^2$ is the Hardy space on the open unit disk). Now $T_f$ is a bounded operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to show this by showing that $f \in H^\infty$. So I try to write $f = G_1 h_1$ and $g = G_2 h_2$ where $G_i$ are outer functions and $h_i$ inner functions. So, what I need to do is if $G_1 G_2$ is in $H^2$ for all $G_2$ outer, then $G_1$ is in $H^\infty$. Does someone have a hint how I could obtain this?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-11T00:39:41.807" LastActivityDate="2010-10-11T00:39:41.807" Title="Bounded operator Hardy space." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6448" PostTypeId="2" ParentId="6310" CreationDate="2010-10-10T18:56:32.583" Score="2" ViewCount="0" Body="&lt;p&gt;At the heart this is really trivial, a  consequence of $\rm\ f(x) = x^n\ \Rightarrow\ f{\:'}(x)\ =\ n\: x^{n-1}\ \equiv\ 0\:\ (mod\ n)\:$. First I give the simple proof, then I explain the viewpoint in terms of derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ n\:|\:(a-b)\frac{a^n-b^n}{a-b}\ \Rightarrow\ n = m\:k,\ \ m\:|\:a-b,\ \ k\:\big|\frac{a^n-b^n}{a-b}\ $ by unique factorization. Thus it suffices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to show $\rm\ \ m\:|\:n,\:a-b\ \Rightarrow\ m\:\big|\frac{a^n-b^n}{a-b}\ =\ a^{n-1}+a^{n-2}\:b+\:\cdots\:+a\:b^{n-2}+b^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, $\rm\: mod\ \:m\:,\ \ a\equiv b\ \ \ \Rightarrow\ \ \ \frac{a^n-b^n}{a-b}\ \equiv\ a^{n-1}+\cdots+a^{n-1} \equiv n\: a^{n-1}\equiv 0\ $ via $\rm\ m|n\ \ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prior line is the special case $\rm\ f = x^n,\ x = b\ $ in this polynomial Taylor series approximant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad  \frac{f(x)-f(a)}{x-a} \: \equiv\ f\:'(a)\ \ \ (mod\ \:x-a)\quad$ for $\rm\ f(x)\in \mathbb Z[x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\ \ $ i.e. $\rm\quad\ f(x)\ =\ f(a)  +\: f\:'(a)\ (x-a) \:+\: (x-a)^2\: g(x)\quad$ for some $\rm\ g(x) \in \mathbb Z[x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus this result about numbers is really nothing but a special case of the following well-known result about functions (here polynomials):  a root $\rm\ x = a\ $ of $\rm\ f(x)\ $ has multiplicity $\rm &gt; 1\ \iff\ f\:'(a) \:=\: 0\:.\:$ In fact, like above, many results about numbers are actually specializations of results about functions. Moreover, because functions have richer structure than numbers - e.g. having derivatives available - we can exploit this structure in the function realm before specializing to numbers. A very powerful example of this is Mason's ABC theorem - which has a trivial high-school level proof for polynomials but is a deep conjecture for numbers. It yields as a consequence a trivial proof of FLT for polynomials. The moral is: to prove a result about numbers, try to interpret it as special case of a result about functions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-10T18:56:32.583" />
  <row Id="6449" PostTypeId="2" ParentId="6445" CreationDate="2010-10-10T19:03:28.097" Score="1" ViewCount="0" Body="&lt;p&gt;Ok, let's try to incorporate your question in a more gneral setting: at least one of the vector spaces in the following is finite-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f\colon U\to U'$ and $g\colon V\to V'$ be linear maps. We easily define the &quot;tensor product&quot; of f and g to be the map $f\otimes g$ such that&#xA;$$ (f\otimes g)(u\otimes v)=f(u)\otimes g(v) $$&#xA;for all $u\in U$ and $v\in V$. So, we obtain a linear map&#xA;$$ \lambda\colon\hom(U,U')\otimes \hom(V,V')\cong \hom(V\otimes U, V'\otimes U')$$&#xA;which is easily checked to be an isomorphism, provided at least one of the pairs $(U,U')$, $(V,V')$, $(U,V)$ consists of finite-dimensional vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an intelligent choice of the spaces, the isomorphism $\lambda$ allows us to prove every sort of useful identity relating tensor product of spaces and their duals: Try putting $U'=V'=\mathbb{K}$ (the underlying field), you'll obtain $(V\otimes U)^\star\cong U^\star\otimes V^\star $ (don't worry for the twist). Try putting $U=V'=\mathbb{K}$, you'll obtain... $\lambda_{UV}\colon V\otimes U^\star\cong \hom(U,V)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $f\colon U\to V$ be a linear map. Using bases for $U$ and $V$, we have&#xA;$$f(u_j) = \sum_i f_i^j v_i$$&#xA;for a family $(f_i^j)$ of scalars. Try proving that&#xA;$$  f = \lambda_{UV}\Big(\sum_{ij} f_i^j v_i\otimes u^j\Big) $$&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-10-10T19:03:28.097" CommentCount="1" />
  <row Id="6450" PostTypeId="2" ParentId="6447" CreationDate="2010-10-10T19:05:05.367" Score="3" ViewCount="0" Body="&lt;p&gt;This isn't the same approach you had in mind, but you can show that $T_f$ is bounded using the closed graph theorem and the fact that evaluation at a point in the open disk is bounded on $H^2$.  You can then show that $f$ is in $H^\infty$ by showing that the complex conjugates of elements of its image on the disk are eigenvalues for the adjoint of $T_f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an elaboration on the last sentence.  For each $w\in\mathbb{D}$, define $k_w:\mathbb{D}\to\mathbb{C}$ by $k_w(z)=\frac{1}{1-\overline{w}z}=\sum_{k=0}^\infty \overline{w}^k z^k$.  Each $k_w$ is in $H^\infty$ and thus in $H^2$.  Using the second expression for $k_w$ and the characterization of the inner product on $H^2$ in terms of the $\ell^2$ sequences of Maclaurin coefficients, notice that $\langle g,k_w\rangle=g(w)$ for all $w\in\mathbb{D}$ and all $g\in H^2$.  It then follows that for all $w$ and $z$ in $\mathbb{D}$, &#xA;$$(T_f^*k_w)(z)=\langle T_f^* k_w,k_z\rangle=\overline{\langle T_f k_z,k_w\rangle}=\overline{f(w)k_z(w)}$$ $$=\overline{f(w)}\overline{\langle k_z,k_w\rangle}=\overline{f(w)}\langle k_w,k_z\rangle=\overline{f(w)}k_w(z).$$  Since $z$ was arbitrary, this shows that $T_f^*k_w=\overline{f(w)}k_w$, so $\overline{f(w)}$ is an eigenvalue for $T_f^*$ with eigenvector $k_w$.  Thus, $\|f\|_\infty\leq \|T_f^*\|=\|T_f\|&amp;lt;\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a standard fact about reproducing kernel Hilbert spaces, and only the particular form of the function $k_w$ is special to the Hardy space. The way I have presented this, it might seem that $k_w$ was summoned by magic, but in fact one could rediscover them without too much work.  The important point is that there exist elements of $H^2$ whose corresponding inner product functionals are point evaluations.  These exist by Riesz's lemma using continuity of the point evaluations, which can be shown by other means.  You don't need to know what these elements are for the argument to carry through.  However, if you did want to discover them, then &quot;working backwards&quot; and considering Maclaurin series would lead you to the second expression for $k_w$ given above.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-10T21:12:58.727" LastActivityDate="2010-10-10T21:12:58.727" CommentCount="5" />
  <row Id="6451" PostTypeId="1" CreationDate="2010-10-10T19:18:37.097" Score="4" ViewCount="94" Body="&lt;p&gt;Hello,&#xA;I'm trying to show that the Möbius strip with boundary circle identified to a point is homeomorphic to P^2 (real projective space). I get geometrically why this is so, but, how does one generally construct maps between these spaces to show that they're homeomorphic? &lt;/p&gt;&#xA;" OwnerUserId="2413" LastActivityDate="2010-10-11T21:12:28.947" Title="Möbius strip with edge identified - constructing map?" Tags="&lt;topology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="6452" PostTypeId="2" ParentId="6445" CreationDate="2010-10-10T19:31:06.850" Score="1" ViewCount="0" Body="&lt;p&gt;Not entirely sure what you're looking for so I'll just throw out some thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $V^{*}$ is the set of linear maps from $V$ to $\mathbb{R}$.  Tensoring with $W$ effectively replaces the $\mathbb{R}$ with $W$.  So the star on $V$ represents that the maps are coming from $V$; the lack of one on $W$ represents that you're getting elements of $W$.  Likewise, I'd expect that elements of $\text{Hom}(V,W)$ would covary with a change of basis of $V$ and contravary with one of $W$ (if you're unfamiliar with these terms, here's &lt;a href=&quot;http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bit more rigor, recall that you have a natural pairing between $V$ and $V^{*}$, which is just a map $$\langle -,- \rangle: V\times V^*\rightarrow\mathbb{R}$$ defined by $\langle v,\alpha\rangle=\alpha(v)$.  As you can check, this is invariant under change of basis (hence the &quot;natural&quot;) -- this actually follows from $V$ and $V$ varying oppositely.  This also has the property that if $v_i,\alpha^i$ are dual bases for $V$ and $V^{*}$, then $\langle v_i,\alpha^i\rangle=1$, and $\langle v_i,\alpha^j\rangle=0$ when $i\ne j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now choose bases $v_i,w_j$ for $V$ and $W$ and a dual basis $\alpha^i$ for $V^{*}$.  Given an element $v=\sum b^iv_i$ of $V$ and an element $f=\sum a_i^jv^i\otimes w_j$ of $V^*\otimes W$, we get&#xA;$$f(v)=\sum_{i,j,k}(a_i^j\alpha^i\otimes w_j)(b^kv_k)=\sum_{i,j,k}(a_i^jb^k\langle v_k,\alpha^i\rangle\otimes w_j).$$&#xA;Since the pairing is only nonzero for $k=i$, this reduces to&#xA;$$f(v)=\sum_{i,j}a_i^jb^i\langle v_i,\alpha^i\rangle\otimes w_j=\sum_{i,j}a_i^jb^iw_j$$&#xA;where the tensor product has disappeared because the output of the pairing is just a real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is just a vector in $W$!  And conversely, given a linear map $V\rightarrow W$, we can pick bases and write the map as a matrix $(a_i^j)$, which then transforms to the above tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.math.binghamton.edu/paul/601-S06/AS5.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; very helpful when I was learning this stuff.  You can change the 5 in the URL to other numbers to read all the notes, though I believe they stop at 9.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-10T19:31:06.850" />
  <row Id="6453" PostTypeId="2" ParentId="6445" CreationDate="2010-10-10T19:31:58.333" Score="1" ViewCount="0" Body="&lt;p&gt;Here's a very &quot;low-tech&quot; answer, in terms of coordinates. With respect to the dual bases (I take $\dim V=\dim W=2$ for simplicity), the linear transformation with matrix $$\begin{pmatrix} a&amp;amp;b \\ c&amp;amp;d \end{pmatrix}$$ can be expanded as&#xA;$$a \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 &amp;amp; 0 \end{pmatrix}&#xA;+ b \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 0 &amp;amp; 1 \end{pmatrix}&#xA;+ c \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 1 &amp;amp; 0 \end{pmatrix}&#xA;+ d \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 &amp;amp; 1 \end{pmatrix}.$$&#xA;Here the column vectors are coordinate vectors for elements in $W$, while the row vectors are coordinate vectors for elements in $V^*$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-10T19:31:58.333" />
  <row Id="6454" PostTypeId="2" ParentId="6451" CreationDate="2010-10-10T19:47:33.887" Score="2" ViewCount="0" Body="&lt;p&gt;I would contruct a homeomorphism from $A/\sim$ to the Mobius strip&#xA;where $A$ is the region on the sphere bounded by two antipodal lines&#xA;of latitude (let's think of them as the arctic and antarctic circles)&#xA;and $\sim$ is the equivelance relation identifying antipodal points.&#xA;Now deforming the artic/antartic regions into the north and south poles,&#xA;induces a homeomorphism from the Mobius strip with edge squeezed to a point&#xA;and ${S^2/\sim}\ = P^2$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-10T19:47:33.887" />
  <row Id="6455" PostTypeId="2" ParentId="6411" CreationDate="2010-10-10T19:49:17.530" Score="2" ViewCount="0" Body="&lt;p&gt;No, this need not be true, unless I am missing some assumption implicit in your use of the phrase &quot;&lt;em&gt;the&lt;/em&gt; induced norm&quot; (emphasis added).  Here is an example where $X=Y=\mathbb{R}$.  All norms on $X\times Y=\mathbb{R}^2$ are equivalent.  The norm $\|(x,y)\|^2=2x^2-2xy+y^2=x^2+(x-y)^2$ does not satisfy your property because, for example $\|(x,x)\|=|x|$ while $\|(x,0)\|=\sqrt{2}|x|$.  (This norm actually comes from an inner product on $\mathbb{R}^2$.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following was a result of misreading the question.  I'll leave it here, for now at least, if for no other reason than leaving Arturo's correction comprehensible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that the projection map $(x,y)\mapsto x$ is continuous.  This implies as a special case the following (which turns out to actually be equivalent to continuity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $\epsilon&gt;0$, there is a $\delta&gt;0$ such that for all $x\in X$ and $y\in Y$, $\|(x,y)\|&amp;lt;\delta$ implies $\|x\|&amp;lt;\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the projection is Lipschitz continuous, and $\delta$ can be taken to be $\epsilon$ divided by the Lipschitz constant.  For the reason Qiaochu Yuan gave in a comment on Ross Millikan's answer, this is the best you could hope for.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-10T20:05:49.500" LastActivityDate="2010-10-10T20:05:49.500" CommentCount="4" />
  <row Id="6456" PostTypeId="1" CreationDate="2010-10-10T20:58:11.020" Score="4" ViewCount="55" Body="&lt;p&gt;Given the binary alphabet {a,b}, I'm trying to find the generating function that distinguishes, for all words of fixed length $n$, the count of blocks of a's and the number of a's. Let $x^p$ count the number of blocks of size $p$ and $y^q$ count the number of a's in a word. As a first example, take the word:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;abbaaabaabaabba&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This has nine a's broken up into a two blocks of length one, two blocks on length two and a block of length three. This would contribute to the generating function a term of $y^9(2x + 2x^2 + x^3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a second example, with $n=3$ there are eight words and the term associated with each of them is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;bbb = 1&#xA;bba = xy&#xA;bab = xy &#xA;baa = (xy)^2&#xA;abb = xy&#xA;aba = 2xy^2&#xA;aab = (xy)^2&#xA;aaa = (xy)^3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the term &lt;code&gt;aba = 2xy^2&lt;/code&gt; is not a typo, there are two blocks of length one. With $A$ as the generating function, this would give: $A_3(x,y) = 1 + 3xy + (2x + 2x^2)y^2 + y^3x^3$. I've been able to come up with a special case of this, when we don't care about the counts of the a's (equivalent to setting all $y=1$). This restricted form $B_n(x) = A_n(x,y=1)$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B_n = B_{n-1} + x^n + \sum_{j=0}^{n-2} ( 2^j x^{n-j-1} -1 + B_j ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $B_0 = 1$. Is there such a formula for $A_n$, or even better, a non-recursive closed form of $A_{npq}$?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-10-10T21:28:38.030" LastActivityDate="2010-10-10T21:28:38.030" Title="Generating function of words in a binary alphabet counting blocks and appearances" Tags="&lt;recursive&gt;&lt;generating-functions&gt;" CommentCount="2" />
  <row Id="6457" PostTypeId="2" ParentId="6414" CreationDate="2010-10-10T21:04:34.010" Score="0" ViewCount="0" Body="&lt;p&gt;First, I am pretty rusty on this stuff, so hopefully somebody&#xA;with more confidence will chime in. I started writing this as another&#xA;comment, but it didn't fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I still think what I said makes sense. If&#xA;$\mathcal{E}_{\mathbb{R}^n}$ is the standard differentiable structure&#xA;on $\mathbb{R}^n$, and $\mathcal{C}_M$ is the sheaf of continuous&#xA;functions on a topological manifold $M$, then a subsheaf $\mathcal{F}$&#xA;of $\mathcal{C}_M$ is a differentiable structure on $M$ if (and only if)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) at each point $p\in M$ there exists a neighbourhood $V$, and&#xA;  $x_1,...x_n \in \mathcal{F}(V)$ such that $q \mapsto (x_1, ...,&#xA;  x_n)$ is a homeomorphism onto an open $U \subset \mathbb{R}^n$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $f \in \mathcal{F}(V)$ iff there exists $F \in&#xA;  \mathcal{E}_{\mathbb{R}^n}(U)$, such that $f = F(x_1,...,x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this definition is derived from Werner (&quot;Foundations on&#xA;differentiable manifolds and Lie groups&quot;), but it has paraphrased by&#xA;me and possibly been distorted through the lens of an absurdly long&#xA;time. In particular, that second condition should probably be reworked&#xA;to be more obviously a purely sheaf theoretic statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, a proper definition of a differentiable structure is&#xA;going to resemble something like this (likewise for holomorphic&#xA;structure, or analytic structure or whatever). And the point is, that&#xA;the \emph{definition} will require that your sheaf of differentiable&#xA;functions on $M$ will be locally isomorphic to the \emph{standard}&#xA;sheaf of differentiable functions on $\mathbb{R}^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should apply even if the topological manifold $M$ is&#xA;$\mathbb{R}^n$. This is why I say that you can't solve the problem&#xA;locally: Locally all differentiable structures are isomorphic to the&#xA;standard one on $\mathbb{R}^n$ by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to my second statement, about an &quot;appropriate&quot; differentiable&#xA;structure on $TM$ defining a unique differentiable structure on $M$,&#xA;this is mostly a guess. However, the qualifier &quot;appropriate&quot; is&#xA;important here. In particular, whatever differentiable structure on&#xA;$TM$ you choose, it has to restrict to the standard structure on&#xA;$\mathbb{R}^n$ on the fibres. In this way, I don't think there is any&#xA;choice left once you've defined the restriction on the zero section.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-10-10T21:04:34.010" CommunityOwnedDate="2010-10-11T02:38:01.230" />
  <row Id="6458" PostTypeId="2" ParentId="6415" CreationDate="2010-10-10T21:12:41.350" Score="0" ViewCount="0" Body="&lt;p&gt;okay ive figured out the first part, now ive been told the chinese remainder theorem for the second part but i dont see how it will give me the extra 2 answers since it only allows the set 1-1000&#xA;I assume the first 6 answers are x ≡ +-3(mod y) where y=7,11,13 yet what about the other 2 answers?&lt;/p&gt;&#xA;" OwnerUserId="2417" LastActivityDate="2010-10-10T21:12:41.350" CommentCount="1" />
  <row Id="6459" PostTypeId="1" AcceptedAnswerId="6480" CreationDate="2010-10-10T21:28:26.737" Score="1" ViewCount="109" Body="&lt;p&gt;I need to write a piece of software that analyzes packet switched computer networks and calculates certain characteristics based on packet loss (the application is Voice over IP quality monitoring).  One approach is to use the following 4-state Markov model (as described &lt;a href=&quot;http://www.voiptroubleshooter.com/indepth/burstloss.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CXMxu.jpg&quot; alt=&quot;4-state markov model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State 1 – packet received successfully&lt;br&gt;&#xA;State 2 – packet received within a burst&lt;br&gt;&#xA;State 3 – packet lost within a burst&lt;br&gt;&#xA;State 4 – isolated packet lost within a gap&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State transitions are bound to certain criteria.  This is how it is expected to work for a stream of network packets:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Loss pattern 000001100101010110110000000000000000000000001000000000&#xA;State        111113322323232332331111111111111111111111114111111111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An example algorithm I found stores the state changes in counter variables (e.g. state change 1-&gt;3 increments counter c13).  So far I get it and I end up with these counters: c14, c11, c13, c22, c23, c33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the actual analysis I need to calculate the state transition probabilities from these counters.  This is where the algorithm chokes.  Besides the issue with zero divisions, I don't see the total number of packets incorporated into the probability calculation (this is from the pseudo code of the algorithm definition, p14 is not included because they added c14 to c11):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p11 = c11/(c11 + c13)&#xA;p13 = 1 - p11&#xA;p31 = c31/(c31 + c32 + c33)&#xA;p32 = c32/(c31 + c32 + c33)&#xA;p33 = 1 - p31 - p32&#xA;p22 = c22/(c22 + c23)&#xA;p23 = 1 - p22&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this even correct?  I don't quite get what the author is doing here.  Am I missing something?  What's the correct way to calculate probabilities p14, p11, p13, p22, p23, p33, p31 and p32?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you understand why I asked this here instead of on &lt;a href=&quot;http://stackoverflow.com&quot;&gt;Stackoverflow&lt;/a&gt;: this is basically just about the calculation of the probabilities, the rest was provided for context.&lt;/p&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="2415" LastEditorDisplayName="" LastEditDate="2010-10-10T23:57:19.910" LastActivityDate="2010-10-11T02:13:20.227" Title="Calculating probabilities for a 4-state Markov model" Tags="&lt;probability&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6460" PostTypeId="1" AcceptedAnswerId="6465" CreationDate="2010-10-10T21:32:17.260" Score="5" ViewCount="149" Body="&lt;p&gt;I'm studiying for an exam I have on combinatorics next week and I'm stuck with the solution to this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have a pentagon and you want to color the corners of the pentagon so that no two adjacent corners have the same color. If you have 'q' colors, how many ways of coloring the pentagon are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say that the answer is $q^2*(q-1)^2*(q-2) + q*(q-1)*(q-2)^2*(q-3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend says the answer is $MyAnswer + q*(q-1)*(q-2)*(q-3)*(q-4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is right? (or are we both wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further if you could point out a resource that explains how to calculate this question for other numbers of corners (say an hexagon) we would really appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering to some comments in the answer, rotations of the pentagon should not be counted. What I mean is that if we have 5 colors (a,b,c,d,e) the pentagon with corners {(1,a),(2,b),(3,c),(4,d),(5,e)} is exactly the same as {(1,c),(2,d),(3,e),(4,a),(5,b)}&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2010-10-11T01:40:53.827" LastActivityDate="2010-10-11T01:40:53.827" Title="Calculate how many ways can you paint the corners of a Pentagon" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6461" PostTypeId="1" AcceptedAnswerId="6469" CreationDate="2010-10-10T21:48:05.170" Score="2" ViewCount="67" Body="&lt;p&gt;Let $A$ and $B$ be compact subsets of a metric space. Denote by $\mathcal{P}(A)$ the set of (Borel) probability measures in $A$, and similarly define $\mathcal{P}(A \cup B)$. They are compact metric. Define $C$ as the set of all elements of $\mathcal{P}(A \cup B)$ which assign probability 1 to $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is there an embedding between $C$ and $\mathcal{P}(A)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, when $A$ and $B$ are finite (and assuming without loss they are also disjoint), the answer is yes because one can identify $\mathcal{P}(A)$ with a closed subset of $\mathcal{P}(A \cup B)$. But I am a little lost on how to (try to) show it in general.&lt;/p&gt;&#xA;" OwnerUserId="2418" LastActivityDate="2010-10-11T00:16:28.717" Title="Existence of an embedding" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6462" PostTypeId="2" ParentId="6421" CreationDate="2010-10-10T22:15:16.057" Score="2" ViewCount="0" Body="&lt;p&gt;The biggest problem here is that you seem to be confusing the chain rule with the power rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power rule is as follows: Suppose you have some real number $n$. Then the function $f(x) =x^n$ has derivative $f'(x)=n\cdot x^{n-1}$. Notice that $n$ is fixed, but $x$ is variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain rule is as follows: Given functions $f$ and $g$, the function $h(x)=(f\circ g)(x) = f(g(x))$ has derivative $h'(x)= (f'\circ g)(x)\cdot g'(x)= f'(g(x))\cdot g'(x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the function you have is $f(x)=10^{-x}$. In this case, your &lt;em&gt;exponent&lt;/em&gt; is variable, but $10$, obviously, is fixed. So the power rule does not apply; that is, we cannot simply bring down the exponent, multiply, and subtract one from the exponent, which is part of what you did in your work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given an exponential function $g(x)=a^x$, the derivative is $g'(x)=a^x\cdot\ln{(a)}$. So why does this rule not apply directly to the function $f(x)=10^{-x}$? Well, notice that the exponent in your $f$ is actually itself a function, namely, $-x$. The exponent in the &lt;em&gt;rule&lt;/em&gt; from your book is merely $x$. That's why we have to apply the chain rule; we technically have a function &quot;inside&quot; another function. From here, follow kahen's solution above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I again want to emphasize that the power rule is not applied to your $f$ because $f$ is exponential: it has a fixed base of $10$ and a variable exponent $-x$ (which is also a function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I should note that the power rule and chain rule &lt;em&gt;as I have written them&lt;/em&gt; are neither rigorous nor precise, but I sacrificed details for readability. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-10T22:15:16.057" CommentCount="2" />
  <row Id="6463" PostTypeId="2" ParentId="6460" CreationDate="2010-10-10T22:21:49.817" Score="3" ViewCount="0" Body="&lt;p&gt;You need to specify the problem better.  Do you count rotations and reflections as different?  For example, imagine a regular pentagon drawn on a sheet of paper with a corner up.  Starting from the top corner and going clockwise, maybe we have 0,1,2,3,4 (using numbers instead of colors).  Is this different from 1,2,3,4,0?  How about from 0,4,3,2,1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest situation is if you say these are all different.  Then the naive approach would be to say you have $q$ choices for the top corner, $q-1$ choices for each of the next three corners (as you can't have neighboring corners that match) and $q-2$ choices for the last, giving $q*(q-1)^3*(q-2)$, but this ignores the fact that the first and fourth corners could be the same color, giving $q-1$ choices for the last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to deal with this situation is to separate into classes based on which corners match the top one.  If we designate the top corner 0 and count clockwise, you can match the top color only in no corner, corner 2, or corner 3.  So matching none, we have $q*(q-1)*(q-2)^3 $ as at corner 1 you can use any color but the color at corner 0, but for each other you have two eliminated.  Matching corner 2, we have $q*(q-1)*1*(q-1)*(q-2)$ and matching corner 3 we have $q*(q-1)*(q-1)*1*(q-2)$.  Add them all up and you have your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the answer has to be a fifth degree polynomial.  For large $q$ the constraint of not matching colors won't reduce the count very much.  So if you count the solutions for six different q (I suggest 0 through 5), you can fit a fifth degree polynomial through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to count rotations and reflections as the same there are two possibilites.  One way is to define a standard position and only count those cases in standard position.  In this case this is easy.  You would say that the smallest number has to be at the top, and that corner 2 is less than corner 3.  A pitfall would be to say the smallest is at the top and corner 1 is less than corner 4.  It could be that corner 1 and 4 are the same, and it could be that one of corner 2 and 3 is the same as corner 0.  But counting how many configurations satisfy this constraint may not be easy.  It is again a fifth degree polynomial.  The other case is to list the configurations of matching corners and see how many times they each get counted.  So if no corners match there are $q*(q-1)*(q-2)*(q-3)*(q-4)$ possibilities, but you have counted each one 10 times, so divide these by 10.  Then you can have one pair or two pair of corners matching.  These are probably easier to count by the standard position approach.  If one pair matches, say they have to be positions 0 and 2.  So we have $q*(q-1)*1*(q-1)*(q-2)$ choices for this.  And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for taking out the asterisks from my expressions, but it rendered in italics and run over itself when they were there.  I think they made it easier to read.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-10T22:29:16.207" LastActivityDate="2010-10-10T22:29:16.207" CommentCount="2" />
  <row Id="6464" PostTypeId="1" AcceptedAnswerId="6476" CreationDate="2010-10-10T22:22:01.990" Score="5" ViewCount="179" Body="&lt;p&gt;In physics there are lots of identities like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - (\nabla \cdot \nabla) A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is an algorithmic algebraic method to prove and/or derive these identities (something like using $e^{i\theta}$ to prove trigonometric identities)?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T03:13:47.490" Title="Identities with Div, Grad, Curl" Tags="&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6465" PostTypeId="2" ParentId="6460" CreationDate="2010-10-10T22:33:08.617" Score="5" ViewCount="0" Body="&lt;p&gt;Here I am going to assume that we are not looking for distinct configurations, that is, we count a rotation or reflection of the pentagon to be a distinct possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You both have incorrect terms. Consider having only $3$ colors, ie $q = 3$. In this condition there must be a vertex which lies between two points which have the same color. There are $3 \times 2 = 6$ ways to color these three points, leaving $2$ ways to color the remaining two points for a total of $12$ possibilities. But $3^2 \times 2^2 \times 1 + 0 + (0) = 36$, which is too many. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: If you are coloring a pentagon, there are three possible cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The pentagon can be colored with $k = 5$ distinct colors chosen from the $q$ possibilities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pentagon can be colored with $k = 4$ distinct colors and one repeat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pentagon can be colored with $k = 3$ distinct colors where two colors appear twice. (We also have that one color could appear three times, but then it would be impossible to color a pentagon without the same color being adjacent.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that it is impossible to color the vertices of a pentagon with only 2 colors such that the same color is not adjacent anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, your friend is accounting for the $q(q-1)(q-2)(q-3)(q-4)$ ways to color a pentagon using $5$ distinct colors. To find the final solution, we need to count how many ways we can color a pentagon in each of the given cases and then we sum all the possibilities together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for each of the three cases, $q$ choose $k$ to fix the k elements you are working with, then consider how many ways you can color a pentagon using those $k$ colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any more assistance will be moving towards an outright solution so I believe I should stop here.&lt;/p&gt;&#xA;" OwnerUserId="557" LastEditorUserId="557" LastEditorDisplayName="" LastEditDate="2010-10-10T22:42:07.160" LastActivityDate="2010-10-10T22:42:07.160" CommentCount="1" />
  <row Id="6466" PostTypeId="1" AcceptedAnswerId="6479" CreationDate="2010-10-10T22:50:19.357" Score="3" ViewCount="181" Body="&lt;p&gt;Hello!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be the fundamental group of the Klein bottle,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = \langle x,y \ ; \ yxy^{-1}=x^{-1} \rangle = {\mathbb Z} \rtimes {\mathbb Z} \ .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the nilpotent subgroups of $G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was only able to find a normal series of abelian subgroups with cyclic quotients in $G$, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\leq \langle y^2 \ ; \ \ \rangle\leq \langle x,y^2 \ ; \ xy^2=y^2x \rangle\leq G \ .$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I'm not an algebraist, I'm sorry if this a silly question. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2412" LastActivityDate="2010-11-28T06:21:27.400" Title="Subgroups - Klein bottle" Tags="&lt;group-theory&gt;&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6467" PostTypeId="1" AcceptedAnswerId="6468" CreationDate="2010-10-10T23:16:36.207" Score="2" ViewCount="93" Body="&lt;p&gt;I am wondering about the quickest way to prove the following from scratch: the lower shift is not similar to a direct sum of other linear transformations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the definitions of the terms used in the question. The lower shift is a mapping from $R^n$ to $R^n$ which maps $(x_1, x_2, \ldots, x_{n-1}, x_n)$ to $(0, x_1, \ldots, x_{n-1})$. Given two linear mappings $T_1: U \rightarrow U, ~T_2: V \rightarrow V$ their direct sum is the map from $U \times V~$ to $~U \times V$ which maps $(u,v)$ to $(T_1 u, T_2 v)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current proof of this appeals to the Cayley-Hamilton theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-10-10T23:49:49.980" Title="The impossibility of decomposing the lower shift operator" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="6468" PostTypeId="2" ParentId="6467" CreationDate="2010-10-10T23:29:18.627" Score="4" ViewCount="0" Body="&lt;p&gt;Tensor by $\mathbb{C}$.  Then the direct sum of two linear transformations has at least two eigenvectors, but the lower shift has only one.  (If you don't want to tensor by $\mathbb{C}$, the corresponding statement over $\mathbb{R}$ is that the direct sum of two linear transformations has at least two one- or two-dimensional invariant subspaces, but the lower shift has only one.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-10T23:29:18.627" CommentCount="1" />
  <row Id="6469" PostTypeId="2" ParentId="6461" CreationDate="2010-10-10T23:35:17.190" Score="1" ViewCount="0" Body="&lt;p&gt;Disclaimer: I am new to this metric on spaces of probability measures, so hopefully there are not silly errors in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\mu\in C$, define $\mu'\in\mathcal{P}(A)$ by $\mu'(E)=\mu(E)$.  The map $\mu\mapsto\mu'$ is surjective, because if $\nu$ is in $\mathcal{P}(A)$, then $\nu=\tilde{\nu}'$, where $\tilde{\nu}\in C$ is defined by $\tilde{\nu}(E)=\nu(E\cap A)$.  The map is also isometric with respect to the &lt;a href=&quot;http://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric&quot; rel=&quot;nofollow&quot;&gt;Lévy–Prokhorov metric&lt;/a&gt;.  To see this, note that although the $\epsilon$-neighborhoods of a Borel subset $E$ of $A$ will be different in $A$ versus in $A\cup B$, the $\mu$-measures of these neighborhoods are the same for all $\mu\in C$, due to the fact that $\mu(B\setminus A)=0$ for such $\mu$.  Hence the sets appearing in the definition of the metric (and therefore also their infima) will be identical whether considering the distance between $\mu$ and $\nu$ in $C$ or between $\mu'$ and $\nu'$ in $\mathcal{P}(A)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-11T00:16:28.717" LastActivityDate="2010-10-11T00:16:28.717" />
  <row Id="6470" PostTypeId="1" AcceptedAnswerId="6493" CreationDate="2010-10-10T23:37:27.447" Score="1" ViewCount="64" Body="&lt;p&gt;Given representations $\rho_1 : G \to \mathrm{GL}(V)$ and $\rho_2 : G \to \mathrm{GL}(W)$, how can we define explicitly the representation of $G$ over $\mathrm{Hom}(V, W) \cong V^* \otimes W$?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-10-11T06:41:50.533" Title="Representation over Hom(V, W)" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="6472" PostTypeId="2" ParentId="6467" CreationDate="2010-10-10T23:49:49.980" Score="3" ViewCount="0" Body="&lt;p&gt;The shift is nilpotent.  If it were nontrivially a direct sum, then each of the summands would be nilpotent on spaces of positive dimension.  The kernel of a nilpotent on a nontrivial space is nontrivial, and the kernel of a direct sum is the direct sum of the kernels, so this would imply that the kernel of the shift has dimension at least 2.  But its kernel is 1-dimensional.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-10T23:49:49.980" CommentCount="1" />
  <row Id="6473" PostTypeId="1" CreationDate="2010-10-11T01:15:21.443" Score="0" ViewCount="102" Body="&lt;p&gt;suppose p,q are two prime number, how to quickly calculate how many numbers x are there such that gcd(x, (q-1)*(p-1)) = 1, without using brute force!&lt;/p&gt;&#xA;" OwnerUserId="2419" LastActivityDate="2010-10-11T02:09:09.537" Title="how to find co-prime numbers" Tags="&lt;cryptography&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6474" PostTypeId="2" ParentId="6410" CreationDate="2010-10-11T01:16:04.503" Score="4" ViewCount="0" Body="&lt;p&gt;Intuitively, the reason that universal quantification can be distributed over conjunction is that universal quantification can already be viewed as conjunction: $(\forall x)\Phi(x)$ can be viwed as the conjunction of $\Phi(d)$ taken over every element $d$ of the domain.  Viewed this way, $(\forall x)(\Phi(x) \land \Psi(x))$ and $(\forall x)\Phi(x) \land (\forall x)\Psi(x)$ are equivalent because they both represent a giant conjunction of every instance of $\Phi(d)$ along with every instance of $\Psi(d)$. This can be made more precise by looking at Tarski's schema for truth in a structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason, existential quantification can be distributed over disjunction, because $(\exists x)\Phi(x)$ can be viewed as the disjunction of every possible substitution instance of $\Phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In very old literature, people actually used $\bigwedge_x$ for $\forall$ and  $\bigvee_x$ for $\exists$, for this reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, continuing this informal viewpoint, the reason that $\forall$ cannot be distributed over disjunction is that a certain distributive rule does not hold:&#xA;$$&#xA;\bigwedge_x\left(\Phi(x) \lor \Psi(x)\right)&#xA;$$&#xA;is not in general the same as &#xA;$$&#xA;(\bigwedge_x \Phi(x)) \lor (\bigwedge_x \Psi(x))&#xA;$$&#xA;Indeed, this rule already fails when there are just two elements in the domain, because the propositional formula&#xA;$(P \lor Q) \land (R \lor S)$ is not equivalent to $(P \land R) \lor (Q \land S)$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-10-11T01:16:04.503" />
  <row Id="6475" PostTypeId="1" AcceptedAnswerId="6477" CreationDate="2010-10-11T01:27:20.067" Score="1" ViewCount="109" Body="&lt;p&gt;Dear all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to prove the following inequality. I have done a plot and it looks ok. $\mu \geq 1$ is an integer and so is $h$, where $0 \leq h \leq \mu-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\mu}\mu^{(\mu+h)}}{(\mu+h)!} \geq \frac{e^{-\mu}\mu^{(\mu-h-1)}}{(\mu-h-1)!}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have rewritten this to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mu-h-1)!\mu^{(\mu+h)} \geq (\mu+h)!\mu^{(\mu-h-1)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however I am not able to go further from there. Can you help me out?&lt;/p&gt;&#xA;" OwnerUserId="2174" LastActivityDate="2010-10-11T01:48:07.920" Title="An inequality involving factorials and powers" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6476" PostTypeId="2" ParentId="6464" CreationDate="2010-10-11T01:35:09.047" Score="4" ViewCount="0" Body="&lt;p&gt;The most straightforward route I know of is through &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_notation&quot; rel=&quot;nofollow&quot;&gt;Einstein tensor notation&lt;/a&gt;, which renders the derivation of such identities largely mechanical by virtue of suppressing distracting notational noise. (I should warn you that I'm largely self-taught on this subject, so the following may contain errors of rigour, but it &quot;works&quot; as long as you're working in Cartesian coordinates with the Euclidean metric.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me work through your example to illustrate. In tensor notation, the $i$th component of the curl of a vector field $v_i$ is given by $\varepsilon_{ijk}\partial_j v_k$, where $\varepsilon_{ijk}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_symbol&quot; rel=&quot;nofollow&quot;&gt;Levi-Civita symbol&lt;/a&gt;, that takes the values $1$, $-1$ and $0$ depending on the order in which the coordinates appear in the subscripts $ijk$. You can think of this as a notational shorthand for the plus and minus signs in the curl formula. So the left-hand side of the desired identity is&#xA;$$\varepsilon_{ijk}\partial_j (\varepsilon_{k\ell m}\partial_\ell A_m)$$&#xA;$$= \varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$&#xA;because derivatives commute with constants and with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, to simplify simplify $\varepsilon_{ijk}\varepsilon_{k\ell m}$, I just looked up the relevant identity of the Levi-Civita symbol, but it should be possible to derive it simply by algebraic manipulation from &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_symbol#Definition&quot; rel=&quot;nofollow&quot;&gt;the purely combinatorial definition of the symbol&lt;/a&gt;. It turns out that $\varepsilon_{ijk}\varepsilon_{k\ell m} = \delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell}$, where $\delta_{ij}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_delta&quot; rel=&quot;nofollow&quot;&gt;Kronecker delta&lt;/a&gt; which is $1$ if and only if $i = j$, and $0$ otherwise. This symbol acts like a substitution operator: $\delta_{ij}v_j = v_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have&#xA;$$\varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$&#xA;$$= (\delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell})\partial_j\partial_\ell A_m$$&#xA;$$= \partial_m\partial_i A_m - \partial_\ell\partial_\ell A_i$$&#xA;$$= \partial_i (\partial_m A_m) - (\partial_\ell\partial_\ell) A_i.$$&#xA;The first term is $\nabla(\nabla\cdot A)$, while the second is $-(\nabla\cdot\nabla)A$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-11T03:13:47.490" LastActivityDate="2010-10-11T03:13:47.490" />
  <row Id="6477" PostTypeId="2" ParentId="6475" CreationDate="2010-10-11T01:48:07.920" Score="5" ViewCount="0" Body="&lt;p&gt;Cancel a factor of $\mu^\mu$ on both sides, move the powers of $\mu$ to the left, and move the factorials to the right to obtain the equivalent statement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu^{2h+1}\geq (\mu+h)(\mu+h-1)(\mu+h-2)\cdots(\mu-h+1)(\mu-h).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break up the right hand side into the middle term, $\mu$, along with $h$ pairs of the form $(\mu+k)(\mu-k)$.  Since $(\mu+k)(\mu-k)=\mu^2-k^2\lt\mu^2$ for $k&gt;0$, it should now be clear.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-11T01:48:07.920" CommentCount="4" />
  <row Id="6478" PostTypeId="2" ParentId="6473" CreationDate="2010-10-11T02:09:09.537" Score="3" ViewCount="0" Body="&lt;p&gt;The answer is $\phi((q-1)(p-1))$, but to compute that you probably need to factor $p-1$ and $q-1$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-11T02:09:09.537" />
  <row Id="6479" PostTypeId="2" ParentId="6466" CreationDate="2010-10-11T02:10:55.320" Score="6" ViewCount="0" Body="&lt;p&gt;Every cyclic subgroup is abelian, hence nilpotent. Every such subgroup is generated by an element of the form $x^ay^b$ with $a$ and $b$ integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, $y^2$ commutes with $x$. Since $y^2$ is central, to determine the commutator of two elements we only need to consider the parity of the exponents of $y$ in their normal forms. Elements of the form $x^ay^{2b}$ and $x^ry^{2s}$ commute; elements of the form $x^ay^{2k+1}$ and $x^r$ commute if and only if $r=0$; and elements of the form $x^ay^{2k+1}$ and $x^ry^{2s+1}$ commute if and only if $a=r$. That gives you lots of abelian subgroups that are isomorphic to $\mathbb{Z}\times\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about higher nilpotency? Suppose you have two noncommuting elements in your subgroup. It is not hard to check that $[x^ay^{2b+1},x^r] = x^{2r}$ and  $[x^ay^{2b+1},x^r y^{2s+1}] = x^{2a-2r}$. But no nontrivial power of $x$ commutes with an element of the form $x^a y^{2b+1}$; and further commutators will just yield further nontrivial powers of $x$ that still do not commute with an element of the form $x^ay^{2b+1}$. So if your subgroup $H$ has an element of the form $h_1=x^a y^{2b+1}$, and $h$ is any nontrivial element of $H$ that is not a power of $h_1$, then $h_1$ and $h$ do not commute, and $h_1$ does not commute with any of $[h_1,h]$, $[h_1,h,h_1]$, $[h_1,h,h_1,h_1],\ldots,[h_1,h,h_1,\ldots,h_1]$, etc. But if $H$ is nilpotent of class $c$, then any commutator of weight $c$ would be central in $H$. Thus, $H$ cannot be nilpotent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only  nilpotent subgroups of $H$ are abelian, and they are given as above.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T06:21:27.400" LastActivityDate="2010-11-28T06:21:27.400" CommentCount="1" />
  <row Id="6480" PostTypeId="2" ParentId="6459" CreationDate="2010-10-11T02:13:20.227" Score="2" ViewCount="0" Body="&lt;p&gt;There is one important rule for assigning probabilities to edges in a Markov model:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $i$ be a node, and $out(i)$ be the set of nodes to which $i$ is connected (via an outgoing edge). Then $\sum_{j \in out(i)}{P_{ij}} = 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider your own model: As an instance, $out(4)=\{1\}$. Therefore, $P_{41}=1$. In other words, if we are in state 4, we will certainly go to state 1 at the next transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider node 1. We have $out(1)= \{1,3,4\}$. Hence $P_{11}+P_{13}+P_{14}=1$. Using the counts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{11} = \frac{C_{11}}{C_{11}+C_{13}+C_{14}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{13} = \frac{C_{13}}{C_{11}+C_{13}+C_{14}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{14} = \frac{C_{14}}{C_{11}+C_{13}+C_{14}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can do the math for other nodes as well.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2010-10-11T02:13:20.227" CommentCount="1" />
  <row Id="6481" PostTypeId="2" ParentId="6466" CreationDate="2010-10-11T03:02:22.747" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a topological version of Arturo's argument.  (I'd leave this as a comment, but I don't have enough rep.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every covering space of the Klein bottle is a surface, and it's fairly easy to convince yourself that the infinite-degree ones have cyclic, hence nilpotent, fundamental group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to consider the finite-degree covering spaces.  Well, these are all closed surfaces of Euler characteristic zero (because Euler characteristic is &lt;strike&gt;nilpotent&lt;/strike&gt; multiplicative), so by the classification of surfaces they are either tori or Klein bottles.  As already noted, the Klein-bottle group itself is not nilpotent, so only the abelian subgroups are left.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-10-11T11:42:54.530" LastActivityDate="2010-10-11T11:42:54.530" CommentCount="2" />
  <row Id="6482" PostTypeId="1" CreationDate="2010-10-11T03:12:27.370" Score="1" ViewCount="199" Body="&lt;p&gt;How would I start proving&#xA;If&#xA;$$\lim_{(x,y)\to(a,b)} f(x,y) = L$$&#xA;and both the $x$ and $y$ limits exist&#xA;that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{x\to a}\left(\lim_{y\to b} f(x,y)\right)  = \lim_{y\to b}\left(\lim_{x\to a} f(x,y)\right) = L$$&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="2319" LastEditorDisplayName="" LastEditDate="2010-10-11T03:40:19.317" LastActivityDate="2010-10-16T17:07:53.413" Title="Limits of 2 variable functions" Tags="&lt;real-analysis&gt;&lt;algebra&gt;&lt;limit&gt;" AnswerCount="2" />
  <row Id="6483" PostTypeId="2" ParentId="6482" CreationDate="2010-10-11T03:23:32.117" Score="6" ViewCount="0" Body="&lt;p&gt;You cannot prove that the two-variable limit equals the iterated limits even if they both exist, since the two-variable limit may fail to exist even if both iterated limits exists and are equal. For example, take $f(x,y) = \frac{xy}{x^2+y^2}$, with $a=b=0$. The iterated limits both exist:&#xA;$$\lim_{x\to 0}\left(\lim_{y\to 0}\frac{xy}{x^2+y^2}\right) = \lim_{x\to 0}\frac{0}{x^2} = 0,$$&#xA;$$\lim_{y\to 0}\left(\lim_{x\to 0}\frac{xy}{x^2+y^2}\right) = \lim_{y\to 0}\frac{0}{y^2} = 0.$$&#xA;But the limit does not exist as $(x,y)\to(0,0)$. If you approach $(0,0)$ along the line $y=x$ you get the limit as $x\to 0$ of $\frac{x^2}{2x^2}$, which goes to $\frac{1}{2}$, not $0$. So the two-variable limit does not exist, even though the two iterated limits both exist and are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that the two-sided limit exists and is equal to $L$, and then you want to prove that the iterated limits will therefore both exist and be equal, then you might have a better shot. Start by thinking about the definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added later:&lt;/strong&gt; And then look at Robin Chapman's example to see that it does not quite work out. So you don't get implications either way.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T17:07:53.413" LastActivityDate="2010-10-16T17:07:53.413" CommentCount="2" />
  <row Id="6484" PostTypeId="1" CreationDate="2010-10-11T03:35:03.267" Score="1" ViewCount="64" Body="&lt;p&gt;why in homological mirror symmetry, we restrict us to a projective variety (calabi-yau)? Because in physics we don't need this condition. What's the general picture for general calabi-yau?&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-11T03:35:03.267" Title="about homological mirror symmetry" Tags="&lt;algebraic-geometry&gt;&lt;physics&gt;" CommentCount="1" />
  <row Id="6485" PostTypeId="1" CreationDate="2010-10-11T03:48:49.973" Score="2" ViewCount="141" Body="&lt;p&gt;Does Slater condition hold trivially (because there are no inequality constraints) for the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min_{x,y} \:\: cx+dy$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^x + e^y = 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I conclude there is a zero duality gap here?&lt;/p&gt;&#xA;" OwnerUserId="2422" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-10T19:41:44.140" LastActivityDate="2010-12-10T19:41:44.140" Title="Does Slater's condition hold for the following problem?" Tags="&lt;optimization&gt;&lt;nonlinear-optimization&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6486" PostTypeId="2" ParentId="6485" CreationDate="2010-10-11T04:21:56.487" Score="-1" ViewCount="0" Body="&lt;p&gt;You can rewrite the equality constraint as $\exp{x} + \exp{y} \leq 1$ and $-\exp{x} -\exp{y} \leq -1$. If slater conditions hold for this new formulation, (since the problem is convex), strong duality (gap zero) would follow.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-11T04:21:56.487" CommentCount="1" />
  <row Id="6487" PostTypeId="2" ParentId="6485" CreationDate="2010-10-11T04:28:25.087" Score="1" ViewCount="0" Body="&lt;p&gt;A Naga says, you can rewrite the equality constraint $h(x)=0$ as a pair of inequality constraints&#xA;$$h(x) \leq 0$$&#xA;$$-h(x) \leq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Slater condition would require&#xA;$$h(x)&amp;lt;0$$&#xA;$$-h(x)&amp;lt;0$$&#xA;which is an impossibility. So the Slater condition cannot be true of a problem with equality constraints. Other regularity conditions may be usable though.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-11T04:28:25.087" />
  <row Id="6488" PostTypeId="2" ParentId="6466" CreationDate="2010-10-11T04:32:52.427" Score="0" ViewCount="0" Body="&lt;p&gt;Is there a shorter (yet elementary) argument showing that $G$ is not nilpotent?&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2010-10-11T04:32:52.427" CommentCount="3" />
  <row Id="6489" PostTypeId="1" AcceptedAnswerId="6490" CreationDate="2010-10-11T04:45:45.797" Score="16" ViewCount="850" Body="&lt;p&gt;As I'm sure many of you do, I read the &lt;a href=&quot;http://xkcd.com/&quot; rel=&quot;nofollow&quot;&gt;XKCD&lt;/a&gt; webcomic regularly. The &lt;a href=&quot;http://xkcd.com/804/&quot; rel=&quot;nofollow&quot;&gt;most recent one&lt;/a&gt; involves a joke about the Axiom of Choice, which I didn't get. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ff9dS.png&quot; alt=&quot;The Banach-Tarski theorem was actually first developed by King Solomon, but his gruesome attempts to apply it set back set theory for centuries.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to Wikipedia to see what the Axiom of Choice is, but as often happens with things like this, the Wikipedia entry is not in plain, simple, understandable language. Can someone give me a nice simple explanation of what this axiom is, and perhaps explain the XKCD joke as well?&lt;/p&gt;&#xA;" OwnerUserId="496" LastEditorUserId="496" LastEditorDisplayName="" LastEditDate="2010-10-11T06:10:32.383" LastActivityDate="2010-10-15T10:53:51.693" Title="Can you explain the &quot;Axiom of choice&quot; in simple terms?" Tags="&lt;set-theory&gt;&lt;axioms&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" />
  <row Id="6490" PostTypeId="2" ParentId="6489" CreationDate="2010-10-11T05:11:47.357" Score="18" ViewCount="0" Body="&lt;p&gt;The joke is really about the Banach-Tarski theorem, which says that you can cut up a sphere into a finite number of pieces which when reassembled give you two spheres of the same size as the original sphere. This theorem is extremely counterintuitive since we seem to be doubling volume without adding any material or stretching the material that we have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof the the theorem requires the Axiom of Choice (AC), which says that if you have a collection of sets then there is a way to select one element from each set. It has been proved that AC cannot be derived from the rest of set theory but must be introduced as an additional axiom. Since AC can be used to derive counterintuitive results such as the Banach-Tarski theorem, some mathematicians are very careful to specify when their arguments depend on AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a formal statement of AC. Suppose we have a set $W$ and a rule associating a set $S_w$ to each $w \in W$. Then AC says that there is a function &#xA;$$f:W \to \bigcup_{w \in W} S_w$$&#xA;such that&#xA;$$f(w) \in S_w$$&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-15T01:00:54.307" LastActivityDate="2010-10-15T01:00:54.307" CommentCount="2" />
  <row Id="6491" PostTypeId="1" AcceptedAnswerId="6520" CreationDate="2010-10-11T05:22:46.290" Score="2" ViewCount="250" Body="&lt;p&gt;Reffered here by &lt;a href=&quot;http://mathoverflow.net/questions/41750/determining-n-in-sigma-x0n&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/41750/determining-n-in-sigma-x0n&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure if this question falls under MathOverflow but neither of my Calculus AP  teachers in high school could help me with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\Sigma_{x=0}^n {f(x)\over2}$ and the output of the summation, how would you find $n$? I've learned how to determine the $n$ given an arithmetic or geometric sequence, but not for an arbitrary function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, when $f(x) = 40 + 6\sqrt{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12 Oct 2010. Edit:&#xA;It seems like I need to explain the entire situation for finding $n$, the number of trapezoids, for trapezoidal rule. It started on a simple review question for Calc AP and a TI-83 program that my calc teacher gave to me to solve the definite integral with trapezoidal rules. Aiming to major in Computer Science, I took it a bit further and completely took apart the program resulting in my original question on StackOverflow: &lt;a href=&quot;http://stackoverflow.com/questions/3886899/determining-the-input-of-a-function-given-an-output-calculus-involved&quot;&gt;http://stackoverflow.com/questions/3886899/determining-the-input-of-a-function-given-an-output-calculus-involved&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there were tumbleweeds for a response, I took it as a personal challenge to reverse engineer the trapezoidal program into an algebraic form with my notes found on my forum: &lt;a href=&quot;http://www.zerozaku.com/viewtopic.php?f=19&amp;amp;t=6041&quot; rel=&quot;nofollow&quot;&gt;http://www.zerozaku.com/viewtopic.php?f=19&amp;amp;t=6041&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reverse engineering the code into some algebra, I derived the formula: &#xA;$$TrapRule(A, B, N) = {(B-A)\over N}({F(A)\over2}+\sum_{k=1}^NF(k)+{F(B)\over2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the values of A and B are constant for the definite integral, I should be able to isolate and solve for $N$. The problem, however, was determining $N$ in $sum_{k=1}^N$ and I came to the conclusion that it was an issue that I called &lt;em&gt;recursive complexity&lt;/em&gt; because it was impossible to determine without recursively adding for the summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, I found MathOverflow and they referred me here. I was hoping only to get help on the issue for a summation because its beyond my skill as a high school student. Now that others have proposed other solutions for my dilemma, I guess I can throw out my thesis Dx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help though, I'll definitely be returning for more.&lt;/p&gt;&#xA;" OwnerUserId="2426" LastEditorUserId="2426" LastEditorDisplayName="" LastEditDate="2010-10-13T01:46:16.893" LastActivityDate="2010-10-13T01:46:16.893" Title="Determining n in sigma ($\Sigma_{x=0}^n$)" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="7" />
  <row Id="6492" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T05:39:36.853" Score="3" ViewCount="0" Body="&lt;p&gt;You are trying to calculate the inverse of the function&#xA;$$S(n)=\sum_{x=0}^n f(x)/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the inverse would not exist (think $f(x)=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the inverse exists, a trivial way to solve your problem would be to start calculating the sum for $n=0,1,2,\ldots$ and stop when the result matches the number provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a formula for $S^{-1}(n)$, there is no reason for there to be such a formula for every $f(x)$. The arithmetic and geometric progression examples you gave were lucky exceptions.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-11T07:57:32.533" LastActivityDate="2010-10-11T07:57:32.533" />
  <row Id="6493" PostTypeId="2" ParentId="6470" CreationDate="2010-10-11T06:41:50.533" Score="4" ViewCount="0" Body="&lt;p&gt;The isomorphism of $\mathrm{Hom}(V,W)$&#xA;with $V^*\otimes W$ only holds when $V$ is finite-dimensional. I'll assume&#xA;both $V$ and $W$ are finite-dimensional. If we write our group actions&#xA;on the right, if $\phi\in\mathrm{Hom}(V,W)$&#xA;then&#xA;$$\phi g:v\mapsto\phi(vg^{-1})g.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-11T06:41:50.533" CommentCount="1" />
  <row Id="6494" PostTypeId="1" AcceptedAnswerId="6496" CreationDate="2010-10-11T06:50:55.073" Score="0" ViewCount="370" Body="&lt;p&gt;Can we have a measurable function $f$, whose inverse is not measurable?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-11T18:46:38.530" Title="Example of a non measurable function!" Tags="&lt;analysis&gt;&lt;measure-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" />
  <row Id="6495" PostTypeId="1" AcceptedAnswerId="6792" CreationDate="2010-10-11T06:57:35.303" Score="2" ViewCount="306" Body="&lt;p&gt;Think of $t$ and $r$ as two independent variables. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose $E$ be a function of $r$ and $V$ be a function of $(t,r)$ such that both go to $0$ at $r=0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There exists a positive function $M(r)$ such that $M(0)=0$ and $V(t,r) = -\frac{M(r)}{R(t,r)}$ where $R$ is another positive function such that $R(0,r)=r$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $p(r) = \frac{E(r)}{V(0,r)}$ be a function regular at $r=0$ such that $p(0) \in (-\infty,1)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also define a function $a$ of $r$ such that, $a(r) = \frac{M(r)}{\frac{4}{3}\pi r^3}$. &#xA;Then $a$ is also a positive definite function with a well-defined value at $r=0$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Define $\alpha = a(0)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now look at this differential equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\dot{R}^2}{2} + V(t,r) = E(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this differential equation has a solution of the form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{t}{t_0} = \sqrt{\frac{\alpha}{a(r)}}\frac{F(p(r))}{F(p(0))} \left [1 - \left ( \frac{R(t,r)}{r} \right)^{\frac{3}{2}}\frac{F\left( \frac{p(r)R(t,r)}{r} \right) }{F(p(r))} \right ] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_0 = \sqrt {\frac{3}{8\pi \alpha}} F(p(0))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the function $F$ is defined over the interval $(-\infty,1)$ as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) = \left\{ \begin{array}{c c}&lt;br&gt;&#xA;                  -\frac{\sqrt{1-x}}{x} - \frac{1}{(-x)^{\frac{3}{2}}} tanh^{-1} \left [ \sqrt{\frac{x}{x-1}} \right ] &amp;amp; x&amp;lt;0 \\&lt;br&gt;&#xA;         \frac{2}{3}  &amp;amp;  x =0 \\&#xA;         \frac{1}{x^{\frac{3}{2}}}tan^{-1} \left [ \sqrt{\frac{x}{1-x}} \right ] - \frac{\sqrt{1-x}}{x}  &amp;amp; 0&amp;lt;x&amp;lt;1 &#xA;             \end {array} \right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one get the above solution? &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-13T20:25:42.500" LastActivityDate="2010-10-14T16:20:19.650" Title="A differential equation " Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6496" PostTypeId="2" ParentId="6494" CreationDate="2010-10-11T07:13:22.073" Score="1" ViewCount="0" Body="&lt;p&gt;I think I have an answer that works now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the problem is trivial if we allow $f$ to be any function between measure spaces (just take the domain to have its powerset as $\sigma$-algebra and/or the trivial $\sigma$-algebra on the codomain), so this must be a question about real functions and the Lebesgue measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be the Cantor set, $Y$ a nonmeasureable set (and hence uncountable) contained in $[0,1]$ and $g: X \to Y$ be any bijective function. Then define &#xA;$$f(x) = \begin{cases}&#xA;  g(x) &amp;amp; \text{if } x \in X \\&#xA;  e^x + 2 &amp;amp; \text{otherwise.}&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $U$ be any measurable set and let us show that $f^{-1}(U)$ is measurable. If $U \cap Y = \emptyset$ then $f^{-1}(U) \subset \mathbb{R} \setminus X$ on which $f$ is continuous. In particular $f^{-1}(U)$ must be measurable. If $U \cap Y \neq \emptyset$ then we split into two more cases. First consider the possibility that $U \cap (2,\infty) = \emptyset$. Then $f^{-1}(U) \subset X$, i.e. it is a null set and hence Lebesgue measurable. And if $U \cap (2,\infty) \neq \emptyset$ then by a similar argument we see that $f^{-1}(U)$ is the disjoint union of a null set contained in $X$ and a measurable subset of $\mathbb{R}\setminus X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f$ is measurable, but its inverse is obviously not measurable since $(f^{-1})^{-1}(X) = f(X) = Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we also require $f$ to be bijective rather than just injective, then I don't know the answer. It seems much harder to construct an example in that case.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-11T09:00:50.990" LastActivityDate="2010-10-11T09:00:50.990" />
  <row Id="6497" PostTypeId="2" ParentId="6494" CreationDate="2010-10-11T08:22:28.533" Score="-5" ViewCount="0" Body="&lt;p&gt;If you're considering only Lebesgue measure, this is Problem 2.2.19 in Problems in Mathematical Analysis III: Integration by Kaczor, Kaczor, and Nowak. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-11T08:22:28.533" CommunityOwnedDate="2010-10-11T08:22:28.533" CommentCount="2" />
  <row Id="6498" PostTypeId="2" ParentId="6489" CreationDate="2010-10-11T09:28:42.423" Score="15" ViewCount="0" Body="&lt;p&gt;It might be a good idea to say something about the connection between the axiom of choice and the Banach-Tarski paradox.  The reason the Banach-Tarski paradox is counterintuitive is that we expect that if you split up a sphere into finitely many pieces, the total &quot;mass&quot; of all the pieces is still the same - so you shouldn't be able to put those pieces together again into something of twice the total &quot;mass.&quot;  The reason this reasoning doesn't apply is that the pieces in question &lt;strong&gt;don't have mass&lt;/strong&gt; at all!  This is not the same as saying that they have zero mass.  It's something much more terrifying: the notion of &quot;mass&quot; (which to a mathematician generally means something precise called &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;measure&lt;/a&gt;) can't be defined for these pieces (in a way that still preserves all the reasonable properties we want of our notion of mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this have to do with the axiom of choice?  Well, it turns out that the axiom of choice is one way we can construct these bizarre pieces (&lt;a href=&quot;http://en.wikipedia.org/wiki/Non-measurable_set&quot; rel=&quot;nofollow&quot;&gt;non-measurable sets&lt;/a&gt;).  Without the axiom of choice, it's not possible to prove that such pieces exist.  With the axiom of choice, we can construct things like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vitali_set&quot; rel=&quot;nofollow&quot;&gt;Vitali set&lt;/a&gt; and like the pieces that occur in the Banach-Tarski paradox because AC greatly increases our ability to write down weird sets.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-11T09:28:42.423" CommentCount="5" />
  <row Id="6500" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T10:26:36.103" Score="0" ViewCount="0" Body="&lt;p&gt;Further to Jyotirmoy's answer, for the $f(x)$ you have given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n)=\sum_{x=0}^{n}\frac{f(x)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\sum_{x=0}^{n}\frac{(40+6\sqrt{x})}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\sum_{x=0}^{n}(20+3\sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\sum_{x=0}^{n}20+3\sum_{x=0}^{n}\sqrt{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second sum is a generalised harmonic number, which doesn't have a nice closed form expression&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2010-10-11T10:40:03.420" LastActivityDate="2010-10-11T10:40:03.420" CommentCount="1" />
  <row Id="6501" PostTypeId="1" AcceptedAnswerId="6504" CreationDate="2010-10-11T10:46:33.987" Score="8" ViewCount="235" Body="&lt;p&gt;So, from what I understand, the axiom of choice is equivalent to the claim that every set can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order&quot; rel=&quot;nofollow&quot;&gt;well ordered&lt;/a&gt;. A set is well ordered by a relation, $R$ , if every subset has a least element. My question is: as anyone constructed a well ordering on the reals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I was going to ask this question about the rationals, but then I realised that if you pick your favourite bijection between rationals and the integers, this determines a well ordering on the rationals through the natural well order on $\mathbb{Z}$ . So it's not the denseness of the reals that makes it hard to well order them. So is it just the &lt;em&gt;size&lt;/em&gt; of $\mathbb{R}$ that makes it difficult to find a well order for it? Why should that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reiterate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a known well order on the Reals?&lt;/li&gt;&#xA;&lt;li&gt;If there is, does a similar construction work for larger cardinalities?&lt;/li&gt;&#xA;&lt;li&gt;Is there a largest cardinality for which the construction works?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-10-11T11:46:23.770" Title="Is there a known well ordering of the reals?" Tags="&lt;set-theory&gt;&lt;large-cardinals&gt;&lt;order-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="6502" PostTypeId="1" CreationDate="2010-10-11T11:06:28.470" Score="2" ViewCount="60" Body="&lt;p&gt;One of the definitions of stochastic continuity is: $\forall a &gt; 0$ and  $\forall s \geq 0$ $$\lim_{t\rightarrow s}\;P(|X(t)-X(s)|&gt;a) = 0.$$ What does it mean intuitively? I know that it implies the jumps but am not sure how and why. &lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-11T15:17:09.860" LastActivityDate="2010-10-11T15:35:24.707" Title="About stochastic continuity" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6503" PostTypeId="2" ParentId="6501" CreationDate="2010-10-11T11:11:52.417" Score="3" ViewCount="0" Body="&lt;p&gt;No, it's not just the size. One can &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartogs_number&quot; rel=&quot;nofollow&quot;&gt;constructively prove&lt;/a&gt; the existence of large well-ordered sets, but for example even when one has the &lt;a href=&quot;http://en.wikipedia.org/wiki/First_uncountable_ordinal&quot; rel=&quot;nofollow&quot;&gt;first uncountable ordinal&lt;/a&gt; in hand, one can't show that it is in bijection with $\mathbb{R}$ without the continuum hypothesis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the difficulty in the problem has to do with what you mean by &quot;constructed.&quot;  If one has a well-ordering on $\mathbb{R}$ then it is possible to carry out the construction of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vitali_set&quot; rel=&quot;nofollow&quot;&gt;Vitali set&lt;/a&gt;, which is a non-measurable subset of $[0, 1]$.  And &lt;a href=&quot;http://en.wikipedia.org/wiki/Solovays_model&quot; rel=&quot;nofollow&quot;&gt;it is known&lt;/a&gt; that the existence of non-measurable subsets of $\mathbb{R}$ is independent of ZF.  In other words, it is impossible to write down a well-ordering of $\mathbb{R}$ in ZF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand given AC one can obviously write down a well-ordering in a non-constructive way (choose the first element, then the second element, then...).  This is probably not what you meant by &quot;construct,&quot; though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-11T11:11:52.417" CommentCount="1" />
  <row Id="6504" PostTypeId="2" ParentId="6501" CreationDate="2010-10-11T11:37:37.577" Score="12" ViewCount="0" Body="&lt;p&gt;I assume you know the general theorem that, using the axiom of choice, every set can be well ordered. Given that, I think you're asking how hard it is to actually define the well ordering. This is a natural question but it turns out that the answer may be unsatisfying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, of course, without the axiom of choice it's consistent with ZF set theory that there is &lt;em&gt;no&lt;/em&gt; well ordering of the reals. So you can't just write down a formula of set theory akin to the quadratic formula that will &quot;obviously&quot; define a well ordering. Any formula that does define a well-ordering of the reals is going to require a nontrivial proof to verify that it's correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is not even a formula that unequivocally defines a well ordering of the reals in ZFC.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The theorem of &quot;Borel determinacy&quot; implies that there is no well ordering of the reals whose graph is a Borel set. This is provable in ZFC.  The stronger hypothesis of &quot;projective determinacy&quot; implies there is no well ordering of the reals definable by a formula in the projective hierarchy. This is consistent with ZFC but not provable in ZFC. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Worse, it's even consistent with ZFC that &lt;em&gt;no&lt;/em&gt; formula in the language of set theory defines a well ordering of the reals (even though one exists). That is, there is a model of ZFC in which no formula defines a well ordering of the reals. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A set theorist could tell you more about these results.  They are in the set theoretic literature but not in the undergraduate literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a positive result. If you work in $L$ (that is, you assume the axiom of constructibility) then a specific formula is known that defines a well ordering of the reals in that context.  However, the axiom of constructibility is not provable in ZFC (although it is consistent with ZFC), and the formula in question does not define a well ordering of the reals in arbitrary models of ZFC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second positive result, for relative definability. By looking at the standard proof of the well ordering principle (Zermelo's proof), we see that there is a single formula $\phi(x,y,z)$ in the language of set theory such that if we have any choice function $F$ on the powerset of the reals then the formula $\psi(x,y) = \phi(x,y,F)$ defines a well ordering of the reals, in any model of ZF that happens to have such a choice function. Informally, this says that the reason the usual proof can't explicitly construct a well ordering is because we can't explicitly construct the choice function that the proof takes as an input. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-11T11:46:23.770" LastActivityDate="2010-10-11T11:46:23.770" />
  <row Id="6505" PostTypeId="1" AcceptedAnswerId="6524" CreationDate="2010-10-11T11:42:59.813" Score="4" ViewCount="161" Body="&lt;p&gt;Online, I have read contradicting opinions on whether &lt;code&gt;intersect&lt;/code&gt; should take precedence over &lt;code&gt;union&lt;/code&gt; (by analogy to logical &lt;code&gt;and&lt;/code&gt; and &lt;code&gt;or&lt;/code&gt;), or whether all set operators should have equal precedence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which way makes more sense and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where does &lt;code&gt;difference&lt;/code&gt; fit in?  (I'd say it should be equal precedence to &lt;code&gt;intersect&lt;/code&gt; because &lt;code&gt;A - B = A intersect B'&lt;/code&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2430" LastEditorUserId="329" LastEditorDisplayName="anon" LastEditDate="2010-11-23T04:05:06.540" LastActivityDate="2010-11-23T04:05:06.540" Title="Precedence of set union, intersect, and difference?" Tags="&lt;notation&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6506" PostTypeId="2" ParentId="6451" CreationDate="2010-10-11T12:36:42.130" Score="1" ViewCount="0" Body="&lt;p&gt;Starting from a flat rectangular Mobius strip $ABA'B'$ with the identification $AB'\cong A'B$. The side $AB$ is to be shrunk to a point and similarly the side $A'B'$ to obtain the identification of the boundary circle to a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct a transformation, we take the side $AB$ is along the x axis and $BA'$ along the y-axis of a Cartesian system. We take the origin of the system at the center of the rectangle and the side  lengths 2a (along the x-axis) and 2b along the (y-axis). ($AB$ is along the line $y=-b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the transformation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = \sqrt {a^2b^2-(a^2-x^2)(b^2-y^2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta = \arcsin{\frac{y}{b}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($r$ and $\theta$ are polar coordinates) maps the Mobius strip to a disc of radius $ab$ with the boundary antipodal points identified (which is homeomorphic to $RP^2$. The identified lines $AB$ $A'B'$ are mapped to  south and the north poles respectively. The lines $BA'$ and $B'A$ are mapped to the right and left boundary semicircles.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-11T12:36:42.130" />
  <row Id="6507" PostTypeId="1" AcceptedAnswerId="6642" CreationDate="2010-10-11T12:40:41.460" Score="2" ViewCount="242" Body="&lt;p&gt;I am of, and I would like to retain a, mindset that mathematics does not have to have numbers as the central object of interest. With that in mind, I have done a fair amount of self-study on topics in undergrad modern algebra, looking for examples to show this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For objects like groups, it is often stated explicitly, at some point, that group elements can be though of as automorphisms on a particular set. I find this pretty interesting, that numbers do not come into play at all here. This makes groups pretty useful for problems that have symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for objects like fields and rings, it seems that their application often ends up dealing with numbers somehow. I'm aware there are probably a handful of cases where rings and fields do not show up as numbers, but those tend to be side-cases that are not of main study. I'm only vaguely aware of Galois theory and it's uses for fields as extensions of groups, so perhaps that is the direction I should follow there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as for rings, aside from the example I just gave, I still come up a bit short on any powerful or broad theorems that make use of those two operations but do not involve numbers in some fundamental way. Perhaps math.stackexchange can prove me wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Answers with really good links to books or other resources will at least get an upvote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An update for those who pointed out my lack of precision.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So admittedly, the choice of words was bad. I did use the phrase &quot;involve numbers in some fundamental way&quot;, which, in my mind, means &quot;does not use facts about numbers&quot;. Perhaps I can  rephrase it better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a group (G, *). There are many examples of groups if I want to let G be the integers, rationals, reals, complex, etc, where the group operation (*) can be addition, multiplication, or some other operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, each of these are &quot;numerical&quot; examples. There are groups which do not behave like numbers, if we let G be the set of all bijections on a set S, and let (*) be the composition, we may not be able to find a group where the set S is a collection of numbers and (*) is a &quot;numerical&quot; operation on them. Our group operation (*) may end just being some random rearrangement of numbers that does not take advantage of the properties of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, and as others often point out, all groups are just specific cases of bijections &amp;amp; composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose I have a ring (R, +, *). The ring of functions on integers, rationals, reals, etc. takes advantage of the properties of numbers when defining (+) and (*). As would the ring of polynomials, or the ring of matrices. Each of these examples of rings, while the set R is not a set of numbers, ends up being an abstraction of numbers, and the operations (+) and (*) make use of this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An example of what I mean&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a example (counterexample?) of what I mean, try Boolean rings. The operations (+) and (*) are isomorphic to the logical operations &quot;xor&quot; and &quot;and&quot;. There is also examples of lattices with join and meet. One could try to squeeze in natural numbers into this with lcd and gcd, but these properties are not unique to numbers so they are not exactly &quot;numerical&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to add more structure, say an additive and multiplicative identity, you can use  Stone's Representation Theorem to show it is isomorphic to a collection of sets with symmetric difference and intersection.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2010-10-12T20:31:23.423" LastActivityDate="2010-10-18T02:32:36.800" CommunityOwnedDate="2010-10-12T20:31:23.423" Title="Are there broad or powerful theorems of rings that do not involve the familiar numerical operations (+) and (*) in some fundamental way?" Tags="&lt;abstract-algebra&gt;&lt;soft-question&gt;&lt;group-theory&gt;&lt;ring-theory&gt;&lt;field-theory&gt;" AnswerCount="6" CommentCount="12" FavoriteCount="1" />
  <row Id="6508" PostTypeId="1" AcceptedAnswerId="6514" CreationDate="2010-10-11T13:01:03.740" Score="0" ViewCount="86" Body="&lt;p&gt;I want to prove that $P(|A+B|&gt;c)\leq P(|A|&gt;c/2)+P(|B|&gt;c/2)$, where $A$ and $B$ are random variables and $c&gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="626" LastActivityDate="2010-10-13T04:20:12.140" Title="how to do i prove this simple probabilty result" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="6509" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T13:05:33.530" Score="0" ViewCount="0" Body="&lt;p&gt;Perhaps you can also work it out by approximating (lower and upper bounding) the sum by an integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^N \sqrt{x} \; dx \le \sum_0^N \sqrt x \le \int_0^N  \sqrt{x+1} \; dx$$&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-11T13:05:33.530" />
  <row Id="6510" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:07:48.627" Score="1" ViewCount="0" Body="&lt;p&gt;If $F$ is a field, then $F[X]$ is a PID.&#xA;If $D$ is a UFD, then so is $D[X]$.&#xA;These are broad and powerful theorems.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-11T13:07:48.627" />
  <row Id="6511" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:21:44.340" Score="0" ViewCount="0" Body="&lt;p&gt;The question is not very well defined. For example, the nullstellensatz shows that the points in $\mathbb{C}^n$ (or any algebraically closed field) are in correspondence with the maximal ideals in $\mathbb{C}[x_1,...,x_n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed this correspondence between geometry and algebra (notice that we can talk about generalized points: prime ideals in $\mathbb{C}[x_1,...,x_n]$) is what's in the &quot;core&quot; of modern algebra. It's at the very least the main reason that we look at rings (arguably the primary reason we look at fields is Galois Theory; and Group Theory has intrinsic value).&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2010-10-11T13:21:44.340" />
  <row Id="6512" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T13:26:18.930" Score="0" ViewCount="0" Body="&lt;p&gt;Following on from leonbloy's answer, one can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler--Maclaurin summation formula&lt;/a&gt; to obtain&#xA;a rather precise approximation of the sum via an integral.  In particular, the error&#xA;term in that formula involves higher derivatives of the function $f(x)$.  Since &#xA;the higher derivatives of your function tend to zero as $x \to \infty$, you should be&#xA;able to obtain reasonably precise approximations by playing around a bit.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-11T13:26:18.930" CommentCount="1" />
  <row Id="6513" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:30:33.143" Score="0" ViewCount="0" Body="&lt;p&gt;Well,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Wedderburn's theorem&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29&quot; rel=&quot;nofollow&quot;&gt;Gauss Lemma,&lt;/a&gt; and Eisentein's Criteria are good ones.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-11T13:37:04.897" LastActivityDate="2010-10-11T13:37:04.897" />
  <row Id="6514" PostTypeId="2" ParentId="6508" CreationDate="2010-10-11T13:31:31.367" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Use the following: If $|A|\leq c/2$ and $|B|\leq c/2$, then $|A+B|\leq c$. It might be useful to negate this statement and then use the subadditivity of the probability measure.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-11T13:31:31.367" />
  <row Id="6516" PostTypeId="2" ParentId="6507" CreationDate="2010-10-11T13:39:38.853" Score="1" ViewCount="0" Body="&lt;p&gt;Rings of &quot;functions&quot; (vs. numbers) are heavily studied - whether formally or truly functionally. Even in a first algebra course you'll find significant theorems on polynomials and rational functions, rings of linear transformations on vector spaces (matrices)  and, perhaps, some nonlinear analogs in commutative algebra (e.g. via Grobner bases). Combinatorics makes heavy use of generating functions (formal power series). Algebraicized analysis employs differential/difference algebra to model elementary functions and their asymptotics (&lt;a href=&quot;http://arxiv.org/abs/0801.4877&quot; rel=&quot;nofollow&quot;&gt;transseries&lt;/a&gt;), e.g. Liouville-Risch theory of integration/summation in finite terms, Rosenlicht's Hardy fields. Differential Galois theory studies solutions of differential equations and similaly for the difference analog (recurrences).  Entire books have been written on rings of continuous functions (e.g. Gillman and Jerison). Abe Robinson's NSA (nonstandard analysis) is another algebraic approach to analysis.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T02:32:36.800" LastActivityDate="2010-10-18T02:32:36.800" />
  <row Id="6517" PostTypeId="2" ParentId="6244" CreationDate="2010-10-11T13:48:07.590" Score="6" ViewCount="0" Body="&lt;p&gt;No transcendental numbers are needed for this question.  Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension.  Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $\sqrt{2}, \sqrt{\sqrt{2}}, \dots, = \cup_{n&gt;0} 2^{2^{-n}} $ is linearly independent over $Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i &amp;lt; n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square roots of the prime numbers are linearly independent over $Q$.  (Proof: this is immediate given the ability to extend the function &quot;number of powers of $p$ dividing $x$&quot; from the rational numbers to algebraic numbers. $\sqrt{p}$ is &quot;divisible by $p^{1/2}$&quot; while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $Q$ unramified at $p$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case.  Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent.  Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-11T14:09:27.680" LastActivityDate="2010-10-11T14:09:27.680" CommentCount="16" />
  <row Id="6518" PostTypeId="2" ParentId="2037" CreationDate="2010-10-11T14:03:30.083" Score="1" ViewCount="0" Body="&lt;p&gt;This isn't a complete answer to your question, but at least it is a closed form. For 5 normal (6-sided), dice, the exact probability of achieving a yahtzee, following the optimum strategy, given $n$ rolls is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1+\frac{53}{13} \left(\frac{5}{6}\right)^{2n+1}+\frac{11 \cdot 5^n}{13 \cdot 2^{n+5} \cdot 3^{3n+1}}  -\frac{5^n}{8 \cdot 3^{2n-2}}  -\frac{7 \cdot 5^{n+1}}{ 2^{n+5} 3^{n-1}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, setting $n=3$ gives $\frac{347897}{7558272}$.&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-10-11T14:03:30.083" />
  <row Id="6519" PostTypeId="1" CreationDate="2010-10-11T14:08:35.300" Score="13" ViewCount="177" Body="&lt;p&gt;Let $M$ be an $n \times n$ matrix whose elements are random reals in [0,1].&#xA;Two questions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; What is the growth rate of the magnitude of the elements of $M^k$ as a function of $k$?  It is definitely&#xA;exponential, but maybe the exponent is known?&lt;/li&gt;&#xA;&lt;li&gt; Is it the case that eventually one element of $M^k$ dominates, as $k \rightarrow \infty$?&#xA;I have some ambiguous experimental evidence that this is the case, but because&#xA;of the exponential growth, exact computation is difficult, rendering my &quot;evidence&quot; tenuous&#xA;at best and perhaps worthless.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One can ask the same question for matrices whose elements are random reals in [-1,1],&#xA;or random 0's and 1's, or random choices among $\lbrace -1, 0, 1\rbrace$, ...&#xA;These question have likely been studied.  Thanks for pointers and/or ideas! &lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-11T19:51:33.727" Title="Powers of random matrices" Tags="&lt;probability&gt;&lt;reference-request&gt;&lt;matrices&gt;&lt;random&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="6520" PostTypeId="2" ParentId="6491" CreationDate="2010-10-11T14:35:37.783" Score="3" ViewCount="0" Body="&lt;p&gt;An approximation to $\displaystyle \sum_{k=1}^{n} \sqrt{k}$ can be found here (on this very site): &lt;a href=&quot;http://math.stackexchange.com/questions/5676/how-closely-can-we-estimate-sum-i0n-sqrti/5728#5728&quot;&gt;http://math.stackexchange.com/questions/5676/how-closely-can-we-estimate-sum-i0n-sqrti/5728#5728&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was shown that $\displaystyle \frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2} -\frac{2}{3} &amp;lt; \sum_{k=1}^{n} \sqrt{k} &amp;lt; \frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle S = \sum_{k=0}^{n} f(x)/2$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} \sqrt{n} = \frac{S - 10n(n+1)}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the approximation above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2} = \frac{S - 10n(n+1)}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fourth degree equation in $\sqrt{n}$ which can be solved exactly (closed formula) in terms of $S$ and would give you a value of $n$ which is close to $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that taking the integer part of the square of an appropriate root will be sufficient to give $n$ or $n-1$ (and so a formula might exist, after all!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not looking for a formula, but a procedure, you can always try binary search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T14:59:40.443" LastActivityDate="2010-10-11T14:59:40.443" CommentCount="1" />
  <row Id="6521" PostTypeId="2" ParentId="6502" CreationDate="2010-10-11T14:57:15.267" Score="2" ViewCount="0" Body="&lt;p&gt;$a$ is a kind of 'tolerance' parameter on $X$. Imagine an adversary of yours setting this parameter at some level, as small as they wish. They also set an upper bound, say $\epsilon$, on the proportion of possible sample paths that may violate the tolerance condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You on the other hand are allowed to choose a time window around the time $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuity then requires that regardless of how tight the two tolerances are set by your adversary, you can always choose a time window narrow enough that within that time window the values of $X$ remain within a distance $a$ of $X(s)$ for no less than $(1-\epsilon)$ proportion of the possible sample paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part here is that you are not required to commit to the sample paths which will violate the tolerance. Depending on the choice of $a$ and $\epsilon$ you may allow different sample paths to 'fail'. All that is required is that the proportion of paths that fail be less than $\epsilon$. This makes your task easier. Therefore this type of continuity is know as &lt;em&gt;weak&lt;/em&gt; continuity. On the other hand we could have required that almost all sample paths be continuous in the ordinary deterministic sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\left(\lim_{t\to s} X(t)=X(s)\right)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which would have been a stronger condition to fulfil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Rasmus points out, these different notions of continuity are a direct counterpart of the more fundamental notions of convergence of random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;['Proportion' above should be interpreted in terms of probability mass.]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-11T15:35:24.707" LastActivityDate="2010-10-11T15:35:24.707" CommentCount="1" />
  <row Id="6522" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T15:14:10.343" Score="3" ViewCount="0" Body="&lt;p&gt;Powers of a matrix are more easily calculated by first diagonalizing it. Let $P, D$ be matrizes, $D$ being a diagonal matrix with $M = PDP^{-1}$, then $$M^k = PD^{k}P^{-1}.$$ The entries in D are the eigenvalues of $M$, so the entries in $M^k$ are growing exponentially with the rate of the logarithm of the largest eigenvalue of $M$, each entry being a linear combination of these eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to point 2: No entry should dominate the others significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only a partial answer.. maybe you can find something about the distribution of eigenvalues of random matrizes. =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-10-11T15:37:00.253" LastActivityDate="2010-10-11T15:37:00.253" CommentCount="2" />
  <row Id="6523" PostTypeId="2" ParentId="6505" CreationDate="2010-10-11T15:39:34.557" Score="6" ViewCount="0" Body="&lt;p&gt;Since there are contradicting opinions, I recommend assuming that whoever is reading your work assumes a different precedence order than you do, and using lots of parentheses. &lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-11T15:39:34.557" CommentCount="1" />
  <row Id="6524" PostTypeId="2" ParentId="6505" CreationDate="2010-10-11T15:43:33.797" Score="6" ViewCount="0" Body="&lt;p&gt;There is no sensible way of preferring one of intersection and union more than the other, since complement switches them.  I don't think you should assume an order at all and you should always use parentheses.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-11T15:43:33.797" />
  <row Id="6525" PostTypeId="2" ParentId="6485" CreationDate="2010-10-11T15:58:19.203" Score="-1" ViewCount="0" Body="&lt;p&gt;Okay. I am pretty sure this problem has zero duality gap. What some other condition can I use?&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2010-10-11T15:58:19.203" CommentCount="1" />
  <row Id="6526" PostTypeId="1" AcceptedAnswerId="6528" CreationDate="2010-10-11T16:08:48.173" Score="3" ViewCount="153" Body="&lt;p&gt;Given an $n &gt;0$ is it possible to partition the set $\mathcal{P} = \{1,2, \cdots, 2n\}$ into $n$ pairs $(a_{i},b_{i})$ such that $a_{i} + b_{i}$ is a prime?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-23T06:45:12.477" LastActivityDate="2010-11-23T06:45:12.477" Title="Partitioning sets such that the sum of 2 elements is Prime" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6528" PostTypeId="2" ParentId="6526" CreationDate="2010-10-11T16:18:48.790" Score="11" ViewCount="0" Body="&lt;p&gt;Yes.  The proof is by strong induction.  The base case is obvious.  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrands_postulate&quot; rel=&quot;nofollow&quot;&gt;Bertrand's postulate&lt;/a&gt; there exists a prime $p$ between $2n+1$ and $4n$, so pick the pairs $\{ p-2n, 2n \}, \{ p-2n+1, 2n-1 \}, ...$ and so forth.  Now it remains to pair up the numbers $\{ 1, 2, ... p-2n-1 \}$, which is possible by the inductive hypothesis.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-11T16:25:14.707" LastActivityDate="2010-10-11T16:25:14.707" />
  <row Id="6529" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T16:33:36.780" Score="2" ViewCount="0" Body="&lt;p&gt;The following analysis treats only the particular case of symmetric matrices $M$, which can be diagonalized by&#xA;an orthogonal transformation. The key observation is that the diagonalizing&#xA;matrix $P$ and the eigenvalue matrix $D$ are statistically independent. The diagonalizing matrix&#xA;elements are $O(n)$ Haar distributed. Now, it is easy to see that every matrix element is a scalar product of&#xA;two $O(n)$ Haar distributed n-vectors weighted by the k-th power of the eigenvalues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At large values of k, the maximal eigenvalue dominates all eigenvalues and the contribution from the diagonalizing matrix is fixed. &#xA;Thus the growth rate of all matrix elements is the natural logarithm of the maximal eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the contribution of the maximal eigenvalue to the $(i,j)$ element is given by $P_{im} D_{mm}^k  P_{jm}$&#xA;($D_{mm}$ is the maximal eigenvalue). Thus the matrix element for which the product $P_{im} P_{jm}$ is maximal dominates all matrix elements of the power matrix.&#xA;This element will be most likely on the diagonal because it is multiplied by the square of a Haar distributed element having a non-zero mean.&#xA;The off-diagonal elements are multiplied by the product of two different Haar distributed elements thus having a zero mean.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-11T16:33:36.780" CommentCount="2" />
  <row Id="6530" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T16:35:03.340" Score="5" ViewCount="0" Body="&lt;p&gt;The best results have been obtained for random matrices with Normally distributed entries, but some of them--like &lt;a href=&quot;http://mathworld.wolfram.com/WignersSemicircleLaw.html&quot; rel=&quot;nofollow&quot;&gt;Wigner's Circular Law&lt;/a&gt;--extend to uniformly distributed entries.  (Independence is crucial.)  Wigner's law applies only to symmetric random matrices.  &lt;a href=&quot;http://mathworld.wolfram.com/GirkosCircularLaw.html&quot; rel=&quot;nofollow&quot;&gt;Girko's Circular Law&lt;/a&gt; holds without the symmetry condition, but afaik requires Normally distributed entries.  It says that for largish $n$ the eigenvalues are approximately uniformly distributed on a disk.  For smaller $n$, before these asymptotics are reached, there is a preference for real eigenvalues extending beyond the disk.  At any rate, these asymptotics will immediately give you the distribution of eigenvalues of positive integral powers of such matrices, especially when you consider that the probability all eigenvalues are distinct (and therefore the matrix is diagonalizable over $\mathbb{C}$) is one.  For example, the largest eigenvalue of $M^k$ will be approximately $n^{k/2} \left( 1 - \frac{1}{2} (3 \pi / (2 n) ) ^ {2/3} \right) ^ k$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-11T16:35:03.340" CommentCount="2" />
  <row Id="6531" PostTypeId="1" CreationDate="2010-10-11T16:39:19.573" Score="1" ViewCount="97" Body="&lt;p&gt;Given n segments of line (into the X axis) with coordinates [li; ri]. You are to choose the&#xA;minimum number of segments that cover the segment [0;M]. All segment are within [0,M].&lt;/p&gt;&#xA;" OwnerUserId="1997" LastActivityDate="2010-10-12T21:09:38.420" Title="line segment covering problem" Tags="&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6532" PostTypeId="1" CreationDate="2010-10-11T16:44:21.370" Score="2" ViewCount="53" Body="&lt;p&gt;Let $k$ be a field, $V$ a finite-dimensional $k$-vectorspace and $M \in End(V)$. How can I determine $Z$, the centralizer of $M \otimes M$ in $End(V) \otimes End(V)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if &#xA;$$M=[[1,0],[0,2]],$$ then $M$ is 6-dimensional, consisting of block matrices of shape 1,2,1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was confused at first, because this seems to be a contradiction to the fact that the centralizer of a subalgebra of the form $A \otimes B$ is just the tensor product of the centralizers of $A$ and $B$; but here we are considering only the element $M \otimes M$, not $A \otimes A$, where $A$ is the subalgebra generated by $M$.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastActivityDate="2010-10-11T18:18:45.940" Title="Centralizer of $M \otimes M$" Tags="&lt;algebra&gt;&lt;tensors&gt;" AnswerCount="1" />
  <row Id="6533" PostTypeId="2" ParentId="4783" CreationDate="2010-10-11T16:45:57.253" Score="5" ViewCount="0" Body="&lt;p&gt;This does not exactly answer the question (which was answered very nicely by Qiaochu), but I am adding this as a curiosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write $\displaystyle \sum_{k=0}^{n} {n \choose k}^{-1}$ as an integral (which can be generalized to other expressions involving the reciprocals of the binomial coefficients).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the following beta integral identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n \choose k}^{-1} = (n+1)\int_{0}^{1} t^{n-k}(1-t)^k \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a nice application of Beta integrals, see this answer by Robin Chapman: &lt;a href=&quot;http://math.stackexchange.com/questions/5058/formula-for-the-harmonic-series-h-n-sum-k1n-1-k-due-to-gregorio-fontana/5234#5234&quot;&gt;Formula for the harmonic series due to Gregorio Fontana.&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} {n \choose k}^{-1} = (n+1) \sum_{k=0}^{n} \int_{0}^{1} t^{n-k}(1-t)^k \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = (n+1)\int_{0}^{1} \sum_{k=0}^{n} t^{n-k}(1-t)^k \ dt$$&#xA;$$ = (n+1)\int_{0}^{1} \frac{t^{n+1} - (1-t)^{n+1}}{2t-1} \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should be able to use the analytical tools available to investigate the properties of this integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, at least two different proofs that the limit is $2$ can be found here: &lt;a href=&quot;http://math.stackexchange.com/questions/6544/different-proofs-of-lim-n-rightarrow-infty-n-int-01-xn-1-xn&quot;&gt;http://math.stackexchange.com/questions/6544/different-proofs-of-lim-n-rightarrow-infty-n-int-01-xn-1-xn&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a different use of this integral formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\displaystyle 2t-1 = y$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} {n \choose k}^{-1} = \frac{n+1}{2^{n+2}} \int_{-1}^{1} \frac{(1+y)^{n+1} - (1-y)^{n+1}}{y} \ dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding out the right side gives us this nice looking formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} {n \choose k}^{-1} = \frac{n+1}{2^n}(\frac{{n+1 \choose 1}}{1} + \frac{{n+1 \choose 3}}{3} + \frac{{n+1 \choose 5}}{5} + \dots)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-12T16:13:56.180" LastActivityDate="2010-10-12T16:13:56.180" />
  <row Id="6534" PostTypeId="2" ParentId="6531" CreationDate="2010-10-11T16:50:36.197" Score="1" ViewCount="0" Body="&lt;p&gt;I think the greedy algorithm will work.  You have to cover 0, so search all segments with a left end of zero and take the longest.  Now search all segments with a left end that you have already covered and take the one with the greatest right end.  Continue until you hit M.  At each stage you have a minimal solution for the range so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If overlap (see the comment below) is not allowed, you need a tree structure.  Make a list of all the segments that have left end zero.  For each one, attach all segments you can, that have the same left end as the existing right end.  If you want, you can prune out paths that have as many or more segments than some other path with the same right endpoint.  Continue to the end.  Then search over all the paths for ones that reach M and take the shortest.  I wouldn't call this a greedy algorithm (as called out in the Math Overflow version) but if overlap is not allowed we can't be sure that any given path will reach M.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-12T21:09:38.420" LastActivityDate="2010-10-12T21:09:38.420" CommentCount="4" />
  <row Id="6535" PostTypeId="1" CreationDate="2010-10-11T16:52:18.427" Score="1" ViewCount="73" Body="&lt;p&gt;So i am reading up on some interesting network topologies for a CS course and what i am reading is $r$-dimensional meshes of tree's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $r$-dimensional mesh of trees is formed by adding complete binary tree's to an $N$-sided $r$-dimensional grid where the grid vertices act as the leafs of each binary tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/s2p0C.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This book says this about the graph $G = (V, E)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|V| = (r+1)N^r - rN^{r-1}$&lt;br /&gt;&#xA;$|E| = 2rN^r - 2rN^{r-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the diameter of this graph (the longest path) is $2rlogN$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they come about getting this result? The book does not explain it at all and i do not see how they calculated it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book i am reading is: Introduction to Parallel Algorithms and Architectures: Arrays, Trees, Hypercubes is any of you are interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br /&gt;&#xA;I have tried to derive this myself by first finding $|V|$ when $r=2$ which would make it a 2-dimensional mesh of trees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this i know there are $2n$ complete binary tree's. Each binary tree have $2^{n-1}-1$ vertices. So this means $2n * 2 ^{n-1} -1$ vetices but we are double counting some because there is a binary tree for every row and column. Therefore we can minus out all the leaves of the column binary trees and they will be accounted in the row binary tree's. So that gives us $n * 2^{n-2}$ vertices to minus from the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n * (2^{n-1} - 1) - n2^{n-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this can i just times it by $n$ for 3 dimensions and $n^2$ for 4 dimensions? Then up to $n^{r-2}$ to find the total in $r$-dimensions? Was i calculating this correct for $r=2$? I read somewhere else that when $r=2$ $|V| = 3n^2 - 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Edit&lt;/strong&gt;&lt;br /&gt;&#xA;Maybe i should use the fact that i know the leafs of each complete binary tree have $n$ vertices which means each binary tree is made out of $n + (n-1)$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $2n$ complete binary tree's which gives me: $2n(n + (n - 1))$ total vertices which contains a few we counted twice. A total of $n^2$ we counted twice. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2n(n + (n-1)) - n^2 = 2n^2 + 2n^2 - 2n - n^2 = 3n^2 - 2n$&lt;br /&gt;&#xA;That looks right, doesn't it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now how would I apply this for 3 dimensions, 4, 5 ect.. all the way to $r$-dimensions?&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T21:02:01.173" LastActivityDate="2010-11-10T21:44:17.957" Title="$r$-dimensional Meshes of Trees" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6536" PostTypeId="2" ParentId="4406" CreationDate="2010-10-11T17:16:40.700" Score="1" ViewCount="0" Body="&lt;p&gt;Morteza, I'd guess you'll get different answers from everyone you ask.  Here is the way I see it (which is, by necessity, somewhat circular):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inference&lt;/strong&gt;, in the narrowest sense, is a single step in a deductive chain.  If I know $P\rightarrow Q$ and $\neg Q$, I can infer $\neg P$ from modus tollens (which itself is sometimes called a &quot;rule of inference&quot;).  More loosely, we can call any conclusion from premises an inference, even if it's not properly deductive (i.e. I look outside, see a clear sky, and infer that it's not raining).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning&lt;/strong&gt; is the mental process of logic -- what goes on inside my head when I use deduction.  Though a lot of people would probably use it as a synonym of &quot;deduction,&quot; I'd say the differences are far more important.  When you reason, you skip steps, explore multiple pathways, and use your intuition, which are all things unavailable to, say, a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deduction&lt;/strong&gt; is the formal process of logic, and an inference is deductive when it follows from an axiom or logical rule.  This is the makeup of most mathematical proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction&lt;/strong&gt; has two meanings.  The first is some sort of mathematical induction (strong, weak, or transfinite), all of which are based on the idea of an infinite deductive chain that can be collapsed into a few steps.  For instance, in normal (weak) induction, you prove the proposition $P(0)$, and you prove that $\forall n:P(n)\rightarrow P(n+1)$.  Modus ponens then gives you $P(1)$, then $P(2)$ using $P(1)$, then $P(3)$ using $P(2)$, and so forth.  In all but the most intuitionist systems of arithmetic/set theory, induction is either provable or taken as an axiom, so its use is properly a deductive inference!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is another, more abstract meaning of induction that contrasts properly with deduction.  The logician Charles Sanders Peirce had an analogy with bags of beans that I find compelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deduction&lt;/strong&gt; has you applying a generalization to a specific case to get a result.  So you know that all the beans in my bag are white (the generalization), and you take a bean from my bag (the case): then that bean must be white (the result).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction&lt;/strong&gt; has you taking a case and a result and generalizing them.  You take a bean from my bag (the case), and you see it is white (the result): therefore, all the beans in my bag are white (generalization).  In this rudimentary form, induction looks like guesswork, and nobody would trust this logic.  &lt;em&gt;But&lt;/em&gt; if you took five or ten beans, and they were all white, your confidence in the generalization would increase.  In fact, everybody uses this form of induction many times a day: you expect the sun to rise, your bed to be in the same place, your food to cook properly, etc. just because they've all happened many times before.  This kind of inference is not deductive at all, yet we still trust it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abduction&lt;/strong&gt; was the third of Peirce's forms of inference.  It is the most suspect of the three, and is best thought of as a sort of &quot;reverse implication.&quot;  The idea is to take a generalization and a result and observe that a case is probably true because it implies the result.  So you have a white bean (the result), and you know that all the beans in my bag are white (the generalization): thus, this bean must be from my bag, for if it were, it would have to be white.  This seems like the sort of reasoning you see a lot on detective shows.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-11T17:16:40.700" />
  <row Id="6537" PostTypeId="1" AcceptedAnswerId="6542" CreationDate="2010-10-11T17:50:17.910" Score="0" ViewCount="128" Body="&lt;p&gt;I need help understanding how to derivate this function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = x^{\arctan(x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="713" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-11T17:52:38.287" LastActivityDate="2010-10-11T18:14:32.160" CommunityOwnedDate="2010-10-11T17:50:17.910" Title="Finding derivative of f(x) = x^(arctan x)" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6538" PostTypeId="1" CreationDate="2010-10-11T17:52:51.900" Score="0" ViewCount="154" Body="&lt;p&gt;It is clear that a finte group of order 105 is not simple since it contains a normal Sylow 7-subgroup or a normal Sylow 5-subgroup. I wonder if it is possible that there exists a Sylow 7-subgroup and a Sylow 5-subgroup ?&lt;/p&gt;&#xA;" OwnerUserId="2438" LastActivityDate="2010-10-18T02:06:19.707" Title="Group of order 105" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6540" PostTypeId="2" ParentId="6494" CreationDate="2010-10-11T17:59:18.483" Score="2" ViewCount="0" Body="&lt;p&gt;My previous answer is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $f(x) = x$ ($f: \mathbb{R} \to \mathbb{R}$) where the domain has the Borel $\sigma$-algebra and the range the trivial $\sigma$-algebra. Then $f$ is clearly measurable but $f$ maps for example $[0,1]$ to $[0,1]$ which is not measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do you want both $\sigma$-algebra's to be Borel? In that case I should modify my first answer (which was wrong because not every subset of a measure zero set is Borel measurable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea would be the following (based on the one below me): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g:[0,1] \to \mathbb{R}$ be the Cantor function. Extend it to all of $\mathbb{R}$ by defining $g(x) = 0$ for $x &amp;lt; 0$ and $g(x) = 1$. Now define $h(x) = x + g(x)$ (this is standard), this function has a lot of nice properties like: $h$ is a bijection and $h(C) = 1$ where $C$ is the Cantor set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that a set of positive measure contains a non-measurable subset. So let $A$ be an non-measurable subset of $h(C)$. So $M = h^{-1}(A)$ is a subset of the Cantor set and has thus zero Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So define $u := h^{-1}$. This function is measurable (Lebesgue/Borel). But $u^{-1}(M) = A$ is non-measurable.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-11T18:46:38.530" LastActivityDate="2010-10-11T18:46:38.530" CommentCount="2" />
  <row Id="6541" PostTypeId="2" ParentId="6537" CreationDate="2010-10-11T18:00:53.413" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \  g^{\:h}\ =\ e^{h\: \log(g)}\:.\ $ Or, take logs, cf. &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_derivative&quot; rel=&quot;nofollow&quot;&gt;logarithmic derivative,&lt;/a&gt; and my &lt;a href=&quot;http://math.stackexchange.com/questions/3773&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-11T18:07:16.773" LastActivityDate="2010-10-11T18:07:16.773" CommunityOwnedDate="2010-10-11T18:00:53.413" CommentCount="2" />
  <row Id="6542" PostTypeId="2" ParentId="6537" CreationDate="2010-10-11T18:03:28.083" Score="3" ViewCount="0" Body="&lt;p&gt;Let $f(x) = x^{\arctan{x}}$ then $\log{f(x)} = \arctan{x} \cdot \log x$. Therefore $$\frac{1}{f(x)} \times f'(x) = \frac{d}{dx} \Bigl[ \arctan{x} \cdot \log x \Bigr] \Longrightarrow f'(x) = f(x) \times  \frac{d}{dx} \Bigl[ \arctan{x} \cdot \log x \Bigr]$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-11T18:14:32.160" LastActivityDate="2010-10-11T18:14:32.160" CommunityOwnedDate="2010-10-11T18:03:28.083" />
  <row Id="6543" PostTypeId="2" ParentId="6482" CreationDate="2010-10-11T18:07:06.367" Score="3" ViewCount="0" Body="&lt;p&gt;This is a rejoinder to Arturo's comment that if the two-variable limit&#xA;exists then both iterated limits exist.&#xA;Define&#xA;$$f(x,y)=\left\{\begin{array}{cl}&#xA;(x^2+y^2)\sin(1/x+1/y)&amp;amp;\textrm{if $xy\ne0$,}\\&#xA;0&amp;amp;\textrm{if $xy=0$.}&#xA;\end{array}\right.$$&#xA;Then certainly $f$ is continuous at $(0,0)$; that is&#xA;$$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$&#xA;But for any nonzero $x$,&#xA;$$\lim_{y\to0}f(x,y)$$&#xA;doesn't exist (the function is wildly oscillatory for $y$ near zero). So&#xA;$$\lim_{x\to0}\left(\lim_{y\to0}f(x,y)\right)$$&#xA;is meaningless.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-11T18:07:06.367" CommentCount="1" />
  <row Id="6544" PostTypeId="1" AcceptedAnswerId="6603" CreationDate="2010-10-11T18:17:18.727" Score="11" ViewCount="314" Body="&lt;p&gt;It can be shown that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n \int_{0}^{1} \frac{x^n - (1-x)^n}{2x-1} \ dx = \sum_{k=0}^{n-1} {n-1 \choose k}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For instance see my answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/6533#6533&quot;&gt;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/6533#6533&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be shown that $$\lim_{n \rightarrow \infty} \  \sum_{k=0}^{n-1} {n-1 \choose k}^{-1} = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For instance see Qiaochu's answer: &lt;a href=&quot;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/4784#4784&quot;&gt;http://math.stackexchange.com/questions/4783/finding-displaystyle-lim-n-to-infty-sum-limits-k0n-n-choose-k/4784#4784&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining those two shows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{n \rightarrow \infty} \ n \int_{0}^{1} \frac{x^n - (1-x)^n}{2x-1} \ dx  = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a different, (preferably analytic) proof of this fact? Please do feel free to add a proof which is not analytic.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-05T08:39:55.590" Title="Different proofs of $\lim_{n \rightarrow \infty} n \int_{0}^{1} (x^n - (1-x)^n)/(2x-1) = 2$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6545" PostTypeId="2" ParentId="6532" CreationDate="2010-10-11T18:18:45.940" Score="1" ViewCount="0" Body="&lt;p&gt;This looks like it can get complicated. Generically, at least over an algebraically&#xA;closed field, a matrix $M$ will have distinct eigenvalues $m_1,\ldots,m_n$,&#xA;and generically $M\otimes M$ will have distinct eigenvalues $m_1^2,\ldots,m_n^2$&#xA;with multiplicity one, and $m_1m_2,m_1m_3,\ldots,m_{n-1}m_n$ with&#xA;multiplicity two. Thus the centralizer will have dimension $n+4{n\choose 2}=2n^2-n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are many degenerate cases: for instance if $M$ has eigenvalues&#xA;$1,a,\ldots,a^{n-1}$ then $M\otimes M$ will have eigenvalues&#xA;$1,a,\ldots,a^{2n-2}$ with multiplicities $1,2,\ldots n-1,n,n-1,\ldots,1$.&#xA;Things can get more complicated still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $M$ might have non-trivial Jordan blocks, and then the real fun starts!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-11T18:18:45.940" />
  <row Id="6546" PostTypeId="1" AcceptedAnswerId="11869" CreationDate="2010-10-11T18:34:51.803" Score="1" ViewCount="129" Body="&lt;p&gt;Motivating factor behind this question is the comments which are given for this question asked some hours ago &lt;a href=&quot;http://math.stackexchange.com/questions/6538/group-of-order-105&quot;&gt;http://math.stackexchange.com/questions/6538/group-of-order-105&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given a group $G$ of order $n$ what are the different methods for  constructing a non abelian group of order $n$. Well, i have seen a method which Herstein uses in his book. He takes a cyclic group of given order and defines an Automorphism on the cyclic group and then places some restrictions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know, whether there are anymore methods for obtaining a Non Abelian group of a given order.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-25T19:24:01.080" Title="Constructing Non Abelian Groups of Given order" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6547" PostTypeId="2" ParentId="6535" CreationDate="2010-10-11T18:38:46.660" Score="1" ViewCount="0" Body="&lt;p&gt;You seem to have mixed up two different formulas for the number of nodes in a complete binary tree. If the &lt;em&gt;height&lt;/em&gt; of the tree is $h$, it has $2^h-1$ nodes. But in this case, $N$ is not the height but the number of &lt;em&gt;leaves&lt;/em&gt;, so the tree has $N-1$ internal nodes, and the total number of nodes is $2n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you get, for $r = 2$, a total of $2N\cdot(2N-1)$ vertices in each tree, less the overcounting of shared leaves, $N\cdot N$, equalling $|V| = 2N(2N-1) - N^2 = 3N^2 - 2N$, which also agrees with the formula $|V| = (r+1)N^r - rN^{r-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way to do it would be to notice that the shared nodes are simply the nodes in the grid, and count them once: $N^r$. The internal nodes of the trees are not shared, there are $N-1$ of them for each tree, and there are $rN^{r-1}$ trees (see below). Add this up and you'll get your solution without worrying about overcounting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of trees is $rN^{r-1}$ because along any direction, a tree connects $N$ nodes, so you need $N^{r-1}$ of them to cover the grid. And the graph contains $r$ copies of this arrangement, one each for the $r$ directions along the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar approach should work for counting the number of edges in the graph -- count the edges in the trees and the edges in the grid separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the diameter, first think about what the shortest path between two nodes is in a traditional mesh (without trees), then consider how the additional tree structure shortens this path. Then it should be clear which pair of nodes remains the farthest apart; their distance is the diameter of the graph.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-11T19:11:16.643" LastActivityDate="2010-10-11T19:11:16.643" />
  <row Id="6548" PostTypeId="2" ParentId="6538" CreationDate="2010-10-11T18:43:37.097" Score="3" ViewCount="0" Body="&lt;p&gt;Of &lt;em&gt;course&lt;/em&gt; there exist Sylow $7$-subgroups and Sylow $5$-subgroups: Sylow's Theorems tell you that the group has Sylow $p$-subgroups for every $p$ (with the Sylow $p$-subgroup being trivial if $p$ does not divide $105$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant to ask &quot;Can the group have &lt;em&gt;normal&lt;/em&gt; Sylow $5$- &lt;em&gt;and&lt;/em&gt; Sylow $7$ subgroup?&quot; then the answer is &quot;yes&quot;: the (unique) abelian group of order $105$ will do (since $105=3\times 5\times 7$ is square free, the only abelian group of order $105$ is the cyclic one). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can &lt;em&gt;always&lt;/em&gt; find a finite group of order $n$ in which all Sylow subgroups are normal: any abelian group of order $n$, and there is always at least one, will do. If any of the primes in the factorization of $n$ is raised to a degree greater than $2$, then you can also find a &lt;em&gt;nonabelian&lt;/em&gt; group in which all the Sylow subgroups are normal, just by taking the direct product of $p$-groups of the correct size, and picking a nonabelian one when the prime is raised to a sufficiently large degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in this case, you can get a nonabelian group in which the Sylow $7$- and the Sylow $5$-subgroups are normal by first constructing a nonabelian group of order $21$ (there are no nonabelian groups of orders $15$ or $35$) which has a normal subgroup of order $7$, and then taking the direct product with with $C_5$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-11T18:43:37.097" />
  <row Id="6549" PostTypeId="1" AcceptedAnswerId="6550" CreationDate="2010-10-11T19:04:38.237" Score="0" ViewCount="40" Body="&lt;p&gt;I have multiple objects in my lab which are fully described in two coordinate systems. Both systems are Cartesian and in meters. If they had a common origin but were rotated relative to each other I could easily reconcile the two. Additionally, if it was just a displacement it would be even easier. However the two are offset in rotation and displacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is find how to convert between the two of them. I would think I would just need the coordinates of the same three objects in the two systems. I know how to do this for two coordinate systems that are rotated OR displaced, but I can't figure out how to do it in a scenario with both rotation and displacement. Anyone got a formula for me or (even better) a link describing how it is done?&lt;/p&gt;&#xA;" OwnerUserId="2440" LastActivityDate="2010-10-11T19:46:07.670" Title="Coordinate substitution between two cartesian systems" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6550" PostTypeId="2" ParentId="6549" CreationDate="2010-10-11T19:46:07.670" Score="0" ViewCount="0" Body="&lt;p&gt;Look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;Affine transformation&lt;/a&gt;&#xA;Basically if you know how to do rotations and translations you can chain them together.  If you have the coordinates of a few objects you can solve the simultaneous equations to get the matrix.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-11T19:46:07.670" />
  <row Id="6551" PostTypeId="1" AcceptedAnswerId="6624" CreationDate="2010-10-11T19:48:45.823" Score="3" ViewCount="111" Body="&lt;p&gt;I'm looking to brush up on SAT math.  (That's basic algebra, probability and trigonometry.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in books and websites.&lt;/p&gt;&#xA;" OwnerUserId="2414" LastEditorUserId="2414" LastEditorDisplayName="" LastEditDate="2010-10-11T23:25:54.640" LastActivityDate="2010-10-19T04:25:12.650" CommunityOwnedDate="2010-10-11T21:14:15.947" Title="Good SAT learning resources?" Tags="&lt;probability&gt;&lt;algebra&gt;&lt;trigonometry&gt;&lt;learning&gt;&lt;self-learning&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" />
  <row Id="6552" PostTypeId="1" AcceptedAnswerId="6561" CreationDate="2010-10-11T19:50:36.893" Score="2" ViewCount="39" Body="&lt;p&gt;I've been trying to find a way to do this: given a point $P(\alpha,\beta)$, a function $f(x)$, and an angle $\theta$, find the area of the sector determined by extending a horizontal line from $P$ to $f$, and then another one from $P$ to $f$ at the angle $\theta$ from the first line. If you look at this picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/ekTqY.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it's a little more clear what I'm talking about. I want to find the area of that sector $S$, which is determined by that triangle $R_1$ and the other region $R_2$. Let the value of $x$ where the horizontal line from $P$ hits $f$ be $q$ (that is, $q=f^{-1}(\beta)$). Now, the line $\overline{PA}$ has the equation $y=\tan\theta(x-\alpha)+\beta$. The intersection of that and $f$, the point $A$ (the first intersection point), is the beginning of the interval determining $R_2$; I'll call the root $p$ of $f(x)=\tan\theta(x-\alpha)+\beta$ that value of $x$, so $p=f^{-1}(\tan\theta(x-\alpha)+\beta)$. The area of $R_1$ is simply the area of the triangle: $R_1=\frac{1}{2}(p-\alpha)(f(p)-\beta)$. The area of $R_2$ is the area under the curve from $p$ to $q$, minus the initial height $\beta$: $R_2=\displaystyle\int_p^q (f(x)-\beta) dx$. Therefore, the total area is $$S=\frac{1}{2}\tan\theta(p-\alpha)^2-\beta(q-p)+\displaystyle\int_p^q f(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the way I did it, but there has to be a simpler way; is there one involving polar equations? Is there a general formula when $f$ is non-invertible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If the function $f$ is translated so that $P$ is now the origin, the area is just $\displaystyle\frac{1}{2}\int_0^\theta r^2 d\phi$ where $r(\phi)$ is $f(x)$ in polar form. However, translating $f$ over to where $P$ is on the origin turns it into $f(x+\alpha)-\beta$. Is there a nice way of putting that into polar form?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-10-11T20:58:20.433" LastActivityDate="2010-10-11T20:58:20.433" Title="Area of a sector between a point and a function determined by an angle" Tags="&lt;geometry&gt;&lt;calculus&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6553" PostTypeId="2" ParentId="6519" CreationDate="2010-10-11T19:51:33.727" Score="1" ViewCount="0" Body="&lt;p&gt;I don't have enough reputation to leave a comment, so here goes....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other have observed, the growth rate of $M^k$ is determined by the largest eigenvalue of $M$. I just want to note that for a nonnegative matrix, the largest eigenvalue is always &#xA;between the largest and smallest column sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, in principle, could allow you to obtain upper and lower bounds by trying to bound how big or small the column sums usually get for the random matrix you described. &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-10-11T19:51:33.727" CommentCount="1" />
  <row Id="6554" PostTypeId="2" ParentId="6544" CreationDate="2010-10-11T19:55:02.453" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a sketch of an elementary analytic demonstration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The integrals all exceed $2/n$ when $x$ exceeds 1: subtract $2/n = \int_0^1{(t^{n-1} + (1-t)^{n-1}) \ dt}$ to obtain a positive integrand (using elementary inequalities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The integrals are bounded by a sequence decreasing to the desired limit.  By looking at the Taylor series of the log of the integrand near 1, for example, and assuming $n \ge 3$, you can easily establish an upper bound of $\exp((n-2)(x-1))$ in the interval $(1/2, 1]$, with a similar bound in the interval $[0, 1/2)$ by symmetry.  The estimate (integrating over one of those two intervals) equals $(1 - \exp(2-n)) / (n-2)$ which, when doubled and multiplied by $n$, converges to $2$.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-11T19:55:02.453" CommentCount="1" />
  <row Id="6555" PostTypeId="2" ParentId="6451" CreationDate="2010-10-11T20:15:16.837" Score="1" ViewCount="0" Body="&lt;p&gt;Maybe you're going to tell me that if all you've got is a hammer, all problems look like nails, but here goes my proposal -following the same strategy as in my answers to &lt;a href=&quot;http://math.stackexchange.com/questions/6357/quotient-topologies-and-equivalence-classes/6362#6362&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, take as a model for the projective plane the disk&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D^2 = \left\{ (x,y) \in \mathbb{R}^2\ \vert \ x^2 + y^2 \leq 1   \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and quotient by the equivalence relation among antipodal points on $S^1$: $(x,y) \sim -(x,y) $ if $x^2 + y^2 = 1$.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathbb{RP}^2 = D^2/\sim \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the Moebius strip inside $\mathbb{RP}^2$ as the vertical strip&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M = \left\{\widetilde{(x,y)} \in \mathbb{RP}^2 \ \vert \ -\frac{1}{2} \leq x \leq \frac{1}{2}   \right\}  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's denote by $M'$ the same set of points inside $D^2$, without any quotient at all. That is, ordinary points of $\mathbb{R}^2$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M = M'/\sim \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you want to prove that, when you quotient out the boundary of the Moebius band&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\partial M = \left\{\widetilde{(x,y)} \in \mathbb{RP}^2 \ \vert \ x = -\frac{1}{2} \ \quad \text{or} \quad  x = \frac{1}{2}   \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get the projective plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M/\partial M \ \cong \ \mathbb{RP}^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, the homeomorphism is easily described in some kind of &quot;eliptical&quot; coordinates: for every point $(x,y) \in D^2$ there is one and only one elipse of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x^2 + \frac{y^2}{b^2} = 1  \qquad \text{for} \qquad -1 \leq b \leq 1 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these elipses have a common major axis, namely the segment $-1\leq x \leq 1, y = 0$. We make the abuse of considering $b=0$ too as the degenerate &quot;elipse&quot; $y=0$ and in order to distingish the &quot;positive&quot; $y\geq 0$ and &quot;negative&quot; $y\leq 0$ branches of the elipse, we  put a sign in $b$: the same as $y$. If all this doesn't sound too rigorous to you, it's ok: look at it as just a motivation. Wait for the resulting formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a point inside the disk is determined once you know its $x$-coordinate and the elipse to which it belongs; that is, its $b$-coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our homeomorphism is going to do the following: points inside the Moebius strip  are going to &quot;travel&quot; along the elipse to which they belong. How far? Well, we want all the points in $\partial M$, those with $x=-1/2$ or $x= 1/2$, to reach points $(-1,0)$ or $(1,0)$, depending on the sign of $x$. Hence, we try something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi }: M \longrightarrow \mathbb{RP}^2 \ , \qquad \widetilde{\varphi} \widetilde{(x,b)}  = \widetilde{(2x, b)} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's write a decent $\varphi$ in Cartesian coordinates. We want our $(2x, b)$ to belong to the $b$-elipse all the time, hence we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(2x)^2 + \frac{y^2}{b^2} = 1 \qquad \Longrightarrow \qquad b = \frac{y}{\sqrt{1-(2x)^2}} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we begin with a map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi : M' \longrightarrow D^2 &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{cases}&#xA;\varphi (x,y) = \left( 2x , \frac{y}{\sqrt{1-(2x)^2}}\right) &amp;amp; \text{if} \ x \neq \frac{\pm 1}{2}  \\&#xA;\varphi (\pm 1/2, y)  = (\pm 1,0) &amp;amp; {}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise.&lt;/strong&gt; Check that $\varphi$ is continuous, surjective and compatible with the antipodal equivalence relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\varphi$ induces a well-defined map&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M \longrightarrow \mathbb{RP}^2 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is continuous, because of the universal property of the quotient topology, and surjective. It is &quot;almost&quot; injective too, except for all the points in the boundary of the Moebius band $x=-1/2$ and $x=1/2$ that go to $\widetilde{(-1,0)} = \widetilde{(1,0)}$. So, if we quotient out them, we get a homeomorphism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\widetilde{\varphi} : M/\partial M \longrightarrow \mathbb{RP}^2 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whit the help, as usual, of the GTET (see the link above, taking into account that, despite being a quotient, $\mathbb{RP}^2$ &lt;strong&gt;is&lt;/strong&gt; a Hausdorff space -see Rotman, &quot;An introduction to Algebraic Topology&quot;, theorem 8.4, &lt;a href=&quot;http://books.google.es/books?id=bApVT-6A43QC&amp;amp;pg=PA180&amp;amp;lpg=PA180&amp;amp;dq=hausdorff+%22projective+plane%22&amp;amp;source=bl&amp;amp;ots=Ge6niTS6G2&amp;amp;sig=SKjgSQKf15psgqzerEDH0OdpXhg&amp;amp;hl=ca&amp;amp;ei=wGOzTJaeIo674gbtrPHDBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CDkQ6AEwBg#v=onepage&amp;amp;q=hausdorff%20%22projective%20plane%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Hausdorff&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-11T21:12:28.947" LastActivityDate="2010-10-11T21:12:28.947" />
  <row Id="6556" PostTypeId="1" AcceptedAnswerId="6560" CreationDate="2010-10-11T20:20:10.803" Score="1" ViewCount="121" Body="&lt;p&gt;Can you help me understand how&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g^{-1}(x) = f^{-1}(x + 2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the inverse of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = f(x) - 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot get my head around how the 2 ends up inside $f$ :/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="713" LastActivityDate="2010-10-11T21:30:02.710" CommunityOwnedDate="2010-10-11T20:27:09.447" Title="Find the inverse of g(x) = f(x) - 2" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6557" PostTypeId="1" AcceptedAnswerId="6558" CreationDate="2010-10-11T20:23:15.133" Score="1" ViewCount="112" Body="&lt;p&gt;Consider I have two sets that neither one is a subset of the other (for example, the set of prime numbers and the set of odd integers). Is there a specific notation that combines the meanings of both ⊄ and ⊅ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there a notation to mean that two sets are completely distinct and share no elements in common (for example, the sets of odd and even integers)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;≠ seems very inappropriate, and I'd like to avoid it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context:&lt;/strong&gt; My area is actually Computer Science. I'm using this refine my vocabulary in order to teach my students about &lt;a href=&quot;http://en.wikipedia.org/wiki/Subnetwork&quot; rel=&quot;nofollow&quot;&gt;subnetting&lt;/a&gt; (which is basically about sets and subsets of IP addresses), and to write some exercises for them. I'll probably use the words &lt;em&gt;incomparable&lt;/em&gt; and &lt;em&gt;disjoint&lt;/em&gt; as suggested.&lt;/p&gt;&#xA;" OwnerUserId="2442" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T04:01:37.307" LastActivityDate="2010-11-23T04:01:37.307" Title="Proper notation for distinct sets" Tags="&lt;notation&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="6558" PostTypeId="2" ParentId="6557" CreationDate="2010-10-11T20:29:06.830" Score="3" ViewCount="0" Body="&lt;p&gt;No, there is no special symbol to denote the situation in which neither set is contained in the other. There &lt;em&gt;is&lt;/em&gt; a special word, though: we say the two sets are &lt;strong&gt;incomparable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also no special symbol to denote the fact that two sets are disjoint; we simply write $A\cap B=\emptyset$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-11T20:29:06.830" />
  <row Id="6559" PostTypeId="2" ParentId="6556" CreationDate="2010-10-11T20:29:32.800" Score="3" ViewCount="0" Body="&lt;p&gt;It's simply $\rm\ g = h\: f\ \Rightarrow\ g^{-1} = (h\:f)^{-1} = f^{-1} h^{-1},\ \ h(x) = x-2\:.\ $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can explicitly check that they're inverses, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ g^{-1}(g(x)) = f^{-1}(g(x)+2) = f^{-1}(f(x)-2 + 2) = x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ g(g^{-1}(x)) = f(g^{-1}(x))-2 = f(f^{-1}(x+2))-2  = x$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-11T20:34:44.497" LastActivityDate="2010-10-11T20:34:44.497" CommunityOwnedDate="2010-10-11T20:29:32.800" />
  <row Id="6560" PostTypeId="2" ParentId="6556" CreationDate="2010-10-11T20:34:57.757" Score="3" ViewCount="0" Body="&lt;p&gt;If you want to complement Bill Dubuque's answer with trying to figure out &lt;em&gt;how&lt;/em&gt; you might have come up with the formula, think about what $g(x)=f(x)-2$ is telling you to do: in order to compute $g(x)$, you first compute $f(x)$, and then you subtract $2$. The inverse of this should be something that &quot;unravels&quot; the process you just went to. So the first thing you need to unravel is the &lt;em&gt;last&lt;/em&gt; thing you did when you computed $g(x)$, namely subtracting $2$. So the first thing you are going to do is add $2$ from whatever they give you. What did you did just prior to subtracting $2$ when you computed $g$? You computed $f$. So the next thing you need to do in the &quot;unraveling&quot; process is to undo the computation of $f$, which is achieved by computing $f^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: to invert (unravel) the process of (i) first compute $f$, then (ii) subtract $2$; you will (I) first add $2$; then (II) compute $f^{-1}$. Symbolically: if you are given $x$ and you want to unravel, first you compute $x+2$, then you compute $f^{-1}$ of that. So if $g(x) = f(x)-2$, then $g^{-1}(x) = f^{-1}(x+2)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-11T20:34:57.757" CommunityOwnedDate="2010-10-11T20:34:57.757" />
  <row Id="6561" PostTypeId="2" ParentId="6552" CreationDate="2010-10-11T20:45:59.213" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, it is easier using polar coordinates. If you have a region in the plane determined by two straight lines $OA$ and $OB$ starting in a point $O$ -like those in your picture- and forming angles $\theta_1$ and $\theta_2$ with respect the $x$-axis ($OB$ doesn't need to be horizontal) and you close your region with a curve which polar equation is $r = r(\theta )$, then you apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Change_of_variables_theorem&quot; rel=&quot;nofollow&quot;&gt;change of variables&lt;/a&gt; theorem, and you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A(D) = \int_D dxdy = \int_{\theta_1}^{\theta_2} \int_0^{r(\theta)} rdrd\theta = \frac{1}{2} \int_{\theta_1}^{\theta_2} r(\theta)^2 d\theta \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $D$ is your sector and $A(D)$ its area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you want to compute the area of a circle of radius $R$ (you already know, but just in case...), $r(\theta ) = R$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A(D) = \frac{1}{2}\int_0^{2\pi} R^2d\theta = \pi R^2 \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazing, isn't it?   :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only problem could be that your data necessarily start with the equation of the curve given in Cartesain coordinates $y = f(x)$ and you have to translate it into polar coordinates.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-11T20:45:59.213" CommentCount="3" />
  <row Id="6562" PostTypeId="2" ParentId="6557" CreationDate="2010-10-11T20:54:33.600" Score="2" ViewCount="0" Body="&lt;p&gt;Let $\rm\:P\: $ be a poset = &lt;a href=&quot;http://en.wikipedia.org/wiki/Partially_ordered_set&quot; rel=&quot;nofollow&quot;&gt;partially-ordered set&lt;/a&gt;, i.e. a set equipped wth a relation $\rm\le$ that is reflexive, antisymmetric and transitive. Then one says that $\rm\: x\: $ and $\rm\: y\: $ are &lt;em&gt;comparable&lt;/em&gt;$\ $ if $\rm\ x \le y\ $ or $\rm\ y \le x\:;\: $ otherwise they are &lt;em&gt;incomparable&lt;/em&gt;, sometimes written $\rm\ x||y\:.\:$ A &lt;em&gt;chain&lt;/em&gt;$\ $ of  $\rm\: P\:$ is a subset in which every two elements are comparable, and dually, an &lt;em&gt;antichain&lt;/em&gt;$\ $ is a subset in which every two elements are incomparable.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-11T20:54:33.600" CommentCount="1" />
  <row Id="6563" PostTypeId="1" AcceptedAnswerId="6564" CreationDate="2010-10-11T21:10:56.173" Score="1" ViewCount="49" Body="&lt;p&gt;&lt;em&gt;This is part of a homework question&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out the perimeter of the black-filled object below (sorry, it's crudely drawn in paint, but I think you get the idea). To find the perimeter, I'm assuming that I should first find the perimeter of the rectangle (12*2+3*2 in this case) then subtract each little triangle (which is suppose to be all the same size by the way). The question mentions only one fact other than the size of each side- that it's trisected. I'm pretty sure the answer to this question lies in here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2sLxC.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2315" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-11T22:54:04.540" LastActivityDate="2010-10-11T22:54:04.540" Title="Properties of a trisected rectangle?" Tags="&lt;geometry&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6564" PostTypeId="2" ParentId="6563" CreationDate="2010-10-11T21:13:26.667" Score="3" ViewCount="0" Body="&lt;p&gt;Well, if each side is trisected, that means it's split into three identical parts. The base of each triangle is one of the parts of the 12 side, and the height of each triangle is one of the parts of the 3 side. You can figure it out from there.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-10-11T21:13:26.667" CommentCount="3" />
  <row Id="6565" PostTypeId="2" ParentId="6321" CreationDate="2010-10-11T21:26:28.743" Score="3" ViewCount="0" Body="&lt;p&gt;The answer to your question is yes. All numbers of the form $x=\sum_{n\ge0}\frac{C_k}{g^{Z(k)}}$ for Z(k) eventually dominating any polynomial are indeed transcendental. As in the question, g and C&lt;sub&gt;k&lt;/sub&gt; are integers with 1&amp;nbsp;&amp;le;&amp;nbsp;C&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;&amp;le;&amp;nbsp;g-1. In fact, the methods used by the paper linked in the question generalize in a quite straightforward way to handle this situation. I don't know of this result appearing in any published paper, but note the following points.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We can say straight-away that x is irrational. This follows from Z(n) eventually dominating any linear function of n, so its base-g expansion is not eventually periodic. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If Z(n+1)/Z(n) is unbounded then, as noted in the comments, x will be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville number&lt;/a&gt; so, by Liouville's theorem, it is transcendental. For any N&amp;nbsp;&amp;gt;&amp;nbsp;0, Z(n+1)&amp;nbsp;&amp;ge;&amp;nbsp;NZ(n) for infinitely many n. The fact that it is a Liouville number follows from taking $p=\sum_{k=1}^nC_kg^{Z(n)-Z(k)}$ and $q=g^{Z(n)}$, giving the rational approximation $|x-p/q|&amp;lt; g^{1+Z(n)-Z(n+1)}\le gq^{-N}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Thue%E2%80%93Siegel%E2%80%93Roth_theorem&amp;amp;oldid=385480595&quot; rel=&quot;nofollow&quot;&gt;Thue–Siegel–Roth theorem&lt;/a&gt;, if Z(n+1)/Z(n)&amp;nbsp;&amp;ge;&amp;nbsp;2+&amp;epsilon; infinitely often (any &amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0) then x will be transcendental. The theorem says that an irrational algebraic number has only finitely many rational approximations $\vert x-p/q\vert\le cq^{-2-\epsilon}$ for any fixed c,&amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0. That x has infinitely many such rational approximations follows in the same way as for point 2 above. Every Z(n+1)&amp;nbsp;&amp;ge;&amp;nbsp;(2+&amp;epsilon;)Z(n) gives a rational approximation $\vert x-p/q\vert&amp;lt; gq^{-2-\epsilon}$, so x cannot be algebraic. This covers the case where Z(n) grows exponentially of rate a&lt;sup&gt;n&lt;/sup&gt; for any a&amp;nbsp;&amp;gt;&amp;nbsp;2, but is not strong enough to cover cases such as Z(n)&amp;nbsp;=&amp;nbsp;2&lt;sup&gt;n&lt;/sup&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If Z(n+1)/Z(n)&amp;nbsp;&amp;gt;&amp;nbsp;1+&amp;epsilon; infinitely often (any &amp;epsilon;&amp;nbsp;&amp;gt;&amp;nbsp;0) then x will be transcendental. This is a consequence of the Roth-Ridout theorem, from the 1957 paper &lt;a href=&quot;http://dx.doi.org/10.1112/S0025579300001182&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Rational approximations to algebraic numbers&lt;/em&gt;&lt;/a&gt; (not free access, but is also quoted in the freely available paper &lt;a href=&quot;http://seminariomatematico.dm.unito.it/rendiconti/cartaceo/53-3.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;An explicit version of the theorem of Roth-Ridout&lt;/em&gt;&lt;/a&gt;, Theorem 2). The Roth-Ridout theorem strengthens the Thue-Siegel-Roth theorem implying, in particular, for irrational and algebraic x, there are only finitely many rational approximations $\vert x-p/q\vert\le cq^{-1-\epsilon}$ when the prime factors of q all belong to some fixed finite set &lt;em&gt;P&lt;/em&gt;. In our case, we can let &lt;em&gt;P&lt;/em&gt; be the set of prime factors of g and the result follows in the same way as for point 3 above. This shows that x is transcendental if Z(n) grows exponentially. (thanks to Mike Bennett over at mathoverflow for pointing out the Roth-Ridout theorem).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A paper by Bugeaud, &lt;a href=&quot;http://rendiconti.math.unipd.it/volumes/vol118.php?lan=english&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;On the b-ary expansion of an algebraic number&lt;/em&gt;&lt;/a&gt; shows that, if x is irrational and algebraic then for large enough n, there are at least (log n)&lt;sup&gt;1+1/(&amp;omega;+4)&lt;/sup&gt; (loglog n)&lt;sup&gt;-1/4&lt;/sup&gt; nonzero digits among the first n digits of the base g expansion. Here, &amp;omega; is the the number of prime divisors of g. This shows that, if Z(n)&amp;nbsp;&amp;ge;&amp;nbsp;exp(cn&lt;sup&gt;&amp;alpha;&lt;/sup&gt;) for large n and any fixed c&amp;nbsp;&amp;gt;&amp;nbsp;0, &amp;alpha;&amp;nbsp;&amp;gt;&amp;nbsp;1/(1+1/(&amp;omega;+4)) then x is transcendental.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After reading through the details of the paper linked in the original question, I note that they do generalize to the base g&amp;nbsp;&amp;ge;&amp;nbsp;2 case. So x is transcendental as long as Z(n) eventually dominates any polynomial. I don't know of any published paper proving this, but posted my proof on &lt;a href=&quot;http://mathoverflow.net/questions/41609/have-all-numbers-with-sufficiently-many-zeros-been-proven-transcendental/41760#41760&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt; where this question was also asked. I have re-read through this proof a few times to be sure, and am now confident that it is correct (modulo small typos, etc). Also, Bugeaud posted an answer to the question agreeing that the method generalizes. Using #(x,n) to denote the number of non-zero digits in the first n digits of the base g expansion of x, the precise statement is as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If x is irrational and satisfies a rational polynomial of degree D then #(x,n)&amp;nbsp;&amp;ge;&amp;nbsp;cn&lt;sup&gt;1/D&lt;/sup&gt; for a positive constant c and all large enough n.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In fact, you can easily remove the &quot;large enough&quot; from this statement, although I find it convenient stated in this way. The proof I wrote out is a generalization of the methods used in the paper linked in the question. There is one change worth noting though. Whereas the paper made use of the Thue-Siegel-Roth theorem at one point (Theorem 3.1), I used &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville's theorem&lt;/a&gt;. This means that the constant c appearing in the statement above is not quite as good (if you go through the proof and work it out explicitly) although, in any case, the paper linked in the question could have obtained a better value by using the Roth-Ridout theorem instead. Using Liouville's theorem does have two advantages though. Firstly, it is elementary. A proof of Liouville's theorem is given in the linked Wikipedia article. Secondly, it is &lt;em&gt;effective&lt;/em&gt;. That is, not only can the constant c be calculated but you can also work out exactly what &quot;large enough&quot; means for n in the statement above (which will depend on the polynomial satisfied by x). The strengthened versions of Liouville's theorem such as Thue-Siegel-Roth and Roth-Ridout are not effective.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-10-11T21:26:28.743" />
  <row Id="6566" PostTypeId="2" ParentId="6556" CreationDate="2010-10-11T21:30:02.710" Score="0" ViewCount="0" Body="&lt;p&gt;$$g(x) = f(x) - 2 = (h \circ f)(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h(x) = x - 2$ and of course $h^\circ(x) = x + 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the inverse $g^{\circ}(x) = (h \circ f)^{\circ}(x) = (f^{\circ}\circ h^{\circ})(x) = f^\circ(x + 2)$ by &lt;a href=&quot;http://math.stackexchange.com/questions/2349/how-to-prove-f-circ-g-1-g-1-circ-f-1/2426#2426&quot;&gt;the usual rules&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-11T21:30:02.710" CommunityOwnedDate="2010-10-11T21:30:02.710" />
  <row Id="6567" PostTypeId="1" AcceptedAnswerId="6600" CreationDate="2010-10-11T21:33:01.970" Score="8" ViewCount="121" Body="&lt;p&gt;The other day I &lt;a href=&quot;http://caicedoteaching.wordpress.com/2008/11/11/275-positive-polynomials/&quot; rel=&quot;nofollow&quot;&gt;came across&lt;/a&gt; the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; A polynomial $f(x,y)$ of degree at most $3$ that vanishes at $8$ of the $9$ points $(x,y)$ with $x, y \in \{-1,0,1\}$ must also vanish at the $9$th point. &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am wondering about how this statement generalizes. Specifically, I am looking for a theorem of the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; Suppose a polynomial $f(x_1, \ldots, x_n)$ of degree $d$ vanishes on a some discrete set $S \subset R^n$, which satisfies _______. Then, defining a discrete set $U$ as ______, the polynomial $f$ must also vanish on $U$.  &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anyone fill in the blanks? &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-10-12T08:50:01.127" Title="The zeros of a multivariable polynomial" Tags="&lt;polynomials&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="6568" PostTypeId="1" AcceptedAnswerId="6569" CreationDate="2010-10-11T21:37:06.213" Score="2" ViewCount="142" Body="&lt;p&gt;I was wondering about the relation between complement of a subset, different between two subsets, union and intersection of subsets. Can we reduce the set of the above operations in a minimal way so that other operations can be represented by these &quot;independent&quot; operations? And how many ways to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the intersection (or union) can be represented by complement and union (or intersection) in the same way as in De Morgan law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can we represent complement or difference in terms of union and intersection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-22T14:36:24.820" LastActivityDate="2010-11-22T18:34:27.377" Title="relation between set operations" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" />
  <row Id="6569" PostTypeId="2" ParentId="6568" CreationDate="2010-10-11T21:46:16.300" Score="2" ViewCount="0" Body="&lt;p&gt;Well, set difference is defined in terms of intersection and complement: $A\setminus B = A\cap B^C$. (Personally, I prefer the notation $A-B$ to $A\setminus B$, but the latter is relatively standard in my experience.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, De Morgan's laws allow us to translate between union and intersection via complements. Beyond this, we must have at least union and complement (or, equivalently, intersection and complement). That is, we cannot define complement solely in terms union and intersection.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-11T21:46:16.300" CommentCount="4" />
  <row Id="6570" PostTypeId="1" AcceptedAnswerId="6571" CreationDate="2010-10-11T21:52:42.047" Score="1" ViewCount="142" Body="&lt;p&gt;(1). According to Wikipedia, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_of_sets&quot; rel=&quot;nofollow&quot;&gt;field of subsets&lt;/a&gt; of X is defined to be a non-empty subset of the power set of X closed under the intersection and union of pairs of sets and under complements of individual sets. I was wondering if there is any redundancy in this definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be defined to be closed only under pairwise intersection and pairwise union? Or only closed under complement and pairwise intersection (or pairwise union)? If yes in each case, how can we show it also closed under the other operation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2). A sigma algebra of subsets of X is defined to be a nonempty subset of the power set of X closed under complementation and under countable union.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be defined to be closed under countable union and countable intersection? If yes, how can we show it is subsequently closed under complement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-22T14:37:08.733" LastActivityDate="2010-11-22T14:37:08.733" Title="Field of sets and Sigma algebra of sets" Tags="&lt;elementary-set-theory&gt;&lt;field-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6571" PostTypeId="2" ParentId="6570" CreationDate="2010-10-11T21:59:31.920" Score="4" ViewCount="0" Body="&lt;p&gt;No, you can't get complements from unions and intersections.  For example, let $X$ be a nonempty set.  Then $\{\emptyset\}$ is nonempty, closed under (arbitrary) intersection and union, but not closed under complements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get intersections from unions and complements using &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Morgan%27s_laws&quot; rel=&quot;nofollow&quot;&gt;De Morgan's laws&lt;/a&gt;.  To get an intersection, just take the complement of the union of the complements.  Similarly, you get unions from intersections and complements.  So yes, the definition of field of subsets was redundant.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-11T21:59:31.920" CommentCount="2" />
  <row Id="6572" PostTypeId="1" CreationDate="2010-10-11T22:11:22.543" Score="0" ViewCount="160" Body="&lt;p&gt;If G is a finite group containing an irreducible character of degree 5, deduce its order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the order of G is divisble by 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the order of G is equal to the sum of the squares of the degrees of irreducible characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I am unable to deduce the order of G with just this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please please please help me determine the exact order of G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&#xA;Sarah&lt;/p&gt;&#xA;" OwnerUserId="2444" LastActivityDate="2010-10-12T18:29:24.150" Title="Finding the order of a group" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6573" PostTypeId="1" AcceptedAnswerId="6574" CreationDate="2010-10-11T23:22:48.207" Score="0" ViewCount="203" Body="&lt;p&gt;Since union of sets is defined to be independent of the number of sets as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Union_%28set_theory%29#Forms&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If M is a set whose elements are themselves sets, then x is an element of the union of M if and only if for at least one element A of M, x is an element of A.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering why for a class of subsets of a set, it being closed under finite union and it being closed under countable union are not same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar question for intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T04:03:19.957" LastActivityDate="2010-11-23T04:03:19.957" Title="Closed under finite union and Closed under countable union" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="6574" PostTypeId="2" ParentId="6573" CreationDate="2010-10-11T23:45:32.610" Score="2" ViewCount="0" Body="&lt;p&gt;Starting from a initial collection of sets being allowed to take countable unions and intersections lets you create many more sets that being allowed to take only finite unions and intersections. Therefore it seems plausible to me that the former can take you out of your starting collection even if the latter does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find examples in any standard measure theory text. Here is a very simple one. Let&#xA;$$A_n=\{1,2,3,\ldots,n\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be the collection of the sets $A_n$ for $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for a finite set of indices, $n_1,n_2,\ldots,n_k$ say, the union&#xA;$$A_{n_1} \cup A_{n_2} \cup \cdots \cup A_{n_k} = A_{\max\{n_1,\ldots,n_k\}}$$&#xA;and hence belongs to $\Omega$. So $\Omega$ is closed under finite unions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the countably infinite union&#xA;$$\cup_{n=1}^{\infty} A_n = \{1,2,3,\ldots\}$$&#xA;and hence does not belong to $\Omega$. So $\Omega$ is not closed under countable unions.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-11T23:45:32.610" CommentCount="0" />
  <row Id="6575" PostTypeId="1" CreationDate="2010-10-11T23:48:35.367" Score="0" ViewCount="90" Body="&lt;p&gt;prove by structural induction that in any tree T, the number of leaves is 1 more than the number of nodes that have right siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proof so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s(n).  in any tree T, the number of leaves(L) is 1 more than the number of nodes(N) that have right siblings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s(1) for a tree of one node, there are no right siblings so n=1. The root itself is a leave so L =1. Thus L=n+1 is true, so the basis is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s(n+1)...Im stuck at the induction part here.&lt;/p&gt;&#xA;" OwnerUserId="2448" LastActivityDate="2010-11-25T05:16:57.633" Title="structural induction" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6576" PostTypeId="1" AcceptedAnswerId="6579" CreationDate="2010-10-11T23:56:39.193" Score="1" ViewCount="87" Body="&lt;p&gt;This is a contest problem. It asks for all the real solutions of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for arbitrary real $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the eq. it's clear that $p$ cannot be negative. By squaring once, splitting the radicals from the other terms and squaring again, I found a formula for $x^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 = 1+\frac{p^2}{16(1-p/2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we have $p&amp;lt;2$ for the second radical to be defined, and by comparing this expression with $p$ we get that $p&amp;lt;2$ makes the first radical be defined as well. Yet solutions of the original equation can only be found when $p\leq 4/3$, and $4/3$ is the (double) root of $9p^2-24p+16=0$ (obtained from comparing $1+\frac{p^2}{16(1-p/2)}$ with $p$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this happens. Perhaps I've done something wrong. If I haven't, I'd appreciate any help in proving that there are no solutions if $p$ is greater than $4/3$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-12T00:51:37.040" Title="Equation with radicals." Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="6577" PostTypeId="1" CreationDate="2010-10-12T00:02:28.387" Score="1" ViewCount="77" Body="&lt;p&gt;Given a certain coordinate frame, I can compute a new one by applying a set of rotations in a given order (what I call Euler Z-Y-X). So I yaw, then pitch then roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that I want to do exactly the opposite: given two coordinate frames (same origin to simplify), how do I find out roll, pitch and yaw angles that were used to transform from one to the other?&lt;/p&gt;&#xA;" OwnerUserId="2449" LastActivityDate="2010-10-12T00:02:28.387" Title="How to extract Euler angles from a a point in a plane?" Tags="&lt;geometry&gt;" CommentCount="4" />
  <row Id="6578" PostTypeId="2" ParentId="6573" CreationDate="2010-10-12T00:13:57.693" Score="2" ViewCount="0" Body="&lt;p&gt;Hmm.  There is a somewhat sneaky and quite false conclusion being made here.  It's true that the Wikipedia definition doesn't mention the number of sets.  However, that doesn't mean that union and intersection are &quot;independent of&quot; the number of sets.  I can't even figure out what this would mean.  To state an obvious counterexample, adding another set to your union will give a set with at least as many elements as before, and adding another set to your intersection will give a set with at most as many elements as before.  (Not even in the sense of cardinality, but just the simple fact that&#xA;$$\left(\bigcup_\alpha A_\alpha\right)\cup B\supset \bigcup_\alpha A_\alpha$$&#xA;and the dual statement for intersections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jyotirmoy's example shows that the countable union of finite sets isn't finite.  Another one comes from topology.  A topology on a set $X$ is &lt;em&gt;defined&lt;/em&gt; to be a set of subsets of $X$ that is closed under unions and finite intersections.  We call such sets &quot;open,&quot; and they're roughly a way of saying what points are close to each other.  A typical example is the Euclidean topology on $\mathbb{R}$, which consists of the open intervals $(a,b)$ and their unions.  This is well-defined because any finite intersection of open intervals is either the empty set or another, smaller open interval.  On the other hand, the set $\bigcap_n (-1/n,1/n)$ is just the point $\lbrace 0\rbrace$, which is definitely not a union of open intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, countable intersections are still kind of important in the same general situation.  A $\sigma$-algebra is a set of subsets of $X$ that is closed under complementation and countable intersections (and thus also countable unions).  You can use such a thing to define a &quot;measure,&quot; which is an assignment of a sort of &quot;size&quot; to each set in the $\sigma$-algebra.  The algebra of Lebesgue-measurable on $\mathbb{R}^n$ is probably the best example, but you can turn any topology into a $\sigma$-algebra, called its Borel algebra, just by iterating countable intersections and countable unions.  A set that's a countable intersection of open sets is called a $G_\delta$, from the German words &quot;gebiet Durchschnitt&quot; meaning &quot;open intersection.&quot;  So we know, for example, that every $G_\delta$ in $\mathbb{R}$, though not necessarily open, has a Lebesgue measure.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T00:13:57.693" />
  <row Id="6579" PostTypeId="2" ParentId="6576" CreationDate="2010-10-12T00:29:52.030" Score="4" ViewCount="0" Body="&lt;p&gt;When solving equations, by doing algebra like squaring etc, one could potentially include more solutions than the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = 1$ implies $x^2 = 1$, by squaring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation also has $x = -1$ as a solution, which is not a solution of the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, when you square etc, you might lose constraints like $x \ge 1$ and $x^2 \ge p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose when you squared, you had the term $x - 2\sqrt{x^2 -1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraint that $x \ge 2 \sqrt{x^2-1}$ (i.e. $x^2 \le 4/3$) gives us the constraint that $p \le 4/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noticed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{p^2}{16(1-p/2)} -p \ge 0$ for all $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{p^2}{16(1-p/2)} \le 4/3$ implies that $0 \le 1 + \frac{p^2}{16(1-p/2)} -p \le 4/3 - p$ which implies $4/3 \ge p$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-12T00:51:37.040" LastActivityDate="2010-10-12T00:51:37.040" CommentCount="5" />
  <row Id="6580" PostTypeId="1" AcceptedAnswerId="6585" CreationDate="2010-10-12T01:35:40.797" Score="2" ViewCount="166" Body="&lt;p&gt;A field is an abelian group under addition and a group under multiplication. But for the power set of a set S, under union and intersection operations and with S and the empty set being their identities, the inverse of any set is not unique for either union or intersection. So is the power set a field of sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Must the inverse of a set for either union or intersection be restricted to the complement of the set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-22T14:37:18.190" LastActivityDate="2010-11-22T14:37:18.190" Title="inverse element in a field of sets" Tags="&lt;elementary-set-theory&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6581" PostTypeId="1" CreationDate="2010-10-12T01:40:22.923" Score="2" ViewCount="270" Body="&lt;p&gt;Hey guys, I have a couple things I want to prove. I'm pretty sure a proof by induction is the best route for these.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First, I need to show that $5^n &amp;lt; n!$ from some $n_{0} &gt; 0$. I'm choosing $n_{0} = 12$ since that's the smallest positive integer where $n! &gt; 5^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: $P(k) : 5^k &amp;lt; k!$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show $P(12)$: $5^{12} &amp;lt; 12!$&lt;/li&gt;&#xA;&lt;li&gt;Assume $P(k)$, show $P(k+1)$:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$P(k+1):$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5^{(k+1)} \leq (k+1)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5*5^k \leq (k+1)*k!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I'm stuck. Since our previous assumption was that $5^k \leq k!$, couldn't we conclude the proof by showing:&#xA;$5 \leq (k+1)$ for $n \geq 12$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second one is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(k): n! \leq n^n$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show $P(1)$: $1 \leq 1$&lt;/li&gt;&#xA;&lt;li&gt;Assume $P(k)$, show $P(k+1)$:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$P (k+1):$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1)! \leq {n+1}^{n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1)*n! \leq {(n+1)} *{(n+1)}^{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we cancel out the common factor (we can do that, right?) $(n+1)$ we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n! \leq (n+1)^n$ which is true for $n \geq1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $n! \leq n^n$ for $n \geq 1 $&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So in conclusion, I'm wondering if these proofs are sufficient (and correct). I kinda feel like I'm missing something in the last few steps - that it's &quot;turtles all the way down&quot;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-12T17:15:16.863" LastActivityDate="2010-10-18T00:44:56.747" Title="Proof of inequalities by induction" Tags="&lt;homework&gt;&lt;induction&gt;" AnswerCount="2" />
  <row Id="6582" PostTypeId="2" ParentId="6580" CreationDate="2010-10-12T01:42:24.527" Score="2" ViewCount="0" Body="&lt;p&gt;One word, two meanings. A 'field of sets' is not a field in the abstract algebra sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; While a 'field of sets' is not a field, a Boolean Algebra (BA) of sets (of which a field of sets is a special case) is indeed an algebra. One can check that a BA is a ring, with symmetric difference $(A \cap B^c) \cup (A^c \cap B)$ as addition and intersection as multiplication. It is then trivially an algebra over the two-element field $\mathbb{F}_2$ by defining $1\cdot A=A$ and $0\cdot A=\emptyset$ and checking that $A+A=\emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Arturo's answer discussed this before me in a more general setting. The edit above is to set right my incorrect assertion in the comments that a BA is not an algebra.]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-12T03:03:29.713" LastActivityDate="2010-10-12T03:03:29.713" CommentCount="10" />
  <row Id="6583" PostTypeId="2" ParentId="6581" CreationDate="2010-10-12T02:29:50.853" Score="4" ViewCount="0" Body="&lt;p&gt;Your first proof is correct.  (If $a\le c$ and $b\le d$ and they're all positive, then $ab\le cd$.)  Your second one needs just a tiny bit more.  How do we know that $n!\le (n+1)^n$?  It follows from your inductive hypothesis, but it's not completely trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general tip, try not to work your proofs backwards.  When you're trying to prove that $P(k)\Rightarrow P(k+1)$, you start out by stating $P(k+1)$ and then deriving a true statement.  It works here because pretty much everything you're doing is reversible (dividing by $k+1$ vs. multiplying), but in later life, this might no longer be true and so it's a good practice to start now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, good work!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T02:29:50.853" CommentCount="1" />
  <row Id="6584" PostTypeId="1" AcceptedAnswerId="6589" CreationDate="2010-10-12T02:57:44.090" Score="1" ViewCount="222" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an amateur mathematician (but I do have degree's in computer science (with mathematics)). Anyways, I have written this paper, where I have proved that for $\zeta(\rho) = 0$ if $\Im(\rho) \to \infty$ then $\Re(\rho) \leq \log_2(3) - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is a big claim, and I am a cranky fella ;) who has a number of withdrawn papers on arxiv. So, chances of me being taken seriously $\to 0$ even though I do put my sweat in for checking any paper I write, before releasing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem is how do I get my paper discussed and verified, before I put it on arxiv (cause I do not want to withdraw it again) and also I do not know any local mathematician working on analytic number theory.&#xA;So here's my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before posting on arxiv are there any&#xA;  other avenues through which I can get&#xA;  my paper checked (on the internet)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-10-12T04:12:07.087" Title="Alternatives to arxiv" Tags="&lt;education&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="6585" PostTypeId="2" ParentId="6580" CreationDate="2010-10-12T02:58:50.503" Score="4" ViewCount="0" Body="&lt;p&gt;A &quot;field of sets&quot; is a collection $\mathcal{F}$ of subsets of a given set $X$ which is closed under binary unions, intersections, and under complementation. Alas, it is not an &quot;field&quot; in the sense of abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; a &quot;subalgebra&quot; of the Boolean algebra of the power set of $X$. Alas, here as well, &quot;Boolean algebra&quot; and &quot;subalgebra&quot; do not have the meaning they have in abstract algebra/ring theory: a &quot;Boolean algebra&quot; is not really an algebra: rather, it is a special kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_%28order%29&quot; rel=&quot;nofollow&quot;&gt;lattice&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, a Boolean algebra is a set $A$ together with two binary operations, $\wedge$ (meet) and $\vee$ (join), one unary operation ${}^c$ (complementation), and two nulary operations (distinguished elements) $0$ and $1$. We require $\vee$ and $\wedge$ to be associative, commutative, and to distribute over each other (this is already different from the case of &quot;algebra&quot; in the ring-theoretic sense, in which only multiplication distributes over sums, and not the other way around); we also have two rules of &quot;absorption&quot; that further describe how $\wedge$ and $\vee$ interact: $x\wedge(x\vee y) = x$ and $x\vee(x\wedge y) = x$. Finally, we require $x\vee x^{c}=1$ and $x\wedge x^{c}=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical example of a Boolean algebra/Boolean lattice is the lattice of subsets of a given set $X$, with $\wedge$ corresponding to intersection, $\vee$ to union, ${}^c$ to complementation, $0$ to $\emptyset$, and $1$ to $X$. Another example is propositional calculus, with $\wedge$ corresponding to conjunction, $\vee$ to disjunction, ${}^{c}$ to negation, $1$ to the class of tautologies, and $0$ to the class of contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for calling it &lt;em&gt;algebra&lt;/em&gt;, in defiance of the meaning of &quot;algebra&quot; from ring theory, is historical: Boole talked about &quot;algebra of thought&quot; or &quot;algebra of logic&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, any Boolean algebra gives rise to a ring. We define $+$ by &quot;symmetric difference&quot;: $x+y = (x\wedge y^{c})\vee(x^{c}\wedge y)$. If you do this, then you get an abelian group, with identity element being $0$, and with every element being its own inverse (that is, every nontrivial element is of additive order $2$). We define $*$ by $x*y=x\wedge y$. The unity of this ring is the $1$ from the Boolean lattice. This ring is an algebra over the field of $2$ elements, since it has characteristic $2$ and a unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask: will this ring be a field? The only time it is a field is if the original Boolean algebra was the trivial algebra, with $0$ and $1$ the only elements. For if $x$ is any other element, $0\neq x$, $x\neq 1$, then if you look at the order induced by the lattice structure ($x\preceq y$ if and only if $x\wedge y = x$) then you have $0\prec x \prec 1$, and $x\wedge y \preceq x$ for all $y$. Therefore, since $x$ is strictly smaller than $1$, then so will $x*y$ for any $y$, so $x$ cannot have a multiplicative inverse in the corresponding ring.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-12T03:07:56.660" LastActivityDate="2010-10-12T03:07:56.660" CommentCount="1" />
  <row Id="6586" PostTypeId="2" ParentId="6581" CreationDate="2010-10-12T03:35:24.027" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Rewrite them as products so to make the induction completely trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely $\quad\quad\displaystyle  \frac{n^n}{n\:!}\ =\ \ \: \frac{n}1\ \ \ \frac{n}2\: \ \ \frac{n}3\ \:\cdots\ \frac{n}n\ \ge\ 1\ \ $ since each factor is $\:\ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad$ and $\rm\quad\quad\displaystyle\ \frac{n\:!}{5^n}\ =\ \frac{12\:!}{5^{12}}\ \frac{13}{5}\ \frac{14}5\ \cdots\ \frac{n}5\ &gt;\ 1\ \ $ since each factor is $\:  &gt; 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need only inductively prove the lemma that a product is $\ge 1\:$ or $&gt; 1\:$ if each factor is - which is completely trivial. Many inductive proofs can be similarly drastically simplified by conceptual preprocessing. Here we've effectively employed &lt;strong&gt;multiplicative telescopy&lt;/strong&gt; to reduce generally intractable exponential inequalities to tractable polynomial inequalities. $\: $ This is a powerful technique with widespread applications. See also the analogous case of &lt;a href=&quot;http://math.stackexchange.com/questions/6814&quot;&gt;&lt;strong&gt;additive telescopy&lt;/strong&gt;&lt;/a&gt; and the fundamental theorem of difference calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T00:44:56.747" LastActivityDate="2010-10-18T00:44:56.747" />
  <row Id="6587" PostTypeId="1" AcceptedAnswerId="6590" CreationDate="2010-10-12T03:46:43.520" Score="2" ViewCount="130" Body="&lt;p&gt;If f is in $L^1(\mathbb{R})$, is it true that $\lim_{h \to 0^+} \frac1{h} \int_x^{x+h} f(t)\mathrm{d}t$ exists and is finite for every x in $\mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to use something along the lines of the following argument: The Lebesgue Differentiation Theorem says that this integral is equal to f(x) a.e., which is finite a.e. if f is in $L^1(\mathbb{R})$. And since the integral doesn't change on a set of measure 0, then the limit itself must be finite a.e. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-10-12T06:20:19.313" LastActivityDate="2010-10-12T06:20:19.313" Title="Finiteness of $\lim_{h \to 0^+} \frac1{h} \int_x^{x+h} f(t)\mathrm{d}t$" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="6588" PostTypeId="2" ParentId="6584" CreationDate="2010-10-12T04:11:05.663" Score="5" ViewCount="0" Body="&lt;p&gt;No one on the internet (or elsewhere) is obliged to check your paper. So you have to catch people's interest and minimize the amount of investment they have to make. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some suggestion based not on personal experience but from being an spectator round the Net:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make your work easy to understand: Present your work in a theorem-lemma format, clearly specifying links to existing literature and pinpointing your own specific contributions.&lt;/li&gt;&#xA;&lt;li&gt;Make your work publicly accessible: Put it up on you website. Or on Google Documents. Or on &lt;a href=&quot;http://vixra.org&quot; rel=&quot;nofollow&quot;&gt;http://vixra.org&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Publicise the key points: Make a newsgroup posting, or create a question here or on Mathoverflow that explains your main insight or innovation and link to your paper. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was a bit hesitant in writing (3) since if you do it wrong you are just creating spam. No one is obliged to read on if you say &quot;I have proved X. Can you please check?&quot;. Much better to say &quot;So far people did not succeed in approach A to problem B because of C. But I think things can be made to work if you do D. Here's an attempt.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-12T04:11:05.663" CommunityOwnedDate="2010-10-12T04:11:05.663" />
  <row Id="6589" PostTypeId="2" ParentId="6584" CreationDate="2010-10-12T04:12:07.087" Score="7" ViewCount="0" Body="&lt;p&gt;There are other places to post papers, but most are either harder to get into (most online journals) or are less reputable than the arXiv (like viXra).  Finding math forums and posting there may be your best bet.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mymathforum.com/&quot; rel=&quot;nofollow&quot;&gt;http://mymathforum.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mersenneforum.org/&quot; rel=&quot;nofollow&quot;&gt;http://mersenneforum.org/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physicsforums.com/&quot; rel=&quot;nofollow&quot;&gt;http://physicsforums.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to name just a few.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On this specific point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your result seems wrong.  Aren't there vertical asymptotes and, in particular, values unbounded above for $\Re(\zeta(x+iy))$ for any fixed $y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have (or think you have) an &lt;em&gt;effective&lt;/em&gt; proof, can you give bounds Y and Z &gt; 0 where, for any $y&gt;Y$, $\Re(\zeta(x+iy)) &amp;lt; Z$?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-12T04:12:07.087" CommentCount="1" />
  <row Id="6590" PostTypeId="2" ParentId="6587" CreationDate="2010-10-12T04:15:41.437" Score="5" ViewCount="0" Body="&lt;p&gt;No.  As you said, it is true almost everywhere, but for example if $f(x)=x^{-1/2}$ on $(0,1)$ (and whatever you like elsewhere), then for $0\lt h\lt 1$, $1/h\int_0^h f(t)dt=2/\sqrt{h}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-12T04:15:41.437" CommentCount="0" />
  <row Id="6591" PostTypeId="2" ParentId="6568" CreationDate="2010-10-12T04:51:25.983" Score="4" ViewCount="0" Body="&lt;p&gt;There is a &quot;minimal&quot; definition of set operations, though it's not really a restriction of the classical ones, and is more a fun trick than a clever insight.  It was discovered by Charles Sanders Peirce, for general Boolean algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A\uparrow B=(A\cap B)^c$.  Then $A\uparrow A=A^c$.  So $(A\uparrow A)\uparrow(B\uparrow B)=(A^c\cap B^c)^c=A\cup B$.  Similarly, $(A\uparrow B)\uparrow(A\uparrow B)=((A\cap B)^c)^c=A\cap B$.  And then you can get difference or whatever else, as expected.  You can do the same thing with $(A\cup B)^c$.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T04:51:25.983" CommentCount="2" />
  <row Id="6592" PostTypeId="2" ParentId="6199" CreationDate="2010-10-12T04:51:32.547" Score="2" ViewCount="0" Body="&lt;p&gt;Here is a somewhat detailed outline of an argument that I think works.  As this is a homework problem, some of the pieces of the argument do need to be filled in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we're in $\mathbb{R}^m$.  For fixed $y \in \Gamma(x)$, the fact that $\Gamma(x)$ has a nonempty interior means that there are $m$ points $z_1, z_2, \ldots, z_m$ in the interior of $\Gamma(x)$ such that these $m$ points and $y$ together are affinely independent.  Thus you can take sufficiently small balls around each of these $m+1$ points such that the balls do not intersect and that any set consisting of one point from each ball is also affinely independent.  Let $z_0 = y$.  Since $\phi$ is lower hemicontinuous, for each $z_i$ there exists a sequence $z_{i_n} \to z_i$ and an $N_i$ such that $z_{i_n} \in \phi(x_n)$ and $z_{i_n}$ is inside that small ball around $z_i$ for all $n \geq N_i$.  For each $n \geq \max \{N_i\}$, construct the convex hull $C_n$ of $\{z_{0_n}, z_{1_n}$, $z_{2_n}, \ldots, z_{i_n}\}$.  Since $\phi$ is convex-valued, $C_n$ is a subset of $\phi(x_n)$.  Do the same thing for each $n$ for the $\psi$ function to obtain sets $D_n$.  Let $S_n = C_n \cap D_n$.  The intersection of the convex hulls of two sets of $m+1$ affinely independent points in $\mathbb{R}^m$ that are pairwise close to each other must be nonempty.  (Consider the supporting hyperplanes.)  Let $y_n$ be the point in $S_n$ closest to $y$.  Since the extreme points of $S_n$ converge to the extreme points of the convex hull of $\{z_0, z_1, z_2, \ldots, z_m\}$, $S_n$ converges as a set to the convex hull of $\{z_0, z_1, z_2, \ldots, z_m\}$.  Thus the point in $S_n$ closest to $y$ $(= z_0)$ must converge to $y$; i.e., $y_n \to y$. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-13T13:29:04.093" LastActivityDate="2010-10-13T13:29:04.093" CommentCount="2" />
  <row Id="6593" PostTypeId="1" AcceptedAnswerId="6693" CreationDate="2010-10-12T05:49:46.957" Score="1" ViewCount="116" Body="&lt;p&gt;If $c$ is a singular 1-cube in $\mathbb{R}^2 \backslash \left\{ (0,0)\right\}$ with $c(0)=c(1)$, show that there is an integer $n$ such that: $c-c_{(1,n)} = \partial (c^2)$ for some $2$-chain $c^2$.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T04:39:00.107" LastActivityDate="2010-10-13T17:56:39.443" Title="Boundary of a 2-chain" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="6594" PostTypeId="1" CreationDate="2010-10-12T06:26:30.657" Score="1" ViewCount="103" Body="&lt;p&gt;I am trying to work through an example from Jordan and Smiths book Nonlinear Ordinary Differential Equations&#xA;Its example 6.1 on page 195&#xA;The question reads: obtain an approximate solution of the family of equations x&quot;+x=ex^3&#xA;with x(e,0)=1  x'(e,0)=0 and error O(e&lt;sup&gt;3&lt;/sup&gt;) uniformly on t&gt;=0, by the method of coordinate pertubation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x&quot;&lt;sub&gt;0&lt;/sub&gt;+x&lt;sub&gt;0&lt;/sub&gt;=0&#xA;So we get: &#xA;x&lt;sub&gt;0&lt;/sub&gt;=A&lt;sub&gt;0&lt;/sub&gt;cost + B&lt;sub&gt;0&lt;/sub&gt;sint&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the conditions given: x&lt;sub&gt;0&lt;/sub&gt;(0)=1 and x'&lt;sub&gt;0&lt;/sub&gt;(0)=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have x&lt;sub&gt;0&lt;/sub&gt;=cost&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation is x&quot;&lt;sub&gt;1&lt;/sub&gt;+x&lt;sub&gt;1&lt;/sub&gt;=x&lt;sub&gt;0&lt;/sub&gt;&lt;sup&gt;3&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which becomes x&lt;sub&gt;1&lt;/sub&gt; + x&lt;sub&gt;1&lt;/sub&gt;=cos&lt;sup&gt;3&lt;/sup&gt;t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So x&quot;&lt;sub&gt;1&lt;/sub&gt;+x&lt;sub&gt;1&lt;/sub&gt; = 3/4cost+ 1/4cos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I assume x&lt;sub&gt;1&lt;/sub&gt;=Acost+Bcos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So x'&lt;sub&gt;1&lt;/sub&gt;=-Asint-3Bsin3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&#xA;ẍ&lt;sub&gt;1&lt;/sub&gt;=-Acost-9Bcos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore ẍ&lt;sub&gt;1&lt;/sub&gt; + x&lt;sub&gt;1&lt;/sub&gt; = -8Bcos3t = 3/4cost+ 1/4cos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore -8B= 1/4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so B = -1/32&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives me x&lt;sub&gt;1&lt;/sub&gt;=Acost - 1/32cos3t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using this and the condition that x&lt;sub&gt;0&lt;/sub&gt;(0)=0 gives me A = 1/32&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however the solution the book gives contains another term : 3/8tsint&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i dont know how to get this term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could someone please help&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditorDisplayName="" LastEditDate="2010-10-12T13:37:52.307" LastActivityDate="2010-10-12T13:37:52.307" Title="differential equations" Tags="&lt;differential-equations&gt;" CommentCount="1" />
  <row Id="6595" PostTypeId="1" CreationDate="2010-10-12T06:51:50.620" Score="1" ViewCount="245" Body="&lt;p&gt;$\displaystyle\lim_{\epsilon \rightarrow 0} \int \frac{1}{x^{1+\epsilon}} \mathrm{d}x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I go about evaluating this integral?&#xA;Does this integral converge to $log_e x $ or to something else?&lt;/p&gt;&#xA;" OwnerUserId="1798" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-12T15:03:31.177" LastActivityDate="2010-10-12T15:03:31.177" Title="How do I evaluate this integral?" Tags="&lt;integral&gt;&lt;limit&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="6596" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T07:03:03.027" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; The integral asked for was $\displaystyle\lim_{\epsilon\to0}\int\frac1{x^{1+\epsilon}}dx$, so it's $=\lim\int x^{-(1+\epsilon)}dx = \lim \frac1{-\epsilon}x^{-\epsilon} = -\mathrm{sign}(\epsilon)\cdot\infty$ (just use $\int x^a dx = \frac1{a+1} x^{a+1}$ for $a\neq -1$).&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-10-12T08:57:19.577" LastActivityDate="2010-10-12T08:57:19.577" CommentCount="9" />
  <row Id="6597" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T07:04:21.230" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, the limit quantity whether its outside the integral or inside the integral doesn't make any sense. So the answer is $$\lim_{\epsilon \to 0} \frac{1}{1+ \epsilon} \cdot \int \frac{1}{x} \ dx = \log{x}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-12T07:04:21.230" CommentCount="2" />
  <row Id="6598" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T07:06:25.783" Score="0" ViewCount="0" Body="&lt;p&gt;Evaluate the integral first using u-substitution and you'll end up with (1/(1+epsilon))ln(u) and the the limit of this as epsilon approaches 0 is indeed ln(u).&lt;/p&gt;&#xA;" OwnerUserId="2302" LastActivityDate="2010-10-12T07:06:25.783" CommentCount="1" />
  <row Id="6599" PostTypeId="1" AcceptedAnswerId="6602" CreationDate="2010-10-12T08:43:11.790" Score="1" ViewCount="121" Body="&lt;p&gt;If $f(x) = \sum_{j=1}^n c_j 1_{A_j}(x)$ ie a simple function. Why is $$\prod_{j=1}^n \exp(t[exp(i(u,c_j))-1]\mu(A_j)) =\exp[t\int_A[\exp(i(u,f(x)))-1]\mu(dx)].$$ The notation (u,c) is dot product, $1_{A}(x)$ is the indicator function and $i=\sqrt{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="626" LastEditorDisplayName="" LastEditDate="2010-10-12T18:16:53.847" LastActivityDate="2010-10-12T18:16:53.847" Title="integrating a simple functions" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6600" PostTypeId="2" ParentId="6567" CreationDate="2010-10-12T08:50:01.127" Score="6" ViewCount="0" Body="&lt;p&gt;The statement you mention is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cayley%E2%80%93Bacharach_theorem&quot; rel=&quot;nofollow&quot;&gt;Cayley-Bacharach theorem&lt;/a&gt;. &#xA;For one way of generalizing the theorem, see the beautiful article &lt;a href=&quot;http://www.ams.org/bull/1996-33-03/S0273-0979-96-00666-0/S0273-0979-96-00666-0.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Cayley-Bacharach theorems and conjectures&quot;&lt;/a&gt; by Eisenbud-Green-Harris.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastActivityDate="2010-10-12T08:50:01.127" />
  <row Id="6601" PostTypeId="2" ParentId="6491" CreationDate="2010-10-12T09:19:02.737" Score="1" ViewCount="0" Body="&lt;p&gt;This was supposed to be a comment to reply to Gio's comment in Matt's answer, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gio, you should have mentioned that little detail about your application to determining the number of panels needed for the trapezoidal rule in your question, because it now looks to me that you're going about it the wrong way. what you actually need here is an estimate of the truncation error of the trapezoidal rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One expression you can use for the error estimate is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{h^2}{12}\left(f^{\prime}(b)-f^{\prime}(a)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ and $b$ are the upper and lower limits, $h=\frac{b-a}{n}$ is the panel size, and $f^{\prime}$ is the derivative of your integrand. To use this, suppose you want the error to be around $10^{-11}$. Equate the error estimate to your desired tolerance, and solve for $n$. This error estimate assumes that $f$ is continuous up to the second derivative ($C^2$); otherwise, the estimate can fail badly. Apart from this, the other possible way the error estimate can fail is if $f^{\prime}$ vanishes at the endpoints (e.g. when integrating a periodic function over one period); for this particular case, the trapezoidal rule is actually more accurate than usual.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-12T09:19:02.737" CommentCount="2" />
  <row Id="6602" PostTypeId="2" ParentId="6599" CreationDate="2010-10-12T09:41:43.043" Score="0" ViewCount="0" Body="&lt;p&gt;Because $e^a \cdot e^b = e^{a+b}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing $t(e^{i\langle u,c_j\rangle}-1)$ does (for the purpose of evaluating the integral) is just to make the manipulations a bit more ugly.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-12T09:41:43.043" />
  <row Id="6603" PostTypeId="2" ParentId="6544" CreationDate="2010-10-12T10:10:43.350" Score="10" ViewCount="0" Body="&lt;p&gt;Let $x=(1+s)/2$, so that the expression becomes&#xA;$$ \frac{n}{2^{n+1}} \int_{-1}^1 \frac{(1+s)^n-(1-s)^n}{s} ds =&#xA;\frac{n}{2^{n}} \int_{0}^1 \frac{(1+s)^n-(1-s)^n}{s} ds. $$&#xA;(The integrand is an even function.)&#xA;Fix some $c$ between 0 and 1, say $c=1-\epsilon$.&#xA;Then the integral from 0 to $c$ will be small in comparison to $2^n$,&#xA;since the integrand is bounded by a constant times $(1+c)^n$,&#xA;so as far as the limit is concerned it is enough to look at the&#xA;integral from $c$ to 1,&#xA;and in that integral we can neglect the term $(1-s)^n$ since it will&#xA;also contribute something much smaller than $2^n$.&#xA;The surviving contribution therefore comes from&#xA;$$ \int_c^1 \frac{(1+s)^n}{s} ds, $$&#xA;which lies between&#xA;$$ \int_c^1 \frac{(1+s)^n}{1} ds $$&#xA;and&#xA;$$ \int_c^1 \frac{(1+s)^n}{c} ds, $$&#xA;that is,&#xA;$$ \frac{2^{n+1} - (1+c)^{n+1}}{n+1} &amp;lt; \int_c^1 \frac{(1+s)^n}{s} ds &amp;lt; \frac{1}{c} \frac{2^{n+1} - (1+c)^{n+1}}{n+1} .$$&#xA;Multiplying by $n/2^n$ and letting $n\to\infty$ shows that the liminf&#xA;is at least 2 and the limsup is at most $2/c$.&#xA;But since this holds for any $c$ between 0 and 1,&#xA;it follows that liminf=limsup=2, hence the limit is 2.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-13T06:41:54.610" LastActivityDate="2010-10-13T06:41:54.610" CommentCount="5" />
  <row Id="6604" PostTypeId="2" ParentId="6595" CreationDate="2010-10-12T10:34:51.367" Score="4" ViewCount="0" Body="&lt;p&gt;As it stands, the limit doesn't exist (see Tobias's answer).&#xA;But if you consider the &lt;em&gt;definite&lt;/em&gt; integral from 1 to $x$,&#xA;then the limit is $\ln x$:&#xA;$$ \int_1^x \frac{dt}{t^{1+\epsilon}} =&#xA;\left[ \frac{t^{-\epsilon}}{-\epsilon} \right]_1^x =&#xA;\frac{1-x^{-\epsilon}}{\epsilon} =&#xA;\frac{1-e^{-\epsilon \ln x}}{\epsilon} =&#xA;\frac{1-(1-\epsilon \ln x+O(\epsilon^2))}{\epsilon}&#xA;\to \ln x$$&#xA;as $\epsilon\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-12T10:34:51.367" CommentCount="3" />
  <row Id="6605" PostTypeId="1" AcceptedAnswerId="6658" CreationDate="2010-10-12T11:56:57.037" Score="1" ViewCount="187" Body="&lt;p&gt;We know that the Mandelbrot set is derived from the iterations of z^2 + c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do anyone know something about magnet Mandelbrot? I found it in the software UltraFractal, and it is much more beautiful than the original Mandelbrot, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know anything about it? What is it iterating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's a picture of a magnet Julia set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/yoZGK.jpg&quot; alt=&quot;Magnet Julia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tZVDM.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-24T10:55:17.007" LastActivityDate="2010-10-24T10:55:17.007" Title="Magnet Mandelbrot Set" Tags="&lt;fractals&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6606" PostTypeId="1" AcceptedAnswerId="6607" CreationDate="2010-10-12T13:34:19.990" Score="1" ViewCount="119" Body="&lt;p&gt;In a boolean algebra, 0 (the lattice's bottom) is the identity element for the join operation $\lor$, and 1 (the lattice's top) is the identity element for the meet operation $\land$. For an element in the boolean algebra, its inverse/complement element for $\lor$ is wrt 1 and its inverse/complement element for $\land$ is wrt 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Boolean algebra can be defined to be a complemented lattice that is also distributive. For a distributive lattice, the complement of x, when it exists, is unique. See Wikipedia (http://en.wikipedia.org/wiki/Lattice_(order)#Complements_and_pseudo-complements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power set of a set $S$ is an example of Boolean algebra. $S$ is the identity for union and $\emptyset$ is the identity for intersection. However, for union, the complement of a set wrt $S$ is not unique; For intersection, the complement of a set wrt $\emptyset$ is not unique either. So is this a contradiction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-10-12T13:51:07.550" Title="identity and inverse/complement elements in a boolean algebra" Tags="&lt;set-theory&gt;&lt;boolean-algebra&gt;" AnswerCount="1" />
  <row Id="6607" PostTypeId="2" ParentId="6606" CreationDate="2010-10-12T13:51:07.550" Score="5" ViewCount="0" Body="&lt;p&gt;By definition, $\rm b\:$ is a complement of $\rm a\:$  if $\rm\ a\vee b = 1,\ a\wedge b = 0\:$. So a unique complement must be a unique solution to &lt;em&gt;both&lt;/em&gt;$\ $ equations (involving &lt;em&gt;both&lt;/em&gt;$\ $ operations), not just a single operation - as you consider above. So there is no contradiction.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-12T13:51:07.550" />
  <row Id="6608" PostTypeId="2" ParentId="6412" CreationDate="2010-10-12T14:59:13.333" Score="0" ViewCount="0" Body="&lt;p&gt;Not an answer to my question, as opposed to a workaround I applied for my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, in my particular case, all $\lvert I \rvert = 2$ for all $I \in \mathbf{I}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sketching, I applied the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;model the weighted graph of non-transitive, symmetric (directionless) equivalences;&lt;/li&gt;&#xA;&lt;li&gt;apply iterations until fixpoint as follows:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;order the edges based on edge weight (assumes a total ordering, and granular ordering such that you don't have to make trivial decisions between equals weights);&lt;/li&gt;&#xA;&lt;li&gt;iterate over the edges in decreasing order, and derive a &quot;strong, transitive equivalence&quot; for the highest weighted edge:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;merge those nodes in the graph, and their edges and edge weights;&lt;/li&gt;&#xA;&lt;li&gt;do not merge nodes which conflict with $\mathbf{I}$;&lt;/li&gt;&#xA;&lt;li&gt;do not merge nodes which have already been involved in a merge in that iteration.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, this is not precisely an answer to my question, but the core idea is to rebuild the equivalences in decreasing order of edge weights, avoiding any equivalences that would equate two identifiers known to be different. You might not end up with a &quot;minimal-cut&quot;, but it's a fairly intuitive, tractable solution (which I hope makes sense).&lt;/p&gt;&#xA;" OwnerUserId="2331" LastActivityDate="2010-10-12T14:59:13.333" />
  <row Id="6609" PostTypeId="1" AcceptedAnswerId="6621" CreationDate="2010-10-12T15:32:18.920" Score="3" ViewCount="404" Body="&lt;p&gt;These operators are written in different forms in Cartesian,&#xA;cylindrical and spherical coordinates. For instance, in spherical&#xA;coordinate system, one has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \overrightarrow{F}=\frac{1}{r^{2}}\frac{\partial }{\partial r}&#xA;\left( r^{2}F_{r}\right) +\frac{1}{r\sin \theta }\frac{\partial }{\partial&#xA;\theta }\left( \sin \theta \cdot F_{\theta }\right) +\frac{1}{r\sin \theta }&#xA;\frac{\partial F_{\varphi }}{\partial \varphi }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Do identities such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \left( \overrightarrow{A}\times \overrightarrow{B}\right) =%&#xA;\overrightarrow{B}\cdot \nabla \times \overrightarrow{A}-\overrightarrow{A}%&#xA;\cdot \nabla \times \overrightarrow{B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hold in general when cylindrical and spherical coordinate systems are used or do they have to be adapted?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: After having read the comments it occured to me that the invariance of&#xA;these identities with regard to the coordinate system is a consequence of&#xA;the definitions of the mentioned operators in terms of integrals, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \overrightarrow{F}=\underset{V\rightarrow 0}{\lim }\frac{1}{V}%&#xA;\underset{S}{\int \int }\overrightarrow{F}\cdot \overrightarrow{n}\;dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is the volume of a bounded closed region $T$, $S$ is the surface&#xA;of $T$, and $\overrightarrow{n}$ the unit outer normal vector to $S$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-26T12:15:21.323" LastActivityDate="2010-12-26T12:15:21.323" Title="Do Vector Calculus Cartesian coordinates identities with Div, Grad, Curl hold in cylindrical and spherical coordinates?" Tags="&lt;calculus&gt;&lt;physics&gt;&lt;vector-calculus&gt;&lt;multivariable-calculus&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6610" PostTypeId="1" AcceptedAnswerId="6611" CreationDate="2010-10-12T15:57:31.830" Score="1" ViewCount="67" Body="&lt;p&gt;Suppose I have a binary vector $v$ that is copied twice by two separate machines $M$ and $N$, resulting in two new vectors $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both machines are faulty in the sense that they both might fail to copy each character correctly. More formally, $M$ has an associated vector $e_M$ of length $|v|$, where $(e_M)_i$ is the probability that bit $i$ in $M$'s input will be inverted in $M$'s output (and $N$ has a similarly defined, but not necessarily identical, associated vector $e_N$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;if the only information I receive is $x$, $y$, and two real numbers $p$ and $q$ describing the probabilities that a mistake has been made in producing $x$ and $y$ (i.e. $p$ is the probability that $M$ made an error when producing $x$ from $v$, and $q$ is the probability that $N$ made an error when producing $y$ from $v$), is there a way to reconstruct, or approximate $e_M$ and $e_N$? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;does the answer to question 1 depend on $|v|$ (which is of course assumed to be at least $2$ for this question to make sense), and if so, how?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(all apologies if the question is trivial, and if I'm not using the right terms)&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-10-12T16:38:58.290" Title="Separate probabilities from &quot;joint&quot; probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="6611" PostTypeId="2" ParentId="6610" CreationDate="2010-10-12T16:38:58.290" Score="0" ViewCount="0" Body="&lt;p&gt;$1-p$ gives you the product of $(1-(e_M)_i)$ over $i$ and similarly for $1-q$.  If you only have one set of $x$ and $y$ and don't know anything about the $(e_M)_i$ and $(e_N)_i$ there is no hope.  Even if you get a lot of $x,y$ pairs but don't get any information of what is the truth all you can tell is the combination of error rates per bit.  The chance of disagreement at a given bit is $e_M+e_N-e_M*e_N$ but you can't sort out any more than that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-12T16:38:58.290" CommentCount="1" />
  <row Id="6613" PostTypeId="1" AcceptedAnswerId="6614" CreationDate="2010-10-12T16:51:30.630" Score="6" ViewCount="103" Body="&lt;p&gt;In MK (Morse-Kelley) set theory life is easy: $\forall X\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)$. If $X=\left\{\right\}$ then $\bigcap X=U$, where $U$ is the universal class. So the (unary) intersection of the empty set is the class that contains all sets as elements. In ZF (Zermelo-Fraenkel) set theory, instead, proper classes are not allowed. So, how can I define $\bigcap X$ in ZF? I tried with the following definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\forall X\left(X\not=\left\{\right\}\rightarrow\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)\right)$. This means that $\bigcap\left\{\right\}$ is undefined, which is not that good.&lt;/li&gt;&#xA;&lt;li&gt;$\forall X\forall y\left(y\in\bigcap X\leftrightarrow X\not=\left\{\right\}\land\forall x\left(x\in X\rightarrow y\in x\right)\right)$. This means that $\bigcap\left\{\right\}=\left\{\right\}$, which is the opposite of MK.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any other valuable definition. Any ideas? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2010-10-12T17:48:01.470" Title="Unary intersection of the empty set" Tags="&lt;set-theory&gt;" AnswerCount="3" />
  <row Id="6614" PostTypeId="2" ParentId="6613" CreationDate="2010-10-12T17:02:59.853" Score="3" ViewCount="0" Body="&lt;p&gt;The fact that proper classes are handled differently in ZF than in MK doesn't really affect the definition of the unary intersection or other operations that can produce proper classes. So in ZF, $\bigcap z$ is defined using the same definition as in MK: &#xA;$$ &#xA;y \in \bigcap z \Leftrightarrow (\forall w \in z) ( y \in w).&#xA;$$&#xA;Just like in MK, $\bigcap \emptyset$ is a proper class in ZF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are treatments of ZF in which the authors don't handle proper classes at all, but there are other treatments in which proper classes are &quot;allowed&quot;. The simplest way to allow them is to restrict to definable proper classes, and silently replace each such class with its definition whenever the class is used. In this way, for any class or set $z$ we have a (possibly proper) class $\bigcap z$ defined as above. We can prove as a theorem in ZF (just as in MK) is that if $z$ is a nonempty set then there is a set $\hat{z}$ such that $\hat{z} = \bigcap z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: I want to emphasize that this does not conflict with the answer given by Arturo Magidin. The definition I stated here (in which $\bigcap \emptyset$ is a proper class) is the one used by Levy, &lt;em&gt;Basic Set Theory&lt;/em&gt;). The definition in the other answer, in which $\bigcap \emptyset = \emptyset$, is used by other authors (I believe). Kunen, Jech, and Halmos all leave $\bigcap \emptyset$ undefined in their well-known texts. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-12T17:48:01.470" LastActivityDate="2010-10-12T17:48:01.470" CommentCount="4" />
  <row Id="6615" PostTypeId="2" ParentId="6613" CreationDate="2010-10-12T17:07:53.237" Score="0" ViewCount="0" Body="&lt;p&gt;The only other way I can think of is to use specification, which is really the only way you know that such a set exists.  Given a universe $U$, define $\bigcap X$ to be the unique $B$ satisfying the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall x(x\in B\Leftrightarrow(x\in U\wedge \exists y(y\in X\wedge x\in y)))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\bigcap\emptyset=U$.  Is this any more or less desirable than having $\bigcap\emptyset$ be a proper class or $\emptyset$ itself?  I don't know.  My topology instructor used it when defining a topology to ensure that the universe was always open, but this is just cosmetic -- there's nothing to be lost by just including this as another axiom.  Ultimately, I suppose everyone but the most hardcore set theorists will just use what works and ignore the axiomatic background.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-12T17:07:53.237" />
  <row Id="6616" PostTypeId="1" AcceptedAnswerId="6618" CreationDate="2010-10-12T17:16:48.733" Score="7" ViewCount="387" Body="&lt;p&gt;While proving that some normed spaces were complete, two questions came to my mind. They relate the topological and the metric structures induced by a norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is it possible to find two equivalent norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space $V \ $ such that $(V \ ,\|\cdot\|_1)$ is complete and $(V \ ,\|\cdot\|_2)$ is not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a vector space $V \ $ and two non-equivalent norms such that $V \ $ is complete relative to both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I'm assuming $V \ $ a vector space over a subfield of $\mathbb{C}$. Also I know that the answer is no if we only consider finite-dimensional vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: I'm considering two norms equivalent if they define the same topology. I think it's the usual notion Jonas referred in his comment.]&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-10-12T19:44:58.100" LastActivityDate="2010-10-12T22:55:23.273" Title="Metric and Topological structures induced by a norm" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="6618" PostTypeId="2" ParentId="6616" CreationDate="2010-10-12T17:24:10.453" Score="11" ViewCount="0" Body="&lt;p&gt;1) No.  It is straightforward to show that equivalent norms yield both the same convergent sequences and the same Cauchy sequences. (Written before Rasmus's answer was posted, but posted afterward.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Yes.  One way to see this is to note that isomorphism classes of vector spaces depend only on linear dimension, so the question amounts to finding 2 nonisomorphic Banach spaces of the same linear dimension.  There are lots of examples of these.  &lt;a href=&quot;http://www.jstor.org/stable/2318458&quot; rel=&quot;nofollow&quot;&gt;Every infinite dimensional separable Banach space has linear dimension $2^{\aleph_0}$&lt;/a&gt;.  However, for example, $\ell^1$ and $c_0$ are separable Banach spaces that are not isomorphic (as Banach spaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, &quot;amounts to&quot; wasn't quite accurate.  It is certainly sufficient that the 2 Banach spaces are not isomorphic, but it is not necessary because you are only asking that one particular map (the identity in the original formulation) is not an isomorphism.  So what I gave above is actually stronger.  To just answer 2), you could just take any infinite dimensional Banach space and induce a new norm via an unbounded linear isomorphism with itself.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer above was assuming that equivalent norms are defined as in &lt;a href=&quot;http://planetmath.org/encyclopedia/EquivalentNorms.html&quot; rel=&quot;nofollow&quot;&gt;this PlanetMath article&lt;/a&gt;.  If instead you meant only that the spaces are homeomorphic in the norm topologies, as Jyotirmoy Bhattacharya suspected, then the examples alluded to above won't work.  However, there are also examples of pairs of Banach spaces that have the same linear dimension but are not homeomorphic, and this will work in either case.  For example, $\ell^\infty$ and $c_0$ are not homeomorphic because $\ell^\infty$ is nonseparable.  Both spaces have linear dimension $2^{\aleph_0}$.  This was already mentioned for $c_0$, and for $\ell^\infty$ it follows because $c_0$ embeds in $l^\infty$ (which gives the lower bound on dimension) and because the cardinality of $\ell^\infty$ is $2^{\aleph_0}$ (which gives the upper bound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm now pretty sure this isn't what you want, based on your edit, but this still gives another example for the actual question as well as an answer to Jyotirmoy's comment.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, another way to see that $2^{\aleph_0}$ is a lower bound for the linear dimensions of $\ell^1$ and friends is to consider the linearly independent set $\{(1,t,t^2,t^3,\ldots):0\lt t\lt 1\}$.  Cardinality of the spaces gives an upper bound.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-12T22:55:23.273" LastActivityDate="2010-10-12T22:55:23.273" CommentCount="6" />
  <row Id="6619" PostTypeId="2" ParentId="6613" CreationDate="2010-10-12T17:31:27.520" Score="2" ViewCount="0" Body="&lt;p&gt;The way I learned it, in ZF, we define the unary union by&#xA;$$\forall y \left(y\in\cup X \Leftrightarrow \exists z(z\in X\wedge y\in z)\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unary intersection is defined using the unary union and the Axiom of Separation:&#xA;$$\cap X = \left\{ y\in\cup X\,|\, \forall z(z\in X\rightarrow y\in z)\right\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this definition, since $\cup\emptyset = \emptyset$, then $\cap\emptyset=\emptyset$ as well. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-12T17:31:27.520" CommentCount="1" />
  <row Id="6620" PostTypeId="2" ParentId="6491" CreationDate="2010-10-12T18:01:07.540" Score="-1" ViewCount="0" Body="&lt;p&gt;If you don't require trapezoids of the same width, you can get a closed form by taking a partition of $[0,n]$ into intervals with endpoints  $n,nq,nq^2,\dots nq^k$ for $q&amp;lt;1$ and $k$ large (or infinite), or by using $(i/n)^2, i=0 \dots n$ as the endpoints.  The first approach leads to the sum of a geometric series and the second is the sum of values of a polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better answers are possible if you specify what is the problem you are ultimately trying to solve.  Is it finding the number of equal-width trapezoids needed to achieve a given accuracy in the integral $\int_0^n f(x) dx$? &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-12T18:06:24.847" LastActivityDate="2010-10-12T18:06:24.847" CommentCount="1" />
  <row Id="6621" PostTypeId="2" ParentId="6609" CreationDate="2010-10-12T18:24:58.650" Score="5" ViewCount="0" Body="&lt;p&gt;[This answer doesn't add much that is new, but it became too painful trying to post is as a comment.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One doesn't need to talk about differential forms, or integral identities, in order to prove the validity of vector calculus formulas in spherical coordinates (although both those points of view are very nice!).  If $\vec{F}(x,y,z)$ is a vector field expressed in Cartesian coordinates, say, and $\vec{F}(r,\theta,\varphi)$ is the same vector field but now expressed in spherical coordinates, then the formula for $\nabla$ in spherical coordinates is &lt;em&gt;determined&lt;/em&gt; by the requirement that $\nabla\cdot \vec{F}(x,y,z)$ and $\nabla\cdot \vec{F}(r,\theta,\varphi)$ correspond to one another as vector fields when you change from Cartesian to shperical coordinates.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $\vec{A}(x,y,z)$ and $\vec{B}(x,y,z)$ are a pair of vector fields expressed in Cartesian coordinates, with $\vec{A}(r,\theta,\varphi)$ and $\vec{B}(r,\theta,\varphi)$ being the same vector fields but now expressed in spherical coordiantes, then $\vec{A}\times \vec{B}$ (expressed in Cartesian coordinates) will correspond to $\vec{A}\times&#xA;\vec{B}$ (expressed in spherical coordinates).  So the left-hand side of the identity to be checked, when computed in Cartesian coordinates, will match (under the change from spherical to Cartesian coordinates) with the same &#xA;expression when computed in spherical coordiantes.  Similarly for the right-hand side of the identity.  Thus the identity is equally valid in Cartesian or spherical coordinates.  (Again, the key point is that the formulas for grad, div, and curl under a coordinate change are &lt;em&gt;defined&lt;/em&gt; so as to make the above argument valid.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-12T18:24:58.650" CommentCount="1" />
  <row Id="6622" PostTypeId="1" AcceptedAnswerId="6632" CreationDate="2010-10-12T18:27:30.800" Score="1" ViewCount="85" Body="&lt;p&gt;I have an inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n \ge C\log^k n \times (A\log(n^r\log^l n) + B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to turn it into an inquality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n &gt; F(A,B,C,k,l,r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that implies the first inequality, while making it $F$ as tight as possible.&#xA;How can I do that?&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2010-10-12T23:22:14.050" Title="how do I transform an inequality on $n$?" Tags="&lt;arithmetic&gt;&lt;natural-number&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6623" PostTypeId="2" ParentId="6572" CreationDate="2010-10-12T18:29:24.150" Score="6" ViewCount="0" Body="&lt;p&gt;I am going to assume that you meant to state that this was a faithful representation, meaning that the map $G \to GL_n(\mathbb{C})$ has no kernel. Otherwise, as Akhil says, the problem is dramatically underconstrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite class of examples is the semidriect product of $S_5$ acting on $(\mathbb{Z}/N)^4$, for $N$ relatively prime to $30$. In other words, look at all permutation matrices whose nonzero entries are $N$-th roots of unity, with product $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another infinite family of examples: Let $G$ be any subgroup of $GL_5(\mathbb{C})$. Let $A$ be an abelian cyclic group, embedded in $GL_5(\mathbb{C})$ as a subgroup of the scalar matrices. Then $A$ is central in $GL_5(\mathbb{C})$, so $GA$ is a finite subgroup of $GL_5(\mathbb{C})$. If $\mathbb{C}^5$ is an irreducible representation of $G$, it is also an irreducible representation of $GA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the rest of my answer says that every example looks like one of these.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a theorem of Jordan which states that there is a bound $J(n)$ such that any subgroup $G$ of $GL_n(\mathbb{C})$ has a normal abelian subgroup $A$ with $|G/A| \leq J(n)$. See &lt;a href=&quot;http://terrytao.wordpress.com/2010/02/18/a-proof-of-gromovs-theorem/#jordan&quot; rel=&quot;nofollow&quot;&gt;Terry Tao's blog&lt;/a&gt; for a nice exposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using modern improvements on Jordan's result, we can get some more precise bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, every $G$ as in your question is either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) An &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_extension&quot; rel=&quot;nofollow&quot;&gt;extension&lt;/a&gt; $0 \to A \to G \to H \to 0$, where $A$ is abelian and $H$ is a subgroup of $S_5$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) A &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_extension_%28mathematics%29#Central_extension&quot; rel=&quot;nofollow&quot;&gt;central extension&lt;/a&gt; $0 \to A \to G \to H \to 0$ where $|H| \leq 25920$ and $A$ is cyclic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think if I knew more about group cohomology I could, in principle, turn this into an infinite  but comprehensible, list of all groups that may occur.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A representation $V$ of $G$ is called primitive it is irreducible and not induced from a proper subgroup $H$ of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt; $\mathbb{C}^5$ is an irreducible, but imprimitive representation, of $G$. Let $\mathbb{C}^5$ be induced from a representation $L$ of $B \subset G$. Then $(\dim L) |G/B| = 5$, so $L$ is one dimensional and $H$ has index $5$ in $G$. Let $A = \bigcap_{g \in G/B} g B g^{-1}$. So $A$ is normal and $G/A$ acts faithfully on the $5$ element set $G/B$, showing that $G/A$ is a subgroup of $S_5$. We must show that $A$ is abelian. Suppose not, and let $C$ be the commutator of $A$. Then $C$ is killed by the map to $GL(L)$, as $L$ is one dimensional. But $C$ is normal in $G$, so it is also killed by the map to $GL(\mathbb{C}^5)$. This contradicts that we are supposed to be dealing with a faithful representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt; $\mathbb{C}^5$ is primitive.&#xA;According to a &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=2334748&quot; rel=&quot;nofollow&quot;&gt;recent paper of Collins&lt;/a&gt;, if $G$ has a a primitive representation of dimension $5$, then $|G/Z(G)| \leq 25920$. (Here $Z(G)$ is the center of $G$.) We take $A=Z(G)$. Since $\mathbb{C}^5$ is irreducible, we know that $Z(G)$ acts by scalars. So $Z(G)$ is a subgroup of $\mathbb{C}^*$, and must be cyclic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find a list of all groups that could occur as $G/Z(G)$, but perhaps you can extract it from Collins' work.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-12T18:29:24.150" CommentCount="1" />
  <row Id="6624" PostTypeId="2" ParentId="6551" CreationDate="2010-10-12T19:06:52.260" Score="0" ViewCount="0" Body="&lt;p&gt;I highly reccomend the Princeton Review series of prep books, they have worked wonders for me. The College Board website also has a lot of useful tools, such as a complete free test that you can take online, and sample questions.&#xA;&lt;a href=&quot;http://sat.collegeboard.com/practice&quot; rel=&quot;nofollow&quot;&gt;http://sat.collegeboard.com/practice&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2254" LastActivityDate="2010-10-12T19:06:52.260" CommunityOwnedDate="2010-10-12T19:06:52.260" CommentCount="2" />
  <row Id="6625" PostTypeId="1" AcceptedAnswerId="7030" CreationDate="2010-10-12T20:08:22.757" Score="3" ViewCount="109" Body="&lt;p&gt;Suppose I'm given a very slowly converging sequence $\sum a_k$. In the literature the Levin's U- transfomation is mentioned as a good universal technique for convergence acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm have difficult understanding how this method actually works and what the transformation actually is. Can anyone help? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After understand how the method works I would like to try it out on a computer. Does anyone know any software packages which have black-boxed this process. I have access to Mathematica, maple, matlab and C++. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-10-17T14:29:54.280" Title="Levin's u-transformation" Tags="&lt;sequences-and-series&gt;&lt;numerical-methods&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6626" PostTypeId="1" AcceptedAnswerId="6627" CreationDate="2010-10-12T20:28:12.213" Score="3" ViewCount="108" Body="&lt;p&gt;Consider the function $Pt(n)$. It tells us how many primitive Pythagorean Triples there are (below $n$) when any argument $n \in \mathbb{N}$ is plugged in. Is there an 'exact formula'; i.e. an elementary function of even a combination of known special functions like the Gamma and Error Function, that describes $Pt(n)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm also interested in the exact value of the limit of $Pt(n)/n$ when $n$ tends to infinity. &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-12T21:35:50.827" LastActivityDate="2010-10-12T21:35:50.827" Title="Number of Pythagorean Triples under a given Quantity" Tags="&lt;number-theory&gt;&lt;analysis&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6627" PostTypeId="2" ParentId="6626" CreationDate="2010-10-12T20:49:17.923" Score="2" ViewCount="0" Body="&lt;p&gt;To answer the sharpened version of the question I suggested (the number of primitive Pythagorean triples with largest element ${\lt}n$ ): by the parametrization of pythagorean triples as sums of two squares, this is (essentially) equal to the question of how many ways there are of expressing all the odd numbers ${\lt}n$ as a sum of two squares.  Mathworld's page on the sum-of-two-squares function at &lt;a href=&quot;http://mathworld.wolfram.com/SumofSquaresFunction.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/SumofSquaresFunction.html&lt;/a&gt; indicates that this is proportional to $n$ (though it might take some work to explicitly work out the constant of proportionality for the odd case), and so in fact your intuition is wrong; the limit you suggest tends to a finite positive value.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-12T20:49:17.923" CommentCount="1" />
  <row Id="6628" PostTypeId="2" ParentId="20" CreationDate="2010-10-12T20:53:23.093" Score="2" ViewCount="0" Body="&lt;p&gt;Going closer to the foundations of mathematics than most ever need or are comfortable with, we can also write down the classic model of $\mathbb{N_0}$ in ZF. This is of course going to be &lt;em&gt;very&lt;/em&gt; bird's eye view and rather devoid of details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There exists a set, $I$, such that $\emptyset \in I$ and $x \in I \Rightarrow x \cup \{x\} \in I$.&quot; This is the axiom of infinity. We've the chosen the letter &quot;I&quot; to represent it since it is infinite and we can do induction over $I$ (thus sets satisfying this axiom are also called &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_set_(axiom_of_infinity)&quot; rel=&quot;nofollow&quot;&gt;inductive sets&lt;/a&gt;&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we construct the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_number&quot; rel=&quot;nofollow&quot;&gt;ordinals&lt;/a&gt; starting with $0 = \emptyset$, $1 = 0 \cup \{0\} = \{\emptyset, \{\emptyset\}\}, 2 = 1 \cup \{1\} \ldots$ Let us not discuss limit ordinals or why &quot;the ordinals&quot; is not a set - it is all rather complicated. Instead let me point out that from the existense of $I$ it's possible to show the existense of the smallest infinite ordinal, $\omega$, the first ordinal with infitely many preceding ordinals. Then we define $\mathbb{N}_0$ to be $\omega$ along with this &quot;simple&quot; definition of sum: $x+0 = x$, $x + 1 = x \cup \{x\}$ and $x + (y \cup \{y\}) = x+y \cup \{x+y\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next couple of steps are easy if you're familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_class&quot; rel=&quot;nofollow&quot;&gt;equivalence classes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z} = \mathbb{N_0}\times\mathbb{N_0}/\sim$, where $(n,m) \sim (n',m')$ iff $(n+m') = (m+n')$. E.g. the integer -2 is the equivalence class consisting of $\{(0,2),(1,3),(2,4),\ldots\}$. Then you show that there is a natural inclusion $\mathbb{N_0} \hookrightarrow \mathbb{Z}$ so it makes sense to write things like $0 \in \mathbb{Z}$ even though we technically defined 0 to be $\emptyset$ and $\emptyset \notin \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Q} = \mathbb{Z}\times\mathbb{Z}\setminus\{0\}/\sim$, where $(a,b) \sim (c,d)$ iff $ad=bc$. Again we have natural inclusions $\mathbb{N}_0 \hookrightarrow \mathbb{Q}$ and $\mathbb{Z} \hookrightarrow \mathbb{Q}$ and each equivalence class is one rational number, i.e. a fraction. E.g. $\tfrac{2}{3}$ is the set $\{(2,3),(4,6),(6,9),\ldots\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we want the real numbers it gets hard. Typically in a first course in analysis you would give the students the reals in the form of an extra axiom:&#xA;There exists a field, $\mathbb{R}$, together with an order (&quot;inequality sign&quot;) $\leq$ such that:&#xA;1) $\mathbb{R}$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordered_field&quot; rel=&quot;nofollow&quot;&gt;ordered field&lt;/a&gt;, and&#xA;2) Every subset of $\mathbb{R}$ that has an upper bound has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Least_upper_bound_property&quot; rel=&quot;nofollow&quot;&gt;least upper bound&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't you start from scratch? Well... first of all there's the issue of time, but also that of mathematical maturity. Any construction of $\mathbb{R}$ is going to require quite a bit of machinery. Some algebra, some topology and being able to manipulate things like &quot;sequences of equivalence classes of sequences of rational numbers&quot; without getting lost. Even formulating the notion of uniqueness of the real numbers can be a bit tricky. We want not just that any two fields satisfying 1) and 2) to be isomorphic (both as fields and w.r.t. order), but also that the isomorphism should be unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what different constructions are there? Quite a lot actually. My 3 personal favourites are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Dedekind cuts. Based on the &quot;obvious&quot; observation that a real number is uniquely determined by which rational numbers it is smaller than and which it is larger than. Pros: Easy to prove that this construction has the least upper bound property. Difficulties: It has to be a field, so we need to define multiplication and division. How do you divide subsets of rational numbers? Especially when the divisor contains $0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Equivalence classes of rational cauchy sequences. Based on the observation that the only way a rational cauchy sequence can &lt;em&gt;not&lt;/em&gt; converge is if the limit is somehow &quot;not there&quot;. So we just set the limit to be the sequence itself! This is of course nonsense, but it can be made precise. Pros: How easy it is to show it's a field: The set of rational cauchy sequences under coordinate-wise addition and multiplication is a ring and the rational sequences that converge to $0$ are a maximal ideal! Difficulties: Defining the order (note that $(\tfrac{1}{n})$ belongs to the same equivalence class as $0$) and then showing that the order is in fact total. Keeping track of sequences of equivalence classes of rational cauchy sequences when you're trying to prove that your field is complete can be a bit tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) &quot;Almost-homomorphisms&quot; of $\mathbb{Z}$. Based on the observation (at least since we got pixel-based displays) that a real number $\alpha$ is in some sense the same as choosing a rule for how to draw a line with slope $\alpha$ on an infinite ($\mathbb{Z}\times\mathbb{Z}$) computer display. Pros and cons? Not sure actually. It could be a fun little project to work out the details I guess.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-13T06:01:26.353" LastActivityDate="2010-10-13T06:01:26.353" CommentCount="2" />
  <row Id="6629" PostTypeId="1" CreationDate="2010-10-12T21:11:27.143" Score="2" ViewCount="217" Body="&lt;p&gt;Suppose I have a variable $s$ that has a geometric distribution with success parameter x. So the probability of success on  trial $s$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_s = (1 - x)^{s - 1} x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following function of $s$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(s) = \delta^s$,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta &amp;lt; 1 $.  Then the expected value of this function is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_ 1^{\infty} V(s) p_s = \frac{\delta x} {1 - \delta (1 - x)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the deviation of V(s) from its mean :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(V(s) - \frac{\delta x} {1 - \delta (1 - x)}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What distribution does this have?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am really interested in is the behavior of the follow ordinary stochastic differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{V} =  \delta^s  - V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stochastic because the law of motion depends on the random variable $s$.  The stochastic approximation technique allows me to focus on the mean dynamics, which are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{V} =  \frac{\delta x} {1 - \delta (1 - x)}  - V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose equilibrium is simply $\hat{V} = \frac{\delta x} {1 - \delta (1 - x)}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to characterize the large deviation properties of the original ODE, IE calculate the exponential likelihood $V$ crosses a particular boundary $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $\delta = .95$ and $x= .01$, $\hat{V} = 0.159664$.  If we set $V_0 = 0.159664$, and let $V_t = V_{t-1} + .2 (\delta^s - V_{t-1})$, and $s$ has the above distribution, how do I calculate the expected time to $V_t$ crossing, say, $c=.4$?  What is the associated rate function?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Editing to give the background to the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the following stochastic dynamic system:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$V_t = V_{t-1} + \gamma (\delta^{S_{t-1}} - V_{t-1})$,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $\delta &amp;lt;1$, $\gamma&amp;lt;1$, both positive, and each $S_t$ is an i.i.d. geometric random variable with success parameter $x$.  What this models is agents who have to wait a geometric length of time to get a reward of value 1, and they have time discount factor $\delta$, and they are learning  the expected value of their reward using a constant gain adaptive  learning procedure, with gain $\gamma$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So considering $S$ as following a Poisson, rather than Geometric, makes little difference to the sense of the problem.  The goal is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the equilibria of the learning dynamics&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Characterize the mean time to escape from this equilibrium, the large deviation properties.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, for large deviation properties, I should be able to use something like Cramer's theorem, I think.  As Mike says below, iterating the dynamics gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_t = (1−γ)^tV_0+γ\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it all depends on $\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$, the discounted sum of some random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Probability that $V_t &gt;c$ is then given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pr(V_t &gt; c ) = Pr(\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k} &gt; \frac{c-(1−γ)^tV_0}{\gamma})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Z:=\frac{c-(1−γ)^tV_0}{\gamma}$.  so we need the distribution of $\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$, which as Mike says is a discounted sum of independent random variables, and so we have that this is approximately normal, for large $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning now to the question of large deviations, if things were not discounted by powers of $(1-\gamma)$, we could appeal directly to Cramer's Theorem; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Pr(\sum \delta^{S_k} &gt; n a) \leq Exp[-n(r^* a - Log(E[Exp(r^* \delta^S)])] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r^* $ is chosen to maximize $r a - Log(E[Exp(r \delta^S)]$.  The problem would be to simply calculate the moment generating function of  $\delta^S$.  But, I don't exactly want $Pr(\sum \delta^{S_k} &gt; n a)$, I need $ Pr(\sum (1-\gamma)^{t-1-k} \delta^{S_k} &gt; a)$; so not an average of $\delta^{S_k}$s, but a discounted sum.  so it is not quite a direct application.&lt;/p&gt;&#xA;" OwnerUserId="2464" LastEditorUserId="2464" LastEditorDisplayName="" LastEditDate="2010-10-19T19:48:58.187" LastActivityDate="2010-10-19T19:48:58.187" Title="Large Deviation Properties of a function of a geometric random variable" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;differential-equations&gt;&lt;approximation&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6630" PostTypeId="2" ParentId="6609" CreationDate="2010-10-12T21:13:00.187" Score="1" ViewCount="0" Body="&lt;p&gt;[I post this answer as suggested in a comment]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Vector Calculus identities with Div, Grad, Curl are independent of the  coordinate system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible explanation is that they may be expressed in terms of integrals whose values do not depend on the particular coordinate system one uses. As an example, one has the following limit for the divergence of the vector field $\overrightarrow{F}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \overrightarrow{F}=\underset{V\rightarrow 0}{\lim }\frac{1}{V}\underset{S}{\int \int }\overrightarrow{F}\cdot \overrightarrow{n}\;dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is the volume of a bounded closed region $T$, $S$ is the surface&#xA;of $T$, and $\overrightarrow{n}$ the unit outer normal vector to $S$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-13T21:00:28.560" LastActivityDate="2010-10-13T21:00:28.560" />
  <row Id="6631" PostTypeId="1" CreationDate="2010-10-12T21:49:20.877" Score="1" ViewCount="101" Body="&lt;p&gt;I have a homework assignment to find the two smallest probable primes with 12 digits, where a probable prime is defined as a number such that a^p-1 = 1 (mod p), where a = 2,3,5,7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm obviously not asking for this to be done for me, but I'm having trouble figuring out a way to do anything other than to brute-force it by checking each of the four congruences for each (odd) 12 digit number n until i find two of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone offer me some hint, relevant theorem, or something that I might be able to use to figure out how to do find these numbers in a more efficient manner? Thanks for any help. &lt;/p&gt;&#xA;" OwnerUserId="2466" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-13T03:23:38.213" LastActivityDate="2010-10-13T03:23:38.213" Title="Finding probable primes with large numbers of digits" Tags="&lt;homework&gt;&lt;prime-numbers&gt;" CommentCount="14" />
  <row Id="6632" PostTypeId="2" ParentId="6622" CreationDate="2010-10-12T22:04:26.930" Score="0" ViewCount="0" Body="&lt;p&gt;I don't think you will get a satisfying bound on n without iteration.  You can expand the right side to $n \ge AC*log^{(k+r)}n+AC*log^kn*log  log^ln+BClog^kn$.  If n is very large, you can even ignore all but the first term as it will dominate.  Using the fact that $log n$ varies slowly with $n$, start with some guess for $log n$, such as 1 or 2.  Plug it in to the right hand side and calculate a new value of n.  Then calculate $logn$.  Continue to convergence, which should be quick.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-12T23:22:14.050" LastActivityDate="2010-10-12T23:22:14.050" CommentCount="2" />
  <row Id="6633" PostTypeId="2" ParentId="6629" CreationDate="2010-10-12T22:09:10.370" Score="1" ViewCount="0" Body="&lt;p&gt;If $\delta &gt; 0$, then the transformations you are applying to the random variable $S$ to get the quantity $T = \delta^S - \frac{\delta x}{1 - \delta(1-x)}$ are all one-to-one.  Since $S$ is discrete, this means you can construct the probability distribution of $T$ by inverting those transformations.  This would yield&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(T = t) = P\left(\delta^S - \frac{\delta x}{1 - \delta(1-x)} = t\right) = P\left(S = \frac{\ln\left(\frac{\delta x}{1 - \delta(1-x)} + t\right)}{\ln \delta}\right) = (1 - x)^{y-1} x,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $$y = \frac{\ln\left(\frac{\delta x}{1 - \delta(1-x)} + t\right)}{\ln \delta},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any value of $t$ such that $y$ is in the set $\{1, 2, \ldots\}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone else can help you with the other questions.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-12T22:09:10.370" CommentCount="6" />
  <row Id="6634" PostTypeId="1" AcceptedAnswerId="6637" CreationDate="2010-10-12T23:01:23.867" Score="2" ViewCount="57" Body="&lt;p&gt;I'm trying to find a factorization for $f(z) = z - c$ in inner and outer functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first I try to find for what $c$, $f$ is cyclic for the unilateral shift $M_z$. There is a theorem that states that if $A$ is a bounded operator on a Hilbert space $H$ then $x \in H$ is cyclic for $A$ iff zero is the only element that is orthogonal to all $A^n x$ for $n = 0,1,2,\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M_z$ be the multiplication operator by $z$ (&quot;the unilateral shift on $H^2$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $g_n(z) = M_z^n (z - c) = -cz^n + z^{n + 1}$ I find that the power series coefficients of $g_n$ (which is given by $g_n(z) = \sum_m a_{n,m} z^m$) are $a_{n,n} = -c$ and $a_{n,n+1} = 1$. If I insert this in the inner product (where $h(z) = \sum \overline{b_n} z^n$), $(g_n, h) = 0$ for all $n = 0, 1, 2, \ldots$. I obtain that $b_{n + 1} - c b_n = 0$ for $n \geq 0$. This gives that $b_n = C \cdot c^n$. But $\sum c^n z^n = 0$ only if $c = 0$. This would imply that $f(z) = z$ is cyclic but a cyclic vector cannot have a zero in the unit disk! What is wrong here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I would find this $c$, then I would find a factorization if the $c$ makes $f$ cyclic because then the inner function is $1$ and the outer function $f$. How to do it if $f$ is not cyclic?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-12T23:37:49.313" LastActivityDate="2010-10-12T23:44:40.983" Title="Factorization inner and outer functions." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6635" PostTypeId="1" AcceptedAnswerId="6703" CreationDate="2010-10-12T23:14:03.337" Score="2" ViewCount="53" Body="&lt;p&gt;Let $f : {\cal S}_+^n \mapsto \mathbb{R}$ be a function defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(Q) := {\rm tr} WQ^{-1}W + {\rm tr} Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W \in {\cal S}_{+}^{n}$ is a symmetric positive definite matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to compute a subdifferential of $f$ at a point $Q \in {\cal S}_+^n$ for which $f(Q) &amp;lt; \infty$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f(Q) = \infty$ for $Q$ such that range of $W$ is not contained within the range of Q. Furthermore, we know that $\min_{Q \in {\cal S}_+^n} f(Q) = 2{\rm tr}W$ which is obtained at $Q = W$.&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-10-13T15:55:11.740" Title="How to compute the subdifferential of a variational represention of the trace norm?" Tags="&lt;matrices&gt;&lt;convex-analysis&gt;&lt;differential&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6636" PostTypeId="2" ParentId="6544" CreationDate="2010-10-12T23:15:05.130" Score="5" ViewCount="0" Body="&lt;p&gt;Here's another idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P_n(x)$ be the polynomial defined by&#xA;$$P_n(x) = \frac{x^n – (1-x)^n}{2x-1},$$&#xD;&#xA;and note that $P_n(x)=(1-x)P_{n-1}(x) + x^{n-1} \qquad (1), $ where $P_1(x)=1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $I_n = \int_0^{1} P_n(x) dx.$ Integrating (1) between 0 and 1 we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \int_0^{1} (1-x)P_{n-1}(x) dx + \frac{1}{n},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and using the symmetry of $P_n(x)$ about 1/2 we note that&#xA;$\int_0^{1} (1-x)P_{n-1}(x) dx = \frac{1}{2}I_{n-1}.$ Hence we obtain the recurrence relation for $n &amp;gt; 1$&#xD;&#xA;$$I_n = \frac{1}{2}I_{n-1} + \frac{1}{n}, \textrm{ where } I_1=1. \qquad (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which it follows that if $\lim_{n \rightarrow \infty} n I_n$ exists then it must equal 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the limit exists we can solve (2) to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \sum_{k=1}^n \frac {1}{k2^{n-k}},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and straightforward algebra shows that $(n-1)I_{n-1} – nI_n &amp;gt; 0$ for sufficiently large $n$ (in fact, $n&gt;6$), and since $nI_n$ is bounded below by 0 the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for completeness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n-1)I_{n-1} – nI_n = \left\lbrace \sum_{k=1}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right\rbrace - \frac{n}{2^{n-1}} \quad \textrm{ for } n&amp;gt;2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing the first term in the summation with the negative term we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{2(n-1)(n-2)} &amp;gt; \frac{n}{2^{n-1}} \quad \textrm{ for } n \ge 13.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\lbrace nI_n \rbrace$ eventually forms a strictly decreasing sequence.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-05T08:39:55.590" LastActivityDate="2010-12-05T08:39:55.590" CommentCount="1" />
  <row Id="6637" PostTypeId="2" ParentId="6634" CreationDate="2010-10-12T23:38:48.157" Score="1" ViewCount="0" Body="&lt;p&gt;What is wrong is the reasoning where you wrote &quot;But $\sum c^n z^n = 0$ only if $c=0$.&quot;  What you need to know is for which choices of $c$ does the condition $b_{n+1}=c b_n$ for all $n$, with $b_0,b_1,\ldots$ being a square summable sequence, force all of the $b_n$'s to be zero.  If $c=0$, this is not the case; just take $h(z)=1$.  The coefficients are not forced to be zero when $|c|&amp;lt;1$, where you can take $h(z)=1 + cz +c^2z^2 +c^3z^3 +\cdots$.  But the condition $b_{n+1}=c b_n$ is impossible for $|c|\geq 1$ unless $h=0$, so your criterion is $|c|\geq1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $f$ is not cyclic, for inner part you can take the holomorphic automorphism of the unit disk that swaps $c$ and 0, and what remains will be an invertible element of $H^\infty$, hence cyclic.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-12T23:44:40.983" LastActivityDate="2010-10-12T23:44:40.983" CommentCount="1" />
  <row Id="6638" PostTypeId="1" AcceptedAnswerId="6697" CreationDate="2010-10-13T00:22:36.790" Score="2" ViewCount="78" Body="&lt;p&gt;Let $p(x)$ be a fixed distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A, C &gt; 0$ be constants.&#xA;Let $\epsilon &gt; 0$. Can we find an example of a distribution $q_{\epsilon}$ such that $\mathrm{KL}(p||q_{\epsilon}) &amp;lt; \epsilon$, but $E[(\log (p(x)/q(x)))^2] \ge A * \epsilon^C$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ and $q_{\epsilon}$ would be a coin toss or a geometric distribution, that would be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I am making myself clear. This is a rather hard question.&#xA;Perhaps I could also get a clue if someone could tell me whether there is a name for the quantity $E[(\log (p(x)/q(x)))^2]$. (KL divergence is defined as $E[\log (p(x)/q(x))]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All expectations are taken with respect to $p$.&#xA;So, for example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E[(\log (p(x)/q(x)))^2] = \sum_x p(x) \left(\log (p(x)/q(x))\right)^2$&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditorDisplayName="" LastEditDate="2010-10-13T11:47:30.240" LastActivityDate="2010-10-13T14:38:54.583" Title="can KL divergence converge to 0, but variance of log odds not converge to 0?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6639" PostTypeId="1" AcceptedAnswerId="6654" CreationDate="2010-10-13T00:37:38.827" Score="2" ViewCount="84" Body="&lt;p&gt;If $x$ is an element in a standard convex linear optimization set constrained by $Ax = b, x \geq 0$, then how can I prove $d$ is a feasible direction only if $Ad=0$ and $di \geq 0$ for every $i$ where $xi=0$?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T01:01:38.007" LastActivityDate="2010-10-13T15:41:02.193" Title="Prove feasible direction" Tags="&lt;linear-programming&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6640" PostTypeId="1" AcceptedAnswerId="6641" CreationDate="2010-10-13T00:48:43.073" Score="2" ViewCount="104" Body="&lt;p&gt;I am looking for algorithms that compute the inverse of Euler's totient function.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-13T03:36:00.213" LastActivityDate="2010-10-13T03:36:00.213" Title="Algorithms for computing inverse of Euler's phi" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="6641" PostTypeId="2" ParentId="6640" CreationDate="2010-10-13T01:23:43.747" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://library.wolfram.com/infocenter/MathSource/696/&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a nice article off of wolfram that gives an efficient algorithm for computing the inverse of the Euler totient function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Download the &lt;strong&gt;invphi.nb&lt;/strong&gt; file at the bottom &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-13T01:30:57.920" LastActivityDate="2010-10-13T01:30:57.920" CommentCount="3" />
  <row Id="6642" PostTypeId="2" ParentId="6507" CreationDate="2010-10-13T01:30:57.120" Score="1" ViewCount="0" Body="&lt;p&gt;You can think of a group (e.g., via Cayley's Theorem) as a collection of bijections that contains the identity and is closed under composition and inverses. So groups are objects that &quot;act on&quot; arbitrary sets (as bijections). That seems to be what you are talking about when you talk about group elements being &quot;automorphisms of a particular set&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jack Schmidt pointed out in &lt;a href=&quot;http://mathoverflow.net/questions/27971/why-is-there-no-cayleys-theorem-for-rings&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt;, every associative unital ring is a subring of the endomorphism ring of its underlying additive group. That is, you can think of any ring as being a collection of group endomorphisms of some group, with the ring addition corresponding to the usual addition of endomorphisms (inherited from the group), and ring multiplication corresponding to composition of functions. This closely parallels the view of &quot;groups as collections of bijections on a set&quot;: rings are &quot;collections of endomorphisms of a group&quot;. If the group has nothing to do with numbers, then the corresponding ring will also have &quot;nothing to do with numbers&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of groups as objects that act on sets as bijections, then you can think of (associative unital) rings as objects that act on abelian groups as endomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this answer is satisfying, but it seems to me to be along the lines you are inquiring.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-13T01:30:57.120" CommunityOwnedDate="2010-10-13T01:30:57.120" CommentCount="1" />
  <row Id="6643" PostTypeId="1" AcceptedAnswerId="6644" CreationDate="2010-10-13T01:38:01.660" Score="2" ViewCount="95" Body="&lt;p&gt;Construct a subset of [0,1] in the same manner as the Cantor set by removing from each remaining interval a subinterval of relative length $\theta$, $0&amp;lt;\theta &amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first statement in a homework exercise. I don't know if my English parsing skills are lacking today or what, but it's unclear to me what this means. I just need help understanding what this construction &quot;looks like.&quot; &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-13T16:57:01.107" Title="Cantor-like construction" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6644" PostTypeId="2" ParentId="6643" CreationDate="2010-10-13T01:47:14.100" Score="3" ViewCount="0" Body="&lt;p&gt;The construction of the Cantor set begins with the removal of the interval $[1/3,2/3]$, an interval of length $1/3$.  The remaining intervals have length $1/3$; you remove intervals of length $1/9$ from each of them.  At stage $k$, you're removing $2^{k-1}$ intervals of length $1/3^k$.  The total length removed is then $$\sum_{k=1}^\infty\frac{2^{k-1}}{3^k}=\frac{1}{3}\frac{1}{1-\frac{2}{3}}=1$$&#xA;so the Cantor set has measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\theta\ne 1/3$.  You can repeat the same construction with $\theta$ instead of $1/3$: at step $k$, you remove $2^{k-1}$ intervals of length $\theta^k$.  This is going to look very much like the Cantor set -- in particular, it has empty interior, which you can prove.  If your homework problem is going where I'm guessing it's going, you should calculate the measure of the set you get for different $\theta$.  Actually, you should do that no matter what the homework says.  The results are quite surprising.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-13T16:57:01.107" LastActivityDate="2010-10-13T16:57:01.107" CommentCount="3" />
  <row Id="6645" PostTypeId="1" AcceptedAnswerId="6651" CreationDate="2010-10-13T02:17:44.817" Score="3" ViewCount="132" Body="&lt;p&gt;How do you prove $|\sqrt[3]{x} - \sqrt[3]{y}|$ $\leq$ $\sqrt[3]{|x-y|}$?&lt;/p&gt;&#xA;" OwnerUserId="2467" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-13T02:39:32.917" LastActivityDate="2010-10-13T02:52:57.243" Title="Cube Root Inequality" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6646" PostTypeId="1" AcceptedAnswerId="6656" CreationDate="2010-10-13T02:18:04.863" Score="2" ViewCount="110" Body="&lt;p&gt;I can't seem to understand Generalized Set Operations. Can someone explain it for me? For example, in Example 1.12, suppose &lt;code&gt;k = 3&lt;/code&gt; then set &lt;code&gt;A(of k) = {3, 4, 5, 6 ...}&lt;/code&gt;, then how is &lt;code&gt;{3,4,5 ...} = {1,2,3 ...}&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NvXFE.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2006" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T02:02:31.353" LastActivityDate="2010-11-21T02:02:31.353" Title="Generalized Set Operation" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6647" PostTypeId="1" AcceptedAnswerId="6650" CreationDate="2010-10-13T02:23:20.670" Score="5" ViewCount="317" Body="&lt;p&gt;I'm facing a practical problem where I've calculated a formula that, with the help of some programming, can bring me to my final answer. However, the numbers involved are so big that it takes ages to compute, and I think it might not be necessary. I have the following formula, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits^{k}_{i=m}(N-i)^{k-i}(\frac{1}{N})^k\frac{k!}{(k-i)!i!} \leq a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which I need to calculate N. The rest of the values are constants, but in the range of the 100,000s. The factorial there is giving me a headache, since the values involved are too large; what simplifications could I make that will loosen the bounds slightly and thereby simplify the calculation? Are there any standard tricks? Or perhaps a way to calculate this in matlab / octave?&lt;/p&gt;&#xA;" OwnerUserId="2471" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-13T03:04:34.030" LastActivityDate="2010-11-25T18:48:17.030" Title="How to simplify or calculate a formula with very big factorials" Tags="&lt;approximation&gt;&lt;binomial-coefficients&gt;&lt;big-numbers&gt;" AnswerCount="3" />
  <row Id="6648" PostTypeId="1" AcceptedAnswerId="6772" CreationDate="2010-10-13T02:28:30.037" Score="4" ViewCount="148" Body="&lt;p&gt;California issues license plates in numeric order (if we turn the letters into numbers).  I have fun noticing the latest plate I have seen.  I am interested in what you can derive from a series of these observations.  I understand that sampling from $\{1,2,3...n\}$ the only useful data is the highest value you have seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's oversimplify the problem.  Assume the highest plate issued is $N_0+n*t$, $n$ in plates/day and $t$ in days.  Assume a similar number of low valued plates come off the road each day.  I don't observe a consistent number of plates each day, but it averages out.  Over a long time, the increase in highest plate seen should give a measure of $n$.  The only other data I have is how frequently I see a new highest plate.  Does that give some measure of how far my highest plate is from the highest issued?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we are asked to cite the source of a question, I made it up.  You probably guessed.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T11:42:35.893" LastActivityDate="2010-10-14T11:45:52.643" Title="License Plate Statistics" Tags="&lt;statistics&gt;&lt;puzzle&gt;&lt;cryptography&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6649" PostTypeId="1" AcceptedAnswerId="6674" CreationDate="2010-10-13T02:45:48.660" Score="6" ViewCount="193" Body="&lt;p&gt;I've been trying to work out the best way to understand why Fourier series converge, and it's a little embarrassing but I don't even know a rigorous proof.  Can someone please help put me on the right track to thinking about these issue's in the proper way?  I am especially interested any geometric ways to think about the convergence issue (something I suppose which takes advantage of the fact that each component $e^{in\theta}$ corresponds to some point along the unit circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T09:48:38.927" LastActivityDate="2010-10-24T13:20:30.903" Title="Geometric intuition behind convergence of Fourier series." Tags="&lt;fourier-series&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="6650" PostTypeId="2" ParentId="6647" CreationDate="2010-10-13T02:46:31.170" Score="7" ViewCount="0" Body="&lt;p&gt;You need &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;.  It is very accurate for large factorials.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-13T02:46:31.170" />
  <row Id="6651" PostTypeId="2" ParentId="6645" CreationDate="2010-10-13T02:52:57.243" Score="7" ViewCount="0" Body="&lt;p&gt;It's not true. Try $x=1$ and $y=-1$.  If you assume $x$ and $y$ have the same sign, and you might as well assume $x\gt y\gt 0$, then it reduces to showing $(x-y)^3\leq x^3-y^3$ (WLOG replacing variables in the original inequality with cubes and cubing both sides).  This is true because $x\gt y\gt 0$ implies $3xy^2&amp;lt;3x^2y$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-13T02:52:57.243" CommentCount="1" />
  <row Id="6653" PostTypeId="1" CreationDate="2010-10-13T03:08:47.753" Score="3" ViewCount="190" Body="&lt;p&gt;I'm trying to prove &quot;the following generalization of Theorem 5 [ Th.5: if $a|bc$ and $(a,b)=1$ then $a | c$ ], which uses the same argument for its proof&quot; (Sierpinski, The Theory of Numbers): if $a$, $b$, and $c$ are integers such that $b | ac$, then $b | (a,b)(b,c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to prove it without any reference to prime numbers (which the author introduces way later), using only divisibility and facts like $(a,b)[a,b]=ab$. Here's what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(a,b)=d_a$, $(b,c)=d_c$, $a=a'd_a$, $c=c'd_c$. Since $b | ac$ and $a | ac$ $\Rightarrow [b,a] | ac$ (this is the argument used in Th5's proof) $\Rightarrow ab/(a,b) | ac \Rightarrow a'b | ac \Rightarrow b | d_a c$. Doing the same for $c$, we get $b | ad_c$. From this we also have $b | (ad_c,d_a c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.            &lt;/p&gt;&#xA;" OwnerUserId="1694" LastEditorUserId="1313" LastEditorDisplayName="" LastEditDate="2010-10-13T03:12:26.200" LastActivityDate="2010-10-19T01:50:34.103" Title="Basic divisibility fact" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="6654" PostTypeId="2" ParentId="6639" CreationDate="2010-10-13T03:23:00.823" Score="2" ViewCount="0" Body="&lt;p&gt;[&lt;strong&gt;EDITED&lt;/strong&gt; after OP's clarification that $i$ is a subscript.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I get the idea of feasible direction right then it means that $x+\lambda d$ should be in the constraint set for some $\lambda&gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose that $Ad \neq 0$. Then $A(x+\lambda d)=b+\lambda Ad \neq b$, violating the equality constraint.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose that there is an $i$ such that $x_i=0$ but $d_i&amp;lt;0$. Then $x_i+\lambda d_i&amp;lt;0$ for all $\lambda&gt;0$ and hence the non-negativity constraint would be violated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-13T15:41:02.193" LastActivityDate="2010-10-13T15:41:02.193" />
  <row Id="6655" PostTypeId="2" ParentId="6653" CreationDate="2010-10-13T03:26:53.747" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: write $(a,b)$ as a linear combination of $a$ and $b$, $(b,c)$ as a linear combination of $b$ and $c$. &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-13T03:26:53.747" CommentCount="2" />
  <row Id="6656" PostTypeId="2" ParentId="6646" CreationDate="2010-10-13T03:31:30.127" Score="7" ViewCount="0" Body="&lt;p&gt;When you're learning set theory, the definitions of union and intersection you're first given are binary: they're defined on a pair of sets.&#xA;\begin{align}A\cup B&amp;amp;=\lbrace x:x\in A\text{ or }x\in B\rbrace\\&#xA;A\cap B&amp;amp;=\lbrace x:x\in A\text{ and }x\in B\rbrace\end{align}&#xA;Since they're also associative (the order you do a chain of consecutive unions or intersections in doesn't matter), you can define the union or intersection of any finite collection of sets.  This is the first pair of equations your book gives above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's always good to generalize your definitions!  In particular, those definitions don't allow us to take a union or intersection of an &lt;em&gt;infinite&lt;/em&gt; collection of sets -- even though the operation makes intuitive sense to us!  (The intersection of sets is just the elements that are in all the sets, for example, and that should make just as much sense with an infinite number of sets as with a finite number of sets.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we do is &lt;em&gt;redefine&lt;/em&gt; union and intersection to apply to a &lt;em&gt;single&lt;/em&gt; set -- this time, a set of sets.  This is $\mathcal{A}$ above, and the second pair of equations your book gives.  There are a couple different notations for this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;$\bigcup\mathcal{A}$&lt;/li&gt;&#xA;&lt;li&gt;$\bigcup(A|A\in\mathcal{A})$ (the one your book uses)&lt;/li&gt;&#xA;&lt;li&gt;$\bigcup_{A\in\mathcal{A}}A$&lt;/li&gt;&#xA;&lt;li&gt;$\bigcup_{i\in I}A_i$, where $I$ is what's called an _indexing set_.  (For example, if $I=\mathbb{N}$, this is equal to $A_1\cup A_2\cup A_3\cup\dotsb$.&lt;/li&gt;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For me, the last notation is the most revealing.  I don't know if you're familiar with cardinality yet.  If not, it's a measure of the size of a set, and one of the most important theorems in set theory is that there is no greatest cardinality -- that is to say, there are multiple sizes of infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the indexing set notation, $I$ itself doesn't really matter as much as its cardinality.  If it is $\mathbb{N}$, you're taking the union of $A_1,A_2,\dotsc$, a countable number of sets.  But it could also be $\mathbb{R}$, in which case you're taking the union of the sets $A_x$ for every real number $x$.  Because $\mathbb{R}$ has a higher cardinality than $\mathbb{N}$, there's no way to write this out in a list!  So thinking of this as $$A_{x_1}\cup A_{x_2}\cup\dotsb$$ is just wrong.  &lt;em&gt;You can't get this union from a generalization of the binary union.&lt;/em&gt;  Hopefully this shows how powerful this generalization is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, I'll explain the example at the bottom of the page.  In this case, the set $A_n=\lbrace n,n+1,n+2,\dotsc\rbrace$: that is, all integers greater than or equal to $n$.  The union $\bigcup_{n\in\mathbb{N}}A_n$ is the set of integers that are in at least one set $A_n$.  But every $n\ge 1$ is in such a set -- for example, $n\in A_n$.  So this union is equal to $\mathbb{N}$.  On the other hand, &lt;em&gt;no integer&lt;/em&gt; is in all sets $A_n$ -- for example, $n\not\in A_{n+1}$.  So $\bigcap_{n\in\mathbb{N}}A_n=\emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.  Ask if you're still confused!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-13T04:51:01.303" LastActivityDate="2010-10-13T04:51:01.303" CommentCount="6" />
  <row Id="6657" PostTypeId="2" ParentId="6653" CreationDate="2010-10-13T03:45:40.113" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ \ \rm bd = ac\ \:\Rightarrow\:\ (a,b)\:(b,c)\: =\ b\: (a,b,c,d)\ $ via GCD distributive, associative laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you know $\rm\ [a,b]\: (a,b)\: =\: ab\ $ then using this LCM*GCD law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have $\rm\ \ \: a,b\:|\:ac\ \:\Rightarrow\ \: [a,b]\:|\:ac\ \ \Rightarrow\:\ ab\:|(a,b)\:ac\ \Rightarrow\ b\:|\:(a,b)\:c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the same to $\rm\:c\:$ yields the result. This appears to be the proof that Sierpinski has in mind since his prior proof is merely the special case where $\rm\ \ (a,b)= 1\:,\: $ and it employs the consequent specialization of the above $\ $ LCM*GCD $\ $ law, explicitly that $\rm\ (a,b) = 1\ \Rightarrow\ [a,b] = ab\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of said LCM*GCD law simpler than Sierpinski's see this &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;one line universal proof  $\:$ [THEOREM here]&lt;/a&gt;. Not only is this proof much simpler but it is also more general - it works in any domain. Note also that the result that you seek is a special case of the powerful $\:$ &lt;strong&gt;Euler four number theorem (Vierzahlensatz)&lt;/strong&gt;,$\:$ or $\:$ &lt;strong&gt;Riesz interpolation&lt;/strong&gt;, or $\:$ &lt;strong&gt;Schreier refinement&lt;/strong&gt;,$\:$ see &lt;a href=&quot;http://groups.google.com/groups?selm=y8z7k9leoku.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;my post here&lt;/a&gt;. For another example of the simplicity of proofs employing the fundamental GCD laws (associative, commutative, distributive, absorptive), see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zej18x733.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post on the Freshman's Dream&lt;/a&gt; $\rm\: (A+B)^n =\: A^n + B^n\ $ for GCDs / Ideals, $\:$ if $\rm\: A+B\ $ is cancellative. It's advantageous to present gcd proofs using these basic laws (vs. the Bezout linear form) since such proofs will generalize better (e.g. to ideal arithmetic) and, moreover, since these laws are so similar to integer arithmetic, we can reuse are well-honed expertise manipulating expressions obeying such laws (associative, commutative, distributive). For examples see said Freshman's Dream post.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-19T01:50:34.103" LastActivityDate="2010-10-19T01:50:34.103" />
  <row Id="6658" PostTypeId="2" ParentId="6605" CreationDate="2010-10-13T04:06:19.407" Score="2" ViewCount="0" Body="&lt;p&gt;I searched Google for &quot;magnetic fractal&quot;, and found the answer on &lt;a href=&quot;http://www.icd.com/tsd/fractals/beginner4.htm&quot; rel=&quot;nofollow&quot;&gt;the first hit&lt;/a&gt;. It quotes &lt;a href=&quot;http://spanky.triumf.ca/www/fractint/magnet_type.html&quot; rel=&quot;nofollow&quot;&gt;the Fractint documentation&lt;/a&gt; (I can't resist mentioning that Fractint is the grand-daddy of freeware fractal-generating software for personal computers -- it had its first release in 1988, and is still being maintained!):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These fractals use formulae derived from the study of hierarchical lattices, in the context of magnetic renormalisation transformations. This kinda stuff is useful in an area of theoretical physics that deals with magnetic phase-transitions (predicting at which temperatures a given substance will be magnetic, or non-magnetic). In an attempt to clarify the results obtained for Real temperatures (the kind that you and I can feel), the study moved into the realm of Complex Numbers, aiming to spot Real phase-transitions by finding the intersections of lines representing Complex phase-transitions with the Real Axis. The first people to try this were two physicists called Yang and Lee, who found the situation a bit more complex than first expected, as the phase boundaries for Complex temperatures are (surprise!) fractals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The formulas for the two fractals are also given there. They are&#xA;$$z \mapsto \left(\frac{z^2 + (c-1)}{2z + (c-2)}\right)^2$$&#xA;for magnet 1, and&#xA;$$z \mapsto \left(\frac{z^3 + 3(c-1)z + (c-1)(c-2)}{3z^2 + 3(c-2)z + (c-1)(c-2) + 1}\right)^2$$&#xA;for magnet 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm by no means knowledgeable on this subject, but I've been looking at some of &lt;a href=&quot;http://math.bu.edu/people/bob/papers.html&quot; rel=&quot;nofollow&quot;&gt;Robert Devaney's papers&lt;/a&gt;, which I came across &lt;a href=&quot;http://www.tetration.org/Fractals/PPEM/index.html&quot; rel=&quot;nofollow&quot;&gt;via tetration.org&lt;/a&gt;. Looking at Devaney's images, I'd guess that the reason why these fractals have the beautiful Sierpinski-gasket-like structures, while the standard quadratic Julia and Mandelbrot sets don't, is that each of the formulas defining these fractals is a &lt;em&gt;rational function&lt;/em&gt;, that is, a ratio of two polynomials, rather than a single polynomial. I believe the field that studies these things is called complex dynamics. My knowledge doesn't extend to &lt;em&gt;how&lt;/em&gt; rational functions with poles give rise to Julia sets with gaskets, but you could try looking for the answer to that in some of Devaney's papers, or in the book &lt;em&gt;Iteration of Rational Functions&lt;/em&gt; by Alan F. Beardon which is cited in &lt;a href=&quot;http://en.wikipedia.org/wiki/Julia_set&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on Julia sets&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-13T04:06:19.407" />
  <row Id="6659" PostTypeId="2" ParentId="6508" CreationDate="2010-10-13T04:20:12.140" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, it directly follows from the union bound.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-13T04:20:12.140" />
  <row Id="6660" PostTypeId="2" ParentId="6507" CreationDate="2010-10-13T04:39:33.907" Score="1" ViewCount="0" Body="&lt;p&gt;I think this is a legitimate observation -- it is possible to do lots of group theory without knowing much about integers or real numbers, but these number systems seem to come up immediately whenever one is dealing with rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the main reason is that every ring has a characteristic, and therefore a ring with unity necessarily includes a copy of either the integers or some $\mathbb{Z}_n$.    Furthermore, any element of a finitely-generated ring can be written as a (possibly noncommutative) polynomial in the generators, where the coefficients of the polynomial are integers (or perhaps some elements of $\mathbb{Z}_n$).  Adding and multiplying these polynomials inherently requires addition and multiplication of integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get past this basic &quot;number-ness&quot;, there are many rings that are otherwise unrelated to either integers or real numbers.  For example, it's possible to develop the theory of polynomial rings quite a bit without using numbers for anything other than coefficients.  The same goes for group rings, whose elements are hardly more number-like than typical group elements.  For a more sophisticated example, the elements of the (integer) cohomology ring of a topological space are essentially geometric objects, not numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fields are the same way.  Every field inherently contains either the rational numbers $\mathbb{Q}$ or a prime field $\mathbb{Z}_p$, but aside from that one could argue that many fields have very little to do with numbers.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-13T04:39:33.907" CommunityOwnedDate="2010-10-13T04:39:33.907" />
  <row Id="6661" PostTypeId="1" AcceptedAnswerId="6662" CreationDate="2010-10-13T04:49:14.667" Score="6" ViewCount="212" Body="&lt;p&gt;Is there a simple explanation of what the Laplace transformation do exactly and how they work?  Reading my math book has left me in a foggy haze of proofs that I don't completely understand. I'm looking for an explanation on lamans terms so that I understand what it is doing as I make these seemingly &quot;magical&quot; transformations. I searched the site and closest to an answer was &lt;a href=&quot;http://math.stackexchange.com/questions/954/inverse-of-laplace-transform&quot;&gt;this&lt;/a&gt; however it doesn't explain things for my simplistic mathematical mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: please edit tags as I don't really know what to put down for this topic&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="1602" LastEditorDisplayName="" LastEditDate="2010-10-20T01:11:09.650" LastActivityDate="2010-10-20T10:03:06.003" Title="Laplace transformations for dummies" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="6662" PostTypeId="2" ParentId="6661" CreationDate="2010-10-13T04:55:38.210" Score="4" ViewCount="0" Body="&lt;p&gt;There are beautiful video lessons at &lt;a href=&quot;http://ocw.mit.edu/index.htm&quot; rel=&quot;nofollow&quot;&gt;MIT Opencourseware&lt;/a&gt;. I'm particularly in love with &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2006/video-lectures/embed19/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; presentation of the Laplace transform.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-13T04:55:38.210" CommentCount="2" />
  <row Id="6663" PostTypeId="1" AcceptedAnswerId="6664" CreationDate="2010-10-13T05:10:08.627" Score="2" ViewCount="97" Body="&lt;p&gt;This is probably a silly question, but a couple of people that I have talked to have had different responses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $C_0$ denote the set of continuous functions with compact support or the set of continuous functions which vanish at infinity?  &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-10-13T06:57:33.737" Title="Definition of $C_0$" Tags="&lt;notation&gt;" AnswerCount="2" />
  <row Id="6664" PostTypeId="2" ParentId="6663" CreationDate="2010-10-13T05:20:10.217" Score="3" ViewCount="0" Body="&lt;p&gt;I have always seen $C_0(X)$ denoting the continuous functions vanishing at infinity, and $C_c(X)$ or $C_{00}(X)$ denoting the continuous functions with compact support, where $X$ is usually a locally compact Hausdorff space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A special case is $c_0$, which is shorthand for $C_0(\mathbb{N})$, and $c_{00}$ means $C_{00}(\mathbb{N})$.  In this case $\mathbb{N}$ has the discrete topology, and &quot;continuous&quot; is redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy, sometimes the compact operators on a Hilbert space are denoted by $B_0(H)$, and the finite rank operators by $B_{00}(H)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;http://eom.springer.de/f/f130270.htm&quot; rel=&quot;nofollow&quot;&gt;Springer Online Reference Works article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-13T05:31:11.717" LastActivityDate="2010-10-13T05:31:11.717" />
  <row Id="6665" PostTypeId="1" CreationDate="2010-10-13T06:41:52.317" Score="5" ViewCount="123" Body="&lt;p&gt;Consider a candy bag that contains &lt;strong&gt;N=100&lt;/strong&gt; candies. There are only two types of candy in the bag. Say the caramel candy and the chocolate candy. Nothing more is known about the contents of the bag.&lt;br&gt;&#xA;Now, you are going to draw (randomly) one candy at a time from the bag until the first caramel appear. Suppose that the first caramel appeared at &lt;strong&gt;k=7th&lt;/strong&gt; drawing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment, what can we say about  the number of caramel candies in the bag?&lt;/p&gt;&#xA;" OwnerUserId="2323" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-13T07:25:54.997" LastActivityDate="2010-10-13T09:48:48.437" Title="The probability theory around a candy bag" Tags="&lt;probability&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6666" PostTypeId="2" ParentId="6663" CreationDate="2010-10-13T06:57:33.737" Score="2" ViewCount="0" Body="&lt;p&gt;The reason people have different responses is that the notation is not completely standardized. For example, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0125850506&quot; rel=&quot;nofollow&quot;&gt;Reed &amp;amp; Simon&lt;/a&gt; use $C_0^{\infty}(X)$ for smooth functions with compact support in a space $X$, and $C_{\infty}(X)$ for continuous functions vanishing at infinity. (But instead of $C_0(X)$ for continuous functions with compact support, they write $\kappa(X)$ for some reason...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you just have to check in every case which convention the text that you're reading uses.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-13T06:57:33.737" CommentCount="1" />
  <row Id="6667" PostTypeId="2" ParentId="6665" CreationDate="2010-10-13T07:23:08.670" Score="0" ViewCount="0" Body="&lt;p&gt;If the number of caramel candies is $a$, then the probability that the first 6 drawn will not be caramel and the 7th drawn will be caramel (assuming that we do not put back the drawn candies) is&#xA;$P(\text{7th}|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}$.  Now, given that this has occurred, the probability $P(a|\text{7th})$ of any particular value of $a$ given that the first caramel is the 7th drawn should be $P(\text{7th}|a)$ for that particular $a$ divided by the sum of all possible $P(\text{7th}|a)$. $\sum_{a=0}^{100}P(\text{7th}|a)=\frac{101}{56}$, so&#xA;$$P(a|\text{7th})=\frac{P(\text{7th}|a)}{\frac{101}{56}}=\frac{56\cdot 93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{101!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bashing out some values and adding things up, the probability that $a\le 19$ is slightly less than 50% (49.673%) and the expected value of $a$ is $\frac{65}{3}=21\frac{2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;edit&lt;/em&gt;&lt;/strong&gt;:  (I've slightly altered my original answer above, mostly in the notation, to better accommodate the work below; I believe that the work above assumed that, without knowing how long it took to draw the first caramel, each possible number of caramels was equally likely.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $P(a)$ is the probability that there are $a$ caramels.  As above, for any particular value of $a$, the probability $P(\text{7th}|a)$ that the first caramel drawn is the 7th candy drawn is&#xA;$$P(\text{7th}|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}.$$&#xA;So, the probability that there are $a$ caramels and that the first caramel drawn is the 7th candy drawn is $P(a\text{ and 7th})=P(a)\cdot P(\text{7th}|a)$.  By Bayes's Theorem: &#xA;$$\begin{align}&#xA;P(a|\text{7th})&amp;amp;=\frac{P(a\text{ and 7th})}{P(\text{7th})}=\frac{P(a\text{ and 7th})}{\sum_{k=0}^{100}P(k\text{ and 7th})}&#xA;\\&#xA;&amp;amp;=\frac{P(a)P(\text{7th}|a)}{\sum_{k=0}^{100}P(k)P(\text{7th}|k)}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $P(a)=\frac{1}{100}$ for all $a$, this yields the results in my original answer.  If $P(a)={100 \choose a}\frac{1}{2^{100}}$ (a binomial distribution with caramel and not equally likely for each individual candy when the bag is originally filled), the expected value of $a$ is 47.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P(a)={100\choose a}p^a(1-p)^{100-a}$ (a binomial distribution where the probability of each single candy being caramel is $p$ when the bag is originally filled), the expected value of $a$ is $1+93p$.  If this expected value of $a$ given that the first caramel drawn was the 7th candy drawn is to equal the expected value of $a$ without having drawn any candies, which is $100p$, then $p=\frac{1}{7}$, so the expected value of $a$ is $\frac{100}{7}=14\frac{2}{7}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-13T09:48:48.437" LastActivityDate="2010-10-13T09:48:48.437" CommentCount="6" />
  <row Id="6668" PostTypeId="1" AcceptedAnswerId="6672" CreationDate="2010-10-13T07:36:45.530" Score="3" ViewCount="92" Body="&lt;p&gt;The field of real numbers has a nice property - for a fixed $n\in \mathbb{N}$, every $r\in \mathbb{R}$ is either an n-th power (if n is odd or if r&gt;0) or -r is an n-th power.&#xA;I want to generalize this - let K be a field extension of F, then $a\in K$ is almost an n-th power if there is $b\in F$ and $c\in K$ such that $ab=c^n$. In other words, if $A=\{ x^n \mid x\in K^\times \}$ then $A F^\times = K^\times$ as groups (In the example above the smaller field is $\mathbb{Q}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is how to find such field extensions $F\leq K$ where every element in K is almost an n-th power for a fixed n (or even better - for all n). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two such examples are when K is algebraically closed field (so every element has an n-th root), or if it is a formally real closed field. Another such example is when K is finite and n is prime to $|K|-1$ (since $K^\times$ is cyclic of order $|K|-1$), though I'm more interested with fields of char 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since my first two examples are fields that are algebraically closed or &quot;almost&quot; closed, I thought that I should start with such fields F where $F(\alpha)$, for $\alpha$ algebraic over F, is algebraically closed (or is formally real closed in the case where F can be ordered). The problem is that besides the real numbers I never encounter such fields. I mean, it is &quot;easy&quot; to start with a field F and ask what are the properties of $F(\alpha)$ but I never saw someone do it in the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my second question is what are the properties of fields that have finite algebraic extensions which are algebraically closed or formally real closed, and specially properties that can help with my first question.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-10-13T07:42:49.650" LastActivityDate="2010-10-13T09:42:29.293" Title="almost n-th power in a field" Tags="&lt;field-theory&gt;" AnswerCount="3" />
  <row Id="6669" PostTypeId="2" ParentId="6647" CreationDate="2010-10-13T07:48:19.200" Score="4" ViewCount="0" Body="&lt;p&gt;You don't need to compute the individual factorials in order to compute $k!/(k-i)!i!$, since that's the binomial coefficient $\binom{k}{i}$. A simple algorithm for computing binomial coefficients can be found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_coefficient#Binomial_coefficient_in_programming_languages&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. A more sophisticated algorithm is due to Goetgheluck (&lt;a href=&quot;http://www.jstor.org/pss/2323099&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt;); implementations can be found &lt;a href=&quot;http://www.numbertheory.org/php/binomial.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.luschny.de/math/factorial/FastBinomialFunction.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, with numbers of the size that you have, this might still not be feasible, and in this case I also recommend Stirling's formula.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-13T07:48:19.200" />
  <row Id="6670" PostTypeId="2" ParentId="6649" CreationDate="2010-10-13T07:50:24.017" Score="2" ViewCount="0" Body="&lt;p&gt;You can write the partial sum $S_n(x)$ as an integral $${1\over 2\pi}\int_{-\pi}^\pi D_n(t) f(x-t)dt,$$&#xA;where the weight function or &quot;kernel&quot; $D_n(t)$ can be easily computed and graphed once and for all. One obtains $$D_n(t)={\sin((n+1/2)t)\over \sin(t/2)}.$$ So $S_n(x)$ is an &quot;average&quot; of $f$-values from the neighborhood of $x$. The essential point is that $D_n(t)$ is heavily concentrated around $t=0$ and oscillates quickly far away from $0$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-10-13T10:04:01.300" LastActivityDate="2010-10-13T10:04:01.300" />
  <row Id="6671" PostTypeId="2" ParentId="6649" CreationDate="2010-10-13T07:59:31.700" Score="2" ViewCount="0" Body="&lt;p&gt;The way I see Fourier series (especially the trigonometric expansions) you simply draw the initial sine and cosine lines at the macro level, and then you start dealing with higher frequencies that correct the smaller details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case of an infinite sum, you always go about correcting a bit more on a smaller scale, and at the limit point you have your original function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm pretty sure I wasn't clear about it, and that I need to wave my arms around, so I've set this CW so if anyone gets the idea and thinks they can clarify it will be easier to do so.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-13T07:59:31.700" CommunityOwnedDate="2010-10-13T07:59:31.700" CommentCount="4" />
  <row Id="6672" PostTypeId="2" ParentId="6668" CreationDate="2010-10-13T08:12:24.727" Score="2" ViewCount="0" Body="&lt;p&gt;Your first question is interesting but I don't know any answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your second question is well-known. If a field $F$ has&#xA;a finite extension which is real-closed then it has a finite extension&#xA;which is algebraically closed (by adjoining $\sqrt{-1}$ to the real-closed&#xA;field). But the only fields which have finite extensions $K$ which are&#xA;algebraically closed are either real-closed or algebraically closed&#xA;already, and the degree of $K/F$ is $2$ or $1$. This is a theorem of&#xA;Artin and Schreier and can be found in the more comprehensive standard&#xA;texts, for instance Lang's &lt;em&gt;Algebra&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-13T08:12:24.727" CommentCount="1" />
  <row Id="6673" PostTypeId="2" ParentId="6329" CreationDate="2010-10-13T08:21:28.233" Score="1" ViewCount="0" Body="&lt;p&gt;Seems like such polyhedra are called &lt;em&gt;not necessarily closed&lt;/em&gt; (NNC) and are usually represented as closed polyhedra with additional dimension $\varepsilon$: every strict inequality $a_{i1}x_1 + \ldots + a_{in}x_n &gt; c_i$ is replaced by $a_{i1}x_1 + \ldots + a_{in}x_n - \varepsilon \geq c_i$ and two additional inequalities $0 \leq \varepsilon \leq 1$ are added to the system. If we call this polyhedron $P'$, the desired polyhedron is the set $\{ (x_1,\ldots,x_n) \mid (x_1,\ldots,x_n,\varepsilon) \in P', \varepsilon &gt; 0 \}$. Such a system of constraints can be converted to representation by generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, they can be characterized directly by three sets of generators $R,P,C \in \mathbb{R}^n$, i.e. every point can be obtained as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_1r_1 + \ldots + \alpha_kr_k + \beta_1p_1 + \ldots + \beta_lp_l + \gamma_1c_1 + \ldots + \gamma_mp_m,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_i \in R, p_i \in P, c_i \in C$ and $\alpha_i, \beta_i, \gamma_i \in \mathbb{R}^+$ and $\sum_{i=1}^l \beta_i + \sum_{i=1}^m \gamma_i = 1$ and there is $1 \leq i \leq l$ such that $\beta_i \neq 0$. The trick here is that points in $C$ don't have to lie within the NNC polyhedron, but its closure, and whenever they appear in the sum, there must also be point from $P$ with nonzero coefficient. This representation can easily be converted to the one mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;br&gt;&#xA;R. Bagnara, P. M. Hill, E. Zaffanella: &lt;em&gt;A New Encoding of Not Necessarily Closed Convex Polyhedra&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.12.5503&quot; rel=&quot;nofollow&quot;&gt;R. Bagnara, E. Ricci, E. Zaffanella, P. M. Hill: &lt;em&gt;Possibly Not Closed Convex Polyhedra and the Parma Polyhedra Library&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like this stuff is used mostly in static analysis/verification and is not interesting to mathematicians, maybe I should have asked on cstheory instead? Any feedback regarding the question welcome, as well as corrections regarding my misuse of terminology/notation.&lt;/p&gt;&#xA;" OwnerUserId="2477" LastActivityDate="2010-10-13T08:21:28.233" CommentCount="1" />
  <row Id="6674" PostTypeId="2" ParentId="6649" CreationDate="2010-10-13T08:38:41.670" Score="3" ViewCount="0" Body="&lt;p&gt;I don't know about a geometric interpretation, but here is a brief sketch of a proof.  First we need to be precise about what we mean by &quot;convergence.&quot;  In the naive sense, Fourier series &lt;em&gt;don't&lt;/em&gt; always converge - that is, pointwise.  (If you change the value of a function at a single point, the Fourier series remains unchanged.)  The sense in which they do always converge is in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space&quot; rel=&quot;nofollow&quot;&gt;Hilbert space&lt;/a&gt; $L^2([0, 1])$, which has inner product defined by $\langle f, g \rangle = \int_0^1 \overline{g(x)} f(x) dx$ inducing a norm, which induces a metric.  In $L^2([0, 1])$ let $X$ be the subspace spanned by the functions $e^{2\pi i nx}, n \in \mathbb{Z}$.  It is fairly straightforward to verify that the functions $e^{2\pi i nx}$ are orthogonal and have norm $1$; generally I think about this in a representation-theoretic way, as a special case of the orthogonality relations for characters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the statement that Fourier series converge is equivalent to the statement that $X$ is dense in $L^2([0, 1])$.  Why?  Given a sequence in $X$ converging to an element of $L^2([0, 1])$ we can compute the Fourier coefficients, which depend continuously on the sequence and hence which converge to a limit.  That these coefficients actually represent the element of $L^2([0, 1])$ is a standard Hilbert space argument and you should take a course in functional analysis if you want to learn this kind of stuff thoroughly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, something else you need to know about $L^2([0, 1])$ is that the subspace $Y$ consisting of all step functions is dense in it.  (If you have trouble believing this, first convince yourself that $Y$ is dense in the continuous functions on $[0, 1]$ and then believe me that the continuous functions are dense in $L^2([0, 1])$.  In fact, $L^2([0, 1])$ can be defined as the completion of $C([0, 1])$ with respect to the $L^2$ norm.)  So to show that $X$ is dense, it suffices to show that the closure of $X$ contains $Y$.  In fact, it suffices to show that $X$ has as a limit point a step function with a single bump, say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(x) = \begin{cases} 0 \text{ if } 0 \le x \le \frac{1}{3}, \frac{2}{3} \le x \le 1 \\ 1 \text{ otherwise} \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to take linear combinations, translations, and dilations of this.  In other words, it suffices to prove convergence for square waves.  But one can do the computations directly here.  There is a &lt;a href=&quot;http://cnx.org/content/m0041/latest/fourier4.png&quot; rel=&quot;nofollow&quot;&gt;standard picture&lt;/a&gt; to stare at, and of course if you have ever actually heard a square wave you should believe that audio engineers, at least, are perfectly capable of approximating square waves by sines and cosines.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-13T08:38:41.670" CommentCount="1" />
  <row Id="6675" PostTypeId="1" AcceptedAnswerId="6699" CreationDate="2010-10-13T08:43:46.943" Score="2" ViewCount="139" Body="&lt;p&gt;Let G be a group with generators and relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in general it is difficult to determine what a group is from its generators and relations. I am interested in learning about techniques for figuring out the order of a group from the given information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I know that if the number of generators exceeds the number of relations then the group has infinite order. If the number of generators equals the number of relations then the group is cyclic or has infinite order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G= [x, y|x^2 = y^3 = (xy)^4 = 1]. My hunch is that G has finite order because (xy)^4 is somehow independent of x^2 and y^3. But if the exponent on xy were bigger, say (xy)^6=1 that relation becomes redundant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is this sort of thinking correct? Furthermore: my method will only tell me if G, or its modification, is finite (or infinite). If G is finite how can I figure out the order of the group? I know that the orders divide the order of the group, but I am looking for a specific number.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-13T14:54:56.393" Title="Order of a Group from its Presentation" Tags="&lt;group-theory&gt;&lt;presentation&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6676" PostTypeId="2" ParentId="6675" CreationDate="2010-10-13T08:53:35.660" Score="3" ViewCount="0" Body="&lt;p&gt;Intuition in these cases is treacherous. Your example $G$ is an&#xA;von Dyck group, a subgroup of index 2 in the more well-known&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle_group&quot; rel=&quot;nofollow&quot;&gt;triangle groups&lt;/a&gt;. The general von Dyck group has presentation&#xA;$$\langle x, y\mid x^m=y^n=(xy)^p=1\rangle$$&#xA;and is finite iff $1/m+1/n+1/p&gt;1$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-13T08:53:35.660" CommentCount="2" />
  <row Id="6677" PostTypeId="2" ParentId="6665" CreationDate="2010-10-13T08:58:59.540" Score="0" ViewCount="0" Body="&lt;p&gt;Since 100 is quite a bit larger than 7 we will assume for the purposes of calculation that the probability of drawing a sweet is not affected by the previous draw (of course it is altered a little, but not too much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the probability of drawing a caramel be $p$ and the probability of drawing a chocolate be $q,$ where $p+q=1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the expected number of draws to draw a caramel is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E= p+2pq+3pq^2+4pq^3+ \cdots  = \frac{1}{p}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we drew a caramel at the seventh draw set $7=1/p,$ so $p = 1/7.$ Thus $100/7 \approx 14$ of the sweets are caramel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we used the correct figures the probability of drawing a caramel at the next draw would go up with each chocolate drawn, which will push the estimated number of caramels down, perhaps to 13, but I don't expect it will alter much since we started with 100 sweets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller the number of the first draw of a caramel, the less that can be reliably said. Imagine if you drew a caramel on the first draw, that could imply they are all caramels which could be wildly wrong.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-13T09:11:12.183" LastActivityDate="2010-10-13T09:11:12.183" CommentCount="1" />
  <row Id="6678" PostTypeId="2" ParentId="6668" CreationDate="2010-10-13T09:11:49.673" Score="1" ViewCount="0" Body="&lt;p&gt;There exist lots of fields $K$ that are closed under taking $n$-th roots for fixed $n$.&#xA;They can be contructed using valuation theory: take a field $K$ which is henselian&#xA;with respect to some valuation $v$ and has the following additional properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n$ is not divisible by the characteristic of the residue field $\overline{K}$.&lt;/li&gt;&#xA;&lt;li&gt;The residue field is algebraically closed.&lt;/li&gt;&#xA;&lt;li&gt;The value group $v(K)$ is divisible by $n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then $K$ is closed under taking $n$-th roots by Hensel's lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2010-10-13T09:11:49.673" />
  <row Id="6679" PostTypeId="2" ParentId="6665" CreationDate="2010-10-13T09:21:00.353" Score="3" ViewCount="0" Body="&lt;p&gt;The question you are asking here is the classical question of &lt;a href=&quot;http://en.wikipedia.org/wiki/Inferential_statistics&quot; rel=&quot;nofollow&quot;&gt;inferential statistics&lt;/a&gt;: &quot;Given the outcome of an experiment, what can be said about the underlying probability distribution?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, for example, give an &lt;a href=&quot;http://en.wikipedia.org/wiki/Estimator&quot; rel=&quot;nofollow&quot;&gt;estimator&lt;/a&gt; for the unknown quantity &quot;number of caramels&quot; (called $a$ from here on). The one most often used (since its easy to calculate) would be the maximum likelyhood estimator, where you estimate $a$ to the the value that maximizes the probability of the outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you'd choose $a$ to maximize $P_a(7)$ (the probability of drawing the first caramel in the seventh draw, assuming there are $a$  of them). A little Excel calculation, along with Isaac's way to calculate $P_a(7)$ results in $a$ to be estimated as 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To judge, what this result is worth, you'd need to calculate the mean squared error of this estimator, which is not as easily done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already had a hypothesis about $a$ (say $a$ &amp;lt; 20), you could use your experimental result to test it, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypothesis_testing&quot; rel=&quot;nofollow&quot;&gt;statistical hypothesis testing&lt;/a&gt;, too.&lt;/p&gt;&#xA;" OwnerUserId="505" LastEditorUserId="505" LastEditorDisplayName="" LastEditDate="2010-10-13T09:28:16.283" LastActivityDate="2010-10-13T09:28:16.283" CommentCount="3" />
  <row Id="6680" PostTypeId="2" ParentId="5248" CreationDate="2010-10-13T09:36:08.967" Score="2" ViewCount="0" Body="&lt;p&gt;Maybe I'm flogging a dead horse, but nobody has mentioned the &lt;strike&gt;standard&lt;/strike&gt; &lt;strong&gt;suspiciously circular (see the comments)&lt;/strong&gt; Fourier analytic proof yet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(t)=1$ for $|t|&amp;lt;1$ and 0 otherwise. Then the Fourier transform is&#xA;$$ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt&#xA;= \int_{-1}^{1} e^{-i\omega t} dt&#xA;= \frac{e^{-i\omega} - e^{i\omega}}{-i\omega}&#xA;= \frac{2\sin\omega}{\omega}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourier's inversion formula states that&#xA;$$ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega $$&#xA;if $f$ is (say) differentiable at $t$.&#xA;In our case, we get in particular that&#xA;$$ 1 = f(0)&#xA;= \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) d\omega&#xA;= \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{2\sin\omega}{\omega} d\omega&#xA;= \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin\omega}{\omega} d\omega.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: Even if this is not really a proof, it's still a good thing to be aware of, since one can use similar ideas to integrate powers of $\sin\omega/\omega$, or integrals &lt;a href=&quot;http://mathoverflow.net/questions/11517/computer-algebra-errors/11607#11607&quot; rel=&quot;nofollow&quot;&gt;like these&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-13T17:42:30.527" LastActivityDate="2010-10-13T17:42:30.527" CommentCount="2" />
  <row Id="6681" PostTypeId="2" ParentId="6668" CreationDate="2010-10-13T09:42:29.293" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/18374/if-l-is-a-field-extension-of-k-how-big-is-l-k&quot; rel=&quot;nofollow&quot;&gt;This question&lt;/a&gt; and its answer is relevant here.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastActivityDate="2010-10-13T09:42:29.293" />
  <row Id="6682" PostTypeId="1" CreationDate="2010-10-13T10:25:17.270" Score="1" ViewCount="346" Body="&lt;p&gt;Show that for any integer n&gt;1 all the numbers n!+2, n!+3, ..., n!+n are composite (i.e. not prime).&lt;/p&gt;&#xA;" OwnerUserId="2401" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-14T20:07:51.007" LastActivityDate="2010-10-15T18:03:49.170" Title="Show that a number is not prime?" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="12" />
  <row Id="6684" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T10:43:40.570" Score="3" ViewCount="0" Body="&lt;p&gt;Hint:&#xA;Try to show, that if for two numbers $a$ and $b$, $a$ is divisible by $d$ and $b$ is divisible by $d$, then so is their sum. Then go looking for such a common divisor in your sums.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-10-13T10:43:40.570" CommentCount="3" />
  <row Id="6685" PostTypeId="1" AcceptedAnswerId="6688" CreationDate="2010-10-13T11:49:46.960" Score="3" ViewCount="83" Body="&lt;p&gt;Given $n$ smooth real functions $f_1, f_2, \dots, f_n$, define a composite function like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \max(f_1(x), f_2(x), \dots, f_n(x)) - \min(f_1(x), f_2(x), \dots, f_n(x))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to say anything useful in general about the shape of this function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems like $f$ will be at least $C^0$ continuous but $f^\prime$ may have arbitrarily many discontinuities. How much would we have to know about the individual $f_k$ to be more specific?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we know that each $f_k$ has $m_k$ extrema, it seems like we should be able to place bounds on both the number of extrema in $f$ and number of discontinuities in $f^\prime$, but I'm having trouble visualising all the possible interactions as $n$ increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've tried to put this in general terms, but for context my particular interest is somewhat related to my &lt;a href=&quot;http://math.stackexchange.com/questions/1632/analysing-an-optics-model-in-discrete-and-continuous-forms&quot;&gt;earlier optics question&lt;/a&gt;. A different but similar imaging process produces something like the above function with constituent $f_k$ of this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_k(x) = \sum_i e_i \sin(x_i + \frac{2k\pi}{n}) P(x - x_i)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, simulations suggest that this process can noticeably improve lateral resolution because of the corners introduced between maxima, but it would be nice to have a more formal characterisation.)&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-10-13T14:40:56.973" Title="Characterising a function involving the max and min of other functions" Tags="&lt;real-analysis&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="6688" PostTypeId="2" ParentId="6685" CreationDate="2010-10-13T13:12:19.057" Score="2" ViewCount="0" Body="&lt;p&gt;Everything you need to know can be deduced from the identity $\text{max}(a, b) = \frac{a+b}{2} + \frac{|a-b|}{2}$ and similarly for $\text{min}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Okay, maybe this is less useful than I thought.  You want to know about the extrema of $f$ and the discontinuities of $f'$.  If each $f_k$ has finitely many extrema, it may still be the case that $f$ has infinitely many discontinuities; consider two functions which are increasing in a &quot;race&quot; in which the victor changes infinitely often.  As a deleted answer says, you want in addition that $f_i - f_j$ has finitely many extrema for every pair $i, j$.  Then outside a bounded range of values of $x$ the relative order of the $f_i$ won't change and everything will be hunky-dorky.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-13T14:40:56.973" LastActivityDate="2010-10-13T14:40:56.973" CommentCount="7" />
  <row Id="6689" PostTypeId="2" ParentId="6661" CreationDate="2010-10-13T13:13:57.487" Score="1" ViewCount="0" Body="&lt;p&gt;I'm going to come at this one from left-field.  In quantum mechanics, we deal with infinite dimensional vector spaces (Hilbert spaces), so I tend to think of integral transforms in those terms.  For instance, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty} K(x,y) f(y) dy = F(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be thought of as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{K} f = F $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $x$ and $y$ from the first equation are the indices of the infinite dimensional vectors and matrix (kernel) $f$, $F$, and $\mathbf{K}$.  Using that interpretation, if $\mathbf{K}$ is unitary then the integral is just a changing the bases of the function (Hilbert) space.  In other words, the integral can be viewed as the decomposition of original vector, $f$, in terms the new basis.  For Fourier transforms the kernel is unitary, and while not true of Laplace transforms, the idea of it being a change of basis still holds.  It should be noted that unlike in the finite case, in the infinite dimensional case care must be taken to ensure that the transform actually converges, but that is another problem entirely.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastEditorUserId="2283" LastEditorDisplayName="" LastEditDate="2010-10-13T13:53:52.283" LastActivityDate="2010-10-13T13:53:52.283" CommentCount="2" />
  <row Id="6691" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T13:23:13.620" Score="-4" ViewCount="0" Body="&lt;p&gt;Take an arbitrary number $n! + k$, where $n &gt; 1, 1 &amp;lt; k \leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is the definition of the factorial function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! = \prod_{i=1}^n i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence every i where $1 \leq i \leq n$ is a factor of $n!$. Since the range k covers is within the range of i, this is clearly true for all k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that k is a factor of $n!$, so let us say $n! = qk, k \in \mathbb{Z}$. Hence $n! + k = qk + k = (q + 1)k$, which implies q + 1 and k are integral factors of $n! + k$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a reasonably rigorous proof. When you become more familiar with the concepts of divisibility and factorials the result should be fairly apparent from first sight.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-10-15T18:03:49.170" LastActivityDate="2010-10-15T18:03:49.170" CommentCount="19" />
  <row Id="6692" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T13:24:50.577" Score="0" ViewCount="0" Body="&lt;p&gt;$8! = 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8$ so combine $k|8!$ and $k|k$ to get $k|8!+k$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: $a|b :\iff \exists k, a k = b.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;: $a|m$ and $a|n$ imply that $a|m+n$.&#xA;&lt;strong&gt;Proof&lt;/strong&gt;: Our hypothesis says that $a k = m$ and $a h = n$, add this and use distributivity to get $a (k + h) = n + m$ which proves that $a|n+m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $k|b$ with ($k$ not $1$ or $b$) then $b$ has a divisor, so it is not prime.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-13T13:47:13.823" CommentCount="3" />
  <row Id="6693" PostTypeId="2" ParentId="6593" CreationDate="2010-10-13T13:38:29.940" Score="1" ViewCount="0" Body="&lt;p&gt;Ok, so, as Jesse says, this is problem 4-24 from Little Spivak. Maybe you should look at problem 4-23, where he explains what $c_{(1,n)}$ is: a parametrization of the circumference of radius $1$, centered at the origin, that goes round the origin $n$ times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;c_{(1,n)} (t) = (\cos 2\pi nt, \sin 2\pi nt) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem says that, given any closed parametrized curve $c(t)$, there is an integer $n$ such that this $c$ together with $c_{(1,n)}$ are the boundary of some $2$-chain $c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm not wrong, this $n$ is the &lt;em&gt;winding number&lt;/em&gt; of $c$; that is, the number of times $c$ goes round the origin. So, if $c$ goes round the origin one time, you need the circumference $c_{(1,1)}$ that goes round the origin once too; if $c$ goes round twice, you need $c_{(1,2)}$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you should try first problem 4-23, where Spivak asks a similar question, but where $c$ is also a circumference (of different radius). There you see too that you need both circumferences to go round the origin the same number of times in order they are the boundary of some $2$-chain. For instance, if $n=1$ (just once round the origin), these circumferences are the boundary of a regular anulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you may try to take a look at &lt;a href=&quot;http://www.ms.uky.edu/~ken/ma570/&quot; rel=&quot;nofollow&quot;&gt;Little Spivak solutions&lt;/a&gt;. They are quite right. For instance, in the solution to 4-24, Ken Kubota (the author of these solutions), begins computing $L = \int_0^1 c^*(d\theta)$, which is the winding number of $c$ multiplied by $2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-13T17:56:39.443" LastActivityDate="2010-10-13T17:56:39.443" />
  <row Id="6694" PostTypeId="1" CreationDate="2010-10-13T13:43:46.387" Score="3" ViewCount="161" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/q/6647/2471&quot;&gt;previous question&lt;/a&gt;, I asked how one would simplify the following equation for the case where the variables are very big:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum\limits^{k}_{i=m}(N-i)^{k-i}(\frac{1}{N})^k\frac{k!}{(k-i)!i!} \leq a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer was basically to use an approximation like Stirling's formula. Having implemented this with some code, it still takes too long to find the maximum value of N so that the inequality holds true. Therefore, I need a direct solution for N. So the new question is, how would you go about solving this equation for N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Some simplifications are acceptable, but I would like to have it as accurate as possible. &#xA;The values this equation will be used for are all in the 100,000-1,000,000 range, except $m$, which is in the 100s range.)&lt;/p&gt;&#xA;" OwnerUserId="2471" LastEditorUserId="2471" LastEditorDisplayName="" LastEditDate="2010-10-13T14:03:37.717" LastActivityDate="2010-12-12T14:44:25.123" Title="How to solve an inequality containing the sum of factorials and powers" Tags="&lt;binomial-coefficients&gt;&lt;big-numbers&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6695" PostTypeId="1" AcceptedAnswerId="6700" CreationDate="2010-10-13T14:00:43.433" Score="15" ViewCount="638" Body="&lt;p&gt;The standard approach for showing $\int \sec \theta  d \theta = \ln |\sec \theta + \tan \theta| + C$ is to multiply by $\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$.  I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta  d \theta$.  Does anyone know of another way to evaluate $\int \sec \theta  d \theta$?&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-15T22:51:37.493" Title="Ways to evaluate $\int  \sec \theta d \theta$" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="6696" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T14:12:00.983" Score="15" ViewCount="0" Body="&lt;p&gt;A useful technique is to use the half angle formulas in terms of $\tan (\theta/2)$ in order to convert trigonometric (rational) functions into rational functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if $t = \tan(\theta/2)$ we have that $\sec \theta = \frac{1+t^2}{1-t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $2dt = (1 + tan^2(\theta/2))\ d\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sec \theta \ d\theta = \int \frac{2dt}{1-t^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can easily be evaluated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \csc \theta \ d\theta = \int \frac{dt}{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $\csc \theta = \frac{1+t^2}{2t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this page out: &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_half-angle_formula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tangent_half-angle_formula&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-13T16:10:15.950" LastActivityDate="2010-10-13T16:10:15.950" CommentCount="4" />
  <row Id="6697" PostTypeId="2" ParentId="6638" CreationDate="2010-10-13T14:38:54.583" Score="0" ViewCount="0" Body="&lt;p&gt;The idea is to modify just one term of a distribution $p$ in a way that changes the distribution very little but alters the variance of the log ratio profoundly.  In order to make this work, the term that gets modified changes at each step, taking us farther and farther into the tail of $p$, in the hope that this will assure the convergence of the modified distributions to $p$ in the sense of the KL divergence between them going to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ be the geometric distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pr(x) = p(x) = 2^{-x}, x = 1, 2, \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for each $n = 1, 2, \ldots$ let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_n(x) = C_n p(x) \quad \text{if}\quad x \ne n$$&#xA;$$= 2^{-f_n} \quad \text{if} \quad x = n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_n$ is a sequence of positive real numbers to be determined and $C_n$ is a sequence of normalizing constants.  The requirement that $q_n$ be a probability function determines $C_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = \sum_x {q(x)} = C_n \sum_{x \ne n} {p(x)} + 2^{-f_n} = C_n (\sum_x {p(x)} - p(n)) + 2^{-f_n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= C_n (1 - p(n)) + 2^{-f_n} = C_n(1 - 2^{-n}) + 2^{-f_n},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_n = (1 - 2^{-f_n}) / (1 - 2^{-n}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now compute the KL divergence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{KL}(p||q_n) = \sum_x {p(x) \log( p(x) / q_n(x))}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking this into a sum over $x \ne n$ and the remaining term as before gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_x {p(x) \log(C_n)} - p(n) \log(C_n) + p(n) \log( 2^{-n} / 2^{-f_n} )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \log(C_n) \left(1 - p(n) \right) + p(n) \left( f_n - n \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \log(C_n) \left(1 - 2^{-n} \right) + 2^{-n} \left( f_n - n \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(using the logarithm base 2 for ease of computation).  Notice that the sequence $\log(C_n)$ converges to $0$ provided $f_n$ diverges, so the first summand goes to $0$.  To get the whole thing to converge to zero we therefore require that $f_n - n = o(2^n)$; &lt;em&gt;i.e.&lt;/em&gt;, $f_n$ should not diverge too rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar computation produces the expectation of the squared log ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_x {p(x) ( \log( p(x) / q_n(x)) )^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_x {p(x) (\log(C_n))^2} - p(n) (\log(C_n))^2 + p(n) \left(\log( 2^{-n} / 2^{-f_n} ) \right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (\log(C_n))^2 \left(1 - 2^{-n} \right) + 2^{-n} \left( f_n - n \right)^2 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again the first summand converges to zero.  However, we can choose $f_n$ (subject to the previous restriction not to diverge too quickly) to make the right summand behave almost as we wish.  Three regimes are exhibited by these choices:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$f_n = n^2$.  The right summand converges to $0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f_n = 2^{n/2} + n$.  The right summand equals a constant $1$, converging to $1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$f_n = (2^{n/2} + n)n$.  The right summand (equal to $n^2$) diverges.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In all cases $f_n - n = o(2^n)$ as required, ensuring that $\mathrm{KL}(p||q_n) \to 0$.  Finally, the variance of $\log( p / q )$ behaves like the expectation of $(\log( p/q ))^2$ because the square of the expectation is converging to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the asymptotic behavior of the variance of the log ratio is essentially independent of the convergence of $q$ to $p$ in the sense of the KL divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$, by the way, is the probability distribution for a simple truncated experiment provided $f_n \ge n$ (which includes all three cases): let $X$ be the number of flips of a fair coin needed to reach heads the first time.  If the number of flips is not $n$, accept $X$.  Otherwise, only in the case $X = n$, generate a uniform random number $U$.  If $U \lt 2^{n - f_n}$, accept the result $X$.  Otherwise, re-run the same experiment recursively until eventually some value is accepted.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-13T14:38:54.583" CommentCount="1" />
  <row Id="6698" PostTypeId="2" ParentId="6694" CreationDate="2010-10-13T14:39:18.403" Score="1" ViewCount="0" Body="&lt;p&gt;Now you are basically into root finding.  I love &lt;a href=&quot;http://www.nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;.  I have had the book over 30 years.  Good discussion of the techniques and code supplied in various languages.   Your formula should be an easy one.  Other numerical analysis books will work, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small thought:  if m is as small as you say, there might be some algebraic simplification in making it 1.  I haven't found it, but others are better at these combinatoric identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are actually doing the sum every time, you might be able to identify the range of i that is the big contribution.  It looks like it should be somewhat below k/2.  You can probably reduce the terms in the sum by a large factor, at least to get close to N.  Then use the full formula for final refinement.  And take the N^(-k) out of the sum.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-13T14:39:18.403" CommentCount="4" />
  <row Id="6699" PostTypeId="2" ParentId="6675" CreationDate="2010-10-13T14:46:11.667" Score="4" ViewCount="0" Body="&lt;p&gt;The group generated by two elements a,b subject only to the relations aa=1, aba=bab has the same number of generators and relations, but is neither cyclic nor infinite (it is the non-abelian group of order 6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful that a relation like a^6 = 1 does not imply a has order 6, only that it has order dividing 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no practical and general method of determining the order of a group given by a presentation.  You may want to look into &quot;quotient methods&quot;, where you can determine the order of nice quotients of your group.  If that order is small, then you can find a finite presentation for the kernel, and try again.  For instance, finding the abelian invariants of a finitely presented group is easy, and if those invariants are small, then finding a presentation of the derived subgroup is easy.  Repeat, until you hit the perfect residuum, the largest perfect subgroup.  Now apply simple group recognition methods to determine the simple quotients of your group.  Find the entire top layer, and then hope beyond hope you can find the kernel.  Now repeat the derived subgroup method on the kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can often use &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt; to compute abelian invariants and derived subgroups fairly easily, as long as the index is kept under a thousand or so.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-10-13T14:46:11.667" />
  <row Id="6700" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T14:49:34.913" Score="15" ViewCount="0" Body="&lt;p&gt;Another way is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sec x dx = \int \frac{\cos x}{1-\sin^2 x} dx =&#xA;\frac{1}{2} \int \left( \frac{1}{1-\sin x} + \frac{1}{1+\sin x} \right) \cos x dx $$ &#xA;$$= \frac{1}{2} \log \left| \frac{1+\sin x}{1-\sin x} \right| + C.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that the answer can appear in many disguises. Another is&#xA;$$\log \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| $$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-13T14:49:34.913" CommentCount="2" />
  <row Id="6701" PostTypeId="2" ParentId="6675" CreationDate="2010-10-13T14:54:56.393" Score="0" ViewCount="0" Body="&lt;p&gt;While somewhat &quot;dated&quot; at this point you might want to look at the book of Coxeter and Moser, Generators and Relations, for work in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/Extras/Coxeter_Moser.html&quot; rel=&quot;nofollow&quot;&gt;http://www-history.mcs.st-and.ac.uk/Extras/Coxeter_Moser.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-13T14:54:56.393" />
  <row Id="6702" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T15:09:45.863" Score="11" ViewCount="0" Body="&lt;p&gt;Using the definitions $$\sec \theta = 1/\cos \theta \quad \text{and} \quad \cos \theta = (\exp(i \theta) + \exp(-i \theta))/2$$ gives  $$\int \sec \theta  d \theta = \int \frac{2 d \theta}{\exp(i \theta) + \exp(-i \theta)}.$$  The only insight needed is to find the substitution $u = \exp( i \theta )$ (what else is there to try?), leading to a multiple of $\int \frac{du}{1+u^2}$, the inverse tangent.  Thus, in an essentially mechanical fashion you obtain the generic solution $$-2 i \arctan(\exp(i \theta)).$$  Unwinding this via the usual algebraic identities between exponential and trig functions not only shows it is equal to the usual solutions, but also reveals why half angles might be involved and where an offset of $\pi /4$ might come from (as in @Derek Jennings' answer): it's a constant of integration, of course.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-14T01:51:09.333" LastActivityDate="2010-10-14T01:51:09.333" CommentCount="5" />
  <row Id="6703" PostTypeId="2" ParentId="6635" CreationDate="2010-10-13T15:55:11.740" Score="1" ViewCount="0" Body="&lt;p&gt;I use the following definition of the  matrix derivative of $f(Q)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\delta f)(Q) = \lim_{t \to 0} f(Q+t \delta Q)-f(Q)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $ \delta Q $ is a symmetric not necessarily positive definite matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to notice that the matrix derivative operator $\delta$ obeys the Leibniz rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in the following,  the cyclicity of the trace, and the identity&#xA;$\delta Q^{-1} = - Q^{-1} \delta Q Q^{-1}$ obtained from the matrix differentiation of $Q Q^{-1} = 1$ are used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first matrix derivative of $f(Q)$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta f(Q) = tr(-W Q^{-1} \delta Q Q^{-1} W + \delta Q )= tr((-Q^{-1} W^2 Q^{-1}+1) \delta Q  )$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the condition $\delta f(Q) = 0$, implies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-Q^{-1} W^2 Q^{-1} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q^2 = W^2 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $Q = W $ is the only solution, because both $Q$ and $W$ are defined to be positive definite. To verify that this solution is a minimum,&#xA;we check the second matrix derivative is positive for an arbitrary $\delta Q$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta^2 f(Q) = tr((Q^{-1} \delta Q Q^{-1} W^2 Q^{-1}+Q^{-1}  W^2 Q^{-1}\delta Q Q^{-1}) \delta Q  ) = 2 tr(Q^{-1} (\delta Q)^2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, the second equality is obtained after the substitution of the solution. &lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-13T15:55:11.740" CommentCount="1" />
  <row Id="6704" PostTypeId="1" AcceptedAnswerId="6711" CreationDate="2010-10-13T16:36:57.360" Score="1" ViewCount="112" Body="&lt;p&gt;I have the following sequences:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = [0, 1, 2, 3, 4, 5]&#xA;b = [1, 2, 4, 7, 13, 26]&#xA;c = [1, 1, 2, 4, 7, 13]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm trying to reverse-engineer an undocumented API and believe that the values in &lt;code&gt;a&lt;/code&gt; can be used to determine the corresponding values in &lt;code&gt;b&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt;. For instance, we could say that third element in &lt;code&gt;b&lt;/code&gt; (&lt;code&gt;4&lt;/code&gt;) is two times the value in &lt;code&gt;a&lt;/code&gt; (&lt;code&gt;2&lt;/code&gt;), while the corresponding value in &lt;code&gt;c&lt;/code&gt; is the same as &lt;code&gt;a&lt;/code&gt;'s. Unfortunately that algorithm doesn't work across all the values f the sequences. Does this make sense? If so, can anyone think of any common series that would give the values of &lt;code&gt;b&lt;/code&gt; and &lt;code&gt;c&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2487" LastActivityDate="2010-10-13T18:58:08.187" Title="Help needed to figure out relationships between a few sequences" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="6705" PostTypeId="2" ParentId="6704" CreationDate="2010-10-13T16:46:58.463" Score="2" ViewCount="0" Body="&lt;p&gt;The sequence $b$ can be obtained from $a$ by the function $$t \mapsto \frac{t^4}{12}-\frac{t^3}{2}+\frac{17 t^2}{12}+1$$ and the sequence $c$ using $$t \mapsto \frac{t^5}{60}-\frac{t^4}{6}+\frac{7 t^3}{12}-\frac{t^2}{3}-\frac{t}{10}+1.$$ I am pretty sure this is not what you want!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-13T16:46:58.463" />
  <row Id="6706" PostTypeId="1" CreationDate="2010-10-13T16:52:37.850" Score="0" ViewCount="43" Body="&lt;p&gt;Using stereoscopic cameras, I track a certain object through it's path in space. For every frame, I compute it's pose in 3D. I can represent it's pose by either a translation vector + rotation matrix or a 6 element parameter vector (X, Y, Z, roll, pitch, yaw).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the output generated by the cameras is noisy, and I'd like to filter some of that noise off by smoothing the poses in 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers on that? What fitting works better for multidimensional problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more information: in my case, the XYZ path can probably be fitted with a 2nd order curve, and roll, pitch and yaw also demonstrate smooth change over samples. &lt;/p&gt;&#xA;" OwnerUserId="2449" LastActivityDate="2010-10-13T16:52:37.850" Title="Fitting object poses in 3D space" Tags="&lt;analysis&gt;" CommentCount="2" />
  <row Id="6707" PostTypeId="1" AcceptedAnswerId="6727" CreationDate="2010-10-13T17:27:22.590" Score="3" ViewCount="104" Body="&lt;p&gt;Let $G = \langle x , y \ | \ x^{12}y=yx^{18} \rangle$. I want to continue a discussion &lt;a href=&quot;http://math.stackexchange.com/questions/6376/normal-closure-in-groups&quot;&gt;here&lt;/a&gt; on the normal closure of $y$ in $G$.&lt;br&gt;&#xA;How can we determine a group presentation for $N$?&lt;/p&gt;&#xA;" OwnerUserId="2424" LastActivityDate="2010-10-14T00:05:35.377" Title="Normal closure in groups II" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6708" PostTypeId="2" ParentId="6682" CreationDate="2010-10-13T17:32:00.737" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad  k\: $ divides $\rm\: k\:m + k\:,\ $ and, of course, $\rm\ k\:$ divides $\rm\: n!\ $ for $\rm\:k =2,3,\:\ldots\:,n\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-13T17:32:00.737" />
  <row Id="6709" PostTypeId="2" ParentId="2234" CreationDate="2010-10-13T17:53:47.030" Score="0" ViewCount="0" Body="&lt;p&gt;Hey guys.... &#xA;There is no correct simple general formula for sum to n terms of the series &#xA;1+1/2+1/3+1/4+ ............. + 1/n &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the following formula will be a good approximation for sum to n-terms of the above series when n&gt;5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;S = log(n+0.5) + 0.5772 + 0.04021/(n*n + 0.8848)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deviation from the actual value fluctuates but remains relatively low.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so i guess this may be a good approximation&lt;/p&gt;&#xA;" OwnerUserId="2488" LastEditorUserId="2488" LastEditorDisplayName="" LastEditDate="2010-10-27T21:17:37.567" LastActivityDate="2010-10-27T21:17:37.567" CommentCount="8" />
  <row Id="6710" PostTypeId="2" ParentId="2234" CreationDate="2010-10-13T18:02:20.463" Score="4" ViewCount="0" Body="&lt;p&gt;A good quick approximation: $\log(n + 0.5) + \gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A yet more accurate approximation: $\log(n)+\gamma+1/2z+1/12z^2-1/120z^4+1/252z^6-1/240z^8+1/132z^{10}-691/32760z^{12}+\cdots$, taking as many terms as is convenient.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Error analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first formula above has maximum error of $1/24n^2$.  uday's first answer has error of about 0.0024 ln n.  Mine is a better approximation for all $n\neq6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first formula is also better than uday's revised formula for all $n\ge46$. The expected error in uday's answer is about 0.000016.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-24T23:58:31.310" LastActivityDate="2010-10-24T23:58:31.310" />
  <row Id="6711" PostTypeId="2" ParentId="6704" CreationDate="2010-10-13T18:31:31.877" Score="1" ViewCount="0" Body="&lt;p&gt;Unfortunately, we don't have enough information here to reverse-engineer your problem. For instance, the Generalized Catalan Numbers have been referenced in the comments (by Qiaochu Yuan) because they appear in your sequence $b.$ This sequence starts $1,1,1,2,4,7,13,26,52,104,\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I could define a “random” sequence by $u_1=1$ and for $n&gt;1$&#xA;$$U_n=2U_{n-1} \quad \textrm{ for } n \equiv 0,1 \textrm{ mod } 4$$&#xA;and&#xA;$$U_n=2U_{n-1}-1 \quad \textrm{ for } n \equiv 2,3 \textrm{ mod } 4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows the stated Catalan sequence exactly, and so contains your sequence $b,$ up until we get 103 instead of 104. I give this as an illustration that with the current information this problem has no definitive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To be fair to the question asker, I'd like to point out that the fact that he/she cannot do what he/she wants to do is often not so clear to people who've been brought up on a diet of school mathematics where one is frequently asked to fill in the missing number in a sequence or find the next number.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-13T18:58:08.187" LastActivityDate="2010-10-13T18:58:08.187" CommentCount="4" />
  <row Id="6712" PostTypeId="2" ParentId="5303" CreationDate="2010-10-13T19:16:20.467" Score="2" ViewCount="0" Body="&lt;p&gt;One essential piece of intuition from theoretical computer science: since P $\neq$ NP (we assume!), there are many statements which are easy to state but hard to prove.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-13T19:16:20.467" />
  <row Id="6713" PostTypeId="1" AcceptedAnswerId="6973" CreationDate="2010-10-13T19:59:06.930" Score="3" ViewCount="91" Body="&lt;p&gt;Rectangles will all be the same size, circle may vary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was discussing this with a friend and he said to just start placing them in the middle and moving outwards but it seems that you could free up more space for rectangles around the edges if you moved that center of mass to one side and maybe put some space between the rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if any of this is confusing or if you have heard of a good method for doing this.&lt;/p&gt;&#xA;" OwnerUserId="2490" LastActivityDate="2010-10-16T19:54:20.263" Title="Is there an algorithm out there that will show you how to fit the most rectangles into a circle?" Tags="&lt;geometry&gt;&lt;calculus&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6714" PostTypeId="1" AcceptedAnswerId="6725" CreationDate="2010-10-13T20:37:00.130" Score="0" ViewCount="69" Body="&lt;p&gt;Suppose x is the solution to a standard linear programming problem ($Ax=b$, $x&gt;=0$) and the set $S$ is every $i$ where $x_{i} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I show this is optimal only where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimize $c'f$ subject to $Af=0, f_{i} \ge 0$, $i$ is in $S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has a cost of zero.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-14T14:33:26.177" LastActivityDate="2010-10-14T14:33:26.177" Title="Convert linear programming problem" Tags="&lt;linear-programming&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6716" PostTypeId="1" AcceptedAnswerId="10476" CreationDate="2010-10-13T20:49:44.203" Score="5" ViewCount="57" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\mathcal{E}$ is a well-pointed elementary topos, that $X$ and $Y$ are objects of $\mathcal{E}$, and that $F$ is a function which maps global elements $p: 1 \to X$ to global elements $F(p): 1 \to Y$ (here $1$ is the terminal object of $\mathcal{E}$). Does there exist a (necessarily unique) arrow $f: X \to Y$ in $\mathcal{E}$ such that $fp = F(p)$ for all $p$? Equivalently, is any object in a well-pointed topos the coproduct over its global elements of $1$? It's easy to show that the answer is &quot;yes&quot; if the coproduct exists since the induced map $\coprod_{p \in \Gamma X} 1 \to X$ is iso. But I don't know whether the coproduct exists in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Could somebody with enough reputation create a &quot;topos-theory&quot; tag and add it to this? Thanks)&lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-13T21:13:00.280" LastActivityDate="2010-11-15T21:31:26.457" Title="Do functions defined on global elements give rise to arrows in a well-pointed topos?" Tags="&lt;category-theory&gt;&lt;topos-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6717" PostTypeId="2" ParentId="6695" CreationDate="2010-10-13T20:55:55.750" Score="3" ViewCount="0" Body="&lt;p&gt;Instead of presenting another way of evaluating this integral, I justify a more general case in an approach which uses partial fractions and trigonometric identities, at the level of a Calculus class, I think:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \dfrac{1}{a+b\cos x}dx=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert &#xA;\dfrac{\sqrt{a+b}+\sqrt{b-a}\tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert \quad a\lt b.\quad (\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+b\cos x=(a-b)+2b\cos ^{2}x/2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{a+b\cos x}=\dfrac{\sec ^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec&#xA;^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec ^{2}x/2}{a+b-(b-a)\tan ^{2}x/2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{a+b-(b-a)\tan ^{2}x/2}=$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{2\sqrt{a+b}}\left( \dfrac{1}{%&#xA;\sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{1}{\sqrt{a+b}+\sqrt{b-a}\tan x/2}%&#xA;\right) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \dfrac{1}{a+b\cos x}dx=$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{2\sqrt{a+b}}\int \left( \dfrac{\sec&#xA;^{2}x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{\sec ^{2}x/2}{\sqrt{a+b}+%&#xA;\sqrt{b-a}\tan x/2}\right) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert \dfrac{\sqrt{a+b}+\sqrt{b-a}%&#xA;\tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we have your particular case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \dfrac{1}{\cos x}dx=\int \dfrac{1}{0+1\cos x}dx=\ln \left\vert \dfrac{%&#xA;1+\tan x/2}{1-\tan x/2}\right\vert . \qquad (\ast\ast)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $\tan \dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}$ and $\sec x+\tan x=\dfrac{1+\sec x+\tan x}{1+\sec x-\tan x}$ it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1+\tan x/2}{1-\tan x/2}=\dfrac{1+\dfrac{\sin x}{1+\cos x}}{1-\dfrac{%&#xA;\sin x}{1+\cos x}}=\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\sec x+\tan x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sec x\; dx=\ln \left\vert \sec x+\tan x\right\vert .$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-15T22:51:37.493" LastActivityDate="2010-10-15T22:51:37.493" />
  <row Id="6718" PostTypeId="1" AcceptedAnswerId="6719" CreationDate="2010-10-13T21:02:11.520" Score="2" ViewCount="98" Body="&lt;p&gt;I'm playing around with a collection of subsets of $\mathbb{R}^n$ let's call them $X_i$. What I want to know is, is the following condition sufficient for some $X_i$ being measurable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all elements of $X_i$ are surrounded by a neighbourhood contained in $X_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is motivated by the idea that non measurable sets are &quot;fractally wiggly&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming standard topology and measure on $\mathbb{R}^n$ but I'm also interested in minimal assumptions regarding topology and measure that keep this a sufficient condition for being measurable.&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-10-14T10:10:34.170" LastActivityDate="2010-10-14T10:10:34.170" Title="Sufficient condition for being measurable?" Tags="&lt;topology&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="6719" PostTypeId="2" ParentId="6718" CreationDate="2010-10-13T21:08:48.463" Score="4" ViewCount="0" Body="&lt;p&gt;I believe this is sufficient, though not necessary.  The set of $x\in X_i$ that are contained in a neighborhood in $X_i$ is the interior of $X_i$, which is open.  If $X_i$ satisfies this condition, then it's an open set union a set of measure zero, both of which are measurable.  This will still be true for any measure defined at least on the Borel $\sigma$-algebra for your space.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-13T21:08:48.463" CommentCount="2" />
  <row Id="6720" PostTypeId="1" CreationDate="2010-10-13T21:16:41.097" Score="0" ViewCount="59" Body="&lt;p&gt;This may be a really basic question but the typical atan2 method isn't working when I write it in a program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a camera object to face a certain target in 3D space.&lt;br&gt;&#xA;I need to calculate the 3 Euler angles (yaw, pitch, roll) for the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/q9SAq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="2493" LastActivityDate="2010-10-13T21:16:41.097" Title="Calculate Euler yaw pitch and roll for a 3d point to face another" Tags="&lt;geometry&gt;" CommentCount="3" />
  <row Id="6721" PostTypeId="2" ParentId="6707" CreationDate="2010-10-13T22:01:37.720" Score="0" ViewCount="0" Body="&lt;p&gt;I don't know the answer myself but you can take a look on Chapter 9 (Presentation of Subgroups) in Johnson's &quot;Presentations of Groups&quot; - I just discovered this very nice book. I'll do that when I get to university.&lt;br&gt;&#xA;Unfortunately not many people participated in the first question, so thanks for continuing the discussion... I'll be voting you up when I can.&lt;/p&gt;&#xA;" OwnerUserId="2342" LastActivityDate="2010-10-13T22:01:37.720" CommentCount="1" />
  <row Id="6722" PostTypeId="1" CreationDate="2010-10-13T22:43:26.157" Score="-1" ViewCount="217" Body="&lt;p&gt;&lt;strike&gt;Tell me everything you know about Euklid, Pythagoras etcetera.&lt;/strike&gt;&#xA;I'm about to begin reading about greek algebra used in calculating geometric figures. After reading I'm going to write a 15-20 pages report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any kind of introduction would be awesome, I'm primarily interested in Euklid and how the theories at that time are important in math today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps. sorry for the bad english, I'm tired, made a 300 words spanish assignment and I'm not a native. &lt;em&gt;Be nice, it's my first time.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2495" LastEditorUserId="2495" LastEditorDisplayName="" LastEditDate="2010-10-17T21:39:26.530" LastActivityDate="2010-10-22T03:57:54.393" Title="Greek Geometry Algebra" Tags="&lt;geometry&gt;&lt;reference-request&gt;&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="6723" PostTypeId="1" CreationDate="2010-10-13T22:44:03.253" Score="1" ViewCount="132" Body="&lt;p&gt;So let's say you have a curved mirror, $y=y(x)$ with this property:&#xA;Whenever a ray of light emanates from the origin, it reflects parallel to the x axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the equation of the mirror.&lt;/p&gt;&#xA;" OwnerUserId="2193" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-17T08:06:12.263" LastActivityDate="2010-10-17T08:06:12.263" Title="Equation of a mirror" Tags="&lt;geometry&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6724" PostTypeId="2" ParentId="6723" CreationDate="2010-10-13T22:48:41.120" Score="1" ViewCount="0" Body="&lt;p&gt;This sounds like homework.  Can you translate the words into an equation?  Think about what happens when the light ray strikes the mirror.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-13T22:48:41.120" />
  <row Id="6725" PostTypeId="2" ParentId="6714" CreationDate="2010-10-13T23:57:36.873" Score="1" ViewCount="0" Body="&lt;p&gt;$f=0$, yielding a cost of $0$, is a feasible point for your second problem. So if the optimal cost is non-zero then it must be negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then $y=x+\lambda f$ would be an improvement over $x$ in the original problem for all $\lambda&gt;0$. Because of the nature of constraints in the second problem $y$ satisfies the equality and binding non-negativity constraints in the first problem. We can choose $\lambda$ small enough so that it also satisfies the non-binding non-negativity constraints. But then $y$ would be superior to $x$ in the original problem, proving that the latter was not an optimal point.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-13T23:57:36.873" />
  <row Id="6726" PostTypeId="1" AcceptedAnswerId="8595" CreationDate="2010-10-13T23:59:52.813" Score="1" ViewCount="76" Body="&lt;p&gt;I read that Smale recast classical mechanics in terms of symplectic geometry. I know a bit about classical mechanics but nothing about symplectic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any writings from Smale on this which I could read?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also what other books or texts are there which cover this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-02T03:06:26.943" Title="References for Smale's Geometry" Tags="&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="6727" PostTypeId="2" ParentId="6707" CreationDate="2010-10-14T00:05:35.377" Score="3" ViewCount="0" Body="&lt;p&gt;This is a much harder question than your last one, and involves two separate parts: finding generators for $N$ and finding relations for $N$.  I'll try to sketch the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, recall from the previous thread that every element of $G$ can be written as a word in $x$ and $y$, and $N$ is the subgroup consisting of all words for which the total power of $x$ is a multiple of $6$.  In particular, $N$ is a subgroup of $G$ with index $6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Generators&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding generators for $N$ involves something called a &lt;strong&gt;&lt;em&gt;Schreier graph&lt;/em&gt;&lt;/strong&gt;.  This is a directed graph defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It has one vertex for each coset of $N$ in $G$ (so six vertices total)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a directed edge labeled $x$ from each coset $aN$ to the coset $xaN$, and a directed edge labeled $y$ from each coset $aN$ to the coset $yaN$.  (Thus there will be $12$ edges total, $6$ labeled $x$ and $6$ labeled $y$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the case you are looking at, the six vertices are $N,xN,x^2N,\ldots,x^5N$.  These six vertices are arranged in a directed hexagon of $x$ edges), and each vertex has a loop labeled $y$:&#xA;&lt;img src=&quot;http://i.imgur.com/TO7fZ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that each path in this graph corresponds to a word in $x$ and $y$, with backwards travel along directed edges corresponding to inverse generators.  In particular, the elements of $N$ are the paths that start and end at the vertex $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the statement that is true is that generators for $N$ are the same as generators for the fundamental group of this graph, using $N$ as the basepoint.  In this case, what that means is that $N$ is generated by the following elements:&#xA;$$&#xA;A=x^6,\quad B_0 = y,\quad B_1 = xyx^{-1},\quad B_2 = x^2yx^{-2},\quad\ldots,\quad B_5 = x^5yx^{-5}&#xA;$$&#xA;(In general, one can obtain generators for the fundamental group of a graph by first choosing a spanning tree for the graph, and then considering all closed paths starting at the basepoint that pass through exactly one edge not in the spanning tree.  In this case, the spanning tree consists of the first five edges of the hexagon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the relations for $N$ come from the relations for $G$.  There is one relation for $G$, namely:&#xA;$$&#xA;x^{12}yx^{-18}y^{-1} = 1.&#xA;$$&#xA;This will lead to six different relations for $N$, obtained by conjugating this relation by representatives for the six cosets of $N$.  Thus, the relations in $N$ can be written as&#xA;$$&#xA;x^{12}yx^{-18}y^{-1} = 1, \quad x(x^{12}yx^{-18}y^{-1})x^{-1} = 1, \quad\ldots,\quad x^5(x^{12}yx^{-18}y^{-1})x^{-5}=1.&#xA;$$&#xA;Of course, these are expressed in terms of $x$ and $y$, and we must rewrite them in terms of the generators $A,B_0,\ldots,B_5$:&#xA;$$&#xA;A^2B_0 A^{-3} B_0^{-1} = 1,\quad A^2B_1 A^{-3} B_1^{-1} = 1,\quad\ldots,\quad A^2B_5A^{-3}B_5^{-1} = 1&#xA;$$&#xA;(The justification for this comes from algebraic topology.  Each group $G$ has an associated 2-complex, which has a single vertex, one loop at the vertex for each generator, and one 2-cell for each relation.  The fundamental group of this 2-complex is equal to $G$.  The finite-index subgroup $N$ corresponds to a finite-sheeted cover of this 2-complex.  This cover has a Schreier graph as its 1-skeleton, and the six 2-cells of the cover are attached along the six words listed above.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Result&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, we have now obtained a presentation of the group $N$:&#xA;$$&#xA;N = \langle A,B_0,\ldots,B_5 \mid A^2B_k = B_kA^3\text{ for each $k$}\rangle&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the group $G$ that you seem to be interested in is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Baumslag-Solitar_group&quot; rel=&quot;nofollow&quot;&gt;Baumslag-Solitar group&lt;/a&gt;, and is usually denoted $B(12,18)$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-14T00:05:35.377" CommentCount="4" />
  <row Id="6728" PostTypeId="1" CreationDate="2010-10-14T00:09:05.827" Score="0" ViewCount="45" Body="&lt;p&gt;what are some &lt;em&gt;lower&lt;/em&gt; bounds (if they exist) on $w^t A w$ in terms of $||w||_2^2$ for $A$ an matrix with all positive values? The lower bound can depend on $A$ and $||w||_2^2$. $w$ is $k \times 1$. Note when $A = I$ then we exactly get the squared norm of $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are no such values, what in the case that $A$ is positive semi-definite? (i.e. $A$ has a $B$ such that $A = BB^T$ and then we are talking $w^tA w = ||Bw||^2_2$ and we want it to be larger than something that depends on $||w||^2_2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2498" LastActivityDate="2010-10-14T07:42:12.417" Title="what are simple lower bounds on w^t A w in terms of ||w||_2^2?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6729" PostTypeId="2" ParentId="6728" CreationDate="2010-10-14T00:22:55.537" Score="2" ViewCount="0" Body="&lt;p&gt;The bounds on $w^TAw$ are&#xA;$$\lambda_{\min} \|w\|_2^2 \le w^TAw \le \lambda_{\max}\|w\|_2^2,$$&#xA;where $\lambda_{\min}$ and $\lambda_{\max}$ are the smallest and largest eigenvalues, respectively, of $\frac{1}{2}\left(A + A^T\right)$. Each bound is attained when $w$ is parallel to the corresponding eigenvector.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-10-14T07:42:12.417" LastActivityDate="2010-10-14T07:42:12.417" />
  <row Id="6730" PostTypeId="2" ParentId="5303" CreationDate="2010-10-14T01:40:33.340" Score="0" ViewCount="0" Body="&lt;p&gt;The question may look almost self-evident, yet when translated into math, it sometimes looses that self-evident feature and so it requires (often fairly complicated) proof. My favorite example is the Jordan Theorem. A closed non-intersecting curve separates the plane into &quot;inner&quot; and &quot;outer&quot; regions. Looks like there isn't anything to prove here. But translate the notion of &quot;closed non-intersecting curve&quot; into the formula, and we no longer have any intuitive feel of &quot;inner region&quot;, &quot;outer region&quot;, and what is that region anyway. The simplest proof requires the machinery of algebraic topology and still quite long.&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-14T01:40:33.340" CommentCount="2" />
  <row Id="6731" PostTypeId="2" ParentId="6340" CreationDate="2010-10-14T02:03:27.597" Score="0" ViewCount="0" Body="&lt;p&gt;Right! Thank you, I didn't think about using H\&quot;older's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;|f(0)| = |\int_{|z|=R} \frac{f(z)dz}{z} | &amp;lt;  (\int_{|z|=R} |f(z)|^p |dz|)^{1/p} (\int_{|z|=R} |z|^{-q} |dz|)^{1/q}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $(\int_{|z|=R} |z|^{-q} |dz|)^{1/q} = R^{-1+1/q} (2\pi)^{1/q} = R^{-1/p} (2\pi)^{1/q}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_0^\infty ((2\pi)^{1/q} |f(0)| R^{1/p})^p dR &amp;lt; || f ||_p^p &amp;lt;\infty&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the only way for this to fly is for $f(0)=0$, for $p\le 1$&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2010-10-14T02:03:27.597" CommentCount="1" />
  <row Id="6732" PostTypeId="1" AcceptedAnswerId="6762" CreationDate="2010-10-14T02:27:53.990" Score="1" ViewCount="87" Body="&lt;p&gt;If I have a bounded set F in N dimensional space and another set G where every element g in G has h'g=c and also must exist in F. H is a vector in the N dimensional space and c is any constant 1x1 matrix (scalar). h is a vector of appropriate dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove every extreme point of G lies on the boundary of F?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say if x and y are extreme points in F then xλ + (1-λ)y = g&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-14T14:34:47.490" LastActivityDate="2010-10-14T14:34:47.490" Title="Show convex combination" Tags="&lt;linear-algebra&gt;&lt;convex-analysis&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="6733" PostTypeId="1" AcceptedAnswerId="6736" CreationDate="2010-10-14T03:11:59.973" Score="1" ViewCount="31" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to simplify this equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y = (¬AB + ¬C)(¬A + ¬D)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;.. into this equation.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y  = ¬A¬C + ¬C¬D + ¬AB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I keep going in circles with expanding and minimizing the booleans. Any tips or advice? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2371" LastActivityDate="2010-10-14T03:31:43.533" Title="Boolean Simplification" Tags="&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6734" PostTypeId="1" AcceptedAnswerId="6735" CreationDate="2010-10-14T03:17:52.503" Score="3" ViewCount="114" Body="&lt;p&gt;The following is a question from the sample GRE Mathematics Subject Test&#xA;found on the ETS website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a $5\times 5$ real matrix.&#xA;Exactly four of the following five conditions on $M$ are equivalent&#xA;to each other.&#xA;Which of the five conditions is equivalent to NONE of the other four?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) For any two distinct column vectors $u$ and $v$ of $M$,&#xA;the set {$u,v$} is linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) The homogeneous system $Mx=0$ has only the trivial solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) The system of equations $Mx=b$ has a unique solution for each real $5\times 1$&#xA;column vector $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(D) The determinant of $M$ is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(E) There exists a $5\times 5$ real matrix $N$ such that $NM$ is the $5\times 5$ identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the correct answer is (A), but I can't figure out why this is true. If $M$ is&#xA;nonsingular, as is implied by statements (B)-(E), then isn't that equivalent to its column vectors being linearly&#xA;independent? And if the 5 column vectors are independent, then I can easily show that&#xA;each pair of vectors are independent. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-10-14T04:46:10.127" Title="GRE linear algebra question" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" />
  <row Id="6735" PostTypeId="2" ParentId="6734" CreationDate="2010-10-14T03:25:43.047" Score="4" ViewCount="0" Body="&lt;p&gt;Linear independence of $n$ vectors, for $n&gt;2$, is &lt;em&gt;not&lt;/em&gt; equivalent to &quot;pairwise&quot; independence. Take three different vectors in the plane, for an example.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-14T03:25:43.047" CommentCount="2" />
  <row Id="6736" PostTypeId="2" ParentId="6733" CreationDate="2010-10-14T03:31:43.533" Score="1" ViewCount="0" Body="&lt;p&gt;Use the following simplification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\neg AB(\neg A + \neg D) = \neg AB\neg A + \neg AB\neg D = \neg AB + \neg AB\neg D = \neg AB$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-14T03:31:43.533" CommentCount="1" />
  <row Id="6737" PostTypeId="1" AcceptedAnswerId="6741" CreationDate="2010-10-14T03:42:15.737" Score="8" ViewCount="298" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; &lt;a href=&quot;http://stackoverflow.com/q/3924602/33686&quot;&gt;This question has been posted on StackOverflow&lt;/a&gt;. I have moved it here because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I am curious about the answer&lt;/li&gt;&#xA;&lt;li&gt;The OP has not shown any interest in moving it himself&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the Communications of the ACM, &lt;a href=&quot;http://delivery.acm.org/10.1145/1790000/1787260/p128-winkler.html?key1=1787260&amp;amp;key2=3588796821&amp;amp;coll=portal&amp;amp;dl=ACM&amp;amp;CFID=108548737&amp;amp;CFTOKEN=59330921&quot; rel=&quot;nofollow&quot;&gt;August 2008 &quot;Puzzled&quot; column&lt;/a&gt;, Peter Winkler asked the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the table before us are 10 dots,&#xA;  and in our pocket are 10 $1 coins.&#xA;  Prove the coins can be placed on the&#xA;  table (no two overlapping) in such a&#xA;  way that all dots are covered. Figure&#xA;  2 shows a valid placement of the coins&#xA;  for this particular set of dots; they&#xA;  are transparent so we can see them.&#xA;  The three coins at the bottom are not&#xA;  needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;http://delivery.acm.org/10.1145/1820000/1810917/p110-winkler.html?key1=1810917&amp;amp;key2=2298796821&amp;amp;coll=portal&amp;amp;dl=ACM&amp;amp;CFID=108548737&amp;amp;CFTOKEN=59330921&quot; rel=&quot;nofollow&quot;&gt;following issue&lt;/a&gt;, he presented his proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We had to show that any 10 dots on a&#xA;  table can be covered by&#xA;  non-overlapping $1 coins, in a problem&#xA;  devised by Naoki Inaba and sent to me&#xA;  by his friend, Hirokazu Iwasawa, both&#xA;  puzzle mavens in Japan.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The key is to note that packing disks&#xA;  arranged in a honeycomb pattern cover&#xA;  more than 90% of the plane. But how do&#xA;  we know they do? A disk of radius one&#xA;  fits inside a regular hexagon made up&#xA;  of six equilateral triangles of&#xA;  altitude one. Since each such triangle&#xA;  has area &lt;code&gt;sqrt(3)/3&lt;/code&gt;, the hexagon&#xA;  itself has area &lt;code&gt;2*sqrt(3)&lt;/code&gt;; since the&#xA;  hexagons tile the plane in a honeycomb&#xA;  pattern, the disks, each with area π,&#xA;  cover &lt;code&gt;π /(2*sqrt(3))&lt;/code&gt; ~ .9069 of the&#xA;  plane's surface.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It follows that if the disks are&#xA;  placed randomly on the plane, the&#xA;  probability that any particular point&#xA;  is covered is .9069. Therefore, if we&#xA;  randomly place lots of $1 coins&#xA;  (borrowed) on the table in a hexagonal&#xA;  pattern, on average, 9.069 of our 10&#xA;  points will be covered, meaning at&#xA;  least some of the time all 10 will be&#xA;  covered. (We need at most only 10&#xA;  coins so give back the rest.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What does it mean that the disks cover&#xA;  90.69% of the infinite plane? The easiest way to answer is to say,&#xA;  perhaps, that the percentage of any&#xA;  large square covered by the disks&#xA;  approaches this value as the square&#xA;  expands. What is &quot;random&quot; about the&#xA;  placement of the disks? One way to&#xA;  think it through is to fix any packing&#xA;  and any disk within it, then pick a&#xA;  point uniformly at random from the&#xA;  honeycomb hexagon containing the disk&#xA;  and move the disk so its center is at&#xA;  the chosen point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand. Doesn't the probabilistic nature of this proof simply mean that in the &lt;strong&gt;majority&lt;/strong&gt; of configurations, all 10 dots can be covered. Can't we still come up with a configuration involving 10 (or&amp;nbsp;less) dots where one of the dots can't be covered?&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-14T06:00:22.397" LastActivityDate="2010-10-14T06:00:22.397" Title="Covering ten dots on a table with ten equal-sized coins: explanation of proof" Tags="&lt;geometry&gt;&lt;probability&gt;&lt;puzzle&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" />
  <row Id="6738" PostTypeId="1" AcceptedAnswerId="6742" CreationDate="2010-10-14T03:50:17.947" Score="2" ViewCount="129" Body="&lt;p&gt;Originally I was just going to ask the problem on my practice math contest, which is asking how many ways there are to write a nine-digit number containing each digit 1-9 so that the first digit is twice as big as the second and the last two digits are odd. (e.g. 846527913).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood that in this case, the number would be in the form of EE(EEEOO)OO or EO(EEOOO)OO where E is an even, O is an odd, the numbers in the parentheses can be ordered in any way, and the only four cases for the first two are 21, 42, 63, and 84. Calculating this gave me 240, which is definitely wrong (I made a program to check, and it says 7680).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are the questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How do I solve this particular problem correctly? Am I on the right track?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How are these solved in general? Given that it's a 9 digit number and each digit 1-9 is used once, how does one handle constraints such as these? This has always confused me (I'm not very strong in combinatorics)&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-10-14T04:10:15.780" Title="Different ways to write a number given digit constraints" Tags="&lt;combinatorics&gt;&lt;probability&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="6739" PostTypeId="2" ParentId="6737" CreationDate="2010-10-14T04:02:11.663" Score="7" ViewCount="0" Body="&lt;p&gt;The key point is that the 90.69% probability is with respect to &quot;the &lt;em&gt;disks&lt;/em&gt; [being] placed randomly on the plane&quot;, not the &lt;em&gt;points&lt;/em&gt; being placed randomly on the plane. That is, the set of points on the plane is &lt;em&gt;fixed&lt;/em&gt;, but the honeycomb arrangement of the disks is placed over it at a random displacement. Since the probability that any such placement covers a given point is 0.9069, a random placement of the honeycomb will cover, on average, 9.069 points (this follows from linearity of expectation; I can expand on this if you like). Now the only way random placements can cover 9.069 points &lt;em&gt;on average&lt;/em&gt; is if &lt;em&gt;some&lt;/em&gt; of these placements cover 10 points -- if all placements covered 9 points or less, the average number of points covered would be at most 9. Therefore, &lt;em&gt;there exists&lt;/em&gt; a placement of the honeycomb arrangement that covers 10 points (though this proof doesn't tell you what it is, or how to find it).&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-14T04:02:11.663" CommentCount="1" />
  <row Id="6740" PostTypeId="2" ParentId="6737" CreationDate="2010-10-14T04:05:05.393" Score="1" ViewCount="0" Body="&lt;p&gt;If you read carefully, this proof is for an &lt;em&gt;arbitrary&lt;/em&gt; placement of dots. So given &lt;em&gt;any&lt;/em&gt; dot arrangement, if we just place the coins randomly (in the honeycomb arrangement,) then on average we will cover slightly more than 9 of the dots. But since we can't cover &quot;part&quot; of a dot (in this problem) then that means that there &lt;strong&gt;exits&lt;/strong&gt; a random placement of the coins that cover all 10 dots. So no matter the configuration of dots, we know that there is always a way to cover the dots with at most 10 coins :) &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-14T04:11:41.460" LastActivityDate="2010-10-14T04:11:41.460" />
  <row Id="6741" PostTypeId="2" ParentId="6737" CreationDate="2010-10-14T04:07:38.023" Score="11" ViewCount="0" Body="&lt;p&gt;Nice! The above proof proves that &lt;em&gt;any&lt;/em&gt; configuration of 10 dots can be covered. What you have here is an example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_method&quot; rel=&quot;nofollow&quot;&gt;probabilistic method&lt;/a&gt;, which uses probability but gives a certain (not a probabilistic) conclusion (an example of &lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_proofs_of_non-probabilistic_theorems&quot; rel=&quot;nofollow&quot;&gt;probabilistic proofs of non-probabilistic theorems&lt;/a&gt;). This proof also implicitly uses the linearity of expectation, a fact that seem counter-intuitive in some cases until you get used to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify the proof: given &lt;em&gt;any&lt;/em&gt; configuration of 10 dots, &lt;strong&gt;fix&lt;/strong&gt; the configuration, and consider placing honeycomb-pattern disks randomly. Now, what is the expected number $X$ of dots covered? Let $X_i$ be 1 if dot $i$ is covered, and $0$ otherwise. We know that $E[X] = E[X_1] + \dots + E[X_{10}]$, and also that $E[X_i] = \Pr(X_i = 1) \approx 0.9069$ as explained above, for all $i$. So $E[X] = 9.069$. (Note that we have obtained this result using linearity of expectation, even though it would be hard to argue about the events of covering the dots being independent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the average over placements of the disks (for the fixed configuration of points!) is 9.069, not &lt;em&gt;all&lt;/em&gt; placements can cover ≤9 dots — at least one placement must cover all 10 dots.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-10-14T04:07:38.023" CommentCount="1" />
  <row Id="6742" PostTypeId="2" ParentId="6738" CreationDate="2010-10-14T04:10:15.780" Score="2" ViewCount="0" Body="&lt;p&gt;You're on the right track.  I'd break the problem up into four cases: where the number starts with 21, 42, 63 or 84.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where you start with 21, you must have 21(EEEOO)OO in your notation. (There are four even digits, so three of them must be in the middle part of the number.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can ask the questions: &#xA;- in how many different places can 4 go? &#xA;- in how many different places can  6 go (given that we've already placed 4?)&#xA;- in how many different places can  8 go (given that we've already placed 4 and 6)? &#xA;The answers are 5, 4, and 3 respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to place the four remaining odd digits among the four remaining places; there are of course $4! = 24$ ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are a total of $5 \times 4 \times 3 \times 24 = 1440$ such numbers beginning with 21.  The same is true if we start with 63 instead of 21. When we start with 42 or 84 the result is different but the method is much the same, and adding up the results from the four cases gives 7680.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (2), there aren't really &quot;general&quot; methods. But one thing that is good to keep in mind is trying to place the digits one at a time (as I've done above) instead of trying to do everything in one fell swoop.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-14T04:10:15.780" CommentCount="1" />
  <row Id="6743" PostTypeId="2" ParentId="4175" CreationDate="2010-10-14T04:12:32.343" Score="2" ViewCount="0" Body="&lt;p&gt;You can also use the following finite calculus formula for alternating binomial transforms.  Let $B(f(k),n)$ denote the alternating binomial transform; i.e., $$B(f(k),n) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k).$$ &#xA;Then (the formula)&#xA;$$B(f(k),n) = -B(\Delta f(k),n-1) + f(0)[n=0].$$&#xA;(Here, $\Delta f(k)$ is the finite difference $f(k+1) - f(k)$, and the expression $[n=0]$ evaluates to $1$ if $n = 0$ and $0$ otherwise.  Also, note that the first argument to $B$ is a function, while the second is a number.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with $f(k) = 1$, we have $\Delta f(k) = 0$.  Since $B(0,n-1)$ is clearly $0$, $B(1,n) = [n=0]$.  The latter is just the known formula &#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} = [n=0].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, continuing to take antidifferences, and using the notation $k^{\underline{m}}$ for the falling factorial $k(k-1)\cdots (k-m+1)$ as well as the power rule for finite differences $\Delta k^{\underline{m}} = m k^{\underline{m-1}}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n (-1)^k \binom{n}{k}k = -B(1,n-1) = - [n-1=0] = -[n=1],$$&#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{2}} = -2B(k,n-1) = 2[n-1 = 1] = 2[n=2],$$&#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{3}} = -3B(k^{\underline{2}},n-1) = -6[n-1 = 2] = -6[n=3],$$&#xA;and so forth, until we get to &#xA;$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{m}} = -mB(k^{\underline{m-1}},n-1) = (-1)^m m![n=m].$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $k(k-1)\cdots (k-m+1) = \frac{k!}{(k-m)!}$, dividing both sides of this last identity by $(-1)^m m!$ proves the OP's identity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The finite calculus formula for alternating binomial transforms and this argument are excerpted from my paper &quot;Combinatorial sums and finite differences,&quot; &lt;em&gt;Discrete Mathematics&lt;/em&gt; 307 (24): 3130-3146, 2007.) &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-14T04:12:32.343" />
  <row Id="6744" PostTypeId="2" ParentId="6734" CreationDate="2010-10-14T04:46:10.127" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ u,v\ $ linear independent $\rm\ \Rightarrow\:\ u + v,\:u+2v,\: u+3v,\:\ldots\: $   pairwise independent&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-14T04:46:10.127" />
  <row Id="6745" PostTypeId="1" AcceptedAnswerId="6761" CreationDate="2010-10-14T04:59:21.177" Score="2" ViewCount="39" Body="&lt;p&gt;It's a well known fact in probability theory that if $\lbrace X_i\rbrace$ is a family of random variables, $\phi : [0,\infty) \to [0,\infty)$ satisfies $$\lim_{x \to \infty} \frac{\phi(x)}{x} = \infty \quad (*) $$ and $\sup_i E\phi(|X_i|) &amp;lt; \infty$, then the family $\lbrace X_i\rbrace$ is uniformly integrable.  I want to know whether there is a single function $\phi$ satisfying (*) such that $\sup_i E\phi(|X_i|) &amp;lt; \infty$ iff $\lbrace X_i\rbrace$ is uniformly integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the answer is no, and that even a family of one absolutely continuous random variable should suffice.  So I conjecture the following: given any $\phi$ satisfying (*), there exists a nonnegative function $f$ such that&#xA;$$\int_0^\infty f(x)dx=1$$&#xA;$$\int_0^\infty x f(x)dx &amp;lt; \infty$$&#xA;$$\int_0^\infty \phi(x)f(x)dx = \infty.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't see offhand what to use for $f$.  Ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nb.  This is not homework.  It was going to be, but I thought I should be able to solve it myself before assigning it to my students :)&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-14T08:13:38.290" Title="No perfect function for uniform integrability" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="6746" PostTypeId="1" AcceptedAnswerId="6747" CreationDate="2010-10-14T05:07:39.977" Score="1" ViewCount="55" Body="&lt;p&gt;Let $T:X \times X \to \mathbb{R}$ be a continuous bilinear operator defined on a normed linear space $X$ s.t. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(\alpha x + \beta y,z) = \alpha T(x,z) + \beta T(y,z)$) and $T(x,y) = T(y,x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a constant $C$ s.t. $||T(x,y)|| \leq C$ $||x||$ $||y|| \forall x,y$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the result is true if $X$ and $Y$ are complete spaces, by using the uniform boundedness principle on $T$ as a continuous function of x for fixed y (and/or the other way around). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure if completeness is necessary, since it is true that a continuous linear operator $T: X \to \mathbb{R}$ has the property $||T(x)|| \leq C ||x|| \forall x$ on any normed linear space $X$ (although linear and bilinear operators are not exactly the same). &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-10-14T13:41:57.733" LastActivityDate="2010-10-14T13:41:57.733" Title="Boundedness of Continuous Bilinear Operators" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="6747" PostTypeId="2" ParentId="6746" CreationDate="2010-10-14T06:09:29.650" Score="2" ViewCount="0" Body="&lt;p&gt;I think you can show this using the same argument as in the continuous linear operators.&#xA;Since T is continuous, then $U=T^{-1}( (-1,1) )$ is open and contains $(0,0)$. find a c&gt;0 small enough such that if $|x|,|y|\leq c$ then $(x,y)\in U$ and then for a general point (x,y) you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|T(x,y)| = |T(\frac{c x}{|x|} \frac {|x|}{c}, \frac{c y}{|y|} \frac {|y|}{c} )| = \frac {|x||y|}{c^2} |T(\frac{c x}{|x|} , \frac{c y}{|y|})| \leq \frac {|x||y|}{c^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if x=0 or y=0 then T(x,y)=0 so you can use the argument above for $x,y \neq 0$. &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-10-14T06:09:29.650" CommentCount="1" />
  <row Id="6748" PostTypeId="1" AcceptedAnswerId="6783" CreationDate="2010-10-14T06:12:06.323" Score="2" ViewCount="158" Body="&lt;p&gt;I'm trying to wrap my brain around adjoint functors. One of the examples I've seen is the categories $\bf IntLE \bf = (\mathbb{Z}, ≤)$ and $\bf RealLE \bf = (\mathbb{R}, ≤)$, where the ceiling functor $ceil : \bf RealLE \rightarrow IntLE$ is left adjoint to the inclusion functor $incl : \bf IntLE \rightarrow RealLE$. I want to check that the following are true, as they seem to be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$floor : \bf RealLE \rightarrow IntLE$ would be right adjoint to $incl$&lt;/li&gt;&#xA;&lt;li&gt;Between the dual categories of $\bf IntGE \bf = (\mathbb{Z}, ≥)$ and $\bf RealGE \bf = (\mathbb{R}, ≥)$, $ceil$ would be right adjoint to $incl$&lt;/li&gt;&#xA;&lt;li&gt;Between $\bf RealGE$ and $\bf IntGE$, $floor$ would be left adjoint to $incl$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is my understanding correct on these points?&lt;/p&gt;&#xA;" OwnerUserId="2378" LastEditorUserId="2378" LastEditorDisplayName="" LastEditDate="2010-10-14T06:25:13.353" LastActivityDate="2010-10-14T20:37:58.503" Title="Adjoint functors" Tags="&lt;category-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="6749" PostTypeId="1" AcceptedAnswerId="6754" CreationDate="2010-10-14T06:18:15.003" Score="2" ViewCount="167" Body="&lt;p&gt;First of all, i'll confess i'm no math geek. I'm from Stackoverflow, but this question seemed more apt here, so i decided to ask you guys :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, i know noone has discovered (or ever will) a Polynomial that generates Prime Numbers. But i've read about Curve Fitting (or Polynomial Fitting) so i was wondering if there was a way, we could have a simple n-degree Polynomial that could generate the first 1000 (or X) primes accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need it to generate all the primes, maybe just upto 1 million, since we already have the data, can we deduce that polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big will be the polynomial for it to be accurate? Could you give an example for the first 100 primes? Am i just plain naive? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in Advance. :)&lt;/p&gt;&#xA;" OwnerUserId="2214" LastActivityDate="2010-10-14T13:02:22.380" Title="The Prime Polynomial : Generating  Prime Numbers." Tags="&lt;prime-numbers&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="6750" PostTypeId="1" AcceptedAnswerId="6756" CreationDate="2010-10-14T06:18:16.373" Score="5" ViewCount="189" Body="&lt;p&gt;What is the difference complete metric space and a closed set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a set be closed but not complete?&lt;/p&gt;&#xA;" OwnerUserId="2504" LastActivityDate="2010-10-14T07:13:14.613" Title="Difference between complete and closed set." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6751" PostTypeId="1" CreationDate="2010-10-14T06:21:55.907" Score="0" ViewCount="54" Body="&lt;p&gt;Let $g_{ij}$ be the components of a symmetric rank-2 positive definite tensor (metric on a Riemannian manifold). Let ${C^i}$ and ${ \beta ^i }$ be components of a vector field on it, the former of which is a functions of a variable $t$ and let $N$ and $M$ be two positive constants. Now let the following inequality hold,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt {g_{ij} \left( \frac{dC^i}{dt} + \beta ^i \right) \left( \frac{dC^j}{dt} + \beta ^j  \right)} &amp;lt; N &amp;lt; M    $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there exists a constant $B$ such that $\sqrt{g_{ij} \beta ^i  \beta ^j}$ is uniformly bounded by $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then apparently the following is true,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt {g_{ij} \frac{dC^i}{dt} \frac{dC^j}{dt} } \leq N + \sqrt{g_{ij} \beta ^i  \beta ^j} \leq  M + B $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this follow?&lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-10-14T06:35:43.323" LastActivityDate="2010-11-13T08:44:19.097" Title="An inequality about norms of vector fields on Riemannian manifolds " Tags="&lt;analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="6754" PostTypeId="2" ParentId="6749" CreationDate="2010-10-14T06:51:09.463" Score="5" ViewCount="0" Body="&lt;p&gt;While Robin was commenting, I cooked up a Mathematica one-liner which gives you the Lagrange interpolation polynomial for the first n primes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[n_] := Sum[Prime[k] Product[x-m,{m,Join[Range[1,k-1], Range[k+1,n]]}] / Product[k-m,{m,Join[Range[1,k-1], Range[k+1,n]]}] ,{k,1,n}] // Expand&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And I don't think anyone would call the results &quot;simple&quot;. For n=100 it's too big to paste here, but n=15 gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4748 - (1752758563*x)/120120 + (71043957851*x^2)/3783780 - &#xA; (4320411427*x^3)/316800 + (378496362427*x^4)/59875200 - &#xA; (7239131749*x^5)/3628800 + (11528263*x^6)/25920 - &#xA; (2075348983*x^7)/29030400 + (5090997277*x^8)/609638400 - &#xA; (977071*x^9)/1382400 + (743507*x^10)/17418240 - (568871*x^11)/319334400 + &#xA; (2207*x^12)/45619200 - (3151*x^13)/4151347200 + (89*x^14)/17435658240&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T06:51:09.463" CommentCount="3" />
  <row Id="6755" PostTypeId="2" ParentId="6340" CreationDate="2010-10-14T07:01:25.227" Score="0" ViewCount="0" Body="&lt;p&gt;$\newcommand\RR{\mathbb{R}}$&#xA;$\newcommand\NN{\mathbb{N}}$&#xA;$\newcommand\e{\varepsilon}$&#xA;$\newcommand\lbg{\lambda}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f: \RR \to \RR$ is an analytic unbounded Lebesgue integrable function, then it must in particular satisfy this property (because it is continuous):&#xA;\[&#xA;   \forall \e,M &gt; 0 \; \exists x_0 : |x| &gt; |x_0| \Rightarrow&#xA;   \lbg(\{x \in \RR : |f(x)| &gt; M\}) &amp;lt; \e,&#xA;\]&#xA;where $\lbg$ is the usual Lebesgue measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just use that $f$ is analytic iff $\forall \text{ compact } K \subset \RR \; \exists C &gt; 0 : x \in K, n \in \NN \Rightarrow \left\vert\frac{\partial^n f}{\partial x^n}(x)\right\vert &#xA;\leq C^{n+1}n!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is easy to see that this is impossible since $|f'|$ will have to be arbitrarily large - or in other words that you can go far enough out to the right on the x-axis that $|f|$ has to go from being larger than some big $M$ to smaller than some small $m$ on a very small interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is missing quite a few details, but it should be pretty easy to fill out.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-14T07:13:28.553" LastActivityDate="2010-10-14T07:13:28.553" />
  <row Id="6756" PostTypeId="2" ParentId="6750" CreationDate="2010-10-14T07:13:14.613" Score="7" ViewCount="0" Body="&lt;p&gt;A metric space is complete if every Cauchy sequence converges.  A subset $F$ of a metric space $X$ is closed if $F$ contains all of its limit points; this can be characterized by saying that if a sequence in $F$ converges to a point $x$ in $X$, then $x$ must be in $F$.  It also makes sense to ask whether a subset of $X$ is complete, because every subset of a metric space is a metric space with the restricted metric.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that a complete subspace must be closed, which essentially results from the fact that convergent sequences are Cauchy sequences.  However, closed subspaces need not be complete.  For a trivial example, start with any incomplete metric space, like the rational numbers $\mathbb{Q}$ with the usual absolute value distance.  Like every metric space, $\mathbb{Q}$ is closed in itself, so there you have a subset that is closed but not complete.  If taking the whole space seems like cheating, just take the rationals in $[0,1]$, which will be closed in $\mathbb{Q}$ but not complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is a complete metric space, then a subset of $X$ is closed if and only if it is complete.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-14T07:13:14.613" CommentCount="2" />
  <row Id="6757" PostTypeId="1" AcceptedAnswerId="6798" CreationDate="2010-10-14T07:45:32.363" Score="4" ViewCount="175" Body="&lt;p&gt;What is the continued fraction for $\displaystyle\sum_{i=1}^n\frac{1}{2^{2^i}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be &quot;almost&quot; periodic, but I can't figure out the exact way to express it.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-10-16T15:56:23.260" Title="Continued Fraction of an Infinite Sum" Tags="&lt;number-theory&gt;&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="6758" PostTypeId="1" AcceptedAnswerId="6765" CreationDate="2010-10-14T07:55:24.650" Score="3" ViewCount="183" Body="&lt;p&gt;I'm still making my way along in Niven's Intro to Number Theory, and the title problem is giving me a little trouble near the end, and I was hoping someone could help get me through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $x^8\equiv 16\pmod{2}$ is solvable with $x\equiv 0\pmod{2}$, so I assume $p$ is an odd prime. From a theorem earlier in the text, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $p$ is a prime and $(a,p)=1$, then the congruence $x^n\equiv a\pmod{p}$ has $(n,p-1)$ solutions or no solution according as $a^{(p-1)/(n,p-1)}\equiv 1\pmod{p}$ or not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So since $(16,p)=1$, the problem reduces to showing that $16^{(p-1)/(8,p-1)}\equiv 1\pmod{p}$ holds for all $p$. I note that $(8,p-1)$ can only take values $2,4,8$. For $2$, the above equivalence is then $4^{p-1}\equiv 1\pmod{p}$, which is true by Fermat's little Theorem. For $4$, it is then $2^{p-1}\equiv 1\pmod{p}$, which again holds by FlT. However, the case where $(8,p-1)=8$ is throwing me off. At best I see that $16^{(p-1)/8}\equiv 2^{(p-1)/2}\pmod{p}$, but I'm not sure how to show this is congruent to $1$ modulo $p$. Maybe there's a more elegant way to do it without looking at cases. Thanks for any insight.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-14T15:13:46.610" Title="Showing $x^8\equiv 16 \pmod{p}$ is solvable for all primes $p$." Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6759" PostTypeId="2" ParentId="98" CreationDate="2010-10-14T08:01:40.487" Score="2" ViewCount="0" Body="&lt;p&gt;The book is Coffee in Memphis by Bollobas. It's the first problem, and there are loads more :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521693950&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Art-Mathematics-Coffee-Time-Memphis/dp/0521693950&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2506" LastActivityDate="2010-10-14T08:01:40.487" />
  <row Id="6760" PostTypeId="2" ParentId="6751" CreationDate="2010-10-14T08:11:13.687" Score="2" ViewCount="0" Body="&lt;p&gt;This is simply the triangle inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lVert\frac{dC}{dt}\rVert = \lVert\frac{dC}{dt} + \beta - \beta\rVert \le \lVert\frac{dC}{dt} + \beta \rVert + \lVert-\beta\rVert &amp;lt; N + \lVert\beta\rVert &amp;lt; M + B$$&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-10-14T08:11:13.687" CommentCount="1" />
  <row Id="6761" PostTypeId="2" ParentId="6745" CreationDate="2010-10-14T08:13:38.290" Score="2" ViewCount="0" Body="&lt;p&gt;By hypothesis, for any positive $k$, there is an $N_k$ such that $\phi(x) &gt; kx$ if $x &gt; N_k$.&#xA;We can of course assume that $(N_k)$ is an increasing sequence, and so it splits the real line into intervals, which we furthermore can assume are of length at least one (say).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $f$ to be piecewise constant: zero if $0 &amp;lt; x &amp;lt; N_1$, and chosen on the other intervals such&#xA;that $\int_{N_k}^{N_{k+1}} x f(x) dx = \frac{c}{k^2}$, where $c$ is just a&#xA;normalization constant (since you require $\int f(x) dx=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is, choose &#xA;$$ f(x) = \frac{c/k^2}{\int_{N_k}^{N_{k+1}} x dx}&#xA;= \frac{2c}{k^2 (N_{k+1}^2-N_k^2)} $$&#xA;for $N_k \le x &amp;lt; N_{k+1}$.&#xA;Then, because we arranged for the intervals to be sufficiently long,&#xA;$\int_0^{\infty} f(x) dx$ reduces to a &lt;em&gt;convergent&lt;/em&gt; sum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$ \int_0^{\infty} \phi(x) f(x) dx \ge \sum_{k\ge 1} \int_{N_k}^{N_{k+1}} kx f(x) dx = \sum_{k\ge 1} k \frac{c}{k^2} = \infty.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T08:13:38.290" CommentCount="1" />
  <row Id="6762" PostTypeId="2" ParentId="6732" CreationDate="2010-10-14T08:25:56.257" Score="1" ViewCount="0" Body="&lt;p&gt;I'll prove that every point of $G$ in the interior of $F$ is not an extreme point of $G$. I'll assume that $N&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;LEMMA.&lt;/em&gt; There is a vector $v\neq 0$ such that $h'v=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Since $N&gt;1$ there is a vector $u$ which is not a multiple of $h$. Let&#xA;$$v=u-\left({{h'u}\over{h'h}}\right)h$$&#xA;Then $h'v=0$. Since $u$ is not a multiple of $h$, $v\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Answer&lt;/em&gt;. Suppose $x \in G$ and $x$ is also in the interior of $F$. Let $v$ be as in the lemma above. Then there is a small enough $\lambda$ such that $(x \pm \lambda v) \in F$. But $h'(x \pm \lambda v)=h'x$ so $(x \pm \lambda v) \in G$. But,&#xA;$$x= {(x + \lambda v) \over 2} + {(x - \lambda v) \over 2}$$&#xA;So $x$ is not an extreme point of $G$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-14T08:25:56.257" />
  <row Id="6763" PostTypeId="1" AcceptedAnswerId="6773" CreationDate="2010-10-14T09:07:02.870" Score="2" ViewCount="319" Body="&lt;p&gt;In measure theory there are three fundamentally related theorems about exchanging limits and integrals: Fatou's lemma, Lebesgue's Dominated Convergence Theorem, and Monotone Convergence Theorem. It is difficult to prove any of these from scratch, but once you have one, the others are easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, for those who have learned these theorems: which one do you prefer to prove first? Difficulty, length, and, perhaps most importantly, how enlightening each path is are the key considerations. I suppose you could also phrase the question: if you were teaching a class in what order would you prove these theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read through all of the proofs and there doesn't seem to be a big difference, but perhaps someone can shed some new light on this question.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-14T13:00:41.277" Title="Theorems in Measure Theory: Fatou's Lemma, Lebesgue DCT, Monotone CT" Tags="&lt;soft-question&gt;&lt;measure-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="6764" PostTypeId="2" ParentId="457" CreationDate="2010-10-14T09:47:58.157" Score="5" ViewCount="0" Body="&lt;p&gt;There's a whole room devoted to physical demonstrations of mathematical concepts in the Museum of Science in Boston. I'll list the ones I can remember, and others who have been there please feel free to add to the list.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A little train that goes around on a track that's on a mobius strip.&lt;/li&gt;&#xA;&lt;li&gt;A &quot;Pachinko&quot; type machine where the balls fall into different slots and form a bell curve as the slots fill up.&lt;/li&gt;&#xA;&lt;li&gt;A pendulum that is free to swing forward and backward as well as left and right. At the bottom is a container of sand with a small hole in it, so as the sand leaks out, it forms &lt;a href=&quot;http://en.wikipedia.org/wiki/Lissajous_curve&quot; rel=&quot;nofollow&quot;&gt;Lissajous curves&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Something about the Riemann Zeta function that I didn't really understand.&lt;/li&gt;&#xA;&lt;li&gt;A straight, solid bar, rotated around an axis that is skew to the bar itself, that fits right through a hyperbola-shaped slot. I'm pretty sure the Exploratorium in San Francisco has this too, and here's a &lt;a href=&quot;http://youtube.com/watch?v=OJGkdO9gcdY&quot; rel=&quot;nofollow&quot;&gt;video of a similar exhibit&lt;/a&gt; thanks to commenter Rahul.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="496" LastEditorUserId="496" LastEditorDisplayName="" LastEditDate="2010-10-18T05:13:16.207" LastActivityDate="2010-10-18T05:13:16.207" CommunityOwnedDate="2010-10-14T09:47:58.157" CommentCount="1" />
  <row Id="6765" PostTypeId="2" ParentId="6758" CreationDate="2010-10-14T09:57:43.250" Score="2" ViewCount="0" Body="&lt;p&gt;One way is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_symbol&quot; rel=&quot;nofollow&quot;&gt;Legendre symbol&lt;/a&gt; identity $2^{(p-1)/2} \equiv (\frac{2}{p}) \equiv (-1)^{(p^2-1)/8} \pmod p$ (for odd primes p), keeping in mind that if (8,p-1)=8 then $p \equiv 1 \pmod 8$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-14T09:57:43.250" CommentCount="2" />
  <row Id="6766" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T10:06:21.260" Score="0" ViewCount="0" Body="&lt;p&gt;I prefer the direction: Dominated convergence theorem -&gt; Beppo-Levi (monotone convergence) -&gt; Fatou.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This direction requires more machinery (like Egoroff and the absolute continuity of the Lebesgue integral) but it is in my opinion more tidy. A book that uses this approach is &lt;a href=&quot;http://books.google.com/books?id=CoSIe7h5mTsC&amp;amp;printsec=frontcover&amp;amp;dq=Bogachev&amp;amp;hl=nl&amp;amp;ei=ENW2TIH0IsqZOozSxOwJ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Bogachev - Measure Theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This way the later theorems are just &quot;corollaries&quot; of DCT, the &quot;biggest&quot; one!&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-14T12:14:34.530" LastActivityDate="2010-10-14T12:14:34.530" />
  <row Id="6767" PostTypeId="2" ParentId="6758" CreationDate="2010-10-14T10:36:53.263" Score="0" ViewCount="0" Body="&lt;p&gt;I usually set this as an exercise when teaching Number Theory.&#xA;My hint is to ask the students: what are the solutions of&#xA;$z^8=16$ in the complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-14T10:36:53.263" />
  <row Id="6768" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T10:41:22.527" Score="0" ViewCount="0" Body="&lt;p&gt;I believe I learned the monotone convergence theorem first, then dominated convergence, then Fatou.  You could argue that this sequence orders the theorems in increasing order of hypothesis complexity.  &lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-14T10:41:22.527" />
  <row Id="6769" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T10:45:08.260" Score="0" ViewCount="0" Body="&lt;p&gt;In fact i would add something more. For me it would be the Bounded Convergence theorem, then the Fatou's lemma and then the Monotone convergence theorem followed by Dominated convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the book by Royden has a very good way proving these theorems&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-14T10:45:08.260" />
  <row Id="6770" PostTypeId="1" CreationDate="2010-10-14T10:49:41.527" Score="1" ViewCount="54" Body="&lt;p&gt;Let $u$ be a fixed vector of length $K$ and $A$ be a matrix of all positive numbers of size $K \times K$. Let $V$ be a set of vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V(\epsilon) \subseteq V$ be a set of vectors of length $K$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \le \sup_{v \in V(\epsilon)} u \cdot v &amp;lt; \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary vector $w \in V$, define $L(w) = \sum_{i=1,j=1}^K (u_i u_j / A_{ij}) \times w_i w_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $\sup_{v \in V(\epsilon)} L(v) &amp;lt; C*\epsilon^r$ for some $C,r &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $L(v)$ can also be rewritten as:&#xA;$L(v) = A \colon (u \otimes v) (u \otimes v)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\otimes$ is Hadamard product and $\colon$ is the Frobenius inner product.&#xA;We also know that the Frobenis inner product $A:B = tr(A^t B)$, therefore we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(v) = tr(A^T (u \otimes v) (u \otimes v)^T)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals (because the reordering of the matrices is a cyclic permutation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(v) = tr((u \otimes v)^T A^T (u \otimes v))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am stuck. Not sure how to deduce the rest, if it is at all possible.&#xA;Basically, if I show the following, then the proof is done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim??: let $A$ be a matrix of positive numbers. let $u$ be a vector of all positive numbers. There is a $C$ and an $r$ such that: let $v \in V$ be a vector. let $w = u \otimes v$ be a vector such that $0 &amp;lt; \sum_i w_i &amp;lt; \epsilon$. then $0 \le tr(w A^T w) \le C\epsilon^r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't think the above claim is true.&#xA;Yet, I have more structure to my problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ in fact has the form $V = ${$ (\log \frac{x}{x^\ast}, \log \frac{(1-x)}{(1-x^\ast)}) \mid x \in [0,1] $} for an $x^\ast \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i.e. $K=2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or for $K = 4$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = ${$ (\log \frac{x}{x^\ast}, \log \frac{(1-x)}{(1-x^\ast)},\log \frac{y}{y^\ast}, \log \frac{(1-y)}{(1-y^\ast)}) \mid y,x \in [0,1] $} for an $x^\ast,y^\ast \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc. (we can have any even $K$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could it be true for this $V$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any insight.&lt;/p&gt;&#xA;" OwnerUserId="2511" LastActivityDate="2010-10-14T10:49:41.527" Title="can we prove a connection between sum of numbers and $L_2$-like norm?" Tags="&lt;linear-algebra&gt;" />
  <row Id="6771" PostTypeId="2" ParentId="2427" CreationDate="2010-10-14T11:32:01.850" Score="4" ViewCount="0" Body="&lt;p&gt;What do we mean by  'sporadic' ?  I suggest that we may find some extension of the&#xA;Chevalley programme (Tohoku J 1950+)  that brings in the sporadics. This is about&#xA;construction - not the classification as we know it.&lt;/p&gt;&#xA;" OwnerUserId="2512" LastActivityDate="2010-10-14T11:32:01.850" CommentCount="1" />
  <row Id="6772" PostTypeId="2" ParentId="6648" CreationDate="2010-10-14T11:39:27.663" Score="1" ViewCount="0" Body="&lt;p&gt;Joseph Gallian has decrypted many of the US state license plate and driver's license codes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=PD0clAlF8O4C&amp;amp;pg=PA27&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=PD0clAlF8O4C&amp;amp;pg=PA27&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think he used Markov chain models.  As whuber mentioned your problem is similar to the German tanks for which the subject reference is &quot;extreme value statistics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T11:45:52.643" LastActivityDate="2010-10-14T11:45:52.643" />
  <row Id="6773" PostTypeId="2" ParentId="6763" CreationDate="2010-10-14T13:00:41.277" Score="2" ViewCount="0" Body="&lt;p&gt;I've generally seen MCT -&gt; Fatou -&gt; DCT.  MCT is nice if the integral is defined as the supremum of the integrals of all simple functions less than $f$.  Fatou points out that you can lose mass when passing to the limit, but cannot gain it.  And DCT is nice to prove with two applications of Fatou, since turning your head upside down shows that you cannot gain mass either positively or negatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree with Jonas's idea that DCT is the &quot;biggest&quot; one, since it doesn't speak about functions not in $L^1$, which the others do; this is often very important.  Also, I see the hypothesis of the DCT as somewhat ad hoc.  To my mind, the &quot;biggest&quot; one is the Vitali convergence theorem, whose hypothesis is uniform integrability, which is necessary and sufficient.  But since it is more complicated it is often skipped.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-14T13:00:41.277" CommentCount="1" />
  <row Id="6774" PostTypeId="1" AcceptedAnswerId="6776" CreationDate="2010-10-14T13:01:13.723" Score="1" ViewCount="67" Body="&lt;p&gt;Let $f: X \rightarrow Y$ be a local homeomorphism with X, Y connected, locally path connected, Hausdorff and with X also compact. Then f is also a covering with finite fibers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to show that the fibers are finite.&#xA;Given that f is a surjection, I know how to show that f is a covering map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I show that f is surjective?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1907" LastActivityDate="2010-10-14T17:05:11.953" Title="Local homeomorphism and coverings" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="6775" PostTypeId="2" ParentId="6749" CreationDate="2010-10-14T13:02:22.380" Score="3" ViewCount="0" Body="&lt;p&gt;From &lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there exists a polynomial  in 10 variables with integer coefficients such that the set of primes equals the set of positive  values of this polynomial obtained as the variables run through all nonnegative integers, although it is really a set of Diophantine equations in disguise (Ribenboim 1991). Jones, Sato, Wada, and Wiens have also found a polynomial of degree 25 in 26 variables whose positive values are exactly the prime numbers (Flannery and Flannery 2000, p. 51).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the primes do not come out in order, so this will not help for what you want.  But it is interesting.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T13:02:22.380" />
  <row Id="6776" PostTypeId="2" ParentId="6774" CreationDate="2010-10-14T13:07:55.590" Score="3" ViewCount="0" Body="&lt;p&gt;I think I have it. The image of X under f is open (use &quot;local homeomorphism&quot;) and closed (use &quot;compact&quot; and &quot;Hausdorff&quot;) in Y and since Y is connected this shows that the image is the whole of Y. &lt;/p&gt;&#xA;" OwnerUserId="1907" LastEditorUserId="1907" LastEditorDisplayName="" LastEditDate="2010-10-14T17:05:11.953" LastActivityDate="2010-10-14T17:05:11.953" />
  <row Id="6777" PostTypeId="1" CreationDate="2010-10-14T13:19:33.000" Score="5" ViewCount="408" Body="&lt;p&gt;What is motivation behind the definition of a complete metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively,a complete metric is complete if they are no points missing from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the definition of completeness (in terms of convergence of cauchy sequences) show that?&lt;/p&gt;&#xA;" OwnerUserId="2504" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-14T16:04:20.357" LastActivityDate="2010-10-15T11:08:07.987" Title="Motivation behind the definition of complete metric space." Tags="&lt;metric-spaces&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="6778" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T13:21:31.397" Score="2" ViewCount="0" Body="&lt;p&gt;When working with metric spaces, we tend to deal with sequences &lt;em&gt;a lot&lt;/em&gt;. Completeness is just the condition that those that &lt;em&gt;should&lt;/em&gt; converge actually converge.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-14T13:21:31.397" CommentCount="6" />
  <row Id="6779" PostTypeId="1" AcceptedAnswerId="6781" CreationDate="2010-10-14T13:21:41.100" Score="0" ViewCount="68" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sum of a set of sphenic numbers can't be equal to the sum of &lt;em&gt;any other&lt;/em&gt; set of sphenic numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By that I meant, Say S is the set of sphenic numbers. Let, S$_1$ $\subset$ S.&#xA;Then, there is no such S$_2$ $\subset$ S so that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S$_1$ $\neq$ S$_2$ AND $\sum S_1$ $=$ $\sum S_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1 : Is this statement correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2 : Is this formulation mathematically right? I mean even if the statement is wrong, is the way that I expressed it conveys mathematical notations/rules etc. Or, how a mathematician would write it if s/he intended to convey the same message?&lt;/p&gt;&#xA;" OwnerUserId="1869" LastActivityDate="2010-10-14T16:18:46.650" Title="Conjecture about the set of Sphenic numbers" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="6780" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T13:23:36.843" Score="3" ViewCount="0" Body="&lt;p&gt;I tend to think of it this way: A Cauchy sequence in any metric space is one such that for all $\varepsilon &gt; 0$ the tail of the sequence is &lt;em&gt;eventually&lt;/em&gt; in some $\varepsilon$-ball (not necessarily around the same point for all $\varepsilon$). In other words the only way a Cauchy sequence can fail to converge is if the limit is somehow &quot;not there&quot;. Hence the choice of the word &quot;complete&quot; because all of the limits that should be there, are in fact there.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-14T16:03:06.003" LastActivityDate="2010-10-14T16:03:06.003" CommentCount="2" />
  <row Id="6781" PostTypeId="2" ParentId="6779" CreationDate="2010-10-14T13:30:08.787" Score="3" ViewCount="0" Body="&lt;p&gt;1) The statement is not correct.  230+1310=231+1309.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphenic_number&quot; rel=&quot;nofollow&quot;&gt;Sphenic numbers&lt;/a&gt; gives the first sequential pair (and implies there are more) and the first run of three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I'm not a mathematician, but the formulation seems fine to me.  You might define Sphenic numbers so people don't have to look it up&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-14T16:18:46.650" LastActivityDate="2010-10-14T16:18:46.650" CommentCount="6" />
  <row Id="6782" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T14:11:13.283" Score="5" ViewCount="0" Body="&lt;p&gt;To &quot;fill the holes&quot; or &quot;add the missing points&quot; would presumably mean embedding the metric space as a subspace of a larger metric space.  To avoid trivialities like placing a line inside the plane, it is required (and it appears to be the only sensible interpretation) that the given space is &lt;em&gt;dense&lt;/em&gt; in the larger space: every neighborhood of a point in the larger space contains points of the smaller space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete metric space is one to which nothing new can be added in this way.  The &quot;no holes&quot; definition of completeness is then equivalent to completeness defined using Cauchy sequences.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more precise term than holes would &quot;punctures&quot;.  Holes also have a topological meaning such as the hole surrounded by an annulus in the plane, the hole in a torus, or the keyhole in a lock.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T17:47:17.680" LastActivityDate="2010-10-14T17:47:17.680" CommentCount="3" />
  <row Id="6783" PostTypeId="2" ParentId="6748" CreationDate="2010-10-14T14:39:47.683" Score="2" ViewCount="0" Body="&lt;p&gt;Given categories $\mathcal{C}$ and $\mathcal{D}$, and functors $\mathbf{F}\colon\mathcal{C}\to\mathcal{D}$ and $\mathbf{U}\colon\mathcal{D}\to\mathcal{C}$, $\mathbf{F}$ is the left adjoint of $\mathcal{U}$ if and only if for every objects $C\in\mathcal{C}$ and $D\in\mathcal{D}$ there is a natural bijection between $\mathcal{C}(C,\mathbf{U}(D))$ and $\mathcal{D}(\mathbf{F}(C),D)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me use $\leq$ and $\geq$ for the relation among reals, and $\preceq, \succeq$ for the relation among integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $ceil$ to be the left adjoint of the inclusion functor, you would need that for all real numbers $r$ and all integers $z$, $\lceil r\rceil \preceq z$ if and only if $r\leq z$. This holds, so you do have an ajunction (this works because in these categories, the morphism set &lt;strong&gt;IntLE&lt;/strong&gt;$(a,b)$ is empty if $a\not\preceq b$, and contains a unique arrow if $a\preceq b$; and similarly with &lt;strong&gt;RealLE&lt;/strong&gt;; so you get a natural bijection between the sets of arrows if and only if they are either both empty or both are singletons at the same time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $floor$ to be a right adjoint to $incl$, you would need that for all real numbers $r$ and all integers $z$, $z\preceq \lfloor r\rfloor$ if and only if $z\leq r$, which again is true; so $floor$ is a right adjoint to the inclusion functor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $ceil$ to be the right adjoint to the inclusion functor in the dual categories, you would need $z\succeq \lceil r \rceil$ if and only if $z\geq r$; and for $floor$ to be the left adjoint, you would need $\lfloor r\rfloor \succeq z$ if and only if $r\geq z$. Both hold, so your assertions 1 through 3 are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Let me second Mariano's suggestion in the comments to keep in mind the case of the underlying set functor and the free group functor for thinking about right and left adjoints. I find myself going back to those two every time I need to remind myself of how things work with adjoints, what adjoints respect or don't respect, and especially when thinking about some of the other equivalent definitions, in particular the one in terms of the unit and co-unit of the adjunction (which are natural transformations between the identity functors and the functors $\mathbf{FU}$ and $\mathbf{UF}$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-14T17:30:14.333" LastActivityDate="2010-10-14T17:30:14.333" CommentCount="2" />
  <row Id="6784" PostTypeId="1" CreationDate="2010-10-14T14:50:01.800" Score="1" ViewCount="182" Body="&lt;p&gt;I need to calculate all the numbers in a certain row of Pascal's triangle. Obviously, this is easy to do with combinatorics. However, what do you do when you need to estimate all the numbers in, say, the 100000th row of Pascal's triangle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to estimate the number so that the costly multiplications and divisions of binomials can be avoided? I'm already estimating factorials with Stirling's formula, but it still takes a number of seconds to calculate only one number - and I need about 100000/2 (since a row is symmetric).&lt;/p&gt;&#xA;" OwnerUserId="2471" LastActivityDate="2010-10-14T17:35:52.343" Title="Estimating a certain row of Pascal's triangle" Tags="&lt;binomial-coefficients&gt;&lt;big-numbers&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="6785" PostTypeId="2" ParentId="6758" CreationDate="2010-10-14T15:01:07.857" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ x^8 - 16\ =\ (x^2 - 2)\: (x^2 + 2)\: (x^4 + 4)\:.\ \:$ If the first two factors have no roots in $\rm\ \mathbb Z/p\ $ then $\:2, -2\:$ are nonsquares so their product $-4\: $ is a square, so $\rm\: i = \sqrt{-1} \in \mathbb Z/p\:$. Thus the third factor has a root since $\rm\ x^4 + 4\ $ has roots $\rm\: \pm 1\pm i\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-14T15:13:46.610" LastActivityDate="2010-10-14T15:13:46.610" CommentCount="2" />
  <row Id="6786" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T15:14:06.207" Score="4" ViewCount="0" Body="&lt;p&gt;Are you computing the factorials and &lt;em&gt;then&lt;/em&gt; dividing?  Don't do that.  You can combine the estimates you get from Stirling's formula into &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n+m \choose n} \approx \sqrt{ \frac{m+n}{2 \pi mn} } \frac{(m+n)^{m+n}}{m^m n^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and even then you can optimize, for example rewriting the second factor as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 1 + \frac{n}{m} \right)^m \left( 1 + \frac{m}{n} \right)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, depending on the relative sizes of $m$ and $n$, applying the approximation $\left( 1 + \frac{x}{n} \right)^n \approx e^x$.  This will work if one of $m$ and $n$ are large compared to the other and in the intermediate case the above powers should be straightforward to compute.  See also Exercises 1 and 2 at &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/&quot; rel=&quot;nofollow&quot;&gt;this blog post by Terence Tao&lt;/a&gt; on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what kind of precision you need you should consider just working with the logarithms and not with the numbers directly.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-14T16:12:27.050" LastActivityDate="2010-10-14T16:12:27.050" CommentCount="1" />
  <row Id="6788" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T15:35:37.850" Score="0" ViewCount="0" Body="&lt;p&gt;Depending on the accuracy you need, this bound might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{n}{k}\right)^k \leq \binom{n}{k} \leq \left(\frac{ne}{k}\right)^k.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-14T15:35:37.850" CommentCount="1" />
  <row Id="6789" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T15:43:43.043" Score="3" ViewCount="0" Body="&lt;p&gt;If you want all the numbers in one particular row, you can get each from the last with one multiply and one divide.  ${n \choose m}={n \choose m-1}*(n-m+1)/m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For individual components, you can use the fact that the distribution is approximately normal with standard deviation $\sqrt n$ and the central coefficient is about $4^n/\sqrt n\pi$.  I'm not sure that is any faster than Qiaochu's suggestion&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T15:43:43.043" CommentCount="1" />
  <row Id="6790" PostTypeId="2" ParentId="6779" CreationDate="2010-10-14T15:44:39.123" Score="2" ViewCount="0" Body="&lt;p&gt;Question 2: Instead of $\sum S_1$ you can write $$\sum_{k \in S_1} k.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T15:44:39.123" CommentCount="3" />
  <row Id="6791" PostTypeId="1" CreationDate="2010-10-14T15:49:08.723" Score="24" ViewCount="470" Body="&lt;p&gt;A high-school student named Erna Fekete made a conjecture to me via email three years ago, &#xA;which I could not answer.  I've since lost touch with her.&#xA;I repeat her interesting conjecture here, in case anyone can provide updated&#xA;information on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how she phrased it.  Let $b(0) = 1$ and $b(n)= \tan( b(n-1) )$.&#xA;In other words, $b(n)$ is the repeated application of $\tan(\;)$ to 1:&#xA;$$\tan(1) = 1.56, \; \tan(\tan(1)) = 74.7, \; \tan^3(1) = -0.9, \; \ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a(n) = \lfloor b(n) \rfloor$.&#xA;Her conjecture is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every integer eventually appears in the $a(n)$ sequence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sequence is not unknown; it&#xA;is &lt;a href=&quot;http://akpublic.research.att.com/~njas/sequences/A000319&quot; rel=&quot;nofollow&quot;&gt;A000319&lt;/a&gt; in Sloane's integer sequences.&#xA;Essentially hers is a question about the orbit of 1 under repeated $\tan(\;)$-applications.&#xA;Her and my investigations at the time led us to believe it was an open problem.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2011-01-06T15:11:25.920" Title="Fekete's conjecture on repeated applications of the tangent function" Tags="&lt;reference-request&gt;&lt;trigonometry&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="6" />
  <row Id="6792" PostTypeId="2" ParentId="6495" CreationDate="2010-10-14T16:20:19.650" Score="3" ViewCount="0" Body="&lt;p&gt;I'm sort of just guessing here (partly based on the solution already found). By explicitly plugging in $V(t,r) = - \frac{M(r)}{R(t,r)}$, you arrive at the ordinary differential equation (for each fixed $r$) for $R$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{R}^2 - \frac{2M}{R} = 2 E $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ and $E$ are constants in time. Now, re-scale the original equation by $t = \lambda s$ and $R = \mu \rho$. Then $\partial_t R = \frac{\mu}{\lambda} \partial_s \rho$. Then you can solve $(\frac{\mu}{\lambda})^2 = \frac{2M}{\mu} = 2E$ to reduce the equation to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{\rho}^2 - \frac{1}{\rho} = 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the weights $\mu$ and $\lambda$ will, roughly speaking, give you the weights $p(r)$ and $a(r)$ in your question). Now note that this scaling degenerates if $E = 0$. In the case that $E = 0$, the equation can be solved by quadrature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{x}^2 = x^{-1} \Rightarrow \sqrt{x} dx = dt \Rightarrow x^{3/2} \sim t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the solution to the homogeneous case. In the inhomogeneous case, you take that as a sort of integrating factor: assume that $\rho^{3/2} f(\rho) \sim t$, this implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{\rho} \left( \rho^{3/2} f(\rho) \right)' = 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we plug this into the equation, which we first re-arrange as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{\rho} = \sqrt{\frac{1}{\frac{\rho}{1+\rho}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we conclude that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{\frac{\rho}{1+\rho}} = \frac{d}{d\rho}( \rho^{3/2} f(\rho)) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you solve this by directly integrating it. I think this $f$ you find should be exactly the $F$ you wrote down above (I didn't check it myself). Note that due to the singular weight $\rho$ which degenerates as $\rho \to 0$, you will have to separately integrate in the regime where $\rho &gt; 0$ and $\rho &amp;lt; 0$. The existence and uniqueness of solution is guaranteed by the theory of Fuchsian ODEs. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-14T16:20:19.650" CommentCount="2" />
  <row Id="6793" PostTypeId="2" ParentId="6748" CreationDate="2010-10-14T16:23:43.427" Score="2" ViewCount="0" Body="&lt;p&gt;Arturo has already posted a nice answer. I'd merely like to emphasize that such universal definitions often enable slick proofs, e.g. see below. For a much more striking example see the theorem in &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;my post here&lt;/a&gt;, which presents a slick one-line proof of the  LCM * GCD law via their universal definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\: \  \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \  $&#xA;for $\rm\ \ n &gt; 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\quad\quad\quad\quad\quad\quad\quad k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\iff\ \ \ \  k\ \le\ \ \lfloor x/(mn)\rfloor $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the above trivial proof to more traditional proofs, e.g. the &lt;a href=&quot;http://chandrumath.wordpress.com/2010/10/07/more-on-the-floor-function/&quot; rel=&quot;nofollow&quot;&gt;special case $\rm\ m = 1\ $ here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-14T20:37:58.503" LastActivityDate="2010-10-14T20:37:58.503" CommentCount="1" />
  <row Id="6794" PostTypeId="1" AcceptedAnswerId="6796" CreationDate="2010-10-14T16:41:23.873" Score="10" ViewCount="425" Body="&lt;p&gt;The Cantor set is uncountable so I expect countably many of them to be able to cover $\mathbb R$, but the set has measure $0$ so countably many of them also has set of measure $0$ and thus can't cover the real line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why/where is my intuition broken?&lt;/p&gt;&#xA;" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-14T16:42:27.290" LastActivityDate="2010-10-25T11:23:19.250" Title="Why can't Cantor sets cover $\mathbb{ R}$?" Tags="&lt;soft-question&gt;&lt;continuum&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6795" PostTypeId="1" CreationDate="2010-10-14T16:41:50.467" Score="1" ViewCount="223" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_number&quot; rel=&quot;nofollow&quot;&gt;perfect number&lt;/a&gt; is a number that is the sum of its proper divisors:&#xA;&lt;code&gt;28 = 1 + 2 + 4 + 7 + 14&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for numbers whose only proper divisors are 1, 2, and N/2? For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 is divisible by 1, 2, and 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 is divisible by 1, 2, and 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;34 is divisible by 1, 2, and 17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for these numbers?&lt;/p&gt;&#xA;" OwnerUserId="2517" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-14T23:53:23.747" LastActivityDate="2010-10-15T18:30:03.327" Title="Is there a name for this kind of number?" Tags="&lt;number-theory&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="6796" PostTypeId="2" ParentId="6794" CreationDate="2010-10-14T16:57:23.117" Score="20" ViewCount="0" Body="&lt;p&gt;You can't cover a complete metric space with a countable union of closed nowhere dense subsets, by Baire's theorem. Cardinality tells us very little.  Even infinite dimensional but separable Banach spaces have the same cardinality as lines, but perhaps here it is more intuitive that you will never cover such a space with countably many lines.  Similarly, you can't cover the plane with countably many lines, polygons, conic sections, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how Baire's theorem is more fundamental than measure here, note that the same is true for &quot;fat&quot; Cantor sets.  That is, you take out smaller intervals to obtain Cantor-like sets with positive measure.  These sets are still closed and nowhere dense, so by Baire's theorem the line is not a countable union of such sets, even though comparing measures wouldn't tell you this.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-14T16:57:23.117" CommentCount="1" />
  <row Id="6797" PostTypeId="2" ParentId="6794" CreationDate="2010-10-14T16:57:54.597" Score="11" ViewCount="0" Body="&lt;p&gt;It's not just that the Cantor set has measure 0, it's also that the reals are a Baire space and the Cantor set is nowhere dense. E.g. take a fat Cantor set, $C$, and look at $C + \mathbb{Z}$ which has infinite measure but is still be nowhere dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: There is a &lt;a href=&quot;http://mathoverflow.net/questions/43478/is-there-a-measure-zero-set-which-isnt-meagre&quot; rel=&quot;nofollow&quot;&gt;related discussion on MO&lt;/a&gt; that could be of interest.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-25T11:23:19.250" LastActivityDate="2010-10-25T11:23:19.250" CommentCount="1" />
  <row Id="6798" PostTypeId="2" ParentId="6757" CreationDate="2010-10-14T17:02:03.437" Score="6" ViewCount="0" Body="&lt;p&gt;We can apply the following general transformation formula of a series into a&#xA;continued fraction, which one can justify (see Addendum 1 and Addendum 2) by comparing the continued fraction fundamental recurrence relations with the series partial sums recurrence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{N}\dfrac{u_{n}}{v_{n}}=\dfrac{u_{1}}{v_{1}+\underset{n=1}{%&#xA;\overset{N-1}{\mathbb{K}}}\left(\left( -\dfrac{u_{n+1}}{u_{n}}v_{n}^{2}\right) /\left( v_{n+1}+\dfrac{u_{n+1}}{u_{n}}v_{n}\right)\right) }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we have $u_{n}=1$, $v_{n}=2^{\left( 2^{n}\right) }$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{N}\dfrac{1}{v_{n}}=\dfrac{1}{4+\underset{n=1}{\overset{N-1}{%&#xA;\mathbb{K}}}\left( \left( -v_{n}^{2}\right) /\left( v_{n+1}+v_{n}\right)\right) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{N}\dfrac{1}{2^{2^{n}}}=\dfrac{1}{4+\underset{n=1}{\overset{N-1}{\mathbb{K}}}\left(\left( -2^{2^{n+1}}\right)/\left( 2^{2^{n+1}}+2^{2^{n}}\right)\right) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{1}{4+}\dfrac{-16}{20+}\cdots \dfrac{-2^{2^{n+1}}}{%&#xA;2^{2^{n+1}}+2^{2^{n}}+}{\cdots }\dfrac{-2^{2^{N}}}{%&#xA;2^{2^{N}}+2^{2^{N-1}}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation of the series into a continued fraction is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty}\dfrac{1}{2^{2^{n}}}=\dfrac{1}{4+\underset{n=1}{\overset{\infty}{\mathbb{K}}}\left(\left( -2^{2^{n+1}}\right)/\left( 2^{2^{n+1}}+2^{2^{n}}\right) \right) }.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 1&lt;/strong&gt;: The series partial sums&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_{n}=\sum_{k=1}^{n}\frac{u_{k}}{v_{k}}=\frac{A_{n}}{B_{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;verify, for $n\geq 2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_{n}=s_{n-1}+\frac{u_{n}}{v_{n}}=\frac{A_{n-1}}{B_{n-1}}+\frac{u_{n}}{v_{n}%&#xA;}=\frac{v_{n}A_{n-1}+u_{n}B_{n-1}}{v^{n}B_{n-1}}=\frac{A_{n}}{B_{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=v_{n}\;A_{n-1}+u_{n}\;B_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=v_{n}\;B_{n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truncated continued fraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{k=1}{\overset{n}{\mathbb{K}}}\left( u_{k}/v_{k}\right) =\frac{%&#xA;A_{n}}{B_{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;verifies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=b_{n}\;A_{n-1}+a_{n}\;A_{n-2}\qquad A_{0}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=b_{n}\;B_{n-1}+a_{n}\;B_{n-2}\qquad B_{0}=1.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 2&lt;/strong&gt;: Detailed algebraic computation. For $n=1$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{u_{1}}{v_{1}}=\frac{a_{1}}{b_{1}}=\frac{A_{1}}{B_{1}}\qquad u_{1}=a_{1}\qquad v_{1}=b_{1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $n-1$ for $n$ in the first recurrence we get for $n\geq 3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n-1}=v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n-1}=v_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which in turn gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=v_{n}\;A_{n-1}+u_{n}\;B_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;\left( v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}\right) +u_{n}\;\left( v_{n-1}\;B_{n-2}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;v_{n-1}\;A_{n-2}+\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;B_{n-2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=v_{n}\;B_{n-1}=v_{n}\;v_{n-1}\;B_{n-2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same substitution in the second recurrence yields (for $n\geq 3$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n-1}=b_{n-1}\;A_{n-2}+a_{n-1}\;A_{n-3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n-1}=b_{n-1}\;B_{n-2}+a_{n-1}\;B_{n-3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining everything we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=b_{n}\;A_{n-1}+a_{n}\;A_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=b_{n}\;\left( v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}\right) +a_{n}\;A_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\left( b_{n}\;v_{n-1}+a_{n}\;\right) \;A_{n-2}+b_{n}\;u_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=b_{n}\;B_{n-1}+a_{n}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=b_{n}\;\left( v_{n-1}\;B_{n-2}\right) +a_{n}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\left( b_{n}\;v_{n-1}+a_{n}\right) \;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing both $A_{n}$ and $B_{n}$ formulae&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=v_{n}\;v_{n-1}\;A_{n-2}+\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{n}=\left( b_{n}\;v_{n-1}+a_{n}\;\right) \;A_{n-2}+b_{n}\;u_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=v_{n}\;v_{n-1}\;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{n}=\left( b_{n}\;v_{n-1}+a_{n}\right) \;B_{n-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one concludes that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{n}\;u_{n-1}+u_{n}\;v_{n-1}=b_{n}\;u_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{n}\;v_{n-1}=b_{n}\;v_{n-1}+a_{n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=v_{n}\;v_{n-1}-b_{n}\;v_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;v_{n-1}-\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;v_{n-1}/u_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=v_{n}\;v_{n-1}-v_{n}\;v_{n-1}-u_{n}\;v_{n-1}\;v_{n-1}/u_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\frac{u_{n}}{u_{n-1}}v_{n-1}^{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}\;u_{n-1}=v_{n}\;u_{n-1}+u_{n}\;v_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}=v_{n}+\frac{u_{n}}{u_{n-1}}v_{n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $n\geq 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=-\frac{u_{n}}{u_{n-1}}v_{n-1}^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}=v_{n}+\frac{u_{n}}{u_{n-1}}v_{n-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T15:56:23.260" LastActivityDate="2010-10-16T15:56:23.260" CommentCount="2" />
  <row Id="6799" PostTypeId="2" ParentId="6784" CreationDate="2010-10-14T17:35:52.343" Score="0" ViewCount="0" Body="&lt;p&gt;The entries in the $n$th row of Pascal's triangle (for large $n$) follow a Gaussian curve quite closely, as a consequence of the central limit theorem. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article about the binomial distribution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-14T17:35:52.343" CommentCount="2" />
  <row Id="6800" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T17:47:13.323" Score="4" ViewCount="0" Body="&lt;p&gt;I'm not going to add nothing directly related to your question and previous answers, but make some propaganda of a theorem I like since I was student and which, I believe, says something stronger than comparing some intuitive notion of completness with its definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat related notion of completeness is the geodesical one. The definition may not be too much appealing unless you're interested in differential geometry, but one of its consequences is easy to explain: if a Riemann manifold is &lt;strong&gt;geodesically complete&lt;/strong&gt;, you can join any two points by a length minimizing geodesic. (But geodesic already implies that it minimizes length, doesn't it? Not quite: just locally. So, for instance, the meridian joining the North Pole with London, but going &quot;backward&quot;, through the Bering Strait and the Pacific Ocean, then the South Pole, Africa and finally London, is a geodesic, but not a length minimizing one blatantly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, $\mathbb{R^2} \backslash \left\{ (0,0)\right\} $ is &lt;strong&gt;not&lt;/strong&gt; geodesically complete, since there is no length minimizing geodesic joining, say, $(-1,0)$ and $(1,0)$, due to the &quot;hole&quot; $(0,0)$. At the same time, as a metric space, $\mathbb{R^2} \backslash \left\{ (0,0)\right\}$ is &lt;strong&gt;not&lt;/strong&gt; complete: the Cauchy sequence $(\frac{1}{n}, 0)$ converges to $(0,0)$, but since $(0,0)$ is not in $\mathbb{R^2} \backslash \left\{ (0,0)\right\}$ it doesn't have a limit there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf-Rinow_theorem&quot; rel=&quot;nofollow&quot;&gt;Hopf-Rinow theorem&lt;/a&gt; tells us that this kind of things always happen together: a &quot;hole&quot; for geodesics is the same as a &quot;hole&quot; for Cauchy sequences, since for a (finite-dimensional) Riemann manifold $M$, both notions agree: $M$ is complete as a metric space if and only if it is geodesically complete.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-15T11:08:07.987" LastActivityDate="2010-10-15T11:08:07.987" CommentCount="3" />
  <row Id="6801" PostTypeId="1" CreationDate="2010-10-14T17:54:40.847" Score="1" ViewCount="124" Body="&lt;p&gt;I'm not quite sure what the mathematical term for what I'm asking is, so let me just describe what I'm trying to figure out.  Let's say that I have two ordered sets of numbers {1, 2} and {3, 4}.  I'm trying to figure out the number of possible ways to combine these two sets into one without breaking the ordering of the two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance, {1, 2, 3, 4}, {3, 4, 1, 2}, and {1, 3, 2, 4} are valid combinations, but {2, 1, 4, 3} isn't.  How do I figure out the number of valid combinations?  This feels like something I should remember from college, but I'm drawing a blank.  It feels somewhere in between a combination and a permutation.  Maybe I'm looking for a partially-ordered permutation (which seems to be a somewhat difficult concept if google is to be believed)?&lt;/p&gt;&#xA;" OwnerUserId="2519" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-14T18:59:56.220" LastActivityDate="2010-10-14T18:59:56.220" Title="How do I determine the possible number of combinations of two ordered sets?" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6802" PostTypeId="2" ParentId="6801" CreationDate="2010-10-14T17:59:21.073" Score="3" ViewCount="0" Body="&lt;p&gt;You are talking about (resulting) tuples I presume (and not sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sets are $S$ with $s$ elements and $T$ with $t$ elements, then the total possible tuples is $\displaystyle {s+t \choose s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you have $s+t$ slots and you pick the slots (say $s$) for one of the sets in $\displaystyle {s+t \choose s}$ ways. Once the slots are chosen, all the $s+t$ numbers can now be filled in only one way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total is $\displaystyle {s+t \choose s}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-14T17:59:21.073" />
  <row Id="6803" PostTypeId="1" CreationDate="2010-10-14T18:07:34.813" Score="1" ViewCount="90" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think the following limits are correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\lim_{d\to\infty}\frac{\sum\limits_{k=1}^{d} {\varphi(N) \choose  k} {d-1 \choose k-1}}{\varphi(N)^d}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\lim_{N\to\infty}\frac{\sum\limits_{k=1}^{d} {\varphi(N) \choose  k} {d-1 \choose k-1}}{\varphi(N)^d}=c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plotted the equations and guessed the results according to the graphs but I could not prove them mathematically by myself. Any hints would be appreciated. Graphs are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://deniz.cs.utsa.edu/plots/&quot; rel=&quot;nofollow&quot;&gt;http://deniz.cs.utsa.edu/plots/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2010-10-14T18:22:54.047" Title="Limit involving the totient function and combination" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;limit&gt;" CommentCount="4" />
  <row Id="6805" PostTypeId="1" CreationDate="2010-10-14T18:09:41.087" Score="1" ViewCount="341" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove using induction that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Base case: P(1) = (1) + (8) + (27) = 36, 36 can be divided by 9 so the base case is valid&lt;/li&gt;&#xA;&lt;li&gt;Inductive step: let p(n) be the statement 9 divides $n^3 + (n+1)^3 + (n+2)^3$ Assume p(k) is true, so 9 divides $k^3 + (k+1)^3 + (k+2)^3$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And this is where I'm stuck. I'm not sure how to demonstrate that 9 divides P(n) when n = k+1. If someone could step me in the right direction that would be awesome. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2520" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:58:19.803" LastActivityDate="2010-10-16T19:21:03.143" Title="Simple Proof by induction: &quot;9 divides  $n^3 + (n+1)^3 + (n+2)^3$&quot;" Tags="&lt;algebra-precalculus&gt;&lt;induction&gt;&lt;proof&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="6806" PostTypeId="2" ParentId="6805" CreationDate="2010-10-14T18:12:25.247" Score="9" ViewCount="0" Body="&lt;p&gt;You want to prove that if $9\mid[n^3+(n+1)^3+(n+2)^3]$&#xA;then $9\mid[(n+1)^3+(n+2)^3+(n+3)^3]$. This boils down to proving&#xA;that $9\mid[(n+3)^3-n^3]$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-14T18:12:25.247" />
  <row Id="6808" PostTypeId="1" CreationDate="2010-10-14T18:23:19.223" Score="1" ViewCount="70" Body="&lt;p&gt;Let's call $S$ the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10.&#xA;Thus, $S = 1234567891011121314151617181920212223242\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any number in $S$ occurs multiple times. The first occurence of 3 is in position 3 of the series, 2nd occurence is in position 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find the position of 3 in 100th occurence? Is there a pattern?&lt;/p&gt;&#xA;" OwnerUserId="2521" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-14T20:23:24.217" LastActivityDate="2010-10-14T20:55:15.457" Title="Finding n-th position of occurence of a number in an infinite series of numbers" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="6809" PostTypeId="2" ParentId="6777" CreationDate="2010-10-14T18:30:19.243" Score="2" ViewCount="0" Body="&lt;p&gt;This answer only applies to the order version of completeness rather than the metric version, but I've found it quite a nice way to think about what completeness means intuitively: consider the real numbers. There the completeness property is what guarantees that the space is connected. The rationals can be split into disjoint non-empty open subsets, for example the set of all positive rationals whose squares are greater than two, and its complement, and the reason this works is because, roughly speaking, there is a &quot;hole&quot; in between the two sets which lets you pull them apart. In the reals this is not possible; there are always points at the ends of intervals, so whenever you partition the reals into two non-empty subsets, one of them will always fail to be open.&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-10-14T18:30:19.243" CommentCount="1" />
  <row Id="6810" PostTypeId="2" ParentId="6801" CreationDate="2010-10-14T18:33:01.463" Score="0" ViewCount="0" Body="&lt;p&gt;Here is another way of visualising Moron's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be the first set, and $T$ be the second set. Now, imagine the elements of $S$ are listed, horizontally, in their order; we will not disturb their order, but will only &quot;place&quot; the elements of $T$ among them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible locations for any element of $T$ are (i) before the first elements of $S$; (ii) between two elements of $S$; and (iii) after the last element of $S$. If $S$ has $s$ elements, then this gives $s+1$ possible locations. We can put more than one element of $T$ in each location, though. For example, if $S=\{1,2,3\}$ and $T=\{a,b,c,d\}$, then we can place elements of $T$ either before $1$, between $1$ and $2$, between $2$ and $3$, or after $3$ (four locations); and we could place one element before $1$, none between $1$ and $2$, two between $2$ and $3$, and one after $3$. But note that once we decide &lt;em&gt;where&lt;/em&gt; to place the elements of $T$, the &lt;em&gt;order&lt;/em&gt; they will appear on is completely determined. If we go the way I just described, you would end up with $a,1,2,b, c, 3, d$, and that is the only way to place elements of $T$ as described while preserving the order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are placing $t$ elements, then, we just need to &quot;select&quot; $t$ locations from $s+1$ possibilities. The order in which we pick them doesn't matter, because in the end we will just put the elements of $T$ in their appropriate order in those locations. And we can select the same location more than once. So we need to compute &quot;combinations with repetitions&quot; (order does not matter, repetitions allowed). The formula for making $n$ selections, with repetitions allowed but where order does not matter, from among $m$ possibilities is $\binom{n+m-1}{n}$ (see for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) so here we have $n=t$ and $m=s+1$, giving $\binom{t+s}{t} = \binom{t+s}{s}$ possibilities. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-14T18:33:01.463" />
  <row Id="6811" PostTypeId="1" AcceptedAnswerId="6816" CreationDate="2010-10-14T19:03:34.217" Score="4" ViewCount="77" Body="&lt;p&gt;This from an exercise in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540390324&quot; rel=&quot;nofollow&quot;&gt;Aigner's book&lt;/a&gt; where one has to evaluate $\sum_{k\ge 0} (-1)^k \binom{n}{k}^2$ using sign reversing involutions. When $n$ is odd, the problem is trivial : let $[n] = \{1,2,\dots,n\}$ consider all pairs of subsets $\{ (A,B) \in 2^{[n]} \times 2^{[n]} : |A| = |B| \}$, and $(A,B)$ has sign $(-1)^{|A|}$ and the sign-reversing involution is $(A,B) \to ([n]-A,[n]-B)$. Any hints on how to approach this problem for even $n$ will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-14T20:07:26.823" LastActivityDate="2010-10-14T20:07:26.823" Title="sign reversing involution proof of a combinatorial identity" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="6812" PostTypeId="2" ParentId="6808" CreationDate="2010-10-14T19:05:09.293" Score="1" ViewCount="0" Body="&lt;p&gt;You could try some careful counting. If mine is careful enough, I make the 100th occurence of 3 to be in the 889th position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the first digit of 333.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I counted like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1 to 99 there are 10 threes in the unit's digit and 10 in the ten's digit, so that's makes 20, and hence there are $3 \times 20=60$ in 1 to 299.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in 300 to 329 there are 30 in the hundred's digit and 3 in the unit's digit, so that's another 33, making 93 so far. Now just look at 330331332333 and you will see that the 100th three is the first digit of 333 which, by some straightforward calculations, is in the 889th position.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-14T20:55:15.457" LastActivityDate="2010-10-14T20:55:15.457" />
  <row Id="6813" PostTypeId="1" AcceptedAnswerId="6817" CreationDate="2010-10-14T19:20:04.697" Score="3" ViewCount="79" Body="&lt;p&gt;I note that the ordinals of L are the same as V, so I guess that it has no $\Pi_1^1$ consequences. On the other hand Wikipedia tells me that it asserts the existance of a $\Delta_2^1$ non-measurable set of reals. &quot;Measurable&quot; involves third-order concepts but I know that there's often a &quot;coding trick&quot; that gets around that sort of thing, so I guess it has some analytic consequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I am guessing -- I am not very good at this stuff. But I am curious. What's the least point in the analytic hierarchy where V=L matters (if any)?&lt;/p&gt;&#xA;" OwnerUserId="2523" LastActivityDate="2010-10-14T20:49:06.120" Title="Where in the analytic hierarchy does V=L start having consequences?" Tags="&lt;real-analysis&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6814" PostTypeId="2" ParentId="6805" CreationDate="2010-10-14T19:24:04.923" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ If $\rm\ 9\:|\:P(n)\ $ then $\rm \ 9\:|\:P(n+1)\ \iff\ 9\ | \ P(n+1)-P(n)\  =\ (n+3)^3 - n^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ This reduction can be viewed as arising from rewriting $\rm\: P(n)\: $ as a &lt;strong&gt;telescoping&lt;/strong&gt; series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm P(n) = (P(n)-P(n-1)) + (P(n-1)-P(n-2)) +\: \cdots\: + (P(1)-P(0))+\ P(0) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\ 9\ $ divides $\rm\ P(n)\ $ if it divides all RHS terms,$\ $ i.e.$\ $ all the differences $\rm \ P(i+1)-P(i)\ $ and $\rm\ P(0)$. Since $\rm\: P\:$ is a polynomial, its difference has lower degree than $\rm\: P\:$, so the problem reduces to a simpler one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such &lt;strong&gt;telescopic&lt;/strong&gt; transformations often simplify inductive proofs. See &lt;a href=&quot;http://math.stackexchange.com/questions/6586&quot;&gt;here for similar multiplicative telescopy.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if you continue to iterate the above reduction then it amounts to showing that the coefficients $\rm \Delta^k P(0)$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;Newton's forward difference formula&lt;/a&gt; are all divisible by 9. Indeed one has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm  P(n)\ =\ 9 + 27\ n + 36\ \binom{n}2 + 18\ \binom{n}3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, one could simply check that four consecutive values of $\rm\:P\:$ are multiples of 9. Analogous remarks hold true for any $\rm\:P(n)\:$ that satisfies a monic linear recurrence with polynomial coefficients (or, much more generally, for proving identities of &lt;a href=&quot;http://www.mathematik.uni-kassel.de/~koepf/Publikationen/Algebra.pdf&quot; rel=&quot;nofollow&quot;&gt;multivariate holonomic functions&lt;/a&gt;). While this viewpoint is overkill here, it is essential when working with nontrivial examples.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-16T17:54:46.930" LastActivityDate="2010-10-16T17:54:46.930" CommentCount="1" />
  <row Id="6815" PostTypeId="2" ParentId="6808" CreationDate="2010-10-14T19:31:06.750" Score="2" ViewCount="0" Body="&lt;p&gt;You can solve this by thinking of some lists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 1-10? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 11-20? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 21-30? -- 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 31-40? -- 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 41-50? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 91-100? -- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how many 3's is there in 1-100?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 101-200?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 201-300?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many 3's is there in 301-400? (obviously sufficient)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when you know the what number comes at the 100:th place you just need count the place - for example by &quot;counting lists&quot; again.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-14T20:22:28.447" LastActivityDate="2010-10-14T20:22:28.447" CommentCount="2" />
  <row Id="6816" PostTypeId="2" ParentId="6811" CreationDate="2010-10-14T19:33:00.507" Score="5" ViewCount="0" Body="&lt;p&gt;Consider the set of pairs $(A,B)$ of subsets of $\{1,\ldots,n\}$&#xA;such that $|A|+|B|=n$ with weight $(-1)^{|A|}$.&#xA;The exceptional pairs where the involution is not defined are those&#xA;where $A=B$. For each other pair $(A,B)$ &#xA;take the smallest element of the symmetric difference of $A$ and $B$&#xA;and move it from one set to the other. The symmetric difference of&#xA;the two new sets is the same, so this is an involution, and it is&#xA;weight-reversing.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-14T19:38:46.673" LastActivityDate="2010-10-14T19:38:46.673" CommentCount="1" />
  <row Id="6817" PostTypeId="2" ParentId="6813" CreationDate="2010-10-14T20:03:14.847" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Absoluteness#Shoenfield.27s_absoluteness_theorem&quot; rel=&quot;nofollow&quot;&gt;Shoenfield's absoluteness theorem&lt;/a&gt; ensures that any $\Sigma^1_2$ statement (and therefore, any $\Pi^1_2$ one) has the same truth value in $L$ as in $V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\Sigma^1_3$ statements differ in general whether they are in $V$ or in $L$. For a silly example, $\exists x\in{\mathbb R}\,(x\notin L)$ can be rewritten as a $\Sigma^1_3$ statement, which is false in $L$ but true in general. Given a $\Sigma^1_2$ formula $\phi(x,y)$, the statement ``$\phi$ defines a well-ordering of the reals'' is $\Pi^1_3$. Again, there is a specific such $\phi$ for which this is true in $L$ and false in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&quot;In general&quot; means here that it can be made false by forcing. There are deeper discrepancies if one allows large cardinals into the picture.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-10-14T20:49:06.120" LastActivityDate="2010-10-14T20:49:06.120" CommentCount="1" />
  <row Id="6818" PostTypeId="1" AcceptedAnswerId="6822" CreationDate="2010-10-14T20:09:30.890" Score="6" ViewCount="302" Body="&lt;p&gt;We know that the &quot;Harmonic Series&quot; $$ \sum \frac{1}{n}$$ diverges. And for $p &gt;1$ we have the result that the series converges $$\sum \frac{1}{n^{p}}$$ converges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can then ask the question of testing the convergence the following 2 Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}, \quad \sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \sin{n}}}$$ where $ k \in (0,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only thing which i have as tool for this problem is the inequality $| \sin{n} | \leq 1$, which i am not sure whether would applicable or not. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-15T07:38:03.877" LastActivityDate="2010-10-15T07:38:03.877" Title="Testing the series $\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6819" PostTypeId="2" ParentId="6808" CreationDate="2010-10-14T20:33:40.640" Score="1" ViewCount="0" Body="&lt;p&gt;Brute force:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[mariano@godot ~]$ ghci&#xA;GHCi, version 6.12.1: http://www.haskell.org/ghc/  :? for help  &#xA;Prelude&amp;gt; [i | (d,i) &amp;lt;- concat (map show [1..]) `zip` [1..], d == '3'] !! 100 - 1&#xA;889&#xA;Prelude&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-14T20:33:40.640" CommentCount="3" />
  <row Id="6820" PostTypeId="2" ParentId="6818" CreationDate="2010-10-14T20:42:53.603" Score="4" ViewCount="0" Body="&lt;p&gt;Here is a handwaving argument that the series diverges for all $k\in(0,2)$:&#xA;Given $k$, there will be a certain density of $n$ for which $k+cos n &amp;lt;1$.  A &quot;random sample&quot; of the harmonic series will diverge by the same argument that shows the harmonic series diverges.  Our series will be greater, term by term, than the harmonic series that has numerator 0 if $k+cos n &gt;1$ and numerator 1 if $k+cos n &amp;lt;1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T20:42:53.603" CommentCount="4" />
  <row Id="6822" PostTypeId="2" ParentId="6818" CreationDate="2010-10-14T20:58:30.637" Score="3" ViewCount="0" Body="&lt;p&gt;Ross' handwaving can be justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use Weyl's theorem on the equidistribution of polynomials (mod 1) with at least one irrational coefficient (see here: &lt;a href=&quot;http://math.stackexchange.com/questions/128/how-do-you-prove-that-pn-xi-for-xi-irrational-and-p-a-polynomial-is-uni/1163#1163&quot;&gt;http://math.stackexchange.com/questions/128/how-do-you-prove-that-pn-xi-for-xi-irrational-and-p-a-polynomial-is-uni/1163#1163&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $\displaystyle p(x) = 2\pi x + \pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle k \in (0,2)$. Let $\displaystyle \delta = k$ if $\displaystyle k \in (0,1)$ and $\displaystyle \delta = k-1$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Weyl's theorem, density of $n$ such that the fractional part of $\displaystyle p(n) &amp;lt; 1- \delta$ is non zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the density (among the set of natural numbers) of the odd numbers $\displaystyle 2n+1$ for which the fractional part of $\displaystyle (2n+1)\pi &amp;lt; 1 - \delta$ is non-zero. Thus the density of $\displaystyle [(2n+1)\pi]$ is non-zero too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such an $\displaystyle n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle N = [(2n+1)\pi]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $\displaystyle (2n+1)\pi - N &amp;lt; 1 - \delta$ and so $\displaystyle \cos(N) - \cos((2n+1)\pi) &amp;lt; 1 - \delta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus $\displaystyle \cos(N) &amp;lt; -\delta$ and so $\displaystyle \cos(N) + k &amp;lt; k - \delta \leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for we have a positive density subset $\displaystyle S$ of $\displaystyle \mathbb{N}$ for which $\displaystyle \cos(N) + k &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{N \in S} \frac{1}{N^{k+\cos(N)}} &gt; \sum_{N \in S} \frac{1}{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle S$ is of positive density, the sum of reciprocals is divergent, for a proof see here: &lt;a href=&quot;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/&quot;&gt;http://math.stackexchange.com/questions/5932/theorem-on-natural-density/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I will leave my earlier attempt here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k \in (0,1)$ consider the below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any $\displaystyle k \in (0,1)$, we can find an infinite number of odd numbers $\displaystyle M$ such that $\displaystyle M\pi - [M\pi] &amp;lt; 1 - k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This easily follows from Vinogradov's theorem that $\displaystyle \{p_{n}\alpha\}$ is equidistributed for irrational $\displaystyle \alpha$ where $\displaystyle p_{n}$ is the $\displaystyle n^{th}$ prime. (See here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistributed_sequence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equidistributed_sequence&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \circ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle n = [M\pi]$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \cos(n) - \cos(M\pi) = \cos(n)  + 1 &amp;lt; 1- k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(using $\displaystyle |\cos x - \cos y| &amp;lt; |x-y|$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $\displaystyle \cos(n) + k &amp;lt; 0$ for an infinite number of $\displaystyle n$ and so the series $\displaystyle \sum \frac{1}{n^{k+\cos n}}$ must be divergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $\displaystyle \sin n$ case, we need the sequence $\displaystyle \{(4k+3)\pi/2\}$ which is an infinite subsequence of $\displaystyle \{p_{n}\pi/2\}$ and so the series  $\displaystyle \sum \frac{1}{n^{k+\sin n}}$ is divergent.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-14T22:55:01.647" LastActivityDate="2010-10-14T22:55:01.647" />
  <row Id="6823" PostTypeId="1" AcceptedAnswerId="6837" CreationDate="2010-10-14T21:38:16.150" Score="0" ViewCount="90" Body="&lt;p&gt;Let $X_1,....,X_n$ be some positive random variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $A_{ij} = \frac{E[X_i*X_j]}{E[X_i]E[X_j]}$ a positive (semi) definite matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that $B_{ij} = Cov(X_i,X_j) = E[X_i*X_j] - E[X_i]E[X_j]$ is PSD because it is a covariance matrix.)&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2010-10-14T23:52:48.807" Title="is the following a positive definite matrix?" Tags="&lt;linear-algebra&gt;&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6824" PostTypeId="2" ParentId="6818" CreationDate="2010-10-14T21:48:15.917" Score="4" ViewCount="0" Body="&lt;p&gt;If $k$ is a real number less than 2 then $\sum_{n=1}^\infty n^{-k-\sin n}$ diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is problem 11162 from the &lt;em&gt;American Mathematical Monthly&lt;/em&gt;, the solution by the Microsoft Research Problems Group appears in the February 2007 issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the solution (if I've transcribed it correctly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k\leq 0$, then the series diverges by comparison with the harmonic series.&#xA;If $0&amp;lt; k &amp;lt;2$, then let $\alpha,\beta$ be such that $\sin\alpha=\sin\beta=1-k$ and&#xA;$\pi/2 &amp;lt; \alpha &amp;lt; \beta &amp;lt; 5\pi/2$. Note that $\sin x &amp;lt; 1-k$ for all $x$ with $\alpha &amp;lt; x&amp;lt; \beta$. Define&#xA;$$ S=(n\in{\mathbb N}:\sin n&amp;lt;1-k )={\mathbb N}\cap\bigcup_{j\in {\mathbb Z}}(2\pi j+\alpha,2\pi j+\beta).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\pi$ is irrational, the sequence $n/(2\pi)$ mod 1 is dense in [0,1], so $S\not=\emptyset$&#xA;and there exist positive integers $a,A,b,B$ with&#xA;$$0 &amp;lt; a-2\pi A &amp;lt; {\beta-\alpha\over2}\quad\mbox{and}\quad -{\beta-\alpha\over2} &amp;lt; b-2\pi B &amp;lt; 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C=\max(a,b)$. Given $s\in S$, we claim that there exists $s^\prime\in S$ with $s &amp;lt; s^\prime\leq s+C$.&#xA;Indeed, there exists $j\in{\mathbb Z}$ with $\alpha &amp;lt; s-2\pi j&amp;lt;\beta$. Compare $s-2\pi j$ to $(\alpha+\beta)/2$.&#xA;If $\alpha &amp;lt; s-2\pi j\leq (\alpha+\beta)/2$, then $\alpha&amp;lt; (s+a)-2\pi(j+A)&amp;lt;\beta$, so we can use&#xA;$s^\prime=s+a$. On the other hand,  if $(\alpha+\beta)/2 &amp;lt; s-2\pi j &amp;lt; \beta$, then&#xA;$a&amp;lt;(s+b)-2\pi(j+B) &amp;lt; \beta$, so we can use $s^\prime=s+b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $S$ is infinite. Index it in increasing order, $s_1 &amp;lt; s_2 &amp;lt; \cdots.$ Now $s_j\leq s_1+C(j-1)$, so &#xA;$$\sum_{n=1}^{s_m}n^{-k-\sin n}\geq \sum_{1\leq n\leq s_m, n\in S} n^{-k-\sin n}&#xA;\geq \sum_{1\leq n\leq s_m , n\in S} {1\over n} \geq \sum_{j=1}^m {1\over s_j}\geq&#xA;\sum_{j=1}^m {1\over s_1+C(j-1)},$$ which diverges as $m\to\infty$ by the integral test.&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-10-14T22:31:34.943" LastActivityDate="2010-10-14T22:31:34.943" CommunityOwnedDate="2010-10-14T21:48:15.917" />
  <row Id="6825" PostTypeId="1" AcceptedAnswerId="6827" CreationDate="2010-10-14T22:02:17.110" Score="0" ViewCount="30" Body="&lt;p&gt;Here's an horrible drawing that tries to explain what I'm asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1164414/SO/Four%20hectagons%20an%20hectagon%20do%20not%20make.png&quot; alt=&quot;Okay, so if you're reading or hearing this you can't see images, so I will take the effort of explaining it with words. Consider four $n$-sided polygons arranged so that their centers are the vertices of a square. The square is exactly as large as the diameter of the smallest circle enclosing each polygon. The polygons do not overlap. Polygons are oriented so that each one has either exactly two sides in common or exactly one side and one vertex in common with the other shapes.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying this with small numbers gives me $f: 4 \to 0, 6 \to 4, 8 \to 4, 10 \to 8, 12 \to 8.$ This suggests that $$f(2n) = 4 \times \left( \lceil \frac{2n}{4} \rceil - 1 \right), 2n &gt; 4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this result be extended for $n \to \infty$? Bonus: What about odd values of $n$?&lt;/p&gt;&#xA;" OwnerUserId="92" LastEditorUserId="92" LastEditorDisplayName="" LastEditDate="2010-10-14T22:19:55.137" LastActivityDate="2010-10-14T22:19:55.137" Title="What is the number of sides of the hollow area generated by juxtaposition of four $2n$-sided regular polygons?" Tags="&lt;geometry&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6826" PostTypeId="2" ParentId="6825" CreationDate="2010-10-14T22:14:33.437" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, and your answer is correct.  Following your figure, assume the polygon has a side at the top and bottom.  If n is even, so the polygons have a multiple of 4 sides, there will be a pair of sides at the left and right.  There will be $n/2-1$ sides between them and four sets of that bounding the central region.  If n is odd, there will be points at the left and right.  From the point to the flat on top is $(n-1)/2$ sides, again multiplied by four to make the boundary of the central region.  This supports your formula, using the ceiling function to combine the cases.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T22:14:33.437" />
  <row Id="6827" PostTypeId="2" ParentId="6825" CreationDate="2010-10-14T22:16:29.590" Score="1" ViewCount="0" Body="&lt;p&gt;There are two clear cases.  When $2n$ is divisible by $4$, each polygon will have a horizontal side and a vertical side touching other polygons.  So the number of sides each polygon contributes to the interior is just the number of sides between a horizontal and a vertical side.  (Each side of the &quot;interior&quot; polygon belongs to exactly one of the exterior ones.)  This is just $(2n-4)/4$, which is then multiplied by $4$ again to give $2n-4$ as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other case, you have the points of tangency being a side and a vertex.  We can reduce this to the other case by &quot;stretching&quot; the polygons so the vertices become edges.  So the number of sides in this case is $(2n+2)-4=2n-2$.  This is your result.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-14T22:16:29.590" />
  <row Id="6828" PostTypeId="1" AcceptedAnswerId="6835" CreationDate="2010-10-14T22:43:46.393" Score="10" ViewCount="224" Body="&lt;p&gt;Can anyone see a nice way to prove the following for $0\le x \le 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{1-x^2}\ge \text{erf}(\sqrt{-\log x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$erf$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$erf(z) = \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{-t^2} \  dt$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-15T01:38:43.720" LastActivityDate="2010-10-15T01:38:43.720" Title="Proving $\sqrt{1-x^2}\ge \text{erf}(\sqrt{-\log x})$" Tags="&lt;inequality&gt;" AnswerCount="1" />
  <row Id="6829" PostTypeId="2" ParentId="6805" CreationDate="2010-10-14T23:01:09.670" Score="4" ViewCount="0" Body="&lt;p&gt;Here's an all-purpose technique for proving your kind of induction problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trying to prove using induction that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ whenever n is a non-negative integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have already demonstrated the base case, so here's the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming it is true for n, is it also true for n+1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means to say: is $(n+1)^3 + (n+2)^3 + (n+3)^3$ divisible by 9?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if $[(n+1)^3 + (n+2)^3 + (n+3)^3] -[n^3 + (n+1)^3 + (n+2)^3]$ (the difference of the inductive step and the formula) is divisible by 9, so is $(n+1)^3 + (n+2)^3 + (n+3)^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? A bit of modular arithmetic here.&#xA;If (x mod 9) - (0 mod 9) = (0 mod 9) then x = (0 mod 9). This means that if the difference is divisible by 9, and one of the numbers in our subtraction is divisible by 9, the other number must be too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to do the algebra on here, but you can verify that $[(n+1)^3 + (n+2)^3 + (n+3)^3] -[n^3 + (n+1)^3 + (n+2)^3]$ simplifies into $9n^2+27n+27$ which you can rewrite as $9(n^2+3n+3)$. Hence the difference is divisible by 9, and so is our inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-10-14T23:01:09.670" />
  <row Id="6830" PostTypeId="1" CreationDate="2010-10-14T23:25:42.310" Score="0" ViewCount="59" Body="&lt;p&gt;I approached this the wrong way apperantly, help please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Consider the word mathematics. In how many ways can you rearrange all the letters so that the vowels are paired and always apart?&quot; worth 6 marks&lt;/p&gt;&#xA;" OwnerUserId="2529" LastActivityDate="2010-10-14T23:30:48.743" Title="Permutations and Combination rearrangement" Tags="&lt;permutations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6831" PostTypeId="2" ParentId="6830" CreationDate="2010-10-14T23:30:48.743" Score="1" ViewCount="0" Body="&lt;p&gt;Given there are four vowels, list all the ways to split them into pairs.  Now for each set of pairs think of each pair as a single character.  How many ways can you permute the resulting collection?  How many of those have the vowel pairs next to each other?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T23:30:48.743" CommentCount="3" />
  <row Id="6832" PostTypeId="1" AcceptedAnswerId="6833" CreationDate="2010-10-14T23:46:04.107" Score="1" ViewCount="352" Body="&lt;p&gt;Problably a simple question and possibly not asked very well. What I want to know is..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In binary, a decimal value of 1 is also 1.&#xA;It can be expressed as  x = 1 X 2º&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(probably the wrong use of a character symbol above, I was trying to represent &quot;to the power of zero&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Why is two to the power of zero equal to one?&#xA;I get that two to the power of one is equal to two, or binary 10, but why is to the power of zero equal to one, is this a math convention? is there a link I could read?&lt;/p&gt;&#xA;" OwnerUserId="2496" LastActivityDate="2010-10-18T00:07:15.683" Title="Why is two to the power of zero equal to binary one?" Tags="&lt;powers&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="6833" PostTypeId="2" ParentId="6832" CreationDate="2010-10-14T23:48:20.563" Score="10" ViewCount="0" Body="&lt;p&gt;Because we want $2^{m+n} = 2^m \cdot 2^n$, and if $n = 0$ this requires that $2^0 = 1$.  More combinatorially, $a^b$ is the number of functions from a set with $b$ elements to a set with $a$ elements, and there is exactly one function from the empty set to any other set (the empty function).  This is the same reason that $0! = 1$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-14T23:48:20.563" CommentCount="1" />
  <row Id="6834" PostTypeId="2" ParentId="6832" CreationDate="2010-10-14T23:48:46.597" Score="4" ViewCount="0" Body="&lt;p&gt;Any number (except zero) to the zero power equals one.  There are many ways to justify it.  Easiest (to me) is to make the laws of exponents work.  We know that $a^b/a^c=a^{b-c}$.  Now put b=c&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-14T23:48:46.597" CommentCount="1" />
  <row Id="6835" PostTypeId="2" ParentId="6828" CreationDate="2010-10-14T23:50:47.453" Score="16" ViewCount="0" Body="&lt;p&gt;Let $y = \sqrt{-\log x}$. Then the inequality reduces to $\text{erf}(y) \leq \sqrt{(1-e^{-2y^2})}$ or equivalently $\text{erf}^2(y) + e^{-2y^2} \leq 1$. Now $\text{erf}^2(y)$ can be written as a double integral $\text{erf}^2(y) = \frac{4}{\pi} \int_{0}^y \int_{0}^{y} e^{- (a^2+b^2)} da db$ (As Qiaochu Yuan points out, the functions involved are well behaved and the double integral is well defined). Replace the area of integration from the square of side $y$ in the first quadrant to a quarter-circle of radius $y\sqrt{2}$ in the first quadrant and switch to polar co-ordinates. This would give the inequality $\text{erf}^2(y) \leq 1-e^{-2y^2}$ which is what we wanted.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-10-15T01:00:18.827" LastActivityDate="2010-10-15T01:00:18.827" CommentCount="5" />
  <row Id="6836" PostTypeId="2" ParentId="6795" CreationDate="2010-10-14T23:52:03.487" Score="5" ViewCount="0" Body="&lt;p&gt;Aside from 8 and arguably 2, these are the even semiprimes, Sloane's &lt;a href=&quot;http://oeis.org/classic&quot; rel=&quot;nofollow&quot;&gt;A100484&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-15T18:30:03.327" LastActivityDate="2010-10-15T18:30:03.327" />
  <row Id="6837" PostTypeId="2" ParentId="6823" CreationDate="2010-10-14T23:52:48.807" Score="3" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we choose arbitrary $\lambda_1,\ldots,\lambda_n$ and form a new random variable&#xA;$$Z={{\lambda_1} \over {E[X_1]}}X_1+\cdots+{{\lambda_n} \over {E[X_n]}}X_n$$&#xA;Then&#xA;$$E[Z^2]=\sum_{i=1}^n\sum_{j=1}^n A_{ij} \lambda_i \lambda_j$$&#xA;But being the expectation of a non-negative variable, $E[Z^2]$ is non-negative for any choice of $\lambda_1,\ldots,\lambda_n$. Which proves that your matrix $A$ is non-negative semi-definite.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-14T23:52:48.807" CommentCount="1" />
  <row Id="6838" PostTypeId="2" ParentId="6832" CreationDate="2010-10-15T00:29:24.427" Score="2" ViewCount="0" Body="&lt;p&gt;The definition $\ 2^0 = 1\ $ is &quot;natural&quot; since it makes the arithmetic of exponents have the same structure as $\mathbb N$ (or $\mathbb Z\:$ if you extend to negative exponents). In more algebraic language: the definition is the canonical extension of the powering homomorphism from $\rm\ \mathbb N_+\: $ to $\rm \mathbb N\ $ (or $\rm\: \mathbb Z\:$),$\ $ viz. $\rm\ 2^n\ =\ 2^{n+0}\ =\ 2^n\ 2^0\ $ $\rm\Rightarrow\ 2^0 = 1\:$. It's just a special case of the fact that the identity element must be preserved by structure preserving maps of certain multiplicative structures (e.g. commutative cancellative monoids).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be viewed as a special case of adjoining an identity element to a commutative semigroup. And it proves very convenient to do so, for the same reason it proves convenient to adjoin the identity element 0 to the positive natural numbers, e.g. it allows every element to be viewed as a sum, so one can write general formulas for sums that work even in extremal cases where an element is indecomposable (e.g. by writing $ 1 = 1 + 0 $ vs. having to separate a special case for the sum-indecomposable element 1 or,$\ $ for $\ 2^\mathbb N\:$, $ 2 = 2\: *\: 1 $). Empty sums and products prove quite handy for naturally terminating recursive definitions.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T00:07:15.683" LastActivityDate="2010-10-18T00:07:15.683" CommentCount="1" />
  <row Id="6839" PostTypeId="1" AcceptedAnswerId="6841" CreationDate="2010-10-15T02:05:43.957" Score="0" ViewCount="60" Body="&lt;p&gt;12, 13, 11, 14, 10, 15, 9, 16... 0, maxNum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with x then alternate on each side of x until you reach 0 on one side and maxNum on the other. Need to do this in some programming code but can't nail it down. How would you write out the formula for that sequence?&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2010-10-15T02:54:33.663" Title="formula for number sequence" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6840" PostTypeId="2" ParentId="6839" CreationDate="2010-10-15T02:18:07.610" Score="2" ViewCount="0" Body="&lt;p&gt;let x denote the start number, the the value of the i'th term can be written:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f_i = x + (-1)^(i+1) * floor((i+1)/2) , if (i &amp;lt;= 2x)&#xA;undefined, otherwise&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2526" LastEditorUserId="2526" LastEditorDisplayName="" LastEditDate="2010-10-15T02:45:28.777" LastActivityDate="2010-10-15T02:45:28.777" CommunityOwnedDate="2010-10-15T02:45:28.777" CommentCount="0" />
  <row Id="6841" PostTypeId="2" ParentId="6839" CreationDate="2010-10-15T02:22:44.080" Score="2" ViewCount="0" Body="&lt;p&gt;Let n start at 0 and x be the initial value for the sequence then take &#xA;$a_{n}(x)=\begin{cases}&#xA;x-\frac{n}{2} &amp;amp; ,n\equiv0\mbox{mod}(2)\\&#xA;x+\frac{n+1}{2} &amp;amp; ,n\equiv1\mbox{mod}(2)\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll just need to add some kind of check statement to notice when you get 0 and some variable to store the maxNum and spit it out at the end.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-10-15T02:28:47.997" LastActivityDate="2010-10-15T02:28:47.997" CommentCount="5" />
  <row Id="6842" PostTypeId="2" ParentId="709" CreationDate="2010-10-15T02:25:24.300" Score="4" ViewCount="0" Body="&lt;p&gt;Already so many answers, but I haven't seen my favorite one posted, so here's another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertex occurs on the vertical line of symmetry, which is not affected by shifting up or down.  So subtract $c$ to obtain the parabola $y=ax^2+bx$ having the same axis of symmetry.  Factoring $y=x(ax+b)$, we see that the $x$-intercepts of this parabola occur at $x=0$ and $x=-\frac{b}{a}$, and hence the axis of symmetry lies halfway between, at $x=-\frac{b}{2a}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-15T02:25:24.300" CommentCount="2" />
  <row Id="6843" PostTypeId="2" ParentId="2324" CreationDate="2010-10-15T02:35:44.463" Score="0" ViewCount="0" Body="&lt;p&gt;As n0vakovic mentioned, it depends on the type of input signal, what you expect the classification to depend on, and the type of noise, if any in the signal. Also, remember that expanding the input to a higher dimensional space also helps a lot, if your classification surface is non complicated (highly nonlinear). This idea is used in nonlinear SVMs and Deep Learning (Geoff Hinton). Now, as to the possible encodings:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Raw input at each data point&lt;/li&gt;&#xA;&lt;li&gt;Scaled input to [0.0,1.0]&lt;/li&gt;&#xA;&lt;li&gt;Take FFT or DCT and use the coefficients as input&lt;/li&gt;&#xA;&lt;li&gt;Wavelet transform and use the coefficients. You have the added advantage of selecting different wavelet bases.&lt;/li&gt;&#xA;&lt;li&gt;Expand into pairwise products of raw input along with the inputs. Now you have approximately 100 + 100*50 inputs&lt;/li&gt;&#xA;&lt;li&gt;Do some denoising (for example in wavelet domain) if you have an idea of what kind of noise it is.&lt;/li&gt;&#xA;&lt;li&gt;Expand the input space by taking moving averages in addition to the raw inputs&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;.... you have many other choices. Also, you can mix and match from the above. i.e., do denoising with wavelets if you think you have some intermittent noise in certain frequencies, and then expand into inputs and pairwise products. You could take pairwise products locally instead of all pairs. Basically, the answer depends, though hopefully the above will give you something to start on.&lt;/p&gt;&#xA;" OwnerUserId="2532" LastEditorUserId="2532" LastEditorDisplayName="" LastEditDate="2010-10-15T18:17:26.450" LastActivityDate="2010-10-15T18:17:26.450" />
  <row Id="6844" PostTypeId="2" ParentId="6839" CreationDate="2010-10-15T02:54:33.663" Score="1" ViewCount="0" Body="&lt;p&gt;WWright and anonymous_21321 have given nice formulas for your sequence.  (Note that x is the zero-th term in those formulas-- you plug in n=0 or i=0 to get x.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you're trying to understand how to write a program for the sequence, a formula may not be the easiest way to go.  What you wrote in your question is an algorithm for the sequence.  To make it a program, you need to make it completely precise and clear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SAGE, I would write:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;print x&#xA;i=1&#xA;while x-i &amp;gt;= 0:&#xA;     print x+i&#xA;     print x-i&#xA;     i = i+1&#xA;print x+i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is a direct translation of your description, written precisely for SAGE.  With a little insight into the sequence, you could write a better program or even a formula for the n-th term, as the other answers do.  If you're just interested in maxNum, there's a simple formula-- you can skip all the other terms in the sequence once you understand what's going on.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-15T02:54:33.663" CommentCount="1" />
  <row Id="6845" PostTypeId="1" CreationDate="2010-10-15T03:04:44.453" Score="8" ViewCount="160" Body="&lt;p&gt;What is the current state of the art in summing (where by 'summing', I mean 'representing in terms of already known constants and whatnot') series such as these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{1+7^{n}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n!}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} e^{-\sqrt{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{1}{n^{3}+\sqrt[7]{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a copy of &lt;a href=&quot;http://en.wikipedia.org/wiki/Konrad_Knopp&quot; rel=&quot;nofollow&quot;&gt;Konrad Knopp&lt;/a&gt;'s book /Theory and Application of Infinite Series/ , but that's fifty years old, and I've been hoping that there have been improvements in the techniques since then. &lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="890" LastEditorDisplayName="" LastEditDate="2010-10-15T04:42:27.857" LastActivityDate="2010-10-15T10:17:15.290" Title="what is the current state of the art in methods of summing &quot;exotic&quot; series?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="6846" PostTypeId="1" CreationDate="2010-10-15T03:38:53.953" Score="0" ViewCount="66" Body="&lt;p&gt;Consider that I have 3 parametric equations as function of time and describe the motion of a body in space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = f(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = g(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;z = h(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These curves are pretty simple and can be modeled within a certain interval by a 2nd order polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it as simple as performing individual fitting on each function or that would break the time relationship between the functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2449" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-17T14:37:57.727" LastActivityDate="2010-10-17T14:37:57.727" Title="How to fit a 3-D parametric equation to datapoints" Tags="&lt;approximation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6847" PostTypeId="2" ParentId="6846" CreationDate="2010-10-15T03:48:44.827" Score="3" ViewCount="0" Body="&lt;p&gt;It should be appropriate to treat them independently since each is a function of time. The case where you have to worry is if f,g,h involve x,y,z. For example, if y=g(t,z,x) you could not treat them independently. Also if x=f(t,x) y=g(t,y) and z=h(t,z) you could still treat the x,y,z independently.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-15T03:48:44.827" />
  <row Id="6848" PostTypeId="1" AcceptedAnswerId="6850" CreationDate="2010-10-15T03:54:53.137" Score="9" ViewCount="230" Body="&lt;p&gt;The problem is to find:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \dfrac{\sin(\cos(x))}{\sec(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rewrite the equation as follows:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \dfrac{\sin(\cos(x))}{\dfrac{1}{\cos(x)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And multiply by $\dfrac{\cos(x)}{\cos(x)}$, producing:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \dfrac{\cos(x)*\sin(\cos(x))}{\dfrac{\cos(x)}{\cos(x)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And rewrite as:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \cos^2(x)\ \dfrac{\sin(\cos(x))}{\cos(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which then becomes:&lt;br&gt;&#xA;$\lim_{x \to 0}\ \cos^2(x) * 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which becomes 1. However, the answer is apparently $\sin(1)$. What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I found a different way to solve this, but I'm still not sure what I did wrong originally.&lt;/p&gt;&#xA;" OwnerUserId="451" LastEditorUserId="451" LastEditorDisplayName="" LastEditDate="2010-10-15T04:20:03.167" LastActivityDate="2011-01-05T17:51:43.657" Title="Finding $\lim_{x \to 0}\ \frac{\sin(\cos(x))}{\sec(x)}$ " Tags="&lt;homework&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="6849" PostTypeId="2" ParentId="5625" CreationDate="2010-10-15T03:58:15.300" Score="1" ViewCount="0" Body="&lt;p&gt;What is n here?&#xA;Is it no .of digits or the value of the number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is the value of the no,, then just convert each digit of base-9 no. in to its base-3 equivalent. &lt;/p&gt;&#xA;" OwnerUserId="2504" LastActivityDate="2010-10-15T03:58:15.300" />
  <row Id="6850" PostTypeId="2" ParentId="6848" CreationDate="2010-10-15T04:07:21.890" Score="9" ViewCount="0" Body="&lt;p&gt;As $\displaystyle x \to 0$, $\displaystyle \cos x \to 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you cannot use the limit $\displaystyle \lim_{h \to 0} \frac{\sin h}{h} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given answer is $\displaystyle \sin(1)$ I presume and not $\displaystyle \sin(0)$ (which is $0$)...&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-15T04:07:21.890" />
  <row Id="6851" PostTypeId="1" AcceptedAnswerId="6969" CreationDate="2010-10-15T04:08:09.287" Score="2" ViewCount="215" Body="&lt;p&gt;Determine and sketch the pairs $(x, y) \in \mathbb{R}^{2}$ that satisfy: $|x \cdot y| \leq 2$&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-15T05:26:10.637" LastActivityDate="2010-10-18T04:45:59.157" Title="Determine and sketch the pairs (x, y) in R x R that satisfy: |x*y| &lt;= 2" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="6852" PostTypeId="1" AcceptedAnswerId="6856" CreationDate="2010-10-15T04:37:03.577" Score="4" ViewCount="111" Body="&lt;p&gt;The following is a lemma I read online, but I don't understand part of the proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d$ be the maximum possible order among integers $a$ prime to $p$. Then for any integer $a$ not divisible by $p$, the order of $a$ divides $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $b$ be an element of order $d$, let $k$ be the order of $a$, and let $g=gcd(k,d)$. Then $a^g$ has order $k/(k,g)=k/g$, and $k/g$ is relatively prime to $d$. Therefore $ba^g$ has order $d\cdot k/g$, which by maximality of $d$ implies that $k/g=1$. Hence $k|d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one part I don't understand is how $k/g$ and $d$ are relatively prime. If I take $d=6$, $k=4$, $g=2$, then $k/g$ and $d$ are not relatively prime. However, I've gone through a few cases for small primes, and such a situation as never arisen, so there must be some constraints on which values $k$ and $d$ can take. Could someone please explain this?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-15T20:52:42.890" Title="Maximum order of integers coprime to a prime $p$." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="6853" PostTypeId="2" ParentId="6851" CreationDate="2010-10-15T05:33:49.443" Score="-2" ViewCount="0" Body="&lt;p&gt;Think graphically it may help you out. So the graph of your curve looks something &#xA; &lt;img src=&quot;http://i.imgur.com/uLMnG.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;like this. The Region you need is the one beneath the rectangular hyperbola. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-16T07:19:19.847" LastActivityDate="2010-10-16T07:19:19.847" CommentCount="1" />
  <row Id="6855" PostTypeId="2" ParentId="6832" CreationDate="2010-10-15T05:49:19.240" Score="2" ViewCount="0" Body="&lt;p&gt;The easy explanation is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2^0 = 1_2$&lt;/li&gt;&#xA;&lt;li&gt;$2^1 = 10_2$ &lt;/li&gt;&#xA;&lt;li&gt;$2^2 = 100_2$&lt;/li&gt;&#xA;&lt;li&gt;$2^3 = 1000_2$&lt;/li&gt;&#xA;&lt;li&gt;$2^4 = 10000_2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we multiply $2^1$ by $2$, we get $2^2$; naturally if we divide $2^1$ by $2$, we should get $2^0$. And it happens that &#xA;${2^1 \over 2} = {2 \over 2} = 1$.&lt;/p&gt;&#xA;" OwnerUserId="2166" LastActivityDate="2010-10-15T05:49:19.240" CommentCount="1" />
  <row Id="6856" PostTypeId="2" ParentId="6852" CreationDate="2010-10-15T05:51:23.997" Score="1" ViewCount="0" Body="&lt;p&gt;Below is the key theorem as it applies to arbitrary finite abelian groups. See my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zr7a0j6us.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for an example of how it is applied to deduce the more general result that  a finite multiplicative subgroup of a domain is cyclic. The lemma is famous as &quot;Herstein's hardest problem&quot; - see the note below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\rm\quad\quad\: maxord(G)\ =\ expt(G)\ $ for a finite abelian group $\rm\: G\:,\ $ i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \ max\ \{ ord(g) : \: g \in G\}\ =\  min\ \{ n&gt;0 : \: g^n = 1\ \:\forall\ g \in G\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \:$ By the lemma below, $\rm\: S\: =\: \{ ord(g) : \:g \in G \}$  is a finite &#xA;set of naturals closed under$\rm\ lcm\:$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence every elt $\rm\ s \in S\:$  is a &#xA;divisor of the max elt $\rm\: m\: $  [else  $\rm\: lcm(s,m) &gt; m\:$],$\ $ so $\rm\ m = expt(G)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ A finite abelian group $\rm\:G\:$ has an  lcm-closed  order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \ X,Y \in G\ \Rightarrow\ \exists\ Z \in G:\  o(Z) = lcm(o(X),o(Y))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;$\ \ $  By induction on  $\rm o(X)\: o(Y)\:.\ $ If it's $\:1\:$ then trivially $\rm\:Z = 1\:$. $\ $ Otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;write  $\rm\ o(X) =\: AP,\: \ o(Y) = BP',\ \ P'|P = p^m &gt; 1\:,\ $  prime $\rm\: p\:$ coprime to $\rm\: A,B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then  $\rm\: o(X^P) = A,\ o(Y^{P'}) = B\:.\ $  By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so  $\rm\ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y))\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ \ $ This lemma was presented as problem 2.5.11, p. 41 in the first edition of Herstein's popular textbook &quot;Topics in Algebra&quot;. In the 2nd edition Herstein added the following note (problem 2.5.26, p.48) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don't be discouraged if you don't get this problem with what you know of group theory up to this stage. I don't know anybody, including myself, who has done it subject to the restriction of using material developed so far in the text.  But it is fun to try. I've had more correspondence about this problem than about any other point in the whole book.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-15T20:52:42.890" LastActivityDate="2010-10-15T20:52:42.890" CommentCount="2" />
  <row Id="6857" PostTypeId="2" ParentId="6832" CreationDate="2010-10-15T05:52:08.487" Score="4" ViewCount="0" Body="&lt;p&gt;I think it's helpful to think about sums for a bit, and then come back to products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With sums… we’re all happy with the idea that you can add up a list of 2 things, or 3 things, or more (so e.g. the sum of the list $(x,y,z)$ is $x+y+z$).  And after moment’s thought, we’re happy with the idea that you can add up a list of 1 thing, eg $(x)$, and get just $x$ itself.  And it takes a moments thought, but it's still pretty intuitive, to decide that you can add up the empty list $()$, and that its sum is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(“I went shopping today, with nothing on my shopping list.  How much did it cost in total?”)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, play exactly the same game with products!  It's just as easy until you get to the empty list.  For some reason, it's much less intuitive at first (at least it was for me, and I think it is for most people) that one can make sense of the product of the empty list, and that it should be $1$.  But every argument I know to justify why the sum of the empty list should be $0$ also works as an argument that its product should be $1$!  (See the arguments about making exponents work nicely, etc, in the earlier comments and answers.)  And, of course, once the lightbulb flashes, it suddenly becomes completely clear and natural that the product is 1, and that that's not just a convention, and how could anyone possibly think otherwise!? :-P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;…unfortunately I don't know an example for products that’s quite as intuitively convincing as the shopping list example is for sums.  The best I can do: at the checkout, they apply a bunch of taxes and/or discounts.  Each one of those is a multiplier: e.g. 7% sales tax multiplies my total by 1.07, a 20% discount is a multiplier of 0.8, and so on.  To get the overall multiplier from all the taxes/discounts currently in effect, you just multiply the list of them together (of course)!  But if there are no taxes or discounts at the store today, what's the overall multiplier on my shopping?&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-10-15T05:52:08.487" CommentCount="1" />
  <row Id="6858" PostTypeId="1" AcceptedAnswerId="6859" CreationDate="2010-10-15T07:14:12.730" Score="-2" ViewCount="205" Body="&lt;p&gt;I've recently volunteered as a tutor for &quot;higher-level mathematics&quot; as a Calculus student because I'm required by my school to achieve at least twenty volunteer hours for service learning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When applying to the NPO as a volunteer, their only position available was for a &quot;higher-level mathematics tutor&quot;. I've been a Calculus AB student and T/A for my Calculus AB teacher since the start of the year (we're on our third chapter) and I've only recently touched upon Calculus so I'm unsure whether my skills are on par with &quot;higher-level mathematics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon after I left their building, the connection between &quot;higher-level education&quot; and &quot;higher-level mathematics&quot; hit me. This NPO is from the elementary to the high school level so I became confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone define &quot;higher-level mathematics&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2426" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-15T07:43:37.310" LastActivityDate="2010-10-15T07:54:55.917" ClosedDate="2010-10-15T18:53:42.897" Title="Define &quot;higher-level mathematics&quot;" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;education&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6859" PostTypeId="2" ParentId="6858" CreationDate="2010-10-15T07:54:55.917" Score="0" ViewCount="0" Body="&lt;p&gt;I think this is a soft question (and have tagged it as such; you can roll it back if you disagree) but I'll try to answer.  First, since the NPO only works up to high school, their definition of &quot;higher-level&quot; probably doesn't correspond to the typical one.  Just think of it in terms of your school.  If it has AP classes, Calc BC is probably the highest math class available, and it's possible a few people are taking basic college-level classes (say, multivariable calc or linear algebra), though they probably won't need tutors.  As a blind guess, I'd say that you'd probably have to tutor kids in calculus, which could be a problem if you're still learning it.  The best solution is to tell them where you are now and ask them or previous tutors what level is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For me, what &quot;higher-level mathematics&quot; really is is the things a math major starts doing around the middle of his or her career -- not linear algebra, multivariable calculus, or basic group theory, but topology, commutative algebra, real/complex analysis, etc. would all be there.  Presumably, mathematicians would find some of this elementary, but there is an objective change in the way the math is done around this time.  You move from calculations and applications of theorems to proving the theorems yourself and learning to work rigorously, axiomatically, intuitively, etc.)&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-15T07:54:55.917" CommentCount="2" />
  <row Id="6860" PostTypeId="1" CreationDate="2010-10-15T08:49:38.467" Score="-2" ViewCount="156" Body="&lt;p&gt;This is a weird logic problem that was posed to me today. This question isn't about the actual logic problem (or the trivial paradox) but in its implications, so read to the end!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A man has been sentenced to death. The executioner comes to his cell on May 31st and tells him that he will die before June 30th. He also tells him that he will not be able to know when he is killed, and that if he is not killed before 9a.m. he will not be killed on that day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On what day(s) can he be killed? The answer can be shown through induction. If the man isn't killed by June 29th, then he knows he will be killed on June 30th, hence He cannot be killed on the 30th because that would violate one of the logical axioms (he cannot know when he will be killed). The inductive step: The 29th now becomes the &quot;last&quot; day of the month, and using similiar logic, he cannot be killed on the 29th. Therefore the man CANNOT be killed without him knowing when it happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logically we have a contradiction in the following statement. $A \land B \Rightarrow \sim A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if we assume we live in a logical world, such a system or situation should not realizable because if we have both $A$ and $\sim A$ true, then we can really prove anything to be true (this is a logic theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question then becomes, how can you reconcile the fact that we can easily do this in the physical world? Find a friend, and tell him that at some point in the next thirty days you will send him an email that says &quot;you're dead&quot;, and that he will not be able to know when it will happen. It works out just fine...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can this be? How can we reconcile the physical realizability of the problem, with its obvious logical paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I don't know the answer, so don't expect one :P&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-10-15T13:34:41.840" ClosedDate="2010-10-15T14:07:03.403" Title="The Hangman's Noose - A Logical Paradox " Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6861" PostTypeId="1" AcceptedAnswerId="6930" CreationDate="2010-10-15T09:39:55.187" Score="2" ViewCount="70" Body="&lt;p&gt;Given two complex varieties over a common base, I can take their fiber product in the category of varieties, or I can take their fiber product in the category of schemes and then take the reduced subscheme. I have heard, that these two operations yield the same.&#xA;Has someone a reference? &lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-10-16T04:41:42.113" Title="Fiber product of varieties vs schemes reference" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" />
  <row Id="6862" PostTypeId="2" ParentId="6845" CreationDate="2010-10-15T10:17:15.290" Score="1" ViewCount="0" Body="&lt;p&gt;Well, much depends on the individual series but one interesting development in the last 50 years is on the acceleration of convergence, which enables us to express some slowly convergent series in terms of one(s) with much faster convergence, or a sum of known constants plus faster converging series. For example, I really like this transformation due to Simon Plouffe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(7) = \frac{19}{56700}\pi^7 – 2\sum_{n=1}^{\infty} \frac{1}{n^7(e^{2\pi n}-1)},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta(7) = \sum_ {n=1}^{\infty} 1/n^7,$ which is given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeta_constant&quot; rel=&quot;nofollow&quot;&gt;this wikipedia page.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrei Markov's 1890 method, which was the basis for Apéry's proof of the irrationality of $\zeta(3),$ has been pushed further by &lt;a href=&quot;http://www.springerlink.com/content/q87617rp46135qwm/&quot; rel=&quot;nofollow&quot;&gt;Mohammed and Zeilbeger&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see some other nice transformations of $\zeta(3)$ &lt;a href=&quot;http://numbers.computation.free.fr/Constants/Zeta3/zeta3.html&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-15T10:17:15.290" CommentCount="4" />
  <row Id="6863" PostTypeId="1" CreationDate="2010-10-15T10:41:19.527" Score="1" ViewCount="75" Body="&lt;p&gt;I'm interested, how is the dual input in a sensor fusioning setup in a Kalman filter modeled? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for instance that you have an accelerometer and a gyro and want to present the &quot;horizon level&quot;, like in an airplane, a good demo of something like this &lt;a href=&quot;http://www.youtube.com/watch?v=AWAFFZ7rPDc&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you actually harvest the two sensors positive properties and minimize the negative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this modeled in the Observation Model matrix (&lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter#Underlying_dynamic_system_model&quot; rel=&quot;nofollow&quot;&gt;usually&lt;/a&gt; symbolized by capital H)?&lt;/p&gt;&#xA;" OwnerUserId="2367" LastEditorUserId="2367" LastEditorDisplayName="" LastEditDate="2010-10-15T10:49:17.733" LastActivityDate="2010-12-20T07:44:25.990" Title="Sensor fusioning in Kalman filter" Tags="&lt;stochastic-processes&gt;&lt;signal-processing&gt;" AnswerCount="1" />
  <row Id="6864" PostTypeId="2" ParentId="6489" CreationDate="2010-10-15T10:53:51.693" Score="3" ViewCount="0" Body="&lt;p&gt;The annotation to &lt;a href=&quot;http://www.irregularwebcomic.net/2339.html&quot; rel=&quot;nofollow&quot;&gt;http://www.irregularwebcomic.net/2339.html&lt;/a&gt; has an explanation of the Banach-Tarski paradox in easy to read language. It does ignore some of the technical details, but it covers most of the idea of the construction nicely.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2010-10-15T10:53:51.693" />
  <row Id="6865" PostTypeId="1" AcceptedAnswerId="6866" CreationDate="2010-10-15T11:57:33.463" Score="1" ViewCount="88" Body="&lt;p&gt;Hi, I'm taking a course on machine learning and came across a model answer which I can't wrap my head around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole model answer is the 2nd exercise here: &lt;a href=&quot;https://noppa.tkk.fi/noppa/kurssi/t-61.3050/viikkoharjoitukset/T-61_3050_proposed_solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;https://noppa.tkk.fi/noppa/kurssi/t-61.3050/viikkoharjoitukset/T-61_3050_proposed_solutions.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I'm specifically having trouble understanding is the last 2 lines of the 2nd page, ie this: &lt;a href=&quot;http://i.imgur.com/SZEPC.png&quot; rel=&quot;nofollow&quot;&gt;http://i.imgur.com/SZEPC.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the left side of the upper equation become just the sum of indicator variables? How does the right side simplify to just lambda*p_i?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastActivityDate="2010-10-15T12:30:18.190" Title="Weird differentation of probabilities, please explain?" Tags="&lt;probability&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="6866" PostTypeId="2" ParentId="6865" CreationDate="2010-10-15T12:30:18.190" Score="2" ViewCount="0" Body="&lt;p&gt;I say there be typos. Particular the suddenly appearing subscript $k$ does not make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Straightforward differentiation gives:&#xA;$${\partial \over {\partial p_i}} L(p|X)={1 \over p_i}\sum_{t=1}^N x^t_i + \lambda$$&#xA;Setting this equal to zero gives us:&#xA;$$\sum_{t=1}^N x^t_i + \lambda p_i=0$$&#xA;Proceeding from which and reading the subscript $k$ in your text as $i$ leads to the same ultimate answer.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-15T12:30:18.190" CommentCount="3" />
  <row Id="6867" PostTypeId="2" ParentId="6860" CreationDate="2010-10-15T13:34:41.840" Score="0" ViewCount="0" Body="&lt;p&gt;While I agree there is no academic consensus on the matter, to me it seems there is a pretty simple resolution. The man goes through all the logical steps to work out that he cannot possibly be hanged, because he would know it was going to happen, so he settles down to wait for the end of June. Then on the day he is actually killed his last thought will be &quot;I didn't know that was going to happen&quot;.&lt;/p&gt;&#xA;" OwnerUserId="397" LastActivityDate="2010-10-15T13:34:41.840" />
  <row Id="6868" PostTypeId="1" AcceptedAnswerId="6888" CreationDate="2010-10-15T13:36:36.477" Score="3" ViewCount="167" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I encounter a problem recently stated as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many pieces of cheese we can obtain from a single thick piece by making five straight slices?(we can't move the cheese when slicing)If we wanna maximize the number of pieces which is denoted by P(n), is there any recurrence relation for P(n)?(n is number of slices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints will be highly appreciated and thank you all in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards.&lt;/p&gt;&#xA;" OwnerUserId="1294" LastActivityDate="2010-10-15T20:08:30.097" Title="How to slice the cheese" Tags="&lt;recursive&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6869" PostTypeId="2" ParentId="6868" CreationDate="2010-10-15T13:54:37.077" Score="12" ViewCount="0" Body="&lt;p&gt;This is a special case of the problem of counting the number of&#xA;regions $\mathbb{R}^n$ is divided into by $k$ hyperplanes &lt;em&gt;in&#xA;general position&lt;/em&gt;. The answer is&#xA;$$\sum_{j=0}^n {k\choose j}.$$&#xA;This is mentioned in Richard Stanley's notes on&#xA;&lt;a href=&quot;http://www-math.mit.edu/~rstan/arrangements/arr.html&quot; rel=&quot;nofollow&quot;&gt;hyperplane arrangements&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-15T13:54:37.077" CommentCount="1" />
  <row Id="6870" PostTypeId="1" CreationDate="2010-10-15T14:56:17.993" Score="7" ViewCount="376" Body="&lt;p&gt;I often see notation/wording like &quot;let f(x) be a continuous function&quot; or &quot;let $f(x) \in C^0(\mathbb{R})$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that sin and $x \mapsto \sin(x)$ are functions, while sin(x) is a real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any correct or best practices in this regard?&lt;/p&gt;&#xA;" OwnerUserId="2043" LastEditorUserId="2043" LastEditorDisplayName="" LastEditDate="2010-10-15T15:26:43.113" LastActivityDate="2010-10-15T17:07:18.370" Title="In written mathematics, is f(x) a function or a number?" Tags="&lt;functions&gt;&lt;notation&gt;" AnswerCount="7" CommentCount="8" />
  <row Id="6871" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T14:59:07.797" Score="-1" ViewCount="0" Body="&lt;p&gt;In Mathematics $f(x)$ denotes a function, like $\sin{x}, \cos{x}, e^{x}$. When the functions are given values like $1,2,3$, then they give you a number. For example, $f(x)=e^{x}$ is a function where as $f(1)=e^{1}=e =2.71$ a number.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-15T14:59:07.797" CommentCount="2" />
  <row Id="6872" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:02:50.963" Score="1" ViewCount="0" Body="&lt;p&gt;It all depends on what you have said $x$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is previously defined as a number (e.g. $x  = 3$) then $f(x)$ is also a number, since you can evaluate $f$ at a certain point $x$ to get the number. (A more common way to notate that would be to say $x_1$ instead of $x$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, $x$ is defined as part of some set (e.g. $x\in[0,1]$) then $f$ is a function defined for input values in that interval. $\sin x$ is defined as a function for $-\infty &amp;lt; x &amp;lt; \infty$, but as a number for $x=\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="161" LastActivityDate="2010-10-15T15:02:50.963" />
  <row Id="6873" PostTypeId="1" AcceptedAnswerId="6880" CreationDate="2010-10-15T15:04:10.787" Score="3" ViewCount="99" Body="&lt;p&gt;This problem was asked in a test, couple of years ago. Looked interesting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a chess tournament, each pair of players plays exactly one game. No game is drawn. Suppose the $i^{th}$ player wins $a_{i}$ games and loses $b_{i}$ games. Show that $$\sum a_{i}^{2} = \sum b_{i}^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of a better title. It would be nice in case someone comes up with a nice title.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-15T16:30:52.090" LastActivityDate="2010-10-15T16:30:52.090" Title="Sum of squares of in-degrees vs out-degrees in a Tournament Graph." Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6874" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:06:01.383" Score="9" ViewCount="0" Body="&lt;p&gt;You are right. Properly speaking, the function is called just $f$, and its value at the point $x$ is denoted by $f(x)$. Speaking of &quot;the function $f(x)$&quot; is what mathematicians call &quot;abuse of notation&quot;, but it is of course often very practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, one usually writes $x \mapsto \sin x$, not $x \to \sin x$; it's &quot;\mapsto&quot; in TeX.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-15T15:06:01.383" CommentCount="1" />
  <row Id="6875" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:06:55.637" Score="1" ViewCount="0" Body="&lt;p&gt;The wording &quot;let $f(x) \in C^0(\mathbb{R})$...&quot; is really sloppy. Of course it doesn't lead to confusion, but there is something to be said for correctness. Ideally whenever you see $f(x)$ it should be because we care about what $f$'s values are rather than $f$ in and of itself as say a member of some space of functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and your example of &quot;$x \to sin(x)$&quot; is IMO better written as &quot;$\mathbb{R} \owns x \mapsto \sin x \in \mathbb{R}$. Obviously not so useful for this particular function, but very smart for things like &quot;$V \times V \owns (x,y) \mapsto \langle x,y \rangle \in \mathbb{C}$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-15T15:06:55.637" />
  <row Id="6876" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T15:07:28.490" Score="10" ViewCount="0" Body="&lt;p&gt;Saying that $f$ or $x\mapsto f(x)$ is the function rather than $f(x)$ is the more common modern perspective in higher mathematics.  I have even seen a precalculus book that stresses that $f$ is a function, whereas $f(x)$ is a value of the function.  The notation $f(x)$ to denote a function remains because it is often more convenient, and it is especially prominent from high school mathematics up to calculus because it is psychologically easier to become accustomed to.  This may lead to ambiguity, because in any case $f(x)$ sometimes &lt;em&gt;does&lt;/em&gt; refer to a value of the function, but the strictest logical correctness of notation isn't for everyone, or for all situations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it off-putting when I read mathematics that was written many decades ago and come across something like, &quot;Let $\varphi(x)$ be a bounded linear functional on $X$....&quot;.  I think to myself, &quot;No, $\varphi$ is the functional!&quot;  Then I take a deep breath and relax, and find that this notation did not hinder the authors from doing and writing great mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ahlfors's complex analysis text has a footnote on page 21 of the 2nd edition, 1966:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Modern students are well aware that $f$ stands for the function and $f(z)$ &#xA;  for a value of the function. However, analysts are traditionally minded and continue to speak of &quot;the function $f(z)$.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-15T15:19:21.327" LastActivityDate="2010-10-15T15:19:21.327" />
  <row Id="6877" PostTypeId="1" CreationDate="2010-10-15T15:16:15.147" Score="5" ViewCount="195" Body="&lt;p&gt;When I first saw the title of &lt;a href=&quot;http://meta.math.stackexchange.com/questions/909/write-an-elevator-pitch-tagline&quot;&gt;this question&lt;/a&gt;, I forgot for a moment I was on meta, and thought it was asking about quick, catchy, attractive, informative one-or-two-liner summaries of various fields of mathematics.  It turned out not to be… so here’s the question I thought it was!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of times when one wants a quick way to explain to someone (family, students, teachers, colleagues) what some field is about, and (hopefully) catch their interest as well — perhaps to come and learn about it themselves, perhaps just to understand why you find it interesting, perhaps to convince them why they should give you funding to work on it…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question linked nicely describes how a good such pitch should work: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This isn't as easy as it sounds. Imagine the user who will never read your FAQ and you have two seconds to grab their attention. It should be catchy but descriptive. It should be thoroughly clear but painfully concise. &lt;strong&gt;Make every... word... count.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A couple of extra thoughts: the level of technicality of the pitch should probably depend on how specialised the field is.  You don't need to describe “algebra” or “calculus” to a professional mathematician; dually, you probably aren't trying to explain “homotopy theory of CDGA’s” to your friend who never took any maths courses (at least, not in one sentence).  This isn’t about any level of maths in particular — whether in recreational maths or cutting-edge research, everyone can benefit from a good slogan sometimes.  Posting a few examples myself, to get the ball rolling&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m not sure to what extent guidelines/conventions for big-list community-wiki questions have solidified on this site, but a couple which work well on MO and SE are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;just one example (tagline) per posted answer, so that they can be voted up/down individually;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;don't be shy of making near-duplicates if you think an idea could have been executed a bit better, nor of suggesting improvements to answers in their comments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Related questions: &lt;a href=&quot;http://math.stackexchange.com/questions/2601/best-intuitive-metaphors-for-math-concepts-of-any-level&quot;&gt;Best intuitive metaphors for math concepts&lt;/a&gt;; &lt;a href=&quot;http://mathoverflow.net/questions/5450/cocktail-party-math&quot; rel=&quot;nofollow&quot;&gt;Cocktail party math&lt;/a&gt; (at MO).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;p.s. I can't see how to make this question community wiki; anyone with enough rep to do so, feel free to do so &amp;amp; delete this request!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="2439" LastEditorDisplayName="" LastEditDate="2010-10-15T16:05:38.630" LastActivityDate="2010-10-21T16:51:04.640" Title="Elevator pitch for a (sub)field of maths?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;intuition&gt;&lt;education&gt;" AnswerCount="8" CommentCount="6" FavoriteCount="2" />
  <row Id="6878" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T15:20:52.687" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Calculus&lt;/strong&gt;: the mathematics of motion and change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not original, of course!)&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-10-15T15:20:52.687" />
  <row Id="6879" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T15:25:48.790" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Category Theory&lt;/strong&gt;:  A birds-eye view of mathematics; a framework for investigating how different areas relate to one another.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-10-15T15:25:48.790" />
  <row Id="6880" PostTypeId="2" ParentId="6873" CreationDate="2010-10-15T15:27:13.670" Score="7" ViewCount="0" Body="&lt;p&gt;$$\sum (a_i^2 - b_i^2) =  \sum (a_i+b_i)(a_i-b_i) = (n-1) \sum (a_i-b_i) $$&#xA;$$= (n-1) \left(\sum a_i - \sum b_i\right) = (n-1)\left(\binom{n}{2} - \binom{n}{2}\right) = 0,$$&#xA;where $n$ is the total number of players.  (Each player plays $n-1$ games, so $a_i+b_i = n-1$, and there are a total of $\binom{n}{2}$ games played, so $\sum a_i = \sum b_i = \binom{n}{2}$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-15T15:41:28.333" LastActivityDate="2010-10-15T15:41:28.333" CommentCount="0" />
  <row Id="6881" PostTypeId="2" ParentId="6873" CreationDate="2010-10-15T15:33:14.407" Score="1" ViewCount="0" Body="&lt;p&gt;Let the number of wins of a random player be A, and let the number of losses of a random player be B. Then A and B have the same mean (half the number of games) and the same variance (since $B = n-A$, where n is the number of players). So $A^2$ and $B^2$ have the same expectation, which is what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, if you write this out more explicitly it'll resemble Mike Spivey's answer quite closely. But this is what went through my head.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-15T15:33:14.407" />
  <row Id="6883" PostTypeId="1" CreationDate="2010-10-15T15:46:25.823" Score="0" ViewCount="139" Body="&lt;p&gt;If a quantity is successively increased by &lt;strong&gt;20%&lt;/strong&gt; for two times and another quantity is successively decreased by &lt;strong&gt;10%&lt;/strong&gt; for three times, the first quantity is &lt;strong&gt;80%&lt;/strong&gt; of the second. What percentage is the original first quantity as compared to the original second quantity?&lt;/p&gt;&#xA;" OwnerUserId="2543" LastEditorUserId="2543" LastEditorDisplayName="" LastEditDate="2010-10-15T16:11:20.583" LastActivityDate="2010-10-15T16:28:10.617" Title="Average--Average---Average" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="6884" PostTypeId="2" ParentId="6873" CreationDate="2010-10-15T16:13:11.720" Score="4" ViewCount="0" Body="&lt;p&gt;Consider flipping the result of one game, say between player 1 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sum (a_i)^2 - \sum (b_i)^2$ changes by $2(a_1 + b_1) - 2(a_2 + b_2) = 0$ as $a_i + b_i = n-1$, where $n$ is the number of players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\sum (a_i)^2 - \sum (b_i)^2$ is the same no matter what the results and thus is same as the extreme case when the winner wins all matches, the runner up all but one etc, for which the sum is 0.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-15T16:13:11.720" />
  <row Id="6885" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T16:18:22.660" Score="2" ViewCount="0" Body="&lt;p&gt;There's the old &quot;&lt;strong&gt;Topology:&lt;/strong&gt; Geometry on a rubber sheet.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both Category Theory and Universal Algebra, I also like &lt;em&gt;The Esperanto of Mathematics/Algebra&lt;/em&gt; (except for the negative connotation of &quot;nobody actually uses it&quot;...).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-15T16:18:22.660" CommunityOwnedDate="2010-10-15T16:18:22.660" CommentCount="2" />
  <row Id="6886" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T16:28:03.537" Score="0" ViewCount="0" Body="&lt;p&gt;(I don't have the points yet to make comment)&#xA;Are you a computer scientist? Because I think your view is more about typed languages.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-10-15T16:28:03.537" />
  <row Id="6887" PostTypeId="2" ParentId="6883" CreationDate="2010-10-15T16:28:10.617" Score="1" ViewCount="0" Body="&lt;p&gt;Say your quantity is $Q$. To get 80% of that, you multiply by $0.8$, so $0.8Q$. How do you get 80% of &lt;em&gt;that&lt;/em&gt;? Well, you multiply by $0.8$ again, so you get $0.8(0.8Q) = (0.8)^2Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you take 80% successively three times? Four times? Five times? $n$ times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you first take $k$% ($k$ between $0$ and $100$), then  you multiply by $\frac{k}{100}$. So if you started with $Q$, you now have $\frac{k}{100}Q$. If you then take $\ell$% (again, $\ell$ between $0$ and $100$), then you need to multiply &lt;em&gt;that&lt;/em&gt; by $\frac{\ell}{100}$. So you would have $\left(\frac{\ell}{100}\right)\left(\frac{k}{100}\right)Q$. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, say that instead of your problem, I wanted to first decrease by 10%, then the result by 20%, then the result of &lt;em&gt;that&lt;/em&gt; by 45%, and finally the result of that by 15%. How would I do it? I would get $(.85)(.55)(.8)(.9)Q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are &lt;em&gt;increasing&lt;/em&gt; by 20%, then instead of multiplying by 0.8, you multiply by $1.2$ (since adding 20% means taking 120% of the original). Same idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you have two quantities, $Q_1$ and $Q_2$; you are doing different things to them. At the end you will have some multiple of $Q_1$ and some multiple of $Q_2$, say $aQ_1$ and $bQ_2$. But you are told that &quot;the first quantity is 80% of the second&quot;. That means that you can write $aQ_1$ as a multiple of $bQ_2$, from which you can &quot;solve for $Q_1$&quot; in terms of $Q_2$. That should tell you exactly what you want to know.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-15T16:28:10.617" />
  <row Id="6888" PostTypeId="2" ParentId="6868" CreationDate="2010-10-15T16:31:19.577" Score="5" ViewCount="0" Body="&lt;p&gt;This was a very old Monthly problem - see below. For an excellent introduction to the general topic see Richard Stanley's paper &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.118.5996&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Hyperplane Arrangements (2004)&lt;/a&gt; and also Renteln's lecture slides &lt;a href=&quot;http://physics.csusb.edu/~prenteln/papers/hyperintro.pdf&quot; rel=&quot;nofollow&quot;&gt;It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements, 2008&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/U07We.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-15T16:39:17.810" LastActivityDate="2010-10-15T16:39:17.810" />
  <row Id="6889" PostTypeId="2" ParentId="6870" CreationDate="2010-10-15T17:07:18.370" Score="5" ViewCount="0" Body="&lt;p&gt;You are quite right: $f$ by itself should denote a function, $f(x)$ by&#xA;itself should denote the element in the codomain of $f$ (in this case,&#xA;the real number) that results when you evaluate $f$ on the element $x$&#xA;of its domain, where $x$ should previously have been defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this rule is honored as much in the breach as the observance:&#xA;there are many situations where it is convenient to break it.  When&#xA;defining a function by a formula, it's hard to avoid a dummy variable,&#xA;and so one likes to say &quot;let $f(x)=e^{-x}$&quot; instead of the more&#xA;correct but awkward &quot;let $f$ be the function $x \mapsto e^{-x}$&quot;.  In&#xA;particular, with functions that have multi-letter symbols like $\sin$,&#xA;I find that people generally prefer to avoid writing them without an&#xA;argument like $\sin x$.  One does not like to talk about $\sin$ as a&#xA;function in itself, so instead of writing something like $\sin'' =&#xA;-\sin$, one would rather say &quot;if $f(x)=\sin x$, then $f'' = -f$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative to a dummy variable that's sometimes used is a dot:&#xA;$\cdot$.  People sometimes write &quot;let $f=g(\cdot + 5)$&quot; to avoid the&#xA;less correct &quot;let $f(x)=g(x+5)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working with functions of several variables, using dummy&#xA;variables often helps keep track of which variable is which.  One&#xA;often writes something like: &quot;let $u(x,t)$ be a solution of the heat&#xA;equation $\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial&#xA;x^2} = 0$&quot;.  Of course, one is really talking about the function $u$&#xA;and not any particular real number of the form $u(x,t)$, but it would&#xA;be much more awkward to write otherwise.  It also reminds you that the&#xA;first argument of $u$ should be interepreted as space and the second&#xA;one as time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: mathematicians are not compilers.  Written mathematics has&#xA;some syntax rules, but they are not quite hard-and-fast, and need not&#xA;be followed at the expense of clarity.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-15T17:07:18.370" CommentCount="2" />
  <row Id="6890" PostTypeId="1" AcceptedAnswerId="7276" CreationDate="2010-10-15T17:19:51.373" Score="1" ViewCount="205" Body="&lt;p&gt;Hi everybody!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to understand the physical meaning of the Feynman's vector potential definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_{m}^{(b)}(x) = e_b \int \delta (xb_{\mu}xb^{\mu})db_m(b), \qquad m=0,1,2,3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(component m of the vector potential of the particle b at the point x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\delta$ is the Dirac delta function&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ xb_{\mu}xb^{\mu} $ is the 4-norm of $x - b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So the $\delta$ function is non-zer $\iff x$ is in the boundary of the light cone of b.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can you give me any clues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-15T18:08:44.357" LastActivityDate="2010-10-21T07:04:42.917" Title="Feynman's vector potential" Tags="&lt;physics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="6891" PostTypeId="1" CreationDate="2010-10-15T19:36:40.820" Score="2" ViewCount="57" Body="&lt;p&gt;A phase change is only possible in a physical system which obeys the laws of statistical mechanics if the infinite series for the partition function of that system converges non-uniformly (i.e. converges point-wise).  This is because in order for a phase change to occur, the partition function must converge to two different continuous functions in different regions of the phase plane; and this can only happen if the number of terms in the partition function is infinite.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be possible, using the analogy with phase changes, to make some general statements about partition functions which converge non-uniformly to two different continuous functions.  This is interesting to me because point-wise convergence is a very weak condition, and the analogy with phase changes seems to suggest a path towards a deeper understanding of this form of convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most interesting subjects in the statistical mechanics of phase transitions is the so-called &quot;renormalization group&quot; (which is not a group) and the critical exponents it predicts (these are non integer numbers which are associated with a &quot;critical point&quot; in the phase plane).  Another powerful result from renormalization group theory is the concept of universality, in which the thermodynamic variables of the system are symmetric across all length scales at a certain critical point.  Universality is directly related to the critical exponents and critical point behavior, but as is fairly standard for theoretical physics, the mathematics used is non-rigorous (I recommend the book &quot;scaling and renormalization in statistical physics&quot; by Cardy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the point-wise convergence of a partition function, can we put the foundations of critical exponents on a rigorous mathematical basis; in the sense that we can somehow predict their values from properties (such as the partial derivatives) of the partition function?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-10-15T19:50:51.087" LastActivityDate="2010-10-15T19:50:51.087" Title="Critical exponents and point-wise convergence." Tags="&lt;analysis&gt;&lt;statistics&gt;" />
  <row Id="6892" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T19:55:37.850" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Algebra:&lt;/strong&gt;&#xA;  The mathematics of structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Set Theory:&lt;/strong&gt; The main foundation of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Logic:&lt;/strong&gt; The main foundation of set theory.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-15T19:55:37.850" />
  <row Id="6893" PostTypeId="1" AcceptedAnswerId="6901" CreationDate="2010-10-15T20:01:39.893" Score="2" ViewCount="123" Body="&lt;p&gt;A loan for $8,000$ must be repaid with 6 year end payments at an annual rate of $11 \%$. What is the annual payment? I know that the present value of an annuity with end payments is $\frac{1-v^n}{i}$ where $v = 1/(1+i)$. Likewise, the future value is $(1+i)^{n}\frac{1-v^n}{i}$. How do I use this to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-18T10:14:29.473" LastActivityDate="2010-10-18T10:14:29.473" Title="Annuities and Loans" Tags="&lt;homework&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="6894" PostTypeId="1" AcceptedAnswerId="6957" CreationDate="2010-10-15T20:03:10.483" Score="2" ViewCount="123" Body="&lt;p&gt;I wanted to prove that $\varphi$ is multiplicative (that for $(a,b)=1$, $\varphi(ab)=\varphi(a)\varphi(b)$) using the following idea:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Define $\varphi'$ by $n = \varphi'(n) + \varphi(n)$.&lt;/li&gt;&#xA;&lt;li&gt;Multiply out $\varphi(a)\varphi(b) = (a - \varphi'(a))(b - \varphi'(b)) = ab - \varphi'(a)b - a \varphi'(b) + \varphi'(a)\varphi'(b)$&lt;/li&gt;&#xA;&lt;li&gt;Use the principle of inclusion exclusion ($|A \cup B| = |A| + |B| - |A \cap B|$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but I could not get it to work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach possible at all? How can it be saved?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So I just need to construct sets $A(a,b)$ and $B(a,b)$ that satisfy the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\varphi'(ab) &amp;amp;=&amp;amp; \varphi'(a) b &amp;amp;+&amp;amp; a \varphi'(b) &amp;amp;-&amp;amp; \varphi'(a)\varphi'(b)\\&#xA;|| &amp;amp; &amp;amp; || &amp;amp; &amp;amp; || &amp;amp; &amp;amp; || \\&#xA;|A \cup B| &amp;amp;=&amp;amp; |A| &amp;amp;+&amp;amp; |B| &amp;amp;-&amp;amp; |A \cap B|&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the proof follows from simple algebra. I just cannot find any way to construct such sets - since the identity is true I imagine they do exist but I am not be completely sure especially since I couldn't construct them.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-16T15:38:21.377" Title="Can this broken proof (of multiplicative of $\varphi$) attempt be fixed?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="6895" PostTypeId="2" ParentId="6868" CreationDate="2010-10-15T20:08:30.097" Score="1" ViewCount="0" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lazy_caterer%27s_sequence&quot; rel=&quot;nofollow&quot;&gt;lazy caterer's sequence&lt;/a&gt;. As others have mentioned, arbitrary dimensional analogues are called hyperplane arrangements.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2010-10-15T20:08:30.097" />
  <row Id="6896" PostTypeId="2" ParentId="6851" CreationDate="2010-10-15T20:19:28.447" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; When you look at the answer of Chandru1 you must think twice. Do you know any algebraic properties of the absolute value? How do you solve an inequality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also like to think of different points $(x,y)=(5,5)$, $(x,y)=(5,-5)$, $(x,y)=(-5,5)$, $(x,y)=(-5,-5)$ -- are they in you set? $(x,y)=(1/5,5)$, $(x,y)=(1/5,-5)$, $(x,y)=(-1/5,5)$, $(x,y)=(-1/5,-5)$ -- are they in you set?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-15T20:19:28.447" />
  <row Id="6897" PostTypeId="1" AcceptedAnswerId="6900" CreationDate="2010-10-15T20:26:48.423" Score="3" ViewCount="65" Body="&lt;h2&gt;The Setting&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be an &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedean_field&quot; rel=&quot;nofollow&quot;&gt;Archimedean field&lt;/a&gt;. TFAE:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$K$ has the least upper bound property.&lt;/li&gt;&#xA;&lt;li&gt;Every &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_sequence#In_topological_groups&quot; rel=&quot;nofollow&quot;&gt;Cauchy sequence&lt;/a&gt; in $K$'s additive group converges.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now proving that 1 implies 2 is easy, but the other direction is slightly harder. Not that that's a problem. Rather the problem is that I can't see a route that doesn't invoke at least &lt;a href=&quot;http://en.wikipedia.org/wiki/Dependent_choice&quot; rel=&quot;nofollow&quot;&gt;dependent choice&lt;/a&gt; at some point.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Strategy 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Starting with a nonempty set $A$ that's bounded above, you could construct a monotonely non-decreasing Cauchy sequence of upper bounds that has the supremum as its limit. Here's a short sketch: Pick an upper bound $B_0$ of $A$. Pick an $a_0 \in A$. Recursively define&#xA;$$ B_{i+1} = \begin{cases}&#xA;\frac{B_i+a_i}{2}, &amp;amp; \text{ if that's an upper bound for } A \\&#xA;B_i,                &amp;amp; \text{ otherwise}&#xA;\end{cases} $$&#xA;and&#xA;$$ a_{i+1} = \begin{cases}&#xA;a_i,  &amp;amp; \text{ if $\tfrac{a_i+B_i}{2}$ is an upper bound for $A$}\\&#xA;\text{choose any } a \in A \text{ s.t. } \frac{a_i+B_i}{2} &amp;lt; a,  &amp;amp; \text{ otherwise.}&#xA;\end{cases} $$&#xA;I can't see a way to get rid of the choice because you really want the $a_i$ to be in $A$ for the argument to go through.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Strategy 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Okay, let's go the long way instead! First we show that $K$ complete implies $[a,b]$ compact. Then we show that &lt;em&gt;that&lt;/em&gt; implies that closed and bounded subsets are compact (&quot;Heine-Borel property&quot;). And finally we show that not(Heine-Borel property) implies not(least upper bound property). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I already get stumped on the first part. Clearly it's easy to show that $K$ complete implies $[a,b]$ is sequentially compact. And from here it'd be nice to use that $K$ is 2nd countable (the intervals with rational endpoints are a basis) to get that $[a,b]$ is in fact compact. So you start with an open covering $U_\alpha$ of $[a,b]$. 2nd countable spaces are Lindelöf... wait... let's make sure and prove that. Let $\{B_i\}$ be a countable basis. Then for each $B_i$ you choose a $U_\alpha$... oh. Choice crept up again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: Does this really require (an admittedly weak form of) choice? Or is there a way to do without?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-15T20:53:31.847" Title="Archimedian field $K$ has LUB property iff it's complete requires DC?" Tags="&lt;real-analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6899" PostTypeId="1" AcceptedAnswerId="6904" CreationDate="2010-10-15T20:46:09.007" Score="2" ViewCount="193" Body="&lt;p&gt;Can we find all positive integers $a,b$ such that $a^{n}+b^{n}$ is an $(n+1)^{th}$ power? I think this question equivalent to solving the statement $$a^{n} + b^{n} = c^{n+1}$$ for $a,b,c \in \mathbb{N}$. But i don't know as to how i can solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted by subsituting $n$ as $n+1$ so that we get $$a^{n+1} + b^{n+1} = c^{n+2} = c \times c^{n+1}=(a^{n}+b^{n}) \cdot c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it seems that this doesn't help!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-16T17:31:37.543" LastActivityDate="2010-10-16T17:31:37.543" Title="Finding $a,b$ such that $a^{n} + b^{n} $ is $(n+1)^{th}$ power" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="6900" PostTypeId="2" ParentId="6897" CreationDate="2010-10-15T20:53:31.847" Score="3" ViewCount="0" Body="&lt;p&gt;It seems to me that you can modify strategy 1 as follows.  Given a nonempty set $A$ that's bounded above, let $x_0$ be a non-upper-bound for $A$ and let $y_0$ be an upper bound for $A$.  Recursively define&#xA;$$&#xA;y_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 &amp;amp; \text{if this is an upper bound for $A$} \\&#xA; y_n &amp;amp; \text{otherwise.}  \end{cases}&#xA;$$&#xA;and&#xA;$$&#xA;x_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 &amp;amp; \text{if this is a non-upper-bound for $A$} \\&#xA; x_n &amp;amp; \text{otherwise.}  \end{cases}&#xA;$$&#xA;Note that $x_1 \leq x_2 \leq \cdots \leq y_2 \leq y_1$. Furthermore, each $y_n$ is an upper bound for $A$, and each $x_n$ is a non-upper-bound for $A$.  It is easy to prove that both sequences are Cauchy sequences, and that they converge to the same limit $L$.  We claim that $L$ is the least upper bound for $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both directions are fairly easy.  If $a \in A$, then $a \leq y_n$ for all $n$, which proves that $a \leq L$.  Thus $L$ is in fact an upper bound for $A$.  Next, if $u$ is any upper bound for $A$, then $u \geq x_n$ for all $n$, and therefore $u \geq L$.  Thus $L$ is the least upper bound for $A$.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-15T20:53:31.847" CommentCount="1" />
  <row Id="6901" PostTypeId="2" ParentId="6893" CreationDate="2010-10-15T21:11:00.633" Score="2" ViewCount="0" Body="&lt;p&gt;We have to find the value of &lt;strong&gt;constant&lt;/strong&gt; (I assume, since nothing is specified on the contrary) payments $A$ given the principal $P$ during $n$ yearly periods and the interest rate $i$. The value of $A$ in the period $k$ is equivalent to the present value $A/\left( 1+i\right) ^{k}$ monetary units, where $i$ is the interest rate in each capitalization period. Summing in $k$, from 1 to $n$, we get the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{k=1}^{n}\dfrac{A}{\left( 1+i\right) ^{k}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a geometric progression with ratio $r=1/(1+i)$ and first term $u_{1}=A/\left( 1+i\right)$ whose sum is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{A}{1+i}\dfrac{\left( \dfrac{1}{1+i}\right)^{n}-1}{\dfrac{1}{1+i}-1}=A\dfrac{\left( 1+i\right) ^{n}-1}{i\left( 1+i\right) ^{n}}=P.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=P\dfrac{i\left( 1+i\right) ^{n}}{\left( 1+i\right) ^{n}-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given problem, the payments will be made during $n=6$ years, with $i=11\%=\dfrac{11}{100}$ and $P=8000$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=8000\times \dfrac{0.11(1.11)^{6}}{(1.11)^{6}-1}\approx 1891$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sum of a geometric progression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{n}\right) -\left(&#xA;ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-r)S=u_{1}-ru_{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\dfrac{u_{1}-ru_{n}}{1-r}=\dfrac{u_{1}-u_{1}r^{n}}{1-r}=u_1\times\dfrac{1-r^{n}}{1-r}=u_1\times\dfrac{r^{n}-1}{r-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T18:45:51.023" LastActivityDate="2010-10-16T18:45:51.023" CommentCount="3" />
  <row Id="6902" PostTypeId="2" ParentId="6899" CreationDate="2010-10-15T21:11:30.927" Score="1" ViewCount="0" Body="&lt;p&gt;A few to start you off:  for n=1 there are many solutions.  a=b=c=2 works for all n.  There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares&lt;/a&gt;.  You can apply that to c^3 to see if all factors of c which are 3 mod 4 come with even exponents you can solve it.  I suspect that is it as there just aren't that many powers to choose from, but don't have a proof.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-15T21:11:30.927" />
  <row Id="6903" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T21:19:06.357" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;b&gt;Geometry&lt;/b&gt;: The ticket to Plato's Academy. :-)&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-15T21:19:06.357" />
  <row Id="6904" PostTypeId="2" ParentId="6899" CreationDate="2010-10-15T21:30:49.183" Score="6" ViewCount="0" Body="&lt;p&gt;When $n = 2$ it's straightforward to give a more-or-less complete solution: $c^3$ is a sum of squares if and only if $c$ is, so write $c = d^2 + e^2$ and the rest is factorization over the Gaussian integers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $n$ there is the family of solutions $a = x(x^n + y^n), b = y(x^n + y^n), c = x^n + y^n$ for integers $x, y$.  But I think finding all solutions is likely to be incredibly hard.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-15T21:30:49.183" CommentCount="1" />
  <row Id="6905" PostTypeId="2" ParentId="6877" CreationDate="2010-10-15T21:41:44.977" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Statistics&lt;/strong&gt;: Making decisions under uncertainty&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-15T21:41:44.977" />
  <row Id="6906" PostTypeId="1" AcceptedAnswerId="7044" CreationDate="2010-10-15T21:47:23.527" Score="4" ViewCount="340" Body="&lt;p&gt;By &quot;Sylvester's Law of Inertia,&quot; I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guess that Sylvester is the name of the dude that proved this theorem, but I have no idea as to why it's called &quot;Law of Inertia.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is this name coming from? I guess it's from physics, but... I just don't see the connection.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-15T21:51:08.800" LastActivityDate="2010-10-17T17:53:28.147" Title="Why is it called Sylvester's Law of Inertia?" Tags="&lt;soft-question&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6907" PostTypeId="2" ParentId="6906" CreationDate="2010-10-15T21:59:38.250" Score="8" ViewCount="0" Body="&lt;p&gt;From Sylvester's &lt;em&gt;On the Theory of the Syzygetic Relations&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This constant number of positive signs which attaches to a quadratic function under all its transformations [...] may be termed conveniently its &lt;em&gt;inertia&lt;/em&gt;, until a better word is found.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-15T22:53:01.527" LastActivityDate="2010-10-15T22:53:01.527" CommentCount="5" />
  <row Id="6908" PostTypeId="1" AcceptedAnswerId="6925" CreationDate="2010-10-15T22:00:28.357" Score="7" ViewCount="168" Body="&lt;p&gt;This is a follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/6828/proving-sqrt1-x2-ge-texterf-sqrt-log-x&quot;&gt;this&lt;/a&gt; question. I came across a tighter potential bound and checked it numerically for $0\le x\le 5$. I think it holds for all positive $x$, can anyone see a proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: using analysis for previous &lt;a href=&quot;http://math.stackexchange.com/questions/6828/proving-sqrt1-x2-ge-texterf-sqrt-log-x&quot;&gt;question&lt;/a&gt; you can show that $1-\exp(-k x^2)$ is an upper bound on $\text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/\pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $\text{erf}(x)^2$ very closely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dashed graph below is $\text{erf}(x)^2$, red is $1-\exp(-k x^2)$ for $k=4/\pi$, other two graphs are for $k=1$ and $k=2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/bounds.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-10-16T00:04:42.547" LastActivityDate="2010-10-16T02:26:03.177" Title="Proving $1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$" Tags="&lt;inequality&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6909" PostTypeId="1" CreationDate="2010-10-15T22:57:35.767" Score="0" ViewCount="103" Body="&lt;p&gt;1.&lt;strong&gt;How many  number of three digit even numbers than can be formed out of the digits 0 to 9 ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question seems confusing since there is no mention of whether repetition is allowed or not ?! since if repetition is allowed then the answer would be &lt;strong&gt;450&lt;/strong&gt; (&lt;code&gt;5*9*10&lt;/code&gt;) where as if not then the answer would be &lt;strong&gt;328&lt;/strong&gt; (&lt;code&gt;9p2+ 4*(8*8) = 328&lt;/code&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct about the question or am missing some point ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.How many calendars that can be prepared for the month of February ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.A golf player wants to put the ball in the hole in 5 shots. He says that he will put the ball in the hole at most by 3 shots and will qualify for the next round. how many number of possibilities of shots he had played ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any clue for this two.&lt;/p&gt;&#xA;" OwnerUserId="2552" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T03:06:42.757" LastActivityDate="2010-10-16T03:06:42.757" Title="Three question on Permutation and combination" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6911" PostTypeId="2" ParentId="6909" CreationDate="2010-10-15T23:14:37.593" Score="1" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I get the same answers you do.  I also agree the question is unclear on repetition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Think of how many ways the pattern of days in February can be different, starting day of week and length&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't understand this question&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-16T00:07:37.430" LastActivityDate="2010-10-16T00:07:37.430" CommentCount="1" />
  <row Id="6913" PostTypeId="2" ParentId="6899" CreationDate="2010-10-15T23:46:35.703" Score="2" ViewCount="0" Body="&lt;p&gt;For $n \ge 3$, the Fermat-Catalan conjecture (http://en.wikipedia.org/wiki/Fermat–Catalan_conjecture) implies that there are only finitely many solutions to your equation. (But the Fermat-Catalan conjecture is just a conjecture!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: as pointed out by yrudoy, this doesn't actually follow. I still think there should be only finitely many solutions, though, based on some probabilistic considerations. See for example &lt;a href=&quot;http://www.thehcmr.org/issue1_1/elkies.pdf&quot; rel=&quot;nofollow&quot;&gt;Noam Elkies, The ABCs of Number Theory&lt;/a&gt;, in vol. 1, issue 1 of the &lt;a href=&quot;http://www.thehcmr.org/&quot; rel=&quot;nofollow&quot;&gt;Harvard College Mathematics Review&lt;/a&gt;, p. 61. (This gives a standard heuristic argument for Fermat's last theorem only having finitely many solutions for each $n \ge 4$, which adapts to this case.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2010-10-16T14:39:01.830" LastActivityDate="2010-10-16T14:39:01.830" CommentCount="1" />
  <row Id="6914" PostTypeId="1" CreationDate="2010-10-15T23:49:12.827" Score="6" ViewCount="254" Body="&lt;p&gt;I'm curious about the rights of ownership regarding problems presented in math textbooks. Can a math problem in general be considered intellectual property of its &quot;inventor&quot;? I'm assuming textual word problems can be &quot;owned&quot; in the same way other literature is by publishers/authors but it becomes more gray with purely numerical expressions. Is it merely a contextual ownership, for example, say no other math author can include such a problem in another textbook or is it more inclusive regarding other mediums like software or blogs etc where money is being made? An &quot;elucidation&quot; here would be much appreciated, thanks. &lt;/p&gt;&#xA;" OwnerUserId="2054" LastActivityDate="2010-10-16T05:30:33.967" Title="Can someone copyright/own a math problem?" Tags="&lt;books&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6916" PostTypeId="1" AcceptedAnswerId="6918" CreationDate="2010-10-16T00:59:14.553" Score="1" ViewCount="105" Body="&lt;p&gt;I've always been taught that one way to look at complex numbers is as a cartesian space, where the &quot;real&quot; part is the x component and the &quot;imaginary&quot; part is the y component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, these complex numbers are like vectors, and they can be added geometrically like normal vectors can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is there a geometric interpretation for the multiplication of two complex numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried out two test ones, $3+i$ and $-2+3i$, which multiply to $-9+7i$.  But no geometrical significance seems to be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a geometric significance for the multiplication of complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-10-16T01:22:26.853" Title="Geometric interpretation of the multiplication of complex numbers?" Tags="&lt;ring-theory&gt;&lt;complex-numbers&gt;&lt;vector-space&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6918" PostTypeId="2" ParentId="6916" CreationDate="2010-10-16T01:16:02.960" Score="6" ViewCount="0" Body="&lt;p&gt;Suppose we multiply the complex numbers $z_1$ and $z_2$. If these numbers are written in the polar form as $r_1 e^{i \theta_1}$ and $r_2 e^{i \theta_2}$, the product will be $r_1 r_2 e^{i (\theta_1 + \theta_2)}$. Equivalently, we are stretching the first complex number $z_1$ by a factor equal to the magnitude of the second complex number $z_2$ and then rotating the stretched $z_1$ counter-clockwise by an angle $\theta_2$ to arrive at the product. There are several websites that expand upon this intuition with graphics and more explanation. See this site for example - &lt;a href=&quot;http://www.suitcaseofdreams.net/Geometric_multiplication.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.suitcaseofdreams.net/Geometric_multiplication.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-16T01:16:02.960" CommentCount="1" />
  <row Id="6919" PostTypeId="2" ParentId="6916" CreationDate="2010-10-16T01:20:16.810" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, there is a simple geometric meaning, but you need to convert to the polar form of the complex numbers to see it clearly.  $3+i$ has magnitude $\sqrt{10}$ and angle about $18^\circ$;&#xA;$-2+3i$ has magnitude $\sqrt{13}$ and angle about $124^\circ$.&#xA;Multiplication of the complex numbers multiplies the two magnitudes, resulting in $\sqrt{130}$,&#xA;and adds the two angles, $142^\circ$.&#xA;In other words, you can view the second number as scaling and rotating the first (or the first scaling and rotating the second).&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-16T01:20:16.810" />
  <row Id="6920" PostTypeId="1" CreationDate="2010-10-16T01:21:09.657" Score="2" ViewCount="75" Body="&lt;p&gt;I'm having difficulty justifying the change of limits in the derivation of the Riemann-Liouville derivative at &lt;a href=&quot;http://www.xuru.org/fc/RiemannLiouville.asp&quot; rel=&quot;nofollow&quot;&gt;xuru.org&lt;/a&gt;. What I don't undestand is how $\int_0^{t_2}$ becomes $\int_{t_1}^x$ in the following statement,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^x \int_0^{t_2} f(t_1) dt_1 dt_2 = \int_0^x \int_{t_1}^x f(t_1) dt_2 dt_1$&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-10-16T02:18:32.623" LastActivityDate="2010-10-16T02:28:04.307" Title="Change of limits in derivation of Riemann-Liouville (Fractional) Derivative" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6921" PostTypeId="2" ParentId="6916" CreationDate="2010-10-16T01:22:26.853" Score="6" ViewCount="0" Body="&lt;p&gt;Add the angles and multiply the lengths.&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-10-16T01:22:26.853" />
  <row Id="6922" PostTypeId="2" ParentId="6920" CreationDate="2010-10-16T01:42:58.927" Score="5" ViewCount="0" Body="&lt;p&gt;Notice that $0\leq t_2\leq x$ and $0\leq t_1\leq t_2$ is equivalent to $0\leq t_1\leq x$ and $t_1\leq t_2\leq x$.  You could shorten the description to $0\leq t_1\leq t_2\leq x$.  Whether you believe me or not, I recommend that you draw a picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that both integrals are over the same triangular region.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-16T02:28:04.307" LastActivityDate="2010-10-16T02:28:04.307" CommentCount="1" />
  <row Id="6923" PostTypeId="1" AcceptedAnswerId="6929" CreationDate="2010-10-16T02:10:25.573" Score="2" ViewCount="77" Body="&lt;p&gt;What is, in the language of Schemes, a G-galois branched cover?&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditorDisplayName="" LastEditDate="2010-10-16T02:21:28.887" LastActivityDate="2010-10-17T13:35:29.860" Title="What is a G-Galois Branched Cover" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6924" PostTypeId="2" ParentId="6899" CreationDate="2010-10-16T02:10:29.363" Score="4" ViewCount="0" Body="&lt;p&gt;The hard case is relatively prime solutions. For $n &gt; 2$ it is conjectured that there are no solutions.  This follows from the more precise conjecture that there are no relatively prime solutions to $x^a + y^b = z^c$ in the hyperbolic case $1/a + 1/b + 1/c &amp;lt; 1$ other than the ten or so examples found by Beukers and Zagier in a computer search (there are large solutions with $(a,b,c)=(2,3,7)$ and a few others) shortly after Wiles announced the proof of Fermat's Last Theorem.   It was proven by Darmon and Granville that the number of relatively prime solutions is finite for any given $n &gt; 2$, and this required the Mordell conjecture (Faltings' theorem) on finiteness of rational points on curves of genus $g \geq 2$.  There is no known elementary approach to these questions, and resolution of the problem for particular values of $(a,b,c)$ uses the full machinery of modern algebraic geometry and number theory. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-16T14:19:38.887" LastActivityDate="2010-10-16T14:19:38.887" />
  <row Id="6925" PostTypeId="2" ParentId="6908" CreationDate="2010-10-16T02:20:52.313" Score="7" ViewCount="0" Body="&lt;p&gt;As before we consider &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$erf(x)^2={4\over \pi}\int_0^x\int_0^x \exp{-(s^2+t^2)}\  ds \ dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare this with the same over the area which is given by the quarter of a circle of radius $\displaystyle \frac{2x}{\sqrt{\pi}}$. The area of this is same as the area of the square of side $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle e^{-(s^2 + t^2)}$ decreases as $\displaystyle s^2 + t^2$ increases, we are done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $\displaystyle s^2 + t^2$ is higher in that region).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T02:26:03.177" LastActivityDate="2010-10-16T02:26:03.177" CommentCount="9" />
  <row Id="6926" PostTypeId="1" CreationDate="2010-10-16T02:53:42.427" Score="0" ViewCount="70" Body="&lt;p&gt;$r$ is a primitive root for the prime $p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1\equiv r^a \pmod{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_2\equiv r^b \pmod{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $gcd(b, p-1)=1$, how can I determine $r$ if $p$, $x_2$, and $b$ are known?&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T03:08:52.533" LastActivityDate="2010-10-16T03:51:35.970" Title="Find $r$ from $x_1\equiv r^a \pmod{p}, x_2\equiv r^b \pmod{p}$ and gcd$(b, p-1)=1$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6927" PostTypeId="2" ParentId="6926" CreationDate="2010-10-16T02:59:29.793" Score="4" ViewCount="0" Body="&lt;p&gt;HINT: If gcd$(b,p-1)=1$, then we can write $1 = tb + s(p-1)$. So $tb\equiv 1 \pmod{p-1}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-16T02:59:29.793" CommentCount="4" />
  <row Id="6928" PostTypeId="2" ParentId="6926" CreationDate="2010-10-16T03:33:58.087" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $The isomorphism $\rm n\to r^n\:$ from the additive group $\rm \mathbb Z/(p-1)$ to the multiplicative group $\rm \mathbb Z/p^*\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;maps $\rm\  b\to x,\ 2b\to x^2,\: \cdots\:,\: nb\to x^n$. You seek $\rm 1\to r\ $  so you need $\rm\ nb\equiv 1\ $ so $\rm\ n\equiv \:\ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-16T03:51:35.970" LastActivityDate="2010-10-16T03:51:35.970" CommentCount="2" />
  <row Id="6929" PostTypeId="2" ParentId="6923" CreationDate="2010-10-16T04:14:09.557" Score="4" ViewCount="0" Body="&lt;p&gt;I don't think this phrase has a single precise translation into the general language of schemes, but it will mean something like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a finite surjective morphism $X \to Y$ of integral schemes such that the corresponding extension of function fields $K(Y) \subset K(X)$ is Galois with Galois group $G$.&#xA;It would also be reasonable to require that $X$ and $Y$ be normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand what it means geometrically, one should imagine that $X$ and $Y$ are projective varieties over some field.  What this will mean then is that $X \to Y$ is surjection with finite fibres, that $G$ acts as a group of automorphisms of $X$ over $Y$,&#xA;and that if we remove the branch locus (i.e. the closed subset of $Y$ along&#xA;which the fibres contain points with multiplicity $&gt; 1$) then each fibre is acted on&#xA;faithfully and transitively by $G$ (so, away from the ramification locus, there&#xA;are $|G|$ sheets of $X$ over $Y$, which are permuted by $G$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example is given by (the projectivization of) the map $(x,y) \mapsto x$&#xA;from the elliptic curve $E$ defined by $y^2 = x^3 - x$ to $\mathbb P^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $G$ is cyclic of order two, acting by $(x,y) \mapsto (x,-y)$, and &#xA;the branch points are precisely the points where $y = 0$ (four of them; three&#xA;three finite points $(0,0), (\pm 1,0)$, and also one at infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, an important result is the Zariski--Nagata purity theorem, which says &#xA;(under mild hypotheses, e.g. that $Y$ and $X$ are smooth, or more generally,&#xA;that $Y$ is smooth and $X$ is normal) that the set of branch points has pure codimension one,&#xA;i.e. is a divisor. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-17T13:35:29.860" LastActivityDate="2010-10-17T13:35:29.860" CommentCount="2" />
  <row Id="6930" PostTypeId="2" ParentId="6861" CreationDate="2010-10-16T04:41:42.113" Score="2" ViewCount="0" Body="&lt;p&gt;Here is a proof of this fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal C$ be the category of (not necessarily irreducible) complex varieties.&#xA;Then $\mathcal C$ can be identified with the category of reduced finite type $\mathbb C$-schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal D$ be the category of all finite type $\mathbb C$-schemes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then obviously $\mathcal C$ is a full subcategory of $\mathcal D$, and the inclusion&#xA;$\mathcal C \subset \mathcal D$ has a right adjoint, namely passage to the underlying reduced subscheme.  General nonsense (i.e. an easy categorical argument) then shows that if $X\to S$ and $Y \to S$ in $\mathcal C$ are two morphisms, the fibre product in $\mathcal C$&#xA;can be computed by first computing the fibre product in the bigger category $\mathcal D$,&#xA;and then applying the right adjoint to the inclusion.  That is,&#xA;the fibre product in the category of varieties over $\mathcal C$ is equal to&#xA;the reduced subscheme of the fibre product in $\mathcal D$ (which coincides with&#xA;the fibre product in the category of all schemes, just because the fibre product&#xA;of morphisms of finite type $\mathbb C$-schemes is again finite type over $\mathbb C$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure of a reference.  Because the proof is easy when you have the right&#xA;categorical framework, it is the kind of thing that is well-known to experts but whose proof&#xA;is not necessarily written down explicitly.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-16T04:41:42.113" CommentCount="1" />
  <row Id="6931" PostTypeId="1" CreationDate="2010-10-16T04:49:51.737" Score="4" ViewCount="93" Body="&lt;p&gt;I understand that the only free groups that are abelian are 1 and Z, hence a difference between 'free abelian groups' and 'abelian free groups'.&lt;br&gt;&#xA;Can someone please tell me what are the solvable free groups?&lt;br&gt;&#xA;How can I construct 'free solvable groups'?&lt;br&gt;&#xA;What is known about free groups in other classes of groups like polycyclic, nilpotent ...?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2554" LastActivityDate="2010-10-16T13:38:11.163" Title="Free groups in some classes?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="6932" PostTypeId="2" ParentId="6931" CreationDate="2010-10-16T05:16:01.410" Score="4" ViewCount="0" Body="&lt;p&gt;I think that you might have to bound the derived length to get precisely the notion that you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notation: For any group $G$, let $G^{(i)}$ denote the $i$th iterated derived subgroup of $G$, so $G^{(0)} = G$ and $G^{(i+1)} = [G^{(i)},G^{(i)}]$).  A group is solvable if &#xA;$G^{(i)}$ is trivial for some $i$, and has derived length $d$ if $G^{(d)}$ is trivial,&#xA;but $G^{(i)}$ is non-trivial for $i &amp;lt; d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $F(X)$ be the free group on a non-empty set $X$ of generators, and consider $F(X)/F(X)^{(d)}$. On the one hand, this group is clearly solvable of derived length $d$. On the other hand, if $G$ is any solvable group of derived length $\leq d$, then giving a map of sets $X \to G$ will be equivalent to giving a map of groups $F(X)/F(X)^{(d)} \to G$.&#xA;Thus $F(X)/F(X)^{(d)}$ is the free object on $X$ in the category of solvable groups of&#xA;derived length $\leq d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can make a similar construction if you consider e.g. the category of nilpotent groups&#xA;of nilpotency class $\leq c$ (i.e. whose central series is of length $\leq c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to consider all solvable groups at once, then you can consider the projective limit $$\lim_{\leftarrow} F(X)/F(X)^{(d)},$$ the so-called &lt;em&gt;pro-solvable&lt;/em&gt; completion of $F(X)$.  Similarly, in the nilpotent setting, you can form the &lt;em&gt;pro-nilpotent&lt;/em&gt; completion of $F(X)$.  But now the &quot;free object&quot; is not literally in the category you want, but is instead is a projective limit of groups from that category.  (This is often okay, though; is you search for &quot;pro-nilpotent completion&quot;, you will see that pro-nilpotent completions of free groups come up a lot, e.g. in arithmetic geometry.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-16T05:16:01.410" CommentCount="0" />
  <row Id="6933" PostTypeId="2" ParentId="6914" CreationDate="2010-10-16T05:24:13.493" Score="3" ViewCount="0" Body="&lt;p&gt;I asked this very question to a copywriter some years ago and he said that you can not copyright protect a formula but he also said that there are copyright protected algorithms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, where is the borderline? I do not know, but I guess this has to be tested in each case. Here is a &lt;a href=&quot;http://imranontech.com/2006/12/04/are-algorithms-copyrightable/&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; on the question, and here is a link to &lt;a href=&quot;http://www.rsa.com/glossary/default.asp?id=1045&quot; rel=&quot;nofollow&quot;&gt;RSA&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a link about &lt;a href=&quot;http://www.math.columbia.edu/~woit/wordpress/?p=590&quot; rel=&quot;nofollow&quot;&gt;plagiarism&lt;/a&gt; in science.  &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-16T05:30:33.967" LastActivityDate="2010-10-16T05:30:33.967" CommentCount="5" />
  <row Id="6934" PostTypeId="2" ParentId="6931" CreationDate="2010-10-16T05:28:39.150" Score="5" ViewCount="0" Body="&lt;p&gt;There is no &quot;free solvable group&quot;, but there &lt;em&gt;are&lt;/em&gt; &quot;free solvable groups of length $n$&quot; for any $n$. There are also free nilpotent groups of class $n$, free Burnside groups of exponent $k$; all groups that are abelian-by-exponent $k$; and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: a &lt;strong&gt;variety of groups&lt;/strong&gt; is a class of groups that is closed under taking isomorphisms, subgroups, quotients, and arbitrary direct products (that is, if $G\in\mathcal{C}$ and $K\cong G$, then $K\in\mathcal{C}$; if $G\in\mathcal{C}$ and $H\lt G$, then $H\in \mathcal{C}$; if $G\in\mathcal{C}$ and $N\triangleleft G$, then $G/N\in\mathcal{C}$; and if $\{G_i\}$ is an arbitrary family of groups with $G_i\in\mathcal{C}$ for every $i$, then $\prod G_i\in\mathcal{C}$). Examples of varieties include &quot;all abelian groups&quot;, &quot;all groups of solvability length at most $n$&quot;; &quot;all nilpotent groups of class at most $c$&quot;; &quot;all groups such that every element is of exponent $k$&quot;; and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathcal{C}$ is a variety of groups, and $G$ is any group, then there is a smallest normal subgroup $\mathcal{C}(G)$ of $G$ such that $G/\mathcal{C}(G)\in\mathcal{C}$ (this normal subgroup is, in fact, not merely normal, but fully invariant).  For example, when $\mathcal{C}$ is the class of all abelian groups, $\mathcal{C}(G)$ is none other than the commutator subgroup of $G$. &lt;em&gt;Proof of the existence of $\mathcal{C}(G)$:&lt;/em&gt; Let $\{N_i\}_{i\in I}$ be the collection of all normal subgroups of $G$ such that $G/N_i\in\mathcal{C}$; the class is nonempty, since $G/G$ is always in any variety. Then consider the obvious map from $G$ to $\prod G/N_i$; the product must be in $\mathcal{C}$ (it is a product of groups in $\mathcal{C}$), so the image of $G$ is in $\mathcal{C}$; the kernel of this map is the group $\mathcal{C}(G)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $X$ is any set, and $F(X)$ is the free group on $X$, then there is a smallest subgroup $\mathcal{C}(F(X))$ such that $F(X)/\mathcal{C}(F(X))\in\mathcal{C}$. The resulting quotient can be seen to have the same universal property &lt;em&gt;relative to groups in $\mathcal{C}$&lt;/em&gt; as $F(X)$ does relative to all groups. So we say that this quotient is the &quot;relatively free $\mathcal{C}$-group on $X$.&quot;  We call the original group $F(X)$ the &quot;absolutely free group on $X$&quot; to distinguish it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Just in case, remember that if $\mathcal{D}$ is a category in which the objects are sets and the arrows are maps of sets, then given a set $X$, the &lt;strong&gt;free $\mathcal{D}$-object on $X$&lt;/strong&gt; is an object $F(X)$ in $\mathcal{D}$, together with a set-theoretic inclusion $i\colon X\to F(X)$, such that for every object $D\in\mathcal{D}$ and every set-theoretic map $f\colon X\to D$ there exists a unique $\mathcal{D}$-morphism $\varphi\colon F(X)\to D$ such that $f=\varphi\circ i$. For the &quot;absolutely free groups&quot;, the category is the category of all groups; for &quot;free abelian groups&quot;, the category is the category of abelian groups. For &quot;relatively free $\mathcal{C}$-group&quot;, the category is the category of all groups in $\mathcal{C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason there is no &quot;free solvable group&quot; is that the class of all solvable groups is &lt;strong&gt;not&lt;/strong&gt; a variety: it is not closed under arbitrary direct products. But if you put a bound on the solvability length then you &lt;em&gt;do&lt;/em&gt; get a variety, and you can construct the corresponding &quot;free solvable group of length $n$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition you can take the pro-$\mathcal{C}$ approach that Matt E mentions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in learning more about varieties, these lie at the intersection of Universal Algebra and Group Theory. I would recommend Hanna Neumann's book &lt;strong&gt;Varieties of Groups&lt;/strong&gt; (a bit old, but still the standard reference); also George Bergman's Universal Algebra notes (available &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-16T13:38:11.163" LastActivityDate="2010-10-16T13:38:11.163" />
  <row Id="6935" PostTypeId="1" AcceptedAnswerId="6937" CreationDate="2010-10-16T05:48:20.043" Score="1" ViewCount="43" Body="&lt;p&gt;I'm trying to understand the approach to calculating an answer to the following question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to do some processing on a list of n items.&#xA;I know that the total processing time takes log2(n) milliseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: What is the maximum number of items that can be processed in 10 seconds (i.e., 10,000 milliseconds) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the right general approach / technique? I should mention that I am a math neophyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that if I have a list of 1,000,000 items it will take 13.8 milliseconds, as log2(1000000) = 13.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's not clear is how to compute the maximum n that can be processed in 10,0000 milliseconds. I know it's a gigantic number, but I am not sure how to calculate it. If the number was small I could probably find it by trial and error, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) It's a giant number such that trial and error seems impractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Even if it was not so huge, I'd really like to understand how to compute it determinsitically rather than by experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much for any pointers.&lt;/p&gt;&#xA;" OwnerUserId="2558" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-16T06:25:25.380" LastActivityDate="2010-10-16T06:26:31.430" Title="Finding the maximum n items that can be processed in x time using a function that gives the time for some n." Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="6936" PostTypeId="2" ParentId="6935" CreationDate="2010-10-16T06:24:25.153" Score="0" ViewCount="0" Body="&lt;p&gt;When I read the headline I thought you were asking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;knapsack problem&lt;/a&gt;, but then I read your question and it turns out you are just asking for an inverse to binary logarithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem asks &quot;find $n$ such that $\log_2 n \geq 10^4$.&quot;&#xA;Note that $\log_2 x \geq y \Leftrightarrow 2^{\log_2 x} = x \geq 2^y$ because $x \mapsto 2^x$ and $\log_2$ are monotonely increasing. This should be enough for you to figure out the answer (and yes, it is huge).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-16T06:24:25.153" CommentCount="1" />
  <row Id="6937" PostTypeId="2" ParentId="6935" CreationDate="2010-10-16T06:26:31.430" Score="2" ViewCount="0" Body="&lt;p&gt;If I understand your question you want to solve an equation of the form: $\log_{2}(x)=100,000$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation#Real_powers&quot; rel=&quot;nofollow&quot;&gt;exponentiation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case we want to do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{\log_{2}(x)}=2^{100,000}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x=2^{100,000}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is quite a large number as you correctly predicted :)&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-16T06:26:31.430" CommentCount="7" />
  <row Id="6938" PostTypeId="1" CreationDate="2010-10-16T08:12:02.700" Score="3" ViewCount="280" Body="&lt;p&gt;Okay, I have this not so pretty 2nd order non-linear ODE I should be able to solve numerically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f''(R) + \frac{2}{R} f'(R)=\frac{0.7}{R} \left( \frac{1}{\sqrt{f(R)}} - \frac{0.3}{\sqrt{1-f(R)}} \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(1)=1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function around the origin is behaving very wildly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of breaking this guy up into a system of two first order ODE's and then solve, but I have no idea how to set this up. What method should I use to set up the system of ODE's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is some other method rather than numerically solving a system of differential equations, please feel welcome to share. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/32jhJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditorDisplayName="" LastEditDate="2010-10-19T15:31:56.740" LastActivityDate="2010-10-19T15:31:56.740" Title="Numerically Solving a Second Order Nonlinear ODE" Tags="&lt;analysis&gt;&lt;differential-equations&gt;&lt;math-software&gt;&lt;numerical-methods&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" />
  <row Id="6939" PostTypeId="1" AcceptedAnswerId="6946" CreationDate="2010-10-16T08:42:17.640" Score="9" ViewCount="317" Body="&lt;p&gt;In the course of playing around with $\sum_{n=1}^{\infty} \frac{1}{\sqrt{1+n^{n}}}$, I used w|α to obtain the power series for $f(x)=\frac{1}{\sqrt{1+x^{x}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{1+x^{x}}} = \frac{1}{\sqrt{2}} - \frac{x\log(x)}{4\sqrt{2}} -\frac{x^{2}\log^{2}(x)}{32\sqrt{2}}+ \frac{5x^{3}\log^{3}(x)}{384\sqrt{2}} + \frac{17x^{4}\log^{4}(x)}{6144\sqrt{2}} - \frac{121x^{5}\log^{5}(x)}{122880\sqrt{2}} - \frac{721x^{6}\log^{6}(x)}{2949120\sqrt{2}} \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I realized that I couldn't really use this to help me with the sum, I found that the denominators (ignoring the $\sqrt{2}$, because all of them have it in common) correspond to &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A047053&quot; rel=&quot;nofollow&quot;&gt;$4^{n}n!$&lt;/a&gt;, what is baffling is that the numerators appear to correspond to the coefficients in the exponential generating function for &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A012259&quot; rel=&quot;nofollow&quot;&gt;$f(x)=e^{\tanh^{-1}(\tan(x))}$&lt;/a&gt; (I believe that the 7th entry should be 1369 and not 6845), and I'm curious what the explanation is, because $f(x)=e^{\tanh^{-1}(\tan(x))}$ is a mighty weird looking function.&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="890" LastEditorDisplayName="" LastEditDate="2010-10-16T20:41:09.013" LastActivityDate="2010-10-18T14:34:06.757" Title="mystery regarding power series of $\frac{1}{\sqrt{1+x^{x}}}$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="6940" PostTypeId="2" ParentId="6938" CreationDate="2010-10-16T09:48:25.527" Score="1" ViewCount="0" Body="&lt;p&gt;The general method to reinterpret a higher-order ODE as a system of first order ODEs is to regard the derivatives of the unknown function as additional unknown functions. In your case, regard $f'$ as a new function $g$. Then the system of first order ODEs consists of two equations, the first being the original equation with $f'$ replaced by $g$ and $f''$ replaced by $g'$, and the second equation being $f'=g$.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-16T09:48:25.527" CommentCount="3" />
  <row Id="6941" PostTypeId="1" AcceptedAnswerId="7113" CreationDate="2010-10-16T10:48:00.313" Score="1" ViewCount="53" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a 2D quadratic Bézier curve which, by construction, is a f(x) function : no loops, a single solution for each defined x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a common mean to convert this curve to a 3rd degree polynomial ? 3 should be enough since there can only be two &quot;bumps&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2561" LastActivityDate="2010-10-18T11:50:51.630" Title="bézier to f(x) polynomial function" Tags="&lt;polynomials&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6942" PostTypeId="1" CreationDate="2010-10-16T11:52:03.363" Score="5" ViewCount="229" Body="&lt;p&gt;The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is the puzzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QGBoB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangle ABC is known to be isosceles, that is, AC=BC. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled &quot;?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients &gt; 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:45:40.083" LastActivityDate="2010-10-16T17:45:40.083" Title="Finding an angle within an 80-80-20 isosceles triangle." Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="6943" PostTypeId="2" ParentId="6851" CreationDate="2010-10-16T12:09:18.733" Score="1" ViewCount="0" Body="&lt;p&gt;My suggestion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Start with the definition of absolute value $\left\vert z\right\vert $, with $z=xy,x,y\in\mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\vert xy\right\vert =xy\qquad \text{if}\ xy\geq 0\iff \left( x\geq 0\wedge y\geq 0\right) \vee \left( x\leq 0\wedge y\leq 0\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\vert xy\right\vert =-xy\qquad \text{if}\ xy&amp;lt;0\iff \left( x\geq 0\wedge y\leq 0\right) \vee \left( x\leq 0\wedge y\geq 0\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) Add the condition  $\left\vert xy\right\vert \leq 2\iff -2\le xy\leq 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) Observe that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $xy\gt 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the 1st and 3rd quadrants; and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)  $xy\lt 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the 2nd and 4th quadrants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Why?)&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T14:26:51.683" LastActivityDate="2010-10-16T14:26:51.683" CommentCount="3" />
  <row Id="6944" PostTypeId="2" ParentId="6890" CreationDate="2010-10-16T12:15:02.180" Score="5" ViewCount="0" Body="&lt;p&gt;Okay, I'm still not very sure about the fact that there's no distance function in the integrand, but if you assume it is somehow absorbed in the definition of $db_m(b)$, you can make some sort of sense out of this expression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you should recall that the classical electromagnetic potentials and their relations to the position and motion of a point particle. The electric potential generated by a charge density $\rho(x)$ is usually written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_0(x) = V(x) = C_1 \int \frac{\rho(y)}{|x - y|} dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the magnetic vector potential generated by a current is usually written as (Biot-Savart law form)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{A}(x) = C_2 \int \frac{\vec{J}(y)}{|x-y|} dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_1,C_2$ are constants which depends on the unit chosen. Then in this case, in units where the speed of light $c = 1$,  $(\rho,\vec{J})$ is a four-vector representing the movement of the particles. (Recall that an object with velocity $\vec{v}$ is traversing $\vec{v}$ spatial units in 1 time unit, so the four-vector corresponding to the tangent vector of its world line will be $(1,\vec{v})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you also realize that the signal (electromagnetic potential) should propagate at the speed of light. So while the above classical notion is valid for small object velocities, they are only ultimately valid in the electrostatics/magnetostatics cases, in which case there is a preferred time direction (the one in which everything stays constant) and we don't have to worry about this propagation); in other words, in those cases the particles on the past light cone can be transported forward using the stationarity to particles on your instantaneous surface and the computations can be done there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By direct analogy then, the proper way to compute the electromagnetic vector four potential is to set $j$ the four-current of your charge distribution (so that conservation of charge demands that the space-time divergence of $j$ vanishes, and we also write $\hat{j}$ for the unit time vector in the $j$ direction) is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_\nu(x) = k\int \frac{j_\nu(y)}{(x-y)_\mu \hat{j}^\mu} \delta((x-y)_\mu(x-y)^\mu) dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x,y$ are space-time events. ($k$ is just some normalizing constant, usually set to $\frac{1}{4\pi\epsilon_0}$). Basically the idea is not to compute $A$'s dependence on $j$, but rather $j$'s contribution to $A$. The influence of $j$ originating from space-time event $y$ will propagate out along the light cone (the $\delta$ function term). In the inertial frame corresponding to the instantaneous velocity $\hat{j}(y)$, the propagation is uniform in directions (because instantaneous at $y$, the distribution is at rest), and decays as $1/r$ where $r$ is the spatial distance, in $\hat{j}(y)$'s rest frame. A direct computation tells you that $r = (x-y)_\mu \hat{j}^\mu$. So the integrand tells you the infinitesmial contribution by $j(y)$ at space-time event $x$ to the four-potential. Then you just add them all up. If you look carefully at the integrand, you'd notice that it is the convolution of the Green's function for the linear wave equation against $j$! So in fact from this expression you recover the differential form of the law for the four-potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla_\mu\nabla^\mu A = 4\pi k j $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is just equation (25.22) in volume 2 of Feynman's lectures. If you just take the formula above, and restrict $j$ to be the tangent vector to the world line of a charged particle, you will recover the equation you wrote down, if you assume $dB_m(b)$ is given by $dy$ times the fraction that appears in the formula above. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-16T12:15:02.180" CommentCount="6" />
  <row Id="6945" PostTypeId="2" ParentId="6942" CreationDate="2010-10-16T12:20:45.313" Score="7" ViewCount="0" Body="&lt;p&gt;The solutions are not as trivial as one would expect from the statement. It's called Langley's problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf&quot; rel=&quot;nofollow&quot;&gt;http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-16T12:20:45.313" CommentCount="2" />
  <row Id="6946" PostTypeId="2" ParentId="6939" CreationDate="2010-10-16T13:14:52.330" Score="15" ViewCount="0" Body="&lt;p&gt;Well,&#xA;$$\frac1{\sqrt{1+x^x}}=\frac1{\sqrt{1+e^y}}$$&#xA;where $y=x\log x$ so it's unsurprising you get a series in terms of $y=x\log x$.&#xA;Then&#xA;$$\frac1{\sqrt{1+e^y}}=\frac1{\sqrt2}\frac1{\sqrt{1+u}}$$&#xA;where $u=\frac12(e^y-1)=o(y)$. This explains the fact that the coefficients&#xA;are rationals over $\sqrt2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $f(x)=\exp(\tanh^{-1}(\tan x))$. Now&#xA;$$\tanh z=\frac{e^{2z}-1}{e^{2z}+1}$$&#xA;so that&#xA;$$e^z=\sqrt{\frac{1+\tanh z}{1-\tanh z}}.$$&#xA;Hence,&#xA;$$\exp(\tanh^{-1} t)=\sqrt{\frac{1+t}{1-t}}.$$&#xA;Putting $t=\tan x$ gives&#xA;$$f(x)=\sqrt{\frac{1+\tan x}{1-\tan x}}=\sum_{n=0}^\infty a_n x^n.$$&#xA;Then&#xA;\begin{eqnarray*}&#xA;&amp;amp;&amp;amp;(1+i)f(ix)+(1-i)f(-ix)\\&#xA;&amp;amp;=&amp;amp;&#xA;2\sum_{m=0}^\infty&#xA;(a_{4m} x^{4m}-a_{4m+1} x^{4m+1}-a_{4m+2} x^{4m+2}+a_{4m+3} x^{4m+3}).&#xA;\end{eqnarray*}&#xA;But&#xA;$$f(ix)=\sqrt{\frac{1+i\tanh x}{1-i\tanh x}}&#xA;=\frac{1+i\tanh x}{\sqrt{1+\tanh^2x}}$$&#xA;and so&#xA;\begin{eqnarray*}&#xA;&amp;amp;&amp;amp;(1+i)f(ix)+(1-i)f(-ix)\\&#xA;&amp;amp;=&amp;amp;&#xA;\frac{2-2\tanh x}{\sqrt{1+\tanh^2 x}}&#xA;=\frac{2\cosh x-2\sinh x}{\sqrt{\cosh^2x+\sinh^2x}}\\&#xA;&amp;amp;=&amp;amp;\frac{2e^{-x}}{\sqrt{(e^{2x}+e^{-2x})/2}}=\frac{2\sqrt2}{\sqrt{1+e^{4x}}}&#xA;\end{eqnarray*}&#xA;which explains why the coefficients in the two series are the same&#xA;up to signs and powers of 4.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-16T13:14:52.330" CommentCount="1" />
  <row Id="6947" PostTypeId="1" CreationDate="2010-10-16T13:39:02.690" Score="2" ViewCount="151" Body="&lt;p&gt;A smooth surface $S$ embedded in $\mathbb{R}^3$ whose metric is inherited from $\mathbb{R}^3$&#xA;(i.e., distance measured by shortest paths on $S$) is a Riemannian 2-manifold: differentiable&#xA;because smooth and with the metric just described.  Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; Are such surfaces a subset of all Riemannian 2-manifolds? Are there Riemannian 2-manifolds&#xA;that are not &quot;realized&quot; by any surface embedded in $\mathbb{R}^3$?  I assume _Yes_.&lt;/li&gt;&#xA;&lt;li&gt; If so, is there any characterization of which Riemannian 2-manifolds are realized by&#xA;such surfaces? In the absence of a characterization, examples would help.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit.&lt;/b&gt; In light of the useful responses, a sharpening of my question occurs to me:&#xA;3. Is the &lt;em&gt;only&lt;/em&gt; impediment embedding vs. immersion? Is every Riemannian 2-manifold realized by&#xA;a surface &lt;em&gt;immersed&lt;/em&gt; in $\mathbb{R}^3$?&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-10-16T16:30:09.707" LastActivityDate="2010-10-16T21:39:19.510" Title="Riemannian 2-manifolds not realized by surfaces in $\mathbb{R}^3$?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="6948" PostTypeId="2" ParentId="6629" CreationDate="2010-10-16T13:50:58.183" Score="0" ViewCount="0" Body="&lt;p&gt;I've thought some more about your problem.  The recursion in your dynamical system yields the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_t = (1- \gamma)^t V_0 + \gamma \sum_{k=0}^{t-1} (1 - \gamma)^{t-1-k} \delta^{S_k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have a sum of discounted random variables.  If they were not discounted the distribution would be approximately normal for large $t$.  It turns out, though, that there is an analogous result for the distribution of a sum of discounted random variables -- a discounted central limit theorem, as it were.  You might have some success using this discounted central limit theorem as an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the specifics.  Let $Y_v = \sum_{k=0}^{\infty} v^k X_k$, where the $X_k$'s are independent and have a common distribution function.  Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_v = \frac{\sqrt{1-v}}{\sigma} \left(Y_v - \frac{\mu}{1-v}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $Z_v$ is asymptotically normal with mean 0 and variance $\frac{1}{1+v}$, for $v \to 1$, where $\mu$ and $\sigma$ are the common mean and standard deviation of the $X_k$'s.  (The expected value of $|X_k - \mu|^3$ must exist, too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there is a bound on the error in this approximation.  Let $F_v(x)$ and $N_v(x)$ be the cdf's of the $Z_v$ random variable and of a normal with mean 0 and variance $\frac{1}{1+v}$, respectively.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|F_v(x) - N_v(x)| \leq \frac{C \rho \sqrt{1-v}}{\sigma^3},$$&#xA;where $\rho = E[|X_k - \mu|^3]$ and $C = 5.4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference is Hans Gerber, &quot;The Discounted Central Limit Theorem and Its Berry-Esseen Analogue,&quot; &lt;em&gt;Annals of Mathematical Statistics&lt;/em&gt; 42(1), pp. 389-392, 1971.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-16T13:50:58.183" />
  <row Id="6949" PostTypeId="2" ParentId="6947" CreationDate="2010-10-16T14:14:43.590" Score="4" ViewCount="0" Body="&lt;p&gt;The hyperbolic plane cannot be smoothly isometrically embedded in $\mathbb{R}^3$. It can be so in $\mathbb{R}^5$. It is open (as far as I know) if it can be embedded in $\mathbb{R}^4$. I believe this is mentioned in Do Carmo's book on curves and surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Not a complete characterization, but Amsler has shown (see below for reference) that any Riemannian surface with constant negative curvature, if attempted to be imbedded in $\mathbb{R}^3$, must have singularities. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Amsler, M.H., Des surfaces a courbure negative constante dans l'espace a trois dimensions et de leurs singularites, Math. Ann. 130, 1955, 234-256&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-16T14:20:06.060" LastActivityDate="2010-10-16T14:20:06.060" CommentCount="2" />
  <row Id="6950" PostTypeId="1" CreationDate="2010-10-16T14:29:32.303" Score="1" ViewCount="137" Body="&lt;p&gt;How to prove that for all positive a,b,c this formula holds true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} &gt; \sqrt{a+b+c}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be invaluable.&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:51:44.213" LastActivityDate="2010-10-16T17:51:44.213" Title="$\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} &gt; \sqrt{a+b+c}$ is true for positive a,b,c" Tags="&lt;algebra-precalculus&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6951" PostTypeId="1" CreationDate="2010-10-16T14:49:45.137" Score="1" ViewCount="208" Body="&lt;p&gt;Are there any odd positive numbers that satisfy the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^2 - b^3 = 4$&#xA;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am certain that there are none but can't prove it. How would you prove that?&lt;/p&gt;&#xA;" OwnerUserId="2562" LastEditorUserId="2562" LastEditorDisplayName="" LastEditDate="2010-10-18T13:28:23.330" LastActivityDate="2010-10-18T13:28:23.330" Title="Odd positive integers that satisfy $a^2 - b^3 = 4$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="6952" PostTypeId="2" ParentId="6951" CreationDate="2010-10-16T14:53:48.037" Score="8" ViewCount="0" Body="&lt;p&gt;Start by rewriting it as $a^2-4=b^3$, and do what comes naturally.&lt;/p&gt;&#xA;" OwnerUserId="625" LastActivityDate="2010-10-16T14:53:48.037" CommentCount="14" />
  <row Id="6954" PostTypeId="2" ParentId="6950" CreationDate="2010-10-16T15:02:05.303" Score="3" ViewCount="0" Body="&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a+b+c=\sqrt{(a+b+c)^2}=\sqrt{a+b+c}\ \sqrt{a+b+c}\iff \dfrac{a+b+c}{\sqrt{a+b+c}}=\sqrt{a+b+c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{\sqrt{a+b}}&gt;\dfrac{a}{\sqrt{a+b+c}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{b}{\sqrt{b+c}}&gt;\dfrac{b}{\sqrt{a+b+c}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{c}{\sqrt{c+a}}&gt;\dfrac{c}{\sqrt{a+b+c}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{\sqrt{a+b}}+\dfrac{b}{\sqrt{b+c}}+\dfrac{c}{\sqrt{c+a}}&gt;\dfrac{a+b+c}{%&#xA;\sqrt{a+b+c}}=\sqrt{a+b+c}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-10-16T15:02:05.303" CommentCount="2" />
  <row Id="6956" PostTypeId="2" ParentId="6942" CreationDate="2010-10-16T15:32:33.430" Score="3" ViewCount="0" Body="&lt;p&gt;Here is the solution I had for this (I had the writeup lying around in an old email I had sent regarding this, so, no tex, also, A and B are interchanged):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9rhDL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that an 80-80-20 triangle is nothing but a part of the triangulation of an 18 sided regular polygon, whose 6 triangles are shown in the the bottom part of the circle above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider an 80-80-20 triangle (ABC on top part of figure) and shoot a light ray from one of the base vertices  (B in the figure) at an angle of 50 degrees to the base (or 30 from one of the equal sides). (See the triangle in the top part of the circle and the red arrows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can show that the ray will reflect twice (first at D then E)  and strike at a 90 degree angle the third time (F) i.e. after  5 refections, the light ray will return back to the vertex! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reflection process can be pictured in another way, by reflecting the triangle each time instead of reflecting the ray (see the red arrows in the 6 triangles in the bottom part)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the point of the third reflection F (i.e. the 90 degree incidence point) is exactly the midpoint of the side on which the ray is incident on. This can be seen by considering the bottom part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the right most B and corresponding triangle CBF. This is a 90-60-30 triangle. Thus CF is half CB which is half CA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Back to triangle at top) i.e. F is midpoint of AC. Thus triangle ACE is isosceles, thus angle CAE = 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we see that angle BDE must be the angle x in the problem, which must be 180-(50+50) = 80. (as DE is BD reflected on AC).  That angle y is 30, follows…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: To get more context about what x and y are, this was the figure when the problem was shown to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2bfkT.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T15:40:54.427" LastActivityDate="2010-10-16T15:40:54.427" />
  <row Id="6957" PostTypeId="2" ParentId="6894" CreationDate="2010-10-16T15:38:21.377" Score="4" ViewCount="0" Body="&lt;p&gt;Oh, I see.  In $\mathbb{Z}/(ab)\mathbb{Z}$ let $A$ be the set of elements not relatively prime to $a$ and let $B$ be the set of elements not relatively prime to $b$.  From here what you want follows by the Chinese Remainder Theorem.  (Of course, you could just use CRT from the get-go.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T15:38:21.377" />
  <row Id="6958" PostTypeId="1" AcceptedAnswerId="6960" CreationDate="2010-10-16T15:39:08.657" Score="3" ViewCount="138" Body="&lt;p&gt;How does the factoring of polynomials over Galois fields work? I cannot seem to understand the basic concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example:&lt;/strong&gt; How do I factorize x&lt;sup&gt;6&lt;/sup&gt; - 1 over GF(3)? I know that the result is (x+1)&lt;sup&gt;3&lt;/sup&gt; (x+2)&lt;sup&gt;3&lt;/sup&gt;, but I'm unable to compute it myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've studied the articles on Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_field&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Galois_field&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Factorization_of_polynomials&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Factorization_of_polynomials&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but I found it very difficult to understand. Is there some algorithm that would help me factorize polynomials like x&lt;sup&gt;n&lt;/sup&gt; - 1 over GF(k)?&lt;/p&gt;&#xA;" OwnerUserId="2411" LastActivityDate="2010-10-16T15:50:02.347" Title="How do I factorize polynomial over Galois field?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6959" PostTypeId="2" ParentId="6958" CreationDate="2010-10-16T15:42:23.690" Score="3" ViewCount="0" Body="&lt;p&gt;Factoring the polynomials $x^n - 1$ is very different from factoring general polynomials, since you already know what the roots are; they are, in a suitably large finite extension, precisely the elements of order dividing $n$.  Since you know that the multiplicative group of a finite field is cyclic, the conclusion follows from here.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T15:42:23.690" />
  <row Id="6960" PostTypeId="2" ParentId="6958" CreationDate="2010-10-16T15:50:02.347" Score="5" ViewCount="0" Body="&lt;p&gt;If $n=pm$ is a multiple of $p$ then over $GF(p)$ one has&#xA;$$x^n-1=(x^m-1)^p$$&#xA;so the problem reduces swiftly to the case where $n$ is coprime to $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is not a factor of $n$ then over an algebraic closure of&#xA;$GF(p)$&#xA;$$x^n-1=\prod_{k=0}^{n-1}(x-\zeta^j)$$&#xA;where $\zeta$ is a primitive $n$-th root of unity.&#xA;One makes this into a factorization over $GF(p)$ by combining conjugate&#xA;factors together. For each $k$, the polynomial&#xA;$$(x-\zeta^k)(x-\zeta^{pk})(x-\zeta^{p^2k})\cdots(x-\zeta^{p^{r-1}k})$$&#xA;has coefficients in, and is irreducible over, $GF(p)$ where $r$&#xA;is the least positive integer with $p^r k\equiv k$ (mod $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, it's easy to work out the degrees of the irreducible&#xA;factors of $x^n-1$, but to find the factors themselves needs a bit more&#xA;work, using for instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Berlekamp%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;Berlekamp's algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-16T15:50:02.347" CommentCount="1" />
  <row Id="6961" PostTypeId="2" ParentId="6947" CreationDate="2010-10-16T17:12:35.663" Score="3" ViewCount="0" Body="&lt;p&gt;The Whitney embedding theorem says you can always embed a smooth $n$-manifold in $\mathbb{R}^{2n}$, and immerse it in $\mathbb{R}^{2n-1}$.  Nonorientable Riemann surfaces, for example, don't embed in $\mathbb{R}^3$, but there are some pretty good immersions (the typical picture of the Klein bottle is a good example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a Riemannian manifold, Nash and Kuiper proved that there's a $C^1$ globally isometric embedding into $\mathbb{R}^{2n+1}$ (and, in fact, that you can arbitrarily closely approximate any metric $C^\infty$ embedding into at least $\mathbb{R}^{n+1}$ by a global isometric $C^1$ embedding).  For a global isometric $C^\infty$ embedding, it looks like the current lower bound is max$(n(n+1)/2+2n,n(n+1)/2+n+5)$.  For a local one, you can do it into $n(n+1)/2+n$-space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that for a globally isometric and analytic embedding of a surface, you might have to go up to $\mathbb{R}^{10}$.  Ew.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-10-16T21:39:19.510" LastActivityDate="2010-10-16T21:39:19.510" CommentCount="1" />
  <row Id="6962" PostTypeId="1" CreationDate="2010-10-16T17:15:39.810" Score="0" ViewCount="66" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/6282/minimizing-gender-regularity-in-a-linear-arrangement-of-boys-and-girls&quot;&gt;Minimizing Gender Regularity in a linear arrangement of boys and girls.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;There are G girl students and B boy students in a class that is about to graduate. You&#xA;need to arrange them in a single row for the graduation. To give a better impression of&#xA;diversity, you want to avoid having too many girls or too many boys seating consecutively.&#xA;You decided to arrange the students in order to minimize the gender regularity. The&#xA;gender regularity of an arrangement is the maximum number of students of the same&#xA;gender (all girls or all boys) that appear consecutively.&#xA;Given G and B, calculate the minimum gender regularity among all possible arrange-&#xA;ments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example&#xA; g=?,b=?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- 10 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- 5 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- 0 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;result &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i want a formula to solve this problem&lt;/p&gt;&#xA;" OwnerUserId="2564" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T17:48:37.073" LastActivityDate="2010-10-16T17:48:37.073" ClosedDate="2010-10-16T19:16:51.747" Title="Minimizing Gender regularity" Tags="&lt;combinatorics&gt;" CommentCount="0" />
  <row Id="6963" PostTypeId="2" ParentId="6939" CreationDate="2010-10-16T17:20:09.453" Score="1" ViewCount="0" Body="&lt;p&gt;It's simple: $\rm\quad\quad  f(x)\ =\  \cos x - \sin x \ =\ \frac{1+i}2 e^{ix} + \frac{1-i}2 e^{-ix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle \Rightarrow\quad\quad\quad\quad\ f(\tan^{-1} x)\ =\ \frac{1-x}{\sqrt{x^2+1}}\quad $ via $\rm\displaystyle\quad e^{\:i\:tan^{-1} x}\ =\ \frac{1+ x\: i}{\sqrt{x^2 + 1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle \Rightarrow\quad f(\tan^{-1}\tanh x)\ =\ \ \frac{\sqrt 2}{e^{4x}+1}\quad\ $ via $\rm\displaystyle\quad\ \tanh{x}\ =\ 1 - \frac{2}{e^{2x}+1}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T14:34:06.757" LastActivityDate="2010-10-18T14:34:06.757" CommentCount="1" />
  <row Id="6964" PostTypeId="1" AcceptedAnswerId="7150" CreationDate="2010-10-16T18:06:31.270" Score="2" ViewCount="89" Body="&lt;p&gt;Let $T_\varphi$ be an analytic Toeplitz operator (meaning $\varphi \in H^\infty$). Further let $K$ be a compact operator that commutes with $T_\varphi$. Now I want to show that the spectrum of $K$ only consists of $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying the following things: $T_\varphi K$ is compact so we can apply the spectral theorem for compact operators. So we know that at least $0$ is in the spectrum of this operator (not of $K$) and that the spectrum is countable with the only limit point $0$. Can I somehow use this? I also know that if $C$ is a compact operator $\|T_\varphi - C\| \geq \|T_\varphi \|$ so $\|T_\varphi(1 - K)\| = \|(1 - K)T_\varphi\| \geq \|T_\varphi \|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions how I can continue?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-18T20:41:05.187" Title="Toeplitz operator commuting with a compact operator." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6965" PostTypeId="1" AcceptedAnswerId="6990" CreationDate="2010-10-16T18:33:41.290" Score="2" ViewCount="172" Body="&lt;p&gt;$$u + v = 2$$&#xA; $$ux + vy = 1$$&#xA; $$ux^2 + vy^2 = - 1$$&#xA; $$ux^3 + vy^3 = -5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a hint for solving this ? I'm kinda stuck. &lt;/p&gt;&#xA;" OwnerUserId="2227" LastActivityDate="2010-10-17T21:37:27.267" Title="system of equations" Tags="&lt;polynomials&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="6966" PostTypeId="2" ParentId="83" CreationDate="2010-10-16T18:45:23.647" Score="2" ViewCount="0" Body="&lt;p&gt;Some of my favorite ways of engaging students from jump street are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Chinese magic square&lt;/li&gt;&#xA;&lt;li&gt;Erostophenes’ way of calculating the circumference of the Earth&lt;/li&gt;&#xA;&lt;li&gt;the “Why are manhole covers always round?” question&lt;/li&gt;&#xA;&lt;li&gt;the Twin-Sisters paradox&lt;/li&gt;&#xA;&lt;li&gt;Morley’s Miracle&lt;/li&gt;&#xA;&lt;li&gt;Pythagorean Triples&lt;/li&gt;&#xA;&lt;li&gt;Euler’s formula for connected planar graphs&lt;/li&gt;&#xA;&lt;li&gt;the reason for the dimensions of A4 paper&lt;/li&gt;&#xA;&lt;li&gt;Kaprekar’s Constant&lt;/li&gt;&#xA;&lt;li&gt;Casting out nines&lt;/li&gt;&#xA;&lt;li&gt;the “Which part of a train is always moving backwards?” question&lt;/li&gt;&#xA;&lt;li&gt;the question about how far the bee flying between two trains goes&lt;/li&gt;&#xA;&lt;li&gt;the red-herring question about “...met a man with seven wives...”&lt;/li&gt;&#xA;&lt;li&gt;the red-herring question about the trains leaving Chicago and Dallas (“When they meet, which one is closer to Dallas?”)&lt;/li&gt;&#xA;&lt;li&gt;the red-herring question about the rowing team that dropped a bottle into the river and row back downstream to retrieve it&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that the curriculum need not be improved or reformed in order to drop one or two of these in at an odd moment or two during class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;- Mike Jones&lt;/p&gt;&#xA;" OwnerUserId="2344" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-16T20:14:28.650" LastActivityDate="2010-10-16T20:14:28.650" CommunityOwnedDate="2010-10-16T18:45:23.647" />
  <row Id="6967" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T18:54:12.107" Score="5" ViewCount="0" Body="&lt;p&gt;One way to do this is using Groebner basis (which is implemented in most serious computer algebra systems, and which you can learn to do by hand in many places, like the amazing book by Cox &lt;em&gt;et al.&lt;/em&gt; called &lt;em&gt;Ideals, Varieties and Algorithms&lt;/em&gt;). For example, using &lt;em&gt;Mathematica&lt;/em&gt; you can compute:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[1]:= GroebnerBasis[{&#xA;  u + v - 2,&#xA;  u x + v y - 1,&#xA;  u x^2 + v y^2 + 1,&#xA;  u x^3 + v y^3 + 5&#xA;  }, {u, v, x, y}&#xA; ]                                                &#xA;                     2&#xA;Out[1]= {2 - 3 y + y , -3 + x + y, -7 + v + 4 y, 5 + u - 4 y}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result is an equivalent system of equations. You'll notice that the first equation&#xA;depends only on $y$, and that the other three allow you to determine $x$, $v$ and $u$, respectively, in terms of $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book I mentioned explains why this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can use the (freely available, great) Macaulay2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Macaulay2, version 1.3.1&#xA;with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,&#xA;               ReesAlgebra, SchurRings, TangentCone&#xA;&#xA;i1 : R = QQ[u,v,x,y,z];&#xA;&#xA;i2 : I = ideal (u + v - 2, u*x + v*y-1, u*x^2+v*y^2+1, u*x^3+v*y^3+5);&#xA;&#xA;o2 : Ideal of R&#xA;&#xA;i3 : gens gb I&#xA;&#xA;o3 = | x+y-3 v+4y-7 u-4y+5 y2-3y+2 |&#xA;&#xA;             1       4&#xA;o3 : Matrix R  &amp;lt;--- R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The polynomials listed in the &lt;code&gt;o3&lt;/code&gt; line are a Groebner basis, &lt;em&gt;i.e.&lt;/em&gt;, an equivalent system of equations, and it allows you to solve it at once.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-16T19:01:33.860" LastActivityDate="2010-10-16T19:01:33.860" CommentCount="1" />
  <row Id="6968" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T19:03:27.880" Score="4" ViewCount="0" Body="&lt;p&gt;An elementary way would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\displaystyle x$ and $\displaystyle y$ are roots of $\displaystyle z^2 + az + b = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that $\displaystyle x^2 + ax + b = 0$ and $\displaystyle y^2 + ay + b= 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the first by $\displaystyle u$, the second by $\displaystyle v$ and adding we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-1 +a + 2b = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mutiplying the first $ux$ and second by $vy$ and adding we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-5 - a + b = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily find $\displaystyle a$ and $\displaystyle b$ now and so can find the possibilities for $\displaystyle x$ and $\displaystyle y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be easy to solve the whole system now.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-16T19:03:27.880" CommentCount="3" />
  <row Id="6969" PostTypeId="2" ParentId="6851" CreationDate="2010-10-16T19:10:38.727" Score="1" ViewCount="0" Body="&lt;p&gt;By taking the seperate cases where $x = 0$ or $y = 0$ and showing that they are in the set, then we can assume $x,y \ne 0$  Therefore we can take $-2 \le xy \le 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$-2/y \le x \le 2/y$&lt;/li&gt;&#xA;&lt;li&gt;$-2/x \le y \le 2/x$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So for all x,y real numbers satisfying the above, and (0, y real number) (x real number, 0), x,y are in R x R.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2010-10-18T04:27:28.247" LastActivityDate="2010-10-18T04:27:28.247" CommentCount="2" />
  <row Id="6970" PostTypeId="2" ParentId="6805" CreationDate="2010-10-16T19:21:03.143" Score="1" ViewCount="0" Body="&lt;p&gt;It's clearer if you write it instead in a symmetric fashion: $(n-1)^3+n^3+(n+1)^3$, which simplifies to $3n(n^2+2)$. If $n$ is a multiple of $3$ then this is a multiple of $9$. Otherwise, $n$ leaves a remainder of $\pm1$ when divided by $3$ and so $n^2+2$ is multiple of $3$, which implies that $3n(n^2+2)$ is a multiple of $9$.&#xA;It's not a proof by induction, though.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-16T19:21:03.143" CommentCount="2" />
  <row Id="6971" PostTypeId="1" AcceptedAnswerId="6995" CreationDate="2010-10-16T19:46:06.293" Score="1" ViewCount="131" Body="&lt;p&gt;Suppose you want to accumulate $12,000$ in a $5 \%$ account by making a level deposit at the beginning of each of the next 9 years. Find the required level payment. So this seems to be an annuity due problem. I know the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle \sum_{k=1}^{n} \frac{A}{(1+i)^{k}} = \frac{A}{1+i} \left[\frac{1- \left(\frac{1}{1+i} \right)^{n}}{1- \left(\frac{1}{1+i} \right)} \right] = P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this problem, we are trying to solve for $P$? Just plug in the numbers? Or do we need to calculate the discount rate $d = i/(i+1)$ since the annuity is being payed at the beginning of the year? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-18T10:16:50.500" Title="Principal and Annuities" Tags="&lt;homework&gt;&lt;finance&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="6972" PostTypeId="2" ParentId="6971" CreationDate="2010-10-16T19:53:45.100" Score="1" ViewCount="0" Body="&lt;p&gt;The problem statement is missing the time when you want to have the 12,000.  If it is at the end of the ninth year, the value of the deposit at the beginning of year n will have increased by 1.05^(10-n).  So if A is the deposit you have $\displaystyle \sum_{k=1}^{n} A*1.05^{(10-k)}=12000$.  Solve for A&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-16T19:53:45.100" CommentCount="2" />
  <row Id="6973" PostTypeId="2" ParentId="6713" CreationDate="2010-10-16T19:54:20.263" Score="2" ViewCount="0" Body="&lt;p&gt;As Rahul's comment indicates, there is no algorithm known to achieve an optimal packing.&#xA;However, some effective heuristics have been developed. For example, there is a 2010 paper by &#xA;Andrea Cassioli and Marco Locatelli entitled&#xA;&quot;A heuristic approach for packing rectangles in convex regions&quot;&#xA;(&lt;a href=&quot;http://www.optimization-online.org/DB_FILE/2010/06/2637.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF download&lt;/a&gt;).&#xA;Here is their abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this paper we propose a heuristic approach for the problem of packing equal rectangles within a convex region. The approach is based on an Iterated Local Search scheme (or, using a terminology employed for continuous problems, a Monotonic Basin Hopping), in which the key step is the perturbation move. Different perturbation moves, both combinatorial and continuous ones, are pro- posed and compared through extensive computational experiments on a set of test instances. The overall results are quite encouraging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This recent paper has 38 references, so this should give you much to pursue and ponder.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-16T19:54:20.263" />
  <row Id="6974" PostTypeId="2" ParentId="6947" CreationDate="2010-10-16T20:13:28.120" Score="2" ViewCount="0" Body="&lt;p&gt;There are pointers to a wealth of information on this question in the responses to the &lt;a href=&quot;http://mathoverflow.net/questions/37708/&quot; rel=&quot;nofollow&quot;&gt;Math Overflow question&lt;/a&gt;, which you mentioned in your comment to Paul VanKoughnett's response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, Deane Yang's response gives a nice summary of the situation, and Bill Thurston's response seems to give a good perspective on the problem of trying to find a characterization of Riemannian manifolds that admit such an embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the third question you mention in an edit. This is essentially a local problem. All this from the same MO question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From BS's response:&#xA;there is not even a local isometric embedding in general:&#xA;&lt;a href=&quot;http://www.springerlink.com/content/m775p64w64351260/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/m775p64w64351260/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from Will Jaggy's response (and Deane Yang's comments on it):&#xA;If the metric is analytic, then you can construct a local isometric embedding.&#xA;Some recent progress on characterizing the requirements when the degree of smoothness is relaxed:&#xA;&lt;a href=&quot;http://arxiv.org/abs/1009.6214&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1009.6214&lt;/a&gt;&#xA;The bibliography for that last one has no shortage of other relevant sounding titles.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-10-16T20:13:28.120" CommentCount="1" />
  <row Id="6975" PostTypeId="1" AcceptedAnswerId="6982" CreationDate="2010-10-16T20:34:46.983" Score="3" ViewCount="130" Body="&lt;p&gt;I'm reading &lt;em&gt;A Computational Introduction to Number Theory and Algebra&lt;/em&gt;, which can be found &lt;a href=&quot;http://www.shoup.net/ntb/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; as a free download. From the book's exercises, I'm stuck with a proof to show that $gcd(a,b)=|a| \iff a | b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current reasoning goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $gcd(a,b) = |a|$,&#xA;we see that&#xA;$aZ + bZ = |a|Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a|b \iff az = b$, for some $z \in Z$,&#xA;we see that both&#xA;$az, b \in |a|Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there exists&#xA;$s, t \in Z$, such that&#xA;$as + bt = az = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is true for $s = 0$ and $t = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q.E.D$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that in the end my reasoning is wrong, but I can't explain it.&lt;/p&gt;&#xA;" OwnerUserId="2566" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-17T00:08:05.583" LastActivityDate="2010-10-18T13:59:03.193" Title="Show that $gcd(a,b)=|a| \iff a | b$" Tags="&lt;number-theory&gt;&lt;abstract-algebra&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6976" PostTypeId="1" AcceptedAnswerId="6980" CreationDate="2010-10-16T20:39:53.287" Score="2" ViewCount="177" Body="&lt;p&gt;I have not been able to solve this problem. Any insights would be appreciated!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $x, n &gt; 1$ be integers. Suppose that for each $k &gt; 1$ there exists an integer $a_{k}$&#xA;such that $x − a_k^n$ is divisible by $k$. Prove that $x = A^{n}$ for some integer $A$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-28T07:00:38.503" LastActivityDate="2010-11-28T07:00:38.503" Title="Proving that an integer is the $n$ th power " Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;arithmetic&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="6977" PostTypeId="2" ParentId="6975" CreationDate="2010-10-16T20:44:38.970" Score="2" ViewCount="0" Body="&lt;p&gt;It is true that the final assertion holds if $s=0$ and $t=1$, but it is not true that  that $as+bt = az = b$ can only hold if $s=0$ and $t=1$ is false. Take, for example, $a=2$ and $b=4$. Then $(-5)a + 3(b) = b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that, by definition, gcd$(x,y)|x$ and gcd$(x,y)|y$. That gives one implication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the converse, remember that $d|x$ and $d|y$ implies $d|$gcd$(x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-16T20:44:38.970" />
  <row Id="6979" PostTypeId="1" AcceptedAnswerId="6984" CreationDate="2010-10-16T20:55:54.397" Score="9" ViewCount="150" Body="&lt;p&gt;Suppose we rotate the graph of $y = f(x)$ about the $x$-axis from $a$ to $b$.  Then (using the disk method) the volume is $$\int_a^b \pi f(x)^2 dx$$ since we approximate a little piece as a cylinder.  However, if we want to find the surface area, then we approximate it as part of a cone and the formula is $$\int_a^b 2\pi f(x)\sqrt{1+f'(x)} dx.$$  But if approximated it by a circle with thickness $dx$ we would get $$\int_a^b 2\pi f(x) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is how come for volume we can make the cruder approximation of a disk but for surface area we can't.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-10-17T01:24:29.310" Title="Areas versus volumes of revolution." Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="6980" PostTypeId="2" ParentId="6976" CreationDate="2010-10-16T20:58:51.893" Score="6" ViewCount="0" Body="&lt;p&gt;This is an old chestnut: an integer which is an $n$-th power modulo&#xA;all primes is an $n$-th power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a sledgehammer proof via the Chebotarev density theorem.&#xA;Suppose for the moment that $n$ is prime.&#xA;Consider $K=\mathbb{Q}(x^{1/n})$ and its Galois closure&#xA;$L=\mathbb{Q}(x^{1/n},\exp(2\pi i/n))$. Then each prime $p$ that is unramified&#xA;in $L$ splits in $K$ into various prime ideals at least one of which has norm $p$.&#xA;So its Frobenius has a fixed point on the permutation representation on the&#xA;$n$-th roots of $x$. By Chebotarev, the Galois group $G$ of $L/\mathbb{Q}$&#xA;has no element of degree $n$ and so must have a fixed point; that is one&#xA;of the $n$-th roots of $x$ must lie in $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure something similar works for any positive integer $n$, but it's&#xA;too late tonight for me to work out the details :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-16T20:58:51.893" CommentCount="2" />
  <row Id="6981" PostTypeId="1" CreationDate="2010-10-16T21:06:19.600" Score="4" ViewCount="189" Body="&lt;p&gt;I'm having trouble wrapping my head around the following issue. My book solves a problem without using complex exponential solution like $C_1  e^{it}$ and using either $A \cos(t) + B \sin(t)$ or $A \cos(t+\phi)$ to a second-order &quot;undamped&quot; linear ODE. I understand that the solution can be purely real or complex. The trouble arises when I use the solution to this ODE inside another function which cubes this solution. Cubing $C_1  e^{it}$ results in another complex sinusoid with 3 times the frequency, which is a very different effect from cubing something like $A \cos(t+\phi)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my work. I've highlighted where I think things have gone awry.&#xA;&lt;img src=&quot;http://i.imgur.com/ORCn9.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tKTRS.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-08T21:06:27.980" LastActivityDate="2011-01-07T21:44:39.953" Title="Using complex exponentials as solution of ODE" Tags="&lt;complex-numbers&gt;&lt;fourier-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6982" PostTypeId="2" ParentId="6975" CreationDate="2010-10-16T21:13:29.813" Score="3" ViewCount="0" Body="&lt;p&gt;Assume that $a,b\in\mathbb{N}$. I use the following definition of $D=\gcd (a,b)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $D$ is a common divisor of $a$ and $b$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Every integer $d\in\mathbb{N}$ which is a common divisor of $a$ and $b$ divides $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition 1: If $\gcd (a,b)=a$, then $a|b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition 2: If $a|b$, then $\gcd (a,b)=a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Since $a|a$ and $a|b$, then any integer $d$ such that $d|a$ and $d|b$&#xA;satisfies also the condition $d|a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $a|a$, $a|b$ and $d|a$ we conclude that $\gcd (a,b)=a$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-16T22:45:02.767" LastActivityDate="2010-10-16T22:45:02.767" CommentCount="4" />
  <row Id="6983" PostTypeId="2" ParentId="6976" CreationDate="2010-10-16T21:20:37.067" Score="4" ViewCount="0" Body="&lt;p&gt;To be a little more concrete, here's the proof for $n = 2$.  We proceed by proving the contrapositive.  Suppose $x$ is not a square, so write $x = k^2 p_1 p_2 ... p_n$ where the $p_i$ are distinct primes, one of which may be $-1$.  By quadratic reciprocity, together with the CRT, there exists an arithmetic progression such that any prime $q$ in that arithmetic progression has the property that $\left( \frac{p_i}{q} \right) = 1$ for $1 \le i \le n-1$ and $\left( \frac{p_i}{q} \right) = -1$ for $i = n$.  A prime $q$ exists in this arithmetic progression by Dirichlet's theorem (a special case of Chebotarev's density theorem!), and $x$ is not a square $\bmod q$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for $n = 3$ one can give a proof using cubic reciprocity, and so on.  In general it should be possible to give a proof along these lines using Artin reciprocity, but if you're using Artin reciprocity and Dirichlet's theorem you might as well use the full strength of Chebotarev.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T21:20:37.067" />
  <row Id="6984" PostTypeId="2" ParentId="6979" CreationDate="2010-10-16T21:21:29.950" Score="6" ViewCount="0" Body="&lt;p&gt;In the case of volume, the error is second order (it is basically revolving a triangle with sides $dx$ and $f'(x)dx$ so is negligible compared to the cylinder which is first order.  This is the same as the 1 dimensional case, where to measure area (the standard integral) you use rectangles but to measure arc length you need to integrate $\sqrt{1+f'(x)^2}dx$, which is the hypotenuse of the triangle.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-17T01:24:29.310" LastActivityDate="2010-10-17T01:24:29.310" CommentCount="1" />
  <row Id="6985" PostTypeId="2" ParentId="6981" CreationDate="2010-10-16T21:25:13.100" Score="1" ViewCount="0" Body="&lt;p&gt;If you cube $cos(t)$ you get $(3cos(t) + cos(3t))/4$ so the triple frequency shows up here, too.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-16T21:25:13.100" CommentCount="5" />
  <row Id="6986" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T21:37:11.830" Score="2" ViewCount="0" Body="&lt;p&gt;Looking for linear terms suggests we write the system as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} 1 &amp;amp; 1 \cr x &amp;amp; y  \cr x^2 &amp;amp; y^2 \cr x^3 &amp;amp; y^3\end{pmatrix}&#xA;\begin{pmatrix} u \cr v\end{pmatrix}&#xA;=\begin{pmatrix} 2 \cr 1 \cr -1 \cr -5 \end{pmatrix}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Row-reduction has to give two entirely zero rows of coefficients,yielding two equations for $x$ and $y$, one of which is linear in $y$.  (It will help to recognize that $x-y$ is a factor of $x^2 - y^2$ and of $x^3 - y^3$.)  That makes it straightforward to find $x$, from which you obtain $y$.  With $x$ and $y$ in hand the coefficients are completely determined, allowing you to solve for $(u, v)$ in the usual ways.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-16T21:37:11.830" />
  <row Id="6987" PostTypeId="2" ParentId="6979" CreationDate="2010-10-16T21:56:53.440" Score="9" ViewCount="0" Body="&lt;p&gt;The problem here is that volume behaves nicely under small deformations of 3D regions in 3D, but surface area does not.  Similarly, area behaves nicely under small deformations of 2D regions in 2D, but circumference / arc length does not.  You can see the essence of the 2D problem, hence the essence of the 3D problem, in the following: the length of the diagonal from $(0, 0)$ to $(1, 1)$ is $\sqrt{2}$, but if we approximate the diagonal by a &quot;staircase&quot; of horizontal and vertical lines of length $\frac{1}{n}$ and let $n \to \infty$ we get a length of $2$ instead.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-16T21:56:53.440" />
  <row Id="6988" PostTypeId="1" CreationDate="2010-10-16T22:04:55.940" Score="3" ViewCount="119" Body="&lt;p&gt;Given two Riemmanian Manifolds of dimension 2, and a point on each. If the scalar curvatures are isomorphic (as functions) in some nbhds of these points, are then the manifolds necessarily locally isometric?&lt;/p&gt;&#xA;" OwnerUserId="2567" LastActivityDate="2010-10-17T03:50:26.637" Title="Is the scalar curvature the only isometric invariant of a Riemannian 2-manifold?" Tags="&lt;differential-geometry&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="6989" PostTypeId="2" ParentId="6988" CreationDate="2010-10-16T22:20:30.820" Score="4" ViewCount="0" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_curvature#Surfaces_of_constant_curvature&quot; rel=&quot;nofollow&quot;&gt;true if the curvatures are constant&lt;/a&gt; (Minding's theorem), but not in general.  &lt;strike&gt;I actually don't know a counterexample, but (edit: I think that)&lt;/strike&gt; Stillwell &lt;a href=&quot;http://books.google.com/books?id=V7mxZqjs5yUC&amp;amp;lpg=PP1&amp;amp;pg=PA344#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;points out on page 344&lt;/a&gt; of &lt;em&gt;Mathematics and its history&lt;/em&gt; that Strubecker's &lt;em&gt;Differentialgeometrie&lt;/em&gt; gives &lt;strike&gt;such an&lt;/strike&gt; a related example in Volume III on page 124, but I'm not 100% sure if that is a counterexample of what you're asking, and I don't currently have access to it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: Okay, here's an example I can access.  In Audin's &lt;em&gt;Geometry&lt;/em&gt;, after the &lt;em&gt;Theorema egregium&lt;/em&gt;, there is a remark that the &quot;converse is wrong; a classical counter-example will be found in Exercise VIII.24.&quot; Thankfully &lt;a href=&quot;http://books.google.com/books?id=U_cTJMCIzdUC&amp;amp;lpg=PP1&amp;amp;pg=PA300#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Exercise VIII.24&lt;/a&gt; is available on Google Books, at least to me.  There is a section with hints and solutions, too, but I can't see whether it includes that exercise.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-16T22:56:53.140" LastActivityDate="2010-10-16T22:56:53.140" CommentCount="2" />
  <row Id="6990" PostTypeId="2" ParentId="6965" CreationDate="2010-10-16T22:23:59.137" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Exploit the innate symmetry: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad  u x^2 + v y^2\ =\ (x+y)\ (ux\ +\ vy)\  -\  xy\ (u+v)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad  u x^3 + v y^3\ =\ (x+y)\ (ux^2+vy^2) -\: xy\ (ux+vy)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yields  $\rm\quad\quad\quad\quad\:\ \ -1 \ =\ \ x+y\ -\ 2\ xy $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\  5\ =\ \ x+y\ \:\ +\:\ \ xy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving yields $\rm\quad\quad\ \  x+y = 3,\ \ \ xy = 2\ \ \Rightarrow\ \ \{x,\:y\}\ =\ \{1,\:2\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ Readers familiar with the  theory of &lt;strong&gt;Lucas-Lehmer sequences&lt;/strong&gt; may recognize this recursion as one of many useful identities satisfied by sums of powers.  Such identities arise in many diverse contexts. Here is another example: Capdegelle's work on &lt;strong&gt;FLT&lt;/strong&gt; (Fermat's Last Theorem) employed the recursion $\rm\ F_{n+3} + S_2\: F_{n+2} + S_1\: F_{n+1} + S_0\: F_n = 0\ $ where $\rm\ F_n = X^n + Y^n - Z^n\ $ is the Fermat polynomial and the polynomials $\rm\ S_k\ $ are the elementary symmetric polynomials $\rm\ S_0 = -XYZ,$ $\rm\ S_1 = XY + XZ + YZ,\ $ $\rm S_2 = -(X+Y+Z)\:.\ $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-17T21:37:27.267" LastActivityDate="2010-10-17T21:37:27.267" CommentCount="10" />
  <row Id="6991" PostTypeId="2" ParentId="6975" CreationDate="2010-10-16T22:41:51.320" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ $ If $\rm\ a\:|\:b\ $ then $\rm\ c\:|\:a,b\ \iff\ c\:|\:a\ \ \ $ [and conversely via $\rm\ c = a\:$]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ a,\:b\ $  and $\rm\:a\:$ have the same common divisors $\rm\:c\:$, hence the same &lt;em&gt;greatest&lt;/em&gt;$\ $ common divisor.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T13:59:03.193" LastActivityDate="2010-10-18T13:59:03.193" />
  <row Id="6992" PostTypeId="2" ParentId="6988" CreationDate="2010-10-16T22:44:06.717" Score="1" ViewCount="0" Body="&lt;p&gt;I'm a little rusty on this stuff, but here's a sketch of a general answer.  Basically the idea is that for any differential-geometric structure on a manifold (Riemannian metric, almost-complex structure, Finsler structure, symplectic structure, jet structure, web structure, etc.) you define the &quot;frame bundle&quot; to be the space of ways of identifying the structure with the canonical one on $\mathbb{R}^n$.  So in the case of a Riemannian metric, point by point you choose an identification of the tangent space with $\mathbb{R}^n$ so that the inner product agrees with the usual one on $\mathbb{R}^n$, and then you allow yourself to post-compose with any element of the orthogonal group $O(n)$ in order to allow for differences in the original choices.  (This is because the orthogonal group is exactly the space of linear maps that preserves the inner product.)  This gives you what is known as a &lt;em&gt;G-structure&lt;/em&gt; on the manifold (here $G=O(n)$).  This is a bundle over the manifold where every fiber is a $G$-torsor, i.e. a space which is topologically the same as $G$ but there's no choice of which point is the identity because there was no canonical choice of how to identify $T_pM$ with $\mathbb{R}^n$.  (So e.g. a real line is a torsor for the group $\mathbb{R}$.)  The next step is to try and make a connection for this bundle, and if I remember correctly, this is the place where we tease out geometric invariants.  There's a Lie-algebra-valued 1-form (with values in $Lie(G)$) that measures the failure of the connection to be trivial(izable), I think, and then you calculate something called the &quot;Spencer cohomology&quot;.  The main point is that this gives &lt;em&gt;all&lt;/em&gt; the invariants, so for example it falls out as an immediate consequence that two manifolds are locally isomorphic if and only if there's a local diffeomorphism that lines up their Gaussian curvatures.  (Something else really cool is that this perspective tells you exactly why we can have the Riemann mapping theorem in dimension 2, but then in all higher dimensions the only conformal maps are Mobius transformations.  Also this can be used to prove Darboux's theorem, and the same method gives the Nijenhaus tensor.)&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-17T03:50:26.637" LastActivityDate="2010-10-17T03:50:26.637" CommentCount="3" />
  <row Id="6993" PostTypeId="1" CreationDate="2010-10-16T22:45:59.347" Score="3" ViewCount="67" Body="&lt;p&gt;During the summer, I did an REU where we focused primarily on one-dimensional dynamics and more specifically kneading theory. One thing that I was always confused about is why the Schwarzian derivatives always seem to pop up in discussions of iterated dynamics on the real line. I understand what a Schwarzian derivative is, but I don't see any intuitive reason that it should show up in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone could explain or provide me with a reference that makes the appearance of Schwarzian derivatives in one-dimensional dynamics on the real line seem natural.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question I have, is there an intuitive motivation for the Schwarzian derivative itself?&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-10-17T00:58:35.490" LastActivityDate="2010-10-17T00:58:35.490" Title="Schwarzian Derivative and One-Dimensional Dynamics - how are they connected?" Tags="&lt;intuition&gt;&lt;reference-request&gt;&lt;dynamical-systems&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="6994" PostTypeId="1" AcceptedAnswerId="7106" CreationDate="2010-10-16T23:46:40.097" Score="10" ViewCount="142" Body="&lt;p&gt;Krull's Hauptidealsatz (principal ideal theorem) says that for a Noetherian ring $R$ and any $r\in R$ which is not a unit or zero-divisor, all primes minimal over $(r)$ are of height 1. How badly can this fail if $R$ is a non-Noetherian ring? For example, if $R$ is non-Noetherian, is it possible for there to be a minimal prime over $(r)$ of infinite height?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The answer is yes. See &lt;a href=&quot;http://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-10-17T23:17:16.380" LastActivityDate="2010-10-18T07:23:06.007" Title="How badly can Krull's Hauptidealsatz fail for non-Noetherian rings?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6995" PostTypeId="2" ParentId="6971" CreationDate="2010-10-17T00:15:35.477" Score="0" ViewCount="0" Body="&lt;p&gt;I get $F = A \frac{(1+i)^{2n}-(1+i)^n}{i(1+i)^n}$. But when I solve for $A$, I get the level payment as being $1088.28$. But the correct answer is $1036.46$. Why is there a discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-17T00:15:35.477" CommentCount="4" />
  <row Id="6996" PostTypeId="1" AcceptedAnswerId="6997" CreationDate="2010-10-17T02:30:41.267" Score="1" ViewCount="54" Body="&lt;p&gt;According to an &lt;a href=&quot;http://www.cs.bu.edu/faculty/crovella/paper-archive/ss-lect-notes.ps&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; I'm studying (&quot;Time series, self-similarity&#xA;and network traffic by Mark Crovella) the expectation of the square of&#xA;a time-scaled Brownian motion process $E[ B(ct)^2 ]$ where $c$ is the time&#xA;scaling is equal to $ct$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate help proving this; i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E[  B(ct)^2 ] = c t$&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-17T02:49:47.730" Title=" Expectation of squared time-scaled Brownian process " Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="6997" PostTypeId="2" ParentId="6996" CreationDate="2010-10-17T02:44:08.583" Score="3" ViewCount="0" Body="&lt;p&gt;This follows from the fact that $B(t)$ has a Gaussian$(0,t)$ distribution.  Therefore, $B(ct)$ has a Gaussian$(0,ct)$ distribution.  Thus $E[B(ct)^2] = Var[B(ct)] + (E[B(ct)])^2 = ct + 0 = ct$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-17T02:49:47.730" LastActivityDate="2010-10-17T02:49:47.730" CommentCount="4" />
  <row Id="6998" PostTypeId="1" AcceptedAnswerId="7009" CreationDate="2010-10-17T03:47:07.083" Score="8" ViewCount="722" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm teaching a group theory course now, and I wanted to give my students a proof that every subgroup of a cyclic group is cyclic. The easiest way I could think to do this is to say that any cyclic group corresponds to a $\mathbb{Z}/n\mathbb{Z}$ and then (implicitly) use the first isomorphism theorem (which they haven't learned yet) to &quot;reduce&quot; it to the theorem that every subgroup of $\mathbb{Z}$ is of the form $m\mathbb{Z}$. The problem is that this latest statement is not simple at all (every subgroup is an ideal, and $\mathbb{Z}$ is a PID because $\mathbb{Z}$ is Euclidean).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all seems way too convoluted for such an easy statement. What's the simplest most elementary proof that I can give?&lt;/p&gt;&#xA;" OwnerUserId="2570" LastActivityDate="2010-10-17T22:34:22.857" Title="How to show every subgroup of a cyclic group is cyclic?" Tags="&lt;group-theory&gt;&lt;education&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="6999" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T03:51:25.963" Score="7" ViewCount="0" Body="&lt;p&gt;If the original group is generated by $a$, and $d$ is the least positive integer such that $a^d$ is in the subgroup, then show using the division algorithm that $a^d$ generates the subgroup.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To elaborate, the division algorithm says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $m$ and $n$ are integers and $n$ is positive, then there are unique integers $q$ and $r$ such that $m=nq+r$ and $0\leq r\lt n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could prove this by taking $r$ to be the least nonnegative integer expressible in the form $m-np$ as $p$ ranges over the integers, which then determines $q$.  (There are details here which I am omitting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $G$ is a cyclic group (finite or infinite) with generator $a$, i.e., $G=\langle a\rangle$.  Let $H$ be a subgroup of $G$.  Let $d$ be the smallest element of the set of positive integers $n$ such that $a^n$ is in $H$.  (Unless $G$ is infinite and $H$ is trivial, this exists.)  If $b$ is an arbitrary element of $H$, then because $b$ is in $G$, $b=a^m$ for some integer $m$.  Apply the division algorithm to write $m=dq+r$ for integers $q$ and $r$ with $0\leq r\lt d$.  Then $b=a^m=(a^d)^q\cdot a^r$.  Since $b$ is in $H$ and $a^d$ is in $H$, it follows that $a^r=a^m\cdot(a^d)^{-q}$ is in $H$.  If $r$ were positive, this would contradict the choice of $d$, so $r=0$, and $b=a^m=(a^d)^q$.  Since $b$ was arbitrary, this shows that $H\subseteq \langle a^d\rangle$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-17T04:32:42.360" LastActivityDate="2010-10-17T04:32:42.360" CommentCount="9" />
  <row Id="7000" PostTypeId="2" ParentId="6976" CreationDate="2010-10-17T04:12:50.847" Score="6" ViewCount="0" Body="&lt;p&gt;If I am not mistaken, the question has a more elementary answer than those provided so far.  I will use the functions $\operatorname{ord}_p: \mathbb{Q}^{\times} \rightarrow \mathbb{Z}$, defined to be the largest power of $p$ appearing in the numerator minus the largest power of $p$ appearing in the denominator.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: A positive rational number $x$ is a rational $n$th power iff for all primes $p$, $\operatorname{ord}_p(x)$ is divisible by $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an easy consequence of unique factorization in $\mathbb{Z}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: A positive integer $x$ is an integral $n$th power iff for all primes $p$, &#xA;$\operatorname{ord}_p(x)$ is divisible by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows easily from Step 1 using the fact that since $\mathbb{Z}$ is a UFD, it is integrally closed.  (Alternately, apply the rational roots theorem to the polynomial $t^n - x$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: As in lhf's comment above, I claim that the given condition on $x$ implies that $x$ is an nth power in $\mathbb{Q}_p$ for all primes $p$.  Indeed, taking $k$ to be a prime power, it implies that for all positive integers $a$, the congruence $t^n -x \equiv 0 \pmod{p^a}$ has a solution, and by a routine application of Hensel's Lemma, this implies that $x$ is an $n$th power in $\mathbb{Q}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4: Since $x$ is an $n$th power in $\mathbb{Q}_p$, the $p$-adic valuation $v_p(x)$ is divisible by $n$.  For a rational number $x$, this means that $\operatorname{ord}_p(x)$ is divisible by $n$.  And we are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As regards the fancy stuff, perhaps people are thinking of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grunwald%E2%80%93Wang_theorem&quot; rel=&quot;nofollow&quot;&gt;Grunwald-Wang Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that if $x$ is an element of a global field $K$ which is an $n$th power in all but finitely many completions at finite places $v$.  Actually, this is &quot;Grunwald's theorem&quot;, i.e., it isn't quite true!  Wang showed that there are counterexamples to this statement, even over $\mathbb{Q}$ (if one uses all but finitely many places, rather than all places): see the wikipedia article for an explanation.  The easy proof that I give above works in any global field which is the fraction field of a PID $R$, with the conclusion that $x$ comes out to be an $n$th power up to a unit of $R$.  (When $R = \mathbb{Z}$, requiring $x$ to be positive fixes the unit ambiguity.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-17T04:12:50.847" CommentCount="3" />
  <row Id="7001" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T04:27:56.593" Score="2" ViewCount="0" Body="&lt;p&gt;See Proposition 5 in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/4400algebra2point5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/4400algebra2point5.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that part of this argument is left as an exercise for the reader.  It &lt;em&gt;is&lt;/em&gt; an incarnation of the same argument that shows that a Euclidean domain is a PID, but one certainly doesn't need this terminology or any prior results.  Namely, one is supposed to show that -- upon identifying the elements of a cyclic group $Z_n$ with the integers $0,1,\ldots,n-1$ as usual -- if $H$ is a nontrivial subgroup of $Z_n$, then it is generated by its smallest nonzero element, say $k$.  Why?  Well, if not, there would exist an element $\ell$ of the subgroup which is not a multiple of $k$, and then by subtracting off a suitable multiple of $k$ -- to be completely explicit, let $a$ be the largest non-negative integer $a$ such that $\ell - ak$ is positive, and subtract $ak$ -- we get an element of the subgroup which is positive but smaller than $k$, contradiction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the more explicitly ring-theoretic proof you have in mind is the one alluded to in the Remark at the bottom of page 2.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-10-17T04:38:53.113" LastActivityDate="2010-10-17T04:38:53.113" />
  <row Id="7002" PostTypeId="1" AcceptedAnswerId="7007" CreationDate="2010-10-17T04:37:46.953" Score="1" ViewCount="133" Body="&lt;p&gt;Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-07T05:36:58.663" Title="Subgroups and finite index" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7004" PostTypeId="1" AcceptedAnswerId="7017" CreationDate="2010-10-17T05:01:18.760" Score="1" ViewCount="123" Body="&lt;p&gt;Suppose $f(x) = \begin{cases} 0 \ \ \text{if} \ x \in \mathbb{R}- \mathbb{Q} \newline \frac{1}{q} \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is $f$ continuous on the irrationals?&#xA;(ii) Is $f$ continuous on the rationals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i) you could use the sequence definition of continuity? Maybe try $a_n = \frac{\sqrt{2}}{n}$ and show that $a_n \to 0$ but $f(a_n) \not \to 0$? So its discontinuous on the irrationals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii) I don't see why there are $1+2+ \cdots + (q-1)$ rational numbers? I know that we need to use this fact to choose an appropriate $\delta$ (e.g  $0&amp;lt; |x-a| &amp;lt; \delta \Rightarrow |f(x)-L| &amp;lt; \epsilon$). &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-17T11:51:47.543" Title="Continuity of function" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="7005" PostTypeId="2" ParentId="7004" CreationDate="2010-10-17T05:11:30.873" Score="1" ViewCount="0" Body="&lt;p&gt;Your $a_n$'s are irrational, so $f(a_n)=0$.  The function is continuous on the irrationals.  You can show this using the $\epsilon$-$\delta$ definition of continuity.  Once you're given an irrational $x$ and an $\epsilon&gt;0$, there is an integer $n&gt;1/\epsilon$, and there are only finitely many rational numbers in, say, $(x-1,x+1)$ having denominator smaller than $n$ is lowest terms.  Thus there is a closest one to $x$, and you can use this to find your $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using similar reasoning, you can show that the limit at each rational is 0, but, as AD. points out, without knowing this there is a quick way to see that the function is discontinuous at each rational number. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-17T05:35:31.370" LastActivityDate="2010-10-17T05:35:31.370" CommentCount="2" />
  <row Id="7006" PostTypeId="2" ParentId="7004" CreationDate="2010-10-17T05:17:08.753" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a picture that might help you see things more clearly: &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thomae%27s_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-17T06:23:34.850" LastActivityDate="2010-10-17T06:23:34.850" />
  <row Id="7007" PostTypeId="2" ParentId="7002" CreationDate="2010-10-17T06:21:49.177" Score="0" ViewCount="0" Body="&lt;p&gt;In fact, the result holds even if you do not assume that we are dealing with finite index, provided you take the equality to be an equality of cardinalities. Just follow the idea Pete Clark suggests, but without assuming that the number of cosets is finite in either case.  And remember that the number of distinct cosets of a subgroup $B$ in a group $A$ is equal, by definition, to the index of $B$ in $A$, $[A:B]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether the $x_i$ are &quot;coset representatives for $H$&quot;, remember that for a collection of elements to be &quot;a set of coset representatives&quot; we require that if $x_iH = x_jH$, then $i=j$, and that for every $g\in G$ there exists $i$ such that $gH = x_iH$ (thinking left cosets; right cosets work the same way). So $x_1,\ldots,x_n$ will not generally be &quot;a coset representative&quot; nor a set of coset representatives, but that is immaterial. Simply show that if $\{x_i\}_{i\in I}$ is a set of coset representatives for $K$ in $G$ (meaning, (i) for every $g\in G$ there exists $i\in I$ such that $gK = x_iK$; and (ii) for all $i,i&amp;#39;\in I$, if $x_iK = x_{i&amp;#39;}K$ then $i=i'$), and $\{y_j\}_{j\in J}$ is a set of coset representatives for $H$ in $K$ (meaning, (1) for every $k\in K$ there exists $j\in J$ such that $kH = y_jH$; and (2) for all $j,j&amp;#39;\in J$, if $y_jH = y_{j&amp;#39;}H$, then $j=j'$); then it follows that $\{x_iy_j\}_{(i,j)\in I\times J}$ is a set of coset representatives for $H$ in $G$ (meaning, you need to prove that (I) for all $g\in G$ there exists $(i,j)\in I\times J$ such that $gH = (x_iy_j)H$; and (II) for all $(i,j),(i&amp;#39;,j&amp;#39;)\in I\times J$, if $(x_iy_j)H = (x_{i&amp;#39;}y_{j&amp;#39;})H$, then $(i,j)=(i',j')$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fact that will no doubt be useful is to remember that for any group $A$ and any subgroup $B$ of $A$, $cB = dB$ if and only if $cB\cap dB\neq\emptyset$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:36:58.663" LastActivityDate="2010-12-07T05:36:58.663" />
  <row Id="7008" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T07:37:07.680" Score="3" ViewCount="0" Body="&lt;p&gt;It is enough to show that every maximal proper subgroup $H$ of $G$ is cyclic, by induction;&#xA;to start the induction, notice that if $G$ is simple, then there is nothing to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $H\subset G$ be a maximal proper subgroup, so that $G/H$ is simple. There exists then a prime $p$ such that $G/H$ is isomorphic to $\mathbb Z/p\mathbb Z$ and then $pG\subset H$. From the exact sequence of abelian groups $$0\to H\to G\to G/H\to 0$$ we get the exact sequence $$0\to H/pG\to G/pG\to G/H\to 0.$$ Now this is an exact sequence of $\mathbb F_p$-vector spaces in an obvious way, and $G/H$ and $G/pG$ are vector spaces of dimension $1$ (for the first, this is obvious; for the second, it is enough to notice that $G/pG$ is cyclic as an abelian group and non-zero because it surjects onto $G/H$) It follows that $H/pG=0$, &lt;em&gt;i.e.&lt;/em&gt;, that $H=pG$. Since $G$ is cyclic, then $H$ is cyclic too.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-17T14:59:25.293" LastActivityDate="2010-10-17T14:59:25.293" CommentCount="4" />
  <row Id="7009" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T07:42:23.057" Score="4" ViewCount="0" Body="&lt;p&gt;If you accept Lagrange's theorem and&#xA;the existence of quotient groups, there's another&#xA;argument in the finite case.&#xA;Let $G=\langle a\rangle$ be a cyclic group of order $n$.&#xA;By Lagrange's theorem if $H$ is a subgroup of $G$ then $m=|H|$&#xA;is a factor of $n$, and is $m\mid n$ then $H_m=\langle a^{n/m}\rangle$&#xA;is a subgroup of order $m$. I claim $G$ has no other order $m$ subgroup.&#xA;The group $G/H$ has order $n/m$ so $x^{n/m}=e$ for all $x\in G/H$.&#xA;This translates to $x^{n/m}\in H$ for all $x\in G$ which means $H_m\le H$.&#xA;These have the same order so $H=H_m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-17T07:42:23.057" CommentCount="3" />
  <row Id="7010" PostTypeId="2" ParentId="6723" CreationDate="2010-10-17T08:05:11.033" Score="0" ViewCount="0" Body="&lt;p&gt;Making Ross's answer more explicit: remember that in the vicinity of the point where the light ray from the origin strikes your mirror, you can replace the curved mirror with the corresponding tangent line. Try deriving a differential equation for your $f(x)$ (hint: you may have to use the formula for the tangent of the difference of two angles) such that the reflected ray has zero slope (i.e., is horizontal).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-17T08:05:11.033" />
  <row Id="7011" PostTypeId="1" AcceptedAnswerId="7037" CreationDate="2010-10-17T09:59:57.057" Score="5" ViewCount="168" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in the process of writing my Master's Thesis on automata theory. The writing must be in English which isn't my mother tongue. &#xA;  So the question is, given that this is my first time long (hundred pages) writing and since English is a second language for me, what is the book you recommend for academic writing style? Ideally the book:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;must address things like how to structure the writing and/or align it with the research &lt;/li&gt;&#xA;&lt;li&gt;address the technical nature of the writing&lt;/li&gt;&#xA;&lt;li&gt;recommend general styling and language rules&lt;/li&gt;&#xA;&lt;li&gt;and as a plus, some hints to those of us for whom English is a second language&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Best regards.&lt;/p&gt;&#xA;" OwnerUserId="982" LastActivityDate="2010-10-17T16:36:44.010" Title="Book about technical and academical writting" Tags="&lt;books&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="2" />
  <row Id="7012" PostTypeId="1" AcceptedAnswerId="7015" CreationDate="2010-10-17T10:08:47.150" Score="1" ViewCount="122" Body="&lt;p&gt;very sorry for the noobish post, I'm studying math on my own and lot of times im very confused about little things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my question, the original problem starts out like this:  Solve the inequality $\vert x -2 \vert \geq \vert 2x -3\vert $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the solved problem from my book &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\vert x -2 \vert \geq \vert 2x -3\vert $  &lt;/li&gt;&#xA;&lt;li&gt;$ (x-2)^2 \geq (2x-3)^2 $  &lt;/li&gt;&#xA;&lt;li&gt;$ 3x^2 - 8x + 5 \leq 0 $&lt;/li&gt;&#xA;&lt;li&gt;$ (x-1)(3x-5) \leq 0$&lt;/li&gt;&#xA;&lt;li&gt;$1 \leq x \leq 5/3$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How does: $(x-1)(3x-5) \leq 0$. (Step 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Become solved as : $1 \leq x \leq 5/3$. (Step 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were separate wouldn't it be:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x \leq 1 \qquad  \text{and} \qquad   x \leq 5/3&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does &lt;strong&gt;$x\leq 1$&lt;/strong&gt; become &lt;strong&gt;$1\leq  x$&lt;/strong&gt;??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much!! =)&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-10-17T15:26:20.360" LastActivityDate="2010-10-17T16:17:19.180" Title="simple inequality problem with modulus" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="7013" PostTypeId="1" AcceptedAnswerId="7135" CreationDate="2010-10-17T10:08:50.980" Score="2" ViewCount="224" Body="&lt;p&gt;How does one evaluate the integral $$\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx \quad \text{for} \ a,b \in (0,1)$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-18T17:05:47.713" Title="Evaluating the improper integral $\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx $" Tags="&lt;analysis&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="7014" PostTypeId="2" ParentId="7012" CreationDate="2010-10-17T10:19:17.273" Score="4" ViewCount="0" Body="&lt;p&gt;Hint: In order for a product to be negative, one of the factors must be positive and the other one negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: It's often helpful to draw a table such as this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         x  |    1    5/3   &#xA;------------+---------------&#xA;       x-1  | -  0  +  +  +&#xA;      3x-5  | -  -  -  0  +&#xA;------------+---------------&#xA;(x-1)(3x-5) | +  0  -  0  + &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On each line you keep track of the signs of the individual factors in the intervals of interest. Then you construct the bottom line by using &quot;minus times minus equals plus&quot;, &quot;zero times anything equals zero&quot;, etc.&#xA;In the bottom line you can then read off that $(x-1)(3x-5) \le 0$ is true exactly when $-1 \le x \le 5/3$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-17T12:01:57.230" LastActivityDate="2010-10-17T12:01:57.230" />
  <row Id="7015" PostTypeId="2" ParentId="7012" CreationDate="2010-10-17T10:19:35.953" Score="1" ViewCount="0" Body="&lt;p&gt;So you want $(x-1) (3x-5) \leq 0$. So either one of the terms has to be positive and the other has to be negative. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;So if $x \geq 5/3$, then both the terms are positive and the product is positive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly if $x \leq 1$, then both the terms are negative and their product is positive. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So observe that you don't want either of these things to happen, that is $x \notin (- \infty,1) \cup (5/3,\infty)$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-17T10:19:35.953" CommentCount="4" />
  <row Id="7016" PostTypeId="2" ParentId="7013" CreationDate="2010-10-17T11:45:35.917" Score="3" ViewCount="0" Body="&lt;p&gt;I think it might be like the proof of $B(x,1-x)=\pi\csc(\pi x)$.  Let $x=\exp(y)$ then evaluate the integral $\int_{-\infty}^\infty \frac{\exp(ay)-\exp(by)}{\exp(y)-1}=\pi(\cot(a\pi)-\cot(b\pi))$ using contour integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave the details as an exercise (that I can't be bothered completing!)&lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2010-10-17T11:45:35.917" CommentCount="0" />
  <row Id="7017" PostTypeId="2" ParentId="7004" CreationDate="2010-10-17T11:51:47.543" Score="0" ViewCount="0" Body="&lt;p&gt;It's an well known function, continuous at all irrationals and discontinuous at all rationals. I will give sketch of the proof below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's see why it's continuous at irrationals. To be continuous at an irrational point $x_0$, we need to show that $\forall\delta&gt;0\exists\epsilon&gt;0:|x-x_0|&amp;lt;\epsilon\Rightarrow |f(x)-f(x_0)|&amp;lt;\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that within any range of length 1, in particular within $[x_0-1/2,x_0+1/2)$, $f(x)$ takes the value 1/2 exactly once, 1/3 twice, 1/4 thrice, and so on, taking 1/k exactly k+1 times. So if we remove these points upto say $\lceil1/\delta\rceil$, we will be removing only finitely many points, and we will be left with points on which $f(x)&amp;lt;1/\lceil1/\delta\rceil\le\delta$. Notice that $x_0$ being irrational was not one of the points that was removed. So find the closest point $a$ to $x_0$ that was removed, and call the distance $|x_0-a|$ as $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, one can argue that the function is discontinuous for rationals. If $x_1$ is a rational point, notice that if we remove all the points of the form $p/q$ upto $0\lt q\le N$, any interval around $x_1$ will contain all points with $f(x)\lt 1/N$. So we can form intervals around $x_1$ with arbitrarily small $f$. Hence $f$ is discontinuous at $x_1$ since $f(x_1)&gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="1310" LastActivityDate="2010-10-17T11:51:47.543" />
  <row Id="7018" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T11:52:02.687" Score="1" ViewCount="0" Body="&lt;p&gt;I found Krantz's &lt;em&gt;A Primer of Mathematical Writing&lt;/em&gt; helpful, though not deep.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-17T11:57:55.433" LastActivityDate="2010-10-17T11:57:55.433" CommunityOwnedDate="2010-10-17T11:52:02.687" />
  <row Id="7019" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T11:54:45.907" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;The Chicago Manual of Style&lt;/em&gt; is one of the standard style guides for English and has lot of useful advice for academic writing. &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-17T11:54:45.907" CommunityOwnedDate="2010-10-17T11:54:45.907" />
  <row Id="7020" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T11:57:20.970" Score="3" ViewCount="0" Body="&lt;p&gt;A set of lecture notes from a course on mathematical writing run by Don Knuth&#xA;&lt;a href=&quot;http://tex.loria.fr/typographie/mathwriting.pdf&quot; rel=&quot;nofollow&quot;&gt;http://tex.loria.fr/typographie/mathwriting.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-17T12:05:31.033" LastActivityDate="2010-10-17T12:05:31.033" CommunityOwnedDate="2010-10-17T11:57:20.970" />
  <row Id="7021" PostTypeId="2" ParentId="7013" CreationDate="2010-10-17T12:20:13.063" Score="2" ViewCount="0" Body="&lt;p&gt;To finish off Simon's answer, see Example 4.3.3 (p. 244–245) in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521534291&quot; rel=&quot;nofollow&quot;&gt;Complex Variables: Introduction and Applications&lt;/a&gt; by Ablowitz &amp;amp; Fokas.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-17T12:20:13.063" />
  <row Id="7022" PostTypeId="1" AcceptedAnswerId="7023" CreationDate="2010-10-17T12:46:11.870" Score="3" ViewCount="135" Body="&lt;p&gt;Young's inequality for convolutions states that if $1 \leq p, q, r \leq \infty$ satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{q} + 1 = \frac{1}{p} + \frac{1}{r}$$&#xA;for all $f \in L^p(G)$ and all $g \in L^r(G)$ where $g$ and $g'$ have the same $L^r$-norm  and $g'(x) = g(x^{-1})$ ($G$ is a topological group) we have that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|f * g\|_q \leq \|g\|_r \|f\|_p.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Grafakos claims we can use this to prove the following inequality due to Hardy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \int_0^\infty \left ( \frac{1}{x} \int_0^x |f(t)| \, dt \right )^p \, dx \right  )^{1/p} \leq \frac{p}{p - 1} \|f\|_{L^p(0, \infty)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint is to consider on the multiplicative group $(\mathbb{R}^+, \frac{dt}{t})$ the convolution of $|f(x)| x^{1/p}$ and ${x^{-1/p'}} 1_{[1, \infty)}$. So if we use this, the RHS is no problem, it is just a direct computation (I can add it if someone wants it for future reference). However, if I compute the convolution I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{x - 1} |f(t)| (y(t - y))^{1/p'} \, dt$$ but I don't see how this is larger (or equal) to the inner integral on the LHS of the inequality. Any suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Willie Wong points out below, the convolution is wrong. It is an multiplicative group, not an additive one.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-17T13:52:17.007" LastActivityDate="2010-10-18T11:13:01.237" Title="An inequality by Hardy." Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="7023" PostTypeId="2" ParentId="7022" CreationDate="2010-10-17T13:45:55.163" Score="4" ViewCount="0" Body="&lt;p&gt;You are doing the convolution wrong. On the multiplicative group $(\mathbb{R}_+, dt/t)$, the convolution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f * g(x) = \int_0^\infty f(y) g(x/y) dy/y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(for harmonic analysis on an abelian group, you need to re-interpret the $+$ and $-$ signs in formulae to be the group binary operator). If you plug in, as $g$, the weight function Grafakos suggested, you should get exactly the LHS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be more general: let $(G,\mu)$ be an Abelian group with an invariant measure $\mu$, where $\cdot$ denotes the group binary operator, then the convolution of two functions $f,g: G\to \mathbb{R}$ is defined as the function $G\to \mathbb{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f*g(y) = \int_G f(x) g(y\cdot x^{-1})\mu(dx) $$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-18T11:13:01.237" LastActivityDate="2010-10-18T11:13:01.237" CommentCount="10" />
  <row Id="7024" PostTypeId="1" CreationDate="2010-10-17T13:49:03.350" Score="1" ViewCount="35" Body="&lt;p&gt;Let $M$ be the mean matrix of a multitype branching process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out whether $A = (I-M)^{-1}$ has any signifiance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think $A_{ij}$ would stand for the number of times that type $j$ would be generated from type $j$, but I can't find any material on that in the web. Anyone knows anything about that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2010-11-16T16:44:18.893" Title="what is the connection between the mean matrix and properties of a branching process?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="7025" PostTypeId="1" AcceptedAnswerId="7060" CreationDate="2010-10-17T13:51:21.977" Score="3" ViewCount="165" Body="&lt;p&gt;We have two functions:&#xA;$F(x)$ and $G(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the improper integral has the value 1.&#xA;$$\int_{1}^{\infty} F(x) G(x) \mathrm{d}x = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we find the value of the integral, where c is a constant?&#xA;$$\int_{1}^{\infty} F(x+c) G(x) \mathrm{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in generalized answers and not specific examples.&#xA;Or, if any links/books' info are given where I can read similar problems, that would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Due to the highly generalized nature of this question, I am giving the examples I am working upon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) = frac(x)$$&#xA;Where frac(x) denotes the fractional part of x, and&#xA;$$G(x) = \frac{sin(xlog(x))}{x^{h+1}}$$&#xA;where constants are $0&amp;lt; h &amp;lt; 1$ and $c = 1/2$&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2010-10-17T14:29:50.863" LastActivityDate="2010-10-18T02:55:15.560" Title="A question on integrals" Tags="&lt;improper-integrals&gt;" AnswerCount="5" />
  <row Id="7027" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:10:52.173" Score="1" ViewCount="0" Body="&lt;p&gt;You asked for references for problems similar to yours.  One such is the &lt;em&gt;convolution&lt;/em&gt; of two functions $f$ and $g$, one form of which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(f*g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t - \tau) d \tau.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convolutions have lots of interesting properties and interpretations.  See MathWorld's article on &lt;a href=&quot;http://mathworld.wolfram.com/Convolution.html&quot; rel=&quot;nofollow&quot;&gt;convolutions&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-18T02:55:15.560" LastActivityDate="2010-10-18T02:55:15.560" CommentCount="1" />
  <row Id="7028" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:13:32.807" Score="0" ViewCount="0" Body="&lt;p&gt;If $c$ is negative, the second integral involves values $F(t)$ for $t&amp;lt;1$, which can be changed arbitrarily without affecting the first integral! But even if you restrict yourself to positive $c$, I guess there's nothing you can say in general without additional information.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-17T14:13:32.807" />
  <row Id="7029" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:21:31.963" Score="3" ViewCount="0" Body="&lt;p&gt;Do you know anything about $F$ and $G$? If not, then certainly no. Let $G(x) = 1$ if $x \leq 2$ and $0$ otherwise. Let $F(x) = 1$ if $x&amp;lt; 2$; we leave $F(x)$ undefined elsewhere for now. Then $\int_1^{\infty} F(x)G(x) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $h(c) = \int_1^\infty F(x+c)G(x)dx$ defined on $0 &amp;lt; c$. I claim that for any continuously differentiable function $H(c)$ such that $H(0) = 1$, I can find an extension to $F(x)$ (i.e. filling in the undefined spots) such that $h(c) = H(c)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Notice that by definition $h(c) = \int_{1+c}^{c+2} F(x) dx$. Then $h'(c) = F(c+2) - F(c+1)$.  So start with $F(x) = 1$ for $1\leq x &amp;lt; 2$, then define $F(x) = H'(x-2) + H'(x - 3) + \ldots + H'(x - k) + F(x - k + 1)$, where $k$ is the integer where $1 \leq x - k + 1 &amp;lt; 2$. The by construction, $h'(c) = H'(c)$ and $h(0) = H(0)$, so the two functions agree. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From the above we can conclude that just knowing $\int_1^\infty F(x) G(x) dx = 1$ tells us absolutely nothing about $\int_1^\infty F(x+c) G(x) dx$; it can be pretty much anything you want it to be. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-17T14:21:31.963" CommentCount="1" />
  <row Id="7030" PostTypeId="2" ParentId="6625" CreationDate="2010-10-17T14:29:54.280" Score="2" ViewCount="0" Body="&lt;p&gt;The article you should be looking at, apart from &lt;a href=&quot;http://dx.doi.org/10.1080/00207167308803075&quot; rel=&quot;nofollow&quot;&gt;David Levin's original paper&lt;/a&gt; is E.J. Weniger's &lt;a href=&quot;http://arxiv.org/abs/math/0306302&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Nonlinear sequence transformations for the acceleration of convergence and the summation of divergent series&lt;/em&gt;&lt;/a&gt;; in there he gives a short FORTRAN routine for implementing Levin's transformations. A more elaborate &lt;a href=&quot;http://doi.acm.org/10.1145/356044.356051&quot; rel=&quot;nofollow&quot;&gt;implementation by Fessler, Ford, and Smith&lt;/a&gt; is &lt;a href=&quot;http://www.netlib.org/toms/602&quot; rel=&quot;nofollow&quot;&gt;available at Netlib&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to show how easy it is to implement the Levin transformation, here's a demonstration program I wrote for the &lt;a href=&quot;http://education.ti.com/educationportal/sites/US/productDetail/us_ti83p.html&quot; rel=&quot;nofollow&quot;&gt;TI-83 Plus calculator&lt;/a&gt; for summing the alternating harmonic series, based on Weniger's short FORTRAN routine (comments are delimited by a backslash):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PROGRAM:XTRPOL&#xA;Prompt N \\ number of terms of the series to use&#xA;1→P:0→S \\ P: alternating sign S: stores partial sums&#xA;For(K,1,N)&#xA;P/K→U \\ K-th term of the series, change to sum a different series&#xA;S+U→S&#xA;1→B \\ adjustable parameter for the Levin transformation&#xA;(B+K-1)U→T \\ required for the Levin u-transform; for t version, remove the (B+K-1)&#xA;prgmLEVINT&#xA;Pause Y&#xA;-P→P&#xA;End&#xA;Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which uses the subroutine&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PROGRAM:LEVINT&#xA;(B+K-1)ֿ¹→W&#xA;W/T→U&#xA;If K=1&#xA;ClrList ∟DL,∟NL&#xA;U→∟DL(K)&#xA;SU→∟NL(K)&#xA;1-W→V&#xA;For(J,K-1,1,-1)&#xA;(B+J-1)W→U&#xA;∟NL(J+1)-U∟NL(J)→∟NL(J)&#xA;∟DL(J+1)-U∟DL(J)→∟DL(J)&#xA;WV→W&#xA;End&#xA;10^(99)→Y&#xA;If abs(∟DL(1))≥10^-99&#xA;∟NL(1)/∟DL(1)→Y&#xA;Y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I haven't bothered to implement the stopping rules described by Fessler, Ford, and Smith in that demo program, but it's doable. Translating that short routine to your language of choice should be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-17T14:29:54.280" CommentCount="4" />
  <row Id="7031" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T14:36:56.637" Score="1" ViewCount="0" Body="&lt;p&gt;In reasonably general situations (using, e.g, distributions if needed) the second integral can be differentiated with respect to $c$ and the derivatives will be $\int F^{(n)}(x)G(x) dx$.  Just as a function is not determined by its value at a point, these derivatives are not determined by the value at $c=0$, so even the local behavior of perturbations of the original integral (the asymptotics of the second integral for small $c&gt;0$) is not determined by the integral with $c=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to think of examples where the second integral is determined, no matter how constrained $F$ and $G$ are.  Consider the situation where $F(x)$ and $1/G(x)$ are polynomials of degrees 2 and 4 with positive coefficients.  This is a very well behaved problem but the variation with respect to $c$ is undetermined by a single piece of information.  If you require integer coefficients or a similar discretization then there is some hope of recovering $F$ and $G$ from the value of the first integral, but this is a Diophantine problem and no longer a question of real analysis.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-17T14:36:56.637" />
  <row Id="7032" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T14:49:23.220" Score="1" ViewCount="0" Body="&lt;p&gt;Here is another: &lt;a href=&quot;http://books.google.com/books?id=9gQd2fJA7Y4C&amp;amp;printsec=frontcover&amp;amp;dq=Nicholas+J.+Higham+-+Handbook+of+Writing+for+the+Mathematical+Sciences&amp;amp;hl=pt-PT&amp;amp;ei=IQy7TJXNJM7rsgaq-PHIDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Handbook of Writing for the Mathematical Sciences - Nicholas J. Higham&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-10-17T14:49:23.220" />
  <row Id="7033" PostTypeId="2" ParentId="7002" CreationDate="2010-10-17T15:09:07.667" Score="1" ViewCount="0" Body="&lt;p&gt;The canonical map $G/H \to G/K$ is surjective. The fiber of $gK$ is $\{gkH : k \in K\}$, which can be identified with $K/H$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-17T15:09:07.667" />
  <row Id="7034" PostTypeId="1" AcceptedAnswerId="7045" CreationDate="2010-10-17T15:31:15.787" Score="11" ViewCount="261" Body="&lt;p&gt;Prove that the sequence $c_{1} = 1$, $c_{(n+1)}= 4/(1 + 5c_{n})$     , $n \geq 1$  is convergent and find its limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok so up to now I've worked out a couple of things.&lt;br&gt;&#xA;$c_1 = 1$&lt;br&gt;&#xA;$c_2 = 2/3$&lt;br&gt;&#xA;$c_3 = 12/13$&lt;br&gt;&#xA;$c_4 = 52/73$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the odd $c_n$ are decreasing and the even $c_n$ are increasing.  Intuitively, it's clear the the two sequences for odd and even $c_n$ are decreasing/increasing less and less.&#xA;Therefore it seems like the sequence may converge to some limit $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sequence has a limit, let $L=\underset{n\rightarrow \infty }{\lim }a_{n}.$   Then $L = 1/(1+5L).$&#xA;So we yield $L = 4/5$ and $L = -1$.  But since the even sequence is increasing and &gt;0, then $L$ must be $4/5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here I am stuck.  I'm not sure how to go ahead and show that the sequence converges to this limit (I tried using the definition of the limit but I didn't manage) and and not sure about the separate sequences how I would go about showing their limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few notes :&#xA;I am in 2nd year calculus.&#xA;This is a bonus question, but I enjoy the challenge and would love the extra marks.&#xA;Note : Once again I apologize I don't know how to use the HTML code to make it nice.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-17T16:32:33.857" LastActivityDate="2010-10-17T23:00:44.417" Title="Prove that the sequence$ c_1 = 1$, $c_{n+1} = 4/(1 + 5c_n) $     , $ n \geq 1$   is convergent and find its limit." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="2" />
  <row Id="7035" PostTypeId="2" ParentId="7024" CreationDate="2010-10-17T15:56:10.150" Score="1" ViewCount="0" Body="&lt;p&gt;The matrix $M$ has $(i,j)$th entry $M_{ij}=E(Z^j_1 | Z_0=e_i)$, that is, $M_{ij}$ is the mean number of type $j$ offspring starting with one type $i$ individual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for the $n$th power of $M$ we get $M_{ij}^n=E(Z^j_n | Z_0=e_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding $A=(I-M)^{-1}=\sum_{n\geq 0} M^n$, we get (ignoring issues of convergence), that  $A_{ij}=E(\sum_{n\geq 0} Z^j_n | Z_0=e_i)$,  the mean number of type $j$ descendants starting with a single type $i$ individual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference is Harris's book &lt;em&gt;The Theory of Branching Processes&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-17T15:56:10.150" />
  <row Id="7036" PostTypeId="2" ParentId="7013" CreationDate="2010-10-17T15:57:02.653" Score="1" ViewCount="0" Body="&lt;p&gt;This looks reminiscent of &lt;em&gt;Frullani's Integrals&lt;/em&gt;; see for example the &lt;a href=&quot;http://arxiv.org/abs/1005.2940&quot; rel=&quot;nofollow&quot;&gt;article:&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-10-17T15:57:02.653" CommentCount="3" />
  <row Id="7037" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T16:02:06.463" Score="2" ViewCount="0" Body="&lt;p&gt;Nicholas Higham's book &lt;em&gt;Handbook of writing for the mathematical sciences&lt;/em&gt; is detailed, clear, and overall very useful. He covers all the four bullet points that you have raised. Moreover, he gives an annotated reference to useful style guides, technical writing guides, dictionaries, and so on. I am sure you won't be disappointed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I guess you might already be familiar with the classic: &lt;em&gt;The Elements of Style&lt;/em&gt; by Strunk &amp;amp; White --- this book is a gem on writing, and is worth reading again and again every few months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here is a paper with very practical advice. This paper really underscores the importance of structure, and offers concrete advice on how to attain good structure: &lt;em&gt;The Science of Scientific Writing&lt;/em&gt;, by G. D. Gopen and J. A. Swan. The&#xA;fundamental thesis of the paper is: Writing to meet the reader’s expectation.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-10-17T16:36:44.010" LastActivityDate="2010-10-17T16:36:44.010" CommentCount="1" />
  <row Id="7038" PostTypeId="2" ParentId="7011" CreationDate="2010-10-17T16:05:41.107" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://retro.seals.ch/digbib/view?rid=ensmat-001:1970:16%3a:59&amp;amp;id=browse&amp;amp;id2=browse2&quot; rel=&quot;nofollow&quot;&gt;How to write mathematics&lt;/a&gt; is a free essay by Paul Halmos which fulfills your first three criteria. It is short, but has several good points.&lt;/p&gt;&#xA;" OwnerUserId="1823" LastActivityDate="2010-10-17T16:05:41.107" CommentCount="2" />
  <row Id="7039" PostTypeId="2" ParentId="6998" CreationDate="2010-10-17T16:09:41.617" Score="1" ViewCount="0" Body="&lt;p&gt;A simple way to present this and related results is to employ concrete fractional models of cyclic groups, i.e. $\rm\ C_n\: =\: \frac{1}n\mathbb Z\ \:(mod\ \mathbb Z)\ =\ \{ 0,\: 1/n,\: 2/n,\: \ldots,\: (n-1)/n\}\:.\ $ Suppose $\rm G$ is a nontrivial subgroup. Since $\rm G\:$ is closed under subtraction, every element of $\rm G\:$ must be an integer multiple of its least element $\rm\:g\:$ (else if $\rm\: h\:$ were the least nonmultiple, then $\rm\: h - g\:$ would be a smaller one). Thus $\rm\:G = \langle g \rangle\:$ is cyclic. $\ $ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-17T22:34:22.857" LastActivityDate="2010-10-17T22:34:22.857" />
  <row Id="7040" PostTypeId="1" AcceptedAnswerId="7048" CreationDate="2010-10-17T16:15:03.007" Score="4" ViewCount="98" Body="&lt;p&gt;Why does $ \text{Var}(Y) = E(\text{Var}(Y|X))+ \text{Var}(E(Y|X))$? What is the intuitive explanation for this? In laymen's terms it seems to say that the variance of $Y$ equals the expected value of the conditional variance plus the variance of the conditional expectation. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-24T09:45:05.033" Title="Conditional  and Total Variance" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="7041" PostTypeId="1" CreationDate="2010-10-17T16:15:13.983" Score="5" ViewCount="161" Body="&lt;p&gt;Hi, I'm a math PhD who has been asked to interview a high school student and determine what he/she is interested in and how strong the student is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually I would want them to talk as much as possible about what classes they've taken and what parts of the classes they like the best. Generally I'd like to allow them to get comfortable talking about what they like rather than what they think I'd like. However, that might not work this time as the student is a non-native english speaker and may get less comfortable as I try to get him/her to talk more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far I'm thinking about watching how he/she solves different problems in different areas of math, but I would like to know if anyone has done this before or has any suggestions for evaluating a student who might not be comfortable talking much.&lt;/p&gt;&#xA;" OwnerUserId="2580" LastEditorUserId="2580" LastEditorDisplayName="" LastEditDate="2010-10-17T17:35:42.207" LastActivityDate="2010-10-17T20:21:17.217" Title="How to assess a non-natively english speaking high-schooler's mathematical ability?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="7042" PostTypeId="1" CreationDate="2010-10-17T16:17:05.293" Score="0" ViewCount="37" Body="&lt;p&gt;Brownian motion is a self-similar process with a Hurst index $H = \frac{1}{2}$. Is it a coincidence that the probability of a positive (forward) or negative (backward) step of the random walk that generates Brownian motion is also $\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-17T16:17:05.293" Title="Is there a connection between the Hurst index and the probabiliy of a random walk that produces Brownian motion?" Tags="&lt;probability&gt;&lt;statistics&gt;" CommentCount="1" />
  <row Id="7043" PostTypeId="2" ParentId="7012" CreationDate="2010-10-17T16:17:19.180" Score="0" ViewCount="0" Body="&lt;p&gt;In the end here when you solve (x-1)(3x-5) &amp;lt;= 0   You are really finding the bounds of what x can be.  In this case, the smallest number is 1, and the greatest number is 5/3.  Therefore x has to be between 1 and 5/3.&#xA;Which is where 1 &amp;lt;= x &amp;lt;= 5/3 comes from.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-10-17T16:17:19.180" CommentCount="1" />
  <row Id="7044" PostTypeId="2" ParentId="6906" CreationDate="2010-10-17T16:32:32.047" Score="2" ViewCount="0" Body="&lt;p&gt;The quote in Mariano's answer is from the introduction to &lt;a href=&quot;http://www.jstor.org/stable/108572&quot; rel=&quot;nofollow&quot;&gt;Sylvester's paper.&lt;/a&gt; Typical of Sylvester's mathematical papers, he used so many nonstandard terms in that paper that he appended a five-page &quot;Glossary of new or unusual Terms, or of Terms used in a new or unusual sense in the preceding Memoir&quot;. There he lists:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Inertia. -- The unchangeable number of integers in the excess of positive over negative signs which adheres to a quadratic form expressed as the sum of positive and negative squares, notwithstanding any real linear transformations impressed upon such form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sylvester did similarly for many mathematical terms, i.e. coined them or used them in a &quot;new or unusual ways&quot; mathematically. You can find many such examples in Jeff Miller's &lt;a href=&quot;http://jeff560.tripod.com/mathword.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Known Uses of Some of the Words of mathematics,&lt;/a&gt; including: allotrious factor, anallagmatic, Bezoutiant, catalecticant, combinant covariant cumulant cyclotomy, cyclotomic, dialytic, discriminant, Hessian, invariant, isomorphic, Jacobian, latent, law of intertia of quadratic forms, matrix, minor, nullity, plagiograph, quintic, Schur complement, sequence, syzygy, totient, tree, umbral calculus, umbral notation, universal algebra, x/y/z-coordinate, zero matrix, zetaic multiplication. Please see each entry for  Sylvester's role - some are major, others are minor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently Sylvester's penchant for colorfully naming mathematical objects arose from his love of language and poetry. Indeed, &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Sylvester.html&quot; rel=&quot;nofollow&quot;&gt;Karen Parshall&lt;/a&gt; wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sylvester's love of poetry and language manifested itself in notable ways even in his mathematical writings. His mastery of French, German, Italian, and Greek was often reflected in the mathematical neologisms - like &quot;meicatecticizant&quot; and &quot;tamisage&quot; - for which he gained a certain notoriety. Moreover, literary illusions, poetic quotations, and unfettered hyperbole spiced his published papers and lectures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sylvester wrote about such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Perhaps I may without immodesty lay claim to the appellation of Mathematical Adam, as I believe that I have given more names (passed into general circulation) of the creatures of mathematical reason than all the other mathematicians of the age combined.-- James Joseph Sylvester, Nature 37 (1888), p. 152.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find a short &lt;a href=&quot;http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Sylvester.html&quot; rel=&quot;nofollow&quot;&gt;Sylvester biography here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-17T17:53:28.147" LastActivityDate="2010-10-17T17:53:28.147" CommentCount="5" />
  <row Id="7045" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T16:47:59.260" Score="2" ViewCount="0" Body="&lt;p&gt;Well done! Your observations are correct and can be completed to give a proof that the sequence is convergent to $\frac{4}{5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tough part in proving your observations it to prove that the odd/even sub-sequences are monotonic and appropriately bounded (what do I mean by this?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $$ c_{2n-1} \ge \frac{4}{5} \ge c_{2n} \ \ \forall n \ge 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try writing $\displaystyle c_{n+2}$ in terms of $\displaystyle c_{n}$ and see if that helps you prove the above bounds (and as a next step, the monotonicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For a simpler proof of the above bound, you can also try showing that if $c_{n} \ge 4/5$ then $c_{n+1} \le 4/5$ and similarly if $c_n \le 4/5$ then $c_{n+1} \ge 4/5$).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-17T18:52:25.757" LastActivityDate="2010-10-17T18:52:25.757" CommentCount="6" />
  <row Id="7046" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T16:52:04.490" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a more explicit version of my hint.  I said it would be easier to show that $d_n = c_n - \frac{4}{5}$ tends to zero as $n \to \infty$.  This gives $d_1 = \frac{1}{5}$ and the recurrence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{n+1} + \frac{4}{5} = \frac{4}{1 + 5(d_n + \frac{4}{5})} = \frac{4}{5} \frac{1}{1 + d_n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{n+1} = - \frac{4}{5} \frac{d_n}{1 + d_n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see what to do from here?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Some more hints.  If the $1 + d_n$ in the denominator were just a $1$, we would be done because then $|d_n|$ would decrease exponentially at least as fast as $\left( \frac{4}{5} \right)^n$.  But it's not.  However, $d_n$ is small, so the denominator should be close enough to $1$ that this argument should still go through.  More precisely, you just need to find a constant $0 &amp;lt; c &amp;lt; \frac{1}{5}$ such that you can prove that $|d_n| \le c$, say for $n \ge 2$.  (Maybe by induction.)  From there it will follow that $|d_n|$ actually decays exponentially at least as fast as $\left( \frac{4}{5(1 - c)} \right)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of a general technique called &quot;bootstrapping,&quot; where you use weak estimates together with relations that a sequence satisfies to get stronger estimates.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-17T23:00:44.417" LastActivityDate="2010-10-17T23:00:44.417" CommentCount="9" />
  <row Id="7047" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T16:59:15.437" Score="0" ViewCount="0" Body="&lt;p&gt;Here's one way to prove it: let $f(x) = 4/(1+5x)$. Say $|x-4/5| \le C$ for some constant $C$. Can you find $C$ and some constant $0 \le k &amp;lt; 1$ so that if $|x-4/5| \le C$, then $|f(x)-4/5| \le k|x-4/5|$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do this, then you can iterate to get $|f^j(x)-4/5| \le k^j |x-4/5|$, for all $j$, and so if you make $j$ large enough then you can get $f^j(x)$ as close to $4/5$ as you like.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-17T16:59:15.437" />
  <row Id="7048" PostTypeId="2" ParentId="7040" CreationDate="2010-10-17T17:25:35.100" Score="6" ViewCount="0" Body="&lt;p&gt;A rigorous proof is &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_total_variance&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;; it relies on the law of total expectation, which says that $E(E(X|Y))=E(X)$.  The intuitive explanation of &lt;em&gt;that&lt;/em&gt; is that $E(X|Y)$ is the expected value of $X$ given a particular value of $Y$, and that $E(E(X|Y))$ is the expected value of that over all values of $Y$.  So $Y$ no longer matters, and we're just looking at $E(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variance law is a bit more difficult to parse, but this is what it says to me.  &quot;How much does $Y$ vary?  We &lt;em&gt;expect&lt;/em&gt; it to vary by the average value of the variances we get by fixing $X$.  But even when we fix $X$, there is some swing in $Y$, and thus swing in $E(Y|X)$.  So we add on the variance of $E(Y|X)$.  The first term is the expected variance &lt;em&gt;from the mean&lt;/em&gt; of $Y|X$; the second is the variance &lt;em&gt;of that mean&lt;/em&gt;.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-17T17:25:35.100" />
  <row Id="7049" PostTypeId="1" CreationDate="2010-10-17T18:28:33.240" Score="2" ViewCount="72" Body="&lt;p&gt;Need to use Fermat's Little Theorem (Let &lt;strong&gt;p&lt;/strong&gt; be a prime number and let &lt;strong&gt;a&lt;/strong&gt; be an integer. Then &lt;strong&gt;a&lt;/strong&gt; ^ &lt;strong&gt;p&lt;/strong&gt; = &lt;strong&gt;a&lt;/strong&gt; mod &lt;strong&gt;p&lt;/strong&gt;. If &lt;strong&gt;p&lt;/strong&gt; does not divide &lt;strong&gt;a&lt;/strong&gt; then &lt;strong&gt;a&lt;/strong&gt; ^ &lt;strong&gt;p&lt;/strong&gt;-1 = 1 mod &lt;strong&gt;p&lt;/strong&gt;.)&#xA;154 is not prime, but 154 = 22*7 and 23 is prime, so a^22 = 1 mod 23.&#xA;a^154 = a^22 * a^7 = 1 mod 23 * 6 mod 23 = 6 mod 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure how to proceed further. Would be grateful for any help.&lt;/p&gt;&#xA;" OwnerUserId="2401" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-17T18:32:42.517" LastActivityDate="2010-10-17T18:38:26.607" Title="Prove that for any integer a  &gt; 1 s.t. gcd(a, 23) = 1, 23 divides a^154 - 1." Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="7050" PostTypeId="2" ParentId="7049" CreationDate="2010-10-17T18:32:09.647" Score="3" ViewCount="0" Body="&lt;p&gt;You solved it! apart from an arithmetic error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a^{mn} = (a^{m})^n$, not $\displaystyle a^{m}* a^{n}$ (which is actually $\displaystyle a^{m+n}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, not sure how you got $a^7 = 6 \mod 23$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-17T18:32:09.647" CommentCount="6" />
  <row Id="7051" PostTypeId="2" ParentId="7049" CreationDate="2010-10-17T18:38:26.607" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ a^{22}\ \equiv 1\ \Rightarrow\ a^{22\: N}\ \equiv\ 1^N\ \equiv\ 1\:.\ $ Now put $\rm\ N = 7\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-17T18:38:26.607" CommentCount="1" />
  <row Id="7052" PostTypeId="2" ParentId="7041" CreationDate="2010-10-17T18:46:26.497" Score="1" ViewCount="0" Body="&lt;p&gt;Well, I'm a non-native English speaker, so I can provide some insight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that one of the first problems would be names and notation in general. As far as I can see, there is no easy way to solve this unless you already know the customs of the country the student is coming from, but your education will probably make it easier to guess what means what. For example in my country we use $log_{base}number$ for general case logarithms, $lg$ for $log_{10}$ and $ln$ for $log_e$. Another example would be popular sets. In some countries 0 is a natural number and in some other countries it isn't. Another example would be popular trigonometric functions. In my country for example secant and cosecant are considered deprecated and aren't used. Same thing with grads as angle measurement units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems with this part can be especially tricky because they can stay undetected for a long time. I've seen some math problems whose solutions depend on whether zero is a natural number or not. Another example here can be integrals. As far as I can see the most popular name in English for operation opposite of finding a derivative is antiderivation. In my area that's called indefinite integration and the indefinite part if often omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still I think that most of the notational problems can be solved with Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next part would be names of fields of mathematics and popular theorems. How big problems this is going to be is closely related to which language terms come from. Sometimes terms are going to be shared between languages and sometimes there will be no direct translation. For example in my language, there is no term for calculus. Instead we use mathematical analysis, which as far as I know is superset of calculus.&lt;br&gt;&#xA;Also in some cultures some theorems are named after famous mathematicians and in some other cultures they are named after what the theorem describes. I can't find any examples right now, but I know I encountered such problems when asking few questions here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem could be non-existence of a concept. For example I was unable to find any mention of &lt;a href=&quot;http://translate.google.com/translate?js=n&amp;amp;prev=_t&amp;amp;hl=sr&amp;amp;ie=UTF-8&amp;amp;layout=2&amp;amp;eotf=1&amp;amp;sl=auto&amp;amp;tl=en&amp;amp;u=http%3A%2F%2Fru.wikipedia.org%2Fwiki%2F%25D0%25A3%25D1%2580%25D0%25B0%25D0%25B2%25D0%25BD%25D0%25B5%25D0%25BD%25D0%25B8%25D0%25B5_%25D0%259C%25D0%25B5%25D1%2589%25D0%25B5%25D1%2580%25D1%2581%25D0%25BA%25D0%25BE%25D0%25B3%25D0%25BE&quot; rel=&quot;nofollow&quot;&gt;Ivan Vsevolodovich Meshcherskiy's equation&lt;/a&gt; &lt;a href=&quot;http://ru.wikipedia.org/wiki/%D0%A3%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5_%D0%9C%D0%B5%D1%89%D0%B5%D1%80%D1%81%D0%BA%D0%BE%D0%B3%D0%BE&quot; rel=&quot;nofollow&quot;&gt;(original article here&lt;/a&gt;) for bodies with variable mass in any English language web-sites. Here it's considered one of the fundamental rocketry equations, right next to Tsiolkovsky's formula. While the example is related more to physics than mathematics, be prepared to find similar examples in mathematics if the student is coming from a different culture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you said that student knows some English, here's my recommendation: Try starting a conversation about some topic in English and see if student is familiar with it. If it doesn't ring a bell, be ready to provide simple examples related to the area. I sometimes had problems finding exact name for some field of mathematics but immediately recognized it as soon as I saw a few simple problems. Another good idea would be to make a list of alternative names for every operation and see if some of them are familiar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another important thing is to have at least some familiarity with mathematics program which the student followed. Sometimes focus of what's studied in classes can be a bit different than what's written on paper. I remember now back when I was in secondary school we had formal mathematical proofs in books and they were supposed to be a part of the curriculum, but teachers in schools just decided that proofs are not important and did not teach them. Also in some schools less attention was payed to proper use of terminology. I know some of my friends who didn't know what was derivative and it's graphical representations, but knew that $x^2{'}=2x$ and same thing for integrals. Another point can be some obvious areas which may have been skipped. When  was in secondary school, we didn't learn almost any probability and statistics and did very little work with percentages. On the other hand I see that in some other countries it's expected that someone who has graduated from primary school can without any problems calculate his interest rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of anything else to add at the moment...&lt;/p&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-10-17T19:14:41.027" LastActivityDate="2010-10-17T19:14:41.027" />
  <row Id="7053" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T18:48:47.777" Score="1" ViewCount="0" Body="&lt;p&gt;As shown by the other answers, there are a few nice ways to approach this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could concentrate only on $C_{2n-1},$ say, since if you establish that $C_{2n-1}$ tends to a limit you automatically nail $C_{2n}$ as well, because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{2n} = \frac{4}{1+5C_{2n-1}} \textrm { so }  \lim C_{2n}  = \lim  \frac{4}{1+5C_{2n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's easy to show $C_{2n-1} &gt; 4/5$ and so expand&#xA;$$(5C_{2n-1}-4)(C_{2n-1}+1)&gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and manipulate (add $20C_{2n-1}$ to both sides and take the 4 to the RHS) to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_{2n-1} &gt; \frac{4+20C_{2n-1}}{21+5C_{2n-1}} = C_{2n+1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $C_{2n-1}$ is bounded below by 4/5 the result follows immediately.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-17T18:48:47.777" CommentCount="2" />
  <row Id="7054" PostTypeId="1" AcceptedAnswerId="7057" CreationDate="2010-10-17T19:00:13.097" Score="2" ViewCount="150" Body="&lt;p&gt;So my brain is frazzled which is probably why this seems like a big deal to me right now, but I just can't get over this reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have F = {All 1-1, increasing functions from N to N}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-1: means that every value of the domain maps to some unique value of the range and every value of the range is equal to f(n) for some n.  Hence, since f is 1-1 AND increasing, the only function that exists is the trivial function, f(n) = n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me why that is wrong :-p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ftr, this isn't the homework question.  I am proving uncountability of F, which is why my brain-fart is bothering me more)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: @Prometheus left this as a comment and then deleted it, probably because he didn't wish to be associated with stupidity as great as mine.  The error is that I'm assuming one-to-one =&gt; onto, which it clearly doesn't. &lt;em&gt;huddles in a ball and cries&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="355" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T02:02:50.073" LastActivityDate="2010-11-21T02:02:50.073" Title="Simple Question on 1-1, Increasing integer functions" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7055" PostTypeId="2" ParentId="7034" CreationDate="2010-10-17T19:07:03.847" Score="0" ViewCount="0" Body="&lt;p&gt;So $c_{1} = 1$ and $c_{n+1} = 4/(1+5c_{n})$ for $n \geq 1$. Let $C(x) = \sum_{n \geq 1} c_{n}x^n$. Then maybe try to express $\sum_{n \geq 1} c_{n+1}x^n$ and $\sum_{n \geq 1} \frac{4x^n}{1+5c_{n}}$ in terms of $C(x)$ to get a closed form. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-17T19:07:03.847" CommentCount="1" />
  <row Id="7056" PostTypeId="2" ParentId="7054" CreationDate="2010-10-17T19:11:52.810" Score="2" ViewCount="0" Body="&lt;p&gt;First, a theorem: $|P(\mathbb{N})| &gt; \aleph_0$ (this is Cantor's theorem applied to the natural numbers. I will not go through the proof here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a second theorem: Let $A$ be the set of all finite or co-finite (that is - the complement is finite) subsets of the natural numbers, then $A$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: The set of all subsets of size $n$ can be mapped to a subset of $\mathbb{N}^n$ (as images of functions from $n \to \mathbb{N}$), since this is a countable set itself, and we have countably many $n$ then we have countably many finite subsets of the natural numbers, and since every one of those can be mapped to its complement - we have the needed proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary: There are uncountably many subsets of $\mathbb{N}$ which are infinite and their complement is also infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each of these subsets we know that we can order it, and thus have a 1-1 function from $\mathbb{N}$ onto it, which is also increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the set of all functions 1-1 and increasing from $\mathbb{N}$ to itself is uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope this is clear, if not - let me know which points needs further expansion.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-17T19:11:52.810" CommentCount="1" />
  <row Id="7057" PostTypeId="2" ParentId="7054" CreationDate="2010-10-17T19:16:12.550" Score="3" ViewCount="0" Body="&lt;p&gt;The question has been answered in the comments.  The functions are not assumed to be onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a hint for an approach that is different from Asaf's: You could consider the sequences of differences $f(n+1)-f(n)$.  It is enough to restrict to functions that only increase by 1 or 2 at each integer.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-17T19:16:12.550" CommentCount="1" />
  <row Id="7058" PostTypeId="2" ParentId="7041" CreationDate="2010-10-17T20:15:36.753" Score="1" ViewCount="0" Body="&lt;p&gt;Seeing textbooks or other material, in any language, from which this person has learned mathematics, would be useful and relatively easy to interpret for an Anglophone mathematician.   If these are not available then equivalents may exist on the internet.  Pointing to material in books in English and seeing if it is familiar may also reveal something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the high end, the International Mathematical Olympiad problems are available online in dozens of languages, probably including that of the student's native country.  This could at least tell you whether the student has heard of some concept used in a problem, and you would have a translation into English so you know you are discussing the same thing.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-17T20:21:17.217" LastActivityDate="2010-10-17T20:21:17.217" />
  <row Id="7060" PostTypeId="2" ParentId="7025" CreationDate="2010-10-17T20:59:05.267" Score="1" ViewCount="0" Body="&lt;p&gt;Now that you have updated your question with the fact that $F(x) = \{x\}$ (the standard notation for the fractional part of $x$), there is quite a bit more that can be said.  In particular, you can give an interpretation to $\int_1^{\infty} \{x + c\} G(x) dx$.  Let $0 &amp;lt; c &amp;lt; 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$\{x + c\} = \{x\} + c, 0 \leq \{x\} &amp;lt; 1 - c;$$ &#xA;$$\{x + c\}= \{x\} + c - 1, 1-c \leq \{x\} &amp;lt; 1.$$&#xA;(Sorry; I can't seem to make the &lt;em&gt;cases&lt;/em&gt; command work right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &#xA;$$\int_1^{\infty} \{x + c\} G(x) dx = \int_{1 \leq x &amp;lt; \infty, 0 \leq \{x\} &amp;lt; 1 - c} (\{x \} + c) G(x) dx + \int_{1 \leq x &amp;lt; \infty, 1-c \leq \{x\} &amp;lt; 1 } (\{x \} + c - 1) G(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \int_1^{\infty} \{x \} G(x) dx + c \int_{1 \leq x &amp;lt; \infty, 0 \leq \{x\} &amp;lt; 1 - c} G(x) dx + (c-1) \int_{1 \leq x &amp;lt; \infty, 1-c \leq \{x\} &amp;lt; 1 } G(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 1 + c \int_{1 \leq x &amp;lt; \infty, 0 \leq \{x\} &amp;lt; 1 - c} G(x) dx - (1-c) \int_{1 \leq x &amp;lt; \infty, 1-c \leq \{x\} &amp;lt; 1 } G(x) dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two remaining integrals constitute an average of sorts, weighted to account for the fact that they are being taken over different percentages of the interval $[1,\infty)$.  The first integral gets weighted by $c$ but includes $1-c$ of the interval $[1,\infty)$, as it is being taken over the set $\cup_{i=1}^{\infty} [i,i+1-c)$.  (Remember that $c$ is a fraction between 0 and 1.)  The second integral gets weighted by $1-c$ but includes $c$ of the interval $[1,\infty)$, as it is being taken over the set $\cup_{i=2}^{\infty} [i-c,i)$.  So $\int_1^{\infty} \{x + c\} G(x) dx$ just shifts the weights on the values of $G(x)$ in $\int_1^{\infty} \{x \} G(x) dx$ in the manner I just described.  The resulting value for $\int_1^{\infty} \{x + c\} G(x) dx$ will be either greater or smaller than 1, depending on whether the larger values of $G(x)$ over $[1, \infty)$ tend to clump just above each integer value of $x$ or just below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than this, I think Willie Wong's answer still applies.  In particular, you still can't get an exact answer for $\int_1^{\infty} \{x + c\} G(x) dx$ -- just an interpretation of it.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-17T21:04:21.277" LastActivityDate="2010-10-17T21:04:21.277" CommentCount="2" />
  <row Id="7061" PostTypeId="1" AcceptedAnswerId="7063" CreationDate="2010-10-17T22:42:08.123" Score="2" ViewCount="167" Body="&lt;p&gt;Prove: If X is NOT divisible by 3 and Y is NOT divisible by 3 Then X^2 - y^2 is Divisible by 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find a good way to prove this.. Tried contrapositive and contradiction but cannot see the answer. Anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2010-10-18T17:46:14.870" Title="Prove: If X is NOT divisible by 3 and Y is NOT divisible by 3 Then X^2 - y^2 is Divisible by 3" Tags="&lt;divisibility&gt;&lt;proof&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7062" PostTypeId="2" ParentId="7061" CreationDate="2010-10-17T22:54:57.743" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ mod 3: $\rm\ x\not\equiv 0\ \Rightarrow\ x \equiv \pm 1\ \Rightarrow\ x^2 \equiv 1\:.\ $ Alternatively, without modular arithmetic,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\ $ 3 divides $\rm\: (x-1)\:x\:(x+1)\ $ but not $\:x\ \Rightarrow\ $ 3 divides $(x-1)(x+1) = x^2 - 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if both $\rm\ x^2-1\ $ and $\rm\ y^2-1\ $ are multiples of 3 then so too is their difference $\rm\ x^2 - y^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\: $ This proof easily generalizes to show that $\:24\:$ divides $\rm x^2-y^2\:$ if $\rm\:x\:$  and $\rm\:y\:$ are coprime to $6\:$. See &lt;a href=&quot;http://math.stackexchange.com/questions/910&quot;&gt;here for proof&lt;/a&gt;, and for generalizations using Carmichael's function (universal exponent of a group).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T17:46:14.870" LastActivityDate="2010-10-18T17:46:14.870" CommentCount="3" />
  <row Id="7063" PostTypeId="2" ParentId="7061" CreationDate="2010-10-17T23:00:10.957" Score="4" ViewCount="0" Body="&lt;p&gt;If $x$ and $y$ are not divisible by 3 then they must be of the form $x = 3k \pm 1$ and $y = 3t \pm 1$. Therefore $x^2 - y^2 = 9k^2 \pm 6k + 1 -9t^2 \mp 6t -1 = 3(3k^2 - 3t^2 \pm 2k \mp 2t).$ So $x^2 - y^2$ is divisible by 3.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-10-18T01:29:02.183" LastActivityDate="2010-10-18T01:29:02.183" CommentCount="2" />
  <row Id="7064" PostTypeId="1" CreationDate="2010-10-17T23:33:06.067" Score="5" ViewCount="201" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_division&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; on long division explains the different notations. I still use the European notation I learned in elementary school in Colombia. I had difficulty adapting to the US/UK notation when I moved to the US. However, I did enjoy seeing my classmates' puzzled faces in college whenever we had a professor that preferred the European notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What long division notation do you use and where did you learn it?&lt;/p&gt;&#xA;" OwnerUserId="2586" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-17T00:14:30.427" LastActivityDate="2010-11-18T21:22:15.223" CommunityOwnedDate="2010-11-18T06:46:45.763" Title="Long division notation (census of nations)" Tags="&lt;big-list&gt;&lt;notation&gt;&lt;arithmetic&gt;&lt;elementary&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="7065" PostTypeId="1" AcceptedAnswerId="7074" CreationDate="2010-10-17T23:40:29.537" Score="6" ViewCount="112" Body="&lt;p&gt;My question is essentially this: Why does the principle of transfinite induction not suffice to show the axiom of choice when the sets to be chosen from are indexed by a well ordered set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that one can prove the axiom of finite choice from simple induction. You induct on the size of the system of sets you are choosing from and pick an element from each set. I understand this. However, my grasp of the details is sketchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why does standard induction alone not suffice to show the axiom of choice for systems of countable sets? Doesn't induction show the truth of the statement for all natural numbers, and therefore for any system of sets that can be indexed by the natural numbers (countable sets)? I know this to be false, but I do not know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Why can't the above &quot;proof&quot; that induction implies the AoC for countable sets not be repaired by using transfinite induction? Isn't this the purpose of transfinite induction, to allow one to induct on sets of infinite size? Shouldn't transfinite induction suffice to prove the axiom of choice for any system of sets indexed by a well-ordered set? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading Jech right now, but my knowledge of ordinals and transfinite induction is very, very poor, so I would greatly prefer answers with a great amount of explanation and hand-holding.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastEditorUserId="2591" LastEditorDisplayName="" LastEditDate="2010-10-18T00:50:18.523" LastActivityDate="2010-10-18T01:02:24.780" Title="Transfinite Induction and the Axiom of Choice " Tags="&lt;set-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7066" PostTypeId="2" ParentId="7064" CreationDate="2010-10-17T23:41:07.507" Score="1" ViewCount="0" Body="&lt;p&gt;I use the same one that is used in the U.S.; it is the one that is (or at least, was up until the 80s) taught in Mexico. I had some classmates in high school who had emigrated form Argentina: they used the one described in the Wikipedia article as &quot;European&quot;, not the one described as &quot;Latin America&quot;. (I'll also note that beginners in Mexico would write the full computation as described in &quot;US notation&quot;; doing the computation mentally was more an 'advanced shortcut' than part of the notation, as I remember).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-17T23:41:07.507" CommunityOwnedDate="2010-10-17T23:41:07.507" />
  <row Id="7067" PostTypeId="1" AcceptedAnswerId="7076" CreationDate="2010-10-18T00:01:20.017" Score="4" ViewCount="143" Body="&lt;p&gt;Can someone give me an explanation targeted to a high school student as to why finding thegcd of two numbers is faster using the euclidean algorithm compared to using factorization, there should be no algorithm efficiency involved, just a general explanation, something my brother in grade 9 can understand.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2010-10-18T06:45:19.593" Title="Euclidean Algorithm vs Factorization" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="7068" PostTypeId="2" ParentId="4990" CreationDate="2010-10-18T00:01:51.407" Score="2" ViewCount="0" Body="&lt;p&gt;The approved answer has caused some risibility at mathoverflow, and I'll elaborate on Robin's&#xA;more reasonable comment (but I'm inclined to attribute the descent argument in this case to&#xA;Euler--at least he wrote it down). The version I give in an undergrad  number theory class is this:  First one develops the standard facts about Z[w] where w^2+w+1=0. (It has unique&#xA;factorization, 2 is prime, the units are 1,-1,w,-w,w^2 and -w^2, any element not 0 or a unit&#xA;has absolute value &gt;1, and each congruence class mod 2 is represented by 0 or a unit). Then one notes that it's enough to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem--There are no a,b,c in R with a+b+c=0, abc a non-zero cube and a=b=1 mod 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the theorem is a reductio. Let H be max(/a/,/b/,/c/) and choose a solution&#xA;a,b,c with minimal H. (H^2 is an integer). a, b and c are evidently pairwise prime. Since&#xA;their product is a non-zero cube, each is (unit)(cube). Since a=b=1 mod 2, a=A^3 and&#xA;b=B^3, and we may assume that A=B=1 mod 2. Since abc is a cube, c=C^3 for some C in R. Since&#xA;2 divides c, 2 divides C and H is at least 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let S=Aw+Bw^2, T=Aw^2+Bw, and U=A+B. Then S+T+U=0 while STU is A^3+B^3=-C^3. Also S=T=1 mod 2, while max(/S/,/T/,/U/) is at most 2(H^(1/3)). This contradicts the minimality assumption.&lt;/p&gt;&#xA;" OwnerUserId="2594" LastActivityDate="2010-10-18T00:01:51.407" />
  <row Id="7069" PostTypeId="2" ParentId="7067" CreationDate="2010-10-18T00:24:41.913" Score="3" ViewCount="0" Body="&lt;p&gt;It boils down to the fact that currently known factorization algorithms are much slower than the fastest known gcd algorithms. It would be difficult to say much more than that that would be comprehensible to a 9'th grade student (in fact one might argue that not much more than that is even known to experts).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-18T00:24:41.913" />
  <row Id="7070" PostTypeId="1" CreationDate="2010-10-18T00:27:05.137" Score="4" ViewCount="74" Body="&lt;p&gt;I am trying to calculate the average distance between any two nodes in an $r$-dimensional mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a $3$-dimensional mesh with $n=3$&#xA;&lt;img src=&quot;http://i.imgur.com/uulgk.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To choose any two points in the mesh there are $\left(n^r \left(n^r - 1\right)\right)/2$ ways of doing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we choose 2 points in the mesh $p\lbrace 1, 2, ..., r \rbrace$ and $q\lbrace 1, 2, ..., r \rbrace$ The distance is calculated as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_geometry&quot; rel=&quot;nofollow&quot;&gt;Manhattan Distance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{r} \left| {q_i - p_i} \right|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i have to find the total distance between all nodes and divide that by the the number of ways of choosing two nodes (given above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do this by looking at this 3-dimensional case which should work on the r-dimensional case also but i am having trouble expressing it in some sort of sum notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how i think i can get the total distance between all nodes, i hope this is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this 3-d mesh let's assume the nodes are labelled $\left( 1, 1, 1 \right)$ to $\left( 3, 3, 3\right)$. Lets start at $\left( 1, 1, 1 \right)$ and sum up the distance between this node and all other nodes. Then we move to $\left( 1, 1, 2 \right)$ and sum up the distances between this node and all other nodes except for $\left( 1, 1, 1 \right)$ because we have already counted the distance between $\left( 1, 1, 1 \right)$ and $\left( 1, 1, 2 \right)$. Then we move onto $\left( 1, 1, 3 \right)$ and sum up all except for $\left( 1, 1, 1 \right)$ and $\left( 1, 1, 2 \right)$. We continue this until we permute through in order $\left( 1, 1, 1 \right)$, $\left( 1, 1, 2 \right)$, $\left( 1, 1, 3 \right)$, $\left( 1, 2, 1 \right)$, $\left( 1, 2, 2 \right)$, ect... Does that make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should give me the total distance without counting anything twice. Then i divide that by the number of ways of choosing 2 nodes and i will have the average distance. Does this sound correct to you? Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-11-17T01:44:19.080" Title="r-dimensional mesh question" Tags="&lt;graph-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7071" PostTypeId="1" CreationDate="2010-10-18T00:54:16.490" Score="-2" ViewCount="244" Body="&lt;p&gt;Is the sum of primes up to p a multiple of p?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e Is 1+2+...+p divisible by p and how would you prove it?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2010-11-10T16:56:26.647" Title="Sum of primes up to p is multiple of p?" Tags="&lt;prime-numbers&gt;" AnswerCount="3" />
  <row Id="7072" PostTypeId="2" ParentId="7070" CreationDate="2010-10-18T00:59:27.587" Score="1" ViewCount="0" Body="&lt;p&gt;This approach is correct, but more work than required.  Using the taxicab metric each dimension is independent.  Taking a cube of side n you can count the vertical segments used as follows:  from the bottom layer to the next layer up, we count $n^2$ (points on the bottom layer) times $n^2*(n-1)$ (points above the bottom layer) = $n^4*(n-1)$.  For the next layer you have $2*n^2*n^2*(n-2)$ and so on.  The $n^4$s distribute out and you can just think about the total distance along a line of length n.  This is $\sum_{i=1}^{i=n}i*(n-i)$.  Then multiply by $n^4$ and you have the total usage in the vertical direction.  Then multiply by 3 for the total usage in each direction.  And divide by the number of pairs of points, which you have calculated already.  The extension to different number of dimensions or non-cubical meshes should be clear.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T00:59:27.587" CommentCount="2" />
  <row Id="7073" PostTypeId="2" ParentId="7071" CreationDate="2010-10-18T01:00:17.543" Score="4" ViewCount="0" Body="&lt;p&gt;What you wrote is not the sum of the primes up to $p$.  1 is not prime, and the meaning of your ellipsis is unclear.  Based on one possible interpretation, a counterexample is that 2 does not divide $1+2$.  Based on another, a counterexample is that 3 does not divide $2+3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $p$ is an odd number, and in particular if it is a prime other than 2, then $p$ divides the sum of the first $p$ positive integers, because this sum is $p\cdot\frac{p+1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-18T01:00:17.543" CommentCount="2" />
  <row Id="7074" PostTypeId="2" ParentId="7065" CreationDate="2010-10-18T01:02:24.780" Score="3" ViewCount="0" Body="&lt;p&gt;The problem is that you also do a &quot;meta&quot; induction when proving choice for a finite number of sets and this &quot;meta&quot; induction fails in the transfinite or countable case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more clear, the proof of choice for a number of sets sets usually goes something like this (I'm assuming choice is &quot;the product of nonempty sets is nonempty&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base case is vacuously true (you're given one nonempty set and want to prove it's nonempty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, assume inductively that for any nonempty sets $X_1,...,X_n$, the product $X_1\times...\times X_n$ is nonempty.  Now, given $n+1$ sets, $X_1,...,X_{n+1}$, you want to show their product is nonempty.  By the induction hypothesis, there is an element $y\in X_1\times....\times X_n$.  Also, there is an element $x_{n+1}\in X_{n+1}$ by assumption.  Then $\{\{y,x_{n+1}\}, x_{n+1}\} = (y, x_{n+1})$ is an element of the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the meta level, what you're saying by unraveling the induction is &quot;I have $n+1$ sets $X_1,...,X_{n+1}$, and I know there is an element (which is itself a set because we're doing set theory!) in each $X_i$ which I'll call $x_i$.  By using the axiom of pairing once, I can form the set $\{x_1,x_2\}$.  By using it again, I can form the set $\{\{x_1,x_2\}, x_1\}$, which is the definition of $(x_1,x_2)$.  Using the axiom of pairing two more times, I can form $(x_1, x_2, x_3)$.  In general (i.e., by induction), by using the axiom of pairing $2(n+1)$ times, I can form the set $(x_1,x_2,...,x_{n+1})$.  This set (element) is a member of $X_1\times...\times X_{n+1}$, and hence this product is nonempty!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What goes wrong as soon as you try to adapt this to the countable or larger setting is on the meta level.  Essentially, you'd have to apply the pairing axiom countably (or more) times to create the set $(x_1,...)$ which &quot;should&quot; be in $X_1\times...\times $.  But naive set theory has taught us that just because we think something &quot;should&quot; be a set, doesn't mean it should (see, for example, Russel's Paradox).  Thus, we at least cannot trust this proof to work in countable or larger settings.  Of course, the failure of this approach doesn't mean there is NO way of proving choice from $ZF$, but it's known from other methods that you cannot prove choice from $ZF$ alone.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-18T01:02:24.780" CommentCount="3" />
  <row Id="7075" PostTypeId="2" ParentId="7071" CreationDate="2010-10-18T01:03:24.933" Score="2" ViewCount="0" Body="&lt;p&gt;Or if by the ellipsis you mean the sum of naturals up to p, it is true and not just for primes.  The sum of naturals up to n is $n*(n+1)/2$, which can be proved by induction, and is divisible by n. (as commented, only if n is odd)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-10T16:56:26.647" LastActivityDate="2010-11-10T16:56:26.647" CommentCount="1" />
  <row Id="7076" PostTypeId="2" ParentId="7067" CreationDate="2010-10-18T01:16:37.730" Score="7" ViewCount="0" Body="&lt;p&gt;The Euclidean algorithm is a definite recipe that tell you exactly what to do at any given step; there is no guessing, no trial and error involved. Trying to factor a number will (even with some of the best methods currently known) involve guesses and trial-and-error; trial division is of course the classic example, but even some of our best methods (elliptic curve factorization and number field sieve, to name two) all involve some &quot;random guessing&quot; and trials to try to find factors. Sure, they are more clever ways of testing than simply trying everything out there, but you still usually end up doing a lot of grunt work along the way that leads nowhere (dividing by a number that is not a factor in trial division; not getting good relations in the number field sieve; performing all computations on an elliptic curve modulo $n$ and not finding any non-invertible elements), or performing good-looking computations that end up with a trivial factor ($1$ or $n$). In essence, this is &quot;wasted effort&quot;, waste that simply does not occur with the Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Note that this is a reflection of our current known factoring methods, and not necessarily (as Bill Dubuque points out, we just don't know either way) an inherent difficulty in factoring. You don't want to compare &quot;factoring&quot; with &quot;Euclidean algorithm&quot;, you want to compare &lt;em&gt;specific ways&lt;/em&gt; of factoring with the Euclidean algorithm. And the ways we know (and the ways the high school students know, which are likely to be trial division plus a handful or two of divisibility tests to make the former simpler) have these drawbacks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps an analogy would be that the Euclidean algorithm is like having a full recipe to prepare a dish, and all the ingredients laid out ready to be used; factorization involves starting to prepare the dish, rummaging through your supplies for ingredients, and possibly realizing part-way through that you don't have the right ingredients, forcing you to start over from scratch with a new dish for which you hope you do have the ingredients. Even if the former situation involves a complex dish while the latter is a series of attempts at very simple and quick dishes, chances are you will spend less total time with the full-recipe-and-all-ingredients-laid-out-method than the let's-try-this-and-hope-we-have-all-the-stuff method.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T03:32:58.980" LastActivityDate="2010-10-18T03:32:58.980" CommentCount="2" />
  <row Id="7077" PostTypeId="2" ParentId="7071" CreationDate="2010-10-18T01:18:09.147" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ \ S\ =\ 1 \ \ \ +\ \ \: 2\ \ \ \ +\ \:\cdots\  +\ p-1\ +\ p $&lt;br&gt;&#xA;$\rm\quad\quad\quad\quad\quad\quad\ S\ =\ p \ +\  p-1\  +\ \cdots\ +\ \  \ \ 2\ \ \ +\ \ \:1 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding $\rm\ \ 2\: S\ =\ p\ (p+1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning&quot; rel=&quot;nofollow&quot;&gt;famous legend&lt;/a&gt; says Gauss used this trick to quickly compute $ 1+2+\:\cdots\:+100\ $ in grade school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This trick of pairing up reflections around the average value is a special case of exploiting innate symmetry - here a reflection or involution. It's a ubiquitous powerful technique, e.g. see my post on &lt;a href=&quot;http://math.stackexchange.com/questions/1865&quot;&gt;Wilson's Theorem&lt;/a&gt; and it's &lt;a href=&quot;http://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1/9345#9345&quot;&gt;group theoretic generalization&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T22:01:57.580" LastActivityDate="2010-11-08T22:01:57.580" CommentCount="4" />
  <row Id="7078" PostTypeId="2" ParentId="7061" CreationDate="2010-10-18T01:20:40.107" Score="4" ViewCount="0" Body="&lt;p&gt;$x^2-y^2 = (x-y)(x+y)$. If neither $x$ nor $y$ are multiples of $3$, then either they have the same remainder and $x-y$ is a multiple of $3$, or they have different remainders ($1$ and $2$) and $x+y$ is a multiple of $3$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-18T01:20:40.107" />
  <row Id="7079" PostTypeId="1" AcceptedAnswerId="7080" CreationDate="2010-10-18T01:37:00.417" Score="-1" ViewCount="120" Body="&lt;p&gt;Let n be a natural number with d(n) = 33 (number of factors are 33)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I prove that n and n + 1 don't have a common factor &gt; 1?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-18T16:07:32.277" LastActivityDate="2010-10-18T16:07:32.277" Title="Number of factors proof" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7080" PostTypeId="2" ParentId="7079" CreationDate="2010-10-18T01:38:35.260" Score="5" ViewCount="0" Body="&lt;p&gt;The first thing you would do is ignore the information that $d(n)=33$, because it is completely irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT: Any common factor of $a$ and $b$ divides $a-b$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-18T01:38:35.260" CommentCount="3" />
  <row Id="7081" PostTypeId="1" CreationDate="2010-10-18T01:41:44.460" Score="3" ViewCount="119" Body="&lt;p&gt;Let G be the group defined by generators a,b and relations $a^4=e$, $a^2b^{-2}=e$, $abab^{-1}=e$. Since the quaternion group of order 8 is generated by elements $a,b$ satisfying the previous relations, there is an epimorphism from $G$ onto $Q_8$. Let $F$ be the free group on $\{a,b\}$ and $N$ the normal subgroup generated by $\{ a^4, a^2b^{-2}, abab^{-1} \}$. How to write down the normal subgroup N and express the group F/N ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T01:56:16.293" LastActivityDate="2010-10-18T10:14:07.423" Title="presentation of quaternion group of order 8" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="7082" PostTypeId="2" ParentId="7081" CreationDate="2010-10-18T01:53:31.450" Score="2" ViewCount="0" Body="&lt;p&gt;To write down the group $F/N$, the best way in this case is to find a normal form. Note that the last relation $abab^{-1}=e$ means that $ba=a^{-1}b = a^3b$ (that is, the element $b^{-1}a^{-3}ba$ is in $N$); so any coset representative that contains the string $ba$ can be replaced with a coset representative in which the string $ba$ is replaced with the string $a^3b$. The fact that $a^2b^{-2}\in N$ means that any coset representative that in which you have a $b^2$ can be replaced with one that has an $a^2$ instead. Proceeding in this way, you can conclude that every coset can be represented by an element of the form $a^i b^j$ with $0\leq i\leq 3$ and $0\leq j\leq 1$ (you can erase any $a^4$ since $a^4\in N$). Thus, the group $F/N$ has at most $8$ elements. You already know it has at least $8$ elements, so that means that each of these elements represent distinct cosets of $N$, so that gives you a way to express $F/N$ using the representatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Writing down&quot; the group $N$ is a bit more difficult, since it is an infinite subgroup in $F$. What exactly do you mean by &quot;writing down $N$&quot;? A way to determine if a given element of $F$ lies in $N$? This can be achieved by finding its &quot;coset representative&quot; from among the special set identified above. If you get $a^0b^0$, then the element was in $N$; if not, then it was not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T02:27:32.537" LastActivityDate="2010-10-18T02:27:32.537" CommentCount="2" />
  <row Id="7083" PostTypeId="2" ParentId="6538" CreationDate="2010-10-18T02:06:19.707" Score="0" ViewCount="0" Body="&lt;p&gt;For any group of order 105, there always exists a normal sylow 7-subgroup and a normal sylow 5-group. &lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-10-18T02:06:19.707" />
  <row Id="7084" PostTypeId="1" AcceptedAnswerId="7087" CreationDate="2010-10-18T02:06:54.557" Score="-3" ViewCount="157" Body="&lt;p&gt;How would I prove that gcd(ab,n) = gcd(a,n)*gcd(b,n)?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2010-10-18T13:49:02.273" Title="GCD Proof Question" Tags="&lt;proof&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="7085" PostTypeId="2" ParentId="7084" CreationDate="2010-10-18T02:13:10.710" Score="0" ViewCount="0" Body="&lt;p&gt;It would be difficult, as it is not true.  Take a=b=n=2&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T02:13:10.710" />
  <row Id="7086" PostTypeId="2" ParentId="7084" CreationDate="2010-10-18T02:17:47.787" Score="1" ViewCount="0" Body="&lt;p&gt;See this recently-asked &lt;a href=&quot;http://math.stackexchange.com/questions/6653/basic-divisibility-fact/6657&quot;&gt;equivalent question&lt;/a&gt;, after imposing the correct hypotheses, e.g. $\rm \gcd(a,b,n) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a proof using said hypothesis and gcd laws (associative, commutative, distributive)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad (a,n)(b,n)\ =\ (ab, n\:(a,b,n))\ =\ (ab,n)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T13:49:02.273" LastActivityDate="2010-10-18T13:49:02.273" />
  <row Id="7087" PostTypeId="2" ParentId="7084" CreationDate="2010-10-18T02:21:08.857" Score="2" ViewCount="0" Body="&lt;p&gt;fmunshi, it isn't true, as it was pointed out. But you can try and see what happens there when $\gcd(a,b)=1$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-18T02:21:08.857" />
  <row Id="7089" PostTypeId="2" ParentId="4844" CreationDate="2010-10-18T02:40:32.613" Score="2" ViewCount="0" Body="&lt;p&gt;The canonical thing to do for a question like this is to look at Neal Sloane's home page. Sure enough, there is a table giving some good arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.research.att.com/~njas/electrons/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www2.research.att.com/~njas/electrons/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was indeed one of the links on the page in wok's answer, but it may be the most complete resource.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-10-18T02:40:32.613" />
  <row Id="7090" PostTypeId="1" AcceptedAnswerId="11796" CreationDate="2010-10-18T02:55:42.190" Score="1" ViewCount="248" Body="&lt;p&gt;I was given a proof of this in class, but it does not make sense to me, can someone provide a better proof, or explain the one I have provided, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\gcd(a,n) = 1$, where $a$ and $n$ are natural numbers. Prove that $0$, $a$, $2a$, $3a,\ldots, (n-1)a$, is complete residue system modulo $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof given in class:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;ka &amp;amp;\equiv ra \pmod{n}\\&#xA;&amp;amp;\Rightarrow  k \equiv r \pmod{n}\\&#xA;&amp;amp;\Rightarrow  \text{$k-r$ divides $n$}\\&#xA;&amp;amp;\Rightarrow \text{$k=r$ because $0\leq k\lt n$, $0\leq r\lt n$}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:59:12.177" LastActivityDate="2010-12-03T22:45:09.197" Title="Complete residue system proof" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="7091" PostTypeId="2" ParentId="4894" CreationDate="2010-10-18T03:35:50.797" Score="3" ViewCount="0" Body="&lt;p&gt;Any two of (a) symplectic structure, (b) almost-complex structure, and (c) Riemannian structure determine the third.  This reflects the fact that $$ U(n) = O(2n) \cap GL(n,\mathbb{C}) \cap Sp(2n),$$ but the intersection of any two is already $U(n)$.  This gives rise to the theory of Kahler manifolds (cf. wikipedia), which are central to mirror symmetry (and the mirror symmetry conjecture).&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-18T20:19:27.963" LastActivityDate="2010-10-18T20:19:27.963" />
  <row Id="7092" PostTypeId="1" AcceptedAnswerId="7093" CreationDate="2010-10-18T03:57:46.073" Score="0" ViewCount="266" Body="&lt;p&gt;What is the highest power of 2 dividing 100!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 multiples of 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25 multiples of 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12 multiples of 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 multiples of 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 multiples of 32&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 multiple of 64&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: giving a highest power of &lt;strong&gt;2^97&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;am I missing anything here?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="2588" LastEditorDisplayName="" LastEditDate="2010-10-18T04:09:04.563" LastActivityDate="2010-10-18T20:02:42.620" Title="What is the highest power of 2 dividing 100!" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="7093" PostTypeId="2" ParentId="7092" CreationDate="2010-10-18T04:06:35.200" Score="2" ViewCount="0" Body="&lt;p&gt;Just look at this page &lt;a href=&quot;http://planetmath.org/encyclopedia/PrimePowerDividingAFactorial.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/PrimePowerDividingAFactorial.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-18T04:06:35.200" CommentCount="1" />
  <row Id="7094" PostTypeId="2" ParentId="7092" CreationDate="2010-10-18T04:07:34.017" Score="1" ViewCount="0" Body="&lt;p&gt;Your list of multiples is correct.  But then the total is 97, not 2^97.  But good to see some work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T04:07:34.017" CommentCount="2" />
  <row Id="7095" PostTypeId="2" ParentId="7092" CreationDate="2010-10-18T04:11:55.470" Score="1" ViewCount="0" Body="&lt;p&gt;You are not missing anything. Using WolframAlpha to factor 100! into primes, we&#xA;get the answer:&#xA;2^97×3^48×5^24×7^16×11^9×13^7×17^5×19^5×23^4×29^3×31^3×37^2×41^2×43^2×47^2×53×59×61×67×71×73×79×83×89×97(239 factors, 25 distinct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore, the highest power of 2 that divides 100! is 2^97.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100!/2^97 = 588971222367687651371627846346807888288472382883312574253249804256440585603406374176100610302040933304083276457607746124267578125&lt;/p&gt;&#xA;" OwnerUserId="1605" LastEditorUserId="1605" LastEditorDisplayName="" LastEditDate="2010-10-18T20:02:42.620" LastActivityDate="2010-10-18T20:02:42.620" CommentCount="8" />
  <row Id="7096" PostTypeId="1" CreationDate="2010-10-18T04:31:00.267" Score="6" ViewCount="97" Body="&lt;p&gt;Kodaira defines a complex analytic family of compact complex manifolds as the data $(E,B,\pi)$, where $E$ and $B$ are complex manifolds, and $\pi$ is a surjective holomorphic submersion such that the preimage $\pi^{-1}(x)$ of any point $x \in B$ is a compact complex submanifold of $E$. (Here complex manifolds are required to be connected.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key property of this definition is that this specifies a differentiable fibre bundle. This fact can almost be obtained from the Ehresmann fibration theorem, which might be stated as follows:&#xA;&quot;Let $X$, $Y$ be differentiable manifolds, and let $f: X \rightarrow Y$ be a proper surjective submersion. Then $f$ is the projection of a differentiable fibre bundle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the map $\pi$ defining a family lacks the properness needed to apply Ehresmann's theorem. It is, however, a corollary of the fact that a family gives a fibre bundle that $\pi$ is indeed proper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more elementary way of seeing that $\pi$ must be proper?&lt;/p&gt;&#xA;" OwnerUserId="2598" LastActivityDate="2010-12-30T19:44:30.230" Title="Proper maps and families of compact complex manifolds" Tags="&lt;differential-topology&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7100" PostTypeId="1" AcceptedAnswerId="7103" CreationDate="2010-10-18T04:58:29.037" Score="0" ViewCount="148" Body="&lt;p&gt;Calculate 1^30 + 2^30 + 3^30 + ... + 17^30 mod 31&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Fermat's Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1^30 = 1 (mod 31) (since 31 is prime)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies the above is congruent to 17 (mod 31)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is correct, right?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2010-10-18T13:25:55.490" Title="Calculate 1^30 + 2^30 + 3^30 + ... + 17^30 mod 31" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7101" PostTypeId="1" CreationDate="2010-10-18T04:58:38.647" Score="-1" ViewCount="108" Body="&lt;p&gt;Prove that for all integers a, b, c, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if (a divides c) and (b divides c) and (gcd (a,b) = d), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then (ab divides cd).&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-18T15:09:52.350" LastActivityDate="2010-10-18T15:09:52.350" Title="Prove that for all integers a, b, c, if a|c and b|c and gcd (a,b) = d, then ab|cd." Tags="&lt;number-theory&gt;&lt;division-algebra&gt;" AnswerCount="2" />
  <row Id="7102" PostTypeId="2" ParentId="7101" CreationDate="2010-10-18T05:06:47.797" Score="1" ViewCount="0" Body="&lt;p&gt;Homework?  You need LCM(a,b)*GCD(a,b)=a*b.  Can you do that?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T05:06:47.797" />
  <row Id="7103" PostTypeId="2" ParentId="7100" CreationDate="2010-10-18T05:16:46.200" Score="5" ViewCount="0" Body="&lt;p&gt;True.  But 1^n=1 mod whatever.  Fermat's Little Theorem says that n^30=1 for all n prime to 31.  So your answer of 17 is correct.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T05:16:46.200" />
  <row Id="7104" PostTypeId="2" ParentId="7101" CreationDate="2010-10-18T05:27:59.447" Score="1" ViewCount="0" Body="&lt;p&gt;This is equivalent to the nontrivial implication in a proof of the identity $\rm\ lcm(a,b)\ =\ ab/gcd(a,b)\:.\ $ See  my &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;post here&lt;/a&gt; for a couple of proofs - including a very simple universal proof that works very generally.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-18T05:37:50.993" LastActivityDate="2010-10-18T05:37:50.993" />
  <row Id="7105" PostTypeId="2" ParentId="7067" CreationDate="2010-10-18T06:45:19.593" Score="1" ViewCount="0" Body="&lt;p&gt;For very small numbers then factorization is quicker than the Euclidean algorithm.&#xA;But the Euclidea algorithm gives more; it allows one to compute reciprocals&#xA;in modular arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger numbers, finding factorizations may be slow or even practically&#xA;impossible, but the Eulcidean algorithm still works well. Also modern&#xA;factorization methods from Pollard rho to the number field sieve, make&#xA;essential use of the Euclidean algorithm.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-18T06:45:19.593" />
  <row Id="7106" PostTypeId="2" ParentId="6994" CreationDate="2010-10-18T07:23:06.007" Score="2" ViewCount="0" Body="&lt;p&gt;Valuation rings demonstrate quite clearly the failure of Krull's principal ideal theorem:&#xA;take a valuation ring O of finite dimension. The prime ideals then form a chain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_0:=0\subset p_1\subset\ldots\subset p_d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that for every $i\in{1,\ldots ,d}$ there exists $r_i\in p_i\setminus p_{i-1}$. Obviously $p_i$ is a minimal prime over $r_iO$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For valuation domains of infinite dimension one has to consider the so-called limit-primes: a prime ideal $p$ of a commutative ring $R$ is called limit-prime if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=\bigcup\limits_{q\in\mathrm{Spec} (R): q\subset p}q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist valuation domains $O$ of infinite Krull dimension such that the maximal ideal $m$ of $O$ is no limit-prime. For example take a valuation ring such that the corresponding value group is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}\times\mathbb{Z}\times\ldots$ (countably many factors ordered lexigraphically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one can find $r\in m$ such that $m$ is minimal over $rO$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2010-10-18T07:23:06.007" />
  <row Id="7107" PostTypeId="1" AcceptedAnswerId="7128" CreationDate="2010-10-18T08:09:32.607" Score="1" ViewCount="101" Body="&lt;p&gt;As part of my solution to a problem, I come to a point where I need to find the solutions to&#xA;$-2\partial_{T}B\left(T\right)+\frac{3}{4}B\left(T\right)\left(A\left(T\right)^{2}+B\left(T\right)^{2}\right)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\partial_{T}A\left(T\right)+\frac{3}{4}A\left(T\right)\left(B\left(T\right)^{2}+A\left(T\right)^{2}\right)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\partial_{T}(f)$ is the derivative with respect to $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that I made a mistake in the steps leading to this because I am supposed to be able to get a not-so-ugly solution for $A(T)$ and $B(T)$. Is there one that exists and I don't see it? I've tried the following:&#xA;&lt;img src=&quot;http://i.imgur.com/yi0U1.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="2256" LastEditorDisplayName="" LastEditDate="2010-10-18T14:45:35.177" LastActivityDate="2010-10-18T14:53:48.223" Title="System of Non-linear ODEs -- Analytic Solution" Tags="&lt;differential-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7108" PostTypeId="1" CreationDate="2010-10-18T08:39:17.287" Score="3" ViewCount="129" Body="&lt;p&gt;What are the conditions for torsion to be zero other than having a plane curve?&#xA;The only thing I can thing of is an equation that have the torsion that cancels out each other.&lt;/p&gt;&#xA;" OwnerUserId="2603" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-19T04:28:33.667" LastActivityDate="2010-10-19T04:28:33.667" Title="Conditions that torsion is zero in a space curve" Tags="&lt;differential-geometry&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7109" PostTypeId="1" AcceptedAnswerId="7116" CreationDate="2010-10-18T09:19:05.287" Score="5" ViewCount="128" Body="&lt;p&gt;$$S = \sum_{k=1}^{\infty} \frac{cos(\theta\log(k))}{k^a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about finding the value of S, given that $\theta \to \infty$ and $0 &amp;lt; a &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any special techniques that might be&#xA;  helpful in calculating this sum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: &#xA;Just to give some background,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was actually trying to figure out&#xA;$$\sum_{k=1}^{\infty} \frac{cos(\theta\log(k))}{k^a} - \sum_{k=1}^{\infty} \frac{cos(\theta\log(k + 0.5))}{(k+0.5)^a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since that expression was a bit complicated, I decided to write the common version...&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2010-10-18T14:56:34.490" LastActivityDate="2010-10-18T14:56:34.490" Title="How to find the sum of this cos series" Tags="&lt;sequences-and-series&gt;&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="7110" PostTypeId="2" ParentId="7081" CreationDate="2010-10-18T10:14:07.423" Score="6" ViewCount="0" Body="&lt;p&gt;Two points about this. As a subgroup of a free group, N is itself a free group. There is an algorithmic procedure for finding a free set of generators of N, known as the Schreier generators. An example of this was presented in detail in a recent item &quot;Normal Closure in groups II&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/6707/normal-closure-in-groups-ii&quot;&gt;http://math.stackexchange.com/questions/6707/normal-closure-in-groups-ii&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a computer program (MAGMA), I got the following set of 9 generators of N:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N.1 = b * a * b * a^-1&#xA;N.2 = b^2 * a^-2&#xA;N.3 = a^-1 * b^2 * a^-1&#xA;N.4 = b^-1 * a * b^-1 * a^-1&#xA;N.5 = a^4&#xA;N.6 = a^2 * b^2&#xA;N.7 = a * b * a * b^-1&#xA;N.8 = a * b^2 * a&#xA;N.9 = a * b^-1 * a * b&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The second point is that the first relation $a^4=e$ in your presentation is actually redundant. The quaternion group is defined by the presentation with two generators $a,b$ and just the two relations $a^2=b^2$ and $abab^{-1} = e$. It is a nice exercise to prove that!&lt;/p&gt;&#xA;" OwnerUserId="2604" LastActivityDate="2010-10-18T10:14:07.423" />
  <row Id="7111" PostTypeId="1" AcceptedAnswerId="7134" CreationDate="2010-10-18T10:28:33.173" Score="1" ViewCount="202" Body="&lt;p&gt;I came across this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d(n)$ denote the number of divisors of $n$. Let $$\nu(z) = \sum\limits_{n=1}^{\infty} d(n) z^{n}$$ Whats the radius of convergence of this power series. We also have to show that $$\nu(z) = \sum\limits_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Regarding the divisor function, we have Dirichlet's formula in hand. But will that help!&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-18T11:02:26.120" LastActivityDate="2010-10-18T16:26:45.870" Title="Radius of Convergence of a given Power series" Tags="&lt;number-theory&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7112" PostTypeId="2" ParentId="7111" CreationDate="2010-10-18T11:35:33.597" Score="3" ViewCount="0" Body="&lt;p&gt;Write your series as&#xA;$$\sum_{n=1}^\infty \sum_{m=1}^\infty z^{mn},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and consider $mn=k,$ say. Does this help you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get stuck look &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_series&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-18T11:41:51.807" LastActivityDate="2010-10-18T11:41:51.807" />
  <row Id="7113" PostTypeId="2" ParentId="6941" CreationDate="2010-10-18T11:50:51.630" Score="0" ViewCount="0" Body="&lt;p&gt;Well, it turns out it's not really possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following Processing code shows my results for a 4th order polynomial. The equations of the parameters are from Maple.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float x1 = 10;&#xA;float y1 = 20;&#xA;float x3 = 90;&#xA;float y3 = 50;&#xA;float d1 = 3;&#xA;float d3 = 1;&#xA;float f = 40;&#xA;&#xA;float x2 = 30;&#xA;float y2 = mouseY;&#xA;&#xA;void draw(){&#xA;background(100);&#xA;   x2 = mouseX;&#xA;   y2 = mouseY;&#xA;&#xA;  noFill();&#xA;  stroke(0, 0, 0);&#xA;  bezier(x1, y1,  x1+f, y1+f*d1, x3-f, y3-f*d3,    x3,y3);&#xA;&#xA;  stroke(255, 102, 0);&#xA;&#xA;  for (float i=x1; i&amp;lt;x3;i++){&#xA;    point(i, approx(i));&#xA;  }&#xA;}&#xA;&#xA;float approx(float time){&#xA;  float a,b,c,d,e;&#xA;&#xA;a = -(8*x1*pow(x3,2)*x2*y1-2*pow(x1,4)*y3+2*pow(x3,4)*y1-pow(x1,2)*d3*pow(x3,3)+2*pow(x1,4)*d3*x3-pow(x1,3)*d3*pow(x3,2)+pow(x1,2)*d1*pow(x3,3)+pow(x1,3)*d1*pow(x3,2)-2*x1*pow(x3,4)*d1-8*x1*pow(x3,2)*x2*y3-8*x3*pow(x1,2)*x2*y3+8*x3*pow(x1,2)*x2*y1-2*pow(x1,3)*x3*x2*d1-2*pow(x3,2)*pow(x1,2)*x2*d1+x3*pow(x1,2)*pow(x2,2)*d1-x3*pow(x1,2)*pow(x2,2)*d3-2*x3*pow(x1,3)*x2*d3+2*x1*pow(x3,3)*x2*d1+x1*pow(x3,2)*pow(x2,2)*d1-x1*pow(x3,2)*pow(x2,2)*d3+2*pow(x1,2)*pow(x3,2)*x2*d3+2*pow(x3,3)*x2*x1*d3+3*pow(x1,3)*pow(x2,2)*d3-2*pow(x1,4)*x2*d3+pow(x1,3)*pow(x2,2)*d1+4*pow(x1,3)*x3*y3+2*pow(x1,4)*y2-2*pow(x3,4)*y2-2*pow(x2,4)*y3+2*pow(x2,4)*y1+4*x1*y3*x3*pow(x2,2)-4*x1*y1*x3*pow(x2,2)+2*pow(x3,4)*x2*d1-3*pow(x3,3)*pow(x2,2)*d1-pow(x3,3)*pow(x2,2)*d3+pow(x2,4)*x3*d3-pow(x2,4)*x1*d3-pow(x2,4)*x1*d1+pow(x2,4)*x3*d1-4*pow(x3,3)*x1*y1+4*pow(x3,3)*x1*y2-4*pow(x3,2)*pow(x1,2)*y1-4*x3*pow(x1,3)*y2-4*pow(x1,2)*pow(x2,2)*y1+4*pow(x1,2)*pow(x2,2)*y3-4*pow(x3,2)*pow(x2,2)*y1+4*pow(x3,2)*pow(x2,2)*y3+4*pow(x3,2)*pow(x1,2)*y3)/((-pow(x3,3)-3*x3*pow(x1,2)+pow(x1,3)+3*x1*pow(x3,2))*(pow(x3,2)*pow(x1,2)-2*x3*pow(x1,2)*x2+pow(x1,2)*pow(x2,2)+4*x1*x3*pow(x2,2)-2*x1*pow(x3,2)*x2-2*x1*pow(x2,3)+pow(x2,4)-2*x3*pow(x2,3)+pow(x3,2)*pow(x2,2)));&#xA;b = (y1*pow(x3,5)-pow(x1,2)*pow(x3,4)*d1+2*pow(x1,3)*pow(x3,3)*d1-2*pow(x1,3)*pow(x3,3)*d3+pow(x1,4)*pow(x3,2)*d3+pow(x1,5)*x3*d3-pow(x1,4)*x3*y3-x1*pow(x3,5)*d1+x1*pow(x3,4)*y1-pow(x1,5)*y3-pow(x3,3)*d3*pow(x2,3)+pow(x3,5)*x2*d1-pow(x3,5)*y2-3*pow(x3,3)*d1*pow(x2,3)+pow(x3,2)*pow(x2,4)*d3-3*y3*x3*pow(x2,4)-3*pow(x1,3)*pow(x3,2)*x2*d1-pow(x1,3)*pow(x3,2)*x2*d3-pow(x1,4)*x3*x2*d3+pow(x1,2)*x3*d1*pow(x2,3)+3*pow(x1,2)*pow(x3,3)*x2*d3+pow(x1,2)*pow(x3,3)*x2*d1-pow(x1,2)*x3*d3*pow(x2,3)+x1*pow(x2,4)*x3*d3-x1*pow(x2,4)*x3*d1+2*pow(x2,4)*pow(x3,2)*d1+pow(x1,5)*y2+3*x3*pow(x2,4)*y1-pow(x1,5)*x2*d3+pow(x1,4)*x3*y2+3*pow(x1,3)*d3*pow(x2,3)+pow(x1,3)*d1*pow(x2,3)-2*pow(x1,2)*pow(x2,4)*d3-pow(x1,2)*pow(x2,4)*d1-3*x1*pow(x2,4)*y3-x1*pow(x3,4)*y2+3*x1*pow(x2,4)*y1+x1*pow(x3,4)*x2*d1-x1*pow(x3,2)*d3*pow(x2,3)+x1*pow(x3,2)*d1*pow(x2,3)+4*pow(x3,2)*y3*pow(x2,3)-12*pow(x1,2)*y3*pow(x3,2)*x2+12*pow(x1,2)*y1*pow(x3,2)*x2+4*x1*x3*pow(x2,3)*y3-4*x1*x3*pow(x2,3)*y1+8*pow(x1,3)*y3*pow(x3,2)-8*pow(x1,3)*pow(x3,2)*y2+4*pow(x1,2)*y3*pow(x2,3)-4*pow(x1,2)*y1*pow(x2,3)-8*pow(x1,2)*pow(x3,3)*y1+8*pow(x1,2)*pow(x3,3)*y2-4*pow(x3,2)*pow(x2,3)*y1)/((-x3+x1)*(pow(x3,2)*pow(x1,4)-2*pow(x1,4)*x3*x2+pow(x1,4)*pow(x2,2)-2*pow(x3,3)*pow(x1,3)+2*pow(x1,3)*x3*pow(x2,2)+2*pow(x1,3)*pow(x3,2)*x2-2*pow(x1,3)*pow(x2,3)+pow(x1,2)*pow(x2,4)+2*pow(x1,2)*x3*pow(x2,3)+pow(x3,4)*pow(x1,2)-6*pow(x1,2)*pow(x2,2)*pow(x3,2)+2*pow(x1,2)*pow(x3,3)*x2+2*x1*pow(x3,3)*pow(x2,2)-2*x1*x3*pow(x2,4)-2*x1*pow(x3,4)*x2+2*x1*pow(x3,2)*pow(x2,3)+pow(x2,4)*pow(x3,2)-2*pow(x3,3)*pow(x2,3)+pow(x3,4)*pow(x2,2)));&#xA;c = -(pow(x1,3)*pow(x3,4)*d1-pow(x1,4)*pow(x3,3)*d3+pow(x1,5)*pow(x3,2)*d3-2*pow(x1,5)*x3*y3-pow(x1,2)*pow(x3,5)*d1+2*pow(x1,4)*d3*pow(x2,3)+2*y1*pow(x3,5)*x1-2*x1*pow(x3,5)*y2+pow(x3,5)*pow(x2,2)*d1-2*d1*pow(x3,4)*pow(x2,3)-2*pow(x1,2)*pow(x2,4)*x3*d1-pow(x1,2)*pow(x2,4)*x3*d3-pow(x1,4)*x3*pow(x2,2)*d3+x1*pow(x3,4)*pow(x2,2)*d1+pow(x3,3)*pow(x1,2)*pow(x2,2)*d1+3*pow(x3,3)*pow(x1,2)*pow(x2,2)*d3-pow(x3,2)*pow(x1,3)*pow(x2,2)*d3-3*pow(x3,2)*pow(x1,3)*pow(x2,2)*d1+2*pow(x3,2)*pow(x2,4)*x1*d3+pow(x3,2)*pow(x2,4)*x1*d1+pow(x3,3)*pow(x2,4)*d1-pow(x1,3)*pow(x2,4)*d3-pow(x1,5)*pow(x2,2)*d3+2*pow(x1,5)*x3*y2-6*x3*x1*pow(x2,4)*y3+6*x3*x1*pow(x2,4)*y1+2*pow(x1,3)*x3*d1*pow(x2,3)+2*pow(x1,3)*x3*d3*pow(x2,3)+2*pow(x1,2)*pow(x3,2)*d1*pow(x2,3)-2*pow(x1,2)*pow(x3,2)*d3*pow(x2,3)-2*x1*pow(x3,3)*d3*pow(x2,3)-2*x1*pow(x3,3)*d1*pow(x2,3)+8*y3*pow(x2,3)*x3*pow(x1,2)+8*y3*pow(x2,3)*x1*pow(x3,2)-8*y1*pow(x2,3)*x3*pow(x1,2)-8*y1*pow(x2,3)*x1*pow(x3,2)-12*pow(x1,2)*pow(x3,2)*pow(x2,2)*y3+12*pow(x1,2)*pow(x3,2)*pow(x2,2)*y1+4*pow(x1,4)*pow(x3,2)*y3-4*pow(x1,4)*pow(x3,2)*y2-4*pow(x1,2)*pow(x3,4)*y1+4*pow(x1,2)*pow(x3,4)*y2)/((-x3+x1)*(pow(x3,2)*pow(x1,4)-2*pow(x1,4)*x3*x2+pow(x1,4)*pow(x2,2)-2*pow(x3,3)*pow(x1,3)+2*pow(x1,3)*x3*pow(x2,2)+2*pow(x1,3)*pow(x3,2)*x2-2*pow(x1,3)*pow(x2,3)+pow(x1,2)*pow(x2,4)+2*pow(x1,2)*x3*pow(x2,3)+pow(x3,4)*pow(x1,2)-6*pow(x1,2)*pow(x2,2)*pow(x3,2)+2*pow(x1,2)*pow(x3,3)*x2+2*x1*pow(x3,3)*pow(x2,2)-2*x1*x3*pow(x2,4)-2*x1*pow(x3,4)*x2+2*x1*pow(x3,2)*pow(x2,3)+pow(x2,4)*pow(x3,2)-2*pow(x3,3)*pow(x2,3)+pow(x3,4)*pow(x2,2)));&#xA;d = (3*x1*pow(x3,2)*pow(x2,4)*y1+2*x3*d3*pow(x1,4)*pow(x2,3)-pow(x3,5)*pow(x1,2)*x2*d1+pow(x3,4)*pow(x1,3)*x2*d1+pow(x3,4)*pow(x1,2)*pow(x2,2)*d1+pow(x3,5)*x1*pow(x2,2)*d1+2*pow(x3,3)*pow(x1,3)*pow(x2,2)*d3-pow(x3,3)*pow(x1,4)*x2*d3-2*pow(x3,3)*pow(x1,3)*pow(x2,2)*d1+pow(x3,3)*pow(x2,4)*x1*d1-pow(x3,2)*pow(x1,4)*pow(x2,2)*d3+pow(x3,2)*pow(x1,5)*x2*d3-x3*pow(x1,5)*pow(x2,2)*d3-3*x3*pow(x1,2)*pow(x2,4)*y3+pow(x3,2)*pow(x1,2)*pow(x2,4)*d3-x3*pow(x1,3)*pow(x2,4)*d3-pow(x3,2)*pow(x1,2)*pow(x2,4)*d1-3*pow(x3,3)*pow(x1,4)*y2-pow(x3,3)*pow(x2,4)*y1-pow(x3,5)*pow(x1,2)*y2+3*pow(x1,3)*pow(x3,4)*y2+pow(x1,3)*pow(x2,4)*y3+pow(x1,5)*pow(x3,2)*y2-2*y3*pow(x1,4)*pow(x2,3)+y3*pow(x1,5)*pow(x2,2)+2*y1*pow(x3,4)*pow(x2,3)-y1*pow(x3,5)*pow(x2,2)-2*y3*pow(x1,5)*x3*x2+y3*x3*pow(x1,4)*pow(x2,2)-2*x1*d1*pow(x3,4)*pow(x2,3)-y1*x1*pow(x3,4)*pow(x2,2)+2*y1*pow(x3,5)*x2*x1+pow(x1,3)*pow(x3,2)*d1*pow(x2,3)-pow(x1,3)*pow(x3,2)*d3*pow(x2,3)+pow(x1,2)*pow(x3,3)*d1*pow(x2,3)-pow(x1,2)*pow(x3,3)*d3*pow(x2,3)+4*pow(x1,4)*pow(x3,2)*y3*x2-4*pow(x1,2)*pow(x3,4)*y1*x2-4*pow(x1,2)*pow(x3,2)*pow(x2,3)*y1+4*pow(x1,2)*pow(x3,2)*y3*pow(x2,3)-4*pow(x3,3)*x1*pow(x2,3)*y1+8*pow(x3,3)*pow(x1,2)*pow(x2,2)*y1+4*pow(x1,3)*x3*pow(x2,3)*y3-8*pow(x1,3)*pow(x3,2)*pow(x2,2)*y3)/((-x3+x1)*(pow(x3,2)*pow(x1,4)-2*pow(x1,4)*x3*x2+pow(x1,4)*pow(x2,2)-2*pow(x3,3)*pow(x1,3)+2*pow(x1,3)*x3*pow(x2,2)+2*pow(x1,3)*pow(x3,2)*x2-2*pow(x1,3)*pow(x2,3)+pow(x1,2)*pow(x2,4)+2*pow(x1,2)*x3*pow(x2,3)+pow(x3,4)*pow(x1,2)-6*pow(x1,2)*pow(x2,2)*pow(x3,2)+2*pow(x1,2)*pow(x3,3)*x2+2*x1*pow(x3,3)*pow(x2,2)-2*x1*x3*pow(x2,4)-2*x1*pow(x3,4)*x2+2*x1*pow(x3,2)*pow(x2,3)+pow(x2,4)*pow(x3,2)-2*pow(x3,3)*pow(x2,3)+pow(x3,4)*pow(x2,2)));&#xA;e = (pow(x3,3)*y1-pow(x1,3)*y3-x1*pow(x3,3)*d1+pow(x1,2)*pow(x3,2)*d1-3*pow(x3,2)*y1*x1-pow(x3,2)*d3*pow(x1,2)+pow(x1,3)*x3*d3+3*x3*y3*pow(x1,2)+x3*d3*pow(x2,3)+pow(x3,3)*x2*d1-2*pow(x3,2)*pow(x2,2)*d1-pow(x3,2)*pow(x2,2)*d3+3*y3*x3*pow(x2,2)-3*y1*x3*pow(x2,2)+x3*d1*pow(x2,3)-pow(x3,3)*y2-2*y3*pow(x2,3)+2*y1*pow(x2,3)-2*pow(x1,2)*x3*x2*d1-pow(x1,2)*x3*x2*d3+x1*pow(x3,2)*x2*d1+2*x1*pow(x3,2)*x2*d3-x1*x3*pow(x2,2)*d3+6*x1*y1*x3*x2-6*x1*y3*x3*x2+x1*x3*pow(x2,2)*d1-x1*d1*pow(x2,3)+3*x1*y3*pow(x2,2)-3*x1*y1*pow(x2,2)+3*x1*pow(x3,2)*y2-pow(x1,3)*x2*d3+2*pow(x1,2)*pow(x2,2)*d3+pow(x1,2)*pow(x2,2)*d1-3*pow(x1,2)*x3*y2-x1*d3*pow(x2,3)+pow(x1,3)*y2)/(-pow(x2,4)*pow(x3,3)+pow(x1,5)*pow(x3,2)-3*pow(x1,4)*pow(x3,3)-pow(x3,5)*pow(x2,2)-pow(x1,2)*pow(x3,5)+3*pow(x1,3)*pow(x3,4)+pow(x2,4)*pow(x1,3)+pow(x1,5)*pow(x2,2)+2*pow(x3,4)*pow(x2,3)-2*pow(x1,4)*pow(x2,3)+3*pow(x2,4)*x1*pow(x3,2)-8*pow(x1,3)*pow(x3,2)*pow(x2,2)+2*pow(x3,5)*x2*x1-3*pow(x2,4)*x3*pow(x1,2)-x1*pow(x3,4)*pow(x2,2)+x3*pow(x1,4)*pow(x2,2)+4*pow(x1,4)*pow(x3,2)*x2-4*pow(x1,2)*pow(x3,4)*x2+8*pow(x3,3)*pow(x1,2)*pow(x2,2)-2*pow(x1,5)*x3*x2+4*pow(x1,3)*x3*pow(x2,3)-4*pow(x3,3)*x1*pow(x2,3));&#xA;  return &#xA;  e*time*time*time*time+&#xA;  a*time*time*time+&#xA;  b*time*time+&#xA;  c*time+&#xA;  d;   &#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'll try with adding the derivative of x2 and adding another point in the middle (with a 5rth order polynomial) but I'm not sure it'll appropriate for my use.&lt;/p&gt;&#xA;" OwnerUserId="2561" LastActivityDate="2010-10-18T11:50:51.630" />
  <row Id="7114" PostTypeId="2" ParentId="2777" CreationDate="2010-10-18T12:25:21.527" Score="0" ViewCount="0" Body="&lt;p&gt;There isn't any simple approximation of the sort you seek. Simply defining an angle $\theta_{k\ell}$ such that $\cos(\theta_{k\ell}) = \tfrac12 \bigl( \cos(\tfrac{2\pi k}n) + \cos(\tfrac{2\pi \ell}n) \bigr)$ is not sufficient. There are infinitely many different angles $\theta_{k\ell}$ which satisfy this equation, after all, which are very different from one another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you want is to obtain &lt;em&gt;some single&lt;/em&gt; value $\theta_{k\ell}$ which you can approximate well using rational multiples of a full rotation, as you describe. This will require mor than just a definition in terms of a periodic function such as cosine.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-10-18T14:36:40.240" LastActivityDate="2010-10-18T14:36:40.240" />
  <row Id="7115" PostTypeId="1" AcceptedAnswerId="7117" CreationDate="2010-10-18T12:32:45.877" Score="4" ViewCount="117" Body="&lt;p&gt;Am I correct to say:&#xA;A difference between an integral and a function is: that an integral can describe an area on a graph, while a function can't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I completely off course here?&lt;/p&gt;&#xA;" OwnerUserId="2605" LastEditorUserId="2605" LastEditorDisplayName="" LastEditDate="2010-10-18T12:52:21.843" LastActivityDate="2010-10-18T13:04:57.277" Title="Can a function describe an area on a graph like an integral?" Tags="&lt;calculus&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="7116" PostTypeId="2" ParentId="7109" CreationDate="2010-10-18T12:53:13.837" Score="5" ViewCount="0" Body="&lt;p&gt;Your value is basically the value of the Riemann zeta function:&#xA;$$&#xA;S = \sum_{k=1}^{\infty} \frac{Re (e^{i \theta \log(k)})}{k^a} = Re( \sum_{k=1}^{\infty} k^{i \theta - a} ) = Re( \zeta ( a - i \theta ))&#xA;$$&#xA;You want to evaluate this on the critical strip $0 &amp;lt; a &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that there is an enormous amount of literatue on the Riemann zeta. The bad news is that this function is nasty on the critical strip.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-10-18T12:53:13.837" CommentCount="4" />
  <row Id="7117" PostTypeId="2" ParentId="7115" CreationDate="2010-10-18T13:00:21.640" Score="2" ViewCount="0" Body="&lt;p&gt;You can view the integral as a function too, assuming it is defined. For instance, if you fix an interval $I=[a,b]$, and consider all functions that are integrable on $I$, you can view the integral as a function that maps each of these functions to a real number, that represents the 'area under the curve'.&#xA;Or, if you fix a function $f$ on the same interval, you can view the integral from $a$ to $x$ (where $a\leq x\leq b$) as a new function of $x$, defined on the interval, and the value at each $x$ would represent the area under the curve up to $x$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-18T13:00:21.640" />
  <row Id="7118" PostTypeId="2" ParentId="7115" CreationDate="2010-10-18T13:04:57.277" Score="5" ViewCount="0" Body="&lt;p&gt;A function $f : A \rightarrow B$ is just an object which, given any $a \in A$ then $f(a) \in B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph of a function $\{(a,f(a))|a\in A\}$ in the case of $f : \mathbb R \rightarrow \mathbb R$ can be viewed as a curve but that's just an illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An anti-derivative, say $F(t) = \int_0^{t} f(x) \mathrm{d}x$ is just another function $F : \mathbb R \rightarrow \mathbb R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is true that say, $F(3)$ gives the area under the curve $f$ defines between $0$ and $3$ on the x-axis.. but this idea of 'area' is just an intuitive guide to help motivate the definitions of integral and similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral $\int_{a}^{b}$ could be thought of as a map that takes $(\mathbb R \rightarrow \mathbb R) \rightarrow \mathbb R$ but that is not usual and it can also just been seen as a notation which only has meaning when completed (given an integrand).&lt;/p&gt;&#xA;" LastActivityDate="2010-10-18T13:04:57.277" />
  <row Id="7120" PostTypeId="2" ParentId="7108" CreationDate="2010-10-18T13:31:13.553" Score="1" ViewCount="0" Body="&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Torsion_of_a_curve#Properties&quot; rel=&quot;nofollow&quot;&gt;states&lt;/a&gt; that &quot;if the torsion of a regular curve is identically zero then this curve belongs to a fixed plane.&quot; By &quot;regular curve&quot; I expect they mean that the curve's first and second derivatives are never zero. I imagine that a planar curve connected via a linear segment to another planar curve lying in a different plane would still have zero torsion everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the limit of my knowledge; I'm posting it as an answer so others can see it and point out if there are any errors.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-18T13:31:13.553" CommentCount="3" />
  <row Id="7121" PostTypeId="2" ParentId="7108" CreationDate="2010-10-18T13:42:06.937" Score="3" ViewCount="0" Body="&lt;p&gt;As has been said, the curve is planar iff the torsion is zero. This is clear if you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas&quot; rel=&quot;nofollow&quot;&gt;Frenet formulas&lt;/a&gt;. For fun, you can get from those a formula that the curve $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ must satisfy. We need&#xA;$$&#xA;\frac{d\mathbf{N}}{ds} + \kappa\mathbf{T} = 0.&#xA;$$&#xA;Assuming $\alpha$ is parameterized by arc length, this could be written&#xA;$$&#xA;\frac{d}{ds}\left(\frac{1}{\kappa}\frac{d^2\alpha}{ds^2}\right) + \kappa\frac{d\alpha}{ds} = 0&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-10-18T13:42:06.937" CommentCount="2" />
  <row Id="7122" PostTypeId="1" AcceptedAnswerId="7124" CreationDate="2010-10-18T13:54:15.243" Score="2" ViewCount="149" Body="&lt;p&gt;Is there a more efficient algorithm besides &lt;a href=&quot;http://en.wikipedia.org/wiki/Gram-Schmidt_process#Algorithm&quot; rel=&quot;nofollow&quot;&gt;Gram-Schmidt&lt;/a&gt; that would produce an orthonormal matrix of rank N, with first row equal to [1 1 1 1 1 ... 1] / sqrt(N)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. for N = 3, the matrix $\begin{align}&#xA;\mathsf A_3 &amp;amp;= \begin{bmatrix}&#xA;  1/\sqrt 3 &amp;amp; 1/\sqrt 3 &amp;amp; 1/\sqrt 3 \\&#xA;  2/\sqrt 6 &amp;amp; -1/\sqrt 6 &amp;amp; -1/\sqrt 6 \\&#xA;  0 &amp;amp; 1/\sqrt 2 &amp;amp; -1/\sqrt 2&#xA;\end{bmatrix}\end{align}$ suffices, but I'm not sure how to generalize.&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-18T14:09:23.647" LastActivityDate="2010-10-18T21:26:16.480" Title="determining orthonormal matrix of rank N with special first row" Tags="&lt;matrices&gt;&lt;algorithms&gt;&lt;orthonormal&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="7123" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T14:04:52.967" Score="1" ViewCount="0" Body="&lt;p&gt;Given the special form, you can probably construct a rotation matrix that would turn one of the standard basis vectors there.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-18T14:04:52.967" CommentCount="2" />
  <row Id="7124" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T14:26:37.320" Score="6" ViewCount="0" Body="&lt;p&gt;There's a reflection swapping $v=(\sqrt{n},0,\ldots,0)$&#xA;with $w=(1,1,\ldots,1)$ which will do what you want.&#xA;It will fix $v+w$ and all vectors orthogonal to $v$ and $w$.&#xA;It will negate $v-w$.&#xA;It will be a symmetric matrix; its first row and column&#xA;is all $1/\sqrt{n}$. The remaining entries $a_{i,j}$ for $i$, $j\ge2$&#xA;will depend only on whether $i=j$ or not. This should be enough&#xA;information to reconstruct it.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-18T14:26:37.320" CommentCount="5" />
  <row Id="7125" PostTypeId="1" AcceptedAnswerId="7129" CreationDate="2010-10-18T14:41:52.890" Score="2" ViewCount="78" Body="&lt;p&gt;Hello, I am a new user!&lt;br&gt;&#xA;I don't understand what is &lt;em&gt;polynomial fitting&lt;/em&gt;.&lt;br&gt;&#xA;Can anyone explain me how to fit a curve to given points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDITED&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="2582" LastEditorDisplayName="" LastEditDate="2010-10-18T16:22:44.563" LastActivityDate="2010-10-18T16:22:44.563" Title="Polynomial fitting - how to fit and what is _polynomial fitting_" Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7126" PostTypeId="1" AcceptedAnswerId="7133" CreationDate="2010-10-18T14:48:14.133" Score="2" ViewCount="138" Body="&lt;p&gt;Here's the Karnaugh map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/NpaAB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer I should be getting from the Karnaugh should be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = R ∙ (CGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm really not seeing how this was arrived at through any simplification methods I've learned thus far.  I can see the answers that are intended are correct (I think), though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, the best answer I can come up with to simplify (only) the Kargnaugh map is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T=RCG'+RCM'+RC'&#xA;  to:&#xA;T=R∙(CG'+CM'+C')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Help appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2609" LastEditorUserId="2609" LastEditorDisplayName="" LastEditDate="2010-10-19T22:09:58.167" LastActivityDate="2010-10-19T22:09:58.167" Title="Boolean Algebra, Simplification: Don't know the method used" Tags="&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7127" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T14:51:09.650" Score="1" ViewCount="0" Body="&lt;p&gt;You should be able to use K-1 rows of Normalized &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadamard_matrix&quot; rel=&quot;nofollow&quot;&gt;Hadamard&lt;/a&gt; matrices of size K, where K is a power of 2 as building blocks. (How to normalize see this: &lt;a href=&quot;http://math.stackexchange.com/questions/4797/what-is-sum-of-rows-of-hadamard-matrix/4798#4798&quot;&gt;http://math.stackexchange.com/questions/4797/what-is-sum-of-rows-of-hadamard-matrix/4798#4798&lt;/a&gt;) You should be able to construct these rows recursively as K is a power of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1 1 ..       1 ]&#xA;[ H_1 | 0  |0 0 ]&#xA;[ 0   | H_2|0 0 ]&#xA;[ 0   | 0  | H_4]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end O(LogN) rows will remain to be filled, which I suppose you can use Gram-Schmidt on.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-18T14:51:09.650" />
  <row Id="7128" PostTypeId="2" ParentId="7107" CreationDate="2010-10-18T14:53:48.223" Score="2" ViewCount="0" Body="&lt;p&gt;You can make the second terms in both equations vanish by multiplying the first by $A(T)$, the second by $B(T)$, and subtracting.  The resulting equation is readily solved for the product $A(T)B(T)$, reducing the system to a single ODE which is directly integrable.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-18T14:53:48.223" CommentCount="2" />
  <row Id="7129" PostTypeId="2" ParentId="7125" CreationDate="2010-10-18T14:59:21.157" Score="8" ViewCount="0" Body="&lt;p&gt;There are two concepts frequently conflated: interpolation and fitting. I'll discuss both since you don't seem to know what you really want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a set of $n$ points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x_i,y_i)\qquad i=1\dots n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpolation is the problem of finding the (unique) polynomial $p(x)$ that passes through all your given points (under the assumption that no two points have the same abscissa), i.e. $p(x_i)=y_i \qquad i=1\dots n$. Your resulting polynomial is of degree $n-1$. The usual techniques for finding the interpolating polynomial are the methods of Lagrange, Newton, and Neville-Aitken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fitting on the other hand assumes your data is contaminated with error, and you want the polynomial that is the &quot;best approximation&quot; to your data. Here polynomial interpolation does not make much sense since you do not want your function to be reproducing the inherent errors in your data as well. Least-squares is a common technique: it finds the polynomial $f(x)$ such that the quantity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n}\left(f(x_i)-y_i\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which measures the departure of your polynomial from the ordinates is minimized (here the assumption is that your abscissas are error-free, and the error in your ordinates is normally distributed). The degree of $f(x)$ can be (and is often) less than $n$. A number of techniques for this are used as well: there's the normal equations, and then there are special matrix decompositions that can be used to efficiently solve this problem.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-18T14:59:21.157" CommentCount="3" />
  <row Id="7130" PostTypeId="1" AcceptedAnswerId="7132" CreationDate="2010-10-18T15:05:38.490" Score="2" ViewCount="123" Body="&lt;p&gt;I'm looking for an explanation on how reducing the Hamiltonian cycle problem to the Hamiltonian path's one (to proof that also the latter is NP-complete). I couldn't find any on the web, can someone help me here? (linking a source is also good).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2610" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-18T15:37:41.550" LastActivityDate="2010-10-19T03:46:36.883" Title="Reduction from Hamiltonian cycle to Hamiltonian path" Tags="&lt;graph-theory&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7131" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T15:08:37.963" Score="0" ViewCount="0" Body="&lt;p&gt;hmm, I think one possibility for orthonormal vectors for N odd is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 1 1 1 1 1 ... 1 ]&#xA;[ 1, cos phi, cos 2*phi, cos 3*phi, ... cos (N-1)*phi ]&#xA;[ 0, sin phi, sin 2*phi, sin 3*phi, ... sin (N-1)*phi ]&#xA;[ 1, cos 2*phi, cos 4*phi, cos 6*phi, ... cos 2*(N-1)*phi ]&#xA;[ 0, sin 2*phi, sin 4*phi, sin 6*phi, ... sin 2*(N-1)*phi ]&#xA;   ...&#xA;[ 1, cos k*phi, cos 2*k*phi, cos 3*k*phi, ... cos k*(N-1)*phi ]&#xA;[ 0, sin k*phi, sin 2*k*phi, sin 3*k*phi, ... sin k*(N-1)*phi ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where k = (N-1)/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'm not sure how to extend to N even.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-10-18T15:08:37.963" />
  <row Id="7132" PostTypeId="2" ParentId="7130" CreationDate="2010-10-18T15:37:17.523" Score="1" ViewCount="0" Body="&lt;p&gt;From Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path_problem&quot; rel=&quot;nofollow&quot;&gt;the Hamiltonian path problem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Hamiltonian path problem for graph G is equivalent to the Hamiltonian cycle problem in a graph H obtained from G by adding a new vertex and connecting it to all vertices of G.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: This answer has been accepted but is incorrect.  Please see Moron's answer instead.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-19T03:46:36.883" LastActivityDate="2010-10-19T03:46:36.883" CommentCount="4" />
  <row Id="7133" PostTypeId="2" ParentId="7126" CreationDate="2010-10-18T16:19:43.280" Score="2" ViewCount="0" Body="&lt;p&gt;Hi Hamster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using sum of products you should be able to derive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RC' + RCG' + RCGM' &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute out the R:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(C' + CG' + CGM')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the identity A + A'B = A + B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(C' + G' + CGM')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That same identity works as a multivariable expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(C' + G' + M')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then apply DeMorgans&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(CGM)'&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-10-18T16:19:43.280" CommentCount="8" />
  <row Id="7134" PostTypeId="2" ParentId="7111" CreationDate="2010-10-18T16:20:34.863" Score="2" ViewCount="0" Body="&lt;p&gt;Elaborating upon Robin's hint about the radius of convergence, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 \le d(n)^{1/n} \le n^{1/n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle n^{1/n} \to 1$, the radius of convergence of the power series is $\displaystyle 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I posted this is because I wanted to mention this quick proof of $\displaystyle  n^{1/n} \to 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle n &gt; 3$ then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge \sqrt[n]{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $AM \ge GM$ on $\displaystyle n-2 \ 1$s and two $\sqrt{n}$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$\displaystyle \frac{n-2 + 2\sqrt{n}}{n} \ge n^{1/n} \ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1- \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $\displaystyle n^{1/n} \to 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-18T16:26:45.870" LastActivityDate="2010-10-18T16:26:45.870" CommentCount="2" />
  <row Id="7135" PostTypeId="2" ParentId="7013" CreationDate="2010-10-18T16:56:06.660" Score="8" ViewCount="0" Body="&lt;p&gt;Here's a sketch of how we can evaluate the integral without complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Let} \quad I(a) = \int_0^\infty \frac{x^{a-1}}{1-x} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split the range into two intervals $(0,1)$ and $(1,\infty)$ then use the substitution $x=1/t$ in the latter part to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(a) = \int_0^1 \frac{x^{a-1}-x^{-a}}{1-x} dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand the integrand as a power series using $(1-x)^{-1} = \sum_{n=0}^\infty x^n$ and integrate to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(a) = \frac{1}{a} + \sum_{n=1}^\infty \left( \frac{1}{a+n} + \frac{1}{a-n} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by differentiating logarithmically the product formula for $\sin x,$&#xA;$$\sin \pi x = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi \cot \pi x = \frac{1}{x} + \sum_{n=1}^\infty \left( \frac{1}{x+n} + \frac{1}{x-n} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$I(a) = \pi \cot(\pi a)$$ and the result follows since the integral in question is $I(a)-I(b).$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-18T17:05:47.713" LastActivityDate="2010-10-18T17:05:47.713" CommentCount="1" />
  <row Id="7136" PostTypeId="1" AcceptedAnswerId="7139" CreationDate="2010-10-18T17:16:10.253" Score="2" ViewCount="241" Body="&lt;p&gt;An Exercise from Apostol which i am not able to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathsf{S_{n}}$ denote the $n$-th partial sum of the series: $$\sum\limits_{r=1}^{\infty} \frac{1}{r(r+1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that for every integer $k&gt;1$, there exists integers $m$ and $n$, such that $$s_{m}-s_{n} =\frac{1}{k}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-19T19:57:40.087" Title="Partial sum of a given series" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="7137" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T17:24:07.467" Score="2" ViewCount="0" Body="&lt;p&gt;First prove that $S_n=1-\frac{1}{n+1}$ by induction.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-18T17:30:59.753" LastActivityDate="2010-10-18T17:30:59.753" />
  <row Id="7139" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T17:40:10.307" Score="3" ViewCount="0" Body="&lt;p&gt;When you've worked out your formula for $s_m-s_n,$ consider $m=(n+1)^2+n.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-18T17:40:10.307" />
  <row Id="7140" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T17:52:36.030" Score="2" ViewCount="0" Body="&lt;p&gt;(EDIT: My original post took $p$ to be the smallest prime that divides $k$.  This is unnecessary.  You can take $p$ to be &lt;em&gt;any&lt;/em&gt; number other than 1 that divides $k$.  Doing so gives you not just one solution but $d(k)-1$ solutions, one for each of the divisors of $k$ other than 1.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(SECOND EDIT: See the answer to &lt;a href=&quot;http://math.stackexchange.com/questions/7173/&quot;&gt;this question&lt;/a&gt; for all solutions in $m$ and $n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First,&#xA;$$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}.$$&#xA;Then $$s_n = \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for any $m,n \geq 1$, $$s_m - s_n = \frac{1}{n+1} - \frac{1}{m+1} = \frac{m-n}{(m+1)(n+1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to find $m,n$ such that this last expression is equal to $\frac{1}{k}$.  Let $p$ be any positive integer other than 1 that divides $k$.  Take $m = (p-1)k-1$.  Clearly $m$ is a positive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we want to show that the resulting $n$ that solves $$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k}$$ is also an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k} \Rightarrow (m-n)k = (m+1)(n+1) $$&#xA;$$\Rightarrow ((p-1)k-1 - n)k = (p-1)k(n+1) \Rightarrow (p-1)k-1 - n = (p-1)n + p-1 $$&#xA;$$\Rightarrow pn = (p-1)k - p \Rightarrow n = \frac{(p-1)k}{p} - 1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that $n$ is an integer because $p|k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have a family of solutions $$m = (p-1)k-1, n = \frac{(p-1)k}{p} - 1,$$&#xA;where $p$ is any positive integer other than 1 that divides $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works even in the case $k = 2$ because then we just get $m = 1$, $n = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-19T19:57:40.087" LastActivityDate="2010-10-19T19:57:40.087" CommentCount="1" />
  <row Id="7141" PostTypeId="2" ParentId="7130" CreationDate="2010-10-18T18:16:19.550" Score="2" ViewCount="0" Body="&lt;p&gt;I don't think Mike's answer is correct. That gives the reduction from Hamiltonian Path &lt;em&gt;to&lt;/em&gt; cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a reduction from Hamiltonian Cycle to Path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a graph $G$ of which we need to find Hamiltonian Cycle, for a single edge $e = \{u,v\}$ add new vertices $u'$ and $v'$ such that $u'$ is connected only to $u$ and $v'$ is connected only to $v$ to give a new graph $G_e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_e$ has a hamiltonian path if and only if $G$ has a hamiltonian cycle with the edge $e=\{u,v\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run the Hamiltonian path algorithm on each $G_e$ for each edge $e \in G$. If all graphs have no hamiltonian path, then $G$ has no hamiltonian cycle. If at least one $G_e$ has a hamiltonian path, then $G$ has a hamiltonian cycle which contains the edge $e$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-18T18:16:19.550" CommentCount="0" />
  <row Id="7142" PostTypeId="1" CreationDate="2010-10-18T18:57:40.590" Score="7" ViewCount="149" Body="&lt;p&gt;I have a problem of the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimize $\|Dx\|_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to  $\|x*x\|_2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x\in\mathbb R^n$, $D$ is a given diagonal matrix of positive entries, and $*$ represents convolution, i.e. $(x*x)_n = \sum_{i+j=n}x_ix_j$ and $x*x\in\mathbb R^{2n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What approach could be used in dealing with this problem numerically? Could this problem be converted to one of the known problem classes that have available solvers?&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2011-01-03T08:44:33.987" Title="A numerical optimization problem with a convolution in the constraint" Tags="&lt;optimization&gt;&lt;nonlinear-optimization&gt;&lt;numerical-optimization&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="7143" PostTypeId="1" AcceptedAnswerId="7147" CreationDate="2010-10-18T19:01:30.843" Score="13" ViewCount="411" Body="&lt;p&gt;The scatter graph at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_triple#Distribution_of_triples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; seems to suggest so. Has anyone attacked this before? Is there a known proof?&lt;/p&gt;&#xA;" OwnerUserId="2612" LastActivityDate="2010-12-08T08:05:14.350" Title="Is it possible to get arbitrarily near any acute angle with Pythagorean triangles?" Tags="&lt;number-theory&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="7144" PostTypeId="1" AcceptedAnswerId="7149" CreationDate="2010-10-18T19:05:22.520" Score="3" ViewCount="143" Body="&lt;p&gt;Let $G=\{x,y|x^2=y^3=(xy)^3=1\}$&#xA;I would like to show that G is isomorphic to A_4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\mathbf{F}_{2} \to G$ be a surjective homomorphism from the free group on two elements to $G$. Let $f$ map $x \to (12)(34)$ and $y \mapsto (123)$. I'm not sure how to show that these elements generate the kernel of $f$. If they do generate the kernel, how do I conclude that the order of $G$ is 12?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I have that the order of the group is of order 12 then I can show that $G$ contains $V$ (the Klein four group) as a subgroup, or that $A_4$ is generated by the image of $x$ and $y$.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-18T19:55:54.920" LastActivityDate="2010-10-19T04:37:55.797" Title="Generators and Relations for A_4" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7146" PostTypeId="2" ParentId="7136" CreationDate="2010-10-18T19:14:46.713" Score="0" ViewCount="0" Body="&lt;p&gt;By partial fractions it &lt;a href=&quot;http://en.wikipedia.org/wiki/Telescoping_series&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;telescopes&lt;/strong&gt;&lt;/a&gt; to a sum matching the RHS of this well-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Egyptian_fraction&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Egyptian fraction&lt;/strong&gt;&lt;/a&gt; sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \frac{1}k\ =\ \frac{1}{k-1}\ -\ \frac{1}{k\:(k-1)}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-18T19:14:46.713" />
  <row Id="7147" PostTypeId="2" ParentId="7143" CreationDate="2010-10-18T19:16:37.730" Score="17" ViewCount="0" Body="&lt;p&gt;If you mean a pythagorean triangle having one of the angles arbitrary close to a given acute angle then the answer is yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle (m^2-n^2, 2mn, m^2+n^2)$ is always a pythagorean triangle for any positive integers $\displaystyle m \gt n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $\displaystyle \tan \theta = \frac{2mn}{m^2-n^2} = \frac{2t}{1-t^2}$ where $\displaystyle t = \frac{n}{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By continuity (and surjectivity) of $\displaystyle \frac{2t}{1-t^2}$ we can find an interval of $\displaystyle t$ (which is infact a subinterval of $\displaystyle (0,1)$)  for which $\displaystyle \frac{2t}{1-t^2}$ is arbitrarily close to any positive number we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rationals are dense, we can find $\displaystyle m,n$ which make the tangent of one the angles arbitrarily close to the tangent of a given acute angle, and hence the angles can be made arbitrarily close.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T08:05:14.350" LastActivityDate="2010-12-08T08:05:14.350" CommentCount="4" />
  <row Id="7149" PostTypeId="2" ParentId="7144" CreationDate="2010-10-18T20:18:49.573" Score="3" ViewCount="0" Body="&lt;p&gt;Abelsh, your definition of $f$ does not make sense: you cannot define the image of $x$ to &lt;strong&gt;be&lt;/strong&gt; $(12)(34)$ unless you already know that $G$ &lt;strong&gt;is&lt;/strong&gt; a group of permutations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, what you want to do is define $f$ from $\mathbf{F}_2$ to $A_4$ by defining it as you want: $f(x) = (12)(34)$ and $f(y)=(123)$. If you do this, then it is straightfoward to check that the images satisfy the given relations, $(f(x))^2 = 1$, $(f(y))^3 = 1$, and since $f(x)f(y) = (12)(34)(123) = (243)$ (I'm composing my permutations right to left), then $(f(x)f(y))^3 = 1$. That means that the kernel of $f$ &lt;em&gt;contains&lt;/em&gt; $x^2$, $y^3$, and $(xy)^3$, and therefore, by van Dyck's Theorem, the homomorphism factors through $G$. That means that $G$ has $A_4$ as a quotient, and thus has at least 12 elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But unless you find a precise description of the kernel, it would be difficult to prove that the kernel is exactly the subgroup generated by $x^2$, $y^3$, and $(xy)^3$. Instead, you might want to find a normal form for elements of $G$ to show that $G$ has &lt;em&gt;at most&lt;/em&gt; 12 elements; if you can do that, you get that $G$ has exactly $12$ elements, and has $A_4$ as a quotient, which gives you what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... do we have a sufficiently nice normal form for elements of $G$? Every element of $F$ can be written as a word in $x$ and $y$; using the relations in $G$ we can replace any power of $x$ with either $x^0$ or $x$, and any power of $y$ with either $y^0$, $y^1$, or $y^2$. So every element of $G$ can be written as a string that alternates $x$s and either $y$'s or $y^2$'s. And $(xy)^3 = 1$ so $(y^2x)^3=1$. That means that you can replace any string of the form $xyx$ with $y^2xy^2$, and replace $xy^2x$ with $yxy$. Note that in each of these replacements, you go from a string with two $x's$ to a string with just one $x$. That means that you can take any string that alternates $xs$ with $y$s or $y^2$s and use those replacements to get it down to one that has at most one $x$ in it. How many such words are there? That gives an upper bound for the number of distinct elements that $G$ can have. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-18T20:18:49.573" />
  <row Id="7150" PostTypeId="2" ParentId="6964" CreationDate="2010-10-18T20:41:05.187" Score="1" ViewCount="0" Body="&lt;p&gt;I think I got the idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma: If $T_\varphi$ is an nonconstant analytic Toeplitz operator then the only invariant finite-dimensional subspace for $T_\varphi$ is $\{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: If $M$ is a finite dimensional subspace them $T_\varphi$ restricted to $M$ has an non-zero eigenvector (because it is compact). But this is not possible for an the Toeplitz operator on the whole space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let $K$ be the compact operator that commutes with $T_\varphi$ and let $\lambda \neq 0$. Define $K_\lambda := \textrm{ker}(K - \lambda)$. Now we see that $K_\lambda$ is an invariant subspace for $T_\varphi$ as follows: $f \in T_\varphi K_\lambda$ then $f = T_\varphi  g$ for some $g \in K_\lambda$, that is $(K - \lambda)g = 0$. Because they commute we have $(K - \lambda) T_\varphi g = (K - \lambda)f = 0$, so $f \in K_\lambda$. Also note that $K_\lambda$ is finite dimensional by the spectral theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $\varphi$ is nonconstant $K_\lambda = 0$. So $\lambda$ is not in the spectrum of $K$. $\{0\}$ is in the spectrum of $K$ by the spectral theorem for compact operators.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-18T20:41:05.187" CommentCount="4" />
  <row Id="7151" PostTypeId="1" AcceptedAnswerId="7156" CreationDate="2010-10-18T20:47:21.373" Score="10" ViewCount="235" Body="&lt;p&gt;What is the least number of comparisons we need to find the median of 6 distinct numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to find the answer to the median of 5 distinct numbers to be 6 comparisons, and it makes sense, however in the case of 6 numbers I can't find an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I was able to do it in by hand was 9 comparisons.  Can that be minimized further?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Median in this case, we are assuming to be the lower median.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-10-19T08:39:31.533" LastActivityDate="2010-10-19T08:39:31.533" Title="Median of distinct numbers" Tags="&lt;algorithms&gt;&lt;sorting&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="7152" PostTypeId="1" AcceptedAnswerId="7157" CreationDate="2010-10-18T20:57:31.250" Score="2" ViewCount="86" Body="&lt;p&gt;I'm familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Householder_transformation&quot; rel=&quot;nofollow&quot;&gt;Householder reflections&lt;/a&gt;; they are a simple transformation that, given a normal vector, describes reflection in the hyperplane perpendicular to that vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do Householder reflections describe the complete space of matrices describing reflection about the origin, or are there other reflections that cannot be described by Householder reflections?&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-10-18T21:05:09.327" LastActivityDate="2010-10-19T04:21:30.230" Title="do Householder reflections describe all reflections?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;reflection&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7153" PostTypeId="2" ParentId="7152" CreationDate="2010-10-18T20:59:47.440" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, they are all reflections.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-18T20:59:47.440" />
  <row Id="7154" PostTypeId="2" ParentId="7122" CreationDate="2010-10-18T21:26:16.480" Score="1" ViewCount="0" Body="&lt;p&gt;I took @Robin Chapman's answer and ran with it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix A that is the Householder reflection of v-w where v = [sqrt(N), 0, 0, 0, ... ] and w = [1 1 1 1 1 ... 1] can be defined by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A(1,i) = A(i,1) = 1/sqrt(N)&#xA;A(i,i) = 1 - K   for i &amp;gt;= 2 &#xA;A(i,j) = -K      for i,j &amp;gt;= 2, i != j&#xA;K = 1/sqrt(N)/(sqrt(N)-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-10-18T21:26:16.480" />
  <row Id="7155" PostTypeId="2" ParentId="7108" CreationDate="2010-10-18T21:31:52.360" Score="1" ViewCount="0" Body="&lt;p&gt;Here's an example of an infinitely differentiable regular curve with identically zero torsion but not contained in any plane.  (By &lt;em&gt;regular&lt;/em&gt;, I mean that the velocity never vanishes.)&#xA;$$&#xA;\alpha(t) = &#xA;\left\{&#xA;\begin{aligned}&#xA;&amp;amp;(t,e^{-1/t},0), &amp;amp; t&gt;0,\\&#xA;&amp;amp;(0,0,0), &amp;amp;t=0,\\&#xA;&amp;amp;(t,0,e^{1/t}), &amp;amp;t&amp;lt;0.&#xA;\end{aligned}&#xA;\right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As Mariano Suárez-Alvarez points out in his comment, the torsion is only defined at points where the curvature is nonzero.  Since the curvature of this curve is zero at the origin, the torsion is not defined there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the argument given by yasmar shows that if the curve is regular, its curvature is nowhere zero, and its torsion is everywhere zero, then it's a plane curve.&lt;/p&gt;&#xA;" OwnerUserId="1421" LastEditorUserId="1421" LastEditorDisplayName="" LastEditDate="2010-10-18T21:56:49.867" LastActivityDate="2010-10-18T21:56:49.867" CommentCount="6" />
  <row Id="7156" PostTypeId="2" ParentId="7151" CreationDate="2010-10-18T21:53:54.173" Score="8" ViewCount="0" Body="&lt;p&gt;It takes at least 8 and can be done in exactly 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lower Bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lower bound to find all the $k$ smallest elements is given here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_algorithm#Lower_bounds&quot; rel=&quot;nofollow&quot;&gt;Lower bounds for Selection&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which says we need at least&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n - k + \sum_{n+1-k &amp;lt; j \le n} \lceil \log_{2} j \rceil $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=6, k=3$, this gives $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 6 - 3 + \lceil \log_{2} 5 \rceil + \lceil \log_{2} 6 \rceil = 9$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus to find the smallest, second smallest and the third smallest, we need at least $9$ compares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now any algorithm which finds the $3^{rd}$ smallest element (say $z$) will have to find the set $S = \{x, y \}$ such that $x &amp;lt; z$ and $y &amp;lt; z$ (otherwise how does it know that $z$ is the third largest?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extra compare between $x$ and $y$ will now give us the smallest, second smallest and third smallest numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus any algorithm which finds the third smallest must use at least 8 comparisons.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper Bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I.J. Kennedy notes, there is an algorithm to find the 3rd smallest element of 6 elements in 8 comparisons and can be found in Knuth's Art of Computer Programming Vol 3, Page 212.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Knuth, this is a result obtained by A. Hadian and M. Sobel [Univ of Minnesota, Dept of Statistics Report 121 (169)] and they show that the $k^{th}$ largest element can be found in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n - k + (k-1) \lceil \log_{2} (n+2-k) \rceil $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comparisons. For $n=6$ and $k=4$ we get the upper bound of $8$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way they do it (Knuth's book has all this information) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First create a knockout tournament on $n-k+2$ items, which takes $n-k+1$ compares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest item (say $L$) of this cannot be a candidate. So of the remaining $k-2$ pick one and follow the path of $L$ up the tree, which gives the the new largest in at most $\lceil \log_{2} n+2-k \rceil$ compares. Remove this new largest, replacing it with one of the remaining and following up the path of this new largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue till the rest of the $k-2$ are exhausted. Finally replace the largest with $-\infty$ and determine the largest of the remaining $n+1-k$ which will be the $k^{th}$ largest as we have removed the largest $k-1$ so far.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf&quot; rel=&quot;nofollow&quot;&gt;http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-19T03:35:59.740" LastActivityDate="2010-10-19T03:35:59.740" CommentCount="15" />
  <row Id="7157" PostTypeId="2" ParentId="7152" CreationDate="2010-10-18T22:32:39.720" Score="2" ViewCount="0" Body="&lt;p&gt;Mariano's answer should be enough, if you read your link to Wikipedia carefully, but maybe I could add some hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do a reflection about a hyperplane containing the origin, you pick a unitary orthogonal vector to that hyperplane, $v$, and then write the vector $x$ you want to reflect about $[v]^\bot$ as a sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = \lambda v + u&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $u \in [v]^\bot$. Now, you want to determine $\lambda \in \mathbb{R}$ in order this to be true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v \cdot x = \lambda v\cdot v + v\cdot u = \lambda  \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, necessarily&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = (v\cdot x) v + u \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if you want to reflect $x$ about $[v]^\bot$, you just substract two times its $v$-component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_v x = x - 2 (v\cdot x) v = x - 2 v (v^t x) = (I - 2vv^t ) x \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any (orthogonal) reflexion about any hyperplane $[v]^\bot$ is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_v = I - 2vv^t \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, a Housholder reflection.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-19T04:21:30.230" LastActivityDate="2010-10-19T04:21:30.230" />
  <row Id="7158" PostTypeId="2" ParentId="7151" CreationDate="2010-10-19T00:06:01.760" Score="5" ViewCount="0" Body="&lt;p&gt;It looks like the answer is 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Knuth Volume Three (1970s&amp;mdash;not as dusty as you think) reports an &lt;strong&gt;upper bound&lt;/strong&gt; of 8, which, paired with Moron's &lt;strong&gt;lower bound&lt;/strong&gt; of 8, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general form of this question is called the Selection Problem. If you google that phrase you will get lots of useful results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Knuth doesn't give an explicit algorithm for finding the median of 6 elements in at most 8 steps (at least in the first edition). However, in exercise 12 of section 5.3.3, he does give the explicit method for finding the median of 7 elements using at most 10 comparisons, which may be of some help.&lt;/p&gt;&#xA;" OwnerUserId="130" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-10-19T03:44:48.620" LastActivityDate="2010-10-19T03:44:48.620" CommentCount="4" />
  <row Id="7159" PostTypeId="1" AcceptedAnswerId="7161" CreationDate="2010-10-19T01:04:35.897" Score="4" ViewCount="136" Body="&lt;p&gt;I'm trying to come up with some math problems (word or otherwise) that get to the meaning of adding zero, but I'm getting stuck because it seems just too simple to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come up with questions like &quot;John is having a birthday party and he invites five friends over but nobody shows up. How many people are at the party?&quot;. While this question gets to the problem at hand, it might be too simple to understand the meaning of adding zero. This problem also suffers from the possibility that first graders will answer like so, &quot;Well there would be three people at the party because John's mom and dad would be there too.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2616" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-19T02:52:29.003" LastActivityDate="2010-10-19T02:52:29.003" Title="What are some deep questions that are applicable to first graders in regards to adding zero?" Tags="&lt;teaching&gt;&lt;elementary&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="7160" PostTypeId="2" ParentId="7143" CreationDate="2010-10-19T01:31:49.140" Score="8" ViewCount="0" Body="&lt;p&gt;In fact, more is true -- consider the triples $(3,4,5), (-7,24,25), (-117,44,125), (-527,-336,625), \ldots$, where the $n$th triple is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (a_n, b_n, c_n) := (Re((3+4i)^n), Im((3+4i)^n), 5^n). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\tan^{-1} 4/3$ is irrational, the sequence of pairs $(a_n/c_n,b_n/c_n)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (3/5, 4/5), (-7/25, 24/25), (-117/125, 44/125), (-527/625, -336/625), \ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistribution_theorem&quot; rel=&quot;nofollow&quot;&gt;equidistributed&lt;/a&gt; on the unit circle. Therefore the sequence $(|a_1/c_1|, |b_1/c_1|), (|a_2/c_2|, |b_2/c_2|)$ obtained by taking absolute values is equidistributed on the arc of the unit circle in the first quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if we can say that the Pythagorean triples (as opposed to some subsequence of them) are equidistributed in the first quadrant in this sense.  Anyone?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-19T01:31:49.140" CommentCount="8" />
  <row Id="7161" PostTypeId="2" ParentId="7159" CreationDate="2010-10-19T01:39:33.367" Score="3" ViewCount="0" Body="&lt;p&gt;I suggest you modify your birthday party idea to include different types of people, such as friends from school, friends from soccer, etc.  Then in general you need to add together the number of attendees from the various categories, and sometimes that will entail adding zero.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-10-19T01:39:33.367" />
  <row Id="7162" PostTypeId="2" ParentId="7159" CreationDate="2010-10-19T01:44:20.093" Score="0" ViewCount="0" Body="&lt;p&gt;Assuming this is a two-student group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give student 1 three pieces. Give student 2 two pieces. Tell student 1 to take all of their pieces in the pot. Ask the students how much are in the pot. Now ask student 2 to put all their pieces in the pot. Ask the students how much are in the pot. Now ask student 1 to take all their pieces and put them in the pot. Now ask the students how much is in the pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last addition to the pot, student 1 didn't have any more pieces, therefore student 1 added zero and the count did not increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution can scale to larger groups of students and the students addition order could be rearranged to show that adding zero at the beginning or the end will not change the outcome.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastActivityDate="2010-10-19T01:44:20.093" />
  <row Id="7163" PostTypeId="1" AcceptedAnswerId="7165" CreationDate="2010-10-19T02:21:45.443" Score="2" ViewCount="50" Body="&lt;p&gt;Is it possible to have 2 different points (non-degenerate) in n-dimensions for any value of n where n&gt;1, share the same cost and both be visited by the simplex method?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-19T04:26:01.903" LastActivityDate="2010-10-19T04:26:01.903" Title="Simplex - 2 different points with the same cost?" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" />
  <row Id="7164" PostTypeId="1" AcceptedAnswerId="7166" CreationDate="2010-10-19T03:25:21.050" Score="2" ViewCount="103" Body="&lt;p&gt;How can I compute the pmf of a random sum(1 to N) of iid bernoulli random variables, where N is distributed geometrically?&lt;/p&gt;&#xA;" OwnerUserId="2617" LastActivityDate="2010-10-19T04:53:31.520" Title="How can I compute the pmf of a random sum?" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7165" PostTypeId="2" ParentId="7163" CreationDate="2010-10-19T03:28:08.130" Score="1" ViewCount="0" Body="&lt;p&gt;No, I think.  At first I thought &quot;Yes&quot; but then convinced myself otherwise by this argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each iteration the simplex method looks for an improving feasible direction.  If there is one, then the next basic feasible solution has a lower cost than the current one (except in the case of degeneracy, which you've already ruled out).  Since the costs of the basic feasible solutions are thus strictly decreasing (again, except for degeneracy) there cannot be two nonoptimal points with the same cost that are visited by the simplex method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there isn't an improving feasible direction, then the simplex method stops, because it has found an optimal solution.  So it wouldn't visit two optimal points with the same cost, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since two such points would both have to be optimal or both be nonoptimal, it appears they cannot exist.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-19T03:28:08.130" />
  <row Id="7166" PostTypeId="2" ParentId="7164" CreationDate="2010-10-19T04:07:00.543" Score="2" ViewCount="0" Body="&lt;p&gt;Let $S = \sum_{i=1}^{N} X_i$ be the random sum that you are interested in. $X_i$ are the iid Bernoulli random variables. To compute the probability that the random sum $S$ equals an integer $s$, use the law of total probability as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(S=s) = \sum_{n=1}^{\infty} P(S=s \mid N=n) P(N=n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we condition on $N = n$, the sum of $n$ iid Bernoulli random variables is easy to compute (the answer will be another familiar distribution). Since $P(N=n)$ follows the geometric law, we can compute both the terms of the product and then compute the infinite sum.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-19T04:07:00.543" CommentCount="2" />
  <row Id="7167" PostTypeId="2" ParentId="6551" CreationDate="2010-10-19T04:25:12.650" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not familiar with SAT, as I'm not from USA, but you might find this useful: &lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this guy explains all of the subjects you might need. It might lack exercises though, but it might be worth checking it out if you need to review a subject.&lt;/p&gt;&#xA;" OwnerUserId="1605" LastActivityDate="2010-10-19T04:25:12.650" CommunityOwnedDate="2010-10-19T04:25:12.650" />
  <row Id="7168" PostTypeId="2" ParentId="7144" CreationDate="2010-10-19T04:37:55.797" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps this answer will use too much technology.  Still, I think it's pretty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $A_4$ as the group of orientation-preserving symmetries of a tetrahedron $S$.  The quotient $X=S/A_4$ is a 2-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbifold&quot; rel=&quot;nofollow&quot;&gt;orbifold&lt;/a&gt;.  Let's try to analyse it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two-dimensional orbifolds have three different sorts of singularities that set them apart from surfaces: cone points, reflector lines and corners where reflector lines meet.  Because $A_4$ acts preserving orientation, all the singularities of $X$ are cone points, and we can write them down: they're precisely the images of the points of $S$ that are fixed by non-trivial elements of $A_4$, and to give $X$ its orbifold structure you just have to label them with their stabilisers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are these points?  There are the vertices of $S$, which are fixed by a rotation of order 3; there are the midpoints of the edges of $S$, which are fixed by a rotation of order 2; and finally, the midpoints of faces, which are fixed by a rotation of order 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fundamental domain for the action of $A_4$ is given by a third of one of the faces, and if you're careful about which sides get identified you can check that $X$ is a sphere with three cone points, one labelled with the cyclic group $C_2$ and the other two labelled with the cyclic group $C_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we can compute a presentation for $A_4$ by thinking of it as the &lt;em&gt;orbifold fundamental group&lt;/em&gt; of $X$ and applying van Kampen's Theorem.  This works just as well for orbifolds, as long as you remember to consider each cone point as a space with fundamental group equal to its label.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complement of the cone points is a 3-punctured sphere, whose fundamental group is free on $x,y$.  The boundary loops correspond to the elements $x$, $y$ and $xy$.  Next, we take account of each cone point labelled $C_n$ by inserting a relation that makes the $n$th power of the appropriate boundary loop equal to $1$.  So we get the presentation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle x,y\mid x^2=y^3=(xy)^3=1\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as required.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-10-19T04:37:55.797" CommentCount="2" />
  <row Id="7169" PostTypeId="1" AcceptedAnswerId="7181" CreationDate="2010-10-19T04:47:10.533" Score="2" ViewCount="139" Body="&lt;p&gt;I was asked to find a solution to&#xA;$$\frac{\sin^2(nx)}{n^2\sin^2(x)}=2^{-1/2}$$&#xA;where $n$ is a fixed integer greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically, there's a solution just above 1/n so I decided to find this one (rather than one of the other roots).  I hoped to find an obvious pattern that I could use to find some sort of closed-form solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the power series of the root jumps out, numerically.  Since x is only used inside a sine, the power series has only odd terms.  The coefficients look like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.0019063576966069771401037205331682344703361047262779848987578751140847/n +&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.466422280109844953629560577622677335783739044717/n^3 +&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.3247444758314010835749/n^5 + ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are annoyingly close to 1, 1/2, 1/3, ... but are clearly distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how to solve this?  This is, I'm sure, well beyond anything the original question-asker cares about, but now my interest is piqued.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-10-20T13:59:09.993" LastActivityDate="2010-10-20T13:59:09.993" Title="Identify this power series / solve this trig equation" Tags="&lt;real-analysis&gt;&lt;trigonometry&gt;&lt;closed-form&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7170" PostTypeId="2" ParentId="7164" CreationDate="2010-10-19T04:53:31.520" Score="2" ViewCount="0" Body="&lt;p&gt;if $N$ is independent of the bernoulli sequence, you have what is called a compound distribution. the pmf can be obtained by conditioning: if $X$ is the [random] sum,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X = k | N = n) = {\rm bin}(k,n,p) = {\binom{n}{k}}p^k(1-p)^{n-k}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $N$ is geometric-$t, P(N = n) = (1- t)t^n, n = 0,1,\dots .$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=k, N=n) = \binom{n}{k}p^k(1-p)^{n-k}(1-t)t^n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now sum out $n$ for $n \ge k$ to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=k) = \sum_{n=k}^\infty \binom{n}{k}p^k(1-p)^{n-k}(1-t)t^n$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= p^kt^k(1-t)\sum_k^\infty \binom{n}{k}[(1-p)t)]^{n-k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= (pt)^k(1-t)\sum_{m=0}^\infty \binom{m+k}{k}[(1-p)t)]^m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(pt)^k(1-t)(1- (1-p)t)^{-(k+1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\frac{1-t}{1-t+pt}[\frac{pt}{1-t+pt}]^k.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $  X \sim {\rm geom}-\frac{pt}{1-t+pt}.$&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-10-19T04:53:31.520" CommentCount="2" />
  <row Id="7171" PostTypeId="2" ParentId="7169" CreationDate="2010-10-19T05:02:15.533" Score="2" ViewCount="0" Body="&lt;p&gt;Let $x=\arccos y$, replace the $\sin^2$ in the numerator by $1-\cos^2$, and use the trigonometric definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev_polynomials&quot; rel=&quot;nofollow&quot;&gt;Chebyshev's polynomials&lt;/a&gt; to transform the equation into a polynomial equation on $y$. Then solve that using any of the usual ways to solve it (numerically!)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-19T05:02:15.533" />
  <row Id="7172" PostTypeId="1" CreationDate="2010-10-19T05:10:38.887" Score="9" ViewCount="319" Body="&lt;p&gt;A function $f: \mathbb{R} \to \mathbb{R}$ is said to have the intermediate-value property if for any $a$,$b$ and $\lambda \in [f(a),f(b)]$ there is a $x \in [a,b]$ such that $f(x)=\lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f$ is injective if $f(x)=f(y) \Rightarrow x=y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is the case that every injective function with the intermediate-value property is continuous. I can prove this using the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An injective function with the intermediate-value property must be monotonic.&lt;/li&gt;&#xA;&lt;li&gt;A monotonic function possesses left- and right-handed limits at each point.&lt;/li&gt;&#xA;&lt;li&gt;For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am not really happy with this proof. Particularly I don't like having to invoke the intermediate-value property twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can there be a shorter or more elegant proof?&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-20T03:18:18.297" LastActivityDate="2010-10-20T03:18:18.297" Title="Injective functions with intermediate value property are continuous. Better proof?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="7173" PostTypeId="1" AcceptedAnswerId="7174" CreationDate="2010-10-19T05:24:33.910" Score="4" ViewCount="89" Body="&lt;p&gt;This is a follow-up to a &lt;a href=&quot;http://math.stackexchange.com/questions/7136/&quot;&gt;question posted recently&lt;/a&gt;.&#xA;Let&#xA;$$s_n = \sum_{r=1}^{n} \frac{1}{r(r+1)},$$&#xA;where we take $s_0 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The problem I am interested in is this: For fixed $k \geq 2$, find all solutions $(m,n)$ in nonnegative integers to the Diophantine equation $$s_m - s_n = \frac{1}{k}.$$   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Current state of knowledge (see the original question):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$s_m - s_n$ can be expressed as &#xA;$$s_m - s_n = \frac{m-n}{(m+1)(n+1)},$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At least some of the solutions are given by taking each divisor $a$ of $k$ such that $a &gt; 1$ and setting $$m = (a-1)k-1,$$ $$n = \frac{(a-1)k}{a} - 1.$$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-20T19:44:35.290" LastActivityDate="2010-10-20T19:44:35.290" Title="Finding all solutions to a Diophantine equation involving partial sums of a given series" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="7174" PostTypeId="2" ParentId="7173" CreationDate="2010-10-19T05:41:18.707" Score="3" ViewCount="0" Body="&lt;p&gt;We are to solve $$\displaystyle \frac{1}{x} - \frac{1}{y} = \frac{1}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be rearranged to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(k+y)(k-x) = k^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each divisor $\displaystyle a &gt; k$ of $\displaystyle k^2$ will give a solution and all solutions are gotten by that. So just by taking divisors of $\displaystyle k$, you might  be missing some solutions for many values of $\displaystyle k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance for $\displaystyle k=15$ you will be missing the solution $\displaystyle m = 9, n = 5$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-19T06:15:55.217" LastActivityDate="2010-10-19T06:15:55.217" CommentCount="1" />
  <row Id="7175" PostTypeId="1" AcceptedAnswerId="7199" CreationDate="2010-10-19T05:49:31.627" Score="4" ViewCount="159" Body="&lt;p&gt;I had already posted this on mathoverflow and was advised to post the same here. So here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X={x,y,z|$x^2+y^2+z^2≤1$ and $ z≥0$} i.e. X is the top half of a 3-Disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z=X/E, where E is the equivalence relation on the the plane z = 0 which is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x,y,0)∼(−x,−y,0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that this space is equivalent to a cone of RP2 (Real Projective Plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the following facts about &quot;Z&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is this a manifold with a boundary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If it is a manifold with a boundary, what are the points of Z that make the boundary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Is it simply connected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) What is the minimum Euclidean dimension in which Z can be embedded in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help. I am new to topology and this problem came up as a part of my project. Any help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. Will.&lt;/p&gt;&#xA;" OwnerUserId="2619" LastActivityDate="2010-10-19T21:56:13.423" Title="Topology of a cone of RP^2." Tags="&lt;algebraic-topology&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7176" PostTypeId="2" ParentId="7175" CreationDate="2010-10-19T06:15:23.160" Score="2" ViewCount="0" Body="&lt;p&gt;First consider just the hemisphere $X'\subseteq X$, and define $Z'= X'/E$.    This is the real projective plane, in one of its many incarnations.  You may have seen $\mathbb{RP}^2$ presented as the space of lines through the origin in $\mathbb{R}^3$.  To get from that presentation to this one, note that each such line intersects $S^2 \subseteq \mathbb{R}^3$ in exactly two points, and for those lines where one point is above the xy-plane and the other is below, we've just chosen the $z&gt;0$ representative.  This is a manifold without boundary, and it is not simply-connected.  Indeed, its universal cover is $S^2$, which is a two-fold cover.  Hence its fundamental group has two elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that in $Z$, we're literally just taking the cone over $Z'$ -- those rays from the origin that end at points in $X'$ that were identified in $Z'$ are themselves identified.  It is a non-trivial (?) fact that $\mathbb{RP}^2$ is not the boundary of any 3-manifold.  I'm pretty sure that if the cone over a space is a manifold, then the boundary is the manifold itself, although I can't instantly tell you why this should be true.  (In fact, $\mathbb{RP}^2$ generates the unoriented cobordism ring in dimension 3.)  Note that the cone over &lt;em&gt;any&lt;/em&gt; space is simply-connected, since it is contractible (to the cone point).  I don't know the answer to 4, though I'd suspect it's either 4 or 5, since 4 is the minimum dimension you need in order to embed $\mathbb{RP}^2$.  In general, the question of minimum dimension for embedding a particular manifold is not so easy, beyond the a priori bounds given by the Whitney embedding theorem.  And this isn't even a manifold...&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-10-19T06:28:03.430" LastActivityDate="2010-10-19T06:28:03.430" CommentCount="8" />
  <row Id="7177" PostTypeId="2" ParentId="7172" CreationDate="2010-10-19T06:27:15.127" Score="3" ViewCount="0" Body="&lt;p&gt;I don't know if this is shorter or elegant, and in particular it does use the hypotheses more than once, but here's something.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your 1., assume WLOG that $f$ is increasing.  Then for all $a\lt b$, $f(a,b)\subseteq (f(a),f(b))$, and thus $(a,b)\subseteq f^{-1}(f(a),f(b))$.  Since each element of $(f(a),f(b))$ has a preimage point in $(a,b)$ and $f$ is injective, $f^{-1}(f(a),f(b))\subseteq (a,b)$.  Combining, $f^{-1}(f(a),f(b))=(a,b)$.  I'll show that this implies that the inverse image of each open interval is open, which implies continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x\lt y$ and suppose that $t$ is in $f^{-1}(x,y)$.  Let $b&gt;t$.  Since $f^{-1}(f(t),f(b))=(t,b)$ and $f(t)\lt y$, there is a $b'$ with $t\lt b' \lt b$ and $f(b')\lt y$.  Thus $f^{-1}(f(t),f(b'))=(t,b')$ is contained in $f^{-1}(x,y)$.  Applying a similar argument on the other side, there is an $a'\lt t$ such that $(a',t)$ is contained in $f^{-1}(x,y)$.  Together this implies that $t$ is in the interior of $f^{-1}(x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-19T06:41:17.417" LastActivityDate="2010-10-19T06:41:17.417" />
  <row Id="7178" PostTypeId="1" AcceptedAnswerId="7179" CreationDate="2010-10-19T07:23:21.860" Score="3" ViewCount="85" Body="&lt;p&gt;Although this is not actually homework, it's an exercise from the book;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to rewrite the variance,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$var[f]=E[(f(x)-E[f(x)])^2]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E[f(x)^2]-E[f(x)]^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how it reduces, and it has had me stumped for a while.  The answer would be nice. Any &lt;em&gt;insight&lt;/em&gt; [a brief, yet, interesting explanation] would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="1774" LastEditorDisplayName="" LastEditDate="2010-10-19T07:37:33.253" LastActivityDate="2010-10-19T07:59:49.867" Title="Expected value and Variance" Tags="&lt;homework&gt;&lt;statistics&gt;&lt;expectation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7179" PostTypeId="2" ParentId="7178" CreationDate="2010-10-19T07:59:49.867" Score="5" ViewCount="0" Body="&lt;p&gt;Let $X = f(x)$ denote the random variable in question. Let $\mu = E(f(x))$. Then, you are looking to prove the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E((X-\mu)^2) = E(X^2)-\mu^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me outline the steps needed to prove this and you can try to fill in the gaps.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Expand the LHS algebraically and use linearity of expectation to split it into $3$ different terms.&lt;/li&gt;&#xA;&lt;li&gt;Note that $\mu$ is a constant and $E(aX)$ is related in a simple way to $E(X)$ when $a$ is a constant.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These two steps combined should get you to the right hand side. &lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-19T07:59:49.867" CommentCount="2" />
  <row Id="7180" PostTypeId="2" ParentId="7172" CreationDate="2010-10-19T08:14:41.090" Score="2" ViewCount="0" Body="&lt;p&gt;As you note, &lt;em&gt;f&lt;/em&gt; is injective and has the intermediate value property =&gt; &lt;em&gt;f&lt;/em&gt; is monotonic.  We may assume &lt;em&gt;f&lt;/em&gt; is increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for any &lt;em&gt;x&lt;/em&gt; and any small* &amp;epsilon; &gt; 0, we have by the IVP &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;c&lt;/em&gt; in [&lt;em&gt;x-1, x&lt;/em&gt;] such that &lt;em&gt;f(c) =  f(x)&lt;/em&gt;-&amp;epsilon; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;d&lt;/em&gt; in [&lt;em&gt;x, x+1&lt;/em&gt;] such that &lt;em&gt;f(d)=f(x)+&lt;/em&gt;&amp;epsilon;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then we choose &amp;delta;=min(&lt;em&gt;x-c,d-x&lt;/em&gt;).  Since &lt;em&gt;f&lt;/em&gt; is increasing, any point within &amp;delta; of &lt;em&gt;x&lt;/em&gt; maps to a point within &amp;epsilon; of &lt;em&gt;f(x)&lt;/em&gt;; thus &lt;em&gt;f&lt;/em&gt; is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*By &quot;small&quot; &amp;epsilon;, I mean &amp;epsilon; smaller than both &lt;em&gt;f(x)-f(x-1)&lt;/em&gt; and &lt;em&gt;f(x+1)-f(x)&lt;/em&gt;, so that the condition of the IVP is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also invoke the IVP more than once.  Perhaps the other Jonas will supply a proof with just one use of the IVP!&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-19T08:14:41.090" />
  <row Id="7181" PostTypeId="2" ParentId="7169" CreationDate="2010-10-19T08:32:04.590" Score="2" ViewCount="0" Body="&lt;p&gt;Just the $x=\arccos\;y$ substitution suggested by Mariano suffices; what you will then have will be the equation $\left(\frac{U_{n-1}(y)}{n}\right)^2=\frac1{\sqrt{2}}$ where $U_n(y)$ is the &lt;a href=&quot;http://dlmf.nist.gov/18.3&quot; rel=&quot;nofollow&quot;&gt;Chebyshev polynomial of the second kind&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-19T08:32:04.590" CommentCount="0" />
  <row Id="7182" PostTypeId="2" ParentId="7172" CreationDate="2010-10-19T09:18:21.980" Score="3" ViewCount="0" Body="&lt;p&gt;[I thought of another proof that uses the IVP and injectiveness once. Putting it as a community wiki answer.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume on the contrary that $f$ is not continuous at $x$. Then there is a sequence $x_n$ converging to $x$ such that $f(x_n)$ does not converge to $f(x)$. Then there is $\epsilon&gt;0$  and a subsequence $x_{n_k}$ such that $f(x_{n_k}) \notin (f(x)-\epsilon,f(x)+\epsilon)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must either be a further subsequence $x_{n_j}$ such that $f(x_{n_j}) \leq f(x)-\epsilon$ or a subsequence $x_{n_q}$ such that $f(x_{n_q}) \geq f(x)+\epsilon$ (or both). Assume without loss of generality the former. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since  $f(x_{n_j}) \leq f(x)-\epsilon &amp;lt; f(x)$, by the IVP for every $j$ there is a $y_j$ such that $x_{n_j}\leq y_j &amp;lt; x$ and $f(y_j)=f(x)-\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $f$ is injective, all the $y_j$ must be the same, say $y$. Because $x_{n_j}$ converges to $x$, $y=x$ by the sandwich theorem. But $f(y)\neq f(x)$. Hence a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-19T14:55:15.543" LastActivityDate="2010-10-19T14:55:15.543" CommunityOwnedDate="2010-10-19T09:18:43.603" />
  <row Id="7183" PostTypeId="1" CreationDate="2010-10-19T11:41:43.753" Score="1" ViewCount="96" Body="&lt;p&gt;There is an intricate connection between Hurwitz Zeta and the (traditional) polygamma function: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_n(z)=(-1)^{n+1}n!\zeta(n+1,z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If to use a generalization for Bernoulli numbers, this can be considered a formula, connecting polygamma and Bernoulli polynomials of negative order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_n(z)=(-1)^{n+1}n!\frac{B_{-n}(x)}n$$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As much as this equality is impressing, it is limited as it only holds for natural n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a couple of unsuccessful attempts to find a more general formula, I encountered a more natural, &quot;balanced&quot; generalization of polygamma function explained in &lt;a href=&quot;http://www.math.tulane.edu/~vhm/papers_html/genoff.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. It turned out that while the old formula still holds for natural n (since the old polygamma and balanced polygamma coincide in integer positive orders), a completely new formula connecting this balanced polygamma with Zeta and Bernoulli numbers can be derived which holds for any z:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(z,q)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,\frac{q}{2}+\frac{1}{2}\right)+\psi \left(z-1,\frac{q}{2}\right)\right)-\psi(z-1,q)\right)}{\ln(2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_z(q) = -\frac{\Gamma (z+1) \left(2^{z-1} \left(\psi\left(-z,\frac{q}{2}+\frac{1}{2}\right)+\psi\left(-z,\frac{q}{2}\right)\right)-\psi(-z,q)\right)}{\ln (2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of them can be expressed completely in terms of balanced polygamma and elementary functions if to notice that $\Gamma(x)=e^{\psi(-1,x)+\frac 12 \ln(2\pi)}$, which allows to get rid of the Gamma function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the target was reached, these expressions still leave a bad impression. I cannot simplify it as no CAS system is capable of operations with the balanced polygamma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I am asking for help on how to simplify the expressions so they could be easier to manage and use. It is also not evident how the letter formulas become the former ones at positive real z.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-10-19T14:09:46.450" LastActivityDate="2010-10-19T14:09:46.450" Title="Connection between Bernoulli polynomials and polygamma function" Tags="&lt;gamma-function&gt;&lt;zeta-function&gt;" CommentCount="2" />
  <row Id="7184" PostTypeId="1" CreationDate="2010-10-19T11:50:55.590" Score="3" ViewCount="87" Body="&lt;p&gt;s and t are two nodes in a graph.&#xA;How to prove that the maximum number of edge-disjoint cut sets which divide s and t equals the length of the shortest path from s to t?&lt;/p&gt;&#xA;" OwnerUserId="2621" LastEditorUserId="2621" LastEditorDisplayName="" LastEditDate="2010-10-19T15:54:08.210" LastActivityDate="2010-10-19T16:58:07.590" Title="Number of cut sets and the length of the shortest path?" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7185" PostTypeId="1" AcceptedAnswerId="7190" CreationDate="2010-10-19T14:19:14.600" Score="1" ViewCount="119" Body="&lt;p&gt;A very simple solution to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffusion_equation&quot; rel=&quot;nofollow&quot;&gt;diffusion equation&lt;/a&gt; is $u(x,t)=x^2+2 t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;br&gt;&#xA;How can this be a solution to the diffusion equation when nothing really diffuses, but just stays the same - see plot: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+x+x%2B2+t&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that it is a solution but can't believe my eyes... Is there some kind of intuition other than: &quot;it satisfies the PDE&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other examples of solutions to well-known equations that behave completely un-intuitive in some situations?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-19T15:13:07.763" Title="Simple Solution to Diffusion equation" Tags="&lt;intuition&gt;&lt;differential-equations&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7186" PostTypeId="1" AcceptedAnswerId="7193" CreationDate="2010-10-19T14:52:09.437" Score="2" ViewCount="48" Body="&lt;p&gt;I'm currently reading Hilton &amp;amp; Stammbach's A First Course in Homological Algebra, and the following point has stumped me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 1.8, they construct co-free modules (&quot;left moodule&quot; over some ring) as essentially coming from the right adjoint to the forgetful functor from $\Lambda$-Modules to Abelian Groups. On the other hand, the free module is constructed as the left adjoint to the forgetful functor from $\Lambda$-modules to Sets. This turns out to be equivalent to requiring free modules to be direct sums of copies of $\Lambda$ considered as a module over itself, and to requiring co-free modules as direct products of the $\Lambda^*=Hom_\mathbb{Z}(\Lambda, \mathbb{Q}/\mathbb{Z})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess my question is: what does the right adjoint to the forgetful functor to Set look like, and why is the right adjoint to the forgetful functor to Abelian Groups more useful?&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-10-19T15:30:19.787" Title="Why is the co-free module defined as the right adjoint to the forgetful functor to Ab rather than Set?" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7187" PostTypeId="1" AcceptedAnswerId="7198" CreationDate="2010-10-19T14:57:14.870" Score="9" ViewCount="332" Body="&lt;p&gt;Quaternions came up while I was interning not too long ago and it seemed like no one really know how they worked. While eventually certain people were tracked down and were able to help with the issue, it piqued my interest in quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading many articles and a couple books on them, I began to know the formulas associated with them, but still have no clue &lt;em&gt;how&lt;/em&gt; they work (why they allow rotations in 3D space to be specific). I back-tracked a little bit and looked at normal complex numbers with just one imaginary component and asked myself if I even understood how they allow rotations in 2D space. After a couple awesome moments of understanding, I understood it for imaginary numbers, but I'm still having trouble extending the thoughts to quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can someone intuitively think about quaternions and how they allow for rotations in 3D space?&lt;/p&gt;&#xA;" OwnerUserId="2623" LastEditorUserId="2623" LastEditorDisplayName="" LastEditDate="2010-10-19T20:58:23.217" LastActivityDate="2010-10-20T12:52:51.393" Title="How can one intuitively think about quaternions?" Tags="&lt;linear-algebra&gt;&lt;complex-numbers&gt;&lt;quaternions&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="7" />
  <row Id="7188" PostTypeId="2" ParentId="7184" CreationDate="2010-10-19T14:58:08.393" Score="1" ViewCount="0" Body="&lt;p&gt;Perhaps this paper might be of use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iis.sinica.edu.tw/page/jise/2001/200103_02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.iis.sinica.edu.tw/page/jise/2001/200103_02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-19T14:58:08.393" CommentCount="1" />
  <row Id="7189" PostTypeId="2" ParentId="7185" CreationDate="2010-10-19T14:59:19.577" Score="0" ViewCount="0" Body="&lt;p&gt;It is difficult to interpret a PDE without knowing its boundary conditions and/or initial values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, at first guess, I'd say you should look at individual time slices of the solution on the same graph and see if that helps your interpretation.  The 3D image is nice for showing everything at once, but that can obscure an interpretation.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2010-10-19T14:59:19.577" />
  <row Id="7190" PostTypeId="2" ParentId="7185" CreationDate="2010-10-19T15:12:14.120" Score="3" ViewCount="0" Body="&lt;p&gt;Here are a few pointers, from the more obtuse to the more direct.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Meeting a friend in a corridor, Wittgenstein said: 'Tell me, why do people always say it was &lt;em&gt;natural&lt;/em&gt; for men to assume that the sun went round the earth, rather than that the earth was rotating?'  His friend said, 'Well, obviously, because it &lt;em&gt;looks&lt;/em&gt; as if the sun is going round the earth.'  To which the philosopher replied, 'Well, what would it have looked like if it had looked as if the earth was rotating?'&quot; Apparently from Tom Stoppard's play &lt;em&gt;Jumpers&lt;/em&gt;. More to the point: what does your intuition think the solution to the diffusion equation with initial conditions $u(x,0) = x^2$ &lt;em&gt;should&lt;/em&gt; have looked like?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$u_1(x,t) = \text{const}$, and $u_2(x,t) = x$ are also solutions of the diffusion equation. Do you find them more or less surprising than the $x^2 + t$ solution?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The standard intuitive interpretation of the diffusion equation is that it &quot;smooths out&quot; variations in the solution. The way it finds these variations is by looking at the second derivative. But the second derivative of $x^2$ is constant everywhere, so what can the diffusion equation do?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Locally, every point &quot;wants to be&quot; equal to the average of the value in its neighbourhood. If the second derivative at a point is positive, the function looks like a valley, so the point increases its value over time; if the second derivative is negative, the point is on a peak, and tries to decrease. With $x^2$, all of $\mathbb{R}$ is a valley.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-19T15:12:14.120" CommentCount="2" />
  <row Id="7191" PostTypeId="2" ParentId="7185" CreationDate="2010-10-19T15:13:07.763" Score="5" ViewCount="0" Body="&lt;p&gt;But there is diffusion! Heat – or whatever quantity $u$ represents – flows from the far left and right ($x\to\pm\infty$, where there's lots of it!) towards the middle ($x=0$). For every fixed point $x$ there is always more heat flowing towards it from the outside than what it emits inwards, and the result is that the heat keeps increasing at each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this might seem unphysical, and that's because it is. For this particular solution, the total amount of heat, $\int_{-\infty}^{\infty} u(x,t) dx$, is infinite for each $t$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-19T15:13:07.763" CommentCount="1" />
  <row Id="7192" PostTypeId="1" CreationDate="2010-10-19T15:27:00.493" Score="2" ViewCount="65" Body="&lt;p&gt;Can any of you guys think of a topological space $(X,\tau)$ and a family of subsets {A&lt;sub&gt;s&lt;/sub&gt;}&lt;sub&gt;${s \in S}$&lt;/sub&gt; of $X$ such that for a certain $x \in X$ you can find a subset $V$ such that $x \in V$ and {$s \in S: V \cap A_{s} \neq \emptyset$} is finite, whereas for every $W \subseteq X$, with $x \in W$, we have that {$s \in S: W \cap \mathrm{cl}(A_{s}) \neq \emptyset$} is never finite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thank you in advance for your replies.&lt;/p&gt;&#xA;" OwnerUserId="2624" LastEditorUserId="2624" LastEditorDisplayName="" LastEditDate="2010-10-19T15:47:17.323" LastActivityDate="2010-10-19T15:47:17.323" Title="A finiteness condition" Tags="&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="7193" PostTypeId="2" ParentId="7186" CreationDate="2010-10-19T15:30:19.787" Score="2" ViewCount="0" Body="&lt;p&gt;If a functor has a right adjoint, it preserves colimits; but the&#xA;forgetful functor from $R$-Mod to Set doesn't. For instance it&#xA;doesn't preserve binary coproducts. So there isn't a right adjoint.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-19T15:30:19.787" />
  <row Id="7194" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T15:37:48.827" Score="3" ViewCount="0" Body="&lt;p&gt;I highly recommend that your read the presentation in Conway and Smith: On quaternions and octonions: Their geometry, arithmetic, and symmetry. Here's an excerpt from John Baez's &lt;a href=&quot;http://www.ams.org/journals/bull/2005-42-02/S0273-0979-05-01043-8/S0273-0979-05-01043-8.pdf&quot; rel=&quot;nofollow&quot;&gt;very informative review:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It follows that the quaternions of norm 1 form a group under multiplication. This group is usually called SU(2), because people think of its elements as 2 × 2 unitary matrices with determinant 1. However, the quaternionic viewpoint is better adapted to seeing how this group describes rotations in 3 and 4 dimensions. The unit quaternions act via conjugation as rotations of the 3d space of &quot;pure imaginary&quot; quaternions, namely those with Re(q) = 0. This gives a homomorphism from SU(2) onto the 3d rotation group SO(3). The kernel of this homomorphism is {±1}, so we see SU(2) is a double cover of SO(3). The unit quaternions also act via left and right multiplication as rotations of the 4d space of all quaternions. This gives a homomorphism from SU(2) × SU(2) onto the 4d rotation group SO(4). The kernel of this homomorphism is {±(1, 1)}, so we see SU(2) × SU(2) is a double cover of SO(4).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These facts are incredibly important throughout mathematics and physics. With their help, Conway and Smith classify the finite subgroups of the 3d rotation group SO(3), its double cover SU(2), the 3d rotation/reflection group O(3), and the 4d rotation group SO(4). These classifications are all in principle &quot;well known'. However, they seem hard to find in one place, so Conway and Smith's elegant treatment is very helpful&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation&quot; rel=&quot;nofollow&quot;&gt;Quaternions and spatial rotation.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-19T15:46:24.207" LastActivityDate="2010-10-19T15:46:24.207" />
  <row Id="7195" PostTypeId="1" CreationDate="2010-10-19T15:39:53.380" Score="7" ViewCount="399" Body="&lt;p&gt;In my attempts to derive the closed form for the arclength of the hyperbola, I wound up with the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{\sqrt{1-m\;\sin^2 u}}{\sin^2 u}\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that such integrals are expressible as combinations of elliptic integrals and trigonometric functions, and &lt;em&gt;Mathematica&lt;/em&gt; does return a result containing elliptic integrals. However, the result I got from &lt;em&gt;Mathematica&lt;/em&gt; seems to suggest that the original integrand was split like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt{1-m\;\sin^2 u}}{\sin^2 u}=\frac{1-m}{\sqrt{1-m\;\sin^2 u}}-\sqrt{1-m\;\sin^2 u}+\frac{\csc^2u-m\;\sin^2 u}{\sqrt{1-m\;\sin^2 u}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then integrated to yield the expression containing the elliptic integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm stuck with is how this splitting was thought of so that the final integral is expressible in terms of the Legendre elliptic integrals. A possible clue lay in the fact that through an appropriate substitution, the original integral can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\mathrm{ds}^2(v|m)\mathrm{d}v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathrm{ds}(v|m)=\frac{\mathrm{dn}(v|m)}{\mathrm{sn}(v|m)}$ is a Jacobian elliptic function, from which this &lt;a href=&quot;http://dlmf.nist.gov/22.16.E25&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; applies. However, I don't know how the formula listed in the DLMF was derived, and I suspect that the answer to my question will also explain that formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any help in understanding why the integrand was split in that manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trivially,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{ds}^2(v|m)=\mathrm{ns}^2(v|m)-m=1-m+\mathrm{cs}^2(v|m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elliptic integral of the second kind is supposed to arise due to the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\mathrm{dn}^2(v|m)\mathrm{d}v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int\mathrm{dn}(v|m)\mathrm{d}(\mathrm{am}(v|m))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int\sqrt{1-m\;\mathrm{sn}^2(v|m)}\mathrm{d}(\mathrm{am}(v|m))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int\sqrt{1-m\;\sin^2(\mathrm{am}(v|m))}\mathrm{d}(\mathrm{am}(v|m))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=E(\mathrm{am}(v|m)|m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I suppose figuring out how to integrate the squares of any of $\mathrm{cs}$, $\mathrm{ns}$, or $\mathrm{ds}$ would do the trick. I suspect something like integration by parts or tricks similar to those used for integrating ratios of trigonometric functions would work here, but I can't seem to find the proper strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another way would be to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int\left(\mathrm{cs}^2(v|m)+\mathrm{dn}^2(v|m)\right)\mathrm{d}v=\mathrm{cs}(v|m)\mathrm{dn}(v|m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating the RHS gives the integrand, but supposing I did not know this expression, how would this integral be evaluated?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T17:00:38.860" LastActivityDate="2010-12-16T17:44:26.080" Title="On deriving the arclength of a hyperbola" Tags="&lt;calculus&gt;&lt;special-functions&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="7196" PostTypeId="2" ParentId="7192" CreationDate="2010-10-19T15:46:58.207" Score="4" ViewCount="0" Body="&lt;p&gt;If I read it correctly you want $V$ to meet only finitely many of the $A_s$, but every set containing $x$ to meet infinitely many of the closures of the $A_s$.  Without some restriction on $V$ and the $A_s$ this is easy.  Just have all the $A_s$ have x as a limit point but not a member.  Then $V$ is just the point $x$ and any $W$ containing $x$ will meet the closure of all the $A_s$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-19T15:46:58.207" />
  <row Id="7197" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T15:47:28.533" Score="1" ViewCount="0" Body="&lt;p&gt;I find the conversion between quaternions and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axis_angle&quot; rel=&quot;nofollow&quot;&gt;axis-angle representation&lt;/a&gt; quite instructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the axis-angle representation, you describe a rotation by specifying the axis of rotation as a unit vector $\omega$ and an angle $\theta$ about which to rotate around this axis. An interesting fact is that any possible rotation can be described in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding quaternion is given simply by $\left(\cos(\theta/2), \omega\sin(\theta/2)\right)$. Here the notation $(a, v)$, where $a$ is a scalar and $v$ a real vector, denotes the quaternion $a + v_xi + v_yj + v_zk$, or $(a,v_x,v_y,v_z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the intuitive interpretation of this. Given a particular rotation axis $\omega$, if you restrict the 4D quaternion space to the 2D plane containing $(1,0,0,0)$ and $(0,\omega_x,\omega_y,\omega_z)$, the unit quaternions representing all possible rotations about the axis $\omega$ form the unit circle in that plane. A rotation of $\theta$ about the axis $\omega$ is the point at an angle $\theta/2$ from $(1,0,0,0)$ on that circle. For example, not rotating at all is $(1,0,0,0)$, rotating 180° is $(0,\omega_x,\omega_y,\omega_z)$, and rotating 360° is $(-1,0,0,0)$, which is the same as not rotating at all (see final paragraph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Multiplying&lt;/em&gt; two quaternions is unintuitive, but I'm not bothered by that, because the composition of two rotations in real life is quite unintuitive in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the quaternions are a &quot;double cover&quot; of the space of rotations, in that any rotation actually has &lt;em&gt;two&lt;/em&gt; quaternions, say $q$ and $-q$, that represent it: $\theta$ and $\theta + 2\pi$ are the same angle, but $\theta/2$ and $(\theta+2\pi)/2$ are not. This is the only &quot;glitch&quot; in the quaternion representation of rotations.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-19T15:47:28.533" CommentCount="1" />
  <row Id="7198" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T15:55:33.217" Score="8" ViewCount="0" Body="&lt;p&gt;Here's one way.  The group of unit quaternions is isomorphic to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_unitary_group&quot; rel=&quot;nofollow&quot;&gt;special unitary group&lt;/a&gt; $\text{SU}(2)$, the group of $2 \times 2$ unitary complex matrices with determinant $1$.  This group acts on $\mathbb{C}^2$ in the obvious way, and so it also acts on &lt;em&gt;lines&lt;/em&gt; in $\mathbb{C}^2$.  (These are complex lines, so they have real dimension $2$.)  The space of lines in $\mathbb{C}^2$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_projective_space&quot; rel=&quot;nofollow&quot;&gt;complex projective line&lt;/a&gt; $\mathbb{CP}^1$, and it turns out there is a natural way to think about this space as a sphere - namely, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere&quot; rel=&quot;nofollow&quot;&gt;Riemann sphere&lt;/a&gt;.  There is a beautiful projection which is pictured at the Wikipedia article which shows this; essentially one thinks of $\mathbb{CP}^1$ as $\mathbb{C}$ plus a &quot;point at infinity&quot; and then projects the latter onto the former in a way which misses one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\text{SU}(2)$ naturally acts on a sphere, and as it turns out it naturally acts by rotations.  This describes the famous 2-to-1 map $\text{SU}(2) \to \text{SO}(3)$ which allows quaternions to describe 3D rotations.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-19T15:55:33.217" CommentCount="3" />
  <row Id="7199" PostTypeId="2" ParentId="7175" CreationDate="2010-10-19T16:02:19.677" Score="2" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's not a manifold either with or without boundary.  For the proof, look at the cone point c.  If U is a small open set around c, then $U-c$ is homeomorphic to $\mathbb{R}P^2 \times (0,1]$.  However, this is not homeomorphic to either $\mathbb{R}^3-pt$ nor a half space minus a point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In fact, it's contractible (the cone over anything is contracitble).  Thus, it's simply connected.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not sure, but it can definitely be topological embedded in R^5.  To see this, let $f$ be an embedding of $\mathbb{R}P^2$ into $\mathbb{R}^4$ (which exists by Whitney's embedding theorem).  Define $G:Z\rightarrow\mathbb{R}^5$ by $G(p, t) = (tf(p),t)$.  This will be a topological embedding.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The minimal possible answer is 4 since that's the smallest in which $\mathbb{R}P^2$ can be embedded into.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-10-19T18:46:53.087" LastActivityDate="2010-10-19T18:46:53.087" CommentCount="8" />
  <row Id="7200" PostTypeId="1" CreationDate="2010-10-19T16:21:35.037" Score="1" ViewCount="81" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/7126/boolean-algebra-simplification-dont-know-the-method-used&quot;&gt;Boolean Algebra, Simplification: Don&amp;#39;t know the method used&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Based on a &lt;a href=&quot;http://math.stackexchange.com/questions/7126/boolean-algebra-simplification-dont-know-the-method-used&quot;&gt;previous question&lt;/a&gt;, the answer I got for the &quot;sum of products&quot; method was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RC' + RCG' + RCGM'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Just for completeness, can someone help walk me through to how this was arrived at?  Thanks again in advance!&lt;/p&gt;&#xA;" OwnerUserId="2609" LastActivityDate="2010-10-19T16:21:35.037" ClosedDate="2010-10-20T05:41:33.297" Title="Boolean Algebra, Simplification: Sum of products" Tags="&lt;boolean-algebra&gt;" CommentCount="2" />
  <row Id="7201" PostTypeId="1" AcceptedAnswerId="10947" CreationDate="2010-10-19T16:41:23.990" Score="1" ViewCount="110" Body="&lt;p&gt;Can someone help direct me towards information on calculating the polynomial remainder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do the basics, but what if I don't have access to the polynomial in expanded form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $\frac{1}{1-x}$ represents the polynomial $\sum_{i=0}^\infty{x^i}$.  I want to be able to find the remainder using an expression like the fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help explain where I can find more resources on this.  I'm not a math expert, so something very readable would be great.  However, I'm willing to learn the math if I need to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me find resources with the equivalent and generating functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to show that this operation is well defined, Dr. Herwig Hauser has written a paper concerning the operation &lt;a href=&quot;http://homepage.univie.ac.at/herwig.hauser/Publications/linear_recurrences_and_powerseries_division.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-10-19T18:43:43.030" LastActivityDate="2010-11-19T09:59:39.423" Title="Polynomial Remainder / Generating Function Remainder" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7202" PostTypeId="2" ParentId="7184" CreationDate="2010-10-19T16:58:07.590" Score="1" ViewCount="0" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First show that the size of any collection of edge-disjoint cut-sets cannot exceed the length of the shortest path.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-19T16:58:07.590" CommentCount="1" />
  <row Id="7204" PostTypeId="1" AcceptedAnswerId="7216" CreationDate="2010-10-19T17:06:09.683" Score="5" ViewCount="196" Body="&lt;p&gt;How does one evaluate show that this limit: $$\lim_{n \to \infty}\sqrt{1 + 2 \sqrt{1+3 \sqrt{1+ \cdots \sqrt{1+(n-1) \sqrt{1+n}}}}}=3$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-19T17:31:53.287" LastActivityDate="2010-10-19T19:53:57.927" Title="Limit of Nested Radical" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7205" PostTypeId="2" ParentId="7204" CreationDate="2010-10-19T17:13:33.360" Score="10" ViewCount="0" Body="&lt;p&gt;This is Ramanujan's famous nested radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information can be found here: &lt;a href=&quot;http://www.isibang.ac.in/~sury/ramanujanday.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.isibang.ac.in/~sury/ramanujanday.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Also: &lt;a href=&quot;http://mathworld.wolfram.com/NestedRadical.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/NestedRadical.html&lt;/a&gt; (number 26).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, this is how he came up with it (sorry, no reference for this claim).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3 = \sqrt{9} = \sqrt{1 + 8} = \sqrt{1 + 2 \cdot 4}$$&#xA;$$ = \sqrt{1 + 2\sqrt{16}} = \sqrt{1 + 2\sqrt{1 + 3 \cdot 5}}$$&#xA;$$ = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25}}} = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + 4 \cdot 6}}}$$&#xA;etc.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-19T17:20:31.400" LastActivityDate="2010-10-19T17:20:31.400" CommentCount="1" />
  <row Id="7206" PostTypeId="2" ParentId="7187" CreationDate="2010-10-19T17:25:04.893" Score="0" ViewCount="0" Body="&lt;p&gt;One way to see the analogy between the transition from real to complex numbers and from complex numbers to quaternions&#xA;is by means of the fact that the spaces of unit complex numbers and the unit quaternions, namely,&#xA;the unit circle $S^1$ and the unit 3-sphere $S^3$ respectively, are Lie groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy can be seen in the group laws, which can be casted in a very similar form as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point on $S^1$ can be represented by the complex number $z = (x, y)$ , $x^2+y^2= 1$.&#xA;A point on $S^3$ can be represented by a quaternion $q = (z, w)$ , where $z$ and $w$ are complex numbers and&#xA;$|z|^2+|w|^2= 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $S^1$ multiplication law has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_1 z_2 = (x_1x_2 - y_1 y_2, x_1y_2 + y_1 x_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $S^3$ multiplication law is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_1 q_2 = (z_1z_2 - w_1 \bar{w_2}, z_1w_2 + w_1 \bar{z_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It easy to verify that these multiplication laws satisfy the group laws. &#xA;Of course the main difference is that the quaternion group law is no longer commutative: since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_2 q_1 = (z_1z_2 - \bar{w_1} w_2, z_1w_2 + \bar{w_1} z_2)$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-10-19T17:25:04.893" />
  <row Id="7207" PostTypeId="1" AcceptedAnswerId="7208" CreationDate="2010-10-19T17:31:20.327" Score="7" ViewCount="90" Body="&lt;p&gt;Given two values $a$ and $b$, how should one go about solving the following inequality for $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \le b^{r}-r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying $log_{b}$ on both sides of the inequality doesn't help me much since that yields the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log_{b}{a} \le log_{b}{(b^{r}-r)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_{z}{x} - \log_{z}{y} = \frac{\log_{z}{x}}{\log_{z}{y}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that doesn't help me to eliminate the exponentiation and solve for $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been ages that I've done algebra, and now I'm back in grad school, and I'm finding this equation in a homework. Can't remember ever seeing a rule of logarithms involving such expressions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual homework (after simplification) involves $9 \le 2^{r} - r$ which is trivially solvable by just eyeballing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there a methodical way to solve $a \le b^{r}-r$ for $r$ given any arbitrary numbers $a$ and $b$?&lt;/p&gt;&#xA;" OwnerUserId="2522" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-19T17:40:21.310" LastActivityDate="2010-10-19T18:05:09.120" Title="How do we solve $a \le b^{r}-r$ for $r$?" Tags="&lt;algebra&gt;" AnswerCount="3" />
  <row Id="7208" PostTypeId="2" ParentId="7207" CreationDate="2010-10-19T17:41:59.800" Score="5" ViewCount="0" Body="&lt;p&gt;The case of equality can be solved in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function#Applications&quot; rel=&quot;nofollow&quot;&gt;Lambert W&lt;/a&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be used to solve equations of the form $$p^{ax+b} = cx + d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(quoted from the wiki page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^{r} -r$ is 'mostly' monotonic, so I suppose solving the equality will be enough to find the solutions to your inequality.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-19T17:41:59.800" CommentCount="1" />
  <row Id="7209" PostTypeId="2" ParentId="7207" CreationDate="2010-10-19T17:45:58.910" Score="2" ViewCount="0" Body="&lt;p&gt;If you don't have the W function, you need an iterative root finding type of solution.  Note that for your example r is somewhere between 3 and 4.  This equation is nicely behaved, so any of them will work.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Root-finding_algorithm&quot; rel=&quot;nofollow&quot;&gt;Root finding&lt;/a&gt; for a start&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-19T17:45:58.910" CommentCount="1" />
  <row Id="7210" PostTypeId="1" AcceptedAnswerId="7229" CreationDate="2010-10-19T17:47:33.297" Score="6" ViewCount="360" Body="&lt;p&gt;A subset $S$ of a metric space $X$ is totally bounded if for any $r&gt;0$, $S$ can be covered by a finite number of $X$-balls of radius $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metric space $X$ is totally bounded if it is a totally bounded subset of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, bounded subsets of $\mathbb{R}^n$ are totally bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any interesting necessary and/or sufficient conditions for a metric space or its subsets to be totally bounded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Background: I was trying to generalize problem 4.8 of baby Rudin which asks you to prove that a real uniformly continuous function on a bounded subset $E$ of the real line is bounded. It seems after a little googling that a more general true statement would require $E$ to be a totally bounded subset of some metric space. But where might we meet such subsets?]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-21T01:32:44.120" LastActivityDate="2010-11-04T12:41:50.667" Title="Which metric spaces are totally bounded?" Tags="&lt;analysis&gt;&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="7211" PostTypeId="2" ParentId="7207" CreationDate="2010-10-19T18:05:09.120" Score="2" ViewCount="0" Body="&lt;p&gt;You can also get a quick-and-dirty approximation for the equality case with the following approach, which should hold in most situations when $a, b &gt; 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange the problem and take logarithms to rewrite it as $$\ln\left(1+ \frac{r}{a}\right) = r \ln b - \ln a.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $\frac{r}{a}$ is less than 1 (and it should be in most cases where $a, b &gt; 1$), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(1+ \frac{r}{a}\right) \approx \frac{r}{a},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives you $$r \approx \frac{- \ln a}{1/a - \ln b}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific problem, this method yields $r \approx 3.78$, whereas the actual answer is closer to $3.66$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can improve this by using the quadratic approximation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(1+ \frac{r}{a}\right) \approx \frac{r}{a} - \frac{r^2}{2a^2},$$&#xA;which requires solving a quadratic equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These approximations come from the Taylor expansion of $\ln (1+x)$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-19T18:05:09.120" />
  <row Id="7212" PostTypeId="2" ParentId="7210" CreationDate="2010-10-19T18:14:30.323" Score="5" ViewCount="0" Body="&lt;p&gt;A metric space is totally bounded if and only if its completion is compact.  (However, a little googling reveals that totally bounded doesn't necessarily imply compact completion without the axiom of choice.)  This gives one explanation of the result in your background, because uniformly continuous functions extend to completions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For subsets of a complete metric space, total boundedness is equivalent to having compact closure, which is what we see for bounded subsets of $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know, is this interesting?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-19T18:14:30.323" CommentCount="5" />
  <row Id="7213" PostTypeId="1" AcceptedAnswerId="7215" CreationDate="2010-10-19T18:14:37.700" Score="3" ViewCount="185" Body="&lt;p&gt;Use the Lagrange remainder to determine how many terms of the Taylor series of cos expanded&#xA;at 0 are needed to estimate $cos(\frac{21\pi}{5})$ . Accuracy to 3 decimal places&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have a lot of rough work on paper. I know math.stackexchange dosn't really answer assignment questions but I am looking for a head start on even where to start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you guys want, i can type up my notes here. Bascically the f'(c) term from the lagrange formula can be 1. I know my inequality has something to do with &amp;lt; 0.001. I need taylor's expansion (obviously to find the n I need). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just can't put it together. This is my first real-analysis course.&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="287" LastEditorDisplayName="" LastEditDate="2010-10-19T18:23:59.620" LastActivityDate="2010-10-19T20:40:57.847" Title="Lagrange remainder and Taylor series" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="3" />
  <row Id="7214" PostTypeId="2" ParentId="7213" CreationDate="2010-10-19T18:21:49.270" Score="1" ViewCount="0" Body="&lt;p&gt;You are right that the derivative term in the error is bounded by 1, so you just need the $\frac{(x-a)^{(n+1)}}{(n+1)!}$ less than .001.  As $21\pi/5$ is rather large, it will take a lot of terms.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-19T18:28:39.420" LastActivityDate="2010-10-19T18:28:39.420" CommentCount="2" />
  <row Id="7215" PostTypeId="2" ParentId="7213" CreationDate="2010-10-19T19:10:43.370" Score="1" ViewCount="0" Body="&lt;p&gt;To cut down the number of terms it seems reasonable to use&#xA;$$\cos \left( \frac {21\pi}{5} \right) = \cos \left( \frac{\pi}{5} \right),$$&#xA;so we look for the smallest $n$ such that&#xA;$$ \frac{1}{(n+1)!} \left( \frac{\pi}{5} \right)^{n+1} &amp;lt; \frac{1}{1000}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-19T19:10:43.370" CommentCount="3" />
  <row Id="7216" PostTypeId="2" ParentId="7204" CreationDate="2010-10-19T19:12:24.983" Score="5" ViewCount="0" Body="&lt;p&gt;This is the special case $\rm\ x,\:n,\:a = 2,\:1,\:0\ $ in Ramanujan's second notebook, chapter XII, entry 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm x + n + a\ =\ \sqrt{ax + (n+a)^2 + x\sqrt{a(x+n) + (n+a)^2 + (x+n) \sqrt{\cdots}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is Ramanujan's solution of the given special case - which was submitted to a journal in April 1911. Note that his solution is incomplete (exercise: why?). For further discussion see this 1935 Monthly article, &lt;a href=&quot;http://www.jstor.org/stable/2301294&quot; rel=&quot;nofollow&quot;&gt;Herschfeld: On infinite radicals.&lt;/a&gt;  It also appeared as Problem A6 on the 27th Putnam competition, 1966. Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $\ \sqrt{a_1 + \sqrt{a_2 +\:\cdots\: +\sqrt{a_n}}}\ \ $ is that $\rm\displaystyle\ \ {\overline \lim}_{n\to\infty}\frac{\log{a_n}}{2^n}\ &amp;lt; \infty\:.\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UBCbs.jpg&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-19T19:53:57.927" LastActivityDate="2010-10-19T19:53:57.927" />
  <row Id="7217" PostTypeId="1" AcceptedAnswerId="7286" CreationDate="2010-10-19T19:32:11.200" Score="7" ViewCount="172" Body="&lt;p&gt;Everything that follows takes place in the Borel $\sigma$-algebra with Lebesgue measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lebesgue mean of $f$ at $x$ is defined as $\displaystyle \lim_{\epsilon\to 0} \int_{x-\epsilon}^{x+\epsilon} \frac{f}{2\epsilon}$. The function defined by $[f](x) =$ &quot;Lebesgue mean of $f$ at $x$&quot; is equal to $f$ almost everywhere so integrals of it will be equal to integrals of $f$, call this property $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the interval $[0,1]$ and removing the middle $1/4$ then the two middle $1/8$-ths then the four middle $1/16$-ths and so on produces the fat Cantor set which has measure $1/2$ but does not contain any interval (Since measures are countably additive this informs us that the set contains uncountably many points). Let $I$ be the indicator function for the fat Cantor set - it is equal to $1$ when applied on a point of the set and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I would have thought that the Lebesgue mean $[I]$ of the indicator for the fat Cantor set would be the zero function. Since that contradicts $P$, it seems more plausible that $[I] = I$ but I cannot prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we construct the function $[I]$?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-20T12:28:17.247" Title="What is the Lebesgue mean of the fat Cantor set?" Tags="&lt;analysis&gt;&lt;continuum&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="7218" PostTypeId="1" CreationDate="2010-10-19T19:35:39.167" Score="1" ViewCount="81" Body="&lt;p&gt;What is the name of the following matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{pmatrix} a &amp; b &amp; 0 \\ &#xA;c &amp; d &amp; 0 \\ &#xA;0 &amp; a &amp; b\\ &#xA;0&amp; c&amp; d&amp; \\ &#xA;b &amp; 0 &amp; a \\&#xA;d &amp; 0 &amp; c\end{pmatrix}\ &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Block_matrix#Block_Toeplitz_matrices&quot; rel=&quot;nofollow&quot;&gt;Block Toeplitz&lt;/a&gt; matrix, but usually one defines those by full shifts by (in this case) 2x2 matrices. In particular i'm interested in solving linear equations of this form. Any reference would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="2629" LastActivityDate="2010-10-20T17:33:22.947" Title="Block Toeplitz Matrix" Tags="&lt;linear-algebra&gt;&lt;numerical-methods&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="7219" PostTypeId="1" CreationDate="2010-10-19T20:11:04.437" Score="1" ViewCount="46" Body="&lt;p&gt;Let $\Sigma$ be a $n-1$ dimensional space-like submanifold of a $n+1$ dimensional space-time $(V,g)$ and let $x \in \Sigma$. Then $(T_x \Sigma)^\perp$ is of dimension $2$ and is time-like. Such a $2$-plane as $(T_x \Sigma)^\perp$ cuts the future lightcone at $x$ along two linearly independent future null directions. Let $l ^+$ and $l^-$ be two future null vectors along these directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $e_a$ with $a \in {2,3,4,5,...}$ be a reference frame on $\Sigma$ and hence orthogonal to $l^+$ and $l^-$. Then the following sequence of identities hold,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K^+_{ab} = K^+(e_a,e_b) = g(l^+,\nabla _{e_a} e_b) = g(e_b,\nabla _ {e_a} l^+) = \nabla _b l^+_a$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By metric compatibility and the fact that $l^+$ and $e_b$ are orthogonal I would have expected a minus between the last but two steps. Why is there no negative sign? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and similarly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K^-_{ab} = \nabla _b l^-_a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ${h_{ab}}$ is the induced metric on the submanifold $\Sigma$ then the {\bf Null Mean Curvature} of $\Sigma$ is defined as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi ^+ = h^{ab}K^+_{ab}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi ^- = h^{ab}K^-_{ab}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $\Sigma$ is an oriented $n-1$ submanifold of a space-like n-submanifold $M$ of $V$. Because of the orientability assumption one can make a continuous choice of unit normal $\nu$ to $\Sigma$ (and tangent to $M$) distinguishing the interior and exterior of $\Sigma$ in $M$. Also define $n$ as the future unit normal to $M$. Choose $\nu$ to be pointing outwards with respect to $\Sigma$ and $n$ to be pointing outwards with respect to $M$. Then one can choose $l^+$ and $l^-$ such that $l^+ = n + \nu$ and $l^- = n - \nu$ ensuring $l^+_\mu l^{-\mu} = -2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric ${h_{ab}}$ defined on $\Sigma$ can be lifted to a metric $h$ (calling it the same by abuse of notation) on $V$ by extending it trivially along the two orthogonal directions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then apparently the following equation is true which I do not understand, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$h^{\mu \nu} = g^{\mu \nu} + \frac{1}{2}(l^{+^\mu} l^{-\nu} + l^{-^\mu} l^{+^\nu})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\mu$ and $\nu$ run over coordinates of $V$ unlike $a$ and $b$ which run over coordinates of $\Sigma$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one see that the lift to $V$ ($h_{\mu \nu}$) of the pulled back metric $h_{ab}$ and the original metric on $V$ ($g_{\mu \nu}$) are related thus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by parallel translation extend $l^+$ and $l^-$ to be tangent field to geodesic congruences. Then the following equation will be true for any $\nu$, &lt;code&gt;$l^{+-\mu}\nabla _{\mu} l^{+-}_\nu = 0$&lt;/code&gt;. Further since these vector fields are also of constant norm one has for any $\nu$, $l^{+-\mu}\nabla &lt;em&gt;{\nu} l^{+-}&lt;/em&gt;\mu = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above one gets the following identity for the expansion $\theta = \nabla _\mu l^{+\mu}$ of the geodesic congruence to which $l^+$ is tangent,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla_\mu l^{+\mu} = g^{\mu \nu}\nabla_\mu l^{+}&lt;em&gt;\nu = h^{\mu \nu} \nabla&lt;/em&gt;\mu l^{+}_\nu$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Now I don't understand why $h^{ab}\nabla_a l^+&lt;em&gt;b = h^{\mu \nu} \nabla&lt;/em&gt;\mu l^+_\nu$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-10-19T20:29:35.247" LastActivityDate="2010-10-19T20:29:35.247" Title="A particular pulling back and lifting of metric " Tags="&lt;differential-geometry&gt;" />
  <row Id="7220" PostTypeId="2" ParentId="7213" CreationDate="2010-10-19T20:40:57.847" Score="1" ViewCount="0" Body="&lt;p&gt;The Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_remainder&quot; rel=&quot;nofollow&quot;&gt;Lagrange remainder&lt;/a&gt; has a good discussion.  The Taylor series is a polynomial that approximates a function near a point, calculated by taking derivatives at the point.  The error term is just that-the difference between the function and the polynomial.  There are various forms, given there, for the error of the Taylor series approximation.  All of them are right, but generally not very useful except as bounds, like you have in this problem.  The difficulty is that you don't know the value of $\xi$, just that it is between $a$ and $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one of the points of the problem is to see that when $(x-a)$ gets large you need a lot of terms for accuracy.  If you calculate the number of terms at $21\pi/5$ and then again at $\pi/5$ it will be evident.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-19T20:40:57.847" />
  <row Id="7221" PostTypeId="1" CreationDate="2010-10-19T20:41:50.637" Score="0" ViewCount="43" Body="&lt;p&gt;If a space has curvature, then the curvature can be seen intrinsically by finding sums of angles in triangles made of geodesics. Under general relativity, space-time is curved on local scales. On global scales, experiments have determined that space-time is almost totally flat.&#xA;If one were to, for each of three points in a gravitational gradient (like near a black hole), find the angle between the sight lines to the other two points and then add up all of the angles, would it add up to 180 (degrees) or perhaps it would show intrinsic curvature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the way of thinking about this good, because it is quite general and conceptual. I don't actually know the equations for general relativity, but this is what I thought of when I was thinking about intrinsic geometric properties.&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2010-10-19T20:41:50.637" Title="Effect of curvature of spacetime on intrinsic geometric properties (under general relativity)" Tags="&lt;differential-topology&gt;&lt;curvature&gt;&lt;geodesy&gt;" CommentCount="2" />
  <row Id="7222" PostTypeId="2" ParentId="5962" CreationDate="2010-10-19T21:02:07.453" Score="0" ViewCount="0" Body="&lt;p&gt;Let me note that &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector-valued_form&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vector-valued_form&lt;/a&gt; seems to be highly relevant. They essentially define the bracket. Note that in the case of a 1-form A valued in $\mathfrak{g}$ you get $[A,A]=[A\wedge A (v \otimes w)]= A(v)\otimes A(w)- A(w)\otimes A(v)$, which has value in  $\mathfrak{g} \otimes \mathfrak{g}$. To combine them you use the bracket in $\mathfrak{g}$, so you get $2 [A(x), A(y)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your definition for $[A, B]$ has a factor of $\frac{1}{(k+l)!}$, which in the case of your vector bundle being over a point does not reproduce the bracket on $\mathfrak{g}$. Maybe it should be $\frac{1}{(k!l!)}$, as in Wikipedia?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Mmm, this was nonsense. Over a point you'd be doing k=l=0, so it would not matter. Anyhow, Wikipedia definition seems to give constants more consistent with the rest. But I can't do arithmetic, so no guarantees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, as suggested by Aaron, you can view A as a matrix of 1-forms, and wedge-multiply the matrices. Then $A\wedge A (x, y) = A(x)A(y)-A(y)A(x)= [A(x), A(y)]$ and indeed, $2 A\wedge A= [A,A]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a bit confused by this wedge notation you give in the question. $A_i$ and $B_j$ differential forms - components of A and B. We then tensor things up, using wedge instead of product, getting a vector in forms valued in the tensor product. We would then need a map $V \otimes V=\mathfrak{g} \otimes \mathfrak{g}$ to $V=\mathfrak{g}$ to get a form with values in $V$ again. Presumably this is the Lie bracket on V, and the result is the 'wedge multiplication of matrices' as above.  The 'issue with skew symmetrization' is that this wedge operation is not the bracket operation on forms, but it's less skewsymmetrized version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cartan equation should work out fine. We have 2-form $\Omega$ and 1-form $\omega$. Their bracket on the triple $a,b,c$ is a sum of 6 terms, with a factor of 1/2 in front. These 6 terms pair up into 3 pairs of 2 equal terms each(as $\Omega$ is antisymmetric). So you get 3 terms with coefficient 1, which we then bracket. On the other hand, wedge multiplying the matrices for $\Omega$ and $\omega$ in different order results in same 3 bracket terms (wedging 2 form and 1 form produces 3 distinct terms). One just needs to write down the formulas. (I'm sure there is a general formula for bracket in terms of antisymmetrization of the wedge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe another point is that $[A,A]=0$ for 'ordinary' $\mathbb{R}$ valued forms is a consequence of the Lie bracket on $\mathbb{R}$ being trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, trace is evidently defined as before - viewing A as a matrix of forms, trace is the summ of the diagonal forms. It is then clear why it commutes with the differential. However, $Tr(A\wedge B) = (-1)^{deg A \times deg B} Tr(B\wedge A)$. This follows from the usual formulas for trace and commuting the forms $A_{i,j}$ past $B_{j,k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the trace of $A\wedge A\wedge A\wedge A$. It is $\Sigma_i A^4_{ii}= \Sigma_i \Sigma_j A^2_{ji} \wedge A^2_{ij}$. $A^2_{kl}=\Sigma_m A_{km} \wedge A_{ml}$ is in general non-zero (just take A  a 3 by 3 matrix with 0 in $A_{12}$ entry and  $A_{2,3}\wedge A_{3,1}  \wedge A_{1,3} \wedge A_{3,2}$ non 0). BUT, you are on a 3-manifold (or you'd be doing a differnt C-S form), so there is no 4-form at all!&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-10-19T21:12:29.033" LastActivityDate="2010-10-19T21:12:29.033" CommentCount="7" />
  <row Id="7223" PostTypeId="1" AcceptedAnswerId="7225" CreationDate="2010-10-19T21:12:18.287" Score="2" ViewCount="73" Body="&lt;p&gt;More specifically, assume d taken from [1, 2^n] and m is taken from [1, n]. What is an upper bound on the probability that d is a multiple of m?&lt;/p&gt;&#xA;" OwnerUserId="2634" LastActivityDate="2010-10-19T22:59:31.537" Title="For an arbitrary positive integer d and random modulus m, what is the probability that d mod m = 0?" Tags="&lt;number-theory&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="7224" PostTypeId="1" CreationDate="2010-10-19T21:15:39.150" Score="1" ViewCount="159" Body="&lt;p&gt;The Complex plane, set of all z=x+iy where x and y are real, surface area equals cross product of x and y equals aleph-something (that's not the question). Projecting the plane onto a sphere, and adding the complex infinity to the set (without a value for arg(z)), gives the riemann sphere, which, being a sphere, is compact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the addition of infinity, as some sort of bounder, make the complex plane compact? is the riemann sphere without infinity compact, since it has an open boundary? also, if the hyperbolic plane can be shown in a disk, what does that mean? The answer to this question might primarily clarify the definitions in my new studies of topology of manifolds, etc. &#xA;thanks.&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2010-10-20T13:36:24.847" Title="Is the riemann sphere compact even though the complex plane isn't?" Tags="&lt;topology&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7225" PostTypeId="2" ParentId="7223" CreationDate="2010-10-19T21:21:28.030" Score="6" ViewCount="0" Body="&lt;p&gt;Fixing $m$, the probability that $d$ is a multiple of $m$ is approximately $1/m$. Now, if you choose $m$ randomly from $[1,n]$, and average $1/m$, you get $\frac{1}{n} \sum_{m=1}^n \frac{1}{m}$, which is approximately $\frac{\log n}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I didn't read the question carefully enough. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least common multiple of the first $n$ numbers grows like $e^n$. So if you take the lcm of the numbers $1 \ldots n \cdot ln 2$, you get a number somewhere around $2^n$ which is going to have as factors all the numbers between $1$ and $n \cdot \ln 2 $. This shows the probability you're interested in is at least $\ln 2$. In fact, this lcm is going to contain as a factor all numbers except prime powers larger than $n \cdot \ln 2$.  Since any prime has at most one power in this interval, then by the prime number theorem, you then get a lower bound for the probability of around $1-1/\ln n$. I expect the upper bound matches, but offhand I don't see how to prove it.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditorDisplayName="" LastEditDate="2010-10-19T22:59:31.537" LastActivityDate="2010-10-19T22:59:31.537" CommentCount="1" />
  <row Id="7226" PostTypeId="2" ParentId="7224" CreationDate="2010-10-19T21:28:22.693" Score="3" ViewCount="0" Body="&lt;p&gt;Riemann sphere without infinity is just the complex plane again, which is not compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that $\mathbb C$ is not compact is because a sequence like $\{1,2,3,\cdots\}$ which tends to infinity does not have a convergent subsequence, whereas any sequence on the Riemann sphere does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of infinity of called the one point compactification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(N.B. what you said about surface area equals cross product of x and y equals aleph-something is probably wrong)&lt;/p&gt;&#xA;" LastActivityDate="2010-10-19T21:28:22.693" />
  <row Id="7227" PostTypeId="2" ParentId="7175" CreationDate="2010-10-19T21:33:20.090" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;This should be a comment somehwere, but got to long.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will, the cone on $P^2_\mathbb R$ is &lt;em&gt;not&lt;/em&gt; a manifold. Consider the long exact sequence for integral reduced homology of the pair $(C,C')$ where $C=C(P^2_\mathbb R)$ is the cone over $P^2_\mathbb R$ and $C'=C\setminus\{a\}$ is the complement of the apex of the cone. Since $C$ is contractible and $C'$ deformation-retracts onto $P^2_\mathbb R$, you get isomorphisms $H^\sharp_2(C,C')\cong H^\sharp_{1}(P^2_\mathbb R)\cong\mathbb Z/2\mathbb Z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If follows that $C$ is not a manifold: in a manifold $M$, for every point $p\in M$ we have that the integral reduced homology $H^\#_\bullet(M,M\setminus\{p\})$ is that of a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Generalizing this reasoning, you get a rather strong condition on a manifold for its cone to be also a manifold. I'm sure the topologists among our fellow M.SEers know of a precise characterization.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, we see that $C$ is not a manifold with boundary, because in such a space $H^\#_\bullet(M,M\setminus\{p\})$ is, for every $p$, either identically zero or that of a sphere.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-19T21:56:13.423" LastActivityDate="2010-10-19T21:56:13.423" CommentCount="3" />
  <row Id="7228" PostTypeId="1" AcceptedAnswerId="7233" CreationDate="2010-10-19T21:35:15.473" Score="1" ViewCount="112" Body="&lt;p&gt;True or False? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle \sum_{n=1}^{\infty}a_{n}^2$ is convergent, and$\displaystyle \sum_{n=1}^{\infty} (\frac{\pi}{2} - b_{n})$ is divergent, then $\displaystyle {a_{n}(\sin(b_{n})-1)}$ is a convergent sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we know that $(a_{n})^2$ is convergent.&#xA;Therefore, $\lim_{n \to \infty} a_{n}^2 = 0$&#xA;I'm not sure what this tells us about or how to relate this to $(a_{n})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we can see that since $-1 \ge \sin(b_{n}) \le 1$&#xA;And therefore, $-2 \ge \sin(b_{n}) - 1 \le 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, again since I don't know what I can conclude about $(a_{n})$, I'm not sure how I can use this fact.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2010-10-19T21:59:47.303" LastActivityDate="2010-10-19T22:18:07.867" Title="Given a convergent $\{(a_{n})^2\}$, divergent $\{b_{n}\}$, prove $\{a_{n}(\sin(b_{n})-1)\}$ converges." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7229" PostTypeId="2" ParentId="7210" CreationDate="2010-10-19T21:35:17.313" Score="6" ViewCount="0" Body="&lt;p&gt;A metric space is totally bounded if and only if every sequence has a Cauchy subsequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Try and prove this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might suspect, this is basically equivalent to what Jonas has said.  The key between these two is provided by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metric space is compact if and only if it is totally bounded and complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, every sequence has a convergent subsequence (compact) if and only if every sequence has a Cauchy sequence (Totally bounded) and every Cauchy sequence converges (complete).&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-11-04T12:41:50.667" LastActivityDate="2010-11-04T12:41:50.667" CommentCount="2" />
  <row Id="7230" PostTypeId="2" ParentId="7224" CreationDate="2010-10-19T21:39:54.003" Score="8" ViewCount="0" Body="&lt;p&gt;For the Riemann sphere, it is the &lt;a href=&quot;http://en.wikipedia.org/wiki/One-point_compactification&quot; rel=&quot;nofollow&quot;&gt;one point compactification&lt;/a&gt; of the plane. So yes, rather by definition it is compact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, a non-compact topological space &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactification&quot; rel=&quot;nofollow&quot;&gt;can be compactified by suitable addition of &quot;points at infinity&quot;&lt;/a&gt;. Particularly useful ones (besides the one-point compactification above) include the Stone-Cech compactification, which in some sense adds the &quot;most points&quot;, and whose existence for any topological space follows from axiom of choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another compactification that often arises is the Penrose/conformal compactification of space-time in general relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A propos your question on hyperbolic plane. Topologically the hyperbolic plane is the same as the normal plane. By considering the Poincare disk model and &quot;adding&quot; the boundary of the disk, you compactify the hyperbolic plane. As mentioned above, any topological space admits some compactification. What's more useful is to consider special compactifications: for example the one-point compactification of the complex plane into the Riemann sphere is conformal relative to the additional &lt;em&gt;geometrical structure&lt;/em&gt; on the two surfaces. Similarly the Penrose compactifications respect the geometrical structure on space-times. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-19T21:39:54.003" />
  <row Id="7231" PostTypeId="1" AcceptedAnswerId="7236" CreationDate="2010-10-19T22:14:10.950" Score="0" ViewCount="121" Body="&lt;p&gt;Based on a &lt;a href=&quot;http://math.stackexchange.com/questions/7126/boolean-algebra-simplification-dont-know-the-method-used&quot;&gt;previous question&lt;/a&gt; (and this ought to be the last of this sequence I promise), I just need to know one more thing.  According to this truth table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9T1x4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told the answer I should be getting by using the &lt;strong&gt;sum of products&lt;/strong&gt; method on the truth table should be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = R ∙ (RCGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Unlike before with the Karnaugh map, I can't see how this was arrived at.  All I can get is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = RC'G'M'+RC'G'M+RC'GM'+RC'GM+RCG'M'+RCG'M+RCGM'&#xA;T = R(C'G'M'+C'G'M+C'GM'+C'GM+CG'M'+CG'M+CGM')&#xA;T = R(C'(G'M'+G'M+GM'+GM)+C(G'M'+G'M+GM'))&#xA;T = R(C'(G'(M'+M)+G(M'+M))+C(G'(M'+M)+GM'))&#xA;T = R(C'(G'+G)+C(G'+GM'))&#xA;T = R(C'+C(G'+M'))&#xA;T = R(C'+G'+M')&#xA;T = R(CGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the right answer I believe, but it's not the same as what I should supposedly have found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again in advance!&lt;/p&gt;&#xA;" OwnerUserId="2609" LastEditorUserId="2609" LastEditorDisplayName="" LastEditDate="2010-10-19T22:34:56.920" LastActivityDate="2010-10-20T19:54:48.807" Title="Boolean Algebra, Simplification: Sum of products from truth table" Tags="&lt;boolean-algebra&gt;" AnswerCount="2" />
  <row Id="7232" PostTypeId="1" CreationDate="2010-10-19T22:14:37.643" Score="0" ViewCount="219" Body="&lt;p&gt;I have a text book which gives two diagrams of beads suspended from strings with different angles, and worked examples to calculate the tension in the strings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first bead is suspended by 2 strings attached to a rod, both strings at angles to the vertical (the bead is at the bottom, the rod is across the top). The text book tells me to label the tensions T1 and T2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the next diagram, one of the threads is now vertical, while the other is at an angle.  This time the text book tells me the tensions are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are they different in one layout, but equal in another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tim&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Agh, sorry - missed some vital info!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diagram, there is a horizontal force applied to the bead to keep it in equilibrium!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I still don't understand why the tensions in the strings would be equal though!)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And some more vital info. I think I'm almost there now.  In the second example, the bead is suspended not by two pieces of string, but by 1! The bead is somewhere in the middle of the string.  The diagrams are so similar, I thought there were two strings in each diagram! (That'll teach me to read the question!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the tensions in the two parts of the string have to be equal because otherwise the string wouldn't be in equilibrium. Or something.  Can anyone confirm that for me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the interested, the text book is the Edexcel AS and A Level Modular Mathematics. The page is 98.)&lt;/p&gt;&#xA;" OwnerUserId="2635" LastEditorUserId="2635" LastEditorDisplayName="" LastEditDate="2010-10-22T06:31:09.973" LastActivityDate="2010-12-21T06:44:26.160" Title="If a bead is suspended by 2 strings, one vertical, the other diagonally, why are the tensions equal?" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7233" PostTypeId="2" ParentId="7228" CreationDate="2010-10-19T22:18:07.867" Score="3" ViewCount="0" Body="&lt;p&gt;In the current form of the question, the condition on $b_n$ is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, $a_n^2\to0$.  This implies that $a_n\to0$.  Then because $|a_n(\sin(b_n)-1)|=|a_n||\sin(b_n)-1|$ and, for the reason you indicated, $|\sin(b_n)-1|\leq2$ for all $n$, the $n^{th}$ term in the sequence has absolute value less than or equal to $2|a_n|$, which converges to 0.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-19T22:18:07.867" CommentCount="5" />
  <row Id="7234" PostTypeId="2" ParentId="6414" CreationDate="2010-10-19T22:35:34.120" Score="0" ViewCount="0" Body="&lt;p&gt;This is actually easier if you use charts. Suppose your smooth structure is given by a maximal atlas C.&#xA;Near any point $p$ in $TU$ you have a neighborhood $V=\pi^{-1} W$ for $W$ - a chart neighborhood of $\pi(p)$, so that on $V$ which there are functions $x_i \cdot\pi$ which are smooth by property (1) and functions $\frac{\partial}{\partial x_i}$, which are smooth by property (2).  But they also define a chart - a map $V$ to $R^{2n}$. So this chart has to be smooth. So we have a cover by charts smooth in both the standard atlas and the maximal atlas C. Hence the two smooth structures coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to do it in sheave seems a bit strange. If your structure sheaf is F, you get $2n$ functions in $F(V)$ and the fact that they define a chart should mean that you can pull back (restriction of) any function f in F to a function on part of $R^{2n}$, so that restriction is in the standard structure sheaf, and hence, (by sheaf axiom) f is too. This seems slightly fishy, even though can probably be fixed, in which case it will be the sheaf translation of the above argument.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-10-19T22:35:34.120" CommentCount="3" />
  <row Id="7235" PostTypeId="1" AcceptedAnswerId="7239" CreationDate="2010-10-19T22:37:05.047" Score="1" ViewCount="51" Body="&lt;p&gt;Suppose we look at $k$-digit strings with digits between $1$ and $n$. How many distinct classes of strings are there when digit identity and order doesn't matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, what is the number of equivalence classes over strings $d_1,\ldots,d_k$ with $d_i\in \{1,\ldots, n\}$ where strings $d,\hat{d}$ are considered equivalent iff there exist bijections $\sigma$ over $\{1,\ldots, k\} $ and $\delta$ over $\{1,\ldots, n\}$ such that the following holds for all $j \in \{1,\ldots,k\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{\sigma(j)}=\delta(\hat{d}_j)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: this seems to correspond to the number of ways to partition integer $k$ into at most $n$ parts&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-10-20T02:31:26.853" LastActivityDate="2010-10-20T02:31:26.853" Title="Number of classes of k-digit strings when digit order and identity doesn't matter" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="7236" PostTypeId="2" ParentId="7231" CreationDate="2010-10-19T23:17:24.630" Score="0" ViewCount="0" Body="&lt;p&gt;The two results are the same.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R(RCGM)'=RR'+RC'+RG'+RM'=RC'+RG'+RM'=R(CGM)'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On the other hand, this means the second &lt;code&gt;R&lt;/code&gt; is kind of redundant, so I don't know why you'd want to write the solution this way rather than your way.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-19T23:17:24.630" />
  <row Id="7237" PostTypeId="2" ParentId="7143" CreationDate="2010-10-19T23:18:34.767" Score="5" ViewCount="0" Body="&lt;p&gt;Although you already have a couple of answers proving the result you ask for, I think it is worth adding a further answer using a completely different approach. There is a simple way of parameterizing all points on an algebraic curve of degree two which, once you know, the answer to your question just drops out really easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: Actually, I didn't realize how close this was to Moron's answer when I posted. Still, my parameterization is a bit more general, so I'll leave it up.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the parameterization: For any Pythagorean triple &lt;em&gt;a&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;&lt;em&gt;c&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; you can divide through by &lt;em&gt;c&lt;/em&gt; to get the equation &lt;em&gt;a&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;1 in rational pairs (&lt;em&gt;a&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;)&amp;nbsp;&amp;isin;&amp;nbsp;Q&lt;sup&gt;2&lt;/sup&gt;. The set of all such rational (&lt;em&gt;a&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;) is nothing other than the rational points in the unit circle S&lt;sup&gt;1&lt;/sup&gt;. Conversely, if you have rational points (&lt;em&gt;a&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;)&amp;nbsp;&amp;isin;&amp;nbsp;S&lt;sup&gt;1&lt;/sup&gt; then you obtain a Pythagorean triple by multiplying through by a common denominator &lt;em&gt;c&lt;/em&gt; (which does not affect the angle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick any rational point on the unit circle (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;) and a nonzero vector (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;), and consider the line t&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tu&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tv&lt;/em&gt;). This must intersect the circle at exactly one other point (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;) -- unless (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;) is tangent to.the circle (in which case we can say that it intersects it twice at (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;) ). In fact, the equation &lt;em&gt;a&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;-&amp;nbsp;1&amp;nbsp;=&amp;nbsp;0 is a quadratic in t with zero constant term (as it is solved by &lt;em&gt;t&lt;/em&gt;&amp;nbsp;=&amp;nbsp;0), so extracting out the factor of &lt;em&gt;t&lt;/em&gt; gives a linear expression. This means that it intersects the curve at a rational point. To be explicit, the equation you get for &lt;em&gt;t&lt;/em&gt; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(u^2+v^2)t^2+2(a_0u+b_0v)t=0\ \Rightarrow\ t=-2(a_0u+b_0v)/(u^2+v^2).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, given any rational point (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;) on the circle distinct from (&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;) you can always take (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;)&amp;nbsp;=&amp;nbsp;(&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;-&amp;nbsp;&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;&amp;nbsp;-&amp;nbsp;&lt;em&gt;b&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;) to see that every such point is reached in this way.&#xA;This parameterizes the rational points on the circle by the nonzero rational vectors (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;)&amp;nbsp;&amp;isin;&amp;nbsp;Q&lt;sup&gt;2&lt;/sup&gt; (up to recaling). In fact, dividing through by &lt;em&gt;u&lt;/em&gt; (if it is nonzero), this parameterizes the rational points on the circle by &lt;em&gt;v&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;Q and the additional point parameterized by (0,1) curresponding to &lt;em&gt;u&lt;/em&gt;&amp;nbsp;=&amp;nbsp;0. This can be seen to be nothing other than &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick any point (&lt;em&gt;x&lt;/em&gt;,&lt;i&gt;y&lt;/i&gt;) on the unit circle and approximate the vector (&lt;em&gt;x&lt;/em&gt;&amp;nbsp;-&amp;nbsp;a&lt;sub&gt;0&lt;/sub&gt;,&lt;em&gt;y&lt;/em&gt;&amp;nbsp;-&amp;nbsp;b&lt;sub&gt;0&lt;/sub&gt;) as closely as you like by rational (&lt;em&gt;u&lt;/em&gt;,&lt;i&gt;v&lt;/i&gt;). The curve t&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;a&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tu&lt;/em&gt;,&lt;i&gt;b&lt;/i&gt;&lt;sub&gt;0&lt;/sub&gt;+&lt;em&gt;tv&lt;/em&gt;) intersects the circle at a rational point, which can be made as close as you like to (&lt;em&gt;x&lt;/em&gt;,&lt;i&gt;y&lt;/i&gt;). That is, &lt;i&gt;the rational points are dense in the unit circle&lt;/i&gt;. So, the angles of pythagorean triples will also be dense in the first quadrant.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-10-20T00:57:03.840" LastActivityDate="2010-10-20T00:57:03.840" />
  <row Id="7238" PostTypeId="1" CreationDate="2010-10-20T00:18:02.650" Score="1" ViewCount="223" Body="&lt;p&gt;So for a writing assignment in one of my classes we are asked to discuss and prove some basic results about compact sets in general topological spaces. I like proving these things, but they dont help me understand what a compact( locally compact, paracompact,...) set in a topological space &quot;looks like.&quot; That said I'm asking for some examples of compact (locally compact, ...) sets in a variety of topological spaces. I'm also interested in some explanation of what extra &quot;benefits&quot; are brought about by singling out these compact sets. For example the p-adic numbers are locally compact and locally compact things (abelian groups to be precise) are a good setting in which to carry out fourier analyis.&lt;/p&gt;&#xA;" OwnerUserId="2636" LastActivityDate="2010-10-20T09:23:13.720" Title="What does a compact set look like?" Tags="&lt;real-analysis&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7239" PostTypeId="2" ParentId="7235" CreationDate="2010-10-20T00:19:00.453" Score="2" ViewCount="0" Body="&lt;p&gt;If I've read this right, you're looking to count k-tuples of integers in {1,..,n} that are inequivalent under the operations (a) permute the coordinates and (b) permute the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, when k=7 and n=2, 1122212 would be equivalent to 2211121, 1112222 and 1111222 (and many others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it would be useful to construct a &lt;em&gt;canonical form&lt;/em&gt; for each class:   In this case, we can take the lexicographic first element in the equivalence class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these would be in canonical form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1111222&#xA;1111122&#xA;1111112&#xA;1111111&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;whereas 1112222 would not be since it belongs to the same class as 1111222.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These canonical forms are equivalent to ordered partitions of k into n parts (some of which can be zero).  For example 1111222 &amp;lt;-&gt; 4+3 since there are four 1's and three 2's.  [If instead n=3, then 1111222 &amp;lt;-&gt; 4+3+0 since there are no 3's in the string.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of these is the number $p_n(k)$ of partitions of k into at most n parts (afterwards, we can append the zeroes to achieve n parts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute $p_n(k)$ using the recurrence relation $p_n(k)=p_n(k-n)+p_{n-1}(k-1)$ (with some appropriate boundary conditions).  [see: Enumerative combinatorics, Volume 1 By Richard P. Stanley p. 28]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-20T00:19:00.453" CommentCount="1" />
  <row Id="7240" PostTypeId="2" ParentId="7238" CreationDate="2010-10-20T01:18:42.860" Score="2" ViewCount="0" Body="&lt;p&gt;I don't know if I understand your question, but here are some elementary examples of compact and non compact spaces. You can find more examples in &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactness&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$[0,1]$ is compact with the subspace topology of the usual, Euclidean topology on $\mathbb{R}$.&lt;/li&gt;&#xA;&lt;li&gt;Neither $(0,1)$, nor $\mathbb{R}$ are compact with the same topology.&lt;/li&gt;&#xA;&lt;li&gt;A set with a finite number of points $A \subset \mathbb{R}$ is always compact. In fact, it is always compact, with no matter which topology.&lt;/li&gt;&#xA;&lt;li&gt;With the usual topology on $\mathbb{R}^n$, $A \subset \mathbb{R}^n$ is compact if and only if it is bounded and closed. Beware: this characterization of compactness is no longer true for arbitrary metric spaces: for instance, with the bounded metric $d(x,y) = \mathrm{min} \left\{ \| x-y\|,1 \right\}$ (which is equivalent to the Euclidean one), every subset $A \subset \mathbb{R}^n$, even the whole $\mathbb{R}^n$, is bounded. So, if the former characterization was still true with this metric, every closed subset of $\mathbb{R}^n$ ($\mathbb{R}^n$ itself included) would be compact.&lt;/li&gt;&#xA;&lt;li&gt;This countable set $\left\{\frac{1}{n}\ \vert \ n \in \mathbb{N} \right\} \cup \left\{ 0\right\} \subset \mathbb{R}$ is compact (with the usual topology).&lt;/li&gt;&#xA;&lt;li&gt;This countable set $\left\{\frac{1}{n}\ \vert \ n \in \mathbb{N} \right\} \subset \mathbb{R}$ is not compact (with the usual topology).&lt;/li&gt;&#xA;&lt;li&gt;This countable set $\mathbb{N} \subset \mathbb{R}$ is not compact (with the usual topology).&lt;/li&gt;&#xA;&lt;li&gt;With the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_complement_topology&quot; rel=&quot;nofollow&quot;&gt;finite complement topology&lt;/a&gt;, $\mathbb{R}$ is compact. In fact, with this topology on the real line, every subset $A \subset \mathbb{R}$ is compact. In fact, every topological space with the finite complement topology is compact.&lt;/li&gt;&#xA;&lt;li&gt;$\mathbb{Q}$ is not compact with the subspace topology of the usual, Euclidean topology on $\mathbb{R}$. &lt;/li&gt;&#xA;&lt;li&gt;Btw, the empty set is always compact.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-20T01:18:42.860" />
  <row Id="7241" PostTypeId="2" ParentId="7143" CreationDate="2010-10-20T01:30:53.873" Score="4" ViewCount="0" Body="&lt;p&gt;The other answers here explain that it is possible to find arbitrarily good approximations to any angle using Pythagorean triples.  Here's a footnote on how to turn this into a constructive procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moron above says to pick a rational number $n/m$ near $t$ where $t$ satisfies $\tan \theta = 2t/(1 - t^2)$.  You could pick $n$ and $m$ using a &lt;a href=&quot;http://www.johndcook.com/blog/2010/10/20/best-rational-approximation/&quot; rel=&quot;nofollow&quot;&gt;Farey algorithm&lt;/a&gt; to find the best rational approximation to $t = (1 - \cos \theta)/\sin\theta$ with a given denominator.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $a = 0$, $b = c = d = 1$.  The number $t$ is between $a/b$ and $c/d$.  Next compute the &quot;mediant&quot; $(a+c)/(b+d)$.  If $t$ is bigger than the mediant, update $a$ and $b$ with $a+c$ and $b+d$.  Otherwise update $c$ with $a+c$ and $d$ with $b+d$. Each iteration of this process produces the best rational approximation to $t$ for a given denominator size.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-10-21T00:10:27.133" LastActivityDate="2010-10-21T00:10:27.133" />
  <row Id="7242" PostTypeId="1" AcceptedAnswerId="7244" CreationDate="2010-10-20T01:31:06.883" Score="0" ViewCount="250" Body="&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the limit of $e^t$ as $t$ approaches infinity is simply infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the limit of $e^{t^2}$ cannot be found because it explodes?  Is this due to the fact that there are differing types of infinity and the second limit produces a non-&quot;countable&quot; infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1/different-kinds-of-infinities&quot;&gt;see referenced question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-20T02:30:38.910" LastActivityDate="2010-10-21T16:22:34.890" Title="Why do some limits &quot;explode&quot; to infinity?" Tags="&lt;limit&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="7243" PostTypeId="2" ParentId="7224" CreationDate="2010-10-20T01:47:41.023" Score="3" ViewCount="0" Body="&lt;p&gt;The complex plane $\mathbb{C}$ is not compact. For instance, here is an open cover that does not admit any finite subcover:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\mathcal U} = \left\{ U_n\right\} \qquad \text{with} \qquad U_n = \left\{ z \in \mathbb{C} \ \vert \ \vert z \vert &amp;lt; n \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why ${\mathcal U}$ does not admit any finite subcover is easy: if you pick just a finite number of $U_n$'s, then their union would be equal to the biggest one of them, which is not the whole $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, thanks to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt;, think of these $U_n$'s inside the Riemann sphere $S^2$. They do not form a cover of $S^2$, since $\infty$ (= the North Pole) is not included in any of them. So, we need to add an open neighborhood of the infinity. For instance, the image of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = \left\{ z \in \mathbb{C} \ \vert \ \vert z \vert &gt; 394 \right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the stereographic projection. Then, we would have an open cover of $S^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;{\mathcal U}' = \left\{ U_n \ \vert \ n \in \mathbb{N}\right\} \cup \left\{ V\right\} \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which admits and open subcover. For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U_{395} \ , \quad V \ .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-20T13:36:24.847" LastActivityDate="2010-10-20T13:36:24.847" />
  <row Id="7244" PostTypeId="2" ParentId="7242" CreationDate="2010-10-20T01:53:28.493" Score="8" ViewCount="0" Body="&lt;p&gt;The limit can be found. What are we saying when we assert that $lim_{t\to\infty} f(t) = \infty$, with $f:\mathbb{R} \to \mathbb{R}$? What does it mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means, &lt;em&gt;by definition&lt;/em&gt;, that given $M &gt;0$ there exists $\varepsilon &gt;0$ such that if $t&gt;\varepsilon$ then $f(t) &gt; M$. Simply this. In (not so rigorous) words, for arbitrary large values of the parameter $t$ the function assumes arbitrary large values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that the limit you want is indeed $\infty$ (in the sense just described). This can be seen noticing that $e^t &amp;lt; e^{t^2}$ for $t&gt;1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this question is not related to the linked one. For me, they are different things. &lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="1178" LastEditorDisplayName="" LastEditDate="2010-10-21T16:22:34.890" LastActivityDate="2010-10-21T16:22:34.890" />
  <row Id="7245" PostTypeId="2" ParentId="7242" CreationDate="2010-10-20T02:13:56.213" Score="1" ViewCount="0" Body="&lt;p&gt;Given your mentioning of the word &quot;countable&quot; (and looking at the link), I think that the &quot;differing types of infinity&quot; that you are thinking of are different infinite cardinalities (i.e. different possible infinite sizes of sets).  This notion of infinity is not the same as the notion of infinity that is used when one says that a limit approaches infinity.  (Of course, they are related at a basic level, in that both allude to the idea of being larger than any finite number, but at a more technical level, they are different notions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ronaldo in his answer has given an explanation of what it means for &#xA;$\displaystyle \lim_{t\to\infty} f(t)$ to equal $\infty$, and this is just as valid for $e^{t^2}$ as it is for $e^t$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dan Ramras notes in one of his comments, the &lt;em&gt;rate&lt;/em&gt; at which $e^{t^2}$ approaches infinity&#xA;is much greater than the rate at which $e^t$ approaches infinty (which is turn much&#xA;greater than the rate at which $t$ approaches infinity, for example).  In terms of Ronaldo's answer, this means that for a given $M$, the $\epsilon$ that you have to choose&#xA;to be sure that $e^{t^2} &gt; M$ when $t &gt; \epsilon$ is smaller than in the corresponding situation for $e^t$.  But both quantities approach infinity as $t$ does.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-20T14:57:44.253" LastActivityDate="2010-10-20T14:57:44.253" CommentCount="2" />
  <row Id="7246" PostTypeId="2" ParentId="7242" CreationDate="2010-10-20T02:37:07.833" Score="5" ViewCount="0" Body="&lt;p&gt;To take a different perspective on it (if this is what you're looking for), there &lt;em&gt;are&lt;/em&gt; limits that can't be found; the limit of $1/x$ as $x\rightarrow 0$ is one example.  The limit doesn't exist because the limit coming from the right and from the left are different: one is $\infty$ and one is $-\infty$.  I guess you could say that $1/x$ &quot;explodes&quot; at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to distinguish the &lt;em&gt;type&lt;/em&gt; of infinity you're talking about, too.  The &quot;countable and uncountable&quot; infinities you mention are &lt;em&gt;cardinalities&lt;/em&gt;: they measure the number of elements in sets.  The infinities you get with limits of functions like these are just limits of real numbers, and don't describe sets at all.  You could call them &quot;uncountable&quot; since $\mathbb{R}$ is uncountable, I suppose, but then there wouldn't be a &quot;countable&quot; infinity in between $\infty$ and the real numbers.  Basically, &quot;infinity&quot; can mean a lot of different things.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-20T02:37:07.833" />
  <row Id="7247" PostTypeId="1" AcceptedAnswerId="7279" CreationDate="2010-10-20T02:38:38.527" Score="4" ViewCount="62" Body="&lt;p&gt;Let $f(x) \in L^p(\mathbb{R})$ and $K \in C^m(\mathbb{R})$. Can I then say that $(f \ast K) (x) = \int_{\mathbb{R}} f(t) K(x-t) dt$ is in $C^m$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is true if $K$ has compact support, but I was wondering if it is possible to have a stronger result (perhaps $K$ vanishing at $\infty$?). &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-10-20T10:01:11.697" Title="Differentiability of Convolutions" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="7248" PostTypeId="1" CreationDate="2010-10-20T02:43:11.423" Score="1" ViewCount="65" Body="&lt;p&gt;I'm attempting to come up with a first order language $L$ that is able to describe vector spaces over fields. I came up with a few sets of nonlogical symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Rs_L=\{Scal,Vec\}$ where $Scal,Vec$ are unary relation symbols which I interpret to be true or false depending on whether $x$ is a scalar or vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $Fs_L=\{+,\cdot,\}$ which are the usual addition and multiplication for both vectors and scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constant symbols, $Cs_L=\{0,1,0_V\}$, where $0$ and $1$ are regular field scalars for the additive and multiplicative identities, respectively, and $0_V$ is the additive identity for vectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these, I was able to formulate axioms for scalar multiplication on a given structure, and I define addition between a scalar and a vector to simply always give the 0 vector and define multiplication between two vectors to always be the 0 vector. This gives a theory such that a structure models it if and only if it is a vector space in the regular algebraic sense. It is easy to see that all linear transformations in the algebraic sense are included in the class of homomorphisms (in the logical sense), but I am having difficulty proving they are the only ones in this class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I want to show that for any $L$-homomorphism $T$, $T(av+w)=aTv+Tw$ for $a$ a scalar and $v,w$ vectors. Since $T$ is a homomorphism, I see that $T(av+w)=Tav+Tw=TaTv+Tw$, so it would be enough to show that all homomorphisms act as the identity on all scalars? Is there a way to show this, or an extra axiom to include?&lt;/p&gt;&#xA;" OwnerUserId="2638" LastActivityDate="2010-10-20T11:00:03.710" Title="First Order Language for vector spaces over fields" Tags="&lt;linear-algebra&gt;&lt;logic&gt;&lt;vector-space&gt;" AnswerCount="2" />
  <row Id="7249" PostTypeId="1" CreationDate="2010-10-20T03:57:21.363" Score="7" ViewCount="116" Body="&lt;p&gt;It's kind of an infamous problem in differential equations to find the correct road surface so that a car with square wheels (and an axle located in the center) keeps its axle level as it drives along.  I hope I won't offend anybody by saying that one smooth piece of the solution (for a wheel with sides of length 2) is $y = -\cosh(x)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually take this solution and describe the position of the axle at any given point, unless I have calculated incorrectly you find that the axle is always positioned directly over the point where the wheel makes contact with the road.  I've been unable to come up with a physical justification of this phenomenon and it seems fairly non-obvious to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a straightforward reason why this must be true?  Is it specific to this wheel shape?&lt;/p&gt;&#xA;" OwnerUserId="1925" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-20T04:38:35.830" LastActivityDate="2010-10-21T04:15:25.823" Title="Question about square-wheeled cars." Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="7250" PostTypeId="2" ParentId="7248" CreationDate="2010-10-20T04:01:43.637" Score="0" ViewCount="0" Body="&lt;p&gt;It's not clear what your motivations are but, in any case, I thought it would be worthwhile to point out that usually the most convenient way to specify a vector space is to view it as a &quot;group with operators&quot;, i.e. for each scalar $c\:$ introduce a unary operation $\ c(v) = c\:v\:$. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-20T04:01:43.637" CommentCount="1" />
  <row Id="7252" PostTypeId="1" CreationDate="2010-10-20T04:10:38.073" Score="0" ViewCount="109" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1&quot;&gt;Sine function dense in $[-1,1]$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;How do i prove this is dense??&lt;/p&gt;&#xA;" OwnerUserId="2639" LastActivityDate="2010-10-20T04:16:43.613" ClosedDate="2010-10-20T04:47:15.867" Title="Is sin(n) for all n in the intergers dense? " Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7253" PostTypeId="2" ParentId="7249" CreationDate="2010-10-20T04:14:55.080" Score="1" ViewCount="0" Body="&lt;p&gt;Quite an exposition is in &lt;a href=&quot;http://poncelet.math.nthu.edu.tw/disk5/js/cardioid/1.pdf&quot; rel=&quot;nofollow&quot;&gt;Stan Wagon's paper&lt;/a&gt;  If you look at Fig 13, it appears that the axle of the ellipse is not over the contact point.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T04:14:55.080" CommentCount="4" />
  <row Id="7254" PostTypeId="2" ParentId="7252" CreationDate="2010-10-20T04:16:43.613" Score="0" ViewCount="0" Body="&lt;p&gt;This was already covered in &lt;a href=&quot;http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1&quot;&gt;http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would vote to close as identical but don't have the rep&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T04:16:43.613" CommentCount="2" />
  <row Id="7255" PostTypeId="1" AcceptedAnswerId="7257" CreationDate="2010-10-20T04:56:40.297" Score="2" ViewCount="100" Body="&lt;p&gt;Trying to remember my high school formulas, and coming up dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have two choices: A and B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(A) = 0.25; P(B) = 0.75&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no conditional probabilities or anything. Each choice is independent of the last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about calculating the probability that I will choose A exactly twice, having chosen ten items?&lt;/p&gt;&#xA;" OwnerUserId="1885" LastActivityDate="2010-10-20T05:36:27.487" Title="Probability of picking an item r times out of n attempts" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="2" />
  <row Id="7256" PostTypeId="2" ParentId="7255" CreationDate="2010-10-20T05:06:18.760" Score="3" ViewCount="0" Body="&lt;p&gt;You want the binomial distribution with $p = 0.25$ and $n = 10$ and $k = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;binomial distribution&lt;/a&gt;, which says,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The probability of getting exactly $k$ successes in $n$ trials [when each trial has success probability $p$] is given by ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \binom{n}{k} p^k (1-p)^{n-k}.&quot;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul VanKoughnett's answer gives you an explanation of why this formula is correct.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-20T05:24:27.557" LastActivityDate="2010-10-20T05:24:27.557" CommentCount="1" />
  <row Id="7257" PostTypeId="2" ParentId="7255" CreationDate="2010-10-20T05:10:33.640" Score="3" ViewCount="0" Body="&lt;p&gt;You want to pick A twice and B eight times, so the probability of doing this in a certain order is $P(A)^2P(B)^8$.  But for you, the order doesn't matter, so you have to divide by the number of ways to choose two things from 10.  This is $\binom{10}{2}=\frac{10!}{8!2!}=45$.  So all in all you get $\frac{(45)3^8}{4^{10}}$, which you can calculate yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the formula for $\binom{n}{r}=\frac{n!}{(n-r)!r!}$, where the exclamation point means &quot;factorial,&quot; $n!=n\times(n-1)\times\dotsb\times 2\times 1$.  Since $n!$ represents the number of arrangements of $n$ things, you can interpret this formula as giving you the number of arrangements of your $n$ things, and then cancelling out the ways you can arrange the $r$ things you want, and the $n-r$ things you don't want.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-20T05:36:27.487" LastActivityDate="2010-10-20T05:36:27.487" CommentCount="3" />
  <row Id="7258" PostTypeId="1" AcceptedAnswerId="7271" CreationDate="2010-10-20T06:29:46.767" Score="0" ViewCount="92" Body="&lt;p&gt;In $\mathbb{R}^2$ specify a linearly independent set that has many elements as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do such sets have in common?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{R}^3$ do the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have an answer in mind, but I would like to know if this wording is clear... and if I have really answered my own question fully and correctly. Thank you.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I removed the 2nd question since people are responding as if it is related or as if I don't know the definition of linear independence.  &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-20T13:31:05.700" LastActivityDate="2010-10-20T13:31:05.700" Title="linearly independent set that has many elements as possible" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" />
  <row Id="7259" PostTypeId="1" AcceptedAnswerId="7260" CreationDate="2010-10-20T06:34:48.980" Score="6" ViewCount="117" Body="&lt;p&gt;Is there an expansion for the following summation?&#xA;$$ (a_1 + a_2 + \cdots + a_k)^n $$&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-10-20T06:44:30.667" Title="Expansion of $ (a_1 + a_2 + \cdots + a_k)^n $" Tags="&lt;binomial-coefficients&gt;" AnswerCount="2" />
  <row Id="7260" PostTypeId="2" ParentId="7259" CreationDate="2010-10-20T06:37:28.610" Score="10" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Multinomial_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what you seek.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-10-20T06:37:28.610" CommentCount="3" />
  <row Id="7261" PostTypeId="2" ParentId="7258" CreationDate="2010-10-20T06:37:42.853" Score="1" ViewCount="0" Body="&lt;p&gt;think about what linear independence means, and make sure you understand the definition of &quot;the subspace generated (or spanned) by the set {x,y,...}.&quot; Once you have done that, it should be clear. If you still have trouble try looking at the vector space $\mathbb{R}$ and do the same. &lt;/p&gt;&#xA;" OwnerUserId="2638" LastActivityDate="2010-10-20T06:37:42.853" CommentCount="1" />
  <row Id="7262" PostTypeId="2" ParentId="7259" CreationDate="2010-10-20T06:44:30.667" Score="1" ViewCount="0" Body="&lt;p&gt;Since you know what $(a+b)^{n}$ is you can take $A = a_{1} + a_{2} + \cdots + a_{k-1}$ and $B=a_{k}$ and try to simplify the big expression by the binomial theorem.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-20T06:44:30.667" CommentCount="2" />
  <row Id="7263" PostTypeId="2" ParentId="7187" CreationDate="2010-10-20T06:53:17.330" Score="0" ViewCount="0" Body="&lt;p&gt;I don't know if the following counts as &quot;intuition&quot;, but perhaps it&#xA;can put things in some perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a theorem that any rotation in $R^n$ can be written as the&#xA;composition of an even number of reflections through hyperplanes.&#xA;To construct a way of doing rotions by algebra (not matrix algebra,&#xA;but rather by multiplying vectors and similar geometric objects), it&#xA;is therefore enough to figure out two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Given a vector $f$, how do we represent the reflection $R_f$&#xA;through the hyperplane orthogonal to $f$? We want $R_f(x)$ to be some&#xA;kind of product involving the vectors $x$ and $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The correspondence constructed in (1) should be such that for two&#xA;vectors $f$ and $g$, the algebraic object $fg$ (whatever that is)&#xA;should represent the composition $R_f \circ R_g$. Then any rotation&#xA;can be written as&#xA;$R_{f_1} \circ R_{f_2} \circ \cdots \circ R_{f_{2m}}$ and therefore represented by the object&#xA;$f_1 f_2 \cdots f_{2m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this to work, we have to give up the idea that the product of two&#xA;vectors must be a vector (as Hamilton famously struggled with for a&#xA;long time). Instead we let the vectors sit inside some bigger space of&#xA;&quot;complex&quot; (compound) objects that can be multiplied with each other&#xA;according to suitable rules (associative, but not commutative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of quaternions, these complex objects consist of a &quot;scalar&#xA;part&quot; plus a &quot;vector part&quot;, and the construction (1) which makes&#xA;everything work is $R_f(x)=-fxf^{-1}$. To see that this is a&#xA;reflection, note that it's linear in $x$, it maps $f$ to $-f$, and it&#xA;maps vectors orthogonal to $f$ to themselves (since they satisfy&#xA;$fx=-xf$ because of the way quaternion multiplication is defined).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do rotations in $R^n$ one constructs a so-called Clifford algebra,&#xA;consisting of complex objects with a &quot;scalar part&quot;, a &quot;vector part&quot;, a&#xA;&quot;bivector part&quot;, a &quot;trivector part&quot;, etc., up to an &quot;n-vector part&quot;,&#xA;and the rules to multiply these objects are generated by the&#xA;requirement that for vectors $x$ and $y$, one wants $xy+yx=2 x\cdot y$&#xA;(or the negative of that, depending on convention). This makes&#xA;orthonal vectors anticommute, so that $R_f(x)=-fxf^{-1}$ gives a&#xA;hyperplane reflection, and everything works as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't go into what a &quot;k-vector&quot; is, but the space of k-vectors&#xA;is $\binom{n}{k}$-dimensional, so that the whole Clifford algebra is&#xA;$2^n$-dimensional.&#xA;The Clifford product of two pure vectors is a scalar plus a bivector,&#xA;which is a little bit different from quaternions where vector times&#xA;vector equals a scalar plus a vector.&#xA;The full Clifford algebra for $R^3$ is 8-dimensional, and it's a bit&#xA;of a &quot;coincidence&quot; that one can actually manage just as well with the&#xA;4-dimensional quaternion algebra if one only wants to work with&#xA;rotations in $R^3$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-20T12:41:03.220" LastActivityDate="2010-10-20T12:41:03.220" />
  <row Id="7264" PostTypeId="2" ParentId="7258" CreationDate="2010-10-20T06:54:17.547" Score="0" ViewCount="0" Body="&lt;p&gt;Just to clarify about the three dimensional case, it is not enough to ask that it won't be a multiple of another vector in the set, you must request for something stronger:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $\{v_1,\ldots ,v_k\}$ which are linear independent, then $v_{k+1}$ can be added to this set while retaining the independence if $v_{k+1}$ cannot be written as the sum: $c_1 v_1 + c_2 v_2 + \cdots + c_k v_k$, where $c_i$ are scalars (in this case real numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The said sum is called &quot;linear combination&quot;, so using this term one can restate that $v_{k+1}$ is linearly independent of $\{v_1,\ldots, v_k\}$ if it cannot be written as a linear combination of $v_1,\ldots ,v_k$.)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-20T06:54:17.547" CommentCount="6" />
  <row Id="7265" PostTypeId="1" AcceptedAnswerId="7267" CreationDate="2010-10-20T06:54:31.660" Score="2" ViewCount="82" Body="&lt;p&gt;Suppose that a rational generating function is given representing the function&#xA;$f(x) = \sum_{i=0}^{2n}{c_i x^i}$ where $c_i \in \mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to determine if the coefficient in the middle, $c_n$, is nonzero or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we could be given $f(x) = \frac{1 - 10x^9 + 9x^{10}}{(x - 1)^2}$, which represents $1 + 2x + 3x^2 + \cdots + 9x^8$.  We are interested, in this case, in the middle coefficient, which happens to be 5.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Original Formulation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The actual value can be determined by evaluating a contour integral.  This is accomplished by first multiplying $f(x)$ by $x^{-n}$.  Then, rewriting the variable $x$ as $x\cdot e^{i t}$ and finally evaluating the integral $\int_C{f(x)dt}$.  The result of all of this is the coefficient in question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Goal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I seek to find alternative formulations to this problem.  Let me give some examples.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 1 : &quot;Tiling&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Start with the original formula $f(x)$.  Make a copy, for example $f(y)$, by rewriting $x$ as $y$.  Then we can subtract the second function from the first, eliminating the first coefficient.  We repeat additions and subtractions (possibly infinitely many times) until all coefficients except for the middle one have been eliminated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 2 : &quot;Division&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We multiply $f(x)$ by a function $g(y)$, where $g(y) = \frac{1}{1-x^{2n+1}}$, essentially making the series repeat forever.  We then &lt;a href=&quot;http://homepage.univie.ac.at/herwig.hauser/Publications/linear_recurrences_and_powerseries_division.pdf&quot; rel=&quot;nofollow&quot;&gt;divide&lt;/a&gt; this by $x^n\cdot g(y)$, and get the remainder.  This remainder is the first half of the original series, and $c_n$ can be easily obtained from this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is desired / Motivations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd simply like to have alternative formulations for the solution.  I'm using this for an algorithm, and I'd like to publish this with anyone who gives a formulation that provides a quick result.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-10-21T00:01:42.607" Title="What is an alternative formulation to a contour integral?" Tags="&lt;soft-question&gt;&lt;sequences-and-series&gt;&lt;big-list&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="7266" PostTypeId="1" AcceptedAnswerId="7275" CreationDate="2010-10-20T07:04:52.720" Score="6" ViewCount="210" Body="&lt;ol&gt;&#xA;&lt;li&gt;Take any number $x$ (edit: x should be positive, heh)&lt;/li&gt;&#xA;&lt;li&gt;Add 1 to it $x+1$&lt;/li&gt;&#xA;&lt;li&gt;Find its reciprocal $1/(x+1)$&lt;/li&gt;&#xA;&lt;li&gt;Repeat from 2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, taking $x = 1$ to start:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1&lt;/li&gt;&#xA;&lt;li&gt;2 (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.5 (the reciprocal)&lt;/li&gt;&#xA;&lt;li&gt;1.5 (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.666... (the reciprocal)&lt;/li&gt;&#xA;&lt;li&gt;1.666... (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.6 (the reciprocal)&lt;/li&gt;&#xA;&lt;li&gt;1.6 (the + 1)&lt;/li&gt;&#xA;&lt;li&gt;0.625&lt;/li&gt;&#xA;&lt;li&gt;1.625&lt;/li&gt;&#xA;&lt;li&gt;0.61584...&lt;/li&gt;&#xA;&lt;li&gt;1.61584...&lt;/li&gt;&#xA;&lt;li&gt;0.619047...&lt;/li&gt;&#xA;&lt;li&gt;1.619047...&lt;/li&gt;&#xA;&lt;li&gt;0.617647058823..&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at just the &quot;step 3&quot;'s (the reciprocals), we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1&lt;/li&gt;&#xA;&lt;li&gt;0.5&lt;/li&gt;&#xA;&lt;li&gt;0.666...&lt;/li&gt;&#xA;&lt;li&gt;0.6&lt;/li&gt;&#xA;&lt;li&gt;0.625&lt;/li&gt;&#xA;&lt;li&gt;0.61584...&lt;/li&gt;&#xA;&lt;li&gt;0.619047...&lt;/li&gt;&#xA;&lt;li&gt;0.617647058823..&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This appears to always converge to 0.61803399... no matter where you start from.  I looked up this number and it is often called &quot;The golden ratio&quot; - 1, or $\frac{1+\sqrt{5}}{2}-1$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there any &quot;mathematical&quot; way to represent the above procedure (or the terms of the second series, of &quot;only reciprocals&quot;) as a limit or series?&lt;/li&gt;&#xA;&lt;li&gt;Why does this converge to what it does for every starting point $x$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;edit: darn, I just realized that the golden ratio is actually 1.618... and not 0.618...; I edited my answer to change what the result is apparently (golden ratio - 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think I could easily make it the golden ratio by taking the +1 &quot;steps&quot; of the original series, instead of the reciprocation steps of the original series:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2&lt;/li&gt;&#xA;&lt;li&gt;1.5&lt;/li&gt;&#xA;&lt;li&gt;1.666...&lt;/li&gt;&#xA;&lt;li&gt;1.6&lt;/li&gt;&#xA;&lt;li&gt;1.625&lt;/li&gt;&#xA;&lt;li&gt;1.61584...&lt;/li&gt;&#xA;&lt;li&gt;1.619047...&lt;/li&gt;&#xA;&lt;li&gt;1.617647058823..&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which does converge to $\frac{1+\sqrt{5}}{2}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explaining either of these series is adequate as I believe that explaining one also explains the other.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-20T07:41:47.950" LastActivityDate="2010-10-20T15:10:31.087" Title="Why does this process, when iterated, tend towards a certain number? (the golden ratio?)" Tags="&lt;limit&gt;&lt;convergence&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="7267" PostTypeId="2" ParentId="7265" CreationDate="2010-10-20T07:10:02.337" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure I understand exactly what you want, but if you differentiate the function $n$ times and then evaluate the result at $x=0$ you get $n! c_n$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-20T07:10:02.337" CommentCount="4" />
  <row Id="7268" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T07:30:21.483" Score="16" ViewCount="0" Body="&lt;p&gt;You are iterating the operation&#xA;$$x\mapsto\frac1{x+1}.$$&#xA;We can represent this in matrix terms. Set&#xA;$$A=\begin{pmatrix}&#xA;0&amp;amp;1\\&#xA;1&amp;amp;1\end{pmatrix}.$$&#xA;If&#xA;$$A&#xA;\begin{pmatrix}&#xA;x\\&#xA;1\end{pmatrix}&#xA;=\begin{pmatrix}&#xA;y_1\\&#xA;z_1\end{pmatrix}$$&#xA;then $1/(x+1)=y_1/z_1$. After $n$ iterations one gets&#xA;$$A^n&#xA;\begin{pmatrix}&#xA;x\\&#xA;1\end{pmatrix}&#xA;=\begin{pmatrix}&#xA;y_n\\&#xA;z_n\end{pmatrix}$$&#xA;and $x_n=y_n/z_n$ is got by applying the map $x\mapsto 1/(x+1)$ $n$ times to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally you won't be surprised to find that the eigenvalues of&#xA;$A$ are $\tau=\frac12(1+\sqrt5)$ and $\tau=\frac12(1-\sqrt5)$.&#xA;The eigenvectors are $v=(1\ \tau)^t$ and $w=(1\ \tau')^t$. We can write&#xA;$$\begin{pmatrix}&#xA;x\\&#xA;1\end{pmatrix}=av+bw$$&#xA;and so&#xA;$$\begin{pmatrix}&#xA;y_n\\&#xA;z_n\end{pmatrix}=a\tau^n v+b\tau'^nw.$$&#xA;Now $\tau&gt;1&gt;|\tau'|$&#xA;so that for large $n$, $y_n$ and $z_n$ are very close to $a\tau^n$and $a\tau^{n+1}$&#xA;so that $x_n\to\infty$ if $a=0$. The only exception is when $a=0$&#xA;which only arises for $x=-\frac12(1+\sqrt5)$. (I bet you didn't test that one!)&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-20T08:10:56.497" LastActivityDate="2010-10-20T08:10:56.497" CommentCount="4" />
  <row Id="7269" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T07:30:41.717" Score="1" ViewCount="0" Body="&lt;p&gt;Define the function $f(x) = \frac{1}{1+x} $ so the process you defined is just going from x to f(x).&#xA;let $x_0\in\mathbb{R}$ and $x_{i+1} = f(x_i)$ then your claim is that $lim x_i = \frac {1+\sqrt{5}}{2} $.&#xA;if this series converges then&#xA;$x = \lim x_i = \lim f(x_{i-1}) = f( \lim x_{i-1} ) = f(x)$ so the limit is $x=\frac{1}{1+x}$ or in other words you have $x^2+x-1=0$. the solutions are $\frac{-1\pm \sqrt{5}}{2}$ &#xA;this isn't exactly the golden ratio. if you take $\frac{1}{x-1}$ instead then you'll have it. you still need to find out for which $x_0$ this sequence converges - this could be done using numerical analysis (sorry, I'm a bit rusty on this, so I'll leave this part to someone else).&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-10-20T07:30:41.717" />
  <row Id="7270" PostTypeId="1" CreationDate="2010-10-20T07:41:00.310" Score="3" ViewCount="80" Body="&lt;p&gt;Thats it! Thats what I want to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If θ is the angle subtended by an arc of length L at its center with radius R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know, θ = L/R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did we get this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't say we got it from generalizing the idea of perimeter of circle.&lt;/p&gt;&#xA;" OwnerUserId="2264" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-20T10:25:00.597" LastActivityDate="2010-10-20T10:26:21.283" Title="θ = (length of arc)/(angle subtended by it).  How?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" />
  <row Id="7271" PostTypeId="2" ParentId="7258" CreationDate="2010-10-20T07:49:26.873" Score="1" ViewCount="0" Body="&lt;p&gt;The first question is clear.  It appears to be a basic question designed to make &lt;em&gt;you&lt;/em&gt; apply the definition of linear independence to familiar objects, the vector spaces $\mathbb{R}^2$ and $\mathbb{R}^3$.  (That's why the responses so far give hints rather than solutions.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note that linear independence is a property of sets of vectors-- so the &quot;sets&quot; in the first question are implied to be sets of vectors.  So both questions are really about sets of vectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asaf Karagila is pointing out that the property in question 2, that &quot;no vector is a constant multiple of another vector in the set,&quot; is a different property than linear independence.  In fact, question 2 has a very different answer than question 1.  If you aren't sure of the answers, explain your thinking and your uncertainty, and I'm sure you'll get a more direct hint.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-20T07:49:26.873" CommentCount="1" />
  <row Id="7272" PostTypeId="2" ParentId="7270" CreationDate="2010-10-20T08:03:56.297" Score="1" ViewCount="0" Body="&lt;p&gt;This is because using &lt;a href=&quot;http://mathworld.wolfram.com/Radian.html&quot; rel=&quot;nofollow&quot;&gt;radian measure&lt;/a&gt; (as opposed to using degrees) in working with things related to circles is mathematically very convenient. In more advanced mathematical work you rarely see people use degrees since the formulae that have the angles in degrees require having to keep around inconvenient factors of $\frac{\pi}{180}$ or its reciprocal.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T08:03:56.297" CommentCount="4" />
  <row Id="7273" PostTypeId="2" ParentId="7238" CreationDate="2010-10-20T08:23:22.880" Score="1" ViewCount="0" Body="&lt;p&gt;You could try reading &lt;a href=&quot;http://terrytao.wordpress.com/2009/02/09/245b-notes-10-compactness-in-topological-spaces/&quot; rel=&quot;nofollow&quot;&gt;Terence Tao's notes&lt;/a&gt; on the subject; I found them very informative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for examples, here is an example and a non-example which I think are informative.  The example is that, by Tychonoff's theorem, $[0, 1]^I$ is compact for any index set $I$.  The non-example is that the closed unit ball in any infinite-dimensional Banach space, say $\ell_1(\mathbb{Z})$, is &lt;em&gt;not&lt;/em&gt; compact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Here is another perspective from which to think about compact &lt;em&gt;Hausdorff&lt;/em&gt; spaces (and LCH spaces).  A compact Hausdorff space $X$ is completely determined by the ring of continuous functions $C(X, \mathbb{R})$; this is a standard exercise which is worked out, for example, &lt;a href=&quot;http://qchu.wordpress.com/2009/11/24/spectra-of-rings-of-continuous-functions/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  One gets the points of $X$ as the maximal ideals of $C(X, \mathbb{R})$ and the topology as the initial topology making every function in $C(X, \mathbb{R})$ continuous.  Slightly generalized, this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfand_representation#Statement_of_the_commutative_Gelfand-Naimark_theorem&quot; rel=&quot;nofollow&quot;&gt;commutative Gelfand-Naimark theorem&lt;/a&gt;, and it says that studying commutative C*-algebras with unit is the same thing as studing compact Hausdorff spaces.  Removing the requirement that we have a unit is the same thing as studying locally compact Hausdorff spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one can write down compact Hausdorff spaces by writing down commutative C*-algebras with unit.  One choice is to take the space $C_b(X, \mathbb{R})$, the space of bounded continuous functions on an arbitrary topological space $X$, with the sup norm.  If $X$ is completely regular Hausdorff, this gives the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification#Construction_using_C.2A-algebras&quot; rel=&quot;nofollow&quot;&gt;Stone-Cech compactification&lt;/a&gt; of $X$.  (Which is hopeless to think about in general; even for $X = \mathbb{N}$ it's very weird, as the Wikipedia article describes.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T09:23:13.720" LastActivityDate="2010-10-20T09:23:13.720" CommentCount="1" />
  <row Id="7274" PostTypeId="2" ParentId="7249" CreationDate="2010-10-20T08:29:49.090" Score="1" ViewCount="0" Body="&lt;p&gt;This is supposed to be in reply to Rahul's second comment, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suppose that the requirement of Hall and Wagon for the axle to have the same horizontal displacement as the contact point of the road and wheel is a generalization of the concept of a circle rolling on a line. In that situation it is obvious that the axle is indeed directly above the point of contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that rolling without slip can be decomposed as a translation, followed by a rotation, and then another translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of the wheel parametrically defined as $(x\qquad y)^T=(\sin\;t\qquad 1-\cos\;t)^T$, the axle is initially at $(0\;1)^T$. If we look at the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bigl(\begin{smallmatrix}t\\0\end{smallmatrix}\bigr)+\bigl(\begin{smallmatrix}\cos\;t&amp;amp;\sin\;t\\-\sin\;t&amp;amp;\cos\;t\end{smallmatrix}\bigr)\cdot\left(\bigl(\begin{smallmatrix}0\\1\end{smallmatrix}\bigr)-\bigl(\begin{smallmatrix}\sin\;t\\1-\cos\;t\end{smallmatrix}\bigr)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in English: translate axle to the point on the circle that would be the future point of contact of the circle with the road, rotate by some angle, and then translate by an amount equal to the length of the arc subtended by said angle horizontally)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we find that this whole mess simplifies to the vector $(t\qquad 1)^T$ which indeed has the same x-coordinate as the current point of contact $(t\qquad 0)$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T08:29:49.090" CommentCount="2" />
  <row Id="7275" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T08:46:54.120" Score="10" ViewCount="0" Body="&lt;p&gt;We want to show that the function $f(x) = \frac{1}{1+x}$ has a unique fixed point to which it converges when iterated on positive $x$.  If $x$ is positive, then $f(x)$ is between $1$ and $0$, so $f(f(x))$ is between $1$ and $\frac{1}{2}$.  It is not hard to see that in fact $f$ fixes the interval $\left[ \frac{1}{2}, 1 \right]$.  Now, for $x, y$ in this interval,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{1}{x+1} - \frac{1}{y+1} \right| = \left| \frac{y-x}{(1+x)(1+y)} \right| \le \frac{4}{9} |x - y|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so on this interval $f$ satisfies the conditions of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach fixed point theorem&lt;/a&gt;.  The unique fixed point to which everything always converges is the unique solution to $f(x) = x$, which you have already found.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T15:10:31.087" LastActivityDate="2010-10-20T15:10:31.087" CommentCount="5" />
  <row Id="7276" PostTypeId="2" ParentId="6890" CreationDate="2010-10-20T09:07:31.323" Score="1" ViewCount="0" Body="&lt;p&gt;The 4-vector density current is&#xA;\begin{equation*}&#xA;j^{\mu} := \rho \frac{\partial x^{\mu}}{\partial \tau} = \rho {\dot{x}}^{\mu} = \rho (c, \mathbf{\dot{x}}) = (\rho c, \mathbf{j})&#xA;\end{equation*}&#xA;where $\tau$ is the proper time.\&#xA;We can ask ${\dot{x}}^{\mu}$ to be a 4-versor, or in other words to be of unit Minkowski norm.&#xA;Usually we define the vector potential $\mathbf{A}$ in terms of $\mathbf{j}$, and the scalar potential $\phi$ in terms of the current density $\rho$&#xA;\begin{equation*}&#xA; \mathbf{A}(\mathbf{x}) = C_1 \int_{space} \frac{\mathbf{j}(\mathbf{y})}{r}d\mathbf{y}, \qquad &#xA; \phi(\mathbf{x}) = C_2 \int_{space} \frac{\rho(\mathbf{y})}{r}d\mathbf{y}&#xA;\end{equation*}&#xA;where $r$ is the spatial distance between $\mathbf{x}$ and $\mathbf{y}$, where $C_1$ and $C_2$ are costants. \&#xA;Let's put all together in the 4-vector $A$&#xA;\begin{equation*}&#xA; A_{\mu}(x) = k \int_{spacetime} \frac{j_{\mu}(y)}{r} dy&#xA;\end{equation*}&#xA;But we want $A$ to be non-zero at the point $\mathbf{x}$ only when the light comes from $\mathbf{y}$ to $\mathbf{x}$, or when the event $x$ os on the boundary of the light cone centered in $y$. To impose this is sufficient to multiply the integrand with $\delta((x-y)^{\nu} (x-y)_{\nu})$&#xA;\begin{equation*}&#xA; A_{\mu}(x) = k \int_{spacetime} \frac{j_{\mu}(y)}{r} \delta((x-y)^{\nu} (x-y)_{\nu}) dy&#xA;\end{equation*}&#xA;We can simplify the expression of $A$, posing&#xA;\begin{equation*}&#xA;\int \delta(f(\alpha,\beta))g(\alpha,\beta) d\beta := \sum_i \frac{g(\alpha,\beta^i)}{| \frac{\partial f}{\partial \beta}(\alpha, \beta^i) |}&#xA;\end{equation*}&#xA;where $\alpha$ is fixed, and the sum is extended to all the $\beta^i$ for which $f(\alpha,\beta^i) = 0$.&#xA;Doing like this we have, speaking about a single particle $b$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation*} \int \delta((x-b)^{\nu} (x-b){\nu})j{\mu}(\beta) d\beta = e_b \int \delta((x-b)^{\nu} (x-b){\nu}) \dot{b}{\mu}(\beta) d\beta = \frac{e_b}{2} \sum_i \frac{j_{\mu}(\beta^i)}{| (x-b)^{\nu} \dot{b}_{\nu} |} \end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we only consider the events $x$ that $(x-b)^{\nu} (x-b)_{\nu} = 0$, we have&#xA;\begin{equation*}&#xA; r = \lvert \mathbf{x-b} \rvert = (x-b)^0&#xA;\end{equation*}&#xA;In the rest frame of the current flux $\dot{b} = (1,0,0,0)$. So we find&#xA;\begin{equation*}&#xA; r = (x-b)^0 = (x-b)^{\nu}{\dot{b}}_{\nu}&#xA;\end{equation*}&#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation*} \int \delta((x-b)^{\nu} (x-b){\nu})j{\mu}(\beta) d\beta = \frac{e_b}{2} \sum_i \frac{j_{\mu}(\beta^i)}{r} \end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we notice $j_{\mu} d\beta = e_b (db_{\mu} / d\beta) d\beta = e_b db_{\mu}$, and so we define the 4-vector potential as follows&#xA;The 4 dimensional vector potential of particle $b$ at the point $x$, with charge $e_b$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\label{A} A_{\mu}^{(b)}(x) := e_b \int \delta((x-b)^{\nu} (x-b){\nu}) db{\mu} \end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we integrate over the whole world line of $b$.&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-21T07:04:42.917" LastActivityDate="2010-10-21T07:04:42.917" CommentCount="2" />
  <row Id="7277" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T09:08:28.263" Score="4" ViewCount="0" Body="&lt;p&gt;This was supposed to be an addendum to Hans's comment, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The iteration Justin considered formally generates a continued fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{1+\cfrac{1}{1+\dots}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as I mentioned in &lt;a href=&quot;http://math.stackexchange.com/questions/3382/continued-fraction-for-frac1e-2/3387#3387&quot;&gt;this answer&lt;/a&gt;, the numerators and denominators of the nth convergent of a continued fraction can be computed recursively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we apply the formula in that answer to this situation, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}C_n\\D_n\end{bmatrix}=\begin{bmatrix}C_{n-1}\\D_{n-1}\end{bmatrix}+\begin{bmatrix}C_{n-2}\\D_{n-2}\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with initial conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}C_{-1}\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix},\qquad \begin{bmatrix}C_{0}\\D_{0}\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, the setup is now similar to Robin's answer, since the two recursions in fact generate the Fibonacci numbers: $C_n=F_n$ and $D_n=F_{n+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your continued fraction then is the limit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty}\frac{F_n}{F_{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the equivalence to Robin's answer is due to the &lt;a href=&quot;http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html&quot; rel=&quot;nofollow&quot;&gt;Binet formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_n=\frac{\varphi^n-\left(-\varphi^{-1}\right)^n}{\sqrt{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which in fact can be derived from Robin's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting that into the limit and evaluating gets you the answer.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T09:08:28.263" />
  <row Id="7278" PostTypeId="2" ParentId="7270" CreationDate="2010-10-20T09:21:28.653" Score="1" ViewCount="0" Body="&lt;p&gt;What is your definition of an angle? The formula you quoted follows directly from the definition of an angle as the length of the arc on the unit circle, combined with the intercept theorems (for “scaling the circle to a unit circle,” if you want).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using angles defined this way is, as J.M. said, much more convenient than degrees. E.g., it connects triangles with trigonometric functions with nice derivatives – from $\sin' = \cos$, $\cos' = -\sin$, $\sin(0) = 0$, and $\cos(0) = 1$, you directly find that $\cos(\pi/2)=0$, i.e., these formulas only hold if the arguments to $\sin$ and $\cos$, traditionally angles, are given in radians.&lt;/p&gt;&#xA;" OwnerUserId="1354" LastActivityDate="2010-10-20T09:21:28.653" />
  <row Id="7279" PostTypeId="2" ParentId="7247" CreationDate="2010-10-20T10:01:11.697" Score="2" ViewCount="0" Body="&lt;p&gt;No. If $K$ does not have compact support, then $f*K$ need not even be bounded. Consider the function $K = 1 / \log( 3 + |x|)$, and $f(x) = (1 + |x|)^{-1/p - \epsilon}$. For $\epsilon &gt; 0$. Then $f\in L^p$, but the convolution integral does not converge anywhere, even when $K$ vanishes at $\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is to require $K$ decays sufficiently rapidly. What you need is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Young%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Young's inequality for convolutions&lt;/a&gt;, which implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sup |f * K| \leq \|f \|_p \|K\|_q $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $1/p + 1/q = 1$. So in particular if $K\in C^m(\mathbb{R}) \cap W^{m,q}(\mathbb{R})$, where $W^{m,q}$ is the Sobolev space of $m$-times weakly differentiable, $q$-integrable functions with $q = p / (p-1)$, you can conclude that the convolution is $C^m$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-20T10:01:11.697" />
  <row Id="7280" PostTypeId="2" ParentId="6661" CreationDate="2010-10-20T10:03:06.003" Score="0" ViewCount="0" Body="&lt;p&gt;Have also a look here - many great resources for the Laplace transform:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-20T10:03:06.003" />
  <row Id="7281" PostTypeId="2" ParentId="7270" CreationDate="2010-10-20T10:12:24.727" Score="1" ViewCount="0" Body="&lt;p&gt;The length of the arc is proportional to the angle by proposition 6.33 in Euclid's &lt;em&gt;&lt;a href=&quot;http://farside.ph.utexas.edu/euclid/Elements.pdf&quot; rel=&quot;nofollow&quot;&gt;Elements&lt;/a&gt;&lt;/em&gt;. We can choose units to make the constant of proportionality equal to $1$ for a circle of radius $1$. Those units are called radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I give the Euclid reference since it uses the notion of angle that might be familiar to you. Nowadays we find it more economical to define the trignometric functions without any appeal to geometry (for eg. via the complex exponential), then define angles by inverting one of these functions, and then derive your result as an exercise in calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple hand-waving argument along those lines. Let OA and OB be two radii of a circle of radius $r$. Then by simple trignometry the length of the line segment AB is $2r\sin(\theta/2)$. For small enough $\theta$, $\sin(\theta/2) \approx \theta/2$ so the length of AB is approximately $r\theta$. In this case the length of AB is also close to the length of the arc AB hence that is also approximately $r\theta$. We can make the approximation as good as we like by making $\theta$ small enough. We can handle large arcs by approximating it by a large number of small segments of this sort.]&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-20T10:26:21.283" LastActivityDate="2010-10-20T10:26:21.283" CommentCount="1" />
  <row Id="7282" PostTypeId="1" AcceptedAnswerId="7283" CreationDate="2010-10-20T10:23:20.667" Score="2" ViewCount="39" Body="&lt;p&gt;For a probability distribution, its quantile function is defined in terms of its distribution function as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Q(p)=F^{-1}(p) = \inf \{ x\in R : p \le F(x) \}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering if, conversely, a quantile function can uniquely determine a distribution and therefore fully describe the probability distribution just as a distribution function does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me be more specific. Because a CDF is nondecreasing, right-continous and limit is 0 when $x \rightarrow -\infty$ and 1 when $x \rightarrow \infty$, its quantile function is nondecreasing, left-continous and a map from (0,1) into $R$. If a function is nondecreasing, left-continous and a map from (0,1) into $R$, can it become a quantile function of some CDF? When it can, is there a way to represent the CDF in terms of the quantile function using infimum or supremum similar as quantile function in terms of CDF?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-20T10:58:43.133" LastActivityDate="2010-10-20T10:58:43.133" Title="Does Quantile Function Uniquely Determine Distribution Function?" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7283" PostTypeId="2" ParentId="7282" CreationDate="2010-10-20T10:34:19.707" Score="2" ViewCount="0" Body="&lt;p&gt;Well, if $Q(p)$ is well-defined and monotonic in the interval $(0,1)$, then certainly.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T10:34:19.707" CommentCount="8" />
  <row Id="7284" PostTypeId="2" ParentId="7248" CreationDate="2010-10-20T11:00:03.710" Score="2" ViewCount="0" Body="&lt;p&gt;If you work in two-sorted first-order logic, it is possible to formalize vector spaces by using one sort for the field of scalars and another sort for the vectors, just as you describe. But there is no way, within the theory itself, to force an embedding of models to act as the identity on the scalars. For example, if you take a structure for your theory, every automorphism of the field will give a homomorphism from the structure to itself.  And every proper subfield of the field of scalars gives rise to a proper substructure in which the vectors remain the same and the scalars are cut down.  This is the motivation for the usual method in which the field is fixed through the addition of a lot of constant symbols to the language. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One response is to simply refuse to consider such things when you work on your proofs. That is, you could simply limit your attention to those homomorphisms that do preserve the field, and those substructures that do not change the field.  It depends on what you are trying to get out of the formalization. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-10-20T11:00:03.710" />
  <row Id="7285" PostTypeId="2" ParentId="2234" CreationDate="2010-10-20T12:18:57.773" Score="0" ViewCount="0" Body="&lt;p&gt;as i have already told you the expression is a theoretical expressions obtained using the curve fitting tool in matlab. &lt;/p&gt;&#xA;" OwnerUserId="2488" LastEditorUserId="2488" LastEditorDisplayName="" LastEditDate="2010-10-20T17:20:28.497" LastActivityDate="2010-10-20T17:20:28.497" CommentCount="1" />
  <row Id="7286" PostTypeId="2" ParentId="7217" CreationDate="2010-10-20T12:28:17.247" Score="1" ViewCount="0" Body="&lt;p&gt;The Lebesgue mean does not exist for &quot;boundary points&quot;. Let $A_n$ be the union of the intervals of step $n$, that is, $A_1=[0,3/8]\cup [5/8,1]$, $A_2=[0,9/64]\cup [15/64,3/8]\cup [5/8,49/64]\cup [55/64,1]$ etc, $A=\cap_n A_n$ is the fat Cantor set. Consider a boundary point $x$ of a set $A_n$. Consider the sequence $\varepsilon_k:=(3/8)^{n+k}$. Then by symmetry we have $I_k=|(x-\varepsilon_k,x+\varepsilon_k)\cap A|=1/4\cdot(1/2)^{n+k}$ for $k$ large enough, thus $I_k/\varepsilon_k \sim (4/3)^{n+k} \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of elements of $A$ as infinite sequences of {Left,Right}. Those sequences which are eventually constant correspond to &quot;boundary points&quot;. My guess is that the points for which the mean is 1 correspond to sequences that are normal in some sense (for example, all patterns of {Left,Right} appear with the same asymptotic frequency).&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-20T12:28:17.247" />
  <row Id="7287" PostTypeId="1" AcceptedAnswerId="7299" CreationDate="2010-10-20T12:44:46.447" Score="3" ViewCount="60" Body="&lt;p&gt;I'm trying to show that if $f$ and $1/f$ are in $H^2$ then $f$ is an outer function. I think I somewhat have a proof (if it is correct) but it doesn't feel like it is the easiest one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that if $f$ is in $H^2$ and non-zero and $B$ is the Blanschke product of its zeros, then there exists a constant $K$ of modulus $1$ and a singular measure $\mu$ on $S^1$ such that $f(z)$ has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = K B(z) \text{exp} \left (-\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \, d\mu(\theta) \right ) \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the first exponential is the singular inner function part. So by uniqueness of this form (up to constants I believe?) and the fact that $1/f$ is in $H^2$ we can write:&#xA;$$\frac{1}{f(z)} = \frac{1}{KB(z)} \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \, d\mu(\theta) \right ) \text{exp} \left (-\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we see that $B(z) = 1$. Further we can see that the singular inner function-part has modulus $&gt;1$ a.e. in the open unit disk if $\mu \neq 0$, and so the limit $r \to 1^-$ is not $1$ because of the maximum modulus principle (maybe this is flawed since $f$ is not analytic on the boundary). So we are left with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(z) = K \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is an outer function. Are there any flaws in this? If not, is there a less &quot;thick&quot; way to show this (I mean, not invoke all the theorems about the form these functions must have and so on) since I'm not really satisfied with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Oh, a nice corollary of this result I found is that if $f$ in $H^\infty$ and $\|f\|_\infty &amp;lt; 1$, then $1 + f$ is outer. (By noting that if $f$ is in $H^\infty$, then $f$ in $H^2$.)&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-20T14:38:36.513" LastActivityDate="2010-10-20T16:03:23.447" Title="$f$, $1/f$ in $H^2$, then $f$ outer." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="7288" PostTypeId="2" ParentId="7187" CreationDate="2010-10-20T12:52:51.393" Score="0" ViewCount="0" Body="&lt;p&gt;Re the last answer, in Geometric Algebra (a Clifford Algebra interpreted geometrically), this is very intuitive, you can even visualize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.geometricalgebra.net/quaternions.html&quot; rel=&quot;nofollow&quot;&gt;http://www.geometricalgebra.net/quaternions.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2646" LastActivityDate="2010-10-20T12:52:51.393" />
  <row Id="7289" PostTypeId="1" AcceptedAnswerId="7292" CreationDate="2010-10-20T13:28:17.123" Score="1" ViewCount="117" Body="&lt;p&gt;In $\mathbb{R}^2$ specify a set of vectors, that has many elements as possible, so that no vector is a constant multiple of another vector in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do such sets have in common?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\mathbb{R}^3$ do the same&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I asked both of these earlier, but people are acting like it is a HW question (giving me &quot;hints&quot; etc.) but I'm not looking for HW help. I made these questions up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are the equivalence classes of fractions for $\mathbb{R}^2$ and for $\mathbb{R}^3$ the equivalence classes of 3-part &lt;em&gt;ordered&lt;/em&gt; ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now that you have heard that answer are there any more interesting answers?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-20T13:42:24.903" LastActivityDate="2010-10-21T01:46:58.110" Title="What do these sets have in common?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7291" PostTypeId="1" CreationDate="2010-10-20T13:42:38.267" Score="5" ViewCount="133" Body="&lt;p&gt;It is well known that the differential calculus has a nice algebraization in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_algebra#Differential_ring&quot; rel=&quot;nofollow&quot;&gt;differential rings&lt;/a&gt; but what about integral calculus? Of course, one sometimes defines an integral in a differential ring $R$ with a derivation $\partial$ as a projection $\pi: R\rightarrow \tilde R$, where $\tilde R$ is a quotient of $R$ w.r.t. the following equivalence relation: $f\sim g$ iff $f-g$ is in the image of $\partial$, but this is not very intuitive and apparently corresponds to the idea of &lt;em&gt;definite&lt;/em&gt; integral over a fixed domain rather than to that of an indefinite one. So my question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there algebraic counterparts for the concept of an &lt;em&gt;indefinite&lt;/em&gt; integral?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1399" LastEditorUserId="1399" LastEditorDisplayName="" LastEditDate="2010-10-20T14:19:36.293" LastActivityDate="2010-10-20T22:52:44.423" Title="Algebraization of integral calculus" Tags="&lt;calculus&gt;&lt;abstract-algebra&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="7292" PostTypeId="2" ParentId="7289" CreationDate="2010-10-20T13:58:21.900" Score="5" ViewCount="0" Body="&lt;p&gt;An easy way to do this is to start with a circle resp. a sphere and delete one of each pair of antipodal vectors.  Any answer can be obtained from this answer by replacing any point with a scalar multiple of itself.  I don't know what else there is to say.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T13:58:21.900" CommentCount="1" />
  <row Id="7293" PostTypeId="2" ParentId="7289" CreationDate="2010-10-20T14:03:46.210" Score="2" ViewCount="0" Body="&lt;p&gt;The intuitive way to answer this, for me, is the following.  Let's start with $\mathbb{R}^2$.  We're going to imagine the vectors as points in $\mathbb{R}^2-{(0,0)}$.  Once you choose one vector, you have basically blocked off the line through the origin that contains it.  So you can choose one vector for every line through the origin.  The same goes for $\mathbb{R}^3$.  There isn't much more to say about this, but if you require the set to be &lt;em&gt;connected&lt;/em&gt;, so that each vector you choose is &quot;close&quot; to other vectors in your set, you've got an embedding of the &quot;real projective plane&quot;, $\mathbb{RP}^{n-1}$, in $\mathbb{R}^n$.  The real projective plane is (equivalently), the space of lines through the origin, the sphere modulo the map sending $x$ to $-x$, or $\mathbb{R}^n$ modulo scalar multiplication.  The last definition is just what you said about equivalence classes of ordered pairs/triples.  Oh, and $\mathbb{RP}^2$ is what you get if you glue a circle to the single edge of a Moebius strip!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-20T14:03:46.210" />
  <row Id="7294" PostTypeId="1" AcceptedAnswerId="7349" CreationDate="2010-10-20T14:18:42.723" Score="0" ViewCount="55" Body="&lt;p&gt;I have some geometry in 3d (like road surface or terrain surface) and there could be several not connected parts of a road. &#xA;So I'd like to know if I may unite this two not connected pieces of road in a term &quot;surface&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I couldn't, then what is the best mathematical term for this?&lt;/p&gt;&#xA;" OwnerUserId="2647" LastEditorUserId="2647" LastEditorDisplayName="" LastEditDate="2010-10-20T15:10:46.030" LastActivityDate="2010-10-20T20:24:32.917" Title="is surface always locally connected area in euklidian 3d ?" Tags="&lt;geometry&gt;&lt;topology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7295" PostTypeId="1" AcceptedAnswerId="16344" CreationDate="2010-10-20T14:27:28.727" Score="5" ViewCount="99" Body="&lt;p&gt;Is there some place (on the internet or elsewhere) where I can find the number and preferably a list of all (isomorphism classes of) finite connected $T_0$-spaces with, say, 5 points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In know that a $T_0$-topology on a finite set is equivalent to a partial ordering, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_order&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; tells me that there are, up to isomorphism, 63 partially ordered sets with precisely 5 elements. However, I am only interested in connected spaces, and I'd love to have a list (most preferably in terms of Hasse diagrams).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2011-01-04T18:33:19.823" Title="Is there a list of all connected $T_0$-spaces with 5 points?" Tags="&lt;topology&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7296" PostTypeId="1" AcceptedAnswerId="7298" CreationDate="2010-10-20T14:35:14.883" Score="7" ViewCount="128" Body="&lt;p&gt;I've seen recently for the first time in &lt;em&gt;Special Functions&lt;/em&gt; (by G. Andrews, R. Askey and R. Roy) the definitions of fractional integral &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(I_{\alpha }f)(x)=\frac{1}{\Gamma (\alpha )}\int_{a}^{x}(x-t)^{\alpha -1}f(t)dt\qquad \text{Re}\alpha &gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and fractional derivative &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^{\nu }w^{\mu }}{dw^{\nu }}=\frac{\Gamma (\mu +1)}{\Gamma (\mu -\nu +1)}w^{\mu -\nu },$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in &lt;em&gt;The Hypergeometric Functions&lt;/em&gt; Chapter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know some applications for Fractional Calculus and/or which results can only be obtained by it, if any.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:27:56.867" LastActivityDate="2010-12-23T05:27:56.867" Title="Applications of Fractional Calculus" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="4" FavoriteCount="4" />
  <row Id="7297" PostTypeId="2" ParentId="7266" CreationDate="2010-10-20T14:45:57.773" Score="1" ViewCount="0" Body="&lt;p&gt;Here is another way of looking at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the special case of starting with $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider what happens when  $\displaystyle x = \frac{f_n}{f_{n+1}}$ where $f_n$ is the $n^{th}$ fibonacci number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get $$\frac{1}{\frac{f_n}{f_{n+1}} + 1} = \frac{f_{n+1}}{f_n + f_{n+1}} = \frac{f_{n+1}}{f_{n+2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle 1 = \frac{f_1}{f_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that after $n$ iterations, $\displaystyle x = \frac{f_{n+1}}{f_{n+2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalized to any other starting value, by using a Fibonacci like sequence, which satisfies the recurrence $\displaystyle a_{n+2} = a_{n+1} + a_{n}$ and choosing appropriate $a_{2}$ and $a_{1}$ so that $\displaystyle \frac{a_1}{a_2}$ is the initial guess for $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n^{th}$ value for $x$ will be given by $\displaystyle \frac{a_n}{a_{n+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general formula for such sequences is given by $a_{n} = A\alpha^n + B\beta^n$ where $\alpha,\beta$ are roots of $\displaystyle z^2 = z + 1$ and thus the limit of $\displaystyle \frac{a_n}{a_{n+1}}$ can be easily found, which will be one of $1/\alpha$ or $1/\beta$ (which you can also see, by assuming there is a limit $1/L$ and setting $\displaystyle 1/L = \frac{1}{1+1/L}$).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-20T14:51:36.180" LastActivityDate="2010-10-20T14:51:36.180" />
  <row Id="7298" PostTypeId="2" ParentId="7296" CreationDate="2010-10-20T14:51:44.417" Score="2" ViewCount="0" Body="&lt;p&gt;Fractional derivatives can be used to establish connections between various special functions.  The book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387488065&quot; rel=&quot;nofollow&quot;&gt;An Atlas of Functions&lt;/a&gt; makes heavy use of this, especially derivatives of order 1/2 and -1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the &lt;em&gt;existence&lt;/em&gt; of fractional derivatives is related to the convergence of Fourier transforms.  For example, if a function has a 1/2 a derivative that means you can multiply its Fourier transform by $x^{1/2}$ and it is still in $L^2$. But I haven't seen much use in actually computing fractional derivatives, only knowing that they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a somewhat related note, see my answer to a &lt;a href=&quot;http://mathoverflow.net/questions/28147/why-do-we-care-about-lp-spaces-besides-p-1-p-2-and-p-infinity/28149#28149&quot; rel=&quot;nofollow&quot;&gt;question on Math Overflow&lt;/a&gt; related to Sobolev spaces.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-10-20T14:58:25.197" LastActivityDate="2010-10-20T14:58:25.197" />
  <row Id="7299" PostTypeId="2" ParentId="7287" CreationDate="2010-10-20T14:53:34.950" Score="1" ViewCount="0" Body="&lt;p&gt;I think your proof i fine. Here is pretty much the same thing as you did, but with less formulas. Write $f(z)= B(z)S(Z)O(z)$, where $B$ is the Blaschke factor, $S$ is the singular inner function of $f$ and where $O$ is the outer part of $f$. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First note that $1/O$ is outer (because $\log\tilde{f}\in L^1$ in your notation, then $\log(1/\tilde{f})=-\log\tilde{f}\in L^1$ too)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since $1/f$ is analytic in the disc, we get $B(z)=1$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S/f=1/O$, is outer so $1/f=(1/S)(1/O)$ is the unique inner outer factorization.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$|S|\le 1$ and $|1/S|\le 1$ means $|S|=1$, so by open mapping $S$ is a unimodular constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-10-20T16:03:23.447" LastActivityDate="2010-10-20T16:03:23.447" CommentCount="3" />
  <row Id="7300" PostTypeId="2" ParentId="7296" CreationDate="2010-10-20T15:00:43.870" Score="1" ViewCount="0" Body="&lt;p&gt;I wouldn't say there are results that can only be obtained through differintegration. It only happens that there are problems whose solutions look neater when we bring in the machinery of differintegrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/bookseries/00765392&quot; rel=&quot;nofollow&quot;&gt;Spanier and Oldham&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471588849&quot; rel=&quot;nofollow&quot;&gt;Miller and Ross&lt;/a&gt; remain useful references on the applications of differintegration. The first reference has a chapter on how certain diffusion problems have a neater formulation when differintegrals are used. For the second reference, the application that jumped out at me was Abel's solution to the so-called &lt;a href=&quot;http://mathworld.wolfram.com/TautochroneProblem.html&quot; rel=&quot;nofollow&quot;&gt;tautochrone problem&lt;/a&gt;: finding the curve such that the time needed for a particle to descend from a given position to the bottom of the curve (assuming there is no friction) is &lt;em&gt;independent&lt;/em&gt; of position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though Huygens and other mathematicians have already obtained this solution long before Abel, he decided to use an integral equation formulation that can then be solved with the help of differintegration. In particular, he arrived at the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{2g}T=\int_0^y\frac{s^{\prime}(\eta)}{\sqrt{y-\eta}}\mathrm{d}\eta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which when reformulated as a differintegral is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\frac{2g}{\pi}}T=\frac{\mathrm{d}^{-\frac12}}{\mathrm{d}y^{-\frac12}}s^{\prime}(y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't spoil the rest of the solution; I'd suggest that you read Miller and Ross if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-20T15:00:43.870" />
  <row Id="7301" PostTypeId="1" AcceptedAnswerId="7312" CreationDate="2010-10-20T15:29:00.573" Score="12" ViewCount="467" Body="&lt;p&gt;Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;fourier transform&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;taylor series&lt;/a&gt; are means to represent functions in a different form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;br&gt;&#xA;What is the connection between these two? Is there a way to get from one to the other (and back again)? Is there an overall, connceting (geometric?) intuition?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T02:07:10.163" Title="Connection between fourier transform and taylor series" Tags="&lt;intuition&gt;&lt;fourier-analysis&gt;&lt;taylor-expansion&gt;&lt;integral-transforms&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="10" />
  <row Id="7302" PostTypeId="1" AcceptedAnswerId="7306" CreationDate="2010-10-20T15:30:58.647" Score="4" ViewCount="156" Body="&lt;p&gt;The Wronskian lets us determine if a set of functions (possibly the solutions to a differential equation) are linearly dependent or not. But, for every example in the book, it is very obvious if one of the functions is a linear combination of the others. The examples in the book use 3-5 functions. What would be an example of a small number of functions where this isn't obvious?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the application of the Wronskian mostly to deal with large sets of functions... where the sheer number makes it hard to tell if they are dependent or not?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-10-20T18:53:03.183" Title="Example of functions where linear dependence isn't obvious." Tags="&lt;linear-algebra&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7303" PostTypeId="1" CreationDate="2010-10-20T15:32:26.977" Score="2" ViewCount="61" Body="&lt;p&gt;After some searching I found:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangulation&quot; rel=&quot;nofollow&quot;&gt;Triangulation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Using theta and &quot;b&quot; to find out the height of a building.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1112" LastActivityDate="2010-10-20T15:36:05.963" Title="What are the uses of trigonometry in architecture?" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7304" PostTypeId="2" ParentId="7303" CreationDate="2010-10-20T15:36:05.963" Score="3" ViewCount="0" Body="&lt;p&gt;Too many to name. Anything that involves the study of the measure of triangles is trigonometry. The Egyptians used Pythagorean triplets to produce right angles using the &quot;rope stretching method.&quot; (a circular rope with 12 knots. stretch so that you have 3, 4, 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it has played a role as long as architecture has been architecture.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-10-20T15:36:05.963" />
  <row Id="7305" PostTypeId="2" ParentId="7301" CreationDate="2010-10-20T15:43:04.507" Score="11" ViewCount="0" Body="&lt;p&gt;A holomorphic function in an annulus containing the unit circle has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Laurent_series&quot; rel=&quot;nofollow&quot;&gt;Laurent series&lt;/a&gt; about zero which generalizes the Taylor series of a holomorphic function in a neighborhood of zero.  When restricted to the unit circle, this Laurent series gives a Fourier series of the corresponding periodic function.  (This explains the connection between the Cauchy integral formula and the integral defining the coefficients of a Fourier series.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's worth mentioning that the Fourier transform is much more general than this and applies in a broad range of contexts.  I don't know that there's a short, simple answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I guess it's also worth talking about intuition.  One intuition for the Taylor series of a function $f(x)$ at at a point is that its coefficients describe the displacement, velocity, acceleration, &lt;a href=&quot;http://en.wikipedia.org/wiki/Jerk_(physics)&quot; rel=&quot;nofollow&quot;&gt;jerk&lt;/a&gt;, and so forth of a particle which is at location $f(t)$ at time $t$.  And one intuition for the Fourier series of a periodic function $f(x)$ is that it describes the decomposition of $f(x)$ into pure tones of various frequencies.  In other words, a periodic function is like a chord, and its Fourier series describes the notes in that chord.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The connection between the two provided by the Cauchy integral formula is therefore quite remarkable; one takes an integral of $f$ over the unit circle and it tells you information about the behavior of $f$ at the &lt;em&gt;origin&lt;/em&gt;.  But this is more a magic property of holomorphic functions than anything else.  One intuition to have &lt;em&gt;here&lt;/em&gt; is that a holomorphic function describes, for example, the flow of some ideal fluid, and integrating over the circle gives you information about &quot;sources&quot; and &quot;sinks&quot; of that flow within the circle.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T21:07:13.843" LastActivityDate="2010-10-20T21:07:13.843" CommentCount="1" />
  <row Id="7306" PostTypeId="2" ParentId="7302" CreationDate="2010-10-20T15:43:10.893" Score="5" ViewCount="0" Body="&lt;p&gt;It can be concealed in various ways.  If the functions are expressed as trig functions the dependence gets hidden easily.  Think of $sin^2(x)$ and $cos(2x)$.  As the functions get messier it gets easier to hide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the Wronskian is that the functions must be sufficiently differentiable and you need to be able to calculate it.  Think of the indicator function on the rationals and the indicator function on the irrationals.  These are dependent, but the Wronskian won't help.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T15:43:10.893" CommentCount="6" />
  <row Id="7307" PostTypeId="1" AcceptedAnswerId="7311" CreationDate="2010-10-20T15:45:14.610" Score="2" ViewCount="114" Body="&lt;p&gt;Perhaps this has a simple answer, but I don't know (I wouldn't be asking if I did). Every set of outer measure zero is Lebesgue measurable with Lebesgue measure zero. Is the converse true? That is, if a set has Lebesgue measure zero, does it necessarily have outer measure zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I suppose I should specify that I'm thinking in $\mathbb{R}^n$. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-20T16:26:50.297" Title="Concerning sets of (Lebesgue) measure zero." Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7308" PostTypeId="2" ParentId="7302" CreationDate="2010-10-20T15:49:04.777" Score="5" ViewCount="0" Body="&lt;p&gt;The cute application of the Wronskian I like to keep in mind is that the invertibility of the Vandermonde matrix implies that the set $\{ e^{\lambda z} : \lambda \in \mathbb{C} \}$ is linearly independent.  The Wronskian also shows up, for example, in the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Variation_of_parameters&quot; rel=&quot;nofollow&quot;&gt;variation of parameters&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T15:49:04.777" CommentCount="2" />
  <row Id="7309" PostTypeId="2" ParentId="7301" CreationDate="2010-10-20T15:49:30.457" Score="9" ViewCount="0" Body="&lt;p&gt;There is a big difference between the Taylor series and Fourier transform-the Taylor series is a local approximation, while the Fourier transform uses information over a range of the variable.  The theorem Qiaochu mentions is very important in complex analysis and is one indication of how restrictive having a derivative in the complex plane is on functions.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T15:49:30.457" />
  <row Id="7310" PostTypeId="2" ParentId="7302" CreationDate="2010-10-20T15:50:05.827" Score="2" ViewCount="0" Body="&lt;p&gt;Have you tried to prove by hand (i.e., only using the definition of linear independence)  that $\sin \theta$ and $ \cos \theta$ are linearly independent? Of course, this can be done with the help of the Wronskian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about $e^{i\theta}$ and $ e^{i (\theta + \frac{\pi}{2})}$? This is geometrically clear, but: can you see the difference between linear independence over the real and complex numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. Just to add still a more elementary example: what about $\sin^2\theta $ and $\cos^2\theta$?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-20T18:53:03.183" LastActivityDate="2010-10-20T18:53:03.183" CommentCount="5" />
  <row Id="7311" PostTypeId="2" ParentId="7307" CreationDate="2010-10-20T15:51:49.773" Score="6" ViewCount="0" Body="&lt;p&gt;If $\mu^{*}$ is an outer measure, then we define the measurable sets in terms of $\mu^{*}$; the measure is then &lt;em&gt;defined&lt;/em&gt; to be the restriction of the outer measure to the measurable sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit: if you have a $\sigma$-algebra and an outer measure $\mu^{*}$ on the algebra, then we say that a set $E$ is $\mu^{*}$-measurable if and only if for every $A$ in the $\sigma$-algebra,&#xA;$$\mu^{*}(A) = \mu^{*}(A\cap E) + \mu^{*}(A\cap E').$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Halmos says in his book &lt;strong&gt;Measure Theory&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;em&gt;It is rather difficutl to get an intuitive understanding of the meaning of $\mu^{*}$-measurability except through familiarity with its implications.&lt;/em&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you have the definiiton of $\mu^{*}$-measurable, then let $S$ be the set of all measurable sets, and you &lt;em&gt;define&lt;/em&gt; the measure $\mu$ on $S$ by $\mu(E) = \mu^{*}(E)$ for all $E\in S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, this holds for the Lebesgue measure: if $E$ is Lebesgue measurable, then the Lebesgue measure of $E$ is equal to the outer measure of $E$, because the Lebesgue measure of $E$ is &lt;em&gt;defined&lt;/em&gt; to be the outer measure of $E$. This holds for any Lebesgue measure, not just for measure $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of the theorem you state earlier is that having outer measure zero &lt;em&gt;implies&lt;/em&gt; that the set is measurable; that's the nontrivial part of the statement (not that the Lebesgue measure of the set will then be zero). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T16:26:50.297" LastActivityDate="2010-10-20T16:26:50.297" CommentCount="3" />
  <row Id="7312" PostTypeId="2" ParentId="7301" CreationDate="2010-10-20T15:52:19.800" Score="10" ViewCount="0" Body="&lt;p&gt;Assume that the Taylor expansion $f(x)=\sum_{k=0}^n a_k x^k$ is convergent for some $|x|&gt;1$. Then f can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^n a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^n a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the &quot;real&quot; function $x\mapsto f(x)$ we started with.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-20T15:52:19.800" />
  <row Id="7313" PostTypeId="2" ParentId="6791" CreationDate="2010-10-20T16:45:52.127" Score="3" ViewCount="0" Body="&lt;p&gt;This isn't a proof, but's too long for a comment, and may just be a &#xA;restatement of the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For contradiction, let k be any integer such that b(n) = k never holds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means k &amp;lt;= a(n) &amp;lt; k+1 never holds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a(n) can't be between k and k+1, the arctan of a(n) can't be either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there's an interval between -Pi/2 and -Pi/2 that a(n) may not &#xA;touch. Let's call in [c,d). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since tan is periodic, a(n) must also avoid m*Pi+[c,d) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Pi is irrational, m*Pi+[c,d) must contain an infinite number of &#xA;integers (pretty sure this is true, but I could be wrong). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there are an infinite number of intervals (approaching &#xA;Pi/2) that a(n) must avoid. Further, a(n) must avoid the arctans of &#xA;these intervals, and the arctans of those intervals, etc. The repeated &#xA;arctan intervals approach 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a(n) also has to avoid those intervals plus any multiple of Pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This non-proof actually applies to any interval a(n) misses, so, if true, shows that a(n) is dense in R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.  &lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2010-10-20T16:45:52.127" />
  <row Id="7314" PostTypeId="2" ParentId="7291" CreationDate="2010-10-20T16:52:06.263" Score="3" ViewCount="0" Body="&lt;p&gt;You might be interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Rota%E2%80%93Baxter_algebra&quot; rel=&quot;nofollow&quot;&gt;Rota-Baxter algebras&lt;/a&gt;.  They are, among other things, an abstraction of how indefinite integration by parts works.  If you are interested, &lt;a href=&quot;http://www.ams.org/bull/1969-75-02/S0002-9904-1969-12156-7/S0002-9904-1969-12156-7.pdf&quot; rel=&quot;nofollow&quot;&gt;Rota's original paper&lt;/a&gt; on the subject is a good read.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T16:52:06.263" CommentCount="1" />
  <row Id="7315" PostTypeId="1" AcceptedAnswerId="7316" CreationDate="2010-10-20T16:57:00.573" Score="5" ViewCount="108" Body="&lt;p&gt;So here is the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solved for a in terms of x:&lt;br&gt;&#xA;$$a^{x} = 10^{2x + 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried:&lt;br&gt;&#xA;$\displaystyle x \cdot \log(a) = (2x+1) \cdot \log\;10 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{x}{2x + 1} = \frac{\log\;10} {\log\;a} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not going in the right direction, the answer according to the book is:&lt;br&gt;&#xA;$$ \frac{1} {\log\;a - 2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Excuse the 'power' tag for this question, there is no logarithm tag&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-20T17:17:41.097" LastActivityDate="2010-10-20T17:55:13.893" Title="Solving $a^{x} = 10^{2x + 1}$" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="7316" PostTypeId="2" ParentId="7315" CreationDate="2010-10-20T17:00:40.740" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: The answer is using $\log_{10}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-20T17:00:40.740" CommunityOwnedDate="2010-10-20T17:01:01.110" CommentCount="14" />
  <row Id="7317" PostTypeId="1" AcceptedAnswerId="7327" CreationDate="2010-10-20T17:11:33.780" Score="3" ViewCount="823" Body="&lt;p&gt;Assume I have only basic math knowledge, what specific areas of math would I need to learn in order to understand the following webpage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Black%E2%80%93Scholes&quot; rel=&quot;nofollow&quot;&gt;Black-Scholes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="2648" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-21T00:07:02.507" LastActivityDate="2010-10-21T15:20:10.657" Title="Understanding Black-Scholes" Tags="&lt;probability&gt;&lt;applications&gt;&lt;stochastic-processes&gt;&lt;finance&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="7318" PostTypeId="2" ParentId="7315" CreationDate="2010-10-20T17:24:06.590" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\ $  Putting $\rm\ a = 10^{\:b}\ $ yields $\rm\ x = 1/(b - 2)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-20T17:54:35.253" LastActivityDate="2010-10-20T17:54:35.253" CommentCount="2" />
  <row Id="7320" PostTypeId="1" AcceptedAnswerId="7333" CreationDate="2010-10-20T17:39:56.413" Score="2" ViewCount="89" Body="&lt;p&gt;How to show that  $$\text{g.c.d} \Biggl( {m \choose k} , {m+1 \choose k } , {m+2 \choose k}, \cdots, {m+k \choose k} \Biggr)=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m,k \in \mathbb{N}$ and $m \geq k $.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-20T17:58:09.693" LastActivityDate="2010-10-20T18:48:43.627" Title="To show that $\text{g.c.d} \Bigl( {m \choose k} , {m+1 \choose k } , {m+2 \choose k}, \cdots, {m+k \choose k} \Bigr)=1$" Tags="&lt;number-theory&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7321" PostTypeId="2" ParentId="7210" CreationDate="2010-10-20T17:49:49.660" Score="3" ViewCount="0" Body="&lt;p&gt;The question asked &quot;where we might meet&quot; total boundedness.   One situation where you do &lt;em&gt;not&lt;/em&gt; meet it is classical analysis or geometry in finite-dimensional spaces; in $R^n$, bounded and totally bounded are equivalent concepts.   So the intuition and conditions for appearance are necessarily more subtle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically, a space fails to be totally bounded if and only if it contains an infinite set of points with all pairwise distances at least $d$ for some $d &gt; 0$.  This is just the negation of the definition of totally bounded: if for some $\epsilon &gt; 0$ there is no finite set of points whose $\epsilon$-neighborhood covers the space, then we can build an infinite set of mutually $\epsilon$-separated points by placing in the set  any point $p_1$ and for $n&gt;1$ inductively adding to the set any point $p_n$ not inside the $\epsilon$-neighborhood of the preceding points.  For finite-dimensional spaces this can happen only for a sequence of points that escapes to infinity.  But in infinite dimensions an infinite set of $d$-separated points can be packed into a ball of finite radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boundedness and total boundedness are both properties of the completion of the space.  A metric space has either boundedness property if and only if its completion does. Considering complete metric spaces is convenient because total boundedness (for a complete metric space) is equivalent to the space being compact.  So one might want a topological characterization of the difference between bounded and totally bounded, for complete spaces.  However, any metric space can be converted into a bounded one (without affecting total boundedness, completeness, or the topology) by replacing the metric $d$ with $d/(1+d)$ or some other bounded monotonic function that preserves the property of being a metric.  So topological characterization suitable for understanding the difference between bounded and totally bounded (complete) spaces is more elusive. Maybe one should consider uniform spaces instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the question of situations where total boundedness appears, there are at least two:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As far as total boundedness requires infinite-dimensionality and a metric, it comes up naturally in functional analysis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It also appears in so-called &quot;constructive analysis&quot;, i.e., analysis using direct and generally negation-free arguments, as in Errett Bishop's book.   The concept of total boundedness first arose from a (classical) logical analysis of the Heine-Borel theorem and what was required to extend it to general metric spaces.   In a situation where nonconstructive arguments, such as determining whether a given sequence has infinitely many positive terms, are not present, spaces like the real numbers are not all that different from general metric spaces, and an analysis of which hypotheses were &quot;really at work&quot; in the classical proofs is useful for building the theory in the more general and direct fashion that Bishop advocated.  So where boundedness might be used in a finite-dimensional classical argument, one might use total boundedness in the constructive one.  This is similar in flavor to the question above in the comments, of what characterizations are true without the axiom of choice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-21T21:34:00.283" LastActivityDate="2010-10-21T21:34:00.283" CommentCount="4" />
  <row Id="7322" PostTypeId="1" AcceptedAnswerId="7375" CreationDate="2010-10-20T17:50:07.480" Score="2" ViewCount="61" Body="&lt;p&gt;We can easily see that if a sequence of inner functions converges then certainly the limit is still inner (using the fact that a countable union of measure zero sets still has measure zero). This implies that the set of inner functions is closed in $H^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is the set of the &lt;em&gt;singular&lt;/em&gt; inner functions closed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that a singular inner function has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(z) = K \text{exp} \left ( -\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i \theta} + z}{e^{i  \theta} - z} \, d\mu(\theta) \right )$$&#xA;where $\mu$ is a positive Borel measure singular with respect to the Lebesgue measure and $K$ is a constant of modulus $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if we take $S_n \to S$ in $H^2$ then $S_n \to S$ uniformly on compact sets. Clearly $S$ is inner because $S_n$ is inner, now we need to show it is singular. My question now is, does anyone have a hint how I show this? I know that if $(\mu_n)$ is a sequence of positive finite Borel measures, then by Riesz and Alaoglu we have that a subsequence converge to a finite positive Borel measure $\mu$. Can I use this?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-20T17:59:51.000" LastActivityDate="2010-10-21T01:08:58.330" Title="Singular inner functions." Tags="&lt;complex-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="7323" PostTypeId="2" ParentId="7315" CreationDate="2010-10-20T17:55:13.893" Score="0" ViewCount="0" Body="&lt;p&gt;HINT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you can find useful to look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithm#Change_of_base&quot; rel=&quot;nofollow&quot;&gt;Logarithm - Change of base&lt;/a&gt;, after solving your equation $\displaystyle \frac{x}{2x+1}=\frac{\text{log} 10}{\text{log}\thinspace a}$. You should finish with something like $x = \displaystyle \frac{1}{\frac{\displaystyle \text{log} \thinspace a}{\displaystyle \text{log} 10}-2}$&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-20T17:55:13.893" />
  <row Id="7324" PostTypeId="1" AcceptedAnswerId="7326" CreationDate="2010-10-20T17:58:43.577" Score="2" ViewCount="79" Body="&lt;p&gt;Suppose $f_{X}(x) = xe^{-x^2/2}$ for $x&gt;0$ and $Y = \ln X$. Find the density function for $Y$. So we want to find $P(Y \leq y)$. This is the same thing as $P(\ln X \leq y)$ or $P(X \leq e^{y})$. Thus $f_{Y}(y) = f_{X}(e^y)$? Or does $f_{Y}(y) = F_{X}(e^y)$ since $P(X \leq x) = F_{X}(x)$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-20T18:05:53.110" Title="Transformation of Random Variables" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="7325" PostTypeId="2" ParentId="7317" CreationDate="2010-10-20T18:01:28.877" Score="0" ViewCount="0" Body="&lt;p&gt;I found &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521497892&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; quite useful.  But it does take the background listed.  The basic result that the equation is the same as one dimensional heat flow and the graphs of roughly how the value flows are not too tough.  There are also free calculators on the web that let you play with it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T18:01:28.877" />
  <row Id="7326" PostTypeId="2" ParentId="7324" CreationDate="2010-10-20T18:05:53.110" Score="2" ViewCount="0" Body="&lt;p&gt;You're correct up to the point where you have $P(Y \leq y) = P(\ln X \leq y) = P(X \leq e^y)$.  The correct next step, though, is that $F_Y(y) = F_X(e^y)$.  To obtain $f_Y(y)$, differentiate both sides of this equation with respect to $y$ (and don't forget to use Leibniz's rule on the right-hand side). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In case you don't remember, Leibniz's rule is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part&quot; rel=&quot;nofollow&quot;&gt;first part of the Fundamental Theorem of Calculus&lt;/a&gt; combined with the chain rule.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-20T18:05:53.110" CommentCount="5" />
  <row Id="7327" PostTypeId="2" ParentId="7317" CreationDate="2010-10-20T18:15:02.630" Score="12" ViewCount="0" Body="&lt;p&gt;The standard low technology argument for Black-Scholes (the famous &quot;binomial tree&quot;) requires only basic material, though there is also a standard medium technology approach using stochastic calculus (informally) and an advanced approach using the rigorous mathematical apparatus of stochastic processes, Brownian motion, and diffusion equations.  I think summation of a finite geometric series, very basic probability, and knowledge of how to take a limit of the binomial distribution to get the Gaussian normal distribution, is enough for the binary tree approach.   This is in almost every elementary text on options pricing, such as the ones used in business schools or quant basic training, where by elementary I mean not starting immediately from stochastic calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This relatively simple derivation applies to the Black-Scholes &lt;em&gt;formula&lt;/em&gt; for pricing the simplest types of options and, in some developments of the theory, also to the Black-Scholes &lt;em&gt;model&lt;/em&gt; giving a more general formalism for pricing arbitrary European options as expected destinations of a random walk.  The binomial tree is not ordinarily presented as a method for producing the Black-Scholes &lt;em&gt;partial differential equation&lt;/em&gt; satisfied by prices in their model, although in theory it could do that.  The differential equation is usually arrived at by a simple heuristic argument using stochastic calculus (given in all finance books that introduce stochastic calculus).  A fully rigorous derivation is quite complicated and does require a lot of mathematical background, but this is not necessary for understanding the formula and its use in finance.  An elementary mathematical derivation of the differential equation not explicitly using the binomial tree was posted at Terry Tao's webpage, though it does not dwell on financial intuitions or interpretations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can be discretized, as pricing of an option to buy a security whose value goes up and down in an (exponentiated) random walk, where there are a finite number of time steps and at each step the price is multiplied by $r$ or $1/r$ with a constant probability of up- or down- motion in the price at each step.  Taking a limit you get the Black-Scholes formula.  The key point is to consider a single time step and show that the price is determined by the relationship between the payoffs for the security and for the option.   The rest is a matter of algebraically propagating this known answer back up through the binary tree (the $2^n$ possible paths for the price of the security between time 0 and time $n$) until you reach the root.  This gives the answer for the finite problem, then taking the continuous limit of the problem &quot;derives&quot; Black-Scholes for the continuous time case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the continuous time theory makes sense mathematically involves more than just suggesting that such a theory might exist as a limit of a well-defined discrete theory.  For that you need to construct the limit theory directly, which requires the machinery of stochastic analysis. To understand the limitations of the finite or continuous theory requires knowledge of finance in addition to the mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[References added:  low-tech approach (Cox-Ross-Rubinstein binomial trees): &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_options_pricing_model&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_options_pricing_model&lt;/a&gt;  ; medium-tech is in most introductory books on derivatives pricing that cover stochastic calculus -- Hull and its competitors;  high-tech is in books on stochastic PDE such as Oksendal or advanced finance books like Shreve.  ]&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-21T01:22:46.387" LastActivityDate="2010-10-21T01:22:46.387" CommentCount="6" />
  <row Id="7328" PostTypeId="1" AcceptedAnswerId="7330" CreationDate="2010-10-20T18:15:11.600" Score="3" ViewCount="51" Body="&lt;p&gt;Let R be a ring with identity. An R-module M is Artinian if it satisfies the descending chain condition on submodules. What is an example of an Artinian module with a proper submodule that is not finitely generated?&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2010-10-20T19:06:33.980" LastActivityDate="2010-10-20T19:06:33.980" Title="Is there an Artinian module with infinitely generated proper submodule?" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="7329" PostTypeId="1" AcceptedAnswerId="7474" CreationDate="2010-10-20T18:24:00.413" Score="0" ViewCount="135" Body="&lt;p&gt;Two players put a dollar in a pot. They decide to throw a pair of dice alternatively. The first one who throws a total of $ 5$ on both dice wins the pot. How much should the player who starts add to the pot to make this a fair game? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my interpretation of this problem is that I first throw a dice and then the other person throws a dice afterwards. E.g. if I throw a 1, and if the other person throws a 4 wins the pot. So we are trying to find the expected payout?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-10-20T18:39:54.987" LastActivityDate="2010-10-22T00:53:42.100" Title="Fair Game and Dice" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="7330" PostTypeId="2" ParentId="7328" CreationDate="2010-10-20T18:24:36.707" Score="5" ViewCount="0" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you can find a concrete example of an Artinian module over the integers which is non-Noetherian and hence has a non-finitely generated submodule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example at hand, the module itself is not finitely generated, and there are also lots of non-finitely generated proper submodules.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-20T18:49:35.843" LastActivityDate="2010-10-20T18:49:35.843" CommentCount="9" />
  <row Id="7331" PostTypeId="1" AcceptedAnswerId="7336" CreationDate="2010-10-20T18:24:40.443" Score="1" ViewCount="131" Body="&lt;p&gt;Maple tells me that $\lim_{n \to \infty} n^2 \log^n(1 - \frac{c \log n}{n}) = 0$ for any constant $c$, but I can't find a way to prove it. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="2649" LastEditorDisplayName="" LastEditDate="2010-10-20T19:15:43.400" LastActivityDate="2010-10-20T19:15:43.400" Title="The limit of $n^2 \log^n(1 - \frac{c \log n}{n})$" Tags="&lt;limit&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7332" PostTypeId="2" ParentId="7296" CreationDate="2010-10-20T18:25:06.823" Score="1" ViewCount="0" Body="&lt;p&gt;Miller and Ross looks very nice indeed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as applications are concerned: &lt;a href=&quot;http://docs.google.com/viewer?a=v&amp;amp;q=cache:aWfxS6fj7WgJ:www.m-hikari.com/ams/forth/rahimyAMS49-52-2010.pdf+An+Introduction+to+the+Fractional+Calculus+and+Fractional+Differential+Equation&amp;amp;hl=en&amp;amp;gl=mx&amp;amp;pid=bl&amp;amp;srcid=ADGEESg0ul80vQSr4JNRvdX8z_LgojseL92CgSyOs19upoGguvZPFjwzTYwtXe-baJIxbksAo2AFdc9hZ2NNOzyS8nmDrkUQJTNZ5a3sCiZml0Cjj3cVCYJPOJKzsI8mQI7XXQSRUNa4&amp;amp;sig=AHIEtbRNtVslJk9HBm6E0JtV-1mAytq9BQ&quot; rel=&quot;nofollow&quot;&gt;Applications of Fractional Diﬀerential Equations&lt;/a&gt;. For some strange reason, the original link is not available. Instead, you can look at it by using Google Docs viewer.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-20T18:25:06.823" />
  <row Id="7333" PostTypeId="2" ParentId="7320" CreationDate="2010-10-20T18:48:43.627" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ if $\rm\ \ f_{k}(m+1)\ \ mod \ \ f_{k}(m)\ =\ f_{k-1}(m)\ \ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $\rm\ \ \gcd(f_k(m),\ f_k(m+1),\:\cdots,\:f_k(m+k)) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad =\ \ \gcd(f_k(m),\ \ f_{k-1}(m),\ \cdots,\ f_0(m))\ \ \ $ [$\rm\ = 1\ $ if $\rm\ f_0(m) = 1\:$ ]&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-20T18:48:43.627" />
  <row Id="7334" PostTypeId="1" CreationDate="2010-10-20T18:59:24.933" Score="0" ViewCount="191" Body="&lt;p&gt;I know that in general, $H_{n}(X)$ counts the number of $n$-cycles that are not $n$-boundaries of a simplicial complex $X$. So for the sphere, $H_{0}(X) \cong \mathbb{Z}$ since it is connected. Also $H_{n}(X) = 0$ for all $n&gt;0$ (e.g. all $1$-cycles are $1$-boundaries, etc..). How do you use this geometric interpretation to deduce that $H_{1}(X) = \mathbb{Z} \times \mathbb{Z}_2$ where $X$ is the Klein bottle? This doesn't seem to correspond to the Betti number. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-20T19:17:42.607" LastActivityDate="2010-10-20T19:42:28.680" Title="Homology of the Klein Bottle" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7335" PostTypeId="2" ParentId="7331" CreationDate="2010-10-20T19:00:40.203" Score="3" ViewCount="0" Body="&lt;p&gt;The Taylor series expansion $- \log (1-x) = \sum_{k \ge 1} \frac{x^k}{k}$ shows that $\log(1 - x) = O(x)$ for small $x$.  So $\log \left( 1 - \frac{c \log n}{n} \right) = O \left( \frac{\log n}{n} \right)$, hence the absolute value of the expression in question is bounded by a constant times $\frac{(\log n)^n}{n^{n-2}}$ which rather clearly goes to zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good example of why it's more flexible to use big-O notation than to deal directly with the definition of a limit.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-20T19:00:40.203" CommentCount="1" />
  <row Id="7336" PostTypeId="2" ParentId="7331" CreationDate="2010-10-20T19:05:07.463" Score="6" ViewCount="0" Body="&lt;p&gt;Well&#xA;$$\log\left(1-c\frac{\log n}{n}\right)=-c\frac{\log n}{n}&#xA;+O\left(\frac{(\log n)^2}{n^2}\right)$$&#xA;and so&#xA;$$n^2\log\left(1-c\frac{\log n}{n}\right)^n=(-c)^n\frac{(\log n)^n}{n^{n-2}}&#xA;\left(1+O\left(\frac{\log n}{n}\right)\right)^n.$$&#xA;Now&#xA;$$n\log\left(1+O\left(\frac{\log n}{n}\right)\right)=O(\log n)$$&#xA;so that&#xA;$$\left(1+O\left(\frac{\log n}{n}\right)\right)^n$$&#xA;grows as at most a polynomial in $n$. The $n^n$ in the denominator&#xA;will swamp everything else...&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-20T19:05:07.463" />
  <row Id="7337" PostTypeId="2" ParentId="7329" CreationDate="2010-10-20T19:14:21.890" Score="1" ViewCount="0" Body="&lt;p&gt;First, find the probability &lt;em&gt;p&lt;/em&gt; that a 5 is thrown and define &lt;em&gt;q&lt;/em&gt; = 1 - &lt;em&gt;p&lt;/em&gt;.  The chance That the first player wins is C=p+pq^2+pq^4+\cdots; calculate this value.  Now set C=(1-C)(1+x) and solve for the extra contribution &lt;em&gt;x&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-20T19:14:21.890" CommentCount="2" />
  <row Id="7338" PostTypeId="2" ParentId="7334" CreationDate="2010-10-20T19:23:02.437" Score="4" ViewCount="0" Body="&lt;p&gt;Think of the bottle as two Möbius strips glued along their edges. A closed loop that circles one of the bands (along its center, say) is not a boundary, but if you follow it twice it becomes one: you can think of it as a loop that follows precisely the edge of one of the original bands, so it's just the boundary of that band. This gives you the &quot;torsion&quot; piece of the homology, and it shows you that homology is a slightly more refined invariant than just counting the number of closed &quot;empty&quot; loops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: perhaps worth to point out that the group-theoretic term &quot;torsion&quot;, which corresponds to elements of finite order, is derived precisely from this geometric picture: the torsion element result from the &quot;twist&quot; in the band.&lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-10-20T19:23:02.437" />
  <row Id="7339" PostTypeId="2" ParentId="7334" CreationDate="2010-10-20T19:42:28.680" Score="3" ViewCount="0" Body="&lt;p&gt;For starters, the Betti numbers are not the homology.  The Betti number is what does the &quot;counting&quot; of cycles you mention, but there's more to the story.  The classification theorem for finitely generated abelian groups says that you can express any such group as a direct sum (equivalently, a product) of copies of $\mathbb{Z}$ and of $\mathbb{Z}_n$ for some values of $n$.  The Betti number is the number of copies of $\mathbb{Z}$ you get, i. e. the largest possible free subgroup (the rest is called the &lt;em&gt;torsion&lt;/em&gt; of the group).  If you've done homology with coefficients yet, a good way to get this is by taking coefficients in $\mathbb{Q}$, which annihilates all the elements of finite order and leaves you with a $\mathbb{Q}$-vector space whose dimension is the Betti number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But back to the main story.  Rather than finding torsion and Betti numbers individually, for simplicial complexes especially, I find it easier to just compute the homology via $H_n=\text{ker}(d_n)/\text{im}(d_{n+1})$.  Let's use the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.sagemath.org/doc/reference/_images/klein.png&quot; alt=&quot;Klein bottle Delta-complex&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a single $0$-simplex, which I'll call $v$; three $1$-simplices, of which the horizontal one will be $a$, the vertical one $b$, and the diagonal one $c$; and two $2$-simplices, of which the upper is $U$ and the lower $L$.  I'm orienting the edges in the direction of their arrows; the faces are oriented so that their boundaries are in the direction of two edges, rather than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $H_1$, you want the $1$-cycles mod those that bound a $2$-cell.  Since each edge is a cycle, the group of $1$-cycles is the free abelian group on $a,b,c$.  The boundary of $U$ is $a+b-c$ and that of $L$ is $c+a-b$.  So we're looking at $\langle a,b,c\rangle/\langle a+b-c,a-b+c\rangle$.  Let's simplify this to $\langle a+b-c,b,c\rangle/\langle a+b-c,2b-2c\rangle=\langle b,c\rangle/\langle 2b-2c\rangle$... and again to $\langle b-c,c\rangle/\langle 2b-2c\rangle$.  Setting $d=b-c$, this is just $\langle d\rangle/\langle 2d\rangle\oplus\langle c\rangle$, which is $\mathbb{Z}\oplus\mathbb{Z}_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're familiar at all with the theory of the fundamental group, there's a cool theorem that says that $H_1(X)$ is the abelianization of $\pi_1(X,x_0)$ for path-connected $X$.  This is another good way to visualize torsion in this specific instance.  I don't know about the Klein bottle, but $\mathbb{RP}^2$ has an element of order $2$ in its fundamental group given by (visualizing it as a disk with its boundary quotiented by the antipodal map) a path from a point on the boundary to the &quot;opposite&quot; point on the boundary.  In the case of $H_1$, you can imagine torsion as the &quot;homology version&quot; of this phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-20T19:42:28.680" />
  <row Id="7340" PostTypeId="1" CreationDate="2010-10-20T19:47:51.307" Score="2" ViewCount="107" Body="&lt;p&gt;Each person from a group of 3 people can choose his dish from a menu of 5 options. Knowing that each person eats only 1 dish what are the number of different orders the waiter can ask the chef?&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T19:49:42.700" LastActivityDate="2010-10-21T06:51:58.133" Title="basic combinatorics question" Tags="&lt;combinatorics&gt;" AnswerCount="4" />
  <row Id="7341" PostTypeId="1" CreationDate="2010-10-20T19:48:53.037" Score="1" ViewCount="62" Body="&lt;p&gt;I would like to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d^2y}{dx^2} - \frac{2}{y^2}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $y(0)=a$ and $y'(0)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $a$ is a known constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="2148" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T19:52:58.610" LastActivityDate="2010-10-20T20:04:02.470" Title="Nonlinear Ordinary differential equation" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7342" PostTypeId="1" CreationDate="2010-10-20T19:51:12.653" Score="1" ViewCount="108" Body="&lt;p&gt;i couldn't do the following question for hours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;minimize $\sum_{i=1}^{n}x_{i}^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t.  $\sum_{i=1}^{n}x_{i}=0$&#xA;      and $\sum_{i=1}^{n}x_{i}^{2}=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Lagrange multiplier rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2653" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-22T19:20:55.463" LastActivityDate="2010-10-23T19:05:59.867" Title="lagrange multiplier for more than 2 equality constraints" Tags="&lt;nonlinear-optimization&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="7343" PostTypeId="2" ParentId="7340" CreationDate="2010-10-20T19:52:08.720" Score="3" ViewCount="0" Body="&lt;p&gt;This is a problem of &lt;em&gt;combinations with repetitions&lt;/em&gt;. See for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;the discussion on Wikipedia&lt;/a&gt; (which lacks a proof right now). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-20T19:52:08.720" CommentCount="1" />
  <row Id="7344" PostTypeId="2" ParentId="7231" CreationDate="2010-10-20T19:54:48.807" Score="0" ViewCount="0" Body="&lt;p&gt;Rather I think the suggestion of $R(RCGM)'$ is a product of sums which -can- be read off directly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R$ for -all- the rows where R is negative&lt;/li&gt;&#xA;&lt;li&gt;$(RCGM)' = R'+C'+G'+M'$ for the single row where all are true &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2010-10-20T19:54:48.807" />
  <row Id="7345" PostTypeId="2" ParentId="7341" CreationDate="2010-10-20T20:04:02.470" Score="8" ViewCount="0" Body="&lt;p&gt;Change variables so that the independent variable is $y$ and the dependent on $p=y'$. Then $p\dot p=y''$ (dots stand for derivatives w.r.t. $y$, primes w.r.t. $x$) and the equation becomes $$p\dot p-\frac2{y^2}=0$$ or $$\frac{d}{dy}(\frac{p^2}2)=\frac2{y^2}.$$ This can be solved by integrating into a first order equation for $y$ as a function of $x$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-20T20:04:02.470" CommentCount="4" />
  <row Id="7346" PostTypeId="1" CreationDate="2010-10-20T20:04:34.370" Score="1" ViewCount="74" Body="&lt;p&gt;Reading &quot;Random number generation and Monte Carlo methods&quot; by James E. Gentle, I encountered an example concerning the Bays-Durham Shuffling algorithm whose result I can't reproduce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, such an example is written in &lt;a href=&quot;http://en.wikibooks.org/wiki/Statistics/Numerical_Methods/Random_Number_Generation#Bays-Durham_Shuffling_of_Uniform_Deviates&quot; rel=&quot;nofollow&quot;&gt;Random Number Generation - Durham Shuffling of Uniform Deviates&lt;/a&gt;. For the full text you can go to &lt;a href=&quot;http://books.google.com/books?id=ERSSDBDcYOIC&quot; rel=&quot;nofollow&quot;&gt;Random number generation and Monte Carlo methods&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I'm trying to reproduce &lt;a href=&quot;http://books.google.com/books?id=ERSSDBDcYOIC&amp;amp;pg=PA23&amp;amp;dq=bays%20durham%20shuffling%20of%20uniform%20deviates&quot; rel=&quot;nofollow&quot;&gt;Figure 1.7&lt;/a&gt; for the linear congruential generator of $x_{i} \equiv 3x_{i-1} \text{mod} 31$. In order to do so, I wrote the following piece of code in Mathematica: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Defines initial variables for the generator:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = 3; mod = 31; x = 9; n = 0; m = {}; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Produce a list with the initial set of numbers (In this case: 27, 19, 26, ... &amp;lt;&amp;lt;27&gt;&gt; ... 1, 3, 9):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;While[n &amp;lt; 30, AppendTo[m, Mod[a x, mod]]; x = Mod[a x, mod]; n++]; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The algorithm generates the first sequence of numbers in a slightly different way using the input value k, so in order to account for this, I had to use this While for the first sequence:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k = 8; i = 1; y = m[[k + 1]];&#xA;&#xA;While[i == 1, choose = Mod[y, 8] + 1; y = l[[choose]]; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Then, for the following sequences:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i = i + 1; m[[choose]] = k; k = k + 2];&#xA;&#xA;While[i &amp;lt; 10000, choose = Mod[y, 8] + 1; y = m[[choose]];&#xA;&#xA;i = i + 1; k++; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given that the congruence relation is congruence modulo 31, the sequence will repeat itself after 30 digits. To avoid looking for the non-existent element m[[31]], I added the conditional:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  If[k &amp;lt; 30, m[[choose]] = m[[k]], k = 1; m[[choose]] = m[[k]]]];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The book says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;'By continuing in this manner to yield 10,000 deviates and plotting the successive pairs, we get Figure 1.7'. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, as you can see, the procedure takes places 10,000 times. The output is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{4, 4, 21, 18, 2, 4, 21, 3, 13, 8, 24, 10, 30, 28, 22, 4, 12, 5, 15, 14, 11, 2, 6, 18, 23, 7, 21, 1, 3, 9}.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Plotting the successive pairs, I get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://miql.zxq.net/modulo.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, even if some of the points look like the ones in Figure 1.7, obviously those images are not the same. Actually, Figure 1.7 contains 122 points (assuming I didn't miscount), however, given the generator, I don't know how you can produce more than 30 points. Actually, the author writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... Remember that there are only 30 different values, however. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-21T02:05:07.187" LastActivityDate="2010-10-21T02:05:07.187" Title="Bays-Durham Shuffling" Tags="&lt;random&gt;" CommentCount="2" />
  <row Id="7347" PostTypeId="2" ParentId="7340" CreationDate="2010-10-20T20:07:45.447" Score="-2" ViewCount="0" Body="&lt;p&gt;For three people and items=5, n=items-2,&#xA;the tetrahedral (or triangular pyramidal) numbers: a(n) = C(n+2,people) = n(n+1)(n+2)/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two people this would become the triangular numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now find the general form for all people and items.&lt;/p&gt;&#xA;" OwnerUserId="2665" LastActivityDate="2010-10-20T20:07:45.447" CommentCount="1" />
  <row Id="7349" PostTypeId="2" ParentId="7294" CreationDate="2010-10-20T20:24:32.917" Score="1" ViewCount="0" Body="&lt;p&gt;The only requirement for a surface is that it be two dimensional. It can be connected or not connected. So it's OK mathematically to call your multi-component terrain object a &quot;surface&quot;, even though in everyday language people might think of it as several surfaces.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2010-10-20T20:24:32.917" />
  <row Id="7350" PostTypeId="1" CreationDate="2010-10-20T20:25:06.457" Score="0" ViewCount="76" Body="&lt;p&gt;What is a good introduction in gradient flows in metric spaces? I know the book Gradient flows: in metric spaces and in the space of probability measures by Luigi Ambrosio, Nicola Gigli and Giuseppe Savaré, but is too hard for an introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone make this a community wiki? I cannot find how you have to do that...&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-21T10:12:38.857" LastActivityDate="2010-10-21T10:12:38.857" Title="Gradient flows in metric spaces." Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;metric-spaces&gt;" CommentCount="6" />
  <row Id="7351" PostTypeId="1" AcceptedAnswerId="7352" CreationDate="2010-10-20T20:25:58.937" Score="0" ViewCount="96" Body="&lt;p&gt;So, I thought of something nerdy...I was making an algebra problem to disguise it so that to see the joke, you'd have to a bit of work, but I think that I might be doing something wrong, because neither &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha&lt;/a&gt; nor &lt;a href=&quot;http://myalgebra.com/&quot; rel=&quot;nofollow&quot;&gt;MyAlgebra&lt;/a&gt; can solve it...I fixed a few things that I had wrong initially, but I don't see anything wrong now...you're supposed to solve for i.&#xA;$$\begin{array}{rcl}&#xA;    (5u)^2-(i+2u)^2 &amp;amp;\gt&amp;amp; 20u^2+4u(i+2u)\\&#xA;    (5u+(i+2u))(5u-(i+2u))&amp;amp;\gt&amp;amp; 4u(5u+(i+2u))\\&#xA;    5u-(i+2u)&amp;amp;\gt&amp;amp; 4u\\&#xA;    3u-i&amp;amp;\gt&amp;amp; 0\\ &#xA;    i&amp;amp;\lt&amp;amp; 3u&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured that you could factor the left side and then divide &lt;code&gt;(5u+(i+2u))&lt;/code&gt; out of both sides, but maybe not?&lt;/p&gt;&#xA;" OwnerUserId="2654" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T20:39:32.727" LastActivityDate="2010-10-20T21:02:40.933" Title="Am I doing something wrong in making this inequality or do algebra tools just usually not factor?" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7352" PostTypeId="2" ParentId="7351" CreationDate="2010-10-20T20:33:26.010" Score="6" ViewCount="0" Body="&lt;p&gt;Your mistake lies in trying to cancel $5u +(i+2u)$ without being careful about whether it is positive, negative, or $0$. &lt;strong&gt;Added:&lt;/strong&gt; There is also a miscalculation in the third line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that &quot;cancelling factors&quot; is &lt;em&gt;really&lt;/em&gt; multiplying by the inverse. But when you multiply an inequality by a number, you have to be careful about whether the number is positive or negative: if you have $a\gt b$ and $c\gt 0$, then $ac\gt bc$; but if $c\lt 0$, then you have $ac\lt bc$ (multiplying by negative numbers reverses the inequality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you try cancelling $5u+(i+2u) = 7u+i$, by not reversing the inequality you are implicitly assuming that $7u+i\gt 0$; that is, that $-7u\lt i$. So in fact, your conclusion is that &lt;strong&gt;if&lt;/strong&gt; $-7u\lt i$ &lt;em&gt;and&lt;/em&gt; the inequality holds, then $i\lt 3u$. But what if $7u+i\lt 0$? Then when you divide both sides by $7u+i$ in order to cancel the factor, you must reverse the inequality, so you actually get $5u -(i+2u) \lt 4u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; At this point you made a slight mistake with the signs. The left hand side is $5u - (i+2u) = 3u-i$; you had $5u - i + 2u$ (missing the parenthesis) so you incorrectly got $7u - i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $7u+i \gt 0$, then you get $3u-i\gt 4u$, or $i\lt -u$. Thus, if $i\gt -7u$, then you must also have $i\lt -u$, or $-7u\lt i \lt -u$. Note that this can only occur if $u\gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $7u+i \lt 0$, then when you cancel that factor you get $3u-i \lt 4u$, or $i\gt -u$; so you have that both $i\lt -7u$ and $i\gt -u$ must hold. That is, $-u\lt i\lt -7u$. Note that this can only occur if $u\lt 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact you have two parts of the solution: you either have $u\gt 0$ and $-7u \lt i\lt -u$, or else you have $u\lt 0$ and $-u\lt i\lt-7u$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-20T21:02:40.933" LastActivityDate="2010-10-20T21:02:40.933" CommentCount="2" />
  <row Id="7353" PostTypeId="1" CreationDate="2010-10-20T21:18:50.490" Score="1" ViewCount="122" Body="&lt;p&gt;I believe I'm right in saying that an isometry of a convex figure will be convex. There are other automorphisms that have this property, for example &quot;stretching along one axis&quot; (e.g. $f(x,y) = (2x,y)$) won't transform a convex shape into a non-convex one or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other maps have this property?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-11-01T02:20:47.147" Title="What automorphisms of R^n preserve convexity?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7354" PostTypeId="1" CreationDate="2010-10-20T21:21:07.340" Score="1" ViewCount="322" Body="&lt;p&gt;Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thought about using the limit theorem or by comparison but am so stuck.  any pointers would be appreciated guys&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-21T08:36:28.480" LastActivityDate="2010-10-21T14:37:58.867" Title="Determine whether this series converges or diverges: $\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="7355" PostTypeId="2" ParentId="7354" CreationDate="2010-10-20T21:22:03.353" Score="2" ViewCount="0" Body="&lt;p&gt;Can you bound the series from above by one that you know converges?  The factorials grow very fast, so you should be able to.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-20T21:22:03.353" />
  <row Id="7356" PostTypeId="1" CreationDate="2010-10-20T21:23:09.400" Score="1" ViewCount="134" Body="&lt;p&gt;Given the coordinates of two rectangles on a coordinate plane, what would be the easiest way to find the coordinates of the intersecting rectangle of the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to do this programatically.&lt;/p&gt;&#xA;" OwnerUserId="2658" LastEditorUserId="2658" LastEditorDisplayName="" LastEditDate="2010-10-24T14:39:37.287" LastActivityDate="2010-10-24T14:39:37.287" Title="How to find rectangle intersection on a coordinate plane" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7357" PostTypeId="2" ParentId="7353" CreationDate="2010-10-20T21:24:05.770" Score="9" ViewCount="0" Body="&lt;p&gt;Any transformation that sends lines to lines will do, since convexity is defined by the existence of certain lines in a figure.  In particular, any transformation of the form $x \mapsto Ax + b$ where $A$ is a matrix and $b$ is a vector works.  (It's not clear to me what you mean by &quot;automorphism&quot; here - in what category? - so depending on your meaning these are the only ones, by which I mean that as an affine space these are the only automorphisms of $\mathbb{R}^n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't trust that explanation, one way to see this is to prove the result for translations (obvious) and then prove it for elementary matrices.  It should be geometrically obvious for all of them, as long as you have a reasonably good handle on &lt;a href=&quot;http://en.wikipedia.org/wiki/Shear_mapping&quot; rel=&quot;nofollow&quot;&gt;shearing&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In fact, I think this condition is necessary.  If a line $L$ is not sent to a line, draw a circle which is split in half by $L$.  Then the image of either one half or the other will not be convex.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-20T21:30:42.027" LastActivityDate="2010-10-20T21:30:42.027" CommentCount="4" />
  <row Id="7358" PostTypeId="2" ParentId="7354" CreationDate="2010-10-20T21:47:48.310" Score="5" ViewCount="0" Body="&lt;p&gt;The series you have is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{1}{3!} + \frac{1}{5!} \ldots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add the even factorial terms, you get an upper bound i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + \frac{1}{3!} + \frac{1}{5!} \ldots &amp;lt; \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}+ \frac{1}{5!} \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be written more compactly as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=0}^\infty \frac{1}{(2n+1)!} &amp;lt; \sum_{n=0}^\infty \frac{1}{n!} = e^1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the series converges.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-20T21:47:48.310" CommentCount="3" />
  <row Id="7359" PostTypeId="2" ParentId="7353" CreationDate="2010-10-20T22:07:45.043" Score="3" ViewCount="0" Body="&lt;p&gt;Search google books for &quot;fundamental theorem of geometry&quot; (it's in a book by M. Berger).&#xA;Using the fact that the only field automorphism of $\mathbb{R}$ is the identity, you get your result (for $n \geq 2$): the affine bijections.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-01T02:20:47.147" LastActivityDate="2010-11-01T02:20:47.147" />
  <row Id="7360" PostTypeId="2" ParentId="7356" CreationDate="2010-10-20T22:21:07.120" Score="0" ViewCount="0" Body="&lt;p&gt;In order to simplify the problem, I will assume that the rectangles are orientated &quot;straight up and down.&quot; By this I mean that each of the rectangles has an edge which is parallel to the x (or y) axis. Judging by the context, this should be a fair assumption. Let me know if this is not the case (in other words, what class is this for?) It might help to first graph the two rectangles whose coordinates are given. In the &quot;simplest&quot; case they will intersect in exactly two places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinates for some intersection will just be the point where &lt;em&gt;&quot;edge a&quot;&lt;/em&gt; of &lt;em&gt;&quot;rectangle 1&quot;&lt;/em&gt; intersects &lt;em&gt;&quot;edge x&quot;&lt;/em&gt; of &lt;em&gt;&quot;rectangle 2.&quot;&lt;/em&gt; That is, the intersection point will have the x value of (edge a or x), and the y value of the other edge (since these rectangles are parallel to the axis the x &lt;strong&gt;or&lt;/strong&gt; y value of an edge will be constant across that edge.) You know the values of these edges by the coordinates given for the corners of your original rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After drawing the rectangles this should be a fairly easy and straight-forward process to work through. If you are trying to write a computer program to do this for you it is slightly more complicated but follows the same logic. Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2010-10-20T22:21:07.120" CommentCount="1" />
  <row Id="7361" PostTypeId="1" AcceptedAnswerId="7365" CreationDate="2010-10-20T22:27:07.853" Score="6" ViewCount="92" Body="&lt;p&gt;Let $I$ be a finitely generated ideal of a commutative ring $R$. Assume every element of $I$ is a zero divisor. Does then exist a $x \neq 0$ in $R$ with $xI=0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true if $0$ is a decomposable ideal, for example if $R$ is noetherian. I wonder if we actually need this. Doesn't it sound plausible? The problem is that we cannot just multiply the elements which kill the generators of $I$, the product can vanish.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-20T23:15:08.443" Title="Ideal consisting of zero divisors " Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="7362" PostTypeId="2" ParentId="7356" CreationDate="2010-10-20T22:30:04.757" Score="1" ViewCount="0" Body="&lt;p&gt;You can think of an axis-aligned rectangle as the set of points $(x,y) \in \mathbb{R}^2$ such that $x_1 \le x \le x_2$ and $y_1 \le y \le y_2$. What can you say about the intersection of two such sets?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-20T22:30:04.757" />
  <row Id="7363" PostTypeId="2" ParentId="7317" CreationDate="2010-10-20T22:34:05.227" Score="1" ViewCount="0" Body="&lt;p&gt;Take a look at the syllabi for the first three &lt;a href=&quot;http://www.beanactuary.org/exams/exam_info.cfm&quot; rel=&quot;nofollow&quot;&gt;actuarial preliminary examinations&lt;/a&gt;. If you study for these tests, in order, you will learn everything you need in order to understand and apply the Black-Scholes formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each exam typically requires three hundred hours of study to do well. If you pace yourself, and you take one exam every six months, you will be studying 11 to 12 hours per week for one and a half years to &lt;i&gt;master&lt;/i&gt; this material. Of course, you could cover this material much faster, if you're more interested in breadth than depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first three exams are 2.5 to 3 hours each. They are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; &lt;a href=&quot;http://www.beanactuary.org/exams/exam1p.cfm&quot; rel=&quot;nofollow&quot;&gt;1/P - The Probability Exam&lt;/a&gt; - A thorough command of calculus and probability topics is assumed.&#xA;&lt;li&gt; &lt;a href=&quot;http://www.beanactuary.org/exams/exam2fm.cfm&quot; rel=&quot;nofollow&quot;&gt;2/FM - Financial Mathematics Exam&lt;/a&gt; - Interest theory (discrete and continuous) and an introduction to derivative securities.&#xA;&lt;li&gt; &lt;a href=&quot;http://www.beanactuary.org/exams/examam.cfm&quot; rel=&quot;nofollow&quot;&gt;3F/MFE - Financial Economics Exam&lt;/a&gt; - Interest rate models, rational valuation of derivative securities, simulation, and financial risk management techniques.&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The syllabi list the textbooks and papers you should study. Additionally, there are &lt;a href=&quot;http://www.actexmadriver.com/&quot; rel=&quot;nofollow&quot;&gt;study guides&lt;/a&gt; that cover each test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exam 3F/MFE covers &lt;b&gt;Black-Scholes&lt;/b&gt;. Specifically, you must be able to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Calculate the value of European and American options using the Black-Scholes option-pricing model.&#xA;&lt;li&gt;Interpret the option &lt;a href=&quot;http://en.wikipedia.org/wiki/Greeks_%28finance%29&quot; rel=&quot;nofollow&quot;&gt;Greeks&lt;/a&gt;.&#xA;&lt;li&gt;Explain the properties of a lognormal distribution and explain the Black-Scholes formula as a limited expected value for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lognormal&quot; rel=&quot;nofollow&quot;&gt;lognormal distribution&lt;/a&gt;.&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-20T22:34:05.227" />
  <row Id="7364" PostTypeId="2" ParentId="7291" CreationDate="2010-10-20T22:52:44.423" Score="4" ViewCount="0" Body="&lt;p&gt;If you find the Rota-Baxter algebra viewpoint of interest then an excellent entry point into the literature on differential-algebraic aspects is the the recent work of Guo, e.g. &lt;a href=&quot;http://arxiv.org/abs/math.RA/0703780&quot; rel=&quot;nofollow&quot;&gt;On differential Rota-Baxter algebras&lt;/a&gt; and &lt;a href=&quot;http://andromeda.rutgers.edu/~liguo/psfiles/guo6.pdf&quot; rel=&quot;nofollow&quot;&gt;Baxter algebras and differential algebras&lt;/a&gt;. See also other papers listed on his &lt;a href=&quot;http://andromeda.rutgers.edu/~liguo/&quot; rel=&quot;nofollow&quot;&gt;home page.&lt;/a&gt; The early papers don't focus so much on these aspects so I would not recommend reading them initially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you may find of interest this paper on classification of related operator identities.&lt;br&gt;&#xA;Freeman, J. M. On the classification of operator identities. Studies in Appl. Math. 51 (1972), 73-84.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-20T22:52:44.423" CommentCount="1" />
  <row Id="7365" PostTypeId="2" ParentId="7361" CreationDate="2010-10-20T23:09:31.460" Score="4" ViewCount="0" Body="&lt;p&gt;No, for a counterexample see Exercises 2-2-6,7 pp. 62-63 in Kaplansky: Commutative Rings (excerpted below). See also the discussion following Theorem 82 p.56.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MMRqu.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/mFafo.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-20T23:15:08.443" LastActivityDate="2010-10-20T23:15:08.443" CommentCount="2" />
  <row Id="7366" PostTypeId="2" ParentId="6791" CreationDate="2010-10-20T23:44:39.037" Score="10" ViewCount="0" Body="&lt;p&gt;I had made the same conjecture as Fekete, apparently around the same time -- mid-2007.  In 2008 I verified that the first twenty million terms do not include 319.  (I actually pushed the verification further, but I can't find the more recent records at the moment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because tan(x) - x = x^3/3 + O(x^5), the function spends a lot of its time in a small neighborhood around 0.  It escapes when it nears $\pi/2$ and quickly returns for many iterations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mostly-unexplained phenomenon presumably related to the above: there are long spans of small numbers followed by short, 'productive' spans with large numbers.  $\tan^k(1)$ is &quot;below 20 or so&quot; (according to a 2008 email I sent) for $360110\le k\le1392490$ but in the next 2000 numbers there are five which are above 20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More theory is needed!&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-01-06T15:11:25.920" LastActivityDate="2011-01-06T15:11:25.920" CommentCount="9" />
  <row Id="7367" PostTypeId="1" CreationDate="2010-10-20T23:58:24.050" Score="4" ViewCount="332" Body="&lt;p&gt;This question and the described solution are copied from a test-paper :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the equation $x^2$ + |x| - 6 = 0 analyze the four statements below for correctness.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;there is only one root&lt;/li&gt;&#xA;&lt;li&gt;sum of the roots is + 1&lt;/li&gt;&#xA;&lt;li&gt;sum of the roots is zero&lt;/li&gt;&#xA;&lt;li&gt;the product of the roots is +4&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answer : (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer Explanation :&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If x &gt; 0 |x| = x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given equation will be $x^2 + x - 6 = 0$⇒ x = 2,- 3 ⇒ x = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If x &amp;lt; 0 |x| = - x.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given equation will b e  $x^2$ - x - 6 = 0 ⇒ x = -2, 3 ⇒x = - 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of roots is 2 - 2 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have a doubt on the statment &quot;&lt;strong&gt;If x &amp;lt; 0 |x| = - x.&lt;/strong&gt;&quot; I think modulus means that |x| is always positive ?! Also I can see that (2) seems to be the correct option isn't ?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please post your views.&lt;/p&gt;&#xA;" OwnerUserId="2663" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-21T16:39:57.530" LastActivityDate="2010-10-21T16:39:57.530" Title="Exploring  the quadratic equation $x^2 + \lvert x\rvert  - 6 = 0$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="7368" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T00:01:12.203" Score="2" ViewCount="0" Body="&lt;p&gt;If x&amp;lt;0, then -x&gt;0, which means that the modulus is indeed positive.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-10-21T00:01:12.203" CommentCount="3" />
  <row Id="7369" PostTypeId="2" ParentId="7265" CreationDate="2010-10-21T00:01:42.607" Score="1" ViewCount="0" Body="&lt;p&gt;If the coefficients are naturals, there is no problem with rounding.  So just divide the polynomials and look at the middle term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather equivalently, you can evaluate it at a number of points and do Lagrange interpolation.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T00:01:42.607" CommentCount="1" />
  <row Id="7370" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T00:13:42.417" Score="6" ViewCount="0" Body="&lt;p&gt;I think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is  obviously a number, so he doesn't have to prove that the series for e converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you can prove the same by using the ratio test $\lim_{n \rightarrow \infty} \displaystyle |\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\displaystyle \frac{1}{(2n+1)!}$ and $a_{n+1}=\displaystyle \frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\lim_{n \rightarrow \infty} \displaystyle \frac{1}{(2n+3)(2n+2)}$ which is 0. According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_tests&quot; rel=&quot;nofollow&quot;&gt;ratio test&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If r &amp;lt; 1, then the series converges. If r &gt; 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the series converges.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-21T01:01:30.610" LastActivityDate="2010-10-21T01:01:30.610" CommentCount="2" />
  <row Id="7371" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T00:15:11.463" Score="1" ViewCount="0" Body="&lt;p&gt;The statement is saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = -x$ (for $x &amp;lt; 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and divide both sides by $x$ to give us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{|x|}{x} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us test a few values to make sure this holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -3$ and $|x| = 3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{|-3|}{-3} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{3}{-3} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1 = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;True.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -5$ and $|x| = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{5}{-5} = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-1 = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;True.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can verify this for all $x &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't believe me, here is a plot of $y = \frac{|x|}{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img716.imageshack.us/img716/4149/wolframalpha20101020192.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-a is the number such that a + (-a) = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;ie,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if a = 3, -a = -3, because $3 + (-3) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if a = -2, -a = 2, because $-2 + 2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can call 3 and -3 &quot;opposites&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite of -2 is 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite of 5 is -5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can translate the original statement then as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If x &amp;lt; 0, |x| is the opposite of x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's test this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = -4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x| = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-4 + 4 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|-4|$ (or, 4) is indeed the opposite of $-4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, the statement says that, if $x &amp;lt; 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x + |x| = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which we can verify as true for all values $x &amp;lt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't believe me, you can look at the plot of $y = x + |x|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img409.imageshack.us/img409/6834/wolframalpha20101020191.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-21T00:28:12.173" LastActivityDate="2010-10-21T00:28:12.173" CommentCount="3" />
  <row Id="7372" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T00:33:15.223" Score="20" ViewCount="0" Body="&lt;p&gt;$\rm\ 0\ =\ x^2 + |x| -6\ =\ (|x| - 2)\:(|x| + 3)\ \Rightarrow\ |x| = 2\ \Rightarrow\ x = \pm2 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-21T00:33:15.223" CommentCount="1" />
  <row Id="7375" PostTypeId="2" ParentId="7322" CreationDate="2010-10-21T01:08:58.330" Score="1" ViewCount="0" Body="&lt;p&gt;You could use &lt;a href=&quot;http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_%28complex_analysis%29&quot; rel=&quot;nofollow&quot;&gt;Hurwitz's theorem&lt;/a&gt;.  For some $R$ with $0\lt R\lt1$, $S$ has no zeros on $\partial B(0,R)$, and $R$ can be taken arbitrarily close to 1.  By Hurwitz's theorem, $S$ has the same number of zeros in $B(0,R)$ as $S_n$ for sufficiently large $n$.  Each $S_n$ is never zero in the open disk, so this shows that $S$ is never zero on $B(0,R)$.  Thus $S$ is never zero on the disk.  This means that $S$ has no Blaschke factor, hence it is singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know whether $(\mu_n)$ converges weak-$*$ to the singular measure of $S$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-21T01:08:58.330" />
  <row Id="7376" PostTypeId="1" AcceptedAnswerId="7509" CreationDate="2010-10-21T01:15:09.850" Score="11" ViewCount="272" Body="&lt;p&gt;Let $C$ be a convex subset of $\mathbb{R}^n$. I've been trying for hours to prove that $\dot{\overline{C}}=\dot{C}$. Somehow my intuition completely fails me. I found a proof in a textbook, but just got stuck on another statement the author considered obvious. Could someone please give a proof that uses little more than elementary linear algebra, topology, and the definition of a convex set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &#xA;The proof mentioned above is from &lt;a href=&quot;http://books.google.de/books?id=1F2B-8ap4wwC&amp;amp;pg=PA32&amp;amp;dq=convex+closure+%22same+boundary%22&amp;amp;hl=de&amp;amp;ei=K_i_TISsCsO54AaN-qS6DA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CDkQ6AEwBA#v=onepage&amp;amp;q=convex%20closure%20%22same%20boundary%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Blackwell and Girshick&lt;/a&gt;:&#xA;Let $y\in\dot{\overline{C}}$ and $T$ be a ball around $y$ contained in $\overline{C}$. Then $C\cap T$ has an inner point, as otherwise $C\cap T$ would be contained in a hyperplane and $\overline{C\cap T}=\overline{T}$ would be contained in the same hyperplane. The problematic statement is &quot;as otherwise $C\cap T$ would be contained in a hyperplane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing: I would be interested in a proof that doesn't use the theorem about separating a convex set from a point by a hyperplane, as I came across this problem in a proof of that very theorem (in the appendix of &lt;a href=&quot;http://books.google.de/books?id=cL-bZSOrqWoC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Stochastic Finance by Föllmer and Schied)&lt;/a&gt;. To be more precise, it occurs in the case of the point in question being in the boundary of $C$, when it is tacitly assumed, that it is also in the boundary of $\overline{C}$. I know this isn't strictly necessary, as I could use another proof, e.g. the one referred to by Mike, but now I'm curious.  &lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditorDisplayName="" LastEditDate="2010-10-21T09:25:40.213" LastActivityDate="2010-10-23T03:47:20.380" Title="Why does a convex set have the same interior points as its closure?" Tags="&lt;topology&gt;&lt;convex-analysis&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="7377" PostTypeId="1" AcceptedAnswerId="7381" CreationDate="2010-10-21T01:31:07.137" Score="10" ViewCount="440" Body="&lt;p&gt;Dear Math.so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, you guys are awesome, and I learn quite a bit just from reading the questions of others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, a friend asked me recently why large primes are important for data security, and I was unable to give him an answer with which I myself was satisfied.  Various wikipedia articles have mostly pointed out an embarrassing paucity in mathematical knowledge on my part, and since this happens to be a very math-related question (and not a programming-related question) I was hoping someone could shed some light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr: question reads as title.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2010-10-22T18:21:23.480" Title="Why are very large prime numbers important in cryptography?" Tags="&lt;prime-numbers&gt;&lt;cryptography&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" />
  <row Id="7378" PostTypeId="2" ParentId="7377" CreationDate="2010-10-21T01:37:45.000" Score="5" ViewCount="0" Body="&lt;p&gt;Some cryptographic algorithms use 2 very large primes (such as 128 bit long) and multiply them together.  The only way we know how to crack that is to try and find the only 2 factors that are available for that number (the 2 large primes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it turns out, it takes A LOT of computer power to be able to find those 2 factors.  Therefore, even if one tries, it'll take so long that the encryption is considered secure enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most numbers have many factors, but a product of 2 primes &lt;em&gt;only&lt;/em&gt; has the 2 primes as factors.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-10-21T07:08:37.677" LastActivityDate="2010-10-21T07:08:37.677" CommentCount="5" />
  <row Id="7379" PostTypeId="2" ParentId="7289" CreationDate="2010-10-21T01:41:21.367" Score="1" ViewCount="0" Body="&lt;p&gt;To gratuitously generalize the question, in $\mathbb{R}^n$ with $n\geq2$ there are continuum-sized subsets each of whose $n$-element subsets is linearly independent.  You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Vandermonde_matrix&quot; rel=&quot;nofollow&quot;&gt;Vandermonde matrices&lt;/a&gt; to see that $\{(1,x,x^2,\ldots,x^{n-1}):x\in\mathbb{R}\}$ is an example of this.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-21T01:46:58.110" LastActivityDate="2010-10-21T01:46:58.110" />
  <row Id="7380" PostTypeId="2" ParentId="7301" CreationDate="2010-10-21T02:07:10.163" Score="4" ViewCount="0" Body="&lt;p&gt;There is a analogy, more direct for fourier series. Both (Fourier series and Taylor series) are a descomposition of a function $f(x)$, which is represented as a linear combination of a (countable) set of functions.  The function results then fully specified by, instead of its values $f(x)$ for each $x$, a sequence of coefficients.  In this sense, both can be called a &lt;strong&gt;transform&lt;/strong&gt; ($f(x) \leftrightarrow \{ a_0, a_1, ...\}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Taylor series (around 0, for simplicity), the set of funtions is $\{1, x , x^2, x^3...\}$. For the Fourier series is $\{1, \sin(\omega x), \cos(\omega x), \sin(2 \omega x), \cos(2 \omega x) ...\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the Fourier series is one the many transformations that uses an orthonomal basis of functions. It is shown that, in that case, the coefficients are obtained by &quot;projecting&quot; $f(x)$ onto each basis function, which amounts to an inner product, which (in the real scalar case) amounts to an integral. This implies that the coefficients depends on a &lt;strong&gt;global&lt;/strong&gt; property of the function (over the full &quot;period&quot; of the function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Taylor series (which does &lt;strong&gt;not&lt;/strong&gt; use a orthonormal basis) is conceptually very different, in that the coeffients depends only in &lt;strong&gt;local&lt;/strong&gt; properties of the function, i.e., its behaviour in a neighbourhood  (it's derivatives).&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-21T02:07:10.163" />
  <row Id="7381" PostTypeId="2" ParentId="7377" CreationDate="2010-10-21T02:52:53.937" Score="19" ViewCount="0" Body="&lt;p&gt;There is a whole class of cryptographic/security systems which rely on what are called &quot;trap-door functions&quot;. The idea is that they are functions which are generally easy to compute, but for which finding the &lt;em&gt;inverse&lt;/em&gt; is very hard (here, &quot;easy&quot; and &quot;hard&quot; refer to how quickly we know how to do it), but such that if you have an extra piece of information, then finding the inverse is easy as well. Primes play a very important role in many such systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such example is the function that takes two integers and multiplies them together (something we can do very easily), versus the &quot;inverse&quot;, which is a function that takes an integer and gives you proper factors (given $n$, two numbers $p$ and $q$ such that $pq=n$ and $1\lt p,q\lt n$).  If $n$ is the product of two primes, then there is one and only one such pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_logarithm&quot; rel=&quot;nofollow&quot;&gt;discrete logarithm&lt;/a&gt;. To consider a simple example, look at the integers modulo, say, $7$. The integers between $1$ and $6$, inclusively, form a group under multiplication, and in fact every number between $1$ and $6$ is a power of $3$. The &quot;discrete logarithm problem&quot; would be, given a number $x$ between $1$ and $6$, to find a number $a$ such that $3^a$ equals $x$ modulo $7$. In this case, you can just try powers of $3$ until you hit the right answer. But if the modulo is very large, then this would take too much time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method for exchanging information over an open channel relies on the fact that we do not have very good methods of finding discrete logarithms in general, but we &lt;em&gt;do&lt;/em&gt; have very good methods for computing modular powers. The idea is: suppose you and I need to exchange information. We want to use some very secure cryptographic system that relies on a complicated key. But, how can we agree on a key? If we have some secure way of communicating so that when we agree on the key nobody will overhear us, then why bother with the entire exercise? We should just communicate using that secure way. So instead we need to communicate at a place where we can be overhead. How can we agree on a &lt;em&gt;secret&lt;/em&gt; key if everyone can hear us? Well, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie-Hellman_key_exchange&quot; rel=&quot;nofollow&quot;&gt;Diffie and Hellman&lt;/a&gt; proposed the following method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a very large prime $p$, and a number $r$ such that every number between $1$ and $p-1$ is a power of $r$ modulo $p$ (such numbers $r$ are known to exist for &lt;em&gt;every&lt;/em&gt; prime; they are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive roots&lt;/a&gt;). Everyone knows $p$ and everyone knows $r$. Then I pick a secret number $a$, and you pick a secret number $b$. I cannot tell you my secret number (it's &lt;em&gt;secret&lt;/em&gt;). But I tell you what $r^a \mod p$ is. Because computing modular powers is easy, I can do this computation easy enough; but because we don't know how to do discrete logarithms easily, we are hoping that nobody will be able to figure out $a$ just from knowing $r^a$... at least, not very quickly. Likewise, you tell me $r^b \mod p$. Now, you know $r^a$, &lt;em&gt;and&lt;/em&gt; you know what $b$ is, so you compute $(r^a)^b \mod p$. By the laws of exponent, you now know (secretly!) the number $r^{ab} \mod p$. I, on the other hand, know $r^b$ (because you told me that number) and I know what $a$ is. So &lt;em&gt;I&lt;/em&gt; compute $(r^b)^a\mod p$. But this is the same as $r^{ab} \mod p$. So now we both have a piece of information, namely the number $r^{ab}\mod p$.  This is going to be our &quot;secret key&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if someone can figure out either $a$ or $b$, then since they also know $r^a$ and $r^b$, they'll be able to figure out our secret key. We hope this is hard, but we certainly need $p$ to be very big: otherwise, they can just try all powers of $r$ until they hit the right one. We need the &quot;search space&quot; to be very big, so we need $p$ to be very big. &lt;strong&gt;Added:&lt;/strong&gt; As jug points out, having $p$ big is not sufficient. There are algorithms for computing discrete logarithms that are particularly good with certain kinds of primes, so we generally also require that $p$ satisfy some additional &quot;good&quot; properties relative to the cryptographic application. You generally want $p$ and $(p-1)/2$ to be both primes, for example. On the other hand, in practice one does not &lt;em&gt;really&lt;/em&gt; need $r$ to be a primitive root. Instead, it is enough that it generate a &quot;large&quot; subgroup of the multiplicative group, which one generally wants to be of prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: figuring out $a$ or $b$ is just &lt;em&gt;one&lt;/em&gt; way in which they could figure out our secret key $r^{ab}$, since everyone knows $p$, $r$, $r^a$, and $r^b$. It is &lt;em&gt;not&lt;/em&gt; known whether this is essentially the only way to break this &quot;key exchange&quot; method; the method really relies on whether one can figure out $r^{ab}$ from knowing $r$, $p$, $r^a$, and $r^b$; this is called the &lt;strong&gt;Diffie-Hellman problem&lt;/strong&gt;; the Diffie-Hellman problem is at most as hard as the Discrete Logarithm Problem, but we do not know if it is &lt;em&gt;just&lt;/em&gt; as hard (it could be easier); and we don't know just how hard the Discrete Logarithm Problem is, we just know that we don't have any easy ways of doing it yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So key exchange is one place where big primes are very important. (Diffie-Hellman is not the only way to do key exchanges). Another place where big primes play a big role is in &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA&quot; rel=&quot;nofollow&quot;&gt;RSA&lt;/a&gt; which is a cryptosystem that also relies on big primes (this time, two big primes $p$ and $q$, and we do arithmetic modulo $n=pq$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Might as well add a quick overview of RSA and how the primes come into play. Here, once again modular exponentiation is part of the process. This is an &quot;public key&quot; system: I will tell everyone how to send me secret messages, which hopefully only I can decode. (In Diffie-Hellman, we did not exchange a message; we agreed on a secret key that we will use with a separate system that requires a secret key; for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Advanced_Encryption_Standard&quot; rel=&quot;nofollow&quot;&gt;AES&lt;/a&gt;). I pick two large primes $p$ and $q$, and compute $n=pq$. I also pick a number $e$ that is relatively prime to $(p-1)(q-1)$ (I can do that because I know $p$ and $q$). Then I use the Euclidean algorithm, which is pretty quick, to find a $d$ such that $ed\equiv 1 \pmod{(p-1)(q-1)}$. Finally, I tell everyone what $n$ and $e$ are. If you want to send me a message, you first convert it to a number $M$ using some standard mechanism. Then you compute $M^e \mod n$, and you tell me what $M^e\mod n$ is. I will take $M^e$ and compute $(M^e)^d = M^{ed}\mod n$. Because $ed\equiv 1 \pmod{(p-1)(q-1)}$, then $M^{ed}\equiv M\pmod{n}$, so that is how I recover $M$. The security of the system relies in hoping that from knowing $n$ and $e$, it is difficult to figure out $d$ (it is easy if I know $p$ and $q$; this is why this is believed to be a &quot;trap-door function&quot; as described in the first paragraph). The problem is at most as hard as factoring $n$, because if you can factor $n$ then you can find $d$ the same way I did; it is not known if the problem of finding $M$ from $n$, $M^e$, and $e$  is at least as hard as factoring (it has been shown that some variants &lt;em&gt;are&lt;/em&gt; at least as hard as factoring), and again we don't know just how hard factoring &lt;em&gt;is&lt;/em&gt;.  But: because we know that if you can factor $n$ then you can read the message, then we want to make $n$ difficult to factor. It only has two factors, but you don't want them to be easy to find, so you want $p$ and $q$ to be large for sure. (Again, there are other conditions one usually puts on $e$, $p$, and $q$ to make sure that certain special attacks do not succeed easily, but at &lt;em&gt;least&lt;/em&gt; we need $p$ and $q$ to be very big).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-22T18:21:23.480" LastActivityDate="2010-10-22T18:21:23.480" CommentCount="15" />
  <row Id="7382" PostTypeId="1" AcceptedAnswerId="7387" CreationDate="2010-10-21T03:44:32.410" Score="2" ViewCount="54" Body="&lt;p&gt;Page 13 of &lt;a href=&quot;http://books.google.com/books?id=4JP7ofjHVh8C&amp;amp;lpg=PP1&amp;amp;ots=tWhNW0T4T4&amp;amp;dq=mikhalkin%20tropical%20algebraic%20geometry&amp;amp;pg=PA13#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Tropical Algebraic Geometry&lt;/a&gt; by Itenberg, Mikhalkin, and Shustin mentions 1-valent vertices, but I haven't been able to find a source that actually defines this term or managed to guess the definition myself.  Either a definition or reference would be greatly appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-21T04:36:16.827" Title="What are valent vertices?" Tags="&lt;terminology&gt;&lt;reference-request&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7383" PostTypeId="2" ParentId="7249" CreationDate="2010-10-21T04:15:25.823" Score="2" ViewCount="0" Body="&lt;p&gt;The physical justification is quite simple. The point of contact between the wheel and the road is instantaneously stationary. Since the wheel moves rigidly, the velocity of any point on the wheel is as though the wheel were rotating about the point of contact. In particular, the velocity of the axle is perpendicular to the line joining it to the point of contact. Since we require that the axle stays level, it velocity must be purely horizontal, so it has to be directly above the contact point.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-21T04:15:25.823" CommentCount="4" />
  <row Id="7384" PostTypeId="2" ParentId="7367" CreationDate="2010-10-21T04:19:33.227" Score="3" ViewCount="0" Body="&lt;p&gt;$f(x)=x^2+|x|-6$ is an even function—that is, $f(x)=f(-x)$ for all $x$, or the graph of $y=f(x)$ is symmetric over the y-axis—so if $f(c)=0$ then $f(-c)=0$, so the sum of the zeros of $f$ must be 0.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-21T04:19:33.227" CommentCount="5" />
  <row Id="7385" PostTypeId="2" ParentId="7376" CreationDate="2010-10-21T04:30:23.893" Score="2" ViewCount="0" Body="&lt;p&gt;One direction is straightforward, so I'll only give the proof in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in \dot{\bar{C}}$.  Then there exists an open ball $B_x$ containing $x$ such that $B_x \subset \bar{C}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x \not\in \dot{C}$.  But $x \in \bar{C}$, which means that $x$ is on the boundary of $C$.   Therefore, $B_x$ contains at least one point $z$ not in $C$.  Since $C$ is convex, there is a hyperplane that separates $z$ and $C$.  But $z \in \dot{\bar{C}}$.  Thus there exists another open ball $B_z$ containing $z$ such that $B_z \subset \bar{C}$.  Now, intersect $B_z$ with the open half-space (from the separating hyperplane) that does not contain $C$ to obtain set $D$.  We have $D \subset \bar{C}$ but $D \cap C = \emptyset$.  Thus $D$ is a subset of the boundary of $C$.  However, $D$ is open because it is the intersection of two open sets.  $D$ thus has the property that it is an open set containing boundary points of $C$ yet fails to contain any points of $C$.  Such a set cannot exist, and we have our contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proof does bring in convexity via the separating hyperplane theorem - the version with a convex set and a point not in the set - rather than via the definition of a convex set.  I'm not sure how to prove this without the separating hyperplane theorem (which is one of the fundamental properties of a convex set).  If you want a proof of the separating hyperplane theorem it can be found on page 47 of &lt;a href=&quot;http://books.google.com/books?id=VdMcrxUfeMUC&amp;amp;pg=PA46&amp;amp;lpg=PA46&amp;amp;dq=separation+theorems+for+convex+sets&amp;amp;source=bl&amp;amp;ots=IQIbpQtnOj&amp;amp;sig=jQLbA7sZol8R6FplrmVJ1Jji0Bc&amp;amp;hl=en&amp;amp;ei=Mba_TP_uGo7CsAOCyfizCQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=8&amp;amp;ved=0CDwQ6AEwBw#v=onepage&amp;amp;q=separation%20theorems%20for%20convex%20sets&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mangasarian&lt;/a&gt;.  I'm also assuming that you're using the usual topology on $\mathbb{R}^n$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-21T04:30:23.893" CommentCount="1" />
  <row Id="7386" PostTypeId="2" ParentId="7342" CreationDate="2010-10-21T04:33:31.637" Score="2" ViewCount="0" Body="&lt;p&gt;Using Lagrange multipliers we get $x_i^2 = \lambda x + \mu$, so that $x_i$ can take only two values, say $X &gt; 0$ and $Y &amp;lt; 0$ (we can't have $X=Y=0$ for obvious reasons). Suppose that $X$ is taken $a$ times, and $Y$ is taken $b = n-a$ times. We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aX + bY = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aX^2 + bY^2 = n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check that the solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^2 = b/a, Y^2 = a/b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective function that we want to minimize is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aX^3 + bY^3 = b\sqrt{b/a} - a\sqrt{a/b}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to make the first summand small and the second big. Note that as we increase $a$, the first summand becomes smaller, and the second one becomes bigger, so we might as well choose $a=n-1$ and $b=1$. So $X = 1/\sqrt{(n-1)}$ and $Y = -\sqrt{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concluding, the optimum is obtained with the sequence $-\sqrt{n-1}, 1/\sqrt{(n-1)} \times (n-1)$, and the optimal value is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ -(n-1)\sqrt{n-1} + 1/\sqrt{n-1} $&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-23T03:30:01.450" LastActivityDate="2010-10-23T03:30:01.450" CommentCount="2" />
  <row Id="7387" PostTypeId="2" ParentId="7382" CreationDate="2010-10-21T04:36:16.827" Score="4" ViewCount="0" Body="&lt;p&gt;In graph theory, the &lt;em&gt;valency&lt;/em&gt; of a vertex is sometimes used to mean its &lt;em&gt;degree&lt;/em&gt;, i.e. the number of edges incident on it. Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Degree_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/VertexDegree.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; mention both terms as synonyms.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-21T04:36:16.827" />
  <row Id="7388" PostTypeId="2" ParentId="7340" CreationDate="2010-10-21T05:52:41.477" Score="3" ViewCount="0" Body="&lt;p&gt;(5+2 choose 3)=(7 choose 3)=35 not (5 choose 3). The 35 different orders orders are {{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 2, 2}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 3}, {1, 3, 4}, {1, 3, 5}, {1, 4, 4}, {1, 4, 5}, {1, 5, 5}, {2, 2, 2}, {2, 2, 3}, {2, 2, 4}, {2, 2, 5}, {2, 3, 3}, {2, 3, 4}, {2, 3, 5}, {2, 4, 4}, {2, 4, 5}, {2, 5, 5}, {3, 3, 3}, {3, 3, 4}, {3, 3, 5}, {3, 4, 4}, {3, 4, 5}, {3, 5, 5}, {4, 4, 4}, {4, 4, 5}, {4, 5, 5}, {5, 5, 5}}.  The notation and formula were from &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A000292&quot; rel=&quot;nofollow&quot;&gt;http://www.research.att.com/~njas/sequences/A000292&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2665" LastActivityDate="2010-10-21T05:52:41.477" />
  <row Id="7389" PostTypeId="2" ParentId="7340" CreationDate="2010-10-21T06:51:58.133" Score="3" ViewCount="0" Body="&lt;p&gt;If we assume that the chef doesn't care which person ordered which dish, just how many of each dish to make, the problem is equivalent to placing 3 balls (each representing a dish the chef needs to make) in 5 urns (representing the 5 possible dishes).  This type of balls/urns problem (or pieces of identical candy to children, donuts of different types, etc.) can be solved with a technique sometimes called &quot;stars and bars&quot;: represent the balls by stars in a line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* * *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, to divide these balls among 5 urns, add 4 bars, denoting the breaking points between different urns, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* * * | | | |   all the balls in the first urn&#xA;* | | * * | |   1 balls in the first urn, 2 in the third urn, none in the rest&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The number of possible ways to do this is the number of ways to rearrange the line of 3 *s and 4 |s, which is the number of ways to choose which 3 of the 7 symbols in the line should be *s, so ${7\choose 3}$.  In general, for $k$ balls into $n$ urns, there are $k$ *s, $n-1$ |s, $k+n-1$ total symbols in the line, so ${k+n-1\choose k}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-21T06:51:58.133" />
  <row Id="7390" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T08:42:12.293" Score="1" ViewCount="0" Body="&lt;p&gt;We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-21T08:42:12.293" CommentCount="1" />
  <row Id="7391" PostTypeId="1" AcceptedAnswerId="8881" CreationDate="2010-10-21T08:48:57.103" Score="4" ViewCount="451" Body="&lt;p&gt;How to solve $$\int_0^{\pi} \sin{2x}\sin{x} dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Sorry! I should have described more. This is not a homework. Recently, Out of the blue I got interest in physics and started reading and solving problems. This is part of a physics problem where I got stuck (because I forgot all high school formulae.). Thanks all of you guys for wonderful solutions. &lt;/p&gt;&#xA;" OwnerUserId="2264" LastEditorUserId="2264" LastEditorDisplayName="" LastEditDate="2010-10-27T18:55:18.683" LastActivityDate="2010-11-04T08:24:33.093" Title="How to solve $ \int_0^{\pi} \sin{2x}\sin{x} dx $? " Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="9" CommentCount="12" FavoriteCount="0" />
  <row Id="7392" PostTypeId="1" AcceptedAnswerId="7409" CreationDate="2010-10-21T08:56:38.103" Score="5" ViewCount="156" Body="&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Does such a ring can be found? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The definition of a noetherian &lt;em&gt;topological space&lt;/em&gt; is similar to that in rings or sets. Every descending chain of closed subsets stops after a finite number of steps &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Question 2:&lt;/strong&gt; is this equivalent to saying that every &lt;em&gt;descending&lt;/em&gt; chain of &lt;em&gt;opens&lt;/em&gt; stops?).&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-10-21T17:00:35.807" Title="A non-noetherian ring with $\text{Spec}(R)$ noetherian" Tags="&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="7393" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T09:01:55.463" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use this formula  $$\cos C - \cos D = 2 \sin\frac{(C+D)}{2} \cdot \sin\frac{(D - C)}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-21T09:01:55.463" CommentCount="9" />
  <row Id="7394" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T09:03:51.727" Score="5" ViewCount="0" Body="&lt;p&gt;I'd suggest using an identity for $\sin 2x$ that rewrites it in terms of $\sin x$ and $\cos x$, then using substitution ($u=\sin x$) on the result.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-21T09:03:51.727" CommentCount="1" />
  <row Id="7395" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T09:06:22.147" Score="5" ViewCount="0" Body="&lt;p&gt;Two more ways: use Euler's formula $\sin ax=(e^{iax}-e^{-iax})/2i$, or integrate by parts twice to get an equality where your sought integral appears twice and can be solved for.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-21T09:06:22.147" CommentCount="3" />
  <row Id="7396" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T09:31:45.847" Score="1" ViewCount="0" Body="&lt;p&gt;To elaborate on the first answer given to this question by Ross Millikan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^\infty \frac{1}{(2n+1)!} = 1 + \sum_{n=1}^\infty \frac{1}{(2n+1)!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&amp;lt; 1 + \sum_{n=1}^\infty \frac{1}{4^n} = \frac{4}{3}, \quad \textrm{ as } &#xA;\frac{1}{(2n+1)!} &amp;lt; \frac{1}{4^n} \textrm{ for } n \ge1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence by &lt;a href=&quot;http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node50.html&quot; rel=&quot;nofollow&quot;&gt;the comparison test&lt;/a&gt; the series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_N &amp;lt; \frac{1}{(2n+3)!} &#xA;\left( 1 +  \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-21T09:31:45.847" />
  <row Id="7397" PostTypeId="2" ParentId="7376" CreationDate="2010-10-21T09:58:15.540" Score="6" ViewCount="0" Body="&lt;p&gt;I understand the problematic statement to mean that if a convex set $X$ has empty interior, then it is contained in a hyperplane. Let me know if I've misunderstood your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sketch of a proof by contrapositive. If $X$ is not contained in a hyperplane, then you can find $n+1$ affinely independent points in $X$. Their convex hull is a simplex contained in $X$, and a nondegenerate simplex has nonempty interior. (I'm pretty sure that last statement is true, but I'd be hard pressed to prove it.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-21T09:58:15.540" CommentCount="2" />
  <row Id="7398" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T10:46:04.467" Score="11" ViewCount="0" Body="&lt;p&gt;Plotting $ \sin{2x}\sin{x} $ in $[0,\pi]$ shows that it's symmetrical with respect to $x=\pi/2$ and so the integral is zero.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-21T10:46:04.467" CommentCount="0" />
  <row Id="7399" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T10:53:13.260" Score="6" ViewCount="0" Body="&lt;p&gt;Have a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29+from+0+to+Pi&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29+from+0+to+Pi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and for all the steps of the derivation (click on &quot;show steps&quot;):&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T10:53:13.260" />
  <row Id="7400" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T11:05:10.310" Score="3" ViewCount="0" Body="&lt;p&gt;Somewhat inspired by lhf's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\pi\sin\;2u\;\sin\;u\;\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0-\frac{\pi}{2}}^{\pi-\frac{\pi}{2}}\sin\left(2\left(u+\frac{\pi}{2}\right)\right)\;\sin\left(u+\frac{\pi}{2}\right)\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\left(\int_{-\frac{\pi}{2}}^0\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\left(-\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we see that the integral is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is convenient here that $\sin\;2u\;\cos\;u$ is an odd function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-21T11:10:48.713" LastActivityDate="2010-10-21T11:10:48.713" CommentCount="4" />
  <row Id="7401" PostTypeId="1" CreationDate="2010-10-21T11:23:12.523" Score="3" ViewCount="75" Body="&lt;p&gt;Suppose there are 9 events, that have a probability of 10%, 20%, 30%, ..., 90% of being a success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find the probability of exactly n number of these events succeeding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For n = 1, I'm thinking it is something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(first succeeds) -&amp;gt; (1-0.9) * 0.8 * 0.7 * 0.6 ... 0.1&#xA;(second succeeds) -&amp;gt; 0.9 * (1-0.8) * 0.7 * 0.6 ... 0.1&#xA;(...)&#xA;(last)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then adding them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probabilities might not always be an an Arithmetic Progression, hoping to find a solution that doesn't depend on that.&lt;/p&gt;&#xA;" OwnerUserId="2670" LastActivityDate="2010-10-21T15:14:38.010" Title="Probability Distribution with different probabilities" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="3" />
  <row Id="7402" PostTypeId="2" ParentId="7317" CreationDate="2010-10-21T11:27:57.567" Score="1" ViewCount="0" Body="&lt;p&gt;Well, it depends if you want to understand esp. &lt;em&gt;this&lt;/em&gt; derivation or &lt;em&gt;an&lt;/em&gt; derivation (so what is going on there).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest and most intuitive derivation I have seen, ever - is this one:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitive Proof of Black-Scholes Formula Based on Arbitrage and Properties of Lognormal Distribution by Alexei Krouglov&lt;br&gt;&#xA;Link: &lt;a href=&quot;http://arxiv.org/abs/physics/0612022&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0612022&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T11:27:57.567" CommentCount="1" />
  <row Id="7403" PostTypeId="2" ParentId="7392" CreationDate="2010-10-21T12:27:41.513" Score="4" ViewCount="0" Body="&lt;p&gt;Answer to question 1: Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a non-noetherian valuation domain $R$ of finite Krull dimension. The spectrum of $R$ is totally ordered under inclusion hence finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;/p&gt;&#xA;" OwnerUserId="2479" LastActivityDate="2010-10-21T12:27:41.513" />
  <row Id="7404" PostTypeId="2" ParentId="7401" CreationDate="2010-10-21T12:49:18.407" Score="1" ViewCount="0" Body="&lt;p&gt;Take the events in order.  After the first, you have 0.1 chance of 1 success and 0.9 chance of 0 successes.  Then add on the second.  The chance of one success after two events is 0.2*the chance of failure on the first + 0.8*the chance of success on the first.  A spreadsheet is your friend for this.  After all the events, you will have a column with the chance of each possible number of successes.  Adding to see if you still get 1 is a good check that your equations are right.  If you use the absolute/relative reference codes correctly, you can copy down and right most of the equations.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T12:49:18.407" />
  <row Id="7405" PostTypeId="1" AcceptedAnswerId="7430" CreationDate="2010-10-21T12:56:48.643" Score="1" ViewCount="53" Body="&lt;p&gt;A quantile function Q is defined in terms of its distribution function F as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Q(p) = \inf \{ x\in R : p \le F(x) \}, p \in (0,1).$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) if the definition of Q can be extended to $[0,1]$ by letting $ x \in R \cup \{\infty, -\infty \}$? if not, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) if it is correct that a distribution function can always be represented in terms of a quantile function defined on $(0,1)$ as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F(x) = \sup \{ p\in (0,1) : Q(p) \le x\}, x \in R-\{x|F(x)=0\text{ or }1\}.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(3) if it is correct that a distribution function can always be represented in terms of a quantile function defined on $[0,1]$ (if (1) is correct) as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F(x) = \sup \{ p\in [0,1] : Q(p) \le x\}, x \in R \cup \{\infty, -\infty\}.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) if the followings are correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F$ is discontiunous at $x_0$ with $F(x_0)=p_0$ and $lim_{x \rightarrow x_0-}F(x)= p_1$ $\Longleftrightarrow$ $Q$ is constant over $[p_1, p_0]$, taking value $x_0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$F$ is constant over $[x_1, x_0]$ taking value $p_0$ $\Longleftrightarrow$ $Q$ is discontiunous at $p_0$ with $Q(p_0)=x_1$ and $lim_{x \rightarrow x_1+}F(x)= x_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-21T13:18:47.350" LastActivityDate="2010-10-21T17:01:45.637" Title="quantile function and distribution function" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="7406" PostTypeId="2" ParentId="7392" CreationDate="2010-10-21T14:03:49.080" Score="7" ViewCount="0" Body="&lt;p&gt;If the non-noetherianness of the ring is hidden inside the nilradical, then $\mathrm{Spec}$ won't see it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k$ be any ring, and let $V$ be a free $k$-module of infinite rank. Consider $R=k\oplus V$, and turn it into a ring by defining $$(a,v)\cdot(b,w)=(ab,aw+bv).$$ (Representation-people call this a &lt;em&gt;trivial extension&lt;/em&gt;) Then $R$ is not-noetherian, because every $k$-submodule of $V$ is an ideal in $R$. Yet $V$ is contained in the nilradical of $R$: if you look at $\mathrm{Spec}\;R$ and at $\mathrm{Spec}\;k$, you'll see that they are very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question: no. If $k$ is an infinite field, then $\mathrm{Spec}\;k[X]$ is noetherian, yet you'll easily find a decreasing chain of open sets init which does not stop. (What examples did you consider before asking the question? :) )&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-21T17:00:35.807" LastActivityDate="2010-10-21T17:00:35.807" CommentCount="1" />
  <row Id="7407" PostTypeId="2" ParentId="7329" CreationDate="2010-10-21T14:05:22.860" Score="0" ViewCount="0" Body="&lt;p&gt;Building on your interpretation, let us define the rules as follows:  A and B alternately throw a single die.  If the sum of the last two throws is 5, the one who just threw wins.  Note that as stated, an initial 5 could win, or a series of throws of a different length than 2.  The analysis would change, but follow the same route.  The below requires precisely two throws to add to 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let p be the probability that the first player wins.  Let q be the probability that the next player to throw wins given that he has received a chance of winning, that is that the last throw is less than 5.  Then if you receive a throw of 5 or 6 your chance of winning is p.  So q=1/6 (that you win on this throw) + (1-p)/3 (that you throw 5 or 6 and then win) +(1-q)/2 (that you throw &amp;lt;5, don't win this throw, but finally win.  p=(1-p)/3 (that you throw 5 or 6 and win)+2(1-q)/3 (that you throw &amp;lt;5 and win).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q=\frac{1}{6}+\frac{1-p}{3}+\frac{1-q}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=\frac{1-p}{3}+\frac{2(1-q)}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have the algebra right, p=15/32 and q=9/16, so the first player should withdraw $2/17. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-21T17:45:25.887" LastActivityDate="2010-10-21T17:45:25.887" />
  <row Id="7408" PostTypeId="2" ParentId="7391" CreationDate="2010-10-21T14:25:15.453" Score="6" ViewCount="0" Body="&lt;p&gt;If you use the identity $\sin{2x} = 2 \cos{x}\sin{x}$ then you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sin{2x} \sin{x} \, dx = 2 \int \sin^2{x} \cos{x} \, dx  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now you can make the substitution $u = \sin{x}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \int u^2 \, du  = \frac{2}{3} u^3 + C = \frac{2}{3} \sin^3{x} + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $$\int_{0}^{\pi} \sin{2x} \sin{x} \, dx = \left ( \frac{2}{3} \sin^3{x} \right )_{0}^{\pi} = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-10-21T14:35:48.983" LastActivityDate="2010-10-21T14:35:48.983" CommentCount="2" />
  <row Id="7409" PostTypeId="2" ParentId="7392" CreationDate="2010-10-21T14:30:48.357" Score="7" ViewCount="0" Body="&lt;p&gt;It is easy to come up with examples when you keep in mind that $X$ and $X_{red}$ are homeomorphic: The topology is not changed when you divide out (all) nilpotent sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $A = k[x_1,x_2,...] / (x_1^2,x_2^2,...)$ satisfies $A/rad(A)=k$, so that $Spec(A)$ consists of exactly one point, namely the maximal ideal $(x_1,x_2,...)$, which is not finitely generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, $Spec(A)$ is a noetherian topological space iff $A$ satisfies the ascending chain condition for &lt;em&gt;radical&lt;/em&gt; ideals.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-10-21T14:30:48.357" CommentCount="1" />
  <row Id="7410" PostTypeId="2" ParentId="7354" CreationDate="2010-10-21T14:37:58.867" Score="2" ViewCount="0" Body="&lt;p&gt;Another way is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} &amp;lt; 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $S_n &amp;lt; 2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-21T14:37:58.867" />
  <row Id="7411" PostTypeId="2" ParentId="7401" CreationDate="2010-10-21T14:41:28.570" Score="1" ViewCount="0" Body="&lt;p&gt;Let the probability that the $i$th event occurs be $p_i = i/10$ and let the probability that exactly $n$ events occur be $P(n).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability that no events occur is $P(0) = \prod_{i=1}^9 (1-p_i) = 3.6288 \times 10^{-4}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S= \lbrace 1,2,3,\ldots,9 \rbrace $ then for $n \ge 1$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(n) = P(0) \times \sum_{i_1,i_2,\ldots,i_n \in S, \quad i_k \textrm { distinct} }&#xA;\frac{p_{i_1} p_{i_2} \ldots p_{i_n} } {(1-p_{i_1})(1-p_{i_2}) \ldots (1-p_{i_n}) },$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where there are ${9 \choose n}$ terms in the summation.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-21T14:41:28.570" />
  <row Id="7412" PostTypeId="1" CreationDate="2010-10-21T14:52:00.710" Score="6" ViewCount="198" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was playing around with &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_carpet&quot; rel=&quot;nofollow&quot;&gt;Sierpinski carpets&lt;/a&gt; (see &lt;a href=&quot;http://stackoverflow.com/questions/3972902/is-it-possible-to-perform-floating-point-operations-on-gpu-when-using-opengl/3973433#3973433&quot;&gt;pretty GPU-produced picture here&lt;/a&gt;), and came up with a variation that I have been unable to find mentioned elsewhere. I'm wondering if anyone can tell me an existing name for it, and help me determine where it fits into existing classifications of fractals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you probably know, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The construction of the &lt;strong&gt;Sierpinski&#xA;  carpet&lt;/strong&gt; begins with a square. The&#xA;  square is cut into 9 congruent&#xA;  subsquares in a 3-by-3 grid, and the&#xA;  [open] central subsquare is removed. The same&#xA;  procedure is then applied recursively&#xA;  to the remaining 8 subsquares, ad&#xA;  infinitum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this variation, &lt;strong&gt;instead of merely removing the central subsquare, we recursively apply the same procedure to it &quot;inversely.&quot;&lt;/strong&gt; (Is there a better term for that?) That is, we treat the central hole as if we want to replace it with the complement of [this variant of] a Sierpinski carpet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we recur on the central hole, we put an &quot;anti-hole&quot; (like a &quot;donut hole&quot; without a donut) in the middle of it. We will then recur on the anti-hole using the normal (positive) procedure, and recur on the 8 subsquares surrounding the anti-hole using the inverse procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks something like this, in its 5th iteration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vVMDk.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think I should post the code for this, I'm happy to, but I didn't think it was very relevant to the math question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been calling it &quot;Sierpinski tablecloth&quot; since I couldn't find another name for it, and because it reminds me of a lace or crocheted tablecloth design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Correction:&lt;/strong&gt; @Rahul pointed out that &quot;the limiting set of the recursive process does not exist.&quot; This is because the central point of the unit square must be part of both the tablecloth and its complement, or neither - which is impossible. (Of course we can still do finite iterations of the constructive process without hitting a contradiction.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix this problem, I tried proposing that we define the tablecloth set T (arbitrarily) to include the central point. The central subsquare is replaced with the &lt;em&gt;union of the central point&lt;/em&gt; with the scaled-down complement of T. As before, the subsquares around the center are replaced with scaled-down copies of T (and therefore the central point of each of the subsquares is contained in T).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another update:&lt;/strong&gt; Unfortunately the same problem is true of any point with an irrational coordinate, and of many rational ones. If we express the coordinates in base 3, as with the Cantor set, a point is a member of T iff the number of 1's in the base-3 expansion of each of its coordinates is even. But this doesn't work for numbers with repeating 1's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we tried to fix this by making all &quot;problem points&quot; belong to T, we'd be left with &quot;almost&quot; a solid unit square, minus a &quot;dust&quot; of rational points with an even number of 1's in their base-3 coordinate expansions. Then T would cease to generate interesting fractal pictures (in the limit). Its area would be 1, like the irrationals$^2$. It's not clear to me that T would be different from the set of the irrationals$^2$ in any interesting way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mind you, we can still draw pictures of the nth iteration of the tablecloth that are fun to look at, but it loses a good deal of its bite if it doesn't exist as a fractal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of a way to save the tablecloth from extinction as a fractal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of this question was written under the misapprehension that the tablecloth exists (is a well-defined set) in the limit.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fractal dimension&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The traditional carpet has &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension#Deterministic_fractals&quot; rel=&quot;nofollow&quot;&gt;Hausdorff dimension&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(8)/log(3) = ~1.8928$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;derived from the fact that it takes 8 self-similar copies at reduction factor 1/3 to cover itself (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractal_dimension#Specific_definitions&quot; rel=&quot;nofollow&quot;&gt;example for Sierpinski triangle&lt;/a&gt;). In this vein I figure that the tablecloth takes 9 self-similar copies at reduction factor 1/3 to cover itself (no smaller copy will cover the middle mini-square), so it would have Hausdorff dimension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(9)/log(3) = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would put it in the same company (in terms of Hausdorff dimension) as space-filling curves like the Sierpinski curve, Hilbert, Peano, etc. Does this sound reasonable so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite comfortable with that, though, because a 1/3-size copy of the tablecloth will not actually cover the whole middle subsquare. In fact I can't see how any finite number of reduced-size copies of the tablecloth would cover the whole middle subsquare. And if we say we need an infinite number, the Hausdorff dimension would be infinite. :-(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative, I considered &lt;a href=&quot;http://en.wikipedia.org/wiki/Multifractal&quot; rel=&quot;nofollow&quot;&gt;multifractals&lt;/a&gt;, since we seem to have different scaling behavior occurring in different parts of the attractor. But that's deeper water than I'm prepared to wade into.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can anyone point me to literature on&#xA;fractals that are defined using&#xA;&lt;em&gt;complements&lt;/em&gt; of reduced-size self-similar sets (as well as the&#xA;usual reduced-size self-similar&#xA;sets)? Is that a class of fractals&#xA;that has been studied? &lt;strong&gt;(Update:)&lt;/strong&gt; Or are such &quot;fractals&quot; doomed to&#xA;be ill-defined (self-contradictory) by their nature?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Similarly to Hausdorff dimension, the &lt;strong&gt;box-counting dimension&lt;/strong&gt; (using the criterion &quot;if the space were divided up into a grid of boxes of size ε, how many boxes of that scale would contain part of the attractor?&quot;) also seems to end up at 2, for much the same reasons. And I'm more comfortable with it because a &quot;box&quot; (as opposed to a reduced copy of the tablecloth) will indeed cover the whole middle subsquare. It seems clear that in the limit, the box-counting dimension converges to 2, because there are no gaps, so every box, at whatever size ε, will always contain part of the fractal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of &lt;strong&gt;area&lt;/strong&gt; (standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;Lebesgue measure&lt;/a&gt;), I'm pretty sure that the tablecloth has zero, like the carpet, because it has no intervals (rectangles) that are entirely in the set. Right? (&lt;strong&gt;Update:&lt;/strong&gt; I stand corrected: the tablecloth seems to have area 1/2.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are several questions above, so to tie them all together: I'm wondering if this tablecloth variant is something new and interesting, or something well-known, or an unknown but trivial variant that can be classified under existing categories. E.g. if there is a variant of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Cantor set&lt;/a&gt; that is defined similarly to the tablecloth (for one dimension), I would appreciate a link to info about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Update:)&lt;/strong&gt; One thing that seems interesting to me about it is that it has points in every interval of the unit square, but it also has holes in every interval. I guess it's like rational numbers in [0,1] in that regard, or like space-filling curves/trees, yet is not a curve or tree (not connected). I can't think of other fractals off-hand that are like that, can you? It seems that this property is connected with the fact that we recur on the inverse instead of leaving holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt; I just read about the term &lt;em&gt;mutually dense&lt;/em&gt; (&lt;a href=&quot;http://www.mathpages.com/home/kmath172.htm&quot; rel=&quot;nofollow&quot;&gt;Interleaving Ad Infinitum&lt;/a&gt;): 'If a set S with linear ordering O can be partitioned into two subsets&#xA;A and B such that between every two distinct elements of A is an &#xA;element of B, and vice versa, then the sets A and B are said to be &#xA;&quot;mutually dense&quot;.' So the tablecloth and its complement are mutually dense, assuming that this term can be generalized to the unit square. At least in any horizontal or vertical linear slice of the unit square, they are mutually dense. &lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T14:02:44.783" LastActivityDate="2010-11-27T14:02:44.783" Title="variant on Sierpinski carpet: rescue the tablecloth!" Tags="&lt;topology&gt;&lt;recursive&gt;&lt;fractals&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="7413" PostTypeId="1" AcceptedAnswerId="7416" CreationDate="2010-10-21T14:52:44.483" Score="6" ViewCount="271" Body="&lt;p&gt;I am currently learning about the concept of convolution  between two functions in my university course. The course notes are vague about what convolution is, so I was wondering if anyone could give me a good explanation. I can't seem to grasp other than the fact that it is just a particular integral of two functions. What is the physical meaning of convolution and why is it useful? Thanks a lot. &lt;/p&gt;&#xA;" OwnerUserId="2673" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-10-26T01:06:54.320" LastActivityDate="2010-10-26T01:06:54.320" Title="Meaning of convolution?" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;convolution&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="7415" PostTypeId="2" ParentId="7413" CreationDate="2010-10-21T14:59:01.713" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution&quot; rel=&quot;nofollow&quot;&gt;The Wikipedia&lt;/a&gt; has some nice graphical explanations.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-10-21T14:59:01.713" CommentCount="1" />
  <row Id="7416" PostTypeId="2" ParentId="7413" CreationDate="2010-10-21T15:02:15.240" Score="4" ViewCount="0" Body="&lt;p&gt;Have a look here:&lt;br&gt;&#xA;&lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20070125163821AA5hyRX&quot; rel=&quot;nofollow&quot;&gt;http://answers.yahoo.com/question/index?qid=20070125163821AA5hyRX&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and lots of good answers here:&lt;br&gt;&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/5892/what-is-convolution-intuitively&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/5892/what-is-convolution-intuitively&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T15:02:15.240" CommentCount="1" />
  <row Id="7417" PostTypeId="2" ParentId="7401" CreationDate="2010-10-21T15:14:38.010" Score="2" ViewCount="0" Body="&lt;p&gt;The generating function for the probabilities $a_n$ for $n$ successes is&#xA;$$\sum_{n=0}^9 a_n t^n=\prod_{k=1}^9(p_k t+(1-p_k))$$&#xA;where $p_k$ is the probability of the $k$-th event occuring, in this&#xA;example $p_k=k/10$. Maybe in this case the product won't simplify&#xA;so nicely, but the method is quite general.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-21T15:14:38.010" CommentCount="1" />
  <row Id="7418" PostTypeId="2" ParentId="7317" CreationDate="2010-10-21T15:20:10.657" Score="1" ViewCount="0" Body="&lt;p&gt;I found this concise paper one of the most accessible ones giving an overview of the traditional approach:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.williams.edu/Mathematics/fmorgan/MorganBlackScholes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.williams.edu/Mathematics/fmorgan/MorganBlackScholes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-21T15:20:10.657" />
  <row Id="7419" PostTypeId="1" AcceptedAnswerId="7420" CreationDate="2010-10-21T15:20:18.527" Score="1" ViewCount="89" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to solve this non-linear differential equation using substitution $\dfrac{y}{x}$ to $t$. However, I can't solve this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this equation and what's the general solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditorDisplayName="" LastEditDate="2010-10-21T16:51:10.150" LastActivityDate="2010-10-22T00:14:37.857" Title="How to solve this non-linear differential equation?" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7420" PostTypeId="2" ParentId="7419" CreationDate="2010-10-21T15:24:49.827" Score="4" ViewCount="0" Body="&lt;p&gt;if you change the dependent variable to $u=y/x$, as you said, the equation becomes $x^2u'=u^3$, which can be immediately be solved by separating variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show us what you tried and we may help you with it.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-21T15:38:02.317" LastActivityDate="2010-10-21T15:38:02.317" CommentCount="2" />
  <row Id="7421" PostTypeId="2" ParentId="7412" CreationDate="2010-10-21T15:29:19.673" Score="2" ViewCount="0" Body="&lt;p&gt;The area, if it exists, is 1/2.  Let the area inside the unit square be A.  Then it is made up of 8 copies at 1/3 linear size, plus the complement of a 1/3 linear size copy (the center square).  So $A=\frac{8}{9}*A+\frac{1-A}{9}$.  There is no problem having an area without a rectangle.  See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fat_Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Fat Cantor Set&lt;/a&gt;.  I believe having positive area also supports dimension 2&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T15:29:19.673" CommentCount="1" />
  <row Id="7422" PostTypeId="2" ParentId="7419" CreationDate="2010-10-21T15:41:14.077" Score="1" ViewCount="0" Body="&lt;p&gt;The substitution $u = \frac{y}{x}$ works. What you have is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = ux$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., $ y' = x u' + u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $x y' = x^2 u' + y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this into your equation, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 u' = u^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be solved using standard methods.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-21T15:41:14.077" CommentCount="2" />
  <row Id="7423" PostTypeId="1" AcceptedAnswerId="7424" CreationDate="2010-10-21T16:04:53.337" Score="2" ViewCount="67" Body="&lt;p&gt;Choose numbers from 1 to 2n uniformly at random.  How many numbers must be chosen, on average, before at least n have been picked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to the coupon-collector problem, but looking for only partial completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Choosing an appropriate meaning of 'random' is part of the question.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-21T16:18:01.997" Title="How long to choose n out of 2n numbers?" Tags="&lt;combinatorics&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="7424" PostTypeId="2" ParentId="7423" CreationDate="2010-10-21T16:15:44.600" Score="4" ViewCount="0" Body="&lt;p&gt;If you stop the sum from the coupon collector problem half-way, you get your answer.  It takes 1 draw on average to get the first different ticket, then $\frac{2n}{2n-1}$ draws for the second, and so on until $\frac{2n}{n+1}$ for the $n^{th}$.  So this is $2n*(H_{2n}-H_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you choose without replacement, the answer is n draws.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T16:15:44.600" CommentCount="1" />
  <row Id="7426" PostTypeId="2" ParentId="7419" CreationDate="2010-10-21T16:23:59.233" Score="1" ViewCount="0" Body="&lt;p&gt;Here is my solution. Is this right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y \\ \\&#xA;\frac{y}{x}=u, \\&#xA;y = \text{ux}, \\&#xA;\text{y$\prime $} = \text{u$\prime $x} + u \\ \\&#xA;x^2\text{u$\prime $} = u^3 \\&#xA;\int \frac{1}{u^3} \, du =\int \frac{1}{x^2} \, dx \\&#xA;-\frac{1}{2u^2} = -\frac{1}{x} + c \\&#xA;u^{2 }= \frac{1}{\frac{2}{x}+c} \\&#xA;y^{2 }= \frac{x^2}{\frac{2}{x}+c} \\&#xA;y = x\left(\frac{2}{x}+c\right)^{-\frac{1}{2}}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="2674" LastEditorUserId="2674" LastEditorDisplayName="" LastEditDate="2010-10-22T00:14:37.857" LastActivityDate="2010-10-22T00:14:37.857" CommentCount="5" />
  <row Id="7427" PostTypeId="2" ParentId="7412" CreationDate="2010-10-21T16:36:45.437" Score="1" ViewCount="0" Body="&lt;p&gt;Regarding your update:  Using &lt;a href=&quot;http://en.wikipedia.org/wiki/Conway_base_13_function&quot; rel=&quot;nofollow&quot;&gt;Conway's base 13 function&lt;/a&gt; or other constructions (I have one I prefer) you can find a function $f: (0,1) \rightarrow (0,1)$ such that for any interval $(a,b) \subset (0,1)$ and $y \in (0,1)$ there is an $x \in (a,b)$ such that $f(x)=y$.  Then for given $y$ the points such that $f(x)=y$ are dense in $(0,1)$.  So you have continuum many disjoint sets, each dense in $(0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T16:36:45.437" CommentCount="2" />
  <row Id="7428" PostTypeId="2" ParentId="6877" CreationDate="2010-10-21T16:47:03.940" Score="4" ViewCount="0" Body="&lt;p&gt;Combinatorics: Counting without your fingers&lt;/p&gt;&#xA;" OwnerUserId="2677" LastActivityDate="2010-10-21T16:47:03.940" />
  <row Id="7429" PostTypeId="2" ParentId="6877" CreationDate="2010-10-21T16:51:04.640" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Numerical Analysis&lt;/strong&gt;: The search for fast, accurate, and stable algorithms for finding numerical solutions.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-21T16:51:04.640" />
  <row Id="7430" PostTypeId="2" ParentId="7405" CreationDate="2010-10-21T16:55:27.533" Score="1" ViewCount="0" Body="&lt;p&gt;I think your last shoud be $Q(p_0)=x_1$ and $lim_{p\rightarrow p_0+} Q(p) =x_0$ with the rest OK&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-21T17:01:45.637" LastActivityDate="2010-10-21T17:01:45.637" />
  <row Id="7431" PostTypeId="2" ParentId="7412" CreationDate="2010-10-21T17:16:17.320" Score="2" ViewCount="0" Body="&lt;p&gt;I wrote a bunch of stuff about the properties of this set assuming it existed, then I realized that it wasn't even obvious that the set was well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the problem: Does the central point belong to the set? If it does, then it does not belong to the middle block, and vice versa, leading to a contradiction. So the limiting set of the recursive process does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, looking at the picture, it feels as though the limit &quot;should&quot; exist in some expanded sense. This could possibly be related to the way that some sequences of functions which do not converge to functions do actually converge to distributions. Or I could just be wrong and the limit simply cannot exist in any sense. &lt;em&gt;Edit:&lt;/em&gt; Well, it turns out that the limiting measure contained in any rectangular region is just half the area of the rectangle. So I suppose it does have a limit in the sense of a measure on $[0,1]^2$, but it's not a very interesting measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious corresponding version of the Cantor set would be to replace every middle third with the complement of the set, rather than leaving it empty. However, the Cantor set has a very nice interpretation in terms of ternary expansions: a real number $x \in [0,1]$ is in the Cantor set if and only if its ternary expansion contains no digit 1 (allowing, for example, $1/3 = 0.1_3$ to be $0.0222\ldots_3$). For this variant, the corresponding condition would have to be that the ternary expansion contains an &lt;em&gt;even&lt;/em&gt; number of ones. This &lt;em&gt;is&lt;/em&gt; well-defined for fractions of the form $a/3^b$ whose ternary expansion terminates, but like above, it is not clear how to deal with rationals like $1/2$, or all the irrationals.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-21T18:03:46.790" LastActivityDate="2010-10-21T18:03:46.790" CommentCount="7" />
  <row Id="7432" PostTypeId="1" CreationDate="2010-10-21T17:30:04.557" Score="0" ViewCount="72" Body="&lt;p&gt;I have a probability space $\mathcal{X} \times \mathcal{Y}$. $\mathcal{X}$ and $\mathcal{Y}$ could be infinite, but they are at worst case countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a matrix $A_{xy}$ ranging over $\mathcal{X}$ for rows $\mathcal{Y}$ for columns, which is either 0 or 1 for each $x,y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the information that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(${$ (x,y) \mid A_{xy} = 1$}$) \le \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to upper bound the probability $P(${$x \mid \exists y A_{xy} = 1$}$)$, where the bound depends on $\epsilon$ and other parameters given in the question? (say, structure of $A$ perhaps... but I'd rather it be $A$-independent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that we can always create an $A$ such that it has $1$ in row (for each $x$ somewhere, therefore the probability that I want to bound would be 1), but still the probability of total 1s would be $\epsilon$. So I am pretty sure I am not approaching the whole problem formulation the right way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe you have an idea what assumptions I can make to get a better bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea that I had is doing the following:&#xA;assume that we have $\alpha = \sup_{x,y} \frac{P(x)}{P(x,y)}$.&#xA;Then, we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(${$x \mid \exists y A_{xy} = 1$}$)  = \sum_x P(x) I(\exists y A_{xy} = 1)\le \sum_{x,y \mathrm{s.t.}\, A_{xy}=1} \alpha P(x,y) \le \alpha \epsilon $ (where $I$ is a 0 1 indicator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that in my case, $\alpha$ could get pretty big if I really define it that way.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2010-11-20T18:44:19.987" Title="can I bound the following probability?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7433" PostTypeId="1" AcceptedAnswerId="7448" CreationDate="2010-10-21T17:51:54.947" Score="0" ViewCount="48" Body="&lt;p&gt;The pdf of the geomtric distribution is the following: $f(x) = (1-p)^{x}p$. Also $E[X] = \frac{1-p}{p}$ and $\text{Var}[X] = \frac{1-p}{p^2}$. The pdf of the negative binomial distribution is: $f(x) = \binom{r+x-1}{x}p^{r}(1-p)^{x}$. Also $E[X] = \frac{r(1-p)}{p}$ and $\text{Var}[X] = \frac{r(1-p)}{p^2}$. So is the geomtric distribution really a special case of the negative binomial? In each case, $X$ is the number of failures until a success occurs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does the geometric distribution have a different version as well? The $\binom{r+x-1}{x}$ seems analogous to $\binom{n+k-1}{k}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-21T19:31:20.567" LastActivityDate="2010-10-21T19:31:20.567" Title="Geometric Distribution versus Negative Binomial Distribution" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7434" PostTypeId="2" ParentId="7432" CreationDate="2010-10-21T17:59:13.067" Score="1" ViewCount="0" Body="&lt;p&gt;You have a two sets of nodes, $X$ and $Y$, with $A_{i,j}=1$ when there is an edge from $i\in X$ to $j \in Y$.  You are given that only a fraction $\epsilon$ of the possible edges are occupied and want the chance that a given $i\in X$ has at least one edge occupied.  If $|Y|$ is large the chance can be quite high.  Your bound would be to recognize that the worst case is if each node in $X$ has only one edge occupied.  Then the bound is $\epsilon * |Y|$ (or 1 if smaller).  The size of $X$ doesn't matter.  Essentially the elements of $X$ are all independent.  If the sets are infinite you need to worry about how you take limits.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T17:59:13.067" />
  <row Id="7435" PostTypeId="1" CreationDate="2010-10-21T18:10:04.110" Score="2" ViewCount="86" Body="&lt;p&gt;In the past couple years, I've read many words pertaining to D-branes without feeling I have really comprehended them. In broad terms, I think I get what they're about: They're supposed to serve as habitats for the ends of open strings and can be viewed as submanifolds (of the target manifold in a sigma model), possibly augmented with a vector bundle, or a sheaf of [somethings], or maybe some other kind of label. (Please correct me if I got that wrong.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hopes of tightening my grasp on the concept, here are some of the questions that have been nagging me during my reading.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What &lt;em&gt;specifically&lt;/em&gt; is the definition of a D-brane, say in the context of a topological field theory? (Or what are the most promising provisional definitions?) What references are most accessible to a mathematical audience?&lt;/li&gt;&#xA;&lt;li&gt;What picture should I have in my head when an author talks about &quot;the moduli space of D-branes&quot;?&lt;/li&gt;&#xA;&lt;li&gt;What is the idea behind the &quot;dynamics of D-branes&quot; that researchers sometimes talk about? (Perhaps when I understand better how to think about these gadgets, it will be easier to conceive of how they should change over time.)&lt;/li&gt;&#xA;&lt;li&gt;What goes into verifying (or at least asserting/conjecturing) that the elements of twisted K-theory classify D-brane charges?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(I considered putting this on MO, but it feels too basic relative to what TFT people talk about there for it to be appropriate.)&lt;/p&gt;&#xA;" OwnerUserId="1347" LastActivityDate="2010-10-21T18:10:04.110" Title="What are D-branes (in a topological field theory)? " Tags="&lt;reference-request&gt;&lt;physics&gt;" CommentCount="2" />
  <row Id="7436" PostTypeId="1" AcceptedAnswerId="7444" CreationDate="2010-10-21T18:11:13.387" Score="3" ViewCount="347" Body="&lt;p&gt;I'm having trouble finding a good explanation of Lebesgue integral. As per the definition, it is the expectation of a random variable. Then how does it model the area under the curve? Lets take for example a function $f(x) = x^2$, how do we find intergral of $f(x)$ under $[0,1]$ using Lebesgue integral?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-21T19:46:48.947" LastActivityDate="2010-10-22T04:12:51.830" Title="Lebesgue integral basics" Tags="&lt;analysis&gt;&lt;probability&gt;&lt;integral&gt;&lt;expectation&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="4" />
  <row Id="7437" PostTypeId="1" AcceptedAnswerId="7439" CreationDate="2010-10-21T18:20:25.740" Score="0" ViewCount="76" Body="&lt;p&gt;Prove(A∩B)’=A’∪B’.&lt;br&gt;&#xA;let x ∈ (A ∩ B)’&lt;br&gt;&#xA;∴ (x ∈ A ∩ B)’&lt;br&gt;&#xA;∴ (x ∈ A ∧ x ∈ B)’&lt;br&gt;&#xA;∴ (x ∈ A)’ ∨ (x ∈ B)’&lt;br&gt;&#xA;∴ x ∈ A’ ∨ x ∈ B’&lt;br&gt;&#xA;∴ x ∈ A’ ∪ B’&lt;br&gt;&#xA;∴ (A ∩ B)’ ⊆ A’ ∪ B’&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the solution provided. Sorry if it seems trivial but I don't understand how&lt;br&gt;&#xA;∴ (x ∈ A ∧ x ∈ B)’&lt;br&gt;&#xA;leads to&lt;br&gt;&#xA;∴ (x ∈ A)’ ∨ (x ∈ B)’&lt;br&gt;&#xA;and not&lt;br&gt;&#xA;∴ (x ∈ A)’ ∧ (x ∈ B)’&lt;/p&gt;&#xA;" OwnerUserId="2679" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T17:54:58.943" LastActivityDate="2010-11-20T17:54:58.943" Title="Proving two sets are equal using bi-directional set inclusion" Tags="&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7438" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T18:26:39.677" Score="1" ViewCount="0" Body="&lt;p&gt;The definition &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; doesn't mention probability or expectation or random variable.  Intuitively, it just says the measure (in $\mathbb R^2$) is the area of the smallest set of rectangles that will cover the set.  Then for an area the Lebesgue integral is just the integral of 1 over the set.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T18:26:39.677" />
  <row Id="7439" PostTypeId="2" ParentId="7437" CreationDate="2010-10-21T18:29:37.233" Score="2" ViewCount="0" Body="&lt;p&gt;This is just one of DeMorgan's Laws.  Not (A and B) is equivalent to (Not A) or (Not B), that is that (A and B) is false if either one is false.  If you go to (Not A) and (Not B) you need both to be false.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T18:29:37.233" CommentCount="1" />
  <row Id="7440" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T18:45:45.153" Score="3" ViewCount="0" Body="&lt;p&gt;The Lebesgue integral is a generalization of the usual Riemann integral taught in basic calculus. If the Riemann integral of a function over a set exists then it equals the Lebesgue integral. So the Lebesgue integral of $x^2$ over $[0,1]$ is just the old $(1/3) 1^3-(1/3)0^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lebesgue integral has the benefit of being defined for many more functions than the Riemann integral. Even more importantly the Lebesgue integral has useful limit properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The expectation of a random variable is a particular application of the Lebesgue integral where the function to be integrated is the random variable (viewed as a function on the sample space) and the integration is with respect to a probability measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to look at one of the many probability and measure books for the details. My own favourites are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pollard, &lt;em&gt;A User's Guide to Measure-Theoretic Probability&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Dudley, &lt;em&gt;Real Analysis and Probability&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Terence Tao has some online lecture notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-21T18:45:45.153" />
  <row Id="7441" PostTypeId="1" AcceptedAnswerId="7463" CreationDate="2010-10-21T18:57:37.303" Score="4" ViewCount="242" Body="&lt;p&gt;For $x&gt;0$ and $y&gt;x+1$, how do we prove that $$\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)} = \frac{x}{y-x-1}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-21T22:50:42.490" Title="Summation of $\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="4" />
  <row Id="7442" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:04:35.317" Score="3" ViewCount="0" Body="&lt;p&gt;One of my graduate school professors, Erhan Cinlar, used to give the following analogy to explain the intuitive difference between the Lebesgue integral and the Riemann integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a pile of coins of different denominations, and you want to know how much money you have.  The Riemann integral is like picking up the coins, one-by-one, and adding the denomination of each to a running total.  The Lebesgue integral is like sorting the coins by denomination first, and then getting the total by multiplying each denomination by how many you have of that denomination and then adding up those numbers.  The methods are different, but you obtain the same result by either method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, when both the Riemann integral and the Lebesgue integral are defined, they give the same value.  As others have said, though, there are functions for which the Lebesgue integral is defined but the Riemann integral is not, and so in that sense the Lebesgue integral is more general than the Riemann.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-22T04:12:51.830" LastActivityDate="2010-10-22T04:12:51.830" CommentCount="2" />
  <row Id="7443" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:05:34.983" Score="0" ViewCount="0" Body="&lt;p&gt;You may also like to refer, these two books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A radical approach to Lebesgue Theory of Integration: Bressoud&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Real Analysis by G.B. Folland.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As Jyotirmoy pointed out, Lebesgue integral is the generalization of Riemann Integral. There are shortcomings of the Riemann Integral, due to which the Lebesgue integral, was discovered. A rigorous definition of the Lebesgue integral needs, you to know what a Simple Function is, and you can read more on this at &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_integration&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lebesgue_integration&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-21T19:05:34.983" />
  <row Id="7444" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:09:34.727" Score="13" ViewCount="0" Body="&lt;p&gt;As has been noted, the usual definition of the Lebesgue integral has little to do with probability or random variables (though the notions of measure theory and the integral can then be applied to the setting of probability, where under suitable interpretations it will turn out that the (Lebesgue) integral of (a certain) functions corresponds to the expectation of (a certain) random variable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not the &lt;em&gt;origin&lt;/em&gt; of the Lebesgue integral. Here is an intuitive idea of what the Lebesgue integral is, as compared to the Riemann integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall from Calculus the idea behind the Riemann integral: the integral $\int_a^b f(x)\,dx$ is meant to represent the net signed area between the $x$-axis, the graph of $y=f(x)$, and the lines $x=a$ and $x=b$. The way we attempt to do this is by breaking up the domain, $[a,b]$, into subintervals $[a=x_0,x_1]$, $[x_1,x_2],\ldots,[x_{n-1},x_n=b]$. Then, on each subinterval $[x_i,x_{i+1}]$ we pick a point $x_i^*$, and we estimate the area under the graph of the function with the rectangle of height $f(x_i^*)$ and base $[x_i,x_{i+1}]$. This leads to the Riemann sums&#xA;$$ \sum_{i=0}^{n-1} f(x_i^*)(x_{i+1}-x_i)$$&#xA;as estimates of the area under the graph. We then consider finer and finer partitions of $[a,b]$ and take limits to estimate the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lebesgue's idea was that instead of partitioning the domain, we will partition the &lt;em&gt;range&lt;/em&gt;; if the function takes values between $c$ and $d$, we can divide the range $[c,d]$ into subintervals $[c=y_0,y_1]$, $[y_1,y_2],\ldots,[y_{m-1},y_m=d]$. Then, we let $E_i$ be the set of all points in $[a,b]$ whose value under $f$ lies between $y_i$ and $y_{i+1}$. That is,&#xA;$$ E_i = f^{-1}([y_i,y_{i+1}]) = \{ x\in[a,b]\,|\, y_i \leq f(x) \leq y_{i+1}\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a way of assigning a &quot;size&quot; to $E_i$, call it its &quot;measure&quot; $\mu(E_i)$, then the portion of the graph of $y=f(x)$ that lies between the horizontal lines $y=y_i$ and $y=y_{i+1}$ will be $A$, where,&#xA;$$ y_i\mu(E_i) \leq A \leq y_{i+1}\mu(E_i).$$&#xA;So Lebesgue suggests to approximate the the area by picking a number $y_i^*$ between $y_i$ and $y_{i+1}$, and considering the sums&#xA;$$ \sum_{i=0}^{n-1} \mu(E_i)y_i^*.$$&#xA;Then consider finer and finer partitions of $[c,d]$, and this gives finer and finer approximations of of the area by these sums. The Lebesgue integral will be the limit of these sums. (The analogy given by Mike Spivey is very apt for the distinction between partitioning the domain and partitioning the range to find the sum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in order for this to make sense, we need to develop a way of measuring fairly intricate subsets of the line, so that we can compute $\mu(E_i)$. So we first develop a way of doing this; turns out that if you accept the Axiom of Choice, then it is impossible to come up with a way of measuring that will (i) assign to an interval $[a,b]$ the &quot;measure&quot; $b-a$; (ii) will be invariant under translation, so so that if $F=E+c = \{e+c | e\in E\}$ then $\mu(F)=\mu(E)$; (iii) will be countably additive: if $E = \cup_{i=1}^{\infty}E_i$ and the $E_i$ are pairwise disjoint, then $\mu(E) = \sum\mu(E_i)$; and (iv) every subset of the line will have a well-defined (possibly infinite) measure. (If you don't accept the Axiom of Choice, then there are models of the reals where we &lt;em&gt;can&lt;/em&gt; achieve this). So one drops the restriction (iv), and constructs a measure for which some sets will be &quot;too weird&quot; to have a measure. We then restrict attention to certain kinds of functions (called the measurable functions), which are the ones for which the sets we get in the process described above are all measurable sets. And then we define the Lebesgue integral for those functions, following the idea described above (but one does not define it exactly that way; instead the usual way is to describe $f$ as a limit of functions for which the integral is easy, and then compute the integral of $f$ as a limit of the integrals that are easy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your function, $f(x)=x^2$, this is fairly easy: the value all lie between $0$ and $1$, so say that we break up the range into subintervals of length $1/n$, so $y_i = i/n$, $i=0,\ldots,n$. Then &#xA;$$f^{-1}([y_i,y_{i+1}]) = f^{-1}([i/n, (i+1)/n]) = [\sqrt{i/n},\sqrt{(i+1)/n}],$$&#xA;so the $n$th estimate, picking $y_i^* = y_i = i/n$ is just&#xA;$$\sum_{i=0}^n (i/n)\left(\sqrt{(i+1)/n} - \sqrt{i/n}\right).$$&#xA;Take the limit as $n\to\infty$, and you will get that the limit is $\frac{1}{3}$, as expected. (I will spare you the details; see the end of this answer for a high-power way of getting the answer similar to the way you do it with the Riemann integral).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that not every function is Lebesgue-integrable, just like not every function is Riemann-integrable. But every function that is Riemann-integrable will &lt;em&gt;also&lt;/em&gt; be Lebesgue integrable, &lt;em&gt;and&lt;/em&gt; the value of its Lebesgue integral will be the same as the value of its Riemann integral. But there are functions that are &lt;em&gt;not&lt;/em&gt; Riemann-integrable but &lt;em&gt;are&lt;/em&gt; Lebesgue-integrable (for example, the characteristic function of the rationals is Lebesgue-integrable, with integral $0$ over any interval, but is not Riemann-integrable). We also have a &quot;Fundamental Theorem of Calculus&quot; for the Lebesgue Integral:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; If $F$ is a differentiable function, and the derivative $F'$ is bounded on the interval $[a,b]$, then $F'$ is Lebesgue integrable on $[a,b]$ and&#xA;$$\int_a^x F'\,d\mu = F(x) - F(a).$$&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the integral is the Lebesgue integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, to &lt;em&gt;finally&lt;/em&gt; answer the question you ask about your example, since $F(x)=\frac{x^3}{3}$ is a differentiable function whose derivative is bounded over any finite interval, in particular over $[0,1]$, then from this theorem you can deduce that the integral over the interval $[0,1]$ of the derivative $F'(x)=x^2$ is equal to $F(1)-F(0)$; that is,&#xA;$$\int_0^1 x^2\,d\mu = \int_0^1 \left(\frac{x^3}{3}\right)'\,d\mu = \frac{1}{3} - \frac{0}{3} =\frac{1}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend the book &lt;em&gt;A Garden of Integrals&lt;/em&gt; by Frank E. Burk (Dolciani Mathematical Expositions 31, MAA, 2007, ISBN 9-780883-853375); it discusses and compares the Cauchy integral, the Riemann integral, the Riemann-Stieltjes integral, the Lebesgue integral, the Lebesgue-Stieltjes integral, and the Henstock-Kurzweil integral; it also discusses the Wiener and Feynman integral. I just finished reading it recently.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-22T03:19:27.760" LastActivityDate="2010-10-22T03:19:27.760" CommentCount="5" />
  <row Id="7446" PostTypeId="1" AcceptedAnswerId="7447" CreationDate="2010-10-21T19:11:38.707" Score="7" ViewCount="262" Body="&lt;p&gt;In &lt;a href=&quot;http://meta.math.stackexchange.com/questions/901/request-for-reopening-equivalence-of-curved-and-straight-lines/903#903&quot;&gt;this answer on meta&lt;/a&gt;, Pete L. Clark said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think the question concerns the idea that a differentiable curve becomes more and more like a straight line segment the closer one zooms in on its graph. (And I must say that I regard part of this confusion as an artifact of badly written recent calculus books who describe this phenomenon as &quot;local linearity&quot;. Ugh!)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, what's wrong with calling it &quot;local linearity&quot;?  (Examples of the specific language from some relatively recent books follow.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From Finney, Demana, Waits, and Kennedy's &lt;em&gt;Calculus: Graphical, Numerical, Algebraic&lt;/em&gt;, 1st ed, p107:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A good way to think of differentiable functions is that they are &lt;strong&gt;locally linear;&lt;/strong&gt; that is, a function that is differentiable at &lt;em&gt;a&lt;/em&gt; closely resembles its own tangent line very close to &lt;em&gt;a&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From Hughes-Hallett, Gleason, et al's &lt;em&gt;Calculus: Single Variable&lt;/em&gt;, 2nd ed, pp138-9:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When we zoom in on the graph of a differentiable function, it looks like a straight line.  In fact, the graph is not exactly a straight line when we zoom in; however, its deviation from straightness is so small that it can't be detected by the naked eye.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following that, there is discussion of the tangent line approximation, then a theorem titled &quot;Differentiability and Local Linearity&quot; (the first time &quot;local linearity&quot;/&quot;locally linear&quot; appears) stating that if a function &lt;em&gt;f&lt;/em&gt; is differentiable at &lt;em&gt;a&lt;/em&gt;, then the limit as &lt;em&gt;x&lt;/em&gt; goes to &lt;em&gt;a&lt;/em&gt; of the quotient of the error in the tangent line approximation and the difference between &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;a&lt;/em&gt; goes to 0.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ostebee and Zorn's &lt;em&gt;Calculus from Graphical, Numerical, and Symbolic Points of View&lt;/em&gt;, 1st ed, p110:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Remarkably, the just-illustrated strategy of zooming in to estimate slope almost always works.  Zooming in on the graph of almost any calculus function $f$, at almost any point $(a,f(a))$, eventually produces what looks like a straight line with slope $f'(a)$.  A function with this property is sometimes called &lt;strong&gt;locally linear&lt;/strong&gt; (or &lt;strong&gt;locally straight&lt;/strong&gt;) at $x=a$. [Margin note: &lt;em&gt;These aren't formal definitions, just descriptive phrases.&lt;/em&gt;]  Local linearity says, in effect, that $f$ &quot;looks like a line&quot; near $x=a$ and therefore has a well-defined slope at $x=a$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(I did not find the term &quot;local linearity&quot; or &quot;locally linear&quot; at a quick glance in Stewart's &lt;em&gt;Calculus: Concepts and Contexts&lt;/em&gt;, 2nd ed, or Leithold's &lt;em&gt;The Calculus 7&lt;/em&gt;; the rest of the calculus books I have on hand predate the inclusion of graphing calculators/software in textbooks, so are not suitable for comparison.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-22T04:30:33.770" Title="Is &quot;locally linear&quot; an appropriate description of a differentiable function?" Tags="&lt;calculus&gt;&lt;education&gt;&lt;teaching&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="7447" PostTypeId="2" ParentId="7446" CreationDate="2010-10-21T19:25:54.090" Score="7" ViewCount="0" Body="&lt;p&gt;None of the books you quoted actually called differentiability &quot;local linearity&quot;; they just used as a good analogy.  It is a good analogy, but it is not a good definition.  A lot of mathematical terminology, especially from topology, uses the word &quot;local,&quot; but it is almost always used with the same purpose.  A locally compact set is one where points have arbitrarily small compact neighborhoods.  A locally connected set is one where points have arbitrarily small connected neighborhoods.  On the other hand, a non-linear differentiable function is &lt;em&gt;not linear on any neighborhood&lt;/em&gt; of a point about which it is differentiable.  It just looks linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, such functions do have tangent lines, which is an equivalent definition of differentiability.  I don't think that a tangent counts as local linearity, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking whether people learning calculus should be taught the words &quot;local linearity&quot; instead of &quot;differentiability,&quot; I mean, that isn't really much more helpful than using the analogy but keeping the terminology.  If you're asking whether we should use the analogy at all, I don't see why not, as long as it's clear that linearity is just an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: &quot;Locally linear&quot; only describes differentiable functions $\mathbb{R}\rightarrow\mathbb{R}$.  &quot;Differentiable&quot; can be extended to functions between arbitrary differential manifolds.  I think it's better to keep the extensible definition.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-21T19:55:22.090" LastActivityDate="2010-10-21T19:55:22.090" CommentCount="7" />
  <row Id="7448" PostTypeId="2" ParentId="7433" CreationDate="2010-10-21T19:29:33.753" Score="1" ViewCount="0" Body="&lt;p&gt;The geometric distribution describes the probability of &quot;x trials are made before a success&quot;, and the negative binomial distribution describes that of &quot;x trials are made before $r$ successes are obtained&quot;, where $r$ is fixed. So you see that the latter is a particular case of the former, namely, when $r=1$.&#xA;As to your last question, it doesn't matter whether you call the variable $x$ or $k$, as long as you keep in mind that it takes on the values $0,1,2...$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-21T19:29:33.753" />
  <row Id="7449" PostTypeId="1" AcceptedAnswerId="7452" CreationDate="2010-10-21T19:30:57.073" Score="1" ViewCount="25" Body="&lt;p&gt;Given that a population (eg. of bacteria) grows exponentially with the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p 2^{\frac{t}{10}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is the initial population and $t$ is the time in minutes. Given that the population consumes (eg. food) at a steady rate, how would you calculate how much the population consumes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that each bacterium consumes $1µg$ of food per minute, how much food would an initial population of &lt;strong&gt;100&lt;/strong&gt; bacteria have consumed after 10 minutes? The solution lies somewhere between the consumption of 100 and 200 bacteria over 10 minutes, thus between $1000µg$ and $2000µg$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured I needed an integral to solve this problem. If I'm not mistaken, the integral of the function above is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{10} p 2^{\frac{t}{10}} dt = \frac{10 p 2^{\frac{t}{10}}}{\ln{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=100$ the result is $2885.39µg$, which is higher than the expected maximum of $2000µg$. Where did I go wrong?&lt;/p&gt;&#xA;" OwnerUserId="2680" LastActivityDate="2010-10-21T19:43:22.280" Title="Consumption of an exponentially growing population" Tags="&lt;integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7451" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T19:33:38.973" Score="1" ViewCount="0" Body="&lt;p&gt;The Riemann integral is pretty good and very intuitive, however the main reason to consider other types of integrals is that &quot;the space of functions that are Riemann integrable&quot;, say $R(I)$ where $I\subset\mathbb{R}^n$ is compact, is to small (it is a linear space in the sense you can add them and multiply by constants). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just look at a piecewise continuous function that vanish outside a bounded region and then you can go on with the Riemann integral. In mathematical analysis we look at various kinds of limits of functions and we would like the limit functions to stay in &quot;the space&quot; (we want the space to be complete). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the best we can do in the Riemann case is to look at uniformly convergent sequences $f_n$ on a compact interval $I\subset\mathbb{R}^n$ - in that case the limit $\lim f_n\in R(I)$  and $\lim\int f_n =\int \lim f_n$. However, uniform convergent is very rare! (Many Fourier series are not continuous even though there partial sums are, etc..).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lebesgue integral can be constructed in several ways (ending up with the same space though). A first try might be to start with norming $R(I)$, $\|f\|=\int|f|$ and then we would get a distance between $f,g\in R(I)$ by $\|f-g\|$, thus turning up with a metric space which we may complete by adding all possible limits - this will not work however because even though $R(I)$ is small it is to large (there are unbounded functions such that $\|f\|=\infty$). A better start would be to look at $C(I)$ = the space of continuous functions on (the compact set) $I$, (certainly each $f\in R(I)$ is a point-wise limit  of $C(I)$ functions) if we norm $C(I)$ in the same manner we would be a normed space and the completion of that space is $L^1(I)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $L^1$ you can sure take limits in norm and moreover, as has already been pointed out in other answers, you have many other better limit theorems such as Lebesgue dominated theorem or the monotone convergence theorem. Also, bounded functions of $R(I)$ do belong to $L^1(I)$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-21T19:33:38.973" />
  <row Id="7452" PostTypeId="2" ParentId="7449" CreationDate="2010-10-21T19:43:22.280" Score="1" ViewCount="0" Body="&lt;p&gt;You have a definite integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{0}^{10} p 2^{\frac{t}{10}} dt = \left[ \frac{10p2^{\frac{t}{10}}}{ln 2} \right]_{0}^{10} = \frac{10p}{ln2} = 1442.69 \mu g $ of food consumed.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-21T19:43:22.280" CommentCount="3" />
  <row Id="7453" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T19:57:10.563" Score="6" ViewCount="0" Body="&lt;p&gt;The sum telescopes since the summand $\rm\ f_n = g_{n+1} - g_n\:$ where $\rm\displaystyle\ g_n = \frac{1-n-y}{y-x-1}\ f_n$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-21T20:21:59.007" LastActivityDate="2010-10-21T20:21:59.007" CommentCount="4" />
  <row Id="7454" PostTypeId="1" CreationDate="2010-10-21T20:04:33.807" Score="15" ViewCount="441" Body="&lt;p&gt;A cohort in a school consists of 75 students who study for 6 years.  Each year, the students are randomly distributed into 3 classrooms of 25 students each.  What is the probability that, after 6 years, each student has at some point been in a classroom with every other student?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally:  Starting with an edgeless (undirected) graph on &lt;em&gt;cn&lt;/em&gt; vertices, a &lt;strong&gt;round&lt;/strong&gt; consists of first randomly partitioning the vertices into &lt;em&gt;c&lt;/em&gt; disjoint sets of &lt;em&gt;n&lt;/em&gt; vertices each, then adding an edge between every pair of not-yet-joined vertices that lie in the same set.  What is the probability that, after &lt;em&gt;y&lt;/em&gt; rounds, the result is a complete graph on &lt;em&gt;cn&lt;/em&gt; vertices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have estimates and solutions to special cases, and it's straightforward to find&#xA;the probability that a single given student sees all the others, but I don't know how to tackle the question in general.  (I do have a very pretty but completely useless expression for the exact answer, which I can supply if there's interest.)  In the case &lt;em&gt;c&lt;/em&gt;=3, &lt;em&gt;n&lt;/em&gt;=25, &lt;em&gt;y&lt;/em&gt;=6 it's clear that the answer is &quot;so close to zero that nobody can tell the difference&quot; but I was hoping for a more precise result.  Any guidance appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-11-06T13:36:21.480" Title="What is the probability that every pair of students studies together at some point?" Tags="&lt;combinatorics&gt;&lt;probability&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="7455" PostTypeId="2" ParentId="7436" CreationDate="2010-10-21T20:18:54.660" Score="1" ViewCount="0" Body="&lt;p&gt;You may want to consider the following sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lebesgue's theory of integration: its origins and development - Thomas Hawkins&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=QnXSqvTiEjYC&amp;amp;printsec=frontcover&amp;amp;dq=The+Calculus+Gallery+Masterpieces+from+Newton+to+Lebesgue&amp;amp;hl=en&amp;amp;ei=Jp3ATPT0I4voOfG25YwM&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCsQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The calculus gallery: masterpieces from Newton to Lebesgue - William Dunham&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=qgiLag9dPpMC&amp;amp;pg=PR3&amp;amp;dq=Carter+M.,+van+Brunt+B.+The+Lebesgue-Stieltjes+integral+-++A+practical+introduction&amp;amp;hl=en&amp;amp;ei=Mp7ATPz6G4SYOruT7ZcM&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCkQ6AEwAA#v=onepage&amp;amp;q=Carter%20M.%2C%20van%20Brunt%20B.%20The%20Lebesgue-Stieltjes%20integral%20-%20%20A%20practical%20introduction&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Lebesgue-Stieltjes integral: a practical introduction - Michael Carter, Bruce Van Brunt&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-10-21T20:18:54.660" />
  <row Id="7456" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T20:33:41.847" Score="1" ViewCount="0" Body="&lt;p&gt;This is not a stand-alone proof like the very nice one provided by Bill, but one can note that the sum is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_function#Special_values_at_z.C2.A0.3D.C2.A01&quot; rel=&quot;nofollow&quot;&gt;this identity for the hypergeometric function ${}_2F_1$&lt;/a&gt;. Let $a=x$, $b=1$, $c=y$, use $\Gamma(z+1)/\Gamma(z)=z$, and subtract 1 to compensate for the fact that the hypergeometric series starts with $n=0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-21T20:33:41.847" />
  <row Id="7457" PostTypeId="1" AcceptedAnswerId="7458" CreationDate="2010-10-21T20:54:14.590" Score="0" ViewCount="70" Body="&lt;p&gt;I'm trying to remember a particular theorem of ZF but unfortunately my memory is quite incomplete.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The theorem is of the form (some set operation) is either (expected answer) or the empty set.&lt;/li&gt;&#xA;&lt;li&gt;If the Axiom of Choice holds, (some set operation) is (expected answer).&lt;/li&gt;&#xA;&lt;li&gt;I believe that if (some set operation) is (expected answer) for all infinite sets, then the Axiom of Choice holds -- that is, the statement is equivalent to AC.&lt;/li&gt;&#xA;&lt;li&gt;The theorem has a formalized proof at &lt;a href=&quot;http://us.metamath.org/&quot; rel=&quot;nofollow&quot;&gt;Metamath&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think it was something like $|S\times S|=|S|$ for infinite sets $S$, or $\bigcup S$ is... something... I'm just not sure.  Does this ring a bell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies for the lack of well-defined information, and all the more for the possibility that some part of this is misleading.  I figured that, short of MathOverflow, this is just about the best place I could come to find my answer.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-10-21T21:02:31.000" Title="Specific equivalent to the Axiom of Choice involving the empty set" Tags="&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" />
  <row Id="7458" PostTypeId="2" ParentId="7457" CreationDate="2010-10-21T20:56:04.380" Score="3" ViewCount="0" Body="&lt;p&gt;I believe what you are looking for is that the Cartesian product of non-empty sets is non-empty.  It is equivalent to AC.  Without AC you cannot be sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tarski's theorem, which you quote in the form of cardinals:  for infinite $S$ there is a bijective map between $S×S$ and $S$ is also an equivalent.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-21T21:02:31.000" LastActivityDate="2010-10-21T21:02:31.000" CommentCount="4" />
  <row Id="7459" PostTypeId="1" AcceptedAnswerId="7507" CreationDate="2010-10-21T21:03:45.427" Score="2" ViewCount="61" Body="&lt;p&gt;I have a polygon P, that may or may not be convex. Is there an algorithm that will enable me to find the collection of points A that are at a distance less than d from P?&#xA;Is A in turn always a polygon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the solution change materially if we try to solve the problem on the surface of a sphere instead of on a Euclidean plane?&lt;/p&gt;&#xA;" OwnerUserId="2684" LastActivityDate="2010-10-22T13:18:27.520" Title="Find all points with a distance less than d to a (potentially not convex) polygon" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="3" />
  <row Id="7460" PostTypeId="2" ParentId="7459" CreationDate="2010-10-21T21:16:10.443" Score="3" ViewCount="0" Body="&lt;p&gt;It will not be a polygon.  If you think about the original polygon being a square of side s, the set A is a square of side s+2d, but with the corners rounded.  The corners become quarter circles with radius d and centered on the original corners of the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general polygon the situation is much the same.  Draw a parallel to the sides offset by d.  Then round the outer corners with a circular arc of radius d centered on the original corners and tangent to the new parallels.  The meeting points will be the intersection of the new line parallel to one side and the extension of the other side of the corner.  The inner corners will stay corners but get less deep and eventually disappear if d is large enough..&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-21T21:16:10.443" />
  <row Id="7461" PostTypeId="2" ParentId="7459" CreationDate="2010-10-21T21:41:37.660" Score="3" ViewCount="0" Body="&lt;p&gt;I heard Ravi Vakil give a series of plenary talks at an MAA conference on this question.  The talks were entitled &quot;The Mathematics of Doodling.&quot;  He generalized the question, though, by asking what happens in the limit when you start with some set in the plane and then iterate the process of finding the set of points at a distance $r$ from the current set.  Some interesting mathematics came out as he showed that the resulting sequence of points gets more and more circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a good picture of him doing this &lt;a href=&quot;http://www.maa.org/news/121407vakil.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; on the MAA website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also has a series of related problems that he developed for the Stanford Math Circle available &lt;a href=&quot;http://math.stanford.edu/circle/notes06f/doodling.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also according to his website, he has an article, &quot;The Mathematics of Doodling,&quot; that will be appearing in the February 2011 issue of the American Mathematical Monthly.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-21T21:57:35.050" LastActivityDate="2010-10-21T21:57:35.050" />
  <row Id="7462" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T21:52:07.323" Score="0" ViewCount="0" Body="&lt;p&gt;Nice observation Hans. I wanted to derive a very similar sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}=\sum \limits_{n=1}^{\infty}\displaystyle  \frac{x^{(n)}}{y^{(n)}}$ where $x^{(n)}$ is the rising factorial which can be written as $x^{(n)} = \displaystyle  \frac{\Gamma(x+n)}{\Gamma(x)}$. Then we have: $\displaystyle  \frac{\Gamma(y)}{\Gamma(x)}\sum\limits_{n=1}^{\infty}\frac{\Gamma(x+n)}{\Gamma(y+n)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, this sum converges to $\displaystyle \frac{\Gamma(x+1)\Gamma(y-x-1)}{\Gamma(y)\Gamma(y-x)}$. After simplication we obtain $\displaystyle  \frac{x}{y-x-1}$. However, I haven't been able to derive the result for  $\displaystyle \sum\limits_{n=1}^{\infty}\frac{\Gamma(x+n)}{\Gamma(y+n)}$&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-21T21:52:07.323" />
  <row Id="7463" PostTypeId="2" ParentId="7441" CreationDate="2010-10-21T22:43:55.433" Score="6" ViewCount="0" Body="&lt;p&gt;This is based on Robert Smith's observation above and Robin Chapman's beta trick in some previous problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\Gamma{(x+n)}\Gamma{(y-x)}}{\Gamma{(y+n)}} = \int_{0}^{1} t^{x+n-1} (1-t)^{y-x-1} dt$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;summing over $n$  we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma{(y-x)}\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \int_{0}^{1} \sum_{n \geq 1} t^{x+n-1} (1-t)^{y-x-1} dt = \int_{0}^{1} t^x (1-t)^{y-x-2}dt, $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{1}{\Gamma(y-x)}\int_{0}^{1} t^{x+1-1}(1-t)^{y-x-1-1}dt = \frac{\Gamma{(x+1)}\Gamma{(y-x-1)}}{\Gamma{(y-x)}\Gamma(y)} = \frac{\Gamma{(x+1)}}{(y-x-1)\Gamma(y)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\Gamma(y)}{\Gamma(x)} \sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{\Gamma(x+1)}{(y-x-1)\Gamma(x)} = \frac{x}{y-x-1}.$&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-21T22:50:42.490" LastActivityDate="2010-10-21T22:50:42.490" CommentCount="2" />
  <row Id="7464" PostTypeId="1" AcceptedAnswerId="7530" CreationDate="2010-10-21T22:48:55.670" Score="2" ViewCount="214" Body="&lt;p&gt;The question &lt;a href=&quot;http://math.stackexchange.com/questions/571/what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff&quot;&gt;what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff&lt;/a&gt; was asked and answered a while back.  This one has a much cleaner answer.  Now you are on a uniform slope, a line through the origin that is not horizontal and you want to throw a stone as far as possible.  As a function of the angle of the slope, what angle should you throw at?  Usual assumptions:  uniform gravity, no friction.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T01:30:58.460" LastActivityDate="2010-10-23T03:09:06.060" Title="What is the optimal angle when throwing a stone on a slope?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;physics&gt;&lt;optimization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7465" PostTypeId="1" CreationDate="2010-10-21T23:49:03.207" Score="1" ViewCount="81" Body="&lt;p&gt;List all the elements of the alternating group A_3 (A sub 3) written in cyclic notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come up with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identity (1)&#xA;Obviously (123)&lt;/p&gt;&#xA;" OwnerUserId="2686" LastActivityDate="2010-10-22T00:08:07.760" Title="Elements of alternating group A_3 (A sub 3)" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" />
  <row Id="7466" PostTypeId="2" ParentId="7465" CreationDate="2010-10-21T23:59:28.463" Score="0" ViewCount="0" Body="&lt;p&gt;Hint: If a group is cyclic, then find an element that is not the identity and find the group that is generated by that element. That is a subset of your group (and even more so a subgroup, albeit unrelated here). And if the group generated is the size of your desired group (you should know the size of $A_3$), then you have all the elements within your group. If you are unfamiliar with cyclic notation please reference the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycle_notation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-10-21T23:59:28.463" CommentCount="1" />
  <row Id="7467" PostTypeId="2" ParentId="7465" CreationDate="2010-10-22T00:08:07.760" Score="1" ViewCount="0" Body="&lt;p&gt;(1) (123) (132)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;got it&lt;/p&gt;&#xA;" OwnerUserId="2686" LastActivityDate="2010-10-22T00:08:07.760" />
  <row Id="7468" PostTypeId="1" AcceptedAnswerId="7472" CreationDate="2010-10-22T00:21:51.960" Score="1" ViewCount="48" Body="&lt;p&gt;Denoted as $\zeta(s,a)$ for a &gt; 0&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do I find topics on the Hurwitz zeta function for a &amp;lt; 0? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any links or resources would be appreciated.&#xA;(Please dont mention wiki or mathworld)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T00:47:31.233" LastActivityDate="2010-10-22T01:19:54.827" Title="On functions similar to Hurwitz zeta function" Tags="&lt;special-functions&gt;&lt;zeta-function&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="7469" PostTypeId="1" CreationDate="2010-10-22T00:23:03.087" Score="1" ViewCount="37" Body="&lt;p&gt;Consider the group of permutation matrices G ={I3, P1, P2, P3, P4, P5}&#xA;For n = 3 the permutation matrices are I3 and the five matrices are:&#xA;P1 = [1,0,0;0,0,1;0,1,0]&#xA;P2 = [0,1,0;1,0,0;0,0,1]&#xA;P3 = [0,1,0;0,0,1;1,0,0]&#xA;P4 = [0,0,1;0,1,0;1,0,0]&#xA;P5 = [0,0,1;1,0,0;0,1,0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write out the elements of a group of permutations that is isomorphic to G, and exhibit an isomorphism from G to this group!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it has to do with Cayley's Theorem.&#xA;With f_a:G-&gt;G defined by f_a(x) = ax for each a that exists in G...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about making a table, but realize I don't know how to since I am dealing with matrices. &lt;/p&gt;&#xA;" OwnerUserId="2686" LastActivityDate="2010-10-22T00:27:36.747" Title="Consider group of permutation matrices and write out elements isomorphic to the group and exhibit it." Tags="&lt;abstract-algebra&gt;" AnswerCount="1" />
  <row Id="7470" PostTypeId="2" ParentId="7469" CreationDate="2010-10-22T00:27:36.747" Score="1" ViewCount="0" Body="&lt;p&gt;HINT: Look at the result of multiplying each of those matrices by the vector $(1,2,3)^T$.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-10-22T00:27:36.747" CommentCount="6" />
  <row Id="7471" PostTypeId="1" CreationDate="2010-10-22T00:44:53.897" Score="1" ViewCount="65" Body="&lt;p&gt;I have a homework question &quot;Show the following is true using theorems. State which theorem you use at each step.&quot; This is just one of many problems I have! So, if you can help me with this one problem than I can apply what I learn to finish the rest. I don't want a handout; I just don't know where to start or what to do. I have looked through the different theorems in the book and I don't see how any would apply to this!&#xA;Am i suppose to prove that the first part  = the second?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D + BC'D' + A'BC + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Should I group two or more and work from there? Since, we are only working with '+' then I should only need these types of theorems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a book that shows the different theorems. But it still doesn't help me tackle this problem. Are we to prove how the first part equals the second part? To me that's expanding and not reducing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can work either side i guess whichever is easier..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of theorems the book provides&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X + 0 = X&#xA;X + 1 = 1&#xA;X + X = X&#xA;X + X' = 1&#xA;X + y = Y + x&#xA;(X + Y) + Z = X +(Y + Z)&#xA;x times (X + Y) = x&#xA;X times Y + X times Y' = x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;these are most of them. there are a few others..&#xA;The book i am using is digital logic - principles and practices.&lt;/p&gt;&#xA;" OwnerUserId="2688" LastEditorUserId="2688" LastEditorDisplayName="" LastEditDate="2010-10-22T02:20:37.667" LastActivityDate="2010-10-22T02:20:37.667" Title="boolean algebra - theorems" Tags="&lt;boolean-algebra&gt;&lt;duality-theorems&gt;" CommentCount="13" />
  <row Id="7472" PostTypeId="2" ParentId="7468" CreationDate="2010-10-22T00:47:00.593" Score="2" ViewCount="0" Body="&lt;p&gt;Well, there's the &lt;a href=&quot;http://dlmf.nist.gov/25.11&quot; rel=&quot;nofollow&quot;&gt;DLMF&lt;/a&gt; and the &lt;a href=&quot;http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta2/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Functions&lt;/a&gt; site... which people really should be checking out first when they encounter an unfamiliar &quot;special function&quot;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-22T00:47:00.593" CommentCount="3" />
  <row Id="7473" PostTypeId="1" CreationDate="2010-10-22T00:47:26.863" Score="1" ViewCount="296" Body="&lt;p&gt;For all $a, m, n \in \mathbb{Z}^+$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$&lt;/p&gt;&#xA;" OwnerUserId="2690" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T13:20:43.707" LastActivityDate="2010-12-04T13:20:43.707" Title="Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7474" PostTypeId="2" ParentId="7329" CreationDate="2010-10-22T00:53:42.100" Score="0" ViewCount="0" Body="&lt;p&gt;Player 1 throws on odd throws and player 2 throws on even throws. This is a geometric distribution. So his distribution is $\left(\frac{8}{9} \right)^{2k} \cdot \frac{1}{9}$ and player 2's distribution is $\left(\frac{8}{9} \right)^{2k-1} \cdot \frac{1}{9}$. So player 1 must give $1/8$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-22T00:53:42.100" />
  <row Id="7475" PostTypeId="1" CreationDate="2010-10-22T00:59:09.117" Score="0" ViewCount="130" Body="&lt;p&gt;Write out the elements of group permutation that is isomorphic to U(14)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G ={I3, P1, P2, P3, P4, P5} For n = 3 the permutation matrices are I3 and the five matrices are:&#xA;P1 = [1,0,0;0,0,1;0,1,0]&#xA;P2 = [0,1,0;1,0,0;0,0,1]&#xA;P3 = [0,1,0;0,0,1;1,0,0]&#xA;P4 = [0,0,1;0,1,0;1,0,0]&#xA;P5 = [0,0,1;1,0,0;0,1,0]&lt;/p&gt;&#xA;" OwnerUserId="2686" LastActivityDate="2010-12-26T20:44:27.977" Title="Find elements of group permutation that is isomorphic to U(14)" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7476" PostTypeId="1" CreationDate="2010-10-22T01:04:23.040" Score="5" ViewCount="78" Body="&lt;p&gt;Let $\mathbb{T}^n = { (z_1,\ldots,z_n) \in \mathbb{C}^n : |z_l| = 1, \; 1 \leq l \leq n }$ denote the $n$-torus, and let $t_1, \ldots, t_n$ be arbitrary real numbers. Then it can be shown that the topological closure $H$ in $\mathbb{T}^n$ of the one-parameter subgroup&#xA;$$H' = {(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in \mathbb{T}^n : y \in \mathbb{R}}$$&#xA;is an $r$-dimensional subtorus of $\mathbb{T}^n$, where $0 \leq r \leq n$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and furthermore that&#xA;$$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{g(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \: dy} = \int_{H}{g(z) \: d\mu_H(z)}$$&#xA;for any continuous function $g : \mathbb{T}^n \to \mathbb{C}$, where $\mu_H$ is the normalised Haar measure on $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, for which sets $B \subset \mathbb{T}^n$ do we have that&#xA;$$\lim_{Y \to \infty} \frac{1}{Y} \int\limits_{{y \in [0,Y] : (e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in B}}{dy} = \mu_H(B).&#xA;$$&#xA;I have a feeling that some sort of approximation by continuous functions should tell you, but I can't seem to make it work for some reason.&lt;/p&gt;&#xA;" OwnerUserId="2689" LastActivityDate="2010-10-22T01:04:23.040" Title="Ergodic flow in tori" Tags="&lt;measure-theory&gt;&lt;fourier-analysis&gt;" FavoriteCount="2" />
  <row Id="7478" PostTypeId="1" CreationDate="2010-10-22T01:30:16.043" Score="1" ViewCount="76" Body="&lt;p&gt;Find an isomorphism from octic group G to group G'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = {e, s, s^2, s^3, b, g, d, t}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e =  (1)  identity &#xA;s =  (1234)&#xA;s^2= (13)(24)&#xA;s^3= (1432)&#xA;b =  (14)(23)&#xA;g =  (24)&#xA;d =  (12)(34)&#xA;t =  (13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G' = {I2, R, R^2, R^3, H, D, V, T}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I2 = [1,0;0,1]  identity matrix&#xA;R =  [0,-1;1,0]&#xA;R^2= [-1,0;0,-1]&#xA;R^3= [0,1;-1,0]&#xA;H =  [1,0;0,-1]&#xA;D =  [0,1;1,0]&#xA;T =  [0,-1;-1,0]&lt;/p&gt;&#xA;" OwnerUserId="2686" LastEditorUserId="2686" LastEditorDisplayName="" LastEditDate="2010-10-22T01:36:58.027" LastActivityDate="2010-10-22T13:31:48.637" Title="Find an isomorphism from octic group G to group G'" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7479" PostTypeId="2" ParentId="7473" CreationDate="2010-10-22T01:51:26.423" Score="2" ViewCount="0" Body="&lt;p&gt;Below is a proof which has the neat feature that it immediately specializes&#xA;to a proof of the integer Bezout identity for $\rm\:x = 1\:$. It allows us to interpret the result as a &lt;strong&gt;q-analog of the integer Bezout identity&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. for $\rm\ m,\:n\ =\ 15,\:21$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\ \frac{x^3-1}{x-1}\ =\ (x^{15} + x^9 + 1)\ \frac{x^{15}-1}{x-1}\ -\ (x^9+x^3)\ \frac{x^{21}-1}{x-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for  $\rm\ x = 1\ $  specializes to  $\ 3\ =\ \ 3\ (15)\ -\ 2\ (21)\:,\ $ i.e. $\ (3)\ =\ (15,21)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DEFINITION&lt;/strong&gt; $\rm\displaystyle \quad n' \: :=\ \frac{x^n - 1}{x-1}\:$. $\quad$ &lt;strong&gt;NOTE&lt;/strong&gt; $\rm\quad  n' = n\ $  for  $\rm\ x = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\quad (m',n')\ =\ ((m,n)')\ $ for naturals $\rm\:m,\:n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PROOF&lt;/strong&gt; $\ $ It is trivially true if $\rm\ m = n\ $  or  if  $\rm\ m = 0\ $  or  $\rm\ n = 0\:$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence suppose  $\rm\:n &gt; m &gt; 0\:$. We proceed by induction on  $\rm\:n + m\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since  $\rm\:\ x^n - 1\  =\  x^r\ (x^m - 1)\  +\  x^r - 1\ \ $  for  $\rm\ \ r = n - m$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we obtain  $\rm\quad\: n'\  =\  x^r\ m'\ +\ r' \quad $   via  dividing the above by  $\rm\: x-1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ \ \ (m', n')\ =\ (m', r')\quad\ \ \ $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad = \  ((m,r)') \quad\ $   via  induction: $\rm\ m+r = n &amp;lt; m+n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad = \  ((m,n)')\quad\:$   via  $\rm\ r = n\ \:(mod\ m)\quad\quad\quad\quad\quad$ &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COROLLARY&lt;/strong&gt; $\ $ Integer Bezout Theorem $\ $ Proof: set $\rm\ x = 1\ $ above: i.e. erase primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A deeper understanding comes when one studies Divisibility Sequences&#xA;and Divisor Theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-22T01:51:26.423" />
  <row Id="7480" PostTypeId="2" ParentId="7232" CreationDate="2010-10-22T02:20:34.060" Score="0" ViewCount="0" Body="&lt;p&gt;Simply put, the tensions are equal because that's the way the problem is set up--it's a given, and you're supposed to use this information to solve for something.&lt;/p&gt;&#xA;" OwnerUserId="2527" LastActivityDate="2010-10-22T02:20:34.060" CommentCount="1" />
  <row Id="7481" PostTypeId="1" CreationDate="2010-10-22T02:23:39.170" Score="4" ViewCount="235" Body="&lt;p&gt;The question is:&#xA;Give an example of a function $f$, continuous on $S=(0,1) \times (0,1)$, such that $f(S)=\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm getting stuck on $f(x)=\tan\left(\pi \left(x-\frac{1}{2}\right)\right)$,&#xA;which covers all of $\mathbb{R}$ but not all of $\mathbb{R}^2$.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-22T02:45:23.580" LastActivityDate="2010-11-08T12:49:21.850" Title="Example of a continuous function from $(0,1) \times (0,1)$ to $\mathbb{R}^2$" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="7482" PostTypeId="2" ParentId="7446" CreationDate="2010-10-22T02:35:32.310" Score="3" ViewCount="0" Body="&lt;p&gt;@Paul and @Issac's comments on a wrong answer I had initially posted made me see why &quot;linear on zooming in&quot; is the right picture of differentiability. Suppose $f$ is some function and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x)=f(a)+m(x-a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a proposed linear approximation to it at a point $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I zoom in on the graph of $f$ and $g$ by a multiple $\lambda&gt;0$ and look at a point $h$ distance apart from $a$ in my zoomed coordinates. Then the vertical error in the approximation in my zoomed coordinates is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e(\lambda)=\lambda[f(a+h/\lambda)-g(a+h/\lambda)]&#xA;=\lambda[f(a+h/\lambda)-f(a)]-mh$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I keep zooming the approximation will look better and better only if $\lim_{\lambda \to \infty} e(\lambda)=0$. This will be the case only if $f$ is differentiable at $a$ and $f'(a)=m$.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-22T02:52:26.653" LastActivityDate="2010-10-22T02:52:26.653" CommunityOwnedDate="2010-10-22T02:52:26.653" />
  <row Id="7484" PostTypeId="2" ParentId="7475" CreationDate="2010-10-22T03:37:02.520" Score="1" ViewCount="0" Body="&lt;p&gt;If U(14) means the group of units modulo 14 under multiplication, then the units are {1,3,5,9,11,13}.  (as on page 2 of this document:  &lt;a href=&quot;http://math.bu.edu/people/rpollack/Teach/541fall09/HW3_Solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.bu.edu/people/rpollack/Teach/541fall09/HW3_Solutions.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  3 generates U(14), and therefore has order |U(14)|=6.  Since the group you're seeking is isomorphic, it is also generated by an element of order 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: there are no elements of order 6 in the symmetric group of order 3 (so 3 by 3 permutation matrices won't work).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-22T03:37:02.520" CommentCount="1" />
  <row Id="7486" PostTypeId="1" CreationDate="2010-10-22T03:48:22.513" Score="0" ViewCount="203" Body="&lt;p&gt;I am trying to make a truth table from an SOP boolean algebra expression. I understand AND, OR, NOT truth tables. I just dont understand these types of tables and there outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the expression&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D  + BC'D' + A'BC + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can use either side whichever is easier.. just let me know what side.&#xA;Would A' be a 1 and the others be a zero? I am also not sure how they get the output?&#xA;I understand the outputs of a AND, OR truth tables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i cant figure out these outputs. would this be considered an OR table since the expression is &quot;+&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would i just construct A, B, D with nots = 1 or zero?&#xA;Than, how do i determine the output?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; -----------------------&#xA; A  | B  | D  |  output&#xA;-----------------------&#xA;| 1 | 0  | 1  |   1?    |  A'BD'&#xA;------------------------&#xA;| 0 | 0  | 0  |   0?    |  BCD&#xA;------------------------&#xA;| 0 | 0  | 1  |   1?    |  ABC'&#xA;-------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i am trying to achieve is how the below expression is true using theorems.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D  + BC'D' + A'BC + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2688" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-22T15:06:37.087" LastActivityDate="2010-10-23T03:24:38.780" Title="how to make a truth table from an boolean expression" Tags="&lt;boolean-algebra&gt;" AnswerCount="4" />
  <row Id="7487" PostTypeId="2" ParentId="6722" CreationDate="2010-10-22T03:57:54.393" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/java/elements/elements.html&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would first really grasp his elements to add math into your report. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for fun check out how many postage stamps were done in his name around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jeff560.tripod.com/stamps.html&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2686" LastActivityDate="2010-10-22T03:57:54.393" CommentCount="3" />
  <row Id="7488" PostTypeId="2" ParentId="7486" CreationDate="2010-10-22T04:02:39.280" Score="1" ViewCount="0" Body="&lt;p&gt;Your truth table will have 2^4=16 lines.  Start with four columns labeled A,B,C,D.  On each line put a different combination of truth values for A,B,C, and D.  Then add more columns for the pieces of your expression.  So to build up the left hand side, make a column A'B and put in the truth values of that combination.  The prime on A is not and the multiply is AND.  So this column would show (NOT A) AND B for the values of A and B on that line.  Continue adding columns for other terms until you have built up the entire left hand side.  Then start on the terms that make up right hand side.  When you have built up the right hand side, compare the left hand side column and the right hand side column.  If they agree on all 16 lines, your equation is correct.  For a simple example, say we want to prove (A+B)'=A'B'  The table would look like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LIcli.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the (A+B)' column and the A'B' column agree everywhere, the equation is correct.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-22T04:02:39.280" CommentCount="1" />
  <row Id="7489" PostTypeId="2" ParentId="7486" CreationDate="2010-10-22T04:02:42.633" Score="0" ViewCount="0" Body="&lt;p&gt;I think the &quot;using theorems&quot; part of the question is intended to prevent you from using the exhaustive &quot;try every possibility&quot; technique (so don't use this technique if it's a homework question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this technique will work fine, you can just run through the $2^4=16$ possible values of (A,B,C,D) and check that both sides of the equation match.  This will be a rigorous proof of the identity -- just not a particularly intelligent one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I understand the notation correctly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;concatenation (i.e. writing the letters next to each other) is short for &quot;and&quot;, for example ABC is &quot;A and B and C&quot;,&lt;/li&gt;&#xA;&lt;li&gt;addition (i.e. +) is short for &quot;or&quot;, for example AB+CD is &quot;(A and B) or (C and D)&quot;&lt;/li&gt;&#xA;&lt;li&gt;adding a dash ' is the notation for negation, i.e. &quot;not&quot;, for example AB' is &quot;A and (not B)&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So A'BD' is the same as (not A) and B and (not D).  In your table, the first row is wrong since (not A) is false (because A is true=1).  The second row is correct since B is false=0 (regardless of what value C has).  The third row is wrong since A is false=0 (again, regardless of what values C has).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-22T04:02:42.633" CommentCount="7" />
  <row Id="7490" PostTypeId="2" ParentId="7446" CreationDate="2010-10-22T04:30:33.770" Score="4" ViewCount="0" Body="&lt;p&gt;The &quot;microscope&quot; or &quot;zooming in&quot; explanation of differentiation is basically nonsense. It contradicts the notion of zooming in actually used in calculus (asymptotic expansion of a function near a point, revealing finer and finer information).  Magnification matches the mathematics only at order 0 and 1, that is, it captures linear information but also destroys the higher order picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f(0)=0$. The $N$-fold magnification of the graph near 0 is the plot of $Nf(x/N)$.  This is the straight line $xf'(0)$ to within an error that shrinks to an invisible size as the picture is &quot;zoomed&quot; by increasing $N$.   This will, literally, linearize any differentiable function.  The higher order approximations -- whose effect is beautifully visible in a &lt;em&gt;fixed scale&lt;/em&gt; picture of the first several Taylor polynomials for $f(x)$ near 0 all superimposed on the same coordinate system -- are made indistinguishable from a straight line by flattening everything with the magnification.  This is a step backward considering that the difference between a tangent line and an osculating circle is easily illustrated with any curve, and many people see this picture in some form before learning calculus.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What one actually wants to display are the effects &lt;em&gt;away&lt;/em&gt; from the point, with the higher approximations being visibly closer to graph over a longer range.   The &quot;zooming&quot; picture may be defensible if presented together with these more informative images, but by itself as an explanation of the derivative as a local linear approximation, it is using diagrams that tell a story opposite to the mathematical concepts involved.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-22T04:30:33.770" />
  <row Id="7491" PostTypeId="1" CreationDate="2010-10-22T07:40:53.127" Score="1" ViewCount="72" Body="&lt;p&gt;If $X\sim \Gamma(a,\sigma_x^2)$ and $Y\sim \Gamma(b,\sigma_y^2)$. What will be the probability density function of R? Where $R=\frac{X+C}{X+Y}$, here $C$ is a positive constant, $\Gamma(.,.)$ denotes standard gamma probability density function and '$\sim$' represents 'distributed as'. X and Y are independent random variables.   &lt;/p&gt;&#xA;" OwnerUserId="1273" LastEditorUserId="1273" LastEditorDisplayName="" LastEditDate="2010-10-22T08:09:24.967" LastActivityDate="2010-10-26T00:43:34.470" Title="what will be the distribution of ratio of  correlated gamma distributed random variables?" Tags="&lt;probability-theory&gt;&lt;statistics&gt;&lt;random-variable&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7492" PostTypeId="1" CreationDate="2010-10-22T07:57:31.737" Score="0" ViewCount="134" Body="&lt;p&gt;Let $S \subset L^2(\mathbb{R})$, and let us define $S^{\perp} =\{f \in L^2(\mathbb R) | \left&amp;lt; f,g\right&gt; =0, g \in S\}$. Show that $S^{\perp}$ is a vector subspace of $L^2$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastActivityDate="2010-10-22T15:50:37.417" Title="Show that $S^{\perp}$ is a vector subspace of $L^2$." Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7493" PostTypeId="1" CreationDate="2010-10-22T08:03:37.243" Score="2" ViewCount="149" Body="&lt;p&gt;Can you solve rudin problem #19, I'm having a lot of difficult with it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've proven the first party namely lim N-&gt;infinity Sum n=1 to N exp(ik(x+nalpha)=1/(2pi)Integral[-pi,pi]… = { 1 if k=0 and 0 otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to prove if $f$ is continuous in $\mathbb{R}$ and $f(x+2\pi)=f(x)$ for all $x$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{N-&gt;\infty} \sum\limits_{n=1}^{N} f(x+n\alpha)=\frac{1}{2\pi} \int\limits_{-\pi}^{\pi}f(t) dt$$ for any $x$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha/\pi$ is irrational&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried writing it as lim n-&gt;infinity 1/N Sum n=1 to N Sum k=0 to N 1/(2pi)integral[-pi,pi]e^(ikt)f(x+nalpha) but that was no help&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-22T08:09:52.783" LastActivityDate="2010-10-22T08:09:52.783" Title="Fejer's Theorem (Problem in Rudin)" Tags="&lt;real-analysis&gt;" CommentCount="6" />
  <row Id="7494" PostTypeId="1" CreationDate="2010-10-22T08:18:48.517" Score="0" ViewCount="256" Body="&lt;p&gt;compute the fourier transform of f(t) = sin(t). Does that converge?&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T08:20:53.283" LastActivityDate="2010-10-22T12:52:50.863" Title="Compute the fourier transform of f(t) = sin(t)?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="7496" PostTypeId="2" ParentId="7494" CreationDate="2010-10-22T09:21:13.797" Score="-1" ViewCount="0" Body="&lt;p&gt;$sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Since, $e^{it}$ and $e^{-it}$ are basis function of the fourier transform. From this easily can be computed using duality property and there will be no convergence issue. &lt;/p&gt;&#xA;" OwnerUserId="1273" LastEditorUserId="1273" LastEditorDisplayName="" LastEditDate="2010-10-22T10:24:50.793" LastActivityDate="2010-10-22T10:24:50.793" CommentCount="2" />
  <row Id="7497" PostTypeId="1" AcceptedAnswerId="7501" CreationDate="2010-10-22T09:28:06.820" Score="4" ViewCount="389" Body="&lt;p&gt;Hi I have just found the proof that 90 equals 95 and was wondering if I have made some mistake. If so, which step in my proof is not true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitions:&lt;/strong&gt;&lt;br&gt;&#xA; 1. $\angle ABC=90^{\circ}$&lt;br&gt;&#xA; 2. $\angle BCD=95^{\circ}$&lt;br&gt;&#xA; 3. $|AB|=|CD|$&lt;br&gt;&#xA; 4. $M:=$ the center of $BC$&lt;br&gt;&#xA; 5. $N:=$ the center of $AD$&lt;br&gt;&#xA; 6. $l:=$ a line perpendicular to $BC$ passing through $M$&lt;br&gt;&#xA; 7. $m:=$ a line perpendicular to $AD$ passing through $N$&lt;br&gt;&#xA; 8. $S:=$ is the cross-section of $l$ and $m$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Based on definitions 1 through 8 we can draw the following image:&lt;/strong&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/CBKzz.jpg&quot; alt=&quot;90 = 95&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Based on the definitions we can derive the following:&lt;/strong&gt;&lt;br&gt;&#xA; 9. $\triangle BSC$ is isosceles (follows from 4, 6 and 8)&lt;br&gt;&#xA; 10. $\triangle ASD$ is isosceles (follows from 5, 7 and 8)&lt;br&gt;&#xA; 11. $|BS|=|CS|$ (follows from 9)&lt;br&gt;&#xA; 12. $|AS|=|DS|$ (follows from 10)&lt;br&gt;&#xA; 13. $\triangle ABS\cong\triangle DCS$ (follows from 3, 11 and 12)&lt;br&gt;&#xA; 14. $\angle ABS=\angle DCS$ (follows from 13)&lt;br&gt;&#xA; 15. $\angle CBS=\angle BCS$ (follows from 9)&lt;br&gt;&#xA; 16. $\angle ABC=\angle ABS - \angle CBS=\angle DCS-\angle BCS=\angle BCD$ (follows from 14 and 15)&lt;br&gt;&#xA; 17. $90^{\circ}=95^{\circ}$ (follows from 1, 2 and 16)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: point $S$ is indeed lying above $BC$. If however it would be below $BC$, then the 'minus' in step 16 would simply have to be changed into a 'plus'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note: The image is not drawn to scale. It only serves as to provide the reader with a intuitive view of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also also note: before posting, I have first investigated if this type of question would be appropriate. Based on &lt;a href=&quot;http://math.stackexchange.com/questions/287/how-come-32-5-31-5&quot;&gt;How come 32.5 = 31.5?&lt;/a&gt; and the following meta &lt;a href=&quot;http://meta.math.stackexchange.com/questions/68/questions-about-math-jokes/70#70&quot;&gt;Questions about math jokes&lt;/a&gt; I have decided to post this question.&lt;/p&gt;&#xA;" OwnerUserId="2262" LastEditorUserId="2262" LastEditorDisplayName="" LastEditDate="2010-10-25T07:21:44.120" LastActivityDate="2010-10-25T07:21:44.120" Title="What is wrong in my proof that 90 = 95? Or is it correct?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;puzzle&gt;&lt;fake-paradoxes&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="7498" PostTypeId="2" ParentId="7494" CreationDate="2010-10-22T10:19:22.380" Score="3" ViewCount="0" Body="&lt;p&gt;If you consider $f:=\sin$ as a function on $\mathbb R$ then it has a Fourier transform only in the distribution sense, as remarked in H.L.'s comment. If, however, you consider $\sin$ as a function on $\mathbb R/(2\pi)$ then its Fourier transform $\hat f$ is a function on $\mathbb Z$, namely the doubly infinite sequence $(c_k)_{k\in\mathbb Z}$ obtained by integration of $\sin$ against the functions $t\mapsto e^{-ikt}$: $$c_k={1\over 2\pi}\int_{-\pi}^\pi \sin(t)e^{-ikt}dt\qquad (k\in\mathbb Z).\qquad(*)$$ As $\sin$ is a nice function Fourier theory guarantees that we can get $\sin (t)$ back using the $c_k$ as follows:  $$\sin (t)=\sum_{k=-\infty}^\infty  c_k e^{ikt}.$$ Now, by Euler's formula, $\sin(t) =(e^{it}-e^{-it})/(2i)$, so we can immediately read off $\hat f(1)=c_1=-i/2$, $\hat f(-1)=c_{-1}= i/2$ and $\hat f(k)=c_k=0$ otherwise, without doing the integration (*).&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-10-22T12:52:50.863" LastActivityDate="2010-10-22T12:52:50.863" CommentCount="2" />
  <row Id="7499" PostTypeId="1" AcceptedAnswerId="9509" CreationDate="2010-10-22T10:40:11.833" Score="2" ViewCount="92" Body="&lt;p&gt;I have a question pertaining to linear separability with hyperplanes in a support vector machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Support_vector_machineclassifier.&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...formally, a support vector machine&#xA;  constructs a hyperplane or set of&#xA;  hyperplanes in a high or infinite&#xA;  dimensional space, which can be used&#xA;  for classification, regression or&#xA;  other tasks. Intuitively, a good&#xA;  separation is achieved by the&#xA;  hyperplane that has the largest&#xA;  distance to the nearest training data&#xA;  points of any class (so-called&#xA;  functional margin), since in general&#xA;  the larger the margin the lower the&#xA;  generalization error of the&#xA;  classifier.classifier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The linear separation of classes by hyperplanes intuitively makes sense to me. And I think I understand linear separability for two-dimensional geometry. However, I'm implementing an SVM using a popular SVM library (libSVM) and when messing around with the numbers, I fail to understand how an SVM can create a curve between classes, or enclose central points in category 1 within a circular curve when surrounded by points in category 2 if a hyperplane in an n-dimensional space V is a &quot;flat&quot; subset of dimension n − 1, or for two-dimensional space - a 1D line.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Here is what I mean:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/wqZMd.png&quot; alt=&quot;circularly enclosed class separation for a 2D binary SVM&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not a hyperplane. That's circular. How does this work? Or are there more dimensions inside the SVM than the two-dimensional 2D input features?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2010-11-09T00:36:34.167" Title="Help me understand linearly separability in a binary SVM" Tags="&lt;algorithms&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7500" PostTypeId="2" ParentId="7492" CreationDate="2010-10-22T10:47:51.643" Score="3" ViewCount="0" Body="&lt;p&gt;I think you mean, show that  $S^\bot = \lbrace f \in L^2| \int_\mathbb{R} fg dx = 0 , \forall g \in S \rbrace$ is a vector subspace of $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f, g  \in S^\bot$ and $\alpha, \beta \in \mathbb{R}$, then for $h \in S$,  $\int_\mathbb{R} (\alpha f + \beta g)h dx = \alpha\int_\mathbb{R}fh dx + \beta\int_\mathbb{R}gh dx = \alpha 0 + \beta 0 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\alpha f + \beta g \in S^\top$, and so $S^\bot$ is a vector subspace.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-10-22T10:47:51.643" />
  <row Id="7501" PostTypeId="2" ParentId="7497" CreationDate="2010-10-22T11:00:35.287" Score="4" ViewCount="0" Body="&lt;p&gt;Your statement 16 depends on the way in which the diagram is drawn to know that ∠DCS = ∠DCB + ∠BCS.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-22T16:00:55.150" LastActivityDate="2010-10-22T16:00:55.150" CommentCount="5" />
  <row Id="7503" PostTypeId="1" CreationDate="2010-10-22T11:56:16.970" Score="-5" ViewCount="205" Body="&lt;p&gt;Use fourier transform and convolution to show that  $$\int_{-\infty}^{\infty}\frac{(\sin(t))^3}{t^3}dt=\frac{3\pi}{4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;convolution is defined by $$(f * g)(t) = \int_{-\infty}^{\infty}(f(s)*g(t-s)ds $$. Fourier transform defined by $$\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-21T06:26:42.560" LastActivityDate="2010-11-21T06:26:42.560" Title="show that  $$\int_{-\infty}^{\infty}\frac{(\sin(t))^3}{t^3}dt=\frac{3\pi}{4}$$" Tags="&lt;calculus&gt;&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7505" PostTypeId="2" ParentId="7486" CreationDate="2010-10-22T13:12:27.797" Score="3" ViewCount="0" Body="&lt;p&gt;A, B, C, and D are boolean variables, meaning that each takes the value&#xA;&quot;true&quot; or &quot;false&quot;.  More complex expressions have value &quot;true&quot; or &quot;false&quot;&#xA;depending on the values of these variables, so for example A'BD' is true if&#xA;A is false, B is true, and D is false, and C is either true or false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that F = G, where F and G are complex expressions, means that, no&#xA;matter what values the boolean variables have, the value of F is the same&#xA;as the value of G (that is, F and G are either both true or both false).&#xA;So, for example, we have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     AB + AC = A(B+C)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;because if A is true and either B or C is true, then both sides are true,&#xA;and in any other case both sides are false---there's no way to assign&#xA;values to A, B, C such that the two side come out differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As previous posters have mentioned, you can always prove (or disprove!)  an&#xA;equality by going through all possible assignments of &quot;true&quot; and &quot;false&quot; to&#xA;the variables.  The goal seems to be to prove &quot;by theorems&quot;, that is, using&#xA;operations previously proven true.  As you say, we can either manipulate&#xA;one side of the equation until it has the form of the other side, or we&#xA;could manipulate both sides and get them into a common form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the first thing to notice is that the right-hand side (RHS)&#xA;is a copy of the LHS with some extra stuff tacked onto the end.  For this&#xA;reason it's simplest to manipulate just the RHS to get rid of the&#xA;surplussage!  We can start with this basic theorem:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  if Q is true whenever P is true, then Q = Q + P&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can prove this theorem by systematically considering all possibilities&#xA;for P and Q.  Or look at it this way:  If Q is true, then both sides of the&#xA;equation are true.  And if Q is false, then P must be false (since, by&#xA;assumption, if P were true Q would be true) hence both sides are false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this theorem we prove:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  XY + X'Z = XY + X'Z + YZ     &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Proof:  Let P be YZ and let Q be XY+X'Z.  Suppose P is true, that is, Y and&#xA;Z are both true.  But if Y and Z are both true, then XY+X'Z must be true.&#xA;(Reason:  If X is true then, since Y is true, XY is true.  And if X is&#xA;false then, since Z is true, X'Z is true.  So in either case XY+X'Z is&#xA;true.)  So we've shown that Q is true whenever P is true, hence by the&#xA;previous theorem Q = Q + P, which in this case is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can prove&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A'D' + AC' = A'D' + AC' + C'D'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is just the same as the previous theorem, putting A for X, D' for Y,&#xA;and C' for Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last theorem implies&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; B(A'D' + AC') = B(A'D' + AC' + C'D')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is the same as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A'BD' + ABC' = A'BD' + ABC' + BC'D'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Given this, we can take the RHS of the original and substitute A'BD' + ABC'&#xA;for A'BD' + ABC' + BC'D', which is to say, we can drop the term BC'D'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With steps similar to the above we can prove these two theorems:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A'BD' + BCD = A'BD' + BCD + A'BC&#xA; BCD + ABC' = BCD + ABC' + ABD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which permit us to drop the final two terms of the RHS of the original,&#xA;completing the proof.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-10-22T13:12:27.797" CommentCount="2" />
  <row Id="7506" PostTypeId="1" AcceptedAnswerId="8192" CreationDate="2010-10-22T13:14:28.230" Score="2" ViewCount="87" Body="&lt;p&gt;Let $A$ be an abelian category. There are various types of full subcategories. I often wonder if it is assumed that these are nonempty, since in most proofs this is used implicitely, but also the empty case then works as a special case. To be sure, I ask you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called thick iff $S$ is closed under subquotients and extensions. Note that $\emptyset$ is thick according to this definition. Is this allowed in the literature you know? Remark that $S$ is nonempty iff $0 \in S$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called topologizing iff $S$ is closed under subquotients and direct sums. The same as above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called Serre iff $S = S^{-}$. Here $S^-$ consists of those objects, such that every nonzero subquotient of it contains a nonzero subobject isomorphic to an object of $S$. Since $0 \in S^{-}$, this always implies that $S$ is nonempty, ok.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S \subseteq A$ is called localizing iff $S$ is thick and $A \to A/S$ admits a right adjoint. I think that the construction $A/S$ makes only sense if $S$ is nonempty. Thus we should add that $S$ is nonempty (i.e. $0 \in S$)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-28T07:44:24.003" Title="Is the empty subcategory thick, localizing, topologizing, etc." Tags="&lt;category-theory&gt;&lt;homological-algebra&gt;" AnswerCount="1" />
  <row Id="7507" PostTypeId="2" ParentId="7459" CreationDate="2010-10-22T13:18:27.520" Score="4" ViewCount="0" Body="&lt;p&gt;What you seek is known as the &lt;em&gt;offset curve&lt;/em&gt; of a polygon.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Offset_curve&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the Wikipedia article.&#xA;This is closely related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_sum&quot; rel=&quot;nofollow&quot;&gt;Minkowski sum construction&lt;/a&gt;, in this case the Minkowski sum of a polygon and a disk.  As others already indicated, the offset of a polygon is not in general&#xA;a polygon.  There is a substantial literature on the topic, which can be reached with these&#xA;key phrases.  There are algorithms; none are easy.  Here is one:&#xA;&quot;An offset algorithm for polyline curves,&quot; &lt;em&gt;Computers in Industry&lt;/em&gt;,&#xA;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1228072&quot; rel=&quot;nofollow&quot;&gt;Volume 58,Issue 3 (April 2007)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-22T13:18:27.520" CommentCount="1" />
  <row Id="7508" PostTypeId="2" ParentId="7478" CreationDate="2010-10-22T13:31:48.637" Score="1" ViewCount="0" Body="&lt;p&gt;Jessica: Preserving the order of all elements isn't sufficient; you have to show that it preserves the group operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a hint though.  $G^\prime$ is the set of symmetries of a square, right?  Try labelling the square's corners.  Then each symmetry is just a permutation of its corners...&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-10-22T13:31:48.637" />
  <row Id="7509" PostTypeId="2" ParentId="7376" CreationDate="2010-10-22T14:18:26.187" Score="2" ViewCount="0" Body="&lt;p&gt;Interior points of a convex set $K$ in $R^n$ are the points in what could be called the &quot;convex interior&quot; of the set: they are inside (the part with positive barycentric coordinates of) at least one nondegenerate &lt;em&gt;n&lt;/em&gt;-dimensional simplex with vertices in $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a point in the interior of the convex set $\overline{C}$, surround it by a nondegenerate $n$-dimensional simplex with vertices in $\overline{C}$.  Because $C$ is dense in its closure, we can perturb the simplex very slightly into one with vertices in $C$, and this simplex will continue to contain the given interior point.  (This is because, for example, the barycentric coordinates are continuous functions of the simplex vertices, as long as the simplex does not degenerate, so a small perturbation will keep the coordinates positive and the point strictly inside the simplex).&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-23T03:47:20.380" LastActivityDate="2010-10-23T03:47:20.380" CommentCount="2" />
  <row Id="7510" PostTypeId="1" CreationDate="2010-10-22T15:54:20.227" Score="0" ViewCount="41" Body="&lt;p&gt;I need to show that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;||(ATA)^-1||2=(||A+||2)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hint:use singular decomposition.I need to proof this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="2706" LastActivityDate="2010-10-22T15:54:20.227" Title="proof with singular value decomposition help me?" Tags="&lt;linear-algebra&gt;" CommentCount="3" />
  <row Id="7511" PostTypeId="1" CreationDate="2010-10-22T16:14:47.553" Score="7" ViewCount="320" Body="&lt;p&gt;Consider (non-trivial) functions that are their own nth derivatives.  For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^2}{\mathrm{d}x^2} e^{-x} = e^{-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^3}{\mathrm{d}x^3} e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2}) = e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}^4}{\mathrm{d}x^4} \sin x = \sin x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_n(x)$ be the function that is it's own nth derivative.  I believe (but I'm not sure) for nonnegative integer n, this function can be written as the following infinite polynomial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n(x) = 1 + \cos(\frac{2\pi}{n})x + \cos(\frac{4\pi}{n})\frac{x^2}{2!} + \cos(\frac{6\pi}{n})\frac{x^3}{3!} + \cdots + \cos(\frac{2t\pi}{n})\frac{x^t}{t!} + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some sense in which this function can be extended to real n using fractional derivatives?  Would it then be possible to graph $z(n, x) = f_n(x)$, and would this function be smooth and continuous on both n and x axes?  Or would it have many discontinuities?&lt;/p&gt;&#xA;" OwnerUserId="2704" LastActivityDate="2010-10-23T18:28:39.117" Title=" Functions that are their Own nth Derivatives for Real n" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="7512" PostTypeId="1" CreationDate="2010-10-22T16:14:54.200" Score="2" ViewCount="331" Body="&lt;p&gt;Hello all,&#xA;  I am a EE grad. student who has had one undergraduate course in real analysis (which was pretty much the only pure math course that I have ever done). I would like to do a self study of some basic functional analysis so that I can be better prepared to take a graduate course in that material in my university. I plan to do that next fall so I do have some time to work through a book fully. Could some one recommend some good books to start working on this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="2707" LastActivityDate="2010-10-26T23:44:06.313" Title="Good book for self study of functional analysis" Tags="&lt;functional-analysis&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="1" />
  <row Id="7513" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T16:20:50.443" Score="18" ViewCount="0" Body="&lt;p&gt;The set of functions $f$ such that $f^{(n)}-f=0$ is a vector space of dimension $n$, spanned by the functions $e^{\lambda t}$ with $\lambda$ an $n$th root of unity. In particular, there are many such functions, not just one: the general such function is of the form $$f(t)=\sum_{k=0}^{n-1}a_ke^{\frac{2\pi i k}{n}t}.$$&#xA;This is explained in every text on ordinary diffential equations; I remember fondly, for example, &lt;em&gt;Theory of Ordinary Differential Equations&lt;/em&gt; by Earl A. Coddington and Norman Levinson, but I am sure you can find more modern expositions in every library.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-22T16:29:26.920" LastActivityDate="2010-10-22T16:29:26.920" CommentCount="2" />
  <row Id="7514" PostTypeId="1" CreationDate="2010-10-22T16:27:49.493" Score="3" ViewCount="234" Body="&lt;p&gt;I sincerely apologize if this question isn't appropriate here. Originally, I wanted to post this question on Overflow because I wanted to hear the opinions of mathematicians who are doing research, but it probably really is not appropriate to post there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, I was looking for alternative ways to carry my batch of papers aside from the laptop, and Kindle came into my mind. I know they recently came out with a new one, and I was wondering if anyone has tried using it for storing, carrying, and reading math article or books, and what their experience was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the Kindle, are there any other electronic tablet or pads, or other devices that you'd recommend for this type of purpose?&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2010-10-22T17:24:21.373" Title="Kindle as a Tool for Mathematicians?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="7515" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T16:28:41.713" Score="3" ViewCount="0" Body="&lt;p&gt;The solutions to a homogeneous linear equation with constant coefficients&#xA;$$ a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = 0$$&#xA;correspond to roots of the &quot;auxiliary polynomial&quot; $a_nt^n + \cdots + a_0$. If the polynomial has roots $r_1,\ldots,r_k$ (in complex numbers), with multiplicities $a_1,\ldots,a_k$, then a basis for the solutions is given by&#xA;$${ e^{r_1x}, xe^{r_1x},\ldots, x^{a_1-1}e^{r_1x},e^{r_2x},\ldots,x^{a_k-1}e^{r_kx}}.$$&#xA;Here, the complex exponential is used, so that if $a$ and $b$ are real numbers and $i$ is the square root of $-1$, we have&#xA;$$e^{a+bi} = e^a(\cos(b) + i \sin(b)).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you are looking at the polynomial $t^n - 1 = 0$, whose roots are the $n$th roots of unity. They are all distinct; so you can either take the complex exponentials, or the real and complex parts. So if $\lambda$ is a primitive $n$th root of $1$, then a basis for the space of solutions is&#xA;$$ e^x, e^{\lambda x}, e^{\lambda^2 x},\ldots,e^{\lambda^{n-1}x}.$$&#xA;The general solution is a linear combination of these with complex coefficients:&#xA;$$f(x) = a_0e^x + a_1e^{\lambda x} + a_2e^{\lambda^2x}+\cdots+a_{n-1}e^{\lambda^{n-1}x},\qquad a_0,\ldots,a_{n-1}\in\mathbb{C}.$$&#xA;If you don't want complex values, you can take a general form as above, and take the real and complex parts separately.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-22T16:28:41.713" CommentCount="3" />
  <row Id="7516" PostTypeId="1" CreationDate="2010-10-22T16:36:35.627" Score="1" ViewCount="91" Body="&lt;p&gt;For example, I have heard of a topological one wherein negation means the interior of the complement (but still would like a reference).&lt;/p&gt;&#xA;" OwnerUserId="1575" LastActivityDate="2010-12-02T23:07:35.427" Title="What are good elementary examples for teaching/introducing/learning about Intuitionistic Logic or Heyting Algebras?" Tags="&lt;logic&gt;&lt;teaching&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7517" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T16:50:53.760" Score="1" ViewCount="0" Body="&lt;p&gt;When you say non-trivial it looks like you mean such that the $m^{th}$ derivative of $f$ is not equal to $f$ for every $m &amp;lt; n$.  Unfortunately, this is still not enough to define $f_n$ uniquely; for any positive integer $1 \le k \le n$ such that $\gcd(k, n) = 1$, any nontrivial linear combination of the $\phi(n)$ functions $e^{ \frac{2\pi i k}{n} x}$ works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to make a consistent choice across all values of $n$ is to take $k = 1$; in that case, you might want to choose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_n(x) = e^{ \frac{2\pi i}{n} x } = \cos \frac{2\pi x}{n} + i \sin \frac{2\pi x}{n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is well-defined for any complex number $n \neq 0$ and all complex numbers $x$, and it is probably true that the $n^{th}$ fractional derivative of $f_n$ is equal to $f_n$.  There is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Essential_singularity&quot; rel=&quot;nofollow&quot;&gt;essential singularity&lt;/a&gt; at $n = 0$ for fixed nonzero $x$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-22T19:32:51.900" LastActivityDate="2010-10-22T19:32:51.900" CommentCount="1" />
  <row Id="7518" PostTypeId="2" ParentId="7514" CreationDate="2010-10-22T16:52:39.887" Score="-1" ViewCount="0" Body="&lt;p&gt;I am also very interested in this question. I am particularly interested in how .pdf files display - are they all good, or do some look wonky, under different magnification levels, etc.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2010-10-22T16:52:39.887" CommentCount="2" />
  <row Id="7519" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T17:06:28.217" Score="1" ViewCount="0" Body="&lt;p&gt;Not to scare you, but list of requirements for a first course in functional analysis is rather long:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Basic theorems of metric spaces including, but not limited to:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Baire category theorem&lt;/li&gt;&#xA;&lt;li&gt;$\ell^p$ is complete&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem&quot; rel=&quot;nofollow&quot;&gt;Arzelà-Ascoli&lt;/a&gt; (how else will you show that an operator is compact?)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Measure theory --- or at least be ready to accept that you have to learn some while reading functional analysis. Because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riesz_representation_theorem&quot; rel=&quot;nofollow&quot;&gt;Riesz representation theorem&lt;/a&gt; essentially says that for a big class of &quot;reasonable&quot; spaces, continuous linear functionals and measures are the same. In other words a lot of the theory will make no sense without at least knowing some measure theory.&lt;/li&gt;&#xA;&lt;li&gt;Topology. If you want to go beyond Banach spaces and study Fréchet spaces. The continuous dual of a Fréchet space that is not a Banach space is not necessarily metrisable --- and you get to work with multiple different topologies on your spaces (weak, strong, weak-*)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If that doesn't scare you off, I can recommend the information-dense &quot;Introduction to Functional Analysis&quot; by Reinhold Meise and Dietmar Vogt. ISBN 0-19-851485-9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when I say dense i mean &lt;em&gt;very&lt;/em&gt; dense. It clocks in at a modest 437 pages, yet in a late undergraduate course in functional analysis we covered less than a third of that book (plus some notes on convexity) in a semester. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Rudin's Real &amp;amp; Complex Analysis: it's a great book, but I don't know if I'd really call it a book on functional analysis. I'd say it's on analysis in general --- hence the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: If you find that you need to brush up on real analysis, Terence Tao has notes for 3 courses on his webpage: Real Analysis &lt;a href=&quot;http://en.wordpress.com/tag/245a-real-analysis/&quot; rel=&quot;nofollow&quot;&gt;245A&lt;/a&gt;  (in progress at the time of writing), &lt;a href=&quot;http://en.wordpress.com/tag/245b-real-analysis/&quot; rel=&quot;nofollow&quot;&gt;245B&lt;/a&gt; and &lt;a href=&quot;http://en.wordpress.com/tag/245c-real-analysis/&quot; rel=&quot;nofollow&quot;&gt;245C&lt;/a&gt;. Actually I think I can highly recommend the entirety of his webpage.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T09:56:25.497" LastActivityDate="2010-10-23T09:56:25.497" CommentCount="6" />
  <row Id="7520" PostTypeId="1" CreationDate="2010-10-22T17:17:05.830" Score="2" ViewCount="337" Body="&lt;p&gt;I am a graduate student of math right now but i was not able to get a topology subject in my undergrad... i just would like to know if you guys know the best one..&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2010-10-22T23:54:22.627" Title="best book for topology?" Tags="&lt;topology&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="2" />
  <row Id="7521" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T17:24:11.843" Score="6" ViewCount="0" Body="&lt;p&gt;I would suggest the following options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Topology by James Munkres&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;General Topology by Stephen Willard&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Basic Topology by M.A. Armstrong&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can take a look at Allen hatcher's webpage for more books on introductory topology. He has a .pdf file containing some very good books. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-22T17:24:11.843" CommunityOwnedDate="2010-10-22T17:24:11.843" CommentCount="5" />
  <row Id="7522" PostTypeId="2" ParentId="7514" CreationDate="2010-10-22T17:24:21.373" Score="1" ViewCount="0" Body="&lt;p&gt;I have the previous-generation Kindle, though I think the software and screen are essentially the same.  I think PDF would probably be the only convenient format and letter-size PDFs are a bit small to read without having to zoom and pan.  Personally, I wouldn't want to deal with a textbook on it—in fact, I wouldn't want to deal with textbooks on any tablet device I've seen thus far because you lose physical features like looking at two books at once and flipping through the book—but I'd certainly consider using it for articles or other shorter works.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-22T17:24:21.373" CommentCount="4" />
  <row Id="7523" PostTypeId="2" ParentId="7516" CreationDate="2010-10-22T17:30:03.603" Score="2" ViewCount="0" Body="&lt;p&gt;The example that you mention in you question is explained in Maclane and Moerdijk  's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540977104&quot; rel=&quot;nofollow&quot;&gt;Sheaves in Geom and Logic&lt;/a&gt; very near the beginning. They discuss this in relation to Heyting algebras for the purpose of Topos Theory. This book is very very nice, I highly recommend it.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-22T17:30:03.603" CommentCount="2" />
  <row Id="7524" PostTypeId="1" AcceptedAnswerId="7546" CreationDate="2010-10-22T17:31:30.777" Score="3" ViewCount="70" Body="&lt;p&gt;I am working with $G_{n,p}$ random graphs. Let $p$ (the probability of an edge) be $c/n$, so $c$ is the expected number of neighbors for a vertex when $n$ is large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $N$ be the number of components with more edges than vertices in the graph. I am trying to find the largest $c$ such that $\mathbb{E}(N)$ tends to zero as $n$ grows large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{E}(N) = \sum_{k=0}^n {n \choose k} \mathbb{E}(X_k)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $X_k$ is the indicator that we have a component with $k$ vertices and $k+1$ edges or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{E}(X_k) = \mathbb{P}(\text{no outgoing edges}) \mathbb{P}(k+1 \text{ edges or more})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{E}(X_k) = (1-p)^{k(n-k)} \mathbb{P}(\text{Binomial}({k \choose 2}, p) \ge k+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am stuck when trying to find an upper bound for this expression. The inequalities I've tried either diverge or cause trouble in the summation. Hoeffding's inequality looked promising but gave me $k^2$ in the exponent which gave me a hard time with the summation. Any tips?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="2649" LastEditorDisplayName="" LastEditDate="2010-10-22T18:38:09.800" LastActivityDate="2010-10-22T20:40:36.427" Title="A component with more edges than vertices in a random graph" Tags="&lt;probability&gt;&lt;graph-theory&gt;&lt;inequality&gt;" AnswerCount="1" />
  <row Id="7525" PostTypeId="2" ParentId="7499" CreationDate="2010-10-22T17:37:00.163" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/3997454/help-me-understand-linear-separability-in-a-binary-svm&quot;&gt;Cross-posted&lt;/a&gt; my question at StackOverflow. More responses there.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2010-10-22T17:37:00.163" />
  <row Id="7526" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T17:38:49.133" Score="1" ViewCount="0" Body="&lt;p&gt;Considering the fact that you have only had one undergraduate course in analysis and will be taking an actual functional analysis class, I don't think you actually want to self-study functional analysis. It would be much more useful for you to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bulk up on your linear algebra &lt;/li&gt;&#xA;&lt;li&gt;review your real analysis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Functional analysis is, for a large part, linear algebra on a infinite dimensional vector space over the real or complex numbers. Having a good intuition from linear algebra is essential: you'll know what is reasonable to expect when the dimensional infinities can be controlled (by some sort of compactness), and when they cannot be controlled, what parts of the argument cannot possibly go wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of real analysis is also helpful because a lot of topological notions are introduced in those books, and familiarity with them is necessary. Furthermore, notions involved in the normed/metric spaces, basic notions of convergence and compactness, and many such are used all the time in functional analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I think you will be better off reviewing the notes for your undergraduate analysis course (or going through Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt;) and studying some linear algebra (unfortunately I can't think of any good book to recommend there). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-22T17:38:49.133" CommentCount="1" />
  <row Id="7527" PostTypeId="1" AcceptedAnswerId="7529" CreationDate="2010-10-22T17:42:20.920" Score="4" ViewCount="202" Body="&lt;p&gt;I have attempting to solve this using the infinite ramsey theorem, with colouring based on whether the sum of two vertices has an even or odd number of distinct prime factors.&#xA;This is leading to an infinite recursion.&#xA;Is this ok? At the end of all time I will be done.&lt;/p&gt;&#xA;" OwnerUserId="2712" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-22T19:21:59.113" LastActivityDate="2010-11-06T19:15:47.177" Title="Find an infinite set of positive integers such that the sum of any two distinct elements has an even number of distinct prime factors." Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="0" />
  <row Id="7528" PostTypeId="1" CreationDate="2010-10-22T17:56:40.600" Score="1" ViewCount="78" Body="&lt;p&gt;Given a continuous function $f$ of period $2\pi$ and $\alpha$ in $\mathbb{R}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can you show for any real $x$ that $\lim_{ N\to \infty} \sum_{n=1}^N f(x+n\alpha) = \lim_{ N\to\infty} \sum_{n=1}^N f(x_n)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, the 'average value' of $f$ diluted by $N$ is the same as the average value of the sequence $x+n\alpha$ since $\alpha/\pi$ is never irrational&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha/\pi$ is irrational&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-21T23:08:43.270" LastActivityDate="2010-12-21T23:08:43.270" Title="Proving a certain sequence is dense in the image of a periodic function" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="7529" PostTypeId="2" ParentId="7527" CreationDate="2010-10-22T18:44:06.123" Score="1" ViewCount="0" Body="&lt;p&gt;Start with $\{5n+1\}$ and construct an infinite complete graph, colouring the edge between $x$ and $y$ with the parity of the number of distinct prime factors of $x+y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the infinite Ramsey theorem, there is an infinite clique of one colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that corresponds to the number of distinct prime factors being even, we are done. Otherwise consider each number multiplied by $5$. Since the sum of any (original) two is not divisible by $5$, after multiplying by $5$, we have that the number of distinct prime factors for each sum must be even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I am guessing there are an infinite number of such sets, such that any two such sets have finite intersection.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-06T19:15:47.177" LastActivityDate="2010-11-06T19:15:47.177" CommentCount="1" />
  <row Id="7530" PostTypeId="2" ParentId="7464" CreationDate="2010-10-22T18:45:57.660" Score="5" ViewCount="0" Body="&lt;p&gt;I am surprised no one seems to have answered this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get $$\pi/4 + \alpha/2$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the line is given by $y = x \tan \alpha$, with $\alpha$ acute and we throw from the origin at an angle $\theta$ from the x-axis, at velocity $1$, then we have that the projectile satisfies, assuming gravity $g=2$ (in appropriate units)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y = t\sin\theta - t^2$, $\displaystyle x = t\cos \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time at which it intersects the line again is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle t\sin\theta - t^2 = t \tan\alpha \cos \theta$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle t = \sin \theta - \tan \alpha \cos \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to maximize the horizontal distance travelled by the projectile, which is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \cos \theta (\sin \theta - \tan \alpha \cos \theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With little manipulation, we need to maximize&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \sin(2\theta - \alpha)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives the result.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-22T21:20:26.790" LastActivityDate="2010-10-22T21:20:26.790" CommentCount="1" />
  <row Id="7531" PostTypeId="1" AcceptedAnswerId="8398" CreationDate="2010-10-22T18:50:55.893" Score="0" ViewCount="133" Body="&lt;p&gt;Actually, i had posted this long ago in MO, and i didn't get a reply to this question as it was unfit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is an exercise, in some textbook ( i think Apostol) and i would be happy if i can receive some answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P(n)$ be the product of positive integers which are $\leq n$ and relatively prime $n$. Prove that $$ \displaystyle P( n) =n^{\phi(n)} \prod\limits_{d \mid n} \biggl(\frac{d!}{d^d} \biggr)^{\mu(n/d)}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-30T20:32:14.653" Title="Proving $P( n) =n^{\phi(n)} \prod\limits_{d \mid n} \biggl(\frac{d!}{d^d} \biggr)^{\mu(n/d)}$" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7532" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T18:59:18.843" Score="1" ViewCount="0" Body="&lt;p&gt;You might look at the answers to this previous MSE question, which had a slightly different&#xA;slant: &lt;a href=&quot;http://math.stackexchange.com/questions/903/choosing-a-topology-text&quot;&gt;&quot;choosing a topology text&quot;&lt;/a&gt;.  Apparently the poser was also interested in self-learning,&#xA;but with less preparation than you.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-22T18:59:18.843" />
  <row Id="7533" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T19:06:17.430" Score="2" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I also like Bourbaki's treatise, but some times it is a bit too logical. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Simmons General topology&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, &quot;A topological picture book&quot; (I don't remember the author though). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;K Jänich Topology.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;J. Kelly General topology.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-22T19:06:17.430" />
  <row Id="7534" PostTypeId="1" AcceptedAnswerId="7540" CreationDate="2010-10-22T19:13:03.593" Score="1" ViewCount="77" Body="&lt;p&gt;Suppose one is given an arbitrary moment generating function $M_{X}(t)$. How would you determine $P(X=k)$ from this? We know that $M_{X}(t) = E[e^{tX}]$ and $M_{X}(0) = 1$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-10-22T19:30:16.523" LastActivityDate="2010-10-22T19:46:43.510" Title="Moment Generating Functions" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7535" PostTypeId="1" AcceptedAnswerId="7543" CreationDate="2010-10-22T19:13:27.703" Score="0" ViewCount="81" Body="&lt;p&gt;Suppose $E$ is a set of edges on graph $G$ over $n$ vertices, $\mathbf{x} \in \{-1,1\}^n$. The following seems to hold if G is not &lt;a href=&quot;http://mathworld.wolfram.com/EulerianGraph.html&quot; rel=&quot;nofollow&quot;&gt;Eulerian&lt;/a&gt;, can anyone see a proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_\mathbf{x} \prod_{ij\in E} x_i x_j=0$$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-10-22T22:03:00.193" LastActivityDate="2010-10-22T22:03:00.193" Title="Sum that is 0 for Eulerian graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="7536" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T19:19:12.377" Score="0" ViewCount="0" Body="&lt;p&gt;There's a book that could fit your actual level perfectly. The book is &lt;a href=&quot;http://books.google.co.cr/books?id=QALoZC64ea0C&amp;amp;pg=PP1&amp;amp;dq=beginning+functional+analysis&amp;amp;hl=es&amp;amp;ei=S-HBTLT6PIbGlQeX1ojTCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Beginning Functional Analysis&lt;/a&gt; by Karen Saxe. It is aimed at undergraduates whose background is a basic course in linear algebra and real analysis. It is pretty well suited for self study since it is very readable (I've done it myself), and the author claims that one of its aims is that the book can be used for self study. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not require any measure theory because it develops the basics of Lebesgue measure in chapter 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend it if you want to read a little bit about functional analysis without having to master more things in order to read more advanced books.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-22T19:19:12.377" CommentCount="1" />
  <row Id="7537" PostTypeId="2" ParentId="7512" CreationDate="2010-10-22T19:19:15.437" Score="1" ViewCount="0" Body="&lt;p&gt;Check out &quot;Introductory Functional Analysis with Applications&quot; by Erwin Kreyszig. I have not read it myself, but I have heard great things. Also, in the preface, he writes that Calculus and a familiarity with Linear Algebra are all that's needed as prerequisites.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-10-22T19:19:15.437" />
  <row Id="7538" PostTypeId="1" CreationDate="2010-10-22T19:38:49.860" Score="3" ViewCount="50" Body="&lt;p&gt;Hi everyone, first time poster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to determine the length of a repeating sequence, with noise.&#xA;I.e. given the sequence 4, 12, 3, 4, 12, 3, 4, 12, 3, ...  the period is apparently 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if the sequence was more complicated. What if each value had added noise?&#xA;I was thinking of something with autocorrelation, or Fourier transform (my intuition tells me the repeating signal has a major frequency component, right? How can I determine which?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It feels like this should be easy, but I'm having trouble implementing a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2714" LastEditorUserId="2714" LastEditorDisplayName="" LastEditDate="2010-10-22T19:45:27.040" LastActivityDate="2010-10-23T01:30:16.613" Title="Determining the length of a repeating sequence, with noise." Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="7539" PostTypeId="1" AcceptedAnswerId="7551" CreationDate="2010-10-22T19:40:44.117" Score="5" ViewCount="230" Body="&lt;p&gt;Given a large univariate polynomial, say of degree 200 or more, is there a procedural way of finding all the complex roots? By &quot;roots&quot;, I mean complex decimal approximations to the roots, though the multiplicity of the root is important. I have access to MAPLE and the closest function I've seen is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;with(RootFinding):&#xA;Analytic(Z,x,-(2+2*I)..2+2*I);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but this chokes if Z is of high degree (in fact it fails to complete even if deg(Z)&gt;15). &lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-22T23:06:07.193" LastActivityDate="2010-10-23T11:11:19.287" Title="Finding all complex zeros of a high-degree polynomial" Tags="&lt;complex-analysis&gt;&lt;numerical-methods&gt;&lt;roots&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="7540" PostTypeId="2" ParentId="7534" CreationDate="2010-10-22T19:46:43.510" Score="1" ViewCount="0" Body="&lt;p&gt;$M_X(\log(t))$ is the probability generating function. Differentiate $k$ times and set $t=0$. Divide the result by $k!$.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-22T19:46:43.510" CommentCount="2" />
  <row Id="7541" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T19:50:45.017" Score="2" ViewCount="0" Body="&lt;p&gt;As an introductory book, &quot;Topology without tears&quot; by S. Morris. You can download PDF for free, but you might need to obtain a key to read the file from the author. (He wants to make sure it will be used for self-studying.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, another great introductory book is Munkres, Topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On graduate level (non-introductory books) are Kelley and Dugunji (or Dugundji?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Munkres said when he started writing his Topology, there wasn't anything accessible on undergrad level, and both Kelley and Dugunji wasn't really undergrad books. He wanted to write something any undergrad student with an appropriate background (like the first 6-7 chapters of Rudin's Principles of Analysis) can read. He also wanted to focus on Topological spaces and deal with metric spaces mostly from the perspective &quot;whether topological space is metrizable&quot;. That's the first half of the book. The second part is a nice introduction to Algebraic Topology. Again, quoting Munkres, at the time he was writing the book he knew very little of Algebraic Topology, his speciality was General (point-set) topology. So, he was writing that second half as he was learning some basics of algebraic topology. So, as he said, &quot;think of this second half as an attempt by someone with general topology background, to explore the Algebraic Topology.&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-22T19:50:45.017" CommentCount="1" />
  <row Id="7542" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T20:02:29.207" Score="1" ViewCount="0" Body="&lt;p&gt;You might consider &lt;em&gt;Topology Now!&lt;/em&gt; by Messer and Straffin. Their idea is to introduce the intuitive ideas of continuity, convergence, and connectedness so that students can quickly delve into knot theory, the topology of surfaces and three dimensional manifolds, fixed points, and elementary homotopy theory. I wish this book had been around when I was a student!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.albion.edu/ram/TopologyNow!/&quot; rel=&quot;nofollow&quot;&gt;http://people.albion.edu/ram/TopologyNow!/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-22T20:02:29.207" CommunityOwnedDate="2010-10-22T20:02:29.207" CommentCount="1" />
  <row Id="7543" PostTypeId="2" ParentId="7535" CreationDate="2010-10-22T20:06:43.183" Score="2" ViewCount="0" Body="&lt;p&gt;if G is Eulerian - as in you have one cyclic route that goes over all the edges, then the degree of each vertex is even so $\prod _{ij\in E} x_i x_j = \prod_{i\in [n]} x_i ^{deg(i)} = 1$ and so the total sum is $2^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if by eulerian you refer to a route that is not a circle then you have exactly two vertices with odd degree so  $\prod _{ij\in E} x_i x_j = x_s x_t$ and it is easy to see that $\sum_x x_s x_t = 0$ &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-10-22T20:06:43.183" CommentCount="1" />
  <row Id="7544" PostTypeId="2" ParentId="7539" CreationDate="2010-10-22T20:12:04.597" Score="1" ViewCount="0" Body="&lt;p&gt;I think this might help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rouche's Theorem:&lt;/strong&gt; Let $D$ be a bounded domain, with piecewise smooth boundary, $\partial{D}$. Let $f(z)$ and $h(z)$ be analytic on $D \cup \partial{D}$. If $|h(z)| &amp;lt; |f(z)$ for $z \in \partial{D}$, then $f(z)$ and $f(z)+h(z)$ have the same number of zero's in $D$, counting multiplicities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; We find the zeros of the function $p(z)=z^{6}+9z^{4}+z^{3}+2z+4$, inside the unit circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For using Rouche's theorem, let $p(z)= f(z) + h(z)$, where $f(z)$ dominates, $h(z)$ inside the unit circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $f(z) = 9z^{4}$, which has four zeros inside the unit circle, all at the origin. $h(z) =z^{6}+z^{3}+2z+4$, which satisfies, $|h(z)| &amp;lt; |f(z)|$ for $|z|=1$. Therefore by Rouche's theorem $p(z)$ also has 4 zeros inside the unit circle. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-22T20:12:04.597" CommentCount="8" />
  <row Id="7545" PostTypeId="2" ParentId="7538" CreationDate="2010-10-22T20:27:33.573" Score="2" ViewCount="0" Body="&lt;p&gt;If the noise is additive, autocorrelation or Fourier transform seem the way to go. In general, time-domain and frequency-domain approaches are the most basic and standard tools in this area. Beware that the spectra of such sequence will have extra peaks corresponding to the &quot;harmonics&quot; of your signal. You can read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Pitch_detection_algorithm&quot; rel=&quot;nofollow&quot;&gt;pitch detection&lt;/a&gt;, which is very related to your problem.&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-22T20:27:33.573" />
  <row Id="7546" PostTypeId="2" ParentId="7524" CreationDate="2010-10-22T20:40:36.427" Score="2" ViewCount="0" Body="&lt;p&gt;The threshold for cycles in $G(n,p)$ is $p=1/n$. The same threshold should work in your case, so the solution is basically $c = 1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-22T20:40:36.427" CommentCount="0" />
  <row Id="7547" PostTypeId="2" ParentId="7528" CreationDate="2010-10-22T20:45:20.473" Score="2" ViewCount="0" Body="&lt;p&gt;If $\alpha/\pi$ is rational, then your claim is false (in general, unless we got lucky), since you're sampling the same points over and over again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $\alpha/\pi$ is irrational, then you're sampling random points of the function (by Weyl's equidistribution theorem), and then with some work you can probably prove that in the limit you do get the mean value of $f$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-22T20:45:20.473" />
  <row Id="7548" PostTypeId="2" ParentId="7535" CreationDate="2010-10-22T20:46:11.753" Score="0" ViewCount="0" Body="&lt;p&gt;I don't think you need the graph to be Eulerian.  You have a bunch of terms $x_i x_j$ for $i,j \in \{1,2,\dots,n\}$ and want the terms to add to zero as the various $x_i$ are $\pm1$.  It is true for n=2 as there are no pairs of $x_i x_j$.  If it is true for n, it will be true for n+1 because all the added terms are of the form $x_i x_{n+1}$ and the ones where $x_{n+1}=+1$ cancel the ones where $x_{n+1}=-1$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-22T20:46:11.753" CommentCount="3" />
  <row Id="7549" PostTypeId="2" ParentId="7539" CreationDate="2010-10-22T20:54:03.177" Score="0" ViewCount="0" Body="&lt;p&gt;I think the following page summarizes the root-finding well, by showing how to reduce it to an appropriate Eigenvalue computation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/PolynomialRoots.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PolynomialRoots.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-10-22T20:54:03.177" />
  <row Id="7550" PostTypeId="1" CreationDate="2010-10-22T22:51:22.107" Score="0" ViewCount="76" Body="&lt;p&gt;Let’s define the codicil of the Geometric Mean – Arithmetic Mean Inequality to be the statement that if the means are equal, then all the terms are equal. Then: I conjecture that most of the GM-AM Inequality is actually in its codicil, in the sense that if you can use the codicil (that is, assume the codicil as part of the hypothesis), then it is a hundred times easier to prove the AM-GM Inequality.&lt;/p&gt;&#xA;" OwnerUserId="2344" LastActivityDate="2010-10-22T22:51:22.107" ClosedDate="2010-10-23T00:07:22.200" Title="Is most of the GM-AM Inequality in its codicil?" Tags="&lt;real-analysis&gt;" CommentCount="1" />
  <row Id="7551" PostTypeId="2" ParentId="7539" CreationDate="2010-10-22T23:04:35.823" Score="4" ViewCount="0" Body="&lt;p&gt;Everyone's first starting point when dealing with the polynomial rootfinding problem should be a peer at &lt;a href=&quot;http://www.yorku.ca/mcnamee/&quot; rel=&quot;nofollow&quot;&gt;J.M. McNamee's excellent bibliography&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=4PMqxwG-eqQC&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is a fact that polynomials of very high degree tend to make most polynomial rootfinders choke. Even the standard blackbox, the Jenkins-Traub algorithm, can choke if not properly safeguarded. Eigenmethods, while they can have nice accuracy, can be very demanding of space and time (O(n²) space and O(n³) operations for a problem with only O(n) inputs!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that unless you are prepared to devote some time and extra precision, this is an insoluble problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having been pessimistic in those last few sentences, one family of methods you might wish to peer at (and I have had personal success with) are the so-called &quot;simultaneous iteration&quot; methods. The simplest of them, (Weierstrass-)Durand-Kerner, is essentially an application of Newton's method to the Vieta formulae, treated as n equations in n unknowns (the assumption taken by (W)DK is that your polynomial is monic, but that is easily arranged).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish for more details and references, the &lt;a href=&quot;http://books.google.com/books?id=4PMqxwG-eqQC&amp;amp;pg=PA67&quot; rel=&quot;nofollow&quot;&gt;book by McNamee&lt;/a&gt; I mentioned earlier is a good start.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-22T23:04:35.823" />
  <row Id="7552" PostTypeId="2" ParentId="7511" CreationDate="2010-10-22T23:38:12.077" Score="2" ViewCount="0" Body="&lt;p&gt;What you are interested here are what &lt;a href=&quot;http://www.sciencedirect.com/science/bookseries/00765392&quot; rel=&quot;nofollow&quot;&gt;Spanier and Oldham&lt;/a&gt; term &quot;cyclodifferential functions&quot;, functions that are regenerated after being differintegrated to the appropriate order (or to put it another way, functions that are eigenfunctions of the differintegration operator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cyclodifferential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${}_0 D_x^{\alpha}y=y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or in more familiar notation, $$\frac{\mathrm{d}^{\alpha}y}{\mathrm{d}x^{\alpha}}=y$$, but the problem with this notation in the setting of differintegrals is that it neglects to take the lower limit that is present in both the Caputo and Riemann-Liouville definitions into account).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you might have seen, $\exp(x)$ is a cyclodifferential function for ${}_0 D_x^1 y$ (i.e. $\exp(x)$ is its own derivative), $\cosh(x)$ and $\sinh(x)$ are cyclodifferentials for ${}_0 D_x^2 y$ (differentiating those two functions twice gives you the originals);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{\pi x}}+\exp(x)\mathrm{erfc}(-\sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\mathrm{erfc}(x)$ is the complementary error function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a cyclodifferential for ${}_0 D_x^{\frac12} y$ (it is its own &lt;em&gt;semiderivative&lt;/em&gt;), and in general&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\alpha-1}\sum_{j=0}^\infty \frac{C^j x^{\alpha j}}{\Gamma(\alpha(j+1))}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\alpha &gt; 0$ and $C$ an appropriate eigenvalue, is a cyclodifferential for ${}_0 D_x^{\alpha}$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-23T08:07:53.183" LastActivityDate="2010-10-23T08:07:53.183" CommentCount="1" />
  <row Id="7553" PostTypeId="2" ParentId="7520" CreationDate="2010-10-22T23:54:22.627" Score="1" ViewCount="0" Body="&lt;p&gt;Singer and Thorpe, &lt;em&gt;Lecture Notes on Elementary Topology and Geometry&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slim book that gives an intro to point-set, algebraic and differential topology and differential geometry. It does not have any exercises and is very tersely written, so it is not a substitute for a standard text like Munkres, but as a beginner I liked this book because it gave me the big picture in one place without many prerequisites.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-22T23:54:22.627" CommentCount="1" />
  <row Id="7554" PostTypeId="1" AcceptedAnswerId="7557" CreationDate="2010-10-23T00:23:42.890" Score="4" ViewCount="137" Body="&lt;p&gt;Is there an algorithm that will allow me to numerically compute the limit of a function f(x) in a principled way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most naive algorithm would be to continue to compute the function for larger values of x. The first problem is how to figure out the 'large' values for x to compute the function for.  How do I know when to stop?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we construct some error bars for this calculation perhaps based on some kind of statistical rationale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any books for further reading will be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-23T01:02:56.900" LastActivityDate="2010-12-24T07:10:56.393" Title="Numerically estimate the limit of a function" Tags="&lt;analysis&gt;&lt;numerical-methods&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="7555" PostTypeId="2" ParentId="7342" CreationDate="2010-10-23T00:42:24.193" Score="3" ViewCount="0" Body="&lt;p&gt;This is a development of my comment, changing the restraints from two to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_{n}=-\sum_{i=1}^{n-1}x_{i}$, the question: find &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \sum_{i=1}^{n}x_{i}^{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{n}x_{i}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^{n}x_{i}^{2}=n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to find &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min f(x_{1},x_{2},\ldots ,x_{n-1}),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(x_{1},x_{2},\ldots ,x_{n-1})=\left(&#xA;\sum_{i=1}^{n-1}x_{i}^{3}\right) -\left( \sum_{i=1}^{n-1}x_{i}\right) ^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s.t. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(x_{1},x_{2},\ldots ,x_{n-1})=0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(x_{1},x_{2},\ldots ,x_{n-1})=\left( \sum_{i=1}^{n-1}x_{i}\right)&#xA;^{2}-n+\sum_{i=1}^{n-1}x_{i}^{2}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions regarding first derivatives are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f(x_{1},x_{2},\ldots ,x_{n-1})-\nabla c(x_{1},x_{2},\ldots&#xA;,x_{n-1})\lambda =0\qquad (\ast ),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f(x)=\begin{pmatrix}\frac{\partial f}{\partial x_{1}} &amp;amp; \ldots  &amp;amp; \frac{\partial f}{\partial x_{n-1}}\end{pmatrix}^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=%&#xA;\begin{pmatrix}&#xA;3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2} &amp;amp; \ldots  &amp;amp; &#xA;3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}%&#xA;\end{pmatrix}%&#xA;^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla c(x)=%&#xA;\begin{pmatrix}&#xA;\frac{\partial c}{\partial x_{1}} &amp;amp; \ldots  &amp;amp; \frac{\partial c}{\partial&#xA;x_{n-1}}%&#xA;\end{pmatrix}%&#xA;^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=%&#xA;\begin{pmatrix}&#xA;2\left( \sum_{i=1}^{n-1}x_{i}\right) +2x_{1} &amp;amp; \ldots  &amp;amp; 2\left(&#xA;\sum_{i=1}^{n-1}x_{i}\right) +2x_{n-1}%&#xA;\end{pmatrix}%&#xA;^{T}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condition $(\ast )$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left(&#xA;\sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{1}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left(&#xA;\sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{n-1}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these equations have the same form (the coefficients of $x_k$, $1\lt k\lt n-1$, are independent of $k$). Thus the solution is atained when all the $n-1$ variables are equal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{1}=\ldots =x_{n-1}=x^{\ast }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have to solve the new system of two equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(x^{\ast })^{2}-3\left( \sum_{i=1}^{n-1}x^{\ast }\right) ^{2}-2\lambda&#xA;\left( \sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0\qquad (\ast&#xA;\ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \sum_{i=1}^{n-1}x^{\ast }\right)&#xA;^{2}-n+\sum_{i=1}^{n-1}(x^{\ast })^{2}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obtain, successively&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(x^{\ast })^{2}-3n+3\sum_{i=1}^{n}(x^{\ast })^{2}-2\lambda \left(&#xA;\sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(x^{\ast })^{2}-3n+3(x^{\ast })^{2}(n-1)-2\lambda x^{\ast }(n-1)-2\lambda&#xA;x^{\ast }=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x^{\ast })^{2}(n-1)-1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\ast }=x_{1}=\ldots =x_{n-1}=\pm \frac{1}{\sqrt{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{n}=-\sum_{i=1}^{n-1}x_{i}=\mp \frac{n-1}{\sqrt{n-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for $n&gt;2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}%&#xA;\right) ^{3}&amp;lt;-\left( \frac{1}{\sqrt{n-1}}\right) ^{3}+\left( \frac{n-1}{%&#xA;\sqrt{n-1}}\right) ^{3},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the minimum is obtained at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{\ast }=x_{1}=\ldots =x_{n-1}=\frac{1}{\sqrt{n-1}},x_{n}=-\frac{n-1}{%&#xA;\sqrt{n-1}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its value is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}%&#xA;\right) ^{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: The original problem in $n$ variables is the solution of $n$ equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f(x_{1},x_{2},\ldots ,x_{n})-\nabla c(x_{1},x_{2},\ldots&#xA;,x_{n})\lambda =0\qquad (\ast\ast\ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus the two conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{1}(x_{1},x_{2},\ldots ,x_{n})=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{2}(x_{1},x_{2},\ldots ,x_{n})=0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla f=%&#xA;\begin{pmatrix}&#xA;\frac{\partial f}{\partial x_{1}} &amp;amp; \ldots  &amp;amp; \frac{\partial f}{\partial&#xA;x_{n}}%&#xA;\end{pmatrix}%&#xA;^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla c=%&#xA;\begin{pmatrix}&#xA;\left( \nabla c_{1}\right) ^{T} &amp;amp; \left( \nabla c_{2}\right) ^{T}%&#xA;\end{pmatrix}%&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda =\begin{pmatrix}&#xA;\lambda _{1} &amp;amp; \lambda _{2}\end{pmatrix}^{T}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The equations would be &lt;em&gt;similar&lt;/em&gt; for $m$ constraints $c_1,\dots ,c_m.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: Since I don't know how to write here matrices with more than one row, I wrote them in the transposed form.  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-23T19:05:59.867" LastActivityDate="2010-10-23T19:05:59.867" />
  <row Id="7556" PostTypeId="2" ParentId="7554" CreationDate="2010-10-23T00:47:15.590" Score="4" ViewCount="0" Body="&lt;p&gt;You have to assume something about f(x).  For the case where the approach to the limit is expected to have an asymptotic expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John Stalker, &lt;em&gt;A convergence speeding algorithm with applications to numerical integration&lt;/em&gt;, Advances in Applied Mathematics Volume 22, Issue 1, January 1999, Pages 119-153 &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-23T00:47:15.590" />
  <row Id="7557" PostTypeId="2" ParentId="7554" CreationDate="2010-10-23T01:07:18.670" Score="4" ViewCount="0" Body="&lt;p&gt;To expand on T's answer, &lt;em&gt;convergence acceleration&lt;/em&gt; methods assume that the error terms of the sequence you are interested in are of a certain form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the general-purpose methods for accelerating the convergence of a sequence is the Shanks transformation. For the case of a sequence of partial sums of a series, the Shanks transformation amounts to constructing a sequence of Padé approximants, which hopefully converge faster to the limit of the sequence of partial sums. (The justly famous Wynn ε algorithm is an efficient realization of the Shanks transformation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another popular technique is Richardson extrapolation, which assumes that the error can be expressed as a power series of some form. This is in effect the application of the usual algorithms for polynomial interpolation to estimate the limit of a sequence. (Richardson extrapolation is the machinery behind the Romberg algorithm for accelerating the convergence of the trapezoidal rule, since the error of the trapezoidal rule is expressible as a series in powers of the panel size.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been deliberately vague here since you have given absolutely no information on the nature of your sequence. There are many convergence acceleration methods to choose from (a good reference is Brezinski and Redivo-Zaglia's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444888144&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Extrapolation Methods: Theory and Practice&lt;/em&gt;&lt;/a&gt;), and the best sequence transformation to use depends a lot on the provenance of your sequence.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-23T01:07:18.670" CommentCount="7" />
  <row Id="7558" PostTypeId="2" ParentId="7538" CreationDate="2010-10-23T01:30:16.613" Score="0" ViewCount="0" Body="&lt;p&gt;In connection with certain algorithms for integer factorization, there are methods due to &lt;a href=&quot;http://doi.acm.org/10.1145/800135.804400&quot; rel=&quot;nofollow&quot;&gt;Floyd&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1007/BF01933190&quot; rel=&quot;nofollow&quot;&gt;Brent&lt;/a&gt; for the so-called &quot;cycle finding&quot; problem. It would look like you can adapt these to your situation, but in the noisy case you will have to use a &quot;fuzzy equality&quot; comparison function, e.g. &lt;code&gt;|x - y| &amp;lt;= tol&lt;/code&gt; where &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; are the two entities being compared, and &lt;code&gt;tol&lt;/code&gt; is a small quantity.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-23T01:30:16.613" />
  <row Id="7559" PostTypeId="2" ParentId="7512" CreationDate="2010-10-23T01:37:46.290" Score="0" ViewCount="0" Body="&lt;p&gt;I am also an EE.   And I think Kreyszig's Funcational Analysis is a good book for your background.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-23T01:37:46.290" CommentCount="1" />
  <row Id="7561" PostTypeId="2" ParentId="7473" CreationDate="2010-10-23T02:21:16.483" Score="2" ViewCount="0" Body="&lt;p&gt;Note $\rm\ f_n\: :=\ a^n-1\ =\ a^{n-m} \: (a^m-1) + a^{n-m}-1\:$,$\ $ i.e. $\rm\ f_n = k\ f_m + f_{n-m}\:$.$\ $ Now apply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\: $  Let $\rm\ f_n\: $ be an integer sequence with $\rm\ f_{\:0} =\: 0\:\ $ and $\rm\ \: f_n\equiv f_{n-m}\ (mod\ f_m)\ $ for $\rm\: n &amp;gt; m\:$.  Then $\rm\:\ (f_n,f_m)\ =\ f_{(n,\:m)}\ \: $ where $\rm\ (i,\:j)\ $ denotes $\rm\ gcd(i,\:j)\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ By induction on  $\rm\:n + m\:$. The theorem is trivially true if $\rm\ n = m\ $ or $\rm\ n = 0\ $ or $\rm\: m = 0\:$.&lt;br&gt;&#xA;So we may assume $\rm\:n &amp;gt; m &amp;gt; 0\:$.$\ $  Note $\rm\ (f_n,f_m)\: =\ (f_{n-m},f_m)\ $ follows from the hypothesis.&lt;br&gt;&#xA;Since $\rm\:\ (n-m)+m \ &amp;lt;\ n+m\: ,\ $ induction yields $\rm\ \ (f_{n-m},f_m)\ =\ f_{(n-m,\:m)}\ =\ f_{(n,\:m)}\ $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T21:00:59.933" LastActivityDate="2010-12-03T21:00:59.933" />
  <row Id="7562" PostTypeId="2" ParentId="7464" CreationDate="2010-10-23T03:09:06.060" Score="2" ViewCount="0" Body="&lt;p&gt;When the solution is so neat, I feel like it should have a high-level explanation &quot;from the book&quot; that makes it immediately obvious. Sadly, I was not able to find one, but I'm posting what I did get in hopes that it might inspire someone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine firing projectiles at all angles from the origin. Assume unit speed and gravity $g = 2$, as in Moron's answer. Consider a frame of reference which starts at the origin and falls freely. There is no gravity in this frame of reference, so all projectiles expand radially outward forming a cone,&#xA;$$x^2 + y^2 = t^{2}.$$&#xA;However, the ground is no longer given by $y = x \tan\alpha$ but rather by&#xA;$$y = x\tan\alpha + t^{2},$$&#xA;which is a parabolic cylinder. The projectiles hit the ground at the points where the cone and the parabolic cylinder intersect. Subtracting the two equations, we find that this intersection satisfies&#xA;$$\left(x^2 + x\tan\alpha\right) + \left(y^2 - y\right) = 0,$$&#xA;&lt;em&gt;i.e.&lt;/em&gt; on the $xy$ plane it projects to a circle. On physical grounds, one can see that the circle passes through the origin, and is tangent to the line $y = x\tan\alpha$ there. The projectile that reaches the farthest corresponds to the point with the largest $x$-coordinate, and the angle of the projectile is simply the angle of the line joining that point to the origin (because projectiles just move radially outward in this frame of reference). I encourage you to draw a little diagram and see that this line bisects the angle between the vertical, at $\pi/2$, and the ground, at $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-23T03:09:06.060" CommentCount="1" />
  <row Id="7563" PostTypeId="2" ParentId="7486" CreationDate="2010-10-23T03:24:38.780" Score="1" ViewCount="0" Body="&lt;p&gt;This is more of an extended remark out of academic curiosity.  I thought I would try to use &lt;a href=&quot;http://www.cs.unm.edu/~mccune/mace4/&quot; rel=&quot;nofollow&quot;&gt;Prover9&lt;/a&gt; to prove the given identity.  [here v denotes &quot;or&quot; and ^ denotes &quot;and&quot;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To test this, I inputted the assumptions into Prover9 (this is the OP's list of axioms from &lt;a href=&quot;http://math.stackexchange.com/questions/7471/boolean-algebra-theorems&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x v x' = y v y'.  % false&#xA;x ^ x' = y ^ y'.  % true&#xA;&#xA;y v (x ^ x') = y.&#xA;y v (x v x') = y.&#xA;x v x = x.&#xA;x v y = y v x.&#xA;(x v y) v z = x v (y v z).&#xA;x ^ (x v y) = x.&#xA;(x ^ y) v (x ^ y') = x.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from which it proved the OP's identity (and if I understand correctly, it was by contradiction):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;((((w' ^ x) ^ z') v ((x ^ y) ^ z)) v ((w ^ x) ^ y')) v ((w ^ x') ^ z) = (((((((w' ^ x) ^ z') v ((x ^ y) ^ z)) v ((w ^ x) ^ y')) v ((w ^ x') ^ z)) v ((x ^ y') ^ z')) v ((w' ^ x) ^ y)) v ((w ^ x) ^ z).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-23T03:24:38.780" CommentCount="1" />
  <row Id="7564" PostTypeId="1" CreationDate="2010-10-23T03:59:37.630" Score="3" ViewCount="226" Body="&lt;p&gt;It's been a while since my last statistics class...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 404 files that went through some automated generation process. I would like to manually verify some of them to make sure that their data is indeed correct. I want to use probability to help me out so that I don't need to check every single file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I calculate what sample size I should use to reach a certain confidence level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I would like to say with 95% confidence that the files are correct, how many of them do I have to check?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an &lt;a href=&quot;http://www.surveysystem.com/sscalc.htm&quot; rel=&quot;nofollow&quot;&gt;online calculator&lt;/a&gt;, but I'm not entirely sure what I should put for the confidence interval. Say I put 20% and leave the confidence factor at 95%. I get a sample size of 23. Let's say now that I tested 23 random files and all of them were fine. Does that mean that &quot;I can be 95% confident that 80% to 100% of the files are correct&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean, then, that for my original question, I would need to use a 99% confidence level with a 4% confidence interval, then I would need to verify that the 291 files (the sample size it gave me) are all correct. And only then I can say with 95% confidence that the files are correct? (99% +- 4% = 95% to 100%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also mentions something about percentages which I'm not quite clear on... does the fact that most (i.e. 100%) of the files I test are valid (since if I found an invalid one, I would stop the whole process and examine my generation process for errors) mean that I can use a smaller sample to get the same confidence factor? If so, how would I calculate it?&lt;/p&gt;&#xA;" OwnerUserId="2718" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T04:17:52.553" LastActivityDate="2010-10-23T08:27:42.933" Title="Calculating a sample size based on a confidence level" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7566" PostTypeId="1" CreationDate="2010-10-23T05:02:36.593" Score="0" ViewCount="40" Body="&lt;p&gt;Is there a way to determine the probability of winning a particular sweepstakes with only the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-the estimated odds of winning are 1 in 13,000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-contestants must be owners (property holders) within the corporation running the sweepstakes and the total number of owners is 222,600&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-each owner can enter the sweepstakes up to 6 times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-there is one grand prize and three runner-up prizes&lt;/p&gt;&#xA;" OwnerUserId="2719" LastActivityDate="2010-10-23T05:44:59.540" Title="Sweepstakes Probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7567" PostTypeId="2" ParentId="7564" CreationDate="2010-10-23T05:03:33.567" Score="2" ViewCount="0" Body="&lt;p&gt;It's not surprising you're a bit confused; understanding what's really going on with confidence intervals can be tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short version: If you don't want to check all the files you have to choose two different percentages: the confidence level (95% in your example), and how far off you're willing to be at that level (20% in your example).  These percentages refer to two different quantities, and so it doesn't make sense to add or subtract them from each other.  Once you've made these choices then I think it is fine to use the online calculator to get a sample size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more detail on what's going on, here's the explanation: You're trying to estimate the true percentage of files that have correct data.  Let's call that percentage $p$.  Since you don't want to calculate $p$ exactly, you have to choose how far off you are willing to be with your estimate, say 20%.  Unfortunately, you can't even be certain that your estimate of $p$ will be within 20%, so you have to choose a level of confidence that that estimate will be within 20% of $p$.  You have chosen 95%.  Then the online calculator gives you the sample size of 23 you need to estimate $p$ to within 20% at 95% confidence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what does that 95% really mean?  Basically, it means that if you were to choose lots and lots of samples of size 23 and calculate a confidence interval from each one, 95% of the resulting confidence intervals would actually contain the unknown value of $p$.  The other 5% would give an interval of some kind that does not include $p$.  (Some would be too large, others would be too small.)  Another way to look at it is that choosing a 95% confidence interval means that you're choosing a method that gives correct results (i.e., produces a confidence interval that actually contains the value of $p$) 95% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Does that mean that 'I can be 95% confident that 80% to 100% of the files are correct'?&quot;  Not precisely.  It really means that you can be 95% confident that the true percentage of correct files is between 80% and 100%.  That's a subtle distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;And only then I can say with 95% confidence that the files are correct? (99% +- 4% = 95% to 100%)?&quot;  No, this is confusing the two kinds of percentages.  The 99% refers to 99% of all confidence intervals constructed if you were to construct lots of them.  The 4% refers to an error margin of $\pm$ 4% for the files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other thing to remember is that the sample size estimator assumes that the population you're drawing from is much, much larger than the size of the sample you end up going with.  Since your population is fairly small you can get away with a smaller-sized sample with the same level of confidence.  The determination of exactly how small, though, is a much more difficult calculation.  It's beyond what you would have seen in a basic statistics class.  I'm not sure how to do it; maybe someone else on the site does.  (EDIT: Even better: take Jyotirmoy Bhattacharya's suggestion and ask on &lt;a href=&quot;http://stats.stackexchange.com/&quot;&gt;Stats Stack Exchange&lt;/a&gt;.)  But this is the only justification for being able to use a smaller sample size than 23 - not the fact that you would abort the confidence interval calculation if you found anything other than 100% for your sample's estimate of the true value of $p$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T05:21:47.637" LastActivityDate="2010-10-23T05:21:47.637" />
  <row Id="7568" PostTypeId="2" ParentId="7566" CreationDate="2010-10-23T05:31:49.187" Score="1" ViewCount="0" Body="&lt;p&gt;If the estimated odds of winning are $1$ in $13,000$ per entrance, then if you enter once, your odds of winning are $1:13000$, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, entering more than once, it's a bit more complicated.  It's easier to calculate the odds of you &lt;em&gt;not&lt;/em&gt; winning, and subtract it from $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The odds of you &lt;em&gt;not&lt;/em&gt; winning once is $12999/13000$.  The odds of you &lt;em&gt;not&lt;/em&gt; winning twice is $(12999/13000)^2$, and three times is $(12999/13000)^3$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could then say we can calculate your odds of not winning $n(x)$, where $x$ is the number of entries, as $n(x) = (12999/13000)^x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you want to calculate the odds of winning $p(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $n(x) + p(x) = 1$ (Think about it.  The sum of the odds of not winning and the odds of winning equal up to 1, because there is a 100% chance that one or the other will happen).  So, $p(x) = 1 - n(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to calculate your odds of winning, you would use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(x) = 1 - (12999/13000)^x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x$ is the amount of times you've entered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you enter 6 times, then your total odds are $\frac{2227329628937465077999}{4826809000000000000000000}$, which is about $0.0004614497$.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-23T05:44:59.540" LastActivityDate="2010-10-23T05:44:59.540" CommentCount="1" />
  <row Id="7569" PostTypeId="1" AcceptedAnswerId="8682" CreationDate="2010-10-23T05:55:16.840" Score="4" ViewCount="163" Body="&lt;p&gt;Are there known sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ for $k &gt; 1$ subject to the constraint $\sum_{p \mid k} \frac{1}{p+1} &amp;lt; 1$? (The factor of +1 in the denominator and the constraint of the sum of inverse &quot;shifted prime divisors&quot; is bounded by 1 are both necessary.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question: Suppose $(p_i)$ is a set of $n$ consecutive primes which minimizes $1 - \sum_{i = 1}^{n} \frac{1}{p_i+1} &gt; 0$ for a given $n &gt; 1$. Are there known bounds for $1 - \sum_{i = 1}^{n} \frac{1}{p_i + 1}$ from below in terms of $n$, e.g., $n^{-\delta n}$ for some fixed $\delta &gt; 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-10-26T14:49:48.390" LastActivityDate="2010-11-02T23:12:17.947" Title="Sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ " Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;asymptotics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7570" PostTypeId="1" CreationDate="2010-10-23T06:00:07.927" Score="2" ViewCount="200" Body="&lt;p&gt;$$\lim_{n\to\infty} \lim_{N\to\infty} a(N,n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the reason I want to know is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a sequence of functions $\{f_n\}$ which converges uniformly to a function $f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I want to say that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n\to\infty} \lim_{N\to\infty} \sum_{k=1}^N \frac{f_k(x_n)}N = \lim_{N\to\infty} \lim_{n\to\infty} \sum_{k=1}^N \frac{f_k(x_n)}N $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here my $a(N,n)$ is that sum ($\sum_{k=1}^N f_k(x_n)/N$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and somehow I want to use that $f_n$ converges uniformly to do this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry.edited&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-23T13:55:42.027" LastActivityDate="2010-12-24T17:23:18.287" Title="When can you interchange limits of a sequence of 2 variables? " Tags="&lt;real-analysis&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7571" PostTypeId="1" CreationDate="2010-10-23T06:18:31.647" Score="1" ViewCount="59" Body="&lt;p&gt;Given a smooth vector field on a closed smooth manifold, does the flow of that vector field exists for all time $t \in R$? This may be a well-known and elementary conclusion, but I just wnat to confirm it to me. &lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2010-10-23T21:26:12.643" Title="flow on a closed manifold" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7572" PostTypeId="2" ParentId="7564" CreationDate="2010-10-23T06:39:05.473" Score="1" ViewCount="0" Body="&lt;p&gt;Let's forget about confidence intervals for a while and try to think of your sample size problem from first principles. This might also answer @Mike's point about finite samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose among your $N$ files $M$ are good and $N-M$ are bad. If you pick a random sample of $n$ files the probability that all of them are good is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(n,M) = {{M \choose n} \over {N \choose n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_distribution&quot; rel=&quot;nofollow&quot;&gt;hypergeometric&lt;/a&gt; distribution. You can calculate it in Excel as &lt;code&gt;HYPGEOMDIST(n,n,M,N)&lt;/code&gt; or in &lt;a href=&quot;http://www.r-project.org/&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt; as &lt;code&gt;dhyper(n,M,N-M,n)&lt;/code&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(n,M)$ will be smaller for larger $n$. A large sample is more likely to uncover bad files even if they are rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(n,M)$ will be larger for larger $M$. Rarer bad files are, the less likely are they to be caught by a sample of a given size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large an $n$ should you choose? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you knew $M$ then you could plot $p(n,M)$ and choose a $n$ large enough to put $p(n,M)$ below the threshold of error you are willing to tolerate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the whole point of the exercise is that you don't know $M$. Now it becomes a question of how optimistic or pessimistic you are. If you are an optimist then you will think that your process will either succeed or fail very badly. So if $M$ is not $N$ then it is a very small number. In this case choosing a small $n$ will let you reach your error tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are a pessimist you will think that if your process fails it will produce a bad file only occasionally, i.e. $M$ will be large but not equal to $N$. In this case you will be forced to take a large sample to reach any given error tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be formal you could try to codify you pre-sampling beliefs about the value of $M$ in terms of a probability distribution. [I have not seen anyone really do it, though I have seen people sometimes pretend to do so.] Let the probability mass function be $f(m)$. Then the probability of a sample of size $n$ having all good files even if some files are bad is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi(n)={1 \over {1-f(N)}}\sum_{m=0}^{N-1} p(n,m)f(m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can choose a large enough $n$ to keep $\pi(n)$ below your error tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I assumed that the 404 files are all you care about. On the other hand if you care about the process that generated the files, then you should model the process as generating bad files with a probability $q$ and instead of $p(n,M)$ have $p(n,q)=(1-q)^n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I assumed from your question that you are interested only in the two possibilities $M=N$ and $M \neq N$. The extension of the approach above when you care about how many files are bad is left as an exercise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;error tolerance&quot; mentioned about should not be a number picked from the air or chosen by convention. It should be the result of considering the costs of inspections, the damages that would result from missing a bad file, and your attitude towards risk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, what about confidence intervals? They are a barbaric relic of the past. In most situations they don't answer any interesting questions.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-23T08:27:42.933" LastActivityDate="2010-10-23T08:27:42.933" CommentCount="1" />
  <row Id="7573" PostTypeId="2" ParentId="7569" CreationDate="2010-10-23T06:40:42.727" Score="0" ViewCount="0" Body="&lt;p&gt;Actually, there is a bound for $\displaystyle \frac{1}{p}$. Perhaps, this should help. You can find this result in a standard Analytic Number theory textbook.(I have taken this from Apostol's &lt;em&gt;Analytic Number Theory&lt;/em&gt; textbook!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition:&lt;/strong&gt; There is a constant $A$ such that $$\sum\limits_{p \leq x} \frac{1}{p}= \log\log{x} + A + \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$$ for all $x \geq 2$, where the constant $A$ is given by $$A = 1 -\log\log{2} + \int\limits_{2}^{\infty} \frac{ R(t)}{t \log^{2}{t}} \ dt$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-23T06:40:42.727" CommentCount="1" />
  <row Id="7574" PostTypeId="2" ParentId="7512" CreationDate="2010-10-23T07:33:26.610" Score="0" ViewCount="0" Body="&lt;p&gt;I like Zeidler's &lt;em&gt;Applied Functional Analysis&lt;/em&gt; (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387944427&quot; rel=&quot;nofollow&quot;&gt;Volume 1&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387944222&quot; rel=&quot;nofollow&quot;&gt;Volume 2&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-23T07:33:26.610" />
  <row Id="7575" PostTypeId="2" ParentId="7512" CreationDate="2010-10-23T08:15:49.810" Score="0" ViewCount="0" Body="&lt;p&gt;One unconventional book is &lt;em&gt;Infinite Dimensional Analysis: A Hitchiker's Guide&lt;/em&gt; by Aliprantis and Border. It's fully rigorous but written for &quot;practical people, such as engineers and economists&quot; rather than math students.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-23T08:15:49.810" />
  <row Id="7577" PostTypeId="2" ParentId="7539" CreationDate="2010-10-23T11:11:19.287" Score="1" ViewCount="0" Body="&lt;p&gt;I think one of the biggest problems is approximating multiple roots. The approach described in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;L.Brugnano, D.Trigiante. &quot;Polynomial Roots: the Ultimate Answer?&quot;, Linear Algebra and its Applications 225 (1995) 207-219&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;relies on the approximation of eigenvalues of a tridiagonal matrix, obtained via the application of Euclid's GCD algorithm to the original polynomial, and seems to work pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find the pdf for the article though, sorry.&lt;/p&gt;&#xA;" OwnerUserId="2691" LastActivityDate="2010-10-23T11:11:19.287" CommentCount="1" />
  <row Id="7578" PostTypeId="1" AcceptedAnswerId="7581" CreationDate="2010-10-23T11:15:36.210" Score="12" ViewCount="388" Body="&lt;p&gt;I'm looking for an example of two metrics that induce the same topology, but so that one metric is complete and the other is not (Since it is known that completeness isn't a topological invariant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any hints or ideas.&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-10-24T03:48:33.183" Title="Two metrics induce the same topology, but one is complete and the other isn't." Tags="&lt;topology&gt;&lt;homework&gt;&lt;metric-spaces&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="3" />
  <row Id="7580" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T11:43:36.807" Score="4" ViewCount="0" Body="&lt;p&gt;Look for a homeomorphism $f: \mathbb{R} \to \mathbb{R}^+$ (you should know a good one). Then take the usual metric on $\mathbb{R}$ and use it to define a metric $d$ on $\mathbb{R}^+$ by $d(x,y) = |f(x)-f(y)|$ (*). Now what can you say about the relationship between $d$ and the usual metric on $\mathbb{R}^+$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Made the question here correct and more illustrative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the identity $\iota: (\mathbb{R}^+,d) \to (\mathbb{R}^+,|\cdot|)$ is a homeomorphism. Now just ask yourself which property, that continuous functions between metric spaces can have, $\iota$ &lt;em&gt;doesn't&lt;/em&gt; have? Can you formulate a theorem about what property a homeomorphism must have to preserve completeness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*): &lt;a href=&quot;http://en.wikipedia.org/wiki/Pullback&quot; rel=&quot;nofollow&quot;&gt;Pullback&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Pushforward&quot; rel=&quot;nofollow&quot;&gt;pushforward&lt;/a&gt; might be of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: &lt;code&gt;\mathbb{R}_+&lt;/code&gt; makes the math renderer choke.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T12:13:36.320" LastActivityDate="2010-10-23T12:13:36.320" CommentCount="0" />
  <row Id="7581" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T11:56:34.643" Score="24" ViewCount="0" Body="&lt;p&gt;The metric space $\{\frac {1}{n} \mid n\in \mathbb{N} \}$ with the usual metric is incomplete (since we don't have zero), and it has the discrete topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the same space with the metric $d(x,y)=1 \iff x\neq y$ also has a discrete topology but is complete, since any cauchy sequence will eventually be constant.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-10-23T11:56:34.643" CommentCount="2" />
  <row Id="7583" PostTypeId="1" AcceptedAnswerId="7591" CreationDate="2010-10-23T13:00:44.417" Score="2" ViewCount="158" Body="&lt;p&gt;Let $\zeta$ be the Riemann- Zeta function. For any integer, $n \geq 2$, how to prove $$\zeta(2) \zeta(2n-2) + \zeta(4)\zeta(2n-4) + \cdots + \zeta(2n-2)\zeta(2) = \Bigl(n + \frac{1}{2}\Bigr)\zeta(2n)$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-24T16:45:49.443" LastActivityDate="2010-10-24T16:45:49.443" Title="An Identity Concerning the Riemann Zeta Function" Tags="&lt;number-theory&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="7586" PostTypeId="1" AcceptedAnswerId="7588" CreationDate="2010-10-23T13:42:13.490" Score="3" ViewCount="125" Body="&lt;p&gt;hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to solve this :$$ z^2 -6z + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book says 'complete the square' so :&lt;br&gt;&#xA;1.$$ (z - 6/2)^2 -36/4 + 25 $$&lt;br&gt;&#xA;2.$$ (z - 3)^2 -9 + 25 $$&lt;br&gt;&#xA;3.$$ (z - 3)^2 + 16 $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how exactly does the above turn into this:&lt;br&gt;&#xA;$$ 3\pm 4\imath $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so Much&lt;br&gt;&#xA;Gideon&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-23T23:43:23.267" LastActivityDate="2010-10-24T15:51:50.350" Title="solve  $ z^2 -6z + 25 $ into complex conjugate" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;&lt;complex-numbers&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="7587" PostTypeId="2" ParentId="7586" CreationDate="2010-10-23T13:44:49.430" Score="3" ViewCount="0" Body="&lt;p&gt;The square roots of -16 are $4i$ and $-4i$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a further hint to anybody who might encounter this sort of thing in the future, a polynomial with complex conjugate roots $\alpha$ and $\alpha^\ast$ will be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2-2(\Re\alpha)x+|\alpha|^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $3^2+4^2=5^2=25$ and that $6=2\times 3$. Thus, once you verify that your quadratic has a negative discriminant, you can almost instantly write down the complex conjugate roots.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-24T05:24:45.643" LastActivityDate="2010-10-24T05:24:45.643" CommentCount="4" />
  <row Id="7588" PostTypeId="2" ParentId="7586" CreationDate="2010-10-23T13:49:12.540" Score="6" ViewCount="0" Body="&lt;p&gt;I think your confusion here stems from the fact you're working with an expression, not an equation. You can only solve equations; it makes no sense to 'solve' an expression!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; You seem to have just edited your post so the first line is an equation. Best to keep it in equation form throughout however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm presuming you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^2 - 6z + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, completing the square:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(z - 6/2)^2 -36/4 + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(z - 3)^2 - 9 + 25 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(z - 3)^2 = -16$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z - 3 = \pm 4i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z = 3 \pm 4i$$&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-10-23T13:49:12.540" CommentCount="3" />
  <row Id="7589" PostTypeId="1" CreationDate="2010-10-23T14:20:46.227" Score="8" ViewCount="244" Body="&lt;p&gt;During my study of plane algebraic curves, I got curious if there is a nontrivial example of a plane algebraic curve that has a &lt;a href=&quot;http://mathworld.wolfram.com/Crunode.html&quot; rel=&quot;nofollow&quot;&gt;node&lt;/a&gt;, a &lt;a href=&quot;http://mathworld.wolfram.com/Cusp.html&quot; rel=&quot;nofollow&quot;&gt;cusp&lt;/a&gt; (for my purposes I do not care which of the two kinds of cusps would the example exhibit), a &lt;a href=&quot;http://mathworld.wolfram.com/Tacnode.html&quot; rel=&quot;nofollow&quot;&gt;tacnode&lt;/a&gt;, and an &lt;a href=&quot;http://mathworld.wolfram.com/Acnode.html&quot; rel=&quot;nofollow&quot;&gt;isolated point&lt;/a&gt;. By &quot;nontrivial&quot; I mean a curve that was not constructed as a chimera of two or more simpler curves, e.g. $(x-y)(x^2+y^2-1)=0$. Of course, it would be a quintic at the very least (i.e. the algebraic degree should be 5 at the minimum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from an explicit example, I would also be interested in a general procedure for constructing algebraic curves with a prescribed number and type of singular points.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After trying out Qiaochu's and T..'s suggestions, I have a follow-up question: does the problem become more difficult if the requirement that the curve be bounded (i.e. one can draw a circle such that the whole curve, including the isolated point, is within the circle) is imposed?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-24T00:28:56.103" LastActivityDate="2010-10-25T05:50:12.987" Title="A plane algebraic curve with all four kinds of double points" Tags="&lt;algebraic-geometry&gt;&lt;plane-curves&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="7590" PostTypeId="2" ParentId="2110" CreationDate="2010-10-23T14:23:05.567" Score="1" ViewCount="0" Body="&lt;p&gt;You can try the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter&quot; rel=&quot;nofollow&quot;&gt;Kalman filter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://calypso.inesc-id.pt/FCUL/psm/docs/kalman-dan-simon.pdf&quot; rel=&quot;nofollow&quot;&gt;http://calypso.inesc-id.pt/FCUL/psm/docs/kalman-dan-simon.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-23T14:23:05.567" CommentCount="2" />
  <row Id="7591" PostTypeId="2" ParentId="7583" CreationDate="2010-10-23T14:29:59.800" Score="7" ViewCount="0" Body="&lt;p&gt;This is too nice an exercise to give away. Your starting point should be:&#xA;$$\frac{t}{e^t-1} = 1 - t/2 + \frac{2 \zeta(2)}{(2 \pi)^2} t^2 - \frac{2 \zeta(4)}{(2 \pi)^4} t^4 + \frac{2 \zeta(6)}{(2 \pi)^6} t^6 - \cdots.$$&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-23T14:29:59.800" CommentCount="1" />
  <row Id="7592" PostTypeId="2" ParentId="7571" CreationDate="2010-10-23T14:49:51.163" Score="3" ViewCount="0" Body="&lt;p&gt;The local existence theorem for ODEs gives a lower bound for the domain of existence of solutions in terms of a Lipschitz constant. On a compact manifold $M$, you can find a &lt;em&gt;global&lt;/em&gt; Lipschitz constant, and then the theorem gives you an $\epsilon&gt;0$ such that for every point $x\in M$, the solution that starts at $x$ is defined at least in $(-\epsilon,\epsilon)$. With this you immediately get existence for all values of $t$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-23T21:26:12.643" LastActivityDate="2010-10-23T21:26:12.643" />
  <row Id="7593" PostTypeId="1" AcceptedAnswerId="7998" CreationDate="2010-10-23T15:36:04.023" Score="10" ViewCount="178" Body="&lt;p&gt;A recent &lt;a href=&quot;http://people.missouristate.edu/lesreid/POW12_0910.html&quot; rel=&quot;nofollow&quot;&gt;Missouri State problem&lt;/a&gt; stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction.  That got me trying to decompose the plane into closed or open intervals.  The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right.  That gets the whole plane except the lattice points.  This seems like it must be a standard problem, but I couldn't find it on the web.  So can the plane be decomposed into unit open intervals?  closed intervals?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-23T19:19:43.510" LastActivityDate="2010-10-27T04:25:29.060" Title="Decomposing the plane into intervals" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;tiling&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="7594" PostTypeId="1" CreationDate="2010-10-23T15:48:47.543" Score="2" ViewCount="124" Body="&lt;p&gt;I am interested in the sum of a Hadamard product of generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are given $n$ functions, where $0 &amp;lt; i \leq n$, of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_i(x_i) = \sum_{j=0}^m{c_{i,j} x_i^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hadamard product is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{j=0}^m{( \prod_{i=0}^n {c_{i,j}})x^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's essentially the same as going through all the generating functions simultaneously and multiplying all the $j$th coefficients together to obtain a new coefficient.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some given information&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have polynomials; I think of them as finite length generating functions.  I know that all of the coefficients are natural numbers.  I'd like the sum of the resulting coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a trick.  Calculating the Hadamard product generates an extremely complicated expression.  I would like to avoid dealing with this expression explicitly, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know ways that I could do this.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T18:11:13.023" LastActivityDate="2010-10-23T18:11:13.023" Title="How can I get this tricky sum?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functions&gt;" CommentCount="5" />
  <row Id="7595" PostTypeId="2" ParentId="7589" CreationDate="2010-10-23T15:55:21.460" Score="9" ViewCount="0" Body="&lt;p&gt;Here is the idea.  Everything depends on what the defining polynomial $f(x, y) = 0$ looks like in local coordinates.  To simplify things we'll take $f(x, y) = y^2 - p(x)$ where $p$ is some polynomial in $x$.  Now, I claim that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $y^2 = x^2 + \text{higher terms}$ then there is a crunode at the origin,&lt;/li&gt;&#xA;&lt;li&gt;If $y^2 = x^3 + \text{higher terms}$ then there is a cusp at the origin,&lt;/li&gt;&#xA;&lt;li&gt;If $y^2 = x^4 + \text{higher terms}$ then there is a tacnode at the origin, and&lt;/li&gt;&#xA;&lt;li&gt;If $y^2 = -x^2 + \text{higher terms}$ then there is an acnode at the origin.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The key point is that in each of these cases we get something locally diffeomorphic to the same guy but without the higher-order terms.  In other words, to get a bunch of double points with arbitrary behavior it suffices to pick a bunch of points $(x_i, 0)$ and place various conditions on the values of $p$ and its derivatives at $x_i$.  In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $p(0) = p'(0) = 0, p''(0) = 1$ then there is a crunode at $(0, 0)$.&lt;/li&gt;&#xA;&lt;li&gt;If $p(1) = p'(1) = p''(1) = 0, p^{(3)}(1) = 1$ then there is a cusp at $(1, 0)$.&lt;/li&gt;&#xA;&lt;li&gt;If $p(2) = p'(2) = p''(2) = p^{(3)}(2) = 0, p^{(4)}(2) = 1$ then there is a tacnode at $(2, 0)$.&lt;/li&gt;&#xA;&lt;li&gt;If $p(3) = p'(3) = 0, p''(3) = -1$ then there is an acnode at $(3, 0)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I claim that this kind of polynomial interpolation is always possible.  In fact, it follows by the Chinese Remainder Theorem!  The conditions are equivalent to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p(x) \equiv x^2 \bmod x^3$&lt;/li&gt;&#xA;&lt;li&gt;$p(x) \equiv (x-1)^3 \bmod (x-1)^4$&lt;/li&gt;&#xA;&lt;li&gt;$p(x) \equiv (x-2)^4 \bmod (x-2)^5$&lt;/li&gt;&#xA;&lt;li&gt;$p(x) \equiv -(x-3)^2 \bmod (x-3)^3.$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course this discussion generalizes.  And it shouldn't be hard for you to convince yourself that you can choose $p$ to be a non-square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  And if you also want to be able to choose the $y$-coordinates arbitrarily, it suffices to replace $y^2$ by a polynomial $q(y)$ which is equal to $y^2 + \text{higher terms}$ in local coordinates at each of the points you're interested in, which can be done by the same method.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-23T19:29:24.753" LastActivityDate="2010-10-23T19:29:24.753" CommentCount="5" />
  <row Id="7596" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T16:24:34.997" Score="3" ViewCount="0" Body="&lt;p&gt;The interval $M:=]-1,1[$ with the usual line element $ds:=|dx|$ is an incomplete metric space: The sequence $x_n:=1-{1\over n} \ (n\to\infty)$ converges in $\mathbb R$, so it is a Cauchy sequence, but it diverges in $M$. On the other hand, the &quot;hyperbolic metric&quot; defined by $ds:=|dx|/(1-x^2)$ induces the same topology on $M$ but is complete. The latter statement needs of course a proof. Suffice it here to say that now the endpoints $\pm1$ are &quot;infinitely far away&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-23T16:24:34.997" />
  <row Id="7597" PostTypeId="2" ParentId="7589" CreationDate="2010-10-23T16:43:22.560" Score="5" ViewCount="0" Body="&lt;p&gt;Everything but the isolated point is a local algebraic requirement that can be satisfied by one component of a rational curve parametrized by polynomials, $(x,y)=(f(z),g(z))$.  For each singularity that requires $k$ segments of the curve to meet in specified ways, write down values of $f$ and $g$ and their first few derivatives at $k$ (or fewer) different values of $z$ that would provide the desired picture.   So this part of the problem reduces to finding polynomial $f$ and $g$ whose coefficients satisfy some linear equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to get two branches to meet we could require $f(1)=f(2)$ and $g(1)=g(2)$; to have the intersection be transversal (resp., a tangency) we need $(f'(1),g'(1))$ different from (resp., equal to) $(f'(2),g'(2))$ and higher tangency conditions can be expressed in the same form.  You can actually draw the curve freely (but unicursally, not lifting the pencil) on paper to define what the pattern of intersections and cusps should be and then choose the values and derivatives at the intersection points to create that behavior in the parametrization.  It can be arbitrarily complicated, such as a point where two branches meet to order $5$ while two other segments meet to order $6$ and are tangent to the first two branches to order $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add an isolated point at $(0,0)$, perform the above construction so that all singularities occur for $z \neq 0$ and the curve does not pass through 0, then replace the parametrization by $x = (1+z^2)f(z), \quad y= (1+z^2)g(z)$.  The algebraic curve you want is the Zariski closure of the parametrized thing.  This is the equation between $x$ and $y$ obtained by elimination of $z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a bounded locus, replace any rational parametrization $(A(t),B(t))$ by $(A,B)/(1+(PAB)^2)$ where $P$ is a polynomial vanishing to high order at all the singular points arranged for $A$ and $B$. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-25T05:50:12.987" LastActivityDate="2010-10-25T05:50:12.987" CommentCount="5" />
  <row Id="7598" PostTypeId="1" AcceptedAnswerId="7599" CreationDate="2010-10-23T16:54:36.140" Score="0" ViewCount="220" Body="&lt;p&gt;I want to know what is the Taylor series of $\sqrt{1+x^2}$  while $x \to 0$. &lt;/p&gt;&#xA;" OwnerUserId="2728" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-23T17:32:51.983" LastActivityDate="2010-10-23T22:37:14.237" Title="Taylor series of $\sqrt{1 + x^2}$" Tags="&lt;sequences-and-series&gt;&lt;homework&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7599" PostTypeId="2" ParentId="7598" CreationDate="2010-10-23T16:58:51.567" Score="3" ViewCount="0" Body="&lt;p&gt;If you just want the answer, you can find it at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=series%28sqrt%281%2Bx%5e2%29%29&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-23T17:40:08.100" LastActivityDate="2010-10-23T17:40:08.100" CommentCount="4" />
  <row Id="7600" PostTypeId="1" CreationDate="2010-10-23T17:25:31.133" Score="0" ViewCount="38" Body="&lt;p&gt;Is there a way to calculate a f test using a beta distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have&#xA;(x1,x2 ..., xn)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &#xA;(y1, y2 ..., yk)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily compute the variance of x and y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how does it relate to a beta distribution?&lt;/p&gt;&#xA;" OwnerUserId="2729" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T17:35:46.550" LastActivityDate="2010-10-23T17:35:46.550" Title="What is the relationship between f test and the beta distribution" Tags="&lt;probability&gt;&lt;statistics&gt;" CommentCount="1" />
  <row Id="7601" PostTypeId="1" CreationDate="2010-10-23T17:41:33.230" Score="0" ViewCount="60" Body="&lt;p&gt;how do you go from (πr²/r)²s² to (2πr)²s² ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;its supposed to be part of propagation of error in radius to circle area&#xA;should be fairly simple but i am newbie :-)&lt;/p&gt;&#xA;" OwnerUserId="2730" LastActivityDate="2010-10-23T17:41:33.230" ClosedDate="2010-10-24T00:48:35.780" Title="how do you go from (πr²/r)²s² to (2πr)²s² ?" Tags="&lt;geometry&gt;" CommentCount="1" />
  <row Id="7602" PostTypeId="2" ParentId="7511" CreationDate="2010-10-23T18:28:39.117" Score="1" ViewCount="0" Body="&lt;p&gt;Sorry to give yet another answer that does not address the issue of fractional $n$ [it seems that fractional derivatives are not such a familiar topic to many research mathematicians; certainly they're not to me], but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a little issue here which has not been addressed.  By the context of the OP's question, I gather s/he is looking for &lt;strong&gt;real-valued&lt;/strong&gt; functions which are equal to their $n$th derivative (and not their $k$th derivative for $k &amp;lt; n$).  Several answerers have mentioned that the set of solutions to $f^{n} = 0$ forms an $n$-dimensional vector space.  But over what field?  It is easier to identify the space of such &lt;strong&gt;complex-valued&lt;/strong&gt; functions, i.e., $f: \mathbb{R} \rightarrow \mathbb{C}$: namely, a $\mathbb{C}$-basis is given by $f(x) = e^{2 \pi i k/n}$ for $0 \leq k &amp;lt; n$.  But what does this tell us about the $\mathbb{R}$-vector space of real-valued solutions to this differential equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it is $n$-dimensional as a $\mathbb{R}$-vector space, though it does not have such an immediately obvious and nice basis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $W$ be the $\mathbb{R}$-vector space of real-valued functions $f$ with $f^{(n)} = 0$ and $V$ the $\mathbb{C}$-vector space of $\mathbb{C}$-valued functions $f$ with $f^{(n)} = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a natural inclusion map $W \mapsto V$.  Phrased algebraically, the question is whether the induced map $L: W \otimes_{\mathbb{R}} \mathbb{C} \rightarrow V$ is an isomorphism of $\mathbb{C}$-vector spaces.  In other words, this means that any given $\mathbb{R}$-basis of $W$ is also a $\mathbb{C}$-basis of $V$.  This is certainly not automatic.  For instance, viewing the Euclidean plane as first $\mathbb{R}^2$ and second as $\mathbb{C}$ gives a map &#xA;$\mathbb{R}^2 \rightarrow \mathbb{C}$ which certainly does not induce an isomorphism upon tensoring with $\mathbb{C}$, since the first space has (real) dimension $2$ but the second space has (complex) dimension $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on this, see Theorem 1.6 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this is actually a problem in &lt;strong&gt;Galois descent&lt;/strong&gt;: according to Theorem 2.14 of the notes of Keith Conrad already cited above, the map $L$ is an isomorphism iff there exists a conjugate-linear involution $r: V \rightarrow V$, i.e., i.e., a map which is self-inverse and satisfies, for any $z \in \mathbb{C}$ and $v \in V$, $r(zc) = \overline{z} r(c)$.&lt;br&gt;&#xA;But indeed we have such a thing: an element of $V$ is just a complex-valued function $f$, &#xA;so we put $r(f) = \overline{f}$.  Note that this stabilizes $V$ since the differential equation $f^{(n)}) = 0$ &quot;is defined over $\mathbb{R}$&quot;: or more simply, the complex conjugate of the $n$th derivative is the $n$th derivative of the complex conjugate.  Thus we have &quot;descent data&quot; (or, in Keith Conrad's terminology, a &lt;strong&gt;G-structure&lt;/strong&gt;) and the real solution space has the same dimension as the complex solution space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a nice exercise to use these ideas to construct an explicit real basis of $W$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-23T18:28:39.117" CommentCount="1" />
  <row Id="7603" PostTypeId="2" ParentId="7593" CreationDate="2010-10-23T18:46:39.030" Score="0" ViewCount="0" Body="&lt;p&gt;Seems pretty easy. We're going to cover the complex plane instead since that's obviously equivalent to covering $\mathbb{R}^2$. Start with the family of unit length line segments $I_\theta = \{a\cdot e^{i\theta} \mid a \in (0,1] \}$, $\theta \in (0,2\pi)$. Then define $I_{\theta,n}$, $n \in \mathbb{N}_0$ to be $I_\theta$ translated a length of $n$ in the direction of $\theta$. This collection covers the entire complex plane except the ray $[0,\infty)$ which we cover with the half-open intervals $[k,k+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Armed with this idea, it should be easy to see how to for any $k$ fill $\mathbb{R}^k$ with intervals of unit length that &quot;point in all directions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI: &lt;code&gt;\mathbb{N}_0&lt;/code&gt; breaks stuff. You have to escape the underscore: &lt;code&gt;\mathbb{N}\_0&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-10-23T18:53:09.080" LastActivityDate="2010-10-23T18:53:09.080" CommentCount="3" />
  <row Id="7604" PostTypeId="1" CreationDate="2010-10-23T19:08:28.073" Score="1" ViewCount="56" Body="&lt;p&gt;I remember hearing somewhere that if $p^n$ is the largest power of a prime $p$ dividing the class number $h_K$ of a number field $K$, then there is unique factorization of the $p^n$-th powers of elements in $\mathcal{O}_K$ into $p^n$-th powers of irreducibles. I can see that there can't be any $p$-torsion for a $p^n$-th power of an ideal class (because $p^n$ is the largest power dividing $h_K$), but I'm confused as to how to translate this into a statement about factorization of elements, and why $p$ is arbitrary. Am I remembering this result correctly (if not, what is the right formulation)? Can someone provide a proof or reference?&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-10-23T20:26:16.397" LastActivityDate="2010-10-23T20:26:16.397" Title="Recovering unique factorization of elements by only looking at $p^n$-th powers?" Tags="&lt;algebraic-number-theory&gt;" CommentCount="5" />
  <row Id="7605" PostTypeId="1" AcceptedAnswerId="7613" CreationDate="2010-10-23T19:28:34.320" Score="3" ViewCount="89" Body="&lt;p&gt;In general, how would one find the distribution of $f(X)$ where $X$ is a random variable? Or consider the inverse problem of finding the distribution of $X$ given the distribution of $f(X)$. For example, what is the distribution of $\max(X_1, X_2, X_3)$ if $X_1, X_2$ and $X_3$ have the same distribution? Likewise, if one is given the distribution of $ Y = \log X$, then the distribution of $X$ is deduced by looking at $\text{exp}(Y)$?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-23T20:42:52.840" Title="Distribution of Functions of Random Variables" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7606" PostTypeId="1" AcceptedAnswerId="7611" CreationDate="2010-10-23T19:30:28.760" Score="1" ViewCount="141" Body="&lt;p&gt;Suppose $p$ is prime, $q|(p-1)$. Prove the congruence $1+x+\cdots+x^{q-1}\equiv 0 \pmod{p}$ has exactly $q$ solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this as a homework assignment and I've having trouble proving it. I'm trying to use Lagrange's theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;If $p$ is prime, an $n$th degree polynomial has at most $n$ solutions mod $p$.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q|(p-1)$ implies $p-1 = qr$ for some $r$ in $\mathbb{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1+x+\cdots+x^{p-1} = (1+x+\cdots+x^{q-1})  h(x)$&#xA;where $h(x)$ is a polynomial of degree $q(r-1)-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(since $x^{q(r-1)-1}x^{q-1}=x^{qr-q-1+q-1}=x^{qr-2}=x^{p-1}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but here I'm stuck. I'm not quite sure where to go from here. I think I can say that all the solutions of $1+x+\cdots+x^{p-1} \mod p$ are either solutions of $1+x+\cdots+x^{q-1}$ or $h(x)$ modilo $p$, but I'm not quite sure if that's true, or if it is, where to go from there. I'd appreciate any help or hints, thanks! &lt;/p&gt;&#xA;" OwnerUserId="2466" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-23T20:32:16.937" LastActivityDate="2010-10-23T21:55:43.290" Title="solutions of polynomials using lagrange's theorem" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7607" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T19:32:56.383" Score="2" ViewCount="0" Body="&lt;p&gt;If $(X,d)$ is a connected, complete metric space and $G$ is a nonempty proper open subset of $X$, then $G$ is not complete with the restricted metric (because it is not closed), but $G$ is homeomorphic to a complete metric space, because the map $x\mapsto(x,1/d(x,X\setminus G))$ embeds $G$ onto a closed subspace of $X\times \mathbb{R}$.  (This is more or less &lt;a href=&quot;http://books.google.com/books?id=zRep59fhoGkC&amp;amp;lpg=PP1&amp;amp;dq=arveson%20invitation&amp;amp;pg=PA61#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Lemma 3.1.1&lt;/a&gt; of Arveson's &lt;em&gt;Invitation&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space that is homeomorphic to a separable complete metric space is called a &lt;a href=&quot;http://books.google.com/books?id=bQwhdmL6IjUC&amp;amp;lpg=PP1&amp;amp;pg=PA195#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Polish space&lt;/a&gt;.  A subspace of a Polish space is Polish if and only if it is a $G_\delta$, while $G_\delta$s in complete metric spaces are not typically complete with the restricted metric.  For example, there is a complete metric inducing the usual topology on the set of irrational numbers.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-23T19:32:56.383" />
  <row Id="7608" PostTypeId="2" ParentId="7578" CreationDate="2010-10-23T19:57:33.527" Score="1" ViewCount="0" Body="&lt;p&gt;Consider one complete normed vector space $(V,\|\cdot\|)$, that is, a Banach space, and let $$B = \{ x \in V : \|x\| &amp;lt; 1 \}$$ be the unit ball centered at the origin. Define $f: V \rightarrow B$ by $x \mapsto x /(1+ \|x\|)$. Then f is continuous and is easy to show that the inverse of $f$ is also continuous, so f is a homeomorphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $B$ isn't complete, since if $x \in V$ and $\| x \| = 1$, then $x_n = (1-1/n)x$ is a cauchy sequence in $B$ that doesn't converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Kahen referred, if we have two metric spaces $X$ and $Y$ with the latter complete and a  uniformly continuous homeomorphism $g : X \rightarrow Y$, then $X$ is complete. To prove this you'll need to observe that since $g$ is uniformly continuous then it sends Cauchy sequences to  Cauchy sequences.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-10-23T19:57:33.527" />
  <row Id="7610" PostTypeId="1" AcceptedAnswerId="7617" CreationDate="2010-10-23T20:01:25.327" Score="-1" ViewCount="142" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which mathematical subjects I have to learn in order to create my own formula about a topic something. For example, I am curious how Einstein could find his final formulas. It is all about logic or he had to know deeply some math rules and subjects in order to get his first formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering how a equation can appear from nothing, and what would be its start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What subjects should I study to create my own formula and test it to see if if its results are plausible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="2732" LastActivityDate="2010-10-23T21:56:59.247" ClosedDate="2010-10-23T21:29:49.093" Title="Which subjects do I have to learn to create a formula from scratch?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="7611" PostTypeId="2" ParentId="7606" CreationDate="2010-10-23T20:21:25.070" Score="3" ViewCount="0" Body="&lt;p&gt;I'm assuming that when you say &quot;congruence&quot;, you mean that you want to show that the congruence &#xA;$$x^{q-1} + \cdots + 1 \equiv 0 \pmod{p}$$&#xA;has exactly $q$ solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are incorrect in saying that $1+x+\cdots + x^{p-1}$ is a multiple of $1+x+\cdots+x^{q-1}$. For example, if $p=7$ and $q=3$ (which divides $7-1=6$), you would need $1+x+x^2$ to divide $1+x+x^2+x^3+x^4+x^5+x^6$. But doing long division, you get&#xA;$$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x^2+x+1)(x^4+x) + 1,$$&#xA;so there is a remainder. Your assertion is false, which is probably why you are stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason I knew you were incorrect in the assertion is: notice that $x^q-1 = (x^{q-1}+\cdots+1)(x-1)$ and $x^p-1 = (x^{p-1}+\cdots+1)(x-1)$. The only solution modulo $p$ to $x^{p}-1$ is $x=1$, because by Fermat's Little Theorem you have $x^p \equiv x \pmod{p}$ for all $x$; if $x^{q-1}+\cdots+1$ divides $x^{p-1}+\cdots + 1$, then $(x^{q-1}+\cdots+1)(x-1)$ divides $(x^{p-1}+\cdots+1)(x-1)$; so every root of $x^{q-1}+\cdots+1$ has to be a root of $x^p-1$. But $x^{q-1}+\cdots+1$ has a lot of roots, and that contradicts the fact that $x^p-1$ only has one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu suggests, consider the polynomial $x^q - 1 = (x-1)(x^{q-1}+\cdots+1)$. Since $p$ is prime, $x^q-1\equiv 0\pmod{p}$ if and only if at least one of $x-1$ and $x^{q-1}+\cdots+1$ are congruent to $0$ modulo $p$; and since $q|p-1$, then $1^{q-1}+1^{q-2}+\cdots+1 = q\not\equiv 0 \pmod{p}$. So $a$ is a solution to the original congruence if and only if $a$ is a solution to $x^q-1 \equiv 0 \pmod{p}$ &lt;strong&gt;and&lt;/strong&gt; $a\not\equiv 1 \pmod{p}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the multiplicative group modulo $p$, which has order $p-1$, and look at a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n&quot; rel=&quot;nofollow&quot;&gt;primitive root modulo $p$.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not know about primitive roots, you can still more or less push through your attempt if you are a little more careful. Instead of claiming that $x^{q-1}+\cdots+1$ divides $x^{p-1}+\cdots + 1$ (which is &lt;em&gt;false&lt;/em&gt;), try to see if it divides $x^{p-1}-1$. By Fermat's Little Theorem, you know exactly how many roots modulo $p$ $x^{p-1}-1$ has. If you can find the $h(x)$ such that $(x^{q-1}+\cdots+1)h(x) = x^{p-1}-1$, then you know that every root of $x^{p-1}-1$ is either a root of $h(x)$ or a root of the polynomial you want. Then show that the two polynomials have no root in common, and use Lagrange's Theorem as you want to give an upper bound for the number of roots of $h(x)$, and hence a lower bound to the number of roots of $x^{q-1}+\cdots+x+1$. Then use Lagrange's Theorem again to get an upper bound.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-23T20:31:48.040" LastActivityDate="2010-10-23T20:31:48.040" CommentCount="9" />
  <row Id="7612" PostTypeId="2" ParentId="7610" CreationDate="2010-10-23T20:22:53.213" Score="0" ViewCount="0" Body="&lt;p&gt;If there had been a simple rule/algorithm that would allow us to derive new things out of the air, everybody would have been an Einstein now.&lt;br /&gt;&lt;br /&gt;&#xA;The background of such formulas is usually a simple idea that the scientist plays with in his/her head and expands it to a more and more complex theory. I personally think that the formulas are the last thing - just a formal formulation of the ideas behind.&lt;/p&gt;&#xA;" OwnerUserId="2733" LastActivityDate="2010-10-23T20:22:53.213" />
  <row Id="7613" PostTypeId="2" ParentId="7605" CreationDate="2010-10-23T20:32:38.323" Score="1" ViewCount="0" Body="&lt;p&gt;Qiaochu is right.  There isn't a magic wand.  That said, there is a set of common procedures that can be applied to certain kinds of transformations.  One of the most important is the cdf (cumulative distribution function) method that you are already aware of.  (It's the one used in your &lt;a href=&quot;http://math.stackexchange.com/questions/7324/&quot;&gt;previous question&lt;/a&gt;.)  Another is to do a change of variables, which is like the method of substitution for evaluating integrals.  You can see that procedure and others for handling some of the more common types of transformations at this &lt;a href=&quot;http://www.math.uah.edu/stat/dist/Transformations.xhtml&quot; rel=&quot;nofollow&quot;&gt;web site&lt;/a&gt;.  (Some of the other examples there include finding maxes and mins, sums, convolutions, and linear transformations.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-23T20:42:52.840" LastActivityDate="2010-10-23T20:42:52.840" />
  <row Id="7614" PostTypeId="1" AcceptedAnswerId="7615" CreationDate="2010-10-23T20:42:25.010" Score="4" ViewCount="114" Body="&lt;p&gt;This is a chug-plug formula given in my book :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.The number of ways in which &lt;strong&gt;mn&lt;/strong&gt; different items can be divided equally int &lt;strong&gt;m groups&lt;/strong&gt;, each containing &lt;strong&gt;n objects and the order of the groups is not important&lt;/strong&gt;,is : $\frac {(mn)!}{(n!)^m} \frac {1}{m!} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.The number of ways in which &lt;strong&gt;mn&lt;/strong&gt; different items can be divided equally int &lt;strong&gt;m groups&lt;/strong&gt;, each containing &lt;strong&gt;n objects and the order of the groups is important&lt;/strong&gt;,is : $\frac {(mn)!}{(n!)^m} \frac {1}{m!} * (m!) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having some troubles in understanding the above two formulas and hence making mistakes while solving problems based on the same, so I am very inquisitive to know the proof.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:17:23.533" LastActivityDate="2010-11-11T15:17:23.533" Title="Grouping items into groups" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7615" PostTypeId="2" ParentId="7614" CreationDate="2010-10-23T20:51:16.010" Score="5" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Imagine first that both the order of the groups &lt;em&gt;and&lt;/em&gt; the order within the groups is important. You have $mn$ items; one way to divide them into $m$ groups of $n$ objects each is to simply list the items. Items 1 through $n$ will go into the first group; items $n+1$ through $2n$ will go into the second group; etc. There are $(mn)!$ ways of listing the items. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in fact, you don't care about the order in which the groups appear on the list: if you take the same list you have, and now list items $n+1$ through $2n$ first, then, say, times $4n+1$ through $5n$, then items $1$ through $n$, etc., you will still get the same division of objects into the same groupings. So any shuffling of the blocks in your list will yield the same distribution of items into the groups. Each distribution into groups can then be described in $m!$ ways, one for each ordering of the $m$ groups. So you have counted each distribution $m!$ times, and you need to divide $(mn)!$ by $m!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are still not done, though, because you &lt;em&gt;also&lt;/em&gt; don't care about the order within each group. Again, say you have a listing: if you take the first $n$ items in the list, and you list them in a different order (say, from last to first) while keeping them in the first $n$ slots, then in the end you will have the same elements separated into the same groups: you don't care who goes into group $1$ first or last, just that they end up in that group. So you have also overcounted by as many ways as there are to list the members of each group in order. In each group, you have $n$ objects, which can be listed in $n!$ different ways. There are $m$ such groups, so each listing of the elements in the groups can be done in $(n!)^m$ ways; so you have &lt;em&gt;also&lt;/em&gt;, separately, overcounted the lists by $(n!)^m$, forcing you to also divide by $(n!)^m$. Putting these two overcount factors together gives you the first formula.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Try to adapt the explanation above to see how to obtain this formula; note that a simpler way of writing it is just as $\frac{(mn)!}{(n!)^m}$, since the two $m!$ will cancel out.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-23T21:07:52.750" LastActivityDate="2010-10-23T21:07:52.750" CommentCount="4" />
  <row Id="7617" PostTypeId="2" ParentId="7610" CreationDate="2010-10-23T20:54:59.410" Score="1" ViewCount="0" Body="&lt;p&gt;Just consider doing an experiment: Let's take a bowling ball up a scaffold, the higher we drop it off the building the harder a dent it will put into the soft earth below. Suppose we did a whole series of experiments and measured each case - when plotting the results we might get a graph that curves up quickly - since graphs are and functions are essentially the same thing - we have a formula too. This is how &lt;a href=&quot;http://en.wikipedia.org/wiki/Willem_%27s_Gravesande&quot; rel=&quot;nofollow&quot;&gt;Willem Gravesande&lt;/a&gt; found the famous formula for kinetic energy $\frac{1}{2}m v^2$. Another fun example of this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_formula#Rydberg_formula_for_hydrogen&quot; rel=&quot;nofollow&quot;&gt;Rydberg formula&lt;/a&gt;, data had been measured and he just tried out lots of different equations until finding one which fit the data nicely. This is the experimental derivation of a physical formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not always necessary to do experiments to discover physical formulae though. Newtons theory of mechanics gives the formula $F = ma$. Using calculus (i.e. purely mathematical reasoning) we can derive the formula $E_k= \frac{1}{2}m v^2$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy#Derivation&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the explicit derivation but the point is just that it is possible to derive new formula as consequences of the mathematical model. This is the mathematical derivation of a physical formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach (you can follow this approach to derive the equations for relativity) is to define some physical principles that you want to model - define a physical theory as a set of postulates. For example &quot;The speed of light is constant for all observers&quot;, etc.. then construct mathematical models of these ideas. Once you have the model you can derive consequences from it and build experiments to check them. Another example: Niels Bohr devised a model of the atom from which he was able to derive the Rydberg formula. I think this is called Mathematical Physics, or maybe just Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, to do this sort of work you need to be a very strong mathematician - but also have a good intuition about reality. If you just study basic physics (like watching the Caltech lectures) you can practice the skills needed.&lt;/p&gt;&#xA;" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-23T21:56:59.247" LastActivityDate="2010-10-23T21:56:59.247" />
  <row Id="7618" PostTypeId="1" AcceptedAnswerId="7620" CreationDate="2010-10-23T21:38:18.373" Score="6" ViewCount="85" Body="&lt;p&gt;I am trying to self-read through Ian Stewart's Galois theory and am stuck at this paragraph. I quote - &quot;Lagrange observed that all methods for solving polynomial equations by radicals involve constructing rational functions of the roots that take a small number of values when the roots $\alpha_j$ are permuted. Prominent among these is the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \prod_{j&amp;lt; k} (\alpha_j - \alpha_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which takes just two values, $\pm \delta$, plus for even permutations and minus for the odd ones. Therefore, $\Delta = \delta^{2}$ is a rational function of the coefficients. This gets us started and it yields a complete solution for the quadratic.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is as follows: Why should an expression that is unchanged by permuting the roots of a polynomial be expressible as a rational function of the polynomial's coefficients? Is this statement even true or have I misunderstood the book? &lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-02T20:16:12.967" LastActivityDate="2010-11-02T20:16:12.967" Title="Expressions unchanged by permuting the roots of a polynomial" Tags="&lt;group-theory&gt;&lt;polynomials&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="7619" PostTypeId="1" CreationDate="2010-10-23T21:42:13.557" Score="2" ViewCount="109" Body="&lt;p&gt;I am trying to solve some problems based on this &lt;a href=&quot;http://math.stackexchange.com/questions/7614/how-to-prove-the-following&quot;&gt;formula&lt;/a&gt;,but am facing some issues in determine whether or not consider ordering as important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In how many ways 15 different books can be divided five heaps of equal number of books ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here according to the answer the ordering of groups is not considered important,hence the answer given is $\frac {15!} {5!(3!)^5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the same thing goes for this one :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In how many ways 3 piles can be formed out of 18 different books&lt;/strong&gt;, which will be $\frac {18!} {3!(6!)^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the following problem &lt;strong&gt;&quot;There are three copies, each of four different books. In how many ways they can be arranged on a self ?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This this is similar to divide 12 books into 4 sets of 3 books each, the answer given here considers the ordering important,why this change ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why in the first two the ordering not and on the third one it is important ? I guess (if the solution is correct) it is something related to copies but am not confident enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example of a problem where ordering should be considered important is say &quot;&lt;strong&gt;In how many ways can 18 different books be divided equally among three students ?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution should be $\frac {18!} {(6!)^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not much confident while in these problems so I will be grateful if somebody helps me to understand when to consider ordering important and when not to.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:18:50.770" LastActivityDate="2010-11-11T15:18:50.770" Title="Dealing with grouping problem in combinatorics" Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="7620" PostTypeId="2" ParentId="7618" CreationDate="2010-10-23T21:42:57.873" Score="8" ViewCount="0" Body="&lt;p&gt;This is a straightforward consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#The_fundamental_theorem_of_symmetric_polynomials&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem on Symmetric Functions&lt;/a&gt;. Indeed, the coefficients of the polynomial are the evaluation of the so called elementary symmetric functions at the roots.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-23T21:42:57.873" CommentCount="3" />
  <row Id="7621" PostTypeId="1" AcceptedAnswerId="7658" CreationDate="2010-10-23T21:46:33.733" Score="8" ViewCount="285" Body="&lt;p&gt;This thing came up in a combinatorics course I am taking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a fixed set of primes $p_1,p_2,\dots,p_k$ and let $A_n$ be number of integers in $\{1,2,\dots,n\}$ which are not divisible by any of the $p_i$'s. $A_n$ is given by $ n - \sum_{1\leq i_1 \leq k} \lfloor \frac{n}{p_{i_1}} \rfloor + \sum_{1\leq i_1 &amp;lt; i_2 &amp;lt; k} \lfloor \frac{n}{p_{i_1}p_{i_2}}\rfloor - \dots $. Now, if $n$ is divisible by each of the $p_i$'s then we have the simpler expression : $A_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$(added later : note I am not assuming $n = \prod_{i=1}^{k}p_i$ here, this holds true whenever $n = c \prod_{i=1}^{k} p_i $ for some integer c, since each $\lfloor \frac{n}{p_{i_1} \dots p_{i_j}} \rfloor = \frac{n}{p_{i_1} \dots p_{i_j}}\ )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another student pointed out that even if $n$ does not have $\prod_{i=1}^{k} p_i$ as a factor, the approximation $B_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$ to $A_n$ is quite close in some specific cases. It is easy to see that $\lim_{n\to\infty}\frac{A_n - B_n}{n}=0$ as $\lim_{n\to\infty}\frac{1}{n}\lfloor \frac{n}{p_{i_1}p_{i_2}\dots p_{i_j}} \rfloor \to \frac{1}{p_{i_1}p_{i_2} \dots p_{i_j}},$ however that is not strong enough to imply that $A_n - B_n$ will be always close.&#xA;Is there any way to analyze how well this approximation will perform in general? I am interested in the worst case for small to moderately sized $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give a feel of how close $A_n$ and $B_n$ can get, assuming the set of primes is $\{2,3,7\}$ we have (assuming my program is correct):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n     A(n)     B(n)&#xA;17    5    4.86  &#xA;27    8    7.71  &#xA;37    11      10.57  &#xA;107   31      30.57  &#xA;1111  318     317.43  &#xA;3001  858     857.43  &#xA;4007  1145    1144.86  &#xA;5000  1429    1428.57 .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-10-25T18:00:58.520" LastActivityDate="2010-10-26T12:19:54.940" Title="Accuracy of approximation to inclusion-exclusion formula in prime sieve" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7622" PostTypeId="2" ParentId="7606" CreationDate="2010-10-23T21:50:32.350" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ Since $\rm\ x^q - 1\ $ divides the prime product $\rm\ x^{p-1} - 1\ =\ (x-1)\:(x-2)\:\cdots\:(x-p+1)\ $ by uniqueness it is uniquely a product of $\rm\:q\:$ of them, so it has precisely $\rm\:q\:$ roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ Products of primes factor uniquely over any integral domain, i.e. any other factorization into irreducibles is the same up to order and unit factors. Such &lt;strong&gt;uniqueness&lt;/strong&gt; of prime factorizations is the trivial part of being a UFD - the nontrivial part is the &lt;strong&gt;existence&lt;/strong&gt; of prime factorizations for all nonzero nonunits.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-23T21:55:43.290" LastActivityDate="2010-10-23T21:55:43.290" />
  <row Id="7623" PostTypeId="2" ParentId="7619" CreationDate="2010-10-23T21:54:13.900" Score="0" ViewCount="0" Body="&lt;p&gt;In the first one, the problem only states that you need to group them, as in, let's say, put the books in a bag, so there could be no order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is very similar. It would be different if it had stated it wanted a stack of books, which would have been something with an order, where you can say: book A is in the bottom, book B is right on top of A, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third, however, asks for an arrangement of books, which would be like putting them all on a shelf. Since you can't distinguish between 2 copies of the same book, swapping them around won't change the order of the books, so that is why you need to divide by the number of copies of the same book, so you don't count the very same arrangement twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I think that in these kind of problems, grouping stuff together means that order is not important (picture putting books in several big bags) and arranging things implies that there is order (imagine putting books in a shelf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your last question, ordering is not important, as you are not asking in which order the books can be given, but which different groups of books can be given to each student. Say we give books A B C D E F to student 1. He would end up with the same books if we had given him B C A E F D, so order is not important.&lt;/p&gt;&#xA;" OwnerUserId="1605" LastEditorUserId="1605" LastEditorDisplayName="" LastEditDate="2010-10-23T21:59:48.140" LastActivityDate="2010-10-23T21:59:48.140" />
  <row Id="7625" PostTypeId="2" ParentId="5921" CreationDate="2010-10-23T22:42:55.797" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a rough sketch of one approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\mathbb{P}[Y\lt X] &#xA;&amp;amp;= \int_{-\infty}^{\infty} \mathbb{P}[Y\lt X|X=x] \cdot f_X(x) \,dx&#xA;\\&#xA;&amp;amp;= \int_{-\infty}^{\infty} F_Y(x) \cdot f_X(x) \,dx&#xA;\\&#xA;&amp;amp;\geq \int_{-\infty}^{\infty} F_X(x) \cdot f_X(x) \,dx&#xA;\\&#xA;&amp;amp;= \left[ \frac{1}{2} \left(F_X(x)\right)^2 \right]_{-\infty}^{\infty}&#xA;\\&#xA;&amp;amp;= \frac{1}{2}.&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="1023" LastActivityDate="2010-10-23T22:42:55.797" />
  <row Id="7626" PostTypeId="2" ParentId="7619" CreationDate="2010-10-23T22:45:58.127" Score="1" ViewCount="0" Body="&lt;p&gt;There are different notions of &quot;ordered&quot; and &quot;unordered&quot; that can arise in relation to set partitions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An unordered set of unordered sets,&lt;/li&gt;&#xA;&lt;li&gt;An unordered set of ordered sets,&lt;/li&gt;&#xA;&lt;li&gt;An ordered set of unordered sets,&lt;/li&gt;&#xA;&lt;li&gt;An ordered set of ordered sets,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;all of which are counted differently.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In how many ways 15 different books can be divided five heaps of equal number of books ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of what is counted here is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{{1,3,8},{2,14,15},{4,5,7},{6,12,13},{9,10,11}}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is an unordered set of unordered sets.  This is what the answer to the first question suggests it's counting.  [Although, the question itself does not resolve the question of order within the heaps.]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three copies, each of four different books. In how many ways they can be arranged on a self ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of what is counted here is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 1 4 1 3 3 2 2 4 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which we can interpret as a set partition counting problem -- place the i-th element in the s(i)-th set, where s(i) denotes the element in the i-th position.  So the above example gives rise to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;({1,4,6},{2,9,10},{3,7,8},{5,11,12})&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is an ordered set of unordered sets.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To highlight the difference between these two situations, for the first question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{{1,3,8},{2,14,15},{4,5,7},{6,12,13},{9,10,11}}&#xA;{{2,14,15},{1,3,8},{4,5,7},{6,12,13},{9,10,11}}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;should be considered the same (this is the same as swapping the positions of the first two heaps), whereas in the second question:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;({1,4,6},{2,9,10},{3,7,8},{5,11,12})&#xA;({2,9,10},{1,4,6},{3,7,8},{5,11,12})&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;should be considered different (this is the same as swapping all the books labelled 1 with all the books labelled 2).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-23T22:45:58.127" />
  <row Id="7627" PostTypeId="1" AcceptedAnswerId="7631" CreationDate="2010-10-23T22:46:41.163" Score="6" ViewCount="111" Body="&lt;p&gt;I want to graph an equilateral triangle. It would be ideal if I had a set of three points: $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ with $x_i, y_i \in \mathbb{Z}$ as the vertices. However, this is impossible. I am willing to &quot;settle&quot; for a triangle that is off by the width of a pencil lead. (That is, if we draw circles of radius $\delta$ around all of the lattice points, the true location of the vertices of the equilateral triangle are inside of these error circles.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For a given $\delta$ what is the smallest such triangle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct is to let the computer guess, is there a more elegant way?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-23T23:52:51.653" LastActivityDate="2010-10-23T23:52:51.653" Title="I want to graph an equilateral triangle on graph paper..." Tags="&lt;number-theory&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7628" PostTypeId="1" AcceptedAnswerId="7847" CreationDate="2010-10-23T23:00:35.083" Score="7" ViewCount="170" Body="&lt;p&gt;It is a theorem that every locally free coherent sheaf on $\mathbb{P}^1$ over an algebraically closed field is isomorphic to a unique sum of sheaves $\mathcal{O}(n)$ for various integers $n$. In particular, the K-ring of locally free coherent sheaves (or all coherent sheaves, $\mathbb{P}^1$ being nonsingular) is isomorphic to $\mathbb{Z}[t, t^{-1}]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topological K-ring of vector bundles on $S^2$ is, by Bott periodicity, isomorphic to $\mathbb{Z}[H]/(H-1)^2$, where $H$ is the canonical bundle. But $S^2$ is homeomorphic to $\mathbb{P}^1_{\mathbb{C}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every locally free sheaf corresponds to a vector bundle on $S^2$. It follows that the map on the K-groups from locally free sheaves to vector bundles is surjective but not injective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What goes wrong? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a version of Bott periodicity for algebraic varieties (or schemes)? (I.e., relating K-groups of $X$ and $X \times \mathbb{P}^1$.) I understand that there is one for the Picard groups. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-10-25T21:00:23.740" Title="Bott periodicity and algebraic geometry" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7629" PostTypeId="1" CreationDate="2010-10-23T23:09:42.143" Score="0" ViewCount="53" Body="&lt;p&gt;I'm wondering if there is other terminology that describes row-completeness, outside the context of a latin square, or if row-complete is actually a general term.&lt;/p&gt;&#xA;" OwnerUserId="2735" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:36:40.500" LastActivityDate="2010-12-19T11:44:26.157" Title="Does the term Row-Complete have any synonyms?" Tags="&lt;combinatorics&gt;&lt;terminology&gt;&lt;latin-square&gt;" AnswerCount="2" />
  <row Id="7630" PostTypeId="2" ParentId="7627" CreationDate="2010-10-23T23:10:16.000" Score="3" ViewCount="0" Body="&lt;p&gt;If you want one of the sides to be along the axis, you can use (0,0),(2x,0),and (x,y).  Then you want $sqrt(y^2+x^2)=2x$ or $y^2=3x^2$.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Pell%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Pell's equation&lt;/a&gt; will help you find solutions in the integers to $y^2=3*x^2\pm1$.  $x=30, y=26$ is within .017 in y.  Once you have one solution (say (a,b)=(1,0)), you can get another one as (2a+3b,a+2b).  These get very close to equilateral.  I don't know if you can do better if you don't put one side along an axis.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-23T23:10:16.000" />
  <row Id="7631" PostTypeId="2" ParentId="7627" CreationDate="2010-10-23T23:17:38.953" Score="5" ViewCount="0" Body="&lt;p&gt;Great question!  Here is an upper bound: let $\frac{p_n}{q_n}$ be the sequence of convergents to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; of $\sqrt{3}$.  It is known that $| \sqrt{3} - \frac{p_n}{q_n} | &amp;lt; \frac{1}{q_n q_{n+1}}$.  Then the vertices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0, 0), (2q_n, 0), (q_n, p_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are within $|\sqrt{3} q_n - p_n| &amp;lt; \frac{1}{q_{n+1}}$ of an equilateral triangle.  Asymptotically I believe that $q_n \sim C \cdot \sqrt{3}^n$ for some constant $C$, which means that for small $\delta$ we can always find a triangle with side length something like $\frac{2}{\delta}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a way to get a lower bound: suppose $(x_i, y_i)$ is a collection of points which are $\delta$-close to an equilateral triangle, and suppose WLOG that $(x_1, y_1) = (0, 0)$.  Then $\frac{x_3 + iy_3}{x_2 + iy_2}$ is a rational approximation to $\frac{1}{2} + \frac{\sqrt{3}}{2} i$, hence taking imaginary parts, $2 \frac{x_2 y_3 - x_3 y_2}{x_2^2 + y_2^2}$ is a rational approximation to $\sqrt{3}$.  It is known that the convergents $\frac{p_n}{q_n}$ give the best rational approximations to $\sqrt{3}$ (in the sense that better ones must have larger denominators), so one should be able to write down how good this approximation is in terms of $\delta$ and use this to give a lower bound on how large $x_2^2 + y_2^2$ must be.  But the details look messy.  In any case, I would be very surprised if one didn't end up with a lower bound of the form $\frac{C}{\delta}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-23T23:27:46.337" LastActivityDate="2010-10-23T23:27:46.337" CommentCount="3" />
  <row Id="7632" PostTypeId="2" ParentId="7586" CreationDate="2010-10-23T23:48:40.123" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/7586/solve-z2-6z-25-into-complex-conjugate/7588#7588&quot;&gt;Noldorin's answer&lt;/a&gt; pretty much covers what I'd say, but I would note that you can proceed in working with the expression until it is factored into linear terms and read off the zeros (the solutions to the equation $(z - 3)^2 + 16=0$) from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing from $(z - 3)^2 + 16$:&#xA;$$\begin{align}&#xA;(z - 3)^2 + 16 &amp;amp;= (z - 3)^2 - (-16)&#xA;\\&#xA;&amp;amp;= (z - 3)^2 - (4i)^2&#xA;\\&#xA;&amp;amp;= ((z-3)-4i)((z-3)+4i)&#xA;\\&#xA;&amp;amp;= (z-(3+4i))(z-(3-4i))&#xA;\end{align}$$&#xA;So, the zeros of the expression are $3+4i$ and $3-4i$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-23T23:48:40.123" CommentCount="1" />
  <row Id="7633" PostTypeId="2" ParentId="7629" CreationDate="2010-10-23T23:55:25.800" Score="0" ViewCount="0" Body="&lt;p&gt;For those who are unfamiliar, a &lt;em&gt;Latin square&lt;/em&gt; $L=(l_{ij})$ is an n by n matrix in which each symbol occurs exactly once in each row an each column, such as the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 1 5 2 4 3&#xA;1 2 0 3 5 4&#xA;5 0 4 1 3 2&#xA;2 3 1 4 0 5&#xA;4 5 3 0 2 1&#xA;3 4 2 5 1 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A Latin square L is called &lt;em&gt;row-complete&lt;/em&gt; if there are n(n-1) distinct tuples of the form $(l_{ij},l_{i(j+1)})$ in L.  The above Latin square is a row-complete Latin square (sourced from &lt;a href=&quot;http://designtheory.org/library/encyc/exs/clsex.html&quot; rel=&quot;nofollow&quot;&gt;http://designtheory.org/library/encyc/exs/clsex.html&lt;/a&gt;) since if we look through it we find each ordered pair (0,1), (0,2), ..., (5,4) exactly once.  These Latin squares (and Latin squares in general) are sometimes used in statistical experiments (or so I'm led to believe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which orders n does there exist a row-complete Latin square is still not completely resolved (although some partial results are known).  In the abstract of J. T. Higham's PhD thesis Construction methods for row-complete Latin squares, row-complete Latin squares are also known as &lt;em&gt;roman squares&lt;/em&gt;.  I haven't heard this term previously myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any cases where &quot;row-complete&quot; has been used in other contexts, and searching through Google and MathSciNet didn't yield anything.  There are probably equivalent notions used in graph theory but under a different name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need an answer to the question, my feeling is you're going to need to ask a bunch of experts in the field.  E.g. you could present your work at a seminar or conference, and ask the audience if they know of an alternative name or usage of &quot;row-complete&quot;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-23T23:55:25.800" />
  <row Id="7634" PostTypeId="1" AcceptedAnswerId="7637" CreationDate="2010-10-23T23:57:36.480" Score="14" ViewCount="496" Body="&lt;p&gt;I use stories like these to develop intuition... or perhaps to destroy it. I have my own answers in mind, but I want to see if I have made any mistakes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are standing at the origin of an &quot;infinite forest&quot; holding an &quot;infinite bb-gun.&quot; The &quot;trees&quot; in this forest are at the lattice points all around you. (The lattice points are like those on graph paper and they align with the cardinal directions: N, S, E, W.) The &quot;forest&quot; is Euclidean in the sense that the trees have no width. To hit a tree with your bb-gun you must aim perfectly at it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would, for example, hit a tree if you fired the gun due north, south east or west. (Your bullets also have no width.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. You fire the gun in an arbitrary direction without bothering to aim. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. You get a new bb-gun and the bullets have a little width to them. ($\delta$?) You fire the gun in an arbitrary direction without bothering to aim. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. All of the trees are removed that have coordinates whose absolute values are not perfect squares. (So, only points such as $(25, 100)$ and $(4,-1600)$ remain.)  Again you use width-less bullets. You fire the gun in an arbitrary direction without bothering to aim. What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. Again, only with perfect squares, but now the bullets have width. What happens?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-24T05:11:15.807" LastActivityDate="2010-10-24T17:55:29.573" Title="You are standing at the origin of an &quot;infinite forest&quot; holding an &quot;infinite bb-gun&quot; " Tags="&lt;number-theory&gt;&lt;intuition&gt;&lt;infinity&gt;&lt;tag-removed&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="7635" PostTypeId="2" ParentId="7629" CreationDate="2010-10-24T00:12:25.630" Score="0" ViewCount="0" Body="&lt;p&gt;Alright, thanks for all the help. ;)  I don't really need to exhaust all possibilities, just wondering (but like you mentioned, it isn't the easiest thing to search for).&lt;/p&gt;&#xA;" OwnerUserId="2737" LastActivityDate="2010-10-24T00:12:25.630" CommentCount="1" />
  <row Id="7636" PostTypeId="1" CreationDate="2010-10-24T00:30:09.093" Score="1" ViewCount="91" Body="&lt;p&gt;Or has it remained a terminal node at the frontier of mathematics?&lt;/p&gt;&#xA;" OwnerUserId="2344" LastActivityDate="2010-10-24T00:30:09.093" Title="Has anything further been done with Morley’s Miracle?" Tags="&lt;euclidean-geometry&gt;" CommentCount="3" />
  <row Id="7637" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T00:31:36.020" Score="16" ViewCount="0" Body="&lt;p&gt;A., C.  The probability that you hit a tree is $0$.  This is equivalent to the statement that the set of angles at which you can see a tree is of measure zero in $[0, 2\pi)$ with the Lebesgue measure, which follows because it is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B.  The probability that you hit a tree is $1$.  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlets_approximation_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's approximation theorem&lt;/a&gt;, for any real $\alpha$ there exist infinitely many pairs $p_n, q_n$ of integers such that $|\alpha - \frac{p_n}{q_n}| &amp;lt; \frac{1}{q_n^2}$.  It follows that the distance between the points $(q_n, q_n \alpha)$ and $(q_n, p_n)$ is at most $\frac{1}{q_n}$, and since the sequence $q_n$ tends to $\infty$ it follows by letting $\alpha = \tan \theta$ where $\theta$ is the angle at which you fired that for any bullet size $\delta &gt; 0$ we will hit a tree at $(q_n, p_n)$ where $q_n &gt; \frac{1}{\delta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D.  The probability that you hit a tree is &lt;strong&gt;still&lt;/strong&gt; $1$.  This is a consequence of the following theorem, which I just learned about &lt;a href=&quot;http://mathoverflow.net/questions/43381/what-numbers-can-be-approximated-pretty-well-by-rationals&quot; rel=&quot;nofollow&quot;&gt;by asking this MO question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eom.springer.de/d/d032580.htm&quot; rel=&quot;nofollow&quot;&gt;Theorem (Khinchin):&lt;/a&gt;  Let $\phi(q) : \mathbb{N} \to \mathbb{R}$ be a monotonically decreasing function.  For almost all real numbers $\alpha$, the number of pairs of positive integers $(q, p)$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| p - q\alpha \right| &amp;lt; \phi(q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is infinite if $\sum \phi(q)$ diverges, and finite if $\sum \phi(q)$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, taking $\phi(q) = \frac{1}{q \ln q}$ (the sum of which diverges, for example by the integral test), we get that for almost all real $\alpha$ there are infinitely many solutions $(q, p)$ to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{p}{q} - \alpha \right| &amp;lt; \frac{1}{q^2 \ln q}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\alpha = \sqrt{\tan \theta}$.  With probability $1$ there will be infinitely many $(q, p)$ satisfying the above condition.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|p^2 - q^2 \tan \theta \right| &amp;lt; \frac{\left| \frac{p}{q} + \sqrt{\tan \theta} \right|}{\ln q}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking $q$ large enough so that the RHS is less than $\frac{\delta}{2}$ it follows that a bullet shot at angle $\theta$ will hit the tree at $(q^2, p^2)$.  (I'm only working in the positive quadrant but the generalization to the other quadrants should be clear.)  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-24T16:56:05.127" LastActivityDate="2010-10-24T16:56:05.127" CommentCount="6" />
  <row Id="7638" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T00:44:38.157" Score="6" ViewCount="0" Body="&lt;p&gt;Interestingly, the &lt;em&gt;fraction&lt;/em&gt; of the lattice points that you could hit (in question A&amp;mdash;not what you asked) from&#xA;the origin is &#xA;$$ 6 / {\pi}^2 $$&#xA;or about 61%.&#xA;See &lt;em&gt;Research Problems in Discrete Geometry&lt;/em&gt;,&#xA;Peter Brass, W. O. J. Moser, János Pach, p.430, &quot;Visibility problems for lattice points.&quot;&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-24T00:44:38.157" CommentCount="8" />
  <row Id="7639" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T00:58:29.737" Score="1" ViewCount="0" Body="&lt;p&gt;B. in the annulus from $r$ to $r+dr$ there are $2*\pi *r*dr$ trees.  If we move the diameter from the bullet to the trees, they subtend an angle $\delta*2*\pi*r*dr/r$.  This reaches $2\pi$ at $r=1/\delta$, so that is about where we would expect to hit a tree.  As the integral diverges, we are guaranteed to hit a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. If we make the tree size δ it subtends an angle $\delta /\sqrt {i^4+j^4}$ Does the sum of this diverge? We need it to exceed $\pi /2$ as I have just covered the first quadrant. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T00:58:29.737" CommentCount="6" />
  <row Id="7640" PostTypeId="1" AcceptedAnswerId="7642" CreationDate="2010-10-24T01:03:37.797" Score="1" ViewCount="49" Body="&lt;p&gt;You have 3 bins.  You can toss balls one at a time to the bins until any one of them has 2 balls in it, and then you stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tosses are independent, and each bin is equally likely to be hit.  It's impossible to not make it in a bin on any given toss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the expected number of tosses so that 1 bin has 2 balls in it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&#xA;Not sure if it's important, but by the pigeon hole principle we know that the maximum number of balls tosses required to do this would be 4.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-10-24T01:14:47.203" Title="Expected number of tosses so that 1 out of 3 bins has 2 balls in it" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="7641" PostTypeId="2" ParentId="7640" CreationDate="2010-10-24T01:07:08.070" Score="0" ViewCount="0" Body="&lt;p&gt;Hint:  So what is the chance that you get success in two throws?  Call it x.  In exactly three throws?  Call it y.  And that tells you the chance that you will need four is 1-x-y.  Then what is the expected number of throws?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T01:07:08.070" />
  <row Id="7642" PostTypeId="2" ParentId="7640" CreationDate="2010-10-24T01:14:47.203" Score="1" ViewCount="0" Body="&lt;p&gt;It can't happen in 1 toss.  In 2 tosses, the first toss doesn't matter (probability 1), but the second must match the first (probability $\frac{1}{3}$), so it happens in 2 tosses with probability $1\cdot\frac{1}{3}=\frac{1}{3}$.  In 3 tosses, the first toss doesn't matter, the second toss must not match the first toss ($\frac{2}{3}$), and the third toss must match one of the first two ($\frac{2}{3}$), so it happens in exactly 3 tosses with probability $1\cdot\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$.  In 4 tosses, the first toss doesn't matter, the second toss must not match the first toss, the third toss must not match either of the first two tosses ($\frac{1}{3}$), and the fourth toss will match one of the first three no matter what (1), so it happens in exactly 4 tosses with probability $1\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot 1=\frac{2}{9}$.  Note that the sum of these three probabilities is 1, which confirms that there is no need to look past 4 tosses.  The expected number of tosses is $2\cdot\frac{1}{3}+3\cdot\frac{4}{9}+4\cdot\frac{2}{9}=\frac{26}{9}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-24T01:14:47.203" />
  <row Id="7643" PostTypeId="1" AcceptedAnswerId="7667" CreationDate="2010-10-24T03:04:24.913" Score="5" ViewCount="206" Body="&lt;p&gt;I remember my professor in college challenging me with this question, which I failed to answer satisfactorily: I know there exists a bijection between the rational numbers and the natural numbers, but can anyone produce an explicit formula for such a bijection?&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T14:21:39.320" LastActivityDate="2010-11-21T14:21:39.320" Title="Produce an explicit bijection between rationals and naturals?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="7644" PostTypeId="1" CreationDate="2010-10-24T03:34:07.280" Score="2" ViewCount="108" Body="&lt;p&gt;The following comes from questions comes from a recent combinatorics paper I attended :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.27 people are to travel by a bus which can carry 12 inside and 15 outside. In how many ways can the party be distributed between inside and outside if 5 people refuse to go outside and 6 will not go inside?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution given C(16,7), I have no clue how they got it ?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.The number of functions f from the set A = {0, 1, 2} into the set B = {1, 2, 3, 4, 5, 6, 7} such that $f(i) \le f(j) $ for $i \lt j $ and $i,j$ belongs to A is&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution given is C(8,3). I didn't really understood this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.The number of ordered pairs $(m, n) m, n $ is in {1 , 2, … , 100} such that $7^m + 7^n$ is divisible by 5 is&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution given is 2500, but how ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.The coefficient of $x^{20}$in the expansion of $(1 + 3x + 3x^2 + x^3)^{20}$, is ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this one elegantly ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5.An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0, 1, …, 9 without replacement. The number of ways in which this can be done is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this one seems impossible for me to solve in 1 mint,or is it ? Given soln is 36(7!)&lt;/p&gt;&#xA;" OwnerUserId="2739" LastActivityDate="2010-10-24T05:30:08.113" Title="Problems on combinatorics" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7645" PostTypeId="2" ParentId="7644" CreationDate="2010-10-24T03:48:05.667" Score="0" ViewCount="0" Body="&lt;p&gt;For number 4 you would use the multinomial theorem.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-10-24T03:48:05.667" />
  <row Id="7646" PostTypeId="2" ParentId="7578" CreationDate="2010-10-24T03:48:33.183" Score="1" ViewCount="0" Body="&lt;p&gt;As many others have pointed out, the way to do this is to think about pairs of metric spaces that are homeomorphic but not isometric.  In many cases, the metric will be complete on one space but not the other.  For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The set $\{1/n \mid n \in \mathbb{N}\}$ is incomplete, but the natural numbers $\mathbb{N}$ are complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ray $(0,\infty)$ and the open interval $(0,1)$ are incomplete, but the real line is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An open disc in the plane is incomplete, but the entire plane is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The punctured plane $\mathbb{R}-\{0\}$ is incomplete, but the infinite cylinder $\mathbb{R} \times S^1$ (where $S^1$ is the circle) is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The real line minus the $x$-axis is incomplete, but a disjoint pair of planes is complete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The twice-punctured plane $\mathbb{R}^2 - \{(-1,0),(1,0)\}$ is incomplete, but the graph of  $$&#xA;z = \frac{1}{(x^2-1)^2 + y^2}&#xA;$$&#xA;in $\mathbb{R}^3$ is complete.  (This graph has asymptotic &quot;cusps&quot; at $(-1,0)$ and $(1,0)$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In each case, the homeomorphism between the two spaces can be used to define a nonstandard metric on the latter space that makes it incomplete.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-10-24T03:48:33.183" />
  <row Id="7647" PostTypeId="1" AcceptedAnswerId="7648" CreationDate="2010-10-24T04:06:57.383" Score="3" ViewCount="87" Body="&lt;p&gt;Does anyone know if there's a client or web app that generates Penrose patterns which can then be converted to a tileable rectangular background image for web site?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this &lt;a href=&quot;http://stephencollins.net/penrose&quot; rel=&quot;nofollow&quot;&gt;http://stephencollins.net/penrose&lt;/a&gt; but I can't seem to get a rectangular shape out of it for web use. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-24T04:12:53.247" LastActivityDate="2010-10-24T04:12:53.247" Title="Penrose Tile generator" Tags="&lt;geometry&gt;&lt;math-software&gt;&lt;tiling&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7648" PostTypeId="2" ParentId="7647" CreationDate="2010-10-24T04:08:36.793" Score="7" ViewCount="0" Body="&lt;p&gt;I believe, since Penrose tilings are aperiodic (lacking translational symmetry), there isn't such a rectangular shape.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-24T04:08:36.793" CommentCount="1" />
  <row Id="7649" PostTypeId="2" ParentId="7644" CreationDate="2010-10-24T04:29:43.297" Score="5" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Five people refuse to go outside, and therefore will go inside. Six people refuse to go inside, so will go outside. That means that you still have $27 - (5+6)=16$ people to accommodate. There are 12 spots for people inside, but five are already &quot;taken&quot; by those who refuse to be outside. That leaves 7 seats inside to assign. So you need to choose which seven people, out of the 16 that are left, will go inside. The number of ways of doing this is precisely $\binom{16}{7}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I misread the question as saying that $f(i)\lt f(j)$ if $i\lt j$. That answer follows: since you require the values of the function to increase (the condition just says that $f(0)\lt f(1)\lt f(2)$), if you know the three values of $f(0)$, $f(1)$, and $f(2)$ you know which one corresponds to which ($f(0)$ is the smallest value, $f(1)$ is the middle value, and $f(2)$ is the largest value). So all you need to do in order to determine a function is to pick three values from $B$. There are $7$ possibilities, you need to pick $3$, so there are $\binom{7}{3}$ ways of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it seems I misread the question. It actually says that $f(i)\leq f(j)$ if $i\lt j$. I gave the number of functions in which all inequalities are strict. There are $\binom{7}{1}$ functions in which all inequalities are actually equalities (just one value). Now, to count the number of functions in which $f(0)=f(1)\lt f(2)$, you just need to pick two values from the set $B$, which there are $\binom{7}{2}$ ways of doing; the same holds for the case in which you have $f(0)\lt f(1)=f(2)$. So the total is $\binom{7}{3}+2\binom{7}{2} + \binom{7}{1}$. Since $\binom{n}{k}+\binom{n}{k-1} = \binom{n+1}{k}$,  this total is equal to &#xA;$$\left(\binom{7}{3} + \binom{7}{2}\right) + \left(\binom{7}{2}+\binom{7}{1}\right) = \binom{8}{3}+\binom{8}{2} = \binom{9}{3}.$$ It seems to me, then, that the answer you give is incorrect, or perhaps you mistyped it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you know the formula for combinations with repetitions, then there is a more direct way of getting this result: simply pick $3$ out of the $7$ possible images, with repetitions allowed; smallest value is $f(0)$, middle value is $f(1)$, largest values $f(3)$ (equality allowed). The total for this is $\binom{7+3-1}{3} = \binom{9}{3}$, as above).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume $m\leq n$; then $7^m+7^n = 7^m(1 + 7^{n-m})$. The product is divisible by $5$ if and only if $1 + 7^{n-m}$ is divisible by $5$ (since $5$ is prime and never divides a power of $7$). For $1+7^{n-m}$ to be divisible by $5$, you need $7^{n-m}$ to have a remainder of $4$ when divided by $5$. If you run over the powers of $7$ and see the remainder when divided by $5$, you will notice that they go $2$, $4$, $3$, $1$, and repeat. So basically, you need $n-m$ to be a multiple of $4$ plus $1$. That is, you need $n-m = 4k+1$. Note in particular that if the pair has $n=m$, then it does not satisfy the condition. So count how many pairs there are where the two differ by a multiple of four plus 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strike&gt;One possibility is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;Multinomial theorem&lt;/a&gt; You would need to figure out all the ways in which you can obtain $x^{20}$ as products of powers of $x$, $x^2$, and $x^3$, and add the appropriate coefficients.&lt;/strike&gt; &lt;strong&gt;Edit:&lt;/strong&gt; But the intended answer is almost certainly the one given by Larry Denenberg.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In order for the number to be divisible by $9$, the digits must add up to a multiple of $9$. The digits $0$ through $9$ add up to $45$, which is a multiple of $9$. So if you omit two of them, they must add up to $9$: thus, if you omit $0$, then you must also omit $9$; if you omit $1$, then you must also omit $8$; etc. So you only have five possible pairs of numbers that you can omit. So pick which of the five pairs you will omit. If you omit $0$ and $9$, then the remaining $8$ digits can be arranged in any order, giving $8!$ possibilities. In all other cases, you cannot place $0$ in the first position, but otherwise can place the rest in any order. That gives $7(7!)$ possible ways of ordering the numbers. Thus, you have one choice that leads to $8!$ numbers, and four choices that lead each to $7(7!)$ numbers. Adding them up gives $8!+(4\times 7)(7!) = 8(7!)+28(7!) = 36(7!)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-24T05:30:08.113" LastActivityDate="2010-10-24T05:30:08.113" CommentCount="2" />
  <row Id="7650" PostTypeId="1" CreationDate="2010-10-24T04:51:02.483" Score="1" ViewCount="91" Body="&lt;p&gt;I am just beginning to read about the use of &quot;Concave Programming&quot; methods and use of the Karush-Kuhn-Tucker conditions to identify the maximum value of a non-linear objective function subject to inequality constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The examples I have seen in the text I have at hand, all involve only linear constraints. Is this method equally applicable to situations where not only are there multiple constraints, but where one or more of those constraints are non-linear ?&lt;/p&gt;&#xA;" OwnerUserId="2740" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-24T19:29:35.807" LastActivityDate="2010-10-24T19:29:35.807" Title="Non-linear programming" Tags="&lt;nonlinear-optimization&gt;&lt;economics&gt;" CommentCount="5" />
  <row Id="7651" PostTypeId="2" ParentId="7644" CreationDate="2010-10-24T04:57:53.557" Score="3" ViewCount="0" Body="&lt;p&gt;For number 4, the multinomial theorem is overkill.  The given expression is just $((x+1)^3)^{20} = (x+1)^{60}$. The coefficient of $x^{20}$ is $60\choose 20$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-10-24T05:03:20.980" LastActivityDate="2010-10-24T05:03:20.980" />
  <row Id="7652" PostTypeId="1" AcceptedAnswerId="7653" CreationDate="2010-10-24T05:51:11.260" Score="0" ViewCount="62" Body="&lt;p&gt;Let $X$ be a Poisson random variable with mean $\lambda$. How do I show that $P[X \geq \lambda] = 1/2$? Also, I was wondering what distributions have this property that the density is concentrated equally around the mean. I see that it's true for Gaussian and Uniform distributions.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-24T06:52:50.887" Title="Concentration of poisson distribution" Tags="&lt;probability&gt;&lt;random-variable&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7653" PostTypeId="2" ParentId="7652" CreationDate="2010-10-24T06:24:03.337" Score="2" ViewCount="0" Body="&lt;p&gt;P[X ≥ &amp;mu;] = 1/2 if and only if the mean (&amp;mu;) and median &quot;coincide&quot;. This is true for Gaussian and uniform distribution, but not always true for Poisson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is exactly one &amp;mu; between every unit interval [n, n+1) such that P[X &amp;ge; &amp;mu;] = 1/2. Some of them are: ln 2, 1.6784…, 2.674…, 3.672…, ...&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-24T06:52:50.887" LastActivityDate="2010-10-24T06:52:50.887" CommentCount="3" />
  <row Id="7654" PostTypeId="2" ParentId="734" CreationDate="2010-10-24T06:30:05.757" Score="2" ViewCount="0" Body="&lt;p&gt;For my experience, the &lt;a href=&quot;http://www.open.ac.uk/&quot; rel=&quot;nofollow&quot;&gt;Open University&lt;/a&gt; is well-respected internationally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note this is different to Open Universities, such as &lt;a href=&quot;http://www.open.edu.au/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; in Australia, which is a collaboration of several universities.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-24T06:30:05.757" CommunityOwnedDate="2010-10-24T06:30:05.757" CommentCount="1" />
  <row Id="7655" PostTypeId="1" AcceptedAnswerId="7661" CreationDate="2010-10-24T06:40:56.027" Score="0" ViewCount="99" Body="&lt;p&gt;If $S$ is a surface which is the complement of finitely many points in a compact surface, and the metric in $S$ is complete, then is Gauss-Bonnet theorem still valid for $S$?&lt;/p&gt;&#xA;" OwnerUserId="2742" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-24T12:35:04.030" LastActivityDate="2010-10-24T14:21:11.990" Title="A question about Gauss Bonnet theorem" Tags="&lt;differential-geometry&gt;" AnswerCount="3" />
  <row Id="7656" PostTypeId="2" ParentId="7454" CreationDate="2010-10-24T07:05:57.753" Score="4" ViewCount="0" Body="&lt;p&gt;As promised in my response to Byron, here is my very partial progress to&#xA;date on this problem.  Though long-winded, it doesn't amount to much.&#xA;I'm still looking for an answer to the general case, or to other special&#xA;cases (like $n=2$ or $c=2$) or even just better estimates.  I'm also&#xA;interested to know about any applicable combinatorial tools even without&#xA;a full answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a caution:  The graphs I'm about to use are not related to the&#xA;graphs in the general formulation of the original question.  Ignore&#xA;those graphs and just think of students and classrooms!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be a graph with a vertex for each student, $cn$ vertices in all.&#xA;Say that a given year's assignment of students to classrooms&#xA;&lt;strong&gt;respects&lt;/strong&gt; $G$ if every edge in $G$ joins two students that are&#xA;assigned to different classrooms.  That is, $G$ encodes constraints on&#xA;assignments; each edge in $G$ represents two students that must be kept&#xA;apart.  Let $R(G)$ be the probability that a random assignment of&#xA;students to classrooms respects $G$.  Note that $R(G)$ does not depend&#xA;on $y$ since the definition of &quot;respects&quot; refers to a single year's&#xA;assignment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $e(G)$ be the number of edges in $G$.  Then the probability that&#xA;after $y$ years every pair of students has at some point shared a&#xA;classroom is exactly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum (-1)^{e(G)} (R(G))^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the sum is taken over all graphs $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result follows from a straightforward application of&#xA;Inclusion/Exclusion.  It's very pretty (at least I think so) but useless&#xA;if you want a real answer.  The problem isn't finding $R(G)$; this is tedious&#xA;but can be automated.  The real difficulty is that you have to handle&#xA;$2^{cn\choose 2}$ graphs, which is far too many.  The graphs come in&#xA;isomorphic bunches (e.g. there are $cn\choose 2$ one-edged graphs that all &#xA;have the probability of being respected) and you don't have to handle most graphs with lots of edges since $R(G) = 0$ for any graph $G$ that is not $c$-partite.  But this doesn't help enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get some use out of this wretched formula by generalizing the&#xA;problem.  Suppose we ask, given the same conditions, for the probability&#xA;that each student &lt;em&gt;in a given group of $p$ students&lt;/em&gt; at some point sees&#xA;each of the other students &lt;em&gt;in that group&lt;/em&gt;.  The formula above gives the&#xA;exact answer if we sum over all graphs with $p$ vertices, each labelled&#xA;with a student in the group.  For example, with $p = 2$ the formula&#xA;gives the probability that two given students at some point see each&#xA;other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 - \left(\frac{n(c-1)}{cn-1}\right)^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course we can get this result by a simpler route.)  For $p=3$ the&#xA;formula gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 - \left(\frac{N-n+1}{N(N-1)}\right)^y(3(N-1)^y - 3(N-n)^y + (N-2n+1)^y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N = cn-1$.  I've fought through the formula for $p$ up to 6, but&#xA;it becomes impossible quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now some approximations.  Define $U_m$ to be the probability that a&#xA;given student $S$ is at some point in a classroom with each student in a&#xA;given set of $m$ students not containing $S$.  Another straightforward&#xA;application of inclusion/exclusion gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_m = \sum_{j=0}^m(-1)^j{m\choose j}{N-j\choose n-1}^y \bigg/ {N\choose n-1}^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N = cn-1$ as before.  $U_N$ is then the probability that a single&#xA;given student will see all the others.  This is a very weak upper bound&#xA;on the probability that &lt;em&gt;all&lt;/em&gt; the students see all the others, though in&#xA;trivial cases one student seeing all the others does imply that they all&#xA;do!  $U_m$ is rather easy to calculate; with the original parameters,&#xA;for example, $U_{74}$ turns out to be a bit less than 1/7792.  Note, BTW,&#xA;that $U_1$ correctly gives the same result as the $p=2$ answer above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;student $A$ sees all the others&quot; were independent of &quot;student $B$&#xA;sees all the others&quot;, then we'd have the answer we seek:  $(U_N)^{cn}$.&#xA;But of course these events are unlikely to be independent.  If many&#xA;students have seen all the others, then surely the probability increases&#xA;that all have seen the others.  &lt;strike&gt;So $(U_N)^{cn}$ may be a lower bound on&#xA;the true answer, if a weak one.&lt;/strike&gt; &lt;strong&gt;Edit:&lt;/strong&gt; Actually, I don't think it can be a lower bound, since I'm pretty sure one could find a case where &lt;em&gt;one&lt;/em&gt; student can see all the others, but it's impossible for &lt;em&gt;all&lt;/em&gt; students to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another way of estimating the answer:  A group of $p$ students&#xA;all see each other if and only if both (a) a particular one of these&#xA;students sees all the other $p-1$ students, and (b) the other $p-1$&#xA;students all see each other.  The probability of (a) is exactly&#xA;$U_{p-1}$.  Let $C_{p-1}$ denote the probability of (b).  If events (a)&#xA;and (b) were independent, then the exact answer to the original question&#xA;would be $U_{p-1}C_{p-1}$.  But by identical reasoning, $C_{p-1} =&#xA;U_{p-2}C_{p-2}$.  Continuing in this way, the exact answer to the original&#xA;question would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_NU_{N-1}U_{N-2}\ldots U_2U_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As above, the events in question are not likely to be independent.  But&#xA;most of the time they're close, right?  With the original parameters,&#xA;the fact that a group of five students all see each other has very&#xA;little impact on whether a sixth student sees each of the five, though&#xA;it should increase that probability slightly.  This would make the&#xA;expression above a lower bound on the actual answer, though I don't know&#xA;how good it is.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-11-06T13:36:21.480" LastActivityDate="2010-11-06T13:36:21.480" CommentCount="3" />
  <row Id="7657" PostTypeId="2" ParentId="7655" CreationDate="2010-10-24T07:51:09.733" Score="0" ViewCount="0" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove a finite set of points from a compact surface, it will no longer be complete. If you change the metric so that you do have a complete surface, you cannot apply the Gauss Bonnet theorem anyway, because the surface isn't compact.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-10-24T07:51:09.733" CommentCount="2" />
  <row Id="7658" PostTypeId="2" ParentId="7621" CreationDate="2010-10-24T08:10:25.360" Score="4" ViewCount="0" Body="&lt;p&gt;To help with the notation I will change your $k$ to an $m$, so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_n = n - \sum \lfloor \frac{n}{p_i} \rfloor + \sum \lfloor \frac{n}{p_i p_j} \rfloor - &#xA;\sum \lfloor \frac{n}{p_i p_j p_k} \rfloor + \cdots, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the summations are over distinct primes in our set $ \lbrace p_1,p_2,\ldots,p_m \rbrace ,$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B_n = n \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $ x \ge \lfloor x \rfloor &gt; x – 1 $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_n \ge n - \sum \frac{n}{p_i} + \sum \frac{n}{p_i p_j} - \binom{m}{2} -&lt;br&gt;&#xA;\sum \frac{n}{p_i p_j p_k} + \sum \frac{n}{p_i p_j p_k p_l} - \binom{m}{4} + \cdots.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that $$\binom{m}{0}+\binom{m}{2}+\binom{m}{4} \cdots = 2^{m-1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we obtain $A_n \ge B_n – 2^{m-1}.$ Similarly, we have $A_n \le B_n + 2^{m-1}$ and hence we can bound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left| \frac{A_n – B_n}{n} \right| \le \frac{2^{m-1}}{n} = O(1/n).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we take $n \equiv -1 \textrm{ mod } \prod_{i=1}^m p_i$ then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n = \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right) = \text{a constant.} \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the difference is a constant infinitely often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Since the differences between the $ \lfloor \frac{n}{p_i p_j \ldots} \rfloor $ and the&#xA;$ \frac{n}{p_i p_j \ldots} $ are a maximum when $n \equiv -1 \textrm{ mod } \prod_{i=1}^m p_i$&#xA;it appears we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n \le \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right), \quad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with equality when $n \equiv -1 \textrm{ mod } \prod_{i=1}^m p_i.$ I do not have a proof of this latter claim, although I have a proof for (1), which I hope to be able to modify to prove (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: I could not find a proof of (2) because, sadly, it's not true! However, (1) is exact for the given $n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a vast improvement on my previous comments, which I believe solves your problem completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n \equiv -r \textrm{ mod } \prod_{i=1}^m p_i, $ where $ 0 \le r &amp;lt; \prod_{i=1}^m p_i$ then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n \le r \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proved this simply by expanding $A_n – B_n$ and doing the summations. Also, I believe the exact formula to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_n – B_n = \delta +  r \prod_{i=1}^m \left( 1- \frac{1}{p_i} \right) - \Phi_P(r),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta = 1$ when $r$ is coprime to all the $p_i$ and $0$ otherwise, and&#xA;$\Phi_P(r)$ is the number of positive integers $\le r$ which are coprime to the $p_i.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $n=711,$ $r=9,$ $\delta = 0$ and $\Phi_P(9) = 2,$&#xA;$P=\lbrace 2,3,5 \rbrace,$&#xA;so the above formula gives&#xA;$$A_{711}-B_{711} = 9 \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right)&#xA;\left( 1 - \frac{1}{5} \right) – 2 = 0.4$$&#xA;which agrees with your $190-189.6 = 0.4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $n=107,$ $r=19,$ $\delta = 1$ and $\Phi_P(19) = 6,$&#xA;$P=\lbrace 2,3,7 \rbrace,$&#xA;so we have&#xA;$$A_{107}-B_{107} = 1 + 19 \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right)&#xA;\left( 1 - \frac{1}{7} \right) – 6 = 0.42857\ldots,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is also in agreement with the rounded figures you've given in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I do not have time to type up the full proof.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-26T12:19:54.940" LastActivityDate="2010-10-26T12:19:54.940" CommentCount="6" />
  <row Id="7659" PostTypeId="1" AcceptedAnswerId="7662" CreationDate="2010-10-24T08:51:40.203" Score="5" ViewCount="138" Body="&lt;p&gt;I've started a little reading on quadratic reciprocity, and a reason for this has eluded me. Here's a little of what I came up with so far. I decided I want to show that for all primes $p$, if $p|x^2-2$, then $p$ does not divide $2y^2+3$. Then, by way of contradiction, if $(x^2-2)/(2y^2+3)$ is an integer, then any $p$ such that $p|2y^2+3$ would have to divide $x^2-2$, a contradiction. I see this is true for $p=2$. I want to find all $p$ such that $x^2\equiv 2\pmod{p}$, and since for any odd $p$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see $(2|p)=1$ iff $p\equiv 1,7\pmod{8}$. So only primes of the form $8k+1$ or $8k+7$ divide $x^2-2$. However, I don't see a way to show that primes of the form $8k+1$ or $8k+7$ do not divide $2y^2+3$, so maybe I'm completely off the mark. Does anyone know how to resolve this, or have a better idea of what to do? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-24T09:37:35.283" LastActivityDate="2010-10-24T09:37:35.283" Title="Why is $(x^2-2)/(2y^2+3)$ never an integer for any integers $x$ and $y$?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7660" PostTypeId="1" AcceptedAnswerId="7665" CreationDate="2010-10-24T08:57:57.737" Score="5" ViewCount="132" Body="&lt;p&gt;Assume you want to prove an identity such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=m+1}^{n}A(k,m)-B(k,m)=S(m)+T(n,m)\qquad\text{for } n,m\in &#xA;\mathbb{Z},n,m\geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: I applied mathematical induction on $m,n$ to prove it. I am unsure because up to now I have seen it applied to properties depending on a single variable only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: does application of two inductive arguments, one on $m$ and the&#xA;other on $n$, guarantee the validity of such a proof? &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-24T10:16:19.987" LastActivityDate="2010-10-24T12:04:54.287" Title="Induction on two integer variables" Tags="&lt;induction&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7661" PostTypeId="2" ParentId="7655" CreationDate="2010-10-24T09:12:22.727" Score="1" ViewCount="0" Body="&lt;p&gt;Sort of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/proc/1982-086-01/S0002-9939-1982-0663893-8/S0002-9939-1982-0663893-8.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/proc/1982-086-01/S0002-9939-1982-0663893-8/S0002-9939-1982-0663893-8.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-10-24T09:18:06.187" LastActivityDate="2010-10-24T09:18:06.187" CommentCount="1" />
  <row Id="7662" PostTypeId="2" ParentId="7659" CreationDate="2010-10-24T09:31:54.707" Score="3" ViewCount="0" Body="&lt;p&gt;Try to show that $2y^2 + 3$ must have at least one prime divisor which is not of the form $8k+1$ or $8k+7$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T09:31:54.707" CommentCount="1" />
  <row Id="7663" PostTypeId="2" ParentId="7040" CreationDate="2010-10-24T09:45:05.033" Score="0" ViewCount="0" Body="&lt;p&gt;Geometrically it's just the Pythagorean theorem. We may measure the &quot;length&quot; of random variables by standard deviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with a random variable Y. E(Y|X) is the projection of this Y to the set of random variables wich may be expressed as a deterministic function of X. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a hypotenuse Y with squared length Var(Y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first leg is E(Y|X) with squared length Var(E(Y|X)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second leg is Y-E(Y|X) with squared length Var(Y-E(Y|X))=...=E(Var(Y|X)).&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-10-24T09:45:05.033" />
  <row Id="7664" PostTypeId="1" CreationDate="2010-10-24T10:48:45.657" Score="1" ViewCount="171" Body="&lt;p&gt;Do somebody knows anything about the Dirac's identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} \label{Dirac}&#xA;    \frac{\partial^2}{\partial x_{\mu}\partial x^{\mu}} \delta(xb_{\mu}xb^{\mu}) =&#xA;    -4\pi  \delta(xb_0)\delta(xb_1)\delta(xb_2)\delta(xb_3)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$xb$, is the 4-vector $x-b$ in Minkowsky spacetime&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\delta$ is the Dirac delta function&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$x_0 = -x^0, \quad x_1 = x^1, \quad x_2 = x^2, \quad x_3, = x^3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you know where can i find some material about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Willie's &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/jk1/lectures/node19.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i've understood that a solution for the linear wave equation&#xA;$$ \square \psi(\mathbf{r},t) = g(\mathbf{r},t) $$&#xA;for a given $g(\mathbf{r},t)$ is&#xA;$$ \psi = \int \int g(r',t')G(r,r',t,t')dV'dt' $$&#xA;where&#xA;$$ G(r,r',t,t') = AG^+(r,r',t,t') + BG^-(r,r',t,t') , \qquad A + B = 1 $$&#xA;and&#xA;$$ G^{\pm}(r,r',t,t') = \frac{\delta(t' - (t \mp | \mathbf{r} - \mathbf{r'} | / c))}{4\pi | \mathbf{r} - \mathbf{r'} | } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Dirac's follow from the solution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \square \psi(r,t) = \delta(\mathbf{r},t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i'm not sure of the details. Can you Willie help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="2354" LastEditorDisplayName="" LastEditDate="2010-10-27T12:50:40.817" LastActivityDate="2010-10-27T12:50:40.817" Title="Dirac's identity" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7665" PostTypeId="2" ParentId="7660" CreationDate="2010-10-24T10:48:56.790" Score="3" ViewCount="0" Body="&lt;p&gt;Here are some induction principles for two variables:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(0,0)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall x,y. P(x,y) \Rightarrow P(x+1,y)$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x,y) \Rightarrow P(x,y+1)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x,y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(0,0)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall x,y. P(x,0) \Rightarrow P(x+1,0)$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x+1,y) \Rightarrow P(x,y+1)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\forall x,y. P(x,y)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2010-10-24T10:48:56.790" CommentCount="4" />
  <row Id="7666" PostTypeId="1" CreationDate="2010-10-24T11:00:50.780" Score="1" ViewCount="110" Body="&lt;p&gt;Let L^2(R)={f:R-&gt;C| \int_{\infty}^{\infty} \abs(f(t))^2 &amp;lt; \infty } and L^1(R)={f:R-&gt;C| \int_{\infty}^{\infty} \abs(f(t)) &amp;lt; \infty }. Give and example of a function such that f \in L^2(R) and f \notin L^1(R).&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-24T12:07:34.937" LastActivityDate="2010-10-24T12:07:34.937" ClosedDate="2010-10-24T12:33:21.173" Title="Give and example of a function such that f \in L^2(R) and f \notin L^1(R)." Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7667" PostTypeId="2" ParentId="7643" CreationDate="2010-10-24T11:05:22.580" Score="8" ViewCount="0" Body="&lt;p&gt;Wikipedia contains several explicit examples:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor pairing function&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Stern-Brocot_tree&quot; rel=&quot;nofollow&quot;&gt;Stern–Brocot tree&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Calkin-Wilf_tree&quot; rel=&quot;nofollow&quot;&gt;Calkin–Wilf tree&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-24T11:05:22.580" />
  <row Id="7668" PostTypeId="2" ParentId="7666" CreationDate="2010-10-24T11:50:27.353" Score="0" ViewCount="0" Body="&lt;p&gt;Try to look at a function of the form $f(x) = x^{-a}$ for some $a &gt; 0$ on an appropriate domain.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-10-24T12:03:24.737" LastActivityDate="2010-10-24T12:03:24.737" CommentCount="3" />
  <row Id="7669" PostTypeId="2" ParentId="7660" CreationDate="2010-10-24T12:04:54.287" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose you are trying to prove a family of statements $P(x, y)$.  This is the same as proving the family of statements $F(x)$, where $F(x) = \forall y : P(x, y)$.  Each statement $F(x)$ can be proven by induction on $y$ (for fixed $x$), and then you can prove $P(x, y)$ by induction on $x$.  You might want to try proving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n+1 \choose k+1} = {n \choose k+1} + {n \choose k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But actually you can be much trickier than this.  Sometimes it suffices to induct on $x + y$, for example.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T12:04:54.287" CommentCount="1" />
  <row Id="7670" PostTypeId="1" AcceptedAnswerId="7671" CreationDate="2010-10-24T12:14:31.650" Score="3" ViewCount="80" Body="&lt;p&gt;How many solutions does the following equation have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k+l+m+n=30,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k,l,m,n\in\mathbb{Z},0\leq k,l,m,n\leq 10.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: this is an exercise in the &lt;em&gt;inclusion-exclusion principle&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2044" LastEditorUserId="2044" LastEditorDisplayName="" LastEditDate="2010-10-24T13:51:39.303" LastActivityDate="2010-10-24T13:51:39.303" Title="Number of integer solutions in [0,10]" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7671" PostTypeId="2" ParentId="7670" CreationDate="2010-10-24T12:36:35.373" Score="7" ViewCount="0" Body="&lt;p&gt;This is the coefficient of $x^{30}$ in $(1 + x + ... + x^{10})^4$.  (To see this, think about what happens when you choose a term from each factor.)  But this is $\frac{(1 - x^{11})^4}{(1 - x)^4}$.  Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1 - x)^4} = \sum_{n \ge 0} {n+3 \choose 3} x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which you'll recognize as the generating function giving the answer to the problem without the range restriction.)  On the other hand, $(1 - x^{11})^4 = 1 - {4 \choose 1} x^{11} + {4 \choose 2} x^{22} - {4 \choose 3} x^{33} + x^{44}$.  Of these terms only the first three contribute to the coefficient of $x^{30}$, making the final answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${33 \choose 3} - {4 \choose 1} {22 \choose 3} + {4 \choose 2} {11 \choose 3} = 286.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the following inclusion-exclusion argument: start with the set of all solutions with no restriction.  Remove all the solutions where one of the four numbers is greater than $10$.  Add back in all the solutions where two of the four numbers are greater than $10$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T12:36:35.373" CommentCount="3" />
  <row Id="7672" PostTypeId="1" AcceptedAnswerId="7750" CreationDate="2010-10-24T12:48:35.310" Score="8" ViewCount="234" Body="&lt;p&gt;Consider commutative semigroup S and its Grothendieck completion group &#xA;G(S).Suppose I insist on defining G(S) as  free abelian group on  basis $[a]$ (with $a\in S$) divided out by the relations $[a+b]-[a]-[b]$. How do I show &lt;em&gt;with that definition&lt;/em&gt; that if the images of $a,b\in S$ become equal in G(S), then necessarily there existed $c\in S$ with a+c=b+c ? I know it true because I can prove it with other construction of  grothendieck group, but I should like  direct proof with above free abelian group construction.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-06T03:32:18.320" LastActivityDate="2010-11-06T03:32:18.320" Title="How to show directly that two elements become equal in Grothendieck group?" Tags="&lt;group-theory&gt;&lt;algebraic-k-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7673" PostTypeId="2" ParentId="7670" CreationDate="2010-10-24T12:49:10.173" Score="5" ViewCount="0" Body="&lt;p&gt;A geometric way of seeing the answer is this: this is the number lattice points in the cube $[0,10]^4$ that intersects the plane $k+l+m+n = 30$. Note that the corners $(0,10,10,10)$ are points on this plane. Since the weights on $k,l,m,n$ are equal, the solutions are in bijection (it would be clear once you draw a picture) with the lattice points in the $k,l,m$ plane that is inside the tetrahedron formed by the vertices $(0,10,10)$, $(10,0,10)$, $(10,10,0)$ and $(0,0,0)$, the total number of which would be the 11th &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedral_number&quot; rel=&quot;nofollow&quot;&gt;tetrahedral number&lt;/a&gt; which is 286. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-24T13:08:43.517" LastActivityDate="2010-10-24T13:08:43.517" />
  <row Id="7674" PostTypeId="1" AcceptedAnswerId="7679" CreationDate="2010-10-24T12:55:55.613" Score="0" ViewCount="111" Body="&lt;p&gt;I didn't get this method.&#xA;$3a^2-9a+6 = 3(a^2-3a+2) = 3(a^2-1a*3+3^2-3^2+2)=$ $= 3[(a-3)^2-7]$, then?&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-24T17:21:42.417" LastActivityDate="2010-10-24T17:21:42.417" Title="Perfect square method" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="7675" PostTypeId="2" ParentId="7664" CreationDate="2010-10-24T13:00:39.173" Score="1" ViewCount="0" Body="&lt;p&gt;Your identity is in fact the expression for the fundamental solution of the linear wave equation in (1+3) dimensions. This should be in most textbooks on electrodynamics or intro to quantum field theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google also tells me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.math.ucsd.edu/~lindblad/110b/l18.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucsd.edu/~lindblad/110b/l18.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.math.ucla.edu/~ralston/266b.1.08w/Funwave.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~ralston/266b.1.08w/Funwave.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;page 15 of &lt;a href=&quot;http://www.math.princeton.edu/~seri/homepage/courses/Analysis2008.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.princeton.edu/~seri/homepage/courses/Analysis2008.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-24T13:00:39.173" CommentCount="15" />
  <row Id="7676" PostTypeId="1" CreationDate="2010-10-24T13:06:17.733" Score="2" ViewCount="106" Body="&lt;p&gt;Is there a specific way to attack congruences with two variables? The reason I ask is because up until now, I had never encountered such a problem. While browsing tonight, I came across this challenge problem in one of my books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(ab,p)=1$ and that $p&gt;2$. Show that the number of solutions $(x,y)$ of the congruence $ax^2+by^2\equiv 1\pmod{p}$ is $p-\left(\frac{-ab}{p}\right)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After experimenting with it for a few hours, I've barely gotten anywhere. So far, I've only managed to break it into some cases. $(ab,p)=1$ implies $(a,p)=(b,p)=1$ and $\left(\frac{-ab}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$, so if&#xA;$\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)=1$ or $-1$, then $\left(\frac{ab}{p}\right)=1$, so I want to show there are $p-(-1)^{(p-1)/2}$ solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $\left(\frac{a}{p}\right)=1$ and $\left(\frac{b}{p}\right)=-1$, I need to show there are $p+(-1)^{(p-1)/2}$ solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, $a$ and $b$ would both be quadratic residues modulo $p$, and hence $c^2\equiv a\pmod{p}$ and $d^2\equiv b\pmod{p}$ for some $c,d$. Then the original problem would come to $(cx)^2+(dy)^2\equiv 1\pmod{p}$, but I'm not sure a sum of squares is any better. Is there a slick solution to a problem like this?, I'm very puzzled over it.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-24T15:50:44.710" Title="Solving congruences in two variables" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="7677" PostTypeId="2" ParentId="6649" CreationDate="2010-10-24T13:20:30.903" Score="0" ViewCount="0" Body="&lt;p&gt;please lisen this vedio lecture &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=3lS5ZMsfUdQ&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=3lS5ZMsfUdQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will understand the geometry behind the convergence of fourier series &lt;/p&gt;&#xA;" OwnerUserId="2745" LastActivityDate="2010-10-24T13:20:30.903" />
  <row Id="7678" PostTypeId="2" ParentId="7676" CreationDate="2010-10-24T13:35:47.353" Score="5" ViewCount="0" Body="&lt;p&gt;The answer to your first question is somewhat long.  Counting solutions to congruences in two variables is equivalent to counting points on (affine) curves over finite fields, which happens to be quite hard in general and is an active field of study.  Even studying cubic polynomials rapidly leads one to deep waters: elliptic curves, modular forms, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadratic case is much simpler.  When working in projective coordinates the underlying curve has genus $0$, hence either has no points or is isomorphic to $\mathbb{P}^1$ (and has $p+1$ points).  So to count points on an affine curve it suffices to determine if any solutions exist and, if so, to find an explicit map from $\mathbb{P}^1$ and see what points it misses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this might just be gibberish to you, so let me illustrate with the example case $a = b = 1$.  It's clear that $x^2 + y^2 = 1$ has nontrivial solutions such as $(1, 0)$.  The existence of these nontrivial solutions allows one to deduce the parameterization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is valid for any $t$ such that $t^2 \not \equiv -1 \bmod p$.  (The argument is roughly as follows: if $(x, y)$ is any other rational solution, then the line from $(x, y)$ to $(1, 0)$ has rational slope, and conversely given any such line it intersects the curve in at most one other point.)  This parameterization misses the solution $(1, 0)$ which corresponds to sending $t$ to $\infty$, so in total there are $p+1$ solutions if $p \equiv 3 \bmod 4$ (so $t$ can be anything) and $p-1$ solutions if $p \equiv 1 \bmod 4$) (so $t$ cannot be the two nontrivial square roots of $-1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For general $a, b$ you get cases depending on whether $a$ and $b$ are quadratic residues.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T13:35:47.353" CommentCount="4" />
  <row Id="7679" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T13:47:21.723" Score="7" ViewCount="0" Body="&lt;p&gt;Ah, so you want to factor the polynomial. Here's how:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3(a^2-2 \frac{3}{2} a + 2) = 3\left(\left(a-\frac{3}{2}\right)^2 - \frac{9}{4} + 2\right) = 3\left(z^2 - \frac{1}{4}\right)$$&#xA;$$= 3\left(z-\frac{1}{2}\right)\left(z+\frac{1}{2}\right) = 3(a-2)(a-1),$$&#xA;where I temporarily used the substitution $z=a-\frac{3}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-24T13:47:21.723" CommentCount="1" />
  <row Id="7680" PostTypeId="1" CreationDate="2010-10-24T14:00:15.597" Score="0" ViewCount="50" Body="&lt;p&gt;Show that the only regular, self-complementary, planar graph is the 5-cycle. Pay special attention to the 4-regular case.&lt;/p&gt;&#xA;" OwnerUserId="2738" LastEditorUserId="2738" LastEditorDisplayName="" LastEditDate="2010-10-24T15:01:06.620" LastActivityDate="2010-10-24T15:01:06.620" Title="Strange Case in Self-Complementary Planarity" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7681" PostTypeId="2" ParentId="7655" CreationDate="2010-10-24T14:14:41.123" Score="0" ViewCount="0" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the sphere $S^2$ and remove one point, giving you a surface $S$ diffeomorphic to $\mathbb{R}^2$.  So we can equip $S$ with a metric making it isometric to $\mathbb{R}^2$, which is certainly complete.  $\mathbb{R}^2$ has zero curvature but its Euler characteristic is $1$, so&#xA;$$ 0 = \int_{\mathbb{R}^2} K dVol \ne \chi(\mathbb{R}^2) = 1.$$&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-10-24T14:21:11.990" LastActivityDate="2010-10-24T14:21:11.990" CommentCount="4" />
  <row Id="7682" PostTypeId="2" ParentId="7680" CreationDate="2010-10-24T14:48:47.757" Score="1" ViewCount="0" Body="&lt;p&gt;Here's something to get you started. Let G be such a graph, with n vertices. If G is k-regular, then its complement is n-1-k-regular, so to be self-complementary, we must have k = n-1-k, so k = (n-1)/2. Now try finding a reason why the graph is nonplanar if k &gt; 2.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-10-24T14:48:47.757" CommentCount="1" />
  <row Id="7683" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T15:00:02.070" Score="1" ViewCount="0" Body="&lt;p&gt;You can guess and then confirm that $a=1$ is a zero of $3a^{2}-9a+6$. Thus&#xA;you can factor it as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3a^{2}-9a+6=3(a-1)(a-x)=3x-3a-3ax+3a^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solve for $x$. Comparing the coefficients of $a^{2},a,a^{0}$, you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3=3,3x=6,-3-3x=-9$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to $x=2$ and $1+x=3$. Hence $x=2$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3a^{2}-9a+6=3(a-1)(a-2)$.&lt;/p&gt;&#xA;" OwnerUserId="2044" LastActivityDate="2010-10-24T15:00:02.070" />
  <row Id="7684" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T15:10:24.017" Score="7" ViewCount="0" Body="&lt;p&gt;There is also a simple way that involves no fraction calculation,it goes like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 3a^{2}-9a+6 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = 3a^{2}-6a-3a+6 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = a(3a-6)-(3a-6) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = (3a-6)(a-1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  = 3(a-1)(a-2) $&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-10-24T15:10:24.017" CommentCount="3" />
  <row Id="7685" PostTypeId="2" ParentId="7676" CreationDate="2010-10-24T15:36:21.123" Score="3" ViewCount="0" Body="&lt;p&gt;What you have is a quadratic form, which makes the problem significantly easier than the general case, and can be split into the following steps.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find at least one solution.&lt;/li&gt;&#xA;&lt;li&gt;Given one solution, use this to parameterize all solutions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is roughly what Qiaochu did in his answer and this method is quite general, applying to quadratic forms in any number of variables and for finding solutions mod p, in the rationals or, in fact, in any field. I also used this method recently in answer to a question about &lt;a href=&quot;http://math.stackexchange.com/questions/7143/is-it-possible-to-get-arbitrarily-near-any-acute-angle-with-pythagorean-triangles/7237#7237&quot;&gt;Pythagorean triples&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me start with (2) first, so we suppose that you already have one solution (x&lt;sub&gt;0&lt;/sub&gt;,y&lt;sub&gt;0&lt;/sub&gt;). The idea is to find other solutions of the form (x,y)&amp;nbsp;=&amp;nbsp;(x&lt;sub&gt;0&lt;/sub&gt;,y&lt;sub&gt;0&lt;/sub&gt;)&amp;nbsp;+&amp;nbsp;(u,v)t. Where you choose nonzero (u,v)&amp;nbsp;&amp;isin;&amp;nbsp;(&amp;#x02124;/(p))&lt;sup&gt;2&lt;/sup&gt; and then solve for t. This will give a quadratic in t (as long as the t&lt;sup&gt;2&lt;/sup&gt; term doesn't cancel) with t&amp;nbsp;=&amp;nbsp;0 as one solution. Extracting out a factor of t gives a linear form with a unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;ax^2+by^2-1=(au^2+bv^2)t^2+2(ux_0+vy_0)t=0\ \Rightarrow\ t=-2\frac{ux_0+vy_0}{au^2+bv^2}\ (*)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works as long as au&lt;sup&gt;2&lt;/sup&gt;+bv&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;&amp;ne;&amp;nbsp;0, and gives us a parameterization of the solutions to the quadratic form in terms of nonzero (u,v)&amp;nbsp;&amp;isin;&amp;nbsp;(&amp;#x02124;/(p))&lt;sup&gt;2&lt;/sup&gt;. The solution we get does not change if (u,v) are rescaled (the projective space &amp;#x2119;&lt;sup&gt;1&lt;/sup&gt; mentioned by Qiaochu is precisely the set of nonzero (u,v) up to rescaling). Rescaling by 1/u, we only look at solutions corresponding to (u,v)&amp;nbsp;=&amp;nbsp;(1,v) and (in case u&amp;nbsp;=&amp;nbsp;0), (u,v)&amp;nbsp;=&amp;nbsp;(0,1). That is p&amp;nbsp;+&amp;nbsp;1 possibilities. The parameterization in terms of (1,v) only fails when a&amp;nbsp;+bv&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;0 so that (bv)&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;-ab. This can only happen if -ab is a quadratic residue, and occurs for bv one of the 2 square roots of -ab. This gives a total number of solutions to the quadratic form as p&amp;nbsp;+&amp;nbsp;1 if -ab is not a quadratic residue and (p&amp;nbsp;+&amp;nbsp;1)&amp;nbsp;-&amp;nbsp;2 if it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at step (1) now. You could try finding a solution by looking at cases depending on whether a,b, ab are quadratic residues, etc. However, there is a quick counting method. For any right hand side c, ax&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;by&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;c can have at most p&amp;nbsp;+&amp;nbsp;1 solutions in x,y (mod p) -- using the above argument. As there are p&lt;sup&gt;2&lt;/sup&gt; possibilities for (x,y), this gives at least p&lt;sup&gt;2&lt;/sup&gt;/(p&amp;nbsp;+&amp;nbsp;1) values of c which occur. As p&lt;sup&gt;2&lt;/sup&gt;/(p&amp;nbsp;+&amp;nbsp;1)&amp;nbsp;&amp;gt;&amp;nbsp;p&amp;nbsp;&amp;minus;&amp;nbsp;1, this shows that you can solve ax&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;by&lt;sup&gt;2&lt;/sup&gt;&amp;nbsp;=&amp;nbsp;c (mod p) for &lt;em&gt;any&lt;/em&gt; value of c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to see that this whole argument carries through almost word-for-word to &lt;em&gt;any&lt;/em&gt; finite field of characteristic &amp;ne;&amp;nbsp;2 (characteristic 2 is a bit different because (*) gives t&amp;nbsp;=&amp;nbsp;0).&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-10-24T15:50:44.710" LastActivityDate="2010-10-24T15:50:44.710" />
  <row Id="7686" PostTypeId="2" ParentId="7586" CreationDate="2010-10-24T15:37:11.300" Score="1" ViewCount="0" Body="&lt;p&gt;You can also use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula&quot; rel=&quot;nofollow&quot;&gt;quadratic formula&lt;/a&gt; to compute the result elegantly, it goes like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^2 - 6z + 25 = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ z = \frac { 6 \pm \sqrt{36 -100} } {2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = \frac{ 6 \pm 8i}{2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=  3 \pm 4i $&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-24T15:51:50.350" LastActivityDate="2010-10-24T15:51:50.350" />
  <row Id="7687" PostTypeId="1" AcceptedAnswerId="7690" CreationDate="2010-10-24T16:02:23.757" Score="-1" ViewCount="149" Body="&lt;p&gt;The shortest median of a right angled triangle ABC is 10 units. If the area of ABC is 96 sq units, what is the length of the longest median&lt;/p&gt;&#xA;" OwnerUserId="2748" LastEditorUserId="2748" LastEditorDisplayName="" LastEditDate="2010-10-25T17:09:21.603" LastActivityDate="2010-10-25T17:09:21.603" Title="how to find length of median in right angled triangle" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7688" PostTypeId="1" AcceptedAnswerId="7722" CreationDate="2010-10-24T16:05:32.087" Score="3" ViewCount="98" Body="&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; The volume of a parallelepiped in $\mathbb{R}^n$ with n sides given by the vectors $(x_{1_1}, x_{1_2} ... x_{1_n}), (x_{2_1}, x_{2_2} ... x_{2_n}) ... (x_{n_1}, x_{n_2} ... x_{n_n})$ and diagonal $(x_{1_1} + x_{2_1}+ \cdots + x_{n_1}, x_{1_2} + x_{2_2} + \cdots + x_{n_2}, ..., x_{1_n} + x_{2_n} + \cdots + x_{n_n})$  Is the determinant of the matrix $A = (x_{j, i})$ where i and J are the double sub-scripts on each $x_{i_j}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case where n= 2 has a nice geometric proof. Shown in this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ybk6O.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u= (a,c)^T$ and $v= (b, d)^T$ then the area of the parallelogram is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ det \left| [ u | v ] \right| = | ad - bc| $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a &quot;nice&quot; geometric proof for n=3?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind the n=2 proof is you take the area of the rectangle and subtract the triangles. So, I'm thinking: take the volume of the rectangular solid and subtract tetrahedrons. How many will I need to subtract? Is it possible to draw this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Suppose that (a,b) and (b, a) are points on the unit circle. Find parallelogram with vertices (0,0), (a,b) (a+b, a+b), (b,a) is where $\theta$ is between the x-axis and (a,b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the answer is $|2cos^2 \theta -1|$. I want to get a sense of how difficult this question is. Could an undergrad taking linear for the first time do this? Is it easier if one is asked to prove that the area is $|2cos^2 \theta -1|$ rather than finding it from scratch? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graphic is from: &#xA;&lt;a href=&quot;http://algebra.math.ust.hk/determinant/01_geometry/lecture1.shtml&quot; rel=&quot;nofollow&quot;&gt;http://algebra.math.ust.hk/determinant/01_geometry/lecture1.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-24T21:21:04.080" LastActivityDate="2010-10-24T21:21:04.080" Title="Nice geometric parallelepiped proof?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;analytic-geometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="7689" PostTypeId="2" ParentId="734" CreationDate="2010-10-24T16:24:38.453" Score="1" ViewCount="0" Body="&lt;p&gt;The Rochester Institute of Technology offers a MS in applied statistics that is completely online (I think, I'm not enrolled yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rit.edu/emcs/ptgrad/_program_detail.php?id=441&quot; rel=&quot;nofollow&quot;&gt;http://www.rit.edu/emcs/ptgrad/_program_detail.php?id=441&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty expensive, since it combines the increased out-of-state tuition with the extra online cost. However an online degree such as this is good for working professionals who &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Can't take classes during the day and&#xA;2) Can get their employer to pay for most of the degree.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-10-24T16:24:38.453" CommunityOwnedDate="2010-10-24T16:24:38.453" />
  <row Id="7690" PostTypeId="2" ParentId="7687" CreationDate="2010-10-24T16:37:06.350" Score="2" ViewCount="0" Body="&lt;p&gt;Let your triangle have sides a, b, and c, with c the hypotenuse and a&gt;b.  The shortest median must be from the right angle to side c (Why?).  If you use this median and drop lines from the midpoint of the hypotenuse to the midpoints of sides a and b you will form four congruent triangles.  So c=20 as the hypotenuse of these four is 10.  Now you have ab=192 and a^2+b^2=400.  This solves as a=16, b=12.  The longest median is the hypotenuse of a 6,16 right triangle=sqrt(36+256)=sqrt(292)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-24T19:00:03.450" LastActivityDate="2010-10-24T19:00:03.450" CommentCount="1" />
  <row Id="7691" PostTypeId="2" ParentId="7674" CreationDate="2010-10-24T16:40:35.117" Score="2" ViewCount="0" Body="&lt;p&gt;If you desire to factor it by way of completing the square then it is simpler to first multiply by 4, namely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\ \ 4\:(a^2-3\: a+2)\ =\ (2\:a-3)^2 - 1\ =\ (2\:a-2)\:(2\:a-4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, divide both sides through by $\rm\ \: 4\ =\ 2\cdot 2$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-24T16:40:35.117" />
  <row Id="7692" PostTypeId="2" ParentId="7688" CreationDate="2010-10-24T16:43:15.617" Score="1" ViewCount="0" Body="&lt;p&gt;For question 2, I think it is easy to find the area if you know a bit of trig, not using the linear algebra.  But maybe the students are too limited by the class at hand to use another approach.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T16:43:15.617" CommentCount="2" />
  <row Id="7693" PostTypeId="1" AcceptedAnswerId="7702" CreationDate="2010-10-24T16:53:33.320" Score="3" ViewCount="82" Body="&lt;p&gt;I'm trying to work out a basic example where formal smoothness should fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering $\mathbb{R} \to \mathbb{R}[x,y]/(x^2-y^2)$.&lt;br&gt;&#xA;The idea is that not every $\mathbb{R}$-homomorphism $\mathbb{R}[x,y]/(x^2-y^2) \to \mathbb{R}$ should lift to a homomorphism $\mathbb{R}[x,y]/(x^2-y^2) \to \mathbb{R}[\varepsilon]/(\varepsilon^2)$. But I can't see that ever being possible: after all, I can just take the exact same homomorphism, with image $\mathbb{R} \subset \mathbb{R}[\varepsilon]/(\varepsilon^2)$; this gives a valid lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to consider something else than $R = \mathbb{R}[\varepsilon]/(\varepsilon^2)$ with $I = (\varepsilon)$ in order to witness the failure of formal smoothness? I would think that was enough, given that it should fully explain lifting of points to tangent vectors.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2010-10-24T17:39:41.923" Title="Have I got the right definition of formal smoothness?" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" />
  <row Id="7694" PostTypeId="2" ParentId="7621" CreationDate="2010-10-24T16:57:10.303" Score="4" ViewCount="0" Body="&lt;p&gt;This might be &lt;em&gt;slightly&lt;/em&gt; off tangent, but it's definitely in the spirit of your post, and is found at the beginning of most texts in sieve theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When your fixed set of primes is the first $k$ primes and $n = p_{k}^2$ then $B_{n}$ is the main term of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_sieve&quot; rel=&quot;nofollow&quot;&gt;sieve of Eratosthenes-Legendre&lt;/a&gt;, which gives an approximation for the number of primes in the interval $[p_{k}, p_{k}^2]$. In this setting &lt;a href=&quot;http://en.wikipedia.org/wiki/Mertens%27_theorems&quot; rel=&quot;nofollow&quot;&gt;Merten's theorem&lt;/a&gt; gives an asymptotic for $B_{n}$. (It's called Merten's Third theorem in the Wikipedia page).&#xA;However, an appeal to the prime number theorem shows that this is a reasonably bad approximation, and that $B_{n}$ overestimates the number of primes by a factor of $2e^{-\gamma}$ where $\gamma$ is the Euler-Mascheroni constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terry Tao has a good &lt;a href=&quot;http://terrytao.wordpress.com/2007/06/05/open-question-the-parity-problem-in-sieve-theory/&quot; rel=&quot;nofollow&quot;&gt;blog posting&lt;/a&gt; on why repeated use of the inclusion-exclusion principle yields non-optimal results.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="592" LastEditorDisplayName="" LastEditDate="2010-10-24T17:06:15.017" LastActivityDate="2010-10-24T17:06:15.017" CommentCount="1" />
  <row Id="7695" PostTypeId="1" AcceptedAnswerId="7700" CreationDate="2010-10-24T16:57:13.780" Score="9" ViewCount="415" Body="&lt;p&gt;I would like to find the apothem of a regular pentagon. It follows from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can this be proved (geometrically or trigonometrically)? &lt;/p&gt;&#xA;" OwnerUserId="2044" LastEditorUserId="2044" LastEditorDisplayName="" LastEditDate="2010-10-25T12:38:31.337" LastActivityDate="2010-10-25T12:38:31.337" Title="How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="7696" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T17:00:15.313" Score="7" ViewCount="0" Body="&lt;p&gt;Look up the &quot;construction of a regular pentagon&quot; using the straight edge and compass. If you keep track each step in this construction you will find that the angle 72 degrees comes in a few places, and this expression follows from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a fun excercise-- you should do it.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-24T17:05:45.467" LastActivityDate="2010-10-24T17:05:45.467" CommentCount="3" />
  <row Id="7697" PostTypeId="1" AcceptedAnswerId="7703" CreationDate="2010-10-24T17:04:35.490" Score="3" ViewCount="118" Body="&lt;p&gt;This question is taken from a contest in India. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prove that we can construct an infinite set of positive integers of the form $2^{n}-3$, where $n \in \mathbb{N}$, such that any two numbers from the set are relatively prime&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to have an answer for this question, and i would also like to know why $2^{n}-3$ has importance here. Can this question be generalized. That is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we construct an infinite set of positive integers of the form $k^{n} -(k+1)$ such that any two numbers from the set are relatively prime?.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-24T20:54:35.507" Title="Construction of an infinite set such that any two number from the set are relatively prime" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7698" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T17:06:14.283" Score="2" ViewCount="0" Body="&lt;p&gt;Or you can go to &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometryAnglesPi5.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt; for $\pi/5$ and use the multiple angle formula&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T17:06:14.283" />
  <row Id="7699" PostTypeId="1" AcceptedAnswerId="7701" CreationDate="2010-10-24T17:08:06.213" Score="0" ViewCount="61" Body="&lt;p&gt;Hello! I have &lt;strong&gt;differential calculus&lt;/strong&gt; problem where I must find the coordinates of a point C on a parabola. There is a point given which is D(8,4) and D is on a line that is intersecting with the parabola at C. Also, the line where there is the point D is perpendicular to the tangent of C.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the equation of the parabola:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = -x^2 + 4x + 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is coordinates of the single point given:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D (8,4) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What are the coordinates of C?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am having trouble with the fact that there is only point given for the intersecting line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for helping out!&lt;/p&gt;&#xA;" OwnerUserId="2751" LastEditorUserId="2751" LastEditorDisplayName="" LastEditDate="2010-10-24T17:24:14.060" LastActivityDate="2010-10-24T17:24:14.060" Title="Finding the coordinates of point on a parabola with a single point on an intersecting line" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7700" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T17:12:18.737" Score="15" ViewCount="0" Body="&lt;p&gt;Since $x := \cos \frac{2 \pi}{5} = \frac{z + z^{-1}}{2}$ where $z:=e^{\frac{2 i \pi}{5}}$, and $1+z+z^2+z^3+z^4=0$ (for $z^5=1$ and $z \neq 1$), $x^2+\frac{x}{2}-\frac{1}{4}=0$, and voilà.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-24T17:45:57.673" LastActivityDate="2010-10-24T17:45:57.673" CommentCount="5" />
  <row Id="7701" PostTypeId="2" ParentId="7699" CreationDate="2010-10-24T17:20:55.863" Score="1" ViewCount="0" Body="&lt;p&gt;Using your differential calculus, you should be able to write the equation for the line tangent to the parabola at a given point on the parabola.  So if $(x,y)$ satisfies $y=-x^2+4x+2$ what is the equation of the tangent line?  Then you should be able to find the equation for the line perpendicular to the tangent at the point on the parabola.  One of these perpendiculars will pass through $(8,4)$.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T17:20:55.863" CommentCount="8" />
  <row Id="7702" PostTypeId="2" ParentId="7693" CreationDate="2010-10-24T17:39:41.923" Score="3" ViewCount="0" Body="&lt;p&gt;Indeed, you need to consider a different artinian ring to witness the failure of formal smoothness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look instead at $\mathbb{R}[\epsilon]/\epsilon^3 \to \mathbb{R}[\epsilon]/\epsilon^2$ and map $\mathbb{R}[x,y]/(x^2-y^2)$ to  $\mathbb{R}[\epsilon]/\epsilon^2$ by $x \mapsto \epsilon$, $y \mapsto 0$. If you wanted to lift this to a map to $\mathbb{R}[\epsilon]/\epsilon^3$, then you would have $x \mapsto \epsilon + a \epsilon^2$ and $y \to b \epsilon^2$, but then $x^2 = \epsilon^2$ and $y^2=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;geometric&quot; intuition here is that the Zariski tangent space to $x^2=y^2$ at $0$ is two dimensional but those tangent vectors which have slope other than $\pm 1$ can not be extended to higher order jets.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-24T17:39:41.923" CommentCount="1" />
  <row Id="7703" PostTypeId="2" ParentId="7697" CreationDate="2010-10-24T17:49:36.740" Score="5" ViewCount="0" Body="&lt;p&gt;Given a set of odd primes $\displaystyle S = \{p_1, p_2, \dots, p_m \}$, we can find an $n$ such that $\displaystyle 2^n - 3 = 1 \mod p_i$, by choosing $n = 2 + \prod (p_i-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus starting with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{2^2 -3,2^3 - 3\}$ we can add a new member which is relatively prime to all the previous members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe a similar argument will hold for other $k$ (when considering $k^n -(k+1)$), since we will have $gcd(p_i,k) =1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-24T17:49:36.740" CommentCount="0" />
  <row Id="7704" PostTypeId="2" ParentId="7688" CreationDate="2010-10-24T17:54:12.657" Score="1" ViewCount="0" Body="&lt;p&gt;Question 2: Since $a=\cos\theta$ and $b=\sin\theta$, it follows from the determinant formula for the area:&#xA;$$ A = \left| \det\begin{pmatrix} a &amp;amp; b \\ b &amp;amp; a \end{pmatrix} \right| = | \cos^2\theta - \sin^2 \theta |.$$&#xA;So I would say this is an easy problem for a linear algebra student.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-24T17:54:12.657" CommentCount="2" />
  <row Id="7705" PostTypeId="2" ParentId="7634" CreationDate="2010-10-24T17:55:29.573" Score="3" ViewCount="0" Body="&lt;p&gt;The problem with trees of finite width at all lattice points is called &quot;Polya's Orchard Problem&quot;.  Other variants such as higher dimensions and different shapes of the orchard are in the literature under the same name, or variations on &lt;em&gt;visibility&lt;/em&gt;, &lt;em&gt;lattice point&lt;/em&gt;, and &lt;em&gt;orchard&lt;/em&gt;.  Sample search result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rose-hulman.edu/mathjournal/archives/2006/vol7-n2/paper9/v7n2-9pd.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.rose-hulman.edu/mathjournal/archives/2006/vol7-n2/paper9/v7n2-9pd.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem with zero-width trees and visibility from the origin (ie., primitive lattice points) I think the $6/\pi^2$ density is attributed to Chebyshev in the 1800's, but it seems like the type of result that could have been known considerably earlier and frequently rediscovered. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-24T17:55:29.573" />
  <row Id="7706" PostTypeId="1" AcceptedAnswerId="7707" CreationDate="2010-10-24T18:09:23.027" Score="0" ViewCount="63" Body="&lt;p&gt;How can you prove that $\Bigl\lfloor{x + \dfrac{1}{2}}\Bigr\rfloor$ is a $\theta(x)$ function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just practicing questions that could come up on an exam and this one was giving me a tough time trying to formally prove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)$ is $\theta(g(x)) \leftrightarrow f(x)$ is $O(g(x))$ and $f(x)$ is $\Omega(g(x))$&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-24T18:12:38.123" LastActivityDate="2010-10-24T18:36:34.350" Title="Proving theta complexity for floor function" Tags="&lt;functions&gt;&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="7707" PostTypeId="2" ParentId="7706" CreationDate="2010-10-24T18:32:53.533" Score="2" ViewCount="0" Body="&lt;p&gt;You can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x - \frac{1}{2} \leq \Bigl\lfloor x+\frac{1}{2}\Bigr\rfloor \leq x+\frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this give you the result?&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-24T18:36:34.350" LastActivityDate="2010-10-24T18:36:34.350" CommentCount="1" />
  <row Id="7708" PostTypeId="1" CreationDate="2010-10-24T18:39:06.733" Score="2" ViewCount="71" Body="&lt;p&gt;If $q$ and $r$ are quaternions, and $p$ is a point, applying $q$ then $r$ to $p$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(qr)p\dfrac{1}{qr}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I want to go the other way? Instead of concatenating rotations, I want to remove them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have $q$ and $m$, and I know that $qx =m$, but I don't know what $x$ is. How can I discover $x$? (How can I do quaternion divison?)&lt;/p&gt;&#xA;" OwnerUserId="2755" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-24T19:50:58.193" LastActivityDate="2010-10-25T00:35:03.980" Title="Quaternion Division" Tags="&lt;quaternions&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="7709" PostTypeId="2" ParentId="7708" CreationDate="2010-10-24T18:43:33.770" Score="0" ViewCount="0" Body="&lt;p&gt;This is described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; under the section Conjugation, the norm, and reciprocal.  You just need the reciprocal of q.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-24T18:43:33.770" />
  <row Id="7710" PostTypeId="2" ParentId="7708" CreationDate="2010-10-24T18:43:52.100" Score="5" ViewCount="0" Body="&lt;p&gt;In algebra, multiplication is denoted by juxtaposition, not by an asterisk.&#xA;The inverse of the operation $p\mapsto qpq^{-1}$ is $p\mapsto q^{-1}pq$.&#xA;The solution of $qx=m$ is $x=q^{-1}m$. Note that&#xA;$$(a+bi+cj+dk)^{-1}=(a^2+b^2+c^2+d^2)^{-1}(a-bi-cj-dk).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this can be found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-24T18:43:52.100" />
  <row Id="7711" PostTypeId="1" AcceptedAnswerId="7725" CreationDate="2010-10-24T19:20:02.490" Score="1" ViewCount="93" Body="&lt;p&gt;Suppose we have two sets of discrete events, $A$ and $B$. Then I think it is true that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sum_{i \in A, j \in B}\Pr[i\ \textrm{AND}\ j] \leq \sum_{i \in A}\Pr[i]+ \sum_{j \in B}\Pr[j] +\sum_{i, j \in A}\Pr[i\ \textrm{AND}\ j] + \sum_{i, j \in B}\Pr[i\ \textrm{AND}\ j]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition for why this should be true is just by analogy to the simple inequality for $a, b \in \mathbb{R}:$&#xA;$$2ab \leq a + b + a(a-1) + b(b-1)$$&#xA;To see why this identity is true, note:&#xA;$$a + b + a(a-1) + b(b-1) = a^2 + b^2 = 2ab + (a-b)^2 \geq 2ab$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a proof for the more complex identity though. Is it true? Why does it hold?&lt;/p&gt;&#xA;" OwnerUserId="2756" LastEditorUserId="2756" LastEditorDisplayName="" LastEditDate="2010-10-24T19:41:33.487" LastActivityDate="2010-10-24T20:57:56.667" Title="Why does this inequality in probability theory hold?" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7712" PostTypeId="1" AcceptedAnswerId="7714" CreationDate="2010-10-24T19:28:58.043" Score="0" ViewCount="47" Body="&lt;p&gt;Hi, if A = { Market research predicates strong demand } and B = { Market demand is strong }, can we reasonably assume that P(A or B) = P(A) * P(B)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I know &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;P(B|A) = 0.8 &lt;/li&gt;&#xA;&lt;li&gt;P(not B | not A) = 0.9, &lt;/li&gt;&#xA;&lt;li&gt;P(B) = 0.2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I need to calculate P(A) and P(not A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, it seems that if P(A or B) != P(A) * P(B), it's impossible to know the answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Actually, I'm drawing a decision tree to determine whether a market research is worth trying.  All the information is listed above.  Without P(A) and P(not A), I can not determine calculate the expected value of the branch of taking the research and can not draw the tree.&lt;/p&gt;&#xA;" OwnerUserId="1618" LastEditorUserId="1618" LastEditorDisplayName="" LastEditDate="2010-10-24T19:50:15.680" LastActivityDate="2010-10-24T19:50:15.680" Title="A problem of probability." Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7713" PostTypeId="1" CreationDate="2010-10-24T19:34:22.303" Score="0" ViewCount="100" Body="&lt;p&gt;Let f(t)=H(t)-H(t-2) and \psi(t)=t*exp(-abs(t)). Compute the wavelet transform of f. Analyse the behavior of the transform when the scaling parameter a approaches zero. H(t) = piecewise( t&gt;0, 1, t&amp;lt;0, 0), W_f(a,b)== \int_{-\infty}^{\infty} f(t)*\psi_{a,b}(t) dt and \psi_{a,b}(t)=a^{-1/2} \psi((t-b)/a)&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-25T09:06:50.430" LastActivityDate="2010-10-25T09:06:50.430" ClosedDate="2010-10-25T01:57:50.297" Title="Let f(t)=H(t)-H(t-2) and \psi(t)=t*exp(-abs(t)). Compute the wavelet transform W_f(a,b) of f. " Tags="&lt;wavelet-transform&gt;" CommentCount="2" />
  <row Id="7714" PostTypeId="2" ParentId="7712" CreationDate="2010-10-24T19:41:20.303" Score="1" ViewCount="0" Body="&lt;p&gt;Your first question is not a mathematical question, but I'd&#xA;say it was reasonable to assume the events are not independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing a Venn diagram of events gives four regions, but&#xA;you only have three equations for the probabilities of each, so&#xA;not enough information to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;With your edit, you introduce a new piece of information.&#xA;I would assign probabilities to the four regions in the Venn&#xA;diagram, say $p_1=P(A \textrm{ and }B)$,&#xA;$p_2=P(A \textrm{ and not }B)$,&#xA;$p_3=P(\textrm{not }A \textrm{ and }B)$ and&#xA;$p_4=P(\textrm{not }A \textrm{ and not }B)$.&#xA;Then you know $p_1+p_2+p_3+p_4=1$, $p_1=0.8(p_1+p_2)$,&#xA;$p_4=0.9(p_3+p_4)$ and $p_1+p_3=0.2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four linear equations in four unknowns!&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-24T19:41:20.303" CommentCount="1" />
  <row Id="7715" PostTypeId="2" ParentId="7527" CreationDate="2010-10-24T20:06:17.533" Score="3" ViewCount="0" Body="&lt;p&gt;For what it's worth, such a sequence constructed by a greedy algorithm begins:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2, 4, 8, 10, 16, 18, 36, 199, 208, 1131, 1347, 3984, 5751, 7310, 27315, 129313, 134101, 169400, 589570&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we start with $A_1 = 2$, and then each $A_n$ is the smallest number greater than $A_{n-1}$ with $A_i + A_n$ having an even number of distinct prime factors for each $n \le 1$. (So, for example, $A_3 = 8$ because $5+4, 6+2, 7+2$ each have an odd number of distinct prime factors, but $8+2$ and $8+4$ both have even numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another such sequence, starting with 1, begins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 5, 9, 13, 35, 39, 286, 290, 381, 385, 866, 4376&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one starting with 3 begins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3, 7, 11, 15, 33, 41, 47, 65, 101, 203, 4102, 6392, 8507, 18608.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these it &lt;i&gt;seems&lt;/i&gt; that these sequences grow roughly exponentially; that is, $A_n \approx k^n$ for some constant $k$. This makes sense. Since approximately half of all integers have an even number of distinct prime factors, once you have $n$ numbers in such a sequence it should take about $2^n$ tries to find the next one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this isn't a proof, but it's at least an argument why such sequences should exist. And judging from the irregularity of the greedily constructed sequences, a greedy method probably isn't the best way to go here even if you want to explicitly construct the sequence.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-24T20:06:17.533" CommentCount="1" />
  <row Id="7716" PostTypeId="1" AcceptedAnswerId="7718" CreationDate="2010-10-24T20:12:57.467" Score="4" ViewCount="136" Body="&lt;p&gt;I was just wondering where the y'/(dy/dx) in implicit differentiation comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^2 + y^2 = 25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d/dx) x^2 + (d/dy) y^2 &lt;strong&gt;(dy/dx)&lt;/strong&gt; = 25 (d/dx)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x + 2y (dy/dx) = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(dy/dx) = -x/y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the bold part come from? Wikipedia says it's a byproduct of the chain rule, but it's just not clicking for me.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2010-10-25T04:32:46.083" Title="Implicit Differentiation" Tags="&lt;calculus&gt;&lt;implicit&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7717" PostTypeId="1" CreationDate="2010-10-24T20:15:21.210" Score="0" ViewCount="315" Body="&lt;p&gt;I've heard all these terms thrown about in proofs and in geometry, but what are the differences and relationships between them? Examples would be awesome! :)&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-25T00:19:54.450" LastActivityDate="2010-10-25T02:56:15.230" Title="Difference between axioms, theorems, postulates, corollaries, and hypothesis" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7718" PostTypeId="2" ParentId="7716" CreationDate="2010-10-24T20:26:06.123" Score="5" ViewCount="0" Body="&lt;p&gt;When you implicitly differentiate $x^2+y^2=25$, you are differentiating with respect to a particular variable—in this case, $x$, so:&#xA;$$\begin{align}&#xA;\frac{d}{dx}(x^2+y^2)&amp;amp;=\frac{d}{dx}25&#xA;\\&#xA;\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)&amp;amp;=0&#xA;\\&#xA;2x+2y\frac{dy}{dx}&amp;amp;=0&#xA;\\&#xA;2y\frac{dy}{dx}&amp;amp;=-2x&#xA;\\&#xA;\frac{dy}{dx}&amp;amp;=-\frac{x}{y}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the 3rd line to the 4th line, $\frac{d}{dx}(y^2)$ is the derivative with respect to $x$ of $y^2$, in which (as in Ryan Budney's comment) we assume that $y$ is some function of $x$, so we apply the chain rule, differentiating $y^2$ with respect to $y$ and multiplying by the derivative of $y$ with respect to $x$ to get $2y\frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: Based on the comments below, I think it might be useful if I introduced a slightly different notation: Let $D_x$ be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_operator&quot; rel=&quot;nofollow&quot;&gt;differential operator&lt;/a&gt; with respect to $x$, which you have previously written as $\frac{d}{dx}$ (and, similarly, $D_y$ is the differential operator with respect to $y$).  When we apply the differential operator to something, we read and write it like a function: $D_x(x^2)=2x$ is &quot;the derivative with respect to $x$ of $x^2$ is $2x$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rewriting the work above in this notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;D_x(x^2+y^2)&amp;amp;=D_x(25)&#xA;\\&#xA;D_x(x^2)+D_x(y^2)&amp;amp;=0&#xA;\\&#xA;2x+D_y(y^2)D_x(y)&amp;amp;=0&#xA;\\&#xA;2x+2yD_x(y)&amp;amp;=0&#xA;\\&#xA;2yD_x(y)&amp;amp;=-2x&#xA;\\&#xA;D_x(y)=\frac{dy}{dx}&amp;amp;=-\frac{x}{y}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, to your question of finding $\frac{dx}{dy}$:&#xA;$$\begin{align}&#xA;D_y(x^2+y^2)&amp;amp;=D_y(25)&#xA;\\&#xA;D_y(x^2)+D_y(y^2)&amp;amp;=0&#xA;\\&#xA;D_x(x^2)D_y(x)+2y&amp;amp;=0&#xA;\\&#xA;2xD_y(x)+2y&amp;amp;=0&#xA;\\&#xA;2xD_y(x)&amp;amp;=-2y&#xA;\\&#xA;D_y(x)=\frac{dx}{dy}&amp;amp;=-\frac{y}{x}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-25T04:32:46.083" LastActivityDate="2010-10-25T04:32:46.083" CommentCount="9" />
  <row Id="7719" PostTypeId="2" ParentId="7717" CreationDate="2010-10-24T20:27:58.157" Score="5" ViewCount="0" Body="&lt;p&gt;In Geometry, &quot;Axiom&quot; and &quot;Postulate&quot; are essentially interchangeable. In antiquity, they referred to propositions that were &quot;obviously true&quot; and only had to be stated, and not proven. In modern mathematics there is no longer an assumption that axioms are &quot;obviously true&quot;. Axioms are merely 'background' assumptions we make. The best analogy I know is that axioms are the &quot;rules of the game&quot;. In Euclid's Geometry, the main axioms/postulates are: (I) Given any two distinct points, there is a line that contains them. (II) Any line segment can be extended to an infinite line. (III) Given a point and a radius, there is a circle with center in that point and that radius. (IV) All right angles are equal to one another. (V) If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. (The &lt;em&gt;parallel postulate&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theorem is a logical consequence of the axioms. In Geometry, the &quot;propositions&quot; are all theorems: they are derived using the axioms and the valid rules. A &quot;Corollary&quot; is a theorem that is usually considered an &quot;easy consequence&quot; of another theorem. What is or is not a corollary is entirely subjective. Sometimes what an author thinks is a 'corollary' is deemed more important than the corresponding theorem. (The same goes for &quot;Lemma&quot;s, which are theorems that are considered auxiliary to proving some other, more important in the view of the author, theorem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;hypothesis&quot; is an assumption made. For example, &quot;If $x$ is an even integer, then $x^2$ is an even integer&quot; I am not asserting that $x^2$ is even or odd; I am asserting that if &lt;em&gt;something&lt;/em&gt; happens (namely, if $x$ happens to be an even integer) then &lt;em&gt;something else&lt;/em&gt; will also happen. Here, &quot;$x$ is an even integer&quot; is the hypothesis being made to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the Wikipedia pages on &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom&quot; rel=&quot;nofollow&quot;&gt;axiom&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorem&quot; rel=&quot;nofollow&quot;&gt;theorem&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Corollary&quot; rel=&quot;nofollow&quot;&gt;corollary&lt;/a&gt;. The first two have many examples.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T00:33:48.633" LastActivityDate="2010-10-25T00:33:48.633" CommentCount="2" />
  <row Id="7720" PostTypeId="1" AcceptedAnswerId="7721" CreationDate="2010-10-24T20:29:47.700" Score="2" ViewCount="74" Body="&lt;p&gt;I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example? &lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2010-10-24T20:35:27.730" Title="Finite group with isomorphic normal subgroups and non-isomorphic quotients?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7721" PostTypeId="2" ParentId="7720" CreationDate="2010-10-24T20:35:27.730" Score="7" ViewCount="0" Body="&lt;p&gt;Take $G = Z_4 \times Z_2$, $H$ generated by $(0,1)$, $K$ generated by $(2,0)$.  Then $H \cong K \cong Z_2$ but $G/H \cong Z_4$ while $G/K \cong Z_2 \times Z_2$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-24T20:35:27.730" CommentCount="2" />
  <row Id="7722" PostTypeId="2" ParentId="7688" CreationDate="2010-10-24T20:53:48.067" Score="1" ViewCount="0" Body="&lt;p&gt;It is a fact of linear algebra that any nonsingular matrix $A$ is a product of &quot;elementary matrices&quot;. An elementary matrix $E$ is either (a) an identity matrix with one diagonal entry replaced by an arbitrary $a\ne 0$ or (b) an identity matrix with one off-diagonal zero replaced by some $b\ne 0$. In case (a) the corresponding linear map multiplies volumes by $|a|$, which coincides with $|\det(E)|$, and in case (b) volumes are unchanged by $E$, which can be verified by looking at a two-dimensional figure. It follows that in case (b) the volume factor is $1=|\det(E)|$ as well. By the multiplicativity of determinants it follows that any nonsingular $A:\mathbb R^n\to\mathbb R^n$ multiplies volumes by the factor $|\det(A)|$. The parallelepiped $P$ considered in the question is the linear image of the unit cube; therefore vol$(P)=|\det(A)|$ where the matrix $A$ has the spanning vectors of $P$ in its columns.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-24T20:53:48.067" />
  <row Id="7723" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T20:53:49.530" Score="8" ViewCount="0" Body="&lt;p&gt;How about &lt;em&gt;combinatorially&lt;/em&gt;?  This follows from the following two facts.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The eigenvalues of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adjacency_matrix&quot; rel=&quot;nofollow&quot;&gt;adjacency matrix&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Path_graph&quot; rel=&quot;nofollow&quot;&gt;path graph&lt;/a&gt; on $n$ vertices are $2 \cos \frac{k \pi}{n+1}, k = 1, 2, ... n$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of closed walks from one end of the path graph on $4$ vertices to itself of length $2n$ is the Fibonacci number $F_{2n}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first can be proven by direct computation (although it also somehow falls out of the theory of quantum groups) and the second is a nice combinatorial argument which I will leave as an exercise.  I discuss some of the surrounding issues in &lt;a href=&quot;http://qchu.wordpress.com/2009/06/07/the-catalan-numbers-regular-languages-and-orthogonal-polynomials/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-24T20:53:49.530" CommentCount="2" />
  <row Id="7724" PostTypeId="2" ParentId="7697" CreationDate="2010-10-24T20:54:35.507" Score="4" ViewCount="0" Body="&lt;p&gt;Further to Moron's answer, consider any sequence of the form x&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;a&lt;sup&gt;n&lt;/sup&gt;&amp;nbsp;+&amp;nbsp;b for fixed integers a,b. Is there an infinite set of the x&lt;sub&gt;n&lt;/sub&gt; for which any two are coprime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, any common factor of a and b will divide every x&lt;sub&gt;n&lt;/sub&gt;. Furthermore, for any n,&#xA;$$&#xA;x_n=(a^{n-1}+a^{n-2}+\cdots+a+1)(a-1)+(b+1).&#xA;$$&#xA;so any common factor of a&amp;nbsp;&amp;minus;&amp;nbsp;1 and b&amp;nbsp;+&amp;nbsp;1 will divide every x&lt;sub&gt;n&lt;/sub&gt;. The answer to the question is no if either a, b or a&amp;nbsp;&amp;minus;&amp;nbsp;1, b&amp;nbsp;+&amp;nbsp;1 have a common prime factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a, b are coprime and every factor of b&amp;nbsp;+&amp;nbsp;1 divides a then the answer is yes.&#xA;Let S&amp;nbsp;=&amp;nbsp;{p&lt;sub&gt;1&lt;/sub&gt;,...,p&lt;sub&gt;m&lt;/sub&gt;} be a finite set of primes not dividing a. Taking n to be a multiple of &amp;prod;&lt;sub&gt;k&lt;/sub&gt;(p&lt;sub&gt;k&lt;/sub&gt;&amp;nbsp;&amp;minus;&amp;nbsp;1) gives x&lt;sub&gt;n&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;1&amp;nbsp;+&amp;nbsp;b&amp;nbsp;&amp;ne;&amp;nbsp;0 (mod p&lt;sub&gt;k&lt;/sub&gt;), so is not a multiple of p&lt;sub&gt;k&lt;/sub&gt;. If S is the set of prime factors of any finite collection of the x&lt;sub&gt;n&lt;/sub&gt;, this shows that we can find a new x&lt;sub&gt;n&lt;/sub&gt; coprime to each of the ones so far and, by induction, we can find an infinite set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the answer is yes in the cases you mention a&amp;nbsp;=&amp;nbsp;2, b&amp;nbsp;=&amp;nbsp;-3 and a&amp;nbsp;=&amp;nbsp;k, b&amp;nbsp;=&amp;nbsp;-(k&amp;nbsp;+&amp;nbsp;1). It does, however, leave open the cases where a,b are coprime and b&amp;nbsp;+&amp;nbsp;1 has prime factors not dividing a or a&amp;nbsp;&amp;minus;&amp;nbsp;1.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-10-24T20:54:35.507" />
  <row Id="7725" PostTypeId="2" ParentId="7711" CreationDate="2010-10-24T20:57:56.667" Score="1" ViewCount="0" Body="&lt;p&gt;If I've understood your notation, $A$ and $B$ are index sets of events denoted $i,j$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequality $2ab\leq a^2+b^2$ applied to the sums $a:=\sum_{i\in A}1_i$ and &#xA;$b:=\sum_{j\in B}1_j$ of indicator functions, followed by integration, gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sum_{i \in A, j \in B}\Pr[i\ \textrm{AND}\ j] \leq  \sum_{i, j \in A}\Pr[i\ \textrm{AND}\ j] + \sum_{i, j \in B}\Pr[i\ \textrm{AND}\ j],$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is stronger than what you asked for. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-24T20:57:56.667" />
  <row Id="7726" PostTypeId="1" AcceptedAnswerId="7732" CreationDate="2010-10-24T21:03:07.057" Score="0" ViewCount="166" Body="&lt;p&gt;Consider the probability space $([0,1]; B[0,1], L)$, where $B[0,1]$ contains the Borel sets&#xA;intersecting $[0,1]$ and $L$ is the Lebesgue measure. How do I find the sigma algebra generated by a random variable defined on this space, $X = 1_{[0,1/2]}$? Secondly, how do I determine whether random variables defined on this space are independent or not, e.g., are $X = 1_{[0,1/2}]$ and $Y = 1_{[1/4,3/4}]$ independent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For finding sigma algebra generated by $X$, I find $X^{-1}(1_{[0,1/2]})$ but what will be this inverse in borel sets? For finding independence, it should be sufficient to show the sigma algebras generated by $X$ and $Y$ are independent right?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-10-24T21:48:03.480" Title="sigma algebra generated by random variable" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7727" PostTypeId="2" ParentId="7695" CreationDate="2010-10-24T21:04:36.103" Score="13" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IvfAH.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a $\triangle ABC$ with $AB=1$, $\mathrm{m}\angle A=\frac{\pi}{5}$ and $\mathrm{m}\angle B=\mathrm{m}\angle C=\frac{2\pi}{5}$, and point $D$ on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$.  Now, $\mathrm{m}\angle CBD=\frac{\pi}{5}$ and $\mathrm{m}\angle BDC=\frac{2\pi}{5}$, so $\triangle ABC\sim\triangle BCD$.  Also note that $\triangle ABD$ is isosceles so that $BC=BD=AD$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x=BC=BD=AD$.  From the similar triangles, $\frac{AB}{BC}=\frac{BC}{CD}$ or $\frac{1}{x}=\frac{x}{1-x}$, so $1-x=x^2$ and $x=\frac{\sqrt{5}-1}{2}$ (the other solution is negative and lengths cannot be negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply the Law of Cosines to $\triangle ABC$:&#xA;$$\begin{align}&#xA;\cos\frac{2\pi}{5}=\cos C&amp;amp;=\frac{a^2+b^2-c^2}{2ab}&#xA;\\&#xA;&amp;amp;=\frac{(\frac{\sqrt{5}-1}{2})^2+1^2-1^2}{2\cdot\frac{\sqrt{5}-1}{2}\cdot 1}&#xA;\\&#xA;&amp;amp;=\frac{\frac{3-\sqrt{5}}{2}}{\sqrt{5}-1}&#xA;\\&#xA;&amp;amp;=\frac{-1+\sqrt{5}}{4}.&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-24T21:04:36.103" CommentCount="3" />
  <row Id="7728" PostTypeId="1" CreationDate="2010-10-24T21:09:52.547" Score="0" ViewCount="118" Body="&lt;p&gt;What is the second derivative of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x^2)y+y(x^2)=6 at point (1,3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;choices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2xsec(xy)   ;   sec(xy) / (x^2)  ;  2xsec(xy)-y  ;    2xsec(xy) / y ;  2xsec(xy)-y / x&lt;/p&gt;&#xA;" OwnerUserId="2758" LastActivityDate="2010-10-25T14:35:15.930" Title="Calculus AB Derivatives" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7729" PostTypeId="1" AcceptedAnswerId="7730" CreationDate="2010-10-24T21:17:37.490" Score="2" ViewCount="64" Body="&lt;p&gt;If $X$ is a random variable with finite mean $\mu$ and variance $\sigma^2$, how do I show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P[\mu − d\sigma &amp;lt; X &amp;lt; \mu + d\sigma] ≥ 1 − 1/d^2~~~~~~\forall d&gt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this in a book but unable to see the proof. Note that $X$ may not be normal.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-24T22:15:56.037" LastActivityDate="2010-10-24T22:15:56.037" Title="lower bound for probability distribution of a random variable" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="7730" PostTypeId="2" ParentId="7729" CreationDate="2010-10-24T21:24:31.817" Score="3" ViewCount="0" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Chebyshev's inequality&lt;/a&gt;, which holds for any probability distribution.  There are two proofs given on the linked Wikipedia page - a measure-theoretic one, and one that uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Markov's inequality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your expression is in a different form, though, than the one on the Wikipedia page.  To see how they are the same, observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P[\mu - d \sigma &amp;lt; X &amp;lt; \mu + d \sigma] \geq 1 - 1/d^2$$&#xA;is equivalent to&#xA;$$P[|X - \mu| &amp;lt; d \sigma] \geq 1 - 1/d^2,$$&#xA;which is equivalent to&#xA;$$P[|X - \mu| &gt; d \sigma] \leq 1/d^2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the one on the Wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-24T21:42:26.383" LastActivityDate="2010-10-24T21:42:26.383" />
  <row Id="7731" PostTypeId="1" CreationDate="2010-10-24T21:38:35.420" Score="1" ViewCount="49" Body="&lt;p&gt;This is one of the descriptions I've seen online:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any even n, say n = 2m, a row complete latin square of order n can be formed by writing down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 2m - 1, 2, 2m - 2, 3,..., m + 1, m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the first row and then developing subsequent rows by adding 1 modulo n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite clear on how that goes, such as what the difference is between 2m and m.  2m stands for n, does m stand for modulo?  I'd like to see and example with an even number or two, such as 4, and 6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2m = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 3, 2, 2... wait, that can't be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should turn out with non repeating sequences, and non recurring pairs between rows.&#xA;I can outline the other methods I've found (but don't understand) if this one is not the best.&lt;/p&gt;&#xA;" OwnerUserId="2759" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:38:00.530" LastActivityDate="2010-11-19T10:38:00.530" Title="How to Form a Row-Complete Latin Square" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" AnswerCount="2" />
  <row Id="7732" PostTypeId="2" ParentId="7726" CreationDate="2010-10-24T21:41:53.210" Score="1" ViewCount="0" Body="&lt;p&gt;The sigma-algebra generated by $1_{[0,1/2]}$ is simply&#xA;$$&#xA;\bigl\{\emptyset,[0,1],[0,1/2],(1/2,1]\bigr\}.&#xA;$$&#xA;It consists of the preimages under the function $1_{[0,1/2]}$ of all Borel sets in the codomaion of the function $1_{[0,1/2]}$, namely, $(\mathbb R,B(\mathbb R))$. (Notice that the preimage $1_{[0,1/2]}^{-1}(M)$ is completely determined by the information of whether 0 and 1 do or do not belong to $M$ respectively.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for $1_{[1/4,3/4]}$ is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The random variables $1_{[0,1/2]}$ and $1_{[1/4,3/4]}$ on $([0,1],B[0,1],L)$ are indeed independent: For this you have to check that $L(A\cap B)=L(A)\cdot L(B)$ for all $A\in 1_{[0,1/2]}^{-1}(B(\mathbb R))$ and $B\in 1_{[1/4,3/4]}^{-1}(B(\mathbb R))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting case is $L([0,1/2]\cap [1/4,3/4])=L([0,1/2])\cdot L([1/4,3/4])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check that both sides are equal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also think about the following question: Are the random variables $1_{[0,1/2]}$ and $1_{[1/4,1]}$ on $([0,1],B[0,1],L)$ also independent?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-10-24T21:48:03.480" LastActivityDate="2010-10-24T21:48:03.480" CommentCount="6" />
  <row Id="7733" PostTypeId="2" ParentId="7731" CreationDate="2010-10-24T21:42:53.803" Score="2" ViewCount="0" Body="&lt;p&gt;If I understand your description correctly, for 2m=4:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  1  3  2&#xA;1  2  0  3&#xA;2  3  1  0&#xA;3  0  2  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For 2m=6:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  1  5  2  4  3&#xA;1  2  0  3  5  4&#xA;2  3  1  4  0  5&#xA;3  4  2  5  1  0&#xA;4  5  3  0  2  1&#xA;5  0  4  1  3  2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: In each case, the first row follows the pattern you gave; in subsequent rows, each number is one more than the number above it, but wrapping around at the size of the square (e.g. in the case where n=2m=6, after 5 comes 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at n=2m=12 (m=6) a bit more closely.  The pattern that you gave said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0, 1, 2m - 1, 2, 2m - 2, 3,..., m + 1, m&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the first row is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0, 1, 12 - 1, 2, 12 - 2, 3,..., 6 + 1, 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0, 1, 12-1=11, 2, 12-2=10, 3, 12-3=9, 4, 12-4=8, 5, 12-5=6+1=7, 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the second row is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0+1=1, 1+1=2, 11+1=12 becomes 0, 2+1=3, 9+1=10, 4+1=5, 8+1=9, 5+1=6, 7+1=8, 6+1=7&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Continuing gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  1 11  2 10  3  9  4  8  5  7  6&#xA;1  2  0  3 11  4 10  5  9  6  8  7&#xA;2  3  1  4  0  5 11  6 10  7  9  8&#xA;3  4  2  5  1  6  0  7 11  8 10  9&#xA;               .&#xA;               .&#xA;               .&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-24T22:10:56.053" LastActivityDate="2010-10-24T22:10:56.053" CommentCount="2" />
  <row Id="7734" PostTypeId="2" ParentId="7728" CreationDate="2010-10-24T22:08:46.420" Score="6" ViewCount="0" Body="&lt;p&gt;The question makes no sense. The derivative &lt;strong&gt;at a point&lt;/strong&gt; is a number (if it exists): the slope of the tangent to the curve at that point; the fact that this is the second derivative is not an issue: you still get the slope of the tangent at the point, in this case the slope of the tangent to the graph of $y'$. None of the choices you give are numbers, so none of them are equal to second derivative of the function at $(1,3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your function as is unclear. It could either mean $x^2y + yx^2 = 6$, in which case you simply have $2yx^2 = 6$, or $y = 3/x^2$; we can verify that $(1,3)$ is on the curve by plugging in. Or else you mean $x^2y(x) + y(x^2)$, where &quot;$y(x)$&quot; means the function $y$ evaluated at $x$ and &quot;$y(x^2)$&quot; means the function $y$ evaluated at $x^2$. Since $y(1) = y(1^2)$, we can verify that the point is on the curve by again substituting $x=1$ and $y=y(x^2)=3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is clear that you have not provided the question accurately, and as such, no answer can be given; in any case, whathever the interpretation of the &lt;em&gt;question&lt;/em&gt;, none of the answers can possibly be the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;strong&gt;if&lt;/strong&gt; the equation is simply $2x^2y = 6$, of $y = 3x^{-2}$, then $y' = -6x^{-3}$ and $y'' = 18x^{-4}$, so the second derivative at $(1,3)$ is $18$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the equation is $xy + y(x^2) = 6$, then we have by implicit differentiation&#xA;$$\begin{array}{rcl}&#xA;\frac{d}{dx}xy + \frac{d}{dx}y(x^2) &amp;amp; = &amp;amp; 0\\&#xA;y + xy'(x) + y'(x^2)(2x) &amp;amp; = &amp;amp; 0\\&#xA;y + xy'(x) + 2xy'(x^2) &amp;amp; = &amp;amp; 0.&#xA;\end{array}$$&#xA;Plugging in $x=1$ and $y=3$ we get&#xA;$3 + y'(1) + 2y'(1) = 0$, or $3y'(1) = -3$, hence $y'(1)=-1$. &#xA;Differentiating again we have:&#xA;$$\begin{array}{rcl}&#xA;y'(x) + y'(x) + xy''(x) + 2y'(x^2) + 2xy''(x^2)(2x) &amp;amp; = &amp;amp; 0\\&#xA;2y'(x) + xy''(x) + 2y'(x^2) + 4x^2y''(x^2) &amp;amp; = &amp;amp; 0.&#xA;\end{array}$$&#xA;Plugging in $x=1$, $y=3$, and $y'(1)=-1$ we get &#xA;$$-2 + y''(1) + 2(-1) + 4y''(1) = 0$$&#xA;or $5y''(1) = 4$. So the second derivative at the point $(1,3)$ is $\frac{d^2y}{dx^2}\Biggm|_{(x,y)=(1,3)} = \frac{4}{5}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T14:35:15.930" LastActivityDate="2010-10-25T14:35:15.930" />
  <row Id="7735" PostTypeId="2" ParentId="7716" CreationDate="2010-10-24T22:12:40.720" Score="6" ViewCount="0" Body="&lt;p&gt;Isaac and Ryan have already answered your question in words.  Now, in symbols, the chain rule gives:&#xA;$$\frac{d(y^2)}{dx} = \frac{d(y^2)}{dy}\frac{dy}{dx} = 2y\frac{dy}{dx}$$&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-10-24T22:12:40.720" CommentCount="1" />
  <row Id="7737" PostTypeId="1" CreationDate="2010-10-24T22:48:52.347" Score="0" ViewCount="208" Body="&lt;p&gt;How do I use Lebesgue Dominated Convergence Theorem to evaluate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \to \infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt(x)}dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what dominating function to use here?&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-24T22:51:43.560" LastActivityDate="2010-10-25T00:50:01.333" Title="Lebesgue Dominated Convergence example" Tags="&lt;probability-theory&gt;&lt;integral&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7738" PostTypeId="1" CreationDate="2010-10-24T22:51:06.243" Score="-1" ViewCount="124" Body="&lt;p&gt;I have trouble understanding Radon-Nikodym theorem. Here's an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the measurable space $(\Omega,F)$, where $\Omega = R$, $F~$ is the $\sigma$-algebra of Borel sets. Let $P[dw] = \frac{1}{\sqrt(2\pi)}e^{-w^2/2}dw$ and $\tilde{P}[dw] = e^{-w}1_{[0,\infty)(w)}dw$ be two probability measures on $\omega$. How to test whether $\tilde{P} &amp;lt;&amp;lt; P$ and $P &amp;lt;&amp;lt; \tilde{P}$. How do we compute Radon-Nikodym derivative $\frac{d\tilde{P}}{dP}$?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-10-25T12:20:09.047" Title="Radon-Nikodym example" Tags="&lt;probability-theory&gt;&lt;integral&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7740" PostTypeId="1" CreationDate="2010-10-24T23:00:08.350" Score="2" ViewCount="94" Body="&lt;p&gt;I want to refer to functions of the form $f(x) = \sum_{i=1}^n a_i x^{\alpha_i}$ where $\alpha_i &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a polynomial, because $\alpha_i$ could be just real arbitrary numbers (though positive ones).&#xA;Is there a name for this class of functions that I can refer to? Wikipedia does claim these are not polynomials.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="30" LastEditorDisplayName="" LastEditDate="2010-10-25T00:45:25.417" LastActivityDate="2010-10-25T00:45:25.417" Title="polynomial of $x$?" Tags="&lt;terminology&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7741" PostTypeId="2" ParentId="7740" CreationDate="2010-10-24T23:07:54.570" Score="3" ViewCount="0" Body="&lt;p&gt;In 1983 they still had no official name, see for example &lt;a href=&quot;http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P16.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Peter Borwein.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-24T23:07:54.570" CommentCount="2" />
  <row Id="7742" PostTypeId="2" ParentId="7737" CreationDate="2010-10-24T23:26:11.120" Score="5" ViewCount="0" Body="&lt;p&gt;Consider this for a solution (it does not use dominated convergence). In $[0,1]$, we have the following inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;0 \leq \frac{n \sin(x)}{1+ n^2 \sqrt{x}} &amp;lt; \frac{n}{n^2 \sqrt{x}} = \frac{1}{n \sqrt{x}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking integral over $[0,1]$ on both sides, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} &amp;lt; \int_{[0,1]} \frac{1}{n \sqrt{x}} = \left[ \frac{2 \sqrt{x}}{n} \right]_0^1&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking limit $n \to \infty$ on both sides, you can show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{n \to \infty} \int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} = 0&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-24T23:26:11.120" CommentCount="1" />
  <row Id="7743" PostTypeId="1" AcceptedAnswerId="8203" CreationDate="2010-10-25T00:07:12.920" Score="16" ViewCount="606" Body="&lt;p&gt;So, I don't like proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me building a proof feels like constructing a steel trap out of arguments to &lt;strong&gt;make true&lt;/strong&gt; what you're trying to assert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oftentimes the proof in the book is something that I get if I study, but hard to come up with on my own.  In other words I can't make steel traps, but I feel fine buying them from others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one acquire the ability to create steel traps with fluency and ease?  Are there any particular reference books that you found helped you really get how to construct a proof fluently?  Or is it just practice?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T22:55:12.823" LastActivityDate="2010-10-28T23:25:21.243" Title="Getting better at proofs" Tags="&lt;soft-question&gt;&lt;proof-strategy&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="7" />
  <row Id="7744" PostTypeId="2" ParentId="7731" CreationDate="2010-10-25T00:10:05.193" Score="0" ViewCount="0" Body="&lt;p&gt;That seems clear.  I guess it was the switch from n to m and modulo that I was confused about.  Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, I meant for this to be a comment, not an answer)&lt;/p&gt;&#xA;" OwnerUserId="2762" LastEditorUserId="2762" LastEditorDisplayName="" LastEditDate="2010-10-25T00:20:48.570" LastActivityDate="2010-10-25T00:20:48.570" />
  <row Id="7745" PostTypeId="1" CreationDate="2010-10-25T00:28:24.627" Score="2" ViewCount="104" Body="&lt;p&gt;Suppose $B_t$ is a Brownian motion. As I understand, $B_2-B_1$ is independent of $B_3-B_2$ from properties of Brownian motion. Does it also mean that $B_1$ and $B_2$ are also independent? Can I use this independence to find the joint density of $B_1+B_2+B_3$ as each Brownian process is a normal process of mean 0 and variance t, it should be trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've another related question. To find the expectation over a Brownian process, can I integrate my stochastic process over the normal density function for Brownian motion (mean 0 and variance t)? I hope this makes sense.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-25T01:31:31.923" LastActivityDate="2010-10-28T16:07:06.293" Title="combination of brownian motion" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="3" />
  <row Id="7746" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T00:32:03.377" Score="22" ViewCount="0" Body="&lt;p&gt;You get better at proofs the same way you get better at basketball or carpentry: lots and lots of practice.  (In particular, like in basketball and carpentry, you can only get so far by reading books.)  Of course, there's good practice and bad practice.  For general experience writing and coming up with proofs I think the same kind of material that people use to prepare for Olympiad-style mathematics is very helpful.  To that end, here are a few references you might find helpful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387982191&quot; rel=&quot;nofollow&quot;&gt;Problem-Solving Strategies&lt;/a&gt;, Engel (somewhat high-level)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Solve_It&quot; rel=&quot;nofollow&quot;&gt;How to Solve It&lt;/a&gt;, Polya&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471135712&quot; rel=&quot;nofollow&quot;&gt;The Art and Craft of Problem Solving&lt;/a&gt;, Zeitz&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For writing proofs specific to a course (e.g. real analysis) you'll generally find that the same basic ideas are being used over and over again, and once you learn to recognize when these ideas will be useful your life will be much easier.  But I think this is a hard skill to teach.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't help that many proofs in textbooks are written in a style that makes it nearly impossible to see how someone could have come up with the proof from first principles.  This is an unfortunate tendency, and you should find a different textbook if this is too much of an issue.  (Alternately, you should see if there are better proofs online, for example on someone's blog.  &lt;a href=&quot;http://gowers.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Tim Gowers&lt;/a&gt; and &lt;a href=&quot;http://terrytao.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Terence Tao&lt;/a&gt; are fond of writing up conceptual proofs of things, and more generally their blogs are a great source of insight into how mathematicians think.)  Once you build up a little problem-solving skill, another way to fix this is to reprove things yourself.  It helps if you can't remember what the proof in the textbook is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm serious about the practice.  $10000$ hours and all that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I also find this &quot;steel trap&quot; analogy very depressing.  While it might seem that way in some textbooks, a properly presented proof should feel more like a poem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit #2:&lt;/strong&gt;  I'd also like to mention that the general outline of Polya's advice is &lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Solve_It#Four_principles&quot; rel=&quot;nofollow&quot;&gt;on Wikipedia&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-28T23:25:21.243" LastActivityDate="2010-10-28T23:25:21.243" />
  <row Id="7747" PostTypeId="2" ParentId="7737" CreationDate="2010-10-25T00:32:21.473" Score="4" ViewCount="0" Body="&lt;p&gt;Note that for $x\in(0,1]$ and $n\geq 1$, we have:&#xA;$$\frac{n\sin(x)}{1+n^2\sqrt{x}} \leq \frac{n}{1+n^2\sqrt{x}} \leq \frac{n}{n^2\sqrt{x}} \leq \frac{1}{n\sqrt{x}} \leq \frac{1}{\sqrt{x}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $g(x) = \frac{1}{\sqrt{x}}$ for $0\lt x\leq 1$ and $g(x)=0$ if $x=0$, then $0\leq f_n(x)\leq g(x)$ almost everywhere in $[0,1]$, where $f_n(x) = \frac{n\sin(x)}{1+n^2\sqrt{x}}$. Since&#xA;$$\int g(x)\,d\mu = \int_{[0,1]}\frac{1}{\sqrt{x}}\,d\mu= 2\sqrt{x}\Biggm|_0^1 = 2\lt \infty$$&#xA;by Lebesgue's Dominated Convergence Theorem you know that if $f(x)$ equals the pointwise limit of the $f_n(x)$ almost everywhere on $[0,1]$, then &#xA;$$\lim_{n\to\infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt{x}}\,d\mu = \int_{[0,1]}f(x)\,d\mu.$$&#xA;So to use Dominated Convergence you now need to figure out an $f(x)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T00:50:01.333" LastActivityDate="2010-10-25T00:50:01.333" CommentCount="1" />
  <row Id="7748" PostTypeId="2" ParentId="7708" CreationDate="2010-10-25T00:35:03.980" Score="0" ViewCount="0" Body="&lt;p&gt;I would note that the situation is similar to matrix multiplication: since multiplication here is noncommutative, you have to distinguish between &quot;premultiplying&quot; with the inverse $p^{-1}q$ and &quot;postmultiplying&quot; with the inverse $qp^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(q\cdot r)^{-1}=r^{-1}\cdot q^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other thing I can say is that what you're essentially doing is the quaternion analog of a &quot;similarity transformation&quot; of a matrix.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T00:35:03.980" />
  <row Id="7749" PostTypeId="2" ParentId="7745" CreationDate="2010-10-25T01:02:09.157" Score="3" ViewCount="0" Body="&lt;p&gt;$B_1$ and $B_2$ are not independent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $B_1$ and $B_2-B_1$ are independent,&#xA;$$0=\mathrm{Cov}(B_1,B_2-B_1)=E[B_1 (B_2-B_1)]=E[B_1 B_2]-E[B_1^2]=E[B_1 B_2] - t_1 \sigma^2$$&#xA;So,&#xA;$$\mathrm{Cov}(B_1,B_2)=E[B_1 B_2]=t_1 \sigma^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this result and the fact that linear combinations of normal variables are normal to calculate the distribution of $B_1+B_2+B_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what exactly you meant by &quot;expectation over a Brownian process&quot;. Can you please give us a little background of the problem you are trying to solve? Your method of taking an expectation with respect to $N(0,\sigma\sqrt{t})$ is fine when you want the expected value of some function of $B_t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/It%C5%8D_calculus&quot; rel=&quot;nofollow&quot;&gt;stochastic integration&lt;/a&gt; &lt;em&gt;with respect to&lt;/em&gt; stochastic processes which might be of use to you.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-25T01:09:49.757" LastActivityDate="2010-10-25T01:09:49.757" CommentCount="3" />
  <row Id="7750" PostTypeId="2" ParentId="7672" CreationDate="2010-10-25T01:03:23.510" Score="8" ViewCount="0" Body="&lt;p&gt;Well your notation is a little bit confusing. Since we are considering free abelian groups, we should avoid additive notations for the commutative semigroup S. Let's talk of $(S, \ast )$. If $F(S)$ is the free abelian group with basis $S$, we can assume $S \subseteq F(S)$. So the set of relations is $$R = \{ (a \ast b) -a - b : a,b \in S \}.$$ Then, the Grothendieck completion is the abelian group $$G(S) = F(S)/ \langle R \rangle.$$ If $a \in S$, I'll denote the coset $a + \langle R \rangle$ by $[a]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof isn't hard. You'll just need to remember that in free abelian groups the expression in terms of a basis is unique. You can find one proof, with an ugly notation, in Rotman's 'Advanced Modern Algebra'. I'll reproduce below the proof on Magum's book 'An Algebraic Introduction to K-theory'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $[a]=[b]$. Then, $a-b \in \langle R \rangle$. That is, $$a-b = \sum_{i=1}^n n_i((a_i \ast b_i) - a_i - b_i),$$ where $n_i = 1$ or $-1$ and $a_i, b_i \in S$. Bringing terms with negative coefficients to the other side, $$ a + \sum_{n_i = -1}(a_i \ast b_i) + \sum_{n_i = 1}(a_i + b_i) = b + \sum_{n_i = 1}(a_i \ast b_i) + \sum_{n_i = -1}(a_i + b_i).$$ Since $S$ is a basis of $F(S)$, the terms on one side of the equation are a permutation of those on the other side. Since $(S, \ast )$ is commutative, it follow that, in S, $$ a \ast \prod_{n_i = -1}(a_i \ast b_i) \ast \prod_{n_i = 1}(a_i \ast b_i) = b \ast \prod_{n_i = 1}(a_i \ast b_i) \ast \prod_{n_i = -1}(a_i \ast b_i). $$ So $a \ast c = b \ast c$, where $$ c = \prod_{i = 1}^n (a_i \ast b_i).$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-10-25T01:30:25.030" LastActivityDate="2010-10-25T01:30:25.030" CommentCount="2" />
  <row Id="7751" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T01:05:59.813" Score="2" ViewCount="0" Body="&lt;p&gt;As Arturo and Qiaochu have pointed out you get better at proving things by practicing a lot, solving exercises and by seeing how other people have proved things. There's a big part in learning how to prove statements by repeating certain strategies that may have worked for particular types of problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, a particular argument that is used countless times in analysis is the &quot;technique&quot; of adding and substracting a quantity in a way that let's you group things in a useful way. And there are lots and lots of tricks such as this one that you will learn during the way by seeing them being used several times (and also by using them yourself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's certainly something about proving things that I must say. I don't know how common this is for people studying at &quot;top universities&quot; around the world were only really talented and smart people are admitted, but in my country it is relatively easy to get accepted in a mathematics program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, I've seen lots of people fail several times in the first mathematics course in my university for mathematics students (where actually students are introduced to proof techniques and basic logic and set theory, basically what in some American universities is referred to as a transition course from the usual calculus courses to proof oriented courses). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for most of them failing such a course from what I've seen, is that some of them just don't seem to get used to the idea of having to prove something rigorously (as a mathematician will expect). I remember very well when one my classmates argued with the professor because he was asked in the test to prove an equality of sets, and my classmate draw a Venn diagram which showed the sets in question. The professor told him repeatedly that as an aid for intuition the drawing was perfectly fine, but that the drawing didn't constitute a proof by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end (as you'll imagine) my classmate didn't win the argument (or extra points for that matter) and he ended up going home frustrated and finally he gave up on the course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that for some reason some people are just better than others at proving things, and I'm not saying that they're just smarter, it's only that the thought processes involved just seem to come more naturally to some persons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there are some books that specifically address the topic of introducing students to the task of proving things. What they usually do is to begin by explaining some basic logic and them they build up some easy facts from set theory, binary relations and functions and in the way they introduce some proof techniques, such as &quot;proof by contradiction&quot;, &quot;proof by contrapositive&quot;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one book that helped me a lot when I was starting is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0817641114&quot; rel=&quot;nofollow&quot;&gt;Proofs and Fundamentals: A First Course in Abstract Mathematics&lt;/a&gt;, but there a lot of other books that I'm pretty sure work really well when making your first steps in proving mathematical statements, such as &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471680583&quot; rel=&quot;nofollow&quot;&gt;How To Read and Do Proofs&lt;/a&gt; by Daniel Solow or &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521675995&quot; rel=&quot;nofollow&quot;&gt;How To Prove It&lt;/a&gt; by Daniel Velleman.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-25T01:05:59.813" />
  <row Id="7752" PostTypeId="2" ParentId="7745" CreationDate="2010-10-25T01:06:21.477" Score="1" ViewCount="0" Body="&lt;p&gt;Expressing $B_1+B_2+B_3=3B_1+2(B_2-B_1)+(B_3-B_2)$ as a sum of independent normal random variables will help you find its density. I'm not sure why you call it a joint density.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-25T01:06:21.477" />
  <row Id="7753" PostTypeId="1" CreationDate="2010-10-25T01:20:57.087" Score="1" ViewCount="49" Body="&lt;p&gt;The usual action of $fg$  on $u⊗v$, where $f,g$  are elements in the Universal Enveloping Algebra $U(G)$ of a Lie algebra $G$ and $u,v$ are elements of a representation $V$  of $G$, is given by $fg(u⊗v)=fgu⊗v+fu⊗gv+gu⊗fv+u⊗fgv$,  right? How to state this fact for $V^{\otimes n}$, i.e. $fg$  acting on $u⊗v$, where $u=⊗_{i=1}^{n-k} u_i$ and $v=\otimes_{i=1}^k v_i$, for each $k=1,...,n−1$? &#xA;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-10-25T03:26:05.367" Title="How to define the action of U(G)  in this situation?" Tags="&lt;universal-algebra&gt;&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="7754" PostTypeId="1" CreationDate="2010-10-25T02:12:35.540" Score="0" ViewCount="117" Body="&lt;p&gt;we have a truck in our game we have a rod against the wall of that truck, and we know the acceleration of truck when the rod start slipping... we just want to know what is acceleration of rod when it touch surface of truck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l2wJa.jpg&quot; alt=&quot;diagram for problem&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2765" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-10-27T05:03:11.427" LastActivityDate="2010-10-27T05:03:11.427" Title="slipping rod on moving truck." Tags="&lt;physics&gt;&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7755" PostTypeId="2" ParentId="7754" CreationDate="2010-10-25T02:35:50.510" Score="1" ViewCount="0" Body="&lt;p&gt;To make the question more precise: there is a max static friction coefficient at points A and B.  Calculating the perpendicular force at A and B is not hard, so the acceleration where it breaks loose is known.  The acceleration increases exceeding this.  So the stick starts to slide aft.  Now you have a sliding friction, which would usually be modeled as a (lower) coefficient of the force on the surface.  You want an equation for the height of A as a function of time. As the acceleration depends upon user input, it seems you want something for short timesteps.  If we could give you delta A=f(acceleration,A,dA/dt previous step), would this meet your needs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rod hits A=0.  This is hard.  One approach would be that it doesn't bounce, but has a certain (different?) sliding friction and continues to move aft.  Another would be that it bounces with a certain recovery fraction, so dA/dt is -(some number times the previous) dA/dt, but there is no more friction at point A.  The sliding friction at B continues unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might even lose contact at A before hitting the deck.  This shouldn't be too hard to detect.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T02:49:39.057" LastActivityDate="2010-10-25T02:49:39.057" CommentCount="3" />
  <row Id="7756" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T02:51:06.327" Score="1" ViewCount="0" Body="&lt;p&gt;I recommend the book by Lay, &quot;Analysis with introduction to proofs&quot;. The first chapter focuses on logic and proofs and personally I found it rather helpful. At least it helped me. Of course to master the art of proofs you have to keep practicing, but knowledge of some basics such that instead of proving the statement &quot;A implies B you can prove contrapositive &quot;not B implies not A&quot;, and how it differs from &quot;Proof by contradiction&quot;, and the fact that negation of &quot;A implies B&quot; is &quot;A and not B&quot;, and how to deal with existential and universal quantifiers, and so forth and so forth. Those are useful techniques one should master. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, as an introduction to analysis, the aforementioned book is quite mediocre. I had a first hand experience with that book, since it was a textbook when I took the &quot;low level introduction to analysis (wasn't sure if I was ready for Rudin's Principles of Analysis). So, this is only the chapter on logic and proof which is worth reading. Of course, purchasing the whole book just for a single chapter is funny, but may be you can find the used one or get one in the library. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that chapter on proofs did help me. When subsequently I took a course on analysis and we used Baby Rudin as a textbook, I was in a very good shape and in fact did quite well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, another suggestion: learn by example. See how other people write the proofs. Rudin's style, for instance, is very terse, you literally have to dig through his proofs, but after a while you start getting used to it, and still after a while you start writing similar proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-25T02:51:06.327" />
  <row Id="7757" PostTypeId="1" AcceptedAnswerId="7762" CreationDate="2010-10-25T02:53:55.737" Score="5" ViewCount="584" Body="&lt;p&gt;I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked &lt;a href=&quot;http://math.stackexchange.com/users/2641/svenkatr&quot;&gt;svenkatr&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/users/242/bill-dubuque&quot;&gt;Bill Dubuque&lt;/a&gt; soution),hence I am removing the combinatorics tag.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-25T14:45:45.310" LastActivityDate="2010-12-06T18:07:50.067" Title="How to prove this  binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ ?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="10" CommentCount="1" FavoriteCount="1" />
  <row Id="7758" PostTypeId="2" ParentId="7717" CreationDate="2010-10-25T02:56:15.230" Score="0" ViewCount="0" Body="&lt;p&gt;Add &quot;lemmas&quot;, &quot;conjectures&quot; and &quot;principles&quot; to your question  and you have the full gamut!&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-25T02:56:15.230" CommentCount="1" />
  <row Id="7761" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:01:05.980" Score="14" ViewCount="0" Body="&lt;p&gt;Here's a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r\binom{n}{r} = n\binom{n-1}{r-1}, \quad r\geq 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here be spoilers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\sum_{r=0}^{n} r\binom{n}{r}&#xA;&amp;amp;= \sum_{r=1}^{n} r\binom{n}{r} &amp;amp; &amp;amp;\text{drop the zero term}&#xA;\\&#xA;&amp;amp;= \sum_{r=1}^{n} n\binom{n-1}{r-1} &amp;amp; &amp;amp;\text{apply the above identity}&#xA;\\&#xA;&amp;amp;= n \sum_{k=0}^{n-1} \binom{n-1}{k} &amp;amp; &amp;amp;\text{replace index $r$ with $k=r-1$} &#xA;\\&#xA;&amp;amp;= n 2^{n-1} &amp;amp; &amp;amp;\text{the sum is just the total number of subsets of $n-1$ elements} &#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="1023" LastEditorUserId="1023" LastEditorDisplayName="" LastEditDate="2010-10-26T06:05:05.743" LastActivityDate="2010-10-26T06:05:05.743" CommentCount="3" />
  <row Id="7762" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:02:53.513" Score="20" ViewCount="0" Body="&lt;p&gt;Take the binomial expansion of $(1+x)^n$. Differentiate both sides with respect to $x$, then substitute $x=1$ and that will give you the identity.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-25T03:02:53.513" CommentCount="6" />
  <row Id="7763" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:04:01.563" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Differentiate $\rm (1+x)^n\:$, use the binomial theorem, then set $\rm\ x = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ Using derivatives, we can pull out of a sum any polynomial function of the index variable, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we have $\rm\:\ k^i\ x^k\ =\ (xD)^i \ x^k\ \ $ for $\rm\ \ D = \frac{d}{dx},\ \ k &gt; 0\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that $\rm\ \sum a_k\: f(k)\: x^k\ =\ f(x\:D) \sum a_k\: x^k\ \ $&#xA; for polynomial $\rm\:f\:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;trick&quot; of representing a numerical series as the value f(1) of a generating function f(x) goes back at least to Euler, who employed it to sum divergent series (among other applications). The power arises from the fact that at the function level one has available much more powerful tools, esp. derivatives, e.g. above. Such techniques come in quite handy when shifting between differential and difference (recurrence) viewpoints, e.g. for hypergeometric functions and their special values.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-25T13:07:53.400" LastActivityDate="2010-10-25T13:07:53.400" />
  <row Id="7764" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T03:07:29.823" Score="1" ViewCount="0" Body="&lt;p&gt;We have $x^{n} = (1+(x-1))^{n}$ and $$[1+(x-1)]^{n} = {n \choose 0 } + {n \choose 1} \cdot (x-1) + {n \choose 2} \cdot (x-1)^{2} + \cdots + {n \choose n} \cdot (x-1)^{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate on both sides w.r.t x and substitute the value $x=2$ in your equation.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-25T03:07:29.823" />
  <row Id="7765" PostTypeId="2" ParentId="7753" CreationDate="2010-10-25T03:26:05.367" Score="1" ViewCount="0" Body="&lt;p&gt;If $V_1,V_2\dots,V_n$ are $U(\mathfrak g)$-modules, then the $U(\mathfrak g)$-module structure on the tensor product $V_1\otimes V_2\otimes\cdots\otimes V_n$ is the unique one such that whenever $x\in\mathfrak g\subset U(\mathfrak g)$ we have $$\begin{aligned}g\cdot v_1\otimes\cdots\otimes v_n&amp;amp;=(x\cdot v_1)\otimes v_2\otimes\cdots\otimes v_n\\&amp;amp;+v_1\otimes(x\cdot v_2)\otimes v_3\otimes\cdots\otimes v_n\\{}&amp;amp;+v_1\otimes v_2\otimes(x\cdot v_3)\otimes v_4\otimes\cdots\otimes v_n\\{}&amp;amp;+\cdots\\{}&amp;amp;+v_1\otimes v_2\otimes\cdots\otimes v_{n-1}\otimes(x\cdot v_n).\end{aligned}$$ Notice that this formula is only valid for elements of $\mathfrak g$, not all elements of $U(\mathfrak g)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know how an arbitrary element of $U(\mathfrak g)$ acts on the tensor product, write it as a sum of products of elements of $\mathfrak g$ and make use of the relation $$xy\cdot v = x\cdot(y\cdot v)$$ valid for all $x$, $y\in U(\mathfrak g)$ and all $v\in V_1\otimes V_2\otimes\cdots\otimes V_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should really pick up a book on Hopf algebras. My favorite is Sweedler's &lt;em&gt;Hopf Algebras&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-25T03:26:05.367" />
  <row Id="7766" PostTypeId="2" ParentId="7743" CreationDate="2010-10-25T03:36:52.323" Score="3" ViewCount="0" Body="&lt;p&gt;It seems you are asking two questions.  The first is how to get better at doing proofs, and the previous answers are better than I can do.  The second is why to do them, which has not been addressed.  From your questions on this site, you seem more an engineer than a mathematician.  This is not a negative-I am educated as a physicist and practicing as an engineer(ing manager), but I enjoy proof-type math.  It just is a different view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to your question &quot;How can I characterize the type of solution vector that comes out of a matrix?&quot; Greg Graviton not only gave an answer of what matrices met your criterion, he proved that matrices that did not met his answer did not work.  This is a major improvement as you know there aren't others out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proofs are designed to make sure you covered all the loose ends.  Ideally, each statement in a proof would be derived from the previous statements through an accepted rule of logic.  But aside from the automated theorem provers, this is unachievable because it would make the proofs too long.  So you only need to display enough steps to convince the audience it can be done.  How big a leap is acceptable varies with the audience.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As physicists/engineers our numbers and functions are better behaved than the ones the mathematicians worry about.  The mathematical world is not uniformly continuous, but ours is.  As such, we interchange limits, integrals, derivatives, and sums freely.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-25T03:36:52.323" />
  <row Id="7767" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T04:43:15.057" Score="6" ViewCount="0" Body="&lt;p&gt;I can't resist giving the standard bijective proof of this fact. Let $[n]$ be the set of integers from $1$ to $n$. Then $r {n \choose r}$ is the number of elements in all the subsets of $[n]$ which have $r$ elements. So your sum is the number of elements in all the subsets of $[n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can pair the sets up into $2^{n-1}$ pairs of the form $S, [n] \setminus S$. Each pair contains $n$ elements, so all the sets together contain $n 2^{n-1}$ elements.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-25T04:43:15.057" />
  <row Id="7768" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T04:45:37.833" Score="16" ViewCount="0" Body="&lt;p&gt;Use &lt;a href=&quot;http://www.jimloy.com/algebra/gauss.htm&quot; rel=&quot;nofollow&quot;&gt;Gauss' trick&lt;/a&gt; for summing a sequence: Combine the sum term-by-term with its reversed-order self, noting that the binomial coefficients are symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = \sum_{r=0}^n r { n \choose r } = \sum_{r=0}^n (n-r) {n \choose n-r }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xA;2 S &amp;amp;=&amp;amp; \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose n-r} \right)\\&#xA; &amp;amp;=&amp;amp; \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose r} \right)\\&#xA;&amp;amp;=&amp;amp; \sum_{r=0}^n \left( n {n\choose r} \right)\\&#xA;&amp;amp;=&amp;amp; n \sum_{r=0}^n {n\choose r}\\&#xA;&amp;amp;=&amp;amp; n 2^n \hspace{0.1in} \text{, by familiar identity}&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-10-25T04:45:37.833" CommentCount="1" />
  <row Id="7769" PostTypeId="1" CreationDate="2010-10-25T05:08:08.093" Score="0" ViewCount="125" Body="&lt;p&gt;How can I find the critical points of a square root function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: What are the critical points of $f(x) = \sqrt{7x^2 + 2x - 2}$?&lt;/p&gt;&#xA;" OwnerUserId="2751" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-25T05:59:24.223" LastActivityDate="2010-10-25T14:33:25.540" Title="Critical Points of a Square Root Function" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7770" PostTypeId="2" ParentId="7589" CreationDate="2010-10-25T05:35:48.277" Score="2" ViewCount="0" Body="&lt;p&gt;I'm setting this answer community wiki so other people can attach their own examples of curves constructed with the methods in this thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Qiaochu's suggestion for building a curve $q(y)=p(x)$ with collinear singular points on the horizontal axis amounts to the construction of an appropriate &lt;a href=&quot;http://mathworld.wolfram.com/HermitesInterpolatingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Hermite interpolation&lt;/a&gt; problem. More explicitly, one wants to find a polynomial (or rational function) whose first few derivatives at preset points vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Mathematica&lt;/em&gt; for instance, the function &lt;code&gt;InterpolatingPolynomial&lt;/code&gt; can be used to generate a Hermite interpolant. (For systems that do not have such a function handy, the Hermite interpolation problem is solved through either an appropriate &lt;a href=&quot;http://books.google.com/books?id=Mp8-z5mHptcC&amp;amp;pg=PA189&quot; rel=&quot;nofollow&quot;&gt;modification of the Newton divided differences scheme&lt;/a&gt; or by solving an associated &lt;a href=&quot;http://dx.doi.org/10.1007/BF01436299&quot; rel=&quot;nofollow&quot;&gt;confluent Vandermonde system&lt;/a&gt;.) The rational interpolant case is a bit tougher, and I am still experimenting with algorithms for the rational Hermite problem so I won't be considering them for now (but might include them in a later edit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, taking the conditions in Qiaochu's answer, here is how one builds a curve with a node at $(-1,0)$, a tacnode at $(0,0)$, and a cusp at $(1,0)$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Expand[InterpolatingPolynomial[{{-1, {0, 0, 1}}, {0, {0, 0, 0, 0, 1}}, {1, {0, 0, 0, 1}}}, x]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result has fractional coefficients, but you can multiply with an appropriate factor so that all the coefficients are integers, resulting in the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x^{11}+2x^{10}-15x^9+21x^7-6x^6-9x^5+4x^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here for instance is a plot of $$y^2=3x^{11}+2x^{10}-15x^9+21x^7-6x^6-9x^5+4x^4$$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n0bhG.png&quot; alt=&quot;monster #1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a more complicated curve, $$y^2-y^3=(x^2+xy+y^2)(3x^{11}+2x^{10}-15x^9+21x^7-6x^6-9x^5+4x^4)$$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/hBmL8.png&quot; alt=&quot;monster #2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to make the prescription for getting an isolated point to work, since the curves generated only manage to have separate branches passing through the desired point, but no isolated points at all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for T..'s suggestion, the &lt;em&gt;Mathematica&lt;/em&gt; code I have requires some serious cleanup, so I shall be editing this answer later to include his parametric construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I am trying to find a curve with four-fold symmetry that has four of each of the types of double points. If I manage to find it, I shall be naming it after Qiaochu and T.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T05:35:48.277" CommunityOwnedDate="2010-10-25T05:35:48.277" CommentCount="1" />
  <row Id="7771" PostTypeId="2" ParentId="7769" CreationDate="2010-10-25T05:51:52.290" Score="5" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take the derivative. This is straightforward here using the chain rule:&#xA;$$\frac{14x + 2}{2\sqrt{7x^2 + 2x - 2}} = \frac{7x+1}{\sqrt{7x^2 + 2x - 2}}.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the points in the domain of $f(x)$ where the derivative does not exist or is equal to zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint the first&lt;/strong&gt;: There are two points that are in the domain of $f(x)$ where the derivative does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint the second&lt;/strong&gt;: A fraction is equal to zero if and only if the numerator is $0$ and the denominator is not zero.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T14:33:25.540" LastActivityDate="2010-10-25T14:33:25.540" />
  <row Id="7772" PostTypeId="1" AcceptedAnswerId="7784" CreationDate="2010-10-25T06:33:36.910" Score="-2" ViewCount="80" Body="&lt;p&gt;A friend of mine who is quite an aggressive Nominalist told me the other day:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Mathematics and numbers are arbitrary; they can accurately predict physical systems in real life only because they are consistent, but any consistent framework will do.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to what other examples of &quot;consistent frameworks&quot; he is talking about.  Perhaps there are alternative definitions of addition?  Of subtraction, or multiplication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different things you can do to numbers that have a &quot;mapping&quot; (isomorphic) to real-world, physical phenomenon?  (Such as mathematical addition isomorphically mapping to, say, putting together two groups of apples, or differentiation of position mapping to velocity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any examples of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might not be; this is somewhat of a vague question and it might end up being very stupid.  But I'm interested in what other completely different frameworks of &quot;math&quot; might exist that are meaningful in describing/modeling physical events.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-10-25T08:24:18.020" LastActivityDate="2010-10-25T08:58:21.333" Title="Alternative, consistent frameworks of mathematics with isomorphic mappings to physical phenomenon" Tags="&lt;abstract-algebra&gt;&lt;math-history&gt;&lt;philosophy&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7773" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T06:35:08.780" Score="6" ViewCount="0" Body="&lt;p&gt;Another way is to consider the following arrangement with $n$ rows and $n+1$ columns&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{pmatrix}&#xA;{n \choose 0} &amp;amp; {n \choose 1} &amp;amp; \dots &amp;amp; {n \choose n} \\&#xA;{n \choose 0} &amp;amp; {n \choose 1} &amp;amp; \dots &amp;amp; {n \choose n} \\&#xA;\vdots &amp;amp; \vdots &amp;amp; \dots &amp;amp; \vdots \\&#xA;{n \choose 0} &amp;amp; {n \choose 1} &amp;amp; \dots &amp;amp; {n \choose n} \\&#xA;\end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we want is the sum of the last $n$ elements of row 1, plus the sum of last $n-1$ elements of row 2 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is same as the sum of the first $n$ elements of row $n$, plus the sum of the first $n-1$ elements of row $n-1$ etc, as ${n \choose r} = {n \choose n-r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the required sum is half the sum of all elements, which is $n2^{n-1}$, as the sum of each row is $2^n$ and there are $n$ rows.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-25T06:35:08.780" />
  <row Id="7774" PostTypeId="1" CreationDate="2010-10-25T06:40:06.330" Score="7" ViewCount="212" Body="&lt;p&gt;Exercise 1.14 of &lt;a href=&quot;http://books.google.ru/books?id=SMiB8VIB5UIC&amp;amp;printsec=frontcover&amp;amp;dq=Rordam+Larsen+Laustsen&amp;amp;source=bl&amp;amp;ots=o_tWQGf66z&amp;amp;sig=IfsFtbHbCz3d-wZc3q7O2ZXQulo&amp;amp;hl=en&amp;amp;ei=CyDFTPPlLMPrOceP0e8L&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the book&lt;/a&gt; Rordam, Larsen and Laustsen &quot;An introduction to K-theory for C*-algebras&quot; asks to prove, that upper triangular matrix with elements from some C*-algebra $A$ is invertible in $M_n(A)$ iff all diagonal entries are invertible in $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to solve this I've found that if $a$ is invertible and $\delta$ is such that $(a^{-1}\delta)^n=0$ then $a+\delta$ is invertible too and its inverse is given by $(a+\delta)^{-1}=\sum_{k=0}^{n} (-a^{-1}\delta)^ka^{-1}$. Using this fact I can show that if diagonal is invertible, then upper-triangular matrix with this diagonal is invertible too, and also that if upper-triangular matrix has upper-triangular inverse, then its diagonal is invertible. So all I need to prove is that if upper-triangular matrix invertible, then its inverse is upper-triangular. I've failed to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there is a hint for this exercise: &quot;Solve the equation $ab=1$ where $a$ is as above [i.e. upper-triangular matrix] and where $b$ is unknown upper triangular matrix&quot;. Solution of this equation follows from my reasoning above, but this doesn't help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update (counterexample attempt):&lt;/strong&gt; I've made one more attempt and it looks for me like I have found a counterexample. However I think there is a mistake in it (because otherwise there is a mistake in &lt;a href=&quot;http://books.google.ru/books?id=SMiB8VIB5UIC&amp;amp;printsec=frontcover&amp;amp;dq=Rordam+Larsen+Laustsen&amp;amp;source=bl&amp;amp;ots=o_tWQGf66z&amp;amp;sig=IfsFtbHbCz3d-wZc3q7O2ZXQulo&amp;amp;hl=en&amp;amp;ei=CyDFTPPlLMPrOceP0e8L&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the book&lt;/a&gt;). Here it is. Let $A=B(l^2(\mathbb{N}))$ --- algebra of bounded operators on sequences $x=\{x_i\}_ {i=1}^ \infty:\|x\|^2=\sum_{x=1}^{\infty}|x_i|^2&amp;lt;\infty$. Let $z\in A$ be defined by $(zx)_ {2n-1}=0$, $(zx)_{2n}=x_n$, and $t\in A$ be defined by $(tx)_{2n-1}=x_n$, $(tx)_ {2n}=0$. Then we have $t^*t=z^ * z=tt^* +zz^* =1$, $t^* z=z^* t=0$. From these we have that&#xA;$$\begin{pmatrix}z&amp;amp;tz^* \\0&amp;amp;t^* \end{pmatrix}\begin{pmatrix}z^* &amp;amp;0\\zt^* &amp;amp;t\end{pmatrix}=\begin{pmatrix}1&amp;amp;0\\0&amp;amp;1\end{pmatrix}$$ and&#xA;$$\begin{pmatrix}z^* &amp;amp;0\\zt^* &amp;amp;t\end{pmatrix}\begin{pmatrix}z&amp;amp;tz^* \\0&amp;amp;t^* \end{pmatrix}=\begin{pmatrix}1&amp;amp;0\\0&amp;amp;1\end{pmatrix}.$$ &#xA;So now my question should say &quot;Where am I wrong?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-25T09:28:31.007" LastActivityDate="2010-10-25T09:28:31.007" Title="Upper-triangular matrix is invertible iff its diagonal is invertible: C*-algebra case." Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;matrices&gt;&lt;c-star-algebras&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="7775" PostTypeId="1" CreationDate="2010-10-25T06:49:27.110" Score="7" ViewCount="118" Body="&lt;p&gt;Consider the scheme of random placing balls into N=1000 cells. We continue the procedure of placing balls as long as a last cell remains empty. The process terminates when a ball is placed into this cell. At this moment several cells (or a certain cell) contain(s) a maximum number of balls among all cells. What is the expectation of this maximum?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an application of this scheme consider N people who enter a lottery game. Each raffle is equivalent to the random placing of a ball into N cells. We take a look at this process as long as each person has won at least once.&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-10-25T16:13:26.320" Title="An occupancy problem" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7776" PostTypeId="1" CreationDate="2010-10-25T06:59:06.740" Score="1" ViewCount="63" Body="&lt;p&gt;I've noticed a type of sum for the number of solutions of a type of quadratic form. For example, the number of solutions to $X^2-Y^2\equiv a\pmod{p}$ for $p$ an odd prime is given by the sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{Y=1}^p(1+\left(\frac{Y^2+a}{p}\right))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\left(\frac{Y^2+a}{p}\right)$ is the Legendre symbol. It seems that the trick is to isolate the $X^2$, add 1 to the RHS, and then let $Y$ range over a complete residue system. It seems that for each value $Y$ takes, there will either be 2 or 0 solutions added to the running total. Is there a simple explanation why this is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that for each $y$ that is a solution, we could take both $x$ and $-x$ to get two solutions $(x,y)$ and $(-x,y)$? My only concern is when $x\equiv -x\pmod{p}$, in which case we would have $x\equiv 0\pmod{p}$. Why does the formula still work in that case as well? Thanks for any explanation.&lt;/p&gt;&#xA;" OwnerUserId="2776" LastEditorUserId="2776" LastEditorDisplayName="" LastEditDate="2010-10-25T07:05:08.777" LastActivityDate="2010-10-25T14:42:39.507" Title="Formula for counting solutions to a quadratic form?" Tags="&lt;number-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7777" PostTypeId="2" ParentId="7776" CreationDate="2010-10-25T07:06:47.200" Score="5" ViewCount="0" Body="&lt;p&gt;This is a general (and well-known) trick. The number of solutions to the&#xA;congruence&#xA;$$x^2\equiv f(y)\pmod{p}$$&#xA;for a polynomial $f$ is&#xA;$$\sum_{y=0}^{p-1}\left[1+\left(\frac{f(y)}{p}\right)\right].$$&#xA;The reason for this is that the congruence $x^2\equiv a$ (mod $p$)&#xA;always has $1+(x/p)$ solutions. Note that when $a\equiv0$ (mod $p$)&#xA;the only solution to $x^2\equiv0$ (mod $p$) is $x\equiv0$ (mod $p$)&#xA;but this is in accord with $(0/p)=0$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-25T08:36:40.903" LastActivityDate="2010-10-25T08:36:40.903" CommentCount="2" />
  <row Id="7778" PostTypeId="1" AcceptedAnswerId="7782" CreationDate="2010-10-25T08:25:48.333" Score="3" ViewCount="328" Body="&lt;p&gt;How to prove this  binomial identity  :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side arises while solving a standard binomial problem the right hand side is given in the solution , I checked using induction that this is true but I am inquisitive to prove it in a rather general way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I have gone through all of the answers posted here,I particularly liked &lt;a href=&quot;http://math.stackexchange.com/users/72/isaac&quot;&gt;Isaac♦&lt;/a&gt;&#xA;answers after which it was not much difficult for me to figure out something  i would rather say an easy and straight algebraic proof, I am posting it here if somebody needs in future :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do always welcome your comments :-)&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-25T19:16:53.647" LastActivityDate="2010-10-26T04:55:39.623" Title="Proving $ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $ " Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="7779" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T08:34:52.840" Score="7" ViewCount="0" Body="&lt;p&gt;If you have proved it via induction, the you have &lt;strong&gt;proved&lt;/strong&gt; it.&#xA;I don't know what you mean by &quot;in a rather general way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note that&#xA;$$1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$$&#xA;is just crying out to be rewritten as&#xA;$$\frac{1\cdot2\cdot3\cdot\cdots\cdot(2n)}&#xA;{2\cdot4\cdot6\cdot\cdots\cdot(2n)}.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-25T08:34:52.840" CommentCount="2" />
  <row Id="7780" PostTypeId="1" AcceptedAnswerId="7785" CreationDate="2010-10-25T08:36:06.483" Score="0" ViewCount="98" Body="&lt;p&gt;i have an object and a 3d direction vector and position for it . I would like to know how do i determine if a certain point X is in the space below the plan determined by my direction ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image that i have drawn to make it more clear . In this image i've made the vector 2d &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bgAYJ.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please Help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, &#xA;Aleandru Badescu&lt;/p&gt;&#xA;" OwnerUserId="2777" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-25T08:41:47.130" LastActivityDate="2010-10-25T09:55:49.553" Title="vector question" Tags="&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="7781" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T08:36:51.850" Score="1" ViewCount="0" Body="&lt;p&gt;Multiply right side by $\frac{n!}{n!}$, and distribute the $2^n$.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-25T08:36:51.850" />
  <row Id="7782" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T08:39:09.597" Score="4" ViewCount="0" Body="&lt;p&gt;Your original identity, ${ 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$, can be rewritten (by multiplying both sides by $(n!)^2$) as $(2n)!=2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!$.  Now, $2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!=$ $1\cdot 3\cdot 5\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n=$ $1\cdot 2\cdot 3\cdots (2n-1)(2n)=(2n)!$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-25T08:39:09.597" CommentCount="1" />
  <row Id="7783" PostTypeId="1" AcceptedAnswerId="7852" CreationDate="2010-10-25T08:53:56.213" Score="10" ViewCount="439" Body="&lt;p&gt;Much effort has been expended on a famous unsolved problem about the [Riemann Zeta function $ \zeta(s) $][1].  Not surprisingly, it&amp;#39;s called the Riemann hypothesis, which asserts:&#xD;&#xA;&#xD;&#xA;$$ \zeta(s) = 0 \Rightarrow \text{Re } s = 1/2 \text{ or Im } s = 0. $$&#xD;&#xA;&#xD;&#xA;Now there are numerical methods for approximating $ \zeta(s) $, but as I understand, no one knows any exact values except at even integers, which include the trivial zeroes for which $ \text{Im } s = 0 $.  So I&amp;#39;ve always wondered:  *For the rest, how does one prove $ \text{Re } s = 1/2 $ &lt;strong&gt;exactly&lt;/strong&gt;?*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All I know that seems vaguely useful is the argument principle, which I'm not sure helps here, but I'd be happy to learn about other techniques that aren't too far advanced.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:  Looking over this again, I found out I missed the closed-form values at &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values&quot; rel=&quot;nofollow&quot;&gt;odd negative integers&lt;/a&gt;.  This doesn't affect the question but seemed worth correcting.  Thanks to the people who contributed.&lt;/p&gt;&#xA;" OwnerUserId="336" LastEditorUserId="336" LastEditorDisplayName="" LastEditDate="2011-01-03T00:36:50.637" LastActivityDate="2011-01-03T01:30:23.843" Title="Proving a known zero of the Riemann Zeta has real part exactly 1/2" Tags="&lt;complex-analysis&gt;&lt;special-functions&gt;&lt;riemann-zeta&gt;&lt;riemann-hypothesis&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="7784" PostTypeId="2" ParentId="7772" CreationDate="2010-10-25T08:58:21.333" Score="4" ViewCount="0" Body="&lt;p&gt;Philosophers of a nominalistic/constructivist bent often come out with&#xA;stuff like this, asserting that maths, physics or whatever is some&#xA;sort of arbitrary, contingent system which could have been totally&#xA;different if history had turned out differently. What these&#xA;philosophers are not very good at doing is proposing credible alternative&#xA;systems. Of course, the burden of proposing alternative&#xA;&quot;constructive frameworks&quot; should be on your chum. Only he can say&#xA;what he has in mind (if anything), not us.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-25T08:58:21.333" CommunityOwnedDate="2010-10-25T08:58:21.333" />
  <row Id="7785" PostTypeId="2" ParentId="7780" CreationDate="2010-10-25T09:55:49.553" Score="1" ViewCount="0" Body="&lt;p&gt;You have a point $x$, vector $v$, and also some vertical vector, pointing down. Lets call it $g$. If I have understood your problem correctly, your plane contains $(0,0,0)$ (or otherwise you can write $x$ in the coordinate system with origin on the plane), vector $v$, and a vector $w=v\times g$ --- horizontal vector, orthogonal to $v$. I want to interpret $x$ as a vector (from the origin to the point $x$). Then for some $\alpha\in\mathbb{R}$ vector $x-\alpha g$ is on the plane and $x$ is below the plane if and only if $\alpha&gt;0$. We have&#xA;$$(x-\alpha g, v\times w)=0,$$&#xA;$$(x-\alpha g, v\times[v\times g])=0,$$&#xA;$$(x-\alpha g, v(v,g)-g(v,v))=0,$$&#xA;$$(x, v)(v,g)-(x,g)(v,v)=\alpha\bigl((g,v)(g,v)-(g,g)(v,v)\bigr),$$&#xA;$$\alpha=\bigl((x, v)(v,g)-(x,g)(v,v)\bigr)/\bigl((g,v)(g,v)-(g,g)(v,v)\bigr).$$&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-25T09:55:49.553" CommentCount="9" />
  <row Id="7786" PostTypeId="1" CreationDate="2010-10-25T10:14:26.367" Score="0" ViewCount="46" Body="&lt;p&gt;When dealing with Dirichlet Processes, according to [Teh, 2007], a DP is defined as by a base Probability H and a scale factor &quot;alpha&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Stick Breaking Construction, the random draws G from a DP:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G~DP(alpha,H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G=sum(pi_k*delta_theta_k) over k from 1 to infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pi_k are ordered draws from a Beta Distribution given the length of an unitary stick&lt;/p&gt;&#xA;&#xA;&lt;p&gt;delta_theta_k is a point mass centered in &quot;theta_k&quot; (theta_k are random draws from the base distribution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have pretty much a clear understanding of all the variables, but I do not know what do they mean by &quot;mass point&quot; is it the probability density of that draw, or is it something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if you could point me in any direction, only a reference would be amazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2010-10-25T10:14:26.367" Title="Mass Point, Dirac Delta in Dirichlet Processes " Tags="&lt;statistics&gt;" CommentCount="1" />
  <row Id="7787" PostTypeId="2" ParentId="7783" CreationDate="2010-10-25T10:32:27.827" Score="12" ViewCount="0" Body="&lt;p&gt;From what I've found, it's mostly presented as a &lt;em&gt;fait accompli&lt;/em&gt;, for instance, the &lt;a href=&quot;http://mathworld.wolfram.com/Riemann-SiegelFunctions.html&quot; rel=&quot;nofollow&quot;&gt;Riemann-Siegel function $Z(t)$&lt;/a&gt;, a function frequently used in the zero-finding, is defined with a $\zeta\left(\frac12+it\right)$ factor. What now one looks for when using constructs like these is if the function ever &quot;turns away&quot; from the horizontal axis without crossing it (if this does happen, the &lt;strong&gt;hypothesis&lt;/strong&gt; is false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why things like &lt;a href=&quot;http://mathworld.wolfram.com/LehmersPhenomenon.html&quot; rel=&quot;nofollow&quot;&gt;Lehmer's phenomenon&lt;/a&gt; are interesting behavior for the ζ function. Put simply, these are sections of the critical strip where there is (very!) nearly no crossing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two traditional ways of visualizing the Lehmer phenomenon: one can look at the graph of the Riemann-Siegel function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l1dHQ.png&quot; alt=&quot;Riemann-Siegel Z(t)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;Plot[RiemannSiegelZ[z], {z, 0, 100}, AspectRatio -&amp;gt; 1/5, Frame -&amp;gt; True]&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the so-called &quot;zeta spiral&quot; $z(t)=\zeta\left(\frac12+it\right)$ in the complex plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tOoxm.png&quot; alt=&quot;zeta spiral&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;ParametricPlot[Through[{Re, Im}[Zeta[1/2 + I t]]], {t, 0, 40}, AspectRatio -&amp;gt; Automatic, Frame -&amp;gt; True]&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is the first instance of the Lehmer phenomenon, seen using both viewpoints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n7rFh.png&quot; alt=&quot;Lehmer for Riemann-Siegel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;Plot[RiemannSiegelZ[z], {z, 7004, 7006}, Frame -&amp;gt; True, PlotRange -&amp;gt; {-2, 2}]&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JUxe8.png&quot; alt=&quot;Lehmer for zeta spiral&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Mathematica&lt;/em&gt; code: &lt;code&gt;ParametricPlot[Through[{Re, Im}[Zeta[1/2 + I t]]], {t, 7004 + 1/2, 7005 + 1/2}, AspectRatio -&amp;gt; Automatic, Frame -&amp;gt; True]&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;hypothesis&lt;/strong&gt; would have been false if for the Riemann-Siegel function, the function displayed a local extremum without crossing the x-axis, or for the zeta spiral, the apparent &quot;cusp&quot; was actually a cusp or did not even pass through the origin. (I won't spoil the fun by posting zoomed-in versions of those last two images, you can do it yourself with &lt;em&gt;Mathematica&lt;/em&gt; or some other computing environment that can evaluate ζ(s) for complex values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two zeroes are in fact quite close: &lt;code&gt;FindRoot[RiemannSiegelZ[x], {x, ##}, WorkingPrecision -&amp;gt; 20] &amp;amp; @@@ {{7005 + 1/50, 7005 + 7/100}, {7005 + 9/10, 7005 + 11/100}}&lt;/code&gt; returns the two approximate zeroes &lt;code&gt;7005.0628661749205932&lt;/code&gt; and &lt;code&gt;7005.1005646726748389&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T10:32:27.827" CommentCount="2" />
  <row Id="7788" PostTypeId="1" AcceptedAnswerId="7790" CreationDate="2010-10-25T10:43:00.373" Score="3" ViewCount="113" Body="&lt;p&gt;As far as i know LU decomposition allow you to calculate matrix determinant following easy and cheap formula: Det[A] = Det[L] Det[U] = Det[U]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying this out in Mathematica 7 gives me correct result by absolute value, i.e. it ignores negative determinants and transforms them to positive ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;matrixA = {{-1, 2, 3, 5}, {-7, -4, 5, 4}, {-89, 7, 8, -6}, {8, 6, -1, 4}};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Det[matrixA]&lt;/code&gt; gives out -2067    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{lu, p, c} = LUDecomposition[matrixA]&#xA;&#xA;u = lu SparseArray[{i_, j_} /; j &amp;gt;= i -&amp;gt; 1, {4, 4}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Det[u]&lt;/code&gt; is 2067&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the question is obvious - how to get correct determinant in Mathematica using LU decomposition?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-25T11:12:47.777" LastActivityDate="2010-10-25T11:12:47.777" Title="Calculating determinant from LUDecomposition" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;mathematica&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7789" PostTypeId="2" ParentId="7788" CreationDate="2010-10-25T11:00:20.960" Score="3" ViewCount="0" Body="&lt;p&gt;From the documentation: &quot;Therefore multiplying L and U recreates the original matrix permuted with the row pivot vector.&quot; So you need to multiply by &lt;code&gt;Signature[p]&lt;/code&gt; to get the correct sign.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-25T11:00:20.960" CommentCount="2" />
  <row Id="7790" PostTypeId="2" ParentId="7788" CreationDate="2010-10-25T11:01:03.643" Score="7" ViewCount="0" Body="&lt;p&gt;Well, this is because you forgot to take into account the &lt;em&gt;permutation matrix&lt;/em&gt; that is output by the LU decomposition routine. You have to remember two facts: 1.) Gaussian elimination generally performs row interchanges for numerical stability reasons, and 2.) the determinant of a matrix changes sign if you interchange rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Mathematica&lt;/em&gt; at least, the function that will rescue you is &lt;code&gt;Signature[]&lt;/code&gt;, which gives the signature of the permutation required to turn the permutation matrix (which &lt;em&gt;Mathematica&lt;/em&gt; outputs as a scrambled list of the numbers from 1 to n, where n is the size of your matrix) into the identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to use your example in &lt;em&gt;Mathematica&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m={{-1, 2, 3, 5}, {-7, -4, 5, 4}, {-89, 7, 8, -6}, {8, 6, -1, 4}};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;we compare&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Det[m]&#xA;-2067&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{lu, p, c} = LUDecomposition[m]&#xA;{{{-1, 2, 3, 5}, {7, -18, -16, -31}, {-8, -11/9, 31/9, 55/9}, {89, 19/2, -963/31, 2067/62}}, {1, 2, 4, 3}, 1}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;code&gt;LUDecomposition[]&lt;/code&gt; here outputs three things: the merged $\mathbf{L}$ and $\mathbf{U}$ matrices, the permutation, and the condition number. We can get the tentative determinant by multiplying together the diagonal elements of &lt;code&gt;lu&lt;/code&gt;, thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Tr[lu, Times]&#xA;2067&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is where &lt;code&gt;Signature&lt;/code&gt; comes in:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Signature[{1, 2, 4, 3}]&#xA;-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that to turn &lt;code&gt;{1, 2, 4, 3}&lt;/code&gt; into &lt;code&gt;{1, 2, 3, 4}&lt;/code&gt;, one needs one swap; namely, to swap the third and fourth elements. Since 1 is odd, the signature is -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Signature[p]*Tr[lu, Times]&#xA;-2067&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-25T11:01:03.643" CommentCount="1" />
  <row Id="7792" PostTypeId="1" AcceptedAnswerId="7794" CreationDate="2010-10-25T11:12:11.570" Score="3" ViewCount="84" Body="&lt;p&gt;One of the most famous 2D-curves are &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_spiral&quot; rel=&quot;nofollow&quot;&gt;logarithmic spirals&lt;/a&gt; (or Spira mirabilis). They can be constructed by using a machinery that ensures a constant angle between the tangent and the radial lines all the time while plotting it.&lt;br&gt;&lt;br&gt; &#xA;&lt;img src=&quot;http://i.imgur.com/yP6mK.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt;&#xA;I can see it in the picture that the spiral arms are getting bigger each turn and I see the math. What I don't understand intuitively is where the logarithm comes from - or put differently: Why does a geometric progression arise just by holding an angle constant. Could you give me some hints? Thank you,&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-25T15:39:13.173" Title="Why does an equiangular spiral become logarithmic (intuitively)?" Tags="&lt;intuition&gt;&lt;plane-curves&gt;&lt;logarithms&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7793" PostTypeId="1" AcceptedAnswerId="7804" CreationDate="2010-10-25T11:43:26.380" Score="1" ViewCount="138" Body="&lt;p&gt;If $x \geq 2$, then how do we prove that $$\int\limits_{2}^{x} \frac{dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-25T14:08:00.517" Title="Dominating the integral $\int\limits_{2}^{x} \frac{dt}{\log^{n}{t}}$" Tags="&lt;number-theory&gt;&lt;analysis&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" />
  <row Id="7794" PostTypeId="2" ParentId="7792" CreationDate="2010-10-25T11:44:03.493" Score="2" ViewCount="0" Body="&lt;p&gt;Much is explained by looking at the polar equation of the spiral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\exp(\theta\cot\alpha)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\alpha$ is the constant angle any tangent to the curve makes with the radius vector (a line segment joining the origin and the point of tangency). This explains the adjective &lt;em&gt;equiangular&lt;/em&gt; (the verification of this property from the defining equation is left as an exercise). As an aside, insects flying towards a point light source like a candle or a light bulb follow the path of an equiangular spiral, since the usual strategy of an insect flying at the daytime to get their bearing is to fly at a constant angle from the sun's rays, and this strategy works against them when encountering man-made light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we have an arithmetic progression of angles $\theta,\theta+\Delta\theta,\theta+2\Delta\theta,\dots$; if we get the corresponding values of the radius vector using the defining equation for the logarithmic spiral (geometrically speaking, this corresponds to a clockwise rotation by $0,\Delta\theta,2\Delta\theta,\dots$ radians), we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp(\theta\cot\alpha),\exp((\theta+\Delta\theta)\cot\alpha),\exp((\theta+2\Delta\theta)\cot\alpha),\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be re-expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp(\theta\cot\alpha),\exp(\theta\cot\alpha)\cdot\exp(\Delta\theta\cot\alpha),\exp(\theta\cot\alpha)\cdot\exp(\Delta\theta\cot\alpha)^2,\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which as you can see is a &lt;em&gt;geometric&lt;/em&gt; progression; that is to say, the &lt;em&gt;logarithms&lt;/em&gt; of the members of this sequence form an arithmetic progression. This is where the &lt;em&gt;logarithmic&lt;/em&gt; adjective arises from.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-25T15:39:13.173" LastActivityDate="2010-10-25T15:39:13.173" CommentCount="2" />
  <row Id="7795" PostTypeId="1" CreationDate="2010-10-25T11:52:04.000" Score="0" ViewCount="238" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;57^46 divide by 17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is best and quickest way to solve such kind of problem manually? Approximate answer will do&lt;/p&gt;&#xA;" OwnerUserId="2784" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-25T11:54:44.200" LastActivityDate="2010-10-25T15:01:22.637" Title="what is 57^46 divide by 17 ??" Tags="&lt;arithmetic&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="7796" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T11:55:47.700" Score="2" ViewCount="0" Body="&lt;p&gt;Euler's theorem gives you the remainder.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-25T11:55:47.700" />
  <row Id="7797" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T12:04:03.937" Score="2" ViewCount="0" Body="&lt;p&gt;You can do it by Fermats little theorem as well.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-25T12:04:03.937" />
  <row Id="7798" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T12:17:40.397" Score="4" ViewCount="0" Body="&lt;p&gt;It really depends what you mean by &quot;approximate answer&quot;. Since you tagged arithmetic...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice 57 = 3 . 19, you can say that $57^{46} / 17 = 3 \cdot 57^{45} \cdot \frac{19}{17}$. So if you know what $57^{45}$ is, you can approximate to within 2% if you multiply that by 3.3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, compute by hand $57^5 = 601692057$, which is less than 0.3% off from $6\cdot 10^8$. Observe that $57^{45} = (57^5)^9$. Using the binomial theorem you see that you can approximate that by $(6\cdot 10^8)^9 = 10 077 696 \cdot 10^{72}$ to within 3%. Rounding off the lower digits won't matter much in the error, so you have $57^{45} \sim 10^{79}$ to 3%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting it all together you have that to within 5%, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 57^{46} \sim 3.3 \cdot 10^{79} $$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-25T12:17:40.397" CommentCount="2" />
  <row Id="7799" PostTypeId="2" ParentId="7738" CreationDate="2010-10-25T12:20:09.047" Score="1" ViewCount="0" Body="&lt;p&gt;$\tilde P &amp;lt;&amp;lt; P$ is true but $P &amp;lt;&amp;lt; \tilde P$ is false. We have&#xA;$$\frac{d\tilde P}{dP}(w)=\sqrt{2\pi}\frac{e^{-w} 1_{[0,\infty)}}{e^{-w^2/2}}$$&#xA;Proving these facts is an exercise that you should really do on your own, as it is not that hard and you would not learn anything if you were just told the proof.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-10-25T12:20:09.047" CommentCount="0" />
  <row Id="7800" PostTypeId="2" ParentId="7793" CreationDate="2010-10-25T12:33:10.047" Score="5" ViewCount="0" Body="&lt;p&gt;Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dx} \frac{x}{\log^n(x)} = \frac{1}{\log^n(x)} - \frac{n}{\log^{n+1}(x)} = \frac{1}{\log^n(x)}\left( 1 - \frac{n}{\log x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sufficiently large $x$, the RHS is bigger than $\frac12 \frac{1}{\log^n(x)}$, and hence we get the asymptotic bound we need. The usual arguments then gives you a sufficiently large constant for the Big O. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-25T12:33:10.047" />
  <row Id="7801" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T12:44:31.223" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm \mod\ 17:\ \ 57\equiv 6,\ \ 6^{46}\equiv 1/6^2,\ \ 1/6\equiv 3\ \ $ which immediately yields the result&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-25T13:10:19.407" LastActivityDate="2010-10-25T13:10:19.407" />
  <row Id="7802" PostTypeId="2" ParentId="7778" CreationDate="2010-10-25T13:51:13.647" Score="4" ViewCount="0" Body="&lt;p&gt;Here's the more general way that you seek. The LHS and RHS are both products of rational functions of $\rm\:n\:$, i.e. they satisfy $\rm\ f(n+1) = r(n)\ f(n)\ $ for some rational function $\rm r(n)\:$. To test whether two such functions are equal you need only test that they have the same value of $\rm\ r(n) = f(n+1)/f(n)\ $ and the same initial condition $\rm\: f(0)$.$\ $ For the above it's trivial: $\rm\ f(0) = 1\:$ and the common value of $\rm\ r(n)\ $ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\displaystyle f(n)\ \ =\  \ {2n \choose n}\ \: =\: \ \frac{(2n)!}{n!^2}\ \ \Rightarrow\ \frac{f(n+1)}{f(n)}\ =\ \frac{2\:(n+1)\:(2n+1)}{\ (n+1)\ (n+1)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\displaystyle f(n)\ =\ 2^n \frac{ 1 \cdot 3  \cdots (2n-1)}{n!} \ \Rightarrow\ \frac{f(n+1)}{f(n)}\ =\ \frac{2\ (2n+1)}{n+1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this method is completely algorithmic - it requires no insight whatsoever. In essence one is employing the &lt;strong&gt;uniqueness theorem&lt;/strong&gt; for first order difference equations  (recurrences) to prove an equality by simply checking that both sides satisfy the same difference / differential equation and initial conditions. Such techniques generalize to much wider classes of multivariate mixed differential - difference equations, e.g.  &lt;strong&gt;holonomic&lt;/strong&gt; functions. For some further examples of this powerful technique see my &lt;a href=&quot;http://math.stackexchange.com/questions/3514&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-26T00:52:42.443" LastActivityDate="2010-10-26T00:52:42.443" CommentCount="7" />
  <row Id="7803" PostTypeId="2" ParentId="7775" CreationDate="2010-10-25T13:56:13.540" Score="2" ViewCount="0" Body="&lt;p&gt;You can get an approximate answer.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupon_collector%27s_problem&quot; rel=&quot;nofollow&quot;&gt;coupon collector's problem&lt;/a&gt; tells you that the expected number of balls is $n*H_n$ or about 7485 for n=1000.  So the average bin will have 7.4 balls.  If you look at the Poisson distribution for 7.485 and find the number where it falls to 1/1000.  I make that 17 or 18.  The motivation is that you have 999 bins with an average occupancy of 7.485.  The highest will be at a probability of about 1/999&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T15:13:22.670" LastActivityDate="2010-10-25T15:13:22.670" CommentCount="2" />
  <row Id="7804" PostTypeId="2" ParentId="7793" CreationDate="2010-10-25T14:08:00.517" Score="7" ViewCount="0" Body="&lt;p&gt;Here is a way to prove it without knowing the answer before hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c &gt; 2$ be a number such that $c &gt; e^{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{c}^{x} \frac{dt}{\log^n t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating by parts, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{c}^{x} \frac{dt}{\log^n t} =  \frac{x}{\log^n x} - K + n \int_{c}^{x} \frac{dt}{\log^{n+1} t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$n\int_{c}^{x} \frac{dt}{\log^{n+1} t} \le \frac{n}{n+1} \int_{c}^{x} \frac{dt}{\log^n t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $\log t \ge n+1$ for $t \ge c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\frac{1}{n+1}\int_{c}^{x} \frac{dt}{\log^n t} &amp;lt; \frac{x}{\log^n x}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{2}^{x} \frac{dt}{\log^n t} = \theta(\frac{x}{\log^n x})$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-25T14:08:00.517" CommentCount="9" />
  <row Id="7805" PostTypeId="2" ParentId="7795" CreationDate="2010-10-25T14:28:50.060" Score="1" ViewCount="0" Body="&lt;p&gt;Why not simply do the computation? The most time-consuming part is to compute the $46$th power of $57$, but that can be done quite fast using &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;repeated squaring&lt;/a&gt; in 6 multiplications. That results in a number with 81 digits. Dividing by 17 using long division is faster, and it takes about 80 easy steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are done, you can check your result with &#xA;&lt;a href=&quot;http://www.google.com/search?q=57%5E46+divided+by+17&quot; rel=&quot;nofollow&quot;&gt;Google&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-25T15:01:22.637" LastActivityDate="2010-10-25T15:01:22.637" CommentCount="3" />
  <row Id="7806" PostTypeId="2" ParentId="7776" CreationDate="2010-10-25T14:42:39.507" Score="0" ViewCount="0" Body="&lt;p&gt;It's a composition of two obvious facts. First, over a finite ring $\rm\:R\:$, any formula that counts the roots of a univariate polynomial extends to bivariate polynomials since the number of roots of $\rm\:f(X,Y)\:$ is simply the sum of the number of roots of $\rm\:f(X,r)\:$ for all $\rm\ r\in R\:$. Second, for quadratic polynomials over some finite field $\rm\:\mathbb F_p\:$ with $\rm\:p\:$ odd, the problem of root counting reduces to testing if the discriminant is a square, hence there exists an elementary closed-form root counting formula expressed in terms of the Legendre symbol. Apply these two observations immediately yields the formula at hand.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-25T14:42:39.507" />
  <row Id="7807" PostTypeId="1" AcceptedAnswerId="7820" CreationDate="2010-10-25T14:44:25.123" Score="5" ViewCount="187" Body="&lt;p&gt;The coordinate transformation (due to Beukers, Calabi and Kolk)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{\sin u}{\cos v}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{\sin v}{\cos u}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;transforms the square domain $0\lt x\lt 1$ and $0\lt y\lt 1$ into the triangle domain $u,v&amp;gt;0,u+v&amp;lt;\pi /2$ (in &lt;em&gt;&lt;a href=&quot;http://books.google.pt/books?id=CuTpgZaLahkC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+from+the+book&amp;amp;source=bl&amp;amp;ots=swepaSdSxh&amp;amp;sig=j1gB1zxabJE8qxEgbFh1ig8RA3Q&amp;amp;hl=pt-PT&amp;amp;ei=Pnj2TKKcCZOKhQfb67yABg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDIQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs from the BOOK&lt;/a&gt;&lt;/em&gt; by M. Aigner and G. Ziegler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the invert transformation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u=\arccos \sqrt{\dfrac{1-x^{2}}{1-x^{2}y^{2}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=\arccos \sqrt{\dfrac{1-y^{2}}{1-x^{2}y^{2}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is easy to see that three of the vertices (although not belonging to the domain) are transformed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y)=(0,0)\mapsto (0,0)=(u,v),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y)=(1,0)\mapsto (\pi /2,0)=(u,v),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,y)=(0,1)\mapsto (0,\pi /2)=(u,v).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt; - But how is the fourth vertex $(x,y)=(1,1)$ transformed? In the plot below of $\dfrac{1-x^{2}}{1-x^{2}y^{2}}$ seems that the following limit does not exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{(x,y)\rightarrow (1,1)}{\lim }\sqrt{\dfrac{1-x^{2}}{1-x^{2}y^{2}}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0Um5U.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt; - As a second question I would like to know how can one &quot;discover&quot; a&#xA;transformation of a square into a triangle such as this. &lt;em&gt;Is there any systematic study of this kind of transformations?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-23T23:46:34.417" LastActivityDate="2010-12-23T23:46:34.417" Title="A limit and a coordinate trigonometric transformation of the interior points of a square into the interior points of a triangle" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;limit&gt;&lt;transformation&gt;&lt;multivariable-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7808" PostTypeId="2" ParentId="7775" CreationDate="2010-10-25T14:49:46.030" Score="0" ViewCount="0" Body="&lt;p&gt;To refine Ross' answer(+1), at the moment when the last (say, 7400) ball fills the last cell, the distribution of the remaining cells should approach that of 999 iid ZERO-TRUNCATED Poisson variables with media u=7399/999. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated: With media = 7.484, this gives $\lambda \approx 7.480 $&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-10-25T16:13:26.320" LastActivityDate="2010-10-25T16:13:26.320" CommentCount="1" />
  <row Id="7809" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T14:53:59.160" Score="1" ViewCount="0" Body="&lt;p&gt;Count all pairs $(x,S)$ where $S \subseteq \{1,2,\dots,n\}$ and $x \in S$ in two ways. Having chosen a $S$ to be an r-element subset, we have $r$ choices for $x$, giving us the left hand side, $\sum_{r \geq 1} r {n \choose r}$. Next having chosen an arbitrary element $x \in \{1,2,\dots,n\}$ we can choose $S$ in $2^{n-1}$ ways, and $x$ can be chosen in $n$ ways giving us the right hand side.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-10-25T14:53:59.160" CommentCount="1" />
  <row Id="7810" PostTypeId="2" ParentId="7757" CreationDate="2010-10-25T14:59:28.327" Score="6" ViewCount="0" Body="&lt;p&gt;There are already some good answers here, but you did say that you wanted something general, so I'll add the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are interested, for some function $f(k)$, in the binomial sum &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(n) = \sum_{k=0}^n \binom{n}{k} f(k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Your problem has $f(k) = k$.)  Then, taking $\Delta f(k) = f(k+1) - f(k)$, denote $A(n)$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n) = \sum_{k=0}^n \binom{n}{k} \Delta f(k).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few years ago I proved the following relationship between $B(n)$ and $A(n)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(n) = 2^n \left(f(0) + \sum_{k=1}^n \frac{A(k-1)}{2^k}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your problem, $f(k) = k$.  So $\Delta f(k) = k+1 - k = 1$.  If you're willing to take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(n) = \sum_{k=0}^n \binom{n}{k} = 2^n,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the formula gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^n \binom{n}{k} k = 2^n \sum_{k=1}^n \frac{2^{k-1}}{2^k} = 2^n \sum_{k=1}^n \frac{1}{2} = n 2^{n-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Even if you're not willing to take $\sum_{k=0}^n \binom{n}{k} = 2^n,$ you can derive that immediately from the expression for $B(n)$ by letting $f(k) = 1$, as $\Delta f(k) = 0$ in that case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for $B(n)$ is Theorem 4 in my paper &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/CombSum.pdf&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Sums and Finite Differences&lt;/a&gt;,&quot; &lt;em&gt;Discrete Mathematics&lt;/em&gt;, 307 (24): 3130-3146, 2007.  This identity is in the paper as well, as an illustration of Theorem 4.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-25T15:56:34.397" LastActivityDate="2010-10-25T15:56:34.397" />
  <row Id="7811" PostTypeId="1" AcceptedAnswerId="7813" CreationDate="2010-10-25T15:13:52.557" Score="7" ViewCount="280" Body="&lt;p&gt;Let $G$ be a group and let $G'$ be the derived subgroup, defined as the subgroup generated by the commutators of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an example of a finite group $G$ where not every element of $G'$ is a commutator? $G'$ is only generated by commutators, but with all of the properties of commutators (ie: what happens under conjugation, exponentation, etc) I can't think of an example.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-10-26T08:16:17.717" Title="Derived Subgroups and Commutators" Tags="&lt;group-theory&gt;" AnswerCount="4" />
  <row Id="7812" PostTypeId="1" AcceptedAnswerId="7817" CreationDate="2010-10-25T15:16:47.880" Score="1" ViewCount="122" Body="&lt;p&gt;I am trying to understand a simple calculation in &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;polar coordinates&lt;/a&gt; - and I am getting totally discombobulated (the original source can be found: &lt;a href=&quot;http://online.redwoods.cc.ca.us/instruct/darnold/CalcProj/Fall07/KableKoch/EquiangularPresentation.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;br&gt;&lt;br&gt;Please have a look at the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EulN4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions&lt;/strong&gt;&lt;br&gt;&lt;br&gt;(1) I don't understand how the length labeled $dr \over d \theta$ can possibly be the change of $r$ with respect to a change in $\theta$. &lt;br&gt;(2) It is then said that $tan \alpha = {r \over {dr \over d \theta}}$. But how can you calculate the tan-function if you don't even have a right angled triangle - and anyway, the formular would suggest that you calculate the tan of the angle that is not labeled in the pic (because $tan={opposite \over adjecent}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, if this is too elementary - but it really bothers me right now... Thank you!&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-25T16:16:09.133" Title="Difficulty with differentiation in polar coordinates" Tags="&lt;calculus&gt;&lt;polar-coordinates&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7813" PostTypeId="2" ParentId="7811" CreationDate="2010-10-25T15:23:23.380" Score="4" ViewCount="0" Body="&lt;p&gt;For a nice source of examples, see &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I. M. Isaacs: Commutators and the Commutator Subgroup. The American Mathematical Monthly. Vol. 84, No. 9 (Nov., 1977), pp. 720-722 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which you can get from &lt;a href=&quot;http://www.jstor.org/stable/2321253&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using GAP I find that the two examples of order 96, which is the minimal possible order, are the groups generated by the following two lists of permutations on $32$ elements.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[ (3,8,6)(4,7,5)(9,27,17)(10,28,18)(11,30,22)(12,29,21)(13,26,23)(14,25,24)(15,31,20)(16,32,19), &#xA;    (1,17,7,23)(2,18,8,24)(3,19,5,21)(4,20,6,22)(9,26,15,32)(10,25,16,31)(11,28,13,30)(12,27,14,29), &#xA;    (1,9,5,13)(2,10,6,14)(3,11,7,15)(4,12,8,16)(17,25,21,29)(18,26,22,30)(19,27,23,31)(20,28,24,32), &#xA;    (1,5)(2,6)(3,7)(4,8)(9,13)(10,14)(11,15)(12,16)(17,21)(18,22)(19,23)(20,24)(25,29)(26,30)(27,31)(28,32), &#xA;    (1,3)(2,4)(5,7)(6,8)(9,11)(10,12)(13,15)(14,16)(17,19)(18,20)(21,23)(22,24)(25,27)(26,28)(29,31)(30,32), &#xA;    (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32) ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[ (3,7,5)(4,8,6)(9,25,17)(10,26,18)(11,31,21)(12,32,22)(13,27,23)(14,28,24)(15,29,19)(16,30,20), &#xA;    (1,17)(2,18)(3,19)(4,20)(5,21)(6,22)(7,23)(8,24)(9,25)(10,26)(11,27)(12,28)(13,29)(14,30)(15,31)(16,32), &#xA;    (1,9)(2,10)(3,11)(4,12)(5,13)(6,14)(7,15)(8,16)(17,25)(18,26)(19,27)(20,28)(21,29)(22,30)(23,31)(24,32), &#xA;    (1,5,2,6)(3,8,4,7)(9,13,10,14)(11,16,12,15)(17,21,18,22)(19,24,20,23)(25,29,26,30)(27,32,28,31), &#xA;    (1,3,2,4)(5,7,6,8)(9,11,10,12)(13,15,14,16)(17,19,18,20)(21,23,22,24)(25,27,26,28)(29,31,30,32), &#xA;    (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32) ]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; I used the following very simple-minded code to find the groups&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;chk := function (g)&#xA;  local comms, sub;&#xA;  comms := Set(List(Cartesian(g, g), p -&amp;gt; p[1]*p[2]*p[1]^-1*p[2]^-1));&#xA;  sub := Set(DerivedSubgroup(g));&#xA;  return Size(comms) = Size(sub);    &#xA;end;;&#xA;examples := AllSmallGroups(96, chk, false);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This works because I remembered there were examples of order 96. To get the permutation representations I did, for example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GeneratorsOfGroup(Image(SmallerDegreePermutationRepresentation(Image(IsomorphismPermGroup(examples[1])))));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-25T16:43:36.557" LastActivityDate="2010-10-25T16:43:36.557" CommentCount="1" />
  <row Id="7814" PostTypeId="2" ParentId="7811" CreationDate="2010-10-25T15:29:34.433" Score="2" ViewCount="0" Body="&lt;p&gt;Keith Dennis has also done some work on this problem.  In the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dennis, R. K.(1-CRNL); Vaserstein, L. N.(1-PAS)&#xA;Commutators in linear groups.&#xA;K-Theory 2 (1989), no. 6, 761–767. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;they consider conditions under which all elements of the commutator subgroup are products of at most 2 commutators.  I seem to recall that Dennis did computer calculations to find the smallest group for which the commutator subgroup contains elements that are not commutators (and my recollection is that the size of the group was around 100 elements) but I can't see anything on-line about that.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-10-25T15:29:34.433" CommentCount="2" />
  <row Id="7815" PostTypeId="1" AcceptedAnswerId="7816" CreationDate="2010-10-25T15:41:22.503" Score="2" ViewCount="83" Body="&lt;p&gt;Let $Z$ be a countable set. Let $f_1,....,f_n$ be a collection of real functions over $Z$. Let $z_1,...,z_m,...$ be an enumeration of elements of $Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $V_1 = \{ (f_1(z),...,f_n(z)) | z \in Z\}$ a set of $n$ dimensional vectors.&#xA;Define $V_2 =\{ (f_i(z_1),...,f_i(z_m),..., )) | i \in\{ 1,...,n\} \}$ a set of infinite dimensional vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $span(V_i)$ be the span (linear span) of $V_i$. Is the linear dimension of $span(V_i)$ the same for $i=1,2$?&lt;/p&gt;&#xA;" OwnerUserId="2789" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T16:15:24.713" LastActivityDate="2010-10-25T16:15:24.713" Title="do two vector spaces have the same linear dimension?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7816" PostTypeId="2" ParentId="7815" CreationDate="2010-10-25T15:53:31.117" Score="1" ViewCount="0" Body="&lt;p&gt;Let me rephrase your question in a way that's less confusing.  Define a matrix $A$ such that $A_{ij} = f_i(z_j)$.  This matrix has finitely many rows but infinitely many columns.  Then the dimension of the span of $V_1$ is the row rank of this matrix, and the dimension of the span of $V_2$ is the column rank of this matrix.  And you want to know if they are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes.  Here's one proof: define $A_m$ to be $A$, but with all columns after the $m^{th}$ column chopped off, and let $a_m$ denote the column rank of $A_m$.  Then it's not hard to see that there exists $m$ such that $a_m$ is the column rank of $A$.  Now apply the theorem that row rank equals column rank for ordinary matrices to $A_m$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-25T15:53:31.117" />
  <row Id="7817" PostTypeId="2" ParentId="7812" CreationDate="2010-10-25T16:09:05.663" Score="2" ViewCount="0" Body="&lt;p&gt;Something is screwy with that picture. I think &lt;a href=&quot;http://beares.net/r/astro/galaxies/galaxies_docs/pitch_angle.htm&quot; rel=&quot;nofollow&quot;&gt;http://beares.net/r/astro/galaxies/galaxies_docs/pitch_angle.htm&lt;/a&gt; has a better picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dr$ should be represented by swinging $r_0$ over to the arm that is denoted by $r$, and looking at the length difference. The $\tan$ is just an approximation: in the infinitesimal limit, the arc that &quot;swings&quot; $r_0$ over to the arm denoted by $r$ approaches a straight line that is perpendicular to the arm at $r$. So $dr/(r\dtheta)$ is the slope, which is just $\tan \alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is sort of understandable. I don't have the software to draw a picture right now, if you need one to illustrate what I wrote in the previous paragraph, I can probably upload a picture in a few hours. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-25T16:09:05.663" CommentCount="3" />
  <row Id="7818" PostTypeId="2" ParentId="7812" CreationDate="2010-10-25T16:10:44.507" Score="2" ViewCount="0" Body="&lt;p&gt;(Willie beat me by a few seconds, but this is what I had written.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No wonder you're confused. That's a horrible picture! You should draw an arc of a circle ($\approx$ a line) at right angles to the side labeled $r_0$; this will have length $r d\theta$ and split the line labeled $r$  into two pieces of lenghts $r_0$ and $dr$ (approximately, for small $d\theta$). Then you have a small right triangle such that $\tan\alpha \approx (r d\theta)/dr$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-25T16:16:09.133" LastActivityDate="2010-10-25T16:16:09.133" />
  <row Id="7819" PostTypeId="2" ParentId="7811" CreationDate="2010-10-25T16:12:59.783" Score="1" ViewCount="0" Body="&lt;p&gt;The &quot;simplest&quot; example is the free group on two generators, but is perhaps not the easiest to play with unless you are comfortable with free groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice survey of the problem of commutators vs. commutators subgroups in finite groups is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Kappe, Luise-Charlotte, and Morse, Robert Fitzgerald. &lt;strong&gt;On commutators in groups.&lt;/strong&gt; &lt;em&gt;Groups St. Andrews 2005, Vol. 2&lt;/em&gt;, pp. 531-558. London Math. Soc. Lecture Notes Ser., &lt;strong&gt;340&lt;/strong&gt;, Cambridge University Press, 2007. MR 2331612 (2008g:20068)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The smallest groups in which the commutator subgroup does not equal the set of commutators have order 96; in fact, there are two non-isomorphic groups of order 96 in which the set of commutators does not equal the commutator subgroup. (This result also appeared in Robert Guralnick's 1977 thesis, &quot;Expressing group elements as products of commutators&quot;, UCLA). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kappe also has an upcoming joint paper in which she discusses the same question, restricted to $p$-groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting item is the Ore Conjecture, which stated that if $G$ is a finite nonabelian simple group, then the set of commutators equals the commutator subgroup. Its proof was recently completed, and it appears in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Liebeck, Martin W.; O'Brien, E. A.; Shalev, Aner; Tiep, Pham Huu. &lt;strong&gt;The Ore conjecture.&lt;/strong&gt; &lt;em&gt;J. Eur. Math. Soc. (JEMS)&lt;/em&gt; 12 (2010), no. 4, 939–1008. MR MR2654085. A PDF preprint can be found at &lt;a href=&quot;http://www.math.auckland.ac.nz/~obrien/research/ore.pdf&quot; rel=&quot;nofollow&quot;&gt;O'Brien's website&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T17:23:51.830" LastActivityDate="2010-10-25T17:23:51.830" CommentCount="2" />
  <row Id="7820" PostTypeId="2" ParentId="7807" CreationDate="2010-10-25T16:27:52.537" Score="4" ViewCount="0" Body="&lt;p&gt;Question 1: Setting $u=\pi/2-v$ in your first set of formulas gives $(x,y)=(1,1)$, so that point corresponds to the whole hypotenuse $u+v=\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: I have no idea. (Luck? Inspiration? Trial and error?)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-25T16:37:27.703" LastActivityDate="2010-10-25T16:37:27.703" CommentCount="1" />
  <row Id="7821" PostTypeId="1" AcceptedAnswerId="7825" CreationDate="2010-10-25T16:45:03.437" Score="2" ViewCount="250" Body="&lt;p&gt;Ok. This question may sound very easy, but actually i am in great need of it. I have been facing trouble in constructing functions, which are only continuous at some particular sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e.g, the standard example of a function which is only continuous at one point, is the function,  $f(x) = x, \ x \in \mathbb{Q}$ and $f(x) = -x, x  \in \mathbb{R} \setminus \mathbb{Q}$. Similarly, i would like to know as to how to construct a function which is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous at exactly $2,3,4$ points.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous exactly at integers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous exactly at Natural numbers&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous exactly at Rationals. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to see many examples (with proof!), so that i can don't struggle when somebody asks me to construct such functions. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-26T03:26:54.163" Title="Constructing Continuous functions at given points." Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="2" />
  <row Id="7822" PostTypeId="1" AcceptedAnswerId="7826" CreationDate="2010-10-25T16:51:22.803" Score="4" ViewCount="118" Body="&lt;p&gt;So this is some I partially understand, I'm not sure what I dont and do understand because most of my understanding is based on assumtions... sorry if I sound a little stupid!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation $6 \cos x - \sin x = 5 $ needs to be turned into the form $ \Re  \cos (x + \alpha) $ then solved for &lt;strong&gt;in the interval&lt;/strong&gt; $ -1/2\pi &amp;lt; x &amp;lt; 1/2\pi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I turned it into this: $ \sqrt{37} \cos( x + 0.165) $ then&lt;br&gt;&#xA; $ \sqrt{37} \cos( x + 0.165)  = 5$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \cos( x + 0.165)  = \frac{5}{\sqrt{37}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x + 0.165= \arccos( \frac{5}{\sqrt{37}} )$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x  = \arccos( \frac{5}{\sqrt{37}} ) - 0.165$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x = 0.44$ YAY! but... the answer is 0.44 &lt;strong&gt;and&lt;/strong&gt; -0.771  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im thinking its asking what other value would of the above equation would end up in 5 also? Correct? How do I do this? &lt;strong&gt;Could someone explain&lt;/strong&gt; what is meant by &lt;em&gt;&quot;solve this equation for that interval&quot;&lt;/em&gt;, and how does one go about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem I think might be related that I JUST cannot get my head around is this one:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The angle made by a wasps wings horizontally is given by the equation $ \theta = 0.4 \sin600t $, where t is time is seconds. How many times a second does its wing oscillate?&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried solving this, honest but I do not know where to begin!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T17:28:18.903" LastActivityDate="2010-10-25T17:28:18.903" Title="understanding intervals in trigonometry" Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7824" PostTypeId="2" ParentId="7821" CreationDate="2010-10-25T17:04:35.487" Score="7" ViewCount="0" Body="&lt;p&gt;Continuous at 2, 3, 4: $f(x)=(x-2)(x-3)(x-4)$ if $x$ is rational, $f(x)=0$ if $x$ is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous at the integers: $f(x)=\sin(\pi x)$ if $x$ is rational, 0 if $x$ is irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous at the natural numbers: $f(x)=\sin(\pi x)$ if $x$ is rational and not a nonpositive integer, 0 if $x$ is irrational, 1 if $x$ is a nonpositive integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous exactly at the rationals: Impossible, because the set of rational numbers is not a $G_\delta$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-25T17:04:35.487" CommentCount="10" />
  <row Id="7825" PostTypeId="2" ParentId="7821" CreationDate="2010-10-25T17:07:30.687" Score="4" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One simple way of constructing a function which is continuous only at a finite number of points, $x=a_1,\ldots,a_n$, is to do a slight modification to the function you give: take a polynomial $p(x)$ that has roots exactly at $x=a_1,\ldots,a_n$ (e.g., $p(x) = (x-a_1)\cdots(x-a_n)$) , and then define&#xA;$$ g(x) = \left\{\begin{array}{ll}&#xA;p(x) &amp;amp; \text{if $x\in\mathbb{Q}$;}\\&#xA;0 &amp;amp; \text{if $x\notin\mathbb{Q}$.}&#xA;\end{array}\right.$$&#xA;The function is continuous at $a_1,\ldots,a_n$, and since $p(x)\neq 0$ for any $x\notin\{a_1,\ldots,a_n\}$ then $g(x)$ is not continuous at any point &lt;em&gt;other than&lt;/em&gt; $a_1,\ldots,a_n$. Other possibilities should suggest themselves easily enough.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function that is continuous exactly at the integers: a similar idea will work: find a function that has zeros exactly at the integers, for example $f(x)=\sin(\pi x)$, and then take&#xA;$$g(x) = \left\{\begin{array}{ll}&#xA;\sin(\pi x) &amp;amp; \text{if $x\in\mathbb{Q}$;}\\&#xA;0 &amp;amp; \text{if $x\notin\mathbb{Q}$.}&#xA;\end{array}\right.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function continuous exactly in the natural numbers: take a function that is continuous at the integers, and redefine it as the characteristic function of the rationals in appropriate places(what happens at $0$ depends on whether you believe $0$ is in the natural numbers or not). Assuming that $0\in\mathbb{N}$, one possibility is:&#xA;$$g(x) = \left\{\begin{array}{ll}&#xA;\sin(\pi x)&amp;amp;\text{if $x\in\mathbb{Q}$ and $x\geq 0$;}\\&#xA;x &amp;amp; \text{if $x\in\mathbb{Q}$ and $-\frac{1}{2}\lt x\leq 0$;}\\&#xA;1 &amp;amp; \text{if $x\in\mathbb{Q}$ and $x\leq -\frac{1}{2}$;}\\&#xA;0 &amp;amp; \text{if $x\notin\mathbb{Q}$.}&#xA;\end{array}\right.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function continuous exactly on the rationals. This one is a bit trickier. There is no such function. This follows because the set of discontinuities of a real valued function must be a countable union of closed sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps then, we might anticipate the next question:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A function that is continuous exactly on the irrationals. An example is the following: let $s\colon\mathbb{N}\to\mathbb{Q}$ be an enumeration of the rationals (that is, a bijection from $\mathbb{N}$ to $\mathbb{Q}$. Define $f(x)$ as follows:&#xA;$$f(x) = \sum_{\stackrel{n\in\mathbb{N}}{s_n\leq x}} \frac{1}{2^n}.$$&#xA;The function has a jump at every rational, so it is not continuous at any rational. However, if $x$ is irrational, let $\epsilon\gt 0$. Then there exists $N$ such that $\sum_{k\geq N}\frac{1}{2^k}\lt \epsilon$. Find a neighborhood of $x$ which excludes every $q_m$ with $m\leq N$, and conclude that the difference between the value of $f$ at $x$ and at any point in the neighborhood is at most $\sum_{k\geq N}\frac{1}{2^k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As I was reminded in the comments by jake, in fact the &quot;standard example&quot; of a function that is continuous at every rational and discontinuous at every rational is &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomae%27s_function&quot; rel=&quot;nofollow&quot;&gt;Thomae's function&lt;/a&gt;. The example I give is a monotone function, and although it is discontinuous at every rational, it is continuous &lt;em&gt;from the right&lt;/em&gt; at every number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T03:26:54.163" LastActivityDate="2010-10-26T03:26:54.163" CommentCount="14" />
  <row Id="7826" PostTypeId="2" ParentId="7822" CreationDate="2010-10-25T17:13:20.227" Score="2" ViewCount="0" Body="&lt;p&gt;As the cosine is periodic, there are many values of $\theta$ which have the same $\cos(\theta)$.  So they are just asking for all the values between $-\pi/2$ and $\pi/2$ that solve the equation.  Your solution got one of them, $\cos(0.605)$ does equal $5/\sqrt(37)$.  But so does $\cos(-.605)$  You were supposed to find that one, too. It leads to the solution -.771 when the .165 is subtracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second problem I presume the &gt; sign is supposed to be *.  How much does the argument of the sine function (the thing you take the sine of) have to increase to go through one cycle?  How much does t have to increase to go through one cycle.  This gives you the period.  The frequency is one divided by this.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-25T17:26:02.823" LastActivityDate="2010-10-25T17:26:02.823" CommentCount="9" />
  <row Id="7827" PostTypeId="1" AcceptedAnswerId="7828" CreationDate="2010-10-25T17:15:10.713" Score="0" ViewCount="140" Body="&lt;p&gt;ok so I did all the revision problems and noted the ones I couldn't do today and Im posting them together, hope thats not a problem with the power that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have exhibit A:&lt;br&gt;&#xA;$e^{-x} -x + 2 $&lt;br&gt;&#xA;So I differentiate to find where the derivative hits 0:&lt;br&gt;&#xA;$-e^{-x} -1 = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now HOW do I figure when this &lt;em&gt;hits&lt;/em&gt; zero!?&lt;br&gt;&#xA;$-1 = e^{-x} $&lt;br&gt;&#xA;$\ln(-1) = \ln(e^{-x})$ ???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More to come ... as one day rests between me and my final exam/attempt at math!&lt;/p&gt;&#xA;" OwnerUserId="2576" LastActivityDate="2010-10-25T18:07:31.470" Title="finding the minima and maxima of some tough functions" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7828" PostTypeId="2" ParentId="7827" CreationDate="2010-10-25T17:20:56.540" Score="3" ViewCount="0" Body="&lt;p&gt;$e^{-x}&gt;0$ for all real $x$. Hence $-e^{-x}&amp;lt;0$ for all real $x$, whence $-e^{-x} - 1 &amp;lt;-1$ for all real $x$. So it nevers &quot;hits&quot; zero. Look at a graph of your function.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T18:07:31.470" LastActivityDate="2010-10-25T18:07:31.470" CommentCount="6" />
  <row Id="7829" PostTypeId="1" CreationDate="2010-10-25T17:48:39.607" Score="1" ViewCount="129" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need calculate ray (line) intersection with torus for my ray-tracing program (I know, its to graphics, but i need math behind it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can solve equation of order x^4, but thats too way slow (Cardano's method). So is there better way, how to calculate this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My torus has center [0,0,0] and R, r diameters&#xA;Line (Ray) is point and its direction vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2791" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T02:46:53.520" LastActivityDate="2010-10-26T02:46:53.520" Title="Annuloid (Torus)-line intersection" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7830" PostTypeId="2" ParentId="7827" CreationDate="2010-10-25T17:49:25.830" Score="0" ViewCount="0" Body="&lt;p&gt;I don't understand why the book wants you to find the max and min.  You have correctly deduced that the derivative is never zero, which says there isn't a max or min.  Looking for the root of this function is not hard.  If you graph it over a reasonable range, say -5 to 5, you will find it close enough to get N correctly&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-25T17:49:25.830" CommentCount="2" />
  <row Id="7831" PostTypeId="2" ParentId="7827" CreationDate="2010-10-25T17:50:03.477" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ e^{-x}\:$ and $\rm\: -x\: $ are both strictly descreasing on $\:\mathbb R\:$, hence so is their sum + 2.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-25T17:50:03.477" />
  <row Id="7832" PostTypeId="2" ParentId="7829" CreationDate="2010-10-25T18:18:08.710" Score="1" ViewCount="0" Body="&lt;p&gt;What kind of speed are you looking for?  In general, smart interval-based solvers work well for this sort of thing: the core idea is that by using interval arithmetic you can bound the values of the 'potential function' (the quartic polynomial whose roots you're trying to find) within any given interval, and then fast-reject if those bounds don't contain zero - that is, if you can confirm that the function is always-positive or always-negative on the interval.  Find starting values of t&lt;sub&gt;min&lt;/sub&gt; and t&lt;sub&gt;max&lt;/sub&gt; by intersecting your ray with the bounding box of the torus, use the interval arithmetic mentioned above to determine whether it's possible for an intersection to appear within that range, and then use binary subdivision (always exploring the 'earlier' - closer to the camera - branch of the subdivision first) to narrow in on the first zero to whatever accuracy is desired.  In practice this isn't much slower than explicitly solving the quartic, and it tends to be more robust, particularly around tangent points/double zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One question I think it's worth asking, though: why are you trying to explicitly handle ray-torus intersection?  Tori don't, in general, make very good primitives for composing a scene; they're &lt;i&gt;too&lt;/i&gt; geometric and not particularly organic enough.  IMHO you're much better off going the pile-of-triangles route and giving yourself more flexibility, rather than tying yourself too closely to complicated geometric primitives.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-10-25T18:18:08.710" CommentCount="7" />
  <row Id="7833" PostTypeId="2" ParentId="7413" CreationDate="2010-10-25T18:48:08.273" Score="0" ViewCount="0" Body="&lt;p&gt;If $X$ and $Y$ are random variables, then their sum $X+Y$ is the convolution of their distributions. &lt;/p&gt;&#xA;" OwnerUserId="2792" LastActivityDate="2010-10-25T18:48:08.273" CommentCount="1" />
  <row Id="7835" PostTypeId="1" AcceptedAnswerId="7841" CreationDate="2010-10-25T19:32:00.323" Score="1" ViewCount="250" Body="&lt;p&gt;I am stuck while proving this identity I verified it using induction but like the other two &lt;a href=&quot;http://math.stackexchange.com/questions/7757/how-to-prove-this-binomial-identity-sum-r0n-r-n-choose-r-n2n-1&quot;&gt;(1)&lt;/a&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/7778/proving-2n-choose-n-2n-frac-1-cdot-3-cdot-5-cdots-2n-1n&quot;&gt;(2)&lt;/a&gt;, I am seeking a more of a general way (algebraic will be much appreciated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!}  $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-10-26T05:11:42.643" Title="Showing  $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!}  $" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="12" />
  <row Id="7837" PostTypeId="1" CreationDate="2010-10-25T20:03:30.947" Score="4" ViewCount="198" Body="&lt;p&gt;Let $r$, $s$ be positive real numbers and $\theta$, $\phi$ real numbers with $|\theta -\phi|&amp;lt;\pi$. Then an argument of $re^{i\theta}+se^{i\phi}$ lies between $\theta$ and $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give a short, clean proof of the statement above that doesn't rely on geometric intuition? Of course the proof may use the not so trivial fact that every nonzero complex number can be written in polar form as well as trigonometrical functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried, but I have no idea how to start, probably because it seems so obvious.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T03:08:52.697" LastActivityDate="2010-10-26T03:08:52.697" Title="Argument of the sum of two complex numbers" Tags="&lt;complex-numbers&gt;&lt;analytic-geometry&gt;" AnswerCount="2" />
  <row Id="7839" PostTypeId="2" ParentId="7835" CreationDate="2010-10-25T20:07:05.583" Score="0" ViewCount="0" Body="&lt;p&gt;Seems easy is $n$ is odd. Multiplying both sides by $n!$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {n \choose 1} + {n \choose 3} + ...  {n \choose n} = \frac{1}{2} \left[  {n \choose 0} + {n \choose 1} + ... {n \choose n}  \right]  = 2^{n-1}$$&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-25T20:07:05.583" />
  <row Id="7840" PostTypeId="1" AcceptedAnswerId="7851" CreationDate="2010-10-25T20:10:21.757" Score="1" ViewCount="134" Body="&lt;p&gt;If I have a doubly stochastic matrix, how can I find the set of all BFS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Doubly_stochastic_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Doubly_stochastic_matrix&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T03:06:45.383" LastActivityDate="2010-10-26T03:06:45.383" Title="Finding all n×n permutation matrices" Tags="&lt;algorithms&gt;&lt;matrices&gt;&lt;linear-programming&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7841" PostTypeId="2" ParentId="7835" CreationDate="2010-10-25T20:15:22.850" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1+x)^{n} - (1-x)^{n} = 2 \Biggl[ {n \choose 1} x + {n \choose 3} x^{3} + \cdots \Biggr]$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-26T04:02:26.553" LastActivityDate="2010-10-26T04:02:26.553" CommentCount="1" />
  <row Id="7842" PostTypeId="2" ParentId="7835" CreationDate="2010-10-25T20:18:02.787" Score="2" ViewCount="0" Body="&lt;p&gt;For even $n$, look at the coefficient of $x^n$ in the expansion of $\sinh^2 x = \frac12(\cosh 2x-1)$. For odd $n$, modify this idea suitably.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-25T20:18:02.787" />
  <row Id="7843" PostTypeId="2" ParentId="7837" CreationDate="2010-10-25T20:29:11.647" Score="4" ViewCount="0" Body="&lt;p&gt;It's enough to do the case $\theta=0$ (just factor out $e^{i\theta}$).  Assume $0&amp;lt;\phi&amp;lt;\pi/2$. Then&#xA;$$0 &amp;lt; \tan\arg(r+s e^{i\phi}) = \frac{s\sin\phi}{r+s\cos\phi} &amp;lt; \frac{s\sin\phi}{s\cos\phi} = \tan\phi,$$&#xA;so the argument of the sum is between $0$ and $\phi$ as desired. Something similar should work when $\phi$ is obtuse.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-25T20:29:11.647" />
  <row Id="7844" PostTypeId="2" ParentId="7837" CreationDate="2010-10-25T20:31:50.457" Score="5" ViewCount="0" Body="&lt;p&gt;You want to show that if you write $re^{i\theta}  + se^{i\phi}$ as $te^{i\chi}$, then $\chi$ lies between $\theta$ and $\phi$. Without loss of generality, assume $\theta\leq\phi$. Factoring out $e^{i\theta}$ you get $re^{i\theta}+se^{i\phi} = e^{i\theta}(r + se^{i(\phi-\theta)})$. Since multiplying by $e^{i\theta}$ is just a rotation by an angle $\theta$, it is enough to consider the case where $\theta=0$ and $0\lt \phi\lt \pi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, you have $r + se^{i\phi} = r + s(\cos(\phi)+i\sin(\phi))$, and you want to express it in the form $t(\cos(\chi) + i \sin(\chi))$. Looking at real and complex parts, you see that $t\sin(\chi)=s\sin(\phi)$, and $r+s\cos(\phi) = t\cos(\chi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume first that $0\lt \phi\lt \frac{\pi}{2}$. Then $\chi$ must also lie in the first quadrant, since we need both $\sin(\chi)$ and $\cos(\chi)$ to be positive (since $t$, $s\sin(\phi)$, and $r+s\cos(\phi)$ are all positive).  If, on the other hand, $\frac{\pi}{2}\lt \phi\lt \pi$, then $\cos(\phi)$ is negative. If $r+s\cos(\phi)\gt 0$, then we need $\cos(\chi)\gt 0$, so $\chi$ is in the first quadrant and automatically smaller than $\phi$ and we are done.  If $r+s\cos(\phi)$ is negative then we need $\chi$ in the second quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former case, $0\lt \chi,\phi\lt \frac{\pi}{2}$, then from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ and since $\sin(x)$ is increasing on $0\leq x\leq\frac{\pi}{2}$, then $0\lt\chi\lt\phi$ if and only if $\frac{s}{t}\lt 1$, if and only if $s\lt t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $t^2 = ||r+se^{i\phi}||^2 = r^2 + s^2 + 2rs\cos(\phi) \gt s^2$, since all of $r$, $s$, and $\cos(\phi)$ are positive. Since $s$ and $t$ are both positive, $s\lt t$, which shows that $0\lt\chi\lt\phi$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other case, where $\frac{\pi}{2}\lt\phi\lt \pi$ and $r+s\cos(\phi)\lt 0$, then we know $\chi$ is also in the second quadrant where $\sin(x)$ is decreasing, so from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ we get that $\frac{\pi}{2}\lt \chi\lt \phi$ if and only if $\frac{s}{t}\gt 1$, if and only if $s\gt t$. Here, since $\cos(\theta)\lt 0$, then you have again $t^2 = ||r+se^{i\phi}||^2 = s^2 + r(r + 2s\cos(\phi)) \lt s^2$ (since $r+s\cos(\phi)\lt 0$ in this situation), so you get $t\lt s$ and hence $\frac{\pi}{2}\lt\chi\lt\phi$, as desired&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-25T20:45:40.010" LastActivityDate="2010-10-25T20:45:40.010" CommentCount="2" />
  <row Id="7845" PostTypeId="1" CreationDate="2010-10-25T20:52:06.173" Score="1" ViewCount="101" Body="&lt;p&gt;Suppose we observe 84 alchoholics with cirrhosis of the liver, of whom 29 have hepatomas, that is, a liver-cell carcinoma. Suppose we know, based on a large sample, that the risk of hepatoma among alcoholics without cirrhosis of the liver is 24%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that we observe exactly 29 alcoholics with cirrhosis of the liver who have hepatomas if the true rate of hepatoma among alcoholics (with or without cirrhosis of the liver) is .24?&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2010-10-25T21:24:42.780" Title="Question from Fundamentals of Biostatistics" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7846" PostTypeId="1" AcceptedAnswerId="7855" CreationDate="2010-10-25T20:54:12.523" Score="1" ViewCount="168" Body="&lt;p&gt;UPDATE: I feel like my original question was too vague and didn't provide enough information (as others have mentioned). So I'm going to restate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Purpose:&lt;/strong&gt; I want to find an equation in order to provide a smooth slope for an animation that I am creating. My animation consists of a lever and a set of numbers on a horizontal line. As you move the lever up and down the numbers will slide down and up the line (opposite of the lever). I want to create a smooth curve so that the user can &quot;feel&quot; the speed increasing the further they get away from the center. Most of the time, the user will stay within the main part of the screen, therefore I want the sliding numbers to go slowly for that space. But once they get closer to the top or bottom I want the increase in speed to be relatively quick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Requirements:&lt;/strong&gt; The variables are the distance from center that the lever is, $x$, and the distance that the numbers should move, $y$. I need $x=y=0$ so there is no movement at the center. $y$ should be positive when $x$ is positive, but $y$ can be positive or negative when $x$ is negative. The graph just needs to be symmetric for $x=0$ or $x=y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I'm using a simple parabola $y = (0.03*x)^2$ but I would like the middle space to be &quot;stretched out.&quot; Therefore, is there a function that I can use that would have some constants (maybe some points) to define what kind of slope is found for, lets say $x=0$ to $x=100$ and then it curving to another point at $x=160$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to use piecewise because unless I use a lot of pieces, the user will be able to &quot;feel&quot; the positions where the equation changes. I think Quadratic Bézier curve might do the trick, but if it can, I would need help in understanding how to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; is the Quadratic Bézier curve what I should be looking into, if not, what do you suggest? If it is, do you have a suggestion of what equation I should use or at least a reference were I can relearn it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Old question:&#xA;&lt;strike&gt;I need to come up with an equation for a curve but I have forgotten most of my calculus by this point. What I need very close to a parabola that opens up. I'm currently using y = (0.03*x)^2 but this does not have a fast enough slope on the ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the slope in the middle to be very gradual, and then near the ends to increase faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing this won't be a standard parabola, but I'm not just sure what I should be looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: Attempting to be more detailed with my &quot;requirements.&quot; This is for a computer animation and a user is sliding a &quot;lever&quot; up and down. As they do so a set of numbers are scrolling the opposite direction on the side. The further away the lever is from center, the faster the numbers should go. Thus, I want the edges (top and bottom) to make the numbers go much faster. But for most of the main section of the screen, they shouldn't go so fast. But I want the transition from slow to fast to be noticeable so that it feels intuitive to the user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I need a function that has a high value for y as x gets larger, but I want the edges to have a greater slope. Bézier curve looks like it might work? But it has been a long time since I've used it. Anyone have suggestions for how to use it in this context?&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="2798" LastEditorUserId="2798" LastEditorDisplayName="" LastEditDate="2010-10-26T16:51:11.110" LastActivityDate="2010-10-26T21:48:28.837" Title="Parabola attributes - custom curve?" Tags="&lt;calculus&gt;&lt;polynomials&gt;&lt;bezier-curve&gt;&lt;parabola&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="7847" PostTypeId="2" ParentId="7628" CreationDate="2010-10-25T21:00:23.740" Score="6" ViewCount="0" Body="&lt;p&gt;It is &lt;strong&gt;not&lt;/strong&gt; true that the Grothendieck ring of coherent sheaves on $\mathbb{P}^1$ is isomorphic to $\mathbb{Z}[t, t^{-1}]$. Although $\mathcal{O} \oplus \mathcal{O}(2)$ is not isomorphic to $\mathcal{O}(1) \oplus \mathcal{O}(1)$, they do have the same class in $K^0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the Grothendieck group of coherent sheaves on a scheme $X$ is that it is generated by isomorphism classes of coherent sheaves, modulo the relation that $[A] + [C] = [B]$ whenever there is a short exact sequence&#xA;$$0 \to A \to B \to C \to 0.$$&#xA;In particular, we have the short exact sequence&#xA;$$0 \to \mathcal{O} \to \mathcal{O}(1)^2 \to \mathcal{O}(2) \to 0,$$&#xA;where the maps are given by $(x \ y)$ and $\binom{-y}{x}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes $K^0$ into $\mathbb{Z}[t, t^{-1}]/(t^2 - 2t +1) \cong \mathbb{Z}[u]/u^2$, just like you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working in the categories of smooth or of topological vector bundles, all short exact sequences split, so you can get away with defining $K$-theory with direct sums. You can't do that in the coherent or the algebraic categories.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-10-25T21:00:23.740" CommentCount="1" />
  <row Id="7848" PostTypeId="2" ParentId="5159" CreationDate="2010-10-25T21:07:05.927" Score="1" ViewCount="0" Body="&lt;p&gt;As you have thought you should solve&#xA;$$\left\{\begin{aligned}3y^2-1=0,\\x^3+y^3-y=0\end{aligned}\right.&#xA;$$&#xA;which gives you 6 ramification points $y=\pm 1/\sqrt{3}, x=\pm e^{2\pi i k/3}\sqrt[3]{2}/\sqrt{3}$ where $k=0,1,2$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-25T21:07:05.927" />
  <row Id="7849" PostTypeId="2" ParentId="7783" CreationDate="2010-10-25T21:16:30.090" Score="8" ViewCount="0" Body="&lt;p&gt;First, one uses the argument principle from complex analysis to compute the number of zeroes with real parts between 0 and 1 and with imaginary parts between 0 and some postive number $T$.  Since the answer is &lt;em&gt;a priori&lt;/em&gt; an integer, one can get a precise answer even though one can't compute the $\zeta$-function exactly; it is &quot;just&quot; a matter of bounding the error in all approximations carefully enough.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, one has to show that this number of zeroes actually lie on the line where $\Re s = 1/2$.&#xA;For this, one factors $\zeta(1/2 + i t)$ as a product of a certain nowhere zero function,&#xA;and a real valued function $Z(t)$ (the function appearing in J.M.'s answer).  (That this can&#xA;be done in a useful way is part of the theory of the $\zeta$-function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count zeroes of $\zeta$ with $\Re s = 1/2$ is now the same as counting zeroes of $Z(t)$,&#xA;which can be determined by sign changes in $Z(t)$ (provided one estimates $Z(t)$ accurately enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that all works out (which it will if RH is true!) one gets a rigorous proof that all zeroes with imaginary part $\leq T$ lie on the line $\Re s = 1/2$.  (In fact, for this to work out, one also needs the zeroes &#xA;of $\zeta$ to be simple, which they are conjectured to be; see T..'s answer&#xA;for an elaboration on this.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-03T01:30:23.843" LastActivityDate="2011-01-03T01:30:23.843" CommentCount="3" />
  <row Id="7850" PostTypeId="2" ParentId="7845" CreationDate="2010-10-25T21:24:42.780" Score="4" ViewCount="0" Body="&lt;p&gt;When you have $n$ people, with probability $p$ for each of them to have hepatoma, and you want to know the probability that exactly $k$ of them have hepatoma, then the formula is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${n \choose k}p^k(1-p)^{n-k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $n$ = 84, $k$ = 29, $p$ = 0.24. This is called the Binomial Distribution. You can learn more about it  here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_distribution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-25T21:24:42.780" />
  <row Id="7851" PostTypeId="2" ParentId="7840" CreationDate="2010-10-25T22:07:13.637" Score="4" ViewCount="0" Body="&lt;p&gt;Don Knuth's Volume 4, Fascicle 2, of &lt;em&gt;The Art of Computer Progamming&lt;/em&gt; has a long section on generating all permutations, including algorithms for doing so.  I found a draft &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CBMQFjAA&amp;amp;url=http%3A%2F%2Fwww-cs-faculty.stanford.edu%2F~uno%2Ffasc2b.ps.gz&amp;amp;rct=j&amp;amp;q=generating%20all%20permutations%20knuth&amp;amp;ei=AvfFTMm_LouksQPf8-DLDQ&amp;amp;usg=AFQjCNEw5tfBK7QyvtTni9utWTNbPAJbIQ&amp;amp;sig2=rwj1e_RcSsJgDYcSpYNI-w&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, going from a permutation to a permutation matrix is fairly straightforward.  For example, suppose you have the permutation 1342 of the numbers 1, 2, 3, and 4.  That can be represented in two-line form as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;1 2 3 4&#xA;1 4 2 3&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;because the permutation sends 1 to the first position, 2 to the fourth position, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the permutation matrix is the matrix with 1's in entries (1,1), (2,4), (3,2), (4,3), and 0's elsewhere; i.e.,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;1 0 0 0 &#xA;0 0 0 1&#xA;0 1 0 0 &#xA;0 0 1 0&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-25T22:18:32.390" LastActivityDate="2010-10-25T22:18:32.390" CommentCount="4" />
  <row Id="7852" PostTypeId="2" ParentId="7783" CreationDate="2010-10-25T22:58:01.737" Score="12" ViewCount="0" Body="&lt;p&gt;As long as there are no multiple zeros on the line up to height  $T$  you can determine, rigorously and using a finite amount of computation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the number of zeros (counted with multiplicity) off the critical line, i.e., the number of counterexamples to the Riemann hypothesis, of height less than $T$.   This is done by contour integrals of $d \log \zeta$ -- the &quot;argument principle&quot; in complex analysis.  For each zero one can locate it to any specified accuracy, and determine its multiplicity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the number of zeros on the critical line, up to height $T$.  For each zero, its ordinate on the line can be calculated to any desired accuracy.  This is done by counting sign changes of real-analytic functions with a $\zeta (1/2 + it)$ factor, on the critical line.  As Matt E mentioned, the ability to write down a real-valued function capturing the behavior of a complex function on a line is a special phenomenon particular to zeta and L-functions, coming from the functional equation relating $\zeta(s)$ to $\zeta(1-s)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If there is a multiple zero of order $n$ at some point on the line, all you can determine by computations of the above type is that a very small neighborhood of that point contains $n$ zeros (counted with multiplicity).  This could mean zeros on the line, counterexamples to the Riemann conjecture, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is part of the package of conjectures surrounding the Riemann hypothesis, that there are no multiple zeros, and that the zeros in fact &quot;repel&quot; each other in a quantifiable sense. There is a lot of numerical and theoretical support for the zero repulsion, especially from random matrix theory.  If the Riemann hypothesis is true but there are multiple zeros, that would be almost as surprising as finding a zero off the line.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-25T23:17:49.193" LastActivityDate="2010-10-25T23:17:49.193" />
  <row Id="7853" PostTypeId="1" CreationDate="2010-10-25T23:17:41.890" Score="3" ViewCount="124" Body="&lt;p&gt;I'm searching for an algorithm that forms a balanced (or quasi-complete) latin square, in which every element is a horizontal neighbor to every other element exactly twice, and a vertical neighbor to every other element exactly twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found one example (for &lt;strong&gt;n = 5&lt;/strong&gt;), but am not clear about a couple of steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;: Write &lt;strong&gt;1 2 ... n&lt;/strong&gt; as the first row&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;: For &lt;strong&gt;i&lt;/strong&gt; even in the first row, fill in the diagonal from upper left to lower right starting with &lt;strong&gt;i&lt;/strong&gt; and alternating with &lt;strong&gt;i - 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;  1 3&#xA;   2&#xA;    1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3&lt;/strong&gt;: For &lt;strong&gt;i&lt;/strong&gt; odd and less than &lt;strong&gt;n&lt;/strong&gt; in the first row fill in the diagonal from upper right to lower left beginning with &lt;strong&gt;i&lt;/strong&gt; and alternating with &lt;strong&gt;i + 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA; 41 3&#xA;3  2&#xA;    1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4&lt;/strong&gt;: Fill in &lt;strong&gt;n&lt;/strong&gt; for the main off diagonal&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA; 4153&#xA;3 52&#xA; 5  1&#xA;5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 5&lt;/strong&gt;: For the last column write &lt;strong&gt;n n-2 n-1 n-3 ... 1 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA; 4153&#xA;3 524&#xA; 5  1&#xA;5   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 6&lt;/strong&gt;: For the even entries of the last column, fill in the diagonal from upper left to lower right beginning with &lt;strong&gt;i&lt;/strong&gt; and alternating with &lt;strong&gt;i - 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 7&lt;/strong&gt;: Complete by symmetry about the main diagonal&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;n = 5&lt;/strong&gt;, the following is obtained:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;24153&#xA;31524&#xA;45231&#xA;53412&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[RE: &lt;strong&gt;Step 5&lt;/strong&gt;] I was able to fill in that portion of the square, but am not sure what would come after &lt;strong&gt;n-3&lt;/strong&gt; in a larger sequence, as it would go &lt;strong&gt;n n-2 n-1 n-3 ??? ... 1 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[RE: &lt;strong&gt;Step 6&lt;/strong&gt;] Are even entries even numbers, or does that refer to even positions in the column? Is &quot;i&quot; in this case the number at the upper left or is it the even entry?  I don't see how the completed square corresponds to the instructions in this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[RE: &lt;strong&gt;Step 7&lt;/strong&gt;] What does &quot;complete by symmetry about the main diagonal&quot; mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if there's a less involved algorithm for balancing a latin square, I'd like to know how it goes.&lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:37:50.607" LastActivityDate="2010-12-19T10:44:26.240" Title="Balancing a Latin Square" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;latin-square&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="7854" PostTypeId="1" CreationDate="2010-10-26T00:09:48.490" Score="1" ViewCount="95" Body="&lt;p&gt;When pricing a european option by monte carlo over 30 days for instance, what's the difference between one big 30 day jump vs 30 one day steps?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-10-26T03:11:49.900" Title="Stochastic/finance monte carlo question" Tags="&lt;finance&gt;&lt;monte-carlo&gt;" AnswerCount="2" />
  <row Id="7855" PostTypeId="2" ParentId="7846" CreationDate="2010-10-26T00:41:37.683" Score="1" ViewCount="0" Body="&lt;p&gt;I think the easiest way may be piecewise—not piecewise-linear, but a piecewise-defined function that is relatively smooth, so there isn't a point where the speed suddenly changes.  Let's say that you want slope $m$ for $-a\le x\le a$ (and $y=0$ when $x=0$), then curving up to the point $(b,c)$.&#xA;$$y=\begin{cases}&#xA;\frac{-a^2 c+a^2 b m}{(a-b)^2}-\frac{\left(2 a c-a^2 m-b^2 m\right) x}{(a-b)^2}+\frac{(-c+b m) x^2}{(a-b)^2} &amp;amp; \text{ if } x&amp;lt;-a&#xA;\\ &#xA;mx &amp;amp; \text{ if } -a\le x\le a&#xA;\\ &#xA;-\frac{-a^2 c+a^2 b m}{(a-b)^2}-\frac{\left(2 a c-a^2 m-b^2 m\right) x}{(a-b)^2}-\frac{(-c+b m) x^2}{(a-b)^2} &amp;amp; \text{ if } x&gt;a&#xA;\end{cases}$$&#xA;(I arrived at this formula by having Mathematica solve for the coefficients of the quadratic polynomial that would go through the point at the end of the linear segment and the other given point and have the same slope as the linear segment at the splice point: &lt;code&gt;a1 x^2 + a2 x + a3 /. &#xA; Solve[a1 a^2 + a2 a + a3 == a m &amp;amp;&amp;amp; 2 a1 a + a2 == m &amp;amp;&amp;amp; &#xA;   a1 b^2 + a2 b + a3 == c, {a1, a2, a3}]&lt;/code&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you wanted a slope of $\frac{1}{4}$ from -100 to 100, then curving up to (180,150), $a=100$, $m=\frac{1}{4}$, $b=180$, $c=150$, and $$y=\begin{cases}&#xA;-\frac{2625}{16}-\frac{97 x}{32}-\frac{21 x^2}{1280} &amp;amp; \text{ if } x&amp;lt;-100&#xA;\\ &#xA;\frac{1}{4}x &amp;amp; \text{ if } -100\le x\le 100&#xA;\\ &#xA;\frac{2625}{16}-\frac{97 x}{32}+\frac{21 x^2}{1280} &amp;amp; \text{ if } x&gt;100&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FW2qS.png&quot; alt=&quot;graph of example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;: even smoother would be to use a cubic polynomial for the end parts and force it to match the first and second derivatives at the splice points (&lt;code&gt;a1 x^3 + a2 x^2 + a3 x + a4 /. &#xA; Solve[a1 a^3 + a2 a^2 + a3 a + a4 == a m &amp;amp;&amp;amp; &#xA;   3 a1 a^2 + 2 a2 a + a3 == m &amp;amp;&amp;amp; 6 a1 a + 2 a2 == 0 &amp;amp;&amp;amp; &#xA;   a1 b^3 + a2 b^2 + a3 b + a4 == c, {a1, a2, a3, a4}]&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{cases}&#xA; \frac{-a^3 c+a^3 b m}{(a-b)^3}-\frac{\left(3 a^2 c-a^3 m-3 a b^2 m+b^3 m\right) x}{(a-b)^3}+\frac{3 a (-c+b m) x^2}{(a-b)^3}-\frac{(c-b m) x^3}{(a-b)^3} &amp;amp; x&amp;lt;-a \\&#xA; m x &amp;amp; -a\leq x\leq a \\&#xA; -\frac{-a^3 c+a^3 b m}{(a-b)^3}-\frac{\left(3 a^2 c-a^3 m-3 a b^2 m+b^3 m\right) x}{(a-b)^3}-\frac{3 a (-c+b m) x^2}{(a-b)^3}-\frac{(c-b m) x^3}{(a-b)^3} &amp;amp; x&gt;a&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you wanted a slope of $\frac{1}{4}$ from -100 to 100, then curving up to (180,150), as above, $a=100$, $m=\frac{1}{4}$, $b=180$, $c=150$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=\begin{cases}&#xA; \frac{13125}{64}+\frac{1639 x}{256}+\frac{63 x^2}{1024}+\frac{21 x^3}{102400} &amp;amp; x&amp;lt;-100 \\&#xA; \frac{x}{4} &amp;amp; -100\leq x\leq 100 \\&#xA; -\frac{13125}{64}+\frac{1639 x}{256}-\frac{63 x^2}{1024}+\frac{21 x^3}{102400} &amp;amp; x&gt;100 &#xA;\end{cases}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Mfp1H.png&quot; alt=&quot;graph of example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;original answer&lt;/em&gt;:&#xA;Would $y=\cot(3x+0.07)$ for $0\le x\le 1$ work?&#xA;Or $y=-\frac{x-0.5}{(x+0.1)(x-1.1)}$?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-26T21:48:28.837" LastActivityDate="2010-10-26T21:48:28.837" CommentCount="8" />
  <row Id="7856" PostTypeId="2" ParentId="7491" CreationDate="2010-10-26T00:43:34.470" Score="1" ViewCount="0" Body="&lt;p&gt;Please see the answer in &lt;a href=&quot;http://mathoverflow.net/questions/43181/what-will-be-the-distribution-of-ratio-of-correlated-gamma-distributed-random-var&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-10-26T00:43:34.470" />
  <row Id="7857" PostTypeId="2" ParentId="7853" CreationDate="2010-10-26T01:29:18.693" Score="1" ViewCount="0" Body="&lt;p&gt;The Latin square L:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;24153&#xA;31524&#xA;45231&#xA;53412&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is isomorphic to the Cayley table of the cyclic group of order 5 via the isomorphism (3,5,4).  That is, if you permute the rows, columns and symbols of L according to the permutation (3,5,4) you will generate&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;12345&#xA;23451&#xA;34512&#xA;45123&#xA;51234&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The n=7 case gives rise to the Latin square&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1234567&#xA;2416375&#xA;3152746&#xA;4627153&#xA;5371624&#xA;6745231&#xA;7563412&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is isomorphic to the Cayley table of the cyclic group of order 7 via the isomorphism (3,7,5,6,4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, my suspicion is that these Latin squares are special cases of those generated in the constructions described by Rosemary Bailey in &lt;a href=&quot;http://www.jstor.org/pss/2345518&quot; rel=&quot;nofollow&quot;&gt;Quasi-complete Latin squares: construction and randomization&lt;/a&gt;, although I haven't gone into the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the smaller questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In papers, by an &lt;em&gt;entry&lt;/em&gt; I exclusively refer to a triple $(i,j,l_{ij})$, where $i$ is the row index, $j$ is the column index and $l_{ij}$ is the symbol in the $(i,j)$-th cell.  For example (2,3,1) would be an entry of the first Latin square.  Importantly, in an entry, the position in the Latin square is important, whereas in a symbol, there is no notion of position (aside from being somewhere in the Latin square).  Consequently, I would use the term &quot;even symbol&quot; (although note that the symbols you use make no difference to whether or not you have a quasi-complete Latin square).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Latin squares L such that $L^T=L$ are called symmetric Latin squares (just as for symmetric matrices).  When giving details of a construction of these types of Latin squares, I find myself writing &quot;...then L admits a unique completion to a symmetric Latin square&quot; or similar.  There's no need to identify every entry since some entries are determined by other entries.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-26T01:29:18.693" CommentCount="3" />
  <row Id="7858" PostTypeId="2" ParentId="7854" CreationDate="2010-10-26T02:24:28.757" Score="1" ViewCount="0" Body="&lt;p&gt;You are in effect replacing many detailed paths by their average effect. This changes the answer since pricing options involves applying the non-linear $\max$ or $\min$ operators. If you are trying to compute the answer corresponding to a continuous time model, then you have to take a large enough number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may help your intuition to draw two binomial tree models, one with one period and one with two periods and solve them by hand.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-26T02:24:28.757" CommentCount="1" />
  <row Id="7859" PostTypeId="2" ParentId="7854" CreationDate="2010-10-26T03:11:49.900" Score="2" ViewCount="0" Body="&lt;p&gt;European options have an efficiently computable deterministic answer in both the continuous time and discrete lattice (binomial tree) settings.  Unless you are outside the Black-Scholes framework, random walk simulations are needed only for path-dependent options.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking about the accuracy of the discretization, it is not only a function of the number of steps.  Both the &quot;time&quot; and &quot;money&quot; coordinates are discretized, and the random walk answer may not converge at all unless the time and price steps are sized correctly relative to each other.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-26T03:11:49.900" />
  <row Id="7860" PostTypeId="1" AcceptedAnswerId="7861" CreationDate="2010-10-26T03:20:40.857" Score="2" ViewCount="70" Body="&lt;p&gt;I am given a lemma and it states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma: Let A be a real n x n matrix.  Then given any e &gt; 0, there is a norm such that &lt;code&gt;norm(A) &amp;lt;= p(A) + e&lt;/code&gt;&lt;/strong&gt;  Where p(A) is the spectral radius of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then go on to state that based on this lemma, &lt;code&gt;if p(A) &amp;lt; 1, then norm(A) &amp;lt; 1&lt;/code&gt; for the correct choice of the norm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for help on is understanding how they came to that conclusion based only on that lemma.  For example, given &lt;code&gt;e = .001&lt;/code&gt;, then the lemma states there is a norm such that &lt;code&gt;norm(A) &amp;lt;= p(A) + .001&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is a &quot;or equal&quot; then this means there is the possibility that &lt;code&gt;norm(A) = p(A) + .001&lt;/code&gt;, and in such a case then &lt;code&gt;norm(A) &amp;gt; p(A)&lt;/code&gt;.  So it seems that based on the lemma, then &lt;code&gt;p(A) &amp;lt; 1&lt;/code&gt; does not necessarily always imply that &lt;code&gt;norm(A) &amp;lt; 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even tried to do a proof of this but still came up with the same result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p(A) &amp;lt; 1 -&amp;gt; &#xA;p(A) + e &amp;lt; 1 + e -&amp;gt; &#xA;norm(A) &amp;lt;= p(A) +e &amp;lt; 1 + e&#xA;norm(A) - e &amp;lt;= p(A) &amp;lt; 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This pretty much says the same thing, if norm(A) - e &amp;lt;= p(A), then p(A) &amp;lt; 1, but if p(A) close to 1 by a difference less than e, then norm(A) &gt; 1 in the &quot;equal to&quot; case where norm(A) - e = p(A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now of course you could pick a smaller e, but the way the lemma reads it says that for a given e, you can find a norm s.t. A &amp;lt;= p(A) + e, so the &quot;or equal&quot; part will still get you using a counter example I gave similar to the e = .001 above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have seen a different proof on wikipedia that shows if p(A) &amp;lt; 1 then norm(A) &amp;lt; 1, so I am not disputing that fact, but that proof gets into looking at entries of the Jordan Normal Form.  So my point is, yes p(A) &amp;lt; 1 then norm(A) &amp;lt; 1, &lt;strong&gt;BUT&lt;/strong&gt; I don't understand how you can conclude that based only on the Lemma I gave at the beginning.  It's these big jumps in rational that confuse me.&lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2010-10-26T03:26:41.473" Title="How does norm(A) &lt;= p(A) + e imply norm(A) &lt; 1 if p(A) &lt; 1 ?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7861" PostTypeId="2" ParentId="7860" CreationDate="2010-10-26T03:26:41.473" Score="3" ViewCount="0" Body="&lt;p&gt;Apply the lemma with $e=\frac{1}{2}(1-p(A))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the fact that it is not strict inequality is insignificant.  Given $e$, you could apply the lemma with $e/2$ to get strict inequality with $e$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-26T03:26:41.473" />
  <row Id="7862" PostTypeId="1" AcceptedAnswerId="7863" CreationDate="2010-10-26T03:51:54.073" Score="4" ViewCount="85" Body="&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1.) Construct a subset of $[0,1]$ in the same manner as the Cantor set, except that at the $k$-th stage, each interval removed has length $\delta 3^{-k}$, $0&amp;lt;\delta &amp;lt;1$. Show that the resulting set is perfect, has measure $1-\delta$, and contains no intervals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that it's perfect is not difficult. The resulting set is an intersection of closed intervals, so is closed. Any point in the set is a limit point of endpoints of intervals (obviously I have more details in my written solution, but that's the idea.) But I don't see how the measure is $1-\delta$... At stage $k$, we remove $2^{k-1}$ intervals of length $\delta 3^{-k}$, so the measure of the resulting set is $$1-\sum\limits_{k=0}^{\infty}{2^{k}\cdot\dfrac{\delta}{3^{k+1}}} = 1-\dfrac{\delta}{3}\cdot\sum\limits_{k=0}^{\infty}{\dfrac{2^k}{3^k}} = 1-\dfrac{2\delta}{3} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2.) Construct a Cantor-type set subset of $[0,1]$ by removing from each interval remaining at the $k$-th stage a subinterval of relative length $\theta_k$, $0&amp;lt;\theta_k&amp;lt;1$. Show that the remainder has measure zero if and only if $\sum{\theta_k}=\infty$. (Use the fact that for $a_k&gt;0$, $\prod\limits_{k=1}^{\infty}{a_k}$ converges, in the sense that $\lim\limits_{N\to\infty}{\prod\limits_{k=1}^N{a_k}}$ exists and is not zero, if only if $\sum\limits_{k=1}^{\infty}{\log{a_k}}$ converges. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand the construction. Suppose we have our $\theta_1$ and we remove an interval of that length. So then we choose $\theta_2$ so that $\theta_2 &amp;lt; \dfrac{1-\theta_1}{2}$, or something like that? Or... ? The phrasing is weird to me. Secondly, the hint does not help at all; it is complete opaque to me, so any insight you can offer would be great. &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-10-26T04:43:44.787" Title="More Cantor-like constructions..." Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7863" PostTypeId="2" ParentId="7862" CreationDate="2010-10-26T04:07:55.533" Score="4" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Your last step in the computation is incorrect. The sum $\sum_{n=0}^{\infty}\frac{2^k}{3^k}$ is:&#xA;$$\sum_{n=0}^{\infty}\frac{2^k}{3^k} = \sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^k = \frac{1}{1-\frac{2}{3}} = 3$$&#xA;so when you multiply it by $\frac{\delta}{3}$ you get $\delta$, exactly what you expect it to be. I'm not sure why you thought you get $1- \frac{2\delta}{3}$. (Recall that if $|x|&amp;lt;1$, then $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Notice that it says &quot;relative length&quot;, not &quot;length&quot;. So at each step, you are taking out an interval which is $\theta_k\cdot\ell$, where $\ell$ is the length of the interval you are looking at. In the Cantor set, you take $\theta_k=\frac{1}{3}$ for every $k$; in the example in (1), you are setting $\theta_k = \frac{\delta}{3}$ for every $k$. If you take out the central $\theta_k\ell$ portion of an interval of length $\ell$, then you are left with two intervals of length $\frac{\ell}{2}(1-\theta_k)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; As per your question in the comments, in &lt;a href=&quot;http://math.stackexchange.com/questions/6643/cantor-like-construction&quot;&gt;this question&lt;/a&gt;, all the $\theta_k$ are constant and equal to $\theta$, much like in the regular Cantor set all $\theta_k$ are equal to $\frac{1}{3}$. To see an example, suppose that you set $\theta_k = \frac{1}{2^{k+1}}$, $k=0,1,\ldots$. First you remove $\theta_0 = \frac{1}{2}$ of the interval, leaving you with $[0,\frac{1}{4}]\cup[\frac{3}{4},1]$. Then you remove $\theta_1=\frac{1}{4}$th of each remaining interval; each interval is of length $\frac{1}{4}$, so you are removing the central $\frac{1}{16}$th part of the interval; this leaves you with $[0,\frac{3}{32}]\cup[\frac{5}{32},\frac{1}{4}]\cup[\frac{3}{4},\frac{27}{32}]\cup[\frac{29}{32},1]$. At the next step, you are removing the central $\theta_2=\frac{1}{8}$th of each interval; the intervals are of length $\frac{3}{32}$, so you are removing the central $\frac{3}{256}$th portion. Etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T04:43:44.787" LastActivityDate="2010-10-26T04:43:44.787" CommentCount="2" />
  <row Id="7864" PostTypeId="1" CreationDate="2010-10-26T04:24:18.303" Score="4" ViewCount="560" Body="&lt;p&gt;Over on &lt;a href=&quot;http://mathoverflow.net/&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt;, there is a popular CW question titled: &lt;a href=&quot;http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics&quot; rel=&quot;nofollow&quot;&gt;Examples of common false beliefs in mathematics&lt;/a&gt;. I thought it would be nice to have a parallel question on this site to serve as a reference for false beliefs within less obscure mathematics. &#xA;That said, it would be good not get bogged down with misconceptions that are generally assumed to be elementary such as: $(x + y)^{2} = x^{2} + y^{2}$.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-26T15:47:19.153" LastActivityDate="2010-10-28T13:03:18.580" CommunityOwnedDate="2010-10-26T20:30:14.583" Title="False beliefs in mathematics (conceptual errors made despite, or because of, mathematical education)." Tags="&lt;big-list&gt;&lt;education&gt;&lt;teaching&gt;&lt;learning&gt;" AnswerCount="9" CommentCount="9" FavoriteCount="1" />
  <row Id="7865" PostTypeId="1" AcceptedAnswerId="7869" CreationDate="2010-10-26T04:29:06.783" Score="3" ViewCount="76" Body="&lt;p&gt;Define a function $F(A, B, C)$ as the number of ways you can roll $B$ $C$-sided dice to sum up to $A$, counting different orderings (rolling a $2$, $2$, and $3$ with three dice is different from rolling a $2$, $3$, and $2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With three $5$-sided dice, the list of $F(A, B, C)$ values in the domain of the possible values of $A$ for $B = 3$ and $C = 5$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(3, 3, 5), F(4, 3, 5), F(5, 3, 5), F(6, 3, 5), ... , F(15, 3, 5)$$ is evaluated to:&#xA;$$1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1$$&#xA;Call this list $L_1$.&lt;br/&gt;&lt;br/&gt;&#xA;Let $s$ be the number of sides on each die, let $n$ be the number of dice, and let $v$ be the total value to roll from the $n$ dice.&#xA;Let $L_2$ be the list of ${v - 1}\choose{v - n}$ in the domain of $v$ values for $n = 3$.&#xA;Then $L_2$ is:&#xA;$${{3 - 1}\choose{3 - 3}}, {{4 - 1}\choose{4 - 3}}, {{5 - 1}\choose{5 - 3}}, {{6 - 1}\choose{6 - 3}}, ... , {{15 - 1}\choose{15 - 3}}$$&#xA;Which is evaluated to:&#xA;$$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91$$&#xA;Comparing $L_1$ with $L_2$, we see that only the first $s$ values of the lists are equal:&#xA;$$1, 3, 6, 10, 15$$&#xA;I have observed that this property holds with other values of $s$, $v$, and $n$, and $A$, $B$, and $C$.&#xA;&lt;br/&gt;&lt;br/&gt;Can someone please explain why $L_1$ and $L_2$ share the first $s$ values?&lt;/p&gt;&#xA;" OwnerUserId="2771" LastEditorUserId="2771" LastEditorDisplayName="" LastEditDate="2010-10-26T04:39:41.280" LastActivityDate="2010-10-26T04:51:26.470" Title="Combinatorics and Rolling Dice Similarity?" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="7866" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T04:34:21.457" Score="3" ViewCount="0" Body="&lt;p&gt;These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-26T04:34:21.457" CommunityOwnedDate="2010-10-26T20:30:52.290" CommentCount="1" />
  <row Id="7867" PostTypeId="1" CreationDate="2010-10-26T04:44:28.173" Score="0" ViewCount="140" Body="&lt;p&gt;I have an expected value say 5(minimum) or 6(minimum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I have 4 values [3, 10, 6, 8]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to compute the percentage on how far the 4 values are from the expected value in terms of percentage. Given 100% means they are all within the minimum and maximum and the farther they are from the expected the lower the percentage would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a mathematical way of computing this. One way is to assign true or false score for each. Meaning either they match or not the minimum or maximum value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3(0.0), 10(0.0), 6(1.0), 8(0.0)] would give 25%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this solution is too crude because I want to give more fine grained computation for how far I am from the expected value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3(0.0), 10(0.0), 6(1.0), 8(0.0)] would give X%&#xA;[1(0.0), 10(0.0), 6(1.0), 9(0.0)] would give Y% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Y% should be lower than X% since their value deviates farther from the mean than X%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do have a boundary of expected values, 1 to 30. I tried to read on statistics like standard deviation but it is the deviation from the mean. Even If i change it to deviation from expected value, How do I get percentage of how far is it from expected value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated:&#xA;I guess I already got an idea on this one. Let say there are N=20 available number to be taken. Then there are X=4 who would share this 20 slots. What I will do is get the average which is A=N/X=5. So I would compute for the summation of the difference of each item from the average. S=sum(abs(Ni-A)) I get the sum of the difference from the maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The percentage would be 1-(S/N)&lt;/p&gt;&#xA;" OwnerUserId="2813" LastEditorUserId="2813" LastEditorDisplayName="" LastEditDate="2010-10-27T04:22:21.003" LastActivityDate="2010-12-27T13:09:44.917" Title="How do you get a Percentage of how far the value is from expected Value" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7868" PostTypeId="1" AcceptedAnswerId="7874" CreationDate="2010-10-26T04:50:46.317" Score="2" ViewCount="81" Body="&lt;p&gt;What is the degree of the polynomial of the following expression ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [x + (x^3 – 1)^{1/2}]^5 + [x – (x^3 – 1)^{1/2}]^5 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am not very wrong the highest power of x is  $\frac {15}{2} $ ?! So the degree is floor(15/2) = 7 ?! The answer is 7 but I am not sure of this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please comment.&lt;/p&gt;&#xA;" OwnerUserId="2814" LastEditorUserId="2814" LastEditorDisplayName="" LastEditDate="2010-10-26T04:56:31.117" LastActivityDate="2010-10-26T05:03:30.440" Title="Degree of polynomial :$ [x + (x^3 – 1)^{1/2}]^5 + [x – (x^3 – 1)^{1/2}]^5 $" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7869" PostTypeId="2" ParentId="7865" CreationDate="2010-10-26T04:51:26.470" Score="5" ViewCount="0" Body="&lt;p&gt;The first $s$ terms of $L_1$ are the compositions of $A$.  They stop being the compositions of $A$ at that point because you hit the limit.  In your example, you miss the composition of 8 as 6+1+1 because the dice only have 5 sides.  The compositions of $n$ into exactly $k$ parts are given by ${{n - 1}\choose{k - 1}}$ as shown &lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T04:51:26.470" />
  <row Id="7870" PostTypeId="2" ParentId="7778" CreationDate="2010-10-26T04:55:39.623" Score="0" ViewCount="0" Body="&lt;p&gt;Here's an enumerative proof. Consider the following procedure for dividing a set $S$ of size $2n$ into two sets $A,B$ of size $n$. The procedure proceeds in $n$ rounds. We start with $A,B$ empty. In each round, let $m = \min S$ and pick some $x \in S \setminus m$. Now decide whether to put $m$ in $A$ and $x$ in $B$, or to put $m$ in $B$ and $x$ in $A$. Remove both $m,x$ from $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the procedure ends with the required partition, and requires $(2n-1)(2)(2n-3)(2)\cdots(1)(2)$ choices. Furthermore, I claim that each partition can be generated in $n!$ ways. Indeed, at each round, either $m \in A$ or $m \in B$; suppose wlog that $m \in A$. This forces our second choice, and for our force choice, we can choose an arbitrary member of $B$. So our total degrees of freedom are $(n)(n-1)\cdots(1)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-26T04:55:39.623" />
  <row Id="7871" PostTypeId="2" ParentId="7867" CreationDate="2010-10-26T04:57:48.873" Score="2" ViewCount="0" Body="&lt;p&gt;Do you mean 5 (minimum) to 6 (maximum)?  You could calculate how many ranges away each item is.  So 3 would get a score of 2 (from being 2 away from 5).  10 would get a score of 4 (10-6).  6 would get a score of 0 for being in range and 8 would get a score of 2.  This doesn't tell you what percentage are in range, but it does tell you how far out the others are.  Then if your range was 5 to 7 the scores would be 1 for 3, 3/2 for 10, and 1/2 for 8.  Maybe you need both figures.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T04:57:48.873" CommentCount="3" />
  <row Id="7872" PostTypeId="2" ParentId="7868" CreationDate="2010-10-26T05:00:46.660" Score="1" ViewCount="0" Body="&lt;p&gt;If you expand the 5th powers, the terms that have square roots in them will cancel.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T05:00:46.660" />
  <row Id="7873" PostTypeId="1" CreationDate="2010-10-26T05:02:16.667" Score="0" ViewCount="138" Body="&lt;p&gt;I'm solving this question, but I don't know what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;============================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a plane $x - y + z = 0$, and a point $b = (1, 2, 0)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Find a basis of this plane, and a basis of orthogonal subspace to this plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Find the closest point $\hat{b}$ on the plane to b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) What is the error between b and $\hat{b}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;============================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a linear algebra problem? What's a basis of orthogonal subspace and closest point $\hat{b}$? What's the error between b and $\hat{b}$??&lt;/p&gt;&#xA;" OwnerUserId="2674" LastActivityDate="2010-10-26T08:47:25.067" Title="Is this a linear algebra problem?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7874" PostTypeId="2" ParentId="7868" CreationDate="2010-10-26T05:03:30.440" Score="4" ViewCount="0" Body="&lt;p&gt;Usually, a polynomial is understood to have only non-negative powers of $x$ (i.e. $x^0=1,x^1=x,x^2$ etc.), so a-priori it's not even clear that this is a polynomial. However, if we open up the binomial theorem, we see that the odd powers cancel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x+y)^n + (x-y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} [y^k + (-y)^{k}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $k$ is odd, $y^k + (-y)^k$ cancels. In your case, the term corresponding to $k = 2l$ has degree $n-2l+3l = n+l$, and so the maximum is achieved for $l=2$ and $n+l = 7$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-26T05:03:30.440" CommentCount="1" />
  <row Id="7875" PostTypeId="2" ParentId="7835" CreationDate="2010-10-26T05:11:42.643" Score="1" ViewCount="0" Body="&lt;p&gt;Multiplying by $n!$, the left-hand side it the number of subsets of $[n]$ of odd size, and the right-hand side is $2^{n-1}$. This is because for each set $S \subseteq [n-1]$, exactly one of $S,S \cup {n}$ has odd size.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-26T05:11:42.643" />
  <row Id="7876" PostTypeId="1" CreationDate="2010-10-26T05:42:22.483" Score="2" ViewCount="76" Body="&lt;p&gt;If in the expansion of $(1 + x)^m \cdot (1 – x)^n $, the coefficients of $ x $ and $ x^2 $are 3 and -6 respectively, then m is ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved it in the following way :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding we get, the coefficient of $ x $ as $ (m-n) = 3 \cdots (1)$ and coefficient of $ x^2 $ as $ \frac{n(n-1)}{2} + \frac{m(m-1)}{2} +- m \cdot n = -6 \cdots (2)$ after substituting and some algebraic treatment I got m = 12 and n = 9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is correct but I am interested if there exists any short procedure such that the entire problem could be done under a mint.&lt;/p&gt;&#xA;" OwnerUserId="2814" LastEditorUserId="2814" LastEditorDisplayName="" LastEditDate="2010-10-26T06:07:35.113" LastActivityDate="2010-10-26T06:17:51.643" Title="How to calculate this efficiently ?" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7877" PostTypeId="2" ParentId="7876" CreationDate="2010-10-26T06:02:51.717" Score="3" ViewCount="0" Body="&lt;p&gt;For the coefficient of $x$ to be 3, $m-n=3$ as you said, so $m&gt;n$ and $m-n&gt;0$ and we can factor the original product:&#xA;$$\begin{align}&#xA;(1+x)^m(1-x)^n&amp;amp;=(1+x)^{m-n+n}(1-x)^n&#xA;\\&#xA;&amp;amp;=(1+x)^{m-n}(1+x)^n(1-x)^n&#xA;\\&#xA;&amp;amp;=(1+x)^{m-n}\left((1+x)(1-x)\right)^n&#xA;\\&#xA;&amp;amp;=(1+x)^{m-n}(1-x^2)^n&#xA;\\&#xA;&amp;amp;=(1+x)^3(1-x^2)^n&#xA;\end{align}$$&#xA;Now, the coefficient of $x^2$ is the sum of the coefficients of $x^2$ in $(1-x^2)^n=1-nx^2+\cdots$ and $(1+x)^3=1+3x+3x^2+x^3$, so $-6=-n+3$ and $n=9$.  Since $m-n=3$, $m=12$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-26T06:17:51.643" LastActivityDate="2010-10-26T06:17:51.643" CommentCount="9" />
  <row Id="7878" PostTypeId="2" ParentId="7873" CreationDate="2010-10-26T06:30:07.107" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, this is a linear algebra problem. Looks like a homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: You are working in $\mathbb{R}^3$. Your plane is given by 1 equation, which strongly suggest that it has dimension 2. Hence you should try to find 2 vectors linear independant and you will have your first basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, 2 vector spaces are orthogonal if any vector of the first one is orthogonal to any vector of the second one (if $v_1 = (x_1, y_1, z_1)^T$ and $v_2 = (x_2, y_2, z_2)^T$, then they are orthogonal if and only if $x_1x_2 + y_1y_2 + z_1z_2 = 0$). When we are talking about the orthogonal subspace in this context, you are supposed to find one with maximal dimension. For example $V = {(0,0,0)^T}$ is always orthogonal to any other subspace, therefore try to find one with dimension &gt; 0. Hint: the dimension of a subspace + dimension of it's orthogonal complement should equals the dimension of the space you are working in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have solved a), then we might help you to solve b) and c).&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-10-26T06:39:53.180" LastActivityDate="2010-10-26T06:39:53.180" />
  <row Id="7879" PostTypeId="1" AcceptedAnswerId="7880" CreationDate="2010-10-26T06:47:25.473" Score="1" ViewCount="25" Body="&lt;p&gt;Let $f(x,y)=1$ if $(x,y)$ is in the unit disk and $f=0$ otherwise. I would like to approximate $f$ by a sequence of smooth functions. The functions need to be evaluated quickly so the results I'm getting from mollification are to awkward to deal with... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance..&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="1794" LastEditorDisplayName="" LastEditDate="2010-10-26T07:03:26.790" LastActivityDate="2010-10-26T07:03:26.790" Title="Sequence of smooth functions approximating a 2d cylinder step function" Tags="&lt;approximation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7880" PostTypeId="2" ParentId="7879" CreationDate="2010-10-26T06:53:31.550" Score="3" ViewCount="0" Body="&lt;p&gt;You might want to try any of formulae 17 to 27 in the &lt;a href=&quot;http://mathworld.wolfram.com/HeavisideStepFunction.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld page for the unit step function&lt;/a&gt;, since the function you're interested in is expressible as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1-H(x^2+y^2-1)$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-26T06:53:31.550" />
  <row Id="7881" PostTypeId="1" AcceptedAnswerId="7903" CreationDate="2010-10-26T07:14:43.230" Score="3" ViewCount="150" Body="&lt;p&gt;For some collection of sets $A$, let $\sigma(A)$ denote the $\sigma$-algebra generated by $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be some collection of subsets of a set $Y$, and let $f$ be a function from some set $X$ to $Y$. I want to prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^{-1}(\sigma(C))=\sigma(f^{-1}(C))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could prove that &#xA;$$\sigma(f^{-1}(C)) \subset f^{-1}(\sigma(C))$$ &#xA;since complements and unions are 'preserved' by function inverse. But how do I go the other way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: One way to go the other way would be to argue that any set in $\sigma(C)$ must be built by &lt;em&gt;repeatedly&lt;/em&gt; applying the complement, union and intersection operations to elements of $C$ and all these operations are preserved when taking the inverse. The problem I am facing with the approach is formalizing the word &quot;repeatedly&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[not-homework]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T14:21:12.830" LastActivityDate="2010-11-21T14:21:12.830" Title="Preimage of generated $\sigma$-algebra" Tags="&lt;elementary-set-theory&gt;&lt;sigma-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7883" PostTypeId="1" AcceptedAnswerId="7891" CreationDate="2010-10-26T07:27:56.463" Score="1" ViewCount="91" Body="&lt;p&gt;This is a problem from Mathematical Analysis by Zorich, from the chapter on continuous functions.&#xA;This is also a homework problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition:&#xA;Let $P_n$ be a polynomial of degree $n$. For a function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already proved the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist a polynomial $P_0(x) = a$ of best approximation of degree 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Q_\lambda(x) = \lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck on proving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there exists a polynomial of best approximation of degree $n$, there also exists a polynomial of best approximation of degree $n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is to prove $\lambda_0 x^{n+1}+P_n$ is the $n+1$ best approximation for $f$. Where $\Delta(\lambda_0 x^{n+1}) =  \min_{\lambda\in \mathbb{R}} \Delta(\lambda x^{n+1})$ and $\Delta(P_n) = E(f(x) - \lambda_0 x^{n+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this approach is right. I'm stuck on how to proceed. &lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-10-26T09:24:03.243" Title="There exist a degree n+1 polynomial of best approximation if there exist a degree n polynomial of best approximation" Tags="&lt;analysis&gt;&lt;polynomials&gt;" AnswerCount="1" />
  <row Id="7884" PostTypeId="2" ParentId="7881" CreationDate="2010-10-26T07:56:13.927" Score="-2" ViewCount="0" Body="&lt;p&gt;Looks like a homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't have time to check the details (I am at work and my boss is watching), however, it seems to me that the following approach should work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: prove that $f^{-1}(\sigma(C))$ is a $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: you should be able to conclude, if you understand what $\sigma(f^{-1}(C))$ means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, you are right, I read your question too quickly. My steps only help you to get the inclusion you already had.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if this new approach is working, but I would:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/ consider $f(\sigma(f^{-1}(C)))$ and try to prove it is a $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ when 1/ is done, prove that it is a $\sigma$-algebra that contains $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3/ If 1/ and 2/ are ok, then that means we have $\sigma(C) \subset f(\sigma(f^{-1}(C)))$ and you can then conclude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that 1 is true, I don't have time to do the details however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ok, me second approach doesn't seem to be working either. I decided to do a little bit of research, and you can actually find the proof online. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S = ${ $B \in \mathcal{P}(Y); f^{-1}(B) \in \sigma(f^{-1}(C)) $}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $S$ is a $\sigma$-algebra that contains $C$. Therefore $\sigma(C) \subset S$ and $f^{-1}(\sigma(C)) \subset f^{-1}(S)$. But, by definition, $ f^{-1}(S) \subset \sigma(f^{-1}(C))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for my &quot;hints&quot;, they were not really useful.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-10-26T11:34:29.793" LastActivityDate="2010-10-26T11:34:29.793" CommentCount="3" />
  <row Id="7885" PostTypeId="2" ParentId="7811" CreationDate="2010-10-26T08:16:17.717" Score="5" ViewCount="0" Body="&lt;p&gt;For any prime $p$ and $n&gt;1$, there are nilpotent groups $G$ of class 2 and order $p^{n(n+1)/2}$ with generators $a_i$ $(1 \le i \le n)$, $b_{ij}$ $(1 \le i &amp;lt; j \le n)$, such that $[a_i,a_j] = b_{ij}$, the $b_{ij}$ are all central in the group, and all generators have order $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $G'$ is the group of order $p^{n(n-1)/2}$ generated by the $b_{ij}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any group, we have $[ax,by] = [a,b]$ when $x,y$ are central in the group, so $G$ has at most $p^{2n}$ distinct elements that are commutators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for any fixed $k&gt;0$, by choosing $n$ sufficiently large we can find $G$ such that not all element of $G'$ are products of at most $k$ commutators.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-10-26T08:16:17.717" CommentCount="1" />
  <row Id="7886" PostTypeId="1" AcceptedAnswerId="7887" CreationDate="2010-10-26T08:23:13.410" Score="3" ViewCount="129" Body="&lt;p&gt;How does one evaluate the limit: $$ \lim_{n \to \infty} \frac{1}{n}\sum\limits_{k=1}^{\lfloor{\frac{n}{2}\rfloor}} \cos\Bigl(\frac{k\pi}{n}\Bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, i recognize this as soon as i saw the problem: $$\int\limits_{0}^{1}f(x) \ dx = \lim_{n \to \infty} \frac{1}{n} \sum\limits_{k=1}^{n} f\Bigl(\frac{r}{n}\Bigr)$$ but the problem is there is $\lfloor{\frac{n}{2}\rfloor}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-26T08:30:14.730" LastActivityDate="2010-10-26T14:49:56.700" Title="Evaluating limit of Summation" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7887" PostTypeId="2" ParentId="7886" CreationDate="2010-10-26T08:35:47.903" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: Consider two separate limits.  The limit of n even going to infinity, and the limit of n odd going to infinity.  If they both converge to the same limit you are done.  Both problems should be easy to solve using your observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one converges to $1/2\int_0^1 cos(\pi x/2) dx$ and the second converges to $\int_0^{1/2}cos(\pi x) dx$&lt;/p&gt;&#xA;" OwnerUserId="780" LastEditorUserId="780" LastEditorDisplayName="" LastEditDate="2010-10-26T09:27:48.477" LastActivityDate="2010-10-26T09:27:48.477" CommentCount="8" />
  <row Id="7888" PostTypeId="2" ParentId="7873" CreationDate="2010-10-26T08:47:25.067" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;(a)&lt;/strong&gt; To find a basis for the plane $x-y+z = 0$, you could solve this equation in terms of $x$ and $z$: $y= x+z$. Then the set of vectors of your plane could be described as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = \left\{ (x, x+z, z) \ \vert \ x,z \in \mathbb{R}   \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this description it's easy to find a basis for your plane $V$: it will have two vectors, say $u,v \in \mathbb{R}^3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = [u,v] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the orthonormal subspace to your plane is a straight line, generated by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_product&quot; rel=&quot;nofollow&quot;&gt;cross product&lt;/a&gt; of $u$ and $v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V^\bot = [u \times v] \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b)&lt;/strong&gt; The closest point $\hat{b}$ to $b$ in the plane $V$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_projection&quot; rel=&quot;nofollow&quot;&gt;orthogonal projection&lt;/a&gt; of $b$ onto $V$. In this case, you don't really need the formulae you'll see in Wikipedia, because it's just the intersection of $V$ with the straight line with direction $u\times v$ that passes through $b$. That is, you have to solve the system of linear equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;x - y +z &amp;amp;= 0    \\&#xA;(x,y,z) &amp;amp;= (1,2,0) + \lambda u\times v  \ .&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(c)&lt;/strong&gt; The &quot;error&quot; between $b$ and $\hat{b}$ is the same as the distance from $b$ to $\hat{b}$; that is, $\| b - \hat{b} \|$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-26T08:47:25.067" />
  <row Id="7889" PostTypeId="1" CreationDate="2010-10-26T09:20:17.003" Score="0" ViewCount="88" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to find the class of all bounded functions $f(w,y)$ which satisfies the following inequality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int\limits_{-\infty}^{+\infty} f(w,y) ~\mathcal{N}(y,w,1) dy &gt;0 ~~~~\forall w&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is assumed that the function $f(w,y)$ is integrable w.r.t. to the Gaussina pdf $\mathcal{N}(y,w,1) $. For instance, the function $I_A(y)-0.5$, where $I_A$ is the indicator function over the set $A={(w,y): w-a\leq y&amp;lt;\infty}$ satisfies the inequality for each scalar $a&gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="2821" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-26T18:59:13.530" LastActivityDate="2010-10-26T18:59:13.530" Title="Functions with constant expected values" Tags="&lt;probability-theory&gt;&lt;integral&gt;&lt;expectation&gt;" />
  <row Id="7890" PostTypeId="1" CreationDate="2010-10-26T09:23:41.137" Score="2" ViewCount="98" Body="&lt;p&gt;According to Mathworld, hyperbolic geometry is well understood in 2 dimensions but not in 3 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/HyperbolicGeometry.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/HyperbolicGeometry.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What isn't well understood about hyperbolic geometry in 3D, that is in 2D or that is in 3D-Euclidean geometry?&lt;/p&gt;&#xA;" OwnerUserId="2822" LastActivityDate="2010-10-26T10:44:17.423" Title="Hyperbolic geometry. 3 dimensions. What is not well understood ?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7891" PostTypeId="2" ParentId="7883" CreationDate="2010-10-26T09:24:03.243" Score="2" ViewCount="0" Body="&lt;p&gt;Determining the best $n$th degree approximation explicitely is a very difficult problem and cannot be done by a simple induction (see the chapter on Tschebyscheff approximation in your favorite textbook of numerical analysis). But existence is easy, it follows from general principles: You have to set up a certain continuous function $\Phi$ on a certain compact set $K$ in a high-dimensional space.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-26T09:24:03.243" CommentCount="2" />
  <row Id="7892" PostTypeId="1" AcceptedAnswerId="9459" CreationDate="2010-10-26T09:29:48.480" Score="5" ViewCount="406" Body="&lt;p&gt;How to calculate without calculator or something like this the values of   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi^{e}$    and    $e^{\pi}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in oder to compare them?&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-08T16:13:04.513" LastActivityDate="2010-11-08T17:31:15.737" Title="Comparing $\pi^{e}$ and $e^{\pi}$" Tags="&lt;real-analysis&gt;&lt;inequality&gt;" AnswerCount="5" />
  <row Id="7893" PostTypeId="1" CreationDate="2010-10-26T09:34:14.160" Score="3" ViewCount="129" Body="&lt;p&gt;Let $\psi$ be a wavelet. Can its Fourier transform $\hat{\psi}$ be also wavelet? produce an example or prove that it is not possible. A wavelet is a function $\psi:\mathbb R\to\mathbb R$ such that (i) $\psi \in L^1(R) \cap L^2(R)$, (ii) $\int_{-\infty}^{\infty} \psi(t) dt = 0$, fourier transform is $\hat f(\omega) =\int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T11:10:24.810" LastActivityDate="2010-11-22T21:42:37.057" Title="Let $\psi$ be a wavelet. Can its Fourier transform $\hat{\psi}$ be also wavelet?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="7894" PostTypeId="2" ParentId="7892" CreationDate="2010-10-26T09:47:53.660" Score="16" ViewCount="0" Body="&lt;p&gt;This is an old chestnut. As a hint, it's easier to consider the more&#xA;general problem: for which positive $x$ is $e^x&gt;x^e$?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-26T09:47:53.660" CommentCount="8" />
  <row Id="7895" PostTypeId="2" ParentId="7892" CreationDate="2010-10-26T09:52:54.003" Score="2" ViewCount="0" Body="&lt;p&gt;Elaborating Robin's answer take $f(x) = \log{x} - \frac{x}{e}$.  We have $$f'(x)= \frac{e-x}{xe}$$ Thus $f'(x)&gt;0$ for $0 &amp;lt; x &amp;lt; e$ and $f'(x) &amp;lt;0$ if $x &gt; e$. Consequently, we have $f(x) &amp;lt; f(e)$ if $x \neq e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise:&lt;/strong&gt; Try to prove this using the same methods: $2^{\sqrt{2}} &amp;lt; e$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-26T10:01:45.413" LastActivityDate="2010-10-26T10:01:45.413" />
  <row Id="7896" PostTypeId="1" AcceptedAnswerId="7898" CreationDate="2010-10-26T10:13:24.387" Score="2" ViewCount="126" Body="&lt;p&gt;I was wondering this today, and my algebra professor didn't know the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are subgroups of finitely generated groups finitely generated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-10-26T10:27:26.653" Title="Subgroups of finitely generated groups" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="7897" PostTypeId="2" ParentId="7896" CreationDate="2010-10-26T10:18:21.240" Score="2" ViewCount="0" Body="&lt;p&gt;No.  The example given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_set_of_a_group#Finitely_generated_group&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; is that the free group $F_2$ contains a subgroup generated by $y^n x y^{-n}, n \ge 1$, which is free on countably many generators.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-26T10:18:21.240" />
  <row Id="7898" PostTypeId="2" ParentId="7896" CreationDate="2010-10-26T10:20:26.220" Score="4" ViewCount="0" Body="&lt;p&gt;It is well-known that the free group $F_2$ on two generators&#xA;has as a subgroup a group isomorphic to a free group on a countably&#xA;infinite set of generators. See Qiaochu's example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a &lt;strong&gt;finite index&lt;/strong&gt; subgroup of a finitely generated&#xA;group is finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-10-26T10:27:26.653" LastActivityDate="2010-10-26T10:27:26.653" />
  <row Id="7899" PostTypeId="2" ParentId="7896" CreationDate="2010-10-26T10:21:18.087" Score="0" ViewCount="0" Body="&lt;p&gt;See this for more details: &lt;a href=&quot;http://planetmath.org/encyclopedia/FinitelyGeneratedSubgroup.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/FinitelyGeneratedSubgroup.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-26T10:21:18.087" />
  <row Id="7900" PostTypeId="1" AcceptedAnswerId="7946" CreationDate="2010-10-26T10:36:25.470" Score="2" ViewCount="106" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Diablo 3 has &lt;em&gt;97 billion&lt;/em&gt; possible skill/trait builds. Per class.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sub&gt;&lt;a href=&quot;http://chat.gaming.stackexchange.com/rooms/3/conversation/97-billion&quot;&gt;LessPop_MoreFizz&lt;/a&gt;, emphasis is mine.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used base two logarithm to claim &quot;97 billion&quot; configurations only are roughly 37 binary choices because $log_2(97×10^9) \simeq 36.5$, but I would like to know if there are more refined techniques to remove the marketing out of these big numbers and see their true meaning.&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2010-10-26T18:15:25.560" Title="Making sense of combinatorics-based marketing hyperboles" Tags="&lt;combinatorics&gt;&lt;big-numbers&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7901" PostTypeId="2" ParentId="7890" CreationDate="2010-10-26T10:44:17.423" Score="1" ViewCount="0" Body="&lt;p&gt;I don't know what &quot;well-understood&quot; means in this context.  It is true that hyperbolic geometry is a very rich area in both dimensions two and three.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dimension two every closed surface admitting one hyperbolic structure admits uncountably many, and these fit together to form moduli space.  Moduli space is an object of intense study from the point of view of algebraic geometry, number theory, dynamical systems, etc.  The &quot;fundamental group&quot; of moduli space is the mapping class group (aka the modular group) and this is again studied by many different groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dimension three we have Mostow rigidity saying that a closed three-manifold admits either zero or essentially one hyperbolic structure.  It follows that the geometric invariants of the hyperbolic structure give topological invariants of the three-manifold, which was a wonderful shock to topologists in the 70's.  (I was recently told a story about Thurston wandering around a conference at Warwick, in the 70's, with a programmable calculator on his belt, saying that he was computing volumes of knots in the three-sphere.  Nobody had any idea what he was saying, apparently.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the moduli spaces of hyperbolic three-manifolds are much more simple.  To get an interesting theory one allows hyperbolic structures that are not complete (or discrete, etc) and then the moduli spaces become very wild and are not well understood.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-10-26T10:44:17.423" CommentCount="2" />
  <row Id="7902" PostTypeId="2" ParentId="7881" CreationDate="2010-10-26T11:05:26.667" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  The original answer that was here is wrong; Carl Mummert's answer contains the correct way to do what I was trying to do.  So instead here is the solution by transfinite induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a sequence of functions $\sigma_i$ where $i$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_number&quot; rel=&quot;nofollow&quot;&gt;ordinal&lt;/a&gt; as follows.  For any subset $C$ of the base set $E$ we take $\sigma_0(C) = C$.  If $i$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Successor_ordinal&quot; rel=&quot;nofollow&quot;&gt;successor ordinal&lt;/a&gt;, take $\sigma_i(C)$ to be the set of all countable unions or complements of elements of $\sigma_{i-1}(C)$; otherwise $i$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_ordinal&quot; rel=&quot;nofollow&quot;&gt;limit ordinal&lt;/a&gt; and we take $\sigma_i(C) = \bigcup_{j &amp;lt; i} \sigma_j(C)$.  For example, $\sigma_{\omega}(C)$ is the set of all sets that can be obtained from $C$ by performing countable union or complement countably many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would like to say that $\sigma(C)$ is the union of the $\sigma_i(C)$ over all ordinals $i$, but the ordinals don't form a set so we can't actually do this.  What I believe is true is that we only need to take ordinals of cardinality at most the cardinality of the powerset of the underlying set.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true, then the proof is as follows.  It's enough to show that $f^{-1}(\sigma_i(C)) = \sigma_i(f^{-1}(C))$ for all $i$.  This is obvious for $i = 0$.  If it's true for all ordinals $j &amp;lt; i$, then it's true for $i$ by the obvious argument.  And now we are done by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfinite_induction&quot; rel=&quot;nofollow&quot;&gt;principle of transfinite induction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me remark that, on the one hand, this is more complicated than Carl Mummert's proof because it requires knowledge of ordinals.  On the other hand, getting into the nitty-gritty like this really shows how complicated $\sigma$-algebras can be.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-26T14:17:45.693" LastActivityDate="2010-10-26T14:17:45.693" CommentCount="4" />
  <row Id="7903" PostTypeId="2" ParentId="7881" CreationDate="2010-10-26T11:16:37.420" Score="7" ViewCount="0" Body="&lt;p&gt;Yes, we &lt;em&gt;can&lt;/em&gt; use transfinite induction to prove this (formalizing the word &quot;repeatedly&quot;). That would be the bottom-up approach. There is also a top-down approach, using the characterization of $\sigma(C)$ as the smallest $\sigma$-algebra containing $C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key fact here is the the preimage operation commutes with all the set algebra operations: if $f \colon X \to Y$ then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$ &lt;/li&gt;&#xA;&lt;li&gt;$f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$ &lt;/li&gt;&#xA;&lt;li&gt;$f^{-1}(A - B) = f^{-1}(A) - f^{-1}(B)$. In particular $f^{-1}(Y - A) = X - f^{-1}(A)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so forth. So $f^{-1}\colon P(Y) \to P(X)$ is a lattice homomorphism on the powerset lattices, to use that terminology.  More importantly for us, the preimage operation even commutes with infinite unions and intersections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $f^{−1}(\sigma(C)) \subseteq \sigma(f^{−1}(C))$, you could follow this strategy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Show that $f^{-1}(C) \in \sigma(f^{-1}(C))$&lt;/li&gt;&#xA;&lt;li&gt;Show that if $f^{-1}(A_i) \in \sigma(f^{-1}(C))$ for $i \in \omega$ then  $f^{-1}(\bigcup A_i) \in \sigma(f^{-1}(C))$&lt;/li&gt;&#xA;&lt;li&gt;Show that if $f^{-1}(A) \in \sigma(f^{-1}(C))$ then $f^{-1}(Y - A) \in \sigma(f^{-1}(C))$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The point here is that, if we let $D$ be the collection of sets $A$ such that $f^{-1}(A) \in \sigma(f^{-1}(C))$, then $D$ is itself a $\sigma$ algebra containing $C$, which means $\sigma(C) \subseteq D$. But by the definition of $D$ this means $f^{-1}(\sigma(C)) \subseteq \sigma(f^{-1}(C))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the three bullets will require taking forward images under $f$. For example, for the third one, let $A$ be as stated. This means $f^{-1}(A)$ is in $\sigma(f^{-1}(C))$, which means that $X - f^{-1}(A)$ is also in $\sigma(f^{-1}(C))$. But $f^{-1}(Y - A)$ is exactly $X - f^{-1}(A)$, so we see that $f^{-1}(Y - A)$ is indeed in $\sigma(f^{-1}(C))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying point here is that the entire proof is algebraic and that a more general theorem is true: you can replace $f^{-1}$ with any other homomorphism of the powerset lattices that preserves countable unions. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-26T11:37:48.353" LastActivityDate="2010-10-26T11:37:48.353" CommentCount="5" />
  <row Id="7904" PostTypeId="2" ParentId="7893" CreationDate="2010-10-26T12:01:27.243" Score="4" ViewCount="0" Body="&lt;p&gt;$f(x)=\sin(x)\cdot\exp(-x^2)$ should do, because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The decay of $f$ ensure $f,\hat{f}\in L^1\cap L^\infty$. &lt;/li&gt;&#xA;&lt;li&gt;$f$ is odd $f(-x)=-f(x)$.&lt;/li&gt;&#xA;&lt;li&gt;$\hat{f}(-\xi)=\int_{-\infty}^\infty e^{ix\xi}f(x)dx=\int_{-\infty}^\infty e^{-i(-x)\xi}f(x)dx=\int_{\infty}^{\infty} e^{-it\xi}f(-t)(-dt)=-\hat{f}(\xi)$&#xD;&#xA;where we used that $f$ is odd.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;If $g$ is integrable and odd then $$\int_{-\infty}^0g(x)dx =\int_{-\infty}^0-g(-x)dx=\int_{+\infty}^0g(t)dt=-\int_0^{+\infty}g(t)dt$$&#xD;&#xA;hence &#xA;$$\int_{-\infty}^\infty g(t)dt = 0.$$&#xD;&#xA;This imply that $$\int_{-\infty}^\infty f dx= \int_{-\infty}^\infty\hat{f}d\xi=0$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-22T21:42:37.057" LastActivityDate="2010-11-22T21:42:37.057" CommentCount="2" />
  <row Id="7905" PostTypeId="2" ParentId="5645" CreationDate="2010-10-26T12:26:52.290" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose that $M$ satisfies all the easy things like extensionality, pairing, etc., plus $\Delta_0$ separation and suppose also that $M$ thinks for every ordinal $\alpha$ that $V_\alpha$ exists (that is, $V_\alpha^M$ exists in $M$). All this is expressible by a single axiom, but when $M$ is a proper class, it ensures that $M$ satisfies the entire Collection scheme, because for any formula $\varphi(x,y)$ and set $A\in M$ for which we want to collect, we may apply Collection in $V$ to find an ordinal $\alpha$ such that whenever $a\in A$ has $\exists y\varphi(a,y)$ in $M$, then such a $y$ may be found in $V_\alpha$ and hence in $V_\alpha^M$. Thus, $V_\alpha^M$ serves as a collection set inside $M$. Similarly, one can get full Separation in $M$ from $\Delta_0$ Separation (or much less, if you care to optimize it), by applying the Reflection theorem, which allows you to bound all the quantifiers by a suitably large $V_\alpha^M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument works only when $M$ is a proper class, however, because it appeals to Collection in $V$, and uses that $M$ grows taller than the resulting collection set. This method would break down completely when $M$ is a set, since the collection set in $V$ could be unbounded in $M$. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-10-26T12:26:52.290" CommentCount="1" />
  <row Id="7906" PostTypeId="2" ParentId="7900" CreationDate="2010-10-26T12:48:45.213" Score="2" ViewCount="0" Body="&lt;p&gt;I think this is a fine approach.  Some of the choices could be three-way, but to get a general idea of how many choices it takes to get that many total possibilities you are spot on.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T12:48:45.213" />
  <row Id="7907" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T13:04:45.623" Score="2" ViewCount="0" Body="&lt;p&gt;I have seen this one time too many&#xA;$$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-26T13:04:45.623" CommunityOwnedDate="2010-10-26T20:31:08.403" CommentCount="2" />
  <row Id="7908" PostTypeId="1" AcceptedAnswerId="7911" CreationDate="2010-10-26T13:40:24.583" Score="0" ViewCount="80" Body="&lt;p&gt;This is NOT a homework, I am over homework stage ages ago :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What function draws semicircle?&#xA;$$y=f(x) = \text{something}$$ &#xA;and I end up with a unit circle.&lt;/p&gt;&#xA;" OwnerUserId="2743" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T14:15:20.433" LastActivityDate="2010-10-26T14:15:20.433" Title="Function in Y semi-circle" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="7909" PostTypeId="2" ParentId="7908" CreationDate="2010-10-26T13:44:09.027" Score="2" ViewCount="0" Body="&lt;p&gt;On the interval $[-1,1]$ you can use the function $\sqrt{1-x^2}$. Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow&quot;&gt;Pythagorean theorem&lt;/a&gt; you can see that the graph of this function is the closed upper semicircle with radius 1 and centre (0,0).&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-26T13:44:09.027" />
  <row Id="7910" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T13:55:05.007" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Every torsion-free Abelian group is free.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This only holds for finitely generated Abelian groups.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-10-26T13:55:05.007" CommunityOwnedDate="2010-10-26T20:31:25.143" CommentCount="2" />
  <row Id="7911" PostTypeId="2" ParentId="7908" CreationDate="2010-10-26T13:58:18.760" Score="5" ViewCount="0" Body="&lt;p&gt;The only &quot;unit semicircles&quot; that can be expressed as $y=f(x)$ are the upper half of the unit circle and the lower half (any other half circle fails the &quot;vertical line test&quot;, so it cannot be expressed as an explicit function of $x$). Since the unit circle consists of all points $(x,y)$ that satisfy $x^2+y^2=1$, solving for $y$ gives you that $y=\sqrt{1-x^2}$ is the upper unit semicircle, and $y=-\sqrt{1-x^2}$ is the lower unit semicircle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are willing to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Parametric_equation&quot; rel=&quot;nofollow&quot;&gt;parametric equations&lt;/a&gt;, you can cover more possibilities. The unit circle can be parametrized as&#xA;$$\begin{array}{rcl}&#xA;x &amp;amp; = &amp;amp; \cos t,\\&#xA;y &amp;amp; = &amp;amp; \sin t;&#xA;\end{array}\qquad 0\leq t\leq 2\pi;$$&#xA;this &quot;traces&quot; the unit circle exactly once, counterclockwise, starting at $0$ radians (the point $(1,0)$). If you want the unit semicircle that starts at $\theta$ radians from the positive $x$-axis (and ends at $\theta+\pi$ radians), then you can obtain it parametrically by $x=\cos(t+\theta)$ and $y=\sin(t+\theta)$, $0\leq t\leq \pi$ (or simply $x=\cos t$, $y=\sin t$ with $\theta\leq t\leq \theta+\pi$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T13:58:18.760" CommentCount="0" />
  <row Id="7912" PostTypeId="1" AcceptedAnswerId="7913" CreationDate="2010-10-26T13:58:44.757" Score="1" ViewCount="166" Body="&lt;p&gt;I have to prove this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2^{k} \,\left|\, {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i did was to get a closed form for this summation, and i haven't been able to get one. Any ideas, or is there any other method for proving this.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T14:01:57.187" LastActivityDate="2010-10-26T19:46:06.853" Title=" $2^{k} \mid {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}$" Tags="&lt;combinatorics&gt;&lt;arithmetic&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7913" PostTypeId="2" ParentId="7912" CreationDate="2010-10-26T14:13:28.157" Score="9" ViewCount="0" Body="&lt;p&gt;The expression on the right just screams &quot;&lt;em&gt;Binomial Theorem! Binomial Theorem!&lt;/em&gt;&quot;. It's a quick matter to figure out &lt;em&gt;what&lt;/em&gt; to apply the Binomial Theorem to...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; So, how can we find out to what to apply the Binomial Theorem? Well, I can tell you what &lt;em&gt;my&lt;/em&gt; thought process was; I'm sure there are better or more direct ones. I'm going to take $(a+b)^{2k}$; $2k$ because the binomial coefficients are $2k$-choose-something. The $a$ should be equal to $1$, because we only have the $3$'s in the expansion. Oh, wait... the power of $3$ that shows up is &lt;em&gt;half&lt;/em&gt; the &quot;choose-something&quot;, and in the binomial theorem, if you take $(1+b)^{2k}$, then the $2i$th term is $\binom{2k}{2i}b^{2i}$... oh, well, let's take $b=3^{1/2}$ then! The exponents will work out. Of course, then now I only have the even powers... but that's okay, because the odd powers are going to have that pesky $\sqrt{3}$ in them. So, I'm going to look at $(1+\sqrt{3})^{2k}$. Let's take it out for a spin and see if that works... Well, what do you know? It does!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $(1+\sqrt{3})^{2k}$; using the binomial theorem, you get&#xA;$$(1+\sqrt{3})^{2k} = \binom{2k}{0}1^{2k} + \binom{2k}{1}1^{2k-1}3^{1/2}+\cdots + \binom{2k}{2k}3^{k}.$$&#xA;The left hand side is $(4 + 2\sqrt{3})^{k} = 2^k(2+\sqrt{3})^{k}$. This can be written uniquely as $x + y\sqrt{3}$ with $x,y\in\mathbb{Z}$, so $x$ is divisible by $2^k$. The right hand side can likewise be written uniquely as $a + b\sqrt{3}$, with $a,b\in\mathbb{Z}$, and $a$ is exactly the sum you have. So $a = x$, and $2^k$ divides $x$, giving the result you want.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T17:32:36.923" LastActivityDate="2010-10-26T17:32:36.923" CommentCount="3" />
  <row Id="7914" PostTypeId="2" ParentId="7892" CreationDate="2010-10-26T14:23:31.827" Score="5" ViewCount="0" Body="&lt;p&gt;Alternatively, we can compare $e^{1/e}$ and $\pi^{1/\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = x^{1/x}$. Then $f'(x) = x^{1/x} (1 - \log(x))/x^2$. Since $\log(x) &gt; 1$ for $x &gt; e$, we see that $f'(x) &amp;lt; 0$ for $e &amp;lt; x &amp;lt; \pi$. We conclude that $\pi^{1/\pi} &amp;lt; e^{1/e}$, and so $\pi^e &amp;lt; e^\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same calculation shows that $f(x)$ reaches its maximum at $e^{1/e}$, and so in general $x^e &amp;lt; e^x$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-26T19:03:40.620" LastActivityDate="2010-10-26T19:03:40.620" CommentCount="1" />
  <row Id="7915" PostTypeId="1" AcceptedAnswerId="7918" CreationDate="2010-10-26T14:26:18.923" Score="1" ViewCount="50" Body="&lt;p&gt;Hello! i have 2 perpendicular vectors in space . How can i determine the plane determined by the 2 vectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Alexandru Badescu&lt;/p&gt;&#xA;" OwnerUserId="2777" LastActivityDate="2010-10-26T14:32:05.057" Title="plane determined by 2 vectors" Tags="&lt;vector-space&gt;" AnswerCount="2" />
  <row Id="7916" PostTypeId="2" ParentId="7864" CreationDate="2010-10-26T14:28:30.360" Score="6" ViewCount="0" Body="&lt;p&gt;I recently caught myself thinking that the formula for the determinant of a 2-by-2 matrix also works for a block matrix, i.e. $det (A B; C D) = det(A)det(D) - det(B)det(C)$.&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-10-26T14:28:30.360" CommunityOwnedDate="2010-10-26T20:30:36.610" CommentCount="1" />
  <row Id="7917" PostTypeId="2" ParentId="7915" CreationDate="2010-10-26T14:30:45.400" Score="1" ViewCount="0" Body="&lt;p&gt;Take their cross product.  This will be perpendicular to the plane they determine.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T14:30:45.400" CommentCount="2" />
  <row Id="7918" PostTypeId="2" ParentId="7915" CreationDate="2010-10-26T14:32:05.057" Score="2" ViewCount="0" Body="&lt;p&gt;The plane determined by two noncollinear vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is the collection of all vectors of the form $\alpha\mathbf{v}_1 + \beta\mathbf{v}_2$, with $\alpha$ and $\beta$ scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by &quot;space&quot; you happen to mean $\mathbb{R}^3$, $\mathbf{v}_1=(a,b,c)$ and $\mathbf{v}_2=(r,s,t)$, then $(a,b,c)\times(r,s,t)$ (the cross product) is perpendicular to both $(a,b,c)$ and $(r,s,t)$, hence perpendicular to the plane they determine, so it will be the normal to the plane. One you have the normal, presumably you know how to get the usual equation of the plane. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T14:32:05.057" />
  <row Id="7919" PostTypeId="1" AcceptedAnswerId="7920" CreationDate="2010-10-26T14:40:16.993" Score="1" ViewCount="83" Body="&lt;p&gt;Venn diagrams captured my attention a few years ago. One night I decided to see if I could come up with some new Venn Diagrams. After a few hours, this is what I came up with. I'm a web developer not a mathematician, so I'm quite proud of this little creation. I've searched the internet and several books to confirm that this construction is unique. Lastly, I wanted to post it here because I know that the this crowd has more resources and knowledge than are available to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nzk3P.png&quot; alt=&quot;http://i55.tinypic.com/303jksh.jpg&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2828" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-26T14:43:44.030" LastActivityDate="2010-10-26T14:48:13.783" Title="Is this 2d five-fold Venn Diagram original and valid?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7920" PostTypeId="2" ParentId="7919" CreationDate="2010-10-26T14:48:13.783" Score="6" ViewCount="0" Body="&lt;p&gt;It seems okay to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to take a look at a very nice paper that appeard in the &lt;em&gt;Notices of the American Mathematical Society&lt;/em&gt; in 2006 about Venn diagrams:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ruskey, Frank; Savage, Carla D.; Wagon, Stan.&#xA;&lt;strong&gt;The search for simple symmetric Venn diagrams.&lt;/strong&gt;&#xA;&lt;em&gt;Notices Amer. Math. Soc.&lt;/em&gt; &lt;strong&gt;53&lt;/strong&gt; (2006), no. 11, 1304–1312.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At least I (an AMS member) can access it &lt;a href=&quot;http://www.ams.org/notices/200611/ea-wagon.pdf&quot; rel=&quot;nofollow&quot;&gt;in PDF format on the web&lt;/a&gt;. I don't know if the link will work for someone who is not a member, though. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T14:48:13.783" CommentCount="2" />
  <row Id="7921" PostTypeId="2" ParentId="7886" CreationDate="2010-10-26T14:49:56.700" Score="2" ViewCount="0" Body="&lt;p&gt;You can probably also use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{m} \cos (\phi + k \alpha) = \frac{\sin((m+1)\alpha/2)\cos(\phi + m\alpha/2)}{\sin \alpha/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be derived using complex numbers, or in a more elementary fashion, using the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2\cos(\phi + k \alpha) \sin(\alpha/2)$  $\displaystyle = \sin (\phi + \alpha(k+1/2)) - \sin(\phi + \alpha(k-1/2))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and telescoping the sum.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-26T14:49:56.700" />
  <row Id="7922" PostTypeId="2" ParentId="7757" CreationDate="2010-10-26T14:50:59.743" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a dirty trick.  $\frac{1}{2^n} \sum_{r \ge 0} r {n \choose r}$ is the expected size of a random subset of an $n$-element set.  But by linearity of expectation, this is $n$ times the probability that any given element is in a subset, which is $\frac{1}{2}$.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2^n} \sum_{r \ge 0} r {n \choose r} = \frac{n}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  And as long as I have this written up somewhere, I might as well use it.  In &lt;a href=&quot;http://math.stackexchange.com/questions/7912/2k-mid-2k-choose-0-cdot-30-2k-choose-2-cdot-31-cdots/7955#7955&quot;&gt;this math.SE answer&lt;/a&gt; I prove the following.  If $a_n, b_n$ are sequences satisfying $b_n = \sum_{k=0}^n {n \choose k} a_k$, and if $A(x) = \sum_{n \ge 0} a_n x^n, B(x) = \sum_{n \ge 0} b_n x^n$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \frac{1}{1 - x} A \left( \frac{x}{1 - x} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example $a_n = n$.  One can prove by various arguments that in this case $A(x) = \frac{x}{(1 - x)^2}$; this is a special case of an identity I use in the above answer.  It follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \frac{1}{1 - x} \left( \frac{ \frac{x}{1-x} }{ \left( 1 - \frac{x}{1-x} \right)^2 } \right) = \frac{x}{(1 - 2x)^2} = \frac{1}{2} \frac{2x}{(1 - 2x)^2} = \frac{1}{2} \sum_{n \ge 0} n \cdot 2^n x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-26T20:22:36.833" LastActivityDate="2010-10-26T20:22:36.833" CommentCount="1" />
  <row Id="7923" PostTypeId="1" CreationDate="2010-10-26T15:10:31.907" Score="5" ViewCount="415" Body="&lt;p&gt;are continuous functions always differentiable?&lt;/p&gt;&#xA;" OwnerUserId="2831" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-26T15:45:31.473" LastActivityDate="2010-10-27T06:21:23.473" Title="Are Continuous Functions Always Differentiable?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="7924" PostTypeId="1" AcceptedAnswerId="7929" CreationDate="2010-10-26T15:18:07.307" Score="4" ViewCount="182" Body="&lt;p&gt;I am learning about various signal processing methods in my university course, and I can't seem to grasp what 'energy' in signals represent. I mean, I know that it is the integral of the absolute value of energy squared (side question: why is absolute value needed?), but I don't know the meaning of what this energy is. Is there an intuitive way to think about these energies? Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="2673" LastActivityDate="2010-10-27T12:53:48.000" Title="How can I interpret &quot;energy&quot; in signals?" Tags="&lt;signal-processing&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7925" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T15:21:01.950" Score="15" ViewCount="0" Body="&lt;p&gt;No. &lt;a href=&quot;http://en.wikipedia.org/wiki/Karl_Weierstrass&quot; rel=&quot;nofollow&quot;&gt;Weierstraß&lt;/a&gt; gave in 1872 the first published example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;contiuous function that's nowhere differentiable&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-26T15:21:01.950" CommentCount="2" />
  <row Id="7926" PostTypeId="1" AcceptedAnswerId="7928" CreationDate="2010-10-26T15:33:03.453" Score="1" ViewCount="126" Body="&lt;p&gt;Notation question (I believe the notes that I'm reading uses pretty common notation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V $ be a vector space over a field $K$. What is $V^\mathbb{N}$? Is it a vector space of infinite dimensions? What are the elements of $V^\mathbb{N}$? Are they infinite length tuples?&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-10-26T16:02:57.067" Title="If $V$ is a vector space, that what is $V^\mathbb{N}$?" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7927" PostTypeId="2" ParentId="7912" CreationDate="2010-10-26T15:36:28.990" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;$\ \ $  Specialize $\rm\ a = 1,\ b = 3 \ $ below and then apply the binomial theorem to the RHS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\ \ \ 2\:|\:a-b\ \Rightarrow\ 2\:|\:(a\pm\sqrt b)^2\ \:in\ \: \mathbb Z[\sqrt b]\ \Rightarrow\ 2^k\ |\ (\:a+\sqrt b)^{2k} + (\:a-\sqrt b)^{2k}\:\ in\ \:\mathbb Z$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the sum is the trace of the $\rm\:$k'th power of an even quadratic integer so it is divisible by $\rm\:2^k\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in your similar recently posted problems the key is to recognize the sum as a slight perturbation of a well-known sum. Here one notices that the given sum is the even bisection of the binomial expansion of power of a quadratic integer, which immediately yields the result as above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-26T16:21:43.467" LastActivityDate="2010-10-26T16:21:43.467" CommentCount="1" />
  <row Id="7928" PostTypeId="2" ParentId="7926" CreationDate="2010-10-26T15:40:31.183" Score="7" ViewCount="0" Body="&lt;p&gt;In my opinion, this notation is ambiguous (edit: but maybe I'm wrong).  It could mean two things: it could mean the space of functions $\mathbb{N} \to V$, which is a vector space in the obvious way.  This is the countable product in the category of vector spaces.  This vector space is very large; in particular, it has uncountable dimension and it is &lt;em&gt;not&lt;/em&gt; spanned by the functions $e_i$ which are equal to $1$ at $i$ and equal to $0$ elsewhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it could mean the space of functions $\mathbb{N} \to V$ which are zero except at finitely many places, which is also a vector space in the obvious way, but countable-dimensional, and this is the vector space spanned by the $e_i$.  This is the countable &lt;em&gt;coproduct&lt;/em&gt; in the category of vector spaces.  (Note that it is not necessary to distinguish between finite products and finite coproducts because they are the same.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to guess, I would guess that it means the former.  Can you give a link to the notes?  It might be possible to figure out which one is meant from context.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-26T16:02:57.067" LastActivityDate="2010-10-26T16:02:57.067" CommentCount="9" />
  <row Id="7929" PostTypeId="2" ParentId="7924" CreationDate="2010-10-26T15:41:34.473" Score="7" ViewCount="0" Body="&lt;p&gt;The energy is the energy carried in the signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute value is probably because signal amplitudes are often represented by &lt;a href=&quot;http://en.wikipedia.org/wiki/Phasor&quot; rel=&quot;nofollow&quot;&gt;phasors&lt;/a&gt;, especially in the context of electromagnetic radiation or in the context of processing via Fourier analysis. And since it uses complex numbers, and the square of a complex number $z^2$ is not the same as its 'absolute value squared' $|z|^2$ (simplest case, take $z = -i$, the imaginary unit, then the former is -1 and the latter 1$), in the formla it is customary to write an absolute value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That we associate energy with the square of the wave amplitude is due to the mechanical model of small amplitude waves of d'Alembert. Essentially the wave medium is thought of as a bunch of springs coupled together. For springs, with Hooke's law we have that the mechanical energy is proportional to the square of the amplitude of displacement. The analogy that the displacement of the media due to the wave should be well approximated by a simple spring leads us to take the infinitesimal mechanical energy associated to an infinitesimal chunck of the medium to be the square of the wave amplitude. So adding them all up you get an integral. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-26T15:41:34.473" CommentCount="5" />
  <row Id="7930" PostTypeId="2" ParentId="7926" CreationDate="2010-10-26T15:46:36.267" Score="2" ViewCount="0" Body="&lt;p&gt;It would be the vector space consisting of an infinite product V&amp;nbsp;&amp;times;&amp;nbsp;V&amp;nbsp;&amp;times;&amp;nbsp;... of copies of V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally: for two sets A and B, A&lt;sup&gt;B&lt;/sup&gt; often denotes the set of functions &lt;em&gt;f&lt;/em&gt;&amp;thinsp;:&amp;thinsp;B&amp;nbsp;&amp;rarr;&amp;nbsp;A. (The only exception I'm aware of is when the sets A and B happen to be ordinals, in set theory; then a different notion of exponentiation is often used.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these two concepts related? Well: as a vector space, V is also a set of vectors. And so a function &lt;em&gt;f&lt;/em&gt;&amp;thinsp;:&amp;thinsp;&amp;#8469;&amp;nbsp;&amp;rarr;&amp;nbsp;V is essentially a block-vector description of a vector &lt;strong&gt;v&lt;/strong&gt;&amp;nbsp;=&amp;nbsp;(&lt;em&gt;f&lt;/em&gt;(0),&amp;nbsp;&lt;em&gt;f&lt;/em&gt;(1),&amp;nbsp;...)&amp;nbsp;&amp;isin;&amp;nbsp;V&amp;nbsp;&amp;times;&amp;nbsp;V&amp;nbsp;&amp;times;&amp;nbsp;...&amp;nbsp;.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-26T15:46:36.267" CommentCount="2" />
  <row Id="7931" PostTypeId="1" AcceptedAnswerId="7933" CreationDate="2010-10-26T15:52:16.840" Score="2" ViewCount="43" Body="&lt;p&gt;in R3 (3d) , having a vector perpendicular with a plane ( so we know where is 'up') , how do we determine if a certain point is below our plane ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Alexandru Badescu&lt;/p&gt;&#xA;" OwnerUserId="2777" LastActivityDate="2010-10-26T16:03:04.443" Title="point below a plane" Tags="&lt;vector-space&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7932" PostTypeId="2" ParentId="7926" CreationDate="2010-10-26T15:55:53.427" Score="1" ViewCount="0" Body="&lt;p&gt;I'd think of it as the set of sequences of elements of $V$, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V^\mathbb{N} = \prod_{n=1}^\infty V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't think there is a really nice description. In the case where $V$ is finite-dimensional, it'd just be $K^\mathbb{N}$ in a clever (and non-canonically isomorphic) disguise, but if $V$ were to be infinite-dimensional, $V^\mathbb{N}$ could be rather ugly. E.g. $\ell_1^\mathbb{N}$: sequences of absolutely summable sequences. Then a sequence of sequences would converge to a sequence iff it converged coordinatewise. Complicated and very confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Munkres' &quot;Topology&quot;, I prefer to use $V^\infty$ for set of sequences in $V$ for which only finitely many elements are nonzero, i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V^\infty = \bigoplus_{n=1}^\infty V = \left\lbrace (v_n) \in V^\mathbb{N} \;{\large\mid}\; \text{$v_n = 0$ from some index onwards}\right\rbrace. $$&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-26T15:55:53.427" />
  <row Id="7933" PostTypeId="2" ParentId="7931" CreationDate="2010-10-26T16:01:26.490" Score="1" ViewCount="0" Body="&lt;p&gt;Isn't this the same as your question &lt;a href=&quot;http://math.stackexchange.com/questions/7780/vector-question&quot;&gt;vector question&lt;/a&gt;?  Take the dot product of the perpendicular and the coordinates of the point.  If it is less than zero, the point is below the plane.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T16:01:26.490" CommentCount="1" />
  <row Id="7934" PostTypeId="2" ParentId="7931" CreationDate="2010-10-26T16:03:04.443" Score="5" ViewCount="0" Body="&lt;p&gt;If $v$ is the vector that points 'up' and $p_0$ is some point on your plane, and&#xA;finally $p$ is the point that might be below the plane, compute the dot product&#xA;$v \cdot (p-p_0)$.  This projects the vector to $p$ on the up-direction.&#xA;This product is $\lbrace -, 0, + \rbrace$ if $p$ is below, on, above the plane, respectively.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-10-26T16:03:04.443" CommentCount="1" />
  <row Id="7938" PostTypeId="1" AcceptedAnswerId="7940" CreationDate="2010-10-26T16:32:56.330" Score="5" ViewCount="224" Body="&lt;p&gt;One observes that $4!+1 =25=5^{2}$, $5!+1=121=11^{2}$ is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is $n!+1$ a perfect square for infinitely many $n$? If yes, then how to prove.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-27T05:05:45.630" LastActivityDate="2010-10-27T05:05:45.630" Title="$n!+1$ being a perfect square" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7939" PostTypeId="2" ParentId="7938" CreationDate="2010-10-26T16:41:33.707" Score="3" ViewCount="0" Body="&lt;p&gt;The sequence of factorials $n!+1$ which are also perfect squares is &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A085692&quot; rel=&quot;nofollow&quot;&gt;here in Sloane&lt;/a&gt;. It contains three terms, and notes that there are no more terms below $(10^9)!+1$, but as far as I  know there's no proof.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-26T16:41:33.707" />
  <row Id="7940" PostTypeId="2" ParentId="7938" CreationDate="2010-10-26T16:43:33.457" Score="10" ViewCount="0" Body="&lt;p&gt;This is Brocard's problem, and it is still open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Brocard%27s_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Brocard%27s_problem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-26T16:43:33.457" CommunityOwnedDate="2010-10-26T16:43:33.457" CommentCount="1" />
  <row Id="7941" PostTypeId="2" ParentId="7938" CreationDate="2010-10-26T16:44:15.780" Score="3" ViewCount="0" Body="&lt;p&gt;My intuition would be that there are very few.  There are just not many squares and even fewer factorials.  &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A025494&quot; rel=&quot;nofollow&quot;&gt;OEIS A025494&lt;/a&gt; lists the squares which are a sum of distinct factorials, which is less restrictive than what you ask and says the list is probably finite.  In particular, there are no more below 31!&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-26T16:44:15.780" />
  <row Id="7942" PostTypeId="2" ParentId="2601" CreationDate="2010-10-26T17:03:34.217" Score="0" ViewCount="0" Body="&lt;p&gt;Using parallel, vertical &quot;gates&quot; to visualize high-dimensional or even infinite cartesian products of intervals. It comes up in topological and measurable spaces (specifically in probability theory), and is nearly essential to intuiting convergence in these spaces.&lt;/p&gt;&#xA;" OwnerUserId="2122" LastActivityDate="2010-10-26T17:03:34.217" CommunityOwnedDate="2010-10-26T17:03:34.217" />
  <row Id="7943" PostTypeId="2" ParentId="7924" CreationDate="2010-10-26T17:11:25.087" Score="3" ViewCount="0" Body="&lt;p&gt;I try to give an explanation in the time and frequency domains, as normaly&#xA;used in Telecomunication Systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a physical process described in the time by an oscillating&#xA;signal $s(t)$ such as an electrical voltage or current, the description&#xA;in the frequency domain is given by its spectrum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overrightarrow{S}(\omega )=\frac{1}{%&#xA;2\pi }\int_{-\infty }^{\infty }s(t)e^{-i\omega t}dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal is $$%&#xA;s(t)=\int_{-\infty }^{\infty }\overrightarrow{S}(\omega )e^{-i\omega&#xA;t}d\omega ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega =2\pi f=\frac{2\pi }{T}$ is the angular frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $\overrightarrow{S}(\omega )$ is the &lt;em&gt;Fourier Transform&lt;/em&gt; of $s(t)$ and $s(t)$ is the inverse transform. Note: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;definition&lt;/a&gt; of the pair $(s(t),\overrightarrow{S}(\omega ))$  with different constant factors is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;energetic activity&quot; during the infinitesimal time interval $%&#xA;\delta t$ is $\delta E=s^{2}(t)\delta t$ (&lt;strong&gt;excluding a constant with physical&#xA;dimensions&lt;/strong&gt;, see note below). The instantaneous power is $$p(t)=\underset{\delta t\rightarrow 0%&#xA;}{\lim }\frac{\delta E}{\delta t}=s^{2}(t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the &quot;energy&quot; of the full process is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=\int_{-\infty }^{\infty }s^{2}(t)dt=2\pi \int_{-\infty }^{\infty&#xA;}S^{2}(\omega )d\omega ,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S(\omega )=\left\vert \overrightarrow{S}(\omega )\right\vert $ and $\overrightarrow{S}(\omega )=S(\omega )e^{i\psi (\omega )}$ is the&#xA;distribution or spectrum with the properties $S(-\omega )=S(\omega )$, $\psi&#xA;(-\omega )=\psi (\omega ).$ The equality between both integrals is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parseval%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note on the  physical dimensions&lt;/strong&gt;. e.g. in &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_network&quot; rel=&quot;nofollow&quot;&gt;electrical circuits&lt;/a&gt; the product of a voltage $V$ by a current $I$ is a power  (energy/time).  By the Ohm's law, in d.c. circuits, $V=RI$ and $VI=RI^2=\dfrac{1}{R}V^2$; in a.c. circuits $| \overrightarrow{V}|| \overrightarrow{I}|=R |\overrightarrow{I}|^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_density&quot; rel=&quot;nofollow&quot;&gt;From the Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The power spectral density (PSD)&#xA;  (...) describes how the power of a&#xA;  signal or time series is distributed&#xA;  with frequency. Here power can be the&#xA;  actual physical power, or more often,&#xA;  for convenience with abstract signals,&#xA;  can be defined as the squared value of&#xA;  the signal, that is, as the actual&#xA;  power dissipated in a load if the&#xA;  signal were a voltage applied to it.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-27T12:53:48.000" LastActivityDate="2010-10-27T12:53:48.000" CommentCount="3" />
  <row Id="7944" PostTypeId="1" AcceptedAnswerId="7948" CreationDate="2010-10-26T18:05:54.607" Score="6" ViewCount="121" Body="&lt;p&gt;I was trying to teach myself some basic Lie theory, and I came across this statement &lt;a href=&quot;http://mathworld.wolfram.com/Commutator.html&quot; rel=&quot;nofollow&quot;&gt;on Mathworld&lt;/a&gt;, relating the commutator of a group, $\alpha\beta\alpha^{-1}\beta^{-1}$, to the commutator of its Lie algebra, $[A,B] = AB-BA$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For instance, let $A$ and $B$ be square matrices, and let $\alpha(s)$ and $\beta(t)$ be paths in the Lie group of nonsingular matrices which satisfy&#xA;  $$\begin{align}&#xA;  \alpha(0)=\beta(0)    &amp;amp;=    I    \\&#xA;  \left.\frac{\partial\alpha}{\partial s}\right|_{s=0}    &amp;amp;=    A    \\&#xA;  \left.\frac{\partial\beta}{\partial s}\right|_{s=0}    &amp;amp;=    B,&lt;br&gt;&#xA;  \end{align}$$&#xA;  then&#xA;  $$\left.\frac{\partial}{\partial s}\frac{\partial}{\partial t}\alpha(s)\beta(t)\alpha^{-1}(s)\beta^{-1}(t)\right|_{(s=0,t=0)}=2[A,B].$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I tried to derive this for myself, using the fact that&#xA;$$\left.\frac{\partial\alpha^{-1}}{\partial s}\right|_{s=0} = \left.-\alpha^{-1}\frac{\partial\alpha}{\partial s}\alpha^{-1}\right|_{s=0} = -A,$$&#xA;I expanded the expression to get&#xA;$$\left.\frac{\partial}{\partial s}\frac{\partial}{\partial t}\alpha\beta\alpha^{-1}\beta^{-1}\right|_{(s=0,t=0)}$$&#xA;$$=\left.\left( \frac{\partial\alpha}{\partial s}\frac{\partial\beta}{\partial t}\alpha^{-1}\beta^{-1} + \alpha\frac{\partial\beta}{\partial t}\frac{\partial\alpha^{-1}}{\partial s}\beta^{-1} + \frac{\partial\alpha}{\partial s}\beta\alpha^{-1}\frac{\partial\beta^{-1}}{\partial t} + \alpha\beta\frac{\partial\alpha^{-1}}{\partial s}\frac{\partial\beta^{-1}}{\partial t} \right)\right|_{(s=0,t=0)}$$&#xA;$$=AB - BA - AB + AB$$&#xA;$$=[A,B].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that the factor of 2 is missing. This seems to agree with &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-755-introduction-to-lie-groups-fall-2004/lecture-notes/&quot; rel=&quot;nofollow&quot;&gt;the lecture notes I found on MIT OCW&lt;/a&gt;, which state (in Ch. 2, PDF 1) that if $X, Y \in \mathfrak{g}$,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\exp(-tX)\exp(-tY)\exp(tX)\exp(tY) = \exp\{t^2[X,Y]+O(t^3)\}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since this is not my area of expertise, I wanted to make sure I got things right before I contacted MathWorld about a typo. Have I done something wrong somewhere, or is the MathWorld statement actually an error?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-26T18:35:23.250" Title="On the relationship between the commutators of a Lie group and its Lie algebra" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="7945" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T18:10:19.370" Score="-2" ViewCount="0" Body="&lt;p&gt;differentiability implies continuity... the other way is not always true.&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2010-10-26T18:10:19.370" />
  <row Id="7946" PostTypeId="2" ParentId="7900" CreationDate="2010-10-26T18:15:25.560" Score="1" ViewCount="0" Body="&lt;p&gt;It's right and wrong.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a certain raw number of possibilities and let's assume from the earlier remarks that 97 billion is the correct number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the hype does not mention is that there is also some statistical distribution to these possibilities. Some are much likelier than others, and some selections with different &quot;genotypes&quot; have similar &quot;phenotypes&quot;: they are almost the same in effect.  The amount of information in one selection from the 97 billion is, on average, a lot less than the number of bits needed to specify a unique integer between 1 and 97000000000.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-26T18:15:25.560" />
  <row Id="7947" PostTypeId="2" ParentId="7924" CreationDate="2010-10-26T18:18:00.633" Score="1" ViewCount="0" Body="&lt;p&gt;The electrical power in a resistive circuit is given by $P = I V$. Frequently one has a fixed resistance (or more in general, impedance), so that voltage and current are variable; taking voltage as independent variable, we have $P = V^2/R$. This gives the instantaneous power (energy over time unit), and hence also gives the average power (or energy over some finite time interval) if the voltage is constant (DC current). When the voltage varies peridocally in time (for example, AC current), the above equation applies to each instant of time ($ P(t) = V(t)^2/R $) , so that, for getting the average power over a period one must integrate (&lt;a href=&quot;http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;). That's why AC energy is typically measured in RMS units: a 120V power outlet gives a sinuosid of approx. 170V peak voltage: that sinusoid &lt;em&gt;has the same energy&lt;/em&gt; as  a constant 120V voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the case of electric signals, but the analogy applies in other fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another (related) motivation comes from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parseval%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's theorem&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-10-26T18:18:00.633" />
  <row Id="7948" PostTypeId="2" ParentId="7944" CreationDate="2010-10-26T18:35:23.250" Score="1" ViewCount="0" Body="&lt;p&gt;Ignoring terms of order 3 and higher, and inverting 1+X using the power series,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1+A)(1+B)(1-A+A^2)(1-B+B^2)&#xA;= (1 + (A+B) + AB) (1 - (A+B) + (A^2 + AB + B^2))&#xA;= 1 - (A+B)^2 + AB + A^2 + AB + B^2 &#xA;= 1 - (A^2 + AB + BA + B^2) + 2AB + A^2 + B^2&#xA;= 1 + AB - BA&lt;br&gt;&#xA;= 1 + [A,B]&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-10-26T18:35:23.250" />
  <row Id="7949" PostTypeId="1" AcceptedAnswerId="13637" CreationDate="2010-10-26T18:42:51.067" Score="9" ViewCount="160" Body="&lt;p&gt;Let $\mathbf C$ be an abelian category admitting projective limits. Let's consider the category whose objects are those of the form&#xA;$$&#xA;0\to A\to B\to C\to 0&#xA;$$&#xA;and whose morphisms are triples of morphisms of $\mathbf C$ such that the diagram&#xA;$$&#xA;\begin{array}{ccccc}&#xA;A&amp;amp;\hookrightarrow &amp;amp; B &amp;amp; \to &amp;amp; C &amp;amp; \newline&#xA;\downarrow &amp;amp;&amp;amp;\downarrow &amp;amp;&amp;amp; \downarrow \newline&#xA;A' &amp;amp;\hookrightarrow &amp;amp; B' &amp;amp;\to &amp;amp;  C'&#xA;\end{array}&#xA;$$&#xA;commutes in all its parts. Call this category $\boldsymbol\Sigma(\mathbf C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could one characterize the limits in $\boldsymbol\Sigma(\mathbf C)$? A little meditation shows that inverse systems are objects of the form&#xA;$$&#xA;\mathcal E_i\colon 0\rightarrow A_i\rightarrow B_i\rightarrow C_i\to 0&#xA;$$&#xA;(every object in the sequence can be thought as an element in a separate inverse system), so the universal property of $\varprojlim_\mathbf J \mathcal E_i$, whatever it turns out to be, must be enjoyed by&#xA;$$&#xA;\textstyle \varprojlim_\mathbf J \mathcal E_i : 0\to \varprojlim_\mathbf J A_i\to \varprojlim_\mathbf J B_i \to \varprojlim_\mathbf J C_i\to 0&#xA;$$&#xA;as soon as one looks to  ${A_i},{B_i},{C_i}$ as three inverse systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What condition(s?) has(ve?) to be imposed on them to make sure the sequence is exact?&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-09T10:22:23.303" Title="Limits in the category of exact  sequences" Tags="&lt;category-theory&gt;&lt;abelian-categories&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="7950" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T18:43:13.230" Score="2" ViewCount="0" Body="&lt;p&gt;For a nice simple example of an everywhere continuous, nowhere differentiable function it's hard to beat &lt;a href=&quot;http://www-formal.stanford.edu/jmc/weierstrass.pdf&quot; rel=&quot;nofollow&quot;&gt;this example&lt;/a&gt; of John McCarthy.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-26T18:43:13.230" CommentCount="1" />
  <row Id="7951" PostTypeId="1" AcceptedAnswerId="7960" CreationDate="2010-10-26T18:50:35.453" Score="1" ViewCount="173" Body="&lt;p&gt;If I have 5 students (A-E) that score 80%, 70%, 70%, 60% and 50% on a test what percentiles do they fall in?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A - 20th percentile (80%)&lt;/li&gt;&#xA;&lt;li&gt;B - 40th percentile (70%)&lt;/li&gt;&#xA;&lt;li&gt;C - 40th percentile (70%)&lt;/li&gt;&#xA;&lt;li&gt;D - 80th percentile (60%)&lt;/li&gt;&#xA;&lt;li&gt;E - 100th percentile (50%)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me expand the question as I wasn't clear earlier: If A is in the top 20th percentile, are B and C in the top 40th percentile or top 60th percentile?&lt;/p&gt;&#xA;" OwnerUserId="2835" LastEditorUserId="2835" LastEditorDisplayName="" LastEditDate="2010-10-27T00:43:23.840" LastActivityDate="2010-10-27T02:45:55.840" Title="How do percentiles work?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7952" PostTypeId="1" AcceptedAnswerId="7957" CreationDate="2010-10-26T19:22:39.827" Score="2" ViewCount="80" Body="&lt;p&gt;This is a question on the homework for my finite fields class. The beginning of the assignment defines the following notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $i\geq1 $, define the following elements of $A=\mathbb{F}_{q}[T]$ : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[i]=T^{q^{i}}-T$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{i}=[i][i-1]\cdots[1]$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_{i}=[i][i-1]^{q}\cdots[1]^{q^{i-1}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and put $D_{0}=L_{0}=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n\geq1$.&#xA;Put $A(n):=\{a \in A\mid \text{deg} (a)&amp;lt; n \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $e_{n}(x):=\prod_{a\in A(n)}\left(x+a\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to prove: $\sum\frac{1}{b}=\frac{(-1)^{n}}{L_{n}}$ where the LHS ranges over all monic $b$ in $A$ of degree equal to $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier on the assignment, we showed $e_{n}(x)+D_{n}=\prod(x+b)$ where $b$ is monic with degree $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I would be able to take the Carlitz analog of the logarithm $\log_{c}(x)$ and then take the formal derivative and evaluate it at zero. But I've realized it is not true that $\log_{c}(xy)=\log_{c}(x)+\log_{c}(y)$. I've been stuck on this problem for some time, and can't think of any other way to approach it. I would appreciate a nudge in the right direction. &lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-26T19:25:55.567" LastActivityDate="2010-10-27T19:11:09.583" Title="Using Carlitz's exponential formula to prove an identity" Tags="&lt;homework&gt;&lt;finite-fields&gt;" AnswerCount="2" />
  <row Id="7954" PostTypeId="1" AcceptedAnswerId="7968" CreationDate="2010-10-26T19:33:27.977" Score="5" ViewCount="163" Body="&lt;p&gt;Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)?&#xA;Are there any interesting examples of this case?&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2010-10-26T20:03:18.730" LastActivityDate="2010-10-26T21:29:40.077" Title="Nil-Radical  equals Jacobson Radical even though not every prime ideal is maximal?" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="7955" PostTypeId="2" ParentId="7912" CreationDate="2010-10-26T19:46:06.853" Score="5" ViewCount="0" Body="&lt;p&gt;Let's talk about some general methods.  This is a special case of the following problem: given that you understand a sequence $a_n$, how can you understand the sequence $b_n = \sum_{k=0}^n {n \choose k} a_k$?  Generating function techniques will come to your rescue.  If you let $A(x) = \sum_{n \ge 0} a_n x^n$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{n \ge 0} b_n x^n = \sum_{n \ge 0} \sum_{k=0}^{n} {n \choose k} a_k x^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we employ one of the simplest but most effective tools in &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;, namely we exchange the order of summation.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{k \ge 0} \sum_{n \ge k} {n \choose k} a_k x^n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we want an answer in terms of $A$, it seems natural to factor $a_k x^k$ out of the inner sum.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{k \ge 0} a_k x^k \sum_{n \ge k} {n \choose k} x^{n-k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a generatingfunctionologist will immediately recognize the inner sum as $\sum_{m \ge 0} {m+k \choose k} x^m = \frac{1}{(1 - x)^{k+1}}$.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \sum_{k \ge 0} a^k \frac{x^k}{(1 - x)^{k+1}} = \frac{1}{1 - x} A \left( \frac{x}{1 - x} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beautiful.  Now, in this problem we are dealing with the sequence $a_{2k} = 3^k, a_{2k+1} = 0$.  This has generating function $A(x) = \frac{1}{1 - 3x^2}$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(x) = \frac{1}{1 - x} \left( \frac{1}{1 - \frac{3x^2}{(1 - x)^2} } \right) = \frac{1 - x}{(1 - x)^2 - 3x^2} = \frac{1 - x}{1 - 2x - 2x^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what we want to prove is that the even coefficients of $B \left( \frac{x}{\sqrt{2}} \right)$ are integral.  To extract the even coefficients we appeal to another general tool in generatingfunctionology: if $B(x) = \sum_{n \ge 0} b_n x^n$ is a generating function you want to extract the even coefficients from, then these are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} b_{2n} x^{2n} = \frac{B(x) + B(-x)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;discrete Fourier transform&lt;/a&gt;.  Once you know this, it is completely mechanical to verify that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{B \left( \frac{x}{\sqrt{2}} \right) + B \left( \frac{-x}{\sqrt{2}} \right)}{2} = \frac{1 - x^2}{1 - 4x^2 + x^4}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has integer coefficients (since the constant term of the denominator is $1$).  This is a bit tedious, but my point here is that once one has picked up a few generating function techniques most of the work in solving problems like these is automatic.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-26T19:46:06.853" CommentCount="1" />
  <row Id="7956" PostTypeId="2" ParentId="7954" CreationDate="2010-10-26T19:49:30.177" Score="4" ViewCount="0" Body="&lt;p&gt;Let $k$ be a field, and let $R=k[x_1,\ldots,x_n]/\mathfrak{I}$, where $\mathfrak{I}$ is an ideal of $k[x_1,\ldots,x_n]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nilradical of $R$ is $\sqrt{\mathfrak{I}}/\mathfrak{I}$, and the Jacobson radical of $R$ is $\mathfrak{M}/\mathfrak{I}$, where $\mathfrak{M}$ is the intersection of all maximal ideals of $k[x_1,\ldots,x_n]$ that contain $\mathfrak{I}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by Hilbert's Nullstellensatz, $\sqrt{\mathfrak{I}}$, the radical of $\mathfrak{I}$, &lt;em&gt;is&lt;/em&gt; the intersection of all maximal ideals of $k[x_1,\ldots,x_n]$ that contain $\mathfrak{I}$, so the nilradical of $R$ equals the Jacobson radical of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in these rings there are generally prime ideals that are not maximal. For example, if $n\geq 3$ and you take $\mathfrak{I}=(x_1)$, then $(x_1,x_2)+\mathfrak{I}$ is prime but not maximal. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-26T19:49:30.177" />
  <row Id="7957" PostTypeId="2" ParentId="7952" CreationDate="2010-10-26T19:49:59.490" Score="3" ViewCount="0" Body="&lt;p&gt;From $e_n(x)+D_n=\prod(x+b)$&#xA;it follows that $\prod b=D_n$ and the $x^1$ coefficient of $e_n(x)$&#xA;is $D_n\sum 1/b$. But the $x^1$ coefficient is equal to $\prod'a$&#xA;where the product is over all nonzero polynomials $a$ of degree $ &amp;lt; n$.&#xA;All you have to do is to prove this product equals $(-1)^nD_n/L_n$.&#xA;The leading coefficient should be easy enough. For the rest you need&#xA;to do a census of the degrees of the irreducible factors of polynomials&#xA;of degree $ &amp;lt; n$. Possibly some earlier results in your course might&#xA;cover this?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-26T19:49:59.490" CommentCount="1" />
  <row Id="7960" PostTypeId="2" ParentId="7951" CreationDate="2010-10-26T20:02:55.093" Score="4" ViewCount="0" Body="&lt;p&gt;You've got them backwards. The 20th percentile would be the level below which only 20% of the observations fall. Since 80% of the scores fall below A's score (four out of 5), then A is in the 80th percentile. Since 20% of the scores fall below D's score, then D is in the 20th percentile, not the 80th. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Percentile&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page&lt;/a&gt; on percentiles (it's rather rough, but the lead will do it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; You edited the question, so here goes the answer to that: &#xA;It seems that you are not really asking about &lt;em&gt;percentiles&lt;/em&gt;, but about &lt;strong&gt;percents&lt;/strong&gt;. They are not the same thing. A is in the top 20 percent, but not &quot;the top 20 percentile&quot;. By &lt;em&gt;definition&lt;/em&gt;, &lt;strong&gt;percentile&lt;/strong&gt; refers to how many quantities are &lt;em&gt;below&lt;/em&gt;. It makes no sense to talk about &quot;top 20 percentile&quot;.  Your intended question is about percents, not 'percentiles'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer that: You have 5 people; A is the top, B and C are tied in second, D is in fourth, and E in fifth. A is certainly in the top 20% of the group, E in the bottom 20%; D and E are in the bottom 40%. A, B, and C are in the top 60% (not the top 40%); B, C, D, and E are in the bottom 80%. But you are using &quot;percentile&quot; wrong. If X is in the 40th percentile, then that means that 40% of the scores are below him; that would mean he is at the bottom of the top 60%. A is in the 80 percentile, B and C are in the 40th percentile, D is in the 20th percentile. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T02:45:55.840" LastActivityDate="2010-10-27T02:45:55.840" />
  <row Id="7961" PostTypeId="1" CreationDate="2010-10-26T20:04:54.647" Score="1" ViewCount="48" Body="&lt;p&gt;Suppose for company A, there is a $60 \%$ chance that no claim is made in the coming year. If one or more claims are made, then the total amount is normally distributed with mean $1000$ and standard deviation $2000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for company B, there is a $70 \%$ chance that no claim is made during the coming year. If one or more claims are made, then the total amount is normally distributed with mean $9000$ and standard deviation $2000$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the total claim amounts are independent. &lt;strong&gt;What is the probability that in the coming year, company B's total claim amount will exceed company A's total claim amount?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_A$ be company A's total claim amount and $X_B$ likewise. We want to find $P(X_B - X_A &gt;0)$. Now $E[X_B-X_A] = 8000$ and $\text{Var}[X_B-X_A] = 2000^2+2000^2$. So what we want to find is:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle P \left[Z \geq \frac{-8000}{\sqrt{2000^2+2000^2}} \right]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that we are given two conditional distributions: $P(X_A|I_A)$ and $P(X_B|I_B)$ where $I_A$ and $I_B$ indicate that more than one claim is made for company A and B respectively.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-10-26T20:31:54.517" LastActivityDate="2010-10-26T20:46:47.243" Title="Normal Distributions and Claim Amounts" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="7962" PostTypeId="2" ParentId="7961" CreationDate="2010-10-26T20:19:05.690" Score="1" ViewCount="0" Body="&lt;p&gt;As the expected difference in claims is 2.8 standard deviations, there is little (how little?) chance that A's claims will exceed B's claims assuming they each get one.  If we ignore it, and assume that the chance of a claim at A is independent of the chance of a claim at B (can we?)  We have probability of no claim either place, so a tie, =60%*70%=42%.  Probability of claim at A, none at B =40%*70%=28% chance of higher claim at A.  Chance of higher claim at B=30%, as if there is a claim it will be higher than A's.  These add to 100%, a good check.  I'll let you figure the chance that they both get a claim and A's is higher than B's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also a little hard to fathom a claim being made on A which is less than zero, but if you are only 1/2 standard deviation low, that will happen.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-26T20:46:47.243" LastActivityDate="2010-10-26T20:46:47.243" CommentCount="4" />
  <row Id="7963" PostTypeId="1" AcceptedAnswerId="7964" CreationDate="2010-10-26T20:43:21.010" Score="0" ViewCount="112" Body="&lt;p&gt;I'm looking for a function that will replicate logical operators. In particular consider the AND operator. Then I am looking for a function z = f(x,y) such that&#xA;$$x,y,z \in \{0,1\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(0,0)=0&#xA;f(1,0)=0&#xA;f(0,1)=0&#xA;f(1,1)=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly I need to generalize over a space where Xi,Yi ∈ {0,1} and I need to do this for other sorts of operators too. (X XOR Y AND Z for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given a vector Xi and a vector Yi I need to be able to generate Zi w&lt;/p&gt;&#xA;" OwnerUserId="2121" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T02:37:29.687" LastActivityDate="2010-10-27T16:44:05.623" Title="How to represent logical operators as functions?" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="7964" PostTypeId="2" ParentId="7963" CreationDate="2010-10-26T20:47:08.930" Score="6" ViewCount="0" Body="&lt;p&gt;If $x, y \in [0, 1]$ you can take $f(x, y) = xy$.  This comes from replacing &quot;true&quot; and &quot;false&quot; with probabilities and from assuming that $x, y$ are probabilities of independent events.  Similarly for NOT take $f(x) = 1 - x$, and for OR take $f(x, y) = 1 - (1 - x)(1 - y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't what you're looking for then you're going to have to be more precise about what you want.  If you want to do something with vectors with entries in $\{ 0, 1 \}$, what's wrong with applying the operations pointwise?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-26T20:47:08.930" CommentCount="6" />
  <row Id="7965" PostTypeId="1" CreationDate="2010-10-26T20:48:16.740" Score="5" ViewCount="99" Body="&lt;p&gt;I'm aware that one can use a Fast Fourier Transform (FFT) to take the cost of multiplication of two polynomials of degree N from O$(N^2)$ to O$(N \ln N)$ (which is an amazing reduction when dealing with large polynomials!). Does a similar transformation procedure exist for multinomials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in the special case where the number of independent variables is only two, ie. $h(x,y) = f(x,y)g(x,y)$, but I'd love to read up on the general procedure. &lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-10-26T20:56:42.473" LastActivityDate="2010-10-26T20:56:42.473" Title="Use of FFT in the multiplication of multinomials" Tags="&lt;algorithms&gt;&lt;polynomials&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="7966" PostTypeId="1" AcceptedAnswerId="9539" CreationDate="2010-10-26T21:16:44.867" Score="2" ViewCount="161" Body="&lt;p&gt;What are good expositions of spectral sequences, which include a thorough introduction to the topic as well as the most important examples of applications - maybe with an emphasis an topological and/or operator algebraic uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the wikipedia article is quite extensive, but I'd like to have something... nicer.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-12-02T06:06:41.990" Title="Reference for spectral sequences" Tags="&lt;reference-request&gt;&lt;algebraic-topology&gt;&lt;homological-algebra&gt;&lt;operator-algebras&gt;" AnswerCount="5" CommentCount="7" />
  <row Id="7967" PostTypeId="2" ParentId="7966" CreationDate="2010-10-26T21:29:15.183" Score="2" ViewCount="0" Body="&lt;p&gt;I really like the last chapter of Aluffi's book &quot;Algebra: Chapter 0.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps one of the canonical references (especially to see tons of applications) is John McCleary's &quot;A User's Guide to Spectral Sequences.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2010-10-26T21:29:15.183" />
  <row Id="7968" PostTypeId="2" ParentId="7954" CreationDate="2010-10-26T21:29:40.077" Score="5" ViewCount="0" Body="&lt;p&gt;There are indeed very many rings in which the nilradical equals the Jacobson radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it.  These rings are called &lt;strong&gt;Hilbert-Jacobson&lt;/strong&gt; rings.  In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed there are very many Hilbert-Jacobson rings.  One useful result in this direction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra.  Then $S$ is also a Hilbert-Jacobson ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some information on this subject, including a proof of the theorem, see Section 8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in particular Theorem 128 on p. 82.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-26T21:29:40.077" />
  <row Id="7973" PostTypeId="2" ParentId="7923" CreationDate="2010-10-26T22:23:47.313" Score="4" ViewCount="0" Body="&lt;p&gt;No, consider the example of $f(x) = |x|$. This function is continuous but not differentiable at $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are even more bizare functions that are not differentiable everywhere, yet still continuous. This class of functions lead to the development of the study of fractals.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-10-26T22:23:47.313" />
  <row Id="7974" PostTypeId="1" AcceptedAnswerId="8043" CreationDate="2010-10-26T22:43:53.690" Score="1" ViewCount="60" Body="&lt;p&gt;I'm trying to sample a lot of points efficiently.  I'm wondering if the following method is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sample points of a function (evaluate the function) mod $n$. I.e. I calculate f(element one), f(element 2)...f(element n).  I do this for all of the points in a ring.  Now I take a second ring (mod $m$, for instance) and calculate the points of a function for all elements of this ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my idea is this: I've sampled only $n+m$ points, but the combinations of these sample points number $n \cdot m$.  I'd like to calculate the sum of (evaluations of) this larger set of points efficiently.  I'm wondering if this is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know as much about this idea as possible.  I'd like to know related ideas, what's possible, or just general ideas on how to do this.  Any pointers would constitute an acceptable answer to me, and an actual method would be absolutely wonderful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this may be more of a numerical method than mathematics, but I'm hoping that everyone will accept this question, since it may allow new methods of integration as a result.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-10-27T13:51:14.493" Title="Efficient Sampling" Tags="&lt;analysis&gt;&lt;intuition&gt;&lt;statistics&gt;&lt;numerical-methods&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="7976" PostTypeId="2" ParentId="7512" CreationDate="2010-10-26T23:44:06.313" Score="1" ViewCount="0" Body="&lt;p&gt;We used Conway (about 3/4th) of the book for functional analysis. It was very dense and I felt like I wanted to stab myself in the guts, but I made it! ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with the other posts that you need to brush up some material before you do this.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-10-26T23:44:06.313" />
  <row Id="7978" PostTypeId="1" AcceptedAnswerId="7985" CreationDate="2010-10-27T00:58:55.023" Score="3" ViewCount="131" Body="&lt;p&gt;Is there a succinct proof for the fact that the rank of a non-zero skew-symmetric matrix ($A = -A^T$)  is at least 2? I can think of a proof by contradiction: Assume rank is 1. Then you express all other rows as multiple of the first row. Using skew-symmetric property, this matrix has to be a zero matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does such a matrix have at least 2 non-zero eigenvalues?&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-27T02:45:27.853" Title="Rank of skew-symmetric matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7979" PostTypeId="2" ParentId="7966" CreationDate="2010-10-27T01:02:18.833" Score="2" ViewCount="0" Body="&lt;p&gt;Allen Hatcher has some textbooks available for free download on his website, including the first 100 pages of an unfinished text on &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/SSAT/SSATpage.html&quot; rel=&quot;nofollow&quot;&gt;Spectral Sequences in Algebraic Topology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should qualify as a thorough first introduction to the topic, but I don't think all of the important examples have made it into the text.  I haven't read it myself, but I remember his more basic text on algebraic topology being very friendly and a good introduction.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-10-27T01:02:18.833" />
  <row Id="7980" PostTypeId="2" ParentId="7864" CreationDate="2010-10-27T01:03:59.343" Score="2" ViewCount="0" Body="&lt;p&gt;The question I've heard on many levels (including the grad level): what is the square root of $a^2$? And everyone says: it's $a$! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it is $|a|$. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-27T01:03:59.343" CommunityOwnedDate="2010-10-27T01:03:59.343" CommentCount="2" />
  <row Id="7981" PostTypeId="1" AcceptedAnswerId="7987" CreationDate="2010-10-27T01:31:23.443" Score="11" ViewCount="1021" Body="&lt;p&gt;I've read so much about it but none of it makes a lot of sense. Also, what's so unsolvable about it?&lt;/p&gt;&#xA;" OwnerUserId="1141" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T05:00:33.927" LastActivityDate="2010-10-28T12:58:38.593" Title="Can someone please explain the Riemann Hypothesis to me... in English?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;complex-analysis&gt;&lt;riemann-hypothesis&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="7982" PostTypeId="2" ParentId="7978" CreationDate="2010-10-27T01:45:49.933" Score="2" ViewCount="0" Body="&lt;p&gt;There exists an invertible matrix $P$ such that $^t P A P$ is diagonal with blocks equal to $\begin{pmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{pmatrix}$ or $0$ (it is a simple exercise in bilinear forms), so that the rank of $A$ is necessarily even.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-10-27T01:45:49.933" />
  <row Id="7983" PostTypeId="1" AcceptedAnswerId="8137" CreationDate="2010-10-27T02:10:11.333" Score="2" ViewCount="115" Body="&lt;p&gt;Suppose I have an optimization problem of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \mathbf{d}^{T}\mathbf{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{M}\mathbf{y} \geq  \mathbf{d}, \qquad \mathbf{y} \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a solution $\mathbf{s}$ which satisfies $\mathbf{M}\mathbf{s} = \mathbf{d}$ and $\mathbf{s} \geq 0$, how can it be shown $\mathbf{s}$ is optimal for the above linear programming problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial thought was to form the dual:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;max p'd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p&gt;=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p'M&amp;lt;=d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus weak duality gives&#xA;d'y &amp;lt;= p'd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the constraint of the dual gives p'M&amp;lt;=d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so : p'My&amp;lt;=dy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus p'd&amp;lt;=dy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the two equalities shows that p'd must equal dy, and thus with strong duality this means that p'd is the optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did I go wrong, and how does being symmetric fit into the problem?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2014" LastEditorDisplayName="" LastEditDate="2010-10-28T02:37:45.800" LastActivityDate="2010-10-28T12:43:41.203" Title="Linear optimization constrained by cost function" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7984" PostTypeId="1" AcceptedAnswerId="8000" CreationDate="2010-10-27T02:31:44.707" Score="7" ViewCount="145" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It is a profoundly erroneous truism, repeated by copybooks and by eminent people when they are making speeches, that we should cultivate the habit of thinking of what we are doing. The precise opposite is the case. Civilization advances by extending the number of important operations which we can perform without thinking about them.&quot; ---Alfred North Whitehead&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;&lt;em&gt;All&lt;/em&gt; math instruction problems are primarily insufficient &lt;em&gt;speed&lt;/em&gt; at a previous step.  If you can't run multiplication in your head, you can't do long division and have it make sense.  If you can't factor numbers in your head, you can't do decent fraction problems.  If math facts aren't instant, basic algebraic manipulations (2x + 3 = 17) take too long to do ... and you never learn the patterns. Algebraic fractions, quadratic factoring, the fundamental theorem of calculus, integration over the complex half-plane ... it's all the same thing.  How fast are you at the prior step?  Can you do it in your head?&quot; ---&lt;a href=&quot;http://aretae.blogspot.com/2010/06/homeschool-curricula.html&quot; rel=&quot;nofollow&quot;&gt;Aretae&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What specific exercises are helpful in internalizing and automatizing computational skills?&lt;/strong&gt; I realize that the obvious answer is &quot;drill and repetition,&quot; but I was hoping the community would have some something more specific to add, so I have a number of related subquestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, are &lt;em&gt;timed&lt;/em&gt; drills helpful? Should one try to do simpler exercises in one's head? Does anyone have tips on how to develop the mental toughness to persist through computations that might be boring or painful in the moment even as one understands that they are ultimately necessary for building understanding? Is there some esoteric metacognitive art of debugging one's own mental algorithms: figuring out the &lt;em&gt;most&lt;/em&gt; efficient way of doing something, and training oneself to do that thing without thinking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After two years of pure autodidacticism in mathematics, I'm taking a differential equations course at the local community college, and it's been more difficult than I expected. Humiliating as it is to admit, I don't seem to yet have the patience or mental toughness for large problem sets, and I worry that I may have developed a few bad study habits. I have a tendency to leisurely examine proofs and casually attempt a few exercises, without (I fear) taking care to establish the firm base of &lt;em&gt;quick and reliable&lt;/em&gt; skills needed for higher understanding. To take one example, I can &lt;em&gt;do&lt;/em&gt; integration by parts, but only after a noticeable hesitation; the operation is not introspectively obvious to me in the way that (say) the distributivity of multiplication over addition is introspectively obvious and hardly even feels like a step. Integration by parts still &lt;em&gt;feels like a step&lt;/em&gt;---and I wonder if perhaps it &lt;em&gt;shouldn't&lt;/em&gt;, supposing one really wants a deep understanding. So while I'm proud of myself for (say) having picked up a lot of cool complex analysis insights from my few months with Mathews and Howell and my quick skim of Tristan Needham, I also feel as if I am missing some fundamental skills that are needed to achieve deeper insight, and I was hoping maybe this community would have some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is a soft question (perhaps too soft?---I dearly apologize), and should therefore be community wiki; however, I don't have the reputation to enable that. Perhaps a moderator could be so kind? I thank you-all in advance for any advice you have to offer, and remain yours.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2010-10-27T04:42:40.877" Title=" Automatizing computational skills" Tags="&lt;soft-question&gt;&lt;learning&gt;&lt;self-learning&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7985" PostTypeId="2" ParentId="7978" CreationDate="2010-10-27T02:45:27.853" Score="6" ViewCount="0" Body="&lt;p&gt;For a skew symmetric (real) matrix, the eigenvalues are all purely imaginary.  This is because if $Av = \lambda v$, then we have $\lambda \langle v,v\rangle = \langle \lambda v, v\rangle = \langle Av,v\rangle = \langle v, -Av \rangle = \langle v, -\lambda v\rangle = -\overline{\lambda} \langle v,v\rangle$, so we conclude that $\lambda = -\overline{\lambda}$, i.e., that $\lambda$ is purely imaginary.  Here, we're using an Hermitian inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a real matrix, complex eigenvalues come in conjugate pairs, so the rank must be even.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-10-27T02:45:27.853" CommentCount="2" />
  <row Id="7986" PostTypeId="1" AcceptedAnswerId="7997" CreationDate="2010-10-27T03:15:00.440" Score="10" ViewCount="212" Body="&lt;p&gt;So I feel a bit strange asking a Calculus question, but this came up today while teaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check that if you start with some integral, which can be see as an &quot;obvious u-substitution problem&quot;, that you can instead use integration by parts, and wind up with the scenario where you have have the original integral on both sides of your equation so you solve for the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Given $I=\int g^n(x)g'(x)dx$ we can clearly use u-substitution, but if we use integration by parts we get the equation $I=-nI+g^{n+1}$. This is nothing exciting or surprising, but it yields the observation that u-sub leads to one of these int by parts equations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Is the opposite true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I mean to say is, if you do integration by parts and you wind up with an equation of this type, does it mean that you could of used some very clever u-substitution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I should know this, but I have thought about it today, and asked a friend or two, and we don't see an immediate proof of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-27T20:51:21.483" Title="Integration by Parts implies U-substitution?" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="7987" PostTypeId="2" ParentId="7981" CreationDate="2010-10-27T03:20:56.997" Score="19" ViewCount="0" Body="&lt;p&gt;The prime number theorem states that the number of primes less than or equal to&#xA;$x$ is approximately equal to $\int_2^x \dfrac{dt}{\log t}.$    The Riemann hypothesis gives a precise answer to how good this approximation is; namely, it states that the difference between the exact number of primes below $x$, and the given integral, is (essentially) $\sqrt{x} \log x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here &quot;essentially&quot; means that one should actually take the absolute value of the difference, and also that one might have to multiply $\sqrt{x} \log x$ by some positive constant.  Also, I should note that the Riemann hypothesis is more usually stated in terms of the location of the zeroes&#xA;of the Riemann zeta function; the previous paragraph is giving an equivalent form, which may be easier to understand, and also may help to explain the interest of the statement.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_hypothesis#Distribution_of_prime_numbers&quot; rel=&quot;nofollow&quot;&gt;the wikipedia entry&lt;/a&gt; for the formulation in terms of counting primes, as well as various other formlations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty of the problem is (it seems to me) as follows: there is no approach currently known to understanding the distribution of prime numbers well enough to establish the desired approximation, other than by studying&#xA;the Riemann zeta function and its zeroes.  (The information about the primes&#xA;comes from information about the zeta function via a kind of Fourier transform.)  On the other hand, the zeta function is not easy to understand; there is no straightforward formula for it that allows one to study its zeroes, and because of this any such study ends up being somewhat indirect.&#xA;So far, among the various possible such indirect approaches, no-one has found&#xA;one that is powerful enough to control all the zeroes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very naive comment, that nevertheless might give some flavour of the problem, is that there are an infinite number of zeroes that one must contend with, so there is no obvious finite computation that one can make to solve&#xA;the problem; ingenuity of some kind is necessarily required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, one can remark that the Riemann hypothesis, when phrased in terms of the location of the zeroes, is very simple (to state!) and very beautiful: it says that all the non-trivial zeros have real part $1/2$.  This suggests that perhaps there is some secret symmetry underlying the Riemann zeta function that would &quot;explain&quot; the Riemann hypothesis.  Mathematicians have had, and continue to have, various ideas about what this secret symmetery might be (in this they are inspired by an analogy with what is called &quot;the function field case&quot; and the&#xA;deep and beautiful theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_conjectures&quot; rel=&quot;nofollow&quot;&gt;the Weil conjectures&lt;/a&gt;), but so far they haven't managed to establish any underlying phenonemon which implies the Riemann hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-27T04:40:47.893" LastActivityDate="2010-10-27T04:40:47.893" CommentCount="1" />
  <row Id="7988" PostTypeId="1" AcceptedAnswerId="7991" CreationDate="2010-10-27T03:24:48.497" Score="4" ViewCount="153" Body="&lt;p&gt;I'm trying to describe a counterexample for a theorem which includes the figure eight or &quot;infinity&quot; symbol, but I'm having trouble finding a good piecewise function to draw it. I need it to be the symbol, except at the &quot;crossing point&quot; the function jumps (not continuous) so that we still have a manifold.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T03:44:18.000" LastActivityDate="2010-10-27T13:56:20.420" Title="The function that draws a figure eight" Tags="&lt;differential-geometry&gt;&lt;plane-curves&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="7989" PostTypeId="2" ParentId="7988" CreationDate="2010-10-27T03:30:20.093" Score="4" ViewCount="0" Body="&lt;p&gt;There are no functions that describe the &quot;lemniscate&quot;, but there are parametric and polar equations for the &lt;a href=&quot;http://mathworld.wolfram.com/Lemniscate.html&quot; rel=&quot;nofollow&quot;&gt;lemniscate of Bernoulli&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/EightCurve.html&quot; rel=&quot;nofollow&quot;&gt;lemniscate of Gerono&lt;/a&gt;, to name two of the more famous lemniscates.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Might as well share this, since the question already has an accepted answer. Here is my favorite way of generating the lemniscate of Bernoulli, as an &lt;a href=&quot;http://mathworld.wolfram.com/Envelope.html&quot; rel=&quot;nofollow&quot;&gt;envelope&lt;/a&gt; of circles centered on a rectangular hyperbola, and passing through the origin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HMyER.png&quot; alt=&quot;lemniscate envelope&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T13:56:20.420" LastActivityDate="2010-10-27T13:56:20.420" CommentCount="2" />
  <row Id="7990" PostTypeId="1" AcceptedAnswerId="7994" CreationDate="2010-10-27T03:30:28.533" Score="0" ViewCount="63" Body="&lt;p&gt;Let $R$ be a ring and $f(X) \in R[x]$ be a non-constant polynomial. We know that the number of roots, of $f(X)$ in $R$ has no relation, to its degree if $R$ is not &lt;em&gt;commutative&lt;/em&gt;, or &lt;em&gt;commutative but not a domain&lt;/em&gt;. But,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;The number of roots, of a non-zero polynomial over commutative integral domain, is at most its degree.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How does one prove above result?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-27T03:54:16.347" Title="Roots of a polynomial in an integral domain." Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7991" PostTypeId="2" ParentId="7988" CreationDate="2010-10-27T03:34:01.600" Score="4" ViewCount="0" Body="&lt;p&gt;You can use the function $$t\in(-\tfrac12\pi,\tfrac32\pi)\mapsto(\cos t,\sin t\cos t)\in\mathbb R^2.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting curve is (I'm omitting a little bit on the left and a little bit on the right from the domain in this picture):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CyNo6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-27T03:47:15.260" LastActivityDate="2010-10-27T03:47:15.260" CommentCount="5" />
  <row Id="7992" PostTypeId="2" ParentId="7988" CreationDate="2010-10-27T03:34:08.873" Score="3" ViewCount="0" Body="&lt;p&gt;Is the &lt;a href=&quot;http://mathworld.wolfram.com/Lemniscate.html&quot; rel=&quot;nofollow&quot;&gt;lemniscate&lt;/a&gt; what you want?  I don't know what you need as a jump at the crossing point, but maybe you can get that with a trigonometric parameterization.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T03:34:08.873" />
  <row Id="7994" PostTypeId="2" ParentId="7990" CreationDate="2010-10-27T03:43:15.973" Score="2" ViewCount="0" Body="&lt;p&gt;This is a consequence of the fact that over a commutative ring with identity $A$, an element $a \in A$ is a zero of a polynomial $f \in A[x]$ if and only if $f(x) = (x - a)q(x)$ for some polynomial $q(x) \in A[x]$.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-10-27T03:43:15.973" />
  <row Id="7995" PostTypeId="2" ParentId="7990" CreationDate="2010-10-27T03:54:16.347" Score="1" ViewCount="0" Body="&lt;p&gt;It follows from the division algorithm, the fact that the evaluation maps gives a homomorphism from $R[x]$ to $R^R$ (functions from $R$ to $R$ with pointwise operations), and that $R$ is a domain. It is the exact same argument as for fields, with the division algorithm suitably restricted to certain kinds of polynomials over $R$ as divisors. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-27T03:54:16.347" />
  <row Id="7997" PostTypeId="2" ParentId="7986" CreationDate="2010-10-27T04:12:51.760" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I interpreted the question as follows: suppose that we use the integration by parts formula,&#xA;$$\int u(x)v'(x)\,dx = u(x)v(x) - \int v(x)u'(x)\,dx$$&#xA;and $\int v(x)u'(x)\,dx = \alpha \int u(x)v'(x)\,dx$ for some constant $\alpha$, so that we can &quot;solve&quot; for the original integral as&#xA;$$\int u(x)v'(x)\,dx = \frac{1}{1+\alpha}u(x)v(x) + C.$$&#xA;Is it the case that we can solve the original integral by doing a direct substitution instead of integration by parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this works: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $v(x)u'(x) = \alpha u(x)v'(x)$ for some $\alpha\neq -1$. (I think this is essentially what you have, since the integral on the left hand side is $\int u(x)v'(x)\,dx$, and the integral on the right is $\int v(x)u'(x)\,dx$). I restrict to $\alpha\neq -1$, because if $\alpha=-1$, then you cannot &quot;solve&quot; for the original integral. But in fact, from the work below we will get that $u=\frac{B}{|v|}$, so that integration by parts will simply result in the true (but useless) $\int u\,dv = B + \int u\,dv$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $u$ or $v$ are zero, then the original integral was the integral of $0$, so we may discard that case. So we get that $\frac{u'}{u} = \alpha \frac{v'}{v}$, and integrating we get $\ln|u|=\alpha\ln|v|+C$, or $|u| = A|v|^{\alpha}$ for some $A\gt 0$; hence $u=B|v|^{\alpha}$ for some $B\neq 0$. So the original integral was&#xA;$$\int u(x)v'(x)\,dx = \int B|v(x)|^{\alpha}v'(x)\,dx$$&#xA;which suggests the substitution $g=v(x)$ to get $B\int |g|^{\alpha}\,dg$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Note that if $\alpha = -1$, we still get $u=B|v|^{-1}$, and&#xA;so again the substitution $g=|v(x)|$ will do the trick. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T16:51:12.063" LastActivityDate="2010-10-27T16:51:12.063" CommentCount="8" />
  <row Id="7998" PostTypeId="2" ParentId="7593" CreationDate="2010-10-27T04:25:29.060" Score="5" ViewCount="0" Body="&lt;p&gt;I posted this to &lt;a href=&quot;http://mathoverflow.net/questions/43611/decomposing-the-plane-into-intervals&quot; rel=&quot;nofollow&quot;&gt;Math Overflow&lt;/a&gt; and Jeff Strom gave the following answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conway and Croft show it can be done for closed intervals and cannot be done for open intervals in the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering a sphere with congruent great-circle arcs. Proc. Cambridge Philos. Soc. 60 1964 787–800.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T04:25:29.060" />
  <row Id="7999" PostTypeId="2" ParentId="4961" CreationDate="2010-10-27T04:38:14.523" Score="2" ViewCount="0" Body="&lt;p&gt;I came across this slick proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Heron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Heron's formula&lt;/a&gt; on artofproblemsolving.com the other day. Heron's formula yields the area of a triangle given the lengths of its three sides:&#xA;$$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $ s = \frac{1}{2}(a+b+c)$.&#xA;The &lt;a href=&quot;http://www.artofproblemsolving.com/Resources/Papers/Heron.pdf&quot; rel=&quot;nofollow&quot;&gt;entire proof&lt;/a&gt;, by high schooler Miles Dillon Edwards, is reproduced here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0PpRN.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $I$ be the center of the incircle of $\triangle ABC$. Let $a = y + z$, $b = x + z$, and&#xA;$c = x + y$ be the lengths of the sides opposite $A$, $B$, and $C$, respectively, and let&#xA;$s = x + y + z$ be the semiperimeter of the triangle. Clearly $2 \alpha + 2 \beta + 2 \gamma = 2\pi$, so&#xA;$\alpha + \beta + \gamma = \pi$. Now notice that&#xA;$$(r + ix)(r + iy)(r + iz) = (u e^{i \alpha})(v e^{i \beta})(w e^{i \gamma}) = u v w e^{i(\alpha+\beta+\gamma)} = u v w e^{\pi i} = −uvw.$$&#xA;Therefore&#xA;$$0 = \text{Im}[(r + ix)(r + iy)(r + iz)] = r^2(x + y + z) − xyz,$$&#xA;so&#xA;$$r = \sqrt{\frac{x y z}{x + y + z}} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$.&#xA;Thus the area of $\triangle ABC$ is&#xA;$$\frac{r a}{2} + \frac{r b}{2} + \frac{r c}{2} = r s =&#xA;\sqrt{s(s-a)(s-b)(s-c)}$$&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-10-27T04:38:14.523" CommunityOwnedDate="2010-10-27T04:38:14.523" />
  <row Id="8000" PostTypeId="2" ParentId="7984" CreationDate="2010-10-27T04:42:40.877" Score="1" ViewCount="0" Body="&lt;p&gt;I too am a self-learner of mathematics working in a field where I have to use math professionally. In my experience speed or automatism has not been important. The trouble in using or learning math comes from not being able to figure out what line of reasoning to follow or how to adequately model a real-world phenomena. Once you have that right being slow or fast only makes a marginal difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practice is important but for a different reason. It is not possible to grasp the relation between different ideas without thinking your way through many concrete instances. Also, every area of mathematics has certain tricks or idioms that are repeated again and again. You get a hang for them only through practice and exposure to good work done by others (good textbooks for beginners like me). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you are being too modest and underplaying your effort, it is a very bad idea to &quot;leisurely examine proofs and casually attempt a few exercises&quot;. Use pen and paper. Know the strategy of the proof well enough to be able to reproduce it on your own. Do at least 90% of the exercises.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-10-27T04:42:40.877" CommentCount="1" />
  <row Id="8001" PostTypeId="1" CreationDate="2010-10-27T05:01:34.877" Score="1" ViewCount="77" Body="&lt;p&gt;I manually arranged two complete latin squares from  ordered pairs of 4 and 6 like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4&#xA;2 4 1 3&#xA;3 1 4 2&#xA;4 3 2 1&#xA;&#xA;1 2 3 4 5 6&#xA;2 4 1 6 3 5&#xA;3 1 5 2 6 4&#xA;4 6 2 5 1 3&#xA;5 3 6 1 4 2&#xA;6 5 4 3 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I like this kind of sequence, but can tell it is going to be more difficult to manually arrange as the number of pairs increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a related question about calculating row-complete latin squares, the following method generated a slightly different arrangement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any even n, say n = 2m, a row complete latin square of order n can be formed by writing down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0, 1, 2m - 1, 2, 2m - 2, 3,..., m + 1, m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the first row and then developing subsequent rows by adding 1 modulo n.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 1 3 2&#xA;1 2 0 3&#xA;2 3 1 0&#xA;3 0 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which can be made column-complete by rearranging the rows (incremented for comparison with the first example)...&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 4 3&#xA;2 3 1 4&#xA;4 1 3 2&#xA;3 4 2 1&#xA;&#xA;1 2 6 3 5 4&#xA;2 3 1 4 6 5&#xA;6 1 5 2 4 3&#xA;3 4 2 5 1 6&#xA;5 6 4 1 3 2&#xA;4 5 3 6 2 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to figure out is how to generate the first pattern with a method similar to that row-complete calculation, or a way to convert the above squares into the first pattern by swapping some values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like they both have their values reversed in rows and columns in the same way, yet their sequences are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes it seem like a simple matter, but I'm not finding a simple (enough) pattern between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example putting one of each square together and looking at it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11 22 63 34 55 46&#xA;22 34 11 46 63 55&#xA;63 11 55 22 46 34&#xA;34 46 22 55 11 63&#xA;55 63 46 11 34 22&#xA;46 55 34 63 22 11&#xA;&#xA;      63 34    46&#xA;   34    46 63   &#xA;63          46 34&#xA;34 46          63&#xA;   63 46    34   &#xA;46    34 63 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does anyone see a way of transposing one to the other, or modifiying the row-complete calculation to match the other row sequence?&lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:37:01.763" LastActivityDate="2010-11-19T10:50:15.237" Title="Finding a Pattern Between Latin Squares" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;latin-square&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8002" PostTypeId="1" AcceptedAnswerId="8005" CreationDate="2010-10-27T05:11:37.780" Score="6" ViewCount="367" Body="&lt;p&gt;Let $\varphi(n)$ be Euler's totient function, the number of positive integers less than or equal to $n$ and relatively prime to $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Challenge: Prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two proofs, one of which is partially combinatorial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm posing this problem partly because I think some folks on this site would be interested in working on it and partly because I would like to see a purely combinatorial proof.  (But please post any proofs; I would be interested in noncombinatorial ones, too.  I've learned a lot on this site by reading alternative proofs of results I already know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll wait a few days to give others a chance to respond before posting my proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The two proofs in full are now given among the answers.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T05:25:23.223" LastActivityDate="2010-11-22T05:25:23.223" Title="Identity involving Euler's totient function: $\sum_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="8003" PostTypeId="2" ParentId="7981" CreationDate="2010-10-27T05:12:37.653" Score="7" ViewCount="0" Body="&lt;p&gt;A direct translation of RH (Riemann Hypothesis) would be very baffling in layman's terms. But, there are many problems that are equivalent to RH and hence, defining them would be actually indirectly stating RH. Some of the equivalent forms of RH are much easier to understand than RH itself. I give what I think is the most easy equivalent form that I have encountered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Riemann hypothesis is equivalent&#xA;  to the statement that an integer has&#xA;  an equal probability of having an odd&#xA;  number or an even number of distinct&#xA;  prime factors. (Borwein page. 46)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-10-27T05:12:37.653" CommentCount="3" />
  <row Id="8004" PostTypeId="1" AcceptedAnswerId="8006" CreationDate="2010-10-27T05:42:01.813" Score="7" ViewCount="176" Body="&lt;p&gt;Hi, I'm just starting to learn some basic topology, and I've mostly encountered definitions. For instance, I read that for a topological space $X$, for any $E\subseteq X$, define its closure $\overline{E}$ as the set of points $p\in X$ such that each neighborhood $N$ of $p$ has a nonempty intersection with $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to verify that this is indeed a closure operation, and here's what I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Suppose $p\in E$, then for any $N$, $p\in N$, so $N\cap E\neq\emptyset$, so $p\in\overline{E}$, and thus $E\subseteq\overline{E}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Suppose $p\in\overline{E\cup F}$. Then for any $N$, $N\cap(E\cup F)\neq\emptyset$, so $(N\cap E)\cup(N\cap F)\neq\emptyset$, and thus either $N\cap E\neq\emptyset$ or $N\cap F\neq\emptyset$, so $p\in\overline{E}$ or $p\in\overline{F}$, and thus $\overline{E\cup F}\subseteq\overline{E}\cup\overline{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) For any $p\in X$, the intersection of any $N$ and $\emptyset$ is empty, so there are no points such that every neighborhood has nonempty intersection with $\emptyset$. Hence $\overline{\emptyset}=\emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) &lt;strong&gt;I'm stuck showing that $\overline{\overline{E}}=\overline{E}$&lt;/strong&gt;. From (1), I know that $\overline{E}\subseteq\overline{\overline{E}}$, but I can't show the other containment. I took $p\in\overline{\overline{E}}$, and so every $N\cap\overline{E}\neq\emptyset$. I want to show $p\in N\cap\overline{E}$, and I think this would be easy if $X$ contains isolated points. But then I found out there are such things as perfect sets or dense-in-itself sets, so that can't work. Is there some trick to show this containment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, I know this is probably very simple, but I'm going a little mad staring at it.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-12-04T12:09:59.680" Title="Verifying a Closure Operation" Tags="&lt;topology&gt;" AnswerCount="3" />
  <row Id="8005" PostTypeId="2" ParentId="8002" CreationDate="2010-10-27T05:42:28.257" Score="10" ViewCount="0" Body="&lt;p&gt;One approach is to use the formula $\displaystyle \sum_{d \mid k} \varphi(d) = k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have that $\displaystyle \sum_{k=1}^{n} \sum_{d \mid k} \varphi(d) = n(n+1)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exchanging the order of summation we see that the $\displaystyle \varphi(d)$ term appears $\displaystyle \left\lfloor \frac{n}{d} \right\rfloor$ times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{d=1}^{n} \left\lfloor \frac{n}{d} \right\rfloor \varphi(d) = n(n+1)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in other words, if we have the $n \times n$ matrix $A$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle A[i,j] = \varphi(j)$ if $j \mid i$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of elements in row $i$ is $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of elements in column $j$ is $\displaystyle \left\lfloor \frac{n}{j} \right\rfloor \varphi(j)$ and the identity just says the total sum by summing the rows is same as the total sum by summing the columns.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T05:58:46.663" LastActivityDate="2010-10-27T05:58:46.663" CommentCount="1" />
  <row Id="8006" PostTypeId="2" ParentId="8004" CreationDate="2010-10-27T05:49:13.517" Score="8" ViewCount="0" Body="&lt;p&gt;From $N\cap\overline{E}\neq\emptyset$ you know that $N$ is a neighborhood of some element of $\overline{E}$, and thus $N\cap E\neq\emptyset$ by your definition of $\overline{E}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-27T05:49:13.517" CommentCount="5" />
  <row Id="8007" PostTypeId="2" ParentId="8004" CreationDate="2010-10-27T05:50:09.420" Score="0" ViewCount="0" Body="&lt;p&gt;You can define the closure operation as taking the smallest closed set which contains $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\bar{E}$ is closed and it contains $\bar{E}$ you have $\bar{\bar{E}} \subseteq \bar{E}$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-10-27T05:50:09.420" CommentCount="3" />
  <row Id="8008" PostTypeId="1" AcceptedAnswerId="8106" CreationDate="2010-10-27T06:00:17.800" Score="2" ViewCount="239" Body="&lt;p&gt;Just wondering if equations used as props in SG1, atlantis and most recently Universe are just random and cool looking, or real/meaningful and cool looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most recent episode of Stargate Universe S02E04 has a &quot;corridor of equations&quot; where dr nick Rush is trying to decipher something. Has anyone familiar with the show got an opinion on this?&lt;/p&gt;&#xA;" OwnerUserId="2496" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T06:02:51.353" LastActivityDate="2010-10-27T21:49:07.530" Title="Do the equations used in Stargate make sense or are they gibberish?" Tags="&lt;popular-math&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="8009" PostTypeId="2" ParentId="7981" CreationDate="2010-10-27T06:16:21.590" Score="1" ViewCount="0" Body="&lt;p&gt;In very layman's terms it states that there is some order in the distribution of the primes (which seem to occur totally chaotic at first sight). Or to say it like Shakespeare: &quot;Though this be madness, yet there is method in 't.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know more there is a new trilogy about that topic where the first volume has just arrived:&#xA;&lt;a href=&quot;http://www.secretsofcreation.com/volume1.html&quot; rel=&quot;nofollow&quot;&gt;http://www.secretsofcreation.com/volume1.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a marvelous and easy to understand book from a number theorist who knows his stuff! &lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-27T06:16:21.590" />
  <row Id="8010" PostTypeId="2" ParentId="7923" CreationDate="2010-10-27T06:21:23.473" Score="0" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_process&quot; rel=&quot;nofollow&quot;&gt;Wiener process&lt;/a&gt; is a continuous everwhere, but differentiable nowhere function (quite an impressive beast by the way...) &lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-10-27T06:21:23.473" />
  <row Id="8011" PostTypeId="2" ParentId="8002" CreationDate="2010-10-27T06:25:49.213" Score="4" ViewCount="0" Body="&lt;p&gt;Look at the problem of counting all the triples (d,m+1,k) where $d\leq m\leq n$ and $gcd(d,m)=\frac{m}{k}$.&#xA;For each (d,m+1) we have exactly one k such that $gcd(d,m)=\frac{m}{k}$, and so we are counting exactly all the pairs (i,j)  where $ i &amp;lt; j \leq n+1 $ which is $\binom{n+1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for each k , if $ gcd(d,m)=\frac{m}{k} $, then $ k \mid m $ and there are $ \left\lfloor \frac{n}{k}\right\rfloor $ such m's, and for each m we have $\varphi(k)$ d's that satisfies this equality, so there are $ \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) $ triples that end with k.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The idea is counting the number of ways to choose two numbers from 1,2,...,n+1 (which is the right side of the equation). Once you choose the larger number m+1, you have m options which is exactly $\sum_{d\mid m} \varphi(d)$ (this is basically what Moron wrote).&#xA;What I wanted to do is to change the counting according to the &quot;relation&quot; between d and m, and the one that worked out is that $\frac{m}{gcd(d,m)}=k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going over triples (d,m+1,k) where $\frac{m}{gcd(d,m)}=k$ and $d\leq m\leq n$ just mean that (d,m+1) is an ordered pair from {1,...,n} and k is decided as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for specific k, if $\frac{m}{gcd(d,m)}=k$ then $k \mid m$ and there are $ \left\lfloor \frac{n}{k}\right\rfloor$ such m's. For each m, if $gcd(d,m)=m/k$ then $gcd(\frac{d}{m/k},\frac{m}{m/k})=gcd(\frac{d}{m}k,k)=1$ and there are $\varphi(k)$ such d's (because $d\leq m$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, or each k there are $ \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) $ options , and summing over k will give the left side of the equation.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-10-28T06:22:09.847" LastActivityDate="2010-10-28T06:22:09.847" CommentCount="2" />
  <row Id="8013" PostTypeId="1" AcceptedAnswerId="8014" CreationDate="2010-10-27T07:10:58.670" Score="7" ViewCount="222" Body="&lt;p&gt;Given a set $S$, one can easily find a set with greater cardinality -- just take the power set of $S$. In this way, one can construct a sequence of sets, each with greater cardinality than the last. Hence there are at least countably infinite many orders of infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But do there exist uncountably infinite orders of infinity?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, does there exist an uncountable set of sets whose elements all have distinct cardinalities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first answer to &lt;a href=&quot;http://math.stackexchange.com/questions/5378/types-of-infinity&quot;&gt;http://math.stackexchange.com/questions/5378/types-of-infinity&lt;/a&gt; suggests the answer is &quot;yes&quot;, but only establishes a countable number of cardinalities (which, to be fair, was what the question was asking about).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been exposed to enough mathematical logic to realize that I'm walking in a minefield; let me know if I've already mis-stepped.&lt;/p&gt;&#xA;" OwnerUserId="1126" LastActivityDate="2010-10-27T16:21:52.773" Title="Are there uncountably infinite orders of infinity?" Tags="&lt;set-theory&gt;&lt;infinity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="8014" PostTypeId="2" ParentId="8013" CreationDate="2010-10-27T07:15:14.833" Score="9" ViewCount="0" Body="&lt;p&gt;The answer is yes.  See &quot;Fact 20&quot; at the end of the following handout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-27T07:15:14.833" CommentCount="1" />
  <row Id="8015" PostTypeId="1" CreationDate="2010-10-27T07:37:21.070" Score="4" ViewCount="100" Body="&lt;p&gt;I'm looking for lightweight, free, Windows, Math software.&lt;br&gt;&#xA;Something I can put an expression and get an answer, or graph it.&lt;br&gt;&#xA;I tried Euler, but it is quiet complicated and HUGE.&lt;br&gt;&#xA;Basic needs:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Expression Based&lt;/li&gt;&#xA;&lt;li&gt;Supports Variables&lt;/li&gt;&#xA;&lt;li&gt;Support Functions, User defined and auto loaded.&lt;/li&gt;&#xA;&lt;li&gt;Supports graphs, 2D.  Not really needing 3D.&lt;/li&gt;&#xA;&lt;li&gt;Supports History.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What do you use?  What do you recommend?&lt;/p&gt;&#xA;" OwnerUserId="2743" LastActivityDate="2010-10-27T09:56:10.793" CommunityOwnedDate="2010-10-27T07:37:21.070" Title="Windows lightweight Math Software" Tags="&lt;math-software&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="8016" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:41:39.840" Score="1" ViewCount="0" Body="&lt;p&gt;I'm also interested in answers to this questions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;http://www.microsoft.com/windowsxp/downloads/powertoys/xppowertoys.mspx&quot; rel=&quot;nofollow&quot;&gt;Power Calculator&lt;/a&gt; from MS, but has a lot to be desired.&lt;/p&gt;&#xA;" OwnerUserId="1099" LastActivityDate="2010-10-27T07:41:39.840" CommentCount="2" />
  <row Id="8017" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:45:06.137" Score="1" ViewCount="0" Body="&lt;p&gt;I haven't personally used &lt;a href=&quot;http://www.wmich.edu/cpmp/CPMP-Tools/index.html&quot; rel=&quot;nofollow&quot;&gt;CPMP-Tools&lt;/a&gt;, but I've heard it referred to as a free CAS.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-27T07:45:06.137" />
  <row Id="8018" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:52:30.353" Score="1" ViewCount="0" Body="&lt;p&gt;Graphmatica does a fairly decent job at plotting functions. Not very powerful, but it can graph a lot of functions and their derivatives and such.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-10-27T09:00:06.583" LastActivityDate="2010-10-27T09:00:06.583" />
  <row Id="8019" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T07:58:25.013" Score="1" ViewCount="0" Body="&lt;p&gt;In your position, I'd just use an interactive &lt;a href=&quot;http://www.python.org/&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt; shell with &lt;a href=&quot;http://matplotlib.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Matplotlib&lt;/a&gt;. (But then, I'm a computer science guy first and foremost.)&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-27T07:58:25.013" CommunityOwnedDate="2010-10-27T07:58:25.013" CommentCount="2" />
  <row Id="8020" PostTypeId="1" AcceptedAnswerId="8063" CreationDate="2010-10-27T08:00:22.360" Score="3" ViewCount="103" Body="&lt;p&gt;Well, in a previous post regarding the explanation of &lt;a href=&quot;http://math.stackexchange.com/questions/7981/can-someone-please-explain-the-riemann-hypothesis-to-me-in-english/7987#7987&quot;&gt;&lt;em&gt;Riemann Hypothesis&lt;/em&gt;&lt;/a&gt; Matt answered that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The prime number theorem states that the number of primes less than or equal to $x$ is approximately equal to $\int_2^x \dfrac{dt}{\log t}.$ The Riemann hypothesis gives a precise answer to how good this approximation is; namely, it states that the difference between the exact number of primes below $x$, and the given integral, is (essentially) $\sqrt{x} \log x$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What i have heard about RH is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The non-trivial Zeros of the &lt;em&gt;Riemann- $\zeta$ -function&lt;/em&gt; have real part as $\frac{1}{2}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me how these two statements are related?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-27T08:08:51.257" LastActivityDate="2010-10-27T15:58:35.210" Title="A Question on RH relating to Prime Number theorem" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;prime-numbers&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="8021" PostTypeId="2" ParentId="8020" CreationDate="2010-10-27T08:10:42.117" Score="5" ViewCount="0" Body="&lt;p&gt;A brief explanation is sufficient I think, since the question is a little vague, because there are many ways to relate the two. Regarding relation between RH and PNT one will have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| \pi(x) - Li(x) | = O({x^\theta}\log{x})$ if and only if $\zeta(s)$ has no non-trivial zeros $\rho$ with $\Re(\rho) &gt; \theta$. Naturally, since the non-trivial zeros of the Riemann Zeta function are symmetric about $\frac{1}{2}$. Hence $| \pi(x) - Li(x) | = O(\sqrt{x}\log{x})$ implies RH.&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-10-27T08:10:42.117" CommentCount="1" />
  <row Id="8022" PostTypeId="2" ParentId="8002" CreationDate="2010-10-27T08:17:18.613" Score="6" ViewCount="0" Body="&lt;p&gt;We can use induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $ \lfloor (n+1)/k \rfloor - \lfloor n/k \rfloor = 1 $ when $n \equiv -1 \textrm{ mod } k,$ (for $k&gt;1$) and $0$ otherwise. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \phi(1) + \sum_{k=2}^n \left( \left\lfloor \frac{n+1}{k} \right\rfloor -&lt;br&gt;&#xA;\left\lfloor \frac{n}{k} \right\rfloor \right) \phi(k) =&#xA;\sum_{k=1, k \textrm{ s.t. } n \equiv -1 \textrm{ mod } k}^n \phi(k)$$&#xA;$$= \sum_{ k | (n+1), k \ne n+1} \phi(k) = (n+1) - \phi(n+1),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $\sum_{d|k} \phi(d) = k.$&#xA;So assuming the result is true for $n,$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n+1} \left\lfloor \frac{n+1}{k} \right\rfloor \phi(k)&#xA;= \phi(n+1) + \sum_{k=1}^n \left( \left\lfloor \frac{n+1}{k} \right\rfloor -&lt;br&gt;&#xA;\left\lfloor \frac{n}{k} \right\rfloor \right) \phi(k) + \frac{n(n+1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \phi(n+1) + \phi(1) + \sum_{k=2}^n \left( \left\lfloor \frac{n+1}{k} \right\rfloor -&lt;br&gt;&#xA;\left\lfloor \frac{n}{k} \right\rfloor \right) \phi(k) + \frac{n(n+1)}{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, using the previous result to substitute for the summation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (n+1) +  \frac{n(n+1)}{2} =  \frac{(n+1)(n+2)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that the result is true when $n=1 \textrm{ and } 2$ completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-27T09:38:06.373" LastActivityDate="2010-10-27T09:38:06.373" CommentCount="1" />
  <row Id="8023" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T08:37:22.050" Score="2" ViewCount="0" Body="&lt;p&gt;As calculator &lt;a href=&quot;http://wxmaxima.sourceforge.net/wiki/index.php/Main_Page&quot; rel=&quot;nofollow&quot;&gt;wxMaxima&lt;/a&gt; (&lt;a href=&quot;http://wxmaxima.sourceforge.net/wiki/index.php/Screenshots&quot; rel=&quot;nofollow&quot;&gt;screenshots&lt;/a&gt;) with export to png and LaTeX syntax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For playing with graphs I use &lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;GeoGebra&lt;/a&gt; with export to png, eps, pdf...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some special cases &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also MS Excel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also a normal calculator - is often faster than PC programs ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And for hard differential/integral/calculus &lt;a href=&quot;http://wood.mendelu.cz/math/maw-html/index.php?lang=en&quot; rel=&quot;nofollow&quot;&gt;Czech Mathematical Assistant on Web&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2839" LastActivityDate="2010-10-27T08:37:22.050" CommunityOwnedDate="2010-11-13T18:39:23.673" />
  <row Id="8024" PostTypeId="1" AcceptedAnswerId="8026" CreationDate="2010-10-27T08:50:11.893" Score="5" ViewCount="204" Body="&lt;p&gt;Lots of people (including myself) face lot of problems in tackling Mathematical Problems, which appear as if we can solve it, but then writing out a solution becomes difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I was asked this question, some time back in an exam. Give an example of a continuous function on $(a,b)$ which is not uniformly continuous. Well, one's obvious choice is $$f(x) = \frac{1}{x-a} \quad \text{or} \ \frac{1}{x-b}$$ I knew this as soon as i saw the problem, and started proving it. One actually has to make an observation that as $x \to a$ then, $\frac{1}{x-a}$ will be larger. But i found that i couldn't actually formally prove it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, to prove that $f(x)=x^{2}$ is not uniformly continuous on $\mathbb{R}$, one again has to play with the quantifiers, to get the contradiction part. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, these are two instances, where i have found the problem, which to me appeared that i could solve it, but writing out a formal solution became difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can students improve upon this? Are there any instances, which happened to you like this!&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-27T15:27:39.633" Title="Difficulty in Mathematical Writing" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="8025" PostTypeId="1" CreationDate="2010-10-27T09:11:16.853" Score="3" ViewCount="116" Body="&lt;p&gt;See &lt;a href=&quot;http://stackoverflow.com/questions/4014090/is-it-safe-to-ignore-the-possibility-of-sha-collisions-in-practice&quot;&gt;this question&lt;/a&gt; on applying probability theory principles in software design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is generally the following: you design some system (say software) and rely on some well-known mathematical concept (say hash function). You know that when this concept is used without caution your system can sometimes fail, however the probability of such failure is extremely low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to evaluate whether you want to alter the design or can just ignore that drawback. Consequences of a failure are usually taken into account when such evaluations are done. For example if a failure leads to a person being mildly offended then it is not that of a problem but if a failure leads to a nuclear power explosion that it is a serious problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;a href=&quot;http://stackoverflow.com/questions/4014090/is-it-safe-to-ignore-the-possibility-of-sha-collisions-in-practice/4014407#4014407&quot;&gt;the accepted answer&lt;/a&gt; goes like this (all numbers here are exaggerated for better perception): probability of Earth colliding with a space rock is 10E-50 and probability of that drawback causing a problem is 10E-100. You see - Earth colliding with a space rock is a gazillion times more likely. So relax, that design is good enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that reasoning correct? Can it be accepted at all times?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastActivityDate="2010-10-27T11:33:30.577" Title="How safe is it to ignore low probability events?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8026" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T09:45:31.357" Score="6" ViewCount="0" Body="&lt;p&gt;I have a (joke) template file for writing papers, which contains in it&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma&lt;/strong&gt; &lt;em&gt;(Main technical lemma)&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let $D$ be a domain in (INSERT SPACE HERE) such that the following properties hold: &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Technical condition 1&lt;/li&gt;&#xA;  &lt;li&gt;Technical condition 2&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Then $D$ is both opened and closed in (INSERT SPACE AGAIN).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What is the point of the above? At some point in your mathematical career you will come to the realisation that the proofs &lt;em&gt;you&lt;/em&gt; personally are going to write are all based on the few small set of technical arguments. For what I do the main tool happens to be the Method of Continuity. By the time you have this realisation, it will also be completely obvious to you how to formulate a given proof to fit the template. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to you come to this realisation? My only suggestion is to read more papers/books/proofs and write more of them yourselfs. Just like a foreign language, the only way to get better and converting your intuitive ideas into formal arguments is through practice and immersion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, from the examples you gave, it is not quite clear whether your difficulty is with implementing $\epsilon$ - $\delta$ s, or with setting up the proof by contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, besides the usual proofs in textbooks, a good place to read up on proof techniques is &lt;a href=&quot;http://en.wikipedia.org/wiki/Proofs_from_THE_BOOK&quot; rel=&quot;nofollow&quot;&gt;Proofs from the BOOK&lt;/a&gt; by Aigner and Ziegler. Try to really figure out the details of each proof so you can explain the idea of it, a few days later, without having the book open. Another good resource for problem solving techniques is &lt;a href=&quot;http://www.tricki.org/&quot; rel=&quot;nofollow&quot;&gt;the Tricki&lt;/a&gt;. Of course, reading up on answers (and providing them) on this website would also help. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A bit on the quantifier issue. First you need to mentally nest the various implications. For example, diagrammatically I think (the following is not formal logic notation) uniform continuity to be something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall \epsilon \to \left( \exists \delta \to ( \forall |x-y| &amp;lt; \delta \to |f(x) - f(y)| &amp;lt; \epsilon ) \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to contradict it by example, you want something that satisfies the hypothesis $\forall \epsilon$ but not the conclusion $\exists \delta \ldots$. Which means that for each $\epsilon$ there &lt;em&gt;cannot exist&lt;/em&gt; a $\delta$ with the requisite property. Which means that for each $\epsilon$ and for each $\delta$ the requisite property must be false. So after step 1 we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall \epsilon, \delta \to \mbox{ not } ( \forall |x-y| &amp;lt; \delta \to |f(x) - f(y)| &amp;lt; \epsilon ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the statement inside the parentheses were to be false, since it is a $\forall$ statement, you just need one example. So you can convert the negation to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \forall \epsilon, \delta \to \left( \exists |x-y| &amp;lt; \delta \mbox{ and } |f(x) - f(y)| &gt; \epsilon \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there, we have converted a &quot;negative statement&quot; which we want to contradict, to a &quot;positive statement&quot; of a property we want our function to have. The rest is picking the mapping $(\epsilon, \delta) \to (x,y)$ for your chosen function $f$ so that the above implication is true. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-27T10:06:46.010" LastActivityDate="2010-10-27T10:06:46.010" CommentCount="4" />
  <row Id="8027" PostTypeId="2" ParentId="8015" CreationDate="2010-10-27T09:56:10.793" Score="0" ViewCount="0" Body="&lt;p&gt;I have also found SpeQ Mathematics.  &lt;a href=&quot;http://www.speqmath.com&quot; rel=&quot;nofollow&quot;&gt;http://www.speqmath.com&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very lightweight, starts quickly and has some good functions.&lt;/p&gt;&#xA;" OwnerUserId="2743" LastActivityDate="2010-10-27T09:56:10.793" />
  <row Id="8028" PostTypeId="2" ParentId="8025" CreationDate="2010-10-27T09:58:13.523" Score="1" ViewCount="0" Body="&lt;p&gt;Maybe do it like the industry does (e.g. with hard drives) and estimate the mean time between failures. If the unsafe operation is used at most once each time interval, and the probability of a failure is $p$, you can expect a time between failures of $\frac{1}{p}$ time intervals. If this time is significantly longer than you expect your system to be used, you are of the safe side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1e-100, is ridiculously low, by any standards. If such an error might occure once every nanosecond, you have a mean time between failures of 3e83 years. Compare that to the estimated age of the universe.... =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-10-27T09:58:13.523" CommentCount="3" />
  <row Id="8029" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T11:00:13.150" Score="1" ViewCount="0" Body="&lt;p&gt;The problem isn't really about writing. It's about being able to do math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the big problems in math is the P = NP problem. Most mathematicians believe that the know that the two aren't equal.&#xA;There's however no accepted proof. As a result a lot of smart people work on the problem. Writing proofs is the main activity of professional mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you 'or'-sign instead of the 'and'-sign because you don't remember with is turned with way, then you have a problem with writing stuff down in a formal matter. &#xA;If you however have no idea about what theorems you need to use to prove the solution then you misstate your problem by saying it's about writing stuff down in a formal matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math centers around finding a way to prove your result. Faith isn't enough.&lt;/p&gt;&#xA;" OwnerUserId="2263" LastEditorUserId="2263" LastEditorDisplayName="" LastEditDate="2010-10-27T11:16:33.130" LastActivityDate="2010-10-27T11:16:33.130" CommentCount="4" />
  <row Id="8030" PostTypeId="1" AcceptedAnswerId="8032" CreationDate="2010-10-27T11:15:31.853" Score="0" ViewCount="135" Body="&lt;p&gt;Problem with this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x)=3sin(x)cos(x/2) a=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been breaking up f(x) into g(x) = 3sin(x) and h(x)cos(x/2), and then multiplying them back (g(x))(h(x)) this was suggested to my by the lecturer. but for the life of me I cant get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is: y = 0.817x+1.398&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-10-27T11:22:16.233" LastActivityDate="2010-11-26T11:44:23.947" Title="Taylor Series Expansions" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8031" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T11:17:57.167" Score="1" ViewCount="0" Body="&lt;p&gt;What makes writing mathematics difficult is lack of confidence.  Someone confident who happens to be wrong will not face difficulties writing up their misapprehensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Willie's argument, getting familiar with lots of proof techniques that together solve most problems in some area, builds confidence: when you see such a problem, your reaction is &quot;I know just what to do with you&quot;.  And if you've seen this technique applied to many different problems, you'll have a good understanding of the difficulties it faced.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, some problems will be tricky.  What happens when your favourite technique comes apart at the seams, and you say to yourself &quot;I'm lost&quot;?  My favourite trick is to look at whatever sublemma I'm trying to prove and to try to doubt it, and once I'm not quite sure it's true, to go looking for a counterexample.  This hopeless search often informs you of the contours of the proof you can then reattempt.  To put into slogan form: &lt;em&gt;It's often hard to provbe what you can't doubt.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polya (1945)'s &lt;em&gt;How to Solve It&lt;/em&gt; has many ideas about techniques to help you keep going with proving things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, being &lt;em&gt;knowledgeable&lt;/em&gt;, as Willie suggests, is one key to being confident, and could call having the resources to deal with problem-solving setbacks being &lt;em&gt;indefatigability&lt;/em&gt;, another key.  Self-help books are full of &quot;you can do it!&quot; advice that will help you write out what you think is a proof, but unfortunately, unlike the above, they reduce the chances that the proof is valid.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="100" LastEditorDisplayName="" LastEditDate="2010-10-27T11:23:10.197" LastActivityDate="2010-10-27T11:23:10.197" />
  <row Id="8032" PostTypeId="2" ParentId="8030" CreationDate="2010-10-27T11:28:39.247" Score="2" ViewCount="0" Body="&lt;p&gt;You apparently only need the 1st-degree Taylor polynomial; note that the 1st-degree Taylor polynomial for a function $f(x)$ at $x=a$ is the same as the tangent line to the graph of $f$ at $a$, $y=f(a)+f'(a)\cdot(x-a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your $g$ and $h$, you can write 1st-degree Taylor polynomials for each of them, then take the product, as you said.  For $g$, $$y=g(1)+g'(1)\cdot(x-1)=3\sin(1)+3\cos(1)\cdot(x-1)\approx 1.621x+0.904.$$  I'll leave $h$ for you to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you've got (for some numbers $a$ and $b$): $$(1.621x+0.904)(ax+b)=1.621ax^2+(0.904a+1.621b)x+0.904b,$$ in which you don't care about the quadratic term (it's not accurate anyway).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given that you only need a 1st-degree Taylor polynomial, it might be easier to just differentiate $f$ as written and compute it directly (not bother with $g$ and $h$ and multiplying).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-10-27T11:28:39.247" CommentCount="1" />
  <row Id="8034" PostTypeId="2" ParentId="8025" CreationDate="2010-10-27T11:33:30.577" Score="2" ViewCount="0" Body="&lt;p&gt;Extremely small probabilities are always suspect.  See &lt;a href=&quot;http://www.johndcook.com/blog/2010/10/18/titanic-effect/&quot; rel=&quot;nofollow&quot;&gt;The Titanic Effect&lt;/a&gt;.  If you think the probability of a software failure is 10^-50 then you're not looking at the most probable source of failure.  If you're hash algorithm, for example, has one chance in 10^50 of failing, then it's more likely that your program, operating system, or hardware has a bug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it's not enough to look at probabilities of failure.  You have to look at probabilities of failure and the consequences of the failures.  In some contexts, a 10% probability of failure is acceptable.  In other contexts, a one-in-a-million chance of failure is quite serious.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-27T11:33:30.577" />
  <row Id="8035" PostTypeId="1" AcceptedAnswerId="8285" CreationDate="2010-10-27T11:40:22.107" Score="7" ViewCount="151" Body="&lt;p&gt;&lt;strong&gt;Statement:&lt;/strong&gt;&#xA;There exist a polynomial P such that |P(x) - cos(x)| &amp;lt;= 10^-6 for all (real) x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer&lt;/strong&gt;: False. All polynomials of a degree n &gt;= 1 are unbound as x tends to infinity. A polynomial of degree n = 0 is bound only when it is in the form y = a (horizontal line) but this will not help because cos(x) varies between [-1,1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; Is my answer reasonable? I am particularly concerned about 'unbound' polynomials as this is the term I just made up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't give me complete answer, I just want to know the flaws in my argument and get some hints to better solutions.&lt;/p&gt;&#xA;" OwnerUserId="2846" LastEditorUserId="2846" LastEditorDisplayName="" LastEditDate="2010-10-27T11:45:38.940" LastActivityDate="2010-10-29T14:27:29.533" Title="Statement about existence of a polynomial - true or false?" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8036" PostTypeId="2" ParentId="7864" CreationDate="2010-10-27T11:47:16.677" Score="6" ViewCount="0" Body="&lt;p&gt;Many well-educated people believe that a p-value is the probability that a study conclusion is wrong.  For example, they believe that if you get a 0.05 p-value, there's a 95% chance that your conclusion is correct.  In fact there may be less than a 50% chance that the conclusion is correct, depending on the context.  Read more &lt;a href=&quot;http://www.johndcook.com/blog/2008/11/18/five-criticisms-of-significance-testing/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-27T11:47:16.677" CommunityOwnedDate="2010-10-27T11:47:16.677" />
  <row Id="8037" PostTypeId="1" CreationDate="2010-10-27T11:54:54.593" Score="0" ViewCount="71" Body="&lt;p&gt;$$&#xA;\begin{pmatrix}&#xA;Y \\ Cb \\ Cr&#xA;\end{pmatrix}&#xA; =&#xA;\mathrm{Round}&#xA;\left\{&#xA;\begin{pmatrix}&#xA;0.2126\times\frac{219}{255} &amp;amp;&#xA;0.7152\times\frac{219}{255} &amp;amp;&#xA;0.0722\times\frac{219}{255} \\&#xA;-\frac{0.2126}{1.8556}\times\frac{224}{255} &amp;amp;&#xA;-\frac{0.7152}{1.8556}\times\frac{224}{255} &amp;amp;&#xA;0.5\times\frac{224}{255} \\&#xA;0.5\times\frac{224}{255} &amp;amp;&#xA;-\frac{0.7152}{1.5748}\times\frac{224}{255} &amp;amp;&#xA;-\frac{0.0722}{1.5748}\times\frac{224}{255} &#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;R \\ G \\ B&#xA;\end{pmatrix}&#xA;\right\}&#xA;+&#xA;\begin{pmatrix}&#xA;16 \\ 128 \\ 128&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to derive formula from above matrix that converts YCbCr to RGB? I am newbie to maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have YCbCr value already available, I want to convert it to RGB.&lt;/p&gt;&#xA;" OwnerUserId="2849" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-27T13:54:05.133" LastActivityDate="2010-10-27T14:51:44.597" Title="YCbCr to RGB conversion with given matrix." Tags="&lt;matrices&gt;" AnswerCount="0" CommentCount="3" />
  <row Id="8039" PostTypeId="1" AcceptedAnswerId="8041" CreationDate="2010-10-27T12:45:36.033" Score="0" ViewCount="84" Body="&lt;p&gt;Let A be and nxn matrix over R or C. Suppose B is an nxn matrix over the same field as A and Bx=Ax for all x \in R^n. Prove that A=B, that is a_{ij}=b_{ij}, for alla i,j&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T12:52:41.870" LastActivityDate="2010-10-27T13:28:07.870" Title="Prove that A=B, that is a_{ij}=b_{ij}, for alla i,j" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8040" PostTypeId="1" AcceptedAnswerId="8047" CreationDate="2010-10-27T13:04:41.467" Score="0" ViewCount="111" Body="&lt;p&gt;I have a problem understanding this equation (I encountered it in a mechanics problem).&#xA;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = x(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q = q(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = dx/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w = dq/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac {\partial v}{ \partial w} = \frac{\partial x}{\partial q}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that true? If yes, why is that true? How can I proof that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Actually it's a passage in the derivation of the Lagrange equations in Analytical Mechanics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = dOP/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{q} = dq/dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v / \partial \dot{q} = \partial OP/ \partial q$&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2010-10-27T13:58:29.073" LastActivityDate="2010-11-25T20:03:39.290" Title="Derivating in respect of a function that's a derivative of another function" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8041" PostTypeId="2" ParentId="8039" CreationDate="2010-10-27T13:28:07.870" Score="1" ViewCount="0" Body="&lt;p&gt;You don't even need to look at the coordinate entries of your matrices. If E and F are vector spaces, and A and B are linear maps from E to F such that for all x in E: Ax=Bx, then by linearity (A-B)x = 0 for all x. Now what can you say about the linear map A-B, and what does that imply about A and B?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-27T13:28:07.870" CommentCount="2" />
  <row Id="8042" PostTypeId="1" AcceptedAnswerId="8171" CreationDate="2010-10-27T13:35:46.393" Score="6" ViewCount="242" Body="&lt;p&gt;How can I prove that for every positive integer $n$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n\pi}{4}−\frac{1}{\sqrt{8n}}&amp;lt;\frac{1}{2}+\sum_{k=1}^{n−1}\sqrt{1−\frac{k^2}{n^2}}$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-28T17:28:42.340" LastActivityDate="2010-10-28T17:28:42.340" Title="An inequality like Riemann sum involving $\sqrt{1-x^2}$" Tags="&lt;sequences-and-series&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8043" PostTypeId="2" ParentId="7974" CreationDate="2010-10-27T13:51:14.493" Score="1" ViewCount="0" Body="&lt;p&gt;I recommend chapters 3 and 5 of &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;.  It sounds like you have a function f which is hard/expensive to evaluate.  You can evaluate it wherever you like, just not too many times.  You then use those values to model f in some way that is much easier to compute and use that model to get values at other points.  What do you know about f?  If you know it is a ninth-degree polynomial (just not which one) you can evaluate it at ten points, do a fit, and have exact values anywhere else (ignoring important issues of numeric accuracy).  If you know it is a finite sum of trig functions, the Nyquist theorem tells you that proper sampling can also give you perfect accuracy on extrapolation.  If you know it is multiplicative, evaluate it at a set of primes.  If you don't know, maybe the best you can do is evaluate it on a grid and use splines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important question is what you will use the values at other points for.  If you need an integral over some region, the formulas will tell you where to evaluate it to get high accuracy &lt;strong&gt;as long as your functional model is correct&lt;/strong&gt;.  For this, you wouldn't need the value at any other points.  If you need derivatives, numeric differentiation is a noise amplifier and dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your f is not in the class of functions you think, things can go badly wrong.  NR has a classic example where poles off the real line can cause problems, even for fitting a real function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to remember that although you can get approximate values at other points, evaluating these values teaches you nothing about f.  They teach you about g, which is the function within the class you selected to approximate f.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T13:51:14.493" />
  <row Id="8044" PostTypeId="1" AcceptedAnswerId="8051" CreationDate="2010-10-27T14:09:03.573" Score="0" ViewCount="110" Body="&lt;p&gt;I have made up this &quot;fun&quot; problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first row of a 3x3 matrix is $(a_{11} a_{12} a_{13})$. The next row consists of the variables $h, g$ and $c$ in any order. The last row are distinct non-zero fixed scalars, such as $2, 1, 7$ or $-4$ etc. They could be any integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an expression for $det(A)$ for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(-c -2h)a_{11} -3ca_{13} + 3ha_{12} + 2ga_{13} + ga_{12}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that a matrix you construct meeting the conditions given and with this determinate will be the original matrix A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ans: 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the last two rows are sets of distinct non-zero integers. What is the probability that a matrix you construct meeting the conditions given will be the original matrix A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example $det(A) = -71a_{11} +13a_{12} +11a_{13}$ how many matrices could produce this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This one seems to be limited to the number of solutions to a Diophantine system, and it's not I type I can solve, unless there is another way...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let the scalars be any real number will the set of possible matrices become infinite?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-27T14:37:54.780" LastActivityDate="2010-10-27T15:12:54.390" Title="Recovering a matrix A from det(A)" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8045" PostTypeId="2" ParentId="8013" CreationDate="2010-10-27T14:09:20.750" Score="7" ViewCount="0" Body="&lt;p&gt;Pete's excellent notes have correctly explained that there is no set containing sets of unboundedly large size in the infinite cardinalities, because from any proposed such family, we can produce a set of strictly larger size than any in that family. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This observation by itself, however, doesn't actually prove that there are uncountably many infinities. For example, Pete's argument can be carried out in the classical Zermelo set theory (known as Z, or ZC, if you add the axiom of choice), but to prove that there are uncountably many infinities requires the axiom of Replacement. In particular, it is actually consistent with ZC that there are only countably many infinities, although this is not consistent with ZFC, and this fact was the historical reason for the switch from ZC to ZFC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it happened was this. Zermelo had produced sets of size $\aleph_0$, $\aleph_1,\ldots,\aleph_n,\ldots$ for each natural number $n$, and wanted to say that therefore he had produced a set of size $\aleph_\omega=\text{sup}_n\aleph_n$. Frankael objected that none of the Zermelo axioms actually ensured that $\{\aleph_n\mid n\in\omega\}$ forms a set, and indeed, it is now known that in the least Zermelo universe, this class does not form a set, and there are in fact only countably many infinite cardinalities in that universe; they cannot be collected together there into a single set and thereby avoid contradicting Pete's observation. One can see something like this by considering the universe $V_{\omega+\omega}$, a rank initial segment of the von Neumann hierarchy, which satisfies all the Zermelo axioms but not ZFC, and in which no set has size $\beth_\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding the Replacement axiom, however, the Zermelo axioms are extended to the ZFC axioms, from which one can prove that $\{\aleph_n\mid n\in\omega\}$ does indeed form a set as we want, and everything works out great. In particular, in ZFC using the Replacement axiom in the form of transfinite recursion, there are huge uncountable sets of different infinite cardinalities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinities $\aleph_\alpha$, for example, are defined by transfinite recursion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\aleph_0$ is the first infinite cardinality, or $\omega$.&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_{\alpha+1}$ is the next (well-ordered) cardinal after $\aleph_\alpha$. (This exists by Hartog's theorem.)&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_\lambda$, for limit ordinals $\lambda$, is the supremum of the $\aleph_\beta$ for $\beta\lt\lambda$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, for any ordinal $\beta$, the set $\{\aleph_\alpha\mid\alpha\lt\beta\}$ exists by the axiom of Replacement, and this is a set containing $\beta$ many infinite cardinals. In particular, for any cardinal $\beta$, including uncountable cardinals, there are at least $\beta$ many infinite cardinals, and indeed, strictly more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinal $\aleph_{\omega_1}$ is the smallest cardinal having uncountably many infinite cardinals below it.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-10-27T16:21:52.773" LastActivityDate="2010-10-27T16:21:52.773" CommentCount="4" />
  <row Id="8046" PostTypeId="1" CreationDate="2010-10-27T14:21:18.437" Score="1" ViewCount="45" Body="&lt;p&gt;&lt;em&gt;Wondering if my following question is an application of information theory:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we have a factory and ship boxes of stuff outside. If a competitor stands outside my factory, observes the stream of boxes coming out of a factory and note their sizes, he can get valuable information about my manufacturing process. So in order to hoodwink him, I manually expand the boxes to sizes different than the original ones. In order to make sure that I am doing a good job at masking the original size distribution from the resultant distribution, I must make sure that I maximize relative entropy between output distribution q and input distribution p, given that I cannot just create a uniform output distribution. Is this premise correct? Do you find any flaw in the question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraint, of course, is that we must use minimum amount of box material. In other words, sum of box sizes for a unit period of time must not exceed a threshold B.&lt;br&gt;&#xA;                 i.e. $\frac{\sum_{j=1}^{i}{BS_i}}{T} \le B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other constraints are:            &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1] Both p and q must be probability distributions i.e. sum up to 1&lt;br&gt;&#xA; 2] $BS_i \le MS$, where MS is the maximum size of a box.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take $BS_i$ as the size of the $i^{th}$ box, then we have an optimization function applying Lagrange multiplier $\lambda$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Lambda(BS_i,\lambda,p,q) = D(q||p) + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B) $&#xA;$                           = \sum_{S=1}^{n}{q(S) \log  \frac{q(S)}{p(S)}} + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$                           = \sum_{S=1}^{n}{q(S) \log q(S)} -\sum_{S=1}^{n}{q(S) \log p(S)} + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To proceed:&lt;/strong&gt;&#xA;Lets say we at the $i^{th}$ iteration. We know the actual box sizes that we already sent out for 1...i-1 iterations, and the actual size of the present box. i.e. $BS_0..BS_i$ are known, and hence p(BS) can be calculated and hence is a known value. We know the output box sizes till the $i-1^{th}$ iteration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Lambda(BS_i,\lambda,q) = \sum_{S=1}^{n}{q(S) \log q(S)} -\sum_{S=1}^{n}{k*q(S) } + \lambda (\frac{\sum_{j=1}^{i}{BS_i}}{T} - B)  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we now solve the optimization function, in order to get the output box size for this $i^{th} iteration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have,&#xA;$\frac{\partial\Lambda}{\partial BS_i} = 0$,&#xA;$\frac{\partial\Lambda}{\partial q(S)} = 0$, and&#xA;$\frac{\partial\Lambda}{\partial \lambda} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pertial differentiation wrt $\lambda$ results in a trivial solution. Since q(S) is dependent on $BS_i$, we will get a factor of $\frac{\partial q(S)}{\partial BS_i}$ in the other two equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I proceed from here? What I am thinking wrong?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-10-27T14:21:18.437" Title="Optimization of relative entropy" Tags="&lt;optimization&gt;&lt;convex-optimization&gt;" />
  <row Id="8047" PostTypeId="2" ParentId="8040" CreationDate="2010-10-27T14:25:10.610" Score="5" ViewCount="0" Body="&lt;p&gt;If you have a parametrized curve $(x,y)=(f(t),g(t))$, then the tangent vector is $(f'(t),g'(t))$. Provided that the tangent vector isn't pointing straight up or straight down (that is, provided that $f'(t)\neq 0$), the curve looks locally like the graph of a function $y=H(x)$, and its slope $dy/dx=H'(x)$ at some point $(x,y)=(a,H(a))$  is just $g'(t_0)/f'(t_0)$ where $t_0$ is the parameter value corresponding to that point on the curve (that is, $a=f(t_0)$ and $H(a)=g(t_0)$). To see this, think of the right triangle with the tangent vector as its hypotenuse; it has $\Delta y=g'(t_0)$ and $\Delta x=f'(t_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what the statement $dy/dx=(dy/dt)/(dx/dt)$ really means is $H'(a)=g'(t_0)/f'(t_0)$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-27T14:25:10.610" />
  <row Id="8048" PostTypeId="1" AcceptedAnswerId="8049" CreationDate="2010-10-27T14:26:35.907" Score="1" ViewCount="85" Body="&lt;p&gt;In this formula for Euler numbers:&#xA;$\begin{align*}A_n = i^{n+1}\sum _{k=1}^{n+1} \sum _{j=0}^k{k\choose{j}} \frac{(-1)^j(k-2j)^{n+1}}{2^ki^kk}\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is $i$? I have to generate the $n-th$ Euler number, and I don't have $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it the imaginary unit, is it?&lt;/p&gt;&#xA;" OwnerUserId="2582" LastActivityDate="2010-10-27T14:39:02.243" Title="Euler numbers formula" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8049" PostTypeId="2" ParentId="8048" CreationDate="2010-10-27T14:39:02.243" Score="7" ViewCount="0" Body="&lt;p&gt;Yes, it is $\sqrt{-1}$.  They will all divide out in the end.  A derivation of your formula is &lt;a href=&quot;http://www.voofie.com/content/117/an-explicit-formula-for-the-euler-zigzag-numbers-updown-numbers-from-power-series/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  The $i$'s come from changing the $\sin(x)$ function to its definition in terms of $\exp(x)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T14:39:02.243" CommentCount="3" />
  <row Id="8051" PostTypeId="2" ParentId="8044" CreationDate="2010-10-27T14:58:28.897" Score="1" ViewCount="0" Body="&lt;p&gt;Letting the entries in the other two rows be $a, b, c, d, e, f$, we are solving a system of Diophantine equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$bf - ce = P, cd - af = Q, ae - bd = R.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($P, Q, R$ are $-71, -13, 11$ in your example.)  This is a system of three equations in six variables so we should expect lots of solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one way to write them down.  Let $Q = \gcd(Q, R) Q', R = \gcd(Q, R) R'$ and fix $a = \gcd(Q, R), d = 2 \gcd(Q, R)$.  Then the last two equations become $f = c - Q', e = R' + 2b$ and the first equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b(c - Q') - c(R' + b) = - bQ' - cR' = P.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\gcd(Q', R') = 1$, this always has a solution by &lt;a href=&quot;http://en.wikipedia.org/wiki/Bezouts_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout's lemma&lt;/a&gt;.  It can easily be found using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;extended Euclidean algorithm&lt;/a&gt;.  In fact, if it has one solution it has infinitely many, since if $(b, c)$ are a pair of solutions, so is $(b + Rt, c - Qt)$ for every $t \in \mathbb{Z}$.  In particular, by choosing $t$ large enough you can arrange for $b$ to be very positive and $c$ to be very negative, which ensures that $a, b, c, d, e, f$ are distinct (although this is not a particularly natural restriction to me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for any choice of $P, Q, R$ with at least two of them nonzero, you get infinitely many solutions, so this &quot;probability&quot; stuff doesn't make sense without more qualification.  If two of them are equal to zero then the vectors $(a, d), (b, e), (c, f)$ are proportional and there are either zero or infinitely many solutions depending on whether the third is zero or nonzero.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-27T15:12:54.390" LastActivityDate="2010-10-27T15:12:54.390" />
  <row Id="8052" PostTypeId="1" CreationDate="2010-10-27T15:00:52.733" Score="1" ViewCount="118" Body="&lt;p&gt;I know it is a very general question, but i'm not sure how to get started and improve number theory related skills related to ACM ICPC and other programming competitions.Any good way or book recommendations ?&lt;/p&gt;&#xA;" OwnerUserId="2860" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T15:34:00.030" LastActivityDate="2010-10-27T17:40:32.573" Title="How to improve number theory related skills ?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8053" PostTypeId="2" ParentId="3978" CreationDate="2010-10-27T15:01:13.520" Score="0" ViewCount="0" Body="&lt;p&gt;OK, after some days of struggling hard with this problem, i think i have found one more solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fermats little theorem, we have $$a^{p} + p  \equiv a ( \text{mod} \ p)$$ and $$b^{p}+p \equiv b (\text{mod} \ p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, $$\frac{b^{p}+p}{a^{p}+p}=\frac{k_{2}p+b}{k_{1}p+a} = C(\text{an integer})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a prime $p$ such that $p \mid a$, then by the above we can see that $p \mid b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we choose a prime such that $p \mid a$, then we have, $$\frac{k_{2}p+b}{k_{1}p+a}=\frac{k_{4}p}{k_{1}p+a} \Longrightarrow p\mid a \ \text{or} \ p=c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $p=c$, then we will have $$\frac{b+1}{a+1}=p \Longrightarrow p\mid (p-1)$$ a contradiction. So $p \mid a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have concluded:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever $p \mid a \Longrightarrow p \mid b$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever $p \mid b \Longrightarrow p \mid a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Does this help!&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-27T15:01:13.520" />
  <row Id="8054" PostTypeId="1" AcceptedAnswerId="8057" CreationDate="2010-10-27T15:06:17.300" Score="2" ViewCount="60" Body="&lt;p&gt;Hello.&#xA;What meaning of &quot;()&quot; in this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_n=\frac1{n+1}\binom{2n}{n}=\frac{(2n)!}{n!(n+1)!}\qquad\mathrm{for}\;n\geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is Catalan number formula (http://en.wikipedia.org/wiki/Catalan_number))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a division?&lt;/p&gt;&#xA;" OwnerUserId="2861" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T15:12:06.033" LastActivityDate="2010-10-27T15:12:06.033" Title="What meaning of () in Catalan number formula?" Tags="&lt;notation&gt;" AnswerCount="1" />
  <row Id="8055" PostTypeId="2" ParentId="8052" CreationDate="2010-10-27T15:06:24.620" Score="0" ViewCount="0" Body="&lt;p&gt;Well, i am not anyone great in answering this question, but atleast on my behalf, i shall tell you some good introductory books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to theory of numbers by David Burton&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to theory of numbers by G.H.Hardy, E.M.Wright&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Introduction to Analytic number theory by Tom Apostol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;104 problems in Number theory by T.Andreescu&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All the above books are getting a very good foundation on the subject. As i said, i am no one to answer, this as i myself struggle in solving problems. But Solving problems are the only way one learns a subject. There are many peers, who would like to answer this question, and you can learn a lot from them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Reading!&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-27T15:06:24.620" />
  <row Id="8057" PostTypeId="2" ParentId="8054" CreationDate="2010-10-27T15:10:42.927" Score="6" ViewCount="0" Body="&lt;p&gt;The notation $n\choose r$ means &quot;the number of ways of choosing $r$ things out of a total of $n$ given things, where order is not important.&quot;  Its value is $n!/r!(n-r)!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given five coins, say a penny, a nickel, a dime, a quarter, and a dollar, there are 10 ways to choose two coins with order unimportant, because ${5\choose 2} = 5!/2!3! = 120/(2)(6) = 10$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-10-27T15:10:42.927" CommentCount="3" />
  <row Id="8058" PostTypeId="1" AcceptedAnswerId="8059" CreationDate="2010-10-27T15:22:27.177" Score="1" ViewCount="80" Body="&lt;p&gt;If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very grateful if somebody show me how to factor this polynomial by hand, as of now I have used to mathematica to get  $(x-2) (2x - 1) (1 + 3x + x^2)$  but I am having trouble to factor this manually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&lt;/p&gt;&#xA;" OwnerUserId="2862" LastActivityDate="2010-10-27T16:48:09.800" Title="How to solve this ?" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8059" PostTypeId="2" ParentId="8058" CreationDate="2010-10-27T15:26:36.907" Score="7" ViewCount="0" Body="&lt;p&gt;You don't need to factor the polynomial to find the value of $x + \frac{1}{x}$.  Note that the given condition implies $\left( 2x^2 + \frac{2}{x^2} \right) + \left( x + \frac{1}{x} \right) - 11 = 0$ by dividing by $x^2$ and collecting symmetric terms.  Now observe that $\left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to factor it by hand anyway, your best friend is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;rational root test&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-27T15:26:36.907" CommentCount="1" />
  <row Id="8060" PostTypeId="2" ParentId="8024" CreationDate="2010-10-27T15:27:39.633" Score="3" ViewCount="0" Body="&lt;p&gt;You need practice in formulating your intutions.  This is a skill that can be learned through practice, but like all skills, you can't begin just by jumping into the deep-end: you need to work consistently at extending your comfort zone, one small step at a time; over time you will then make substantial improvements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What form should your practice take?  Well, you have to practice writing careful and complete proofs.   (The word &quot;writing&quot; here is important; it is not enough just to think aboutit and convince yourself that you &lt;em&gt;could&lt;/em&gt; write a correct answer --- rather, sit down with pen and paper and carefully write out the full and correct answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a practical suggestion of where to begin (in light of your specific problem): rather than beginning directly with a concept like (failure of) uniform continuity, which involves a rather large number of quantifiers, be sure you can completely prove some simpler statements, such as &quot;$1/(x-a)$ is unbounded on $(a,b)$&quot; or &quot;$x^2$ is unbounded on $\mathbb R$&quot;.   Once you can comfortably prove such statements, you can build up to carefully proving analytic statements with more quantifiers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing: When trying to prove statements (using precise defintions)&#xA;which you feel that you have a good intuitive understanding for, try to self-conciously refelect on how your intuition accords with the formal definitions.  For example, ask yourself &quot;why do I think that $x^2$ is unbounded on $\mathbb R$?&quot; Try to analyze your own intuition and extract some statement out of it.  Then compare this with the formal defintion of unbounded and see if they are the same.  If so, great (!) ; you've got good intuition.  If they dont' quite agree, try to understand why; what was missing in your intuition?  what extra aspect of the situation is the formal definition trying to capture?  Again, reflecting on and analyzing your mathematical intuition is a skill that can be practiced and which you can improve --- but, like all skills, you have to practice it if you want to get better.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-27T15:27:39.633" CommentCount="3" />
  <row Id="8062" PostTypeId="1" CreationDate="2010-10-27T15:45:14.707" Score="1" ViewCount="56" Body="&lt;p&gt;If a,b,c are in arithmetic progression., p,q,r in harmonic progression and ap,bq,cr are in geometric progression., then $\frac{p}{r}+\frac{r}{p} = $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I have tried to use the basic/standard properties of the respective progressions to get the desired result, but I am not yet successful.&lt;/p&gt;&#xA;" OwnerUserId="2862" LastEditorUserId="2862" LastEditorDisplayName="" LastEditDate="2010-10-27T15:56:56.163" LastActivityDate="2010-10-27T18:09:38.983" Title="A problem on progression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="8063" PostTypeId="2" ParentId="8020" CreationDate="2010-10-27T15:49:18.110" Score="4" ViewCount="0" Body="&lt;p&gt;The point is that there is an explicit formula (due to Riemann) relating&#xA;$\pi(x)$ to the zeroes of the zeta-function.  (The proof is via a kind of Fourier transform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rough shape is that&#xA;$$\pi(x) = \mathrm{Li}(x) + \sum_{\rho} \mathrm{Li}(x^{\rho}) + \text{ lower order terms},$$&#xA;where the sum is over zeroes $\rho$ of $\zeta(s)$ in the critical strip (i.e. with real&#xA;parts between $0$ and $1$).&#xA;(See &lt;a href=&quot;http://en.wikipedia.org/wiki/Explicit_formula&quot; rel=&quot;nofollow&quot;&gt;the wikipedia entry&lt;/a&gt; for a more precise statement; this is the same link as in Qiaochu's comment above.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the (simple but) key fact to remember is that&#xA;$| x^{\rho}| = x^{\Re \rho}$, for a positive real number $x$.&#xA;So to get asymptotics on $\pi(x)$ from this, one has to give &#xA;upper bounds on $\Re \rho$.  For example, to get the prime number theorem, one has to show that $\Re \rho &amp;lt; 1$ for all $\rho$ (i.e. that $\zeta(s)$&#xA;has no zeroes on the line $\Re s = 1$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best possible estimate comes if you assume RH.  Then $\Re \rho = 1/2$&#xA;for all $\rho$, so $| x^{\rho}| = x^{1/2}$, and (careful) estimates&#xA;give the error term $\sqrt{x} \log x$ for the difference between&#xA;$\pi(x)$ and $\mathrm{Li}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-27T15:58:35.210" LastActivityDate="2010-10-27T15:58:35.210" />
  <row Id="8064" PostTypeId="2" ParentId="8062" CreationDate="2010-10-27T15:49:57.823" Score="0" ViewCount="0" Body="&lt;p&gt;If a,b,c are in arithmetic progression, then you can write them as a, a+d, a+2d.  Maybe all you need is 2b=a+c.  Similarly for geometric progression, you can write them as ap, apz, apz^2 where z is the common ration (often we use r, but that was taken).  Again, maybe (bq)^2=ap*cr is all you need.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T15:49:57.823" CommentCount="1" />
  <row Id="8065" PostTypeId="2" ParentId="8062" CreationDate="2010-10-27T16:08:50.487" Score="3" ViewCount="0" Body="&lt;p&gt;Let's represent your arithmetic progression as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a,a+h,a+2h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h=b-a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and your harmonic progression as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p,\frac{p}{l+1},\frac{p}{2l+1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(following &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_progression_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the condition that the products of corresponding terms of the arithmetic and harmonic sequences form a geometric sequence implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a+h}{a(l+1)}=\frac{a+2h}{a+h}\frac{l+1}{2l+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{r}+\frac{r}{p}=2l+1+\frac1{2l+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we solve for $l$, which has two solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave the rest to you. ;)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-27T16:08:50.487" />
  <row Id="8066" PostTypeId="1" AcceptedAnswerId="8068" CreationDate="2010-10-27T16:25:24.443" Score="0" ViewCount="99" Body="&lt;p&gt;And if not, what are the most common hypotheses (on open sets it induces, for example) under which distance is a continuous function?&lt;/p&gt;&#xA;" OwnerUserId="2866" LastActivityDate="2010-10-27T16:36:59.843" Title="Is the function distance continuous?" Tags="&lt;analysis&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8067" PostTypeId="2" ParentId="8066" CreationDate="2010-10-27T16:33:40.327" Score="2" ViewCount="0" Body="&lt;p&gt;Yes.  The standard definition of the topology induced by a metric ensures this; in fact it's not hard to see that it's the coarsest topology such that $d(x, y)$ is continuous for fixed $x$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-27T16:33:40.327" CommentCount="1" />
  <row Id="8068" PostTypeId="2" ParentId="8066" CreationDate="2010-10-27T16:36:59.843" Score="0" ViewCount="0" Body="&lt;p&gt;As Qiaochu points out $d(x,y)$ is continuous for fixed $x$. You may like to see this as well, as this is a familiar result in Topology.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A$ is a non empty closed subset of a metric space $(X,d)$ then show that the function f on X given by $f(x)=d(x,A)$ is continuous.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because $$| f(x) - f(y) | = | d(x,A) - d(x,B) | \leq d(x,y)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $f$ is uniformly continuous (use $\delta = \epsilon$ in any point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ and $y$ be points in $X$, and $p$ any point of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $d(x,p) \leq d(x,y) + d(y,p)$ (triangle inequality)&#xA;and so $d(x,A) \leq d(x,y) + d(y,p)$ (as $d(x,A)$ is the infimum).&#xA;But then $d(y,p) \geq d(x,A) - d(x,y)$ (for all $p$)&#xA;so that $d(y,A) \geq d(x,A) - d(x,y)$ (as $d(y,A)$&#xA; is the infimum).&#xA;So : $d(x,A) - d(y,A) \leq d(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now reverse the roles of $x$ and $y$ to get&#xA;$d(y,A) - d(x,A) \leq d(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is taken from &lt;a href=&quot;http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2004;task=show_msg;msg=1323.0001&quot; rel=&quot;nofollow&quot;&gt;http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2004;task=show_msg;msg=1323.0001&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-27T16:36:59.843" />
  <row Id="8069" PostTypeId="2" ParentId="7963" CreationDate="2010-10-27T16:44:05.623" Score="1" ViewCount="0" Body="&lt;p&gt;In one of the standard approaches to &lt;a href=&quot;http://en.wikipedia.org/wiki/Fuzzy_logic&quot; rel=&quot;nofollow&quot;&gt;fuzzy logic&lt;/a&gt;--used in household appliances everywhere!--the truth values are taken to be real numbers in the unit interval, where $0$ represents &lt;em&gt;false&lt;/em&gt; and $1$ represents &lt;em&gt;true&lt;/em&gt;, and the Boolean connectives are defined by the following equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(not)   $\neg p = 1-p$&lt;/li&gt;&#xA;&lt;li&gt;(and)   $p\wedge q= \text{min}(p,q)$&lt;/li&gt;&#xA;&lt;li&gt;(or)    $p\vee q= \text{max}(p,q)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These values agree with the classical values when restricted to $0$ and $1$, and thus agree with Qiaochu's answer on the classical values, but they disagree on non-classical values. In particular, this version of fuzzy logic gives the same truth value to $p$ as $p\wedge p$, but this isn't the case with the probabilistic formulas, which (inappropriately) interprets $p\wedge p$ as a conjunction of independent events. Similarly, fuzzy logic also says $p$ is equivalent to $p\vee p$, whereas the probabilistic formula does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fun exercise to show that fuzzy logic doesn't work with classical tautologies very well. For example, $p\vee \neg p$ is not uniformly value $1$, although it is always at least $\frac 12$. Indeed, a statement is a classical tuatology if and only if it gets fuzzy value at least $\frac 12$ on all input. There are numerous other such fun problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also numerous variations on the fuzzy theme, however, with many different fuzzy logics, and the Wikipedia page appears extensive.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-10-27T16:44:05.623" />
  <row Id="8070" PostTypeId="1" CreationDate="2010-10-27T16:45:58.630" Score="1" ViewCount="89" Body="&lt;p&gt;2x ≡ 1 (mod p) where p is an odd prime&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really stuck on this one.&lt;/p&gt;&#xA;" OwnerUserId="2868" LastActivityDate="2010-10-27T17:15:24.893" Title="Solving 2x ≡ 1 (mod p) where p is an odd prime" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8071" PostTypeId="1" AcceptedAnswerId="8076" CreationDate="2010-10-27T16:47:24.417" Score="8" ViewCount="454" Body="&lt;p&gt;At the Blizzcon 2010, StarCraft II multiplayer panel, this stuff was supposed to explain the ladder matchmaking system. I look at this and go eh? what!?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CYV2b.png&quot; alt=&quot;StarCraft II: Ladder math&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is any of this real? or are they just messing with me, because I can't make heads or tails of this humongous equation. Can anyone break it down for me? or at least explain the principle behind this, I don't get how these integrals, derivatives, Euler-functions and multiplication sums yield anything meaningful...&lt;/p&gt;&#xA;" OwnerUserId="2867" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T02:40:37.200" LastActivityDate="2010-10-28T02:40:37.200" Title="StarCraft II: Ladder math" Tags="&lt;calculus&gt;&lt;statistics&gt;&lt;applications&gt;&lt;tag-removed&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="8072" PostTypeId="2" ParentId="8058" CreationDate="2010-10-27T16:48:09.800" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad \bigg(a\ \bigg(x+\frac{1}x\bigg)^2 + b\ \bigg(x+\frac{1}x\bigg)+c\bigg)\ x^2\: \ =\ \ a\ (x^4 + 1) + b\ (x^3+x) + (c+2a)\ x^2 $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above allows you to determine the values of $\rm\:a,\:b,\:c\:$ by inspection. The same works for arbitrary even degree symmetric polynomials: $\ $ simply subtract $\rm\ a\ (x^2 +1)^k$ and recurse on what remains.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially a special case of Gauss's algorithm for rewriting symmetric polynomials in terms of elementary symmetric polynomials. It's one of the earliest known occurrences of Grobner basis rewriting, e.g. see p. 314 of Cox, Little, O'Shea:$\ $ Ideals, Varieties and Algorithms. In chapter 7 they present a generalization to compute the ring of invariants of a finite matrix group $\rm\: G \subset GL(n,k)\:$. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-27T16:48:09.800" />
  <row Id="8073" PostTypeId="1" AcceptedAnswerId="8074" CreationDate="2010-10-27T17:05:25.377" Score="0" ViewCount="77" Body="&lt;p&gt;I need to do a series expansion of this forumla&#xA;$$z = \sqrt{\frac{x+1}{y+1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get Zx Zxx and so on. but I really dont know where to start with this this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;just a push in the right direction should be enough.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-10-27T18:52:58.270" LastActivityDate="2010-10-27T18:52:58.270" Title="partial derivatives." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8074" PostTypeId="2" ParentId="8073" CreationDate="2010-10-27T17:10:07.260" Score="1" ViewCount="0" Body="&lt;p&gt;If you know how to differentiate already, remember that partial differentiation with respect to a certain independent variable treats the other independent variables as constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use $(x+y)^2$ as an example, $\partial_x (x+y)^2$ is just $2(x+y)$ (chain rule). Now try this operation on your function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-27T17:10:07.260" CommentCount="1" />
  <row Id="8075" PostTypeId="2" ParentId="8070" CreationDate="2010-10-27T17:15:24.893" Score="1" ViewCount="0" Body="&lt;p&gt;The trick proposed by Qiaochu in the comment is actually a very special case of &lt;a href=&quot;http://math.stackexchange.com/questions/3230&quot;&gt;Gauss's algorithm&lt;/a&gt; for computing inverses $\rm\:(mod\ p)\:$ for $\rm\:p\:$ prime. This special case of the Euclidean algorithm works simply by repeatedly scaling the fraction so to reduce the denominator until it reaches one. See my linked post for further details.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-27T17:15:24.893" />
  <row Id="8076" PostTypeId="2" ParentId="8071" CreationDate="2010-10-27T17:24:16.150" Score="14" ViewCount="0" Body="&lt;p&gt;It looks like an ordinary statistical calculation.  The numerator with $\Pi_{g=1}^G$ is a likelihood or probability density, presumably of some outcomes for games $1$ to $G$.  The denominator with $\int \Pi \dots$ is the integral of the numerator over all outcomes; it is a normalization constant to ensure the total probability of all results is $1$.  Everything in the formulas is a calculation of (z-scores in) independent normal distributions, so they have a fairly simple probability model for how a player's ranking parameters drive the game outcomes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of the calculation might be to calculate a player's set of ranking parameters $y$ (a vector of numbers measuring strength, speed, skill, wins, or whatever interpretation the quantities have for the game) that maximize the conditional probability $P(g_j | y)$ of having observed the game outcomes $g_i$ for $i = 1$ to $G$.  In other words, Maximum Likelihood Estimation of a player's parameters from game data. I can't read everything in the formula -- can you post a larger magnification? -- but the $\theta_{1,g} - \theta_{2,g}$ look like a measure of how one side of the game performed compared relative to the other, such as a difference in number of points, or a measure of how the sides were &lt;em&gt;expected&lt;/em&gt; to perform relative to each other, given their ratings.  Alternatively, $P(g_j | y)$ could be a Bayesian &quot;posterior&quot; distribution on $y$ in light of the game outcomes, so that the formula is a rule for updating the rankings given some game results. Here $\Phi(\theta_0 + \gamma_i + \psi_{i,0})$ can be understood as implying an initial rating, where the distribution of skills in the player population is assumed to be normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also infer from the formula that either they are doing the wrong calculation (after $G$ new games), or the big formula is actually a summary of what has happened after $G$ separate re-estimation steps, one after each game (so that in a single step there is no product involved, and the formula would involve only the ratings just before the game, and the game outcome).  The probabilies they are computing for the $G$ games are of the form &quot;what is the chance the player had performance at least $x$ in game 1, and at least $y$ in game 2, $z$ or better in game 3, ...&quot;.  This is not the correct way to assess the probability that the whole set of $G$ game results is, collectively, above a certain level of performance. But if the parameters are re-estimated after every game, and the older game outcomes forgotten, then &quot;chances of a victory at least $X$ big&quot; is the only thing you can do, so this would account for the shape of their formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as the old joke goes:  &quot;and by the way, what's StarCraft?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-27T20:02:45.843" LastActivityDate="2010-10-27T20:02:45.843" CommentCount="5" />
  <row Id="8077" PostTypeId="1" CreationDate="2010-10-27T17:25:44.387" Score="1" ViewCount="78" Body="&lt;p&gt;I've got a function whose general form is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{1}{x^\alpha}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x &gt; 0$ and $\alpha &gt; 0$. I would like to show that if $0&amp;lt; \alpha &amp;lt; 1$ $f(x)$ has slow decay and if $\alpha &gt; 1$ the $f(x)$ has rapid decay. (I've already verified these properties of $f(x)$ using a graphing application.)&lt;/p&gt;&#xA;" OwnerUserId="1752" LastActivityDate="2010-10-27T22:54:54.007" Title="How to show that a function has slow or rapid decay" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8078" PostTypeId="1" AcceptedAnswerId="8083" CreationDate="2010-10-27T17:34:55.917" Score="3" ViewCount="106" Body="&lt;p&gt;Suppose $B \subset X$ where $X$ is a vector space. $B$ is called balanced if $\alpha B \subset B$ for every $\alpha \in \Phi$ with $|\alpha| \leq 1$. Note that $\Phi = \textbf{R}$ or $\Phi = \textbf{C}$. What is the intuition behind balanced sets? Why do we define them? Is it &quot;good&quot; for a vector space to have more balanced sets? Why do we require $| \alpha| \leq 1$?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-27T17:39:38.470" LastActivityDate="2010-10-27T20:26:22.227" Title="Intuition Behind Balanced Sets" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8079" PostTypeId="2" ParentId="8052" CreationDate="2010-10-27T17:40:32.573" Score="1" ViewCount="0" Body="&lt;p&gt;For the ACM you should probably focus on their problem sets and could also look into problems from discrete mathematics. I really don't think they ask questions that rely on a graduate level understanding of number theory so reading books could be overkill for your purposes and would only confuse you if you rush it. It should take several weeks to read several chapters in a good number theory book - assuming you're not too busy (but you're in cs, right?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: Go and ask a math professor for some problem sets.&lt;/p&gt;&#xA;" OwnerUserId="2738" LastActivityDate="2010-10-27T17:40:32.573" />
  <row Id="8081" PostTypeId="1" AcceptedAnswerId="8082" CreationDate="2010-10-27T17:44:11.203" Score="0" ViewCount="109" Body="&lt;p&gt;Does any compact subset belong to Borel Sigma Algebra? &#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here Borel Sigma algebra is defined to be the sigma algebra generated by all open subsets in the topological space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that there are two kinds of definitions for a Borel sigma algebra: one generated by all open subsets, and the other by all compact subsets.&#xA;These two definitions are equivalent for most typical spaces, including any locally compact separable metrizable space (or more generally any σ-compact space), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in the definition for inner regular measure, is it correct that all the compact subsets are required to be Borel measurable, where the Borel sigma algebra is defined to be generated by all open sets instead of compact sets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw in Wikipedia, inner regular measure is defined to be on Hausdorff space. Is a Haursdorff space a locally compact separable metrizable space or a σ-compact space? Or in a Hausdorff space, the sigma algebra generated by all compact subsets is a proper subset of the sigma algebra generated by all open subsets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-10-27T19:07:35.443" LastActivityDate="2010-10-27T19:07:35.443" Title="Does any compact subset belong to Borel Sigma Algebra ?" Tags="&lt;topology&gt;&lt;sigma-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8082" PostTypeId="2" ParentId="8081" CreationDate="2010-10-27T17:48:16.183" Score="11" ViewCount="0" Body="&lt;p&gt;If your space is Hausdorff then any compact subset is closed (hence the complement of something open), in which case the answer is &lt;em&gt;yes&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your set is a finite set with more than one element and with the indiscrete topology (just the whole set and the empty set are open),&#xA;then every subset is compact, but proper subsets are not Borel.  So in complete generality the answer is &lt;em&gt;no&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-27T17:48:16.183" CommentCount="1" />
  <row Id="8083" PostTypeId="2" ParentId="8078" CreationDate="2010-10-27T17:52:48.830" Score="5" ViewCount="0" Body="&lt;p&gt;[Added in response to the comments below:] Despite what is written below,&#xA;&lt;em&gt;balanced&lt;/em&gt; needn't imply convex; it simply doesn't follow from the definition.  So one should take the following discussion as rather an explanation about &lt;em&gt;convex&lt;/em&gt; balanced sets, for whatever that is worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Original answer follows:]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Balanced&lt;/em&gt; means that the set is symmetric around the origin (i.e. invariant under $v \mapsto -v$) and is convex.  These are nice properties, generalizing some useful features of balls around the origin in $\mathbb R^n$.  (Incidentally, if you didn't have the condition $|\alpha| = 1$, then any balanced subset with non-empty interior (in $\mathbb R^n$ say) would be all of $\mathbb R^n$, so that wouldn't be such an interesting condition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any vector space will have plenty of such sets; you can form them just by taking the convex hull of a collection of points invariant under $v \mapsto -v$.  What is more relevant in functional analysis is whether the origin of a topological vector space contains a &lt;em&gt;basis&lt;/em&gt; of balaced neighbourhoods.  The topological vector space is then called &lt;em&gt;locally convex&lt;/em&gt; and this has important consequences for the space (especially for&#xA;its duality theory).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that balls around the origin in a normed space are automatically balanced, so normed spaces are locally convex; so again, in the topological setting, one sees that this is a way of capturing some the geometric aspects&#xA;of normed vector spaces, without insisting on the existence of a norm.&#xA;(So, while normed spaces are not closed under certain operations, such as&#xA;forming infinite direct sums, locally convex spaces are --- any direct sum&#xA;(just to take one example) of a collection of locally convex spaces has &#xA;a natural locally convex topology of its own.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-10-27T20:26:22.227" LastActivityDate="2010-10-27T20:26:22.227" CommentCount="3" />
  <row Id="8084" PostTypeId="2" ParentId="8078" CreationDate="2010-10-27T17:53:49.747" Score="0" ViewCount="0" Body="&lt;p&gt;In &lt;em&gt;Complex analysis in locally convex spaces&lt;/em&gt; by Seán Dineen, it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;Balanced open sets arise as the natural domain of convergence of Taylor series expansion at the origin of &lt;i&gt;holomorphic functions.&lt;/i&gt;&lt;/blockquote&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T18:42:35.153" LastActivityDate="2010-10-27T18:42:35.153" CommentCount="8" />
  <row Id="8085" PostTypeId="1" AcceptedAnswerId="8087" CreationDate="2010-10-27T18:01:11.143" Score="7" ViewCount="126" Body="&lt;p&gt;Suppose $A$ is a set of finite measure.   Is it possible that $A$ can be an &lt;strong&gt;uncountable&lt;/strong&gt; union of disjoint subsets of $A$, each of which has positive measure?  &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-10-27T18:04:51.227" Title="Set of Finite Measure: Uncountable disjoint subsets of non-zero measure. " Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8086" PostTypeId="1" AcceptedAnswerId="8089" CreationDate="2010-10-27T18:02:12.203" Score="3" ViewCount="173" Body="&lt;p&gt;How is a Borel measure on Euclidean space defined? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean the Borel measure on the Borel sigma algebra generated by usual toplogy on an Euclidean space, which coincides with the normal definition of volume of subsets in the Euclidean space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Lebesgue measure on an Euclidean space, it is defined in terms of the outer measure on the power set of the Euclidean space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Borel measure can be defined independent of the Lebesgue measure, then the Lebesgue measure can be defined in terms of the Borel measure by completion of measure. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-06T03:37:25.243" LastActivityDate="2010-11-06T03:37:25.243" Title="How is a Borel measure on Euclidean space defined?" Tags="&lt;topology&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="8087" PostTypeId="2" ParentId="8085" CreationDate="2010-10-27T18:04:51.227" Score="8" ViewCount="0" Body="&lt;p&gt;No.  Suppose $A$ is an uncountable disjoint union of measurable subsets $A_i, i \in I$ with positive measure.  Then $I$ is a countable union of the sets of indices $i$ such that $\mu(A_i) &gt; \frac{1}{n}, n \in \mathbb{N}$, so it follows that one of these sets must be uncountable.  In particular a countable union of some subcollection of the $A_i$ has arbitrarily large measure; contradiction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-27T18:04:51.227" CommentCount="6" />
  <row Id="8088" PostTypeId="2" ParentId="8062" CreationDate="2010-10-27T18:09:38.983" Score="0" ViewCount="0" Body="&lt;p&gt;Notice that $\rm\:\ \quad\displaystyle \frac{p}r+\frac{r}p\ =\ \frac{(p+r)^2}{pr} - 2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we have that$\rm\quad\displaystyle p\:r\ =\ \frac{(bq)^2}{ac}\ \ $ via $\rm\ ap,\:bq,\:cr\ $ geometric  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have $\rm\quad\ \ \displaystyle p+r\ =\ \frac{2pr}q\quad\ \ \ $ via $\rm\ p,q,r\ $ harmonic&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-27T18:09:38.983" />
  <row Id="8089" PostTypeId="2" ParentId="8086" CreationDate="2010-10-27T18:15:47.440" Score="3" ViewCount="0" Body="&lt;p&gt;The Borel $\sigma$-algebra on $\mathbb{R}$ is generated by the ring of finite disjoint unions of half-open intervals.  On this ring one can define a countably additive measure in the obvious way (although one should prove that it is countably additive).  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Carath%C3%A9odorys_extension_theorem&quot; rel=&quot;nofollow&quot;&gt;Carathéodory's extension theorem&lt;/a&gt; this measure extends to the entire Borel $\sigma$-algebra, and by &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynkin_system#Dynkin.27s_.CF.80-.CE.BB_Theorem&quot; rel=&quot;nofollow&quot;&gt;Dynkin's lemma&lt;/a&gt; this extension is unique.  Then one can take products to get a Borel measure on $\mathbb{R}^n$ and complete to get Lebesgue measure on the Lebesgue $\sigma$-algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this isn't essentially different - the proof of Caratheodory that I've seen uses outer measure anyway.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-27T18:21:51.950" LastActivityDate="2010-10-27T18:21:51.950" />
  <row Id="8090" PostTypeId="2" ParentId="8077" CreationDate="2010-10-27T18:24:27.363" Score="1" ViewCount="0" Body="&lt;p&gt;If you start at 1 (your sum is not defined at x=0), you can bound the sums as integrals: $ \int_{x=2}^{\infty}x^{-\alpha} dx \lt \sum_{x = 1}^\infty x^{-\alpha} \lt \int_{x=1}^{\infty}x^{-\alpha} dx$.  One will solve it for $\alpha \le 1$ and one for $\alpha \gt 1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T18:24:27.363" />
  <row Id="8091" PostTypeId="2" ParentId="8078" CreationDate="2010-10-27T18:25:23.080" Score="0" ViewCount="0" Body="&lt;p&gt;It's like balancing a tire, you spin the set around! (A bit tongue in cheek, but that's how I've always felt about it...)&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-10-27T18:25:23.080" />
  <row Id="8092" PostTypeId="2" ParentId="8077" CreationDate="2010-10-27T18:30:59.417" Score="1" ViewCount="0" Body="&lt;p&gt;Another approach is to use&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_condensation_test&quot; rel=&quot;nofollow&quot;&gt;Cauchy's condensation test&lt;/a&gt;&#xA;which states that a series $\sum a_n$, where the terms $a_n$&#xA;are positive and decreasing, is convergent if and only if&#xA;$\sum 2^m a_{2^m}$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example the line $\mathbf{Re}(s)=1$ is the&#xA;edge of the region of ansolute convergence of the usual&#xA;formula $\sum_{n=1}^\infty 1/n^s$ for the Riemann zeta function.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-27T18:30:59.417" CommentCount="1" />
  <row Id="8093" PostTypeId="2" ParentId="7952" CreationDate="2010-10-27T19:11:09.583" Score="0" ViewCount="0" Body="&lt;p&gt;In class today we asked the professor about this problem and he told us to look up logarithmic derivatives. This approach leads to the following solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the previous homework problem we know that: $e_{n}(x)+D_{n}=\prod(x+a)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\frac{f'}{f}$ where we are formally differentiating with respect to x. This transforms the LHS to $\frac{e\prime_{n}(x)}{e_{n}(x)+D_{n}}$ where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ e\prime_{n}(x)=\frac{d}{dx}\sum_{i=0}^{n-1}x^{q^{i}}\frac{D_{n}}{D_{i}L_{n-i}}=(-1)^{n}\frac{D_{n}}{L_{n}}+\sum_{i=1}^{n-1}(-1)^{n-i}q^{i}\frac{D_{n}}{D_{i}L_{n-i}^{q}}x^{q^{i-1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $x=0$, $e\prime_{n}(x)=(-1)^{n}\frac{D_{n}}{L_{n}}$ and so the LHS becomes $\frac{(-1)^{n}}{L_{n}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the RHS gets transformed to $\sum\frac{1}{x+b}$ since $f'=\sum_{b}\frac{\prod(x+a)}{x+b}$ and $f=\prod(x+a)$. At $x=0$ this becomes $\sum\frac{1}{b}$ which is what we wanted.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-27T19:11:09.583" />
  <row Id="8094" PostTypeId="1" CreationDate="2010-10-27T19:38:46.403" Score="1" ViewCount="124" Body="&lt;p&gt;I am not a math expert nor a native English speaker, so maybe I am using wrong terminology... In this case, help me to improve my question :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my set of equations, I have to calculate values $x_n$, $y_n$ and $z_n$. I know the values  of A, B, C, Total and sums of parameters ($S_1$ = $x_1$ + $x_2$ + $x_3$, $S_2$ = $y_1$  + $y_2$ + $y_3$, $S_3$ = $z_1$ + $z_2$ + $z_3$). It is not necessary, that each equation gives us exactly the same result, but each equation should have its solution it's close to others as possible - in perfect case this would be Total/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1A + y_1B + z_1C = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_2A + y_2B + z_2C = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_3A + y_3B + z_3C = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured it out, this is a set of diophantine equations. I could solve this with a genetic algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First randomly assign values $x_n$, $y_n$ and $z_n$ (just watch for the sum of parameters)&lt;/li&gt;&#xA;&lt;li&gt;Then mutate this numbers so, that the solution of each equation is as close to others as possible. By mutating I mean to increase one parameter while decreasing another and calculate fitness function.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the first solution should work (have to try it out though). But it was a simplification of my original problem, which is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i(x_1A + y_1B + z_1C) = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j  (x_2A + y_2B + z_2C) = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k  (x_3A + y_3B + z_3C) = \text{Total}/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem is, that I cannot just randomly increase/decrease parameters, because $i$, $j$ and $k$ could be different and hence the set of these equations could yield different solution as previous mutation. Also - i, j and k are not given, I have to calculate them by my self.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to solve my two problems?&lt;/p&gt;&#xA;" OwnerUserId="2873" LastEditorUserId="2873" LastEditorDisplayName="" LastEditDate="2010-10-28T07:59:17.083" LastActivityDate="2010-12-27T08:44:28.197" Title="How to calculate this set of diophantine equations" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8095" PostTypeId="1" AcceptedAnswerId="8098" CreationDate="2010-10-27T19:49:27.543" Score="0" ViewCount="62" Body="&lt;p&gt;For each $n&gt;0$, how do we prove that $$\Gamma'(n+1)&gt; \log{n} \cdot \Gamma(n+1)$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I had spent about half an hour on this question, but just could find any way of proceeding for the solution. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wikipedia page gave me an interesting identity $$\Gamma'(n+1)= n! \cdot \Biggl( - \gamma + \sum\limits_{k=1}^{n} \frac{1}{k}\Biggr)$$ But i don't know how it can be applied here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-27T20:00:35.377" Title="Inequality concerning the Gamma function" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;special-functions&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8096" PostTypeId="1" CreationDate="2010-10-27T19:51:33.937" Score="2" ViewCount="93" Body="&lt;p&gt;Let $X$ be integral scheme and $\mathcal K$ sheaf of rational functions on $X$. For any &#xA;point $y\in X$ different of generic point we know that fiber of $\mathcal K$ (defined as usual as $\mathcal K _y / \mathcal m_y \mathcal K_y$) is zero. I'll be very gratefull if you explain intuitively why this is so, in language of restriction of $\mathcal K$ to reduced subscheme $Y=\overline{\{y \} }$. I have difficulty because many rational functions on $X$ can be restricted to nonzero rational functions on  $Y$ . How is that compatible with fiber of $\mathcal K$ equals zero at $y$?  &lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-27T19:57:23.583" LastActivityDate="2010-10-27T19:57:23.583" Title="Intuition for fiber of rational function sheaf" Tags="&lt;algebraic-geometry&gt;" CommentCount="7" />
  <row Id="8097" PostTypeId="2" ParentId="7475" CreationDate="2010-10-27T19:59:31.807" Score="0" ViewCount="0" Body="&lt;p&gt;I made a table using the labels {I3, p1, p2, p3, p4, p5} instead of filling the table with the matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;---  I3  p1  p2  p3  p4  p5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I3  I3  p1  p2  p3  p4  p5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p1  p1  I3  p3  p2  p5  p4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p2  p2  p5  I3  p4  p3  p1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p3  p3  p4  p1  p5  p2  I3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p4  p4  p3  p5  p1  I3  P2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p5  p5  p2  p4  I3  p1  p3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f_I3 (x) = I3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then looking at a matrix mapping looking at p1 line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = I3 p1 p2 p3 p4 p5&#xA;    p1 I3 p3 p2 p5 p4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p1*I3 = p1  p1*p1 = I3  this brings it back to start so &#xA;(I3, p1)&#xA;p1*p2 = p3  p1*p3 = p2 brings it back to start so &#xA;(I3, p1),(p2, p3)&#xA;p1*p4 = p5  p1*p5 = p4 brings it back to start so &#xA;(I3, p1),(p2, p3),(p4, p5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f_p1 (x) = (I3, p1),(p2, p3),(p4, p5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then looking at a matrix mapping looking at p2 line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = I3 p1 p2 p3 p4 p5&#xA;    p2 p5 I3 p4 p3 p1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f_p2 (x) = (I3, p2), (p1, p5), (p3, p4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on... &lt;/p&gt;&#xA;" OwnerUserId="2686" LastActivityDate="2010-10-27T19:59:31.807" CommentCount="1" />
  <row Id="8098" PostTypeId="2" ParentId="8095" CreationDate="2010-10-27T20:00:35.377" Score="6" ViewCount="0" Body="&lt;p&gt;Well, if we rearrange the inequality to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)} &gt; \log n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)}=-\gamma+\sum_{j=1}^n \frac1{j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you should be able to easily deduce the inequality.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-27T20:00:35.377" CommentCount="2" />
  <row Id="8099" PostTypeId="1" CreationDate="2010-10-27T20:07:39.767" Score="1" ViewCount="81" Body="&lt;p&gt;Given octic group G = {e, s, s^2, s^3, b, g, d, t}   (sigma, betea, delta... )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find subgroup of G that has order 2 and is a normal subgroup of G.&#xA;Find subgroup of G that has order 2 and is NOT a normal subgroup of G.&lt;/p&gt;&#xA;" OwnerUserId="2686" LastActivityDate="2010-10-27T20:07:39.767" Title="Find subgroup of octic group that is normal and NOT normal" Tags="&lt;abstract-algebra&gt;&lt;homework&gt;" CommentCount="6" />
  <row Id="8100" PostTypeId="2" ParentId="58" CreationDate="2010-10-27T20:20:10.033" Score="11" ViewCount="0" Body="&lt;p&gt;Some of the data generated by Math Stack Exchange itself (and, presumably, by similar sites) ought to approximately follow Benford's law.  These would include the distribution of first digits of &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Frequency of tag use,&lt;/li&gt;&#xA;&lt;li&gt;Number of votes for questions,&lt;/li&gt;&#xA;&lt;li&gt;User reputation,&lt;/li&gt;&#xA;&lt;li&gt;Number of views for questions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is because Benford's law applies to exponentially growing quantities, and the total number of all of these quantities ought to be growing exponentially.  It's only approximate because of artifacts in the way that some of these quantities are determined and because you need several orders of magnitude in order to see Benford's law really kick in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I decided to test this for the first three.  The first digits weren't actually that hard to compile because the site allows you to sort the first three from highest to lowest.  (Unfortunately, it does not allow you to sort the fourth that way, and I don't feel like wading through 2200+ questions to collect the data.) Here are the results I got on the first three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tag Use&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Data collected October 25, 2010.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/njTkU.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a bad fit, especially when you consider that there are only three orders of magnitude represented in this measure.  There are a disproportionately large number of tags that were created but only used once, which explains the larger frequency of 1 as a first digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Votes Per Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Data collected October 26, 2010.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/zC1pM.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also not a bad fit, especially since there are currently only two orders of magnitude represented in this measure.  Also, I threw out the questions with 0 and negative numbers of votes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;User Reputations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Data collected October 25 and 26, 2010.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kJYsl.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the worst fit of the three, as the frequency of 1 as a first digit is so much larger than the others.  However, there are a very large number of users who have never posted a question or an answer and so have a rep of 1.  And, for reasons unknown to me, there are also a large number of users who have a reputation of 101, despite never having asked or answered a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove 1 as a possible first digit and then rescale the Benford law probabilities to consider only 2 through 9 as possible first digits then the picture looks like the following, which is a much better Benford fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZM83s.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, none of these data sets would pass Pearson's chi-square test for goodness-of-fit with respect to the Benford probabilities at a reasonable level of significance.  However, given some of the artifacts in the data and the fact that there are relatively few orders of magnitude represented, the fit with Benford's law is really not that bad.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-27T20:20:10.033" CommentCount="3" />
  <row Id="8101" PostTypeId="1" AcceptedAnswerId="8119" CreationDate="2010-10-27T20:24:36.073" Score="6" ViewCount="150" Body="&lt;p&gt;Polynomial $P$ satisfies $P(n)&gt;n$ for all positive integers $n$. Every positive integer $m$ is a factor of some number of the form $P(1),P(P(1)),P(P(P(1))),\ldots $. Prove that $P(x)=x+1$.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-27T20:28:24.170" LastActivityDate="2010-10-28T00:03:54.260" Title="Iterated polynomial problem" Tags="&lt;number-theory&gt;&lt;polynomials&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="8102" PostTypeId="2" ParentId="7986" CreationDate="2010-10-27T20:51:21.483" Score="3" ViewCount="0" Body="&lt;p&gt;This expands on a comment that might not have been perfectly clear (perhaps because of its obviousness, for which I apologize).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have in hand some magic method of integration (such as the one in the question).  By this I mean &lt;em&gt;any&lt;/em&gt; procedure that takes a functional &lt;em&gt;expression&lt;/em&gt; $f(x)$ and returns some &lt;em&gt;expression&lt;/em&gt; $F(x)$ for its indefinite integral. (I also assume you accept $F(x)$ as a valid solution to your integration; &lt;em&gt;e.g.&lt;/em&gt;, some people might not accept an elliptic function, or imaginary values are &lt;em&gt;verboten,&lt;/em&gt; or they might not like power series, &lt;em&gt;etc&lt;/em&gt;.  I leave the criterion of acceptability to you as a matter of taste.  All I require is that you know how to differentiate $F$ and are able to compare that result to $f$ in order to check the validity of your magic method.)  A trivial application of the Fundamental Theorem of Calculus asserts that the &quot;very clever&quot; substitution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = F(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will enable you to perform the integral.  This happens, of course, because you can calculate that $du = F'(x)dx = f(x)dx$, whence the substitution converts the original integral into $\int{du}$ with general solution $u + C = F(x) + C$ by back-substitution.  In other words, the question as posed merely asks whether the FTC is true in a special case.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-10-27T20:51:21.483" CommentCount="5" />
  <row Id="8103" PostTypeId="2" ParentId="8035" CreationDate="2010-10-27T20:54:37.780" Score="0" ViewCount="0" Body="&lt;p&gt;You could write $y(x)=a$ instead of $y=a$ to point out you are meaning a constant function, as $y(a)=a$ is not constant. Part (horizontal line) is not necessary. &lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-27T20:54:37.780" CommentCount="0" />
  <row Id="8104" PostTypeId="1" AcceptedAnswerId="8129" CreationDate="2010-10-27T21:01:13.140" Score="3" ViewCount="70" Body="&lt;p&gt;If i have a standard LP problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \mathbf{d}^T \mathbf{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{B}\mathbf{x}=\mathbf{f},\qquad \mathbf{x} \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{y}$ is the optimal solution and $\mathbf{z}$ is the optimal solution to the dual problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the same cost function $\mathbf{d}$, if $\mathbf{f}$ is replaced by $\mathbf{b}$ then $\mathbf{x}$ becomes the new optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{z}^T (\mathbf{b}-\mathbf{f}) \leq \mathbf{d}^T (\mathbf{x}-\mathbf{y})$$&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T01:55:05.023" LastActivityDate="2010-10-28T01:55:05.023" Title="Prove property of dual Linear Programming problem" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8105" PostTypeId="1" AcceptedAnswerId="8216" CreationDate="2010-10-27T21:27:08.070" Score="1" ViewCount="90" Body="&lt;p&gt;I'm quite positive that I saw before an encyclopedia (more accurately, dictionary/glossary) on Elsevier website. I remember I looked up the definition of a hyperplane there, few months ago. But I can't find it through Google anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the link?&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T01:50:36.107" LastActivityDate="2010-10-29T19:29:25.217" Title="Website for Elsevier Encyclopedia (or glossary)?" Tags="&lt;reference-request&gt;&lt;online-resources&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8106" PostTypeId="2" ParentId="8008" CreationDate="2010-10-27T21:41:32.567" Score="6" ViewCount="0" Body="&lt;p&gt;I can't believe this question is not closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As near as I can tell, Chloe solved the problem which has stymied Dr. Rush by successfully integrating a constant. Apparently Rush's expertise in ancient technology does not cover first year calculus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kv6EZ.jpg&quot; alt=&quot;Before&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Z4lZ7.jpg&quot; alt=&quot;After&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: Actually, looking over it, I might be mistaken. I interpreted the question as $-\int_0^t g \sin \gamma dt$ and Chloe's answer as $-\ddot{r} t \sin \gamma$, which is integrating a constant and making a gratuitous substitution of $\ddot{r}$ for $g$. I can't make any sense at all of the middle line in Chloe's answer.&lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2010-10-27T21:49:07.530" LastActivityDate="2010-10-27T21:49:07.530" CommentCount="1" />
  <row Id="8107" PostTypeId="1" AcceptedAnswerId="8198" CreationDate="2010-10-27T22:27:44.467" Score="2" ViewCount="57" Body="&lt;p&gt;Imagine I have a company that makes widgets, where each widget costs me A dollars to make. Each month I can allocate money toward research and development with the aim of finding a new process that will allow me to build widgets for a cost of A/B dollars. Presume that I know that for each C dollars I spend on research and development there's a D% chance of finding a breakthrough. Of course, spending money on research and development means that I have less to spend on building widgets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a monthly budget of E dollars. This budget is not directly tied to my profit margin, but it is safe to say that it my profit margins influence future budgets (i.e., if I make no widgets for three straight months b/c I do all research and development, it's likely that my budget will be reduced, whereas if I discover a breakthrough the first month my profits will skyrocket and I'll likely see that budget grow over time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case that is too abstract, here's the real world scenario I'm interested in solving (although I'd like a more general approach, as well):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A = 15 dollars&lt;/li&gt;&#xA;&lt;li&gt;B = 3&lt;/li&gt;&#xA;&lt;li&gt;C = 5 dollars&lt;/li&gt;&#xA;&lt;li&gt;D = 2.75%&lt;/li&gt;&#xA;&lt;li&gt;E = 30 dollars&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That is, today widgets cost me 15 dollars to build but if I can find a breakthrough I know I can make them at 1/3 the cost (5 dollars). For each 5 dollars I spend on research and development there is a 2.75% chance I'll find the breakthrough. However, I have only 30 dollars to spend each month. If I spend it all on research and development and have no success then I have made no widgets for sale. If I spend it all on widget construction I have no chance of finding a breakthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some statistical distribution or formula that can let me plug in these variables and see some sort of breakdown that gives me an idea of whether it's a good idea to spend any money on research and development each month and, if so, how much?&lt;/p&gt;&#xA;" OwnerUserId="2876" LastEditorUserId="2876" LastEditorDisplayName="" LastEditDate="2010-10-28T15:43:09.940" LastActivityDate="2010-10-28T16:51:58.243" Title="How to analyze risk vs. reward for spending on research and development work?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8108" PostTypeId="2" ParentId="8077" CreationDate="2010-10-27T22:36:43.830" Score="0" ViewCount="0" Body="&lt;p&gt;I'm probably talking nonsense but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_1^\infty \frac{1}{x^\alpha} = \left. \frac{x^{1 - \alpha}}{1 - \alpha} \right|_1^\infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral diverges if $1 - \alpha &gt; 0$ i.e. if $\alpha &amp;lt; 1$, but converges if $1 - \alpha &amp;lt; 0$ i.e. $\alpha &gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My difficulty now is in showing that the integral is a bound on the sum, which from what I've studied can only be done if the terms of the sum are taken as follows&#xA;&lt;img src=&quot;http://i.imgur.com/OniXk.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;instead of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4J5KC.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is more natural and intuitive.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-10-27T22:54:54.007" LastActivityDate="2010-10-27T22:54:54.007" CommentCount="1" />
  <row Id="8109" PostTypeId="2" ParentId="7391" CreationDate="2010-10-27T22:37:16.643" Score="0" ViewCount="0" Body="&lt;p&gt;To supplement lhf's observation of the symmetry of the graph, note that if $f(x)=\sin(2x)\sin(x)$, then the symmetry observed is that $f(\pi/2+x)=-f(\pi/2-x)$, which means that the integral over $[0,\pi/2]$ exactly cancels the integral over $[\pi/2,\pi]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetry can be seen algebraically by noting that $\sin(\pi/2+x)=\sin(\pi/2-x)$, and that $$\sin(2(\pi/2+x))=\sin(\pi+2x)=-\sin(2x)=\sin(-2x)=-\sin(\pi-2x)=-\sin(2(\pi/2-x)),$$ with each equality being evident from the unit circle definition of $\sin$.  Multiplying yields $f(\pi/2+x)=-f(\pi/2-x)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-27T22:37:16.643" />
  <row Id="8110" PostTypeId="1" CreationDate="2010-10-27T22:47:10.720" Score="5" ViewCount="55" Body="&lt;p&gt;&lt;code&gt;Given three containers of specified volume, how many different volumes can you measure ?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we have cans with capacities 2, 3, 4 litres.&#xA;We can measure :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;1 litre by filling the 3 litre can and pouring it in the 2 litre can. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt; 2, 3, 4 litres are trivial.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt; 5 litres by filling both the 2 and 3 litre cans. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;6 litres by filling the 4 litre can with 3 litres and filling the 3 litre can fully or filling the 2 and 4 litre cans. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt; 7 litres by filling the 3 and 4 litre cans.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;8 litres by getting 6 litres and filling the 2 litre can.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;9 litres by filling all the cans. &lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for this case is 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general way to answer this for three capacities a, b, c?&lt;/p&gt;&#xA;" OwnerUserId="2865" LastActivityDate="2010-10-28T19:58:26.160" Title="Measure different volumes based on specified capacities" Tags="&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="8111" PostTypeId="1" AcceptedAnswerId="8120" CreationDate="2010-10-27T22:58:37.800" Score="3" ViewCount="198" Body="&lt;p&gt;Is there a short way to say $f(f(f(f(x))))$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you can use recursion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x,y)=\begin{cases}&#xA;f(g(x,y-1)) &amp;amp; \text{if } y &amp;gt; 0, \&#xA;\newline x &amp;amp; \text{if } y = 0.&#xA;\end{cases}$&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-10-27T23:20:24.920" LastActivityDate="2010-10-27T23:34:25.377" Title="A short way to say f(f(f(f(x))))" Tags="&lt;functions&gt;&lt;notation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="8112" PostTypeId="2" ParentId="8094" CreationDate="2010-10-27T23:00:22.363" Score="1" ViewCount="0" Body="&lt;p&gt;It looks like you have six equations, the three you list in the post and $x_1+x_2+x_3=y_1+y_2+y_3=z_1+z_2+z_3=S$, where $S$ is given to you.  If we add the three equations in your post together we find $S(A+B+C)=Total$.  As all the terms in this equation are givens, it had better be true.  This will not apply to Part 2.  So you have five equations in nine unknowns and it should not be hard to satisfy.  If you just do a solution of linear equations you should find that there is a four parameter solution set and all the terms are rational multiples of each other.  The only thing that can go wrong is for things to go negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Part 2, it looks like $i, j, k$ are givens as well.  You still have more unknowns than equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have misunderstood something, please help me out.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T23:00:22.363" CommentCount="3" />
  <row Id="8113" PostTypeId="1" CreationDate="2010-10-27T23:05:49.600" Score="3" ViewCount="201" Body="&lt;p&gt;&lt;strong&gt;1.How many proper subset of {1,2,3,4,5,6,7} contain the numbers 1 and 7 ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets consider {1,7} as a single element then the number of possible subset is $2^6$ and hence the number of proper subset is &lt;strong&gt;62&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.A survey show that 63% of the Americans like cheese where as 73% like apples.If x% of Americans like both cheese and apples, then we have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) $x \ge 39 $   (B) $x \le 63 $ (C) $39 \le x \le 63 $ (D) None of these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if a% and b%  like only cheese and only apples then we have, $ a + x + x + b = 100 $ ,$ a + x  = 63 $ and $ b + x  = 63 $ solving we get $x = 39%$. So (D) is my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct ?&lt;/p&gt;&#xA;" OwnerUserId="2877" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T18:04:16.467" LastActivityDate="2010-11-20T18:20:02.643" Title="Two basic question on set theory" Tags="&lt;algebra-precalculus&gt;&lt;elementary-set-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="8114" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:09:21.813" Score="0" ViewCount="0" Body="&lt;p&gt;You should define it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;\text{iterate}_0(f) &amp;amp;:=&amp;amp; id \\&#xA;\text{iterate}_{n + 1}(f) &amp;amp;:=&amp;amp; \text{iterate}_{n}(f) \circ f&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then write $\text{iterate}_4(f)(x)$.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-27T23:09:21.813" CommentCount="4" />
  <row Id="8115" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:11:16.223" Score="2" ViewCount="0" Body="&lt;p&gt;Some will say $f^4(x)$.  But it becomes confused with the fourth power  or fourth derivative of $f(x)$.  I'm not sure what you mean here by &quot;piecewise&quot;.  To me &quot;piecewise&quot; would be something like a step function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=1$ if $x\gt 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=0$ if $x\le 0$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T23:11:16.223" CommentCount="13" />
  <row Id="8116" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:11:18.800" Score="0" ViewCount="0" Body="&lt;p&gt;See &quot;function powers&quot; in Wikipedia &quot;Function composition&quot;. &lt;/p&gt;&#xA;" OwnerUserId="2878" LastActivityDate="2010-10-27T23:11:18.800" CommentCount="2" />
  <row Id="8117" PostTypeId="2" ParentId="8113" CreationDate="2010-10-27T23:15:47.827" Score="4" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No, the subsets must have 1 and 7.  The other five elements are optional, but you can't have them all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have $a + b + x +  \text{those who like neither cheese nor apples} =100%$.  Don't count x twice when adding to 100%.  But you are right that $a+x=63, b+x=39$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T03:16:04.797" LastActivityDate="2010-10-28T03:16:04.797" CommentCount="6" />
  <row Id="8118" PostTypeId="2" ParentId="8110" CreationDate="2010-10-27T23:19:57.077" Score="1" ViewCount="0" Body="&lt;p&gt;If gcd(a,b,c) is not 1, you can only get multiples of this gcd.  You can't get more than a+b+c.  I think if you play around you can convince yourself you can get any multiple of the gcd, but you would need to define the allowable activities carefully to be able to prove it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-27T23:19:57.077" CommentCount="1" />
  <row Id="8119" PostTypeId="2" ParentId="8101" CreationDate="2010-10-27T23:34:06.347" Score="2" ViewCount="0" Body="&lt;p&gt;Denote the iterates by $x_0 = 1, x_{n+1} = P(x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the coefficients of $P$ are integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at some point $P(x_n) &gt; 2x_n$, then I claim that $m = P(x_n)-x_n$ does not divide any iterate. First, $x_n &amp;lt; m$, so $x_0,\ldots,x_n$ cannot be divisible by $m$. Second, we prove by induction that for $k \geq n$, $x_k \equiv x_n \pmod{m}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{k+1} = P(x_k) \equiv P(x_n) \equiv x_n \pmod{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_n &amp;lt; m$, we see that $m$ doesn't divide any of the iterates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that always $P(x_n) \leq 2x_n$. Thus $P(x) = ax+b$ with $a \leq 2$. On the one hand $P(1) &gt; 1$, and on the other hand $P(1) \leq 2$. Thus $P(1) = 2$, and therefore either $a = 1$ or $a = 2$. If $a = 2$ then $P(x) = 2x$, and we generate only powers of $2$. Thus $a = 1$ and $P(x) = x + 1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-28T00:03:54.260" LastActivityDate="2010-10-28T00:03:54.260" CommentCount="3" />
  <row Id="8120" PostTypeId="2" ParentId="8111" CreationDate="2010-10-27T23:34:25.377" Score="8" ViewCount="0" Body="&lt;p&gt;I personally prefer &#xA;$f^{\circ n} = f \circ f^{\circ n-1} = \dotsb = \kern{-2em}\underbrace{f \circ \dotsb \circ f}_{n-1\text{ function compositions}}$&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-27T23:34:25.377" CommentCount="4" />
  <row Id="8121" PostTypeId="1" AcceptedAnswerId="8122" CreationDate="2010-10-27T23:50:53.587" Score="6" ViewCount="225" Body="&lt;p&gt;I'm asking because numerical tests seem to give nonsensical answers, and I thought I would check if there was an analytic way of checking for divergence, but I couldn't think of one offhand.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-10-28T01:17:23.090" Title="does $\sum_{n=1}^{\infty} \frac{1}{\zeta(1+\frac{1}{n})}$ diverge or converge?" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="8122" PostTypeId="2" ParentId="8121" CreationDate="2010-10-28T00:11:32.677" Score="12" ViewCount="0" Body="&lt;p&gt;$(s-1)\zeta(s) = 1 + a(s-1) + b(s-1)^2 + \dots$ is analytic near $1$ (in fact entire, but we don't need that for this problem).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $s=1+1/n$ this gives $\zeta(1+1/n)=n + a + b/n + \dots  =n + O(1)$, so the sum diverges.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T00:47:47.343" LastActivityDate="2010-10-28T00:47:47.343" CommentCount="0" />
  <row Id="8123" PostTypeId="2" ParentId="8121" CreationDate="2010-10-28T00:20:48.543" Score="12" ViewCount="0" Body="&lt;p&gt;You could use an integral comparison to get a bound on $\zeta(1+1/n)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(1+1/n)\lt 1+\int_1^\infty x^{-1-\frac{1}{n}}=1+n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $0\lt a\lt 1$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{a}=\int_1^\infty x^{-1-a}\lt\zeta(1+a)\lt1+\int_1^\infty x^{-1-a}=1+\frac{1}{a}&amp;lt;\frac{2}{a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $a_1,a_2,\ldots$ is a sequence of positive numbers converging to $0$, then&#xA; $\sum_{n=1}^{\infty} \frac{1}{\zeta(1+a_n)}$ converges if and only if $\sum_{n=1}^\infty a_n$ does.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-28T01:17:23.090" LastActivityDate="2010-10-28T01:17:23.090" />
  <row Id="8124" PostTypeId="2" ParentId="8105" CreationDate="2010-10-28T00:29:22.713" Score="1" ViewCount="0" Body="&lt;p&gt;What you are looking for doesn't seem to exist (if you go to Elsevier site and type any combination of what you are looking for, like math glossary, you don't get any relevant answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to check Wolfram Mathworld: &lt;a href=&quot;http://mathworld.wolfram.com/&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1605" LastEditorUserId="1605" LastEditorDisplayName="" LastEditDate="2010-10-29T19:29:25.217" LastActivityDate="2010-10-29T19:29:25.217" CommentCount="1" />
  <row Id="8125" PostTypeId="1" AcceptedAnswerId="8127" CreationDate="2010-10-28T01:32:31.730" Score="3" ViewCount="139" Body="&lt;p&gt;A union of finite sets can always be converted to a union of finite disjoint sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if a union of countable sets can always be converted to a union of countable disjoint sets? If yes how to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update with Paul's question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I wonder about the natural extension of this question to uncountable unions. Can we always refine an arbitrary set {Aα} to {Bα} (i.e. with Bα⊂Aα for each α) such that ⋃αBα=⋃αAα? The answer below uses countability pretty blatantly and I can't think of a way to do it without. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T17:55:22.470" LastActivityDate="2010-11-20T17:55:22.470" Title="convert union of sets into union of disjoint sets" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="8126" PostTypeId="1" AcceptedAnswerId="8166" CreationDate="2010-10-28T01:33:29.357" Score="16" ViewCount="203" Body="&lt;p&gt;I'm trying to think of an example of a homomorphism of commutative rings $f:A\rightarrow B$ and ideals $I,J$ of $B$ such that $f^{-1}(I)+f^{-1}(J)$ is not a preimage of any ideal of $B$.  I can't seem to come up with one... anyone know one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To clear up some basic facts / head off some mistakes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Arturo points out, we can assume $f$ is an inclusion.  Perhaps I should have written the question in terms of inclusions in the first place, but, eh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, $f^{-1}(I)+f^{-1}(J)$ is not equal to $f^{-1}(I+J)$ in general.  A counterexample would be the inclusion of $\mathbb{C}$ in $\mathbb{C}[x]$; consider $(x)$ and $(1-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show an ideal $K\subseteq A$ is not a preimage of any ideal of $B$, it suffices to show that it's not equal to $f^{-1}(Bf(K))$.&lt;/p&gt;&#xA;" OwnerUserId="2884" LastEditorUserId="2884" LastEditorDisplayName="" LastEditDate="2010-10-28T08:02:05.840" LastActivityDate="2010-10-28T14:44:47.643" Title="Basic counterexample re: preimages of ideals" Tags="&lt;abstract-algebra&gt;&lt;algebra&gt;&lt;commutative-algebra&gt;&lt;counter-examples&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="8127" PostTypeId="2" ParentId="8125" CreationDate="2010-10-28T01:39:43.593" Score="12" ViewCount="0" Body="&lt;p&gt;Given $A_1,A_2,\ldots$, let $B_1=A_1$, and for $n\gt1$, $B_n=A_n\setminus\cup_{k=1}^{n-1}A_k$.  Then for $m\lt n$, $B_m$ is contained in $A_m$ while $B_n$ is contained in $A_n\setminus A_m$, so $B_m$ and $B_n$ are disjoint.  You can check that $\cup_{k=1}^n B_k=\cup_{k=1}^n A_k$ for all $n$, and $\cup_{k=1}^\infty B_k=\cup_{k=1}^\infty A_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a collection of sets indexed by an arbitrary set, then assuming the axiom of choice you can put a well-order on the index set.  Given $\{A_i\}_{i\in I}$, where $(I,\lt)$ is a well-ordered set, we can define $\{B_i\}_{i\in I}$ as follows.  Let $B_{i_0}=A_{i_0}$, where $i_0$ is the least element of $I$, and for $i\gt i_0$, $B_i=A_i\setminus\cup_{j\lt i}A_j$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-28T02:22:01.230" LastActivityDate="2010-10-28T02:22:01.230" CommentCount="3" />
  <row Id="8128" PostTypeId="2" ParentId="8113" CreationDate="2010-10-28T01:48:02.377" Score="2" ViewCount="0" Body="&lt;p&gt;On #1, you are over counting. You have thought of {1,7} as a single elements but your answer has included the possibility that it is not in there. Think of it as we are setting 1 and 7 to the side. Now for each remaining element, we can either include it or exclude it from a subset. If you still don't see how to obtain the answer, I would recommend the following short explanation of the &lt;a href=&quot;http://www.icoachmath.com/SiteMap/MultiplicationPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;multiplication principle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On #2, you are correct in thinking that each person must fall into 1 of four groups: people who like apples and cheese, people who just like cheese, people who like just apples, and people who like neither. If you think abut the information you are given, you should see an upper bound for x. For a lower bound on x, assume that everyone either likes apple or cheese and use the &lt;a href=&quot;http://users.csc.tntech.edu/~srini/DM/chapters/review3.6.html&quot; rel=&quot;nofollow&quot;&gt;inclusion exclusion principle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-28T01:48:02.377" CommentCount="4" />
  <row Id="8129" PostTypeId="2" ParentId="8104" CreationDate="2010-10-28T01:51:41.503" Score="2" ViewCount="0" Body="&lt;p&gt;You are almost there.  The feasible region of the dual for the new problem is exactly the same as that of the dual for the original problem.  (Write down the two duals to see that.)  Since &lt;strong&gt;z&lt;/strong&gt; is feasible for the dual of the original problem, it is also feasible for the dual of the new problem.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-28T01:51:41.503" />
  <row Id="8131" PostTypeId="1" CreationDate="2010-10-28T01:58:04.443" Score="0" ViewCount="66" Body="&lt;p&gt;What is the easiest way to find the maximum of 9cos(t) - 8cost(3t)?&lt;/p&gt;&#xA;" OwnerUserId="2885" LastEditorUserId="2885" LastEditorDisplayName="" LastEditDate="2010-10-28T02:38:26.140" LastActivityDate="2010-10-28T02:38:26.140" Title="What is the easiest way to find the maximum of 9cos(t) - 8cos(3t)?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8132" PostTypeId="2" ParentId="8131" CreationDate="2010-10-28T02:06:32.160" Score="2" ViewCount="0" Body="&lt;p&gt;Express $\cos(3t)$ in terms of $\cos(t)$ and differentiate with respect to $t$ or $\cos(t)$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-10-28T02:06:32.160" />
  <row Id="8133" PostTypeId="1" CreationDate="2010-10-28T02:18:05.457" Score="6" ViewCount="119" Body="&lt;p&gt;Of course, Euler proved that the Riemann zeta function can be defined as the analytic continuation of a product over all primes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(s) = \prod_{p \in \mathbb{P}}\frac1{1-p^{-s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known (but not something I understand) that the positions of zeros of the zeta function allows one to make inferences about the asymptotic behavior of primes.  Is this a general phenomenon? Does Euler's transform generalize to products over other subsets of the natural numbers $\mathbb{A}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha(s) = \prod_{a \in (\mathbb{A} \subset \mathbb{N})}\frac1{1-a^{-s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can one then reverse Euler's transform and derive the generating subset $\mathbb{A}$ completely from the new function's zero set?  More generally, how do properties of the derived function's zeros translate to properties of the generating subset?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, specifically for the standard Riemann zeta function, if it was shown that exactly one single zero existed off the critical line, what would its position say about the distribution of primes?&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-28T02:45:47.927" LastActivityDate="2010-10-30T00:43:37.700" Title="An inverse for Euler's zeta function product formula" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;riemann-zeta&gt;&lt;riemann-hypothesis&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="8134" PostTypeId="1" AcceptedAnswerId="8139" CreationDate="2010-10-28T02:49:56.470" Score="12" ViewCount="261" Body="&lt;p&gt;Let $z_1, z_2 \ldots z_m$ be complex numbers, $m \in \mathbb{N}$. Can anybody tell me how to prove the following inequality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you're wondering, this is asserted without proof in a paper by Von Neumann, about infinite tensor products of Hilbert spaces.&lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T17:42:01.470" LastActivityDate="2010-10-28T17:42:01.470" Title="Complex-number inequality  $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$" Tags="&lt;complex-numbers&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="8135" PostTypeId="1" AcceptedAnswerId="8136" CreationDate="2010-10-28T03:00:42.727" Score="0" ViewCount="82" Body="&lt;p&gt;Define the relation $\sim$ on $\mathbb{Z}$ by $a \sim b$ if and only if $5a \equiv 2b \pmod{3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $\sim$ is an equivalence relation on $\mathbb{Z}$.&#xA;Identify the distinct equivalence classes on $\mathbb{Z}/\sim$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove something like this? Do I need to prove something for reflexiveness, symmetry, and transitivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T03:07:52.963" LastActivityDate="2010-10-28T04:14:18.693" Title="Define the relation $\sim$on $\mathbb{Z}$ by $a \sim b$ if and only if $5a \equiv 2b \pmod{3}$" Tags="&lt;modular-arithmetic&gt;&lt;proof&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8136" PostTypeId="2" ParentId="8135" CreationDate="2010-10-28T03:10:57.157" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, you need to &quot;prove something&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $\sim$ is an equivalence relation, you need to prove that it is reflexive, symmetric, and transitive. That is, you need to prove that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For every $a\in\mathbb{Z}$, $a\sim a$ is true. Since &quot;$x\sim y$&quot; means &quot;$5x\equiv 2y\pmod{3}$&quot;, then that means that you need to prove that for every $a\in\mathbb{Z}$, $5a\equiv 2a\pmod{3}$ is true. &lt;/li&gt;&#xA;&lt;li&gt;For every $a,b\in\mathbb{Z}$, if $a\sim b$ is true, then $b\sim a$ is true. If you are unsure what this means, then &quot;unpack&quot; the meaning of &quot;$a\sim b$&quot; and of &quot;$b\sim a$&quot; the way I did above.&lt;/li&gt;&#xA;&lt;li&gt;For every $a,b,c\in\mathbb{Z}$, if $a\sim b$ and $b\sim c$ are both true, then $a\sim c$ is true.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That will prove that $\sim$ is an equivalence relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need to think about the equivalence classes. That is, given $a\in\mathbb{Z}$, what is $[a]=\{b\in\mathbb{Z}\mid a\sim b\}$, the set of all integers that are related to $a$? The different sets you get this way are the equivalence classes of $\mathbb{Z}/\sim$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T04:14:18.693" LastActivityDate="2010-10-28T04:14:18.693" />
  <row Id="8137" PostTypeId="2" ParentId="7983" CreationDate="2010-10-28T03:31:17.993" Score="3" ViewCount="0" Body="&lt;p&gt;You can go this route: You know &lt;strong&gt;Ms&lt;/strong&gt; = &lt;strong&gt;d&lt;/strong&gt;.  Taking transposes, you have &lt;strong&gt;s&lt;/strong&gt;$^T$ &lt;strong&gt;M&lt;/strong&gt;$^T$ = &lt;strong&gt;d&lt;/strong&gt;$^T$.  But (the key step) since &lt;strong&gt;M&lt;/strong&gt;$^T$ = &lt;strong&gt;M&lt;/strong&gt;, this means that &lt;strong&gt;s&lt;/strong&gt; is feasible for the dual problem.  Since &lt;strong&gt;s&lt;/strong&gt; is feasible for the primal and for the dual, and since the primal and dual have the same objective function, weak duality shows you that &lt;strong&gt;s&lt;/strong&gt; must be optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;strong&gt;M&lt;/strong&gt;$^T \neq $ &lt;strong&gt;M&lt;/strong&gt; then &lt;strong&gt;s&lt;/strong&gt; might not be optimal.  For example, suppose we have the following linear program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min x + y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2}x + y \geq 1,$$&#xA;$$\frac{1}{2}x + 2y \geq 1,$$&#xA;$$x,y \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective and right-hand side coefficient vectors are the same, as in the problem statement.  Solving the inequalities at equality gives you $x = 2$, $y=0$.  However, this is not optimal.  The solution $x = 0$, $y = 1$ yields a smaller objective function value (and is actually the optimal solution).&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-28T12:43:41.203" LastActivityDate="2010-10-28T12:43:41.203" />
  <row Id="8138" PostTypeId="2" ParentId="8134" CreationDate="2010-10-28T03:37:03.900" Score="-1" ViewCount="0" Body="&lt;p&gt;EDIT: This answer is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all boils down to the inequality $|xy-1| \leq |x-1|+|y-1|$, which I expect to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this inequality, prove by induction that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|z_1\cdots z_m - 1| \leq |z_1 - 1| + \cdots + |z_m - 1|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use $e^x \geq 1 + x$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-10-28T04:20:39.117" LastActivityDate="2010-10-28T04:20:39.117" CommentCount="2" />
  <row Id="8139" PostTypeId="2" ParentId="8134" CreationDate="2010-10-28T04:01:38.487" Score="17" ViewCount="0" Body="&lt;p&gt;The inequality in question bounds how far you can get from $1$ by multiplying several complex numbers that may individually not be far from $1$. So it makes sense to try to derive a bound for the product of just &lt;em&gt;two&lt;/em&gt; complex numbers, and then proceed by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Suppose $\lvert z_1 - 1\rvert = \alpha_1$ and $\lvert z_2 - 1\rvert = \alpha_2$. Then $\lvert z_1 z_2 - 1\rvert \le (1 + \alpha_1)(1 + \alpha_2) - 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; We know $\alpha_1\alpha_2 = \lvert z_1z_2 - z_1 - z_2 + 1\rvert$. By triangle inequality on the three points $z_1 z_2$, $z_1 + z_2 - 1$, and $1$, we have&#xA;$$\begin{align}&#xA;\lvert z_1 z_2 - 1\rvert &amp;amp;\le \lvert z_1 z_2 - z_1 - z_2 + 1\rvert + \lvert z_1 + z_2 - 2\rvert \\&#xA;&amp;amp;\le \alpha_1\alpha_2 + \alpha_1 + \alpha_2 \\&#xA;&amp;amp;= (1 + \alpha_1)(1 + \alpha_2) - 1.&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for several numbers,&#xA;$$\begin{align}&#xA;\lvert z_1 z_2 \cdots z_m - 1\rvert &amp;amp;\le (1 + \alpha_1)(1 + \alpha_{2,\ldots,m}) - 1 \\&#xA;&amp;amp;\le (1 + \alpha_1)(1 + \alpha_2)(1 + \alpha_{3,\ldots,m}) - 1 \\&#xA;&amp;amp;\vdots \\&#xA;&amp;amp;\le (1 + \alpha_1)(1 + \alpha_2)\cdots(1 + \alpha_m) - 1,&#xA;\end{align}$$ where $\alpha_{2,\ldots,m}$, for example, is my hopefully transparent abuse of notation to denote $\lvert z_2\cdots z_m - 1\rvert.$ Finally, since $1+x \le e^x$ for real $x$, the desired inequality follows.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-28T04:01:38.487" CommentCount="7" />
  <row Id="8140" PostTypeId="1" AcceptedAnswerId="15069" CreationDate="2010-10-28T04:12:39.473" Score="1" ViewCount="408" Body="&lt;p&gt;I have a graph $G$ with a list of edges $E$, all the edges can form cycles, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1Ux3b0FtuIMpVGX4e-AbblH05h2gIZII9u8pACRzymPE&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the coordinates for the vertex $V$ &lt;strong&gt;are known&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wat is the algorithm to find all the faces $F$? For the example above, there are 4 such faces, as indicated by the looping arrow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Why is it that the algorithm to find the face in the case without coordinate is different from the one with coordinates? Simply because by specifying the coordinates there is a unique set of faces, and without coordinates there could be sets of faces that are topologically equal. See the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1wAxp6gJgN7wAkr5xVpSfBv7_8Gfd2FHIFX6CY5bdgdc&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update 2: Clear up the terminology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPdate 3: &lt;a href=&quot;http://math.stackexchange.com/questions/8140/find-all-cycles-faces-in-a-graph/15069#15069&quot;&gt;It is suggested&lt;/a&gt; that I construct a cyclically linked list at each vertex, sorted according to the angles made between the incoming edge and the outgoing edge. Then one would just have to continue along the edge, but I afraid this algorithm doesn't work with the so called butterfly graph. Consider this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1QF2JouvziCYaPCkERy5kld1No8NORf63O7XwKN22nXY&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that we start at $e_1$, and the arrow denotes the direction of the movement. If we take the left turn as our orientation then we would be visiting $v_1$ twice before completing a cycle-- and the cycle is not really a cycle. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-22T02:43:27.190" LastActivityDate="2010-12-24T18:49:30.593" Title="Find All Cycles (Faces) In a Graph" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="8141" PostTypeId="1" CreationDate="2010-10-28T04:18:00.483" Score="0" ViewCount="64" Body="&lt;p&gt;This animation from wikipedia shows basically what I want to accomplish, however - I'm hoping to have it flipped around, where it starts progressing more towards the destination and &quot;up&quot; (in this image), and then arcs more directly to the end point. However, I only have access to a starting point and ending point, what I am hoping to do is be able to determine the other points by specifying a &quot;height&quot; (or width, whatever you want to call it), to determine how high the arc actually goes.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Bezier_3_big.png&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Bezier_3_big.png&lt;/a&gt; (unable to post image due to no rep)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help or direction would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2886" LastActivityDate="2010-10-28T04:18:00.483" Title="How to calculate a bezier curve with only start and end points?" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;bezier-curve&gt;" CommentCount="4" />
  <row Id="8142" PostTypeId="1" AcceptedAnswerId="8146" CreationDate="2010-10-28T04:18:41.040" Score="7" ViewCount="162" Body="&lt;p&gt;Let $C$ be a circle of radius $r$ with $n$ points. Prove that there is a point on the circle such that the product of the distances from this point to the other $n$ points is greater than $r^n$. So we seem to be looking at chords that are of length $\leq 2r$. To show the existence of such a point, would you use a constructive argument? Or more of an indirect argument? Likewise, to show that the inequality holds, would we invoke certain &quot;famous&quot; inequalities such as AM-GM or Cauchy-Schwartz? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-28T17:25:16.443" LastActivityDate="2010-10-28T20:16:14.783" Title="Lower bound on product of distances from points on a circle" Tags="&lt;geometry&gt;&lt;inequality&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8143" PostTypeId="1" AcceptedAnswerId="8148" CreationDate="2010-10-28T04:50:33.747" Score="2" ViewCount="64" Body="&lt;p&gt;How many distinct integral value of $n$ satisfies the equation $2^{2n} - 3 \cdot (2^{n+2}) + 32 = 0 $ ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-10-28T04:57:37.300" Title="How many distinct integral value of $n$ satisfies the equation $2^{2n} - 3 \cdot (2^{n+2}) + 32 = 0 $ ?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="8144" PostTypeId="2" ParentId="8143" CreationDate="2010-10-28T04:53:58.980" Score="1" ViewCount="0" Body="&lt;p&gt;You should be able to prove a fairly small limit on n.  Think about the fact that 3&amp;lt;2^2.  Then you can just try them all up to there.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-28T04:53:58.980" />
  <row Id="8145" PostTypeId="2" ParentId="8142" CreationDate="2010-10-28T04:55:10.010" Score="3" ViewCount="0" Body="&lt;p&gt;We can take $r=1$ by scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between $e^{i\alpha}$ and $e^{i\beta}$ is $2|\sin\frac{\alpha-\beta}{2}|$. Denoting the given points by $e^{i\alpha_i}$ and the required point by $e^{i\beta}$, the total distance is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \prod_i 2|\sin\frac{\alpha_i-\beta}{2}| $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take a random angle $\beta$. The average multiplicative value of sine on $[0,\pi]$ is $1/2$, see &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Identities_without_variables&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. Therefore the multiplicative expected value of the product is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be more formal, replace &quot;multiplicative&quot; by taking of the logarithm.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-28T04:55:10.010" />
  <row Id="8146" PostTypeId="2" ParentId="8142" CreationDate="2010-10-28T04:55:58.220" Score="11" ViewCount="0" Body="&lt;p&gt;Let $r&gt;0$, $D=\{z\in\mathbb C:|z|&amp;lt;r\}$, and $S=\{z\in\mathbb C:|z|=r\}$. Pick $n\in\mathbb N$ and distinct points $z_1,\dots,z_n\in S$, and consider the polynomial $f(z)=\prod_{i=1}^n(z_i-z)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is holomorphic on $\mathbb C$ and non-constant, its absolute value attains its maximum on the closed disc $D\cup S$ at some point $p$ of $S$ and not inside $D$. Since $|f(0)|=r^n$, it follows that $|f(p)|&amp;gt; r^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is precisely what you wanted.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-28T05:19:17.437" LastActivityDate="2010-10-28T05:19:17.437" CommentCount="2" />
  <row Id="8148" PostTypeId="2" ParentId="8143" CreationDate="2010-10-28T04:57:37.300" Score="7" ViewCount="0" Body="&lt;p&gt;HINT: make the substitution $x=2^{n}$ This will reduce your equation to a quadratic.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-10-28T04:57:37.300" CommentCount="2" />
  <row Id="8149" PostTypeId="1" AcceptedAnswerId="8156" CreationDate="2010-10-28T05:23:59.040" Score="1" ViewCount="97" Body="&lt;p&gt;We know that the points of discontinuity of a monotone function on an interval $[a,b]$ are countable.Using this can we prov that: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any bijection $f: \mathbb{R} \to [0,\infty)$ has infinitely many points of discontinuity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If yes, how or otherwise how to prove the above result?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-28T05:55:33.890" Title="Points of discontinuity of a bijective function $f:\mathbb{R} \to [0,\infty)$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8150" PostTypeId="1" AcceptedAnswerId="8153" CreationDate="2010-10-28T05:24:34.213" Score="2" ViewCount="57" Body="&lt;p&gt;Is the localization of a reduced ring (no nilpotents) still reduced?&lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-10-28T05:46:06.727" Title="Does localization preserve reducedness?" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8151" PostTypeId="2" ParentId="8150" CreationDate="2010-10-28T05:30:50.917" Score="-1" ViewCount="0" Body="&lt;p&gt;Yes. Why don't you see wiki for more: &lt;a href=&quot;http://en.wikipedia.org/wiki/Reduced_ring&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Reduced_ring&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-28T05:30:50.917" />
  <row Id="8152" PostTypeId="2" ParentId="8150" CreationDate="2010-10-28T05:32:15.087" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT: This argument is incorrect but I feel others could learn from Mariano Suárez-Alvarez's comments so I've made the post CW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly you are asking does reduce ring imply reduced localization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argue by contrapositive, assume that localization is not reduced, i.e. contains nilpotents. Since elements of the localization of our ring are of the form $\frac{r}{s}$ where s is a subset of our ring that does not contain 0. Then choose a nilpotent element in the localization $(\frac{r}{s})^{n}=0$ Since 0 is not in S it must be the case that $r^{n}=0$, i.e. r is nilpotent. Hence the original ring is nilpotent.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2010-10-28T05:46:06.727" LastActivityDate="2010-10-28T05:46:06.727" CommunityOwnedDate="2010-10-28T05:46:06.727" CommentCount="4" />
  <row Id="8153" PostTypeId="2" ParentId="8150" CreationDate="2010-10-28T05:32:23.867" Score="5" ViewCount="0" Body="&lt;p&gt;Let $A$ be a ring, $S\subset A$ a multiplicatively closed subset, and suppose that $0\neq a/b\in A_S$ is nilpotent. Then there exists $n$ such that $(a/b)^n=0$, &lt;em&gt;i.e.&lt;/em&gt;, such that there exists $t\in S$ with $ta^n=0$. But then $ta$ is nilpotent in $A$. If it is zero, then $a/b=0$ in $A_S$, which it isn't.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-28T05:32:23.867" />
  <row Id="8154" PostTypeId="2" ParentId="8113" CreationDate="2010-10-28T05:45:30.033" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Say {1,7} is a single element, then total number of subsets possible = $2^6$ Now, how many subsets among these &lt;strong&gt;don't have {1,7}&lt;/strong&gt; ? It will be $2^5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,number of subsets &lt;strong&gt;having {1,7}&lt;/strong&gt; is  $2^6$ -  $2^5$ = 32.But since you have asked for proper subset so the answer would be 31.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-28T05:52:01.060" LastActivityDate="2010-10-28T05:52:01.060" CommentCount="2" />
  <row Id="8155" PostTypeId="1" AcceptedAnswerId="8168" CreationDate="2010-10-28T05:50:17.987" Score="0" ViewCount="125" Body="&lt;p&gt;I want to write the equation of a line in $\mathbb C^n$ passing through a point $(z_1,z_2,...,z_n)$. Actually I have a set of points and I suspect they all lie in the same line which passes through this point and I want a convenient way to check it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="2890" LastActivityDate="2010-10-29T18:40:28.157" Title="How to write the equation of a line in $\mathbb C^n$" Tags="&lt;algebra&gt;&lt;linear&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8156" PostTypeId="2" ParentId="8149" CreationDate="2010-10-28T05:55:33.890" Score="4" ViewCount="0" Body="&lt;p&gt;Suppose to the contrary that $f$ has a finite number $n$ of discontinuities, at $x_1, x_2, \ldots x_n$. Then $\mathbb{R} - ( x_1, \ldots x_n )$ is a union of open intervals $I_1, \ldots I_{n + 1}$. $f$ restricted to each $I_m$ is continuous and injective, and therefore monotone, so each image $f|_{I_m}(I_m)$ is an open (in $\mathbb{R}$) interval $J_m$. The $J_m$ are non-empty, pairwise disjoint and contained in $(0, \infty)$; suppose that $J_{m_1} &amp;lt; J_{m_2} &amp;lt; \ldots &amp;lt; J_{m_{n + 1}}$. Then between each $J_{m_p}$ and $J_{m_{p + 1}}$ there is at least one point, which must be the image of some $x_q$. But this exhausts all the $x$'s, so that $0$ is not in the image of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, can anybody tell me how to get the forum to display braces in math mode?&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-10-28T05:55:33.890" CommentCount="6" />
  <row Id="8157" PostTypeId="1" AcceptedAnswerId="8161" CreationDate="2010-10-28T05:59:19.727" Score="1" ViewCount="81" Body="&lt;p&gt;If a set contain $(2n+1)$ elements and if the number of subsets which contain at most n elements is 4096, then what is the value of $n$?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T02:02:57.343" LastActivityDate="2010-11-21T02:02:57.343" Title="If a set contain $(2n+1)$ elements and if the number of subsets which contain at most n elements is 4096, then what is the value of $n$?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8159" PostTypeId="2" ParentId="8157" CreationDate="2010-10-28T06:11:57.187" Score="0" ViewCount="0" Body="&lt;p&gt;Use the identity $(1+1)^n= n_{C_0}+n_{C_1}+\dots+n_{C_n}$ and the fact that $n_{C_r}=n_{C_{n-r}}$.&lt;/p&gt;&#xA;" OwnerUserId="2890" LastActivityDate="2010-10-28T06:11:57.187" />
  <row Id="8160" PostTypeId="1" AcceptedAnswerId="8163" CreationDate="2010-10-28T06:14:21.163" Score="4" ViewCount="66" Body="&lt;p&gt;In a compact hyperbolic Reimann surface without boundary is length minimizing geodesic between two points $p$ and $q$ unique.&lt;/p&gt;&#xA;" OwnerUserId="2893" LastActivityDate="2010-10-28T06:49:46.053" Title="Uniqueness of length minimizing geodesic" Tags="&lt;geometry&gt;&lt;riemann-surface&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8161" PostTypeId="2" ParentId="8157" CreationDate="2010-10-28T06:23:44.687" Score="4" ViewCount="0" Body="&lt;p&gt;There are $2^{2n + 1}$ subsets in total. Now note that the function which takes a subset to its complement is a bijection, and that a subset has $n$ or fewer elements if, and only if, its complement has more than $n$ elements. Thus the number of subsets with $n$ or fewer elements is equal to the number with more than $n$; thus the number of each is $2^{2n + 1}/2 = 2^{2n}$. Now solve for $n$.&lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-10-28T06:23:44.687" />
  <row Id="8162" PostTypeId="2" ParentId="8155" CreationDate="2010-10-28T06:40:23.527" Score="0" ViewCount="0" Body="&lt;p&gt;How much do you know about vectors?  You can take two points and find the difference between them, as a vector; this is the analogue of the slope of a line in $\mathbb{R}^2$.  If the vector is $\mathbf{v}$ and one point is $a$, then points on the line are of the form $a+\lambda\mathbf{v}$, where $\lambda$ is real.  (Unless you're talking about a &quot;complex line,&quot; which is really a plane.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you're just uncomfortable with $\mathbb{C}^n$, you can just rewrite all your points as points in $\mathbb{R}^{2n}$, like $(Re(z_1),Im(z_1),\dotsc,Re(z_n),Im(z_n))$.  This is an isomorphism of the real vector space structure of $\mathbb{C}^n$, so in particular it preserves lines and addition.  The only thing you lose is complex multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: If you're looking for a complex line, just let $\lambda$ be complex in the above vector expression.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-28T09:03:23.223" LastActivityDate="2010-10-28T09:03:23.223" CommentCount="2" />
  <row Id="8163" PostTypeId="2" ParentId="8160" CreationDate="2010-10-28T06:47:46.877" Score="6" ViewCount="0" Body="&lt;p&gt;Think about a Riemann surface of genus 2 say. This is a 2-holed doughnut.&#xA;It has a curve winding about one of its holes. (Or more mathematically speaking&#xA;it has an embedded circle representing a nonzero homology class.)&#xA;We can deform this curve so that it is length-minimizing in its homotopy&#xA;class. Then the curve becomes a geodesic $C$. Then let $p$ and $q$ be &quot;opposite&quot;&#xA;points of $C$, that is the distance from $p$ to $q$ along $C$ is a half of&#xA;the length of $C$. Then $C$ splits into two equal-length geodesics between&#xA;$p$ and $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose there may be even shorter geodesics between $p$ and $q$, but&#xA;I think that choosing the homology class to minimize the length of $C$&#xA;shoudl sort that out.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-28T06:47:46.877" />
  <row Id="8164" PostTypeId="2" ParentId="8160" CreationDate="2010-10-28T06:49:46.053" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not a hyperbolic geometer, but I'm going to say &lt;strong&gt;no&lt;/strong&gt;.  Here is my reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be your compact hyperbolic Riemann surface, so it is the quotient of its universal cover, the hyperbolic plane, by a cocompact Fuchsian group $\Gamma$.  Let $\pi: \mathbb{H} \rightarrow \Gamma \backslash \mathbb{H} = C$ be the covering map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have points $P$ and $Q$ on $C$.  Fix one lift $\tilde{P}$ of $P$ and consider all possible lifts &lt;code&gt;$\tilde{Q}_{\gamma}$&lt;/code&gt; of $Q$: this will be a full orbit of $\Gamma$.  For &#xA;each $\gamma \in \Gamma$, there is a unique geodesic in $\mathbb{H}$ from $P$ to $\tilde{Q}_{\gamma}$ and its image under $\pi$ is a geodesic from $P$ to $Q$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So certainly there are lots of geodesics in question.  The ones of minimal length are going to correspond to the choices of $\gamma$ such that the hyperbolic distance from $P$ to $\tilde{Q}_{\gamma}$ is minimized.  But it seems pretty clear that we can choose $P$ so as to be equidistant between two such points: choose any point $Q_1$, and then choose $Q_2$ to be distinct from $Q_1$, to lie in the same $\Gamma$-orbit as $Q_1$ and to be closest (not necessarily uniquely closest!) to $Q_1$ among all such points.  Let $P$ be the midpoint of the geodesic arc from $Q_1$ to $Q_2$.  Then so long as $P$ is at least as far away from any other point in the $\Gamma$-orbit of $Q$ as it is from $Q_1$ and $Q_2$, on the quotient we have two length-minimizing geodesics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure you can arrange for this to happen with a suitable choice of $\Gamma$.  For instance, if $\Gamma$ is such that the fundamental domain is a regular hyperbolic octagon with edges identified in the usual way to get a genus $2$ quotient, then I think it works to take $Q_1$ and $Q_2$ to be adjacent vertices of the octagon.  See for instance the picture towards the bottom of this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.cornell.edu/~mec/Winter2009/Victor/part4.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~mec/Winter2009/Victor/part4.htm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-10-28T06:49:46.053" />
  <row Id="8165" PostTypeId="2" ParentId="7864" CreationDate="2010-10-28T07:24:28.983" Score="1" ViewCount="0" Body="&lt;p&gt;I have seen this many times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 + a^3 = a^5$$&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-28T07:24:28.983" CommunityOwnedDate="2010-10-28T07:24:28.983" CommentCount="1" />
  <row Id="8166" PostTypeId="2" ParentId="8126" CreationDate="2010-10-28T07:30:53.207" Score="4" ViewCount="0" Body="&lt;p&gt;Unless I'm missing something, this is the preimage of I+J (I missed something... this is not correct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $I,J$ are ideals in $B$ then $f^{-1}(I),f^{-1}(J)\subseteq f^{-1}(I+J)$ so $f^{-1}(I)+f^{-1}(J)\subseteq f^{-1}(I+J)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $x\in f^{-1}(I+J)$ then $f(x)=i+j,\; i\in I,j\in J$. Let $a\in f^{-1}(I)$ such that $f(a)=i$, and $b\in f^{-1}(J),\;f(b)=j$ then $f(a+b)=i+j=f(x)$ so $a+b=x+k$ where $k \in ker(f)$. $ker(f)\subseteq f^{-1}(I)+f^{-1}(J)$ and $a,b\in  f^{-1}(I)+f^{-1}(J)$ so also $x \in f^{-1}(I)+f^{-1}(J)$ and we get that $f^{-1}(I+J)\subseteq f^{-1}(I)+f^{-1}(J)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ok, another try&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $f:\mathbb{Z}[x]\rightarrow \mathbb{Q}[x]$ be the inclusion function. If $I=(x-3)\mathbb{Q}[x],\; J=x \mathbb{Q}[x]$ then their preimage is $(x-3)\mathbb{Z}[x], \; x\mathbb{Z}[x]$ (I used here Gauss lemma). The sum of the preimages is not all of $\mathbb{Z}[x]$ and it contains 3. If it is in itself a preimage of K then it  $3\in K$ in $\mathbb{Q}[x]$ which is invertible and so this is all of $\mathbb{Q}[x]$ and we get a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this one is ok....&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-10-28T08:41:19.400" LastActivityDate="2010-10-28T08:41:19.400" CommentCount="3" />
  <row Id="8168" PostTypeId="2" ParentId="8155" CreationDate="2010-10-28T07:51:31.100" Score="1" ViewCount="0" Body="&lt;p&gt;Since $\mathbb{C}$ is 2-dimensional, as a real vector space,&#xA;a one-dimensional subspace of $\mathbb{C^n}$ is actually a complex plane, and is therefore not an ordinary line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, such subspaces/&quot;lines&quot; can be expressed as in the real case, as&#xA;a solution to $n-1$ equations of the form&#xA;$\lambda_1z_1+\lambda_2z_2+\dots+\lambda_nz_n = c$&#xA;where c is 0 if you want a subspace, any complex number if you want any set that &quot;looks&quot; like $\mathbb{C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Fixed error.&lt;/p&gt;&#xA;" OwnerUserId="934" LastEditorUserId="934" LastEditorDisplayName="" LastEditDate="2010-10-29T13:52:32.743" LastActivityDate="2010-10-29T13:52:32.743" CommentCount="3" />
  <row Id="8170" PostTypeId="1" AcceptedAnswerId="8197" CreationDate="2010-10-28T09:46:39.330" Score="6" ViewCount="247" Body="&lt;p&gt;I'm trying to understand basic tensor analysis. I understand the basic concept  that the valency of the tensor determines how it is transformed, but I am having trouble visualizing the difference between different valencies when it comes to higher order tensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this picture in my mind for the lower order tensors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^i = \left(\begin{array}{x} x^1 \\ x^2 \\ x^3\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_i = \left(\begin{array}{ccc} x_1 &amp;amp; x_2 &amp;amp; x_3\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^i_j = \left(\begin{array}{ccc} x^1_1 &amp;amp; x^1_2 &amp;amp; x^1_3 \\&#xA; x^1_2 &amp;amp; x^2_2 &amp;amp; x^3_2 \\ x^1_3 &amp;amp; x^2_3 &amp;amp; x^3_3\end{array} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $X^{ij}$ and $X_{ij}$ they are represented in the same 2d array, but the action on a vector isn't defined in the same way as with matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am having trouble with is intuitively understanding the difference between $X^{ijk}$, $X_{k}^{ij}$, $X_{jk}^{i}$, and $X_{ijk}$ (other permutations of the valence $(2,1)$ and $(1,2)$ omitted for brevity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED&lt;/strong&gt;&#xA;After reading the responses and their comments I came up with this new picture in my head for higher order tensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I am somewhat comfortable with tensor products in quantum mechanics, I can draw a parallel with the specific tensor space I'm used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a rank-5 tensor with a valence of (2,3) then we can consider it in the braket notation as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \langle \psi_i \mid \otimes \ \langle \psi_j \mid \otimes \ \langle \psi_k \mid \otimes \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle = X_{ijk}^{lm} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we operate with this tensor on rank-3 contravariant tensor, we are-left with a constant (from the inner product) and a rank-2 contravariant tensor, unmixed tensor product &#xA;$\begin{eqnarray}(\langle \psi_i \mid \otimes \ \langle \psi_j \mid \otimes \ \langle \psi_k \mid \otimes \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle)(\mid \Psi_i \rangle \ \otimes \mid \Psi_j \rangle \ \otimes \mid \Psi_k \rangle) &amp;amp;=&amp;amp;  c \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle \\ &amp;amp;=&amp;amp; X_{ijk}^{lm}\Psi^{ijk} = cX'^{lm}\end{eqnarray}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to further operate with a rank-2 covariant tensor (from the right, per convention that a covector and vector facing each other is an implied direct product) we would simply get a number out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I am confused about though, is that in one of the answer to this question there was a point made that we are taking tensor products of a Vector space with itself (and possibly it's dual), however in the quantum mechanics picture (although I didn't rely on it in this example) we often take tensor products between different, often disjoint, subspaces of the enormous Hilbert space that describes the quantum mechanical universe. Does the tensor picture change in this case? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments on my example would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-10-28T20:19:31.977" LastActivityDate="2010-10-28T20:19:31.977" Title="Intuitive way to understand covariance and contravariance in Tensor Algebra" Tags="&lt;intuition&gt;&lt;tensors&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8171" PostTypeId="2" ParentId="8042" CreationDate="2010-10-28T09:55:36.467" Score="8" ViewCount="0" Body="&lt;p&gt;Write the inequality as&#xA;$$\frac{\pi}{4} &amp;lt; \frac{1}{2n} + \frac{1}{n} \sum_{k=1}^{n-1} \sqrt{1-\left(\frac{k}{n}\right)^2} + \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$&#xA;The left-hand side $\pi/4$ is the area of the part of the unit circle that&#xA;lies in the first quadrant (below the curve $y=f(x)=\sqrt{1-x^2}$).&#xA;We want to interpret the right-hand side as the area of a region $D$ which&#xA;covers that quarter circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $f$ is concave, so that its graph lies below any tangent line.&#xA;Thus the trapezoid bounded by the lines $x=a-\epsilon$ and $x=a+\epsilon$&#xA;and by the $x$ axis and the tangent line through $(a,f(a))$ will cover the&#xA;corresponding part of the circle:&#xA;$$\int_{a-\epsilon}^{a+\epsilon} f(x) dx &amp;lt; 2\epsilon f(a).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, taking $D$ to be the union of the following pieces does the trick:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A rectangle of height 1 between $x=0$ and $x=1/2n$.&lt;/li&gt;&#xA;&lt;li&gt;Trapezoids as above, of width $\frac{1}{n}$ and centered at $x=k/n$ for $k=1,\ldots,n-1$.&lt;/li&gt;&#xA;&lt;li&gt;A trapezoid as above, of width $\frac{1}{2n}$ and centered at $x=1-1/4n$. This last one has area&#xA;$$\frac{1}{2n} f(1-1/4n) = \frac{1}{2n} \sqrt{\frac{1}{2n} - \frac{1}{16n^2}} &amp;lt;   \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-28T09:55:36.467" />
  <row Id="8172" PostTypeId="2" ParentId="8170" CreationDate="2010-10-28T10:03:00.327" Score="3" ViewCount="0" Body="&lt;p&gt;I prefer to think of them as maps instead of matrices.  When you move to tensor bundles over manifolds, you won't have global coordinates, so it might be preferable to think this way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $x_i$ is a map which sends vectors to reals.  Since it's a tensor, you're only concerned with how it acts on basis elements.  It's nice to think of them in terms of dual bases: then $x_i(x^j)=\delta_{ij}$, which is defined as $1$ when $i=j$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, $x^i$ is a map which sends covectors to reals, and is defined by $x^i(x_j)=\delta_{ij}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have more indices, then you're dealing with a tensor product $V^*\otimes\dotsb\otimes V^*\otimes V\otimes\dotsb\otimes V$, say with $n$ copies of the vector space and $m$ copies of the dual.  An element of this vector space takes in $m$ vectors and gives you back $n$, again in a tensorial way.  So, for example, $X_{ijk}$ is a trilinear map; $X^{ijk}$ is a trivector (an ordered triple of vectors up to linearity); $X_{ij}^k$ is a bilinear map taking two vectors to one vector; and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth thinking about these in terms of the tensors you've seen already.  The dot product, for example, is your basic (0,2)-tensor.  The cross product is a (1,2)-tensor.  If you study Riemannian manifolds, it turns out you can use the metric to &quot;raise and lower indices&quot;; so the Riemannian curvature tensor, for example, is alternately defined as a (1,3)-tensor and a (0,4)-tensor, depending on the author's needs.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2010-10-28T10:39:57.533" LastActivityDate="2010-10-28T10:39:57.533" CommentCount="10" />
  <row Id="8173" PostTypeId="1" AcceptedAnswerId="8184" CreationDate="2010-10-28T10:27:06.037" Score="3" ViewCount="139" Body="&lt;p&gt;What is an example of a lambda-system that is not a sigma algebra?&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T19:10:48.930" LastActivityDate="2010-10-28T19:10:48.930" Title="What is an example of a lambda-system that is not a sigma algebra?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8174" PostTypeId="2" ParentId="8173" CreationDate="2010-10-28T10:38:26.213" Score="4" ViewCount="0" Body="&lt;p&gt;Take the $\lambda$-system generated by the open intervals $\{ (a, b) : a &amp;lt; \frac{1}{2} &amp;lt; b \}$ in $[0, 1]$.  Every element of this $\lambda$-system either does or does not contain an interval about $\frac{1}{2}$; in particular, $\{ \frac{1}{2} \}$ is not in this $\lambda$-system, so this $\lambda$-system is not closed under countable intersection.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-28T11:16:32.027" LastActivityDate="2010-10-28T11:16:32.027" CommentCount="0" />
  <row Id="8175" PostTypeId="2" ParentId="7864" CreationDate="2010-10-28T11:13:26.270" Score="3" ViewCount="0" Body="&lt;p&gt;Both my students and some of my colleagues (!) believe that the graph of a function cannot cross a horizontal asymptote.  Obviously this implies that they misunderstand the definition of an asymptote.  More worryingly (in my eyes), it also seems to imply that they don't understand &lt;em&gt;why&lt;/em&gt; we even care about asymptotes.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-10-28T11:13:26.270" CommunityOwnedDate="2010-10-28T11:13:26.270" CommentCount="1" />
  <row Id="8176" PostTypeId="1" AcceptedAnswerId="8201" CreationDate="2010-10-28T12:11:21.717" Score="5" ViewCount="216" Body="&lt;p&gt;This is probably an obvious parallel that most people are aware of, but I only just noticed it the other day and it made me quite excited. The &lt;strong&gt;determinant&lt;/strong&gt;  in algebra has a lot in common with the &lt;strong&gt;integral&lt;/strong&gt; in analysis. For example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They are both applied to functions, the integral to integrable functions, the determinant to linear transformations $T:V \rightarrow V$.&lt;/li&gt;&#xA;&lt;li&gt;They are both &quot;sums of products.&quot;&lt;/li&gt;&#xA;&lt;li&gt;They both &lt;em&gt;can&lt;/em&gt; be used to give a scalar result. (Not always, of course, but this is how they are first developed.)&lt;/li&gt;&#xA;&lt;li&gt;They are both important major structures in algebra and analysis.&lt;/li&gt;&#xA;&lt;li&gt;They are both defined in ways that feel 'backwards'-- the formal definition isn't always useful for calculating them-- then they come to represent &lt;strong&gt;multiple&lt;/strong&gt; important concepts acting as a fulcrum in their fields.  (ie. AREA is connected to ANTI-DERIVATIVES... or that SOLUTIONS TO AX=B are connected to LINEAR TRANSFORMATIONS.)&lt;/li&gt;&#xA;&lt;li&gt;They can both be used to give area and volume. (under a curve, or of a parallelepiped)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Question: &lt;strong&gt;What mathematical structure encompasses both?&lt;/strong&gt; (If the answer is category theory, please go slowly with me, I don't understand that stuff yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else could we add to this list? &lt;strong&gt;Are there any problems or proofs that bring these parallels in to the light?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any other mathematical structures that follow the pattern established by these two structures&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Were they developed independently (what I suspect) or is the determinant in some way patterned after the integral or vice versa?&lt;/strong&gt; (I know my math history and have not come across anything about this.)&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-10-28T15:43:44.263" LastActivityDate="2010-10-28T17:24:52.717" Title="The determinant is the integral of algebra. The integral is the determinant of analysis." Tags="&lt;linear-algebra&gt;&lt;analysis&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="8177" PostTypeId="2" ParentId="7981" CreationDate="2010-10-28T12:58:38.593" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a very simply description of the Riemann Hypothesis that requires nothing more than a 3rd grade education to understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2323497&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2323497&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a beautiful proof linking the Farey sequence of fractions to the Riemann hypothesis by Jerome Franel.  It's only three pages long and should be able to be understood by any undergraduate mathematics major.&lt;/p&gt;&#xA;" OwnerUserId="2899" LastActivityDate="2010-10-28T12:58:38.593" />
  <row Id="8178" PostTypeId="2" ParentId="7864" CreationDate="2010-10-28T13:03:18.580" Score="2" ViewCount="0" Body="&lt;p&gt;To generalize a few of the answers, for pretty much any function, someone somewhere will make the mistake of treating it as if it is linear in all of its variables. Thus we get:&#xA;$e^a + e^b = e^{a+b}$, $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$, $a/(b+c) = a/b + a/c$, ...&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-10-28T13:03:18.580" CommunityOwnedDate="2010-10-28T13:03:18.580" />
  <row Id="8179" PostTypeId="2" ParentId="8126" CreationDate="2010-10-28T13:06:31.280" Score="2" ViewCount="0" Body="&lt;p&gt;Let $k$ be field of caracteristic $\neq 2$, $A=k[x,y], B=k[x,y,x^{-1},y^{-1}]$ and take $f:A\hookrightarrow B$ be inclusion. Example for what you want is $I=(x+y)B, J=(x-y)B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f^{-1}I=(x+y)A, f^{-1}J=(x-y)A$ and $f^{-1}I+f^{-1}J=(x,y)A$ . Since $(x,y)B=B$,  sum &#xA;$f^{-1}I+f^{-1}J$ can not be $f^{-1}$ of ideal in $B$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-28T14:44:47.643" LastActivityDate="2010-10-28T14:44:47.643" CommentCount="2" />
  <row Id="8180" PostTypeId="1" CreationDate="2010-10-28T13:49:35.613" Score="1" ViewCount="56" Body="&lt;p&gt;Short question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an element in the Kalman gain vector ever be over one?&lt;/p&gt;&#xA;" OwnerUserId="2367" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-20T07:04:59.780" LastActivityDate="2010-11-20T07:04:59.780" Title="Kalman gain over one" Tags="&lt;signal-processing&gt;" AnswerCount="1" />
  <row Id="8181" PostTypeId="1" AcceptedAnswerId="8183" CreationDate="2010-10-28T14:04:04.380" Score="2" ViewCount="103" Body="&lt;p&gt;I remember seeing something like the following problem in the past and would like to know if it has a solution (or if I can find a source for it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt; Given a finite set of points in the plane in general position (no three are collinear), does there exist a line that intersects no points?&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T19:02:46.863" LastActivityDate="2010-10-28T19:49:30.153" Title="Existence of lines not containing given points in general position." Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;discrete-geometry&gt;&lt;projective-geometry&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="8182" PostTypeId="2" ParentId="8142" CreationDate="2010-10-28T14:09:02.360" Score="1" ViewCount="0" Body="&lt;p&gt;Using complex numbers, but a more elementary (and constructive) approach than the Maximum Modulus theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the unit circle. It is enough if we show that the product of distances is greater than $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we will show that the product of distances of some point is at least $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the points be $\displaystyle z_1, z_2, \dots z_n$. Rotate the circle so that $\displaystyle (-1)^n \prod_{j=1}^{n} z_j = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle P(z) = \prod_{j=1}^{n}(z-z_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product of distances from $\displaystyle  z$ to $\displaystyle z_j$ is given by $\displaystyle |P(z)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\displaystyle \omega_j$ be the $\displaystyle n$ $n^{th}$ roots of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for $\displaystyle 1 \le k &amp;lt; n$, we have that $\displaystyle \sum_{j=1}^{n} (\omega_j)^k = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{j=1}^{n} P(w_j) = 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \sum_{j=1}^{n} |P(w_j)| \ge |\sum_{j=1}^{n} P(w_j)| = 2n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence there is some $j$ for which $|P(w_j)| \ge 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have shown that there is a point whose product of distances from the $n$ points is at least $2r^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I believe we can also show that if for all points, the product of distances is  $\leq 2r^n$, then the $n$ points must be equally spaced! (I will leave that for you :-)).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-28T20:16:14.783" LastActivityDate="2010-10-28T20:16:14.783" />
  <row Id="8183" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T14:09:30.550" Score="5" ViewCount="0" Body="&lt;p&gt;Yes.  Choose any direction in the plane.  There are continuum many lines parallel to that direction.  Only finitely many hit one of the points.  So continuum many of these lines do not hit any points.  This argument works as well for countably many points.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-28T14:09:30.550" CommentCount="1" />
  <row Id="8184" PostTypeId="2" ParentId="8173" CreationDate="2010-10-28T14:26:13.380" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a somewhat more natural example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\Omega, \mathcal{F})$ be a measurable space, and let $P,Q$ be two probability measures on $\mathcal{F}$.  It is a good exercise to verify that $$\mathcal{L} := \{ A \in \mathcal{F} : P(A) = Q(A) \}$$ is a $\lambda$-system.  (This is a common application of the $\pi$-$\lambda$ theorem : if one can show that $P$ and $Q$ agree on a $\pi$-system that generates $\mathcal{F}$, then $P$ and $Q$ must be the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\mathcal{L}$ need not be a $\sigma$-algebra.  For instance, consider a sample space consisting of two coin flips: $$\Omega = \{ HH, HT, TH, TT \}, \quad \mathcal{F} = 2^\Omega.$$ Let $P$ be the probability measure under which the coins are independent and unbiased, and let $Q$ be the measure under which the first coin is unbiased but the second coin is stuck to the first so that they always come up the same.  Explicitly,&#xA;$$P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}$$&#xA;$$Q(HH)=Q(TT)=\frac{1}{2}; \quad Q(HT)=Q(TH)=0.$$&#xA;Then one can check that the events on which $P$ and $Q$ agree are those which only look at one of the coins (or none), so that&#xA;$$\mathcal{L} = \{ \{HH,HT\}, \{HH,TH\}, \{HT,TT\}, \{TH,TT\}, \emptyset, \Omega\}.$$&#xA;This is not a $\sigma$-algebra since it is not closed under intersections.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-10-28T14:26:13.380" CommentCount="4" />
  <row Id="8185" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T14:35:23.490" Score="1" ViewCount="0" Body="&lt;p&gt;Ross' argument with equations says that, if you choose a Cartesian coordinate system in the plane, every straight line can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = ax + b \ , \qquad \text{or} \qquad x = 0 \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some real numbers $a, b$. Now, choosing a direction means picking up some $a \in \mathbb{R}$. Everyone of your finite (or countable) number of points $(x_n, y_n)_{n=1,2, \dots}$ determines a straight line: you just solve the equation for $b$ in terms of $y_n, a, x_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y_n = a x_n + b \qquad \Longleftrightarrow \qquad b = y_n - ax_n \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you just need to choose any $b \neq y_n - ax_n, \ n = 1,2,\dots$, which you can do, since there is just a finite (or countable) number of them, but an uncountable quantity of real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could fix the height $b$ instead of the slope $a$; for instance, $b=0$, and the points $(x_n, y_n)_{n=1,2,\dots}$ would give you a finite (or countable) number of slopes of straight lines from the origin $(0,0)$ to each of your points, using the same equation as above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a = \dfrac{y_n}{x_n} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In case you have $x_n = 0$, you just have the straight line $x=0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you pick up any $a\neq\dfrac{y_n}{x_n} $, which you can do for the same reason as before, and you are done.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-28T14:35:23.490" />
  <row Id="8186" PostTypeId="1" CreationDate="2010-10-28T14:51:18.943" Score="5" ViewCount="269" Body="&lt;p&gt;It is well-known that given two primes &lt;em&gt;p&lt;/em&gt; and &lt;em&gt;q&lt;/em&gt;, &lt;em&gt;pZ + qZ = Z&lt;/em&gt;  where &lt;em&gt;Z&lt;/em&gt; stands for all integers.&#xA;It seems to me that the set of natural number multiples, i.e. &lt;em&gt;pN + qN&lt;/em&gt; also span all natural numbers that are large enough. That is, there exists some &lt;em&gt;K&gt;0&lt;/em&gt;, such that&#xA;&lt;em&gt;pN + qN = [K,K+1,...)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, given &lt;em&gt;p&lt;/em&gt; and &lt;em&gt;q&lt;/em&gt;, can we get a upper bound on &lt;em&gt;K&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2901" LastActivityDate="2010-11-07T00:54:29.040" Title="When do the multiples of two primes span all large enough natural numbers? " Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;cryptography&gt;&lt;number&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8187" PostTypeId="2" ParentId="8186" CreationDate="2010-10-28T14:55:05.867" Score="13" ViewCount="0" Body="&lt;p&gt;$K = pq + 1$ if $\mathbb{N} = \{ 1, 2, 3, ... \}$, and $K = pq - p - q + 1$ if $\mathbb{N} = \{ 0, 1, 2, 3, ... \}$.  This is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;coin problem&lt;/a&gt;, or Frobenius problem (and you only need $p, q$ relatively prime).  It frequently appears on middle- and high-school math competitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I completely misremembered how hard the proof is.  Here it is.  If $n$ is at least $pq+1$, then the positive integers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n-p, n-2p, ... n-qp$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have distinct residue classes $\bmod q$, so one of them must be divisible by $q$.  On the other hand, it's not hard to see that $pq$ itself cannot be written in the desired way.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-07T00:01:15.903" LastActivityDate="2010-11-07T00:01:15.903" CommentCount="6" />
  <row Id="8188" PostTypeId="1" AcceptedAnswerId="8189" CreationDate="2010-10-28T14:56:53.023" Score="3" ViewCount="129" Body="&lt;p&gt;For a ring $R$, an $R$-right-module $M$, an $R$-left-module $N$, and an abelian group $P$, one can use the universal property of the tensor product to construct maps&#xA;$$&#xA;M\otimes_R N\to P.&#xA;$$&#xA;It concrete cases, it is often easy to see that the constructed map is surjective by just writing down pre-images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be harder to verify that the map is injective, because then one has to consider general sums of elementary tensors in $M\otimes_R N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any tricks to avoid this, and to achieve injectivity more elegantly?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More concretely, an example I have in mind is the following: I have a pre-additive category $C$ with finitely many objects. I consider modules over this category (that is, functors from $C$ to Abelian groups; or, equivalently, modules over the category ring of $C$). Now, I consider the &quot;free $C$-right-module $Q_Y$ over an object $Y$ in $C$&quot; which is the hom functor $C(\bullet,Y)$. I want to show that for an arbitrary left-module $M$:&#xA;$$&#xA;Q_Y\otimes_C M\cong M(Y)&#xA;$$&#xA;as abelian groups. Using the module structure of $M$, I can accomplish a natural epimorphism $Q_Y\otimes_C M\to M(Y)$ which should be injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I think that the formula&#xA;$$&#xA;M(Y)\to Q_Y\otimes_C M;\quad m\mapsto\mathrm{id}_Y\otimes m&#xA;$$&#xA;gives indeed a (two-sided) inverse. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-02T12:48:07.210" LastActivityDate="2010-11-02T12:48:07.210" Title="Is there a trick to prove injectivity for maps out of tensor products?" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8189" PostTypeId="2" ParentId="8188" CreationDate="2010-10-28T15:26:34.100" Score="1" ViewCount="0" Body="&lt;p&gt;If the image Q is easy to find and well understood, then you can try to construct an inverse from Q→M⊗N. If Q is well understood, then you can define homomorphisms from it (say by defining the images of some generators, and checking that all the relations are sent to 0 in M⊗N).  Then you just check that the composition is the identity (say by sending generators of Q into M⊗N and then back into P, and checking relations to check equality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This depends heavily on the image being well understood.  It works well to prove an abstract tensor product is equal to some concretely understood module.  If the image is confusing, then checking that the homomorphism is well defined (or even just defining it) can be hard, and even checking equality of elements in P can be hard in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other tricks using embeddings and exact sequences, but they also have limited applicability (usually when the entire problem has exact sequences floating around).  Feel free to give some more concrete examples for more concrete tricks.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-10-28T15:26:34.100" CommentCount="2" />
  <row Id="8190" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T16:01:06.853" Score="6" ViewCount="0" Body="&lt;p&gt;What is this general position stuff?  Take any line outside the convex hull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you're thinking of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem&quot; rel=&quot;nofollow&quot;&gt;Sylvester-Gallai theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-28T16:01:06.853" />
  <row Id="8191" PostTypeId="2" ParentId="7745" CreationDate="2010-10-28T16:07:06.293" Score="1" ViewCount="0" Body="&lt;p&gt;I can not post comments, so I will post an answer..  Using $S_0 + B_t$ to model a stock price is not a very good idea, because $S_t$ may well turn negative. Have you considered using &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;Geometric Brownian Motion&lt;/a&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="731" LastActivityDate="2010-10-28T16:07:06.293" />
  <row Id="8192" PostTypeId="2" ParentId="7506" CreationDate="2010-10-28T16:21:05.780" Score="1" ViewCount="0" Body="&lt;p&gt;Normally subcategories of Abelian Categories are assumed nonempty for the simple reason that they need contain the zero object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want your subcats to be abelian, then they need this zero. For this reason, I would say no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is at least the viewpoint in NCAG.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-10-29T07:28:12.907" LastActivityDate="2010-10-29T07:28:12.907" CommentCount="2" />
  <row Id="8193" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T16:22:36.730" Score="2" ViewCount="0" Body="&lt;p&gt;Such statements are often easier to see in the projectively dual version.&#xA;The dual statement would be that given any finite set of lines in the projective plane (which secretely we are thinking of the dual plan of lines in&#xA;the projective plane where our original points lay), and a point $P$ not on them (this is the point in the dual plane corresponding to the line at infinity&#xA;in our original plan), there is a point $Q$ distinct from $P$, and also not lying on any of the lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just stating it again without the parenthetical explanations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Given any finite set of lines in the projective plane, and a point $P$ not lying on them any of them, there exists a point $Q$ distinct from $P$ and not lying on any of them.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will (pretty clearly) be true just provided we are working over an&#xA;infinite field.  (Thus we don't need that $\mathbb R$ is uncountable;&#xA;the fact that it is infinite is enough.  This is made clear in Agusti Roig's answer, of which the above is slighty more theoretical reformulation.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-28T16:22:36.730" CommentCount="1" />
  <row Id="8194" PostTypeId="1" AcceptedAnswerId="8196" CreationDate="2010-10-28T16:32:25.517" Score="3" ViewCount="90" Body="&lt;p&gt;Consider a function $f : \mathcal X \times \mathcal Y \mapsto \mathbb R$. I want to define $g_x(y) = f(x,y) : \mathcal Y \mapsto \mathbb R$. I want to say that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$g_x$ is a &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;/strong&gt; of function $f$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the appropriate word for &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;____&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2904" LastActivityDate="2010-10-28T16:49:26.260" Title="Function of two variables as a function of a function" Tags="&lt;terminology&gt;&lt;functions&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="8195" PostTypeId="2" ParentId="8194" CreationDate="2010-10-28T16:35:32.497" Score="4" ViewCount="0" Body="&lt;p&gt;This is sometimes called &lt;a href=&quot;http://en.wikipedia.org/wiki/Currying&quot; rel=&quot;nofollow&quot;&gt;currying&lt;/a&gt;.  It is closely related to the notion of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_object&quot; rel=&quot;nofollow&quot;&gt;exponential object&lt;/a&gt;.  But you don't really need to use either of these terms to perform this construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Ah, I was assuming you were varying $x$.  If $x$ is fixed, you might want to call $g_x$ a &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_(mathematics)#Restrictions_and_extensions&quot; rel=&quot;nofollow&quot;&gt;restriction&lt;/a&gt; of $f$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-28T16:45:30.393" LastActivityDate="2010-10-28T16:45:30.393" CommentCount="2" />
  <row Id="8196" PostTypeId="2" ParentId="8194" CreationDate="2010-10-28T16:39:19.737" Score="6" ViewCount="0" Body="&lt;p&gt;I've seen your $g_x$ called the $x$-&lt;strong&gt;section&lt;/strong&gt; of $f$.  E.g. Folland's &lt;em&gt;Real Analysis&lt;/em&gt;, section 2.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Another notation that's often useful is to write $f(x, \cdot)$ instead of $g_x$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-10-28T16:49:26.260" LastActivityDate="2010-10-28T16:49:26.260" CommentCount="1" />
  <row Id="8197" PostTypeId="2" ParentId="8170" CreationDate="2010-10-28T16:47:27.600" Score="4" ViewCount="0" Body="&lt;p&gt;Since you asked for an intuitive way to understand covariance and contravariance, I think this will do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, remember that the reason of having covariant or contravariant tensors is because you want to represent the same thing in a different coordinate system. Such a new representation is achieved by a transformation using a set of partial derivatives. In tensor analysis, a good transformation is one that leaves invariant the quantity you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we consider the transformation from one coordinate system $x^1,...,x^{n}$ to another $x^{'1},...,x^{'n}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{i}=f^{i}(x^{'1},x^{'2},...,x^{'n})$ where $f^{i}$ are certain functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at a couple of specific quantities. How do we transform coordinates? The answer is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dx^{i}=\displaystyle \frac{\partial x^{i}}{\partial x^{'k}}dx^{k}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Every quantity which under a transformation of coordinates, transforms like the coordinate differentials is called a contravariant tensor.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we transform some scalar $\Phi$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{\partial \Phi}{\partial x^{i}}=\frac{\partial \Phi}{\partial x^{'k}}\frac{\partial x^{'k}}{\partial x^{i}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Every quantity which under a coordinate transformation, transforms like the derivatives of a scalar is called a covariant tensor.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, a reasonable generalization is having a quantity which transforms like the product of the components of two contravariant tensors, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^{ik}=\displaystyle \frac{\partial x^{i}}{\partial x^{'l}}\frac{\partial x^{k}}{\partial x^{'m}}A^{'lm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is called a contravariant tensor of rank two. The same applies to covariant tensors of rank n or mixed tensor of rank n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having in mind the analogy to coordinate differentials and derivative of a scalar, take a look at this picture, which I think will help to make it clearer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8ZQfK.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;The contravariant components of a vector are obtained by projecting onto the coordinate axes. The covariant components are obtained by projecting onto the normal lines to the coordinate hyperplanes.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, you may want to read: &lt;a href=&quot;http://en.wikipedia.org/wiki/Orthogonal_coordinates#Basis_vectors&quot; rel=&quot;nofollow&quot;&gt;Basis vectors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I don't recommend to rely blindly in the picture given by matrices, specially when you are doing calculations.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-10-28T16:47:27.600" CommentCount="2" />
  <row Id="8198" PostTypeId="2" ParentId="8107" CreationDate="2010-10-28T16:51:58.243" Score="2" ViewCount="0" Body="&lt;p&gt;Since your question is about seeing how your different options play out, and you have a small number of them (the point of my previous question), you can use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Decision_tree&quot; rel=&quot;nofollow&quot;&gt;decision tree&lt;/a&gt;.  (From Wikipedia: &quot;In decision analysis, a 'decision tree'... is used as a visual and analytical decision support tool, where the expected values (or expected utility) of competing alternatives are calculated.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want the first level of branches of the tree to be the various decisions you have (spend 0 dollars on research, spend 5 dollars on research, etc.).  The second level of branches of the tree will be whether or not the breakthrough is found.  Based on the outcomes and probabilities for the second level, you can calculate an expected value for each option in the first level.  Presumably you want to choose the option with the highest expected value (or expected utility, if you are comfortable with utility functions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think you need some more information to get a really good answer.  In particular, your example doesn't quantify exactly how your budget would change over time.  That needs to be included somehow.  In addition, perhaps going with the highest expected monthly value isn't what you really want, as your description of the scenario implies that you might be willing to go three months without profit in order to increase your chance of a breakthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: I think a decision tree is what you want (and there are lots of resources out there besides the Wikipedia page that can help you with that), but you need to quantify some of the other aspects of your problem before you can get a good answer.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-28T16:51:58.243" />
  <row Id="8199" PostTypeId="2" ParentId="8176" CreationDate="2010-10-28T16:52:35.823" Score="5" ViewCount="0" Body="&lt;p&gt;I guess one very natural answer to the question is provided by differential forms.  On a smooth $n$-manifold, an $n$-form is a type of &quot;field&quot; whose value at each point is a determinant on the tangent space at that point.  (I say &quot;a determinant&quot; because it's only on $\mathbb R^n$ -- or, more generally, an inner product space -- that there's a natural way to define &lt;em&gt;the&lt;/em&gt; determinant.)  An $n$-form is exactly the type of field that can be integrated over an $n$-manifold in a coordinate-independent way.  More generally, a $k$-form on an $n$-manifold is a field that yields a determinant on each $k$-dimensional subspace of each tangent space, and it is the type of field that can be integrated over $k$-dimensional submanifolds.  (You can find more about this in my book &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="1421" LastActivityDate="2010-10-28T16:52:35.823" CommentCount="3" />
  <row Id="8200" PostTypeId="2" ParentId="8186" CreationDate="2010-10-28T16:56:17.387" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ p\ (p^{-1}\: mod\ q) + q\ (q^{-1}\: mod\ p)\ =\ pq+1\ $ since it's $1$ mod $\rm\:p,q\:$ and $\rm &gt;2\:$ and $\rm &amp;lt; 2\:pq$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To represent largers numbers keep adding $\rm\ 1 = ap+bq\ $ while, if need be, adding $\rm\:\pm (q ,-p)\:$ to the coefficients to keep them positive.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-28T16:56:17.387" />
  <row Id="8201" PostTypeId="2" ParentId="8176" CreationDate="2010-10-28T17:18:03.240" Score="4" ViewCount="0" Body="&lt;p&gt;It's an interesting question. If there were any strong and formalizable analogy it probably would have been developed a long time ago and inscribed in the textbooks.  A few observations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The linear algebra analogue of integration is a trace.   Determinants are an exponentiated trace. For example $\det \exp A = \exp Tr(A)$ for matrices.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you view integration as solution of differential equations rather than measure, then determinant appears as the Wronskian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Integration (as measure) and determinants are closely related in the theory behind the change of variables formula in integrals: differential forms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The integral is a trace on an infinite-dimensional space (the commutative algebra of functions on, e.g., a closed interval or the real line) while the determinant is specifically finite-dimensional.  The Lebesgue measure used in ordinary $n$-dimensional integrals is in some sense defined using determinants (volume), which is why it does not generalize well to infinite dimensional spaces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thinking about integrals and determinants in terms of formal properties they satisfy leads to &quot;K-theory&quot;, specifically $K_1$, but I don't think this produces any deep or striking analogies between the concepts.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-10-28T17:24:52.717" LastActivityDate="2010-10-28T17:24:52.717" />
  <row Id="8202" PostTypeId="1" CreationDate="2010-10-28T18:28:01.187" Score="0" ViewCount="185" Body="&lt;p&gt;I have a function that is difficult to integrate.  So I elect to work with power series representations.  Suppose the power series representation for this function is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=M}^N{c_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempt to integrate each term of the power series seperately.  For instance, looking at a few terms, we can integrate this to obtain $g(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \cdots + c_0 x^0 + c_1 x^1 + c^2 x^2 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \cdots + c_0 x^1 + \frac{1}{2}c_1 x^2 + \frac{1}{3}c_2 x^3 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect, we take $c_i x_i \mapsto \frac{1}{i+1}c_i x^{i+1}$ integrating terms individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its possible to do this using generating function methods, which is fairly routine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Problem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two terms are problematic.  The term $c_{-1} x^{-1}$ should integrate to $\log(x)$.  The constant of integration is not known, so this is the second problem term.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to perhaps solve this problem.  For instance, finding a way to approximate the two problem terms would be a good start.  In the end, I want to get an approximation to the integral within an error bounds.  I really want to consider this particular method, though.  I'd be interested in seeing related literature.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some Ideas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We may be able to use approximations to similar integrals to solve for approximations to the two problem terms.  I'm interested in trying to get an approximation (to the original integral) within a certain amount of error.  If this is possible, I'd consider the problem solved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An Example:  Particular integrals that I'm using this on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a description of the function(s) I want to integrate.  I take:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x^b \frac{\prod_{i=0}^m{(\pm x^{c_i} \pm 1)}}{\prod_{j=0}^n{(\pm x^{d_j} \pm 1)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that both the quotient and the denominator consist of products of the form $\pm x^r \pm 1$.  Here, $r \in \mathbb{N}$ and $-(2^s) \leq r \leq 2^s$ for some natural $s$.  Also, $b$ is similarly restricted/defined as $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take this formula and make the substitution $x \mapsto e^{i t}$ and integrate from $t=-\pi$ to $t=\pi$.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-10-29T16:25:17.917" LastActivityDate="2010-10-29T16:25:17.917" Title="How effective is this alternative to integration?" Tags="&lt;complex-analysis&gt;&lt;intuition&gt;&lt;integral&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" CommentCount="7" />
  <row Id="8203" PostTypeId="2" ParentId="7743" CreationDate="2010-10-28T19:16:44.633" Score="23" ViewCount="0" Body="&lt;p&gt;I'd like to second one part of Qiaochu Yuan's answer: the recommendation to read Polya's book.  Unlike many other books I've seen (albeit none of the others recommended above), it actually does contain guidance on how to construct a proof &quot;out of nothing&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's one problem with the &quot;practise, practise, practise&quot; mantra.  Practise what?  Where are the lists of similar-but-not-quite-identical things to prove to practise on?  I can find lists of integrals to do and lists of matrices to solve, but it's hard coming up with lists of things to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, practise is correct.  But just as with anything else in mathematics, there's guidelines to help get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to realise is that reading others proofs is not guaranteed to give you any insight as to how the proof was developed.  A proof is meant to convince someone of a result, so a proof points to the theorem (or whatever) and knowing how the proof was constructed does not (or at least, &lt;em&gt;should not&lt;/em&gt;) lend any extra weight to our confidence in the theorem.  Proofs can be written in this way, and when teaching we should make sure to present some proofs in this way, but to do it every time would be tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what are the guidelines for constructing a proof?  You'll probably get different answers from different mathematicians so these should be construed as being my opinion and not a(n attempt at a) definitive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation is that you take the statement that you want to prove and apply the following steps to it as often as you can:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Expand out unfamiliar terms.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Replacing generic statements by statements about generic objects.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Including implicit information.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you've done all that, the hope is that the proof will be much clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Original statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The composition of linear transformations is again linear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace generic statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S$ and $T$ are two composable linear transformations then their composition, $S T$, is again linear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important to be precise here.  The word &quot;composable&quot; could have been left out, as the statement only makes sense if $S$ and $T$ are composable, but until you are completely familiar with this kind of process, it is better to be overly precise than otherwise.  In this case, leaving in the word &quot;composable&quot; reminds us that there is a restriction on the domains and codomains which will be useful later.  (However, one has to draw the line somewhere: even the word &quot;composable&quot; is not quite enough since it leaves open the question as to whether it is $S T$ or $T S$!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Include implicit information:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S \colon V \to W$ and $T \colon U \to V$ are linear transformations then $S T \colon U \to W$ is again linear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's where remembering that $S$ and $T$ are composable in the previous step helps keep things clear.  As $S$ and $T$ are composable, we only need $3$ vector spaces.  Then, since we explicitly have the vector spaces the fact that $S$ and $T$ are composable is plain, though some may prefer to keep that fact in the statement.  Also, some may like to have the fact that $U$, $V$, and $W$ are &lt;strong&gt;vector&lt;/strong&gt; spaces explicitly stated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Expand out definitions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S \colon V \to W$ and $T \colon U \to V$ are such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$, then $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$ for all $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that I have been careful not to repeat myself with the newly introduced symbols.  It would be technically alright to reuse $u_1$ and $u_2$ in place of $x_1$ and $x_2$ since these are &lt;em&gt;local&lt;/em&gt; declarations (restricted by the phrases &quot;for all ...&quot;).  However, humans are not good at differentiating between local and global declarations so it is best not to reuse symbols unless the scope is very clear.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace generic statements:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $S \colon V \to W$ and $T \colon U \to V$ are such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$, then whenever  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$,  $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up to now, the rephrasing has not taken into account the fact that there is a &lt;em&gt;conclusion&lt;/em&gt; and a &lt;em&gt;hypothesis&lt;/em&gt;.  This rephrasing modifies a part of the conclusion to turn it from a generic statement &quot;$P(p)$ is true for all $p \in Q$&quot; to a statement about a generic object &quot;whenever $p \in Q$ then $P(p)$ is true&quot;.  We do not do this for the similar statements in the hypothesis.  This is because these two pieces are treated differently in the proof.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace generic statements, and reorganise to bring choices to the fore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $S \colon V \to W$ and $T \colon U \to V$ be such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$.  Let  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.  Then  $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this form, the distinction between &lt;em&gt;hypothesis&lt;/em&gt; and &lt;em&gt;conclusion&lt;/em&gt; is all the clearer.  Parts of the hypothesis use the word &quot;Let&quot;, parts of the conclusion use the word &quot;Then&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With this formulation, the proof essentially writes itself.  With all it's gory details:&lt;/p&gt;&#xA;&#xA;Proof&#xA;&#xA;&lt;p&gt;Let $S \colon V \to W$ and $T \colon U \to V$ be such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$.  Let  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.[^quick]  Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S T(x_1 + \eta x_2) = S \big( T(x_1) + \eta T(x_2)\big)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the hypothesis on $T$ as $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.  So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the hypothesis on $S$ as $T(x_1), T(x_2) \in V$ and $\eta \in \mathbb{R}$.  Hence the conclusion is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&#xA;1. This could be condensed, but the important thing here is how to &lt;em&gt;find&lt;/em&gt; it, not what the final form should be.&#xA;2. Notice that I wrote &quot;as $x_1, x_2 \in U$&quot; rather than &quot;with $u_1 = x_1$ and $u_2 = x_2$&quot;.  This is partly style, and partly because in the statement of linearity, $u_1$ and $u_2$ are &lt;em&gt;placeholders&lt;/em&gt; into which we put $x_1$ and $x_2$.   So saying $u_1 = x_1$ is semantically incorrect as it equates a virtual vector with an actual vector.  This is a very minor point, though.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, I would like to disagree with one part of Qiaochu's answer.  I actually like the imagery of a steel trap.  A proof is a bit like a trap: we want to capture the theorem in a trap so that it can't wriggle out.  We construct the proof so that there is no possibility of escape.  Eventually, yes, we want the proof to be beautiful but when it's first constructed we just want it to do the job.  Only once the theorem is caught can we spend a little time decorating the cage to make it look pretty and set it off to its best advantage.  So build the trap because theorems can be dangerous!  An escaped theorem can do untold damage, rampaging across the countryside, laying waste like an unchecked viking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Okay, not quite finally.  The step-by-step proof above was taking from a page I wrote for my students on the nature of proof.  The original can be found &lt;a href=&quot;http://syvert.math.ntnu.no:2500/tma4145/show/proof&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-10-28T19:16:44.633" CommentCount="3" />
  <row Id="8204" PostTypeId="1" AcceptedAnswerId="8208" CreationDate="2010-10-28T19:31:33.153" Score="2" ViewCount="182" Body="&lt;p&gt;Is it true that for all proofs $\forall x \exists y : R(x, y)$, then $y = y(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A while back I remember reading a book on functional programming that was leading into some questions about what exactly mathematical proofs were and how they were related to functions. In fact, I just found the book and page: here on PDF page 16 (marked page 3):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cs.kent.ac.uk/people/staff/sjt/TTFP/ttfp.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.kent.ac.uk/people/staff/sjt/TTFP/ttfp.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the author mentions that if we have a proof that  $\forall x \exists y : R(x, y)$, then, in a sense, because x is bound before y, then y is dependent on x, and therefore $y = y(x)$. There may be counterexamples. Perhaps y is a constant itself, but then y(x) can also just be a constant function. I'm trying to understand under what circumstances this is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f : \mathbb{R}^n \to \mathbb{R}$ be continous at a point $\vec{v_0}$ and $f(\vec{v_0}) = A &amp;lt; 0$. Prove that there exists a number $\delta &gt; 0$ such that if $\| \vec{v} - \vec{v_0} \| &amp;lt; \delta $ then $f(\vec{v}) &amp;lt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My argument was simply to use the statement of continuity at $\vec{v_0}$ to prove this. That is, since f is continuous, the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall \epsilon &gt; 0 \exists \delta &gt; 0 : \| \vec{v} - \vec{v_0} \| &amp;lt; \delta \rightarrow \| f(\vec{v}) - f(\vec{v_0}) \| &amp;lt; \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as referred to above, we must have $\delta = \delta(\epsilon)$. Then we can make the replacement $\delta = \delta(|A|)$, and we have the necessary $\delta$ to prove that $f(\vec{v}) &amp;lt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one main question (the one that closes this), and one side question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Under what conditions can we assume that $\forall x \exists y : R(x, y)$, then $y = y(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there indirect ways to prove a proposition of the form $\forall x \exists y : R(x, y)$ where y is somehow not dependent on x and also not a constant? Perhaps it varies on R, but that is given, I suppose. Certainly all $\epsilon-\delta$ proofs that I've come across have been constructive in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a canonical way to prove the proposition above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually struggled with a possible proof for a while before I came up with the technique above. I used this technique in a HW assignment and on my latest Advanced Calc test. The HW was marked as correct although with the note that it was &quot;indirect&quot;, while the test was marked as wrong. Granted, the HW was done by the TA, the test by my prof. I personally don't see a more &quot;direct&quot; way to prove this result.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2010-10-28T19:50:28.730" LastActivityDate="2010-10-29T16:21:57.180" Title="Is it true that for all proofs of the statement, $\forall x \exists y : R(x, y)$, then we can say $y = y(x)$? (Example given)" Tags="&lt;real-analysis&gt;&lt;soft-question&gt;&lt;logic&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="8205" PostTypeId="2" ParentId="8204" CreationDate="2010-10-28T19:47:37.667" Score="0" ViewCount="0" Body="&lt;p&gt;Some texts define a relation as simply a set of ordered pairs and a function as a set of ordered pairs in which every first element is unique. So if for all x there is a unique y such that R(x,y), then you can say you have a function y(x).  You have not specified that y has to be unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by &quot;prove a relation&quot; under question 1.  Your proof seems fine to me.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-28T19:47:37.667" />
  <row Id="8206" PostTypeId="2" ParentId="8181" CreationDate="2010-10-28T19:49:30.153" Score="1" ViewCount="0" Body="&lt;p&gt;For a very nice treatment of the combinatorics of point and line configurations and the symmetry of such configurations, look at Branko Grünbaum's book: Configurations of Points and Lines, American Mathematical Society, 2009.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-28T19:49:30.153" CommentCount="1" />
  <row Id="8207" PostTypeId="1" AcceptedAnswerId="8209" CreationDate="2010-10-28T19:50:04.783" Score="2" ViewCount="160" Body="&lt;p&gt;Simple example in order to clarify what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequence - {2 , -1, 4, 2, 9 }&#xA;Sum and product combination: { 16, -144 }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Is a combination of sum and product unique for any* sequence?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;any sequence&quot; except that one which consists only of zeros.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPD:&lt;/strong&gt; The answer to my initial question is &quot;no&quot;. Thanks.&#xA;But what's the answer if we restrict ourselves to sequences of the same length and made of non-negative integers? Do two or more non-negative integer sequences exist, each of which has the same {sum,product} combination and the same length?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPD:&lt;/strong&gt; Both answers are &quot;no&quot;. Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="1934" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-10-31T01:43:11.697" LastActivityDate="2010-10-31T01:43:11.697" Title="Is a combination of sum and product of any sequence unique?" Tags="&lt;sequences-and-series&gt;&lt;algebra&gt;&lt;proof&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="8208" PostTypeId="2" ParentId="8204" CreationDate="2010-10-28T19:50:27.060" Score="8" ViewCount="0" Body="&lt;p&gt;You seem to be asking: if for every $x$ there is a $y$ for which $R(x,y)$ holds, can we then define a function $f$ (in the usual sense) such that for every $x$, $R(x,f(x))$ holds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a set-theoretic point of view, that statement is in fact none other than the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_Choice&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Axiom of Choice has many forms, but one form is the following: given a nonempty family of nonempty sets, $S_i$ with $i\in I$ (&quot;nonempty family&quot; means that $I$ is not empty, so we have some sets; &quot;nonempty sets&quot; means that each $S_i$ is not empty), there exists a &quot;choice function&quot;: a way to select for each $i\in I$ an element of the corresponding $S_i$. Formally, there exists a function $f\colon I\to \cup S_i$ from the index set to the union of the sets, with the property that $f(i)\in S_i$ for each $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that by assumption, for every $x$ in our universe $X$, the set $S_x = \{y\mid R(x,y)\}$ is nonempty, and you have a function $f$ with the desried property. The function $f$ will be a &quot;choice function&quot;: a function  $f:X \to \cup_{x\in U}S_x$ such that for each $x\in X$, $f(x)\in S_x$. Conversely, a choice function for the family $\{S_x\}_{x\in X}$ will yield and $f$ with the desired property. The existence of such functions for &lt;em&gt;every&lt;/em&gt; nonempty family of nonempty sets is precisely the Axiom of Choice. Conversely, suppose that the Axiom of Choice fails, and let $\{S_x\}_{x\in X}$ be a nonempty family of nonempty sets for which no choice function exists. Then let $R(x,y)$ be &quot;$y$ is in $S_x$&quot;. For every $x$ there exists a $y$ such that $R(x,y)$, since $S_x\neq\emptyset$. But since we are assuming that the family has no choice function, there does not exist any function $f$ such that for every $x$, $f(x)$ is in $S_x$. Thus, the question not only depends on the Axiom of Choice, but is &lt;em&gt;equivalent&lt;/em&gt; to the Axiom of Choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some circumstances, the exact nature of the sets $S_x$ mean that you do not need to invoke the Axiom of Choice, because we have some &quot;natural way&quot; of defining the choice. For instance, if every set $S_x$ is a subset of the natural numbers, then you can define $f(x)$ to be &quot;the least element of $S_x$&quot;.  Another standard example is when $S_x$ is a singleton for every $x$; that is, for each $x$ there exists one &lt;strong&gt;and only one&lt;/strong&gt; $y$ for which $R(x,y)$ holds. In that case, you can define your function $f$ to be &quot;$f(x)$ is the unique $y$ for which $R(x,y)$ holds&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in general, the Axiom of Choice is, in some sense, &quot;inherently non-constructive&quot;.  And the thing is, the Axiom of Choice is &lt;em&gt;independent&lt;/em&gt; of the usual axioms of set theory. Like the parallel postulate in geometry, you can accept it or not at your taste (the consequences of accepting and of not accepting it are different; for example, if you accept the Axiom of Choice, then that means that there is a way to slice a ball of radius 1 into a finite number of parts, and then rearrange the parts without changing their shape or size, into &lt;em&gt;two&lt;/em&gt; balls of radius of 1; this is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach-Tarski_paradox&quot; rel=&quot;nofollow&quot;&gt;Banach-Tarski paradox&lt;/a&gt;; on the other hand, if you want all your &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;vector spaces&lt;/a&gt; to have &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_%28linear_algebra%29&quot; rel=&quot;nofollow&quot;&gt;bases&lt;/a&gt;, then you &lt;em&gt;must&lt;/em&gt; accept the Axiom of Choice. In short, it has good consequences and weird/bad consequences). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In the situation you have, it is possible to describe a function that will work; however, the &quot;description&quot; is a bit like cheating, in the sense that you can give an explicit rule for what the value of the function will be, but in practice it would be difficult to actually find the value. Namely, for every $\epsilon\gt 0$ you know that the set of $\delta\gt 0$ for which the implication $||v-v_0||\lt\delta\Rightarrow ||f(v)-f(v_0)||\lt\epsilon$ holds is nonempty. In particular, by the Archimedean property, the set $A_{\epsilon}=\{n\in\mathbb{N}| ||v-v_0||\lt\frac{1}{n} \Rightarrow ||f(v)-f(v_0)||\lt\epsilon\}$ is nonempty. This is a nonempty set of positive integers, so it has a least element. You can then define your function $\mathbf{f}\colon (0,\infty)\to(0,\infty)$ by the &quot;formula&quot; $\mathbf{f}(\epsilon)=1/\min(A_{\epsilon})$, the reciprocal of the smallest $n$ in the set. This definition would not require you to invoke the Axiom of Choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question 2: you almost had it, I think, but you should not invoke (implicitly) the Axiom of Choice: pick a specific $\epsilon$ (for example, $|A|/2$). You &lt;em&gt;know&lt;/em&gt; that there exists a $\delta\gt 0$ that works for &lt;em&gt;that&lt;/em&gt; epsilon, so you pick &lt;em&gt;any&lt;/em&gt; $\delta_0$ that works and use &lt;em&gt;that&lt;/em&gt;. That would be the standard way of proving the proposition at hand. If you assume you have some kind of &quot;black-box function&quot; that will hand you a $\delta$ whenever you hand it an $\epsilon$, you are invoking the Axiom of Choice and you are making your proof somewhat non-constructive. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-29T16:21:57.180" LastActivityDate="2010-10-29T16:21:57.180" CommentCount="8" />
  <row Id="8209" PostTypeId="2" ParentId="8207" CreationDate="2010-10-28T19:53:04.940" Score="8" ViewCount="0" Body="&lt;p&gt;Not in general.  Knowing the sum and product of a sequence $r_1, ... r_n$ only fixes two coefficients of the polynomial $(x - r_1)...(x - r_n) = x^n + a_{n-1} x^{n-1} + ... + a_0$, namely $a_0$ and $a_{n-1}$, and for $n &gt; 2$ one can choose the other coefficients arbitrarily and solve for the complex roots of the resulting polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With certain restrictions, this might sometimes be possible, for example if you restrict the sequence to be integers.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-28T19:53:04.940" CommentCount="2" />
  <row Id="8210" PostTypeId="2" ParentId="8110" CreationDate="2010-10-28T19:58:26.160" Score="2" ViewCount="0" Body="&lt;p&gt;For the case where we have cans with $2,3,4$ liters, consider the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;(1+x^2)^3 (1+x^3)^2(1+x^4)^2\left(1+\frac{1}{x^2}\right)^2\left(1+\frac{1}{x^3}\right) \left(1+\frac{1}{x^4}\right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the terms $x^k$ where $k\geq 1$. I believe all those powers of $x$ are the capacities you can measure using the three cans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any term of the form $1+x^a$ allows for two choices - You either fill the can of volume $a$ or you don't fill it. If we multiplied three terms and wrote $(1+x^2)(1+x^3)(1+x^4)$, this will allow you to find out how many different volumes you can measure if you are allowed to fill each can at most once and if a can once filled is never emptied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the case of cans of size $2,3$ and $4$, it is possible to fill the 2 liter can once, empty it into the 4 liter can, fill the 2 liter can again, pour it into the 4 liter can again and fill the 2 liter can a third time. This is why we have $(1+x^2)^3$. Similarly for the 3 litre can. The 4 litre can could be emptied into the 2 and 3 liter cans combined and the refilled, so we have the term $(1+x^4)^2$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should also allow for subtracting of volumes - This implies that terms of the form $(1+\frac{1}{x^a})$ should be present. If we look at the 2 liter can, you could fill it (and empty it) from the 4 liter can twice. Therefore, we have a term $(1+1/x^2)^2$. Similarly we get the term $(1+1/x^3)$. The 4 liter can could also be emptied once after filling it from the other two cans. Combining all of this, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;(1+x^2)^3 (1+x^3)^2(1+x^4)^2\left(1+\frac{1}{x^2}\right)^2\left(1+\frac{1}{x^3}\right) \left(1+\frac{1}{x^4}\right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea can be extended to other values of $a,b$ and $c$. I tried it for a few other values (using Wolfram Alpha to expand the expressions) and it seems to work. &lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-28T19:58:26.160" />
  <row Id="8212" PostTypeId="2" ParentId="8207" CreationDate="2010-10-28T20:19:08.007" Score="1" ViewCount="0" Body="&lt;p&gt;Let's say we have a finite tuple (not set) of positive integers, $A = (a_1,a_2,\dotsb,a_n)$, with $\sum a_i = s$, $\prod a_i = p$. Consider the simplest prime factorization $p = p_1^{n_1} \dotsb p_k^{n_k}$ and construct $B = (1,\dotsb,1,p_1,\dotsb,p_1,p_2,\dotsb,p_k,\dotsb,p_k)$ where each $p_i$ occurs $n_i$ times and we fill out with enough $1$s to make $\sum B = s$ (why can we do this?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving that this produces a $B$ with different elements than $A$ in &quot;most&quot; cases is left as an exercise for the reader. As is extending it to the case where $A$ may contain negative integers. In that case you want to fill up with a clever combination of $1$s and $-1$s.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-10-28T20:19:08.007" CommentCount="1" />
  <row Id="8213" PostTypeId="1" AcceptedAnswerId="8214" CreationDate="2010-10-28T22:17:39.343" Score="2" ViewCount="87" Body="&lt;p&gt;Hi all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quick formulation of the problem:&lt;/strong&gt;&lt;br&gt;&#xA;Given two points: $(x_l, y_l)$ and $(x_u, y_u)$&lt;br&gt;&#xA;with: $x_l &amp;lt; x_u$ and $y_l &amp;lt; y_u$,&lt;br&gt;&#xA;and given lower asymptote=0 and higher asymptote=1,&#xA;what's the logistic function that passes through the two points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanatory image:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/QoQix.png&quot; alt=&quot;logistic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other details:&lt;/strong&gt;&lt;br&gt;&#xA;I'm given two points in the form of Pareto 90/10 (green in the example above) or 80/20 (blue in the example above), and I know that the upper bound is one and the lower bound is zero.&lt;br&gt;&#xA;How do I get the formula of a sigmoid function (such as the logistic function) that has a lower asymptote on the left and higher asymptote on the right and passes via the two points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T08:36:38.767" LastActivityDate="2010-10-31T08:36:38.767" Title="Logistic function passing through two points?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8214" PostTypeId="2" ParentId="8213" CreationDate="2010-10-28T22:29:57.373" Score="7" ViewCount="0" Body="&lt;p&gt;I believe you're looking for constants $a$ and $b$ so that $f(x_\ell) = y_\ell$ and $f(x_u) = y_u$ where $f(x) = \exp(a + bx) / (1 + \exp(a + bx))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the linear system $a + b x_\ell = g(y_\ell)$ and $a + b x_u = g(y_u)$ where $g(y) = f^{-1}(y) = \log(y/(1-y))$.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-10-28T22:29:57.373" CommentCount="1" />
  <row Id="8215" PostTypeId="2" ParentId="3590" CreationDate="2010-10-28T22:32:10.090" Score="1" ViewCount="0" Body="&lt;p&gt;Since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(z)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,1\right)+\psi \left(z-1,\frac{1}{2}\right)\right)-\psi(z-1,1)\right)}{\ln(2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\psi(x,z)$ is the generalized polygamma following Espinosa's &lt;a href=&quot;http://www.math.tulane.edu/~vhm/papers_html/genoff.pdf&quot; rel=&quot;nofollow&quot;&gt;generalization&lt;/a&gt;, whatever we say about Zeta function we can also say about the right hand part of this identity. It consists only of Gamma function, its (fractional) derivatives and integrals.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2010-12-05T03:25:14.210" LastActivityDate="2010-12-05T03:25:14.210" CommentCount="2" />
  <row Id="8216" PostTypeId="2" ParentId="8105" CreationDate="2010-10-28T22:33:38.210" Score="1" ViewCount="0" Body="&lt;p&gt;You may be confusing Elsevier with Springer. &lt;a href=&quot;http://eom.springer.de/&quot; rel=&quot;nofollow&quot;&gt;The Springer Online Encyclopaedia of Mathematics&lt;/a&gt; is quite an impressive reference.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-28T22:33:38.210" />
  <row Id="8217" PostTypeId="1" AcceptedAnswerId="8218" CreationDate="2010-10-28T22:53:36.287" Score="2" ViewCount="54" Body="&lt;p&gt;I am working on a game program.&#xA;I have an array of 8 points in 3d space (X,Y,Z) that are the 8 corners of a cube whose W=H=D. The 8 points are listed in no particular order. For the sake of determining this algorithm, The cube's center is at the world origin, (0,0,0) but the cube's orientation is not aligned with the Cartesian axis (it's arbitrarily rotated). I'd like to create 6 sets of 4 points each that represent the faces of the cube but I'm stumped as to how to proceed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea of an algorithm that would help me? &lt;/p&gt;&#xA;" OwnerUserId="2915" LastActivityDate="2010-10-28T22:58:27.233" Title="Determine which of 8 points make up the 4 corners of a cube's face" Tags="&lt;linear-algebra&gt;&lt;linear-programming&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="8218" PostTypeId="2" ParentId="8217" CreationDate="2010-10-28T22:58:27.233" Score="4" ViewCount="0" Body="&lt;p&gt;Pick a point $p$, calculate the distance from that point to all the others, three will be the same length and smaller than the other lengths. Call these three points $x_1$, $x_2$, $x_3$. Do the same for $x_1$, $x_2$, $x_3$. Each pair of those will share two points, one will be $p$ the other will be $y_{12}$, $y_{13}$, $y_{23}$ respectively. You just defined three sides by points. Now, pick the only remaining point that hasn't been named. Call it $q$. Run the above algorithm. Done.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-28T22:58:27.233" CommentCount="4" />
  <row Id="8220" PostTypeId="1" CreationDate="2010-10-28T23:44:55.900" Score="7" ViewCount="133" Body="&lt;p&gt;This was suggested by a problem on FreeNode's #math a little while ago...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct a directed graph $\Gamma$ with vertex set the set of compact convex sets in $\mathbb R^2$, and an arrow $A\to B$ if $A$ and $B$ are disjoint and there is a point $a$ in $A$ and a $t&gt;0$ such that $a+(t,0)$ is in $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess there are no oriented cycles in $\Gamma$, so it is a directed acyclic graph. Can you give a non-ugly proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; Rahul's example below shows that this is too much to hope for. In fact, what I &lt;em&gt;really&lt;/em&gt; wanted is to know if the subgraph spanned by every finite set of disjoint convex sets has &lt;em&gt;no&lt;/em&gt; cycles.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-10-30T14:05:49.563" LastActivityDate="2010-10-30T14:05:49.563" Title="A tree of convex sets?" Tags="&lt;convex-sets&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="8221" PostTypeId="1" CreationDate="2010-10-29T00:02:54.290" Score="1" ViewCount="128" Body="&lt;p&gt;How does one go about solving the following quadratic congruence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4x^2 \equiv 2 \ (\text{mod} \ 7)$&lt;/p&gt;&#xA;" OwnerUserId="2868" LastEditorUserId="2868" LastEditorDisplayName="" LastEditDate="2010-10-29T00:24:42.700" LastActivityDate="2010-10-29T05:10:36.363" Title="Solving a quadratic congruence: $4x^2 \equiv 2 \ (\text{mod} \ 7)$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8222" PostTypeId="2" ParentId="8221" CreationDate="2010-10-29T00:14:41.687" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Multiply both sides by 2.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-29T00:14:41.687" />
  <row Id="8223" PostTypeId="2" ParentId="8221" CreationDate="2010-10-29T00:28:32.990" Score="1" ViewCount="0" Body="&lt;p&gt;Although not as useful in general, with a small modulus like $7$, one can let $x$ run through all possible congruence classes modulo $7$. Using Bill Dubuque's hint will make the mental calculation easier to see which $x$ actually satisfy the congruence.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-29T00:28:32.990" />
  <row Id="8224" PostTypeId="1" CreationDate="2010-10-29T00:36:59.367" Score="0" ViewCount="193" Body="&lt;p&gt;In a survey of political preference, 78% of those asked were in favor of at-least one of the proposals: I, II and III. 50% of those asked favored proposal I, 30% favored proposal II, and 20% favored proposal III. If 5% favored all the three proposals, what % of those asked favored more than one of the three proposals?&lt;/p&gt;&#xA;" OwnerUserId="2917" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-28T06:58:32.987" LastActivityDate="2010-11-28T16:16:04.023" Title="A homework problem involving counting using a special case of  inclusion/exclusion " Tags="&lt;combinatorics&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8225" PostTypeId="2" ParentId="8220" CreationDate="2010-10-29T00:39:04.527" Score="5" ViewCount="0" Body="&lt;p&gt;Am I misunderstanding the conditions of the problem or is this a counterexample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lFHHH.png&quot; alt=&quot;Two ellipses in an X, and two circles between them&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; I think the answer to your updated question is yes -- or rather, no, any subgraph induced by disjoint convex sets has no cycles. (However, that would make it a &lt;em&gt;directed acyclic graph&lt;/em&gt; or DAG, and not a tree, unless that's computer science terminology.) I have a sketch of what I think should be a valid proof, but filling in the details is a little ugly. Unfortunately, I think there is no completely non-ugly proof, for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The power of convexity is a little lost because separating hyperplanes are of no help here. If $A \rightarrow B \rightarrow C$, there is not necessarily a hyperplane separating $C$ from $A \cup B$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The proposition is not valid in more than 2 dimensions, so any proof must fundamentally depend on the 2D nature of the problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That said, here's the fairly ugly proof I got. Suppose for the sake of contradiction that there is a cycle of disjoint compact convex sets, $A_1 \rightarrow A_2 \rightarrow \cdots \rightarrow A_n \rightarrow A_1$. Each set $A_i$ contains a pair of points $p_i$ and $q_i$ such that $q_{i-1} \rightarrow p_i$ and $q_i \rightarrow p_{i+1}$ (where $i+1$ and $i-1$ are modulo $n$, of course). Therefore, it suffices to prove the impossibility of a cycle on the line segments $p_1 q_1 \rightarrow p_2 q_2 \rightarrow \cdots \rightarrow p_n q_n \rightarrow p_1 q_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the rightmost &quot;fringe&quot; of the first $i$ segments, as a function of $y$; that is, let $f_{i}(y) = \sup \{x : (x,y) \in p_1 q_1 \cup p_2 q_2 \cup \cdots \cup p_i q_i\}$. While $f_i$ is not continuous, I think you can prove by induction that if $p_1 q_1 \rightarrow p_2 q_2 \rightarrow \cdots \rightarrow p_i q_i \rightarrow p_{i+1} q_{i+1}$, then the discontinuities of $f_i$ are not &quot;visible&quot; to $p_{i+1}$. What I mean is that $f_i$ jumps leftwards rather than rightwards as you move in $y$ away from $p_{i+1}$, so there is no way for $q_{i+1}$ to sneak in to the left of $f_i$ without the segment $p_{i+1}q_{i+1}$ intersecting $f_i$ somewhere. Therefore, every $p_{i+1}$ and $q_{i+1}$ is to the right of $f_i$. In particular, $q_n$ is to the right of $f_{n-1}$. But $p_1$ is on or to the left of $f_{n-1}$, so $q_n \rightarrow p_1$ is impossible.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-30T07:28:44.270" LastActivityDate="2010-10-30T07:28:44.270" CommentCount="3" />
  <row Id="8226" PostTypeId="1" CreationDate="2010-10-29T00:53:43.227" Score="0" ViewCount="100" Body="&lt;p&gt;The question reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathfrak{p}$ be a prime ideal of $A$. Show that $A_\mathfrak{p}$ has a unique maximal ideal, consisting of all elements $a/s$ with $a \in \mathfrak{p}$ and $s \notin \mathfrak{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;$A_\mathfrak{p}$&quot; is $A/\mathfrak{p}$, then this is equivalent to saying $A$ is entire $\Rightarrow A$ has a unique maximal ideal, which is false, no? (consider $\mathbb{Z}$).&lt;/p&gt;&#xA;" OwnerUserId="654" LastActivityDate="2010-10-29T00:53:43.227" Title="What does this notation mean, in Lang's Algebra Ch2 Problem3?" Tags="&lt;abstract-algebra&gt;&lt;notation&gt;&lt;books&gt;" CommentCount="2" />
  <row Id="8227" PostTypeId="2" ParentId="6228" CreationDate="2010-10-29T01:05:57.240" Score="1" ViewCount="0" Body="&lt;p&gt;Following the advice of Weltschmerz, from 'The Classical Theory of Fields':&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;The total charge present in all of space is equal to the integral $\int \rho dV$ over all space. We can write this integral in four-dimensional form:&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, you have the equality you wrote: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int{\rho}dV = \displaystyle \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;where the integral is taken over the entire four-dimensional hyperplane perpendicular to the $x^{0}$ axis (clearly this integration means integration over the whole three dimensional space). Generally, the integral &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{1}{c}\int j^{i}dS_{i}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;over an arbitrary hypersurface is the sum of the charges whose world lines pass through this surface. &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then, $dS_{i}$ is not a surface in the conventional sense. Instead, is a hyper-surface. &lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-29T01:14:36.893" LastActivityDate="2010-10-29T01:14:36.893" />
  <row Id="8229" PostTypeId="2" ParentId="8140" CreationDate="2010-10-29T01:09:52.947" Score="3" ViewCount="0" Body="&lt;p&gt;Consider your graph is planarly embedded in the plane. And think of edge unoriented edge as a pair of oriented arcs in opposite directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick one oriented arc $a$. There is a unique face whose boundary, when drawn in counterclockwise direction, &quot;contains&quot; $a$, and in that boundary $a$ is followed by a well-defined arc $b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To construct the faces, it is enough to be able to decide for each $a$ what  the corresponding $b$ is. This can be done easily by sorting the edges incident to each vertex cyclically.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-29T01:09:52.947" CommentCount="2" />
  <row Id="8230" PostTypeId="1" AcceptedAnswerId="8238" CreationDate="2010-10-29T01:17:43.783" Score="1" ViewCount="61" Body="&lt;p&gt;I'd like to get the exact value of the following sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{\lceil \frac{k}{2} \rceil}{({k \choose i}\cdot i)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know the asymptotic limits of the function.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-10-29T04:13:44.660" Title="What is the sum of half of the choices (weighted)?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="8231" PostTypeId="2" ParentId="8224" CreationDate="2010-10-29T01:19:21.417" Score="0" ViewCount="0" Body="&lt;p&gt;hmm, I got 17, not 27...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we count |A|+|B|+|C|, then we overcount intersections of any 2 sets twice, and intersections of all three 3 times. Then, if we subtract $|A \cup B \cup C|$, we get the intersections of any 2 sets once and 3 sets 2 times. (most easily seen by drawing a Venn diagram). Subtracting the 3-set intersection once more gives 100-78-5=17 .&lt;/p&gt;&#xA;" OwnerUserId="654" LastActivityDate="2010-10-29T01:19:21.417" />
  <row Id="8233" PostTypeId="1" CreationDate="2010-10-29T01:30:28.827" Score="1" ViewCount="54" Body="&lt;p&gt;why does this algorithm fail to factor polynomials over Z/2Z, what would I need to do to fix this? &lt;/p&gt;&#xA;" OwnerUserId="2143" LastActivityDate="2010-10-29T01:30:28.827" Title="Cantor-Zassenhaus split" Tags="&lt;number-theory&gt;&lt;number&gt;" />
  <row Id="8234" PostTypeId="1" AcceptedAnswerId="8320" CreationDate="2010-10-29T02:26:27.343" Score="1" ViewCount="154" Body="&lt;p&gt;While reading Munkres' Topology section about integers and reals (Chapter 1, Section 4), he defines the set $\mathbb{Z}_+$ as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Definition: A subset $A$ of the real&#xA;  numbers is said to be inductive if it&#xA;  contains the number 1, and if for&#xA;  every $x$ in $A$, the number $x+1$ is&#xA;  also in $A$. Let $\cal{A}$ be the&#xA;  collection of all inductive subsets of&#xA;  $\mathbb{R}$. Then the set&#xA;  $\mathbb{Z}_+$ of positive integers is&#xA;  defined by the equation $\bigcap _{A&#xA;  \in \cal{A}} A$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Isn't it possible to define $\mathbb{Z}_+$ as the set defined this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\in B \wedge (j \in B, n-1 = j) \Rightarrow n\in B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $B = \mathbb{Z}_+$&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-10-29T22:52:41.660" Title="Defining $\mathbb{Z}_+$" Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8235" PostTypeId="1" AcceptedAnswerId="8574" CreationDate="2010-10-29T02:29:35.393" Score="9" ViewCount="317" Body="&lt;p&gt;Inspired by &#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/8186/when-do-the-multiples-of-two-primes-span-all-large-enough-natural-numbers/8187#8187&quot;&gt; this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be a subset of ${\mathbb Z}^d$ that generates the whole additive group ${\mathbb Z}^d$, and let $S$ be the additive semigroup generated by $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that there is a $d$-dimensional convex cone $C\subset {\mathbb R}^d$ so that&#xA;all elements of $C \cap {\mathbb Z}^d$ with sufficiently large norm are contained in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Feel free to retag!&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-04T11:07:01.237" Title="A multi-dimensional Frobenius problem  " Tags="&lt;discrete-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8236" PostTypeId="2" ParentId="4054" CreationDate="2010-10-29T03:11:06.173" Score="2" ViewCount="0" Body="&lt;p&gt;See Needham's &lt;a href=&quot;http://usf.usfca.edu/vca/&quot; rel=&quot;nofollow&quot;&gt;Visual Complex Analysis&lt;/a&gt; for lots of geometric intuition and much more.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-29T03:11:06.173" />
  <row Id="8237" PostTypeId="2" ParentId="8221" CreationDate="2010-10-29T03:21:15.660" Score="0" ViewCount="0" Body="&lt;p&gt;Since $2 \equiv 9 \pmod 7$, you have $(2x-3)(2x+3) = 4x^2-9 \equiv 0 \pmod 7$. Now use that $7$ is prime.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2010-10-29T05:10:36.363" LastActivityDate="2010-10-29T05:10:36.363" CommentCount="2" />
  <row Id="8238" PostTypeId="2" ParentId="8230" CreationDate="2010-10-29T04:13:44.660" Score="3" ViewCount="0" Body="&lt;p&gt;Use $\binom{k}{i} i = k \binom{k-1}{i-1}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=0}^{\lceil k/2 \rceil} \binom{k}{i} = k\sum_{j=0}^{\lceil k/2 \rceil - 1} \binom{k-1}{j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = 2m$ is even then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k\sum_{j=0}^{m-1} \binom{2m-1}{j} = k 2^{2m-2} = k2^{k-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = 2m+1$ is odd then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k\sum_{j=0}^m \binom{2m}{j} = k \left(2^{2m-1} +\frac{1}{2} \binom{2m}{m}\right) = k\left(2^{k-2} + \frac{1}{2} \binom{k-1}{(k-1)/2}\right)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the asymptotics, $\frac{k}{2} \binom{k-1}{(k-1)/2} \approx 2^{k-1} \sqrt{\frac{k}{2\pi}}$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-29T04:13:44.660" CommentCount="1" />
  <row Id="8239" PostTypeId="1" CreationDate="2010-10-29T04:20:10.690" Score="2" ViewCount="134" Body="&lt;p&gt;This is a problem I'm stuck on that our professor gave us for additional practice (not homework, but its recommended that we understand how to prove it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know X and Y are complete metric spaces, and we need to show that $X \times Y$ is complete. I'm really lost on the proof technique. We were given an outline as follows, but I could only fully figure out (1). Part 3 is what we've been really stuck on though.  I was wondering whether someone could give an proof for say a more specific space where $X = \mathbb{R}, Y = \mathbb{R}$, so I could understand the principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outlined: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Show that $d_{X \times Y} ( (a_1,b_1) , (a_2,b_2)) = \max \left\{ d_X (a_1,a_2) , d_Y (b_1, b_2) \right\}$ is the product metric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Prove that this gives the product topology on $X \times Y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Prove that  if  $a_n, b_n$ are Cauchy sequences, where $a_n \in X$ and $b_n \in Y$, then $(a_n,b_n )$ is Cauchy.&lt;/p&gt;&#xA;" OwnerUserId="2919" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-29T08:42:11.853" LastActivityDate="2010-10-29T08:55:23.043" Title="Cartesian Product of Two Complete Metric Spaces is Complete" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;topology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="8240" PostTypeId="2" ParentId="8239" CreationDate="2010-10-29T04:33:06.273" Score="2" ViewCount="0" Body="&lt;p&gt;For 3) if you are given an $\epsilon$, you can find $N$ such that $|a_n-a_m|&amp;lt;\epsilon \text{ if } n,m&gt;N$ and similarly for $M$ and the $b$ series.  Doesn't $\max{(N,M)}$ give a limit that $|(a_n,b_n)-(a_m,b_m)|&amp;lt;\epsilon$?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-29T04:33:06.273" />
  <row Id="8241" PostTypeId="1" AcceptedAnswerId="8246" CreationDate="2010-10-29T04:34:11.173" Score="4" ViewCount="86" Body="&lt;p&gt;Given are some positive integers $t_1, ..., t_n$ such that $\mathrm{gcd}(t_1, ..., t_n) = 1$. Now we create a linear combination of these: $k_1 t_1 + ... + k_n t_n$ where $k_1, ..., k_n$ are nonnegative integers that you can choose to generate different integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest $N$ such that all integers greater than or equal to $N$ can be generated by the linear combination?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $k_1 3 + k_2 4$ can generate $0, 3, 4, 6, 7, 8, 9, ...$. So here $N = 6$ and $k_1 = 2, k_2 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T15:49:17.480" LastActivityDate="2010-10-29T15:49:17.480" Title="Generating integers from a linear combination of integers" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8242" PostTypeId="2" ParentId="8239" CreationDate="2010-10-29T04:34:37.210" Score="2" ViewCount="0" Body="&lt;p&gt;If ${a_n}$, ${b_n}$ are Cauchy, then for some N, $d_X(a_n,a_m) &amp;lt; \epsilon$, $d_Y(b_n,b_m) &amp;lt; \epsilon$ when $n,m \geq N$. Because of this, $d_{X \times Y} ((a_n,b_n),(a_m,b_m)) &amp;lt; \epsilon$ for $n,m \geq N$. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-10-29T04:34:37.210" />
  <row Id="8243" PostTypeId="1" AcceptedAnswerId="8245" CreationDate="2010-10-29T04:43:20.223" Score="1" ViewCount="188" Body="&lt;p&gt;If $\gcd(a,b) = 1$, $x \equiv 0 \pmod{a}$, $x \equiv 0 \pmod{b}$ then prove that $x \equiv 0 \pmod{ab}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at answering the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;x &amp;amp;\equiv 0 \pmod{a}\\&#xA;&amp;amp;\Longrightarrow x\text{ is divisible by $a$}\\&#xA;&amp;amp;\Longrightarrow x = ma\text{ for some integer $m$}\\&#xA;\ \\&#xA;x &amp;amp;\equiv 0 \pmod{b}\\&#xA;&amp;amp;\Longrightarrow  x\text{ is divisible by $b$}\\&#xA;&amp;amp;\Longrightarrow x = mb\text{ for some integer $m$}\\&#xA;\ \\&#xA;x^2 &amp;amp;= (ma)(mb)\\&#xA;x^2 &amp;amp;= (m^2)(ab)\\&#xA;x &amp;amp;= \sqrt{m^2ab}\\&#xA;x &amp;amp;= m\sqrt{a}\sqrt{b}&#xA;\end{align*}&#xA;Let $m$ be $k\sqrt{a}\sqrt{b}$. Then&#xA;\begin{align*}&#xA;x &amp;amp;= kab\\&#xA;&amp;amp;\Longrightarrow x \equiv 0 \pmod{ab}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct, if not can someone point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T13:12:34.463" LastActivityDate="2010-12-04T13:12:34.463" Title="If $\gcd(a,b) = 1$, $x \equiv 0 \pmod{a}$, $x \equiv 0 \pmod{b}$ then prove that $x \equiv 0 \pmod{ab}$" Tags="&lt;number-theory&gt;&lt;proof&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="8244" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T04:52:38.480" Score="1" ViewCount="0" Body="&lt;p&gt;No. This is not correct.&lt;br&gt;&#xA;1. Integers $m_1$ and $m_2$ in $x=m_1 a$ and $x=m_2 b$ can be different.&lt;br&gt;&#xA;2. In the beginning of your proof all numbers are integers. However when you write &quot;let m be k*sqrt(a)*sqrt(b)&quot;, you don't know, that k is integer too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right direction: use, the fact, that each number $y$ can be uniquely represented as $p_1^{\alpha_1} p_2^{\alpha_2}\dots p_n^{\alpha_n}$, where $p_i$ are primes and $\alpha_i$ are integers.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-29T04:52:38.480" CommentCount="1" />
  <row Id="8245" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T04:57:40.680" Score="6" ViewCount="0" Body="&lt;p&gt;You are given that $a$ divides $x$. Therefore, you can write $x = ma$ where m is an integer. You are also given that $b$ divides $x$. This implies that $b$ divides $ma$. But $b$ and $a$ are coprime. Therefore $b$ must divide $m$. So you can write $m = kb$ where $k$ is another integer. Therfore, you have $x = kab$.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-10-29T04:57:40.680" CommentCount="1" />
  <row Id="8246" PostTypeId="2" ParentId="8241" CreationDate="2010-10-29T05:07:16.700" Score="4" ViewCount="0" Body="&lt;p&gt;This is a famous problem, sometimes called the coin problem of Frobenius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=2$ the answer to your question is known to be $N = t_1 t_2 - (t_1 + t_2) + 1$.  A proof of this can be found in the answer to this &lt;a href=&quot;http://math.stackexchange.com/questions/8186/&quot;&gt;recent question&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For three or more integers, there is no known closed-form solution for $N$.  There are some bounds on the values of $N$ in the $n = 3$ case, as well as some algorithms for determining $N$.  For more information and references, see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/CoinProblem.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt; pages on the Frobenius coin problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the problem is considered solved when $n = 2$, partially solved (because of the bounds and algorithms) when $n = 3$, and unsolved when $n \geq 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Guy's &lt;em&gt;Unsolved Problems in Number Theory&lt;/em&gt; says, &quot;The case $n = 3$ was first solved explicitly by Selmer and Beyer, using a continued fraction algorithm.&quot;  So I guess the $n=3$ case has been solved.  I suppose you have would have to dig up their paper (it's in the MathWorld references) to see their solution. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-29T15:46:37.763" LastActivityDate="2010-10-29T15:46:37.763" CommentCount="4" />
  <row Id="8247" PostTypeId="1" CreationDate="2010-10-29T05:19:58.367" Score="0" ViewCount="115" Body="&lt;p&gt;&lt;strong&gt;Suppose n is an integer, such that the sum of the digits of n is 2, and $10^{10} \lt n \lt 10^{11} $. The number of different values for n is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to list them :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1) 11000000000&#xA;(2) 10100000000&#xA;(3) 10010000000&#xA;...&#xA;(10)10000000001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I am getting 10 possible values but the answer is 11.What  am I missing here ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I was missing 20000000000.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-29T06:06:27.750" LastActivityDate="2010-10-29T06:06:27.750" Title="Simple number theory" Tags="&lt;number-theory&gt;" CommentCount="4" />
  <row Id="8248" PostTypeId="1" CreationDate="2010-10-29T07:02:12.220" Score="0" ViewCount="162" Body="&lt;p&gt;By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.  What is the 10001st prime number?&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2010-10-29T09:43:42.307" Title="How to find next prime number?" Tags="&lt;prime-numbers&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="8249" PostTypeId="1" CreationDate="2010-10-29T07:15:37.680" Score="2" ViewCount="52" Body="&lt;p&gt;Define $f$ on $[-1,1]$ by &#xA;$$f(x) = \left\{ \begin{array}{ll}&#xA;0 &amp;amp; \mbox{if $x\lt 0$;}\\&#xA;0 &amp;amp; \mbox{if $x=0$;}\\&#xA;1 &amp;amp; \mbox{if $x\gt 0$.}&#xA;\end{array}\right.$$&#xA;Let the integrator $a$ be defined by &#xA;$$a(x) = \left\{\begin{array}{ll}&#xA;0 &amp;amp; \mbox{if $x\lt 0$;}\\&#xA;0 &amp;amp; \mbox{if $x=0$;}\\&#xA;1 &amp;amp; \mbox{if $x\gt 0$.}&#xA;\end{array}\right.$$&#xA;Show that $f$ is Riemann-Stieltjes integrable on  $[-1,1]$ even though &#xA;$$\lim_{||P||\to 0} S(P,f,a)$$&#xA;does not exist.&lt;/p&gt;&#xA;" OwnerUserId="2920" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-29T16:30:45.890" LastActivityDate="2010-10-29T18:49:51.933" Title="to show a function is RS integrable even when lim S(P,f,a) does not exist" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8250" PostTypeId="1" CreationDate="2010-10-29T07:47:18.270" Score="9" ViewCount="424" Body="&lt;p&gt;Here is a statement of &lt;a href=&quot;http://en.wikipedia.org/wiki/Farkas_lemma&quot; rel=&quot;nofollow&quot;&gt;Farkas Lemma&lt;/a&gt; from the Wikipedia. Let $A$ be an $m \times n$ matrix and $b$ an $m$-dimensional vector. Then, exactly one of the following two statements is true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There exists an $x \in \mathbb{R}^n$ such that $Ax = b$ and $x \geq 0$.&lt;/li&gt;&#xA;&lt;li&gt;There exists a $y \in \mathbb{R}^m$ such that $A^T y \geq 0$ and $b^T y &amp;lt; 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This result has a simple geometric interpretation: either $b$ lies in the cone formed by the columns of $A$ or it is possible to find a vector $y$ such that $y$ forms an acute angle with all columns of $A$ and an obtuse angle with $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was wondering if there is a way to make the result intuitive purely at the algebraic level of solving linear equations and inequalities?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lemma is important in finance and the geometric intuition is not much help there since the vectors are payoffs and prices of financial assets which have no natural geometric meaning. Gale's &lt;em&gt;Theory of Linear Economic Models&lt;/em&gt; has a purely algebraic proof, but that is an opaque induction argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A little more about my application. We have $m$ assets and $n$ possible future states of nature. $A_{ij}$ is the payoff by asset $i$ in state $j$. $b_i$ is the price of asset $i$. $x_j$ is the value of one dollar in state $j$. $y_i$ is the amount of asset $i$ in a portfolio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Farkas' lemma tells us that either (1) there is a way of assigning a non-negative price to a dollar in each state in a way such that the price of each asset is just the sum total of the value of its payoffs, or (2) there is a portfolio whose price is negative, so you get paid for holding it, but whose payoffs are non-negative, which means that you do not have to pay anything back. I want to understand, in terms which make economic sense, why (1) and (2) should be mutually exclusive. Acute and obtuse angles are no help here.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastEditorUserId="1195" LastEditorDisplayName="" LastEditDate="2010-10-29T18:12:43.890" LastActivityDate="2010-11-05T22:18:18.113" Title="Farkas’ lemma: purely algebraic intuition" Tags="&lt;convex-analysis&gt;&lt;finance&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="8251" PostTypeId="2" ParentId="8239" CreationDate="2010-10-29T08:55:23.043" Score="2" ViewCount="0" Body="&lt;p&gt;As for (2), it suffices to prove that the balls produced by the distance $d_{X\times Y}$ are a basis for the product topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's write down what it is a ball for the distance $d_{X\times Y}$ with center $(a,b)$ and radius $\varepsilon &gt; 0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B_{X\times Y}((a,b); \varepsilon) = \left\{ (x,y) \in X\times Y \ \vert \ d_{X\times Y} ((a,b), (x,y))= \max  \left\{ d_X(a,x) , d_Y (b,y) \right\}   &amp;lt; \varepsilon \right\} \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\max  \left\{ d_X(a,x) , d_Y (b,y) \right\}   &amp;lt; \varepsilon \quad \Longleftrightarrow \quad d_X(a,x) &amp;lt;\varepsilon \ \text{and} \ d_Y(b,y) &amp;lt; \varepsilon \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B_{X\times Y }((a,b); \varepsilon)\  = \ B_X (a;\varepsilon ) \times B_Y (b; \varepsilon ) \ .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: a ball for the distance $d_{X \times Y}$ is the same as a product of balls for the distances $d_X$ and $d_Y$. Which means that balls for the distance $d_{X\times Y}$ form a basis for the product topology.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-29T08:55:23.043" />
  <row Id="8252" PostTypeId="2" ParentId="8248" CreationDate="2010-10-29T08:58:40.983" Score="2" ViewCount="0" Body="&lt;p&gt;It is 104743, according to &lt;a href=&quot;http://primes.utm.edu/lists/small/100000.txt&quot; rel=&quot;nofollow&quot;&gt;http://primes.utm.edu/lists/small/100000.txt&lt;/a&gt; and the related list of the first 10,000 primes ;-)&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-10-29T08:58:40.983" CommentCount="1" />
  <row Id="8253" PostTypeId="1" AcceptedAnswerId="8304" CreationDate="2010-10-29T09:05:35.963" Score="0" ViewCount="127" Body="&lt;p&gt;In the proof of Rouche's theorem I saw in a book, there are two points I failed to understand, or failed to prove myself. (if you aren't familiar with the theorem, please try to look at the two statements here below and explain them regardless of the theorem):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Consider the complex plane. It's not difficult to show that the index of a curve C with respect to a vector field v is equal to the sum of indices of singular points, i.e., those at which v(z) = 0.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;If f(z) is a polynomial, and let v(z) = f(z), where v is a vector field, then the index of the singular point z0 is equal to the multiplicity of the root z0 of f.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am having trouble understanding this, can anyone please help?&lt;/p&gt;&#xA;" OwnerUserId="1214" LastEditorUserId="1214" LastEditorDisplayName="" LastEditDate="2010-10-29T09:19:01.707" LastActivityDate="2010-10-29T18:56:59.080" Title="Polynomials, Rouche's theorem and index of vector fields" Tags="&lt;complex-analysis&gt;&lt;algebra&gt;&lt;polynomials&gt;&lt;plane-curves&gt;&lt;roots&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8255" PostTypeId="2" ParentId="8248" CreationDate="2010-10-29T09:43:42.307" Score="1" ViewCount="0" Body="&lt;p&gt;I assume you want to program this? Then the question belongs in a CS forum. However, I will try to motivate how to program this using properties of prime numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For a number you &lt;em&gt;could&lt;/em&gt; try to divide it by every single positive integer smaller than the respective number (except for 1). This is stupid, as you obviously could just search through the odd numbers (checking 2 first). But can you do even better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Do you have to stop at n-1, or can you stop way earlier? You can. Find out where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now count primes.&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-10-29T09:43:42.307" CommentCount="3" />
  <row Id="8256" PostTypeId="1" CreationDate="2010-10-29T09:48:24.633" Score="2" ViewCount="134" Body="&lt;p&gt;I was recently asked to help someone with the following question on their first year analysis course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that  $\mathbb{N}$ is the set of all positive integers. Use the principle of induction to show that $n \ge 1$ for all $ n \in \mathbb{N}.$ [ Hint: Let $ S = \lbrace n \in \mathbb{N} | n \ge 1 \rbrace.$ ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am concerned that the first sentence of the question assumes that we already know exactly what the positive integers are, which renders the rest of the question redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer to see the question written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\mathbb{N}$ is the set of $ x \in \mathbb{R}$ such that $x$ is a member of every inductive set. [ $P$ is an inductive set if (a) $ 1 \in P $ and (b) $ x \in P \Rightarrow x+1 \in P.$ ] Then we can answer as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $S$ as in the hint, by definition $ S \subseteq \mathbb{N}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 \in S$ since $1 \in \mathbb{N},$ as it is an inductive set.&#xA;For $ n \in S, $ $n+1 &gt; n \ge 1,$ therefore $n+1 \in S.$&#xA;Hence $S$ is an inductive set, and so  $ \mathbb{N} \subseteq S.$ Therefore $ S=\mathbb{N}.$&#xA;Therefore $1$ is the least element of $\mathbb{N}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being way too picky, or is the question flawed as it stands?&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-29T13:00:46.093" LastActivityDate="2010-10-29T18:09:14.360" Title="Is this analysis problem involving induction flawed?" Tags="&lt;analysis&gt;&lt;induction&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8257" PostTypeId="1" CreationDate="2010-10-29T10:23:16.027" Score="0" ViewCount="75" Body="&lt;p&gt;&lt;strong&gt;How many even positive integers are there that are divisors of 720 ?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to compute the number of divisors but how to compute the number of &lt;strong&gt;even or odd positive divisors of a number&lt;/strong&gt; ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we list the divisors of 720 (using mathematica) : &lt;strong&gt;{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45,48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 720}&lt;/strong&gt; among these only &lt;strong&gt;24&lt;/strong&gt; are even,I am looking for some tricks that can be used in solving similar kinds of problems during exam (under a minute solution).&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-10-29T10:58:37.557" Title="How many even positive integers are there that are divisors of 720 ?" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8258" PostTypeId="2" ParentId="8257" CreationDate="2010-10-29T10:28:28.007" Score="1" ViewCount="0" Body="&lt;p&gt;You have an even number given, so just cut it in half and count the divisors of this.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-29T10:28:28.007" CommentCount="4" />
  <row Id="8259" PostTypeId="1" AcceptedAnswerId="8276" CreationDate="2010-10-29T10:39:05.123" Score="0" ViewCount="56" Body="&lt;p&gt;Hi everybody!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please can you explain me why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla^2 G_{\omega}(R) = f(R) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{R} \frac{d^2(RG_\omega(R))}{d R^2} = f(R) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G_\omega(R)$ is the Fourier transform of $G$&lt;/li&gt;&#xA;&lt;li&gt;$G$ is the Green's function&lt;/li&gt;&#xA;&lt;li&gt;We suppose that $G$ and $f$ are spherically symmetric&lt;/li&gt;&#xA;&lt;li&gt;$R = \sqrt{x^2 + y^2 + z^2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2354" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T10:46:56.343" LastActivityDate="2010-10-29T13:15:27.627" Title="Laplacian and spherical symmetry" Tags="&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="8260" PostTypeId="2" ParentId="8257" CreationDate="2010-10-29T10:39:52.423" Score="0" ViewCount="0" Body="&lt;p&gt;1*2*2*2*2*3*3*5. Odd divisors: (1,3,3,5) =&gt; (1,3,5,9,15,45). &#xA;Let A = (1,3,5,9,15,45). Then all factors are, A, 2*A, 4*A, 8*A, 16*A. &lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-10-29T10:39:52.423" CommentCount="1" />
  <row Id="8261" PostTypeId="2" ParentId="4054" CreationDate="2010-10-29T10:50:27.307" Score="2" ViewCount="0" Body="&lt;p&gt;I don't have enough reputation to add a comment, but this is just a minor add-on to KCd's answer. He points out that the whole thing reduces to computing integrals of the form $\int z^m dz$, where &lt;em&gt;m&lt;/em&gt; is an integer. Certainly one can compute these integrals by polar coordinates, but the following way to think about this computation is in my opinion closer to geometric intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, when $m \neq -1$ the function $z^m$ has a single-valued antiderivative  $z^{m+1}/(m+1)$ on $\mathbb C \setminus \{0\}$, and so the integral of $z^m$ over &lt;em&gt;any&lt;/em&gt; closed loop in $\mathbb C \setminus \{0\}$ vanishes, even though the punctured plane is not simply connected. The only &lt;em&gt;m&lt;/em&gt; for which one fails to get a single-valued antiderivative is $m=-1$, in which case one gets the multi-valued function $\log z$. The value of $\log z$ for two ''adjacent'' branches always differs by exactly $2\pi i$, so the integral of $1/z$ over a closed loop in $\mathbb C \setminus \{0\}$ just gives you $2\pi i$ times the winding number of the loop around the origin, i.e. the total number of times it has ''jumped'' to another branch in the counterclockwise direction. In the assumptions of your theorem the boundary &lt;em&gt;C&lt;/em&gt; will traverse the singularity once counterclockwise, which gives the result.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-10-29T10:50:27.307" />
  <row Id="8262" PostTypeId="2" ParentId="8257" CreationDate="2010-10-29T10:58:37.557" Score="2" ViewCount="0" Body="&lt;p&gt;An even number $2m$ is a factor of $720$ iff $m$ is a factor of $360$.&#xA;So it's the same problem as counting divisors of $360$. For&#xA;that it helps to consider the prime factorization of $360$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-29T10:58:37.557" />
  <row Id="8263" PostTypeId="1" CreationDate="2010-10-29T11:01:17.350" Score="0" ViewCount="141" Body="&lt;p&gt;1.What should be added to $15^{81}$ to make it divisible by 7 ? Suggest a general technique to counter problems like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.For how many two-digit numbers, the sum of the digits is greater than the product of the digits?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-10-29T11:07:28.390" LastActivityDate="2010-12-22T14:44:13.733" Title="Some problems in Number theory" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8264" PostTypeId="1" AcceptedAnswerId="8271" CreationDate="2010-10-29T11:06:31.703" Score="0" ViewCount="104" Body="&lt;p&gt;I'm a student and I'm struggling to understand the basics of discrete maths. If such a noob question wont offend you, please help me understand the equation and how to prepare a truth table for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p \rightarrow (\neg q \lor \neg r) \land \neg p \land q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use any substitution or such, explain how you derived them too .&lt;/p&gt;&#xA;" OwnerUserId="2925" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T11:15:26.450" LastActivityDate="2010-10-29T12:46:04.857" Title="Provide truth table" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8265" PostTypeId="1" AcceptedAnswerId="8267" CreationDate="2010-10-29T11:15:11.473" Score="1" ViewCount="78" Body="&lt;p&gt;Hi, I have the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that no uncountable subset of $P(\omega)$ is well-ordered by the inclusion relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think they want me to do by embedding it in a separable complete dense linear order (like $\mathbb{R}$), which I have already proved cannot have any uncountable well-ordered subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints or ideas are welcome.&lt;/p&gt;&#xA;" OwnerUserId="2926" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-29T11:27:06.643" LastActivityDate="2010-10-29T11:53:06.993" Title="Showing that well-ordered subsets of $P(\omega)$ are countable" Tags="&lt;set-theory&gt;&lt;order-theory&gt;" AnswerCount="2" />
  <row Id="8266" PostTypeId="2" ParentId="8263" CreationDate="2010-10-29T11:20:22.480" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; For the first one use the fact that $15^{2} \equiv 1 \ (\text{mod} \ 7)$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-29T11:20:22.480" CommentCount="7" />
  <row Id="8267" PostTypeId="2" ParentId="8265" CreationDate="2010-10-29T11:36:34.403" Score="2" ViewCount="0" Body="&lt;p&gt;Say $X$ is a well-ordered subset of $P(\omega)$.&#xA;For each $x \in X$ which is not the biggest element of $X$, let $y$ be the smallest amongst the elements of $X$ which are strictly bigger than $x$, and choose $f(x) \in y \setminus x$.&#xA;That way you get an injective ($x$ is the biggest element of $X$ not containing $f(x)$) map from $X$ (possibly minus one element) to $\omega$, which is countable.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-10-29T11:36:34.403" />
  <row Id="8268" PostTypeId="2" ParentId="8264" CreationDate="2010-10-29T11:40:47.053" Score="0" ViewCount="0" Body="&lt;p&gt;When first doing truth tables, it's best to break a sentence into its constituent parts. Here the propositions are $p,q,r$, so have the first three columns of your truth table be possible truth values for these. Then the remaining columns should be things such as $\neg q$, $\neg r$, $\neg q \vee\neg r$, $\neg p\wedge q$, $p\to (\neg q\vee\neg r)$, finally building up to $p\to (\neg q\vee\neg r)\wedge\neg p\wedge q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get started, the possible triples for $(p,q,r)$ are $(T,T,T),(T,T,F),(T,F,T),(F,T,T),(T,F,F),(F,T,F),(F,F,T),(F,F,F)$, and the rest of the table should follow easily from there. and the sentence should be easier to understand in retrospect, now that you see how each part affects each other part.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-29T12:46:04.857" LastActivityDate="2010-10-29T12:46:04.857" />
  <row Id="8269" PostTypeId="2" ParentId="8263" CreationDate="2010-10-29T11:52:30.457" Score="1" ViewCount="0" Body="&lt;p&gt;A good approach to these is to reduce your base number according to the modulus. So since $15\equiv 1\pmod{7}$, you would have $15^{89}\equiv 1^{89}\equiv 1\pmod{7}$. In other words, $15^{89}$ is 1 greater than a multiple of $7$, or 6 less than a multiple of $7$. So to get a complete solution set, you could add any integer of the form $6+7k$, or $-1+7k$, and these two sets coincide. So you can take either what must be subtracted or added to get the number to a multiple of your modulus. For example, for $29^2=841$, you could either add 4 to get 845, or subtract 1 to get 840, both of which are divisible by 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, as was noted in the comments, $29^2\equiv(-1)^2\equiv 1\pmod{5}$, and you should be able to get a complete solution set by the same reasoning. I just mentioned $29\equiv -1$ since $(-1)^2$ is nicer to deal with than $4^2$, and you would have to reduce 16 modulo 5 anyway.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-10-29T11:52:30.457" />
  <row Id="8270" PostTypeId="2" ParentId="8265" CreationDate="2010-10-29T11:53:06.993" Score="1" ViewCount="0" Body="&lt;p&gt;I assume by &quot;inclusion&quot; that you mean the relation A&amp;nbsp;&amp;sube;&amp;nbsp;B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a well-ordered subset &amp;sigma;&amp;nbsp;&amp;sube;&amp;nbsp;P(&amp;omega;). Clearly, with respect to inclusion, &amp;sigma; has some ordinal type; the question is whether this type is countable. For any a,b&amp;nbsp;&amp;isin;&amp;nbsp;&amp;sigma; such that a&amp;nbsp;&amp;sub;&amp;nbsp;b, we have |b&amp;nbsp;\&amp;nbsp;a|&amp;nbsp;&amp;ge;&amp;nbsp;1. Thus, the cardinality of V&amp;nbsp;=&amp;thinsp;[&amp;nbsp;&amp;cup;&amp;nbsp;&amp;sigma;&amp;nbsp;]&amp;thinsp;&amp;sube;&amp;nbsp;&amp;omega; is at least that of the order type of &amp;sigma;, and at most that of &amp;omega;.&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-29T11:53:06.993" />
  <row Id="8271" PostTypeId="2" ParentId="8264" CreationDate="2010-10-29T12:20:42.320" Score="4" ViewCount="0" Body="&lt;p&gt;This answer is meant to elaborate on yunone's earlier answer. I also assume an order of precedence so that &quot;IF a THEN b&quot; takes lower precedence than &quot;a AND b&quot; or &quot;a OR b&quot; &amp;mdash; that is, I assume that the expression you must evaluate is p&amp;nbsp;→&amp;nbsp;((¬q&amp;nbsp;∨&amp;nbsp;¬r)&amp;nbsp;∧&amp;nbsp;¬p&amp;nbsp;∧&amp;nbsp;q).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw out the following table (drawn below in monospaced unicode): spaces in between columns are just to indicate separation because ASCII double-lines look ugly.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; p | q | r    ¬q | ¬r | ¬q ∨ ¬r    ¬p | ¬p ∧ q | (¬q ∨ ¬r) ∧ ¬p ∧ q    p → ((¬q ∨ ¬r) ∧ ¬p ∧ q)&#xA;===|===|===  ====|====|=========  ====|========|====================  ==========================&#xA; F | F | F       |    |               |        |&#xA; F | F | T       |    |               |        |&#xA; F | T | F       |    |               |        |&#xA; F | T | T       |    |               |        |&#xA; T | F | F       |    |               |        |&#xA; T | F | T       |    |               |        |&#xA; T | T | F       |    |               |        |&#xA; T | T | T       |    |               |        |&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The columns pre-filled with T and F describe, in each row, a different possible setting of your variables p, q, and r. From these, you can evaluate other expressions such as ¬q and ¬r, use these to evaluate other expressions such as (¬q&amp;nbsp;∨&amp;nbsp;¬r), and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps, introduce auxiliary variables for expressions that you have to evaluate along the way. For instance, you can define A&amp;nbsp;=&amp;nbsp;¬q&amp;nbsp;∨&amp;nbsp;¬r&amp;nbsp; and &amp;nbsp;B&amp;nbsp;=&amp;thinsp;&amp;thinsp;¬p&amp;nbsp;∧&amp;nbsp;q&amp;nbsp;, which will help you to evaluate the expression A&amp;nbsp;∧&amp;nbsp;B&amp;thinsp;&amp;thinsp;=&amp;thinsp;&amp;thinsp;(¬q&amp;nbsp;∨&amp;nbsp;¬r)&amp;nbsp;∧&amp;nbsp;¬p&amp;nbsp;∧&amp;nbsp;q&amp;nbsp;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, just break down the expression into pieces that you can manage, and re-assemble the pieces to get the whole. (Which is a high-level description of mathematical problem solving in general, of course)&lt;/p&gt;&#xA;" OwnerUserId="439" LastActivityDate="2010-10-29T12:20:42.320" />
  <row Id="8272" PostTypeId="1" CreationDate="2010-10-29T12:35:28.253" Score="0" ViewCount="58" Body="&lt;p&gt;Let $G$ be a topological group and $G^{[0,1]}$ the set of paths $[0,1]\rightarrow G$ with the compact-open topology. This should make $G^{[0,1]}$ a topological group. I'd like to be able to refer to a simple proof that multiplication $p\ast q(t)=p(t)q(t)$ of paths is continuous without having to give all the details. I'm not so much interested in the proof itself as I am in a reference. Any suggestions for texts that this (or a generalization) shows up in?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastActivityDate="2010-10-29T12:35:28.253" Title="Continuity of multiplication in Map([0,1],G)" Tags="&lt;topology&gt;&lt;reference-request&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="8273" PostTypeId="1" AcceptedAnswerId="8282" CreationDate="2010-10-29T12:37:25.743" Score="3" ViewCount="117" Body="&lt;p&gt;Hello, I'm learning about continued fractions, and I've enjoyed them so far, but I'm unsure if I've done the following correctly. I have no real experience with analysis, so I'm not sure if my reasoning is formal enough, or correct. Any feedback would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\xi$ be an irrational number with continued fraction expansion $\langle a_0,a_1,a_2,a_3\dots\rangle$. Let $b_1,b_2,b_3,\cdots$ be any sequence of positive integers, either finite or infinte. Prove that $\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n,b_1,b_2,b_3\dots\rangle=\xi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I let $r_n=\langle a_0,a_1,\dots,a_n\rangle$ and $\xi'=\langle b_1,b_2,b_3,\dots\rangle$, and let $\beta_n=\langle a_0,a_1,a_2,a_3\dots,a_n,\xi'\rangle$. So* &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta_n-r_n=\beta_n-\frac{h_n}{k_n}=\frac{\xi'h_n+h_{n-1}}{\xi'k_n+k_{n-1}}-\frac{h_n}{k_n}$$&#xA;$$=\frac{-(h_nk_{n-1}-h_{n-1}k_n)}{k_n(\xi'k_n+k_{n-1})}=\frac{(-1)^n}{k_n(\xi'k_n+k_{n-1})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ${k_n}$ is a positive increasing series, $\xi'$ is a positive real number, so as $n$ approaches $\infty$, the denominator goes to $\infty$ while the numerator alternates between $-1$ and $1$, so the fraction tends to $0$. Hence we have that $\lim_{n\to\infty}(\beta_n-r_n)=0$, so $$\lim_{n\to\infty}\beta_n=\lim_{n\to\infty}r_n=\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n\rangle=\langle a_0,a_1,a_2\dots\rangle=\xi.$$ Hence $\lim_{n\to\infty}\langle a_0,a_1,a_2,\dots,a_n,b_1,b_2,b_3\dots\rangle=\xi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this the correct route to go? As a small side question, how does it make sense to have integers $b_i$ at the end of this sequence, if the sequence is infinite? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*If it's not well known, $\{h_n\}$ is the sequence defined by $h_{-2}=0,h_{-1}=1,h_i=a_ih_{i-1}+h_{i-2}$ and $\{k_n\}$ is defined as $k_{-2}=1,k_{-1}=-1, k_i=a_ik_{i-1}+k_{i-2}$, and $r_n=\langle a_0,a_1,\dots,a_n\rangle$, for any sequence of integers $a_0,a_1,a_2\dots$ all positive except perhaps $a_0$. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-10-30T02:23:07.807" LastActivityDate="2010-10-30T18:19:01.090" Title="Adding integers to an infinite continued fraction expansion doesn't change the value?" Tags="&lt;number-theory&gt;&lt;continued-fractions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8275" PostTypeId="1" AcceptedAnswerId="8281" CreationDate="2010-10-29T13:03:29.780" Score="0" ViewCount="46" Body="&lt;p&gt;Suppose $S^n$ is the n-sphere with basepoint $x$. If based map $f:S^n\rightarrow S^n$ is such that the pull-back $f^{-1}(S^{n}-x)$ is connected do we necessarily have that $deg(f)$ is either -1,0, or 1?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastEditorUserId="2423" LastEditorDisplayName="" LastEditDate="2010-10-29T13:59:02.127" LastActivityDate="2010-10-29T14:02:40.330" Title="Degrees of certain maps of spheres" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8276" PostTypeId="2" ParentId="8259" CreationDate="2010-10-29T13:15:27.627" Score="3" ViewCount="0" Body="&lt;p&gt;If $H(R)$ is spherically symmetric, we can write the Laplacian of $H$ in spherical coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace_operator&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = R^2\frac{d^2H}{dR^2} + 2R \frac{dH}{dR}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other way, $\frac{d}{dR}(RH(R)) = R\frac{dH}{dR} + H$&#xA;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2}{dR^2}(RH(R)) = R\frac{d^2H}{dR^2} + 2\frac{dH}{dR}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \frac{1}{R^2}\frac{d}{dR}\bigg(R\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply it to $G_{\omega}$.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-10-29T13:15:27.627" CommentCount="1" />
  <row Id="8277" PostTypeId="1" AcceptedAnswerId="8278" CreationDate="2010-10-29T13:18:14.783" Score="2" ViewCount="105" Body="&lt;p&gt;How to prove that if you take any 400 consecutive Octobers then exactly 14 % of those years have five Fridays and Sundays?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-10-29T14:13:12.973" LastActivityDate="2010-10-29T16:47:36.007" Title="Five Fridays and Sundays on October" Tags="&lt;recreational-mathematics&gt;" AnswerCount="2" />
  <row Id="8278" PostTypeId="2" ParentId="8277" CreationDate="2010-10-29T13:30:35.807" Score="2" ViewCount="0" Body="&lt;p&gt;To have five Fridays and Sundays, October must start on Friday.  A span of 400 years has 146097 days in it-remember the years that are multiples of 100 but not 400 are not leap years.  This is a multiple of 7.  So the pattern repeats.  Count how many of the 400 start October with Thursday or Friday.  You should find 56.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrected-used to say start on Thursday or Friday. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-10-29T16:47:36.007" LastActivityDate="2010-10-29T16:47:36.007" CommentCount="4" />
  <row Id="8279" PostTypeId="2" ParentId="8273" CreationDate="2010-10-29T13:39:42.047" Score="3" ViewCount="0" Body="&lt;p&gt;Your work looks good to me.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does make sense to have the $b_n$ at the end of the sequence as each stage of the sequence is finite.  This is an important point.  At each stage, the list of $a$'s is finite, so it makes sense to have more numbers (of any sort) at the end.  But as you let $n \rightarrow \inf $ they get pushed farther and farther out.  The way continued fractions work, they have less and less impact, as you showed.  So, &quot;in the limit&quot;, they don't matter.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-30T18:19:01.090" LastActivityDate="2010-10-30T18:19:01.090" CommentCount="1" />
  <row Id="8280" PostTypeId="1" AcceptedAnswerId="8283" CreationDate="2010-10-29T13:55:44.033" Score="0" ViewCount="52" Body="&lt;p&gt;If 72% percent of new apps are reviewed within the last 7 days, then there is a 72% chance that any app is reviewed in the last week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$72\%  = \sum\limits_{i = 0}^{i = 6} {x\left( {1 - x} \right)^i }$$&lt;br&gt;&#xA;          where $x$ represents the probability that an app is reviewed each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p = \left( {1 - x} \right)$, then the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\eqalign{&#xA;  &amp;amp; 72\%  = x + xp + xp^2  + xp^3  + xp^4  + xp^5  + xp^6   \cr &#xA;  &amp;amp; 72\%  = x\left( {1 + p + p^2  + p^3  + p^4  + p^5  + p^6 } \right)  \cr &#xA;  &amp;amp; 72\%  = 1 - p^7   \cr &#xA;  &amp;amp; p^7  = 1 - 72\%   \cr &#xA;  &amp;amp; p^7  = 28\%   \cr &#xA;  &amp;amp; \left( {1 - x} \right)^7  = 28\%   \cr &#xA;  &amp;amp; 1 - x \approx 0.8337  \cr &#xA;  &amp;amp; x \approx 16.63\%  \cr} &#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my work correct?&lt;/p&gt;&#xA;" OwnerUserId="2929" LastActivityDate="2010-10-29T14:09:11.293" Title="Apple App Store Review Rate" Tags="&lt;probability&gt;&lt;expectation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="8281" PostTypeId="2" ParentId="8275" CreationDate="2010-10-29T14:02:40.330" Score="3" ViewCount="0" Body="&lt;p&gt;Look at $S^2$ as the result of adding $\infty$ to $\mathbb C$, and let $f:S^2\to S^2$ be the map $z\mapsto z^2$. Then the preimage of $S^2-z$ is connected for all $z\in S^2$ (because the preimage of $z$ is finite). The degree is, of course, 2.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-10-29T14:02:40.330" CommentCount="1" />
  <row Id="8282" PostTypeId="2" ParentId="8273" CreationDate="2010-10-29T14:03:30.247" Score="4" ViewCount="0" Body="&lt;p&gt;You are on the right track, but some statements do not make any sense at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;Since $\xi'$ is a sequence of positive integers&quot; : This does not make any sense. $\xi'$ is a constant real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You are using confusing notation. $\beta = \langle a_0, a_1, \dots, a_n, \xi' \rangle$. This varies with $n$, so you need to talk about $\beta_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &quot;Hence we have $\lim_{n \to \infty} (\xi' -r_n) = 0$&quot; is not right. This implies that any two real numbers are equal (think about it)! You need to consider $\lim_{n \to \infty} \beta_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to you side question, it is not one sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a sequence of sequences!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0, a_1, b_1, b_2, b_3, \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0, a_1, a_2, b_1, b_2, b_3, \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0, a_1, a_2, \dots, a_n, b_1, b_2, b_3 \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that in each sequence, we only had a finite number of the $a_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each sequence corresponds to a real number (the $\beta_n$ above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we get a new sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta_1, \beta_2, \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the problem is asking you to prove that $\lim_{n \to \infty} \beta_n = \xi$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-10-29T14:37:43.720" LastActivityDate="2010-10-29T14:37:43.720" CommunityOwnedDate="2010-10-29T14:18:54.353" CommentCount="2" />
  <row Id="8283" PostTypeId="2" ParentId="8280" CreationDate="2010-10-29T14:09:11.293" Score="2" ViewCount="0" Body="&lt;p&gt;Looks good.  You have made the assumption, probably incorrect, that the probability of an app being reviewed is independent of the time since release and independent of whether it has already been reviewed.  But for math problems (as opposed to the real world) we often assume such.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-10-29T14:09:11.293" />
  <row Id="8284" PostTypeId="1" CreationDate="2010-10-29T14:16:09.110" Score="1" ViewCount="34" Body="&lt;p&gt;&lt;strong&gt;Formulation of the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B_t$ denote the standard Brownian Motion on the interval $[0,1]$. For a given confidence level $\alpha \in (0,1)$ the confidence band on $[0,1]$ is a function $u$ with the property $P(\omega; |W(\omega)_t| &amp;lt; u(t) \quad \forall t\in [0,1])=\alpha$. In other words, the probability that a path of the Brownian Motion stays within a confidence band is $\alpha$. Additionally the boundary hitting position for those paths leaving the band must be uniformly distributed on $[0,1]$. This condition can be stated using the stopping time $\tau(\omega) = \inf [ t \in [0,1], W_t(\omega)=u(t) ]$. $\tau $ is the position of the first hitting, and is uniform on $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;References and links to literature or papers considering this or similar problems&lt;/li&gt;&#xA;&lt;li&gt;Thoughts, ideas, discussion&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The context of the problem is a rather boring one, so will not state it here. The problem itself seem to be non-trivial and interesting.&lt;/p&gt;&#xA;" OwnerUserId="731" LastActivityDate="2010-10-29T14:16:09.110" Title="Confidence band for Brownian Motion with uniformly distributed hitting position" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;stochastic-processes&gt;" CommentCount="2" />
  <row Id="8285" PostTypeId="2" ParentId="8035" CreationDate="2010-10-29T14:27:29.533" Score="0" ViewCount="0" Body="&lt;p&gt;Thanks everyone! The model answer I was given is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;False. Roughly speaking, any polynomial can be made as large as you like by taking x to be very large (provided it is of degree greater than zero), whereas | cos x| &amp;lt;= 1 . There is obviously no polynomial of degree zero (i.e. no constant number) for which the statement holds.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from &lt;em&gt;Advanced Problems in Core Mathematics&lt;/em&gt; by Siklos. I am not very worried about the details, I am more than happy to be broadly right which I am on this occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again, everyone!&lt;/p&gt;&#xA;" OwnerUserId="2846" LastActivityDate="2010-10-29T14:27:29.533" />
  <row Id="8286" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T14:30:10.300" Score="2" ViewCount="0" Body="&lt;p&gt;Since $\rm\ \gcd(a,b) = 1\:,\ $ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's Lemma,&lt;/a&gt; $\rm\ \ a\:|\:b\:(x/b)\ \Rightarrow\ a\:|\:x/b\ \Rightarrow\ ab\:|\:x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively $\rm\ \ b,a\:|\:x\ \Rightarrow\ ab\: |\: ax,bx\ \Rightarrow\ ab\ |\ gcd(ax,bx)\ =\ x\ gcd(a,b)\ =\ x $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the special case $\rm\ gcd(a,b) = 1\ $ of $\rm\ gcd(a,b)\ lcm(a,b)\ =\ ab\ $ which has a &lt;a href=&quot;http://math.stackexchange.com/questions/3149&quot;&gt;similar proof.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T06:38:50.247" LastActivityDate="2010-12-03T06:38:50.247" />
  <row Id="8287" PostTypeId="2" ParentId="8277" CreationDate="2010-10-29T14:52:00.510" Score="0" ViewCount="0" Body="&lt;p&gt;As people have said, the best way to do this is brute force, once you know that the calendar repeats with period 400 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd note that to have five Fridays and five Sundays, October must &lt;i&gt;start&lt;/i&gt; on a Friday (like it does this year), and therefore run from Friday, October 1 to Sunday, October 31. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it feels like cheating to use date rules that somebody has already computed, you could (for example) note that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;this year, October 1 was a Friday, the sixth day of the week;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if October 1, year $n$ is the $k$th day of the week, then October 1, year $n+1$ is either the $k+1$st or $k+2$nd day of the week, depending on whether $n+1$ is an ordinary year or a leap year. (Of course the $8$th and $9$th days of the week can be converted to the $1$st and $2$nd days.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, it's easy to see that the answer should be &quot;approximately 1/7&quot; because the calendar rules don't make reference to any particular days of the week, although of course this is not a proof.  This is why the claim that went around the Internet earlier this month that an October like the current one only occurs every 823 years was so laughable; see &lt;a href=&quot;http://www.geekosystem.com/5-fridays-saturdays-sundays-not-823-years/&quot; rel=&quot;nofollow&quot;&gt;here for a debunking&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-29T14:52:00.510" />
  <row Id="8288" PostTypeId="1" AcceptedAnswerId="8330" CreationDate="2010-10-29T15:35:17.317" Score="3" ViewCount="118" Body="&lt;p&gt;Is it possible to divide an equilateral triangle into 5 equal (i.e., obtainable&#xA;from each other by a rigid motion) parts?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-30T01:54:01.897" Title="Splitting equilateral triangle to 5 parts" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="8289" PostTypeId="1" AcceptedAnswerId="8294" CreationDate="2010-10-29T15:35:34.140" Score="2" ViewCount="130" Body="&lt;p&gt;I need to find the number of natural numbers between 1 and 1000 that are divisible by 3, 5 or 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that given a set of numbers, 1 ... n, the number of numbers divisible by d is given by $\lfloor \frac{n}{d} \rfloor$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we let $A$ be the set of numbers divisible by 3, $B$ be the set of numbers divisible by 5 and $C$ be the set of numbers divisible by 7, then the size of the union of those sets is going to be the number required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to work out $A \cup B \cup C$, I need to know $A \cap B$ (and $A \cap C$ etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only thought was that to work out $A \cap B$, I could use $\lfloor \frac{1000}{3 \times 5} \rfloor$, but I am unsure if this misses out some numbers. Is that the correct way to do it or not? If no, can anyone suggest the correct way.&lt;/p&gt;&#xA;" OwnerUserId="108" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-29T15:47:10.620" LastActivityDate="2010-10-29T15:49:12.350" Title="How many numbers between 1 and 1000 can be divided by  both $a$ and $b$" Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8290" PostTypeId="1" CreationDate="2010-10-29T15:39:06.807" Score="2" ViewCount="103" Body="&lt;p&gt;A triple of positive integers $(a,b,c)$ is an $abc$-triple if $a$ and $b$ are coprime and $c = a + b$. Define the &lt;em&gt;quality&lt;/em&gt; or &lt;em&gt;power&lt;/em&gt; of an $abc$-triple as $P(a,b,c) = \frac{\log c}{\log \text{rad}(abc)}$, where $\text{rad}(k)$ denotes the product of distinct prime divisors of $k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One version of the $abc$-Conjecture is that for each $\varepsilon &gt; 0$, there are finitely many $abc$-triples such that $P(a,b,c) &gt; 1 + \varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are finitely many known triples satisfying $P &gt; 1.4$, the so called &lt;em&gt;good triples&lt;/em&gt;,  and the largest (quality) is $P(2,3^{10} \cdot 109, 23^{5}) = 1.629911684 \dots$ (discovered by E. Reyssat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there an upper bound for $P(a,b,c)$ simply in terms of $c$ and absolute constants (sharper than $\log_{2} c$)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there an absolute upper bound for $P(a,b,c)$ so that no triple has higher quality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Best Answer:&lt;/strong&gt; If there were such a bound, then asymptotic FLT would be in hand. (Thanks Ace of Base on MO!)&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T03:55:05.947" LastActivityDate="2010-11-04T03:55:05.947" Title="Upper bound for the quality of an $abc$-triple" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;diophantine-equations&gt;&lt;asymptotics&gt;" FavoriteCount="2" />
  <row Id="8291" PostTypeId="2" ParentId="8289" CreationDate="2010-10-29T15:40:53.577" Score="5" ViewCount="0" Body="&lt;p&gt;That's right. A number is divisible by 3 and 5 if and only if it is divisible by 15.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-29T15:40:53.577" />
  <row Id="8293" PostTypeId="1" AcceptedAnswerId="8296" CreationDate="2010-10-29T15:48:34.763" Score="1" ViewCount="91" Body="&lt;p&gt;In a problem my teacher did the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5a - 2b = 3m &#xA;5b - 2c = 3n    +&#xA;-----------------&#xA;5a - 2c + 3b = 3m + 3n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I tried solving for b in the first equation and then plugging it in to the second but could not get the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is adding full equations like this valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2010-10-30T16:39:57.753" Title="When can you add full equations together?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8294" PostTypeId="2" ParentId="8289" CreationDate="2010-10-29T15:49:12.350" Score="5" ViewCount="0" Body="&lt;p&gt;In general, you have to use $\left\lfloor\frac{1000}{[a,b]}\right\rfloor$, where $[a,b]$ is the least common multiple of $a$ and $b$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-10-29T15:49:12.350" />
  <row Id="8295" PostTypeId="1" CreationDate="2010-10-29T16:34:30.903" Score="1" ViewCount="81" Body="&lt;p&gt;I want to solve a simple stochastic problem. Imagine there are n agents who want to access a resource, with a probability p at a given time t. What ist the probability that the resource will be free and not accessed by an agent. How does the result change with 2n and 2 available resources?&#xA;How do i solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="2931" LastActivityDate="2010-12-28T18:44:29.343" Title="n agents accessing a resource" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8296" PostTypeId="2" ParentId="8293" CreationDate="2010-10-29T16:50:57.093" Score="1" ViewCount="0" Body="&lt;p&gt;If A = B and C = D then A+C = B+D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if $f$ is any function, then $x = y$ implies $f(x) = f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we have used the function f(x) = x + C. to go from A = B to A + C = B + C as well as g(x) = B + x to go from C = D to B + C = B + D. Then we paste the two equations together (transitivity) to get A + C = B + D.&lt;/p&gt;&#xA;" LastActivityDate="2010-10-29T16:50:57.093" />
  <row Id="8297" PostTypeId="2" ParentId="8243" CreationDate="2010-10-29T17:22:33.513" Score="2" ViewCount="0" Body="&lt;p&gt;gcd(a,b)=1 so there are integers s and t such that sa+tb=1, whence sax+tbx=x. But a divides x so x=ax' for some integer x'. As b divides x, x=bx&quot; for some integer x&quot;. Substituting on the left-hand side of the preceding equation yields sabx&quot;+tabx'=x. Factor out the commn ab on the left to get ab(ax&quot;+tx&quot;)=x, so that ab divides x.&lt;/p&gt;&#xA;" OwnerUserId="2933" LastActivityDate="2010-10-29T17:22:33.513" />
  <row Id="8298" PostTypeId="2" ParentId="8288" CreationDate="2010-10-29T17:42:17.143" Score="0" ViewCount="0" Body="&lt;p&gt;I -think- there's a proof related to sums of angles that says this can't be done.&lt;/p&gt;&#xA;" OwnerUserId="1474" LastActivityDate="2010-10-29T17:42:17.143" CommentCount="1" />
  <row Id="8299" PostTypeId="1" CreationDate="2010-10-29T17:53:42.420" Score="2" ViewCount="59" Body="&lt;p&gt;Is there a relation between the max of a Gaussian random walk of 10 &#xA;steps vs the max of 10 Gaussian random walks? Specifics (in &#xA;Mathematica notation): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;(* a Gaussian random walk with standard deviation 1 *) &#xA;a[0] := 0 &#xA;a[n_] := a[n-1] + RandomReal[NormalDistribution[0, 1]] &#xA;&#xA;(* the max of the walk over 10 steps *) &#xA;b := Max[Table[a[i],{i,1,10}]] &#xA;&#xA;(* calculate max many times to get good sample set *) &#xA;(* Mathematica &quot;magic&quot; insures we're not using the same random #s each time *) &#xA;c = Table[b,{i,1,10000}] &#xA;&#xA;(* distribution isn't necessarily normal, but we can still compute mu + SD *) &#xA;Mean[c] (* 3.66464 *) &#xA;StandardDeviation[c] (* 1.61321 *) &#xA;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, consider 10 people doing a Gaussian random walk of 1 step and we &#xA;take the max of these 10 values. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &#xA;(* max of 10 standard-normally distributed numbers *) &#xA;d := Max[Table[RandomReal[NormalDistribution[0, 1]],{i,1,10}]] &#xA;&#xA;(* get a good sample set *) &#xA;f = Table[d,{i,10000}] &#xA;&#xA;(* and now the mean and SD *) &#xA;Mean[f] (* 1.54843 *) &#xA;StandardDeviation[f] (* 0.580593 *)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The two means/SDs are obviously different, but I sense they're related &#xA;somehow, perhaps by Sqrt[10], since the sum (not max) of 10 random &#xA;walks is normal with SD of Sqrt[10], and I sense that somehow the &#xA;cumulative sum of the first 9 somehow cancel out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these known distributions? &lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2010-12-29T01:44:29.003" Title="Max of multistep Gaussian walk vs max of multiple single-step Gaussian walks" Tags="&lt;random&gt;&lt;standard-deviation&gt;" AnswerCount="1" />
  <row Id="8300" PostTypeId="2" ParentId="8256" CreationDate="2010-10-29T18:09:14.360" Score="1" ViewCount="0" Body="&lt;p&gt;I agree that if we define $\mathbb{N}$ as a &quot;set of positive integers&quot;, the question is redundant. However if we say that $\mathbb{N}$ is a set of natural numbers (and, after all, that is exactly what $\mathbb{N}$ stands for), defined by Peano axioms, then the question is no longer redundant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, calling $\mathbb{N}$ a &quot;set of positive integers&quot; is, IMHO, a bad style. Unfortunately, there is no standard convention for &quot;set of positive integers&quot;. Many books use $\mathbb{Z}^{+}$ for that, however at least one book I'm familiar with uses $\mathbb{Z}^{+}$ for &quot;group of integers under the regular addition&quot;.  May be $\mathbb{Z}_{+}$ is better (?)&lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-29T18:09:14.360" />
  <row Id="8301" PostTypeId="2" ParentId="8295" CreationDate="2010-10-29T18:38:12.747" Score="2" ViewCount="0" Body="&lt;p&gt;The resource will be free with probability $(1-p)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there are $2n$ agents and $2$ resources, one of the resources will be free with probability $(1-p)^{2n} + 2np(1-p)^{2n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you want to calculate the tail of a Binomial distribution. When $p = c/n$ and $n$ is large, the distribution is approximately Poisson.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-29T18:38:12.747" />
  <row Id="8302" PostTypeId="2" ParentId="8155" CreationDate="2010-10-29T18:40:28.157" Score="0" ViewCount="0" Body="&lt;p&gt;It doesn't matter if you work with complex, real numbers, or elements of any field $\mathbb{K}$: if you have a point $p = (z_1, \dots , z_n) \in \mathbb{K}^n$, or any $\mathbb{K}$-vector space $V$, an equation for a straight line in $\mathbb{K}^n$ (or in $V$) passing through $p$ may always be written, for instance, as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p + \lambda v \ ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $v= (v_1, \dots , v_n) \in \mathbb{K}^n$ (or $v\in V$) and $\lambda \in \mathbb{K}$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-29T18:40:28.157" />
  <row Id="8303" PostTypeId="2" ParentId="8249" CreationDate="2010-10-29T18:49:51.933" Score="1" ViewCount="0" Body="&lt;p&gt;HINT: Take a sequence of partitions which alternate between including $0$ and not including $0$ while having their size go to $0$ (the most natural choice will do, since $0$ is the midpoint of the interval); pick your &quot;tag&quot; appropriately in the ones that do not include $0$ so that the sum gives you one value, and see what you get when the partition does include $0$. With some appropriate choices, you'll get a sequence that does not converge, showing that the limit does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the function is Riemman-Stieltjes integrable relative to $a$, break up the integral into two interals, possibly redefining $f$ and $a$ at a single point in one or both of them, so that you can use the standard formula for Riemman-Stieltjes when the functions are &quot;nice&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-29T18:49:51.933" CommentCount="2" />
  <row Id="8304" PostTypeId="2" ParentId="8253" CreationDate="2010-10-29T18:56:59.080" Score="2" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;If by index of the curve C they mean the integral:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2 \pi i} \int_\gamma \frac{v'(z)}{v(z)} dz $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is what I am guessing that they mean, then this follows from the residue theorem applied to the complex valued function $v(z)$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The index of a vector field at $z_0$ is defined to be the winding number you get for taking a small circle about $z_0$ of winding number $1$ and computing the winding number of its image about $0$. This can be computed as:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2 \pi i} \int_\gamma \frac{v'(z)}{v(z)} dz $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is a parametrization of a sufficiently small circle which winds around $z_0$ once. For $v(z) = f(z)$ and $f(z_0) = 0$, we can write $f(z) = (z-z_0)^n g(z)$ where $n$ is maximal such that $g(z_0) \neq 0$. Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f'(z) = n (z-z_0)^{n-1} g(z) + (z-z_0)^n g'(z) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{f'(z)}{f(z)} = \frac{n}{z-z_0} + (z-z_0)^n \frac{g'(z)}{g(z)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term above is analytic in this region if we make $\gamma$ a small enough circle so that $g(z)$ is non-zero in the interior. Thus, when we integrate we just get that the index is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2 \pi i} \int_\gamma \frac{n}{z - z_0} dz = n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the multiplicity of $z_0$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-29T18:56:59.080" />
  <row Id="8305" PostTypeId="1" AcceptedAnswerId="8311" CreationDate="2010-10-29T19:09:46.660" Score="1" ViewCount="144" Body="&lt;p&gt;If we have a uniform continuous function $f: X \to Y$, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ takes a cauchy sequence in $X$ to a cauchy sequence in $Y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now is this statement true: $f$ is uniformly continuous iff,  given $\epsilon &gt; 0$, there is an $N &gt; 0$, such that for every $x,y \in I$,where $I$ is an interval ($x \neq y$) we have $$ \Biggl| \frac{f(x)-f(y)}{x-y} \Biggr| &gt; N \ \Longrightarrow |f(x)-f(y)|&amp;lt; \epsilon.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, how to prove it?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-29T20:24:57.473" Title="Necessary and sufficient condition for uniform continuity" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8306" PostTypeId="2" ParentId="8299" CreationDate="2010-10-29T19:29:19.043" Score="0" ViewCount="0" Body="&lt;p&gt;I don't have an answer but only some trivial observations about your question that I post below. It got too big to be left as a comment. If I understand your question correctly, you want to compare the following two problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Let $X_i$ be i.i.d unit normals and you are interested in $X^{*} = \max(X_i, 1 \leq i \leq 10)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Let $Y_i$ be i.i.d unit normals. Define $S_0 = 0$ and $S_i = S_{i-1} + Y_i$ for $i &gt; 0$ and $S^{*} = \max(S_i, 1 \leq i \leq 10)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that $P(X^{*} \leq x) = P(X_i \leq x)$ for all $1 \leq i \leq 10$ and therefore the distribution of $X^{*}$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_{X^{*}}(x) = 10 \Phi(x)^9 \phi(x)$ where $\phi$ and $\Phi$ are the pdf and cdf of the standard normal. From this, you can technically compute the mean of $X^{*}$ though I don't know if the messy integration yields a nice solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of determining the distribution of $S^{*}$ seems much trickier. I found some references &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.aoap/1174323252&quot; rel=&quot;nofollow&quot;&gt;online&lt;/a&gt; that study the asymptotic behavior for long random walks and even that seems very hard. You can argue that $P(S^{*} \leq s) = P(S_i \leq s)$ for all $1 \leq i \leq n$ but the $S_i$ are dependent random variables. Their joint distribution is easy to derive but once again, I don't know if the integration is manageable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you are only interested in comparing the first moment of $X^{*}$ and $S^{*}$, there might be a clever way to do it that avoids all the integrations and such. If there exists such a proof, I would love to learn it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I found some links that give the asymptotics for these random variables as the number of random variables $n \rightarrow \infty$. See &lt;a href=&quot;http://www.springerlink.com/content/lt45q14201550468/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.pjm/1102911985&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. As best as I can determine, $E(X^{*})$ grows as $\sqrt{2\ln(n)}$ and $E(S^{*})$ grows as $\sqrt{\frac{n}{\pi}}$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-10-30T00:47:18.573" LastActivityDate="2010-10-30T00:47:18.573" CommentCount="3" />
  <row Id="8307" PostTypeId="1" CreationDate="2010-10-29T19:35:40.070" Score="4" ViewCount="79" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual approach is to induct on the length of the arithmetic progression, which is difficult to simplify directly to the case of two colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a different approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastActivityDate="2010-10-29T19:35:40.070" Title="Is there a simpler proof of Van Der Waerden's Theorem when there are only two colors?" Tags="&lt;graph-theory&gt;" />
  <row Id="8308" PostTypeId="1" AcceptedAnswerId="8310" CreationDate="2010-10-29T19:36:20.327" Score="0" ViewCount="91" Body="&lt;p&gt;I'm trying to understand the intuitions behind working with very large numbers.  Specifically, I'm talking about numbers of the form $a^b$ where $a &gt; 10,000$ and $b &gt; 10,000$, and in general $a$ and $b$ are small, but $a^b$ has millions of digits.  Obviously, computing the number is infeasible, but I still would like to determine things about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am particularly interested in ideas or rules about evaluating which of two large numbers is larger.  I believe I want to use logarithms somehow, but they've always confused me a little bit, so I can't quite see where to go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, I'd love to hear more generic discussion of how one can work with such large numbers easily, sacrificing precision, but not accuracy.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2010-10-29T19:58:05.103" Title="Working With Large Exponents" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="8309" PostTypeId="2" ParentId="8250" CreationDate="2010-10-29T19:56:53.153" Score="-1" ViewCount="0" Body="&lt;p&gt;First suppose both conditions are true. From the first of them we know $x\geq 0$, from the second --- $y^TA\geq 0$. Therefore $y^Tb=y^TAx\geq 0$. Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that both conditions are false. Let $\{e_k\}_{k=1}^n$ be the standard basis in $\mathbb{R}^n$. Note that the set $\{Ax\mid x\geq 0\} = \{\sum_kx_k Ae_k\mid x_k\geq 0\}$ is closed and take the nearest point to $b$ in it. From minimality of $\|b-Ax\|^2$ for any $k$ we have either $x_k&gt;0$ and $\partial_{x_k}\|b-Ax\|^2=0$ either $x_k=0$ and $\partial_{x_k}\|b-Ax\|^2\geq 0$. Thus&#xA;$$\left[\begin{aligned}x_k&amp;amp;&gt;0,&amp;amp;(b-Ax,Ae_k)&amp;amp;=0,\\x_k&amp;amp;=0,&amp;amp; (b-Ax,Ae_k)&amp;amp;&amp;lt;0\end{aligned}\right.\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $y=Ax-b$. From (1) we have $(y,Ae_k)\geq 0$ and hence $A^Ty\geq 0$. Finally&#xA;$$b^Ty=(y,b)=(Ax-b,b)=(Ax-b,b-Ax)+(Ax-b,Ax)=$$&#xA;$$-\|b-Ax\|^2-\sum_kx_k (b-Ax,Ae_k)=-\|b-Ax\|^2&amp;lt;0.$$&#xA;So the second condition should be true. Contradiction.&lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-10-29T19:56:53.153" />
  <row Id="8310" PostTypeId="2" ParentId="8308" CreationDate="2010-10-29T19:58:05.103" Score="7" ViewCount="0" Body="&lt;p&gt;Since logarithms are monotonically increasing, you can determine if $$a^b &gt; c^d$$ by instead checking if $$\log(a^b) &gt; \log(c^d)$$ which can be rewritten (using basic properties of logarithms) as $$b\log(a) &gt; d\log(c)$$ This results in the comparison of two numbers on the order of those given in the input (that is, both numbers will be within an order of magnitude or so of $a,b,c,d$), which is definitely feasible on a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use logs to calculate other properties, as long as you remember to convert your solution back to the original numbers (by means of an exponential) when you're done. For example, to compute whether $${a_1}^{b_1}{a_2}^{b_2} &gt; {c_1}^{d_1}{c_2}^{d_2}$$, you can take the log of both sides to get &#xA;\begin{align}&#xA;\log({a_1}^{b_1}{a_2}^{b_2}) &amp;amp;&gt; \log({c_1}^{d_1}{c_2}^{d_2}) \\&#xA;\log({a_1}^{b_1}) + \log({a_2}^{b_2}) &amp;amp;&gt; \log({c_1}^{d_1}) + log({c_2}^{d_2}) \\&#xA;b_1\log(a_1) + b_2\log(a_2) &amp;amp;&gt; d_1\log(c_1) + d_2\log(c_2) \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is again computationally feasible.&lt;/p&gt;&#xA;" OwnerUserId="1507" LastActivityDate="2010-10-29T19:58:05.103" CommentCount="2" />
  <row Id="8311" PostTypeId="2" ParentId="8305" CreationDate="2010-10-29T20:24:57.473" Score="2" ViewCount="0" Body="&lt;p&gt;Well, the standard definition of uniform continuous on an interval $I$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall \epsilon &gt; 0,  \exists \delta &gt; 0 \text{ such that } \forall x, y \in I,  |x-y| &amp;lt; \delta \Rightarrow |f(x)-f(y)| &amp;lt; \epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to prove is that this is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall \epsilon &gt; 0,  \exists N &gt; 0 \text{ such that } \forall x, y \in I,  \frac{|f(x)-f(y)|}{|x-y|} &gt; N \Rightarrow |f(x)-f(y)| &amp;lt; \epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{|f(x)-f(y)|}{|x-y|} &gt; N \iff |x-y| &amp;lt; \frac{|f(x) - f(y)|}{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof (of your question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ : take $N$ such that $\delta=\frac{ \sup( f(x)) - \inf(f(y))}{N}$. The supremum and infimum are well-defined because of uniform continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftarrow$ : take $\delta := \frac{\epsilon}{N}$&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-29T20:24:57.473" />
  <row Id="8312" PostTypeId="1" AcceptedAnswerId="8313" CreationDate="2010-10-29T20:25:00.587" Score="0" ViewCount="66" Body="&lt;p&gt;Let $f\in\mathbb{Z}[x]$ be monic and irreducible, let $K=$ splitting field of $f$ over $\mathbb{Q}$. What can we say about the relationship between $disc(f)$ and $\Delta_K$? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't? &lt;/p&gt;&#xA;" OwnerUserId="264" LastActivityDate="2010-10-29T20:39:51.217" Title="Discriminant of a monic irreducible integer polynomial vs. discriminant of its splitting field" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8313" PostTypeId="2" ParentId="8312" CreationDate="2010-10-29T20:32:34.303" Score="3" ViewCount="0" Body="&lt;p&gt;The two are the same if the roots of $f$ form an integral basis of the ring of integers of the splitting field (e.g. if $f$ is a cyclotomic polynomial) because then, well, they're defined by the same determinant (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field#Examples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;), but in general they don't.  In the general case $\mathbb{Z}[\alpha_1, ... \alpha_n]$ is an order in $\mathcal{O}_K$ so one can write the $\alpha_i$ as an integer linear combination of an integral basis, so the matrices whose determinants define the two discriminants should be related by the square of a matrix with integral entries, hence integral determinant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact if I'm not totally mistaken, the quotient of the two discriminants should be precisely the index of $\mathbb{Z}[\alpha_1, ... \alpha_n]$ in $\mathcal{O}_K$ as lattices, or maybe its square...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, since the discriminant of the field is defined in terms of $\mathcal{O}_K$ it is the &quot;right&quot; choice for carrying information about, for example, ramification.  One can see this even in the quadratic case: if $d \equiv 1 \bmod 4$ then the discriminant of $x^2 - d$ is $4d$ but the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$, and the latter is the &quot;right&quot; choice because $2$ doesn't ramify in $\mathbb{Z} \left[ \frac{1 + \sqrt{d}}{2} \right]$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-29T20:39:51.217" LastActivityDate="2010-10-29T20:39:51.217" />
  <row Id="8314" PostTypeId="1" AcceptedAnswerId="8316" CreationDate="2010-10-29T20:45:59.703" Score="2" ViewCount="164" Body="&lt;p&gt;I know a few proofs over this theorem (where $\mathfrak{c}$ is the cardinality of $[0,1]$ and $\aleph_0$ is the cardinality of $\mathbb{N}$) where they construct two injections and then use Schröder-Bernstein (via the Cantor set or something like that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I was wondering if I could do something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $s: \{0,1\}^\mathbb{N} \to [0,1]$ by&#xA;$$s(x) = \sum_{i = 1}^\infty \frac{x(i)}{2^i}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is clearly a surjection because this is just a binary expansion of numbers in $[0,1]$, but not injective because these expansions are not unique. Is there a way to make this work? Can I use this to construct a bijection?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-21T14:21:57.040" LastActivityDate="2010-11-21T14:21:57.040" Title="$\mathfrak{c} = 2^{\aleph_0}$." Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="8315" PostTypeId="1" CreationDate="2010-10-29T20:49:21.623" Score="2" ViewCount="152" Body="&lt;p&gt;Consider a function $f_1$ defined by $f_1(x)=1-x+o(x)$ and $f_1(2x)=f_1(x)^2 + 0$. It's simple to find that $f_1(x)=e^{-x}$ (for example by writing series near $x=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a function $f_2$ defined by $f_2(x)=2-x^2+o(x^2)$ and $f_2(2x)=f_2(x)^2-2$. It can be proven that $f_2(x)=2 \cos(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any formula (probably with use of special functions) for the generalization of this, i.e. function $f_n$ defined by $f_n(x)=2^{n-1}-x^n+o(x^n)$ and $f_n(2x)=f_n(x)^2-2^{2n-2}+2^{n-1}$?&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="695" LastEditorDisplayName="" LastEditDate="2010-10-30T07:18:58.207" LastActivityDate="2010-10-30T07:18:58.207" Title="Generalization of cos: is this function known?" Tags="&lt;functions&gt;&lt;special-functions&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8316" PostTypeId="2" ParentId="8314" CreationDate="2010-10-29T20:53:46.353" Score="7" ViewCount="0" Body="&lt;p&gt;Yes, but in my opinion, it's really not worth it.  Binary expansions are unique for all real numbers except dyadic rationals $\frac{k}{2^n}$, since they can end with a string of zeroes or a string of ones.  But there are only countably many dyadic rationals.  So define $s : \{ 0, 1 \}^{\mathbb{N}} \to [0, 1]$ to be what you said for all strings that don't end with a string of zeroes or a string of ones, and for the countably many exceptions pick any bijection you like.  For example, if a sequence $a_i \in \{ 0, 1 \}^{\mathbb{N}}$ ends with a string of ones, send it to $0.1 a_1 a_2 a_3 ...$, and if it ends with a string of zeroes, send it to $0.0 a_1 a_2 a_3 ...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really, there's no point in not using Schroeder-Bernstein.  Explicit bijections are highly overrated; in general it is much easier to construct an injection and a surjection.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-29T20:53:46.353" CommentCount="3" />
  <row Id="8318" PostTypeId="1" CreationDate="2010-10-29T21:32:04.870" Score="0" ViewCount="75" Body="&lt;p&gt;I'm reading through &lt;a href=&quot;http://en.wikipedia.org/wiki/LFSR#Fibonacci_LFSRs&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; about Linear Feedback Shift Registers (Specifically Fibonacci LFSRs) and the only restrictions it mentions about the taps are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The LFSR will only be maximum-length if the number of taps is even; just 2 or 4 taps can suffice even for extremely long sequences.&lt;/li&gt;&#xA;&lt;li&gt;The set of taps must be relatively prime, and share no common divisor to all taps.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It has two other bullet points basically saying that there can be more than one working set of taps, and if you find one set of taps, you can use that to determine another set of taps. Is this all it takes to find taps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these the only restrictions to finding taps in a Fibonacci LFSR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are plenty of reference tables to look up taps, but I'm trying to understand the fundamentals of an LFSR better.&lt;/p&gt;&#xA;" OwnerUserId="757" LastActivityDate="2010-10-30T18:08:44.893" Title="How are the taps of an LFSR found?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="8319" PostTypeId="1" AcceptedAnswerId="8381" CreationDate="2010-10-29T21:44:35.780" Score="8" ViewCount="198" Body="&lt;p&gt;Why $\frac{1}{\cos x}=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi }{x^2-\left (n-\frac{1}{2}\right )^2\pi^2}$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-31T06:44:08.973" Title="Series representation for 1/cos x" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="8320" PostTypeId="2" ParentId="8234" CreationDate="2010-10-29T22:52:41.660" Score="0" ViewCount="0" Body="&lt;p&gt;Quoting Adrián Barquero:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This doesn't determine B as the set of&#xA;  positive integers. Basically you're&#xA;  just saying that B is an inductive&#xA;  set. In fact, if you look at the&#xA;  definition you gave from Munkres's&#xA;  book, the set of positive integers&#xA;  being defined as the intersection of&#xA;  all inductive sets is just saying that&#xA;  the positive integers constitute a&#xA;  minimal inductive set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-10-29T22:52:41.660" CommunityOwnedDate="2010-10-29T22:52:41.660" />
  <row Id="8321" PostTypeId="1" CreationDate="2010-10-29T23:24:28.970" Score="2" ViewCount="92" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you don't know what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading Farkas and Kra's &quot;Riemann Surfaces,&quot; and they talk briefly about these guys in section 0.2. Please tell me if I am misunderstanding something or saying something wrong or funny. Let $\wp$ be the Weierstrass function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, $\wp$ maps the torus to $CP^1 \times CP^1$ in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \mapsto (\wp(a),\wp'(a)) = (z,w)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the image of this map lies on the zero set of the polynomial $P(z,w) = 4(z-e_1)(z-e_2)(z-e_3) - w^2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is the description of the inverse to this map. It says that the integral of the differential form $\frac{dz}{w}$ from $\infty$ to a point $Q$ along a path $c$ is the inverse of this map.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I don't understand what they mean by $\infty$ here.&lt;/li&gt;&#xA;&lt;li&gt;I don't understand why this is the inverse to the map.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Heuristically, I see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_\infty^Q \frac{dz}{w} = \int_0^z \frac{\wp'(u)}{\wp'(u)} du = \int_0^a du = a&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but unfortunately, I have no clue what I just did here. I mean, I guess I did something like a pull back? (Set $z=\wp$ and $w=\wp'$. Isn't this a map into the complex plane, and not the Torus?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a related question, but why is infinity the branch point?  In general, when you have $w^2 = p(z)$, with degree of $p$ odd, why is infinity one of the branch points? Why is it when the degree of $p$ is even, it is not a branch point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time!&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-30T13:04:06.677" LastActivityDate="2010-10-30T13:04:06.677" Title="A Torus and the Weierstrass P function??" Tags="&lt;complex-analysis&gt;&lt;riemann-surface&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8322" PostTypeId="1" CreationDate="2010-10-29T23:51:12.830" Score="4" ViewCount="79" Body="&lt;p&gt;Given an infinite dimensional manifold modeled on a Banach space, what does it mean for it to have a Riemannian metric? Does it necessarily mean that it is actually a Hilbert manifold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that Hilbert Spaces have inner products, whereas Banach Spaces just have norms. If a Banach manifold has a Riemannian metric, it means that at each tangent space, which is a Banach space, it has an inner product... so wouldn't that make any Banach manifold a Hilbert Manifold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm obviously completely confused and mixing up something.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2010-10-30T19:13:54.393" Title="A Banach Manifold with a Riemannian Metric?" Tags="&lt;soft-question&gt;&lt;inner-product-space&gt;&lt;manifold&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8323" PostTypeId="1" AcceptedAnswerId="8340" CreationDate="2010-10-30T00:42:41.487" Score="8" ViewCount="355" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that the number of digits of a natural number n&gt;0, which represent by d (n) is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d(n)= 1 + floor(log n)   (I)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;log indicates logarithm base 10&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well .. the classical approach to the Stirling factorial natural number n&gt;1 is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n! ~ f(n) = [(2.n.pi) ^ (1 / 2)]. [(n / e) ^ n]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The symbol ~ indicates &quot;approximately.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of digits n!, according to equality (I), is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d(n!) = 1 + floor(log n!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It seems to me that for all natural n&gt; 1, log n! and log [f (n)] have the same floor:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;floor[log(n!)] = floor[log(f(n))]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's my big question!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we could write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d (n!) = 1 + floor[log(f(n))]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope someone has some little time for the theme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Sergio Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-01T19:39:18.293" LastActivityDate="2010-11-02T13:50:06.387" Title="Does Stirling's formula give the correct number of digits of n!?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="8324" PostTypeId="2" ParentId="8133" CreationDate="2010-10-30T00:43:37.700" Score="4" ViewCount="0" Body="&lt;p&gt;I'll just answer your last question for now.  It is not possible for the zeta function to have a single zero off the critical line.  The functional equation for the Riemann zeta function immediately shows that the zeros in the critical strip are symmetric about the line Re z=1/2.  Hence, if there is one zero, there must be two.  Plus, $\overline{\zeta(\bar s)}=\zeta(s)$.  So, if you have one zero, you have four.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-10-30T00:43:37.700" />
  <row Id="8325" PostTypeId="1" CreationDate="2010-10-30T00:59:37.000" Score="0" ViewCount="73" Body="&lt;p&gt;How can we find out the topology of an open manifold (a manifold with boundaries)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given a cylinder with open ends(cartesian product of $S^1$ and $(0,1)$), what is the genus of this shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a cylinder with one end closed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2939" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-10-30T02:41:34.057" LastActivityDate="2010-10-30T02:41:34.057" Title="Topology of an open manifold?" Tags="&lt;topology&gt;&lt;manifold&gt;" CommentCount="3" />
  <row Id="8326" PostTypeId="1" CreationDate="2010-10-30T01:05:42.380" Score="2" ViewCount="204" Body="&lt;p&gt;Is there a good quick introduction to knot theory?  I am relatively mathematically savvy so any level is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="355" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-30T01:18:30.303" LastActivityDate="2010-11-02T21:45:05.527" Title="A good quick introduction to Knot Theory?" Tags="&lt;recreational-mathematics&gt;&lt;knot-theory&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="1" />
  <row Id="8327" PostTypeId="2" ParentId="8315" CreationDate="2010-10-30T01:25:00.610" Score="5" ViewCount="0" Body="&lt;p&gt;Assuming Willie Wong's correction, by a method of undetermined coefficients it's not too hard to generate a series for  $f_3$, and it begins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4-x^3+(1/56)x^6-(1/14112)x^9+(13/115379712)x^{12}-(53/629973227520)x^{15}+(17413/495415985907548160)x^{18} + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it's not clear what to do with this. For example, the denominators have large-ish prime factors: $495415985907548160 = 2^{14} 3^4 5^1 7^5 13^1 31^1 73^1 151^1$. This seems to suggest that there's not going to be an easy third-order differential equation, whereas your $f_1$ and $f_2$ satisfy first- and second-order differential equations.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-10-30T01:25:00.610" />
  <row Id="8328" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T01:39:09.663" Score="5" ViewCount="0" Body="&lt;p&gt;As Michael comments, Colin Adams has a well regarded text called &quot;The Knot Book&quot;. Adams has also written a comic book about knot theory called &quot;Why Knot?&quot;. It's very humorous but is a genuine introduction to the mathematics involved. This comic book comes with a plastic &quot;rope&quot; that can be knotted, unknotted, and twisted into different shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &quot;Why Knot?&quot; qualifies well as a &quot;good, quick&quot; introduction to the topic. Well worth tracking down.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-30T01:39:09.663" CommunityOwnedDate="2010-10-30T01:39:09.663" CommentCount="2" />
  <row Id="8329" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T01:47:30.627" Score="1" ViewCount="0" Body="&lt;p&gt;Two books I would recommend are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knots and Surfaces, David Farmer and Theodore Standford (American Mathematical Society; 1996)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knots and Links, Peter Cromwell (Cambridge U. Press; 2004).&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-30T01:47:30.627" />
  <row Id="8330" PostTypeId="2" ParentId="8288" CreationDate="2010-10-30T01:54:01.897" Score="2" ViewCount="0" Body="&lt;p&gt;You might want to look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.michaelbeeson.com/research/papers/TriangleTiling.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.michaelbeeson.com/research/papers/TriangleTiling.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the references given there.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-30T01:54:01.897" CommentCount="2" />
  <row Id="8331" PostTypeId="1" AcceptedAnswerId="8335" CreationDate="2010-10-30T02:17:21.750" Score="2" ViewCount="176" Body="&lt;p&gt;I've just started learning about schemes, so maybe I'm missing something basic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exercise I-24(a):&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Take Z = Spec$\mathbb{C}[x]$, let $X$ be the result of identifying the two closed points (x) and (x-1) of |Z|, and let $\phi: Z \to X$ be the natural projection. Let $\mathcal{O}$ be $\phi_* \mathcal{O}_Z$, a sheaf of rings on&#xA;$X$. Show that $(X, \mathcal{O})$ satisfies condition (i) above for all elements $f \in \mathcal{O}(X) = \mathbb{C}[x]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition (i) referred to: For any $f \in \mathbb{C}[x]$ define $U_f \subset X$ as the set of points $x \in X$ such that $f$ maps to a unit of the stalk $\mathcal{O}_x$. (i) means that $\mathcal{O}(U_f) = \mathbb{C}[x][f^{-1}]$ for all f. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But how can this be? Put f = x. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U_f = X \setminus \{(x)\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi^{-1}(U_f) = Z \setminus \{ (x), (x-1) \}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{O}(U_f) = \mathcal{O}_Z(\phi^{-1}(U_f)) = \mathbb{C}[x][ ((x)(x-1))^{-1} ]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that is not $\mathbb{C}[x][f^{-1}]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Regarding the answer and comments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;evgeniamerkulova's answer reassures me that I'm not out of my mind, but obviously Matt E and Mariano know what they're talking about, so I don't know what to think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Mariano and Matt E imply that $\mathcal{O}(X)$ is not $\mathbb{C}[x]$, but that seems obviously wrong (and contradicts the book itself). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my reasoning, spelled out. O(X) is C[x]. This is because $\mathcal{O}_Z(\phi^{-1}(X)) = \mathcal{O}_Z(Z) = \mathbb{C}[x]$. In order for the condition to be satisfied, we need $\mathcal{O}(U) = \mathbb{C}[x,x^{-1}]$ for some open U in X. So we need $\mathcal{O}_Z(\phi^{-1}(U)) = \mathbb{C}[x,x^{-1}]$. For that to happen we need $\phi^{-1}(U) = Z \setminus \{ (x) \}$. But all the inverse images of sets in X either include both (x) and (x-1) or neither of them, so this can never happen. &lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2010-10-30T15:08:25.887" LastActivityDate="2010-10-30T15:08:25.887" Title="Exercise from Eisenbud &amp; Harris's The Geometry of Schemes" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="8332" PostTypeId="1" CreationDate="2010-10-30T03:02:11.513" Score="5" ViewCount="154" Body="&lt;p&gt;Can we always break an arbitrary field extension $L/K$ into an extension $F/K$ in which the only roots of unity of $F$ are those in $K$, followed by an extension $L/F$ which is of the form $L=F(\{\omega_i\})$ where the $\omega_i$ are roots of unity? What if we restrict to $L/K$ separable, or finite?&lt;/p&gt;&#xA;" OwnerUserId="264" LastEditorUserId="264" LastEditorDisplayName="" LastEditDate="2010-11-07T01:08:48.857" LastActivityDate="2010-11-07T01:08:48.857" Title="Roots of unity and field extensions" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" CommentCount="1" FavoriteCount="2" />
  <row Id="8333" PostTypeId="2" ParentId="8321" CreationDate="2010-10-30T03:17:24.580" Score="5" ViewCount="0" Body="&lt;p&gt;It is better to think of mapping to $\mathbb C P^2$ rather than a product of $\mathbb C P^1$s.  The point $\infty$ is then the unique point at infinity on the cubic $y^2 = 4(x-e_1)(x-e_2)(x-e_3).$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the integral depends on the path, not just the endpoint; the ambiguity in the value of the integral if you just specify the endpoints is given by integrating over closed loops, i.e. over representatives of the 1st homology.  These integrals span a lattice $\Lambda$ in $\mathbb C$ (the original lattice with respect to which you defined $\wp$), and so the integrals really take values in $\mathbb C/\Lambda$, not in $\mathbb C$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-30T03:17:24.580" CommentCount="3" />
  <row Id="8334" PostTypeId="1" AcceptedAnswerId="8685" CreationDate="2010-10-30T04:44:38.530" Score="4" ViewCount="98" Body="&lt;p&gt;I find confusing some examples I have seen. Maybe you can help me to determine what is going on with them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Generalized Feedback Shift Register (GFSR) sequence defines a sequence $\{W_{i}\}$ satisfying the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_{k+p}=c_{0}W_{k}\bigoplus c_{1}W_{k+1}\bigoplus...\bigoplus c_{p-1}W_{k+p-1} \qquad \qquad (1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\bigoplus$ is the binary exclusive-or operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the polynomial $f(x)=c_{0}+c_{1}x+c_{2}x^{2}+...+c_{p-1}x^{p-1}+x^{p}$ is a primitive polynomial over $GF(2)$, then the sequence $\{W_{i}\}$ will have maximal sequence $2^{p}-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1: Let's consider the trinomial $1+x+x^{4}$ and a bit sequence $1, 0, 1, 0$. For the polynomial we have $c_{0}=1$, $c_{1}=1$, $c_{2}=0$, $c_{3}=0$ and $p=4$. Therefore, the equation $(1)$ should be $W_{k+4}=W_{k}\bigoplus W_{k+1}$. According to this, we calculate the values for $W_{5}, W_{6},...$ etc (since we already know that $W_{1}=1, W_{2}=0, W_{3}=1, W_{4}=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure generates the following sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1,0,1,0,1,1,1,1,0,0,0,1,0,0,1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the example takes 4 bit chunks (changing to decimal representation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1010=10, 1111=15, 0001=1, 0011=3, 1010+1111=0101=5, 0001+0011=0010=2$ and so on. So a '4-wise decimation' using the recurrence yields the numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10, 15, 1, 3, 5, 2, ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a standard way to generate a bigger sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;By the using the bit stream from the trinomial $1+x+x^{4}$ and the starting sequence $1,0,1,0$, and... forming 4-bit words by putting the bits into a fixed binary position with a delay of 3 between binary positions, we have&#xA;    $$1010=10, 1110=14, 0011=3, 0101=5, 1111=15, 0001 = 1, 0010=2, 0111=7,...$$&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, both examples are dealing with exactly the same problem. However, they lead to different sequences. I don't even know how the second example generates its sequence (it looks like it is taking the first bit sequence $1, 0, 1, 0$ and applying the binary exclusive-or operation for the first two terms $1 \bigoplus 0 = 1$ which is the first term of the following bit sequence, then take the second and third terms $0 \bigoplus 1=1$ which is the second term of the new sequence and so on). However, I don't know how it gets the last term. Such a pattern works for all the sequences of the Example 2, which makes me thing that I'm not seeing the full picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-30T16:06:18.307" LastActivityDate="2011-01-01T20:44:30.010" Title="Generalized Feedback Shift Registers" Tags="&lt;cryptography&gt;&lt;random&gt;" AnswerCount="1" />
  <row Id="8335" PostTypeId="2" ParentId="8331" CreationDate="2010-10-30T08:47:33.240" Score="2" ViewCount="0" Body="&lt;p&gt;Everything you say is right and Eisenbud and  Harris are false. I don't inderstand Matt E ' comments either because you did no mistake:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) That &quot;x&quot; is not function on $X$ has nothing to do with problem, and you never said it was function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) He writes &quot;To compute correctly on $X$, you need to figure out what $\mathcal O(X)$ is&quot; : you have computed that it is   $\mathbb C[x]$ and you are right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness stalk $\mathcal O_a$ of $\mathcal O$ at  quotient point $a\in X$  corresponding to $0,1$ [you write x, but you may not because x is already polynomial] is  ring $S \subset \mathbb C(x)$ of all fractions $f(x)/g(x)$  ($f(x), g(x) \in \mathbb C [x])$)  such that $g(0)\neq 0$ and $ g(1)\neq 0$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-10-30T08:47:33.240" CommentCount="3" />
  <row Id="8336" PostTypeId="1" AcceptedAnswerId="8370" CreationDate="2010-10-30T09:41:47.130" Score="0" ViewCount="96" Body="&lt;p&gt;There was a problem in &lt;em&gt;Apostol's&lt;/em&gt; book namely, to prove that:&#xA;$$\sum\limits_{k=1}^{n} \Biggl\lfloor{\frac{k}{2}\Biggr\rfloor} = \Biggl\lfloor{\frac{n^{2}}{4}\Biggr\rfloor}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which i could solve. The following probelm is eluding me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $a =1,2, \cdots, 7$, prove that there exists an integer $b$ depending on $a$ such that $$\sum\limits_{k=1}^{n} \Biggr\lfloor{\frac{k}{a}\Biggr\rfloor} =\Biggl\lfloor{\frac{(2n+b)^{2}}{8a}\Biggr\rfloor} $$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-30T18:01:15.967" Title="proving $\sum\limits_{k=1}^{n} \Bigl\lfloor{\frac{k}{a}\Bigr\rfloor} =\Bigl\lfloor{\frac{(2n+b)^{2}}{8a}\Bigr\rfloor} $" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8337" PostTypeId="1" CreationDate="2010-10-30T10:04:43.110" Score="13" ViewCount="629" Body="&lt;p&gt;As I have heard people did not trust Euler when he first discovered the formula&#xA;$$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$&#xA;However, Euler was Euler and he gave other proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe many of you know some nice proofs of this, can you please share it with us?!&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T08:32:00.257" LastActivityDate="2010-11-03T18:52:08.830" Title="Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$." Tags="&lt;sequences-and-series&gt;&lt;complex-analysis&gt;&lt;big-list&gt;&lt;fourier-analysis&gt;&lt;math-folklore&gt;" AnswerCount="8" CommentCount="10" FavoriteCount="7" />
  <row Id="8338" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T10:17:24.640" Score="4" ViewCount="0" Body="&lt;p&gt;I love the book &quot;On Knots&quot; by Louis Kauffman. It's got a playful style, yet he develops a lot of deep mathematics. I read this in high school, and I got quite a lot out of it, and as my mathematical knowledge progressed, I got more and more out of it.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-10-30T10:17:24.640" />
  <row Id="8339" PostTypeId="1" AcceptedAnswerId="8407" CreationDate="2010-10-30T10:39:18.517" Score="0" ViewCount="256" Body="&lt;p&gt;Suppose A and B are similar matrices. Show that A and B have the same eigenvalues with the same geometric multiplicities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar matrices; Suppose A and B are $n\times n$ matrices over $\mathbb R$ or $\mathbb C$. We say A and B are similar, or that A is similar to B, if there exists a matrix P such that $B = P^{-1}AP$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-10-30T10:42:23.983" LastActivityDate="2010-10-30T21:36:00.027" Title="Suppose A and B are similar matrices. Show that A and B have the same eigenvalues with the same geometric multiplicities." Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="8340" PostTypeId="2" ParentId="8323" CreationDate="2010-10-30T10:52:30.393" Score="4" ViewCount="0" Body="&lt;p&gt;Here are some thoughts on the conjecture that may lead one to suspect that it is true. This is not a proof that it is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to know whether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\lfloor \log_{10} n! \right\rfloor = \left\lfloor \log_{10} \lfloor f(n) \rfloor \right\rfloor$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is true for all $n &gt; 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note that this would NOT be true if the interval&#xA;$I_n = ( \log_{10} n!,\log_{10} \lfloor f(n) \rfloor)$ contains an integer but is true otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's look at the length of $I_n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling series&lt;/a&gt; we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) &amp;lt; &#xA; \log_{10} n! -  \log_{10} f(n) &amp;lt; &#xA;\frac{1}{\log_{e}10} \frac{1}{12n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so taking the integer part of $f(n)$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) &amp;lt; &#xA; \log_{10} n! -  \log_{10} \lfloor f(n) \rfloor &amp;lt; &#xA;\frac{1}{\log_{e}10} \left( \frac{1}{12n} + \frac{1}{ \lfloor f(n) \rfloor } \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where to achieve the RHS we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log_{10} f(n) - \log_{10} \lfloor f(n) \rfloor  &amp;lt; \frac{1}{ \lfloor f(n) \rfloor \log_e 10 }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&#xA;$$\text{Length}(I_n) &amp;lt; \frac{1}{\log_{e}10} \left( \frac{1}{360n^3} +&#xA;\frac{1}{ \lfloor f(n) \rfloor } \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can verify that the conjecture holds for $n=2,3,\ldots,10,$ so summing up the remaining lengths of the $I_n$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=11}^\infty \text{Length}(I_n) &amp;lt; \frac{1}{360 \log_e 10}&#xA;\left( \sum_{n=11}^\infty \frac{1}{n^3} +  \sum_{n=11}^\infty \frac{1}{ \lfloor f(n) \rfloor } \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $ \lfloor f(n) \rfloor  \ge (n-1)! $ and so, doing some calculations (replacing the&#xA;$ \lfloor f(n) \rfloor $ on the RHS by $ (n-1)! $ , we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=11}^\infty \text{Length}(I_n) &amp;lt; \frac{1}{360 \log_e 10}&#xA;\left( 0.00452492 + 0.00000030 \right) &amp;lt; 5.5 \times 10^{-6}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that an integer falls in any of the intervals is very small.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-10-30T16:49:09.177" LastActivityDate="2010-10-30T16:49:09.177" CommentCount="2" />
  <row Id="8341" PostTypeId="1" AcceptedAnswerId="8343" CreationDate="2010-10-30T11:09:29.820" Score="2" ViewCount="99" Body="&lt;p&gt;What is the name for I can only describe as the circumference of a sphere. I mean like the equator on a planet. What is the line called which goes around the entire sphere, it can be anywhere (so not just in the middle of the top and the bottom, if your talking about a planet or something) and the length of it equals 2πr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to describe it is if you put a plane through sphere at any angle so that 2 equal halves are created, what is the name of the line where the plane intersect the sphere's surface.&lt;/p&gt;&#xA;" OwnerUserId="2075" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-30T13:06:54.600" LastActivityDate="2010-10-30T13:06:54.600" Title="&quot;circumference&quot; of a sphere" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="1" />
  <row Id="8342" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T11:12:53.913" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \det (P^{-1}P - P^{-1}A P) = \det( P^{-1}(I - A)P ) = \det P^{-1} \cdot \det (I - A) \cdot \det P $$ &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-10-30T11:12:53.913" CommentCount="2" />
  <row Id="8343" PostTypeId="2" ParentId="8341" CreationDate="2010-10-30T11:14:51.287" Score="5" ViewCount="0" Body="&lt;p&gt;A &quot;great circle&quot;, so called because they are the largest possible circles on a sphere. If you're on a sphere (say, Earth) and you want to go between two points using the shortest path, then a great circle is the path you should follow. This makes them like straight lines for spheres.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-10-30T11:14:51.287" CommentCount="1" />
  <row Id="8345" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T11:20:29.667" Score="0" ViewCount="0" Body="&lt;p&gt;It's basically something like:  If $v_1,\dots,v_k$ is a basis for the eigenspace of A corresponding to eigenvalue $\lambda$, then $P^{-1}v_1,\dots,P^{-1}v_k$ is a basis for the eigenspace of B corresponding to the same eigenvalue.  There are some details to be filled in, like showing that any vector $w$ with $Bw=\lambda w$ can be written as a linear combination of the $P^{-1}v_i$ by using that the same holds for the matrix $A$ and the vectors $v_i$ by assumption.&lt;/p&gt;&#xA;" OwnerUserId="2942" LastActivityDate="2010-10-30T11:20:29.667" />
  <row Id="8346" PostTypeId="1" CreationDate="2010-10-30T11:48:18.783" Score="2" ViewCount="55" Body="&lt;p&gt;What requirements should a CRC polynomial of a given degree satisfy to make the CRC catch a maximum of errors?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;br&gt;&#xA;I'm talking about &lt;strong&gt;GF(2) polynomials&lt;/strong&gt;.&lt;br&gt;&#xA;As an example of the kind of requirements I'm looking for: I can imagine (but don't know for sure) that a prime polynomial catches more errors than a composite polynomial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematician, so please type slowly :-)&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-01T12:17:04.637" LastActivityDate="2010-11-01T12:17:04.637" Title="What requirements should a CRC polynomial satisfy?" Tags="&lt;computer-science&gt;&lt;finite-fields&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8347" PostTypeId="1" AcceptedAnswerId="8414" CreationDate="2010-10-30T12:21:58.610" Score="0" ViewCount="100" Body="&lt;p&gt;I hate min and max questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a function&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G(x, y) = 1/x + 4/y + 9/(4-x-y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to prove that G(x,y) has a minimum value of 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've partialy derived them for x and y giving&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(1) 1/x^2 + 9/(4-x-y)^2 = 0&#xA;(2) 4/y^2 + 9/(4-x-y)^2 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is where I get stuck I'm not entirely sure how to continue. I need to find x = ? and y = ? but this simultaneous equation stumps me.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastActivityDate="2010-10-31T04:39:48.330" Title="Two Var Function." Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="8348" PostTypeId="1" CreationDate="2010-10-30T12:22:30.330" Score="4" ViewCount="279" Body="&lt;p&gt;It is a rather surprising fact (to me, at least) that $\displaystyle \binom{14}{4} = 1001$; $\displaystyle \binom{14}{5} = 2002$; $\displaystyle \binom{14}{6} = 3003$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, this is the only instance where three consecutive binomial coefficients are in the ratio $\displaystyle 1 : 2 : 3$. I found it quite interesting to investigate under what conditions consecutive members of a row of Pascal’s triangle can be in the ratio $\displaystyle a : b : c$, where $\displaystyle a,b,c$ are positive integers with  $\displaystyle \mathrm{gcd}(a,b,c) = 1$ and $\displaystyle a &amp;lt; b &amp;lt; c$, except where otherwise stated. That is, only the left-hand side of the triangle will be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{n}{k} : \binom{n}{k+1} : \binom{n}{k+2} = a : b : c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancelling and rearranging,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle b(k + 1) = a(n - k)$......….....[1]&lt;br&gt;&#xA;$\displaystyle c(k + 2) = b(n - k - 1)$.........[2]               &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \frac{a(b + c) + c(a + b)}{(b^2 - ac)}$$                                  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = \frac{a(b + c)}{b^2 - ac} - 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore n and k are integers iff $\displaystyle b^2-ac$ divides both $\displaystyle a(b + c)$ and $\displaystyle c(a + b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle n &gt; 0$ implies $\displaystyle b^2 &gt; ac$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle k\ge 0$ implies $\displaystyle c \ge \frac{b(b - a)}{2a}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence a third condition is $\displaystyle \frac{b^2}{a} &gt; c \ge \frac{b(b - a)}{2a}$        &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most interesting special case is $\displaystyle c = a + b$, when for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a,b &amp;lt; 100$ there are only five solutions. Namely,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (1,2,3)$ gives $\displaystyle \{n,k\} = \{14,4\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (3,5,8)$ gives $\displaystyle \{n,k\} = \{103,38\} $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (8,13,21)$ gives $\displaystyle \{n,k\} = \{713,271\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (21,34,55)$ gives $\displaystyle \{n,k\} = \{4894,1868\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (55,89,144)$ gives $\displaystyle \{n,k\} = \{33551,12814\}$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, there are solutions only when $\displaystyle (a,b,c) = (F(2m), F(2m + 1), F(2m + 2)) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m = 1,2,3...,$ where $\displaystyle F(m)$ is the $\displaystyle m^{th}$ Fibonacci number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \{n,k\} = \{F(2m + 2)F(2m + 3) - 1, F(2m)F(2m + 3) - 1\} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All solutions satisfy $\displaystyle F(2m)n = F(2m+2)k + F(2m+1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where possible I have been able to derive formulae for $\displaystyle \{n,k\}$ for all special cases I could think of (eg. $\displaystyle a,b,c$ in arithmetic progression) except for the case $\displaystyle c = a^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle a,b &amp;lt; 3000$ there are only three solutions:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (1,2,1)$ gives $\displaystyle \{n,k\} = \{2,0\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (2,3,4)$ gives $\displaystyle \{n,k\} = \{34,13\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (a,b,c) = (13,47,169)$ gives $\displaystyle \{n,k\} = \{1079,233\}$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $\displaystyle c = a^2$ and dividing equation [2] by [1] leads to&#xA;$\displaystyle a(k + 1)(k + 2) = (n - k)(n - k - 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, all solutions satisfy $\displaystyle n^2 - (2k + 1)n - (k + 1)[(a - 1)k + 2a] = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discriminant $\displaystyle D$ of the above quadratic is $\displaystyle 4a(k + 1)(k + 2) + 1$, and a necessary and sufficient condition for $\displaystyle n$ to be an integer is that this expression be a perfect square.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = 1, k = 0$ gives $\displaystyle D = 9 = 3^2$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = 2, k = 13$ gives $\displaystyle D = 1681 = 41^2$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = 13, k = 233$ gives $\displaystyle D = 2859481 = 1691^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my question is: can one prove (or disprove) that there are no more solutions?&lt;/p&gt;&#xA;" OwnerUserId="1770" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-10T19:40:34.187" LastActivityDate="2011-01-09T19:44:40.163" Title="$\binom{n}{k} : \binom{n}{k+1} : \binom{n}{k+2} = a : b : c$" Tags="&lt;binomial-coefficients&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="8349" PostTypeId="1" CreationDate="2010-10-30T13:07:06.367" Score="1" ViewCount="134" Body="&lt;p&gt;Is there any way to decipher, manually of course , whether a number (may be large enough) is a prime?&lt;/p&gt;&#xA;" OwnerUserId="2943" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T15:08:37.630" LastActivityDate="2010-12-02T15:08:37.630" Title="How do you determine if a number is prime or composite?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" />
  <row Id="8350" PostTypeId="2" ParentId="5248" CreationDate="2010-10-30T13:50:53.383" Score="4" ViewCount="0" Body="&lt;p&gt;Let's consider the integrals&lt;br&gt;&#xA;$$I_1(t)=\int_t^{\infty}\frac{\sin(x-t)}{x}dx\qquad\mbox{ and }\qquad&#xA;I_2(t)=\int_0^{\infty}\frac{e^{-tx}}{1+x^2}dx,\qquad t\geq 0.$$&#xA;A direct calculation shows that $I_1(t)$ and $I_2(t)$ satisfy the ordinary differential equation&#xA;$$y''+y=\frac{1}{t},\qquad t&gt;0.$$&#xA;Therefore, the difference $I(t)=I_1(t)-I_2(t)$ satisfy the homogeneous differential equation&#xA;$$y''+y=0,\qquad t&gt;0,$$&#xA;hence it should be of the form &#xA;$$I(t)=A\sin (t+B) $$&#xA;with some constants $A$, $B$. But  $I_1(t)$ and $I_2(t)$ both converge to $0$ as $t\to\infty$. This implies that $A=0$ and $I_1(t)=I_2(t)$ for all $t\geq 0$. Finally, we have that&#xA;$$\int_0^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\arctan(\infty)-\arctan(0)=\frac{\pi}{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-10-30T23:35:44.327" LastActivityDate="2010-10-30T23:35:44.327" CommentCount="2" />
  <row Id="8352" PostTypeId="2" ParentId="8349" CreationDate="2010-10-30T14:29:50.953" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primality_test&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Primality_test&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-10-30T14:29:50.953" CommentCount="1" />
  <row Id="8353" PostTypeId="2" ParentId="8337" CreationDate="2010-10-30T14:32:30.077" Score="16" ViewCount="0" Body="&lt;p&gt;OK, here's my favorite. I thought of this after reading a proof from the book &quot;Proofs from the book&quot; by Aigner &amp;amp; Ziegler, but later I found more or less the same proof as mine in &lt;a href=&quot;http://homepage.univie.ac.at/josef.hofbauer/02amm.pdf&quot; rel=&quot;nofollow&quot;&gt;a paper&lt;/a&gt; published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9&#xA;(EDIT: ...which is actually the proof that I read in Aigner &amp;amp; Ziegler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $0 &amp;lt; x &amp;lt; \pi/2$ we have $0&amp;lt;\sin x &amp;lt; x &amp;lt; \tan x$ and thus&#xA;$$\frac{1}{\tan^2 x} &amp;lt; \frac{1}{x^2} &amp;lt; \frac{1}{\sin^2 x}.$$&#xA;Note that $1/\tan^2 x = 1/\sin^2 x - 1$.&#xA;Split the interval $(0,\pi/2)$ into $2^n$ equal parts, and sum&#xA;the inequality over the (inner) &quot;gridpoints&quot; $x_k=(\pi/2) \cdot (k/2^n)$:&#xA;$$\sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k} - \sum_{k=1}^{2^n-1} 1 &amp;lt; \sum_{k=1}^{2^n-1} \frac{1}{x_k^2} &amp;lt; \sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k}.$$&#xA;Denoting the sum on the right-hand side by $S_n$, we can write this as&#xA;$$S_n - (2^n - 1) &amp;lt; \sum_{k=1}^{2^n-1} \left( \frac{2 \cdot 2^n}{\pi} \right)^2 \frac{1}{k^2} &amp;lt; S_n.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although $S_n$ looks like a complicated sum, it can actually be computed fairly easily. To begin with,&#xA;$$\frac{1}{\sin^2 x} + \frac{1}{\sin^2 (\frac{\pi}{2}-x)} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x \cdot \sin^2 x} = \frac{4}{\sin^2 2x}.$$&#xA;Therefore, if we pair up the terms in the sum $S_n$ except the midpoint $\pi/4$ (take the point $x_k$ in the left half of the interval $(0,\pi/2)$ together with the point $\pi/2-x_k$ in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into $2^{n-1}$ parts. And the midpoint $\pi/4$ contributes with $1/\sin^2(\pi/4)=2$ to the sum. In short,&#xA;$$S_n = 4 S_{n-1} + 2.$$&#xA;Since $S_1=2$, the solution of this recurrence is&#xA;$$S_n = \frac{2(4^n-1)}{3}.$$&#xA;(For example like this: the particular (constant) solution $(S_p)_n = -2/3$ plus the general solution to the homogeneous equation $(S_h)_n = A \cdot 4^n$, with the constant $A$ determined by the initial condition $S_1=(S_p)_1+(S_h)_1=2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have&#xA;$$ \frac{2(4^n-1)}{3} - (2^n-1) &amp;lt;  \frac{4^{n+1}}{\pi^2} \sum_{k=1}^{2^n-1} \frac{1}{k^2}  &amp;lt; \frac{2(4^n-1)}{3}.$$&#xA;Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. This squeezes the partial sums between two sequences both tending to $\pi^2/6$. Voilà!&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-10-30T17:44:03.363" LastActivityDate="2010-10-30T17:44:03.363" CommentCount="1" />
  <row Id="8354" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T14:54:19.533" Score="4" ViewCount="0" Body="&lt;p&gt;Rolfsen's textbook &quot;Knots and links&quot; is quite nice. It assumes a 1st course in algebraic topology, and is pleasant self-learner text.  Plenty of nice exercises. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the higher-end of the knots textbook world, Burde and Zieschang's &quot;Knots&quot; covers quite a lot of ground in much detail.   Kawauchi's &quot;A survey of knot theory&quot; covers much more ground but in less detail.  Hillman's &quot;Algebraic invariants of links&quot; is more specialized and tends to focus on ideas such as Alexander modules, but it goes into them in more detail than I've seen anywhere outside of Jerry Levine's papers. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-10-30T14:54:19.533" />
  <row Id="8355" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T14:56:26.740" Score="5" ViewCount="0" Body="&lt;p&gt;The matrices $A$ and $B$ describe the same linear transformation $L$ of some vector space $V$ with respect to different bases. The eigenspaces $E_\lambda:=\lbrace x\in V\ |\ Lx=\lambda x\rbrace$ of $L$ have an invariant geometrical meaning and a well defined dimension. So there is actually nothing to prove.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-10-30T14:56:26.740" />
  <row Id="8356" PostTypeId="1" CreationDate="2010-10-30T14:56:41.467" Score="1" ViewCount="108" Body="&lt;p&gt;If we have two Random Variables X and Y, what is the interpretation for the three cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a] p(X=x,Y=y) = $p_1$(X=x) * $p_2$(Y=y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b] p(X=x,Y=y) &amp;lt; $p_1$(X=x) * $p_2$(Y=y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c] p(X=x,Y=y) &gt; $p_1$(X=x) * $p_2$(Y=y)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where p -&gt; joint pdf of X and Y,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_1$ -&gt; marginal pdf of X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_2$ -&gt; marginal pdf of Y&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-10-31T04:47:57.580" LastActivityDate="2010-12-31T03:44:30.187" Title="Interpretation of joint pdf/product of marginal pdfs" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="8357" PostTypeId="1" AcceptedAnswerId="8358" CreationDate="2010-10-30T15:29:36.377" Score="1" ViewCount="431" Body="&lt;p&gt;Give &lt;code&gt;a &amp;gt; b&lt;/code&gt;, where &lt;code&gt;a,b ∈ ℝ&lt;/code&gt;, is it always true that &lt;code&gt;a² &amp;gt; b²&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2946" LastActivityDate="2010-11-08T16:06:32.617" Title="If a &gt; b is a² &gt; b² ?" Tags="&lt;inequality&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" />
  <row Id="8358" PostTypeId="2" ParentId="8357" CreationDate="2010-10-30T15:30:39.303" Score="5" ViewCount="0" Body="&lt;p&gt;no its not. When $a,b$ are positive it happens. Consider a=-2 and b =-3. notice that inequality reverses.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-30T15:30:39.303" />
  <row Id="8359" PostTypeId="1" AcceptedAnswerId="8372" CreationDate="2010-10-30T15:37:53.723" Score="0" ViewCount="106" Body="&lt;p&gt;Could I calculate the answer to either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\left[ x^{2} y \right ] - \sum xy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum \left[ \left ( x - \sum xy \right )^{2} y \right ]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had any of these variables&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum xy $; $ \sum x $; $ \sum y $; $ \sum x^2 $; $ \sum y^2 $;&lt;br&gt;&#xA;and &lt;code&gt;n&lt;/code&gt; ( the amount of items in the list )&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;adding more info&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate the &lt;strong&gt;variance&lt;/strong&gt; of a discrete probability distribution on my calculator. There is no function $\sum \left[ \left ( x - \sum xy \right )^{2} y \right ]$ on my model, and I was wondering if I could get there using the other &lt;strong&gt;functions&lt;/strong&gt;. ( those listed above )&lt;/p&gt;&#xA;" OwnerUserId="2945" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-10-30T18:28:02.420" LastActivityDate="2010-10-30T18:28:02.420" Title="Calculating  $\sum\left[ x^{2} y \right ]$" Tags="&lt;sequences-and-series&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8361" PostTypeId="2" ParentId="8356" CreationDate="2010-10-30T16:13:55.797" Score="0" ViewCount="0" Body="&lt;p&gt;Two random variables $X$ and $Y$ are independent if $P(X=x, Y=y) = P(X=x)P(Y=y)$, i.e. their joint pdf factorizes. E.g. $X$ = getting heads on the first toss of a coin, $Y$ = getting heads on the second toss of the coin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for random variables $X$ and $Y$, $P(X = x, Y = y) \leq P(X=x)$ and $P(X = x, Y=y) \leq P(Y=y)$. This implies that (c) can't hold true. There's no specific interpretation to (b).&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-10-30T16:13:55.797" CommentCount="5" />
  <row Id="8362" PostTypeId="2" ParentId="8357" CreationDate="2010-10-30T16:21:47.373" Score="20" ViewCount="0" Body="&lt;p&gt;If $\rm\ a &gt; b\  $ then $\rm\ a^2 - b^2 = (a-b)\:(a+b) &gt; 0\ \iff\ a+b &gt;0 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-30T16:21:47.373" CommentCount="1" />
  <row Id="8363" PostTypeId="2" ParentId="8293" CreationDate="2010-10-30T16:39:57.753" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\ \: A = A',\ \ \: B = B'\ \Rightarrow\ A+B\  =\:\ A'+B'$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ 0 = A-A' + B-B' \: =\ (A+B) - (A'+B')$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-30T16:39:57.753" CommentCount="3" />
  <row Id="8364" PostTypeId="2" ParentId="123" CreationDate="2010-10-30T16:44:29.150" Score="3" ViewCount="0" Body="&lt;p&gt;Many kinds of nonlinear PDE have wave solutions explicitly expressed using hyperbolic tangents and secants: shock-wave profiles, solitons, reaction-diffusion fronts, and phase-transition fronts, for starters.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-10-30T16:44:29.150" />
  <row Id="8365" PostTypeId="2" ParentId="8322" CreationDate="2010-10-30T17:02:23.970" Score="1" ViewCount="0" Body="&lt;p&gt;Lang has written several (very similar) books that discuss these topics.  Basically, you're right: Riemannian metrics only make sense when the manifold is modeled on a Hilbert space.  There is, however, a notion of psuedo-Riemannian manifold, which makes sense when the manifold is modeled on a self-dual Banach space.  Take a look at Chapter 7 of Lang's book, visible through Google Books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=VfxGB5nYv1MC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=VfxGB5nYv1MC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1392" LastActivityDate="2010-10-30T17:02:23.970" CommentCount="1" />
  <row Id="8366" PostTypeId="1" CreationDate="2010-10-30T17:17:52.210" Score="0" ViewCount="60" Body="&lt;p&gt;I want to count integral of f(x)=x^2 in range for example 0-10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I read to do it by myself and apply it to other funcs ?&lt;/p&gt;&#xA;" OwnerUserId="1755" LastActivityDate="2010-10-30T18:37:55.423" Title="How to use integral - x^2" Tags="&lt;integral&gt;" AnswerCount="1" />
  <row Id="8367" PostTypeId="2" ParentId="8366" CreationDate="2010-10-30T17:20:52.737" Score="2" ViewCount="0" Body="&lt;p&gt;First learn differential calculus, then integral calculus. The formula to be used here is :$$\int_{a}^{b} x^{n} \ dx = \bigg[\displaystyle \frac{x^{n+1}}{n+1}\bigg]^{b}_{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the answer will be $\frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}$ &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-30T18:37:55.423" LastActivityDate="2010-10-30T18:37:55.423" CommentCount="4" />
  <row Id="8368" PostTypeId="2" ParentId="8336" CreationDate="2010-10-30T17:35:33.423" Score="1" ViewCount="0" Body="&lt;p&gt;It seems that as of now, doing individual cases might be a good way to get a feel for things. However, the outline below is motivated by 1.2.4 from TAOCP Volume 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that&#xA;$$\sum_{k=1}^n c_k = n c_n - \sum_{k=1}^{n-1} k (c_{k+1}-c_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this identity for the case where $c_k = \lfloor k/2\rfloor$, and it rapidly leads to the alleged proof. I have not worked out the details, but it seems that the other cases that you mention above should also follow similarly, noting for example that when $a=3$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{k+1}-c_k = \left\lfloor \frac{k+1}{3}\right\rfloor - \left\lfloor \frac{k}{3}\right\rfloor,$$ which is zero for all $k$, except for $k=3j-1$, as $j=1,2,\ldots$, in which case it equals $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these ideas, the latter sum in the identity above can be greatly simplified, which should help prove your claims.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-10-30T17:35:33.423" />
  <row Id="8369" PostTypeId="1" CreationDate="2010-10-30T17:57:58.307" Score="0" ViewCount="59" Body="&lt;p&gt;Convex optimization problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimize,&#xA; $\sum{|y-x|^2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that,&lt;br&gt;&#xA;a] K-L divergence between probability distributions for x and y, is maximum i.e. D(p(y)||q(x)) is maximum    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b] $\sum{y} &amp;lt; \alpha$, where $\alpha$ is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c] x,y &amp;lt; $\beta$&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-10-30T17:57:58.307" Title="convex optimization problem" Tags="&lt;optimization&gt;&lt;convex-optimization&gt;" />
  <row Id="8370" PostTypeId="2" ParentId="8336" CreationDate="2010-10-30T18:01:15.967" Score="2" ViewCount="0" Body="&lt;p&gt;You should be able to use the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle n = aj+r$, where $ 0 \le r &amp;lt; a$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=1}^{n} \left\lfloor \frac{k}{a} \right\rfloor = aj(j-1)/2 + rj$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying and dividing the RHS by $8a$ and using $aj = n-r$ would help, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I haven't worked out the remaining details myself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-10-30T18:01:15.967" CommentCount="1" />
  <row Id="8371" PostTypeId="2" ParentId="8318" CreationDate="2010-10-30T18:08:44.893" Score="1" ViewCount="0" Body="&lt;p&gt;The taps, including an extra &quot;zero&quot; tap, form a polynomial $P(X)$ over $GF(2)$, we should satisfy two properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$P(X)$ must be irreducible. Otherwise, the sequence decomposes (in some way) as a &quot;direct sum&quot; of several different LFSRs (try an example to see what this means).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The order of $X$ modulo $P(X)$ is the period of the LFSR sequence. If the period is maximal ($2^{\deg P} - 1$), then the polynomial is called primitive. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the number of taps is odd then $P(X)$ will have an even number of elements, so $P(1) = 0$. This implies that $P(X)$ is divisible by $X+1$, and so is reducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the taps have a common factor $k$, then taking $Y=X^k$ you can recover LFSR sequences of a smaller polynomial $Q(Y) = P(X)$ (look at the output at intervals of $k$), and this implies that the period of the original sequence can't be maximal. Try and see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not difficult to calculate (using Moebius inversion) the number of irreducible/primitive polynomials, and there turn out to be lots of them. There are also ways to test if a polynomial is irreducible (in general, to factor it) and whether it's primitive (this requires factorizing $2^{\deg P}-1$ and then becomes trivial). So in practice, you can just &quot;do it yourself&quot; - try several polynomials, and soon enough you'll hit upon a nice one.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-30T18:08:44.893" CommentCount="2" />
  <row Id="8372" PostTypeId="2" ParentId="8359" CreationDate="2010-10-30T18:22:24.923" Score="2" ViewCount="0" Body="&lt;p&gt;No, you cannot recover the variance of the distribution from the information you have. Here is a simple counterexample:&#xA;$$A = \{(0,1), (1,0), (2,0), (3,1)\},$$&#xA;$$B = \{(0,0), (1,1), (2,1), (3,0)\}.$$&#xA;Both $A$ and $B$ have the same $n = 4$, $\sum x = 6$, $\sum y = 2$, $\sum x^2 = 14$, $\sum y^2 = 2$, and $\sum xy = 3$, but $\sum x^2 y$ is $9$ for $A$ and $5$ for $B$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-10-30T18:22:24.923" CommentCount="2" />
  <row Id="8373" PostTypeId="2" ParentId="8337" CreationDate="2010-10-30T18:26:36.583" Score="6" ViewCount="0" Body="&lt;p&gt;I have two favorite proofs.  One is the last proof in Robin Chapman's collection; you really should take a look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is a proof that generalizes to the evaluation of $\zeta(2n)$ for all $n$, although I'll do it &quot;Euler-style&quot; to shorten the presentation.  The basic idea is that meromorphic functions have infinite partial fraction decompositions that generalize the partial fraction decompositions of rational functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular function we're interested in is $B(x) = \frac{x}{e^x - 1}$, the exponential generating function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt; $B_n$.  $B$ is meromorphic with poles at $x = 2 \pi i n, n \in \mathbb{Z}$, and at these poles it has residue $2\pi i n$.  It follows that we can write, a la Euler,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{e^x - 1} = \sum_{n \in \mathbb{Z}} \frac{2\pi i n}{x - 2 \pi i n} = \sum_{n \in \mathbb{Z}} - \left( \frac{1}{1 - \frac{x}{2\pi i n}} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can expand each of the terms on the RHS as a geometric series, again a la Euler, to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{e^x - 1} = - \sum_{n \in \mathbb{Z}} \sum_{k \ge 0} \left( \frac{x}{2\pi i n} \right)^k = \sum_{k \ge 0} (-1)^{n+1} \frac{2 \zeta(2n)}{(2\pi )^n} x^{2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because, after rearranging terms, the sum over odd powers cancels out and the sum over even powers doesn't.  (This is one indication of why there is no known closed form for $\zeta(2n+1)$.)  Equating terms on both sides, it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_{2n} = (-1)^{n+1} \frac{2 \zeta(2n)}{(2\pi)^n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} (2\pi)^n}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.  To compute $\zeta(2)$ it suffices to compute that $B_2 = \frac{1}{6}$, which then gives the usual answer.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-30T18:26:36.583" CommentCount="1" />
  <row Id="8374" PostTypeId="1" CreationDate="2010-10-30T18:51:34.797" Score="1" ViewCount="47" Body="&lt;p&gt;I would like a family of functions $f_a(x)$ so that $f_a$ is $a\in\mathbb{R}$ fractionally differentiable but not $a+\epsilon$ fractionally differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know such functions which are easily described? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-10-31T20:03:40.423" Title="Family of function with fractional derivatives" Tags="&lt;functions&gt;" AnswerCount="1" />
  <row Id="8375" PostTypeId="1" AcceptedAnswerId="8392" CreationDate="2010-10-30T18:54:52.700" Score="1" ViewCount="61" Body="&lt;p&gt;While reading a probability paper titled &lt;a href=&quot;http://www.springerlink.com/content/j83542h24q561674/&quot; rel=&quot;nofollow&quot;&gt;Gröbner bases and factorisation in discrete probability and Bayes&lt;/a&gt; (can't find free version, sorry),  I came across the explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of indicator functions $I_i$ form a Boolean algebra that is in one-to-one correspondence with the quotient of the ring $\mathbb{R}[x_1, \ldots, x_n]$ with respect to the relationships $x_i(x_i - 1)=0$ where $x_i$ represents $I_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_i$ is defined to be the indicator function $I_i(\omega) = 1 \;\textrm{if}\;  \omega \in A_i \;\textrm{else}\; 0$, $i = 1, \ldots, n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a rudimentary understanding of rings, but can you please explain the correspondence shown here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: In the rest of the paper, the authors don't seem to limit themselves to $\mathbb{F}_2$. In fact, I can't see a need for the use of the minus sign in $x_i(x_i-1)$ if the field is $\mathbb{F}_2$. I can't verify if the coefficients of the polynomials are disregarded yet.&lt;/p&gt;&#xA;" OwnerUserId="2949" LastEditorUserId="2949" LastEditorDisplayName="" LastEditDate="2010-10-31T11:57:13.757" LastActivityDate="2010-10-31T11:57:13.757" Title="Quotient ring and Boolean algebra correspondence" Tags="&lt;algebra&gt;&lt;ring-theory&gt;&lt;boolean-algebra&gt;" AnswerCount="2" />
  <row Id="8376" PostTypeId="2" ParentId="8375" CreationDate="2010-10-30T18:59:57.810" Score="2" ViewCount="0" Body="&lt;p&gt;I assume $i$ is an element of $\{ 1, 2, ... n \}$.  The correspondence comes from identifying $x_i$ with the indicator function of $\{ i \}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the sentence you quote is imprecise.  If you wanted a Boolean algebra (more precisely, a Boolean ring), you would take the base field to be $\mathbb{F}_2$, not $\mathbb{R}$.  If you wanted real multiples of indicator functions, you wouldn't say &quot;the set of indicator functions.&quot;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-30T18:59:57.810" />
  <row Id="8377" PostTypeId="1" AcceptedAnswerId="8441" CreationDate="2010-10-30T19:00:10.603" Score="0" ViewCount="87" Body="&lt;p&gt;Greetings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides Un+1 - Un ?&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2950" LastEditorDisplayName="" LastEditDate="2010-10-30T23:55:34.450" LastActivityDate="2010-10-31T11:03:26.620" Title="Another way to calculate monotonicity of a sequence" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8378" PostTypeId="2" ParentId="8337" CreationDate="2010-10-30T19:01:54.847" Score="8" ViewCount="0" Body="&lt;p&gt;We can use the function $f(x)=x^{2}$ with $-\pi \leq x\leq \pi $ and find&#xA;its expansion into a trigonometric Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}\cos nx+b_{n}\sin x),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is periodic and converges to $f(x)$ in $-\pi \leq x\leq \pi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing that $f(x)$ is even, it is enough to determine the coefficients &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }f(x)\cos nx\;dx\qquad n=0,1,2,3,...,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;because &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }f(x)\sin nx\;dx=0\qquad n=1,2,3,... .$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=0$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{0}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }x^{2}dx=\dfrac{2}{\pi }\int_{0}^{\pi&#xA;}x^{2}dx=\dfrac{2\pi ^{2}}{3}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for $n=1,2,3,...$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{n}=\dfrac{1}{\pi }\int_{-\pi }^{\pi }x^{2}\cos nx\;dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\dfrac{2}{\pi }\int_{0}^{\pi }x^{2}\cos nx\;dx=\dfrac{2}{\pi }\times \dfrac{%&#xA;2\pi }{n^{2}}(-1)^{n}=(-1)^{n}\dfrac{4}{n^{2}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int x^2\cos nx\;dx=\dfrac{2x}{n^{2}}\cos nx+\left( \frac{x^{2}}{%&#xA;n}-\dfrac{2}{n^{3}}\right) \sin nx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\dfrac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\dfrac{1}{n^{2}}%&#xA;\cos nx\right) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f(\pi )=\pi ^{2}$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\dfrac{4}{%&#xA;n^{2}}\cos \left( n\pi \right) \right) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\left( (-1)^{n}(-1)^{n}%&#xA;\dfrac{1}{n^{2}}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi ^{2}=\dfrac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=\dfrac{\pi ^{2}}{4}-\dfrac{\pi ^{2}}{12}=%&#xA;\dfrac{\pi ^{2}}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Second method (available on-line a few years ago) by Eric Rowland. From&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1-t)=-\sum_{n=1}^{\infty}\dfrac{t^n}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and making the substitution $t=e^{ix}$ one gets the series expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=\text{Log}(1-e^{ix})=-\sum_{n=1}^{\infty }\dfrac{e^{inx}}{n}=-\sum_{n=1}^{%&#xA;\infty }\dfrac{1}{n}\cos nx-i\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose radius of convergence is $1$. Now if we take the imaginary part of both sides, the RHS becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im w=-\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the LHS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Im w=\arg \left( 1-\cos x-i\sin x\right) =\arctan \dfrac{-\sin x}{%&#xA;1-\cos x}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arctan \dfrac{-\sin x}{1-\cos x}=-\arctan \dfrac{2\sin \dfrac{x}{2}\cdot \cos \dfrac{x}{2}}{2\sin ^{2}\dfrac{x}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\arctan \cot \dfrac{x}{2}=-\arctan \tan \left( \dfrac{\pi }{2}-\dfrac{x}{2}%&#xA;\right) =\dfrac{x}{2}-\dfrac{\pi }{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following expansion holds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi }{2}-\frac{x}{2}=\sum_{n=1}^{\infty }\dfrac{1}{n}\sin nx.\qquad&#xA;(\ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the identity $(\ast )$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi }{2}x-\dfrac{x^{2}}{4}+C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\cos&#xA;nx.\qquad (\ast \ast )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $x=0$, we get the relation between $C$ and $\zeta (2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=-\zeta (2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for $x=\pi $, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta (2)=2\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}}{n^{2}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we deduce&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi ^{2}}{4}+C=-\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\cos n\pi&#xA;=\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}}{n^{2}}=\dfrac{1}{2}\zeta (2)=-\dfrac{1}{%&#xA;2}C.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=-\dfrac{\pi ^{2}}{6},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we thus prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta (2)=\dfrac{\pi ^{2}}{6}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: this 2nd method can generate all the zeta values $\zeta (2n)$ by integrating repeatedly $(\ast\ast )$. This is the reason why I appreciate it. Unfortunately it does not work for $\zeta (2n+1)$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-10-31T22:19:36.730" LastActivityDate="2010-10-31T22:19:36.730" />
  <row Id="8379" PostTypeId="1" AcceptedAnswerId="8402" CreationDate="2010-10-30T19:13:45.920" Score="1" ViewCount="232" Body="&lt;p&gt;Well, this is an exercise problem from &lt;em&gt;Herstein&lt;/em&gt; which sounds difficult:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does one prove that if $|G|&gt;2$, then $G$ has non-trivial automorphism?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only thing, i know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the &lt;em&gt;Inner- Automorphism&lt;/em&gt; of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-10-30T19:24:11.587" LastActivityDate="2010-10-30T21:07:41.440" Title="$|G|&gt;2$ implies $G$ has non trivial automorphism." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="8380" PostTypeId="2" ParentId="8322" CreationDate="2010-10-30T19:13:54.393" Score="2" ViewCount="0" Body="&lt;p&gt;There are two notions of what it means for a(n infinite dimensional) manifold to have a Riemannian structure.  A &lt;strong&gt;strong&lt;/strong&gt; Riemannian structure means a (smooth) choice of inner product on each tangent space which induces an isomorphism of each tangent space with its corresponding cotangent space.  A &lt;strong&gt;weak&lt;/strong&gt; Riemannian structure simply means a (smooth) choice of inner product on each tangent space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong Riemannian structures only exist if the manifold is modelled on a Hilbert space, and even then they have to be chosen correctly (so the usual $L^2$-metric on the space of $L^{1,2}$-loops is not a strong structure, even though the manifold is Hilbertian).  Weak Riemannian structures exist much more widely.  For a weak Riemannian structure you only need to know that the manifold admits smooth partitions of unity and that the model spaces admit continuous inner products.  So, for example, continuous loops in a smooth manifold admit a weak Riemannian structure but not a strong one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although strong Riemannian structures are very good for generalising much of ordinary differential geometry to infinite dimensions, there are occasions where the requirement of having a Hilbert manifold is too strong, and one can get away with merely having a weak Riemannian structure.  I've written an article where having a weak Riemannian structure on the space of smooth loops was an essential step and where the construction &lt;em&gt;would not have worked&lt;/em&gt; on a Hilbertian manifold (though actually it was a &lt;em&gt;co&lt;/em&gt;-Riemannian structure that I needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Declaration of interests: I've actually proposed a refinement of the &quot;weak/strong&quot; classification as I found it too harsh.  See my article &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Research/Preprints/constructdirac.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for this, and the above-mentioned result, and a load of examples of spaces with different types of Riemannian structure.)&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-10-30T19:13:54.393" CommentCount="2" />
  <row Id="8381" PostTypeId="2" ParentId="8319" CreationDate="2010-10-30T19:18:07.930" Score="4" ViewCount="0" Body="&lt;p&gt;You can find such partial fraction expansions given as examples in many texts on complex analysis. For example, you will find a derivation of this series in Example 1 at &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;lpg=PA64&amp;amp;dq=partial%20fraction%20series%20csc%20cot%20sec%20complex&amp;amp;pg=PA59#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; to Markushevich and Silverman's book.  It uses &quot;Cauchy's theorem on partial fraction expansions&quot;, given a few pages earlier at &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;lpg=PA64&amp;amp;dq=partial%20fraction%20series%20csc%20cot%20sec%20complex&amp;amp;pg=PA56#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2 gives the series for $\cot$, although from the preview I can't see whether $\tan$ and $\csc$ are included. The analogous series for $\tan$ is derived on &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_fractions_in_complex_analysis#Example&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.  Each of these series can be useful for evaluating numerical series.  For example, taking $x=0$ in your series you get &#xA;$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.$$&#xA; And although it is not directly related to your question, I thought I'd add in light of the &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n1-infty-frac1n2&quot;&gt;recent question&lt;/a&gt; on evaluating the sum of the reciprocals of the squares that if you take the partial fraction series &#xA;$$\tan(x) = \sum_{k=0}^{\infty} \frac{-2x}{x^2 - \left(k + \frac{1}{2}\right)^2\pi^2},$$&#xA;divide by $x$ and let $x$ go to $0$, rearranging yields &#xA;$$\frac{\pi^2}{8}=1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots,$$&#xA;which in turn (because the sum over the evens is $\frac{1}{4}$ the total sum) leads to &#xA;$$\frac{\pi^2}{6}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots.$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-31T06:44:08.973" LastActivityDate="2010-10-31T06:44:08.973" />
  <row Id="8382" PostTypeId="2" ParentId="8326" CreationDate="2010-10-30T19:20:28.033" Score="1" ViewCount="0" Body="&lt;p&gt;This answer is a little less popular of a suggestion, but it was how I first learned about knots, and I really enjoyed it. This book is primarily focused on Vassiliev Diagrams, and is currently unpublished, but available through the authors webpages. Yay! free stuff!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to &lt;a href=&quot;http://www.math.ohio-state.edu/~chmutov/preprints/&quot; rel=&quot;nofollow&quot;&gt;the CDbook&lt;/a&gt; by Chmutov, Duzhin, and Mostovoy. It is the first link on the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally found the focus on invariants very useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second favorite would be Rolfsen, as Ryan suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-10-30T19:20:28.033" />
  <row Id="8383" PostTypeId="2" ParentId="8346" CreationDate="2010-10-30T19:21:01.450" Score="3" ViewCount="0" Body="&lt;p&gt;Typically one considers the Hamming distance for the possible lengths of the messages, the larger the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.5.5027&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Koopman &amp;amp; Chakravarty&lt;/em&gt;, Cyclic Redundancy Code (CRC) Polynomial Selection For Embedded Networks&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="210" LastActivityDate="2010-10-30T19:21:01.450" />
  <row Id="8384" PostTypeId="1" CreationDate="2010-10-30T19:26:31.030" Score="2" ViewCount="65" Body="&lt;p&gt;Given the following game,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is a hider and a seeker, each game consists of the hider randomly selecting one of two places to hide that is A with probability p and B with 1 - p, and the seeker randomly choses to search in A with probability q and B with 1 - q. If seeker finds hider then game ends, else both randomly pick again. This continues each time both randomly picking where to hide and search, until seeker finds hider.&#xA;I want to know how to formulate the average number of game iterations to expect using theory...that is using p and q, how can i express how long i should expect it to take for seeker to find hider?&lt;/p&gt;&#xA;" OwnerUserId="2952" LastActivityDate="2010-10-30T20:18:45.320" Title="how do i theoretically find the expected length of the following game?" Tags="&lt;game-theory&gt;" AnswerCount="2" />
  <row Id="8385" PostTypeId="1" AcceptedAnswerId="8395" CreationDate="2010-10-30T19:30:08.840" Score="3" ViewCount="198" Body="&lt;p&gt;Using n_th root of unity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$(e^{\frac{2ki\pi}{n}})^{n} = 1$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2951" LastEditorUserId="2951" LastEditorDisplayName="" LastEditDate="2010-10-30T19:45:00.543" LastActivityDate="2010-11-01T03:47:26.787" Title="Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$" Tags="&lt;trigonometry&gt;&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="8386" PostTypeId="2" ParentId="8348" CreationDate="2010-10-30T19:33:40.837" Score="0" ViewCount="0" Body="&lt;p&gt;Sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\binom{n}{k},\binom{n}{k+1},\binom{n}{k+2}$ are in ratio $1:2:3$. Then $\frac{n-k}{k+1}=2$ and $\frac{n-k-1}{k+2}=\frac{3}{2}$. Solving these two equations gives $n=14$ and $k=4$.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-10-30T19:33:40.837" CommentCount="2" />
  <row Id="8387" PostTypeId="2" ParentId="8384" CreationDate="2010-10-30T19:39:00.053" Score="5" ViewCount="0" Body="&lt;p&gt;The probability that the game doesn't end in a particular round is $x = p(1 - q) + q(1 - p)$, so the probability that the game lasts for exactly $n$ rounds is $x^{n-1}(1 - x)$.  This means that the expected length of a game is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 1} n x^{n-1} (1 - x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sum can be found by various methods, depending on what you're comfortable with; for now I'll just say that the answer works out to be $\frac{1}{1 - x}$.  In other words, it's the reciprocal of the probability that the game ends.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-10-30T19:39:00.053" CommentCount="3" />
  <row Id="8388" PostTypeId="2" ParentId="3087" CreationDate="2010-10-30T19:39:05.890" Score="1" ViewCount="0" Body="&lt;p&gt;In high school in the 1970s I got tremendous benefit out of a one-term elective course called Mathematical Logic, which was nothing fancy, just basic prepositional stuff.  Understanding the basic structure of logical arguments empowers one to learn anything more efficiently. If college-bound students could ditch half of their AP Calculus for a course in &lt;b&gt;Logical Reasoning&lt;/b&gt; that includes basic material from Daniel Velleman's book &lt;i&gt;How to Prove It&lt;/i&gt; and some basic facts about statistics, they would be in a far better position to pursue many subjects, not just math.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-10-30T19:39:05.890" CommunityOwnedDate="2010-10-30T19:39:05.890" CommentCount="1" />
  <row Id="8389" PostTypeId="1" CreationDate="2010-10-30T19:39:40.910" Score="2" ViewCount="135" Body="&lt;p&gt;Let G be a finite p-group and has a nontrivial normal subgroup H, then the intersection of H and its center is not trivial.&lt;/p&gt;&#xA;" OwnerUserId="2953" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T04:24:48.123" LastActivityDate="2010-10-31T17:55:41.333" Title="A question about p-group" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8390" PostTypeId="2" ParentId="8389" CreationDate="2010-10-30T19:54:38.257" Score="1" ViewCount="0" Body="&lt;p&gt;Let $a_{1}, . . . , a_{k}$ be representatives of the conjugacy classes of $G$, ordered such that $a_{m} \in H$ and $a_{m+1}, \cdots , a_{k} \notin H$. The conjugacy class $C(a_{i})$ have either $C(a_{i}) \subset H$ or $C(a_{i}) \cap H = {e}$. First arrange the ${a_{1}, . . . , a_{m}}$ so that the first $r$  represent conjugacy classes of size 1, (i.e. elements in $H \cap Z$) and the latter $m − r$ represent classes of size larger than 1. Then we can write the&#xA;class equation for $H \cap Z = H$ as: $$|H| = \sum\limits_{i=1}^{m} |C(a_{i} \cap H| = |H \cap Z| + \sum\limits_{i=r}^{m} |C(a_{i})| =  |H \cap Z| + \sum\limits_{i=r}^{m}\frac{|G|}{|N(a_{i})|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $|H| &amp;lt; p^{n}$ every term in the sum is divisible by $p$ so $|H \cap Z|$ from above is divisible by $p$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-30T19:54:38.257" />
  <row Id="8391" PostTypeId="2" ParentId="3087" CreationDate="2010-10-30T20:03:32.207" Score="4" ViewCount="0" Body="&lt;p&gt;I strongly believe that even if something can be done by calculator, a high school student should understand some nuts and bolts of how it's being done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent some time doing volunteer tutoring at CRLS (Cambridge Rindge and Latin School). I still remember the girl at 8th or 9th grade who was studying inequalities, yet she couldn't compute $2-1$ without the help of calculator. The most frustrating case was the kid in 9th or 10th grade who didn't know multiplication tables, he couldn't compute, say, 3 times 7 in his head. I was supposed to help him with square roots. Well, how do you explain that $\sqrt{8} = 2\sqrt{2}$? First of, you have to realize that $8 = 2 \times 2 \times 2$, and that's exactly what he couldn't do. I ended up spending some extra time with him, afterward been told that he passed the test or exam and was very happy with my help - but I wasn't! I felt we were wasting lots of time and just because he didn't know multiplication tables, he couldn't break down numbers into factors, and calculator wasn't really help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line: I think in school students &lt;em&gt;still&lt;/em&gt; should learn basic arithmetic before moving to more advanced stuff. I think it's a pure idiocy to introduce the kids to set theory while having them relying on their calculators to compute $1 + 1$. Honestly, the advanced stuff can wait till college. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-10-30T20:03:32.207" CommunityOwnedDate="2010-10-30T20:03:32.207" />
  <row Id="8392" PostTypeId="2" ParentId="8375" CreationDate="2010-10-30T20:08:16.737" Score="0" ViewCount="0" Body="&lt;p&gt;A nice exposition of Stone's cryptomorphism between Boolean algebras and Boolean rings is in Burris and Sankappanavar's textbook &lt;a href=&quot;http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra.pdf#page=154&quot; rel=&quot;nofollow&quot;&gt;A Course in Universal Algebra&lt;/a&gt;. There you'll also find a presentation of Boolean products - a powerful tool that facilitates the transfer of Boolean algebra results to other algebraic structures - leading to very general representation theories.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-30T20:08:16.737" />
  <row Id="8393" PostTypeId="1" AcceptedAnswerId="8399" CreationDate="2010-10-30T20:09:06.007" Score="1" ViewCount="109" Body="&lt;p&gt;Simple question-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for a linear programming optimization problem possible to have a degenerate optimal solution whereas the dual has a unique optimal solution? I can't find a scenario where this is possible but intuitively it seems that it is possible.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-10-30T21:55:28.157" LastActivityDate="2010-10-30T21:55:28.157" Title="Primal- degenerate optimal, Dual - unique optimal" Tags="&lt;optimization&gt;&lt;linear-programming&gt;" AnswerCount="1" />
  <row Id="8395" PostTypeId="2" ParentId="8385" CreationDate="2010-10-30T20:18:15.243" Score="8" ViewCount="0" Body="&lt;p&gt;$$P=\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=(2i)^{1-n}e^{-i\pi n(n-1)/(2n)}\prod_{k=1}^{n-1}(e^{2ik\pi/n}-1)=$$&#xA;$$(-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1)=2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k),$$&#xA;where $\xi=e^{2i\pi/n}$. Now note, that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, thus cancelling $x-1$ we have $\prod_{k=1}^{n-1} (x-\xi^k) =\sum_{k=0}^{n-1}x^k$. Substituting $x=1$ we have $\prod_{k=1}^{n-1} (1-\xi^k)=n$. Therefore $P=n2^{1-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;In order to note that $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, note that $1,\xi,\dots,\xi^{n-1}$ are roots of $x^n-1$. Therefore by &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;polynomial reminder theorem&lt;/a&gt; we have $x^n-1=Q(x) \prod_{k=0}^{n-1} (x-\xi^k)$. Comparing degrees of lhs and rhs we can find, that $Q(x)$ has degree 0. Comparing highest coefficients we can conclude $Q(x)=1$.&lt;/p&gt;&#xA;" OwnerUserId="695" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T03:47:26.787" LastActivityDate="2010-11-01T03:47:26.787" CommentCount="10" />
  <row Id="8396" PostTypeId="2" ParentId="8384" CreationDate="2010-10-30T20:18:45.320" Score="6" ViewCount="0" Body="&lt;p&gt;You can sometimes answer such questions using the idea of &quot;conditioning&quot;. Students often prefer this method as there are no infinite sums to calculate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things can happen on the first round of this game: either the seeker finds the hider (probability = $pq+(1-p)(1-q)$), or the game starts over (probability = $p+q-2pq$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $t$ is the average number of iterations that the game lasts, then averaging over  these two scenarios gives $$t=(1)(pq+(1-p)(1-q))+(1+t)(p+q-2pq).$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $t$ gives $t=1/(pq+(1-p)(1-q))$.   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-10-30T20:18:45.320" />
  <row Id="8398" PostTypeId="2" ParentId="7531" CreationDate="2010-10-30T20:32:14.653" Score="0" ViewCount="0" Body="&lt;p&gt;Success finally!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f( n) = \sum_{(k,n)=1;1\leq k\leq n} \log\Bigl(\frac{k}{n}\Bigr)$$ therefore we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{d|n}f(d) =\log\Bigl(\frac{1}{n}\Bigr)+...+\log\Bigl(\frac{n}{n}\Bigr)=\log\left(\frac{n!}{n^n}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by Moebius Inversion Formula: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = \sum_{d|n}\log\left(\frac{d!}{d^d}\right)\cdot \mu\left(\frac{n}{d}\right) = \log\left(\prod_{d|n}\left(\frac{d!}{d^d}\right)^{\mu\left(\frac{n}{d}\right) }\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(k)} -\phi(n)\cdot \log( n) = \log(P(n))-\log(n^{\phi(n)})$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-10-30T20:32:14.653" />
  <row Id="8399" PostTypeId="2" ParentId="8393" CreationDate="2010-10-30T20:55:13.407" Score="2" ViewCount="0" Body="&lt;p&gt;The answer is yes, but only if there are other optimal solutions than the degenerate one.  For example, suppose the primal problem is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max x_1 + x_2$$&#xA;subject to&#xA;$$x_1 \leq 1,$$&#xA;$$x_1 + x_2 \leq 1,$$&#xA;$$x_1, x_2 \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution $(1,0)$ is optimal and degenerate, but every solution $(a,1-a)$, for $0 \leq a \leq 1$ is also optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min y_1 + y_2$$&#xA;subject to&#xA;$$y_1 + y_2 \geq 1,$$&#xA;$$y_2 \geq 1,$$&#xA;$$y_1, y_2 \geq 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual has the unique (degenerate) optimal solution $(0,1)$.  So we do have a situation with a degenerate optimal solution in the primal but a unique dual optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the degenerate optimal solution is unique, then there &lt;em&gt;must&lt;/em&gt; be multiple optimal solutions in the dual.  The following table is from Sierksma's &lt;em&gt;Linear and Integer Programming: Theory and Practice&lt;/em&gt;, Volume 1, page 144.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Primal Optimal Solution                     Dual Optimal Solution&#xA;(a) Multiple                     implies    Degenerate&#xA;(b) Unique and nondegenerate     implies    Unique and nondegenerate&#xA;(c) Multiple and nondegenerate   implies    Unique and degenerate&#xA;(d) Unique and degenerate        implies    Multiple&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-30T20:55:13.407" />
  <row Id="8400" PostTypeId="2" ParentId="8389" CreationDate="2010-10-30T21:03:12.257" Score="6" ViewCount="0" Body="&lt;p&gt;Perhaps a slightly less computationally-intensive argument is to simply note that since $H$ is normal, it must be a union of conjugacy classes. Each conjugacy class of $G$ has $p^i$ elements for some $i$; since $H$ contains at least one conjugacy class with $p^0 = 1$  elements (the class of the identity), and $|H|\equiv 0 \pmod{p}$, it must contain other classes with just one element, which must be classes of central elements of $G$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-30T21:03:12.257" CommentCount="6" />
  <row Id="8401" PostTypeId="2" ParentId="8379" CreationDate="2010-10-30T21:05:58.573" Score="9" ViewCount="0" Body="&lt;p&gt;If $G$ is not abelian, then conjugation by a noncentral element will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is abelian, then $x\mapsto x^{-1}$ is an automorphism. It will be nontrivial unless every element of $G$ equals its inverse, that is, if every element of $G$ is of exponent $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If every element of $G$ is of exponent $2$, then $G$ is a vector space over the field of $2$ elements, so it is isomorphic to a (possibly infinite) sum of copies of $C_2$, the cyclic group of two elements. Since $|G|\gt 2$, there are at least two copies, so the linear transformation that swaps two copies of $C_2$ is a nontrivial automorphism.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-10-30T21:05:58.573" CommentCount="16" />
  <row Id="8402" PostTypeId="2" ParentId="8379" CreationDate="2010-10-30T21:07:41.440" Score="5" ViewCount="0" Body="&lt;p&gt;As you note in the question, the group of inner automorphisms is isomorphic to $G/Z(G)$. In particular, it's non-trivial if and only if $Z(G)=G$.  So there is a non-trivial (inner) automorphism unless $G=Z(G)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice that, by definition, $Z(G)=G$ if and only if $G$ is abelian; so we have reduced to the abelian case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is abelian then $g\mapsto -g$ is an automorphism, and it is non-trivial unless $g=-g$ for all $g\in G$.  But $g=-g$ if and only if the order of $g$ divdes two.  So we have now reduced to the case in which $2g=0$ for all $g\in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $G$ is a vector space over the field $\mathbb{Z}/2$.  As&#xA;$|G|$ is equal to 2 raised to the power of the $\mathbb{Z}/2$-dimension of $G$,&#xA;the hypothesis that $|G|&gt;2$ implies that $\mathrm{dim}_{\mathbb{Z/2}} G&gt;1$.  But now we can write down lots of linear automorphisms of $G$. For instance, you could fix any basis $g_1,g_2,\ldots$ and take the automorphism $g_1\mapsto g_2$, $g_2\mapsto g_1$ and $g_i\mapsto g_i$ for every $i&gt;2$.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-10-30T21:07:41.440" />
  <row Id="8403" PostTypeId="1" AcceptedAnswerId="8406" CreationDate="2010-10-30T21:19:09.143" Score="0" ViewCount="93" Body="&lt;p&gt;I'm sorry if the English isn't correct, I'm not native.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say for example these two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$((-1)^n+1/n)^2$$&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2950" LastEditorDisplayName="" LastEditDate="2010-10-31T14:24:00.877" LastActivityDate="2010-10-31T14:24:00.877" Title="How to check if a sequence is bounded from above, bounded from below or bounded?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8404" PostTypeId="2" ParentId="8002" CreationDate="2010-10-30T21:29:49.207" Score="1" ViewCount="0" Body="&lt;p&gt;In case anyone is interested, here are the full versions of my two proofs.  (I constructed the combinatorial one from my original partially combinatorial one after I posted the question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The non-combinatorial proof&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Derek Jennings observes, $\lfloor \frac{n+1}{k} \rfloor - \lfloor \frac{n}{k} \rfloor$ is $1$ if $k|(n+1)$ and $0$ otherwise.  Thus, if $$f(n) = \sum_{k=1}^n \left\lfloor\frac{n}{k} \right\rfloor \varphi (k),$$&#xD;&#xA;then $$\Delta f(n) = f(n+1) - f(n) = \sum_{k|(n+1)} \phi(k) = n+1,$$&#xD;&#xA;where the last equality follows from the well-known formula Moron cites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$\sum_{k=1}^n \left\lfloor\frac{n}{k} \right\rfloor \varphi (k) = f(n) = \sum_{k=0}^{n-1} \Delta f(k) = \sum_{k=0}^{n-1} (k+1) = \frac{n(n+1)}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The combinatorial proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sides count the number of fractions (reducible or irreducible) in the interval (0,1] with denominator $n$ or smaller.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the right side, the number of ways to pick a numerator and a denominator is the number of ways to choose two numbers with replacement from the set $\{1, 2, \ldots, n\}$.  This is known to be &#xA;$$\binom{n+2-1}{2} = \frac{n(n+1)}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the left side.  The number of irreducible fractions in $(0,1]$ with denominator $k$ is equal to the number of positive integers less than or equal to $k$ and relatively prime to $k$; i.e., $\varphi(k)$.  Then, for a given irreducible fraction $\frac{a}{k}$, there are $\left\lfloor \frac{n}{k} \right\rfloor$ total fractions with denominators $n$ or smaller in its equivalence class.  (For example, if $n = 20$ and $\frac{a}{k} = \frac{1}{6}$, then the fractions $\frac{1}{6}, \frac{2}{12}$, and $\frac{3}{18}$ are those in its equivalence class.)  Thus the sum &#xA;$$\sum_{k=1}^n \left\lfloor\frac{n}{k} \right\rfloor \varphi (k)$$&#xD;&#xA;also gives the desired quantity.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T05:15:18.223" LastActivityDate="2010-11-22T05:15:18.223" />
  <row Id="8405" PostTypeId="1" CreationDate="2010-10-30T21:31:55.070" Score="2" ViewCount="71" Body="&lt;p&gt;In a 365 day year, Joe, Greg &amp;amp; Dean visit the park multiple times for a walk.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Joe visits every 3rd day&lt;/li&gt;&#xA;&lt;li&gt;Greg visits every 5th day&lt;/li&gt;&#xA;&lt;li&gt;Dean visits every 7th day&lt;/li&gt;&#xA;&lt;li&gt;All three visit the park on the first day of the year&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the year:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On how many days is Greg alone in the park?&lt;/li&gt;&#xA;&lt;li&gt;On how many days will Dean meet Joe?&lt;/li&gt;&#xA;&lt;li&gt;On how many days will all three meet?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can work this out brute force by laying down the multiples of 3,5,7 from 1 to 365. I'm looking for an efficient way of formalizing the calculation step by step. What'd be the best way?&lt;/p&gt;&#xA;" OwnerUserId="2956" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2010-10-30T21:50:54.800" LastActivityDate="2010-10-30T21:54:23.353" Title="Simple Sequence Problem: Walk in the park" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8406" PostTypeId="2" ParentId="8403" CreationDate="2010-10-30T21:33:41.123" Score="2" ViewCount="0" Body="&lt;p&gt;If you just want the bounded property, you can simply find crude upper and lower bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. if $(-1)^n+1/n$ is bounded, then so is $((-1)^n+1/n)^2$ (so we can essentially disregard the power of 2).  Then we note that both $(-1)^n$ and $1/n$ are between $-1$ and $1$ for all $n \geq 1$ (I told you it was crude).  Hence $-2 \leq (-1)^n+1/n \leq 2$ and $((-1)^n+1/n)^2 \leq 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first sequence $\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$ can be resolved by similar techniques.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-30T21:33:41.123" CommentCount="2" />
  <row Id="8407" PostTypeId="2" ParentId="8339" CreationDate="2010-10-30T21:36:00.027" Score="0" ViewCount="0" Body="&lt;p&gt;$B = P^{-1}AP \ \Longleftrightarrow \ PBP^{-1} = A$. If $Av = \lambda v$, then $PBP^{-1}v = \lambda v \ \Longrightarrow \ BP^{-1}v = \lambda P^{-1}v$. so, if $v$ is an eigenvector of $A$, with eigenvalue $\lambda$, then $P^{-1}v$ is an eigenvector of $B$ with the same eigenvalue. So, every eigenvalue of $A$  is an eigenvalue of $B$ and since you can interchange the roles of $A$ and $B$ in the previous calculations, every eigenvalue of $B$ is an eigenvalue of $A$ too. Hence, $A$ and $B$ have the same eigenvalues. Geometrically, in fact, also $v$ and $P^{-1}v$ are the same vector, written in different coordinate systems.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-10-30T21:36:00.027" />
  <row Id="8408" PostTypeId="2" ParentId="8405" CreationDate="2010-10-30T21:54:23.353" Score="4" ViewCount="0" Body="&lt;p&gt;Lets say its the $x$th day. Then, Greg is alone if and only if $x$ is a multiple of $5$ but not of $3$ or $7$. Dividing 365 by $5$ we get that Greg will go to the park 73 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe will go to the park every 3 times that Greg goes to the park. 73 = 24*3 + 1, so will be 24 occasions in which Joe is also in the park. Also, as 73 = 10*7 + 3, there will be 10 occasions when Dean is also in the park. Finally they are all in the park if $x$ is a multiple of $3*5*7=105$, which is only on 3 days. Thus, tallying it all up Greg is alone 73 - 24 - 10 + 3 = 42 days out of the year (the reason I add 3 is because in subtracting 24 and 10 I have also included days when they are both there twice, but we are only counting how many days he is alone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Dean will meet Joe if and only if $x$ is both a multiple of 3 and 7, or if and only if $x$ is a multiple of 21. As $365 = 17*21 + 8$, they will meet on $17$ occasions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically to attack these problems just think about dividing the 365 days and using the quotient, as it is equal to the number of days a given person goes to the park.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-30T21:54:23.353" CommentCount="1" />
  <row Id="8409" PostTypeId="1" CreationDate="2010-10-30T22:29:43.837" Score="4" ViewCount="201" Body="&lt;p&gt;I'm self-studying field extensions. I ran over an exercise which I can't completely solve. (I haven't yet started studying Galois theory, and I think this exercise isn't meant to be solved using it, just in case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is:&#xA;a) Prove $\sqrt{2}+\sqrt[3]{5}$ is algebraic over $\mathbb{Q}$ of degree 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Done: I know it has degree $\leq 6$ because $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ which has degree 6; then I explicitly found the polynomial by solving a 6-equation linear system, and Wolfram Alpha proved it irreducible (btw: how can I prove it by hand?). The polynomial is $t^6-6t^4-10t^3+12t^2-60t+17$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) What's its degree over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this part b) which I can't solve. Of course its degree is $\leq 6$ in both cases, but I don't know what else to do.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-10-31T02:50:19.037" Title="Degree of $\sqrt{2}+\sqrt[3]{5}$ over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="8410" PostTypeId="2" ParentId="8374" CreationDate="2010-10-30T22:33:04.320" Score="3" ViewCount="0" Body="&lt;p&gt;x^a is fractionally differentiable at $x=0$ with $b$-fractional derivative proportional to $x^{a-b}$ for $b \leq a$. However, if $b &gt; a$, then the fractional derivative blows up at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a collection of functions which have $a$-fractional derivatives but such that the $b$-fractional derivatives for $b&gt;a$ simply do not exist, consider the $a$-fractional anti-derivatives of $|x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are easy to find, because $|x| = x$ for $x&gt;0$, and $|x| = -x$ for $x \leq 0$, and so you can just use the formula for the $k$th anti-derivative of $x$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-10-31T20:03:40.423" LastActivityDate="2010-10-31T20:03:40.423" CommentCount="1" />
  <row Id="8411" PostTypeId="2" ParentId="8409" CreationDate="2010-10-30T22:39:57.727" Score="4" ViewCount="0" Body="&lt;p&gt;Bruno: The degree over ${\mathbb Q}(\sqrt2)$ is 3: Consider $(x-\sqrt 2)^3-5$. Since $\root 3\of 5$ is in ${\mathbb Q}(\sqrt 2)(\sqrt 2+\root 3\of 5)$, the only other option is that it already belongs to ${\mathbb Q}(\sqrt 2)$. But this is impossible, since its minimal polynomial is $x^3-5$ over ${\mathbb Q}$ and ${\mathbb Q}(\sqrt 2)$ is an extension of degree 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ By the way, one can use this and the tower law, to show that ${\mathbb Q}(\sqrt 2+\root 3\of 5)$ has degree 6 over ${\mathbb Q}$. The irreducibility of your polynomial then follows for free. ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the degree over ${\mathbb Q}(\root 3\of 5)$ is 2.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-10-30T22:39:57.727" CommentCount="8" />
  <row Id="8413" PostTypeId="2" ParentId="123" CreationDate="2010-10-30T23:05:47.493" Score="6" ViewCount="0" Body="&lt;p&gt;The catenary has been mentioned a number of times, but apparently not the corresponding surface of revolution, the &lt;a href=&quot;http://mathworld.wolfram.com/Catenoid.html&quot; rel=&quot;nofollow&quot;&gt;catenoid&lt;/a&gt;. It and the plane are the only surfaces of revolution that have zero mean curvature (i.e. they are &lt;em&gt;minimal surfaces&lt;/em&gt;). This surface is the form a soap bubble (approximately) takes when it is stretched across two rings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8Zo0d.jpg&quot; alt=&quot;catenoid bubble&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(image from &lt;a href=&quot;http://www.soapbubble.dk/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-30T23:05:47.493" />
  <row Id="8414" PostTypeId="2" ParentId="8347" CreationDate="2010-10-30T23:33:53.137" Score="4" ViewCount="0" Body="&lt;p&gt;As always, to find the maximum and minimum of a function, you have to calculate first the critical points, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial G(x,y)}{\partial x}=0$$&#xA;$$\frac{\partial G(x,y)}{\partial y}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the respective equations are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial G(x,y)}{\partial x}=-\frac{1}{x^{2}}+\frac{9}{(4-x-y)^{2}}$$&#xA;$$\frac{\partial G(x,y)}{\partial y}=-\frac{4}{y^{2}}+\frac{9}{(4-x-y)^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying you should obtain $x=1-\displaystyle \frac{y}{4}$ and $\displaystyle \frac{5y}{2}=4-x$ which lead to the solution $(x,y)=\bigg\{\displaystyle \frac{2}{3},\displaystyle \frac{4}{3}\bigg\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you substitute those values in your original function $G(\frac{2}{3},\frac{4}{3})=9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, such a point is just a critical point (which means, it could be a local minimum, a local maximum or a saddle point). In order to figure out, we have to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_partial_derivative_test&quot; rel=&quot;nofollow&quot;&gt;Second partial derivative test&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the details in the Wikipedia article. As you already know what you want to prove (that $(\frac{2}{3},\frac{4}{3})$ is a local minimum), we expect that $M(\frac{2}{3},\frac{4}{3})&gt;0$ and $f_{xx}(\frac{2}{3},\frac{4}{3})&gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'll leave it there. You can do the rest of the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the aforementioned procedure works to obtain only a local minimum, maximum or saddle point, not global extrema.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-10-31T04:39:48.330" LastActivityDate="2010-10-31T04:39:48.330" />
  <row Id="8415" PostTypeId="1" CreationDate="2010-10-30T23:57:11.313" Score="3" ViewCount="98" Body="&lt;p&gt;The origin of this question is the identity&#xA;$$\sum_{k=0}^n \binom{n}{k} H_k = 2^n \left(H_n - \sum_{k=1}^n \frac{1}{k 2^k}\right),$$&#xA;where $H_n$ is the $n$th harmonic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $2^n$, we have&#xA;$$2^{-n} \sum_{k=0}^n \binom{n}{k} H_k = H_n - \sum_{k=1}^n \frac{1}{k 2^k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum on the left can now be interpreted as a weighted average of the harmonic numbers through $H_n$ -- where the weights, of course, are the binomial coefficients.  Thus the difference between $H_n$ and its &quot;binomial average&quot; (I'm guessing there's no term for this) is&#xA;$$H_n - 2^{-n} \sum_{k=0}^n \binom{n}{k} H_k =  \sum_{k=1}^n \frac{1}{k 2^k}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum on the right is known to converge to $\ln 2$ as $n \to \infty$.  (Substitute $-\frac{1}{2}$ into the Maclaurin series for $\ln (1+x)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to my question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we classify nonnegative functions $f(n)$ for which &#xA;  $$\lim_{n \to \infty} \left(f(n) - 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k) \right)$$ is finite and nonzero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would seem that if $f$ increases sufficiently rapidly, then the limit would be $\infty$.  This is the case with both $f(n) = a^n$ and $f(n) = n$.  If $f$ decreases or is constant, then the limit is zero.  If $f$ has basically logarithmic growth, then it seems the limit would behave as $H_n$.  But can this be proved?  And what about other sublinear, increasing functions?&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-12T01:32:51.243" Title="Asymptotic difference between a function and its binomial average" Tags="&lt;binomial-coefficients&gt;&lt;asymptotics&gt;&lt;average&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="8416" PostTypeId="2" ParentId="8409" CreationDate="2010-10-31T00:31:53.580" Score="1" ViewCount="0" Body="&lt;p&gt;For part b, we can find the minimal polynomial in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $a =\sqrt{2} + \sqrt[3]{5}$ over $\mathbb{Q}(\sqrt[3]{5})$. Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (x - \sqrt[3]{5})^2 -2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a degree two polynomial in $\mathbb{Q}(\sqrt[3]{5})[x]$ that $a$ is a root of. We know that $a \notin \mathbb{Q}(\sqrt[3]{5})$, so the degree of its minimal polynomial must be greater than 1, and now we have a polynomial of degree 2 that it is a root of, so its minimal polynomial must be that one above. Therefore $a$ has degree two over $\mathbb{Q}(\sqrt[3]{5})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider $a$ over $\mathbb{Q}(\sqrt{2})$. Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (x - \sqrt{2})^3 - 5 = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a degree 3 polynomial in $\mathbb{Q}(\sqrt{2})[x]$ that $a$ is a root of, and hence the minimal polynomial, which must divide this polynomial, can have degree either 2, or 3. But, if it has degree two, then it is attained from the above one by dividing by $x - a$. We get by polynomial division that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (x - \sqrt{2})^3 - 5 = (x - \sqrt{2} - \sqrt[3]{5})(-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2 \notin \mathbb{Q}(\sqrt{2})[x]$ because the $x$ coefficient is not in $\mathbb{Q}(\sqrt{2})$. If it were, then as $2\sqrt{2} \in \mathbb{Q}(\sqrt{2})$, we would get that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \sqrt{2} + (-2\sqrt{2} + \sqrt[3]{5}) = \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is impossible. Thus that quadratic polynomial cannot be the minimal polynomial for $a$. Hence, the minimal polynomial of $a$ has degree 3.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-10-31T02:50:19.037" LastActivityDate="2010-10-31T02:50:19.037" CommentCount="2" />
  <row Id="8417" PostTypeId="1" CreationDate="2010-10-31T00:54:43.520" Score="1" ViewCount="98" Body="&lt;p&gt;Hi all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had recently asked the question:&lt;br&gt;&#xA;&lt;a href=&quot;http://math.stackexchange.com/q/8213/2916&quot;&gt;http://math.stackexchange.com/q/8213/2916&lt;/a&gt;&lt;br&gt;&#xA;which received a very helpful answer. However, I'd like to ask your help again as the formulation of the problem has slightly changed and i'm lost...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;br&gt;&#xA;Given two points: $(x_l, y_l)$ and $(x_u, y_u)$&#xA;with: $x_l &amp;lt; x_u$ and $y_l &amp;lt; y_u$,&lt;br&gt;&#xA;and given that the higher asymptote is one ($\lim_{x\to+\infty}f(x)=1$)&lt;br&gt;&#xA;what's the logistic function that passes through the two points &lt;strong&gt;and the origin $(0, 0)$&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="2916" LastEditorDisplayName="" LastEditDate="2010-10-31T10:40:40.093" LastActivityDate="2010-10-31T10:40:40.093" Title="Logistic function passing through two points AND the origin?" Tags="&lt;statistics&gt;" CommentCount="8" />
  <row Id="8418" PostTypeId="2" ParentId="8337" CreationDate="2010-10-31T01:28:49.853" Score="6" ViewCount="0" Body="&lt;p&gt;Here is one more nice proof, I learned it from Grisha Mikhalkin: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $Z$ be a complex curve in $\mathbb{C}^2$. Let $R(Z) \subset \mathbb{R}^2$ be the projection of $Z$ onto its real parts and $I(Z)$ the projection onto its complex parts. If these projections are both one to one, then the area of $R(Z)$ is equal to the area of $I(Z)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; There is an obvious map from $R(Z)$ to $I(Z)$, given by lifting $(x_1, x_2) \in R(Z)$ to $(x_1+i y_1, x_2 + i y_2) \in Z$, and then projecting to $(y_1, y_2) \in I(Z)$. We must prove this map has Jacobian $1$. WLOG, translate $(x_1, y_1, x_2, y_2)$ to $(0,0,0,0)$ and let $Z$ obey $\partial z_2/\partial z_1 = a+bi$ near $(0,0)$. To first order, we have $x_2 = a x_1 - b y_1$ and $y_2 = a y_1 + b x_1$. So $y_1 = (a/b) x_1 - (1/b) x_2$ and $y_2 = (a^2 + b^2)/b x_1 - (a/b) x_2$. So the derivative of $(x_1, x_2) \mapsto (y_1, y_2)$ is $\left( \begin{smallmatrix} a/b &amp;amp; - 1/b \\ (a^2 + b^2)/b &amp;amp; -a/b \end{smallmatrix} \right)$ and the Jacobian is $1$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the curve $e^{-z_1} + e^{-z_2} = 1$, where $z_1$ and $z_2$ obey the following inequalities: $x_1 \geq 0$, $x_2 \geq 0$, $-\pi \leq y_1 \leq 0$ and $0 \leq y_2 \leq \pi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a point on $e^{-z_1} + e^{-z_2} = 1$, consider the triangle with vertices at $0$, $e^{-z_1}$ and $e^{-z_1} + e^{-z_2} = 1$. The inequalities on the $y$'s states that the triangle should lie above the real axis; the inequalities on the $x$'s state the horizontal base should be the longest side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Projecting onto the $x$ coordinates, we see that the triangle exists if and only if the triangle inequality $e^{-x_1} + e^{-x_2} \geq 1$ is obeyed. So $R(Z)$ is the region under the curve $x_2 = - \log(1-e^{-x_1})$. The area under this curve is&#xA;$$\int_{0}^{\infty} - \log(1-e^{-x}) dx = \int_{0}^{\infty} \sum \frac{e^{-kx}}{k} dx = \sum \frac{1}{k^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, project onto the $y$ coordinates. Set $(y_1, y_2) = (-\theta_1, \theta_2)$ for convenience, so the angles of the triangle are $(\theta_1, \theta_2, \pi - \theta_1 - \theta_2)$. The largest angle of a triangle is opposite the largest side, so we want $\theta_1$, $\theta_2 \leq \pi - \theta_1 - \theta_2$, plus the obvious inequalities $\theta_1$, $\theta_2 \geq 0$. So $I(Z)$ is the quadrilateral with vertices at $(0,0)$, $(0, \pi/2)$, $(\pi/3, \pi/3)$ and $(\pi/2, 0)$ and, by elementary geometry, this has area $\pi^2/6$.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-10-31T12:11:28.040" LastActivityDate="2010-10-31T12:11:28.040" CommentCount="4" />
  <row Id="8419" PostTypeId="1" AcceptedAnswerId="8420" CreationDate="2010-10-31T01:38:08.503" Score="1" ViewCount="247" Body="&lt;p&gt;Please suggest a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-10-31T01:52:36.230" LastActivityDate="2010-10-31T16:21:56.547" Title="What is the remainder when $4^{44}$ is divided by 44 ?" Tags="&lt;number-theory&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="8420" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T01:58:43.080" Score="8" ViewCount="0" Body="&lt;p&gt;Well... $4^{44}$ is obviously divisible by 4, that is, $4^{44} \equiv 0 \pmod 4$.  So you only need to calculate $4^{44} \pmod {11}$, which can be done using &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's Little Theorem&lt;/a&gt; (since 4 and 11 are coprime), then use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese Remainder Theorem&lt;/a&gt; to combine the two congruences.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-31T01:58:43.080" CommentCount="3" />
  <row Id="8421" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T02:03:57.967" Score="4" ViewCount="0" Body="&lt;p&gt;$4^{44}=4^4\times4^{40}=4^4\times(4^{10}-1+1)^4$. The second factor leaves remainder 1 one divided by 11. This is because of Fermat's little theorem, but can be seen directly: $4^{10}-1=(2^{10}-1)(2^{10}+1)=11\times93\times1025$, so $4^{10}$ is a multiple of 11 plus 1, and so is $4^{40}$. Finally, $4^{44}=4^4\times(11k+1)\equiv 4^4=256=44\times5+36\equiv 36\pmod{44}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to consider the prime decomposition of 44 and treat each factor separately.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-10-31T02:03:57.967" />
  <row Id="8422" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T02:13:36.383" Score="2" ViewCount="0" Body="&lt;p&gt;Another method is to simply compute $4^{44} \pmod {44}$, which could be done in any self-respecting computer algebra system (or you could write your own code for it, watching out for overflow errors).  For example:  you could use &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warning:  If this is a homework question, finding the answer computationally is likely to annoy the lecturer.  However, it's still a good way to check your answer.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-31T02:13:36.383" CommentCount="2" />
  <row Id="8423" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T02:41:54.237" Score="1" ViewCount="0" Body="&lt;p&gt;As previously stated, since $4^{44} \equiv 0 \pmod 4$, all you need to find is $4^{44} \mod 11$ (which makes sense if you think about the fact that if $4^{44}$ is evenly divisible by one factor of 44, it won't affect the result of modding, so you only need to consider the other factor, 11).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A faster way to do this is note that $4^{44}=2^{88}=2^{85}\cdot8=32^{17}\cdot8$, and that $32\equiv -1 \pmod{11}$, which turns it into $(-8)\mod 11$. However, it is important to note that this is still in mod 11, so the answer must be turned into $(-8)\mod 44$. (which is the same thing, as we noted earlier)&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-10-31T08:16:12.957" LastActivityDate="2010-10-31T08:16:12.957" CommentCount="0" />
  <row Id="8424" PostTypeId="1" AcceptedAnswerId="8428" CreationDate="2010-10-31T03:06:53.563" Score="2" ViewCount="107" Body="&lt;p&gt;I was browsing through Wikipedia for some reason or another, and it says that there is a way to do Galois theory without worrying about whether or not polynomials are irreducible.  That's a bit surprising to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Wikipedia wrong (as is often the case when I read something surprising)?  If not, where can I hear more about it?  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-10-31T15:19:50.647" Title="Galois theory with non-irreducible polynomials" Tags="&lt;galois-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8425" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T03:08:28.627" Score="3" ViewCount="0" Body="&lt;p&gt;One way to do this is to figure out that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4^6 \equiv 64^2 \equiv 20^2 \equiv 4 \mod 44 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4^{44} \equiv (4^6)^7 * 4^2 \equiv 4^7 * 4^2 \equiv 4^4 \equiv 36 \mod 44 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the answer is $36$, and all you had to do is calculate a few powers of $4$ mod $44$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-31T03:08:28.627" />
  <row Id="8426" PostTypeId="1" AcceptedAnswerId="8427" CreationDate="2010-10-31T03:14:07.963" Score="2" ViewCount="75" Body="&lt;p&gt;I'm attempting to implement an equation (for calculating &lt;a href=&quot;http://dx.doi.org/10.1109/TMAG.2010.2049026&quot; rel=&quot;nofollow&quot;&gt;magnetic forces between coils&lt;/a&gt;, eqs (22–24) in the linked paper) that requires the use of elliptic integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately these equations require the evaluation of the elliptic integrals far outside their standard parameter range of $0\le m\le 1$ and the numerical implementations I have available to evaluate them give inconsistent results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that Mathematica is correct in its answer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;EllipticF[ArcSin[Sqrt[1/c5]], c5] //. c5 -&amp;gt; 817.327&#xA;=&amp;gt; 0.054961 - 1.17196*10^-17 i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Whereas Matlab's MuPad engine gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mfun('EllipticF',asin(sqrt(1/817.327)),sqrt(817.327))&#xA;=&amp;gt; 0.054961 - 0.000707i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Mma's function takes parameter $m=k^2$ whereas MuPad takes modulus $k$, explaining the &lt;code&gt;sqrt&lt;/code&gt;)&#xA;While I can use Mathematica for my own work, my colleagues only have Matlab available and I'd like them to be able to use this code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a pretty unfamiliar with the theory behind elliptic integrals, but &lt;a href=&quot;http://www.gutenberg.org/ebooks/31076.bibrec.html&quot; rel=&quot;nofollow&quot;&gt;Baker's ‘Elliptic Functions’&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We shall see later on that the quantity $k^2$ [...] can always be considered real and less than unity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which leads me to ask: can the arguments to these elliptic integrals be re-stated in terms of an input of $k&gt;1$, such as I seem to require above?&lt;/p&gt;&#xA;" OwnerUserId="2961" LastEditorUserId="2961" LastEditorDisplayName="" LastEditDate="2010-11-01T00:12:18.057" LastActivityDate="2010-11-01T00:12:18.057" Title="Elliptic integrals with parameter outside 0&lt;m&lt;1" Tags="&lt;numerical-methods&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8427" PostTypeId="2" ParentId="8426" CreationDate="2010-10-31T03:45:05.133" Score="2" ViewCount="0" Body="&lt;p&gt;What you seem to require here are the so-called &lt;a href=&quot;http://dlmf.nist.gov/19.7#ii&quot; rel=&quot;nofollow&quot;&gt;&quot;reciprocal-modulus transformations&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific case, upon applying the reciprocal-modulus transformation, your elliptic integral can actually be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{c_5}}F\left(\frac{\pi}{2}\vert \frac1{c_5}\right)=\frac1{\sqrt{c_5}}K\left(\frac1{c_5}\right)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K(m)$ is the complete elliptic integral of the first kind with parameter $m$ (though I understand MuPAD uses $k=\sqrt{m}$ as argument, in which case you seem to know the conversion formulae already).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are similar formulae for the elliptic integrals of the second and third kinds, and I direct you to the DLMF link I gave above for them.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-31T03:45:05.133" CommentCount="2" />
  <row Id="8428" PostTypeId="2" ParentId="8424" CreationDate="2010-10-31T05:25:45.047" Score="5" ViewCount="0" Body="&lt;p&gt;There is a very big difference between &quot;you can do Galois theory without worrying about whether or not a polynomial is irreducible&quot; and &quot;one can do Galois theory while working with polynomials that are not irreducible&quot;. The former implies that the results will hold regardless of whether the polynomial is irreducible or not; the latter merely says that one can adjust the work so as to deal with reducible polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x)$ is not irreducible, but factors as $f(x)=p_1(x)^{a_1}\cdots p_k(x)^{a_k}$ with $p_j(x)$ irreducible and separable, then you can certainly still talk about the Galois group of the splitting field of $f(x)$, and it will have all the usual properties: it would simply be compositum of the splitting fields of the different $p_i$, and so the Galois group of $f$ would have normal subgroups corresponding to the splitting fields of each $p_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the polynomials are inseparable one runs into greater difficulties; the usual technique is to separate the extension into a separable extension and a purely inseparable extension. For the separable ones, the usual techniques of Galois Theory work; for the purely inseparable ones, we have special results that apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the quote from Wikipedia is rather poor, since separability is a property of the &lt;strong&gt;irreducible factors&lt;/strong&gt; of the polynomials. Repeated factors do not affect separability or lack thereof. But even if we were to take it at face value, it does not claim that one can do Galois theory &quot;without worrying&quot;; it would say that if you are doing Galois theory with non-irreducible polynomials (in the way that one usually does: by splitting the polynomial into irreducible factors), then you would not worry about whether the polynomial itself has repeated roots or not, because repeated roots could come from repeated &lt;em&gt;separable&lt;/em&gt; irreducible factors. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-10-31T15:19:50.647" LastActivityDate="2010-10-31T15:19:50.647" CommentCount="4" />
  <row Id="8429" PostTypeId="1" AcceptedAnswerId="8437" CreationDate="2010-10-31T06:08:17.423" Score="0" ViewCount="47" Body="&lt;p&gt;I was always thinking about this questions when it comes to lottery. lets say you have a number range from 1 - 45. You need 5-6 numbers to win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let says you have 2 options&#xA;Draw 3 sets of numbers with 10 numbers on each set.&#xA;or &#xA;1 set of numbers with 19 numbers for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which chance will have higher of winning?&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2010-10-31T09:36:32.240" Title="More choices lesser draw or more draw lesser choices? " Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8430" PostTypeId="2" ParentId="392" CreationDate="2010-10-31T06:09:01.513" Score="5" ViewCount="0" Body="&lt;p&gt;This is related to &lt;a href=&quot;http://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sinx-and-cosx/395#395&quot;&gt;Justin L.'s answer&lt;/a&gt;, because I have basically the same interpretation, but whereas that answer (as I'm interpreting it) gives a great intuitive &lt;em&gt;check&lt;/em&gt; that the derivatives are correct, I intend to present how one might actually (somewhat intuitively) arrive at the derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $s(x)=(\cos(x),\sin(x))$ gives the point on the unit circle after traveling an arclength $x$ from the point $(1,0)$, oriented counterclockwise.  Parametrization with respect to arclength is precisely the condition that guarantees that the curve has unit speed, i.e., $|s'(x)|\equiv 1$.  Because $s$ also has constant length, the product rule can be used to show that $s'(x)$ is perpendicular to $s(x)$ for all $x$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s\cdot s \equiv 1 \Rightarrow s'\cdot s + s\cdot s'\equiv 0 \Rightarrow s\cdot s'\equiv 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for each $x$, $s'(x)$ is a unit length vector perpendicular to $s(x)$.  This leaves 2 possibilities: either a counterclockwise or clockwise rotation by $\frac{\pi}{2}$ from $s(x)$.  But now, because $s'$ tells us how $s$ is changing, it must point in the direction of motion of $s$, namely counterclockwise.  Thus $s'(x)$ is a counterclockwise rotation by $\frac{\pi}{2}$ from $s(x)$, which means &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s'(x)=s\left(x+\frac{\pi}{2}\right)=\left(\cos\left(x+\frac{\pi}{2}\right),\sin\left(x+\frac{\pi}{2}\right)\right)=(-\sin(x),\cos(x)).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we also have $s'(x)=(\cos'(x),\sin'(x))$, so matching coordinates yields $\cos'=-\sin$ and $\sin'=\cos$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-10-31T06:09:01.513" CommentCount="2" />
  <row Id="8431" PostTypeId="1" AcceptedAnswerId="8432" CreationDate="2010-10-31T06:22:47.240" Score="1" ViewCount="83" Body="&lt;p&gt;Let G=(V,E) be a connex graph. Color it's edges randomly with red/blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-prove that there exists an Eulerian circuit, without any two adjacent edges of the same color.. only if for any &lt;code&gt;v&lt;/code&gt; vertex of G the number of incident blue edges is equal to the number of red ones.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-if G is complete and &lt;code&gt;x,y,z&lt;/code&gt; are three distinct vertices prove that, if there is a path without consecutive edges of the same color from &lt;code&gt;x&lt;/code&gt; to &lt;code&gt;y&lt;/code&gt; passing through &lt;code&gt;z&lt;/code&gt;,  then there is a path with the same property that has &lt;code&gt;xz&lt;/code&gt; as the first edge or &lt;code&gt;zy&lt;/code&gt; as the last edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me by at least providing some suitable references that may help me solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2010-10-31T06:34:32.473" Title="Eulerian graphs proof" Tags="&lt;homework&gt;&lt;graph-theory&gt;&lt;proof&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8432" PostTypeId="2" ParentId="8431" CreationDate="2010-10-31T06:34:32.473" Score="1" ViewCount="0" Body="&lt;p&gt;These probably should be comments but I don't have sufficient reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for problem 1: If there are an equal number of red and blue edges at each vertex, what does that tell you about the parity of the degree of each vertex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for problem 2: If you have a path from x to y passing through z, can you use the fact that the graph is complete to find another path that fits the alternating condition?&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-10-31T06:34:32.473" CommentCount="2" />
  <row Id="8433" PostTypeId="1" AcceptedAnswerId="8438" CreationDate="2010-10-31T07:18:27.503" Score="0" ViewCount="83" Body="&lt;p&gt;pseudocode:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;void recursive('k'){ // 'k' and 'i' vertices&#xA;  sumA = 0;&#xA;  sumB = 0;&#xA;  for each non visited 'i' neighbor do{&#xA;     recursive('i');&#xA;     sumA = sumA + b['i'];&#xA;     sumB = sumB + max(a['i'], b['i']);&#xA;     }&#xA;  a['k'] = 1 + sumA;&#xA;  b['k'] = sumB;&#xA;  }&#xA;&#xA;&#xA;void main(){&#xA; a = b = 0; //initialize tables with 0 (zeros)&#xA; recursive('X');  //let 'X' be an arbitrary root&#xA; cout&amp;lt;&amp;lt;max(a['X'], b['X']);&#xA; }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;need proof that &lt;code&gt;max(a['X'], b['X'])&lt;/code&gt; is the cardinal of the maximum independent set in the tree.&#xA;What am I missing ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2010-10-31T09:52:27.187" Title="Maximum independent set in a tree. Review algorithm, need proof." Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8434" PostTypeId="2" ParentId="8419" CreationDate="2010-10-31T07:59:57.717" Score="0" ViewCount="0" Body="&lt;p&gt;$\rm\ 4^5\equiv 2^{10}\equiv 1\ \:(mod\ 11)\ \Rightarrow\ 4^{4+5\cdot8}\equiv 4^4\equiv 4\cdot 20\equiv -8\ (mod\ 11)\:$ and $\rm (mod\ 4)\:$, so $\rm (mod\ 44)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-10-31T16:21:56.547" LastActivityDate="2010-10-31T16:21:56.547" CommentCount="1" />
  <row Id="8435" PostTypeId="1" AcceptedAnswerId="8471" CreationDate="2010-10-31T08:00:09.887" Score="3" ViewCount="97" Body="&lt;p&gt;Suppose n is an integer. What kind of bounds do we know for how close the closest prime p &gt; n will be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd especially appreciate an answer that pushes me in the right direction of proving a good bound for this without showing the proof fully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context: I ran into this while coming up to a solution for &lt;a href=&quot;http://stackoverflow.com/questions/4058172/tricky-algorithm-interview-question&quot;&gt;http://stackoverflow.com/questions/4058172/tricky-algorithm-interview-question&lt;/a&gt;. Depending on the bound, the algorithm I come up with may or may not be deterministic.&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-10-31T22:07:13.360" Title="Bounds on prime number spacing" Tags="&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8436" PostTypeId="2" ParentId="8429" CreationDate="2010-10-31T09:00:17.110" Score="1" ViewCount="0" Body="&lt;p&gt;Hopefully I understand your question properly. To make sure, here's how I understand it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lottery, where 5 distinct random numbers will be chosen from 1-45 inclusive. If on your lottery card, you have all 5, you win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Options:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a) draw 3 cards with 10 random numbers per card&lt;/li&gt;&#xA;&lt;li&gt;b) draw 1 card with 19 random numbers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which option is best?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) The probability of winning if you drew one card with 10 numbers is 10/45 * 9/44 * 8/43 * 7/42 * 6/41 = 4/19393. Your probability of not winning is 1-4/19393. Hence your probability of winning with at least one card is 1-(1-4/19393)^3, which is approximately equal to 0.0006187.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) On the other hand, your probability of winning with the 1 card option is 19 / 45 * 18/44 * 17/43 * 16/42 * 15/41 = 1292/135751, which is approximately equal to 0.009517.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So clearly the second option is the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;strong&gt;A more intuitive (less numerical) explanation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your chances of winning with a card with 10 numbers is very low since half of the numbers on the card half to be winning numbers. Clearly with 19 numbers the chances are greater. If you crunch the numbers, it turns out that 3 10-number cards are not sufficient to overcome the advantage held by the 19-number card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can extend the calculations above to show how many 10-number cards you'd need before that becomes the better option...?&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-10-31T09:00:17.110" />
  <row Id="8437" PostTypeId="2" ParentId="8429" CreationDate="2010-10-31T09:36:32.240" Score="3" ViewCount="0" Body="&lt;p&gt;We have $45$ numbers $1,2,\ldots,45$ of which $6$ are winners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice is between (a) buying three cards, each of which contains $10$ distinct numbers chosen from the $45$  (the same number can appear on different cards) and (b) buying one card which contains $19$ distinct numbers chosen from the $45.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To win we must have at least $5$ of the winning numbers on a card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the probability of at least $5$ winning number on a card of size $10$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \binom{6}{5} \binom{39}{5} +  \binom{6}{6} \binom{39}{4} }{ \binom{45}{10} }&#xA;= \frac{1}{902}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the probability of $ \le 4 $ winners is $ 1 – 1/902 = 901/902.$ Hence the probability of at least one of our three cards of size $10$ winning is ($1$ minus the probability that they all lose)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 - \left( \frac{901}{902} \right)^3 \approx 0.003322.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of at least five winners on a card of size $19$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \binom{6}{5} \binom{39}{14} +  \binom{6}{6} \binom{39}{13} }{ \binom{45}{19} }&#xA;= \frac{5491}{135751} \approx 0.040449.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $0.040449 &gt; 0.003322$ you have a better chance of winning by buying one card of size $19.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-10-31T09:36:32.240" CommentCount="3" />
  <row Id="8438" PostTypeId="2" ParentId="8433" CreationDate="2010-10-31T09:52:27.187" Score="0" ViewCount="0" Body="&lt;p&gt;The element $a[i]$ is the size of the maximal independent set in the subtree rooted in $i$ which contains $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The element $b[i]$ is the size of the maximal independent set in the subtree rooted in $i$ which doesn't contain $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm computes these recursively, and then choose the maximal of the two at the root.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-10-31T09:52:27.187" CommentCount="2" />
  <row Id="8439" PostTypeId="1" AcceptedAnswerId="8440" CreationDate="2010-10-31T10:06:53.540" Score="2" ViewCount="154" Body="&lt;p&gt;$$\cos A \cdot \cos 2A  \cdot \cos 2^{2}A  \cdot \cos 2^{3}A  ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin  A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very much inquisitive to see how this trigonometrical identity can be proved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:I am not much of interested about an inductive proof.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-10-31T11:05:39.540" LastActivityDate="2010-10-31T14:54:24.227" Title="Prove: $\cos A \cdot \cos 2A  \cdot \cos 2^{2}A  \cdot \cos 2^{3}A  ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin  A } $" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8440" PostTypeId="2" ParentId="8439" CreationDate="2010-10-31T10:39:33.043" Score="9" ViewCount="0" Body="&lt;p&gt;The proof is just repeated application of the double-angle formula for the sine function. For example, the case $n=3$:&#xA;$$\sin 8A = 2 \sin 4A \cos 4A = 2 (2 \sin 2A \cos 2A) \cos 4A = 2 (2 (2 \sin A \cos A) \cos 2A) \cos 4A.$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-31T10:39:33.043" CommentCount="4" />
  <row Id="8441" PostTypeId="2" ParentId="8377" CreationDate="2010-10-31T11:03:26.620" Score="1" ViewCount="0" Body="&lt;p&gt;I'll write something so this question doesn't remain unanswered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes it can be quite difficult to prove that a sequence is increasing.  For example, Smetaniuk gave a proof that the number of Latin squares $L_n$ (Sloane's &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A002860&quot; rel=&quot;nofollow&quot;&gt;A002860&lt;/a&gt;) is increasing (actually, he proved $L_{n+1} \geq (n+1)!L_n$), which is one of the best results around regarding the mysterious $L_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smetaniuk, Bohdan&#xA;A new construction of Latin squares. II. The number of Latin squares is strictly increasing. &#xA;Ars Combin. 14 (1982), 131–145.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-10-31T11:03:26.620" />
  <row Id="8442" PostTypeId="2" ParentId="5892" CreationDate="2010-10-31T11:49:06.070" Score="1" ViewCount="0" Body="&lt;p&gt;this link provides some good resources ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;check it up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://entrance-exam.net/gre-papers/&quot; rel=&quot;nofollow&quot;&gt;http://entrance-exam.net/gre-papers/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2969" LastActivityDate="2010-10-31T11:49:06.070" CommunityOwnedDate="2010-10-31T11:49:06.070" />
  <row Id="8443" PostTypeId="1" CreationDate="2010-10-31T12:56:48.943" Score="4" ViewCount="75" Body="&lt;p&gt;I just read this document - &lt;a href=&quot;http://www.math.ohio-state.edu/~econrad/Jacobi/sumofsq/sumofsq.html&quot; rel=&quot;nofollow&quot;&gt;Jacobi's Four Square Theorem&lt;/a&gt;. It shows how to count the number of representations of a number as the sum of four squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can follow the proof but currently it just appears like a brilliant coincidence, so I am wondering what else can you count with these?&lt;/p&gt;&#xA;" LastActivityDate="2010-10-31T19:46:34.110" Title="What else can the elliptic integral count?" Tags="&lt;fourier-series&gt;&lt;elliptic-curves&gt;&lt;counting&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8444" PostTypeId="1" CreationDate="2010-10-31T13:49:03.427" Score="1" ViewCount="116" Body="&lt;p&gt;What could be the possible way to find the &lt;code&gt;Nth term&lt;/code&gt; of following series where the sign toggles after each triangular number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;1 -2 -3 4 5 6 -7 -8 -9 -10 11 12 13 14 15 -16 -17 ....&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series cannot be in a Geometric Progression because there are 4 distinct triangular numbers in the above series.&lt;/p&gt;&#xA;" OwnerUserId="2971" LastActivityDate="2010-11-01T09:07:10.250" Title="Nth term of the series where sign toggles after a triangular  number" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8445" PostTypeId="2" ParentId="3087" CreationDate="2010-10-31T14:07:26.733" Score="1" ViewCount="0" Body="&lt;p&gt;Whatever you think the high school mathematics curriculum &quot;should&quot; be, in the United States a curriculum has been put in place, known as the Common Core Standards (CCS), which will significantly - I believe - change American mathematics education. (The CCS extend to all K-12 mathematics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.corestandards.org/the-standards/mathematics&quot; rel=&quot;nofollow&quot;&gt;http://www.corestandards.org/the-standards/mathematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think these changes are for the better. A short summary of my personal views are available here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-10-31T14:07:26.733" CommunityOwnedDate="2010-10-31T14:07:26.733" />
  <row Id="8446" PostTypeId="2" ParentId="3087" CreationDate="2010-10-31T14:43:33.380" Score="1" ViewCount="0" Body="&lt;p&gt;I recall the Asimov short story about humanity that has completely lost the knowledge of how to do simple arithmetic without use of a computer, called &quot;The Feeling of Power&quot;. The point is, despite computers being able to do these things for us, it is still important and valuable to know how to do simple operations. At times, we are amazingly close to that point in this very day, certainly in the current crop of kids growing up now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic mathematical skills are tremendously valuable in a variety of places. Just because you CAN factor a polynomial by computer does not mean that at least understanding the tools to do so are no longer necessary for 99% of the people in the world.&lt;/p&gt;&#xA;" OwnerUserId="1608" LastActivityDate="2010-10-31T14:43:33.380" CommunityOwnedDate="2010-10-31T14:43:33.380" />
  <row Id="8447" PostTypeId="2" ParentId="1116" CreationDate="2010-10-31T14:48:27.290" Score="7" ViewCount="0" Body="&lt;p&gt;I had a student work on this. We came up with a complicated recurrence relation, &#xA;indicated below. The first few answers are 1, 2, 6, 25, 128, 758, 5014, 36194, 280433, 2303918, 19885534, 179028087, 1671644720, 16114138846, 159761516110, 1623972412726, &#xA;16880442523007, 179026930243822, 1933537655138482, 21231023519199575, &#xA;236674460790503286, 2675162663681345170, 30625903703241927542, &#xA;354767977792683552908, 4154708768196322925749, &#xA;49152046198035152483150, 587011110939295781585102, &#xA;7072674305834582713614923. Note that we are counting rotations and reflections as distinct tilings. An interesting pattern is that mod 2, there is an 8-fold periodicity which we don't understand and can't prove in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of the cases n=1,2,3,4, with 1,2,6,25 tilings in each case.The way to systematically generate these is to &quot;push in&quot; a vertical line from the right to all previously constructed tilings in all possible ways. That's how the recurrence relation is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OftaF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, here is the recurrence:&#xA;Let $a_{\ell,j,m}$ be the number of distinct tilings with $\ell$ tiles, $j$ edges that meet the right-hand side of the square and $m$ 4-valent vertices.&#xA;$$a_{\ell,j,m}=\sum_{p=1}^\ell(-1)^{p+1}\sum_{i=0}^m\sum_{n=1}^{\ell-1}\sum_{k=0}^{n-1}\alpha_{n,k,i,\ell,j,m,p} a_{n,k,i}$$&#xA;where, letting $t=m-i, s=\ell-n-p-t$ and $r=k+s+t+p-j$,&#xA;$$\alpha_{n,k,i,l,j,m,p}=\binom{r-1}{p-1}\binom{k-r+2}{p}\binom{s+r-1}{r-1}\binom{r-p}{t}.$$&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2010-11-02T09:35:40.290" LastActivityDate="2010-11-02T09:35:40.290" CommentCount="4" />
  <row Id="8449" PostTypeId="1" CreationDate="2010-10-31T15:41:43.990" Score="2" ViewCount="150" Body="&lt;p&gt;Consider the set $F_n: = [n]^{[n]}$ of all functions $f:[n] \rightarrow [n]$, $[n] = \lbrace 1,2,...,n\rbrace$. It is well known that $|F_n| = n^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Let two functions $f, g$ in $F_n$ be of the same isomorphism type ($f\sim g$) iff there exists a permutation $\pi$ such that $f\pi = \pi g $&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the number of isomorphism types of functions $f:[n] \rightarrow [n]$, i.e. what is $|F_n/_{\sim}|$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$|F_2| = 4$, $|F_2/_{\sim}| = 3$ &lt;/li&gt;&#xA;&lt;li&gt;$|F_3| = 27$, $|F_3/_{\sim}| = 7$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-10-31T23:19:24.577" LastActivityDate="2010-11-01T03:52:00.047" Title="Number of isomorphism types of functions f:[n]-&gt;[n]" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8450" PostTypeId="1" AcceptedAnswerId="8454" CreationDate="2010-10-31T16:00:15.657" Score="0" ViewCount="120" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part1(Solved)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to create a function that maps x to y as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y 0 1 2 3 4 5 4 3 2 1  0  1  2  3  4  5 ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, this is what I have been able to come up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;|5 sin ((1/10)*pi*x)|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which will give the following y's for x = 0 to 10:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0.00000&#xA; 1.5451&#xA; 2.9389&#xA; 4.0451&#xA; 4.7553&#xA; 5.0000&#xA; 4.7553&#xA; 4.0451&#xA; 2.9389&#xA; 1.5451&#xA; 0.00000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This sequence will repeat every 10 steps, with the peak (5) at step 5. This is getting close to what I want so all that is left is to somehow adjust the function so that the required integers appear (1,2,3,4), but I don't know how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the function is cycling through a series of numbers over and over, I figured that it might be possible to write the function in terms of a periodic function and not piece by piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about writing a function that produce the following plot:(x is an integer &gt;=0 , the pattern repeats after x = 8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7tXVu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this plot be expressed using a single formula or must it be defined piecewise?&#xA;  Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="2973" LastEditorDisplayName="" LastEditDate="2010-11-20T16:06:40.383" LastActivityDate="2010-11-20T16:06:40.383" Title="Creating a sine function with specific parameters" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8451" PostTypeId="2" ParentId="8449" CreationDate="2010-10-31T16:08:00.060" Score="2" ViewCount="0" Body="&lt;p&gt;If you want to count the number of functions up to renamings on the domain and codomain, then the number is $p_n$, the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;partitions&lt;/a&gt; of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Later:&lt;/strong&gt; now that you've made precise what you wanted... This is counted &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001372&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, with references.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-01T03:52:00.047" LastActivityDate="2010-11-01T03:52:00.047" CommentCount="2" />
  <row Id="8452" PostTypeId="1" AcceptedAnswerId="8460" CreationDate="2010-10-31T16:30:26.610" Score="2" ViewCount="220" Body="&lt;p&gt;How do I solve for $x$ from this equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-1/x^2 + 9/(4-x-y)^2 = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get this into $x=$&quot;blah&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also. How do you guys get that funky formatting? &lt;/p&gt;&#xA;" OwnerUserId="2874" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-19T00:11:40.827" LastActivityDate="2010-11-19T00:11:40.827" Title="Make x subject of $-1/x^2 + 9/(4-x-y)^2 = 0$." Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="8453" PostTypeId="2" ParentId="8450" CreationDate="2010-10-31T16:40:54.990" Score="3" ViewCount="0" Body="&lt;p&gt;You ask for a function. A function is a mapping from one set to&#xA;another set but you neglect to specify which sets. Your list of&#xA;$x$-values consists of small nonnegative integers together with&#xA;an ellipsis. From this, one might surmise that your intended&#xA;domain is $\mathbb{N}_0$, the set of nonnegative integers. Likewise&#xA;you do not specify a codomain, but let's say for argument's sake&#xA;that it is $\mathbb{N}_0$, the same as the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are uncountably many functions from $\mathbb{N}_0$&#xA;to $\mathbb{N}_0$ containing your specified values. But you mention&#xA;it has to be a &quot;sine&quot; function. The term sine refers to a particular&#xA;function on the real numbers, taking all velues between $-1$ and $1$&#xA;and having period $2\pi$. Your function cannot have all of these&#xA;properties. Perhaps by a &quot;sine&quot; function you mean a periodic function?&#xA;If so here is one defined on $\mathbb{N}_0$ which extends your table&#xA;of values:&#xA;$$f(n)=\left\{&#xA;\begin{array}{cl}&#xA;0&amp;amp;\textrm{if $n\equiv0$ (mod 10),}\\&#xA;1&amp;amp;\textrm{if $n\equiv\pm1$ (mod 10),}\\&#xA;2&amp;amp;\textrm{if $n\equiv\pm2$ (mod 10),}\\&#xA;3&amp;amp;\textrm{if $n\equiv\pm3$ (mod 10),}\\&#xA;4&amp;amp;\textrm{if $n\equiv\pm4$ (mod 10),}\\&#xA;5&amp;amp;\textrm{if $n\equiv5$ (mod 10).}&#xA;\end{array}&#xA;\right.$$&#xA;Of course, unlike the acutal sine function, this never takes negative&#xA;values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps as in Hans's comment, you want a function on the reals&#xA;restricting to your table of values. Here is a piecewise linear&#xA;example as Hans suggests:&#xA;$$g(x)=\left\{&#xA;\begin{array}{cl}&#xA;10\{x/10\}&amp;amp;\textrm{if $\{x/10\}&amp;lt;1/2$,}\\&#xA;10-10\{x/10\}&amp;amp;\textrm{if $\{x/10\}\ge 1/2$}\\&#xA;\end{array}&#xA;\right.$$&#xA;where $\{t\}$ denotes the fractional part of a real number $t$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-10-31T16:40:54.990" />
  <row Id="8454" PostTypeId="2" ParentId="8450" CreationDate="2010-10-31T16:41:34.857" Score="9" ViewCount="0" Body="&lt;p&gt;$\frac{5}{\pi}\arccos\left(\cos\left(\frac{\pi x}{5}\right)\right)$ seems to do what you want, unless you need something smoother.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-10-31T16:41:34.857" CommentCount="6" />
  <row Id="8455" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:17:42.423" Score="1" ViewCount="0" Body="&lt;p&gt;When you have fractions involved, the usual trick is to put everything on the same side, so you have &quot;= 0&quot; on the other side (This step is already ok in your case). Then, you put all the fractions on the same denominator and you multiply by this denominator (which vanishes since you have 0 on the other side). Usually it's then easier, since you don't have fractions anymore (well, it depends on the situation, but this is a good idea in general).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{-1}{x^2} + \frac{9}{(4-x-y)^2} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{-(4-x-y)^2}{x^2(4-x-y)^2} + \frac{9x^2}{x^2(4-x-y)^2} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-(4-x-y)^2 + 9x^2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can extand $(4-x-y)^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(4-x-y)^2 = ((4-y)-x)^2 = (4-y)^2 - 2(4-y)x + x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus your equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-(4-y)^2 + 2(4-y)x + 8x^2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a standard equation, the same as $ax^2 + bx + c = 0$, that can be solved for $x$ (consider $y$ as if it was any number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, don't forget to check if the solution you obtain is ok with your beginning equation (no division by 0 for example).&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-10-31T17:17:42.423" />
  <row Id="8456" PostTypeId="2" ParentId="3087" CreationDate="2010-10-31T17:30:22.860" Score="2" ViewCount="0" Body="&lt;p&gt;Students should learn &lt;em&gt;math&lt;/em&gt;. You know, the stuff that most of us &lt;em&gt;do&lt;/em&gt;. When we use the word &quot;math&quot; we certainly don't have in our minds adding, multiplying, factoring polynomials, or even calculus. So why do we teach children that this is what math is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This view has been in several published places like Devlin's &lt;em&gt;The Math Gene&lt;/em&gt;. He makes a good argument that the concept of a group isn't any harder than other stuff done in grade school. The problem is that it is introduced later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing I've read about math education is &lt;em&gt;A Mathematician's Lament&lt;/em&gt; by Paul Lockhart.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-10-31T17:30:22.860" CommunityOwnedDate="2010-10-31T17:30:22.860" />
  <row Id="8457" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:31:46.183" Score="1" ViewCount="0" Body="&lt;p&gt;If you're working on the same problem as asked about in &lt;a href=&quot;http://math.stackexchange.com/q/8347/1242&quot;&gt;this question&lt;/a&gt;, the answer is that you don't need to right away, since you can first use the other equation to get a relation between $x$ and $y$ that will simplify things a little. See my comment there.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-10-31T17:31:46.183" CommentCount="1" />
  <row Id="8458" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:45:24.073" Score="3" ViewCount="0" Body="&lt;p&gt;Put $z = 4-x-y\:.\ $ Multiplying by $(xz)^2$ yields $$0 = 9x^2-z^2 = (3x-z)\:(3x+z) = (4x+y-4)(2x-y+4)$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-31T17:45:24.073" />
  <row Id="8460" PostTypeId="2" ParentId="8452" CreationDate="2010-10-31T17:56:40.003" Score="5" ViewCount="0" Body="&lt;p&gt;If you add $\frac{1}{x^{2}}$ to both sides of your equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{x^{2}}+\frac{9}{\left( 4-x-y\right) ^{2}}=0\qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get this equivalent one (provided that $\frac{1}{x^{2}}$ is finite,&#xA;i.e $x\neq 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{x^{2}}=\frac{9}{\left( 4-x-y\right) ^{2}}.\qquad (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is satisfied if the square root of one side is equal or&#xA;symmetric to the square root of the other side: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{x}=\pm \frac{3}{4-x-y}.\qquad (3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation $(3)$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=\pm \left( 4-x-y\right) \qquad (4)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided that $x\neq 0$ and $4-x-y\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation $(4)$ represents two equations. One is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=4-x-y,\qquad (5)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x=4-y\Leftrightarrow x=1-\frac{1}{4}y\qquad (6)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=-\left( 4-x-y\right), \qquad (7)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3x=-4+x+y\Leftrightarrow x=-2+\frac{1}{2}y.\qquad (9)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $(1)$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=1-\frac{1}{4}y\qquad (10)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=-2+\frac{1}{2}y,\qquad (11)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided that $y\neq 4$ because the conditions $x\neq 0$ and $4-x-y\neq 0$&#xD;&#xA;correspond to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\neq 0\Leftrightarrow 1-\frac{1}{4}y\neq 0\Leftrightarrow y\neq 4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\neq 0\Leftrightarrow -2+\frac{1}{2}y\neq 0\Leftrightarrow y\neq 4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4-x-y\neq 0\Leftrightarrow 4-\left( 1-\frac{1}{4}y\right) -y\neq&#xD;&#xA;0\Leftrightarrow y\neq 4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4-x-y\neq 0\Leftrightarrow 4-\left( -2+\frac{1}{2}y\right) -y\neq&#xD;&#xA;0\Leftrightarrow y\neq 4.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-18T21:41:35.687" LastActivityDate="2010-11-18T21:41:35.687" />
  <row Id="8461" PostTypeId="1" AcceptedAnswerId="8538" CreationDate="2010-10-31T18:25:54.540" Score="1" ViewCount="95" Body="&lt;p&gt;claims arrive in an insurance portfolio, according to a homogenous Poission process X(t), t&gt;=0. Assume that each of the 12 months in a year has exactly 30 days. Calculate the value of the following conditional probability:&#xA;P{X(2) = 3, X(3) = 5 | X(4) = 5}&#xA;where X(k) is the number of claims during the first k months of the year&#xA;can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-11-01T17:56:54.300" Title="claims in an insurance portfolio" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8462" PostTypeId="1" CreationDate="2010-10-31T19:08:48.297" Score="1" ViewCount="99" Body="&lt;p&gt;I have a three pairs of points namely : p{1,1} p{1,2} p{2,1} p{2,2} p{3,1} p{3,2}, I want to find p such that : |p-p{i,1}| = |p-p{i,2}| for all i. I know that these represent three planes, but I am not sure whether these planes do meet in a single point. My simple question is even if the three planes are not parallel do they always meet in a single point. Intuitively I can visualize three planes which are not parallel but don't meet at a single point but how do I know that mathematically ?&lt;/p&gt;&#xA;" OwnerUserId="2978" LastActivityDate="2010-10-31T20:23:24.737" Title="How to determine a point in R3 which is which is equidistant to three pairs of point ?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8463" PostTypeId="2" ParentId="8462" CreationDate="2010-10-31T19:26:29.970" Score="5" ViewCount="0" Body="&lt;p&gt;The planes described by the equations $z=0$, $z=y$, and $z=2y$ have the $x$-axis in common, so you may have infinitely many points.  The planes described by the equations $z=0$, $z=y$, and $z=1-y$ have no points in common, but no two are parallel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to determine whether the given planes will have a unique point of intersection is using linear algebra.  Given the equations $a_1x+b_1y+c_1z=d_1$, $a_2x+b_2y+c_2z=d_2$, and $a_3x+b_3y+c_3z=d_3$ describing the planes, there is a unique solution to the system of equations, corresponding to the point where the planes meet, if and only if the coefficient matrix&#xA;$$A= \begin{bmatrix}&#xA;  a_1 &amp;amp; b_1 &amp;amp; c_1 \\&#xA;  a_2 &amp;amp; b_2 &amp;amp; c_2 \\&#xA;  a_3 &amp;amp; b_3 &amp;amp; c_3&#xA;\end{bmatrix}$$&#xA;has an inverse, in which case the solution is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}&#xA;  x\\&#xA;  y \\&#xA;  z&#xA;\end{bmatrix}&#xA;=A^{-1}&#xA;\begin{bmatrix}&#xA;  d_1\\&#xA;  d_2 \\&#xA;  d_3&#xA;\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-10-31T20:23:24.737" LastActivityDate="2010-10-31T20:23:24.737" CommentCount="1" />
  <row Id="8464" PostTypeId="2" ParentId="8443" CreationDate="2010-10-31T19:46:34.110" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;What else can you count with these?&quot;: Solutions to the Fermat equation $x^n + y^n = z^n$.  (If this doesn't make immediate sense, I will add more material later, when I have more time.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-10-31T19:46:34.110" />
  <row Id="8465" PostTypeId="1" AcceptedAnswerId="8474" CreationDate="2010-10-31T20:07:31.527" Score="1" ViewCount="131" Body="&lt;p&gt;Greetings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=x}^{2x} 5 - \sum_{k=2x+1}^{5x+2} 5 = -25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I discover the value of X?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Turns out it was 5x+2, not 5x+1. My bad, apologies for the trouble.&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2950" LastEditorDisplayName="" LastEditDate="2010-10-31T20:40:09.020" LastActivityDate="2010-11-01T00:23:09.960" Title="Solving summation problem" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8466" PostTypeId="2" ParentId="8465" CreationDate="2010-10-31T20:13:52.160" Score="0" ViewCount="0" Body="&lt;p&gt;Use the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=x}^{2x} 5 = 5 \cdot (2x - x) = 5x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=2x+1}^{5x+1} 5 = 5 \cdot (5x+1-2x-1) = 15x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5x - 15x = -25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is easy to solve, you just get $x = 25/10$ or in simpler terms $x = 5/2$&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-31T20:13:52.160" CommentCount="6" />
  <row Id="8467" PostTypeId="2" ParentId="8435" CreationDate="2010-10-31T20:23:51.460" Score="2" ViewCount="0" Body="&lt;p&gt;There are many things that are known about this. Most of which I believe come from something called the Prime Number Theorem, which you can read about on wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something that might be helpful with regards to your question are the formulas in this section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proofs for most of these theorems are really tricky and require a lot of knowledge of analysis though, so be warned!&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-10-31T20:23:51.460" />
  <row Id="8468" PostTypeId="1" AcceptedAnswerId="8469" CreationDate="2010-10-31T21:34:46.363" Score="0" ViewCount="72" Body="&lt;p&gt;Hello.&#xA;$$\sum_{i=1}^{x} x + \sum_{i=1}^{x} 2 = 8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to discover $x$ here, so I made:&#xA;$$x(x-1) + 2(x-1) = 8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which should give me the $x$, but instead of an integer I get:&#xA;$$x = \frac{-1\pm\sqrt{41}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the answer should be $x=2$, according to the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="2950" LastActivityDate="2010-10-31T21:49:10.947" Title="Summation problem not giving expected answer" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8469" PostTypeId="2" ParentId="8468" CreationDate="2010-10-31T21:43:10.330" Score="2" ViewCount="0" Body="&lt;p&gt;You have $x$ integers between $1$ and $x$. Look: $1, 2, 3, \ldots, x-1, x$. So you are just not counting properly.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-10-31T21:43:10.330" />
  <row Id="8471" PostTypeId="2" ParentId="8435" CreationDate="2010-10-31T21:54:34.807" Score="6" ViewCount="0" Body="&lt;p&gt;The keyword you want is &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_gap&quot; rel=&quot;nofollow&quot;&gt;prime gap&lt;/a&gt;.  The best unconditional result I can see on that page appears to be $O(n^{3/4 + \epsilon})$ for any $\epsilon &gt; 0$.  With the Riemann hypothesis you get $O(\sqrt{n} \log n)$, but a famous conjecture asserts that you should be able to do considerably better, something like $O((\log n)^2)$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-10-31T22:07:13.360" LastActivityDate="2010-10-31T22:07:13.360" CommentCount="2" />
  <row Id="8472" PostTypeId="2" ParentId="392" CreationDate="2010-10-31T21:55:42.093" Score="1" ViewCount="0" Body="&lt;p&gt;One important fact that ought to be mentioned explicitly is that &#xA;$$\frac{d \sin x}{dx} = \cos x$$&#xA;&lt;em&gt;only when $x$ is measured in radians.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general angle measure, the derivative of $\sin x$ is some scalar multiple of $\cos x$.  In fact, it could be argued that this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Radian#Advantages_of_measuring_in_radians&quot; rel=&quot;nofollow&quot;&gt;major reason for the usefulness of radians&lt;/a&gt;: The radian is the angle measure that makes that scalar multiple equal to 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something that not a lot of people realize.  For example, here's a quote from a Math Overflow answer to the question &lt;a href=&quot;http://mathoverflow.net/questions/40082/&quot; rel=&quot;nofollow&quot;&gt;&quot;Why do we teach calculus students the derivative as a limit?&quot;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to point out a simple question that very few calculus students and even teachers can answer correctly: Is the derivative of the sine function, where the angle is measured in degrees, the same as the derivative of the sine function, where the angle is measured in radians. In my department we audition all candidates for teaching calculus and often ask this question. So many people, including some with Ph.D.'s from good schools, couldn't answer this properly that I even tried it on a few really famous mathematicians. Again, the difficulty we all have with this question is for me a sign of how badly we ourselves learn calculus. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To see why radians are crucial, look at the slopes of the graphs of $\sin x$ at $x = 0$ when $x$ is measured in radians and when $x$ is measured in degrees.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, when $x$ is in radians:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4AsQJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope appears to be close to 1.  (And, of course, we know that it is 1.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, when $x$ is in degrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1I74Q.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope is much, much smaller than 1.  So the derivative of $\sin x$ at $x = 0$ when $x$ is in degrees cannot be $\cos(0) = 1$.    The correct answer, if $x$ is in degrees, is that the derivative of $\sin x$ is $\frac{\pi}{180}\cos x$ (via the chain rule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, all of the answers to the OP's question given here implicitly assume that $x$ is being measured in radians.  (It might be an interesting exercise for students reading this to go through each of the other arguments to see exactly where that assumption is being made.)  However, as the Math Overflow quote points out, this is something that a lot of people don't realize.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-10-31T21:55:42.093" />
  <row Id="8473" PostTypeId="2" ParentId="8337" CreationDate="2010-10-31T22:05:39.633" Score="3" ViewCount="0" Body="&lt;p&gt;Here is an other one which is more or less what Euler did in one of his proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $\sin(x)$ where $x\in\mathbb{R}$ is zero exactly at $x=n\pi$ for each integer $n$. If we factorized it as an infinite product we get&#xA;$$\sin(x)= \cdots(1+\frac{x}{3\pi})(1+\frac{x}{2\pi})(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(1-\frac{x}{2\pi})(1-\frac{x}{3\pi})\cdots =$$&#xA;$$\qquad x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^2\pi^2})(1-\frac{x^2}{3^2\pi^2})\cdots $$&#xA;We can also represent $\sin(x)$ as a Taylor series at $x=0$&#xA;$$\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$&#xA;Multiplying the product and identifying the coefficient of $x^3$ we see that&#xA;$$\frac{x^3}{3!}=x(\frac{x^2}{\pi^2} + \frac{x^2}{2^2\pi^2}+ \frac{x^2}{3^2\pi^2}+\cdots)$$&#xA;or &#xA;$$\frac{\pi^2}{3!}=\sum_{n=1}^\infty\frac{1}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two interesting links: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.math.dartmouth.edu/~euler/date-publish.html&quot; rel=&quot;nofollow&quot;&gt;Euler's papers&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://www.southernct.edu/~sandifer/Ed/History/Preprints/Talks/NYU%20Basel%20Problem%20Paper.PDF&quot; rel=&quot;nofollow&quot;&gt;Euler’s Solution of the Basel Problem – The Longer Story&lt;/a&gt;&lt;/em&gt; an essay on the subject written by Ed Sandifer. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-10-31T22:05:39.633" />
  <row Id="8474" PostTypeId="2" ParentId="8465" CreationDate="2010-10-31T22:33:14.637" Score="1" ViewCount="0" Body="&lt;p&gt;You asked a similar question, so I think you didn't understand Derek's comment. Here we need to count how many integers there are between $x$ and $2x$ and between $2x+1$ and $5x+2$. Since you are not familiar with these things, try to write all possible numbers. You'll get something like $$x,\ x+1, \ldots, \ x+ (x-1), \ 2x$$ and $$2x+1, \ldots, \ 2x + (x-1), \ 3x, \ 3x+1, \ldots, \ 3x+ (x-1), \ 4x, $$ $$4x+1, \ldots, \ 4x +(x-1), \ 5x, \ 5x+1, \ 5x+2.$$ Now it's easy to see that you have $x+1$ numbers between $x$ and $2x$ and $3x+2$ between $2x+1$ and $5x+2$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-01T00:23:09.960" LastActivityDate="2010-11-01T00:23:09.960" />
  <row Id="8475" PostTypeId="1" CreationDate="2010-10-31T22:37:30.213" Score="1" ViewCount="63" Body="&lt;p&gt;I need a proof of this well known fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a group  $G$  is given as a quotient of a free group $F/R$ then its pro-finite&#xA;completion is given by the quotient $\hat{F}/ \bar{R}$ where $\bar{R}$ is &#xA;the closure of $R$ in $\hat{F}$. (since $F$ embeds into $\hat{F}$ densely &#xA;we can think of $R$ inside $\hat{F}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2980" LastActivityDate="2010-11-01T00:32:27.777" Title="Pro-finite presentations" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8476" PostTypeId="1" CreationDate="2010-10-31T22:59:18.900" Score="1" ViewCount="129" Body="&lt;p&gt;What I'm attempting to do is rearrange the formula for the sum of a geometric series so as to find the value of $r$. I've tried several different methods, all of which have failed; though I can't understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was my last attempt using logs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n=\frac{a(1-r^n)}{1-r}$$&#xA;$$S_n(1-r) = a(1-r^n)$$&#xA;$$S_n-S_nr = a - ar^n$$&#xA;$$\frac{S_n}{a}-1 = \frac{S_n}{a}r-r^n$$&#xA;$$\log{\frac{S_n}{a}} = \log{\frac{S_n}{a}}+\log{r}-n\log{r}$$&#xA;$$0 = (\log{r})(n-1)$$&#xA;$$r = 10^{\frac{0}{n-1}}$$&#xA;$$r=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand where I'm going wrong. Any advice would be great. Also, if you know a formula to find $r$ using $S_n$ &amp;amp; $a$ then that would be equally fantastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ben.&lt;/p&gt;&#xA;" OwnerUserId="2981" LastActivityDate="2010-11-01T16:01:15.017" Title="Rearrange the formula for the sum of a geometric series to find &quot;r&quot;?" Tags="&lt;sequences-and-series&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8477" PostTypeId="2" ParentId="8476" CreationDate="2010-10-31T23:16:42.447" Score="1" ViewCount="0" Body="&lt;p&gt;No need to solve the full polynomial as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{n+1} - rS_n = a$$ (by def. of geometric series)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I would not leave this as an own entry if I had enough points to comment everywhere, sorry about that)&lt;/p&gt;&#xA;" OwnerUserId="2900" LastEditorUserId="2900" LastEditorDisplayName="" LastEditDate="2010-10-31T23:48:20.410" LastActivityDate="2010-10-31T23:48:20.410" />
  <row Id="8478" PostTypeId="2" ParentId="8476" CreationDate="2010-10-31T23:21:33.190" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Consider $\ S_{n+1} - r\ S_n $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-10-31T23:21:33.190" />
  <row Id="8479" PostTypeId="1" CreationDate="2010-10-31T23:30:53.227" Score="0" ViewCount="124" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8323/does-stirlings-formula-give-the-correct-number-of-digits-of-n&quot;&gt;Does Stirling&amp;#39;s formula give the correct number of digits of n!?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(n) = [(2.n.pi) ^ (1 / 2)]. [(n / e) ^ n]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f(n) is the classical approach of Stirling's factorial natural number n&gt;1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;floor[log(f(n)] = floor(log n!), whatever the natural number n&gt;1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(log indicates logarithm to base 10.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I STILL CAN NOT GET NO COUNTEREXAMPLE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already, many thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-01T09:53:47.973" LastActivityDate="2010-11-01T09:53:47.973" ClosedDate="2010-11-01T13:15:39.687" Title="About Stirling's formula" Tags="&lt;number-theory&gt;" CommentCount="5" />
  <row Id="8480" PostTypeId="2" ParentId="8476" CreationDate="2010-10-31T23:38:27.110" Score="3" ViewCount="0" Body="&lt;p&gt;Added: I assume you don't know both $S_n$ and $S_{n+1}$. Also I thought you had a geometric progression, i.e. not an infinite series. If you mean a series then you have to apply limits to $S_n$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{S_{n}}{a}-1=\frac{S_{n}}{a}r-r^{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you cannot deduce&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log \frac{S_{n}}{a}=\log \frac{S_{n}}{a}+\log r-n\log r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (x+y)\neq \log x+\log y.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct property of the $\log $ function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (xy)=\log x+\log y.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebraic identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{n}-c^{n}=(x-c)(x^{n-1}+cx^{n-2}+c^{2}x^{n-3}+\cdots +c^{n-2}x+c^{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives for $c=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+x^{n-3}+\cdots +x+1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{S_{n}}{a}=\frac{1-r^{n}}{1-r}=r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to find $r$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1-\frac{S_{n}}{a}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this polynomial equation has no algebraic solution, in general, for $n\ge 6$. You need a different method (a numerical one).&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-01T16:01:15.017" LastActivityDate="2010-11-01T16:01:15.017" CommentCount="2" />
  <row Id="8481" PostTypeId="2" ParentId="6234" CreationDate="2010-10-31T23:46:38.267" Score="0" ViewCount="0" Body="&lt;p&gt;What you're basically looking for is a function $f(x)$ such that $f(2x)=f^2(x)-1$ and $f(0)=\phi$, from there:&#xA;\begin{align}&#xA;2f'(2x)&amp;amp;=2f(x)f'(x)\\&#xA;\frac{f'(2x)}{f'(x)}&amp;amp;=f(x)\\&#xA;\frac{f'(x)}{f'(x/2)}&amp;amp;=f(x/2)\\&#xA;\frac{f'(x)}{f'(x/2^n)}&amp;amp;=\prod_{k=1}^n f(x/2^k)&#xA;\end{align}&#xA;and, given a value $x_0$ such that $f(x_0)=1$,&#xA;\begin{align}&#xA;\Phi&amp;amp;=\prod_{k=1}^{\infty} \frac{f(x_0/2^k)}{\phi}\\&#xA;    &amp;amp;=\lim_{n\rightarrow\infty}\phi^{-n} \prod_{k=1}^n f(x_0/2^k)\\&#xA;    &amp;amp;=\lim_{n\rightarrow\infty}\phi^{-n} \frac{f'(x_0)}{f'(x_0/2^n)}\\&#xA;    &amp;amp;=\lim_{h\rightarrow0}h^\alpha \frac{f'(x_0)}{f'(hx_0)}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha=\frac{ln(\phi)}{ln(2)}$.  Unfortunately, I have no idea how to get $f(x)$, and the fact that $f(x)=1+O(x^{1+\alpha})$ does not make finding this function look easy.&lt;/p&gt;&#xA;" OwnerUserId="2957" LastActivityDate="2010-10-31T23:46:38.267" CommentCount="3" />
  <row Id="8482" PostTypeId="2" ParentId="8461" CreationDate="2010-11-01T00:27:06.320" Score="1" ViewCount="0" Body="&lt;p&gt;(Major rewrite: The following is a much simpler and cleaner answer than my previous one involving binomial distributions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by considering what $X(4) = 5$ actually means.  It means that through the first four months of the year there have been 5 claims.  Given that, you're looking for the probability that all three of the following events occurred:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Exactly 3 claims in the first two months.  &lt;/li&gt;&#xA;&lt;li&gt;Exactly 2 claims in the third month.  &lt;/li&gt;&#xA;&lt;li&gt;Exactly 0 claims in the fourth month.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each claim is equally likely to fall in each month.  So, out of 5 total items, you're trying to fit 3 into a category with a probability of $\frac{1}{2}$, 2 into a category with a probability $\frac{1}{4}$, and 0 into a category with probability $\frac{1}{4}$.  This situation is exactly what a &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_distribution&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;multinomial&lt;/em&gt; distribution&lt;/a&gt; models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability is &#xA;$$\frac{5!}{3! 2! 0!} \left(\frac{1}{2}\right)^3 \left(\frac{1}{4}\right)^2 \left(\frac{1}{2}\right)^0 = \frac{5}{64}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-01T04:11:09.677" LastActivityDate="2010-11-01T04:11:09.677" CommentCount="2" />
  <row Id="8483" PostTypeId="2" ParentId="8475" CreationDate="2010-11-01T00:32:27.777" Score="3" ViewCount="0" Body="&lt;p&gt;The profinite completion is the inverse limit of the $G/H$ for all normal subgroups $H$ of $G$ of finite index, that is the inverse limit of the $F/I$ for all normal subgroups $I$ of $F$ which are of finite index and contain $R$.&#xA;So you have a continuous morphism $\hat{F} \rightarrow \hat{G}$, it is surjective and its kernel is the intersection of the above $I$, and it is easily seen to be $\bar{R}$.&#xA;Thus you have a continuous bijective morphism $\hat{F}/\bar{R} \rightarrow \hat{G}$, and since $\hat{F}$ is compact (it is profinite) it is a homeomorphism, so it's an isomorphism of topological groups.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-01T00:32:27.777" />
  <row Id="8484" PostTypeId="1" AcceptedAnswerId="8486" CreationDate="2010-11-01T00:34:32.947" Score="5" ViewCount="142" Body="&lt;p&gt;Find all positive integers n, such that $$\sum\limits_{k=1}^{n}(k+2)^n=(n+3)^n $$&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T02:26:46.420" LastActivityDate="2010-11-01T02:26:46.420" Title="Find all positive integers n, such that..." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8485" PostTypeId="1" AcceptedAnswerId="8534" CreationDate="2010-11-01T01:32:41.677" Score="2" ViewCount="116" Body="&lt;p&gt;The Hoeffding's inequality is $P(S_n - E[S_n] \geq \epsilon) \leq e^{-2\epsilon^2/k'}$, where $S_n = \sum_{i=1}^{n} X_i$, $X_i$'s are independent bounded random variables, and $k'$ depends on the random variables.&#xA;In the proof of Hoeffding's inequality, an optimization problem of the form is solved:&#xA;$$\min_{s} \ \ e^{-s\epsilon}e^{ks^2}$$ subject to $s &gt; 0$, to obtain a tight upper bound (which in turn yields the Hoeffding's inequality). &#xA;It turns out that $s = \epsilon/2k$ is the value that obtains the Hoeffding's inequality. I don't see how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Note that $k &gt; 0, \epsilon &gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastEditorUserId="1527" LastEditorDisplayName="" LastEditDate="2010-11-01T03:48:24.260" LastActivityDate="2010-11-01T16:18:05.860" Title="Proof of Hoeffding's inequality " Tags="&lt;probability-theory&gt;&lt;convex-optimization&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8486" PostTypeId="2" ParentId="8484" CreationDate="2010-11-01T02:13:15.420" Score="6" ViewCount="0" Body="&lt;p&gt;$k^n/(n+3)^n = \left(1-\frac{n+3-k}{n+3}\right)^n$, so using that $\log (1+x) \leq x$ for all $x$,&#xA;\[ \frac{\sum_{k=3}^{n+2} k^n}{(n+3)^n} \leq \sum_{j=1}^n e^{-jn/(n+3)} \leq \frac{1}{e^{n/(n+3)}-1} \]&#xA;This last term is decreasing, and for $n=7$ it is strictly less than $1$, so you are left with only $n \leq 6$ to check.&#xA;It is probably possible to sharpen the inequalities so that there would be less to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you get $n=2$ or $3$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-01T02:19:10.877" LastActivityDate="2010-11-01T02:19:10.877" />
  <row Id="8487" PostTypeId="1" AcceptedAnswerId="8488" CreationDate="2010-11-01T02:29:09.993" Score="1" ViewCount="84" Body="&lt;p&gt;Please imagine that we have a circular segment with some arc length 's' and chord length 'a' (using notation from &lt;a href=&quot;http://mathworld.wolfram.com/CircularSegment.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/CircularSegment.html&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided only 'a' and 's', and placing the left-hand-side point of the chord at the origin of the Euclidean plane (or a more convenient point), is there sufficient information to write an expression for the height of the circular segment (i.e. the y-axis/&quot;vertical&quot; distance between the chord on the x-axis and the circular arc) as a function of a position on the chord?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a simple matter to express the chord length in terms of the arc length and 'theta': a = (s) * 2(Sin[theta/2])/theta, or an expression for the arc length in terms of the chord length and theta: s = a/(2(Sin[theta/2])/theta).  And one can write an express for the maximum height as: h = (R - (1/2)*((-a)^2+4*R^2)^(1/2)), where the radius of the circle, 'R' is related to theta as: R = 1/2 Sqrt[a^2/(-1 + Cos[theta/2]^2)]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is insufficient information to accomplish the above, I would love to have an intuitive explanation for why this is so. &lt;/p&gt;&#xA;" OwnerUserId="5009" LastEditorUserId="5009" LastEditorDisplayName="" LastEditDate="2010-11-01T02:54:52.730" LastActivityDate="2010-11-01T02:54:52.730" Title="Calculating the height of a circular segment at all points provided only chord and arc lengths" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8488" PostTypeId="2" ParentId="8487" CreationDate="2010-11-01T02:52:44.400" Score="3" ViewCount="0" Body="&lt;p&gt;Using the notation of the figure you have linked to, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;R \sin \frac{\theta}{2} = \frac{a}{2}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can also write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\theta = \frac{s}{R} = \frac{2 s \sin \theta/2}{a}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation, you can solve for $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have solved for $\theta$, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;h = R - R \cos(\theta/2)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $R = a/(2 \sin \theta/2)$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;h = \frac{a}{2 \sin \theta/2} \left( 1 - \cos\left(\frac{s \sin\theta/2}{a}\right)\right)&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-01T02:52:44.400" CommentCount="2" />
  <row Id="8489" PostTypeId="1" AcceptedAnswerId="8492" CreationDate="2010-11-01T02:53:14.920" Score="2" ViewCount="57" Body="&lt;p&gt;Let ${a_k}$, ${b_k}$ be two sequences in $\mathbb{R}$. Suppose ${a_k}$ converges to $a$ s.t. $\liminf_{k \to \infty} {a_k} = \lim_{k \to \infty} {a_k} = a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it then true that $\liminf_{k \to \infty} (a_k - b_k) = \lim_{k \to \infty} a_k + \liminf_{k \to \infty} -b_k$?&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-01T03:45:16.013" Title="$\liminf$ of difference of two sequences" Tags="&lt;analysis&gt;" AnswerCount="3" />
  <row Id="8490" PostTypeId="2" ParentId="392" CreationDate="2010-11-01T03:08:29.877" Score="1" ViewCount="0" Body="&lt;p&gt;As an addendum, you can download the Mathematica notebook from &lt;a href=&quot;http://demonstrations.wolfram.com/GraphingDerivatives/&quot; rel=&quot;nofollow&quot;&gt;Graphing Derivatives&lt;/a&gt;, which allows you to play a little bit with $\text{sin}(x)$, $\text{cos}(x)$ and another function. I think it shows a very obvious but interesting construction of those trigonometric functions.&#xA;In case you don't want to download or install anything, I posted an amateurish &lt;a href=&quot;http://www.youtube.com/watch?v=hBJfoLLZJ8A&quot; rel=&quot;nofollow&quot;&gt;screencast&lt;/a&gt;, so you can see the demonstration. Basically, you draw the $\text{sin}(x)$ function, and in each point $(x,y)$ you calculate/draw the slope. The value of the slope corresponds to the value of the $y$ coordinate of the derivative of the function (in this example, $\text{cos}(x)$), keeping the same $x$ coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a wonderful exercise to plot some random function, and drawing the derivative of that function based on this procedure, then take a look at the 'true' derivative and see how much your drawing resembles the derivative.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-01T03:49:15.053" LastActivityDate="2010-11-01T03:49:15.053" />
  <row Id="8491" PostTypeId="2" ParentId="8489" CreationDate="2010-11-01T03:11:51.170" Score="1" ViewCount="0" Body="&lt;p&gt;Yes.  Here's an approach to proving this, without full details.  Given $\epsilon\gt0$, take $N$ sufficiently large so that $a_k-b_k$ is within $\epsilon$ of $a-b_k$ for all $k\geq N$.  Then the $\liminf$ of the sequence $a_N-b_N,a_{N+1}-b_{N+1},\ldots$ must lie within $\epsilon$ of that of the sequence $a-b_N,a-b_{N+1},\ldots$.  Removing a finite number of terms from the sequence doesn't change the $\liminf$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-01T03:11:51.170" />
  <row Id="8492" PostTypeId="2" ParentId="8489" CreationDate="2010-11-01T03:19:58.080" Score="3" ViewCount="0" Body="&lt;p&gt;Yes. In general you only have subadditivity of the limit inferior (that is, in general you only have $\liminf (a_n+b_n) \geq \liminf(a_n) + \liminf(b_n)$) but if at least one sequence converges, then you have equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why you have equality, think of $\liminf$ as the infimum of all limits of converging subsequences. If $a_{j_k}+b_{j_k}$ is a converging subsequence of $\{a_n+b_n\}$, then $b_{j_k}$ must also converge (since $a_{j_k}$ converges: every subsequence of a converging sequence converges); and likewise, if $b_{j_k}$ is a converging subsequence of $b_n$, then $a_{j_k}+b_{j_k}$ is a converging subsequence of $a_n+b_n$. Moreover, if $b_{j_k}$ converges to $s$, then $a_{j_k}+b_{j_k}$ converges to $a+s$. Conversely, if $a_{j_k} + b_{j_k}$ converges to $a+t$, then $b_{j_k}$ must converge to $t$. So if $L$ is the set of all points that are limits of subsequences of $b_n$, then the set of all points that are limits of subsequences of $a_n+b_n$ is $a+L = \{a+t \mid t\in L\}$. Therefore&#xA;\[ \liminf (a_n+b_n) = \inf (a+L) = a + \inf L = a + \liminf b_n = \lim_{n\to\infty}a_n + \liminf b_n.\]&#xA;To get the difference, just apply this to $a_n$ and $(-b_n)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or you can try the same argument with $\limsup$, and use the fact that $\liminf -b_n = -\limsup b_n$; you would think of $\limsup$ as the supremum of all limits of converging subsequences of $b_n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now see if you can figure out why you only have an inequality in the general case...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-01T03:45:16.013" LastActivityDate="2010-11-01T03:45:16.013" CommentCount="3" />
  <row Id="8493" PostTypeId="2" ParentId="8356" CreationDate="2010-11-01T03:31:54.993" Score="0" ViewCount="0" Body="&lt;p&gt;Since I can't make comments (haven't got enough points yet), I writing them in the form of answer. The word &quot;pdf&quot; stands for &quot;probability density function&quot;, not for &quot;probability distribution function&quot;. In your particular case, of course, you mean some discrete distributions, which are referred to as &quot;pmf&quot; (probability mass function). So, please &lt;em&gt;do&lt;/em&gt; use the correct terminology. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-01T03:31:54.993" />
  <row Id="8494" PostTypeId="2" ParentId="8489" CreationDate="2010-11-01T03:32:35.180" Score="0" ViewCount="0" Body="&lt;p&gt;Yes. One possible proof outline:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If there is a $N$ such that $c_k \geq d_k$ for all $k &gt; N$ then $\liminf c_k \geq \liminf d_k$&lt;/li&gt;&#xA;&lt;li&gt;If $C$ is a constant then $\liminf (C+a_k)=C+\liminf a_k$&lt;/li&gt;&#xA;&lt;li&gt;Let $\lim a_k=a$. For any $\epsilon &gt; 0$ we can find a $N$ such that for all $k &gt; N$&#xA;$$a-\epsilon - b_k \leq a_k - b_k \leq a+\epsilon -b_k$$&#xA;Apply (1) and (2), then make use of the fact that the result is true for all $\epsilon &gt; 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-11-01T03:32:35.180" />
  <row Id="8495" PostTypeId="1" AcceptedAnswerId="8499" CreationDate="2010-11-01T03:59:34.213" Score="1" ViewCount="269" Body="&lt;p&gt;The series in question is $\sum\limits_{n=2}^{\infty}\frac{(-1)^n n^2}{( \ln(n))^n}$&#xA;I think this diverges.  So using the divergence test, I am trying to show that limit of the general term of the series is not 0.&#xA;By a few algebraic manipulations we get :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\lim (-1)^n(n^2)( \ln(n))^{-n}$&lt;/li&gt;&#xA;&lt;li&gt;$\lim (-1)^n(n)/ \ln(n)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to proceed here.  If you get rid of the alternating, the limit is $\infty$.&#xA;However, I am not sure how to manipulate this to show that (I suspect) the limit does not exist.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T04:31:19.870" LastActivityDate="2010-11-04T02:15:55.493" Title="Is $\sum\limits_{n=2}^{\infty} \frac{(-1)^n n^2}{( \ln(n))^n}$ divergent?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="8496" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:17:19.370" Score="3" ViewCount="0" Body="&lt;p&gt;If $a_n$ is a sequence, then $\lim_{n\to\infty}a_n = 0$ if and only if $\lim_{n\to\infty} |a_n| = 0$. So if the absolute value of the terms does not go to zero, then the terms don't go to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if you know that $\lim \frac{n}{\ln (n)}\neq 0$, then it follows that $\lim\frac{(-1)^n n}{\ln n}\neq 0$. (because the former is the limit of the absolute values of the latter). So if your reduction is accurate, and your proof that the limit without the $(-1)^n$ factor is not zero is accurate, then you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I'm not sure how you manage to get to that limit. If you had $(\ln(n^n))^{-1}$, then I would agree; but how do you manage to get rid of the exponent and cancel an $n$ if you have&#xA;\[ \frac{(-1)^n n^2}{(\ln n)^n} \text{?}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Well, looks like the beans have been spilt; your &quot;few algebraic manipulations&quot; were incorrect. Remember that while $\ln(a^k) = k\ln(a)$, it is &lt;strong&gt;not&lt;/strong&gt; true that $(\ln a)^k = k\ln a$. $\ln(a^k)$ is the logarithm of the $k$th power of $a$; $(\ln a)^k$ is the $k$th power of the logarithm of $a$; they are usually different.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-01T04:30:05.960" LastActivityDate="2010-11-01T04:30:05.960" />
  <row Id="8497" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:25:02.233" Score="2" ViewCount="0" Body="&lt;p&gt;The series converges by the alternating series test.  Take logs of the absolute values of the terms to get $2\ln(n)-n\ln(\ln(n))$.  Consider the function $f(x)=2\ln(x)-x\ln(\ln(x))$, with derivative $f'(x)=\frac{2}{x}-\ln(\ln(x))-\frac{1}{\ln(x)}$.  Since $f$ goes to $-\infty$ and $f'$ is eventually always negative, continuity and monotonicity of the exponential function tell us that the terms of your sequence have absolute value eventually decreasing to 0, so that the alternating series test applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second expression you give for the limit does not go to zero, and it is not clear where it came from.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-01T04:25:02.233" />
  <row Id="8498" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:26:02.290" Score="1" ViewCount="0" Body="&lt;p&gt;You want to prove that $n^2 (\ln (n))^{-n}$ is monotonically decreasing and use Liebniz' test.  As long as $\ln (n)$ is greater than 1, the exponential will dominate the polynomial.  So pick your favorite starting place and fix ln(n) to ln(a) for some a and argue that the actual term is smaller in absolute value than this.  Now show that the version with ln(a) also goes to zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-01T04:26:02.290" />
  <row Id="8499" PostTypeId="2" ParentId="8495" CreationDate="2010-11-01T04:36:25.760" Score="7" ViewCount="0" Body="&lt;p&gt;Applying the root test you can see that the series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \lim_{n\to\infty} |a_{n}|^{1/n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L&amp;lt;1$, then the series is absolutely convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L&gt;1$ the series is divergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L=1$ then the test is inconclusive. However, if $|a_{n}|^{1/n}\geq 1$ for infinitely many distinct values of n, then the series $\sum_{n=1}^{\infty} a_{n}$ diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your series, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \lim_{n\to\infty}\bigg|\displaystyle \frac{(-1)^{n}n^{2}}{\text{ln}^{n}(n)}\bigg|^{\frac{1}{n}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $L=\lim_{n\to\infty}\bigg|\displaystyle\frac{(-1)n^\frac{2}{n}}{\text{ln}(n)}\bigg|=0$ (because $\lim_{n\to\infty}\ln(n)=\infty$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $L&amp;lt;1$ and the series convergences. &lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-04T02:15:55.493" LastActivityDate="2010-11-04T02:15:55.493" CommentCount="6" />
  <row Id="8500" PostTypeId="1" AcceptedAnswerId="8533" CreationDate="2010-11-01T04:40:03.430" Score="0" ViewCount="49" Body="&lt;p&gt;Let G be a graph and δ(G) the minimum degree of a peak. Describe an algorithm in pseudocode that, for a given tree T with k&amp;lt;= δ(G) edges, should be build (in polinomial time) a sub-graph H of G so that H is isomorphic with T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I even start ? &lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2010-11-01T16:12:31.527" Title="The isomorphic subgraph problem." Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="8501" PostTypeId="1" CreationDate="2010-11-01T04:52:13.290" Score="1" ViewCount="364" Body="&lt;p&gt;This may sound silly, but... Suppose an aspiring amateur mathematician wanted to plan to move to another city...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some cities that are home to some of the largest number of the brightest mathematicians? I'm sure this may depend on university presence, or possibly industry presence, or possibly something surprising. Wondering where the best place to take a non-faculty job at a university and try to make friends with some sharp minds in the computer lab or at the nearby pub might be.&lt;/p&gt;&#xA;" OwnerUserId="2984" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T04:56:54.690" LastActivityDate="2010-11-05T07:29:42.070" Title="Best Cities for Mathematical Study" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8502" PostTypeId="1" CreationDate="2010-11-01T05:47:08.130" Score="1" ViewCount="116" Body="&lt;p&gt;$$\sum_{n=1}^\infty\frac{e^\frac{1}{n}}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the limit comparison test with $1/n$, which diverges. So I divide the $n^{th}$ term of my series by $1/n$, which is the same as multiplying by $n$, and all you're left with is $\lim_{n\to\infty}e^\frac{1}{n}$, which is $e^0$, which is $1$, and since $1$ is a constant greater than $0$, and $1/n$ diverges, then the original series diverges as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were my steps for this problem, but I got 0 points for this (not even partial credit!). What did I do wrong??&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-01T06:01:39.240" LastActivityDate="2010-11-01T06:01:39.240" ClosedDate="2010-11-01T06:32:22.010" Title="Convergence/Divergence using Limit Comparison Test" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" CommentCount="3" />
  <row Id="8503" PostTypeId="1" AcceptedAnswerId="8507" CreationDate="2010-11-01T06:14:12.360" Score="1" ViewCount="145" Body="&lt;p&gt;Determine all &#xA;    $ N, |N| \le 9 $ such that $ x^2-94y^2=N $ is solvable in positive integers.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T03:54:03.220" LastActivityDate="2010-11-04T03:54:03.220" Title="Pell's equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8504" PostTypeId="1" AcceptedAnswerId="8505" CreationDate="2010-11-01T06:58:06.847" Score="3" ViewCount="337" Body="&lt;p&gt;Is the subset consisting of all integrable (or square integrable) smooth functions of the set of all integrable (or square integrable) functions dense under the usual Euclidean or integral of absolute difference metric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By smooth I mean derivatives of all orders exist.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-02T14:13:12.030" LastActivityDate="2010-11-03T19:54:01.873" Title="Are the smooth functions dense in either L^2 or L^1?" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8505" PostTypeId="2" ParentId="8504" CreationDate="2010-11-01T07:06:57.990" Score="7" ViewCount="0" Body="&lt;p&gt;Yes.  In fact, by the Stone-Weierstrass theorem and the existence of smooth bump functions, smooth functions with compact support are uniformly dense in the space of continuous functions with compact support.  Uniform density implies $L^2$ and $L^1$ density for functions with compact (and therefore finite measure) support, and since continuous functions with compact support are dense in $L^2$ and $L^1$, the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to see this more directly, you can go through the iterations of approximating an arbitrary ($L^1$ or $L^2$) function with a bounded function with bounded support, then with a simple function, then with a step function, and finally approximate the step function with a smooth function using &lt;a href=&quot;http://en.wikipedia.org/wiki/Bump_function&quot; rel=&quot;nofollow&quot;&gt;bump functions&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-01T15:07:18.127" LastActivityDate="2010-11-01T15:07:18.127" CommentCount="7" />
  <row Id="8506" PostTypeId="2" ParentId="8444" CreationDate="2010-11-01T09:07:10.250" Score="1" ViewCount="0" Body="&lt;p&gt;Using the formula for the triangular numbers we note that if&#xA;$m \in I = [2n^2+n+1,2n^2+3n+1]$ for some $n=0,1,2,\ldots$ then $f(m)=m,$ otherwise&#xA;$f(m)=-m.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible choice of $n$ is $ \lfloor \sqrt{m/2} \rfloor,$ since if we write&#xA;$l(n) =  2n^2+n+1$ and $u(n) =  2n^2+3n+1$ by writing $\sqrt{m/2} = N + r,$ where $N$ is an integer and $0 \le r &amp;lt; 1$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u \left( \lfloor \sqrt{m/2} \rfloor – 1 \right) = 2N^2 – N &amp;lt; 2N^2+4Nr+r^2 &amp;lt; m,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $m \notin I.$ Similarly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l \left( \lfloor \sqrt{m/2} \rfloor + 1 \right) &gt; m,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $m \notin I.$ Hence we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(m) = m \textrm{ when } m \in [2t^2+t+1,2t^2+3t+1] \textrm{ for }&#xA;t =  \lfloor \sqrt{m/2} \rfloor,$$&#xA;otherwise $f(m)=-m.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-01T09:07:10.250" />
  <row Id="8507" PostTypeId="2" ParentId="8503" CreationDate="2010-11-01T09:20:33.417" Score="4" ViewCount="0" Body="&lt;p&gt;You are asking for which $N$ there is an element of norm&#xA;$N$ in the quadratic ring $R=\mathbb{Z}[\sqrt{94}]$.&#xA;Now $R$ is the full ring of integers in the field&#xA;$\mathbb{Q}(\sqrt{94})$ which has class number one.&#xA;The problem is almost the same as determining for which&#xA;$N$ there is a norm $N$ ideal in $R$. There is a norm&#xA;$N$ ideal in $R$ iff either $N$ or $-N$ is a norm from $R$.&#xA;But we can't have both $N$ and $-$ being norms, since $-1$&#xA;is not a norm from $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question reduces to&#xA;(i) for which $N$ is there an ideal in $R$ of norm $N$?&#xA;(ii) if there is such an ideal, is the corresponding solution to&#xA;Pell's equation with $+N$ or $-N$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i) $N=p_1^{r_1}\cdots p_k^{r_k}$ is a norm from $R$&#xA;iff $r_j$ is even whenever $p_j$ is inert in $R$. Now&#xA;$p$ is inert iff the Legendre symbol $(94/p)=-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii) we can sometimes use congruence conditions.&#xA;From (i) for instance $x^2-94y^2=\pm3$ is soluble, but&#xA;thinking modulo $8$ we see that $x^2-94y^2=-3$ is insoluble.&#xA;So $x^2-94y^2=3$ is soluble.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-01T09:20:33.417" CommentCount="1" />
  <row Id="8508" PostTypeId="1" AcceptedAnswerId="8551" CreationDate="2010-11-01T09:22:34.170" Score="3" ViewCount="142" Body="&lt;p&gt;Hi,&#xA;I define $X_i$ as a random variable that is uniformly distributed between (0,1). What is the expected number of such variables I require to make the sum go just higher than 1. &#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2988" LastActivityDate="2010-11-01T22:42:00.193" Title="Expected number of (0,1) distributed continuous random variables required to sum upto 1." Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="8510" PostTypeId="1" AcceptedAnswerId="8514" CreationDate="2010-11-01T09:51:53.300" Score="4" ViewCount="97" Body="&lt;p&gt;Consider the sphere $S^2 = \lbrace (x,y,z) :\ x^2 + y^2 + z^2 = 1 \rbrace$. This is a smooth manifold in $\mathbb{R}^3$, and for a given point $s \in S^2$, one can consider its coordinate neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to put a smooth structure on $S^2$, but all require at least two coordinate neighborhoods. One simple way is to make use of the local coordinates $\theta, \phi$ (azimuthal angle and inclination angle). For $s \in U = S^2 \setminus {(0,0,1), (0,0,-1)}$, define a coordinate map by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g : (0,2\pi) \times (0,\pi) \rightarrow U $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with formula given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g(\theta, \phi) = (\sin(\phi) \cos(\theta), \sin(\phi) \sin(\theta), \cos(\phi)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This map is one-to-one, onto, and bicontinuous, so it is a homeomorphism. One can construct a similar map onto $V = S^2 \setminus {(1,0,0),(-1,0,0)}$, so the whole sphere is covered. The fact that this defines a differentiable structure is easy to work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I want to ask is, are there any non-constant solutions to the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f : S^2 \rightarrow R$ is a smooth function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, what sorts of boundary conditions are necessary in order to ensure uniqueness.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-01T10:47:38.780" LastActivityDate="2010-11-01T11:49:12.557" Title="Harmonic functions on $S^2$" Tags="&lt;real-analysis&gt;&lt;differential-equations&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8511" PostTypeId="1" CreationDate="2010-11-01T11:24:55.190" Score="0" ViewCount="36" Body="&lt;p&gt;If you can help me in any of these two questions, I can get the answer to the other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a discrete random variable x and the data pertaining to it in X(n), how in matlab can we determine the prob. mass function pmf(X)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I use CVX to minimize a function log(pmf(X)) ?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-11-01T11:24:55.190" Title="CVX for matlab question" Tags="&lt;matlab&gt;&lt;convex-optimization&gt;" />
  <row Id="8512" PostTypeId="1" CreationDate="2010-11-01T11:34:17.557" Score="-2" ViewCount="106" Body="&lt;p&gt;Let $A$ be and $n\times n$ matrix over $\mathbb{C}$. If $A$ is diagonalizable and $k$ is a positive integer, prove that $A^k$ is diagonalizable. Matrix $A$ $n\times n$ is diagonal, if $a_{ij}=0$, when $i \neq j$. $n\times n$ matrix $A$ is diagonalizable if $A$ is similar to some diagonal matrix. In other words, a matrix is diagonalizable if there is a diagonal matrix $D$ and an invertible matrix $P$ such that $P^{-1}AP$= $D$. How are the eigenvalues and eigenvectors of $A^k$ related to those of $A$?&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-01T12:04:34.160" LastActivityDate="2010-11-01T12:04:34.160" Title=" prove that $ A^k$  is diagonalizable" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8513" PostTypeId="2" ParentId="8510" CreationDate="2010-11-01T11:46:57.393" Score="6" ViewCount="0" Body="&lt;p&gt;If I understand your question correctly, you are aware that $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2}$ is not actually the Laplacian on a sphere, but you want to know whether there are any nontrivial smooth functions on $S^2$ which are harmonic with respect to this modified Laplacian. The answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f(\theta,\phi)$ which satisfies $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$ is really a harmonic function on the rectangle $[0,2\pi] \times [0,\pi]$, which is mapped onto the sphere via $g^{-1}$. Continuity on the sphere requires that $f(\theta,0) = \text{const}$ (continuity at the north pole), $f(\theta,\pi) = \text{const}$ (continuity at the south pole), and $f(0,\phi) = f(2\pi,\phi) = h(\phi)$ for some $h:[0,\pi]\rightarrow \mathbb{R}$ (continuity across the prime meridian). For any $h$, the boundary conditions are independent of $\theta$, so $f$ is a function of $\phi$ alone. But then $\frac{\partial^2 f}{\partial \phi^2} = 0$ implies that $f$ is linear in $\phi$. If the slope is nonzero, $f$ on the sphere fails to be smooth at the poles. Therefore, for $f$ to be smooth on the sphere and satisfy $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$, it has to be constant.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-01T11:46:57.393" CommentCount="2" />
  <row Id="8514" PostTypeId="2" ParentId="8510" CreationDate="2010-11-01T11:49:12.557" Score="6" ViewCount="0" Body="&lt;p&gt;Since $f$ is a smooth function from $S^2\to \mathbb{R}$, it is a smooth function on your coordinate chart. It is easy to see that by periodicity, you also have that $f$ is a smooth function from $S^1\times(0,\pi)\to\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ is smooth on $S^2$, the limits&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\phi \to 0, \pi} \frac{1}{\sin \phi} \frac{\partial}{\partial \theta} f $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;need to exist and be finite: this is the requirement that $f$ is smooth in &quot;the other chart&quot;. Therefore you have that $\partial_\theta f \to 0$ as $\phi$ approach either boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation, however, requires that $\partial_\theta f$ be a harmonic function on $S^1\times (0,\pi)$, a compact domain with boundary. Using the maximum principle, you immediately have that $\partial_\theta f = 0$ everywhere. Plugging back into the equation, this implies $\partial^2f/\partial \phi^2 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $f$ must be a linear function in $\phi$. And we now apply the smoothness constraint yet again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the north pole, smoothness will require that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\phi \to 0} \partial_\phi f(\theta,\phi) = \lim_{\phi \to 0} \frac{1}{\sin \phi} \partial_\theta f(\theta + \pi/2, \phi) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies that the constant slope of the linear function $f(\phi)$ is 0. And hence $f$ must be the constant function. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-01T11:49:12.557" />
  <row Id="8515" PostTypeId="2" ParentId="8512" CreationDate="2010-11-01T12:02:49.357" Score="2" ViewCount="0" Body="&lt;p&gt;I think robinhoode and Robin are right: is it so difficult to ask things nicely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, here you have some hints:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let $v$ be an eigenvector of eigenvalue $\lambda$. That is, $Av = \lambda v$. Compute $A^k v$.&lt;/li&gt;&#xA;&lt;li&gt;Compute $(P^{-1}AP)^k$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-01T12:02:49.357" CommentCount="1" />
  <row Id="8516" PostTypeId="2" ParentId="8337" CreationDate="2010-11-01T12:32:20.437" Score="3" ViewCount="0" Body="&lt;p&gt;This is not really an answer, but rather a long comment prompted by &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n1-infty-frac1n2/8418#8418&quot;&gt;David Speyer's answer&lt;/a&gt;.&#xA;The proof that David gives seems to be the one in &lt;a href=&quot;http://arxiv.org/abs/math/0701039&quot; rel=&quot;nofollow&quot;&gt;How to compute $\sum 1/n^2$ by solving triangles&lt;/a&gt; by Mikael Passare,&#xA;although that paper uses a slightly different way of seeing that&#xA;the area of the region $U_0$ (in Passare's notation)&#xA;bounded by the positive axes and the curve $e^{-x}+e^{-y}=1$,&#xA;$$\int_0^{\infty} -\ln(1-e^{-x}) dx,$$&#xA;is equal to $\sum_{n\ge 1} \frac{1}{n^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings me to what I really wanted to mention, namely &lt;em&gt;another&lt;/em&gt; curious&#xA;way to see why $U_0$ has that area; I learned this from&#xA;&lt;a href=&quot;http://www.math.chalmers.se/~wastlund/&quot; rel=&quot;nofollow&quot;&gt;Johan Wästlund&lt;/a&gt;.&#xA;Consider the region $D_N$ illustrated below for $N=8$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6CGRB.png&quot; alt=&quot;A shape with area = sum of reciprocal squares&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's not immediately obvious,&#xA;the area of $D_N$ is $\sum_{n=1}^N \frac{1}{n^2}$.&#xA;&lt;em&gt;Proof&lt;/em&gt;: The area of $D_1$ is 1. To get from $D_N$ to $D_{N+1}$ one removes the boxes along the&#xA;top diagonal, and adds a new leftmost column of rectangles of width $1/(N+1)$&#xA;and heights $1/1,1/2,\ldots,1/N$,&#xA;plus a new bottom row which is the &quot;transpose&quot; of the new column,&#xA;plus a square of side $1/(N+1)$ in the bottom left corner.&#xA;The $k$th rectangle from the top in the new column&#xA;and the $k$th rectangle from the left in the new row (not counting the&#xA;square) have a combined area which exactly matches the $k$th box in the removed diagonal:&#xA;$$ \frac{1}{k} \frac{1}{N+1} + \frac{1}{N+1} \frac{1}{N+1-k} = \frac{1}{k} \frac{1}{N+1-k}. $$&#xA;Thus the area added in the process is just that of the square, $1/(N+1)^2$.&#xA;Q.E.D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apparently this shape somehow comes up in connection with the &quot;random&#xA;assignment problem&quot;, where there's an expected value of something which&#xA;turns out to be $\sum_{n=1}^N \frac{1}{n^2}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now place $D_N$ in the first quadrant, with the lower left corner at the origin.&#xA;Letting $N\to\infty$ gives nothing but the region $U_0$:&#xA;for large $N$ and for $0&amp;lt;\alpha&amp;lt;1$,&#xA;the upper corner of column number $\lceil \alpha N \rceil$ in $D_N$ lies at&#xA;$$ (x,y) =&#xA;   \left(&#xA;    \sum_{n=\lceil (1-\alpha) N \rceil}^N \frac{1}{n},&#xA;    \sum_{n=\lceil \alpha N \rceil}^N \frac{1}{n}&#xA;   \right)&#xA;   \sim&#xA;   \left(\ln\frac{1}{1-\alpha}, \ln\frac{1}{\alpha}\right),$$&#xA;hence (in the limit) on the curve $e^{-x}+e^{-y}=1$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-01T13:18:49.870" LastActivityDate="2010-11-01T13:18:49.870" CommentCount="1" />
  <row Id="8517" PostTypeId="2" ParentId="8504" CreationDate="2010-11-01T12:58:28.340" Score="4" ViewCount="0" Body="&lt;p&gt;Jonas's argument is good.  Another proof is: given $f \in L^p$ (here $p=1,2$), take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution&quot; rel=&quot;nofollow&quot;&gt;convolution&lt;/a&gt; of $f$ with a sequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mollifier&quot; rel=&quot;nofollow&quot;&gt;mollifiers&lt;/a&gt; $\eta_\epsilon$.  Using properties of convolutions, it's easy to check that $f * \eta_\epsilon$ is a smooth function, and that $f * \eta_\epsilon \to f$ in $L^p$ as $\epsilon \to 0$.  This has the advantage of being a little more direct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; For a reference, see Folland's &lt;em&gt;Real Analysis&lt;/em&gt;, section 8.2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smoothness of $f * \eta_\epsilon$ is Proposition 8.10 and comes from differentiating under the integral sign in the convolution (with justification!), and choosing to put the derivative on $\eta_\epsilon$.  Intuitively, it comes from the idea that convolution is an &quot;averaging&quot; operation and tends to smooth, smear, or blur rough areas of $f$ together, and so should be a smoothing operation.  (The wikipedia article has a nice animation illustrating this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that $f * \eta_\epsilon \to f$ in $L^p$ is Folland's Theorem 8.14 (a), and it's pretty elementary.  He also has Proposition 8.17 which proves that $C^\infty_c$ is dense in $L^p$, but it sort of inexplicably starts by using the fact that $C_c$ is dense in $L^p$.  I suppose this is used to get a compactly supported function, so that you can approximate $f \in L^p$ by functions which are not only smooth (which $f * \eta_\epsilon$ is) but also compactly supported (which $f * \eta_\epsilon$ need not be, although $\eta_\epsilon$ is).  But an easier argument would be to first approximate $f$ in $L^p$ norm by a function $g$ which is compactly supported but not necessarily continuous; for example, $g = f 1_{[-N,N]}$ for large $N$ (this works by dominated convergence), and then apply mollifiers to $g$.  Unless, of course, there is some subtlety that I've missed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit 2&lt;/b&gt;: Indeed there is.  Folland's 8.14 (a) relies upon the fact that translation is strongly continuous in $L^p$, which uses the density of $C_c$.  So apparently it is not so easy to bypass this step, and that destroys a lot of the &quot;directness&quot; of my argument.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-03T19:54:01.873" LastActivityDate="2010-11-03T19:54:01.873" CommentCount="18" />
  <row Id="8518" PostTypeId="1" AcceptedAnswerId="8521" CreationDate="2010-11-01T13:09:50.147" Score="2" ViewCount="109" Body="&lt;p&gt;Given a continuous function $g$ on $\mathbb R$. Is it possible to decompose $\mathbb R$ as the union of a countable collection of intervals $I_n=]a_n, a_{n+1}]$ so that $g$ is monotone on each $I_n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same question may be asked for a continuous function defined on a bounded interval.&lt;/p&gt;&#xA;" OwnerUserId="2989" LastActivityDate="2010-11-01T18:08:17.563" Title="Decomposition of a continuous function with monotone functions" Tags="&lt;real-analysis&gt;" AnswerCount="3" />
  <row Id="8519" PostTypeId="2" ParentId="8508" CreationDate="2010-11-01T13:26:22.327" Score="1" ViewCount="0" Body="&lt;p&gt;I don't know the exact value, but by &lt;a href=&quot;http://en.wikipedia.org/wiki/Wald%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Wald's equation&lt;/a&gt; it is between 2 and 4.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-01T13:26:22.327" />
  <row Id="8520" PostTypeId="2" ParentId="8518" CreationDate="2010-11-01T13:41:54.037" Score="5" ViewCount="0" Body="&lt;p&gt;Whenever someone comes with a guess as to whether continuous functions are &quot;nice&quot; in some way, the answer is almost inevitably &quot;no&quot;. For example this &lt;a href=&quot;http://math.stackexchange.com/questions/7923/are-continuous-functions-always-differentiable/7925&quot;&gt;other question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;Weierstraß function&lt;/a&gt; is likewise a counterexample to your question. Note that it is not of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bounded_variation&quot; rel=&quot;nofollow&quot;&gt;bounded variation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Made the opening line less... confrontational.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-01T14:00:47.477" LastActivityDate="2010-11-01T14:00:47.477" />
  <row Id="8521" PostTypeId="2" ParentId="8518" CreationDate="2010-11-01T13:42:10.293" Score="1" ViewCount="0" Body="&lt;p&gt;You might think about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_function&quot; rel=&quot;nofollow&quot;&gt;Weierstrass function&lt;/a&gt;.  I think you can prove it is not monotone over any interval by altering the proof that it is not differentiable.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-01T13:42:10.293" />
  <row Id="8522" PostTypeId="1" AcceptedAnswerId="8537" CreationDate="2010-11-01T14:01:00.613" Score="0" ViewCount="116" Body="&lt;p&gt;Hello members of MSE,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what the expansion series of the function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = -\frac{1}{x^3} \cdot \frac{1}{\Gamma(x) \cdot \Gamma(-(\exp(\frac{2}{3}\pi\cdot i))x) \cdot \Gamma(-(\exp(\frac{4}{3}\pi \cdot i))x)} $$ is, at $x = 0$. I'm also interested in the method behind computing the series expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think: &quot;why don't you just paste this equation in wolframalpha and then see what the expansion series is?&quot; Well, I did exactly that, but wolframalpha couldn't compute it! Does that mean the expansion series doesn't exist or is it more likely that it's just too 'hard' for the computational knowledge engine to find the series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT For more information:  this function is  function (27) at&lt;/em&gt; &lt;a href=&quot;http://mathworld.wolfram.com/InfiniteProduct.html&quot; rel=&quot;nofollow&quot;&gt; this page&lt;/a&gt;, &lt;em&gt;when n=3. I offer my apologies for not stating the question well at first.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-01T17:01:32.820" LastActivityDate="2010-11-01T17:25:33.480" Title="What is the Series Expansion of the following function?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="8523" PostTypeId="2" ParentId="8518" CreationDate="2010-11-01T14:21:17.857" Score="2" ViewCount="0" Body="&lt;p&gt;The function $f(x) = x \sin(1/x)$ is continuous on $\mathbb{R}$ and is not monotone on any nontrivial interval containing $0$.  For that matter, the same holds for $f_a(x) = x^a \sin(1/x)$ for any positive integer $a$, and this gives a (simple!) example which is highly differentiable: i.e., $k$ times differentiable for each fixed $k$ and all sufficiently large $a$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-01T18:08:17.563" LastActivityDate="2010-11-01T18:08:17.563" CommentCount="3" />
  <row Id="8524" PostTypeId="1" CreationDate="2010-11-01T14:27:06.023" Score="10" ViewCount="310" Body="&lt;p&gt;I am a soon-to-be high school graduate with inclination towards mathematics. I enjoy doing math, and am relatively good at it, but I dislike the way I am being taught. I feel like I am being taught methods for solving a lot of problems, but I don't see how any of this fits together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was considering pursuing a career in mathematics, but I realized I don't really know much about it as a whole as opposed to how to solve specific problems that will appear on some test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good books out there about the &quot;math mindset&quot;, or whatever it is that it's called?&lt;/p&gt;&#xA;" OwnerUserId="2993" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-01T14:32:00.730" LastActivityDate="2010-11-01T22:57:47.963" CommunityOwnedDate="2010-11-01T15:53:50.130" Title="On the Math Mindset" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="4" />
  <row Id="8525" PostTypeId="2" ParentId="8522" CreationDate="2010-11-01T14:33:58.057" Score="2" ViewCount="0" Body="&lt;p&gt;Up to a constant and a power of $x$, your function is $\frac1{x\Gamma(x)}$. Mathematica tells me that the Taylor series at zero is &#xA;$$&#xA;1+\gamma  x+\left(\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) x^2+\frac{1}{12} x^3 \left(2 \gamma ^3-\gamma  \pi ^2-2 \psi ^{(2)}(1)\right)+\frac{x^4 \left(60 \gamma ^4-60 \gamma ^2 \pi ^2+\pi ^4-240 \gamma  \psi ^{(2)}(1)\right)}{1440}+\dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\gamma$ is Euler's constant, and the $\psi$'s are poly-gamma functions.)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-01T14:33:58.057" CommentCount="9" />
  <row Id="8526" PostTypeId="1" AcceptedAnswerId="8531" CreationDate="2010-11-01T14:36:21.550" Score="1" ViewCount="87" Body="&lt;p&gt;Suppose $f(x,y) = c$ for $0\lt y\lt x\lt 1$ and $0$ outside. What is $P(X+Y \leq 1)$? What is $P(X^2+Y^2 \leq 1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $P(X+Y \leq 1) = \int_{0}^{1} \int_{0}^{1-x} 2 \ dy \ dx$? Likewise, $P(X^2+Y^2 \leq 1) = \int_{0}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \ dy \ dx$? This is assuming that $c=2$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-11-01T14:42:42.980" LastActivityDate="2010-11-01T15:32:55.273" Title="Probability and Integrals" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8528" PostTypeId="2" ParentId="8526" CreationDate="2010-11-01T14:56:18.817" Score="1" ViewCount="0" Body="&lt;p&gt;No, $P(X+Y \leq 1 ) = \int_0^1 \int_0^{\min (x,1-x)} 2 dy dx = \int_0^{1/2} \int_0^x 2 dy dx + \int_{1/2}^1 \int_0^{1-x} 2 dy dx = 1/4+1/4=1/2$&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-01T14:56:18.817" />
  <row Id="8529" PostTypeId="1" AcceptedAnswerId="8530" CreationDate="2010-11-01T14:58:25.730" Score="1" ViewCount="75" Body="&lt;p&gt;Is there a formula that can get sunrise at a particular latitude and longitude (and elevation from sea level)? If so, what it is? (Why does it work?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please tag as &quot;earth-science&quot; and any other appropriate tags. &lt;/p&gt;&#xA;" OwnerUserId="2414" LastActivityDate="2010-11-01T15:29:08.210" Title="Formula for getting time of sunrise at a particular location?" Tags="&lt;math-software&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8530" PostTypeId="2" ParentId="8529" CreationDate="2010-11-01T15:29:08.210" Score="4" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sunrise_equation&quot; rel=&quot;nofollow&quot;&gt;sunrise equation&lt;/a&gt; is what you're looking for. This gives the number of hours of daylight as a function of your latitude and the Sun's declination; the declination is itself a function of the day of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that one could figure out the time of sunrise (using the longitude and time zone for correcting for the position in the time zone), assuming that noon is exactly halfway between sunrise and sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ignoring time zones and daylight savings time, the time halfway between sunrise and sunset is not noon! On &lt;i&gt;average&lt;/i&gt; it's noon, but sometimes the sun runs slow and sometimes it runs fast. The correction for this is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_time&quot; rel=&quot;nofollow&quot;&gt;&quot;equation of time&quot;&lt;/a&gt;; the Wikipedia article gives an approximate formula good to within 45 seconds.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-01T15:29:08.210" />
  <row Id="8531" PostTypeId="2" ParentId="8526" CreationDate="2010-11-01T15:32:55.273" Score="2" ViewCount="0" Body="&lt;p&gt;Since the joint density is constant on the support region, you can get your answers by considering areas.  The answer in each case is the area that corresponds to the event in question divided by the total area of the support region.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to calculate $P(X + Y \leq 1)$, you want the area of the lower triangular region below (i.e., the region for which $x + y \leq 1$), divided by the total area of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Cq1RV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is $$\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, to calculate $P(X^2 + Y^2 \leq 1)$, you want the area of the circle sector divided by the total area of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IdpYU.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is $$\frac{\frac{\pi}{8}}{\frac{1}{2}} = \frac{\pi}{4}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-01T15:32:55.273" CommentCount="5" />
  <row Id="8532" PostTypeId="1" CreationDate="2010-11-01T15:35:08.983" Score="14" ViewCount="352" Body="&lt;p&gt;Many young, and not so young, mathematicians struggle with how to spend their time. Perhaps this is due to the 90%-10% rule for mathematical insight: 90 pages of work yield only 10 pages of useful ideas. A venerable mathematician once described his career to me as constantly stumbling around in the dark. Of course, this struggle is largely personal...perhaps it is an evolutionary struggle to find one's own way to find something relevant to contribute to mathematics as literature. Such an evolutionary struggle necessitates the artist's turmoil and requires an acknowledgement of &quot;l'importance d'être seul&quot;. Even when I have successfully done some reasonable mathematics, it behooves me to look for a more efficient way to proceed. This sort of searching turns up various nuggets of advice on how to do mathematics, like the following (paraphrased) ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Read a paper of a master for a year, and you will get something&lt;/em&gt;- advice given to young mathematicians by Israel Gelfand at Harvard during his 90th birthday conference on the unity of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Try to imagine a proof, no matter how vague&lt;/em&gt;-Gowers's online essay on the philosophy of mathematics and our relationship to formalism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I try to colorize it in my mind, to try to see what it's really getting at, rather than simply what it says&lt;/em&gt;-Thurston's online response to Ashley Reiter on reading mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also recall Grothendieck's images of a softening nut and the rising sea...online interviews/documents with suggestions of Atiyah, Singer, Connes, Gromov...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other such nuggets, many of which can be found on posts on this site. These all provide very inspiring images, but I would like to be more blunt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is your method? How do you go&#xA;  about doing mathematics on a day to&#xA;  day basis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditorDisplayName="" LastEditDate="2010-11-01T17:39:35.800" LastActivityDate="2010-11-12T19:28:24.213" CommunityOwnedDate="2010-11-01T15:53:01.393" Title="What is your method?" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;meta-math&gt;&lt;mathematicians&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="6" />
  <row Id="8533" PostTypeId="2" ParentId="8500" CreationDate="2010-11-01T16:12:31.527" Score="0" ViewCount="0" Body="&lt;p&gt;Scan the tree $T$ in some arbitrary way, meanwhile tracing out the subgraph $H$ (start at an arbitrary vertex of $G$). Take care not to close cycles. The condition on the minimum degree ensures that you'll never &quot;run out of edges&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-01T16:12:31.527" />
  <row Id="8534" PostTypeId="2" ParentId="8485" CreationDate="2010-11-01T16:18:05.860" Score="1" ViewCount="0" Body="&lt;p&gt;Since $\exp$ is monotone increasing, your problem is equivalent to minimizing the quadratic $ks^2 - \epsilon s = k(s - \epsilon/2k)^2 - \epsilon^2/4k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-01T16:18:05.860" />
  <row Id="8535" PostTypeId="1" CreationDate="2010-11-01T16:33:42.977" Score="3" ViewCount="61" Body="&lt;p&gt;Suppose $R=S/I$ is the homomorphic image of a local Cohen-Macaulay ring $(S,m)$.&#xA;I know the completion of $R$ as an $S$-module is just&#xA;$R \otimes \hat S$, where $\hat S$ is the $m$-adic completion of $S$.&#xA;Is this also the completion of $R$ as a ring?&#xA;(I think this comes down to showing that the inverse limit of $S/(m^i +I)$&#xA;is the same as the (inverse limit of $S/m^i$) tensored with $S/I$.)&lt;/p&gt;&#xA;" OwnerUserId="2998" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-01T18:41:29.997" LastActivityDate="2010-11-02T04:14:29.040" Title="Completion of the homomorphic image of a Cohen-Macaulay ring." Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="8536" PostTypeId="1" CreationDate="2010-11-01T17:04:40.697" Score="3" ViewCount="130" Body="&lt;p&gt;I've wasted a few days trying to prove Thue's lemma with no success! The proof is absolutely easy if you just count. What 'red flags' are useful to give you a warning that counting might be more productive than other approaches?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can start with one point, but it's more of a conceptual one and requires you to see the theorem in a bigger context: If you know that you cannot prove an existential effectively then counting is probably going to be involved. (You might notice that effective solution would imply fast algorithm for factoring or similar).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T01:06:52.163" Title="Learning When to count?" Tags="&lt;proof-strategy&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8537" PostTypeId="2" ParentId="8522" CreationDate="2010-11-01T17:25:33.480" Score="2" ViewCount="0" Body="&lt;p&gt;Mathematica can handle (rather easily) that function. For example, the expansion at x=0 to order 5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = -1-2 \gamma  x-2 \gamma ^2 x^2+\displaystyle \frac{1}{6} x^3 \left(-8 \gamma ^3-\psi ^{(2)}(1)\right)-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\displaystyle\frac{1}{3} x^4 \left(\gamma  \left(2 \gamma ^3+\psi ^{(2)}(1)\right)\right)+\frac{1}{60} x^5 \left(-16 \gamma ^5-20 \gamma ^2 \psi ^{(2)}(1)+\psi ^{(4)}(1)\right)+O\left(x^6\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(same notation than the one used in Mariano's post).&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-11-01T17:25:33.480" CommentCount="2" />
  <row Id="8538" PostTypeId="2" ParentId="8461" CreationDate="2010-11-01T17:56:54.300" Score="1" ViewCount="0" Body="&lt;p&gt;Hi Mike, thanks for your re-edit...now it makes much more sense and you are right, much cleaner!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I was last working on the problem, I have come up with the following alternative solution which yields the same numerical answer as you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P{X(2) = 3, X(3) = 5 | X(4) = 5}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= P{X(2) = 3, X(3) = 5, X(4) = 5} / P{X(4) = 5}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= P{X(2) = 3, X(3) - X(2) = 2, X(4) - X(3) = 0} / P{X(4) = 5} = (using independent increments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= P{X(2) =3}.P{X(3) - X(2) = 2}.P{X(4) - X(3) = 0} / P{X(4) = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= {exp((1/3!)-2u(2u)^3).exp((1/2!)-u(u)^2).exp(-u)} / exp((1 / 5!)-4u(4u)^5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 5!(2)^3 / 3!2!1!(4)^5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= (4 . 5 . 5) / 4^5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 5/64&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what do you think of this approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, sorry if it looks messy everyone, i just dont know how to get fancy formatting using the text entry, can anyone enlighten me on this?&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-11-01T17:56:54.300" CommentCount="2" />
  <row Id="8539" PostTypeId="1" AcceptedAnswerId="8542" CreationDate="2010-11-01T18:01:32.527" Score="0" ViewCount="75" Body="&lt;p&gt;$$\sum_{n\geq0} ({\frac{2}{3}})^{2n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to find that this is a geometric series with common ratio of $4/9$, therefore convergent. That means it's possible to calculate the sum of the series. Can I get any tip of how to do the sum of any kind of geometric series that converge? I've been researching, but all articles seem very confusing, specially with English, which I'm not native on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-01T18:40:29.360" LastActivityDate="2010-11-01T18:40:29.360" Title="Convergence of a series and respective sum" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;&lt;convergence&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8540" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T18:10:36.317" Score="8" ViewCount="0" Body="&lt;p&gt;Unfortunately, a very popular approach to teaching mathematics at the grade school level is to de-emphasize the concepts and focus more on the rote, procedural aspects of the subject.  I don't think many people really like this type of education but it's one of the simpler approaches to teaching. On the positive side, by focusing on the procedural you have a systematic way with which you can judge a students' progress.   In some sense this is a mode of teaching that is dictated largely by the desire for a convienient means of evaluation, rather than being dictated by holding the students' interest, or teaching them valuable things.  The hope is that by learning enough procedures by rote, the underlying concepts somehow sink-in, as if by magic.  I think this does work for many students but by-and-large it's unsatisfactory to almost all involved.  To teach mathematics in any other way with similar objective measures of success tends to be much more time and energy consuming for all involved, as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complement Sam Nead's very nice suggestions, I enjoyed reading Devaney's &quot;Introduction to Chaotic Dynamical Systems&quot;.  I don't remember when exactly I picked it up, might have been grade 12 or 1st year of university.  Parts of it were over my head at the time, but I was able to digest much of the contents, slowly.  In that regard I think it's useful to take a non-linear approach to learning.  It's useful to have as a goal to understand some high-level topic, and to have a mental roadmap for how you're going to get there. You don't want to be too aggressive on this, though.  Around grade 12, Devaney's text was very useful for me.  It's a book that can be used both for graduate courses and for upper-level undergraduate courses.  So you'll probably see it as several steps ahead of what you're used to.  But it's well-written and well-motivated.  Digest what you can, slow down and think about aspects that appear hard, maybe come back to them a few months later.  Attempt some of the more fun-looking problems at the end of sections.  Do some computer experiments.  That kind of thing. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-01T18:10:36.317" CommunityOwnedDate="2010-11-01T18:10:36.317" CommentCount="1" />
  <row Id="8541" PostTypeId="2" ParentId="8539" CreationDate="2010-11-01T18:11:31.900" Score="2" ViewCount="0" Body="&lt;p&gt;$$S = 1 + x + x^2 + x^3 + \cdots, \quad |x|&amp;lt;1. \quad (1)$$&#xA;$$xS = x + x^2 + x^3 + \cdots \quad (2)$$ &#xA;$$(1) - (2) \Rightarrow S-xS = 1.$$&#xA;$$S = \frac{1}{1-x}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-01T18:11:31.900" CommentCount="1" />
  <row Id="8542" PostTypeId="2" ParentId="8539" CreationDate="2010-11-01T18:13:54.353" Score="3" ViewCount="0" Body="&lt;p&gt;$$ (\frac{2}{3})^{2n-1} = \frac{3}{2}(\frac{4}{9})^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a geometric progression with common ratio $r$, and sum from 0 to infinity the sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s = \sum_{i=0}^\infty ar^i = \frac{a}{1-r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $|r|&amp;lt;1$ and where $a$ is the &quot;common factor&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-01T18:13:54.353" />
  <row Id="8543" PostTypeId="2" ParentId="8539" CreationDate="2010-11-01T18:30:07.780" Score="1" ViewCount="0" Body="&lt;p&gt;$\displaystyle\rm\ \ \sum_{n\ge 0}\ r^{\:a+b\:n}\ =\ \frac{r^a}{1-r^b}\ \ if\ \ |r| &amp;lt; 1\ \ $ e.g. see the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_series&quot; rel=&quot;nofollow&quot;&gt;geometric series.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-01T18:30:07.780" CommentCount="1" />
  <row Id="8544" PostTypeId="1" CreationDate="2010-11-01T19:17:44.723" Score="2" ViewCount="87" Body="&lt;p&gt;There is a 5 by 5 matrix of points on a plane. How many triangles can be formed using points on this matrix?&lt;/p&gt;&#xA;" OwnerUserId="3004" LastActivityDate="2010-11-01T23:22:10.033" Title="There is a 5 by 5 matrix of points on a plane. How many triangles can be formed using points on this matrix?" Tags="&lt;combinatorics&gt;&lt;puzzle&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8545" PostTypeId="2" ParentId="8544" CreationDate="2010-11-01T19:35:23.473" Score="-1" ViewCount="0" Body="&lt;p&gt;by total do u mean the total number of ways in which any 3 points can be connected?&lt;/p&gt;&#xA;" OwnerUserId="2969" LastActivityDate="2010-11-01T19:35:23.473" CommentCount="1" />
  <row Id="8546" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T19:39:50.333" Score="5" ViewCount="0" Body="&lt;p&gt;I have spent a TON of time in the math library at Rutgers just thumbing through all the books, and I never really found a book about the &quot;math mindset&quot; that was half decent.  You could do a lot worse than to read stack exchange every day and become a regular on this forum; everyone here is really helpful and you shouldn't be afraid of posting a &quot;silly&quot; question out of fear of being embarrassed; the only thing that upsets people here is when someone doesn't put a sufficient amount of thought/effort into the problem before asking, or else when the question being asked is clearly off topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that talking with mathematicians about specific problems you are working on will give you a far deeper insight into how mathematicians think than reading some low level book for a popular audience will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing in my opinion is not the book you finally choose, but rather the attitude that you should always try to solve EVERY SINGLE problem in whatever math book you read.  You should spend more time solving the problems than reading the sections.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you pick up &quot;A survey of modern algebra&quot; by Birkhoff and Mclane.  The difficulty level is pretty low, and the problems are numerous and instructive.  One reason I so highly recommend this particular book for you is that &quot;modern algebra&quot; is not only fundamentally important for contemporary mathematics, it also is not taught in high schools generally, not to mention that many of the results are beautiful.  Another great thing about this specific book is that the author's are famous mathematicians, and you can figure out how they think about math just by seeing the organization of how the material is presented and how the results are derived.  The first chapter in this book deals with number theory, but actually contains a very nice introduction to writing logical and rigorous proofs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you should just read through books on specific subjects, and pick the subject you will study based on your current interests and level of knowledge.  Some subjects to start with are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;set theory, metric spaces, complex analysis, group theory, number theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you might want to try is just to go ahead and check out some books by the great masters.  For example, there is a great set theory book by Frankel (who has the axioms of set theory named after him), and a great book on number theory by Sierpinski; and I recommend both of these not only because they are really well written and basically at the high school level, but because this will give you an insight into how the great masters thought about math.  Finally, if you are feeling confident go ahead and check out Disquisitions Arithmetic by Gauss, just to try your hand at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also mention a compilation of writings for a popular audience by Henri Poincare called &quot;The Value of Science&quot;.  A large portion of this book deals with physics, but there are two particular passages you should pay attention to for understanding the &quot;math mindset&quot;: the first is the discussion of induction, and the second is a fairly famous section where Poincare discusses the internal psychological process he experienced when solving a highly technical problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of Luck!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-01T22:10:37.913" LastActivityDate="2010-11-01T22:10:37.913" CommunityOwnedDate="2010-11-01T19:39:50.333" />
  <row Id="8547" PostTypeId="1" CreationDate="2010-11-01T19:49:40.877" Score="1" ViewCount="82" Body="&lt;p&gt;I've recently come upon the following (seemingly) simple observation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt;: A positive integer $k = d d^{\prime} &gt; 1$ has the opposite parity of $\text{gcd}(d+1,d^{\prime}+1)$ for any pair of distinct divisors $d$ and $d^{\prime}$ of $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are substantially simple proofs (see comments), and one proof follows from the formula: &#xA;\begin{eqnarray}&#xA;\text{gcd}(a,b) = a + b - ab + 2 \sum_{k=1}^{a-1} \lfloor \tfrac{kb}{a} \rfloor.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary&lt;/strong&gt;: Given a multiplicative arithmetic function $f \colon \mathbb{N} \to \mathbb{N}$ and a pair of coprime integers $a$ and $b$, &#xA;\begin{eqnarray}&#xA;f(ab) + 1 \equiv \text{gcd}(f(a)+1,f(b)+1) \mod 2.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Do either of these simple results have any neat use(s) or interesting application(s)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(NB: The purpose of this post is not so much about finding simple (or simpler) proofs, but rather more about determining possible applications.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-01T22:50:52.103" LastActivityDate="2010-11-01T22:50:52.103" Title="Parity of $k = d d^{\prime} &gt; 1$ vs $\text{gcd}(d+1,d^{\prime} +1)$ for distinct divisors $d$ and $d^{\prime}$." Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="8548" PostTypeId="2" ParentId="8547" CreationDate="2010-11-01T20:02:52.717" Score="2" ViewCount="0" Body="&lt;p&gt;I don't have an answer for the applications question (so feel free to vote me down, I guess?), but the proof is much simpler conceptually if you do it in cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $k = ab$, and if a is odd and b is even, then the product is even and the divisors swap parity when you add one, so they don't have a factor of two in common. If they're both even, then the product is even but $a+1$ and $b+1$ are odd, so the gcd is odd. If they're both odd, then the product is odd but $a+1$ and $b+1$ are both even so the gcd is even.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2010-11-01T20:20:49.713" LastActivityDate="2010-11-01T20:20:49.713" CommentCount="1" />
  <row Id="8550" PostTypeId="1" CreationDate="2010-11-01T20:08:32.717" Score="1" ViewCount="186" Body="&lt;p&gt;The definition of a Cumulative Distribution Function $(CDF)$ says that&#xA;$$P(X \le x) = F(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then my text book gives the following theorem without proof:&#xA;$$P(X \lt x) = F(x-)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says that the proof is easy, but with my rusted calculus skills, I have trouble to even intuitively understand this theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to sketch a proof, to improve my understanding:&#xA;$$F(x-) = \lim_{n \to \infty} F(x - \frac{1}{n}) = \lim_{n \to \infty} P(X &amp;lt; x - \frac{1}{n}) = P(X &amp;lt; x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume we can't say $P(X \le x - \frac{1}{n})$, because then $X(\omega) \ge x$, for some $\omega \in \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct reasoning?&lt;/p&gt;&#xA;" OwnerUserId="2566" LastEditorUserId="2566" LastEditorDisplayName="" LastEditDate="2010-11-01T20:16:21.020" LastActivityDate="2010-11-03T17:06:19.060" Title="Proof of $P(X &lt; x) = F(x−)$" Tags="&lt;calculus&gt;&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8551" PostTypeId="2" ParentId="8508" CreationDate="2010-11-01T20:14:03.857" Score="4" ViewCount="0" Body="&lt;p&gt;I asked a related question in MathOverflow a while back. Here is the &lt;a href=&quot;http://mathoverflow.net/questions/10567/number-of-uniform-rvs-needed-to-cross-a-threshold&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;. Let $N$ be the number of $U(0,1)$ random variables needed for the sum to cross $1$. It is in fact possible to derive the pmf of $N$ and compute $E(N)$ as David Bar Moshe has indicated. But there is a nifty derivation if we are only interested in $E(N)$ that involves conditioning on the very first random variable $U_1$ that we pick. I have sketched an outline below. Let me know if any part of the derivation isn't clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N(x)$ be expected number of $U(0,1)$ random variables needed for the sum to cross $x$ where $0 \leq x \leq 1$. Then, a recursion can be derived for $f(x) = E(N(x))$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \int_{0}^{1} E(N(x) \mid U_1 = y) dy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This integral naturally splits into two - that between $0$ and $x$ and that between $x$ and $1$. If $U_1 &amp;lt; x$, $E(N(x))$ is simply $1 + f(x-U_1)$. If $U_1 &gt; x$, $E(N(x))$ is just $1$. This will give us an integral equation for $f(x)$ that can be solved to get the solution (with suitable initial condition) as $f(x) = e^x$. &lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-11-01T20:14:03.857" />
  <row Id="8552" PostTypeId="1" AcceptedAnswerId="8572" CreationDate="2010-11-01T20:39:30.730" Score="4" ViewCount="121" Body="&lt;p&gt;I am interested in a system of differential equations that is non-linear, but it doesn't seem to be too crazy. I'm not very good at non-linear stuff, so I thought I'd throw it out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual equations I'm looking at have several parameters that'd I'd like to tweak eventually.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;q' = k - m / r&#xA;r' = i - n r - j q&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;i, j, k, m and n are all real-valued constants. I'm guessing that this system would be cyclical in nature, but I'm not sure if it has any explicit solution, so I have produced a version of it with the constants removed to see if that can be solved:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;q' = 1 - 1 / r&#xA;r' = 1 - r - q&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Anyone know if either of these are solvable and what kind of techniques would be needed to solve them if so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation is based on a polar coordinate system where Q (or theta) is the angle and r is radius, and I've made a number of simplifications to make it somewhat tractable.&lt;/p&gt;&#xA;" OwnerUserId="2247" LastActivityDate="2010-11-01T22:37:15.817" Title="Explicit solutions to this nonlinear system of two differential equations" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8553" PostTypeId="2" ParentId="8550" CreationDate="2010-11-01T20:49:05.263" Score="3" ViewCount="0" Body="&lt;p&gt;The intuition is that the CDF is just &quot;adding probabilities up&quot;, a little bit at a time.  Remember, the integral is a &lt;b&gt;s&lt;/b&gt;um; that's why $\int$ is an elongated 's'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think of your expression $F(x-\frac 1 n)$ as &quot;adding a little bit more&quot;; as $n$ gets larger, you added in the difference&#xA;$$&#xA;\int_{-\infty}^{x-\frac 1 {n+1}} f(x) - \int_{-\infty}^{x-\frac 1 n} f(x) = \int_{x-\frac 1 n}^{x-\frac 1 {n+1}} f(x).&#xA;$$&#xA;Since $f(x) \geq 0$ for all $x$, this difference is also $\geq 0$. The limit is then the sum of the area under the curve $f(x)$ between $x-1/n$ and $x-1/(n+1)$:&#xA;$$&#xA;\sum_{n=1}^\infty \int_{x-\frac 1 n}^{x-\frac 1 {n+1}} f(x).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $X$ is a continuous random variable, this sum &lt;i&gt;will&lt;/i&gt; converge to $F(x)$. However, if $X$ has &quot;discrete pieces&quot; things can go wrong and the CDF can jump: see the third picture in &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aDiscrete_probability_distribution_illustration.png&quot; rel=&quot;nofollow&quot;&gt;this diagram&lt;/a&gt;.  In general, however, the sum will converge to $\lim_{b\to x^-} F(b)$.&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-01T20:49:05.263" CommentCount="7" />
  <row Id="8554" PostTypeId="2" ParentId="8552" CreationDate="2010-11-01T20:56:37.987" Score="4" ViewCount="0" Body="&lt;p&gt;I don't think there's very much hope for an explicit solution. However, you can easily draw a phase portrait and see qualitatively how the solutions behave (in your case, the solutions in the upper half plane $r&gt;0$ spiral towards the stable equilibrium $(q,r)=(0,1)$. &lt;a href=&quot;http://math.rice.edu/~dfield/dfpp.html&quot; rel=&quot;nofollow&quot;&gt;Here's a Java applet by John Polking&lt;/a&gt; for drawing phase portraits (including nullclines); it can also find and analyze equilibria.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-01T20:56:37.987" CommentCount="6" />
  <row Id="8555" PostTypeId="2" ParentId="8550" CreationDate="2010-11-01T20:57:59.013" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT: I rewrite my answer, since it appears it was not clear enough. I will try to go in more details even though I think this is not really interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $F(x) := P(X \leq x)$ and we want to prove $F(x-) = P(X &amp;lt; x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a sequence $(a_n)$ such that $\lim_{n \rightarrow \infty} a_n = x$, $a_n &amp;lt; x$ for all $n$. For the sake of simplicity, let's admit $a_n$ is increasing ($a_{n+1} \geq a_n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's consider the sequence $(b_n)$ given by $b_n := F(a_n)$. By definition of $F$, we have $b_n = P(X \leq a_n)$. Thus $(b_n)$ is a sequence in [0,1]. Since $b_{n+1} = P(X \leq a_{n+1}) \geq P(X \leq a_n) = b_n$, we have that the sequence $b_n$ is also increasing. Hence, it converges. Now let's calculate the limit (let's call it $b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $P(X \leq a_n) \leq P(X &amp;lt; x)$, so $b_n \leq  P(X &amp;lt; x)$ for all $n$. Thus this should remain true for the limit $b$. You might argue that you are not sure why this should remain true for the limit and why we could not have $b = P(X \leq x)$. Here comes the $\varepsilon$ argument. If  $P(X \leq x) = P(X &amp;lt; x)$ then there is no problem, so suppose $P(X \leq x) \ne P(X &amp;lt; x)$. In that case, let $\varepsilon := \frac{1}{2}\left(P(X \leq x) - P(X &amp;lt; x)\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of the limit, you should find infinite members of the sequence in $[b- \varepsilon, b + \varepsilon]$. But this can't be, because all the $b_n$ are smaller than $P(X &amp;lt; x)$ and thus at distance at least 2 $\varepsilon$ from the limit. This proves that if $P(X \leq x) \ne P(X &amp;lt; x)$, then $b \ne P(X \leq x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so we know $b \leq P(X &amp;lt; x)$. We still need to prove there is an equality here. I will not go into further technical details here, the intuition should tell you that since $a_n \rightarrow x$, you can come as near to $P(X &amp;lt; x)$ as you want with the sequence $b_n$ (consider $\varepsilon = \frac{P(X &amp;lt; x) -b}{2}$ and see what this implies bla-bla-bla).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the beginning I said we could consider $a_n$ to be increasing. Suppose it is not. Then you can construct a new sequence $c_n$, such that $\lim c_n = x$, $c_n \leq a_n$ for all $n$ and $c_n$ is increasing. Then you compare the sequence $d_n := F(c_n)$ with the sequence $b_n$ and bla-bla-bla. This really becomes technical, but it works. So $F(x-) = P(X &amp;lt; x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the way, sorry if I can't write my sequences properly, I can't get { } in math mode, even with 1 backslash, 2 backslaches or 3 backslaches before the {. This really annoys me.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-03T17:06:19.060" LastActivityDate="2010-11-03T17:06:19.060" CommentCount="2" />
  <row Id="8556" PostTypeId="1" CreationDate="2010-11-01T21:07:39.770" Score="1" ViewCount="125" Body="&lt;p&gt;If I have a stochastic matrix X- the sum of each row is equal to 1 and all elements are non-negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this property, how can I show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x'X=x', x&gt;=0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has a non-zero solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this has something to do with proving a feasible dual, but not may be wrong..&lt;/p&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2010-11-03T01:32:29.707" Title="duality with a stochastic matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="8557" PostTypeId="1" CreationDate="2010-11-01T21:18:19.470" Score="2" ViewCount="207" Body="&lt;p&gt;$$\int_1^{10}x^xdx$$&#xA;I figured it with a relative error of 1% and I have a response $$≈0.3*10^{10}$$ &#xA;But I don´t know how to accurately calculate it ...&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="2779" LastEditorDisplayName="" LastEditDate="2010-11-09T07:35:01.870" LastActivityDate="2010-11-09T07:35:01.870" Title="Calculate the integral $\int_1^{10}x^xdx$" Tags="&lt;analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="8558" PostTypeId="1" AcceptedAnswerId="8560" CreationDate="2010-11-01T21:19:41.560" Score="1" ViewCount="83" Body="&lt;p&gt;I'm trying to use inverse cosine, in the cosine rule to find an angle of a triange when you know 3 sides. I know this formula and have it written down. However I've left my calculator and I'm having to use Google:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.co.uk/webhp?hl=en#sclient=psy&amp;amp;hl=en&amp;amp;site=webhp&amp;amp;source=hp&amp;amp;q=arccos((5.9%5E2+%2B+4.1%5E2+-+3.2%5E2)+%2F+(2+&quot; rel=&quot;nofollow&quot;&gt;http://www.google.co.uk/webhp?hl=en#sclient=psy&amp;amp;hl=en&amp;amp;site=webhp&amp;amp;source=hp&amp;amp;q=arccos((5.9%5E2+%2B+4.1%5E2+-+3.2%5E2)+%2F+(2+&lt;/a&gt;&lt;em&gt;+5.9+&lt;/em&gt;+4.1))++in+degrees&amp;amp;aq=f&amp;amp;aqi=&amp;amp;aql=&amp;amp;oq=&amp;amp;gs_rfai=&amp;amp;pbx=1&amp;amp;qscrl=1&amp;amp;fp=3dd9e71350957573&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it doesn't return the correct answer which makes think theres a difference between arccos and inverse cos.&lt;/p&gt;&#xA;" OwnerUserId="2075" LastActivityDate="2010-11-01T21:37:03.323" Title="Difference between arccos and Cos^-1" Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="8559" PostTypeId="1" CreationDate="2010-11-01T21:23:36.407" Score="0" ViewCount="64" Body="&lt;p&gt;If i have $Q = \left\{ y \in \mathbb{R}^n \ \vert \ By=c, y \geq 0 \right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y_{j} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove there exists some $p$ where $p'A \geq 0'$, $p'b = 0$ and $(p'A)_{j} &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At face value this looks like a dual problem, but I'm confused how to proceed without an objective function. Can I arbitrarily assign one, or is there another way to proceed?&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-02T04:32:59.017" LastActivityDate="2010-11-02T04:32:59.017" Title="matrix with null variable" Tags="&lt;linear-programming&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="8560" PostTypeId="2" ParentId="8558" CreationDate="2010-11-01T21:30:15.623" Score="2" ViewCount="0" Body="&lt;p&gt;Wolfram Alpha gives about 31.21 degrees.  My calculator gives 31.20576  When I first pasted it into Google, it didn't bring up the calculator, then when I put it into Alpha it lost the s off arccos.  Also between the numbers in the denominator I get spaces instead of asterisks, so maybe it is not multiplying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people distinguish between the inverse cosine function (which has the range restricted so there is only one value) and arccos not having the range restricted so it is multivalued, but that doesn't seem to be your problem.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-01T21:37:03.323" LastActivityDate="2010-11-01T21:37:03.323" CommentCount="4" />
  <row Id="8561" PostTypeId="1" AcceptedAnswerId="8708" CreationDate="2010-11-01T21:42:41.777" Score="5" ViewCount="188" Body="&lt;p&gt;This is the first exercise from Sierpinski's Elementary Theory of Numbers. He gives a proof using induction and I was wondering if this approach was correct as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a!b!|(a+b)! \iff \exists c \in \mathbb{N} \text{ such that } (a+b)! = c(a!b!)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming without loss of generality that $a \leq b$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a!b! = \displaystyle\prod\limits_{n=1}^a n^2 \displaystyle\prod\limits_{n=a+1}^b n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we define the set S:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = \{n \in \mathbb{N} :n &amp;lt; a^2 \wedge \not \exists m \in \mathbb{N}\text{ such that }m^2=n)  \} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c = \displaystyle\prod\limits_{n \in D}n \displaystyle\prod\limits_{n=b+1}^{a+b}n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $(a+b)! = c(a!b!)$&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T02:48:48.593" LastActivityDate="2010-11-03T01:24:07.513" Title="Proving that if a and b are natural numbers, then a!b! | (a+b)!" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="1" />
  <row Id="8564" PostTypeId="1" AcceptedAnswerId="8571" CreationDate="2010-11-01T21:51:11.843" Score="4" ViewCount="161" Body="&lt;p&gt;Okay, so I'm trying to find $ \int \frac1{\cos x}\mathrm{d}x$ using the substitution $t = \tan\left(\frac{x}{2}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sub in the trig identity for $\sec$ as $\frac{1+t^2}{1-t^2}$ and then rearrange and substitute $\frac{\mathrm{d}t}{\mathrm{d}x} = \frac12 \left(1+ \tan^2\left(\frac{x}{2}\right)\right)$ so I am left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac2{1-t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then used partial fractions to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac2{1-t^2} = \frac1{1+t} + \frac1{1-t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore integrating I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \frac1{\cos x}dx = \ln(t+1) - \ln(t-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But subbing in $t = \tan\left(\frac{x}{2}\right)$ doesn't seem to get me anywhere close to the solution that I want to find, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \int \frac1{\cos x}\mathrm{d}x = \ln(\sec x + \tan x) + C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help on this would be greatly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3011" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-02T07:09:36.390" LastActivityDate="2010-11-02T13:05:47.003" Title="Integral of $\frac1{\cos x}$ using t substitution" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8565" PostTypeId="2" ParentId="8556" CreationDate="2010-11-01T22:00:38.217" Score="2" ViewCount="0" Body="&lt;p&gt;If your matrix $X$ is $n\times n$, then $X$ is the transition matrix of a Markov chain with state space $\lbrace  1 , 2 , \dots , n \rbrace$. The vector $x$ that you are looking for is a stationary measure for the Markov chain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of such a non-zero $x$ can be proven using probability theory, but a purely linear approach falls under &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem&quot; rel=&quot;nofollow&quot;&gt;Perron-Frobenius theory&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the averages ${1\over n}\sum_{k=1}^n X^k$ of the powers of $X$ converge to a matrix $M$ (this statement needs proof!). Then, it is easy to see that any row of $M$ can serve as $x'$ and solves $x'X=x'$. Note that the row sums of $M$ all equal 1, so $x\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you suggest, maybe there is also an approach via linear programming. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-01T22:00:38.217" />
  <row Id="8566" PostTypeId="2" ParentId="8557" CreationDate="2010-11-01T22:03:12.223" Score="5" ViewCount="0" Body="&lt;p&gt;One way to solve this problem is to expand in terms of a power series as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^x = e^{\ln(x) x} = \sum_{n=0}^\infty \frac{(\ln(x) x)^n}{n!} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating on both sides from $1$ to $10$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_1^{10} x^x dx = \sum_{n=0}^\infty \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $x^n ln(x)^n$ is monotonically increasing, so the integral inside the sum is bounded by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx &amp;lt; \frac{9}{n!} \cdot \ln(10)^n \cdot 10^n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want the sum to whatever precision, you can use that expression to determine how small the remainder term is. Next, how on earth do you compute that integral? Well it looks ugly but there is a recursive formula. Let $k, l$ be positive integers. Then using integration by parts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \int_1^{10} x^k \ln(x)^l dx = \frac{x^{k+1} ln(x)^l}{k+1} \Bigr \vert_1^{10} - \frac{l}{k+1} \int_1^{10} x^k \ln(x)^{l-1} dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you chose the limits of integration to be from $1$ to $10$ you might have to take quite a few terms in this series in order to compute an accurate value, but that is to be expected as $x^x$ blows up so quickly.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-01T22:03:12.223" CommentCount="1" />
  <row Id="8567" PostTypeId="2" ParentId="8564" CreationDate="2010-11-01T22:18:49.200" Score="2" ViewCount="0" Body="&lt;p&gt;Try writing $\sec x + \tan x$ in terms of $t$ and see if you can reorganize the two answers to agree.  You may need to tweak the value of $C$ to do this.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-01T22:18:49.200" CommentCount="1" />
  <row Id="8568" PostTypeId="1" CreationDate="2010-11-01T22:29:25.430" Score="0" ViewCount="63" Body="&lt;p&gt;Let $(X,x_0),(Y,y_0)$ be pointed spaces. There is a canonical continuous bijection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(X,x_0) \vee (Y,y_0) \to X \times \{y_0\} \cup \{x_0\} \times Y$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which sends $x$ to $(x,y_0)$ and $y$ to $(x_0,y)$. Here we regard the space on the right as a subspace of the product $X \times Y$. Actually this map is also open and thus an homeomorphism. But the proof I know uses a distinction of cases whether the open $U$ in the wedge contains the base point or not. The same problem when you want to show that the space satisfies the universal property of a coproduct. Is there a shorter proof which avoids this distinction?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-01T22:29:25.430" Title="wedge sum embeds into product" Tags="&lt;topology&gt;" CommentCount="3" />
  <row Id="8569" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T22:29:30.917" Score="3" ViewCount="0" Body="&lt;p&gt;I think certain parts of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691118809&quot; rel=&quot;nofollow&quot;&gt;The Princeton Companion to Mathematics&lt;/a&gt; might be just what you need.  While many of the topics will go over your head at the moment, there is a very lucid introduction to the history and basic points of view of modern mathematics in Parts I and II, as well as a lot of interesting historical overviews and essays in Parts VI and VIII.  And of course much of the book is filled with fascinating mathematics.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-01T22:29:30.917" CommunityOwnedDate="2010-11-01T22:29:30.917" />
  <row Id="8570" PostTypeId="2" ParentId="8508" CreationDate="2010-11-01T22:32:41.747" Score="3" ViewCount="0" Body="&lt;p&gt;This problem is well-known. For an answer, see, for example, the first part of Section 2 in &lt;a href=&quot;http://myweb.facstaff.wwu.edu/curgus/Papers/27Unexpected.pdf&quot; rel=&quot;nofollow&quot;&gt;http://myweb.facstaff.wwu.edu/curgus/Papers/27Unexpected.pdf&lt;/a&gt;, or Equations (7)-(10) in &lt;a href=&quot;http://mathworld.wolfram.com/UniformSumDistribution.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/UniformSumDistribution.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-01T22:42:00.193" LastActivityDate="2010-11-01T22:42:00.193" />
  <row Id="8571" PostTypeId="2" ParentId="8564" CreationDate="2010-11-01T22:32:51.420" Score="2" ViewCount="0" Body="&lt;p&gt;The problem equivocates to proving that $\frac{t+1}{t-1} = secx + tanx$ since $ln(t+1) - ln(t-1) = ln(\frac{t+1}{t-1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that $tan\frac{x}{2} = \frac{1-cosx}{sinx}$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{t+1}{t-1} = \frac{1+sinx-cosx}{1-sinx-cosx} = \frac{secx+tanx-1}{secx-tanx-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, $(secx+tanx)(1+secx-tanx) = secx + tanx +(sec^2x-tan^2x) = secx+tanx+1$, so that the result follows.&lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorUserId="702" LastEditorDisplayName="" LastEditDate="2010-11-02T13:05:47.003" LastActivityDate="2010-11-02T13:05:47.003" CommentCount="1" />
  <row Id="8572" PostTypeId="2" ParentId="8552" CreationDate="2010-11-01T22:37:15.817" Score="4" ViewCount="0" Body="&lt;p&gt;Taking that second question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r' = i - nr - jq$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and differentiating gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r'' = -nr' - jq' = -nr' - j(k-\frac{m}{r})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in other words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r'' + ar' + \frac{b}{r} = c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a much simpler differential equation only one variable. I think that you could probably solve this with power series or clever guessing, but it needs to be worked out.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-01T22:37:15.817" CommentCount="6" />
  <row Id="8573" PostTypeId="2" ParentId="8524" CreationDate="2010-11-01T22:57:47.963" Score="3" ViewCount="0" Body="&lt;p&gt;Here are some further classics that you may find inspiring and enlightening: Polya: How to solve it; Rademacher and Toeplitz: The enjoyment of mathematics; Kac and Ulam: Mathematics and Logic. &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-01T22:57:47.963" CommunityOwnedDate="2010-11-01T22:57:47.963" />
  <row Id="8574" PostTypeId="2" ParentId="8235" CreationDate="2010-11-01T23:01:49.427" Score="5" ViewCount="0" Body="&lt;p&gt;Take $\mathcal{E}=(e_1,\ldots,e_d) \in A^d$ that is a basis of $\mathbb{Q}^d$.&#xA;We choose the norm $|| \cdot ||$ to be $|| \cdot ||_{\infty,\mathcal{E}}$ (sup-norm of the coefficients in the basis $\mathcal{E}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}e_1 + \ldots + \mathbb{Z}e_d$ contains $N \mathbb{Z}^d$ for some positive integer $N$.&#xA;Let $C_0$ be the cone spanned by the $e_i$'s.&#xA;Take $C$ to be the cone spanned by $f_1,\ldots,f_n$ where $f_i = e_i +  \epsilon \sum_{j \neq i} e_j$, choosing $\epsilon&gt;0$ small enough so that $(f_1,\ldots,f_n)$ is free.&#xA;There exists $\eta&gt;0$ (depending on $\epsilon$) such that for every $x \in C$, the coefficients of $x$ in the basis $\mathcal{E}$ are all $\geq \eta ||x||$, so $B(x,\eta ||x||) \subset C_0$ (draw a picture!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now choose $s_1,\ldots,s_k \in S$ exhausting $\mathbb{Z}^d/N\mathbb{Z}^d$.&#xA;Let $M=\max_i ||s_i||$.&#xA;Take $x \in C \cap \mathbb{Z}^d$, with $||x||&gt;M/\eta$.&#xA;There exists $i$ such that $x-s_i \in N \mathbb{Z}^d$, and $x-s_i \in B(x,M) \subset C_0$, so the coefficients of $x-s_i$ in $\mathcal{E}$ are non-negative integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, $x=s_i+(x-s_i)$ is in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to add rigor to the statement with the &quot;picture&quot;: if $x = \sum_i x_i f_i \in C$, and if $x_{i_0} = \max_i x_i$, then $x = \sum_i y_i e_i$ with $y_{i_0} = \max_i y_i \leq (1+(d-1)\epsilon ) x_{i_0}$ and $y_i \geq \epsilon x_{i_0}$ for each $i$, so that $y_i \geq \epsilon / (1+(d-1)\epsilon) ||x||$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-04T11:07:01.237" LastActivityDate="2010-11-04T11:07:01.237" CommentCount="2" />
  <row Id="8575" PostTypeId="1" AcceptedAnswerId="8584" CreationDate="2010-11-01T23:03:41.073" Score="2" ViewCount="59" Body="&lt;p&gt;So suppose I want a path from 0 to $c&gt;0$ on the real line, and I am going to use the function $S(t)$ to get there in (discrete) time $T$.  That is, my position at time 0 is 0, my position at time $T$ is $c$, and my position $P_t$ changes in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_t = (1-\gamma) P_{t-1} + \gamma S(t) &#xA;$$&#xA;where $\gamma&amp;lt;1$ limits fast I can move.  Which can be written&#xA;$$&#xA;\frac{P_t - P_{t-1}}{\gamma} = - P_{t-1} +S(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, I have a drift towards zero of $-P$, and $S$ has to drive me away from zero, towards $c$.  So $S(t)$ is the &quot;size&quot; of my push towards $c$ at date $t$.  The &lt;i&gt;cost&lt;/i&gt; of $S$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C(T,S(t)) = \sum_1^{T} (S(t))^2&#xA;$$&#xA;So I am penalized for large movements in my position, and &lt;i&gt;big&lt;/i&gt; pushes towards $c$.  I would like to calculate &#xA;$$&#xA;\min_{T,S(\cdot)} C(T,S(t)) &#xA;$$&#xA;subject to $P_0 =0$, $P_T = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if I am free to choose the number of steps  it takes to get there, and the path can be whatever I want, what is the cost minimizing route from 0 to $c$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, you could set $S(1) =  c / \gamma$, so I arrive immediately, at cost $( c/\gamma )^2$.  So that is an upper bound on the cost.  I could do it in two steps;  If I step $x$ the first period,  I will need to step &#xA;$$\frac{(c-(1-\gamma) \gamma x)}{\gamma}$$&#xA;the second period, for a total cost of of&#xA;$$&#xA;x^2 + (\frac{(c-(1-\gamma) \gamma x)}{\gamma})^2,&#xA;$$&#xA;Which is minimized for $x=\frac{c (1-\gamma)}{\gamma (2-2 \gamma+\gamma^2 )}$, which therefore costs &#xA;$$&#xA;\frac{c^2 (1-\gamma)^2}{\gamma^2 (2-2 \gamma+\gamma^2)^2}+\frac{(c+\frac{c (1-\gamma) (-1+\gamma)}{2-2 \gamma+\gamma^2})^2}{\gamma^2}&#xA;$$&#xA;which is a lower cost than the one-step option, since $\gamma&amp;lt;1$.  It is not clear to me how to extend this analysis to large $T$ - is there a simple recursive form for the cost minimizing $k$-step path, and the problem reduces to choosing the best $k$?&lt;/p&gt;&#xA;" OwnerUserId="2464" LastActivityDate="2010-11-02T01:38:44.770" Title="Minimizing the cost of a path in a dynamic system" Tags="&lt;calculus&gt;&lt;differential-equations&gt;&lt;optimization&gt;&lt;dynamical-systems&gt;" AnswerCount="1" />
  <row Id="8576" PostTypeId="1" AcceptedAnswerId="8644" CreationDate="2010-11-01T23:19:35.157" Score="2" ViewCount="172" Body="&lt;p&gt;Let &lt;strong&gt;n&lt;/strong&gt; be a natural number &lt;strong&gt;k&lt;/strong&gt; digits. Show that the quantity &lt;strong&gt;Q&lt;/strong&gt; of digits required to write the natural numbers from &lt;strong&gt;1&lt;/strong&gt; to &lt;strong&gt;n&lt;/strong&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q = k(n+1) - 111 ... 11&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(111 ... 11 contains k digits equal to 1.)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-02T10:02:47.987" LastActivityDate="2010-11-02T13:53:06.450" Title="Number of digits from 1 to n" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8577" PostTypeId="2" ParentId="8576" CreationDate="2010-11-01T23:21:42.813" Score="0" ViewCount="0" Body="&lt;p&gt;Start with the one digit numbers. How many digits are needed for them.  Do you include zero?  How many digits to go from 10 through 99?  And so on...&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-01T23:21:42.813" CommentCount="4" />
  <row Id="8578" PostTypeId="2" ParentId="8544" CreationDate="2010-11-01T23:22:10.033" Score="4" ViewCount="0" Body="&lt;p&gt;This is the obvious approach (to me, at least), and is mostly a matter of bookkeeping.  To begin with, there are ${{5^2} \choose 3}$ sets of three points.  However, some of these are colinear, which we don't want to count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $2 \times 5 \times {5 \choose 3}$ horizontal or vertical sets of three colinear points.  E.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; .xxx.   .....&#xA; .....   ...x.&#xA; .....   ...x.&#xA; .....   .....&#xA; .....   ...x.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are $3+3+3+3$ sets of three colinear points with rise/run $\in {\pm 2, \pm 1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; x....   ..x..&#xA; ..x..   .....&#xA; ....x   ...x.&#xA; .....   .....&#xA; .....   ....x&#xA;&#xA; ...x.   .....&#xA; .....   ....x&#xA; ..x..   ..x..&#xA; .....   x....&#xA; .x...   .....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are $2 \times {5 \choose 3}$ sets of three colinear points with rise/run $\in {\pm 1}$ along the main diagonal or main anti-diagonal.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; x....   .....&#xA; .x...   ...x.&#xA; ..x..   ..x..&#xA; .....   .....&#xA; .....   x....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are $4 \times {4 \choose 3}$ sets of three colinear points along the &quot;shunted&quot; main diagonal and main anti-diagonal (is there a better name for these diagonals?).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; .x...   ...x.   .....   .....&#xA; .....   ..x..   x....   .....&#xA; ...x.   .....   .x...   ...x.&#xA; ....x   x....   ..x..   ..x..&#xA; .....   .....   .....   .x...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Finally, there are these $4$ remaining:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ..x..   ..x..   .....   .....&#xA; .x...   ...x.   .....   .....&#xA; x....   ....x   x....   ....x&#xA; .....   .....   .x...   ...x.&#xA; .....   .....   ..x..   ..x..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So there are \[{{5^2} \choose 3}-2 \times 5 \times {5 \choose 3}-(3+3+3+3)-2{5 \choose 3}-4{4 \choose 3}-4=2148\] triangles that can be formed.  I also checked this answer with some code in &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Here, I have assumed that you want to count congruent triangles separately.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-01T23:22:10.033" CommentCount="2" />
  <row Id="8579" PostTypeId="1" AcceptedAnswerId="8580" CreationDate="2010-11-01T23:32:44.143" Score="0" ViewCount="78" Body="&lt;p&gt;I'm having a hard time. I want to generate an equation system with two equations from a second degree equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, get two first degree equations from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;y''=(-k/m)*abs(y')*y'-g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(k, m and g are constants)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, it is a homework, I just can't figure out how to do it.&lt;/p&gt;&#xA;" OwnerUserId="3012" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-01T23:37:29.923" LastActivityDate="2010-11-01T23:37:29.923" Title="How to transform a second degree differential equation into two first degree linear equations?" Tags="&lt;homework&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="8580" PostTypeId="2" ParentId="8579" CreationDate="2010-11-01T23:35:55.417" Score="2" ViewCount="0" Body="&lt;p&gt;Set v = y'. Then you get the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v' = (-k/m)*abs(v)*v-g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = y'&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-01T23:35:55.417" />
  <row Id="8581" PostTypeId="1" AcceptedAnswerId="8583" CreationDate="2010-11-01T23:48:56.573" Score="0" ViewCount="61" Body="&lt;p&gt;Forgive the total beginner question, it's been a few years since I needed to do this stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the expansion of (1 + x)^n?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: substituted expansion for simplification.&lt;/p&gt;&#xA;" OwnerUserId="3013" LastEditorUserId="3013" LastEditorDisplayName="" LastEditDate="2010-11-01T23:56:23.313" LastActivityDate="2010-11-02T00:04:03.243" Title="Simplify polynomial" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8582" PostTypeId="2" ParentId="8576" CreationDate="2010-11-01T23:57:31.990" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Chopping off the unit digit is a 10-to-1 map from k+1 digit onto k-digit naturals. Therefore, since there are 9 naturals with 1-digit, there are 90 with 2-digits, and 900 with 3-digits, etc.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-01T23:57:31.990" CommentCount="4" />
  <row Id="8583" PostTypeId="2" ParentId="8581" CreationDate="2010-11-02T00:04:03.243" Score="2" ViewCount="0" Body="&lt;p&gt;$$1 + n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 2n + n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 3n + 3n^2 + n^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 4n + 6n^2 + 4n^3 + n^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 5n + 10n^2 + 10n^3 + 5n^4 + n^5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + 6n + 15n^2 + 20n^3 + 15n^4 + 6n^5 + n^6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if you can spot a pattern!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-02T00:04:03.243" CommentCount="2" />
  <row Id="8584" PostTypeId="2" ParentId="8575" CreationDate="2010-11-02T01:38:44.770" Score="3" ViewCount="0" Body="&lt;p&gt;For fixed $T$, or $k$ which is the same thing, it turns out that you are just minimizing a quadratic function with a linear constraint, and a simple solution exists. However, there is no best $k$ -- you can make the cost arbitrarily small by taking longer and longer to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $s$ be the vector with components $s_i = S(i)$ for $i = 1, \ldots, k$. Then if you expand out the recursion in your first equation $P_k = \gamma s_k + (1-\gamma) P_{k-1}$, you will find that&#xA;$$\begin{align}&#xA;P_k &amp;amp;= \sum_{i=1}^k \gamma(1-\gamma)^{k-i} s_i + (1-\gamma)^k P_0 \\&#xA;&amp;amp;= b^Ts + 0&#xA;\end{align}$$ where $b$ is a vector whose components are $b_i = \gamma(1-\gamma)^{k-i}$. So your problem is that of minimizing $C(s) = \sum_{i=1}^k s_i^2 = s^Ts$, subject to the constraint $b^Ts = c$. Using Lagrange multipliers, you can get a straightforward solution. However, you will find that the cost decreases asymptotically with $k$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-02T01:38:44.770" CommentCount="1" />
  <row Id="8585" PostTypeId="2" ParentId="8501" CreationDate="2010-11-02T01:51:54.247" Score="1" ViewCount="0" Body="&lt;p&gt;I imagine that you could sit in on a lot of classes at Berkeley if you ask the professors. I doubt many would mind if you are very serious about mathematics. &lt;/p&gt;&#xA;" OwnerUserId="2919" LastActivityDate="2010-11-02T01:51:54.247" />
  <row Id="8586" PostTypeId="1" AcceptedAnswerId="8589" CreationDate="2010-11-02T01:52:38.033" Score="3" ViewCount="133" Body="&lt;p&gt;I would really appreciate if someone could just write down for me one example of a second order, or higher, differential equation for which it is known that there are no smooth solutions; and it's fine if it's a partial differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought it would be easy to either come up with an example or else find one by searching google/wiki/arxiv; but now I am not so sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a thing for non-smooth functions, and it just bothers me that I don't even know a single example of this type of differential equation.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-02T15:32:04.277" Title="What is an example of a second order differential equation for which it is known that there are no smooth solutions?" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8587" PostTypeId="2" ParentId="8561" CreationDate="2010-11-02T02:09:41.200" Score="0" ViewCount="0" Body="&lt;p&gt;Try $a=b=2$. Then $(a+b)!/a!/b! = 6$ and so $\prod_{n=b+1}^{a+b} n = 12$ cannot divide your $c$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-02T02:09:41.200" CommentCount="1" />
  <row Id="8588" PostTypeId="1" AcceptedAnswerId="8594" CreationDate="2010-11-02T02:10:48.250" Score="0" ViewCount="45" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following formula:&#xA;$\exists y \forall x ((x \geq y) \wedge \neg(y \geq x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This essentially boils down to:&#xA;$\exists y \forall x (x &gt; y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to check whether this applies to certain universes. But my result depends on whether I can assume that x can also equal y, because in that case the condition will never be true (how could a number be smaller than and at the same time equal to another number?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a hint?&lt;/p&gt;&#xA;" OwnerUserId="3016" LastEditorUserId="3016" LastEditorDisplayName="" LastEditDate="2010-11-02T02:19:28.807" LastActivityDate="2010-11-02T02:34:04.440" Title="Predicate logic: Two variables - same value allowed?" Tags="&lt;homework&gt;&lt;logic&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="8589" PostTypeId="2" ParentId="8586" CreationDate="2010-11-02T02:20:42.327" Score="8" ViewCount="0" Body="&lt;p&gt;There are already &lt;em&gt;first order&lt;/em&gt; linear partial differential equations with smooth coefficients which do not admit any smooth solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hans Lewy produced the first &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewy%27s_example#CITEREFMizohata1962&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt; of such a PDE. The equation reads &#xA;$$\left[-i\partial_x+\partial_y-2(x+iy)\partial_z\right]u(x,y,z)=f(x,y,z),\qquad(x,y,z)\in\mathbb R^{3}.$$&#xA;The equation does not have distribution solutions in any neighbourhood of any point in $\mathbb R^3$ provided $f=f(x,y,z)$ is not a real analytic function (it can be smooth though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original paper by Lewy is nice, clear and less than 4 pages long (freely available &lt;a href=&quot;http://www.math.ucla.edu/~yhhuang/266.09w/lewy.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).   &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-02T15:32:04.277" LastActivityDate="2010-11-02T15:32:04.277" CommentCount="1" />
  <row Id="8590" PostTypeId="2" ParentId="8586" CreationDate="2010-11-02T02:23:55.457" Score="4" ViewCount="0" Body="&lt;p&gt;Consider the partial differential equations associated to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nash_embedding_theorem&quot; rel=&quot;nofollow&quot;&gt;isometric embedding problem&lt;/a&gt; of the hyperbolic plane into Euclidean 3-space. In $C^1$ there exists a solution by Nash-Kuiper theorem, but it is known classically that there cannot be solutions that are twice or more continuously differentiable. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-02T02:23:55.457" CommentCount="2" />
  <row Id="8591" PostTypeId="1" AcceptedAnswerId="8593" CreationDate="2010-11-02T02:24:42.080" Score="1" ViewCount="135" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to understand why polynomial multiplication works the way it is defined.  For example, we know that $(x+1)^2 = x^2+2x+1$, but how can we prove that this actually works?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, how did we came up with the method used to multiply polynomials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2797" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-02T02:34:34.273" LastActivityDate="2010-11-02T04:50:58.950" Title="Proof that polynomial multiplication works" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8592" PostTypeId="2" ParentId="8591" CreationDate="2010-11-02T02:31:48.593" Score="2" ViewCount="0" Body="&lt;p&gt;If you have a set $X$ and two real valued functions $f$, $g:X\to\mathbb R$, then there is a standard way to multiply $f$ and $g$: the product is the function $h:X\to\mathbb R$ such that $$h(p)=f(p)g(p)$$ for all elements $p$ of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X=\mathbb R$ and if $f$ and $g$ are polynomials, then the function $h$ turns out also to be a polynomial, given by the usual formula. This completely determines the multiplication rule.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-02T02:31:48.593" CommentCount="7" />
  <row Id="8593" PostTypeId="2" ParentId="8591" CreationDate="2010-11-02T02:32:29.730" Score="8" ViewCount="0" Body="&lt;p&gt;We define the property that multiplication has to distribute over addition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is $A(B+C) = AB + AC$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With polynomials we try to do something similiar, analogous to the real numbers we all know so well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x+1)^2 = (x+1)(x+1) = (x+1)x + (x+1)1 = (x^2 + x) + (x + 1) = x^2 + 2x + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, by the distributive law of multiplication you can see how the defined operation makes sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You CAN come up with some binary operation that combines polynomials and doesnt distribute over addition, however this operation would not be called multiplication.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-02T03:04:59.687" LastActivityDate="2010-11-02T03:04:59.687" CommentCount="0" />
  <row Id="8594" PostTypeId="2" ParentId="8588" CreationDate="2010-11-02T02:34:04.440" Score="3" ViewCount="0" Body="&lt;p&gt;I'll put the answer here so the question can be marked as answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For every $x$&quot; means &lt;strong&gt;for every&lt;/strong&gt; $x$, period.  So, yes, there are no restrictions on $x$ not being equal to $y$; any such restrictions would have to be given as predicates (a clause $x\neq y$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll note (as I did in the comments) that whether the transformation from $(x\geq y) \wedge \neg(y\geq x)$ is equivalent to $x\gt y$ (and whether $\gt$ is areflexive) depends on the interpretation of $\geq$ and $\gt$ on the universe in question. The specification of the universe should not be only the set, but also the meaning of the relational and functional symbols like $\geq$. Only if there is some working convention can you simply assume that the meaning will be &quot;the usual one&quot;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-02T02:34:04.440" />
  <row Id="8595" PostTypeId="2" ParentId="6726" CreationDate="2010-11-02T03:06:26.943" Score="0" ViewCount="0" Body="&lt;p&gt;I'd recommend Foundations of Mechanics by Abraham and Marsden. It's a huge book, with a very rigorous treatment of mechanics. Altough Smale is not an author, he's heavily cited in the book (by the way, the book provides an &lt;em&gt;excellent&lt;/em&gt; bibliography).&lt;/p&gt;&#xA;" OwnerUserId="1034" LastActivityDate="2010-11-02T03:06:26.943" />
  <row Id="8596" PostTypeId="1" AcceptedAnswerId="8618" CreationDate="2010-11-02T03:24:21.730" Score="2" ViewCount="169" Body="&lt;p&gt;Is category theory worth learning for the sake of learning it? Can it be used in applied mathematics/probability? I am currently perusing &lt;em&gt;Categories for the Working Mathematician&lt;/em&gt; by Mac Lane. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-11-02T03:50:09.653" LastActivityDate="2010-11-02T08:09:14.857" Title="What are the prerequisites for learning category theory?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8597" PostTypeId="2" ParentId="8591" CreationDate="2010-11-02T03:24:36.707" Score="5" ViewCount="0" Body="&lt;p&gt;In addition to crasic's symbolic answer, one can also look at the geometric interpretation of multiplication of two quantities (the area of the rectangle formed with sides of the given length) and find that it is consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://1.bp.blogspot.com/_PnLYRqe0k9g/S0nZ-dZsl9I/AAAAAAAAARE/MtAEo96oqp0/s320/multiply-two-binomials.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x+a) * (x+b) = x^2 + ax + bx + ab$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As pointed out by Douglas, the geometric interpretation is only &lt;em&gt;defined&lt;/em&gt; at all when $x$, $a$, and $b$ are positive, so care must be taken to keep algebraic self-consistency when generalizing to the real numbers.  Geometric &quot;proofs&quot; should only be used as an &quot;intuition check&quot;, in any case.&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="51" LastEditorDisplayName="" LastEditDate="2010-11-02T04:50:58.950" LastActivityDate="2010-11-02T04:50:58.950" CommentCount="2" />
  <row Id="8598" PostTypeId="1" AcceptedAnswerId="8601" CreationDate="2010-11-02T03:50:03.023" Score="2" ViewCount="89" Body="&lt;p&gt;A group $G$ is is &lt;em&gt;abelian&lt;/em&gt; if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ab = ba $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all pairs of elements $a,b \in G$, i.e., if multiplication commutes.  Another way of stating this condition is that the &lt;em&gt;identity map&lt;/em&gt; $\mathrm{id}$ gives an isomorphism between left- and right- multiplication.  In other words, we can think of left-and right- multiplication as group actions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L: G \rightarrow \mathrm{Sym}(G);\ L(a)\cdot b := ab, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R: G \rightarrow \mathrm{Sym}(G);\ R(a)\cdot b := ba. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an abelian group we have $L(a)\cdot b = ab = ba = R(a)\cdot b$.  Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(\mathrm{id}(a))=R(a).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might consider a weaker condition: there is &lt;em&gt;some&lt;/em&gt; isomorphism $\varphi:G \rightarrow G$ (not necessarily the identity) such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(\varphi(a))=R(a).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $a \in G$.  (In other words, if we &lt;em&gt;really&lt;/em&gt; didn't like multiplying on the right, we could always multiply on the left by $\varphi(a)$ instead without losing anything.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard name for this weaker condition?  When can it be satisfied (if ever)?  For some standard examples (e.g., the quaternions) I can work out that no such isomorphism exists, but I don't know what to say about the general case.&lt;/p&gt;&#xA;" OwnerUserId="1612" LastEditorUserId="1612" LastEditorDisplayName="" LastEditDate="2010-11-02T04:05:07.693" LastActivityDate="2010-11-02T04:30:13.420" Title="Can nonabelian groups still have isomorphic left- and right- multiplication?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8599" PostTypeId="1" CreationDate="2010-11-02T03:54:42.983" Score="3" ViewCount="128" Body="&lt;p&gt;Let '$f$' be a continous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function ? &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-02T04:04:20.653" LastActivityDate="2010-11-02T08:03:42.223" Title="Let 'f' be a continous but nowhere differentiable function. Is 'f' convolved with mollifier, a smooth function ?" Tags="&lt;analysis&gt;&lt;convolution&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8600" PostTypeId="2" ParentId="8596" CreationDate="2010-11-02T03:57:12.573" Score="2" ViewCount="0" Body="&lt;p&gt;I do not know of any specific applications to probability (&quot;applied mathematics&quot; is a bit too wide a label, though I don't know any specific ones either). I view Category Theory as an interesting unifying language that allows you to speak about ideas that may seem to be isolated from each other in different subjects with the same notions, recognizing the connections between them. The results that I most often actually find useful have to do with adjuctions (specifically, I seem to find very useful the fact that left adjoints commute with colimits and right adjoints commute with limits...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that it is good to know at least the most general ideas, enough so that you can communicate and recognize those connections; of course, I'm a big fan of knowing stuff &quot;just for the sake of knowing it&quot; so I may not be the best advice giver in that respect. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-02T03:57:12.573" CommunityOwnedDate="2010-11-02T03:57:12.573" />
  <row Id="8601" PostTypeId="2" ParentId="8598" CreationDate="2010-11-02T03:58:32.643" Score="4" ViewCount="0" Body="&lt;p&gt;In particular, you would have $L(\varphi(a))\cdot e = R(a)\cdot e$, that is, $\varphi(a)=a$.  So $\varphi=\mathrm{id}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-02T03:58:32.643" CommentCount="1" />
  <row Id="8602" PostTypeId="2" ParentId="8535" CreationDate="2010-11-02T04:14:29.040" Score="2" ViewCount="0" Body="&lt;p&gt;If $S$ is any local ring with maximal ideal $m_S$, and $R = S/I$ is a quotient of $S$, then the maximal ideal $m_R$ of $R$ is $m_S/I$, and its $i$th power (for any&#xA;$i \geq 0$) is equal to $m_S^i + I /I$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $m_R$-adic completion of $R$ is equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;inverse limit of $R/m_R^i$ = inverse limit of $S/(I + m_S^n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The completion of $R$ as an $S$-module is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;inverse limit of $R\otimes_S (S/m_S^i)$ = inverse limit of $S/(I +m_s^i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which by the Artin--Rees theorem is the same as $R\otimes_S \hat{S}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the two notions of completion are indeed the same (and there is no need&#xA;to assume that $S$ is Cohen--Macaulay).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-02T04:14:29.040" />
  <row Id="8603" PostTypeId="1" AcceptedAnswerId="8613" CreationDate="2010-11-02T04:18:28.563" Score="2" ViewCount="49" Body="&lt;p&gt;Prove that the Markov-Hurwitz equation $x^2+y^2+z^2=dxyz$ is solvable in positive integers iff d= 1 or 3.  Of course the reverse direction is easy, just set x=y=z=1, d=3.  But I really have no idea how to attack the forward direction.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-11-02T06:36:45.403" Title="Markov-Hurwitz equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" />
  <row Id="8604" PostTypeId="2" ParentId="8596" CreationDate="2010-11-02T04:23:39.853" Score="3" ViewCount="0" Body="&lt;p&gt;This issue has been discussed multiple times in MO. For instance &lt;a href=&quot;http://mathoverflow.net/questions/903/resources-for-learning-practical-category-theory&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, with a view towards programming, or &lt;a href=&quot;http://mathoverflow.net/questions/8543/learning-to-think-categorically/8586#8586&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, where I proposed a selection of Mac Lane's book.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-02T04:23:39.853" CommentCount="1" />
  <row Id="8605" PostTypeId="2" ParentId="8598" CreationDate="2010-11-02T04:30:13.420" Score="0" ViewCount="0" Body="&lt;p&gt;Note that $L$ is a one-to-one homomorphism from $G$ to $S_G$, while $R$ is a one-to-one homomorphism from $G^{\mathrm{op}}$ to $S_G$ ($R(ab) = R(b)\circ R(a)$, so you have an anti-homomorphism if you work with $G$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could ask more generally if these two subgroups of $S_G$ are conjugate. That is, do there exist permutations $\pi$ and $\psi$ of $G$ such that for every $g\in G$, $\pi^{-1}\circ L(g) \pi = R(\psi(g))$ ? In fact, they always are: take $\pi=\psi$ to be the map that sends every element $x$ to $x^{-1}$ (which is, after all, an isomorphism from $G$ to $G^{op}$!):&#xA;\[ \pi^{-1}(L(g)(\pi(x))) = \pi^{-1}(L(g)(x^{-1})) = \pi^{-1}(gx^{-1}) = xg^{-1}) = R(g^{-1})(x).\]&#xA;So $L(G)$ and $R(G)$ are always conjugate in $S_G$, whether $G$ is abelian or not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-02T04:30:13.420" />
  <row Id="8606" PostTypeId="2" ParentId="8603" CreationDate="2010-11-02T04:36:36.130" Score="0" ViewCount="0" Body="&lt;p&gt;For starters, d must be odd.  For if d is even, either all of x,y, and z are even or two are odd and one is even.  If they are all even, there is also a solution $(x/2)^2 + (y/2)^2 + (z/2)^2 = (2d)(x/2)(y/2)(z/2)$ with smaller x,y,z.  If only one of x,y,z is even and d is even, the LHS is 2 mod 4 and the RHS is 0 mod 4.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-02T04:36:36.130" />
  <row Id="8607" PostTypeId="2" ParentId="4667" CreationDate="2010-11-02T04:37:37.407" Score="3" ViewCount="0" Body="&lt;p&gt;It makes sense to consider some sets of combinatorial objects as spaces (or &lt;a href=&quot;http://en.wikipedia.org/wiki/Polytope&quot; rel=&quot;nofollow&quot;&gt;polytopes&lt;/a&gt;) and therefore discuss dimensionality (e.g. the set of n by n (-1,0,+1)-matrices).  Although, perhaps the word &quot;dimension&quot; could better be described as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29&quot; rel=&quot;nofollow&quot;&gt;degrees of freedom&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mathematically, degrees of freedom is&#xA;  the dimension of the domain of a&#xA;  random vector, or essentially the&#xA;  number of 'free' components: how many&#xA;  components need to be known before the&#xA;  vector is fully determined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect that it will be difficult to discuss dimensionality in many combinatorial settings.  For example, imagine constructing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt;, starting from an empty matrix, placing symbols one-at-a-time in a non-clashing manner.  After placing (say) half of the symbols, we might find: (a) there are still many completions of this partial Latin square, (b) there are no completions of this partial Latin square or (c) there is a unique completion of this partial Latin square.  This seems to go against the notion of dimensionality -- the number of &quot;components&quot; required to determine the Latin square is not fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could think of the set of partitions of a set of n elements as having dimension n.  You require n pieces of information to determine the partition (which set each element is in).  But as Qiaochu Yuan points out, who cares?  There's no point in having a notion of &quot;dimension&quot; unless you can use it for something.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-02T04:37:37.407" />
  <row Id="8608" PostTypeId="2" ParentId="8550" CreationDate="2010-11-02T04:49:00.263" Score="1" ViewCount="0" Body="&lt;p&gt;For $x$ fixed, define events $A_n$ by $A_1 = { X \leq x-1 }$ and, for $n \geq 2$, $A_n = { x-\frac{1}{{n - 1}} &amp;lt; X \leq x-\frac{1}{n} }$. Since the disjoint union $A_1 \cup A_2 \cup \ldots$ gives the event ${ X &amp;lt; x }$, we have ${\rm P}(X &amp;lt; x)= {\rm P}(A_1 \cup A_2 \cup \ldots) = \sum\nolimits_{k = 1}^\infty {{\rm P}(A_k )}$. Noting that $\sum\nolimits_{k = 1}^n {{\rm P}(A_k )} = {\rm P}(X \leq x - \frac{1}{n})$, we thus obtain the desired result: ${\rm P}(X &amp;lt; x) = \lim _{n \to \infty } {\rm P}(X \le x - \frac{1}{n}) = \lim _{n \to \infty } F(x - \frac{1}{n}) = F(x - )$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-02T17:51:00.887" LastActivityDate="2010-11-02T17:51:00.887" CommentCount="2" />
  <row Id="8609" PostTypeId="1" CreationDate="2010-11-02T05:02:37.767" Score="9" ViewCount="142" Body="&lt;p&gt;A divisor $d$ of $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}}$ is &lt;em&gt;unitary&lt;/em&gt; if and only if $d = p_{1}^{\varepsilon_{1}} \cdots p_{n}^{\varepsilon_{n}}$, where each exponent $\varepsilon_{i}$ is either $0$ or $r_{i}$. Let $D_{k} = ${ $d$ } be the subset of unitary divisors of an integer $k &gt; 1$ satisfying $\omega(d) = \omega(k) - 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;. A positive integer $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}} &gt; 1$ is &lt;em&gt;hyperbolic&lt;/em&gt; if and only if $\sum_{i = 1}^{n} p_{i}^{-r_{i}} &amp;lt; 1$ or, equivalently, $\sum_{d \in D_{k}} d &amp;lt; k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $3$, $10$ and $20$ are hyperbolic, but $30$ and $510510 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$ are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming my calculations are correct, I'll state the following with some confidence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, many positive integers are hyperbolic. Of the first $10^{8}$ integers, $70334760$ are hyperbolic. Non-trivial prime powers, squares or higher, are also hyperbolic. If $k$ is a hyperbolic integer, then so are its proper, non-trivial unitary divisors; however, the same cannot be inferred for all divisors (consider the hyperbolic integer $900$ and its non-hyperbolic divisor $30$). In fact, an arbitrary product of any number of non-trivial unitary divisors of a hyperbolic integer is again hyperbolic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of hyperbolic integers is closed under &lt;em&gt;exponentiation&lt;/em&gt;, but not under addition (e.g., $10 + 20 = 30$) or under multiplication (e.g., $3 \times 10 = 30$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &lt;em&gt;Prime zeta function&lt;/em&gt;, $\zeta_{P}(s) = \sum_{p \text{ prime}} p^{-s}$, which converges absolutely for $\mathsf{Re}(s) &gt; 1$. Recall that the multiplicity of a prime divisor $p$ of $k$ is the largest exponent $r$ such that $p^{r}$ divides $k$ but $p^{r+1}$ does not. If the minimum multiplicity of an integer $k$ is $2$ or greater, then $k$ is hyperbolic as can be seen by the elementary bound&#xA;\begin{eqnarray}&#xA;\sum_{i = 1}^{n} p_{i}^{-r_{i}} &amp;lt; \zeta_{P}(2) \approx 0.452247 ....&#xA;\end{eqnarray}&#xA;Thus, the question of hyperbolicity is non-trivial only for integers with minimum multiplicity $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical evidence suggests that the natural density of the hyperbolic integers is greater than $0.988284 \dots$, and I conjecture that &lt;em&gt;almost all&lt;/em&gt; integers are indeed hyperbolic (i.e., the natural density is 1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is anything presently known about such integers? (References welcome!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is there a simple proof showing (or refuting) that almost all integers are hyperbolic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-02T23:38:18.897" LastActivityDate="2010-11-02T23:38:18.897" Title="Positive integers $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}} &gt; 1$ satisfying $\sum_{i = 1}^{n} p_{i}^{-r_{i}} &lt; 1$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;asymptotics&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="8610" PostTypeId="1" AcceptedAnswerId="8805" CreationDate="2010-11-02T05:09:11.257" Score="9" ViewCount="178" Body="&lt;p&gt;Prove that continued fraction of tan(1)=[1;1,1,3,1,5,1,7,1,9,1,11,...].  I tried using the same sort of trick used for finding continued fractions of quadratic irrationals and trying to find a recurrence relation, but that didn't seem to work.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-11-04T02:51:17.377" Title="Continued Fraction expansion of tan(1)" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8611" PostTypeId="1" AcceptedAnswerId="8912" CreationDate="2010-11-02T05:49:23.853" Score="3" ViewCount="193" Body="&lt;p&gt;How can I compute the number of common divisors of two naturals ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,if we consider &lt;code&gt;(12,24)&lt;/code&gt; the answer is &lt;strong&gt;6&lt;/strong&gt; i.e &lt;code&gt;{1,12,2,6,3,4}&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT : I got an answer &lt;a href=&quot;http://books.google.co.in/books?id=di0WSsXV74wC&amp;amp;pg=PA8&amp;amp;lpg=PA8&amp;amp;dq=%22Number+of+common+divisors%22&amp;amp;source=bl&amp;amp;ots=Wg7wRQKNul&amp;amp;sig=djqKheSaFypwJ_yqmzk_g7yVtO0&amp;amp;hl=en&amp;amp;ei=caTPTNbbH4yivgPYy4DFBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CDIQ6AEwBg#v=onepage&amp;amp;q=%22Number%20of%20common%20divisors%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.The solution boils down to finding the number of divisor of the GCD of the two numbers.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-02T16:06:23.743" LastActivityDate="2010-11-11T20:11:11.437" Title="Number of common divisors between two given numbers" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8612" PostTypeId="2" ParentId="3510" CreationDate="2010-11-02T06:25:38.380" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;:&#xA;Consider the function $f(t) = e^{-it}(\cos t + i \sin t)$ for $t \in \mathbb{R}$. By the quotient rule &#xA;\begin{eqnarray}&#xA;f^{\prime}(t) = e^{-i t}(i \cos t - \sin t) - i e^{-i t}(\cos t + i \sin t) = 0&#xA;\end{eqnarray}&#xA;identically for all $t \in \mathbb{R}$. Hence, $f$ is constant everywhere. Since $f(0) = 1$, it follows that $f(t) = 1$ identically. Therefore, $e^{it} = \cos t + i \sin t$ for all $t \in \mathbb{R}$, as claimed.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-24T23:30:12.673" LastActivityDate="2010-11-24T23:30:12.673" />
  <row Id="8613" PostTypeId="2" ParentId="8603" CreationDate="2010-11-02T06:36:45.403" Score="4" ViewCount="0" Body="&lt;p&gt;See Theorem 6.1 of &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf&lt;/a&gt;. &#xA;By the way, the point of the case d = 1 and d = 3 is not that there are integral solutions (that is trivial) but that they can be systematically derived from small obvious solutions by a recursive method. See the start of Section 6 at that link for a discussion of that issue and a reference for more information.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-11-02T06:36:45.403" CommentCount="1" />
  <row Id="8614" PostTypeId="2" ParentId="8503" CreationDate="2010-11-02T06:48:58.930" Score="5" ViewCount="0" Body="&lt;p&gt;Putting the bound on N at 9 seems quite non-random. The continued fraction method for solving x^2 - dy^2 = 1 also works on x^2 - dy^2 = N as long as |N| &amp;lt; sqrt(d), which corresponds exactly to your bound on N when d = 94. Therefore your question almost sounds like it was set by someone who knew continued fractions. How did you come up with your question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, your questions would be better posed if you explain where they come from instead of writing them in the &quot;Solve this&quot; format, which makes them look suspiciously like homework tasks.&lt;/p&gt;&#xA;" OwnerUserId="619" LastActivityDate="2010-11-02T06:48:58.930" />
  <row Id="8615" PostTypeId="2" ParentId="8586" CreationDate="2010-11-02T07:21:26.830" Score="3" ViewCount="0" Body="&lt;p&gt;How about you take the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{dy}{dx} = |x|  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a linear non-homogeneous differential equation, whose solution is $C^1$ but not smooth at $x=0$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-02T07:21:26.830" CommentCount="3" />
  <row Id="8616" PostTypeId="2" ParentId="8611" CreationDate="2010-11-02T07:48:16.513" Score="0" ViewCount="0" Body="&lt;p&gt;can u post your algorithm here debanjan , i will be interested to have a looka t it&lt;/p&gt;&#xA;" OwnerUserId="2969" LastActivityDate="2010-11-02T07:48:16.513" CommentCount="3" />
  <row Id="8617" PostTypeId="2" ParentId="8599" CreationDate="2010-11-02T08:03:42.223" Score="3" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that when you (as you say in the comments) get the two scenarios:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f \star (D g) = D(f \star g) = (D f) \star g&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then &lt;strong&gt;you get to choose which!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $D f$ doesn't make sense, then you can ignore it and choose to use the identity $D(f \star g) = f \star (D g)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this to the extreme, you get the - bizarre, in my opinion - result that if $p$ is a polynomial then $f \star p$ is always a polynomial.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-11-02T08:03:42.223" CommentCount="14" />
  <row Id="8618" PostTypeId="2" ParentId="8596" CreationDate="2010-11-02T08:09:14.857" Score="2" ViewCount="0" Body="&lt;p&gt;It depends on whether you are talking about &lt;em&gt;Category Theory&lt;/em&gt; as a topic in mathematics (on a par with &lt;em&gt;Geometry&lt;/em&gt; or &lt;em&gt;Probability&lt;/em&gt;) or &lt;em&gt;Category Theory&lt;/em&gt; as a viewpoint on mathematics as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the former, the main prerequisite is that you should have encountered a situation where you wanted to move from one type of &quot;thing&quot; to another type of &quot;thing&quot;: say from a group to its group ring, or from a space to its ring of functions, or from a manifold to its differential graded algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the latter, then there are no prerequisites and it is a Very Good thing to do!  But if the latter, then reading Mac Lane isn't necessarily the best way to go.  However, I'm not sure if there is a textbook (or other) that tries to teach elementary mathematics (of any flavour) from a categorical viewpoint.  I try to teach this way, but I've not written a textbook!  I wrote a bit more on this in response to a question on MO, I copied my answer &lt;a href=&quot;http://www.math.ntnu.no/~stacey/CountingOnMyFingers/TeachingCategories.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-11-02T08:09:14.857" CommentCount="3" />
  <row Id="8619" PostTypeId="2" ParentId="8556" CreationDate="2010-11-02T08:10:27.600" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a hint for you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $v = (1, 1, 1, \dots, 1)'$, the vector with only ones. By definition of your Matrix, you have $Xv = v$. Thus, $v$ is an eigenvector with eigenvalue 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now change the first basis vector of your vector space to $v$ and keep the other unchanged. In this new basis, what will $X$ looks like?&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-02T08:10:27.600" CommentCount="2" />
  <row Id="8620" PostTypeId="1" AcceptedAnswerId="8628" CreationDate="2010-11-02T09:06:42.280" Score="3" ViewCount="102" Body="&lt;p&gt;The standard definition of a ring is an abelian group that is a monoid under multiplication (with distributivity). However there are some books that have a weaker definition in that a ring only has to be closed under multiplication (no identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a problem in my algebra book asking me to prove that if a ring (defined in the second way) has $p$ elements, where $p$ is prime. If the multiplication is not trivial (i.e. sending everything to 0) then the ring is forced to have a multiplicative identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its seems like a trivial proof, but I just can't see what I'm missing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have so far:&#xA;Given $R$ is an ring with p elements&#xA;R is an abelian group of prime order, therefore it is cyclically generated, and of characteristic $p$ and isomorphic to $Z/pZ$. Essentially it boils down to showing $Z/pZ$ is forced to have a multiplicative identity, but I just can't see where this comes from (every resource I found seems to take this as fact). Since this is a requirement regardless of multiplicative structure I can't just use the fact that $Z/pZ - 0$  is a group under the typical multiplication.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2010-11-02T09:25:57.020" LastActivityDate="2010-11-02T17:28:42.347" Title="Finite rings of prime order must have a multiplicative identity" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="4" />
  <row Id="8621" PostTypeId="1" CreationDate="2010-11-02T09:19:57.960" Score="1" ViewCount="88" Body="&lt;p&gt;Interesting problem... I tried solving it through isosceles triangles, and I've proved it doesn't work on equilateral triangles. Can anyone give me some hints towards a geometric proof?&lt;/p&gt;&#xA;" OwnerUserId="2131" LastActivityDate="2010-11-02T10:25:42.527" Title="Can two angle bisectors of a triangle be perpendicular?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8622" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T09:22:23.590" Score="0" ViewCount="0" Body="&lt;p&gt;Go on a hunt for all the ways in which &lt;em&gt;R&lt;/em&gt; can &lt;em&gt;fail&lt;/em&gt; to be isomorphic to &amp;#8484;/p&amp;#8484;, and demonstrate the failure of any of them to hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the subset &lt;em&gt;Z&lt;/em&gt;&amp;nbsp;&amp;sube;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; of elements such that left-multiplication by z&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt; &lt;strong&gt;does&lt;/strong&gt; map everything to zero. We may easily show that &lt;em&gt;Z&lt;/em&gt; is an ideal in &lt;em&gt;R&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For y,z&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt;, we have (y+z)r = yr + zr = 0+0 = 0 for any r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;, so y+r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt;;&lt;/li&gt;&#xA;&lt;li&gt;For any z&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt; and a,r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;, we have (az)r = a(zr) = 0 and (za)r = z(ar) = 0; so az,&amp;thinsp;za&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;Z&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, really, we only needed the first of these two criteria, because it shows that (&lt;em&gt;Z&lt;/em&gt;,+) must be a subgroup of the abelian group (&lt;em&gt;R&lt;/em&gt;,+), which is of prime order. We know that &lt;em&gt;Z&lt;/em&gt;&amp;nbsp;&amp;ne;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;; so it follows that &lt;em&gt;Z&lt;/em&gt;&amp;nbsp;=&amp;nbsp;{0}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay then. If none of the non-zero elements map &lt;em&gt;everything&lt;/em&gt; to zero via multiplication, perhaps some of them map &lt;em&gt;some&lt;/em&gt; things to zero? Select a&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt;&amp;nbsp;\&amp;nbsp;{0} arbitrary, and consider the set &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;{b&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; | ab&amp;nbsp;=&amp;nbsp;0}. Well, again we find that &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; is an abelian subgroup of &lt;em&gt;R&lt;/em&gt;, and again we know that &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; is not allowed to be all of &lt;em&gt;R&lt;/em&gt;; so we know that &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;{0}. That is, &lt;em&gt;R&lt;/em&gt; contains no zero divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, are there any non-zero elements such that multiplication is not invertible? Well, no: any a,b&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; such that ar&amp;nbsp;=&amp;nbsp;br for some non-zero r&amp;nbsp;&amp;isin;&amp;nbsp;&lt;em&gt;R&lt;/em&gt; would have the property (a&amp;nbsp;&amp;minus;&amp;nbsp;b)r&amp;nbsp;=&amp;nbsp;0, so that a&amp;nbsp;=&amp;nbsp;b; so multiplication by elements of &lt;em&gt;R&lt;/em&gt;&amp;nbsp;\&amp;nbsp;{0} yield bijections on  &lt;em&gt;R&lt;/em&gt;&amp;nbsp;\&amp;nbsp;{0}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should now enable you to answer your question.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-11-02T10:11:51.240" LastActivityDate="2010-11-02T10:11:51.240" CommentCount="0" />
  <row Id="8623" PostTypeId="1" AcceptedAnswerId="8642" CreationDate="2010-11-02T09:38:35.583" Score="3" ViewCount="87" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Find the distinct number of permutations of four letters of the word MATHEMATICS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 11P4/(2!)^3 = 990&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? This question along with the answer is from a module from my course, so this is all the details that I am able to provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of this question is how many distinct 4 letter words can be formed from the word 'MATHEMATICS'. Therefore, a word like 'MAME' is valid. I know that using 11P4 alone will not suffice because it will say that there are 4 ways to create the 4 letter word MAME from the word 'MATHEMATICS'. So, we I will need to adjust the output of 11P4, and this is where I am stuck. The answer divides 11P4 by (2!)^3. I think this represents the 3 non-distinct letters in the word 'MATHEMATICS' (i.e. M,A,T) which have two copies each. Using Sage, the answer I have is 2,454 which is not the same as the answer provided. Here is the code I wrote:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#Input according to question&#xA;inputQ='M A T H E M A T I C S'.split(' ')&#xA;&#xA;#Same input, but each letter in input is distinct&#xA;inputTest='M1 A1 T1 H E M2 A2 T2 I C S'.split(' ')&#xA;&#xA;print inputQ&#xA;print inputTest&#xA;&#xA;#Number of letters to form from the given input&#xA;choice=4&#xA;&#xA;ansQ=Permutations(inputQ,choice).list()&#xA;ansTest=Permutations(inputTest,choice).list()&#xA;&#xA;print 'Total distinct permutations of ' + str(choice) + ' letter word(s) from the word ' + ''.join(inputQ) +':' ,len(ansQ)&#xA;print 'If each letter in '+ ''.join(inputQ) + ' were distinct: ', len(ansTest)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What if I only want 1 letter? Then the answer would be 8, which I got by manually listing down all possibilities. But what it the general formula for this sort of problem?&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="2973" LastEditorDisplayName="" LastEditDate="2010-11-02T13:30:50.023" LastActivityDate="2010-11-02T14:55:04.923" Title="Distinct permutations of x items from n items" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="8624" PostTypeId="1" CreationDate="2010-11-02T09:38:36.730" Score="-1" ViewCount="159" Body="&lt;p&gt;Is pi mean 22/7? If so, how circumference of the circle will be 2*22/7 if radius would be 1.&lt;/p&gt;&#xA;" OwnerUserId="2997" LastActivityDate="2010-11-02T22:34:54.543" Title="How the letter 'pi' came in mathematics?" Tags="&lt;pi&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="8625" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T09:42:16.053" Score="3" ViewCount="0" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi#The_letter_.CF.80&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The constant is named &quot;$\pi$&quot; because &quot;$\pi$&quot; is the first letter of the Greek word περίμετρος (perimeter)...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions that can be avoided with a quick look at a wikipedia article are typically frowned upon here.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-11-02T09:42:16.053" CommentCount="3" />
  <row Id="8626" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T09:43:55.013" Score="1" ViewCount="0" Body="&lt;p&gt;$\pi \neq \frac{22}{7}$. First of all $\frac{22}{7}$ is just an approximate value of $\pi$.  Note that $\pi$ is irrational so it cannot be expressed in the form $\frac{p}{q}$ where $p$ and $q \neq 0$ are integers. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For more information about $\pi$ i would suggest you to read the wiki article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pi&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-02T09:43:55.013" CommentCount="4" />
  <row Id="8627" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T09:59:20.437" Score="2" ViewCount="0" Body="&lt;p&gt;From &lt;a href=&quot;http://books.google.co.uk/books?id=QFPvAAAAMAAJ&amp;amp;q=Pi%3a+a+biography+of+the+world%27s+most+mysterious+number&amp;amp;dq=Pi%3a+a+biography+of+the+world%27s+most+mysterious+number&amp;amp;hl=en&amp;amp;ei=oN7PTIKkItTR4gaM6-SiBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CC0Q6AEwAABlockquote&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Pi: A Biography of the World's Most Mysterious Number&lt;/em&gt;&lt;/a&gt; by  Posamentier and  Lehmann:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the well-known mathematics historian Florian Cajori (1859-1930), the symbol $\pi$ was first used in mathematics by William Oughtred (1575-1660) in 1652 when he referred to the ratio of the circumference of a circle to its diameter as $\frac{\pi}{\delta}$, where $\pi$ represented the periphery of a circle and $\delta$ represented the diameter...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In 1706 William Jones (1675-1749) published his book &lt;em&gt;Synopsis palmariorum matheseos&lt;/em&gt;, in which he used $\pi$ to represent the ratio of the circumference of a circle to its diameter. This is believed to have been the first time that $\pi$ was used as it is defined today... But not until he [&lt;em&gt;Leonhard Euler - A.R.&lt;/em&gt;] used the symbol $\pi$ in his famous book &lt;em&gt;Introductio in analysin infinitorum&lt;/em&gt; did the use of $\pi$ to represent the ratio of the circumference of a circle to its diameter become widespread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-02T10:26:28.073" LastActivityDate="2010-11-02T10:26:28.073" CommentCount="1" />
  <row Id="8628" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T10:10:14.013" Score="3" ViewCount="0" Body="&lt;p&gt;Let $x$ be a nonzero element of the ring. Then $R=\{0,x,2x,3x,\ldots,(p-1)x\}$&#xA;where $2x$ means $x+x$ etc. Then $x^2=jx$ where $1\le j\le p-1$.&#xA;Moreover $(ax)(bx)=abx^2=(abk)x$. All you need to do is to prove that for&#xA;some $a$, $(abk)x=bx$ for all $b$. (It's surely enough to do this for $b=1$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-02T10:10:14.013" CommentCount="2" />
  <row Id="8629" PostTypeId="2" ParentId="8621" CreationDate="2010-11-02T10:25:42.527" Score="9" ViewCount="0" Body="&lt;p&gt;The reason you cannot find a geometric proof is because it is not true.&#xA;Suppose you have a triangle ABC such that two angle bisectors are perpendicular, say the angle bisectors of A and B. Suppose they intersect at some point D, then we have a right-angled triangle ADB. If in ABC the angle at A is a and the angle at B is b, then in ADB the angle at A is a/2 and the angle at B is b/2. ADB right-angled triangle implies a/2+b/2=90, which implies a+b=180, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2712" LastActivityDate="2010-11-02T10:25:42.527" />
  <row Id="8630" PostTypeId="1" CreationDate="2010-11-02T11:12:01.610" Score="2" ViewCount="98" Body="&lt;p&gt;Given two integers A and B , how to efficiently compute the sum of all the digits of every number int the set  $\{N | A \le N \le B \} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;strong&gt;A = 100&lt;/strong&gt;  and &lt;strong&gt;B = 777&lt;/strong&gt;, then the required answer is &lt;strong&gt;8655&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in deriving an formula/efficient algorithm for the same.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-02T11:24:20.617" LastActivityDate="2010-11-02T13:06:36.980" Title="How to (efficiently) compute the  sum of  the  digits  of evey number  lying between two given numbers ?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" />
  <row Id="8631" PostTypeId="2" ParentId="8623" CreationDate="2010-11-02T11:49:39.860" Score="3" ViewCount="0" Body="&lt;p&gt;After the edit by OP,here is a similar &lt;a href=&quot;http://sg.answers.yahoo.com/question/index?qid=20100302091836AA9Cf9K&quot; rel=&quot;nofollow&quot;&gt;problem and solution&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-02T13:58:08.813" LastActivityDate="2010-11-02T13:58:08.813" CommentCount="2" />
  <row Id="8632" PostTypeId="1" CreationDate="2010-11-02T11:51:07.943" Score="-1" ViewCount="57" Body="&lt;p&gt;I have a problem. I wish to understand how following can be proven to hold; $\langle P_S(v), u_m \rangle = \sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2}  \langle u_j,u_m \rangle = \langle v , u_m \rangle $, where $P_S(v) =\sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2} u_j $ and Suppose V is a complex inner product space and $B={u_1, u_2,..., u_n }$ is orthogonal set in $V$ with $u_j&amp;lt;&gt;0$ for all j. also prove that $P_S(v)$ is a linear transformation. &lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-02T12:24:21.463" LastActivityDate="2010-11-02T12:48:20.923" Title="explain that  $\langle P_S(v), u_m \rangle = \sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2}  \langle u_j,u_m \rangle = \langle v, u_m \rangle$" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8633" PostTypeId="1" AcceptedAnswerId="8663" CreationDate="2010-11-02T12:17:48.590" Score="2" ViewCount="130" Body="&lt;p&gt;Fair warning: I am not a math expert (that's why I'm here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be able to calculate the probability of rolling a certain side on a die with &lt;em&gt;n&lt;/em&gt; sides where any number of those sides has an unfair advantage over the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: On a 6-sided die, we could assume that the 1 - 6 sides are longer than the other sides due to wear and tear (or someone shaved the 1 or 6) and, therefore, more likely to come face up on a roll. I know that, on a perfect die, the roll is uniformly random and each side has a 1/side chance of coming up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the probability of each side if sides 1 and 6 are longer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if sides 1 and 6 are longest, 2 and 5 are second longest, and 3 and 4 are shortest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I'm using a 12-sided die? 10-sided? 20-sided?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a formula that I can plug numbers into and I'd really, really like an explanation of how it all works, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://books.google.com/books?id=GJP08Lv7vpYC&amp;amp;lpg=PA87&amp;amp;ots=bydhFz2oEh&amp;amp;dq=simulate%2520shaved%2520dice&amp;amp;pg=PA86#v=twopage&amp;amp;q&amp;amp;f=false%20%22this%20link%22&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; that talks about using Mathematica to calculate one side shaved on a 6-sided die, but I don't know how this changes when you increase the number of dice, or the syntax being used. I feel like the equation below (from the link) makes sense somewhat (in that &amp;sigma; represents the &lt;em&gt;increase&lt;/em&gt; in probability of a 1 or 6, but I would like to know how to calculate &amp;sigma;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = { 1/6 + &amp;sigma;, 1/6 - &amp;sigma;/6, 1/6 - &amp;sigma;/6, 1/6 - &amp;sigma;/6, 1/6 - &amp;sigma;/6, 1/6 + &amp;sigma; };&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I use the same formula to represent a 20 sided die?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = {1/20 + &amp;sigma;, 1/20 - &amp;sigma;/20, ...  ... ... ..., 1/20 - &amp;sigma;/20, 1/20 + &amp;sigma;}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I took an intro to Statistics in college, but that's all the exposure I've had. I would like to understand more, but this is where I am now. I appreciate any help you can give me and any resources you can point me to.&lt;/p&gt;&#xA;" OwnerUserId="3023" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-02T13:03:56.427" LastActivityDate="2010-11-02T16:50:16.397" Title="Probability concerning unfair dice with N sides" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;physics&gt;&lt;applications&gt;" AnswerCount="3" />
  <row Id="8634" PostTypeId="2" ParentId="8632" CreationDate="2010-11-02T12:48:20.923" Score="5" ViewCount="0" Body="&lt;p&gt;The first equality follows by definition. Now notice that $$\langle u_j, u_m \rangle = 0$$ if $i\neq j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the sum reduces to $$\frac{\langle v, u_m \rangle}{||u_m||^2}\langle u_m, u_m \rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is the definition of the norm?&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-02T12:48:20.923" CommentCount="2" />
  <row Id="8636" PostTypeId="2" ParentId="8633" CreationDate="2010-11-02T12:52:36.513" Score="1" ViewCount="0" Body="&lt;p&gt;There are 2 ways to consider your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probabilistic way is just to say we have an unfair dice. This means we give probabilites $p_1$, $p_2$, ..., $p_6$. The conditions are that $p_1 + p_2 + \dots + p_6 = 1$ and that $p_i \geq 0$ for $i = 1, 2, ..., 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we assign &quot;Probability that the dice will show face number $i$&quot; is $p_i$. For example, a fair dice would have $p_1 = p_2 = \dots = p_6 = \frac{1}{6}$. An unfaire dice like you want (face 1 and 6 bigger, 2 and 5 normal, 3 and 4 smaller) could have $p_1 = \frac{1}{4}, p_2 = \frac{1}{6}, p_3 = \frac{1}{12}, p_4 = \frac{1}{12}, p_5 = \frac{1}{6}, p_6 = \frac{1}{4}$. This is the probabilist approach. You &lt;em&gt;decide&lt;/em&gt; the probabilities you want, and then you do some calculations (for example, you could be interested what is the distribution of the sum of 2 such dices, and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistician has a dice and knows/supposes that it is unfair. He &lt;em&gt;cannot&lt;/em&gt; know the exact value $p_i$ for each face. Hovewer, he can try to guess thoses values and create tests to see if these values are probable or not. The usual way to guess the values is to throw the dice many times (say 1000 times) and to count of much 1's, 2's, ... we get. For example, if you count 100 1's, you would say that the probability of having a 1 could be $\frac{100}{1000} = \frac{1}{10}$. After that, you can build tests to verify this guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I answered your question. But it truly depends on what you want to do. According to your question, you have the unfair dice, and you want to guess the probabilities. This would more be the statistic approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you have a model (like the one you give with $p_i = \frac{1}{n} + \sigma$) and you must create a test to estimate $\sigma$. Basically, you will have to throw the dice several times anyway.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-02T12:52:36.513" CommentCount="1" />
  <row Id="8637" PostTypeId="2" ParentId="8633" CreationDate="2010-11-02T12:54:53.423" Score="2" ViewCount="0" Body="&lt;p&gt;There are two distinct questions posed here. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How to work with probabilities of dice that are biased in some way.  For an $n$ sided die this means that there are some numbers $p_1, p_2, \dots p_n$ with $p_1 + p_2 + \dots p_n = 1$, where $p_i$ is interpreted as the probability of landing on face $i$.  Then the standard methods of probability theory give recipes for calculating the probabilities of outcomes of random processes based on such dice: sums of several dice tosses, results of games, etc.  Calculations are easier for the case where all $p_i$ are equal -- known as a &quot;uniform distribution&quot; or &quot;fair dice&quot; -- but there is nothing fundamentally different about this case compared to that of unequal probabilities.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to compute the probabilities $p_i$ from the physics of a die and the dice-tossing process.  For coins there is some recent analysis of &quot;dynamical bias in coin tossing&quot; starting from the article with that title by Diaconis, Holmes and Montgomery.  For rectangular dice with uniform mass density (not loaded toward some faces), a simple and not necessarily physically accurate prescription is to assign each face a probability proportional to its surface area, or equivalently, probability equal to its share of the surface area of the entire die.  This can be viewed as the $k=0$ approximation to the probabilities for a thrown die that rolls onto at most $k$ additional faces after landing. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-02T16:02:52.117" LastActivityDate="2010-11-02T16:02:52.117" />
  <row Id="8638" PostTypeId="2" ParentId="8630" CreationDate="2010-11-02T13:06:36.980" Score="2" ViewCount="0" Body="&lt;p&gt;You should find the formula $f(n)$ for the sum of digits for numbers from 1 to n, then do $f(B)-f(A)$.  It is not very clean.  The sum of digits of all k digit numbers, that is from $10^{(k-1)} \text { to } 10^k-1$, is $45*(10^k-1)/9$.  Stopping part way is harder, but you can do it by recursion.  If you want the sum of digits of numbers up to 7655, say, you can do sum of digits up to 999, plus 1000*6*5/2 (for the thousands digits up through 6) + 6*sum of digits up to 999 (for the last three digits for 1000 through 6999) + 7*656 (for the thousands in the 7s) + sum of digits up to 655.  The last has one less digit, so you just call your subroutine with that.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-02T13:06:36.980" CommentCount="2" />
  <row Id="8639" PostTypeId="1" AcceptedAnswerId="8641" CreationDate="2010-11-02T13:17:54.543" Score="0" ViewCount="68" Body="&lt;p&gt;What is the probability to get $n$ times the same element $k$ choosing randomly from a set $A$ knowing the cardinality $|A| = c$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance,&lt;br&gt;&#xA;rubik&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-02T15:06:51.400" LastActivityDate="2010-11-02T15:06:51.400" Title="Probability to get always the same number choosing randomly from a set of $c$ elements" Tags="&lt;probability&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="8641" PostTypeId="2" ParentId="8639" CreationDate="2010-11-02T13:23:49.903" Score="2" ViewCount="0" Body="&lt;p&gt;It depends: is $k$ fixed in advance? if so then the probability is given by $\frac{1}{c^n}$. If you just ask for the probability to have $n$ times the same element, then it is $\frac{1}{c^{n-1}}$ (since the first time you can have any object). By the way, I supposed you are doing exactly $n$ tries. Of course, the probability changes if you are doing more than $n$ tries.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-02T13:52:26.183" LastActivityDate="2010-11-02T13:52:26.183" CommentCount="3" />
  <row Id="8642" PostTypeId="2" ParentId="8623" CreationDate="2010-11-02T13:42:50.310" Score="1" ViewCount="0" Body="&lt;p&gt;After you edited your question, here is the new answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to solve this question, you should separate into different cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's try to find all the 4-letter words that have different letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Debanjan pointed out, this leads to 8*7*6*5 = 1680 words. The idea for this is just to consider you have 8 possibilities for the first letter, but for the second letter you only have 7 possibilities since it must be different from the first one, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's count the word that have 2 &quot;M&quot;, but all other letters are different. You have 7 choices for another letter, and 6 for the last letter. However, the 2 &quot;M&quot; might be anywhere in the word, thus you multiply this by the number of permutation, which is given by $\frac{4!}{2!2!} = 6$ Thus, there are 7*6*6 = 252 such words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ EDIT: Here I need to explain something, as Ross commented out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only count 252 such words and not 504, because in the 42 choices of letters, I can choose &quot;S&quot; and &quot;C&quot;, but I can also choose &quot;C&quot; and then &quot;S&quot;. So either you have to divide the 7*6 by 2, or you only allow permutations that don't change the order of the first other letter chosen and the second other letter chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I didn't want to explain this in details, but it is true it is needed, in order to avoid confusion. ]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also 252 words with 2 &quot;A&quot; and all other letters different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also 252 words with 2 &quot;T&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are 6 words with 2 &quot;A&quot; and 2 &quot;M&quot;, 6 words with 2 &quot;A&quot; and 2 &quot;T&quot; and 6 words with 2 &quot;M&quot; and 2 &quot;T&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total: 1680 + 252 + 252 + 252 + 6 + 6 + 6&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-02T14:44:53.227" LastActivityDate="2010-11-02T14:44:53.227" CommentCount="5" />
  <row Id="8643" PostTypeId="2" ParentId="8323" CreationDate="2010-11-02T13:50:06.387" Score="0" ViewCount="0" Body="&lt;p&gt;My numerical work and computing skills are not to be trusted, but the first &quot;near miss&quot; that I recorded was $n=252544447$. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-02T13:50:06.387" CommunityOwnedDate="2010-11-02T13:50:06.387" CommentCount="3" />
  <row Id="8644" PostTypeId="2" ParentId="8576" CreationDate="2010-11-02T13:53:06.450" Score="1" ViewCount="0" Body="&lt;p&gt;As this question is now 14 hours old, I will give an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First sum the total number of digits of all the natural numbers $ \le k-1.$ Call this sum $S.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $9 \times 10^{i-1}$ natural numbers of length $i,$ so the total number of digits of  natural numbers of length $i$ is $9i \times 10^{i-1}.$ Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S=9(1+2.10 + 3.10^2 + \cdots + (k-1)10^{k-2}) = (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first natural number with $k$ digits is $10^{k-1}$ and so there are $n- 10^{k-1}+1$ natural numbers from $10^{k-1}$ to $n$ inclusive. Each of these numbers has $k$ digits and so the number of digits required to write the natural numbers from $1$ to $n$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  k( n- 10^{k-1}+1) + S =  k( n- 10^{k-1}+1) + (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = k(n+1) - \frac{10^k-1}{9}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-02T13:53:06.450" CommentCount="1" />
  <row Id="8645" PostTypeId="2" ParentId="8639" CreationDate="2010-11-02T13:59:28.307" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming that you draw $n$ times uniformly at random from the set of $c$ elements, it is $\frac{1}{c^{n-1}}$ as Djaian says. You are rolling a $c$-sided die.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-02T13:59:28.307" CommentCount="1" />
  <row Id="8646" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T14:04:54.690" Score="0" ViewCount="0" Body="&lt;p&gt;If $R$ is a ring of order $p$, then it injects into the ring $\text{End}(\mathbb{Z}/p\mathbb{Z})$ of endomorphisms of the abelian group $\mathbb{Z}/p\mathbb{Z}$.    How many elements does this ring have?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-02T14:20:56.830" LastActivityDate="2010-11-02T14:20:56.830" CommentCount="2" />
  <row Id="8647" PostTypeId="2" ParentId="8623" CreationDate="2010-11-02T14:13:18.903" Score="3" ViewCount="0" Body="&lt;p&gt;Note: This answer presumes that the question is &quot;How many four-letter words can be formed using the letters of the word 'MATHEMATICS'.&quot;  So MAME should be counted as a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The provided answer, ${}_{11}P_4 / (2!)^3$, is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's temporarily change the question to &quot;How many distinct rearrangements are the of the letters of MATHEMATICS?&quot;, that is, how many 11-letter words can we form.  If the letters were all distinct, the answer would be 11!, that is, ${}_{11}P_11$.  But the letters are not all different, so we must adjust.  We can do so as in the original problem:  First, attach subscripts to identical letters, so that the letters are in fact all distinct, making 11! &quot;words&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to erase subscripts.  First take the Ms.  There are two of them, M1 and M2, and swapping these does not change a &quot;word&quot; when we're ignoring subscripts.  So there are 11!/2 words if we ignore the subscripts on M.  Proceeding similarly, we find that there are $11!/(2*2*2)$ distinct words when all the subscripts are erased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need factorials when we have more than two of a letter.  Suppose there were three Xs.  Then there are 3! ways of rearranging the Xs, and all these ways lead to the same word when subscripts are erased.  So the adjustment would be to divide by 3!.  (Really this is happening with the Ms too, since 2! = 2.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general:  Suppose we have a total of $n$ letters, including $c_1$ copies of one letter, $c_2$ copies of the second letter, . . . , and $c_i$ copies of the $i^{\rm th}$ letter.  Then the total number of words we can make is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n! \over { c_1!c_2!\ldots c_i! }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this formula works even for letters that aren't replicated, since 1! = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good.  The trouble is that the original problem does not ask for words formed from &lt;em&gt;all&lt;/em&gt; the letters, just from four at a time.  And here the trick doesn't work.  We do need to adjust, but we can't simply divide by 2! when there are two copies of a letter, because some of the words don't contain all the copies of that letter!  The adjusting trick, as described, works only when all copies of the letter are necessarily in the word, i.e., when each word contains all letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this even more clearly, suppose I had asked for words of &lt;em&gt;two&lt;/em&gt; letters formed from the letters of MATHEMATICS.  Would the answer be ${}_{11}P_2/(2!)^3$?&#xA;Certainly not, since this expression is not even an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way of phrasing the question so that the provided answer is correct?  I don't see it, and I find it doubtful anyway, for the same reason as in the preceding paragraph.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-11-02T14:55:04.923" LastActivityDate="2010-11-02T14:55:04.923" />
  <row Id="8648" PostTypeId="1" CreationDate="2010-11-02T14:14:07.610" Score="-3" ViewCount="77" Body="&lt;p&gt;explain $\langle \sum_{j=1}^n \langle v,u_j \rangle u_j, \sum_{k=1}^n \langle w,u_k \rangle u_k \rangle =\sum_{j=1}^n \langle v,u_j \rangle \sum_{k=1}^n \langle u_j, \langle w, u_k \rangle u_k \rangle$, V is complex inner product space with orthonormal basis $R={u_1, u_2,..., u_n}$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-02T14:40:19.720" LastActivityDate="2010-11-02T20:17:44.127" Title="explain below part of the proof of Parseval's relation" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8649" PostTypeId="2" ParentId="8648" CreationDate="2010-11-02T14:32:49.783" Score="2" ViewCount="0" Body="&lt;p&gt;The inner product is sesquilinear, and each $\langle v,u_j\rangle$ and $\langle w,u_k\rangle$ are scalars. Rewrite $\alpha_j = \langle v,u_j\rangle$ and $\beta_k = \langle w,u_k\rangle$. Then the left hand side is&#xA;$$\left\langle \sum_{j=1}^n\langle v,u_j\rangle u_j, \sum_{k=1}^{n}\langle w,u_k\rangle u_k\right\rangle = \left\langle \sum_{j=1}^n \alpha_ju_j,\sum_{k=1}^n\beta_k u_k\right\rangle.$$&#xA;So by additivity and homogeneity in the first component you have:&#xA;\begin{eqnarray*}&#xA;\left\langle \sum_{j=1}^n\langle v,u_j\rangle u_j, \sum_{k=1}^n\langle w,u_k\rangle u_k\right\rangle &amp;amp; = &amp;amp; \left\langle \sum_{j=1}^n \alpha_ju_j,\sum_{k=1}^n\beta_k u_k\right\rangle\\&#xA;&amp;amp; = &amp;amp; \sum_{j=1}^{n}\left\langle \alpha_j u_j, \sum_{k=1}^n\beta_k u_k\right\rangle\\&#xA;&amp;amp; = &amp;amp; \sum_{j=1}^n\left( \alpha_j\left\langle u_j, \sum_{k=1}^n\beta_k u_k\right\rangle\right).&#xA;\end{eqnarray*}&#xA;Then applying linearity on the second component you get the given expression.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T20:17:44.127" LastActivityDate="2010-11-02T20:17:44.127" CommentCount="3" />
  <row Id="8650" PostTypeId="1" AcceptedAnswerId="8666" CreationDate="2010-11-02T15:10:15.733" Score="5" ViewCount="103" Body="&lt;p&gt;what is the number of permutations of a set say {1, 2, 3, 4, 5} that keep at least one integer fixed ?&lt;/p&gt;&#xA;" OwnerUserId="3027" LastEditorUserId="3027" LastEditorDisplayName="" LastEditDate="2010-11-02T15:30:57.347" LastActivityDate="2010-11-02T17:00:59.453" Title="Permutations of a set that keep at least one integer fixed" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8651" PostTypeId="1" CreationDate="2010-11-02T15:25:13.740" Score="4" ViewCount="180" Body="&lt;p&gt;The 2nd part of the &quot;Fundamental Theorem of Calculus&quot; has never seemed as earth shaking or as &lt;strong&gt;fundamental&lt;/strong&gt; as the first to me. Why is it &quot;fundamental&quot; -- I mean, the &lt;em&gt;mean value theorem&lt;/em&gt;, and the &lt;em&gt;intermediate value theorems&lt;/em&gt; are both pretty exciting by comparison. And after the joyful union of integration and the derivative that we find in the first part, the 2nd part just seems like a yawn. So, what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear I'm talking about this:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let ƒ be a real-valued function defined on a closed interval [a, b] that admits an antiderivative F on [a,b]. That is, ƒ and F are functions such that for all x in [a, b],&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = F'(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ƒ is integrable on [a, b] then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_a^b f(x)dx = F(b) - F(a).$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've been through the proof a few times. It makes sense to me.  But, it didn't help me to see the light. To me it just looks like &quot;OK here is how you do the definite integral.&quot; Which dosen't seem like such a big deal, especially when indefinite integrals can be more interesting. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-02T15:31:52.220" LastActivityDate="2010-11-02T19:38:05.627" Title="The 2nd part of the &quot;Fundamental Theorem of Calculus.&quot; " Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8652" PostTypeId="1" CreationDate="2010-11-02T15:40:33.820" Score="13" ViewCount="1296" Body="&lt;p&gt;I'm a senior undergrad at a top-ish(say, top 15) math school. I'm a solid, not stellar, student. This year I'm taking the qualifying exam grad courses in algebra and analysis and have been taken aback by the &quot;pressure cooker&quot; atmosphere among grad students here. That is, even moreso than in the undergraduate program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm self driven, could going to a &quot;less prestigious&quot; school afford me more space(I mean in a psychological sense) to produce a more solid contribution to math? By &quot;less prestigious&quot;, I mean a school &quot;ranked&quot; significantly lower than the range of schools that I could comfortably get into. For me, &quot;less prestigious&quot; would be ranked around 40-60 on, say, USNews or NRC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is that at such a school, I would be more able to learn the fundamentals at my own pace, as opposed to a pace dictated to me by the program. I know I want to do math, and I think my learning style may be better suited to going at my own pace. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="3030" LastActivityDate="2010-11-04T04:49:05.350" Title="Does it ever make sense NOT to go to the most prestigious graduate school you can get into?" Tags="&lt;soft-question&gt;" AnswerCount="9" CommentCount="9" FavoriteCount="10" />
  <row Id="8653" PostTypeId="2" ParentId="8651" CreationDate="2010-11-02T15:42:17.037" Score="0" ViewCount="0" Body="&lt;p&gt;This is the part of the fundamental theorem that allows you to &lt;em&gt;compute&lt;/em&gt; integrals; then you can compute areas, and with more theory even volumes, surfaces and so on. Exciting enough?&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-11-02T15:42:17.037" CommentCount="3" />
  <row Id="8654" PostTypeId="1" AcceptedAnswerId="8657" CreationDate="2010-11-02T15:43:38.890" Score="4" ViewCount="108" Body="&lt;p&gt;Dear MSE-Community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me summarize the method by which L. Euler solved the Basel Problem and how he found the exact value of $\zeta(2n)$ up to $n=13$. Euler used the infinite product  $$ \displaystyle   f(x) = \frac{sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2}\Big) , $$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities&quot; rel=&quot;nofollow&quot;&gt; Newton's identities&lt;/a&gt; and the (Taylor) &lt;a href=&quot;http://www.wolframalpha.com/input/?i=sin(x)/x&quot; rel=&quot;nofollow&quot;&gt; Series Expansion&lt;/a&gt; (at $x=0$) of the sine function divided by 'x' to arrive at $$ 1 - \frac{x^2}{\pi^2} \cdot (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ... + \frac{1}{n^2}) + x^4(...) = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ... $$ .&#xA;Upon subtracting 'one' from both sides, equating the $x^2$ terms to each other and multiplying both sides by $ - \pi^2$, one finds that $$\zeta(2)=\frac{\pi^2}{6}$$.&#xA;When I first saw this proof and the way it was extended to find the values of the other even zeta-constants, I couldn't help myself thinking: &quot;How could this method be strenghtened to find the values of the &lt;em&gt;odd&lt;/em&gt; zeta-constants?&quot; (And, a little while later, &quot;why hasn't this been done before?&quot;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started looking for a similar-looking infinite product, only now I focussed on one of the form $$ \displaystyle   f(x) = \prod_{n=1}^{\infty} \Big(1-\frac{x^a}{k^3 \cdot q}\Big) $$ (for some $ a \in \mathbb{N} , q \in \mathbb{R} $). A little while later I stumbled upon &lt;a href=&quot;http://mathworld.wolfram.com/InfiniteProduct.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; website and fixated my eyeballs on equation (27). If we take $n=3$, Prudnikov et al. tell us that $$ \prod_{n=1}^{\infty} \Big(1-\frac{x^3}{k^3}\Big) = - \frac{1}{x^3} \cdot \prod_{k=1}^{2} \frac{1}{\Gamma(-e^{2/3 \pi i k} \cdot x)} $$ . Now, I thought that if we could use Newton's Identities again on the left side of the equation and find out what the Taylor Series Expansion of the right-hand side would be, we could find out what the exact value of Apery's Constant and other odd zeta-constants would be. In this &lt;a href=&quot;http://math.stackexchange.com/questions/8522/what-is-the-series-expansion-of-the-following-function&quot;&gt;answer&lt;/a&gt; by Robert Smith, I was told the Series Expansion. So we have $$ 1 - x^3(1 + \frac{1}{8} + \frac{1}{27} + ... + \frac{1}{n^3}) = -1 - 2 \cdot \gamma x - 2 \gamma^2 x^2 + \frac{1}{6}x^3(-8\gamma^3 - \psi^{(2)} (1)) - x^4(...)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that on the left side we only have 'one minus a term with an x^3 coefficient', while on the other side we see 'minus one plus x, x^2, x^3 coefficients with their terms'. This is important, because it probably answers the question why the following will not work, but I don't know why and I really would like to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you know what I will attempt to do now. We equate the x^3 terms with each other, set $x=1$, multiply by minus one and 'find' that $$\zeta(3) = \frac{1}{6}(8\gamma^3 + \psi^{(2)} (1)) $$.&#xA;By combining this with the already known result $$ \zeta(3) = -\frac{1}{2} \psi^{2}(1)$$, we 'find' that $$\zeta(3) '=' \gamma^3$$. Obviously, this is wrong. Apery's constant is larger than one, and this value is clearly smaller than one. Could someone please elaborate one &lt;em&gt;where&lt;/em&gt; I went wrong? And does anybody have any sugguestions and/or ideas related to the discussion from above using which we could find &quot;better&quot; values for Apery's Constant and the other odd zeta constants? (For example by pointing out a similar infinite product relation, and by showing that that infinite product has a nicer Series Expansion?) Or could someone point out to me why this approach to finding nicer closed-form representations for these constants clearly won't lead to any results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Moderators: If you find any spelling mistakes or errors grammar errors, feel free to correct them. To the rest: $\gamma$ is the Euler-Mascheroni Constant, and it amounts to approximately $0.5772$. The $\psi^{(2)}(x)$ stands for the second logarithmic deriviative of the Gamma-function. As usual, Wikipedia is a pretty good reference for this sort of things.)&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-02T16:21:13.477" LastActivityDate="2010-11-02T16:21:13.477" Title="Why is the following evaluation of Apery's Constant wrong and do you have suggestions on how, if at all, this method could be improved?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="8655" PostTypeId="2" ParentId="8651" CreationDate="2010-11-02T15:49:27.373" Score="8" ViewCount="0" Body="&lt;p&gt;It's natural that the Fundamental Theorem of Calculus has two parts, since morally it expresses the fact that differentiation and integration are mutually inverse processes, and this amounts to two statements: (i) integrating and then differentiating and (ii) differentiating and then integrating get us (essentially) back where we started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people have noticed that the two parts are not completely independent though: e.g. if $f$ is continuous, then (ii) follows easily from (i).  However, for discontinuous -- but Riemann integtrable -- $f$, the theorem still holds, and this is what requires an additional argument.  See page 8 of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/243integrals1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/243integrals1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some discussion of this point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't tell from your question how squarely this answer addresses it.  If yes, and you have further concerns, please let me know.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-02T15:55:31.993" LastActivityDate="2010-11-02T15:55:31.993" CommentCount="1" />
  <row Id="8657" PostTypeId="2" ParentId="8654" CreationDate="2010-11-02T16:08:42.200" Score="4" ViewCount="0" Body="&lt;p&gt;Your previous question was about the wrong function.  Instead of $\Gamma(x)$ in the denominator you should have $\Gamma(-x)$.  If you fix this you'll probably end up with the same polygamma identity you already knew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, proving anything about $\zeta(2k+1)$ is known to be quite hard.  If anything simple worked, Euler would have done it, or somebody in the last few centuries anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me also mention that Prudnikov's identity is, if you are willing to accept Euler-style manipulations, trivial.  It is equivalent to a product formula for $\frac{1}{\Gamma(x)}$ which follows (again if you are willing to accept Euler-style manipulations) from an investigation of its roots and does not really tell you anything deep about zeta values.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-02T16:15:45.773" LastActivityDate="2010-11-02T16:15:45.773" CommentCount="6" />
  <row Id="8658" PostTypeId="1" CreationDate="2010-11-02T16:25:51.580" Score="4" ViewCount="187" Body="&lt;p&gt;$f(x) = \frac{1}{\cos x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f'(x) = \frac{\sin(x)}{\cos^2(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f''(x) = \frac{2\sin^2(x)+\cos^2(x)}{\cos^3(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(3)}(x) = \frac{6\sin^3(x)+5\cos^2(x)\sin(x)}{cos^4(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(n)}(x) = \frac{ ?}{cos^{n+1}(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these are easy: &lt;a href=&quot;http://darkwing.uoregon.edu/~jcomes/251exn.pdf&quot; rel=&quot;nofollow&quot;&gt;http://darkwing.uoregon.edu/~jcomes/251exn.pdf&lt;/a&gt;&#xA;Others are not. Why?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T03:08:00.100" LastActivityDate="2010-11-03T03:08:00.100" Title="$f(x) = \frac1{\cos x}$ then $f^{(n)}(x)$ is ...?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8659" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T16:38:55.017" Score="20" ViewCount="0" Body="&lt;p&gt;I think that there are several important points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you will learn just as much from your fellow&#xA;graduate students, especially in the initial years, than&#xA;you will from your professors, and far more from your&#xA;fellow students if there is a good atmosphere for it. Thus,&#xA;I suggest that you try to land at a school with a&#xA;supportive and studious student environment. You can visit&#xA;the places to find out, and talk to graduate students&#xA;there. From this perspective, it is very helpful to be at a&#xA;place that has many talented graduate students---they will&#xA;help you rise to their level (or you will all rise&#xA;together). Since the top schools tend to have stronger&#xA;students, this can be a good reason to go to a top school&#xA;when it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, most of the pressure on graduate students is&#xA;self-imposed, instantiating their drive to do well&#xA;mathematically. Every school, including the top programs,&#xA;have some students that proceed at a different pace. So you&#xA;can often resist whatever external pressure you imagine is&#xA;there. (One important exception to this is at a school&#xA;where financial support might be withdrawn for slow&#xA;progress---so look into that at the places in which you are&#xA;interested.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third---and actually I find this to be the most important&#xA;point---you shouldn't look at the school only and make a&#xA;such an important life decision based only on mathematics&#xA;and prestige. Rather, look at all aspects of how the choice&#xA;of a school will affect your life. You must consider the&#xA;city and region as well. For example, do you prefer city&#xA;living or country living? If you like the mountains, hiking&#xA;and snow, then don't go to Florida, and similarly, if big&#xA;city living is your preference, then don't go to small&#xA;town. I find that this kind of consideration is oddly often&#xA;neglected among mathematics students.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-11-02T16:38:55.017" CommentCount="2" />
  <row Id="8660" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T16:41:12.643" Score="12" ViewCount="0" Body="&lt;p&gt;There are circumstances that could lead one to be admitted to more prestigious graduate program X and less prestigious graduate program Y and choose to attend Y.  For instance, there was a graduate student at UGA a few years ago who was doing well in the program but at some point revealed to me that he had to care for a sick family member in a different state, so moved to that state and enrolled in a distinctly less prestigious program there.  Of course I don't disagree with his decision -- indeed I admire it -- but there was no question that this was being done for higher reasons than his mathematical well-being and future career.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treating this question as advice, I would say &lt;strong&gt;No&lt;/strong&gt; -- you should attend the most prestigious graduate program you can get into.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, graduate programs are not linearly ranked: i.e., holding up two graduate programs to each other, one may not be able to identify one as more prestigious than the other.  When this is possible, it is usually the case that one is &lt;em&gt;significantly&lt;/em&gt; more prestigious.  So, to refine the advice of the previous paragraph: you should choose a maximal element of the set of graduate schools you get admitted to, partially ordered by prestige.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Included in this are considerations such as the following: you strongly believe that you want to work with a particular faculty member at a program which is overall less prestigious than some other.  This is a tricky decision that I don't want to enter into here in detail, except to say: if there is any doubt in your mind that you will complete a PhD under Professor X, be very skeptical about trading more than a little in the prestige of the school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the OP's case, s/he seems to want to choose a non-maximal element in order to take some of the pressure off.  Although I understand the sentiment, I advise strongly against this.  Two points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Graduate school is a high-pressure, hard-slog situation for everyone.  It has to be.  If you're not fully prepared for that, perhaps you're not fully prepared for a PhD program: think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Much of graduate school &lt;em&gt;does&lt;/em&gt; proceed at your own pace.  However, the quality of the students and faculty in your program sets an example and a standard, in my experience the higher the better.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-02T16:41:12.643" />
  <row Id="8661" PostTypeId="2" ParentId="8658" CreationDate="2010-11-02T16:45:03.753" Score="11" ViewCount="0" Body="&lt;p&gt;This is asking for the $n$-th derivative of the secant function.&#xA;As the derivative of $\sec$ is $\sec\tan$ and that of $\tan$ is&#xA;$\sec^2=1+\tan^2$ then the $n$-th derivative of $\sec$ is&#xA;$\sec f_n(\tan)$ where $f_0(t)=1$ and $f_{n+1}(t)=tf_n(t)+(t^2+1)f_n'(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably too much to hope to find a nice formula for the&#xA;coefficients of the $f_n$. The constant coefficients of $f_{2m}$&#xA;are essentially the Euler numbers.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-02T16:45:03.753" />
  <row Id="8662" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T16:50:11.197" Score="7" ViewCount="0" Body="&lt;p&gt;I really appreciate this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people still hold the out-dated belief that you learn mathematics in the classroom. Sure, you may learn classical formulas, the statement of famous theorems and some folklore surrounding its solver or length of openness (if there is none), but you'll learn &quot;how to be a mathematician&quot; at home or in the library (with [insert your favorite book here]) or &lt;strong&gt;alone&lt;/strong&gt; at the board wrestling with problems. This is the time when you develop your problem solving skills and gauge your limitations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is nice about the better schools is having a (sound) sounding board for your ideas. If your colleagues are sharp, they will provide quick and clean jugular shots to your questions. However, sometimes it's better to think about questions on your own time and settle them yourself with no interference. This simple and time-honored exercise develops your &quot;internal compass&quot; much more than having access to slick solutions from others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, it is sometimes (perhaps often) the case that we do not have the best vantage point to judge the right course of action in a given (perhaps crucial) point in our lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I think: &lt;em&gt;Postpone your question for a little while. Go to the best school that you can get into, and find an advisor that suits you and your tempo. Be honest with him/her about your strengths and weaknesses. Develop an on-going dialogue with your colleagues. Work hard but don't go crazy doing so. Ask questions if you get stuck, but not too many. Be respectful. Listen to people that you hold in high regard (and some that you may not). Think for yourself. Find an interesting problem. Write a good thesis. Graduate. Return to the issue of &quot;solid contribution&quot; when you have a better view of the landscape of your chosen field.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T04:49:05.350" LastActivityDate="2010-11-04T04:49:05.350" CommentCount="2" />
  <row Id="8663" PostTypeId="2" ParentId="8633" CreationDate="2010-11-02T16:50:16.397" Score="1" ViewCount="0" Body="&lt;p&gt;Answering to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I initially started writing a software program that would simulate uniformly random dice rolls, but then thought about how that wouldn't be realistic because dice get tumbled and smoothed before reaching the consumer (which is why some people have &quot;favorite dice&quot; because they roll 20s often). I realized that I could approximate and fake out the realism, but I got ridiculously curious about how it all actually worked. So, to answer, I'm not measuring real physical objects, but trying to determine how to realistically simulate the rolling of virtual dice.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you could do is the following. Start with a perfect dice ($p_1 = p_2 = \dots = p_6$). Then, each time you roll it, you can alter the probabilities $p_i$ with an error $\varepsilon_i$. Thus, the probabilities would not be the same in the future. This would simulate the damaging of the dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $p_1 = p_2 = \dots = p_6 = \frac{1}{6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generate 5 random errors $\varepsilon_1, \varepsilon_2, \dots, \varepsilon_5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute $ \varepsilon_6 = - \varepsilon_1 - \varepsilon_2 - \varepsilon_3 - \varepsilon_4 - \varepsilon_5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set new probabilities to $p_1 + \varepsilon_1, p_2 + \varepsilon_2, \dots, p_5 + \varepsilon_5, p_6 + \varepsilon_6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roll the dice with this new probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute new errors...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I strongly suggest that you choose the way of generating the $\varepsilon_i$ so that they will always be 0 or as near to 0 as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fine tuning purpose, it would be good if it was not always $p_6$ that gets the difference of all other errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method can easily be adjusted to dices with any number of faces.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-02T16:50:16.397" CommentCount="2" />
  <row Id="8664" PostTypeId="2" ParentId="8658" CreationDate="2010-11-02T16:51:11.867" Score="1" ViewCount="0" Body="&lt;p&gt;That depends on what you mean by &quot;easy.&quot;  As far as the examples in the .pdf you link to, I claim that the following are true for any reasonable definition of &quot;easy&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives of powers and logarithms.  &lt;/li&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives of solutions to homogeneous linear ODEs.&lt;/li&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives of $x^n f(x)$ if it is easy to compute the iterated derivatives of $f(x)$.&lt;/li&gt;&#xA;&lt;li&gt;It is easy to compute the iterated derivatives $f(x) + g(x)$ if it is easy to compute the iterated derivatives of $f(x)$ and of $g(x)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think that covers all of them.  Otherwise, there is no general reason to expect that it should be easy to compute iterated derivatives.  Even at $x = 0$ they may define complicated and highly nontrivial sequences, for example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt; (which are related to this problem).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-02T16:51:11.867" />
  <row Id="8665" PostTypeId="2" ParentId="8561" CreationDate="2010-11-02T16:58:24.710" Score="0" ViewCount="0" Body="&lt;p&gt;It is known that (a+b)!/a!b! represents the number of combinations of (a + b) elements, taken &quot;b to b&quot;. Therefore, it is an integer.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-02T16:58:24.710" CommentCount="1" />
  <row Id="8666" PostTypeId="2" ParentId="8650" CreationDate="2010-11-02T17:00:59.453" Score="7" ViewCount="0" Body="&lt;p&gt;This is equal to $n!$ minus the number of permutations which keep no elements fixed.  These are known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow&quot;&gt;derangements&lt;/a&gt; and you can find all the relevant formulas at the Wikipedia article.  You can count them, for example, using inclusion-exclusion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want the number of derangements of an $n$-element set, then a nice way to compute it is to round $\frac{n!}{e}$ to the nearest integer.  For example, when $n = 5$ we want to round $\frac{120}{e} \approx 44.1$, so there are $44$ derangements, hence $120 - 44 = 76$ permutations which fix at least one element.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-02T17:00:59.453" CommentCount="1" />
  <row Id="8667" PostTypeId="1" AcceptedAnswerId="8728" CreationDate="2010-11-02T17:05:42.457" Score="10" ViewCount="135" Body="&lt;p&gt;Suppose $X$ is a space and $A_1\subseteq A_2\subseteq A_3\subseteq ...\subset X$ is a sequence of subspaces each of which is closed in $X$ and such that $X\cong \varinjlim_{n}A_n$ (i.e. $U$ is open in $X$ if and only if $U\cap A_n$ is open in $A_n$ for each $n$). This topology on $X$ has many names (direct limit, inductive limit, weak topology, maybe more) but I can't seem to find much dealing with separation properties in this general setting. Specifically, I am asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A_n$ is Hausdorff for each $n$, then must $X$ also be Hausdorff?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastActivityDate="2010-11-03T19:16:50.850" Title="Separation in direct limits of closed inclusions" Tags="&lt;general-topology&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="8668" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T17:14:25.197" Score="4" ViewCount="0" Body="&lt;p&gt;There are many good comments above, I'd like to add a few other points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prestige is a fairly abstract, borderline meaningless word.  It's meant to be an amalgamam of other, &lt;em&gt;actually important&lt;/em&gt; ideas.  And those concepts are the things one really ought to pay attention to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when you go to grad school, one thing to keep in mind is &quot;will university X give me the time it takes to do substantial research?&quot;.  It's a good thing to go to a school that can promise you long-term funding with enough teaching work so that you get familiar with the process, but not so much that it distracts substantially from your graduate studies.  Depending on how strictly you use this criterion, you'll find this may narrow down the list of potential graduate schools to something pretty close to the traditional &quot;prestigious&quot; university list.  The above is concerned with the component of prestige associated to &quot;does the university have ample research funding for grad students?&quot; Even among the &quot;top&quot; universities you'll find this varies pretty widely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to keep in mind is much of the point of going to grad school is to be exposed to as many ideas from diverse areas of mathematics as possible, as most undergraduate programs rarely get beyond the 19th century in mathematical understanding. Smaller universities tend to have less diverse faculty members, less grad students and postdocs, sometimes their teaching load is higher.  The net effect is that if you're at a smaller, more isolated place you're less likely to go to some wonderful talk on a random topic you didn't know anything about.  It's amazing how quickly one can pick up new ideas by going to good random talks, ideas that affect how you look at other subjects, the interconnections between fields, applications to other areas of life, an outsider's perspective on &quot;your&quot; field, and so on.  So going to a &quot;big&quot; place is very useful in that regard. So this is the component of prestige associated to being a large department/university. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course other components to prestige, but the above are two fairly substantial ones.&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-02T19:23:08.137" LastActivityDate="2010-11-02T19:23:08.137" />
  <row Id="8669" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T17:17:44.173" Score="4" ViewCount="0" Body="&lt;p&gt;I think realistically people are going to be trying really hard at even a mid-level PhD program; they just will not have gotten into better schools. They might not be &lt;em&gt;as&lt;/em&gt; intense as those at higher-ranked places, but I seriously doubt things will be at all slow-paced. And I think if you are better prepared than them you may have trouble relating to the other students when going through the various travails of graduate school, which may make it harder to bond with them. And if they're not as strong they won't be useful people to learn from. Not to mention the result of having a thesis advisor at a lesser place, and the greater difficulty being on the job market from a lower-ranked place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion would be to go to one of the best places you get into, and then try to work your life around being a grad student there. Find friends who are less miserable, try to not get too stressed out over exams, and so on. Also, once exams are over, no matter where you go things tend not to be that stressed. You just are more acquainted with the ones who are going through that process.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-02T17:17:44.173" CommentCount="5" />
  <row Id="8670" PostTypeId="2" ParentId="8620" CreationDate="2010-11-02T17:28:42.347" Score="0" ViewCount="0" Body="&lt;p&gt;$\rm R = \{0,a,2a,\cdots,(p-1)a\}\ $ for $\rm\ 0 \ne a\in R\:$. So $\rm\ a^2 = k\:a\ $ and $\rm\ k \ne 0\:$ via multiplication nontrivial. Now $\rm\ j = 1/k\ \:mod\ p\ \Rightarrow\ ia\ ja\ = \ ijka\ =\ ia\ \Rightarrow\ ja = 1\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-02T17:28:42.347" />
  <row Id="8671" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T17:47:16.517" Score="7" ViewCount="0" Body="&lt;p&gt;Maybe you would like an answer from someone who is there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to undergrad at a &quot;Top 100&quot; school. I did extremely well there, but utterly failed to get into the level of grad schools I was aiming at, and am now a grad student at a &quot;Top 50&quot; school. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only answer based on what I've seen at these schools, so maybe things are different elsewhere. But to me, the situation is kind of sad here. The students know so little and don't seem to really have the right attitude. I think I'm missing a big component of interacting with peers who are learning the same material. I can honestly say I've never had any sort of study session or even just a discussion about mathematics with fellow students that was productive for me. (Although this may more reflect my experience at my undergrad school... I mostly talked to the grad students there, and this was true there, but I'm still new at my graduate school and don't know many people, so maybe this will change.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never been to one of these prestigious schools, but seeing people online talk about them, it seems like a completely different world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, it's nice because I can stand out here. But I can stand out here without even trying very hard. I think this leads to more of a complacent attitude on my part, which is not good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder, what would happen if I were at a top 10 school? I honestly don't know. I might get massacred. I certainly have a much weaker background than entering students at those schools seem to have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the career issue. As far as I can tell, very few students graduating here end up in a tenure track job at the kind of university I would like to be at. It seems like a big challenge to overcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not all negative. I suppose it is nice to not have so much pressure. There are some strong professors who do interesting work I might like to work with. But if you're ambitious (and why get a PhD in math if you're not ambitious?), I wouldn't recommend coming to this type of school if you have other options. &lt;/p&gt;&#xA;" OwnerUserId="3036" LastActivityDate="2010-11-02T17:47:16.517" CommentCount="1" />
  <row Id="8672" PostTypeId="1" AcceptedAnswerId="8700" CreationDate="2010-11-02T17:51:45.140" Score="5" ViewCount="370" Body="&lt;p&gt;Hello.&lt;br&gt;&#xA;Lets say I have a 2 x 2 Matrix (Actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Structure_tensor&quot; rel=&quot;nofollow&quot;&gt;Structure Tensor&lt;/a&gt; of a Discrete Image - I):&#xA;$$ \begin{bmatrix}&#xD;&#xA; \frac{\partial I}{\partial x}\frac{\partial I}{\partial x} &amp;amp; \frac{\partial I}{\partial x}\frac{\partial I}{\partial y} \\\ &#xD;&#xA;    \frac{\partial I}{\partial y}\frac{\partial I}{\partial x} &amp;amp; \frac{\partial I}{\partial y}\frac{\partial I}{\partial y}&#xD;&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has 2 properties:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Symmetric.&lt;/li&gt;&#xA;&lt;li&gt;Positive Semidefinite.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given those properties, What would be the easiest method to compute (Numerical) its Eigen Vectors (Orthogonal) and Eigen Values?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditorDisplayName="" LastEditDate="2010-12-25T09:57:56.743" LastActivityDate="2010-12-25T09:57:56.743" Title="Eigenvalues and Eigenvectors of 2 x 2 Matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8673" PostTypeId="2" ParentId="8672" CreationDate="2010-11-02T18:03:44.650" Score="5" ViewCount="0" Body="&lt;p&gt;Since you work with a $2\times2$ matrix, the corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_polynomial&quot; rel=&quot;nofollow&quot;&gt;characteristic  polynomial&lt;/a&gt; is quadratic so the eigenvalues can be expressed in closed form in terms of the matrix elements.&#xA;As soon as you get the eigenvalues all you have to do is to solve two systems of linear equations with two unknowns (i.e. the coordinates of the corresponding eigenvector in the standard basis). Again, everything is nice and explicit, no need to apply a fancy numerical method.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-02T18:16:36.883" LastActivityDate="2010-11-02T18:16:36.883" CommentCount="3" />
  <row Id="8674" PostTypeId="1" AcceptedAnswerId="8686" CreationDate="2010-11-02T18:25:20.070" Score="5" ViewCount="149" Body="&lt;p&gt;Consider a situation with a bag with infinity number of balls. Each ball is of some color. Number of colors is finite but it is not known. Balls are drawn from the bag one by one and checked for the color.&#xA;We want to stop drawing balls when there is small probability that we will find color that is not already drawn.&#xA;Or more exact, we want to be very sure (with large probability) that there is small portion of balls in bag with colors that are not already drawn.&#xA;Is there a statistical model to describe when to stop?&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2010-11-03T00:54:20.703" Title="Bag with infinite number of colored balls" Tags="&lt;probability-theory&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8675" PostTypeId="2" ParentId="5166" CreationDate="2010-11-02T18:37:10.093" Score="-1" ViewCount="0" Body="&lt;p&gt;Answer given above is totally wrong&lt;/p&gt;&#xA;" OwnerUserId="3039" LastActivityDate="2010-11-02T18:37:10.093" CommentCount="2" />
  <row Id="8676" PostTypeId="2" ParentId="8658" CreationDate="2010-11-02T18:41:24.647" Score="1" ViewCount="0" Body="&lt;p&gt;You might want to check out Faa di Bruno's formula: &lt;a href=&quot;http://mathworld.wolfram.com/FaadiBrunosFormula.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/FaadiBrunosFormula.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you have $f(x) = g(h(x))$ where $g(x) = 1/x$ and $h(x) = \cos(x)$.  The answer is not simple unfortunately, and involves some combinatorics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, notice $f(x) \cos(x) = 1$, and thus by taking the $m$th derivative of both sides we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^m {m \choose k} f^{(k)}(x) \left( \frac{d^{m-k}}{dx^{m-k}} \cos(x) \right) = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $m \geq 1$.  Thus, you can view the vector $(f(x),f'(x),\dots,f^{(n)}(x))$ as the solution to the system of linear equations above with $m=0,1,\dots,n$.  You can invert this matrix numerically, if that if your interest.  A closed form might be possible too. &lt;/p&gt;&#xA;" OwnerUserId="2515" LastActivityDate="2010-11-02T18:41:24.647" />
  <row Id="8677" PostTypeId="2" ParentId="8651" CreationDate="2010-11-02T18:44:29.200" Score="4" ViewCount="0" Body="&lt;p&gt;The names &quot;first&quot; and &quot;second&quot; for the two parts of the theorem are meaningless.  More correct names would be &lt;em&gt;existence&lt;/em&gt; and &lt;em&gt;uniqueness&lt;/em&gt;.  It also is not unreasonable to separate the uniqueness statement from the formula relating definite integrals to antiderivatives, which is an algebraic &lt;em&gt;consequence&lt;/em&gt; of the (analytic) uniqueness statement.  The formula could be considered as a third part of the theorem, but numbering pieces of a theorem in a particular order is an uninformative nomenclature -- as is calling theorems &quot;fundamental&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamental theorem of calculus asserts existence and uniqueness of antiderivatives (solutions of the differential equation $y' = f(x)$ with given value of $y(x_0)$ at one point).  Apart from purely logical considerations there are several reasons the uniqueness theorem is important.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Indefinite integrals of the form $\int_p^x f(t) dt$, which are what appear in most presentations of the existence part of the theorem, in some cases do not account for all antiderivatives of $f(x)$ as the basepoint $p$ is varied over all real numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the more precise presentation $y(x) = y(a) + \int_a^x f(t) dt$ there is still the possibility that other processes, even more magical than integration, might be related to anti-differentiation.  So it is of interest to either find these exotic species, or show that integrals give everything. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An explicit analysis of uniqueness becomes more pressing when integrating functions with singularities, as in $\int dx/|x|^p$ for $p=1$ and $p=1/2$ (the number of integration constants changes, so this is needed for writing down solution formulas in full generality).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the algebraic formula implied by uniqueness, $\int_a^b f = F(b)-F(a)$, is important both as a means of computing integrals and as the basis of the notation supporting changes of integration variable (substitutions).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-02T19:38:05.627" LastActivityDate="2010-11-02T19:38:05.627" />
  <row Id="8678" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T18:48:22.187" Score="9" ViewCount="0" Body="&lt;p&gt;I happened to have gone to a toppish school (look up my profile if you really care which one). Even though I didn't realize it at the time, there are distinct advantages from the mathematical environment there. But I also think you shouldn't base your decision on graduate schools purely on artifical rankings of prestige. If you are deciding between schools, &lt;em&gt;you have to visit them to get a sense of the environment there&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike the undergraduate degree, if you do a graduate degree at a reasonably strong place, you will be spending 4, 5, or maybe 6 years with a close-knit group of people that number from 20-30 to 100+. It is where you will get to know a lot of future colleagues in your field, it is where you pick up new ideas from passing discussions, and it is where you learn how to do research. And the most important factor that goes into your decision of which graduate school to go to is &lt;strong&gt;the people&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to find out if you can work with the faculty there: is there anyone there in your field? Is his or her interest something you want to study? Are they taking students? How do they treat their students? All these are things you should talk to people, especially current students, about when you visit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to find out if you have fellow students: you don't want to be the only student in the department doing (for example) PDEs when everyone else does algebraic geometry. The students whose research interests are closely related to yours will be the ones you discuss most with (and see most of); they will be the ones you ask for help and they will be your future colleagues. Find out if they have informal student seminars. Arrange your trip so that you can sit in on some of them. Having a couple more senior students who are willing to give you sound advice when you are just starting out will be very useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really must make a decision without first hand experience visiting, then so be it: higher ranked schools tend to have a larger, more diverse pool of faculty. They also tend to attract the better students. So that's where I will put my money were I to bet blindly. But I cannot stress enough the importance of finding out whether the school and environment matches you first hand. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A little side remark about &quot;more room psychologically&quot;: I would argue that that is actually counter-productive. When I was in graduate school there was significant amount of down time where the research just doesn't go well. Looking back, I am really glad to have had my large and rowdy office: at least one of us will have made some progress at any given time. And with seminars happening now and then, one can at least learn something useful through discussions when the thesis project is stuck. Were not for all these &quot;distractions&quot;, I fear that the ennui caused by the glacial pace of my thesis research may have led to psychological problems. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-02T18:48:22.187" CommentCount="2" />
  <row Id="8679" PostTypeId="2" ParentId="8609" CreationDate="2010-11-02T18:53:40.690" Score="4" ViewCount="0" Body="&lt;p&gt;Say $k$ has the prime factorization $p_1^{r_1} p_2^{r_2} \cdots p_n^{r_n}$. Then let $H(k)$ be the &quot;hyperbolicity&quot; of this integer, $H(k) = \sum_{i=1}^n p_i^{-r_i}$. Hyperbolic integers are those integers $k$ that have $H(k) &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $H(k)$ for some &quot;random&quot; integer $k$ looks like a sum of independent random variables $X_p$, one for each prime $p$, where $P(X_p = p^{-k})$ is the probability that a random integer is divisible by $p^k$ but not $p^{k+1}$. Thus $P(X_p = p^{-k}) = (p-1)/p^{k+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular &#xA;$$ E(X_p) = \sum_{k \ge 1} p^{-k} P(X_p = p^{-k}) = \sum_{k \ge 1} {p-1 \over p^{2k+1}} = {1 \over p^2 + p} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get a bit tricky in finding the distribution of $\sum_p X_p$; in particular a central limit theorem won't work, because the contribution of each of the small primes is too large.  But there is some limiting distribution, and I suspect it at least bears some resemblance to the limiting distribution of the actual hyperbolicities of integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, my tests seem to point to a natural density of hyperbolic integers somewhere in the neighborhood of 99 percent. I'd conjecture that integers with hyperbolicity less than $k$ have a natural density.  There are similar results on the distribution of $\sigma(n)/n$ where $\sigma(n)$ is the sum of divisors of $n$; see for example Marc Deleglise, &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.em/1048515661&quot; rel=&quot;nofollow&quot;&gt;Bounds for the density of abundant integers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-02T18:53:40.690" CommentCount="1" />
  <row Id="8681" PostTypeId="2" ParentId="8652" CreationDate="2010-11-02T19:39:50.453" Score="11" ViewCount="0" Body="&lt;p&gt;Maybe this is just getting redundant, but I think this is a slightly different point from what people has said above:  while there do exist compelling reasons to go to grad schools that don't maximize prestige, it really doesn't sound to me like you have put forth any.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think on the whole grad students are pretty naive about how big an influence where they go to grad school has on their opportunities after graduating.  Of course, lots of people do manage to go to less fancy schools and go on to have good careers, but if you look at the schools in the 40-60 range you mention, most of their faculty went to more prestigious places.  For example, at Oregon where I'm a professor (US News rank #56), fully half the professors who got Ph.D.s in the US got them at Harvard, MIT, Berkeley or Stanford (all in the top 5 on the USN ranking) and most of the rest at institutions in the low 2 digits (Texas, Maryland, Wisconsin, Penn).  Some of this is for unfair reasons (bias, etc.), some of this is selection reasons, and some is because going to a better school give you more opportunities to be a better mathematician (more exposure to new ideas, more visitors, higher quality peers).  I wouldn't give up on those things because of a nebulous fear of not liking the environment.  Try it out; it's much easier to go to a Berkeley or Michigan for a year and then transfer somewhere else than the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Just as an addendum, I thought I should emphasize here that like Pete said, prestige isn't really well ordered.  There are lots of pairs where it's hard to say really conclusively which is more prestigious, in which case you should go with your gut.  So I'm not talking about things like comparing Chicago and Columbia, and choosing Columbia because you'd rather be in New York, but rather the difference between going to graduate school in the 40-60 range (in USNWR or NRC), and something in, say, the top 20.&lt;/p&gt;&#xA;" OwnerUserId="3041" LastEditorUserId="3041" LastEditorDisplayName="" LastEditDate="2010-11-03T14:16:38.503" LastActivityDate="2010-11-03T14:16:38.503" />
  <row Id="8682" PostTypeId="2" ParentId="7569" CreationDate="2010-11-02T19:42:54.723" Score="0" ViewCount="0" Body="&lt;p&gt;My answer on MO:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments &lt;a href=&quot;http://mathoverflow.net/questions/43326/sharp-upper-bounds-for-sums-of-the-form-sum-p-mid-k-frac1p1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), for any set of $n$ positive integers {$a_{1}, \dots, a_{n}$} such that $\sum_{i = 1}^{n} \frac{1}{a_{i}} &amp;lt;1$, we have&#xA;\begin{eqnarray}&#xA;\sum_{i = 1}^{n} \frac{1}{a_{i}} \leq \sum_{i = 1}^{n} \frac{1}{d_{i}} = \frac{d_{n+1} - 2}{d_{n+1} -1} &amp;lt; 1,&#xA;\end{eqnarray}&#xA;where $d_{i}$ is the $i^{\text{th}}$-Euler number, which satisfies the quadratic recurrence $d_{i} = d_{1} \cdots d_{i-1} + 1$ with $d_{1} = 2$. The first few terms of the sequence are 2, 3, 7, 43, 1807....  (A000058). It is relatively straightforward to show that the aforementioned recurrence is equivalent to $d_{i} = d_{i-1}(d_{i-1}-1) + 1$, and it is known from the recurrence that $d_{i} = \lfloor \theta^{2^{i}} + \frac{1}{2} \rfloor$, where $\theta \approx 1.2640$.... Thus,&#xA;\begin{eqnarray}&#xA;\sum_{p \mid k} \frac{1}{p + 1} \leq \sum_{i = 1}^{\omega(k)} \frac{1}{d_{i}} = \frac{d_{\omega(k) + 1} - 2}{d_{\omega(k)+1} - 1} = \frac{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 2}{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 1} &amp;lt; 1.&#xA;\end{eqnarray}&#xA;One can also show that $d_{i} - 1 = \lfloor \vartheta^{2^{i-1}} - \tfrac{1}{2} \rfloor$, where where $\vartheta \approx 1.5979$...., so we have&#xA;\begin{eqnarray}&#xA;1 - \sum_{p \mid k} \frac{1}{p + 1} \geq 1 - \frac{d_{\omega(k)+1} - 2}{d_{\omega(k)+1} - 1} = \frac{1}{d_{\omega(k)+1} - 1} = \lfloor \vartheta^{2^{\omega(k)}} - \tfrac{1}{2} \rfloor^{-1} &gt; 0.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt; E. Deutsche points out that the sequence {$d_{i} - 1$} (A007018) counts the number of ordered rooted trees with out-going degree up to 2 with all leaves at top level.  &lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-02T23:12:17.947" LastActivityDate="2010-11-02T23:12:17.947" />
  <row Id="8683" PostTypeId="1" AcceptedAnswerId="8696" CreationDate="2010-11-02T19:48:21.317" Score="5" ViewCount="78" Body="&lt;p&gt;A closed curve on a surface can be represented by the boundary it intersects. Suppose the original surface is created by glueing a to A, b to B etc. Pick an orientation, each time it intersects a boundary, add the boundary to the representation. For example in the image, it's &#xA;&lt;img src=&quot;http://i.imgur.com/uNKZX.png&quot; alt=&quot;alt text&quot;&gt;&#xA;The black dot is the starting point to trace the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of letters is called a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of a word is defined as the amount of letters in the word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A word W represents a simple closed curve if there exist a simple closed curve that can be represented by W. Let's call those words simple words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given surface, are there algorithms to find simple words with length up to L in polynomial time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive algorithm is to generate every word and test which ones are simple, but that is $O(c^L)$ for some constant c. The amount of simple word is $O(L^k)$ for some constant k. As shown in &lt;a href=&quot;http://www.math.princeton.edu/~mmirzakh/data/Growth.pdf&quot; rel=&quot;nofollow&quot;&gt;Growth of the number of simple closed geodesics on hyperbolic surfaces&lt;/a&gt; by Maryam Mirzakhani. The bound given in that paper can be converted into the bound on words.&lt;/p&gt;&#xA;" OwnerUserId="96" LastActivityDate="2010-11-03T09:42:54.867" Title="Algorithms to find simple closed curves on a surface" Tags="&lt;topology&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8684" PostTypeId="1" AcceptedAnswerId="8704" CreationDate="2010-11-02T19:48:25.807" Score="3" ViewCount="113" Body="&lt;p&gt;I know how to generate solutions to $x^2-Dy^2=\pm 1$ from the fundamental solution to $x^2-Dy^2=\pm 4$.  But how do we know that &lt;strong&gt;all&lt;/strong&gt; the solutions are generated from this fundamental solution?&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="2505" LastEditorDisplayName="" LastEditDate="2010-11-02T21:33:08.523" LastActivityDate="2010-11-03T01:56:28.973" Title="Generalized Pell's Equation: Why is 4 special?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8685" PostTypeId="2" ParentId="8334" CreationDate="2010-11-02T20:10:16.857" Score="1" ViewCount="0" Body="&lt;p&gt;I just read &quot;Generalized Feedback Shift Register Pseudorandom Number Algorithm&quot; by T. G. Lewis and W. H. Payne. &#xA;I think that paper settles the question I was raising (going to the source, right?). In essence, the question is &#xA;&quot;What is the correct procedure to use the Generalized Feedback Shift Register Algorithm (GFSR)?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.- Start with a sequence and a primitive polynomial $x^{p}+x^{q}+1$. &#xA;For example, $a_{0}=a_{1}=a_{2}=a_{3}=a_{4}=1$ and $x^{5}+x^{2}+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.- Elements of the sequence follow $a_{k}=a_{k-p+q}\bigoplus a_{k-p}$ with $k=p, p+1,...$. &#xA;In this example, since we have the first 5 elements of the sequence and according to the polynomial, we are given that $p=5, q=2$.&#xA;Therefore, we can know the next elements of the sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix} a_{6}=a_{3}\bigoplus a_{1}=0 \\ a_{7}=a_{4}\bigoplus a_{2}=0 &#xA;\\ a_{8}=a_{5}\bigoplus a_{3}=1 \\ a_{9}=a_{6}\bigoplus a_{4}=1 \\ ... \\ \end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this way we construct the rest of the sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a_{i}\}_{0}^{30}={1111100011011101010000100101100}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to produce a better random sequence, we apply Kendall's algorithm. Although there are several variations of Kendall's algorithm, the point is to shift the original sequence $1111100011011101010000100|101100$ forwards by 6 bits, that is, $1011001111100011011101010|000100$.&#xA;And again three times more (until we are back with the original sequence). This process gives the following sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;\text{Key} &amp;amp; \text{Sequence} \\&#xA;0 &amp;amp; \|11111\|00011011101010000100|101100\\&#xA;1 &amp;amp; 1011001111100011011101010|000100\\&#xA;2 &amp;amp; 0001001011001111100011011|101010\\&#xA;3 &amp;amp; 101010000100101100111100|011011\\&#xA;4 &amp;amp; 0110111010100001001011001|111100&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we take n-tuples (in this example, 5-tuples are used) which are positioned as the columns of a new array:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix}&#xA;W_{0}: &amp;amp; \|1\|1010 &amp;amp; W_{10}: &amp;amp; 01001&amp;amp; W_{20}: &amp;amp; 00111\\&#xA;W_{1}: &amp;amp; \|1\|0001 &amp;amp; W_{11}: &amp;amp; 10000&amp;amp; W_{21}: &amp;amp; 01111\\&#xA;W_{2}: &amp;amp; \|1\|1011 &amp;amp; W_{12}:&amp;amp; 10110&amp;amp; W_{22}: &amp;amp; 10010\\&#xA;W_{3}: &amp;amp; \|1\|1100 &amp;amp; W_{13}:&amp;amp; 10100&amp;amp; W_{23}: &amp;amp; 01100\\&#xA;W_{4}: &amp;amp; \|1\|0011 &amp;amp; W_{14}:&amp;amp; 01110&amp;amp; W_{24}: &amp;amp; 00101\\&#xA;W_{5}: &amp;amp; 00001 &amp;amp; W_{15}:&amp;amp; 11111&amp;amp; W_{25}: &amp;amp; 10101\\&#xA;W_{6}: &amp;amp; 01101 &amp;amp; W_{16}:&amp;amp; 00100&amp;amp; W_{26}: &amp;amp; 00011\\&#xA;W_{7}: &amp;amp; 01000 &amp;amp; W_{17}:&amp;amp; 11000&amp;amp; W_{27}: &amp;amp; 10111\\&#xA;W_{8}: &amp;amp; 11101 &amp;amp; W_{18}:&amp;amp; 01011&amp;amp; W_{28}: &amp;amp; 11001\\&#xA;W_{9}: &amp;amp; 11110 &amp;amp; W_{19}:&amp;amp; 01010&amp;amp; W_{29}: &amp;amp; 00110&#xA;\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $W_{i}$ is called a 'word'. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Since each column obeys the recurrence $a_{k}=a_{k-p+q}\bigoplus a_{k-p}$, each word must also obey&#xA;    $W_{k}=W_{k-p+q}\bigoplus W_{k-p}$.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know, that's the correct procedure for using the GFSR algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections or comments will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-02T20:25:20.667" LastActivityDate="2010-11-02T20:25:20.667" />
  <row Id="8686" PostTypeId="2" ParentId="8674" CreationDate="2010-11-02T20:12:34.703" Score="3" ViewCount="0" Body="&lt;p&gt;This problem often comes up in Biology when one is interested in estimating the number of species in an area based on some survey. There are a wide variety of methods developed to suit the specific application in hand. For example, it is very hard to come up with a good estimator for the number of biological species in a rain forest since there are too many &quot;extremely rare&quot; species to take into account. There is no one unique estimator that is globally optimal and you would have to customize your own estimator based on your needs. A good starting point is this &lt;a href=&quot;http://biomet.oxfordjournals.org/content/63/3/435.abstract&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; and references therein.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-11-02T20:12:34.703" CommentCount="1" />
  <row Id="8687" PostTypeId="2" ParentId="8684" CreationDate="2010-11-02T20:33:33.727" Score="3" ViewCount="0" Body="&lt;p&gt;I had written out an answer previously, but to make it work I need to assume a little more.  Namely, let me assume that $D$ is &lt;strong&gt;not&lt;/strong&gt; congruent to $1 \pmod 4$.  (I'll think a little more about what's going on in general, and someone else is of course welcome to jump in and help me out.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the running assumptions of the Pell equation is that $D$ is squarefree.  In particular, it is not $0 \pmod 4$.  So if you reduce the equation $x^2 - Dy^2 = \pm 4$ &#xA;modulo $4$, you get $x^2 - Dy^2 \equiv 0 \pmod 4$.  Since the squares mod $4$ are $0$ and $1$, running through the four possibilities for $x^2$ and $y^2$ mod $4$ one sees that the only solutions are when $x^2 \equiv y^2 \equiv 0$, i.e., when $x$ and $y$ are even.  [N.B.: here I am using that $D \not \equiv 1 \pmod 4$; otherwise $1^2 - D \cdot 1^2 \equiv 0 \pmod 4$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, every integral solution to $x^2 - Dy^2 = \pm 4$ occurs with $x$ and $y$ even, so we may write $x = 2X$, $y = 2Y$ and substitute, getting &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2X)^2 - D (2Y)^2 = 4(X^2 - DY^2) = \pm 4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $4$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^2 - DY^2 = \pm 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process can also be reversed: if $X^2 - DY^2 = \pm 1$, then $(2X)^2 - D(2Y)^2 = \pm 4$.  Thus we have found a bijection between the sets of solutions to $x^2 - Dy^2 = \pm 4$ and $X^2 - DY^2 = \pm 1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is the most direct answer to your question.  One can still wonder whether this occurs for integers other than $4$, but for other $k &gt; 1$ it can happen that there are solutions to $x^2 - Dy^2 = \pm k$ in which $x$ and $y$ are &lt;em&gt;not&lt;/em&gt; both divisible by $k$.  Try it and see.  So $4$ is special here because the theory of quadratic equations modulo $4$ is especially simple.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: See &lt;a href=&quot;http://sites.google.com/site/tpiezas/008&quot; rel=&quot;nofollow&quot;&gt;http://sites.google.com/site/tpiezas/008&lt;/a&gt; for some further information on going from $x^2 - dy^2 = \pm 4$ to $x^2 - dy^2 = \pm 1$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-03T01:21:00.360" LastActivityDate="2010-11-03T01:21:00.360" CommentCount="1" />
  <row Id="8688" PostTypeId="1" AcceptedAnswerId="8729" CreationDate="2010-11-02T20:50:16.143" Score="1" ViewCount="149" Body="&lt;p&gt;I am taking my first real-analysis course this year and have a midterm on friday. I have NO idea what is going on in this class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if anyone could clarify what the following means.&#xA;The book starts off with talking about Archimedes and his principle which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The value of an infinite series, if it&#xA;  exists, is the number T such that&#xA;  given any rational numbers L and M&#xA;  such that L &amp;lt; T &amp;lt; M, all of the finite&#xA;  sums from some point on will be&#xA;  strictly contained in the interval&#xA;  between L and M.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book then goes on about talking about series that do not converge (ie diverges and how hard it is to get a value for that). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the book has the following definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An infinite series converges if there is a target value T so that for any L &amp;lt; T and any   M &gt; T, all of the partial sums from some point on are strictly between L and M.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I have no idea what this means. What does the interval mean? I am more confused about L and M and the interval then I am about the series. Why are they giving me an interval?&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T21:07:11.630" LastActivityDate="2010-11-03T05:01:45.280" Title="Infinite series - Archimedean principle." Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8689" PostTypeId="2" ParentId="8688" CreationDate="2010-11-02T21:06:41.630" Score="2" ViewCount="0" Body="&lt;p&gt;&quot;The interval between $L$ and $M$&quot; means all the numbers that are between $L$ and $M$; we call $L$ and $M$ the &quot;endpoints&quot; of the interval. If you imagine the real numbers as the points on the real number line, with $L$ and $M$ located somewhere, the &quot;interval&quot; would be the segment of the line that is between the points $L$ and $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, an interval may or may not contain (either or both) of its endpoints. In your case, you are looking at the interval that does &lt;strong&gt;not&lt;/strong&gt; contain either $L$ and $M$ (that is the &quot;strictly between $L$ and $M$&quot; condition in the second quote). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concreteness, say $L=0$ and $M=1$. Then the interval of numbers strictly between $L$ and $M$ would be all real numbers that are greater than $0$ and smaller than $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number $x_0$, if $a\lt x_0$ and $x_0\lt b$, then we think of the interval determined by $a$ and $b$ (all real numbers strictly between $a$ and $b$: this includes $x_0$, but will include other things as well) as being a &quot;neighborhood&quot; of $x_0$: they are all points that are within a certain distance from $x_0$, either below $x_0$ or above $x_0$ (no further than $a$ is if they are below $x_0$, and no further than $b$ is if they are above). The definition you quote is basically saying: if you specify how far from $T$ you are willing to tolerate your total (no further than $L$ is from $T$ if the total is less than $T$, and no further than $M$ is from $T$ if the total is more than $T$), then you can always ensure that from some point forward the partial sums will have a total that you is within your margin of tolerance (a total that you can tolerate). This will hold no matter how tolerant or intolerant you are (you are very tolerant if you allow totals that are far away from $T$; that is, both $L$ and $M$ are &quot;far away&quot; from $T$; you are being very intolerant if your $L$ and your $M$ are really, really, really, close to $T$). You can be as intolerant as you wish, but you are required to allow &lt;em&gt;some&lt;/em&gt; leeway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, $T$ is what you think the sum of the series &quot;should&quot; be (if it made sense to add infinitely many numbers). The definition says that the sums you get a long the way will get &quot;as close as you care to specify&quot; to that total $T$, provided you add enough terms of your series, and that once you get there, any further refinments (by adding more terms of your series) will still keep you within your specified 'neighborhood' of $T$. It is very similar to the $\epsilon-\delta$ definition of a limit, if you saw that, or to the &quot;intuitive notion&quot; of limit: as long as you stay close enough to the point (for series, as long as you add up enough terms), the value of the function (for series, the running total) will be as close as you care to &quot;pre-specify&quot; from the limit (for series, as close as you care to &quot;pre-specify&quot; from $T$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; To respond to your query in the comments: &quot;If we have a target value, why add the tolerance?&quot; Keep in mind that we can only add finitely many numbers at a time. We can add two numbers, and then using that we can add three, four, five, etc. But we don't really know what it means to add infinitely many numbers, like the series says we &quot;should&quot; or &quot;want&quot;. So we cannot &lt;em&gt;actually&lt;/em&gt; perform the act of adding up all the (infinitely many) terms. Rather, we can only add up to a certain number of the terms; these are the partial sums. We can (in principle, at least) compute a partial sum to any term we want, but we cannot actually add up all of them (just like we can, in principle, count as high as we want, but we cannot actually count through &quot;all&quot; the natural numbers). So what we have actual access to are the partial sums and only the partial sums. We think that if we &lt;em&gt;could&lt;/em&gt; add up all the terms of the series, then it would make sense to call that sum $T$; so what we do is we compare those partial sums to our target, $T$. But of course, $T$ is supposed to be what we would get after adding up &lt;em&gt;all&lt;/em&gt; the terms of the series (if we could do that), and what we are comparing is &lt;strong&gt;not&lt;/strong&gt; the result of adding all the terms, but only some of them. So the total we have in hand (the partial sum) has no reason to be exactly the same as the target (just like, if you add the first 100 numbers in a list of a thousand numbers, the running total has no reason to be the same as the final total). So we need to add some tolerance to account for the fact that we are not &lt;em&gt;actually&lt;/em&gt; adding up all the terms, but only some of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I've never seen what you quote as the &quot;Archimedean Principle&quot; called that, but that may very well be just me...)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-02T23:39:36.523" LastActivityDate="2010-11-02T23:39:36.523" CommentCount="3" />
  <row Id="8690" PostTypeId="1" CreationDate="2010-11-02T21:07:00.230" Score="14" ViewCount="539" Body="&lt;p&gt;Hi, I encountered a problem today to prove that (X_n) with X_n = cos(n!) does not have a limit (when n approaches infinity). I have no idea how to do it formally. Could someone help? The simpler the proof (by that i mean less complex theorems are used) the better. Thanks&lt;/p&gt;&#xA;" OwnerUserId="3046" LastEditorUserId="3046" LastEditorDisplayName="" LastEditDate="2010-11-02T21:29:32.567" LastActivityDate="2010-11-03T01:17:56.027" Title="Is there a limit of cos (n!)?" Tags="&lt;analysis&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="6" />
  <row Id="8691" PostTypeId="2" ParentId="8688" CreationDate="2010-11-02T21:16:50.400" Score="1" ViewCount="0" Body="&lt;p&gt;This is really quite a convoluted way to define the limit of a sequence.  I can sympathize with your confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to note is that the limit of an infinite series is the limit of its sequence of partial sums.  So, we just need to understand limits of sequences.  The reason for the &lt;em&gt;interval&lt;/em&gt;, is that the series may have positive and negative terms.  So, while a series is busy converging, it may jump to the left or right of its limit.  But the main point is that for any &lt;em&gt;error tolerance&lt;/em&gt; (no matter how small) all the partial sums will &lt;em&gt;eventually&lt;/em&gt; be within this error tolerance of the limit.  Being within a small error tolerance means that you are in a small interval containing the limit (hence the interval business).  Here is how we quantify &lt;em&gt;error tolerance&lt;/em&gt; and &lt;em&gt;eventually&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sequence $(s_n)$ converges to a limit $s$ if for any $\epsilon&gt;0$ there exists an index $N$ such that $s_n \in (s-\epsilon, s+\epsilon)$ for all $n \geq N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think of $\epsilon$ as the &lt;em&gt;error tolerance&lt;/em&gt;, and the $N$ (which depends on $\epsilon$) as a measure of the &lt;em&gt;eventually&lt;/em&gt;.  This definition is mathematically equivalent to the one you gave where $s-\epsilon$ corresponds to $L$ and $s+\epsilon$ corresponds to $M$, but this one is superior for a couple of reasons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is no reason to insist that $L$ and $M$ are rational.  We &lt;em&gt;can&lt;/em&gt; insist that they are rational because rationals approximate real numbers arbitrarily well, but it is confusing to include it in the definition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, there is no reason to specify &lt;em&gt;two&lt;/em&gt; error tolerances $\epsilon_1$ and $\epsilon_2$ which is what the confusing definition is doing.  Instead of saying we eventually stray by at most $\epsilon_1$ to the left of the limit and by at most $\epsilon_2$ to the right of the limit, we may as well specify a &lt;em&gt;single&lt;/em&gt; error tolerance $\epsilon:=\min(\epsilon_1, \epsilon_2)$. We can then insist that the sequence eventually strays by at most $\epsilon$ (in either direction) from the limit.  &lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditorDisplayName="" LastEditDate="2010-11-03T00:06:01.797" LastActivityDate="2010-11-03T00:06:01.797" CommentCount="2" />
  <row Id="8692" PostTypeId="2" ParentId="8690" CreationDate="2010-11-02T21:30:27.503" Score="19" ViewCount="0" Body="&lt;p&gt;I suspect that you have miscopied the problem, or your source has an error. I will argue that, while your statement is extremely likely to be true, we probably do not know enough to prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is why. Suppose, hypothetically, that &#xA;$$\frac{1}{2 \pi} = \sum_{i \geq 0} \frac{d_i}{i!}$$&#xA;where all the $d_i$ are integers and, for $i$ sufficiently large, each $d_i$ is either $0$ or $1$. This is extraordinarily unlikely. However, if we knew how to prove that this didn't happen, we would probably know enough to show that every digit occurs infinitely often in the decimal expansion of $\pi$, and I believe that is an open problem. So, while this scenario is almost surely false, I don't think we know enough to prove that it is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in this scenario,&#xA;$$\cos n! = \cos (n! 2 \pi \sum_{i \geq 0} \frac{d_i}{i!}) = \cos (2 \pi N + 2 \pi \sum_{i \geq n+1} \frac{d_i }{(n+1)(n+2) \cdots i})$$&#xA;where $N$ is a very large integer. By the periodicity of $\cos$, this is just&#xA;$$\cos (2 \pi\sum_{i \geq n+1} \frac{d_i }{(n+1)(n+2) \cdots i}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, once $n$ is large enough that $d_i$ is $0$ or $1$ for $i$ larger than $n$, we have&#xA;$$\sum_{i \geq n+1} \frac{d_i n!}{i!} \leq \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \cdots = \frac{1}{n}.$$&#xA;So, in our unlikely scenario, $\lim_{n \to \infty} \sum_{i \geq n+1} \frac{d_i n!}{i!} =0$ and $\lim_{n \to \infty} \cos n! = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I may try to extract some basic lessons here, questions about trig functions evaluated at integers almost always turn into questions about $\pi$. Often they can be resolved simply by noting that $\pi$ is irrational; occasionally it is important to show that $\pi$ is not well approximated by rationals and sometimes, as in this example, they turn into claims about $\pi$ which are beyond current knowledge. The key point is to work out what they are saying about $\pi$, and then find out whether that fact is known.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise:&lt;/strong&gt; Show that, for any $a$ between $-1$ and $1$, there is a $\theta$ such that $\lim_{n \to \infty} \cos (n! \theta)$ is $a$.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-11-02T21:30:27.503" CommentCount="1" />
  <row Id="8693" PostTypeId="2" ParentId="8326" CreationDate="2010-11-02T21:34:09.280" Score="0" ViewCount="0" Body="&lt;p&gt;V.O. Manturov,  Knot theory. The first chapter gives an elementaty introduction. To continue, you have to know basical topological concepts, e.g. foundamental group.&lt;/p&gt;&#xA;" OwnerUserId="3047" LastActivityDate="2010-11-02T21:34:09.280" />
  <row Id="8694" PostTypeId="2" ParentId="8326" CreationDate="2010-11-02T21:45:05.527" Score="0" ViewCount="0" Body="&lt;p&gt;My intro to knot theory graduate course used &quot;An Introduction to Knot Theory&quot; by Lickorish.  The early chapters on Seifert surfaces and polynomials are quite nice.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-11-02T21:45:05.527" />
  <row Id="8695" PostTypeId="1" AcceptedAnswerId="8702" CreationDate="2010-11-02T21:55:10.973" Score="5" ViewCount="114" Body="&lt;p&gt;Let $A$ be a stochastic matrix.  Thus $A$ has nonnegative entries, and the sum of the elements in each row is 1. This implies that the vector $\begin{bmatrix} 1 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 1 \end{bmatrix}^T$ is an eigenvector corresponding to the eigenvalue 1.  Is it true that there exists a vector $b$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(A - I)x \geq b$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;has no solutions in $x$?  If so, is there a simple proof?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I've been trying to construct an answer to &lt;a href=&quot;http://math.stackexchange.com/questions/8556/&quot;&gt;another question&lt;/a&gt; using linear programming duality (as the OP implies he is interested in).  If my reasoning is correct, this is the only step I need to complete the argument.  I feel like this should be an easy question to answer, but I've been working on it for a while with no success.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-03T00:39:15.747" Title="No solutions to a matrix inequality?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8696" PostTypeId="2" ParentId="8683" CreationDate="2010-11-02T22:32:25.980" Score="3" ViewCount="0" Body="&lt;p&gt;Fix a triangulation $T = (t_i)_{i=0}^N$ of the surface $S$.  An arc $\alpha$ embedded in a triangle $t_i$ is &lt;em&gt;normal&lt;/em&gt; if the two points $\partial \alpha$ lie in distinct edges of $t_i$. A single triangle $t_i$ admits three different normal arcs (up to isotopy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple closed curve $\alpha \subset S$ is &lt;em&gt;normal&lt;/em&gt; with respect to $T$ if and only if $\alpha \cap t_i$ is a collection of normal arcs for each $i$.  Exercise: every essential simple curve may be isotoped to be normal.  We may replace any normal curve by a vector of length $3N$ by counting the number of each type of normal arc.  These vectors have non-negative integer entries and satisfy certain &quot;matching equations&quot;.  Conversely, any such vector represents a simple closed multicurve.  (Note that a single isotopy class of curve may have many different representations as a normal curve!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matching equations cut a cone out of the positive orthant.  Thus to enumerate all curves up to a fixed length (where here I count the number of normal arcs as the length) it suffices to find a Hilbert basis for this cone and take all combinations up to a fixed size, and so on, and so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is an algorithm that enumerates all essential simple closed curves up to length $N$ in time polynomial in $N$.  However the precomputation (of the Hilbert basis for the cone, etc) is no joke. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a &lt;a href=&quot;http://www.math.wisc.edu/~rkent/normal.ps&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; starting on page 13 of Cameron Gordon's notes for a course on normal surfaces in three-manifolds.  I'll end with a remark: in the case of the (once-holed) torus you can get away with exactly two triangles, the cone is very simple, the basis is very pretty, and everything can be done by hand.  (In fact you can this way rediscover for yourself the classification of simple closed curves in the torus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'll add one more remark.  You are representing curves on the surface via &quot;cutting sequences&quot; which are very closely related to writing out words in terms of the fundamental group.  When working with &lt;em&gt;simple&lt;/em&gt; curves (and the simplicity is crucial here) it can be exponentially more efficient to use the &quot;normal coordinates&quot; I described.  These are very similar to Thurston's &quot;train track&quot; coordinates for simple closed curves.  There is a third way to represent simple curves: Fix a small collection of curves (ie the very short curves) and then act on those by the mapping class group of the surface, as generated by Dehn twists (say).  These &quot;mapping class group coordinates&quot; can again be exponentially more efficient than using cutting sequences.  &lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-11-03T09:42:54.867" LastActivityDate="2010-11-03T09:42:54.867" />
  <row Id="8697" PostTypeId="2" ParentId="8624" CreationDate="2010-11-02T22:34:54.543" Score="0" ViewCount="0" Body="&lt;p&gt;Citing &lt;em&gt;Ed Sandifer&lt;/em&gt; preprint on L. Euler and the &lt;a href=&quot;http://www.southernct.edu/~sandifer/Ed/History/Preprints/Talks/NYU%20Basel%20Problem%20Paper.PDF&quot; rel=&quot;nofollow&quot;&gt;Basel Problem&lt;/a&gt; (that was published in &lt;a href=&quot;http://books.google.se/books?id=tK_KRmTf9nUC&amp;amp;pg=PA115&amp;amp;lpg=PA115&amp;amp;dq=Ludolf+Kuelen&amp;amp;source=bl&amp;amp;ots=lum0ovQKcQ&amp;amp;sig=7X4HZ1NrHoeNFM5VWquGShwJQHc&amp;amp;hl=sv&amp;amp;ei=aJHQTODdMsOBOuOupfEE&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CB0Q6AEwAQ#v=onepage&amp;amp;q=Ludolf%20Kuelen&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Euler at 300: an appreciation&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“It is clear that I employ here the letter $\pi$ to indicate the number of Ludolf of Kuelen, $3.14159265$, etc.” By this time, Euler has been using p to denote that constant for several years, but the convention will take many years more before it is universally adopted.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-02T22:34:54.543" />
  <row Id="8698" PostTypeId="1" AcceptedAnswerId="8725" CreationDate="2010-11-02T22:57:23.537" Score="4" ViewCount="87" Body="&lt;p&gt;What is the smallest square which contains 11 non-overlapping (except boundary) unit squares?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is open but I would like to know a method to verify the best known answer at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading paper at &lt;a href=&quot;http://www.emis.de/journals/EJC/Volume_10/PDF/v10i1r8.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; . In figure 2, there has been said that the best known packing has side length about 3.8772 and tilt angle about 40.182 degrees. How can I verify those values, as another source (Which Way did the Bicycle Go page 105) says that the side length is about 3.877083? I think I have to denote the tilt angle by $\alpha$, a side of the square by $x$ and form two equations but I don't see how to make those equations.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-11-03T01:40:30.507" LastActivityDate="2010-11-03T04:21:01.943" Title="Eleven unit squares inside a larger square" Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="8699" PostTypeId="2" ParentId="8683" CreationDate="2010-11-02T22:58:41.270" Score="0" ViewCount="0" Body="&lt;p&gt;An algorithm to determine whether a given element of the fundamental group has a simple representative on surfaces with boundary was described by Birman and Series in their paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;amp;co4=AND&amp;amp;co5=AND&amp;amp;co6=AND&amp;amp;co7=AND&amp;amp;dr=all&amp;amp;pg4=AUCN&amp;amp;pg5=AUCN&amp;amp;pg6=PC&amp;amp;pg7=ALLF&amp;amp;pg8=ET&amp;amp;review_format=html&amp;amp;s4=birman&amp;amp;s5=series&amp;amp;s6=&amp;amp;s7=&amp;amp;s8=All&amp;amp;vfpref=html&amp;amp;yearRangeFirst=&amp;amp;yearRangeSecond=&amp;amp;yrop=eq&amp;amp;r=5&amp;amp;mx-pid=744104&quot; rel=&quot;nofollow&quot;&gt;An algorithm for simple curves on surfaces&lt;/a&gt;. &lt;em&gt;J. London Math. Soc.&lt;/em&gt; (2) 29 (1984), no. 2, 331–342&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the MathSciNet review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The given algorithm is purely mechanical and easy to check (for finite words). For other algorithms, also for closed surfaces, see the bibliography in the paper and also B. Reinhart [Ann. of Math. (2) 75 (1962), 209--222; MR0150740 (27 #727)].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When writing this answer, I hadn't noticed the 'polynomial time' stipulation.  This won't give you that.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-11-03T00:09:59.997" LastActivityDate="2010-11-03T00:09:59.997" CommentCount="5" />
  <row Id="8700" PostTypeId="2" ParentId="8672" CreationDate="2010-11-02T23:20:27.903" Score="7" ViewCount="0" Body="&lt;p&gt;As a two-by-two matrix, applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobi_eigenvalue_algorithm&quot; rel=&quot;nofollow&quot;&gt;Jacobi&lt;/a&gt; in fact gives the answer at once!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that given a two-by-two matrix $\mathbf{A}$, one can construct an orthogonal matrix $\mathbf{V}=\bigl(\begin{smallmatrix}c&amp;amp;s\\-s&amp;amp;c\end{smallmatrix}\bigr)$ such that $\mathbf{V}^T\mathbf{A}\mathbf{V}$ is diagonal, where the two numbers $c$ and $s$ satisfy $c^2+s^2=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the off-diagonal elements are not zero (why?), computing $c$ and $s$ is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\frac{a_{22}-a_{11}}{2a_{12}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\frac{\mathrm{sgn}(\tau)}{|\tau|+\sqrt{1+\tau^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\frac1{\sqrt{1+t^2}}\qquad s=ct$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eigendecomposition easily follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See Golub and Van Loan's excellent &lt;a href=&quot;http://books.google.com/books?id=mlOa7wPX6OYC&amp;amp;pg=PA426&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; for further details.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The point was apparently missed that this method returns an eigensystem with orthogonal eigenvectors, so I suppose I should spell out in code how Jacobi works here, with a &lt;em&gt;Mathematica&lt;/em&gt; example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = N[{{3, -1}, {-1, 2}}, 20];&#xA;&#xA;tau = (a[[2, 2]] - a[[1, 1]])/a[[1, 2]]/2;&#xA;&#xA;t = Sign[tau]/(Abs[tau] + Sqrt[1 + tau^2]);&#xA;&#xA;{c, s} = {1, t}/Sqrt[1 + t^2]&#xA;{0.85065080835203993218, 0.5257311121191336060}&#xA;&#xA;{l1, l2} = {a[[1, 1]] - t a[[1, 2]], a[[2, 2]] + t a[[1, 2]]}&#xA;{3.6180339887498948482, 1.3819660112501051518}&#xA;&#xA;Eigenvalues[a] == {l1, l2}&#xA;True&#xA;&#xA;Eigenvectors[a] == {{c, -s}, {s, c}}&#xA;True&#xA;&#xA;a.{c, -s} == l1 {c, -s}&#xA;True&#xA;&#xA;a.{s, c} == l2 {s, c}&#xA;True&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T00:19:14.947" LastActivityDate="2010-11-04T00:19:14.947" CommentCount="2" />
  <row Id="8701" PostTypeId="1" AcceptedAnswerId="8715" CreationDate="2010-11-02T23:31:18.547" Score="0" ViewCount="62" Body="&lt;p&gt;I need to figure out the Probability that Y &gt; 23 for a chi-square distribution with degrees of freedom = 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says to do this with some applet. I don't know what it's talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I do this on my Ti-89 (preferable) or on some website somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT/UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a program on my ti-89 for chi-square distribution. However it asks me for lower and upper bound and degrees of freedom. I put in 7 for degrees of freedom but i'm not sure how to translate probability Y &gt; 23 into an upper and lower bound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put in lower = 0, upper = 23, degrees of freedom = 7 and I said to auto-scale and I got .998295. Is that the right probability?&lt;/p&gt;&#xA;" OwnerUserId="3049" LastEditorUserId="3049" LastEditorDisplayName="" LastEditDate="2010-11-03T00:03:25.807" LastActivityDate="2010-11-03T02:17:13.953" Title="chi-square distribution with v = 7 degrees of freedom Ti-89?" Tags="&lt;probability&gt;&lt;homework&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="8702" PostTypeId="2" ParentId="8695" CreationDate="2010-11-02T23:49:33.017" Score="5" ViewCount="0" Body="&lt;p&gt;Your inequality $(A-I)x \ge b$ has no solutions in $x$ as soon as $b&gt;0$. Indeed, any potential solution would have to satisfy $A x \ge  x + b$ and, since rows of $A$ are nonnegative and sum to one, each element of vector $Ax$ is a convex combination of the components of $x$, which must be less than $x_{max}$, the largest component of $x$. On the other hand, at least one element of $x+b$ is greater than $x_{max}$, which proves the impossibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Farkas%27_lemma&quot; rel=&quot;nofollow&quot;&gt;Farkas' Lemma&lt;/a&gt; to this impossible system shows that the following always admits solutions in y&#xA;$$y^T (A-I) = 0, y \ge 0 \text{ and } y^T b &gt; 0$$&#xA;which expresses the fact that $A$ necessarily admits a nonnegative left eigenvector with eigenvalue $1$ (the last inequality ensures that $y$ is nonzero). &lt;/p&gt;&#xA;" OwnerUserId="55" LastEditorUserId="55" LastEditorDisplayName="" LastEditDate="2010-11-03T00:39:15.747" LastActivityDate="2010-11-03T00:39:15.747" CommentCount="1" />
  <row Id="8703" PostTypeId="2" ParentId="8690" CreationDate="2010-11-02T23:58:06.797" Score="0" ViewCount="0" Body="&lt;p&gt;I tried the following approach where I assumed the limit existed and attempted to get a contradiction. However, the best I could do was to get the same answer as David above that if (in the extremely unlikely case that) the limit exists, it will be 1. I just wanted to make a record of the approach here in case someone sees any potential to this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x_{n+1} = \cos((n+1)n!)$, it can be written as a $(n+1)$th degree polynomial in $\cos(n!)$. Should the limit exist, it will be a fixed point of this polynomial. Using standard multiple angle formulas (found &lt;a href=&quot;http://mathworld.wolfram.com/Multiple-AngleFormulas.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for ex.), we can show that the limit if it exists should equal $1$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-11-02T23:58:06.797" CommentCount="7" />
  <row Id="8704" PostTypeId="2" ParentId="8684" CreationDate="2010-11-03T00:35:39.200" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The usual descent argument works: $\: $ if $\rm\ x^2-D\:y^2 = 1\ $ has fundamental solution $\rm\:\epsilon^k\:$ and $\rm\:\epsilon^n&gt;0\:$ is another solution then $\rm\:k|n\:$ (else $\rm\:\epsilon^{n\ mod\ k}\:$ is a solution smaller than the fundamental one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hint is not detailed enough then you can find full details of this standard argument e.g. in section 1.3 of Jacobson and Williams: Solving the Pell equation. There you'll also find the standard table giving the value of $\rm\ k$ (viz. $\rm\:k|6\:$ and depends on the parity of $\rm x$ and $\rm y$ and the sign of $\pm4)$. Note in particular that the complete proof is not as trivial as the special case in Pete L. Clark's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why &quot;4 is special&quot;, i.e. why the equation $\rm\ x^2 - D\:y^2 = \pm 4\ $ proves more fundamental than $\rm\ x^2 - D\:y^2 = \pm 1\ $ is simply because the former equation is the one that arises when checking that an integer has norm $\pm 1$, i.e. $\rm\ N((x+y\sqrt{D})/2) = \pm1$ iff $\rm\ x^2-D\:y^2 = \pm 4\:$; i.e. the 4 arises from clearing &quot;denominators&quot; in the norm equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T01:56:28.973" LastActivityDate="2010-11-03T01:56:28.973" CommentCount="1" />
  <row Id="8705" PostTypeId="2" ParentId="8674" CreationDate="2010-11-03T00:54:20.703" Score="0" ViewCount="0" Body="&lt;p&gt;If you assume there are N colored balls, with equal frequency p=1/N, then this should be pretty do-able.  Of course, you're desired confidence interval will affect the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without working out all the details, consider that you have drawn the point where you see a few colors 2 or 3 times.  By considering how many balls you have drawn, and how many uniques have been witnessed, you can make an estimate of that N within a confidence bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give the details if you like, but if these assumptions are acceptable you can probably take it from here?&lt;/p&gt;&#xA;" OwnerUserId="2677" LastActivityDate="2010-11-03T00:54:20.703" />
  <row Id="8706" PostTypeId="2" ParentId="8690" CreationDate="2010-11-03T01:17:56.027" Score="0" ViewCount="0" Body="&lt;p&gt;Above posters seem to be right, I thought I had this solved for the limit = 1, but realized I was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its of any use, here is my faulty reasoning. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;As we get more and more factors we will only get closer and closer to a factor that is a multiple of pi.  We know that n! is even for any n &gt;=2, so we get a abritarily closer to 2*pi|n! as n approaches infinity.&#xA;    Sorry of the code is out of form, but:&lt;/p&gt;&#xA;    &#xA;    &lt;blockquote&gt;&#xA;      &lt;p&gt;eps = .00001&lt;br&gt;&#xA;      for i in xrange(2,800000,2):&lt;br&gt;&#xA;      ...     if i*math.pi-math.floor(i*math.pi) &amp;lt; eps: print i&lt;br&gt;&#xA;      ...&lt;br&gt;&#xA;      431230&lt;br&gt;&#xA;      530762&lt;br&gt;&#xA;      630294  &lt;/p&gt;&#xA;      &#xA;      &lt;p&gt;math.cos(math.floor(431230*math.pi)) =&#xA;      0.99999999997167566&lt;br&gt;&#xA;      So any n! for n &gt; 431230*pi will be at least this close, or closer, to 1  &lt;/p&gt;&#xA;    &lt;/blockquote&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is that product of the factors NOT close to pi is going to grow faster than that epsilon shrinks.  And obviously no integer is going to be an exact multiple of pi (as it is irrational).  Therefore, I am led to agree with other posters that this limit is unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS - I know this isn't rigorous by any means, but was hoping maybe better minds would have some insightful comments.&lt;/p&gt;&#xA;" OwnerUserId="2677" LastActivityDate="2010-11-03T01:17:56.027" />
  <row Id="8707" PostTypeId="1" AcceptedAnswerId="8709" CreationDate="2010-11-03T01:18:33.007" Score="3" ViewCount="79" Body="&lt;p&gt;Prove or disprove that closed 3-manifolds which are not simply connected cannot be embedded in three-dimensional Euclidean spaces. I am not a mathematics major and I am taking introductory topology this semester. But I need to apply this result for my research. Any help is much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Srikanth.&lt;/p&gt;&#xA;" OwnerUserId="3050" LastActivityDate="2010-11-03T01:57:51.563" Title="embedding of closed 3-manifolds." Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8708" PostTypeId="2" ParentId="8561" CreationDate="2010-11-03T01:24:07.513" Score="1" ViewCount="0" Body="&lt;p&gt;Per fmartin's request, I've collected my above comments into an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proposed approach doesn't work. The 2nd last equation is $c = de$, where $e = (b+1)\cdots (a+b)$. Therefore $(a+b)! = d(b+1)\cdots (a+b)a!b! = da!(a+b)!$ implies $1 = d a!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Sierpinski's inductive proof is expressed much more clearly by explicitly mentioning the underlying binomial identity that enables the descent, viz. $\binom{a+b}a = \binom{a+b-1}a + \binom{a+b-1}b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For various integrality proofs of binomial coefficients see also &lt;a href=&quot;http://math.stackexchange.com/questions/2192&quot;&gt;this thread&lt;/a&gt;. There you'll find a very simple proof I discovered that shows how to write a binomial coefficient as a product of fractions whose denominators are all coprime to any given prime p.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-03T01:24:07.513" />
  <row Id="8709" PostTypeId="2" ParentId="8707" CreationDate="2010-11-03T01:25:11.523" Score="4" ViewCount="0" Body="&lt;p&gt;If your manifold is closed you must be using some definition that ensures it's compact?  So if $f : M \to \mathbb R^3$ is an embedding, isn't the image simultanously compact and open? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit in response to your comment:  if $f : M \to \mathbb R^3$ is an embedding, let $B \subset M$ be an open subset of $M$ which is homeomorphic to an open ball in $\mathbb R^3$.  You need to argue that $f(B)$ is open in $\mathbb R^3$.  If your embedding is smooth there's a big theorem from calculus that gives you the result.  If you're talking about topological embeddings you're going to need a tool. Have you studied the &quot;invariance of dimension&quot; theorem? &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-03T01:57:51.563" LastActivityDate="2010-11-03T01:57:51.563" CommentCount="2" />
  <row Id="8710" PostTypeId="2" ParentId="8556" CreationDate="2010-11-03T01:32:29.707" Score="1" ViewCount="0" Body="&lt;p&gt;Here's the linear programming approach (with considerable help from another user, Fanfan).  Consider the LP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min 0^T y$$&#xA;subject to&#xA;$$(X - I)y \geq 1,$$&#xA;where $0$ and $1$ are, respectively, vectors containing all 0's and all 1's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fanfan's &lt;a href=&quot;http://math.stackexchange.com/questions/8695/&quot;&gt;answer to my question&lt;/a&gt; this LP is infeasible.  Thus its dual,&#xA;$$\max 1^T x$$ &#xA;subject to &#xA;$$x^T (X-I) = 0^T,$$&#xA;$$x \geq 0,$$&#xA;is either infeasible or unbounded.  But $x = 0$ is a solution, and so this dual problem is feasible.  Thus it must be unbounded.  But that means there must be a nonzero solution $x_1$ in its feasible region.  Thus we have $x_1^T X = x_1^T$ with $x_1 \geq 0$, $x_1 \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Fanfan's answer to my question also includes another answer to your question - one that uses Farkas's Lemma rather than LP duality.  It ends up being quite similar to my answer here, as, of course, Farkas's Lemma and LP duality are basically equivalent.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-03T01:32:29.707" />
  <row Id="8711" PostTypeId="1" AcceptedAnswerId="8712" CreationDate="2010-11-03T01:34:59.780" Score="5" ViewCount="202" Body="&lt;p&gt;Can a derivative operation commute over an integral operation irrespective of the properties of the function under the integral ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-11-03T02:02:30.277" Title="Derivative commuting over integral." Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8712" PostTypeId="2" ParentId="8711" CreationDate="2010-11-03T01:57:00.190" Score="8" ViewCount="0" Body="&lt;p&gt;Not in general.  I recommend Gelbaum and Olmsted's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=cDAMh5n4lkkC&amp;amp;printsec=frontcover&amp;amp;source=gbs_atb#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Counterexamples in Analysis&lt;/a&gt;&lt;/em&gt;, which is where I turned to find a counterexample to your question.  Namely, example 15 on page 123 is titled&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A function $f$ for which $d/dx\int_a^b f(x,y)dy\neq\int_a^b[\partial/\partial x f(x,y)]dy$, although each integral is proper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The example is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x,y) = \left\{&#xA;     \begin{array}{lr}&#xA;       \frac{x^3}{y^2}e^{-x^2/y} &amp;amp; : y&gt;0, \\&#xA;       0 &amp;amp; : y=0, &#xA;     \end{array}&#xA;   \right.&#xA;$$&#xA;integrated with respect to $y$ from $0$ to $1$.  Actually, differentiating under the integral sign works here except where $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function and its partial derivative are not jointly continuous.  When they are jointly continuous, differentiation and integration commute.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-03T02:02:30.277" LastActivityDate="2010-11-03T02:02:30.277" CommentCount="7" />
  <row Id="8713" PostTypeId="1" AcceptedAnswerId="8714" CreationDate="2010-11-03T02:08:43.687" Score="7" ViewCount="137" Body="&lt;p&gt;This operation is similar to discrete convolution and cross-correlation, but has binomial coefficients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n)\star g(n)=\sum_{k=0}^n \binom{n}{k}f(n-k)g(k) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^n\star b^n=(a+b)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;following binomial theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wonder if there is a name for such operation and where I can read about its properties.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-11-03T03:41:32.010" Title="How this operation is called?" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="8714" PostTypeId="2" ParentId="8713" CreationDate="2010-11-03T02:15:10.590" Score="11" ViewCount="0" Body="&lt;p&gt;It's called a &lt;em&gt;binomial convolution&lt;/em&gt; in Graham, Knuth, and Patashnik's &lt;em&gt;Concrete Mathematics&lt;/em&gt;.  I don't have that text in front of me (but I bet someone here can give you a page number), but here's a reference on the &lt;a href=&quot;http://fermatslasttheorem.blogspot.com/2006/10/generating-function-for-bernoulli.html&quot; rel=&quot;nofollow&quot;&gt;Fermat's Last Theorem blog&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also be worth checking out Section 2.3 of Wilf's &lt;em&gt;Generatingfunctionology&lt;/em&gt;.  This is on exponential generating functions.  The property of interest is that if $F(x)$ and $G(x)$ are the exponential generating functions of $f(n)$ and $g(n)$, respectively, then $F(x)G(x)$ is the exponential generating function of $f(n) \star g(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(FYI: You can download the second edition of &lt;em&gt;Generatingfunctionology&lt;/em&gt; from &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;Wilf's web site&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-03T03:41:32.010" LastActivityDate="2010-11-03T03:41:32.010" CommentCount="2" />
  <row Id="8715" PostTypeId="2" ParentId="8701" CreationDate="2010-11-03T02:17:13.953" Score="3" ViewCount="0" Body="&lt;p&gt;There's an online calculator &lt;a href=&quot;http://www.fourmilab.ch/rpkp/experiments/analysis/chiCalc.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Plugging $\chi^2 = 23$ and $df=7$, we get $.0017$. Your calculator calculated the probability that $0 \leq \chi^2 \leq 23$, which is $1-.0017 = .9983$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I commented above, you could tell that you needed to take $1-p$ instead of $p$ since $\chi^2 &gt; 23$ is an unlikely event, given that the expectation is $7$ and the standard deviation is $\sqrt{14} \approx 4$ (Chebyshev and even Markov would give you that $p &amp;lt; 1/2$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-03T02:17:13.953" CommentCount="2" />
  <row Id="8716" PostTypeId="2" ParentId="8713" CreationDate="2010-11-03T02:18:33.313" Score="0" ViewCount="0" Body="&lt;p&gt;It corresponds to multiplying exponential generating functions (check &lt;em&gt;generatingfunctionology&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-03T02:18:33.313" />
  <row Id="8717" PostTypeId="1" CreationDate="2010-11-03T02:26:13.460" Score="0" ViewCount="49" Body="&lt;p&gt;Most of what I've read defines the support of a distribution topologically if it defines support precisely; e.g the smallest closed subset that has full measure (i.e. the intersection of all closed subsets of full measure). I've seen allusions to purely measure-theoretic definitions, and I think I even came across one once, but I can't find it. Is there a measure-theoretic definition of support that everyone agrees on?&lt;/p&gt;&#xA;" OwnerUserId="2122" LastActivityDate="2010-11-03T12:31:45.490" Title="Non-topological notion of &quot;support&quot; of a probability distribution" Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="8718" PostTypeId="1" AcceptedAnswerId="8721" CreationDate="2010-11-03T02:43:42.140" Score="0" ViewCount="97" Body="&lt;p&gt;Let $P \in \mathbb{Z}[x], P(x) = \displaystyle\sum\limits_{j=0}^n a_j x^j, a_n \neq 0$ and $a_0 \neq 0$; if $p/q$ is a root of P (with p and q coprimes) then $p|a_0$ and $q|a_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've managed to prove the first part ($p|a_0$) and I suppose I'm not far from proving the second, though I'd really like some feedback since I'm just starting with making proofs of my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(x) = a_n(x-p/q)\displaystyle\prod\limits_{j=2}^n (x-r_j)$, with $r_j$ being the other n-1 roots of P(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $a_0 = a_n(-p/q)\displaystyle\prod\limits_{j=2}^n (-r_j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $-p/q|a_0$ and obviously $p/q|a_0$. Rephrasing, $a_0 = l\frac{p}{q} = \frac{l}{q} p$ with $l \in \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies $p|a_0$ if $l/q \in \mathbb{Z}$, but this is trivial since $q|lp$ and q and p are coprimes, so $q|l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $p|a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second part, we want to see that $q_i|a_n \forall i \leq n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define $d$ as the least common multiple of $\{q_1, q_2,...,q_n\}$. Then, $q_i|a_n \forall i \leq n \iff d|a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it follows that $d|\displaystyle\prod\limits_{j=1}^n q_j$, so we want to see that $a_n = l \displaystyle\prod\limits_{j=1}^n q_j$ with $l \in \mathbb{Z}$. Here's where I have my doubts with the proof as I have no way to show that l is indeed an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging the previously given equation for $a_0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = a_0 \displaystyle\prod\limits_{j=1}^n \frac{-q_j}{p_j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the previous reasoning, as $p_i|a_0 \forall i \leq n$, then $a_0 = k \displaystyle\prod\limits_{j=1}^n p_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $a_0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = k \displaystyle\prod\limits_{j=1}^n -q_j$, which is equivalent to $q|a_n$ as shown earlier.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-11-03T04:59:31.007" Title="Proving Gauss' polynomial theorem" Tags="&lt;number-theory&gt;&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8719" PostTypeId="2" ParentId="8713" CreationDate="2010-11-03T02:47:13.480" Score="3" ViewCount="0" Body="&lt;p&gt;Remark that one can employ the powerful umbral calculus to compute closed forms for many such &lt;strong&gt;binomial convolutions&lt;/strong&gt; of special functions, e.g. from p. 161 of Roman: Umbral Calculus:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/Wt0Ii.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-03T02:47:13.480" CommentCount="3" />
  <row Id="8720" PostTypeId="1" AcceptedAnswerId="8722" CreationDate="2010-11-03T02:49:15.527" Score="2" ViewCount="106" Body="&lt;p&gt;I am looking for a simple proof to show that given a countable set of natural numbers $C$ that is closed under addition and whose gcd is 1, there exists two elements $c_1, c_2 \in C$ such that $\gcd(c_1, c_2)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the Chinese remainder theorem, but my proof is about half a page and I was wondering if there is a shorter one.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-11-03T03:04:17.730" LastActivityDate="2010-11-03T04:25:11.403" Title="Chinese remainder theorem" Tags="&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8721" PostTypeId="2" ParentId="8718" CreationDate="2010-11-03T02:58:16.417" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\displaystyle\rm\ \ 0 = q^n\ P(\frac{p}q)\ =\ a_n\ p^n + q \ (\cdots)\ \Rightarrow\ q\:|\: a_n\:p^n\ \Rightarrow\ q\:|\:a_n\ $ by Euclid's Lemma.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $  This result is not usually called Gauss' polynomial theorem but, rather, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test / Theorem&lt;/a&gt; - which is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_lemma%28polynomial%29&quot; rel=&quot;nofollow&quot;&gt;Gauss's Lemma&lt;/a&gt; for polynomials. Generally if $\rm\:\alpha\:$ is an algebraic number with primitive minimal polynomial $\rm\: f(x)\in \mathbb Z[x]\:$ and if $\rm\: g(\alpha) = 0\:$ for $\rm\: g(x)\in \mathbb Z[x]\:$ then  $\rm\ f(x)|g(x)\ in\ \mathbb Z[x]$.  The above is the special case $\rm\ \alpha = p/q,\ \ f(x) = q\:x - p,\ \ \gcd(p,q) = 1\:$. Namely, $\rm\ (q\:x-p)\:|\:g\in \mathbb Z[x]\ \Rightarrow\ q|g_n,\ p|g_0\ in\ \mathbb Z\: $.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T04:59:31.007" LastActivityDate="2010-11-03T04:59:31.007" CommentCount="1" />
  <row Id="8722" PostTypeId="2" ParentId="8720" CreationDate="2010-11-03T03:15:59.653" Score="3" ViewCount="0" Body="&lt;p&gt;Because $\gcd(C)=1$ there are $a_1,a_2,\ldots,a_n\in C$ such that $\gcd(a_1,\ldots,a_n)=1$.  Thus there are integers $m_1,\ldots m_n$ such that $m_1a_1+\cdots m_na_n=1$.  Let $\{i_1,\ldots,i_p\}\subset\{1,\ldots,n\}$ be those indices corresponding to positive coefficients $m_{i_k}$ and $\{j_1,\ldots,j_q\}$ those corresponding to negative coefficients $m_{j_k}$.  Because $C$ is closed under addition, $c_1=\sum_{k=1}^pm_{i_k}a_{i_k}$ and $c_2=\sum_{k=1}^q-m_{j_k}a_{j_k}$ are in $C$, and $c_1-c_2=1$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-03T03:15:59.653" CommentCount="2" />
  <row Id="8723" PostTypeId="2" ParentId="8718" CreationDate="2010-11-03T03:23:00.520" Score="1" ViewCount="0" Body="&lt;p&gt;If you know how to prove the first part, just apply it to the polynomial $t^n P(t^{-1})$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-03T03:23:00.520" CommentCount="2" />
  <row Id="8724" PostTypeId="1" CreationDate="2010-11-03T03:24:28.007" Score="3" ViewCount="124" Body="&lt;p&gt;The Riemann Zeta Function is defined as $ \displaystyle \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{1}{n^s}$. It is not absolutely convergent or conditionally convergent for $\text{Re}(s) \leq 1$. Using analytic continuation, one can derive the fact that $\displaystyle \zeta(-s) = -\frac{B_{s+1}}{s+1}$ where $B_{s+1}$ are the Bernoulli numbers. Can one obtain this result without resorting to analytic continuation? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-11-03T15:11:23.513" LastActivityDate="2010-11-03T15:11:23.513" Title="Riemann Zeta Function and Analytic Continuation" Tags="&lt;complex-analysis&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="8725" PostTypeId="2" ParentId="8698" CreationDate="2010-11-03T04:15:05.770" Score="0" ViewCount="0" Body="&lt;p&gt;It's a pity this example isn't really describe anywhere. Assuming that there's a triple intersection at $(1,2)$, we get the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k - 3 - \sin(\alpha) - \cot(\alpha)&lt;em&gt;(2 + 2&lt;/em&gt;\cos(\alpha) - k) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you can get more equations by assuming that the tilted squares touch each other and corners of the straight squares, but these seem not to be satisfied by the given $k,\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How to find more equations: the tilted squares come in three groups (two, two and one), and all seem to be just touching each other. For each of the groups, one vertex has one known coordinate (the middle group might have a known vertex at $(1,2)$, but I don't know if that's true). You can get equations from the fact that these groups touch certain corners of the straight squares, and from the (apparent) fact that they just touch each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to get these equations is to define vectors $s,t$ for the sides of the tilted square (they can be expressed in terms of $\alpha$), and now all these touchings amount to pairs of points $p,q$ which are parallel to either $s$ or $t$. So $p-q$ is proportional to $s$ (or $t$), and we get an equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, pursuing this route, I get conflicting equations (conflicting in the sense that when I substitute $k$ and $\alpha$, they are contradictory). Maybe you can try and tell us if it works out for you.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-03T04:21:01.943" LastActivityDate="2010-11-03T04:21:01.943" />
  <row Id="8726" PostTypeId="2" ParentId="8720" CreationDate="2010-11-03T04:25:11.403" Score="0" ViewCount="0" Body="&lt;p&gt;(as requested)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose on the contrary that $t_1 \equiv t_2 \equiv 0 \pmod p$ where $p = \gcd(t_1, t_2) &gt; 1$.&#xA;Then, there must be a $t_3 \nequiv 0 \pmod p$ otherwise all elements are divisible by $p &gt; 1$, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $t_1 = pq$ and $t_2 = pr$.&#xA;By the Chinese remainder theorem, for all $s \ge qr$, there are $c_1, c_2 \in \mathbb{Z}$ such that &#xA;$&#xA;s = c_1q + c_2r&#xA;$.&#xA;Let $s = p^a$ where $a \in \mathbb{N}$ is chosen so that $p^a \ge qr$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&#xA;$&#xA;c_1t_1 + c_2t_2 = c_1pq + c_2pr = p(c_1q + c_2r) = ps  = p^{a+1}.&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\gcd(c_1t_1 + c_2t_2, t_3) = 1$ since the prime factorization of $c_1t_1 + c_2t_2$ consists of $p^{a+1}$ and $t_3$ is relatively prime to $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $C$ is closed under addition, $c_1t_1 + c_2t_2\in C$.  Thus, it and $t_3$ are the two elements that have g.c.d. 1.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-11-03T04:25:11.403" />
  <row Id="8727" PostTypeId="1" CreationDate="2010-11-03T04:28:29.360" Score="2" ViewCount="126" Body="&lt;p&gt;How would I start off proving that S= (the set of symmetric nxn matrices) is a manifold. &#xA;I tried using the definition directly by saying&#xA;Mn = the space of all nxn matrices&#xA;For every A in Mn there exists open sets U=V=Mn and a bijection F: U-&gt;V by F(A)= A-A^T&#xA;Therefore we have&#xA;F(U ∩ S) = F(S) since S is a subset of Mn&#xA;={0} ∩ Mn&#xA;this is where I get stuck.&#xA;Also, I know that the set of all symmetric nxn matrices is (n^2 + n)/2, therefore that is the dimension of the manifold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition: A set M (subset of R^n) is a k-dimensional manifold if for every x in M there exists open sets U, V and a bijection h:U-&gt;V with &#xA;x in U and &#xA;H(U ∩ M = V ∩ (R^k X {c^(k+1),...,c^n}) for all c's constants&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="2319" LastEditorDisplayName="" LastEditDate="2010-11-03T04:42:11.097" LastActivityDate="2010-11-03T07:16:39.937" Title="The set of symmetric matrices as a manifold" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="8728" PostTypeId="2" ParentId="8667" CreationDate="2010-11-03T04:32:21.173" Score="6" ViewCount="0" Body="&lt;p&gt;The answer is no. &lt;a href=&quot;http://cms.math.ca/cmb/v12/cmb1969v12.0337-0338.pdf&quot; rel=&quot;nofollow&quot;&gt;H. Herrlich&lt;/a&gt;  showed, in 1969, that even if you consider each $A_n$ a completely regular space, the direct limit may fail to be Hausdorff. However if all $A_n$ are T$_4$ - spaces then $X$ is a T$_4$ - space (it's not hard to prove this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment about the definition of direct limit. Usually, in category theory, we call &lt;a href=&quot;http://en.wikipedia.org/wiki/Direct_limit&quot; rel=&quot;nofollow&quot;&gt;direct limit&lt;/a&gt; a colimit of a directed family of objects. Using this terminology it's well known that the category of Hausdorff spaces isn't closed under direct limits. You can find some examples in Dugundji's 'Topology' (a shame it's out of print). The definition you are using is very particular, so Herrlich's example is special.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://dml.cz/bitstream/handle/10338.dmlcz/700787/Toposym_03-1971-1_36.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper D. Hajek and G. Strecker exhibit sufficient conditions for the Hausdorff property to be preserved under direct limits.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-03T19:16:50.850" LastActivityDate="2010-11-03T19:16:50.850" CommentCount="1" />
  <row Id="8729" PostTypeId="2" ParentId="8688" CreationDate="2010-11-03T04:52:21.843" Score="4" ViewCount="0" Body="&lt;p&gt;What a spectacularly rubbish way to define convergence. Here's how to think of it in a simple, yet rigorous way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we have a sequence $S_{n} = \frac{1}{2^n}$. We know that $S_{0} = 1, S_{1} = \frac{1}{2}, S_{2} = \frac{1}{4}$, and so on and so forth, just as you would have defined it in Calculus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might now be interested in the sum of $\sum_{n=0}^{\infty} S_{n} = 1 + \frac{1}{2} + \frac{1}{4} + ...$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may remember that this series will converge to 2. How should we express this analytically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we cannot actually sum up an infinite number of terms, we'll never actually get to 2 if we sum this up by hand. However, suppose I only wanted to get a value CLOSE to 2. How close? Let's say we want our sum to be within 0.5 of 2. That is, we want the sum to be between $2-0.5$ and $2+0.5$. This is the interval $(1.5,2.5)$. I don't have to sum up all of the terms to get in this interval. In fact, I need to sum up only 3 terms: $1 + \frac{1}{2} + \frac{1}{4} = 1.75$ to be in this interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $\sum_{n=0}^{3} S_{n}$ is in the interval. We can, if we want to, sum up more than 3 terms, and we will still remain in the interval. The point is that any number of terms more than 3 will still keep us in the interval. We can also say that we need at least the partial sum $S_{2}$ (which is our third term).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we wanted our interval to be smaller, between $2-0.01$ and $2+0.01$, which is the interval (1.99,2.01). It turns out you need 8 terms for our series to get within this interval. After these eight terms (i.e. after the partial sum $S_7$), your sequence stays in the interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our above examples, we wanted to get within 0.5 or 0.01 of the point of convergence. These are our epsilons. That is $(2-\epsilon, 2 + \epsilon)$ is our interval, and we can set $\epsilon$ to be anything we like. We chose to set epsilon to 0.5 or 0.01, but we could have chosen a smaller number.And therein lies the idea of convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see, no matter how small I chose my epsilon (as long as it is greater than $0$), I can still find a point where my sequence enters the interval $(2-\epsilon, 2 + \epsilon)$, and never leaves it afterwards. When epsilon was 0.5, we needed $S_2$, when epsilon was 0.01, we needed $S_7$. If we can demonstrate that no matter HOW SMALL you make your epsilon, there is a $S_n$ that is in that interval (to stay), you have shown convergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I actually do this? Well, $\sum_{n=0}^{k} \frac{1}{2^n}$ is a geometric series equal to $2^{-k}(2^{k+1}-1)$. There is a formula for this in your textbook, I presume.&#xA;If we want $2^{-k}(2^{k+1}-1) &gt; 2 - \epsilon$ we can solve algebraically to get that $k&gt; -\frac{\log{\epsilon}}{\log{2}}$. If we set $\epsilon=0.01$, we'll get that $k&gt;6.64...$. This means that we need at least $S_7$ as I mentioned above. Either way, the mere existence of this function demonstrates that we can always find a sufficiently large k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this suffice, conceptually? In order for you to argue that the sequence doesn't converge to a particular point, you need to show that the sequence is always some distance away from my point. Yet, no matter how small you make that distance, it is not small enough, for the sequence gets into that interval, and remains there. &lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="647" LastEditorDisplayName="" LastEditDate="2010-11-03T05:01:45.280" LastActivityDate="2010-11-03T05:01:45.280" CommentCount="5" />
  <row Id="8730" PostTypeId="1" AcceptedAnswerId="8733" CreationDate="2010-11-03T05:04:02.890" Score="2" ViewCount="97" Body="&lt;p&gt;I presume that it is because one needs to take a limit to define the derivative, but is there another reason?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-03T06:08:10.527" Title="Why are holomorphic functions defined in reference to an open set?" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="8731" PostTypeId="1" AcceptedAnswerId="8790" CreationDate="2010-11-03T05:46:54.333" Score="3" ViewCount="122" Body="&lt;p&gt;What are some algorithms used for finding the shortest path between 2 points on a (riemannian) manifold (the manifold may have a smooth boundary)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've had 3 ideas, none of which seem that good:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) (Rubber band principle) Find any path between the two points, then draw the path taut like a stretched rubber band with some damping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) (Ray-tracing analog) Shoot out geodesic rays from one point in all directions. If a ray hits the target, we're done. If the ray hits the boundary and is not close to tangent, ignore it. If the ray hits the boundary approximately tangent, then follow the geodesic on the boundary in that direction, shooting tangent rays off occasionally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) (Discretization) Triangulate/tetrahedralize/etc the manifold, and construct a weighted graph where the nodes correspond to the centerpoints of each tetrahedron, the edges correspond to tetrahedra that touch, and the weights correspond to the distance between tetrahedra centers. Then compute the shortest path in the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the stuff I've seen in the literature is concerned with the theoretical existence of the path rather than actually computing it. How do you find the shortest path in practice?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-11-05T00:00:47.273" LastActivityDate="2010-11-05T00:00:47.273" Title="Algorithm for shortest path on manifold" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;manifold&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8732" PostTypeId="1" CreationDate="2010-11-03T06:05:30.183" Score="3" ViewCount="121" Body="&lt;p&gt;The following two identities comes from my trigonometry module without any sort of proof,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A + B + C = \pi $ then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan A + \tan B + \tan C = tan A \cdot tan B \cdot tan C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:I am not much sure about whether the first one is fully correct or not, so if not please suggest the correct one and also I will be grateful if somebody suggest a suitable method (may be using mathematica) to verify an identity like this prior to proving.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-11-03T11:19:19.183" Title="How to prove this trignometrical Identities ?" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8733" PostTypeId="2" ParentId="8730" CreationDate="2010-11-03T06:08:10.527" Score="6" ViewCount="0" Body="&lt;p&gt;You may want to take a look at &lt;em&gt;When is a Function that Satisfies the Cauchy-Riemann Equations Analytic?&lt;/em&gt; by J. D. Gray and S. A. Morris, The American Mathematical Monthly, Vol. 85, No. 4, (Apr., 1978), pp. 246-256. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, in a sense, we want open domains so limits can be computed, but we have a notion of directional limit, so this is not much of a reason; what I mean is, some justification is needed that we actually lose something otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper above gives examples showing how the Cauchy-Riemann equations alone do not suffice for analyticity, &lt;em&gt;even in the presence of continuity&lt;/em&gt;, and proceeds to look for weak additions to these two requirements. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-03T06:08:10.527" />
  <row Id="8734" PostTypeId="1" AcceptedAnswerId="8745" CreationDate="2010-11-03T06:27:04.517" Score="4" ViewCount="299" Body="&lt;p&gt;After learning about the binary number system(only 2 symbols i.e. 0 and 1), I just thought why did we adopt the decimal number system (10 symbols) after all?&lt;br&gt;&#xA;I mean if you go to see, it's rather inefficient when compared to the octal(8 symbols) and the hexadecimal(16 symbols)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tag this appropriately.&lt;/p&gt;&#xA;" OwnerUserId="3062" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T07:05:01.787" LastActivityDate="2010-11-03T22:05:08.627" Title="why have we chosen our number system to be decimal (base 10)" Tags="&lt;number-systems&gt;" AnswerCount="5" CommentCount="8" />
  <row Id="8735" PostTypeId="2" ParentId="8732" CreationDate="2010-11-03T06:32:24.073" Score="1" ViewCount="0" Body="&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prelude.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. If $\alpha+\beta+\gamma=\pi$, then $\gamma=\pi-\alpha-\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. $\tan(\pi-\theta)=-\tan(\theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. $\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d. $\tan\left(\frac{\pi}{2}-\theta\right)=\frac1{\tan(\theta)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Act I.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(\alpha)+\tan(\beta)-\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=-\tan(\alpha)\tan(\beta)\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the left-hand side be made to look like the right-hand side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Act II.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)+\left(\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)\right)\left(\frac{\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)}{1-\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)}\right)^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplify the above expression.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T11:19:19.183" LastActivityDate="2010-11-03T11:19:19.183" CommentCount="1" />
  <row Id="8737" PostTypeId="2" ParentId="8727" CreationDate="2010-11-03T07:16:39.937" Score="1" ViewCount="0" Body="&lt;p&gt;If you don't want to explicitly construct the manifold from charts, you could instead use the regular value theorem. Consider the &quot;how antisymmetric is this matrix&quot; map, f:Rnxn -&gt; R, f(A)=||A-A^t||. If zero is a regular value of f, then the regular value theorem says that the inverse image of 0 (the set of matrices that are symmetric) must be a submanifold of Rnxn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one could bound Df(A)h from below with the reverse triangle inequality:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(||A+sh-(A+sh)^t||-||A-A^t||)/s &#xA;&gt;= (||A-A^t|| - s||h-h^t|| - ||A-A^t||)/s&#xA;= ||h-h^t||&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(or ... &amp;lt;= -||h-h^t|| depending on sign)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By judiciously choosing the matrix h, you can make this any real number, so Df(A) is onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the result is even stronger since we didn't use the fact that A is symmetric - one could imagine the sets of &quot;equally antisymmetric&quot; matrices (set of all A where ||A-A^t||=c) as submanifolds layered on top of each other that foliate Rnxn.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-03T07:16:39.937" />
  <row Id="8738" PostTypeId="2" ParentId="8732" CreationDate="2010-11-03T07:16:43.713" Score="4" ViewCount="0" Body="&lt;p&gt;If $A+B+C= \pi \Longrightarrow \tan(A+B) = \tan(\pi -C) =-\tan(C)$. So we have $$\tan(A+B)= \frac{\tan(A) + \tan(B)}{1 - \tan(A)\cdot \tan(B)} = -\tan(C) $$ $$\Longrightarrow \tan(A)+\tan(B) = -\tan(C) \cdot \Bigl[1 - \tan(A)\tan(B)\Bigr]$$ from which the first one follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the second one, we have $\displaystyle\frac{A}{2} + \frac{B}{2} =\frac{\pi}{2}- \frac{C}{2} \Longrightarrow \tan\Bigl(\frac{A+B}{2}\Bigr)= \cot\Bigl(\frac{C}{2}\Bigr)$ Now expanding we have $$\tan\Bigl(\frac{A+B}{2}\Bigr)= \frac{\tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr)}{1- \tan\Bigl(\frac{A}{2}\Bigr)\cdot \tan\Bigl(\frac{B}{2}\Bigr)} = \cot\Bigl(\frac{C}{2}\Bigr)$$ Multiplying both sides by $\tan\frac{C}{2}$ we have $$\tan\Bigl(\frac{C}{2}\Bigr) \cdot \Bigl[ \tan\Bigl(\frac{A}{2}\Bigr) + \tan\Bigl(\frac{B}{2}\Bigr) \Bigr] = 1 \cdot \Bigl[ 1 - \tan\Bigl(\frac{A}{2}\Bigr) \cdot \tan\Bigl(\frac{B}{2}\Bigr)\Bigr]$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-03T10:45:48.867" LastActivityDate="2010-11-03T10:45:48.867" CommentCount="3" />
  <row Id="8739" PostTypeId="1" AcceptedAnswerId="8741" CreationDate="2010-11-03T08:13:20.627" Score="6" ViewCount="88" Body="&lt;p&gt;While reading about general topology, I started looking at the metric topology. At one point, the interior of a topological space $X$ is defined as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${E}^{\circ}=\{p\in E\ | \ B_p(\epsilon)\subseteq E, \epsilon&gt;0\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $B_p(\epsilon)$ is a ball with center $p$ and radius $\epsilon$. I tried verifying that this met the general properties of an interior, and didn't have much trouble showing ${E}^{\circ}\subseteq E$, $(E\cap F)^{\circ}={E}^{\circ}\cap {F}^{\circ}$, and ${X}^{\circ}=X$. However, I don't see how ${{E}^{\circ}}^{\circ}={E}^{\circ}$. I see that ${{E}^{\circ}}^{\circ}\subseteq {E}^{\circ}$, but tried without success to see the other containment. Is there a reason why this follows simply from the axioms of what a metric is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore, apparently another way to approach this topology is to define the family of open sets $\mathcal{T}$ as those which are the union of a family of balls. Again, I tried verifying this for myself. I see that for any $\mathcal{U}\subseteq\mathcal{T}$, $\cup\mathcal{U}$ is a union of sets which each may be represented as the union of a family of balls, so the whole union again may be represented as the union of a family of balls, namely those in each set in $\mathcal{U}$. Also, $X$ may be represented as the union of balls where I take one ball with center $p$ for all $p\in X$. Finally $\emptyset$ is just an empty union of balls. However, I was unable to show that for finite $\mathcal{U}\subseteq\mathcal{T}$, $\cap\mathcal{U}$ is again open. Is there some way to see that the finite intersection could be represented as a union of balls from the properties of the metric?&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-03T08:30:07.950" Title="Verifying some basics of the metric topology" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="2" />
  <row Id="8740" PostTypeId="2" ParentId="8739" CreationDate="2010-11-03T08:30:05.797" Score="3" ViewCount="0" Body="&lt;p&gt;For the first problem, show that for every point in $E^{\circ}$ there is a ball around it which is also in $E^{\circ}$.  In other words, show that interiors are open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second problem, do the same: show that for every point in the intersection there is a ball around it which is also in the intersection.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-03T08:30:05.797" CommentCount="1" />
  <row Id="8741" PostTypeId="2" ParentId="8739" CreationDate="2010-11-03T08:30:07.950" Score="2" ViewCount="0" Body="&lt;p&gt;A hint for your first question: show that if $p \in E^{\circ}$, then there exists an $\epsilon &gt; 0$ such that $B_p(\epsilon)$ is contained in $E^{\circ}$. This shows that $p$ lies in the interior of $E^{\circ}$, i.e., $p \in (E^{\circ})^{\circ}$.  This comes down to showing that if you have a point $p$ in an open ball $B$, you can find a smaller opne ball $B'$ around $p$ which is entirely contained in $B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint for your second question: it's enough to show that if $B_1$ and $B_2$ are two open balls and $p \in B_1 \cap B_2$ (i.e., $p$ lies in both balls), then there is a small ball $B_3$ around $p$ such that $B_3$ is contained in both $B_1$ and $B_2$.  This is actually quite similar to the exercise of the previous paragraph.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-03T08:30:07.950" CommentCount="1" />
  <row Id="8742" PostTypeId="1" AcceptedAnswerId="8757" CreationDate="2010-11-03T10:20:03.050" Score="2" ViewCount="122" Body="&lt;p&gt;It's a really simple question. However I didn't see it in books and I tried to find the answer on the web but failed.&#xA;If I have two independent random variables, X1 and X2, then I define two other random variables Y1 and Y2, where Y1 = f1(X1) and Y2 = f2(X2).&#xA;Intuitively, Y1 and Y2 should be independent, and I can't find a counter example, but I am not sure. Could anyone tell me whether they are independent? Does it depend on some properties of f1 and f2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2010-11-03T14:07:55.147" Title="Are functions of independent variables also independent?" Tags="&lt;probability&gt;&lt;random-variable&gt;" AnswerCount="2" />
  <row Id="8743" PostTypeId="2" ParentId="8742" CreationDate="2010-11-03T10:45:41.317" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, they are independent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are studying rigorous probability course with sigma-algebras then you may prove it by noticing that the sigma-algebra generated by $f_{1}(X_{1})$ is smaller than the sigma-algebra generated by $X_{1}$, where $f_{1}$ is borel-measurable function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are studying an introductory course - then just remark that this theorem is consistent with our intuition: if $X_{1}$ does not contain info about $X_{2}$ then $f_{1}(X_{1})$ does not contain info about $f_{2}(X_{2})$.&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-11-03T10:45:41.317" CommentCount="1" />
  <row Id="8744" PostTypeId="2" ParentId="8732" CreationDate="2010-11-03T10:56:05.480" Score="0" ViewCount="0" Body="&lt;p&gt;More generally, we have from &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Moivre%27s_formula&quot; rel=&quot;nofollow&quot;&gt;De Moivre's theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(\alpha_1+\alpha_2+\cdots+\alpha_n)=&#xA;\text{Re}\prod_{k=1}^n(\cos \alpha_k + i \sin \alpha_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(\alpha_1+\alpha_2+\cdots+\alpha_n)=&#xA;\text{Im}\prod_{k=1}^n(\cos \alpha_k + i \sin \alpha_k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(\alpha_1+\alpha_2+\cdots+\alpha_n)=&#xA;\frac{ \text{Im}\prod_{k=1}^n ( 1 + i \tan \alpha_k) }{  \text{Re}\prod_{k=1}^n(1 + i \tan\alpha_k) }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case $n=3.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-03T11:07:10.867" LastActivityDate="2010-11-03T11:07:10.867" />
  <row Id="8745" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T10:58:22.657" Score="4" ViewCount="0" Body="&lt;p&gt;Expanding on the comment by J.M., let me quote from the (highly recommended) book by Georges Ifrah &lt;a href=&quot;http://books.google.com/books?id=FMTI7rwevZcC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Universal History of Numbers&lt;/em&gt;&lt;/a&gt; (Wiley, 2000, pp. 21-22):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Traces of the anthropomorphic origin of counting systems can be found in many languages. In the Ali language (Central Africa), for example, &quot;five&quot; and &quot;ten&quot; are respectively &lt;em&gt;moro&lt;/em&gt; and&#xA;  &lt;em&gt;mbouna&lt;/em&gt;: &lt;em&gt;moro&lt;/em&gt; is actually the word for &quot;hand&quot; and &lt;em&gt;mbouna&lt;/em&gt; is a contraction of &lt;em&gt;moro&lt;/em&gt; (&quot;five&quot;) and &lt;em&gt;bouna&lt;/em&gt;, meaning &quot;two&quot; (thus &quot;ten&quot;=&quot;two hands&quot;).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is therefore very probable that the Indo-European, Semitic and Mongolian words for the first ten numbers derive from expressions related to finger-counting. But this is an unverifiable hypothesis, since the original meanings of the names of the numbers have been lost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ifrah then goes on to explain that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the hand makes the two complementary aspects of integers entirely intuitive. It serves as an instrument permitting natural movement between cardinal and ordinal numbering. If you need to show that a set contains three, four, seven or ten elements, you raise or bend &lt;em&gt;simultaneously&lt;/em&gt; three, four, seven or ten fingers, using your hand as cardinal mapping. If you want to count out the same things, then you bend or raise three, four, seven or ten fingers in &lt;em&gt;succession&lt;/em&gt;, using the hand as an ordinal counting tool.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-03T11:15:30.873" LastActivityDate="2010-11-03T11:15:30.873" CommentCount="4" />
  <row Id="8746" PostTypeId="1" CreationDate="2010-11-03T11:15:51.843" Score="1" ViewCount="45" Body="&lt;p&gt;Consider a cube, with standard topological faces: 6 faces, 12 edges and 8 vertices.&#xA;Define a &lt;em&gt;cube part&lt;/em&gt; as set of faces (possible of different dimension) so that no 2 set elements share same vertex.&#xA;E.g. there is no cube part consisting of 3 2-dimensional faces, because there have to be 2 faces that share a vertex.&#xA;Note, cube part doesn't have to contain all cube vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many different cube parts there are, counting topologically isomorphic cube parts as same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is application of Redfield–Pólya Theorem, but I'm not familiar with topic and some hints will be usefull.&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2010-11-03T11:15:51.843" Title="Number of sets of cube topological faces" Tags="&lt;combinatorics&gt;" CommentCount="2" />
  <row Id="8747" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T11:41:34.840" Score="1" ViewCount="0" Body="&lt;p&gt;Because it makes the metric system so much simpler :).&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-11-03T11:41:34.840" CommentCount="2" />
  <row Id="8748" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T12:01:16.870" Score="6" ViewCount="0" Body="&lt;p&gt;I think the answer here might be, that the guys who thought base 10 was a good idea had the largest sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one trusts the wikipedia, the Babylonians had a base 60 system, which can still be felt today with this &quot;60 minutes in an hour&quot; nonesense, and a (related) base 12 system was widely in use too. There are still unique words for &quot;eleven&quot; and &quot;twelve&quot;, as well as expressions as &quot;a dozen&quot;. After all, you can count to twelve using a single hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there was the base 1 latin system, and (wikipedia again) a base 20 system for the mayan, who obvsiously had no shoes, since they could count their toes, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something as easy as &quot;base 10 is natural for humans&quot; does not explain it all. =)&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-11-03T12:01:16.870" CommentCount="3" />
  <row Id="8749" PostTypeId="2" ParentId="8250" CreationDate="2010-11-03T12:11:21.250" Score="8" ViewCount="0" Body="&lt;p&gt;If you want an intuition that explains &lt;em&gt;why&lt;/em&gt; Farka's Lemma should be true, you will have to use the geometric interpretation; there's no way around that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an intuition that shows &lt;em&gt;what&lt;/em&gt; Farka's Lemma achieves (but not why it should be true), there is indeed an algebraic explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, Farka's Lemma answers the following question: what is a sufficient and necessary condition for the system of equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n &amp;amp;=&amp;amp; b_1 \\&#xA;a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n &amp;amp;=&amp;amp; b_2 \\&#xA;&amp;amp;\vdots&amp;amp; \\&#xA;a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n &amp;amp;=&amp;amp; b_m&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to have a positive solution $x_i\ge 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oviously, if one of the equations has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_1x_1 + c_2x_2 + \dots + c_nx_n = b $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where all the $c_i$ are positive $c_i &gt; 0$, but the right-hand side $b$ is negative $b&amp;lt;0$, then it's clearly impossible to choose all the $x_i$ positive and have them satisfy the equation, because the left-hand side would always be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was only for a single equation, but it is equally obvious that if any &lt;em&gt;linear combination&lt;/em&gt; of the equations of your system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\sum y_i a_{i1}\right)x_1 + \left(\sum y_i a_{i2}\right)x_2 + \dots + \left(\sum y_i a_{in}\right) x_n = \sum y_i b_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has this form with positive coefficients on the left- and a negative number on the right-hand side, then you can't solve it for $x_i \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what Farka's Lemma says is that this is in fact the &lt;strong&gt;only bad thing&lt;/strong&gt; that can happen. If the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax = b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;doesn't have a solution $x\ge0$, then this must be because there exists a nasty linear combination of some equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(y^TA)x = y^Tb$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that obstructs solvability because it satisfies $y^TA \ge 0$ but $y^Tb \le 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, Farka's Lemma shows that the non-solvability of the system can be &quot;certified&quot;; the vector $y$ is a &quot;certificate&quot; for the fact that the system cannot be solved. Such certificates, commonly called &lt;strong&gt;obstructions&lt;/strong&gt;, are common in other branches of mathematics, like algebraic geometry (&lt;a href=&quot;http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Nullstellensatz&lt;/a&gt;) or algebraic topology (homology, cohomology).&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-11-03T12:11:21.250" />
  <row Id="8750" PostTypeId="1" CreationDate="2010-11-03T12:28:58.783" Score="11" ViewCount="150" Body="&lt;p&gt;The &lt;a href=&quot;http://mathworld.wolfram.com/ReuleauxPolygon.html&quot; rel=&quot;nofollow&quot;&gt;Reuleaux polygons&lt;/a&gt; are analogs of the regular polygons, except that the &quot;sides&quot; are composed of circle arcs instead of lines. It is known that for an odd number of sides, e.g. the Reuleaux triangle, the polygon has constant width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading the paper &lt;a href=&quot;http://www.maa.org/pubs/sampMMA.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Roads and Wheels&lt;/em&gt;&lt;/a&gt; by Stan Wagon and Leon Hall, I got curious on how one might construct the appropriate &quot;road&quot; for Reuleaux wheels; i.e., finding the curve such that when a Reuleaux polygon rolls on it, the axle at the centroid of the polygon experiences no vertical displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that it does not seem straightforward, at least to me, how to construct the corresponding differential equation for the road, as presented in the paper. Since circles roll on horizontal lines, and equiangular spirals roll on inclined lines, I would suppose that the road needed for a rolling Reuleaux would not be piecewise linear. &lt;a href=&quot;http://demonstrations.wolfram.com/ARollingReuleauxTriangle/&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; demonstrates that the &quot;road&quot; cannot be a horizontal line for a Reuleaux triangle, as the axle does not remain level when the curve is rolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does one construct the road? A solution for just the Reuleaux triangle would be fine, but a general solution is much better.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-03T12:28:58.783" Title="Reuleaux Rollers" Tags="&lt;geometry&gt;" CommentCount="11" FavoriteCount="3" />
  <row Id="8751" PostTypeId="2" ParentId="8717" CreationDate="2010-11-03T12:31:45.490" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a comment made by Dmitri Pavlov for &lt;a href=&quot;http://mathoverflow.net/questions/42310/when-is-l2x-separable/42314#42314&quot; rel=&quot;nofollow&quot;&gt;this MathOverFlow question&lt;/a&gt;.&#xA;Maybe it will be helpful to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The support of a measure $m$ on a measurable space is defined as follows. Consider the Boolean algebra of all measurable sets modulo null sets (i.e., two measurable sets are equivalent if their symmetric difference is a measure $0$ set). This Boolean algebra is complete, as described in the link. Now take the supremum of all elements $p$ of this algebra such that $m$ vanishes on $p$. The complement of this supremum is the support of $m$. It is extremely important to factor out the null sets, otherwise the above procedure doesn't make sense. – Dmitri Pavlov Oct 16 at 15:09&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-03T12:31:45.490" CommunityOwnedDate="2010-11-03T12:31:45.490" />
  <row Id="8752" PostTypeId="1" CreationDate="2010-11-03T12:35:37.730" Score="11" ViewCount="458" Body="&lt;p&gt;Four Color Theorem is equivalent to the statement: &quot;Every cubic planar bridgeless graphs is 3-edge colorable&quot;. There is computer assisted proof given by Appel and Haken. Dick Lipton in of his &lt;a href=&quot;http://rjlipton.wordpress.com/2009/04/24/the-four-color-theorem/&quot; rel=&quot;nofollow&quot;&gt;beautiful blogs&lt;/a&gt;&#xA;&lt;a href=&quot;http://rjlipton.wordpress.com/2009/04/24/the-four-color-theorem/&quot; rel=&quot;nofollow&quot;&gt;&lt;/a&gt; posed the following open problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there non-computer based proofs of the Four Color Theorem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, While I was reading this paper, &#xA;Anshelevich and Karagiozova, &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1250790.1250849&quot; rel=&quot;nofollow&quot;&gt;Terminal backup, 3D matching, and covering cubic graphs&lt;/a&gt;&#xA;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1250790.1250849&quot; rel=&quot;nofollow&quot;&gt;&lt;/a&gt;, the authors state that Cahit  proved that &quot;every 2-connected cubic planar graph is edge-3-colorable&quot; which is equivalent to the Four Color Theorem (I. Cahit, Spiral Chains: The Proofs of Tait's and Tutte's Three-Edge-Coloring Conjectures. arXiv preprint, math CO/0507127 v1, July 6, 2005).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does Cahit's proof resolve the open problem in Lipton's blog by providing non-computer based proof for the Four Color Theorem? Why isn't Cahit's proof widely known and accepted?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cross posted on cstheory.stackexchange.com as &lt;a href=&quot;http://cstheory.stackexchange.com/questions/2597/human-checkable-proof-of-the-four-color-theorem-closed&quot;&gt;Human checkable proof of the Four Color Theorem?&lt;/a&gt;&#xA;&lt;a href=&quot;http://cstheory.stackexchange.com/questions/2597/human-checkable-proof-of-the-four-color-theorem-closed&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2459" LastActivityDate="2010-12-04T16:44:25.170" Title="Human checkable proof of the Four Color Theorem?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="8753" PostTypeId="1" AcceptedAnswerId="8778" CreationDate="2010-11-03T12:54:54.727" Score="3" ViewCount="144" Body="&lt;p&gt;Let $$\nu(x) = \frac{1}{2} \log{2\pi x} - x + \int\limits_{1}^{x} \frac{\lfloor{t\rfloor}}{t} \ dt$$ then what is $\displaystyle \lim_{x \to \infty} x \cdot \nu(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know to proceed. Tried something using the expansions of $\log$ and seeing how the integral behaves, but couldn't figure out. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-03T23:32:08.490" Title="Evaluating a given limit" Tags="&lt;number-theory&gt;&lt;analysis&gt;" AnswerCount="1" />
  <row Id="8755" PostTypeId="1" AcceptedAnswerId="8756" CreationDate="2010-11-03T13:22:36.367" Score="3" ViewCount="74" Body="&lt;p&gt;Working on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Birthday_problem&quot; rel=&quot;nofollow&quot;&gt;generalized birthday problem&lt;/a&gt;, where you draw with replacement from $\{1,2,3, \ldots,d\}$ and look for the number of draws $n$ for which you have greater than $1/2$ chance of a match I said&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d!}{d^d(d-n)!} \lt \frac{1}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Stirling's approximation  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^d \exp (d-n)} {\exp(d) d^n (d-n)^{(d-n)}} \lt \frac{1}{2}$&lt;br&gt;&#xA;$\frac{1}{\exp(n)}(1+\frac{n}{d-n})^{(d-n)}\lt \frac{1}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking logs and using the Taylor expansion  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(d-n) \ln(1+\frac{n}{d-n})-n \lt -\ln (2)$&lt;br&gt;&#xA;$(d-n) (\frac{n}{d-n}-\frac{n^2}{2(d-n)^2})\lt -\ln(2)$&lt;br&gt;&#xA;$n^2&gt;2 \ln(2) (d-n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the n in comparison with d in the last term, because if we wanted it we should keep the next term in the expansion of ln, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n&gt;\sqrt{2 \ln (2) d}$  which agrees with the published result&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is fine, but like much math writing it obscures the difficulties of finding the solution.  Before taking logs I wanted to say &#xA;$(1+\frac{n}{d-n})^{(d-n)} \rightarrow \exp(n)$ but that left me with the unhelpful $1 \lt \frac{1}{2}$.  Seeing what happened after taking logs, I need the next term in the expansion as the first one cancels.  I know how to expand a Taylor series as far as I want, but I don't know what the next term in this one is.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-03T13:36:28.990" LastActivityDate="2010-11-03T14:03:02.883" Title="Next term in $(1+a/n)^n \rightarrow \exp (n)$" Tags="&lt;approximation&gt;&lt;taylor-expansion&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8756" PostTypeId="2" ParentId="8755" CreationDate="2010-11-03T13:31:34.247" Score="3" ViewCount="0" Body="&lt;p&gt;$$\left(1+\frac{x}{n}\right)^{n}=e^x\left(1-\frac{x^2}{2n}+\frac{x^3(8+3x)}{24n^2}+\cdots\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.maa.org/pubs/Calc_articles/ma052.pdf&quot; rel=&quot;nofollow&quot;&gt;note&lt;/a&gt; might be helpful.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-03T13:50:08.433" LastActivityDate="2010-11-03T13:50:08.433" CommentCount="3" />
  <row Id="8757" PostTypeId="2" ParentId="8742" CreationDate="2010-11-03T13:42:20.710" Score="0" ViewCount="0" Body="&lt;p&gt;For any two (measurable) sets $A_i$, $i=1,2$, $Y_i \in A_i$ if and only if $X_i \in B_i$, where $B_i$ are the sets { $s : f_i (s) \in A_i$ }. Hence, since the $X_i$ are independent, ${\rm P}(Y_1 \in A_1 , Y_2 \in A_2) = {\rm P}(Y_1 \in A_1) {\rm P}(Y_2 \in A_2)$. Thus, the  $Y_i$ are independent (which is intuitively clear anyway). [We have used here that random variables $Z_i$, $i=1,2$, are independent if and only if ${\rm P}(Z_1 \in C_1 , Z_2 \in C_2) = {\rm P}(Z_1 \in C_1) {\rm P}(Z_2 \in C_2)$ for any two measurable sets $C_i$.]&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-03T14:07:55.147" LastActivityDate="2010-11-03T14:07:55.147" CommentCount="2" />
  <row Id="8758" PostTypeId="2" ParentId="8337" CreationDate="2010-11-03T14:00:17.530" Score="2" ViewCount="0" Body="&lt;p&gt;Another variation. We make use of the following identity (proved at the bottom of this note):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \cot^2 \left( \frac {2k-1}{2n} \frac{\pi}{2} \right) = 2n^2 – n. \quad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $1/\theta &gt; \cot \theta &gt; 1/\theta - \theta/3 &gt; 0$ for $0&amp;lt; \theta&amp;lt; \pi/2 &amp;lt; \sqrt{3}$ and so&#xA;$$ 1/\theta^2 – 2/3 &amp;lt; \cot^2 \theta &amp;lt; 1/\theta^2. \quad (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\theta_k = (2k-1)\pi/4n,$ summing the inequalities $(2)$ from $k=1$ to $n$ we obtain   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2n^2 – n &amp;lt; \sum_{k=1}^n \left( \frac{2n}{2k-1}\frac{2}{\pi} \right)^2 &amp;lt; 2n^2 – n + 2n/3.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\pi^2}{16}\frac{2n^2-n}{n^2} &amp;lt; \sum_{k=1}^n \frac{1}{(2k-1)^2} &amp;lt;&#xA;\frac{\pi^2}{16}\frac{2n^2-n/3}{n^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the limit as $n \rightarrow \infty$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = \frac{\pi^2}{8},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the result for $\sum_{k=1}^\infty 1/k^2$ follows easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove $(1)$ we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos 2n\theta = \text{Re}(\cos\theta + i \sin\theta)^{2n} = &#xA;\sum_{k=0}^n (-1)^k {2n \choose 2k}\cos^{2n-2k}\theta\sin^{2k}\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\cos 2n\theta}{\sin^{2n}\theta} =  \sum_{k=0}^n (-1)^k {2n \choose 2k}\cot^{2n-2k}\theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so setting $x = \cot^2\theta$ we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \sum_{k=0}^n (-1)^k {2n \choose 2k}x^{n-k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has roots $x_j = \cot^2 (2j-1)\pi/4n,$ for $j=1,2,\ldots,n,$ from which $(1)$ follows since&#xA;${2n \choose 2n-2} = 2n^2-n.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-03T14:00:17.530" />
  <row Id="8759" PostTypeId="2" ParentId="8755" CreationDate="2010-11-03T14:03:02.883" Score="2" ViewCount="0" Body="&lt;p&gt;$$\rm\quad \bigg(1+\frac{a}n\bigg)^{\:n}\ =\ e^{n\log(1+a/n)}\ =\ e^a\ e^{n\log(1+a/n)-a}\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm n\:\log \bigg(1+\frac{a}n\bigg)-a\ =\ \frac{a^2}{2\:n} + \frac{a^3}{3\:n^2} +\: \cdots $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm e^{n\log(1+a/n)-a}\ =\ 1 - \frac{a^2}{2\:n} + \frac{3a^4 + 8a^3}{24\: n^2} +\: \cdots $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-03T14:03:02.883" />
  <row Id="8760" PostTypeId="1" CreationDate="2010-11-03T14:37:32.057" Score="0" ViewCount="164" Body="&lt;p&gt;What are the three cube roots of -1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if this is a trick question, But I have been asked this. one of the ansers is -1, what are the other 2?&lt;/p&gt;&#xA;" OwnerUserId="3069" LastEditorUserId="120" LastEditorDisplayName="" LastEditDate="2010-11-03T14:52:29.903" LastActivityDate="2010-11-03T22:33:41.657" Title="What are the three cube roots of -1?" Tags="&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8761" PostTypeId="1" CreationDate="2010-11-03T14:40:48.463" Score="0" ViewCount="60" Body="&lt;p&gt;This is a strange question, it might be easy for all you math wizards or it even may be impossible. If you don't understand what I mean let me know so I can change the way I post the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Product A can be built using Product B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get Product B I have to transform product C or Product D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 50% chance to get product B by transforming product C and 25% to get product B by transforming product D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to predict the cost of production for Product A knowing the price for Product C and D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&#xA;There are no transformation costs, only the costs for the materials &lt;/p&gt;&#xA;" OwnerUserId="3068" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-03T16:29:10.010" LastActivityDate="2010-11-03T16:29:10.010" Title="Price Calculation based on probabilities" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="8762" PostTypeId="2" ParentId="8760" CreationDate="2010-11-03T14:44:57.043" Score="2" ViewCount="0" Body="&lt;p&gt;Write $-1$ in polar form as $e^{i\pi}$. In general, the cube roots of $r e^{i\theta}$ are given by $r^{1/3}e^{i\theta/3}$, $r^{1/3}e^{i(\theta/3 + 2\pi /3)}$ and $r^{1/3}e^{i(\theta/3 + 4\pi /3)}$. In your case $r = 1$ and $\theta = \pi$, so your cube roots are $e^{i\pi / 3}$, $e^{i\pi}$, and $e^{i 5\pi/ 3}$. Put back into rectangular form, they are ${1 \over 2} + i{\sqrt{3} \over 2}$, $-1$, and ${1 \over 2} - i{\sqrt{3} \over 2}$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-03T14:44:57.043" CommentCount="1" />
  <row Id="8763" PostTypeId="2" ParentId="8760" CreationDate="2010-11-03T14:45:11.680" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Let $\rm\ \ x\ \to\ -x\ \ $ in $\rm\displaystyle\ \ \frac{1-x^3}{1-x}\ =\ 1+x+x^2\:.\ $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally suppose $\rm\:f(x)\:$ is a polynomial over a field with roots $\rm\: a \ne b\:$. Then $\rm\ f(x) = (x-a)\ g(x)\ $  hence $\rm\: f(b) = 0\: \Rightarrow\ (a-b)\:g(b) = 0\ \Rightarrow\ g(b) = 0\ $ i.e. $\rm\:b\:$ is a root of $\rm\ f(x)/(x-a)\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a factorization perspective, the reason that this works is because, over a domain, monic linear polynomials are prime, so the linear factors of a polynomial are unique, i.e. the roots and their multiplicity are unique. e.g. see my &lt;a href=&quot;http://math.stackexchange.com/questions/7622&quot;&gt;post here&lt;/a&gt;. This fails over coefficient rings that are not domains, i.e. have zero-divisors, e.g. $\rm\ x^2-1 = (x-1)(x+1) = (x-4)(x+4)\ $ over $\ \mathbb Z/15\:$. Here, although $4 \ne 1$ is a root of $\rm\ x^2 - 1$ it is not true that 4 is a root of $\rm\ (x^2-1)/(x-1) = x+1\:$. For the example at hand we have $\rm\ x^3 + 1 = (x+1)(x+9)(x-10) = (x+16)(x+22)(x-38)\ $ over $\ \mathbb Z/91\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T16:00:28.907" LastActivityDate="2010-11-03T16:00:28.907" />
  <row Id="8764" PostTypeId="1" CreationDate="2010-11-03T14:58:36.130" Score="1" ViewCount="49" Body="&lt;p&gt;I'm working to understand some noise on some of our analog-to-digital converter signals. I've analyzed a log of 2^18 samples in MATLAB, and can make the following observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the noise consists of large single-sample spikes on top of my expected signal&lt;/li&gt;&#xA;&lt;li&gt;the FFT spectrum of the noise looks white (hard to say since there is some signal there as well, but all the peaks seem to be at the fundamental + harmonics of my signal)&lt;/li&gt;&#xA;&lt;li&gt;amplitude of the spikes seems equally likely to be positive or negative&lt;/li&gt;&#xA;&lt;li&gt;the distribution of spike amplitudes seems to be approx exp(-K*|x|): that is, if I do either an ascending or descending sort of the data, and plot it on a semilog scale (log-x, linear y) then at the low end of the sort (i.e. the outliers) the distribution looks linear on the semilog plot.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do any of you know of a random process that might meet/approximate the above criteria? I am looking for some insight into what type of noise this is, so I might better guess where it is coming from.&lt;/p&gt;&#xA;" OwnerUserId="120" LastActivityDate="2010-11-03T22:07:16.063" Title="characterizing noise PDF" Tags="&lt;stochastic-processes&gt;&lt;noise&gt;" AnswerCount="1" />
  <row Id="8765" PostTypeId="1" CreationDate="2010-11-03T15:04:05.663" Score="1" ViewCount="95" Body="&lt;p&gt;((n+1)H&lt;sub&gt;n&lt;/sub&gt; - (n+1)) + H&lt;sub&gt;n+1&lt;/sub&gt; = (n+2)H&lt;sub&gt;n+1&lt;/sub&gt; - (n+2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove the above statement.&lt;/p&gt;&#xA;" OwnerUserId="3071" LastEditorUserId="3071" LastEditorDisplayName="" LastEditDate="2010-11-03T15:37:08.557" LastActivityDate="2010-11-03T16:38:07.713" Title="I would like a hint on this problem regarding harmonic numbers." Tags="&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;induction&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="8766" PostTypeId="1" AcceptedAnswerId="8767" CreationDate="2010-11-03T15:04:28.713" Score="1" ViewCount="97" Body="&lt;p&gt;I want to find all the roots of a polynomial and decided to compute the eigenvalues of its companion matrix.&#xA;How do I do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have this polynomial: $4x^3 - 3x^2 + 9x - 1$, I compute the companion matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix} 0&amp;amp;0&amp;amp;\frac{3}{4} \\ 1&amp;amp;0&amp;amp;-\frac{9}{4} \\ 0&amp;amp;1&amp;amp;\frac{1}{4} \end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how can I find the eigenvalues of this matrix?&lt;br&gt;&#xA;Thanks in advance,&lt;br&gt;&#xA;rubik&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="2582" LastEditorDisplayName="" LastEditDate="2010-11-03T15:30:00.983" LastActivityDate="2010-11-03T15:30:00.983" Title="Eigenvalues of matrix" Tags="&lt;matrices&gt;&lt;polynomials&gt;&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="8767" PostTypeId="2" ParentId="8766" CreationDate="2010-11-03T15:16:30.933" Score="4" ViewCount="0" Body="&lt;p&gt;Typically, one uses the &lt;a href=&quot;http://dx.doi.org/10.1007/BF02163331&quot; rel=&quot;nofollow&quot;&gt;QR algorithm&lt;/a&gt; on a matrix with a preliminary reduction to upper Hessenberg form (all zeroes below the subdiagonal) for determining the eigenvalues. However, since the Frobenius companion matrix is already Hessenberg to begin with, it can easily be fed into an eigenroutine without preliminary transformation (though a preliminary &lt;a href=&quot;http://dx.doi.org/10.1007/BF02165404&quot; rel=&quot;nofollow&quot;&gt;balancing&lt;/a&gt; might be necessary if the coefficients of the polynomial vary widely in magnitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is a rather wasteful algorithm: the Frobenius companion matrix is sparse, and subjecting it to the QR algorithm causes the zeroes above the subdiagonal to be filled in quickly. However, there are now &lt;a href=&quot;http://scg.ece.ucsb.edu/publications/oadv2.pdf&quot; rel=&quot;nofollow&quot;&gt;new methods&lt;/a&gt; that exploit the sparsity of the Frobenius companion matrix; if you're sure you do not want to investigate &lt;a href=&quot;http://books.google.com/books?id=4PMqxwG-eqQC&quot; rel=&quot;nofollow&quot;&gt;other methods&lt;/a&gt;, you should look into this new development.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-03T15:16:30.933" CommentCount="1" />
  <row Id="8768" PostTypeId="2" ParentId="8766" CreationDate="2010-11-03T15:17:08.653" Score="3" ViewCount="0" Body="&lt;p&gt;You may want to take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;Companion matrix&lt;/a&gt; to be sure that you have the right companion matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the eigenvalues of a square matrix, you need to solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{det}(A-\lambda I)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is the matrix you are interested in and I is the identity matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a couple of examples &lt;a href=&quot;http://en.wikipedia.org/wiki/Eigenvalue,_eigenvector_and_eigenspace#Computation_of_eigenvalues.2C_and_the_characteristic_equation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-11-03T15:17:08.653" CommentCount="7" />
  <row Id="8769" PostTypeId="1" CreationDate="2010-11-03T15:24:41.263" Score="-2" ViewCount="327" Body="&lt;p&gt;I am a student in Iran. I must write a paper in algebraic geometry in next twoyears. But I have not any idea and I dont know how can I start studing?&#xA;So do you know any easy field or idea?&lt;/p&gt;&#xA;" OwnerUserId="3073" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-03T16:01:07.223" LastActivityDate="2010-11-04T17:32:59.463" ClosedDate="2010-11-04T06:27:55.803" Title="How can I write a paper in algebraic geometry?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="8770" PostTypeId="2" ParentId="7454" CreationDate="2010-11-03T15:31:07.193" Score="2" ViewCount="0" Body="&lt;p&gt;The only help I can give is to suggest a lower bound.  You have cn students and the study together graph could then have cn(cn-1)/2 edges.  Each session you pick $\frac{cn(n-1)}{2}$ edges to color in and you ask whether after y sessions all the edges are colored.  If you ignore the class grouping, you can do the same problem randomly choosing $\frac{ycn(n-1)}{2}$ edges independently with replacement to color.  I think your case will color the whole graph with higher probability, as no edge can claim more than $y$ of the colorings, but it should be close.  Now this is a nice Poisson distribution.  Each edge is colored with probability $1-\exp(-\lambda)$, where $\lambda$ is the average number of colorings each edge receives, here $\frac{ycn(n-1)}{cn(cn-1)}$, or just about $\frac{y}{c}$.  The chance that all edges are colored is then $(1-\exp(-\lambda))^{\frac{cn(cn-1)}{2}}$, which as you say is very small.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-03T15:31:07.193" CommentCount="2" />
  <row Id="8771" PostTypeId="2" ParentId="8765" CreationDate="2010-11-03T15:44:28.530" Score="6" ViewCount="0" Body="&lt;p&gt;Hint: Your identity is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(n+1)(H_{n+1}-H_{n})=1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is something you can deduce by a simple algebraic manipulation.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-03T15:44:28.530" CommentCount="1" />
  <row Id="8772" PostTypeId="2" ParentId="8761" CreationDate="2010-11-03T15:51:21.637" Score="2" ViewCount="0" Body="&lt;p&gt;You haven't given any information about the cost of turning A into B or turning C or D into B.  Can we assume these are zero?  In any case, it's not too hard to compute the probability distribution of costs of A by just working out similarly to what follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you go with the strategy &quot;buy C and attempt a transformation to B until you get B&quot;.  Then with probability 1/2, this will cost C plus whatever it costs to turn C into B, which I'll call $B_C$, with probability 1/4, cost $2C+B_C$, with probability $1/2^n$ cost $nC+B_C$.  Thus the expected price will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_C+C\sum_{n=1}^\infty \frac{n}{2^n}=B_C+2C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the strategy &quot;buy D and attempt a transformation to B until you get B&quot; will have expected price&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_D+D\sum_{n=1}^\infty \frac{n}{4^n}=B_D+4D/9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll want to compare these two and see which is minimal and run with that strategy.&lt;/p&gt;&#xA;" OwnerUserId="826" LastActivityDate="2010-11-03T15:51:21.637" CommentCount="2" />
  <row Id="8773" PostTypeId="2" ParentId="8337" CreationDate="2010-11-03T15:56:10.233" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a complex-analytic proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $z\in D=\mathbb{C}\backslash${$0,1$}, let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(z)=\sum\frac{1}{\log^2 z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the sum is taken over all branches of the logarithm. Each point in $D$ has a neighbourhood on which the branches of $\log(z)$ are analytic. Since the series converges uniformly away from $z=1$, $R(z)$ is analytic on $D$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a few observations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Each term of the series tends to $0$ as $z\to0$. Thanks to the uniform convergence this implies that the singularity at $z=0$ is removable and we can set $R(0)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The only singularity of $R$ is a double pole at $z=1$ due to the contribution of the principal branch of $\log z$. Moreover, $\lim_{z\to1}(z-1)^2R(z)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) $R(1/z)=R(z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By (i) and (iii) $R$ is meromorphic on the extended complex plane, therefore it is rational. By (ii) the denominator of $R(z)$ is $(z-1)^2$. Since $R(0)=R(\infty)=0$, the numerator has the form $az$. Then (ii) implies $a=1$, so that&#xA;$$R(z)=\frac{z}{(z-1)^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, setting $z=e^{2\pi i w}$ yields&#xA;$$\sum\limits_{n=-\infty}^{\infty}\frac{1}{(w-n)^2}=\frac{\pi^2}{\sin^2(\pi w)}$$&#xA;which implies that $$\sum\limits_{k=0}^{\infty}\frac{1}{(2k+1)^2}=\frac{\pi^2}{8},$$&#xA;and the identity $\zeta(2)=\pi^2/6$ follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is due to T. Marshall (&lt;em&gt;American Mathematical Monthly&lt;/em&gt;, Vol. 117(4), 2010, P. 352).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-03T18:52:08.830" LastActivityDate="2010-11-03T18:52:08.830" />
  <row Id="8774" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T15:56:10.447" Score="1" ViewCount="0" Body="&lt;p&gt;It is believed that the decimal system evolved mainly due to anthropomorphic reasons (5 digits on each hand) and is thought to be a simplification of the Babylonian sexagesimal (base 60) counting method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this analogy precise, note that the normal hand has 4 fingers (excluding the thumb) with 3 segments, along with 5 digits on the other hand to be used as segment pointers. This gives 3 x 4 x 5 = 60 unique configurations.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-03T17:09:48.730" LastActivityDate="2010-11-03T17:09:48.730" CommentCount="4" />
  <row Id="8775" PostTypeId="1" AcceptedAnswerId="8786" CreationDate="2010-11-03T16:14:18.130" Score="7" ViewCount="257" Body="&lt;p&gt;Start with the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^n i^3 = \left( \sum_{i = 1}^n i \right)^2 = \left(\frac{n(n+1)}{2}\right)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate the left-most term with respect to $i$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{di} \sum_{i=1}^n i^3 = 3 \sum_{i = 1}^n i^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate the right-most term with respect to $n$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dn} \left(\frac{n(n+1)}{2}\right)^2 = \frac{1}{2}n(n+1)(2n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equate the derivatives, obtaining&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is known to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any neat reason why this method happens to get lucky and work for this case?&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-11-04T00:11:05.833" Title="An incorrect method to sum the first $n$ squares which nevertheless works" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="8776" PostTypeId="2" ParentId="8769" CreationDate="2010-11-03T16:22:27.850" Score="1" ViewCount="0" Body="&lt;p&gt;Hi Hamid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can sense your anxiety, so I'll be happy to offer some advice. You'll first need to read some algebraic geometry and digest some of its results. Finding a reasonable problem to work on should follow from an extension of some basic exercises. After a few months of determining what is known (and what isn't) about a given problem/topic, it'll be clearer how to proceed -- the problem may even define itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What category does your paper fall under: a Ph.D. thesis, a minor thesis or a review article?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basic results like &lt;em&gt;Bezout's Theorem&lt;/em&gt; or elementary &lt;em&gt;Intersection Multiplicity&lt;/em&gt; might be appropriate for the introduction section in all three. This might be a good place to start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T17:32:59.463" LastActivityDate="2010-11-04T17:32:59.463" CommentCount="2" />
  <row Id="8777" PostTypeId="2" ParentId="8765" CreationDate="2010-11-03T16:28:58.883" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If $\rm\:f(n)\:$ satsifies a first-order recurrence $\rm\ \ f(n+1) = a\ f(n) + b\ $ then it may be used as a rewrite rule to reduce $\rm\ c_0\ f(n) + c_1\ f(n+1) + \:\cdots\: + c_k\ f(n+k)\ $ to the form $\rm\ d\ f(n) + e\:.\ $ Said in operator form, if $\rm \ S^2 = a\ S + b\ $ then $\rm\ f(S)\ \:mod\:\ (S^2-a\ S-b)\ =\ c\ S + d\ $ has degree 1, since one can eliminate $\rm\:S^n,\ n&gt;2\ $ via the rewrite rule $\rm\: S^2 \to a\ S + b\:$. In essence one is employing the division / Euclidean algorithm to compute a remainder in a polynomial ring of operators. Precisely the same method works for solutions of any (monic) linear recurrence or differential equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-03T16:38:07.713" LastActivityDate="2010-11-03T16:38:07.713" />
  <row Id="8778" PostTypeId="2" ParentId="8753" CreationDate="2010-11-03T16:44:24.363" Score="4" ViewCount="0" Body="&lt;p&gt;[The answer is given at the end.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let $n = \left\lfloor x \right\rfloor$. We have $\int_k^{k + 1} {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t}  = k[\log (k + 1) - \log (k)]$, $k=1,\ldots,n-1$, and $\int_n^x {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t}  = n[\log (x) - \log (n)]$. From this it follows straightforwardly that &#xA;$$&#xA;\nu(x) = \frac{1}{2}\log (2\pi x) - x - \log (n!) + n\log (x).&#xA;$$&#xA;Next, let $\delta_x = x - n$. Thus, $x = n + \delta_x$. Substituting this for $x$ in the above expression, and using &#xA;$$&#xA;\log (M + a) = \log (M) + \frac{a}{M} - \frac{{a^2 }}{{2M^2 }} + O\bigg(\frac{1}{{M^3 }}\bigg), &#xA;$$&#xA;as $M \to \infty$, leads to&#xA;$$&#xA;\nu (x) = \frac{1}{2}\log (2\pi n) - n - \log (n!) + n\log (n) + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),&#xA;$$&#xA;or&#xA;$$&#xA;\nu (x) = \log \Bigg[\frac{{(\sqrt {2\pi n} )e^{ - n} n^n }}{{n!}}\Bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),&#xA;$$&#xA;as $n \to \infty$. Then, from the well-known estimate&#xA;$$&#xA;n! = (\sqrt {2\pi n} )e^{ - n} n^n \bigg(1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg)&#xA;$$&#xA;(and noting that $\log(1/\xi)=-\log(\xi)$), it follows that&#xA;$$&#xA;\nu (x) =  - \log \bigg[1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),&#xA;$$&#xA;and in turn,&#xA;$$&#xA;\nu (x) =  - \frac{1}{{12n}} + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg).&#xA;$$&#xA;Finally, since $x = n + \delta_x$, we obtain&#xA;$$&#xA;x \nu(x) = - \frac{1}{{12}} + \frac{{\delta _x }}{{2}} - \frac{{\delta _x^2 }}{{2}} + O\bigg(\frac{1}{{n}}\bigg).$$&#xA;We conclude that $x \nu(x)$ does not have a limit as $x \to \infty$. Rather, since $\delta_x /2 - \delta_x^2 / 2$ is oscillating between $0$ and $1/8$, we have&#xA;$$&#xA;\mathop {\lim \inf }\limits_{x \to \infty } [x\nu (x)] =  - \frac{1}{{12}} \;\;\; {\rm and} \;\;\; \mathop {\lim \sup }\limits_{x \to \infty } [x\nu (x)] =  \frac{1}{{24}}.&#xA;$$&#xA;Nevertheless, we have&#xA;$$&#xA;\mathop {\lim }\limits_{n \to \infty } n\nu (n) =  - \frac{1}{{12}},&#xA;$$&#xA;as $n \to \infty$ (integer).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-03T23:32:08.490" LastActivityDate="2010-11-03T23:32:08.490" CommentCount="9" />
  <row Id="8779" PostTypeId="1" AcceptedAnswerId="8782" CreationDate="2010-11-03T16:53:02.370" Score="3" ViewCount="126" Body="&lt;p&gt;Prove that $f_n\to f$ in measure on $E$ if and only if given $\varepsilon&gt;0$, there exists $K$ such that |{$x\in E : |f(x)-f_k(x)|&gt;\varepsilon$}|$&amp;lt;\varepsilon$ for $k\ge K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;only if&quot; direction of this is immediate from the definition of convergence in measure, but the other direction is less obvious to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, we suppose that given $\varepsilon &gt;0$, there is a $K$ such that |{$x\in E : |f(x)-f_k(x)|&gt;\varepsilon$}|$&amp;lt;\varepsilon$ for $k\ge K$. My initial thought was to bound the measure of the set in question by, say, $\frac{1}{k}$. But I'm not sure I can do that because $\varepsilon$ not only bounds the measure of the set, but the set also depends on the choice of $\varepsilon$. To show something convergence in measure, I need to show that &lt;em&gt;for every $\varepsilon$&lt;/em&gt; the limit as $k\to\infty$ of the measures of those sets is zero...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Subquestion: is the use of |$\cdot$| standard for denoting Lebesgue measure? I had never seen it until this course. I had always seen $m(-)$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Sub-subquestion: is there any particular reason that set brackets don't display in math mode? The commands \ { and \ } didn't do anything...]&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2010-11-03T17:16:34.853" LastActivityDate="2010-11-03T17:16:34.853" Title="Characterization of convergence in measure" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8780" PostTypeId="1" AcceptedAnswerId="8783" CreationDate="2010-11-03T16:57:06.667" Score="6" ViewCount="125" Body="&lt;p&gt;I am wondering if there is a general coordinate-independent way to define a Partial Differential Equation on a Smooth manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is definitely true that in each coordinate neighborhood you could define a function to satisfy a differential equation, but when you change coordinates then the differential equation will most likely have a different form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in $\mathbb{R}^3$ one says a function is harmonic if it satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial^2}{\partial x^2} f + \frac{\partial}{\partial y^2} f + \frac{\partial}{\partial z^2} f = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it is &lt;strong&gt;not&lt;/strong&gt; true that in spherical coordinates a function is harmonic if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial^2}{\partial r^2} f + \frac{\partial}{\partial \theta^2} f + \frac{\partial}{\partial \phi^2} f = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of variables to spherical coordinates gives you a much different PDE. So basically, how can you define a differential operator which is coordinate independent on a smooth manifold, so that you can have some notion of a PDE on a manifold.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-03T19:01:01.040" Title="PDEs on Manifolds" Tags="&lt;analysis&gt;&lt;differential-equations&gt;&lt;differential-geometry&gt;" AnswerCount="2" FavoriteCount="5" />
  <row Id="8782" PostTypeId="2" ParentId="8779" CreationDate="2010-11-03T17:10:53.703" Score="4" ViewCount="0" Body="&lt;p&gt;Convergence in measure means that for all $\varepsilon\gt0$, $|\{x\in E : |f(x)-f_k(x)|&gt;\varepsilon\}|$ goes to $0$, which means that for all $\varepsilon\gt0$, for all $\delta\gt0$, there exists $K$ such that $|\{x\in E : |f(x)-f_k(x)|&gt;\varepsilon\}|&amp;lt;\delta$ for all $k\geq K$.  Taking $\delta=\varepsilon$ gives the &quot;only if&quot;.  To see &quot;if&quot;, given positive $\varepsilon$ and $\delta$, take $K$ such that $k\geq K$ implies that $|\{x\in E : |f(x)-f_k(x)|&gt;\min(\varepsilon,\delta)\}|&amp;lt;\min(\varepsilon,\delta)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-03T17:10:53.703" CommentCount="4" />
  <row Id="8783" PostTypeId="2" ParentId="8780" CreationDate="2010-11-03T17:15:40.003" Score="5" ViewCount="0" Body="&lt;p&gt;You could view the concept of a &quot;Jet bundle&quot; to be precisely what you need to make partial differential equations coordinate-independent on manifolds, as points in jet bundles represent precisely the Taylor polynomials of functions in local coordinates.  Take a look at Chapter 4 in Hirsch's &quot;Differential Topology&quot; textbook for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying problem of course is that the tangent bundle of a manifold is not trivial in general so you have to do something new to make sense of partial differential equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the simplest would be to view a vector field as a 1st order differential equation via the directional derivative operator.  From there you can take mixed directional derivatives.  This isn't ideal because you can't express &quot;natural&quot; differential equations like the Laplacian in this manner (for example, think of $S^2$ -- every vector field has a zero...).  edit: as pointed out, on $S^2$ there &lt;em&gt;is&lt;/em&gt; a formulation of the Laplacian in terms of iterated 1st order DE's. I suppose I should have phrased things as &quot;even if you can express a PDE as iterated 1st order DEs on a manifold, it may not be desirable due to the breaking of symmetries brought on by bringing in vector fields.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most direct approach is to think of partial differential equations as functions on jet bundles.  But you can frequently express interesting differential equations in other manners -- having a connection on a manifold is one of the simplest ways of turning 2nd derivatives into tangential data.  The relevant formulation is called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ehresmann_connection&quot; rel=&quot;nofollow&quot;&gt;Ehresmann Connection&lt;/a&gt;. But there's all kinds of other structures on manifolds where PDEs can lurk.  Differential forms give you a standard language -- exterior derivative and Hodge star operators on Riemann manifolds give you a language where you can express the Laplacian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a very good question in that to a large extent the language you need to express a PDE on a manifold very much informs on what you actually know geometrically about the PDE.  So it's a key step in getting insight into the nature of a PDE.  Having a PDE at the jet bundle level basically means you know nothing about it, other than it's a PDE.&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-03T18:47:53.680" LastActivityDate="2010-11-03T18:47:53.680" CommentCount="13" />
  <row Id="8785" PostTypeId="2" ParentId="8775" CreationDate="2010-11-03T17:34:18.277" Score="8" ViewCount="0" Body="&lt;p&gt;Consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's formula&lt;/a&gt; which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_1^n i^m = \frac{1}{m+1} \sum_0^m {m+1 \choose k} B_k n^{m+1-k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_k$ is the k'th Bernoulli number. Take the derivative with respect to $n$ on the right hand side, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dn} \sum_1^n i^m = \frac{1}{m+1} \sum_0^m \frac{(m+1)\cdot m!}{(m+1 - k)\cdot (m -1 + 1 - k)! k!}\cdot (m+1-k) B_k n^{m-1 + 1-k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_0^m { m \choose k} B_k n^{m-k} = m \sum_1^n i^{m-1} + B_m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the difference between formally taking the derivative w.r.t. $i$ as you did, and formally taking the derivative w.r.t. $n$, is exactly the Bernoulli constant for the power $m$. Now Bernoulli's constant is equal to 0 for all odd $m$ that is greater than 1. For those cases your observation is also true: the two &quot;derivatives&quot; are equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, some random guessing, since I am not an analytic number theorist. The deep connection to Bernoulli numbers probabaly has something to do with analytic number theory and harmonic analysis that I do not know myself. That the formulae work for odd powers is possibly something related to the method of descent, where analytical results in even dimensions can be arrived from results in one higher dimension via some sort of trace. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T17:34:18.277" />
  <row Id="8786" PostTypeId="2" ParentId="8775" CreationDate="2010-11-03T17:37:20.363" Score="16" ViewCount="0" Body="&lt;p&gt;Let $f_k(n)=\sum_{i=1}^n i^k$. We all know that $f_k$ is actually&#xA;a polynomial of degree $k+1$. Also $f_k$ can be characterised by the two&#xA;conditions:&#xA;$$f_k(x)-f_k(x-1)=x^k$$&#xA;and&#xA;$$f_k(0)=0.$$&#xA;Differentiating the first condition gives&#xA;$$f_k'(x)-f_k'(x-1)=k x^{k-1}.$$&#xA;Therefore the polynomial $(1/k)f_k'$ satisfies the &lt;strong&gt;first&lt;/strong&gt; of the two&#xA;conditions that $f_{k-1}$ does. But it may not satisfy the second. But then&#xA;$(1/k)(f_k'(x)-f_k'(0))$ does. So&#xA;$$f_{k-1}(x)=\frac{f_k'(x)-f_k'(0)}k.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mysterious numbers $f_k'(0)$ are related to the Bernoulli numbers,&#xA;and when $k\ge3$ is odd they obligingly vanish...&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-03T17:37:20.363" CommentCount="2" />
  <row Id="8787" PostTypeId="1" AcceptedAnswerId="8789" CreationDate="2010-11-03T17:44:30.137" Score="0" ViewCount="75" Body="&lt;p&gt;I'm interested in finding a recursion or simple representation for a &quot;Hadamard product&quot; of two power series.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Hadamard Product&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Hadamard product is defined on generating functions $f(x)$ and $f(y)$ as $f(z)$ below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=0}^\infty{c_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(y) = \sum_{i=0}^\infty{d_i y^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z) = \sum_{i=0}^\infty{c_i \cdot d_i z^i}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My Particular Interest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now suppose that we work on the series above with only the first N terms as possibly nonzero, and the rest will be considered to be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the coefficients of $f(x)$ to be the reciprocals of the first $N$ naturals, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=1}^N{\frac{1}{i} x^{i+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $f(y)$ is in the form of an expression, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(y) = \frac{1-(2y)^{N-1}}{1 - 2y}$ represents the first N terms of the form $(2y)^N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(y) = \frac{1-(n+1)x^N+Nx^{N+1}}{(x-1)^2}$ represents the first N naturals in series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or some similar function...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can we find a simple expression, such as a recursion or formula, to represent the &quot;Hadamard product&quot; of these new series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could simply write out the $N$ terms in the resulting function $f(z)$.  However, I'm wondering if there is a more compact way of representing this.  I'm particularly interested when $f(x)$ is the series of reciprocals mentioned above.  However, I am interested in finding this result for almost any $f(y)$.  Is there a way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to consider both the cases when coefficients of $f(y)$ take on real values and when it is complex-valued.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-03T17:52:38.863" Title="Is an easier formulation possible?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;intuition&gt;" AnswerCount="1" />
  <row Id="8789" PostTypeId="2" ParentId="8787" CreationDate="2010-11-03T17:52:38.863" Score="1" ViewCount="0" Body="&lt;p&gt;It's $z \int_0^z \frac{f(y)}{y^2} dy$ after you remove the constant and linear terms.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-03T17:52:38.863" CommentCount="5" />
  <row Id="8790" PostTypeId="2" ParentId="8731" CreationDate="2010-11-03T18:16:20.580" Score="2" ViewCount="0" Body="&lt;p&gt;There are many algorithms for computing shortest paths on polyhedral 2-manifolds.&#xA;With a student I computed the shortest paths shown below with one, the Chen-Han algorithm.&#xA;The algorithms fan out shortest paths to a &lt;em&gt;frontier&lt;/em&gt;, mimicking the structure&#xA;(but not the details) of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;Dijkstra's algorithm&lt;/a&gt; for shortest paths in a graph.  Sometimes the method is called the &quot;continuous Dijkstra&quot; method.  These algorithms (all in $\mathbb{R}^3$) are described in many places, including &#xA;the book &lt;em&gt;&lt;a href=&quot;http://www.gfalop.org&quot; rel=&quot;nofollow&quot;&gt;Geometric Folding Algorithms: Linkages, Origami, Polyhedra&lt;/a&gt;&lt;/em&gt;, Chapter 24.&#xA;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/DCiki.jpg&quot; alt=&quot;Polyhedron&quot;&gt;&#xA;&lt;br /&gt;&#xA;I would assume the same approach will work for triangulated manifolds in arbitrary dimensions, but of course it will be significantly more complicated to implement.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-11-03T18:16:20.580" CommentCount="3" />
  <row Id="8791" PostTypeId="1" AcceptedAnswerId="8796" CreationDate="2010-11-03T18:59:27.593" Score="1" ViewCount="146" Body="&lt;p&gt;I got this problem from one of my mates, and i rearranged them, and got it in a summable form. From here could anyone tell me as to how i can sum up this interesting series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one sum the given series: $$S = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{1}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{1}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{1}{5^{3}} \cdot \frac{1}{7} + \cdots + \text{ad inf}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-04T10:10:46.787" Title="Summing a given series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="8792" PostTypeId="2" ParentId="8780" CreationDate="2010-11-03T19:01:01.040" Score="7" ViewCount="0" Body="&lt;p&gt;Ryan already gave a very good answer. I just want to give a few clarifications on how one should think about PDEs in general. Most importantly: you should &lt;em&gt;not&lt;/em&gt; think a differential operator as defined by &quot;a formula&quot;. What do I mean by that? In your question, you wrote that the Laplacian in Euclidean coordinates is $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$, but in Spherical coordinates it is not $\frac{\partial^2}{\partial r^2} + \frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial \phi^2}$. But why should one have thought that the two expressions are the same? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate, think of the function on the plane $f(x,y) = x^2 + y^2$. This represents the square of the distance of a point from the origin. In radial coordinates, the same function is written as $f(r,\theta) = r^2$. &lt;em&gt;Not&lt;/em&gt; $r^2 + \theta^2$, which is a very different beast.  Just like how the direct replacement $x \to r$ and $y\to \theta$ changes the &lt;em&gt;function&lt;/em&gt; you are considering, the replacements $\frac{\partial}{\partial x} \to \frac{\partial}{\partial r}$ etc changes the &lt;em&gt;operator&lt;/em&gt; you are considering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: to think geometrically (on manifolds), a (real valued) function is an assignment of a number to each point on the manifold. Then in any coordinate system there is a representation of the function by a formula. Since we like to work with formulas, we can interpret the notion of a function as a &quot;map&quot;. This map takes as its input a coordinate system, and outputs a formula that represents your function in that coordinate system. (If you are familiar with category theory, this is similar to how one can think in terms of arrows instead of dots.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a partial differential operator can be thought of as an object in itself. Our usual way of writing it as partial derivatives in some coordinate system with coefficients is just a convenient representation of the real object upstairs. Therefore, analogous to how a function is merely a &quot;map&quot; from coordinate system to the formulaic representation, you can think of a partial differential operator also as a &quot;map&quot; which takes as its input a coordinate system and outputs an expression which is a sum of partial derivatives with some coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jet bundle formulation is just a sophisticated, rigorous way of formulating this idea. The main difficulty is making sure the intuition above is compatible with the change of variables formulas. Suppose we are given a coordinate system $A$, and an expression in terms of partial derivatives which we'll call $L_A$. The question becomes: does there exist some abstract partial differential operator $L$ such that its representation in the coordinate system $A$ is precisely $L_A$? And if there is such an operator, is the fact that, for some function $f$ (written in coordinate system $A$ as $f_A$), $L_A f_A = 0$ invariant under change of coordinate system? (That given any other coordinate system $B$, $L_Bf_B = 0$ also.) The tools of jet bundles allows you to make such a consistent definition/description of partial differential equations. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T19:01:01.040" CommentCount="1" />
  <row Id="8793" PostTypeId="1" CreationDate="2010-11-03T19:14:30.710" Score="3" ViewCount="96" Body="&lt;p&gt;Let $B \subset [0,2\pi]$ be a Lebesgue measurable set. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{B} \cos(nx) dx = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK I did this assuming B is an open interval, this is pretty easy using the fact that the sine function is bounded by 1. Now I'm stuck in the general case, I'm somewhat confused by &quot;Lebesgue measurable&quot; I know this means that the measure is given by the outer measure i.e the infimum of the sum of all measures of open covers of B. But I'm having trouble writing it, I get confused when working with Lebesgue measurable sets. Can you please help me?&lt;/p&gt;&#xA;" OwnerUserId="3079" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-03T19:19:47.510" LastActivityDate="2010-11-03T20:23:01.670" Title="Compute integral of a Lebesgue measurable set" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8794" PostTypeId="1" AcceptedAnswerId="8803" CreationDate="2010-11-03T19:18:53.260" Score="7" ViewCount="77" Body="&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; In odd dimensions, solutions to the wave equation: $u_{tt}(x,t)=\nabla u(x,t)$, $u_t(x,0)=0$, $u(x,0)=f(x)$, ($t\geq 0, x\in \mathbb{R}^n$) have the nice property that the value of $u(x,t)$ only depends on the values $f(y)$ with $|y-x|=t$. For even dimensions, the value $u(x,t)$ depend on all the values $f(y)$ with $|y-x|\leq t$. A consequence of this is, that when you switch a light bulb on and then off (in 3D), there will be a light wave traveling with the speed on light (D'oh) and behind the wave, there will be total darkness. But when you throw a rock into a pond (with a 2D surface), there will be lots of wave traveling outwards from where the rock hit the water and, in theory, the water will never a still again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can anyone give an intuitive explanation of this difference between odd and even dimensions?&lt;/p&gt;&#xA;" OwnerUserId="3078" LastActivityDate="2010-11-03T20:52:15.877" Title="Intuitive explanation of the difference between waves in odd and even dimensions" Tags="&lt;intuition&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="8795" PostTypeId="1" AcceptedAnswerId="8801" CreationDate="2010-11-03T19:24:17.623" Score="4" ViewCount="110" Body="&lt;p&gt;The topological $K$-theory of a complex compact manifold $X$ is the commutative monoid $K(X)$ of isomorphism classes of complex vector bundles. Two classes $[E]$ and $[F]$ are equivalent in $K$-theory if they are &lt;i&gt;stably equivalent&lt;/i&gt;: &#xA;there exists a trivial bundle $G$ with &#xA;$$&#xA;(E \oplus G) \cong (F \oplus G).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In what sense is this the correct notion of equivalence in $K(X)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, what does stable equivalence buy you?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="3008" LastEditorDisplayName="" LastEditDate="2010-11-03T19:42:25.907" LastActivityDate="2010-11-04T23:04:39.120" Title="Why is stable equivalence necessary in topological K-theory?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8796" PostTypeId="2" ParentId="8791" CreationDate="2010-11-03T19:25:24.503" Score="7" ViewCount="0" Body="&lt;p&gt;I presume this is $\sum_{n=0}^\infty a_n 5^{-n}$&#xA;where $(2n+1)a_n=\sum_{k=0}^n 1/(2k+1)$.&#xA;Let&#xA;$$f(x)=\sum_{n=0}^\infty a_n x^{2n+1}$$&#xA;so that $f(0)=0$ and&#xA;$$f'(x)=\sum_{n=0}^\infty x^{2n}\sum_{k=0}^n\frac{1}{2k+1}.$$&#xA;Therefore&#xA;$$f'(x)=\frac{1}{1-x^2}\frac{\tanh^{-1}x}{x}.$$&#xA;Your sum equals&#xA;$$\int_0^{1/5}\frac{\tanh^{-1}x}{x(1-x^2)}dx.$$&#xA;If we let $x=\tanh y$ then $dx=(1-x^2)dy$ and we get&#xA;$$\int_0^t y\coth y\ dy$$&#xA;where $t=\tanh^{-1}(1/5)$.&#xA;I'm getting a nasty feeling that $\int y\coth y\ dy$&#xA;is one of those integrals that can't be done elementarily. :-(&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-03T19:25:24.503" CommentCount="2" />
  <row Id="8797" PostTypeId="2" ParentId="3633" CreationDate="2010-11-03T19:26:25.703" Score="1" ViewCount="0" Body="&lt;p&gt;Introduce a new coordinate system with a fixed point of $f$ as origin, e.g. the point $\omega:=e^{\pi i/3}$. Writing $x=\omega+\xi$ with a new independent variable $\xi$  one has $\phi(\omega+\xi)=\omega +\psi(\xi)$ for a new unknown function $\psi$ with $\psi(0)=0$. This function $\psi$ satisfies in a neighbourhood of $\xi=0$ the functional equation $\psi(\psi(\xi))=2\omega\xi+\xi^2$. Now you can recursively determine the Taylor coefficients of $\psi$. If you are lucky the resulting formal series actually defines an analytical function in a neighbourhood of $\xi=0$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-11-03T20:25:07.503" LastActivityDate="2010-11-03T20:25:07.503" />
  <row Id="8798" PostTypeId="2" ParentId="8791" CreationDate="2010-11-03T19:27:05.610" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If you massage the odd bisection of the generating function of the harmonic numbers then you should obtain a closed form in terms of (di)logs evaluated in $\:\mathbb Q(\sqrt{5})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; $\ $ See Andrey's later answer for some further details of the method I hinted above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-04T01:09:36.053" LastActivityDate="2010-11-04T01:09:36.053" CommentCount="2" />
  <row Id="8799" PostTypeId="2" ParentId="8793" CreationDate="2010-11-03T19:30:50.457" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re-write the integral as the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} \chi_B(x) \cos(nx) dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\chi_B$ is the characteristic function of the set $B$ (so that it equals 1 on $B$ and 0 else where). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you've already proven the case where $B$ is an open interval. Now take a sequence of decreasing coverings for $B$ by finitely many disjoint open intervals. For each covering, the result is true by what you've already shown. Take the limit using dominated convergence theorem (if $B \subset C$, then $|\chi_B \cos| \leq |\chi_C \cos|$). &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T19:30:50.457" />
  <row Id="8800" PostTypeId="2" ParentId="8793" CreationDate="2010-11-03T19:36:33.147" Score="1" ViewCount="0" Body="&lt;p&gt;Each Lebesgue integrable function can be approximated by a finite step function in norm. That is if $f\in L^1$ and $\varepsilon&gt;0$ then there are intervals $I_1,\ldots,I_N$, and scalars $a_1,\ldots,a_N$ such that &#xA;$$\int |f-\sum_{n=1}^N a_n\chi_n|&amp;lt;\varepsilon.$$&#xA;Where $\chi_n$ is the characteristic function on $I_n$, that is $\chi_n(x)=1$ for $x\in I_n$ and $\chi_n(x)=0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on each bounded interval $I$ you have already proved that $\int_I \cos(nx)dx\to 0$, hence given $\varepsilon&gt;0$ there is a step function as above and we get&#xA;$$\limsup\left|\int_B\cos(kx)dx\right|= \limsup\left|\int (\chi_B-\sum_{n=1}^N a_n\chi_n +\sum_{n=1}^N a_n\chi_n)\cos(kx)dx\right|$$&#xA;$$\leq \limsup\int |\chi_B-\sum_{n=1}^N a_n\chi_n|dx +\limsup\left|\int\sum_{n=1}^N a_n\chi_n\cos(kx)dx\right|$$&#xA;$$\le\varepsilon + \sum_{n=1}^N |a_n|\cdot \limsup\left|\int\chi_n\cos(kx)dx\right| =\varepsilon +0.$$&#xA;From which you conclude the result since this holds for any $\varepsilon&gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works not only for $\chi_B$, but for any $f\in L^1$ - it is called the Riemann-Lebesgue Lemma. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-03T19:56:59.550" LastActivityDate="2010-11-03T19:56:59.550" />
  <row Id="8801" PostTypeId="2" ParentId="8795" CreationDate="2010-11-03T19:40:07.923" Score="4" ViewCount="0" Body="&lt;p&gt;It comes down to whether or not you want $K(X)$ to be a group.  If all you care about is isomorphism classes of vector bundles, that's a perfectly good monoid.  People wouldn't call the monoid $K$-theory because $K$-theory of stuff generally means you've performed some kind of group-completion of a monoid or a category or something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's two ways to respond to your question.  The simplest would be &quot;why don't vector bundles have inverses?&quot; which on the most superficial level has the answer not all vector bundles are $0$-dimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to respond would be, what if we performed the formal group-completion of the monoid of isomorphism classes of vector bundles?  Well, you get the same thing.  General nonsense says you have a map from the group completion to $K(X)$ defined your way, since bundles over reasonable spaces have complementary bundles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this roughly an answer to your question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit in response to your comment: Say $X$ is a finite-dimensional CW-complex, and $\epsilon$ is a finite-dimensional vector bundle over $X$.  Then there is a classifying map for $\epsilon$, this is a map $X \to G_{\infty,k}$ where $k$ is the dimension of $\epsilon$.  $G_{\infty,k}$ is the Grassmannian of $k$-dimensional subspaces of $\mathbb R^\infty$.  A basic theorem about $G_{\infty,k}$ is that any map of a finite-dimensional CW-complex into $G_{\infty,k}$ is homotopic to a map into $G_{m,k}$ for $m$ perhaps very large.  $G_{m,k}$ is canonically isomorphic to $G_{m,m-k}$ (take orthogonal complements).  The bundle classified by the correponding map $X \to G_{m,m-k}$ is the &lt;em&gt;complementary bundle&lt;/em&gt; to $\epsilon$, denote it $\epsilon'$.  By design, $\epsilon \oplus \epsilon'$ is a trivial bundle as it's the pull-back of the tangent bundle to $\mathbb R^m$.  The fact that maps $X \to G_{\infty,k}$ are homotopic to maps $X \to G_{m,k}$ has many proofs -- one is from the Schubert cell decomposition of $G_{m,k}$ together with cellular approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the idea.  I believe complementary bundles exist for more than bundles over finite-dimensional spaces but that's the most convienient argument that comes to mind (have to run off to teach a class now). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd edit:  Let $\mathcal V(X)$ be the group completion of the monoid $\mathcal M(X)$ of isomorphism classes of finite-dimensional vector bundles over $X$.  Let $K(X)$ be the stable isomorphism classes of vector bundles over $X$.  $K(X)$ is a group for a fairly broad class of spaces, by the arguments above.  So by group-completion formal nonsense there is a map $\mathcal V(X) \to K(X)$.   The question is whether this map has an inverse or not. Given an element $[\alpha] \in K(X)$, $\alpha \in \mathcal M(X)$ is a vector bundle over $X$ and is well-defined up to stable equivalenc.  To $\alpha$ there is an associated element of $\mathcal V(X)$ and the question is, is the map &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(X) \to \mathcal V(X)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;well-defined, where you map $[\alpha]&lt;em&gt;{K(X)} \longmapsto [\alpha]&lt;/em&gt;{\mathcal V(X)}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, okay, I see the problem now.  This isn't well-defined.  But things are off only by a tiny thing.  $\mathcal V(X)$ contains a copy of the integers as a direct-summand.  This is because $\mathcal M(X) \to \mathbb Z$ given by taking the dimension of a bundle is a homomorphism of monoids (using addition in $\mathbb Z$).  There is a map back $\{0,1,2,\cdots \} \to \mathcal M(X)$ given by taking trivial bundles over $X$.  So in $\mathcal V(X)$ this gives a splitting $\mathcal V(X) = \mathbb Z \oplus \overline{\mathcal V(X)}$ where $\overline{\mathcal V(X)}$ is the &quot;reduced&quot; group-completion of isomorphism types of vector bundlex over $X$.  There &lt;em&gt;is&lt;/em&gt; a well-defined map $K(X) \to \overline{\mathcal V(X)}$ since you can think of $\overline{\mathcal V(X)}$ as being $\mathcal V(X)$ modulo trivial bundles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense now?   I haven't thought about this stuff in much detail for years. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-04T23:04:39.120" LastActivityDate="2010-11-04T23:04:39.120" CommentCount="6" />
  <row Id="8802" PostTypeId="2" ParentId="8791" CreationDate="2010-11-03T19:49:38.050" Score="1" ViewCount="0" Body="&lt;p&gt;$$S(x) = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{x^2}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{x^4}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{x^6}{5^{3}} \cdot \frac{1}{7} + \cdots$$&#xA;$$=\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{x}{\sqrt 5}\right)^{2k} \cdot \frac{1}{2k+1}$$&#xA;$$=\frac{1}{x}\int_{0}^x\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{y}{\sqrt 5}\right)^{2k}dy$$ &#xA;$$=\frac{\sqrt 5}{x}\int_{0}^{x/\sqrt 5}\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) t^{2k+1}\frac{dt}{t}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let&#xA;$$F(t)=\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) t^{2k+1},$$&#xA;then &#xA;$$F(t)+\frac{t}{2}F(t)=H(t):= \sum_{n=0}^{\infty} \Bigl(1 + \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\Bigr) t^n.$$&#xA;$H(t)$ is the generating function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;harmonic numbers&lt;/a&gt; and is well known:&#xA;$$H(t)=\frac{-\ln(1-t)}{1-t}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining it all together we get that&#xA;$$S(x)=\frac{\sqrt 5}{x}\int_{0}^{x/\sqrt 5} \frac{2\ln(1-t)}{t(t-1)(t+2)}dt$$&#xA;which probably can be calculated in elementary functions.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-03T19:49:38.050" CommentCount="3" />
  <row Id="8803" PostTypeId="2" ParentId="8794" CreationDate="2010-11-03T20:14:36.753" Score="4" ViewCount="0" Body="&lt;p&gt;I had this question sometime ago and was shown the explanation in Balazs 1954 paper &quot;&lt;a href=&quot;http://iopscience.iop.org/0370-1298/68/6/307&quot; rel=&quot;nofollow&quot;&gt;Wave propagation in even and odd dimensional spaces&lt;/a&gt;&quot;.  The singularities of the integrands in the solution vary based on the dimension giving rise to this effect.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When n is even, the integrand has a&#xA;pole in the complex plane, so the&#xA;contour has to be deformed, and the&#xA;result refers to only one time value -- that of the pole.&lt;/li&gt;&#xA;&lt;li&gt;When n is odd, the integrand has a&#xA;branch point, and the contour cannot&#xA;encircle the singularity. &#xA;Consequently the solution contains&#xA;contributions from all times along&#xA;the branch cut.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2010-11-03T20:14:36.753" />
  <row Id="8804" PostTypeId="2" ParentId="8793" CreationDate="2010-11-03T20:23:01.670" Score="0" ViewCount="0" Body="&lt;p&gt;Thanks to both. I'm still confused because books define Lebesgue measurable sets in different ways (Caratheodory extension, outer measure, etc). What is the exact definition you are using for Lebesgue measurable set? I'm sure once I understand how it is defined I will be able to understand fully your hint(s). Thanks again.&lt;/p&gt;&#xA;" OwnerUserId="3082" LastActivityDate="2010-11-03T20:23:01.670" CommentCount="3" />
  <row Id="8805" PostTypeId="2" ParentId="8610" CreationDate="2010-11-03T20:50:42.383" Score="7" ViewCount="0" Body="&lt;p&gt;We  use the formula given here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction_of_Gauss#Applications&quot; rel=&quot;nofollow&quot;&gt;Gauss' continued fraction&lt;/a&gt; for $\tan z$ and see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(1) = \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5 -\dots}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To transform $$\cfrac{1}{a - \cfrac{1}{b - \cfrac{1}{c - \dots}}}$$ to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-2 + \dots}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get the expansion for $\displaystyle \tan(1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above expansion for $\tan(1)$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cfrac{1}{1-1 + \cfrac{1}{1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  = 1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}$$&#xA;$$= 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{5 + \dots}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the transformation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $\displaystyle x = b - \cfrac{1}{c - \dots}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$&#xA;$$ = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-1 + \cfrac{1}{c - \dots}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the identity again to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{b-1 + \cfrac{1}{c - \dots}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-1 + \cfrac{1}{d - \dots}}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying again to $\cfrac{1}{c-1 + \cfrac{1}{d - \dots}}$ etc gives the required CF.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-04T02:51:17.377" LastActivityDate="2010-11-04T02:51:17.377" CommentCount="1" />
  <row Id="8806" PostTypeId="2" ParentId="8794" CreationDate="2010-11-03T20:52:15.877" Score="3" ViewCount="0" Body="&lt;p&gt;This is a very good question. &lt;a href=&quot;http://www.mathpages.com/home/kmath242/kmath242.htm&quot; rel=&quot;nofollow&quot;&gt;John Baez once gave a very detailed answer to it&lt;/a&gt; which I will not bother to reproduce here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A not-exactly precise way to think about it is this: a wave propagating in $n$ dimensions can be thought of as a wave propagating in $n+1$ dimensions, but one degree of freedom removed. More precisely: if $f$ solves the wave equation $(\partial_t^2 - \triangle) f(t,x) = 0$ on $\mathbb{R}^d$, then  it also solves the equation $(\partial_t^2 - \partial_y^2 - \triangle) f(t,y,x) = 0$ on $\mathbb{R}^{d+1}$, if you assume that $f(t,y,x) = f(t,0,x) = f(t,x)$ for all $y$. (It is constant in the $y$ direction, so any derivative in that direction is 0.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if a wave were to propagate in, say, 5 dimensions with the property you mentioned. Then in 4 dimensions, the points &quot;inside the light cone&quot; is reachable in 5 dimensions as on the light cone. Or, in other words, assume $f(t,x)$ is a solution to wave equation in 4 dimensions. And let $f(t,y,x)$ be the trivial extension with one dimension added. So $f(t,y,x)$ solves the wave equation in 5 dimensions. Suppose at time $t = 0$, a light bulb turns on at the origin $x = 0$ in 4 dimension space. This corresponds to a bank of lightbulbs turning on in 5 dimensional space along the $y$ axis. The five dimensional principle says that a point at coordinates $(y,x)$ will be illuminated by a lightbulb at $(y_0,0)$ at time $t$ if $t^2 = x^2 + (y-y_0)^2$. But for every $x, y$ such that $|x| &amp;lt; |t|$, you can find two values of $y_0$ such that $t^2 = x^2 + (y-y_0)^2$. And so some light must lag behind. (In the study of partial differential equations, this is also known as the method of descent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by this argument, there should still be some light lagging behind for 3 dimensions also! The claim, however, is this: due to the nature of the propagation, there is a destructive interference when you drop two dimensions. That you have destructive interference when you drop two dimensions, but not such problem when you drop only one dimension, is not at all obvious intuitively (at least to me). It however does fall out of the expression for the fundamental solution to the wave equation. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-03T20:52:15.877" />
  <row Id="8807" PostTypeId="1" AcceptedAnswerId="8810" CreationDate="2010-11-03T21:02:57.527" Score="4" ViewCount="47" Body="&lt;p&gt;I have a semigroup $G$ (a set with associative binary operation) such that for all $a,b\in G$ there exists $x,y\in G$ such that $ax=ya=b$. Is this property enough to show that $G$ is a group, and if so, how?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-03T21:11:54.550" Title="Semigroup with &quot;transitive&quot; operation is a group?" Tags="&lt;group-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8808" PostTypeId="1" AcceptedAnswerId="8809" CreationDate="2010-11-03T21:04:20.787" Score="11" ViewCount="296" Body="&lt;p&gt;A lot of mathematics as far as I know is interested in the study of Euclidean and locally Euclidean spaces (manifolds).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the special feature of Euclidean spaces that makes them interesting?&lt;/li&gt;&#xA;&lt;li&gt;Is there a field that studies spaces that are neither globally nor locally Euclidean spaces?&lt;/li&gt;&#xA;&lt;li&gt;If such a field exists, are there any practical uses for it (as in physically existing models)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2949" LastActivityDate="2010-12-08T20:00:31.727" Title="Why the interest in locally Euclidean spaces?" Tags="&lt;geometry&gt;&lt;differential-geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="3" />
  <row Id="8809" PostTypeId="2" ParentId="8808" CreationDate="2010-11-03T21:06:55.873" Score="12" ViewCount="0" Body="&lt;p&gt;One of the main reasons for interest is that manifolds are &lt;em&gt;homogeneous&lt;/em&gt;.  So from a geometric point of view, we study manifolds for much the same reason we study groups.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put in less abstract ways, there is the &quot;classical problem&quot; of whether or not the earth is flat.  The &quot;flat earth problem&quot; deals with the possibility that on small scales something may look linear, but macroscopically it need not be.  Manifolds are the abstract manifestation of the flat earth problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People do study large families of objects that are locally modelled on spaces that aren't Euclidean spaces.  There's infinite-dimensional manifolds, fractal manifolds, orbifolds (locally modelled on euclidean space mod a finite group action), etc... Fibre bundles are maps that are locally projection maps.  This idea resurfaces in mathematics in many different forms and many different ways. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-03T21:19:37.703" LastActivityDate="2010-11-03T21:19:37.703" CommentCount="2" />
  <row Id="8810" PostTypeId="2" ParentId="8807" CreationDate="2010-11-03T21:11:54.550" Score="6" ViewCount="0" Body="&lt;p&gt;Yes. This is a standard exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let $a\in G$ be arbitrary. Then there exists $e\in G$ such that $ae=a$ (using the first condition with $a=b$). Now let $b\in G$. Then there exists $z\in G$ such that $za=b$. Therefore, $be=(za)e=z(ae) = za = b$. Thus, $e$ is a right identity for $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the fact that $ax=e$ is always solvable it follows that every element has a right inverse. Since your semigroup has a right identity and right inverses, then it is a group.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-03T21:11:54.550" />
  <row Id="8811" PostTypeId="1" AcceptedAnswerId="8835" CreationDate="2010-11-03T21:30:21.803" Score="5" ViewCount="269" Body="&lt;p&gt;Américo Tavares pointed out in his answer to &lt;a href=&quot;http://math.stackexchange.com/questions/8476/rearrange-the-formula-for-the-sum-of-a-geometric-series-to-find-r&quot;&gt;this question&lt;/a&gt; that finding the ratio of a geometric progression only from knowledge of the sum of its first $n+1$ terms $S = 1+x+x^2+\cdots+x^n$ amounts to solving a polynomial of degree $n$. This suggested to me that there might be up to $n$ real solutions of $x$ for a given sum, but I could not find any. In fact, it turned out that the following fact is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n \ge 1$ and $S \in \mathbb{R}$, the polynomial equation $x^n + x^{n-1} + \cdots + x + 1 = S$ has at most two real solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A corollary is that if $n$ is odd, there is exactly one real solution. I was only able to prove this using a rather contrived geometric argument based on the shape of the graph of $y = x^{n+1}$. Is there a simple, direct (and ideally, intuitive) proof of this fact?&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-05T23:00:57.123" Title="Why does the polynomial equation $1 + x + x^2 + \cdots + x^n = S$ have at most two solutions in $x$?" Tags="&lt;sequences-and-series&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="8812" PostTypeId="2" ParentId="8672" CreationDate="2010-11-03T21:34:16.997" Score="4" ViewCount="0" Body="&lt;p&gt;Despite other answers, I thought it might benefit the impatient to see the explicit answer below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$M = &#xA;\left(&#xA;\begin{array}{cc}&#xA; a &amp;amp; b \\&#xA; b &amp;amp; c&#xA;\end{array}&#xA;\right),$$&#xA;be the input matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the discriminant: $\Delta = \sqrt{a^2+4 b^2-2 a c+c^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the eigenvalues of $M$ are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda_1 = 0.5(a+c-\Delta)$ and $\lambda_2 = 0.5(a+c+\Delta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can find a matrix $V$ such that &#xA;$$M = V^{-1} \begin{pmatrix}&#xA;\lambda_1 &amp;amp; 0\\&#xA;0 &amp;amp; \lambda_2&#xA;\end{pmatrix}V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica says that the matrix $V$ is given by&#xA;$$&#xA;V = \begin{pmatrix}&#xA;\frac{a-c-\Delta}{2b} &amp;amp; 1\\&#xA;\frac{a-c+\Delta}{2b} &amp;amp; 1&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for orthogonal $V$, then the above calculations need some changes.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-03T21:34:16.997" CommentCount="2" />
  <row Id="8813" PostTypeId="2" ParentId="8811" CreationDate="2010-11-03T21:36:58.410" Score="9" ViewCount="0" Body="&lt;p&gt;The roots are also roots of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{n+1} - Sx + S - 1 = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we get by multiplying your equation by $x-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This polynomial ($x^{n+1} - Sx + S-1$), as we move from $x = -\infty$ to $x = \infty$ is either &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Monotonically increasing, and thus has at most one real root.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Monotonically decreasing, and then monotonically increasing and hence can have at most two real roots.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Monotonically increasing, then decreasing and then again increasing (happens only when $n$ is even). In which case there are at most three real roots, one of which is $1$. So for $S \ne n+1$, the original equation does not have more than two solutions. If $S=n+1$ and $n$ is even, then the turning points are $-1$ and $1$ and the value of the polynomial at $-1$ is positive. So the only roots are $1$ and a root which is $&amp;lt; -1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This can be seen by looking at its derivative, which is an increasing function for odd $n$, and for even $n$, it is positive, then possibly negative (depending on $S$) and then positive again, as we move from $x = -\infty$ to $x = \infty$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-03T22:28:49.673" LastActivityDate="2010-11-03T22:28:49.673" CommentCount="5" />
  <row Id="8814" PostTypeId="1" CreationDate="2010-11-03T21:39:01.623" Score="9" ViewCount="933" Body="&lt;p&gt;Here is a funny exercise &#xA;$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$&#xA;(If you prove it don't publish it here please).&#xA;Do you have similar examples?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-27T06:58:45.720" CommunityOwnedDate="2010-11-04T06:29:02.260" Title="Funny identities" Tags="&lt;big-list&gt;&lt;trigonometry&gt;&lt;fun-problem&gt;" AnswerCount="18" CommentCount="9" FavoriteCount="4" />
  <row Id="8815" PostTypeId="2" ParentId="6339" CreationDate="2010-11-03T21:48:52.257" Score="0" ViewCount="0" Body="&lt;p&gt;A trivial case when $B$ is semidef is say when the original matrix is elementwise non-negative (and you define sign(0) = 1). In that case, $B=ee^T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it might be hard to give a non-trivial set of sufficient conditions on $A$ so that $B$ is semidefinite.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-03T21:48:52.257" />
  <row Id="8816" PostTypeId="2" ParentId="8811" CreationDate="2010-11-03T21:57:58.027" Score="6" ViewCount="0" Body="&lt;p&gt;A powerful algorithmic way to handle such problems is to employ &lt;a href=&quot;http://en.wikipedia.org/wiki/Sturm%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Sturm's Theorem&lt;/a&gt;. If you work out the details you will see that it is quite simple for this example. This in turn is a special case of the CAD (cylindrical algebraic decomposition) algorithm  - an effective implementation of Tarski's  quantifier elimination for the first order theory of the reals. The general ideas behind these  methods prove helpful in solving a variety of problems. It's well worth the effort to learn the general methods rather than ad-hoc techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per Rahul's request, here are further details of applying Sturm's algorithm to the example at hand. We desire to prove that $\rm\ \ g(x) =\ x^n +\:\cdots\: + x^2 + x + 1 - s\ \ $ has at most two distinct real roots. Consider $\rm\ f(x) = (x-1)\ g(x) =\ x^{n+1}-1-s\ (x-1)\:.\ $ Since $\rm\ \ f\:' = (n+1)\ x^n - s\ \ $ we have $\rm\ f\ mod\ f\:'\: =\ f\: - \: x/(n+1)\ f\:'\ =\ a\ x + b\ $ for some $\rm\:a,\:b\in \mathbb R\:$. So the euclidean remainder sequence in the calculation of $\rm\ gcd(f,\:f\:')\ $ has length at most $4$, viz. $\rm\ f,\ f\:',\ a\ x + b,\ c\ $. Thus it has at most $3$ sign changes at any point, so Sturm's theorem implies that $\rm\ f(x)\ $ has at most $3$ distinct real roots. So if $\rm\: x = 1\:$ isn't a multiple root of $\rm \:f(x)\:$ then $\rm\ g(x) = f(x)/(x-1)\ $ has at most $2$ distinct real roots. Else $\rm\ x-1\:|\:gcd(f,\:f\:')\ \Rightarrow\ x-1\:|\:c\ \Rightarrow\ c=0\:$. So in this case  the remainder sequence has length at most $3$, so at most $2$ sign changes, so $\rm\:f\ $ has at most $2$ distinct real roots, therefore ditto for $\rm\:g\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Sturm's theorem is slightly more work here than Rolle's theorem, it has the added benefit that it allows one to compute the precise number of roots in any interval (versus only bounds using Rolle's theorem).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-05T23:00:57.123" LastActivityDate="2010-11-05T23:00:57.123" CommentCount="3" />
  <row Id="8817" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:01:42.733" Score="7" ViewCount="0" Body="&lt;p&gt;The Frobenius automorphism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x + y)^p = x^p + y^p$$&lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T22:01:42.733" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8818" PostTypeId="2" ParentId="8734" CreationDate="2010-11-03T22:05:08.627" Score="2" ViewCount="0" Body="&lt;p&gt;I don't believe you understand the notion of efficiency in terms of encoding. Informally speaking, you have to keep it mind there are two factors involved: (i) cost of having different symbols (in case of base 10 there as 10 different symbols, in case of base 16 there are 16 different symbols etc) and the length of the resulting string to encode a particular number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you consider both factors and apply some basic information theory to it, the answer may look a bit surprising: the most efficient encoding has a base $e$ (yes, that very $e = 2.718\dots$). Since we'd rather have some natural number as a base, the best we can get is base 3, and the next is base 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why, then, computers use base 2 (0 and 1) rather than base 3 (say, -1, 0, and 1)? The answer is that it is simple to design the circuits that distinguish between two (rather than three) states. (I do remember reading some of the earlier computers did use base 3, but I can't recall all the details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with respect to octals and hexes, those are simply convenient ways to record the binary strings. If you did some machine-level debugging, you probably had a chance to read what's known as &quot;hexadecimal dump&quot; (contents of a memory). Surely it's easy to read than if it were written as binary dump. But what's lurking underneath of that is base 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The answer on &quot;why do we use base 10&quot; has been answered elsewhere.)  &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-03T22:05:08.627" CommentCount="1" />
  <row Id="8819" PostTypeId="2" ParentId="8532" CreationDate="2010-11-03T22:06:21.463" Score="5" ViewCount="0" Body="&lt;p&gt;The method is &quot;keep trying&quot;. Try reading the paper again, try reading a different paper, try stopping reading papers and just work it out yourself. Try talking to other people, try struggling on your own, then try talking to people again. Try a different problem, then try going back to the same problem. Try the same method again and again, then try different methods, then go back to the first method and try it one more time. Try just writing out some calculations even if you don't see them going anywhere, try a stupid example, try drawing some pictures. Finally try taking a break, and then also try not taking a break. Try everything once, then try it again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But enough of that, here's some more practical advice from my own experience (I'm a postdoc by the way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For a hard problem, work intensely on it for a few weeks or months, until you feel like you've completely hit a wall. During this period you want to be producing as many calculations, lemmas, pictures, examples, etc as possible. They don't all have to be super-relevant -- your goal is maximum output of material. Then when you finally feel like you're going insane, take a deep breath and move on to something else. After a break of a few weeks or months, which may be spent either on work of some other kind or even on vacation, come back to the problem and see if the fresh perspective helps. It may take multiple cycles like this for a really hard problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Look for a variation of your problem which is easier or which your methods can better handle. Even if the variation is not as strong of a result as what you are ultimately aiming for, it may still &quot;count&quot; and end up being a nicer paper than you realize at first. It's good for your morale to register progress in this way, and if you can get a publication it's good for your CV too. Here is where advisors and mentors can help a lot -- they can help you identify which variation is meaningful, doable, and &quot;interesting&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Remember that almost nothing ever works. One of the worst feelings to me is when I find a mistake in something I thought I had figured out, and all of a sudden weeks or months of progress begin to unravel. This can be extremely depressing. Try not to take it too hard, and over time you can rebuild and recover. Try also to appreciate the progress you are making, even if it is much less than you wish it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="3087" LastActivityDate="2010-11-03T22:06:21.463" CommunityOwnedDate="2010-11-03T22:06:21.463" CommentCount="3" />
  <row Id="8820" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:07:04.300" Score="11" ViewCount="0" Body="&lt;p&gt;$$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-03T22:07:04.300" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8821" PostTypeId="2" ParentId="8764" CreationDate="2010-11-03T22:07:16.063" Score="2" ViewCount="0" Body="&lt;p&gt;This seems to be Laplace noise. Have a look at: &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Laplace_distribution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is handy for modeling such spikey, sparse noise.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-03T22:07:16.063" />
  <row Id="8822" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:12:30.473" Score="2" ViewCount="0" Body="&lt;p&gt;$$ \infty! = \sqrt{2 \pi} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes from the zeta function.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-03T22:12:30.473" CommunityOwnedDate="2010-11-04T06:29:02.260" CommentCount="3" />
  <row Id="8823" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:18:49.217" Score="2" ViewCount="0" Body="&lt;p&gt;\begin{eqnarray}&#xA;\sum_{i_1 = 0}^{n-k} \, \sum_{i_2 = 0}^{n-k-i_1} \cdots \sum_{i_k = 0}^{n-k-i_1 - \cdots - i_{k-1}} 1 = \binom{n}{k}&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-03T22:18:49.217" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8824" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:23:47.103" Score="3" ViewCount="0" Body="&lt;p&gt;$$ \sum\limits_{n=1}^{\infty} n =  1 + 2 +3 + \cdots \text{ad inf.} = - \frac{1}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now doesn't this sound funny. You can also see many more on &lt;a href=&quot;http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-03T22:23:47.103" CommunityOwnedDate="2010-11-04T06:29:02.260" CommentCount="5" />
  <row Id="8825" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:26:57.677" Score="1" ViewCount="0" Body="&lt;p&gt;\begin{eqnarray}&#xA;\zeta(0) = \sum_{n \geq 1} 1 = -\frac{1}{2}&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-03T22:26:57.677" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8826" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:33:13.333" Score="6" ViewCount="0" Body="&lt;p&gt;\begin{eqnarray}&#xA;1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2}&#xA;\end{eqnarray}&#xA;More generally, let $D_{k} = ${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then &#xA;\begin{eqnarray}&#xA;\sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2}&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\mathsf{d}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-23T00:42:18.930" LastActivityDate="2010-11-23T00:42:18.930" CommunityOwnedDate="2010-11-04T06:29:02.260" />
  <row Id="8827" PostTypeId="2" ParentId="8760" CreationDate="2010-11-03T22:33:41.657" Score="0" ViewCount="0" Body="&lt;p&gt;You may take a look at this previous &lt;a href=&quot;http://math.stackexchange.com/questions/3315/what-is-the-square-root-of-i/3346#3346&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-03T22:33:41.657" />
  <row Id="8828" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:33:44.283" Score="8" ViewCount="0" Body="&lt;p&gt;$$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-03T22:33:44.283" CommunityOwnedDate="2010-11-03T22:35:34.187" />
  <row Id="8829" PostTypeId="2" ParentId="8814" CreationDate="2010-11-03T22:40:28.237" Score="3" ViewCount="0" Body="&lt;p&gt;Two related integrals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-03T22:40:28.237" CommunityOwnedDate="2010-11-03T22:40:28.237" CommentCount="2" />
  <row Id="8830" PostTypeId="1" AcceptedAnswerId="8849" CreationDate="2010-11-03T22:46:20.100" Score="1" ViewCount="90" Body="&lt;p&gt;I am working in the field of machine learning, and I have come across a few papers that show relationships between Gröbner bases and discrete probability. So I come here for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please explain how can Gröbner bases used to describe discrete probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at Gröbner bases and I understand the general concepts (and used Maple to calculate a few examples). So it is the link that is missing for me.&lt;/p&gt;&#xA;" OwnerUserId="2949" LastActivityDate="2010-11-04T00:47:21.133" Title="How can Gröbner bases used to describe discrete probability ?" Tags="&lt;probability&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8831" PostTypeId="1" AcceptedAnswerId="8833" CreationDate="2010-11-03T22:56:40.540" Score="2" ViewCount="108" Body="&lt;p&gt;Hello everybody.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basktball player shoots basket for 100 times. On the first throw the ball is always in, on the second - always out. For all consequent shoots the probability of hit is equal to precentage of all scored shoots diveded by total number of shoots done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability the he will score total 50 balls out of 100?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a homework, I'm just anxious about solution of this problem. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My ruminitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are random events $E_1, E_2, \dots, E_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_1$ always happens, $E_2$ - never.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $E_n$ with $n &gt; 2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(E_n) = \frac{\sum_{k=1}^{i-1}{x_k}}{i-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${x}$ - random variables, that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i = 1$ if event is happened, $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;$P (\sum_{i=1}^{100}{x_i} = 50) = ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is it true that this problem can be solved with Limit Theorem?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequent events are dependant on outcoimes ov previous, can Limit Theorem handle that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Could you point on how can i solve this problem?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanx! &#xA;=)&lt;/p&gt;&#xA;" OwnerUserId="3089" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T00:36:15.723" LastActivityDate="2010-11-04T00:41:48.083" Title="[Probability Theorem Problem] Limit Theorem applicability" Tags="&lt;probability-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="8832" PostTypeId="2" ParentId="8652" CreationDate="2010-11-03T22:59:39.887" Score="7" ViewCount="0" Body="&lt;p&gt;It depends largely on what your career goals are.  If your goal is to be a professor at a top 100 research school, then you should certainly go to the most prestigious option available as explained by Pete and Ben (and with their caveats that &quot;more&quot; means &quot;significantly more.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you have different goals (for example, you want a job at a more teaching oriented institution) then it seems less clear to me that the increase in prestige is always for the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I was a graduate student a professor moved from a top 25 school to Berkeley bringing several of his students along with him.  Those students had the experience of having the same advisor at two different schools, and they found the experience at the two schools to be substantially different.  In particular, at the previous school the professor was not very busy and spent hours talking with his students.  Whereas at the new school there were lots and lots of visitors/postdocs/other profs around all the time and the prof had the opportunity to talk to those people rather than his students.  (He was still a responsible advisor who met regularly with his students.)  So, if having an advisor with lots of extra time who actually seeks you out to talk to you is really important to you then going to a lower ranked school actually can make a significant difference in experience.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-11-03T22:59:39.887" CommentCount="2" />
  <row Id="8833" PostTypeId="2" ParentId="8831" CreationDate="2010-11-03T23:24:44.047" Score="3" ViewCount="0" Body="&lt;p&gt;The probability is $1/99$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote by $P(a,b)$ the probability that after $a+b$ shots, exactly $a$ were in and $b$ were out. Then $P(1,1)=1$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(a,b) = \frac{a-1}{a+b-1} P(a-1,b) + \frac{b-1}{a+b-1} P(a,b-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We prove by induction on $a+b$ that $P(a,b) = 1/(a+b-1)$. When $a+b = 2$, this is true. Otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(a,b) = \frac{a+b-2}{a+b-1} \cdot \frac{1}{a+b-2} = \frac{1}{a+b-1}$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-03T23:24:44.047" CommentCount="1" />
  <row Id="8834" PostTypeId="2" ParentId="8831" CreationDate="2010-11-03T23:28:04.510" Score="1" ViewCount="0" Body="&lt;p&gt;This is problem B1 from the 2002 Putnam exam.  It is useful to generalize and allow for any number of free throws.  We can then prove the stronger result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;. For any number of free throws $n$, and any $i \in [n-1]$ the probability of hitting exactly $i$ free throws after $n$ attempts is $\frac{1}{n-1}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; We proceed by induction on $n$.  The base case $n=3$ is trivially trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inductive step, assume the result for $n-1$, and let $i \in [n-1]$.  There are two ways to make exactly $i$ shots after $n$ attempts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;make $i-1$ shots after $n-1$ attempts and clutch up on your final attempt, or &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;make $i$ shots after $n-1$ attempts and epic fail the final attempt. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By induction, the required probability is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{n-2} \frac{i-1}{n-1} + \frac{1}{n-2} \frac{n-1-i}{n-1} = \frac{1}{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your particular question is $\frac{1}{99}$.  &lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditorDisplayName="" LastEditDate="2010-11-03T23:35:49.600" LastActivityDate="2010-11-03T23:35:49.600" CommentCount="2" />
  <row Id="8835" PostTypeId="2" ParentId="8811" CreationDate="2010-11-03T23:45:12.717" Score="8" ViewCount="0" Body="&lt;p&gt;In between the previous two answers - one very specific and one very general, there is the following one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rolle%27s_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Rolle's theoerm&quot;&gt;Rolle's theorem&lt;/a&gt; states that between any two zeros of the function lies a zero of the derivative. So each time you differentiate the number of real roots can not decrease by more than 1. This actually holds with multiplicity as well - if a root is of multiplicity k, then it is of multiplicity (k-1) for the derivative, and there are still roots of the derivative strictly between consecutive distinct roots of the original function. So $x^{n+1}−Sx+S−1$ has at most one root more than $(n+1)x^n-S$ and at most two more than $n(n+1)x^{n-1}$, so not more than three total (counted with multiplicity). One of those is $x=1$, so there are no more than two for the $1+x+\ldots+x^n=S$. QED.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-03T23:45:12.717" CommentCount="5" />
  <row Id="8837" PostTypeId="1" AcceptedAnswerId="8840" CreationDate="2010-11-03T23:53:05.037" Score="5" ViewCount="115" Body="&lt;p&gt;I recall seeing a hand-drawn picture of spec of a ring (maybe of Z?) that had been passed around in the early days of the Zariski topology.  Does anyone know where I can find a copy?&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-04T00:29:59.503" Title="Picture of spec?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="8838" PostTypeId="1" AcceptedAnswerId="8848" CreationDate="2010-11-03T23:58:20.123" Score="8" ViewCount="480" Body="&lt;p&gt;Is solving problems quickly an important trait for a mathematician to have? Is solving textbook/olympiad style problems quickly necessary to succeed in math? To make an analogy, is it better to be a sprinter or a marathon runner in mathematics? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-04T06:50:27.667" Title="Is speed an important quality in a mathematician? " Tags="&lt;soft-question&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="5" />
  <row Id="8840" PostTypeId="2" ParentId="8837" CreationDate="2010-11-04T00:09:13.743" Score="6" ViewCount="0" Body="&lt;p&gt;I think you mean &lt;a href=&quot;http://www.neverendingbooks.org/wp-content/uploads/2008/12/MumfordsTreasureMap.jpg&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt; by Mumford. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-11-04T00:09:13.743" CommentCount="2" />
  <row Id="8841" PostTypeId="2" ParentId="8775" CreationDate="2010-11-04T00:11:05.833" Score="3" ViewCount="0" Body="&lt;p&gt;This is quite obvious if expressed most naturally in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_polynomials&quot; rel=&quot;nofollow&quot;&gt;Bernoulli polynomials&lt;/a&gt;, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad\quad\quad\quad\rm\ \ \ \sum n^k\ = \:\frac{1}{k+1} (B_{k+1}(n+1) - B_{k+1}(0))\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\rm\quad\quad\ B_k^{\:'}(x)\ =\ k\ B_{k-1}(x)\ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus $\ \rm\ (\sum n^k)'\ =\ k\ \sum n^{k-1} + B_{\:k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.ams.org/journals/mcom/1993-61-203/S0025-5718-1993-1197512-7/S0025-5718-1993-1197512-7.pdf&quot; rel=&quot;nofollow&quot;&gt;Knuth's exposition&lt;/a&gt; these identities go all the way back to Jacobi - with special cases known to Faulhaber. Note that the identities in Robin Chapman's post are simply special cases of these well-known identities for Bernoulli polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bernoulli polynomials are a special-case of polynomials amenable to study by the powerful techniques of the Umbral Calculus. This is the best way to understand the genesis of their many interesting properties. For a nice introduction see Steven Roman's book.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T00:11:05.833" />
  <row Id="8842" PostTypeId="2" ParentId="8837" CreationDate="2010-11-04T00:12:16.347" Score="2" ViewCount="0" Body="&lt;p&gt;You are thinking of the picture of Spec($Z[X]$) in Mumford's Red Book of Varieties and Schemes.   A reproduction is downloadable at the link posted in the other answer, and other (less famous) pictures of prime spectra appear in the Eisenbud and Harris introductory book on schemes.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-04T00:18:14.920" LastActivityDate="2010-11-04T00:18:14.920" />
  <row Id="8843" PostTypeId="2" ParentId="8831" CreationDate="2010-11-04T00:18:34.030" Score="2" ViewCount="0" Body="&lt;p&gt;Some additional references may be of interest. This result is a &#xA;well-known feature of Polya's urn model starting with one ball&#xA;each of two different colours.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, this problem is Exercise 5.9 in the &#xA;first edition of &#xA;&lt;em&gt;Introduction to Stochastic Processes&lt;/em&gt; by Lawler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a detailed analysis in section 2.4 of &#xA;&lt;em&gt;Markov Chains and Mixing Times&lt;/em&gt; by Levin, Peres, and Wilmer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect that it can be found in other books on probability and Markov chains. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-04T00:18:34.030" CommunityOwnedDate="2010-11-04T00:18:34.030" />
  <row Id="8844" PostTypeId="2" ParentId="8837" CreationDate="2010-11-04T00:29:59.503" Score="3" ViewCount="0" Body="&lt;p&gt;Be sure not to miss this very interesting paper (in English) on &quot;Mumford's treasure map&quot; and related concepts: Lieven Le Bruyn: &lt;a href=&quot;http://matrix.cmi.ua.ac.be/DATA2/Alev60.pdf&quot; rel=&quot;nofollow&quot;&gt;Un dessins d'enfants&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T00:29:59.503" />
  <row Id="8846" PostTypeId="2" ParentId="8831" CreationDate="2010-11-04T00:41:48.083" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an enumerative proof. We can think of the process as an urn with balls of two colors, say black and white. In the beginning, there is one ball of each color. In each round we pick one of the balls and duplicate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us count the number of ways to get $a+1,b+1$ balls after $a+b$ rounds. This involves adding $a$ black balls and $b$ white balls. The first time a black ball is added, there is no choice. The next time, we can choose any of the two existing balls, and so on. In total, there are $a!$ choices for black balls, and $b!$ choices for white balls. We also have to choose in which round we add which color - there are $\binom{a+b}{a} = (a+b)!/a!b!$ such choices. In total, we get $(a+b)!$ choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've shown that the number of ways to get $a+1,b+1$ balls depends only on $a+b$. In fact, we can directly calculate the probability by calculating the total number of possible moves. In the first move we pick one of two balls, then one of three balls, and so on. So in $a+b$ rounds there are $(a+b+1)!$ possible moves. We conclude that each outcome has probability $1/(a+b+1)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-04T00:41:48.083" />
  <row Id="8847" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T00:42:56.070" Score="8" ViewCount="0" Body="&lt;p&gt;It is not speed but depth that counts.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-04T00:42:56.070" />
  <row Id="8848" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T00:44:05.303" Score="17" ViewCount="0" Body="&lt;p&gt;To some degree speed matters. If you are going to sit there thinking &quot;hmmmm ... what &lt;em&gt;is&lt;/em&gt; a subspace?&quot; every single time you see the word &quot;subspace&quot; you may never get through even the simplest of linear algebra proofs. So, some knowledge must be there automatically, that's why we all practice so much. That's why we don't just learn definitions but &lt;em&gt;memorize&lt;/em&gt; them. That's the point of knowing a proof by heart (and I do mean &lt;em&gt;knowing&lt;/em&gt; not just memorizing it in that case) so that it falls off of your tongue with ease. That way, when you need to prove something similar you aren't struggling with the mechanics. You have room in your brain to think about the big picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, the process of doing math is all about doing things more quickly. Think of how long it took you to solve a quadratic when you were in grade school. Now you just look at it and you know the solutions are there, and you can probably see the factors, or write the quadratic formula without thinking about it. What once was a big problem is now just a tiny step in much larger problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That all said, I do think that at times lay people mistake being fast with being good at mathematics. Some of the mathematicians I admire most are careful... even plodding in their thinking, going over each step in a way that seems almost naive. --yet they never seem to &lt;em&gt;miss&lt;/em&gt; anything and in their careful way of reasoning they make many small discoveries that someone who was rushing to the answer might miss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These mathematicians whom I admire stop now and then and think carefully about even the solutions to quadratics, they are looking for something deeper in each detail. And they are &lt;em&gt;not self-conscious&lt;/em&gt; about the speed at which they work. &lt;strong&gt;Rushing to impress others is a great source of errors.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm as guilty as anyone else of this. But, I will try to emulate my mentors and not care if it takes me a long time. As long as in the end what I say is logical... and correct.  &lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-04T00:44:05.303" CommentCount="1" />
  <row Id="8849" PostTypeId="2" ParentId="8830" CreationDate="2010-11-04T00:47:21.133" Score="2" ViewCount="0" Body="&lt;p&gt;You can find a nice introduction here: &amp;#56256;Pistone, Giovanni and Riccomagno, Eva and Wynn, Henry, Pistone, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pistone, Giovanni; Riccomagno, Eva; Wynn, Henry&lt;br&gt;&#xA;&lt;a href=&quot;http://alexandria.tue.nl/repository/books/542984.pdf&quot; rel=&quot;nofollow&quot;&gt;Computational commutative algebra in discrete statistics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T00:47:21.133" CommentCount="1" />
  <row Id="8850" PostTypeId="1" AcceptedAnswerId="8852" CreationDate="2010-11-04T01:28:39.633" Score="0" ViewCount="41" Body="&lt;p&gt;I have a branching process of the form p0=0.1, p1 = 0.6, p2 = 0.3. (any other pn = 0). Z0, the original population is 1. Z1 is the population after 1 timestep, Z2 is the population after 2 timesteps, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I compute the probability mass function of Z3? It would suffice to have an expression I could expand on my TI-89.&lt;/p&gt;&#xA;" OwnerUserId="2617" LastEditorUserId="2617" LastEditorDisplayName="" LastEditDate="2010-11-04T01:35:44.900" LastActivityDate="2010-11-04T01:59:35.763" Title="Given probabilities for a branching process, how do I compute the probability mass function at a time?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8851" PostTypeId="1" AcceptedAnswerId="8855" CreationDate="2010-11-04T01:56:32.073" Score="0" ViewCount="50" Body="&lt;p&gt;Given that X_i is a partition of the probability space, and Y,Z are events in the probability space such that P[Y intersect X_i] &gt; 0 for all i, how can I prove that P[Z|Y] = sum(from i=1 to infinity) P[Z|Y intersect X_i] * P[X_i | Y] ?&lt;/p&gt;&#xA;" OwnerUserId="2617" LastActivityDate="2010-11-04T02:34:26.073" Title="How can I prove this law of &quot;generalized&quot; total probability?" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="8852" PostTypeId="2" ParentId="8850" CreationDate="2010-11-04T01:59:35.763" Score="3" ViewCount="0" Body="&lt;p&gt;The probability of having $k$ individuals after 3 steps is the coefficient of $x^k$ in the expansion of $f(f(f(x)))$, where $f(x) = 0.1 + 0.6x + 0.3x^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generalizes a lot: you can have different types of individuals, each with their own branching probabilities, which can change from step to step. &lt;/p&gt;&#xA;" OwnerUserId="2032" LastActivityDate="2010-11-04T01:59:35.763" CommentCount="1" />
  <row Id="8853" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T02:08:02.747" Score="13" ViewCount="0" Body="&lt;p&gt;Depth of thought is much more important than speed. A good illustration of this is the work of Grigori Perelman, who proved the Poincaré Conjecture. In their article about him published in The New Yorker, Sylvia Nasar and David Gruber wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'At Leningrad University, which Perelman entered in 1982, at the age of sixteen, he took advanced classes in geometry and solved a problem posed by Yuri Burago, a mathematician at the Steklov Institute, who later became his Ph.D. adviser. “There are a lot of students of high ability who speak before thinking,” Burago said. “Grisha was different. He thought deeply. His answers were always correct. He always checked very, very carefully.” Burago added, “He was not fast. Speed means nothing. Math doesn’t depend on speed. It is about deep.”'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tinyurl.com/3xga4f5&quot; rel=&quot;nofollow&quot;&gt;http://tinyurl.com/3xga4f5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2010-11-04T02:08:02.747" CommentCount="3" />
  <row Id="8854" PostTypeId="1" CreationDate="2010-11-04T02:25:52.593" Score="2" ViewCount="279" Body="&lt;p&gt;The classic text of Deligne and Rapoport giving a model for the mod p reduction of the modular curve is found in a &quot;Lecture Notes in Mathematics&quot; volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has it been translated from the original french and/or LaTeX-ed so that it can be searched more easily?&lt;/p&gt;&#xA;" OwnerUserId="2580" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T02:10:16.417" LastActivityDate="2010-12-23T00:44:25.973" Title="Has Deligne-Rapoport been translated?" Tags="&lt;reference-request&gt;&lt;modular-forms&gt;" AnswerCount="2" />
  <row Id="8855" PostTypeId="2" ParentId="8851" CreationDate="2010-11-04T02:34:26.073" Score="4" ViewCount="0" Body="&lt;p&gt;Because $X_i$ are a partition,&#xA;$$P(Z \cap Y) = \sum_i P(Z \cap Y \cap X_i)$$&#xA;Now apply the definition of conditional probability, once to the left-hand side and twice to the right-hand side, to get your result.&lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-11-04T02:34:26.073" />
  <row Id="8856" PostTypeId="1" AcceptedAnswerId="8861" CreationDate="2010-11-04T02:56:39.660" Score="9" ViewCount="210" Body="&lt;p&gt;My class had a good time redeveloping the unit circle under the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_geometry&quot; rel=&quot;nofollow&quot;&gt;taxicab metric&lt;/a&gt;. Now some of them want to do it again with another similar metric. I want to give this to some of my &quot;honors&quot; high-school* calculus I students to work on independently. As such, it can't be too difficult or require too many advanced concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is another metric that I could introduce them to where one can describe a &quot;unit circle&quot; and try to make sense of the sine, cosine and tangent functions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I have a group of high school students enrolled in a college course. The course, as planned, is &lt;em&gt;far&lt;/em&gt; too easy for them. (At the same time, it is quite hard for the college students enrolled. Go figure.) After the high school students do what has been assigned by the online learning system I have them work on other stuff-- which I allow to be pretty self-guided. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-13T13:15:21.310" LastActivityDate="2010-11-13T18:00:17.880" Title="Developing the unit circle in geometries with different metrics: beyond taxi cabs." Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;big-list&gt;&lt;metric-spaces&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="5" />
  <row Id="8857" PostTypeId="2" ParentId="8856" CreationDate="2010-11-04T03:06:50.743" Score="8" ViewCount="0" Body="&lt;p&gt;Any norm defines a distance, by $d((a,b),(c,d)) = ||(a-c,b-d)||$. Some common norms:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are all the $p$-norms: $||(x,y)||_p = \sqrt[p]{|x|^p + |y|^p}$ (the usual norm occurs with $p=2$); you can do them for any $p$, $0\lt p\lt \infty$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sup norm: $||(x,y)||_{\infty} = \max\{|x|,|y|\}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can take a positive linear combination of norms to create a new one. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given any linear transformation $A\colon\mathbb{R}^2\to\mathbb{R}^2$, and any norm $||\cdot||$, you can define the norm that maps $(x,y)$ to $||A(x,y)||$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is also the discrete metric, though that will create a rather nasty &quot;unit circle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually nice to treat the sup norm, the $p$-norms, and the taxi-cab norm together. If you draw the &quot;unit circles&quot; for all of them, the limit of the $p$ norms as $p\to 0^+$ is the taxicab norm, while the limit as $p\to\infty$ is the sup norm. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T03:47:17.560" LastActivityDate="2010-11-04T03:47:17.560" CommentCount="6" />
  <row Id="8859" PostTypeId="1" AcceptedAnswerId="8873" CreationDate="2010-11-04T03:36:25.853" Score="1" ViewCount="49" Body="&lt;p&gt;This is somewhat of a follow-up question to &lt;a href=&quot;http://math.stackexchange.com/q/8426/2961&quot;&gt;‘Elliptic integrals with parameter outside 0&amp;lt;m&amp;lt;1’&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an equation that I'm attempting to simplify that has terms that look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\Re\bigl( -i (c-1) \mathrm K(c) -2i \mathrm E(c) \bigr)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c&gt;1$. I.e., they're being evaluated outside their standard parameter range of $0\le m\le 1$. I know the equation has been derived with Mathematica, and I'm suspicious that the $\Re(i\mathrm K(c))$ terms, etc., are more complex than they need to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've taken a look at &lt;a href=&quot;http://dlmf.nist.gov/19.7#i&quot; rel=&quot;nofollow&quot;&gt;Legendre's Relation&lt;/a&gt; which seems to be related to splitting up these integrals into real and imaginary parts, and it doesn't seem to click—I suspect that the expression above is about as simple as I can get it (i.e., even if I manage to replace the integrals by alternate versions that have only real components, the resulting expression will be &lt;em&gt;more&lt;/em&gt; complex than that above). Is there anything I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="2961" LastEditorUserId="2961" LastEditorDisplayName="" LastEditDate="2010-11-04T04:57:13.467" LastActivityDate="2010-11-04T12:57:46.513" Title="Separate elliptic integrals into real and imaginary parts" Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8860" PostTypeId="1" AcceptedAnswerId="8862" CreationDate="2010-11-04T03:49:40.243" Score="2" ViewCount="58" Body="&lt;p&gt;My number theory library of choice doesn't implement the logarithmic integral for complex values.  I thought that I might take a crack at coding it, but I thought I'd ask here first for algorithmic advice and/or references.  I'm sure there are better methods than naively calculating the integral.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T03:52:10.553" LastActivityDate="2010-11-04T04:23:03.603" Title="Efficiently calculating the logarithmic integral with complex argument" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;special-functions&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="8861" PostTypeId="2" ParentId="8856" CreationDate="2010-11-04T03:57:06.740" Score="7" ViewCount="0" Body="&lt;p&gt;Two funny distances on $\mathbb{R}^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The elevator distance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;d((x_1,y_1),(x_2,y_2)) = &#xA;\begin{cases}&#xA;\vert y_1 - y_2 \vert                  &amp;amp; \text{if}\ x_1 = x_2 \\&#xA;\vert y_1 \vert + \vert x_1 - x_2 \vert + \vert y_2\vert &amp;amp; \text{otherwise} &#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The mail office distance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;d(p,q) = &#xA;\begin{cases}&#xA;0            &amp;amp; \text{if}\ p = q \\&#xA;\|p\| + \|q\|    &amp;amp; \text{otherwise}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good question to ask: why are they called &quot;elevator&quot; and &quot;mail office&quot; distances, respectively?  :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. As Arturo Magidin points out, with these distances balls centered at the origin are not particularly interesting: you have to try with balls NOT centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MORE EDIT. Don: have you seen Arturo Magidin's comment about making balls centered at the origin &quot;interesting&quot;?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-04T06:35:56.103" LastActivityDate="2010-11-04T06:35:56.103" CommentCount="7" />
  <row Id="8862" PostTypeId="2" ParentId="8860" CreationDate="2010-11-04T04:23:03.603" Score="5" ViewCount="0" Body="&lt;p&gt;For this answer, I'm assuming the definition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{li}(z):=\mathrm{PV}\int_0^z\frac{\mathrm{d}u}{\ln\;u}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we assume the Cauchy principal value (the more common definition in number theory that has a lower limit of 2 differs merely by a constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the first thing you have to note is the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{li}(z)=\mathrm{Ei}(\ln\;z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathrm{Ei}(z)$ is an &lt;a href=&quot;http://dlmf.nist.gov/6.2&quot; rel=&quot;nofollow&quot;&gt;exponential integral&lt;/a&gt;. (Again, I repeat my advice to people who encounter strange functions: you would really do well to check the DLMF first for identities and references.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathrm{Ei}(z)$ is a slightly more tractable beastie to numerically evaluate, since the singularity at $z=0$ can be confined to a logarithmic part; to wit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Ei}(z)=\gamma+\ln\;z+\int_0^z \frac{\exp(u)-1}{u}\mathrm{d}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the last portion is an entire function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, depending on which of the left or right half-planes should the exponential integral be evaluated, your strategy will differ (it is a common fact that most special function routines are &lt;em&gt;polyalgorithms&lt;/em&gt;, since their behavior can markedly differ in different regions of the complex plane). I will be vague for the rest of this answer since you did not clarify your region of interest. Suffice it to say that one usually uses a &lt;a href=&quot;http://dlmf.nist.gov/6.9&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; for arguments in the left half-plane, a power series for small to medium-sized arguments, and &lt;a href=&quot;http://dlmf.nist.gov/6.12&quot; rel=&quot;nofollow&quot;&gt;asymptotic expansions&lt;/a&gt; for large arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If implementing it yourself is starting to sound daunting (because it is), I will have to point out this &lt;a href=&quot;http://dx.doi.org/10.1145/78928.78934&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; and the corresponding &lt;a href=&quot;http://www.netlib.org/toms/683&quot; rel=&quot;nofollow&quot;&gt;FORTRAN subroutine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps a bit.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-04T04:23:03.603" CommentCount="7" />
  <row Id="8863" PostTypeId="2" ParentId="8808" CreationDate="2010-11-04T04:24:07.533" Score="7" ViewCount="0" Body="&lt;p&gt;In my opinion (because it really is an opinion), the main reason to care about euclidean spaces is that they come equipped with a lot of particularly nice structures.  In particular, they are finite-dimensional Hilbert spaces.  This means that they are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Finite-dimensional vector spaces (so we can talk about addition and scaling)&lt;/li&gt;&#xA;&lt;li&gt;Complete metric spaces (so we can talk about distances and limits, and limits behave as we would like them to)&lt;/li&gt;&#xA;&lt;li&gt;Inner product spaces (so we can talk about the notion of &quot;angle,&quot; and thereby do all sorts of geometric things.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And really, there's just so much geometric intuition that comes along with these ideas -- not to mention an entire calculus apparatus.  After all, there are plenty of topological spaces where the above notions are not defined or otherwise fail to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason (at least to me) to study locally euclidean spaces is that we want to study spaces that are more general than euclidean spaces, yet still retain many of their nice features.  In particular, we want a place where calculus makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Areas of math which study more abstract spaces include topology and algebraic geometry.  Admittedly, I'm not very well-versed in either just yet, but I'm sure practical uses (and physical models) have been found in both.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-11-04T04:24:07.533" CommentCount="3" />
  <row Id="8864" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T04:46:07.663" Score="2" ViewCount="0" Body="&lt;p&gt;Speed is great for brute force calculations and computer simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For proofs that rely on abstract ideas and arguments, speed is inconsequential.&lt;/p&gt;&#xA;" OwnerUserId="3017" LastActivityDate="2010-11-04T04:46:07.663" CommentCount="1" />
  <row Id="8865" PostTypeId="1" AcceptedAnswerId="8868" CreationDate="2010-11-04T04:53:44.743" Score="5" ViewCount="268" Body="&lt;p&gt;I'm a bit curious as to how &quot;real&quot; mathematicians would solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Find the least prime number greater than 2000.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I can always go brute force:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;IF N IS PRIME, OUTPUT N&#xA;ELSE INCREASE N BY 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But that's no fun. (Especially when the number is really high.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any algorithms/tricks/etc that can help me solve this quickly, and efficiently, especially if I were given large values of N?&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-11-04T22:50:52.830" LastActivityDate="2010-11-04T22:50:52.830" Title="The least prime greater than 2000" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="8867" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T05:11:31.497" Score="7" ViewCount="0" Body="&lt;p&gt;If you want to find several primes at once starting from some point, you can use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Eratosthenes_sieve&quot; rel=&quot;nofollow&quot;&gt;sieve&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-04T05:11:31.497" CommentCount="1" />
  <row Id="8868" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T05:57:27.523" Score="11" ViewCount="0" Body="&lt;p&gt;I expect there are more powerful tools available; but for a simple approach I would apply the Prime Number Theorem (PNT) and a sieve (as Yuval has already suggested).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One useful consequence of the PNT is that around a number &lt;em&gt;N&lt;/em&gt;, approximately one out of every &lt;em&gt;log(N)&lt;/em&gt; numbers is prime.  (By '&lt;em&gt;log&lt;/em&gt;,' number theorists always mean the natural log '&lt;em&gt;ln&lt;/em&gt;'.)  So around 2000, about 1 out of every 7.6 numbers is prime.  Let's just look among the numbers 2001 to 2060 for our next prime-- I'm leaving extra space in case a big prime gap happens to show up around 2000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use the sieve-- we can throw out all even numbers in our list, since they are certainly not primes.  Next, we can throw out everything divisible by 3, then everything divisible by 5, by 7, by 11,...  When should we stop sieving?  We can stop when we get to $\sqrt{2060} \approx 45$, since if any number in our list can be factored as &lt;em&gt;M = ab&lt;/em&gt;, then one of the factors must be less than or equal to the square root of the biggest number in our list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a computer would be happy to do this process for you.  But, I just wrote down the list and went through my basic sieve (primes up to 43); and I got the prime numbers 2003, 2011, 2017, 2027, 2029, 2039, 2053.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-04T05:57:27.523" CommentCount="4" />
  <row Id="8870" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T06:45:22.180" Score="1" ViewCount="0" Body="&lt;p&gt;An integer $p$ is prime if and only if the following equation holds:&#xA;\begin{eqnarray}&#xA;\sum_{n \geq 1} \left \lfloor \frac{p}{n} \right \rfloor - \left \lfloor \frac{p-1}{n} \right \rfloor  = 2&#xA;\end{eqnarray}&#xA;NB: The infinite number of terms of sum is there to accommodate an arbitrarily large integer $p$, but the sum eventually produces a string of $0$s. (See Crandall and Pomerance, &lt;em&gt;Prime Numbers: A Computational Perspective&lt;/em&gt;, New York: Springer, ISBN 0-387-04777-9)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After observing that $2001$ and $2002$ both violate the test early on but $2003$ does not, then you can use any number of primality tests (or choose to prove the sum equals 2 to establish primality directly). Otherwise, you may try something like Wilson's theorem, which states: An integer $p$ is prime if and only if $(p-1)! \equiv -1 \mod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica gives the correct result in less than 1 second: $2003$ is the first prime greater than $2000$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T07:49:17.273" LastActivityDate="2010-11-04T07:49:17.273" CommentCount="5" />
  <row Id="8871" PostTypeId="2" ParentId="8838" CreationDate="2010-11-04T06:50:27.667" Score="0" ViewCount="0" Body="&lt;p&gt;What do YOU think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mathematics, intelligence is needed.  And intelligence is instantaneous. And something that is instantaneous has no notion of speed; speed requires time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed as a measure cannot be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For arithmetic, speed can be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is an almost universal confusion between arithmetic and mathematics. Mathematics is the process of digging into the unknown.  Arithmetic is the process of playing with the artefacts that the process of mathematics has uncovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is termed mathematics in school is usually nothing more than arithmetic.  It depends not upon the material but upon the teaching.  If the student is actively enquiring,  with the teacher guiding this enquiry only when needed,  then mathematics will ensue.   If the teacher is intent upon stuffing the syllabus into the head of the poor student, the best that can be hoped for is arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="3096" LastActivityDate="2010-11-04T06:50:27.667" />
  <row Id="8873" PostTypeId="2" ParentId="8859" CreationDate="2010-11-04T07:00:36.917" Score="3" ViewCount="0" Body="&lt;p&gt;Well, for the simplified case that you now presented in your current version of the question (again, as I said, the complete elliptic integral of the third kind requires a different sort of analysis), the identities in &lt;a href=&quot;http://dlmf.nist.gov/19.7.E3&quot; rel=&quot;nofollow&quot;&gt;formula 19.7.3 in the DLMF&lt;/a&gt; can be rewritten in parameter form as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(m)=\frac1{\sqrt{m}}\left(K\left(\frac1{m}\right)-i K\left(1-\frac1{m}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(m)=\sqrt{m}\left(E\left(\frac1{m}\right)-\left(1-\frac1{m}\right)K\left(\frac1{m}\right)+i\left(E\left(1-\frac1{m}\right)-\frac1{m} K\left(1-\frac1{m}\right)\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we apply both to your expression above, it can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\sqrt{c}E\left(1-\frac1{c}\right)-\frac{1+c}{\sqrt{c}}K\left(1-\frac1{c}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T12:57:46.513" LastActivityDate="2010-11-04T12:57:46.513" CommentCount="3" />
  <row Id="8875" PostTypeId="1" CreationDate="2010-11-04T07:14:18.410" Score="2" ViewCount="109" Body="&lt;p&gt;I came across an interesting problem but unable to see how to approach it. How do I use the dominated convergence theorem (LDCT), to show that first derivative of characteristic function of probability distribution at t = 0, $\phi^′(0)=iE[X]$? Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory&lt;/a&gt;)&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="957" LastEditorDisplayName="" LastEditDate="2010-11-04T07:42:05.157" LastActivityDate="2010-11-28T22:46:44.740" Title="derivative of characteristic function" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8876" PostTypeId="1" CreationDate="2010-11-04T07:22:58.900" Score="3" ViewCount="107" Body="&lt;p&gt;What does this expression mean - $\lim_{n\rightarrow\infty} E|X_n-X|=0$? $X_n$ is a sequence of random variables and $X$ is a random variable. What does this expression imply? Can I say that the sequence $X_n$ converges to $X$ in probability and almost sure convergence?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-11-04T08:33:38.020" Title="convergence of sequence of random variables" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8877" PostTypeId="2" ParentId="8876" CreationDate="2010-11-04T07:33:05.227" Score="2" ViewCount="0" Body="&lt;p&gt;This is called &lt;em&gt;convergence in the mean&lt;/em&gt; or convergence in the $L^{1}$-norm. In general, if &#xA;\begin{eqnarray}&#xA;\lim_{n \to \infty} \mathbb{E}(| X_{n} - X|^{p}) = 0,&#xA;\end{eqnarray}&#xA;then $X_{n}$ is said to converge to $X$ in the $L^{p}$-norm (provided that $\mathbb{E}(|X_{n}|^{p})$ is finite for all $n \geq 1$). Analytically, there are nice implications of such convergence. For example, convergence in an $L^{p}$-norm implies convergence in an $L^{q}$-norm if $p \geq q$. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_random_variables&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Convergence_of_random_variables&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markov's inequality states&#xA;\begin{eqnarray}&#xA;\mathbb{P}(|X_{n} - X| &gt; \epsilon) \leq \epsilon^{-p} \, \mathbb{E}(|X_{n} - X|^{p}).&#xA;\end{eqnarray}&#xA;Thus, $L^{p}$-norm convergence implies convergence in probability.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T07:46:04.093" LastActivityDate="2010-11-04T07:46:04.093" CommentCount="1" />
  <row Id="8878" PostTypeId="2" ParentId="8876" CreationDate="2010-11-04T07:58:49.120" Score="0" ViewCount="0" Body="&lt;p&gt;When $\lim _{n \to \infty } E|X_n - X| = 0$, we say that $X_n$ converges in mean to $X$. It is very well known that this implies that $X_n$ converges in probability to $X$, but not that $X_n$ converges almost surely to $X$. Consider a sequence $(X_n)$ of independent random variables such that &#xA;$P(X_n  = 0) = 1 - 1/n$ and $P(X_n  = 1) = 1/n$. Then $E|X_n|=1/n$, and hence $X_n$ converges in mean to $0$. However, since $\sum\nolimits_{n = 1}^\infty  {P(X_n  = 1)}  = \infty $ and the $X_n$ are independent, almost surely the sequence $(X_n)$ contains infinitely many $1$'s; in particular $X_n$ does not converge to $0$ in the almost sure sense.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-04T08:33:38.020" LastActivityDate="2010-11-04T08:33:38.020" />
  <row Id="8879" PostTypeId="2" ParentId="8875" CreationDate="2010-11-04T08:04:14.123" Score="1" ViewCount="0" Body="&lt;p&gt;You need to show that if $(a_n)$ is a sequence of nonzero numbers&#xA;which tends to zero, that $(\phi(a_n)-\phi(0))/a_n\to iE[X]$. Now&#xA;$$\frac{\phi(a)-\phi(0)}{a}=E\left(\frac{e^{iaX}-1}{a}\right).$$&#xA;You need some hypothesis on $X$ for the result to work, for&#xA;instance that $|X|$ has finite expectation. Certainly $(e^{iaX}-1)/a\to iX$&#xA;so to apply dominated convergence we need a function $f(X)$ with finite&#xA;expectation and with $|(e^{iaX}-1)/a|\le f(X)$ at least for $a$ in a&#xA;deleted neighbourhood of $0$. Does $f(X)=|X|$ work?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-04T08:04:14.123" />
  <row Id="8880" PostTypeId="2" ParentId="7142" CreationDate="2010-11-04T08:07:15.277" Score="0" ViewCount="0" Body="&lt;p&gt;I think this can easily be formulated as a &lt;a href=&quot;http://en.wikipedia.org/wiki/QCQP&quot; rel=&quot;nofollow&quot;&gt;QCQP&lt;/a&gt;&#xA;that with a Positive Definite matrix. Therefore the problem is convex and can be solved using interior point methods.&lt;/p&gt;&#xA;" OwnerUserId="3097" LastActivityDate="2010-11-04T08:07:15.277" CommentCount="1" />
  <row Id="8881" PostTypeId="2" ParentId="7391" CreationDate="2010-11-04T08:24:33.093" Score="1" ViewCount="0" Body="&lt;p&gt;There is no need for trigonometric identities, complex exponentials or the like. Observe that&#xA;\begin{eqnarray}&#xA;\int_{0}^{\pi} \sin(2 x) \sin (x) dx = \int_{0}^{\pi/2}  \sin(2 x) \sin (x) dx  + \int_{\pi/2}^{\pi}  \sin(2 x) \sin (x) dx &#xA;\end{eqnarray}&#xA;By a change of variables ($x \to \pi - x$) and the oddness of the integrand on the interval $[0,\pi]$, you find that &#xA;\begin{eqnarray}&#xA; \int_{\pi/2}^{\pi}  \sin(2 x) \sin (x) dx = - \int_{0}^{\pi/2}  \sin(2 x) \sin (x) dx ,&#xA;\end{eqnarray}&#xA;which implies that the original integral vanishes identically.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-04T08:24:33.093" />
  <row Id="8882" PostTypeId="2" ParentId="8814" CreationDate="2010-11-04T08:36:14.877" Score="1" ViewCount="0" Body="&lt;p&gt;$$ \sin \theta \cdot \sin \bigl(60 - \theta \bigr) \cdot \sin \bigl(60 + \theta \bigr) = \frac{1}{4} \sin 3\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos \theta \cdot \cos \bigl(60  - \theta \bigr) \cdot \cos \bigl(60 + \theta \bigr) = \frac{1}{4} \cos 3\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tan \theta \cdot \tan \bigl(60  - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \frac{1}{4} \tan 3\theta $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-04T08:36:14.877" CommunityOwnedDate="2010-11-04T08:36:14.877" CommentCount="4" />
  <row Id="8883" PostTypeId="2" ParentId="8814" CreationDate="2010-11-04T08:37:19.310" Score="5" ViewCount="0" Body="&lt;p&gt;$$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-04T08:37:19.310" CommunityOwnedDate="2010-11-04T08:37:19.310" />
  <row Id="8884" PostTypeId="1" AcceptedAnswerId="8891" CreationDate="2010-11-04T08:53:05.593" Score="9" ViewCount="299" Body="&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/7583/an-identity-concerning-the-riemann-zeta-function&quot;&gt;post&lt;/a&gt;, David Speyer, actually, gave an expression for $\displaystyle  \frac{t}{e^{t}-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is can we sum the given series, using that expression, if not how does one sum this series. $$\sum\limits_{n=1}^{\infty} \frac{n}{e^{2\pi n}-1}=\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-12T04:07:07.623" Title="Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8885" PostTypeId="2" ParentId="8856" CreationDate="2010-11-04T09:04:47.440" Score="4" ViewCount="0" Body="&lt;p&gt;One closely related to the two in Agusti's answer is what I call the &lt;strong&gt;bus metric&lt;/strong&gt; (though I change the name to reflect the name of the local bus company - which seems to change each time I teach about metric spaces!).  This is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;d(p,q) = \begin{cases}&#xA;\|p - q\|_2 &amp;amp; \text{if} p, q \text{ are collinear} \\&#xA;\|p\|_2 + \|q\|_2 &amp;amp; \text{otherwise}&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In Trondheim (where I am) then the majority of bus routes are radial, so this does link to the students' intuition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The students could do a nice animation of what happens to a ball of unit length centred at a point $(x,0)$ as $x$ ranges from, say, $-2$ to $2$.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastActivityDate="2010-11-04T09:04:47.440" CommentCount="2" />
  <row Id="8886" PostTypeId="2" ParentId="8814" CreationDate="2010-11-04T09:07:00.950" Score="5" ViewCount="0" Body="&lt;p&gt;Machin's Formula:&#xA;\begin{eqnarray}&#xA;\frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T08:54:51.073" LastActivityDate="2010-11-05T08:54:51.073" CommunityOwnedDate="2010-11-04T09:07:00.950" />
  <row Id="8889" PostTypeId="1" CreationDate="2010-11-04T10:04:05.877" Score="2" ViewCount="77" Body="&lt;p&gt;Let $\mathbf{x} =$ {$x_{i}$} be a set of $n$ positive reals. In every good book on inequalities, one finds the classical result &#xA;\begin{eqnarray}&#xA;AM(\mathbf{x}) \geq GM(\mathbf{x}) \geq HM(\mathbf{x}),&#xA;\end{eqnarray}&#xA;where $AM(\mathbf{x}) = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$ is the arithmetic mean, $GM(\mathbf{x}) = \sqrt[n]{x_{1} \cdots x_{n}}$ is the geometric mean and $HM(\mathbf{x}) = n (\sum_{i = 1}^{n} \frac{1}{x_{i}})^{-1}$ is the harmonic mean of $\mathbf{x}$, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: I'm curious about sharp bounds of the form: &#xA;\begin{eqnarray}&#xA;HM(\mathbf{x}) \geq f(\mathbf{x}) AM(\mathbf{x}) + g(\mathbf{x}),&#xA;\end{eqnarray}&#xA;where $f$ and $g$ are some functions &lt;em&gt;which do not imply the classical result above or render the inequality trivial&lt;/em&gt;. Do such results exist in the literature (or mathematical folklore)? (References are welcome.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-04T17:26:25.460" LastActivityDate="2010-11-04T17:26:25.460" Title="Sharp (Reverse) Harmonic-Arithmetic Mean Bounds" Tags="&lt;real-analysis&gt;&lt;algebra&gt;&lt;reference-request&gt;&lt;inequality&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="8890" PostTypeId="1" CreationDate="2010-11-04T10:17:57.210" Score="-5" ViewCount="81" Body="&lt;p&gt;for $z, w \in C$, prove that $2|zw| &amp;lt;= |z|^2+|w|^2$, and deduce that $|z+w|^2 &amp;lt;= 2(|z|^2+|w|^2)$. where $|v|$ is modulus of vector $v \in C$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-22T01:14:43.220" LastActivityDate="2010-11-22T01:14:43.220" Title="for $z, w \in C$, prove that $2|zw| &lt;= |z|^2+|w|^2$" Tags="&lt;complex-numbers&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8891" PostTypeId="2" ParentId="8884" CreationDate="2010-11-04T10:54:51.530" Score="20" ViewCount="0" Body="&lt;p&gt;What you require here are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eisenstein_series&quot; rel=&quot;nofollow&quot;&gt;Eisenstein series&lt;/a&gt;. In particular the evaluation of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at $\tau = i. $ Rearrange to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_series&quot; rel=&quot;nofollow&quot;&gt;Lambert series&lt;/a&gt; for additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_ 2(\tau) = \zeta(2) \left(&#xD;&#xA;1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)&#xD;&#xA;=\zeta(2)E_2(\tau)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;satisfies the quasimodular transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =&#xD;&#xA;(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as given in the comment to the question by Slowsolver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very nice generalisation of the sum in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For odd $ m &gt; 1 $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_k$ are the Bernoulli numbers defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{z}{e^z - 1} = \sum_{k=0}^\infty \frac{B_k}{k!} z^k \quad \textrm{ for }&#xD;&#xA;|z| &amp;lt; 2 \pi.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T04:07:07.623" LastActivityDate="2010-11-12T04:07:07.623" CommentCount="8" />
  <row Id="8892" PostTypeId="1" CreationDate="2010-11-04T11:16:53.877" Score="1" ViewCount="74" Body="&lt;p&gt;I need to find the number of undirected trees on $n$ vertices such that the edges (and not the vertices) are labeled and exactly one label appears twice (i.e. there are $n-2$ possible labels and they all appear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning goes like this: From Cayley's formula we know there are $n^{n-2}$ trees with labeled vertices. Erase the label $n$ from its vertex and for every other vertex, move the label from the vertex itself to the edge in the direction of the $n$ vertex. This way we get a tree with $n-1$ labels for the edges. However, this argument is flawed, since for example 0--1--2 and 0--2--1 give the same labeled tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that I managed to fix the problem and find the number of edge-labeled trees with $n-1$ edges, and let's denote it by $k$; my next idea is choosing one of the $n-2$ first labels and relabeling $n-1$ to what I choose. This causes double-counting so I think I need to divide by 2 here, but nothing more. Hence the final number is $(n-2)k/2$. Am I correct in this reasoning - and how can I find k? Quite obviously it's not $n^{n-2}$ since otherwise $(n-2)k/2$ might not even be an integer (if $n$ is odd).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-04T12:03:08.360" LastActivityDate="2011-01-03T12:44:34.183" Title="Number of undirected trees with labeled edges, one repeating" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;trees&gt;&lt;enumeration&gt;" AnswerCount="1" />
  <row Id="8893" PostTypeId="1" AcceptedAnswerId="8896" CreationDate="2010-11-04T11:25:11.120" Score="7" ViewCount="266" Body="&lt;p&gt;I am having trouble expressing the titular question as iterated integrals over a given region. I have tried narrowing down the problem, and have concluded that the simplest way to approach this is to integrate over the XZ plane in the positve octant and multiply by 8, but I am having trouble identifying the bounding functions.&lt;/p&gt;&#xA;" OwnerUserId="3102" LastActivityDate="2010-11-04T14:47:19.917" Title="Volume bounded by cylinders x^2 + y^2 = r^2 and z^2 + y^2 = r^2" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8894" PostTypeId="2" ParentId="8892" CreationDate="2010-11-04T11:32:36.813" Score="0" ViewCount="0" Body="&lt;p&gt;The correspondence between edge- and vertex-labellings may work better starting from the degree 1 vertices instead of edge $n$ which would be at a random location inside the tree.   For the degree 1 vertices we have a unique choice of label to assign.  Remove these vertices and edges and repeat the process until the tree is a single vertex (0 edges) or a single edge.   These cases correspond to enumeration of trees on labelled vertices, with the &quot;center&quot; of the tree being a vertex or an edge, respectively.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-04T12:02:02.463" LastActivityDate="2010-11-04T12:02:02.463" />
  <row Id="8895" PostTypeId="1" AcceptedAnswerId="8897" CreationDate="2010-11-04T11:40:52.730" Score="1" ViewCount="55" Body="&lt;p&gt;This one comes from my module without proper explanation.Please suggest how to approach this problem ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:How do you write degree (in symbol) in math-mode ?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-11-04T11:52:56.930" LastActivityDate="2010-11-04T11:54:47.810" Title="In right triangle PQR, ∠Q = $90 ^\circ$ and PR = 2PQ, then ∠R is equal to ?" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8896" PostTypeId="2" ParentId="8893" CreationDate="2010-11-04T11:46:53.200" Score="9" ViewCount="0" Body="&lt;p&gt;The solid lies above the region $D$ in the $xy$-plane bounded by the circle $x^{2} + y^{2} = r^{2}$, so the volume is given by the integral $$\int\int\limits_{D} f(x,y) \ dA = \int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \ dx dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the required volume of the solid is: $$\int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} 2\sqrt{r^{2}-y^{2}} \ dx dy = \frac{16}{3}r^{3}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-04T11:46:53.200" CommentCount="1" />
  <row Id="8897" PostTypeId="2" ParentId="8895" CreationDate="2010-11-04T11:54:47.810" Score="2" ViewCount="0" Body="&lt;p&gt;As Djaian pointed out use the fact that $$\sin\angle{R} = \frac{PQ}{PR}=\frac{PQ}{2PQ}=\frac{1}{2}$$ So $\angle{R} = ?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also i hope you are aware of the fact that in a right angled triangle $$\sin\theta =\frac{\text{Opposite Side}}{\text{Hypotenuse}}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-04T11:54:47.810" />
  <row Id="8898" PostTypeId="1" CreationDate="2010-11-04T13:14:05.177" Score="0" ViewCount="104" Body="&lt;p&gt;Let A and B be nxn matrices over C. How to prove $(AB)^{\ast}=B^{\ast}A^{\ast}$? $A^{\ast}$ is complex conjugate transpose( if matrix $A$ is real then $A^{\ast} = A^{t}$. This is homework and is in the area of linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-04T13:20:48.720" LastActivityDate="2010-11-04T13:20:48.720" Title="How to prove $(AB)^{\ast}=B^{\ast}A^{\ast}$?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8899" PostTypeId="2" ParentId="8898" CreationDate="2010-11-04T13:16:54.960" Score="1" ViewCount="0" Body="&lt;p&gt;Do you have the theorem that $(AB)^t=B^tA^t$ for real matrices?  Can you follow that proof through for the complex case (using conjugates)?  Without going back to look, I think it works.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T13:16:54.960" CommentCount="8" />
  <row Id="8900" PostTypeId="1" CreationDate="2010-11-04T13:22:23.917" Score="0" ViewCount="63" Body="&lt;p&gt;Hi all,&#xA;I need some online tool for diagonalizing 2x2 matrices or at least finding the eigenvectors and eigenvalues of it. I don't like to download any stuf because I'm not able to, some online tool will do the job.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3107" LastActivityDate="2010-11-04T14:12:56.287" Title="Online tool for diagonalizing matrices ?" Tags="&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8901" PostTypeId="2" ParentId="8865" CreationDate="2010-11-04T13:42:34.250" Score="6" ViewCount="0" Body="&lt;p&gt;Sieving, alone, is a poor approach to the problem unless n is small (which, in your example, it is).  A better approach for large numbers: choose an appropriate interval,* sieve out small primes, test remaining numbers with a probable-prime test, then (if desired) prove primality with ECPP or a similar algorithm (for small primes, APR-CL is better).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many programs can do this search for you, though they generally omit the last step (Pari and Mathematica at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.* If the interval is chosen at random, length 4.5 log x is 95% likely to hold at least one prime, since $4.5e^{-4.5}\approx0.05$.  If your interval is adversarially chosen to have few primes, you could choose an interval as long as about $2e^{-\gamma}\log^2x$.  If performance is critical you can determine the optimal sieving distance based on time and other resources taken for both parts.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-11-04T13:42:34.250" CommentCount="2" />
  <row Id="8902" PostTypeId="1" AcceptedAnswerId="8913" CreationDate="2010-11-04T13:44:08.290" Score="0" ViewCount="52" Body="&lt;p&gt;Let M, K, and L be isomorphic normal subgroups of a group G. Suppose $M \cap K \leq M \cap L$. Can we conclude anything about the index of MK in G and the index of ML in G? I would like it to be the case that MK has smaller index than ML, but I'm having a hard time convincing myself either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: I initially left out the word &quot;normal&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="759" LastEditorUserId="759" LastEditorDisplayName="" LastEditDate="2010-11-04T14:30:46.760" LastActivityDate="2010-11-04T14:36:39.557" Title="Can we say something about the index of these subgroups?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8903" PostTypeId="1" CreationDate="2010-11-04T13:49:11.100" Score="-3" ViewCount="157" Body="&lt;p&gt;How to prove polarization identity?&#xA;For any $u, v \in V$ &#xA;$4 \langle T(u), v \rangle = \langle T(u+v), u+v \rangle - \langle T(u-v), u-v \rangle + i\langle T(u+iv), u+iv \rangle - i\langle T(u-iv), u-iv \rangle$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-04T15:50:40.333" LastActivityDate="2010-11-21T04:07:39.877" Title="How to prove polarization identity?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8904" PostTypeId="1" AcceptedAnswerId="8915" CreationDate="2010-11-04T13:59:55.313" Score="3" ViewCount="241" Body="&lt;p&gt;From this &lt;a href=&quot;http://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbbq/6250#6250&quot;&gt;post&lt;/a&gt; we see that $\mathbb{R}$ over $\mathbb{Q}$ is infinite dimensional. Similarly $\mathbb{C}$ over $\mathbb{Q}$ is also infinite dimensional, and i rememeber having solved a problem that $\mathbb{R} \cong \mathbb{C}$ when considered as vector spaces over $\mathbb{Q}$. So i would like to extend this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $V$ and $W$ are two vector spaces, over the field $\mathbb{Q}$. Can we say that $V \cong W$ if they have the same dimension over $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know how to prove or disprove this!&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-04T15:06:16.370" LastActivityDate="2010-11-13T13:21:53.110" Title="Isomorphism of Vector spaces over $\mathbb{Q}$" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="8905" PostTypeId="2" ParentId="8893" CreationDate="2010-11-04T14:04:09.590" Score="7" ViewCount="0" Body="&lt;p&gt;This is one of those results in calculus which were anticipated by&#xA;Archimedes. He gave a correct formula for the volume but it is not known exactly &#xA;how Archimedes solved this problem. There is, however, a simple way to&#xA;obtain the answer without much calculus. Let me quote from late Gardner's &lt;a href=&quot;http://books.google.co.uk/books?id=PtBPcAAACAAJ&amp;amp;dq=unexpected+hanging&amp;amp;hl=en&amp;amp;ei=57rSTJywFNSP4ga9j8jcDw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCwQ6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Unexpected Hanging and Other Mathematical Diversions&lt;/em&gt;&lt;/a&gt; (Gardner considers the case $r=1$ but this is not essential, of course):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine a sphere of unit radius inside the volume common&#xA;  to the two cylinders and having as its center the point&#xA;  where the axes of the cylinders intersect. Suppose that the&#xA;  cylinders and sphere are sliced in half by a plane through&#xA;  the sphere's center and both axes of the cylinders. &#xA;  The cross section of the volume common to the cylinders will be a square. &#xA;  The cross section of the sphere will be a circle that fills the square.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/SwXzR.jpg&quot; alt=&quot;alt text&quot;&gt;        &lt;img src=&quot;http://i.imgur.com/vaASe.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now suppose that the cylinders and sphere are sliced by a&#xA;  plane that is parallel to the previous one but that shaves off&#xA;  only a small portion of each cylinder (&lt;em&gt;have a look at the picture on the left&lt;/em&gt;).&#xA;  This will produce parallel tracks on each cylinder,&#xA;  which intersect as before to form a square cross section of&#xA;  the volume common to both cylinders. Also as before, the&#xA;  cross section of the sphere will be a circle inside the square.&#xA;  It is not hard to see (with a little imagination and pencil&#xA;  doodling) that any plane section through the cylinders, parallel&#xA;  to the cylinders' axes, will always have the same result:&#xA;  a square cross section of the volume common to the cylinders,&#xA;  enclosing a circular cross section of the sphere.&#xA;  Think of all these plane sections as being packed together&#xA;  like the leaves of a book. Clearly, the volume of the sphere&#xA;  will be the sum of all circular cross sections, and the volume&#xA;  of the solid common to both cylinders will be the sum of all&#xA;  the square cross sections. We conclude, therefore, that the&#xA;  ratio of the volume of the sphere to the volume of the solid&#xA;  common to the cylinders is the same as the ratio of the area&#xA;  of a circle to the area of a circumscribed square. A brief calculation&#xA;  shows that the latter ratio is $\pi/4$. This allows the&#xA;  following equation, in which $x$ is the volume we seek:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\frac{4\pi r^3/3}{x}=\frac{\pi}{4}.$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The $\pi$'s drop out, giving $x$ a value of $16r^3/3$. The radius in&#xA;  this case is 1, so the volume common to both cylinders is&#xA;  $16/3$. As Archimedes pointed out, it is exactly $2/3$ the volume&#xA;  of a cube that encloses the sphere; that is, a cube with&#xA;  an edge equal to the diameter of each cylinder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-04T14:47:19.917" LastActivityDate="2010-11-04T14:47:19.917" CommentCount="2" />
  <row Id="8906" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T14:05:08.567" Score="0" ViewCount="0" Body="&lt;p&gt;All finite dimensional vector spaces of the same dimension over a field $k$ are isomorphic.  Fix a basis of $V$ and $W$; they each have $n$ elements if $V$ and $W$ are $n$-dimensional.  Then the map that sends basis elements $e_i$ of $V$ to basis elements $f_i$ of $W$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We usually say that if $V$ and $W$ are vector spaces over $k$ of the same dimension, they are isomorphic, but not canonically.  This is because to exhibit an isomorphism requires a &lt;i&gt;choice&lt;/i&gt; of basis.&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-04T14:05:08.567" CommentCount="5" />
  <row Id="8907" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T14:12:23.487" Score="2" ViewCount="0" Body="&lt;p&gt;Two vector spaces whose dimensions over a field are of the same cardinality are easily seen to be isomorphic by mapping one basis onto the other; i.e. if $V$ has basis ${e_i}_{i\in I}$ and $W$ has basis ${f_j}_{j\in J}$ where $I$ and $J$ are of the same cardinality, pick a bijection $\phi:I\rightarrow J$ and define a linear map by sending $e_i\mapsto f_{\phi(i)}$. Since the $e_i$ map bijectively to the $f_j$, this must be an isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="1306" LastActivityDate="2010-11-04T14:12:23.487" CommentCount="3" />
  <row Id="8908" PostTypeId="2" ParentId="8900" CreationDate="2010-11-04T14:12:56.287" Score="1" ViewCount="0" Body="&lt;p&gt;Wolframalpha has an option. Try this: &lt;a href=&quot;http://www.wolframalpha.com/examples/Matrices.html&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/examples/Matrices.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-04T14:12:56.287" />
  <row Id="8909" PostTypeId="1" CreationDate="2010-11-04T14:16:23.030" Score="10" ViewCount="176" Body="&lt;p&gt;I'm interested to know if anyone can point me to a non-calculus way of seeing that the volume of a pyramid = 1/3 x area of base x height. Yes, I've googled.&lt;/p&gt;&#xA;" OwnerUserId="3110" LastActivityDate="2010-11-05T00:00:19.863" Title="How to intuitively see that the volume of a pyramid = 1/3 x area of base x height" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="8910" PostTypeId="2" ParentId="8909" CreationDate="2010-11-04T14:22:58.393" Score="5" ViewCount="0" Body="&lt;p&gt;The easiest intuitive one I know is to take a cube and divide it into three pyramids.  Take one corner and form the pyramid with each of the three faces that don't touch that corner.  They are congruent, so have volume 1/3*base*height as the base is a square the side of the cube and the height is the side of the cube as well.  Then appeal to affine transformations to say this applies to all pyramids.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T14:22:58.393" CommentCount="2" />
  <row Id="8911" PostTypeId="1" CreationDate="2010-11-04T14:26:32.297" Score="-1" ViewCount="147" Body="&lt;p&gt;Suppose that for all $u \in V, \langle T(u), u \rangle =0$. How to prove that T is the 0 operator, that is $T(v)=0$ for all $v \in V$? Hint: Apply information on my last question about polarization identity to obtain that $\langle T(u), v \rangle = 0$ for all $u, v \in V$ and then let $v=T(u)$. this is a homework in the area of linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-04T15:10:10.670" LastActivityDate="2010-11-05T12:59:11.310" Title="How to prove that T is the 0 operator, that is  T(v)=0 for all $u, v \in V$?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8912" PostTypeId="2" ParentId="8611" CreationDate="2010-11-04T14:30:43.547" Score="4" ViewCount="0" Body="&lt;p&gt;user2969 asked for the full answer so I'll post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to take the gcd of the two numbers.  This can be done with the Euclidean algorithm and is fast: the naive implementation takes quadratic time, which is good enough.  Call this number g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to factor g.  This is the hardest step if g is large.  Basic approach: divide g by the primes up to some fixed limit, say 10^5, or until what remains is less than the square of the current prime.  At this point test what remains for primality (first with a Miller-Rabin test then, if desired, with the test of your choice, APR-CL, ECPP, etc.).  If the number is composite, use Pollard's rho and/or Shanks' SQUFOF.  (Whenever you find a factor, check it and its cofactor for primality and return to this step if composite.)  If those fail to find a factor, use ECM to search for a small factor.  At an appropriate crossover point you may wish to use MPQS/SIQS to factor the number instead.  If the number is larger, increase the bounds on ECM.  If no factor is found to ~40 digits, consider using GNFS to factor what remains.  This will take at least a day, depending on the size of the remaining number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the factorization it's easy to count the number of divisors: start an accumulator at 1, and for each prime multiply the accumulator by one more than the exponent of that prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for this problem you may be able to short-circuit the factorization for numbers of an appropriate size by trial-dividing up to the cube root of the number and showing that what remains is composite and not a square.  In this case the number of divisors is the number of divisors of the factored part times 4, since the unfactored part is a squarefree semiprime.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are some scripts to calculate the value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pari.math.u-bordeaux.fr/&quot; rel=&quot;nofollow&quot;&gt;Pari/GP&lt;/a&gt;: &lt;code&gt;common(m,n) = numdiv(gcd(m,n))&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica: &lt;code&gt;Common[m_,n_] := DivisorSigma[0,GCD[m,n]]&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-04T19:52:26.270" LastActivityDate="2010-11-04T19:52:26.270" CommentCount="1" />
  <row Id="8913" PostTypeId="2" ParentId="8902" CreationDate="2010-11-04T14:36:39.557" Score="6" ViewCount="0" Body="&lt;p&gt;It's true for finite groups, since $|MK| = |M| |K|/|M \cap K|$. It will not be true in general for infinite groups. The example that springs to mind is $G={\mathbb Z}^{\infty}$, where it is easy to find subgroups $M,K,L$ all isomorphic to $G$ with $M \cap K = M \cap L = 1$, $ML=G$ and $|G:MK|$ infinite.&lt;/p&gt;&#xA;" OwnerUserId="2604" LastActivityDate="2010-11-04T14:36:39.557" CommentCount="1" />
  <row Id="8914" PostTypeId="2" ParentId="8911" CreationDate="2010-11-04T14:51:24.957" Score="1" ViewCount="0" Body="&lt;p&gt;You've left out some important information.  It's not enough to guarantee T(v) is the zero operator just to know T(v) orthogonal to v for all v.  (Think about a right angle rotation in the plane.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if T(v) is orthogonal to u for all u, then (in particular) T(v) must be zero since &amp;lt; T(v),T(v)&amp;gt; = 0 implies that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So look for the missing extra condition(s)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited per Robin Chapman's comments, to incorporate the OP's earlier question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/8903/how-to-prove-polarization-identity&quot;&gt; How to prove polarization identity?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polarization identity expresses &amp;lt;T(v),u&amp;gt; for any v,u in a complex inner product space as a linear combination of terms of the form &amp;lt;T(w),w&amp;gt;.  Thus if all of the latter are zero, so must &amp;lt;T(v),u&amp;gt; be always zero, and the result follows.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-11-05T12:59:11.310" LastActivityDate="2010-11-05T12:59:11.310" CommentCount="1" />
  <row Id="8915" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T14:53:56.173" Score="6" ViewCount="0" Body="&lt;p&gt;It is not hard to prove that for &lt;em&gt;any&lt;/em&gt; vector space $\mathbf{V}$, if $\beta$ and $\beta'$ are two bases for $\mathbf{V}$, then the cardinality of $\beta$ equals the cardinality of $\beta'$. For finite dimensional spaces this is standard. For infinite dimensional, for each $x\in\beta$ let $\beta'_x$ be a finite subset of $\beta'$ such that $x\in\mathrm{span}(\beta'_x)$. Then $\cup\beta'_x = \beta'$, so $|\beta'|\leq \aleph_0|\beta|=|\beta|$. Now repeat the process for $y\in\beta'$ to get the other inequality. &lt;strong&gt;Edit:&lt;/strong&gt; Note that this assumes the Axiom of Choice, so that we can talk about the cardinals of $\beta$ and $\beta'$. In the absence of AC, it could be that $\beta$ and $\beta'$ are &quot;incomparable&quot; (no injections going either way), so we cannot conclude that any two bases have the same cardinality without AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a field $F$, and let $\mathbf{V}$ and $\mathbf{W}$ be vector spaces over $F$. Assuming the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_Choice&quot; rel=&quot;nofollow&quot;&gt;Axiom of Choice&lt;/a&gt; we can prove that every vector space has a basis (in fact, &quot;Every vector space over any field has a basis&quot; is equivalent to the Axiom of Choice). So let $\beta=\{v_i\}_{i\in I}$ be a basis for $\mathbf{V}$, and let $\gamma=\{w_j\}_{j\in J}$ be a basis for $\mathbf{W}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{V}$ is isomorphic to $\mathbf{W}$, then let $\varphi\colon\mathbf{V}\to\mathbf{W}$ be an isomorphism. Then $\varphi(\beta)$ is a basis for $\mathbf{W}$, and since $\varphi(\beta)$ and $\gamma$ are both bases of $\mathbf{W}$, and $\varphi$ is a bijection, you have $|\beta|=|\varphi(\beta)|=|\gamma|$. So if $\mathbf{V}$ and $\mathbf{W}$ are isomorphic, then they have bases of the same cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $|\beta|=|\gamma|$. Let $f\colon\beta\to\gamma$ be a bijection, and let $\varphi\colon \mathbf{V}\to\mathbf{W}$ be the unique linear transformation that extends $f$ (that is, extend $f$ linearly to all of $\mathbf{V}$). Since $\varphi(\mathbf{V}) = \varphi(\mathrm{span}(\beta)) = \mathrm{span}(\varphi(\beta)) = \mathrm{span}(f(\beta)) = \mathrm{span}(\gamma)=\mathbf{W}$, then $\varphi$ is onto. It is straightforward to verify that $\varphi$ is one-to-one (it takes a basis to a basis). So $\varphi$ is an isomorphism. Thus, if $\mathbf{V}$ and $\mathbf{W}$ have bases of the same cardinality, then $\mathbf{V}$ is isomorphic to $\mathbf{W}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, two vector spaces over the same field are isomorphic if and only if they have bases of the same cardinality, if and only if they have the same dimension (assuming the Axiom of Choice). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as vector spaces over $\mathbb{Q}$, since they both have dimension $2^{\aleph_0}$. But you have to specify over what field you are working: $\mathbb{R}$ and $\mathbb{C}$ are &lt;strong&gt;not&lt;/strong&gt; isomorphic as vector spaces over $\mathbb{R}$ (dimensions 1 and 2, respectively)!. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T03:20:18.040" LastActivityDate="2010-11-05T03:20:18.040" CommentCount="7" />
  <row Id="8916" PostTypeId="2" ParentId="8909" CreationDate="2010-11-04T14:57:49.293" Score="12" ViewCount="0" Body="&lt;p&gt;I actually find it easiest to take a cube and divide it into 6 pyramids: just connect all of the corners of a cube to the midpoint, and for each of the 6 faces of the cube, you get a pyramid having that face as its base. If the cube has a side length s, then each pyramid has an s by s base and height s/2, and its volume is 1/6 the volume of the cube, which is 1/3 base times height.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-04T14:57:49.293" />
  <row Id="8917" PostTypeId="1" CreationDate="2010-11-04T15:43:05.433" Score="2" ViewCount="82" Body="&lt;p&gt;How can we show that every automorphism of S4 is an inner automorphism ?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-11-04T15:50:19.247" Title="Automorphism of S4" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8918" PostTypeId="2" ParentId="8917" CreationDate="2010-11-04T15:50:19.247" Score="1" ViewCount="0" Body="&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Any automorphism preserves the conjugacy class of transpositions.( By counting size of conjugacy classes)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any automorphism preserves whether two transpositions share an element (by looking at the order of their product)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any automorphism permutes the four classes of transpositions {those that move 1}, {those that move 2}, {those that move 3}, {those that move 4}&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This gives you a permutation on {1,2,3,4} and it is not hard to show that your automorphism is conjugation by this permutation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-04T15:50:19.247" />
  <row Id="8919" PostTypeId="2" ParentId="8752" CreationDate="2010-11-04T16:06:58.457" Score="1" ViewCount="0" Body="&lt;p&gt;Any proof of the 4CT is human-checkable so long as that human has enough time on their hands.&lt;/p&gt;&#xA;" OwnerUserId="2994" LastActivityDate="2010-11-04T16:06:58.457" CommentCount="2" />
  <row Id="8920" PostTypeId="1" CreationDate="2010-11-04T16:37:00.167" Score="1" ViewCount="105" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://math.stackexchange.com/questions/8623/distinct-permutations-of-x-items-from-n-items&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the 11-letter word &quot;MATHEMATICS&quot;. The general question is to find how many words it is possible to have if we can make 4 permutations of letters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-question a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each letter has a position in the word, thus when I say 1 permutation, we exchange the position of 2 letters. For example, with 1 permutation it is possible to have &quot;AMTHEMATICS&quot;, but not &quot;SMATHEMATIC&quot;.&#xA;How much word is it possible to make with 4 permutations, if ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a.1) we want exactly 4 permutations and it is forbidden to have permutation $p$ and its inverse permutation. But it is possible to permute an &quot;M&quot; with another &quot;M&quot;.&lt;/li&gt;&#xA;&lt;li&gt;a.2) same as a.1) but it is forbidden to permute &quot;M&quot; with the other &quot;M&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;sub-question b) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that by permutations, we mean changing the letters in the word in such a way that 4 permutations could lead to &quot;TICSMATHEMA&quot; (when a letter enter a new position, the other letters are pushed).&#xA;How much word is it possible to make with 4 permutations, if we want exactly 4 permutations, that means 4 letters changing positions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that there is no trivial solution. At least I don't see it. Of course, it would be interesting to have a general solution for the case of a word with $n$ letters (possibly not all different) with $k$ permutations.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-04T16:53:52.650" Title="Permutation of 4 letters in a 11-letter word : how many different words can we make?" Tags="&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8921" PostTypeId="1" CreationDate="2010-11-04T16:43:33.180" Score="5" ViewCount="66" Body="&lt;p&gt;Assume Q is a convex central symmetric curve, whose area is $\displaystyle S$.&#xA;The area of the maximum parallelogram inside Q is $\displaystyle S'$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove the conjecture that $\displaystyle \frac{S'}{S} \ge \frac{2}{\pi}=0.6366\dots$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, If Q is an ellipse, $\displaystyle S'=2ab$, $\displaystyle S=\pi ab$. &#xA;If Q is a regular hexagon, $\displaystyle \frac{S'}{S}= \frac{2}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's trivial that $\displaystyle \frac{S'}{S} \ge \frac{1}{2}$, and I know how to prove $\displaystyle \frac{S'}{S} \ge \frac{4}{4+\pi}=0.56\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From many reason, I believe this conjecture is true.&#xA;  Denote MAP=&quot;Maximum Area Parallelogram&quot;:&#xA;  For any $Q$ and any direction $\theta$, let $P(Q,\theta)$ be the area of MAP which have a corner in this direction.     $S'=\max\{P(Q,\theta)\}$.      In order to make $\frac{S'}{S}$ smallest, We need keep the largest one of $\{P(Q,\theta)\}$ small while S is a constant. Ellipse just keeps everyone in  $\{P(Q,\theta)\}$ average. This is very special, I don't think there will be other curve having this property. On the other hand, distribute equally always lead to the min-max in our knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the $\frac{4}{4+\pi}$ lowerbound, the idea is as follows:&#xA;First, use polar function $r(\theta)$ to describe the curve. The condition is that $r(a)r(b)\sin|a-b|\leq C$, and we want to bound is $S=\int_{\theta}{r(\theta)}^2$. &#xA;Second, Without lose of generality, We assume $r(0)=r(90)=1,C=1$, and assume $Q$ is in the boundary of $Z=\{(x,y)|-1\leq x,y\leq 1\}$.&lt;br&gt;&#xA;Third, let $a=r(\theta)$ and $b=r(\theta+90)$ and find a bound for $(a^2+b^2)$ by Cauchy-Inequality. and it will give a bound for the area $S$.&lt;/p&gt;&#xA;" OwnerUserId="3109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T19:19:18.823" LastActivityDate="2010-11-04T22:05:57.670" Title="A conjecture of parallelogram inside convex and central symmetric curve" Tags="&lt;geometry&gt;&lt;convex-optimization&gt;&lt;conjectures&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8922" PostTypeId="2" ParentId="8920" CreationDate="2010-11-04T16:53:52.650" Score="1" ViewCount="0" Body="&lt;p&gt;For a.1, there are 11*10/2=55 ways to pick the first permutation.  If you just prohibit inverses(the inverse is the same as the swap), there are then 55*54*53*52/24 ways to pick four permutations.  But this ignores the fact that you may move the same letter twice.  If you want eight letters to move in four swaps, you have 55*36*21*10/24&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T16:53:52.650" />
  <row Id="8923" PostTypeId="1" AcceptedAnswerId="8997" CreationDate="2010-11-04T17:02:24.447" Score="3" ViewCount="69" Body="&lt;p&gt;Let $X$ denote a metric space. One can assume that $X$ is Polish if that helps, but I was trying to avoid to assume that $X$ is compact. Let $P(X)$ denote the set of Borel probability measures on $X$. The weak*-topology on $P(X)$ is defined as usual: a net $(p_{\alpha})$ in $P(X)$ converges to $p \in P(X)$ iff $|\int fdp_{\alpha}-\int fdp|$ converges to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Is there any metric that metrizes the weak*-topology on $P(X)$ &lt;strong&gt;and&lt;/strong&gt; has convex open balls?  &lt;/p&gt;&#xA;" OwnerUserId="3117" LastActivityDate="2010-11-05T00:16:07.353" Title="Metrization of the weak*-topology on a set of probability measures" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8925" PostTypeId="1" CreationDate="2010-11-04T17:08:32.910" Score="2" ViewCount="110" Body="&lt;p&gt;How to show that $e^{x} \geq \left (1 + \frac{x}{n} \right) ^{n}$ holds for each non-negative real $x$ and each integer $n \geq 1$ ? I tried series and induction but got stuck. Can you please help?&lt;/p&gt;&#xA;" OwnerUserId="3118" LastActivityDate="2010-11-05T06:37:15.240" Title="Help inequality involving exponential function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8927" PostTypeId="1" AcceptedAnswerId="8956" CreationDate="2010-11-04T17:18:57.030" Score="5" ViewCount="147" Body="&lt;p&gt;Wikipedia articles on &quot;unit sphere&quot; and &quot;unit circle&quot; say the radius is 1. Articles on the &quot;unit square&quot; and &quot;unit cube&quot; say the length of the side is 1. Would you expect a unit torus to have major &lt;em&gt;radius&lt;/em&gt; 1 or major &lt;em&gt;diameter&lt;/em&gt; 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, a torus is a different animal than a sphere, but... It feels most natural to me that the &quot;unit&quot; length should apply to the (major) radius, not the major diameter. Yet I recently came across open source code where someone generated a &quot;unit torus&quot; of major diameter 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that &quot;wrong enough&quot; that I should change it (in a package of related changes that I'm already preparing to submit)? Can you give me a more solid mathematical basis for advocating that change? Or should I accept the status quo as just a different but legitimate interpretation of &quot;unit torus&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see from search hits like the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cis.upenn.edu/~ngns/docs/References/CDC09%20RGG%20March06.pdf&quot; rel=&quot;nofollow&quot;&gt;Spectral Analysis of Virus Spreading in Random Geometric Graphs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.scientificcommons.org/29544141&quot; rel=&quot;nofollow&quot;&gt;Unconditional Proof of the Boltzmann-Sinai Ergodic Hypothesis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1496770.1496776&quot; rel=&quot;nofollow&quot;&gt;The cover time of random geometric graphs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache%3acuTivzCr9Y0J%3aannals.princeton.edu/annals/2006/163-3/annals-v163-n3-p06-p.pdf&quot; rel=&quot;nofollow&quot;&gt;Dimers and amoebae&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;that the term &quot;unit torus&quot; is used in some fields, like dynamical systems and discrete algorithms. But I'm unable to tell from these papers or abstracts what the authors mean exactly by &quot;unit torus&quot;. &lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache%3acuTivzCr9Y0J%3aannals.princeton.edu/annals/2006/163-3/annals-v163-n3-p06-p.pdf&quot; rel=&quot;nofollow&quot;&gt;Dimers and amoebae&lt;/a&gt; actually gives this definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the unit torus T&lt;sub&gt;2&lt;/sub&gt; = {(z,w) ∈ C&lt;sub&gt;2&lt;/sub&gt; : |z|&#xA;  = |w| = 1}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This definition appears to give a definite size. But if it's in the two-dimensional vector space over the complex numbers, I don't know how to apply it to $\mathbb{R}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;unit torus&quot; (in $\mathbb{R}^3$) actually means something that &lt;em&gt;does not have any particular size&lt;/em&gt;, then that would be important to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is really not one of programming, but of what this term means in mathematics... including, to what degree is it actually defined (or not) in math?&#xA;I will base my software decisions on that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Would tag this &quot;torus&quot; if I could create the tag.)&lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="2656" LastEditorDisplayName="" LastEditDate="2010-11-04T20:15:05.630" LastActivityDate="2010-11-04T20:32:52.213" Title="what size is a &quot;unit torus&quot;?" Tags="&lt;geometry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="19" FavoriteCount="1" />
  <row Id="8928" PostTypeId="2" ParentId="5052" CreationDate="2010-11-04T17:25:25.570" Score="1" ViewCount="0" Body="&lt;p&gt;If you accept enumerative combinatorics as part of number theory, here is a nice application of the Legendre polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Count the number of ways to insert paired, grammatical and non-repeating parentheses enclosing at least two letters in a word of length $n &amp;gt; 1$. Define $a_1 = 1$ and ignore outmost paratheses. For $n = 2$, there is only one way $(ab) = ab$. For $n = 3$, we have $abc, (ab)c, a(bc)$. For $n = 4$, we have $abcd$, $(ab)cd$, $a(bc)d$, $ab(cd)$, $(ab)(cd)$, $a(bcd)$, $a(b(cd))$, $a((bc)d)$, $(abc)d$, $(a(bc))d$, and $((ab)c)d$. And, so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence continues {$1$, $1$, $3$, $11$, $45$, $197$, $903$, $4279 \dots$ } (A001003) and the terms are called &lt;em&gt;Super Catalan&lt;/em&gt; numbers or &lt;em&gt;Little Schroeder&lt;/em&gt; numbers. &#xA;In general,&#xA;\begin{eqnarray}&#xA;a_n = \tfrac{1}{n} \sum_{k =0}^{n-2} \binom{2n-k-2}{n-1} \binom{n-2}{k} = \tfrac{1}{4n}(3P_{n-1}(3) - P_{n-2}(3)) + \tfrac{1}{2} \epsilon(n),&#xA;\end{eqnarray}&#xA;where $\epsilon$ is the unit arithmetic function, which is $1$ when the argument is $1$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there are at least $10$ different combinatorial problems enumerated by this sequence, involving restricted lattice paths, integer compositions and even matrix row-sums. The Online Encyclopedia of Integer Sequences has the present state of knowledge about these fascinating numbers; see there reference section of &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A001003&quot; rel=&quot;nofollow&quot;&gt;http://www.research.att.com/~njas/sequences/A001003&lt;/a&gt; for relevant papers and books.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T21:16:22.587" LastActivityDate="2010-11-15T21:16:22.587" CommentCount="1" />
  <row Id="8929" PostTypeId="2" ParentId="8925" CreationDate="2010-11-04T17:26:16.123" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Consider $\rm\ e^z\ \ge\ 1 + z,\ \ z\ =\ x/n $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T17:26:16.123" CommentCount="1" />
  <row Id="8931" PostTypeId="2" ParentId="8925" CreationDate="2010-11-04T18:06:30.607" Score="0" ViewCount="0" Body="&lt;p&gt;For each fixed nonnegative $x$, $(1+\frac{x}{n})^n$ increases with $n$.  To see why this is true, note that $(1+\frac{x}{n})^n$ is the amount of money in an account one year after $1$ unit of money is invested at a yearly interest rate of $100x\%$, compounded $n$ times per year.  As $n$ goes to $\infty$, this increasing sequence converges to $e^x$, corresponding to continuously compounded interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I do not claim that this is a rigorous proof.  It is merely intended as an explanation.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-04T18:06:30.607" />
  <row Id="8933" PostTypeId="1" AcceptedAnswerId="8941" CreationDate="2010-11-04T18:25:12.023" Score="4" ViewCount="243" Body="&lt;p&gt;How can I prove that there is no closed form solution to the equation $2^x + 3^x = 10$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-12-15T03:17:33.547" Title="Proof that there is no closed form solution" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" />
  <row Id="8934" PostTypeId="2" ParentId="8904" CreationDate="2010-11-04T18:25:17.700" Score="2" ViewCount="0" Body="&lt;p&gt;This answer was motivated by some comments of Djaian. As Arturo said, two vector spaces over the same field are isomorphic if and only if they have bases of the same cardinality. The infinite dimensional case is valid in much general context. So the important property of fields is that it is true for finite-dimensional spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To justify this we'll talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_module&quot; rel=&quot;nofollow&quot;&gt;free modules over a ring with identity&lt;/a&gt;. Roughly speaking a module is  a generalization of vector space to arbitrary rings and the word free means that the module has a basis. We have the following result:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we consider a ring $R$ with identity and a free $R$-module $F$ with an infinite basis $X$, then every basis of $F$ has the same cardinality as $X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof is based on the Cantor–Schröder-Bernstein theorem and is similar to the one Arturo gave for vector spaces. It needs the Axiom of Choice too. Thus if $F_1$ and $F_2$ are two free modules with infinite bases $X_1$ and $X_2$, respectively, then $F_1 \cong F_2$ if and only if $|X_1| = |X_2|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how badly it can be, there is a ring $R$ such that, as a free left $R$-module, $R$ has a basis of $n$ elements, for each positive integer $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that a ring with identity has &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariant_basis_number&quot; rel=&quot;nofollow&quot;&gt;invariant basis number&lt;/a&gt;  (IBN) if for every free $R$-module $F$, any two bases of $F$ have the same cardinality. The cardinal number of any basis of $F$ is called the rank (or dimension) of $F$ over $R$. It's easy to see that if $R$ is a ring with IBN and $F_1$ and $F_2$ are two free modules over $R$, then $F_1 \cong F_2$ if and only if they have the same rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any nonzero commutative ring has IBN. Also any division ring has IBN. Other examples include finite rings, left-Noetherian rings, local rings and group rings.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-13T13:21:53.110" LastActivityDate="2010-11-13T13:21:53.110" CommentCount="6" />
  <row Id="8935" PostTypeId="1" CreationDate="2010-11-04T18:26:52.007" Score="2" ViewCount="69" Body="&lt;p&gt;I am trying to understand the interpretation of &lt;strong&gt;$\sin \frac{A}{2}$&lt;/strong&gt; and &lt;strong&gt;$\cos \frac{A}{2}$&lt;/strong&gt; in terms of  &lt;strong&gt;$\sin A$&lt;/strong&gt; from my book, here is how it is given :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $ \bigl( \sin \frac{A}{2} + \cos \frac{A}{2} \bigr)^{2} = 1 + \sin A $ and $ \bigl( \sin \frac{A}{2} - \cos \frac{A}{2} \bigr)^{2} = 1 - \sin A $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding and subtracting we have $ 2 \cdot \sin \frac{A}{2} = \pm \sqrt{  1 + \sin A } \pm \sqrt{  1 - \sin A } $ ---- (1) and $ 2 \cdot \cos \frac{A}{2} = \mp \sqrt{  1 + \sin A } \mp \sqrt{  1 - \sin A } $ ---(2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have understood upto this far well, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now they have broke the them into quadrants :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 1st quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&#xA; $$ 2 \cdot \cos \frac{A}{2} =  \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 2nd quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&#xA; $$ 2 \cdot \cos \frac{A}{2} =  \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 3rd quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&#xA;  $$ 2 \cdot \cos \frac{A}{2} =  \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 4th quadrant :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cdot \sin \frac{A}{2} =  - \sqrt{  1 + \sin A } - \sqrt{  1 - \sin A } $$&#xA; $$ 2 \cdot \cos \frac{A}{2} =   - \sqrt{  1 + \sin A } + \sqrt{  1 - \sin A } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, In knew the ALL-SINE-TAN-COSINE rule but still I am not able to figure out how the respective signs are computed in these (above) cases.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-04T18:51:53.047" LastActivityDate="2010-11-04T18:56:49.053" Title="Expressions of    $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of  $\sin A$" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8936" PostTypeId="1" CreationDate="2010-11-04T18:37:01.020" Score="2" ViewCount="51" Body="&lt;p&gt;Let X be a measure space and let $f_{n}$ be a sequence of functions which converge pointwise to a function f in $L^{p}(X)$ where $p&gt;1$ and suppose $g_{n}$ is a sequence of functions which converge pointwise to a function $g$ in $L^{\frac{p}{p-1}}(X)$. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{X} f_{n}(x) g_{n}(x) dx = \int_{X} f(x)g(x) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea how to proceed. Any help? I guess somewhere we need Holder and DCT.&lt;/p&gt;&#xA;" OwnerUserId="3118" LastActivityDate="2010-11-04T19:00:37.907" Title="Convergence of integrals in L^p and L^(p/(p-1))." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8937" PostTypeId="1" CreationDate="2010-11-04T18:44:36.380" Score="5" ViewCount="109" Body="&lt;p&gt;I am currently engaged in independent study of algebraic geometry, using Dan Bump's book. One of the exercises in it outlines a proof of the Krull Intersection Theorem, which [here] is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$, and let $M$ be the intersection of all of the $\mathfrak{m}^n$. Then $M = 0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hints direct me to use the Artin-Rees lemma to show that $\mathfrak{m} M = M$, then use Nakayama's lemma to show that $M = 0$ (this second step is easy). I showed this to a professor and he accused the book of using big machinery for no reason, arguing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathfrak{m} M = \mathfrak{m} \bigcap_{n \ge 0} \mathfrak{m}^n = \bigcap_{n \ge 1} \mathfrak{m}^n = M.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this argument work? Does Bump apply Artin-Rees because that argument works in some broader context where the above argument fails? &lt;/p&gt;&#xA;" OwnerUserId="3120" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-04T18:56:05.200" LastActivityDate="2010-12-04T19:44:24.970" Title="Why is the Artin-Rees lemma used here?" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8938" PostTypeId="2" ParentId="8936" CreationDate="2010-11-04T18:54:22.233" Score="3" ViewCount="0" Body="&lt;p&gt;Not true: Let $f_n(x)$ be the a constant times the characteristic function of $[1/n,2/n]$ and $g_n(x)$ be another constant times the characteristic function of $[1/n,2/n]$, the constants defined that the $L^p$ and $L^{p \over p-1}$ norms of&#xA;$f_n$ and $g_n$ respectively are both 1. Then they both converge pointwise to zero, so that the right hand side is zero. But the left integral is always 1. Thus what you're saying isn't even true if $f_n$ and $g_n$ are in $L^p$ and $L^{p \over p-1}$ respectively.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-04T19:00:37.907" LastActivityDate="2010-11-04T19:00:37.907" CommentCount="1" />
  <row Id="8939" PostTypeId="2" ParentId="8935" CreationDate="2010-11-04T18:56:49.053" Score="4" ViewCount="0" Body="&lt;p&gt;The easiest way of computing the signs is to make them match; we know that sin x &amp;gt; 0 if 0 &amp;lt; x &amp;lt; &amp;pi; and that cos x &amp;gt; 0 if -&amp;pi;/2 &amp;lt; x &amp;lt; &amp;pi;/2.  Knowing whether sin A is greater than 0 or less than zero tells you whether $\sqrt{1-\mathrm{sin} A}$ is greater or less than $\sqrt{1+\mathrm{sin} A}$; that in turn lets you figure out what the overall sign on all of the right-hand terms is, and each quadrant corresponds to one of the four positive/negative pairs on the right-hand terms.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-04T18:56:49.053" />
  <row Id="8940" PostTypeId="1" AcceptedAnswerId="8942" CreationDate="2010-11-04T19:06:20.780" Score="1" ViewCount="180" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like a demonstration of the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original question:&lt;/strong&gt;  If a square matrix of order $n\gt 1$ has a row (column) that is linear combination of rows (columns) parallel to it, then the determinant of this matrix is null.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Actual intended question:&lt;/strong&gt; If the determinant of a square matrix of order $n\gt 1$ is zero, then this matrix has some row (column) that is linear combination of rows (columns) parallel to it.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T21:13:03.783" LastActivityDate="2010-11-05T03:15:39.630" Title="Determinant equal to zero" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="8941" PostTypeId="2" ParentId="8933" CreationDate="2010-11-04T19:07:38.330" Score="2" ViewCount="0" Body="&lt;p&gt;There is a solution to this equation, but I don't know that you can make a closed-form solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\lim_{x\rightarrow -\infty} 2^x + 3^x = 0$ and $\lim_{x \rightarrow \infty} 2^x + 3^x = \infty$ and because $2^x + 3^x$ is continuous there must be a point where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^x + 3^x = 10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the intermediate value theorem.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-04T19:07:38.330" CommentCount="1" />
  <row Id="8942" PostTypeId="2" ParentId="8940" CreationDate="2010-11-04T19:13:08.450" Score="2" ViewCount="0" Body="&lt;p&gt;If you know the determinant is multilinear (linear in each row/linear in each column), then express the the determinant as a sum of determinants of matrices that have repeated rows/columns. Then show that the determinant of a matrix with two identical rows/columns is always zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do &lt;em&gt;not&lt;/em&gt; know the determinant is multilinear in the rows/columns, and you know the definition in terms of sums of products over all permutations, then the same argument works to show the determinant is a sum of determinants of matrices with repeated rows/columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your only definition of determinant is inductively (by minors), then prove it by induction, expanding along a row or column that is &lt;em&gt;not&lt;/em&gt; a linear combination of other rows/columns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; You've edited your question, and now the question is the &lt;em&gt;precise converse&lt;/em&gt; of what you originally posted. This is a bit of bad form, since any answers already posted would automatically look completely wrong-headed. In the future, consider editing to &lt;em&gt;add&lt;/em&gt; the new question to the old, indicating this is an edit or addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now you want to show that &lt;em&gt;if&lt;/em&gt; the determinant is zero, then a row/column is a linear combination of other rows/columns. Looking at the tranpose if necessary it suffices to consider the case of rows. Note that applying elementary row operations does not change whether the determinant is zero or nonzero (an elementary row operation will either multiply the determinant by a nonzero constant, or will not change the determinant). So the determinant of the matrix is zero if and only if the determinant of its row-echelon form is zero; the determinant of the row-echelon form is zero if and only if there is a row of zeros. Tracing back the elementary row operations will express the original row as a linar combination of the other rows. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T21:33:03.290" LastActivityDate="2010-11-04T21:33:03.290" CommentCount="1" />
  <row Id="8943" PostTypeId="1" CreationDate="2010-11-04T19:23:44.610" Score="1" ViewCount="93" Body="&lt;p&gt;Consider a number say N = 345&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Permutations are :{{3, 4, 5}, {3, 5, 4}, {4, 3, 5}, {4, 5, 3}, {5, 3, 4}, {5, 4, 3}}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GCD of all of them is &lt;strong&gt;3&lt;/strong&gt; which is the answer.Please suggest a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-04T19:36:51.683" LastActivityDate="2010-11-05T13:10:07.867" Title="Computing GCD of all permutations (of the digits) of a given number." Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8944" PostTypeId="1" AcceptedAnswerId="8948" CreationDate="2010-11-04T19:25:58.087" Score="8" ViewCount="256" Body="&lt;p&gt;I want to prove that the series&#xA;$$ \sum_{n=1}^{\infty}(-1)^n\left(2^{1/n}-1\right)$$&#xA;converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fairly certain that this converges.  Using the ratio or root test does not seem to work.  Therefore the other tests I am left with are &lt;strong&gt;alternating series test&lt;/strong&gt;, &lt;strong&gt;intergral test&lt;/strong&gt;, &lt;strong&gt;comparison test&lt;/strong&gt; and &lt;strong&gt;limit comparison test&lt;/strong&gt; (that we know about).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like it would be a messy integral and that would not work or I would not be able to solve it.&lt;br&gt;&#xA;I think I could show this using the alternating series test IF the limit of the series is 0.  My problem is formally showing that the limit of this series is 0.  I can't quite get past that.&lt;br&gt;&#xA;My only option left if the alternating series test does not work is the limit comparison test, since the series is not &gt; 0 for large n, and therefore we cannot use the comparison test.  If this is the case I honestly don't have a clue what other series I should compare this with.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for reading!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2010-11-04T22:41:17.023" LastActivityDate="2010-11-13T21:21:56.407" Title="Prove that $\sum\limits_{n=1}^{\infty}(-1)^n(2^{1/n} - 1)$ is convergent." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="8945" PostTypeId="1" AcceptedAnswerId="8955" CreationDate="2010-11-04T19:30:05.963" Score="2" ViewCount="119" Body="&lt;p&gt;How to evaluate this integral: $$\int_{0}^{\infty} \biggl\lfloor{\frac{n}{e^{x}}\biggr\rfloor} \ dx, $$where $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same integral when asked to evaluate for $n=2$ (say) i can do it by splitting the limits from $x = 0$ to $x = \log{2}$ where $e^{x}$ takes the value 1, and then from $x= \log{2}$ to $x = \infty$. But how to do this for the general case $n$. I thought of two ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Using induction on $n$. This will not work!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Splitting the limits from $x =0$, to $x = \log{n}$ also seems to cause some problems for me. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T20:04:53.450" LastActivityDate="2010-11-04T20:04:53.450" Title="Evaluating the integral $\int_{0}^{\infty} \Bigl\lfloor{\frac{n}{e^{x}}\Bigr\rfloor} \ dx $" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="8946" PostTypeId="1" CreationDate="2010-11-04T19:31:46.867" Score="2" ViewCount="110" Body="&lt;p&gt;I am confused on how to apply Ito's formula on certain problems, especially where expectations are involved. For example, if $W_t$ is a Weiner process and $X_t$ satisfies a below SDE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$    dX_t = (X_t-\mu)dt + \sigma\sqrt{X_t}dW_t,~~~~~~ X_0 = x_o$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find $\partial_t \phi$ or $\partial_\xi \phi$ where $\phi(t,\xi)=E[e^{i\xi X_t}]$ is characteristic function of $X_t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't quite understand how to approach this problem. Should I first solve the SDE for $X_t$, then compute the expectation $E[e^{i\xi X_t}]$, and then apply Ito's Lemma to find $\partial_t\phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking it a step further, how would I compute $\partial_t\psi$ where $\psi(t,\xi)=\ln\phi(t,\xi)$ and solve resulting SDE for $\psi(t,\xi)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/It%C5%8D&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/It%C5%8D&lt;/a&gt;'s_lemma&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T19:42:13.760" LastActivityDate="2010-12-07T21:10:49.700" Title="application of Ito's lemma" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8947" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T19:33:16.537" Score="0" ViewCount="0" Body="&lt;p&gt;I believe that the alternating series test is the right way to go.  All you have to do is prove that $\lim (2^{1 \over n} - 1)$ is zero.  That is, to employ the test, you &lt;em&gt;don't&lt;/em&gt; need to show that the limit of the &lt;em&gt;series&lt;/em&gt; is zero, but merely that the limit of the &lt;em&gt;summands&lt;/em&gt; is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ETA: Arturo reminds us that one must also show that the summands strictly decrease.  But the AST is still the correct approach.]&lt;/p&gt;&#xA;" OwnerUserId="1801" LastActivityDate="2010-11-04T19:33:16.537" CommentCount="3" />
  <row Id="8948" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T19:33:42.573" Score="12" ViewCount="0" Body="&lt;p&gt;This is an alternating series, so it suggests (nay, &lt;em&gt;screams&lt;/em&gt;) for the use of the Alternating Series Test. You need to show that $\lim\limits_{n\to\infty}(2^{1/n}-1) = 0$, and that for all sufficiently large $n$, $2^{\frac{1}{n+1}}-1 \lt 2^{1/n} - 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is equivalent to showing that $2^{\frac{1}{n+1}}\leq 2^{\frac{1}{n}}$; you might want to consider taking logarithms, say base $2$, which respect the order (that is, if $0\lt a\lt b$, then $\log_2(a)\lt\log_2(b)$). As to the former, it amounts to showing that $\lim\limits_{n\to\infty}2^{1/n} = 1$. Since $a^b = e^{b\ln(a)}$, this is equivalent to showing that $\lim\limits_{n\to\infty}e^{\frac{\ln(2)}{n}} = 1$; and since the exponential is continuous, this amounts to checking what happens to the exponent as $n\to\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that none of the other tests that you mention are even applicable! The integral test can only be applied to a series $\sum a_n$ in which all terms $a_n$ are nonnegative (which is not the case here), and where you can find a positive, decreasing, and continuous function $f(x)$ such that $f(n)=a_n$ for all $n$ (or at least all sufficiently large $n$). So it is not applicable here. The Comparison Test requires the series to have positive terms, as does the Limit Comparison Test. You &lt;em&gt;could&lt;/em&gt; combine them with the Absolute Convergence Test by checking the convergence of $\sum|a_n|$, and then deducing the convergence of $\sum a_n$ if the series of absolute values converges, and apply one of these tests to the series $\sum|a_n|$. But you cannot apply any of the Integral, Comparison, or Limit Comparison Tests to this series directly.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T20:06:52.183" LastActivityDate="2010-11-13T20:06:52.183" CommentCount="6" />
  <row Id="8949" PostTypeId="2" ParentId="8937" CreationDate="2010-11-04T19:34:59.593" Score="4" ViewCount="0" Body="&lt;p&gt;There is a proof due to Herstein that is &quot;elementary&quot; in the sense of avoiding the Artin-Rees lemma. Below is Kaplansky's presentation of this proof, from his &quot;Commutative Rings&quot;. A proof using primary decomposition (as in Krull's original proof) can be found in Zariski and Samuel.&#xA;&lt;img src=&quot;http://i.imgur.com/IIOGX.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/sDQoD.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-04T19:34:59.593" />
  <row Id="8950" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T19:36:26.733" Score="1" ViewCount="0" Body="&lt;p&gt;Yes the &lt;strong&gt;Alternating Test&lt;/strong&gt; will work. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-04T19:36:26.733" />
  <row Id="8951" PostTypeId="1" CreationDate="2010-11-04T19:36:49.760" Score="1" ViewCount="53" Body="&lt;p&gt;Suppose $X$ and $Y$ are independent random variables. The mgf for $X$ is $M_{X}(t) = (1-t)^{-1}, \ t&amp;lt;1$. The mgf for $Y$ is $M_{Y}(t) = (1-2t)^{-1}, \ t&amp;lt; 0.5$. Two other random variables $U$ and $V$ are defined by $U = \frac{1}{2}(X+Y)$ and $V = \frac{1}{2}(X-Y)$. Calculate $\text{Cov}(U,V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\text{Cov}(U,V) = E(UV)-E(U)E(V)$. This is the same thing as $\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y))$. Expanding out, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y)) = \frac{1}{4} \text{Var}(X) -\frac{1}{4} \text{Cov}(X,Y)+\frac{1}{4} \text{Cov}(Y,X)-\frac{1}{4} \text{Var}(Y)$ which equals $\frac{1}{4}(\text{Var}(X)-\text{Var}(Y))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easier way to obtain this without going through all this work? Could one use the fact $U+V = X$ and $U-V = Y$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-04T19:36:49.760" Title="Covariance and Random Variables" Tags="&lt;probability&gt;" CommentCount="1" />
  <row Id="8952" PostTypeId="2" ParentId="8943" CreationDate="2010-11-04T19:37:34.960" Score="1" ViewCount="0" Body="&lt;p&gt;In the case you have given you just need to note that the sum of the digits of 345 is divisible by 3. More generally, my first thoughts are that there is no straightforward answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course lots of cases such as numbers like, 198, 5505005, 777777777, 4848448, etc where we can say something more easily.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-04T19:37:34.960" />
  <row Id="8953" PostTypeId="2" ParentId="8943" CreationDate="2010-11-04T19:43:52.263" Score="3" ViewCount="0" Body="&lt;p&gt;This is Sloane's &lt;a href=&quot;http://oeis.org/classic/A069652&quot; rel=&quot;nofollow&quot;&gt;A069652&lt;/a&gt;. The best way I can think of is to actually permute the digits and calculate -- though checking for 2, 5, and 3/9 first in the obvious way (all digits even, all digits 0 or 5, all digits add to a multiple of 3/9) in hopes that the rest of the number will have gcd 1 and terminate early that way.  Storing numbers in &lt;A href=&quot;http://en.wikipedia.org/wiki/Binary-coded_decimal&quot; rel=&quot;nofollow&quot;&gt;BCD&lt;/a&gt; (or as a vector of digits, or as a byte array) would make this fastest, I imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Building on Gabe's answer, pick two distinct digits &lt;em&gt;m&lt;/em&gt; and &lt;em&gt;n&lt;/em&gt; in the number, if they exist. (If not, return the number.)  The gcd divides 9(m-n). Together with the 2/3/5/9 trick, this should determine most numbers quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, continue choosing digits until you're gone through all combinations.  The hardest case is numbers where all digits are divisible by 3 and the sum of digits is divisible by 9; you must actually work with the numbers in this case to determine what power of 3 divides the number (at least 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case here's some naive Math'ca code in case someone wants to check their implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;f[n_] := GCD @@ FromDigits /@ Permutations[IntegerDigits[n]]&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-05T13:10:07.867" LastActivityDate="2010-11-05T13:10:07.867" CommentCount="2" />
  <row Id="8955" PostTypeId="2" ParentId="8945" CreationDate="2010-11-04T19:55:09.173" Score="7" ViewCount="0" Body="&lt;p&gt;The integrand will only take on integral values m, and only for a finite measure of x values. The interval of x values in which the integrand equals exactly m is $ (- \ln \frac{m+1}{n} , - \ln \frac{m}{n} ] $. Split the integral over these intervals, evaluate (upper bound of interval minus lower bound times m) and turn them into a summation as $$ \sum_{m=1}^n m \left( \ln \frac{m+1}{m} \right) = \ln \prod_{m=1}^n \left( 1 + \frac{1}{m} \right)^m $$&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-11-04T19:55:09.173" CommentCount="3" />
  <row Id="8956" PostTypeId="2" ParentId="8927" CreationDate="2010-11-04T19:55:53.780" Score="2" ViewCount="0" Body="&lt;p&gt;My best guess is that the unit $d$-torus is the quotient $\mathbb{R}^d/\mathbb{Z}^d$, endowed with some combination of the following additional structures depending on the field of mathematics in consideration:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The structure of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_group&quot; rel=&quot;nofollow&quot;&gt;topological group&lt;/a&gt; (with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quotient_space&quot; rel=&quot;nofollow&quot;&gt;quotient topology&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quotient_group&quot; rel=&quot;nofollow&quot;&gt;quotient group&lt;/a&gt; structure),&lt;/li&gt;&#xA;&lt;li&gt;The structure of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemannian_manifold&quot; rel=&quot;nofollow&quot;&gt;Riemannian manifold&lt;/a&gt; (with the induced &lt;a href=&quot;http://en.wikipedia.org/wiki/Flat_manifold&quot; rel=&quot;nofollow&quot;&gt;flat metric&lt;/a&gt;; this means that it is not curved, by contrast with the induced metric on the usual torus in $\mathbb{R}^3$)&lt;/li&gt;&#xA;&lt;li&gt;The structure of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_space&quot; rel=&quot;nofollow&quot;&gt;probability space&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;None of these structures depend on an embedding into $\mathbb{R}^n$.  The naming is by analogy with the case $d = 1$, where one gets a description of the usual topological group structure on the circle but, again, &lt;em&gt;without a preferred embedding into $\mathbb{R}^n$.&lt;/em&gt;  None of these structures suggest a good definition of &quot;unit torus&quot; in $\mathbb{R}^3$.  In particular, as Willie notes in the comments, $\mathbb{R}^2/\mathbb{Z}^2$ with the flat metric can't even isometrically embed into $\mathbb{R}^3$&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-04T20:32:52.213" LastActivityDate="2010-11-04T20:32:52.213" CommentCount="7" />
  <row Id="8957" PostTypeId="2" ParentId="8933" CreationDate="2010-11-04T20:06:46.057" Score="-5" ViewCount="0" Body="&lt;p&gt;x=log_2(10)/(1+log_2(3))&lt;/p&gt;&#xA;" OwnerUserId="3125" LastActivityDate="2010-11-04T20:06:46.057" CommentCount="3" />
  <row Id="8958" PostTypeId="1" AcceptedAnswerId="8962" CreationDate="2010-11-04T20:23:03.687" Score="2" ViewCount="53" Body="&lt;p&gt;Let $f \in End_{F}(V)$ be an Endomorphism of a vector spaces $V$ over the field $F$ with the property $f \circ f=id_{V}$. Show that if $char(F) \neq 2$, the following equotation is true: $$V = Ker(f-id_{V}) \oplus Ker(f+id_{V})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: $\oplus$ is the direct sum, $id_{V}$ is the identity map, Ker is the kernel and $\circ$ is the function composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts:&#xA;I think you have to show that every Element in V can be expressed with the direct sum. It is obvious that the direct sum is always an element in V because the Kernel will return an element in V and the sum has to be in V again. But I have no clue where you need that $char(F) \neq 2$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-11-04T20:41:08.360" Title="Equality for subspaces with a linear map" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8959" PostTypeId="2" ParentId="8943" CreationDate="2010-11-04T20:29:02.900" Score="3" ViewCount="0" Body="&lt;p&gt;I've gotta run, so I can't flesh this out quite as much as I want, but here's a start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that gcd(a, b) = gcd(a, b-a). Thus, if a and b are close to each other, they must have a small gcd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, consider a number N. Let a consist of all of the digits of N, sorted in ascending order. Now switch the last digit with the largest digit that is different from it, and call this new number b. Then b-a will be &quot;small&quot;. I suspect it's possible to refine this to get a quick answer.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-04T20:29:02.900" CommentCount="5" />
  <row Id="8960" PostTypeId="1" AcceptedAnswerId="8963" CreationDate="2010-11-04T20:32:06.953" Score="0" ViewCount="48" Body="&lt;p&gt;I have a set of quadratic equations of the form..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 2S_0q_0(q_0S_0 + q_1S_1 + ...... + q_iS_n)+k_0 = 0$&lt;br&gt;&#xA;$ 2S_1q_1(q_0S_0 + q_1S_1 + ...... + q_iS_n)+k_1 = 0$&lt;br&gt;&#xA;   .&lt;br&gt;&#xA;   .&lt;br&gt;&#xA;   .&lt;br&gt;&#xA;$ 2S_nq_n(q_0S_0 + q_1S_1 + ...... + q_nS_n)+k_n = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have n equations and n variables $q_0,q_1,q_2.....q_n$. The k and S values are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I analytically reach a solution for this system?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T20:38:41.247" LastActivityDate="2010-11-04T20:48:06.637" Title="Solving this set of quadratic equations" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8961" PostTypeId="1" CreationDate="2010-11-04T20:32:23.367" Score="5" ViewCount="192" Body="&lt;p&gt;I have been thinking about the differentials that we use in derivatives and integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have an equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\int{w}{dr}} = other stuff$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The context for this strange equation was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q'(t) = a - \frac{b}{r(t)}, r'(t) = c - d r(t) - e q(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r''(t)= d r'(t) - e q'(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r''(t)= d r' - e a + \frac{eb}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w=r', w' = d w - ea + \frac{eb}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dw}{dr} \frac{dr}{dt}= d w - ea + \frac{eb}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w \frac{dw}{dr} = dw - ea + \frac{eb}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate both sides with respect to r:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{w^2}{2} = \int{d*w dr} - ea r + eb log(r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shift the integral to the left hand side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int d * w dr = - \frac{w^2}{2} - ea * r - eb * log(r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide by d:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int w dr = \frac{w^2}{2d} - \frac{ear}{d}-\frac{eb}{d} log(r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I get to that w dr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I can replace w with ${\frac{dr}{dt}}$, and differentials can be multiplied like normal, I would get ${\frac{dr^2}{dt}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I have the idea of multiplying by ${\frac{dt}{dt}}$. Can I go ahead and do this? That would give me &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\int{\frac{dr^2}{dt^2} dt}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which looks like ${w^2}$ to me. But can I do that? &lt;/p&gt;&#xA;" OwnerUserId="2247" LastEditorUserId="2247" LastEditorDisplayName="" LastEditDate="2010-11-05T12:27:36.487" LastActivityDate="2010-11-05T13:17:27.587" Title="To what extent can you manipulate differentials like dy and dt like actual values?" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="8962" PostTypeId="2" ParentId="8958" CreationDate="2010-11-04T20:32:35.677" Score="3" ViewCount="0" Body="&lt;p&gt;You want to show that two things happen:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Every element of $V$ can be written as a sum $u+w$, where $u\in\mathrm{ker}(f-\mathrm{id}_V)$ and $v\in\mathrm{ker}(f+\ker{id}_v)$; and&lt;/li&gt;&#xA;&lt;li&gt;The intersection of $\mathrm{ker}(f-\mathrm{id}_V)$ and $\mathrm{ker}(f+\mathrm{id}_V)$ is trivial (contains only the zero vector).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One reason why you need $\mathrm{char}(F)\neq 2$ is that if $\mathrm{char}(F)=2$, then $1=-1$, so $f-\mathrm{id}_V = f+\mathrm{id}_V)$. And the sum of a subspace with itself cannot be a direct sum unless the subspace is trivial (which in this case would require $V$ to be trivial). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part is fairly easy: suppose $x$ is in the intersection. Then $(f+\mathrm{id}_V)(x) = \mathbf{0}$ and $(f-\mathrm{id}_V)(x) = \mathbf{0}$; the first equation tells you that $f(x)+x=\mathbf{0}$, the second that $f(x)-x=\mathbf{0}$. Now, since $2\neq 0$, you can conclude that $x=\mathbf{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you &quot;just&quot; need to show that every element of $V$ can be expressed as a sum of vectors in the appropriate place. Now, notice that so far we have not used the fact that $f\circ f=\mathrm{id}_V$, so presumably we are going to need to use it somehow. For one thing, what happens if you take an arbitrary vector $x$, and you look at $x-f(x)$? Then $f(x-f(x)) = f(x)-f(f(x)) = f(x)-x = -(x-f(x))$. So, does $x-f(x)$ lie in one of the two subspaces? What about $x+f(x)$? And, can you express $x$ as a linear combination of $x+f(x)$ and $x-f(x)$ (again, the fact that $2\neq 0$ is going to rear its ugly head...)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-04T20:41:08.360" LastActivityDate="2010-11-04T20:41:08.360" CommentCount="1" />
  <row Id="8963" PostTypeId="2" ParentId="8960" CreationDate="2010-11-04T20:48:06.637" Score="5" ViewCount="0" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $X = q_0S_0 + q_1S_1 + \dots + q_n S_n$ and add up all the equations.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-04T20:48:06.637" CommentCount="1" />
  <row Id="8964" PostTypeId="2" ParentId="8944" CreationDate="2010-11-04T20:50:55.290" Score="2" ViewCount="0" Body="&lt;p&gt;In many cases expanding the exponential works, in this case:&#xA;$2^{\frac{1}{n}} = \exp ( \frac{\ln 2}{n} ) = 1 + \frac{\ln 2}{n} + c_n $. Where $ \lim_{n \to \infty} n^2 c_n  =  (\ln 2)^2 \times ( \exp ( \frac{\ln 2}{n} ) - 1 - \frac{\ln 2}{n} ) /( \frac{\ln 2}{n})^2 = \frac{(\ln 2)^2}{2}.$ So, $c_n$ converges absolutely.&#xA;Now, $\sum_{ n \geq 1} (-1)^n (2^{\frac{1}{n}} - 1) = \sum_{n \geq 1} (-1)^n \frac{\ln 2}{n} + \sum_{n \geq 1} (-1)^n c_n$, which clearly converges.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-11-04T20:50:55.290" />
  <row Id="8965" PostTypeId="1" AcceptedAnswerId="8972" CreationDate="2010-11-04T20:55:20.050" Score="1" ViewCount="30" Body="&lt;p&gt;Let $(X, \prec)$ and $(Y, &amp;lt;)$ be two well-ordered sets. Now, I want to understand the proof that $X$ is isomorphic to an initial segment of $Y$ or $Y$ is isomorphic with an initial segment of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, they define $\mathcal{F}$ to be the set of all mappings with as domain an initial segment of $X$ and $Y$ as codomain and $F: \mathcal{F} \to Y \cup \{Y\}$ as $F(f) = \text{min}(Y \setminus \text{ran} f)$ if $f$ is not surjective and equal to $Y$ if $f$ is surjective. Now, they want to apply the recursion principle to $F$ to get a mapping from $X$ to $Y \cup \{Y\}$ that satisfies $f(x) = F(f|_{\hat{x}})$ for all $x$. Now I wonder how this is possible with the Recursion principle because the codomain is different. (Arturo solved this (very easy...) question in the comments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further they begin $x \prec y$ in $X$ and $f(y) \neq Y$ then $f(x) \lt f(y)$, so I assume the above but then I see that (strict) $Y \setminus \text{ran} f|_{\hat{y}} \subset Y \setminus \text{ran} f|_{\hat{x}}$ but why would could the minimum not be equal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-04T21:18:23.363" LastActivityDate="2010-11-04T21:26:52.087" Title="Initial segments order-isomorphic." Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8966" PostTypeId="1" CreationDate="2010-11-04T20:57:51.820" Score="1" ViewCount="271" Body="&lt;p&gt;I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some trouble to do that and I'd glad with any help I may get.&lt;/p&gt;&#xA;" OwnerUserId="3128" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-05T02:17:07.430" LastActivityDate="2010-12-08T08:11:57.290" Title="Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$" Tags="&lt;algebra&gt;" AnswerCount="5" CommentCount="8" />
  <row Id="8967" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T21:01:28.490" Score="3" ViewCount="0" Body="&lt;p&gt;Cube both sides and collect terms. You should get&#xA;\begin{eqnarray}&#xA;512 - 3264 x + 8856 x^2 - 13457 x^3 + 12702 x^4 - 7794 x^5 + &#xA; 3136 x^6 - 844 x^7 + 120 x^8 = 0&#xA;\end{eqnarray}&#xA;which factorizes into&#xA;\begin{eqnarray}&#xA;(8 - 17 x + 6 x^2) (64 - 272 x + 481 x^2 - 456 x^3 + 258 x^4 - &#xA;   84 x^5 + 20 x^6) = 0.&#xA;\end{eqnarray}&#xA;Since this is an $8^{\text{th}}$-degree polynomial equation, you can use Mathematica or a calculator to determine six of the eight solutions (by solving the $6^{\text{th}}$-degree polynomial equation numerically) and the other two by solving the quadratic equation: $\frac{1}{12}(17 \pm \sqrt{97})$. The 6 solutions are 3 pairs of complex conjugates:&#xA;\begin{eqnarray}&#xA;&amp;amp; &amp;amp; 0.647522 \dots \pm i 2.21209 \dots  \&#xA;&amp;amp; &amp;amp; 0.657656 \dots \pm i 0.218497 \dots  \&#xA;&amp;amp; &amp;amp; 0.794822 \dots \pm i 0.788971 \dots &#xA;\end{eqnarray}&#xA;Of course, you'll have to check that these solutions work. At most one real solution and two complex solutions are spurious.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T06:30:19.867" LastActivityDate="2010-11-05T06:30:19.867" CommentCount="7" />
  <row Id="8968" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T21:02:19.890" Score="2" ViewCount="0" Body="&lt;p&gt;Note that $\frac{1}{12} \left(17+\sqrt{97}\right)$ is not a solution. It just appeared because of cubing.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-11-04T21:22:50.000" LastActivityDate="2010-11-04T21:22:50.000" CommentCount="6" />
  <row Id="8969" PostTypeId="1" CreationDate="2010-11-04T21:06:52.010" Score="1" ViewCount="75" Body="&lt;p&gt;If I have two different discrete distributions of random variables X and Y, such that their probability mass functions are related as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=x_i) = \lambda\frac{P (Y=x_i)}{x_i} $   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what can I infer from this equation? Any observations or interesting properties that you see based on this relation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if,&lt;br&gt;&#xA;  P($X=x_i$) = $\lambda\sqrt{\frac{P (Y=x_i)}{x_i} }$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, $\lambda$ is a constant.&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-04T23:26:01.930" LastActivityDate="2010-11-05T02:48:44.750" Title="Relationship between these two probability mass functions." Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8970" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T21:12:40.333" Score="0" ViewCount="0" Body="&lt;p&gt;Really? Is it impossible solve it by factorize or trick of algebra? Sorry, I´m just a high school grade studant.&lt;/p&gt;&#xA;" OwnerUserId="3128" LastActivityDate="2010-11-04T21:12:40.333" CommentCount="3" />
  <row Id="8971" PostTypeId="1" CreationDate="2010-11-04T21:21:37.640" Score="1" ViewCount="53" Body="&lt;p&gt;I hope this question is not too general, but I am not completely sure yet how to phrase it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we all know that when we have two sequences of random variables $X_n$ and $Y_n$ for $n \ge 1$, that converge to $X$ and $Y$ (using some measure, almost surely, or with probability, in distribution, etc.) then we have all kind of nice properties such as $X_n + Y_n \rightarrow X + Y$ and $X_n Y_n \rightarrow X Y$ so on (with some regularity conditions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what if we have a general operator on $X$ and $Y$? Are there some mild/not mild conditions on an operator $g(X_n,Y_n)$ that returns a random variable $Z_n$ such that $g(X_n,Y_n) \rightarrow g(X,Y)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the continuous mapping theorem would state that if $g$ is continuous, then we have that this holds. Are there milder conditions?&lt;/p&gt;&#xA;" OwnerUserId="3129" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-05T02:26:02.457" LastActivityDate="2010-11-05T02:26:02.457" Title="Sufficient conditions for convergence of functions of random variables." Tags="&lt;probability&gt;&lt;statistics&gt;" />
  <row Id="8972" PostTypeId="2" ParentId="8965" CreationDate="2010-11-04T21:26:52.087" Score="2" ViewCount="0" Body="&lt;p&gt;Let $z$ be the least element of $X$ strictly larger than $x$; it exists, since you have at least $y$ among the upper bounds. You cannot have $f(z)=Y$, because you do not have $f(y)=Y$; so $f(z)$ is the least element which is not the image of any element of $\{a\in X\mid a\preceq x\}$. If $z=y$, then this shows that $f(y)$ cannot be equal to $f(x)$ (since $f(y)$ is the least element of the complement of a set that contains $f(x)$); if $z\neq y$, then $f(z)\in \mathrm{ran}f|_{\hat{y}}$; so the minimum of the complement of the range of $f|_{\hat{y}}$ cannot be equal to the minimum of the complement of the range of $f|_{\hat{x}}$: the latter &lt;em&gt;is&lt;/em&gt; $f(z)$, and this is &lt;em&gt;not&lt;/em&gt; in the complement of the former. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-04T21:26:52.087" />
  <row Id="8973" PostTypeId="2" ParentId="8961" CreationDate="2010-11-04T21:37:32.563" Score="4" ViewCount="0" Body="&lt;p&gt;None of your examples make any sense to me. The manipulations with&#xA;differentials which are allowed in calculus are basically only those&#xA;that follow from the chain rule. In particular, an integral can be&#xA;transformed using the change-of-variables formula, but that's it. (OK,&#xA;maybe I'm over-simplifying a little here, but the basic advice is that&#xA;if you're unsure whether something is allowed or not, don't do it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think much of the confusion surrounding this comes from the fact&#xA;that the notation $dy/dx$ doesn't show at which point the derivative&#xA;is supposed to be evaluated. For example, the derivative of an inverse&#xA;function is given by the rule $dx/dy=1/(dy/dx)$ in this notation,&#xA;which is suggestive and useful, but not very precise. What this really&#xA;says is that if $(a,b)$ is a point on the graph $y=f(x)$ and if the&#xA;slope at that point is $f'(a)=s$, then the point $(b,a)$ lies on the&#xA;graph of the inverse function, $y=f^{-1}(x)$, and the slope of that&#xA;graph &lt;em&gt;at that point&lt;/em&gt; is $1/s$:&#xA;$$ (f^{-1})'(b)=1/s=1/f'(a).$$&#xA;There's no &quot;magic&quot; here, just a simple&#xA;geometric fact which can be illustrated in a picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example of a similar flavour, see &lt;a href=&quot;http://math.stackexchange.com/q/8040/1242&quot;&gt;this&#xA;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I was a bit too quick. There is something behind your manipulations after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you correctly, you are looking at a second order ODE of the form $d^2 r/dt^2 = f(r,dr/dt)$; let's call this equation (A). This is equivalent to the first order system&#xA;$dr/dt=w$, $dw/dt=f(r,w)$, which I'll call (B). Any solution $r=R(t)$ of (A)&#xA;corresponds to a solution $(r,w)=(R(t),R'(t))$ of (B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix one such solution for the moment.&#xA;(We're trying to solve the equation, so we probably don't know such a solution&#xA;yet, but we can pretend that we have one and see where that leads.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $(r,w)=(R(t),R'(t))$ is a parametrized curve in the $(r,w)$-plane&#xA;(the &lt;em&gt;phase plane&lt;/em&gt; of the system).&#xA;Provided this curve doesn't have vertical slope, it looks (at least locally)&#xA;like the graph of a function, $w=H(r)$ say.&#xA;This function $H$ satisfies a first-order ODE,&#xA;$$H'(r)=\frac{f(r,H(r))}{H(r)},$$&#xA;which is the useful fact that you learned from Mariano in that other thread.&#xA;[It follows from &quot;$dw/dr=(dw/dt)/(dr/dt)=f(r,w)/w$&quot;&#xA;as explained in the answer that I linked to above.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral you asked about is&#xA;$$ \int_{r_0}^{r_1} H(r) dr.$$&#xA;(I've attached some limits $r_0$ and $r_1$ just because that makes it easier for me to think.)&#xA;Making the change of variables $r=R(t)$, where $R$ is the solution of (A)&#xA;that we used in order to define the function $H$, we get&#xA;$$ \int_{t_0}^{t_1} H(R(t)) R'(t) dt$$&#xA;where $t_0$ and $t_1$ are suitable time values corresponding to $r_0$ and $r_1$&#xA;[that is, $r_0=R(t_0)$ and $r_1=R(t_1)$].&#xA;By the definition of $H$ we have $H(R(t))=R'(t)$,&#xA;since $w=H(r)$ defines a point on the curve $(w,r)=(R(t),R'(t))$.&#xA;Thus the integral becomes&#xA;$$ \int_{t_0}^{t_1} (R'(t))^2 dt,$$&#xA;just like you (and Dan in his answer) said, although expressed in a more&#xA;precise way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, whether this leads to something useful, I don't know.&#xA;But at least it should be a bit clearer what the manipulations &lt;em&gt;mean&lt;/em&gt;, I hope.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-05T13:17:27.587" LastActivityDate="2010-11-05T13:17:27.587" CommentCount="5" />
  <row Id="8974" PostTypeId="2" ParentId="8969" CreationDate="2010-11-04T21:51:22.067" Score="1" ViewCount="0" Body="&lt;p&gt;I don't have enough reputation yet, or this would be a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A random variable can have only one probability mass function, so it is not clear what you are asking.  Where do these equations come from?&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-04T21:51:22.067" CommentCount="1" />
  <row Id="8975" PostTypeId="1" AcceptedAnswerId="9018" CreationDate="2010-11-04T22:05:29.067" Score="0" ViewCount="139" Body="&lt;p&gt;Can someone explain the following definitions of a limit and derivatives. Issac did a good job explaining it in the previous thread: &lt;a href=&quot;http://math.stackexchange.com/questions/8688/infinite-series-archimedean-principle&quot;&gt;http://math.stackexchange.com/questions/8688/infinite-series-archimedean-principle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a limit is defined as the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$lim_{x \to a} f(x) = T$, what we actually mean is if we take any number M &gt; T, then we can force $f(x) &amp;lt; M$ by taking $x$ sufficiently close to $a$. Similarly, if we take any $L &amp;lt; T$, then we can force $f(x) &gt; L$ by taking $x$ sufficiently close to $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a derivative is defined as the following:&#xA;The derivative of $f$ at $a$ is the value denoted by $f'(a)$, such that for any $L &amp;lt; f'(a)$ and any $M &gt; f'(a)$, we can force $L &amp;lt; \frac{f(x)-f(a)}{x - a} &amp;lt; M$ simply by taking $x$  sufficiently close  to (but not equal to $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first real analysis course and I am very confused. If you choose to use episilon delta, please try to make it as simple as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="287" LastActivityDate="2010-11-05T07:54:54.450" Title="limit and derivatives - precise definitions" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8976" PostTypeId="2" ParentId="8921" CreationDate="2010-11-04T22:05:57.670" Score="3" ViewCount="0" Body="&lt;p&gt;You might find this paper of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fulton C.M., Stein S.K. (1960) Parallelograms inscribed in convex curves. Amer Math Monthly 67: 257–258&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-04T22:05:57.670" CommentCount="1" />
  <row Id="8977" PostTypeId="1" CreationDate="2010-11-04T22:19:46.550" Score="3" ViewCount="195" Body="&lt;p&gt;How can I calculate the arc between two circles? The arc must be tangent to the two points on the circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture illustrating it. I'm trying to code and calculate the orange arc and the blue arc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8GsDV.png&quot; alt=&quot;circle arcs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The points B, F, D, and E can be changed by the user and dynamically update.&lt;/p&gt;&#xA;" OwnerUserId="3132" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-04T22:23:44.223" LastActivityDate="2010-11-05T12:52:21.887" Title="Calculating a tangent arc between two points on two circles." Tags="&lt;geometry&gt;&lt;circle&gt;&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="8978" PostTypeId="1" CreationDate="2010-11-04T22:43:13.373" Score="9" ViewCount="226" Body="&lt;p&gt;Suppose that you have an e.g.f.: $E(z)=\sum_{n=0}^{\infty} \frac{a_{n}z^{n}}{n!}$, and that the definition of $a_{n}$ can be reasonably extended to noninteger arguments. (the Catalan numbers $C_{n}$, would be written in terms of the Gamma function thusly: $C_{n} = \frac{\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+1)}$, for instance), what then is the combinatorial significance of this integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U(z)=\int_{0}^{\infty} \frac{a_{v}z^{v}dv}{\Gamma(v+1)}$$&#xA;?&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-11-11T15:27:51.120" Title="what is the combinatoric significance of an integral related to the e.g.f." Tags="&lt;combinatorics&gt;&lt;generating-functions&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="8979" PostTypeId="1" AcceptedAnswerId="8986" CreationDate="2010-11-04T22:44:53.453" Score="0" ViewCount="58" Body="&lt;p&gt;Let f be an increasing function defined on [a,b] and let&lt;br&gt;&#xA;a &amp;lt; x1 &amp;lt; x2 &amp;lt; ... &amp;lt; xn &amp;lt; b be n points in the interior of [a,b]&lt;br&gt;&#xA;Show that the sum of [f(xk+) - f(xk-)] (with k ranging from 1 to n)&lt;br&gt;&#xA;is less than or equal to f(b-) - f(a+)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof I have is very wordy and sketchy. I know it would be easier if I could find a telescoping sum but I've got a road block in my mind. :/&lt;/p&gt;&#xA;" OwnerUserId="2319" LastActivityDate="2010-11-04T23:22:04.853" Title="Inequality on an increasing function" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="8980" PostTypeId="1" CreationDate="2010-11-04T22:52:25.637" Score="4" ViewCount="155" Body="&lt;p&gt;Can someone show mathematically how gimbal lock happens when doing matrix rotation with Euler angles for yaw, pitch, roll?  I'm having a hard time understanding what is going on even after reading several articles on Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the only work-around to use quaternions?&lt;/p&gt;&#xA;" OwnerUserId="3134" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T11:19:45.247" LastActivityDate="2010-11-05T11:19:45.247" Title="Euler angles and gimbal lock." Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;&lt;quaternions&gt;" AnswerCount="2" />
  <row Id="8981" PostTypeId="2" ParentId="8975" CreationDate="2010-11-04T23:00:54.627" Score="0" ViewCount="0" Body="&lt;p&gt;I think it's easier if you think about it visually. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With loss of generality, we can assume as we get closer to 'a' our function is increasing, it could be constant, decreasing, or whatever.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have X1 &amp;lt; a &amp;lt; X2 on the X axis with the corresponding f(X1)=L &amp;lt; T &amp;lt; M=f(X2)&lt;br&gt;&#xA;[Notice that f(a) doesn't necessarily equal T]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start squeezing X1 and X2 closer together and hence closer to a, we are forcing f(X1) and f(X2) closer to T. Therefore for x &amp;lt; X2, f(x) &amp;lt; f(X2)=M.&lt;br&gt;&#xA;Similarly, for x &gt; X1, f(x) &gt; f(X1)=L  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hoped this helped. When in doubt, draw a picture.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastActivityDate="2010-11-04T23:00:54.627" />
  <row Id="8982" PostTypeId="2" ParentId="8961" CreationDate="2010-11-04T23:01:22.460" Score="3" ViewCount="0" Body="&lt;p&gt;Your calculation looks to me like it produces the right answer provided one interprets it correctly. The only sensible meaning one can attach to $dr^2/dt^2$ should be $(dr/dt)^2$. So it seems your calculation shows that $\int w dr = \int w^2 dt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience from my own undergraduate calculus is that: (i) any calculation I did by multiplying or dividing differentials like this would produce a meaningful and correct answer in the end; (ii) whatever answer was found in the end could also be justified by only using the normal chain rule. In this case, for instance, the change-of-variables formula tells you that $dr = (dr/dt)dt = wdt$, which gives the above formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I realised afterwards that point (i) above might be misleading when one takes a course on calculus in several variables, since $dx \neq \frac{\partial x}{\partial y} dy$ if $x$ depends on more variables than $y$. In principle I would guess that multiplication and division of differentials is still possible as a heuristic but one needs to be very careful to distinguish between d's and $\partial$'s.&lt;/p&gt;&#xA;" OwnerUserId="677" LastEditorUserId="677" LastEditorDisplayName="" LastEditDate="2010-11-05T06:52:06.187" LastActivityDate="2010-11-05T06:52:06.187" />
  <row Id="8983" PostTypeId="1" AcceptedAnswerId="8987" CreationDate="2010-11-04T23:11:10.203" Score="1" ViewCount="121" Body="&lt;p&gt;My question is how do I reduce A'B'C'+AB'C'+ABC' (note that (') stands for a bar over the letter). To get (A+B')C'. I'm so lost just been trying to get it for awhile only using the 10 boolean simplification rules.&lt;/p&gt;&#xA;" OwnerUserId="3135" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-05T02:24:34.327" LastActivityDate="2010-11-05T02:24:34.327" Title="Boolean Simplification of A'B'C'+AB'C'+ABC'" Tags="&lt;boolean-algebra&gt;" AnswerCount="2" />
  <row Id="8984" PostTypeId="2" ParentId="8980" CreationDate="2010-11-04T23:16:23.273" Score="6" ViewCount="0" Body="&lt;p&gt;&quot;Euler Angles&quot; you can think of as a function $(S^1)^3 \to SO_3$ or $\mathbb R^3 \to SO_3$.  The derivative of this function does not always have rank 3, so you have degenerate submanifolds where the function is many-to-one.  In this special case that's called &quot;gimbal lock&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One formalism that avoids this is quaternions.  You can of course use other formalisms, and many other formalisms are naturally related to the quaternion version, so people tend to gravitate to the quaternion version.  One version that's closely related to quaternions would be to use the exponential map for the unit quaternion group.  But this also has &quot;gymbal lock&quot; but of a different kind.  But it does have the rather appealing interpretation as rotations about arbitrary axis -- this is perhaps more useful if you're only interested in rotations that differ from the identity matrix (or some given matrix) by a small amount, they're very natural coordinates on &quot;small scales&quot; in $SO_3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any special properties you'd like for coordinates on $SO_3$? That might give a sense for where you want to go with this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit in response to 1st comment: There's various conventions for Euler angles. Let's use what Wikipedia calls &quot;Proper Euler Angles&quot;   &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_angles&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euler_angles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a unit vector $\vec v \in \mathbb R^3$ and a number $\theta \in \mathbb R$, let $R_{\vec v, \theta}$ be the rotation matrix which fixes $\vec v$ and which rotates counter-clockwise by an angle $\theta$ in the plane orthogonal to $\vec v$.  One version of &quot;Proper Euler angles&quot; would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{(1,0,0), \theta_1} R_{(0,1,0), \theta_2} R_{(1,0,0), \theta_3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Tait-Bryan&quot; variant would use the three distinct vectors, notice above we only used the standard vectors for the xy-plane.   The Tait-Bryan variant is the one appropriate to pitch/roll/yaw of an aeroplane.  Wikipedia has an excellent picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PcWFU.png&quot; alt=&quot;Pitch/Roll/Yaw&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case you'd have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{(1,0,0),\theta_1} R_{(0,1,0), \theta_2} R_{(0,0,1),\theta_3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_3$ would be the roll, $\theta_2$ the pitch and $\theta_1$ the yaw. To relate my coordinates to the picture, $(1,0,0)$ is the direction the plane is pointing. $(0,1,0)$ is the direction of the left wing. $(0,0,1)$ is the direction of the yellow axis sticking out of the top of the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write out the matrices quite explicitly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R_{(1,0,0), \theta} = &#xA; \pmatrix{ 1 &amp;amp; 0 &amp;amp; 0 \cr &#xA;           0 &amp;amp; \cos \theta &amp;amp; -\sin \theta \cr&#xA;           0 &amp;amp; \sin \theta &amp;amp; \cos \theta }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, one occurance of &quot;gimbal lock&quot; is $\theta_2 = 0$.  In the Tait-Bryan variant it would be when the aeroplane is either pointing straight up or down, which is $\theta_2 = \pm \pi/2$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-05T00:02:03.897" LastActivityDate="2010-11-05T00:02:03.897" CommentCount="2" />
  <row Id="8985" PostTypeId="2" ParentId="8983" CreationDate="2010-11-04T23:18:19.713" Score="0" ViewCount="0" Body="&lt;p&gt;First use the distributive law to pull out the C', then work on the As and Bs&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-04T23:18:19.713" />
  <row Id="8986" PostTypeId="2" ParentId="8979" CreationDate="2010-11-04T23:22:04.853" Score="6" ViewCount="0" Body="&lt;p&gt;First, define $x_0 = a$ and $x_{n+1}=b$.&#xA;Then, for any $k=1,\ldots,n$, we have&#xA;$$&#xA;f(x_k +)-f(x_k -) \leq f\bigg(\frac{{x_k  + x_{k + 1} }}{2}\bigg) - f\bigg(\frac{{x_{k - 1}  + x_k }}{2}\bigg)&#xA;$$&#xA;(since $f$ is increasing). Thus,&#xA;$$&#xA;\sum\limits_{k = 1}^n {[f(x_k  + ) - f(x_k  - )]}  \le f\bigg(\frac{{x_n  + x_{n + 1} }}{2}\bigg) - f\bigg(\frac{{x_0  + x_1 }}{2}\bigg).&#xA;$$&#xA;Now, since $(x_n  + x_{n + 1} )/2 = (x_n  + b)/2 &amp;lt; b$ and $(x_0  + x_1 )/2 = (a  + x_1)/2 &gt; a$, we get&#xA;$$&#xA;\sum\limits_{k = 1}^n {[f(x_k  + ) - f(x_k  - )]}  \le f(b-) - f(a+).&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-04T23:22:04.853" CommentCount="1" />
  <row Id="8987" PostTypeId="2" ParentId="8983" CreationDate="2010-11-04T23:27:46.040" Score="3" ViewCount="0" Body="&lt;pre&gt;&lt;code&gt;A'B'C'+AB'C'+ABC'&#xA;C'(A'B'+AB'+AB)&#xA;C'(A'B'+A(B'+B))&#xA;C'(A'B'+A)&#xA;C'(B'+A)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It's that last step that used to trip me up.  &lt;code&gt;A'+AB = A'+B&lt;/code&gt;  Forget what that law is called (identity?).&lt;/p&gt;&#xA;" OwnerUserId="3136" LastEditorUserId="3136" LastEditorDisplayName="" LastEditDate="2010-11-04T23:41:36.120" LastActivityDate="2010-11-04T23:41:36.120" CommentCount="4" />
  <row Id="8988" PostTypeId="1" AcceptedAnswerId="8991" CreationDate="2010-11-04T23:30:14.883" Score="1" ViewCount="123" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like a demonstration of the fact below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being given real numbers &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; (nonzero) and a positive integer &lt;strong&gt;n&lt;/strong&gt;, the order &lt;strong&gt;p&lt;/strong&gt;, that occupies the maximum term (in absolute value)&#xA; of the development of power (a+b)^ n, according to decreasing powers of &lt;strong&gt;a&lt;/strong&gt; is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;p = 1 + integer part of [|b|(n+1)/(|a|+|b|)]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;strong&gt;n&lt;/strong&gt; is integer, there are maximum two terms: those of order &lt;strong&gt;p&lt;/strong&gt; and &lt;strong&gt;p-1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;|a|&lt;/strong&gt; and &lt;strong&gt;|b|&lt;/strong&gt; are the modules of the numbers &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already, very grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-05T16:23:24.267" LastActivityDate="2010-11-05T16:23:24.267" Title="Maximum term of (a + b) ^ n" Tags="&lt;binomial-coefficients&gt;" AnswerCount="1" />
  <row Id="8989" PostTypeId="2" ParentId="8966" CreationDate="2010-11-04T23:33:14.993" Score="3" ViewCount="0" Body="&lt;p&gt;Once someone gives the answer, it becomes easier!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume you are only looking for real roots and that $\displaystyle (5x-x^3)^{1/3}$ is the unique &lt;em&gt;real&lt;/em&gt; number whose cube is $\displaystyle 5x - x^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x-2)^3 + 5x - x^3 = -6x^2 + 17x -8$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $\displaystyle a = (5x-x^3)^{1/3}$ and $\displaystyle b = x-2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle -2x^2(x-2) - (-6x^2 - 17x - 8) = 2x^2(5x-x^3)^{1/3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle -2x^2 b - (a^3 + b^3) = 2x^2a$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (a+b)(a^2 - ab + b^2 + 2x^2) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for any real numbers $\displaystyle a,b,x$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a^2 -ab + b^2 \ge 0$ and  $\displaystyle 2x^2 \ge 0$ and thus we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a + b = 0$ or $\displaystyle 2x^2 = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle a = -b$ can be cubed to give $\displaystyle 6x^2 - 17x + 8 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x^2 = 0$ can be easily eliminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the transformations we did were equivalent, and so &lt;em&gt;both&lt;/em&gt; roots of $\displaystyle 6x^2 - 17x + 8 = 0$ are also roots of the original equation, given the definition of cuberoot at the top of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define $\displaystyle z^{1/3}$ using the principal branch of $\log z$, then the above assumption of $\displaystyle a$ being real is valid only if $\displaystyle 5x - x^3 \ge 0$, which eliminates $\displaystyle \dfrac{17 + \sqrt{97}}{12}$ as a root.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T08:11:57.290" LastActivityDate="2010-12-08T08:11:57.290" CommentCount="5" />
  <row Id="8990" PostTypeId="1" CreationDate="2010-11-04T23:41:00.410" Score="0" ViewCount="69" Body="&lt;pre&gt;&lt;code&gt;AB+CD+A’BD+A’BC+AB’D+AB’C&#xA;AB+CD+A’B(D+C)+AB’(D+C)&#xA;CD+(AB+A'B(D+C)+AB’(D+C))&#xA;CD+(AB+B(D+C)+A(D+C))&#xA;CD+AB+(B(D+C)+A(D+C))&#xA;CD+AB+(A+B)(D+C)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is what I got, but I'm not sure if it's correct.  It's as good as I can get it algebraically.  Can it be simplified further?&lt;/p&gt;&#xA;" OwnerUserId="3136" LastActivityDate="2010-11-05T05:38:12.020" Title="Boolean Algebra: Can this be simplified further?" Tags="&lt;boolean-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8991" PostTypeId="2" ParentId="8988" CreationDate="2010-11-04T23:45:55.807" Score="5" ViewCount="0" Body="&lt;p&gt;Let $f(k)=C_n^k |a|^{n-k} |b|^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(k+1)/f(k)=\frac{|b|}{|a|} \frac{n-k}{k+1} &gt; 1$ iff $k&amp;lt;\frac{|b|(n+1)}{(|b|+|a|)}-1$, so $f(k)$ is increasing until $k=I\left(\frac{|b|(n+1)}{(|b|+|a|)}\right)$ where $I$ denotes the integral part.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-04T23:45:55.807" CommentCount="4" />
  <row Id="8992" PostTypeId="1" CreationDate="2010-11-04T23:51:06.830" Score="2" ViewCount="111" Body="&lt;p&gt;Hello. I'm trying to compute the following limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{0}^{\infty} (1+\frac{x}{n})^{(-n)} \sin(x/n) dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the integrand converges to 0 pointwise so the trick is to find the dominating function. So for simplicity let ${f_{n}}$ denote the sequence of functions of the integrand: it is not hard to see that $|f_{n}| \leq (1+x/3)^{-3}x$ for all $n\geq 3$ and the latter function is integrable on $(0,\infty)$. But doesn't Lebesgue dominated convergence theorem requires to bound it for all $n\in\mathbb{N}$? What happens if we can't obtain a dominating function for $n=1$ or $n=2$ ? Can you please clarify this? I have read in a book that you can drop some terms, say $n=1,2,3$ and the bound the rest , but why is this possible? doesn't this violates the hypothesis of Lebesgue's theorem?&lt;/p&gt;&#xA;" OwnerUserId="3139" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T03:45:26.800" LastActivityDate="2010-11-05T04:30:50.677" Title="Understanding dominating functions" Tags="&lt;measure-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8993" PostTypeId="1" CreationDate="2010-11-04T23:55:06.997" Score="4" ViewCount="154" Body="&lt;p&gt;The Poincaré conjecture is listed as one of the Millennium Prize Problems and has received significant attention from the media a few years ago when Grigori Perelman presented a proof of this conjecture. But why is this interesting at all? What practical and interesting questions can be answered now that this conjecture is proven?&lt;/p&gt;&#xA;" OwnerUserId="3138" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T00:30:05.953" LastActivityDate="2010-11-05T00:31:42.750" Title="What is the importance of the Poincaré conjecture?" Tags="&lt;topology&gt;&lt;manifold&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8995" PostTypeId="2" ParentId="8909" CreationDate="2010-11-05T00:00:19.863" Score="0" ViewCount="0" Body="&lt;p&gt;Put together 6 pyramids to make a cube. You will see that volume of whole cube can be considered as sum of 2 equal smaller volumes (area by half-height of cube). Its also the height of pyramid multiplieds by area. The small volume is also the sum of 3 pyramids, because 3 is half of 6. So single pyramid is 1/3rd of this multiplication.&lt;/p&gt;&#xA;" OwnerUserId="3140" LastActivityDate="2010-11-05T00:00:19.863" CommentCount="2" />
  <row Id="8996" PostTypeId="1" AcceptedAnswerId="9001" CreationDate="2010-11-05T00:15:48.463" Score="2" ViewCount="80" Body="&lt;p&gt;Hi. This is an exercise from Rotman, An Introduction to Homological Algebra, which I've been thinking now and then for a few days and I haven't solved it yet. I've decided to ask here because it is bugging me and I don't have any friends also studying this subject (in case it is of any use, I'm an undergraduate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a commutative noetherian ring. If $A$, $B$ are finitely generated $R$-modules, then $Hom_R(A,B)$ is a finitely generated $R$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've thought as of yet (maybe the problem is that I haven't been trying to apply the useful theorems...):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, since $R$ is noetherian, it suffices to inject $Hom_R(A,B)$ into some finitely generated module (in noetherian rings, submodules of f.g. are f.g.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $R$ is commutative (or since $R$ is noetherian), a finitely generated $R$-module is a quotient of $R^n$ for some $n$. Then let us write $A\simeq \frac{R^n}{I}$ and $B\simeq \frac{R^m}{J}$. Now we've got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Hom_R(A,B)\simeq Hom_R\left(\frac{R^n}{I},\frac{R^m}{J}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had $Hom_R\left(\frac{R^n}{I},\frac{R^m}{J}\right) \hookrightarrow Hom_R(R^n, R^m)$ (this is what I tend to think is the wrong direction, but a interesting question-rising path nevertheless) then it would be over since $Hom_R(R^n, R^m)\simeq R^{nm}$ which is finitely generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask myself: can I see $\frac{R^n}{I}$ as a submodule of $R^n$? Because if I could, then doing the same thing for $R^m$ and passing it to the hom, then it would be over. Now, this would be true if the sequence $$0\to I\hookrightarrow R^n\to \frac{R^n}{I} \to 0$$ split. But this (of course) doesn't always happen. But if I somehow had the third module to be projective, or the first one to be injective, then it would happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why would, for example, be $\frac{R^n}{I}$ be projective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where I got stuck. I'm sorry if this is overly detailed, but I remember reading that it is always good, when asking about a textbook question, to add what you've thought up to that moment. Also, even if my thoughts don't lead to the solution, I would like to know if they are correct!&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2010-11-05T02:00:49.527" LastActivityDate="2010-11-05T02:00:49.527" Title="Hom of finitely generated modules over a noetherian ring" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="1" />
  <row Id="8997" PostTypeId="2" ParentId="8923" CreationDate="2010-11-05T00:16:07.353" Score="3" ViewCount="0" Body="&lt;p&gt;On a separable metric space $(X,d)$, weak convergence of probability measures is equivalent to convergence with respect to the Kantorovich-Rubinshtein metric defined by &#xA;$$ \beta(\mu,\nu) = \sup \left( \left|\int_X f\ d(\mu-\nu)\right| : f\in {\rm Lip}(X),\  \sup_{x\in X}|f(x)|\leq 1 \right),$$&#xA;where &#xA;$${\rm Lip}(X):=\left( f:X\to{\mathbb R}, |f(x)-f(y)|\leq d(x,y) {\rm \ for\ all\ } x,y\in X   \right). $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, see Theorem 11.3.3 of Dudley's &lt;em&gt;Real Analysis and Probability&lt;/em&gt; or Section 8.3 of Bogachev's measure &lt;em&gt;Theory Volume 2&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-05T00:16:07.353" CommentCount="3" />
  <row Id="8998" PostTypeId="2" ParentId="8993" CreationDate="2010-11-05T00:31:42.750" Score="13" ViewCount="0" Body="&lt;p&gt;While I'm sure an expert could give a much more informative answer, let me give a naive one.  In mathematics we are always interested in classification results.  If we want to understand an object that occurs in some problem and we have a classification result for that type of object, we can use it to gain traction on the problem.  For example we use the classification of semisimple Lie algebras or the classification of finite simple groups all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Poincare conjecture is part of a similar classification effort, but for closed 3-manifolds.  Now, closed 2-manifolds have a well-understood classification in a few senses; there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface#Classification_of_closed_surfaces&quot; rel=&quot;nofollow&quot;&gt;topological classification&lt;/a&gt;, and there is also a &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniformization_theorem#Geometric_classification_of_surfaces&quot; rel=&quot;nofollow&quot;&gt;geometric classification&lt;/a&gt;.  These classification results allow us to tackle many problems involving surfaces, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_surface&quot; rel=&quot;nofollow&quot;&gt;Riemann surfaces&lt;/a&gt; which occur when one analytically continues a holomorphic function.  So it is natural to look for a corresponding classification result in higher dimensions.  (In science journalism this is often justified as saying that topologists are trying to understand the shape of the universe, or something like that; you should feel free to take this justification or leave it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topological classification of surfaces shows in particular that a surface is determined up to homeomorphism by its &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;integral homology&lt;/a&gt;, so it's natural to ask whether the same is true for 3-manifolds.  Unfortunately, it's not; there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_sphere#Poincar.C3.A9_homology_sphere&quot; rel=&quot;nofollow&quot;&gt;famous example&lt;/a&gt; of a 3-manifold with the same homology as a sphere but which is not homeomorphic to it.  One way to prove this is to show that it has a nontrivial &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_group&quot; rel=&quot;nofollow&quot;&gt;fundamental group&lt;/a&gt;, so now the natural question arises as to whether this is enough to uniquely identify the 3-sphere.  (If it is, then the problem of deciding whether a 3-manifold is the 3-sphere is in some sense purely algebraic, and in topology it is always desirable to reduce topological questions to algebraic ones since the latter tend to be easier.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem became famous in part because it was extremely hard and in part because many mathematicians gave incorrect proofs.  This is often how problems become famous; if a problem in a field is hard that indicates that the tools of the field are not adequate to easily address it, so the problem spurs people to improve those tools.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perelman ended up proving the Poincare conjecture by proving a stronger result, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrization_conjecture&quot; rel=&quot;nofollow&quot;&gt;geometrization conjecture&lt;/a&gt;, which is the analogue for 3-manifolds of the uniformization theorem.  So work on the Poincare conjecture has led to a deeper understanding of 3-manifolds in general.  Not only that, but Perelman's work introduced important techniques which can now be used on other problems.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-05T00:31:42.750" />
  <row Id="8999" PostTypeId="2" ParentId="8992" CreationDate="2010-11-05T00:43:29.407" Score="5" ViewCount="0" Body="&lt;p&gt;The limit of a sequence $\{ a_{n} : n \geq 1\}$ is the same as that of $\{ a_{n+2} : n \geq 1 \}$.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T04:30:50.677" LastActivityDate="2010-11-05T04:30:50.677" />
  <row Id="9000" PostTypeId="1" CreationDate="2010-11-05T00:52:34.717" Score="0" ViewCount="59" Body="&lt;p&gt;Suppose $G_1$ and $G_2$ are Lie groups with isomorphic Lie algebras.  Then from standard Lie theory we know that there is a simply connected Lie group $G$ such that $G/H_i = G_i$ where $H_i$ is a discrete subgroup of the center of $G$.  I am curious if there is a nice condition on $H_1,H_2$ and how they sit in $G$ that implies $G_1$ and $G_2$ are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally I guess, when is $G/H = G/K$ for a general group $G$ and normal subgroups $H$ and $K$ (maybe having $H$ and $K$ be in the center makes things easier?).&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-11-05T00:52:34.717" Title="When is $G/H = G/K$?  In particular, when are two Lie groups with isomorphic Lie algebras isomorphic?" Tags="&lt;group-theory&gt;&lt;lie-groups&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="9001" PostTypeId="2" ParentId="8996" CreationDate="2010-11-05T01:28:10.663" Score="5" ViewCount="0" Body="&lt;p&gt;You're nearly there: with your notations, $\mathrm{Hom}_R(A,B)$ injects in $\mathrm{Hom}_R(R^n,B) \simeq B^n$, and $B^n$ is finitely generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed $\mathrm{Hom}_R(A,B) = \left\{f \in \mathrm{Hom}_R(R^n,B)\ |\ I \subset \ker f \right\}$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-05T01:28:10.663" CommentCount="2" />
  <row Id="9002" PostTypeId="1" AcceptedAnswerId="9003" CreationDate="2010-11-05T01:34:38.660" Score="0" ViewCount="57" Body="&lt;p&gt;I was experimenting with binary operators, and devised a commutative operator with function $f$ such that the operator mapped $f \colon \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{R}$ in order to make up some problem (e.g. if $f(3,10)=30$ and $f(4,3)=10$, find $f(4,6)$, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are there any functions $f$ other than $a \otimes b = f(a,b) = qab+r(a+b)$ where $q,r \in \mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it always true that the operator is also distributive if $a \otimes 0 = 0$?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-11-05T01:41:52.723" Title="Some stuff with binary operators." Tags="&lt;abstract-algebra&gt;&lt;fun-problem&gt;" AnswerCount="1" />
  <row Id="9003" PostTypeId="2" ParentId="9002" CreationDate="2010-11-05T01:41:52.723" Score="3" ViewCount="0" Body="&lt;p&gt;Every such $f$ has the form $f(a, b) = g(a, b) + g(b, a)$ for some arbitrary function $g : \mathbb{Z} \times \mathbb{Z} \to \mathbb{R}$.  And no.  For example, take $f(a, b) = a^2 b^2$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-05T01:41:52.723" CommentCount="3" />
  <row Id="9004" PostTypeId="2" ParentId="8969" CreationDate="2010-11-05T01:44:33.147" Score="1" ViewCount="0" Body="&lt;p&gt;First, you don't have the freedom to choose $\lambda$.  (Maybe you already realize that?)  In order for $P(X = x_i)$ to be a true probability distribution, you must have &#xA;$$\lambda = \left(\sum_{x_i} \frac{P(Y = x_i)}{x_i}\right)^{-1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, if $Y$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_distribution&quot; rel=&quot;nofollow&quot;&gt;geometric&lt;/a&gt; with success probability $p$, and $P(X = x_i) = \lambda P(Y = x_i)/x_i$ then $X$ does have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_distribution&quot; rel=&quot;nofollow&quot;&gt;logarithmic distribution&lt;/a&gt; with parameter $q = 1-p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why:  $P(Y = k) = (1-p)^{k-1}p$, for $k = 1, 2, \ldots$.  Thus &#xA;$$P(X = k) = \lambda \frac{P(Y = k)}{k} = \lambda \frac{q^{k-1} p}{k} = \frac{\lambda p}{q} \frac{q^k}{k}.$$&#xA;The expression on the right is the logarithmic probability mass function with $\lambda = \frac{-q}{p \ln p}$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-05T01:44:33.147" />
  <row Id="9005" PostTypeId="1" AcceptedAnswerId="9006" CreationDate="2010-11-05T01:56:09.620" Score="0" ViewCount="25" Body="&lt;p&gt;I have a piecewise 2-input function that I would like to repeatedly tile diagonally across a grid. See this image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pLaFa.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, I know the equations of the pink, blue and green areas in the range 0 - 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that i'm having trouble finding a way to tile the function diagonally across like shown in the image by transforming the inputs. I tried using mod (x1 = x1 mod 2, x2 = x2 mod 2) to transform the input, and it worked somewhat, but it was wrong at grids (2-3, 1-2), (1-2, 2-3), (4-5, 3-4) and (3-4, 4-5). Any help is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2010-11-05T02:17:51.743" Title="Repeating Tiled Piecewise Function Help" Tags="&lt;geometry&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9006" PostTypeId="2" ParentId="9005" CreationDate="2010-11-05T02:17:51.743" Score="3" ViewCount="0" Body="&lt;p&gt;Let $F$ be the original functions, $f$ the new one. Define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) = F(x-m,y-m)$ where $m = \min(\lfloor x \rfloor, \lfloor y \rfloor)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T02:17:51.743" CommentCount="1" />
  <row Id="9007" PostTypeId="1" CreationDate="2010-11-05T02:28:29.910" Score="3" ViewCount="109" Body="&lt;p&gt;The &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_mean&quot; rel=&quot;nofollow&quot;&gt;harmonic mean&lt;/a&gt;&lt;/em&gt; of a finite set of positive real numbers $\{x_1, x_2, \ldots, x_n\}$ is defined to be $$H(\{x_1, x_2, \ldots, x_n\}) = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_mean&quot; rel=&quot;nofollow&quot;&gt;logarithmic mean&lt;/a&gt;&lt;/em&gt; of two distinct positive real numbers $a$ and $b$ is defined to be $$L(a,b) = \frac{b - a}{\ln b - \ln a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the first applications of integration that students often see is the extension of the arithmetic mean of a finite set of real numbers to the arithmetic mean of a function $f(x)$ on a continuous interval $[a,b]$ via $\frac{1}{b-a} \int_a^b f(x) dx.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, you can extend the harmonic mean so that it applies to a positive function $f(x)$ over a continuous interval $[a,b]$.  You get &#xA;$$\frac{b-a}{\int_a^b \frac{dx}{f(x)}} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you take $f(x) = x$ you obtain the harmonic mean of the continuous interval $[a,b]$.  This is &#xA;$$H([a,b]) = \frac{b-a}{\int_a^b \frac{1}{x}  dx} = \frac{b-a}{\ln b - \ln a} = L(a,b).$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is this: Is there an intuitive reason why $H([a,b]) = L(a,b)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For comparison purposes, note that if $A$ denotes the arithmetic mean, then $A([a,b]) = A(a,b) = \frac{a+b}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-05T04:02:30.157" Title="Harmonic mean and logarithmic mean" Tags="&lt;real-analysis&gt;&lt;intuition&gt;&lt;average&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9008" PostTypeId="1" CreationDate="2010-11-05T02:43:20.417" Score="6" ViewCount="172" Body="&lt;p&gt;Let F be a strictly increasing function on S, a subset of the real line. If you know that F(S) is closed, prove that F is continuous.&lt;/p&gt;&#xA;" OwnerUserId="2319" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-05T04:28:40.897" LastActivityDate="2010-11-05T22:23:32.930" Title="If F is strictly increasing with closed image, then F is continuous" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9009" PostTypeId="2" ParentId="8969" CreationDate="2010-11-05T02:48:44.750" Score="2" ViewCount="0" Body="&lt;p&gt;if you rewrite the equation as $P(Y = x) = xP(X = x)/\lambda$, then the distribution of $Y$ is called the length-biased distribution for $X$. it arises, for example, if one has a bunch of sticks in a bag and reaches in and selects one at random - where the probability a particular stick is selected is proportional to its length. if the lengths of the sticks are realizations of the random variable $X$, the distribution of the length of the selected stick is that of $Y$.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-11-05T02:48:44.750" />
  <row Id="9010" PostTypeId="1" AcceptedAnswerId="9020" CreationDate="2010-11-05T02:57:52.910" Score="4" ViewCount="223" Body="&lt;p&gt;I hear people use these words relatively interchangeably.  I'd believe that any differentiable manifold can also be made into a variety (which data, if I understand correctly, implicitly includes an ambient space?), but it's unclear to me whether the only non-varietable manifolds should be those that don't admit smooth structures.  I'd hope there's more to it than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard too that affine schemes are to schemes local coordinates are to manifolds, so maybe my question should be about schemes instead -- I don't even know enough to know...&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-05T15:02:45.500" Title="What is the difference between a variety and a manifold?" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="9011" PostTypeId="1" AcceptedAnswerId="9015" CreationDate="2010-11-05T02:58:44.103" Score="0" ViewCount="52" Body="&lt;p&gt;I've been wondering if/when it's possible to &quot;truncate&quot; a series.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the closed form for the series of naturals is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{(x-1)^2}$ = $1 + 2z + 3x^2 + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;truncated&quot; form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1-(n+1)x^n+n\cdot x^{n+1}}{(x-1)^2}$ = $1 + 2z + 3x^2 + \cdots + n\cdot x^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The closed form of integers of the form $2^n$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{1-2x}$ = $1 + 2x + 4x^2 + \cdots + (2x)^n + \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;truncated&quot; form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1 - (2x)^n}{1 - 2x}$ = $1 + 2x + 4x^2 + \cdots + (2x)^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Search&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could anyone help with finding appropriate resources that address &quot;truncating&quot; a series?  It would be great if there is some work that may include math that relates to finding truncated versions of series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a very good background in math, so anything that doesn't require a lot of background would be very beneficial.  However, I am very interested in learning, and ANY resource that touches upon this topic would be very beneficial and interesting for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd just like to know where I can learn more about this and related topics.  Again, I'm very interested in being able to produce a closed form or recurrence for a &quot;piece&quot; of a series.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T03:42:59.063" LastActivityDate="2010-11-05T03:42:59.063" Title="Constructing finite versions of arbitrary series" Tags="&lt;sequences-and-series&gt;&lt;intuition&gt;&lt;recursive&gt;" AnswerCount="1" />
  <row Id="9012" PostTypeId="2" ParentId="9008" CreationDate="2010-11-05T03:08:52.760" Score="3" ViewCount="0" Body="&lt;p&gt;Here's an approach by contraposition.  Let $f$ be a strictly increasing function discontinuous at $x\in S$.  Then $f(x)\lt\lim_{y\to x+}f(y)$ or $f(x)\gt\lim_{y\to x-}f(y)$ (or both).  Suppose $f(x)\lt\lim_{y\to x+}f(y)$.  Then you can show that $\lim_{y\to x+}f(y)$ is in $\overline{f(S)}\setminus f(S)$, so $f(S)$ is not closed.  To see that the limit is in the closure of $f(S)$ is a straightforward unwinding of definitions. It's not in $f(S)$ because for every $z\lt x$, $f(z)\lt f(x)\lt\lim_{y\to x+}f(y)$, and for every $z\gt x$, $\lim_{y\to x+}f(y)\lt f(z)$.    (Similarly on the other side.  It may help to keep in mind that $\lim_{y\to x-}f(y)=\sup_{y\lt x}f(y)$ and $\lim_{y\to x+}f(y)=\inf_{y\gt x}f(y)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a way that doesn't use contraposition (although there is a bit of contradiction).  Let $x$ be an element of $S$, and let $x_1,x_2,\ldots$ be an increasing sequence in $S$ converging to $x$.  Then $f(x_1),f(x_2),\ldots$ is an increasing sequence bounded above by $f(x)$, and hence it converges.  Since $f(S)$ is closed, there is a $z\in S$ such that $f(x_n)\to f(z)$ as $n\to \infty$.  I claim that $z=x$.  If $z$ were bigger than $x$, then we'd have $f(x_n)\leq f(x)\lt f(z)$ for all $n$, making the convergence impossible.  If $z$ were smaller than $x$, we'd have $z$ smaller than $x_n$ for some $n$, so $f(z)\lt f(x_n)\leq f(x_{n+1})\leq\cdots$, again making the convergence impossible.  So $z=x$ as claimed.  This implies that the left-hand limit of $f$ at $x$ exists and equals $f(x)$.  Similarly on the right, so $f$ is continuous.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-05T03:58:39.427" LastActivityDate="2010-11-05T03:58:39.427" CommentCount="8" />
  <row Id="9013" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T03:09:17.540" Score="2" ViewCount="0" Body="&lt;p&gt;If you are French, a variety &lt;em&gt;is&lt;/em&gt; a manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, the connection is a bit more subtle.  The correct analogue of a manifold is a scheme of finite type over a field.  That is, it has a compatible covering by ring spectra (which is similar to the requirement that a manifold have a covering by compatible copies of euclidean space), and it lives over a field (the base space is an honest &quot;point&quot;). Here, I am being intentionally vague about &quot;compatibility&quot;, since this would lead me into a discussion of &lt;em&gt;sheaves&lt;/em&gt;, which I would rather not discuss unless you feel it's necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical notion of a variety much more resembles the classical notion of a manifold (see, for instance, Milnor's book &quot;Topology from a differentiable viewpoint&quot;).  The classical notion of a manifold is simply a locally closed subset of $\mathbf{P}^n(\mathbf{R})$ (or just $\mathbf{R}^n$, although looking at locally-closed subsets of projective space makes the analogy that much stronger), and the notion of a variety (quasi-projective, I guess) comes from considering locally closed subsets of a projective space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, there is a theorem of Serre called GAGA (Géométrie algébrique et géométrie analytique), which says that in nice cases, we can &quot;replace&quot; complex varieties with complex-analytic manifolds to &quot;do&quot; cohomology and get a better picture of the real underlying space.  (Here, by &lt;em&gt;complex variety&lt;/em&gt;, we simply mean finite-type schemes over $\mathbf{C}$)&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-05T15:02:45.500" LastActivityDate="2010-11-05T15:02:45.500" CommentCount="8" />
  <row Id="9014" PostTypeId="2" ParentId="8940" CreationDate="2010-11-05T03:15:39.630" Score="1" ViewCount="0" Body="&lt;p&gt;Think of this square n x n matrix as a transformation T on a vector space V. This transformation sends $V^{n} \rightarrow V^{n}$. Now, if the determinant of a matrix is equal to 0, we know it isn't invertible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since our vector space is finite-dimensional, a transformation that has a non-invertible matrix representation is neither onto nor one-to-one. However, if all our columns were linearly independent vectors in $V^n$, they would span $V^n$ and T would be onto. One of the columns must therefore be a linear combination of the others. &lt;/p&gt;&#xA;" OwnerUserId="647" LastActivityDate="2010-11-05T03:15:39.630" />
  <row Id="9015" PostTypeId="2" ParentId="9011" CreationDate="2010-11-05T03:21:36.543" Score="3" ViewCount="0" Body="&lt;p&gt;Truncating to the first $n$ terms is the same as taking the Hadamard product with $\frac{1 - x^n}{1 - x}$, so if you can compute the corresponding integral then you're home free.  In the special case that the original series is rational you can also do it using partial fraction decomposition, and then you are actually done by using appropriate generalizations of the two formulas you've already listed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general I see no reason for truncations to have nice closed forms.  In particular I see no reason for the truncations of a series like $- \ln (1 - x)$ to have nice closed forms, since that would imply that the harmonic series has a nice closed form.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-05T03:21:36.543" CommentCount="2" />
  <row Id="9016" PostTypeId="2" ParentId="8966" CreationDate="2010-11-05T03:36:04.160" Score="0" ViewCount="0" Body="&lt;p&gt;The algebraic $\frac{1}{12}(17 + \sqrt{97})$ is not a root of the equation&#xA;\begin{eqnarray}&#xA;-2 x^3 + 10 x^2 - 17 x + 8 = (2 x^2) (5 x - x^3)^{1/3}&#xA;\end{eqnarray}&#xA;Plugging it in, you find that the left hand side is real and equal to &#xA;\begin{eqnarray}&#xA;\tfrac{1}{216}(-149 - 37 \sqrt{97}) = -2.37689 \dots&#xA;\end{eqnarray}&#xA;The right side is&#xA;\begin{eqnarray}&#xA;\tfrac{1}{432} (\tfrac{1}{2}( 595 - 61 \sqrt{97})^{1/3} (17 + \sqrt{97})^2 = 1.18844 \dots + i 2.05845 \dots&#xA;\end{eqnarray}&#xA;Note: $595 &amp;lt; 61 \sqrt{97}$. I think the ambiguity lies in the fact that we have not used the third-roots of unity. Numerical computations aside, just plot the two functions. The RHS is a positive function defined only in the I and II quadrants. The LHS is cubic. There is only one real intersection point.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T04:09:41.780" LastActivityDate="2010-11-05T04:09:41.780" CommentCount="4" />
  <row Id="9017" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T03:37:44.330" Score="2" ViewCount="0" Body="&lt;p&gt;A variety is usually defined to be the zero set of some polynomials. I don't know much about this, but I think people usually call manifolds that can be realized as varieties &quot;algebraic manifolds&quot;. Take for example the polynomial $x^2+y^2-1$, then the zero set of this is $x^2+y^2=1$, or the circle $S^1$. So that is an algebraic manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be varieties that are not manifolds, for instance, $y^2-x^2(x+1)=0$ is a &quot;nodal cubic&quot; and so it has a singularity at $(0,0)$. It can't be a manifold because it looks like &quot;X&quot; a cross at the origin so is not homeomorphic locally to $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very useful theorem, sometimes called the Regular Level Set Theorem, which says that if you have a smooth map $M\to N$, then the level set of a regular value is a smooth manifold. So in this particular case, our polynomial $f:\mathbb{R}^n \to \mathbb{R}$ is a smooth map, and we care about whether or not zero is a regular value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This amounts to checking whether or not the matrix $(\frac{\partial f}{\partial x_1}(c), \ldots , \frac{\partial f}{\partial x_n}(c))$ has rank 1 at all $c$ such that $f(c)=0$. i.e. if at least one of the partial derivatives is non-zero. This can easily be extended to work if your variety is defined by a bunch of polynomials $\mathbb{R}^n\to\mathbb{R}^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a name, called the Jacobian criterion, and for varieties it says that if the Jacobian has full rank, then the variety is non-singular. But these two things are equivalent (Jacobian criterion for non-singular variety and the level set theorem to be a smooth manifold). Thus every non-singular variety is a smooth manifold (Just to be clear, this was considering &quot;variety&quot; to mean a zero set of a collection of polynomials over $\mathbb{R}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to reiterate, some varieties are manifolds (if the defining polynomials satisfy a certain condition on partial derivatives) and some are not. Is every smooth manifold a variety? I think not, but it seems harder. A quick Google search turns up a result that seems to say every smooth compact manifold is algebraic.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-11-05T03:37:44.330" CommentCount="1" />
  <row Id="9018" PostTypeId="2" ParentId="8975" CreationDate="2010-11-05T03:48:55.767" Score="3" ViewCount="0" Body="&lt;p&gt;Here's how I would explain limits:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the function $f(x) = x^{2}$. We might want to ask ourselves: &quot;What is $\lim_{x \rightarrow 3} f(x)$?&quot; Put simply: As $x$ gets near 3, where happens to our function? Clearly, our function goes to 9, which is $3^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must understand, however, that a limit is a much more subtle argument. In our case, we could just &quot;plug in&quot; 3. But suppose we wanted to take $\lim_{x \rightarrow 0} \frac{1}{x}$ or $\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$. In these cases, we cannot simply &quot;plug in&quot; our value for x, and we must understand what a limit really is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrow $\rightarrow$ in the limit sign means &quot;approaching.&quot; In our first example, we wanted to find out what happens as x approaches 3. Let's look at this ourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=1$, then $x^2=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2$, then $x^2=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2.5$, then $x^2=6.25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2.9$, then $x^2=8.41$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=2.99$, then $x^2=8.9401$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As x gets closer and closer to 3, $x^2$ (our function) gets closer and closer to 9.&#xA;We can also come from the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=5$, then $x^2=25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=4$, then $x^2=16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=3.5$, then $x^2=12.25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=3.1$, then $x^2=9.61$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x=3.01$, then $x^2=9.0601$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let me introduce the terminology of epsilons $\epsilon$ and deltas $\delta$. The delta is how far away we are from our chosen limit value for x. In this case, the delta was how far away we were from 3. When we set x=3.5, delta was equal to 0.5. When x=3.1, delta was equal to 0.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correspondingly, epsilon was how far away we were from the value of the limit: 9. When we set x=3.1, and our delta was 0.1, we got $f(x) = 9.61$, so the epsilon was 0.61.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And therein lies the fundamental idea of a limit. We can make the epsilon smaller simply by shrinking the delta. The closer your x is to 3, the closer your f(x) is to 9. If you wanted to be within 0.001 of 9 (i.e. $\epsilon = 0.001$), you'll need to be really close to x, and have a really small $\delta$. To be precise, $\delta = 0.00017$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We therefore say that the limit of $f(x) = x^2$ as x approaches 3 is equal to 9, because we can make the output of the function be as close to 9 as we want, simply by constricting the input to be close to 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Bael9.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above picture provides a great visual description. We have some curvy function $f(x)$, and we want to say that the limit at $x=a$ is equal to $L$. This can be shown because if we narrow the lines $a-\delta$ and $a+\delta$, thereby narrowing our input, we narrow the lines $L+\epsilon$ and $L-\epsilon$, constricting our output. We therefore must conclude that the limit is $L$, because the value of our function approaches $L$ and gets closer and closer to it the closer we get to $x=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this explanation has helped. &lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="647" LastEditorDisplayName="" LastEditDate="2010-11-05T04:56:10.767" LastActivityDate="2010-11-05T04:56:10.767" CommentCount="4" />
  <row Id="9019" PostTypeId="2" ParentId="8992" CreationDate="2010-11-05T03:49:50.617" Score="3" ViewCount="0" Body="&lt;p&gt;There are a couple of ways to think about it; as acarchau says, since you are concerned only about the limit, and the &quot;tail&quot; has the same limit, it is enough to consider the tail. Equivalently: &lt;em&gt;the limit of a sequence depends only on what it is doing &quot;eventually&quot;&lt;/em&gt;. Any amount of &quot;weird&quot; stuff that happens up to a finite bound $N$ is irrelevant (that is, if the sequence is &quot;nice&quot; for all $n\geq N$, then that is enough, because as far as the limit is concerned, we don't care what happens at $n=1$, $n=2,\ldots,n=N-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, consider the sequence of functions $\{g_n\}$, where $g_n(x) = f_{N+n}(x)$. Then the hypothesis of Lebesgue's Theorem hold exactly as written for $g_n(x)$, so we know that we can get the conclusion for the $g_n$. Now rewrite $g_n$ as $f_{N+n}$, and do a change of variable $m=N+n$; then $m\to\infty$ as $n\to\infty$, so the limit of $f_{N+n}$ as $n\to\infty$ is the same as the limit of $f_m$ as $m\to\infty$, giving the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all in all, it's always the same moral: what happens at finitely many terms &quot;at the beginning&quot; does not affect the end behavior of a sequence. Only the &quot;eventual&quot; behavior matters.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-05T03:49:50.617" CommentCount="1" />
  <row Id="9020" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T03:50:24.993" Score="4" ViewCount="0" Body="&lt;p&gt;In English (as opposed to French, in which language variety and manifold are synonyms) the word &lt;em&gt;variety&lt;/em&gt; is short for &lt;em&gt;algebraic variety&lt;/em&gt;.  The main differences, then, between (algebraic) varieties and (smooth) manifolds are that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Varieties are cut out in their ambient (affine or projective) space as the zero loci of polynomial functions, rather than simply as the zero loci of smooth functions.   This gives them a more rigid structure.  (Here I am thinking just of quasi-projective varieties; there are objects that people call varieties which can't be immersed into projective space, but there is no need to think about them when you are just learning the subject.  Also, a manifold need not be regarded as lying in an ambient Eulcidean space, but can always be immersed into one, and can then be thought of as being cut out as the zero locus of smooth&#xA;functions.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rigidity of varieties is reflected in the definition of isomorphism: we define two varieties to be isomorphic if we can find polynomial maps giving rise to mutually inverse bijections from one to the other, while two manifolds are isomorphic (i.e. diffeomorphic) if we can find smooth maps giving rise to mutually inverse bijections between them.   It turns out, for example, that&#xA;the only diffeomorphism invariant of a compact connected surface is its genus $g$, while if we look at smooth connected  projective curves over the complex numbers (which, when we forget the variety structure and just think of them as manifolds, &lt;em&gt;are&lt;/em&gt;&#xA;compact connected surfaces --- note that one complex dimension gives two real&#xA;dimensions) then the genus is not a complete invariant.  For a fixed genus $g \geq 2$,&#xA;there is a $6g-6$-dimensional family of non-isomorphic curves of genus $g$.&#xA;(When $g = 1$ there is a $2$-dimensional family, and when $g = 0$, the&#xA;variety structure is actually uniquely determined.  Also, by &quot;dimension&quot; here&#xA;I mean real dimension; but these families have their own natural algebraic variety structures, of half the dimension --- i.e. there is a $3 g - 3$ dimensional variety parameterizing isomorphism classes of genus $g$ curves&#xA;when $g \geq 2$.  Again the halving of dimension reflects the difference between real and complex dimension.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)  Varieties can admit singularities, whereas we stipulate that manifolds be non-singular (i.e. locally Euclidean).   Here it is useful to think about the fact that the critical locus of a (collection of) smooth function(s) can be pretty nasty, and so if we consider the zero loci of smooth functions and allow singularities, we will allow extremely nasty singularities.  On the other hand,&#xA;the critical locus of a (collection) of polynomial(s) is not so bad (e.g. it is always of codimension at least one in the zero locus), and so allowing singularities in the theory turns out to be okay (and in fact to be more than okay; it turns out to be one of the more powerful features of algebraic geometry).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-05T03:50:24.993" CommentCount="16" />
  <row Id="9021" PostTypeId="2" ParentId="8990" CreationDate="2010-11-05T03:54:38.587" Score="1" ViewCount="0" Body="&lt;p&gt;The function is majority (with tie-breaking): the expression is true iff at least two of $A,B,C,D$ are true. This follows from your last term by opening up the parentheses, and can also be checked directly.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T03:54:38.587" />
  <row Id="9022" PostTypeId="2" ParentId="9007" CreationDate="2010-11-05T04:02:30.157" Score="3" ViewCount="0" Body="&lt;p&gt;The harmonic mean can be generalized to an arbitrary invertible function $w$: define the mean of $x_1,\ldots,x_n$ to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w^{-1}\left(\frac{w(x_1) + \cdots + w(x_n)}{n}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logarithmic mean can also be generalized re its mean value interpretation for an arbitrary function $f$ such that $f'$ is invertible: define the mean of $x,y$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f')^{-1}\left(\frac{f(x) - f(y)}{x - y}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can generalize your observation by taking $w = f'$. The corresponding mean, before taking $w^{-1}$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{y-x}\int_x^y f'(t) \, dt = \frac{f(y)-f(x)}{y-x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the (former) mean of the interval with weight $f'$ is the same as the (latter) mean with respect to $f$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T04:02:30.157" CommentCount="5" />
  <row Id="9023" PostTypeId="1" CreationDate="2010-11-05T04:18:43.593" Score="2" ViewCount="64" Body="&lt;p&gt;This may be a question whose answer is lost in the mists of time, but why is the elliptical integral of the first kind denoted as $F(\pi/2,m)=K(m)$ when that of the second kind has $E(\pi/2,m)=E(m)$? It's not very consistent! Aside from convention, is there anything stopping us from rationalising these names a little?&lt;/p&gt;&#xA;" OwnerUserId="2961" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T11:09:47.847" LastActivityDate="2010-11-05T11:09:47.847" Title="Inconsistent naming of elliptic integrals" Tags="&lt;soft-question&gt;&lt;notation&gt;&lt;special-functions&gt;&lt;math-history&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9024" PostTypeId="2" ParentId="9023" CreationDate="2010-11-05T04:22:13.090" Score="2" ViewCount="0" Body="&lt;p&gt;This discrepancy seems to come from the various conventions in defining the nome $q$ of complex lattice $\Lambda_{\tau} = \mathbb{Z} \oplus \tau \mathbb{Z}$ as one of the four $e^{2 \pi i \tau}$, $e^{\pi i \tau}$, $e^{2 i \tau}$ or $e^{i \tau}$, where $\tau \in \mathbb{H}$ is the lattice parameter. Investigate the vast literature on theta functions, and you'll see exactly what I mean by the problem of too many &quot;&lt;em&gt;conventions&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T07:26:25.257" LastActivityDate="2010-11-05T07:26:25.257" CommentCount="1" />
  <row Id="9025" PostTypeId="2" ParentId="8977" CreationDate="2010-11-05T04:32:39.590" Score="0" ViewCount="0" Body="&lt;p&gt;I'm going to ignore arc AC and address what determines arcs FD and BE based on the points A, B, C, D, E, and F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two circles with centers X and Y are tangent at P, then X, Y, and P are collinear.  So, line AF and line CD both pass through the center of the circle containing arc FD and line AB and line CE both pass through the center of the circle containing arc BE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have coordinates for all of these points, you can write equations of the line and algebraically find their points of intersection.  Given the coordinates of the centers of the circles, you can determine their radii.  Given the coordinates of the centers of the circles and the endpoints of the arcs, you can determine the measure of the central angles that subtend the arcs.  From the radii and central angles, you can determine the arc lengths.  (It's hard to be more specific without introducing a bunch of variables, but I can add more detail if you are not sure how to carry out any of these steps.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-11-05T04:32:39.590" />
  <row Id="9026" PostTypeId="1" CreationDate="2010-11-05T04:58:40.620" Score="6" ViewCount="174" Body="&lt;p&gt;I tried to do some math in &lt;a href=&quot;http://www.geekality.net/?p=751&quot; rel=&quot;nofollow&quot;&gt;a blog post&lt;/a&gt; of mine and came to one with a floor function. I wasn't sure how to deal with it so I just ignored it, and then added the ceiling function in my final equation as that seemed to give me the result I wanted. I'm wondering what is the correct way of handling these functions in equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;G(n) &amp;amp;= \left\lfloor n\log{\varphi}-\dfrac{\log{5}}{2}\right\rfloor+1 \\&#xA;n\log{\varphi} &amp;amp;= G(n)+\dfrac{\log{5}}{2}-1 \\&#xA;n &amp;amp;= \left\lceil\dfrac{G(n)+\dfrac{\log{5}}{2}-1}{\log\varphi}\right\rceil&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should have done this in a correct way? How do I work with the ceiling and floor functions when I shuffle around with equations?&lt;/p&gt;&#xA;" OwnerUserId="3147" LastActivityDate="2010-11-05T17:51:20.293" Title="How do you handle the floor and ceiling function in an equation?" Tags="&lt;algebra&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9027" PostTypeId="2" ParentId="9026" CreationDate="2010-11-05T05:26:29.833" Score="5" ViewCount="0" Body="&lt;p&gt;You can replace $\lfloor x \rfloor$ with $x - \theta$, where $\theta \in [0,1)$ is some unknown quantity. Similarly, $\lceil x \rceil = x + \theta$ (a different $\theta$ within the same range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another helpful identity is $\lfloor x \rfloor + n = \lfloor x + n \rfloor$ for any integer $n$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T05:26:29.833" />
  <row Id="9028" PostTypeId="2" ParentId="9026" CreationDate="2010-11-05T05:29:54.760" Score="6" ViewCount="0" Body="&lt;p&gt;Your final expression gives you the number you want.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to your blog post, you're looking for the smallest integer $n$ (i.e., the &quot;first Fibonacci number with 1000 digits&quot;) that satisfies &#xA;$$G(n) = \left\lfloor n \log \varphi - \frac{\log 5}{2} \right\rfloor + 1.$$&#xA;There may, of course, be more than one integer $n$ for which this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of the floor function, the values of $n$ that satisfy this are the values that satisfy $$G(n) - 1 \leq n \log \varphi - \frac{\log 5}{2} &amp;lt; G(n),$$&#xA;which, since $\log \phi &gt; 0$, are the values that satisfy&#xA;$$\frac{G(n) + \frac{\log 5}{2}}{\log \varphi} - \frac{1}{\log \varphi} \leq n  &amp;lt; \frac{G(n) + \frac{\log 5}{2}}{\log \varphi}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\frac{1}{\log \varphi} \approx 4.78$, there are either four or five integers in this interval.  But the smallest one is obtained by taking the ceiling of the lower endpoint of the interval; i.e.,&#xA;$$\left\lceil\frac{G(n) + \frac{\log 5}{2} - 1}{\log \varphi}\right\rceil.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this argument also apparently shows that there are either four or five Fibonacci numbers that have a given number of digits.  (Except in the single-digit case, where there are six (not counting 0).  But your formula for $G(n)$ doesn't hold when $n=1$, so we shouldn't expect this calculation to be true in the single-digit case anyway.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-05T17:51:20.293" LastActivityDate="2010-11-05T17:51:20.293" />
  <row Id="9029" PostTypeId="2" ParentId="8990" CreationDate="2010-11-05T05:38:12.020" Score="2" ViewCount="0" Body="&lt;p&gt;To go along with my comment above, the minimal form is what you arrived at. Using a Karnaugh map, you will get $AB+CD+AC+AD+BC+BD$, which factors to your answer.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-05T05:38:12.020" />
  <row Id="9030" PostTypeId="2" ParentId="9026" CreationDate="2010-11-05T05:41:31.643" Score="2" ViewCount="0" Body="&lt;p&gt;Observe that&#xA;\begin{eqnarray}&#xA;G(n) = \left \lfloor n \log \varphi - \log \sqrt{5} \right \rfloor + 1 = \left \lceil n \log \varphi - \log \sqrt{5} \right \rceil&#xA;\end{eqnarray}&#xA;and write&#xA;\begin{eqnarray}&#xA;\left \lceil \frac{G(n)}{\log \varphi} + \log_{\varphi} \sqrt{5} \right \rceil &amp;amp; = &amp;amp; \left \lceil \tfrac{1}{\log \varphi} \left \lceil n \log \varphi - \log \sqrt{5}  \right \rceil + \log_{\varphi} \sqrt{5} \right \rceil = n.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T06:09:30.330" LastActivityDate="2010-11-05T06:09:30.330" />
  <row Id="9031" PostTypeId="2" ParentId="8975" CreationDate="2010-11-05T05:56:06.867" Score="-1" ViewCount="0" Body="&lt;p&gt;We say that the limit of $f$ at $a$ is $\ell$ means: For any $x$ in the domain of $f$ other than $a$ itself, $$\forall (\varepsilon &gt;0) \exists (\delta &gt;0) d(x,a)&amp;lt;\delta \Rightarrow d(f(x),\ell)&amp;lt;\varepsilon,$$ where $d$ is the distance function for the metric space $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, for any open neighborhood of $U$ of $\ell$, we can find neighborhood $V$ of $a$ such that  $f(V)\subseteq U$.  Indeed, there is some subtlety here about the domain of $f$ (in particular, $f$ need not be defined &lt;em&gt;at&lt;/em&gt; $a$, only on a subspace whose closure contains $a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, let $X$ and $Y$ be topological spaces, and let $\Omega\subseteq X$ be a subspace such that $a\in Cl_X(\Omega-\{a\})$.  Then we say that $\ell$ is &lt;strong&gt;a&lt;/strong&gt; limit point of $f$ at $a$ (in this generality, limits need not be unique) provided that for any open neighborhood $U$ of $\ell$, there exists an open neighborhood $V\subseteq X$ of $a$ such that $f(V\cap \Omega)\subseteq U$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that no matter how small a neighborhood we give around $\ell$, we can always find a small enough neighborhood of $a$ such that $f$ maps this neighborhood of $a$ into the neighborhood of $\ell$.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-05T07:54:54.450" LastActivityDate="2010-11-05T07:54:54.450" CommentCount="4" />
  <row Id="9032" PostTypeId="1" CreationDate="2010-11-05T06:11:44.960" Score="3" ViewCount="127" Body="&lt;p&gt;Suppose I have two tables, each of unknown size, and I'd like to estimate the average of their true sizes. I hire 2 contractors: one guarantees good precision (i.e., her measurement normally-distributed about the true value, with a standard deviation of 5mm), while the other is a dimwit (i.e., his measurement is also unbiased, but with a standard deviation of 100mm). What's the optimal way to combine the two measurements to form a final estimate of their true average size? The formal way to ask this question is, &quot;Given a sample from a normally-distributed random variable with known variance, and another sample from a second normally-distributed random variable with known (but potentially different) variance, what is the best guess as to the expected value of the average of the two random variables?&quot; Is the answer &quot;Just average them.&quot;? Ideally, I want to prove the answer, whatever it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm not sure if it was clear, but one contractor measures table 1, while the other contractor measures table 2. &lt;/p&gt;&#xA;" OwnerUserId="3298" LastEditorUserId="3298" LastEditorDisplayName="" LastEditDate="2010-11-09T04:54:36.630" LastActivityDate="2010-11-09T05:20:47.153" Title="Optimally combining samples to estimate averages" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;statistics&gt;&lt;optimization&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="9035" PostTypeId="1" AcceptedAnswerId="9046" CreationDate="2010-11-05T06:57:16.197" Score="9" ViewCount="141" Body="&lt;p&gt;Im giving a small talk for a combinatorics class on the &lt;a href=&quot;http://garden.irmacs.sfu.ca/?q=op/erdos_szekeres_conjecture&quot; rel=&quot;nofollow&quot;&gt;Erdos-Szekeres conjecture&lt;/a&gt; regarding the happy ending problem (the paper is focused on recent work regarding the conjecture).  I always find that when the audience is not up for a long technical proof, and they aren't familiar with the field (in this case combinatorial geometry) that it is important to provide strong motivation for why the problem is important, and perhaps mention a couple interesting related results that have applications in other fields, or to other problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Happy Ending problem, I am aware of connections to the $n$-hole problem for points in general position, Ramsey theory (via the original paper by erdos and szekeres in 1935), and the generalization of the E&amp;amp;S theorem for convex bodies.  But Im not sure if these are good things to mention since the class does not know any Ramsey theory and convex bodies are probably too abstract to mention offhand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: does anyone know of other interesting related results to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Happy_Ending_problem&quot; rel=&quot;nofollow&quot;&gt;Happy Ending Problem, and the corresponding theorem and conjecture.&lt;/a&gt; Or, can you tell me succinctly why or why not this conjecture is important in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will accept anything along the lines of similar geometrically or combinatorially, directly related, or mentioned in a paper involving the E&amp;amp;S conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original paper from 1935 is a goldmine of things like the ordered pigeonhole principle, but Im interested in things I havent read yet.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-11-05T11:23:15.153" Title="Interesting results related to The Happy Ending Problem" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;discrete-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9036" PostTypeId="2" ParentId="8501" CreationDate="2010-11-05T07:12:59.970" Score="3" ViewCount="0" Body="&lt;p&gt;Without sounding biased in any way, I would say Cambridge/Boston is a good choice for you. In no particular order there is Harvard University, MIT, Boston University, Boston College, Northeastern University, Brandeis University, Tufts University, Bently University, University of Massachusetts at Boston, Curry College, Eastern Nazarene College, Pine Manor College, Hellenic College, Lesley University, Wheelock College, Lasell College, Simmons University, Cambridge College and Bunker Hill Community College (and many, many more) within the metropolis. See &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_colleges_and_universities_in_metropolitan_Boston&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_colleges_and_universities_in_metropolitan_Boston&lt;/a&gt; for a complete list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number of these institutions offer extension programs (with open enrollment and classes in the evening or weekends) suitable for life-long learners and aspiring amateur mathematicians. For example, the &lt;em&gt;Masters for Mathematics Teaching Program&lt;/em&gt; at Harvard University offers courses in all major mathematics subject areas, taught by many instructors which hold separate positions in the university (like adjunct/junior faculty, preceptors, senior lecturers, post-doctoral or teaching fellows and even a senior graduate student).&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-05T07:29:42.070" LastActivityDate="2010-11-05T07:29:42.070" />
  <row Id="9037" PostTypeId="1" CreationDate="2010-11-05T07:31:37.510" Score="2" ViewCount="143" Body="&lt;p&gt;Is there a surjective map between the (class of) ordinal numbers On and the set No (Conway's surreal numbers) and is it constructable, In Conway's system we have for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_0 = &amp;lt; 0,1,2,3,... | &gt; $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon = &amp;lt; 0 | 1, 1/2, 1/4, 1/8, ... &gt; $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\epsilon$ is not the first uncountable ordinal, but the &quot;reciprocal&quot; of $\omega_0$). My question is thus: can you device this for every ordinal, or does Conway's system On eventually run &quot;out of space&quot;).&lt;/p&gt;&#xA;" OwnerUserId="3152" LastEditorUserId="3152" LastEditorDisplayName="" LastEditDate="2010-11-05T08:12:22.033" LastActivityDate="2010-11-05T22:40:45.337" Title="surreal and ordinal numbers" Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="9038" PostTypeId="2" ParentId="9037" CreationDate="2010-11-05T07:52:09.437" Score="3" ViewCount="0" Body="&lt;p&gt;I''m not entirely sure what your question is. In Conway's&#xA;notation &lt;strong&gt;On&lt;/strong&gt; denotes the ordinal numbers (and &lt;strong&gt;No&lt;/strong&gt; denotes the set of&#xA;all surreal Numbers). Basically the elements of &lt;strong&gt;On&lt;/strong&gt; are just von Neumann&#xA;ordinals. In von Neumann's construction an ordinal $\alpha$ is just&#xA;the set $\{\beta:\beta&amp;lt;\alpha\}$ of all preceding ordinals and this&#xA;corresponds to the Conway ordinal $\langle\beta:\beta&amp;lt;\alpha\mid\ \rangle$.&#xA;So the von Neumann and Conway ordinals naturally correspond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe your question is whether there is a one-to-one correspondence&#xA;between the ordinals and the whole class of surreal numbers. These&#xA;are both proper classes with respect to ZFC. There are subtleties associated&#xA;to defining bijections between proper classes, which I don't really&#xA;understand, but given a strong version of AC I believe that any&#xA;two proper classes can be paired off. (Maybe an experienced set theorist&#xA;could say more?)&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-05T08:55:34.173" LastActivityDate="2010-11-05T08:55:34.173" CommentCount="2" />
  <row Id="9039" PostTypeId="2" ParentId="9008" CreationDate="2010-11-05T08:05:17.230" Score="3" ViewCount="0" Body="&lt;p&gt;Let $f$ be any strictly increasing &lt;s&gt;(not necessarily strictly)&lt;/s&gt; function on $S$.  To show that $f$ is continuous on $S$, it is enough to show that it is continuous at $x$ for every $x \in S$.  If $x$ is an isolated point of $S$, every function is continuous at $x$, so assume otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that monotone functions can only be discontinuous in a very particular, and simple, way.  Namely, the one-sided limits $f(x-)$ and $f(x+)$ always exist (or rather, the first exists when $x$ is not left-isolated and the second exists when $x$ is not right-isolated): it is easy to see for instance that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x-) = \sup_{y &amp;lt; x, \ y \in S} f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore a discontinuity occurs when $f(x-) \neq f(x)$ or $f(x+) \neq f(x)$.  In the first case we have that for all $y &amp;lt; x$, $f(y) &amp;lt; f(x-)$ and for all $y \geq x$, $f(y) &gt; f(x-)$.  Therefore $f(x-)$ is not in $f(S)$.  But by the above expression for $f(x-)$, it is certainly a limit point of $f(S)$.  So $f(S)$ is not closed.  The other case is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other nice, related properties of monotone functions include: a monotone function has at most countably many points of discontinuity and a monotone function is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Regulated_function&quot; rel=&quot;nofollow&quot;&gt;regulated function&lt;/a&gt; in the sense of Dieudonné.  In particular the theoretical aspects of integration are especially simple for such functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: As Myke notes in the comments below, the conclusion need not be true if $f$ &#xA;is merely increasing (i.e., $x_1 \leq x_2$ implies $f(x_1) \leq f(x_2)$).  A counterexample &#xA;is given by the characteristic function of $[0,\infty)$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-05T22:23:32.930" LastActivityDate="2010-11-05T22:23:32.930" CommentCount="5" />
  <row Id="9040" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T08:15:36.600" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Exotic Spheres&lt;/strong&gt;. Milnor's proof that there exists 28 distinct differentiable $7$-manifolds that are homeomorphic, but not diffeomorphic, to a standard $7$-sphere.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-05T08:15:36.600" CommunityOwnedDate="2010-11-05T08:15:36.600" />
  <row Id="9041" PostTypeId="2" ParentId="8980" CreationDate="2010-11-05T08:15:55.693" Score="5" ViewCount="0" Body="&lt;p&gt;This refers to Ryan's answer, but was too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was getting confused for a while too, but I think I understand now.&#xA;To avoid misunderstandings, the Tait–Bryan angles should be called&#xA;heading/elevation/bank instead of yaw/pitch/roll. The heading angle&#xA;refers to rotations about the $z$ axis in space, &lt;em&gt;not&lt;/em&gt; about the yaw&#xA;axis which by definition is perpendicular to the &quot;body-wing-plane&quot; of&#xA;the airplane, so that &quot;excellent picture&quot; is in fact a bit misleading&#xA;in this context! The gimbal lock &lt;a href=&quot;http://en.wikipedia.org/wiki/Gimbal_lock#Gimbal_lock_in_three_dimensions&quot; rel=&quot;nofollow&quot;&gt;illustrated on&#xA;Wikipedia&lt;/a&gt;&#xA;occurs when the plane is pointing straight up or down (elevation = 90&#xA;degrees), so that changing the bank is the same thing as changing the&#xA;heading. (That is, there is momentarily no way to change &lt;em&gt;the yaw&lt;/em&gt; by&#xA;manipulating the heading/elevation/bank angles!)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-05T08:15:55.693" CommentCount="1" />
  <row Id="9042" PostTypeId="2" ParentId="9032" CreationDate="2010-11-05T08:38:03.997" Score="2" ViewCount="0" Body="&lt;p&gt;Just averaging them does not account for the different variances, and therefore does not yield an optimal result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best (as in minimum variance, unbiased) estimation of the expected value $\mu$ of those two distributions $X_1, X_2$ with variances $\sigma^2_1,\sigma^2_2$ is $$\hat\mu(x_1, x_2) = \frac{\sigma_1^2x_2+\sigma_2^2x_1}{\sigma_1^2+\sigma_2^2}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct my notaion. I've learned this stuff with German names. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The joint distribution of $X_1, X_2$ forms an exponential family in $T(x_1, x_2) = \sigma_1^2x_2+\sigma_2^2x_1$, since its probability density function has the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x_1, x_2) = \frac{1}{\sqrt{2\pi\sigma_1^2}\sqrt{2\pi\sigma_2^2}}\exp({-\frac{(x_1-\mu)^2}{2\sigma_1^2}-\frac{(x_2-\mu)^2}{2\sigma_2^2}})$$ $$= h(x)A(\mu)\exp({\frac{\mu(\sigma_1^2x_2+\sigma_2^2x_1)}{2\sigma_1^2\sigma_2^2}}) $$ with $h(x)A(\mu)$ being all the uninteresting stuff. =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $T$ is sufficient, $\mu$ depends only on $T$ and is unbiased and the Lehmann–Scheffé theorem together with some mumbling about completeness concludes the proof.&lt;/p&gt;&#xA;" OwnerUserId="505" LastActivityDate="2010-11-05T08:38:03.997" />
  <row Id="9043" PostTypeId="2" ParentId="8814" CreationDate="2010-11-05T08:46:53.767" Score="2" ViewCount="0" Body="&lt;p&gt;\begin{eqnarray}&#xA;\sum_{k = 0}^{\lfloor q  - q/p) \rfloor} \left \lfloor \frac{p(q - k)}{q} \right \rfloor = \sum_{k = 1}^{q} \left \lfloor \frac{kp}{q} \right \rfloor&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-05T08:46:53.767" CommunityOwnedDate="2010-11-05T08:46:53.767" />
  <row Id="9044" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T08:58:16.993" Score="1" ViewCount="0" Body="&lt;p&gt;I have had two results in the last year that surprised me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One appeals to my intuition in physics, although it may seem more obvious to those more versed mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of pauli matrices (with identity) when multiplied by $i$ form the group of unitary quaternions (under matrix multiplication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The was surprising to me how you can connect such a physical concept as electron spin to an abstract algebraic concept. Its what led me to dive into group and representation theory as it applies to physics. Now I understand that we can consider spin symmetry as SU(2) , which is injective into SO(3) the group of symmetry of $R^3$ of which the quaternions can be thought of as a representation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second result, which I had to prove in an algebra assignment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a field $F$ of characteristic $p$ where $p$ is prime $(x+y)^p = x^p + y^p$ for $x,y \in F$. Lovely little result that spits in the face of everything that our grade school teachers taught us in algebra. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-11-05T08:58:16.993" CommunityOwnedDate="2010-11-05T08:58:16.993" />
  <row Id="9045" PostTypeId="1" AcceptedAnswerId="10662" CreationDate="2010-11-05T10:14:07.173" Score="-5" ViewCount="80" Body="&lt;p&gt;How to prove that an nxn matrix A is unitary if and only if $||Az||=||z||$ for all $z \in C^n$? Suggestion: Apply previous question on &quot;How to prove that T is the 0 operator, that is $T(v)=0$ for all $u,v \in V$ ?&quot; to &#xA;$A^{\ast}A-I$, this is a homework in the area of linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastActivityDate="2010-11-17T06:17:16.627" Title="How to prove that an nxn matrix A is unitary if and only if $||Az||=||z||$ for all $z \in C^n$?" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9046" PostTypeId="2" ParentId="9035" CreationDate="2010-11-05T11:23:15.153" Score="4" ViewCount="0" Body="&lt;p&gt;You might look at this for some recent related work, and investigate its 14 references:&#xA;&quot;Every Large Point Set contains Many Collinear Points or an Empty Pentagon.&quot; &lt;a href=&quot;http://www.informatik.uni-trier.de/~ley/db/conf/cccg/cccg2009.html#AbelBBCDHKLPW09&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;CCCG&lt;/em&gt; 2009, 99-102&lt;/a&gt;. The main result is exactly what is stated in the title.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-11-05T11:23:15.153" CommentCount="3" />
  <row Id="9048" PostTypeId="1" CreationDate="2010-11-05T11:48:43.557" Score="1" ViewCount="102" Body="&lt;p&gt;OK, so I got my ships moving along the bezier in time T, now I need to factor in acceleration, but I don't really know how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the DeCasteljau Algorithm, that's what my code looks like:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function linp(d, a, b, t) {&#xA;    d.x = a.x + (b.x - a.x) * t;&#xA;    d.y = a.y + (b.y - a.y) * t;&#xA;}&#xA;&#xA;// a, b, c, d are the 4 controls points, dest will contain the x/y coordinates and t is the time value from 0.0 to 1.0&#xA;Ship.prototype.bezier = function(dest, a, b, c, d, delta) {&#xA;    var ab = {x:0, y: 0}, bc = {x:0, y: 0}, cd = {x:0, y: 0};&#xA;    var abbc = {x:0, y: 0}, bccd = {x:0, y: 0};&#xA;    linp(ab, a, b, t);&#xA;    linp(bc, b, c, t);&#xA;    linp(cd, c, d, t);&#xA;    linp(abbc, ab, bc, t);&#xA;    linp(bccd, bc, cd, t);&#xA;    linp(dest, abbc, bccd, t);&#xA;};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the question is, what is a simple way to factor in the acceleration from the starting speed to the end speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&lt;br&gt;&#xA;Here's a small video how the thing looks when it's not working. I hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=EIJg1WFbjh4&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=EIJg1WFbjh4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3154" LastEditorUserId="3154" LastEditorDisplayName="" LastEditDate="2010-11-05T13:26:56.943" LastActivityDate="2011-01-10T02:21:16.190" Title="Moving along a bezier curve in time T, factoring in acceleration from X to Y" Tags="&lt;bezier-curve&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="9049" PostTypeId="1" AcceptedAnswerId="9056" CreationDate="2010-11-05T11:58:44.093" Score="2" ViewCount="90" Body="&lt;p&gt;On what basis the numbering less than 20, the fraction 17/6 (base 10) does not generate a regular tithe?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-05T12:00:34.417" LastActivityDate="2010-11-05T13:11:43.997" Title="Numbering in base other than 10" Tags="&lt;number-systems&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="9050" PostTypeId="2" ParentId="9010" CreationDate="2010-11-05T12:24:31.177" Score="1" ViewCount="0" Body="&lt;p&gt;This is answer to 978...answer (I have not enough reputation to comment).&#xA;He writes &quot;Also, we can be jerks and do something silly: Every compact manifold admits the structure of an affine scheme, by Hochster's Theorem&quot;&#xA;This is not true because Hochster theorem writes that quasicompact opens sets must be base of opens sets and that is completely false for usual manifolds.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-05T12:24:31.177" CommentCount="3" />
  <row Id="9051" PostTypeId="2" ParentId="9037" CreationDate="2010-11-05T12:38:49.377" Score="8" ViewCount="0" Body="&lt;p&gt;The existence of a bijection between the class of ordinals $On$ and the class of surreal numbers $No$ is independent of the axioms of set theory. There are several interesting possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If ZFC is consistent, then there is a model of ZFC in which there is a definable such bijection. This is true in Goedel's constructible universe $L$, for example, for in $L$ there is a definable well-ordering of the universe, and we can use this well-ordering to well-order the surreals, which provides the desired bijection.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More generally, there is a first-order definable bijection between $On$ and $No$ if and only if the axiom known as $V=HOD$ holds. For the one direction, if $V=HOD$ holds, then there is a definable well-ordering of the universe and hence in particular a definable well-ordering of the surreals. Conversely, under ZFC if there is a definable bijection between $On$ and $No$, then there is a definable well-ordering of $No$. This allows us to construct a definable well-ordering of the class of sets of ordinals, since any set of ordinals determines a transfinite binary sequence of some ordinal length, and we can interpret this sequence as a $\pm 1$ sequence, which determines a unique surreal number by climbing through the tree of left-right cuts. Thus, we can well-order the class of sets of ordinals. But in ZFC every set is coded by a set of ordinals, and so we can construct a well-ordering of the entire universe, by looking for the least ordinal mapping to a surreal whose $\pm 1$ representation codes that set. So in this case, V=HOD holds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another way to summarize this argument is to say that if you can well-order $No$---and this is what your bijection to $On$ amounts to---then you can well-order every class.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you drop the requirement that the bijection be definable, then we should move to the Goedel-Bernays context, in order to treat classes. The assertion that there is a bijection between $On$ and $No$ is equivalent over ZFC+GB to the axiom of Global Choice, which asserts that there is a well-ordering of the universe. This is by the same argument as above. (Note, we need AC for sets in order to make the last step of the argument; the class bijection in effect allows us to sew the set sized well-orderings together into a class well-ordering.) Thus, the theory ZFC+GB+(your bijection) is equivalent to GBC.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because of this, if ZFC is consistent, then there are models of ZFC that have no bijection between $On$ and $No$, either definable or definable-from-parameters or otherwise. This is because it is known that ZFC does not imply global choice. One can construct such models by performing a class forcing iteration, adding a Cohen subset to every regular cardinal. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Meanwhile, every model of ZFC has a class forcing extension in which there &lt;em&gt;is&lt;/em&gt; a class well-ordering of the universe, simply by forcing to add a global well-ordering, and this forcing extension adds no new sets, only classes. In this sense, it is compatible with every model of ZFC set theory to have the desired bijection as a class, without adding any new sets. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Further, every model of ZFC has a class forcing extension in which there is a &lt;em&gt;definable&lt;/em&gt; bijection between $On$ and $No$, since we can force $V=HOD$. (This forcing, however, does add new sets.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Lastly, upon reading your question again, I see that you asked for a surjection from $On$ onto $No$, rather than a bijection. But these are equivalent, since if there is a surjection, then we can remove the redundant ordinals from the domain by only using the least ordinal that maps to a given surreal, and this gives a bijection from a proper class of ordinals to $No$. But every proper class of ordinals is bijection with $On$ simply by collapsing to the order type of the predecessors.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-11-05T22:40:45.337" LastActivityDate="2010-11-05T22:40:45.337" />
  <row Id="9052" PostTypeId="2" ParentId="8814" CreationDate="2010-11-05T12:41:36.190" Score="4" ViewCount="0" Body="&lt;p&gt;Well, i don't know whether to classify this as funny or surprising,  but ok it's worth posting.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $(X,\tau)$ be a topological space and let $A \subset X$ . By iteratively applying operations of closure and complemention, one cannot produce more than 14 disjoint sets. It's called as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kuratowski%27s_closure-complement_problem&quot; rel=&quot;nofollow&quot;&gt;Kuratowski's Closure complement problem&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-05T12:41:36.190" CommunityOwnedDate="2010-11-05T12:41:36.190" />
  <row Id="9053" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T12:43:29.277" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kuratowski%27s_closure-complement_problem&quot; rel=&quot;nofollow&quot;&gt;Kuratowski's Complement problem&lt;/a&gt;, is the one which i came across recently, and i was clearly flabbergasted.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-05T12:43:29.277" CommunityOwnedDate="2010-11-05T12:43:29.277" />
  <row Id="9054" PostTypeId="2" ParentId="8977" CreationDate="2010-11-05T12:52:21.887" Score="1" ViewCount="0" Body="&lt;p&gt;As others have mentioned in comments, your control points cannot be independent. Nevertheless, if we assume that a given configuration has the properties you want, we can analyze the geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll consider the orange arc, $BE$, and I'll assume that both circles $A$ and $C$ overlap the interior of the orange circle, which I'll further assume has not degenerated into a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a = |AB|$, $c = |CE|$, and $x=|AC|$; all of these can be considered known quantities. Let the (unknown) radius of the orange circle be $r = |PB| = |PE|$, where $P$ is the circle's center. Because radii $AB$ and $PB$ are perpendicular to a common tangent line at $B$, these segments lie on the same line; likewise for $CE$ and $PE$; consequently, $P$ lies at the intersection of the two extended radii $AB$ and $CE$, so that the angle $BPE$ is congruent to the angle between the &lt;em&gt;vectors&lt;/em&gt; $AB$ and $CE$. Call the measure of that angle $\theta$; it, too, can be considered a known quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, triangle $APC$ has sides of length $x$, $r-a$, and $r-c$ (the last two because of the assumed overlap of circles), with angle $\theta$ between the last two. By the Law of Cosines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 = (r-a)^2 + (r-c)^2 - 2 (r-a)(r-c) \cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this quadratic equation for $r$, and you can calculate whatever else you need to know: arc length, location of $P$, equation of the orange circle, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The equation of the orange circle can be expressed in a form that degenerates into a line equation as $r$ approaches infinity. Note that, in such a degenerate case, $\theta = 0$.)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-05T12:52:21.887" />
  <row Id="9055" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T13:09:52.503" Score="1" ViewCount="0" Body="&lt;p&gt;1) Exotic structures on $\mathbb{R}^4$ probably puzzles anyone learning about differentiable topology. Even more, the fact that it is &lt;em&gt;only&lt;/em&gt; for $n=4$ is quite remarkable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)$S^n$ not being a Lie group for all $n$ ...&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-05T13:09:52.503" CommunityOwnedDate="2010-11-05T13:09:52.503" CommentCount="2" />
  <row Id="9056" PostTypeId="2" ParentId="9049" CreationDate="2010-11-05T13:11:43.997" Score="2" ViewCount="0" Body="&lt;p&gt;The terminating decimals (though that term should be reserved for base 10, but I have no better) in any base are precisely the fractions whose denominator has every prime factor represented in the factorization of the base.  So in base 10, it is fractions with denominators of the form 2^x*5^y, where x and y are allowed to be zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T13:11:43.997" CommentCount="2" />
  <row Id="9057" PostTypeId="1" CreationDate="2010-11-05T14:00:38.513" Score="1" ViewCount="122" Body="&lt;p&gt;I am CS major, please be patient if my question is not well-stated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dataset is quantitative mass spectrometry (MS) data. By labeling proteins of two different samples A and B, we get the relative abundance of 100 to thousands of proteins in A/B. Alongside with this ratio, we can estimate its variance based on the signal intensities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wanted: A list of proteins significantly different from the set of all protein ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most proteins remain unchanged between A and B. The population of log-ratios distributes around 1. The histogram shows a bell shape with fat tails. Two-term Gaussian mixture model has been found to provide a good fit to experimental noise. I suppose it would work good for this data - think of experimental and biological noise.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to test for significantly different ratios on such a two-term Gaussian mixture model?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for your responses!&lt;/p&gt;&#xA;" OwnerUserId="3156" LastEditorUserId="3156" LastEditorDisplayName="" LastEditDate="2010-11-08T11:37:50.390" LastActivityDate="2011-01-07T11:44:40.127" Title="How to do hypothesis testing on Gaussian mixture model?" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9058" PostTypeId="1" AcceptedAnswerId="9067" CreationDate="2010-11-05T14:05:38.057" Score="1" ViewCount="151" Body="&lt;p&gt;How do I solve the following question?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given the identity    $x^2-ax+144\equiv(x-b)^2$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Work out the values of $a$ and $b$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question appears in AQA 43005/1H.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="3157" LastEditorDisplayName="" LastEditDate="2010-11-06T16:49:58.863" LastActivityDate="2010-11-06T16:49:58.863" Title="Work out the values of a and b from the identity `x² - ax + 144 ≡ (x - b)²`" Tags="&lt;algebra&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="9059" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T14:06:21.393" Score="0" ViewCount="0" Body="&lt;p&gt;Posted by &lt;a href=&quot;http://in.answers.yahoo.com/my/profile;_ylt=AuGKDZFCaIla44dugPaZqxXaHwx.;_ylv=3?show=31f743411a42c3a8de21b1c9c1b15871aa&quot; rel=&quot;nofollow&quot;&gt;Pi R Squared&lt;/a&gt; on &lt;a href=&quot;http://in.answers.yahoo.com/question/index?qid=20090510094003AA5Xzgh&quot; rel=&quot;nofollow&quot;&gt;Yahoo Answers&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a=24$ and $b=12$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$x^2-24x+144=(x-12)^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="3157" LastEditorDisplayName="" LastEditDate="2010-11-05T15:46:28.930" LastActivityDate="2010-11-05T15:46:28.930" CommentCount="6" />
  <row Id="9060" PostTypeId="1" AcceptedAnswerId="9066" CreationDate="2010-11-05T14:12:05.613" Score="3" ViewCount="199" Body="&lt;p&gt;Consider the set &#xA;$$S = [0,1/2] \cup [3/4, 7/8] \cup [15/16, 31/32] \cup \cdots$$&#xA;and the set &#xA;$$T = (0,1/2) \cup  (3/4, 7/8) \cup (15/16, 31/32) \cup \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $S$ open, closed or neither? Is $T$ open, closed, or neither?&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T16:53:41.850" LastActivityDate="2010-11-05T19:32:32.223" Title="open and closed sets" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9061" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T14:14:31.230" Score="4" ViewCount="0" Body="&lt;p&gt;You have to note first that $12^2=144$. After this, you have to recall the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-b)^2=x^2-2bx+b^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare coefficients to get the equations you need to solve for $a$ and $b$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-05T14:14:31.230" CommentCount="1" />
  <row Id="9062" PostTypeId="2" ParentId="9057" CreationDate="2010-11-05T14:33:54.243" Score="0" ViewCount="0" Body="&lt;p&gt;When you say the data fits one distribution or another, what are you plotting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One hypothesis would be that the two samples are drawn from the same distribution.  Then you would expect A/B to be the same over all proteins.  If the distribution is the frequency of A/B for all the different proteins you observe you can test the hypothesis if you have a model of the errors in A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you model the data should be dependent upon how you will use the model.  It is not clear to me what the model is for.  You can certainly fit any curve through the data.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T14:33:54.243" CommentCount="1" />
  <row Id="9063" PostTypeId="1" AcceptedAnswerId="9065" CreationDate="2010-11-05T14:45:52.733" Score="7" ViewCount="131" Body="&lt;p&gt;Given a $\sigma$-finite measure $\mu$ on a set $X$ is it possible to formulate a topology on the space of functions $f:X \rightarrow \mathbb{R}$ that gives convergence $\mu$-almost everywhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to find any way to write this and am suspecting that no such topology exists! Is this true?  If so, is there some generalisation of a topological space where one can make sense of convergence without having open sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments, references or tips would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="780" LastActivityDate="2010-11-05T15:49:14.060" Title="Characterization of Almost-Everywhere convergence" Tags="&lt;analysis&gt;&lt;measure-theory&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="9064" PostTypeId="2" ParentId="9060" CreationDate="2010-11-05T15:10:34.630" Score="1" ViewCount="0" Body="&lt;p&gt;If we have the standard topology T is the countable union of disjoint open sets and so T is an open set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complement of S is $(-\infty, 0) \cup (1/2, 3/4) \cup (7/8, 15/16) \cup \cdots$, etc. if this is open then S is closed. Is it open? Why?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T16:01:26.063" LastActivityDate="2010-11-05T16:01:26.063" CommentCount="3" />
  <row Id="9065" PostTypeId="2" ParentId="9063" CreationDate="2010-11-05T15:11:36.710" Score="9" ViewCount="0" Body="&lt;p&gt;Given I understand you correctly (topologize almost everywhere convergence), we can show that this is not possible. If we have a topological space, then we have convergence of a sequences if and only if every subsequence has a further convergent subsequence and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So pick a sequence that converges in measure but not almost everywhere. There is a theorem that states that every subsequence of this also converges in the same measure. So it has a subsequence that converges almost everywhere (by another theorem). But the original sequence does not converge almost everywhere so we cannot have convergence in a topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I add more detail?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-05T15:49:14.060" LastActivityDate="2010-11-05T15:49:14.060" CommentCount="1" />
  <row Id="9066" PostTypeId="2" ParentId="9060" CreationDate="2010-11-05T15:12:09.777" Score="5" ViewCount="0" Body="&lt;p&gt;For the sake of the answer I am going to assume this is the standard topology of the real numbers, but I'll try to be as general as possible so it will be easier to generalize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A union of any number of open sets is still open, so $T$ is open.&#xA;A union of countably many closed sets does not have to be closed, this sort of set is called $F_\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While some $F_\sigma$ are open sets (in fact in a metric space, such as the real numbers with their standard topology, every open set is $F_\sigma$) this one is not. One of the requirements from an open set (in any topology) is that it will have an open neighborhood surrounding each point inside of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case above that means there will be an interval around each point, which is a property not being satisfied by $S$, as the point $\frac{1}{2}$ is in $S$ but there is no interval around it contained in $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is still the case where $S$ could be closed as well, however if we look at the sequence $a_n = \frac{2^{2n-1}-1}{2^{2n-1}}$ (the endpoints of each closed interval within the union) then it is clear that $a_n$ approaches $1$ but since $1 \notin S$ then we have that $S$ is not closed.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-11-05T19:19:13.920" LastActivityDate="2010-11-05T19:19:13.920" CommentCount="2" />
  <row Id="9067" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T15:42:19.557" Score="0" ViewCount="0" Body="&lt;h3&gt;Find the Value of $b$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$x^2-ax+144\equiv(x-b)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand the right-hand side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-ax+144\equiv{x^2-2bx+b^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get the coefficients $(x=0)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b^2=-144$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b=\pm12$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Find the Value of $a$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Substitute $b=\pm12$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-ax+144\equiv{x^2\pm24x+144}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduce the expanded fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cancel{x^2}-ax\cancel{+144}\equiv\cancel{x^2}\pm24x\cancel{+144}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the coefficients $(x=1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a=\pm{24}$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Answer&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$a=24, b=12$ &lt;strong&gt;or&lt;/strong&gt; $a=-24, b=-12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figured out from @J. M. and @Djaian's answers and comments.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="3157" LastEditorDisplayName="" LastEditDate="2010-11-06T16:09:20.287" LastActivityDate="2010-11-06T16:09:20.287" CommentCount="3" />
  <row Id="9068" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T15:43:49.273" Score="5" ViewCount="0" Body="&lt;p&gt;The prior answer presumes that you already know the fact that two polynomials functions are identical (if and) only if they have identical coefficients. But the above problem might be posed as motivation before this general fact is proved. In that case you can prove it directly without using this more general fact as follows. Evaluating the identity at $\rm\ x = 0\ $ yields $\rm\ (-b)^2 = 144 = 12^2\ $. So $\rm\ (x-b)^2 = x^2 -2b + 144 \equiv x^2 -ax + 144\:.\ $ RHS $-$ LHS yields $\rm\ (2b-a)\:x \equiv 0\:$. Evaluating this at $\rm\ x = 1\ $ yields $\rm\ a = 2b\:.\ $ And, of course, $\rm\ 0 = 12^2 - (-b)^2 = (12+b)\ (12-b)\ \Rightarrow\ b = \pm12\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above proof yields$\ \ $ (1) the equation is true for all $\rm\: x\ \Rightarrow\ $ (2) the equation is true for $\rm x = 0,\ and\ x = 1\ $ $\rm\ \Rightarrow\ $ (3) $\rm\ a = 2b,\ b^2 = 144\:.\ $ But (3) implies that both sides of the equation have equal coefficients, so it is true for all $\rm\:x\:$, implying (1).$\ $ So all three statements are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact generally it is true over $\mathbb C$ that two polynomial functions of degree $\rm\:n\:$ are identical iff they have identical coefficients iff they have equal values at $\rm\:n+1\:$ points (or $\rm\:n\:$ points if they have the same leading coefficient - as above). In fact, this is true not only for polynomials with coefficients in $\mathbb C$ but in any integral domain, i.e any ring without zero-divisors, i.e. $\rm\ ab = 0 \Rightarrow\ a=0\ or\ b=0\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-06T16:27:28.487" LastActivityDate="2010-11-06T16:27:28.487" CommentCount="1" />
  <row Id="9069" PostTypeId="1" AcceptedAnswerId="9096" CreationDate="2010-11-05T15:45:22.533" Score="1" ViewCount="112" Body="&lt;p&gt;The probability that a fire will occur is $0.001$. If there is a fire, the amount of damage, $X$, will have a Pareto distribution given by $P(X&gt;x) = \left(\frac{2(10)^6}{2(10)^6+x} \right)^2$. An insurance will pay the excess of the loss over a deductible of $100,000$. For this coverage the one-time insurance premium will be $110 \%$ of the expected payment. Calculate the premium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected payment is $E[W]$ where $W$ denotes the payment. Then $E[W] = E[W| \text{fire}]P(\text{fire})+E[W| \text{no fire}]P(\text{no fire})$. To calculate $E[W| \text{fire}]$, we could use $\int_{0.1}^{\infty} [1-F(x)] \ dx$? This would be: $\int_{0.1}^{\infty} 1-\left[1-\left(\frac{2(10)^6}{2(10)^6+x} \right)^2\right] \ dx$ which equals $\int_{0.1}^{\infty} \left(\frac{2(10)^6}{2(10)^6+x} \right)^2 \ dx$?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-05T20:36:59.363" Title="Probability of Fire" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="9070" PostTypeId="2" ParentId="2949" CreationDate="2010-11-05T16:10:13.553" Score="2" ViewCount="0" Body="&lt;p&gt;To me the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_surface#27_lines_on_a_cubic_surface&quot; rel=&quot;nofollow&quot;&gt;Cayley-Salmon theorem&lt;/a&gt; is an example of a result that still strikes me as rather surprising.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem(Cayley-Salmon)&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A smooth cubic surface $\mathcal{S} \subset \mathbb{P}^{3}_{k}$ contains exactly 27 lines, where $k$ is an algebraically closed field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/PrintHT/Cubic_surfaces.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to some history about it. There's a really nice treatment of this result in chapter 5 of Klaus Hulek's book &lt;a href=&quot;http://books.google.co.cr/books?id=uI_21kSf0iAC&amp;amp;printsec=frontcover&amp;amp;dq=Klaus+hulek&amp;amp;hl=es&amp;amp;ei=fCrUTN-wJoOclge-68XaBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Elementary Algebraic Geometry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-05T16:10:13.553" CommunityOwnedDate="2010-11-05T16:10:13.553" />
  <row Id="9071" PostTypeId="1" CreationDate="2010-11-05T16:14:49.140" Score="2" ViewCount="72" Body="&lt;p&gt;This comes up in a paper by Beskos and Roberts on exact simulation of diffusions. I'm sure it's easy, but I can't work out what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a differentiable function $\alpha$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha(u) \geq k_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha'(u) \leq k_2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constants $k_1$ and $k_2$, show that there exist constants $c_1$ and $c_2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_1 \leq \alpha^2(u) + \alpha'(u) \leq c_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly one condition implies alpha is bounded below and one implies that alpha has at most linear growth, but I don't see how this is enough for the conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit - I'm a fool. They wrote $k_1 \leq \alpha(u), \alpha'(u) \leq k_2$, but there was a line break after the comma. I was reading them as two seperate conditions on $\alpha$ and $\alpha'$. Sorry everyone, thanks for the help. If it wasn't for the counterexamples it would have taken a lot longer to realise that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3162" LastEditorUserId="3162" LastEditorDisplayName="" LastEditDate="2010-11-05T16:33:41.573" LastActivityDate="2010-11-05T16:33:41.573" Title="simple functional inequality" Tags="&lt;analysis&gt;&lt;inequality&gt;" CommentCount="5" />
  <row Id="9072" PostTypeId="1" CreationDate="2010-11-05T16:39:59.227" Score="1" ViewCount="184" Body="&lt;p&gt;Can someone explain how to picture or construct the Monster Group?&lt;/p&gt;&#xA;" OwnerUserId="3163" LastActivityDate="2010-11-06T11:56:25.063" Title="What is the simplest way to fathom the Monster Group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="9073" PostTypeId="2" ParentId="9069" CreationDate="2010-11-05T16:51:57.140" Score="1" ViewCount="0" Body="&lt;p&gt;TO calculate $E(W|fire)$, you first need to calculate the pdf. You can get this by taking derivative of $1-P(X&gt;x)$. Call this function $p(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the insurer pays out any amount greater than the deductible of $100000$, the expected payment is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{100000}^\infty (x - 100000) p(x) dx&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The premium should be 110% of the value of the integral.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-05T16:51:57.140" CommentCount="2" />
  <row Id="9074" PostTypeId="2" ParentId="9060" CreationDate="2010-11-05T16:52:35.593" Score="4" ViewCount="0" Body="&lt;p&gt;To be able to answer these questions you need to know what &quot;open&quot; and &quot;closed&quot; mean. And that depends on context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context you are looking at, in which you are dealing with subsets of the real line, without any further information we usually assume that we are dealing with the &lt;em&gt;standard&lt;/em&gt; meanings of &quot;open&quot; and &quot;closed&quot;. The meanings are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A set is $\mathcal{O}\subseteq\mathbb{R}$ is &lt;strong&gt;open&lt;/strong&gt; if and only if for every $x\in \mathcal{O}$ there is an $\epsilon\gt 0$ (which may depend on $x$) such that $(x-\epsilon,x+\epsilon)\subseteq \mathcal{O}$ (where $(x-\epsilon,x+\epsilon)$ is the set of all real numbers $y$ such that $x-\epsilon \lt y \lt x+\epsilon$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set $\mathcal{C}\subseteq\mathbb{R}$ is &lt;strong&gt;closed&lt;/strong&gt; if and only if its complement $\mathbb{R}-\mathcal{C}$ is open; if and only if for every point $x\notin \mathcal{C}$, there exists $\epsilon\gt 0$ (which may depend on $x$) such that $(x-\epsilon,x+\epsilon)\cap \mathcal{C}=\emptyset$. Alternatively, a set $\mathcal{C}\subseteq\mathbb{R}$ is &lt;strong&gt;closed&lt;/strong&gt; if and only if for every converging sequence of real numbers $\{x_n\}_{n=1}^{\infty}$, if each $x_n$ is in $\mathcal{C}$, and $\lim\limits_{n\to\infty}x_n = x$, then $x\in\mathcal{C}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some consequences of these definitions are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Both the sets $\mathbb{R}$ and $\emptyset$ are open and closed.&lt;/li&gt;&#xA;&lt;li&gt;Every open interval is an open set. Every closed interval is a closed set.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathcal{O}_1$ and $\mathcal{O}_2$ are both open, then their intersection $\mathcal{O}_1\cap\mathcal{O}_2$ is also open (given $x$ in the intersection, find $\epsilon_1\gt 0$ and $\epsilon_2\gt 0$ so that $(x-\epsilon_i,x+\epsilon_i)\subseteq\mathcal{O}_i$, and then let $\epsilon=\min\{\epsilon_1,\epsilon_2\}$ be the witness for the intersection). As a consequence, using De Morgan's laws, if $\mathcal{C}_1$ and $\mathcal{C}_2$ are closed, then $\mathcal{C}_1\cup\mathcal{C}_2$ is closed.&lt;/li&gt;&#xA;&lt;li&gt;Using induction, from 3 it follows that any intersection of &lt;em&gt;finitely many&lt;/em&gt; open sets is open, and any union of &lt;em&gt;finitely many&lt;/em&gt; closed sets is closed.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathcal{O}_i$ is any family (of any size, finite or infinite) of open sets, then $\cup\mathcal{O}_i$ is also open (pick $x$ in the union; then it belongs to some $\mathcal{O}_j$, and since $\mathcal{O}_j$ is open and $x\in\mathcal{O}_j$, there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq\mathcal{O}_j\subseteq \cup\mathcal{O}_i$). Consequently, if $\mathcal{C}_i$ is any family (of any size, finite or infinite) of closed sets, then $\cap\mathcal{C}_i$ is closed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, looking back at your examples, it should be clear that $T$ must be open, since it is a union of open intervals; open intervals are open sets, and an arbitrary union of open sets is always open. What is not clear is whether $S$ is closed or not, and that is because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In general, an arbitrary intersection of open sets is not necessarily open; it &lt;em&gt;may&lt;/em&gt; be open, or it may be closed (e.g., $\mathop{\cap}\limits_{n=1}^{\infty}(-\frac{1}{n},1+\frac{1}{n}) = [0,1]$), or it may be neither open nor closed. &lt;/li&gt;&#xA;&lt;li&gt;In general, an arbitrary union of closed sets is not necessarily closed; it &lt;em&gt;may&lt;/em&gt; be closed, but it may also be open, or neither.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So... what about $S$? It is somewhat difficult at first thought to try to determine if it is closed using the definition of &quot;complement is open&quot;; but it is not too bad to verify that it does not satisfy the condition I labeled as &quot;equivalent&quot;: if we have a sequence of points in $S$ that converges in $\mathbb{R}$, will the limit be in $S$? Well, take $x_n = 1 - \frac{1}{2^{2n+1}}$; that is $x_1=\frac{7}{8}$, $x_2 = \frac{31}{32}$, $x_3=\frac{127}{128}$, etc. All of these are in $S$, and the sequence converges to $1$. Is $1$ in $S$? $1$ is in $S$ if and only if it is in at leat one of the intervals in the union... is it? Well, no. So $S$ is not closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $S$ open, then? Well, if it were open, then since $\frac{1}{2}\in S$, there would have to be some $\epsilon\gt 0$ such that $(\frac{1}{2}-\epsilon,\frac{1}{2}+\epsilon)\subseteq S$. Is there such an $\epsilon$? Well, no; any such interval would contain points smaller than $\frac{1}{2}$, and there are no points smaller than $\frac{1}{2}$ in $S$. So $S$ is not open either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $S$ is not open, but is not closed either... then it's neither.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Asaf points out, countably infinite unions of closed sets &lt;em&gt;do&lt;/em&gt; show up a lot, so even though they may not be closed, we give them a special name because we often will want to say &quot;this may be neither open nor closed, but &lt;em&gt;is&lt;/em&gt; a countable union of closed sets.&quot; We call them $\mathcal{F}_{\sigma}$ (pronounced &quot;eff-sigma&quot;) sets; $\mathcal{F}$ comes from the French &lt;em&gt;fermé&lt;/em&gt;, for &quot;closed&quot;; $\sigma$ is a letter often used in mathematics to denote that a sum or union is taken a countable, but possibly infinite, number of times (from the French &lt;em&gt;somme&lt;/em&gt; for sum). Dually, we are often interested in sets that are countable intersections of open sets; these sets may be open, closed, or neither, but we call them $\mathcal{G}_{\delta}$ (&quot;gee-delta&quot;); this time, in a fit of internationalism, it comes from the &lt;strong&gt;German&lt;/strong&gt;, specifically &lt;em&gt;Gebiet&lt;/em&gt; (open) and &lt;em&gt;Durchshnitt&lt;/em&gt; (intersection). Your set $S$ is an $\mathcal{F}_\sigma$ set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, one talks about &lt;strong&gt;Topological spaces&lt;/strong&gt;, generalizing these notions. A topological space is an ordered pair $(X,\mathcal{O})$, where $X$ is any set, and $\mathcal{O}$ is a collection of subsets of $X$; basically, we explicitly specify what sets are going to be the &quot;open&quot; sets. We require that the collection $\mathcal{O}$ satisfy some of the same conditions that the open sets in $\mathbb{R}$ satisfy, namely, we require that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\emptyset$ and $X$ are both &quot;open&quot; sets (elements of $\mathcal{O}$);&lt;/li&gt;&#xA;&lt;li&gt;If $A$ and $B$ are both &quot;open&quot; sets, then $A\cap B$ is an open set (that is, if $A$ and $B$ are both elements of $\mathcal{O}$, then $A\cap B$ is an element of $\mathcal{O}$); and&lt;/li&gt;&#xA;&lt;li&gt;If $\{A_i\}_{i\in I}$ is a family, finite or infinite, of &quot;open&quot; sets, then $\mathop{\cup}\limits_{i\in I}A_i$ is also an &quot;open&quot; set (if each $A_i$ is an element of $\mathcal{O}$, then so is their union). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then we define a subset of $X$ to be closed if and only if its complement is open, and go from there. &lt;em&gt;Caveat&lt;/em&gt;: the description of closed sets via limits of sequences is no longer true; one &lt;em&gt;can&lt;/em&gt; define the notion of convergence, but in arbitrary topological spaces the properties are not as nice as they are in the real numbers (there are spaces where the properties &lt;em&gt;are&lt;/em&gt; nice; they get special names for that). One needs to consider instead a more general notion (called &quot;nets&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the same set may have many &quot;topologies&quot;, and many notions of what an open or closed set may mean. Which is why people are asking you if it is the &quot;standard&quot; topology or not (assuming you've gotten this far in my response, anyway...) That is, is it the notion I described above, or some other version of a topology? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all part of the very interesting and deep field of Topology.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T19:32:32.223" LastActivityDate="2010-11-05T19:32:32.223" CommentCount="3" />
  <row Id="9075" PostTypeId="1" AcceptedAnswerId="9079" CreationDate="2010-11-05T16:55:29.467" Score="4" ViewCount="197" Body="&lt;p&gt;Q: $\displaystyle \int \frac{1}{\sin x\cos x} dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the correct answer can be obtained by doing:&lt;br&gt;&#xA;$\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, doing the following gets a completely different answer:&#xA;\begin{eqnarray*}&#xA;\int \frac{1}{\sin x\cos x} dx&lt;br&gt;&#xA;&amp;amp;=&amp;amp;\int \frac{\sin x}{\sin^2(x)\cos x} dx\\&#xA;&amp;amp;=&amp;amp;\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.&#xA;\end{eqnarray*}&#xA;let $u=\cos x, du=-\sin x dx$; then&#xA;\begin{eqnarray*}&#xA;\int \frac{1}{\sin x\cos x} dx&lt;br&gt;&#xA;&amp;amp;=&amp;amp;\int \frac{-1}{(1-u^2)u} du\\&lt;br&gt;&#xA;&amp;amp;=&amp;amp;\int \frac{-1}{(1+u)(1-u)u}du\\ &#xA;&amp;amp;=&amp;amp;\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\&#xA;&amp;amp;=&amp;amp;-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C&#xA;\end{eqnarray*}  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?&lt;/p&gt;&#xA;" OwnerUserId="2869" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-05T17:08:30.747" LastActivityDate="2010-11-05T17:48:23.537" Title="Integration of 1/(sinx*cosx)" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="9077" PostTypeId="1" AcceptedAnswerId="9108" CreationDate="2010-11-05T17:09:57.703" Score="1" ViewCount="114" Body="&lt;p&gt;I am working on a &lt;a href=&quot;http://www.chem.uoa.gr/applets/appletsmooth/appl_smooth2.html&quot; rel=&quot;nofollow&quot;&gt;signal-smoothing algorithm&lt;/a&gt; for personal interest. I understand the basic concept of the Savitzky-Golay algorithm but I would like to understand how the &lt;a href=&quot;http://www.statistics4u.info/fundstat_eng/cc_savgol_coeff.html&quot; rel=&quot;nofollow&quot;&gt;coefficients&lt;/a&gt; were discovered. Can someone please explain the the mathematics behind it, so I could calculate the coefficients on my own?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the matrix look? For example &lt;a href=&quot;http://www.statistics4u.com/fundstat_eng/cc_filter_savgol_math.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; states: &quot;For n=5 and p=3, the filtering coefficients are b= [-0.0857 0.3429 0.4857 0.3429 -0.0857].&quot; However, it does not show the steps to achieve this. If n equal the number of points to sample and p equals the power of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=8mAZYv5wIcE&quot; rel=&quot;nofollow&quot;&gt;Least Squares Tutorial&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.vias.org/tmdatanaleng/cc_filter_savgol_math.html&quot; rel=&quot;nofollow&quot;&gt;Applying Least Squares&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3165" LastEditorUserId="3165" LastEditorDisplayName="" LastEditDate="2010-11-06T01:55:30.217" LastActivityDate="2010-11-06T01:55:30.217" Title="Calculating the Savitzky-Golay Coefficients " Tags="&lt;algorithms&gt;&lt;signal-processing&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9078" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:16:54.823" Score="2" ViewCount="0" Body="&lt;p&gt;The second method gives the same answer as the first. By the first method, the answer you get is $-\log(\cos x) + \log(\sin x)$. The first term is the same as what you get by the second method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to show is that $\log(\sin x) = \frac{1}{2}\log(1-\cos x) + \frac{1}{2}\log(1+\cos x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\begin{split}&#xA;\frac{1}{2}\log(1-\cos x) + \frac{1}{2}\log(1+\cos x) &amp;amp;= \frac{1}{2}\left( \log(2 \sin^2 \frac{x}{2}) + \log(2 \cos^2 \frac{x}{2})\right) =  \log(2 \sin \frac{x}{2} \cos \frac{x}{2})&#xA;\end{split}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-05T17:16:54.823" />
  <row Id="9079" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:17:07.770" Score="3" ViewCount="0" Body="&lt;p&gt;If I take the derivative of your second answer (call it $g(x)$), I get:&#xA;\begin{eqnarray*}&#xA;\frac{dg}{dx}&#xA;&amp;amp; = &amp;amp; -\frac{-\sin x}{\cos x} + \frac{\sin x}{2(1-\cos x)} + \frac{-\sin x}{2(1+\cos x)}\\&#xA;&amp;amp; = &amp;amp; \frac{\sin x\left(1-\cos^2 x + \frac{1}{2}\cos x(1+\cos x) - \frac{1}{2}\cos x(1-\cos x)\right)}{\cos x(1-\cos x)(1+\cos x)}\\&#xA;&amp;amp; = &amp;amp; \frac{\sin x\left( 1- \cos^2 x + \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{2}\cos^2 x\right)}{\cos x(1-\cos^2 x)}\\&#xA;&amp;amp; = &amp;amp; \frac{\sin x}{\cos x\&gt;\sin^2 x} = \frac{1}{\cos x\sin x}.&#xA;\end{eqnarray*}&#xA;So I'm not sure why Mathematica says the second method is not &quot;the right answer&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-05T17:17:07.770" />
  <row Id="9080" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:17:13.013" Score="3" ViewCount="0" Body="&lt;p&gt;This may be an easier method $$\int\frac{1}{\sin{x} \cdot \cos{x}} \ dx = \int\frac{\sec^{2}{x}}{\tan{x}} \ dx$$ by multiplying the numerator and denominator by $\sec^{2}{x}$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-05T17:17:13.013" CommentCount="1" />
  <row Id="9081" PostTypeId="2" ParentId="9032" CreationDate="2010-11-05T17:24:41.663" Score="0" ViewCount="0" Body="&lt;p&gt;You can do away with the normality if you restrict yourself to linear estimates. Given $\mu, \sigma_1^2, \sigma_2^2, \sigma_{12}$,, you have random variables $X_1,X_2$, with $E(X_1) = E(X_2) = \mu $, and $\text{Var}(X_1) = \sigma_1^2$ and $\text{Var}(X_2) = \sigma_2^2$, and $\text{Cov}(X_1,X_2) = \sigma_{12}$. You want $l_0,l_1,l_2$ such that $E(l_0+l_1 X_1 + l_2 X_2) = \mu $ (unbiased) and $\text{Var}(l_0 + l_1 x_1 + l_2 X_2)$  is as small as possible (minimum variance unbiased estimator). From the first equation you get $l_0 + l_1 \mu + l_2 \mu = \mu $ for all $\mu$ i.e., $l_0 = \mu ( 1 - l_1 - l_2)$ for all $\mu$. Which is possible iff $l_0 = 0$  and $l_1 + l_2 = 1$. So the second condition reduces to minimizing $\text{Var}(l_1 X_1 + l_2 X_2)^2$ over all $l_1,l_2$ subject to $l_1 + l_2 = 1$. Simple calculus of lagrangian multipliers applied to $\text{Var}(l_1 X_1 + l_2 X_2) - \lambda(l_1 + l_2 - 1)$  leads to&lt;br&gt;&#xA;$&#xA;l_1 \sigma_1^2 + l_2 \sigma_{12} - \lambda = 0&#xA;$&lt;br&gt;&#xA;$&#xA;l_1 \sigma_{12} + l_2 \sigma_{2}^2 - \lambda = 0&#xA;$&lt;br&gt;&#xA;$&#xA;l_1  + l_2  = 1$&lt;br&gt;&#xA;from which we get our estimator as&lt;br&gt;&#xA;$\frac{\alpha_1}{\alpha_1 + \alpha_2} X_1 + \frac{\alpha_2}{\alpha_1 + \alpha_2} X_2$ where $\alpha_1 = \frac{\sigma_2^2 - \sigma_{12}}{\sigma_1^2\sigma_2^2 -\sigma_{12}^2}$ and $\alpha_2 = \frac{\sigma_1^2 - \sigma_{12}}{\sigma_1^2\sigma_2^2 -\sigma_{12}^2}$. Note, if $\sigma_{12} = 0$ then it reduces to $\frac{\sigma_1^2 X_1 + \sigma_2^2 X_2}{\sigma_1^2 + \sigma_2^2}$ as noted above.&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-11-05T17:24:41.663" CommentCount="1" />
  <row Id="9082" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:27:00.863" Score="2" ViewCount="0" Body="&lt;p&gt;Taking log of $\rm\ sin^2(x)\ =\ 1 - cos^2(x)\ = (1-cos(x))\ (1+cos(x))\ $ shows both answers identical&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-05T17:27:00.863" />
  <row Id="9083" PostTypeId="2" ParentId="9075" CreationDate="2010-11-05T17:42:13.597" Score="0" ViewCount="0" Body="&lt;p&gt;Another way to integrate would be to let $\displaystyle t = \frac{\sin{x}}{\cos{x}}$, then $\displaystyle \frac{dt}{dx} = \frac{1}{\cos^2{x}} \Rightarrow dx = \cos^2{x}\;{dt}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $ \displaystyle I = \int\frac{\cos^2{x}}{\sin{x}}\;{dt} = \int\frac{\cos{x}}{\sin{x}}\;{dt} = \int\frac{1}{t}\;{dt} = \ln{t}+k = \ln\frac{\sin{x}}{\cos{x}}+k$. &lt;/p&gt;&#xA;" OwnerUserId="3167" LastEditorUserId="3167" LastEditorDisplayName="" LastEditDate="2010-11-05T17:48:23.537" LastActivityDate="2010-11-05T17:48:23.537" CommentCount="1" />
  <row Id="9084" PostTypeId="2" ParentId="9032" CreationDate="2010-11-05T18:18:06.713" Score="0" ViewCount="0" Body="&lt;p&gt;More generally, the minimal-variance unbiased linear combination of estimators is their reciprocal-variance-weighted average.&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-11-05T18:18:06.713" />
  <row Id="9085" PostTypeId="1" AcceptedAnswerId="9087" CreationDate="2010-11-05T18:59:15.770" Score="1" ViewCount="118" Body="&lt;p&gt;I am wondering about particle trajectories for solutions of the Navier-Stokes equation.  Is it possible that there is a Manifold $M$ for which fluid particles move along geodesic's or &quot;straight lines on the manifold&quot;?  If so, would it be fair to say that the Navier-Stokes equations are partial differential equations which &quot;embed&quot; this manifold in $R^3$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be somewhat analogous to the way that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_disc_model&quot; rel=&quot;nofollow&quot;&gt;Poincare Disk Model&lt;/a&gt; transforms straight lines in the hyperbolic plane into &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow&quot;&gt;arcs of circles that are orthogonal to the boundary circle&lt;/a&gt;.  Except in this case, the &quot;straight lines&quot; are transformed into fluid particle trajectories in $R^3$.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-05T19:14:25.333" LastActivityDate="2010-12-24T05:45:48.790" Title="Manifold interpretation of Navier-Stokes equations" Tags="&lt;differential-equations&gt;&lt;differential-geometry&gt;&lt;manifold&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="9086" PostTypeId="1" CreationDate="2010-11-05T19:05:51.967" Score="0" ViewCount="119" Body="&lt;p&gt;The rational root theorem states that, given $P \in \mathbb{Z}[x], P = \displaystyle\sum_{j=0}^n a_j x^j$ with $a_0 \neq 0$ and $a_n \neq 0$, if $P(p/q)=0, p\perp q$ then $p|a_0$ and $q|a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/8718/proving-gauss-polynomial-theorem&quot;&gt;this question&lt;/a&gt;, I'd like to know how can I prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0 = a_n \displaystyle\prod_{j=1}^n (-r_j)$, with $r_j$ being roots of P(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same as saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0 = a_n \frac{-p}{q} \displaystyle\prod_{j=2}^n (-r_j)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging this, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_n = q\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove that $\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)} \in \mathbb{Z}$, without using the rational root theorem (actually, I want to prove the rational root theorem using this identity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've tried to prove it by contradiction, assuming $a_0 = k(-p)\displaystyle\prod_{j=2}^n (-r_j) + \epsilon$ with $\epsilon \in (0, (-p)\displaystyle\prod_{j=2}^n (-r_j))$ but I don't seem to find anything weird rearranging that equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you manage to find a proof please just post a small hint. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-11-05T19:28:29.913" LastActivityDate="2010-12-09T05:44:25.243" Title="Proving a statement involving the rational root theorem." Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="9087" PostTypeId="2" ParentId="9085" CreationDate="2010-11-05T19:28:03.270" Score="7" ViewCount="0" Body="&lt;p&gt;For the Navier Stokes equation, I am rather doubtful, because the presence of the viscosity term makes the equation not time-symmetric. And the geodesic flow is necessarily time-symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove the viscosity term, you are down to the Euler equation, for which there is a well-known &lt;a href=&quot;http://terrytao.wordpress.com/2010/06/07/the-euler-arnold-equation/&quot; rel=&quot;nofollow&quot;&gt;characterisation due to V I Arnold&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-05T19:28:03.270" />
  <row Id="9088" PostTypeId="1" AcceptedAnswerId="9091" CreationDate="2010-11-05T19:42:42.627" Score="1" ViewCount="59" Body="&lt;p&gt;If you've got multiple arrays like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(24,36,28,28,16,27)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(38,38,45,57,35,50)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;every array being 6 integers, each integer in range [0,60]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find the distance between those 2 arrays.  The problem is related to computer science, however, I would like to interpret these arrays as &quot;planes&quot; so as to find the distance between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to find the distance between those 2 arrays?  Can we interpret them as 6 points on a plane?  Or maybe 3 coordinates?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-11-05T19:51:28.813" Title="Problem interpretation - Distance Formula?" Tags="&lt;geometry&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9089" PostTypeId="2" ParentId="9058" CreationDate="2010-11-05T19:44:51.207" Score="2" ViewCount="0" Body="&lt;p&gt;The RHS of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{2}-ax+144\equiv (x-b)^{2}\qquad (1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be expanded into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{2}-2bx+b^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So identity $(1)$ is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{2}-ax+144\equiv x^{2}-2bx+b^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2b-a)x+144-b^{2}\equiv 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;which can only be an identity&lt;/em&gt; (&lt;em&gt;i.e. independent of the value of&lt;/em&gt; $x$) &lt;em&gt;if&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=2b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^{2}=144.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system of equations has two solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=-12,a=-24$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=12,a=24.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-05T23:17:58.583" LastActivityDate="2010-11-05T23:17:58.583" />
  <row Id="9090" PostTypeId="2" ParentId="9057" CreationDate="2010-11-05T19:46:07.927" Score="0" ViewCount="0" Body="&lt;p&gt;If I understand properly, you have a histogram of A/B values for various proteins.  You don't need any model to find the most extreme, just find the 2.5 percentile and 97.5 percentile by counting.  The proteins outside these bounds are the 5% most extreme.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T19:46:07.927" CommentCount="1" />
  <row Id="9091" PostTypeId="2" ParentId="9088" CreationDate="2010-11-05T19:51:28.813" Score="4" ViewCount="0" Body="&lt;p&gt;You can interpret them any way you want, each array as three points in a plane (but using which order), two points in space, one point in six-dimensional space.  There are various distances available.  If you define the points as $a=(a_1,a_2,a_3,a_4,a_5,a_6)$ and similarly for $b$ you can use  $\sum |a_i-b_i|$ or $\sum (a_i-b_i)^2$, for example.  Note I didn't use the dimension at all.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-05T19:51:28.813" CommentCount="5" />
  <row Id="9092" PostTypeId="1" CreationDate="2010-11-05T20:01:16.257" Score="6" ViewCount="189" Body="&lt;p&gt;How can I compute the following limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{a \to 0^+} \left(a \int_{1}^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\mathrm dx\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints you can please give?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="3170" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T17:51:52.373" LastActivityDate="2010-12-02T18:41:42.373" Title="Compute $\lim_{a \to 0^+} \left(a \int_1^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\mathrm dx\right)$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9094" PostTypeId="2" ParentId="9092" CreationDate="2010-11-05T20:13:05.327" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: try a change of the independent variable $y:=ax$.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-05T20:13:05.327" CommentCount="4" />
  <row Id="9095" PostTypeId="1" AcceptedAnswerId="9097" CreationDate="2010-11-05T20:30:02.633" Score="1" ViewCount="82" Body="&lt;p&gt;Hi! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having some problem wrapping my head around an assignment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to find out how many additions, subtractions, multiplications and divisions are used while solving an NxM linear equation, while using Gauss Jordan elimination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2078" LastActivityDate="2010-11-05T20:41:27.907" Title="Gauss Jordan elimination - count of steps for NxM equation" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" />
  <row Id="9096" PostTypeId="2" ParentId="9069" CreationDate="2010-11-05T20:36:59.363" Score="0" ViewCount="0" Body="&lt;p&gt;So $E[W| \text{fire}] = \int_{0.1}^{\infty} [1-F(x)] \ dx = 4 \int_{0.1}^{\infty} (2+x)^{-2} \ dx = 4/2.1$. Thus $E[W] = (4/2.1)(0.001)+0 = 0.00190476$. So the premium is $(0.0019047)(10^6)(1.1) =2,095$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-05T20:36:59.363" />
  <row Id="9097" PostTypeId="2" ParentId="9095" CreationDate="2010-11-05T20:41:27.907" Score="3" ViewCount="0" Body="&lt;p&gt;It depends on how efficient you want to be, and it also depends on the rank of the matrix. But for a &quot;generic&quot; matrix, you can expect the rank to be $\min(N,M)$, and for &quot;most&quot; entries to be nonzero. (And, do you count multiplying zero by a constant as a &quot;multiplication&quot;? Adding two zeros as an &quot;addition&quot;?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpret your question as asking for the number of multiplications and additions of entries, rather than adding rows, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time you multiply a row by a constant, you are performing $M$ multiplications (minus the number of entries that are zero if they don't count); every time you add a row to another row you are performing $M$ additions/subtractions (minus the number of entries that are zero in both).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, generically, let us assume that the only zeros you will encounter are the ones you create throught Gauss-Jordan elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you get started? First you pick a pivot; then you use that pivot to eliminate the other entries in the column; there are $N-1$ entries to be eliminated; each elimination requires (i) multiplying your pivot row by a constant; and (ii) adding the pivot row to another row. So you are performing $M$ products and $M$ sums in each step, and you are doing this $N-1$ times. Then, you multiply the first row appropriately to get a $1$ in the pivot, for another $M$ products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you pick your next pivot, and you need to eliminate the entries below it on the column; there are $N-2$ entries; each entry requires (i) multiplying your pivot row by a constant, which requires $M-1$ products (first entry is $0$, you can ignore it), and then (ii) adding to the other row, requiring $M-1$ sums (first entry in both is $0$, you can ignore it). You do this $N-2$ times, and then multiply the first row by a constant ($M-1$ products again) to get a $1$ in the pivot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you pick your next pivot... etc. Figure out how many products/sums we are doing in each step, how many steps, add them all up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be an &lt;em&gt;upper bound&lt;/em&gt;, since in practice you could end up with lots of zeros, or a matrix that has rank strictly smaller than $\min(M,N)$, so that you get a row of zeroes early on, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading your title, though, you might be counting &quot;add a row to another row&quot; as a single step, and &quot;multiply a row by a constant&quot; as a single step; in that case, proceed as above but you only count the number of row operations instead of the number of individual numerical operations, and the number of common zeros between the rows is irrelevant.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-05T20:41:27.907" CommentCount="1" />
  <row Id="9098" PostTypeId="1" AcceptedAnswerId="9104" CreationDate="2010-11-05T21:21:42.770" Score="0" ViewCount="47" Body="&lt;p&gt;Dear Friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like, if possible, to obtain a proof of the theorem below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Being given real numbers &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;, with |a|&gt;|b|, and &lt;strong&gt;m&lt;/strong&gt; is a positive integer, the order &lt;strong&gt;p&lt;/strong&gt;, which occupies the maximum term (in absolute value) the development of power &lt;strong&gt;(a+b)^(-m)&lt;/strong&gt; , according to decreasing powers of &lt;strong&gt;a&lt;/strong&gt;, is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;p = 1 + integer part of [|b|(m-1)/(|a|-|b|)]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the expression &lt;strong&gt;|b|(m-1)/ (|a|-|b|)&lt;/strong&gt; is an integer, then there are maximum two terms (in absolute value): they are of order &lt;strong&gt;p&lt;/strong&gt; and &lt;strong&gt;p-1&lt;/strong&gt;. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already, most grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rio de Janeiro, Brazil&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-05T22:40:42.567" Title="Order of the maximum term of (a +b)^(-m)" Tags="&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="9099" PostTypeId="1" CreationDate="2010-11-05T21:22:26.263" Score="4" ViewCount="140" Body="&lt;p&gt;I have an equation system of the form Aix + Biy + Ciz = Di, where (x,y,z) is a unit vector, and (Ai, Bi, Ci, Di) are sets of measurements from a noisy system (with typically 3-5 independant readings). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first intuition to solve this problem was to pose this as an overdetermined linear equation system AX = B where X = (x,y,z), and to solve for X. However, with that approach, I have no way to enforce that the solution for vector X is a unit vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an elegant (or standard) solution to that problem, or should I simply dive into non-linear equation solving solutions?&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2010-11-06T02:17:56.420" Title="Solve a linear equation system, while enforcing a unit vector solution." Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9100" PostTypeId="2" ParentId="9099" CreationDate="2010-11-05T22:04:39.523" Score="3" ViewCount="0" Body="&lt;p&gt;Unfortunately, the problem is almost certainly innately non-linear, since your solution space is inherently so.  Pragmatically, the best approach is probably to treat your overdetermined linear system as a minimization-problem, least-squares style; this gives you a (positive definite) function to be minimized.  Find the (exact) minimum value in $\mathbb{R}^3$, project that down to $S^2$ (i.e., normalize your vector X) to find the closest point on the sphere, and use that as a starting point for some descent-based method (i.e., find the component of the gradient of your goodness-of-fit function tangent to the sphere, add that to your current point on $S^2$ and renormalize down to the sphere again, and keep doing that until you meet whatever convergence criterion you want to use.  There's no guarantee that this will work if your data is particularly ill-conditioned, but in practice it's likely to be the most straightforward method.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-05T22:04:39.523" />
  <row Id="9101" PostTypeId="2" ParentId="9099" CreationDate="2010-11-05T22:11:11.257" Score="1" ViewCount="0" Body="&lt;p&gt;Your objective is to solve $Ax=b$ subject to the constraint $||x||=1$. One way to do this is to recognize that the solution to $Ax=b$ is an affine space (translated subspace). This means that out of all the solutions to $Ax=b$, there is a unique &quot;smallest&quot; one, i.e. where $||x||$ is as small as possible. To find this solution, you can solve the minimium-norm optimization problem, whose solution is:&#xA;$$&#xA;x_0 = \lim_{\lambda \to 0^+} (A^TA+\lambda I)^{-1}A^Tb&#xA;$$&#xA;In the case where $A$ is skinny and full-rank, this simplifies to $(A^TA)^{-1}A^Tb$. In general, the solution can be found by taking the SVD of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have verified that $x_0$ is indeed a solution, the next thing to do is check that $||x_0|| \leq 1$, for if that were not the case, then you could confidently assert that there is no solution to $Ax=b$ with norm 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to find a basis for the nullspace of $A$, i.e. the set of vectors $x$ such that $Ax=0$. This can be done using the SVD again. If $N$ is a matrix whose columns form a basis for the nullspace of $A$, then the set of all solutions to $Ax=b$ is:&#xA;$$&#xA;x = x_0 + Ny&#xA;$$&#xA;where $y$ is an arbitrary vector of appropriate dimension. The key point here is that $x_0$ is orthogonal to all vectors in the nullspace of $A$ (because $x_0$ is the solution point closest to the origin). Thus, you may compute the norm of $x$ using the Pythagorean theorem:&#xA;$$&#xA;||x||^2 = ||x_0||^2 + ||Ny||^2&#xA;$$&#xA;Now we see that the if we want $||x||=1$, we simply need to choose $y$ so that&#xA;$$&#xA;||Ny|| = \sqrt{1 - ||x_0||^2}&#xA;$$&#xA;We may always choose the columns of $N$ to be orthonormal, so let's assume we have done that. Then the set of desired $y$ is precisely the sphere with radius $\sqrt{1 - ||x_0||^2}$. By choosing any $y$ on this sphere, we parametrize all solutions to your problem. If you only care about a particular solution, then you can choose $y$ of the form $[\alpha,0,...,0]$, where $\alpha = \sqrt{1 - ||x_0||^2}$.f&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditorDisplayName="" LastEditDate="2010-11-06T02:17:56.420" LastActivityDate="2010-11-06T02:17:56.420" CommentCount="7" />
  <row Id="9103" PostTypeId="2" ParentId="9099" CreationDate="2010-11-05T22:30:19.093" Score="3" ViewCount="0" Body="&lt;p&gt;You could also solve it as a quadratically constrained quadratic program (&lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratically_constrained_quadratic_program&quot; rel=&quot;nofollow&quot;&gt;QCQP&lt;/a&gt;).  The standard approach for solving an overdetermined linear system is least-squares (as with Steven Stadnicki's answer), which gives you a quadratic objective function.  Then your restriction that $X$ be a unit vector is equivalent to $$x^2 + y^2 + z^2 = 1,$$ which gives you a very nice quadratic constraint.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrices in your QCQP formulation will all be $2I$, where $I$ is the identity matrix, and so they will all be positive definite.  This means your QCQP problem can easily be solved using interior-point methods.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-05T22:30:19.093" />
  <row Id="9104" PostTypeId="2" ParentId="9098" CreationDate="2010-11-05T22:40:42.567" Score="0" ViewCount="0" Body="&lt;p&gt;The expansion is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{p=0}^{\infty} \binom{-m}{p} a^{-m-p} b^p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio between the absolute values of the terms corresponding to $p+1$ and to $p$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m+p}{p+1} \cdot \frac{|b|}{|a|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $p+1$st term is larger than or equal to the $p$th term if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m+p \geq (p+1)\frac{|a|}{|b|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \left(1 - \frac{|a|}{|b|}\right) \geq \frac{|a|}{|b|} - m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $|a| &amp;lt; |b|$, this is the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \leq \frac{|a|/|b|-m}{1-|a|/|b|} = \frac{|a|-m|b|}{|b|-|a|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p + 1 \leq \frac{(1-m)|b|}{|b|-|a|} = \frac{(m-1)|b|}{|a|-|b|}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the $p$th term is maximal if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \leq \frac{(m-1)|b|}{|a|-|b|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p+1 \geq \frac{(m-1)|b|}{|a|-|b|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the fraction is an integer $N$, then both $p = N$ and $p = N - 1$ are solutions. Otherwise, $p$ is the floor of the fraction, and I'm off by one wrt your solution.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-05T22:40:42.567" />
  <row Id="9105" PostTypeId="1" AcceptedAnswerId="9107" CreationDate="2010-11-05T22:50:11.627" Score="1" ViewCount="96" Body="&lt;p&gt;I have the following series of equations (n+2 equations n+2 variables):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_0q_0+\lambda q_0  + c_0 = 0$&lt;br&gt;&#xA;$k_1q_1+\lambda q_1  + c_1 = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_nq_n+\lambda q_n  + c_n = 0$&lt;br&gt;&#xA;$q_1+q_2+....+q_n = 1$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables are $q_0,q_1,.....,q_n$ and $\lambda$. k and c are series of constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I solve this?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-06T12:22:54.890" LastActivityDate="2010-11-06T12:22:54.890" Title="Solving Series of equations" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functional-equations&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9106" PostTypeId="2" ParentId="9105" CreationDate="2010-11-05T22:57:31.847" Score="2" ViewCount="0" Body="&lt;p&gt;It's not linear, first of all: you have two variables $\lambda$ and $q_i$ multiplied together. Luckily, your set of equations is algebraic; &lt;a href=&quot;http://mathworld.wolfram.com/GroebnerBasis.html&quot; rel=&quot;nofollow&quot;&gt;Gröbner&lt;/a&gt; may have a shot here.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-05T22:57:31.847" CommentCount="2" />
  <row Id="9107" PostTypeId="2" ParentId="9105" CreationDate="2010-11-05T23:10:06.827" Score="4" ViewCount="0" Body="&lt;p&gt;$\displaystyle q_i = \frac{-c_i}{k_i + \lambda}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in the last equation gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{n} \frac{-c_i}{k_i + \lambda} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be made into a polynomial equation in $\displaystyle \lambda$, which you should be able to solve by standard numerical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you find a root of the above polynomial, substituting $\displaystyle \lambda$ with the value of the root in $\displaystyle q_i = \frac{-c_i}{k_i + \lambda}$ will give the other variables.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-05T23:10:06.827" CommentCount="3" />
  <row Id="9108" PostTypeId="2" ParentId="9077" CreationDate="2010-11-05T23:11:47.427" Score="2" ViewCount="0" Body="&lt;p&gt;This was supposed to be a comment, but it got too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm personally not too enamored of the NR routine for Savitzky-Golay myself; it has to use Gaussian elimination (and thus a two-dimensional array) for what I think is a well-structured problem. Peter Gorry's proposal in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac00205a007&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; is a better choice, IMHO (a proper implementation merely requires a few one-dimensional arrays). In any event, looking at the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac60214a047&quot; rel=&quot;nofollow&quot;&gt;original article&lt;/a&gt; by Savitzky and Golay should help you a great deal, as well as F.B. Hildebrand's &lt;a href=&quot;http://books.google.com/books?id=hqucruPBheQC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Introduction to Numerical Analysis&lt;/em&gt;&lt;/a&gt;; Hildebrand does not explicitly refer to the Savitzky-Golay coefficients, but he gives the &lt;a href=&quot;http://books.google.com/books?id=hqucruPBheQC&amp;amp;pg=PA357&quot; rel=&quot;nofollow&quot;&gt;general idea&lt;/a&gt; of using least-squares fit polynomials for smoothing data.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-05T23:11:47.427" CommentCount="4" />
  <row Id="9109" PostTypeId="1" CreationDate="2010-11-05T23:22:43.680" Score="1" ViewCount="93" Body="&lt;p&gt;How to find a function which satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^x=\lim_{h\to\infty} \left( f_a \left(f_a^{[-1]}(x)+\frac xh\right)\right)^{[h]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f^{[n]}(x)$ is the number if iterations of a function (if n=-1 it's inverse function)?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T00:54:40.290" LastActivityDate="2010-11-06T00:54:40.290" Title="How to find a function which satisfies such functional equation?" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;functional-equations&gt;&lt;dynamical-systems&gt;" CommentCount="2" />
  <row Id="9110" PostTypeId="1" CreationDate="2010-11-05T23:35:48.137" Score="0" ViewCount="50" Body="&lt;p&gt;Is there a simple/efficient way of enumerating the elements of a free distributive lattice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm doing some computations with them, and listing their elements in what's probably the least efficient way possible...)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-05T23:42:29.520" LastActivityDate="2010-11-06T00:12:14.807" Title="Enumerating the elements of free distributive lattices" Tags="&lt;lattices&gt;" AnswerCount="1" />
  <row Id="9111" PostTypeId="2" ParentId="5008" CreationDate="2010-11-06T00:11:30.410" Score="2" ViewCount="0" Body="&lt;p&gt;It is probably worth mentioning that another name for this is the category of abstract (not necessarily separated) algebraic $k$-varieties.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-06T08:31:51.333" LastActivityDate="2010-11-06T08:31:51.333" CommentCount="2" />
  <row Id="9112" PostTypeId="2" ParentId="9110" CreationDate="2010-11-06T00:12:14.807" Score="1" ViewCount="0" Body="&lt;p&gt;It's equivalent to enumerating monotone Boolean functions, e.g. see Fidytek et al. &lt;a href=&quot;http://www.ii.uni.wroc.pl/~lorys/IPL/article79-6-4.pdf&quot; rel=&quot;nofollow&quot;&gt;Algorithms counting monotone Boolean functions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-06T00:12:14.807" CommentCount="1" />
  <row Id="9113" PostTypeId="2" ParentId="9092" CreationDate="2010-11-06T00:25:38.313" Score="1" ViewCount="0" Body="&lt;p&gt;Dear Andrey, Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I end up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{a \to 0+} \int _{a}^{\infty} e^{-y} cos \left( \frac{2\pi}{1+(y/a)^{2}}\right) dy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I'm stuck. How do I apply DCT here?&lt;/p&gt;&#xA;" OwnerUserId="3176" LastActivityDate="2010-11-06T00:25:38.313" CommentCount="1" />
  <row Id="9114" PostTypeId="2" ParentId="9105" CreationDate="2010-11-06T01:55:59.887" Score="0" ViewCount="0" Body="&lt;p&gt;Define $c_{n+1} = 1$ and $q_{n+1} = 0$. Observe that we can write the above system in the following $(n+2) \times (n + 2)$-matrix form: $K_{\lambda} \mathbf{q} = - \mathbf{c}$, where $K_{\lambda}$ is $\text{diag}(k_{i} + \lambda)$ for the upper $(n+1) \times (n+1)$-submatrix, ($0$ $1$ $1$ $\cdots$ $1$ $0$) for the bottom row and ($0$ $0$ $0$ $\cdots$ $0$ $1$)$^{\top}$ for the right most column. (By enlarging the matrix, we have encoded the constraint $\sum_{i = 1}^{n} q_{i} = 1$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\lambda$ is unknown, we can proceed presupposing that $K_{\lambda}$ is not singular (i.e., $\det K_{\lambda} \neq 0$), and solve the constrained system by matrix inversion: $\mathbf{q} =  - K_{\lambda}^{-1} \mathbf{c}$. The result is a vector $\mathbf{q}$ in terms of the unknown $\lambda$ and given constants. To find the allowed values for $\lambda$, numerically (or analytically) solve the polynomial $\mathbf{q} \cdot \mathbf{1} - 1 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-06T02:35:38.043" LastActivityDate="2010-11-06T02:35:38.043" />
  <row Id="9115" PostTypeId="1" AcceptedAnswerId="9117" CreationDate="2010-11-06T02:47:57.693" Score="5" ViewCount="165" Body="&lt;p&gt;So, I'm relearning Group Theory.  And I got the axioms down, I think.  So let's make a concrete example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The collection of numbers the positive integers less than 7:  1,2,3,4,5,6&lt;/li&gt;&#xA;&lt;li&gt;The • operation will be multiplication mod 7.  &lt;/li&gt;&#xA;&lt;li&gt;Associativity holds.&lt;/li&gt;&#xA;&lt;li&gt;The Identity e is 1.&lt;/li&gt;&#xA;&lt;li&gt;Every element has an inverse: &#xA;&lt;ul&gt;&#xA;&lt;li&gt;1*? mod 7 = 1 --&gt; 1&lt;/li&gt;&#xA;&lt;li&gt;2*? mod 7 = 1 --&gt; 4&lt;/li&gt;&#xA;&lt;li&gt;3*? mod 7 = 1 --&gt; 5&lt;/li&gt;&#xA;&lt;li&gt;4*? mod 7 = 1 --&gt; 2&lt;/li&gt;&#xA;&lt;li&gt;5*? mod 7 = 1 --&gt; 3&lt;/li&gt;&#xA;&lt;li&gt;6*? mod 7 = 1 --&gt; 6&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;!  What is the &lt;em&gt;order&lt;/em&gt; of the group?!  I thought the order would be 7.  But there are 6 elements!  So maybe I was wrong and 0 should be in the group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 0 does not have an inverse!  There is no x such that 0*x mod 7 = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what am I misunderstanding here?  Is it the definition of order?  Is it some other trick about groups?&lt;/p&gt;&#xA;" OwnerUserId="3179" LastActivityDate="2010-11-06T02:57:08.880" Title="What happens to the 0 element in a Finite Group?" Tags="&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9116" PostTypeId="1" AcceptedAnswerId="9170" CreationDate="2010-11-06T02:48:00.370" Score="3" ViewCount="123" Body="&lt;p&gt;I am seeking conditions on the distribution of the step sizes that guarantee that a  random walk on the 2D lattice will return to the origin (with probability 1).  Essentially, under what conditions can Pólya's theorem be proved?  Certainly it holds if the steps are of size 1 (and equally probable in all directions), and I believe if the variance of the step sizes is finite, the return probability is still 1.&#xA;But what about distributions with infinite variance, like the &lt;a href=&quot;http://en.wikipedia.org/wiki/L%C3%A9vy_distribution&quot; rel=&quot;nofollow&quot;&gt;Lévy distribution&lt;/a&gt;?&#xA;Does &quot;Lévy flight&quot; return to the origin?  And more generally, are there conditions on&#xA;the distribution that guarantee this return?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is likely all well known, and if so, pointers to the relevant literature would be appreciated.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-11-06T13:13:47.677" LastActivityDate="2010-11-06T15:27:01.460" Title="Conditions for 2D random walk to return to origin" Tags="&lt;probability-theory&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="9117" PostTypeId="2" ParentId="9115" CreationDate="2010-11-06T02:51:15.143" Score="9" ViewCount="0" Body="&lt;p&gt;The only error is your belief that the order &quot;should&quot; be 7. The order of a finite group is just the number of elements in the group. Your group consists of the positive integers that are smaller than, and relatively prime to, 7. There are six of them, so your group has order 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not sure why you thought the order should be 7...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, you cannot add $0$ to the mix and still have a group. If you consider the numbers $0,1,\ldots,6$ under multiplication modulo $7$ you do not get a group, you get a &lt;a href=&quot;http://en.wikipedia.org/wiki/Semigroup&quot; rel=&quot;nofollow&quot;&gt;semigroup&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Ah, Jonas Meyer's reply suggests what is going on; since you say you are &lt;em&gt;relearning&lt;/em&gt; Group Theory, you might have vague memories of the &quot;group of integers modulo $n$&quot; as having order $n$. The group of integers modulo $n$ &lt;em&gt;under addition&lt;/em&gt; has order $n$; but the &lt;strong&gt;multiplicative group&lt;/strong&gt; modulo $n$ consists of the positive integers less than, and relatively prime to, $n$, with the operation being multiplication modulo $n$, and has $\varphi(n)$ elements (Euler's phi function). When $n=7$ (the case you are looking at), the group has $\varphi(7)=6$ elements, as you observed.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-06T02:57:08.880" LastActivityDate="2010-11-06T02:57:08.880" CommentCount="1" />
  <row Id="9118" PostTypeId="2" ParentId="9115" CreationDate="2010-11-06T02:51:25.720" Score="7" ViewCount="0" Body="&lt;p&gt;You're right, the group has order 6 because it has six elements.  You can make {0,1,2,3,4,5,6} a group with addition mod 7.  This would be a group of order 7.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-06T02:51:25.720" />
  <row Id="9119" PostTypeId="2" ParentId="9115" CreationDate="2010-11-06T02:51:28.473" Score="5" ViewCount="0" Body="&lt;p&gt;Good thinking.  0 is not in the group, so the order is 6.  0 is in another group in the multiplicative $\mathbb{Z}/\mathbb{Z}_7$, which is the trivial group (one element).  The groups are disjoint.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-06T02:51:28.473" />
  <row Id="9120" PostTypeId="1" AcceptedAnswerId="9121" CreationDate="2010-11-06T03:15:30.023" Score="0" ViewCount="46" Body="&lt;p&gt;If we can say that we're able to determine the weather at 10ms intervals, and if we observe that over a whole second (1000 samples) that it's raining at every sample, what is the accuracy rate within which we've calculated the weather for the second? I'm sure this sounds like homework to some, but it's not. I was reading a statement about sampling rates that claimed a sample every 10ms, and that the result was within 10% accuracy. I am not sure how they came up with this guarantee.&lt;/p&gt;&#xA;" OwnerUserId="3180" LastActivityDate="2010-11-06T03:34:41.717" Title="How can you calculate the accuracy rate of the following set of statements?" Tags="&lt;statistics&gt;" AnswerCount="1" />
  <row Id="9121" PostTypeId="2" ParentId="9120" CreationDate="2010-11-06T03:34:41.717" Score="0" ViewCount="0" Body="&lt;p&gt;That depends whether you calculated that it would be raining or not.  The accuracy is 100% if yes, 0% if no.  Flip response aside, it is hard to claim an accuracy rate for a single digital event.  10% accuracy should either refer to an analog prediction, in which case you should be within 10% of the measured value, or a many event digital prediction, in which case the observed rate should be within 10% of your prediction.  But in the popular press, &quot;10% accuracy&quot; has no numeric content, it just means &quot;pretty good, but not perfect&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-06T03:34:41.717" />
  <row Id="9122" PostTypeId="1" AcceptedAnswerId="9124" CreationDate="2010-11-06T05:29:21.267" Score="3" ViewCount="159" Body="&lt;p&gt;I have a generating function that I'm trying to create.  It has a general form to it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + cx + c(c-1)x^2 + c(c-1)(c-2)x^3 + \dots + c!x^c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see a closed form for this function.  I'd like something other than a summation.  I'm particularly interested if someone could walk through how to derive the form for me.  I was hoping that someone could carry on with what I've done and show me how to solve what's left.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My attempt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I try to create a function $A(x)$ (which will be the closed form) using recurrences.  Here I'm using techniques (and especially notation) from &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;Wilf's &lt;em&gt;Generatingfunctionology&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I set $a_0=1$.  I then proceed to multiply by $(c-n)$.  My guess is that the math should be something like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1}=c \cdot a_n - \frac{d A(x)}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to say that the next term should be $c$ times the previous term minus $n$ times the previous term.  The latter is represented by the derivative, I believe.  So I guess that this could be correct.  Unfortunately, I don't know for sure, and so I'm kind of lost at this point.  I would greatly appreciate if someone could take it from here and explain the rest in detail.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Differential Equation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that the last part is correct, the function becomes a differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A(x)-1}{x}=c \cdot A(x) - \frac{d A(x)}{dx}$&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-06T05:40:20.210" LastActivityDate="2010-11-08T05:30:16.450" Title="Help with a generating function and differential equation..." Tags="&lt;sequences-and-series&gt;&lt;homework&gt;&lt;differential-equations&gt;&lt;recursive&gt;&lt;generating-functions&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="9123" PostTypeId="1" AcceptedAnswerId="9125" CreationDate="2010-11-06T05:37:46.653" Score="1" ViewCount="129" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In maths, you can use something as&#xA;  simple as statistical analysis to&#xA;  intuit the theory, and in comp-science&#xA;  you can use simulators.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the above statement true for maths ?&lt;/p&gt;&#xA;" OwnerUserId="3027" LastActivityDate="2010-11-06T07:21:40.307" Title="Is the above statement true for maths ?" Tags="&lt;soft-question&gt;&lt;statistics&gt;&lt;math-history&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9124" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T06:20:22.180" Score="3" ViewCount="0" Body="&lt;p&gt;I believe that your initial attempt is incorrect. If I solve the suggested differential equation, I get&#xA;\begin{eqnarray}&#xA;A(x) = -\frac{1}{cx} \left( 1 - e^{cx} \right) = \sum_{k \geq 1} \frac{(cx)^{k-1}}{k!},&#xA;\end{eqnarray}&#xA;where I have chosen the initial condition to ensure $1$ is the first term of the series expansion. The generating function that you are seeking is the following:&#xA;\begin{eqnarray}&#xA;e^{1/x} \left( x^{c} \Gamma(c+1, \tfrac{1}{x}) - c (-x)^{c} \Gamma(0, \tfrac{1}{x}) (1 -c)^{(c)} \right) = 1 + \sum_{k = 1}^{c} c(c-1) \cdots (c-k+1) \ x^{k},&#xA;\end{eqnarray}&#xA;which may be simplified to&#xA;\begin{eqnarray}&#xA;e^{1/x} \left( \frac{(-x)^{c} \ \Gamma(0, \frac{1}{x})}{\Gamma(-c)} + x^{c} \ \Gamma(c + 1, \tfrac{1}{x}) \right) = 1 - \sum_{k = 1}^{c} c (1 - c)^{(k-1)} \ (-x)^{k},&#xA;\end{eqnarray}&#xA;where $\Gamma(n,x)$ is the incomplete gamma function defined as&#xA;\begin{eqnarray}&#xA;\Gamma(n,x) = \int_{x}^{\infty} t^{n-1} e^{-t} dt = (n-1)! e^{-x} \sum_{k = 0}^{n-1} \frac{x^{k}}{k!}&#xA;\end{eqnarray}&#xA;and $(x)^{(n)}$ is the Pochhammer symbol or rising factorial, $x(x+1)\cdots (x+n-1)$. This form makes no assumption on the integrality of $c$. However, if $c$ is a positive integer, then the formula simplifies greatly. For example, if $c = 3$, the left side specializes to &#xA;\begin{eqnarray}&#xA;e^{1/x} x^3 \Gamma(4, \tfrac{1}{x}) = 1 + 3 x + 6 x^{2} + 6 x^{3}. &#xA;\end{eqnarray}&#xA;NB: There is no extra term. The last exponent of $x$ is indeed $c$ with coefficient $c!$. In fact, the last coefficient is repeated because $c(c-1) \cdots 2 = c(c-1) \cdots 2 \cdot 1 = c!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the generating function for integer $c$ is &#xA;\begin{eqnarray}&#xA;e^{1/x} x^{c} \ \Gamma(c+1, \tfrac{1}{x}).&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-08T05:30:16.450" LastActivityDate="2010-11-08T05:30:16.450" CommentCount="5" />
  <row Id="9125" PostTypeId="2" ParentId="9123" CreationDate="2010-11-06T07:21:40.307" Score="2" ViewCount="0" Body="&lt;p&gt;If I'm understanding the question correctly, I suppose the OP is aiming to posit that in mathematics we &lt;em&gt;verify&lt;/em&gt; a mathematical truth in nature by experimentation and statistical data analysis versus how we verify a computational truth by seeing it in practice through simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to be overly pedantic, but to me a mathematical or computational result is true when it is the conclusion of a rigorous application of sound thought through a sequence of logical steps from an initial set of assumptions within a particular (logical) framework. I see no distinction between mathematical or computational truths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, when we try to apply mathematical or computational rules to nature, the results aren't always so easy to interpret, because &lt;em&gt;truth&lt;/em&gt; isn't well-defined. This is why statistics is so applicable -- it gives us a way to quantify the grey area. Through simulations we can test ideas in an ideal setting, so the grey area is as big as we allow it to be. &lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-06T07:21:40.307" />
  <row Id="9126" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T08:05:26.663" Score="2" ViewCount="0" Body="&lt;p&gt;The general form of your sum can be expressed succinctly as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum (-c)_k (-x)^k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(a)_k$ is a Pochhammer symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We recognize at once that this is a hypergeometric series; specifically, it is a ${}_2 F_0$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${}_2 F_0 \left(-c,1;;-x\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as a &lt;a href=&quot;http://dlmf.nist.gov/13.6.vi&quot; rel=&quot;nofollow&quot;&gt;Tricomi confluent hypergeometric function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{x}U\left(1,c+2,\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be re-expressed as an &lt;a href=&quot;http://dlmf.nist.gov/8.5.E3&quot; rel=&quot;nofollow&quot;&gt;incomplete gamma function&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^c\exp\left(\frac1{x}\right)\Gamma\left(c+1,\frac1{x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the differential equation, it can be obtained from &lt;a href=&quot;http://functions.wolfram.com/07.33.13.0011.01&quot; rel=&quot;nofollow&quot;&gt;this formula&lt;/a&gt; to give:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}-\left(c-2-\frac1{x}\right)x\frac{\mathrm{d}y}{\mathrm{d}x}-cy=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the solution of interest satisfies the initial conditions $y(0)=1$ and $y^{\prime}(0)=c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave others to flesh out the details.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T09:58:24.620" LastActivityDate="2010-11-06T09:58:24.620" CommentCount="5" />
  <row Id="9127" PostTypeId="1" CreationDate="2010-11-06T08:06:29.767" Score="2" ViewCount="72" Body="&lt;p&gt;Say I have a function    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V(x)=A(x-x1)(x-x2)(x-x3)&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;x1&lt;/strong&gt;, &lt;strong&gt;x2&lt;/strong&gt;, &lt;strong&gt;x3&lt;/strong&gt; are the three roots in increasing order and &lt;strong&gt;A&lt;/strong&gt; is positive. Clearly &lt;strong&gt;V(x)&lt;/strong&gt; is positive at large &lt;strong&gt;x&gt;x3&lt;/strong&gt;, negative between &lt;strong&gt;x2&lt;/strong&gt; and &lt;strong&gt;x3&lt;/strong&gt;, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say I wish to   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Integrate Sqrt[-V(x)]&lt;/strong&gt; between the limits &lt;strong&gt;x2&lt;/strong&gt; and &lt;strong&gt;x3&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I do it?&lt;/p&gt;&#xA;" OwnerUserId="3181" LastActivityDate="2010-11-06T08:24:47.603" Title="Integrals of the square root of a cubic polynomial" Tags="&lt;integral&gt;" AnswerCount="2" />
  <row Id="9128" PostTypeId="2" ParentId="9127" CreationDate="2010-11-06T08:16:01.663" Score="3" ViewCount="0" Body="&lt;p&gt;This is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_integral&quot; rel=&quot;nofollow&quot;&gt;elliptic integral&lt;/a&gt; and can be expressed&#xA;in terms of the complete elliptic integral of the second kind.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-06T08:16:01.663" />
  <row Id="9129" PostTypeId="2" ParentId="9127" CreationDate="2010-11-06T08:18:13.450" Score="3" ViewCount="0" Body="&lt;p&gt;Well, you can't integrate it in terms of elementary functions like the arctangent or the logarithm in general (unless there is some special configuration of your roots). Having an integrand that contains the square root of a cubic will result in the use of so-called &quot;&lt;a href=&quot;http://dlmf.nist.gov/19&quot; rel=&quot;nofollow&quot;&gt;elliptic integrals&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't give further details, but I would suggest looking at Abramowitz and Stegun, particularly &lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/page_596.htm&quot; rel=&quot;nofollow&quot;&gt;this section&lt;/a&gt;, or any other book that discusses elliptic integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387053182&quot; rel=&quot;nofollow&quot;&gt;Byrd-Friedman&lt;/a&gt; would probably be the best reference on this; they have a comprehensive listing of formulae for reducing elliptic integrals to the Legendre-Jacobi forms.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T08:24:47.603" LastActivityDate="2010-11-06T08:24:47.603" />
  <row Id="9130" PostTypeId="1" AcceptedAnswerId="9166" CreationDate="2010-11-06T08:29:24.350" Score="3" ViewCount="119" Body="&lt;p&gt;Let $a_n$ be a sequence of real numbers such that $\exp(it\cdot a_n)$ converges to 1 for all real t.&#xA;Show that $a_n$ converges to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can show this by letting $X_n$ be degenerate random variables which take value $a_n$ with prob 1 and then using results about characteristic functions and convergence.&#xA;But this seems like overkill and it doesn't provide any intuition for what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like there should be a simpler proof based on basic properties of continuous functions and convergent sequences.&#xA;Any ideas ?&lt;/p&gt;&#xA;" OwnerUserId="3182" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T14:51:31.320" LastActivityDate="2010-11-06T17:41:56.290" Title="Elementary solution of convergence problem ?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="9131" PostTypeId="1" AcceptedAnswerId="9136" CreationDate="2010-11-06T09:29:29.877" Score="3" ViewCount="87" Body="&lt;p&gt;Hi! What do I miss in the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a commutative Noetherian ring with unit. A map $f:M\to N$ of $R$-modules is injective/surjective iff the associated map $f_p:M_p\to N_p$ on the localization is injective/surjective for every prime ideal $p$ of the ring $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an isomorphism $M_p=M\otimes_R R_p$. A sequence&#xA;$$0\to K\xrightarrow{k} M\xrightarrow{f} N\to 0$$&#xA;is exact iff $k$ is injective and $s$ is surjective. Hence the above sequence is exact iff for every prime ideal $p$ of $R$ the sequence&#xA;$$0\to K\otimes_R R_p\xrightarrow{k'} M\otimes_R R_p\xrightarrow{f'} N\otimes_R R_p\to 0$$&#xA;is exact. But $R_p$ is not flat over $R$ in general, is it? What am I missing? &#xA;Is the sequence still exact if I tensor with the residue field $R_p/m_p$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3183" LastActivityDate="2010-11-06T16:33:34.480" Title="Flatness of local rings" Tags="&lt;algebra&gt;&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="9132" PostTypeId="1" AcceptedAnswerId="9155" CreationDate="2010-11-06T09:34:04.653" Score="1" ViewCount="54" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any known bound on sum of independent but not identically distributed geometric random variables?&#xA;I have to show that the tail of the sum drops exponentially (like in the Chernoff bounds for the sum of iid geom. variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, if $X_i \sim Geom(p_i)$, $X = \sum_{i=1}^n X_i$, and $E[X]=\Theta(n)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to show that $\Pr(X &amp;lt; 2E[X]) &gt; 1 - \delta ^n$, where $\delta &amp;lt; 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance,&#xA;Michael.&lt;/p&gt;&#xA;" OwnerUserId="3184" LastEditorUserId="3184" LastEditorDisplayName="" LastEditDate="2010-11-08T10:35:08.367" LastActivityDate="2010-11-08T10:35:08.367" Title="Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v. " Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="9133" PostTypeId="1" CreationDate="2010-11-06T09:41:50.567" Score="1" ViewCount="73" Body="&lt;p&gt;I was asked to prove the next formula - &#xA;(A+BC'+C)C' = ABC'+AB'C'+A'BC'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i need to show all the stages of the simplification, i have all of the rules/identities.&lt;br&gt;&#xA;i have tried many times and i did not make it, i always get something else &#xA;thank you&lt;/p&gt;&#xA;" OwnerUserId="3185" LastActivityDate="2010-11-06T11:50:41.527" Title="Boolean algebra,  ABC'+AB'C'+A'BC' = (A+BC'+C)C', can anyone show it please " Tags="&lt;boolean-algebra&gt;" AnswerCount="1" />
  <row Id="9134" PostTypeId="1" AcceptedAnswerId="9135" CreationDate="2010-11-06T09:54:49.203" Score="2" ViewCount="115" Body="&lt;p&gt;Please suggest suitable approach for this problem&lt;/p&gt;&#xA;" OwnerUserId="3187" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T11:00:10.720" LastActivityDate="2010-11-06T11:00:10.720" Title="Find the maximum and minimum value of $ \sin ^4 \theta + \cos ^4 \theta $ ?" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9135" PostTypeId="2" ParentId="9134" CreationDate="2010-11-06T10:03:11.197" Score="4" ViewCount="0" Body="&lt;p&gt;Note that&#xA;$$1 = (\sin^2\theta + \cos^2\theta)^2$$&#xA;and use $\sin2\theta = 2\sin\theta\cos\theta.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-06T10:03:11.197" CommentCount="1" />
  <row Id="9136" PostTypeId="2" ParentId="9131" CreationDate="2010-11-06T10:08:29.520" Score="1" ViewCount="0" Body="&lt;p&gt;Your characterisation of an exact sequence is missing exactness at $M$.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-11-06T10:08:29.520" CommentCount="1" />
  <row Id="9137" PostTypeId="1" AcceptedAnswerId="9165" CreationDate="2010-11-06T10:23:43.480" Score="0" ViewCount="30" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;x    |     0     1     2     3     4     5&#xA;------------------------------------------&#xA;f(x) |    .7    .2   .05   .03   .01   .01&#xA;F(x) |    .7    .9   .95   .98   .99     1&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where f(x) is the density function for x and F(x) is the cumulative distribution function for x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;And I am being asked to use F to verify that P[X=3]=0.03&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that I should start with P[X=3]=F(2) ... something ... something, since F(x)=P(X&amp;lt;=x) but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="713" LastActivityDate="2010-11-06T14:40:23.590" Title="How to use the cumulative dist. fun to verify an exact density (Probability)" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="9138" PostTypeId="1" AcceptedAnswerId="9185" CreationDate="2010-11-06T10:38:57.543" Score="5" ViewCount="326" Body="&lt;p&gt;Hello alltogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dealing with a physics Problem I get the following equation to solve for $\alpha$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this in Mathematica gives the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a==2 \text{ArcTan}\left[\frac{1-m}{1+m}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I am unable to get this result myself. No matter what I try normal equation transformations or rewriting the equation with the complex e-Function ..., everything fails. Even going the other Direction from Mathematica's Solution to my original equation resulted in nothing sensible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help of how to do do this transformation is very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T17:06:00.067" LastActivityDate="2010-11-11T17:06:00.067" Title="Solve trigonometric equation: $1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$" Tags="&lt;trigonometry&gt;" AnswerCount="7" CommentCount="0" />
  <row Id="9139" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T10:48:30.203" Score="5" ViewCount="0" Body="&lt;p&gt;So you have $1 = m\cos{\alpha} + \sin{\alpha}$. Now you have $1+m = m(\cos{\alpha}+1) + \sin\alpha$ and $1-m = m(\cos\alpha -1) + \sin{\alpha}$. So then you have $$\frac{1-m}{1+m} = \frac{m(\cos\alpha-1) + \sin\alpha}{m(\cos\alpha +1) + \sin\alpha} = \frac{-m \tan\frac{\alpha}{2} + 1}{m\cot\frac{\alpha}{2} + 1}$$ using the half angle formulae's. Then cross multiplying and solving you should be able to get the value of $\alpha$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-07T09:39:50.823" LastActivityDate="2010-11-07T09:39:50.823" CommentCount="7" />
  <row Id="9140" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T10:49:16.317" Score="6" ViewCount="0" Body="&lt;p&gt;Consider the equation&#xA;$$a\cos t+b\sin t=c$$&#xA;as an equation for $t$. Here $a$ and $b$ are real numbers, not both zero.&#xA;The trick to solve this is to divide by $\sqrt{a^2+b^2}$ to get&#xA;$$\frac{a}{\sqrt{a^2+b^2}}\cos t+\frac{b}{\sqrt{a^2+b^2}}\sin t&#xA;=\frac{c}{\sqrt{a^2+b^2}},$$&#xA;equivalently&#xA;$$a'\cos t+b'\sin t=c'$$&#xA;where now $a'^2+b'^2=1$. This means that $(a',b')$ lies on the unit circle.&#xA;There is then an angle $u$ so that $(a',b')=(\cos u,\sin u)$.&#xA;The equation becomes&#xA;$$\cos u\cos t+\sin u\sin t=c',$$&#xA;that is&#xA;$$\cos(t-u)=c'.$$&#xA;We see there is a solution for $t$ iff $|c'|\le1$. In general when $|c'|&amp;lt;1$&#xA;there are exactly two solutions for $t$ modulo $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All of this is standard A-level maths.)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-06T10:49:16.317" CommentCount="6" />
  <row Id="9141" PostTypeId="1" AcceptedAnswerId="9144" CreationDate="2010-11-06T11:44:47.157" Score="-1" ViewCount="160" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone provide, please, a proof of the theorem below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Being &lt;strong&gt;x&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; integers greater than 1, which can not be represented as powers of the same basis(positive integer) and &lt;strong&gt;integer exponent&lt;/strong&gt;, then the logarithm of &lt;strong&gt;x&lt;/strong&gt;, in base &lt;strong&gt;b&lt;/strong&gt;, is an irrational number.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well... Assuming that the logarithm is a rational number p/q (p and q are relatively prime integers), I know I can write &lt;strong&gt;x^q&lt;/strong&gt; = &lt;strong&gt;b^p&lt;/strong&gt;, but I can not conclude from this fact that &lt;strong&gt;b&lt;/strong&gt; e &lt;strong&gt;x&lt;/strong&gt; are powers of a same integer (and with integer exponent).&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-07T11:33:33.857" LastActivityDate="2010-11-07T11:33:33.857" Title="About irrational logarithms" Tags="&lt;logarithms&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9142" PostTypeId="2" ParentId="9133" CreationDate="2010-11-06T11:50:41.527" Score="1" ViewCount="0" Body="&lt;p&gt;opening the brackets is done as with &quot;regular&quot; algebra, which yields:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;AC' + BC'C' + CC'&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from your identites:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;C'C' = C'  &#xA;CC' = 0  &#xA;X + 0 = X  &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which imply that the above expression can be simplified to:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;AC' + BC'&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from here you should look to expand the expression by the following identities:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;X*1 = X  &#xA;X + X' = 1  &lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which yields:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;AC'(B + B') + BC'(A + A')&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;expand the brackets as usual, change the order of multiplication with the identity XY = YX, cancel common terms based on X + X = X, and youre done...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;someone should add a h/w tag to this question. i dont have any points...&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2010-11-06T11:50:41.527" />
  <row Id="9143" PostTypeId="2" ParentId="9072" CreationDate="2010-11-06T11:56:25.063" Score="6" ViewCount="0" Body="&lt;p&gt;I think the answer really depends on what you mean by &quot;fathom&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a short construction of the Monster, there is a sketch by Conway in one of the later chapters of &lt;em&gt;Sphere Packings, Lattices, and Groups&lt;/em&gt; (occasionally available on Google Books).  The construction there goes through a nice progression of increasingly complicated exceptional objects, like the Golay code and the Leech lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to understand a &quot;natural&quot; object on which the Monster acts by symmetries, you should read up about vertex algebras.  At our current state of knowledge (and depending on who you ask), the most natural object on which the monster acts is the monster vertex algebra $V^\natural$ (also known as the moonshine module, or the monster VOA), which is a graded vector space, together with some extra structure like a rather complicated multiplication operation $V^\natural \otimes V^\natural \to V^\natural((z))$.  The construction of $V^\natural$ is given in the book &lt;em&gt;Vertex operator algebras and the Monster&lt;/em&gt; by I. Frenkel, Lepowsky, and Meurman.  A string theorist might say that it is given by orbifolding the Leech lattice CFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to consider some basic facts about the Monster, you can have a look at the ATLAS of finite groups, or play around with the software GAP.  Both have the character table, and orders of centralizers of elements, etc.  Wilson showed that the monster is a Hurwitz group, so Ryan Budney's comment about acting on a Riemann surface holds for the minimum possible genus (about $10^{52}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to understand finer points about the structure of the Monster group, you're pretty much out of luck.  It's big enough that there are plenty of explicit questions whose answers we don't know.  For example, the conjugacy classes of homomorphisms from $\mathbb{Z} \times \mathbb{Z}$ (i.e., pairs of commuting elements) are not classified, and $H^4$ with coefficients in $\mathbb{Z}$ is still unknown (very annoying).&lt;/p&gt;&#xA;" OwnerUserId="3189" LastActivityDate="2010-11-06T11:56:25.063" />
  <row Id="9144" PostTypeId="2" ParentId="9141" CreationDate="2010-11-06T11:57:00.697" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\log_b\;x$ were a rational number $\frac{p}{q}$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^q=b^p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, go back to your initial assumption...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add: &lt;a href=&quot;http://www.southalabama.edu/mathstat/personal_pages/feldvoss/logintirrat.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; might be of interest.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-06T12:08:48.683" LastActivityDate="2010-11-06T12:08:48.683" CommentCount="2" />
  <row Id="9145" PostTypeId="1" AcceptedAnswerId="9856" CreationDate="2010-11-06T12:16:30.500" Score="0" ViewCount="63" Body="&lt;p&gt;Hey guys, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider n loaded dice with differing probability distributions. I have constructed an equation that tells me the probability mass, the number of times a certain number is expected to be thrown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i want to extend the numerical range from a line onto a plane. How do i get the expected number of times a number/number combination is thrown now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two dicerolls are independent. Can i compute the probability mass for the combination using my equation for computing the probability mass for the line? Since the probability mass is not a probability i cannot just multiply those. What will i have to change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the advice&lt;/p&gt;&#xA;" OwnerUserId="2362" LastActivityDate="2010-11-11T14:05:45.213" Title="Gaussian Distributions and probability mass" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9146" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T12:28:34.407" Score="0" ViewCount="0" Body="&lt;p&gt;I found a differential equation which gives the correct result under &lt;em&gt;Mathematica 7.0.1&lt;/em&gt;.  The differential equation is the same one I derived in the question (see my comment on adding x):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{A(x)-1}{x}=c \cdot A(x) - \frac{d A(x)}{dx}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result it gives as the solution (besides the series) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{-1/x}x^{-c}(C_1 - x^{c-1}E_c(-\frac{1}{x}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E_c(z)$ is the Exponential Integral Function, for which the following equation holds: $-x^{c-1}E_c(b)$ = $b^c x^c \Gamma[1-c,b]$ and $b=-\frac{1}{x}$. $C_1$ is the constant of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note my derivation assumed $c \in \mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-06T12:28:34.407" />
  <row Id="9147" PostTypeId="1" AcceptedAnswerId="9153" CreationDate="2010-11-06T12:42:24.743" Score="1" ViewCount="80" Body="&lt;p&gt;This is an exercise from Apostol's number theory book. How does, one prove that $$ \frac{\sigma(n)}{n} &amp;lt; \frac{n}{\varphi(n)} &amp;lt; \frac{\pi^{2}}{6} \frac{\sigma(n)}{n} \quad \text{if} \ n \geq 2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of using the formula $$\frac{\varphi(n)}{n} = \prod\limits_{p \mid n} \Bigl(1 - \frac{1}{p}\Bigr)$$ but couldn't get anything further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sigma(n)$ stands for the sum of divisors&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\varphi(n)$ stands for the Euler's Totient Function.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-06T12:53:46.740" LastActivityDate="2010-11-06T13:52:57.037" Title="Proving $ \frac{\sigma(n)}{n} &lt; \frac{n}{\varphi(n)} &lt; \frac{\pi^{2}}{6} \frac{\sigma(n)}{n}$" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9148" PostTypeId="1" AcceptedAnswerId="9151" CreationDate="2010-11-06T13:04:10.253" Score="0" ViewCount="55" Body="&lt;p&gt;&lt;a href=&quot;http://stackoverflow.com/questions/4113133/changing-a-positive-value-to-a-negative-one&quot;&gt;http://stackoverflow.com/questions/4113133/changing-a-positive-value-to-a-negative-one&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why? is this just a computer science problem , or is this a mathematical property.&lt;/p&gt;&#xA;" OwnerUserId="3191" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-11-06T18:50:32.200" LastActivityDate="2010-11-06T18:50:32.200" Title="Why does 0-N invert's the sign. " Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="9149" PostTypeId="1" CreationDate="2010-11-06T13:04:14.807" Score="1" ViewCount="38" Body="&lt;p&gt;For a mapping algorithm I'm working on, I'm trying out the effect of sigmoid weightings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I'm using $$y = \frac{1}{1+x^n}$$ where n is the steepness of the sigmoid function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph shows values of x between 0 and 256 and the following values of n: 2, 4 and 32. X values have been scaled to a range of -10 to 10. Also plotted is equivalent $y=x$ line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Fu0Ev.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect I would like to achieve is a function that lies somewhere between the $y=x$ and the $n=2$ functions. When I set n to be between 1 and 2 however, the location of the asymptotes changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I use values of n such that $1 &amp;lt; n &amp;lt; 2$ and keep the asymptotes at 0 and 1?&lt;/p&gt;&#xA;" OwnerUserId="3190" LastActivityDate="2010-11-06T15:11:51.273" Title="Is there a general (logistic?) function for sigmoids over a given range?" Tags="&lt;functions&gt;" AnswerCount="2" />
  <row Id="9150" PostTypeId="1" AcceptedAnswerId="9181" CreationDate="2010-11-06T13:14:10.350" Score="1" ViewCount="84" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach's fixed point theorem&lt;/a&gt; gives us a sufficient condition for a function in a complete metric space to have  a fixed point, namely it needs be a contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in how to calculate the limit of the sequence $x_0 = f(x), x_1 = f(x_0), \ldots, x_n = f(x_{n-1})$ for a fixed $x$. I couldn't figure out a way to do this limit with ordinary limits calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I have at my disposal is the proof of the theorem, from which we see that the sequence $x_n$ is a Cauchy sequence; from this, I'm able to say, for example, that $\left|f(f(f(x))) - f(f(f(f(x))))\right| \leq \left|f(x_0)-f(x_1)\right| ( \frac{k^3}{1-k})$, where $k$ is the contraction constant, but I can't get any further in the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My question is&lt;/em&gt;: how should I procede to calculate this limit exactly? If there are non-numerical (read: analytical) way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt;: I'm interested in functions $\mathbb{R} \rightarrow \mathbb{R}$ (as it can be seen from my use of the euclidean metric in $\mathbb{R}$)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2010-11-06T17:33:23.030" Title="Finding the fixed points of a contraction" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9151" PostTypeId="2" ParentId="9148" CreationDate="2010-11-06T13:18:23.310" Score="4" ViewCount="0" Body="&lt;p&gt;I'm not sure what level of answer you're looking for, so (at the risk of appearing stupid) here's my explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 is nothing. A negative number is the absence of something, a deficit. Subtraction is the removal of one quantity from another quantity. Therefore subtracting something from nothing results in a deficit, and absence, a negative number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put visually:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-5  -4  -3  -2  -1   0   1   2   3   4   5&#xA;         ^-----------^+++++++++++^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If I want to make 3 a negative number, I can subtract 3 from 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a bonus, you can also make a number negative by multiplying it by -1.&lt;/p&gt;&#xA;" OwnerUserId="3190" LastActivityDate="2010-11-06T13:18:23.310" />
  <row Id="9152" PostTypeId="1" CreationDate="2010-11-06T13:19:39.173" Score="1" ViewCount="181" Body="&lt;p&gt;Possibly not a Math problem in the true sense of the word, but here goes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three Nines - and various math operators = can be turned into various simple numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9-9)*9 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9/9)^9 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9+9)/9 = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt (9+9-9) = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sqrt(9) +9/9) = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sqrt(9!) - 9/9 = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sqrt (9*9) - sqrt (9) = 6     (or 9 - (9/Sqrt (9) ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 I am stumped on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 - (9/9) = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 * (9/9) = 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 +(9/9) = 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;99/9 = 11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9 + (9/sqrt(9)) = 12&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANy answers for 7, and anything for 13 and beyond?&lt;/p&gt;&#xA;" OwnerUserId="3193" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-06T14:53:35.557" LastActivityDate="2010-11-07T07:08:25.120" Title="The Three Nines" Tags="&lt;puzzle&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="9153" PostTypeId="2" ParentId="9147" CreationDate="2010-11-06T13:28:32.520" Score="8" ViewCount="0" Body="&lt;p&gt;Writing $n=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ for primes $p_i$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma (n) = \prod_{i=1}^k \frac{p_i^{a_i+1} – 1}{p_i – 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma(n)\phi(n) = n^2 \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right) \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which both inequalities follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the RH inequality we note that the expansion of the reciprocal of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \prod_{i=1}^k \left( 1 - \frac{1}{p_i^{a_i+1}} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is $&amp;lt; \sum_{r=1}^\infty 1/r^ 2 = \pi^2/6.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma(n) = \prod_{i=1}^k (1+p_i +p_i^2 + \cdots + p_ i^{a_i}),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is where the formula for $\sigma(n)$ comes from and to obtain $(1)$ we've just multiplied this by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(n) = n \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and factored out all the $p_i^{a_i}.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-06T13:52:57.037" LastActivityDate="2010-11-06T13:52:57.037" />
  <row Id="9154" PostTypeId="2" ParentId="9152" CreationDate="2010-11-06T13:44:22.137" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a quick cheat for 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7 = 9 - 9^0 - 9^0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what functions one's allowed they are easy to come by. Here are some more examples for $7.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d(9)! + \log_9 9 = \lfloor \sqrt {\sigma(9)} \rfloor ! + \frac{9}{9}$$&#xA;$$= \lfloor \log \sqrt{9!} \rfloor +  \frac{9}{9} =&#xA;\lfloor \sqrt{ \log 9^9 } \rfloor + \sqrt{9} = 7.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, $d(n)$ is the number of divisors of $n$&#xA;and $\sigma(n)$ is the sum of the divisors of $n.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-06T19:55:14.677" LastActivityDate="2010-11-06T19:55:14.677" CommentCount="4" />
  <row Id="9155" PostTypeId="2" ParentId="9132" CreationDate="2010-11-06T13:44:45.177" Score="0" ViewCount="0" Body="&lt;p&gt;Hoeffding's inequality can be used but the random variables need to be bounded to get meaningful bounds. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hoeffding%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hoeffding%27s_inequality&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-06T13:44:45.177" CommentCount="1" />
  <row Id="9156" PostTypeId="1" CreationDate="2010-11-06T13:45:44.290" Score="0" ViewCount="56" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have curve, that is created from circles in axis X (only x coordinate changes, y is still 0). My curve is given by this equation:&#xA;(x + R * sin(a))^2 + y^2 = (r * sin(a))^2 .... a = parameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Basicly, it creates Spiric sections)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i have got parametric (on any other kind of equtaion) of line&#xA;X = X0 + D * t   ..... t = parameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate intersection of &quot;curve&quot; x line. How can i do that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2791" LastActivityDate="2010-11-06T13:45:44.290" Title="Parametric curve vs line intersection" Tags="&lt;geometry&gt;" CommentCount="2" />
  <row Id="9157" PostTypeId="1" AcceptedAnswerId="9202" CreationDate="2010-11-06T13:59:24.387" Score="1" ViewCount="71" Body="&lt;p&gt;hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem has me scared to my wits, mostly because there are more like it!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;The mean of a random sample of n observations drawn from a N $ ( \mu ,\sigma^{2} ) $ distribution is denoted by $  \overline{\chi} $. Given that P $ (| \overline{\chi} - \mu| &gt; 0.5\sigma ) \lt 0.05 $ &lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find the smallest value of n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried:&lt;br&gt;&#xA;Since $\overline{\chi}$ ~ N $ \left ( \mu ,\frac{\sigma^2}{n} \right )  $&lt;br&gt;&#xA;P $ (| \overline{\chi} - \mu| &gt; 0.5\sigma ) \lt 0.05 $  Should be:&lt;br&gt;&#xA;$$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 0.05$$&#xA;$$1-\phi\left(\frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}}\right) \lt 0.05 $$&lt;br&gt;&#xA;($\phi$ is the normal distribution function, I lookup a value from the normal dist. table.)&#xA;$$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 1.645$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now where do I go from here!?&#xA;This is totally confusing me, especially that darn modulus. How do I use my result so far with this?&lt;br&gt;&#xA;P $ (| \overline{\chi} - \mu| &gt; 0.5\sigma )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&#xA;Gideon&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="2576" LastEditorDisplayName="" LastEditDate="2010-11-07T03:27:48.550" LastActivityDate="2010-11-07T03:27:48.550" Title="finding n with N $\left ( \mu ,\frac{\sigma^2}{n} \right ) $ with given probability" Tags="&lt;homework&gt;&lt;probability-theory&gt;&lt;statistics&gt;&lt;central-limit-theorem&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9158" PostTypeId="2" ParentId="9150" CreationDate="2010-11-06T14:00:47.250" Score="5" ViewCount="0" Body="&lt;p&gt;The limit is the fixpoint, so you &quot;just&quot; need to solve the equation $x=f(x)$. Whether this can be done in closed form or not depends on what $f$ looks like.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-06T14:00:47.250" CommentCount="8" />
  <row Id="9159" PostTypeId="1" AcceptedAnswerId="9162" CreationDate="2010-11-06T14:08:23.033" Score="4" ViewCount="80" Body="&lt;p&gt;I'm trying to parametrize the space curve determined by the boundary of a standard orange peel: for example, the one on &lt;a href=&quot;http://www.dreamstime.com/stock-photo-orange-peel-image14340&quot; rel=&quot;nofollow&quot;&gt;this photo&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the ideal curve would be inside the unit cube; have only one point of intersection with every horizontal plane $z=k$, when $k\in [-1,1]$; would start in (0,0,-1) and end in (0,0,1), wrapping itself around them; and touch the boundary of the cube when z=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sort of a standard helix, compressed. I hope I was clear.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-06T14:25:13.817" Title="How to parameterize an orange peel" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="9160" PostTypeId="2" ParentId="9150" CreationDate="2010-11-06T14:11:40.027" Score="1" ViewCount="0" Body="&lt;p&gt;In addition to what Hans Lundmark has said about solving $x = f(x)$, you could also try writing a simple computer programme to read a number $n$ and a starting value $x_0$, and compute the result of applying f to $x_0$ $n$ times, thus delivering an approximation to the root that you are seeking. The value of $n$ may have to be fairly large in some cases. &lt;/p&gt;&#xA;" OwnerUserId="3195" LastEditorUserId="3195" LastEditorDisplayName="" LastEditDate="2010-11-06T14:19:07.163" LastActivityDate="2010-11-06T14:19:07.163" CommentCount="2" />
  <row Id="9161" PostTypeId="1" AcceptedAnswerId="9173" CreationDate="2010-11-06T14:16:30.070" Score="4" ViewCount="215" Body="&lt;p&gt;Reading the Wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;Random Walk&lt;/a&gt;, I was wondering what is the definition for a general random walk as a random process, so that all the concepts such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Lattice_random_walk&quot; rel=&quot;nofollow&quot;&gt;random walk on $\mathbb{Z}^d$&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Gaussian_random_walk&quot; rel=&quot;nofollow&quot;&gt;Gaussian random walk&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Random_walk_on_graphs&quot; rel=&quot;nofollow&quot;&gt;random walk on a graph&lt;/a&gt; can fall into this general definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In Ross's book for stochastic process, he defined a random walk to be the sequence of partial sums of a sequence of i.i.d. real-valued random variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is a Gaussian random walk the sequence of partial sums of a sequence of i.i.d. real-valued random variables with a common Gaussian distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does a random walk on a graph fit into Ross's definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) I was thinking maybe the concept of random walk is equivalent to that of discrete-time Markov process, where the state space can be either continuous or discrete or even non-numeric (such as for a random walk on a graph)? Is it correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can I say the graph in &quot;a random walk on a graph&quot; is just the &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphical_model&quot; rel=&quot;nofollow&quot;&gt;graphical model&lt;/a&gt; for the distribution of the random walk as a stochastic process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the graph in &quot;a random walk on a graph&quot; has nothing to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_graph&quot; rel=&quot;nofollow&quot;&gt;random graph&lt;/a&gt;, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) if you can come up with a definition for random walk, please also explain how various special kinds of random walks fit into the definition you provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Dr Schmuland for the reply! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I have chosen the best answer, I am still not clear about the following questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Is a random walk (including random walk on a group such as $\mathbb{Z}^d$ and random walk on a graph) equivalent to a discrete-time Markov process? If not, what makes random walks different from general discrete time Markov processes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) For a random walk defined on a group such as $\mathbb{Z}^d$, does it really require the increments between every two consecutive indices be i.i.d. (seems to be the definition in Ross's Stochastic processes) or just independence will be enough and identical distribution is not necessary? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reply to Update of Dr Schmuland&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Dr Schmuland, for providing a big picture that I have not clearly realized before! It took me some while to understand. Learning more makes me feel better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally when the state space is additive group, are these two properties - markov property and increment-independence - equivalent? Must time-homogeniety be defined only for Markov process? If it can be define for a general process, is it equivalent to the property that the increments over same length period have the same distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wonder if, for a random walk on a graph, at each vertex, the transition probabily must be uniform? If yes, is the transition probability uniform at a vertex over a set of the neighbours and itself or just over a set of its neighbours (itself excluded, which means it is impossible to stay at the vertex)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your last exercise, I think time-homogeniety and identical-increment-distribution are equivalent, and Markov property and increment-independence are equivalent, so every time-homogeneous Markov chain will be a random walk on the group {1,2,3}. But my answer will be different for it to be a random walk on the graph derived for {1,2,3} as you mentioned, the transition probability from a vertex to other vertices has to be uniform.&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-09T16:06:43.700" LastActivityDate="2010-11-09T16:06:43.700" Title="Concept of Random Walk" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9162" PostTypeId="2" ParentId="9159" CreationDate="2010-11-06T14:25:13.817" Score="6" ViewCount="0" Body="&lt;p&gt;Well, you seem to have a lot of options; there are a number of spherical spirals that would do. The &lt;a href=&quot;http://mathworld.wolfram.com/SphericalSpiral.html&quot; rel=&quot;nofollow&quot;&gt;loxodrome&lt;/a&gt; is one (the spherical analogue of the equiangular spiral), and &lt;a href=&quot;http://mathworld.wolfram.com/SeiffertsSphericalSpiral.html&quot; rel=&quot;nofollow&quot;&gt;Seiffert's spiral&lt;/a&gt; is another.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-06T14:25:13.817" CommentCount="1" />
  <row Id="9163" PostTypeId="1" AcceptedAnswerId="9174" CreationDate="2010-11-06T14:32:56.923" Score="1" ViewCount="83" Body="&lt;p&gt;Quoted from &lt;a href=&quot;http://en.wikipedia.org/wiki/Wiener_process&quot; rel=&quot;nofollow&quot;&gt;Wikepedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Wiener process can be constructed as the scaling limit of a random walk, or other discrete-time stochastic processes with stationary independent increments. This is known as Donsker's theorem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the scaling limit of any &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot; rel=&quot;nofollow&quot;&gt;random walk&lt;/a&gt; is always a Wiener process, or just the scaling limits of some special kinds of random walks are, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk#Gaussian_random_walk&quot; rel=&quot;nofollow&quot;&gt;Gaussian random walk&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-11-06T16:25:35.493" Title="Wiener Process and Random Walk" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="9164" PostTypeId="2" ParentId="9152" CreationDate="2010-11-06T14:37:52.433" Score="7" ViewCount="0" Body="&lt;p&gt;$7=(\sqrt{9})!+\frac{9}{9}$.  Also for 5 you need to move the ! outside the parens.&lt;br&gt;&#xA;$18=\sqrt{(\sqrt{9})!!+9}-9=\sqrt{9}^{\sqrt{9}}-9$&lt;br&gt;&#xA;and you can sqrt the last 9 and add and subtract to get a few more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13=\frac{9}{.9}+\sqrt{9}$&lt;br&gt;&#xA;If you allow the overline for a repeating decimal or the floor function  (floor(sqrt(9)) you can convert a 9 into a 1.  This gives new flexibility&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-06T15:30:26.957" LastActivityDate="2010-11-06T15:30:26.957" />
  <row Id="9165" PostTypeId="2" ParentId="9137" CreationDate="2010-11-06T14:40:23.590" Score="2" ViewCount="0" Body="&lt;p&gt;The CDF $F(x)$ is the sum of the probabilities $f(y)$ for $y \leq x$.  Since $F(x+1) - F(x) = f(x+1)$, you can verify that $0.98 - 0.95 = 0.3$. &lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-06T14:40:23.590" CommentCount="1" />
  <row Id="9166" PostTypeId="2" ParentId="9130" CreationDate="2010-11-06T14:42:22.830" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt; The sequence $(a_n)$ cannot be bounded, if it doesn't converge to 0. Suppose the contrary, then up to a subsequence $(a_{\sigma(n)})$ converges to some value $A$. Let $t = \pi/A$. Then $\exp i t a_\sigma(n) \to -1$ contradicting convergence assumption. (Another way to see this step is to note that if $(a_n)$ were bounded, then the sequence $(\exp i t a_n)$ is equicontinuous, and on any compact interval the convergence must be uniform.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt; The sequence cannot be unbounded. Assume the contrary, then up to a subsequence we can choose $a_{\sigma(n+1)} &gt; 3 a_{\sigma(n)}$. Without loss of generality we can assume the subsequence is positive. Let $t_n = 2\pi a_{\sigma(n)}^{-1}$. Define a sequence of closed intervals recursively. Let $I_1 = [t_1/4, 3t_1/4]$. By construction $\exists m_2 \in \mathbb{Z}$ such that $m_2t_2, (m_2+1)t_2 \in I_1$. Let $I_2 = [ (m_2 + 1/4) t_2, (m_2 + 3/4) t_2]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each step $I_n$ has length $t_n / 2$. Hence there exists $m_{n+1}$ such that $m_{n+1}t_{n+1}, (m_{n+1}+1)t_{n+1} \in I_n$. And so we construct $I_{n+1}$ analogously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we have $I_n \supset I_{n+1} \supset I_{n+2} \ldots$. Furthermore, by construction, $\Re e^{it a_{\sigma(n)}} \leq 0$ for $t \in I_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for $t\in \cap_1^\infty I_n$, we have that $\exp i t a_j$ is bounded away from 1 on a subsequence, and hence contradicts the assumption of convergence. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-06T14:42:22.830" CommentCount="3" />
  <row Id="9167" PostTypeId="2" ParentId="9149" CreationDate="2010-11-06T14:43:21.347" Score="2" ViewCount="0" Body="&lt;p&gt;I think it would help if you quantified what you mean by the &lt;em&gt;location&lt;/em&gt; of the asymptote.  In your graph, it &lt;em&gt;looks&lt;/em&gt; like $y$ takes on the values 0 or 1 at different places for different values of $n$, but $y \ne 0$ for any $x &gt; 0$ and $y \ne 1$ for any finite $x$ no matter the value of $n&gt;0$. You could define the location of the asymptotes, for example, as the values of $x$ where $y$ equals 0.001 and 0.999.  Or you could define the locations of the asymptotes as places where the derivative $y'$ equals 0.001, for example.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-06T14:43:21.347" />
  <row Id="9168" PostTypeId="2" ParentId="9149" CreationDate="2010-11-06T15:11:51.273" Score="1" ViewCount="0" Body="&lt;p&gt;You can also use $y = \frac{1}{1+ax^n}$ to give yourself another parameter to play with.  It will still go from 0 to 1 as x goes from -infinity to infinity, but you can change how quickly it gets close.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-06T15:11:51.273" CommentCount="1" />
  <row Id="9169" PostTypeId="1" CreationDate="2010-11-06T15:21:26.193" Score="0" ViewCount="116" Body="&lt;p&gt;I am trying to evaluate this integral or at least get bounds for its aboslute value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have where $\tau \to \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty} f(t) \frac{\tau \sin(\tau\log t)}{t^{\sigma+1}} dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(t) = (t + 1/2) - \lfloor t + 1/2 \rfloor$ (where $\lfloor.\rfloor$ is the floor function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating by parts gives me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty} \frac{f(t)}{t^\sigma} d(-\cos(\tau \log t)) = \left[ -\frac{f(t)}{t^\sigma}\cos(\tau \log t) \right]_{1}^{\infty} + B$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B = \int_{1}^{\infty} \cos(\tau\log t) g'(t) dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(t) = f(t)/t^\sigma$ and is piecewise continuous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Riemann-Lebesgue lemma we have $B = 0$ as $\tau \to \infty$, giving us,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{1}^{\infty} \frac{f(t)}{t^\sigma} d(-\cos(\tau \log t)) = \left[ -\frac{f(t)}{t^\sigma}\cos(\tau \log t) \right]_{1}^{\infty}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since, $0 \leq f(t) \leq 1$, I think it makes the absolute value of the integral $\leq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are these steps correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any further insights will be highly appreciated...&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2320" LastEditorDisplayName="" LastEditDate="2010-11-06T15:41:00.173" LastActivityDate="2010-11-06T15:41:00.173" Title="Is this the right way to evaluate this integral?" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;integral&gt;" CommentCount="3" />
  <row Id="9170" PostTypeId="2" ParentId="9116" CreationDate="2010-11-06T15:27:01.460" Score="2" ViewCount="0" Body="&lt;p&gt;You are correct. The &quot;crude&quot; criterion for recurrence of a 2d random walk is $\mu=0$ and $\sigma^2&amp;lt;\infty$ for the jump distribution. The jump sizes are otherwise unrestricted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;detailed&quot; criterion involves the characteristic function $\phi$ of the jump distribution, i.e., its Fourier transform. It says that 2d random walk is transient or recurrent as the real part of $(1-\phi(\theta))^{-1}$ is Lebesgue integrable on a neighborhood of the origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results are from Section 8 of Spitzer's &lt;em&gt;Principles of Random Walk (2e)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spitzer gives a detailed example of &lt;strong&gt;symmetric&lt;/strong&gt; one-dimensional random walks, and shows that their recurrence or transience depends on the size of the tail of the jump distribution. That is, he supposes that &#xA;$$0&amp;lt;\lim_{|x|\to\infty} |x|^{1+\alpha}P(0,x)=c&amp;lt;\infty,$$ and concludes that this walk is recurrent when $\alpha\geq 1$ and transient when $\alpha&amp;lt;1$.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, somehat unexpectedly, there exist symmetric transient random walks in one dimension. Their jump distribution has such large tails that the walk leaps back and forth with large jumps and satisfies $\liminf_n X_n=-\infty$ and $\limsup_n X_n=+\infty$ without a guarantee of returning to the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be possible to modify his arguments to the two dimensional case.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-06T15:27:01.460" CommentCount="3" />
  <row Id="9171" PostTypeId="1" CreationDate="2010-11-06T15:51:20.450" Score="0" ViewCount="132" Body="&lt;p&gt;What is the minimum for this function of $x_1,x_2,...,x_n$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{i=1}^{n}{c_i \log x_i} + \lambda (\sum_{i=1}^{n}{d_i x_i})$&#xA;   , where $\lambda$, c and d series are positive constants, $x_i \in (0.02,1]$ and $\sum x_i = 1$.&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-06T16:47:18.867" LastActivityDate="2011-01-05T17:44:36.090" Title="Minimum for this function" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="9172" PostTypeId="2" ParentId="9122" CreationDate="2010-11-06T15:51:29.630" Score="1" ViewCount="0" Body="&lt;p&gt;If you know about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Incomplete_gamma_function&quot; rel=&quot;nofollow&quot;&gt;incomplete Gamma function&lt;/a&gt; then the solution is immediate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\rm\quad\quad\quad\quad\quad\ \  \Gamma(s,x)\ =\ (s-1)\ \Gamma(s-1,x) + x^{s-1}\ e^{-x} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence $\rm\displaystyle\quad e^x\ \Gamma(s+1,x)\ =\ \sum_{k=0}^s\ \frac{s!}{k!}\: x^k\ \ $ for $\rm\:s\in \mathbb N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your method of summing the lower-factorical coefficient recurrence $\rm\ c_{(k+1)}\ = \ (c-k)\ c_{(k)}\ $ will yield a hypergeometric differential equation with the above solution - presuming that you know how to solve such hypergeometric differential equations. But that's overkill compared to the above.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-06T15:51:29.630" />
  <row Id="9173" PostTypeId="2" ParentId="9161" CreationDate="2010-11-06T15:53:27.803" Score="4" ViewCount="0" Body="&lt;p&gt;It's a good question. The naive answer is that the two concepts are different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random walks on ${\mathbb Z}^d$ or ${\mathbb R}$, etc. are examples of a random walk on an additive group $G$. There is a fixed distribution $\mu$ on $G$, and at each time you randomly select an element of the group (using distribution $\mu$) and add it to the current state.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, for a random walk on a (locally finite) graph, you randomly choose an edge  uniformly and follow that edge to the next state (node). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the concepts may be related at a deeper level. The first chapter of Woess's &lt;em&gt;Random Walks on Infinite Graphs and Groups&lt;/em&gt; explains how a random walk on a graph is related to a random walk on a quotient of its automorphism group, and also how a random walk on a group is related to a walk on its Cayley graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't read the book in detail, but my impression is that, nevertheless, the two concepts are not exactly equivalent.  &lt;hr&gt;&#xA;&lt;strong&gt;Update (Nov. 7 2010)&lt;/strong&gt; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any stochastic process $(X_n)$ taking values in a countable state space&#xA;can be built up out of the following pieces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) an initial distribution $P(X_0=x)$ that tells us&#xA;  how to choose the starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) transition kernels that tell us how to choose the next&#xA;state given the history of the process up to the present, that is,&#xA; $P(X_{n+1}=x_{n+1} | X_n=x_n, \dots, X_0=x_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a process is called Markov if the kernels only depend on $n$, $x_{n+1}$ and $x_n$,&#xA;that is,&#xA;$P(X_{n+1}=x_{n+1} | X_n=x_n, \dots, X_0=x_0) = P(X_{n+1}=x_{n+1} | X_n=x_n)$.&#xA;This is a big assumption and a huge simplification. But enough models of&#xA;real world random behaviour have this &quot;memoryless&quot; property to make Markov&#xA;processes wide spread and extremely useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further simplification is to assume that the Markov process is time homogeneous,&#xA;so that the transition rules only depend on the states and not on time. That is,&#xA;we assume that $P(X_{n+1}=y | X_n=x)=P(X_1=y | X_0=x)$ for all $x,y$.&#xA;The crucial point here is that $p(x,y):=P(X_{n+1}=y | X_n=x)$ is the same for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now assume that the state space is an additive group.&#xA;The transition kernels for a general process can be rewritten in terms of increments:&#xA;$$P(X_{n+1}=x_{n+1} | X_n=x_n, \dots, X_0=x_0)$$&#xA;can be written &#xA;$$P(X_{n+1}-X_n=x_{n+1}-x_n | X_n-X_{n-1}=x_n-x_{n-1}, \dots, X_0=x_0).$$&#xA;If the increments are independent, then this becomes&#xA;$$P(X_{n+1}-X_n=x_{n+1} -x_n)=:\mu_n(x_{n+1}-x_n).$$&#xA;In this case, the process is automatically Markov and&#xA;the transition kernels are an arbitrary  sequence $(\mu_n)$ of probabilities on the state space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for a random walk we assume &lt;strong&gt;both&lt;/strong&gt; time homogeneity and independent increments.&#xA;Thus, the transition kernel $p(x,y)$ is represented by a single measure: $p(x,y)=\mu(y-x)$.&#xA;This imposes  a kind of homogeneity in space and in time.&#xA;We are down to an extremely special, yet still&#xA;interesting process, that only depends on the initial distribution and&#xA;the single measure $\mu$. This is why Ross insists that for random walks the increments are&#xA;identically distributed; just being independent is not enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because random walks are so special, they can be analyzed by &#xA; mathematical tools that don't work for most Markov processes. In particular,&#xA; tools from harmonic analysis are very fruitful here.&#xA;A standard reference, though a little old, is  &lt;em&gt;Principles of Random Walk (2nd ed)&lt;/em&gt;&#xA;by Frank Spitzer. A more recent treatment is given in &lt;em&gt;Random Walk: A Modern Introduction&lt;/em&gt;&#xA; by Gregory Lawler and Vlada Limic (2010).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an exercise to convince you that random walks are only a small part of&#xA;all Markov chains. Take the state space to be the corners of a triangle, labelled $\{ 0, 1, 2\}$, &#xA;considered as the group ${\mathbb Z}$ mod 3. Any 3 by 3 stochastic matrix&#xA; (non-negative entries and row sums all equal to 1) defines a time homogeneous Markov&#xA;chain on this state space. Which of these are random walks?&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-11-07T20:32:08.440" LastActivityDate="2010-11-07T20:32:08.440" CommentCount="4" />
  <row Id="9174" PostTypeId="2" ParentId="9163" CreationDate="2010-11-06T16:25:35.493" Score="3" ViewCount="0" Body="&lt;p&gt;In light of your &lt;a href=&quot;http://math.stackexchange.com/questions/9161/concept-of-random-walk&quot;&gt;other question&lt;/a&gt;, one should say there are a lot of things referred to by the name &quot;random walk&quot;.  In its simplest form, Donsker's theorem is about a process $X_n$ in $\mathbb{R}^d$ whose increments $X_n - X_{n-1}$ are iid with any distribution that has zero mean and finite variance.  In particular, the increments do not have to be Gaussian.  It then asserts that the scaling limit of the process $X_n$ is $d$-dimensional Brownian motion (aka Wiener process).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of other stochastic processes that have Brownian motion as their scaling limits, and still other processes that have a scaling limit that is not Brownian motion.  Is there something specific you're interested in?&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-06T16:25:35.493" CommentCount="2" />
  <row Id="9175" PostTypeId="2" ParentId="9171" CreationDate="2010-11-06T16:32:29.493" Score="1" ViewCount="0" Body="&lt;p&gt;Let $ x_i=1/m $ certainly $ x_i \in (0,1] $ for all $i$ and all natural numbers $m$.&#xA;Then your function becomes&#xA;$-\log m \sum_{i=1}^{n}{c_i} + \frac{\lambda}{m} (\sum_{i=1}^{n}{d_i})$ which can be written as $-m A + B/m $ where $A,B$ are positive constants.&#xA;This can be made arbitrarily large and negative, so there is no minimum.&#xA;If this is part of some work on Lagrangian multipliers you need to think again about your restrictions on $\lambda$.&lt;/p&gt;&#xA;" OwnerUserId="2712" LastActivityDate="2010-11-06T16:32:29.493" CommentCount="4" />
  <row Id="9176" PostTypeId="2" ParentId="9131" CreationDate="2010-11-06T16:33:34.480" Score="1" ViewCount="0" Body="&lt;p&gt;Exact sequence of $R=\mathbb Z$ modules $0\to \mathbb Z \to \mathbb Q$ becomes not exact after choice $p=2\mathbb Z$  and tensor with $R_p/m_p =\mathbb Z /(2) $ because it becomes    $0 \to  \mathbb Z /(2)\to 0$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-06T16:33:34.480" />
  <row Id="9178" PostTypeId="1" AcceptedAnswerId="9179" CreationDate="2010-11-06T16:40:23.447" Score="1" ViewCount="137" Body="&lt;p&gt;I'm looking to strengthen my understanding of the math that is directly useful to practical computer science, as opposed to unsolved computer science problems. In other words, the kind of math that would be useful in day to day algorithmic analysis, and come up during senior level interviews at software companies. My math knowledge is very rudimentary. I want to say it's almost at a 9th or 10th grade level. Will I need to go back and refresh on algebra, or can I jump right into combinatorics? What are the prerequisites?&lt;/p&gt;&#xA;" OwnerUserId="3180" LastActivityDate="2010-11-06T22:07:45.527" Title="What are the prerequisites for combinatorics?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9179" PostTypeId="2" ParentId="9178" CreationDate="2010-11-06T16:53:35.007" Score="2" ViewCount="0" Body="&lt;p&gt;In my opinion the &quot;prerequisite&quot; knowledge for Combinatorics/Discrete mathematics is primarily basic logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people would say that this falls under discrete math, but I think the main things that you will need to understand are Truth Tables, De Morgan, Induction, set theory and such. Many discrete books cover these topics, so don't be too afraid. If you really want to get another book to cover this material, the first chapter of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471368792&quot; rel=&quot;nofollow&quot;&gt;Herstein&lt;/a&gt; does this very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as algorithm analysis, you will probably want to understand some basic asymptotics, big O, and recursion, all of these things can be found in most introductions to discrete math, such as the often recommended &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are particularly worrying about Geometry, Trig and Calc, I wouldn't. A few things from PreCalc/Calc will be useful, but I don't thing overwhelmingly so. I don't expect you to see much Trig, and Geometry will mostly used to provide examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other question I linked in the comment will also be useful probably.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-06T16:53:35.007" />
  <row Id="9180" PostTypeId="2" ParentId="9178" CreationDate="2010-11-06T16:53:41.613" Score="1" ViewCount="0" Body="&lt;p&gt;To &quot;strengthen your understanding of the math that is directly useful to practical computer science&quot; I think your best bet is simply to master the material in  Graham; Knuth; Patashnik. Concrete Mathematics and Rosen: Discrete mathematics and its applications.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-06T16:53:41.613" CommentCount="1" />
  <row Id="9181" PostTypeId="2" ParentId="9150" CreationDate="2010-11-06T17:33:23.030" Score="3" ViewCount="0" Body="&lt;p&gt;@Andy (in reply to your comment/question &quot;Could you provide some example that has a closed form and explain if (and how) it is possible to find the fixed point without solving x = f(x) but trying to calculate the limit of x_n?&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that you would be hard-pressed to achieve this, since your function $f$ is a continuous function (being a contraction map in the first place); and if you then take limits of both sides of $x_n = f(x_{n-1})$, you will get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} f(x_{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which (by continuity) leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \rightarrow \infty} x_n = f (\lim_{n \rightarrow \infty} x_{n-1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l = f(l)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $l = \lim_{n \rightarrow \infty} x_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you will have to solve $l = f(l)$, which was what you wanted to avoid in the first place!&lt;/p&gt;&#xA;" OwnerUserId="3195" LastActivityDate="2010-11-06T17:33:23.030" CommentCount="2" />
  <row Id="9182" PostTypeId="1" AcceptedAnswerId="9184" CreationDate="2010-11-06T17:37:29.217" Score="2" ViewCount="63" Body="&lt;p&gt;In Hatcher's &lt;em&gt;Algebraic Topology,&lt;/em&gt; section 3.2, during the computation of the cohomology ring of a $n$-torus, the following assertion is made. Let $Y$ be a space and $R$ a commutative ring. Then the natural map in cohomology&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H^n(I \times Y, R) \to H^n( \partial I \times Y, R)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a split injection. This is likely straightforward, and I believe it follows directly from the Kunneth theorem. However, I don't see a direct proof, which is probably what Hatcher is indicated. Can someone explain to me what I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-06T17:43:27.117" Title="Why is $H^n(I \times Y, R) \to H^n( \partial I \times Y, R)$ a split injection?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="9184" PostTypeId="2" ParentId="9182" CreationDate="2010-11-06T17:43:27.117" Score="6" ViewCount="0" Body="&lt;p&gt;Isn't $I$ just the closed unit interval? Then $I\times Y$ is homotopy&#xA;equivalent to&#xA;$Y$ and $\partial I\times Y$ is just two copies of $Y$. It's&#xA;clear that your map then is just the diagonal map from&#xA;$H^n(Y)$ to $H^n(Y)\oplus H^n(Y)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-06T17:43:27.117" CommentCount="1" />
  <row Id="9185" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T17:44:15.123" Score="8" ViewCount="0" Body="&lt;p&gt;Added: As explained in the comments, certain trigonometric equations such as the linear equations in $\sin x$ and $\cos x$ can be solved by a resolvent quadratic equation. One method is to write the $\sin x$ and $\cos x$ functions in terms of the same trigonometric function. Since all [direct] trigonometric functions of the simple angle can be expressed rationally as a function of the $\tan$ of the half-angle, such a conversion is adequate for these equations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since $$\cos \alpha =\frac{1-\tan ^{2}\frac{\alpha }{2}}{1+\tan ^{2}\frac{%&#xA;\alpha }{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \alpha =\frac{2\tan \frac{\alpha }{2}}{1+\tan ^{2}%&#xA;\frac{\alpha }{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m\cos \alpha +\sin \alpha =1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m-m\tan ^{2}\frac{\alpha }{2}+2\tan \frac{\alpha }{2}=1+\tan ^{2}\frac{\alpha }{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may set $x=\tan \frac{\alpha }{2}$ ($\alpha =2\arctan x$), and thus get&#xA;the quadratic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 1+m\right) x^{2}-2x+1-m=0.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its solutions are: $x=\frac{1}{m+1}\left( -m+1\right) $ (if $m\neq -1$) or $%&#xA;x=1$ (if $m=-1$), which gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) If $m\neq -1$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha =2\arctan x=2\arctan \frac{1-m}{m+1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) If $m=-1,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha =2\arctan 1=\frac{\pi }{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different technique to solve a linear equation in $\sin \alpha $ and $\cos&#xA;\alpha $ is to use an auxiliary angle $\varphi $. If you set $m=\tan \varphi &#xA;$, your equation takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \alpha +\tan \varphi \cdot \cos \alpha =1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin (\alpha +\varphi )=\cos \varphi =\frac{1}{\pm \sqrt{1+\tan ^{2}\varphi &#xA;}}=\pm \sqrt{\frac{1}{1+m^{2}}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha =\pm \arcsin \sqrt{\frac{1}{1+m^{2}}}-\arctan m.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Detailed derivation: from $m\cos \alpha +\sin \alpha =1$ and $m=\tan \varphi &#xA;$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \alpha +\tan \varphi \cdot \cos \alpha =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \alpha +\dfrac{\sin \varphi }{\cos \varphi }\cdot \cos \alpha =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\sin \alpha \cdot \cos \varphi +\sin \varphi \cdot \cos \alpha }{\cos&#xA;\varphi }=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\sin \left( \alpha +\varphi \right) }{\cos \varphi }=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \left( \alpha +\varphi \right) =\cos \varphi $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos \varphi =\pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be obtained as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin ^{2}\varphi +\cos ^{2}\varphi =1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\sin ^{2}\varphi }{\cos ^{2}\varphi }+1=\dfrac{1}{\cos ^{2}\varphi }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan ^{2}\varphi +1=\dfrac{1}{\cos ^{2}\varphi }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cos ^{2}\varphi =\dfrac{1}{1+\tan ^{2}\varphi }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin \left( \alpha +\varphi \right) =\pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }%&#xA;}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha +\varphi =\arcsin \left( \pm \sqrt{\dfrac{1}{1+\tan ^{2}\varphi }}%&#xA;\right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($m=\tan \varphi $, $\varphi =\arctan m$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha +\arctan m=\arcsin \left( \pm \sqrt{\dfrac{1}{1+m^{2}}}\right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha =\arcsin \left( \pm \sqrt{\dfrac{1}{1+m^{2}}}\right) -\arctan m.$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-07T12:50:06.913" LastActivityDate="2010-11-07T12:50:06.913" CommentCount="10" />
  <row Id="9186" PostTypeId="1" CreationDate="2010-11-06T17:51:29.743" Score="1" ViewCount="137" Body="&lt;p&gt;This is a follow-up (refinement?) of &lt;a href=&quot;http://math.stackexchange.com/questions/9010/what-is-the-difference-between-a-variety-and-a-manifold&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In learning some algebraic topology, I've learned to think of an affine scheme as spec $R$.  (I've been told that this is a legitimate use of terminology because it provides an embedding of $Rings^{op}$ into a larger category.)  One construction we can do, for example, is create the &quot;tangent scheme&quot;, which is obtained by localizing and completing.  The examples I've been looking at are $\hat{\mathbb{G}}_a$, $\hat{\mathbb{G}}_m$, $(\mathbb{Z}[[t]],F)$ (FGLs (1-dimensional, commutative) more generally), $T_1C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_m)\cong C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_a)$ (&quot;commutative $k$-variate FGLs satisfying the 2-cocycle condition&quot;), etc.  We then have exponential maps, which e.g. in the last case is $exp:C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_a) \rightarrow C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_m)$ given by $g\mapsto 1+g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to test the waters, here is my question (although please feel free to push it further or in a different direction).  Presumably these exponential maps are not always injective.  Once we apply this picture to a particular ring (or perhaps even before?), can we translate differential-geometric ideas like conjugate points, geodesics, &quot;$exp$ is a local isomorphism&quot;, etc. into algebro-geometric language?  When we can, which theorems for manifolds carry over to schemes and which must we discard?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-06T21:01:40.413" LastActivityDate="2010-12-26T09:31:35.473" Title="To what extent do the stories on manifolds carry over to schemes?" Tags="&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9187" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T17:55:15.110" Score="4" ViewCount="0" Body="&lt;p&gt;Because you're familiar with complex exponentials, all you needed to succeed is to observe that by setting $w = m - i$ your equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w \exp(i \alpha) + \bar{w} \exp(-i \alpha) = 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edit&lt;/strong&gt; What might motivate this choice of $w$?  Well, if you're thinking of complex numbers you're thinking that $\cos(\alpha) + i \sin(\alpha) = \exp(i \alpha)$.  To obtain $m \cos(\alpha) + \sin(\alpha)$ from this you need to multiply $\cos(\alpha)$ by $m$ and $i \sin(\alpha)$ by $-i$.  Thus it's automatic to set out by computing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(m - i)(\cos(\alpha) + i \sin(\alpha)) = \left(m \cos(\alpha) + \sin(\alpha)\right) + \left(\text{real stuff}\right) i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real part of this expression is exactly what you need, whence the original equation reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = \mathbb{Re}\left(w \exp(i \alpha))\right) = \frac{w \exp(i \alpha) + \overline{w \exp(i \alpha)}}{2} = \frac{w \exp(i \alpha) + \bar{w} \exp(-i \alpha)}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by 2 to clear the fraction.  &lt;strong&gt;End of edit&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $z = \exp(i \alpha)$ to simplify the notation yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w z^2 - 2 z + \bar{w} = 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with two generic solutions by the quadratic formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z = \frac{1 + \sqrt{1 - |w|^2}}{w}, \quad z = \frac{1 - \sqrt{1 - |w|^2}}{w}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\sqrt{1 - |w|^2} = \pm m i$, the solutions simplify to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z = \frac{i + m}{1 + m i}, \quad z = i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Once again we see a connection to fractional linear transformations, as per &lt;a href=&quot;http://math.stackexchange.com/questions/6695/ways-to-evaluate-int-sec-theta-d-theta/6702#6702&quot;&gt;@Qiaochu Yuan's comment&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $\alpha$ is the argument (polar coordinate) of a solution $z$.  Writing $\tan(\theta) = m$ (which defines $\theta$ up to an integral multiple of $\pi$), the argument of the denominator is $\theta$ while the argument of the numerator is equal to $\pi/2 - \theta$.  Whence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \pi/2 - 2\theta \quad \text{or} \quad \alpha = \pi/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(both modulo an integral multiple of $2 \pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer, in the first case you may write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \pi/2 - 2\arctan(m).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second case obviously gives the trivial solutions $1 = m \cos(\pi/2 + 2 n \pi) + \sin(\pi/2 + 2 n \pi)$ $= m 0 + 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first solution appears to be a new one, but it's equivalent to the others that have been obtained.  As a check, we may assume $\theta$ is not an odd multiple of $\pi/2$, whence $\alpha$ is not an odd multiple of $\pi/2$, and compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \cos(\alpha) + \sin(\alpha)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \tan(\theta) \cos(\pi/2 - 2 \theta) + \sin(\pi/2 - 2\theta) = \tan(\theta) \sin(2 \theta) + \cos(2 \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{\sin(\theta)}{\cos(\theta)}\left(2 \sin(\theta) \cos(\theta)\right) + \left(\cos(\theta)^2 - \sin(\theta)^2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sin(\theta)^2 + \cos(\theta)^2 = 1.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Although for a problem like this I would ordinarily take the route described by @Robin Chapman, I like this complex number approach because it produces a simpler solution (which actually is defined for $m = -1$, btw) by means of straightforward techniques and it explicitly produces &lt;em&gt;all&lt;/em&gt; solutions.  When $m \lt -1$ and we use the standard branch of the arctangent it differs by $2 \pi$ from the one &lt;em&gt;Mathematica&lt;/em&gt; obtained, but that's of no consequence.  (It's curious that &lt;em&gt;Mathematica&lt;/em&gt; cannot tell these two solutions are the same, even when you restrict the domain to real nonnegative $m$.  A plot of their difference will remove any lingering doubts.)&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-11-06T20:45:46.643" LastActivityDate="2010-11-06T20:45:46.643" CommentCount="3" />
  <row Id="9188" PostTypeId="1" AcceptedAnswerId="9196" CreationDate="2010-11-06T18:23:03.233" Score="0" ViewCount="335" Body="&lt;p&gt;In this &lt;a href=&quot;http://math.stackexchange.com/questions/8904/isomorphism-of-vector-spaces-over-mathbbq&quot;&gt;post&lt;/a&gt; we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In know these as $\mathbb{Q}$-Vector spaces, are isomorphic from the  linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-06T19:47:59.120" LastActivityDate="2010-11-07T01:01:30.787" Title="Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="9189" PostTypeId="1" AcceptedAnswerId="9211" CreationDate="2010-11-06T18:26:40.497" Score="0" ViewCount="153" Body="&lt;p&gt;Hi, I was reading John Lee's Introduction to Smooth manifolds, and I came across this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $M$ be a smooth manifold, and let $\delta : M \rightarrow \mathbb{R}$ be a positive continuous function. Using a partition of unity, show that there is a smooth function $\tilde{\delta} : M \rightarrow \mathbb{R}$ such that $0 &amp;lt; \tilde{\delta}(x) &amp;lt; \delta(x)$ for all $x \in M$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I thought about it for a while, and I'm pretty stuck on it. Does anyone have any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit: It has been pointed out that one can basically assume that $M = \mathbb{R}$, because the proof should be the same in both cases. If you do not know anything about smooth manifolds, feel free to do this!)&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-06T20:40:46.790" LastActivityDate="2010-11-06T22:22:01.867" Title="Question on Partitions of Unity" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9190" PostTypeId="2" ParentId="9188" CreationDate="2010-11-06T18:37:54.027" Score="6" ViewCount="0" Body="&lt;p&gt;To prove this: Suspect the fields are not isomorphic, then we can attempt to find a property which holds inside one and does not the other - but whose truth is preserved by isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the field $\mathbb{Q}(\sqrt{2})$ there is an element which satisfies the field property $x^2=2$. There is no element in $\mathbb{Q}(\sqrt{3})$ which satisfies this, but suppose for a contradiction that there was an isomorphism $\psi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{3})$ we would have $\psi(x^2) = \psi(2)$ which is equivalent to $\psi(x)^2 = \psi(1)+\psi(1)$ and since $\psi(1) = 1$ we have an element of $\mathbb{Q}(\sqrt{3})$ which, squared, is 2.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-06T18:37:54.027" CommentCount="8" />
  <row Id="9191" PostTypeId="1" CreationDate="2010-11-06T18:50:57.950" Score="0" ViewCount="142" Body="&lt;p&gt;Assume function $f$ has a discontinuity (jump discontinuity) at $x_0$. This discontinuity should be buried to level $k$, meaning (let $f$ is modified to $g$ after an operation) $g$, $g'$, $g'',\ldots g^{(k-1)}$ ($g^{(i)}$ means $i$th derivative of $g$) should all be continuous at $x_0$. What is the operation on $f$ by which this an be achieved (for example convolving with a suitable function, etc.,) ? For a slightly different question, assume $f$ is continous at $x_0$ but not differentiable.   &lt;strong&gt;EDIT&lt;/strong&gt;: $f$ has a compact support or atleast belong to $L^1$ or $L^2$. same should be the case with $g$.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-06T20:56:57.453" LastActivityDate="2010-11-07T02:41:46.037" Title="How to bury a discontinuity to a higher level ?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="27" />
  <row Id="9192" PostTypeId="1" AcceptedAnswerId="9194" CreationDate="2010-11-06T18:55:29.543" Score="1" ViewCount="57" Body="&lt;p&gt;I have a velocity vector on X, Y and Z axes. The values are represented as M/S (meters per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What algorithm should I use to convert this vector into a general MPH (miles per hour) value that takes into account all axes of the vector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may (or may not) be a very simple question. But I am far from being a genius at math! So apologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much.&lt;/p&gt;&#xA;" OwnerUserId="3198" LastActivityDate="2010-11-06T19:04:04.773" Title="Converting a velocity vector into MPH?" Tags="&lt;vector-space&gt;" AnswerCount="1" />
  <row Id="9193" PostTypeId="2" ParentId="9188" CreationDate="2010-11-06T18:59:11.353" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Compare discriminants: $\rm\displaystyle\ \big\{\big(\alpha-\alpha'\big)^2:\ \alpha \in \mathbb Q(\sqrt 2)\big\}\: =\ 2\ \mathbb Q^{\:2}\ \ $ vs. $\rm\ \ 3\ \mathbb Q^{\:2}\ $ for $\rm\ \mathbb Q(\sqrt 3)\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $\rm\ \alpha,\: \alpha'\ \not\in\mathbb Q\ $ are conjugate then they remain so under any field isomorphism since their minimal polynomial $\rm\ (x-\alpha)\ (x-\alpha')\ $ is in $\rm\:\mathbb Q[x]\:$ so it is fixed by any isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact quadratic fields are characterized uniquely be their &lt;strong&gt;discriminant&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-06T23:09:37.403" LastActivityDate="2010-11-06T23:09:37.403" CommentCount="4" />
  <row Id="9194" PostTypeId="2" ParentId="9192" CreationDate="2010-11-06T19:04:04.773" Score="2" ViewCount="0" Body="&lt;p&gt;Multiply each coordinate by 2.23693629.  In general, to convert units of a vector, you just convert units of each coordinate.  This works because scalar multiplication by positive constants preserves direction, and because lengths of vectors and scalar multiplication are related by $\|c v\|=|c|\|v\|$.  Scalar multiplication in this case just means multiplying each coordinate by the conversion factor.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-06T19:04:04.773" CommentCount="7" />
  <row Id="9195" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T19:11:24.167" Score="7" ViewCount="0" Body="&lt;p&gt;Since we are adding different proofs, here is a geometric one, for the special case when $m &gt; 0$ and $\alpha$ is acute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_tangents&quot; rel=&quot;nofollow&quot;&gt;Law of Tangents&lt;/a&gt; which states, that in a triangle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gkGVS.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{a-b}{a+b} = \frac{\tan\frac{\alpha - \beta}{2}}{\tan\frac{\alpha+\beta}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the below triangle (pardon the crude figure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1ivPA.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\displaystyle \angle{BCA} = \alpha$, $\displaystyle  \angle{CBA} = \pi/2$ and $\displaystyle \angle{CAB} = \pi/2 - \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ABC is a right angled triangle with $\displaystyle AC = 1 = m\cos\alpha + \sin \alpha$. Point $\displaystyle D$ is chosen on $\displaystyle AC$ such that $\displaystyle AD = \sin \alpha$ and so $\displaystyle DC = m\cos \alpha$. We also have $\displaystyle BC = \cos \alpha$ and $AB = \sin \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle AD = \sin \alpha = AB$ and so $\triangle ABD$ is isoceles and hence $\displaystyle \angle{ABD}  = \angle{ADB} = \pi/4 + \alpha/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x = \angle{CDB} = 3\pi/4 - \alpha/2$ and $\displaystyle  y = \angle{CBD} = \pi/4-\alpha/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply the law of tangents to $\displaystyle \triangle BCD$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\cos \alpha - m\cos\alpha}{\cos\alpha + m\cos\alpha} = \frac{\tan (x/2 - y/2)}{\tan(x/2 + y/2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{\tan(\pi/4)}{\tan(\pi/2 - \alpha/2)} = \tan(\alpha/2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1-m}{1+m} = \tan(\alpha/2)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-06T19:16:55.993" LastActivityDate="2010-11-06T19:16:55.993" CommentCount="3" />
  <row Id="9196" PostTypeId="2" ParentId="9188" CreationDate="2010-11-06T19:33:29.927" Score="7" ViewCount="0" Body="&lt;p&gt;More generally: suppose $d$ and $d'$ are both squarefree integers, both different from $1$, and consider $F_1 = \mathbb{Q}(\sqrt{d})$ and $F_2 = \mathbb{Q}(\sqrt{d'})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are both isomorphic as $\mathbb{Q}$-vector spaces, since they are both of dimension $2$; or more explicitly, every element of $F_1$ can be written uniquely as $a+b\sqrt{d}$ with $a,b\in\mathbb{Q}$ (unique because  $\sqrt{d}\notin\mathbb{Q}$), and every element of $F_2$ can be written uniquely as $x+y\sqrt{d'}$ with $x,y\in\mathbb{Q}$. The map $f\colon F_1\to F_2$ given by $f(a+b\sqrt{d}) = a + b\sqrt{d'}$ is additive and $\mathbb{Q}$-homogeneous, clearly bijective, so $F_1$ and $F_2$ are isomorphic as vector spaces over $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they are never isomorphic as fields; clearly, $d$ is a square in $F_1$. I claim $d$ can only be a square in $F_2$ if $d=d'$. Indeed, if$(x+y\sqrt{d'})^2 = d$. That means that $x^2 + d'y^2 + 2xy\sqrt{d'} = d$, hence $2xy = 0$ and $x^2+d'y^2=d$. If $x=0$, then $d=d'y^2$, so clearing denominators you get $da^2 = d'b^2$ for some $a,b\in\mathbb{Z}$, $\gcd(a,b)=1$; since both $d$ and $d'$ are squarefree, it follows that $|a|=|b|=1$, so $d=d'$. If $y=0$, then $d=x^2$, so $d$ is the square of a rational, contradicting the fact that it is a squarefree integer different from $1$. Thus, of $d$ is a square in $F_2$, then $d=d'$. Hence, if $F_1\cong F_2$, then $d=d'$ (converse is immediate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since every quadratic extension of $\mathbb{Q}$ is equal to $\mathbb{Q}(\sqrt{d})$ for some squarefree integer $d$ different from $1$, you conclude that any two quadratic extensions are either identical or not isomorphic as fields. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-06T20:08:49.083" LastActivityDate="2010-11-06T20:08:49.083" CommentCount="6" />
  <row Id="9197" PostTypeId="1" CreationDate="2010-11-06T19:55:03.010" Score="2" ViewCount="82" Body="&lt;p&gt;Let $f$ be in $L^{1}(\mathbb{R})$, $\mathbb{R}$ the real numbers. Show that for every $\varepsilon &gt; 0$ there exists $A \subseteq R$ , measurable, such that $m(A) &amp;lt; \infty$ , $f$ is bounded on $A$ and $ \int_{\mathbb{R}} |f|  &amp;lt; \int_{A} |f| + \varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take $A$ as the support of the simple function which approximates $f$ in the $L^{1}$ norm then this has finite measure and it satisfies the other conditions. But I don't see why $f$ must be bounded on it. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3176" LastActivityDate="2010-11-06T21:46:08.253" Title="Measurable set on which a function is bounded" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9198" PostTypeId="2" ParentId="9197" CreationDate="2010-11-06T20:05:09.190" Score="4" ViewCount="0" Body="&lt;p&gt;$|f|I_{|f|\leq n} $ increases to $|f|I_{|f| &amp;lt; \infty} = |f|$ (a.e.) as $f$ is in $L^{1}(\mathcal{R})$, the result follows from monotone convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update : This answer is incomplete, please see Jonas's answer.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-11-06T21:46:08.253" LastActivityDate="2010-11-06T21:46:08.253" CommentCount="1" />
  <row Id="9199" PostTypeId="1" AcceptedAnswerId="9203" CreationDate="2010-11-06T20:09:25.667" Score="7" ViewCount="273" Body="&lt;p&gt;This is motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/8933/proof-that-there-is-no-closed-form-solution&quot;&gt;this question&lt;/a&gt; and the fact that I have no access to the Timothy Chow's paper &lt;a href=&quot;http://www.jstor.org/pss/2589148&quot; rel=&quot;nofollow&quot;&gt;What Is a Closed-Form Number?&lt;/a&gt; indicated there by&#xA;Qiaochu Yuan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an equation $f(x)=0$ has no closed form solution, what does it normally&#xA;mean? Added: $f$ may depend (and normally does) on parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me this is equivalent to say that one cannot solve it for $x$ in&#xA;the sense that there is no elementary expression $g(c_{1},c_{2},\ldots&#xA;,c_{p})$ consisting only of a finite number of polynomials, rational&#xA;functions, roots, exponentials, logarithmic and trigonometric functions,&#xA;absolute values, integer and fractional parts, such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(g(c_{1},c_{2},\ldots ,c_{p}))=0$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-07T14:46:49.140" LastActivityDate="2010-11-07T14:46:49.140" Title="What does closed form solution usually mean?" Tags="&lt;soft-question&gt;&lt;elementary&gt;&lt;closed-form&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="9200" PostTypeId="2" ParentId="9197" CreationDate="2010-11-06T20:11:57.560" Score="4" ViewCount="0" Body="&lt;p&gt;You can do this as a &quot;$2\epsilon$-proof&quot; (or $\epsilon/2$ if you prefer).  First, since $(\int_{-n}^n|f|)_{n=1}^\infty$ converges to $\int_{\mathbb{R}}|f|$, there is an $n$ such that $\int_{\mathbb{R}}|f|\lt\int_{-n}^n|f|+\epsilon/2$.  Then, since $(\int_{-n}^n|f|\cdot\chi_{\{|f|\leq m\}})_{m=1}^\infty$ converges to $\int_{-n}^n|f|$, there is an $m$ such that $\int_{-n}^n|f|\lt\int_{-n}^n|f|\cdot\chi_{\{|f|\leq m\}}+\epsilon/2$.  Take $A=[-n,n]\cap\{x:|f(x)|\leq m\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than using simple functions to show this, I would use this as a first step to showing that $f$ can be approximated by simple functions, because now $f$ can be uniformly approximated by simple functions on $A$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-06T20:34:56.377" LastActivityDate="2010-11-06T20:34:56.377" CommentCount="4" />
  <row Id="9201" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T20:13:10.737" Score="5" ViewCount="0" Body="&lt;p&gt;Yet another approach, which is very specific to this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;1 - \sin\alpha &amp;amp;=&amp;amp; m \cos\alpha \\&#xA;\left( 1 - \sin\alpha \right)^2 &amp;amp;=&amp;amp; m^2 \cos^2\alpha \\&#xA;\left( 1 - \sin\alpha \right)^2 &amp;amp;=&amp;amp; m^2 ( 1 - \sin\alpha )( 1 + \sin\alpha ) \\&#xA;\left( 1 - \sin\alpha \right)\left((1 - \sin\alpha ) - m^2(1+\sin\alpha)\right) &amp;amp;=&amp;amp; 0 \\&#xA;\left( 1 - \sin\alpha \right)\left( (1-m^2) - ( 1 + m^2 )\sin\alpha \right) &amp;amp;=&amp;amp; 0&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\sin\alpha=1$ or $\sin\alpha=\frac{1-m^2}{1+m^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, if $\sin\alpha=1$, then your original equation reduces to the $m$-less tautology, &quot;$0=0$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, a little manipulation shows relations between $m$ and other trig functions of $\alpha$. For instance, you could substitute back into your original equation to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = m\cos\alpha + \frac{1-m^2}{1+m^2} \;\;\; \Longrightarrow \;\;\; \cos\alpha=\frac{2m}{1+m^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which you can get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1-m^2}{2m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha} = \frac{(1-m)^2}{1-m^2}=\frac{1-m}{1+m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whichever you prefer.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-06T20:13:10.737" CommentCount="3" />
  <row Id="9202" PostTypeId="2" ParentId="9157" CreationDate="2010-11-06T20:33:23.493" Score="2" ViewCount="0" Body="&lt;p&gt;This is not a good question because it actually has nothing to do with sampling theory (nor with the Chebyshev inequality, for that matter): it's quite artificial.  Nevertheless, you have grasped its intent well.  A good way to work through it is to reason in rather basic terms, rather than relying solely on mathematical manipulation; to wit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The standard deviation of $\bar{\chi}$ is $1/\sqrt{n}$ times the SD of the parent distribution, $\sigma$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore we should express the cutoff $0.5 \sigma$ in terms of the SD of $\bar{\chi}$ itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.5 \sigma = 0.5 \sqrt{n} \left(\sigma / \sqrt{n}\right) = 0.5 \sqrt{n} SD_{\bar{\chi}}.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We know (or can look up in a table) that $5$% of the time any normally distributed variate is further than $1.96$ times its standard deviation from its mean.  (The value $1.645$ corresponds to a probability of $10$%: the absolute value requires you to look at both tails, not just the right hand tail.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus the question is merely asking you to solve the equation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$0.5 \sqrt{n} \ge 1.96.$$&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-11-06T20:33:23.493" CommentCount="1" />
  <row Id="9203" PostTypeId="2" ParentId="9199" CreationDate="2010-11-06T20:39:27.777" Score="9" ViewCount="0" Body="&lt;p&gt;I would say it very much depends on the context, and what tools are at your disposal. For instance, telling a student who's just mastered the usual tricks of integrating elementary functions that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{\exp{u}-1}{u}\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\sqrt{(u+1)(u^2+1)}\mathrm{d}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have no closed form solutions is just the fancy way of saying &quot;no, you can't do these integrals yet; you don't have the tools&quot;. To a working scientist who uses exponential and elliptic integrals, however, they do have closed forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar vein, when we say that nonlinear equations, whether algebraic ones like $x^5-x+1=0$ or transcendental ones like $\frac{\pi}{4}=v-\frac{\sin\;v}{2}$ have no closed form solutions, what we're really saying is that we can't represent solutions to these in terms of functions that we know (and love?). (For the first one, though, if you know hypergeometric or theta functions, then yes, it has a closed form.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is fair to say that for as long as we haven't seen the solution to an integral, sum, product, continued fraction, differential equation, or nonlinear equation frequently enough in applications to give it a &lt;em&gt;standard&lt;/em&gt; name and notation, we just cop out and say &quot;nope, it doesn't have a closed form&quot;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T02:09:12.013" LastActivityDate="2010-11-07T02:09:12.013" CommentCount="6" />
  <row Id="9204" PostTypeId="1" AcceptedAnswerId="9219" CreationDate="2010-11-06T21:07:12.483" Score="1" ViewCount="129" Body="&lt;p&gt;You are allowed to use +, -, / and * (plus, minus, division and multiplication) signs and bracketing. These signs you can put between the numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1963&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to form mathematical expressions. You must put at least one sign between two numbers and – cannot be used as “negative”, thus -1+9+6+3 is not allowed, but 1-9+6+3 is allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: what is the smallest natural number that cannot be expressed in this way? How to show (elegantly, without computer) that it is impossible to get 14?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found myself the following representations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0=1*9-6-3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1=(1/9)*(6+3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2=1+9/(6+3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3=1*9/(6-3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4=1+9/(6-3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5=(1+9)/(6/3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6=1*9-6+3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7=1+9-6+3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8=1+9-6/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9=1*(9-6)*3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10=1+(9-6)*3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11=1*9+6/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12=1+9+6/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13=1+9+6-3&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-11-07T11:30:22.800" LastActivityDate="2010-11-07T11:30:22.800" Title="A problem related to the number 1963" Tags="&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9205" PostTypeId="2" ParentId="9138" CreationDate="2010-11-06T21:30:58.060" Score="3" ViewCount="0" Body="&lt;p&gt;Another:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{eqnarray}&#xA;\frac{1-\sin\alpha}{\cos\alpha} &amp;amp;=&amp;amp; m \\&#xA;\frac{1-\cos\left(\frac{\pi}{2}-\alpha\right)}{\sin\left(\frac{\pi}{2}-\alpha\right)} &amp;amp;=&amp;amp;m \\&#xA;\tan\frac{\frac{\pi}{2}-\alpha}{2} = \tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) &amp;amp;=&amp;amp; m \\&#xA;\frac{\tan\frac{\pi}{4}-\tan\frac{\alpha}{2}}{1+\tan\frac{\pi}{4} \tan\frac{\alpha}{2}} = \frac{1-\tan\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}} &amp;amp;=&amp;amp; m \\&#xA;\Longrightarrow \tan\frac{\alpha}{2} &amp;amp;=&amp;amp; \frac{1-m}{1+m}&#xA;\end{eqnarray}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-06T21:30:58.060" CommentCount="1" />
  <row Id="9206" PostTypeId="1" AcceptedAnswerId="9215" CreationDate="2010-11-06T21:41:30.660" Score="2" ViewCount="57" Body="&lt;p&gt;The following inequality is from the proof that the $L^p$ norm is Gâteaux differentiable for $ 1 &amp;lt; p&amp;lt;\infty$ (from &quot;Analysis&quot; by Lieb and Loss). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$, $b\in\mathbb{C}$ and $-1\leq t\leq 1$, $t\not=0.$ Then &#xA;$$|a|^p-|a-b|^p\leq\frac{1}{t}(|a+tb|^p-|a|^p) \leq |a+b|^p-|a|^p.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to prove the second inequality for positive $t$ by writing $a+tb=(1-t)a+t(a+b)$and using the convexity of $\cdot^p$. From this the first inequality follows for negative $t$ by substituting $-b$ for $b$. The same trick would finish the proof, if I could prove either the second inequality for negative $t$ or the first inequality for positive $t$.   &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2010-11-06T23:35:14.570" Title="The complex numbers inequality $(|a+tb|^p-|a|^p)/t \leq |a+b|^p-|a|^p$" Tags="&lt;analysis&gt;&lt;inequality&gt;" AnswerCount="1" />
  <row Id="9207" PostTypeId="1" CreationDate="2010-11-06T21:44:42.513" Score="0" ViewCount="148" Body="&lt;p&gt;Hi, I need to prove or disprove whether non-singular matrices commute. Can you help me and tell me how? What should i use cos i really don't know, I am beginner in this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="3203" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T00:24:48.680" LastActivityDate="2010-11-07T00:24:48.680" Title="Nonsingular matrices commute - prove or disprove" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9208" PostTypeId="2" ParentId="9178" CreationDate="2010-11-06T22:07:45.527" Score="0" ViewCount="0" Body="&lt;p&gt;Graham et al. and Rosen are both excellent books but they are also very comprehensive. If you are looking for something that is shorter (definitely not comprehensive) and more like an &quot;outline&quot; you might find this small paperback useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essential Discrete Mathematics for Computer Science by Todd Feil and Joan Krone, Prentice-Hall, 2003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chapters are somewhat independent and include pseudo-code for some algorithms. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-06T22:07:45.527" />
  <row Id="9209" PostTypeId="1" AcceptedAnswerId="9214" CreationDate="2010-11-06T22:09:26.510" Score="1" ViewCount="47" Body="&lt;p&gt;According to the definition of stochastic process in &lt;a href=&quot;http://en.wikipedia.org/wiki/Stochastic_process#Definition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a probability space $(\Omega, \mathcal{F}, P)$, a stochastic process (or random process) with state space $X$ is a collection of $X$-valued random variables indexed by a set $T$ (&quot;time&quot;). That is, a stochastic process $F$ is a collection&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\{ F_t : t \in T \}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where each $F_t$ is an $X$-valued random variable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does that mean for all values of the index $t$, the state space $X$ and the $\sigma$-algebra on it must be same? For a stochastic process, Can it be allowed that the state space and the $\sigma$-algebra on it is different for different value of the index?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct that, as a stochastic process, &quot;a random walk on a graph&quot; seems to have different state spaces at different stages/times? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-06T22:48:13.013" LastActivityDate="2010-11-06T22:48:13.013" Title="state space of stochastic process" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="9210" PostTypeId="1" CreationDate="2010-11-06T22:16:02.890" Score="7" ViewCount="120" Body="&lt;p&gt;Is there a general formula to determine the probability of unbounded, cumulative dice rolls hitting a specified number? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Example, with a D6 and 14:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 + 2 + 3 + 4 = 14 : success&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 1 + 1 + 6 + 5 + 4 = 17 : failure &lt;/p&gt;&#xA;" OwnerUserId="3204" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-06T23:13:35.827" LastActivityDate="2010-11-07T01:09:01.290" Title="Probability of cumulative dice rolls hitting a number" Tags="&lt;number-theory&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="9211" PostTypeId="2" ParentId="9189" CreationDate="2010-11-06T22:22:01.867" Score="1" ViewCount="0" Body="&lt;p&gt;Have to do it with 2 coverings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering  1: For each $m\in M$ chose open $V_m \subset M$ and $\epsilon_m &gt;0$ with $\epsilon_m &amp;lt; inf \delta(V_m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering 2: Take locally finite covering $(U_i)_{i\in I}$ finer than preceding covering. For each  $i\in I$ chose $m\in M$ with  $U_i\subset V_m$ and put $\delta_i =\epsilon_m$  . Remark that on $U_i$ we have $\delta_i &amp;lt; inf  \delta (U_i).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then associate partition of unity $(\phi_i)_{i\in I}$ with covering $(U_i)_{i\in I}$ and wanted function is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde \delta=\sum \delta_i \phi_i $$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-06T22:22:01.867" CommentCount="3" />
  <row Id="9212" PostTypeId="2" ParentId="9199" CreationDate="2010-11-06T22:36:32.030" Score="3" ViewCount="0" Body="&lt;p&gt;To better understand closed forms, you may want to familiarize yourself with what's called Differential Algebra.  Just as number theory relies on abstract structures such as rings, fields, ideals, etc to express roots algebraic equations using elementary numbers, similarly there is a parallel apparatus for expressing functions (i.e. solutions of differential equations) using differential rings, fields, ideals  called &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_algebra&quot; rel=&quot;nofollow&quot;&gt;Differential Algebra&lt;/a&gt;.  It is this underlying mechanism that defines which functions can be expressed as &quot;closed forms&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallels:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Similar to splitting fields for&#xA;algebraic equations, there is a&#xA;parallel Galois theory with&#xA;Picard-Vessiot extensions and what&#xA;not.&lt;/li&gt;&#xA;&lt;li&gt;Similar to correspondence between&#xA;subfields of number fields and&#xA;Galois subgroups, on the&#xA;differential side, there is a correspondence between differential subfields and subgroups of algebraic groups.&lt;/li&gt;&#xA;&lt;li&gt;Just as algebraic equations can be&#xA;determined to be solvable by&#xA;radicals, similarly linear&#xA;differential equations can be&#xA;determined to be solvable by&#xA;exponentials, Liouvillian functions,&#xA;etc.  There is an ascending tower of differential fields which can be built &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is more...I am no expert in this differential algebra field but if you want some freely available references, see &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Seiler &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.47.6596&quot; rel=&quot;nofollow&quot;&gt;Computer Algebra and&#xA;differential equations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Van der Put &lt;a href=&quot;http://iwi.eldoc.ub.rug.nl/FILES/root/1999/JSymbCompvdPut/1999JSymbCompvdPut.pdf&quot; rel=&quot;nofollow&quot;&gt;Galois theory of&#xA;differential equations, algebraic&#xA;groups and Lie algebras&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Papers by &lt;a href=&quot;http://www4.ncsu.edu/~singer/ms_papers.html&quot; rel=&quot;nofollow&quot;&gt;Michael F Singer&lt;/a&gt; are good. See for example Galois theory of linear differential equations.&lt;/li&gt;&#xA;&lt;li&gt;Check the &lt;a href=&quot;http://www.sci.ccny.cuny.edu/~ksda/posted.html&quot; rel=&quot;nofollow&quot;&gt;Kolchin seminar in&#xA;Differential Algebra&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2010-11-06T22:36:32.030" />
  <row Id="9213" PostTypeId="2" ParentId="9210" CreationDate="2010-11-06T22:42:39.250" Score="8" ViewCount="0" Body="&lt;p&gt;Assuming the order matters (i,e 1+2 is a different outcome from 2+1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of getting the sum $n$ with dice numbered $1,2,\dots,6$ is the coefficient of $x^n$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=0}^{\infty}(\frac{x+x^2+x^3+x^4+x^5+x^6}{6})^j = \frac{6}{6-x-x^2-x^3-x^4-x^5-x^6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing it as partial fractions (using roots of $6-x-x^2-x^3-x^4-x^5-x^6=0$)  or using Cauchy's integral formula to find the coefficient of $x^n$, Taylor series, etc should work.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-06T22:50:50.050" LastActivityDate="2010-11-06T22:50:50.050" CommentCount="5" />
  <row Id="9214" PostTypeId="2" ParentId="9209" CreationDate="2010-11-06T22:44:40.297" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, the measurable space $(X, \mathcal{M})$ ($M \subset 2^X$ being a $\sigma$-algebra) should be the same for all $t$ in the index set.  If you want the process to occupy different sets $X_t$ at different times, then the right things to do is to take $X$ to be the disjoint union of all the $X_t$.  There is no requirement that $F_t$ be surjective, so there is no harm if some of the state space is not used at some times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a random walk on a graph, the state space $X$ is just the vertex set of the graph, typically with the discrete $\sigma$-algebra.  Then for each $t \in T$, $F_t$ is some random element of $X$, i.e. a random vertex of the graph.  Formally, if your underlying probability space is $(\Omega, \mathcal{F}, P)$, then $F$ is a map from $\Omega \times T$ into $X$, with the property that for each $t \in T$, the map $\Omega \ni \omega \mapsto F(\omega, t) \in X$, denoted $F_t$, is measurable.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-06T22:44:40.297" />
  <row Id="9215" PostTypeId="2" ParentId="9206" CreationDate="2010-11-06T23:35:14.570" Score="1" ViewCount="0" Body="&lt;p&gt;The same basic idea works for negative values of $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to prove is equivalent to $|a + tb|^p - |a|^p \geq t|a + b|^p - t|a|^p$. This in turn is the same as $(1 - t)|a|^p \leq |a + tb|^p + (-t)|a + b|^p$, which in turn is equivalent to $|a|^p \leq {1 \over 1 - t} |a + tb|^p + {-t \over 1 - t} |a + b|^p$. Note that $a = {1 \over 1 - t} (a + tb) + {-t \over 1 - t} (a + b)$, so you can use convexity as you used before.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-06T23:35:14.570" />
  <row Id="9216" PostTypeId="2" ParentId="9207" CreationDate="2010-11-06T23:40:57.100" Score="-1" ViewCount="0" Body="&lt;p&gt;This is one of those problems where one example that contradicts the claim is enough to disprove it (also known as giving a counterexample).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phrase non-singular matrices indicates we are talking about square matrices, and presumably you want them to be of equal size (since otherwise their product is not defined in either order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as the commentators have suggested, we should try a couple of &quot;random&quot; two-by-two matrices, as these will &quot;usually&quot; not commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a pair of non-commuting non-singular matrices:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  0&#xA;0 -1&#xA;&#xA;1  1&#xA;0  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Multiplying by the first of these matrices on the left scales the second row by a factor of -1 (and otherwise leaves the entries of the first row unchanged), while multiplying by it on the right scales the second column by a factor of -1 (and otherwise leaves the entries of the first column unchanged).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try and you'll quickly see that the pair of matrices above do not commute (but both are non-singular, e.g. because their determinants are nonzero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-06T23:40:57.100" CommentCount="2" />
  <row Id="9217" PostTypeId="2" ParentId="8186" CreationDate="2010-11-07T00:04:55.157" Score="0" ViewCount="0" Body="&lt;p&gt;This follows easily by Bézout's lemma (as Qiaochu notes, x and y only need to be coprime to generate the unit ideal (indeed, this is the proper generalization of the term &quot;coprime&quot;)).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can ever so-slightly strengthen this to show that $\mathbf{Z}$ is a PID (this is the true content of Bézout's lemma).&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-07T00:54:29.040" LastActivityDate="2010-11-07T00:54:29.040" />
  <row Id="9219" PostTypeId="2" ParentId="9204" CreationDate="2010-11-07T00:17:17.140" Score="0" ViewCount="0" Body="&lt;pre&gt;&#xA;from fractions import Fraction&#xA;&#xA;def Jaska(L):&#xA;    if len(L) == 1:&#xA;        yield L[0], Fraction(L[0])&#xA;    else:&#xA;        for l in range(len(L) - 1):&#xA;            for e1, v1 in Jaska(L[:l+1]):&#xA;                for e2, v2 in Jaska(L[l+1:]):&#xA;                    yield '(%s + %s)' % (e1, e2), v1 + v2&#xA;                    yield '(%s - %s)' % (e1, e2), v1 - v2&#xA;                    yield '(%s * %s)' % (e1, e2), v1 * v2&#xA;                    if v2 != 0:&#xA;                        yield '(%s / %s)' % (e1, e2), v1 / v2&#xA;&#xA;def JaskaAll(L):&#xA;    solutions = dict([])&#xA;    for e, v in Jaska(L):&#xA;        if v.denominator == 1:&#xA;            v = v.numerator&#xA;        if v not in solutions:&#xA;            solutions[v] = []&#xA;        solutions[v].append(e[1:-1])&#xA;    return solutions&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can check &lt;pre&gt;14 in JaskaAll([1,9,6,3])&lt;/pre&gt; to make sure.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T00:17:17.140" CommentCount="1" />
  <row Id="9221" PostTypeId="2" ParentId="9210" CreationDate="2010-11-07T01:09:01.290" Score="3" ViewCount="0" Body="&lt;p&gt;The short answer is not really, but it doesn't matter.  If you're interested in computing the exact probability there is a way to do it reasonably quickly by writing down a certain matrix and using binary exponentiation, but I'll let you in on a secret: for large $n$, the answer is exponentially close to $\frac{1}{6}$ (that is, it's exactly what you would naively expect).  A very similar result is proven in the MO thread Michael linked to.  So if that's all you wanted to know, then you know it.  If you really wanted to estimate how close it was to $\frac{1}{6}$ then you'd need to estimate the next largest root of the characteristic polynomial $x^6 = \frac{x^5 + ... + 1}{6}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-07T01:09:01.290" />
  <row Id="9222" PostTypeId="1" AcceptedAnswerId="9234" CreationDate="2010-11-07T01:14:00.053" Score="3" ViewCount="104" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble with the following exercise in do Carmo's Riemannian geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ and $Y$ be differentiable vector fields on a Riemannian manifold $M$. Let $p \in M$ and let $c: I \to M$ be an integral curve of $X$ through $p$, i.e. $c(t_0) = p$ and $\frac{dc}{dt} = X(c(t))$. &#xA;Prove that the Riemannian connection of $M$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\nabla_XY \ )(p) = \frac{d}{dt} (P^{-1}_{c,t_0,t}(Y(c(t)))\ |_{t=t_0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P^{-1}_{c,t_0,t}: T_{c(t_0)}M \to T_{c(t)}M$ is the parallel transport along $c$, from $t_0$ to $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess, I don't have enough understanding of how to handle the parallel transport (since it is only given as the unique solution to a differential equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;S. L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Do Carmo first introduced the notion of an affine connection &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla: \text{Vect}(M) \times \text{Vect}(M) \to \text{Vect}(M)$,   $(X,Y) \mapsto \nabla_XY$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\nabla_{fX + gY}Z = f\nabla_XZ + g \nabla_YZ$&lt;/li&gt;&#xA;&lt;li&gt;$\nabla_X(Y+Z) = \nabla_XY + \nabla_XZ$&lt;/li&gt;&#xA;&lt;li&gt;$\nabla_X(fY) = f\nabla_XY + X(f)Y$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;for $X,Y,Z \in \text{Vect}(M)$ and $f,g \in C^\infty(M)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then showed that there is a unique correspondence which associates to a vector field $V$ along the differentiable curve $c: I \to M$ another vector field $\frac{DV}{dt}$ along c, called covariant derivative of $V$ along $c$, with three more properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\frac{D}{dt}(V+W) =  \frac{D}{dt}V +  \frac{D}{dt}W$&lt;/li&gt;&#xA;&lt;li&gt;$ \frac{D}{dt}(fV) =  \frac{df}{dt}V +  f\frac{D}{dt}V$&lt;/li&gt;&#xA;&lt;li&gt;If $V$ is induced by a vector field $Y \in \text{Vect}(M), then  \frac{D}{dt}V = \nabla_{dc/dt}Y$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then he showed existence and uniqueness of the parallel transport along a curve, and went on to prove existence and uniqueness of Levi-Civita connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes things clearer? Thanks for the quick answer!&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="3208" LastEditorDisplayName="" LastEditDate="2010-11-07T01:45:05.440" LastActivityDate="2010-11-07T19:14:11.043" Title="Expression for Levi-Civita Connection" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="9223" PostTypeId="1" AcceptedAnswerId="9224" CreationDate="2010-11-07T01:48:02.990" Score="1" ViewCount="93" Body="&lt;p&gt;The table of integrals says that&#xA;$\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\arctan\frac{x}{a}+C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with my proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;y &amp;amp;= \arctan\frac{x}{a} \\&#xA;a\tan y&amp;amp;=x \\&#xA;\tan y &amp;amp;= \frac{x}{a} \\&#xA;\frac{d}{dx} \tan y &amp;amp;= \frac{d}{dx} \frac{x}{a} \\&#xA;\sec^{2}y \frac{dy}{dx} &amp;amp;= \frac{1}{a} \\&#xA;\frac{dy}{dx} &amp;amp;= \frac{1}{a} \frac{1}{\sec^{2}y} \\&#xA;&amp;amp;= \frac{1}{a} \frac{1}{1 + \tan^{2} y} \\&#xA;\frac{dy}{dx}&amp;amp;= \frac{1}{a} \frac{1}{1 + (\frac{x}{a})^{2}} \\&#xA;\int \frac{dy}{dx}\,dx &amp;amp;= \int \frac{1}{a} \frac{1}{1 + (\frac{x}{a})^{2}}\,dx\\&#xA;\arctan\frac{x}{a} &amp;amp;= \frac{1}{a} \int \frac{1}{1 + (\frac{x}{a})^{2}}\,dx \\&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right so far?&#xA;Any help appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T01:54:21.160" LastActivityDate="2010-11-07T03:42:45.530" Title="Proving an Integral" Tags="&lt;trigonometry&gt;&lt;integral&gt;&lt;proof&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9224" PostTypeId="2" ParentId="9223" CreationDate="2010-11-07T01:58:41.343" Score="3" ViewCount="0" Body="&lt;p&gt;Other than having unnecessary steps (why did you move $a$ to the left hand side in the second step, only to bring it back to the right hand side in the next step?), and missing the constant of integration on the left hand side in the last equation (as well as the missing &quot;$dx$&quot; which I fixed), nothing wrong so far. Now perform the operation inside the integral to get $\frac{a^2}{a^2+x^2}$, pull out the constant, simplify, and move it the left hand side to get the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if what you want is to show that the integral of $\frac{1}{a^2+x^2}$ is really equal to $\frac{1}{a}\arctan(\frac{x}{a}) + C$, then why not simply differentiate the latter and see if you get the integrand?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-07T01:58:41.343" CommentCount="3" />
  <row Id="9225" PostTypeId="1" CreationDate="2010-11-07T02:40:09.467" Score="1" ViewCount="136" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;On a small island there are 25&#xA;  inhabitants. One of these inhabitants,&#xA;  named Jack, starts a rumor which&#xA;  spreads around the isle. Any person&#xA;  who hears the rumor continues&#xA;  spreading it until he or she meets&#xA;  someone who has heard the story&#xA;  before. At that point, the person&#xA;  stops spreading it, since nobody likes&#xA;  to spread stale news.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) Do you think all 25 inhabitants will eventually hear the rumor or will the rumor die out before that happens? Estimate the proportion of inhabitants who will hear the rumor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would one go about calculating the probability? I would try to block it out, but there are so many potential branches. This problem is in the simulations portion of our statistics book, but I'm curious if there is any other way that doesn't involve simulating the situation.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T02:50:09.783" LastActivityDate="2010-11-07T04:28:57.070" Title="Probability of the propagation of a rumor problem." Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="9226" PostTypeId="2" ParentId="9191" CreationDate="2010-11-07T02:41:46.037" Score="3" ViewCount="0" Body="&lt;p&gt;If $f(x)$ has a jump discontinuity at $x_0$, then $g_k(x) = f(x)(x-x_0)^{k+1}$ has a continuous $k$th derivative at $x_0$ but is not $(k+1)$ times differentiable at $x_0$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-07T02:41:46.037" CommentCount="3" />
  <row Id="9227" PostTypeId="2" ParentId="9222" CreationDate="2010-11-07T02:44:07.607" Score="3" ViewCount="0" Body="&lt;p&gt;This is a cultural comment, rather than an answer; but it is a bit long for the comment box, which is why I am writing it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My comment is the following: giving a connection, and giving the associated parallel transport, are essentially the same thing, and so it should be possible to establish the formula you are trying to prove by reasonably conceptual, high-level, thinking, rather than mucking about too much with complicated differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate a little: how would you define the derivative of a vector field along a curve?  Well, the idea is that you want to form the usual Newton quotient&#xA;$$ \dfrac{V\bigl(c(t+\epsilon)\bigr) - V\bigl(c(t)\bigr)}{\epsilon}$$&#xA;and then let $\epsilon$ go to zero, to compute that derivative of $V$ at the&#xA;point $c(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is that the subtraction in this formula doesn't make sense,&#xA;because the tangent vectors being subtracted are based at different points:&#xA;one is based at $c(t+\epsilon)$ and the other at $c(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter parallel transport: suppose we have an agreed upon way to shift, or &lt;em&gt;transport&lt;/em&gt;, tangent&#xA;vectors along a curve.  Then we can apply this so as to transport &#xA;$V\bigl(c(t+\epsilon)\bigr)$ from $c(t+\epsilon)$ to $c(t)$, and then form the&#xA;above Newton quotient, and proceed to compute a derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An agreed upon method for shifting vectors along curves is formally referred to as a choice of &lt;em&gt;parallel transport&lt;/em&gt;, because if you have such a method, and if you define a vector field along a curve $c(t)$ by choosing some $V_0$ at&#xA;$c(t_0)$ and then using the agreed upon parallel transport to define&#xA;$V\bigl(c(t)))$ by tranporting $V_0$ along $c$ to $c(t)$, you obtain a vector&#xA;field along $c(t)$ whose derivative at each point will be zero (by construction!).  Thus this vector field does not change along $c(t)$,&#xA;and so consists of parallel vectors, hence the name &lt;em&gt;parallel tranport&lt;/em&gt;.&#xA;(Of course, the notions of &lt;em&gt;change&lt;/em&gt; and &lt;em&gt;parallel&lt;/em&gt; are not intrinsict to&#xA;the vector field; they depend upon our particular choice of parallel transport.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, our choice of parallel transport should satisfy some axioms (it should be linear; it should be smooth (so that the vector fields along curves&#xA;that it gives rise to are smooth); and so on).  If you look in the right text,&#xA;you will find these axioms written down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we've seen, a choice of parallel transport on our manifold gives a way&#xA;of differentiating vector fields along curves.  But since a tangent vector&#xA;is just an infinitesimal curve, one sees that we in fact have a way of &#xA;differentiating a vector field in the direction of a given vector field: extend that vector field to a curve, and then differentiate along that curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a choice of parallel transport determines an affine connection, i.e. a way&#xA;of differentiating vector fields along tangent vectors.  Conversely, this is enough information to determine the parallel transport: we parallel transport a tangent vector along a curve by making sure that its derivative (using the given affine connection) at every point of the curve, in the tangent direction of the curve, vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have gone full circle, from parallel transport, to affine connection, back to parallel transport.  The text you are reading is spelling out the second half of this circle, but is perhaps a little scanty on details regarding the first half; in fact, the exercise you are trying to solve is exactly about the first half of the circle, and its goal is to verify that you truly are going around in a circle: i.e. that you are ending up where you started from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really is something to verify here (i.e. the exercise is not trivial if you have not done it before, and are learning these ideas for the first time), but I hope that the above discussion may help shed some light on its meaning,&#xA;and also make it seem less daunting (and perhaps more conceptual and less computational) than it might otherwise appear.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-07T19:00:21.193" LastActivityDate="2010-11-07T19:00:21.193" CommentCount="2" />
  <row Id="9228" PostTypeId="2" ParentId="9225" CreationDate="2010-11-07T03:00:20.303" Score="0" ViewCount="0" Body="&lt;p&gt;Take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logistic_function&quot; rel=&quot;nofollow&quot;&gt;logistic function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-07T03:00:20.303" />
  <row Id="9229" PostTypeId="2" ParentId="704" CreationDate="2010-11-07T03:13:54.083" Score="1" ViewCount="0" Body="&lt;p&gt;I encountered this situation when dealing with permutations; we phrased it as follows.  If $\sigma$ is a permutation of $\mathbb{Z}_n$ and $d$ is a divisor of $n$, then it is impossible for precisely two of the following statements to hold for some $i,j \in \mathbb{Z}_n$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$i \equiv j \pmod d$,&lt;/li&gt;&#xA;&lt;li&gt;$\sigma(i) \equiv \sigma(j) \pmod d$ and&lt;/li&gt;&#xA;&lt;li&gt;$\sigma(i)-i \equiv \sigma(j)-j \pmod d$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I suspect a modification of this would suffice in most cases.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-07T03:13:54.083" />
  <row Id="9230" PostTypeId="1" CreationDate="2010-11-07T03:25:39.717" Score="2" ViewCount="167" Body="&lt;p&gt;I see people like Terry Tao and others take extensive notes. But is this really necessary? When I do this I feel like I am rewriting a textbook.&lt;/p&gt;&#xA;" OwnerUserId="3209" LastActivityDate="2010-11-07T04:04:52.273" Title="What are examples of mathematicians who dont take many notes?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9231" PostTypeId="1" CreationDate="2010-11-07T03:31:38.067" Score="0" ViewCount="73" Body="&lt;p&gt;I've read a few articles explaining the way to use the Minkowski reduced basis in a lattice in order to measure the uniformity of the output of a random number generator. However, I can't prove a conclusion drawn from the following statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a set of linearly independent vectors $\{v_{1}, v_{2}, ... ,v_{d}\}$ in $\mathbb{R^{d}}$, a lattice is the set of vectors $w$ of the form $\sum_{i=1}^{d}z_{i}v_{i}$, where $z_{i}$ are integers. The set of vectors $\{v_{i}\}$ is a basis for the lattice. The basis is a Minkowski reduced basis if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$||v_{k}||\leq||\sum_{i=1}^{d}z_{i}v_{i}||, \qquad \text{for} \quad 1\leq k \leq d,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all sets of $d$ integers $z_{i}$ such that the greatest common divisor of the set $\{z_{k},z_{k+1},...,z_{d}\}$ is $1$. Accordingly, this implies $|v_{1}|\leq |v_{2}| \leq ... \leq |v_{d}|$. In the condition for a Minkowski reduced basis, $||v||$ denotes the Euclidean norm of the vector $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can we infer that $|v_{1}|\leq |v_{2}| \leq ... \leq |v_{d}|$)?. That's what I'd like to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, $||v_{1}||$ (for example), differs only from $||v_{2}||$ in having the restriction $\text{gcd}(\{z_{1},z_{2}\})=1$, while the second vector has $\text{gcd}(\{z_{2}\})=1$ in $\mathbb{R^{2}}$, however, I don't know how can I reach the conclusion from that fact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-07T04:48:59.150" LastActivityDate="2010-12-07T05:44:25.140" Title="Proof for Minkowski reduced basis" Tags="&lt;random&gt;&lt;lattices&gt;" AnswerCount="1" />
  <row Id="9232" PostTypeId="2" ParentId="9230" CreationDate="2010-11-07T03:36:11.457" Score="5" ViewCount="0" Body="&lt;p&gt;Everyone learns a bit differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people learn better by listening; others by writing. I find that the best way for me to understand someone else's proof is to write it out in careful detail, rather than simply reading it.  For me, taking notes is a way to help me keep up with the material while digesting and internalising it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other people don't learn that way. George Bergman (my advisor), for example, throughly dislikes lecturing (in the traditional way) because he says that it is a method for transferring information from the lecturer's notes to the student's notebook without passing through the brain of either (an opinion he has held since he took his first graduate course; so clearly he derived little benefit from taking notes the way you describe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all you are doing is writing and this process does not help you think about what you are writing, then it is probably not in your best interest to write notes. Better to keep the pad for any sudden ideas or insights you might want to come back later, and trust the textbook to provide the information you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... whether it is &quot;necessary&quot; is not an absolute, but rather a function of your own personal preference. In short: &quot;whatever floats &lt;em&gt;your&lt;/em&gt; boat&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T04:04:07.910" LastActivityDate="2010-11-07T04:04:07.910" CommentCount="4" />
  <row Id="9233" PostTypeId="2" ParentId="9223" CreationDate="2010-11-07T03:42:45.530" Score="1" ViewCount="0" Body="&lt;p&gt;A better way to prove it would be to let $\displaystyle x = a\tan{\theta}$, then $\displaystyle \frac{dx}{d\theta} = a\sec^2{\theta} \Rightarrow dx = a\sec^2{\theta}\;{d\theta}$. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int\frac{1}{x^2+a^2}\;{dx} = \int\frac{a\sec^2{\theta}}{a^2\tan^2{\theta}+a^2}\;{d\theta} = \frac{1}{a}\int\frac{a\sec^2{\theta}}{a(\tan^2{\theta}+1)}\;{d\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  = \frac{1}{a}\int\frac{a\sec^2{\theta}}{a\sec^2{\theta}}\;{d\theta} = \frac{1}{a}\int\;{d\theta} = \frac{1}{a}\theta+k = \frac{1}{a}\arctan\frac{x}{a}+k$.&lt;/p&gt;&#xA;" OwnerUserId="3207" LastActivityDate="2010-11-07T03:42:45.530" CommentCount="4" />
  <row Id="9234" PostTypeId="2" ParentId="9222" CreationDate="2010-11-07T03:46:44.623" Score="3" ViewCount="0" Body="&lt;p&gt;Let $\{e_i\}\subseteq T_{c(t_0)}M$ be a basis.  Define $E_i(t)$ as the parallel translation of $e_i$ along $c(t)$.  Prove to yourself that $\{E_i(t)\}$ forms a basis of $T_{c(t)}$ for all $t$ (hint, use the uniqueness part of solving linear ODEs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can write $Y(c(t)) = \sum_i a_i(t) E_i(t)$.  Then, since the $E_i(t)$ are parallel, and since $P_{c,t_0,t}$ is linear, we have $P^{-1}_{c,t_0, t}(Y(c(t)) = \sum_i a_i(t) e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, by actually computing the limit, it's not too hard to see that $\frac{d}{dt} P^{-1}_{c,t_o,t}Y(c(t))|_{t=t_0} = \sum_i a_i'(t_0)e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the goal is to show that $\nabla_X Y(p)$ can also be written like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\nabla_X Y(p) = \nabla_X \sum_i a_i(t)E_i(t)|_{t=t_0} = \sum_i a_i(t_0) \nabla_X E_i(t)|_{t=t_0} + \sum_i a_i'(t_0) E_i(t_0)$.  (The second equality is just the Leibniz rule every connection must satisfy).  However, $\nabla_X E_i(t)$ is $\frac{D}{dt} E_i(t)$ and this is 0 since the $E_i$ are parallel.  Hence, the first term of the sum vanishes, so we get the desired result.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-07T19:14:11.043" LastActivityDate="2010-11-07T19:14:11.043" CommentCount="1" />
  <row Id="9235" PostTypeId="2" ParentId="9230" CreationDate="2010-11-07T03:56:08.727" Score="2" ViewCount="0" Body="&lt;p&gt;When I go to a lecture I try to discern the general idea and condense the talk into a small summary -- sometimes half a page, sometimes just a couple of lines.  Sometimes I get technical details out of a talk, sometimes it's more of a big picture.  Sometimes I miss the point. Sometimes I latch onto an idea I don't understand and spend the entire talk trying to work that out, or I try to prove the speaker wrong.  But as Arturo mentions, different people have different styles. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-07T04:04:52.273" LastActivityDate="2010-11-07T04:04:52.273" />
  <row Id="9236" PostTypeId="2" ParentId="9225" CreationDate="2010-11-07T04:00:14.737" Score="3" ViewCount="0" Body="&lt;p&gt;An &lt;a href=&quot;http://en.wikipedia.org/wiki/Epidemic_model#The_SIR_Model&quot; rel=&quot;nofollow&quot;&gt;SIR model&lt;/a&gt; is also appropriate because of this aspect of the problem: &quot;Any person who hears the rumor continues spreading it until he or she meets someone who has heard the story before.  At that point, the person stops spreading it, since nobody likes to spread stale news.&quot;  Here, susceptible = hasn't heard rumor, infected = has heard rumor but is still willing to spread it, and recovered = has heard rumor but is no longer willing to spread it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see you're in 9th grade, and so the differential equations in the SIR model may not be something you've seen yet.  Still, you may get something out of reading the explanation of the model on the Wikipedia page.  Also, the standard SIR model doesn't quite fit this scenario; the equations for the infected and recovered populations for your scenario should be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dI}{dt} = \beta SI - \gamma RI,$$&#xA;$$\frac{dR}{dt} = \gamma RI.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you said you only wanted hints, but I'm giving you a fuller explanation of the SIR model because, according to &lt;a href=&quot;http://www.math.vt.edu/people/gboquet/papers/discrete.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, &quot;While the formal model is simple, in general it cannot be solved analytically.&quot;  So this may just be one of those problems for which simulation is the best solution we have now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O.K., I have to add this: For fun, you should also check out the &lt;a href=&quot;http://cse.ucdavis.edu/~chaos/courses/nlp/Projects2009/DanielSpielman/Report.pdf&quot; rel=&quot;nofollow&quot;&gt;SIRZ&lt;/a&gt; (susceptible-infected-recovered-zombie) model.  We can all be thankful that the authors say, &quot;Most of the analysis concluded that, under reasonable circumstances, a zombie apocalypse is unlikely.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-07T04:28:57.070" LastActivityDate="2010-11-07T04:28:57.070" />
  <row Id="9237" PostTypeId="2" ParentId="9231" CreationDate="2010-11-07T04:02:56.460" Score="1" ViewCount="0" Body="&lt;p&gt;Just consider the sets $\{z_1,\ldots,z_d\} = \{0,\ldots,0,1,0,\ldots 0 \}$ in the definition of Minkowski reduced bases, where for $v_k$ you put a 1 at the $k+1$-st place.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-07T04:02:56.460" CommentCount="8" />
  <row Id="9238" PostTypeId="2" ParentId="9152" CreationDate="2010-11-07T04:05:30.523" Score="1" ViewCount="0" Body="&lt;p&gt;I recommend &lt;a href=&quot;http://mrob.com/pub/ries/index.html&quot; rel=&quot;nofollow&quot;&gt;ries&lt;/a&gt; which finds solutions to this problem, related problems with different digits and operators, and in general finds operations that end up with a specified number.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-11-07T04:05:30.523" />
  <row Id="9239" PostTypeId="1" AcceptedAnswerId="9246" CreationDate="2010-11-07T04:22:57.923" Score="4" ViewCount="93" Body="&lt;p&gt;A powder can be compressed by packing it down. Each time it is packed down it loses $\frac{1}{2}$ then $\frac{1}{4}$ then $\frac{1}{8}$ ... etc. of it's total volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This powder is placed in to a container of unit volume and packed down. Then the remaining space is filled with fresh powder and packed down again. &lt;strong&gt;The packing action acts on the powder multiple times.&lt;/strong&gt; So, for example, here is the volume of powder after the first few packings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12) = \frac12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12)(\frac34) + (\frac12)(\frac12) =  \frac58$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12)(\frac34)(\frac78) + (\frac12)(\frac12)(\frac34) + (\frac38)(\frac12) = \frac{45}{64}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\frac12)(\frac34)(\frac78)(\frac{15}{16}) + (\frac12)(\frac12)(\frac34)(\frac78) + (\frac38)(\frac12)(\frac34) + (\frac{19}{64})(\frac12)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would like to know how much powder is in the container after it has been packed n times, in terms of n.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-07T04:44:56.917" LastActivityDate="2010-11-07T05:49:36.717" Title="Packing powder series, suspect it is easy, yet..." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9240" PostTypeId="2" ParentId="9225" CreationDate="2010-11-07T04:23:44.853" Score="4" ViewCount="0" Body="&lt;p&gt;Mike Spivey's remark that there are only three classes of people is quite relevant if you want to simulate this.  At the start, there are 24 susceptible people and one infected.  Call the number of each category S, I, and R.  If two I meet, they both become R, so I goes down by 2 and R goes up by 2.  If an S meets an I, the S becomes I, so S goes down by 1 and I goes up by 1.  If R meets I, I goes down by 1 and R goes up by 1.  Other meetings have no result.  My guess is that somebody doesn't hear, because as S falls more and more meetings deplete the I's.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T04:23:44.853" CommentCount="1" />
  <row Id="9241" PostTypeId="2" ParentId="9239" CreationDate="2010-11-07T04:40:22.983" Score="2" ViewCount="0" Body="&lt;p&gt;Not much help, but the limiting packing density is a factor 3.462746619.  The &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html&quot; rel=&quot;nofollow&quot;&gt;inverse symbolic calculator&lt;/a&gt; finds the compression (limiting volume of the initial unit volume) as 0.2887880950866024 with the correct formula.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T04:40:22.983" CommentCount="1" />
  <row Id="9242" PostTypeId="1" AcceptedAnswerId="9268" CreationDate="2010-11-07T04:52:37.307" Score="5" ViewCount="166" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about it for a while and consulting other students, no one seems to be able to find an example of the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the PDE &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\partial f}{\partial x} = 0 \quad $      on $U = { (x,y) \in \mathbb R^2 ; y&gt;0, 1 &amp;lt; x^2 + y^2 &amp;lt; 4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a solution $f$ that does not only depend on $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exercise is taken form Lee's &quot;Introduction to smooth manifolds&quot;, p. 517 at the end of the chapter on the Frobenius theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: According to the errata, the condition on $U$ is $y &gt; 0$, not $x &gt; 0$ as your copy of the book might state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-07T13:58:39.843" LastActivityDate="2010-11-07T15:25:39.687" Title="Counterintuitive PDE" Tags="&lt;differential-equations&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="9243" PostTypeId="1" AcceptedAnswerId="9247" CreationDate="2010-11-07T04:56:42.657" Score="1" ViewCount="65" Body="&lt;p&gt;$f$ is smooth (derivatives of all orders exist) except at $x_1$,$x_2$,$x_3$ where derivatives exist only upto orders of $k_1$,$k_2$,$k_3$ respectively, where $k_i$ is a natural number. Same is the case with another function $g$ which is smooth except at $t_1$,$t_2$,$t_3$ where derivatives exist only upto orders of $l_1$,$l_2$,$l_3$ respectively. Both $f$,$g$ belong to either $L^1$ or $L^2$ or have compact support. What is the behaviour of the function $h=f*g$&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-07T05:39:21.700" LastActivityDate="2010-11-07T06:00:39.117" Title="behaviour of discontinuities  after convolution" Tags="&lt;calculus&gt;&lt;convolution&gt;" AnswerCount="1" />
  <row Id="9244" PostTypeId="2" ParentId="9239" CreationDate="2010-11-07T05:03:04.837" Score="2" ViewCount="0" Body="&lt;p&gt;The best I can come up with is horribly recursive (not so bad if you store intermediary values, as mine does below), but it would work to use Mathematica or some other software to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(n)=\displaystyle\sum_{i=0}^{n-1}\bigg(\displaystyle\prod_{j=1}^{n-i}(1-\frac{1}{2^j})\bigg)(1-f(i))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f[0] = 0;&#xA;f[n_] := f[n] = Sum[Product[(1 - 1/2^j), {j, 1, n - i}] (1 - f[i]), {i, 0, n - 1}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-07T05:16:11.620" LastActivityDate="2010-11-07T05:16:11.620" />
  <row Id="9245" PostTypeId="1" AcceptedAnswerId="9257" CreationDate="2010-11-07T05:03:15.347" Score="7" ViewCount="141" Body="&lt;p&gt;Imagine that we have something like the rubik's cube on $U=I^{3}=[0,1]^{3}$, and that we say that all the points on the xy, yz, and zx faces are black, and that all the points on the other three faces are white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a legal move to be a rotation by an integral multiple of $\frac{\pi}{4}$ around one of the lines going through the center of each pair of opposite faces, for any square subset of $U$ orthogonal to the respective line. Note that there are an uncountable number of them on each axes. Points retain their color after legal moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a sequence of legal moves that will produce a circle -- some points one color and some points another color? -- on one of the faces of $U'$? (where $U'$ is meant to denote $U$ after said sequence of moves has been performed?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;addition&lt;/strong&gt;: I strongly suspect that no single countably infinite set of legal moves works, just on the basis of cardinality arguments -- a circle would have a continuum of points, and any countably infinite sequence of legal moves would only be able to move a countably infinite set of points.&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="890" LastEditorDisplayName="" LastEditDate="2010-11-07T08:30:57.717" LastActivityDate="2010-11-07T12:40:01.403" Title="continuous rubik's cube circle drawing" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="9246" PostTypeId="2" ParentId="9239" CreationDate="2010-11-07T05:49:36.717" Score="2" ViewCount="0" Body="&lt;p&gt;The amount of powder remaining at each point is $(1-1/2),(1-1/4),\ldots$, and so the constant you're looking for (in the limit) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\prod_{n=1}^{\infty} (1 - 2^{-n}) \approx 0.288788095086602$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also the limiting probability that an $n \times n$ matrix over $GF(2)$ be regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the exact result for $n$ by $R_n$. We want to estimate the error $R_n - R_\infty$. Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = \frac{R_\infty}{\prod_{m=n+1}^{\infty} (1 - 2^{-m})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logarithm of the denominator is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\Theta(\sum_{m=n+1}^\infty 2^{-m}) = -\Theta(2^{-n})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the denominator itself is $1 - \Theta(2^{-n})$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = R_\infty (1 + \Theta(2^{-n})) = R_\infty + \Theta(2^{-n})$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The product can be converted to a sum (exercise!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_\infty = \sum_{n=2}^\infty \frac{(-1)^n}{\prod_{k=2}^n (2^k-1)} = \frac{1}{3} - \frac{1}{3\cdot 7} + \frac{1}{3 \cdot 7 \cdot 15} - \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this sum, it's easy to prove that this constant is irrational (hint: same as the usual proof that $e$ is irrational using its series representation $\sum_{n=0}^\infty 1/n!$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also obtain excellent rational approximations this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1/3, 2/7, 13/45, 188/651, 5731/19845, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the series is Leibniz, you know that the error after summing $k$ terms is at most the next term. For example, $5731/19845$ is at most $1/78129765$ too large than the actual constant.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T05:49:36.717" CommentCount="2" />
  <row Id="9247" PostTypeId="2" ParentId="9243" CreationDate="2010-11-07T06:00:39.117" Score="1" ViewCount="0" Body="&lt;p&gt;Assume $f$ and $g$ has compact support, and put $h=f*g$, then there is no problem for the derivatives to be integrable. If $D^{n_1}f,D^{n_2}g\in L^1$ then we have&#xA;$$D^n h = D^{n_1}f*D^{n_2}g, \qquad \text{where } n_1+n_2=n.$$&#xA;Hence we certainly have that $h$ is differentiable of order $n=n_1+n_2$ where $n_1=\min(k_1,k_2,k_3)$ and $n_2= \min(l_1,l_2,l_3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the discontinuity  at $x_i$ the integrability of $D^if$ might be lost ($i=1,2,3$) and the same holds for $g$, so we can not do better than this.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-07T06:00:39.117" CommentCount="1" />
  <row Id="9248" PostTypeId="2" ParentId="9242" CreationDate="2010-11-07T07:00:02.550" Score="0" ViewCount="0" Body="&lt;p&gt;$f$ can be locally but not globally independent of $x$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $|y| &amp;lt; 1$ the function can have different values on the left and right sides of the annulus.  The restriction of the function to the left or the right side depends only on $y$.  For example, $f = {\rm sgn(x)}(1-y^2)$ for $|y| \leq 1$ and $0$ in the rest of the region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(restored following correction of problem.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-07T15:25:39.687" LastActivityDate="2010-11-07T15:25:39.687" />
  <row Id="9249" PostTypeId="2" ParentId="9152" CreationDate="2010-11-07T07:08:25.120" Score="5" ViewCount="0" Body="&lt;p&gt;As Derek points out, we can blow the question wide open by allowing operations not included on basic calculators. In fact, even a single nine could be quite productive if we allow (say) floor, ceiling, factorial, and the natural logarithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor\ln\ln9\rfloor=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln\ln9\rceil=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor\ln 9\rfloor=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln 9\rceil=3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln\ln\lceil\ln(9!)\rceil!\rceil=4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor\ln\ln\lceil\ln\lceil\ln\lceil\ln(9!)\rceil!\rceil!\rceil!\rfloor&#xA;= 5$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln 9\rceil!=6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lceil\ln\lceil\ln9\rceil!!\rceil= 7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vdots$$&lt;/p&gt;&#xA;" OwnerUserId="2474" LastActivityDate="2010-11-07T07:08:25.120" CommentCount="1" />
  <row Id="9250" PostTypeId="2" ParentId="9242" CreationDate="2010-11-07T08:15:10.680" Score="7" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;Mean Value Theorem&lt;/a&gt; implies that any function f differentiable on U must be constant w.r.t. x for any fixed y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an error in the text -- it should say y&gt;0 instead of x&gt;0.  (see &lt;a href=&quot;http://www.math.washington.edu/~lee/Books/Smooth/errata.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.washington.edu/~lee/Books/Smooth/errata.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-07T08:15:10.680" CommentCount="2" />
  <row Id="9251" PostTypeId="1" CreationDate="2010-11-07T09:24:17.217" Score="0" ViewCount="65" Body="&lt;p&gt;I have Ubuntu 10.10 installed and need to plot 3D graphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have installed several free applications available in the repository, like QtiPlot and GNU Octave. I have found out and created the following graphic.&#xA;&lt;img src=&quot;http://i.imgur.com/88XMt.png&quot; alt=&quot;Gaussian function&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have to show in the same graphic the position of my experiment results, which consist of elements with three parameters: X, Y and Z coordinates which had been calculated with the same function as above graphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea to do that?  Would be better if you propose solution in free apps,because there are several proprietary apps like Maple or MATLAB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="2698" LastActivityDate="2010-11-07T09:24:17.217" Title="Gaussian function" Tags="&lt;functions&gt;&lt;applications&gt;" CommentCount="2" />
  <row Id="9252" PostTypeId="1" AcceptedAnswerId="9360" CreationDate="2010-11-07T10:47:44.250" Score="6" ViewCount="170" Body="&lt;p&gt;Happened to get this from a friend.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does one show that for all pairs of real numbers $x$ and $y$, there is $n \in \mathbb{N}$ such that $$ x &amp;lt; \frac{\varphi(n+2)}{\varphi(n)} &amp;lt; y$$ where $\varphi$ is the Euler's totient function.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-08T02:57:11.390" Title="Showing $x &lt; \frac{\varphi(n+2)}{\varphi(n)} &lt; y$ for $x,y \in \mathbb{R}$." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" />
  <row Id="9253" PostTypeId="2" ParentId="154" CreationDate="2010-11-07T11:04:57.770" Score="4" ViewCount="0" Body="&lt;p&gt;Even though quite a few people have already contributed towards answering your question, I would like to offer some thoughts of mine that may help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not going to give you ways of showing that complex numbers are necessary and meaningful, rather an idea about why I think people find them &lt;em&gt;meaningless&lt;/em&gt; and how this can be resolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am prompted by the part of your question that says: &quot;...most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning.&quot; I have thought about this a lot in the past and have come to the conclusion that the problem of finding complex numbers meaningful lies in the meaning we have attached to the number systems &quot;preceding&quot; them in the hierachical chain: $\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}$. The meaning we have attached is that of quantity! Numbers all the way up to and including the reals are scalar quantities: We use them for distance, area, volume, weight, speed, intensity and so on. However, when you get to the complex numbers all that has to go away. There is no such thing as $2 + 3i$ kilogrammes or $-4i$ dollars or $1-5i$ centimetres... Yet we ask the learner to call them numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the learner is faced with an apparent contradiction in terms: They are asked to think of these strange entities as &lt;em&gt;numbers&lt;/em&gt; while at the same time these new additions to the realm don't behave as the numbers of old. And you should take into account that the older numbers have been around for a long time. That's where I think the problem lies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To circumvent this, I think that what one needs to do is explain to the learner that - from now on - numbers are going to take on a much wider role than before: They are going to be used not only to denote quantities, but also to denote &lt;em&gt;directions&lt;/em&gt;, which is precisely what complex numbers do on the two-dimensional plane. Every complex number can be represented by a vector and this automatically suggests a direction. The &quot;older&quot; meaning is not lost, since a complex number has a modulus in addition to its real and imaginary parts and these three are &lt;em&gt;quantities&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one might say &quot;But real numbers denote directions as well: +1 denotes a unit displacement to the right of 0 along the x-axis while -1 denotes a displacement to the left&quot;. This is indeed true, however, I sincerely doubt that the untrained mind of a learner, who encounters complex numbers for the firs time, will delve in that direction. And, if it does, then that's fine, because it shows that the direction concept was already lurking in the number hierarchy ever since negative integers were introduced!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope all this is of help to you.&lt;/p&gt;&#xA;" OwnerUserId="3195" LastActivityDate="2010-11-07T11:04:57.770" CommentCount="1" />
  <row Id="9254" PostTypeId="2" ParentId="3405" CreationDate="2010-11-07T11:37:32.217" Score="0" ViewCount="0" Body="&lt;p&gt;I think these observations are probably already in other answers, or even too simple to have been listed, but I wanted to add my working anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p(2n+2)=(n+1)/n(p(2n)-2p(n)/(2n+1))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus if k(n)= np(n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k(2n+2)=(n+1)^2/n^2/(2n+1)((2n+1)k(2n)-4k(n))&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2010-11-07T11:37:32.217" />
  <row Id="9255" PostTypeId="1" CreationDate="2010-11-07T12:01:43.480" Score="0" ViewCount="97" Body="&lt;p&gt;How can I render the &lt;em&gt;connected&lt;/em&gt; sum of $n$ topological spaces (or other quite strange topological operations) in a &quot;product-like&quot; manner? What I'm looking for is writing, for example&#xA;$$&#xA;{\Large{*}}_{i=1}^n G_i&#xA;$$&#xA;for the free product of n groups. How can I render this with the # operation in topological spaces? (Yes, I'm studying Hatcher's AT...)&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-11-07T12:56:02.210" LastActivityDate="2010-11-08T10:08:38.463" Title="Define strange operators" Tags="&lt;soft-question&gt;&lt;latex&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9256" PostTypeId="1" CreationDate="2010-11-07T12:13:36.553" Score="1" ViewCount="133" Body="&lt;p&gt;Prove that, where a,b,...,e are real numbers and a≠0,&#xA; if ax + by = c&#xA;has the same solution set as&#xA; ax+dy=e&#xA;then they are the same equation. What if a=0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If a≠0 then the solution set of the first equation is {(x,y)|x=c-by/a}&lt;/p&gt;&#xA;" OwnerUserId="3215" LastActivityDate="2010-11-08T20:02:35.127" Title="Prove that ax + by = c has same solution set as ax+dy=e" Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9257" PostTypeId="2" ParentId="9245" CreationDate="2010-11-07T12:21:16.377" Score="5" ViewCount="0" Body="&lt;p&gt;What a great question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution. I am aiming for a black circle inscribed on a&#xA;white face (where the boundary edge points are invisible,&#xA;as mentioned in the comments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick one of the white faces to be the desired final face F.&#xA;Now, consider any given horizontal row on that face, which&#xA;is all white. We want to add two black dots to such a row (except for top/bottom/center, which are handled already by the edge point color condition). These rows come in upper/lower symmetric dual&#xA;pairs. Call such a pair the&#xA;current working pair of rows. We make a quarter turn to&#xA;each of these working rows to the side. Now, operating on&#xA;that side face S, we make two moves parallel to F, which&#xA;will not upset F, in order to make the working rows on S&#xA;have exactly two black dots each in the correct positions.&#xA;This is possible since the squares containing the working&#xA;rows have not been moved previously, and the corresponding&#xA;columns containing the desired black dots have also not&#xA;been moved previously, since those dots will only be moved&#xA;for this pair of working rows. That is, it is precisely the&#xA;black dot opposite the dual white dot that we desire, and&#xA;by working in dual pairs, we put all four black dots into&#xA;position with two moves. (Is it clear?) Thus, on the side&#xA;face, we have the two working rows looking exactly like we&#xA;want. So we turn them each a quarter turn back to the main&#xA;face F, and proceed with another pair of working rows. The&#xA;center row and top/bottom rows do not need adjusting,&#xA;because of the invisible edge points. So this seems to do&#xA;it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are problematic issues in setting up your&#xA;problem. This kind of problem, where one wants to describe&#xA;a task involving infinitely many steps, is known as a&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supertask&quot; rel=&quot;nofollow&quot;&gt;supertask&lt;/a&gt;, and is&#xA;open to a number of interesting paradoxes and problematic&#xA;issue, some of which I describe in &lt;a href=&quot;http://mathoverflow.net/questions/22033/infinite-cpu-clock-rate-and-hotel-hilbert/22038#22038&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow&#xA;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this problem, we can imagine carrying out my solution in&#xA;a transfinite sequence of moves. The fact that each row on the desired face is&#xA;handled with finitely many moves makes my particular&#xA;solution less problematic than other situations one can&#xA;imagine. In general, for example, if a square has been&#xA;moved infinitely many times, what position is it in at the&#xA;limit? You haven't said, and it isn't clear what it should&#xA;be. But solutions in which every given square is moved only&#xA;finitely often seem to avoid this problem, particularly when the order that the finite sequences of moves are made in doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit.&lt;/em&gt; I notice now that you didn't say that the diameter of the circle should be the same as the diameter of the square. You can easily modify my solution for any size circle on any face. In this case, you do have to handle the top/bottom/middle rows of the circle, but this is no problem and is handled the same as the other rows. The top/bottom rows of the circle form a dual pair of working rows just like the others, and the center row has no dual, and is handled by itself.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-11-07T12:40:01.403" LastActivityDate="2010-11-07T12:40:01.403" />
  <row Id="9258" PostTypeId="2" ParentId="9256" CreationDate="2010-11-07T12:29:58.453" Score="0" ViewCount="0" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ax + by = c$ has the same solution set as $ax + dy = e$ then any pair $(x,y)$ solving the first equation solves the second too. Let $(x,y)$ and $(x',y')$ with $y \neq y'$ be solutions. Subtracting the two we have $(b - d)y = c - e$ and $(b - d)y' = c - e$ but the only way for $(b - d)y = (b - d)y'$ to hold is when $b = d$, because $y$ is not equal to $y'$. Clearly then, $0 = c - e$ and this proves that $c = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this proof does not go through when $a = 0$ is because there are no solutions with $y$ distinct.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T10:16:34.537" CommentCount="2" />
  <row Id="9259" PostTypeId="1" AcceptedAnswerId="9305" CreationDate="2010-11-07T12:32:21.163" Score="-1" ViewCount="207" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this problem, I can not figure it out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the sum of all the multiples of 3 or 5 below 1000.&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-07T18:02:06.653" LastActivityDate="2010-11-07T19:03:23.657" Title="Find the sum of all the multiples of 3 or 5 below 1000." Tags="&lt;number-theory&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9260" PostTypeId="2" ParentId="9259" CreationDate="2010-11-07T12:36:00.893" Score="3" ViewCount="0" Body="&lt;p&gt;The multiples of 3 are 3,6,9,12,15,18,21,24,27,30,....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multiples of 5 are 5,10,15,20,25,30,35,40,45,....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intersection of these two sequences is 15,30,45,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the first numbers 1+2+3+4+...+n is n(n+1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the first few multiples of k, say k+2k+3k+4k+...+nk must be kn(n+1)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can just put these ingredients together to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To find n use 1000/3 = 333 + remainder, 1000/5 = 200 + remainder, 1000/15 = 66 + remainder and then sum multiples of 3: $3\cdot 333(333+1)/2 = 166833$. multiples of 5: $5\cdot 200(200+1)/2 = 100500$ and subtract multiples of 15 $15\cdot 66(66+1)/2 = 33165$ to get 234168.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T12:36:00.893" />
  <row Id="9261" PostTypeId="2" ParentId="275" CreationDate="2010-11-07T12:37:26.820" Score="1" ViewCount="0" Body="&lt;p&gt;The Shape of Space by Jeffrey Weeks is really great.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-07T12:37:26.820" CommunityOwnedDate="2010-11-07T12:37:26.820" />
  <row Id="9262" PostTypeId="1" AcceptedAnswerId="9263" CreationDate="2010-11-07T12:38:26.633" Score="0" ViewCount="168" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a simple formula for the n-th derivative of some expression. I'm suppose to prove that it's true for all N, by using induction. I set n=1, and show that the base case works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step, were I assume that it works a K is fine. But how is that shown on paper? And the that I don't understand at all, is how to show algebraically that this is true. Do you just &quot;plug-in&quot;, replace all the k's with k+1 and then what? Or? Well as you see, I don't get it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for any input :)  &lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-08T19:19:36.513" Title="Induction, not what, but how it's done? " Tags="&lt;homework&gt;&lt;induction&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="9263" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T12:53:28.080" Score="6" ViewCount="0" Body="&lt;p&gt;The principle of induction says that if you have some $\mathbb N$ indexed family of propositions $P(n)$, that you can prove the theorem $\forall n \in \mathbb N, P(n)$ just by proving:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(1)$&lt;/li&gt;&#xA;&lt;li&gt;$\forall n \in \mathbb N. P(n) \Rightarrow P(n+1)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Intuitively this is quite simple, for any given number say 3051 the statement P(3051) must be true, since P(0) is true, P(1) is true, P(2) is true ... all the way up to P(3051). Mathematically this is just reflecting the nature of $\mathbb N$ so it's a tautology.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is probably not clear what a proposition is though, because nobody ever talks about them. Well if you are familiar with functions, like $f(x) = x^2-1$, given any number say 3 you find that f(3) is a number too: 8. A proposition is like a function except that it computes a mathematical statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could define the following proposition: $S(x) :\iff \exists n, n^2 = x$ and that would hold for all square numbers for example $S(9) \iff \exists n, n^2 = 9$ and so we can prove $S(9)$ by proving $\exists n, n^2 = 9$ by exhibiting the witness 3 then showing that $3^2 = 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more interesting example might be the proposition $\text{Strong}_P(n)$ which is defined like so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(1) :\iff P(1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(2) :\iff P(1) \text{ and } P(2)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(3) :\iff P(1) \text{ and } P(2) \text{ and } P(3)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\text{Strong}_P(n+1) :\iff \text{Strong}_P(n+1) \text{ and } P(n+1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is relatively easy to verify that $\text{Strong}_P(n) \iff (\forall k \leq n, P(k))$ so this really is the principle of &quot;strong&quot; induction. Furthermore you can prove the definition holds for all $n$ using normal induction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A very beautiful theorem is that the sum of cubes is the square of the sum, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1^3 + 2^3 + 3^3 + 4^3 + 5^3 = (1 + 2 + 3 + 4 + 5)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $P(n)$ be the statement &quot;sum of cubes up to n = square of sum up to n&quot;. Obviously it holds for P(1) since 1 = 1, now we would like to prove $\forall n \in \mathbb N. P(n) \Rightarrow P(n+1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all n, given that the sum of cubes up to n equals the square of the sum of to n, then &quot;the same statement for n+1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have the assumption $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ and need to prove $1^3 + 2^3 + \cdots + n^3 + (n+1)^3 = (1 + 2 + \cdots + n + n+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use algebra, the right hand side is $(1 + 2 + \cdots + n)^2 + 2(n+1)(1 + 2 + \cdots + n) + (n+1)^2$. So all we have to show now is that $(n+1)^3 = 2(n+1)(1 + 2 + \cdots + n) + (n+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying out some things $n^3 + 3 n^2 + 3 n + 1 = 2(n+1)(1 + 2 + \cdots + n) + n^2 + 2 n + 1$ cancelling some things gives: $n (n + 1)^2 = 2(n+1)(1 + 2 + \cdots + n)$ and we quickly notice it being equivalent to the formula $(1 + 2 + \cdots + n) = n (n + 1)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shall we try to prove this by induction also! Let $T(n)$ be the statement that $(1 + 2 + \cdots + n) = n (n + 1)/2$. Clearly T(1) holds, since 1 = 1. Now we should like to prove that for any given $n$ the hypothesis $T(n)$, that is $(1 + 2 + \cdots + n) = n (n + 1)/2$ implies that $(1 + 2 + \cdots + n + n+1) = (n + 1) (n + 2)/2$. Algebraically we can just substitute the hypothesis in to get $n (n + 1)/2 + n+1 = (n + 1) (n + 2)/2$ doubling this $n (n + 1) + 2 n + 2 = (n + 1) (n + 2)$ and this is obviously true by multiplying out both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this was a stupid example though beacuse you can just check the first 4 cases and not do any induction).&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T19:19:36.513" CommentCount="9" />
  <row Id="9264" PostTypeId="1" AcceptedAnswerId="9272" CreationDate="2010-11-07T12:53:47.580" Score="-1" ViewCount="106" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have this series of numbers &lt;code&gt;2, 5, 10, 17&lt;/code&gt;. Now somebody told me that next number is 26. He used this function for that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(n) = n*n + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain how does that function solve the problem to come up with 26?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T13:45:13.070" LastActivityDate="2010-11-07T13:59:23.520" Title="How to solve number series with f(n) = n*n + 1" Tags="&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="9265" PostTypeId="1" CreationDate="2010-11-07T13:01:56.213" Score="0" ViewCount="100" Body="&lt;p&gt;I understand that we use the chain rule to differentiate the $y$ part of it. Like if it stood $xy$, I would use the product rule, that say that the derivative of $y$ is $dy/dx$. But what if it's just $y$ alone or with a constant in front off it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T07:12:42.610" LastActivityDate="2010-11-14T07:57:46.033" Title="Question about implicit differentiation" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9268" PostTypeId="2" ParentId="9242" CreationDate="2010-11-07T13:38:37.183" Score="4" ViewCount="0" Body="&lt;p&gt;I will use the corrected version mentioned by Douglas, i.e.  $U$ will be the domain defined by $y&gt;0$ and $1&amp;lt; x^2 +y^2&amp;lt; 4$. Consider a $C^\infty$ function $\phi (y)$ which is equal to $1$ for negative $y$, $0$ for $y\geq 1/2$ and strictly decreasing for $0 &amp;lt; y &amp;lt; 1/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required function $f$ is then defined by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) =-\phi (y)$ if $(x,y)\in U$ and $x\leq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,y) =+\phi (y)$ if $(x,y)\in U$ and $x\geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not only depend on $y$ since $f(-3/2,1/4)&amp;lt;0$ and $f(+3/2,1/4)&gt;0$. Nevertheless we do have $\dfrac {\partial f}{\partial x}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS&lt;/strong&gt; The Mean Value Theorem doesn't apply because the segment joining the two points $(-3/2,1/4)$ and $(+3/2,1/4)$ (for example) is &lt;em&gt;not&lt;/em&gt; entirely included in $U$.&lt;/p&gt;&#xA;" OwnerUserId="3217" LastEditorUserId="3217" LastEditorDisplayName="" LastEditDate="2010-11-07T14:32:03.167" LastActivityDate="2010-11-07T14:32:03.167" CommentCount="5" />
  <row Id="9269" PostTypeId="1" AcceptedAnswerId="9288" CreationDate="2010-11-07T13:47:11.603" Score="0" ViewCount="68" Body="&lt;p&gt;If I have a quadratic form like $y^2 - x^2 - x = k$ none of the techniques I know work because of the nasty $x$. Note that homogenizing doesn't work because a solution of $Y^2 - X^2 - X Z = k Z^{(2)}$ does not lead to a solution of the original equation in integers, at least as far as I have been able to determine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one understand the solution set of these equation? How can I solve them?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T18:42:35.650" Title="How to solve inhomogeneous quadratic forms in integers?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;&lt;quadratic-forms&gt;" AnswerCount="2" />
  <row Id="9270" PostTypeId="2" ParentId="275" CreationDate="2010-11-07T13:49:39.430" Score="1" ViewCount="0" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://mathoverflow.net/questions/8609/favorite-popular-math-book&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/8609/favorite-popular-math-book&lt;/a&gt; for other opinions&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2010-11-07T13:49:39.430" CommunityOwnedDate="2010-11-07T13:49:39.430" />
  <row Id="9271" PostTypeId="2" ParentId="9269" CreationDate="2010-11-07T13:52:38.537" Score="4" ViewCount="0" Body="&lt;p&gt;These all reduce to equations&#xA;$$f(x,y)=k$$&#xA;where $x$ and $y$ are integers, possibly satisfying some&#xA;congruence conditions, and $f$ is a homogeneous&#xA;quadratic form. In two variables this basically&#xA;comes down to solving norm equations in the rings&#xA;of integers in quadratic fields.&#xA;For your example you get&#xA;$$4y^2-4x^2-4x-1=4k-1$$&#xA;that is&#xA;$$y'^2-x'^2=4k-1$$&#xA;where $x'=2x-1$ is odd and $y'=2y$ is even. This is atypically&#xA;easy, since it factorizes as $(y'-x')(y'+x')=4k-1$. More typically&#xA;you'd get a Pell-type equation.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-07T18:42:35.650" LastActivityDate="2010-11-07T18:42:35.650" CommentCount="3" />
  <row Id="9272" PostTypeId="2" ParentId="9264" CreationDate="2010-11-07T13:59:23.520" Score="4" ViewCount="0" Body="&lt;p&gt;In general given any length $n$ sequence of numbers there exists a polynomial $p(x)$ of degree $n$ which gives this sequence. It's easy to find such a polynomial so I will not discuss how to do this. Note that as a consequence of this, if we are given a length $n$ sequence we can find a polynomial for the length $n+1$ sequence which gives any value we want after giving the values in the initial sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore, there are many different types of function other than polynomials - many of which can fit any given sequence. For this reason it is impossible to really be &quot;right&quot; when answering a question of the form &quot;What's the next number ...?&quot;. Any prediction is just as (in)valid without putting some constraints on it.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T13:59:23.520" CommentCount="2" />
  <row Id="9273" PostTypeId="2" ParentId="275" CreationDate="2010-11-07T14:35:21.353" Score="2" ViewCount="0" Body="&lt;p&gt;If you are willing to invest a bit of time, a lot of Conway's books are very good. I especially recommend&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathematical_Plays&quot; rel=&quot;nofollow&quot;&gt;Winning ways for your mathematical plays&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://plus.maths.org/content/symmetries-things&quot; rel=&quot;nofollow&quot;&gt;Symmetries of things&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A bit less &quot;story&quot; but still quite a lot of fun is Robert Lang's &lt;a href=&quot;http://www.langorigami.com/pubs/books/books.php4&quot; rel=&quot;nofollow&quot;&gt;Origami design secrets&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also recommend James R. Newman's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486432688&quot; rel=&quot;nofollow&quot;&gt;The world of mathematics&lt;/a&gt;. But be warned, while the content is not very technical (indeed, many articles in the collection are from public lectures of famous scientists), it can get a little bit dry at times. If you are patient enough for it, it is a very good companion to Courant and Robbins' &lt;em&gt;What is mathematics&lt;/em&gt;. (It is sort of like the Princeton Companion, but older and slightly more down-to-earth). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, you can also try the various books and articles by &lt;a href=&quot;http://bit-player.org/&quot; rel=&quot;nofollow&quot;&gt;Brian Hayes&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-07T14:35:21.353" CommunityOwnedDate="2010-11-07T14:35:21.353" />
  <row Id="9274" PostTypeId="1" CreationDate="2010-11-07T15:19:57.743" Score="0" ViewCount="439" Body="&lt;p&gt;Hi, I need to find an isomorphism between [0,1] to (0,1) can you help me with this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks. benny&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T21:22:48.343" LastActivityDate="2010-11-28T09:19:44.817" Title="Isomorphism between [0,1] to (0,1)" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="7" CommentCount="2" />
  <row Id="9275" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T15:27:16.357" Score="1" ViewCount="0" Body="&lt;p&gt;Hello Benny,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you mean by &quot;isomorphism&quot;? Are you considering these sets with a given group or ring structure (if then, which one?), or as topological spaces (in which case you'd be looking for a &quot;homeomorphism&quot;), or just as sets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of a canonical group structure on either set, but if you want your map between the two to be continuous you might have some luck by looking at the notion of compactness and how it behaves under continuous functions.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-07T15:27:16.357" CommentCount="3" />
  <row Id="9276" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T15:36:04.413" Score="0" ViewCount="0" Body="&lt;p&gt;The bijection 1/x turns (0,1) into (1,$\infty$) which you can chop up into countably many &lt;code&gt;(]&lt;/code&gt;'s. Also you can split &lt;code&gt;[]&lt;/code&gt; into &lt;code&gt;[)(]&lt;/code&gt; and these can be divide countably many times to bisect with the &lt;code&gt;(]&lt;/code&gt;'s but it flips the ordering around.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-07T15:36:04.413" />
  <row Id="9277" PostTypeId="1" AcceptedAnswerId="9290" CreationDate="2010-11-07T15:38:31.720" Score="0" ViewCount="66" Body="&lt;p&gt;I thought of writing this question &lt;a href=&quot;http://math.stackexchange.com/questions/9171/minimum-for-this-function&quot;&gt;Minimum for this function&lt;/a&gt; in a different way, if it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to minimize,&#xA; $\sum_{i=1}^{n}{a_ix_i} +\nu \sum_{i=1}^{n}{b_i 2^{x_i}}$&lt;br&gt;&#xA;where,&lt;br&gt;&#xA;$a_i \epsilon [0,1],\quad b_i \epsilon (0,\infty),\quad x_i \epsilon [xmin,0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;br&gt;&#xA;$\sum_{i=1}^{n}(2^{x_i}) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;xmin, $\nu$, $a_i$  and  $b_i$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the tricky part is to minimize the function while ensuring all $2^{x_i}$ sum upto 1 for these n variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your patience and help.&lt;/p&gt;&#xA;" OwnerUserId="2858" LastActivityDate="2010-11-07T16:44:53.333" Title="Minimum for this function" Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="9278" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T15:42:18.020" Score="0" ViewCount="0" Body="&lt;p&gt;muad's answer is clear, but one could emphasize a couple of points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical induction is essentially a &lt;strong&gt;logical chain reaction&lt;/strong&gt;. &#xA;It helps to visualize the goal ``$\forall n\ P(n)$'' as an infinite number of separate assertions $P(1), P(2), \ldots$  The aim of the induction step is to establish &lt;em&gt;logical links&lt;/em&gt; between successive assertions:&#xA;$$&#xA;P(1)\Rightarrow P(2)\Rightarrow \ldots \Rightarrow P(17)\Rightarrow P(18)\Rightarrow\ldots&#xA;$$ &#xA;That is, $\forall n\in\mathbb N\ P(n)\Rightarrow P(n+1)$. These links are &lt;em&gt;purely hypothetical inferences&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The handout &lt;a href=&quot;http://www.math.cmu.edu/~bobpego/21131/lcr.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; provides an expanded explanation with examples.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-07T15:42:18.020" CommentCount="2" />
  <row Id="9279" PostTypeId="1" CreationDate="2010-11-07T15:43:46.393" Score="2" ViewCount="62" Body="&lt;p&gt;I have $X_1,...$ Bernoulli variables with probability $p$ of success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get an $n$ such that with probability $\delta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\sum_{i=1}^n X_i \ge k) \ge \delta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $n$ would of course depend on $k$, $p$ and $\delta$. Can I get an upper bound for that $n(k,\delta,p)$? (meaning, I want some $n(k,\delta,p)$ for which the above holds, the smaller $n$ is, the better.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3227" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-07T16:04:08.640" LastActivityDate="2010-11-07T17:06:55.270" Title="can I get an upper bound on the tail of a binomial variable?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" />
  <row Id="9280" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T15:55:21.937" Score="5" ViewCount="0" Body="&lt;p&gt;You can identify a countably infinite collection of points starting with 0 and converging to some limit less than 1/2.  Just shift each point one down the line.  Now repeat from the top end.  All points not involved go to themselves.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T15:55:21.937" CommentCount="2" />
  <row Id="9281" PostTypeId="1" CreationDate="2010-11-07T15:59:15.477" Score="2" ViewCount="142" Body="&lt;p&gt;Let $k$ be an algebraic closed field. Let $x$ be a point in $X=P_k^1$. What is $O_{X,x}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have $x=(t-a)\in \text{Spec }k[t]$. Looking $x$ inside $P_k^1$, does $O_{X,x}=k[t]_{(t-a)}$? I'm confused when I have to deal with the sheaf of rings.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-01-06T22:44:39.320" Title="Stalks on Projective Scheme" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;&lt;projective-schemes&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9282" PostTypeId="2" ParentId="9279" CreationDate="2010-11-07T16:08:31.830" Score="3" ViewCount="0" Body="&lt;p&gt;Consult a discussion of Chernoff bounds like &lt;a href=&quot;http://en.wikipedia.org/wiki/Chernoff_bound&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chernoff_bound&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-07T16:08:31.830" />
  <row Id="9283" PostTypeId="2" ParentId="9277" CreationDate="2010-11-07T16:18:55.593" Score="0" ViewCount="0" Body="&lt;p&gt;I don't think rewriting this way helps. In the first place, how would you show that if $y$ is the optimal solution of $\min_y f(y)$, then $x = 2^y$ is the optimal solution to $\min_x f(\log x)$?&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-07T16:18:55.593" />
  <row Id="9284" PostTypeId="2" ParentId="9281" CreationDate="2010-11-07T16:20:13.097" Score="2" ViewCount="0" Body="&lt;p&gt;The stalk of the generic point is, of course, the rational function field $k(t)$. Now $\mathbb{P}^1$ is covered by two copies of $\mathbb{A}^1$. If a closed point $x$ corresponds to the maximal ideal generated by $(t-a)$ for some $a \in k$, then the stalk is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_{\mathbb{A}^1,x} = k[t]_{(t-a)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is isomorphic to $k[t]_{(t)}$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-07T16:20:13.097" CommentCount="2" />
  <row Id="9285" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T16:27:06.080" Score="7" ViewCount="0" Body="&lt;p&gt;The maps $(0,1) \to [0,1], x \mapsto x$ and $[0,1] \to (0,1), x \mapsto 1/4 + x/2$ are injective. The theorem of Cantor-Schröder-Bernstein implies that there is a bijection. It is actually possible to work out the proof of the theorem in these simple examples and get an explicit bijection, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g : [0,1] \to (0,1), x \mapsto 1/2 \pm 1/2^{n+2}$ if $x = 1/2 \pm 1/2^{n+1}$ for some $n$, and $x \mapsto x$ else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://matheplanet.com/matheplanet/nuke/html/article.php?sid=873&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; if you know german. ;)&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-07T16:27:06.080" />
  <row Id="9286" PostTypeId="1" AcceptedAnswerId="9292" CreationDate="2010-11-07T16:33:19.933" Score="2" ViewCount="470" Body="&lt;p&gt;How to prove&#xA; $$\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-07T17:23:58.123" LastActivityDate="2010-11-22T01:21:02.573" Title="Proving $\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="9287" PostTypeId="2" ParentId="9274" CreationDate="2010-11-07T16:37:05.817" Score="1" ViewCount="0" Body="&lt;p&gt;If you just want a bijection there are many ways to do it. Here's a rather contrived one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\alpha$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_ordinal&quot; rel=&quot;nofollow&quot;&gt;limit ordinal&lt;/a&gt; of the same cardinality as $(0,1)$. By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor-Bernstein-Schröder theorem&lt;/a&gt; there exists a bijective map $f: (0,1) \to \alpha$. Let $S$ denote the &lt;a href=&quot;http://en.wikipedia.org/wiki/Successor_ordinal&quot; rel=&quot;nofollow&quot;&gt;successor &quot;function&quot;&lt;/a&gt; (I know it's not a function - hence the quotes). Note that $\alpha$ does not have a maximal element, so it makes sense to define $\phi: [0,1] \to (0,1)$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(x) = \begin{cases}&#xA;(f^{-1} \circ S \circ S \circ f)(x) &amp;amp; \text{ for $x \in (0,1)$}, \\&#xA;f^{-1}(1) &amp;amp; \text{ for $x = 1$}, \\&#xA;f^{-1}(0) &amp;amp; \text{ for $x = 0$}.&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that $\phi$ is a bijection is an easy exercise. It's also completely useless and nonconstructive. &lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-07T16:37:05.817" CommentCount="1" />
  <row Id="9288" PostTypeId="2" ParentId="9269" CreationDate="2010-11-07T16:43:57.677" Score="3" ViewCount="0" Body="&lt;p&gt;Over 200 years ago Lagrange solved the general binary quadratic Diophatine equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm a\ x^2 + b\ xy + c\ y^2 + d\ x + e\ y + f = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reduces to a Pell equation: put $\rm\ D = b^2-4ac,\ E = bd-2ae,\ F = d^2-4af\:.\ $ Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm  D\ Y^2\ =\ (D\ y + E)^2 + D\ F - E^2,\quad\quad Y\ =\ 2ax + by + d $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if we put $\rm\quad\ \ X = D\ y + E,\quad\ \ N = E^2 - D\ F\quad\ \ $ we have the Pell equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm X^2 - D\ Y^2\ =\ N $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dario Alpern has a web page &lt;a href=&quot;http://www.alpertron.com.ar/QUAD.HTM&quot; rel=&quot;nofollow&quot;&gt;Quadratic two integer variable equation solver&lt;/a&gt; that will solve any such equation - with complete descriptions of the methods involved. For some  recent optimizations of Lagrange's algorithm see this paper H. C. Williams et al. &lt;a href=&quot;http://www.springerlink.com/content/58614q7346xg2650/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;A new look at an old equation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-07T16:43:57.677" />
  <row Id="9289" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T16:44:09.573" Score="0" ViewCount="0" Body="&lt;p&gt;So you get it conceptually (assume $P$ true for an arbitrary $n$, show that that implies  $P$ true for $n+1$) (and yes that seems weird but if you parenthesize the formula with quantifiers, it works out that you're not really assuming the consequent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in practice...that's what you want to know how to do it. Examples are best. Let's try a simple one that you've probably seen before but specify exactly what is happening. For&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n-1} 2k+1 = n^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the inductive case, assume the above is true, then show that the LHS with $n+1$ substituted for $n$ equals the RHS with the same substitution. That is, show &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n} 2k+1 = (n+1)^2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strategy is to do algebraic manipulation so that you see where you can apply the induction hypothesis. You can do the manipulation on either side, but you have to follow the rules (here specifically when proving an equational identity, i.e. 'LHS = RHS', you have to make sure you keep them separate, that is show a chain of manipulation leading from one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start from either side: the RHS obviously expands to $n^2 + 2n + 1$. Now use your pattern matching skills to find something that looks like something in the induction hypothesis. Here it would be the $n^2$ part. This looks like the RHS of the induction hypothesis; you can replace it with the LHS (of the induction hypothesis), to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sum_{k=0}^{n-1} 2k+1)  + 2n + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(extra parens put in to show what got replaced) And the extra part outside of the summation, the $2n+1$, is the same as extending the top of the range of the summation to $n$ (here you have to be familiar with the summation notation). So this last part can be converted immediately into &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n} 2k+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But  that is the end of the equational chain, going from  $(n+1)^2$ to $\sum_{k=0}^{n} 2k+1$, maintaining equality the  the whole way, most of the time just doing the algebraic manipulation, but necessarily using the induction hypothesis somewhere along the line, that is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(n+1)^2 = ... = ... = \sum_{k=0}^{n} 2k+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suspect a lot of the difficulty here is just keeping separate what you know already and what you have to prove (what you don't actually know yet -in- the proof). And ...yeah...just keeping track of all the symbols.&lt;/p&gt;&#xA;" OwnerUserId="1919" LastActivityDate="2010-11-07T16:44:09.573" />
  <row Id="9290" PostTypeId="2" ParentId="9277" CreationDate="2010-11-07T16:44:53.333" Score="2" ViewCount="0" Body="&lt;p&gt;First let $x = (x_1, \dots, x_n)$ and write $f(x) = \sum_i^n a_i x_i + \nu\sum_i^n b_i 2^{x_i}$ and $g(x) = \sum_i^n 2^{x_i} - 1$. Now the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers&quot; rel=&quot;nofollow&quot;&gt;Lagrange's multipliers&lt;/a&gt; tells you that you have a solution $x'$ iff it is an extremal point of the function $f(x) + \lambda g(x)$ with $\lambda$ a parameter to be determined. So you get equations $a_i + (\nu b_i + \lambda) 2^{x'_i} \log{2} = 0$ for all $1 \leq i \leq n$. It's now easy to express $x'_i$ as a function of lambda and plugging this into the remaining constraint $g(x') = 0$ to find the lambda and complete the solution. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-07T16:44:53.333" CommentCount="4" />
  <row Id="9292" PostTypeId="2" ParentId="9286" CreationDate="2010-11-07T16:47:10.920" Score="15" ViewCount="0" Body="&lt;p&gt;This is an old favorite of mine.&lt;br&gt;&#xA;Define $$I=\int_{-\infty}^{+\infty} e^{-x^2} dx$$ &#xA;Then $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$  &#xA;$$I^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$  &#xA;Now change to polar coordinates&lt;br&gt;&#xA;$$I^2=\int_{0}^{+2 \pi}\int_{0}^{+\infty}e^{-r^2} rdrd\theta$$  &#xA;The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$  &#xA;$$I^2=2\pi\int_{0}^{+\infty}e^{-u}du/2=\pi$$ &#xA;So $$I=\sqrt{\pi}$$ and your integral is half this by symmetry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T23:43:16.607" LastActivityDate="2010-11-20T23:43:16.607" CommentCount="12" />
  <row Id="9293" PostTypeId="1" AcceptedAnswerId="9294" CreationDate="2010-11-07T16:54:51.377" Score="0" ViewCount="36" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R\subset A\times A$ a binary relation, and note $a\sim b$ for $(a,b)\in R$. We define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a,b)\in R\circ R\Leftrightarrow  \exists b\in A : a\sim b,  b\sim c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $R^n=R\circ\dots\circ R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My books says that $(a,b)\in R^n$ iff there is a path of length $n$ in the graph of $R$. Isn't it false ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $(a,b)\in R^3$, there exists $c\in A$ such that $(a,c)\in R^2$, $(c,b)\in R^2$ $\Leftrightarrow$ there exists $d, f\in A$ such that &#xA;$$a\sim d\sim c\sim f\sim b$$&#xA;which is a path of length 4, not 3.&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2010-11-07T16:56:45.730" Title="Binary relation powers and path length" Tags="&lt;graph-theory&gt;&lt;discrete-mathematics&gt;&lt;relations&gt;" AnswerCount="1" />
  <row Id="9294" PostTypeId="2" ParentId="9293" CreationDate="2010-11-07T16:56:45.730" Score="1" ViewCount="0" Body="&lt;p&gt;Your book is right. Note that $R^3 = R \circ R \circ R = R^2 \circ R$, not $R^2 \circ R^2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T16:56:45.730" CommentCount="1" />
  <row Id="9295" PostTypeId="2" ParentId="9279" CreationDate="2010-11-07T17:06:55.270" Score="1" ViewCount="0" Body="&lt;p&gt;The following observation, which is straightforward to derive, may be very useful here (also with regard to Naga's answer):&#xA;$$&#xA;{\rm P}\bigg(\sum\limits_{i = 1}^n {(1 - X_i ) \ge n - k} \bigg) \le 1 - \delta &#xA;$$&#xA;implies &#xA;$$&#xA;{\rm P}\bigg(\sum\limits_{i = 1}^n {X_i  \ge k} \bigg) \ge \delta. &#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-07T17:06:55.270" />
  <row Id="9297" PostTypeId="1" CreationDate="2010-11-07T17:45:30.773" Score="0" ViewCount="53" Body="&lt;p&gt;I need to turn the equation L=12/x + x into a quadratic forumla to solve. i have been given the answers of 0.619m 0r 19.38m if L = 20m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what iv go so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0= 12/x + x -20&#xA;but i dont know how to turn 12/x to a x^2 value&lt;/p&gt;&#xA;" OwnerUserId="3230" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-07T18:00:12.563" LastActivityDate="2010-11-07T18:00:12.563" Title="Quadratic equation rearranging " Tags="&lt;algebra-precalculus&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="9298" PostTypeId="2" ParentId="9256" CreationDate="2010-11-07T17:46:50.500" Score="2" ViewCount="0" Body="&lt;p&gt;Follow the supplied hint: evaluate $\rm\ x = (c-b\: y)/a = (e-d\:y)/a\ $ at $\rm\ y = 0\ $ and $\rm\ y = -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, see also this closely related &lt;a href=&quot;http://math.stackexchange.com/questions/9058/work-out-the-values-of-a-and-b-from-the-identity-x-ax-144-x-b/9068#9068&quot;&gt;recent question.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T20:02:35.127" LastActivityDate="2010-11-08T20:02:35.127" />
  <row Id="9299" PostTypeId="2" ParentId="9297" CreationDate="2010-11-07T17:49:51.547" Score="2" ViewCount="0" Body="&lt;p&gt;Hi Robin. Multiply the equation by $x$ and you'll get the polynomial you want.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-11-07T17:49:51.547" />
  <row Id="9300" PostTypeId="2" ParentId="9286" CreationDate="2010-11-07T17:52:16.620" Score="4" ViewCount="0" Body="&lt;p&gt;A variation on Ross Millikan's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can start again with the observation&#xA;$$\left(\int_{-\infty}^{\infty}e^{-x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-y^2}dy\right)=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)=V.$$&#xA;Now $V$ is simply the volume of the body &#xA;$$-\infty &amp;lt; x,y &amp;lt; \infty,\qquad 0 &amp;lt; z &amp;lt; e^{-(x^2+y^2)},$$&#xA;or, equivalently,&#xA;$$0 &amp;lt; x^2+y^2 &amp;lt; -\ln z,\qquad  0 &amp;lt; z &amp;lt; 1.$$&#xA;This implies that the body is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_of_revolution&quot; rel=&quot;nofollow&quot;&gt;solid of revolution&lt;/a&gt;. Using the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Disk_integration&quot; rel=&quot;nofollow&quot;&gt;disk integration&lt;/a&gt; formula, we have&#xA;$$V=\int_{0}^{1}\pi(-\ln z)dz=[\pi(z-z\ln z)]_{0}^1=\pi.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-07T20:10:12.667" LastActivityDate="2010-11-07T20:10:12.667" CommentCount="4" />
  <row Id="9301" PostTypeId="2" ParentId="9262" CreationDate="2010-11-07T18:07:45.717" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a very simple illustrative example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad D^n\ e^{c\:x}\ =\ c^n\ e^{c\:x }\quad$ for all $\rm\:n\in \mathbb N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Base step:&lt;/strong&gt; $\ $ It is true for $\rm\: n = 0\ $ since $\rm\ D^0 = I, \ \ c^0 = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction step:&lt;/strong&gt; $\ $ If it is true for $\rm\: n\:$ then differentiating both sides yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad \quad D^{n+1}\ e^{c\:x}\ =\ D^n\: (c^n\ e^{c\:x })\ =\ c^{n+1}\ e^{c\:x}  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence it is true for $\rm\: n+1\:$. Therefore it is true for all $\rm\:n\in \mathbb N\:$ by induction.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-07T18:07:45.717" />
  <row Id="9302" PostTypeId="1" AcceptedAnswerId="9303" CreationDate="2010-11-07T18:28:45.753" Score="2" ViewCount="163" Body="&lt;p&gt;Say I have a symmetric matrix. I have the concept of norm as defined on wikipedia. Now I want to prove (disprove?) that the norm of a symmetric matrix is maximum absolute value of its eigenvalue. I would really appreciate if this can be done only using simple concepts of linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite new to mathematics. &lt;/p&gt;&#xA;" OwnerUserId="1674" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2011-01-03T12:42:05.763" LastActivityDate="2011-01-03T13:33:23.740" Title="Norm of a symmetric matrix?" Tags="&lt;linear-algebra&gt;&lt;norm&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9303" PostTypeId="2" ParentId="9302" CreationDate="2010-11-07T18:34:42.670" Score="5" ViewCount="0" Body="&lt;p&gt;Given a symmetric matrix, you have a complete set of eigenvalues and eigenvectors.  So any vector can be represented as a linear combination of the eigenvectors.  Multiply your matrix by an arbitrary unit vector decomposed into the eigenvectors.  Then note that the maximum length of the resultant vector is achieved when the input vector is along the eigenvector associated with the largest eigenvalue.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-07T18:34:42.670" CommentCount="5" />
  <row Id="9304" PostTypeId="2" ParentId="9286" CreationDate="2010-11-07T18:55:11.317" Score="2" ViewCount="0" Body="&lt;p&gt;An alternative derivation is to show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=I,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is your integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\int_{0}^{\infty}e^{-x^2}\; dx,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then evaluate $I^2$ by reversing the order of integration. If $x&gt;0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=x\int_{0}^{\infty}e^{-{(xy)}^2}\; dy=x\int_{0}^{\infty}e^{-u^2}\dfrac{du}{x}=\int_{0}^{\infty}e^{-u^2}\; du=I.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I^2=\int_{0}^{\infty}e^{-x^2}\; dx\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=\displaystyle\int_{0}^{\infty}dy\int_{0}^{\infty}xe^{-x^2}e^{-x^{2}y^{2}}\; dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int_{0}^{\infty}dy\int_{0}^{\infty}xe^{-x^{2}(1+y^2)}\; dx=\int_{0}^{\infty}dy\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int_{0}^{\infty }\dfrac{1}{2\left( 1+y^{2}\right) }\; dy=\dfrac{1}{2}\left[ \arctan y\right] _{y=0}^{\infty }=\dfrac{\pi}{4}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\dfrac{\sqrt{\pi}}{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-07T21:54:53.343" LastActivityDate="2010-11-07T21:54:53.343" CommunityOwnedDate="2010-11-07T18:55:11.317" CommentCount="0" />
  <row Id="9305" PostTypeId="2" ParentId="9259" CreationDate="2010-11-07T18:58:11.493" Score="4" ViewCount="0" Body="&lt;p&gt;The previously posted answer isn't correct. The statement of the problem is to sum the multiples of 3 and 5 below 1000, not up to and equal 1000. The correct answer is&#xA;\begin{eqnarray}&#xA;\sum_{k_{1} = 1}^{333} 3k_{1} + \sum_{k_{2} = 1}^{199} 5 k_{2} - \sum_{k_{3} =1}^{66} 15 k_{3} = 166833 + 99500 - 33165 = 233168,&#xA;\end{eqnarray}&#xA;where we have the used the identity&#xA;\begin{eqnarray}&#xA;\sum_{k = 1}^{n} k = \tfrac{1}{2} n(n+1).&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-07T19:03:23.657" LastActivityDate="2010-11-07T19:03:23.657" />
  <row Id="9306" PostTypeId="1" AcceptedAnswerId="9307" CreationDate="2010-11-07T19:09:15.907" Score="2" ViewCount="126" Body="&lt;p&gt;According to the solution manual:&#xA;$\int \frac{x}{\sqrt{1-x^{4}}}dx = \frac{1}{2}\arcsin x^{2}+C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution doesn't seem to be working. I know another way of solving it (setting $u=x^{2}$) but the fact that this way of solving it doesn't work bothers me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{set }u=1-x^{4}\text{ so } dx=\frac{du}{-4x^{3}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;\int \frac{x}{\sqrt{1-x^{4}}}dx &amp;amp;= \int \frac{x}{\sqrt{u}}dx  \\&#xA;&amp;amp;= \int \frac{xdu}{-4x^{3}\sqrt{u}} \\&#xA;&amp;amp;= -\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} \\&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{set } v=\sqrt{u} \text{ so }du=2\sqrt{u}\,dv&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;-\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} &amp;amp;= -\frac{1}{2} \int \frac{dv}{x^{2}}  \\&#xA;&amp;amp;= -\frac{1}{2} \int \frac{dv}{\sqrt{1-v^{2}}} \\&#xA;&amp;amp;= -\frac{1}{2} \arcsin (v) + C \\&#xA;&amp;amp;= -\frac{1}{2} \arcsin (\sqrt {1-x^{4}}) + C \\&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be happy to clarify any steps I took. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T19:13:50.270" LastActivityDate="2010-11-07T19:28:15.543" Title="Can't Solve an Integral" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="9307" PostTypeId="2" ParentId="9306" CreationDate="2010-11-07T19:22:38.563" Score="6" ViewCount="0" Body="&lt;p&gt;Your solution is &lt;em&gt;an&lt;/em&gt; antiderivative of the original function. You can always check whether your solution is correct by taking its derivative. This also implies that the book solution and your solution differ by a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this specific problem, imagine the right triangle with sides $x^2$ and $\sqrt{1-x^4}$ and hypotenuse $1$. Then $\arcsin\sqrt{1-x^4} = \frac{\pi}{2} - \arcsin x^2$, and it should be easy to see from there how both solutions are related.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-07T19:28:15.543" LastActivityDate="2010-11-07T19:28:15.543" />
  <row Id="9311" PostTypeId="1" AcceptedAnswerId="9315" CreationDate="2010-11-07T19:42:17.930" Score="10" ViewCount="262" Body="&lt;p&gt;This should be an easy exercise: Given a finite odd abelian group $G$, prove that $\prod_{g\in G}g=e$. Indeed, using Lagrange's theorem this is trivial: There is no element of order 2 (since the order must divide the order of $G$, but it is odd), and so every element except $e$ has a unique inverse which is different from it. Hence both the element and its inverse participate in the product and cancel each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is simple - I need to solve this &lt;strong&gt;without&lt;/strong&gt; Lagrange's theorem. So either there's a smart way to prove the nonexistance of an element of order 2 in an odd abelian group, or I'm missing something even more basic...&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-08T01:38:39.527" Title="Product of all elements in an odd finite abelian group is 1" Tags="&lt;group-theory&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="9312" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T19:55:03.817" Score="0" ViewCount="0" Body="&lt;p&gt;Let $G$ be a finite abelian group of odd order. Then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\prod_{g \in G}g \prod_{g \in G} g = &#xA;\prod_{g \in G} g \prod_{g \in G} g^{-1} = \prod_{g \in G} gg^{-1} = e.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $x:= \prod_{g \in G}\;g$ is its own inverse. Now you just need to show that the only element that's its own inverse in $G$ is $e$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-07T19:55:03.817" CommentCount="3" />
  <row Id="9313" PostTypeId="1" AcceptedAnswerId="9317" CreationDate="2010-11-07T20:09:40.097" Score="2" ViewCount="101" Body="&lt;p&gt;Say I have a binomial coefficient $y=\binom{5n+3}{n+2}$ or $y=\binom{n^2+4}{3n}$ something of the sorts in terms of the variable $n$. How can I determine $f$ so that $y = O(f)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general method for this sort of thing.  I have only ever seen $O$-notation in a computer science setting and its treatment was not concerned with deriving anything except the bounds for &quot;simple&quot; functions like logarithms and polynomials.  As a side question, what would be a good resource to learn further methods to find $O$ of more complex functions?  I understand the formal definition, but Im unsure how to use it practically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies if this is a stupid question. Its not homework.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2010-11-07T23:10:32.693" LastActivityDate="2010-11-07T23:10:32.693" Title="How can I determine asymptotic growth of binomial coefficients?" Tags="&lt;binomial-coefficients&gt;&lt;asymptotics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9314" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:12:39.857" Score="1" ViewCount="0" Body="&lt;p&gt;Here is one argument (although surely there are simpler ones): since $G$ is abelian, the elements of order dividing 2 form a subgroup $H$ of $G$.  On the other hand, an abelian group every element of which is of order dividing $2$ can be thought of as a vector space over the field of 2 elements, and so (since $H$ is finite, and hence has finite dimension) we see that either $H$ is trivial (if its dimension over the field of 2 elements is zero), or else that the number of elements in  $H$ is even (if its dimension is positive).  One element of $H$ is the identity $e$, and so either $H$ is trivial, i.e. $G$ contains no elements of exact order $2$, or else the number of elements in $H \setminus \{e\}$, i.e. the number of elements of exact order 2 in $G$, is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is easy to see that when $G$ is odd, the number of elements of exact order 2 is even.  (Count the elements of $G$ by thinking about the orbits of the map $g \mapsto g^{-1}$.)  Thus when $G$ is odd, the group $H$ must indeed be trivial, and so $G$ contains no elements of order 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, we have avoided an appeal to Lagrange's theorem by instead appealing to a somewhat coarser counting argument, together with linear algebra over the field of 2 elements.  Whether or not this is absurd, the readers can decide!&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-07T20:12:39.857" />
  <row Id="9315" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:14:15.273" Score="8" ViewCount="0" Body="&lt;p&gt;If don't exist and element with order 2 than you are done. Supose $g\in G$ such that $g^2=e$. Since ${g_1,\ldots, g_n}={gg_1,\ldots,gg_n}$ (the brackets should be here), than $\prod g_i = g^n \prod g_i$ and $g^n=e$. Putting $n=2k+1$, $e=g^{2k+1}=g^{2k}g=g$.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="1971" LastEditorDisplayName="" LastEditDate="2010-11-07T20:19:31.137" LastActivityDate="2010-11-07T20:19:31.137" CommentCount="5" />
  <row Id="9316" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:17:24.693" Score="2" ViewCount="0" Body="&lt;p&gt;It's kind of easy to prove Lagrange's theorem in this case. For any $h \in G$,  $h^{|G|}\prod_{g \in G} g$ $= \prod_{g \in G} hg$. Since $g \rightarrow hg$ permutes the elements of the group, $= \prod_{g \in G} hg = \prod_{g \in G} g$ and&#xA;thus $h^{|G|} = e$. It's a short step from here to say that the order of $h$ divides $|G|$. In any event if $|h|$ were 2, one could write $1 = |G| - 2k$ for some $k$ and get that $h^1 = h^{|G|} (h^2)^{-k} = e$. Surely a Lagrange-free proof wouldn't be much simpler...&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-07T20:17:24.693" />
  <row Id="9317" PostTypeId="2" ParentId="9313" CreationDate="2010-11-07T20:18:06.520" Score="5" ViewCount="0" Body="&lt;p&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt; in $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-07T20:18:06.520" />
  <row Id="9318" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T20:19:41.143" Score="1" ViewCount="0" Body="&lt;p&gt;By the classification of finite abelian groups, any abelian group $G$ can be written as $Z_{n_1} \oplus \cdots \oplus Z_{n_k}$ where the $n_i$ are prime powers.  For odd abelian groups, none of the $n_i$ are powers of two. Write each element of the group $G$ as a $k$-tuple $(m_1,\ldots, m_k)$ where $0 \le m_i &amp;lt; n_i$.  Let $N = n_1 n_2 \cdots n_k$ be the order of the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the $j$th component of $\sum_{g \in G} g$ is &#xA;$$ {N \over n_j} \left( \sum_{m_j=0}^{n_j-1} m_j \right) $$&#xA;and the sum is of course $n_j(n_j-1)/2$. This simplifies to $N(n_j-1)/2$. Since $n_j$ is odd this is a multiple of $N$ (as an integer); therefore it's a multiple of $n_j$ (as an integer) and thus $0$ (as an element of $Z_{n_j}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't have an algebra text at hand; does the proof of the classification use Lagrange's theorem?)&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-07T20:19:41.143" CommentCount="3" />
  <row Id="9319" PostTypeId="1" AcceptedAnswerId="9333" CreationDate="2010-11-07T20:31:36.627" Score="4" ViewCount="238" Body="&lt;p&gt;We were given a challenge by our calculus professor and I've been stuck on it for a while now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the set of subsequence limits of $A_n=\sin(n)$ is $[-1, 1]$ (another way to phrase this would be: $\forall r\in [-1,1]$ there exists a subsequence of $A_n=\sin(n)$ that converges to $r$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a good way to start?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T21:33:21.763" LastActivityDate="2010-11-08T20:28:43.747" Title="sin(n) subsequence limits set" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="19" FavoriteCount="1" />
  <row Id="9320" PostTypeId="2" ParentId="9313" CreationDate="2010-11-07T20:33:47.087" Score="2" ViewCount="0" Body="&lt;p&gt;The answer to your second question is to read Graham, Knuth, and Patashnik's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;, at least for simpler practical examples.  For complicated sequences it can be extremely difficult and in many cases open to find asymptotics, e.g. for Ramsey-theoretic sequences, so you'll have to narrow the scope of your question a little to get a reasonable answer.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-07T20:33:47.087" CommentCount="1" />
  <row Id="9321" PostTypeId="1" CreationDate="2010-11-07T20:50:55.397" Score="1" ViewCount="111" Body="&lt;p&gt;Hi. I'm reading page number 4 here. In particular the section where it deals with the case $p=\infty$, that is , showing that $L^{\infty}$ is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.core.org.cn/NR/rdonlyres/Mathematics/18-125Fall2003/5E3917E2-C212-463B-9EDB-671486133388/0/18125_lec15.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.core.org.cn/NR/rdonlyres/Mathematics/18-125Fall2003/5E3917E2-C212-463B-9EDB-671486133388/0/18125_lec15.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why is the convergence uniform? where it says &quot;for $x \in N^{c}$ , $f_{n}$ is a Cauchy sequence of complex numbers. Thus $f_{n} \rightarrow f$ uniformly. Clearly we have pointwise convergence but why is it uniform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I don't see why $||f_{n} - f||_{\infty} \rightarrow 0$. Can you please explain this step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3176" LastActivityDate="2010-11-07T21:22:03.697" Title="Understanding proof of completeness of $L^{\infty}$" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9322" PostTypeId="2" ParentId="9321" CreationDate="2010-11-07T21:22:03.697" Score="3" ViewCount="0" Body="&lt;p&gt;The key is that we are working in $N^c$, where we have, morally speaking, defined $N$ to be the set of points where things go wrong. In detail, $N$ is the set of points $x$ where either 1) $f(x)$ is bigger than the limsup $\|f\|$, or 2) the series $(f_n)$ is not Cauchy at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside of $N$, the terms $f_n$ are bounded and the series $(f_n)$ is Cauchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question 1), use the fact that $(f_n)$ is (uniformely!) Cauchy outside of $N$ and that $\mathbb C$ is complete to define a limit function $f$ on the set $N^c$. It should follow, pretty much by definition of $f$, that $f_n \to f$ uniformely on $N^c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question 2), then we know from question 1 that $\|f_n - f\|\infty \to 0$ on $N^c$. Next, extend the limit function $f$ to the whole set by setting it equal to 0 on $N$. What is the measure of $N$? How does that play into the definition of the $\| \cdot \|_\infty$ norm, and thus the question about limits?&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-07T21:22:03.697" />
  <row Id="9323" PostTypeId="1" AcceptedAnswerId="9326" CreationDate="2010-11-07T21:27:43.913" Score="1" ViewCount="108" Body="&lt;p&gt;Suppose that $S_1$ and $S_2$ are the vanishing sets of a system of polynomial equations in n variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[X_1,\dots,X_n]$) and a system of polynomial equations in m variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[Y_1,\dots,Y_m]$). We can give $S_1$ and $S_2$ the Zariski topology by letting all algebraic subsets of $S_1$ and $S_2$ to be closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, if $Y_j\circ f\colon S_1\to\mathbb{k}$ is a polynomial in $\mathbb{k}[X_1,\dots,X_n]/I(S_1)$, for every $j$, then $f$ is continuous in the Zariski topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition that $Y_j\circ f$ be polynomial, however, seems too strict: as long as the vanishing set of $Y_j\circ f$ were equal to the vanishing set of a polynomial function, $f$ would be continuous, which means that the usual morphisms from $S_1$ to $S_2$ are not the same as the continuous maps between them as (Zariski-)topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question then is of two parts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Is it true that there exist Zarsiki-continuous maps that are not polynomial? &lt;/li&gt;&#xA;  &lt;li&gt;If yes, then what is the geometric way of thinking about the extra structure that polynomial maps preserve, i.e. continuous maps squash subsets of $S_1$ in such a way that non-algebraic subsets never get squashed into algebraic ones, but some algebraic ones may get squashed into non-algebraic ones, but polynomials also give what features to the geometric squashing?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-11-07T21:40:46.947" Title="Are there Zariski-continuous maps between algebraic sets that are not polynomial maps?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9324" PostTypeId="1" AcceptedAnswerId="9330" CreationDate="2010-11-07T21:28:10.080" Score="1" ViewCount="66" Body="&lt;p&gt;I've been trying to google the answer to this question, but have had no luck, so I'm asking this question here. Let's say the origin is at (0, 0), body 1 with mass m1 is at (-r1, 0), and body 2 with mass m2 is at (r2, 0). How can I calculate the initial velocities v1 and v2 for the bodies so that they trace circular orbits around each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I'm asking is because I've only been able to get the answer for the specific problem where m1 = m2, but I can't explain it very well, and because I can't explain it very well I can't generalize it to bodies of different masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any assistance!&lt;/p&gt;&#xA;" OwnerUserId="3240" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-07T21:30:27.973" LastActivityDate="2010-11-07T21:59:40.740" Title="two-body problem circular orbits" Tags="&lt;physics&gt;&lt;classical-mechanics&gt;" AnswerCount="2" />
  <row Id="9325" PostTypeId="1" CreationDate="2010-11-07T21:30:57.333" Score="2" ViewCount="53" Body="&lt;p&gt;I was wondering if equilibrium distribution, steady-state distribution, stationary distribution and limiting distribution mean the same, or there are differences between?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learn them in the context of Discrete-time Markov Chain as far as I know. Or they also appear in other situations of stochastic processes and probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Markov_chain&quot; rel=&quot;nofollow&quot;&gt;Markov Chain&lt;/a&gt;, there seems not very clear to me in defining and using these concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-11-07T21:30:57.333" Title="equilibrium distribution,   steady-state distribution,  stationary distribution and limiting distribution" Tags="&lt;stochastic-processes&gt;" />
  <row Id="9326" PostTypeId="2" ParentId="9323" CreationDate="2010-11-07T21:36:41.907" Score="4" ViewCount="0" Body="&lt;p&gt;If I understand your question correctly, then (1) yes, there are going to be lots of Zariski continuous maps that are not polynomial (although, if one asks that all higher products of the map preserve the Zariski topology on the corresponding sources and targets, this is a much stronger restriction, I think --- does anyone know the details?); and (2) the usual extra structure that one adds is the sheaf of regular functions on source and target; polynomial maps pullback regular functions on the target to regular functions on the source.  Thus varieties are naturally thought of as being objects&#xA;in the category of locally ringed spaces (i.e. spaces equipped with a certain structure sheaf of rings) rather than just in the category of topological spaces.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-07T21:36:41.907" />
  <row Id="9327" PostTypeId="2" ParentId="9323" CreationDate="2010-11-07T21:37:33.480" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the affine line and Look for an example there yourself.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-07T21:37:33.480" />
  <row Id="9328" PostTypeId="2" ParentId="9323" CreationDate="2010-11-07T21:40:46.947" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, there exists Zariski continous maps between algebraic varieties that are not polynomials. For example all  bijection between $\mathbb A ^1$ and itself is Zariski continous but not polynomial in general: think of bijection that permutes $0$ and $1$ and leaves other points fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometric way of thinking is algebraic subsets must be preserved . Only use polynomials and rationnal functions.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-07T21:40:46.947" CommentCount="5" />
  <row Id="9329" PostTypeId="2" ParentId="9324" CreationDate="2010-11-07T21:53:04.220" Score="0" ViewCount="0" Body="&lt;p&gt;Hello Bob,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silly question, but have you read the Wikipedia page on the two-body problem (http://en.wikipedia.org/wiki/Two-body_problem)? If I understand your problem correctly, then it seems that it is fairly well treated on that page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you want to determine the set of initial vectors $v_1$ and $v_2$ such that the two bodies trace circular orbits around each other. If $x_1(t)$ and $x_2(t)$ denote the positions of the bodies at time $t$, then do you mean that you want the distance $|x_1(t) - x_2(t)|$ between the two to be equal to some fixed radius $r$? If so, then this seems to be treated in the &quot;Displacement vector motion&quot; section of the Wikipedia page (if the radius $r$ is constant, then its derivative w/r/t time is ...?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WP page seems to solve two one-body problems to arrive at a complete description of the system of solutions. In contrast, you only ask for a partial description of the system (i.e. what inital vectors give circular orbits). An alternative solution in your case could be to redefine the origin of your coordinate system to be given by $x_1(t)$ at time $t$; then your job is just to determine which initial vectors give circular orbits of one mass in orbit around another fixed one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second method actually sounds like it's more instructive and fun to do than working through the WP page. I'd think of Isaac Newton and try the second method to start with, then turn to the WP page (or here) if I got stuck. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-07T21:53:04.220" CommentCount="1" />
  <row Id="9330" PostTypeId="2" ParentId="9324" CreationDate="2010-11-07T21:59:40.740" Score="2" ViewCount="0" Body="&lt;p&gt;I will assume the phrase &quot;circular orbits around each other&quot; means that you are looking for a solution where the two bodies remain constant distance apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest thing to do is to work in a coordinate system adapted to the centre of mass. The total separation between the two masses is $R = r_1 + r_2$. The centre of mass sits at $(R_0,0)$ where $R_0 = (-m_1 r_1 + m2 r_2)/(m_1 + m_2)$. Without loss of generality, we'll assume that the centre of mass remains immobile (else we can subtract the velocity of the centre of mass from both bodies and use the fact that Galilean motions leave the mechanics invariant). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the distance from $m_1$ to the centre of mass is $|-r_1 - R_0| = R m_2 / (m_1 + m_2)$, similarly the distance from $m_2$ to the centre of mass will be $R m_1/(m_1 + m_2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us write down the equations of motion. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The assumption that the centre of mass remains immobile implies that the total momentum is 0. This means that $m_1 v_1 + m_2 v_2 = 0$. Furthermore, that $R$ is constant requires that the initial velocities be such that the two masses start out moving perpendicular to the $x$ axis. So hereon we can assume that $v_1, v_2$ are scalars, which represent the initial speed in the $y$ direction for the masses respectively. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Relative to the (immobile) centre of mass, the first mass is travelling in a circular orbit of radius $R m_2 / (m_1 + m_2)$. This requires force $ m_1 v_1^2 / (R m_2 / (m_1 + m_2) ) = v_1^2 \cdot \frac{m_1 (m_1 + m_2)}{R m_2}$. We have also a similar expression for the force acting on $m_2$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The gravitational attraction between the two masses is $G m_1 m_2 / R^2$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So now we solve the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_1 v_1 + m_2 v_2 = 0 $$&#xA;$$ \frac{G m_1 m_2}{R^2} = v_1^2 \frac{m_1 (m_1 + m_2)}{m_2 R} = v_2^2 \frac{m_2(m_1 + m_2)}{m_1R} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_1 = \sqrt{ \frac{G m_2^2}{R(m_1 + m_2)} }, v_2 = \sqrt{ \frac{G m_1^2}{R(m_1 + m_2)} }$$&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-07T21:59:40.740" CommentCount="1" />
  <row Id="9332" PostTypeId="2" ParentId="9281" CreationDate="2010-11-07T22:02:36.203" Score="2" ViewCount="0" Body="&lt;p&gt;For topological space $X$, open subset $U \subset X$ and point $x\in U$ we have for all sheaf $F$ on $X$ : $F_x = (F|U)_x$. Apply to $X=\mathbb P ^1, U=\mathbb A^1, x=(t-a)$&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-07T22:02:36.203" CommentCount="3" />
  <row Id="9333" PostTypeId="2" ParentId="9319" CreationDate="2010-11-07T22:04:56.157" Score="0" ViewCount="0" Body="&lt;p&gt;Following Qiaochu hint, I'll try to elaborate a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This is a complete rewrite of the proof to fix a flaw pointed out by Qiaochu and make it overall clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First some notation. Let $S^1 = [0, 2\pi)$ with 0 and $2\pi$ identified. For two points $a, b \in S^1$ we'll denote by $a \oplus b = a + b \mod 2\pi$ (and likewise for $\ominus$) and let $A_n = \{n \mod 2\pi\}_{n=1}^{\infty}$. Also denote by $U_{\varepsilon}(p)$ a punctured $\varepsilon$-neighborhood of $p \in S^1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our strategy will be to show that $\{A_n\}$ has at least one limit point in $S^1$. From this we will conclude that 0 is also a limit point and finally we will use this fact to show that every point in $S_1$ is a limit point of $A_n$ (this means that $\{A_n\}$ is dense in S^1). Having established this, we will use continuity of $\sin$ to finally resolve the problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a preparation we will note the relation $A_{n+m} = A_n \oplus A_m \quad (1)$. As a corollary of this we have that the sequence $\{A_n\}$ is injective (for if not, there would exist $n, m \in \mathbb{N}, k \in \mathbb{Z} \quad n&gt;m$  such that $A_n = A_m$ and so $n - m = 2 k \pi$, a contradiction with irrationality of $\pi$). This implies simple but crucial fact that the image of the sequence contains infinitely many points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to establish the density of ${A_n}$ in $S^1$ we will first show that there exists at least one limit point $p \in S^1$. This is established by a standard argument: consider intervals $I = [0, \pi]$ and $J = [\pi, 2\pi]$ and take the one which contains infinitely many points of $\{A_n\}$ (if both do, take the &quot;bottom&quot; one, i.e. $I$). Call this interval $I_0$. Now again divide it in two intervals of same length and let $I_1$ be the one with infinitely many points. Continue in this way to obtain an infinite sequance of intervals $I_0 \supset I_1 \supset \cdots$. Then the set  $K = \cap_{n=0}^{\infty} I_n$ is non-empty (this should be covered in standard calculus course, I hope) and any point $p \in K$ is surely a limit point (by construction of $\{I_n\}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the above we now know that for each $\varepsilon &gt; 0$ there exist infinitely many points in $U_\varepsilon(p)$, i.e. infinitely many $n, m \quad n &gt; m$ such that $|A_n - A_m - 2k\pi| &amp;lt; 2\varepsilon$ for some $k \in \mathbb{Z}$. But from (1) we get that $|A_{n-m} - 2k\pi| &amp;lt; 2\varepsilon$. Because we identify 0 and $2\pi$ in $S^1$ we can see that these differences are concentrated around 0. Therefore we have shown that 0 is also a limit point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we will show that every $p \in S^1$ is a limit point. Suppose we are given $\varepsilon &gt; 0$. We just take any $A_k$ from $U_{\varepsilon}(0)$ and note that for $l = \lfloor {p \over A_k} \rfloor$ we have $A_{lk} \in U_{\varepsilon}(p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude the answer we will show that any $r \in [-1, 1]$ is indeed a limit point of $\{\sin(n)\}$. Take $s = \arcsin (r) \mod 2\pi$. Then by the above, we know that $s$ is a limit point of $\{A_n\}$ so there is a subsequence $\{A_{n_k}\} \to s$ and observe by continuity of $\sin$ that $\{\sin(A_{n_k})\} = \{\sin(n_k)\} \to sin(s) = r$.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-08T18:55:36.713" LastActivityDate="2010-11-08T18:55:36.713" CommentCount="15" />
  <row Id="9335" PostTypeId="1" CreationDate="2010-11-07T22:28:11.550" Score="1" ViewCount="133" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove $\limsup(\{A_n \cup B_n\}) = \limsup(\{A_n\}) \cup \limsup(\{B_n\})$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3242" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-08T20:16:58.143" LastActivityDate="2010-11-08T20:16:58.143" Title="Proof: Limit superior intersection" Tags="&lt;homework&gt;&lt;limit&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="9336" PostTypeId="2" ParentId="9335" CreationDate="2010-11-07T22:33:20.690" Score="3" ViewCount="0" Body="&lt;p&gt;Use the definition, and double inclusion; that is, show that every element of $\limsup(A\cup B)$ must be either an element of $\limsup(A)$ or of $\limsup(B)$; then show that every element of $\limsup(A)$ must be in $\limsup(A\cup B)$ and that every element of $\limsup(B)$ must be in $\limsup(A\cup B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one must assume that you mean your &quot;$A$&quot; to be a sequence of sets and your &quot;$B$&quot; to likewise be a sequence of sets... Otherwise, what you write does not really make much sense  (limit superior and limit inferior of a single set is not usually defined).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-07T22:42:52.880" LastActivityDate="2010-11-07T22:42:52.880" CommentCount="4" />
  <row Id="9337" PostTypeId="2" ParentId="9335" CreationDate="2010-11-07T22:46:54.280" Score="0" ViewCount="0" Body="&lt;p&gt;Another nice way is to use characteristic functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map $\chi : \mathcal{P}(\Omega) \to \{0,1\}^\Omega$ assigns to every subset of $\Omega$ its characteristic function.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\chi$ is bijective.&lt;/li&gt;&#xA;&lt;li&gt;$\chi$ is continuous, i.e. $\chi_{\lim\sup_{n\to\infty} A_n} = \lim\sup_{n\to\infty}\, \chi_{A_n}$ (pointwise limit)&lt;/li&gt;&#xA;&lt;li&gt;$\chi$ is a homomorphism, i.e. $\chi_{A \cup B} = \chi_A + \chi_B - \chi_A \chi_B$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now your question reduces to the computation of an ordinary limit.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2010-11-07T22:46:54.280" />
  <row Id="9338" PostTypeId="1" CreationDate="2010-11-07T22:49:02.610" Score="1" ViewCount="78" Body="&lt;p&gt;Say I have the following numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10&#xA;10&#xA;9&#xA;5&#xA;4&#xA;4&#xA;4&#xA;1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How do i go about calculating the percentile for each score?  Is there a standard formula for figuring this out?&lt;/p&gt;&#xA;" OwnerUserId="2490" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T05:56:11.100" LastActivityDate="2010-12-08T05:56:11.100" Title="How do you calculate percentile?" Tags="&lt;statistics&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9339" PostTypeId="2" ParentId="9338" CreationDate="2010-11-07T22:54:12.370" Score="2" ViewCount="0" Body="&lt;p&gt;The $x$th percentile is the point where $x$ percent of the data lies below. Here you have 8 data points; $1$ is in the bottom $12.5\%$, so it is the 12.5-percentile. $4$ is the (top of the) bottom $50\%$ so it corresponds to the 50 percentile. $5$ is at the top of the bottom $62.5\%$, so it corresponds to the 62.5 percentile. Etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-07T22:54:12.370" />
  <row Id="9340" PostTypeId="1" CreationDate="2010-11-07T23:09:10.670" Score="0" ViewCount="115" Body="&lt;p&gt;Hello, I am trying to reproduce the following bound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{1\leq m\leq x, m\neq n}\frac{1}{|\log(m/n)|}=O(x\log(x))$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $x\geq 2$ and some $n$, $1\leq n\leq x$ (the implicit constant shouldn't depend on $n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to bound each term by $\frac{1}{|\log(1\pm 1/x)|}$, but this doesn't seem to be good enough. Splitting the sum at $\log(x)$ also didn't help. Any ideas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your comments or hints.&lt;/p&gt;&#xA;" OwnerUserId="3244" LastEditorUserId="3244" LastEditorDisplayName="" LastEditDate="2010-11-08T00:39:43.287" LastActivityDate="2011-01-07T01:44:39.067" Title="Bounding the series $\sum_{m\leq x,m\neq n}\frac{1}{|\log(m/n)|}$" Tags="&lt;sequences-and-series&gt;&lt;analytic-number-theory&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9341" PostTypeId="1" CreationDate="2010-11-07T23:19:28.893" Score="-4" ViewCount="235" Body="&lt;p&gt;I hear about the 10,0000 hour rule all the time. I also hear about people studying the night before and acing an exam. Is that just a romantic image? Do most people really need to study ~10,000 hours to become an expert at some field? &lt;/p&gt;&#xA;" OwnerUserId="3246" LastActivityDate="2010-11-08T00:09:49.003" ClosedDate="2010-11-19T13:12:09.857" Title="Do most people study?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9342" PostTypeId="2" ParentId="9341" CreationDate="2010-11-07T23:47:33.460" Score="4" ViewCount="0" Body="&lt;p&gt;In my experience as a high school teacher, here's what I've noticed: students who cram the night before may, in fact, pass (or even ace) an exam, but they are unable to retain any of the knowledge or skills the exam is meant to have assessed.  They may achieve passing grades, but they don't end up becoming experts in the field.  Their lack of expertise becomes especially apparent when they are asked questions they haven't seen before but that someone with mastery over the topic could answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a very simplistic example, I often teach algebra students how to graph the solution to an inequality on the real number line, and the students will practice this skill.  Then on the test, I'll give them a graph that represents the solution to an unknown inequality and ask them to produce that inequality.  If they've mastered the material, this is an easy exercise, but if they've simply crammed the night before without any real understanding of the connection between the inequality and the graph, this becomes more difficult.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-11-07T23:47:33.460" CommentCount="4" />
  <row Id="9343" PostTypeId="2" ParentId="9340" CreationDate="2010-11-07T23:49:16.007" Score="1" ViewCount="0" Body="&lt;p&gt;This answer corresponds to the original question, where $n$ was assumed fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the corresponding integral $\int_a^x {\frac{{du}}{{\log (u/n)}}}  = n\int_{a/n}^{x/n} {\frac{{du}}{{\log (u)}}} $ (where $a$ is some constant) and the fact that the logaritmic integral function li(x) is $O(x/\log(x))$ as $x \to \infty$, your bound should be $O(x/\log(x))$ (and not $O(x\log(x))$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-08T00:54:02.823" LastActivityDate="2010-11-08T00:54:02.823" CommentCount="1" />
  <row Id="9344" PostTypeId="2" ParentId="9302" CreationDate="2010-11-07T23:50:35.250" Score="3" ViewCount="0" Body="&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices&quot; rel=&quot;nofollow&quot;&gt;spectral theorem&lt;/a&gt; to obtain an orthogonal basis of eigenvectors for the matrix is probably the best approach, similar to what Ross said, given your criterion of &quot;simple concepts of linear algebra&quot;.  Here is another approach I thought worth mentioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest of the absolute values of the eigenvalues of a matrix $A$ is called the spectral radius of $A$, which I'll denote $r(A)$.  In general one allows complex eigenvalues (in part because there are real matrices with no real eigenvalues, like &#xA;$\begin{bmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{bmatrix}$), but fortunately it turns out that real symmetric matrices can have only real eigenvalues, so it makes no difference here.  The inequality $r(A)\leq \|A\|$ is easy, because if $\lambda$ is an eigenvalue for $A$ and $v$ is a corresponding (nonzero) eigenvector, then $\|Av\|=\|\lambda v\| = |\lambda|\|v\|$, which shows that $\|A\|\geq |\lambda|$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remarkably, there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_radius#Theorem_.28Gelfand.27s_formula.2C_1941.29&quot; rel=&quot;nofollow&quot;&gt;formula&lt;/a&gt; for the spectral radius in terms of the norm, $r(A)=\lim_{k\to\infty}\|A^k\|^{1/k}$.  I'll note before going on that everything I've said so far is independent of the norm on the space on which matrices act (that is, independent of which &lt;a href=&quot;http://en.wikipedia.org/wiki/Operator_norm&quot; rel=&quot;nofollow&quot;&gt;operator norm&lt;/a&gt; you choose for the matrices).  But now, assuming the Euclidean norm (which is implicit in your question), the matrix norm has the property that $\|A^2\|=\|A\|^2$ for a real symmetric matrix $A$.  Inductively you get $\|A^{2^k}\|=\|A\|^{2^k}$, and applying the spectral radius formula yields $r(A)=\lim_{k\to\infty}\|A^{2^k}\|^{1/2^k}=\lim_{k\to\infty}\|A\|=\|A\|$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-07T23:59:01.480" LastActivityDate="2010-11-07T23:59:01.480" />
  <row Id="9345" PostTypeId="2" ParentId="9311" CreationDate="2010-11-07T23:50:41.807" Score="4" ViewCount="0" Body="&lt;p&gt;If finite abelian group $\rm G$ has an elt $\rm\: j\: $ of order $\:2\:$ then $\rm\:\ \:g\ \to\ j\: g\ \ $ pairs its elts so $\rm G$ has even order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the often useful fact that the cardinalities of a finite set and its fixed-point set under an involution have equal parity, since the non-fixed points are paired by the involution. Hence, as above, when there are no fixed points ($\rm\: j\ne 1\: \Rightarrow\: j\:g\ne g\: $) the set has even cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such simple symmetries often lie at the heart of elegant proofs, e.g. the famous &lt;a href=&quot;http://mathoverflow.net/questions/31113/zagiers-one-sentence-proof-of-fermats-theorem&quot; rel=&quot;nofollow&quot;&gt;Heath-Brown-Zagier proof&lt;/a&gt; that every prime $\rm\:\equiv 1\ (mod\ 4)\ $ is a sum of two squares.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T01:38:39.527" LastActivityDate="2010-11-08T01:38:39.527" />
  <row Id="9346" PostTypeId="2" ParentId="9341" CreationDate="2010-11-07T23:50:43.367" Score="2" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was at the mathematical school, where the master taught his pupils after a method scarce imaginable to us in Europe. The proposition and demonstration were fairly written on a thin wafer, with ink composed of a cephalick tincture. This the student was to swallow upon a fasting stomach, and for three days following eat nothing but bread and water. As the wafer digested, the tincture mounted to his brain, bearing the proposition along with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://books.google.co.uk/books?id=3KTxJd9j3zIC&amp;amp;printsec=frontcover&amp;amp;dq=gulliver%27s+travels+jonathan+swift&amp;amp;hl=en&amp;amp;ei=DDvXTI-jE4yM4gba0Zn4Bw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCwQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this source&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-07T23:50:43.367" CommunityOwnedDate="2010-11-08T00:03:04.257" CommentCount="4" />
  <row Id="9348" PostTypeId="2" ParentId="9274" CreationDate="2010-11-08T00:36:13.643" Score="0" ViewCount="0" Body="&lt;p&gt;The easiest way (if you can apply Cantor-Bernstein-Schroder theorem) is to find a bijection from $[0,1]$ onto a subset, say $[1/4,3/4]$; that is easy to do. Since the identity map from $(0,1)$ to $[0,1]$ is a bijection, it follows from the theorem that the two sets are isomorphic.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-08T00:36:13.643" CommentCount="1" />
  <row Id="9349" PostTypeId="1" CreationDate="2010-11-08T00:40:58.947" Score="1" ViewCount="80" Body="&lt;p&gt;Hi. Can someone simply explain to me how to calculate linear integral linke below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{L} 5y \mathrm{d}L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where L is line segment from (0;0) to (0,2;0,2).&lt;/p&gt;&#xA;" OwnerUserId="3250" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-08T05:38:28.693" LastActivityDate="2010-11-08T05:38:28.693" Title="Liner integral ?" Tags="&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9350" PostTypeId="1" CreationDate="2010-11-08T00:45:52.497" Score="1" ViewCount="75" Body="&lt;p&gt;Is there some increasing function $f(n)$ that grows slower than $n^{c}$ for some $c &gt; 1$ such that $\sum_{n=1}^{\infty} \frac{1}{f(n)}$ converges? &lt;/p&gt;&#xA;" OwnerUserId="3252" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T00:50:21.790" LastActivityDate="2010-11-08T00:52:31.807" Title="Smallest function whose inverse converges" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9351" PostTypeId="1" CreationDate="2010-11-08T00:47:00.327" Score="2" ViewCount="97" Body="&lt;p&gt;This is a bit of a silly question, but I've been puzzled on how this can be done, so I ask here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given an implicit Cartesian equation like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4 x^4-4 x^3+8 y^2 x^2-27 x^2+12 y^2 x+4 y^4-27 y^2+27=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^6+3 y^2 x^4-3 x^4+3 y^4 x^2+21 y^2 x^2+3 x^2+y^6-3 y^4+3 y^2-1=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and are told that you are not allowed to graph the equation (which makes the determination of geometric properties a bit difficult), but are allowed to do manipulations to the equation. &lt;strong&gt;Is it possible to determine the symmetry of the graph of a curve only from its equation?&lt;/strong&gt; Or to be more precise, can one determine $n$ in the sentence &quot;the curve with implicit Cartesian equation so-and-so has n-fold symmetry&quot; just from the implicit equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some curves, one quick way of determining symmetry is to transform to polar coordinates: for instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x^2+y^2)^2=x^2-y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;transforms into the polar equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r^2=\cos\;2\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and due to the periodicity of the cosine, it is seen that it has two-fold symmetry (i.e., invariant under rotations of $\frac{2\pi}{2}=\pi$ radians.) However, we all know that not all algebraic curves can undergo such a neat conversion, and that this method presumes that the origin is &quot;in the vicinity&quot; of the curve. (On the other hand, if you determine through the usual methods that the curve has the origin as a singular point, conversion to polar coordinates is especially attractive!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I would like to see if a more general method is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tangentially, we know that only the circle has infinitely many axes of symmetry, so if the method can also demonstrate that only the circle has this property, that would be a plus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not the least (and since I feel it does not deserve to be asked as a separate question): is there a bounded algebraic curve that has no axes of symmetry whatsoever? Most of the examples I see have at least two-fold symmetry.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-08T01:18:13.343" Title="Determining the symmetry of a plane algebraic curve from its Cartesian equation." Tags="&lt;algebraic-geometry&gt;&lt;plane-curves&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9352" PostTypeId="2" ParentId="9350" CreationDate="2010-11-08T00:49:38.407" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, $f(n)=n(\log(n))^2$.  (Of course you can ignore the first term where the reciprocal is undefined, and the finitely many terms where it may be bigger than $n^c$ for $c$ close to $1$.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-08T00:49:38.407" />
  <row Id="9353" PostTypeId="2" ParentId="9350" CreationDate="2010-11-08T00:52:31.807" Score="4" ViewCount="0" Body="&lt;p&gt;The answer to your title question, however, is no.  That is, given an increasing function $f(n) : \mathbb{N} \to \mathbb{R}$ such that $\sum \frac{1}{f(n)}$ converges, there exists an increasing function $g(n)$ such that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sum \frac{1}{g(n)}$ converges, and&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{n \to \infty} \frac{g(n)}{f(n)} = 0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(Unfortunately the proof escapes me at the moment.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-08T00:52:31.807" CommentCount="4" />
  <row Id="9354" PostTypeId="1" AcceptedAnswerId="9371" CreationDate="2010-11-08T01:04:26.440" Score="4" ViewCount="66" Body="&lt;p&gt;In several &quot;proof-theory-for-dummies&quot;-type texts I have read, $I\Delta_0+Exp$ or a theory equivalent to it shows up as a &quot;base&quot; theory, though in what sense it is minimal is not clearly addressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise this is a vague question but if I had the knowledge to make it more specific, I wouldn't need to ask it in the first place.&lt;/p&gt;&#xA;" OwnerUserId="3254" LastActivityDate="2010-11-08T03:39:55.447" Title="What is special about elementary recursive arithmetic?" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="9355" PostTypeId="1" AcceptedAnswerId="10079" CreationDate="2010-11-08T01:06:52.317" Score="8" ViewCount="260" Body="&lt;p&gt;Consider a random walk on an undirected, non-bipartite graph. Let $\pi$ be the stationary distribution of this process, and let the hitting time $H(s,t)$ be the expected time until a walk beginning at node $s$ reaches node $t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned from &lt;a href=&quot;http://www.cs.yale.edu/publications/techreports/tr1029.pdf&quot; rel=&quot;nofollow&quot;&gt;Random walks on graphs: a survey,&lt;/a&gt; by L. Lovasz, that the quantity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{t} \pi(t) H(s,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is independent of $s$. In the Lovasz survey, this falls out as a byproduct of a lengthy calculation. I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is there a simple (calculation-free, combinatorial) proof of this statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) In what generality does this statement hold? Random walks on undirected graphs are a very particular kind of Markov process. What conditions on the probability transition matrix of the process do you need for the above to hold?&lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2010-11-13T21:59:40.457" Title="About a weighted sum of  hitting times for random walks on graphs" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9356" PostTypeId="2" ParentId="9351" CreationDate="2010-11-08T01:12:46.820" Score="2" ViewCount="0" Body="&lt;p&gt;I'll assume we're talking about Euclidean symmetries.  Then we're looking at transformations of the form $x \mapsto ax + by + c, y = -bx + ay + d$ where $a^2 + b^2 = 1$ (or the analogous set of equations for orientation-reversing symmetries).  The condition $k f(x, y) = f(ax + by + c, -bx + ay + d)$ for some nonzero $k$ (which can be determined in terms of $a, b, c, d$ by evaluating both sides at a point where $f(x, y) \neq 0$), when equating coefficients, into a system of polynomial equations in $a, b, c, d, k$.  Put this together with the equation $a^2 + b^2 = 1$ and you have a system you can plug into a CAS which will, using Grobner basis methods, tell you what the solutions look like.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am assuming that $f$ has a certain nondegeneracy property here, namely that $f$ generates the ideal of polynomial functions vanishing on $f(x, y) = 0$ in $\mathbb{R}[x, y]$.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-08T01:18:13.343" LastActivityDate="2010-11-08T01:18:13.343" CommentCount="1" />
  <row Id="9357" PostTypeId="1" AcceptedAnswerId="9358" CreationDate="2010-11-08T01:54:17.537" Score="4" ViewCount="177" Body="&lt;p&gt;Further to my understanding &lt;a href=&quot;http://math.stackexchange.com/questions/7743/getting-better-at-proofs&quot;&gt;What makes a proof&lt;/a&gt;, I'm doing exercises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this question is to compare the proof a certain textbook provides with the one that is &lt;em&gt;logical to me&lt;/em&gt;, that I made up.  If my proof is weak, &lt;em&gt;why is it weak&lt;/em&gt;?  Why is the proof from the book acceptable (other than it finds a way to involve $\lim_{h \to 0}$ :)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without further ado the question:  Prove &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_x(\int_a^x f(t)dt) = f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proof is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\int f(t)dt = F(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $D_x( \int_a^x{f(t)dt} ) = D_x( \left[ F(t) \right]_a^x )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= D_x( F(x) - F(a) )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since F(a) is constant with respect to x, it goes to 0 when derived:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= F'(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= f(x) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that good?  If not, specifically what is wrong with it?  I know proofs have to do with &quot;your audience&quot;, but a lot of proofs are written in assembly language -- they're so low level!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The proof in the book I'm using says&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let $h(x) = \int_a^x{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h(x + \Delta x) - h(x) = \int_a^{x+\Delta x}{f(t)dt} - \int_a^x{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \int_a^x{f(t)dt} + \int_x^{x+\Delta x}{f(t)dt} - \int_a^x{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \int_x^{x+\Delta x}{f(t)dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \Delta x (f(x*))$ for some x* between x and x + $\Delta$ x by the mean value theorem for integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, dividing both sides by $\Delta x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{h(x + \Delta x) - h(x)}{\Delta x} = f(x*)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $D_x(\int_a^x f(t)dt) = D_x(h(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \lim_{\Delta x \to 0} \frac{h(x + \Delta x) - h(x)}{\Delta x} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \lim_{\Delta x \to 0} f(x*) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as $\Delta x \to 0$, $x + \Delta x \to x$ so $x* \to x$ (since $x*$ is between $x$ and $x + \Delta x$).  Since f is continuous, $\lim_{x \to 0} f(x*) = f(x)$&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-11-08T13:05:59.287" Title="Prove $D_x(\int_a^x f(t)dt) = f(x)$" Tags="&lt;calculus&gt;&lt;proof&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9358" PostTypeId="2" ParentId="9357" CreationDate="2010-11-08T02:02:06.640" Score="7" ViewCount="0" Body="&lt;p&gt;Your first line has a serious problem: $\int f(t)dt$ is not a function of $t$; the indefinite integral is a &lt;em&gt;family&lt;/em&gt; of functions, not a single function. So saying &quot;Let $\int f(t)dt= F(t)$&quot; does not make sense. So as soon as you try to get started, you are already in trouble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can say, &quot;well, &lt;em&gt;pick&lt;/em&gt; some antiderivative instead of taking the whole family.&quot; But if you try to define $F$ as &quot;pick some antiderivative of $f$&quot;, then your problem is that you have no way to guarantee that there &lt;em&gt;is&lt;/em&gt; such a thing in the first place! The whole point of this result is showing that there &lt;em&gt;is&lt;/em&gt; an antiderivative, so you cannot assume there is one to start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you assume the First Fundamental Theorem of Calculus &lt;em&gt;and&lt;/em&gt; that $f(t)$ has an antiderivative, then this result is very easy, exactly as you do it: $\int_a^x f(t)dt = F(x)-F(a)$, so the derivative with respect to $x$ is $F'(x)=f(x)$. But this assumes that there is an antiderivative for $f$ and that the FTC (Part 1) holds; but since this theorem is often used to prove Part 1, that could also make your argument circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, really, the major flaw is that you are assuming that there is an antiderivative for $f$ in the first place (you can prove FTC (Part 1) without this result, so the circularity problem is not fatal). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T05:13:29.953" LastActivityDate="2010-11-08T05:13:29.953" />
  <row Id="9359" PostTypeId="2" ParentId="9349" CreationDate="2010-11-08T02:14:00.127" Score="3" ViewCount="0" Body="&lt;p&gt;Assuming you mean to calculate the line integral $\int_L {5y \, {\rm d}s}$, where $L$ is the segment from $(0,0)$ to $(0.2,0.2)$, then define a bijective parametrization $r:[0,0.2] \to L$ by $r(t)=(t,t)$. Also let $f(x,y)=5y$. Then, &#xA;$$&#xA;\int_L {f\,{\rm d}s}  = \int_0^{0.2} {f(r(t))|r'(t)|\,{\rm d}t}. &#xA;$$&#xA;Here, $f(r(t)) = f(t,t) = 5t$ and $|r'(t)| = |(1,1)| = \sqrt{2}$.&#xA;Thus, $\int_L {5y \, {\rm d}s} = \sqrt 2 \int_0^{0.2} {5t \, {\rm d}t} $.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-08T02:42:18.763" LastActivityDate="2010-11-08T02:42:18.763" CommentCount="1" />
  <row Id="9360" PostTypeId="2" ParentId="9252" CreationDate="2010-11-08T02:17:24.157" Score="8" ViewCount="0" Body="&lt;p&gt;Okay, here is an answer assuming plausible conjectures about linear equations in primes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; For any real numbers $(a,b)$ with $0 &amp;lt; a &amp;lt; b &amp;lt; 1$, there is a positive integer $n$ such that $\phi(n)/n \in (a,b)$. Moreover, for any integer $k$, we can arrange that $n$ is relatively prime to $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $N$ be larger than $k$, and large enough that $b(1-1/N)&gt;a$. Consider the product $\prod_{p &gt;N} (1-1/p)$, where the product runs over primes greater than $N$. This infinite product is $0$, because $\sum 1/p=\infty$. So there is some $M$ such that $\prod_{N &amp;lt; p &amp;lt; M} p$ is less than $b$. Choose the first such $M$ for which this happens. Then $\prod_{N &amp;lt; p &amp;lt; M} p &gt; b (1-1/M) &gt; a$ by the choice of $M$. Take $n=\prod_{N&amp;lt; p &amp;lt; M} p$. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer your question. Choose intervals $(a_1, b_1)$ and $(a_2, b_2)$ with $0 &amp;lt; a_i &amp;lt; b_i &amp;lt; 1$ such that $x &amp;lt; a_2/b_1 &amp;lt; b_2/a_1 &amp;lt; y$. Using the lemma, find $n_1$ and $n_2$ such that $\phi(n_i)/n_i \in (a_i, b_i)$, the $n_i$ are both odd, and $GCD(n_1, n_2) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you believe the &lt;a href=&quot;http://arxiv.org/abs/math/0606088&quot; rel=&quot;nofollow&quot;&gt;generalized Hardy-Littlewood conjecture&lt;/a&gt; then there are infinitely many pairs  $(q_1, q_2)$ of prime numbers such that $n_1 q_1 +2 = n_2 q_2$. For $q_1$ large enough, $n:=n_1 q_1$ has the desired property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this equation is one of the ones to which Green and Tao's results do not apply. (In their terminology, $\psi_1$ and $\psi_2$ are linearly dependent.) You might be able to dodge this issue by asking for $q_1$ and $q_2$ to be almost prime rather than prime. (Almost prime meaning a product of two large primes.) I'm not enough of an expert to know whether that trick would work in this setting; if you seriously care about the question, you might try e-mailing Terry Tao.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-11-08T02:57:11.390" LastActivityDate="2010-11-08T02:57:11.390" />
  <row Id="9361" PostTypeId="1" AcceptedAnswerId="9362" CreationDate="2010-11-08T02:21:09.180" Score="1" ViewCount="100" Body="&lt;p&gt;I had hoped that this would be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have power series that I'm using.  I want to know what simple structure describes the operations that I'm performing on them.  It would be better if I could find something that non-mathematicians (actually computer scientists) would understand, but I'd like to know the technical terminology too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me define three power series, $f(x)$, $g(x)$, and $h(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \sum_{i=m_1}^{n_1}{c_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{i=m_1}^{n_1}{d_i x^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h(y) = \sum_{i=m_2}^{n_2}{e_i y^i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the two operations that I perform are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Subtraction $r(x) = f(x) - g(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(x) = \sum_{i=m_1}^{n_1}{c_i - d_i}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Multiplication $r(x,y) = f(x)h(y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(x,y) = \sum_{i=m_1}^{n_1}{\sum_{j=m_2}^{n_2}{c_i \cdot e_j}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the coefficients and variables are integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm writing a paper, and I'm searching for a more &quot;definite&quot; description than what I have.  It is somewhat ackward to describe, and I want to convey this information exactly and concisely.  Plus, I don't want the reader to have any doubts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been considering that I should just write out the math, since it seems to leave less doubt.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-08T03:23:03.410" LastActivityDate="2010-11-08T03:33:45.303" Title="What structure is this?" Tags="&lt;calculus&gt;&lt;soft-question&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9362" PostTypeId="2" ParentId="9361" CreationDate="2010-11-08T02:24:32.317" Score="0" ViewCount="0" Body="&lt;p&gt;The first is $\rm\ (f-g)(1)\ =\ f(1)-g(1)\ $ and the second is $\rm\ (f\:h)(1)\ =\ f(1)\ h(1)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T03:33:45.303" LastActivityDate="2010-11-08T03:33:45.303" CommentCount="5" />
  <row Id="9363" PostTypeId="2" ParentId="154" CreationDate="2010-11-08T03:05:26.850" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a possible line of reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, we start with natural numbers, $\mathbb{N}$, then add $0$, making them the whole numbers $\mathbb{N}_0$, then add negative numbers, making them integers $\mathbb{Z}$, then expand them to rationals $\mathbb{Q}$, and finally to reals $\mathbb{R}$. In some respect, we keep filling up the line, till there are no gaps left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question: why do we call them &quot;numbers&quot;? What are the properties we expect from some object so we can call it a number. In fact, just a few: addition and multiplication must be defined for all of them, and we want these operations to be commutative and associative.  The distributive law of multiplication over addition must hold. It's nice to have a notion of ordering, so we can say $a \lt b$, and multiplication by some positive number must be compatible with that notion: if $a \lt b$ and $x$ is positive, then $ax \lt bx$. Those are the minimal requirements, met by $\mathbb{N}$. As we expand our numbering system, we get more and more useful properties, and the structure (both algebraic and topological) gets more complicated: from semigroup in case of $\mathbb{N}$ to complete ordered field in case of $\mathbb{R}$, but those basic ones always hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the next step in our &quot;quest for expanding our system&quot; would be to look at $\mathbb{R}^2$, $\mathbb{R}^3$ etc. In general, the objects (&quot;points&quot;) populating the space $\mathbb{R}^n$ are called &quot;vectors&quot; and they differ from numbers in one important respect: we can't define vector multiplication so it meets our requirements for numbers (commutativity, associativity and having multiplication distributed over addition). So we don't treat vectors in $\mathbb{R}^n$ as the numbers, in particular, we don't think of them as atomic objects, but rather in terms of the components or coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one important exception, however: i is $\mathbb{R}^2$. In fact we &lt;em&gt;can&lt;/em&gt; define multiplication of two objects (points, vectors) in $\mathbb{R}^2$ so it meets all our requirements and we can call these objects &quot;numbers&quot;. To be precise, all except one: the ordering. Multiplication we define is no longer compatible with the ordering. So we agree that probably not having that feature isn't such a big deal. Once we introduced multiplication, we end up with complete field (but not ordered field) of entities which we have a right to call the numbers. So we call them the complex numbers and designate them as $\mathbb{C}$. There are some similarities between $\mathbb{R}^2$ and $\mathbb{C}$, but there are also some difference, the most important one being the fact that multiplication is defined so $\mathbb{C}$ is a field rather than vector space. In fact we can treat the elements of $\mathbb{C}$ as the scalars and use them to build a vector space, a complex matrices, etc. Well, this is roughly, why we can call complex numbers &quot;numbers&quot;. They exhibit the same behavior (except for ordering) as the numbers we're familiar with do. One of my favorite examples: when we define the derivative in vector analysis, the formulas look quite different from those we got used in Calculus. But when we define the derivative for complex-valued functions, the formula is exactly the same, because multiplication and hence division) is defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why complex numbers are important is a different question. What I wanted to focus is the fact that we have a right to call them &quot;numbers&quot;. But, just like we can think of rational numbers as a special case of reals, and when move from reals to rationals, we loose the important property of completeness, we &lt;em&gt;can&lt;/em&gt; think of real numbers as a special case of complex numbers, and when we move from complex to reals we loose some nice properties as well. In many respects, studying complex numbers helps us to better understand real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a sidenote, again, when we move from $\mathbb{R}$ to $\mathbb{C}$, we loose the ordering. Complex numbers form a complete field, but this is not an ordered field. (There is only one complete ordered field, up to isomorphism.) Now suppose we want to explore $\mathbb{R}^n$ further. That's the next best? $\mathbb{R}^4$. We can define multiplication of objects (vectors) in $\mathbb{R}^4$ which is associative, but - alas - not commutative. It's kind of hard to think of numbers whose multiplication is not commutative. So we no longer call them (quaternions, $\mathbb{H}$) numbers. Finally, the next interesting space is $\mathbb{R}^8$, where we still can define multiplication, but it is no longer associative (in addition to being non-commutative). It's &lt;em&gt;really&lt;/em&gt; hard to deal with something which is not associative, and certainly it's hard to think of those entities as numbers. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-08T03:05:26.850" />
  <row Id="9365" PostTypeId="1" CreationDate="2010-11-08T03:19:06.777" Score="1" ViewCount="78" Body="&lt;p&gt;In a Cartesian system, I've got the slope, start point and distance of a line segment.  What's the formula to find the endpoint?&lt;/p&gt;&#xA;" OwnerUserId="3259" LastActivityDate="2010-11-08T04:07:50.827" Title="Endpoint of a line knowing slope, start and distance" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="9366" PostTypeId="1" AcceptedAnswerId="9368" CreationDate="2010-11-08T03:20:52.150" Score="1" ViewCount="61" Body="&lt;p&gt;Suppose $A$ is a set of $2n$ positive integers { $a_{i}$ } with $a_{i} \geq 2$ which can be partitioned into at most $r \leq n$ coprime pairs. The following fact is true: For all $A$ and every $n \geq 1$,&#xA;$$&#xA;\prod_{i = 1}^{2n} (a_{i} - 1) \equiv 0 \pmod{2^{r}}. &#xA;$$&#xA;For example, the set {$2,3,5,6$} can be partitioned into two coprime pairs $(2,3)$ and $(5,6)$, so $2^{2} = 4$ divides $1 \cdot 2 \cdot 4 \cdot 5 = 40$. The set {$2,4,8,16$} cannot be partitioned into coprime pairs (i.e., $r = 0$), so the result above is not particularly interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can this result be sharpened? References are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Update&lt;/strong&gt;) I think the above result can be generalized to the following:&#xA;Suppose $A$ is a set of $2n$ positive integers { $a_{i}$ } with $a_{i} \geq 2$ which can be partitioned into at most $r \leq n$ pairs with &lt;em&gt;odd greatest common divisor&lt;/em&gt;. Then $2^{r}$ divides $\prod_{i = 1}^{2n} (a_{i} - 1)$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-08T04:32:07.207" LastActivityDate="2010-11-08T04:32:07.207" Title="Power of $2$ Divisibility Problem" Tags="&lt;number-theory&gt;&lt;number&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="9367" PostTypeId="2" ParentId="9365" CreationDate="2010-11-08T03:22:44.073" Score="2" ViewCount="0" Body="&lt;p&gt;If the point is $(a,b)$, then the distance from $(a,b)$ to $(x,y)$ is&#xA;$$\sqrt{(x-a)^2 + (y-b)^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the point is $(a,b)$, then the points that lie on the line through $(a,b)$ with slope $m$ are the points of the form&#xA;$$(a,b) + k(1,m)$$&#xA;where $k$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the two together, if you know the start point $(a,b)$, and the slope $m$, and the distance $d$, then find the (two) values of $k$ that will give you a distance of $d$ by plugging in and solving for $k$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T03:29:17.250" LastActivityDate="2010-11-08T03:29:17.250" CommentCount="1" />
  <row Id="9368" PostTypeId="2" ParentId="9366" CreationDate="2010-11-08T03:25:40.500" Score="3" ViewCount="0" Body="&lt;p&gt;Each pair contains at most one even number, since the two numbers have odd gcd. That means that at least one $a_i-1$ for each pair is even, so the product is divisible at least $2^s$, where $s$ is the number of pairs with odd gcd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; As far as sharpening the result, take a set of $2n$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_numbers&quot; rel=&quot;nofollow&quot;&gt;Fermat numbers&lt;/a&gt;, numbers of the form $F_k=2^{2^k}+1$. If $k\neq\ell$, then $F_k$ and $F_{\ell}$ share no common factor, so if you take $2n$ of them, say $F_{k_1},\ldots,F_{k_{2n}}$ with $k_1\lt k_2\lt\cdots\lt k_{2n}$, then you can divide them into $n$ coprime pairs; your product will equal $2^{2^{k_1}+\cdots+2^{k_{2n}}}$. So you can make this product divisible by arbitrarily large powers of $2$ for a fixed value of $n$ and $r$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T04:08:42.257" LastActivityDate="2010-11-08T04:08:42.257" CommentCount="4" />
  <row Id="9370" PostTypeId="1" CreationDate="2010-11-08T03:39:10.250" Score="1" ViewCount="105" Body="&lt;p&gt;In one of the programming work I am doing, I encountered a set of linear eqns. with modulo. I am putting it in simple format with only 2 variables,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a11x + a12y) mod 8 = b1&lt;br&gt;&#xA;(a21x + a22y) mod 8 = b2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of simple eqns. with out modulo I can use: AX=B, X=inv(A)*B. But how to handle this modulo? Any clue or pointers will be very helpful.&lt;/p&gt;&#xA;" OwnerUserId="3260" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-11-08T05:06:39.147" LastActivityDate="2010-11-08T05:06:39.147" Title="how to solve linear equations involving modulo?" Tags="&lt;algebra&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="9371" PostTypeId="2" ParentId="9354" CreationDate="2010-11-08T03:39:55.447" Score="4" ViewCount="0" Body="&lt;p&gt;The main use of this theory is that it is strong enough to prove many truly basic facts about natural numbers, while still being weak enough that it carries little &quot;baggage&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\Delta_0 + \text{Exp}$ is strong enough to define functions that encode and decode finite sequence of numbers (Goedel numbering) and, similarly, it can encode and decode finite sets of numbers as single numbers. This makes it possible to formalize many basic number-theoretic and proof-theoretic results in this theory. Those results, in turn, make it possible to use $I\Delta_0 + \text{Exp}$ as a base over which we can prove stronger statements are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Harvey Friedman has made a somewhat polemical &quot;&lt;a href=&quot;http://cs.nyu.edu/pipermail/fom/1999-April/003014.html&quot; rel=&quot;nofollow&quot;&gt;grand conjecture&lt;/a&gt;&quot; that &lt;em&gt;every&lt;/em&gt; fact of elementary number theory proven in &lt;em&gt;Annals of Mathematics&lt;/em&gt; can be proven in $I\Delta_0 + \text{Exp}$. This includes, for example, Fermat's Last Theorem. This is not to say that the original proofs will work - the proofs in $I\Delta_0 + \text{Exp}$ may be much more lengthy and complex. But the fact that this conjecture is even plausible speaks to the strength of $I\Delta_0 + \text{Exp}$. (The status of FLT here is not known.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\Delta_0 + \text{Exp}$ is very weak in other senses. Its proof theoretic ordinal is $\omega^3$, which is extremely smaller than the ordinal of Peano arithmetic and even much smaller than the ordinal of the &quot;next&quot; canonical theory, $I\Sigma_1$. This is philosophically relevant in certain instances where we wish to keep our assumptions relatively weak (in terms of their proof-theoretic ordinal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\Delta_0 + \text{Exp}$ is not &quot;minimal&quot; in a formal sense. We could work with &lt;em&gt;ad hoc&lt;/em&gt; theories that are weaker, but still include the facts of number theory we need (perhaps as axioms). However, $I\Delta_0 + \text{Exp}$ has a relatively concrete and natural definition, which makes it more attractive to work with than some concocted system.  (Of course, it may be that $I\Delta_0 + \text{Exp}$ is equivalent to some axiom $T$ over some even weaker base theory, in which case $I\Delta_0 + \text{Exp}$ would be minimal as an extension of the base theory that proves $T$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another commonly used weak theory is $I\Sigma_1$. This theory extends $I\Delta_0 + \text{Exp}$ (modulo some definitional extensions) but $I\Sigma_1$ is still relatively weak (its ordinal is $\omega^\omega$). Many basic facts of number theory are easier to prove in $I\Sigma_1$, making it more convenient than $I\Delta_0 + \text{Exp}$ for various purposes, at the cost of a larger proof theoretic ordinal. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-08T03:39:55.447" />
  <row Id="9372" PostTypeId="1" AcceptedAnswerId="9379" CreationDate="2010-11-08T03:41:17.643" Score="5" ViewCount="388" Body="&lt;p&gt;I am teaching a Calc II course and came across the following series when finding the interval of convergence for the Taylor series of $f(x)=\sqrt{x}$ centered at $x=1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_{n=2}^\infty \frac{1\cdot 3\cdot 5\cdot 7\cdots (2n-3)}{2^nn!}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it doesn't converge, but how do you show it?  I would like to be able to use only the technology of a Calc II course, but any answer would be enlightening.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-11-08T16:08:10.400" Title="Why Doesn't This Series Converge?" Tags="&lt;calculus&gt;&lt;teaching&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="0" />
  <row Id="9373" PostTypeId="2" ParentId="9370" CreationDate="2010-11-08T03:50:31.663" Score="2" ViewCount="0" Body="&lt;p&gt;You can do Gauss-Jordan elimination performing operations modulo $8$. Just be mindful that &quot;division&quot; is only defined for odd integers (you are really multiplying by the inverse modulo $8$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-08T03:50:31.663" />
  <row Id="9375" PostTypeId="2" ParentId="9365" CreationDate="2010-11-08T04:07:50.827" Score="2" ViewCount="0" Body="&lt;p&gt;An equivalent way to Arturo's answer is as follows: from the slope $m$, you can determine the cosine and the sine of the angle from the horizontal axis of a line with that slope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\frac{1}{\sqrt{1+m^2}} \qquad s=\frac{m}{\sqrt{1+m^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(exercise: verify that they are the cosine and sine of a certain angle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this construction, you can easily determine the two points at a distance r from your starting point $(h,k)$ as $(h,k)\pm r(c,s)$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-08T04:07:50.827" />
  <row Id="9376" PostTypeId="2" ParentId="9370" CreationDate="2010-11-08T04:09:22.520" Score="1" ViewCount="0" Body="&lt;p&gt;You can apply Cramer's rule if the determinant is odd (so invertible mod 8). For an introduction to linear algebra over commutative rings see Wm. C. Brown: Matrices over commutative rings.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T04:09:22.520" />
  <row Id="9377" PostTypeId="1" AcceptedAnswerId="9380" CreationDate="2010-11-08T04:20:55.520" Score="1" ViewCount="123" Body="&lt;p&gt;Suppose a set S has 11 elements. How many subsets of S have an even&#xA;number of elements? Express your answer as a summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure how to approach this. I think the general formula for this type of question is n!/k!(n-k)! where n is 11 and k is the number we're choosing. So would we just take the summation starting at 0 to 5 of 11!/2i!(n-2i)! Or what?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:07:37.737" LastActivityDate="2010-11-28T04:07:37.737" Title="Combinations and Permutations Question" Tags="&lt;combinatorics&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9378" PostTypeId="1" CreationDate="2010-11-08T04:30:10.903" Score="5" ViewCount="253" Body="&lt;p&gt;Where, or do there exist, good visualizations of sheaves, stalks, stacks, and/or schemes? I'm a better visual thinker than I am a symbolic thinker, and it would be easier for me to follow some of the algebraic geometry papers that have no pictures whatsoever if I had some kind of visual intuition of what these objects look like. &lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T05:17:23.877" LastActivityDate="2010-11-13T12:52:00.460" Title="Visualizations of some of the abstractions of algebraic geometry." Tags="&lt;algebraic-geometry&gt;&lt;visualization&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="9379" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T04:32:11.280" Score="16" ViewCount="0" Body="&lt;p&gt;I don't think the series diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \frac{1 \cdot 3 \cdot 7 \cdots (2n-3)}{2 \cdot 4 \cdot 6 \cdots 2n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{{2n \choose n}}{(2n-1)4^n} =  \frac{1}{(2n-1)\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ {2n \choose n}}{4^n} = \frac{1}{\sqrt{\pi n}}(1 + O(\frac{1}{n}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the series must converge! Perhaps you have a mistake in your computation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more elementary proof of convergence, see the end of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe you should be able to compute it using the series expansion for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{1-x^2}} = \sum_{k=0}^{\infty} \frac{{2k \choose k} x^{2k}}{4^k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(you will need to subtract some terms, divide by $x^2$ and integrate).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An elementary proof of convergence.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $\displaystyle S_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}$ then&#xA;$\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves the partial term of your series, which is $\displaystyle \sum_{k=2}^{n} \frac{S_k}{2k-1} &amp;lt; \sum_{k=2}^{\infty} \frac{1}{(2k-1)\sqrt{k+1}} &amp;lt; C$ (for some constant $C$), and thus is bounded above. Since the series is monotonically increasing and bounded above, it is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will prove that $\displaystyle S_n \le \frac{1}{\sqrt{n+1}}$ by induction on $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ it is clearly true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $S_{n+1} = S_n \frac{2n+1}{2n+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\displaystyle  1 - \frac{2n+1}{2n+2} = \frac{1}{2n+2}  \ge \frac{1}{\sqrt{n+2}(\sqrt{n+2} + \sqrt{n+1})} = \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S_{n} \le \frac{1}{\sqrt{n+1}} \Rightarrow S_{n+1} \le \frac{1}{\sqrt{n+2}}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-08T15:23:06.247" LastActivityDate="2010-11-08T15:23:06.247" CommentCount="11" />
  <row Id="9380" PostTypeId="2" ParentId="9377" CreationDate="2010-11-08T04:34:53.200" Score="6" ViewCount="0" Body="&lt;p&gt;Hi fprime. If you can use the identity: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k = 0}^{n} (-1)^{k} {n \choose k} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you can show that the cardinality of the set of subsets with even cardinality is equal to the cardinality of the set of subsets with odd cardinality.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-11-08T04:34:53.200" CommentCount="8" />
  <row Id="9381" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T04:49:46.853" Score="2" ViewCount="0" Body="&lt;p&gt;Using Stirling formula $n!\sim \sqrt {2\pi n}(n/e)^n$, I also get that&#xA;the $n$th term of the series is $O(1/n^{3/2})$. So the series is convergent. In the derivation I used that $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=e^{-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-08T04:49:46.853" />
  <row Id="9382" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T04:50:18.210" Score="2" ViewCount="0" Body="&lt;p&gt;In fact the sum converges. You can find a closed form by massaging the formula that I posted on &lt;a href=&quot;http://math.stackexchange.com/questions/4744/formal-power-series-coefficient-multiplication/4750#4750&quot;&gt;Hadamard products&lt;/a&gt; viz. $\rm\ (1-4x)^{-1/2} \;=\; \sum\ \binom{2n}{n}\ x^n\:.\ $ See also &lt;a href=&quot;http://math.stackexchange.com/questions/7778/proving-2n-choose-n-2n-frac-1-cdot-3-cdot-5-cdots-2n-1n/7802#7802&quot;&gt;this post&lt;/a&gt;.  Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/GktjF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T15:52:02.743" LastActivityDate="2010-11-08T15:52:02.743" CommentCount="1" />
  <row Id="9384" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T05:06:18.447" Score="7" ViewCount="0" Body="&lt;p&gt;You can use Taylor approximations here. Note that the ratio between consecutive terms is ${2n - 3 \over 2n} = \exp(\ln(1 - 3/2n)) = \exp(-{3 \over 2n} + O(1/n^2))$. So the product is comparable to $\exp(-{3 \over 2} \sum_{i = 2}^n {1 \over n} + O(1/n))$, which in turn is comparable to $\exp(-{3 \over 2} \ln(n))$ or $n^{-{3 \over 2}}$. Thus the series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: Seeing others computing the actual value... once you know it converges (absolutely) you can use the original power series to get the actual value. For $|x| &amp;lt; 1$, $(1 - x)^{1 \over 2} = 1 - {x \over 2} - {1 \over 2} \sum_{n=2}^{\infty} A_nx^n$, where $A_n$ is the $n$ term of the sum. Since all terms of the sum are positive and it converges for $x = 1$, if you take limits as $x$ goes to 1 you get the sum $S$ we're looking at. In other words, 0 = 1 - 1/2 - ${S \over 2}$ or $S = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And actually if you think about it.. if the series did diverge, taking limits of the series as $x$ goes to 1 from below would give infinity since all terms are positive. Thus just taking limits of both sides as $x$ goes to 1 gives both convergence and the correct value. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-08T16:04:04.973" LastActivityDate="2010-11-08T16:04:04.973" CommentCount="1" />
  <row Id="9385" PostTypeId="2" ParentId="9338" CreationDate="2010-11-08T05:18:42.350" Score="3" ViewCount="0" Body="&lt;p&gt;well, if they come sorted like that, you can:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;calculate the length of the list (in this case, 8)&lt;/li&gt;&#xA;&lt;li&gt;start going up in ascending order (from 1, for example)&lt;/li&gt;&#xA;&lt;li&gt;once you hit a number bigger than the number you want, count how many numbers are before it (if you're looking for 5, that would be 5 numbers because you stopped at 9)&lt;/li&gt;&#xA;&lt;li&gt;divide what you found in #3 by the length of the list in #1, multiply it by 100, and there's your percentile.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;otherwise, just count how many numbers are &amp;lt;= the number you're looking for, then divide it by the total. the above is just faster for when they come sorted.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-11-08T05:18:42.350" />
  <row Id="9386" PostTypeId="1" AcceptedAnswerId="9388" CreationDate="2010-11-08T05:20:49.353" Score="1" ViewCount="85" Body="&lt;p&gt;If x,y,z are natural numbers, how many solutions are there to x+y+z=25?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I figure this out? I cant even begin dissecting this problem. Where do I begin?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-08T05:33:52.950" LastActivityDate="2010-11-08T06:03:43.980" Title="Number of possible solutions for an equation" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="9387" PostTypeId="2" ParentId="9386" CreationDate="2010-11-08T05:26:01.077" Score="1" ViewCount="0" Body="&lt;p&gt;This is called the &quot;partition function&quot; of 25. See here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_function_%28number_theory%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Partition_function_%28number_theory%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an easy-to-compute-directly function - the best bet is simply writing a small script that counts solutions using a double loop.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-08T05:26:01.077" CommentCount="5" />
  <row Id="9388" PostTypeId="2" ParentId="9386" CreationDate="2010-11-08T05:33:01.157" Score="5" ViewCount="0" Body="&lt;p&gt;Suppose that &quot;natural number&quot; doesn't include 0. Take 25 balls, and put a wall between any two. A partition $a+b+c=25$ is the same as a choice of two walls, and there are $\binom{24}{2} = 276$ of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If zeroes are allowed, then $(x+1)+(y+1)+(z+1) = 28$, and so the answer is $\binom{27}{2} = 351$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T05:33:01.157" CommentCount="6" />
  <row Id="9389" PostTypeId="1" AcceptedAnswerId="9390" CreationDate="2010-11-08T05:40:26.280" Score="2" ViewCount="55" Body="&lt;p&gt;let $f$:&lt;strong&gt;R&lt;/strong&gt;-&gt;&lt;strong&gt;N&lt;/strong&gt; such that given any $x_1$,$x_2$ belong to &lt;strong&gt;R&lt;/strong&gt; such that $x_1 &amp;lt; x_2$ then there exists a $x_3$ belongs to $(x_1,x_2)$ such that $f(x_3)&gt; max(f(x_1),f(x_2))$. How many such mappings are possible ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-11-08T05:48:05.797" Title="a Mapping problem" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="1" />
  <row Id="9390" PostTypeId="2" ParentId="9389" CreationDate="2010-11-08T05:48:05.797" Score="2" ViewCount="0" Body="&lt;p&gt;There are $\aleph_0^\aleph$ such mappings. The upper bound is trivial. For the lower bound, pick any function $f\colon \mathbb{R} \longrightarrow \mathbb{N}$ such that $f(\pm \frac{p}{q}) = q$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T05:48:05.797" CommentCount="7" />
  <row Id="9391" PostTypeId="1" AcceptedAnswerId="9393" CreationDate="2010-11-08T05:48:34.370" Score="3" ViewCount="132" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Trinomial_expansion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the number of terms in (x+y+z)^30 is 496. I'm assuming this is before like terms are added up. How many terms would there be if like terms were combined? How would I go about figuring that out?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2010-11-08T05:58:32.397" LastActivityDate="2010-11-08T23:16:09.737" Title="Number of terms in a trinomial expansion" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9392" PostTypeId="2" ParentId="9391" CreationDate="2010-11-08T05:57:47.280" Score="4" ViewCount="0" Body="&lt;p&gt;First note that every $x^ay^bz^c$ with $a + b + c = 30$ appears. One way of seeing this is observing ${\partial^{30}(x+y+z)^{30} \over \partial_x^a\partial_y^b\partial_z^c} $ at $(0,0,0)$ is $30!$, which is not zero. So for a given value of $a$, every possible $(b,c)$ with $b + c = 30 - a$ can happen. Since $b$ goes from $0$ to $30 - a$, there are $31 - a$ possibilities for $b$, each of which forces $c$ to have the single value $30 - a - b$. Thus for a given $a$ there are &#xA;$31 - a$ different possible $(b,c)$. Adding this over all $a$ from 0 to 31 this gives the sum of $31 + 30 + ... + 0 = 496$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-08T23:16:09.737" LastActivityDate="2010-11-08T23:16:09.737" CommentCount="2" />
  <row Id="9393" PostTypeId="2" ParentId="9391" CreationDate="2010-11-08T05:59:12.863" Score="8" ViewCount="0" Body="&lt;p&gt;No, the 496 is the number of terms after like terms are combined.  Before like terms are combined there are $3^{30}$ terms.  This is because you have 30 different factors, and so the number of terms you get before combining is the number of ways to choose 30 elements when there are three choices for each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zaricuse's answer is hinting at how to derive the formula on the Wikipedia page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another way to look at the formula on the Wikipedia page: The number of terms in the expansion of $(x+y+z)^n$ after combining is the number of ways to choose $n$ elements &lt;em&gt;with replacement&lt;/em&gt; (since you can choose $x,y,z$ more than once) in which order does not matter from a set of 3 elements.  This formula is known to be &#xA;$$\binom{3+n-1}{n} = \binom{n+2}{n} = \frac{(n+1)(n+2)}{2}.$$ &#xA;See, for example, MathWorld's entry on &lt;a href=&quot;http://mathworld.wolfram.com/BallPicking.html&quot; rel=&quot;nofollow&quot;&gt;Ball Picking&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-08T22:06:29.063" LastActivityDate="2010-11-08T22:06:29.063" CommentCount="4" />
  <row Id="9394" PostTypeId="2" ParentId="9386" CreationDate="2010-11-08T06:03:43.980" Score="-1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ (x,y,z)\ \to\ \{x,\ x+y\}\ $ bijects solutions with two elt subsets of $\{1,2,\cdots,24\} $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T06:03:43.980" />
  <row Id="9395" PostTypeId="1" AcceptedAnswerId="9401" CreationDate="2010-11-08T06:34:06.817" Score="0" ViewCount="110" Body="&lt;p&gt;What is the cardinality of set of all smooth functions belonging to $L^1$ or $L^2$ ? What is that of set of all integrable or square integrable functions ? &lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-11-08T12:32:07.567" Title="what is the cardinality of set of all smooth functions in $L^1$ ?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="17" FavoriteCount="1" />
  <row Id="9396" PostTypeId="2" ParentId="9378" CreationDate="2010-11-08T06:44:43.767" Score="3" ViewCount="0" Body="&lt;p&gt;Mumford's The Red Book of Varieties and Schemes is quite famous for its pictures and the emphasis on geometric intuition.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-08T06:44:43.767" />
  <row Id="9397" PostTypeId="2" ParentId="9377" CreationDate="2010-11-08T06:49:58.933" Score="4" ViewCount="0" Body="&lt;p&gt;Here is a very easy way of seeing that the number of sets of even cardinality is equal to the number of sets of odd cardinality: if the total number of elements is odd (like in your case) then the map that sends a set to its complement establishes a bijection between sets of even and of odd cardinality. If the total number of elements is even, then there is the same bijection between the sets of even cardinality not containing the element 1 and the sets of odd cardinality not containing 1. Also, there is the obvious bijection between sets of even cardinality not containing 1 and those of odd cardinality containing 1, so again you have a bijection between odd and even sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you just have to compute the total number of subsets, which I am sure you can do.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-08T07:04:24.240" LastActivityDate="2010-11-08T07:04:24.240" />
  <row Id="9398" PostTypeId="1" AcceptedAnswerId="9399" CreationDate="2010-11-08T06:53:09.550" Score="0" ViewCount="124" Body="&lt;p&gt;I have a function given implicitly, you know. X og Y on both sides. Than it says, assume y=y(x). That's fine. And find y'(0), that I should be able to do. But what about y''(0)? How do you treat the dy/dx parts when taking the second derivative? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I would also like to follow the tip in the book, that says when I'm after actual values. We can just insert the value instead solving for dy/dx. &lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-08T23:13:35.200" Title="Find the second derivative of some implicit function? " Tags="&lt;homework&gt;&lt;implicit&gt;" AnswerCount="2" />
  <row Id="9399" PostTypeId="2" ParentId="9398" CreationDate="2010-11-08T06:56:03.710" Score="5" ViewCount="0" Body="&lt;p&gt;You simply replace all the $\frac{\mathrm{d}y}{\mathrm{d}x}$ terms in your second derivative with the expression you got for $\frac{\mathrm{d}y}{\mathrm{d}x}$ through implicit differentiation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an explicit example, suppose we wanted to find $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ for the expression $x^2+y^2-r=0$ ($r$ here is a constant). We find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and differentiating again gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we substitute the expression we got for the first derivative into the second derivative, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{1+\left(-\frac{x}{y}\right)^2}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{x^2+y^2}{y^3}$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T07:04:16.367" LastActivityDate="2010-11-08T07:04:16.367" />
  <row Id="9400" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T07:00:21.383" Score="4" ViewCount="0" Body="&lt;p&gt;Let's start with the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=1}^\infty \frac{\binom{2n}{n}}{(2n-1)4^n}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{\binom{2n}{n}}{2n-1} = \frac{(2n)!}{(2n-1)n!n!} = 2 \frac{(2n-2)!}{n!(n-1)!} = 2C_{n-1},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_n$ is the $n$th Catalan number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $C_n$ counts the number of paths of length $2(n+1)$ that reach the origin for the first time at their end, always being to the &lt;i&gt;right&lt;/i&gt; of the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a one-dimensional random walk will get back to the origin w.p. $1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=1}^\infty \frac{C_{n-1}}{4^n} = \frac{1}{2}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the original sum equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=1}^\infty \frac{2C_{n-1}}{4^n} = 1.$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T07:00:21.383" />
  <row Id="9401" PostTypeId="2" ParentId="9395" CreationDate="2010-11-08T07:29:32.983" Score="3" ViewCount="0" Body="&lt;p&gt;A continuous function is determined by its value on the rational points, so there are at most $\aleph^{\aleph_0} = \aleph$ of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, it's not difficult to find $\aleph$ smooth (integrable) functions in $L_1 \cap L_2$, just take any such non-zero function $f$ and consider $\{rf : r \in \mathbb{R}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For definiteness, $x \mapsto e^{-x}$ is $C^\infty$ and in $\bigcap_{p&gt;0} L_p$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-08T07:51:12.397" LastActivityDate="2010-11-08T07:51:12.397" CommentCount="7" />
  <row Id="9402" PostTypeId="1" CreationDate="2010-11-08T07:40:58.583" Score="5" ViewCount="252" Body="&lt;p&gt;Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form?&#xA;$$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$&lt;/p&gt;&#xA;" OwnerUserId="2323" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-08T17:33:10.077" LastActivityDate="2010-11-08T17:33:10.077" Title="Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}$ without using complex analysis" Tags="&lt;real-analysis&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="9403" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T07:53:15.237" Score="-1" ViewCount="0" Body="&lt;p&gt;You can try using the Taylor expansion of $\cos$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T07:53:15.237" />
  <row Id="9404" PostTypeId="1" CreationDate="2010-11-08T08:00:37.030" Score="2" ViewCount="137" Body="&lt;p&gt;Let $f \colon (\mathbb{C}^{n},\mathbf{0}) \to (\mathbb{C},0)$ be a complex analytic function with isolated critical point at the origin. Define the singular hypersurface $V_{f, \kappa} = f^{-1}(\kappa)$ for small $\kappa &gt; 0$. Let $\pi \colon \tilde{V} \to V$ be a resolution of $V = V_{f,0}$, and let $p_{g} = \dim H^{n-2}(\tilde{V}, \mathcal{O})$ denote the corresponding &lt;em&gt;geometric genus&lt;/em&gt; of $V$, which is independent of the chosen resolution. (This is the same geometric genus in the Yau-Durfee Theorem: $\mu \geq n! \ p_{g}$ for $n &gt; 2$, See &lt;a href=&quot;http://www.intlpress.com/JDG/archive/1993/37-2-375.pdf&quot; rel=&quot;nofollow&quot;&gt;Y93&lt;/a&gt; or &lt;a href=&quot;http://www.mathjournals.org/mrl/2006-013-006/2006-013-006-006.pdf&quot; rel=&quot;nofollow&quot;&gt;Y06&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the map $\varphi_{f} = f/\| f \| \colon S_{\epsilon}^{2n-1} \setminus V_{f, \kappa} \to S^{1}$, where $\epsilon &gt; 0$ is sufficiently small. The intersection of $V_{\kappa}$ with a small sphere $S_{\epsilon}^{2n-1}$ is an algebraic link. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the Brieskorn-Pham polynomial $f = \sum_{i = 1}^{n} z_{i}^{a_{i}}$ with $a_{i} \geq 2$. In particular, for $f = z_{1}^{p} + z_{2}^{q}$, the intersection above is a torus knot $T_{p,q}$ if $p$ and $q$ are coprime. In 1968, Milnor conjectured that the unknotting number $u(T_{p,q})$ is related to the dimension of the local algebra $A_{f} = \mathbb{C}[x,y] / \langle x^{p-1}, y^{q-1} \rangle$, which is also the rank of the middle homology group $H_{1}(F_{f,0}; \mathbb{Z})$ of the corresponding fiber $F_{f,0} = \varphi_{f}^{-1}(e^{i \theta})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1928, Brauner proved $\mu(f) = \text{rank} \ H_{1}(F_{f,0}, \mathbb{Z}) = (p-1)(q-1)$ for $f = x^{p} + y^{q}$. In 1994, Kronheimer and Mrowka proved the Thom conjecture, and as a consequence, the Milnor conjecture followed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edited) &lt;strong&gt;Question(s)&lt;/strong&gt;: Is the genus $g$ of the fiber $F_{f,0}$ related to the geometric genus $p_{g}$ of $V$ for $n &gt; 2$? Is there a geometric genus-like object for $n = 2$? If so, how might it relate to invariants of the torus knots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also posted on MO.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-08T20:49:09.397" LastActivityDate="2010-11-08T20:49:09.397" CommunityOwnedDate="2010-11-08T20:49:09.397" Title="The Milnor Conjecture on the Unknotting Number of a Torus Knot" Tags="&lt;complex-analysis&gt;&lt;algebraic-geometry&gt;&lt;algebraic-curves&gt;" CommentCount="7" FavoriteCount="2" />
  <row Id="9405" PostTypeId="1" CreationDate="2010-11-08T08:02:36.410" Score="0" ViewCount="64" Body="&lt;p&gt;Let G be the group U(24) and H the cyclic subgroup generated by the element 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exhibit all of the distinct cosets of H in G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am coming up with this.&#xA;(I created the Cayley table for U(24) - using the elements 1,5,7,11,13,17,19,23.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;lt;1&gt; = {1}&#xA;&amp;lt;5&gt; = (5*5 = 1)  {5}&#xA;&amp;lt;7&gt; = (7*7 = 1)  {7}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so forth since each element times itself is equal to the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? there are 8 elements and therefore 8 distinct cosets for U(24).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Next piece of the problem I'm working thru.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is H a normal subgroup of G?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not quite sure what H is... exactly to compare &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing &amp;lt;7&gt; = H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so by looking at the left and right cosets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;lt;1&gt;H  &amp;lt;-&gt;  H&amp;lt;1&gt;&#xA;&amp;lt;5&gt;H (5*7 = 11) &amp;lt;-&gt; H&amp;lt;5&gt; (7*5 = 11)&#xA;&amp;lt;7&gt;H (7*7 = 1) &amp;lt;-&gt; H&amp;lt;7&gt; (7*7 = 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then I would stop there and say it was normal subgroup of G.&#xA;with order 1 and the index of H in G is 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Lagrange Theorem states the order of G = (order of H)* index of H in G... this doesn't end up working in this case (Makes me wonder if I am doing this part wrong)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Exhibit distinct elements of the quotient group of G/H... (So I can construct a Cayley Table for G/H.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am just starting to work on this... some help to get me started would be nice :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3256" LastActivityDate="2010-11-08T08:16:02.950" Title="Exhibiting the distinct cosets of a cyclic subgroup H in group G?" Tags="&lt;abstract-algebra&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="9406" PostTypeId="2" ParentId="9405" CreationDate="2010-11-08T08:16:02.950" Score="1" ViewCount="0" Body="&lt;p&gt;The group $H$ has two elements $\{7, 1\}$ and therefore its index in $G$ is 4. Also note that right cosets coincide with left cosects so that $gH = Hg$ (this comes from the fact that G is abelian and also answers your second question). So e.g. $5 \{7, 1\} = \{11, 5\} = 11 \{7, 1\}$. You'll get 4 cosets like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third question is really the same thing as the first question (thanks to $H$ being normal).&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-08T08:16:02.950" CommentCount="6" />
  <row Id="9407" PostTypeId="1" CreationDate="2010-11-08T08:16:41.640" Score="3" ViewCount="106" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what are some &lt;em&gt;natural&lt;/em&gt; examples of integral domains $A$ together with sequences of elements satisfying $... | f_2 | f_1 | f_0$, but so that $f_i$ divides no power of $f_{i+1}$? So in the spectrum there will be an ascending chain of basic-open subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we could just take the union of the $k[y^{-n} x,y]$ in $k(x,y)$ or the union of the $\mathbb{Z}[2^{-n} x]$ in $\mathbb{Q}(x)$, or a valuation domain whose value group has arbitrary small positive elements, but I don't call these examples natural. What about rings of holomorphic functions on a connected open subset of $\mathbb{C}$?&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-08T21:27:49.347" Title="Examples of infinite divisor sequences" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9408" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T08:18:20.660" Score="1" ViewCount="0" Body="&lt;p&gt;The previous answer is not correct. If you use the Taylor expansion of cosine and integrate termwise you consider integrals of the following form:&#xA;\begin{eqnarray}&#xA;\int_{0}^{\infty} \frac{x^{a} \ dx}{1 + x^{2}} = \tfrac{\pi}{2} \sec (\tfrac{\pi a}{2})&#xA;\end{eqnarray}&#xA;which is only well-defined if $-1 &amp;lt; a &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-08T08:18:20.660" />
  <row Id="9409" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T08:18:54.573" Score="10" ViewCount="0" Body="&lt;p&gt;This can be done by the useful technique of differentiating under the integral sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is exercise 10.23 in the second edition of &quot;Mathematical Analysis&quot; by Tom Apostol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the brief sketch (as laid out in the exercise itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle F(y) = \int_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx$ for $y &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle F''(y) - F(y) + \pi/2 = 0$ and hence deduce  that $\displaystyle F(y) = \frac{\pi(1-e^{-y})}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this to deduce that for $y &gt; 0$ and $a &gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \int_{0}^{\infty} \frac{\sin xy}{x(x^2 + a^2)} \ dx = \frac{\pi(1-e^{-ay})}{2a^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \int_{0}^{\infty} \frac{\cos xy}{x^2 + a^2} \ dx = \frac{\pi e^{-ay}}{2a}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-08T08:18:54.573" CommentCount="5" />
  <row Id="9410" PostTypeId="2" ParentId="9407" CreationDate="2010-11-08T08:30:08.633" Score="4" ViewCount="0" Body="&lt;p&gt;In rings of analytic functions (real or complex), you could take any function with an infinite product expansion, and remove one factor at a time.  E.g., you could take $f_0(z)=\sin(\pi z)$, and $f_k(z)=\frac{\sin(\pi z)}{(z-1)(z-2)\cdots(z-k)}$.  As indicated in this example, you don't really need to think about infinite products.  You can take any function with infinitely many zeros, and divide out by factors corresponding to those zeros (including the multiplicity of each zero, so that taking powers won't add them back in).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It might be nicer to take $f_0(z)=\sin(\pi z)$ and $f_k(z)=\frac{\sin(\pi z)}{\left(1-\frac{z^2}{1^2}\right)\left(1-\frac{z^2}{2^2}\right)\cdots\left(1-\frac{z^2}{k^2}\right)}$, so that you're actually removing factors from the product expansion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this on any connected open subset of $\mathbb{C}$ or $\mathbb{R}$ (connected because you want an integral domain), but another &quot;natural&quot; class of examples that comes to mind is the set of &lt;a href=&quot;http://eom.springer.de/b/b016630.htm&quot; rel=&quot;nofollow&quot;&gt;Blaschke products&lt;/a&gt; with infinitely many zeros on the open unit disk.  In this case, you could take&#xA;$$f_k(z)=\frac{f_0(z)}{(\phi_{a_1}(z))^{m_1}(\phi_{a_2}(z))^{m_2}\cdots(\phi_{a_k}(z))^{m_k}},$$&#xA;where $a_1,a_2,\ldots$ are the distinct zeros of the infinite Blaschke product $f_0$, $m_k$ is the multiplicity of $a_k$, and $\phi_{a_k}$ is a holomorphic automorphism of the disk that sends $a_k$ to $0$ (unique up to a constant multiple of modulus $1$).  Thus each $f_k$ is a Blaschke product and has one less zero than $f_{k-1}$.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-08T21:27:49.347" LastActivityDate="2010-11-08T21:27:49.347" CommentCount="1" />
  <row Id="9411" PostTypeId="1" CreationDate="2010-11-08T09:09:42.717" Score="0" ViewCount="66" Body="&lt;p&gt;If I am given a Cayley Table of 8 elements (an octic table). How would I go about finding a subgroup that satisfies the following conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Find H is a normal subgroup of G&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) G is a normal subgroup of octic group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) G is NOT normal subgroup of octic group&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1) If aH = Ha for every a in G, then H is said to be a normal subgroup. So I must prove this is a fact, by trial and error picking out a subgroup of order 2 that will satisfy this statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 &amp;amp; 3) Not sure what it means to show that G is a subgroup of the octic group. Isn't G the actual octic group?&lt;/p&gt;&#xA;" OwnerUserId="3256" LastActivityDate="2010-11-08T09:09:42.717" Title="If I am given a Cayley Table of  8 elements how can I derive some facts about it?" Tags="&lt;abstract-algebra&gt;&lt;homework&gt;" CommentCount="5" />
  <row Id="9412" PostTypeId="1" AcceptedAnswerId="9418" CreationDate="2010-11-08T09:54:28.433" Score="1" ViewCount="96" Body="&lt;p&gt;Hi there, In a calculus course I just completed I struggled to understand the Taylor and Maclaurin series. I got the basics but not some of the harder concepts. Could anyone point me in the right direction? Books, tutorials, etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="2897" LastActivityDate="2010-11-08T12:12:06.810" Title="Learning Taylor/Maclaurin Series" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="9413" PostTypeId="2" ParentId="9255" CreationDate="2010-11-08T10:08:38.463" Score="1" ViewCount="0" Body="&lt;p&gt;This question was asked over at tex.se, &lt;a href=&quot;http://tex.stackexchange.com/questions/5042/define-strange-operators&quot;&gt;Define strange operators&lt;/a&gt;, with an accepted answer by frabjous:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;code&gt;\mathop{..}&lt;/code&gt; command from amsmath makes its argument behave like a &quot;large operator&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with some example code.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-08T10:08:38.463" CommunityOwnedDate="2010-11-08T10:08:38.463" />
  <row Id="9414" PostTypeId="1" AcceptedAnswerId="9415" CreationDate="2010-11-08T10:37:53.200" Score="0" ViewCount="82" Body="&lt;p&gt;I have posted about induction on it's own so I think I understand that part. But I'm slightly confused when it comes to using induction to show that n+1 is also the derivative. The base case if fine, I just set n=1. And show that is equal to the derivative not using the formula. But how about k+1? Can I just choose k=the-basecase. And than prove k+1 by just finding f''(x)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f&lt;sup&gt;(n)&lt;/sup&gt;(x) = a&lt;sup&gt;n-1&lt;/sup&gt;e&lt;sup&gt;ax&lt;/sup&gt;(ax+n) &lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-08T17:30:42.227" Title="Induction to prove a simple formula for calculating the n-th derivative? " Tags="&lt;calculus&gt;&lt;induction&gt;" AnswerCount="2" />
  <row Id="9415" PostTypeId="2" ParentId="9414" CreationDate="2010-11-08T11:06:48.040" Score="3" ViewCount="0" Body="&lt;p&gt;We have $f^n(x)= a^{n-1}e^{ax}(ax+n)$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dx} f^n(x) = a^{n-1} \frac{d}{dx} \lbrace e^{ax}(ax+n) \rbrace&#xA;= a^{n-1} \left( ae^{ax}(ax+n) + ae^{ax} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= a^n e^{ax}(ax + n+1) = f^{n+1}(n).$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-08T11:06:48.040" />
  <row Id="9416" PostTypeId="1" CreationDate="2010-11-08T11:27:55.503" Score="1" ViewCount="104" Body="&lt;p&gt;Say I pass 512 samples into my FFT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My microphone spits out data at 10KHz,  so this represents 1/20s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(So the lowest frequency FFT would pick up would be 40Hz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FFT will return an array of 512 frequency bins&#xA; - bin 0: [0 - 40Hz)&#xA; - bin 1: [40 - 80Hz)&#xA;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if my original sound contained energy at say 115Hz,  how can I accurately retrieve this frequency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is going to lie in bin #2,  but very close to bin #3.   so I would expect both bins to contain something nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: how about the bins either side of this? Would they be guaranteed to be zero if there are no other frequencies close in the original signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main question:  is there some algorithm for deciphering the original frequency given the relative bin strengths?&lt;/p&gt;&#xA;" OwnerUserId="3096" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T11:36:27.200" LastActivityDate="2010-12-22T09:33:42.713" Title="Extracting exact frequencies from FFT output" Tags="&lt;fourier-analysis&gt;&lt;signal-processing&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9417" PostTypeId="1" CreationDate="2010-11-08T11:41:35.493" Score="3" ViewCount="141" Body="&lt;p&gt;Please find a closed form for partial sum of a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=a^{1/x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want it to be expressed in terms of bounded number of elementary functions and/or well known special functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No computer algebra systems I have tried so far could find a satisfactory solution. I believe that the expression can exist in the terms of incomplete Gamma function or its generalizations because indefinite integral  of this function can be expressed in terms of incomplete Gamma function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int f(x) dx= x\sqrt[x]{a}-\operatorname{Ei}\left(\frac{\ln a}{x}\right)\ln a$$&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-11-09T11:19:41.607" Title="Find a closed form for $\sum_{k=1}^{x-1} a^{1/k}$" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="9418" PostTypeId="2" ParentId="9412" CreationDate="2010-11-08T12:12:06.810" Score="6" ViewCount="0" Body="&lt;p&gt;I think it's great that you're seeking a deeper understanding of an important topic-- even after the class is over!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't done so, go back to your calculus text and carefully read the sections on Taylor and MacLaurin series.  (I know math textbooks often seem obtuse and unhelpful, but it's a valuable skill to be able to read slowly and learn with just the text as your guide.)  Pay close attention to the examples, because they are acutely chosen to help you see the ideas and connections you need.  If there's a gap you can't fill in, ask!  --the more detailed your question, the more direct our help will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that solving some problems by hand would make you more comfortable working with power series.  Here are some problems to try.  (It may take a lot of writing, but as long as you have one idea of what to do next, try it!  There's a lot to keep track of with power series, and practice is important.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove, by expanding as MacLaurin series, that $e^{ix} = \cos x + i \sin x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start with the MacLaurin series for $(\cos x)$ and $(\sin x)$, square them, then add to show that $\cos^2 x + \sin^2 x = 1$.  (To start, just work up to $x^4$, but go further if you can.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start with the MacLaurin series for $e^x$ and $\ln (1-x)$ and show that $e^{\ln (1-x)} = 1-x$.  (Again, just work up to $x^4$ or so.  You could also try the same thing with $e^{-\ln(1-x)}$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove, using MacLaurin series that $\frac{d}{dx} e^x = e^x$, $\frac{d}{dx} \sin x = \cos x$, and $\frac{d}{dx} \cos x = - \sin x$.  (For these, don't stop at $x^4$.  You can use summation notation to work with all terms, from 0 to infinity, at the same time.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Write the MacLaurin series for $\frac{1}{1-x}$, then take the integral of the series.  (What do you get?)  Take the derivative of the series.  (What do you get?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the MacLaurin series for $(\arctan x)$ by taking its derivative, finding the MacLaurin series expansion, then integrating that expansion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't have other references for you, but power series expansions can be used in solving limits (replacing L'Hospital's rule), in differential equations, in combinatorics (generating functions), and in complex analysis.  If any of those topics piques your interest, I'm sure people here or in your department can give some references to check out.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-08T12:12:06.810" CommentCount="1" />
  <row Id="9419" PostTypeId="2" ParentId="9395" CreationDate="2010-11-08T12:32:07.567" Score="2" ViewCount="0" Body="&lt;p&gt;The second part of the question asks for the cardinality of the set of integrable or square-integrable functions.  I will assume you actually mean &quot;functions&quot; rather than &quot;equivalence classes of functions under the relation of equality almost everywhere&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\beta$ be the cardinality of the set of all functions from $\mathbb{R}$ to $\mathbb{R}$; by standard set theory this is the same as the cardinality of the powerset of the real numbers: $\beta = 2^{|\mathbb{R}|} = 2^{2^{\aleph_0}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly the set of integrable functions, and the set of square integrable functions, can have cardinality no more than $\beta$. It turns out this is exactly the cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be a Cantor set; the key properties are that $|E| = |\mathbb{R}|$ and the measure of $E$ is $0$. Consider the set of all functions that are $0$ for every $x$ that is not in $E$. All of these functions are both integrable and square integrable, because the measure of $E$ is zero. The cardinality of this set is the cardinality of the set of functions from $E$ to $\mathbb{R}$, which is exactly $\beta$ because $|E| = |\mathbb{R}|$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-08T12:32:07.567" />
  <row Id="9420" PostTypeId="1" AcceptedAnswerId="9474" CreationDate="2010-11-08T12:40:32.783" Score="2" ViewCount="163" Body="&lt;p&gt;Suppose $h$ is a hash function, $h$ : { 0, 1 } &lt;sup&gt;*&lt;/sup&gt; $\rightarrow$ { 0, 1 } &lt;sup&gt;n&lt;/sup&gt; and for all $w_1$, $w_2$ it holds: $h(w_1 \oplus w_2) =&#xA;h(w_1) \oplus h(w_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;$\oplus$ is the XOR operation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't $h$ a cryptographically good hash function?&lt;/p&gt;&#xA;" OwnerUserId="2411" LastActivityDate="2010-11-09T23:47:11.487" Title="Why is hash function $h$ ($h(w_1 \oplus w_2) =  h(w_1) \oplus h(w_2)$) not good?" Tags="&lt;homework&gt;&lt;cryptography&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="9421" PostTypeId="2" ParentId="7481" CreationDate="2010-11-08T12:49:21.850" Score="1" ViewCount="0" Body="&lt;p&gt;In what follows I assume that a map $f:\ ]0,1[\to \mathbb R^2$ is wanted. $$ $$ There are continuous surjective maps $p:[0,1]\to[0,1]^2$; they are called Peano curves. One may choose $p$ in such a way that the &quot;curve&quot; begins at $(0,0)$ and ends at $(1,1)$. Therefore it is possible to construct a continuous map $f:\ ]0,1]\to\mathbb R^2$ such that the restrictions of $f$ to the intervals $[{1\over n+1},{1\over n}]$ are Peano curves covering an increasing sequence of squares $Q_n$ whose union is $\mathbb R^2$. At the end, the point $1$ may safely be omitted from the domain of $f$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-08T12:49:21.850" />
  <row Id="9422" PostTypeId="1" AcceptedAnswerId="9444" CreationDate="2010-11-08T12:53:07.197" Score="6" ViewCount="192" Body="&lt;p&gt;Could you provide a geometric explanation of taylor expansion? &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-11-08T15:32:30.663" Title="Intuition explanation of taylor expansion?" Tags="&lt;intuition&gt;&lt;taylor-expansion&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="9423" PostTypeId="1" CreationDate="2010-11-08T12:58:43.300" Score="3" ViewCount="47" Body="&lt;p&gt;I read a little remark in a topology text that the &lt;em&gt;direct&lt;/em&gt; image of an open set under a continuous mapping is not necessarily open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the definition of &lt;em&gt;direct&lt;/em&gt; image in this case? I tried googling to no avail, and only found references to sheaf theory on wikipedia, which I don't believe to be what I'm looking for. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2010-11-08T13:00:32.597" Title="Definition of direct image?" Tags="&lt;terminology&gt;&lt;definition&gt;" AnswerCount="1" />
  <row Id="9424" PostTypeId="2" ParentId="9423" CreationDate="2010-11-08T13:00:32.597" Score="3" ViewCount="0" Body="&lt;p&gt;They just mean &quot;image&quot;. An example here is the image of the set $(0, 2\pi)$ under the map $\sin(x)$, which is $(0, 1]$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-08T13:00:32.597" CommentCount="2" />
  <row Id="9425" PostTypeId="1" AcceptedAnswerId="9428" CreationDate="2010-11-08T13:05:32.693" Score="0" ViewCount="130" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine all positive integers &lt;strong&gt;x&lt;/strong&gt; and &lt;strong&gt;y&lt;/strong&gt; that satisfy the equation &#xA;&lt;strong&gt;x^2 + 2 = 5y&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T13:42:12.597" LastActivityDate="2010-11-08T14:45:14.250" Title="x^2 + 2 = 5y (being x and y positive integers)" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;diophantine-equations&gt;" AnswerCount="2" />
  <row Id="9426" PostTypeId="2" ParentId="9357" CreationDate="2010-11-08T13:05:59.287" Score="1" ViewCount="0" Body="&lt;p&gt;The integral is a priori defined as the area beneath a curve (assuming Riemann-Stieltjes integral).&#xA;Hence, you should not assume that it is the anti-derivative of a function,&#xA;that is $\int f dx $ has nothing AT ALL to do with $\int_a^b f dx$&#xA;the first expression means anti-derivative, the second means area under a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proof is supposed to give a connection between the two notions, antiderivative, and area under graph.&lt;/p&gt;&#xA;" OwnerUserId="934" LastActivityDate="2010-11-08T13:05:59.287" />
  <row Id="9427" PostTypeId="1" CreationDate="2010-11-08T13:08:51.010" Score="1" ViewCount="63" Body="&lt;p&gt;Given an $R$-algebra $S$, is $Hom_{R-Alg}(\prod_{i=1}^n R, S) = Hom_{S-Alg}(\prod_{i=1}^n S, S)$?&lt;/p&gt;&#xA;" OwnerUserId="3267" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-11-09T03:26:19.663" LastActivityDate="2010-11-09T03:26:19.663" Title="$Hom_{R-Alg}(\prod R, S)$" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="9428" PostTypeId="2" ParentId="9425" CreationDate="2010-11-08T13:13:31.150" Score="-1" ViewCount="0" Body="&lt;p&gt;Hint: When is $5y - 2$ a square? Modulo 5 this is $5y - 2 = 3$, but squares mod 5 are always 0, 1 or 4 so the equation has no solutions.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T14:10:17.227" CommentCount="1" />
  <row Id="9430" PostTypeId="2" ParentId="9378" CreationDate="2010-11-08T13:36:11.437" Score="4" ViewCount="0" Body="&lt;p&gt;David Einsenbud and Joe Harris' book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387986375&quot; rel=&quot;nofollow&quot;&gt;The Geometry of Schemes&lt;/a&gt; has a large number of really nice pictures which could help you gain a little bit of geometric intuition. That being said, it is also an excellent book which you should check out if you are interested in schemes.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-08T13:36:11.437" />
  <row Id="9431" PostTypeId="1" AcceptedAnswerId="9439" CreationDate="2010-11-08T13:36:35.807" Score="1" ViewCount="166" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-09T04:47:42.287" LastActivityDate="2010-11-09T04:47:42.287" Title="$n^2 + 3n +5$ is not divisible by $121$" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;modular-arithmetic&gt;&lt;divisibility&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9432" PostTypeId="1" CreationDate="2010-11-08T13:36:38.130" Score="1" ViewCount="74" Body="&lt;p&gt;The Clifford product of a pair of vectors $a,b$ is an associative operation defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ab = a \cdot b + a \wedge b.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sufficiently low dimensions I am used to being able to define the Clifford product on arbitrary $k$-vectors by repeatedly applying the vector definition.  For instance, suppose that I build a Clifford algebra over $\mathbb{R}^3$ with the usual (positive) Euclidean inner product.  Then I can easily write out the Clifford product of &lt;em&gt;any&lt;/em&gt; pair of basis bivectors.  For instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e_{12}e_{13} = e_1 e_2 e_1 e_3 = -e_2(e_1 e_1)e_3 = -e_2 (1) e_3 = -e_{23}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In four dimensions I get stuck, because it's possible that two of the indices don't &quot;cancel,&quot; and then I don't know how to apply the product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{12}e_{34} = e_1 e_2 e_3 e_4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do I go from here?  I strongly suspect that this equals just $e_{1234}$, but I don't know how to show (in an explicit, pedantic, algebraic way) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_1 e_2 e_3 e_4 = e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3268" LastActivityDate="2010-11-08T18:43:09.847" Title="How do I evaluate the Clifford product in dimensions greater than 3?" Tags="&lt;exterior-algebra&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="9434" PostTypeId="1" CreationDate="2010-11-08T13:56:16.213" Score="1" ViewCount="102" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working on a Stock Market Analysis website as a college project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was keen on knowing some algorithms which I could potentially use for analysing trends in the stock market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be glad if some one could help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You.&lt;/p&gt;&#xA;" OwnerUserId="3269" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T14:03:17.140" LastActivityDate="2010-11-08T23:21:36.050" Title="Examples for Algorithms that can be used for Stock Market Analysis" Tags="&lt;algorithms&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9435" PostTypeId="1" AcceptedAnswerId="9440" CreationDate="2010-11-08T14:00:00.710" Score="-3" ViewCount="104" Body="&lt;p&gt;I admit, the question is a little provocative, but is asked in all earnestness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is inspired by this one: &lt;a href=&quot;http://math.stackexchange.com/questions/11/does-99999-1&quot;&gt;Does .99999... = 1?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the arguments here point to the fact that 0.999... and 1 are arbitrarily close together. Put differently this means that &quot;there is no gap in-between the two&quot; - which concludes that these numbers are just the same one number: 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another field, concerning the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcendental_number&quot; rel=&quot;nofollow&quot;&gt;transcendental numbers&lt;/a&gt;, it is proved that they are complete. So again put differently there are &quot;no gaps in the number-line&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough this is not seen as a proof that all transcendental numbers are only just one number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I can't find a good explanation why &quot;no gaps&quot; here means &quot;is equal&quot; and there it means &quot;they are all different&quot; (=complete). Could anybody please enlighten me?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-08T14:38:27.567" Title="Does the &quot;no gaps&quot;-property with transcendental numbers mean that there is only &quot;one number&quot;?" Tags="&lt;transcendental-numbers&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="9436" PostTypeId="2" ParentId="9427" CreationDate="2010-11-08T14:03:28.343" Score="3" ViewCount="0" Body="&lt;p&gt;If $R$ is a commutative ring and $T$ an $R$-algebra, a morphism $\phi:R\times \cdots\times R\to T$ of $R$-algebras is in particular left $R$-linear, so there exist $a_1$, $\dots$, $a_n\in T$ such that $\phi(x_1,\dots,x_n)=x_1a_1+\cdots x_ny_n$. Since $\phi$ is unitary, $\phi(1,\dots,1)=1$, so $a_1+\cdots+a_n=1$. Let $e_i=\phi(0,\dots,0,1,0,\dots,1)$, with the one in the $i$th position. Then  multiplicativity implies that $e_i^2=e_i$ for all $i$ and $e_ie_j=0$ if $i\neq j$. This gives a concrete description of the set $\hom_{R\mathrm{-alg}}(R\times \cdots\times R, T)$, and you should be able to use it to answer your question.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-08T14:09:26.607" LastActivityDate="2010-11-08T14:09:26.607" />
  <row Id="9437" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:08:32.167" Score="5" ViewCount="0" Body="&lt;p&gt;I give it a try: If you want to know where you will be in x time driving a car you can find out by separating the different components: position at the moment, speed, acceleration, jolt and so on and add them all together.&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-08T14:08:32.167" CommentCount="1" />
  <row Id="9438" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:12:41.093" Score="6" ViewCount="0" Body="&lt;p&gt;Think of a Taylor series not as one entity but as a sequence of approximations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term gives a constant approximation: f(x + h) is approximately f(x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two terms give a linear approximation: f(x + h) is approximately f(x) plus a trend term, h f'(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first three terms include a constant approximation, a linear trend, and a curvature term to account for the change in the linear trend: f(x + h) is approximately f(x) + h f'(x) + h^2 f''(x)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you add a term to account for the change in the curvature, etc.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-08T14:12:41.093" />
  <row Id="9439" PostTypeId="2" ParentId="9431" CreationDate="2010-11-08T14:15:57.893" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ m\ =\ n^2 + 3\:n+5\ \equiv\ (n-4)^2\ \:(mod\ 11)\ \Rightarrow\ n\ =\ 4+11\:k \ \Rightarrow\ m = \ldots\ (mod\ 11^2)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T14:15:57.893" />
  <row Id="9440" PostTypeId="2" ParentId="9435" CreationDate="2010-11-08T14:22:31.000" Score="4" ViewCount="0" Body="&lt;p&gt;The property of the real numbers which formalizes these informal statements about gaps is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Least-upper-bound_property&quot; rel=&quot;nofollow&quot;&gt;least upper bound property&lt;/a&gt;.  It states that any set of real numbers that has an upper bound has a least upper bound, i.e. a supremum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $\{ .9, .99, .999, ... \}$ has an upper bound, so it has a least upper bound.  The least upper bound cannot be less than $1$, so it must be exactly $1$.  On the other hand, for any reasonable definition of $.999...$, this number is also an upper bound of the same set, and it is less than or equal to $1$.  So the two are equal (because least upper bounds are unique).  The &quot;gap&quot; which does not exist here is the nonexistent gap between two least upper bounds, and its nonexistence is a purely formal property given the existence of least upper bounds, so the nonexistence of this gap has nothing to do with completeness or the least upper bound property; it's the existence of the least upper bound that is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The least upper bound property implies that Cauchy sequences of real numbers converge, which is usually what people mean when they say that there are no &quot;gaps&quot; in the number line.  Another way to formalize this statement is that the real line is &lt;a href=&quot;http://en.wikipedia.org/wiki/Connected_space&quot; rel=&quot;nofollow&quot;&gt;connected&lt;/a&gt;, which implies, among other things, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate_value_theorem&quot; rel=&quot;nofollow&quot;&gt;intermediate value theorem&lt;/a&gt;.  (Compare with the case of $\mathbb{Q}$, which is totally disconnected.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-08T14:22:31.000" />
  <row Id="9441" PostTypeId="2" ParentId="8357" CreationDate="2010-11-08T14:36:03.183" Score="0" ViewCount="0" Body="&lt;p&gt;If $a &gt; b &gt; 0$ then $a^2 &gt; b^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a &gt; b$ means there is a positive $k$ such that $a = b + k$. Squaring this equation we have $a^2 = b^2 + (2bk + k^2)$ but $2bk + k^2$ is just another positive so $a^2 &gt; b^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we know $2bk + k^2$ is positive is because of the ordered field axioms, one says if $x$ and $y$ are positive so is $xy$ and another says that $x+y$ is positive. That is why we need $b$ to be positive.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T14:36:03.183" />
  <row Id="9442" PostTypeId="2" ParentId="9435" CreationDate="2010-11-08T14:38:27.567" Score="2" ViewCount="0" Body="&lt;p&gt;1/ To say that a topological set is &quot;complete&quot; has a mathematical meaning. For example, one definition of completeness is that all Cauchy-sequences converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number is not complete (or at least I have never heard of such a definition for numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that the real numbers are complete, because they satisfy the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ When we have a sequence, we can (sometimes) see if this sequence converge or not. Again, the notion of convergence has a mathematical meaning, and suppose that the sequence obey a certain definition. When a sequence converges, we name an element &quot;limit&quot;, this is an element that obey a certain definition too. It is straightforward from the definitions of convergence and of limit, that if a limit exists, it is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example we can study the sequence given by 0, 0.9, 0.99, 0.999, ... This sequence converges and has limit 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we can study another sequence $\frac{p_n}{q_n}$ given by the conditions $q_n = 10^n$ and $\left(\frac{p_n}{q_n}\right)^2 \leq 2 \leq \left(\frac{p_n+1}{q_n}\right)^2$. This sequence will also converge, and the limit will be $\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if observe that this last example is a sequence composed only by elements in $\mathbb{Q}$, we might be surprised to see that the limit is actually not in $\mathbb{Q}$. Mathematicians don't like the fact of not having the limit within the working topological set. Therefore they build $\mathbb{R}$ which is the set all (classes of equivalence of) Cauchy-sequences of elements in $\mathbb{Q}$. Hence $\mathbb{R}$ satisfy the definition of completeness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, saying that $\mathbb{R}$ is complete is just a matter of definition. It has nothing to do with the word &quot;gap&quot;. As far as I know, &quot;gap&quot; has no mathematical meaning.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-08T14:38:27.567" CommentCount="1" />
  <row Id="9443" PostTypeId="2" ParentId="9425" CreationDate="2010-11-08T14:45:14.250" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \  x^2 \equiv\ \ldots\ \:(mod\ 5)\ $ versus $\rm\ \mathbb Z^2\ \equiv\ \{0,\:\pm1,\:\pm2\}^2\ \equiv\ \ldots \ \:(mod\ 5)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T14:45:14.250" />
  <row Id="9444" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:48:13.990" Score="2" ViewCount="0" Body="&lt;p&gt;We know that the higher the degree of an equation, the more &quot;turning points&quot; it &lt;em&gt;may&lt;/em&gt; have. For example, a parabola has one &quot;turning point.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/D6fEV.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;(A parabola has an equation of the form $y=ax^2 + bx +c$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cubic of the form $y=ax^3 + bx^2 +cx +d$ can have up to two &quot;turning points,&quot; though it may have fewer. In general, an equation of degree $n$ may have up to $n-1$ turning points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/35376.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here is the polynomial $f(x) = 2x^4 - x^3 -3x^2 + 7x - 13$. It is degree 4 and it has the maximum number of turning points, 4-1=3. But, keep in mind, some degree 4 polynomials have only one or two turning points. The degree gives us the MAXIMUM number as $n=1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is important because, if you want to use a polynomial to approximate a function, you will want to use a polynomial of high enough degree to match the &quot;features&quot; of the function. The Taylor series will let you do this with functions that are &quot;infinitely differentiable&quot; since it uses the derivatives of the function to approximate the functions behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are Taylor polynomials of increasing degree and the sine curve. Notice how they are &quot;wrapping around&quot; the sine curve, giving an approximation that fits better and better over more of the curve as the degree of the Taylor polynomial increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/j9caf.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source for this image: &lt;a href=&quot;http://202.38.126.65/navigate/math/history/Mathematicians/Taylor.html&quot; rel=&quot;nofollow&quot;&gt;http://202.38.126.65/navigate/math/history/Mathematicians/Taylor.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sine curve has &lt;em&gt;so many&lt;/em&gt; turning points it is easy to see that to match all of the features of the sine curve we will need to take the limit of the $n^{th}$ degree Taylor polynomial as $n \rightarrow \infty$.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the intuition behind the Taylor series. The higher the degree, the better the &quot;fit.&quot;  Why? Because higher degree curve have more &quot;turning points&quot; so they can better match the shape of things like the sine function. (As long as the function we are approximating is differentiable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Side note: A function may have only a few turning points and still need infinitely many terms of the Taylor polynomial. Take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;nofollow&quot;&gt;catenary&lt;/a&gt;, for example, which only has one turning point since it looks like a parabola. The Taylor series for the catenary will not have any terms where the coefficients are zero, since the derivatives of the catenary are hyperbolic sinusoidal functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, even with the catenary, higher degree polynomials give a better approximation.   &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-08T15:01:34.370" LastActivityDate="2010-11-08T15:01:34.370" CommentCount="1" />
  <row Id="9446" PostTypeId="1" CreationDate="2010-11-08T14:57:14.103" Score="6" ViewCount="154" Body="&lt;p&gt;Why is every group the quotient of a free group by a normal subgroup?&lt;/p&gt;&#xA;" OwnerUserId="3272" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T15:43:10.620" LastActivityDate="2010-11-08T19:09:51.407" Title="Every group is the quotient of a free group by a normal subgroup" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9447" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T14:59:33.993" Score="0" ViewCount="0" Body="&lt;p&gt;We are approximating a function by polynomials at a point.As a first approximation, we give a give polynomial whose value at that point is same as the functions. In the second step, we make the first derivative equal too. In the third step,the second derivative is made equal and so on... &lt;/p&gt;&#xA;" OwnerUserId="3271" LastActivityDate="2010-11-08T14:59:33.993" />
  <row Id="9448" PostTypeId="2" ParentId="9432" CreationDate="2010-11-08T15:13:46.473" Score="2" ViewCount="0" Body="&lt;p&gt;It's not easy to give a proof without knowing exactly what your&#xA;definitions of the Clifford and exterior algebras are, so the&#xA;following argument is still a little handwaving. But let's say that we&#xA;somehow have defined what the exterior algebra of a vector space $V$&#xA;is; we know that its elements are multivectors, and the rule which&#xA;generates everything is that $x \wedge y + y \wedge x=0$ if $x$ and&#xA;$y$ are vectors in $V$. The Clifford algebra has the same&#xA;elements as the exterior algebra, and the same linear space structure,&#xA;but the multiplication is different: it is generated by the rule&#xA;$xy+yx=Q(x,y)$ where $Q$ is the inner product on $V$. Since&#xA;&quot;orthogonal&quot; means that $Q(e_i,e_j)=0$, it shouldn't be too hard to&#xA;believe that the Clifford and exterior multiplications agree for&#xA;orthogonal vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe &lt;a href=&quot;http://en.wikipedia.org/wiki/Clifford_algebra#Relation_to_the_exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;this section on Wikipedia&lt;/a&gt; can be of some help too?&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-08T16:04:53.183" LastActivityDate="2010-11-08T16:04:53.183" />
  <row Id="9449" PostTypeId="2" ParentId="9422" CreationDate="2010-11-08T15:32:30.663" Score="1" ViewCount="0" Body="&lt;p&gt;Predict global while computing local!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Taylor expansion of a function $f$ around some value $x_0$ is similar to a prediction of the function at a neighboring value $x$ knowing progressively more about the variation of $f$ at the point $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First step: easiest prediction: nothing changed, that is, $f(x) = f(x_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second step: we know the first derivative, so we predict the function was linear between $x_0$ and $x$ : $f(x) = f(x_0) + (x-x_0)f'(x_0)$. See, everything still local as the derivate is given at $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step give a generalization of this predictions for higher derivatives. The different forms give bounds to the error or more knowledge of the residual.&lt;/p&gt;&#xA;" OwnerUserId="1090" LastActivityDate="2010-11-08T15:32:30.663" />
  <row Id="9450" PostTypeId="2" ParentId="9446" CreationDate="2010-11-08T15:42:52.493" Score="6" ViewCount="0" Body="&lt;p&gt;Given any group G, let F(G) be the free group whose generators are the elements of G. Define a group homomorphism f from F(G) to G in the obvious way: by sending the generator g of F(G) to the element g of G. This is a surjective homomorphism, so G is isomorphic to F(G) / ker f.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, let M be the set of words of the form $g_1 g_2 h^{-1}$ where $g_1 g_2 = h$ in G -- in other words, M encodes the multiplication table for G. Let N be the normal closure of M. Then G is isomorphic to F(G) / N.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-08T15:42:52.493" />
  <row Id="9451" PostTypeId="2" ParentId="7892" CreationDate="2010-11-08T15:43:35.800" Score="4" ViewCount="0" Body="&lt;p&gt;From &lt;em&gt;&lt;a href=&quot;http://books.google.co.uk/books?id=Kx2cjyzTIYkC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+without+words&amp;amp;hl=en&amp;amp;ei=7RnYTLrRHZnc4wbR-on9Bw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDEQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs without Words&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0SBs6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-08T16:11:57.947" LastActivityDate="2010-11-08T16:11:57.947" CommentCount="1" />
  <row Id="9452" PostTypeId="2" ParentId="9446" CreationDate="2010-11-08T15:47:13.820" Score="9" ViewCount="0" Body="&lt;p&gt;It is a consequence of the universal property of free groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is a set, then the &lt;strong&gt;free group on&lt;/strong&gt; $X$, $F(X)$, is a group $F(X)$ that contains $X$ as a subset, and such that for every group $G$ and every set-theoretic map $f\colon X\to G$ there exists a unique group homomorphism $\mathcal{F}\colon F(X)\to G$ such that $\mathcal{F}(x) = f(x)$ for all $x\in X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is a group, let $X$ be the underlying set of $G$ (the set of all elements of $G$, considered merely as a set), and let $F(X)$ be the free group on $X$. The set-theoretic map $i\colon X\to G$ that maps each element of $G$ to itself induces a homomorphism $\mathcal{F}\colon F(X)\to G$ such that $\mathcal{F}(x)=i(x)$ for all $x\in X$. Since the image of $\mathcal{F}$ includes at least the image of $i$, the map $\mathcal{F}$ is onto. By the First Fundamental Theorem of Homomorphisms, the image of $\mathcal{F}$ (which is $G$) is isomorphic to $F(X)/\mathrm{ker}(\mathcal{F})$. So $G$ is isomorphic to the quotient of a free group by a normal subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can replace the underlying set of $G$ with any generating set of $G$ and the proof goes through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Of course, this answer requires that you know that a free group exists on any set, and that it has the appropriate universal property. A good exposition, with motivation and three different constructions of the free group, is in &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;Chapter 2 of George Bergman's book&lt;/a&gt;, &lt;em&gt;An invitation to General Algebra and Universal Constructions&lt;/em&gt;. Here's a link to a &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/Ch.2.ps&quot; rel=&quot;nofollow&quot;&gt;Postscript version&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T17:14:41.037" LastActivityDate="2010-11-08T17:14:41.037" />
  <row Id="9453" PostTypeId="2" ParentId="8357" CreationDate="2010-11-08T16:06:32.617" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, when &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; are positive real numbers. In this case, we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&gt;b =&gt; a-b&gt;0 =&gt; (a+b).(a-b)&gt;0 =&gt; (a^2)-(b^2)&gt;0&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-08T16:06:32.617" />
  <row Id="9454" PostTypeId="2" ParentId="9372" CreationDate="2010-11-08T16:08:10.400" Score="5" ViewCount="0" Body="&lt;p&gt;Further to the other answers posted, not only does the sum converge, but you can conclude that it converges precisely because of the way it was derived. As you state, the terms $C_n=\frac{1\cdot3\cdot5\cdot7\cdots(2n-3)}{2^nn!}$ arise as the Taylor expansion of $f(x)=\sqrt{x}$ about $x=1$,&#xA;$$&#xA;\sqrt{1-x}=1-\frac{x}{2}-\sum_{n=2}^\infty C_nx^n.&#xA;$$&#xA;As $\sqrt{1-x}$ is a well-defined analytic function for $\Vert x\Vert &amp;lt; 1$, this has radius of convergence 1. Also, as the square root is well defined at 0, &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Monotone_convergence_theorem&amp;amp;oldid=388896065#Convergence_of_a_monotone_series&quot; rel=&quot;nofollow&quot;&gt;monotone convergence&lt;/a&gt; gives&#xA;$$&#xA;\sum_{n=2}^\infty C_n=\lim_{x\to1}\sum_{n=2}^\infty C_nx^n=\lim_{x\to1}\left(1-\frac{x}{2}-\sqrt{1-x}\right)=\frac12.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-11-08T16:08:10.400" />
  <row Id="9455" PostTypeId="1" CreationDate="2010-11-08T16:27:02.170" Score="0" ViewCount="83" Body="&lt;p&gt;If G is a finite nilpotent group, then every minimal normal subgroup of $G$ is contained in the center of $G$ and has prime order.&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-08T16:59:36.290" LastActivityDate="2010-11-08T17:02:28.897" Title="A question about nilpotent group" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9456" PostTypeId="2" ParentId="9402" CreationDate="2010-11-08T17:01:36.917" Score="4" ViewCount="0" Body="&lt;p&gt;Since &#xA;$$\frac{x}{1+x^2}=\int_{0}^{\infty}e^{-y}\sin (xy)dy,$$&#xA;we have that&#xA;$$I=\int_{0}^{\infty}\frac{\cos bx}{1+x^2}dx=\int_{0}^{\infty}\frac{\cos bx}{x}dx\int_{0}^{\infty}e^{-y}\sin (xy)dy.$$&#xA;Changing the order of integration (which can be justified by the standard method) yields&#xA;$$I=\int_{0}^{\infty}e^{-y}dy\int_{0}^{\infty}\frac{\sin xy}{x} \cos bx dx.$$&#xA;The calculation of the integral (a.k.a. the discontinuous Dirichlet factor)&#xA;$$\int_{0}^{\infty}\frac{\sin xy}{x} \cos bx dx=\begin{cases}0, &amp;amp; 0 &amp;lt; y &amp;lt; b&#xA;\\ \ \\ \pi/2, &amp;amp; 0 &amp;lt; b &amp;lt; y, \end{cases}$$&#xA;can be easily reduced to the calculation of the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_integral&quot; rel=&quot;nofollow&quot;&gt;Dirichlet integral&lt;/a&gt;.&#xA;Therefore,&#xA;$$I=\frac{\pi}{2}\int_{b}^{\infty}e^{-y}dy=\frac{\pi}{2}e^{-b}.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-08T17:01:36.917" CommentCount="3" />
  <row Id="9457" PostTypeId="2" ParentId="9455" CreationDate="2010-11-08T17:02:28.897" Score="3" ViewCount="0" Body="&lt;p&gt;A finite nilpotent group is a product of $p$-groups. So you can do a very quick computation to show that you can reduce to the case where $G$ is a $p$-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then look at &lt;a href=&quot;http://math.stackexchange.com/questions/8389/a-question-about-p-group/8400#8400&quot;&gt;this question&lt;/a&gt; to answer Steve D's comment query. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-08T17:02:28.897" CommentCount="2" />
  <row Id="9458" PostTypeId="2" ParentId="9414" CreationDate="2010-11-08T17:30:42.227" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad\quad\ \  (f_0\:e^{\:g})^{(n)}\ \ \ \ =\ \ f_n\ e^g\ \: $ when $\rm\: \ g'\ f_n + f_n'\: =\ f_{n+1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Base case:&lt;/strong&gt; $\rm\quad\ (f_0\:e^{\:g})^{(0)}\ \ \ \ =\ \ f_0\ e^g\ \ $ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction:&lt;/strong&gt; $\rm\quad (f_0\:e^{\:g})^{(n+1)} = \ (f_n\ e^g)'\ \:=\  (g'\ f_n + f_n')\ e^g \ =\ f_{n+1}\ e^g\quad\ $  &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\rm\ g = ax,\ f_{n+1} = a^n\:(ax+n+1)\ =\ a\ (a^{n-1} (ax + n))+ a^n\ =\ g'\ f_n + f_n'$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-08T17:30:42.227" />
  <row Id="9459" PostTypeId="2" ParentId="7892" CreationDate="2010-11-08T17:31:15.737" Score="3" ViewCount="0" Body="&lt;p&gt;Another proof uses the fact that $\displaystyle \pi &gt; e$ and that $e^x &gt; 1 + x$ for $x &gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $$e^{\pi/e -1} &gt; \pi/e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{\pi/e} &gt; \pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{\pi} &gt; \pi^e$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-08T17:31:15.737" />
  <row Id="9460" PostTypeId="1" AcceptedAnswerId="9490" CreationDate="2010-11-08T17:37:54.460" Score="2" ViewCount="135" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g: \mathbb{C}\times[a,b]\to\mathbb{R}$ a continuous function and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(t)=g(h_0,t)=\lim_{h\to h_0} g(h,t) \ \forall t\in \mathbb{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following result/reasoning correct ? If we have a sequence of functions, we need uniform convergence to pass the limit inside the integral, but it seems it works everytime here. Or there is a flaw ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{h\to h_0} \int_a^b g(h,t)dt= \int_a^b g(t)dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Idea 1:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\forall t\in[a,b], \ \forall&#xA;  \epsilon &gt;0, \exists \delta(t)&gt;0\text{&#xA;  such that }|g(h,t)-g(t)|&amp;lt;\epsilon \&#xA;  \forall |h-h_0|&amp;lt;\delta(t)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Then if&#xA;  $|h-h_0|&amp;lt;\delta=\inf_{t\in[a,b]}&#xA;  \delta(t)$, &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$|g(h,t)-g(t)|&amp;lt;\epsilon \ \forall&#xA;  |h-h_0|&amp;lt;\delta(t), \ \forall&#xA;  t\in[a,b]$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\left|\int_a^b&#xA;  (g(h,t)-g(t))dt\right|\le \int_a^b&#xA;  |g(h,t)-g(t)|dt\le (b-a)\epsilon$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Doesn't work if $\inf_{t\in[a,b]}&#xA;  \delta(t)=0$, see George's comment.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;~&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Idea 2:&lt;/strong&gt; (added after George's comment)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let $\overline{B(h_0, \gamma)}\subset&#xA;  C$ a closed ball. Resctrict $g$ to $&#xA;  \overline{B(h_0, \gamma)}\times[a,b]$.&#xA;  Because it is continuous on a compact,&#xA;  this function will be uniformly&#xA;  continuous. Thus&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\forall \epsilon &gt;0 \ \exists&#xA;  \delta&gt;0 \ : \&#xA;  |g(h,t)-g(h_0,t)|&amp;lt;\epsilon \ \forall&#xA;  |h-h_0|&amp;lt;\min(\delta, \gamma)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;and&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\left|\int_a^b&#xA;  (g(h,t)-g(t))dt\right|\le \int_a^b&#xA;  |g(h,t)-g(t)|dt\le (b-a)\epsilon$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;if $|h-h_0|&amp;lt;\min(\delta, \gamma)$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Does this one work ;) ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help :)&lt;/p&gt;&#xA;" OwnerUserId="524" LastEditorUserId="524" LastEditorDisplayName="" LastEditDate="2010-11-08T19:50:02.707" LastActivityDate="2010-11-08T21:59:50.420" Title="Limit of integral" Tags="&lt;real-analysis&gt;&lt;integral&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9461" PostTypeId="1" AcceptedAnswerId="9465" CreationDate="2010-11-08T18:08:21.630" Score="1" ViewCount="106" Body="&lt;p&gt;A set M $\in$ $\mathbb{Z}^2$ is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Rule 1) $(0, 1)$ $\in$ M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Rule 2) If $(x, y)$ $\in$ M, then $(x + 1, y + 2x + 3)$ $\in$ M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) (2 points) Starting with $(0, 1)$, write out the six pairs with the smallest first coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) (2 points) State the (simple) relationship that holds between the first and second coordinates of all pairs in M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok for a I got:&#xA;$(0,1)$,$(1,4)$,$(2,9)$,$(3,16)$,$(4,25)$,$(5,36)$,$(6,49)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for b, I cant see the relationship. Can anyone give me any hints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok I found the relation: $y=(x+1)^2$. However, I must now prove this claim using structural induction. I can get up to the inductive hypothesis, but after that I'm not really sure what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base case: $(0,1), 1=(0+1)^2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive Hypothesis: For any element $x,y$ if $x\in M$ then $y=(x+1)^2$. We must show that $y+1=(x+1)^2 + 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure that my IH is correct either.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2010-11-08T18:45:48.473" LastActivityDate="2010-11-08T19:33:58.363" Title="Find a pattern in a recursive definition" Tags="&lt;number-theory&gt;&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;number&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9462" PostTypeId="2" ParentId="9446" CreationDate="2010-11-08T18:23:09.907" Score="4" ViewCount="0" Body="&lt;p&gt;This is one of the most intuitive observations in all of group theory, and it illustrates the quotient operation in the most fundamental way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll provide two separate answers. The first is fully intuitive; the second is a formalized version of the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First answer:&lt;/strong&gt; Take a group $G$. A relation on $G$ is an equation satisfied by some of the elements. For instance, $eg = g$ where $e$ is the identity is a relation satisfied by all group elements $g \in G$. Because we can always multiply by inverses in a group, we can rewrite this relation as $egg^{-1} = gg^{-1} = e$, i.e., $e = e$. This can be applied to any relation. If $G$ is abelian, then $ab = ba$ for all $a,b \in G$, and we can rewrite this as $aba^{-1}b^{-1} = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a relation asserts that some product of group elements coincides with the identity, so the only information we need to understand the relation is the product which occurs on the left side of the equals sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now every group has a few relations which are implied directly by the group axioms. $aa^{-1} = e$ is one of them. We can ask whether the group has any &lt;em&gt;extra&lt;/em&gt; relations which are not implied by the group axioms. If no such relations exist, i.e., if the only relations which hold are those which must hold by virtue of the group axioms, then the group is said to be &lt;strong&gt;free&lt;/strong&gt;; the group is &quot;free of additional relations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a group $G$, one natural thing to do is to introduce new relations into it and to thereby create new groups. But you can't just introduce completely random relations because (a) the relations can't contradict each other or pre-exising relations and (b) the resulting structure must again be a group. Now we saw earlier that a relation can be specified as a product of group elements. In order that the relations satisfy (a) and (b), it turns out it is necessary and sufficient that the corresponding products form a &lt;em&gt;normal subgroup&lt;/em&gt; $N$. The result of introducing the collection of relations $N$ into the group $G$ is the quotient $G/N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any group $G$ can be obtained in this manner. You start with the free group $F$ whose generators are elements of $G$ &lt;em&gt;considered as a set&lt;/em&gt;. And then you look at all the additional relations satisfied by elements of $G$ and assemble them into a normal subgroup $N$. Then $G = F/N$ by the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second answer:&lt;/strong&gt; Given any set $S$, the free group on $S$ is that group $F(S)$ for which every function $f : S \rightarrow G$ from $S$ to an arbitrary group $G$ extends to a unique &lt;em&gt;homomorphism&lt;/em&gt; $\tilde{f} : F(S) \rightarrow G$. There are various ways of constructing $F(S)$ explicitly. For instance, you may take $F(S)$ to consist of words over the alphabet whose letters are elements of $S$ and $S'$, where $S'$ has the letter $s^{-1}$ (a symbol at the moment) for each symbol $s \in S$. It's important to notice that $F(S)$ actually contains &lt;em&gt;equivalence classes&lt;/em&gt; of words, because we introduce the obvious cancellation rules; e.g., $abb^{-1}c$ can be reduced via cancellation to $ac$. It must be proved that all possible algorithms for reduction yield the same reduced word; I'll omit that step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to prove that this group $F(S)$ satisfies the stated universal property. I won't prove this in detail, but it is more or less intuitive. Since $\tilde{f}$ has to be a homomorphism, we find, for instance, that $\tilde{f}(ab) = \tilde{f}(a) \tilde{f}(b) = f(a)f(b)$. In general, since $f$ is defined for all elements of $S$, $\tilde{f}$ is thereby defined uniquely for all elements of $F(S)$. [It is via similar reasoning that you may determine that it is sufficient to know the values of a linear operator on the elements of a basis of a vector space.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we start with our group $G$ which we would like to write as a quotient of a free group. Which free group? That free group whose generators are the symbols from $G$. So we pick $F(G)$. Now we need to introduce the needed relations in order to collapse $F(G)$ into $G$. How do we carry it out? By the first answer, we could easily accomplish this if only we knew the normal subgroup $N$ of relations, but it seems that in this general case we don't really know $N$ concretely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we can figure out $N$ as follows. We can take the identity map $f : G \rightarrow G$ and extend it to a homomorphism $\tilde{f} : F(G) \rightarrow G$. The extension $\tilde{f}$ is in general not injective, and its kernel is precisely the group of relations $N$! (Formally this is an application of one of the standard theorems on homomorphisms.) Then $G = F(G)/N$ as before.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-11-08T19:09:51.407" LastActivityDate="2010-11-08T19:09:51.407" CommentCount="2" />
  <row Id="9463" PostTypeId="2" ParentId="9432" CreationDate="2010-11-08T18:43:09.847" Score="2" ViewCount="0" Body="&lt;p&gt;It appears that you are thinking in terms of &quot;geometric algebra&quot;.&#xA;For those unfamliar with this doctrine, it's a way of regarding&#xA;the Clifford algebra on an inner product space $(V,Q)$ and the exterior algebra&#xA;on the $V$ as the same set, but with different product operations.&#xA;Let's stick to a ground field of characteristic zero. Then the Clifford&#xA;algebra $C$ of $(V,Q)$ is generated by the elements $v\in V$ with relations&#xA;$vv=Q(v)$. Now for $v_1,\ldots,v_k\in V$ one can define wedge product&#xA;$$v_1\wedge v_2\wedge\cdots\wedge v_k&#xA;=\frac1{k!}\sum_{\pi\in S_k}(-1)^{\text{sgn}(\pi)}v_{\pi(1)}v_{\pi(2)}\cdots&#xA;v_{\pi(k)}\in C.$$&#xA;This wedge product identifies $C$ with the exterior algebra $\bigwedge(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this definition, Hans's comments are absolutely right. If&#xA;$v_1,\ldots,v_k$ are pairwise orthogonal then the Clifford product&#xA;$v_1v_2\cdots v_k$ equals the exterior product&#xA;$v_1\wedge v_2\wedge\cdots\wedge v_k$, and this is certainly the case&#xA;for orthogonal basis vectors: $e_1 e_2 e_3 e_4&#xA;=e_1\wedge e_2\wedge e_3\wedge e_4$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-08T18:43:09.847" />
  <row Id="9464" PostTypeId="2" ParentId="9427" CreationDate="2010-11-08T18:44:35.647" Score="1" ViewCount="0" Body="&lt;p&gt;Tensor products commute with finite products (of modules, but the same isom. gives it for rings), in particular $(\prod_{i=1}^{n} R) \otimes_R S = \prod_{i=1}^{n} S$.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-08T18:44:35.647" CommentCount="4" />
  <row Id="9465" PostTypeId="2" ParentId="9461" CreationDate="2010-11-08T18:47:43.383" Score="1" ViewCount="0" Body="&lt;p&gt;Your inductive step is to show that if $(x,(x+1)^2) \in M$, then $(x+1,(x+2)^2) \in M$.  So apply Rule 2 to $(x,(x+1)^2)$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-08T19:04:11.107" LastActivityDate="2010-11-08T19:04:11.107" CommentCount="3" />
  <row Id="9466" PostTypeId="2" ParentId="9461" CreationDate="2010-11-08T18:52:51.660" Score="1" ViewCount="0" Body="&lt;p&gt;Your inductive hypothesis is wrong. You should be showing that if $(x,(x+1)^2)\in M$, then $(x+1, ((x+1)+1)^2) \in M$. From there you should be able to see how to get the result (just use Rule 2).&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-08T18:58:25.373" LastActivityDate="2010-11-08T18:58:25.373" />
  <row Id="9467" PostTypeId="2" ParentId="9461" CreationDate="2010-11-08T19:04:33.643" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ f\ (x,\:y)\ = \ (x+1,\ y+2x+3)\ \Rightarrow\ f\ (x,\ (x+1)^2)\ =\ (x+1,\ (x+2)^2) $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-08T19:33:58.363" LastActivityDate="2010-11-08T19:33:58.363" />
  <row Id="9468" PostTypeId="1" AcceptedAnswerId="9506" CreationDate="2010-11-08T19:12:31.090" Score="5" ViewCount="154" Body="&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/CatalansConstant.html&quot; rel=&quot;nofollow&quot;&gt;Quite&lt;/a&gt; a  &lt;a href=&quot;http://mathworld.wolfram.com/MertensConstant.html&quot; rel=&quot;nofollow&quot;&gt;few&lt;/a&gt; mathematical constants are known to arise in several branches of mathematics (more &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_product&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). I have no doubt that they are useful to the understanding of some mathematical structures and add to the whole body of knowledge of mathematics as a whole. Sometimes, however, I find myself looking at these constants and I feel unsatisfied. I suspect that more constants can be expressed in more 'fundamental' constants like $\pi, e$ and $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Euler probably came close to be a living one, mathematicians of the past did not have computers. We can now use computers to determine te value of the afore mentioned constants with great precision. We can also try to conjecture the exact value of these constants by finding the value of some arithmetic combination of $\pi, e$ and $\gamma$ and the real numbers that corresponds to one of the 'unevaluated' constants with high precision. My question is: have experimental mathematicians 'found' the value(s) of some of the constants I described (by the the method I described or a similar one)?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. If someone knows of a reference to a paper/book that summarizes some results in this subield of a field, I would be grateful to him/her.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-19T15:46:30.380" LastActivityDate="2010-11-19T16:12:56.710" Title="Suggesting closed-form representations of mathematical constants by means of experimental mathematics?" Tags="&lt;number-theory&gt;" AnswerCount="5" CommentCount="7" />
  <row Id="9470" PostTypeId="2" ParentId="9468" CreationDate="2010-11-08T19:18:58.530" Score="2" ViewCount="0" Body="&lt;p&gt;Have you tried the &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html&quot; rel=&quot;nofollow&quot;&gt;Inverse Symbolic Calculator&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T19:18:58.530" CommentCount="7" />
  <row Id="9471" PostTypeId="2" ParentId="457" CreationDate="2010-11-08T19:28:31.727" Score="1" ViewCount="0" Body="&lt;p&gt;Well, one Stokes theorem surely deserves to be viewed in physical terms. It's hard not to think about these matters in terms of physical flows (be it fluid, or electrical fields), sinks, sources, etc. and this can give intuition when dealing with arbitrary differential forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another physical concept often used in abstract mathematics is conservation of energy, or more generally integrals of motion. Used all the time in analysis of PDE, as far as I know.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-08T19:28:31.727" CommunityOwnedDate="2010-11-08T19:28:31.727" />
  <row Id="9472" PostTypeId="2" ParentId="9468" CreationDate="2010-11-08T19:33:26.010" Score="3" ViewCount="0" Body="&lt;p&gt;The rational constant for an Apery like series for $$\zeta(4)$$ was found experimentally, using continued fractions.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T19:33:26.010" CommentCount="2" />
  <row Id="9473" PostTypeId="2" ParentId="9468" CreationDate="2010-11-08T19:38:19.547" Score="0" ViewCount="0" Body="&lt;p&gt;Gauss discovered the AGM/elliptic integral connection through the value $$\frac{\omega}{\pi}$$ by experimental mathematics.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-08T19:38:19.547" />
  <row Id="9474" PostTypeId="2" ParentId="9420" CreationDate="2010-11-08T19:52:26.433" Score="2" ViewCount="0" Body="&lt;p&gt;The function $h$ violates the one-wayness property of cryptographically good hash functions, because it's not computationally infeasible to recover the message $w$ from the hash $h(w)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can generate (obtain) a set of message-hash pairs, then we can use the XOR operations on some of the hashes to get the $h(w)$ hash. If we use the same XORing procedure on the corresponding messages, we can recover the message $w$ as well. The whole process is computationally feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Kudos to &lt;a href=&quot;http://math.stackexchange.com/users/232/qiaochu-yuan&quot;&gt;Qiaochu Yuan&lt;/a&gt; for the help.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(be free to comment or edit)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2411" LastActivityDate="2010-11-08T19:52:26.433" />
  <row Id="9475" PostTypeId="1" AcceptedAnswerId="9477" CreationDate="2010-11-08T20:03:18.823" Score="4" ViewCount="138" Body="&lt;p&gt;The usual symbol for the cardinality of the continuum (i.e. the real numbers) is Fraktur $\mathfrak{c}$. However, I recall some sources also using $\aleph$ (with no subscript). This usage is not mentioned in Wikipedia or Mathworld, but I found some support for it over Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the $\aleph$ notation standard?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-08T20:31:24.067" Title="Symbol for the cardinality of the continuum" Tags="&lt;set-theory&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="9477" PostTypeId="2" ParentId="9475" CreationDate="2010-11-08T20:12:22.290" Score="2" ViewCount="0" Body="&lt;p&gt;I have seen the use of $\aleph$ and $\mathfrak{c}$ as well the explicit $2^{\aleph_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're uncertain, it's best just to add &quot;We denote the cardinality of the continuum by ...&quot;.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-11-08T20:12:22.290" />
  <row Id="9479" PostTypeId="2" ParentId="9319" CreationDate="2010-11-08T20:28:43.747" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose you want to find positive integers $n, k$ such that $|n - 2 \pi k - \arcsin r| &amp;lt; \frac{1}{m}$ for some large positive integer $m$ and $r \in [-1, 1]$.  Then I claim that some value of $n$ less than $2\pi m$ works.  This is the pattern I was trying to get you to see, although I wasn't doing a very good job of it (I should've asked you to plot $k$ instead of $n$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I think this is a little hard to prove except when $r = 0$; fortunately, using the second half of Marek's argument you can prove the same result with a weaker bound on $n$ for general $r$ using the case of $r = 0$, so I would encourage you to try that case first.  To do this, try plotting the fractional parts of the numbers $\frac{n}{2\pi}$ in $[0, 1)$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-08T20:28:43.747" CommentCount="1" />
  <row Id="9480" PostTypeId="2" ParentId="9475" CreationDate="2010-11-08T20:31:24.067" Score="3" ViewCount="0" Body="&lt;p&gt;I have never seen $\aleph$ used without a subscript in any treatise of Set Theory.&lt;/p&gt;&#xA;" OwnerUserId="3195" LastActivityDate="2010-11-08T20:31:24.067" />
  <row Id="9481" PostTypeId="2" ParentId="6981" CreationDate="2010-11-08T20:33:25.327" Score="1" ViewCount="0" Body="&lt;p&gt;The complex exponential solutions make sense for linear ODEs because you can get a characteristic polynomial since the exponential is the eigenfunction of the derivative operator. The reason why a complex solution times any constant is still a complex solution is due to the linearity in the linear ODE. If $x(t)$ is a solution and $i*y(t)$ is a solution then any linear combination is also a solution. Additivity and homogeneity are not properties of the solutions of a non-linear ODE and so it makes no sense to assume that the solution is of the form $e^{st}$ .&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-11-08T20:33:25.327" />
  <row Id="9482" PostTypeId="1" AcceptedAnswerId="9485" CreationDate="2010-11-08T20:44:09.120" Score="1" ViewCount="87" Body="&lt;p&gt;The implicit expression $(b-a)=(a+b)^3$ looks like it could be written explicitly for $a$ as a function of $b$. The only region of interest is for $a,b&gt;0$ Here is what the plot looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jPQ4Z.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2256" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-08T21:17:07.270" LastActivityDate="2010-11-08T21:20:11.760" Title="Express implicit equation explicitly" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9483" PostTypeId="1" AcceptedAnswerId="9805" CreationDate="2010-11-08T20:44:39.367" Score="2" ViewCount="92" Body="&lt;p&gt;Let $(X, \prec)$ and $(Y, &amp;lt;)$ be two well-ordered sets. I want to show that if $X$ is isomorphic to a subset of Y then $X$ is isomorphic with an initial segment of $Y$. (The other direction is of course true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some hints. Does this require the recursion theorem to construct the isomorphism?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-11T02:40:05.730" Title="Order-isomorphic with a subset iff order-isomorphic with an initial segment." Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="9485" PostTypeId="2" ParentId="9482" CreationDate="2010-11-08T21:05:42.730" Score="5" ViewCount="0" Body="&lt;p&gt;Writing $p=a+b$ you have the cubic equation,&#xA;$$&#xA;p^3+p-2b=0.&#xA;$$&#xA;This is already in &quot;depressed cubic&quot; form (no $p^2$ term), so it can be solved directly by &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Cubic_function&amp;amp;oldid=395526018#Roots_of_a_cubic_function&quot; rel=&quot;nofollow&quot;&gt;standard methods&lt;/a&gt;. The coefficient of p is positive, so it is strictly increasing and there will be a single real root.&#xA;$$&#xA;p = \sqrt[3]{\sqrt{b^2+1/27}+b}-\sqrt[3]{\sqrt{b^2+1/27}-b}&#xA;$$&#xA;or,&#xA;$$&#xA;a = \sqrt[3]{\sqrt{b^2+1/27}+b}-\sqrt[3]{\sqrt{b^2+1/27}-b}-b.&#xA;$$&#xA;Alternatively, using the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Cubic_function&amp;amp;oldid=395526018#Trigonometric_.28and_hyperbolic.29_method&quot; rel=&quot;nofollow&quot;&gt;hyperbolic method&lt;/a&gt;,&#xA;$$&#xA;a=\frac{2}{\sqrt{3}}\sinh\left(\frac13\sinh^{-1}\left(3\sqrt{3}b\right)\right)-b.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-08T21:13:18.303" LastActivityDate="2010-11-08T21:13:18.303" CommentCount="13" />
  <row Id="9488" PostTypeId="1" AcceptedAnswerId="9497" CreationDate="2010-11-08T21:44:51.287" Score="2" ViewCount="259" Body="&lt;p&gt;Inspired by the recent post &quot;Limit of integral&quot;, I propose the following problem (hoping it will not turn out to be too easy). Suppose that $g:[0,1] \times [0,1] \to {\bf R}$ is continuous in both variables separately. Is it true that, for all $x_0 \in [0,1]$,&#xA;$$&#xA;\mathop {\lim }\limits_{x \to x_0 } \int_0^1 {g(x,y)\,{\rm d}y}  = \int_0^1 {g(x_0 ,y)\,{\rm d}y} .&#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-12T06:30:18.900" Title="Limit of integral - part 2" Tags="&lt;real-analysis&gt;&lt;integral&gt;&lt;limit&gt;" AnswerCount="3" />
  <row Id="9490" PostTypeId="2" ParentId="9460" CreationDate="2010-11-08T21:59:50.420" Score="1" ViewCount="0" Body="&lt;p&gt;So, the second idea was the right one: considering $g$ on a compact to have it uniformly continuous. See my message above.&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2010-11-08T21:59:50.420" CommentCount="0" />
  <row Id="9491" PostTypeId="1" AcceptedAnswerId="9492" CreationDate="2010-11-08T22:02:14.007" Score="4" ViewCount="99" Body="&lt;p&gt;If $a$ and $b$ are positive real numbers, and if $f(x)$ has the following asymptotic property&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then is the following true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{a}{b} + O(\frac{1}{\sqrt{x}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might look like homework but it isn't.&lt;/p&gt;&#xA;" OwnerUserId="3282" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-08T22:49:05.267" LastActivityDate="2010-11-08T22:49:05.267" Title="function asymptotic where $f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$" Tags="&lt;asymptotics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9492" PostTypeId="2" ParentId="9491" CreationDate="2010-11-08T22:18:41.580" Score="5" ViewCount="0" Body="&lt;p&gt;Yes.  One way to see this is to actually do the long division (like the kind you learned in elementary school)!  Unfortunately, typesetting that in full on this forum will overtax my LaTeX powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, dividing $b + O\left(\frac{1}{\sqrt{x}}\right)$ into $a + O\left(\frac{1}{\sqrt{x}}\right)$ yields $\frac{a}{b}$ with a remainder of $O\left(\frac{1}{\sqrt{x}}\right)$.  So we have &#xA;$$\frac{a + O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + \frac{O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + O\left(\frac{1}{\sqrt{x}}\right),$$&#xA;since $b + O\left(\frac{1}{\sqrt{x}}\right) = O(1).$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-08T22:18:41.580" />
  <row Id="9493" PostTypeId="1" CreationDate="2010-11-08T22:25:47.307" Score="0" ViewCount="66" Body="&lt;p&gt;I have a table in my database that records, 48 times per day, the total users, total active users within 2 weeks, total users logged in within the past 24 hours , and total users that logged in within the last 24 hours  that had visited the site prior to the last 24 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use this data to calculate the churn rate of my users, but in some cases I am getting a negative number so I assume I am doing it incorrectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(Yesterday's active in past 24 hours&#xA;  [returning] - Todays's active in past&#xA;  24 hours [returning] + Change in&#xA;  Active Users) / Todays's active in&#xA;  past 24 hours [returning]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume I have the correct data somewhere with all of the data I am saving, I'm just not using it the right way.&lt;/p&gt;&#xA;" OwnerUserId="3284" LastActivityDate="2010-11-08T22:25:47.307" Title="How to calculate daily user churn rate?" Tags="&lt;analysis&gt;&lt;computer-science&gt;" CommentCount="3" />
  <row Id="9494" PostTypeId="2" ParentId="9491" CreationDate="2010-11-08T22:28:16.750" Score="6" ViewCount="0" Body="&lt;p&gt;It is true.  In the spirit of epsilon/delta, you are challenged to prove that $|f(x)-\frac{a}{b}| \lt \frac{M}{\sqrt{x}}$ for $x\gt x_0$ where your challenger gives M and you have to find an $x_0$ that works.  But you get to challenge back saying the numerator should be within $\frac{N}{\sqrt{x}}$ of $a$ and similarly the denominator should be within $\frac{P}{\sqrt{x}}$ of $b$.  So take $N=\frac{M}{2b}$ and $P=\frac{aM}{2b^2}$ and take the larger of the $x_0$'s that come back.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-08T22:28:16.750" CommentCount="4" />
  <row Id="9495" PostTypeId="2" ParentId="9488" CreationDate="2010-11-08T22:34:43.707" Score="1" ViewCount="0" Body="&lt;p&gt;This is true if there is a (measurable) function $f(y)$ such that $$|g(x,y)|\le f(y), \mbox{ for }0\le x\le 1 $$  and $$\int_0^1 f(y)\,dy&amp;lt;\infty.$$&#xA;This follows from the Lebesgue dominated convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Every continuous function is measurable.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-08T23:45:20.760" LastActivityDate="2010-11-08T23:45:20.760" CommentCount="4" />
  <row Id="9497" PostTypeId="2" ParentId="9488" CreationDate="2010-11-08T22:44:33.180" Score="5" ViewCount="0" Body="&lt;p&gt;The answer is no. In fact, it is related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_integral_rule&quot; rel=&quot;nofollow&quot;&gt;Leibniz's rule&lt;/a&gt;. The counterexample below is from 'Counterexamples in Analysis' and was recently posted &lt;a href=&quot;http://math.stackexchange.com/questions/8711/derivative-commuting-over-integral/8712#8712&quot;&gt;here&lt;/a&gt; by Jonas Meyer. Consider the function &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(x,y) = \left\{&#xA;     \begin{array}{lr}&#xA;       \left (\frac{3x^2}{y^2} - \frac{2x^4}{y^3} \right)e^{-x^2/y} &amp;amp;  y&gt;0, \\&#xA;       0 &amp;amp; y=0. &#xA;     \end{array}&#xA;   \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that $g$ is continuous in both variables separately. We have that $\int_{0}^1 g(0,y) dy = 0 $ and $$ \int_{0}^1 g(x,y) dy = e^{-x^2}(1-2x^2)$$ if $ x \ne 0$. So $$\lim_{x \rightarrow 0} \int_{0}^1 g(x,y) dy = 1 \ne \int_{0}^1 g(0,y) dy.$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-08T23:37:40.667" LastActivityDate="2010-11-08T23:37:40.667" CommentCount="6" />
  <row Id="9498" PostTypeId="2" ParentId="7499" CreationDate="2010-11-08T23:04:06.657" Score="0" ViewCount="0" Body="&lt;p&gt;Check out Andrew Ng's website for a tutorial on SVMs.&#xA;www.stanford.edu/class/cs229/notes/cs229-notes3.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yeah, in this case I believe that the feature space and what you are representing are two different dimensions.&lt;/p&gt;&#xA;" OwnerUserId="3287" LastActivityDate="2010-11-08T23:04:06.657" />
  <row Id="9499" PostTypeId="2" ParentId="9398" CreationDate="2010-11-08T23:13:35.200" Score="0" ViewCount="0" Body="&lt;p&gt;If you can use partial derivatives, then you can do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you find $dy/dx$, say $$\frac{dy}{dx}=g(x,y).$$ Then by chain rule &#xA;$$\frac{d^2y}{dx^2}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}\frac{dy}{dx}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}g(x,y).$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-08T23:13:35.200" />
  <row Id="9500" PostTypeId="2" ParentId="9488" CreationDate="2010-11-08T23:15:05.283" Score="4" ViewCount="0" Body="&lt;p&gt;I think I have another counterexample.  Define $f(x)=\int_{0}^x e^{-1/t}dt$ for $x\gt0$.  This is chosen because it goes to zero as $x$ goes to zero from the right and because experimentation led me to the differential equation $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$ as a sufficient condition for the following to work.  Define &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x,y) = \left\{&#xD;&#xA;     \begin{array}{lr}&#xD;&#xA;       \frac{xy}{f(x)+y^2} &amp; \text{if } x\gt0, \\&#xD;&#xA;       0 &amp; \text{if } x=0, &#xD;&#xA;     \end{array}&#xD;&#xA;   \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $(x,y)$ in $[0,1]\times[0,1]$, and let $x_0=0$.  The right hand side of your tentative equation is obviously $0$.  The left hand side is &#xA;$$\lim_{x\to0+}\frac{x}{2}\log\left(1+\frac{1}{f(x)}\right),$$&#xA;which comes out to $\frac{1}{2}$ after $2$ applications of l'Hôpital's rule, if I did it correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I started by playing with the standard example of a discontinuous but separately continuous function on $\mathbb{R}^2$, $f(x,y)=\frac{xy}{x^2+y^2}$ when $x$ or $y$ is nonzero, $f(x,y)=0$ when $x=y=0$. Then I tried to see how the $x^2$ in the denominator could be modified to give a counterexample here, by replacing it with an unknown $f(x)$ that goes to $0$ at $0$ and seeing what further properties of $f(x)$ would make it work.  As I mentioned, this led in particular to the sufficient condition $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$, and hence to this example.  Unfortunately, I can't offer any real intuition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to look at this a little more, and came up with an example simpler than the other one I gave.  With $$g(x,y)=\frac{-y}{\log(x/2)(x+y^2)}$$ &#xA;for $x\gt0$ and $0$ otherwise, the same result as above holds.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-12T06:30:18.900" LastActivityDate="2010-11-12T06:30:18.900" CommentCount="8" />
  <row Id="9501" PostTypeId="2" ParentId="9434" CreationDate="2010-11-08T23:21:36.050" Score="2" ViewCount="0" Body="&lt;p&gt;Just a comment up front: Obviously, you're not going to find the perfect algorithm to predict stock prices.  Such a thing would the Holy Grail of Wall Street. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, during my brief stint as a part-time business school prof here are the basic methods we discussed in the forecasting unit.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Moving average&lt;/li&gt;&#xA;&lt;li&gt;Weighted moving average&lt;/li&gt;&#xA;&lt;li&gt;Exponential smoothing&lt;/li&gt;&#xA;&lt;li&gt;Exponential smoothing with trend&lt;/li&gt;&#xA;&lt;li&gt;Least-squares linear regression (with a nod to BBischof!)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each has pros and cons.  The first four are discussed in the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_smoothing&quot; rel=&quot;nofollow&quot;&gt;exponential smoothing&lt;/a&gt;, although exponential smoothing with trend is called &quot;double exponential smoothing&quot; there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me emphasize again that these are &lt;em&gt;basic&lt;/em&gt; methods.  If you want something fancier, you could try &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;geometric Brownian motion&lt;/a&gt;.  Be aware, though, that there are LOTS of other algorithms out there.  You can probably imagine how many finance companies are trying to come up with a better way of predicting stock prices.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-08T23:21:36.050" />
  <row Id="9502" PostTypeId="1" CreationDate="2010-11-08T23:33:34.510" Score="1" ViewCount="58" Body="&lt;p&gt;Hello, I am interested in a method to find the roots of the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;f(t) = \sum_{i=1}^n \alpha_i e^{\beta_i t} + \gamma t + \delta = 0.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my application, coefficients $\alpha_i$, $\beta_i$, $\gamma$, and $\delta$ are real.  $n$ is typically a small integer, say 10.  In particular I am interested in the smallest positive real root of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those interested, this equation arises when attempting to compute the point of intersection between the solution to the linear ODE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\dot x(t) &amp;amp;= Ax(t) + b &amp;amp; (A = A^T) \\ &#xA;x(0) &amp;amp;= x_0&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the boundary of a set of linear constraints&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;Cx(t) \ge d.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial point is always feasible $(Cx_0\ge d)$.  For my purposes, all matrices and vectors are real.&lt;/p&gt;&#xA;" OwnerUserId="3285" LastActivityDate="2010-11-08T23:52:09.063" Title="solving a univariate equation with a sum of exponentials" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;&lt;differential-equations&gt;&lt;roots&gt;" AnswerCount="1" />
  <row Id="9503" PostTypeId="2" ParentId="9378" CreationDate="2010-11-08T23:41:25.957" Score="1" ViewCount="0" Body="&lt;p&gt;I've just started with this stuff, so take this for what it's worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's been working for me is to imagine a sheaf as the sheaf of continuous functions from the reals to the reals. The restriction maps are just the normal restrictions of functions to open sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For presheaves I think of the presheaf of bounded continuous functions from reals to reals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stalks are just the germs thereof. I draw an element of a stalk by drawing coordinate axes, choosing a point x on the horizontal axis, and then draw the graph of a continuous function on a tiny neighborhood of x. The neighborhood is supposed to be &quot;infinitesimal.&quot; The stalk is the set of all such graphs on that neighborhood. &lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-11-08T23:41:25.957" CommentCount="4" />
  <row Id="9504" PostTypeId="2" ParentId="9502" CreationDate="2010-11-08T23:52:09.063" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming $\alpha_i, \beta_i, \gamma$, and $\delta $ are givens, any numeric root finder should make quick work of this.  There won't be an algebraic solution.  There is good info at &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; chapter 9.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-08T23:52:09.063" CommentCount="3" />
  <row Id="9505" PostTypeId="1" AcceptedAnswerId="9515" CreationDate="2010-11-09T00:23:57.653" Score="8" ViewCount="791" Body="&lt;p&gt;We know that $2^4 = 4^2$ and $(-2)^{-4}$ = $(-4)^{-2}$. Is there another pair of integers $(x, y)$, with different $x$ and $y$, which satisfies the equality  $x^y = y^x$?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T01:54:53.973" LastActivityDate="2010-11-19T19:49:10.333" Title="$x^y = y^x$ (x and y are integers)" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="4" />
  <row Id="9506" PostTypeId="2" ParentId="9468" CreationDate="2010-11-09T00:24:39.610" Score="4" ViewCount="0" Body="&lt;p&gt;I can't resist giving a couple of examples of (contrived) numerical coincidences (both are from &lt;a href=&quot;http://books.google.co.uk/books?id=csfnE14Kj-cC&amp;amp;printsec=frontcover&amp;amp;dq=experimentation+in+mathematics+computational+paths+to+discovery&amp;amp;hl=en&amp;amp;ei=A5TYTJu1E8mG4Aap0OXqCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDMQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Experimentation in Mathematics&lt;/em&gt;&lt;/a&gt; by J.M. Borwein, D.H. Bailey and R. Girgensohn).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}\cos(2x)\prod_{k=1}^{\infty}\cos(x/k)\ dx=\frac{\pi}{8}-\epsilon,$$&#xA;where &#xA;$0&amp;lt;\epsilon&amp;lt;10^{-41}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{k=1}^{\infty}e^{-(k/10)^2}\approx5\sqrt\pi-\frac{1}{2}=8.362269254527580...$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, they agree through 427 (four hundred twenty seven) digits yet they are not equal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A moral. Make sure you understand the context and use your inverse symbolic calculator with caution.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-09T05:41:37.797" LastActivityDate="2010-11-09T05:41:37.797" CommentCount="4" />
  <row Id="9507" PostTypeId="2" ParentId="9505" CreationDate="2010-11-09T00:34:10.720" Score="19" ViewCount="0" Body="&lt;p&gt;For every integer $n$, $x = y = n$ is a solution. So assume $x \neq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x \mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n &amp;lt; m$) we must have $n &amp;lt; e$, i.e. $n = 1$ or $n = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 2$ then $n^m$ is a power of $2$, and so (since $m &gt; 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} &amp;gt; 3$, and so an easy induction shows that $k \leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Up till now we considered $n,m&amp;gt;0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=0$ and $m \neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &amp;gt; 0$ and $m &amp;lt; 0$ then $0 &amp;lt; n^m \leq 1$ whereas $|m^n| \geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n,m &amp;lt; 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-16T18:28:27.290" LastActivityDate="2010-11-16T18:28:27.290" CommentCount="4" />
  <row Id="9508" PostTypeId="1" AcceptedAnswerId="9512" CreationDate="2010-11-09T00:35:25.027" Score="5" ViewCount="159" Body="&lt;p&gt;I need to write a coupon code system but I do not want to save each coupon code in the database. (For performance and design reasons.) Rather I would like to generate codes subsequent that are watermarked with another code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They should like kind of fancy and random. Currently they look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: AKFCU2, 2: AKFDU2, 3: AKFDW2, 4: AKHCU2, 5: AKHCW2, ..., 200: CLFSU2, 201: CLFSW2, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's obvious that subsequent codes look very similar as I just converted my code ingredients (the watermark and the integer in the front) to binary base and permutated the order by a fixed scheme. But to prevent people to easily guess another valid code or even accidently enter another valid code (thus making another code invalid by using it) I would prefer something more chaotic, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: FIOJ32, 2: X9NIU2, 3: SIUUN7, 4: XTVV4S, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end the problem is to find a bijective, discrete function on the domain {0,1}^27 (or alternatively {0,1,2,3,4, ..., [10^(8.5)]}) that is far away from being continuous. Also it should be as simple as possible to implement. (EDIT: I also need to implement the reverse function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions for such a function?&lt;/p&gt;&#xA;" OwnerUserId="3290" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-09T08:36:55.387" LastActivityDate="2010-11-10T02:53:58.737" Title="Looking for a bijective, discrete function that behaves as chaotically as possible" Tags="&lt;discrete-mathematics&gt;&lt;computer-science&gt;&lt;cryptography&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9509" PostTypeId="2" ParentId="7499" CreationDate="2010-11-09T00:36:34.167" Score="2" ViewCount="0" Body="&lt;p&gt;As mentioned, the kernel tricks embeds your original points to a higher dimensional space (in fact, in some cases infinite dimensional - but of course the linear subspace generated by your actual points is finite dimensional).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, using the embedding $(x,y) \mapsto (x,y,x^2,y^2)$ (that actually corresponds to a quadratic kernel, I think) then the equation of an arbitrary ellipse becomes linear.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-09T00:36:34.167" />
  <row Id="9510" PostTypeId="1" CreationDate="2010-11-09T00:38:51.410" Score="13" ViewCount="178" Body="&lt;p&gt;Winding yarn into a ball suggests some mathematical questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/neiZH.jpg&quot; alt=&quot;Yarn&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Under some natural model, what paths should the yarn follow to achieve the most dense ball?&#xA;One model is that used by Henryk Gerlach and Heiko von der Mosel in their paper&#xA;&quot;On sphere-filling ropes&quot; &lt;a href=&quot;http://arxiv.org/abs/1005.4609&quot; rel=&quot;nofollow&quot;&gt;arXiv:1005.4609v1 (math.GT)&lt;/a&gt;.&#xA;(This is the same model I suggested in a &lt;a href=&quot;http://mathoverflow.net/questions/26525/coiling-rope-in-a-box&quot; rel=&quot;nofollow&quot;&gt;MO question&lt;/a&gt;.)&#xA;I expect it would be difficult to extend the optimal solutions of the above paper to the multiple layers for a ball of yarn. (Below is shown part of Fig.6 from their paper.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0kn2O.jpg&quot; alt=&quot;Rope&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What paths should the yarn follow to achieve the least dense ball?&#xA;Here I imagine layers forming a grid that suspends the yarn above as much empty space as possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Random winding.&#xA;Typical instructions for how to do this by hand say, &quot;Change directions every once in a while while you are winding,&quot; or &quot;As you wrap, slowly rotate the ball counterclockwise to keep the distribution even.&quot; What density is achieved by random winding?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I ask these questions primarily out of curiosity. Perhaps there is an analogous process (winding the interior of a golf ball or baseball?) that has been studied mathematically.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="237" LastEditorDisplayName="" LastEditDate="2010-11-09T00:55:46.600" LastActivityDate="2010-11-09T00:55:46.600" Title="Optimal yarn balls" Tags="&lt;geometry&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="9511" PostTypeId="1" AcceptedAnswerId="9514" CreationDate="2010-11-09T00:43:08.677" Score="2" ViewCount="54" Body="&lt;p&gt;The unit doublet is a symbolic object whose convolution with a differentiable function is supposed to give the derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x * u_1)(t) = \frac{dx(t)}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Unit_doublet&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Unit_doublet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this equation be made rigorous, in a way that it would be valid for all $x\in C^1(\mathbb R)$?  I.e., is there a definition of the unit doublet as a distribution or a measure, and a version of the convolution operation that would make sense for all $C^1(\mathbb R)$ functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to my question on MO: &lt;a href=&quot;http://mathoverflow.net/questions/2969/&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/2969/&lt;/a&gt; , but the theorems referred in the answers to that question include only a small fraction of $C^1(\mathbb R)$.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-09T01:02:10.617" LastActivityDate="2010-11-09T01:38:31.383" Title="Rigorous definition of convolution with the unit doublet " Tags="&lt;fourier-analysis&gt;&lt;signal-processing&gt;" AnswerCount="1" />
  <row Id="9512" PostTypeId="2" ParentId="9508" CreationDate="2010-11-09T00:47:24.937" Score="3" ViewCount="0" Body="&lt;p&gt;Take any odd $a$ and calculate $x \mapsto ax + b \pmod{2^{27}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's a more sophisticated suggestion. The following functions are all invertible and easy to implement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt; Multiplication by an odd number modulo $2^{27}$.&#xA;&lt;li&gt; Addition of an arbitrary number modulo $2^{27}$.&#xA;&lt;li&gt; XOR of an arbitrary 27-bit number.&#xA;&lt;li&gt; Rotation of the bits (can be implemented using two shifts).&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you compose them repeatedly they become harder to break (but probably still easy for someone who already knows the general form of your cipher).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By composing, I mean you apply several of them in succession (you can apply a given mapping more than once with same or different parameters).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another suggestion is to use a random permutation. It's not difficult to generate a random permutation, and given a permutation to calculate its inverse. You can store both permutations in a file and load them to memory (if you have enough - it's 1GB for both) each time your main program starts.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-09T07:36:53.853" LastActivityDate="2010-11-09T07:36:53.853" CommentCount="7" />
  <row Id="9513" PostTypeId="1" AcceptedAnswerId="9518" CreationDate="2010-11-09T00:57:58.813" Score="2" ViewCount="63" Body="&lt;p&gt;I am trying to prove the following fact for a homework assignment in algebraic geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let R be a local ring, and X a prescheme. Show that there is a one-one correspondence between R-valued points on X and pairs (x, g) of points x on X and local homomorphisms $g: O_{X, x} \to R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given f: Spec R -&gt; X, and writing x for f(M) where M is the maximal ideal of R, the induced map on stalks gives us a map $g: O_{X,x} \to R_M = R$. Then it remains to show that given a point x of X and a map $g: O_{X,x} \to R$, that there is a map f: Spec R -&gt; X such that f(M) = x and such that the induced map is g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't had any luck yet with associating a map f with a given g. Here are my thoughts and some of the things I've tried:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When R is a field, then Spec R has only a single point, and thus the map f is completely determined by where it sends the zero ideal. This induces a local homomorphism from $O_{X,x}$ to k, which is to say that it induces a map from $O_{X,x} / M_x \to k$ since the maximal ideal $M_x$ of $O_{X,x}$ is contained in the kernel of the local homomorphism. Anyway, as long as there is some local homomorphism from $O_{X,x}$ to k, then the map f sending (0) to x is a well-defined morphism. Now, when R is again just a local ring, it seems unlikely that f is completely determined by where it sends M, and I can't see how to work with this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I've tried extending the map g to a map of sheaves, thinking that maybe from there I'd be able to find my map of preschemes. But here I get a little lost, and I'm not sure how to make it work.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We know a lot about the topology of Spec R, and this forces some properties of X. For example, the only open neighborhood of M in Spec R is the whole space. Therefore, f(Spec R) covers any open neighborhood of f(M). I haven't pursued this way very far yet, as it seems a little further afield.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I just need a little hint to get me going again -- I've been thinking about this one for a while and I can't seem to get unstuck.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-13T11:40:11.503" Title="How to find a bijection between R-valued points of X and local ring homomorphisms?" Tags="&lt;homework&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9514" PostTypeId="2" ParentId="9511" CreationDate="2010-11-09T01:28:24.360" Score="3" ViewCount="0" Body="&lt;p&gt;For $C^1(\mathbb{R})$ it is a bit delicate. But for $C^1(\mathbb{R})$ with compact support (or rapid decay a la Schwartz functions), you can define $u_1$ as (up to a negative sign) the distributional derivative of the Dirac distribution. It is a distribution of compact support, and so also a tempered distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that it has compact support, you should be able to convolve it against any distribution (see Hormander, Analysis of Linear Partial Differential Operator, Volume 1, Chapter 4). A function in $C^1$ is locally integrable, and hence a distribution. So the object $x*u_1$ is well defined as a distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least distributionally the equation you wrote make sense: that acting on any smooth function with compact support, the two sides act the same way as linear functionals. (Since for any test function $\psi \in C^\infty_c(\mathbb{R})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle(x*u_1),\psi\rangle = -\langle x,(\psi*u_1)\rangle = - \langle x,\psi'\rangle = \langle x',\psi\rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you see that the two sides are equally defined as distributions, using that a convolution of a test function against a distribution is always well defined and also a smooth function.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equation you wrote down is meaningful in the sense of distributions. And hence you can take $dx/dt$ as a continuous representative (in the equivalent class of distributions) of $x*u_1$. Which I think is about as rigorous as you need it. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-09T01:38:31.383" LastActivityDate="2010-11-09T01:38:31.383" CommentCount="1" />
  <row Id="9515" PostTypeId="2" ParentId="9505" CreationDate="2010-11-09T01:40:19.037" Score="11" ViewCount="0" Body="&lt;p&gt;This is a classic (and well known problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general solution of $x^y = y^x$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = (1+1/u)^u, y = (1+1/u)^{u+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, see here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey&quot; rel=&quot;nofollow&quot;&gt;http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/x%5Ey-x%5Ey&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.qbyte.org/puzzles/p048s.html&quot; rel=&quot;nofollow&quot;&gt;http://www.qbyte.org/puzzles/p048s.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T05:03:27.990" LastActivityDate="2010-11-09T05:03:27.990" CommentCount="1" />
  <row Id="9516" PostTypeId="1" CreationDate="2010-11-09T02:06:18.103" Score="1" ViewCount="67" Body="&lt;p&gt;Is there a way of enumerating necklaces, in the spirit in which the factorial number system ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Factorial_number_system&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Factorial_number_system&lt;/a&gt; ) can be used to enumerate permutations?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-11-09T02:34:07.540" Title="Enumerating necklaces" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9517" PostTypeId="2" ParentId="9516" CreationDate="2010-11-09T02:34:07.540" Score="2" ViewCount="0" Body="&lt;p&gt;There is code here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Recn/Necklaces/&quot; rel=&quot;nofollow&quot;&gt;http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Recn/Necklaces/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3293" LastActivityDate="2010-11-09T02:34:07.540" />
  <row Id="9518" PostTypeId="2" ParentId="9513" CreationDate="2010-11-09T03:12:57.397" Score="3" ViewCount="0" Body="&lt;p&gt;For any point $x \in X$, there is a canonical morphism $\mathrm{Spec} \mathcal{O}_{X, x} \to X$. To prove this, first take an affine neighborhood of $x$. You may also want to look at the definition of a morphism of locally ringed spaces / schemes, specifically, the requirement on stalks.&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2010-11-09T03:12:57.397" CommentCount="1" />
  <row Id="9519" PostTypeId="2" ParentId="6437" CreationDate="2010-11-09T03:33:41.100" Score="1" ViewCount="0" Body="&lt;p&gt;To connect all computers to each other you need n-1 connections, i.e., 4. To connect 3 printers you need exactly 3 connections. So the number connections required is 7. &lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-11-09T03:33:41.100" />
  <row Id="9520" PostTypeId="1" CreationDate="2010-11-09T03:46:56.433" Score="4" ViewCount="124" Body="&lt;p&gt;Is this a known special function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 a^p(1-a)^{1-p}\,b^{1-p}\,(1-b)^p dp\qquad ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really only interested in maximizing this over $(a,b)$ in $[0,1] \times [0,1]$, so a pointer to a nice numerical evaluation is appreciated as much or more so than an unstable exact formula.&lt;br&gt;&#xA;Thanks for any help&lt;/p&gt;&#xA;" OwnerUserId="3295" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-09T03:57:26.503" LastActivityDate="2010-11-09T04:46:34.663" Title="Is this a known special function?" Tags="&lt;integral&gt;&lt;optimization&gt;&lt;special-functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9521" PostTypeId="2" ParentId="9355" CreationDate="2010-11-09T03:54:15.427" Score="0" ViewCount="0" Body="&lt;p&gt;To answer (ii), I think it needs to be ergodic, i.e., aperiodic and recurrent.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-11-09T03:54:15.427" />
  <row Id="9522" PostTypeId="1" AcceptedAnswerId="9535" CreationDate="2010-11-09T03:55:38.047" Score="6" ViewCount="165" Body="&lt;p&gt;Given a row of 16 houses where 10 are red and 6 are blue, what is the expected number of neigbors of a different color?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-11-10T17:52:51.113" Title="Expected number of neighbors" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;expectation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9523" PostTypeId="1" AcceptedAnswerId="14092" CreationDate="2010-11-09T03:58:12.667" Score="4" ViewCount="170" Body="&lt;p&gt;My text gives a much more complicated proof of this result, which makes me wonder if the argument I have in my head for this has something wrong with it.  Does this work, or have I made a bad assumption somewhere along the line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U \subseteq \mathbb{C}$ be open, $\overline{D}(z_0, r) \subset U$, and $f : U \rightarrow \mathbb{C}$ be holomorphic with a zero of order $n$ at $z_0$ and no other zeroes in $U$. Taking a power series expansion at $z_0$,&#xA;$$f(z) = a_n(z-z_0)^n + o((z-z_0)^{n+1}).$$&#xA;Differentiating,&#xA;$$f'(z) = n a_n(z-z_0)^{n-1} + o((z-z_0)^n),$$&#xA;so we have&#xA;$$\lim_{z \rightarrow z_0} \frac{(z-z_0) f'(z)}{f(z)} = \lim_{z \rightarrow z_0} \frac{n a_n(z-z_0)^n + o((z-z_0)^{n+1})}{a_n(z-z_0)^n + o((z-z_0)^{n+1})} = n.$$&#xA;Define the function $g : U \rightarrow \mathbb{C}$ by&#xA;$$g(z) = \begin{cases}&#xA;(z-z_0) f'(z)/f(z) &amp;amp; \text{if } z \neq z_0 \\&#xA;n &amp;amp; \text{otherwise}&#xA;\end{cases}$$&#xA;By the Cauchy integral formula,&#xA;$$n = g(z_0) = \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{g(z)}{z - z_0}  dz =  \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{(z-z_0)f'(z)}{(z-z_0)f(z)} dz=  \frac{1}{2 \pi i} \oint_{\partial\overline{D}(z_0, r)} \frac{f'(z)}{f(z)} dz$$&#xA;as desired.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastEditorUserId="3296" LastEditorDisplayName="" LastEditDate="2010-11-09T05:44:52.713" LastActivityDate="2010-12-12T22:14:13.440" Title="Proof that $\frac{1}{2\pi i}\oint f'(z)/f(z) \, dz = n$" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="9524" PostTypeId="1" AcceptedAnswerId="14902" CreationDate="2010-11-09T04:01:42.107" Score="10" ViewCount="485" Body="&lt;p&gt;What are some good ways to motivate the material on infinite series that appears at the end of a typical American Calculus II course?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My students in this course are generally from biochemistry, computer science, economics, business, and physics (with a few humanities folks taking the course for fun) - not just math majors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have struggled some in the past to motivate the infinite series material to these students.  For one, it doesn't fit with the rest of Calc II, which is on the integral.  Over the years I have &quot;converged&quot; on telling them that the main point of the unit is Taylor series and that the rest of the material is there primarily so that we have the tools we need in order to understand Taylor series.  Then I illustrate some of the many uses of Taylor series (mainly function approximation, at this level).  This approach works better than anything I've come up with thus far with respect to getting my students to care about infinite series, but I feel a little like I'm selling the rest of the material short by subordinating it to Taylor series.  Does anyone have other ways of motivating infinite series that they would like to share?  (Again, only a small percentage of the students in my class are math majors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: The material in this unit typically consists of sequences, basic series (like geometric and telescoping ones), a slew of tests for convergence (e.g., integral test, ratio test, root test), an introduction to power series, Taylor and Maclaurin series, and maybe binomial series.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-19T23:33:36.320" LastActivityDate="2011-01-06T23:22:44.117" Title="Motivating infinite series" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;teaching&gt;" AnswerCount="9" CommentCount="15" FavoriteCount="3" />
  <row Id="9525" PostTypeId="1" AcceptedAnswerId="9568" CreationDate="2010-11-09T04:05:48.133" Score="0" ViewCount="118" Body="&lt;p&gt;This is in relation to a problem dealing with the three-dimensional analogue of Pell's Equation.  I would like to factor&#xA;$ x^3+Dy^3+D^2z^3-3Dxyz $ into $\frac{1}{2}(x+Dy+D^2y)$ and another factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like just a hint on how to go about doing this as I'm pretty much stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I think the factor I was looking for is $\frac{1}{2}(x+Ny+N^2z)$ where $N=D^3$.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="2505" LastEditorDisplayName="" LastEditDate="2010-11-09T23:22:51.400" LastActivityDate="2010-11-09T23:32:53.793" Title="Factoring Multiple Variable  Polynomials" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9527" PostTypeId="2" ParentId="9523" CreationDate="2010-11-09T04:09:47.367" Score="5" ViewCount="0" Body="&lt;p&gt;If $f$ has a zero of order $n$ at $z_0$, then there is an function $h$ holomorphic near $z_0$ and such that $f(z)=(z-z_0)^nh(z)$ near $z_0$ and $h(z_0)\neq0$. Computing, $$\frac{f'}{f}=\frac{n}{z-z_0}+\frac{h'(z)}{h(z)}.$$ If you use the Cauchy formula now over a sufficiently small circle around $z_0$ you get (a corrected version of) what you want.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-09T04:18:26.970" LastActivityDate="2010-11-09T04:18:26.970" CommentCount="3" />
  <row Id="9530" PostTypeId="2" ParentId="9520" CreationDate="2010-11-09T04:17:31.080" Score="8" ViewCount="0" Body="&lt;p&gt;You can get a closed-form answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 a^p (1-a)^{1-p} b^{1-p} (1-b)^p dp = b(1-a) \int_0^1 \left(\frac{a(1-b)}{b(1-a)}\right)^p dp = \left. \frac{b(1-a)}{\ln \frac{a(1-b)}{b(1-a)}} \left(\frac{a(1-b)}{b(1-a)}\right)^p \right|_0^1 $$&#xA;$$= \frac{a(1-b) - b(1-a)}{\ln a + \ln (1-b) - \ln b - \ln (1-a)} = \frac{a-b}{\ln a + \ln (1-b) - \ln b - \ln (1-a)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds if $a \neq b$ and if neither of $a$ or $b$ is 0 or 1.  If $a = b$, then instead we have $$a(1-a) \int_0^1 dp = a - a^{2}.$$&#xA;And, of course, if $a$ or $b$ is 0 or 1 then the value of the integral is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as far as maximizing, you can use the usual approach of finding where both partial derivatives are 0.  I haven't worked through the calculations, but I strongly suspect that because of the symmetry in $a$ and $b$ that the maximum value will occur at $a = b$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-09T04:46:34.663" LastActivityDate="2010-11-09T04:46:34.663" CommentCount="5" />
  <row Id="9531" PostTypeId="2" ParentId="9524" CreationDate="2010-11-09T04:19:16.330" Score="3" ViewCount="0" Body="&lt;p&gt;While I somewhat struggle to relate to the problem (series was possibly the most interesting section of Calc II to me), I can share some insight on how my teacher was able to make them seem interesting. She used Mathematica demonstrations throughout every section of our calculus classes (I also had her for Calc III), and I found that the visualization of terms and partial sums helped to visually see the asymptotic behavior of each towards zero and the sum, respectively. Taylor and Maclaurin series were also very &quot;unimportant&quot; to me (in the sense that I did not at the time grasp their usefulness), until I was shown how the polynomials converged to the function. Some of her files are available on demonstrations.wolfram.com:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://demonstrations.wolfram.com/SeriesAFewExamples/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/SeriesAFewExamples/&lt;/a&gt;&#xA;&lt;a href=&quot;http://demonstrations.wolfram.com/SeriesStepsOnANumberLine/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/SeriesStepsOnANumberLine/&lt;/a&gt;&#xA;&lt;a href=&quot;http://demonstrations.wolfram.com/GraphsOfTaylorPolynomials/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/GraphsOfTaylorPolynomials/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a key idea that has always stuck with me, is that we only know how to perform the basic arithmetic operations, and more complex functions like $e^x$, $\sin x$, etc. can all be computed using just addition, subtraction, multiplication, and division.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-09T04:19:16.330" CommentCount="5" />
  <row Id="9532" PostTypeId="1" AcceptedAnswerId="9533" CreationDate="2010-11-09T04:27:54.823" Score="4" ViewCount="271" Body="&lt;p&gt;arising out of comment made by Yuval Filmus in &lt;a href=&quot;http://math.stackexchange.com/questions/9395/what-is-the-cardinality-of-set-of-all-smooth-functions-in-l1&quot;&gt;what is the cardinality of set of all smooth functions in $L^1$ ?&lt;/a&gt; I got this idea (forgive me for my ignorance for if it is nothing but an elementary definition/result in real analysis). The idea is like this                        &lt;em&gt;Let $f:X\to Y$ is a mapping, where $X$ is a complete metric space (not sure if its strictly needed or whether a looser condition would do). If $f$ is a continuous mapping then $f$ is uniquely specified by a mapping $g:E\to Y$ where $E$ is a dense subset of $X$.&lt;/em&gt; What are the condition under which it is valid ? also  the validity of the converse statement.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T20:24:20.237" LastActivityDate="2010-11-13T20:37:23.970" Title="Link between a Dense subset and a Continuous mapping" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="9533" PostTypeId="2" ParentId="9532" CreationDate="2010-11-09T04:41:16.033" Score="11" ViewCount="0" Body="&lt;p&gt;If $Y$ is a Hausdorff topological space, then the value of a continuous function $f\colon X\to Y$ is completely determined by the value of $f$ on a dense subset $E$ of $X$; to see this, suppose $f$ and $g$ are two functions that agree on a dense subset $E$ of $X$, and let $u\in X\setminus E$. If $f(u)\neq g(u)$, then there are open neighborhoods $U$ and $V$ of $f(u)$ and $g(u)$, respectively, such that $U\cap V=\emptyset$. Then $f^{-1}(U)$ is an open neighborhood of $u$, as is $g^{-1}(V)$. Their intersection is an open neighborhood of $u$, and therefore must contain elements of $E$; but then any $e\in E$ in the intersection has $f(e)=g(e)$, with $f(e)\in U$ and $g(e)\in V$, contradicting that $U\cap V=\emptyset$. Therefore, $f(u)=g(u)$, hence $f=g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, it does not matter if $X$ is a complete metric space (or even a metric space); the key is $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the key is $Y$, consider the extreme case in which $Y$ has the indiscrete topology (the only open sets are the empty set and the set $Y$). Then &lt;em&gt;any&lt;/em&gt; function into $Y$ is continuous, so you can make your $X$ anything you want, and have two functions that agree on any subspace you care to specify and yet differ somewhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Note that any metric space is necessarily Hausdorff, so if your maps are between metric spaces, then the property holds (as in the case of Yuval's answer). This because you have the property that $d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$. Thus, given $x,y\in Y$, $x\neq y$, let $\epsilon=d(x,y)\gt 0$. Then $B(x,\frac{\epsilon}{2})$ and $B(y,\frac{\epsilon}{2})$ are open neighborhoods of $x$ and $y$ (respectively) that are disjoint: if $z$ were in the intersection, then $d(x,z)\lt \frac{\epsilon}{2}$, $d(z,y)\lt\frac{\epsilon}{2}$, and by the triangle inequality we would conclude that $d(x,y)\lt\epsilon$, a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final Addition:&lt;/strong&gt; See &lt;a href=&quot;http://math.stackexchange.com/questions/9825/does-this-property-characterize-a-space-as-hausdorff&quot;&gt;here&lt;/a&gt; for a discussion of the converse. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; A result towards a possible converse: certainly we need some separation on $Y$. Suppose that there exist points $x,y\in Y$, $x\neq y$, such that every open subset that contains $x$ also contains $y$ (so $Y$ could be $T_0$, but cannot be $T_1$). Let $X$ be the Sierpinski space, $X=\{a,b\}$ with topology $\tau = \{\emptyset, \{b\}, X\}$. Let $f,g\colon X\to Y$ be the following maps: $f$ is the constant function that maps everything to $x$; $g$ is the function that maps $a\mapsto x$, $b\mapsto y$. The constantfunction is certainly continuous. For $g$, if $U$ is an open subset that contains $y$ but not $x$, then $g^{-1}(U)=\{b\}$, which is open; so $g$ is continuous. And $g$ and $f$ agree on the dense set $\{a\}$, but are not equal. &#xA;I'm still trying to figure out the $T_1$ case (for all $x,y\in Y$, $x\neq y$, there exist open sets $U,V$ such that $x\in U-V$ and $y\in V-U$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 3:&lt;/strong&gt; Another step: $T_1$ is not sufficient; a colleague came up with this one (I kept trying the cofinite topology on $\mathbb{N}$ and not getting anywhere): take two copies of the real line and identify every point except the origin; this is $Y$. The result is a $T_1$ space, but not Hausdorff since no neighborhoods of the two copies of the origin are disjoint.  Now let $X$ be the real line, and let $E = (-\infty,0)\cup(0,\infty)$ be the dense subset. The two obvious injections, one mapping $0$ to the first copy in $Y$ and the other mapping it to the other copy, are both continuous and agree on $E$ but not on all of $X$, so $T_1$ does not suffice for the property. At least, then, in the hierarchy of $T$-spaces, the first level at which we are guaranteed the property is Hausdorff. This does not, however, establish whether the converse property characterises Hausdorff-ness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 4.&lt;/strong&gt; I almost have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conjecture.&lt;/strong&gt; Assuming the Axiom of Choice, the following are equivalent for a topological space $Y$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Y$ is Hausdorff.&lt;/li&gt;&#xA;&lt;li&gt;For every topological space $X$, every dense subset $E$ of $X$, and every pair of continuous maps $f,g\colon X\to Y$, if $f$ and $g$ agree on $E$, then $f=g$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Argument.&lt;/em&gt; Suppose that $Y$ is not Hausdorff. If there exist points $x,y\in Y$, $x\neq y$, such that every open neighborhood of $x$ contains $y$, then the map from the Sierpinski space indicated above shows that $Y$ does not have property 2. So we may assume that $Y$ is at least a $T_1$ space. But since $Y$ is not Hausdorff, there exist points $s,t\in Y$, $s\neq t$, such that for every neighborhoods $U$ of $s$ and $V$ of $t$ such that $s\in U-V$ and $t\in V-U$, we have $U\cap V\neq \emptyset$. Let $\mathfrak{U}\_s$ be the family of open neighborhoods of $s$ that do not contain $t$, and let $\mathfrak{V}\_t$ be the family of open neighborhoods of $t$ that do not contain $s$. Let $P=\mathfrak{U}\_s\times \mathfrak{V}\_t$, and partially order $P$ by letting $(U,V)\leq (U&amp;#39;,V&amp;#39;)$ if and only if $U&amp;#39;\subseteq U$ and $V&amp;#39;\subseteq V$. This makes $P$ into a directed partially ordered set (given any $(U,V),(R,S)\in P$, there exists $(U&amp;#39;,V&amp;#39;)\in P$ such that $(U,V)\leq (U&amp;#39;,V&amp;#39;)$ and $(R,S)\leq (U&amp;#39;,V&amp;#39;)$. Now, for each $(U,V)\in P$, we are assuming that $U\cap V\neq\emptyset$, so using the Axiom of Choice let $y_{(U,V)}\in Y$ be an element of $U\cap V$. Note that $y_{(U,V)}\neq s$ and $y_{(U,V)}\neq t$ for all $(U,V)\in P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $X = \{y_{(U,V)}\mid (U,V)\in P\}\cup\{s,t\}$, and give $X$ the induced topology from $Y$, so that the inclusion maps $X\hookrightarrow Y$ is continuous. Note that the set $E=\{y_{(U,V)}\mid (U,V)\in P\}$ is dense in $X$: for every open neighborhood $B$ of $s$ in $X$, there exists an open set $\mathcal{O}\_B\in Y$ such that $\mathcal{O}\_B\cap X = B$; in particular, $\mathcal{O}\_B$ is a neighborhood of $s$; let $\mathcal{V}$ be any open neighborhood of $t$ that does not contain $S$, and let $B&amp;#39;=\mathcal{V}\cap X$; let $\mathcal{U}$ be any open neighborhood of $s$ that does not contain $t$. Then $\mathcal{U}\cap\mathcal{O}\_B$ is open, hence $B&amp;#39;=\mathcal{U}\cap\mathcal{O}\_B\cap X$ is an open subset of $X$ that is contained in $B$. Consider now $y_{(\mathcal{U}\cap\mathcal{O}\_B,\mathcal{V})}$. This is in $\mathcal{U}\cap\mathcal{O}_B\cap\mathcal{V}\cap X \subseteq B&amp;#39;$, and is plainly in $E$. In particular, in $X$ we have that $B&amp;#39;\cap E\neq\emptyset$, and hence $B\cap E\neq \emptyset$. Thus, every open neighborhood of $s$ in $X$ contains points of $E$, hence $x$ lies in the closure of $E$. A symmetric argument holds for $t$. Thus, $E$ is dense in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the maps $f,g\colon X\to Y$ defined as follows: $f$ and $g$, restricted to $E$, are the identity; $f(s)=f(t)=s$; and $g(s)=g(t)=t$. I claim that $f$ and $g$ are both continuous. Indeed, let $\mathcal{O}$ be an open set in $Y$. If $\mathcal{O}\cap\{s,t\}=\emptyset$ or $\{s,t\}\subseteq\mathcal{O}$, there is nothing to do: the inverse image under both $f$ and $g$ is just the intersection with $X$, hence open in $X$. So assume without loss of generality that $s\in\mathcal{O}$ but $t\notin \mathcal{O}$. Note that $g^{-1}(\mathcal{O}) = (\mathcal{O}-\{s\})\cap X$, and since $Y$ is $T_1$ removing a single point from an open set results in an open set, so $g^{-1}(\mathcal{O})$ is open. So we just need to show that $f^{-1}(\mathcal{O}) = (\mathcal{O}\cap X)\cup\{t\}$ is open in $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that is where I am a bit stuck at present. Can anyone either verify or falsify this?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T20:37:23.970" LastActivityDate="2010-11-13T20:37:23.970" CommentCount="8" />
  <row Id="9534" PostTypeId="2" ParentId="9524" CreationDate="2010-11-09T05:03:40.453" Score="2" ViewCount="0" Body="&lt;p&gt;I believe that power series were the main motivation for the formal study of infinite series; for instance, lots and lots of Euler's work consists of just multiplying out formal power series without any explicit concern for convergence.  Perhaps demonstrating e.g. Euler products in number theory would show why we'd be interested in having a firm basis on which to rest such an argument.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-09T05:03:40.453" CommentCount="4" />
  <row Id="9535" PostTypeId="2" ParentId="9522" CreationDate="2010-11-09T05:12:59.257" Score="8" ViewCount="0" Body="&lt;p&gt;The answer is 7.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be the number of neighbors of a different color.  Let $X_i$ be 1 if houses $i$ and $i+1$ are different colors and 0 otherwise.  Then $X = \sum_{i=1}^{15} X_i.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected number of neighbors of a different color is &#xA;$$E\left[\sum_{i=1}^{15} X_i\right] = \sum_{i=1}^{15} E[X_i] = \sum_{i=1}^{15} P(X_i = 1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue}) $$&#xA;$$ = \frac{10}{16} \frac{6}{15} + \frac{6}{16} \frac{10}{15} = \frac{120}{16(15)} = \frac{1}{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$E[X] = \sum_{i=1}^{15} \frac{1}{2} = 7.5.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, using indicator variables is often an effective way to calculate expected values.  The linearity of the expectation operator allows you to sidestep all the nasty dependencies you would otherwise be forced to deal with.  For lots of interesting examples using this approach, see Section 7.2 of Sheldon Ross's &lt;em&gt;A First Course in Probability&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the answer comes out so nicely I would not be surprised if there is a cleaner argument than mine.  If someone comes up with one I would love to see it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Moron's generalization has inspired me to generalize mine. :)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $m$ red houses and $n$ blue houses.  Then the expected number of neighbors of a different color is $$\frac{2mn}{m+n}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to my argument above,&#xA;$$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue}) $$&#xA;$$ = \frac{m}{m+n} \frac{n}{m+n-1} + \frac{n}{m+n} \frac{m}{m+n-1} = \frac{2mn}{(m+n)(m+n-1)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$E[X] = \sum_{i=1}^{m+n-1} \frac{2mn}{(m+n)(m+n-1)} = \frac{2mn}{m+n}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-10T17:52:51.113" LastActivityDate="2010-11-10T17:52:51.113" CommentCount="2" />
  <row Id="9537" PostTypeId="2" ParentId="9524" CreationDate="2010-11-09T05:27:19.150" Score="3" ViewCount="0" Body="&lt;p&gt;I think series do fit in with the integral quite nicely - the integral can be seen a &quot;series on a continuous set of summands&quot; while series are the discrete analog. They are both generalizations of finite everyday sums. The techniques used in the study of infinite series are similar to the ones used for improper integrals, and the integral convergence test is another reason why knowing integrals before studying series can be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why studying series is important, I think you should use motivating examples from the fields your students know, if possible. There is one basic &quot;global&quot; example I can think of - the paradoxes of Zeno.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-09T05:27:19.150" CommentCount="2" />
  <row Id="9538" PostTypeId="2" ParentId="9086" CreationDate="2010-11-09T05:35:03.803" Score="2" ViewCount="0" Body="&lt;p&gt;Given $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$ has roots $r_1, r_2, \ldots r_n$, we have&#xA;$$a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = a_n (x - r_1)(x - r_2) ... (x-r_n);$$&#xA;dividing through by $a_n$ gives&#xA;$$x^n + \frac{a_{n-1}}{a_n} x^{n-1} + \ldots + \frac{a_0}{a_n} = (x - r_1)(x - r_2) ... (x-r_n).$$&#xA;Multiplying out the RHS, the constant term is $\prod_{i=1}^n (-r_i)$.  This must match up with the constant term on the left, $a_0 / a_n$, which establishes the identity you wanted to come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for actually proving the rational root theorem, it's not hard, but this isn't the way to go about it.  Just assume that $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$ as before and that $f(p/q) = 0$, where $p/q$ is in lowest terms (i.e., where $p$ and $q$ are relatively prime.)  Then we have&#xA;$$0 = f(p/q) = a_n \frac{p^n}{q^n} + a_{n-1} \frac{p^{n-1}}{q^{n-1}} + \ldots + a_0$$&#xA;Multiplying each side by $q^n$, we have&#xA;$$0 = a_n p^n + a_{n-1} p^{n-1} q + \ldots + a_0 q^n;$$&#xA;rearranging,&#xA;$$- a_n p^n = q(a_{n-1} p^{n-1} +  a_{n-2} p^{n-2} q +\ldots + a_0 q^{n-1}).$$&#xA;Since the RHS is divisible by $q$, so is the LHS.  But $p$ is relatively prime to $q$, which means that $a_n$ must be divisible by $q$.  Going back and rearranging things in a different way,&#xA;$$p(a_n p^{n-1} + a_{n-1} p^{n-2} q + \ldots a_1) = - a_0 q^n;$$&#xA;since the LHS is divisible by $p$, the RHS must be as well.  Again, $p$ is relatively prime to $q$, so $a_0$ is divisible by $p$.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-09T05:35:03.803" CommentCount="2" />
  <row Id="9539" PostTypeId="2" ParentId="7966" CreationDate="2010-11-09T06:16:17.710" Score="2" ViewCount="0" Body="&lt;p&gt;I really like the way Bott &amp;amp; Tu develop spectral sequences in &quot;Differential Forms in Algebraic Topology&quot;.  It would be an especially good choice if you happen to have some familiarity with de Rham cohomology; they begin by just sort of easing you into the whole idea by proving the isomorphism of de Rham cohomology and Cech cohomology by constructing a double complex whose cohomology is isomorphic to both of those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, they only give uses for spectral sequences in topology.  I think they mainly introduce them to be able to give the spectral sequences for the homology and cohomology of the total space of a fibration.  In this application there is particularly nice geometric intuition for the algebraic machinery, which I found very useful for trying to get to the bottom of what these things really do.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-09T06:16:17.710" CommentCount="1" />
  <row Id="9540" PostTypeId="1" CreationDate="2010-11-09T06:23:34.190" Score="3" ViewCount="30" Body="&lt;p&gt;This question was firstly posted on &lt;a href=&quot;http://mathoverflow.net/questions/45252/does-order-isomorphism-of-linear-extensions-of-two-partially-ordered-sets-imply-o&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;. Two answers are pretty interesting. The potential counter-example given in the second answer is really interesting, but it is not surely a counter-example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two partially ordered sets $A = {a&amp;lt; b,a&amp;lt; c}$, $B={x&amp;lt; z,y&amp;lt; z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their linear extensions (here we allow equality in linear extensions) for $A, B$ are &#xA;$A_L={A_1={a&amp;lt; b&amp;lt; c}, A_2={a&amp;lt; b= c}, A_3={a&amp;lt; c&amp;lt; b}}$&#xA;and&#xA;$B_L = { B_1 = {x&amp;lt; y&amp;lt; z}, B_2={y&amp;lt; x&amp;lt; z}, B_3 ={x= y&amp;lt; z}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may define $f_1: A_1\to B_1$ by $f_1(a)=x, f_1(b)=y, f_1(c)=z$ and $f_2:A_3\to B_2$ by $f_2(a)=y, f_2(b)=x, f_2(c)=z$, but there are no bijective mappings from $A_2$ to $B_3$ s.t. $t&amp;lt; s\Longleftrightarrow f_3(t)&amp;lt; f_3(s)$ and $t=s \Longleftrightarrow f_3(t)=f_3(s)$.  (And it's easy to see that no other pairing of the $A_i$ with the $B_j$ will allow such maps to be chosen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no order isomorphisms from $A$ to $B$, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following conjecture true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any partially ordered sets $A,B$, if there exist isomorphisms from $A$'s linear extensions to $B$'s, then there exists an isomorphism from $A$ to $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, could you give me a counter example? &lt;/p&gt;&#xA;" OwnerUserId="3303" LastEditorUserId="3303" LastEditorDisplayName="" LastEditDate="2010-11-09T06:31:18.890" LastActivityDate="2010-11-09T06:31:18.890" Title="Does order isomorphism of linear extensions of two partially ordered sets imply order isomorphism of themselves?" Tags="&lt;combinatorics&gt;&lt;homological-algebra&gt;&lt;order-theory&gt;" FavoriteCount="1" />
  <row Id="9541" PostTypeId="1" CreationDate="2010-11-09T07:51:04.227" Score="2" ViewCount="85" Body="&lt;p&gt;Given $xy=C$ where $x, y$ are integer variables  and $C$ is integer constant. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the most efficient algorithm that finds $x,y$ such that $x+y$ is minimum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Providing references is highly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Input integers are reasonably encoded.&lt;/p&gt;&#xA;" OwnerUserId="2459" LastEditorUserId="2459" LastEditorDisplayName="" LastEditDate="2010-11-09T08:00:20.923" LastActivityDate="2010-11-09T08:00:20.923" Title="Hardness of perimeter minimization?" Tags="&lt;computer-science&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="9542" PostTypeId="1" AcceptedAnswerId="9555" CreationDate="2010-11-09T08:03:10.800" Score="2" ViewCount="133" Body="&lt;p&gt;I began reading Paul Halmos' &quot;Naive Set Theory&quot;, and encountered the &quot;Axiom of Specification&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;Blockquote&gt; To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.&#xA;&lt;/Blockquote&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier in the same section, I learned that statements in set theory should be &quot;sentences&quot;.  A sentence was defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;Blockquote&gt; There are two basic types of sentences, namely, assertions of belonging, &#xA;$x \in A$,&#xA;and assertions of equality,&#xA;$A = B$;&#xA;all other sentences are obtained from such &lt;i&gt;atomic&lt;/i&gt; sentences by repeated applications of the usual logical operators...&lt;/Blockquote&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete definition of a sentence follows, which can be read on Google Books here: &lt;a href=&quot;http://goo.gl/XvK2B&quot; rel=&quot;nofollow&quot;&gt;http://goo.gl/XvK2B&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to translate the axioms and theorems in the book into sentences, but it seems like the Axiom of Specification is not a sentence.  It refers to &quot;every condition&quot;, but I have no way to build a sentence that refers to &quot;every condition&quot; because the atomic sentences only refer to sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Axiom of Specification a sentence?  If not, does that mean that statements about set theory do not need to be sentences?  What other sorts of statements are allowed?  (I'm using &quot;statement&quot; colloquially since I don't know the technical term.)&lt;/p&gt;&#xA;" OwnerUserId="3305" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-09T14:51:15.590" LastActivityDate="2010-11-09T14:51:15.590" Title="How can I write the Axiom of Specification as a sentence?" Tags="&lt;set-theory&gt;&lt;logic&gt;&lt;axioms&gt;" AnswerCount="2" />
  <row Id="9543" PostTypeId="2" ParentId="9542" CreationDate="2010-11-09T08:17:09.713" Score="-2" ViewCount="0" Body="&lt;p&gt;$$\forall A, \forall S, \exists B ; x \in A, x \in S \Rightarrow x \in B \text{ and } x \in B \Rightarrow x \in A, x \in S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read it as: &quot;For all $A$, for all $S$, there exists $B$ such that, if $x$ is in $A$ and $x$ is in $S$, then $x$ is also in $B$ and, if $x$ is in $B$, then $x$ is in $A$ and $x$ is in $S$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Saying that a property $S(x)$ is true is usually translated as $x \in S$). &lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-09T08:17:09.713" CommentCount="5" />
  <row Id="9544" PostTypeId="2" ParentId="9508" CreationDate="2010-11-09T08:25:07.537" Score="0" ViewCount="0" Body="&lt;p&gt;you could make sth inspired by RSA, where you do not need two keys. like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;calculate everything modulo $p$, where $p$ is prim. that will then be the GF[$p$]  &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_field&quot; rel=&quot;nofollow&quot;&gt;(GaloisField)&lt;/a&gt;.&#xA;now instead of finding a generator $g$ for that GF[$p$] and calculating anything in powers of $g$, you could use any large numbers for multiplication and addition. this allone will then be very unsafe, but it will be very chaotic in combination with some bit-operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the encoding and decoding functions can be combined out of simple unsafe functions. the combination makes them more unpredictable, especially with the bit-operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(calculate once the reciprocal ($m^{-1}\equiv m^{p-2}$) of every $m$, that you use to multiply; you'll need it for decryption.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;let's define functions enc1 and dec1 based on multiplication:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc1_m(x)=x·m$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec1_m(y)=y·m^{-1}$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and another pair based on addition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc2_a(x)=x+a$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec2_a(y)=y-a$ (mod p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and another pair based on bit-shuffling, but that need to be invertible and never generate a value $\geq p$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $(\odot,\oplus,\otimes,\neg)$ be the bitwise (and,or,xor,not) operations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $permutateAndXorLowestNBits_{s}(x,n)$ change the lowest n bits of x so that it is invertible. in other words, this condition holds:&#xA;$$\forall x,s,n: permutateAndXorLowestNBits_{s}^{-1}(permutateAndXorLowestNBits_{s}(x,n),n)=x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let N(x) be the number of lowest bits of x that can be &quot;safely&quot; changed so that after changing those bits, N will result in the same number and the value never exceeds p. in other words, these two conditions hold: $$x \oplus (2^{N(x)}-1) &amp;lt; p$$   $$\forall y. x \odot \neg(2^{N(x)}-1) \leq y \leq x \oplus (2^{N(x)}-1) \Rightarrow N(y)=N(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc3_s(x) = permutateAndXorLowestNBits_{s}(x,N(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec3_s(y) = permutateAndXorLowestNBits_{s}^{-1}(y,N(y))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, the resulting key will be (m,a,s) and the functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$enc_{m,a,s}(x)=enc1_m(enc2_a(enc3_s(x)))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dec_{m,a,s}(y)=dec3_s(dec2_a(dec1_m(y)))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but any other more complec combination would be valid, too. remember, that any combination of several enc1s and enc2s can be expressed as one single combination of enc1 and enc2; so, alternate it with enc3.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;i do not know, how hard it will be to break this if only p is known. without the bit-operations, a linear equation would break this, but with those bit-operations it will be very chaotic. does anyone know that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;otherwise, you could still make sth like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Feistel_cipher&quot; rel=&quot;nofollow&quot;&gt;Feistel cipher&lt;/a&gt;, but i do not know wether it would be more secure or how to measure that.&lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="3226" LastEditorDisplayName="" LastEditDate="2010-11-10T02:53:58.737" LastActivityDate="2010-11-10T02:53:58.737" CommentCount="2" />
  <row Id="9545" PostTypeId="1" CreationDate="2010-11-09T10:01:37.287" Score="1" ViewCount="62" Body="&lt;p&gt;Let $\Sigma$ be an alphabet of size $s$, with which we build strings of length $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Hamming ball&lt;/strong&gt; of radius $d$ centered at $x\in\Sigma^n$ is the set $B(x, d)$ of words in $\Sigma^n$ that differ from $x$ in &lt;em&gt;at most&lt;/em&gt; $d$ positions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the &lt;strong&gt;Hamming circle&lt;/strong&gt; of radius $d$ centered at $x\in\Sigma^n$ is the set $C(x, d)$ of words in $\Sigma^n$ that differ from $x$ in &lt;em&gt;exactly&lt;/em&gt; $d$ positions (note: &quot;Hamming ball&quot; is a standard term, but I do not know whether the expression &quot;Hamming circle&quot; exists at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there closed expressions for $|B(x, d_1)\cap B(y, d_2)|$ and $|C(x, d_1)\cap C(y, d_2)|$, given any two distinct strings $x,y\in\Sigma^n$ and any two $d_1,d_2\in\mathbb{N}$?&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-11-09T14:21:55.200" Title="Intersections of Hamming balls and &quot;circles&quot;" Tags="&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="9546" PostTypeId="1" AcceptedAnswerId="9557" CreationDate="2010-11-09T10:05:55.067" Score="0" ViewCount="63" Body="&lt;p&gt;Let $A \in \mathbb{R}^{n \times n}$ be a positive definite matrix and let $a &gt; 0$ be some constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the following quantity&#xA;$$&#xA;\hat x = {\rm arg} \max_{x \in R^n}\ x' \left( \frac{1}{a + \|x\|^2_2} A \right) x &#xA;$$&#xA;Is it possible to relate $\|\hat x\|_2^2$ to the eigenvalue and $\hat x$ to the eigenvector of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, suppose that $E \in \mathbb{R}^{p \times p}$ is such that $A + E$ is still positive definite and $\|E\|$ is small in some norm (say, spectral). Consider &#xA;$$&#xA;\hat y = {\rm arg} \max_{y \in R^n}\ y' \left( \frac{1}{a + \|y\|^2_2} (A+E) \right) y&#xA;$$&#xA;Are there known results that can be used to quantify $| \|\hat x\|_2^2 - \|\hat y\|_2^2 |$? I suppose the norms should be close if the perturbation $E$ is small, but don't know how to quantify this notion. &lt;/p&gt;&#xA;" OwnerUserId="2384" LastEditorUserId="2384" LastEditorDisplayName="" LastEditDate="2010-11-09T11:58:30.370" LastActivityDate="2010-11-09T12:20:23.417" Title="How to relate the following problem to known results on the eigenvectors?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;perturbation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9547" PostTypeId="1" AcceptedAnswerId="9551" CreationDate="2010-11-09T10:47:17.603" Score="0" ViewCount="52" Body="&lt;p&gt;This is from Calculus I test, that I failed to solve correctly. This really bugs me, because for &lt;code&gt;Solve[y == Log[E, 3 - 2x] / Sqrt[x + 1], x]&lt;/code&gt;, the Mathematica is telling me that: &quot;The equations appear to involve the variables to be solved for in an essentially non-algebraic way.&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Eyjzo.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the answer is $x \in \mathbb{R}$&lt;/p&gt;&#xA;" OwnerUserId="2317" LastEditorUserId="2317" LastEditorDisplayName="" LastEditDate="2010-11-09T11:11:58.967" LastActivityDate="2010-11-09T11:24:32.947" Title="Finding Range of: $y=\log_e(3-2x)/\sqrt{x+1}$?" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="9548" PostTypeId="1" AcceptedAnswerId="9565" CreationDate="2010-11-09T10:53:11.240" Score="4" ViewCount="168" Body="&lt;p&gt;f(a,b) = f(a-1, b) + f(a-1, b-1) + f(a, b-1), ab != 0 &lt;br&gt;&#xA;f(a,b) = 1, ab = 0 &lt;br&gt;&#xA;So what is f(a, b)?&lt;/p&gt;&#xA;" OwnerUserId="3307" LastActivityDate="2010-11-09T21:33:36.843" Title="General expression of f(a, b) if f(a, b)=f(a-1,b) + f(a, b-1) + f(a-1, b-1)?" Tags="&lt;recursive&gt;" AnswerCount="4" />
  <row Id="9549" PostTypeId="2" ParentId="9417" CreationDate="2010-11-09T11:12:52.663" Score="2" ViewCount="0" Body="&lt;p&gt;Just for fun here are some bounds for $\sum_{k=1}^n a^{1/k}$ for $ a \ge 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{1/k} = 1 + \frac{\log a}{1! k} + \frac{(\log a)^2}{2! k^2}&#xA;+ \frac{(\log a)^3}{3! k^3} + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n a^{1/k} = \zeta_n(0) + (\log a)\zeta_n(1) + \frac{(\log a)^2}{2!}\zeta_n(2) +&#xA;\frac{(\log a)^3}{3!}\zeta_n(3) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta_n(r) = \sum_{k=1}^n 1/k^r.$&#xA;Thus $\zeta_n(0)=n$ and $\zeta_n(1)=H_n = 1 + 1/2 + 1/3 + \cdots + 1/n.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $H_n= \log(n + 1/2) + \gamma + \epsilon(n),$ where $0&amp;lt; \epsilon(n)&amp;lt; 1/24n^2$ and $\gamma$ is the Euler-Mascheroni constant, &#xA;and (by comparing the sum with the integral of $1/x^r$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(r-1)(n+1)^{r-1}} &amp;lt; \sum_{k=n+1}^\infty \frac{1}{k^r} &amp;lt;&#xA; \frac{1}{(r-1)n^{r-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(r-1)(n+1)^{r-1}} &amp;lt; \zeta(r) - \zeta_n(r) &amp;lt;&#xA; \frac{1}{(r-1)n^{r-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ l(a) \le \sum_{k=1}^n a^{1/k} \le u(a) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma \right)&#xA;+\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n} \right)$$&#xA;$$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2n^2} \right)&#xA;+\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3n^3} \right) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma + \frac{1}{24n^2} \right)&#xA;+\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n+1} \right)$$&#xA;$$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2(n+1)^2} \right)&#xA;+\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3(n+1)^3} \right) + \cdots.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Providing $a$ is not enormous and $n$ is sufficiently large, we only need a few terms for a good approximation to our sum.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-09T11:19:41.607" LastActivityDate="2010-11-09T11:19:41.607" CommentCount="4" />
  <row Id="9550" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T11:17:45.587" Score="5" ViewCount="0" Body="&lt;p&gt;I don't know about a formula for $f(a,b)$, but the generating function is&#xA;$$\sum_{a,b=0}^\infty f(a,b)x^ay^b=\frac1{1-x-y-xy}.$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-09T11:17:45.587" />
  <row Id="9551" PostTypeId="2" ParentId="9547" CreationDate="2010-11-09T11:24:32.947" Score="2" ViewCount="0" Body="&lt;p&gt;Just observe that at $\lim_{x\to -1^+} y(x) = \infty$ and $\lim_{x \to 3/2^-} y(x) = -\infty$ and $y(x)$ is continuous on $x \in (-1, 3/2)$. Therefore the range is indeed $(-\infty,\infty)$.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-09T11:24:32.947" />
  <row Id="9552" PostTypeId="1" AcceptedAnswerId="9558" CreationDate="2010-11-09T11:29:17.660" Score="0" ViewCount="145" Body="&lt;p&gt;So, &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_space&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; has a page about &quot;affine space&quot;, where such a space is an object (set, topological space, blah) with a free transitive (blah, continuous, blah) action by the additive group of a free module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least when the base ring is an algebraically closed field, we can identify the top. sp. of $\mathbf{A}^n_k$, with such an affine space (I don't care enough to check whether or not the topology is compatible).  (The point sets agree by the Nullstellensatz)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case that k is a field but not alg. closed, is there any way to make sense of the discrepancy between the definitions, or is calling the affine variety $\mathbf{A}^n_k$ &quot;affine n-space&quot; just a relic of the ultraclassical algebraically closed case where the identification actually makese sense?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-19T01:09:12.577" LastActivityDate="2010-11-19T04:11:35.230" Title="Axiomatic affine n-space and the (classical) affine variety $A^n_k$ only agree for K an alg. closed field?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9553" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T11:32:08.457" Score="3" ViewCount="0" Body="&lt;p&gt;I doubt there's a nice closed form in both entries.  For example, the diagonal entries $f(n, n)$ have generating function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=0}^{\infty} f(n, n) x^n = \frac{1}{\sqrt{1 - 6x - x^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be proven using the methods described &lt;a href=&quot;http://qchu.wordpress.com/2009/10/07/extracting-the-diagonal/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but suggests that $f(n, n)$ doesn't have a product formula (for example like the Catalan numbers do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fixed $n$, it turns out that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{m=0}^{\infty} f(m, n) x^m = \frac{(1 + x)^n}{(1 - x)^{n+1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which translates into the binomial identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(m, n) = \sum_{k=0}^{n} {n \choose k} {m + n-k \choose n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this identity isn't terribly useful for large $n$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-09T14:02:43.353" LastActivityDate="2010-11-09T14:02:43.353" CommentCount="2" />
  <row Id="9554" PostTypeId="1" AcceptedAnswerId="9556" CreationDate="2010-11-09T12:02:18.400" Score="0" ViewCount="249" Body="&lt;p&gt;I'd heard of propositional logic for years, but until I came across &lt;a href=&quot;http://math.stackexchange.com/questions/4043/what-are-good-resources-for-learning-predicate-logic-predicate-calculus&quot;&gt;this question&lt;/a&gt;, I'd never heard of predicate logic.  Moreover, the fact that &lt;em&gt;Introduction to Logic: Predicate Logic&lt;/em&gt; and &lt;em&gt;Introduction to Logic: Propositional Logic&lt;/em&gt; (both by Howard Pospesel) are distinct books leads me to believe there are significant differences between the two fields.  What distinguishes predicate logic from propositional logic?&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-11-09T15:11:06.870" Title="What's the difference between predicate and propositional logic?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9555" PostTypeId="2" ParentId="9542" CreationDate="2010-11-09T12:02:50.743" Score="6" ViewCount="0" Body="&lt;p&gt;The axiom of specification is not a sentence. It's an &quot;axiom scheme&quot;, which is to say that it is a family of sentences. This is one of the things that becomes more clear when you move to axiomatic set theory, instead of naive set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each sentence $S(x)$ that does not mention $B$, the axiom of specification includes the axiom&#xA;$$&#xA;\forall A \exists B \forall x ( x \in B \Leftrightarrow x \in A \land S(x)). &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that axiom says, informally, is that given a set $A$ and a definition $S$ of a subset of $A$, that subset actually exists.   The scheme is slightly more general than my previous formula, because the scheme allows sentences with &quot;parameters&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The restriction that $S$ does not mention $B$ is to avoid paradoxes. Otherwise we would have as an axiom (letting $A = \{0\}$ and letting $S$ be &quot;$x \not \in B$&quot;)&#xA;$$&#xA;\exists B \forall x ( x \in B \Leftrightarrow x \in \{0\} \land x \not \in B).&#xA;$$&#xA;That set is paradoxical - it contains 0 if and only if it doesn't contain $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that we cannot quantify over sentences is that set theory is formalized using the logical system of &quot;first order logic&quot;. That system is not able to quantify over sentences. This isn't an arbitrary choice; the inability to quantify over sentences is a necessary result of certain logical properties of first-order logic that are desirable. There are other logics in which one can quantify over sentences, but these logics do not have nice properties (and some have argued these logics themselves include set theory). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is explained, in great detail, in books on axiomatic set theory. One reasonable book is Levy's &lt;em&gt;Basic set theory&lt;/em&gt;. The standard graduate textbook is Kunen's &lt;em&gt;Set theory: an introduction to independence proofs&lt;/em&gt;, and it can be used to learn axiomatic set theory, but it is somewhat terse at the beginning and is better as a second book on axiomatic set theory in my opinion. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-09T12:17:02.713" LastActivityDate="2010-11-09T12:17:02.713" CommentCount="1" />
  <row Id="9556" PostTypeId="2" ParentId="9554" CreationDate="2010-11-09T12:10:50.753" Score="2" ViewCount="0" Body="&lt;p&gt;Propositional logic (also called sentential logic) is the logic the includes sentence letters (A,B,C) and logical connectives, but not quantifiers. The semantics of propositional logic uses truth assignments to the letters to determine whether a compound propositional sentence is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicate logic is usually used as a synonym for first-order logic, but sometimes it is used to refers to other logics that have similar syntax. Syntactically, first-order logic has the same connectives as propositional logic, but it also has variables for individual objects, quantifiers, symbols for functions, and symbols for relations. The semantics include a domain of discourse for the variables and quantifiers to range over, along with interpretations of the relation and function symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many undergrad logic books will present both propositional and predicate logic, so if you find one it will have much more info. A couple well-regarded options that focus directly on this sort of thing are Mendelson's book or Enderton's book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This set of &lt;a href=&quot;http://www.math.psu.edu/simpson/courses/math557/logic.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt; by Stephen Simpson is free online and has a nice introduction to the area. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-09T12:10:50.753" />
  <row Id="9557" PostTypeId="2" ParentId="9546" CreationDate="2010-11-09T12:20:23.417" Score="2" ViewCount="0" Body="&lt;p&gt;Let $\lambda$ be the biggest eigenvalue of A and $x_{\lambda}$ a corresponding eigenvector. Since $x' A x \leq \lambda \left\| x \right\|^2$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x' \left( \frac{1}{a + \left\| x \right\|_2^2} A \right) x&#xA;\leq \frac{\lambda \left\| x \right\|^2}{a + \left\| x \right\|^2} &amp;lt; \lambda&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for $x = t x_{\lambda}$ and $t \to \infty$, the expression goes to $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the maximum is never attained.&lt;/p&gt;&#xA;" OwnerUserId="960" LastActivityDate="2010-11-09T12:20:23.417" />
  <row Id="9558" PostTypeId="2" ParentId="9552" CreationDate="2010-11-09T12:44:34.593" Score="3" ViewCount="0" Body="&lt;p&gt;Affine space means a vector space in which the origin is not preferred.  Its automorphisms are the group of affine linear transformations over $k$ (the usual linear transformations with respect to some (arbitrary) choice of origin, together with translations).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In algebraic geometry, one often encodes geometric objects in slightly complicated ways, and the preferred method of encoding is time-dependent, e.g. in that what Weil meant by a variety, and what it typically meant now, are not literally the same mathematical object, although the are both supposed to refer to the same Platonic reality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, one shouldn't think about $\mathbb A^n_k$ too literally in terms of its points when comparing with the Wikipedia discussion of affine space (since what its points are will depend on what foundations you use!).  It does encode the geometric concept of affine space (as in the linked Wikipedia article) in the context of algebraic geometry.  For example, its automorphisms are precisely the group of affine linear transformations over $k$. [Correction: As Mariano points out in a comment below, there are many non-linear automorphisms of $\mathbb A^n$; what I should have written is that the automorphisms of $\mathbb A^n$ which extend to $\mathbb P^n$ --- so these are the automorphisms of $\mathbb A^n$ that extend to the hyperplane at infinity in some reasonable sense --- are precisely the affine linear automorphisms.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do insist on making the connection precise on the level of points, then&#xA;one way to say it is that if $\Omega$ is &lt;em&gt;any&lt;/em&gt; extension field of $k$, then&#xA;the $\Omega$-valued points of $\mathbb A^n_k$ are naturally an $n$-dimensional affine space over $\Omega$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-19T04:11:35.230" LastActivityDate="2010-11-19T04:11:35.230" CommentCount="4" />
  <row Id="9561" PostTypeId="2" ParentId="9554" CreationDate="2010-11-09T14:08:37.047" Score="1" ViewCount="0" Body="&lt;p&gt;Propositional logic is an axiomatization of Boolean logic.  As such predicate logic includes propositional logic.  Both systems are known to be consistent, e.g. by exhibiting models in which the axioms are satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propositional logic is decidable, for example by the method of truth tables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_table&quot; rel=&quot;nofollow&quot;&gt; [Truth table -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &quot;complete&quot; in that every tautology in the sentential calculus (basically a Boolean expression on variables that represent &quot;sentences&quot;, i.e. that are either True or False) can be proven in propositional logic (and conversely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicate logic (also called predicate calculus and first-order logic) is an extension of propositional logic to formulas involving terms and predicates.  The full predicate logic is undecidable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/First-order_logic&quot; rel=&quot;nofollow&quot;&gt; [First-order logic -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &quot;complete&quot; in the sense that all statements of the predicate calculus which are satisfied in every model can be proven in the &quot;predicate logic&quot; and conversely.  This is a famous theorem by Gödel (dissertation,1929):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del_completeness_theorem&quot; rel=&quot;nofollow&quot;&gt; [Gödel's completeness theorem -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hardmath&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-11-09T15:11:06.870" LastActivityDate="2010-11-09T15:11:06.870" />
  <row Id="9562" PostTypeId="2" ParentId="7966" CreationDate="2010-11-09T14:09:41.900" Score="1" ViewCount="0" Body="&lt;p&gt;I like the exposition in the appendix to Eisenbud's &lt;em&gt;Commutative Algebra,&lt;/em&gt; which does everything in a kind of friendly way via exact couples. It &lt;em&gt;doesn't&lt;/em&gt; do the Grothendieck spectral sequence, though, but that's in many other places (e.g. Lang's &lt;em&gt;Algebra&lt;/em&gt;), and essentially follows (after some homological stuff) from the spectral sequence of a double complex. (I should confess that I've not myself needed to know too much about spectral sequences at the moment (beyond the Leray spectral sequence and the Cech-to-derived functor one in sheaf cohomology), and my knowledge of them is correspondingly limited.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EGA III has several applications of spectral sequences in algebraic geometry. The very first result, that the higher quasi-coherent cohomology vanishes on an affine scheme, implicitly uses the Cech-to-derived functor spectral sequence since the computation given is that of the Cech cohomology (but he doesn't actually go through the details--I think they're in Godement's &lt;em&gt;Theorie de faisceaux&lt;/em&gt;, but the deduction of it as a special case of the Grothendieck's spectral sequence is in Milne's online notes on etale cohomology). Another application in EGA III is the proof of the proper mapping theorem: that a proper map of noetherian schemes preserves coherence. The argument is very pretty, and starts in the standard case by doing this for a &lt;em&gt;projective&lt;/em&gt; morphism (which is essentially the calculation, as in Hartshorne, of the cohomology of the standard line bundles on $\mathbb{P}^n$). But to deduce it for a proper morphism, the real fun starts when you &quot;approximate&quot; a proper $f: X \to Y$  morphism by a projective one (using Chow's lemma) and then show that there is a &quot;significant&quot; sheaf on $X$ which pushes forward to $Y$ coherently using the projective morphism and the Grothendieck spectral sequence between various push-forwards. (There's also a devissage argument that I'm skipping here.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-09T14:09:41.900" />
  <row Id="9563" PostTypeId="2" ParentId="9545" CreationDate="2010-11-09T14:21:55.200" Score="2" ViewCount="0" Body="&lt;p&gt;To get started, let us work on the circle intersection where the distances are the same:&#xA;$|C(x,d)\cap C(y,d)|$.  Let the distance between $x$ and $y$ be $m$.  Then we pick $i$ places of the disagreement where we agree with $x$, have to have $i$ of the disagreement that we agree with $y$, and $d-i$ of the others where we have to disagree with both to get the correct distance.  Each of those $d-i$ can have $s-2$ choices, as they have to disagree with $x$ and $y$, as can the $m-2i$ that we didn't choose to agree with x or y.  So $$|C(x,d)\cap C(y,d)|=\sum_{i=0}^{\frac{m}{2}} \binom{m}{i} \binom{m-i}{i} \binom{n-m}{d-i} (d+m-3i)^{(s-2)}$$  If you can sum this, you are better than I.  For the ball, we can just sum over $m$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-09T14:21:55.200" />
  <row Id="9564" PostTypeId="1" CreationDate="2010-11-09T14:23:51.660" Score="2" ViewCount="27" Body="&lt;p&gt;How can I prove that the Cartier dual of $\alpha_p$ is again $\alpha_p$ (using the Yoneda lemma)? It should be something like $\alpha_p(R) \to (\alpha_p(R) \to \mu_p(R)), x \mapsto (y \mapsto exp_{p-1}(x + y)$, where $exp_{p-1}$ is the truncated exponential sequence. My problem is that this isn't a homomorphism.&lt;/p&gt;&#xA;" OwnerUserId="3267" LastActivityDate="2010-11-09T14:23:51.660" Title="Cartier dual of alpha_p" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="9565" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T14:24:02.127" Score="5" ViewCount="0" Body="&lt;p&gt;These numbers actually have a name.  They're called &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A008288&quot; rel=&quot;nofollow&quot;&gt;Delannoy numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link here is to the On-Line Encyclopedia of Integer Sequences.  Generally, when you have a sequence of integers that you want to learn more about the OEIS is a great place to start looking.  For instance, both Robin Chapman's and Qiaochu Yuan's formulas are given at the link provided, as well as some other formulas and a large number of references and related problems.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-09T14:35:42.677" LastActivityDate="2010-11-09T14:35:42.677" />
  <row Id="9567" PostTypeId="1" CreationDate="2010-11-09T14:55:04.790" Score="0" ViewCount="102" Body="&lt;p&gt;how to find the center of an arc given start and end points&lt;/p&gt;&#xA;" OwnerUserId="3310" LastActivityDate="2010-11-10T22:35:41.070" Title="how to calculate an arc center" Tags="&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9568" PostTypeId="2" ParentId="9525" CreationDate="2010-11-09T15:25:21.630" Score="7" ViewCount="0" Body="&lt;p&gt;You've got the factor wrong.  There is a well-known identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2} \left( a + b + c \right) \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you were probably thinking of, but to get the correct factorization you should take $a = x, b = \sqrt[3]{D} y, c = \sqrt[3]{D^2} z$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-09T15:25:21.630" CommentCount="1" />
  <row Id="9569" PostTypeId="1" CreationDate="2010-11-09T15:55:56.270" Score="0" ViewCount="95" Body="&lt;p&gt;Ok. That is a fact. There is a theorem that proof that. As it says in &lt;a href=&quot;http://math.stackexchange.com/questions/6148/if-a-graph-can-be-colored-with-max-4-colors-is-it-planar&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just cant understand how maths can proof such a thing, what does exactly the theorem says?? how mathematically proofs it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diego&lt;/p&gt;&#xA;" OwnerUserId="3314" LastActivityDate="2010-11-09T15:55:56.270" Title="Every planar graph can be colored with 4 colors max." Tags="&lt;graph-theory&gt;" CommentCount="5" />
  <row Id="9570" PostTypeId="2" ParentId="9522" CreationDate="2010-11-09T16:00:52.453" Score="3" ViewCount="0" Body="&lt;p&gt;The chances of a particular neighbour pair being the same colour is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{{14 \choose 8} + {14 \choose 10}}{{16 \choose 6}} = \frac{4004}{8008} = \frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the answer is $\displaystyle 7.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is happening because $\displaystyle {14 \choose 8}, {14 \choose 9}, {14 \choose 10}$ are in arithmetic progession: The number of ways of being same colour is $\displaystyle {14 \choose 8} + {14 \choose 10}$ and the number of ways of being different is $\displaystyle 2{14 \choose 9}$. The probability is $\displaystyle \frac{1}{2}$ if these two are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, $\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression if and only if $\displaystyle n+2$ is a perfect square and $\displaystyle r$ is given by $\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$ (see the end of the answer for a proof).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance, the whole bunch of problems:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;15 red, 10 blue&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;21 red, 15 blue&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;etc give rise to this neat probability of being $\displaystyle \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that n+2 is a perfect square&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression iff&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2{n \choose r+1} = {n \choose r} + {n \choose r+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2 = \frac{r+1}{n-r} + \frac{n-r-1}{r+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing some manipulations gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (n-2r-2)^2 = n+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which has an integer solution iff $\displaystyle n+2$ is a perfect square.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T18:24:03.647" LastActivityDate="2010-11-09T18:24:03.647" CommentCount="3" />
  <row Id="9571" PostTypeId="1" CreationDate="2010-11-09T16:06:24.540" Score="1" ViewCount="72" Body="&lt;p&gt;From Wikipedia's triangle center article: &quot;Thus every point is potentially a triangle center. However the vast majority of triangle centers are of little interest, just as most continuous functions are of little interest. The Encyclopedia of Triangle Centers is an ever-expanding list of interesting ones.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperbolic Barycentric Coordinates are used in:&#xA;&lt;a href=&quot;http://ajmaa.org/searchroot/files/pdf/v6n1/v6i1p18.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ajmaa.org/searchroot/files/pdf/v6n1/v6i1p18.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes a triangle center interesting, and why would hyperbolic triangle centers be interesting ? Wouldn't they just be copies of the euclidean triangle centers ?&lt;/p&gt;&#xA;" OwnerUserId="3316" LastEditorUserId="3316" LastEditorDisplayName="" LastEditDate="2010-11-10T10:05:33.940" LastActivityDate="2010-11-10T10:05:33.940" Title="Triangle centers" Tags="&lt;geometry&gt;" />
  <row Id="9572" PostTypeId="1" CreationDate="2010-11-09T16:09:16.430" Score="1" ViewCount="70" Body="&lt;p&gt;During my studies of real analysis I've come across the definition of a field (or algebra) of sets. My question is: What is the connection between this structure and the structure of a field or algebra known from abstract algebra?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-11-09T20:51:37.433" Title="Field of sets versus a field as an algebraic structure" Tags="&lt;real-analysis&gt;&lt;abstract-algebra&gt;&lt;definition&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="9573" PostTypeId="1" AcceptedAnswerId="9581" CreationDate="2010-11-09T16:13:07.893" Score="2" ViewCount="76" Body="&lt;p&gt;We have a perfect binary tree on 2^k-1 nodes. Every node in the tree is marked with probability 1/2, and a node is either marked or unmarked. We want to find a marked node and return it. Our algorithm is as follows&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;procedure find(v);&#xA; if (v is marked)&#xA;   print(v)&#xA; else if (v is not a leaf)&#xA;   v1 is the left child of v&#xA;   v2 is the right child of v&#xA;   find(v1) ( recursive call )&#xA;   find(v2) ( recursive call )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So the algorithm basically starts from the top. Searches down the tree in a depth first fashion, and returns at least one marked node if there are any. Note that the algorithm can easily return multiple marked nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I want to do is to say something about the expected number of nodes the algorithm will consider on a random tree (1/2 probability of marking nodes), when the algorithm is run on the root of the tree.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="3317" LastEditorDisplayName="" LastEditDate="2010-11-09T17:32:29.010" LastActivityDate="2010-11-09T23:54:41.857" Title="Expected time of tree search algorithm on random input" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;algorithms&gt;&lt;random&gt;&lt;expectation&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9574" PostTypeId="2" ParentId="9572" CreationDate="2010-11-09T16:32:30.607" Score="6" ViewCount="0" Body="&lt;p&gt;The term &quot;field&quot; in &quot;field / algebra of sets over A&quot; has no relation to the term &quot;field&quot; as used in ring theory. Rather it denotes a Boolean subalgebra of the power set of A, i.e. a collection of subsets of A closed under union, intersection and complement. The term is well-known due to its frequent use in the famous Stone representation theorem: $\ $every Boolean algebra is isomorphic to a field of sets. For further information see &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_of_sets&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt; and &lt;a href=&quot;http://groups.google.com/group/sci.logic/browse_frm/thread/20373f3ea47d79c0/cceccb1e3469a031&quot; rel=&quot;nofollow&quot;&gt;this sci.math thread&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Boolean algebras can be considered as &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_ring&quot; rel=&quot;nofollow&quot;&gt;Boolean rings&lt;/a&gt;, i.e. rings where every element is idempotent, i.e. $\: x^2 = x\:$. However, except for the field of two elements, such rings are never fields: $\ 0 = x^2 - x = x\ (x-1)\ \Rightarrow\ x = 0\ \ {\rm or}\ \ x = 1\ $ in a field, i.e. fields have only trivial idempotents.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-09T20:51:37.433" LastActivityDate="2010-11-09T20:51:37.433" />
  <row Id="9575" PostTypeId="1" AcceptedAnswerId="9576" CreationDate="2010-11-09T16:42:44.653" Score="2" ViewCount="220" Body="&lt;p&gt;If there is none, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the other side, what about open set (0, 1) to closed set [0, 1] with a continuous function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3318" LastActivityDate="2010-11-09T16:53:02.680" Title="Is there a continuous function from [0, 1] onto (0, 1)?" Tags="&lt;analysis&gt;&lt;topology&gt;" AnswerCount="2" />
  <row Id="9576" PostTypeId="2" ParentId="9575" CreationDate="2010-11-09T16:44:49.640" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; For the first one use the fact that, Continuous image of a compact set is compact.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-09T16:44:49.640" CommentCount="4" />
  <row Id="9578" PostTypeId="2" ParentId="9575" CreationDate="2010-11-09T16:53:02.680" Score="3" ViewCount="0" Body="&lt;p&gt;For the other side consider $f: (0,1) \to [0,1]$ defined as $f(x)= |\cos(2\pi x)|^{2}$&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-11-09T16:53:02.680" CommentCount="1" />
  <row Id="9579" PostTypeId="2" ParentId="9573" CreationDate="2010-11-09T17:09:21.853" Score="0" ViewCount="0" Body="&lt;p&gt;Idea: calculate recursively the expected number of nodes the algorithm visits on a complete binary tree of height $n$ &lt;i&gt;assuming at least one node is marked&lt;/i&gt;. Note that this assumptions alters slightly the probabilities of relevant events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This idea only makes sense if you stop after the first marked node is found, in which case JeffE's solution is much much better.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-09T19:06:52.483" LastActivityDate="2010-11-09T19:06:52.483" />
  <row Id="9580" PostTypeId="1" CreationDate="2010-11-09T17:18:59.470" Score="3" ViewCount="106" Body="&lt;p&gt;if $X_n$ converges to $X$ and $Y_n$ converges to $Y$ in distribution, what about $X_n + Y_n $ would that converge to $X+Y$ in distribution ?&#xA;any ideas how i could prove or disprove this &lt;/p&gt;&#xA;" OwnerUserId="3321" LastActivityDate="2010-11-09T18:40:17.147" Title="Linearity of convergence in distribution of random variables" Tags="&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="9581" PostTypeId="2" ParentId="9573" CreationDate="2010-11-09T17:20:36.680" Score="1" ViewCount="0" Body="&lt;p&gt;Let's assume as Yuval suggests that 2^k - 1 nodes are meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can get a recursive expression for the expected number of nodes N(k), where the basis case is N(1) = 1, i.e. that if we have one node, we have to visit it (whether or not it turns out to be marked).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If k &gt; 1, then with probability 1/2 the top node or root of the binary tree will be marked, and thus only the top node will be visited.  In the alternative we have to visit both left and right subtrees.  Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N(k+1) = 1/2 + (1/2)*[N(k) + N(k)] = 1/2 + N(k)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution of this recurrence relation + initial condition is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N(k) = (k+1)/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regardless of whether there are or are not marked nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Correction:&lt;/b&gt;  As Ross points out (see comment), the recurrence is N(k) = 1 + N(k), with solution N(k) = k, due to the fact that we will always visit the top node (before finding out whether it's marked).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-11-09T23:54:41.857" LastActivityDate="2010-11-09T23:54:41.857" CommentCount="1" />
  <row Id="9583" PostTypeId="1" CreationDate="2010-11-09T17:29:01.487" Score="2" ViewCount="93" Body="&lt;p&gt;I am looking for an algorithm that &quot;describes&quot; the following situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15 - 22 people are combined into groups of  4 or 5 based upon stated preferences of who they wish to be grouped with.  Each person will be grouped with at least one of their preferences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't &quot;homework&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3322" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T21:03:42.897" LastActivityDate="2010-11-11T20:14:07.673" Title="Looking for an algorithm for grouping based on preferences" Tags="&lt;combinatorics&gt;&lt;algorithms&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="9584" PostTypeId="2" ParentId="9573" CreationDate="2010-11-09T17:32:34.017" Score="2" ViewCount="0" Body="&lt;p&gt;Assuming nodes are marked by independent coin flips, then the order that you visit nodes has no effect whatsoever.  So the expected number of nodes visited is exactly the expected number of fair coin flips to get heads: 2.&lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2010-11-09T17:32:34.017" CommentCount="2" />
  <row Id="9587" PostTypeId="2" ParentId="9580" CreationDate="2010-11-09T17:55:04.670" Score="1" ViewCount="0" Body="&lt;p&gt;Convergence in distribution is a pretty weak concept.. Suppose you consider probability distributions on $[0,1]$. Let $X = Y = X_n$ for all $n$ have a density supported on $[0,1/2]$ alone, and let $Y_n$ be the same distribution except shifted to the right by $1/2$. Then all these random variables have the same distribution, so convergence in distribution is automatic. But also each $X_n + Y_n$ is the same, but different from $X + Y$, so you won't get convergence in distribution. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-09T17:55:04.670" />
  <row Id="9588" PostTypeId="2" ParentId="1890" CreationDate="2010-11-09T17:55:45.697" Score="8" ViewCount="0" Body="&lt;p&gt;Here is a solution found at &lt;a href=&quot;http://books.google.com/books?id=Xw5T-a-CaUkC&amp;amp;lpg=PP1&amp;amp;pg=PA189#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KIqSX.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-09T17:55:45.697" CommunityOwnedDate="2010-11-09T17:55:45.697" />
  <row Id="9589" PostTypeId="2" ParentId="6428" CreationDate="2010-11-09T17:59:53.627" Score="4" ViewCount="0" Body="&lt;p&gt;There's a canonical example of fibrations used in algebraic topology which are not fibre bundles.  That is, given a continuous function $f : X \to Y$ there is a homotopy-equivalence $\phi : X' \to X$ and a fibration $f' : X' \to Y$ such that $f \circ \phi$ is homotopic to $f'$.  I believe this idea goes back to Serre (or perhaps earlier).  The fibre of $f'$ is called the homotopy-fibre of $f$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common usage of this construction is with the Postnikov Tower of a space.  In his dissertation Jean-Pierre Serre used this (and some &quot;closure&quot; observations of the Serre spectral sequence of a fibration, then it was called Serre $\mathcal C$-theory, nowadays this technology is subsumed in localization) to show that most of the homotopy-groups of the spheres are finite.  &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-09T17:59:53.627" />
  <row Id="9590" PostTypeId="2" ParentId="2120" CreationDate="2010-11-09T18:25:33.667" Score="2" ViewCount="0" Body="&lt;p&gt;Here is a solution found at &lt;a href=&quot;http://www.imc-math.org.uk/imc1999/prob_sol1.pdf&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/z4ZFY.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-09T18:25:33.667" CommunityOwnedDate="2010-11-09T18:25:33.667" />
  <row Id="9591" PostTypeId="2" ParentId="9580" CreationDate="2010-11-09T18:31:25.663" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose that $X_n$ converges in distribution to $X$ where $X$ is a symmetric random variable, say $X \sim N(0,1)$. Then, trivially, $X_n$ also converges in distribution to $-X$ (since $X$ and $-X$ are identically distributed). However, $X_n + X_n$ does not converge in distribution to $X+(-X)=0$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-09T18:40:17.147" LastActivityDate="2010-11-09T18:40:17.147" />
  <row Id="9592" PostTypeId="1" AcceptedAnswerId="9599" CreationDate="2010-11-09T18:37:19.987" Score="5" ViewCount="100" Body="&lt;p&gt;I have an interesting math question I've developed while playing cards with my family and wonder if anyone has a way to approach the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have n people who each have a deck of cards. Each deck has a different number of cards in it. We are shuffling the cards for a game and in the end we all want to have the same number of cards to shuffle. Each individual can do one of two actions per round of shuffling. They can shuffle their deck, or they can split their deck in half and swap one half of their deck with half of another person's deck. My question is, is there an equation based on n to describe a method of swapping that will eventually give all n players the same amount of cards? Furthermore, how many rounds does it take? (In my mind, I define a round as a period of time where everyone performs one action. A player can only perform one action per round and the round ends when all players have performed one action)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the problem simpler, you can assume that each individual can only do one action: split their deck and swap with any other player.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is trivial when n is a power of 2. For example if we have 2 people, they split their decks, swap, and they have the same number of cards. It takes one round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4 people, you split them into 2 groups of 2 people and perform the swap internally in the group. Then you make 2 groups with one person from each group in round one and have them swap. This takes 2 rounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see if there is an simple elegant solution when n is not a power of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can also assume that each deck splits evenly every time, which is a big assumption but should give a good basis for how to solve the problem. A similar problem would be to replace the cards with cups of liquid which gets rid of passing half of cards around.)&lt;/p&gt;&#xA;" OwnerUserId="3324" LastEditorUserId="3324" LastEditorDisplayName="" LastEditDate="2010-11-09T20:43:46.027" LastActivityDate="2010-11-09T22:34:06.730" Title="Card shuffling to Even Decks" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="9593" PostTypeId="2" ParentId="9592" CreationDate="2010-11-09T18:44:12.747" Score="2" ViewCount="0" Body="&lt;p&gt;It depends on the number of cards each person holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance with 3 people, one holding 8 cards, one 16 and the other 32 you can never get each to have the same number, as the total is not divisible by 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I misunderstood your intent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cards were &quot;liquid&quot; (or basically arbitrary real numbers), even then, there is no procedure (independent of the cards each person holds), which will end in finite number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you start with 1,1,2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a finite number of steps each will have a finite representation when written in base 2, but $4/3$ does not have a finite representation in base 2.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T19:09:46.077" LastActivityDate="2010-11-09T19:09:46.077" CommentCount="6" />
  <row Id="9594" PostTypeId="1" CreationDate="2010-11-09T18:54:51.987" Score="3" ViewCount="184" Body="&lt;p&gt;I am writing a program to solve a Rubik's cube, and would like to know the answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 12 ways to make one move on a Rubik's cube. How many ways are there to make a sequence of six moves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my project's specification: up to six moves may be used to scramble the cube. My job is to write a program that can return the cube to the solved state. I am allowed to use up to 90 moves to solve it. Currently, I can solve the cube, but it takes me over 100 moves (which fails the objective)... so I ask this question to figure out if a brute force method is applicable to this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of ways to make six moves is not overly excessive, I can just make six random moves, then check to see if the cube is solved. Repeat if necessary.&lt;/p&gt;&#xA;" OwnerUserId="3325" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-11T02:18:18.260" LastActivityDate="2011-01-10T02:44:42.150" Title="How many ways can I make six moves on a Rubik's cube?" Tags="&lt;probability&gt;&lt;counting&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9595" PostTypeId="1" AcceptedAnswerId="9601" CreationDate="2010-11-09T18:56:47.543" Score="1" ViewCount="102" Body="&lt;p&gt;I'm considering a parametrized &lt;a href=&quot;http://en.wikipedia.org/wiki/Sigmoid_function&quot; rel=&quot;nofollow&quot;&gt;sigmoid&lt;/a&gt; function such as the following logistic:&lt;br&gt;&#xA;$$f(x)=\frac{e^{a+bx}}{1+e^{a+bx}}$$&lt;br&gt;&#xA;And I'm interested only in the interval $\displaystyle x &gt;= 0$ and $\displaystyle x &amp;lt; x_{max}$ (with a given $x_{max}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two tiny questions regarding its properties:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is the &lt;strong&gt;sum of sigmoid functions always a sigmoid function itself&lt;/strong&gt;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is the log function (i.e. $\displaystyle f(x) = log_{a}(bx^{c})$ or similar) &lt;strong&gt;a particular case of a sigmoid function&lt;/strong&gt;? In other words, can I pick parameters for the logistic function above so that it behaves like a log function in the given interval?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="2916" LastEditorDisplayName="" LastEditDate="2010-11-09T21:40:16.653" LastActivityDate="2010-11-10T00:58:17.697" Title="Properties of sigmoid functions" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9596" PostTypeId="1" CreationDate="2010-11-09T19:01:47.023" Score="3" ViewCount="224" Body="&lt;p&gt;I have always taken for granted that expected value is a linear operator. For &lt;em&gt;any&lt;/em&gt; random variables $X$ and $Y$: $E(aX + bY) = aE(X) + bE(Y)$. Can anyone point me to a rigorous proof of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I know that generally median $Med()$ is not a linear operator, meaning $Med(aX + bY)$ might not be equal to $a Med(X) + b Med(Y)$. Are there absolute criteria / rules when $Med$ is a linear operator, and when it is not?&lt;/p&gt;&#xA;" OwnerUserId="3326" LastEditorUserId="3326" LastEditorDisplayName="" LastEditDate="2010-11-09T22:37:03.357" LastActivityDate="2010-11-11T21:58:06.740" Title="Expected value is a linear operator? Under what conditions is median also a linear operator?" Tags="&lt;probability-theory&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="9597" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T19:03:00.513" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the definition of $$ E(X) = \sum\limits_{x} x P(X)$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-09T19:03:00.513" CommentCount="2" />
  <row Id="9598" PostTypeId="1" AcceptedAnswerId="9600" CreationDate="2010-11-09T19:47:53.720" Score="1" ViewCount="165" Body="&lt;p&gt;How does one prove this identity: $$\lim_{n \to \infty} 2^{2n-1} \sqrt{n} \frac{ \Gamma(n)^{2}}{\Gamma(2n)} = \sqrt{\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from &lt;em&gt;Gamelin : Complex Analysis&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-10T02:17:06.250" Title="Proving $\lim_{n \to \infty} 2^{2n-1} \sqrt{n} \frac{ \Gamma(n)^{2}}{\Gamma(2n)} = \sqrt{\pi}$" Tags="&lt;complex-analysis&gt;&lt;special-functions&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9599" PostTypeId="2" ParentId="9592" CreationDate="2010-11-09T19:48:09.663" Score="6" ViewCount="0" Body="&lt;p&gt;There is a standard problem (I often assign it in linear algebra, and I was assigned the problem in representation theory); seems like it might be related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $n$ people sitting in a circular table, each of them with two neighbors. They each have a bowl of porridge in front of them, and a spoon in each hand. Unhappy with their individual portions, they all simultaneously grab half the porridge from their left-hand neighbor, half the porridge from the right-hand neighbor, and they put it in their bowl (which has been emptied by his neighbors doing the same). What happens to the distribution of porridge over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: More explicit, applied to the cards.&lt;/em&gt; So, let's consider the case of three eaters/shufflers. I'm assuming for convenience now that the cards are, as Moron suggests, &quot;liquid&quot;; we'll get back to the discrete case presently. Let $x_i(n)$ be the number of cards that the $i$th person has after $n$ steps, with $\mathbf{x}(0) = (a_0,b_0,c_0)$ being the initial distribution of cards (so $a_0,b_0,c_0\geq 0$). Then the procedure can be represented as&#xA;$$\begin{array}{rcrcrcl}&#xA;x_1(n+1) &amp;amp;=&amp;amp;  &amp;amp; &amp;amp;\frac{1}{2}x_2(n) &amp;amp; + &amp;amp; \frac{1}{2}x_3(n)\\&#xA;x_2(n+1) &amp;amp; = &amp;amp; \frac{1}{2}x_1(n) &amp;amp; &amp;amp; &amp;amp;+&amp;amp; \frac{1}{2}x_3(n)\\&#xA;x_3(n+1) &amp;amp; = &amp;amp; \frac{1}{2}x_1(n) &amp;amp; + &amp;amp; \frac{1}{2}x_2(n)&#xA;\end{array}$$&#xA;So we can view this as the linear system $\mathbf{x}(n+1) = A\mathbf{x}(n)$, where&#xA;$$A = \left(\begin{array}{lll}&#xA;0 &amp;amp; 0.5 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0.5 &amp;amp; 0&#xA;\end{array}\right).$$&#xA;The characteristic polynomial of $A$ is $-t(t-0.5)^2$. Since the rows all add up to $1$, then $(1,1,1)^T$ is an eigenvector of $\lambda=1$ (that is, if everyone starts with $1$, then after applying the procedure they all end up again with $1$: $(1,1,1)^T$ is mapped to one times itself). A basis for the eigenspace corresponding to $\lambda=-\frac{1}{2}$ is $(1,-1,0)^T$ and $(1,0,-1)^T$. If, for instance, player one began with one card, and player two with one &quot;anti-card&quot; (a card made of antimatter, say), then after the first step, player three got half a card from 1 and half an anti-card from 2, which vanished in a puff of smoke; player 2 now has half a card he got from 1, and player 1 has half an anti-card he got from 2. So $(1,-1,0)^T$ is mapped to $(-\frac{1}{2},\frac{1}{2},0)$. Similarly with $(1,0,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because $(1,1,1), (1,-1,0), (1,0,-1)$ is a basis for $\mathbb{R}^3$, our original distribution $\mathbf{x}(0)$ can be written uniquely as&#xA;\[ \mathbf{x}(0) = a(1,1,1) + b(1,-1,0) + c(1,0,-1),\]&#xA;for some $a$, $b$, and $c$. Note that the number of cards among all three players is $3a$. If we apply the procedure $n$ times to get $\mathbf{x}(n)$, we will get:&#xA;\begin{align*}&#xA;\mathbf{x}(n) &amp;amp;= A^n\mathbf{x}(0)\\&#xA;&amp;amp;= A^n(a(1,1,1)^T + b(1,-1,0)^T + c(1,0,-1)^T)\\&#xA;&amp;amp;= (a,a,a)^T + (-0.5)^n(b,-b,0)^T + (-0.5)^n(c,0,-c)^T\\&#xA;&amp;amp;= (a + (-0.5)^n(b+c), a-(0.5)^nb, a-(0.5)^nc)^T.&#xA;\end{align*}&#xA;Now, in reality we are interested in the floor of these quantities (since we are really dealing with actual cards, no negative and fractional cards). As soon as $(0.5)^nb$ and $(0.5)^nc$ are both smaller than $\frac{1}{4}$, the given distribution will be essentially the even distribution $(a,a,a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explicit example: suppose that we start with Player 1 having $30$ cards, Player $2$ with $16$ cards, and player $3$ with $22$ cards (I picked the numbers at random). First, we write $(30,16,22)$ as a linear combination of $(1,1,1)$, $(1,-1,0)$, and $(1,0,-1)$. Solving the corresponding system we get:&#xA;$$(30,16,22) = \frac{68}{3}(1,1,1) + \frac{20}{3}(1,-1,0) + \frac{2}{3}(1,0,-1).$$&#xA;So after $n$ applications of the procedure, we will have&#xA;$$&#xA;\mathbf{x}(n) = \frac{68}{3}(1,1,1) + \frac{(-1)^n20}{3(2^n)}(1,-1,0) + \frac{(-1)^n2}{3(2^n)}(1,0,-1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each player will have exactly one third of the cards, modified by a bit. &#xA;If $n=2$, then the coefficient of $(1,-1,0)$ is $1.67$ and the coefficient of $(1,0,-1)$ is $0.167$; so you would expect the first player to have the nearest integer to  $\frac{68}{3}+1.67$ cards, namely about $24$; the second player should have the nearest integer to $\frac{68}{3} -1.67$ cards, so about $21$ cards; and the third player the nearest integer to $\frac{68}{3}-0.167$ cards; this is almost $22.5$, so he'll have $23$ cards (to complete the tally). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=3$, you get about $22$ cards for the first player, $23$ for the second, and $23$ for the third). If $n=4$, the first player should have approximately $23$ cards; the second player should have approximately $22$ cards; and the third player should have approximately $23$ cards. After this point, you'll just be shuffling around who is one card short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar phenomenon occurs with any odd number of people. This may not be the fastest procedure for dividing them among an odd number of players, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to do this with $n=4$ people, then the matrix you get is:&#xA;$$ A = \left(\begin{array}{llll}&#xA;0 &amp;amp; 0.5 &amp;amp; 0 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0 &amp;amp; 0.5 &amp;amp; 0\\&#xA;0 &amp;amp; 0.5 &amp;amp; 0 &amp;amp; 0.5\\&#xA;0.5 &amp;amp; 0 &amp;amp; 0.5 &amp;amp; 0&#xA;\end{array}\right).$$&#xA;This time, the characteristic polynomial is $t^2(t-1)(t+1)$, so you have one eigenvalue $\lambda=1$ (with corresponding eigenvector $(1,1,1,1)$, which is associated to &quot;the cards are evenly distributed&quot;); you get one eigenvalue $\lambda=-1$, with corresponding eigenvector $(1,-1,1,-1)$; and you get two eigenvectors mapping to $0$; for instance $(0,1,0,-1)$ and $(1,0,-1,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you express your original distribution as a linear combination of these,&#xA;\[ (x_1(0),x_2(0),x_3(0),x_4(0)) = \alpha(1,1,1,1) + \beta(1,-1,1,-1) + \gamma(0,1,0,-1) + \delta(1,0,-1,0),\]&#xA;the last two components don't matter after the first go around, so you end up with &#xA;\[ \mathbf{x}(n) = \bigl( \alpha + (-1)^n\beta, \alpha - (-1)^n\beta, \alpha+(-1)^n\beta, \alpha-(-1)^n\beta\bigr).\]&#xA;So depending you are going to have a certain portion of the cards evenly distributed (corresponding to $\alpha$), and a certain portion of the cards (corresponding to $\beta$) swapping between even- and odd-numbered players at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: Another way of looking at the $n=4$ case:&lt;/em&gt; A better way of seeing what is happening in this case is to replace the vector $(1,-1,1,-1)$ with the vector $(2,0,2,0)$ (the sum of the eigenvector and $(1,1,1,1)$); we still have a basis, so we can write any initial distribution as&#xA;$$(x_1(0),x_2(0),x_3(0),x_4(0)) = \alpha(1,1,1,1) + \beta(2,0,2,0) + \gamma(0,1,0,-1)$ + \delta(1,0,-1,0).$$&#xA;But now, when you apply $A$, the vector $(2,0,2,0)$ transforms into the vector $(0,2,0,2)$, and this vector is in turn transformed back into $(2,0,2,0)$. The total number of cards is $4(\alpha+\beta)$. Of these, $4\alpha$ of the cards will eventually end up evenly distributed among the players, but the remaining $4\beta$ cards will end up evenly distributed between two non-adjacent players, and then swapped over to the other two players, then swapped back. If you imagine you are in a bridge table, first the North-South partnership has the extra $4\beta$ cards, $2\beta$ cards each, then they go to East-West, then back to North-South, then back ot East-West, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar phenomenon occurs with any even number of players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; &lt;em&gt;Back to the discrete case.&lt;/em&gt; Okay, above we were dealing with the &quot;liquid cards&quot;/&quot;porridge&quot; case, where the cards are divisible any way we want. What happens if we are playing actual cards that we don't want to split up? Notice that each player is only required to divide his cards in half; so the only problem arises if a player has an odd number of cards. In that case, the player can just give the extra card to whichever of his neighbors he likes the most, or randomly; the final distribution in this case is very close to the ideal &quot;continuous&quot; distribution: each of his neighbors was supposed to get $x$-and-a-half cards from him, but got either $x$ or $x+1$; so the most that any one person can be off from the &quot;ideal&quot; result of applying the procedure is $1$ card (half a card from one neighbor, half a card from the other). This vector will be fairly close to the &quot;ideal&quot; continuous vector, and so the result of applying the procedure to this distribution will also result in a distribution which is fairly close to what the ideal distribution is. (Linear maps are continuous, and in this case the eigenvalues are all of absolute value less than or equal to $1$, so the approximation will not get worse as time goes on; it will either remain the same, or get better). You can see this in the example I worked out for $n=3$. If you start with $(30,16,22)$, then after the first step you get $(19,26,23)$; at the next step, the &quot;predicted&quot; distribution is $(24,21,23)$. The actual distributions you get are either $(24,22,22)$, $(25,21,22)$, $(24,21,23)$, or $(25,20,23)$, depending on how the extra cards get distributed. The sup-distance to the predicted distribution is 1 in all cases. Do it a third time, and depending on how things break you will end up in either $(22,24,22)$, $(21,24,23)$, $(22,23,23)$, $(21,23,24)$, $(23,23,22)$, or $(21,25,22)$. The &quot;worst&quot; distribution (the last one) is still better than the worst distribution in the previous step. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary:&lt;/strong&gt; the &quot;give half your cards to each of your neighbors and get half of each of their cards&quot; will yield an even distribution after a fairly small-ish number of steps if there is an odd number of players, but will not necessarily yield and even distribution if there is an even number of players. In the latter case, part of the deck will get evenly distributed, while the other part ends up swapping between adjacent &quot;teams&quot;.  One advantage of this, at least for an odd number of players, is that it is a single procedure that is repeated a number of times (much like riffle-shuffling a deck) in order to achieve uniformity, rather than a more complex algorithm that involves different actions at different steps.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-09T22:34:06.730" LastActivityDate="2010-11-09T22:34:06.730" CommentCount="11" />
  <row Id="9600" PostTypeId="2" ParentId="9598" CreationDate="2010-11-09T19:50:50.147" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;Stirling's approximation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-09T19:50:50.147" />
  <row Id="9601" PostTypeId="2" ParentId="9595" CreationDate="2010-11-09T19:58:39.573" Score="4" ViewCount="0" Body="&lt;p&gt;I am assuming that a sigmoid function is bounded, and its graph has exactly one inflection point (among other properties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (1), then, the answer is no.  If $f(x)=\frac{e^{3x}}{1+e^{3x}}$ and $g(x)=\frac{e^{2x-5}}{1+e^{2x-5}}$, then $h(x)=f(x)+g(x)$ can be seen to have two points of inflection, so h(x) is not sigmoid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2), your function is unbounded on $0\le x \le x_{\mbox{max}}$, but sigmoid functions are bounded.  So these functions have decidedly different behavior.  Though, it does depend somewhat on what you mean by &quot;like&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-09T19:58:39.573" CommentCount="5" />
  <row Id="9602" PostTypeId="1" AcceptedAnswerId="9603" CreationDate="2010-11-09T20:10:48.530" Score="2" ViewCount="66" Body="&lt;p&gt;What are the open sets of a Stone space of a Boolean algebra B?&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stone_space&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;, they are the ultrafilters on B that contain an element of B.  However, this cannot be the case since an ultrafilter is a set of subsets of B and contains no elements of B.  Using the alternate definition of the Stone space as the space of homomorphisms from B to {0,1} doesn't offer much help, either.&lt;/p&gt;&#xA;" OwnerUserId="3327" LastActivityDate="2010-11-09T20:28:45.933" Title="Topology on Stone Spaces of Boolean Algebra" Tags="&lt;topology&gt;&lt;logic&gt;&lt;boolean-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9603" PostTypeId="2" ParentId="9602" CreationDate="2010-11-09T20:18:14.460" Score="5" ViewCount="0" Body="&lt;p&gt;I think you are confused by the two meanings of ultrafilter. In the context of a Boolean algebra $B$, an ultrafilter is just a maximal proper filter, that is, a subset of $B$ that is closed under $\wedge$, upward closed, and does not contain $0$. The Stone space $M(B)$ consists of all such maximal filters, and its usual topology is generated by basic open sets of the form&#xA;$$&#xA;N_p = \{ f \in M(B) : p \in f \}&#xA;$$&#xA;with one such basic open set $N_p$ for each $p \in B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a second, incompatible meaning of ultrafilter: an ultrafilter on an arbitrary set $S$ is a maximal filter in the above sense on the lattice of subsets of $S$. This is what people mean when they say &quot;an ultrafilter on $\omega$&quot;, for example.  When $S$ is already being treated as a lattice, the former sense is usually what is meant, but you can only tell by context. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-09T20:28:45.933" LastActivityDate="2010-11-09T20:28:45.933" />
  <row Id="9605" PostTypeId="1" AcceptedAnswerId="9609" CreationDate="2010-11-09T20:48:42.753" Score="1" ViewCount="72" Body="&lt;p&gt;As an exercise in Python programming I've written a script that finds amicable pairs, and I've had fun doing so. But I'm wondering if amicable pairs have a use, in mathematics or elsewhere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a reminder:&lt;br&gt;&#xA;Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number. (A proper divisor of a number is a positive integer divisor other than the number itself. For example, the proper divisors of 6 are 1, 2, and 3.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the smallest pair of amicable numbers is (220, 284); for the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110, of which the sum is 284; and the proper divisors of 284 are 1, 2, 4, 71, and 142, of which the sum is 220.&#xA;(from the Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Amicable_number&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first few amicable pairs, according to &lt;em&gt;Math World&lt;/em&gt;, are:&#xA;(220, 284), (1184, 1210), (2620, 2924) (5020, 5564), (6232, 6368), (10744, 10856), (12285, 14595), (17296, 18416), (63020, 76084)&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-09T20:58:56.720" Title="Amicable pairs: any use for them yet?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="9606" PostTypeId="1" CreationDate="2010-11-09T20:54:06.547" Score="1" ViewCount="115" Body="&lt;p&gt;I have a set of functions that map square matrices with $n$ rows and columns to square matrices with $k &amp;lt; n$ rows and columns.  Is there a name for this property?  I know that 'projection' would be the wrong word.  Also I don't mean that the function &lt;em&gt;only&lt;/em&gt; changes the size, it might also do other things.&lt;/p&gt;&#xA;" OwnerUserId="3282" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-09T21:59:35.210" LastActivityDate="2010-11-11T20:22:07.407" Title="What is the name for a function of a matrix that changes the matrix size?" Tags="&lt;terminology&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9608" PostTypeId="1" AcceptedAnswerId="9643" CreationDate="2010-11-09T20:55:28.983" Score="1" ViewCount="141" Body="&lt;p&gt;I have a stochastic variable x with this property: if it's measured at &#xA;t1 and again at t2, then x(t2)-x(t1) has a normal distribution with &#xA;mean 0 and standard deviation Sqrt[t2-t1]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the distribution of the maximum value this variable &#xA;reaches between t1 and t2, or confirm my intuition that this is not &#xA;well-defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach: break [t1,t2] into multiple intervals and integrate the &#xA;following (PDF[NormalDistribution[x,y]] = normal probability &#xA;distribution with mu of x and standard deviation of y): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PDF[NormalDistribution[0,Sqrt[1/4]]][x0]* &#xA;PDF[NormalDistribution[x0,Sqrt[1/4]]][x1]* &#xA;PDF[NormalDistribution[x1,Sqrt[1/4]]][x2]* &#xA;PDF[NormalDistribution[x2,Sqrt[1/4]]][x3] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each xi is integrated from -Infinity to m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This specific example computes the probability that the maximum on &#xA;[0,1] is less than m by breaking [0,1] into 4 parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking [0,1] into more parts should yield more accurate results, &#xA;although I slightly suspect that the limit diverges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica slows to a crawl even breaking [0,1] into 5 or more parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried replacing the normal distribution with others (uniform, &#xA;DeltaDirac, C/(1+x^2), etc), with no better luck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling yields many results (this appears to be a &quot;Wiener Process&quot;), &#xA;but I can't find the actual distribution of the maximum anywhere (nor &#xA;does it say anywhere that such a maximum doesn't exist). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimate goal is to price box options:&#xA;&lt;a href=&quot;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&quot;&gt;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2010-11-10T18:00:28.353" Title="Determining distribution of maximum of dependent normal variables" Tags="&lt;integral&gt;&lt;normal-distribution&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9609" PostTypeId="2" ParentId="9605" CreationDate="2010-11-09T20:58:56.720" Score="5" ViewCount="0" Body="&lt;p&gt;Apparently, John Conway says no.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, John Conway says they are not very useful mathematically:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The only application or use for these numbers is the original one -&#xA;     you insert a pair of amicable numbers into a pair of amulets, of &#xA;     which you wear one yourself and give the other to your beloved!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://mathforum.org/library/drmath/view/52560.html&quot; rel=&quot;nofollow&quot;&gt;Dr. Math&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-09T20:58:56.720" />
  <row Id="9610" PostTypeId="2" ParentId="9598" CreationDate="2010-11-09T20:59:23.833" Score="5" ViewCount="0" Body="&lt;p&gt;You can also use &lt;a href=&quot;http://en.wikipedia.org/wiki/Wallis_product&quot; rel=&quot;nofollow&quot;&gt;Wallis product&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Qxr9R.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be shown by considering $\displaystyle \int_{0}^{\pi/2} \sin^n x$ (as Wallis originally did) or using the product expansion for $\sin x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sin x = x\prod_{n=1}^{\infty} (1- \frac{x^2}{\pi^2 n^2})$ and setting $\displaystyle x=\pi/2$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-09T20:59:23.833" />
  <row Id="9611" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T21:13:02.193" Score="0" ViewCount="0" Body="&lt;p&gt;For the second question, we want to show that $\text{Med}(aX+bY)$ might not equal $a \text{Med}(X)+ b \text{Med}(Y)$. Suppose $f(x)$ is the density function of $X$ and $g(y)$ is the density function of $Y$. Let $m_1$ be the median of $X$ and $m_2$ be the median of $Y$. Then $m_1$ is the value for which $\int_{0}^{m_1} f(x) \ dx = 0.5$. Likewise, $m_2$ is the value for which $\int_{0}^{m_2} g(y) \ dy = 0.5$. I guess you could use this fact to show that in general, $\text{Med}(aX+bY) \neq a \text{Med}(X)+ b \text{Med}(Y)$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-09T21:13:02.193" />
  <row Id="9612" PostTypeId="1" CreationDate="2010-11-09T21:18:38.870" Score="2" ViewCount="122" Body="&lt;p&gt;How to solve the recursion: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p[n,m] = p[n-1,m-1] + p[n+1,m-1] + p[n,m-1]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally in general, but if you need base cases: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p[n,0] = 0 \text{ (for } n \neq 0),$$ &#xA;$$p[0,0] = 1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've asked a similar question previously and believe $n^{m-2}$ is &#xA;somehow involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would turning this into a DFQ help? &lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-10T18:15:16.377" LastActivityDate="2010-11-10T18:15:16.377" Title="Solve recursion $p[n,m] = p[n-1,m-1] + p[n+1,m-1] + p[n,m-1]$" Tags="&lt;recursive&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="9613" PostTypeId="2" ParentId="9606" CreationDate="2010-11-09T21:20:07.137" Score="2" ViewCount="0" Body="&lt;p&gt;Well, &lt;em&gt;projection&lt;/em&gt; could be the right word, depending on what you want to do with the matrices. Note that matrices with $n$ rows and $m$ columns form a vector space of dimension $n \times m$. If you now restrict to matrices of $k &amp;lt; n$ rows, you can treat them as a $k \times m$-dimensional subspace of the original vector space, having entries in rows $i &gt; k$ set to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the dual operation of reducing the number columns might be called a &lt;em&gt;restriction&lt;/em&gt;. This comes from the fact that you can treat $n \times m$ dimensional matrix as a representation of an operator mapping $m$-dimensional vector space to a $n$-dimensional vector space and reducing number of columns restricts this operator to a subspace of the original vector space.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-09T21:20:07.137" />
  <row Id="9614" PostTypeId="2" ParentId="9612" CreationDate="2010-11-09T21:31:00.293" Score="2" ViewCount="0" Body="&lt;p&gt;See OEIS &lt;a href=&quot;http://www.research.att.com/~njas/sequences/A027907&quot; rel=&quot;nofollow&quot;&gt;A027907&lt;/a&gt; and references on trinomial coefficients.  If you just start making the triangle in a spreadsheet and type the numbers into OEIS you will find it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-09T21:31:00.293" />
  <row Id="9615" PostTypeId="2" ParentId="9548" CreationDate="2010-11-09T21:33:36.843" Score="0" ViewCount="0" Body="&lt;p&gt;Following Robin Chapman's answer, $\sum_{a,b \geq 0} f(a,b)x^a y^b = \frac{1}{1-x-y-xy} = \sum_{n \geq 0} (x+y+xy)^n = \sum_{n \geq 0} \sum_{i+j+k=n} \frac{n!}{i!j!k!} x^i y^j (xy)^k$&#xA;so $f(a,b)=\sum_{k=0}^{\min (a,b)} \frac{(a+b-k)!}{(a-k)!(b-k)!k!}$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-09T21:33:36.843" />
  <row Id="9616" PostTypeId="1" AcceptedAnswerId="9618" CreationDate="2010-11-09T21:42:44.793" Score="0" ViewCount="118" Body="&lt;p&gt;Take a Multiplicative group G, defined by a prime p.  Now take an element t from it.  Its inverse is t&lt;sup&gt;-1&lt;/sup&gt;.  To calculate the inverse you can use the extended Euclidian algorithm or this trick, &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_multiplicative_inverse#Direct_Modular_Exponentiation&quot; rel=&quot;nofollow&quot;&gt;which I found on wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t&lt;sup&gt;-1&lt;/sup&gt; = t&lt;sup&gt;p-2&lt;/sup&gt; mod p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this makes some sense to me, but the following, which I found in some code, does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take t, and raise it to u (also in G):  t&lt;sup&gt;u&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(t&lt;sup&gt;u&lt;/sup&gt;)&lt;sup&gt;-1&lt;/sup&gt; = t&lt;sup&gt;p-1-u&lt;/sup&gt; mod p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this trick called, and why does it work?&lt;/p&gt;&#xA;" OwnerUserId="3179" LastActivityDate="2010-11-10T00:15:09.273" Title="Why does this trick to find the inverse of an element in a Multiplicative Cyclic Group work?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9617" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T21:43:40.207" Score="3" ViewCount="0" Body="&lt;p&gt;First question: By definition, on a probability space ($\Omega,\mathcal{F},P$), the expected value of a random variable $X:\Omega\to \mathbb{R}$ is defined as&#xA;$$E(X)=\int_\Omega X(\omega) dP(\omega).$$ Note that it is only well-defined if the integral converges absolutely, i.e. $$\int_\Omega |X(\omega)| dP(\omega)&amp;lt;\infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The integrals above are Lebesgue integrals. If $X$ is discrete, $P$ is the point measure and this integral turns into a sum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if the expected values of both $X$ and $Y$ exist, then $E(X+Y)$ exists (triangle inequality) and, for $a,b\in \mathbb{R}$, we can use the linearity of the Lebesgue integral to conclude&#xA;$$E(aX+bY)=\int_\Omega aX +bYdP=a\int_\Omega X dP+b\int_\Omega YdP=aE(X)+bE(Y).$$&lt;/p&gt;&#xA;" OwnerUserId="3244" LastActivityDate="2010-11-09T21:43:40.207" CommentCount="2" />
  <row Id="9618" PostTypeId="2" ParentId="9616" CreationDate="2010-11-09T21:46:10.070" Score="3" ViewCount="0" Body="&lt;p&gt;The &quot;trick&quot; is that $\displaystyle t^{p-1} = 1 \mod p$ always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\displaystyle t^{p-1-u} \cdot t^u = 1 \mod p$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle t^u$ and $\displaystyle t^{p-1-u}$ are inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for a general group $G$ and $a, b \in G$, $a^b$ does not make much sense as a power of $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Arturo points out, it in fact is used to denote the element $b^{-1}ab$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-09T21:56:22.383" LastActivityDate="2010-11-09T21:56:22.383" CommentCount="5" />
  <row Id="9619" PostTypeId="2" ParentId="9616" CreationDate="2010-11-09T21:52:43.223" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad\quad  A^N = 1\ \ \Rightarrow\ \ A^{-K}\ =\ \frac{1}{A^K}\ =\ \frac{A^N}{A^K}\ =\ A^{N-K} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not really a &quot;trick&quot; nor does it have any standard name. It's simply a consequence of the cyclic structure, viz $\rm\ -K \ \equiv\ N-K\ \:(mod\ N)\:$. This is already quite familiar to everyone in the form of clock arithmetic, e.g. 1:50  = 10 before 2,$\ $ i.e. $\rm\: -10\ \equiv\ 50\ \ (mod\ 60)\:.\:$  Note that, as a consequence, there can be no concept of sign (positive/negative) in cyclic groups, i.e. they cannot be ordered. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-10T00:15:09.273" LastActivityDate="2010-11-10T00:15:09.273" />
  <row Id="9620" PostTypeId="1" CreationDate="2010-11-09T22:07:10.980" Score="2" ViewCount="137" Body="&lt;p&gt;I get confused in the following problem. Actually I don't know how to derive the derivative of the Jacobian. Could anybody help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a smooth vector field $\vec{b}$ on $R^{n}$, let $\vec{X}(s)=\vec{X}(s,x,t)$solves the ODE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{\vec{X}}=\vec{b}(\vec{X})(s \in R)$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{X}(t)=\vec{X}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the Jacobian $J(s,x,t)=det D_{x}\vec{X}(s,x,t)$,  derive Euler's formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_{s}=(div\vec{b}(\vec{X}))J$&lt;/p&gt;&#xA;" OwnerUserId="2855" LastActivityDate="2011-01-08T22:44:40.600" Title="A problem in Evan's PDE 2nd edition. (Problem 6 on Pg 162)" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9621" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T22:14:31.650" Score="4" ViewCount="0" Body="&lt;p&gt;EXTENDED/REVISED ANSWER&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some general points concerning the second question. By definition, $m$ is a median of $X$ if ${\rm P}(X \ge m) \geq 1/2$ and ${\rm P}(X \le m) \geq 1/2$. While a median is uniquely determined for any common example of a continuous random variable, it is not uniquely determined in general. For example, any number $m \in [-1,1]$ is a median for random variable $X$ with ${\rm P}(X=1) = {\rm P}(X=-1) = 1/2$. Hence my previous answer to this question (see below), where I assumed that $m(X)=0$ since $X$ is symmetric, should be revised. This is done simply as follows. We define $X$ and $Y$ exactly as before, and introduce another random variable $\tilde X$ defined to be equal to $X$ with probability $1-1/n$ and to $0$ with probability $1/n$. It is immediately checked that the symmetric random variables $\tilde X$ and $Y$ have a unique median, equal to $0$. Thus $m(\tilde X) + m(Y) = 0$, as required. On the other hand, one easily verifies that ${\rm P}(\tilde X + Y = 1) \to 3/4$ as $n \to \infty$ (cf. my previous answer), which implies that $\tilde X + Y$ has a unique median, equal to $1$. So, $m(\tilde X + Y) \neq m(\tilde X) + m(Y)$, as required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In view of this example, we now give a counterexample for the case where $X$ and $Y$ are independent. Let $X$ and $Y$ be i.i.d. random variables with common probability mass function given by $p(2)= p(-1) = \frac{1}{2}(1 - \frac{1}{n})$, $p(0) = \frac{1}{n}$. Then, $X$ and $Y$ have a unique median, equal to $0$. On the other hand, one verifies that both ${\rm P}(X+Y \geq 1)$ and  ${\rm P}(X+Y \leq 1)$ tend to $3/4$ as $n \to \infty$; hence, $X + Y$ has a unique median, equal to $1$. So, $m(X + Y) \neq m(X) + m(Y)$, as required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In view of the preceding examples, we finally consider the case where $X$ and $Y$ are both symmetric and independent. Assuming both $X$ and $Y$ have a unique median, it must obviously be equal to $0$. For any fixed numbers $a$ and $b$, $aX + bY$ is also symmetric. Moreover, $aX + bY$ has a unique median, equal to $0$. This can be carried out straightforwardly, upon observing that ${\rm P}(X \in (-\varepsilon,\varepsilon), Y \in (-\varepsilon,\varepsilon)) &gt; 0$ for any $\varepsilon &gt; 0$. Hence, $m(aX+bY)=am(X)+bm(Y)=0$. From this, it easy to establish the following generalization. Suppose that $X$ and $Y$ are independent and symmetric around arbitrary points, say $m_1$ and $m_2$, respectively. Assume that both $X$ and $Y$ have a unique median (these medians are necessarily given by $m(X)=m_1$ and $m(Y)=m_2$). Then, for any fixed numbers $a$ and $b$, $m(aX+bY)$ has a unique median, equal to $am(X)+bm(Y)=am_1+bm_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PREVIOUS ANSWER&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, let us show that even if $X$ and $Y$ are symmetric random variables, then $m(X+Y)$ might be different from $m(X)+m(Y)$ (where $m$ denotes median). Suppose that ${\rm P}(X=1) = {\rm P}(X=-1) = 1/2$; hence $X$ is symmetric. Define $Y$ as follows: if $X=1$ then $Y=0$, whereas if $X=-1$ then $Y=2$ or $-2$ with probability $1/2$ each. Then, ${\rm P}(Y=0)=1/2$, ${\rm P}(Y=2)=1/4$, and ${\rm P}(Y=-2)=1/4$. Hence $Y$ is symmetric and, in turn, $m(X)+m(Y)=0+0=0$. However, ${\rm P}(X+Y=1) = 3/4$ (and ${\rm P}(X+Y=-3) = 1/4)$. In particular, $m(X+Y)=1$ (since ${\rm P}(X+Y=1) \geq 1/2$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-11T21:58:06.740" LastActivityDate="2010-11-11T21:58:06.740" CommentCount="3" />
  <row Id="9622" PostTypeId="1" AcceptedAnswerId="9729" CreationDate="2010-11-09T22:18:04.183" Score="1" ViewCount="96" Body="&lt;p&gt;Reading a book on fractional calculus reminded me that I'd like to know more on the following method/idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an integral: $\displaystyle \int_{-\pi}^\pi{\frac{\displaystyle\sum_{j=a}^b{e^{i\cdot c_j \cdot t}}}{\displaystyle\sum_{k=a}^b{e^{i \cdot d_k \cdot t}}} dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use recurrences and generating functions to find all of the derivatives.  Using all of the derivatives (evaluated at a point), I can essentially create another function (for example, the function at point $n$ is the $n$th derivative).  The function brings up a curiousity:  how accurate could it be at predicting the integral itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;  An additional thought:  If this method fails, perhaps the derivatives could still be used to aid in a very good spline extrapolation using only a few points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering what problems arise when trying to extrapolate/interpolate an integral of the form above using derivatives.  I'd like to know any studies of this problem if they exist, and what I can expect.  I realize I'm being a bit vague, but I'm not exactly sure what the good questions are regarding this issue.  I would like to explore this possibility as much as possible, and hope that answers arrive with hope and how to evaluate the integral this way.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-09T23:40:34.020" LastActivityDate="2010-11-10T16:21:59.640" Title="Using derivatives to evaluate an integral of exponentials" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9623" PostTypeId="1" AcceptedAnswerId="9628" CreationDate="2010-11-09T22:20:03.730" Score="5" ViewCount="158" Body="&lt;p&gt;I know the general Black-Scholes formula for Option pricing theory (for calls and puts), however I want to know the other solutions to the Black-Scholes PDE and its various boundary conditions. Can someone start from the B-S PDE and derive its various solutions based on different boundary conditions? Even if you could provide some links/sources where it is done, I'll appreciate that. The point is that I want to know various other solutions and their boundary conditions which are derived from Black-Scholes PDE. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="754" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T15:25:12.367" LastActivityDate="2010-11-10T16:45:47.060" Title="Black Scholes PDE and its many solutions" Tags="&lt;differential-equations&gt;&lt;finance&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9624" PostTypeId="1" CreationDate="2010-11-09T22:28:28.493" Score="0" ViewCount="52" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/9606/what-is-the-name-for-a-function-of-a-matrix-that-changes-the-matrix-size&quot;&gt;What is the name for a function of a matrix that changes the matrix size?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;What are some examples of named functions that change the size of a matrix?  For example, I guess a 'projection' does this under certain circumstances?  What are other examples?&lt;/p&gt;&#xA;" OwnerUserId="3282" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-09T22:51:00.357" LastActivityDate="2010-11-09T22:51:00.357" ClosedDate="2010-11-11T11:08:05.473" Title="names of functions which change the size of a matrix" Tags="&lt;terminology&gt;&lt;big-list&gt;&lt;matrices&gt;" CommentCount="1" />
  <row Id="9625" PostTypeId="2" ParentId="9583" CreationDate="2010-11-09T22:34:47.303" Score="1" ViewCount="0" Body="&lt;p&gt;This appears to be a variation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stable_marriage_problem&quot; rel=&quot;nofollow&quot;&gt;marriage problem&lt;/a&gt;. You might want to consider work related to this one.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-09T22:34:47.303" />
  <row Id="9626" PostTypeId="2" ParentId="9620" CreationDate="2010-11-09T22:41:57.293" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of the determinant can be evaluated using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant#Derivative&quot; rel=&quot;nofollow&quot;&gt;adjugate&lt;/a&gt;. Then it remains to compute &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{ds} D_x X(s,x,t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which you need to use the ODE.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-09T22:41:57.293" CommentCount="1" />
  <row Id="9627" PostTypeId="1" CreationDate="2010-11-09T22:58:07.070" Score="4" ViewCount="91" Body="&lt;p&gt;Say I have two locally free sheaves $F,G$ on projective variety $X$. I know the cohomology groups $H^i(X,F)$ and $H^i(X,G)$. Is this enough to give me information about $H^i(X,F\otimes G)$? In particular, if $H^i(X,F)=0$, what conditions on $G$ guarantee that also $H^i(X,F\otimes G)=0$?&lt;/p&gt;&#xA;" OwnerUserId="3332" LastActivityDate="2010-11-09T22:58:07.070" Title="Cohomology of a tensor product of sheaves" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="9628" PostTypeId="2" ParentId="9623" CreationDate="2010-11-09T23:05:46.703" Score="7" ViewCount="0" Body="&lt;p&gt;The Black-Scholes &lt;em&gt;formula&lt;/em&gt; involving the standard normal distribution is specific to call or put options.  The Black-Scholes &lt;em&gt;formalism&lt;/em&gt;, relating the prices to random walks and PDE, works for pricing a European option with arbitrary payoff.  For any boundary condition (except some artificial ones with  incredibly rapid growth that makes the random walk expectations diverge) the price is the expected value of the option value at the time of maturity, the expectation taken over price paths of the underlying security whose &quot;risk-adjusted&quot; drift and volatility are calculated from the parameters assumed to govern the &quot;physical&quot; price movement.  So one does not need the whole theory of random walks, just the probability density of where the risk-adjusted path will land at the time of maturity, and integrating the European option payoff against this density will give the answer.  The Black-Scholes theory shows that the result satisfies the PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, Black Scholes is, after a change of variables, equivalent to a &quot;backwards&quot; heat diffusion equation -- one whose time parameter is $(-t)$, or better, $(T_{m} - t)$ where $T_m$ is the maturity time of the option.  So your question is the same as asking for all solutions of the heat equation. I assume that means heat diffusion with arbitrary boundary conditions.  As long as the boundary conditions do not grow extremely fast the expectation of a random walk gives the solution. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-10T16:45:47.060" LastActivityDate="2010-11-10T16:45:47.060" />
  <row Id="9629" PostTypeId="1" AcceptedAnswerId="9630" CreationDate="2010-11-09T23:11:26.617" Score="1" ViewCount="86" Body="&lt;p&gt;Let M  be an (n-1)-manifold in R^n . Let M(e)  be the set of end-points of normal vectors (in both directions) of length e  and suppose e  is small enough so that M(e) is also an (n-1)-manifold. Show that M(e)  is orientable (even if M  is not)&lt;/p&gt;&#xA;" OwnerUserId="3333" LastActivityDate="2010-11-09T23:19:02.167" Title="Orientation on a Manifold" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9630" PostTypeId="2" ParentId="9629" CreationDate="2010-11-09T23:19:02.167" Score="2" ViewCount="0" Body="&lt;p&gt;Each point $p$ of $M(\epsilon)$ comes from a point $q$ in $M$. Consider the vector $X$ field on $M(\epsilon)$ which on $p$ takes the value $\vec{qp}$. Then $N$ is a non-zero normal field on $M(\epsilon)$ (maybe one needs to consider the projection of $N$ onto the normal line to $M(\epsilon)$ at each point, but that projection is surely non-zero and, moreover, to normalize this projection)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-09T23:19:02.167" />
  <row Id="9631" PostTypeId="2" ParentId="9525" CreationDate="2010-11-09T23:32:53.793" Score="0" ViewCount="0" Body="&lt;p&gt;Thanks to Qiaochu Yuan's identity I was able to factor into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{2}(x+Ny+N^2z)[(x-Ny)^2+(Ny-N^2z)^2+(N^2z-x)^2] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows that if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x^3+Dy^3+D^2z^3-3Dxyz=1 $ and $ x,y,z $ are very large positive, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x-Dy $ and $ y-Dz $ must be small.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2010-11-09T23:32:53.793" />
  <row Id="9632" PostTypeId="2" ParentId="9612" CreationDate="2010-11-09T23:41:55.893" Score="1" ViewCount="0" Body="&lt;p&gt;Given your boundary conditions, if we open up the recurrence relation then we discover an $m$-step process, where at each step we change $n$ by $0$ or $\pm 1$. For every walk that ends up at $0$, we &quot;score&quot; one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many such walks exist? There is an equal number $k$ of $+1$s and $-1$s, and so we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=0}^{\lfloor m/2 \rfloor} \binom{m}{k,k} = \sum_{k=0}^{\lfloor m/2 \rfloor} \frac{m!}{k!k!(m-2k)!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to estimate it, let's calculate the largest coefficient, assuming $m$ is divisible by $3$. It is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{m!}{(m/3)!} \approx \frac{\sqrt{2\pi m}(m/e)^m}{\sqrt{2\pi (m/3)}^3 (m/3e)^m} = \frac{3^m}{(2\pi/3^{1.5})m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly, we expect about $O(\sqrt{m})$ of the summands to be of comparable size, so a conjectured upper bound would be $O(3^m/\sqrt{m})$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-09T23:41:55.893" />
  <row Id="9633" PostTypeId="2" ParentId="9420" CreationDate="2010-11-09T23:47:11.487" Score="0" ViewCount="0" Body="&lt;p&gt;In WEP, encryption is done by XORing some stream cipher (RC4), and CRC is used as a signature (can't remember if CRC is applied before or after encryption). Since both operations commute, you can modify a packet by XORing and then fix the signature, without ever knowing the key (or the parts of the packet you didn't modify).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CRC is often used as a hash function, although nowadays MD5 is more common. CRC is linear, and WEP is a case where that's a problem (even checksum would've been better).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-09T23:47:11.487" />
  <row Id="9634" PostTypeId="2" ParentId="9596" CreationDate="2010-11-09T23:56:40.117" Score="3" ViewCount="0" Body="&lt;p&gt;Intuitive (and somewhat informal) justification for the linearity of expectation. Suppose that random variables $X$ and $Y$ have expectation ${\rm E}(X)$ and ${\rm E}(Y)$, respectively. Suppose also that $(X_n,Y_n)$ is a sequence of i.i.d. random vectors having the (joint) distribution of $(X,Y)$. Then, in particular, $X_n$, $Y_n$, and $aX_n + bY_n$ are sequences of i.i.d. random variables having the distribution of $X$, $Y$, and $aX + bY$, respectively. Thus, from&#xA;$$&#xA;a{\rm E}(X) + b{\rm E}(Y) \stackrel{{\rm a.s.}}{=} a\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n} + b\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {Y_i } }}{n},&#xA;$$&#xA;we conclude that&#xA;$$&#xA;{\rm E}(aX + bY) \stackrel{{\rm a.s.}}{=} \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {(aX_i  + bY_i )} }}{n}  \stackrel{{\rm a.s.}}{=} a{\rm E}(X) + b{\rm E}(Y),&#xA;$$&lt;br&gt;&#xA;where $\stackrel{{\rm a.s.}}{=}$ stands for `almost surely equal', and where we have used the strong law of large numbers, namely, if $Z_n$ is a sequence of i.i.d. random variables having the distribution of a random variable $Z$ with finite expectation, then ${\rm E}(Z) \stackrel{{\rm a.s.}}{=} \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\nolimits_{i = 1}^n {Z_i } }}{n}$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-09T23:56:40.117" />
  <row Id="9635" PostTypeId="2" ParentId="9595" CreationDate="2010-11-10T00:58:17.697" Score="0" ViewCount="0" Body="&lt;p&gt;It is incorrect to speak of &quot;the sigmoid function&quot;.  The Wikipedia link shows that there are many to choose from.  For (2), note that $\displaystyle f(x) = \log_{a}(bx^{c})=c\log_{a}{x}+\log_{a}{b}=d\ln{x}+e$ by absorbing the constants into the change of base of log.  This is definitely not sigmoidal.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T00:58:17.697" />
  <row Id="9636" PostTypeId="1" CreationDate="2010-11-10T02:00:53.490" Score="0" ViewCount="113" Body="&lt;p&gt;Let $L^{\infty}$ denote the set of all essentially bounded functions and suppose that&#xA;$f \in L^{\infty}$ and $g \in L^{\infty}$. Then the product $fg$ is in $L^{\infty}$. Now the question is: prove or disprove: $\|fg\|_\infty \leq \|f\|_\infty \|g\|_\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="3337" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T16:15:01.740" LastActivityDate="2010-11-10T16:15:01.740" Title="Prove or disprove $L^{\infty}$ inequality" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9637" PostTypeId="2" ParentId="9598" CreationDate="2010-11-10T02:17:06.250" Score="0" ViewCount="0" Body="&lt;p&gt;Wolfram|Alpha shows an &lt;a href=&quot;http://www.wolframalpha.com/input/?i=limit+as+n+goes+to+infinity+of+%28+%28power%282%2C2n-1%29%29+%28sqrt%28n%29%29+%28power%28gamma%28n%29%2C2%29+%2F+gamma%282n%29%29+%29&quot; rel=&quot;nofollow&quot;&gt;expansion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3293" LastActivityDate="2010-11-10T02:17:06.250" />
  <row Id="9638" PostTypeId="1" CreationDate="2010-11-10T02:24:24.893" Score="2" ViewCount="133" Body="&lt;p&gt;Our math prof gave us this and we are not sure what to make of the notation, can someone give us a hand? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/r5Fvf.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3338" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T05:36:11.640" LastActivityDate="2010-11-10T05:36:11.640" Title="Notation help. What do $\mathcal{P}(A)$ and $\chi_{-}$  mean here?" Tags="&lt;homework&gt;&lt;notation&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9639" PostTypeId="2" ParentId="9636" CreationDate="2010-11-10T02:26:47.593" Score="1" ViewCount="0" Body="&lt;p&gt;What Qiaochu said, and use the definition of the $L^\infty$-norm along with a similar result regarding absolute values.&lt;/p&gt;&#xA;" OwnerUserId="2527" LastActivityDate="2010-11-10T02:26:47.593" CommentCount="3" />
  <row Id="9640" PostTypeId="2" ParentId="9638" CreationDate="2010-11-10T02:32:30.580" Score="3" ViewCount="0" Body="&lt;p&gt;$\mathcal{P}(A)$ is the power set of $A$, i.e., the set of subsets of $A$, and $2^A$ is the set of functions from $A$ to $\{0,1\}$.  The characteristic function $\chi_S$ of a subset $S$ of $A$ is the function $\chi_S:A\to\{0,1\}$ defined by $\chi_S(x)=1$ if $x$ is in $S$ and $\chi_S(x)=0$ if $x$ is not in $S$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it is a good idea to ask your professor to clarify notation on the homework if it is not easy to find in the notes or text.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T02:32:30.580" CommentCount="4" />
  <row Id="9641" PostTypeId="2" ParentId="9638" CreationDate="2010-11-10T02:38:02.560" Score="5" ViewCount="0" Body="&lt;p&gt;$\mathcal{P}(A)$ is the power set of $A$; it is a set whose elements are precisely the subsets of $A$. For example, if $A=\{1,2,3\}$, then $\mathcal{P}(A) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any two sets $A$ and $B$, $A^B$ represents the set of all functions from $B$ to $A$. So $2^A$, with &quot;$2$&quot; meaning the set $\{0,1\}$ means the set of all functions $f\colon A\to \{0,1\}$; all functions whose domain is $A$, and which take only the values $0$ and $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is a set and $S$ is a subset, then $\chi_S$ is the characteristic function of $S$. It is a function $\chi_S\colon A\to \{0,1\}$ defined by:&#xA;$$\chi_S(a) = \left\{\begin{array}{ll}&#xA;0 &amp;amp;\mbox{if $a\not\in S$,}\\&#xA;1 &amp;amp; \mbox{if $a\in S$.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;$\chi_{-}$&quot; he means the function that takes each subset $S$ to the characteristic function $\chi_S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no new notation in part (b).&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-10T02:38:02.560" />
  <row Id="9642" PostTypeId="1" CreationDate="2010-11-10T02:38:17.190" Score="2" ViewCount="75" Body="&lt;p&gt;Assume a capacitated network flow problem with graph $G = (N,A)$ and capacities $u_{ij}$ (finite), costs for each path of $c_{ij}$ and I/O of $b_{i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it be shown there is an equivalent transportation problem with a source node for every arc $(i,j)$ in $N$ having supply $u_{ij}$, and a sink node with demand $\sum _{k | (i,k) \in A}  u_{ik}-b_{i}$ for every node $i$ in $N$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At every supply node $(i,j)$ there are two outgoing infinite capacity arcs: one goes to demand node $i$, and its cost coefficient is 0; the other goes to demand node $j$ and its cost coefficient is $c_{ij}$.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-14T17:22:11.077" LastActivityDate="2010-11-14T17:33:41.537" Title="Convert capacitated network flow problem" Tags="&lt;network-flow&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="9643" PostTypeId="2" ParentId="9608" CreationDate="2010-11-10T02:43:41.063" Score="5" ViewCount="0" Body="&lt;p&gt;You have a stochastic process $\lbrace{X(t):t \geq 0\rbrace}$ with the property that $X(t_2)-X(t_1) \sim N(0,t_2 - t_1)$, which is quite obviously supposed to be Brownian motion (BM). Suppose first that you want to find the distribution function of the running maximum $M(t)=\mathop {\max }\limits_{0 \le s \le t} X(s)$ (the maximum exists, since BM has continuous sample paths). There is a very simple formula for that, namely:&#xA;$$&#xA;{\rm P}(M(t) \le x) = \sqrt {\frac{2}{{\pi t}}} \int_0^x {e^{ - u^2 /(2t)} {\rm d}u}, \;\; x \geq 0. &#xA;$$&#xA;The situation is a little more complicated if you want to find the distribution function of $&#xA;M(t_1 ,t_2 ) = \mathop {\max }\limits_{t_1  \le s \le t_2 } X(s)$ (i.e., the maximum of $X$ over the time interval $[t_1,t_2]$). For this purpose, we need to condition on the initial value $X(t_1)$. Since $X(t_1) \sim N(0,t_1)$, it has density function $f(u;t_1) = \frac{1}{{\sqrt {2\pi t_1 } }}e^{ - u^2 /(2t_1 )}$, and by the law of total probability we have&#xA;$$&#xA;{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty  {{\rm P}(M(t_1 ,t_2 ) \le x|X_{t_1 }  = u)f(u;t_1) {\rm d}u}. &#xA;$$&#xA;Now, as follows from basic properties of BM, conditioned on $X_{t_1}=u$, $M(t_1 ,t_2 )$ can be replaced by $u + M(0,t_2 - t_1)$, i.e. by $u + M(t_2 - t_1)$ (more precisely, by $u$ plus an independent copy of $M(t_2 - t_1)$, which is independent of $X_{t_1}$). This leads to&#xA;$$&#xA;{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty  {{\rm P}(M(t_2  - t_1 ) \le x - u)f(u;t_1 ){\rm d}u}. &#xA;$$&#xA;Finally, since $M(t_2  - t_1 )$ cannot be negative, we have to integrate only from $-\infty$ to $x$. That is,&#xA;$$&#xA;{\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^x {{\rm P}(M(t_2  - t_1 ) \le x - u)f(u;t_1 ) {\rm d}u}, \;\; x \in {\bf R}.&#xA;$$&#xA;So, we have a double integral with an elementary integrand. Maybe one can simplify it. Also, maybe one can find the result in the literature. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-10T17:54:31.510" LastActivityDate="2010-11-10T17:54:31.510" CommentCount="6" />
  <row Id="9644" PostTypeId="2" ParentId="9636" CreationDate="2010-11-10T03:14:57.393" Score="0" ViewCount="0" Body="&lt;p&gt;Isn't it true that $$  ||fg||_{\mathrm{sup}} \leq ||f||_{\mathrm{sup}}||g||_{\mathrm{sup}} $$ for any two bounded function f and g? I do not seem to understand the subtlety here. Doesn't this prove the question (since f and g can be unbounded only over a measure zero set)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T04:51:27.783" LastActivityDate="2010-11-10T04:51:27.783" CommentCount="1" />
  <row Id="9645" PostTypeId="1" AcceptedAnswerId="9646" CreationDate="2010-11-10T03:34:48.483" Score="1" ViewCount="114" Body="&lt;p&gt;When I first learned linear algebra, both the professor and the book used brackets like &lt;code&gt;[&lt;/code&gt; and &lt;code&gt;]&lt;/code&gt; to enclose matrices. However, in my current differential equations textbook, matrices are enclosed by parenthesis, and I suddenly realize everybody else are using them too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are brackets/parenthesis for enclosing matrices always totally interchangeable?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T03:54:57.467" LastActivityDate="2010-11-10T04:05:32.813" Title="Parenthesis vs brackets for matrices" Tags="&lt;matrices&gt;&lt;notation&gt;" AnswerCount="2" />
  <row Id="9646" PostTypeId="2" ParentId="9645" CreationDate="2010-11-10T03:44:22.780" Score="5" ViewCount="0" Body="&lt;p&gt;It is just a question of notation and what you are comfortable with. The only thing you might want to be aware is |A| stands for determinant of the matrix A and hence cannot be used in place of () or [] to represent the matrix&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T03:44:22.780" />
  <row Id="9647" PostTypeId="2" ParentId="9645" CreationDate="2010-11-10T04:05:32.813" Score="3" ViewCount="0" Body="&lt;p&gt;They're interchangeable. I think a lot of people tend to use parentheses just because they're easier to write by hand.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-10T04:05:32.813" CommentCount="3" />
  <row Id="9648" PostTypeId="1" AcceptedAnswerId="9674" CreationDate="2010-11-10T04:26:19.850" Score="9" ViewCount="155" Body="&lt;p&gt;To motivate my question, recall the following well-known fact: Suppose that $p\equiv 1\pmod 4$ is a prime number. Then the equation $x^2\equiv -1\pmod p$ has a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can show this as follows: Consider the following polynomial in ${\mathbb Z}_p[x]$: $x^{4k}-1$, where $p=4k+1$. The roots of this polynomial are precisely the elements of ${\mathbb Z}_p^*$, each one with multiplicity 1. The polynomial factors as $(x^{2k}-1)((x^k)^2+1)$, and it follows that if $a$ is any element of ${\mathbb Z}_p^*$ with $a^{2k}\ne 1$ (and there are precisely $(p-1)/2$ possible such $a$), then $b=a^k$ satisfies $b^2\equiv -1\pmod p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are other arguments, but I am interested in pursuing this line of reasoning. My question is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the quadratic reciprocity law, we&#xA;  have that $x^2\equiv -2\pmod p$ has a&#xA;  solution iff $p\equiv 1$ or $3\pmod&#xA;  8$. Is there a proof of the&#xA;  right-to-left implication using some&#xA;  polynomial and appropriate counting of&#xA;  roots, as in the case shown above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3340" LastActivityDate="2010-11-10T16:38:25.087" Title="A particular case of the quadratic reciprocity law" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="9649" PostTypeId="1" AcceptedAnswerId="9727" CreationDate="2010-11-10T04:46:11.093" Score="2" ViewCount="128" Body="&lt;p&gt;This is a real-analysis homework question so I of course have to be very precise and justify anything or any theorem I use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b_n$ is a bounded sequence and $\lim(a_n) = 0$, show that $\lim(a_nb_n) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, since $b_n$ is bounded, then sup($b_n$) is some finite number and therefore we can take an $N$ natural number as large as we need such that for all $n\gt N$ $b_na_n$ approaches $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought to use the limit theorems, but since $a_n$ is not bounded, the general limit theorems do not reply. (I am referring to $\lim(X + Y) = \lim X + \lim Y$ for $X,Y$ sequences etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking then to use the definition of the limit somehow to show that since $b_n$ is bounded we can take as intuitively stated above $N$ large enough to show the statement is true.  I'm not sure how to proceed with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your replies in advance!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T04:52:40.623" LastActivityDate="2010-11-10T16:20:13.650" Title="If $b_n$ is a bounded sequence and $\lim a_n = 0$, show that $\lim(a_nb_n) = 0$" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="9650" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:51:57.030" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Let $N(\epsilon)$ be such that $\frac{1}{N(\epsilon)}&amp;lt;\frac{\epsilon}{M}$, where $|b_{n}| \leq M$ ( since $b_{n}$ is bounded.) Then think of $|a_{n}b_{n} - 0|$ for $n \geq N(\epsilon)$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-10T04:51:57.030" />
  <row Id="9651" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:52:34.737" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: If $|a_n|\leq\varepsilon$, then $|a_nb_n|\leq\varepsilon |b_n|\leq \varepsilon \sup_k|b_k|$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T04:52:34.737" />
  <row Id="9652" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:53:39.920" Score="2" ViewCount="0" Body="&lt;p&gt;Since the $b_n$ are bounded, there is a constant $C$ such that $-C\lt b_n\lt C$ for all $n$. Fix $n$. Suppose first $a_n\gt 0$. Then $-C a_n&amp;lt; a_nb_n &amp;lt; Ca_n$. Similarly, if $a_n\lt 0$ then $Ca_n &amp;lt; a_n b_n &amp;lt; -C a_n$. In short: $|a_n b_n|\lt C|a_n|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for any sequence $c_n$, $c_n\to 0$ iff $|c_n|\to 0$. Right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $C|a_n|\to 0$ since $a_n\to 0$. Since $|a_nb_n|\le C|a_n|$, we must also have $|a_n b_n|\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is just a sketch. Let me know if you need me to clarify something.)&lt;/p&gt;&#xA;" OwnerUserId="3340" LastActivityDate="2010-11-10T04:53:39.920" />
  <row Id="9653" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T04:55:52.213" Score="3" ViewCount="0" Body="&lt;p&gt;You write *... but since $a_n$ is not bounded...*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you mean that you have not been explicitly told that the sequence $a_n$ is bounded. That is true; you have not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as it happens, the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $c_n$ be any sequence of real numbers. If $c_n$ converges, then it is bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, why is this true? Well, for large enough $n$, the values of $c_n$ must be close to the limit $L$; so only the &quot;early&quot; values of $c_n$ can get away from the limit, and since there are only finitely many of them, they can only go so far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, try to prove this carefully, so you can conclude that $a_n$ &lt;strong&gt;is&lt;/strong&gt; bounded, so that the theorems you want to apply &lt;em&gt;can&lt;/em&gt; be applied.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-10T04:55:52.213" CommentCount="2" />
  <row Id="9654" PostTypeId="1" CreationDate="2010-11-10T04:59:16.333" Score="0" ViewCount="117" Body="&lt;p&gt;Suppose $\mu(X) &amp;lt; \infty$ and $0 &amp;lt; p&amp;lt; q \leq \infty$. I'm reading the following proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $q=\infty$ then $L^{q}(\mu) \subseteq L^{p}(\mu)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $q=\infty$ then $\int |f|^{p} \leq (||f||_{\infty})^{p}  \int 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why we have the above inequality?? &lt;/p&gt;&#xA;" OwnerUserId="3337" LastActivityDate="2010-11-10T06:45:42.513" Title="Please explain inequality" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="9655" PostTypeId="1" AcceptedAnswerId="9658" CreationDate="2010-11-10T05:00:02.030" Score="4" ViewCount="98" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a compact, connected smooth manifold. If $p, q$ are points in $M$, is there always a geodesic that goes from $p$ to $q$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is certainly not true if $M$ is not compact, but I couldn't find a counterexample for the compact case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody help me out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2010-12-07T20:37:25.857" Title="Geodesics on Compact Manifolds" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9656" PostTypeId="2" ParentId="9654" CreationDate="2010-11-10T05:03:53.357" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Note that $-||f|\|_{\infty} \leq f(x)\leq ||f||_{\infty}$ almost everywhere. Hence...&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-10T05:03:53.357" />
  <row Id="9657" PostTypeId="2" ParentId="9654" CreationDate="2010-11-10T05:05:54.870" Score="1" ViewCount="0" Body="&lt;p&gt;Let's just write $\|f\|$ for the infinity norm. We have that $\|f\|$ is defined as the essential sup of the values that $f$ takes. But then $|f(x)|^p\le\|f\| ^p$ for any $x\in X$ except in a set of measure 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, what you have is $\int|f|^p\le \int\|f\| ^p$, because the set where the inequality between $|f|$ and $\|f\|$ is reversed is negligible. And, since $\|f\| ^p$ is a constant, we can move it out of the integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note $\|f\|$ could very well be infinite, but in that case the inequality you need is obvious, because anything is $\le\infty$. So, only the case where $\|f\| &amp;lt;\infty$ needs arguing.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ok, I was using the wrong norm. You have $\|f\|$ defined as the infimum of $\|h\|_ {\rm sup}$ where $h$ ranges over all (measurable) functions that agree with $f$ a.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix any such $h$. Then $|f|=|h|$ a.e., so $|f|^p=|h|^p\le \|h\|^ p_ {\rm sup}$ a.e. So $\int |f|^p=\int |h|^p\le \|h\|^p_ {\rm sup}\int 1$, as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the infimum of the right hand side.&lt;/p&gt;&#xA;" OwnerUserId="3340" LastEditorUserId="3340" LastEditorDisplayName="" LastEditDate="2010-11-10T05:28:32.113" LastActivityDate="2010-11-10T05:28:32.113" CommentCount="4" />
  <row Id="9658" PostTypeId="2" ParentId="9655" CreationDate="2010-11-10T05:08:10.353" Score="7" ViewCount="0" Body="&lt;p&gt;Yes. This is part of the classical Hopf-Rinow theorem, &lt;em&gt;q.v.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-10T05:08:10.353" CommentCount="1" />
  <row Id="9659" PostTypeId="1" AcceptedAnswerId="9664" CreationDate="2010-11-10T05:37:42.280" Score="4" ViewCount="340" Body="&lt;p&gt;I am a bit confused about the concept of proper subsets,precisely whether to include one or both of the &lt;strong&gt;void set&lt;/strong&gt; and &lt;strong&gt;the set itself&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extract from my module goes like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Obviously,every set is the subset of itself and the void set $\phi$ is the subset of every set. These two subsets are called improper subsets.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also includes a theorem which states that &quot;&lt;strong&gt;&lt;em&gt;Let A be a finite set having n elements. Then the total number of subsets of A is ($2^n$) and the number of proper subsets of A is ($2^{n}-1).$&lt;/em&gt;&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again in a sample solution of this problem &quot;&lt;strong&gt;&lt;em&gt;If A = {a,b,c},then the number of proper subsets of A is ?&lt;/em&gt;&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Total no of subsets of {a,b,c} = $2^3$ = 8.&#xA;But each set have two improper subset, so number of improper subsets are 6.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this solution correct ? If so please explain the concept.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-10T05:59:24.867" LastActivityDate="2010-12-22T20:41:47.783" Title="Is the void set (∅) a proper subset of every set ? " Tags="&lt;terminology&gt;&lt;elementary-set-theory&gt;" AnswerCount="6" />
  <row Id="9660" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:43:44.180" Score="1" ViewCount="0" Body="&lt;p&gt;A set A is a subset of a set B if every element of A is also an element of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set A is a proper subset of a set B if A is a subset of B and there is at least one element of B that's not an element of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the void set is a subset of all sets, and it's a proper subset of every set except itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, notice that we notate the void set using $\emptyset$, not $\phi$.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-11-10T05:43:44.180" />
  <row Id="9662" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:43:59.967" Score="0" ViewCount="0" Body="&lt;p&gt;The null set is a proper subset of any set other than itself. A set $A$ is a proper set of $B$ if $\forall x \in A, x \in B$ and $\exists x \in B$ such that $x \notin A$. Clearly $\phi$ satisfies this provided the superset is not $\phi$ itself&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T05:43:59.967" />
  <row Id="9664" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:45:01.770" Score="8" ViewCount="0" Body="&lt;p&gt;Calling $\emptyset$ and $A$ &quot;improper&quot; subsets of a set $A$ is not universal, and it is confusing in this case, because the meaning of &quot;proper&quot; is not the same.  It is standard to say that $S$ is a proper subset of $A$ if (and only if) every element of $S$ is an element of $A$, but $S$ is not equal to $A$, i.e., at least one element of $A$ is not in $S$.  Under this definition, $\emptyset$ is a proper subset of every &lt;em&gt;nonempty&lt;/em&gt; set, even though it is &quot;improper&quot; according to the convention you were also given.  Just remember that mathematical terminology varies and isn't always logical.  Here &quot;improper&quot; does not mean &quot;not proper&quot;.  (For this reason I would personally not use the convention of calling the empty set &quot;improper&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When finding all proper subsets, you should count the empty set.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T05:45:01.770" CommentCount="12" />
  <row Id="9665" PostTypeId="2" ParentId="9659" CreationDate="2010-11-10T05:46:23.977" Score="2" ViewCount="0" Body="&lt;p&gt;Given a set $A$, a proper subset is any set $B$ such that $B\subseteq A$ and $B\neq A$; that is, $B$ is contained in $A$ but is not equal to $A$. This is denoted by $B\subset A$ in some texts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while $A$ is a subset of itself, it is &lt;em&gt;not&lt;/em&gt; a proper subset of itself. And this is true for &lt;em&gt;any&lt;/em&gt; set, even the empty set (or void set, as you call it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of which, the empty set $\emptyset$ is not only a subset of any set, but also a proper subset of any non-empty set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to include a solution to OP's new question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the answer in front of you. If $A$ has cardinality $n$, then the number of subsets is $2^n$ and the number of proper subsets is $2^n-1$, because the only set we have to &quot;throw out&quot; is $A$ itself in order to get all the proper subsets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $A=\{a,b,c\}$, then there are $2^3-1=7$ proper subsets. We can just list them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a,b\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a,c\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{b,c\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{a\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{b\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{c\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\emptyset$&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="2313" LastEditorDisplayName="" LastEditDate="2010-11-10T05:52:15.913" LastActivityDate="2010-11-10T05:52:15.913" CommentCount="1" />
  <row Id="9666" PostTypeId="2" ParentId="9654" CreationDate="2010-11-10T05:48:15.610" Score="0" ViewCount="0" Body="&lt;p&gt;Note that $|f| \leq ||f||_{\infty}$ a.e. This follows from the definition of $||f||_{\infty}$. Hence, $\int |f|^{p} d \mu \leq \int (||f||_{\infty})^{p} d \mu = \int ||f||_{\infty}^{p} \int d \mu$. Thus it follows.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-10T06:45:42.513" LastActivityDate="2010-11-10T06:45:42.513" />
  <row Id="9667" PostTypeId="1" CreationDate="2010-11-10T05:54:34.423" Score="1" ViewCount="66" Body="&lt;p&gt;Working on probability theory in an economics class. This was on of the practice problems that I'm struggling to back up with math. Any help would be appreciated! :)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An accident insurance company has reserves to cover claims on no more than 3 accidents. If all accidents have a probability of 0.01 of occurring, if all accidents are independent, and if the insurance company writes 100 policies, what&#xA;is the probability that they will not have enough reserves?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3343" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-10T06:03:44.977" LastActivityDate="2010-11-10T12:10:15.203" Title="Economics Practice Homework Problem" Tags="&lt;probability&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9668" PostTypeId="1" CreationDate="2010-11-10T06:18:01.713" Score="1" ViewCount="68" Body="&lt;p&gt;As usual denote $L^p$ the quotient space where two integrable functions are identified if they are equal almost everywhere. So I'm using the definition written here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lp_space&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have the following result: for each $p \geq 1$ we have $L^{\infty}(X) \subseteq L^{p}(X)$ where X is a finite measure space and $L^{\infty}$ denotes the set of all essentially bounded functions endowed with the $||f||_{\infty}$ pseudonorm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I took $f \in L^{\infty}(X)$ then by definition there is some bounded function $g$ such that $g=f$ a.e. But then $f=g$. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int |f|^{p} = \int |g|^{p} \leq \int (||g||_{\sup})^{p} &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: Is the above correct? Why do we need $p \geq 1$. Why wouldn't $p&gt;0$ work? Is it because we need $p \geq 1$ in the case q is not $\infty$ or where exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3337" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-10T06:36:07.727" LastActivityDate="2010-11-10T06:36:07.727" Title="Quotients , L^p" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9669" PostTypeId="1" AcceptedAnswerId="9671" CreationDate="2010-11-10T06:22:58.947" Score="1" ViewCount="51" Body="&lt;p&gt;Suppose $X$ is a Banach space and $T: X \to X$ is a bounded linear operator. If $T$ is onto, is it necessarily 1-1? &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-10T06:29:20.170" Title="Surjectivity implying injectivity for bounded linear operators  " Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9670" PostTypeId="2" ParentId="9668" CreationDate="2010-11-10T06:26:21.723" Score="1" ViewCount="0" Body="&lt;p&gt;If p&amp;lt;1, then the integral $(\int |f|^p d \mu)^{\frac{1}{p}}$ no longer defines a norm. That is why we need $p \geq 1$ typically. However, for the conclusion you want to draw, I guess it will still work if $p \geq 0$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T06:26:21.723" CommentCount="15" />
  <row Id="9671" PostTypeId="2" ParentId="9669" CreationDate="2010-11-10T06:29:20.170" Score="3" ViewCount="0" Body="&lt;p&gt;No.  This is true in the finite dimensional case, where a linear transformation is one-to-one iff it is onto iff it is invertible.  Some counterexamples for your question are the &quot;backward shifts&quot; on sequence spaces like $c_0$ or $\ell^p$ with $1\leq p\leq \infty$.  Let $X$ be one of these spaces, and let $T(a_0,a_1,a_2,\ldots)=(a_1,a_2,a_3,\ldots)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-10T06:29:20.170" />
  <row Id="9672" PostTypeId="2" ParentId="9594" CreationDate="2010-11-10T06:37:16.893" Score="0" ViewCount="0" Body="&lt;p&gt;I would leave this as a comment, but I do not have the reputation ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you consider a move? The normal way of understanding the Rubik's cube, is that you can do 18 or 27 moves at each step. That is; you can move each side 90  degrees in two different directions, or make a 180 degree move. In total 3 moves per side, and you have 6 or 9 sides depending on whether you count the &quot;middle slices&quot; of the cube as valid sides or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you arrive at 12 legal moves, do you only consider the 90 degree moves on the &quot;outer&quot; sides as legal?&lt;/p&gt;&#xA;" OwnerUserId="1823" LastActivityDate="2010-11-10T06:37:16.893" CommentCount="2" />
  <row Id="9673" PostTypeId="1" AcceptedAnswerId="9675" CreationDate="2010-11-10T06:56:56.627" Score="3" ViewCount="117" Body="&lt;p&gt;How do I show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2 \log(\Gamma(z))}{\mathrm{d}z^2} = \displaystyle \sum_{n = 0}^{\infty} \frac{1}{(z+n)^2}$$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma(z)$ is the gamma function.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T16:08:10.767" LastActivityDate="2010-11-10T19:46:51.240" Title="$\frac{\mathrm{d}^2 \log(\Gamma (z))}{\mathrm{d}z^2} = \displaystyle \sum_{n = 0}^{\infty} \frac{1}{(z+n)^2}$" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="0" />
  <row Id="9674" PostTypeId="2" ParentId="9648" CreationDate="2010-11-10T07:25:40.830" Score="10" ViewCount="0" Body="&lt;p&gt;If $p=8k+1$ then consider $a$ with $a^{4k}\equiv-1$ (mod $p$).&#xA;Let $b=a^k-a^{7k}$. Then&#xA;$$b^2=a^{2k}-2a^{8k}+a^{14k}\equiv a^{2k}(1+a^{4k})-2&#xA;\equiv-2\pmod{p}.$$&#xA;For $p\equiv3$ (mod $8$) this argument will also work but you need to use the&#xA;Galois field $GF(p^2)$ rather than the integers modulo $p$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-10T07:25:40.830" CommentCount="1" />
  <row Id="9675" PostTypeId="2" ParentId="9673" CreationDate="2010-11-10T07:59:14.597" Score="6" ViewCount="0" Body="&lt;p&gt;Use the hadamard product formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left( 1 + \frac{z}{k} \right)^{-1} e^{z/k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d \log(\Gamma(z))} {dz} = \frac{\Gamma'(z)}{\Gamma(z)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an infinite product, there is an easy way to compute this expression. If&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(z) = \prod f_n(z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then it is not hard to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{f'(z)}{f(z)} = \sum \frac{f_n'(z)}{f_n(z)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;applying this to the Gamma function gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\Gamma'(z)}{\Gamma(z)} = -\gamma - \frac{1}{z} + \sum_{k=1}^\infty \frac{-1}{k(1 + z/k)} + \frac{1}{k} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have to take one more derivative to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{d^2 \log(\Gamma(z))} {dz^2} = \frac{1}{z^2} + \sum_{k=1}^\infty \frac{1}{(k + z)^2} = \sum_{n=0}^\infty \frac{1}{(z+n)^2} $&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-10T07:59:14.597" />
  <row Id="9677" PostTypeId="2" ParentId="9673" CreationDate="2010-11-10T08:06:05.890" Score="3" ViewCount="0" Body="&lt;p&gt;Let's assume the Gauss formula &#xA;$$\frac{\Gamma'(a)}{\Gamma(a)}+\gamma=\int_{0}^{1}\frac{1-t^{a-1}}{1-t}dt$$&#xA;holds (where $\gamma$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler–Mascheroni constant&lt;/a&gt;). Integrating the identity &#xA;$$\frac{1-t^{a-1}}{1-t}=\sum_{k=0}^{\infty}(t^{k}-t^{a+k-1})$$&#xA;yields the series&#xA;$$\frac{d\ln \Gamma(a)}{da}+\gamma=\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{a+k}\right)$$&#xA;which converges uniformly on finite intervals $a\in[0,A]$. Now we can differentiate the latter series in $a$ to obtain that&#xA;$$\frac{d^2\ln\Gamma(a)}{da^2}=\sum_{k=0}^{\infty}\frac{1}{(a+k)^2}.$$&#xA;The differentiation is valid since the resulting series converges uniformly for $a\geq 0.$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation of the Gauss formula.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the basic properties of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_function&quot; rel=&quot;nofollow&quot;&gt;beta function&lt;/a&gt;, we get&#xA;$$\Gamma(b)-B(a,b)=\Gamma(b)-\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\frac{b\Gamma(b)(\Gamma(a+b)-\Gamma(a))}{b\Gamma(a+b)}$$&#xA;$$=\frac{\Gamma(b+1)}{\Gamma(a+b)}\cdot\frac{\Gamma(a+b)-\Gamma(a)}{b}.$$&#xA;Passing to the limit $b\to0$ yields&#xA;$$\frac{d\ln \Gamma(a)}{da}=\frac{\Gamma'(a)}{\Gamma(a)}=\lim\limits_{b\to 0}(\Gamma(b)-B(a,b)),$$&#xA;or &#xA;$$\frac{\Gamma'(a)}{\Gamma(a)}=\lim\limits_{b\to 0}\int_{0}^{\infty}x^{b-1}\left(e^{-x}-\frac{1}{(1+x)^{a+b}}\right)dx=\int_{0}^{\infty}\left(e^{-x}-\frac{1}{(1+x)^{a}}\right)\frac{dx}{x}.\qquad(1)$$&#xA;Identity (1) can be used to define the Euler constant $\gamma$&#xA;$$\qquad\qquad\qquad\qquad\qquad\qquad-\gamma:=\frac{\Gamma'(1)}{\Gamma(1)}=\int_{0}^{\infty}\left(e^{-x}-\frac{1}{1+x}\right)\frac{dx}{x}.\qquad\qquad\qquad\qquad\qquad\qquad(2)$$&#xA;Subtracting (2) from (1) and using the substitution $t=\frac{1}{1+x}$ we obtain that&#xA;$$\frac{\Gamma'(a)}{\Gamma(a)}-\frac{\Gamma'(1)}{\Gamma(1)}=\int_{0}^{1}\frac{1-t^{a-1}}{1-t}dt.$$&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-10T19:46:51.240" LastActivityDate="2010-11-10T19:46:51.240" />
  <row Id="9679" PostTypeId="1" AcceptedAnswerId="9682" CreationDate="2010-11-10T09:14:24.257" Score="0" ViewCount="56" Body="&lt;p&gt;I am doing some pre study/revision in maths before i take my next education steps and am working through a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201624427&quot; rel=&quot;nofollow&quot;&gt;hefty engineering maths book&lt;/a&gt; from the beginning. The first chapter is a review of algebraic techniques using indices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to simplify this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\frac{2xy^2}{(2xy)^2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solution given is this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\frac{1}{2x}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what are the intermediate steps to arrive at this solution?&lt;/p&gt;&#xA;" OwnerUserId="676" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T14:20:11.963" LastActivityDate="2010-11-10T14:20:11.963" Title="What are the steps to perform this algebraic simplification" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="9680" PostTypeId="1" CreationDate="2010-11-10T09:20:57.673" Score="0" ViewCount="81" Body="&lt;p&gt;what i cos^2 (value) equal to? (what is anytrigono^2(theta) equal to?)&lt;/p&gt;&#xA;" OwnerUserId="3344" LastActivityDate="2010-11-10T11:36:05.593" Title="what i cos^2 (value) equal to? (what is anytrigono^2(theta) equal to?)" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9682" PostTypeId="2" ParentId="9679" CreationDate="2010-11-10T09:33:57.943" Score="2" ViewCount="0" Body="&lt;p&gt;$$\frac{2xy^{2}}{\left( 2xy\right) ^{2}}=\frac{2xy^{2}}{2^{2}x^{2}y^{2}}=%&#xA;\frac{2xy^{2}/\left( 2xy^{2}\right) }{2^{2}x^{2}y^{2}/\left( 2xy^{2}\right) }%&#xA;=\frac{1}{2x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided that $2xy^{2}\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Justification&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( 2xy\right) ^{2}=2^{2}x^{2}y^{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2^{2}x^{2}y^{2}}{2xy^{2}}=2x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2xy^{2}}{2xy^{2}}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2xy^{2}}{\left( 2xy\right) ^{2}}=2xy^{2}\left( 2xy\right)&#xA;^{-2}=2xy^{2}2^{-2}x^{-2}y^{-2}=2^{1-2}x^{1-2}y^{2-2}=2^{-1}x^{-1}y^{0}=\frac{1}{2x}$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-10T09:33:57.943" CommentCount="1" />
  <row Id="9683" PostTypeId="1" CreationDate="2010-11-10T09:45:13.173" Score="1" ViewCount="79" Body="&lt;p&gt;Hello..is it possible to calculate shadow areas of buildings or simulate shadows of buildings in a city, using the heights of these buildings and the sun angle and azimuth? the basic light tracing concept using trigonometry is known. But the Problem is how to find away to come up with simulated shadow of group of buildings in one place.  &lt;/p&gt;&#xA;" OwnerUserId="3346" LastActivityDate="2010-11-10T11:51:44.910" Title="shadow simulation from buildings" Tags="&lt;geometry&gt;&lt;algorithms&gt;&lt;computer-science&gt;&lt;matlab&gt;&lt;geometric-construction&gt;" AnswerCount="2" />
  <row Id="9684" PostTypeId="1" AcceptedAnswerId="9704" CreationDate="2010-11-10T09:52:38.190" Score="0" ViewCount="68" Body="&lt;p&gt;Suppose we are given two real numbers: $a, b \in R$, $b &gt; a$. Find $m \in [1, 2, 2.5, 5]$ and $k \in Z$, satisfying the following condition: $\frac{b-a}{m10^k} \in [5; 10]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this problem without using brute force?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastActivityDate="2010-11-10T14:24:49.677" Title="How to solve this problem programmatically?" Tags="&lt;nonlinear-optimization&gt;&lt;convex-optimization&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9685" PostTypeId="1" AcceptedAnswerId="9689" CreationDate="2010-11-10T10:05:29.043" Score="2" ViewCount="68" Body="&lt;p&gt;When f is a Lebesgue measurable function, is $\Vert f \Vert_p$ continuous in p, when p is in a set such that f belongs to $L_p$? If yes, how to show that? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes. It is an exercise taken from Rudin's book. I have spent quite a while thinking but got no idea.&lt;/p&gt;&#xA;" OwnerUserId="3348" LastEditorUserId="3348" LastEditorDisplayName="" LastEditDate="2010-11-10T10:22:29.097" LastActivityDate="2010-11-10T10:38:51.277" Title="$\Vert f \Vert_p$  is continuous in p ?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9687" PostTypeId="2" ParentId="9685" CreationDate="2010-11-10T10:23:36.693" Score="1" ViewCount="0" Body="&lt;p&gt;So you're asking about $p \mapsto \left( \int_X |f|^p d\mu \right)^{1/p}$. Isn't this clearly a composition of continuous functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First $p \mapsto |f|^p$ (no hints, but this should be pretty easy), then the Lebesgue integral (hint: it's a linear map between the Banach space $L^p(X,\mu)$ and $\mathbb{R}$. Which linear maps are continuous?), and finally $x \mapsto x^{1/p}$. &lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-10T10:38:25.030" LastActivityDate="2010-11-10T10:38:25.030" CommentCount="1" />
  <row Id="9688" PostTypeId="1" AcceptedAnswerId="9739" CreationDate="2010-11-10T10:29:21.157" Score="0" ViewCount="72" Body="&lt;p&gt;For a Lebesgue measurable function f, $\Vert f \Vert_p$ converges to $\Vert f \Vert_\inf$ as $p$ goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if $\Vert f \Vert_p$ converges to $\exp(\int \log |f|)$ as $p$ goes to 0 for probability measure? How to prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-10T15:57:00.157" LastActivityDate="2010-11-10T17:19:50.370" Title="convergence of $\Vert f \Vert_p$ with respect to p" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9689" PostTypeId="2" ParentId="9685" CreationDate="2010-11-10T10:38:51.277" Score="1" ViewCount="0" Body="&lt;p&gt;This may not be the answer you are looking for, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Holder's inequality you can show that the map $p^{-1}\mapsto \|f\|_p$ is &lt;strong&gt;log-convex&lt;/strong&gt;. Now just use the fact that any convex function on an interval is continuous. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-10T10:38:51.277" CommentCount="1" />
  <row Id="9690" PostTypeId="1" CreationDate="2010-11-10T10:43:41.600" Score="3" ViewCount="110" Body="&lt;p&gt;In a parallelogram ABCD. M is the midpoint of CD. Line BM intersects AC at L and it also intersects AD extended at E. Prove that EL=2BL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This is not a homework problem. I was solving geometry for fun. I'm unable to solve this. :(&lt;/p&gt;&#xA;" OwnerUserId="2264" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T15:12:59.007" LastActivityDate="2010-11-11T15:12:59.007" Title="A parallelogram and a line joining a vertex to the midpoint of opposite side." Tags="&lt;geometry&gt;&lt;high-school&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="9691" PostTypeId="2" ParentId="9683" CreationDate="2010-11-10T10:44:45.607" Score="3" ViewCount="0" Body="&lt;p&gt;I'll assume that the buildings are made up of polygons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Project each polygon onto the ground to find its shadow.&lt;/li&gt;&#xA;&lt;li&gt;Merge these shadow polygons (e.g. using &lt;a href=&quot;http://en.wikipedia.org/wiki/Weiler%E2%80%93Atherton_clipping_algorithm&quot; rel=&quot;nofollow&quot;&gt;Weiler-Atherton&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Find the area of the merged polygon.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3350" LastActivityDate="2010-11-10T10:44:45.607" CommentCount="1" />
  <row Id="9692" PostTypeId="1" AcceptedAnswerId="9741" CreationDate="2010-11-10T10:52:20.273" Score="2" ViewCount="163" Body="&lt;p&gt;As we know that $L_p \subseteq L_q$ when $0 &amp;lt; p &amp;lt; q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastActivityDate="2010-11-10T23:28:06.377" Title="When $L_p = L_q$ ?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="9693" PostTypeId="2" ParentId="9680" CreationDate="2010-11-10T11:36:05.593" Score="-2" ViewCount="0" Body="&lt;p&gt;Are you asking what &quot;^2&quot; means? It means to take the square of the element, id est to multiply it by itself. For example, cos2(x)=cos(x)⋅cos(x). – Djaian 1 hour ago&lt;/p&gt;&#xA;" OwnerUserId="3344" LastActivityDate="2010-11-10T11:36:05.593" CommentCount="1" />
  <row Id="9694" PostTypeId="2" ParentId="9692" CreationDate="2010-11-10T11:49:30.433" Score="2" ViewCount="0" Body="&lt;p&gt;For probability measure $ L_p \subset L_q $ .&#xA;Equality is unreachable, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lp_space&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1035" LastEditorUserId="1035" LastEditorDisplayName="" LastEditDate="2010-11-10T11:55:49.233" LastActivityDate="2010-11-10T11:55:49.233" />
  <row Id="9695" PostTypeId="2" ParentId="9683" CreationDate="2010-11-10T11:51:44.910" Score="-1" ViewCount="0" Body="&lt;p&gt;if you have the data on the buildings, model the whole scene as tall boxes in OpenGL, render it with a suitable light source, see the shadow.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-10T11:51:44.910" CommentCount="2" />
  <row Id="9696" PostTypeId="2" ParentId="9688" CreationDate="2010-11-10T12:03:51.017" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate by non-negative step functions (if $f$ vanishes on a set of positive measure, using a Holder inequality you can argue that $\|f\|_p \leq C |\mathrm{supp}(f)|^{\frac1p - 1} \searrow 0$). For step functions, that $\|f\|_p\geq \exp \int \log |f|$ is a consequence of the arimetic-geometric mean inequality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AM-GM inequality becomes an equality when all its terms are equal. For a non-negative step function $f$, as $p\to 0$, you have that $f \to 1$. By continuity you can quantify the difference between AM-GM. So you have that the convergence is true for step functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finish off you use the usual diagonal argument to extract a subsequence from the approximating sequence of step functions, together with a suitable sequence of $p$'s. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-10T12:03:51.017" CommentCount="1" />
  <row Id="9697" PostTypeId="2" ParentId="9667" CreationDate="2010-11-10T12:10:15.203" Score="3" ViewCount="0" Body="&lt;p&gt;Let's $X_i$ be a Bernoulli variable that is 1 if an accident occured with the policy number $i$, 0 otherwise. Thus we have:&#xA;$$P(X_i = 1) = 0.01, P(X_i = 0) = 0.99$$&#xA;This is true for all $i$ since we assume independance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y = X_1 + X_2 + \dots + X_{100}$ the sum of all $X_i$. $Y$ counts the number of accidents that occured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means you are looking for $P(Y &gt; 3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you are having probability class, you should have seen what a Bernoulli variable is, and what the distribution of a variable that is the sum of independant Bernoulli variables is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know, look &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that according to your question I had to made the assumption that the probability you give (0.01), is the probability that for 1 policy there will be 1 accident and that there cannot be more than 1 accident for the same policy. Usually, we give a probability that in a given amount of time there will be an accident. &lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2010-11-10T12:10:15.203" />
  <row Id="9698" PostTypeId="2" ParentId="9692" CreationDate="2010-11-10T12:17:07.890" Score="5" ViewCount="0" Body="&lt;p&gt;Equality only when $p = q$. Suppose $p &amp;lt; q$, and let $r = \frac{p+q}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B_n$ be a sequence of disjoint measurable subsets of your probability space with the property that their measures $|B_n| = m_n \leq 2^{-n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\chi_n$ denote the characteristic function of $B_n$. Consider the sequence of functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_n = \sum_1^n \frac{1}{m_k^{1/r}} \chi_k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to check that this sequence is bounded in $L^p$ and Cauchy, so converges to a limit function $f$. But the sequence is unbounded in $L^q$. Hence the limit function $f$ is an element of $L^p$ that is not in $L^q$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-10T23:28:06.377" LastActivityDate="2010-11-10T23:28:06.377" CommentCount="4" />
  <row Id="9699" PostTypeId="2" ParentId="9690" CreationDate="2010-11-10T12:17:31.063" Score="4" ViewCount="0" Body="&lt;p&gt;Let BD and AC intersects at point O.&#xA;We have triangle BCD. BO=OD because ABCD is parallelogram.&#xA;In this triangle CO and BM are medians.&#xA;Medians intersects in one point and divide each other in relation 2:1.&#xA;So  BL = 2LM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also BM = ME thus triangle MBC = triangle MDE because they are congruent and DM = CM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So BL = 2/3 BM =&gt; BL = 2EL.    &lt;/p&gt;&#xA;" OwnerUserId="1035" LastActivityDate="2010-11-10T12:17:31.063" />
  <row Id="9700" PostTypeId="1" CreationDate="2010-11-10T12:36:54.133" Score="-2" ViewCount="143" Body="&lt;p&gt;requires a decision&lt;/p&gt;&#xA;" OwnerUserId="3352" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T14:39:15.947" LastActivityDate="2010-11-10T15:15:31.100" Title="$\sum\limits_{n=1}^{\infty} \lbrace (-1)^n-(n-1)^3/(n^4+3n^2+2) \rbrace $" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="9701" PostTypeId="1" CreationDate="2010-11-10T12:45:34.250" Score="0" ViewCount="92" Body="&lt;p&gt;A Steinmetz solid is the intersection of two cylinders. My question is how to find the surface area of one with both cylinders of radius 1, without parameterizing the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I parameterized it; that is, used $r(u,v)=\langle \sqrt{1-v^2}, u \sqrt{1-v^2}, v \rangle$, $u,v \in [-1,1]$, and integrated the magnitude of the  cross product of the partials: $$\int_{-1}^1 \int_{-1}^1 |r_u \times r_v| dR$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I make the solid into a function $z=f(x,y)$ rather than parameterizing it?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-11-10T14:55:25.087" Title="Surface Area of a Steinmetz Solid" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="9702" PostTypeId="2" ParentId="9700" CreationDate="2010-11-10T13:47:43.973" Score="3" ViewCount="0" Body="&lt;p&gt;I'm assuming you want to know about the convergence of&#xA;$$\sum_{n=1}^\infty \lbrace (-1)^n - (n-1)^3/(n^4+3n^2+2) \rbrace.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: If $\sum a_n $ is convergent then $a_n \rightarrow 0$ as $ n \rightarrow \infty.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-10T13:47:43.973" />
  <row Id="9703" PostTypeId="1" AcceptedAnswerId="9788" CreationDate="2010-11-10T14:22:38.950" Score="10" ViewCount="855" Body="&lt;p&gt;My daughter is stuck on the concept that 2&lt;sup&gt;0&lt;/sup&gt; = 1, having the intuitive expectation that it be equal to zero. I have tried explaining it, but I guess not well enough. How would you explain the concept to a child, other than the teachers &quot;that is just the rule&quot; approach?&lt;/p&gt;&#xA;" OwnerUserId="3356" LastActivityDate="2010-11-28T04:03:09.460" Title="How do I explain 2 to the power of zero equals 1 to a child" Tags="&lt;arithmetic&gt;&lt;teaching&gt;&lt;elementary&gt;" AnswerCount="12" CommentCount="11" FavoriteCount="4" />
  <row Id="9704" PostTypeId="2" ParentId="9684" CreationDate="2010-11-10T14:24:49.677" Score="2" ViewCount="0" Body="&lt;p&gt;This is a &quot;normalization&quot; type operation.  For simplicity we can replace the two parameters a,b with the single one c = b - a &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the interpretation proposed by Eric above, you want c*(10^-k) to fall into an interval m*[5,10) where m is in {1,2,2.5,5}.  In other words we want to normalize c by a power of 10 so that it belongs to (at least) one of these intervals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[5,10), [10,20), [12.5,25), [25,50)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the middle pair of these intervals has nonempty overlap, the union of all four intervals is [5,50) and covers exactly one order of magnitude (power of 10).  It follows that the exponent k is unique, although the factor m may or may not be unique.  Note that this is equivalent to pigeonholing c/5 into one of four subintervals covering [1,10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find these &quot;without using brute force&quot; ?  The best programmatic approach depends on what language and hardware is available.  The IEEE standard for floating-point arithmetic (IEEE-754) allows some fairly portable assumptions to be made, but the challenge is to get a &quot;leading digit&quot; and characteristic exponent in base 10 when the usual representation of floating-point numbers is base 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the &quot;common logarithm&quot; log_10(c/5) and break it into an integer part and a fractional part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;log_10(c/5) = iPart + fPart  where fPart in [0,1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now iPart here is the same value we want for k in your original notation for the problem, and we can determine m from fPart by doing a bit of testing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if ( fPart &amp;lt; log_10(2) ) then m = 1&lt;br&gt;&#xA;else if ( fPart &amp;lt; log_10(2.5) ) then m = 2&lt;br&gt;&#xA;    else if ( fPart &amp;lt; log_10(5) ) then m = 2.5&lt;br&gt;&#xA;        else m = 5  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-10T14:24:49.677" />
  <row Id="9705" PostTypeId="1" CreationDate="2010-11-10T14:28:52.747" Score="2" ViewCount="122" Body="&lt;p&gt;How many subgroups with index 2 are there of a free group on two generators? What are their generators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I know is that the subgroups should have 2*2 + 1 - 2 = 3 generators. &lt;/p&gt;&#xA;" OwnerUserId="3358" LastActivityDate="2010-11-10T15:40:17.790" Title="Finding subgroups of a free group with a specific index" Tags="&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9706" PostTypeId="2" ParentId="9705" CreationDate="2010-11-10T14:30:58.613" Score="2" ViewCount="0" Body="&lt;p&gt;Let $F$ be free on $a$ and $b$, and let $N$ be a subgroup of index $2$. Then $N$ is normal, and there is a map $\phi:F\to\mathbb Z/2\mathbb Z$ with kernel precisely $N$. What are the possible images of $a$ and $b$? For each possible choice you can describe the corresponding generators of $N$ as a normal subgroup, and with a little extra work, generators as subgroup.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-10T14:46:59.597" LastActivityDate="2010-11-10T14:46:59.597" />
  <row Id="9707" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T14:32:27.887" Score="0" ViewCount="0" Body="&lt;p&gt;I would appeal to the laws of exponents.  Show for $a,b&gt;0$ that $2^a* 2^b=2^{a+b}$, then extend it to negative $b$, then set $a=b$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T14:32:27.887" CommentCount="1" />
  <row Id="9709" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T14:34:20.040" Score="4" ViewCount="0" Body="&lt;p&gt;This is a question answered all over the Internet, here's an OK answer:&#xA;&lt;a href=&quot;http://mathforum.org/dr.math/faq/faq.number.to.0power.html&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/dr.math/faq/faq.number.to.0power.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-10T14:34:20.040" CommentCount="1" />
  <row Id="9710" PostTypeId="1" AcceptedAnswerId="9715" CreationDate="2010-11-10T14:47:06.003" Score="1" ViewCount="113" Body="&lt;p&gt;I received some homework (calculus), which I can't prove:&#xA;&lt;code&gt;f(n) * g(n) &amp;gt;= f(n) + g(n)&lt;/code&gt; for some n&gt;=1 is always true &lt;code&gt;(f(n),  g(n) &amp;gt; = 1)&lt;/code&gt;, I think that this is true, so I need to prove that there is some &lt;code&gt;n&amp;gt;=1&lt;/code&gt; (n is a natural number) for which this inequality will be true always, thanks in advance for any idea&lt;/p&gt;&#xA;" OwnerUserId="3359" LastEditorUserId="3359" LastEditorDisplayName="" LastEditDate="2010-11-10T14:56:08.020" LastActivityDate="2010-11-11T19:31:40.807" Title="prove or refute" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9711" PostTypeId="2" ParentId="9705" CreationDate="2010-11-10T14:51:05.960" Score="1" ViewCount="0" Body="&lt;p&gt;I like to approach this sort of problem using graphs. The free group on two generators is the fundamental group of a wedge of two circles $R_2$, which I picture as a red oriented circle and a black oriented circle. A subgroup of index $ k$ corresponds to a covering map $G\to R_2$ of index $k$. $G$ can be pictured as a (Edit: basepointed) $k$-vertex connected graph with red and black oriented edges such that at every vertex there is one incoming and one outgoing edge of each color. In the case $k=2$, it's not hard to write down all such graphs. I count three myself.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastEditorUserId="2941" LastEditorDisplayName="" LastEditDate="2010-11-10T15:17:04.470" LastActivityDate="2010-11-10T15:17:04.470" CommentCount="9" />
  <row Id="9712" PostTypeId="1" AcceptedAnswerId="10401" CreationDate="2010-11-10T14:52:40.500" Score="2" ViewCount="32" Body="&lt;p&gt;Let $C$ be a convex subset of a Banach space $X$ and $T:C\to C$ a (norm) continuous affine map, i.e. $$T(tx+(1-t)y)=tT(x)+(1-t)T(y)$$ for $0\le t\le 1$. Is $T$ weakly continuous, i.e continuous as a map from $(C,\tau)$ to $(C,\tau)$ where $\tau$ is the topology induced by the weak topology of $X$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-15T14:29:30.687" Title="Weak Continuity of  Affine maps" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;&lt;convex-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9713" PostTypeId="1" CreationDate="2010-11-10T14:54:08.113" Score="6" ViewCount="151" Body="&lt;p&gt;I'm teaching first semester calculus and trying to find a way to explain why each hypothesis in l'Hostpial's rule is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ and $g$ are real differentiable functions on an interval containing a point $c$ then there are three hypotheses one needs to check in order to apply l'Hospitals rule to compute $\lim_{x \to c} f(x)/g(x)$ :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$g'(x) \neq 0$ on a neighborhood of $c$, (the wikipedia page misses this)&lt;/li&gt;&#xA;&lt;li&gt;$\lim_{x \to c} f'(x)/g'(x)$ exists (in the extended sense including $\pm \infty$),&lt;/li&gt;&#xA;&lt;li&gt;Either $\lim_{x \to c} g(x) = \pm \infty$ or $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is a pair of functions $f$ and $g$ so that 2 and 3 both hold and you can reasonably compute $\lim_{x \to c} f(x)/g(x)$ but get a different answer than computing the limit of the quotient of the derivatives. Bonus points if you can make $\lim_{x \to c} f(x) = \pm \infty$ as well in the case that $g(x) \to \pm \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was shown an example like this a long time ago in my analysis sequence but it would be lost on the students I think:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = 2x + \sin(2x)$ and let $g(x) = (2x+\sin(2x))e^{-\sin(x)}$. Then $f/g = e^{\sin x}$ on $(0,\infty)$ so $\lim_{x \to \infty} f(x)/g(x)$ does not exist. Both functions will go to $\infty$ as $x \to \infty$ and the limit of the quotients of the derivatives actually goes to zero. The issue is that $g'(x)$ has infinitely many zeros as $x \to \infty$ so you cannot use l'Hospital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clue on how to make this more inviting to a freshman calculus student?&lt;/p&gt;&#xA;" OwnerUserId="3341" LastActivityDate="2010-11-13T16:00:24.710" Title="Non-vanishing of derivative hypothesis in l'Hospital's rule" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="9714" PostTypeId="2" ParentId="9701" CreationDate="2010-11-10T14:55:25.087" Score="2" ViewCount="0" Body="&lt;p&gt;Taking the axes of the cylinders along $x$ and $y$, you have $z=\sqrt{1-(\max{(|x|,|y|)})^2}$.  So you have to integrate $$\int_{-1}^1 \int_{-x}^x f(x) dx dy + \int_{-1}^1 \int_{-y}^y f(y) dx dy$$  where $f$ is the area element.  The first integral gets x&gt;y and the second gets y&gt;x.  By symmetry they are the same.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T14:55:25.087" />
  <row Id="9715" PostTypeId="2" ParentId="9710" CreationDate="2010-11-10T14:57:13.747" Score="4" ViewCount="0" Body="&lt;p&gt;Hint: $2.25 &amp;lt; 3$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-10T14:57:13.747" CommunityOwnedDate="2010-11-10T14:57:13.747" CommentCount="4" />
  <row Id="9716" PostTypeId="2" ParentId="9705" CreationDate="2010-11-10T14:58:56.360" Score="3" ViewCount="0" Body="&lt;p&gt;Mariano's answer is quite right, but relies on the 'accident' that all subgroups of index two are normal.  If you wanted to find all the subgroups of index three, say, you would need to try another approach. Here's one idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free group of rank two, $F$, is the fundamental group of the figure-eight graph $\Gamma$, which has one vertex and two edges.  Fix the vertex as the base point.  The subgroups of index $k$ correspond, via covering-space theory, to connected covering spaces $\widehat{\Gamma}\to\Gamma$ of degree $k$, with a choice of base vertex $\hat{v}\in\widehat{\Gamma}$.  (If, on the other hand, you only wanted to count &lt;em&gt;conjugacy classes&lt;/em&gt; of subgroups, you could forget about $\hat{v}$.)  Because the covering map $p:\widehat{\Gamma}\to\Gamma$ has degree $k$, the graph $\widehat{\Gamma}$ has exactly $k$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's decorate the graph $\Gamma$ so that we can see $F=\langle a,b\rangle$ more clearly.  Label each edge with the corresponding generator $a$ or $b$.  Furthermore, orient each edge to indicate in which direction the generator goes around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the covering map $p$ to pull the labels and orientations back from $\Gamma$ to $\widehat{\Gamma}$.  That is, if $\hat{e}$ is an edge of $\widehat{\Gamma}$ and $p(\hat{e})=e$ is labelled $a$, then you should label $\hat{e}$ with $a$ also, and orient $\hat{e}$ so that $p$ sends the orientation on $\hat{e}$ to the orientation on $e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little thought, it's not too hard to see that this decoration on $\widehat{\Gamma}$---the labels and orientation---are enough to reconstruct the map $p$: they tell you where to send each edge, and there's only one choice of where to send each vertex. The statement that $p$ is  a covering map translates into a nice condition on the decoration of $\widehat{\Gamma}$: for each vertex $\hat{u}$, you see exactly one edge with label going into $\hat{u}$, and exactly one edge with each label going out of $\hat{u}$.  I'll call this condition 'the covering condition'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This discussion turns counting the subgroups of index $k$ into the combinatorial problem of counting the connected, decorated, based graphs $\widehat{\Gamma}$ with $k$ vertices that satisfy the covering condition.  With a little thought, one can write down a formula for this.  &lt;a href=&quot;http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;amp;co4=AND&amp;amp;co5=AND&amp;amp;co6=AND&amp;amp;co7=AND&amp;amp;dr=all&amp;amp;pg4=AUCN&amp;amp;pg5=AUCN&amp;amp;pg6=PC&amp;amp;pg7=ALLF&amp;amp;pg8=ET&amp;amp;review_format=html&amp;amp;s4=hall%2C%20m%2a&amp;amp;s5=&amp;amp;s6=&amp;amp;s7=&amp;amp;s8=All&amp;amp;vfpref=html&amp;amp;yearRangeFirst=&amp;amp;yearRangeSecond=&amp;amp;yrop=eq&amp;amp;r=181&amp;amp;mx-pid=28836&quot; rel=&quot;nofollow&quot;&gt;Marshall Hall Jr&lt;/a&gt; did just this (though I don't think he thought in terms of the covering spaces of graphs) and came up with the following formula. Let $N(k,r)$ be the number of subgroups of index $k$ in the free group of rank $r$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(k,r)=k(k!)^{r-1}-\sum_{i=1}^{k-1}((k-1)!)^{r-1}N(i,r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some related things, I wrote a blog post about proving theorems about free groups by thinking about covering spaces of graphs &lt;a href=&quot;http://ldtopology.wordpress.com/2008/12/01/in-memoriam-j-r-stallings-topology-of-finite-graphs/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you don't like covering spaces, an equivalent point of view is to count transitive actions by permutation groups with $r$ generators on based sets of size $k$.  This turns out to be the same combinatorial problem.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-11-10T15:40:17.790" LastActivityDate="2010-11-10T15:40:17.790" CommentCount="3" />
  <row Id="9717" PostTypeId="2" ParentId="9648" CreationDate="2010-11-10T15:02:14.187" Score="9" ViewCount="0" Body="&lt;p&gt;Similar arguments work for other cases as well. For example, if $p = 1 \mod 3k$, find $a$ such that $a^{3k}=1$ and $a^k \neq 1$. Set $z=a^k$, then $z^2+z+1=0$ and we deduce that $(z - z^2)^2 = -3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a larger example, if $p=5k+1$, then there is an $a$ in $\mathbb{F}_p$ with $a^{5k}=1$ and $a^k \neq 1$. Let $z=a^k$, so $z^4+z^3+z^2+z+1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b = z - z^2 - z^3 + z^4$. Then &#xA;$$b^2 = z^8 -  2 z^7 - z^6 + 4 z^5 - z^4 - 2 z^3 + z^2=$$&#xA;$$-z^4 - z^3 - z^2 - z + 4 = 5.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $p \equiv 1 \mod 5$ implies that $\left( \frac{5}{p} \right)=1$. Just as in Robin's answer, you can also prove that $p \equiv 4 \mod 5$ implies $\left( \frac{5}{p} \right)=1$ in this manner, but it is more difficult: the element $z$ will live in $\mathbb{F}_{p^2}$, and you will need to check that $b$ lives in $\mathbb{F}_{p}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for every squarefree $M$, let $\Delta$ be $|M|$ if $M \equiv 1 \mod 4$ and $\Delta=4 |M|$ otherwise. Then you can prove in this manner the statement &quot;if $p = 1 + \Delta k$ then $\left( \frac{\Delta}{p} \right) = 1$.&quot; The first steps are the same in each case: find $a$ in $\mathbb{F}_p$ such that $a^{\Delta k}=1$ but $a^{d k} \neq 1$ for any proper divisor $d$ of $\Delta$. Let $z=a^k$, so $z$ is a primitive $\Delta$ root of unity in $\mathbb{F}_p$. At this point, some magic polynomial $g_M$ is introduced, such that $g_M(z)^2 = M$. If you want to prove the full quadratic reciprocity law, you can do so using this approach as well, but you will need to work in finite fields of non-prime order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often thought it would be fun to assign a class the job of finding the magic polynomials $g_M$, for various values of $M$, and see if they could guess the general pattern. Hint: $g$ stands for &quot;Gauss sum&quot;.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-11-10T16:16:29.227" LastActivityDate="2010-11-10T16:16:29.227" CommentCount="5" />
  <row Id="9718" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T15:02:16.320" Score="36" ViewCount="0" Body="&lt;p&gt;I will give a different answer than the answer I gave in the other thread which tries to appeal to intuition.  I am sure your daughter has no problem accepting that $2*0 = 0$.  Intuitively this is because if you add $2$ to itself zero times, you get zero.  Or, to be concrete, if someone gives you two apples zero times, you have zero apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For repeatedly adding $2$, talking about collections of apples is a good model.  But for repeatedly multiplying by $2$, it isn't necessarily, since you can't multiply apples and apples (at least, not in a way that makes sense to a child).  But you can multiply apples by numbers; that is, you can start with $1$ apple, then double the number of apples you have to get $2$ apples, then double the number of apples you have to get $4$ apples, and so forth.  In general if you double your apples $n$ times, you have $2^n$ apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if you double your apples zero times?  Well, that means you haven't started doubling them yet, so you still have $1$ apple.  If you want your notation to be consistent, then you should say $2^0 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a subtly different argument from the argument I gave before.  It's intuitive what it means to add different amounts of apples, and it's intuitive what it means to have zero apples.  But the twos I am now working with aren't numbers of apples, they're just abstract numbers; in other words, they're unitless, so it's harder to get a grip on them.  What $2^n$ really represents above is an endomorphism of the free commutative monoid on an apple, which is much less concrete than an apple.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a way to gain intuition here which sort of involves units, but I don't know if you can convince your daughter that it makes sense.  One way to interpret $2^n$ is that it is the &quot;size&quot; of an $n$-cube of side length $2$ in dimension $n$.  For example, the length of a segment of side length $2$ is $2$, the area of a square of side length $2$ is $4$, and so forth.  One way to say this is that $2^n$ is the number of $n$-cubes of side length $1$ that fit into an $n$-cube of side length $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a meaningful interpretation of the above when $n = 0$ we need to decide what $0$-dimensional objects are.  Well, if $2$-dimensional space is a plane and $1$-dimensional space is a line, then $0$-dimensional space must be... a point.  In particular, a $0$-cube, of any side length, is a point, and so exactly one $0$-cube of side length $1$ fits into a $0$-cube of side length $2$.  Hence $2^0 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm really curious what her response to this argument will be, actually.  Could you report back on this?)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-10T15:02:16.320" CommentCount="7" />
  <row Id="9719" PostTypeId="1" AcceptedAnswerId="9721" CreationDate="2010-11-10T15:12:18.483" Score="1" ViewCount="72" Body="&lt;p&gt;I have proved this using Venn diagram but when I am trying to prove this using the rule that &lt;strong&gt;&lt;em&gt;&quot;If $ A \subset B \text{ and } B \subset A $ then $ A = B $&quot;&lt;/em&gt;&lt;/strong&gt;, I am having some problems with my understanding of the same,here is how I  did so far :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x \in A \cap (B-c) \Rightarrow x \in A  \text{ and }  x \in (B-C) \Rightarrow x \in A \text{ and } (x \in B \text{ and } x \notin C)  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to proceed next ? Since if I am do something like this : &lt;strong&gt;$ x \in A \text{ and } x \in B \text{ and } x \in A \text{ and } x \notin C $&lt;/strong&gt;,it's not giving the correct results, what exactly I am missing here ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-10T15:17:13.220" Title="Proving : $A \cap (B-C) = (A \cap B) - (A \cap C) $ " Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="9720" PostTypeId="2" ParentId="9700" CreationDate="2010-11-10T15:15:31.100" Score="3" ViewCount="0" Body="&lt;p&gt;Considering your first coment you certainly want to know about the convergence of &#xA;$\sum\limits_{n=1}^{\infty}(-1)^{n-1} \frac{(n-1)^3}{n^4+3n^2+2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&#xA;$\frac{(n-1)^3}{n^4+3n^2+2} \to 0$ when $n \to \infty$, so...&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-11-10T15:15:31.100" CommentCount="3" />
  <row Id="9721" PostTypeId="2" ParentId="9719" CreationDate="2010-11-10T15:17:13.220" Score="6" ViewCount="0" Body="&lt;p&gt;Hint: $x \notin C \Rightarrow x \notin D \cap C$ for any set $D$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-10T15:17:13.220" CommentCount="1" />
  <row Id="9722" PostTypeId="2" ParentId="9710" CreationDate="2010-11-10T15:34:39.840" Score="0" ViewCount="0" Body="&lt;p&gt;If f and g are both constant functions, then it's easy to see that the claim isn't true in general. Let's assume that g is not constant, and since this problem came from a calculus class, let's assume that f and g are differentiable. Compare the derivative of f(x) * g(x) with the derivative of f(x) + g(x). What can you say about the relative size of the derivatives? What does that tell you about the functions?&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-10T15:34:39.840" CommentCount="3" />
  <row Id="9723" PostTypeId="1" AcceptedAnswerId="9725" CreationDate="2010-11-10T15:38:30.200" Score="1" ViewCount="55" Body="&lt;p&gt;Find a necessary and sufficient condition for the linear equation Ax = b to have&#xA;no solution. (hint: Use duality to find a strong alternative to Ax = b).&lt;/p&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2010-11-10T15:53:26.217" Title="Use duality to find a strong alternative" Tags="&lt;linear-programming&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="9724" PostTypeId="1" AcceptedAnswerId="9744" CreationDate="2010-11-10T15:42:04.527" Score="1" ViewCount="119" Body="&lt;p&gt;Which of the following is the unbounded set ?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$ X = \{ x\mid x = \frac{1}{n},n \in \mathbb{N} \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ Y = \{ x\mid x = \frac{1}{2^n},n \in  \mathbb{N} \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ Z = \{ x\mid x = 2^n,n \in \mathbb{N} \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ W = \{ x\mid x \in \mathbb{N}, x \lt 4532 \} $ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After reading the definition from &lt;a href=&quot;http://en.wikipedia.org/wiki/Bounded_set&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; , I can only convince myself that it can't be W but am not getting which is correct from the other three ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T05:50:57.830" LastActivityDate="2010-11-17T05:50:57.830" Title="A question on Unbounded set" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="9725" PostTypeId="2" ParentId="9723" CreationDate="2010-11-10T15:48:25.317" Score="1" ViewCount="0" Body="&lt;p&gt;Solution exists iff $b$ is orthogonal to the null space of $A^{*}$. This is typically referred to as the Fundamental Theorem of Linear Algebra.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-10T15:53:26.217" LastActivityDate="2010-11-10T15:53:26.217" CommentCount="4" />
  <row Id="9726" PostTypeId="2" ParentId="9724" CreationDate="2010-11-10T16:13:36.853" Score="5" ViewCount="0" Body="&lt;p&gt;A set in $\mathbb{R}$ is bounded if you can find two numbers, $a$ and $b$ such that all elements of the set are between $a$ and $b$.  If you want to claim a set is unbounded, whatever $a$ and $b$ I give you, you should be able to find at least one element outside the interval $(a,b)$.  Which of your sets satisfies that?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T16:13:36.853" CommentCount="1" />
  <row Id="9727" PostTypeId="2" ParentId="9649" CreationDate="2010-11-10T16:20:13.650" Score="2" ViewCount="0" Body="&lt;p&gt;You want to show that the absolute value of (b_n)(a_n) can be made arbitrarily small. But, for all n, the absolute value of b_n does not exceed some positive constant M since the sequence b is bounded. Thus |b_na_n| does not exceed M|a_n|. Since the sequence a converges to 0, this is all you should need. The argument is basically the Squeeze Theorem from elementary calculus.&lt;/p&gt;&#xA;" OwnerUserId="2933" LastActivityDate="2010-11-10T16:20:13.650" CommentCount="1" />
  <row Id="9728" PostTypeId="1" CreationDate="2010-11-10T16:21:35.273" Score="3" ViewCount="204" Body="&lt;p&gt;Let $g:[0,1] \rightarrow \mathbb{R}$ be twice differentiable with $g&amp;#39;&amp;#39;(x)\gt 0$ for all $x\in[0,1]$. If $g(0)&amp;gt;0$ and $g(1)=1$, show that $g(d)=d$ for some point $d\in(0,1)$ if and only if $g&amp;#39;(1)\gt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there exits $d\in(0,1)$ such that $g(d)=d$. Then by the MVT applied on $[d,1]$, $f'(c)(d-1)=g(d)-g(1) $for some $c\in(d,1)$. But $g(d)=d$ and $g(1)=1$ so $f&amp;#39;(c)=1$. Now, since $g&amp;#39;&amp;#39;(x)\gt 0$ for $x\in[0,1]$, $g&amp;#39;$ is increasing on this interval, therefore $1\gt d$ implies $g&amp;#39;(1)-g&amp;#39;(d)\gt 0$. &#xA;Assume $g&amp;#39;(1)\gt 1$ and let $f(x)=g(x)-x$. Then $f(0)=g(0)-0\gt 0$ by hypothesis. I want to show $f(x)\lt 0$ for some $x$ in a neighborhood of $1$. $g&amp;#39;(1)=  \lim\limits_{x\to1}\frac{g(x)-g(1)}{x-1}$ so there exists $\delta\gt 0$ such that given $\epsilon\gt 0$ with $x\in(1-\delta,1)$ implies $|g&amp;#39;(1)-\frac{g(x)-1}{x-1}|\lt \epsilon$. If $\epsilon=g&amp;#39;(1)-1\gt 0$, then $g&amp;#39;(1)-\epsilon\lt \frac{g(x)-1}{x-1}$ which implies $g(x)\lt x$ for $x\in(1-\delta,1)$. The result follows by the IVT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments, corrections or different solutions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="2467" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T17:45:54.180" LastActivityDate="2010-11-17T17:45:54.180" Title="Twice differentiable function, show there is a fixed point." Tags="&lt;real-analysis&gt;&lt;homework&gt;" CommentCount="1" />
  <row Id="9729" PostTypeId="2" ParentId="9622" CreationDate="2010-11-10T16:21:59.640" Score="1" ViewCount="0" Body="&lt;p&gt;If you can calculate all the derivatives at a point, say t=0, you can create a Laurent series around zero.  As long as all the poles of the function are outside a circle of radius $\pi$, this will converge to the function.  You can then integrate term by term to get the integral.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T16:21:59.640" />
  <row Id="9730" PostTypeId="1" AcceptedAnswerId="9734" CreationDate="2010-11-10T16:22:49.767" Score="0" ViewCount="42" Body="&lt;p&gt;Imagine I had two sets of samples. The only information I have about them is their&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Size&lt;/li&gt;&#xA;&lt;li&gt;Average&lt;/li&gt;&#xA;&lt;li&gt;Standard deviation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it possible to calculate the standard deviation of a third set that is composed by the union of all samples in both sets?&#xA;And what if the original sets had the same size?&lt;/p&gt;&#xA;" OwnerUserId="3364" LastEditorUserId="3364" LastEditorDisplayName="" LastEditDate="2010-11-10T16:58:56.460" LastActivityDate="2010-11-10T16:58:56.460" Title="Is it possible to calculate the standard deviation without the samples?" Tags="&lt;statistics&gt;&lt;standard-deviation&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9731" PostTypeId="1" AcceptedAnswerId="9743" CreationDate="2010-11-10T16:30:51.340" Score="4" ViewCount="183" Body="&lt;p&gt;I'm stuck with this induction proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, far, given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T(1)  = 2&lt;br&gt;&#xA;T(n)  = 2T(n/2)+2&lt;br&gt;&#xA;      = 2(2T(n/[2^2])+2) + 2&lt;br&gt;&#xA;      = [2^2]T(n/[2^2]) + [2^2] + 2&lt;br&gt;&#xA;      = 2^2 + [2^2] + 2&lt;br&gt;&#xA;      = [2^3]T(n/[2^3]) + [2^3] + [2^2] + 2&lt;br&gt;&#xA;      = [2^3]T(n/[2^3]) + 2{[2^2] + [2^1] + 1} &#xA;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;      = [2^k]T(n/[2^k]) + 2{2^(k) - 1}&#xA;How then do I show this to be correct (the proof). So far I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let (n/[2^k]) = 1&lt;br&gt;&#xA;=&gt; n = 2^k&lt;br&gt;&#xA;So, T(n) = nT(1) + 2(n  - 1)&#xA;T(n) = 4n - 2&#xA;//This is where I'm stuck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof (by induction):&#xA;When n = 1, T(1) = 2.&#xA;Assume T(k) is true [T(n) = 4n - 2]&#xA;//THIS IS WHERE I'M STUCK&lt;/p&gt;&#xA;" OwnerUserId="3366" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-10T23:27:34.727" LastActivityDate="2010-11-10T23:27:34.727" Title="Prove by induction (T(n) = 2T(n/2) + 2)" Tags="&lt;induction&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9732" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T16:32:05.263" Score="1" ViewCount="0" Body="&lt;p&gt;Why not just take out a calculator, let her punch in any number she likes, and then let her hit square root over and over.  The limit is 1, and to show $2^0=1$ will take some explaining, but i think it's the best way to think about this in the long term.  Actually this sequence will allow you to later on introduce $log$ and $e$ in a simple way.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-10T16:32:05.263" />
  <row Id="9733" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T16:32:35.240" Score="13" ViewCount="0" Body="&lt;p&gt;How about this: There's always an implicit 1 in the expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^3 = 2 * 2 * 2 * 1 = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^2 = 2 * 2 * 1 = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^1 = 2 * 1 = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^0 = 1 = 1&lt;/p&gt;&#xA;" OwnerUserId="3367" LastActivityDate="2010-11-10T16:32:35.240" CommentCount="1" />
  <row Id="9734" PostTypeId="2" ParentId="9730" CreationDate="2010-11-10T16:37:25.803" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, it is possible.  The equation should be in your book, or you can look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_deviation&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page&lt;/a&gt; under the heading Population-based statistics&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T16:37:25.803" />
  <row Id="9735" PostTypeId="2" ParentId="9648" CreationDate="2010-11-10T16:38:25.087" Score="4" ViewCount="0" Body="&lt;p&gt;If I remember correctly, this approach with a Gauss sum (associated to an 8th root of 1 in a finite field) is used in the first page or two of Serre's &lt;em&gt;Cours d'Arithmetique&lt;/em&gt; to determine the &quot;supplementary law&quot; for $\left( \frac{2}{p} \right)$.  The magical algebraic identity in this case is that if $a^4 = -1$, then $(a \pm 1/a)^2 = \pm 2$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-11-10T16:38:25.087" />
  <row Id="9736" PostTypeId="2" ParentId="9730" CreationDate="2010-11-10T16:41:27.543" Score="0" ViewCount="0" Body="&lt;p&gt;As Ross suggested, from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard deviations of non-overlapping (X ∩ Y = ∅) sub-populations can be aggregated as follows if the size (actual or relative to one another) and means of each are known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/math/9/5/1/951ab01cabd025deadd5bf497bae971f.png&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/math/9/5/1/951ab01cabd025deadd5bf497bae971f.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3364" LastActivityDate="2010-11-10T16:41:27.543" />
  <row Id="9737" PostTypeId="2" ParentId="9731" CreationDate="2010-11-10T16:50:41.003" Score="1" ViewCount="0" Body="&lt;p&gt;If you calculate $T(n)$ you can observe that $T(n)=4n-2$.  To prove that by induction, observe that it is true for $n=1$ as $T(1)=2=4*1-2$.  Then assume it is true for $T(n)$ and prove that it is true for $T(2n)$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-10T17:32:24.167" LastActivityDate="2010-11-10T17:32:24.167" CommentCount="2" />
  <row Id="9738" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T17:08:18.947" Score="7" ViewCount="0" Body="&lt;p&gt;I would try to explain it in terms of exponents&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2=2^{1}=2^{1+0}=2^{1}\times 2^{0}=2\times 2^{0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by a division&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2}{2}=\dfrac{2}{2}\times 2^{0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1=1\times 2^{0}=2^{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Applying the same argument to any number different from $0$ gives the same result.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-10T18:59:22.567" LastActivityDate="2010-11-10T18:59:22.567" />
  <row Id="9739" PostTypeId="2" ParentId="9688" CreationDate="2010-11-10T17:19:50.370" Score="2" ViewCount="0" Body="&lt;p&gt;As my approach is different to that of Willie Wong, I give it here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $f$ is in &lt;em&gt;some&lt;/em&gt; $L^{p_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(p_n)$ be such that $0 &amp;lt; p_n &amp;lt; p_0$ and $p_n \downarrow 0$. Define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_n(x) = \frac{1}{p_0} (|f(x)|^{p_0} - 1) - \frac{1}{p_n} (|f(x)|^{p_n} - 1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\frac{1}{p} (x^p - 1) \downarrow \log x$ as $p \downarrow 0$ for all $x &gt; 0$. So let the $g_n \to g$. Then by the Monotone Convergence Theorem we have that&#xA;$$\int g_n \uparrow \int g.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that &#xA;$$\int \frac{1}{p_n} (|f|^{p_n} -1) \downarrow \int \log |f|.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by Jensen's inequality we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{exp} \left ( \int \log |f| \right ) \leq \left ( \int |f|^{p_n} \right )^\frac{1}{p_n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by the inequality $|t| \leq \text{exp}(|t| - 1)$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \int |f|^{p_n} \right )^\frac{1}{p_n} \leq \text{exp} \left ( \int \frac{1}{p_n} (|f|^{p_n} - 1) \right ).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this together with our previous claim proves the theorem.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-10T17:19:50.370" />
  <row Id="9740" PostTypeId="2" ParentId="9608" CreationDate="2010-11-10T17:30:15.913" Score="0" ViewCount="0" Body="&lt;p&gt;OK, I think I've found a closed form (someone double check my math here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the CDF of x (for x&gt;0) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{erf}\left(\frac{x}{\sqrt{2} \sqrt{t}}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the PDF (for x&gt;0) is clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t}}}{\sqrt{t}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the probability that the maximum between t1 and t2 is exactly x, we compute the probability that x[t1] == y (normal distribution w SD of Sqrt[t1]):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\frac{y^2}{2 \text{t1}}}}{\sqrt{2 \pi } \sqrt{\text{t1}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then multiply the probability that the max will be x-y between t1 and t2 (only valid for y&amp;lt;=x):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{(x-y)^2}{2&#xA;(\text{t2}-\text{t1})}}}{\sqrt{\text{t2}-\text{t1}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we integrate this wrt y from -Infinity to x:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\frac{(x-y)^2}{2 (\text{t2}-\text{t1})}-\frac{y^2}{2\text{t1}}}}{\pi  \sqrt{\text{t1}} \sqrt{\text{t2}-\text{t1}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{e^{-\frac{x^2}{2 \text{t2}}} \left(\text{erf}\left(\frac{x\sqrt{\frac{1}{\text{t1}}-\frac{1}{\text{t2}}}}{\sqrt{2}}\right)+1\right)}{\sqrt{2 \pi } \sqrt{\text{t2}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize all of this was in the previous post, but I wanted to play around with LaTex.&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2469" LastEditorDisplayName="" LastEditDate="2010-11-10T18:00:28.353" LastActivityDate="2010-11-10T18:00:28.353" CommentCount="1" />
  <row Id="9741" PostTypeId="2" ParentId="9692" CreationDate="2010-11-10T17:31:24.697" Score="6" ViewCount="0" Body="&lt;p&gt;This is essentially Exercise 6.5 in Folland's &lt;em&gt;Real Analysis&lt;/em&gt;.  Let $(X,\mathcal{F},\mu)$ be a measure space, and let $$m = \inf\{\mu(A) : A \in \mathcal{F}, \mu(A) &gt; 0\}$$&#xA;$$M = \sup\{\mu(A) : A \in \mathcal{F}, \mu(A) &amp;lt; \infty\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $0 &amp;lt; p &amp;lt; q &amp;lt; \infty$, it is a fact that $L^p(\mu) \subset L^q(\mu)$ iff $m &gt; 0$, and $L^q(\mu) \subset L^p(\mu)$ iff $M &amp;lt; \infty$ (which in particular holds when $\mu$ is finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a finite measure $\mu$, it is necessary and sufficient that for some $m &gt;  0$ every set either has measure 0 or has measure at least $m$.  This is going to force your space to be &quot;effectively&quot; finite in some sense.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-10T17:31:24.697" />
  <row Id="9742" PostTypeId="2" ParentId="9731" CreationDate="2010-11-10T17:31:45.553" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ From the first few values we guess $\rm\ T(2^n)\ =\ 2^{n+2}-2\ $ and induction confirms it:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm T(2^{n+1})\ =\ 2\ T(2^n) + 2 \ =\ 2\ (2^{n+2}-2) +\ 2\ =\ 2^{n+3} - 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can extend $\rm\:T\:$ to $\:\mathbb N\:$ by defining $\rm\ T(2k+1) = 2\ T(k+1)-2 $ and now one easily proves by induction that $\rm\ T(k) = 4\:k-2\ $ since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad T(2k+1)\ =\ 2\ T(k+1)-2\ =\ 2\ (4k+2)-2\ =\ 4(2k+1)-2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\ T(2k)\ =\ 2\ \ \ \ T(k)\ \ +\ \ \: 2\ =\ 2\ (4k-2) + 2\ =\ 4 (2k) - 2 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-10T19:16:05.943" LastActivityDate="2010-11-10T19:16:05.943" />
  <row Id="9743" PostTypeId="2" ParentId="9731" CreationDate="2010-11-10T17:36:19.217" Score="1" ViewCount="0" Body="&lt;p&gt;Here is how I attempt this from the point you left off:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that &lt;code&gt;T(1) = 2&lt;/code&gt;. We are trying to prove that &lt;code&gt;T(n) = 4n-2&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is trivially true for n = 1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(1) = 4(1) - 2&#xA;     = 4 - 2&#xA;     = 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Assume&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(k) = 4k - 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From the original definition:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(k+1) = 2T( [k+1] / 2 ) + 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;//since we assumed up to T(k) is correct and (k+1)/2 is less than T(k), we substitute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, We have &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2( 4( (k+1) /2) - 2  ) + 2&#xA;= 4(k+1)- 4 + 2&#xA;= 4(k+1) - 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Proven.&lt;/p&gt;&#xA;" OwnerUserId="3371" LastEditorUserId="3371" LastEditorDisplayName="" LastEditDate="2010-11-10T20:01:19.530" LastActivityDate="2010-11-10T20:01:19.530" CommentCount="3" />
  <row Id="9744" PostTypeId="2" ParentId="9724" CreationDate="2010-11-10T18:01:51.110" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; When do you say that a set is bounded. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;A set $X \subset \mathbb{R}$ is said to be bounded if you can find a $M \in \mathbb{R}$ such that $|x| \leq M$ for all $x \in X$.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now consider your cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$X = \{ x | x = \frac{1}{n} ; \ n \in \mathbb{N}\}$ *&lt;em&gt;NOTE:&lt;/em&gt;* $\frac{1}{n} \leq 1$ for all $n \in \mathbb{N}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly the Second one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now , the third one is unbounded. (Why?). Assume $2^{n} \leq M$ for all $n \in \mathbb{N}$ and obtain a contradiction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fourth one follows from the definition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-17T02:56:52.863" LastActivityDate="2010-11-17T02:56:52.863" CommentCount="1" />
  <row Id="9745" PostTypeId="1" CreationDate="2010-11-10T18:32:59.247" Score="2" ViewCount="78" Body="&lt;p&gt;Let $f \in L^{1}(\mathbb{R})$ such that $f(x) &gt;0$ for almost all $x \in \mathbb{R}$. Let A be a Lebesgue measurable set such that $\int_{A} f =0$. Prove $m(A) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wlog we can assume $f(x)&gt;0$ everywhere define $A_{n}={x \in A: f(x) &gt; \frac{1}{n}}$&#xA;so since $f(x) &gt; 0$ everywhere, $\int_{A_n} f dm \leq \int_{A} fdm = 0$. Then it is easy to see that A is a countable union of sets of measure 0, so A has measure 0 in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: why do we need A to be Lebesgue measurable? doesn't the above proof works for any measurable set? I still don't see the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3373" LastActivityDate="2010-11-10T18:43:29.333" Title="Integral, Lebesgue measure" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9746" PostTypeId="1" AcceptedAnswerId="9806" CreationDate="2010-11-10T18:43:28.880" Score="1" ViewCount="93" Body="&lt;p&gt;It is claimed in some logic books that the compactness theorem of first-order logic can be proven using Tychnoff's theorem from topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to me this feels very strange because I consider logic as something &quot;lower&quot; in the sense that the theorems of topology follow from the axioms and theorems in logic. How can we avoid circularities then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another small question I have is that if we have $A \iff B$ and we prove both directions with contraposition, does there exist a direct proof? You could say we could just follow the steps backwards, but is this always possible?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-11T02:47:55.600" Title="Tychonoff and compactness (logic) and another small logic question." Tags="&lt;topology&gt;&lt;logic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9747" PostTypeId="2" ParentId="9745" CreationDate="2010-11-10T18:43:29.333" Score="1" ViewCount="0" Body="&lt;p&gt;When they write $L^1(\mathbb{R})$, the measure on $\mathbb{R}$ hasn't been specified, so it's assumed to be Lebesgue measure.  If you had $L^1(\mu)$ for an arbitrary measure $\mu$ your argument would work just as well.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-10T18:43:29.333" CommentCount="1" />
  <row Id="9748" PostTypeId="1" CreationDate="2010-11-10T19:03:33.667" Score="0" ViewCount="47" Body="&lt;p&gt;Let X be a sigma finite measure space with $p &gt; 0$. The book states the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists a map $f \in L^{p}(X)$ such that $f&gt;0$ and f is bounded by $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $C_{n}$ be a sequence of disjoint sets of finite measure such that X is the union&#xA;of these sets. Now just put $f(x) = (\frac{2^{-n}}{1+\mu(C_{n})})^{\frac{1}{p}}$ whenever $x \in C_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pretty clear the boundedness of f. How do we show that $f \in L^{p}(X)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because $\int_{X} f \leq \mu(C_{n}) &amp;lt; \infty$ ?&lt;/p&gt;&#xA;" OwnerUserId="3373" LastEditorUserId="3373" LastEditorDisplayName="" LastEditDate="2010-11-10T19:47:18.573" LastActivityDate="2010-11-10T19:47:22.303" Title="Sigma finite and construction of map" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9749" PostTypeId="1" AcceptedAnswerId="9751" CreationDate="2010-11-10T19:08:25.143" Score="5" ViewCount="235" Body="&lt;p&gt;What are some interesting applications of the Mean Value Theorem for derivatives (both the 'extended' or 'non-extended' versions as seen &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; are of interest)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've seen some trivial applications like finding the number of roots of a polynomial equation. What are some more interesting applications of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking this as I'm not exactly sure why MVT is so important - so examples which focus on explaining that would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="181" LastActivityDate="2010-11-13T12:34:04.600" Title="Applications of the Mean Value Theorem" Tags="&lt;calculus&gt;&lt;applications&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="1" />
  <row Id="9750" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:20:24.837" Score="2" ViewCount="0" Body="&lt;p&gt;There are applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an important one, Taylor series proof relies on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other application I like is to quickly come up with and prove inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1)&#xA;$\displaystyle |\cos x - \cos y| \le |x - y|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2)&#xA;$\displaystyle \frac{1}{2\sqrt{n+1}} &amp;lt; \sqrt{n+1} - \sqrt{n} &amp;lt; \frac{1}{2\sqrt{n}}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-10T19:20:24.837" />
  <row Id="9751" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:23:49.383" Score="4" ViewCount="0" Body="&lt;p&gt;There are several applications of the Mean Value Theorem. It is one of the most important theorems in analysis and is used all the time. I've listed $5$ important results below. I'll provide some motivation to their importance if you request.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1)$ If $f: (a,b) \rightarrow \mathbb{R}$ is differentiable and $f'(x) = 0$ for all $x \in (a,b)$, then $f$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2)$ &lt;strong&gt;Leibniz's rule:&lt;/strong&gt; Suppose $ f : [a,b] \times [c,d] \rightarrow \mathbb{R}$ is a continuous function with $\partial f/ \partial x$ continuous. Then the function $F(x) = \int_{c}^d f(x,y)dy$ is derivable with derivative $$ F'(x) = \int_{c}^d \frac{\partial f}{\partial x} (x,y)dy.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3)$ &lt;strong&gt;L'Hospital's rule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4)$ If $A$ is an open set in $\mathbb{R}^n$ and $f:A \rightarrow \mathbb{R}^m$ is a function with continuous partial derivatives, then $f$ is differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5)$ &lt;strong&gt;Symmetry of second derivatives:&lt;/strong&gt; If $A$ is an open set in $\mathbb{R}^n$ and $f:A \rightarrow \mathbb{R}$ is a function of class $C^2$, then for each $a \in A$, $$\frac{\partial^2 f}{\partial x_i \partial x_j} (a) = \frac{\partial^2 f}{\partial x_j \partial x_i} (a)$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-10T20:06:45.493" LastActivityDate="2010-11-10T20:06:45.493" />
  <row Id="9752" PostTypeId="2" ParentId="9748" CreationDate="2010-11-10T19:27:51.897" Score="1" ViewCount="0" Body="&lt;p&gt;Note that $X = \displaystyle \cup_{n=1}^{\infty} C_n$ such that $\mu(C_n) &amp;lt; \infty$. (This is because $X$ is a sigma finite measure space under $\mu$. Also, note that we can choose $C_n$ such that $C_n$'s are mutually disjoint. So, we can assume that $C_i \cap C_j = \phi$. (We need this to define $f(x)$ uniquely. You could of course circumvent this by defining $f$ uniquely on the intersection incase they are not disjoint)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = (\frac{2^{-n}}{1+\mu(C_n)})^{1/p}$, whenever $x \in C_n$. (If $C_n$'s are not disjoint, define $f$ to be the infimum of $(\frac{2^{-n}}{1+\mu(C_n)})^{1/p}$ on the intersection of the sets). Clearly, $f(x) &gt; 0$ (Since each of the $\mu(C_n) &amp;lt; \infty$). Also, $f(x)$ is bounded above by $1$. (By construction of $f(x)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\int_X |f|^p d\mu = \int_{ \displaystyle \cup_{n=1}^{\infty} C_n} |f|^p d\mu \leq \displaystyle \sum_{n=1}^{\infty} \int_{C_n} |f|^p d\mu = \displaystyle \sum_{n=1}^{\infty} \int_{C_n} (\frac{2^{-n}}{1+\mu(C_n)}) d\mu$&#xA;$ = \displaystyle \sum_{n=1}^{\infty} (\frac{2^{-n}}{1+\mu(C_n)}) \mu(C_n) &amp;lt; \displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n} = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f \in \mathcal{L}^p(X)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-10T19:47:22.303" LastActivityDate="2010-11-10T19:47:22.303" CommentCount="4" />
  <row Id="9753" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:28:12.967" Score="3" ViewCount="0" Body="&lt;p&gt;Some more applications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If the derivative of a function $f$ is everywhere strictly positive, then f is a strictly increasing function&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Suppose $f$ is differentiable on whole of $\mathbb{R}$, and $f'(x)$ is a constant. Then $f$ is linear.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mean Value theorem plays an important role in the proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem of Calculus&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Suppose $f$ is continuous on $[a,b]$ and $f'$ exists and is bounded on the interior, then $f$ is of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bounded_variation&quot; rel=&quot;nofollow&quot;&gt;Bounded Variation&lt;/a&gt; on $[a,b]$&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-10T19:43:08.847" LastActivityDate="2010-11-10T19:43:08.847" />
  <row Id="9754" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T19:47:12.880" Score="1" ViewCount="0" Body="&lt;p&gt;The generalization of the Mean Value Theorem known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;Cauchy's Mean Value Theorem&lt;/a&gt; can be used to prove L'Hopital's Rule. See for instance the following &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#Proof_of_l.27H.C3.B4pital.27s_rule&quot; rel=&quot;nofollow&quot;&gt;proof&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-10T19:47:12.880" />
  <row Id="9755" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T19:52:45.870" Score="1" ViewCount="0" Body="&lt;p&gt;I think it's quiet simple to understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base point is 1 (I'm not talking about the base... read on...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from this initial point of 1, you are applying the base to the result n times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^3 = 1 * (2 *2 *2) = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^-3 = ((( 1 / 2) /2) / 2)  = 1/8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base point is always 1 so&#xA;when you are in the 0 point (n^0) you don't have to multiply the very basic 1 so you have &quot;1&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the starting point is the &quot;1&quot; not the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I've just noticed that Zarkonnen already gave this answer, yet I think that the way I presented the answer is easier for a child to understand- So Zarkonnen, I hope you are ok with this post ;)&lt;/p&gt;&#xA;" OwnerUserId="3375" LastEditorUserId="3375" LastEditorDisplayName="" LastEditDate="2010-11-10T20:03:24.203" LastActivityDate="2010-11-10T20:03:24.203" />
  <row Id="9756" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T20:12:16.600" Score="4" ViewCount="0" Body="&lt;p&gt;I'd demonstrate this using a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^3 = 8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^2 = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^1 = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^0 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{-1} = 1/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{-2} = 1/4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you decrease the exponent, you divide by 2. So, when you go from 2&lt;sup&gt;1&lt;/sup&gt; to 2&lt;sup&gt;0&lt;/sup&gt;, of course you divide by 2, which gives you 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, you can segue into negative exponents, if you'd like. Just keep dividing by 2.&lt;/p&gt;&#xA;" OwnerUserId="2994" LastActivityDate="2010-11-10T20:12:16.600" CommentCount="1" />
  <row Id="9757" PostTypeId="2" ParentId="9746" CreationDate="2010-11-10T20:14:45.803" Score="1" ViewCount="0" Body="&lt;p&gt;For the second question, suppose that in some proof system you can prove $p(x) \Leftrightarrow q(x)$ for some expressions $p,q$, and only $p(x)\Rightarrow q(x)$ requires proof by contradiction. Then to prove $p(x) \land q(y) \Leftrightarrow q(x) \land p(y)$ you probably need proof by contradiction on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only know that $p(x) \Rightarrow q(x)$ requires proof by contradiction ($p(x) \Leftarrow q(x)$ not necessarily being true), then $p(x) \Rightarrow p(x) \land q(x)$ also probably requires proof by contradiction, and this time both sides are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-10T20:14:45.803" CommentCount="2" />
  <row Id="9758" PostTypeId="1" CreationDate="2010-11-10T20:17:32.560" Score="3" ViewCount="136" Body="&lt;p&gt;What is an intuitive explanation of a positive-semidefinite matrix? Or a simple example which gives more intuition for it rather than the bare definition. Say $x$ is some vector in space and $M$ is some operation on vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $n$ × $n$ Hermitian matrix M is called &lt;em&gt;positive-semidefinite&lt;/em&gt; if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{*} M x \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $x \in \mathbb{C}^n$ (or, all $x \in \mathbb{R}^n$ for the real matrix), where $x^*$ is the conjugate transpose of $x$.&lt;/p&gt;&#xA;" OwnerUserId="915" LastEditorUserId="915" LastEditorDisplayName="" LastEditDate="2010-11-10T20:29:49.630" LastActivityDate="2010-11-10T23:50:20.640" Title="Intuitive explanation of a positive-semidefinite matrix" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="9759" PostTypeId="2" ParentId="9749" CreationDate="2010-11-10T20:21:34.317" Score="1" ViewCount="0" Body="&lt;p&gt;If $f$ and $g$ are two differentiable functions on an interval $I$ and for&#xA;all $x\in I$, $f^{\prime }(x)=g^{\prime }(x)$, then the difference $f(x)-g(x)&#xA;$ is constant on $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical interpretation (assuming the movement is uninterruptible): If the average speed of a car between two locations was $V$ km/h, then there was at least one instant where the speed indicator displayed $V$ km/h. &lt;/p&gt;&#xA;" OwnerUserId="2044" LastActivityDate="2010-11-10T20:21:34.317" />
  <row Id="9760" PostTypeId="1" CreationDate="2010-11-10T20:22:58.653" Score="13" ViewCount="265" Body="&lt;p&gt;We have three counters, $i, j, k$, all initialized to zero. Each step consists of adding or subtracting one from one of the counters, so $(\Delta i, \Delta j, \Delta k)$ is selected among $(\pm1, 0, 0), (0, \pm1, 0), (0, 0, \pm1)$, each with probability 1/6. What is the expected number of steps before all the counters are N modulo 2N at the same time?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-12T04:33:41.310" LastActivityDate="2010-11-13T20:45:14.273" Title="Expected number of steps before three counters reach N modulo 2N at the same time" Tags="&lt;probability&gt;&lt;expectation&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="9" />
  <row Id="9761" PostTypeId="2" ParentId="9758" CreationDate="2010-11-10T20:24:34.173" Score="3" ViewCount="0" Body="&lt;p&gt;One intuitive definition is as follows. Multiply any vector with a positive definite matrix. The angle between the original vector and the resultant vector will always be less than $\frac{\pi}{2}$. In some sense the positive definite matrix tries to keep the vector within a certain half space of the vector. This is analogous to what a positive number does to a real variable. Multiply it and it only stretches or contracts the number but never reflects it about the origin.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-10T20:24:34.173" CommentCount="1" />
  <row Id="9762" PostTypeId="1" AcceptedAnswerId="9764" CreationDate="2010-11-10T20:25:23.327" Score="1" ViewCount="73" Body="&lt;p&gt;Injective functions are not the same thing as injective &lt;em&gt;correspondences&lt;/em&gt;, correct? Injective functions are a subset of the injective &lt;em&gt;correspondences&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in $y^2 = x$, y is not a function of x, but this is still a correspondence between $\mathbb{R} \rightarrow \mathbb{R}$. Can we say that the correspondence, $C:X \rightarrow Y$ is injective? (Since, if $C(x_1) = C(x_2)$ then $x_1 = x_2$.) Can we say it is also a surjective &lt;em&gt;correspondence&lt;/em&gt; since every y-value has at least one (mostly two) pre-images?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I way off base applying these definitions to things that are not functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last question. &quot;Mappings&quot; are, as the Wikipedia suggests, the same thing as functions. (For some reason I thought mappings were correspondences. In fact, it is in my old notes. My old notes are wrong, right? )&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-10T20:37:03.127" LastActivityDate="2010-11-10T20:44:05.773" Title="Injective functions vs. correspondences." Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9763" PostTypeId="2" ParentId="9758" CreationDate="2010-11-10T20:40:17.200" Score="2" ViewCount="0" Body="&lt;p&gt;First I'll tell you how I think about Hermitian positive-definite matrices.  A Hermitian positive-definite matrix $M$ defines a sesquilinear inner product $\langle Mv, w \rangle = \langle v, Mw \rangle$, and in fact every inner product on a finite-dimensional inner product space $V$ has this form.  In other words it is a way of computing angles between vectors, or a way of projecting vectors onto other vectors; over the real numbers it is the key ingredient to doing Euclidean geometry.  An inner product can be recovered from the norm $\langle Mv, v \rangle = \langle v, Mv \rangle$ it induces, and a norm in turn can be recovered from its unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$.  This unit sphere is a distorted version of the usual unit sphere; the distortions will occur along axes corresponding to the eigenvectors of $M$, and the amount of distortion corresponds to the (inverses of the) corresponding eigenvalues.  For example when $\dim V = 2$ it is an ellipse and when $\dim V = 3$ it is an ellipsoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Hermitian positive-semidefinite matrix $M$ no longer describes an inner product because it is not necessarily positive-definite, but it still defines a sesquilinear form.  It also defines a function $\langle Mv, v \rangle$ which is no longer a norm because it is not necessarily positive-definite; some people call these &quot;pseudonorms,&quot; I think.  The corresponding unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$ might now be lower-dimensional than the usual unit sphere, depending on how many eigenvalues are equal to zero; for example if $\dim V = 3$ it might be an ellipsoid, or an ellipse, or two points.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-10T20:40:17.200" CommentCount="3" />
  <row Id="9764" PostTypeId="2" ParentId="9762" CreationDate="2010-11-10T20:44:05.773" Score="2" ViewCount="0" Body="&lt;p&gt;What you're describing is universally known as a &quot;relation.&quot;  It is true that you can have an injective relation which is not a function.  Unfortunately, there are different uses for the term &quot;correspondence&quot; -- used alone, it can denote a general relation, but &quot;one-to-one correspondence&quot; can also denote a bijection.  The safest thing to do is probably to avoid using the word altogether.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-10T20:44:05.773" CommentCount="2" />
  <row Id="9765" PostTypeId="2" ParentId="9758" CreationDate="2010-11-10T20:44:56.280" Score="3" ViewCount="0" Body="&lt;p&gt;Let's consider the set $E$ of all vectors $y=Mx$, where $x\in\mathbb R^n$ belongs to the unit sphere (i.e. $\|x\|=1$). In other words, $E$ is the image of the unit sphere under the linear transformation. If the matrix is non-degenerate, $E$ is an $n$-dimensional ellipsoid. But if we assume additionally that $M$ is symmetric then we can say much more about the structure of $E$. Namely, the directions of the axes of the ellipsoid are pairwise orthogonal and represented by the eigenvectors of the matrix. Moreover, the lengths of the semiaxes are given by the corresponding eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-11-10T23:50:20.640" LastActivityDate="2010-11-10T23:50:20.640" />
  <row Id="9766" PostTypeId="1" CreationDate="2010-11-10T20:56:28.677" Score="1" ViewCount="62" Body="&lt;p&gt;Hello&#xA;i have to solve the following problem&#xA;find an $\displaystyle f(k)$ where $\displaystyle S_k=\theta(f(k))$&#xA;where $\displaystyle S_k =\sum_{n=1}^{k^2-1} \sqrt{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried first of all to calculate or &quot;limit&quot; my sum using integral rule so i came up with&#xA;$\displaystyle \frac{2(k^2-1)^{3/2}}{3} \leq S_k \leq \frac{2(k^3-1)}{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but after that i am in a dead end as i do not know $\displaystyle S_k$ so i can not simplify anything&#xA;can anyone help how to proceed this problem ?&#xA;thanks &lt;/p&gt;&#xA;" OwnerUserId="3382" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-10T21:02:28.970" LastActivityDate="2010-11-10T21:28:37.123" Title="Problem calculating theta notation " Tags="&lt;asymptotics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9767" PostTypeId="2" ParentId="9766" CreationDate="2010-11-10T21:04:28.237" Score="4" ViewCount="0" Body="&lt;p&gt;As far as I see you already have the solution there. In Theta-notation only the fastest growing terms are important. And that is the same for the upper and lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Theta(k^3) \leq S_k \leq \Theta(k^3).$$&lt;/p&gt;&#xA;" OwnerUserId="915" LastEditorUserId="915" LastEditorDisplayName="" LastEditDate="2010-11-10T21:28:37.123" LastActivityDate="2010-11-10T21:28:37.123" CommentCount="5" />
  <row Id="9768" PostTypeId="1" CreationDate="2010-11-10T21:10:06.663" Score="4" ViewCount="82" Body="&lt;p&gt;This may be embarassingly simple, but I can't see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be a Riemannian manifold of dimension $n$; fix $x \in M$, and let $B(x,r)$ denote the geodesic ball in $M$ of radius $r$ centered at $x$.  Let $V(r) = \operatorname{Vol}(B(x,r))$ be the Riemannian volume of $B(x,r)$.  It seems to be the case that for small $r$, $V(r) \sim r^n$, i.e. $V(r)/r^n \to c$ with $0 &amp;lt; c &amp;lt; \infty$.  How is this proved, and where can I find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a neighborhood $U \ni x$ and a chart $\phi : U \to \mathbb{R}^n$, certainly $\phi$ has nonvanishing Jacobian, hence (making $U$ smaller if necessary) bounded away from 0.  So $\operatorname{Vol}(\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))) \sim r^n$.  But I do not see how to relate the pullback $\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))$ of a Euclidean ball to a geodesic ball in $M$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-10T22:43:59.873" Title="Volume of a geodesic ball" Tags="&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="9769" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T21:21:09.797" Score="0" ViewCount="0" Body="&lt;p&gt;When I was a kid, I had to teach this concept (and fractions) to some kids in a lower grade. The best method is to work backward with division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^3 = 8/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^2 = (2^3)/2 = 8/2 = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^1 = (2^2)/2 = 4/2 = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^0 = (2^1)/2 = 2/2 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. This is also the only way to wrap your head around negative powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the best way to teach fractions is by having pizza for dinner.&lt;/p&gt;&#xA;" OwnerUserId="3385" LastEditorUserId="3385" LastEditorDisplayName="" LastEditDate="2010-11-10T23:47:51.813" LastActivityDate="2010-11-10T23:47:51.813" CommentCount="1" />
  <row Id="9770" PostTypeId="1" AcceptedAnswerId="9783" CreationDate="2010-11-10T21:27:10.203" Score="5" ViewCount="223" Body="&lt;p&gt;Greetings!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand what it means to have an imaginary number in an exponent. What does $x^{i}$ where $x$ is real mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read a &lt;a href=&quot;http://mathforum.org/library/drmath/view/52251.html&quot; rel=&quot;nofollow&quot;&gt;few&lt;/a&gt;  &lt;a href=&quot;http://www.newton.dep.anl.gov/newton/askasci/1995/math/MATH057.HTM&quot; rel=&quot;nofollow&quot;&gt;pages&lt;/a&gt; on this issue, and they all seem to boil down to the same thing:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any real number $x$ can be written as $e^{\ln{x}}$ (seems obvious enough.)&lt;/li&gt;&#xA;&lt;li&gt;Mumble mumble mumble&lt;/li&gt;&#xA;&lt;li&gt;This is equivalent to $e^{\cos{x} + i\sin{x}}$ &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Clearly I'm missing something in step 2. I understand (at least I think I do) how the complex number $\cos{x} + i\sin{x}$ maps to a point on the unit circle in a complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am missing, I suppose, is how this point is related to the natural log of $x$. Moreover, I don't understand what complex exponentiation &lt;em&gt;is&lt;/em&gt;. I can understand integer exponentiation as simple repeated multiplication, and I can understand other things (like fractional or negative exponents) by analogy with the operations that undo them. But what does it mean to repeat something $i$ times?&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-11-11T11:39:49.023" Title="Understanding imaginary exponents" Tags="&lt;trigonometry&gt;&lt;complex-numbers&gt;&lt;logarithms&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="9771" PostTypeId="2" ParentId="9768" CreationDate="2010-11-10T21:49:54.190" Score="1" ViewCount="0" Body="&lt;p&gt;A Google search for the title of the question finds &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.mmj/1029001150&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; where the first five coefficients in the series expansion of the volume (in powers of $r$) are computed. The first term is the same as the Euclidean volume (proportional to $r^n$, in other words); then come higher-order corrections depending on the curvature.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-10T21:49:54.190" CommentCount="2" />
  <row Id="9772" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T21:54:27.420" Score="1" ViewCount="0" Body="&lt;p&gt;It is a way to extend your real variable function into the complex domain in a &quot;nice&quot; (read nice as holomorphic) way. You might ask the question why do we need to even extend in the first place? The need for extension comes from the fact that a $n^{th}$ order polynomial with real co-efficients need not have all its roots as real numbers. So you end up with these complex numbers. Once we find that we have these new numbers, we begin to wonder why should we not try to feed these numbers into functions which are defined on the real variables. But then we find that we have a zillion ways to define these functions for these new complex numbers. A &quot;nice&quot; way to extend these functions is to do in a way that the extended function is holomorphic. That is why you extend $e^{i \theta}$ to the complex plane as $cos(\theta) + i sin(\theta)$. Once you do this extension then all the other things related to logarithms and exponentiation follows in a logically coherent way. For instance, we can now define $x^i$ (say x&gt;0 or else rewrite $x$ as $-y$ where $y&gt;0$ and write -1 as $e^{i \pi}$) as $(e^{log (x)})^{i} = e^{ilog(x)} = cos(log(x)) + i sin(log(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I understand it. Kindly correct me if I am wrong.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-10T22:31:19.493" LastActivityDate="2010-11-10T22:31:19.493" CommentCount="2" />
  <row Id="9773" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T21:56:18.660" Score="2" ViewCount="0" Body="&lt;p&gt;Lets say you want to figure out what $x^{a + ib}$ is, then like you mentioned, you start by writing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{a + ib} = e^{a \ln(x) + i b \ln(x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be split up though as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ e^{a \ln(x) + i b \ln(x)} = e^{a \ln(x)} e^{i b \ln(x)} = x^a e^{i b \ln(x)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then you can use euler's formula to take care of the imaginary exponent, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{a + ib} = x^a \cos(b \ln(x)) + i x^a \sin(b \ln(x)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula does not provide much intuition as to what is really happening though. You mentioned that you can understand integer exponentiation as simple repeated multiplication, but I don't think that that is the right way to look at it in complex analysis. I think it is much better to look at exponentiation geometrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When exponentiating a real number by a complex value, we found that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{a + ib} = x^a e^{i b \ln(x)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so exponentiating $x$ by $a + ib$ gives you the point in the plane with magnitude $x^a$ and angle $b \ln(x)$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-10T21:56:18.660" CommentCount="2" />
  <row Id="9774" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T22:07:37.547" Score="0" ViewCount="0" Body="&lt;p&gt;The quantity x^i is well-defined if x is a &lt;b&gt;positive&lt;/b&gt; real number.  For negative (or complex) values of x, the evaluation is open to multiple interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the following.  The natural logarithm of positive real numbers is well-defined, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^i = exp(i*ln(x))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that for real x &gt; 0, all one knows about ln(x) is that it is real.  It might be positive, negative, or zero depending on the comparison of x to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the function exp can be defined by a power series that converges everywhere in the complex plane.  As it turns out, that definition of the exponential function implies the following when the argument is purely imaginary (as here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^i = exp(i*ln(x)) = cos(ln(x)) + i*sin(ln(x))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt; [Euler's formula -- Wikipedia]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-10T22:07:37.547" CommentCount="3" />
  <row Id="9775" PostTypeId="2" ParentId="9594" CreationDate="2010-11-10T22:10:19.700" Score="1" ViewCount="0" Body="&lt;p&gt;12^6 is just under 3 million.  So it would probably not work to randomly try six unscrambles.  But it wouldn't be too hard to make a data file of all the positions and their unscramble twists if you can find a reasonable way to search it, like some hash function on a description of the position.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T22:10:19.700" CommentCount="2" />
  <row Id="9776" PostTypeId="1" AcceptedAnswerId="9778" CreationDate="2010-11-10T22:10:39.067" Score="2" ViewCount="177" Body="&lt;p&gt;How do I calculate the outcome of taking one complex number to the power of another, ie  (a + bi)^(c + di)?&lt;/p&gt;&#xA;" OwnerUserId="3367" LastActivityDate="2010-11-11T11:02:25.487" Title="A complex number to the power of another complex number" Tags="&lt;complex-numbers&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="9777" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T22:15:11.397" Score="5" ViewCount="0" Body="&lt;p&gt;The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+\sin y)$$ &#xD;&#xA;&#xD;&#xA;which agrees with the real exponential function when $y=0$. &#xD;&#xA;&#xD;&#xA;The [principal logarithm][1] of $z$ is the complex number &#xD;&#xA;&#xD;&#xA;$$w=\text{Log }z=\log |z|+i\arg z$$&#xD;&#xA;&#xD;&#xA;so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$. &#xD;&#xA;&#xD;&#xA; &#xD;&#xA;&#xD;&#xA;The complex power is &#xD;&#xA;&#xD;&#xA;$$z^w=e^{w\text{ Log} z}.$$ &#xD;&#xA;&#xD;&#xA;In your example $z=x,w=i$ is therefore $x^i=e^{i \text{ Log}x}$. &#xD;&#xA;&#xD;&#xA;If $x&gt;0$, $\text{Log }x=\log x$. If $x&amp;lt;0$, $\text{Log }x=\log |x|+i\pi$. &#xD;&#xA;&#xD;&#xA;Examples:&#xD;&#xA;&#xD;&#xA;$(-1)^i=e^{i\text{Log }(-1)}=e^{i(i\pi)}=e^{-\pi}$.&#xD;&#xA;&#xD;&#xA;$2^i=e^{i\text{Log }(2)}=e^{i\log 2}=\cos (\log 2)+i\sin (\log 2)$.&#xD;&#xA;&#xD;&#xA;$(-2)^i=e^{i\text{Log }(-2)}=e^{i(\log 2+i\pi)}=e^{i\log 2}e^{-\pi}=(\cos (\log 2)+i\sin (\log 2))e^{-\pi}.$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-11T11:39:49.023" LastActivityDate="2010-11-11T11:39:49.023" CommentCount="2" />
  <row Id="9778" PostTypeId="2" ParentId="9776" CreationDate="2010-11-10T22:16:55.910" Score="7" ViewCount="0" Body="&lt;p&gt;First you need to realize that this is a multi-valued function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us choose the principal logarithm for convenience. So our argument $\theta$ will lie between $(-\pi,\pi]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then write $a+ib = r e^{i \theta} = e^{(ln(r) + i \theta)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we get $(a+ib)^{c+id} = e^{(ln(r) + i \theta)(c+id)}$. Do the necessary algebraic manipulations in the exponent to get $e^{(cln(r) - d \theta) + i(d ln(r) + c \theta)}$. You might also want to take a look at the previous question asked on a similar &lt;a href=&quot;http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents&quot;&gt;topic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-10T23:54:07.967" LastActivityDate="2010-11-10T23:54:07.967" />
  <row Id="9779" PostTypeId="2" ParentId="9776" CreationDate="2010-11-10T22:21:01.890" Score="5" ViewCount="0" Body="&lt;p&gt;Well, assuming principal values of the complex logarithm (otherwise much craziness ensues):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+bi)^{c+di}=\exp((c+di)\ln(a+bi))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\exp((c+di)(\ln|a+bi|+i\arg(a+bi)))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\exp((c\ln|a+bi|-d\arg(a+bi))+i(c\arg(a+bi)+d\ln|a+bi|))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\exp((c\ln|a+bi|-d\arg(a+bi)))\exp(i(c\arg(a+bi)+d\ln|a+bi|))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I'll let you finish it at this point, using the fact that $\exp(ix)=\cos\;x+i\sin\;x$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-10T22:21:01.890" />
  <row Id="9780" PostTypeId="2" ParentId="9567" CreationDate="2010-11-10T22:35:41.070" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LENum.png&quot; alt=&quot;circles through 2 points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Américo Tavares and J. M. said in their comments, two points do not uniquely determine an arc.  Shown above are several circles that pass through two given points.  However, the center of any circle that passes through those two points is on the perpendicular bisector of the line segment with those points as endpoints.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-11-10T22:35:41.070" CommentCount="1" />
  <row Id="9781" PostTypeId="2" ParentId="9768" CreationDate="2010-11-10T22:43:59.873" Score="1" ViewCount="0" Body="&lt;p&gt;It's simple, all right.  As I realized not long after posting (and as Hans also suggested), the key is the exponential map.  The tangent space $T_x M$ gets an inner product space structure from the Riemannian metric; we can isometrically identify it with $\mathbb{R}^n$.  Now $\exp_x : \mathbb{R}^n \to M$ is a diffeomorphism on some small ball $B_{\mathbb{R}^n}(0,\epsilon)$; on this ball, straight lines map to length-minimizing geodesics (see Do Carmo, &lt;em&gt;Riemannian Geometry&lt;/em&gt;, Proposition 3.6), and thus Euclidean balls map to geodesic balls of the same radius.  Taking $\epsilon$ smaller if necessary, we can assume the Jacobian of $\exp_x$ is bounded away from $0$ and $\infty$ on $B_{\mathbb{R}^n}(0, \epsilon)$; thus for $r &amp;lt; \epsilon$ we have that $\operatorname{Vol}(B(x,r))$ is comparable to $\operatorname{Vol}(B_{\mathbb{R}^n}(0,r)) \sim r^n$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-10T22:43:59.873" />
  <row Id="9782" PostTypeId="2" ParentId="9690" CreationDate="2010-11-10T22:44:19.577" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a different way to look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make two copies of the parallelogram, sharing the side CD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/czbHh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second copy is A'B'DC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will show that B'=E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\triangle{B'AB}$ and $\triangle{B'DM}$. These are similar and since B'D = AD, we have that DM = AB/2. M is the midpoint of CD and thus B' = E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus B'L' = BL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $\triangle{B'DL'}$ and $\triangle{B'AL}$. These are similar and so B'L' = LL', as B'D = AD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus EL = EL' + LL' = 2B'L' = 2BL.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-10T23:10:27.057" LastActivityDate="2010-11-10T23:10:27.057" />
  <row Id="9783" PostTypeId="2" ParentId="9770" CreationDate="2010-11-10T22:58:32.723" Score="5" ViewCount="0" Body="&lt;p&gt;Consider a real number $A$, and take it to the power $i$.  If our system of complex numbers is to be consistent, then $A^i$ must be a complex number; in other words, there must be two real numbers $x$ and $y$, which depend on $A$, such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A^i=x+iy$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we can write $A^{-i}=x-iy$ for the same $x$ and $y$.  Hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+y^2=(x+iy)(x-iy)=A^iA^{-i}=A^{i-i}=A^0=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that for any real number $A$, $|A^i|=1$, and therefore $A^i$ corresponds to a complex number which lies some angle $\theta$ along the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the sine and cosine functions for extremely small angles $\epsilon$.  A tiny angle $\epsilon$ cuts out a slice of the unit circle, and the curvature of the circumference over this small angle is negligible.  We can therefore think of this slice as a right triangle with angle $\epsilon$, and the hypotenuse and adjacent sides are both length one since they correspond to the radius of the unit circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula for the arc length of a circle, it's easy to determine that in the right triangle formed by the small angle approximation, the length of the side opposite to the angle $\epsilon$ is equal to $\epsilon$.  We can read off the $(x,y)$ coordinates from this diagram (which are $(cos(\epsilon),sin(\epsilon))$), and therefore we conclude that for very small angles $\epsilon$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sin(\epsilon) \approx \epsilon \hspace{10mm} cos(\epsilon) \approx 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore $cos(\epsilon) + isin(\epsilon) \approx 1+i\epsilon$, and hence for real numbers $A$ which are extremely close to one (so that $lnA$ is small), the complex number $A^i$ lies approximately at an angle $lnA$ along the unit circle, since $A^i=e^{ilnA}\approx 1+i(lnA)$.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-10T23:34:26.873" LastActivityDate="2010-11-10T23:34:26.873" CommentCount="4" />
  <row Id="9784" PostTypeId="1" CreationDate="2010-11-10T23:05:35.173" Score="7" ViewCount="134" Body="&lt;p&gt;If A and B are two points on the earth, how could I find any arbitrary point between them along the shortest distance side of their great circle path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points are in radians&lt;br&gt;&#xA;longitude = 0 to 2pi&lt;br&gt;&#xA;latitude = 0 to pi, 0 being at north pole&lt;br&gt;&#xA;Points are not antipodal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I desire something where I specify a range 0.0 to 1.0, with 0.0 being point A and 1.0 being point B and 0.5 being the midpoint between them, with all other values being their corresponding points.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  This is not homework.  I'm 41yrs old and this is for a personal project I'm working on.&lt;/p&gt;&#xA;" OwnerUserId="3390" LastActivityDate="2010-11-11T00:19:17.247" Title="Spherical geometry:  Arbitrary point between two points" Tags="&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="9785" PostTypeId="1" CreationDate="2010-11-10T23:12:41.093" Score="2" ViewCount="139" Body="&lt;p&gt;Let $\Omega = $ {$\omega_{i}$} be an ordered set of $n$ positive reals in the unit interval, $\omega_{1} \leq \cdots \leq \omega_{n} \leq 1$. Define the $n$-simplex $\Delta(\Omega; (\mathbb{R}^{+})^{n})$ by the non-negative points $(x_{1}, \dots, x_{n}) \subset (\mathbb{R}^{+})^{n}$ which satisfy the inequality &#xA;\begin{eqnarray}&#xA;\omega_{1} x_{1} + \cdots + \omega_{n} x_{n} \leq 1.&#xA;\end{eqnarray}&#xA;Let $X$ be a non-trivial subset of the integers $\mathbb{Z}^{n}$. Define $\Delta(\Omega, X) = X \cap \Delta(\Omega, (\mathbb{R}^{+})^{n})$. It is well-known that&#xA;\begin{eqnarray}&#xA;|\Delta(\Omega, \mathbb{N}^{n})| \leq \frac{1}{n!} \prod_{i = 1}^{n} \frac{1}{\omega_{i}} \quad \text{and} \quad |\Delta(\Omega, (\mathbb{Z}^{+})^{n})| \leq \frac{1}{n!} \left(1 + \sum_{i = 1}^{n} \omega_{i} \right)^{n} \prod_{i = 1}^{n} \frac{1}{\omega_{i}},&#xA;\end{eqnarray}&#xA;where $\mathbb{Z}^{+}$ denotes the set of non-negative integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question(s)&lt;/strong&gt;: For the given bounds above, are any sharper bounds known? Given the similarity in form, are there formulas for other $X$ sets, say for integers greater than some arbitrary integer $c$ or integers satisfying some congruence condition (e.g., $a \equiv b$ mod $d$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Update&lt;/em&gt;) The theory of Ehrhart polynomials is relevant to the question above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Suppose I'd like to use the Ehrhart machinery to count the number of non-negative integer solutions of $a_{1} x_{1} + \cdots + a_{n} x_{n} \leq r$ for a non-negative integer $r$ and positive integers {$a_{i}$}. How does one proceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-16T21:48:26.607" LastActivityDate="2010-12-16T22:44:26.017" Title="Upper Bounds on the Number of Lattice Points in an $n$-Simplex" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="9786" PostTypeId="1" AcceptedAnswerId="9787" CreationDate="2010-11-10T23:24:03.673" Score="1" ViewCount="37" Body="&lt;p&gt;I think the following picture helps explain it easier than words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lggXy.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the coordinates of point a, which is at a corner of the triangle, as well as the coordinates of point b, which is somewhere along the edge of the opposite side.  I also know the y-coordinate of point c, and I'm trying to find its x-coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've got so far is that as the y-coordinate of b moves up to c, the x coordinate of c approaches the x coordinate of b.  And as the y-coordinate of b moves down to a, the x coordinate of c moves to half-way between the x coordinates of a and b.  But I don't know the direct relationship.&lt;/p&gt;&#xA;" OwnerUserId="3392" LastActivityDate="2010-11-10T23:35:17.233" Title="How to solve for the x-coordinate of the top of an isosceles triangle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9787" PostTypeId="2" ParentId="9786" CreationDate="2010-11-10T23:35:17.233" Score="3" ViewCount="0" Body="&lt;p&gt;Assuming the base is horizontal as you have drawn it, we have enough information.  Let us call the coordinates $(xa,ya), (xb,yb),$ and $(xc,yc)$.  Then $$\frac{bx-cx}{by-cy}=-\frac{ax-cx}{ay-cy}$$ because the slopes have to be negatives of each other.  You have five of the six variables.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-10T23:35:17.233" CommentCount="2" />
  <row Id="9788" PostTypeId="2" ParentId="9703" CreationDate="2010-11-10T23:43:58.760" Score="6" ViewCount="0" Body="&lt;p&gt;I want to extend the answer by @Qiaochu Yuan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the kid accepts $2*0 = 0$. In other terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Some number &lt;em&gt;times&lt;/em&gt; 0 yields the no-changer of &lt;em&gt;plus&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analoguously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Some number &lt;em&gt;to the power&lt;/em&gt; 0 yields the no-changer of &lt;em&gt;times&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By no-changer I refer, of course, to the unit element. That this can be added/multiplied to anything without resulting in a change should be accepted. I am unsure wether this approach helps understanding the hierarchy of arithmetic operators or wether you need the hierarchy for understanding the approach.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-10T23:43:58.760" CommentCount="2" />
  <row Id="9789" PostTypeId="2" ParentId="9785" CreationDate="2010-11-10T23:55:12.863" Score="0" ViewCount="0" Body="&lt;p&gt;If $\omega_1 = \cdots = \omega_n = 1/k$ then the equation translates to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_1 + \cdots + x_n \leq k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the number of non-negative integer solutions is $\binom{n+k-1}{k}$, which is roughly $n^k/k!$, compared to your estimate $k^n/n!$ (that's assuming $k$ is much smaller; if we assume $n$ is much smaller, we get your estimate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, how do you derive your own upper bound? I assume you start with the volume of the simplex $1/n!$ and then blow up dimension $i$ by $1/\omega_i$, getting your expression. How do you then relate it to the number of lattice points? You can easily construct a convex polytope of arbitrarily small area with arbitrarily many lattice points.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-10T23:55:12.863" />
  <row Id="9790" PostTypeId="2" ParentId="9784" CreationDate="2010-11-11T00:06:44.403" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://williams.best.vwh.net/avform.htm#Dist&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; aviation website has the information that you were looking for. The formula presented there returns the latitude and longitude of a point that is a fraction $f$ between points A and B except when they are antipodal just as you mentioned in the question.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-11-11T00:19:17.247" LastActivityDate="2010-11-11T00:19:17.247" CommentCount="1" />
  <row Id="9791" PostTypeId="1" AcceptedAnswerId="10301" CreationDate="2010-11-11T00:18:06.203" Score="7" ViewCount="189" Body="&lt;p&gt;The definition of small is that they have O(lg n) bits. One way is just to test the integers 2,3,... for primality and keep the first n primes, but this takes at least O(n log n) time (times the cost of primality testing) using a naive algorithm. Is it possible to do it in linear time?&lt;/p&gt;&#xA;" OwnerUserId="2634" LastActivityDate="2010-11-15T18:07:32.193" Title="How can one efficiently generate n small relatively prime integers?" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="0" />
  <row Id="9792" PostTypeId="1" CreationDate="2010-11-11T00:18:22.383" Score="1" ViewCount="49" Body="&lt;p&gt;I am looking for an algorithm to compute the N shortest lattice vectors in R^2. The problem statement is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a lattice $L: \{ m \vec{u}+n\vec{v} \} \in \mathcal{R}^2$, a norm $\left\lVert\cdot\right\rVert$, and an integer $N&gt;0$, compute a set of $N$ shortest lattice vectors $S$ such that if $\vec{x}\in S$, then for all $\vec{y} \notin S$, $\left\lVert\vec{y}\right\rVert \ge \left\lVert\vec{x}\right\rVert$. One may assume that $\vec{u}$ and $\vec{v}$ are reduced basis vectors returned from Gauss' algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, the algorithm can take an existing $S$ and expand it by adding in the next &quot;ring&quot; of vectors with the next greatest norm. I have a rough idea of how to go about doing this for the Euclidean norm; you can move along the perimeter of the already-found vectors and add in new vectors of minimal norm, but I am wondering if it is possible to generalize this to any norm. Even generalizing to $L_1$ or $L_\infty$ norms would be good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Description of my algorithm for the $L_2$ norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm seeks to find the next ring of points with the next largest norm (radius from the origin) given an existing ring. We accomplish this by scanning along the &quot;perimeter&quot; of the existing ring, in only one of the dimensions since the other length in the other dimension is roughly known based on the equation of a circle.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume you are given the reduced lattice basis vectors $\vec{u}$ and $\vec{v}$, integer coefficients $(a,b)$ such that $a \vec{u} + b\vec{v}$ is one of the points on the existing ring of radius $r$.&lt;/li&gt;&#xA;&lt;li&gt;Determine the maximum extent we need to scan in the $\vec{u}$ direction:&#xA;Compute $m&gt;0$ such that $\left\lVert m\vec{u}+n\vec{v} \right\rVert = r$ has no solution for any $n$. This is a simple quadratic equation. Round $m$ away from zero by one to be safe. The amount of time it will take to complete the scan is proportional to $m$.&lt;/li&gt;&#xA;&lt;li&gt;Initialize $r_b = \infty$, the radius of the best next candidate ring. Initialize the set $S$ of next ring points to empty.&lt;/li&gt;&#xA;&lt;li&gt;For each $i$ in $-m \le i \le m$, determine $j$ so that $r_c = \left\lVert i\vec{u}+j\vec{v} \right\rVert &gt; r$. If $r_c \lt r_b$, then we found an even smaller ring: set $r_b = r_c$ and set $S = \{i,j\}$. Otherwise if $r_c = r_b$ we found another point on the ring and add $\{i,j\}$ to $S$. Otherwise, if $r_c \gt r_b$, do nothing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A little thought should convince you that this works. This only works for the $L_2$ norm since it explicitly makes use of the shape of the equi-norm contours. For a general norm, the shape is not known ahead of time.&lt;/p&gt;&#xA;" OwnerUserId="398" LastEditorUserId="398" LastEditorDisplayName="" LastEditDate="2010-11-11T06:35:59.310" LastActivityDate="2010-11-11T06:35:59.310" Title="The N-shortest lattice vectors problem in R^2" Tags="&lt;geometry&gt;&lt;lattices&gt;" CommentCount="4" />
  <row Id="9793" PostTypeId="2" ParentId="9776" CreationDate="2010-11-11T00:19:27.077" Score="4" ViewCount="0" Body="&lt;p&gt;I transcribe part of &lt;a href=&quot;http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents/9777#9777&quot;&gt;my answer&lt;/a&gt; to &lt;a href=&quot;http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+\sin y)$$ &#xD;&#xA;&#xD;&#xA;which agrees with the real exponential function when $y=0$. The [principal logarithm][3] of $z$ is the complex number &#xD;&#xA;&#xD;&#xA;$$w=\text{Log }z=\log |z|+i\arg z$$&#xD;&#xA;&#xD;&#xA;so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$. &#xD;&#xA;&#xD;&#xA;&#xD;&#xA;The complex power is &#xD;&#xA;&#xD;&#xA;$$z^w=e^{w\text{ Log} z}.$$ &#xD;&#xA;&#xD;&#xA;In your case you have: $z=a+bi,w=c+di$&#xD;&#xA;&#xD;&#xA;$$(a+bi)^{c+di}=z^{w}=e^{w\text{Log }z}=e^{(c+di)\text{Log }(a+bi)}$$&#xD;&#xA;&#xD;&#xA;$$=e^{(c+di)(\log |a+bi|+i\arctan \frac{d}{c}))}$$&#xD;&#xA;&#xD;&#xA;Since&#xD;&#xA;&#xD;&#xA;$$(c+di)\left(\log |a+bi|+i\arctan \frac{d}{c}\right)=(c+di)\log |a+bi|+(c+di)i\arctan \frac{d}{c}$$&#xD;&#xA;&#xD;&#xA;$$=c\log |a+bi|-d\arctan \frac{d}{c}+i\left( d\log |a+bi|+c\arctan \frac{d}{c}%&#xD;&#xA;\right) $$&#xD;&#xA;&#xD;&#xA;we obtain&#xD;&#xA;&#xD;&#xA;$$e^{(c+di)\left(\log |a+bi|+i\arctan \frac{d}{c})\right)}=e^{c\log |a+bi|-d\arctan &#xD;&#xA;\frac{d}{c}}e^{i\left( d\log |a+bi|+c\arctan \frac{d}{c}\right) }$$&#xD;&#xA;&#xD;&#xA;and&#xD;&#xA;&#xD;&#xA;$$e^{i\left( d\log |a+bi|+c\arctan \frac{d}{c}\right) }=$$&#xD;&#xA;&#xD;&#xA;$$=\cos \left( d\log |a+bi|+c\arctan \frac{d}{c}\right)+i\sin \left( d\log |a+bi|+c\arctan \frac{d}{c}\right).$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-11T11:02:25.487" LastActivityDate="2010-11-11T11:02:25.487" CommentCount="1" />
  <row Id="9794" PostTypeId="2" ParentId="9048" CreationDate="2010-11-11T00:30:41.167" Score="1" ViewCount="0" Body="&lt;p&gt;OK, got it figured out my self, before including the acceleration I first had to parameterize the curve itself, you can read all the stuff over at my gamedev question:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gamedev.stackexchange.com/questions/5373/moving-ships-between-two-planets-along-a-bezier-missing-some-equations-for-accel&quot;&gt;http://gamedev.stackexchange.com/questions/5373/moving-ships-between-two-planets-along-a-bezier-missing-some-equations-for-accel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3154" LastActivityDate="2010-11-11T00:30:41.167" />
  <row Id="9795" PostTypeId="1" CreationDate="2010-11-11T00:38:32.687" Score="0" ViewCount="89" Body="&lt;p&gt;$G$ and $G \times G$ where $G = Z\_2 \times Z\_2 \times Z\_2 \times Z\_2 \times  Z\_2 \times \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer says yes but I cannot figure out what homomorphism function I could use.&lt;/p&gt;&#xA;" OwnerUserId="3395" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T00:59:01.083" LastActivityDate="2010-11-11T01:11:43.607" Title="Are they isomorphic?" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9796" PostTypeId="2" ParentId="9795" CreationDate="2010-11-11T00:45:45.567" Score="3" ViewCount="0" Body="&lt;p&gt;Think of $$G = Z\_{2_1}\times Z\_{2_2} \times Z\_{2_3} \times Z\_{2_4} \times Z\_{2_5} \times \ldots$$ and $$G \times G= (Z\_{2_1}\times Z\_{2_3} \times Z\_{2_5} \times \ldots) \times (Z\_{2_2}\times Z\_{2_4} \times Z\_{2_6} \times \ldots)$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-11T01:11:43.607" LastActivityDate="2010-11-11T01:11:43.607" CommentCount="4" />
  <row Id="9797" PostTypeId="1" AcceptedAnswerId="9798" CreationDate="2010-11-11T01:14:13.043" Score="0" ViewCount="67" Body="&lt;p&gt;Ok this is a really silly question and I should know this, but I cant seem to figure something out:&lt;img src=&quot;http://i.imgur.com/F0U3t.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the last step, how do they know that 0&amp;lt;=x&amp;lt;=4? If we use the minimum value of theta, which is $-pi/2$, and plug that into $x=cos(theta)$, then we get 0, and same for pi/2.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-11T03:46:28.687" Title="Parametric Equation Question" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="9798" PostTypeId="2" ParentId="9797" CreationDate="2010-11-11T01:18:28.440" Score="3" ViewCount="0" Body="&lt;p&gt;Because over $-\pi /2\leq \theta \leq \pi /2, 0\leq\cos{\theta}\leq1$.  At $\theta =0, \cos{\theta}=1$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T01:18:28.440" />
  <row Id="9799" PostTypeId="1" AcceptedAnswerId="9801" CreationDate="2010-11-11T02:16:42.727" Score="6" ViewCount="190" Body="&lt;p&gt;I know that Banach Tarski Paradox gives that &quot;A three-dimensional Euclidean ball is equidecomposable with two copies of itself&quot; and I have read that doubling the ball can be accomplished with five pieces. My question is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Is it possible to construct (mathematically) these five sets or is the proof more of an existence result and not a construction?&quot; If it is possible to construct, I would really appreciate if someone can show the construction here.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-11T03:23:57.847" Title="Banach Tarski Paradox" Tags="&lt;set-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9800" PostTypeId="2" ParentId="9760" CreationDate="2010-11-11T02:21:17.550" Score="4" ViewCount="0" Body="&lt;p&gt;This question in its full generality is probably best tackled by some generating function approach as Qiaochu points out. I toyed around with the $N=1$ case and got some interesting results that I am sharing below. For the $N=1$ case, we don't have to distinguish between the $+1$ and the $-1$s. In fact, a nice visualization is to think of a row of bulbs one of which is randomly toggled at each stage. Question: Given that they are all off initially, how long does it take on an average for all of them to light up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_k$ be the expected number of steps for all bulbs to light up starting from $k$ lighted bulbs out of a total of $n$ bulbs. Then, we have the recursions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0 = 1 + f_1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_k = 1 + \frac{k}{n} f_{k-1} + \frac{n-k}{n} f_{k+1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_n = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to solve for $f_0$. I cheated on solving the recursion by numerically solving it for various $n$ and appealing to OEIS. The answer turns out to be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0 = n \sum_{j=0}^{n-1} \frac{2^j}{j+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $3$ bulbs, the answer is $10$. I don't know how to extend this approach to the case of arbitrary $N$ without making it messy. I am not very comfortable with generating function based solution approaches. If someone were to post one for this problem, I would love to learn the techniques of &quot;generating functionology&quot; through that.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-11-11T02:26:34.103" LastActivityDate="2010-11-11T02:26:34.103" CommentCount="2" />
  <row Id="9801" PostTypeId="2" ParentId="9799" CreationDate="2010-11-11T02:23:05.170" Score="9" ViewCount="0" Body="&lt;p&gt;No, it is not possible to explicitly construct the pieces. The pieces are, by necessity, non-measurable. This means that they cannot be Borel sets, which covers essentially every &quot;explicit construction&quot; technique you might think of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there are models of ZF set theory (without the axiom of choice) in which the Banach-Tarski paradox fails. So the construction must, necessarily, make use of some form of the axiom of choice.  This means that an even wider range of construction techniques - those that can be carried out in ZF - are insufficient to form the decomposition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After clarification, it seems that one part of the question is to find an explicit proof that (using the axiom of choice) it is possible to get a decomposition using exactly 5 pieces. A proof of this is given by Francis Su's thesis on the paradox &lt;a href=&quot;http://www.math.hmc.edu/%7Esu/papers.dir/banachtarski.pdf&quot; rel=&quot;nofollow&quot;&gt;(PDF)&lt;/a&gt;. Theorem 20 gives the proof of the five-piece decomposition, and by tracing back the previous results you can work out exactly what the pieces are. That thesis is, by the way, a wonderful reference for many other aspects of the paradox as well. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-11T03:23:57.847" LastActivityDate="2010-11-11T03:23:57.847" CommentCount="10" />
  <row Id="9802" PostTypeId="1" CreationDate="2010-11-11T02:35:00.560" Score="1" ViewCount="79" Body="&lt;p&gt;If | | denotes the absolute value then is it always true $|x^{s}|=|x|^{s}$ whenever s is any &lt;em&gt;positive&lt;/em&gt; real number? I know this holds for any positive integer. What about positive reals?&lt;/p&gt;&#xA;" OwnerUserId="3398" LastActivityDate="2010-11-11T03:12:00.123" Title="Elementary question power absolute values" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9803" PostTypeId="1" AcceptedAnswerId="9857" CreationDate="2010-11-11T02:35:11.860" Score="0" ViewCount="160" Body="&lt;p&gt;Let V be a vector space with a non-trivial subspaces U, W. $ U \neq W$. Consider $u_0 \in V$, with $u_0 \neq 0$ and $w_0 \in V$ with $w_0 \neq 0$ then we have $u_0 + U \in V/U$ and $w_0 + W \in V/W$. Is there any sensible way to define:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(u_0 + U) + (w_0 + W)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking &quot;no&quot; $(2,4) + (2,6,7)$ makes no sense. So, why should this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, we would need to find some new vector space that had all of the elements in $V/U$ and in $V/W$ and their sums and scalar products. Maybe $V/U + V/W$ -- but then we need both $V/U$ and $V/W$ to be subspaces of some larger vector space. Is there a way to make sense of this at all?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-11T15:24:40.260" LastActivityDate="2010-11-11T15:24:40.260" Title="Bad ideas: Adding vectors from different vector spaces?" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9804" PostTypeId="2" ParentId="9703" CreationDate="2010-11-11T02:36:19.667" Score="3" ViewCount="0" Body="&lt;p&gt;I think that it is worth pointing out that, strictly speaking, explaining “2&lt;sup&gt;0&lt;/sup&gt; equals 1” is not really what people do in this situation.  The best we can do is to convince a child of the following facts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If 2&lt;sup&gt;0&lt;/sup&gt; is any number, it makes more sense to consider that 2&lt;sup&gt;0&lt;/sup&gt;=1 than considering 2&lt;sup&gt;0&lt;/sup&gt; as any other numbers (such as 0).&lt;/li&gt;&#xA;&lt;li&gt;It is more interesting to consider 2&lt;sup&gt;0&lt;/sup&gt; to be 1 than giving up.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some of the other answers provide good ways to convince a child of these facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the reason that 2&lt;sup&gt;0&lt;/sup&gt; equals 1 rather than 2&lt;sup&gt;0&lt;/sup&gt; is undefined is really conventions and experiences: it is much more convenient to define 2&lt;sup&gt;0&lt;/sup&gt;=1 than leaving 2&lt;sup&gt;0&lt;/sup&gt; undefined.  I do not think that it is possible to convince a child of this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to the following.  Let &lt;i&gt;f&lt;/i&gt;(&lt;i&gt;x&lt;/i&gt;) = sin &lt;i&gt;x&lt;/i&gt; / &lt;i&gt;x&lt;/i&gt;.  What is &lt;i&gt;f&lt;/i&gt;(0)?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If &lt;i&gt;f&lt;/i&gt;(0) were any number, it would make more sense to consider that &lt;i&gt;f&lt;/i&gt;(0)=1 than considering &lt;i&gt;f&lt;/i&gt;(0) as any other values.&lt;/li&gt;&#xA;&lt;li&gt;It would be more interesting to consider &lt;i&gt;f&lt;/i&gt;(0) to be 1 than giving up.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, these “evidences” do not make &lt;i&gt;f&lt;/i&gt;(0)=1 under the usual way of mathematical writing.  I think that the only things that differentiate these two cases are conventions and experiences.&lt;/p&gt;&#xA;" OwnerUserId="926" LastActivityDate="2010-11-11T02:36:19.667" />
  <row Id="9805" PostTypeId="2" ParentId="9483" CreationDate="2010-11-11T02:40:05.730" Score="3" ViewCount="0" Body="&lt;p&gt;Jonas: What you need to show is that if $Y$ is well-ordered, then any subset $Z$ of $Y$ is isomorphic to an initial segment of $Y$. How would you define such an isomorphism $f$? You do not have many possibilities: The least element of $Z$ must be mapped by $f$ to the smallest element of $Y$. The next element of $Z$ must be mapped by $f$ to the next element of $Y$, and so on. You just need to check that this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more detail, say that $g$ is an &lt;em&gt;approximation&lt;/em&gt; iff its domain is an initial segment of $Z$, its range is an initial segment of $Y$, and $g$ is an order isomorphism from its domain onto its range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One checks (by considering least disagreements) that any two approximations coincide on the common part of their domains, so we can let $f$ be the common function resulting from pasting all the approximations together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We just need to see that $f$ exhausts $Z$. For this, one can check that $f(x)\le x$ for all $x$, so one &quot;does not run out of room.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a slightly more elegant (and formal) way of presenting this argument, by appealing to the recursion theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x$ in $Z$, define $f(x)=\sup_Y({\rm succ}_Y f(y)\mid y\in Z, y\lt x)$, where ${\rm succ}_Y a$ means the successor in $Y$ of $a$. Then $f$ is precisely the map we wanted. (One still needs a little argument to see that it works).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trickier question for another day is whether the recursion theorem is &lt;em&gt;needed&lt;/em&gt; for the argument. This may be formalized by asking whether Zermelo set theory proves the result (so we do not have access to replacement), but there may be finer formulations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The above was written quickly. Let me know of infelicities or plain falsehoods and I'll try to edit.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-11T02:40:05.730" CommentCount="2" />
  <row Id="9806" PostTypeId="2" ParentId="9746" CreationDate="2010-11-11T02:47:55.600" Score="2" ViewCount="0" Body="&lt;p&gt;When we study mathematical logic, we use the same mathematical methods that would be used to study any other area (algebra, analysis, topology, etc.).  This includes the use of set-theoretical techniques.  Otherwise, we would be needlessly hamstringing ourselves.  In principle, the results that we obtain could be built up entirely from primitive axioms, just like the rest of mathematics; but just like the rest of mathematics complete formalization is not usually the goal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even ignoring those practical considerations, it's not hard to see that none of the axioms of point set topology or set theory used in topology presuppose that the compactness theorem of logic holds. So there is no circularity if we use topological methods to prove the compactness theorem. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-11T02:47:55.600" />
  <row Id="9807" PostTypeId="1" CreationDate="2010-11-11T02:48:13.867" Score="2" ViewCount="107" Body="&lt;p&gt;The question is: Given a licence plate that can have either 2 or 3 letters followed by either 2 or 3 numbers, how many different license plates can be printed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My math is as such $(26^2 + 26^3)*(9^2+9^3)$ which comes to 14,784,120 but the book says the answer is 20,077,200. What am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="3071" LastActivityDate="2010-11-22T08:11:11.890" Title="Issue with permutation problem" Tags="&lt;homework&gt;&lt;discrete-mathematics&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9808" PostTypeId="2" ParentId="9791" CreationDate="2010-11-11T02:49:36.860" Score="-1" ViewCount="0" Body="&lt;p&gt;Check out Chapter 1 of Paul Pollack's book &quot;&lt;a href=&quot;http://books.google.com/books?id=MI7zfW7CkzIC&amp;amp;lpg=PP1&amp;amp;ots=4uNSVP2yUZ&amp;amp;dq=pollack%20%22not%20always%20buried%20deep%22&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Not Always Buried Deep&lt;/a&gt;&quot;.  He lists several elementary proofs for the infinitude of primes, some of which can be adapted to form simple iterative algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, I can't get Latex to work here.   Isn't it like MathOverflow?  Anyway..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let S = {p1...pk} be the set of previously generated coprime integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And take M = product of integers in S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now do any of the following to get new integer coprime to the set S.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;(Euclid)&lt;/em&gt;: M + 1 &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Stieltjes)&lt;/em&gt;: Factor M as A.B in some way, and take A + B&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Euler)&lt;/em&gt; Totient(M)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Braun and Metrod)&lt;/em&gt;: N = M/p1 + ... + M/p_k&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(Goldbach)&lt;/em&gt;: N = 2 + M &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is also Saidak's proof from the same book:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take N1=n; N2 = N1(N1+1); N3 =&#xA;N2(N2+1)... Nk=Nk(Nk+1) and so on. &#xA;This a relative prime sequence.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am probably missing a couple of other proofs.  Anyway, check the book!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1 : **I am going to retain the answer despite the negative votes, in the hope that it might spur someone else to invent a better algorithm ! *&lt;/p&gt;&#xA;" OwnerUserId="1558" LastEditorUserId="1558" LastEditorDisplayName="" LastEditDate="2010-11-11T14:13:13.260" LastActivityDate="2010-11-11T14:13:13.260" CommentCount="7" />
  <row Id="9809" PostTypeId="1" AcceptedAnswerId="9813" CreationDate="2010-11-11T02:50:53.067" Score="3" ViewCount="83" Body="&lt;p&gt;I've been reading Vakil's notes on algebraic geometry (on my own -- this is not part of a class), and I'm stuck on one problem (number 6.1.H).  It goes as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a scheme.  Prove that $X$ is quasicompact and quasiseparated if and only if $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to show one direction, namely that if $X$ is quasicompact and quasiseparated then it has a cover of the indicated form.  It's also not hard to prove that if $X$ has a cover of the indicated form, then $X$ is quasicompact.  I'm having difficulty with the &quot;quasiseparated&quot; part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for any help.&lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2010-11-12T18:04:59.373" Title="Quasiseparated if finitely covered by affines in appropriate way" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9810" PostTypeId="2" ParentId="9802" CreationDate="2010-11-11T02:54:15.763" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, even when $x \in \mathbb{C}$.  To see this, write $x = |x| e^{i \theta}$, in which case $$|x^s| = ||x|^s e^{i s \theta}| = |x|^s |e^{i s \theta}| = |x|^s.$$&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-11T02:54:15.763" CommentCount="1" />
  <row Id="9811" PostTypeId="2" ParentId="9802" CreationDate="2010-11-11T03:02:56.210" Score="0" ViewCount="0" Body="&lt;p&gt;This is a fairly standard argument proceeding along the lines of first extending the relation from $\mathbb{Z}$ to $\mathbb{Q}$ and finally to $\mathbb{R}$ (of course only for the values where the relation makes sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first extension use the fact that every rational $r \in \mathbb{Q}$ can be written as $r = p/q$ with $p, q \in \mathbb{Z}$. For the second, use the continuity of the norm and density of $\mathbb{Q}$ in $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-11T03:02:56.210" />
  <row Id="9812" PostTypeId="2" ParentId="9802" CreationDate="2010-11-11T03:12:00.123" Score="2" ViewCount="0" Body="&lt;p&gt;Assuming you're talking about real $x$, the only time you would be concerned with whether the equality holds is if $x$ is negative.  (Otherwise, $|x|=x$ and $|x^s|=x^s$, so there's nothing to show.)  Now, how do you define $x^s$ if $x$ is negative and $s$ is irrational? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One answer is given in more generality by user3296.  You can use complex numbers, in which case it works out well.  However, if you only want to work with real values, then trying to take irrational powers of negative numbers is best avoided.  To see this, think first about how $x^s$ can be defined when $x$ is positive.  One way is by taking rational numbers $p/q$ approaching $s$, and taking a limit of $x^{p/q}=\sqrt[q]{x}^p$.  But this definition won't work if $x$ is negative, because for some choices of $p/q$ the result will be imaginary, for others negative, and for others positive.  Another way is using the definition $x^s=e^{s\ln{x}}$, where $\ln$ is the natural logarithm defined on the positive reals.  To extend this definition to negative $x$, you need a way of defining a logarithm of negative numbers.  This actually works (although there is not a unique choice), but the result is that you end up with complex numbers rather than real numbers.  For example, you can define $\log$ on $(-\infty,0)$ by $\log(x)=\ln(-x)+\pi i$, so that $x^s=e^{s(\ln(-x)+\pi i)}=e^{s\pi i}(-x)^s$.  But this is not real when $s$ is not an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you have no problem using complex exponentials (and logarithms, at least implicitly), then the answer is yes, as user3296 says.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-11T03:12:00.123" />
  <row Id="9813" PostTypeId="2" ParentId="9809" CreationDate="2010-11-11T03:14:13.943" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: Starting from your open affine cover of $X$, construct an open affine cover&#xA;of $X \times X$, and then show that the preimage of each member of that open affine cover is a union of finitely many affines.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hint, more directly relevant to Vakil's definition of quasi-separated:&#xA;I am implicitly cheating somewhat, because I am still secretely thinking about quasi-separated as meaning that the diagonal map is quasi-compact.  If I do that,&#xA;then the exercise at hand translates into the following: show that if a morphism&#xA;$f$ has the property that the target has an open affine cover, the preimage of each member of which is quasi-compact, then $f$ is quasi-compact (i.e. each quasi-compact in the target has quasi-compact preimage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should certainly remind yourself how to prove this fact, because the tools used there will be the same tools you need in your present exercise.  (Even though you may not be working explicitly with the formulation in terms of the diagonal morphism, the underlying argument will have to be essentially the same.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. first convince yourself that you only have to check that the intersection of any two open affines is quasi-compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, an important point in all these kinds of arguments: if $f: X \to Y$&#xD;&#xA;with $X$ and $Y$ affine, say $X =$ Spec $B$ and $Y =$ Spec $A$, and $a \in A$, so that Spec $A_a$ is a distiguished open, then the inverse image of Spec $A_a$&#xD;&#xA;is a distinguished open affine in $X$ (it is Spec $B_a$).  This a fundamental tool: it is the one way that you have to construct arbitrarily small open affine neighbourhoods in $Y$ whose preimages in $X$ are again affine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will want to use the idea of the preceding paragraph, but now &quot;inverse image&quot; will be replaced by &quot;intersection&quot;.  (When you take inverse images of product neighbourhoods along the diagonal map, you are computing intersections,&#xA;so this replacement makes sense; but again, you won't have to think about the diagonal map explicitly if you dont' want to.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's probably enough of a rambling hint for now.  If you have those ideas in mind and put everything you know together, you hopefully can find your way to a complete proof.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-12T18:04:59.373" LastActivityDate="2010-11-12T18:04:59.373" CommentCount="3" />
  <row Id="9814" PostTypeId="2" ParentId="9760" CreationDate="2010-11-11T03:26:05.783" Score="9" ViewCount="0" Body="&lt;p&gt;Still working on a general solution, but here are a few exact solutions for small $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N = 1..8 $ the exact values are 10, $\frac{416}{5}$, $\frac{10294}{35}$, $\frac{18481664}{25823}$, $\frac{235115486250}{165493097}$, $\frac{1690479776164}{681135455}$, $\frac{25081308000562315334439801586}{6314587098176165651428105}$, $\frac{282872889746351048352862015165111214080}{47431617359482015647230177875265569}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical values: &lt;code&gt;10, 83.2, 294.114, 715.706, 1420.7, 2481.86, 3971.96, 5963.8.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were computed &quot;the hard way&quot; by making a large state diagram, exploiting a lot of symmetry, and then solving a system of linear equations. There are 165 states, for example, in the case of $N=8$, which means solving for 165 equations in 165 unknowns. There is undoubtedly a more clever way to compute this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I offer the bound $c \cdot N^3$ without proof; I believe the expected number of steps is asymptotic to this value with $c \approx 12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 13-Nov-2010&lt;/strong&gt;:&#xA;The following table provides strong evidence for the $12N^3$ asymptotic bound given earlier.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; N        E(N)    12(N^3)&#xA;&#xA; 1        10.0        12&#xA; 2        83.2        96&#xA; 3       294.1       324&#xA; 4       715.7       768&#xA; 5      1420.7      1500&#xA; 6      2481.9      2592&#xA; 7      3972.0      4116&#xA; 8      5963.8      6144&#xA; 9      8530.2      8748&#xA;10     11743.8     12000&#xA;11     15677.6     15972 &#xA;12     20404.2     20736&#xA;13     25996.5     26364&#xA;14     32527.3     32928&#xA;15     40069.3     40500&#xA;16     48695.3     49152&#xA;17     58478.1     58956&#xA;18     69490.5     69984&#xA;19     81805.3     82308&#xA;20     95495.3     96000&#xA;21    110633.2    111132&#xA;22    127291.9    127776&#xA;23    145544.0    146004&#xA;24    165462.6    165888&#xA;25    187120.2    187500&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Edit 11-Nov-2010:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the system of linear equations for $N=3$. Ostensibly there are $6^3 = 216$ states but by symmetry this can be reduced to $20$. The three subscripts on each variable represent the distance from the goal state in each dimension. In accordance with the problem statement we start at the farthest node from the goal. Thus we are solving for $x_{3,3,3}$ here. (The answer is $\frac{10294}{35}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I show this tedium in hopes someone can figure out how to generalize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{0,0,0}=0 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,0,1}=\frac{1}{6} x_{0,0,0}+\frac{1}{6} x_{0,0,2}+\frac{2}{3} x_{0,1,1}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,0,2}=\frac{1}{6} x_{0,0,1}+\frac{1}{6} x_{0,0,3}+\frac{2}{3} x_{0,1,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,0,3}=\frac{1}{3} x_{0,0,2}+\frac{2}{3} x_{0,1,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,1,1}=\frac{1}{3} x_{0,0,1}+\frac{1}{3} x_{0,1,2}+\frac{1}{3} x_{1,1,1}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,1,2}=\frac{1}{6} x_{0,0,2}+\frac{1}{6} x_{0,1,1}+\frac{1}{6} x_{0,1,3}+\frac{1}{6} x_{0,2,2}+\frac{1}{3} x_{1,1,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,1,3}=\frac{1}{6} x_{0,0,3}+\frac{1}{3} x_{0,1,2}+\frac{1}{6} x_{0,2,3}+\frac{1}{3} x_{1,1,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,2,2}=\frac{1}{3} x_{0,1,2}+\frac{1}{3} x_{0,2,3}+\frac{1}{3} x_{1,2,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,2,3}=\frac{1}{6} x_{0,1,3}+\frac{1}{3} x_{0,2,2}+\frac{1}{6} x_{0,3,3}+\frac{1}{3} x_{1,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{0,3,3}=\frac{2}{3} x_{0,2,3}+\frac{1}{3} x_{1,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,1,1}=\frac{1}{2} x_{0,1,1}+\frac{1}{2} x_{1,1,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,1,2}=\frac{1}{3} x_{0,1,2}+\frac{1}{6} x_{1,1,1}+\frac{1}{6} x_{1,1,3}+\frac{1}{3} x_{1,2,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,1,3}=\frac{1}{3} x_{0,1,3}+\frac{1}{3} x_{1,1,2}+\frac{1}{3} x_{1,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,2,2}=\frac{1}{6} x_{0,2,2}+\frac{1}{3} x_{1,1,2}+\frac{1}{3} x_{1,2,3}+\frac{1}{6} x_{2,2,2}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,2,3}=\frac{1}{6} x_{0,2,3}+\frac{1}{6} x_{1,1,3}+\frac{1}{3} x_{1,2,2}+\frac{1}{6} x_{1,3,3}+\frac{1}{6} x_{2,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{1,3,3}=\frac{1}{6} x_{0,3,3}+\frac{2}{3} x_{1,2,3}+\frac{1}{6} x_{2,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{2,2,2}=\frac{1}{2} x_{1,2,2}+\frac{1}{2} x_{2,2,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{2,2,3}=\frac{1}{3} x_{1,2,3}+\frac{1}{3} x_{2,2,2}+\frac{1}{3} x_{2,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{2,3,3}=\frac{1}{6} x_{1,3,3}+\frac{2}{3} x_{2,2,3}+\frac{1}{6} x_{3,3,3}+1 $ &lt;br /&gt;&lt;br /&gt;&#xA;$x_{3,3,3}=x_{2,3,3}+1 $&#xD;&lt;/p&gt;&#xA;" OwnerUserId="130" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-11-13T20:45:14.273" LastActivityDate="2010-11-13T20:45:14.273" />
  <row Id="9815" PostTypeId="1" AcceptedAnswerId="9826" CreationDate="2010-11-11T03:33:28.383" Score="2" ViewCount="162" Body="&lt;p&gt;This is a homework question so specifically I am looking for a direction (help). Not an absolute answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how to solve 2nd order nonhomogeneous (and I think 3rd order is similiar) recurrence relations. Whats kinking me up with this one is a term with n.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;h[n] = 4h[n- 1] - 4h[n-2] + 3n + 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here I am using square brackets to indicate subscripts.&#xA;Now I can reorganize this equation into:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 = h[n] - 4h[n-1] + 4h[n-2] -3n-1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From which I can derive the characteristic equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 = x^3 - 4x^2 + 4x - 3n -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Normally at this point I would solve the roots and be on my way, but here I have that extra n. How do I deal with this?&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-11T04:51:01.967" LastActivityDate="2010-11-11T14:56:08.317" Title="How to solve 3rd order nonhomogeneous recurrence relation with n in equation?" Tags="&lt;combinatorics&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9816" PostTypeId="2" ParentId="9797" CreationDate="2010-11-11T03:46:28.687" Score="0" ViewCount="0" Body="&lt;p&gt;Well, y^2 has to be positive, so the minimum that y^2/25 could be is 0.  So, x^2/16 is 1 at the maximum =&gt; x≤4.  The same argument applies for x = 0&lt;/p&gt;&#xA;" OwnerUserId="3401" LastActivityDate="2010-11-11T03:46:28.687" CommentCount="1" />
  <row Id="9817" PostTypeId="2" ParentId="9815" CreationDate="2010-11-11T04:11:52.497" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Find a particular polynomial solution $\rm\:h[n]\:$ by undetermined coefficients (notice that the degree can be at most 1).$\ $ The general solution is then the sum of any particular solution and the general solution of the associated homogeneous equation $\rm\ h[n] -4\ h[n-1] + 4\ h[n-2]\ =\ 0\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-11T14:56:08.317" LastActivityDate="2010-11-11T14:56:08.317" CommentCount="2" />
  <row Id="9819" PostTypeId="1" CreationDate="2010-11-11T05:09:15.967" Score="1" ViewCount="73" Body="&lt;p&gt;A triangle on a sphere is comprised of points A, B and C. How to determine its area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the formula:&lt;br&gt;&#xA;A = E * R^2, where R is radius of sphere, and E is the excess angle of (a + b + c - pi), but how to determine the angles between ABC, ACB and BAC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! :-)&lt;/p&gt;&#xA;" OwnerUserId="3402" LastActivityDate="2010-11-11T05:58:22.240" Title="Area of a spherical triangle" Tags="&lt;spherical-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9820" PostTypeId="1" CreationDate="2010-11-11T05:32:31.367" Score="2" ViewCount="102" Body="&lt;p&gt;According to the book the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be a measurable function and suppose $|f| \leq h$ where $h$ is a real-valued function&#xA;and $h$ is integrable. Then f is integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: suppose instead we have the inequality $ |f| \leq h$ but almost everywhere and of course $h$ assumed to be integrable. Is it still true that f is integrable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this follow because the sets of measure zero &quot;don't count&quot; when integrating?&lt;/p&gt;&#xA;" OwnerUserId="3398" LastActivityDate="2010-11-11T19:53:19.313" Title="Doubt inequality almost everywhere" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9821" PostTypeId="2" ParentId="9820" CreationDate="2010-11-11T05:43:23.833" Score="3" ViewCount="0" Body="&lt;p&gt;Yes on both counts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be set of measure zero such that $|f|\leq h$ on $X-E$. Then let $g=f\chi_{X-E}$ (that is, redefine $f$ to be zero in $E$). Then $|g|\leq h$ everywhere, so $g$ is integrable. Then you have&#xA;\begin{align*}&#xA;\int_X g\,d\mu &amp;amp;= \int_{(X-E)\cup E}g\,d\mu = \int_(X-E)g\,d\mu + \int_E g\,d\mu\\&#xA;&amp;amp;= \int_{X-E}f\,d\mu + 0 = \int_{X-E}f\,d\mu = \int_{X-E}f\,d\mu + \int_E f\,d\mu\\&#xA;&amp;amp;= \int_{X}f\,d\,\mu,&#xA;\end{align*}&#xA;because $\int_E fd\,\mu$ exists and is equal to $0$, since $\mu(E)=0$ (so any sequence of simple functions that converges to $f$ will always have integral equal to $0$ over $E$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:53:19.313" LastActivityDate="2010-11-11T19:53:19.313" CommentCount="10" />
  <row Id="9822" PostTypeId="2" ParentId="9819" CreationDate="2010-11-11T05:44:47.017" Score="2" ViewCount="0" Body="&lt;p&gt;The angle between two great circles is equal to the angle between the planes containing them. Such a plane always passes through the centre of the sphere, say $O$. So, for example, angle $a$ is the angle between two planes, one of which passes through $A$, $B$, and $O$, and the other passes through $A$, $C$, and $O$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle between two planes, in turn, equals the angle between their normals, and you can determine the normal from the three points by using the cross product.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-11T05:49:53.063" LastActivityDate="2010-11-11T05:49:53.063" />
  <row Id="9823" PostTypeId="2" ParentId="9819" CreationDate="2010-11-11T05:58:22.240" Score="3" ViewCount="0" Body="&lt;p&gt;The area of a spherical triangle can also be calculated using the lengths of its sides, as in&#xA;this &lt;a href=&quot;http://mathforum.org/library/drmath/view/65316.html&quot; rel=&quot;nofollow&quot;&gt;Dr.Math link.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-11T05:58:22.240" />
  <row Id="9824" PostTypeId="1" AcceptedAnswerId="9832" CreationDate="2010-11-11T06:08:30.027" Score="2" ViewCount="94" Body="&lt;p&gt;You have three bags, A, B, and C. Bags A and B hold x red and y blue marbles each; bag C holds 2x red marbles and 2y blue marbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say that we pull out z marbles from bag A, z marbles from bag B, and 2z marbles from bag C. Will the expected number of red marbles pulled from bag A and bag B together be the same as the expected number of marbles pulled from bag C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is yes, but I am having trouble figuring out why.&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-15T22:26:41.697" LastActivityDate="2010-11-15T22:26:41.697" Title="Probability question" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" />
  <row Id="9825" PostTypeId="1" AcceptedAnswerId="10083" CreationDate="2010-11-11T06:15:04.560" Score="26" ViewCount="427" Body="&lt;p&gt;As a result of &lt;a href=&quot;http://math.stackexchange.com/questions/9532/link-between-a-dense-subset-and-a-continuous-mapping&quot;&gt;this question&lt;/a&gt;, I've been thinking about the following condition on a topological space $Y$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For every topological space $X$, $E\subseteq X$, and continuous maps $f,g\colon X\to Y$, if $E$ is dense in $X$, and $f$ and $g$ agree on $E$ (that is, $f(e)=g(e)$ for all $e\in E$), then $f=g$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $Y$ is Hausdorff, then $Y$ satisfies this condition. The question is whether the converse holds: if $Y$ satisfies the above condition, will it necessarily be Hausdorff? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Y$ is not at least $T_1$, then $Y$ does not have the property: if $u,v\in Y$ are such that $u\neq v$ and every open neighborhood of $u$ contains $v$, then let $X$ be the Sierpinski space, $X=\{a,b\}$, $a\neq b$, with topology $\tau=\{\emptyset,\{b\},X\}$, $E=\{b\}$, let $f,g\colon X\to Y$ be given by $f(a)=f(b)=v$, and $g(a)=u$, $g(b)=v$. Then both $f$ and $g$ are continuous, agree on the dense subset $E$, but are distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume $Y$ is $T_1$ but not $T_2$; let $u$ and $v$ be witnesses to the fact that $Y$ is not $T_2$, let $\mathcal{U}\_s$ and $\mathcal{V}\_t$ be the collection of all open nbds of $s$ that do not contain $t$, and all open nbds of $t$ that do not contain $s$, respectively. Construct a net with index set $\mathcal{U}\_s\times\mathcal{V}\_t$ (ordered by $(U,V)\leq (U&amp;#39;,V&amp;#39;)$ if and only if $U&amp;#39;\subseteq U$ and $V'\subseteq V$) by letting $y_{(U,V)}$ be a point in $U\cap V$ (this is where AC comes in). Let $E=\{y_{(U,V)}\mid (U,V)\in\mathcal{U}\_s\times\mathcal{V}\_t\}$, and let $X=E\cup\{s\}$. Give $X$ the induced topology; let $f\colon X\to Y$ be the inclusion map, and let $g\colon X\to Y$ be the map that maps $E$ to itself identically, but maps $s$ to $t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is I cannot quite prove that $g$ is continuous; the difficulty arises if I take an open set $\mathcal{O}\in \mathcal{V}_t$; the inverse image under $g$ is equal to $((\mathcal{O}\cap X)-\{t\})\cup\{s\}$, and I have not been able to show that this is open in $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the condition above characterize Hausdorff spaces? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If not, I would appreciate a counterexample. If it &lt;em&gt;does&lt;/em&gt; characterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; A little digging turned up  this question raised in the Problem Section of the &lt;strong&gt;American Mathematical Monthly&lt;/strong&gt; back in 1964 by Alan Weinstein. The &lt;a href=&quot;http://www.jstor.org/stable/2315068&quot; rel=&quot;nofollow&quot;&gt;solution by Sim Lasher&lt;/a&gt; gives a one paragraph proof that does not require one to consider $T_1$ and non-$T_1$ spaces separately. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T21:08:25.210" LastActivityDate="2010-11-14T21:08:25.210" Title="Does this property characterize a space as Hausdorff?" Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="3" />
  <row Id="9826" PostTypeId="2" ParentId="9815" CreationDate="2010-11-11T06:27:30.523" Score="2" ViewCount="0" Body="&lt;p&gt;Your 'characteristic equation' doesn't work, as you noted, because of the presence of the variable $n$; it leaves the equation making no sense.  While the particular-solution method works well, there's another general technique that can be used (in principle) to solve recurrence relations that are linear in previous terms and polynomial in $n$: generating functions.  Consider the generating function $H(x) = \sum_{n=0}^{\infty} h[n]x^n$ and use your equation for $h[n]$ to come up with a corresponding equation for $H(x)$. Note that the generating function for $h[n-1]$ is simply $xH(x)$; the 'generating function' for a constant $c$ is $c\over 1-x$, and the generating function for $n$ is $x\over(1-x)^2$ (as can be found by taking the derivative of the generating function for 1 and multiplying by $x$).  Once you come up with a formula you can use it to solve for $H(x)$ and then use the usual partial-fraction methods to turn this back into an explicit formula for $h[n]$.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-11T06:27:30.523" />
  <row Id="9827" PostTypeId="2" ParentId="9824" CreationDate="2010-11-11T06:33:57.250" Score="2" ViewCount="0" Body="&lt;p&gt;Yes.  The expected number of red marbles from A is $\frac{xz}{x+y}$ and the same for B.  The expected number from C is $\frac{4xz}{2x+2y}$  The 4 in the numerator is 2*2, one from the number of pulls, one from the number of red marbles.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T06:33:57.250" />
  <row Id="9828" PostTypeId="2" ParentId="457" CreationDate="2010-11-11T06:37:42.627" Score="1" ViewCount="0" Body="&lt;p&gt;I recall what I now remember as &quot;the Metereological theorem&quot; from college:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At any point in time there exist two&#xA;  antipodal points on the equator that&#xA;  have the same temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trivial to prove. Only requires the mean value theorem and continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could demonstrate it by stating that you can not draw a simple closed curve around a circle without having two points 180 degrees apart being exactly the same distance from the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true of any continuous physical property like pressure, humidity, etc. Also, it can be extended to any three metereological properties on a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's rather amazing what you can do with just continuity and the mean value theorem. &lt;/p&gt;&#xA;" OwnerUserId="3403" LastEditorUserId="3403" LastEditorDisplayName="" LastEditDate="2010-11-11T06:48:32.227" LastActivityDate="2010-11-11T06:48:32.227" CommunityOwnedDate="2010-11-11T06:37:42.627" />
  <row Id="9829" PostTypeId="2" ParentId="9803" CreationDate="2010-11-11T06:43:45.077" Score="0" ViewCount="0" Body="&lt;p&gt;Certainly you can add them as elements of $V$.  They should all have the same number of components as the dimension of $V$.  But if $u_0 \in U$ why isn't $u_0+U$ just $U$?  $U$ is a vector space, so is closed under addition.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T06:43:45.077" CommentCount="2" />
  <row Id="9830" PostTypeId="1" AcceptedAnswerId="9855" CreationDate="2010-11-11T06:46:09.450" Score="3" ViewCount="100" Body="&lt;p&gt;$K$ numbers are chosen with replacement form the numbers $1,2,3, \cdots N$.In how many ways we can choose the numbers so that the maximum number chosen is exactly $R \le N $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation given in my module is like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Since the maximum number is $R$, each of $K$ numbers can be chosen in $R$ ways.so, the number be can be chosen in $R^K$ ways.These includes the $(r-1)^k$ choices in which none of the $K$chosen numbers is $R$. Hence, the required number is $R^K - (r-1)^k $.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not getting this explanation,and after reading &lt;a href=&quot;http://math.stackexchange.com/users/742/arturo-magidin&quot;&gt;Arturo Magidin&lt;/a&gt;'s answer I am somewhat more confused now.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T15:11:28.467" LastActivityDate="2010-11-11T15:11:28.467" Title="Number of ways of choosing K numbers from $\mathbb{N}$ which doesn't exceed a maximum" Tags="&lt;combinatorics&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="9831" PostTypeId="2" ParentId="9830" CreationDate="2010-11-11T06:50:16.607" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; My original answer was incorrect even with the interpretation I had in mind. Here is the correction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I originally gave a formula for the number of ways to pick $K$ numbers so that the maximum is &lt;em&gt;at most&lt;/em&gt; $R$, rather than exactly $R$. Moreover, I interpreted the question as asking for the number of different selections, so that if you first picked $1$ and then $2$, this was the same &quot;pick&quot; as first picking $2$ and then $1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we interpret it that way, so that we are asking for different multisets of choices, then we are asking about combinations with repetitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways of choosing $r$ objects from among $n$ objects with replacement is $\binom{n+r-1}{r}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of ways to pick $K$ from among the $R$ that are less than or equal to $R$ is $\binom{R+K-1}{K}$. However, some of them will not include the number $R$; how many? Well, $\binom{(R-1)+K-1}{K}=\binom{R-K-2}{K}$, since these are the number of choices from among the first $R-1$ instead. Thus, the total with this interpretation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation*}&#xA;\binom{R+K-1}{K} - \binom{R+K-2}{K} = \binom{R+K-2}{K-1}.&#xA;\end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, reserve one spot for for $R$, and then pick the remaining $K-1$ any way you want from among the first $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the explanation posted tells us that in fact they are thinking of the &lt;em&gt;order&lt;/em&gt; of the choices as important: so picking $1$ first and $2$ second is different from picking $2$ first and $1$ second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, we can pick any of the first $R$ numbers at each step, $K$ steps, so that gives $R^K$ choices. In some of them we never picked $R$, however; how many? Well, if we can only pick from among $1,\ldots,R-1$, then by the same argument we have $(R-1)^K$ possibilities. So the total is&#xA;$R^K - (R-1)^K$ as explained there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Since, as you say, you've never seen the formula for combinations with repetitions, and since it is dear to my heart (I figured it out from scratch during my first semester, when considering the question of how many different rolls with $n$ $r$-sided dice there are by constructing a big table, eventually recognizing it as Pascal's Triangle on its side, then using the combinatorial formula for the entry to get the formula, and finally coming up with a proof of it) (and that was probably more than you needed to know) here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $n$ elements; without loss of generality, assume they are the numbers $1$ through $n$. Now, we want to pick $r$ elements form them with repetitions, so we will have $b_1,\ldots,b_r$. Order them so that $b_1\leq b_2\leq\cdots\leq b_r$. Now consider the vector $(b_1,b_2+1,b_3+2,\ldots,b_{r}+r-1)$. Then the entries of these vector are all distinct, and each entry is a number between $1$ and $n+r-1$. Different choices lead to different vectors, so the number of possible choices with repetitions is at most the number of ways to pick $r$ objects from $n+r-1$ possibilities &lt;em&gt;without&lt;/em&gt; repetition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose you pick $r$ numbers from among $1,2,\ldots,n+r-1$ without repetition, $a_1,a_2,\ldots,a_r$. Order them so that $a_1\lt a_2\lt\cdots\lt a_r$. Then consider $(a_1, a_2-1, a_3-2,\ldots,a_r-r+1)$. Since $a_i\geq i$, then $a_i-(i-1)\geq 1$, and $a_i\lt a_{i+1}$ so $a_i-i+1 \leq a_{i+1}-i$. Thus, this new vector corresponds, as before, to a choice of $r$ numbers from among $1,2,\ldots,n$ &lt;em&gt;with&lt;/em&gt; repetitions allowed.  Different choices lead to different vectors, so the number of possible ways to pick $r$ objects from $n+r-1$ possibilities without repetition is at most the number of ways to pick $r$ objects from among $n$ possibilities &lt;em&gt;with&lt;/em&gt; repetions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the two quantities are equal. Since the number of ways to pick $r$ objects from among $n+r-1$ possibilities without repetition is $\binom{n+r-1}{r}$, the formula follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(When I proved it to myself, I took my original set and added symbols &quot;R1&quot;, &quot;R2&quot;,$\ldots$,R(r-1), which stood for &quot;repeat position 1&quot;, &quot;repeat position 2&quot;, etc. Then I picked from among $1,2,\ldots,n$,R1,R2,$\ldots$,R(r-1) without repetition, and wrote them in order, with $1\lt 2\lt\cdots\lt n\lt$ R1$\lt\cdots\lt$ R(r-1); this then gave me a &quot;pick with repetitions&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T07:26:28.857" LastActivityDate="2010-11-11T07:26:28.857" CommentCount="5" />
  <row Id="9832" PostTypeId="2" ParentId="9824" CreationDate="2010-11-11T06:59:11.680" Score="2" ViewCount="0" Body="&lt;p&gt;The expected number of red balls pulled from A is $z\frac{x}{x+y}$. This is because the probability that any of the $z$ individual balls is red is $\frac{x}{x+y}$. To see that, consider the experiment of taking $z$ balls. Number each ball you've taken - the first ball number $1$, the second ball $2$, and so on. Each of the numbers on its own is uniformly distributed, so the probability that the $k$th ball is red is exactly $\frac{x}{x+y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is an easy calculation, as in Ross's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the number of red balls is distributed according to a Hypergeometric distribution (q.v.).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-11T06:59:11.680" CommentCount="4" />
  <row Id="9833" PostTypeId="1" AcceptedAnswerId="9834" CreationDate="2010-11-11T07:00:41.563" Score="2" ViewCount="141" Body="&lt;p&gt;If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm &amp;lt; $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking in terms of a converse, if $T$ is any bounded linear operator defined on $X$, then does the existence of a bounded inverse $S=(I-T)^{-1}$ imply that $S$ can be represented as $S=\sum_{n=0}^\infty T^n$?   &lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditorDisplayName="" LastEditDate="2010-11-11T07:08:04.890" LastActivityDate="2010-11-11T20:55:07.647" Title="If $(I-T)^{-1}$ exists, can it always be written in a series representation?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9834" PostTypeId="2" ParentId="9833" CreationDate="2010-11-11T07:23:20.593" Score="11" ViewCount="0" Body="&lt;p&gt;No, not even in the finite-dimensional case. If $T$ is a linear map&#xA;from $\mathbb{R}^n$ to itself with all eigenvalues $&gt;1$ in absolute&#xA;value, then $I-T$ is invertible, but $\sum T^n$ certainly does not&#xA;converge.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-11T07:23:20.593" CommentCount="1" />
  <row Id="9835" PostTypeId="1" AcceptedAnswerId="9870" CreationDate="2010-11-11T07:31:19.450" Score="1" ViewCount="76" Body="&lt;p&gt;i was reading wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Naive_Bayes_classifier&quot; rel=&quot;nofollow&quot;&gt;Naive Bayes classifier&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but i didn't understand some of the calculations.&#xA;can anybody explain. how they have calculated this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(height|male) = 1.5789 (A probability distribution over 1 is OK. It is the area under the bell curve that is equal to 1.)&#xA;P(weight|male) = 5.9881e-06&#xA;P(foot size|male) = 1.3112e-3&lt;/p&gt;&#xA;" OwnerUserId="1093" LastActivityDate="2010-11-11T17:22:39.470" Title="Naive Bayes classifier" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9836" PostTypeId="1" AcceptedAnswerId="9839" CreationDate="2010-11-11T07:48:58.937" Score="2" ViewCount="78" Body="&lt;p&gt;How does someone approximate something like &#xA;$$\frac{1}{n}\sum_{k=1}^n e^{-k I(k)^2}$$&#xD;&#xA;where I(k) is equidistributed on $[0,1]$. Ideally I would like to show its about $K/\sqrt{n}$ for some constant $K$. This is not homework, but a question I stumbled upon &quot;in nature&quot; and I don't really have any ideas on how to tackle it.&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-11-11T08:24:28.610" Title="Summing something with random and non-random parts." Tags="&lt;probability&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9837" PostTypeId="2" ParentId="9833" CreationDate="2010-11-11T07:56:32.453" Score="3" ViewCount="0" Body="&lt;p&gt;Well there are slightly weaker condition that improves the result you cite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $T$ is a bounded operator with spectral radius $r(T)&amp;lt; 1$&#xA;Then by the spectral radius formula we have &#xA;$$\lim_{n\to\infty}\|T^n\|^{1/n} =\inf_n\|T^n\|^{1/n}=r(T)&amp;lt; 1$$ &#xD;&#xA;which ensure the convergence of &#xD;&#xA;$$\sum_{n=0}^\infty\|T^n\|$$&#xD;&#xA;which in turn, by the triangle inequality, bounds &#xD;&#xA;$$\left\|\sum_{n=0}^\infty T^n\right\|$$ &#xD;&#xA;Now, it is a standard exercise to show that  &#xD;&#xA;$$\sum_{n=0}^\infty T^n=(I-T)^{-1}$$&#xD;&#xA;If there is any doubt at all do not hesitate to ask..&#xD;&#xA;&#xD;&#xA;**Edit:** By the Banach algebra inequality we have&#xD;&#xA;$\|T^n\|\le\|T\|^n$, which means that $$r(T)=\inf_k\|T^k\|^{1/k}\le\|T^n\|^{1/n}\le\|T\|$$&#xD;&#xA; Hence $\|T\|&amp;lt;1$ implies not only $r(T)&amp;lt;1$, but also $r(T)\le\|T\|&amp;lt;1$. &#xD;&#xA;&#xD;&#xA;Also, this is not the case in the example of Robin above, because we also have $r(T)=\sup{|\lambda|:\lambda\in\sigma(T)}$ where $\sigma(T)$ is the spectrum of $T$ (the set of all $\lambda\in\mathbb{C}$ such that $\lambda I-T$ is not invertible) and the eigenvaules of $T$ is certainly in the spectrum. (Note that $r(T)$ is the radius of the smallest closed disc that contain $\sigma(T)$ - hence the name &lt;em&gt;spectral radius&lt;/em&gt;). &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-11T20:55:07.647" LastActivityDate="2010-11-11T20:55:07.647" CommentCount="10" />
  <row Id="9838" PostTypeId="1" CreationDate="2010-11-11T08:00:33.563" Score="0" ViewCount="45" Body="&lt;p&gt;In a proposition in Bourbaki, it is said that for a filter $\mathcal{F}$ and subset $A$ on a topological space $X$, the trace $\mathcal{F}_A$ of $\mathcal{F}$ is a filter iff each set of $\mathcal{F}$ meets $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the definition of the trace of a filter? I couldn't find it mentioned anywhere else online.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2010-11-11T08:00:33.563" Title="What is the trace of a filter?" Tags="&lt;topology&gt;&lt;terminology&gt;&lt;definition&gt;" CommentCount="1" />
  <row Id="9839" PostTypeId="2" ParentId="9836" CreationDate="2010-11-11T08:24:28.610" Score="6" ViewCount="0" Body="&lt;p&gt;To get a rough idea for your sum, I tried the following, don't know if it is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First compute the expected value of $e^{-k x^2}$ as&#xA;$$\int_0^1 e^{-k x^2} dx = \frac{\sqrt{\pi}\mbox{erf}(\sqrt{k})}{2\sqrt{k}}$$&#xD;&#xA;&#xD;&#xA;Now, I used the cheap bound that $\mbox{erf}(\sqrt{k}) &amp;lt; 1$, to see that in expected value each term of your sum is approximately $c/(2\sqrt{k})$, for some constant $c$. Thus, the your sum is approximately:&#xD;&#xA;$$\sum_{k=1}^n \frac{c}{2\sqrt{k}} \approx K\sqrt{n},$$&#xD;&#xA;&#xD;&#xA;so that after using the normalizing $1/n$ outside the sum, the approximate value of your sum is indeed $K/\sqrt{n}$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-11T08:24:28.610" CommentCount="1" />
  <row Id="9840" PostTypeId="1" AcceptedAnswerId="9852" CreationDate="2010-11-11T08:54:22.300" Score="1" ViewCount="172" Body="&lt;p&gt;The formula for finding the roots of a polynomial is as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac {-b \pm \sqrt{ b^2 - 4ac }}{2a} $$&#xD;&#xA;&#xD;&#xA;what happens if you want to find the roots of a polynomial like this simplified one&#xD;&#xA;&#xD;&#xA;$$ 3x^2 + x + 24 = 0 $$&#xD;&#xA;&#xD;&#xA;then the square root value becomes&#xD;&#xA;&#xD;&#xA;$$ \sqrt{ 1^2 - 4\cdot3\cdot24 } $$&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;$$ = \sqrt{ -287 } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the square root of a negative number, which isn't allowed. What do you do in this case? I know there are other methods, i.e. factorisation and completing the square, but does this mean that this formula can only be used in specialised cases or have i gone wrong somewhere along the path?&lt;/p&gt;&#xA;" OwnerUserId="676" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-11T10:34:28.400" LastActivityDate="2010-11-12T03:20:11.593" Title="Finding roots of polynomials, negative square root" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9841" PostTypeId="1" AcceptedAnswerId="9847" CreationDate="2010-11-11T08:57:03.860" Score="0" ViewCount="65" Body="&lt;p&gt;Is the average of i.i.d. random variables with zero mean and finite variance convergent to 0 in probability or $L^2$? How do I show this? I am just beginning to learn convergence of random variables, so an example would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-11-11T11:30:33.433" Title="convergence of average of iid random variables " Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;random-variable&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="9842" PostTypeId="2" ParentId="9840" CreationDate="2010-11-11T08:59:48.753" Score="3" ViewCount="0" Body="&lt;p&gt;Yeah so have you read about complex numbers. The root will be &#xA;$$x = \frac{-1 \pm{i} \sqrt{287}}{2 \times 3}$$&#xD;&#xA;&#xD;&#xA;where $i=\sqrt{-1}$. Read more about complex numbers and when a polynomial can have complex roots, that happens when the *Discriminant* factor $b^{2}-4ac &amp;lt;0$. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-11T08:59:48.753" CommentCount="4" />
  <row Id="9843" PostTypeId="2" ParentId="9840" CreationDate="2010-11-11T09:04:37.770" Score="4" ViewCount="0" Body="&lt;p&gt;Hi Aran,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100% correct, and good observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this, we define $\sqrt{-1}=i$ where $i$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Imaginary_unit&quot; rel=&quot;nofollow&quot;&gt;imaginary unit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\sqrt{-287}=\sqrt{287}i$, and we can solve as per the general quadratic formula. Numbers of the form $a+bi$ are known as complex numbers and are extremely useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the term $b^2-4ac$ is known as the discriminant of the quadratic equation. It should be clear that if $b^2-4ac&gt;0$ there exists two real solutions, if $b^2-4ac=0$ there is one solution (the repeated root) and if &#xA;$$b^2-4ac \lt 0$$ there are two complex solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadratic formula is the most general way to solve the quadratic equation - so you are doing the right thing.&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2010-11-12T03:20:11.593" LastActivityDate="2010-11-12T03:20:11.593" CommentCount="4" />
  <row Id="9844" PostTypeId="1" CreationDate="2010-11-11T10:44:42.673" Score="0" ViewCount="66" Body="&lt;p&gt;Suppose $h$ is a function on $Z$ that is periodic with period N, that is $h(n+N)=h(n)$ for all $n$. How to prove: for any $m \in Z$, &#xA;$\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$. In other words, any sum over an interval of length $N$ yields the same result.&#xA;This is problem from the area of linear algebra and is a homework&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-11T11:26:27.087" LastActivityDate="2010-11-11T13:58:18.350" Title="How to prove: for any $m \in Z$, $\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9845" PostTypeId="2" ParentId="9841" CreationDate="2010-11-11T10:52:55.617" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;Law of large numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-11T10:52:55.617" />
  <row Id="9846" PostTypeId="1" CreationDate="2010-11-11T10:56:53.273" Score="3" ViewCount="79" Body="&lt;p&gt;One way to tessellate a 3D sphere is by iterated subdivision of an icosahedron. I am wondering whether this method gives a homogeneous surface density of vertices. To the eye, it seems to do so, and logic indicates that too (each face has the same are in the icosahedron, and faces in each subdivision are created of equal area), but is there some bias that I'm not thinking of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, what tessellation method can yield a constant density of vertices?&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-11T15:05:30.220" Title="Which tessellation of the sphere yields a constant density of vertices?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="9847" PostTypeId="2" ParentId="9841" CreationDate="2010-11-11T10:58:23.457" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with ${\rm E}(X_i)=0$ and ${\rm Var }(X_i) = \sigma^2$. Then, by the strong law of large numbers, the average $\bar X_n  = \frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $0$ almost surely, hence, in particular, in probability. However, this can also be shown directly as follows. By definition of converges in probability, we need to show that, for any $\varepsilon &gt; 0$, ${\rm P}(|\bar X_n - 0| \geq \varepsilon) \to 0$ as $n \to \infty$. We have ${\rm E}(\bar X_n) = 0$ and ${\rm Var}(\bar X_n) = \sigma^2/n$. Hence, by Chebyshev's inequality, &#xA;$$&#xD;&#xA;{\rm P}(|\bar X_n - 0|  \ge \varepsilon ) \le \frac{{{\rm Var}(\bar X_n )}}{{\varepsilon ^2 }} \to 0.&#xD;&#xA;$$&#xD;&#xA;The convergence in probability is thus proved. As for the convergence to $0$ in $L^2$, this follows straight from the definition. Indeed,&#xD;&#xA;$$&#xD;&#xA;{\rm E}(|\bar X_n  - 0|^2 ) = {\rm Var}(\bar X_n ) \to 0.&#xD;&#xA;$$  &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-11T11:30:33.433" LastActivityDate="2010-11-11T11:30:33.433" CommentCount="1" />
  <row Id="9848" PostTypeId="2" ParentId="9844" CreationDate="2010-11-11T11:07:11.267" Score="1" ViewCount="0" Body="&lt;p&gt;Let $m \in \{0 \dots, N-1\}$. Then we have&#xA;$$\sum_{n=m}^{m+N-1} h(n) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n + N) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n) = \sum_{n=0}^{N-1} h(n)$$&#xD;&#xA;&#xD;&#xA;Now for $m \in \mathbb{Z}$ write $m = kN + r$ with $k \in \mathbb{Z}$ and $r \in \{0, \dots, N-1\}$. We clearly have $$\sum_{n=kN+r}^{k(N+1) + r -1} h(n) = \sum_{n=r}^{N+r-1} h(n+kN) = \sum_{n=r}^{N+r-1} h(n)$$&#xD;&#xA;Now the last expression equals $\sum_{n=0}^{N-1} h(n)$ by the first part of the argument.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-11T11:38:22.503" LastActivityDate="2010-11-11T11:38:22.503" CommentCount="3" />
  <row Id="9849" PostTypeId="1" CreationDate="2010-11-11T11:12:37.300" Score="0" ViewCount="67" Body="&lt;p&gt;How to prove that $N^{-1/2} W_N$ is an unitary matrix( Suggestion: Use lemma 2.2 and lemma 1.105 ii.) \&#xA;Lemma 1.105 &#xA;Let $A$ be an $nxn$ matrix over $C$. Then the following statements are equivalent: &#xA;i. A is unitary&#xA;ii. The columns of A form an orthonormal basis for $C^n$&#xA;iii. The rows of A form an orthonormal basis for $C^n$&#xA;iv. A preserves inner products, that is, $\langle Az, Aw \rangle =\langle z, w \rangle $ for all $z,w \in C^n$&#xA;v. $||Az|| = ||z||$, for all $z \in C^n$ \&#xA;lemma 2.2&#xA;The set ${E_0,...,E_{N-1} }$ is an orthonormal basis for $l^2(Z_N)$.&#xA;Also $W_N = \left(\begin{array}{ccccccc}&#xD;&#xA;1 &amp; 1 &amp; 1 &amp; 1 &amp; . &amp; . &amp; 1  \\&#xD;&#xA;1 &amp; \omega_N &amp; \omega_N^2 &amp; \omega_N^3 &amp; . &amp; . &amp; \omega_N^{N-1} \\&#xD;&#xA;1 &amp; \omega_N^2 &amp; \omega_N^4 &amp; \omega_N^6 &amp; . &amp; . &amp; \omega_N^{2(N-1)} \\&#xD;&#xA;1 &amp; \omega_N^3 &amp; \omega_N^6 &amp; \omega_N^9 &amp; . &amp; . &amp; \omega_N^{3(N-1)} \\&#xD;&#xA;. &amp; . &amp; . &amp; . &amp; . &amp; . &amp; . \\&#xD;&#xA;. &amp; . &amp; . &amp; . &amp; . &amp; . &amp; . \\&#xD;&#xA;1 &amp; \omega_N^{N-1} &amp; \omega_N^{2(N-1)} &amp; \omega_N^{3(N-1)} &amp; . &amp; . &amp; \omega_N^{(N-1)(N-1)} \\&#xD;&#xA;\end{array}\right)$, &#xA;Where $\omega_N = e^{-2 \pi i/N}$ then &#xA;$\omega_N^{mn} = e^{-2 \pi i mn/N}$ &#xA;This a homework in the area of linear algebra. And I have no idea how to procede in the proof, because it might become quite technical proof if you would just make straight computation, but I hope it is done by lemma 1.105 and lemma 2.2.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-11T11:33:08.173" LastActivityDate="2010-12-11T14:44:24.980" Title="How to prove that $N^{-1/2} W_N$ is an unitary matrix?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9850" PostTypeId="2" ParentId="9844" CreationDate="2010-11-11T11:28:13.733" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand&#xA;$$\sum_{n=m}^{m+N-1}h(n)$$&#xD;&#xA;What do you know about $h$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-11T11:28:13.733" />
  <row Id="9851" PostTypeId="1" CreationDate="2010-11-11T11:39:20.567" Score="11" ViewCount="243" Body="&lt;p&gt;If for a function $f(x)$ only its absolute value $|f(x)|$ and the absolute value $|\tilde f(k)|$ of its Fourier transform $\tilde f(k)=N\int f(x)e^{-ikx} dx$ is known, can $f(x) = |f(x)|e^{i\phi(x)}$ and thus the phase function $\phi(x)$ be extracted? (with e.g. $N=1/(2\pi)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Marek already &lt;a href=&quot;http://math.stackexchange.com/questions/9851/can-the-phase-of-a-function-be-extracted-from-only-its-absolute-value-and-its-fou/9854#9854&quot;&gt;stated&lt;/a&gt;, this is even not uniquely possible for $f(x)=c\in\mathbb C$, since the global phase cannot be re-determined. So please let me extend the question to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under what circumstances is the phase-retrieval (up to a global phase) uniquely possible, and what ambiguities could arise otherwise?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-11-12T09:36:31.587" LastActivityDate="2010-12-22T09:17:55.817" Title="Can the phase of a function be extracted from only its absolute value and its Fourier transform's absolute value?" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;&lt;fourier-analysis&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="9852" PostTypeId="2" ParentId="9840" CreationDate="2010-11-11T11:56:29.627" Score="9" ViewCount="0" Body="&lt;p&gt;The other answers are nice, so I won't reiterate them.  It's worth pointing out, though, that when you say &quot;I know there are other methods, i.e. factorisation and completing the square, but does this mean that this formula can only be used in specialised cases&quot;, you seem to be implying that these other methods might yield different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't the case.  If the quadratic formula fails to give you real-valued answers, then the other two methods (and every applicable method) will also fail, because the quadratic polynomial itself simply doesn't &lt;em&gt;have&lt;/em&gt; any real-valued roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visualize this easily: as you may know, the graph of a quadratic polynomial is a parabola, and the real roots of a polynomial correspond to its graphs' zeros, i.e. points where the graph crosses the $x$-axis.  So if you imagine a parabola with its vertex above the $x$-axis and opening upwards, you can see that it will &lt;em&gt;never&lt;/em&gt; cross the $x$-axis -- it has no real zeros at all!&#xA;&lt;img src=&quot;http://i.imgur.com/wNzW3.png&quot; alt=&quot;alt graph of quadratic polynomial&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly the situation with $y = 3x^2 + x + 24$ and every other quadratic polynomial for which $b^2 - 4ac$ (the discriminant) is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly (since it seems as though this may be one of your first introductions to complex numbers) let me clarify something that confuses my students greatly: don't let the words &quot;real&quot; and &quot;imaginary&quot; mislead you into thinking that real numbers are somehow genuine or actual and imaginary numbers are somehow fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like a &quot;negative&quot; number isn't sad, cynical, or irritated, and just as an &quot;irrational&quot; number isn't my crazy ex-girlfriend, &quot;real&quot; and &quot;imaginary&quot; are technical terms, and not meant to carry with them their English connotations.  Although they earned those names due to prior misunderstandings about numbers, today we know that imaginary numbers are just as actual and valid as real numbers, and in fact complex (not &quot;complicated&quot;) numbers have huge real-world applications and consequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along these same lines: you say that taking the square root of a negative number &quot;isn't allowed&quot;.  More accurate would be to say that the square root of a negative number isn't a real number (&quot;real&quot;, again, in the technical sense).  If you require the result of taking the square root to be a real number, then you're right, taking the square root of a negative isn't allowed.  But otherwise it's perfectly fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy: you're allowed to divide the whole number 6 by the whole number 3, and the result is the whole number 2.  But if you divide the whole number 1 by the whole number 2, then the result, $\frac{1}{2}$, isn't a whole number.  If you require your operations on a certain class of numbers to produce a result from that same class of numbers, then lots of operations (like division, and subtraction too, for that matter) will have situations in which they aren't &quot;allowed&quot; to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This requirement is sometimes very desirable, and there are certainly contexts in which you will want to impose it.  But you should understand that imposing this requirement is a choice that you make because it fits the context of the problems you're working on.  It's not divine law.  If you find yourself in a situation in which you would like to take the square root of a negative number and produce the corresponding imaginary number, you are definitely &quot;allowed&quot; to do so.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2010-11-12T02:58:22.973" LastActivityDate="2010-11-12T02:58:22.973" CommentCount="3" />
  <row Id="9853" PostTypeId="2" ParentId="9844" CreationDate="2010-11-11T13:28:47.857" Score="1" ViewCount="0" Body="&lt;p&gt;The important point here is that any set of $N$ consecutive integers is a complete residue system modulo $N$. See &lt;a href=&quot;http://mathrefresher.blogspot.com/2008/10/complete-residue-system.html&quot; rel=&quot;nofollow&quot;&gt;complete residue system.&lt;/a&gt; We consider the index modulo $N$ since $h(n)$ has period $N.$ i.e. the numbers&#xA;$ \lbrace m,m+1,m+2,\ldots,m+N-1 \rbrace $ modulo $N$ are $ \lbrace 0,1,2,\ldots,N-1 \rbrace $ in some order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your identity follows immediately from this fact.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-11T13:58:18.350" LastActivityDate="2010-11-11T13:58:18.350" />
  <row Id="9854" PostTypeId="2" ParentId="9851" CreationDate="2010-11-11T13:45:48.963" Score="5" ViewCount="0" Body="&lt;p&gt;The short answer is no. Take constant function $f(x) \equiv C \in \mathbb{C}$. Disregarding normalization, we have $\hat{f} = C \delta$ (in the sense of distributions). Clearly, there is no way to recover the phase of $C$ once we take the absolute value on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this a little more explicit, consider a lot easier version of the problem on the group $G = \mathbb{Z} / N\mathbb{Z}$. Its dual is $\hat{G} = G$. If you'll write out the Fourier transform equations (i.e. $\hat{f}(k) = \sum_{n=0}^{N-1} f(n) \exp(-{i k n \over 2 \pi})$), you'll obtain $2N$ real equations for $2N$ coefficients ($N$ Fourier phases and $N$ original phases). The properties of this system of equations are not clear to me, but the case $N=1$ (this is the same as in the first paragraph, but here we don't need to talk about distributions) already shows that the solutions need not be unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope someone else can provide more information, I'd be also interested to see what conditions on $f$ one needs to assume to get a unique solution. Even for the case $G = \mathbb{Z} / N\mathbb{Z}$ this looks interesting enough.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-11T14:00:17.033" LastActivityDate="2010-11-11T14:00:17.033" CommentCount="5" />
  <row Id="9855" PostTypeId="2" ParentId="9830" CreationDate="2010-11-11T14:04:26.663" Score="2" ViewCount="0" Body="&lt;p&gt;The explanation in your module is a common technique in combinatorics.  If you want to count how many there are of something, count some set that includes all of them and some more, then subtract out the members of the set that don't belong.  We can work a small example (unfortunately the numbers get large quickly). Let K=4, R=2.  N doesn't matter in this problem as we just throw away all the numbers $\{R+1, R+2, R+3,\ldots,N\}$.  There are $2^4=16$ strings of four numbers with each number being 1 or 2.  You can list them-they look like the binary numbers from 0 to 15 with the zeros changed to twos.  But one =1^4 of them is all ones, so we shouldn't count it.  The answer is $15=2^4-1^4=R^K-(R-1)^K$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T14:04:26.663" CommentCount="1" />
  <row Id="9856" PostTypeId="2" ParentId="9145" CreationDate="2010-11-11T14:05:45.213" Score="3" ViewCount="0" Body="&lt;p&gt;What you want is called the &quot;Joint Distribution Function&quot; of the dice.  That is, for two dice with random variables $X$ and $Y$ measuring the number showing on the faces, it is $P(X=x,Y=y)$.  Since the dice are independent, the JDF &lt;i&gt;is&lt;/i&gt; the product of the individual distribution functions of $X$ and $Y$ (what you call the mass function):&#xA;$$P(X=x,Y=y) = P(X=x)P(Y=y).$$&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-11-11T14:05:45.213" />
  <row Id="9857" PostTypeId="2" ParentId="9803" CreationDate="2010-11-11T14:25:07.493" Score="1" ViewCount="0" Body="&lt;p&gt;Well, let's go back to the definitions. I am sorry if I go too far into the details, but it is difficult to judge what you exactly understand based only on your question. So maybe I explain too much, but don't take this as an offense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is $V/U$ ? I will explain this a little, since it might be the source of the problem. $V/U$ is the set of the class of equivalences of vectors from $V$. The equivalence relation is given by $x \sim y :\iff x-y \in U$. This set can naturally be considered as a vector space if you define:&#xA;$$[x] + [y] := [x+y] \text{ and } \lambda[x] := [\lambda x]$$&#xD;&#xA;&#xD;&#xA;Let's take an example: $V = \mathbb{R}^4$, $U = \mathbb{R}^2$, $W = \mathbb{R}$. First, what does it mean to consider $\mathbb{R}^2$ and $\mathbb{R}$ as subspaces of $\mathbb{R}^4$ ? Well, we need to actually consider $\mathbb{R}^2 \times \{(0,0)\}$ and $\mathbb{R} \times \{(0,0,0)\}$. Thus, actually $U = \{ x = (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | x_3 = x_4 = 0\}$ and $W = \{ x = (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | x_2 = x_3 = x_4 = 0\}$.&#xD;&#xA;&#xD;&#xA;So a first equivalence relation $\sim_A$ can be defined for $V/U$:&#xD;&#xA;$$x \sim_A y :\iff x_3 = y_3 \text{ and } x_4 = y_4$$&#xD;&#xA;&#xD;&#xA;Same for $W/U$:&#xD;&#xA;$$x \sim_B y :\iff x_2 = y_2, x_3 = y_3, x_4 = y_4$$&#xD;&#xA;&#xD;&#xA;Thus let's write, $V/U = \{[x]_A \}$ and $V/W = \{[x]_B \}$, where obviously $[x]_A = \{y \in \mathbb{R}^4 | y_3 = x_3, y_4 = x_4\}$ and $[x]_B = \{y \in \mathbb{R}^4 | x_2 = y_2, x_3 = y_3, y_4 = x_4\}$.&#xD;&#xA;&#xD;&#xA;Note that $V/U$ has dimension 2 (a base is given by $\{[(0,0,1,0)]_A , [(0,0,0,1)]_A\}$) and that $V/W$ has dimension 3 (a base is given by $\{[(0,1,0,0)]_B , [(0,0,1,0)]_B, [(0,0,0,1)]_B\}$ ).&#xD;&#xA;&#xD;&#xA;Now, how can we define an addition $[u]_A + [v]_B$ ? This would be trying to add sets. You can try to define $[u]_A + [v]_B$ in a natural way but you won't manage it: simply saying $[u]_A + [v]_B := (u_1+v_1, u_2 + v_2 , u_3 + v_3, u_4 + v_4)$ won't work, the addition will not be well-defined (because if you change the representant of $[u]_A$, $u_1$ and $u_2$ can be anything, and if you change representant $[v]_B$, $v_1$ can be anything).&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;Furthermore, your intuition should tell you that since both dimensions of $V/U$ and $V/W$ are 2 and 3, such an addition can only be defined properly in a vector space of at least dimension 5 (otherwise you lose information). Too bad $\mathbb{R}^4$ has only dimension 4. So you should be convinced that it is impossible to define a natural addition properly in $V$ for this example.&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;(Caution: I don't say it is impossible to define an addition inside $V$. Actually, it is possible to define several additions, but it won't be &quot;natural&quot;)&#xD;&#xA;&#xD;&#xA;&#xD;&#xA; &#xD;&#xA;Hence, the only way to do it properly is to consider $V/U \oplus V/W$. Of course, in special cases, where $\dim(V/U) + \dim(V/W) \leq \dim(V)$, then it is possible to rearrange things so that you consider $V/U \oplus V/W$ as a subspace of $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the long answer, that depending on your level of understanding might not be very clear.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-11T14:33:11.090" LastActivityDate="2010-11-11T14:33:11.090" CommentCount="2" />
  <row Id="9858" PostTypeId="2" ParentId="9849" CreationDate="2010-11-11T14:34:25.097" Score="1" ViewCount="0" Body="&lt;p&gt;Following Slowsolver,note that the entries along a row or column are in geometric progression.  So just write down $W_N^*W_N$ from the definition of a matrix multiplication.  It will be a sum of N terms in geometric progression, which you can sum.  The powers in each sum off the main diagonal will form a permutation of $\{0,1,2,\ldots,N-1\}$.  As they are all the $N^{th}$ roots of unity, they sum to zero.  Along the main diagonal you will get all 1's, which sum to N.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T14:34:25.097" />
  <row Id="9859" PostTypeId="2" ParentId="9846" CreationDate="2010-11-11T14:58:04.783" Score="2" ViewCount="0" Body="&lt;p&gt;See the links below. They're not about tesselation, but about evenly distributing points on a sphere (from which you can extract a tesselation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://local.wasp.uwa.edu.au/~pbourke/geometry/spherepoints/&quot; rel=&quot;nofollow&quot;&gt;http://local.wasp.uwa.edu.au/~pbourke/geometry/spherepoints/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.vanderbilt.edu/~esaff/texts/161.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.vanderbilt.edu/~esaff/texts/161.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.research.att.com/~njas/icosahedral.codes/&quot; rel=&quot;nofollow&quot;&gt;http://www2.research.att.com/~njas/icosahedral.codes/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sitemason.vanderbilt.edu/page/hmbADS#code&quot; rel=&quot;nofollow&quot;&gt;http://sitemason.vanderbilt.edu/page/hmbADS#code&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eqsp.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;http://eqsp.sourceforge.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/index/spheres.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.niu.edu/~rusin/known-math/index/spheres.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-11T14:58:04.783" />
  <row Id="9860" PostTypeId="2" ParentId="9846" CreationDate="2010-11-11T15:05:30.220" Score="1" ViewCount="0" Body="&lt;p&gt;I've got my answer from one of the links (http://www.math.vanderbilt.edu/~esaff/texts/161.pdf) given by lhf, in the last paragraph of page 9 and first of page 10:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's more, they are not&#xA;  asymptotically uniformly distributed.&#xA;  To see this, it is enough to observe&#xA;  that after the first step, the points&#xA;  are centered at 80 spherical triangles&#xA;  which do not all have the same area&#xA;  (the projection process in- creases&#xA;  the areas of the &quot;middle triangles&quot;&#xA;  more than the rest). As the further&#xA;  steps in the process yield the same&#xA;  number of points in each of the 80&#xA;  triangles, asymptotic uniform&#xA;  distribution cannot hold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for the link!&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-11T15:05:30.220" CommentCount="1" />
  <row Id="9861" PostTypeId="1" CreationDate="2010-11-11T15:11:07.510" Score="2" ViewCount="42" Body="&lt;p&gt;I have two sets (a,b) of data and I am restricted to using the &#xA;values in the sets once and only once when they are combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the 2nd highest value possible from &#xA; a data set of a{3,2} and b{2,0} is a1+b2=3. &#xA;There are two possible ways of combining these two data sets: &#xA; (1) a1+b1 and a2+b2 which results in values of 5 and 2, and &#xA; (2) a1+b2 and a2+b1 which results in values of 3 and 4. &#xA;The (2) combination results in the highest (or maximum) 2nd-high value &#xA;possible of 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ergo, the 3rd highest value possible from &#xA; a data set of a{3,2,1} and b{4,3,0} is a1+b3=3. &#xA;There are six possible ways of combining these two data sets: &#xA;(1) a1+b1 followed by a2+b2  and a3+b3, which results in values of 7, 5 and 1, and &#xA;(2) a1+b1 followed by a2+b3  and a3+b2, which results in values of 7, 2 and 4, and &#xA;(3) a1+b2 followed by a2+b1  and a3+b3, which results in values of 6, 6 and 1, and &#xA;(4) a1+b2 followed by a2+b3  and a3+b1, which results in values of 5, 2 and 5, and &#xA;(5) a1+b3 followed by a2+b1  and a3+b2, which results in values of 3, 6 and 4, and &#xA;(6) a1+b3 followed by a2+b2  and a3+b1, which results in values of 3, 5 and 5. &#xA;The (5) and (6) combinations both result in a maximum 3rd-highest value of 3, &#xA;  from a1+b3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By brute force and analogy, I have determined the maximum 1st through nth highest combination of ai+bi in two data sets can be determined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum             Formula &#xA;1st-high            a1+b1 &#xA;2nd-high            min(a1+b2, a2+b1) &#xA;3rd-high            min(a1+b3, a2+b2, a3+b1) &#xA;4th-high            min(a1+b4, a2+b3, a3+b2, a4+b1) &#xA;... &#xA;8th-high            min(a1+b8, a2+b7, a3+b6, a4+b5, a5+b4, a6+b3, a7+b2, a8+b1) &#xA;... &#xA;nth-high            min(a1+bn, a2+b(n-1), ... , a(n-1)+b2, an+b1) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a handle on the formula, what I want to know is &#xA;(a) where the answer to this question has been published, and&#xA;(b) what it is formally described by in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure this is not the first time this has come up.&#xA;Is it addressed in a text book or in the literature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3411" LastActivityDate="2010-11-11T15:27:42.270" Title="Maximum combined 1st through nth highest combinations of ai + bi" Tags="&lt;algebra-precalculus&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9862" PostTypeId="2" ParentId="9861" CreationDate="2010-11-11T15:27:42.270" Score="1" ViewCount="0" Body="&lt;p&gt;Let $a = 10,9,0$ and $b = 10,8,0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First highest is $a_1+b_1 = 20$, second highest is $a_2+b_1 = 19$, third highest is $a_1+b_2 = 18$. Third highest invalidates your formula.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-11T15:27:42.270" CommentCount="1" />
  <row Id="9863" PostTypeId="1" AcceptedAnswerId="9897" CreationDate="2010-11-11T15:35:29.257" Score="1" ViewCount="106" Body="&lt;p&gt;I am trying to understand (simplify) the following space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z = S^2 \times S^2 / E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(a_1 , a_2 , a_3) \in S^2$ and $(n_1 , n_2 , n_3) \in S^2$. Here ${a_1}^2 + {a_2}^2 + {a_3}^2 = 1$ and ${n_1}^2 + {n_2}^2 + {n_3}^2 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence relations $E$ are the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(\pi; a_1,a_2,a_3)$ is a rotation of angle $\pi$ along the axis $(a_1, a_2, a_3)$. So the point $(\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)$ is the mirror image of $(n_1, n_2, n_3)$ in the plane whose normal is $(a_1, a_2, a_3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to simplify the equivalence relation (2) by a homeomorphism of $S^2 \times S^2$. For example, the approach I am trying is to rotate the second $S^2$ such that the mirror plane becomes horizontal (XY plane). Since the normal of the mirror plane is $(a_1, a_2, a_3)$, the amount by which the second $S^2$ has to be rotated depends on the $(a_1, a_2, a_3)$ it corresponds to. And for this transformation to be a homeomorphism the rotation operations should vary continuously. The transformation should not complicate the first equivalence relation; one way this condition could be imposed is by ensuring that the rotation operation corresponding to $(a_1, a_2, a_3)$ and $(-a_1, -a_2, -a_3)$ are be the same. I am not able to come up with such a transformation. All my attempts to simplify (2) end up complicating (1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific, I want to get a homeomorphism between the space $Z$ and the following space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z^' = S^2 \times S^2 / E^' $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence relations $E^'$ are the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (n_1, n_2, -n_3)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas are appreciated. Please let me know if you have any questions.&lt;/p&gt;&#xA;" OwnerUserId="3050" LastEditorUserId="3050" LastEditorDisplayName="" LastEditDate="2010-11-11T18:02:00.360" LastActivityDate="2010-11-11T19:56:08.360" Title="Equivalence relations on $S^2 \times S^2$" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="9864" PostTypeId="2" ParentId="8814" CreationDate="2010-11-11T15:53:30.517" Score="3" ViewCount="0" Body="&lt;p&gt;$\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-11-11T15:53:30.517" CommunityOwnedDate="2010-11-11T15:53:30.517" />
  <row Id="9865" PostTypeId="2" ParentId="457" CreationDate="2010-11-11T16:50:36.673" Score="1" ViewCount="0" Body="&lt;p&gt;I think one that comes up often in discussions like this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Brouwers Fixed point theorem&lt;/a&gt; which I think can be demonstrated by putting a map of your country on a table and asserting that one of the points on the map is directly above the corresponding point in your country (ie where you are right now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way i think is simpler is to take two sheets of paper, crumple one up and place it on top of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember someone telling me that when you stir a cup of coffee, one point in the fluid will always end up at the same place as it started by this theorem, but im not convinced this is really a continuous map since the particles in the fluid are discrete objects.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-11-11T16:50:36.673" CommunityOwnedDate="2010-11-11T16:50:36.673" />
  <row Id="9866" PostTypeId="2" ParentId="457" CreationDate="2010-11-11T16:59:54.103" Score="1" ViewCount="0" Body="&lt;p&gt;Physical interpretation of the Mean Value Theorem: If the average speed of a car between two locations was V km/h, then there was at least one instant where the speed indicator &#xA;displayed V km/h.&lt;/p&gt;&#xA;" OwnerUserId="2044" LastActivityDate="2010-11-11T16:59:54.103" CommunityOwnedDate="2010-11-11T16:59:54.103" />
  <row Id="9867" PostTypeId="1" CreationDate="2010-11-11T17:12:35.117" Score="0" ViewCount="140" Body="&lt;p&gt;it is a question Convergence/Divergence of calculus II! Please give me a hand!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine convergence or divergence using any method covered so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n = 1}^{\infty} \sin (1/n)$$&lt;/p&gt;&#xA;" OwnerUserId="3414" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T17:44:32.653" LastActivityDate="2010-11-11T17:44:32.653" Title="Convergence/Divergence of $\sum_{n=1}^{\infty} \sin(1/n)$" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9868" PostTypeId="1" CreationDate="2010-11-11T17:13:44.007" Score="0" ViewCount="48" Body="&lt;p&gt;Hi. Suppose given two manifolds $X$ and $Y$, both orientable of dimension $n$, and a map $f:X\to Y$. Is there a relationship between the degree of $f$ calculated with respect to homology (by which I mean the induced map on the top homology groups) and the degree of $f$ calculated with respect to cohomology (by which I mean the induced map on the top cohomology groups)? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2272" LastActivityDate="2010-11-11T17:13:44.007" Title="Computing degree of map" Tags="&lt;algebraic-topology&gt;" CommentCount="3" />
  <row Id="9869" PostTypeId="2" ParentId="9867" CreationDate="2010-11-11T17:17:48.817" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: $\sin(x) / x \to 1$ as $x \downarrow 0$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-11T17:17:48.817" />
  <row Id="9870" PostTypeId="2" ParentId="9835" CreationDate="2010-11-11T17:22:39.470" Score="1" ViewCount="0" Body="&lt;p&gt;These are the probability densities of the assumed distributions, evaluated at the sample data. In this case, the distributions of height, weight and foot size are assumed to follow independent Gaussian distributions whose parameters are given in the second table. If you don't know the probability density function of a Gaussian distribution, look &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for the height, the assumed distribution is a Gaussian with mean $\mu = 5.855$ and variance $\sigma^2 = 3.5033\times 10^{-2}$. As the sample's value is $x = 6$, you can plug that in to the probability density function of the distribution to get &lt;a href=&quot;http://www.wolframalpha.com/input/?i=f+%3D+e%5E%28-%28x-mu%29%5E2%2F%282+sigma%5E2%29%29%2F%28sqrt%282+pi+sigma%5E2%29%29%2C+x+%3D+6%2C+mu+%3D+5.855%2C+sigma%5E2+%3D+3.5033e-2&quot; rel=&quot;nofollow&quot;&gt;$f(x) = 1.5789$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-11T17:22:39.470" />
  <row Id="9871" PostTypeId="2" ParentId="9867" CreationDate="2010-11-11T17:23:08.227" Score="2" ViewCount="0" Body="&lt;p&gt;You need to determine convergence for $\sum_{n=1}^\infty\sin(1/n)$. The series diverges. The two hints below may guide you when trying to justify this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 1: $ \lim_{\theta\to0}\sin(\theta)/\theta=1$ and $1/n\to 0$ as $n\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: $\sin(1/n)$ is positive. So you may attempt a limit comparison test.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-11T17:40:23.970" LastActivityDate="2010-11-11T17:40:23.970" CommentCount="4" />
  <row Id="9872" PostTypeId="1" CreationDate="2010-11-11T17:40:52.367" Score="1" ViewCount="124" Body="&lt;p&gt;I think it is desirable to have that $M_{m\times n}\left(\mathbb{K}\right)\not=M_{m'\times n'}\left(\mathbb{K}\right)$ if $m\not=m'$ or $n\not=n'$. In other words, the set of all $m\times n$ matrices on $\mathbb{K}$ should be different from the set of all $m'\times n'$ matrices on the same field when $m\not=m'$ or $n\not=n'$. But if I define $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^n\right)^m$ then $M_{0\times n}\left(\mathbb{K}\right)=M_{0\times n'}\left(\mathbb{K}\right)$ for every $n$ and $n'$. That is, two matrices with zero rows are equal, no matter how many columns they have, because $X^0$ is the set containing the empty tuple, for every set $X$. I could have defined $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^m\right)^n$ instead, but then the problem is with $M_{m\times 0}\left(\mathbb{K}\right)$ and $M_{m'\times 0}\left(\mathbb{K}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Am I defining matrices correctly?&lt;/li&gt;&#xA;&lt;li&gt;Are $m\times 0$ or $0\times n$ matrices that important?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastEditorUserId="2128" LastEditorDisplayName="" LastEditDate="2010-11-11T17:49:44.727" LastActivityDate="2010-11-11T19:58:36.507" Title="$0\times n$ matrices" Tags="&lt;matrices&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9873" PostTypeId="1" AcceptedAnswerId="9874" CreationDate="2010-11-11T17:52:28.437" Score="1" ViewCount="77" Body="&lt;p&gt;How to convert the following partial differential equation (pde) &#xA;$$\frac{\partial V}{\partial t}=aV-as\frac{\partial V}{\partial s}-b^2s^2\frac{\partial^2 V}{\partial s^2}$$&#xD;&#xA;into a pde of the form&#xD;&#xA;$$\frac{\partial u}{\partial t}=x^2\frac{\partial^2 u}{\partial x^2}+cx\frac{\partial u}{\partial x}$$&#xD;&#xA;by some change of variables?&#xD;&#xA;Here $a,b,c$ are constants. (A hint should be enough.)&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-11T17:57:22.127" Title="Change of variables in a partial differential equation" Tags="&lt;analysis&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="9874" PostTypeId="2" ParentId="9873" CreationDate="2010-11-11T17:57:22.127" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you send $t\mapsto -t$, it will fix the sign. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An exponential integrating factor factor of $e^{At}$ will get rid of the $aV$ term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just need to send $s\mapsto\lambda x$ and compute the correct scaling factors. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-11T17:57:22.127" CommentCount="1" />
  <row Id="9875" PostTypeId="1" AcceptedAnswerId="9894" CreationDate="2010-11-11T18:31:17.747" Score="2" ViewCount="84" Body="&lt;p&gt;Let $$\cdots \rightarrow A_{n+1}\rightarrow^{f_{n+1}} A_n \rightarrow^{f_{n}} A_{n-1}\rightarrow \cdots $$ be an inverse system of finite dimensional vector spaces with the property that the $A_i$ are 'eventually constant', i.e., there is an $m$ such that the maps $f_i$ are isomorphisms for every $i\ge m$. Does it follow that $$\lim_{\leftarrow} A_n \simeq A_m?$$&lt;/p&gt;&#xA;" OwnerUserId="3332" LastActivityDate="2010-11-11T20:12:24.623" Title="inverse limit of isomorphic vector spaces" Tags="&lt;commutative-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9876" PostTypeId="1" CreationDate="2010-11-11T18:32:01.383" Score="2" ViewCount="33" Body="&lt;p&gt;Let $A,B \in \mathbb{R}^{n \times n}$ be two positive semi-definite matrices and let $a &gt; 0$ be a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following maximization problem&#xA;$$&#xD;&#xA;\max_{x \in \mathbb{R}^n, \gamma}\ x^T \left(\frac{\gamma}{a + \gamma} A - \gamma B \right)x&#xD;&#xA;\qquad \text{subject to } \gamma &gt; 0, \|x\| = 1&#xD;&#xA;$$&#xD;&#xA;and suppose that we know that $\hat x$ and $\hat \gamma$ maximize the expression. &#xD;&#xA;In this case, we have that $\hat x$ is the largest eigenvector of the matrix $\frac{\hat \gamma}{a + \hat \gamma} A - \hat \gamma B$. &#xD;&#xA;&#xD;&#xA;Suppose that we perturb the matrix A by another matrix $E$, which is supposed to be small in some norm (say operator). Consider again the following optimization problem&#xD;&#xA;$$&#xD;&#xA;\max_{x \in \mathbb{R}^n, \gamma}\ x^T \left(\frac{\gamma}{a + \gamma} (A+E) - \gamma B \right)x&#xD;&#xA;\qquad \text{subject to } \gamma &gt; 0, \|x\| = 1&#xD;&#xA;$$&#xD;&#xA;and denote $\tilde x$ and $\tilde \gamma$ the maximizers. &#xD;&#xA;&#xD;&#xA;The question is how to bound the difference $|\hat \gamma - \tilde \gamma|$ as some function of $E$? Intuitively, if $E$ is small then $\hat \gamma$ and $\tilde \gamma$ should also be close. Any pointers to the relevant literature would be also appreciated. &lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-11T18:32:01.383" Title="How to bound the change between optimal points when perturbing an objective function?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;optimization&gt;&lt;perturbation-theory&gt;" FavoriteCount="1" />
  <row Id="9877" PostTypeId="1" AcceptedAnswerId="9878" CreationDate="2010-11-11T18:37:09.800" Score="2" ViewCount="124" Body="&lt;p&gt;I understand that for a number $a$ to have a multiplicative inverse in mod $n$, it must be coprime to $n$; therefore, all numbers that do not have a multiplicative inverse mod $n$ must share a factor with $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my prediction was that if $n$ is even, then the sum of all numbers that do not have a multiplicative inverse mod $n$ is $\frac{n}{2}\bmod{n}$, and if $n$ is odd, then the sum is $0\bmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally my proof was constructed something like this: if $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then the sum of all the elements of this array:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\begin{matrix}&#xD;&#xA;p_1 &amp;amp; p_2 &amp;amp; \cdots &amp;amp; p_k \\ 2p_1 &amp;amp; 2p_2 &amp;amp; \cdots &amp;amp; 2p_k \\ \vdots &amp;amp; \vdots &amp;amp; \cdots &amp;amp; \vdots \\ n-2p_1 &amp;amp; n-2p_2 &amp;amp; \cdots &amp;amp; n-2p_k \\ n-p_1 &amp;amp; n-p_2 &amp;amp; \cdots &amp;amp; n-p_k \end{matrix}&#xD;&#xA;\end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would cancel out in the correct way. However, while trying to prove this, I hit a roadblock  because I wasn't sure how to account for the multiplicity of the factors of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand &lt;a href=&quot;http://en.wikipedia.org/wiki/Coprime#Properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; correctly, this is the same as finding the sum of all numbers that are not units in the ring of integers modulo $n$, $\mathbb{Z}/n\mathbb{Z}$. Is there any way to do this? If the result is so nice, I feel like there is also a nice way to prove this.&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T15:38:13.110" LastActivityDate="2010-11-12T15:55:10.930" Title="Prove the sum of all numbers that do not have a multiplicative inverse mod $n$" Tags="&lt;algebra&gt;&lt;ring-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" />
  <row Id="9878" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T18:41:20.217" Score="4" ViewCount="0" Body="&lt;p&gt;If $k$ is not coprime with $n$, then $n-k$ is not coprime with $n$ either.&#xA;So you can group those $k$ by pairs $(k,n-k)$, and if $n$ is even and $\neq 2$ (and only in this case), $n/2$ is not coprime with $n$ and &quot;alone&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-11T18:41:20.217" CommentCount="1" />
  <row Id="9879" PostTypeId="1" AcceptedAnswerId="9887" CreationDate="2010-11-11T18:42:05.940" Score="1" ViewCount="150" Body="&lt;p&gt;I wanted to use this fact in a proof, it seems obvious, but I should probably prove this as well.  Any proof will do.  Also for the future is there a way to search for these proofs google does not work very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Ah, it may also be good to show that the n-th root of any real number a s.t. $0\lt a\lt1$ is also $\lt1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:33:45.953" LastActivityDate="2010-11-11T20:46:02.327" Title="Prove that if $a$ is a real number, and $0\lt a \lt 1$, then $\sqrt{a} \gt a$" Tags="&lt;real-analysis&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="9880" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T18:46:50.270" Score="3" ViewCount="0" Body="&lt;p&gt;Note that if $(a,n) = 1$, then $(n-a,n)=1$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-11T18:46:50.270" />
  <row Id="9882" PostTypeId="2" ParentId="9872" CreationDate="2010-11-11T18:48:52.370" Score="1" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No,I guess the only sensible way of defining $\mathbb{K}^0$ when $\mathbb{K}$ is a field is by $\mathbb{K}^0=0$, since it should be a 0-dimensional vector space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No, I have never seen $m\times 0$-matrices used anywhere, other than here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3332" LastActivityDate="2010-11-11T18:48:52.370" />
  <row Id="9883" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T18:49:14.720" Score="2" ViewCount="0" Body="&lt;p&gt;Depending on the level of the audience, it might be seen as obvious.  I would just argue that $\sqrt{x}$ is monotonic and $\gt 0$ on $(0,1)$, so $\sqrt{a}\lt1$ then multiply both sides by $\sqrt{a}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T18:49:14.720" />
  <row Id="9886" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T19:05:17.877" Score="3" ViewCount="0" Body="&lt;p&gt;We argue by contradiction. Suppose $a \in (0,1)$ and $\sqrt{a} \leq a$. Then &#xA;\begin{eqnarray}&#xA;\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a} \leq 1, &#xA;\end{eqnarray}&#xA;which implies that $\sqrt{a} \geq 1$ and therefore $a \geq 1$, which is absurd by construction.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-11T19:05:17.877" />
  <row Id="9887" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T19:13:33.983" Score="4" ViewCount="0" Body="&lt;p&gt;If you know that whenever $a\lt b$ and $c\gt 0$ then $ac\lt bc$, then for any $x$ with $0\lt x\lt 1$, multiplying through by $x$ gives &#xA;\begin{equation*}&#xA;0 \lt x^2 \lt x \lt 1.&#xA;\end{equation*}&#xA;(The last inequality is &quot;inherited&quot; from the fact that you already know that $x\lt 1$). Repeat to get&#xA;\begin{equation*}&#xA;0\lt x^3 \lt x^2 \lt x \lt 1.&#xA;\end{equation*}&#xA;And so on; you get $0\lt x^n \lt x^{n-1}\lt\cdots \lt x \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working the other way, if $1\lt y$, then multiplying through by $y$ you get &#xA;\begin{equation*}&#xA;1\lt y\lt y^2&#xA;\end{equation*}&#xA;(with the first inequality &quot;inherited from your original assumption) and repeating this process leads to&#xA;\begin{equation*}&#xA;1\lt y\lt y^2 \lt\cdots \lt y^n&#xA;\end{equation*}&#xA;So: if $0\lt z\lt 1$, then $0\lt z^n\lt 1$. If $1\lt z$, then $1\lt z^n$. And of course, if $z=1$ then $z^n = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, start with $0\lt a\lt 1$. Then $\sqrt[n]{a}$ cannot be greater than $1$ (then $a = \left(\sqrt[n]{a}\right)^n \gt 1$ by the above), so $\sqrt[n]{a}\lt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now replace $x$ with $\sqrt[n]{a}$ to get the desired inequality.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:31:57.623" LastActivityDate="2010-11-11T19:31:57.623" CommentCount="3" />
  <row Id="9888" PostTypeId="2" ParentId="9872" CreationDate="2010-11-11T19:27:30.790" Score="6" ViewCount="0" Body="&lt;p&gt;As Laurent S points out, the sensible way to define $\mathbb{K}^0$ is as the one element set (either $\{0\}$ or $\{\emptyset\}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could likewise argue that since you want $M_{m\times n}(\mathbb{K})$ to correspond to linear transformations from $\mathbb{K}^n$ to $\mathbb{K}^m$, and since $\mathbb{K}^0$ is a perfectly good vector space, then we &lt;em&gt;should&lt;/em&gt; define the matrices $M_{0\times n}(\mathbb{K})$ and $M_{m\times 0}(\mathbb{K})$. Then you are certainly forced to define both as the trivial set, since only the trivial linear transformation exists in either case (the zero map for $m=0$, and the trivial inclusion for $n=0$).   In that respect, they would fit with the general theory and not require exceptions when describing vector spaces (where you would need to explicitly exclude the zero-dimensional vector space, for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, then, are $M_{0\times n}(\mathbb{K})$ and $M_{0\times n'}(\mathbb{K})$ both equal even when $n\neq n'$, if the first &quot;is&quot; a linear transformation from $\mathbb{K}^n$ to $\{0\}$ and the second &quot;is&quot; a linear transformation from $\mathbb{K}^{n'}$ to $\{0\}$, and hence different? The key is the &lt;code&gt;&quot;&lt;/code&gt;. They are both &lt;em&gt;identified&lt;/em&gt; with those maps, but as a vector space itself they are in fact isomorphic, so there is no horrible universe-shattering error that crept in by having the &quot;matrix&quot; rings be equal. Of course, now your theorem that $M_{m\times n}(\mathbb{K})$ is uniquely determined (in some way) by the pair $(m,n)$ is no longer true: you need to say that they are equal if and only if either $(m,n)=(m',n')$ or $mn=m'n'=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such is life: there is often a special case that needs to be dealt with separately, and your choice is whether you exclude it by design or you exclude it by mention. Here, your choice is between specifying &quot;nonzero&quot; in a lot of theorems about vector spaces, or specifying &quot;or $mn=m'n'=0$&quot; in the theorem about matrices. The latter is usually less disruptive. (In set theory, one often has to deal with the emptyset in a special manner; that corresponds to $0$ here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to their interest: they have very little structure to them, so they are not generally interesting in and of themselves. They can show up as part of a general scheme (as above, fitting into the role of representing linear transformations between the standard finite dimensional vector spaces) so that exceptions don't need to be made, but as such they play more the role of &quot;exception-avoiders&quot; than &quot;interesting in and of themselves.&quot;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T19:58:36.507" LastActivityDate="2010-11-11T19:58:36.507" />
  <row Id="9889" PostTypeId="2" ParentId="9710" CreationDate="2010-11-11T19:31:40.807" Score="0" ViewCount="0" Body="&lt;p&gt;I know its been a while since I got to exercise the old brain matter, but to disprove this equation, all you need to do is find one example which makes it false.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take any n in which f(n) or g(n) returns 1.  And take any n in which f(n) or g(n) does not return 1, say it returns x.  Which would make the equation false, 1*x &amp;lt; 1+x.&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2010-11-11T19:31:40.807" />
  <row Id="9890" PostTypeId="1" AcceptedAnswerId="9892" CreationDate="2010-11-11T19:33:25.737" Score="3" ViewCount="53" Body="&lt;p&gt;I am learning the basics of category theory, so this question is probably obvious to anyone who knows the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resources I've seen all take the following approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0) A category is a collection of objects and morphisms between those objects that satisfy some rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A functor is a morphism in the category of categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A natural transformation is a morphism in the category of functors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they all stop right there. What about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) the morphisms in the category of natural transformations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Or the &quot;morphisms in the category of the morphisms in the category of natural transformations&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these uninteresting? Why does the &quot;meta-ness&quot; stop at 2 levels deep?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-11T19:40:01.667" Title="Morphisms in the category of natural transformations ?" Tags="&lt;category-theory&gt;&lt;morphism&gt;" AnswerCount="1" />
  <row Id="9891" PostTypeId="1" AcceptedAnswerId="9988" CreationDate="2010-11-11T19:34:17.147" Score="1" ViewCount="82" Body="&lt;p&gt;Could someone provide a good reference for learning matrix calculus? I've recently moved to a more engineering-oriented field where it's commonly used and don't have much experience with it.&lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2010-11-12T09:53:04.770" Title="Reference for matrix calculus" Tags="&lt;calculus&gt;&lt;matrices&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9892" PostTypeId="2" ParentId="9890" CreationDate="2010-11-11T19:40:01.667" Score="5" ViewCount="0" Body="&lt;p&gt;They are not uninteresting. Just look up the term &lt;em&gt;n-category&lt;/em&gt; (e.g. &lt;a href=&quot;http://math.ucr.edu/home/baez/n_categories/index.html&quot; rel=&quot;nofollow&quot;&gt;Baez's introduction&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it's true that functors and natural transformations already suffice for most common ideas and constructions used across mathematics, so this is why one usually stops there.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-11T19:40:01.667" />
  <row Id="9893" PostTypeId="2" ParentId="9891" CreationDate="2010-11-11T19:41:32.247" Score="1" ViewCount="0" Body="&lt;p&gt;Gene Howard Golub, Charles F. Van Loan book on &quot;Matrix Computations&quot; is regarded as the &quot;Bhagavad Gita&quot; for Matrix Algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=mlOa7wPX6OYC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=mlOa7wPX6OYC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another book by &quot;Gene Howard Golub, Gerard Meurant&quot; on &quot;Matrices, moments, and quadrature with applications&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=IZvkFET3LlwC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=IZvkFET3LlwC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &quot;Numerical Linear Algebra&quot; by Trefethen and Bau is well-written and easy to read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=bj-Lu6zjWbEC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=bj-Lu6zjWbEC&amp;amp;printsec=frontcover&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would highly recommend Trefethen and Bau since I have read it completely. I feel it is ideal for self-study or for a one quarter course. Once you are done with this you can take a look at Golubs' book. Golubs' book is really good for reference.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T19:44:06.050" LastActivityDate="2010-11-11T19:44:06.050" CommentCount="1" />
  <row Id="9894" PostTypeId="2" ParentId="9875" CreationDate="2010-11-11T19:51:03.713" Score="2" ViewCount="0" Body="&lt;p&gt;Remember that if your directed set has a maximum, then the inverse limit is just the maximum; here this is &quot;essentially&quot; what you have, since after a certain point you are not really getting anything new: you can identify all $A_i$ with $i\geq m$, collapsing the &quot;left tail&quot; of your inverse system into a terminating one, so the inverse system is just that maximum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this works whether you are working with finite dimensional vector spaces or any kind of structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, &quot;essentially&quot; is not the same as &quot;exactly.&quot; So you use the universal property. I claim that $A_m$ has the desired property. Let  $f_{ij}$ with $i\geq j$ be defined as the identity if $i=j$, and as the composition $f_i\circ f_{i-1}\circ\cdots\circ f_j$ if $i\gt j$. Then the projection maps $\pi_j\colon A_m\to A_j$ are defined by $\pi_j=f_{mj}$ if $j\leq m$, and as $\pi_j = f_{jm}^{-1}$ if $j\gt m$; note that this makes sense since $f_{jm}$ is a composition of isomorphisms when $j\gt m$, so it is itself an isomorphism and has an inverse. Note that for any $i\gt j$ we have $f_{ij}\circ \pi_j = \pi_i$, since you have $f_{ij}\circ f_{jk} = f_{ik}$ whenever $i\geq j \geq k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $P$ be any object together with maps $p_j\colon P\to A_j$ such that for all $i\gt j$ you have $f_{ij}\circ p_i = p_j$. Then $p=p_m\colon P\to A_m$ has the appropriate property, that is, $p_j = \pi_j\circ p$ for all $j$: if $j\leq m$, then this because $\pi_j = f_{mj}$, so $\pi_j\circ p = f_{mj}\circ p_m = p_j$, and if $j\gt m$ then $f_{jm}\circ p_j = p_m$, so pre-composing with $f_{jm}^{-1}$ we get $p_j = (f_{jm})^{-1}\circ p_m = \pi_j\circ p$. The map is unique: if $f\colon P\to A_m$ has the same property, then $f=f_m = p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $(A_m, \{\pi_j\})$ has the desired universal property, and so &quot;is&quot; the inverse limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll note that the fact that we are dealing with finite dimensional vector spaces is completely immaterial; what matters is the properties of the functions in play. This is typical of universal constructions.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-11T20:12:24.623" LastActivityDate="2010-11-11T20:12:24.623" CommentCount="2" />
  <row Id="9895" PostTypeId="2" ParentId="8611" CreationDate="2010-11-11T19:54:01.793" Score="1" ViewCount="0" Body="&lt;p&gt;In response to user2969 :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first you need to computed the &lt;a href=&quot;http://en.wikipedia.org/wiki/Greatest_common_divisor&quot; rel=&quot;nofollow&quot;&gt;gcd&lt;/a&gt;, it is always best to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Euclidean algorithm&lt;/a&gt; for this,my C++ implementation for the same :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int E_GCD(int a,int b){&#xA;   return b ? gcd(b,a%b):a;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now we need to compute the number of divisors of &lt;code&gt;N = E_GCD(A,B)&lt;/code&gt;, this can be done efficiently by first computing the factorization using any fast algorithm like &lt;a href=&quot;http://en.wikipedia.org/wiki/Pollard%27s_rho_algorithm&quot; rel=&quot;nofollow&quot;&gt;Pollard rho integer factorization&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Lenstra_elliptic_curve_factorization&quot; rel=&quot;nofollow&quot;&gt;ECM&lt;/a&gt; then using the standard number theory $\text{trick}^*$ to for finding the number of factors,however in this problem if the values of &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt;  are small (say:$0 \lt A,B \le 10^6$),then we don't need really need to use any of such sophisticated algorithms,instead you could build up your own algorithm simply based on the fact the the factors of a number occur in pair that is if say &lt;code&gt;i&lt;/code&gt; is a factor of &lt;code&gt;N&lt;/code&gt; then so is $N/i$ using this key observation I build up the algorithm,here is my C++ implementation of this idea :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  N = E_GCD(A,B);&#xA;  int ans = 0;&#xA;  int sqt = (int)sqrt(N);&#xA;  for(int i = 1 ; i &amp;lt;= sqt; i++){&#xA;      if(N % i == 0){&#xA;          ans += 2 ;&#xA;          if(i == N/i) ans--;&#xA;      }&#xA;  }&#xA;  printf(&quot;%d\n&quot;,ans);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice the fact that a number cannot a have a divisor which is greater than the square root of the number,rest of the code I believe is self explanatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*$\text{If }N = a^p \cdot b^q  \cdot c^r \cdots \text{,a,b,c ... are primes,number of factors of N is :}(p+1) \cdot (q+1)\cdot(r+1)\cdots$ &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-11T20:11:11.437" LastActivityDate="2010-11-11T20:11:11.437" />
  <row Id="9896" PostTypeId="1" AcceptedAnswerId="9908" CreationDate="2010-11-11T19:55:55.553" Score="6" ViewCount="321" Body="&lt;p&gt;In naieve set theory, you quickly run into existence trouble if you try to do meta-things things like take &quot;the set of all sets&quot;. For example, does the set of all sets that don't contain themselves contain itself? (see, Russell)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, people have no trouble casually talking about the &quot;category of all categories&quot;, etc. How do we know there isn't a contradiction lurking somewhere here - especially since many books define categories in terms of sets?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-16T00:49:04.417" Title="Category of all categories vs. Set of all sets" Tags="&lt;set-theory&gt;&lt;category-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="9897" PostTypeId="2" ParentId="9863" CreationDate="2010-11-11T19:56:08.360" Score="1" ViewCount="0" Body="&lt;p&gt;This isn't quite what you asked for, but notice that (1) simply gives us $\mathbb{RP}^2 \times S^2$.  I'm not sure I understand your notation for (2); it seems like it doesn't depend at all on $\pi$.  If it's true that you're just reflecting $\vec{n}$ through the plane perpendicular to $\vec{a}$, then you're just getting a copy of $D^2$ over every point of $\mathbb{RP}^2$.  This is a fiber bundle, and if you think through it I think you'll find that it's the disk bundle of the tangent bundle $T\mathbb{RP}^2$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-11T19:56:08.360" CommentCount="8" />
  <row Id="9898" PostTypeId="1" CreationDate="2010-11-11T20:01:02.567" Score="1" ViewCount="77" Body="&lt;p&gt;Given a diffusion $dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$, if one applies the transform&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \int_a^x \frac{1}{\sigma(u)}du$, one can use Ito's lemma to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$df(X_t) = \left( \frac{\mu(X_t)}{\sigma(X_t)} - \frac{\sigma'(X_t)}{2} \right)dt + dW_t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to explicitly invert this transform in the general case, assuming $\sigma$ is sufficiently nice?&lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2010-11-11T20:53:52.640" Title="How do I invert this transform?" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;&lt;transformation&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9899" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T20:01:29.290" Score="0" ViewCount="0" Body="&lt;p&gt;Since you were explicitly looking for a different proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle S_n$ be the sum of all numbers $k$ such that $\displaystyle 1 \leq k$ &amp;lt; $\displaystyle n$ and $\displaystyle (n,k) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that $\displaystyle S_2 = 1, S_3 = 3 = 0 \mod 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is that $\displaystyle S_n = 0 \mod n$ if $\displaystyle n &amp;gt; 2$ (which can easily be seen by pairing $\displaystyle k$ with $\displaystyle n-k$ as noted by other answers, but we will give a different proof).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this differently, notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S_n + \sum_{d \mid n, d &amp;gt; 1} (d \times S_{n/d}) = n(n-1)/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is got by partitioning the numbers $\displaystyle \{1,2,\cdots, n-1 \}$ by their gcd with $\displaystyle n$ and adding up the individual partitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partition $\displaystyle \{k : (k,n) = 1\}$ sums to $\displaystyle S_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partition $\displaystyle \{k : (k,n) = d\}$ is same as the set $\displaystyle \{k : (k/d, n/d) = 1\}$ and so sums to $\displaystyle  d \times S_{n/d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also notice that $\displaystyle n(n-1)/2 = 0 \mod n$ if $\displaystyle n$ is odd, and $\displaystyle n(n-1)/2 = n/2 \mod n$ if $\displaystyle n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by total induction on $\displaystyle n$, $S_n = 0 \mod n$ if $\displaystyle n &amp;gt; 2$ as $\displaystyle S_{n/d} = 0 \mod n/d$ implies $\displaystyle d \times S_{n/d} = 0 \mod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what you are looking for is $\displaystyle \sum_{d \mid n, d &amp;gt; 1} (d \times S_{n/d})$ which is $\displaystyle 0 \mod n$ if $\displaystyle n$ is odd, and $\displaystyle n/2 \mod n$ if $\displaystyle n$ is even: the only non-zero term $\displaystyle \mod n$ corresponds to $\displaystyle n/d = 2$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T15:12:45.317" LastActivityDate="2010-11-12T15:12:45.317" />
  <row Id="9901" PostTypeId="2" ParentId="9896" CreationDate="2010-11-11T20:05:07.710" Score="1" ViewCount="0" Body="&lt;p&gt;You can talk about &quot;the category of all categories&quot; with the proviso &quot;in a given universe&quot;. &quot;Our universe&quot; is big enough to include the set of natural numbers and all the usual operations you do with sets (union, intersection...) See Mac Lane's &quot;Categories for the working mathematician&quot;, first chapter, for the details. &lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-11T20:05:07.710" />
  <row Id="9902" PostTypeId="2" ParentId="6387" CreationDate="2010-11-11T20:08:36.847" Score="3" ViewCount="0" Body="&lt;p&gt;Make the observation that $|x| = \theta(x) x - \theta(-x) x$ for all real $x$, where $\theta$ is the Heaviside function (evaluating to $1$ if the argument is positive and $0$ otherwise). It is known that&#xA;\begin{eqnarray}&#xA;\int \theta(x) \ x \ dx = \theta(x) \frac{x^{2}}{2} + C \quad \text{and} \quad \int \theta(-x) \ x \ dx = \theta(-x) \frac{x^{2}}{2} + C^{\prime},&#xA;\end{eqnarray}&#xA;where $C$ and $C^{\prime}$ are constants of integration. The identity for $n = 1$ follows by subtraction and the representation of $|x|$ above. With $n$ integrations, we have&#xA;\begin{eqnarray}&#xA;\int \cdots \int |x| \ dx = \theta(x) \int \cdots \int x \ dx  - \theta(-x) \int \cdots \int x \ dx = \frac{|x| x^{n}}{(n+1)!} + P_{n},&#xA;\end{eqnarray}&#xA;where $P_{n}$ is a polynomial in $x$, as claimed.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-11T20:08:36.847" />
  <row Id="9903" PostTypeId="2" ParentId="9877" CreationDate="2010-11-11T20:09:48.447" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It's a special case of Wilson's theorem for groups - see &lt;a href=&quot;http://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1/9345#9345&quot;&gt;my answer here&lt;/a&gt; - which highlights the key role played by symmetry (here a negation reflection / involution). Namely, since your set is closed under negation, its non-fixed points $\rm -k\not\equiv k\:$ pair up and contribute zero to the sum, leaving only the sum of the fixed points $\rm - k\equiv k\ \iff\ 2\:k\equiv 0\:,\ $ so $\rm\ k\equiv 0\ $ if $\rm\: n\:$ is odd, else $\rm\ k \equiv 0,\ n/2\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-12T15:55:10.930" LastActivityDate="2010-11-12T15:55:10.930" />
  <row Id="9904" PostTypeId="2" ParentId="9606" CreationDate="2010-11-11T20:22:07.407" Score="0" ViewCount="0" Body="&lt;p&gt;If the space of nxn and kxk matrices are viewed as vector spaces, then functions mapping between them would be called &quot;operators&quot;. That's probably not as specific as you want since it doesn't take into account the matrix structure, only the vector space structure.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-11T20:22:07.407" />
  <row Id="9905" PostTypeId="2" ParentId="9896" CreationDate="2010-11-11T20:24:19.073" Score="12" ViewCount="0" Body="&lt;p&gt;In general there are two ways to make foundations of category theory precise. The first one is to make use of &lt;a href=&quot;http://en.wikipedia.org/wiki/Grothendieck_universe&quot; rel=&quot;nofollow&quot;&gt;Grothendieck universes&lt;/a&gt;, which is just a way to limit the number of sets we are considering. So when we say the category of sets (groups, modules, ...) is implicit that we are restricting ourselves to some universe which is a set in the usual sense. This approach isn't all nice, since we cannot prove in ZFC that every set $S$ belongs to some universe $\mathcal{U}$, so we need to take it as an axiom. However people seems to prefer this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Von_Neumann-Bernays-Godel_set_theory&quot; rel=&quot;nofollow&quot;&gt;Neumann–Bernays–Gödel set theory&lt;/a&gt; (NBG). It is an axiomatic set theory where class is the primitive concept. Then we say that a class $S$ is a set if there is a class $C$ such that $A \in C$. Thus a set is a particular kind of class and a class that is not a set is called proper class. Using this axiomatic, the collection of all sets is a proper class. So when we say the category of sets (rings, topological spaces, ...) we have a proper class of objects. However we cannot speak of the class of all classes and we introduce the name small category to refer to categories with a &lt;strong&gt;&lt;em&gt;set&lt;/em&gt;&lt;/strong&gt; of objects. Then we can speak of the category of all small categories and functors between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NBG does not mess the other things we already have in mathematics since it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Conservative_extension&quot; rel=&quot;nofollow&quot;&gt; conservative extension &lt;/a&gt; of ZFC. Meaning that all theorems we already know in ZFC are still valid in NBG and if we prove a theorem in NBG using only the language of ZFC then it is a theorem in ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you don't need to worry too much with these things. In my opinion they are very technical for a first course and in most cases they are not really important. Just treat your categories in the same naive way almost everyone treats sets and remember that things can be done precisely.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-16T00:49:04.417" LastActivityDate="2010-11-16T00:49:04.417" />
  <row Id="9906" PostTypeId="2" ParentId="9825" CreationDate="2010-11-11T20:27:27.203" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a sketch of my idea for proving that $g$ is continuous:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Show that $s$ and $t$ are witnesses that $Y$ is not $T_2$ iff any eventually nonconstant net converging to $s$ or $t$ converges to &lt;em&gt;both&lt;/em&gt; $s$ and $t$.&lt;/li&gt;&#xA;&lt;li&gt;Observe that $g$ is continuous on $E$ (why?).&lt;/li&gt;&#xA;&lt;li&gt;Show that $g$ is continuous at $s$ by using the convergent net definition as follows: Let $(x_\alpha)$ be a net that converges to $s$. We split into cases:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(x_\alpha)$ is eventually constant. Here we need to use that $Y$ is $T_1$. Because of this, $x_\alpha$ is eventually equal to $s$, so $g(x_\alpha)$ is eventually equal to $t$ and hence $g(x_\alpha) \to t = g(s)$. &lt;/li&gt;&#xA;&lt;li&gt;$(x_\alpha)$ is not eventually constant and converges to $s$. By the first subresult, we have $g(x_\alpha) \to t = g(s)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(NB! My topology class used Munkres' &quot;Topology&quot; which outside of a few exercises never mentions nets. So it might be wrong and full of holes to fill out)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a lot of time trying to figure out some cleverer $X$ than the one you came up with, but it was really hard. It seems obvious though that it has to be constructed from $Y$ somehow. Some of my ideas involved starting with $2^Y$, i.e. the set of indicator functions on $Y$, giving it the product topology and identifying with $\mathcal{P}(Y)$ and from there you can pass to a topology on $E = \lbrace U \cap V \mid U \in \mathcal{U}_s, V \in \mathcal{V}_t \rbrace$, but that effectively boils down to what you did. It doesn't even avoid choice as you still have to use that to define $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea I tried was to consider only the case where $Y$ is infinite and then constructing something using the cofinite topology, but that too didn't pan out.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-11T20:27:27.203" CommentCount="2" />
  <row Id="9907" PostTypeId="2" ParentId="9879" CreationDate="2010-11-11T20:46:02.327" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm \ \sqrt a - a \ =\ \sqrt a\ \:(1 - \sqrt a)\ &gt; 0\ $ since both factors are $&gt; 0$  on the interval $(0,1)$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-11T20:46:02.327" />
  <row Id="9908" PostTypeId="2" ParentId="9896" CreationDate="2010-11-11T20:50:07.333" Score="6" ViewCount="0" Body="&lt;p&gt;If you base your mathematics on Set Theory, then you run into the same foundational issues in Category Theory as you do in Set Theory (since Categories will have to be modeled &quot;with&quot; Set Theory in some sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can go the other way, though: you can base your mathematics with Category Theory, and develop the Category of Sets through it, as Lawvere does &lt;a href=&quot;http://www.tac.mta.ca/tac/reprints/articles/11/tr11abs.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Then your primitive notions, the bedrock of your theory, is not Set Theory but Categories. There are precise ways in which you can talk about &quot;Category of all categories&quot;, analogous to the way in which you can talk about &quot;the class of all sets&quot; if you start with Set Theory instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know there are no problems lurking in this viewpoint? Well, we don't, just like we don't know that there are no serious problems lurking in Set Theory (e.g., that it is not consistent). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-11T20:50:07.333" CommentCount="1" />
  <row Id="9909" PostTypeId="2" ParentId="9898" CreationDate="2010-11-11T20:53:52.640" Score="2" ViewCount="0" Body="&lt;p&gt;if you take $\mu(u)=\sigma(u)=\alpha+u^2$ then the image diffusion is the same whatever $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="2905" LastActivityDate="2010-11-11T20:53:52.640" />
  <row Id="9910" PostTypeId="2" ParentId="154" CreationDate="2010-11-11T21:03:43.360" Score="2" ViewCount="0" Body="&lt;p&gt;I usually say: &quot;Believe it or not, electricity and radio waves actually &lt;strong&gt;do&lt;/strong&gt; behave like complex numbers. You don't see that in high school, but electric and electronics engineers do.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2612" LastActivityDate="2010-11-11T21:03:43.360" />
  <row Id="9911" PostTypeId="1" AcceptedAnswerId="9913" CreationDate="2010-11-11T21:20:10.393" Score="4" ViewCount="141" Body="&lt;p&gt;We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s&gt;1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was curious to see till what extent I can push the denominator so that it will still diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s&gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the same argument as previous). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't make sense to me though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were to converge, then I should be able to find a $s_1 &gt; 1$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I make sense of (or) interpret this result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am assuming that my convergence and divergence conclusions are right).&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-11T21:54:40.240" LastActivityDate="2010-11-11T23:37:56.930" Title="$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n\log^sn}$" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="9913" PostTypeId="2" ParentId="9911" CreationDate="2010-11-11T21:36:09.030" Score="5" ViewCount="0" Body="&lt;p&gt;Yes $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^s}$ is convergent if $\displaystyle s &gt; 1$, we can see that by comparing with the corresponding integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your other question, if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does not imply $\log n$ grows faster than a power of $\displaystyle n$. You cannot compare them term by term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that the first &quot;few&quot; terms of the series dominate (the remainder goes to 0). For a small enough $\displaystyle \epsilon$,  we have that $\log n &gt; n^{\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\log n$ to dominate the other.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T22:08:08.337" LastActivityDate="2010-11-11T22:08:08.337" CommentCount="1" />
  <row Id="9915" PostTypeId="2" ParentId="9911" CreationDate="2010-11-11T22:01:23.423" Score="3" ViewCount="0" Body="&lt;p&gt;This is a special case of classic general logarithmic convergence tests. Some early work in asymptotics was motivated by attempts to determine the &quot;boundary of convergence&quot; in terms of various functions, e.g. log-exp functions. Below is a a very interesting excerpt from Hardy's classic &quot;A course in pure mathematics&quot;. For further work see his &quot;Orders of infinity&quot; and see especially this very interesting &lt;a href=&quot;http://www.gfisher.org/fisher%20-%20the%20infinite%20and%20infinitesimal%20quantities%20of%20du%20bois-reymond%20and%20their%20reception.pdf&quot; rel=&quot;nofollow&quot;&gt;paper by G. Fisher&lt;/a&gt; on Paul du Bois-Reymond's work on the boundary between convergence and divergence (where he discovered &lt;strong&gt;diagonalization&lt;/strong&gt; before Cantor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note in particular the following very general result that is presented in the excerpt: the following series and integral are convergent if $\rm\ s &gt; 1\ $ and divergent if $\rm\: s\le 1\ $ where $\rm\: n_0,\ a\ $ are any numbers large enough to ensure positivity of $\rm\ log_k x := \log \log \cdots \log x\:,\ $ iterated $\rm\:k\:$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \sum_{n_0}^\infty \frac{1}{n \log n\ \log_2 n\ \cdots\ \log_{k-1} n\ (\log_k n)^s } $$&#xD;&#xA;&#xD;&#xA;$$\rm \int_{a}^\infty \frac{dx}{x \log x\ \log_2 x\ \cdots\ \log_{k-1} x\ (\log_k x)^s } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cChMu.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/5Ge4P.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/VUS3e.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/S5njR.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/FEDXp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-11T23:37:56.930" LastActivityDate="2010-11-11T23:37:56.930" />
  <row Id="9916" PostTypeId="1" AcceptedAnswerId="9920" CreationDate="2010-11-11T22:25:03.430" Score="2" ViewCount="117" Body="&lt;p&gt;I know this means that there are two cosets. I also know that one must be H itself.&#xA;This means that the remaining coset(s) must be equal for right and left. Also since there is only one possible coset, this means that for all elements, a,b in G, then aH = bH and Ha = Hb, since each element in G acting on the subgroup must produce the same set. This means aH = Ha I believe. From there H = aHa-1, which given the information I am not sure if this good or how to use it. Any help would good.&lt;/p&gt;&#xA;" OwnerUserId="3417" LastActivityDate="2010-11-11T22:49:28.630" Title="H is a subgroup of index 2 in finite group G, then every left coset of H is a right coset as well" Tags="&lt;group-theory&gt;&lt;group&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="9917" PostTypeId="2" ParentId="9911" CreationDate="2010-11-11T22:30:52.047" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;This is an expansion of my comment above.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{a}^{+\infty }\frac{1}{x(\log x)^{p}}dx$ is convergent for $a&gt;1$, $p&gt;1$&#xA;and divergent for $a&gt;1$, $p\leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$I(t)=\int_{a}^{t}\frac{1}{x(\log x)^{p}}dx=\frac{(\log t)^{1-p}-(\log&#xD;&#xA;a)^{1-p}}{(1-p)(\log a)^{p-1}}.$$&#xD;&#xA;&#xD;&#xA;(The integral can be evaluate by parts or by the substitution $u=\log x$.)&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;If $p&gt;1$ and $a&gt;1$, $I(t)\rightarrow 0$, as $t\rightarrow \infty $.&#xD;&#xA;Therefore $\int_{a}^{+\infty }\frac{1}{x(\log x)^{p}}dx$ is convergent. If $%&#xA;p\leq 1$ and $a&gt;1$, $I(t)\rightarrow \infty $, as $t\rightarrow \infty $ and &#xD;&#xA;$\int_{a}^{+\infty }\frac{1}{x(\log x)^{p}}dx$ is divergent.&#xD;&#xA;&#xD;&#xA;The series&#xD;&#xA;&#xD;&#xA;$$\sum_{n=2}^{\infty }\frac{1}{n(\log n)^{s}}$$&#xD;&#xA;&#xD;&#xA;behaves like $\int_{2}^{+\infty }\frac{1}{x(\log x)^{s}}dx$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The improper integral is exercise 2 on page 640 0f Angus Taylor's Advanced Calculus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0NIrW.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-11T23:24:15.903" LastActivityDate="2010-11-11T23:24:15.903" />
  <row Id="9918" PostTypeId="2" ParentId="9916" CreationDate="2010-11-11T22:32:38.427" Score="2" ViewCount="0" Body="&lt;p&gt;You say &quot;Also since there is only one possible coset, this means that for all elements, a,b in G, then aH = bH and Ha = Hb, since each element in G acting on the subgroup must produce the same set.&quot; but this is not true.  Let a be the identity and b not be in H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But earlier you say &quot;I know this means that there are two cosets. I also know that one must be H itself.&quot;  And if all cosets are the same size, what can you say about the one that isn't H?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-11T22:32:38.427" />
  <row Id="9919" PostTypeId="2" ParentId="9911" CreationDate="2010-11-11T22:35:09.970" Score="3" ViewCount="0" Body="&lt;p&gt;Du Bois-Reymond was interested in this question, positing some sort of series &quot;between&quot; all convergent series and all divergent series. Hausdorff didn't like that idea, and showed that given any countable sequences of convergence series converging &quot;ever more slowly&quot; and divergent series diverging &quot;ever more slowly&quot;, you can find a convergent and a divergent series &quot;in between&quot;. Here the order relation is the one given by the (conclusive) ratio test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bill's remark is related to the recursive Elias codes. The first few codes correspond to the convergent series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{n(\log n)^2}, \frac{1}{n\log n (\log \log n)^2}, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then diagonalizes to obtain his $\omega$ code (Elias omega code, if you want to look it up), which corresponds to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{n\log n\log \log n \cdots (\log^* n)^2},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\log^* n$ is the number of times you need to apply $\log$ until you get below some constant (the other logs continue up to this constant); confusingly, coding theorists use $\log^* n$ to mean the $\log n \log\log n \cdots$ part. We can continue the iteration like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{n\log n\log\log n \cdots \log^*n (\log \log^* n)^2}, \frac{1}{n\log n\log\log n \cdots \log^*n \log \log^* n (\log\log \log^* n)^2}, \cdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on. We can diagonalize again to obtain an $\omega + \omega$ code, and so on at least for $n \omega + m$. We an probably continue even to $\omega^2$ and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding divergent series are&#xA;$\displaystyle \frac{1}{n\log n}, \frac{1}{n\log n\log\log n}, \ldots, \frac{1}{n\log n\log\log n \cdots \log^* n}, \frac{1}{n\log n\log\log n \cdots \log^* n \log\log^* n}, \ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can ask whether there is a \emph{scale} for convergent series, that is a chain of (mutually comparable) series converging more and more slowly, such that for each convergent series there's one in the chain converging even more slowly; you can ask the same question about divergent series. Surprisingly, the existence of these is independent of ZFC (they exist given CH, and models in which they don't exist can be constructed using forcing).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-11T22:35:09.970" CommentCount="2" />
  <row Id="9920" PostTypeId="2" ParentId="9916" CreationDate="2010-11-11T22:41:17.443" Score="3" ViewCount="0" Body="&lt;p&gt;The (left) cosets of $H$ partition the group $G$; since there are two cosets and one of them is $H$, that means that, whatever the other one is (call it $\mathcal{C}$), &lt;em&gt;as a set&lt;/em&gt;, you have that $H\cap\mathcal{C}=\emptyset$ and $H\cup&#xD;&#xA;\mathcal{C}=G$. These two together tell you that the other one &lt;strong&gt;must&lt;/strong&gt; be equal, as a &lt;em&gt;set&lt;/em&gt;, to the complement of $H$ inside of $G$; that is, $\mathcal{C}=G-H$. So the two cosets are $H$ and $G-H$. Of course, $G-H$ can be written as $xH$ for &lt;em&gt;some&lt;/em&gt; $x$; in fact, for any $x\notin H$ you have $xH = G-H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the exact same argument holds for &lt;em&gt;right&lt;/em&gt; cosets: one of them is $H$, and the other one must be, as a set, the complement of $H$. So the two right cosets are $H$ and $G-H$, and $G-H$ can be written as $Hy$ for some $y$; in fact, for any $y\notin H$ you have $Hy=G-H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that in both cases, the cosets are:&#xA;    1. $H$, the coset of the elements of $H$; and&#xA;    2. $G-H$, the coset of the elements not in $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each of the two left cosets is also a right coset and vice-versa.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-11T22:49:28.630" LastActivityDate="2010-11-11T22:49:28.630" CommentCount="1" />
  <row Id="9922" PostTypeId="1" CreationDate="2010-11-11T23:22:42.497" Score="3" ViewCount="136" Body="&lt;p&gt;Find all continuous functions on R satisfying f(x), that f(x)=f(sinx) for all x.&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-11T23:32:13.287" LastActivityDate="2010-11-12T07:37:50.637" Title="Find all continuous functions on R satisfying f(x), that f(x)=f(sinx) for all x." Tags="&lt;functional-equations&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="9923" PostTypeId="2" ParentId="9922" CreationDate="2010-11-11T23:28:19.657" Score="2" ViewCount="0" Body="&lt;p&gt;For each $x$, $\sin(\sin(\sin...\sin(x)...))$ (n-times) goes to zero as $n$ goes to infinity, so by continuity &lt;strike&gt; f(x)=0 &lt;/strike&gt; $f(x)=f(0)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-11T23:28:19.657" CommentCount="2" />
  <row Id="9924" PostTypeId="2" ParentId="9922" CreationDate="2010-11-11T23:28:38.517" Score="9" ViewCount="0" Body="&lt;p&gt;$f(x)$ must be constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must have that if $x_1 = x$ and $x_{n+1} = \sin x_n$, then $f(x_{n+1}) = f(x_n) = f(x_{n-1}) \dots = f(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $x_n \to 0$ as $n \to \infty$. Thus $f(x) = f(0)$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-11T23:28:38.517" />
  <row Id="9925" PostTypeId="2" ParentId="154" CreationDate="2010-11-11T23:36:09.973" Score="2" ViewCount="0" Body="&lt;p&gt;I have just discovered this web page and your question.  This is probably a late response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people think that at some point someone came along and simply asserted &quot;yes there is a square root of -1 after all&quot; and called it 'i'. That's a common misconception, and, unfortunately, seems to be what is often &quot;taught&quot; in &quot;high schools&quot; in the US.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of Real Numbers that you are used to is an example of what mathematicians call a &quot;field&quot;. That means you can add, subtract, multiply, and divide according to familiar properties. The field C contains the Reals: it is just the two-dimensional plane endowed with a very natural addition (vectors) and a really cool multiplication, which takes a little effort to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complex number is just a point on the plane, an ordered pair. The absolute value of a complex number is its distance to the origin, and let's call its &quot;angle&quot; the angle it forms with the positive x-axis. The Real numbers are just the x-axis, and &quot;i&quot; is just (0,1). So the real number 1 is (1,0) and -1 is (-1,0). Then multiplying complex numbers multiplies their absolute values and adds the angles. That's why (0,1) times itself is -1 = (-1,0). 90 degrees plus 90 degrees = 180 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is essentially no other way to imbed R (the reals) in a field in which all numbers have square roots. If you want to be technical, any other such field is isomorphic to C. Furthermore, there is no way to make any higher R^n (R^3 = space for example) into a field in a natural way. The quaternions are a way to make R^4 into a division ring, close but not a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in short, the field of complex numbers is extremely real and concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last March Steven Strogatz wrote a column about C = the field of complex numbers in the New York Times. You might enjoy this exchange of emails I had with someone a few months later, after Strogatz's article appeared in the New York Times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://home.bway.net/lewis/Complex.htm&quot; rel=&quot;nofollow&quot;&gt;http://home.bway.net/lewis/Complex.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1837" LastActivityDate="2010-11-11T23:36:09.973" />
  <row Id="9927" PostTypeId="1" CreationDate="2010-11-12T00:23:58.440" Score="0" ViewCount="60" Body="&lt;p&gt;Hi all,&#xA;I wanted help in problem 7.4 of  Gilbarg and Trudinger . &#xA;The question says( in my language) that prove the following product formula for differentiation&lt;br&gt;&#xA;D(uv) = uDv + vDu for all u , v  in W^1{\omega} such that uv, uDv + vDu are locally integrable.&#xA;Some help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3423" LastActivityDate="2010-11-12T00:23:58.440" Title="product rule ( sobolev functions)" Tags="&lt;real-analysis&gt;" CommentCount="1" />
  <row Id="9928" PostTypeId="1" CreationDate="2010-11-12T00:47:19.327" Score="2" ViewCount="73" Body="&lt;p&gt;In the derived category $D(C)$ of an abelian category $C$, one formally inverts quasi-isomorphisms.  In the context of model categories, one inverts weak equivalences.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does one gain by doing so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a big-picture way to think about what is being done?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-12T00:47:19.327" Title="Yoga of localization in categories?" Tags="&lt;category-theory&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="9929" PostTypeId="2" ParentId="8536" CreationDate="2010-11-12T01:06:52.163" Score="1" ViewCount="0" Body="&lt;p&gt;The short answer, in my opinion, is that counting is one of the first things you should try when you approach a problem and it looks even remotely possible to use.  If counting doesn't work, &lt;em&gt;then&lt;/em&gt; you should try all your other tools.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T01:06:52.163" />
  <row Id="9930" PostTypeId="2" ParentId="8415" CreationDate="2010-11-12T01:32:51.243" Score="1" ViewCount="0" Body="&lt;p&gt;Here's an idea that might lead somewhere.  Let $F(x) = \sum_{n \ge 0} \frac{f(n)}{n!} x^n$, let $g(n) = \frac{1}{2^n} \sum_{k=0}^{n} {n \choose n-k} f(k)$, and let $G(x) = \sum_{n \ge 0} \frac{g(n)}{n!} x^n$.  Then it's not hard to see that $G(x) = e^{\frac{x}{2}} F \left( \frac{x}{2} \right)$, so we are looking for functions $F$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) - e^{\frac{x}{2}} F \left( \frac{x}{2} \right) = A(x)$$&#xD;&#xA;&#xD;&#xA;for some $A(x)$ with Taylor coefficients tending to some nonzero limit.  Let $H(x) = e^x F(x)$.  Now we want &#xD;&#xA;&#xD;&#xA;$$H(x) - H \left( \frac{x}{2} \right) = e^{-x} A(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition is easier to work with on the left now but harder to work with on the right.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T01:32:51.243" CommentCount="1" />
  <row Id="9931" PostTypeId="1" CreationDate="2010-11-12T01:52:46.237" Score="0" ViewCount="44" Body="&lt;p&gt;Let $G = \hbox{proj.lim.}_{\alpha} \{ G_{\alpha} , \varphi^{\alpha}_{\beta} \}$ be a projective limit of simple groups (i.e., each $\varphi^{\alpha}_{\beta}\colon G_{\alpha}\to G_{\beta}$ is a surjective group homomorphism between simple groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that if $\varphi^{\alpha}_{\beta}$ is not the trivial map ($x\mapsto 1$), then $\varphi^{\alpha}_{\beta}$ is an isomorphism. I believe $G$ is always a simple group. But I don't know how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="3426" LastActivityDate="2010-11-12T04:23:51.993" Title="projective limit of simple groups" Tags="&lt;group-theory&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9933" PostTypeId="1" AcceptedAnswerId="9942" CreationDate="2010-11-12T02:11:02.340" Score="6" ViewCount="395" Body="&lt;p&gt;Someone recently asked me why a negative * a negative is positive, and why a negative * a positive is negative, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went ahead and gave them a proof by contradiction like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume (-x) * (-y) = -xy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divide both sides by (-x) and you get (-y) = y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have a contradiction, then our first assumption must be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing I did something wrong here...since the conclusion of (-x) * (-y) = (xy) is hard to derive from what I wrote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better way to explain this?  Is my proof incorrect?  Also, what would an intuitive way to explain the negation concept be, if there is one?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-11-15T06:41:49.017" Title="Why negative times negative = positive?" Tags="&lt;teaching&gt;" AnswerCount="11" CommentCount="4" />
  <row Id="9934" PostTypeId="1" CreationDate="2010-11-12T02:14:17.193" Score="7" ViewCount="170" Body="&lt;p&gt;Cyclic groups have at most one subgroup of any given finite index. Can we describe the class of all groups having such property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3427" LastActivityDate="2010-11-28T15:01:03.837" Title="Groups having at most one subgroup of any given finite index" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="9935" PostTypeId="1" CreationDate="2010-11-12T02:21:20.957" Score="5" ViewCount="116" Body="&lt;p&gt;I know a proof that $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof goes as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a prime $p$ divides $(15n)!(10n)!(6n)!$, then the power of the prime dividing $(15n)!(10n)!(6n)!$ is lesser than the $(30n)!n!$. It is relatively easy to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{My question is there a counting argument to prove that this is an integer?}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this, I mean does $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ count something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have a gut feeling that this should count something but I have not thought in depth about this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance $\frac{n!}{r!(n-r)!}$ counts the number of ways of choosing $r$ objects out of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there other examples similar to this? (I tried searching for other such examples in vain)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-12T02:46:47.287" Title="Does $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ count something?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="9936" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:34:40.947" Score="0" ViewCount="0" Body="&lt;p&gt;This come from ring axioms. (As Qiaochu pointed out, we do not need the Field Axioms for this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First prove $0 \times y = 0$, $\forall y$ in the ring and where $0$ is the additive identity in the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then prove $(-x) \times y = - (x \times y)$, $\forall x,y$ in the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use these two and then manipulate to get the result $(-x) \times (-y) = x \times y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you run into any difficulty.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-12T02:51:27.670" LastActivityDate="2010-11-12T02:51:27.670" CommentCount="2" />
  <row Id="9937" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:36:43.920" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a proof.  First, for all $x$, $x\cdot 0=x\cdot(0+0)=x\cdot 0 +x\cdot 0$.  Subtracting $x\cdot0$ from each side, $x\cdot0=0$.  Now, for all $x$ and $y$, $0=x\cdot0=x\cdot(-y+y)=x\cdot(-y)+x\cdot y$.  Subtracting $x\cdot y$ from both sides, $x\cdot(-y)=-(x\cdot y)$.  Applying this twice along with the identity $-(-a)=a$, $(-x)\cdot(-y)=-(-x)\cdot y=-(-(x\cdot y))=x\cdot y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible.  These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(x\cdot y)$ rather than simply $-xy$ to not be confused with $(-x)\cdot y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being &quot;$5$ in the other direction&quot;, and so multiplying by $-5$ means &quot;multiply by $5$ and switch direction&quot;, i.e., sign.  This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-12T02:36:43.920" CommentCount="1" />
  <row Id="9938" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:37:50.750" Score="1" ViewCount="0" Body="&lt;p&gt;I think the &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; get in the way a bit; you can see the crucial steps using just 1 and -1.  What you've really shown is that (-1)(-1)=-1 leads to a contradiction.  If we divide by -1, we get -1=1, which is not true!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the right answer, (-1)(-1)=1, uses a couple more steps:  First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0).  Now, we simplify the parts we know to get (-1)(-1)+(-1)=0.  Solve for (-1)(-1), and you get (-1)(-1)=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One physical explanation people often like for negative*negative=positive is multiplying rates.  You can film someone walking forwards (positive rate) or walking backwards (negative rate).  Now, play the film back, but in reverse (another negative rate).  What do you see if you play a film backwards of someone walking backwards?  You see the person walking forwards, because negative*negative=positive!&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-12T02:37:50.750" />
  <row Id="9939" PostTypeId="1" AcceptedAnswerId="9953" CreationDate="2010-11-12T02:38:24.027" Score="8" ViewCount="139" Body="&lt;p&gt;$H$ is a subgroup of $G$ and $a,b$ are in $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was rather lost for a while but I think I may have&#xA;actually proved it, though I am unsure. . Does this make sense? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$aH = bH$ &#xA;$\Rightarrow$ $b$ is an element of $aH$ &#xA;$\Rightarrow$ there exists an $h$ in $H$ s.t. $ah = b$&#xA;$\Rightarrow$ $a^{-1}ah = a^{-1}b$ &#xA;$\Rightarrow$ $h = a^{-1}b$&#xA;$\Rightarrow$ $hb^{-1} = a^{-1}bb^{-1}$&#xA;$\Rightarrow$ $hb^{-1} = a^{-1}$&#xA;This means $a^{-1}$ is an element of $Hb^{-1}$, so they are equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wasn't quite sure I could do those operations.&#xA;Can anyone tell me where to learn to do the special symbols, like quantifiers and relations b/c it would make it simpler.&lt;/p&gt;&#xA;" OwnerUserId="3417" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T03:37:36.290" LastActivityDate="2010-11-12T04:57:21.617" Title="If $aH = bH$ then  $Ha^{-1} = Hb^{-1}$, prove or find a counter example." Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9940" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:40:03.480" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ x\ y\ =\ (-x)\ (-y)\ $ by rewriting $\rm\ x\ y + x\ (-y) + (-x)\ (-y)\ $ in two different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, $\rm\ x\ y,\ (-x)\ (-y)\ $ are both inverses of $\rm\ x\ (-y)\ $ so they're equal by &lt;strong&gt;uniqueness of inverses&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I've &lt;a href=&quot;http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint/3514#3514&quot;&gt;emphasized before&lt;/a&gt;, &lt;strong&gt;uniqueness theorems&lt;/strong&gt; are powerful tools for proving equalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally the &lt;strong&gt;Law of Signs&lt;/strong&gt; holds for any &lt;strong&gt;odd&lt;/strong&gt; functions &#xA;under composition, e.g. polynomials with all terms having odd power. Indeed we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ (-f)\ (-g)\  =\ -(f(-g))\ =\ f(g)\ \iff\ f(-g)\ = -(f(g))\ =\ (-f)(g) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm \phantom{\ \ (-f)\ (-g)\  =\ -(f(-g))\ =\ f(g)\ }\iff \ f(-x)\ = -f(x)\:,\ $ ie. $\rm\:f\:$ odd, $\ $ via $\rm\ g = x = $ identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally such functions enjoy only a weaker &lt;a href=&quot;http://en.wikipedia.org/wiki/Nearring&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;near&lt;/em&gt;-ring&lt;/a&gt; structure. &#xA;In above ring case, distributivity implies that multiplication &#xA;is linear hence odd (here one views the ring in Cayley-style &#xA;as the ring of endormorphisms of its abelian additive group, &#xA;i.e. represent each ring elt $\rm\ r\ $ by the linear map  $\rm\ x \to r\ x\:,\ $ &#xA;i.e. as a 1-dim matrix).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-12T03:00:44.017" LastActivityDate="2010-11-12T03:00:44.017" CommentCount="1" />
  <row Id="9941" PostTypeId="1" CreationDate="2010-11-12T02:40:48.850" Score="2" ViewCount="98" Body="&lt;p&gt;I'm trying to find a general solution for $\displaystyle\iint_{S}[[x+y]] dA$ where $S=[a,b] \times [c,d]$ and $[[x]]$ is the greatest integer function. I know that if $a,b,c,d \in \mathbb{Z}$, then the answer is just $\frac{(d-c)(b-a)(a+b+c+d-1)}{2}$ (proven below), but how do I account for the extra volume when $a,b,c,d\in \mathbb{R}$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By looking at the function itself, it's clear that the volume of the region above $S=[a,a+1]\times[b,b+1]$ is $a+b+\frac{1}{2}$ (the volume of the rectangular prism is $a+b$ while the volume of the triangular prism on top of it is $\frac{1}{2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the total volume is:&#xA;$$\displaystyle\sum_{y=c}^{d-1}\sum_{x=a}^{b-1} x+y+\frac{1}{2}=\sum_{y=c}^{d-1}\frac{b^2-a^2}{2}+(b-a)y=\frac{(d-c)(b-a)(a+b+c+d-1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other question I'd like to confirm is that $\displaystyle\underset{S}{\int \cdots \int} [[x_1+x_2+\cdots+x_n]] dx_1 \cdots dx_n$ over $S=[i_1,i_2] \times [i_3,i_4] \times \cdots \times [i_{2n-1},i_{2n}]$ is:&#xA;$$\displaystyle\frac{1}{2}\left(\sum_{j=1}^{2n}i_j\right)\left(\prod_{k=1}^{n}(i_{2k}-i_{2k-1})\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true if $i_k\in\mathbb{Z}$ for all $k\in[0,n]$? Can this, too, be extended to all reals?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-11-12T03:45:49.650" LastActivityDate="2010-11-13T20:32:13.310" Title="How to find a closed form solution to $\displaystyle\iint_{S}[[x+y]] dA$" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="9942" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T02:42:58.317" Score="4" ViewCount="0" Body="&lt;p&gt;Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you'll have to try to picture what could (-2)*(-3) mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this will help but it's the only way I can think of this in some intuitive sense.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-12T02:42:58.317" />
  <row Id="9943" PostTypeId="2" ParentId="9935" CreationDate="2010-11-12T02:46:47.287" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/26336/integer-valued-factorial-ratios&quot; rel=&quot;nofollow&quot;&gt;This MO thread&lt;/a&gt; has a lot of good information.  I think we should be pessimistic.  Pietro Majer's answer and the comments therein suggest that this would be hard, at least in the sense that one should not expect a proof anything like the proof for binomial coefficients.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T02:46:47.287" CommentCount="1" />
  <row Id="9944" PostTypeId="2" ParentId="9939" CreationDate="2010-11-12T03:06:39.307" Score="4" ViewCount="0" Body="&lt;p&gt;$aH=bH$ does not imply that $a$ or $b \in H$.  It just implies they are in the same coset.  It does imply that $b^{-1}a \in H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the nice formatting, enclose your formulas in dollar signs.  If you right click on a formula, there is a menu item Show Source that will show you the text (without the dollar signs) that created it.  Double dollar signs get display mode.  When you are typing, an interpreted version is visible below.  Handy references for LaTeX are &lt;a href=&quot;http://web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.artofproblemsolving.com/Wiki/index.php/LaTeX%3aCommands&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T03:06:39.307" />
  <row Id="9945" PostTypeId="1" CreationDate="2010-11-12T03:09:03.317" Score="1" ViewCount="77" Body="&lt;p&gt;Why is the following inequality true: if $x \geq 0$ then $(1-\frac{x}{n})^{n} \leq e^{-x}$ ? here $n$ is a positive integer. Is there a quick way to see this?&lt;/p&gt;&#xA;" OwnerUserId="3398" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T04:05:03.053" LastActivityDate="2010-11-12T04:05:03.053" Title="$(1-\frac{x}{n})^n\lt \exp(-x)$" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="9946" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:15:12.207" Score="0" ViewCount="0" Body="&lt;p&gt;You can look at it this way:&#xA;$(1 - \frac{x}{n})^n = \sum_{k=0}^{n} C(n,k) (-1)^k \frac{x^k}{n^k}$&#xA;and&#xA;$e^{-x} = \sum_{k=0}^{+\infty} (-1)^k \frac{x^k}{k!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it clearer now ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(hint: one of the sums is finite, the other not, and one of the terms grows quicker than the other)&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-11-12T03:15:12.207" CommentCount="2" />
  <row Id="9947" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:20:50.500" Score="1" ViewCount="0" Body="&lt;p&gt;You might want to prove this for $0 \leq x \leq n$. Else it might not be always true. (For instance, consider the case when $x$ is really large and $n$ is even)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $f(y) = e^{y} \times (1-y)$. Note that $f(0) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now prove that $f(y)$ is a decreasing function $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{df(y)}{dy} = e^{y} - ye^{y} - e^{y} = -ye^{y} \leq 0$, $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $f(y)$ is a decreasing function $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have $f(y) \leq f(0) = 1$, $\forall y \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $e^{y} \times (1-y) \leq 1$, which implies $(1-y) \leq e^{-y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace $y = \frac{x}{n}$ and raise both sides to the $n^{th}$ power where $n \in \mathbb{R}^{+}$. The inequality remains the same since $0 \leq x \leq n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we get $(1-\frac{x}{n})^n \leq e^{-x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also useful to see that $\displaystyle \lim_{n \rightarrow \infty} (1-\frac{x}{n})^n = e^{-x}$ from below.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-12T03:40:13.557" LastActivityDate="2010-11-12T03:40:13.557" CommentCount="1" />
  <row Id="9948" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:24:33.913" Score="1" ViewCount="0" Body="&lt;p&gt;consider the case where $x = 4$ and $n=2$. Here, LHS &gt; RHS. So the inequality does not always hold.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-12T03:24:33.913" />
  <row Id="9949" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T03:25:31.387" Score="3" ViewCount="0" Body="&lt;p&gt;The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 X (-40) = -120$ is added to your account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite of being billed would be billing someone else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you bill 3 people for 40 dollars each, $(-3) X (-40)$ is added to your account. This value should be positive since it results in you receiving money. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-12T03:25:31.387" />
  <row Id="9950" PostTypeId="2" ParentId="9939" CreationDate="2010-11-12T03:25:45.770" Score="1" ViewCount="0" Body="&lt;p&gt;You're on the right track. $aH = bH$ is equivalent to $a^{-1}b \in H$, and $Ha^{-1} = Hb^{-1}$ is equivalent to $a^{-1} (b^{-1})^{-1} = a^{-1}b \in H$, which is the same thing. One approach that should work would be to turn your arguments into statements of this nature. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T03:25:45.770" />
  <row Id="9951" PostTypeId="1" CreationDate="2010-11-12T03:29:40.600" Score="-1" ViewCount="143" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/9141/about-irrational-logarithms&quot;&gt;About irrational logarithms&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Please help proving that $\log_{10}(2)$ is irrational.&lt;/p&gt;&#xA;" OwnerUserId="3430" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T03:39:38.677" LastActivityDate="2010-11-14T20:15:10.407" ClosedDate="2010-11-15T07:48:25.783" Title="prove $\log_{10} (2)$ is irrational .. " Tags="&lt;logarithms&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9952" PostTypeId="2" ParentId="9945" CreationDate="2010-11-12T03:32:02.597" Score="0" ViewCount="0" Body="&lt;p&gt;First thing: you need $n$ to be even for this if  1 - x/n is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement is the same as $1 - {x \over n} \leq e^{-{x \over n}}$. Letting $t = {x \over n}$ this follows from $1 - t \leq e^{-t}$. This can be proven using calculus.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T03:32:02.597" CommentCount="2" />
  <row Id="9953" PostTypeId="2" ParentId="9939" CreationDate="2010-11-12T03:37:03.403" Score="2" ViewCount="0" Body="&lt;p&gt;For formatting, see &lt;a href=&quot;http://meta.math.stackexchange.com/questions/107/what-should-go-in-the-math-stackexchange-faq/255#255&quot;&gt;here&lt;/a&gt; and some of the links there. You need to know a bit of LaTeX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your argument, it is (mostly) correct. My only quibbles would be of presentation. From $aH=bH$ you conclude that there must exist $h\in H$ such that $b=ah$. That's fine. Your next step should be that &quot;there exists $h\in H$ such that $a^{-1}b=a^{-1}ah = h$&quot;. Then &quot;There exists $h\in H$ such that $a^{-1}=a^{-1}bb^{-1}=hb^{-1}$&quot;, hence $a^{-1}$ is in the coset $Hb^{-1}$. Your final statement &quot;hence they are equal&quot; is technically wrong, since $a^{-1}$ is not equal to the coset $Hb^{-1}$; what you &lt;em&gt;meant&lt;/em&gt; to say is that the since $a^{-1}$ is in $Hb^{-1}$, then the coset $Ha^{-1}$ is equal to the coset $Hb^{-1}$. Other that these quibbles, the argument seems correct to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to see if you understand it well, try to figure out which, if any, of your implications are reversible, to see if the &lt;em&gt;converse&lt;/em&gt; also holds.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T04:57:21.617" LastActivityDate="2010-11-12T04:57:21.617" />
  <row Id="9954" PostTypeId="2" ParentId="9951" CreationDate="2010-11-12T03:40:35.663" Score="5" ViewCount="0" Body="&lt;p&gt;Hint: if $10^{a/b} = 2$ then $10^a = 2^b$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-12T03:40:35.663" />
  <row Id="9956" PostTypeId="2" ParentId="9951" CreationDate="2010-11-12T03:41:37.940" Score="0" ViewCount="0" Body="&lt;p&gt;Prove by contradiction. So start by assuming&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{10}{2}=\frac{m}{n}$$, where $m$ and $n$ are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10^\frac{m}{n}=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10^m=2^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and derive a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-12T03:41:37.940" CommentCount="2" />
  <row Id="9957" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T03:47:56.043" Score="3" ViewCount="0" Body="&lt;p&gt;Quite a good explanation is that one wants the distributive law to work in general with &lt;em&gt;positive&lt;/em&gt; quantities when you add (smaller) negative ones:&#xA;If $x&gt;a\ge0$ and $y&gt;b\ge0$ then&#xA;$$&#xD;&#xA;(x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b)&#xD;&#xA;$$&#xA;For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-12T03:47:56.043" />
  <row Id="9958" PostTypeId="1" CreationDate="2010-11-12T03:49:28.350" Score="6" ViewCount="389" Body="&lt;p&gt;I would like this to be a community wiki. But, I don't know how to do that here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know sources, and examples of good &quot;challenge&quot; problems for students who have studied pre-calculus and some calculus. (differentiation and the very basics of integration.) Topics could be related to things such as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Taylor Series.&lt;/li&gt;&#xA;&lt;li&gt;Product Rule, Quotient Rule, Chain Rule.&lt;/li&gt;&#xA;&lt;li&gt;Simple limits.&lt;/li&gt;&#xA;&lt;li&gt;Delta Epsilon Proofs.&lt;/li&gt;&#xA;&lt;li&gt;Induction proofs for the sum of the first n, integer, squares, etc.&lt;/li&gt;&#xA;&lt;li&gt;Integration by substitution.&lt;/li&gt;&#xA;&lt;li&gt;Other topics...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What I have found so far are too many problems that are just a bit too difficult. The problem can have a &quot;trick&quot; but it needs to be something a freshman could do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one problem that I thought was just at the right level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x) = \frac{x}{x+\frac{x}{x+ \frac{x}{x+ \vdots}}}$, find $f'(x)$*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*To be honest this problem makes me a little nervous. Still, I like it.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-12T19:26:59.030" CommunityOwnedDate="2010-11-12T08:33:04.083" Title="Interesting calculus problems of medium difficulty?" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;&lt;big-list&gt;" AnswerCount="16" CommentCount="2" FavoriteCount="5" />
  <row Id="9959" PostTypeId="2" ParentId="9934" CreationDate="2010-11-12T03:54:05.687" Score="13" ViewCount="0" Body="&lt;p&gt;Since $G$ has exactly one subgroup of each finite index, and the index of a conjugate of $H$ equals the index of $H$, then every subgroup of finite index is normal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is finite, then &lt;em&gt;every&lt;/em&gt; subgroup is normal, so the group must be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_group&quot; rel=&quot;nofollow&quot;&gt;Dedekind group&lt;/a&gt; (also known as Hamiltonian groups). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All such groups that are nonabelian are of the form $G = Q_8 \times B \times D$, where $Q_8$ is the quaternion group of $8$ elements, $B$ is a direct sum of copies of the cyclic group of order $2$, and $D$ is an abelian group of odd order. Any of the factors may be missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Q_8$ contains several subgroups of index $4$, if a factor of $Q_8$ appears then $G$ would have several subgroups of the same index, hence $G$ must in fact be an abelian group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $G$ is finite and abelian, it is isomorphic to a direct sum of cyclic groups, $G = C_{a_1}\oplus\cdots\oplus C_{a_k}$, where $1\lt a_1|a_2|\cdots|a_k$. If $k\gt 1$, then $G$ contains at least two subgroups of order $a_{k-1}$; thus $k=1$ so $G$ is in fact cyclic. So the only finite groups with the desired property are the cyclic groups.&#xA;If $G$ is infinite, you can have other possibilities. One example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pr%C3%BCfer_group&quot; rel=&quot;nofollow&quot;&gt;Prufer group&lt;/a&gt;, &lt;strong&gt;Added:&lt;/strong&gt; but only by vacuity: it has no proper subgroups of finite index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $H$ if a subgroup of finite index in $G$ then $H$ is normal, as above, and $G/H$ also has the desired property and is finite; thus, $G/H$ is cyclic for every subgroup of finite index by the argument above. I'm sure there's more to be said, but I'll think about it a bit first...&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T02:32:26.113" LastActivityDate="2010-11-13T02:32:26.113" CommentCount="11" />
  <row Id="9960" PostTypeId="1" AcceptedAnswerId="9973" CreationDate="2010-11-12T03:54:37.590" Score="1" ViewCount="80" Body="&lt;p&gt;How does one show for complex numbers u and v, and for p&gt;1 that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$||u|^{p-1}u - |v|^{p-1}v|\leq c_p |u-v|(|u|^{p-1}+|v|^{p-1})$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_p$ is some constant dependent on p.  My intuition is to use some version of the mean value theorem with $F(u) = |u|^{p-1}u$, but I'm not sure how to make this work for complex-valued functions.  Plus there seems to be an issue with the fact that $F$ may not smooth near the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, this shows up in Terry Tao's book Nonlinear Dispersive Equations: Local and Global Analysis on pg. 136, where it is stated without proof as an &quot;elementary estimate&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1692" LastActivityDate="2010-11-12T05:34:05.237" Title="Complex inequality $||u|^{p-1}u - |v|^{p-1}v|\leq c_p |u-v|(|u|^{p-1}+|v|^{p-1})$" Tags="&lt;complex-numbers&gt;&lt;inequality&gt;" AnswerCount="2" />
  <row Id="9961" PostTypeId="2" ParentId="9951" CreationDate="2010-11-12T03:57:01.947" Score="0" ViewCount="0" Body="&lt;p&gt;We argue by contradiction. Suppose $\log_{10} 2 = \frac{p}{q}$ is rational with $q &amp;gt; p &amp;gt; 0$. Then $ 2^{q} = 10^{p} = 2^{p} \ 5^{p}$, so $2^{q - p} = 5^{p}$. Since $q &amp;gt; p$ and $p &amp;gt; 0$, it follows that $2^{q - p} \equiv 0 \mod 5$, which is impossible since no power of $2$ ends in $0$ or $5$. Hence $\log_{10} 2$ is irrational. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method generalizes easily to prove the irrationality of many reals of the form $\log_{b} \ a$ for $a, b \in \mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-14T20:15:10.407" LastActivityDate="2010-11-14T20:15:10.407" CommentCount="2" />
  <row Id="9963" PostTypeId="2" ParentId="9931" CreationDate="2010-11-12T04:14:49.053" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose that for every $\beta$, there is $\alpha \geq \beta$ such that $\phi_{\beta}^{\alpha}$ is trivial.  Then the projective limit is trivial.&#xA;(Because if $(x_{\beta})$ is an element of the projective limit, then&#xA;for each $\beta$, we have $x_{\beta} = \phi_{\beta}^{\alpha}(x_{\alpha}) = 1,$&#xA;if we choose $\alpha$ for the given $\beta$ so that $\phi_{\beta}^{\alpha}$ is&#xA;trivial --- as we may, by assumption.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose that for some $\beta$, the maps $\phi_{\beta}^{\alpha}$&#xA;are isomorphisms for all $\alpha \geq \beta$.  If $A$ is the directed set indexing the projective limit, then replacing $A$ by its subset $A_{\geq \beta}&#xD;&#xA;:= \{\alpha \in A \, | \ \alpha \geq \beta\}$ doesn't change the projective&#xA;limit (since $A_{\geq \beta}$ is cofinal in $A$), and now all the transition&#xA;maps $\phi_{\gamma}^{\alpha}$ for $\alpha, \gamma \in A_{\geq \beta}$ are isomorphisms. (Since each of $\phi_{\beta}^{\gamma}$ and  $\phi_{\beta}^{\alpha} = \phi_{\beta}^{\gamma}\circ \phi_{\gamma}^{\alpha}$ is an isomorphism, by assumption.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the projective limit of the $G_{\alpha}$ (for $\alpha \in A_{\geq &#xD;&#xA;\beta}$) has all its transition maps being isomorphisms, and so is isomorphic to $G_{\beta}$,&#xA;and hence is simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the projective limit is either trivial or simple, depending on whether one is in the first or second case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added in response to Arturo Magidin's comment below: I hadn't paid attention to the fact that the transition maps are assumed surjective.  This implies that in my first case, all the $G_{\beta}$ are trivial (!), since they are (by assumption) the image of some $G_{\alpha}$ under the trivial map.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unless all the $G_{\beta}$ are trivial, then one must be in the second case, when the $\phi^{\alpha}_{\beta}$ are all eventually isomorphisms.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-12T04:23:51.993" LastActivityDate="2010-11-12T04:23:51.993" CommentCount="6" />
  <row Id="9964" PostTypeId="1" AcceptedAnswerId="9969" CreationDate="2010-11-12T04:23:26.490" Score="3" ViewCount="53" Body="&lt;p&gt;Given a sequence $(x_n)$ in a metric space $M$, call it &lt;em&gt;uniformly separated&lt;/em&gt; if all pairwise distances $d(x_n,x_m)$ between distinct terms are uniformly bounded away from zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that a given sequence $(x_n)$  has the property that it has no Cauchy subsequence. Must it have a uniformly separated subsequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of my students asserted this in homework without watertight proof, and proof or disproof is eluding me. Clearly any counterexample cannot be in ${\mathbb R}^n$. Enjoy!&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-12T04:33:05.353" Title="If no Cauchy subsequence exists, must a uniformly separated subsequence exist?" Tags="&lt;convergence&gt;&lt;metric-spaces&gt;" AnswerCount="1" />
  <row Id="9965" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T04:24:02.363" Score="1" ViewCount="0" Body="&lt;p&gt;I recommend the following sources (big advantage: free access):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Mathematical Reflections: &lt;a href=&quot;http://awesomemath.org/mathematical-reflections/&quot; rel=&quot;nofollow&quot;&gt;http://awesomemath.org/mathematical-reflections/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Through the problem column, we challenge students to develop their creative problem solving and reasoning skills by devising solutions to the proposed questions. Exceptional solutions will be published, with the intent of encouraging students to formally write out and submit their work to be showcased in print&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) School Science and Mathematics: &lt;a href=&quot;http://ssmj.tamu.edu/problems.php&quot; rel=&quot;nofollow&quot;&gt;http://ssmj.tamu.edu/problems.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This section offers readers an opportunity to exchange interesting mathematical problems and solutions. Proposals are always welcomed...&#xA;The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-12T04:24:02.363" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9966" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T04:27:51.170" Score="2" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://people.missouristate.edu/lesreid/POTW.html&quot; rel=&quot;nofollow&quot;&gt;Missouri State problems&lt;/a&gt; are often very good&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T04:27:51.170" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9967" PostTypeId="2" ParentId="9760" CreationDate="2010-11-12T04:29:17.103" Score="3" ViewCount="0" Body="&lt;p&gt;Here's another direction of attack.  The expected number of steps before a single, given counter reaches $N$ is exactly $3N^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in I. J. Kennedy's approach, let $x_k$ denote the expected time before a given counter reaches $N$ given that the counter starts $k$ steps away from $N$.  Then the $x_k$'s satisfy, for $1 \leq k \leq N-1,$&#xA;$$x_k = \frac{1}{6} (1 + x_{k+1}) + \frac{4}{6} (1 + x_k) + \frac{1}{6} (1 + x_{k-1}),$$&#xA;with boundary conditions&#xA;$x_0 = 0$ and $x_N = \frac{2}{3} (1 + x_N) + \frac{1}{3}(1 + x_{N-1}),$ the second of which simplifies to $x_N = x_{N-1} + 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference equation reduces to $$x_{k+1} - 2x_k + x_{k-1} = -6,$$ and thus is linear, nonhomogeneous, and second-order.  The solution technique is very similar to that for the corresponding differential equation.  The characteristic equation for the homogeneous difference equation is $r^2 -2r + 1 = 0$, which means we have the double root $r = 1$.  Thus the general solution to the homogeneous equation is $y_k = A(1)^k + B k (1)^k = A + Bk$.  We then use the method of undetermined coefficients with $z_k = Ck^2$ as our candidate to find a particular solution to the nonhomogeneous equation.  This yields $C = -3$.  Thus the general solution to the difference equation is &#xA;$$x_k = A + Bk - 3k^{2}.$$&#xA;Now, we use the boundary conditions to find $A$ and $B$.  These yield $A = 0$ and &#xA;$$A + BN - 3N^2 = A + B(N-1) - 3(N-1)^2 + 3.$$&#xA;Solving this, we have $B = 6N$.  Thus our solution is &#xA;$$x_k = 6Nk - 3k^2,$$&#xA;which means &#xA;$$x_N = 3N^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have time right now to pursue this further, but maybe someone else can use this to find a reasonable upper bound for the time until all three counters hit $N$ simultaneously like the OP wants.  In particular, though, if it takes on the order of $N^2$ steps on average for a given counter to reach $N$ for the first time it seems that it should take more than the order of $N^3$ steps on average for all three of them to hit $N$ together.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-12T04:29:17.103" />
  <row Id="9968" PostTypeId="1" CreationDate="2010-11-12T04:29:19.270" Score="4" ViewCount="74" Body="&lt;p&gt;Let we have a competitive survival game in which a player has choice between different resources to earn. The question here is which resource should he prefer to maximize the chance of survival. I tried to find out if there already established measures of such sort, but so far without success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some thoughts led me to the following measure of the resource' value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=\int_{t_0}^{t_1} e^{\int_0^t \log (p_1(u))du}\log \frac{p_1(t)}{p_0(t)} dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_0(t)$ is the probability density of survival without the resource, $p_1(t)$ is the probability density of survival with the resource and $(t_0,t_1)$ is the period of time through which the resource affects the probability of survival (the time for which the value is evaluated considered t=0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if there any similar measures already proposed and about the possible drawbacks of this proposed value. I also wonder to which extent this can be applied to real economics.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2010-11-12T10:32:37.817" LastActivityDate="2010-11-12T10:32:37.817" Title="Measure of value of resources in a competitive game" Tags="&lt;probability&gt;&lt;game-theory&gt;&lt;economics&gt;" CommentCount="1" />
  <row Id="9969" PostTypeId="2" ParentId="9964" CreationDate="2010-11-12T04:33:05.353" Score="3" ViewCount="0" Body="&lt;p&gt;A proof can be inferred from the answers to an earlier question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/7210/which-metric-spaces-are-totally-bounded/&quot;&gt;Which metric spaces are totally bounded?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-11-12T04:33:05.353" CommentCount="1" />
  <row Id="9970" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T04:37:19.490" Score="1" ViewCount="0" Body="&lt;p&gt;Variational calculus is always fun (for example, derive from Newton's law the shape of a chain left hanging from its two ends, you end up with cosh) and is a nice introduction to analytical mechanics. Or pursuit problems (http://mathworld.wolfram.com/PursuitCurve.html) were a real challenge, but were quite enjoyable. However, those a more dealing with differential equations. &lt;/p&gt;&#xA;" OwnerUserId="1034" LastActivityDate="2010-11-12T04:37:19.490" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9971" PostTypeId="2" ParentId="9760" CreationDate="2010-11-12T05:04:31.830" Score="5" ViewCount="0" Body="&lt;p&gt;Here's a sketch of another derivation of the hitting time for one counter (re Mike's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the usual $\pm 1$ random walk on the integer line. It is well known that the expected time it takes to reach distance $N$ is $N^2$. In our case reaching distance $N$ and reaching the value $N$ modulo $2N$ is the same, since we can't get to $\pm 3N$ without moving through $\pm N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We move the specified counter in expectation once every three steps, and therefore it's $3N^2$ and not $N^2$ (because the direction of movement and this &quot;waiting for our turn&quot; are independent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument shows that the case of one counter is rather special.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T05:04:31.830" CommentCount="2" />
  <row Id="9972" PostTypeId="2" ParentId="9960" CreationDate="2010-11-12T05:18:07.613" Score="1" ViewCount="0" Body="&lt;p&gt;Without loss, you can assume $|u|\le|v|$ and replace $u$ by $uv$ to reduce the problem to the situation $v=1$ and $|u|\le1$. Unless you need $c_p$ explicitly, then it's clear, yes?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2010-11-12T05:18:07.613" />
  <row Id="9973" PostTypeId="2" ParentId="9960" CreationDate="2010-11-12T05:34:05.237" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose without loss of generality that $|u| \geq |v| &gt; 0$. Then you can divide the equation through by $|v|^p$ and your task it to prove $||w|^{p-1}w - 1| \leq c_p|w - 1|(|w|^{p-1} + 1)$, where $w = u/v$. Note that &#xA;$$||w|^{p-1}w - 1| =  ||w|^{p-1}w - |w|^{p-1} + |w|^{p-1} - 1| $$&#xA;$$\leq ||w|^{p-1}w - |w|^{p-1}| + ||w|^{p-1} - 1|$$&#xA;Note the first term is $|w|^{p-1}|w - 1|$ is automatically bounded by your right hand side. So you're left trying to show that $||w|^{p-1} - 1|$ is bounded by your right hand side. For this it suffices to show that&#xA;$$||w|^{p-1} - 1| \leq c_p||w| - 1|| (|w|^{p-1} + 1)$$&#xA;Since $|w| \geq 1$ by the assumption that $|u| \geq |v|$, it suffices to show&#xA;that for all real $r \geq 1$ one has&#xA;$$r^{p-1} - 1 \leq c_p(r - 1)(r^{p-1} + 1)$$&#xA;Now use the mean value theorem as you originally wanted to.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T05:34:05.237" CommentCount="1" />
  <row Id="9974" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:41:12.580" Score="3" ViewCount="0" Body="&lt;p&gt;I like this problem, and my students seem to like it also. Technically, they should use density of the rationals, but you can skirt around this.(At least the solution I am thinking of).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and related:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Show a function is linear iff $f(x)=f(x-a)+f(x-b)-f(x+a+b)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-12T05:41:12.580" CommunityOwnedDate="2010-11-12T08:33:04.083" CommentCount="2" />
  <row Id="9975" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:44:45.697" Score="2" ViewCount="0" Body="&lt;p&gt;Another one that is kinda silly, but a little fun is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^{\pi/4}f\left(x\right)dx=\int_0^{\pi/8}f\left(x\right)dx=\int_0^{\pi/16}f\left(x\right)dx=\int_0^{\pi/32}f\left(x\right)dx=\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f\left(\sqrt[3]{t}e^t\right)=-\left(3t\right)f\left(\sqrt[3]{t}e^t\right)$ for infinately many $t$.&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-12T05:44:45.697" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9976" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:48:36.097" Score="1" ViewCount="0" Body="&lt;p&gt;You got me started...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume $f\left(x\right)$ is a function ith a continuous derivative and $f\left(1\right)=0$. Show $\exists c\in\mathbb{R}$ such that $c^2f'\left(c\right)+2cf\left(c\right)=0.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-11-12T05:48:36.097" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9977" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T05:51:59.607" Score="1" ViewCount="0" Body="&lt;p&gt;Ok last one for a while, I promise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably all know this one as it is quite famous:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:\left[0,1\right]\rightarrow\left[0,1\right]$, then define &lt;em&gt;a point of period n,&lt;/em&gt; for $n\in\mathbb{N}$ to be a point $x_1$ such that &#xA;            \begin{equation}&#xA;                f^n\left(x_1\right)=\underbrace{f\circ f\circ\dots\circ f}_{n-times}\left(x_1\right)=x_1 ,&#xA;            \end{equation}&#xA;        and that for no $m$ less than $n$, &#xA;$f^m\left(x_1\right)=x_1$. That is, $n$ is the smallest natural number such that $f^n\left(x_1\right)=x_1$. So a point, $x_1$, has period $2$ if  $f\left(f\left(x_1\right)\right)=f\left(x_2\right)=x_1$ and $x_1\neq x_2$. Now define a &lt;em&gt;fixed point of f&lt;/em&gt; to be a point $x_0$ such that $f\left(x_0\right)=x_0$; i.e. it has period one.   Show that if $f$ has a point of period $2$, then it has a fixed point.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-12T11:27:20.750" LastActivityDate="2010-11-12T11:27:20.750" CommunityOwnedDate="2010-11-12T08:33:04.083" />
  <row Id="9978" PostTypeId="1" AcceptedAnswerId="9979" CreationDate="2010-11-12T06:02:41.353" Score="0" ViewCount="162" Body="&lt;p&gt;I have been sick for my real analysis class and I am a having a tough time getting the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain the what open, closed and compact sets are and how they are related with uniform continuity? I think the hinging point was a poor understanding of accumulation points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3417" LastActivityDate="2010-11-12T21:00:39.260" Title="Real Analysis Continuity" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9979" PostTypeId="2" ParentId="9978" CreationDate="2010-11-12T06:25:23.400" Score="5" ViewCount="0" Body="&lt;p&gt;I assume you're talking about open, closed, and compact subsets of a metric space.  If you're just talking about the real line $\mathbb{R}$, then ignore the commentary about abstract metric spaces and just replace the term &quot;neighborhood&quot; with &quot;open interval.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In metric spaces, a set $O$ is &lt;strong&gt;open&lt;/strong&gt; if every point in $O$ is an interior point of $O$.  In other words, every point of $O$ can be surrounded by an open ball (neighborhood) which itself lies in $O$.  So, for instance, the open intervals $(a,b)$ are open subsets of $\mathbb{R}$, and discs $\{x^2 + y^2 \lt R^2 \}$ are open subsets of $\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one thing to keep in mind is that in more abstract metric spaces, the open sets can be a little strange-looking.  For instance, there are metric spaces where every subset is an open set!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, a set $F$ is &lt;strong&gt;closed&lt;/strong&gt; if every accumulation point of $F$ is in $F$.  Accumulation points are points $x$ such that every open neighborhood around $x$ must intersect the set in question.  So, for instance, the accumulation points of the open disc $\{x^2 + y^2 \lt R^2\}$ are precisely the set of points in the closed disc $\{x^2 + y^2 \leq R^2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about closed sets is that they are closed under the operation of taking limits.  That is, if one were to take a sequence of points in a closed set, then their limit would still have to be in the set.  (This can be made rigorous and should be proven at some point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with open sets, the closed sets of a metric space can still be rather strange-looking, and again, there are examples of spaces in which every subset is closed.  Nevertheless, it is always true that finite subsets of metric spaces are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note also that if $(X,d)$ is a metric space, then the sets $\emptyset$ and $X$ are always going to be both open and closed.  In general, the subsets of a metric space may be open, closed, both or neither.  (So just because a set is not open does not make it closed!)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Compact sets are a little less intuitive.  The rigorous definition is that a set $K$ is &lt;strong&gt;compact&lt;/strong&gt; if every open cover of $K$ has a finite subcover.  Since I don't want to babble for too long, I won't go into what this means in detail.  However, I've always intuited compact sets as being &quot;super closed&quot; sets -- they're like closed sets, but even stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fact (whose proof you should learn or create) that every compact subset of a metric space is closed and bounded, but the converse is not necessarily true.  In the euclidean spaces $\mathbb{R}^n$, the converse is, in fact, true, but this is not obvious, and is known as the Heine-Borel Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike open and closed sets, in a metric space $(X,d)$, the set $X$ may not be compact.  Therefore, it makes sense to talk about &quot;compact metric spaces,&quot; whereas it doesn't make sense to talk about &quot;open&quot; or &quot;closed&quot; metric spaces (since they're all open and closed subsets of themselves).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uniform continuity&lt;/strong&gt; is a stronger condition than continuity.  That is, every uniformly continuous function is continuous, but not necessarily conversely.  It is a theorem (Heine-Cantor) that on compact sets, the converse will hold.  So on compact sets, continuity is equivalent to uniform continuity.  Again, this is an important theorem whose proof you should learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, uniform continuity of a function $f$ on a set $E$ can be thought of as saying that the &quot;degree of continuity&quot; is the same for every point $x$ in the set $E$.  This can be made slightly more precise by saying that the $\delta &gt; 0$ that one gets in the $\epsilon-\delta$ definition of continuity can be chosen to be the same for all $x$ in $E$ -- that is, it will depend only on $\epsilon$, and not on the point $x$.  For plain old continuous functions, the $\delta$ will depend on both $\epsilon$ and $x$.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-11-12T06:40:04.380" LastActivityDate="2010-11-12T06:40:04.380" CommentCount="3" />
  <row Id="9980" PostTypeId="2" ParentId="9978" CreationDate="2010-11-12T06:28:42.787" Score="1" ViewCount="0" Body="&lt;p&gt;Difficult one to answer, given the numerous definitions of open sets. Also - more information on what you know, etc, would be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try not to be overly formal - someone will probably be more so than me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a point $x$ in a set. If for every point $x$ we can find another point arbitrarily close then this is an open set. From the standard wikipedia definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset $U$ of the Euclidean n-space $\mathbb{R}^n$ is called open if, given any point $x$ in $U$, there exists a real number $\epsilon &gt; 0$ such that, given any point $y$ in $\mathbb{R}^n$ whose Euclidean distance from $x$ is smaller than $\epsilon$, $y$ also belongs to $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (again see Wikipedia) you can replace the Euclidean space with any metric space, and come up with an equivalent definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closed sets are sets where the complement is open. One of the most confusing aspects is there exists sets that are both open and closed. These are known as Clopen sets. The definition of closed sets you are after may be the one involving accumulation points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition: $x$ is an accumulation points of $U$ if every $\epsilon$ neighborhood of $x$ contains a points ($\ne x$) in $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset of $U$ is closed if if it contains all its accumulation points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see Jesse has posted a nice answer - I will leave this as an alternative reference. &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-12T06:28:42.787" />
  <row Id="9981" PostTypeId="2" ParentId="9978" CreationDate="2010-11-12T06:31:19.930" Score="2" ViewCount="0" Body="&lt;p&gt;So I'm going to try to explain what you asked for, though you should probably get your book ASAP or reference a &lt;a href=&quot;http://kr.cs.ait.ac.th/~radok/math/mat6/calc12.htm#1.6%20Further%20Discussion%20of%20the%20Concept%20of%20Limit&quot; rel=&quot;nofollow&quot;&gt;free analysis book&lt;/a&gt; and Wikipedia, and probably most importantly, get someone to take notes for you in your class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point p is an accumulation/limit point of a set E if &lt;strong&gt;every&lt;/strong&gt; neighborhood around that point has an element of the set E not equal to p. Think of a neighborhood as an ball around a point. A neighborhood in $\mathbb{R}$ (the real line) is just an interval. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is closed if every limit point of a set E is inside E. An example of where you could have a limit point that is not an element of a set is given by the set $\mathbb{Q}$ (the rationals). You can build a series of rational numbers that converge to say for example $\pi$, and hence $\pi$ is a limit point of $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say if you have an any interval around $\pi$, you'll have an element of $\mathbb{Q}$. However, pi is not an element of $\mathbb{Q}$, hence the rationals aren't a closed set (Actually every irrational number is a limit point of $\mathbb{Q}$... you may want to think about why).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is open if every point of that set is an interior point. A point is an interior point if there exists at least one open ball around that point such that that open ball is contained entirely in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your usage, assuming that you'll be studying $\mathbb{R}$, a compact set is one which is closed and bounded. A set is bounded if you can but an open ball at its center that covers the entire set. Think [-1,1], with the ball centered at 0, and radius 1.&#xA;Note that (-1,1) is not closed, since both endpoints are limit points of the set, but not elements of the set (ie. you can have an infinite series that converges to 1, and has elements that are in (-1,1), meaning 1 is a limit point, yet 1 is not an element of the set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between compactness and uniform continuity is given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem&quot; rel=&quot;nofollow&quot;&gt;Heine-Cantor&lt;/a&gt;.&#xA;Hope that helps. &lt;/p&gt;&#xA;" OwnerUserId="2919" LastEditorUserId="2919" LastEditorDisplayName="" LastEditDate="2010-11-12T21:00:39.260" LastActivityDate="2010-11-12T21:00:39.260" CommentCount="1" />
  <row Id="9982" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T06:51:09.063" Score="7" ViewCount="0" Body="&lt;p&gt;If $C_0 + C_1/2 + \ldots + C_n/(n+1) = 0$, where each $C_i \in \mathbb{R}$, then prove that the equation $C_0 + C_1x + \ldots + C_nx^n = 0$ has at least one real root between 0 and 1.  (Rudin, Ch 5 Exercise 4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|f(x)| \leq |x|^2$ for all $x \in \mathbb{R}$, then prove that $f$ is differentiable at $x = 0$.  (Spivak?) (OK, so this one is easier than &quot;medium&quot; perhaps, but I like it anyway.  It illustrates nicely that growth conditions can have an impact on smoothness.)&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-11-12T06:51:09.063" CommunityOwnedDate="2010-11-12T08:33:04.083" CommentCount="1" />
  <row Id="9983" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T07:02:00.047" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose $\displaystyle f:\mathbb{R} \to \mathbb{R}$ and $\displaystyle g:\mathbb{R} \to \mathbb{R}$ are differentiable functions such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f'(x) = 3f(x) + 7g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g'(x) = 2f(x) + 3g(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find all such $f$ and $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: Consider $\displaystyle h(x) = 2g(x) - f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the level they are at, I suppose this is the right level of difficulty you are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little more advanced, they can differentiate again and construct a second order differential equation in $\displaystyle f$ (or $\displaystyle g$), which will work with arbitrary coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More advanced than that, they can consider a matrix differential equation.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-12T07:02:00.047" CommunityOwnedDate="2010-11-12T07:04:58.953" />
  <row Id="9984" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T07:04:11.247" Score="2" ViewCount="0" Body="&lt;p&gt;Since they know &quot;basic&quot; integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the integer part of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{10000} \dfrac{1}{\sqrt[4]{n}}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-12T07:04:11.247" CommunityOwnedDate="2010-11-12T07:05:14.337" CommentCount="1" />
  <row Id="9985" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T07:15:53.637" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose $\displaystyle f:[0,1] \to \mathbb{R}$ is a function such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\displaystyle g(t) = \lim_{x \to t} f(x) $ exists $\displaystyle \forall t \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\displaystyle f(x) + g(x) = 2x \ \ \forall x \in [0,1]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find all such $\displaystyle f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to realize that $\displaystyle g$ must be continuous (perhaps this is near the medium/difficult border).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extra credit, they can try showing that 1) is enough to show that there is at least one point where $\displaystyle f$ is continuous. (See this: &lt;a href=&quot;http://math.stackexchange.com/questions/3777/&quot;&gt;Is there a function with a removable discontinuity at every point?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it can be shown that if 1) is true, the set of discontinuities of $\displaystyle f$ is countable (See the same question above).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T07:21:08.537" LastActivityDate="2010-11-12T07:21:08.537" CommunityOwnedDate="2010-11-12T07:15:53.637" />
  <row Id="9986" PostTypeId="2" ParentId="9922" CreationDate="2010-11-12T07:37:50.637" Score="3" ViewCount="0" Body="&lt;p&gt;For those of you who wonder why the sequence $x_{n+1} = \sin x_n$ tends to zero, note that after the first application of sine, we get a number in $\pm 1$, and from there on, the numbers keep their sign, and monotonically approach zero (since $|\sin x| \leq |x|$). So the sequence must approach a limit, which is also a fixed point of the sine function. The only fixed point of sine is zero.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T07:37:50.637" />
  <row Id="9987" PostTypeId="1" CreationDate="2010-11-12T09:09:17.797" Score="1" ViewCount="109" Body="&lt;p&gt;I apologize for posting such an untechnical question, but with responses it could surely be posed in a better form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a math noob, but I've seen (as we all have) a few examples of &quot;connections&quot; between different areas of math by way of an equality or some other relating device.  An example is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e^iπ + 1 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These five constants serve different functions in a variety mathematical tasks.  It seems to me that any constant is a &quot;measurement&quot; of some kind, which then begs the question &quot;What is it a measurement of?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of some places to start?  If you have a &quot;bridge&quot; you like, post it.&lt;/p&gt;&#xA;" OwnerUserId="1825" LastActivityDate="2010-11-12T16:42:11.993" Title="Making meaning of mathematical &quot;bridges&quot;" Tags="&lt;philosophy&gt;&lt;meta-math&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9988" PostTypeId="2" ParentId="9891" CreationDate="2010-11-12T09:53:04.770" Score="1" ViewCount="0" Body="&lt;p&gt;Actually the books cited above by Sivaram are excellent for numerical stuff. If you want &quot;matrix calculus&quot; then the following books might be helpful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Matrix Differential Calculus with Applications in Statistics and Econometrics by Magnus and Neudecker&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Functions of Matrices by N. Higham&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculus on Manifolds by Spivak&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some classic, but very useful material can also be found in&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Introduction to Matrix Analysis by Bellman.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As a simple example, the books will teach (unless you already know it) how to compute, say, the derivative of $f(X) = \log\det(X)$ for an invertible matrix $X$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-12T09:53:04.770" />
  <row Id="9989" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T10:14:41.033" Score="1" ViewCount="0" Body="&lt;p&gt;This is a classic one due to Arnold (A Mathematical Trivium):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1 \over 2\pi} \int_{0}^{2\pi} \sin^{100}(x) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They should also try to find a numeric value (to within one-percent error, say).&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-12T10:14:41.033" CommunityOwnedDate="2010-11-12T10:14:41.033" />
  <row Id="9990" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T11:13:51.697" Score="1" ViewCount="0" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best site I have come across when it comes to these kind of problems, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/index.php&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="3438" LastActivityDate="2010-11-12T11:13:51.697" CommunityOwnedDate="2010-11-12T11:13:51.697" CommentCount="1" />
  <row Id="9991" PostTypeId="1" CreationDate="2010-11-12T11:54:06.670" Score="4" ViewCount="110" Body="&lt;p&gt;A graph G is planar if and only if xxx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can xxx be substituted for? Note that this is from a topological POV so a graph is a 1-dim cw complex and I guess the fundamental group should be used somehow. &lt;/p&gt;&#xA;" OwnerUserId="3439" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-12T12:02:16.113" LastActivityDate="2010-11-12T14:33:01.243" Title="When is a graph planar? " Tags="&lt;topology&gt;&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9992" PostTypeId="2" ParentId="9991" CreationDate="2010-11-12T11:59:44.587" Score="3" ViewCount="0" Body="&lt;p&gt;Please refer Kuratowski's theorem on Planar graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kuratowski:&lt;/strong&gt; A graph $G$ is &lt;em&gt;planar&lt;/em&gt; iff $G$ does not contain a sub division of $K_{5}$ or $K_{3,3}$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-12T12:07:04.483" LastActivityDate="2010-11-12T12:07:04.483" CommentCount="2" />
  <row Id="9993" PostTypeId="1" CreationDate="2010-11-12T12:22:46.150" Score="4" ViewCount="100" Body="&lt;p&gt;Solving a heat equation with central symmetry i got the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to find a sphere center temperature vs. time given that the surface of the sphere is kept constant at $$T_1$$ and at initial moment t=0 the temperature inside the sphere is distributed uniformly with value $$T_0$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution is following:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(t)=T_1+2 (T_0-T_1)\sum_{k=1}^\infty (-1)^{k+1} e^{-t(\frac{k\pi a}{r})^2}$$  &#xA;Here r is the radius of the sphere and a=const is a coefficient of thermal conductivity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the solution must satisfy the initial condition $$T(0)=T_0$$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(0)=T_1+2 (T_0-T_1)\sum_{k=1}^\infty (-1)^{k+1} $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is satisfied only if the following holds:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-1+1-1+1-1+ ... =1/2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we found this divergent sum  through physical arguments? Is this appropriate for the pure mathematician?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets go a step further and differentiate the solution by t:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {d T(t)}{dt}=-2 (T_0-T_1)(\frac{\pi a}{r})^2\sum_{k=1}^\infty (-1)^{k+1} k^2 e^{-t(\frac{k\pi a}{r})^2}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, at initial moment t=0 we have:     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {d T(0)}{dt}=-2 (T_0-T_1)(\frac{\pi a}{r})^2\sum_{k=1}^\infty (-1)^{k+1} k^2 $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously  $$\frac {d T(0)}{dt}=0$$  because  the maximum speed in the universe is limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the following must be valid:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1^2 - 2^2 + 3^2 - 4^2 + ... =0  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can say what has been said previously, i think. But at the moment i'm not sure about the following divergent sum:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - 2 + 3 - 4 + ... &lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-11-12T20:27:01.177" Title="Summing divergent series based on physics" Tags="&lt;sequences-and-series&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9994" PostTypeId="1" CreationDate="2010-11-12T12:30:13.440" Score="1" ViewCount="119" Body="&lt;p&gt;I am not good with vector spaces so I would be grateful for any help.&lt;br&gt;&#xA;As I've been told I need to take $v \in \mathrm{Im}(T)$, $v\neq 0$, and show that if $T(v) = \mathbf{0}$ then $T^2 = 0$. But if $T(v) \neq \mathbf{0}$, then need to show that if $\{e_1,\ldots,e_{n-1}\}$ is a basis of $\mathrm{Ker}(T)$, then $\{e_1,\dots,e_{n-1},v\}$ is a basis of $V$. Then deduce that $T$ is diagonalisable.  I don't really know how to do that though...&lt;/p&gt;&#xA;" OwnerUserId="2401" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T15:32:28.460" LastActivityDate="2010-11-12T15:32:28.460" Title="Let $V$ be a $k$-vector space of dimension $n$ and $T: V \to V$ a linear map of rank 1. Show that either $T^2 = 0$ or that $T$ is diagonalisable?" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="4" />
  <row Id="9995" PostTypeId="2" ParentId="9993" CreationDate="2010-11-12T12:35:23.260" Score="0" ViewCount="0" Body="&lt;p&gt;It was good enough for Euler and Ramanujan apparently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%C2%B7_%C2%B7_%C2%B7&quot; rel=&quot;nofollow&quot;&gt;1-2+3-4+...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1+2+3+4+...&quot; rel=&quot;nofollow&quot;&gt;1+2+3+4+...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-12T12:35:23.260" />
  <row Id="9996" PostTypeId="1" AcceptedAnswerId="9998" CreationDate="2010-11-12T12:46:38.293" Score="1" ViewCount="88" Body="&lt;p&gt;I'm trying to prove that $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is a divisible $\mathbb{Z}$-module (p is prime, and the direct sum and direct product are taken over the set of all primes). It is an exercise from Rotman, An Introduction to Homological Algebra. Here's what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigoplus \mathbb{Z_p}$ is the torsion submodule of $\prod \mathbb{Z_p}$, so the quotient is torsion-free. Since $\mathbb{Z}$ is a PID, then the quotient is flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can flatness help me prove divisibility? Well, since $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is flat then $Hom_\mathbb{Z} \left( \frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}, \frac{\mathbb{Q}}{\mathbb{Z}}\right)$, the character module, is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I don't know how to continue. I don't think this is the way to go, but it's what I've tried. Another thing I've observed is that again, since $\mathbb{Z}$ is a PID, then $\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is divisible iff it is injective, but once again, I don't know what to do with this.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-12T13:17:13.413" Title="$\frac{\prod \mathbb{Z_p}}{\bigoplus \mathbb{Z_p}}$ is a divisible abelian group" Tags="&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9997" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T13:10:41.460" Score="1" ViewCount="0" Body="&lt;p&gt;A great source of problems I have used in the past is, &quot;Problems in calculus of one variable&quot;, by I A Maron. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-12T13:10:41.460" CommunityOwnedDate="2010-11-12T13:10:41.460" />
  <row Id="9998" PostTypeId="2" ParentId="9996" CreationDate="2010-11-12T13:15:47.520" Score="7" ViewCount="0" Body="&lt;p&gt;what about the direct approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have $(a_i)\in \prod \mathbb{Z}_p$ and an integer n, then modulo $\bigoplus \mathbb{Z_p}$ you may assume that $a_p=0$ for all the primes dividing n. n is invertible in all the rest of $\mathbb{Z}_p$ so you can find $\frac{a_p}{n}$ in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So define $b_p = 0 $ for $p\mid n$ and $b_p = \frac{a_p}{n}$ for $p\not\mid n$ and then&#xA;$n (b_p)+\bigoplus \mathbb{Z_p} = (a_n) + \bigoplus \mathbb{Z_p}$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-12T13:17:13.413" LastActivityDate="2010-11-12T13:17:13.413" CommentCount="1" />
  <row Id="9999" PostTypeId="1" AcceptedAnswerId="10001" CreationDate="2010-11-12T13:17:52.120" Score="11" ViewCount="595" Body="&lt;p&gt;The standard way (other than generating up to $N$) is to check if $(5*N*N + 4)$ or $(5*N*N - 4)$ is a perfect square.  What is the mathematical logic behind this? Also, is there any other way for checking the same?  &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T21:41:17.843" LastActivityDate="2010-12-26T23:51:24.940" Title="Checking if a number is a Fibonacci or not?" Tags="&lt;number-theory&gt;&lt;fibonacci-number&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="2" />
  <row Id="10000" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T13:33:49.017" Score="0" ViewCount="0" Body="&lt;p&gt;EDITED after Prometheus's comment to use the characterization using sum of dimension of eigenspaces instead of counting eigenvalues. This argument should be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $\dim V = n \geq 1$. Consider $T$'s restriction to its image. This is a linear map on a 1-dimensional space, so we can write $T\vert_{T(V)}(v) = cv$ for some $c \in k$. There are now two possibilities. Either $c = 0$ or $c \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$c=0$: Then $T^2 = 0$ (why?) and all of $T$'s eigenvalues are $0$ since we must have $0 = T(T(v))  =T(\lambda v) = \lambda T(v) = \lambda^2 v$ for any eigenvalue $\lambda$ and any eigenvector $v$. We conclude that $T$ can't be diagonalizable since all of its eigenvalues are $0$, but the eigenspace corresponding to $0$ is $\ker T$ which has dimension $n-1$.&lt;/li&gt;&#xA;&lt;li&gt;$c \neq 0$: First note that clearly $T^2 \neq 0$. We show that the sum of the dimensions of $T$'s eigenspaces is $n$, so $T$ must be diagonalizable [&lt;a href=&quot;http://en.wikipedia.org/wiki/Diagonalizable_matrix&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;]. And this is very easy: $T(V)$ is a 1-dimensional eigenspace with eigenvalue $c$, and $\ker T$ is an n-1-dimensional eigenspace with eigenvalue $0$. Thus the sum of the dimensions of the eigenspaces is at least $n$, but it can't be greater than $n$, so it must be exactly $n$ (not 100% sure on this... LinAlg is a long time ago...)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-12T15:14:05.630" LastActivityDate="2010-11-12T15:14:05.630" CommentCount="2" />
  <row Id="10001" PostTypeId="2" ParentId="9999" CreationDate="2010-11-12T13:42:31.977" Score="9" ViewCount="0" Body="&lt;p&gt;Binet's formula tells you that $F_n$ is asymptotic to $\frac{1}{\sqrt{5}} \phi^n$, where $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio.  So to check if a number $x$ is a Fibonacci number you should compute $n = \frac{\log (\sqrt{5} x)}{\log \phi}$.  If this number is very close to an integer then $x$ should be very close to the Fibonacci number $F_{[n]}$ where $[n]$ denotes the closest integer to $n$.  In fact I think $x$ is a Fibonacci number if and only if the error is less than about $\frac{\log \phi}{5x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you don't trust that computation, you can compute $F_{[n]}$ in $O(\log n)$ time, for example by using &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponentiation_by_squaring&quot; rel=&quot;nofollow&quot;&gt;binary exponentiation&lt;/a&gt; to compute powers of the matrix $\left[ \begin{array}{cc} 1 &amp;amp; 1 \\\ 1 &amp;amp; 0 \end{array} \right]$, and then you can check whether this number equals $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first question can be answered using the theory of (generalized) &lt;a href=&quot;http://en.wikipedia.org/wiki/Pells_equation&quot; rel=&quot;nofollow&quot;&gt;Pell equations&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-12T14:33:36.313" LastActivityDate="2010-11-12T14:33:36.313" CommentCount="12" />
  <row Id="10002" PostTypeId="2" ParentId="9999" CreationDate="2010-11-12T13:58:53.067" Score="9" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ n\:$ is a Fibonacci number iff the interval $\rm\ [\phi\ n - 1/n,\ \phi\ n + 1/n]\ $ contains a positive integer (the next Fibonacci number for $\rm\ n &gt; 1\:$). This follows from basic properties of continued fractions. This is probably the most efficient test since it requires only a few multiplications and additions. $\ $ For example, $\rm\ 2 \to [2.7,\ 3.7],\ \ 3\to [4.5,\ 5.2],\ \ 5 \to [7.9,\ 8.3]\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-12T14:41:09.520" LastActivityDate="2010-11-12T14:41:09.520" CommentCount="4" />
  <row Id="10003" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T14:03:13.503" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose that $0\neq v\in Im(T)$, then $T(v)\in Im(T)$. $dim(Im(T))=1$ so $Im(T)=span(v)=F\cdot v$ and you get that there is a $\lambda$ such that $T(v)=\lambda v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T(v)=0$ then $T^2(V) \subseteq T(Im(T)) = T(F\cdot v)=0$, so $T^2 = 0$. Assume now that $T(v)\neq 0$, then $Im(T) \cap \ker(T) = {0}$ so a basis for the kernel (which is of dimension n-1) is a set of eigenvectors of eigenvalue 0 and $v$ is the n-th eigenvector of eigenvalue $\lambda\neq 0$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-12T14:03:13.503" CommentCount="7" />
  <row Id="10004" PostTypeId="2" ParentId="9991" CreationDate="2010-11-12T14:05:58.497" Score="2" ViewCount="0" Body="&lt;p&gt;There are lots of characterizations of planar graphs, e.g. Kuratowski's&#xA;theorem as already mentioned. Another is Whitney's theorem that a finite&#xA;graph $G$ is planar if and only if the dual matroid to the matroid of $G$&#xD;&#xA;is graphic (also comes from a graph).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-12T14:05:58.497" />
  <row Id="10005" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T14:24:40.770" Score="1" ViewCount="0" Body="&lt;p&gt;First, if the rank of T is 1, then the dimension of its kernel is n-1. That means 0 is an eigenvalue of order n-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as Prometheus said, you just have to find out whether you will find that last eigenvalue or not. To find it out, just take a vector $v \notin Ker(T)$ (thus, $v \in Im(T)$), and look at $T(v)$. Either you'll again fall on $0$. What does it mean ?&#xA;You have taken a vector from the only dimension not belonging to the kernel of $T$. It means it belongs to $Im(T)$. Any other vector there can be obtained by multiplying $v$ by some coefficient $t \in \mathbb{R}$ (or $\mathbb{C}$, ...). Since T is linear, the same will go for the image of $v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $T(v) = 0$, all the vectors from that dimension will fall on $0$ when going through $T$. So for any vector $u \in Im(T)$, $T(u) = 0$. That means that for any vector $v \in V$, $T(T(v)) = 0$, that is, $T^2(v) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $T(v)$ isn't zero, there will be some vector $u \in V$ (but different from $0$) such that $T(v) = u$. Thus, for any $t \in \mathbb{R}$, $T(t \cdot v) = t \cdot T(v) = t \cdot u$. And $u$ is in $Im(T)$ thus colinear to $v$. Now you have v being an eigenvector of $T$. Taking a basis of $Ker(T)$ and adding $v$ to it provides you a basis of $V$ on which $T$ is diagonalisable.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-11-12T14:24:40.770" CommentCount="4" />
  <row Id="10006" PostTypeId="2" ParentId="9991" CreationDate="2010-11-12T14:33:01.243" Score="1" ViewCount="0" Body="&lt;p&gt;A very good treatment of when a graph can be embedded in the plane and more generally into other surfaces is given by the excellent book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphs on Surfaces by Bojan Mohar and Carsten Thomassen (John Hopkins U. Press, 2001)&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-12T14:33:01.243" />
  <row Id="10007" PostTypeId="1" AcceptedAnswerId="10012" CreationDate="2010-11-12T15:02:12.513" Score="3" ViewCount="64" Body="&lt;p&gt;This question is a follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/9996/frac-prod-mathbbz-p-bigoplus-mathbbz-p-is-a-divisible-abelian-group&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm trying to prove that the torsion submodule of $\prod \mathbb{Z}_p$ is not a direct summand of $\prod \mathbb{Z}_p$. I found this interesting because it's an example that over a PID, a non-finitely generated module may fail to be decomposed as torsion $\oplus$ something free, showing even that it can't be decomposed as torsion $\oplus$ anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is easily proven that the torsion submodule of $\prod \mathbb{Z}_p$ is $\bigoplus \mathbb{Z}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we could decompose $\prod \mathbb{Z}_p$ as torsion $\oplus$ something it would have to be $\prod \mathbb{Z}_p = \bigoplus \mathbb{Z}_p \oplus \frac{\prod \mathbb{Z}_p}{\bigoplus \mathbb{Z}_p}$ (this is correct, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using the question linked above, we know the right summand is divisible ($\iff$ injective), besides it is torsion-free ($\iff$ flat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think) I've learned my lesson from my previous question, so I also tried proving that there is a member in $\prod \mathbb{Z}_p$ which can't be written as a sum of something in $\bigoplus \mathbb{Z}_p$ plus something in $\frac{\prod \mathbb{Z}_p}{\bigoplus \mathbb{Z}_p}$, but I couldn't advance any further in this direction either.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-12T15:25:59.360" Title="The torsion submodule of $\prod \mathbb{Z}_p$ is not a direct summand of $\prod \mathbb{Z}_p$" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;homological-algebra&gt;" AnswerCount="2" />
  <row Id="10008" PostTypeId="1" CreationDate="2010-11-12T15:02:18.363" Score="0" ViewCount="72" Body="&lt;p&gt;I have a problem that I cannot solve regarding interest payments, and payment holidays.&#xA;The scenario is this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I purchase a product which is paid for over a 12 month period, I need to calculate the monthly interest payments. However, the complexity here, is that I have the option to take one month payment holiday (is optional, but can be any month over the 12 month period).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;annual interest (a) = (e.g for 5%, a=5) &#xA;&#xA;monthly interest (m) = ((1 + a/100)^(1/12) - 1) x 100&#xA;&#xA;balance = balance*(1+m) - instalment&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the instalment, especially when including payment holiday?&lt;/p&gt;&#xA;" OwnerUserId="3442" LastActivityDate="2011-01-11T15:44:43.030" Title="Instalment APR Calculation Formula" Tags="&lt;finance&gt;" AnswerCount="1" />
  <row Id="10009" PostTypeId="2" ParentId="9994" CreationDate="2010-11-12T15:16:34.123" Score="1" ViewCount="0" Body="&lt;p&gt;Prometheus' answer is essentialy correct. I'd just like to point something out to you so that you can gain a little intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the rank of $T$ is one-dimensional, which means that there exists ($n-1$)-dimensional subspace $W \subset V$ such that $T(W) = 0$ (imagine this in three dimensions as a plane that is being sent to zero). Now pick any $u \in V$ not lying in $W$. We have $T(u) = v \neq 0$ (because otherwise the rank of $T$ would be zero). So what can we say about $v$? Well, it either lies in $W$ and therefore for any $u&amp;#39; \in V$, with decomposition $u&amp;#39; = u + w&amp;#39;$ for some $w&amp;#39; \in W$, we have $T^2(u&amp;#39;) = T^2(u + w&amp;#39;) = T^2(u) + T^2(w&amp;#39;) = T(v) + 0 = 0$ (because both $v$ and $w&amp;#39;$ are in $W$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume that $v \notin W$. But then we know $T(v) = \lambda v$ because the span of $v$ is one-dimensional. Therefore we can write $T = \lambda E_v \oplus 0_W$ which is the desired diagonalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try thinking about it in two dimensions and matrices and convince yourself that this problem is all about this:&#xA;$$\pmatrix{0&amp;amp;1\cr0&amp;amp;0} \quad \pmatrix{\lambda &amp;amp; 0\cr 0 &amp;amp;0}$$&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-12T15:16:34.123" CommentCount="1" />
  <row Id="10010" PostTypeId="1" AcceptedAnswerId="10015" CreationDate="2010-11-12T15:22:06.720" Score="7" ViewCount="104" Body="&lt;p&gt;By the mean value theorem it's easy to show that $|a_{n+1}-a_{n}| \leq \frac{5}{6}|a_{n}-a_{n-1}|$ for every n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I thought of saying $|a_{n+1}-a_{n}| \leq ... \leq (\frac{5}{6})^{n}|a_{1}| \to 0$ and somehow show that ** if $M_{n}$ is the closed interval whose end points are $a_{n}$ and $a_{n-1}$ then $a_{n+1} \in M_{n}$ which implies $M_{n+1} \subseteq M_{n}$ and then to finish with &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_intersection_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor's intersection theorem&lt;/a&gt; that gives us convergence of $a_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not even sure if ** is correct and I haven't even used the fact that $a_{0} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Following the tip and some more thought I've come up with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every $m\gt n$: $|a_{m}-a_{n}|\leq|a_{m}-a_{m-1}+a_{m-1}-...+a_{n-1}-a_{n}|\leq$ &lt;br&gt;&#xA;$\leq\sum_{k=n}^{m-1}|a_{k+1}-a_{k}|\leq|a_{1}|\sum_{k=n}^{m-1}(\frac{5}{6})^{k}\le$ &lt;br&gt;&#xA;$\le|a_{1}|\sum_{k=n}^{\infty}(\frac{5}{6})^{k}=|a_{1}|\frac{(\frac{5}{6})^{n}}{\frac{1}{6}}=6|a_{1}|(\frac{5}{6})^{n} \to 0$ and from here it's easy to show that the sequence is Cauchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I made an error.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T18:25:51.330" LastActivityDate="2010-11-12T18:25:51.330" Title="Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10011" PostTypeId="2" ParentId="10007" CreationDate="2010-11-12T15:24:14.150" Score="3" ViewCount="0" Body="&lt;p&gt;Ask yourself this question: if $x \in \prod_p \mathbb{Z}_p$ is such that $x \in n \prod_p \mathbb{Z}_p$ for each $n \in \mathbb{Z} \setminus \{0\}$, then what can we say about $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to conclude that there is no embedding from $\frac{\prod_p\mathbb{Z}_p}{\otimes_p \mathbb{Z}_p}$ to $\prod_p \mathbb{Z}_p$&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-12T15:24:14.150" />
  <row Id="10012" PostTypeId="2" ParentId="10007" CreationDate="2010-11-12T15:25:59.360" Score="4" ViewCount="0" Body="&lt;p&gt;Let $G=\prod_p Z_p$ and $H=\bigoplus_p Z_p$ and regard $H$&#xD;&#xA;as a subgroup of $G$. You want to know if $H$ is a direct factor&#xA;of $G$. If it was there would be a subgroup $K$ of $G$&#xD;&#xA;isomorphic to $G/H$. There isn't. You know that $G/H$&#xD;&#xA;is divisible, but there are no nonzero elements $x\in G$ having the property that&#xA;$x=p y$ for some $y\in G$ for all primes $p$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-12T15:25:59.360" CommentCount="0" />
  <row Id="10013" PostTypeId="2" ParentId="10008" CreationDate="2010-11-12T15:31:53.483" Score="0" ViewCount="0" Body="&lt;p&gt;There will not be a closed form, as it will vary with the month of the holiday.  If you want to come out exactly even you have to choose in advance the month of the holiday.  One approach is to assume the holiday is the first month (which will lead to the highest payment-do you see why?).  Then you can apply one month's interest to the original balance and start an 11 month repayment with no holiday allowed.  You should be able to use your usual formulas for this.  Then as long as no holiday is taken, apply the interest to the current balance and take off the payment as with repaying a loan in variable installments.  If no holiday is taken the eleventh month payment will be slightly smaller and no payment will be due in month twelve. Alternately you could bill the first month assuming the holiday will be the second month, then if no holiday is taken in the first bill the second as if the holiday will be the third, etc.  This will be quite messy.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T15:31:53.483" />
  <row Id="10014" PostTypeId="1" AcceptedAnswerId="10016" CreationDate="2010-11-12T15:40:08.477" Score="1" ViewCount="61" Body="&lt;p&gt;Ok, so this might not be a permutation or combination but I'm curious as to what the answer is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: There are 3 groups and 5 Objects in each group. How many combinations are there if your choosing one from each group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes this is in reference to Call of Duty: Black Ops perks lol.&lt;/p&gt;&#xA;" OwnerUserId="3444" LastActivityDate="2010-11-12T15:48:50.387" Title="Simple Combination/Permutation Question" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10015" PostTypeId="2" ParentId="10010" CreationDate="2010-11-12T15:42:42.120" Score="4" ViewCount="0" Body="&lt;p&gt;You presumably want $|f&amp;#39;(x)|\le 5/6$. It's not the case that $a_{n+1}$&#xD;&#xA;need lie in the interval between $a_{n-1}$ and $a_n$. What can you say&#xA;about $|a_n-a_m|$? (The value of $a_0$ isn't relevant).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-12T15:42:42.120" CommentCount="1" />
  <row Id="10016" PostTypeId="2" ParentId="10014" CreationDate="2010-11-12T15:43:47.817" Score="3" ViewCount="0" Body="&lt;p&gt;You have five choices from the first group, times five from the second, times five from the third, for a total of 125 choices total.  To see why this is the case, imagine writing out the possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,2,1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-11-12T15:43:47.817" CommentCount="1" />
  <row Id="10017" PostTypeId="2" ParentId="10014" CreationDate="2010-11-12T15:45:38.527" Score="1" ViewCount="0" Body="&lt;p&gt;How many combinations are there if there is only one group?  For each of those, how many ways can you choose from the second group?  And the third?  Multiplication is your friend.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-12T15:45:38.527" />
  <row Id="10019" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T16:08:38.130" Score="0" ViewCount="0" Body="&lt;p&gt;Here are some interesting problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int\limits_{0}^{\frac{\pi}{2}} \sin{x}^{\cos{x}} \Bigl[ \cos{x}\cot{x} - \log(\sin{x})^{\sin{x}}\Bigr] \ dx$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\displaystyle \int \tan{x} \cdot \tan{2x} \cdot \tan{3x} \ dx$ : Why i say this is interesting because, many people would attempt to solve it directly but one can use the fact that $\tan(3x)=\tan(x + 2x)$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are many resources available in the web. The &lt;em&gt;Harvard - MIT Mathematics&lt;/em&gt; tournament problems should be challenging.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-12T16:08:38.130" CommunityOwnedDate="2010-11-12T16:08:38.130" />
  <row Id="10020" PostTypeId="1" AcceptedAnswerId="10027" CreationDate="2010-11-12T16:15:11.483" Score="0" ViewCount="50" Body="&lt;p&gt;I'm trying to show that the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_i^2 \mid \omega \mid^{-\beta}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\mid \omega \mid = ( \omega_1^2 + \omega_2^2)^{\frac{1}{2}}$ and $i = [1, 2]$) is defined at the origin under certain conditions. Here's my attempt. I'd like to know if its correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\mid \omega \mid^2 = \omega_1^2 + \omega_2^2 \Rightarrow \omega_i^2 \le \mid \omega \mid^2$ for any $i$. Therefore, $\omega_i^2 \mid \omega \mid^{-\beta} \le \mid \omega \mid^2 \mid \omega \mid^{-\beta} = \mid \omega \mid^{2 - \beta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the expression is defined at the origin if $2 - \beta &amp;gt; 0$ or $\beta &amp;lt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: in addition can it be shown that because $\omega_i \le \mid \omega \mid$ (I assume this is obvious), $\omega_i^3 \le \mid \omega \mid^3$ and by combining both inequalities it is possible to show in general that $\omega_i^n \le \mid \omega \mid^n$ for odd and even values of $n$.&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="1752" LastEditorDisplayName="" LastEditDate="2010-11-12T17:14:20.347" LastActivityDate="2010-11-12T17:53:30.603" Title="For what values of $\beta$ does $\omega_i^2 \mid \omega \mid^{-\beta}$ not blow up at the origin?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10021" PostTypeId="1" CreationDate="2010-11-12T16:23:53.707" Score="-2" ViewCount="158" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/11/does-99999-1&quot;&gt;Does .99999&amp;hellip; = 1?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Help solve an argument: is 1 greater than 0.9 (recurring)?&lt;/p&gt;&#xA;" OwnerUserId="3446" LastActivityDate="2010-11-12T16:46:33.907" ClosedDate="2010-11-12T18:05:34.510" Title="Is 1 greater than 0.9 (recurring)?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10022" PostTypeId="1" AcceptedAnswerId="10032" CreationDate="2010-11-12T16:28:49.770" Score="3" ViewCount="93" Body="&lt;p&gt;We are playing a game with my friend and trying to determine who is the better player.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The game has an luck element to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current score is 9-7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following questions have risen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How many games we should play to get statistically significant result that one is better than the other? We can assume p=9/16 or p=0,5 and something standard for z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What can we infer from the current game score? We tried &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Normal_approximation_interval&quot; rel=&quot;nofollow&quot;&gt;binomial proportion confidence interval&lt;/a&gt; (we used Wilson score, because N&amp;lt;30). &lt;a href=&quot;http://bit.ly/aM259u&quot; rel=&quot;nofollow&quot;&gt;Here are the results&lt;/a&gt;. Not sure how to interpret them.. does it mean that with 95%, the probability, at which the game score is 9-7, is within 34%-76%?&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2010-11-12T21:06:32.370" Title="How many games should we play to determine the better player statistically significantly?" Tags="&lt;probability&gt;&lt;card-games&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10024" PostTypeId="2" ParentId="9987" CreationDate="2010-11-12T16:42:11.993" Score="1" ViewCount="0" Body="&lt;p&gt;This is discussed at length in the book &lt;a href=&quot;http://en.wikipedia.org/wiki/Where_Mathematics_Comes_From&quot; rel=&quot;nofollow&quot;&gt;Where Mathematics Comes From&lt;/a&gt; by Lakoff and Núñez.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T16:42:11.993" />
  <row Id="10025" PostTypeId="2" ParentId="10021" CreationDate="2010-11-12T16:46:33.907" Score="-3" ViewCount="0" Body="&lt;p&gt;$1 - 0.9 = 0.1 &amp;gt; 0$.&#xA;You're welcome.&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2010-11-12T16:46:33.907" />
  <row Id="10026" PostTypeId="1" AcceptedAnswerId="10028" CreationDate="2010-11-12T17:45:54.217" Score="2" ViewCount="55" Body="&lt;p&gt;I want to prove that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then $L$ is also a subsequential limit of $a_{n}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I came up with the following: Let $\epsilon\gt0$; if $(x_{k})\to L$ then we simply pick $x_{i}\in(L-\epsilon, L+\epsilon)$ and because $x_{i}$ is a subsequential limit of $a_{n}$ we know that in every neighborhood of $L$ there are infinite elements of $a_{n}$ and we conclude that $L$ is also a subsequential limit of $a_{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems a bit clumsy, is there a better way to show this? Perhaps with Bolzano-Weistrass?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T17:53:56.597" LastActivityDate="2010-11-12T19:23:19.670" Title="If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then..." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10027" PostTypeId="2" ParentId="10020" CreationDate="2010-11-12T17:53:30.603" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, with your edit, your proof is fine. I think what you are trying to show is that the function $f : \omega \mapsto \omega_i^2\lvert\omega\rvert^{-\beta}$ has a well-defined and finite limit at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make your proof fully rigorous, all you need to say is that $f$ is bounded between two continuous functions $g : \omega \mapsto 0$ and $h : \omega \mapsto \lvert\omega\rvert^{2-\beta}$ whose gap tends to zero at the origin, therefore $f$ must also be continuous at the origin. And that's a perfectly good way to show that a function is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also show that the condition is necessary, i.e. that if $\beta \ge 2$ then $f$ is not continuous at the origin, simply by taking limits approaching the origin along two different directions, for example one along $\omega_1 = 0$ and the other along $\omega_2 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-12T17:53:30.603" CommentCount="2" />
  <row Id="10028" PostTypeId="2" ParentId="10026" CreationDate="2010-11-12T18:21:28.057" Score="1" ViewCount="0" Body="&lt;p&gt;I'm as mystified as Jonas Meyer on why you think this is &quot;clumsy&quot;. It follows exactly along the intuition: I can get arbitrarily close to $L$ using the $x_i$, and I can find subsequence of $(a_i)$ that gets arbitrarily close to the $x_i$, so I can find subsequences that get arbitrarity close to things that get arbitrarily close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps what you want is some idea of which subsequence that might be? Well, we can get it done as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an $N_1$ such that if $k\geq N_1$, then $|x_k-L|\lt 1$. And since $x_k$ is the limit of a subsequence of $(a_n)$, there is an $n_1$ such that $|a_{n_1}-x_{N_1}|\lt 1$. In particular, $|a_{n_1}-L|\lt 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is an $N_2\gt N_1$ such that for all $k\geq N_2$, $|x_k-L|\lt\frac{1}{2}$. Since $x_{N_2}$ is the limit of a subsequence of $(a_n)$, there is an $n_2$, $n_2\gt n_1$, such that $|a_{n_2}-x_{N_2}|\lt \frac{1}{2}$; in particular, $|a_{n_2}-L|\lt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue this way; assume that we have found $N_k$, $N_k\gt\cdots\gt N_1$ such that $|x_{N_i}-L|\lt \frac{1}{2^{i-1}}$, and $n_1\lt n_2\lt\cdots\lt n_k$ with $|x_{n_i}-x_{N_i}|\lt \frac{1}{2^{i-1}}$, so $|x_{n_i}-L|\lt \frac{1}{2^{i-2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is an $N_{k+1}\gt N_k$ such that for all $j\geq N_{k+1}$, $|x_{j}-L|\lt \frac{1}{2^k}$. Since $x_{N_{k+1}}$ is the limit of a subsequence of $(a_n)$, there is an $n_{k+1}\gt n_k$ such that $|a_{n_{k+1}}-x_{N_{k+1}}|\lt \frac{1}{2^k}$, and in particular $|a_{n_{k+1}}-L|\lt \frac{1}{2^{k-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductively, we get a subsequence $(a_{n_k})$ of $(a_n)$. I claim this subsequence converges to $L$. Let $\epsilon\gt 0$; find $k$ such that $0\lt \left(\frac{1}{2}\right)^{k-2}\lt \epsilon$. Then for all $\ell\geq k$ we have&#xA;\begin{equation*}&#xA;|a_{n_{\ell}} - L|\lt \frac{1}{2^{\ell-2}} \lt \frac{1}{2^{k-2}}\lt \epsilon.&#xA;\end{equation*}&#xA;Thus, the sequence converges to $L$, as claimed. QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I don't think this is particularly &quot;elegant&quot;, but I don't think it is clumsy either. It is exactly the intuition: get very close to $L$ using the $x_i$, then get very close to $x_i$ using some $a_j$, and this gives you an $a_j$ that is very close to $L$. Just keep doing it and you get a subsequence converging to $L$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-12T19:23:19.670" LastActivityDate="2010-11-12T19:23:19.670" />
  <row Id="10029" PostTypeId="1" AcceptedAnswerId="10035" CreationDate="2010-11-12T18:28:11.403" Score="6" ViewCount="354" Body="&lt;p&gt;I along with one of my friends were just discussing some basic things in group theory, when this question came up:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the &lt;em&gt;fundamental results&lt;/em&gt; in group theory?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We happened to list out some:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Fundamental Theorem of Group Homomorphism:&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Cayley's Theorem&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Sylow's Theorem&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There may be many more, but as far as my little knowledge is concerned, i think these are very important. Then we started explaining why each one of the above results were more powerful. I could explain as to how the &lt;em&gt;Fundamental Theorem of Group Homomorphism&lt;/em&gt;  can be used to derive some good results, in group theory, and also i could show my friends the power of the &lt;em&gt;Sylow Theorems&lt;/em&gt; just by considering groups of order $pq$. But i could never illustrate him as to how powerful Cayley's theorem is. Can anyone explain the significance of &lt;em&gt;Cayley's theorem&lt;/em&gt; and why it plays a central role in group theory. I am also curious to know whether any important results proved in Group theory using &lt;strong&gt;Cayley's theorem.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-12T19:22:51.807" Title="Importance of  Cayley's theorem" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="10030" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T18:42:13.657" Score="1" ViewCount="0" Body="&lt;p&gt;It is central to the theory of group representations. It allows to prove that any group $G$ of cardinal $n$ has a representation (a vector space $E$ such that $G$ is isomorphic to the linear group of $E$) of dimension $n$.&lt;/p&gt;&#xA;" OwnerUserId="441" LastActivityDate="2010-11-12T18:42:13.657" CommentCount="3" />
  <row Id="10031" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T18:43:08.323" Score="4" ViewCount="0" Body="&lt;p&gt;It shows that the axiomatization of transformation groups is correct.  Any transformation group is an axiomatic group, and any axiomatic group -- a structure, or more precisely a &quot;model&quot;, satisfying the group axioms -- is a group of transformations.  Historically, concrete groups of transformation appeared before abstract groups defined by algebraic axioms, so there is a question whether the algebra captures all (or possibly, more than) the intended examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Cayley's theorem with its initially strange but retroactively instinctive idea of considering a structure's internal &quot;action on itself&quot;, promoting symmetries of an object to the status of an object in their own (collective) right, is an early example of the formal style typical of later algebra and it helps prepare the ground psychologically for working in an abstract or axiomatized mode that was new in the 19th century.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-12T18:49:52.543" LastActivityDate="2010-11-12T18:49:52.543" CommentCount="5" />
  <row Id="10032" PostTypeId="2" ParentId="10022" CreationDate="2010-11-12T18:52:44.270" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure I understand your notation, but if we assume that one player wins 9/16 of the time, and we want a confidence of 99.5% (p=0.5%), then we can write the probability that the ratio of 9/16 would arise by chance as a function of the number of games, then solve for the number of games that gives a probability of 0.5%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I wrote a function to calculate this for 9/16 and p=5%.  The result is that it takes 265 trials (149 wins by one player) to exceed confidence of 95%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (2), you can infer that you can't yet say who is better with 95% confidence.  All you can say with 95% confidence is that the &quot;true&quot; outcome is somewhere between 34% and 76%.  As long as your confidence interval includes 50%, you can't say who's better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my sample code in Python (you need the scipy.stats package):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = 9&#xA;n = 16&#xA;&#xA;while binom_test(s, n) &amp;gt; 0.05:&#xA;    n += 1&#xA;    s += 9.0/16&#xA;&#xA;print n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3378" LastEditorUserId="3378" LastEditorDisplayName="" LastEditDate="2010-11-12T21:06:32.370" LastActivityDate="2010-11-12T21:06:32.370" CommentCount="1" />
  <row Id="10033" PostTypeId="1" AcceptedAnswerId="10403" CreationDate="2010-11-12T18:59:57.760" Score="3" ViewCount="108" Body="&lt;p&gt;In Stein-Shakarchi's book Fourier Analysis (p. 170), the solution of the Black-Scholes equation&#xA;$$\frac{\partial V}{\partial t}+rs\frac{\partial V}{\partial s}+\frac{\sigma^2s^2}{2}\frac{\partial^2 V}{\partial s^2}-rV=0$$ for &#xA;$0\lt t\lt T$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with boundary condition $V(s,T)=F(s)$, is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(s,t)=\frac{e^{-r(T-t)}}{\sqrt{2\pi\sigma^2(T-t)}}\int_0^\infty e^{-\frac{(\log(s/s^*)+(r-\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}} F(s^*)\,ds^*.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own derivation shows that $ds^*$ in the formula should be replaced by $$\frac{ds^*}{s^*}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a typo in the book?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T17:22:18.757" LastActivityDate="2010-11-15T17:22:18.757" Title="Black-Scholes equation" Tags="&lt;analysis&gt;&lt;differential-equations&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10034" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T19:05:08.840" Score="9" ViewCount="0" Body="&lt;p&gt;Cayley's Theorem was &lt;em&gt;very&lt;/em&gt; important historically. Groups originally arose from considering groups of permutations (specifically, the action of some functions on the roots of some polynomials). Every group was really a collection of permutations of some set of specific objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Cayley introduced the notion of an &quot;abstract&quot; group; the idea that you simply had some &lt;em&gt;things&lt;/em&gt; that you could &quot;compose&quot; in some way giving you an associative operation, with an &quot;identity&quot; and inverses. He points out that all the things that people had been considering up to that point were &quot;groups&quot; in this sense. But he also wanted to make the point that he was not introducing a new class of objects, but that every object that satisfied his definition would also be a &quot;group&quot; in the old sense. That is, all he was doing was to recast the old notion, rather than expanding the class (at least, as applied to &lt;em&gt;finite&lt;/em&gt; groups; for infinite groups there was no consensus on what &quot;permutation of an infinite set&quot; meant [whether it meant what we now think of as &quot;bijection&quot;, or whether it meant what we now call &quot;a bijection with finite support&quot;, that is, that only moves finitely many objects]). What Cayley's Theorem says is &quot;Every &lt;em&gt;thing&lt;/em&gt; that satisfies this definition can be thought of as a &quot;group&quot; in the old sense of a collection of permutations that is closed under composition, inverses, and includes the identity.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the proof turns out to be useful in other contexts as well, as indicated by T. So I would classify Cayley's theorem as more historically important than important today.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-12T19:05:08.840" />
  <row Id="10035" PostTypeId="2" ParentId="10029" CreationDate="2010-11-12T19:17:50.293" Score="5" ViewCount="0" Body="&lt;p&gt;Well, it gives (part of) a proof of the first of Sylow's theorems: it is quite easy to prove that if $G$ is a finite group that admits a $p$-Sylow subgroup $S$, and if $H$ is a subgroup of $G$, then $H$ also admits a $p$-Sylow subgroup ($H$ acts on $G/S$, whose cardinal is prime to $p$, so there must be an orbit of cardinal prime to $p$, and $\mathrm{Stab}(gS)=H \cap gSg^{-1}$, so you find that the $\mathrm{Stab}$ of any element of this orbit is a $p$-Sylow of $H$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using Cayley's theorem, you can embed any finite group $H$ in $G=\mathrm{GL}_n(\mathbb{Z}/p \mathbb{Z})$ (think permutation matrices), where $n= \mathrm{card} H$, and $G$ admits a $p$-Sylow: the subgroup of upper-triangular unipotent matrices (recall $\mathrm{card} G = (p^n-1)(p^n-p) \ldots (p^n-p^{n-1})=p^{n(n-1)/2} m$ with $m$ prime to $p$).&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-12T19:22:51.807" LastActivityDate="2010-11-12T19:22:51.807" CommentCount="1" />
  <row Id="10036" PostTypeId="2" ParentId="9958" CreationDate="2010-11-12T19:26:59.030" Score="1" ViewCount="0" Body="&lt;p&gt;Let $m$ be an integer $\geq 2$. Define $u_0=e^{1/m}$, and $u_{n+1}=\frac{u_n}{u_n-I(u_n)}$ where $I$ denotes integral part.&#xA;Prove that $I(u_n)=mn$ for $n \geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-12T19:26:59.030" CommunityOwnedDate="2010-11-12T19:26:59.030" CommentCount="3" />
  <row Id="10037" PostTypeId="2" ParentId="8532" CreationDate="2010-11-12T19:28:24.213" Score="0" ViewCount="0" Body="&lt;p&gt;I think in complement to the above answer, it is useful to consider the comments of Littlewood in &lt;a href=&quot;http://www.springerlink.com/content/558154440wkt1833/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/558154440wkt1833/fulltext.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what the current opinion of this is?&lt;/p&gt;&#xA;" OwnerUserId="941" LastActivityDate="2010-11-12T19:28:24.213" CommunityOwnedDate="2010-11-12T19:28:24.213" />
  <row Id="10038" PostTypeId="1" AcceptedAnswerId="10056" CreationDate="2010-11-12T20:07:31.840" Score="2" ViewCount="111" Body="&lt;p&gt;One proof of the halting problem goes by contradiction like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume there is a Turing machine $H$ that can decide the halting problem, then construct a Turing machine $Q$ that takes as input a Turing machine $M$ and an input $I$ to $M$, such that if $H$ decides that $M(I)$ halts, then $Q$ doesn't halt, and vice versa. Then $Q(Q)$ should not halt if $Q(Q)$ halts. A contradiction, then $H$ couldn't have existed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is a small technical details. The input to $Q$ required to generate this contradiction is not well-founded. That is, $Q$ takes a machine $Q$ and it's input $I$ (i.e. a pair $(Q,I)$). Evidently, $I$ is a also pair $(Q,I&amp;#39;)$, where $I&amp;#39;=(Q,I&amp;#39;&amp;#39;)$, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, let $I=(Q,I&amp;#39;)$ be equivalently written as a set $I={(1,Q),I&amp;#39;}={(1,Q),{(1,Q),I&amp;#39;&amp;#39;}}=...$; showing that $I$ is not well-founded. So does the contradiction in the proof arises because $H$ couldn't exist or because $I$ is not well-founded ?&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-11-13T07:58:32.290" Title="Does this proof for the undecidability of the halting problem violate the axiom of regularity ?" Tags="&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="10039" PostTypeId="1" CreationDate="2010-11-12T20:16:54.383" Score="2" ViewCount="109" Body="&lt;p&gt;Find all functions $f:\mathbb{R} \to \mathbb{R}$, that are continuous at the point $x=0$ and satisfy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f (x+y)=f (x)+f (y)+xy (x+y) \ \ \forall x,y \in \mathbb{R} $$ &lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T20:42:15.420" LastActivityDate="2010-11-12T20:44:04.750" Title="Solve the functional equation $f (x+y)=f (x)+f (y)+xy (x+y)$, $f$ continuous at $0$" Tags="&lt;functions&gt;&lt;functional-equations&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10040" PostTypeId="2" ParentId="10039" CreationDate="2010-11-12T20:26:52.597" Score="7" ViewCount="0" Body="&lt;p&gt;Let $\displaystyle g(x) = f(x) - x^3/3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $\displaystyle g(x+y) = g(x) + g(y)$ which by continuity at $\displaystyle 0$ implies $\displaystyle g(x) = kx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $\displaystyle f(x) = x^3/3 + kx$ which satisfies the original equation for any $k$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a proof that $\displaystyle g(x+y) = g(x) + g(y)$ and $\displaystyle g$ continuous at $\displaystyle 0$ implies $\displaystyle g(x) = kx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that $\displaystyle g(0) = 0$ and that continuity at $\displaystyle 0$ implies continuity everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by induction, we can prove that for any integer $\displaystyle n$, $\displaystyle g(n) = ng(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can then be extended to the rationals: $\displaystyle g(r) = rg(1) \ \ \forall r \in \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any real $\displaystyle x$, pick a sequence of rationals $\displaystyle \{r_n\}$ converging to $\displaystyle x$ and use the continuity of $\displaystyle g$ at $\displaystyle x$ and that $\displaystyle g(r_n) = r_n g(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a classic and you should find plenty of literature on it.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T20:44:04.750" LastActivityDate="2010-11-12T20:44:04.750" CommentCount="3" />
  <row Id="10041" PostTypeId="2" ParentId="9993" CreationDate="2010-11-12T20:27:01.177" Score="3" ViewCount="0" Body="&lt;p&gt;(Background for other readers: the derivation of the solution can be found&#xA;&lt;a href=&quot;http://books.google.com/books?id=eHANhZwVouYC&amp;amp;pg=PA89&quot; rel=&quot;nofollow&quot;&gt;on Google books&lt;/a&gt;,&#xA;see equations (6.18) and (6.19).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good and very natural question, but your reasoning is flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution formula that you quote for $T(t)$ is only valid for $t&amp;gt;0$,&#xA;and it does satisfy $\lim_{t\to 0} T(t) = T_0$;&#xA;it's just that taking the limit $t\to 0$ termwise isn't justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That the limit is $T_0$ can be seen by tracing all the steps in&#xA;the derivation carefully and using knows facts about convergence of&#xA;Fourier series. Unfortunately, explaining this would require drawing&#xA;more figures than I can do here with a reasonable amount of effort...&#xA;Someone who knows more than I about theta functions can probably&#xA;say something about how to compute the limit directly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument &quot;because the maximum speed in the universe is limited&quot; is useless,&#xA;since the heat equation is not the real world; in fact the heat equation&#xA;admits &lt;em&gt;infinite&lt;/em&gt; speed of propagation!&#xA;(A solution with compact support at $t=0$ will not have compact support for any $t&amp;gt;0$.)&#xA;Moreover, taking the limit $t\to 0$ termwise is even less justified in the&#xA;differentiated sum than it was in the original sum.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-12T20:27:01.177" />
  <row Id="10042" PostTypeId="1" AcceptedAnswerId="10296" CreationDate="2010-11-12T20:39:07.693" Score="4" ViewCount="84" Body="&lt;p&gt;Take projective real space $\mathbb P_n (\mathbb R)$ of ODD dimension. It is easy to proof that all his Stiefel-Whitney numbers are zero . So according Thom theorem&#xA;there must exists manifold $M$ with boundary such that boundary is &#xA;$\partial M= \mathbb P_n (\mathbb R)$. I should like  to see directly such $M$, without using Thom Theorem . For example if $n=1$  evident choice is $M=$ closed disk.&lt;br&gt;&#xA;I have no idea in general case. Can some one help please ?&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-13T14:29:15.803" LastActivityDate="2010-11-14T20:47:31.820" Title="For what manifold is  boundary given odd-dimensional projective space?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10043" PostTypeId="2" ParentId="10038" CreationDate="2010-11-12T20:46:30.423" Score="3" ViewCount="0" Body="&lt;p&gt;You got the proof wrong in the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us denote with $\uparrow$ resp $\downarrow$ nontermination resp. termination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $H$ is the Turing machine (i.e. its code) that can solve the halting problem, that is $H(M,x) = 1 \Leftrightarrow M(x) \downarrow$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $Q(M) = \begin{cases}1, &amp;amp;H(M,M) = 0 \\\\ \uparrow, &amp;amp;H(M,M)=1\end{cases}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $Q(Q) \downarrow\ \Leftrightarrow\ Q(Q) \uparrow$ which yields the contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, everything is well defined, no problem at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Please see Arturo's answer for a more verbose explanation how you should understand arguments in $\mathbb{N}^k, k&amp;gt;1$, for Turing machines.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastEditorUserId="3330" LastEditorDisplayName="" LastEditDate="2010-11-13T07:55:41.480" LastActivityDate="2010-11-13T07:55:41.480" CommentCount="1" />
  <row Id="10044" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T21:02:12.210" Score="0" ViewCount="0" Body="&lt;p&gt;Perhaps some intuition can be gained by plotting each number's position on the number line.  Taking the inverse of any number is visualized by taking the mirror-image of the original plot.  So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero).  Likewise, the inverse of a negative number is a positive number.  If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.&lt;/p&gt;&#xA;" OwnerUserId="3378" LastActivityDate="2010-11-12T21:02:12.210" />
  <row Id="10045" PostTypeId="1" AcceptedAnswerId="10050" CreationDate="2010-11-12T21:11:19.807" Score="2" ViewCount="75" Body="&lt;p&gt;For what value of $\alpha$ is $W_t^{\alpha}$ a martingale where $\alpha \in \mathbb{Z}$? ($W_t$ standard Brownian motion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove for $\alpha = 0$ and $\alpha = 1$, it is a martingale. (rather it is obvious). But my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove that $W_t^{\alpha}$ is not a martingale for $\alpha \in \mathbb{Z}\cap\{0,1\}^{c}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question was asked in an phone interview. They were only interested in the answer which I told correctly.)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-12T21:33:37.730" LastActivityDate="2010-11-12T21:43:57.250" Title="For what value of $\alpha$ is $W_t^{\alpha}$ a martingale?" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="10046" PostTypeId="1" AcceptedAnswerId="10064" CreationDate="2010-11-12T21:12:26.630" Score="6" ViewCount="271" Body="&lt;p&gt;In exercise 36 &lt;em&gt;Miscellaneous Taylor Coefficients using Bernoulli&#xA;numbers&lt;/em&gt; on pages 88-89 of Louis Comtet's &lt;em&gt;&lt;a href=&quot;http://books.google.pt/books?id=C0HPgWhEssYC&amp;amp;printsec=frontcover&amp;amp;dq=Louis+Comtet+Advanced+Combinatorics&amp;amp;source=bl&amp;amp;ots=rEhBo3Xuv8&amp;amp;sig=rTARizLlaQKIHQNZcHAlHKRUKUc&amp;amp;hl=pt-PT&amp;amp;ei=jaHdTJnVI5OKhQeWqpiJDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBkQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Advanced Combinatorics&lt;/a&gt;&lt;/em&gt;, 1974,&#xA;one is asked to obtain the following explicit formula for the Bernoulli numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_{2n}=(-1)^{n-1}\dfrac{1+\left[ \varphi _{n}\right] }{2(2^{2n}-1)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varphi _{n}=\dfrac{2(2^{2n}-1)(2n)!}{2^{2n-1}\pi ^{2n}}\displaystyle\sum_{k=1}^{3n}\dfrac{1}{k^{2n}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with $\displaystyle\sum_{n\geq 0}B_{n}\dfrac{t^{n}}{n!}=\dfrac{t}{e^{t}-1}$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to prove, among other sums, that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{\dbinom{2n}{n}}=\dfrac{1}{3}+\dfrac{2\pi\sqrt{3}}{27}\qquad (\ast ).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alfred van der Poorten wrote &lt;a href=&quot;http://www.springerlink.com/content/3866u5593p65l016/&quot; rel=&quot;nofollow&quot;&gt;here (section 10)&lt;/a&gt;: &lt;em&gt;seeing that&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{n=1}^{\infty}\dfrac{x^{2n}}{n^{2}\dbinom{2n}{n}}=2\arcsin^{2}\left( \dfrac{x}{2}\right) \qquad (\ast \ast )$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(...) &lt;em&gt;formula&lt;/em&gt; [ $(\ast )$ ] &lt;em&gt;become&lt;/em&gt;[s] &lt;em&gt;quite accessible to proof&lt;/em&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to show formula $(\ast \ast )$ neither how it can be used to&#xA;prove $(\ast )$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Could you provide (a) more detailed hint(s) on how and/or&#xA;different ways in which formula $(\ast )$ can be derived?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: For information the other sums are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n\dbinom{2n}{n}}=\dfrac{\pi \sqrt{3}}{9},\quad$ $\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{2}\dbinom{2n}{n}}=\dfrac{\pi ^{2}}{18},\quad$ $\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{4}\dbinom{2n}{n}}=\dfrac{17\pi ^{4}}{3240}.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-13T14:55:51.003" LastActivityDate="2010-11-14T23:14:47.127" Title="Exercise from Comtet's Advanced Combinatorics: prove $27\sum_{n=1}^{\infty }1/\binom{2n}{n}=9+2\pi \sqrt{3}$" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;binomial-coefficients&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="10047" PostTypeId="1" AcceptedAnswerId="10052" CreationDate="2010-11-12T21:16:36.953" Score="4" ViewCount="128" Body="&lt;p&gt;Let $S$ be a sheaf over $X$ and $r$ an element in $S_x$ for some $x$ in $X$. Must there exist a section $s$ in $S(X)$ such that such that $s$ equals $r$ when mapped to $S_x$ by the canonical map?  &lt;/p&gt;&#xA;" OwnerUserId="3453" LastActivityDate="2010-11-13T22:04:06.697" Title="Can every element in the stalk be represented by a section in the top space?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="10048" PostTypeId="2" ParentId="10046" CreationDate="2010-11-12T21:21:50.720" Score="6" ViewCount="0" Body="&lt;p&gt;$1/\binom{2n}{n} = (2n+1) \int_0^1 x^n(1-x)^n dx$.  Sum these over $n$ to get the integral of a rational function.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-11-12T21:21:50.720" CommentCount="4" />
  <row Id="10049" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T21:22:10.070" Score="0" ViewCount="0" Body="&lt;p&gt;There's also the somewhat dissatisfactory answer &quot;because the definition of an ordered field says so&quot;. Recall that an &lt;em&gt;ordered field&lt;/em&gt; is a field $K$ such that there exists $P \subset K$ such that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $x,y \in P$ then $x+y \in P$ and $xy \in P$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x \in K$ *exactly one* of the following holds. $x \in P$, $-x \in P$ or $x = 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And the elements of $P$ are called positive. &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordered_field&quot; rel=&quot;nofollow&quot;&gt;Obligatory Wikipedia link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you show that a negative real number times a negative real number is positive? Well either you give the student that $\mathbb{R}$ is an ordered field (typical for a first course in real analysis) and that essentially boils down to &quot;negative times negative is positive &lt;em&gt;because I said so, da#&amp;amp;it&lt;/em&gt;&quot;, or you go the long and arduous route and construct the reals - typically using Dedekind cuts or equivalence classes of rational Cauchy sequences - and show that they are in fact an ordered field.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-12T21:27:19.080" LastActivityDate="2010-11-12T21:27:19.080" CommentCount="2" />
  <row Id="10050" PostTypeId="2" ParentId="10045" CreationDate="2010-11-12T21:31:26.570" Score="1" ViewCount="0" Body="&lt;p&gt;First, the question doesn't even make sense if $\alpha$ is not an integer, because $W^\alpha$ is undefined as a real valued process when W goes negative. So, in that case, I'll assume that W is started from a positive value and stopped when it hits zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two methods: Use Ito's formula for $X=W^\alpha$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;dX = \alpha W^{\alpha-1}\,dW + \frac12\alpha(\alpha-1) W^{\alpha-2}\,dt.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term on the right is a local martingale. So the only way that X can be a martingale is for $\alpha(\alpha-1)\int W_t^{\alpha-2}\,dt$ to be a local martingale. It is a standard result that continuous finite variation processes cannot be a local martingale unless they are constant, giving $\alpha(\alpha-1)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, use Jensen's inequality. As $f(x)=x^\alpha$ is strictly convex for $\alpha\not\in[0,1]$ and strictly concave for $\alpha\in(0,1)$ (restricting to the positive reals), the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t\mapsto\mathbb{E}[W_t^\alpha]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is strictly increasing for $\alpha\not\in[0,1]$ and strictly decreasing for $\alpha\in(0,1)$. So, $W^\alpha$ is not a martingale in these cases. In fact, it is a submartingale for $\alpha \not\in[0,1]$ and a supermartingale for $\alpha\in(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, Jensen's inequality is the simplest method. However, Ito's formula applies much more generally, to any twice differentiable function of a semimartingale.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-12T21:43:57.250" LastActivityDate="2010-11-12T21:43:57.250" CommentCount="1" />
  <row Id="10051" PostTypeId="2" ParentId="10045" CreationDate="2010-11-12T21:39:35.487" Score="1" ViewCount="0" Body="&lt;p&gt;Assume that $\alpha&amp;gt;0$. &#xA;Apply the optional stopping theorem to the process $W^\alpha_t$ using the hitting time $T$ of $\{-1 , 2\}$. If $W^\alpha_t$ were a martingale, we'd have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=E(W^\alpha_0)=E(W^\alpha_T)=(2/3)(-1)^\alpha+(1/3)2^\alpha.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that  $2^{\alpha-1}=1$, so $\alpha=1$.   &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-12T21:39:35.487" CommentCount="6" />
  <row Id="10052" PostTypeId="2" ParentId="10047" CreationDate="2010-11-12T21:39:56.247" Score="10" ViewCount="0" Body="&lt;p&gt;No: take sheaf of holomorphic functions $S=\mathcal O$ on $X=\mathbb C$.Then stalk at $x=1$ of $1/z$ does not come from $\mathcal O(\mathbb C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Arturo: no you are not misremembering. Here is an example. Take $X=\mathbb P_n (\mathbb C)$ and $S=\mathcal O (-1)$= tautologic bundle.Then every stalk $S_x$ is isomorphic with $\mathcal O_x$ so is not zero but $S(\mathbb P_n (\mathbb C))=0$. (By GAGA principle of Serre you can chose to think in holomorphic way or algebraic way in this example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Robin Chapman: Yes, there are sheaves $S$ with all morphisms $S(X) \to S_x$ surjective but that are not flabby. For example, take $X=\mathbb R$ and $S=C$ = sheaf of continous functions. Then  for $x\in \mathbb R$ and $f_x \in C_x$ take bump function $\phi$ equal $1$ in vicinity of $x$ and support smaller than domain of definition of $f$. Then $\phi f$ (extended by zero) is global function in $C(\mathbb R)$ and his germ at $x$ is equal $f_x$. But of course $C$ is not flabby.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T22:04:06.697" LastActivityDate="2010-11-13T22:04:06.697" CommentCount="2" />
  <row Id="10053" PostTypeId="2" ParentId="9933" CreationDate="2010-11-12T21:40:33.130" Score="0" ViewCount="0" Body="&lt;p&gt;It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-12T21:40:33.130" />
  <row Id="10054" PostTypeId="1" CreationDate="2010-11-12T21:45:56.390" Score="-2" ViewCount="99" Body="&lt;p&gt;I learn less while writing as opposed to not writing. In other words, when I do problems, it seems like I can solve them quicker by not writing out all the details. &lt;/p&gt;&#xA;" OwnerUserId="3428" LastActivityDate="2010-11-12T22:14:14.310" ClosedDate="2010-11-12T22:19:26.810" Title="Is this unusual......?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10055" PostTypeId="2" ParentId="10054" CreationDate="2010-11-12T21:56:02.743" Score="0" ViewCount="0" Body="&lt;p&gt;It's not unusual, because it's naturally easier to solve problems when you don't work out all the details. If you do work out all the details in your head, then writing those details down shouldn't affect the rate at which you solve the problems (since the details are already solved as well). &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-12T21:56:02.743" />
  <row Id="10056" PostTypeId="2" ParentId="10038" CreationDate="2010-11-12T22:04:48.327" Score="3" ViewCount="0" Body="&lt;p&gt;This is essentially Raphael's answer expanded, and taking into account what I was (probably unsuccessfully) trying to communicate in the comments. As I recall, the reason the Universal Machine is important is that it is through the UTM that we can actually explain how the &quot;construction&quot; of the modified machines come about; but I could have messed that up and if so I apologize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I guess the point is that you are technically correct that the proof you present, which glosses over some important details, is incorrect. The machine $Q$ you give cannot be asked about the behavior of $Q$ with $Q$ as an input, given the way you describe $Q$. Technically the reason is that if you try to formalize it along the lines I give below, you end up unable to formalize the input you should put into $Q$. The argument is &quot;morally&quot; correct in that a few technical (yet standard) tricks will solve that issue by using a slight modification of $Q$ instead of $Q$ directly. Then again, it is clear that the proof is not trying to be technically accurate, but only &quot;morally&quot; accurate, so one might forgive slight blurrings. It's similar to what happens in the usual description of Cantor's 2nd proof of uncountability of the reals, where the issue of distinct representations of reals is glossed over and can lead to a techncially incorrect argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turing machines each have a code $C$, and take numbers as input. (We can use, for example, Goedel numbering to translate specific sentences or problems into numbers). Let us write $C(m)$ to mean the giving the machine with cod $C$ the input $m$.We can take a pair $(C,m)$, where both $C$ and $m$ are numbers, and obtain a number $N$ that &quot;corresponds&quot; (in a precise and well-defined way) with a pair $(C,m)$. Let us write that as $N\sim (C,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a machine $H$ that solves the Halting Problem; we can tweak it so that it will accept any in put and, when given an input $N$, will do the following:&#xA;\begin{equation*}&#xA;H(N) = \left\{\begin{array}{ll}&#xA;1 &amp;amp; \mbox{if $N\sim (C,m)$, and $C$ halts when $m$ is its input;}\\&#xA;0 &amp;amp; \mbox{if $N\sim (C,m)$ and $C$ does not halt when $m$ is its input;}\\&#xA;0 &amp;amp; \mbox{if $N\not\sim(C,m)$ for any $C$ and $m$.}&#xA;\end{array}\right.&#xA;\end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $H$ exists, then we can &quot;construct&quot; (that is, there also exists) a Turing Machine $Q$ such that:&#xA;\begin{equation*}&#xA;Q(N) = \left\{\begin{array}{ll}&#xA;1 &amp;amp; \mbox{if $H(M)=0$, where $M\sim(N,N)$;}\\&#xA;\mbox{does not halt} &amp;amp; \mbox{if $H(M)=1$, where $M\sim(N,N)$.}&#xA;\end{array}\right.&#xA;\end{equation*}&#xA;Note that $M$ is well-defined, since it is just the code for the pair $(N,N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that $Q$ does not take as an input a pair of the form $(C,k)$; it takes a single number $k$, and then figures out what $H(k,k)$ would have been. It's much more restrictive machine than one that &quot;does the opposite&quot; of what $H$ does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $Q$ being a Turing machine has a specific code $C$. So the question is, what is $Q(C)$? By definition, $Q(C) = 1$ if and only if $H(M)=0$, where $M\sim(C,C)$; but $H(M)=0$ means either that $M$ is not the code of a valid pair, or else that $C$ does not halt when given the number $C$ as an input. However, $M$ *is* the code of a pair, because that is how we defined it. So that means that $Q(C)=1$ if and only if the machine with code $C$ does not halt when given $C$ as an input. But the machine with code $C$ *is* $Q$, so $Q(C)=1$ if and only if the result of doing $Q(C)$ is non-terminating. Therefore, we cannot have $Q(C)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is $Q(C)\neq 1$? $Q(C)\neq 1$ if and only if $Q(C)$ does not halt; and $Q(C)$ does not halt if and only if $H(M)=1$, where $M\sim(C,C)$. But $H(M)=1$ in this case if and only if the machine with code $C$ halts when it is given the input $C$. But the machine with code $C$ *is* $Q$, so $Q(C)$ fails to halt if and only if $Q(C)$ halts, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the key here is two-fold: first, that the inputs to Turing Machines are just numbers, rather than actual pairs or sets. We are identifying what in the meta-language corresponds to a pair with a single number in order to do the input, so we cannot run into problems of foundation (inputs are not pairs of pairs of pairs... etc, they are just &lt;em&gt;numbers&lt;/em&gt;). The issue &lt;em&gt;could&lt;/em&gt; arise if we could not define the number because it requires a recursive process that &quot;goes on indefinitely&quot; and leads to an &quot;infinite number&quot;. And &lt;strong&gt;second&lt;/strong&gt;: you do not construct $Q$ so that it will halt if $H$ says 'doesn't halt' with that same input, or will halt if $H$ says 'it &lt;em&gt;does&lt;/em&gt; halt' with the same input. Instead, $Q$ transforms its input in a precise and well defined way (in this case, taking $N$ to the code for $(N,N)$) and only &lt;em&gt;then&lt;/em&gt; asking whether $H$ will halt with &lt;em&gt;that&lt;/em&gt; new input or not. That is, $Q$ need not be quite as general as you make it out to be in your post, which is why we can actually use $Q$ to ask about itself. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T07:58:32.290" LastActivityDate="2010-11-13T07:58:32.290" CommentCount="4" />
  <row Id="10057" PostTypeId="2" ParentId="10046" CreationDate="2010-11-12T22:05:32.033" Score="4" ViewCount="0" Body="&lt;p&gt;You want to get rid of the factor of $n^2$ in the denominator.  So differentiate with respect to $x$, that gets rid of one factor.  If you modify what you get you can differentiate again to get rid of the other.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-12T22:05:32.033" CommentCount="1" />
  <row Id="10059" PostTypeId="1" AcceptedAnswerId="10063" CreationDate="2010-11-12T22:13:35.230" Score="1" ViewCount="103" Body="&lt;p&gt;I'm sorry for this kind of specific question, I'd love if you could link to resources (prime lists, etc) that can answer similar questions more generically.&lt;/p&gt;&#xA;" OwnerUserId="3455" LastActivityDate="2010-11-12T22:32:25.297" Title="What is the largest prime less than 2^31?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;online-resources&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10060" PostTypeId="2" ParentId="10054" CreationDate="2010-11-12T22:14:14.310" Score="3" ViewCount="0" Body="&lt;p&gt;With certain kinds of problems, sometimes one can start writing up without actually visualising where it all will end and how, and just keep taking steps until you reach the end. Think of it as walking in the dark with a small flashlight: you can illuminate a couple of steps ahead but not the entire path. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But sometimes we need to visualize the whole path before we start walking; or at least the major turning points without focusing too much on every little stone along the way. &quot;Visualizing&quot; the path will likely take a lot less time than actually walking the walk, of course. But one must still walk the path to reach the destination, though, and that requires taking &lt;em&gt;each&lt;/em&gt; step, not just describing the major turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps your experience is that you run into fewer dead-ends and can understand what is going on better with the second method, trying to visualize those major turns only and seeing how one might reach the destination, rather than starting to walk without really knowing where you are going. That's certainly a possibilitiy; but remember you must still get there, so you must still walk it (that is, you still need to write it up with the details). It is very common to &lt;em&gt;think&lt;/em&gt; that you can get there by visualizing the entire trip and not realize that there is a serious problem along the way, because you aren't thinking about the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I personally find that even when using the former flashlight-a-few-steps-at-a-time method to &lt;em&gt;solve&lt;/em&gt; the problem, I want to step back at the end and look at the entire path that was traversed, to see if I got there the best possible way, and to get a good feel for the entire thing. That's always good practice &lt;em&gt;anyway&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end, assuming you want to do things in a good way, I would say that you do not save or lose time either way: you're just changing the order of what you are doing. Either try working out all the details a few steps at a time first, and then look back and try to take in the view; or look around to get a feel for the entire problem, see how you are going to get from beginning to end, and only &lt;em&gt;then&lt;/em&gt; start writing up all the details. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-12T22:14:14.310" />
  <row Id="10061" PostTypeId="2" ParentId="10059" CreationDate="2010-11-12T22:19:10.117" Score="5" ViewCount="0" Body="&lt;p&gt;It is $2^{31}-1$. You might want to check Mersenne prime for similar details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mersenne_prime&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mersenne_prime&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-12T22:19:10.117" CommentCount="5" />
  <row Id="10062" PostTypeId="2" ParentId="9524" CreationDate="2010-11-12T22:31:40.350" Score="2" ViewCount="0" Body="&lt;p&gt;I want to give a more elaborate example for computer scientists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a context free grammar, we can derive (in well-behaved cases) its structure function with a (often) simple calculation. This function's coefficients $a_n$ are the number of derivation trees for words of length $n$ (or, for unambiguous grammars, the number of words). With additional tricks, we can get a number of properties of the generated language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple example: Let a grammar with the set of rules $S \rightarrow aS | bS | \varepsilon$. This directly translates to $S(z) = 2zS(z) + 1$. Solving this equation is simple; we obtain $S(z) = (1-2z)^{-1}$. The coefficients $[z^n]S(z) = 2^n$ are, of course, the number of different words of length $n$ this (unambiguous) grammar can generate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This directly yields a convenient tool to check wether a grammar is unambiguous if we know the number of words (and we can solve the equation system and obtain the series expansion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many further extensions and applications.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-12T22:31:40.350" />
  <row Id="10063" PostTypeId="2" ParentId="10059" CreationDate="2010-11-12T22:32:25.297" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.prime-numbers.org/prime-number-2147480000-2147485000.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.prime-numbers.org/prime-number-2147480000-2147485000.htm&lt;/a&gt; tells you that it's 2147483647 (about 2/3rds of the way down, third column).  This website seems like a good resource if you're looking for lots of primes.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2010-11-12T22:32:25.297" CommentCount="1" />
  <row Id="10064" PostTypeId="2" ParentId="10046" CreationDate="2010-11-12T22:50:35.830" Score="7" ViewCount="0" Body="&lt;p&gt;Another one, which has been mentioned multiple times on this site (search for Wallis' product) is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{\pi/2} \sin^{2n-1} x \ dx = \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} \ \ n \gt 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\displaystyle \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} = \dfrac{2^{2n}}{2n{2n \choose n}} = \dfrac{2^{2n-1}}{n{2n \choose n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{\pi/2} n\left(\dfrac{\sin x}{2}\right)^{2n-1} \ dx = \dfrac{1}{{2n \choose n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the sum can be found easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula by Alfred Van der Pooten can be proved using this approach too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof of Wallis Product is given here: &lt;a href=&quot;http://crypto.stanford.edu/pbc/notes/pi/wallis.xhtml&quot; rel=&quot;nofollow&quot;&gt;http://crypto.stanford.edu/pbc/notes/pi/wallis.xhtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a (closely related) problem on this site using this, see here: &lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-12T23:04:00.223" LastActivityDate="2010-11-12T23:04:00.223" CommentCount="1" />
  <row Id="10065" PostTypeId="1" AcceptedAnswerId="10067" CreationDate="2010-11-12T23:08:57.350" Score="6" ViewCount="334" Body="&lt;p&gt;I started by showing that $1\leq a_{n} \leq n$ (by induction) and then $\frac{1}{n}\leq \frac{a_{n}}{n} \leq 1$ which doesn't really get me anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a different path I showed that $a_{n} \to \infty$ but can't see how that helps me.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-11-12T23:18:15.990" LastActivityDate="2010-11-15T18:19:39.677" Title="Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim_{n\to\infty}\frac{a_{n}}{n}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;homework&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="1" />
  <row Id="10066" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:16:03.930" Score="8" ViewCount="0" Body="&lt;p&gt;Now that the homework is solved, here is some more investigation into this interesting sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can show that $$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$ for some constant $\displaystyle C \gt 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By $\displaystyle x_n \sim y_n$ I mean $\displaystyle \lim_{n \to \infty} (x_n - y_n) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $b_n = a_{n}^2 - 2n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $\displaystyle b_{n+1} = b_n + \dfrac{1}{b_n + 2n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $b_3 \gt 0$ and thus for sufficiently large $\displaystyle n$, $\displaystyle b_n \gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also easy to show that $\displaystyle b_n \lt 2n$. In fact, we can easily show that $b_n \lt \log n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have that, for sufficiently large $\displaystyle m,n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  b_{m+1} - b_n = \sum_{k=n}^{m} \dfrac{1}{b_k + 2k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle 0 \lt b_k \lt \log k$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k}(1- \dfrac{b_k}{2k})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here we used $\displaystyle \dfrac{1}{1+x} \gt \ \ 1-x, 1 \gt x \gt 0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Since $b_k \lt \log k$, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k} - \sum_{k=n}^{m} \dfrac{\log k}{4k^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that $\displaystyle H_m - H_n = \log(\dfrac{m+1}{n}) + O(\dfrac{1}{n} - \dfrac{1}{m})$, where $\displaystyle H_n = \sum_{k=1}^{n} \dfrac{1}{k}$ is the $\displaystyle n^{th}$ harmonic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $c_n = b_n - \dfrac{\log n}{2}$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle O(\dfrac{1}{n} -\dfrac{1}{m}) \gt c_{m+1} - c_n \gt O(\dfrac{1}{n} -\dfrac{1}{m}) -\sum_{k=n}^{m} \dfrac{\log k}{4k^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \sum_{k=1}^{\infty} \dfrac{\log k}{k^2}$ is convergent and so by the Cauchy convergence criteria, we have that $\displaystyle  c_n$ is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the sequence $\displaystyle a_{n}^2 - 2n - \dfrac{\log n}{2}$ converges and hence, for some $\displaystyle C$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in other words&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle a_{n} \sim \sqrt{2n + \dfrac{\log n}{2} - C}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick (possibly incorrect) computer simulation seems to show a very slow convergence to $\displaystyle C = 1.47812676429749\dots$ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The previous answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: Consider $(a_n)^2$ and try to apply similar reasoning as you did for $a_n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T18:19:39.677" LastActivityDate="2010-11-15T18:19:39.677" CommentCount="18" />
  <row Id="10067" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:16:36.757" Score="11" ViewCount="0" Body="&lt;p&gt;Use $a_{n+1} = \frac{1}{a_n} + \frac{1}{a_{n-1}} + \cdots + 1$ and $a_n \longrightarrow \infty$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-12T23:16:36.757" CommentCount="4" />
  <row Id="10068" PostTypeId="1" AcceptedAnswerId="10137" CreationDate="2010-11-12T23:18:01.070" Score="0" ViewCount="146" Body="&lt;p&gt;Let $f(x)$ continuous function on $R$ wich can be in different signs. Prove, that there is exists an arithmetic progression $a, b, c   (a&amp;lt;b&amp;lt;c)$,  such that $f(a)+f(b)+f(c)=0$.&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-13T09:19:56.870" LastActivityDate="2010-11-13T15:58:16.427" Title="Continuous function of one variable" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10069" PostTypeId="1" CreationDate="2010-11-12T23:22:06.260" Score="2" ViewCount="87" Body="&lt;p&gt;Say we have the function $$G(z) = \sum_{n \geq 0} g_n z^n.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for the transform T defined so that $$(T(G))(z) = \sum_{n \geq 0} g_n z^{n^2}?$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything known about this kind of transformation?&lt;/p&gt;&#xA;" OwnerUserId="3456" LastActivityDate="2010-11-12T23:22:06.260" Title="What is known about the transformation of a power series in which $z^n$ is replaced with $z^{n^2}$?" Tags="&lt;transformation&gt;&lt;power-series&gt;" CommentCount="10" FavoriteCount="1" />
  <row Id="10070" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:26:53.727" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: prove by induction that $a_n \leq 2\sqrt{n}$.&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-11-12T23:26:53.727" />
  <row Id="10071" PostTypeId="2" ParentId="10065" CreationDate="2010-11-12T23:28:11.217" Score="5" ViewCount="0" Body="&lt;p&gt;As $a_{n+1}^2 = a_n^2 + 2 + \frac{1}{a_n^2}$, we know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ a_{n+1}^2 \leq a_n^2 + 3 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_n^2 \leq 3n$, then $a_{n+1}^2 \leq 3(n+1) $, and since $a_1^2 = 1 \leq 3$, by the induction hypothesis $a_n^2 \leq 3n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\frac{a_n^2}{n^2} \leq 3/n$ for all $n$, and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \rightarrow \infty} \frac{a_n^2}{n^2} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from this it follows though that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \rightarrow \infty} \frac{a_n}{n} = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-11-12T23:28:11.217" CommentCount="0" />
  <row Id="10072" PostTypeId="1" AcceptedAnswerId="10075" CreationDate="2010-11-12T23:30:25.213" Score="2" ViewCount="88" Body="&lt;p&gt;I have a book that uses the notation $l^\infty(\mathcal{F})$, where $\mathcal{F}$ is a set of measurable functions, $f \in \mathcal{F}$ then $f \colon \mathcal{X} \rightarrow \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure what $l^\infty(\mathcal{F})$ is. I know what $l^\infty$ is by itself, the set of bounded sequences (Wikipedia), but not sure about this one.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-12T23:39:00.287" LastActivityDate="2010-11-13T00:18:58.537" Title="what could this notation mean?" Tags="&lt;real-analysis&gt;&lt;probability&gt;&lt;statistics&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10073" PostTypeId="1" CreationDate="2010-11-12T23:40:09.507" Score="2" ViewCount="206" Body="&lt;p&gt;In complex analysis, we care about the functions of a complex variable. Are such functions just special case of the complex valued function defined in Rudin's Principles of Mathematical Analysis, since they have many similar properties such as uniformly convergence in series. Moreover, I wonder the relationship between the functions of two real variables and the functions of a complex variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T09:16:38.353" LastActivityDate="2010-11-25T09:16:38.353" Title="A question about complex analysis" Tags="&lt;soft-question&gt;&lt;complex-analysis&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="10074" PostTypeId="2" ParentId="10068" CreationDate="2010-11-12T23:43:33.747" Score="3" ViewCount="0" Body="&lt;p&gt;Try defining the following function on $R \times R^+$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g(x,h) = f(x-h) + f(x) + f(x+h) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argue that if $f$ can take positive and negative values, then so can $g$.  Then, use the intermediate value theorem.  &lt;/p&gt;&#xA;" OwnerUserId="2515" LastEditorUserId="2515" LastEditorDisplayName="" LastEditDate="2010-11-12T23:51:18.037" LastActivityDate="2010-11-12T23:51:18.037" />
  <row Id="10075" PostTypeId="2" ParentId="10072" CreationDate="2010-11-13T00:18:58.537" Score="5" ViewCount="0" Body="&lt;p&gt;Did you check page 34 (which I got by checking the List of Symbols on page 506)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=seH8dMrEgggC&amp;amp;pg=PA34&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=seH8dMrEgggC&amp;amp;pg=PA34&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/QlpEf.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-13T00:18:58.537" />
  <row Id="10076" PostTypeId="2" ParentId="10073" CreationDate="2010-11-13T00:21:39.810" Score="7" ViewCount="0" Body="&lt;p&gt;$\newcommand{\Hol}{\mathcal{H}}$&#xD;&#xA;Complex analysis (usually) refers specifically to questions involving differentiable functions starting with the deceptively simply looking case of functions of one varible on open subsets of the complex plane. Already here there are many differences from the theory of differentiable functions $f: (a,b) \to \mathbb{R}$. Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $G \subset \mathbb{C}$ is open and $f: G \to \mathbb{C}$ is once differentiable or has an antiderivative, it is infinitely often differentiable. Such functions are called &quot;holomorphic&quot; or &quot;holomorphic on $G$&quot; if one wants to be precise (sometimes written $f \in \Hol(G)$). This is in stark contrast to the real case where a differentiable function may fail to have even a continuous derivative.&lt;/li&gt;&#xA;&lt;li&gt;If $f \in \Hol(G)$, then for any $z \in G$ there exists an open ball around $z$ contained in $G$ on which $f$ is equal to its own Taylor series at $z$. Even smooth real functions are not this nice. Wikipedia has an article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&quot; rel=&quot;nofollow&quot;&gt;the standard example&lt;/a&gt; - though I also like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fabius_function&quot; rel=&quot;nofollow&quot;&gt;Fabius function&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Cauchy's integral formula which roughly states that if you know the values of a holomorphic function on the boundary of a disk, you know its values everywhere inside the disk.&lt;/li&gt;&#xA;&lt;li&gt;Liouville's theorem which states that if $f \in \Hol(\mathbb{C})$ is bounded, then $f$ is constant. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Essentially: When you're wondering whether the theory of holomorphic functions is &quot;nice&quot; in some way, the answer is usually yes. Whereas the theory of even smooth functions is a lot more messy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not provided links to Wikipedia's articles on the relevant theorems. Instead I recommend &lt;a href=&quot;http://www.math.ku.dk/noter/filer/koman-10.pdf&quot; rel=&quot;nofollow&quot;&gt;these free notes&lt;/a&gt; by Christian Berg from the University of Copenhagen. They are in English and have been in use in the university's first course in complex analysis for close to a decade (to the best of my knowledge) and have had multiple updates and corrections over the years.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-13T00:21:39.810" CommentCount="1" />
  <row Id="10077" PostTypeId="2" ParentId="10073" CreationDate="2010-11-13T00:51:55.360" Score="1" ViewCount="0" Body="&lt;p&gt;For the second part of your question, one can think of complex functions as continuous functions of two real variables that satisfy some differential equations. The differential equations are the Cauchy-Riemann equations mentioned in the answer Jonas Meyer linked to: &lt;a href=&quot;http://math.stackexchange.com/questions/5108/how-is-mathbbc-different-than-mathbbr2/5111#5111&quot;&gt;How is $\mathbb{C}$ different than $\mathbb{R}^2$?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives straight away a reason why all functions of two real variables are not functions of a single complex variable, since not all functions of two real variables satisfy those differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat deeper understanding of what is going on can be obtained by looking at the complex coordinates themselves. In two real variables we often denote our coordinates by $(x,y)$. Let's agree to identify the complex plane $\mathbb C$ with $\mathbb R^2$ in the usual way (i.e. via $z = x + iy$). Then we can perform a coordinate change; instead of looking at $(x,y)$, we will consider the coordinates $(z,\bar z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very subtle change in point of view, and merits your attention. Please take the time to convince yourself that this amounts to a change of basis of the $\mathbb R$-vector space $\mathbb R^2$: if $e_1 = (1,0)$ and $e_2 = (0,1)$ is the usual basis, then the new basis is given by $e_1 + e_2 = (1,1)$ and $e_1 - e_2 = (1,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that any function $f$ of the real variables $(x,y)$ can be expressed as a function of the new variables $(z,\bar z)$. Now one can say that just as in the case of real analysis we start by considering the case of one real variable, then we want to consider one complex variable to begin with in the case of complex analysis. This leads us to consider the functions $f$ which only depend on the variable $z$ (or $\bar z$, the choice amounting to a choice of a square root of $-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we consider those functions $f$ of the variables $(z,\bar z)$ which do not depend on $\bar z$. Another way to express this property is to say that we require $f$ to satisfy the differential equation $\partial f/\partial \bar z = 0$, but once you write out what this means in the original coordinates $(x,y)$ this equation turns out to be equivalent to the Cauchy-Riemann equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, when we investigate what it means for a function to depend on a single complex variable, we stumble upon the Cauchy-Riemann equations, and thus upon the definition of a holomorphic function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wholeheartedly recommend the first chapter of Greene and Krantz's &quot;Function theory of one complex variable&quot; for a more in depth discussion of this point of view. They take some lovely examples with polynomials in $(x,y)$ and then in $z$ which show quite explicitly what is going on.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-13T00:51:55.360" CommentCount="1" />
  <row Id="10078" PostTypeId="1" CreationDate="2010-11-13T00:56:09.783" Score="0" ViewCount="74" Body="&lt;p&gt;If I have a distribution mass function p(Y) , and values of $X_0,X_1,...,X_i$, I want to assign values to $X_{i+1},...,X_n$ such that the series of X follows p(Y) distribution where Y is the set of all distinct values of X. Is there any technique to do this?&lt;/p&gt;&#xA;" OwnerUserId="2858" LastEditorUserId="2858" LastEditorDisplayName="" LastEditDate="2010-11-13T03:18:33.493" LastActivityDate="2010-11-18T01:31:52.727" Title="Creating data set to match a particular distribution." Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10079" PostTypeId="2" ParentId="9355" CreationDate="2010-11-13T01:29:49.997" Score="6" ViewCount="0" Body="&lt;p&gt;This holds for &lt;em&gt;any&lt;/em&gt; irreducible finite state Markov chain. I'm not sure about a pure combinatorial proof, but it can be shown quite quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M(t)$ be the expected time the chain takes to return to its initial state, if started at $t$. As it spends, on average, a time $M(t)-1$ outside the state $t$ after each visit to $t$, the fraction of its time spent at $t$ is $1/M(t)$. So, the stationary distribution is $\pi(t)=1/M(t)$ (this is standard, see &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Markov_chain&amp;amp;oldid=394586440#Steady-state_analysis_and_limiting_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $X_n$ be a Markov chain with the given transition matrix and, for a fixed state $t$, consider the process $H(X_n,t)$. This is just the expected time remaining before it next hits $t$. On average, at each step, this will decrease by 1 when $X_n\not=t$ and increase by $M(t)-1=1/\pi(t)-1$ when $X_n=t$. So, $Y_n\equiv\sum_tH(X_n,t)\pi(t)$ stays the same on average across each step, regardless of the value of $X_n$. That, is it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Martingale_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;martingale&lt;/a&gt;. Consider what happens when $Y_n$ reaches its maximum value. As it can't increase further and its expected value remains constant, $Y_n$ must remain constant. So, $\sum_tH(s,t)\pi(t)$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The bit above the line answers your question. However, the calculations in the linked paper follow through for all irreducible Markov chains. In particular the formula (3.3) holds. This is the product of the lengthy calculation that you mention. I would like to show this now, which will involve going through some calculations. Let $M_{ij}=\mathbb{P}(X_{n+1}=j\vert X_n=i)$ be the transition matrix, $H_{ij}=H(i,j)$ be the hitting time matrix, $I$ be the identity matrix, $J_{ij}=1$ be the matrix of ones, and $F$ be the diagonal matrix with entries being the mean return times $F_{ii}=1/\pi(i)$. The description above of the average change in $H(X_n,t)$ across each time step can be written in matrix form &#xA;$$&#xD;&#xA;(M-I)H=F-J.\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xD;&#xA;Note that $F\pi=J\pi={\bf 1}$. So, (1) gives $(M-I)H\pi=0$, which is just the martingale property described above. The only eigenvalues of M with eigenvector 1 are proportional to ${\bf 1}$. So, $H\pi$ is proportional to ${\bf 1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve (1). Let $M^*={\bf 1}\pi^{\rm T}$. Then $MM^*=M^*M=M^*$ and $M^*F=M^*J=J$. Trying $A=(I-M+M^*)^{-1}(J-F)$ in place of H in (1),&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;(M-I)A &amp;amp;=(I-M+M^*)^{-1}(M-I)(J-F)\\\\&#xD;&#xA;&amp;amp;=(I-M+M^*)^{-1}(M-M^*-I)(J-F)\\\\&#xD;&#xA;&amp;amp;=F-J.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;Also, the only square matrices satisfying $(M-I)A=0$ are the ones with constant columns, so of the form ${\bf 1}v^{\rm T}$ for a vector v. So, (1) has the general solution&#xA;$$&#xD;&#xA;H = (I-M+M^*)^{-1}(J-F)+{\bf 1}v^{\rm T}&#xD;&#xA;$$&#xD;&#xA;and, using the condition $H_{ii}=0$ gives&#xA;$$&#xD;&#xA;v_i=\left((1-M+M^*)^{-1}(F-J)\right)_{ii}.&#xD;&#xA;$$&#xD;&#xA;Next, $J\pi=F\pi={\bf 1}$ so,&#xA;$$&#xD;&#xA;H\pi = (v^{\rm T}\pi){\bf 1}&#xD;&#xA;$$&#xD;&#xA;has constant entries of value&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;v^{\rm T}\pi &amp;amp;= \sum_i\left((I-M+M^*)^{-1}(F-J)\right)_{ii}\pi_i\\\\&#xD;&#xA;&amp;amp;= {\rm Tr}\left[(I-M+M^*)^{-1}(F-J)F^{-1}\right]\\\\&#xD;&#xA;&amp;amp;= {\rm Tr}\left[(I-M+M^*)^{-1}(I-M^*)\right].&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;If the eigenvalues of M are $\lambda_1,\ldots,\lambda_n$ with $\lambda_1=1$, then it can be seen that the corresponding eigenvalues of $(I-M+M^*)^{-1}(I-M^*)$ are $0,(1-\lambda_2)^{-1},(1-\lambda_3)^{-1},\ldots,(1-\lambda_n)^{-1}$ giving&#xA;$$&#xD;&#xA;(H\pi)_i=\sum_{j=2}^n\frac{1}{1-\lambda_j}.&#xD;&#xA;$$&#xD;&#xA;This is independent of i, and is the same as equation (3.3) in the linked paper.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-13T21:59:40.457" LastActivityDate="2010-11-13T21:59:40.457" CommentCount="1" />
  <row Id="10080" PostTypeId="2" ParentId="10065" CreationDate="2010-11-13T01:47:51.817" Score="0" ViewCount="0" Body="&lt;p&gt;You might also want to take another sequence $b_n = \frac{a_n}{n}$ and use the recurrence to try to prove that $\displaystyle \lim_{n \rightarrow \infty} b_n = 0$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-13T01:47:51.817" />
  <row Id="10082" PostTypeId="1" CreationDate="2010-11-13T02:29:30.847" Score="0" ViewCount="87" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/51/the-cow-in-the-field-problem-intersecting-circular-areas&quot;&gt;The cow in the field problem (intersecting circular areas)&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;A farmer ties a cow with a rope to a pole at the edge of a round field of radius &lt;strong&gt;r&lt;/strong&gt;. How long must the rope be compared to &lt;strong&gt;r&lt;/strong&gt; if the farmer wants to ensure that the cow can roam exactly half of the field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The point is to find a &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed-form_expression&quot; rel=&quot;nofollow&quot;&gt;closed-form expression&lt;/a&gt;, not solve this with numerical analysis etc. Is that possible?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's a nice depiction of the problem here:&#xA;_http://www11.speedyshare.com/files/25162122/download/Cow.png&lt;/p&gt;&#xA;" OwnerUserId="3461" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T04:58:10.480" LastActivityDate="2010-11-13T04:58:10.480" ClosedDate="2010-11-13T19:08:43.160" Title="Is it possible to express the area of the intersection of 2 circles as a closed-form expression?" Tags="&lt;geometry&gt;" CommentCount="2" />
  <row Id="10083" PostTypeId="2" ParentId="9825" CreationDate="2010-11-13T03:12:34.047" Score="3" ViewCount="0" Body="&lt;p&gt;[This answer moved from &lt;a href=&quot;http://math.stackexchange.com/questions/9532/link-between-a-dense-subset-and-a-continuous-mapping&quot;&gt;this question&lt;/a&gt; on Arturo's advice]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the following goes a long way towards proving a converse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(Y, T)$ be any T1 topological space with at least two points and let $a$ and $b$ be distinct points in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X = Y\setminus\{b\}$.&#xA;Let $f: X \to Y$ be the inclusion of $X$ in $Y$.&#xA;Let $g: X \to Y$ agree with $f$ on $X\setminus\{a\}$ and $g(a) = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, define the topology on $X$ to be the coarsest topology that makes both $f$ and $g$ continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these assumptions it turns out that $X\setminus\{a\}$ is dense in $X$ if and only if $a$ and $b$ do not have disjoint neighbourhoods in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this is true, let us construct a base of the topology on $X$.&#xA;To make $f$ continuous we only need to take the subspace topology. Since&#xA;X is open in Y this is $S_1 = \{ G \in T \mid b \notin G \}$. To also make $g$ continuous we need to add the open neighbourhoods of $b$, with $b$ replaced by $a$. This gives $S_2 = \{ (H\setminus\{b\} \cup \{a\} \mid H \in T, b \in H \}$.&#xA;Now $S_1 \cup S_2$ is a subbase of the topology on $X$.&#xA;Since $S_1$ and $S_2$ are already closed under finite intersection, and each covers $X$, we can say that&#xA;$B = \{ G \cap H \mid G \in S_1, H \in S_2 \}$ is a base of the topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then (remembering that finite sets are closed in Y) we find that the following are all equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X\setminus\{a\}$ is not dense in $X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\{a\}$ is open in $X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\{a\} \in B$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;there are $G \in S_1, H \in S_2$ such that $G \cap H = \{a\}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;there are $G \in S_1, H \in S_2$ such that $a \in G$ and $G \cap (H\setminus\{a\}\cup\{b\}) = \oslash$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$a$ and $b$ have disjoint neighbourhoods in $(Y, T)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3457" LastEditorUserId="3457" LastEditorDisplayName="" LastEditDate="2010-11-14T09:26:26.920" LastActivityDate="2010-11-14T09:26:26.920" CommentCount="1" />
  <row Id="10084" PostTypeId="1" CreationDate="2010-11-13T03:54:15.760" Score="-2" ViewCount="143" Body="&lt;p&gt;I cant study for long periods of time (e.g. hours and hours straight). Is this bad?&lt;/p&gt;&#xA;" OwnerUserId="3463" LastActivityDate="2010-11-13T06:32:42.100" ClosedDate="2010-11-13T07:29:02.103" Title="Am I at a disadvantage?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10085" PostTypeId="1" CreationDate="2010-11-13T03:55:34.867" Score="12" ViewCount="409" Body="&lt;p&gt;I am not a math major but more of an amateur mathematician working in computational math. So the question I ask may not make complete sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am comfortable with the different sizes of infinities and Cantor's &quot;diagonal argument&quot; to prove that the set of all subsets of an infinite set has cardinality strictly greater than the set itself. So if we have a set $\Omega$ and $|\Omega| = \aleph_i$, then the cardinality of $2^{\Omega}$ is $|2^{\Omega}| = \aleph_{i+1} &amp;gt; \aleph_i$ and we have $\aleph_{i+1} = 2^{\aleph_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am fine with these argument. (I am assuming what I have written down makes sense!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why should there be a smallest $\infty$? Is there a proof that there exists a smallest $\infty$ and that this is what we address as &quot;countably infinite&quot;? or to rephrase the question why can't I find a set whose power set gives us $\mathbb{N}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I am asking this question is in &quot;some sense&quot; $\aleph_i = \log_2(\aleph_{i+1})$. I do not completely understand why this process should stop when $\aleph_i = \aleph_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Though coming to think about it I can feel that if I take an infinite set with $log_2 \aleph_0$ elements I can still put it in one-to-one correspondence with the Natural number set. So Is $log_2 \aleph_0 = \aleph_0$ (or) am I just confused? If $n \rightarrow \infty$, then $2^n \rightarrow \infty$ faster while $\log (n) \rightarrow \infty$ slower and $\log (\log (n)) \rightarrow \infty$ even slower and $\log(\log(\log (n))) \rightarrow \infty$ even &quot;more&quot; slower and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a clean (and relatively elementary) way to explain this to me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am totally fine if you direct me to some paper (or) webpage. I tried googling but could not find an answer to my question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I apologize if this question is too trivial or doesn't make sense.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-13T04:47:55.273" LastActivityDate="2010-11-13T04:47:55.273" Title="Why is $\aleph_0$ the smallest $\infty$?" Tags="&lt;set-theory&gt;&lt;logic&gt;&lt;philosophy&gt;&lt;infinity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" />
  <row Id="10086" PostTypeId="1" CreationDate="2010-11-13T03:57:54.960" Score="1" ViewCount="105" Body="&lt;p&gt;If my understanding is correct, naive set theory needs to be restricted in order to avoid paradoxes including the Russell paradox. Typically, the restriction is expressed in terms of size. For example, the set of all sets must be excluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall that I came across a paper in Arxiv some time ago which explained that a useful restriction may be expressed in terms of symmetry conditions rather than size. Can anyone explain to me the concept and/or provide a link to the paper in question?&lt;/p&gt;&#xA;" OwnerUserId="2949" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-11-13T07:18:37.563" LastActivityDate="2010-11-13T07:18:37.563" Title="Limiting set theory using symmetry" Tags="&lt;set-theory&gt;&lt;reference-request&gt;" CommentCount="1" />
  <row Id="10087" PostTypeId="2" ParentId="10085" CreationDate="2010-11-13T04:26:16.483" Score="11" ViewCount="0" Body="&lt;p&gt;First, let me clear up a misunderstanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:  Is $2^\omega = \aleph_1$?  More generally is $2^{\aleph_\alpha} = \aleph_{\alpha+1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer of &quot;yes&quot; to the first question is known as the continuum hypothesis (CH), while an answer of &quot;yes&quot; to the second is known as the generalized continuum hypthothesis (GCH).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:  Both are undecidable using ZFC.  That is, Godel has proven that if you assume the answer to CH is &quot;yes&quot;, then you don't add any new contradictions to set theory.  In particular, this means it's impossible to prove the answer is &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, Cohen showed that if you assume the answer is &quot;no&quot;, then you don't add any new contradictions to set theory.  In particular, this means it's impossible to prove the answer is &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for GCH is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is just to say that while you are allowed to assume an answer of &quot;yes&quot; to GCH (which is what you did in your post), there is no way you can prove that you are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that out of the way, let me address your actual question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there is a proof that $\omega$ is the smallest infinite cardinality.  It all goes back to some very precise definitions.  In short, when one does set theory, all one really has to work with is the &quot;is a member of&quot; relation $\epsilon$.  One defines an &quot;ordinal number&quot; to be any transitive set $X$ such that $(X,\epsilon)$ is a well ordered set.  (Here, &quot;transitive&quot; means &quot;every element is a subset&quot;.  It's a weird condition which basically means that $\epsilon$ is a transitive relation).  For example, if $X = \emptyset$ or $X=\{\emptyset\}$ or $X = \{\{\emptyset\},\emptyset\}$, then $X$ is an ordinal.  However, if $X=\{\{\emptyset\}\}$, then $X$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 important facts about ordinals.  First, EVERY well ordered set is (order) isomorphic to a unique ordinal.  Second, for any two ordinals $\alpha$ and $\beta$, precisely one of the following holds:  $\alpha\epsilon\beta$ or $\beta\epsilon\alpha$ or $\beta = \alpha$.  In fact, it turns out the the collection of ordinals is well ordered by $\epsilon$, modulo the detail that there is no set of ordinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, cardinal numbers are simply special kinds of ordinal numbers.  They are ordinal numbers which can't be bijected (in, perhaps NOT an order preserving way) with any smaller ordinal.  It follows from this that the collection of all cardinal numbers is also well ordered.  Hence, as long as there is one cardinal with a given property, there will be a least one.  One example of such a property is &quot;is infinite&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me just point out that for FINITE numbers (i.e., Finite natural numbers), one usually cannot find a solution $m$ to $m = log_2(n)$.  Thus, as least from an inductive reasoning point of view, it's not surprising that there are infinite cardinals which can't be written as $2^k$ for some cardinal $k$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-11-13T04:26:16.483" />
  <row Id="10088" PostTypeId="2" ParentId="10085" CreationDate="2010-11-13T04:30:28.747" Score="9" ViewCount="0" Body="&lt;p&gt;Suppose $A$ is an infinite set. In particular, it is not empty, so there exists a $x_1\in A$. Now $A$ is infinite, so it is not $\{x_1\}$, so there exists an $x_2\in A$ such that $x_2\neq x1$. Now $A$ is infinite, so it is not $\{x_1,x_2\}$, so there exists an $x_3\in A$ such that $x_3\neq x_1$ and $x_3\neq x_2$. Now $A$ is infinite, so it is not $\{x_1,x_2,x_3\}$, so there exists an $x_4\in A$ such that $x_4\neq x_1$, $x_4\neq x_2$  and $x_4\neq x_3$... And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was you can construct a sequence $(x_n)_{n\geq1}$ of elements of $A$ such that $x_i\neq x_j$ if $i\neq j$. If you define a function $f:\mathbb N\to A$ so that $f(i)=x_i$ for all $i\in\mathbb N$, then $f$ is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, then, the cardinal of $A$ is greater or equal to that of $\mathbb N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathbb N$ is itself infinite, this shows that the smallest infinite cardinal is $\aleph_0=|\mathbb N|$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-13T04:30:28.747" CommentCount="2" />
  <row Id="10089" PostTypeId="2" ParentId="10085" CreationDate="2010-11-13T04:36:18.180" Score="3" ViewCount="0" Body="&lt;p&gt;If you assume both the continuum hypothesis and the (weaker) axiom of countable choice, then no infinite set can have cardinality strictly smaller than that of $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, assume the continuum hypothesis, that there is no set of cardinality strictly between $\aleph_{0} = | \mathbb{N} |$ and $2^{\aleph_{0}} = \aleph_{1}$. Suppose $X$ is an infinite set of cardinality strictly smaller than $\aleph_{0} = |\mathbb{N}|$. Then the corresponding power sets must satisfy $|\mathcal{P}(X)| = 2^{|X|} \leq 2^{|\mathbb{N}|} = \aleph_{1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|\mathcal{P}(X)| = \aleph_{1}$, then it follows that $2^{|X|} = 2^{|\mathbb{N}|}$, hence $|X| = |\mathbb{N}|$, which is impossible by construction. Thus, $|\mathcal{P}(X)|$ &amp;lt; $\aleph_{1}$, and by CH, we have $|\mathcal{P}(X)|$ &amp;lt; $\aleph_{0}$. Certainly, $|X|$ &amp;lt; $|\mathcal{P}(X)|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the set $X$ has cardinality strictly less than a set (its power set) that has cardinality strictly less than $\aleph_{0}$. Continuing in this fashion, we find an infinite sequence of decreasing transfinite cardinals, and $X$ is an infinite set with arbitrarily small transfinite cardinality. (If we also assume the generalized continuum hypothesis, then there are no infinite sets with cardinality between these sets either.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axiom of countable choice says that there is a least transfinite cardinal, and no such pathological set $X$ can therefore exist.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-13T04:36:18.180" CommentCount="15" />
  <row Id="10090" PostTypeId="2" ParentId="10068" CreationDate="2010-11-13T05:29:14.563" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Think about the intermediate value theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think about some $x$ and some $y$ with $f(x)\gt0$ and $f(y)\lt0$. &lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-13T14:30:54.847" LastActivityDate="2010-11-13T14:30:54.847" CommentCount="2" />
  <row Id="10091" PostTypeId="2" ParentId="10084" CreationDate="2010-11-13T06:32:42.100" Score="0" ViewCount="0" Body="&lt;p&gt;Everyone is like this.  Just study a hard 1 hour then take a 30 minute break and repeat.  This is especially nice before exams.  Find whatever works for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: You are at a disadvantage of about 1% of people.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-11-13T06:32:42.100" />
  <row Id="10092" PostTypeId="1" AcceptedAnswerId="10093" CreationDate="2010-11-13T07:38:36.263" Score="4" ViewCount="101" Body="&lt;p&gt;Are all dense open sets in R cocountable?&#xA;That is, are all dense open sets in R such that their complements are at most countable?&#xA;It would seem like they must be since the closed sets that are uncountable are all intervals, so their complements are not dense.&lt;/p&gt;&#xA;" OwnerUserId="3327" LastActivityDate="2010-11-28T20:09:30.240" Title="Dense Open Sets and Cocountability" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="10093" PostTypeId="2" ParentId="10092" CreationDate="2010-11-13T07:48:14.050" Score="6" ViewCount="0" Body="&lt;p&gt;No, the cardinality of their complements can be as big as that of all of the real numbers.  For an explicit example, the complement of the Cantor set is dense and open.  It consists of all of the &quot;middle third&quot; open intervals removed from $[0,1]$ together with the unbounded intervals $(-\infty,0)$ and $(1,\infty)$.  The Cantor set is a closed uncountable set whose connected components are points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complement of a dense open set can be &quot;big&quot;in respects other than cardinality.  Let the rational numbers be listed as $r_1,r_2,r_3,\ldots$, and for each $k$ let $I_k$ be an open interval of length $1/2^k$ containing $r_k$.  Then $U=I_1\cup I_2\cup I_3\cup\cdots$ is a dense open subset of $\mathbb{R}$, but $U$ is a union of intervals the sum of whose lengths is $1/2 + 1/2^2 + 1/2^3+\cdots=1$.  So in a sense the complement of this dense open set must take up &quot;most of the space&quot; in the real numbers.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-13T08:03:23.457" LastActivityDate="2010-11-13T08:03:23.457" CommentCount="0" />
  <row Id="10094" PostTypeId="2" ParentId="10092" CreationDate="2010-11-13T07:49:27.310" Score="3" ViewCount="0" Body="&lt;p&gt;No; the complement of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_Set&quot; rel=&quot;nofollow&quot;&gt;Cantor set&lt;/a&gt; is a dense open set in $\mathbb{R}$ that is not cocountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cantor set is closed and uncountable, so its complement is open. By the by, the Cantor set is a closed set that is uncountable but not an interval; in fact, the Cantor set has empty interior, so it cannot contain any intervals. Every open neighborhood of every point in the Cantor set contains both points of the Cantor set and points not in the Cantor set, so the complement is dense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-13T07:49:27.310" />
  <row Id="10095" PostTypeId="1" AcceptedAnswerId="10101" CreationDate="2010-11-13T08:58:48.923" Score="1" ViewCount="111" Body="&lt;p&gt;Hi, I'm very n00b in maths, and I don't understand what I can do with a generating function.&lt;br&gt;&#xA;For example, if I want to compute the first spread polynomial, how can I do that from the generating function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TBMoJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the exponential one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bCswn.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;rubik&lt;/p&gt;&#xA;" OwnerUserId="2582" LastEditorUserId="2582" LastEditorDisplayName="" LastEditDate="2010-11-14T08:47:48.170" LastActivityDate="2010-11-14T08:47:48.170" Title="How to find an explicit formula from a generating function" Tags="&lt;polynomials&gt;&lt;generating-functions&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="10096" PostTypeId="1" CreationDate="2010-11-13T09:03:46.283" Score="3" ViewCount="117" Body="&lt;p&gt;One of the answers to &lt;a href=&quot;http://math.stackexchange.com/questions/2502/what-are-large-cardinals-for&quot;&gt;this question&lt;/a&gt; indicates that large cardinals are useful for destructive testing of set theory. That aside, and not assuming GCH, are there any sets known that have a cardinality of $\aleph_{n}$, $n&amp;gt;0$, or that of any of the large cardinals? There are a few examples of sets that have these cardinalities on &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;the wikipedia page&lt;/a&gt;, but they are meager and few compared to the examples on the page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Beth_number&quot; rel=&quot;nofollow&quot;&gt;Beth cardinalities&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-13T14:08:37.633" LastActivityDate="2010-11-13T14:30:01.057" Title="Examples of sets whose cardinalities are $\aleph_{n}$, or any large cardinal. (not assuming GCH)" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;large-cardinals&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10097" PostTypeId="1" CreationDate="2010-11-13T09:30:09.040" Score="1" ViewCount="239" Body="&lt;p&gt;Let $f:\mathbb{R} \to \mathbb{R}$ is a function with these special properties. $f$ is continuous everywhere. $f$ is not smooth (not infinitely differentiable). $f$ is differentiable only finitely many times everywhere. $f$ belongs to $L^p$. For any $k$ belongs to $\mathbb{N}$ let $E$ be the set of all points where $f$ is differentiable exactly $k$ times then $E$ is not dense in any open subset of $\mathbb{R}$.Is such a function feasible? I need suggestions on ways to construct such a special function.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the final version. not going to change it further. I apologize for the inconvinience&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-13T11:03:52.330" LastActivityDate="2010-11-17T18:14:55.360" Title="Construction of a special  function" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="10098" PostTypeId="2" ParentId="154" CreationDate="2010-11-13T10:05:31.230" Score="4" ViewCount="0" Body="&lt;p&gt;I think some confusion comes from calling the numbers that only have imaginary part &quot;imaginary numbers&quot;. I believe if mathematicians had other name to give to them (but I am really not thinking on a better name), there would be less confusion.&lt;/p&gt;&#xA;" OwnerUserId="3467" LastActivityDate="2010-11-13T10:05:31.230" CommentCount="1" />
  <row Id="10100" PostTypeId="2" ParentId="10097" CreationDate="2010-11-13T10:26:30.700" Score="2" ViewCount="0" Body="&lt;p&gt;I deleted this temporarily while I was working on an update. Also... whoever upvoted this, please remove your upvote as this answer is incorrect. The older answers are below this update. As for why it took me so long to edit this to reflect that it's wrong, that's because there has been water in the telephone cable or something. This area has been without phones and internet for about 48 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of trying to answer this question, I'll try and point out some examples of odd behaviour. The essential problem is that an exhibit of a function with similar properties as those you stated is probably going to be as a limit of differentiable functions. But uniform limits and differentiability don't play that nicely together - q.v.  [&lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence#to_Differentiability&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start simple. Consider $f_1 \in C^1([-1,1])$ and define $f_n$ as follows:&#xA;$$ f_n(x) = \begin{cases}&#xD;&#xA;g_n(x) &amp;amp; \text{ if } -\tfrac{1}{n} \lt x \lt \tfrac{1}{n} \\\\&#xD;&#xA;f_{n-1}(x) &amp;amp; \text{ otherwise,}&#xD;&#xA;\end{cases} $$&#xD;&#xA;where $g_n \in C^n([-\tfrac{1}{n},\tfrac{1}{n}])$ such that $g_n^{(k)}(\tfrac{\pm 1}{n}) = f_{n-1}^{(k)}(\tfrac{\pm 1}{n})$, $k = 0,\ldots n-1$ and $\sup_{x \in [-\tfrac{1}{n},\tfrac{1}{n}]} \lvert g_n (x)- f_{n-1}(x) \rvert &amp;lt; 2^{-n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By construction this is a Cauchy sequence in $C[-1,1]$, so the limit, $f$ is continuous. Moreover for any $X \subset [-1,1]$ that doesn't contain a neighbourhood of $0$, the sequence $(f_n|_X)$ is eventually constant. So we know everything about $f$'s behaviour with respect to differentiability everywhere except at $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly it's possible for $f$ to be smooth at $0$ - for example if $(f_n)$ is eventually constant on some neighbourhood of $0$. But $g_n$ can have pretty much any behaviour on $[-\tfrac{1}{2n},\tfrac{1}{2n}]$, so $(g_n^{(k)}(0))$ can be &lt;em&gt;any&lt;/em&gt; sequence of real numbers for each $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This demonstrates that limits of real functions can be truly bizarre. The $f_n$ get nicer and nicer around $0$, but the limit can have any behaviour at all at $0$: Not differentiable, exactly $k$ times differentiable or smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure this next idea says anything at all about the problem (because the limit might not be differentiable at all), but I find it a bit of a waste to not write it down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall say that a function $f$ is &quot;$F^k$ on $U$&quot; if $f$ is continuous and $k$ times differentiable on $U$ and nowhere $k+1$ times differentiable in $U$. Obviously the requirement that $f$ be continuous isn't needed for $k \neq 0$ ($f$ differentiable $\Rightarrow$ $f$ continuous), but stating it twice isn't a problem. We are mostly going to be concerned about intervals, so for endpoints of intervals we shall in addition require that the one-sided derivatives exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $f_0$ be an $F^0$ function on $[0,1]$, asumme that we have defined the sequence up to $f_{n-1}$. We define $f_{n}$ in the following way: For each $k = 0, \ldots n-1$ and for each maximal open subinterval where $f_{n-1}$ is $F^k$, set $f_n$ to be an $F^{k+1}$ function on the two outer thirds of the interval such that the $k$ derivatives of $f_{n-1}$ and $f_n$ agree on the endpoints and such that $|f_n(x) - f_{n-1}(x)| &amp;lt; 2^{-n}$ for each $x$ in the interval (so the sequence is Cauchy). On the inner one third of the intervals we set $f_n = f_{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simple ASCII diagram should help a bit with grokking the idea:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;     0                             F^0                             1&#xA;f_0: |-------------------------------------------------------------|&#xA;              F^1                  F^0                  F^1&#xA;f_1: |-------------------)(-------------------)(-------------------|&#xA;       F^2    F^1    F^2    F^1    F^0    F^1    F^2    F^1    F^2 &#xA;f_2: |-----)(-----)(-----)(-----)(-----)(-----)(-----)(-----)(-----]&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If this makes you think of the Cantor set and the Towers of Hanoi, it's not a coincidence - that's exactly how I got the idea. Anyway as I said, the limit of the $f_n$ might not be differentiable anywhere, but it seems to me that it's at least a starting point for constructing a sequence which has a limit that's $F^k$ on a nowhere dense set in $[0,1]$ for all $k &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This answer is based on a misreading of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; There exists no $f \in C(\mathbb{R})$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f \notin C^\infty(\mathbb{R})$,&lt;/li&gt;&#xA;&lt;li&gt;If $k \in \mathbb{N}_{0}$, then $E_k := \lbrace x \in \mathbb{R} \mid f \text{ is exactly } k \text{ times differentiable at } x\rbrace$ is nowhere dense.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; Consider $\cup_{k=0}^\infty E_k$. This has to be all of $\mathbb{R}$ since $f$ is continuous. But then we'd have written $\mathbb{R}$ as a countable union of nowhere dense sets contradicting that $\mathbb{R}$ is a Baire space.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is my second answer which is also wrong&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; There exists no function, $f$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f \in C(\mathbb{R})$,&lt;/li&gt;&#xA;&lt;li&gt;for all $k \in \mathbb{N}$ the set of points where $f$ is exactly $k$ times differentiable is nowhere dense,&lt;/li&gt;&#xA;&lt;li&gt;$f$ is nowhere smooth.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof.&lt;/strong&gt; Recall that if $f$ is differentiable at a point $x_1$, there exists an entire open interval (UPDATE: no there doesn't. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_function&quot; rel=&quot;nofollow&quot;&gt;Cantor function&lt;/a&gt; is a counterexample), $I_1$ around $x_1$ where $f$ is differentiable. Find a closed interval, $J_1$ containing $x_1$ inside $I_1$. Since the set of points where $f$ is exactly $k$ times differentiable is nowhere dense, there must be a point $x_2 \in J_1$ where $f$ is twice differentiable. Like before we get a closed interval $J_2$ contained in $J_1$ where $f$ is twice differentiable. Proceed by induction and we get a nested sequence of closed intervals, $J_1 \supset J_2 \supset \dotsb$. By the nested interval theorem, $\cap_{\mathbb{N}} J_n$ is nonempty and it's easy to see that $f$ has to be smooth on all the points in the intersection, contradicting our assumption that $f$ was nowhere smooth.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-17T18:14:55.360" LastActivityDate="2010-11-17T18:14:55.360" CommentCount="12" />
  <row Id="10101" PostTypeId="2" ParentId="10095" CreationDate="2010-11-13T10:32:56.997" Score="4" ViewCount="0" Body="&lt;p&gt;If you know what Taylor expansion is, then you should know that $S_n(s)$ in e.g.f. is just its $n$th derivative at zero. For the ordinary g.f., it's almost the same, except that you have to divide the derivative by $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if you want to do more with g.f. than this, I strongly advise you to at least look at this &lt;a href=&quot;http://en.wikipedia.org/wiki/Generating_function&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; page and perhaps even into the &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt; book, which in my opinion does a pretty good job of introducing g.f. to a beginner.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastEditorUserId="3223" LastEditorDisplayName="" LastEditDate="2010-11-13T12:40:59.810" LastActivityDate="2010-11-13T12:40:59.810" CommentCount="1" />
  <row Id="10102" PostTypeId="1" AcceptedAnswerId="10132" CreationDate="2010-11-13T10:44:00.693" Score="10" ViewCount="234" Body="&lt;p&gt;A friend of mine is a differential geometer, and keeps insisting that he doesn't need the axiom of choice for the things he does. I'm fairly certain that's not true, though I haven't dug into the details of Riemannian geometry (or the real analysis it's based on, or the topology, or the theory of vector spaces, etc...) to try and find a theorem or construction that uses the axiom of choice, or one of its logical equivalences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do you know of any result in Riemannian geometry that needs the axiom of choice? They should be there somewhere, I particularily suspect that one or more is hidden in the basic topology results one uses.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-13T16:18:41.227" Title="Where do we need the axiom of choice in Riemannian geometry?" Tags="&lt;topology&gt;&lt;logic&gt;&lt;differential-geometry&gt;&lt;axiom-of-choice&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="10103" PostTypeId="2" ParentId="10097" CreationDate="2010-11-13T10:44:53.003" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a construction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the operator $T:X\to Y$ where $X=L^1\cap L^\infty$ defined by &#xA;$$T(f)(x)=f*g(x)$$ where $g(x)=e^{-x^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some properties of $Y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Young's inequality on convolutions we see that $T(f)\in L^r$ where $1/r=1/p+1/q-1$ provided $f\in L^p$ and $g\in L^q$ -  in our case this is true for all $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $T(f)$ is smooth (we talked about that in a previous thread of yours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we by some reason wish to destroy the smoothness, we can if wish use $g(x)=e^{-x^2}h(x)$ where $h\in C^n\cap L^\infty$ instead of the previous $g$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-13T10:44:53.003" CommentCount="6" />
  <row Id="10104" PostTypeId="1" AcceptedAnswerId="10117" CreationDate="2010-11-13T11:24:23.543" Score="1" ViewCount="39" Body="&lt;p&gt;Let any Amicable Pair be represented by (a, b), with a &amp;lt; b. In a search for Amicable Pairs, let (n, m) with n &amp;lt; m, be a candidate to be tested for being an amicable pair. Given n, is there way to calculate both the least upper bound and greatest lower bound for the m that would make (n, m) an Amicable Pair?&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-13T13:03:59.263" Title="Amicable Pair (a, b): given a, what are limits on size of b?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10105" PostTypeId="2" ParentId="10047" CreationDate="2010-11-13T11:29:34.290" Score="1" ViewCount="0" Body="&lt;p&gt;This is only true in pathological cases. Neither in complex analysis (see above) nor in algebraic geometry: If we take the structure sheaf $\mathcal{O}_{Spec(A)}$ of an affine scheme, then the question is whether the localzation maps $A \to A_{\mathfrak{p}}$ are surjective, which is almost never true.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-13T11:29:34.290" CommentCount="2" />
  <row Id="10106" PostTypeId="2" ParentId="10047" CreationDate="2010-11-13T11:38:08.403" Score="1" ViewCount="0" Body="&lt;p&gt;This raises another question. If the restriction maps from $S(X)$ to $S(U)$&#xD;&#xA;are surjective for all open sets $U$ (a sheaf with this&#xA;property is called &lt;em&gt;flasque&lt;/em&gt; or &lt;em&gt;flabby&lt;/em&gt;) then certainly the maps&#xA;from $S(X)$ to each stalk $S_x$ are surjective. So are there sheaves&#xA;$S$ with the property that each map $S(X)\to S_x$ is surjective but which are&#xA;not flabby? I expect there are, but cannot see an example immediately.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-13T11:38:08.403" CommentCount="1" />
  <row Id="10107" PostTypeId="2" ParentId="9513" CreationDate="2010-11-13T11:40:11.503" Score="0" ViewCount="0" Body="&lt;p&gt;Let $s$ be the special point of $Spec(R)$, i.e. the one which corresponds to the unique maximal ideal of $R$. Then every point $p$ of $Spec(R)$ generalizes to $s$, i.e. $s \in \overline{\{p\}}$. If $f : Spec(R) \to X$ is a continuous map, we thus have $f(s) \in \overline{\{f(p)\}}$, i.e. every open neighborbood of $f(s)$ contains $f(p)$. In other words, if $U$ is an open neighborhood of $f(s)$, the map $f$ actually factors as $Spec(R) \to U \to X$. If $f$ is a morphism of schemes, this is also a factorization in the category of schemes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a morphism $f : Spec(R) \to X$ mapping $s \mapsto x$ corresponds to a equivalence class of morphisms $Spec(R) \to U$ mapping $s \mapsto x$, where $U$ is an affine open neighboord of $x$. Such a morphism corresponds to a homomorphism $\mathcal{O}_X(U) \to R$, and since we consider equivalence classes, we get a homomorphism $\mathcal{O}_{X,x} \to R$, whose locality corresponds to the condition $s \mapsto x$ above.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-13T11:40:11.503" />
  <row Id="10108" PostTypeId="2" ParentId="9749" CreationDate="2010-11-13T11:50:39.437" Score="0" ViewCount="0" Body="&lt;p&gt;An application which is used a lot in any calculus course:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of a differentiable function vanishes at an extremum(ie maximum or minimum) point.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-13T11:50:39.437" />
  <row Id="10109" PostTypeId="1" AcceptedAnswerId="10111" CreationDate="2010-11-13T12:07:51.213" Score="1" ViewCount="47" Body="&lt;p&gt;When operating with sigma algebras, what does it mean when we talk about a countably infinite set? And, what are then closed countably infinite interesections?&lt;/p&gt;&#xA;" OwnerUserId="3468" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-13T14:00:17.587" LastActivityDate="2010-11-13T14:00:17.587" Title="What's a closed countably infinite intersection? " Tags="&lt;measure-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10110" PostTypeId="2" ParentId="10073" CreationDate="2010-11-13T12:21:29.893" Score="0" ViewCount="0" Body="&lt;p&gt;Holomorphicity is a stronger condition compared to differentiability in real variables. Here are some theorems of complex analysis, which are not true in real analysis.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If two holomorphic functions agree on an open set, they have to agree everywhere. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any bounded function holomorphic on all of $\mathbb C$ is necessarily constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A nonconstant holomorphic function can miss at most one value.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can see from these theorems that one has much more control over a holomorphic function, as opposed to a differentiable function of two real variables.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-13T12:21:29.893" />
  <row Id="10111" PostTypeId="2" ParentId="10109" CreationDate="2010-11-13T12:21:34.263" Score="4" ViewCount="0" Body="&lt;p&gt;I expect that you have read that a sigma algebra $\mathcal{A}$ has&#xA;to be &quot;closed under countably infinite intersections&quot;. What this means&#xA;is that if you have a family $(A_n)$ of elements of $\mathcal{A}$&#xD;&#xA;indexed by the natural numbers $\mathbb{N}$, then the intersection&#xA;$\bigcap_{n\in\mathbb{N}}A_n$ of all the $A_n$ is also an element of $\mathcal{A}$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-13T12:27:45.993" LastActivityDate="2010-11-13T12:27:45.993" CommentCount="0" />
  <row Id="10112" PostTypeId="1" AcceptedAnswerId="10135" CreationDate="2010-11-13T12:31:42.080" Score="3" ViewCount="181" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, since I do a job, and work on mathematics in my spare time, I do not have any connection with the academia. Secondly, the internet is the only way I can interact and reach out to other mathematicians or math enthusiasts like me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I need to discuss my paper with someone, where do I do so over the internet? (This is not about peer review, or about publishing my paper.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats my question, more clearly:&#xA;&lt;strong&gt;Where do I find people on the internet who can advise me on the papers that I write?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where I can discuss my papers on the internet?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically to give some background: I recently released a paper about a new zero-free region for Riemann Zeta Function &lt;a href=&quot;http://arxiv.org/pdf/0911.5572v14&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/0911.5572v14&lt;/a&gt;. I wish to get feedback about that. Do you know of any forums/groups, etc. where I can get a good and guiding response? Any links, forums, groups relevant to this topic would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roupam&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-11-14T05:48:01.437" Title="About finding advisors on the internet" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10113" PostTypeId="2" ParentId="9749" CreationDate="2010-11-13T12:34:04.600" Score="1" ViewCount="0" Body="&lt;p&gt;MVT is very important. In calculus and analysis, of course. But it's important in other areas too, like applied mathematics an even number theory. For example, for showing Liouville numbers are transcendental.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2010-11-13T12:34:04.600" />
  <row Id="10114" PostTypeId="2" ParentId="9378" CreationDate="2010-11-13T12:40:53.943" Score="0" ViewCount="0" Body="&lt;p&gt;The books &quot;Geometry of Schemes&quot; and &quot;The red book of Varieties and Schemes&quot; are great to get intuition about this. But here's a thing that you can do by yourself: try to produce your own examples and figures. That's not easy, needs training. But you should try.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2010-11-13T12:40:53.943" />
  <row Id="10115" PostTypeId="2" ParentId="10065" CreationDate="2010-11-13T12:42:44.087" Score="2" ViewCount="0" Body="&lt;p&gt;Lets rewrite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_{n+1}-a_n}{n+1-n}=\frac{1}{a_n}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as n goes to infinity lets denote $$a_n=f$$ .Then approximately:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d f}{dn}=\frac{1}{f}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After integrating:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f^2}{2}=n+c=&amp;gt;\frac{f^2}{n^2}=\frac{2}{n}+\frac{c}{n^2}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-11-13T12:42:44.087" CommentCount="1" />
  <row Id="10116" PostTypeId="2" ParentId="9378" CreationDate="2010-11-13T12:52:00.460" Score="0" ViewCount="0" Body="&lt;p&gt;Algebraic varieties are a lot like complex analytic spaces, which are special kinds of manifolds. So if you develop intuition about geometrical objects equipped with complex analytic functions, then it would help you a great deal in intuition on varieties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to get intuition on curves, which are simpler than the rest. And smooth complex projective curves are like compact Riemann surfaces, which can be visualized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could again look at pictures such as that of $Spec\ \mathbb Z$ and $Spec\ \mathbb Z[x]$ given in Mumford's Red book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The functor of points approach is also quite a nice way to think. Roughly, a scheme can be specified by its $R$-valued points for every ring $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example of a stalk is the notion of germs of holomorphic functions, which might be familiar from complex analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-13T12:52:00.460" />
  <row Id="10117" PostTypeId="2" ParentId="10104" CreationDate="2010-11-13T13:03:59.263" Score="1" ViewCount="0" Body="&lt;p&gt;An Amicable pair of numbers $(m,n)$ is such that each is equal to the sum of the proper divisors of the other, so if you know $n$ then $m=\sigma(n)-n,$ where $\sigma(n)$ is the usual sum of divisors function. Hence you are essentially asking for bounds on $\sigma(n).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\sigma(n)-n=1$ when $n$ is prime and so for a lower bound $m \ge 1,$ since $\sigma(n) \ge n+1$ for all $n \ge 2.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following upper bound for $\sigma(n)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma(n) &amp;lt; &#xD;&#xA;e^\gamma  n \log \log n + \frac{ 0.6482 n}{\log \log n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;was given on &lt;a href=&quot;http://mathoverflow.net/questions/33755/erdos-kac-for-sum-of-divisors&quot; rel=&quot;nofollow&quot;&gt;mathoverflow.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-13T13:03:59.263" CommentCount="4" />
  <row Id="10118" PostTypeId="2" ParentId="10102" CreationDate="2010-11-13T13:56:33.363" Score="7" ViewCount="0" Body="&lt;p&gt;It is very possible that your colleague generally avoids using the axiom of choice, because manifolds are relatively concrete objects.  Assuming that your friend is studying only separable manifolds, or better yet only compact ones, that makes it possible to do many constructions very explicitly, without using the axiom of choice. This is similar to the way that various principles of analysis that require the axiom of choice in general do not require the axiom of choice when they are applied to Euclidean spaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you're right that it may be easier to find AC in the background results.  The trouble is that many of these background results are studied in far more generality than they are used. For example, suppose an analyst uses the fact that $[0,1]\times[0,1]$ is compact. This fact follows from Tychonoff's theorem, which for general topological spaces does require the axiom of choice. But we could prove the compactness of the unit square more directly, avoiding the axiom of choice altogether (this relies on the separability of the square, in particular). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So sometimes the axiom of choice is used for convenience, by the invocation of a very general result, but it could be avoided if necessary. If we were to study only smooth manifolds that had already been embedded into Euclidean space, I suspect that we would be able to do pretty much everything in Zermelo-Fraenkel set theory without the axiom of choice. But it would take careful attention in the proofs to make sure that we replace choice-based techniques with alternative methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this isn't a direct answer to the question, but I think it's relevant since it explains a caveat with possible answers: just because a general result requires AC doesn't mean that AC is required for all consequences of that result. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-13T13:56:33.363" CommentCount="3" />
  <row Id="10119" PostTypeId="1" CreationDate="2010-11-13T14:14:52.020" Score="2" ViewCount="107" Body="&lt;p&gt;During lesson we were given a starter activity which was to try and factorise polynomials and see what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomials were&#xA;$x^3-8$,&#xA;$x^3-3x^2+ x -3$&#xD;&#xA;and&#xA;$x^4 - 16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not work out what happened to them, and it's bugging me. If anyone can explain then it'd be helpful! Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1968" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-13T15:09:54.000" LastActivityDate="2010-11-14T03:41:15.440" Title="Factorising certain polynomials." Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10120" PostTypeId="1" AcceptedAnswerId="10130" CreationDate="2010-11-13T14:16:16.450" Score="0" ViewCount="85" Body="&lt;p&gt;Hi, I've got a solution to a economy class calculation, but I can't figure out how they came to the answer. Could you help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-5p + 80 = 8p +-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.......? = ......? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;120 = 12p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p = 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra: do you know which rule they are using to get from the first line to the 3d?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math beginner,&#xA;Frank&lt;/p&gt;&#xA;" OwnerUserId="2605" LastActivityDate="2010-11-14T05:25:22.077" Title="quick basic math equation question" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10121" PostTypeId="1" AcceptedAnswerId="10127" CreationDate="2010-11-13T14:20:18.443" Score="1" ViewCount="50" Body="&lt;p&gt;This is actually a generalized version I wrote of a homework question that intrigued me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f$ be continuous in $[0,1]$ and&#xA;  $\forall x\in [0,1], \ f(x)=f(1-x)$.&#xA;  If $r\in [0,1]$ and $1-r$ are the only&#xA;  sub limits of $a_{n}$ then $f(a_{n})$&#xD;&#xA;  converges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(the original question states $r=\frac{1}{3}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved the original question using a method shown in class and tried to implement it to prove the generalized version (see my own answer). I was wondering if there's a better way.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-13T15:21:45.737" Title="if $r\in (0,1), 1-r$ are the only sub limits of $a_{n}$ then $f(a_{n})$ converges when..." Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10122" PostTypeId="2" ParentId="10121" CreationDate="2010-11-13T14:21:17.053" Score="0" ViewCount="0" Body="&lt;p&gt;Here's my own attempt at solving this (comments are appreciated!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We denote $b_{n}=f(a_{n})$ and by Bolzano-Weistrauss, $b_{n_{k}}\to b$.&#xA;The subsequence $a_{n_{k}}$ might not converge, but again by Bolzano-Weistrauss there exists $a_{n_{k_{m}}}$ that converges to either $r$ or $1-r$ (because those are the only sublimits of $a_{n}$).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if $a_{n_{k_{m}}}\to r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(r)$ (because $f$ is continuous in $[0,1]$).&lt;/li&gt;&#xA;&lt;li&gt;Similarly, if $a_{n_{k_{m}}}\to 1-r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(1-r)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But since $\forall x\in [0,1], \ f(x)=f(1-x)$, we get $f(r)=f(1-r)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finish the argument we note that $b_{n_{k_{m}}}$ is a subsequence of $b_{n_{k}}\to b$, thus $b_{n_{k_{m}}}\to b$ and we conclude that $b=f(r)=f(1-r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $b$ was chosen arbitrarily we get that $b_{n}$ has a single sublimit which implies $b_{n}\to f(r)$ &lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-13T14:21:17.053" />
  <row Id="10123" PostTypeId="2" ParentId="10096" CreationDate="2010-11-13T14:23:34.303" Score="3" ViewCount="0" Body="&lt;p&gt;There are very few examples where you directly prove in ZFC that a certain set must have size $\aleph_n$. This is because most of the sets we construct are defined in terms of power sets. This means that we can compute the size of these sets in terms of ℶ numbers pretty easily, but we can't compute them in terms of ℵ numbers. The difficulty is related to the unprovability of the continuum hypothesis. It turns out that ZFC can say very, very little about the cardinalities of ℶ numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to get sets of a fixed cardinality is to talk directly about well orderings. For example, $\aleph_1$ is exactly the set of order types of well orderings of $\omega$ (regardless of what $\beth_1$ is). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large cardinals, there is no way to explicitly compute their cardinality. For example, any inaccessible cardinal number $\kappa$ will have the property that $|\kappa| = \aleph_\kappa$, so you will not be able to make progress by trying to compute its ℵ number. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-13T14:30:01.057" LastActivityDate="2010-11-13T14:30:01.057" CommentCount="3" />
  <row Id="10124" PostTypeId="1" AcceptedAnswerId="10213" CreationDate="2010-11-13T14:23:42.650" Score="5" ViewCount="148" Body="&lt;p&gt;Suppose I have a set $\mathcal{S}$ of $N$ distinct items. Now consider the set $\mathcal{P}$ of all possible pairs that I can draw from $S$. Naturally, $|\mathcal{P}| = \binom{N}{2}$. Now when I draw $k$ items (pairs) from $\mathcal{P}$ with a uniform distribution, what is the expected number of distinct items from $S$ in those $k$ pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: I also asked this question over at &lt;a href=&quot;http://stats.stackexchange.com/questions/4453/&quot;&gt;stats&lt;/a&gt;, but got no answers so far, so I am trying here. Thanks for your time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; I pick the pairs without replacement.&lt;/p&gt;&#xA;" OwnerUserId="3470" LastEditorUserId="3470" LastEditorDisplayName="" LastEditDate="2010-11-13T22:47:51.273" LastActivityDate="2010-11-14T16:14:21.260" Title="How to calculate the expected number of distinct items when drawing pairs?" Tags="&lt;combinatorics&gt;&lt;expectation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" />
  <row Id="10125" PostTypeId="2" ParentId="10120" CreationDate="2010-11-13T14:25:10.233" Score="2" ViewCount="0" Body="&lt;p&gt;Add $5p+50$ to both sides of the = sign, and then simplify each side.&#xA;(If you wish you can do this in two steps: Add $5p$ and then 50 to each side of the equality sign.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &#xA;The above leads to the equation $130 = 13p$ which can be solved directly by dividing both sides of the equality sign by 13: $130/13 = 13p/13$, which after simplification gives us&#xA;$10=p$ -- which is the same as saying $p=10$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: To reach the equation $120=12p$, that is contained in the original question, is not necessary in order to solve the equation.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-14T05:25:22.077" LastActivityDate="2010-11-14T05:25:22.077" CommentCount="4" />
  <row Id="10126" PostTypeId="1" CreationDate="2010-11-13T14:49:19.913" Score="1" ViewCount="64" Body="&lt;p&gt;This is a follow-up to the question &lt;a href=&quot;http://mathoverflow.net/questions/45378/undecidability-in-conways-game-of-life&quot; rel=&quot;nofollow&quot;&gt;Undecidability in Conway’s Game of Life&lt;/a&gt; I posted at mathoverflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some cellular automata it can be proven that they can simulate a Turing machine, normally by explicit construction. For some of them it can be proven that they can &lt;em&gt;not&lt;/em&gt; simulate a Turing machine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there computable sufficient conditions on a cellular automaton to be Turing-complete (besides of being one of the known cases)? (My guess is: No.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If not: Why not?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If not: Can &lt;em&gt;non&lt;/em&gt;-computable sufficient conditions be named?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What about necessary conditions?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What's the &quot;standard&quot; way of showing that a cellular automaton is &lt;em&gt;not&lt;/em&gt; Turing-complete?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastActivityDate="2010-11-13T14:49:19.913" Title="(Semi-)Decidability of Turing-completeness of cellular automata" Tags="&lt;computability&gt;" CommentCount="2" />
  <row Id="10127" PostTypeId="2" ParentId="10121" CreationDate="2010-11-13T14:54:06.850" Score="0" ViewCount="0" Body="&lt;p&gt;We claim that $f(a_n) \to \xi = f(r) = f(1-r)$. If $r = \tfrac{1}{2}$ then $(a_n)$ must be convergent (why?). Since $f$ is continuous $f(a_n)$ converges too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume that $r \neq \tfrac{1}{2}$. WLOG we can assume that $r \lt \tfrac{1}{2}$. Consider the subsequence $(a_{n_k})$ consisting of the elements of $(a_n)$ that are less than $\tfrac{1}{2}$ and the subsequence $(a_{n_l})$ consisting of the elements of $(a_n)$ that are greater than or equal to $\tfrac{1}{2}$ and convince yourself that they are in fact subsequences and that they exhaust the $a_n$. In particular $a_{n_k} \to r$ and $a_{n_l} \to 1-r$. Now all that's left to do is to observe that since $f$ is continuous, $f(a_{n_k})$ and $f(a_{n_l})$ are eventually in any open interval around $\xi$. Since they exhaust $(a_n)$ we must have $f(a_n) \to \xi$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-13T14:54:06.850" />
  <row Id="10128" PostTypeId="2" ParentId="10119" CreationDate="2010-11-13T15:07:20.353" Score="1" ViewCount="0" Body="&lt;p&gt;It's not clear exactly what's your problem, but the last polynomial factorizes like this:&#xA;$$x^4-16=(x^2)^2-4^2=(x^2+4)(x^2-4)=(x^2+4)(x+2)(x-2).$$&#xD;&#xA;For the first one, it's easy to find the root $x=2$, which means that you can factor out $x-2$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-13T15:07:20.353" CommentCount="1" />
  <row Id="10129" PostTypeId="2" ParentId="10124" CreationDate="2010-11-13T15:10:43.257" Score="4" ViewCount="0" Body="&lt;p&gt;You are picking edges from a complete graph and looking for the expected number of vertices that get picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the expected number of vertices that &lt;em&gt;don't&lt;/em&gt; get picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that a vertex gets picked in 1 try is $\dfrac{2}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it does not get picked in all tries is $(1-\dfrac{2}{n})^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the expected number of vertices that don't get picked is $n(1-\dfrac{2}{n})^k$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus the expected number of vertices that get picked is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(1 - (1-\dfrac{2}{n})^k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross points out, this assumes you pick the pairs with replacement.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-13T15:41:23.857" LastActivityDate="2010-11-13T15:41:23.857" CommentCount="4" />
  <row Id="10130" PostTypeId="2" ParentId="10120" CreationDate="2010-11-13T15:11:27.653" Score="2" ViewCount="0" Body="&lt;p&gt;-5p+80=8p-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80=5p+8p-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80=13p-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80+50=13p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;130=13p (to get 120=12p at this step, you can multiply both sides by 12/13, but it's not necessary to arrive at the final answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p=10&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-13T15:11:27.653" CommentCount="1" />
  <row Id="10131" PostTypeId="2" ParentId="10102" CreationDate="2010-11-13T15:14:22.003" Score="5" ViewCount="0" Body="&lt;p&gt;Your friend will probably weasel out of this example by claiming that it isn't strictly &quot;Riemannian&quot; geometry, but it's definitely differential geometry.  The example is the Hodge theorem, which asserts that every cohomology class of a Riemannian manifold is represented by a unique harmonic form (where &quot;harmonic&quot; is with respect to the Laplace-Beltrami operator).  I highly doubt that a proof can be crafted without some reasonably serious functional analysis; the standard approach uses elliptic theory and Sobolev theory which requires the Banach-Alaoglu theorem which in turn requires the Tychonoff theorem.  In general I would wager that any result which involves geometric PDE theory (including some results in, say, minimal surface theory which genuinely are Riemmannian) is going to demand the axiom of choice at some level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, you might check out the work of Alexander Nabutovsky.  He has obtained some really serious results about the structure of geodesics and on the moduli space of Riemannian metrics using techniques from logic and computability theory - I wouldn't be surprised if AC is hiding somewhere.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2010-11-13T15:14:22.003" />
  <row Id="10132" PostTypeId="2" ParentId="10102" CreationDate="2010-11-13T15:21:19.093" Score="6" ViewCount="0" Body="&lt;p&gt;It looks to me like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem&quot; rel=&quot;nofollow&quot;&gt;Arzel&amp;agrave;--Ascoli theorem&lt;/a&gt; needs at least some weak form of choice.  (I have started &lt;a href=&quot;http://mathoverflow.net/questions/45928/does-arzela-ascoli-require-choice&quot; rel=&quot;nofollow&quot;&gt;an MO question&lt;/a&gt; to clarify this.)  One often uses this in geometry; for example, to guarantee the existence of minimizing geodesics connecting pairs of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: See Andres Caicedo's answer on MO (at above link).  The answer is affirmative.  Also, the database list of equivalents he mentions contains some very innocuous-looking statements that I bet your friend has never thought twice about using.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-13T16:18:41.227" LastActivityDate="2010-11-13T16:18:41.227" CommentCount="1" />
  <row Id="10133" PostTypeId="2" ParentId="10121" CreationDate="2010-11-13T15:21:45.737" Score="0" ViewCount="0" Body="&lt;p&gt;It is quite obvious that if $r$ and $1-r$ are the only sublimits of $\{a_n\}$ in the interval $[0,l]$, then for any $\delta \gt 0$, $(r-\delta, r+\delta)\cup (1-r-\delta,1-r+\delta)$ contains all but finitely many terms of the sequence. &#xA;Let $\epsilon&amp;gt;0$. Choose $\delta&amp;gt;0$ such that $|f(x)-f(r)|\lt \epsilon$ for all $x\in (r-\delta, r+\delta)$. Then by symmetry of $f$,&#xA;also $|f(x)-f(r)|\lt \epsilon$ for all $x\in (1-r-\delta, 1-r+\delta)$. It follows that&#xA;$$|f(a_n)-f(r)|\lt \epsilon$$ for all but finitely many $n$'s. This proves that $\lim_{n\to\infty} f(a_n)=f(r)$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-13T15:21:45.737" />
  <row Id="10134" PostTypeId="1" CreationDate="2010-11-13T15:23:21.523" Score="0" ViewCount="129" Body="&lt;p&gt;Let $f:[0,1] \rightarrow [0,\infty)$ be a measurable function such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu (\{x \in [0,1]: f(x) &amp;gt; t \}) \leq \frac{1}{t(ln(t))^{2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holds for each $t&amp;gt;3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show $f$ is an integrable map.&lt;/p&gt;&#xA;" OwnerUserId="3471" LastActivityDate="2010-11-19T11:00:14.900" Title="Prove f is an integrable map" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10135" PostTypeId="2" ParentId="10112" CreationDate="2010-11-13T15:24:16.420" Score="3" ViewCount="0" Body="&lt;p&gt;I think randomly emailing professors is not going to earn any replies. Professors usually have a lot on their plate and they get hundreds of emails each day. Unsolicited emails are usually ignored. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility you can explore is trying to establish email contact with a few graduate students in related areas. Once you have done this, you may send copies of your work to some of them and try to get feedback. You are more likely to find graduate students who are responsive and some of the advanced ones are actually used to reviewing papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I found another useful piece on this subject by Henry Cohn&#xA;&lt;a href=&quot;http://research.microsoft.com/en-us/um/people/cohn/Thoughts/advice.html&quot; rel=&quot;nofollow&quot;&gt;http://research.microsoft.com/en-us/um/people/cohn/Thoughts/advice.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T05:48:01.437" LastActivityDate="2010-11-14T05:48:01.437" CommentCount="1" />
  <row Id="10136" PostTypeId="1" CreationDate="2010-11-13T15:50:34.480" Score="1" ViewCount="94" Body="&lt;p&gt;How do I show that a monomorphism $F \rightarrow G$ of sheaves induces an injection on stalks? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When showing that monomorphism is an injection on sets one uses the maps $x \mapsto a$ and $x \mapsto b$ where $a,b$ are some fix elements such that they have the same image under the monomorphism. Then by using the monomorphism property we must have $a = b$. But since we are dealing with sheaves we are supposed to construct a skyscraper sheaf and two maps that map &quot;almost all&quot; section of the skyscraper sheaf to $a$ and $b$ (respectively), or at least elements that are &quot;connected&quot; to $a$ and $b$ (since these are representations of elements in the stalk there are other options depending on which open set we are taking). When I say &quot;almost all&quot; and &quot;connected&quot; it is just my visualization of how it should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not able to neither define the correct skyscraper sheaf nor the maps to $F$ (especially such that the maps commute with the restriction maps). I was hoping for &lt;a href=&quot;http://math.stackexchange.com/questions/10047/can-every-element-in-the-stalk-be-represented-by-a-section-in-the-top-space&quot;&gt;this&lt;/a&gt; being true which might have helped me but now I am stuck and need some help with this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I am still interested in how the problem will be solved by using a skyscraper sheaf. &lt;/p&gt;&#xA;" OwnerUserId="3473" LastEditorUserId="3473" LastEditorDisplayName="" LastEditDate="2010-11-17T17:20:18.467" LastActivityDate="2010-11-17T17:20:18.467" Title="Monomorphisms of sheaves gives an injection of stalks" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="10137" PostTypeId="2" ParentId="10068" CreationDate="2010-11-13T15:58:16.427" Score="0" ViewCount="0" Body="&lt;p&gt;Let's ponder like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point $x$ $f(x)&amp;gt;0$, therefore, in the vicinity of this point there is an increasing arithmetic progression&#xA;$a_{0}, \ b_{0}, \ c_{0}$ that, $f(a_{0})+f(b_{0})+f(c_{0})&amp;gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this one will be found increasing arithmetic progression of &#xA;$a_{1}, \ b_{1}, \ c_{1}$ that, $f(a_{1})+f(b_{1})+f(c_{1})&amp;lt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all values of parametr $t[0,1]$ сonsider the arithmetic progression &#xA;$a(t), \ b(t), \ c(t)$, where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(t)=a_{0}(1-t)+a_{1}t$, &lt;br&gt; &#xA;$b(t)=b_{0}(1-t)+b_{1}t$, &lt;br&gt;&#xA;$c(t)=c_{0}(1-t)+c_{1}t$. &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Function $F(t)=f(a(t))+f(b(t))+f(c(t))$ continuously depends on $t$, at $t=0 \ F(t)&amp;gt;0$, and at $t=1 \ F(t)&amp;lt;0$. It means that in some $t \ F(t)=0$ and the corresponding progress  $a(t), \ b(t), \ c(t)$  is required. &lt;/p&gt;&#xA;" OwnerUserId="2779" LastActivityDate="2010-11-13T15:58:16.427" />
  <row Id="10138" PostTypeId="2" ParentId="9713" CreationDate="2010-11-13T16:00:24.710" Score="1" ViewCount="0" Body="&lt;p&gt;Strictly speaking we cannot satisfy (2) exists lim&lt;sub&gt;x→c&lt;/sub&gt;f′(x)/g′(x), even in the extended sense, unless (1) g'(x) is nonzero in a (deleted) neighborhood of x = c.  That is required for f'(x)/g'(x) to be defined there (in the deleted neighborhood).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such gaps in the definition of f'(x)/g'(x) only matter if they accumulate at x = c, since we can otherwise work around them by shrinking the neighborhood to avoid them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can entertain the possibility that those gaps all turn out to be removable discontinuities in f'(x)/g'(x), so that a continuous extension of f'(x)/g'(x) as x approaches c is uniquely determined.  For that to happen we would need f'(x) to have roots of equal or greater multiplicity as g'(x) at those gaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this amounts to &quot;multiplying top and bottom by the same thing&quot;, in the sense that f'(x) and g'(x) would have a common factor containing all the roots of g'(x) sufficiently near x = c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proposed example f(x) = 2x + sin(2x) and g(x) = f(x)e&lt;sup&gt;-sin(x)&lt;/sup&gt; works out in this way.  The quotient f'(x)/g'(x) is undefined when g'(x) = 0, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g'(x) = [f'(x) - cos(x) f(x)] e&lt;sup&gt;-sin(x)&lt;/sup&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f'(x) = 2(1 + cos(2x)) = 4 cos&lt;sup&gt;2&lt;/sup&gt;(x)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common factor containing all the roots of g'(x) for sufficiently large x is then cos(x).  If we &quot;cancel&quot; that factor, removing roots in the denominator, the resulting &quot;simplified&quot; f'(x)/g'(x) has a bounded numerator and a denominator that tends to -&amp;infin; as x grows (since f(x) tends to &amp;infin;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently we cannot blithely patch over removable singularities in f'(x)/g'(x) for the sake of l'Hopital's rule.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-13T16:00:24.710" />
  <row Id="10139" PostTypeId="2" ParentId="10119" CreationDate="2010-11-13T16:03:37.417" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ x-a \ $ divides $\rm\ x^n - a^n\ $ handles the 1st and 3rd. The 2nd is $\rm\ x^2\ (x-3) + (x-3)\ =\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T16:03:37.417" />
  <row Id="10140" PostTypeId="2" ParentId="10136" CreationDate="2010-11-13T16:23:50.923" Score="2" ViewCount="0" Body="&lt;p&gt;There is a nice proof of this on page 7 of Kiran Kedlaya's notes (and also that surjectivity carries over to stalks)&#xA;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/MIT18_726s09_lec03_sheaves.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/MIT18_726s09_lec03_sheaves.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-13T16:23:50.923" CommentCount="1" />
  <row Id="10141" PostTypeId="2" ParentId="10136" CreationDate="2010-11-13T16:27:13.813" Score="4" ViewCount="0" Body="&lt;p&gt;In fact, more is true. If $U$ is an open set, then $F(U) \to G(U)$ is a monomorphism for $F \to G$ a monomorphism of sheaves. To see this, use the constant sheaf $\mathbb{Z}_U$ over $U$. To hom from $\mathbb{Z}_U$ into a sheaf $H$ is the same thing as giving a section of $H$ over $U$. This implies injectivity on the stalks because filtered colimits are exact in the category of abelian groups (or modules over a ring).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-13T16:27:13.813" CommentCount="1" />
  <row Id="10142" PostTypeId="2" ParentId="10095" CreationDate="2010-11-13T16:40:21.133" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad\quad $ With $\rm\ \ \alpha\ =\ 1 - 2\ s + 2\ \sqrt{s^2 - s}\ \ $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \rm \frac{s\ x\ (1+x)}{(1-x)^3 + 4\ s\ x\ (1-x)}\ =\ \frac{\alpha}{4\ (x-\alpha)} +  \frac{\alpha&amp;#39;}{4\ (x-\alpha&amp;#39;)} - \frac{1}{2\ (x-1)} $$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T16:40:21.133" CommentCount="0" />
  <row Id="10144" PostTypeId="2" ParentId="10095" CreationDate="2010-11-13T16:48:06.413" Score="1" ViewCount="0" Body="&lt;p&gt;Here it's easier to work with the exponential generating function.  You just need to find the Taylor series of each term on the RHS.  You already know the Taylor series of one term.  To find the Taylor series of the other term, use the fact that $\cos t = \frac{e^{it} + e^{-it}}{2}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-13T16:48:06.413" />
  <row Id="10145" PostTypeId="2" ParentId="9941" CreationDate="2010-11-13T17:22:29.380" Score="1" ViewCount="0" Body="&lt;p&gt;Although this is not an answer, I thought that the following identity (pulled out of the book &quot;Concrete Mathematics&quot;) might be useful for your analysis (or rather its proof actually that should be useful):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1\cdots\int_0^1 f(\lfloor x_1+\cdots+x_n\rfloor)dx_1\cdots dx_n = \sum_k \genfrac{\langle}{\rangle}{0pt}{}{n}{k} \frac{f(k)}{n!},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\genfrac{\langle}{\rangle}{0pt}{}{n}{k}$ is the Euler-number (first-kind).&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-13T17:22:29.380" />
  <row Id="10146" PostTypeId="1" AcceptedAnswerId="10149" CreationDate="2010-11-13T17:30:20.840" Score="1" ViewCount="154" Body="&lt;p&gt;Often i have wondered about this question, but today i had a chance to recollect it and hence i am posting it here. During &lt;em&gt;high-school&lt;/em&gt; days one generally learns &lt;em&gt;Integration&lt;/em&gt; and i still loving doing problems on &lt;em&gt;Integration.&lt;/em&gt; But i have never understood this idea of &lt;em&gt;integration by parts&lt;/em&gt;. Why do we adopt the &lt;strong&gt;ILATE&lt;/strong&gt; or &lt;strong&gt;LIATE&lt;/strong&gt; rule to do problems and what was the reason behind thinking of such a rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Integration_by_parts&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Integration_by_parts&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-13T19:32:36.610" Title="Intuition behind the ILATE rule" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" />
  <row Id="10147" PostTypeId="1" AcceptedAnswerId="10152" CreationDate="2010-11-13T17:39:31.003" Score="2" ViewCount="205" Body="&lt;p&gt;How to prove that finite additivity follows from countable additivity!!?&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T01:45:56.867" LastActivityDate="2010-11-20T01:45:56.867" Title="Finite additivity follows from countable additivity!" Tags="&lt;measure-theory&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10149" PostTypeId="2" ParentId="10146" CreationDate="2010-11-13T17:49:15.727" Score="7" ViewCount="0" Body="&lt;p&gt;The way I see it, when you differentiate an inverse trigonometric function, you don't get another inverse trigonometric function. Instead you get &quot;simpler&quot; funtions like $1/(1 + x^2)$ or $1/\sqrt{1-x^2}$. This does not typically happen with the antiderivative of such functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when you differentiate a logarithmic funtion, the logarithm disappears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when using integration by parts $\int u dv = uv - \int v du$, it makes sense to select the inverse trigonometric or logarithmic function to be the one that is the u term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of algebraic, trigonometric and exponential functions both integration and differentiation don't change the nature of the function, so they come later in the ILATE order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is just intuition and there are examples where you can violate this so called rule and still integrate by parts without any problems&lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2010-11-13T19:32:36.610" LastActivityDate="2010-11-13T19:32:36.610" CommentCount="1" />
  <row Id="10150" PostTypeId="2" ParentId="10146" CreationDate="2010-11-13T17:50:41.463" Score="4" ViewCount="0" Body="&lt;p&gt;As a technique for explicitly integrating functions given by formulas as usually seen in calculus classes, integration by parts works because $u&amp;#39;v$ can be easier to integrate than $uv&amp;#39;$.  There are multiple ways an integrand can be considered as a product of the form $u&amp;#39;v$, but some choices lead to $uv&amp;#39;$ being of no use, perhaps even harder to integrate.  The ILATE mnemonic you mention (or whatever it is) gives some rules that approximate the best guesses as to what will work well.  Better than this mnemonic (which I've never thought about) is to just have enough experience doing integrals to be able to see what will work and why (at least for the type of integrals one typically sees in a calculus class).&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-13T17:50:41.463" />
  <row Id="10151" PostTypeId="2" ParentId="8856" CreationDate="2010-11-13T18:00:17.880" Score="5" ViewCount="0" Body="&lt;p&gt;It is a lovely result of Hermann Minkowski that any plane centrally symmetric convex set can serve as the &quot;unit ball&quot; of a distance function.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-13T18:00:17.880" />
  <row Id="10152" PostTypeId="2" ParentId="10147" CreationDate="2010-11-13T18:02:16.723" Score="10" ViewCount="0" Body="&lt;p&gt;I guess this question arises because in the definition of countable additivity it is assumed that there are infinitely many events.  But if there are finitely many events, you can just throw in countably many empty events.  That is, given $A_1,\ldots,A_n$, define $A_{n+1}=\emptyset$, $A_{n+2}=\emptyset$, and so on, then apply countable additivity.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-13T18:02:16.723" />
  <row Id="10153" PostTypeId="2" ParentId="10147" CreationDate="2010-11-13T18:20:49.417" Score="2" ViewCount="0" Body="&lt;p&gt;I assume that you want to show that finite additivity follows from countably infinite additivity. In general, though countable means both finite and countably infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given that $\mu(\displaystyle \cup_{i=1}^\infty A_i) = \displaystyle \sum_{i=1}^\infty \mu(A_i)$ whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we now want to show $\mu(\displaystyle \cup_{i=1}^n A_i) = \displaystyle \sum_{i=1}^n \mu(A_i)$, $\forall n \in \mathbb{N}$, whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $A_{i} = \emptyset$, $\forall i &amp;gt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way we have defined, note that $A_i \cap A_j = \emptyset$, $\forall i \neq j$. (Note: $A \cap \emptyset = \emptyset$, $\forall A$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By countably infinite additivity, $\mu(\displaystyle \cup_{i=1}^\infty A_i) = \displaystyle \sum_{i=1}^\infty \mu(A_i)$ and note that $\mu(A_i) = 0$, $\forall i &amp;gt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that $\displaystyle \cup_{i=1}^\infty A_i = \displaystyle \cup_{i=1}^n A_i$, since $A_i = \emptyset$, $\forall i &amp;gt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\mu(\displaystyle \cup_{i=1}^n A_i) = \displaystyle \sum_{i=1}^n \mu(A_i)$ whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-13T19:13:14.060" LastActivityDate="2010-11-13T19:13:14.060" CommentCount="6" />
  <row Id="10154" PostTypeId="1" AcceptedAnswerId="10158" CreationDate="2010-11-13T18:35:57.950" Score="0" ViewCount="95" Body="&lt;p&gt;I've written Haskell function, that helps me to compute $x^{-1} \mod m$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;invM x m = invMhelp x 0 m; --modular multiplicative inverse x^-1 (mod m) = inv x m  &#xA;invMhelp x i m = if (x*i `mod` m == 1) then i else (invMhelp x (i+1) m);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Main&amp;gt; 13 `invM` 2109&#xA;649&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That means $13^{-1} = 649 \mod 2109$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I compute $x^{-2} \mod m$?&lt;/strong&gt; ($x^{-2} \mod m = (x^{-1})^2 \mod m$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can I simply use the power of two of the multiplicative inverse and then apply the modulo m again?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $13^{-1} = 649 \mod 2109$, $649^2 \mod 2109 = 1510$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $13^{-2} \mod 2109$ equal to $1510$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the noob question, but I'm not sure about this. Thanks for answers.&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="2411" LastEditorDisplayName="" LastEditDate="2010-11-13T18:41:47.897" LastActivityDate="2010-11-13T21:18:23.977" Title="How do I compute x^(-2) mod m?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="10155" PostTypeId="1" AcceptedAnswerId="10157" CreationDate="2010-11-13T18:39:21.067" Score="0" ViewCount="160" Body="&lt;p&gt;Please observe the triangle shown in the figure below(In red).I consider the angle between 5 and 4.The base should be 5 ,hypotenuse 4 and perpendicular 3,but according to the solution manual the base is 4 and the hypotenuse 5.Please tell me how.Thanks!&lt;img src=&quot;http://i.imgur.com/pdKkI.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2857" LastEditorUserId="2857" LastEditorDisplayName="" LastEditDate="2010-11-13T18:53:15.313" LastActivityDate="2010-11-16T14:07:18.193" Title="How to solve this triangle?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10156" PostTypeId="2" ParentId="10154" CreationDate="2010-11-13T18:42:40.800" Score="3" ViewCount="0" Body="&lt;p&gt;Yes. That will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \equiv b \pmod{m} \Rightarrow  a^k \equiv b^k \pmod{m}$, whenever $k \in \mathbb{N}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-13T18:49:50.810" LastActivityDate="2010-11-13T18:49:50.810" CommentCount="2" />
  <row Id="10157" PostTypeId="2" ParentId="10155" CreationDate="2010-11-13T18:51:48.500" Score="6" ViewCount="0" Body="&lt;p&gt;This is not a complete answer to your question, but here's a hint. (I cannot post comments yet).&#xA;(3,5,4) is not a Pythagorean triplet &lt;em&gt;i.e.&lt;/em&gt; there are no right angled triangle with base 5, perpendicular 3 and hypoteneuse 4.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-13T18:51:48.500" />
  <row Id="10158" PostTypeId="2" ParentId="10154" CreationDate="2010-11-13T18:55:58.620" Score="5" ViewCount="0" Body="&lt;p&gt;$\rm\: a\equiv b^{-1} \Rightarrow\ a^2 \equiv b^{-2}\: $ via &lt;strong&gt;congruences are preserved by multiplication&lt;/strong&gt; (so too by squaring). If you don't yet know that property then you can instead prove the sought result as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm a\equiv b^{-1}\ \:(mod\ m)\ \Rightarrow\ m\:|\: ab-1\ \Rightarrow\ m|(ab-1)(ab+1)\ =\ a^2 b^2 - 1\ \Rightarrow\ a^2 \equiv b^{-2}\ \:(mod\ m)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, here is the proof that congruences are preserved by mutliplication&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \:(mod\ m)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ m\: |\: A-a,\:\:\ B-b\ \Rightarrow\ m\ |\ (A-a)\ B + a\ (B-b)\ =\ AB - ab $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;strong&gt;congruence product rule&lt;/strong&gt; is at the heart of many other similar product rules, for example Leibnitz's product rule for derivatives in calculus, e.g. see my &lt;a href=&quot;http://math.stackexchange.com/questions/3101/derivative-of-a-product-and-derivative-of-quotient-of-functions-theorem-i-dont/3178#3178&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-13T21:18:23.977" LastActivityDate="2010-11-13T21:18:23.977" CommentCount="1" />
  <row Id="10159" PostTypeId="1" CreationDate="2010-11-13T19:33:52.253" Score="1" ViewCount="87" Body="&lt;p&gt;Can somebody please explain continuity and discontinuity in fourier series?&lt;/p&gt;&#xA;" OwnerUserId="3478" LastActivityDate="2010-11-14T08:17:27.197" Title="Continuity and discontinuity in fourier series?" Tags="&lt;fourier-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10160" PostTypeId="1" CreationDate="2010-11-13T19:42:26.177" Score="4" ViewCount="116" Body="&lt;p&gt;I watched the excellent interview with Richard Feynman: &lt;a href=&quot;http://www.youtube.com/watch?v=PsgBtOVzHKI&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=PsgBtOVzHKI&lt;/a&gt; In the interview Feynman mention that he at young age re-invented operator calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched for &quot;operator calculus&quot; and have not found any accessible references that introduce the topic. Maybe operator calculus go under another name today, than at the time of the interview?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a reasonably simple explanation of operator calculus, and also give some references to books on the subject?&lt;/p&gt;&#xA;" OwnerUserId="3479" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-13T19:55:36.393" LastActivityDate="2010-11-13T20:56:21.267" Title="What is operator calculus?" Tags="&lt;reference-request&gt;&lt;operator-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="10161" PostTypeId="2" ParentId="10160" CreationDate="2010-11-13T20:11:16.703" Score="11" ViewCount="0" Body="&lt;p&gt;At the heart the key idea is quite simple. Namely, one views functions, endomorphisms, etc. as &quot;numbers&quot; and manipulates them as such (where valid). E.g. for a derivative $\rm D = d/dx\ $ we have that $\rm D^2 - c^2\ =\ (D - c)(D + c) $ just as for numbers, as long $\rm\:D\:$ commutes with $\rm\:c\:$, i.e. $\rm\:c\:$ is constant. So we can solve constant coefficient linear differential / difference equations by simply factoring their operator (&quot;characteristic&quot;) polynomials into linear factors over $\mathbb C$. One can also similarly perform transcendental operator manipulations such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm f(D)\ e^{t\:x}\ =\ f(t)\ e^{t\ x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm e^{\:t\ D}\ f(x)\ =\ f(x+t) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm t^{x\ D}\ f(x)\ =\ f(t\: x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Generalized Leibniz Rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm g(D)\ f(x)\ = \ \sum_{n=0}^\infty\ \frac{f^{(n)}(x)\ g^{(n)}(D)}{n!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details see Roman: Umbral Calculus and Rainville: Special Functions and especially Rota: Finite operator calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-13T20:56:21.267" LastActivityDate="2010-11-13T20:56:21.267" CommentCount="1" />
  <row Id="10162" PostTypeId="2" ParentId="10159" CreationDate="2010-11-13T20:19:51.840" Score="7" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have a removable discontinuity at a point, the Fourier series will converge to the limit of the function at the point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have a jump discontinuity at some point, then the Fourier series will converge to the average of the values of left and right limits. But the higher harmonics are significant, resulting in the &quot;Gibbs phenomenon&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For more complicated discontinuities, we really cannot say anything. You can have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_Fourier_series&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; for more details.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The best one can say about the convergence of Fourier series is that it always converges almost everywhere, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Carleson's theorem&lt;/a&gt;. But this kind of theorems involve more sophisticated levels of mathematical analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-11-14T08:17:27.197" LastActivityDate="2010-11-14T08:17:27.197" CommentCount="3" />
  <row Id="10163" PostTypeId="2" ParentId="9941" CreationDate="2010-11-13T20:32:13.310" Score="1" ViewCount="0" Body="&lt;p&gt;Moving the rectangle $j$ units to the right changes the integral by $j|S|$. So we may assume $a+c\geq0$. Parametrize $S$ by $x=(u-v)/2$, $y=(u+v)/2$. The corresponding domain $S&amp;#39;$ in the $(u,v)$-plane is a rectangle with $45^\circ$ slopes of its sides. Cut up $S&amp;#39;$ by two vertical lines into two triangles and a parallelogram. The integral in question is a linear combination of integrals of the following kind:&#xA;$\int_0^q \lfloor u\rfloor du$, $ \int_0^q u \lfloor u\rfloor du$.&#xA; These integrals have to be precomputed, and in the end some simplifications should be possible.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-13T20:32:13.310" />
  <row Id="10164" PostTypeId="1" AcceptedAnswerId="10192" CreationDate="2010-11-13T20:44:10.220" Score="0" ViewCount="118" Body="&lt;p&gt;Let $g(t)$ be a real function of a real variable such that &#xA;$$\sup |p(t)g(t)|\lt \infty$$ for any polynomial $p$. Is it true that for any nonngetive&#xA;integer $k$, one has $$\sup_{(x,y)\in\mathbb{R}^2}(\frac{|x|}{1+|y|})^k |g(x-y)|\le A_k, $$&#xD;&#xA;for some positive constant $A_k\lt \infty$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-13T21:55:29.343" LastActivityDate="2010-11-14T01:24:10.357" Title="A real function question" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10165" PostTypeId="2" ParentId="10164" CreationDate="2010-11-13T21:07:35.003" Score="1" ViewCount="0" Body="&lt;p&gt;Hint 1: What kind of function must $g(t)$, satisfying the first inequality, be? Can you think of some simple such function yourself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: What can you say about the function $y \mapsto \left({1 \over 1 + |y|}\right)^k$? Can it spoil the second inequality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 3: Let $x = y + t$. Triangle inequality is your friend.&lt;/p&gt;&#xA;" OwnerUserId="3223" LastActivityDate="2010-11-13T21:07:35.003" CommentCount="4" />
  <row Id="10166" PostTypeId="1" CreationDate="2010-11-13T21:17:55.507" Score="0" ViewCount="72" Body="&lt;p&gt;I was wondering if this question should be posted in this forum or in Stack Exchange, but reading both &quot;What kind of questions can I ask here?&quot;, in Math I found&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Software that mathematicians use&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I guess it should fit here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a TI-89 (a --powerful, I guess-- Texas Instruments calculator) which I've had since high-school days; I very rarely use it anymore. Of course, for more advanced mathematical computations there is great software for the PC, but maybe some of it (the most simple of it) has been done for the TI-89, and I thought it might be cool finding some use for it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most notably while studying Galois theory lately, I've been using Wolfram Alpha to quickly determine the discriminant of a polynomial of degree &gt;= 3. Of course one could just use the formula (it's not that cumbersome for degrees 3 and 4), but I thought there might be a program that would do it for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is: what good, useful &quot;advanced&quot; (advanced undergraduate and forth) math software do you know for the TI-89?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, do you know of one to compute the discriminant of a polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the question is well-posed and suited for this forum. If not, please redirect me.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-13T21:23:56.123" Title="Advanced math software for the TI-89 (discriminant of a general polynomial?)" Tags="&lt;math-software&gt;&lt;calculator&gt;" AnswerCount="1" />
  <row Id="10167" PostTypeId="1" AcceptedAnswerId="10175" CreationDate="2010-11-13T21:19:23.213" Score="2" ViewCount="147" Body="&lt;p&gt;Let $A$ be a finitely generated Abelian group. Let $tA$ denote the torsion subgroup. Prove that $A$ has a subgroup isomorphic to $A/tA$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $A/tA$ is torsion free, so my thinking so far has been to take the non-torsion elements of $A$ and consider the subgroup generated by them. Then show that subgroup and $A/tA$ are isomorphic. (I'm doing all this in additive notation, just so you know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A&amp;#39;=\{a\in A\,|\, na\neq 0 \text{ for all } n\in \mathbb{Z}^+\}$ and consider the subgroup $B=\langle A&amp;#39;\rangle$. Since $A$ is finitely generated, so is $B$. Let the generators be $\{a_1, \ldots , a_n\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restrict the natural projection $\pi: A\to A/tA$ to a map $\varphi: B\to A/tA$. Since $\pi$ is a well-defined homomorphism, so is $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'd like to show that $\varphi$ is bijective, but I'm a little stuck. First of all, I haven't convinced myself it's even true. Secondly, for injectivity, does it suffice to only consider two generators $a_i$ and $a_j$ and suppose their images are distinct and go from there? Or do I have to take two arbitrary elements of $B$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be great, thanks!&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-15T17:36:14.650" Title="Quotient of an Abelian group with its torsion subgroup" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="10168" PostTypeId="2" ParentId="8944" CreationDate="2010-11-13T21:21:56.407" Score="1" ViewCount="0" Body="&lt;p&gt;A general suggestion: There are a few problems asking for convergence of divergence of series of the form $\sum_n (-1)^ng(n)$ or $\sum_n g(n)$ for some function $g$ for which one can easily rewrite $g(n)$ as&#xA;$$f(1/n)-f(0)$$ for some $f$ differentiable at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all these cases one can try to replace $g(n)=f(1/n)-f(0)$ with $f&amp;#39;(0)/n$ and use elementary estimates and inequalities for the remainder error, if needed (sometimes a limit comparison test will already suffice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the problem here, $g(n)=2^{1/n}-1$ so blatantly we can try $f(x)=2^x$. We have $f&amp;#39;(0)=\ln 2$, which is the estimate that appears in acarchau's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example, $\sum_n\sin(1/n)$ diverges: Take $f(x)=\sin(x)$. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-13T21:21:56.407" />
  <row Id="10169" PostTypeId="2" ParentId="10166" CreationDate="2010-11-13T21:23:56.123" Score="1" ViewCount="0" Body="&lt;p&gt;The discriminant can be computed as a (Sylvester) determinant (or resultant), e.g. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Discriminant&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; Thus you need merely find a program that computes determinants.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T21:23:56.123" CommentCount="4" />
  <row Id="10170" PostTypeId="2" ParentId="10167" CreationDate="2010-11-13T21:25:22.903" Score="1" ViewCount="0" Body="&lt;p&gt;Unfortunately the whole group will be generated by the nontorsion elements if the group is infinite. One thing you can do is use the classification theorem for finitely generated abelian groups which says that a finitely generated abelian group is of the form $Z_{m_1} \times ... \times Z_{m_n} \times Z \times ... \times Z$, with finitely many $Z$'s appearing on the right. In this case $tA$ is $Z_{m_1} \times ... \times Z_{m_n}$ and $A$ / $tA$ will be the $Z \times ... \times Z$ which is clearly a subgroup of $A$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-13T21:25:22.903" CommentCount="2" />
  <row Id="10171" PostTypeId="1" CreationDate="2010-11-13T21:26:04.963" Score="0" ViewCount="48" Body="&lt;p&gt;If I understand correctly, the Euclidean Dirac operator is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D=\sum_{i=1}^n e_i \frac{\partial}{\partial x_i},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e_i$ are bases for $Cl_{0,n}(\mathbb{R})$, i.e., the $n$-dimensional Clifford algebra with negative-definite signature over the reals (so $e_i^2=-1$), and $x_i$ are the corresponding coordinates.  Several sources state that $D^2 = -\Delta_n$ where $\Delta_n$ is the standard Euclidean Laplace operator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_n = \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I write out $D^2 f$ explicitly for some function $f:\mathbb{R}^n \rightarrow \mathbb{R}$, scalar terms from the Laplacian certainly appear, e.g.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_1 \frac{\partial}{\partial x_1}\left( e_1 \frac{\partial}{\partial x_1} f \right) = e_1 \left( e_1 \frac{\partial^2}{\partial x_1^2}f + (\frac{\partial}{\partial x_1}e_1)\frac{\partial}{\partial x_1}f \right)=e_1^2 \frac{\partial^2}{\partial x_1^2}f = -\frac{\partial^2}{\partial x_1^2}f.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I also end up with bivector cross terms that shouldn't be there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_1 \frac{\partial}{\partial x_1}\left( e_2 \frac{\partial}{\partial x_2} f \right) = e_1 \left( e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f + (\frac{\partial}{\partial x_1}e_2)\frac{\partial}{\partial x_2}f \right)=e_1 e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f = e_{12}\frac{\partial^2}{\partial x_1 \partial x_2}f.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I only be considering the scalar part of $D^2$, or am I simply doing something wrong here?&lt;/p&gt;&#xA;" OwnerUserId="2330" LastActivityDate="2010-11-13T21:26:04.963" Title="How do you show that the Laplacian is the square of the (Euclidean) Dirac operator?" Tags="&lt;differential-geometry&gt;&lt;vector-calculus&gt;" CommentCount="2" />
  <row Id="10172" PostTypeId="1" AcceptedAnswerId="10226" CreationDate="2010-11-13T21:36:54.673" Score="1" ViewCount="46" Body="&lt;p&gt;What are some families of graphs with known &lt;a href=&quot;http://en.wikipedia.org/wiki/Tree_decomposition&quot; rel=&quot;nofollow&quot;&gt;tree width&lt;/a&gt; k?&#xA;For instance, for tree k=1, for cycle k=2, for series-parallel graph which is not a tree k=2, what are some other examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for some easy to construct families (or famous graphs which I can find in Mathematica's GraphData or on the internet) for testing tree decomposition code.&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-11-14T07:55:38.470" Title="Families of graphs with treewidth k" Tags="&lt;graph-theory&gt;&lt;mathematica&gt;" AnswerCount="1" />
  <row Id="10173" PostTypeId="1" AcceptedAnswerId="10176" CreationDate="2010-11-13T21:41:16.417" Score="0" ViewCount="101" Body="&lt;p&gt;All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any number, how can we retrieve each digit of that number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. For number = 4321, how can I get individual digits 4,3,2 and 1?&lt;/p&gt;&#xA;" OwnerUserId="2659" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-13T22:09:22.273" LastActivityDate="2010-11-14T01:34:17.143" Title="Extract each digit of a number" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10174" PostTypeId="2" ParentId="10173" CreationDate="2010-11-13T21:46:43.350" Score="2" ViewCount="0" Body="&lt;p&gt;1) Take N modulo 10.&#xA;2) Set N = [N/10] ([] is the greatest integer funtion)&#xA;3) GO back to step 1) until N = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I haven't misunderstood your question here.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-13T21:46:43.350" />
  <row Id="10175" PostTypeId="2" ParentId="10167" CreationDate="2010-11-13T21:47:40.087" Score="4" ViewCount="0" Body="&lt;p&gt;In fact more is true. Every finitely generated abelian group is a direct sum of its torsion part and the quotient by its torsion part. The structure theorem of finitely generated abelian groups is even stronger than this. &#xA;You can find a proof of the weaker result in Peter May's notes (Theorem 1.11)&#xA;&lt;a href=&quot;http://www.math.uchicago.edu/~may/TQFT/Lecture2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uchicago.edu/~may/TQFT/Lecture2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T22:12:46.967" LastActivityDate="2010-11-13T22:12:46.967" CommentCount="4" />
  <row Id="10176" PostTypeId="2" ParentId="10173" CreationDate="2010-11-13T21:48:22.630" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad 3\ =\ \bigg\lfloor \frac{4321}{100}\bigg\rfloor \ mod\ 10 \ =\ 43\ mod\ 10\ \ $ i.e. rightshift then chop off least significant digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if you need all digits &quot;unwind&quot; the Horner representation $\rm\ N\ =\ d_0 + 10\ (d_1 + 10\ (d_2 + \:\cdots\: ))\:$ by using division by 10 with remainder, viz $\rm\ d_0 = N\ mod\ 10,\ \ d_1 = (N-d_0)/10\ mod\ 10,\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-13T22:11:55.923" LastActivityDate="2010-11-13T22:11:55.923" />
  <row Id="10177" PostTypeId="1" AcceptedAnswerId="10248" CreationDate="2010-11-13T21:52:21.503" Score="3" ViewCount="136" Body="&lt;p&gt;I'm trying to follow a proof about immersing/embedding $\mathbb{RP}^n$ into $\mathbb{R}^{n+1}$, which goes roughly as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $\tau=T\mathbb{RP}^n$.  The normal bundle $\nu$ has rank 1, so its Steifel-Whitney class is $w(\nu)=1$ or $w(\nu)=1+x$.  In every case, we need $w(\nu)\cdot w(\tau) = w(\nu \oplus \tau) = w(\epsilon^{n+1})=1$.  If $w(\nu)=1$, then $w(\tau)=(1+x)^{n+1}=1$, so $n+1=2^r$.  If $w(\nu)=1+x$, then similarly $(1+x)^{n+2}=1$ so $n+2=2^r$.  &lt;strong&gt;If the immersion is an embedding, the former case must hold.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this true?  I feel like there should be an easy reason, but none of the people I talked with were able to nail down anything precise.  This could be wrong, but it seems like this is tacitly saying that a codimension-1 embedding of a closed manifold must be in fact of an orientable manifold, which is the same as saying that the the normal line bundle has trivial $w_1$ (since line bundles are totally classified by their orientability, i.e. by $w_1$).  Is this true?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-14T05:46:38.250" LastActivityDate="2010-11-14T20:23:22.630" Title="Does having a codimension-1 embedding of a closed manifold $M^n \subset \mathbb{R}^{n+1}$ require $M$ to be orientable?" Tags="&lt;differential-topology&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="10178" PostTypeId="2" ParentId="10173" CreationDate="2010-11-13T21:52:34.700" Score="2" ViewCount="0" Body="&lt;p&gt;For the k'th digit (counting left from the unit's place), divide the number by $10^k$ and call the remainder r. Then, $[r/(10^{k-1})]$ gives the k'th digit.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T01:34:17.143" LastActivityDate="2010-11-14T01:34:17.143" />
  <row Id="10179" PostTypeId="2" ParentId="10155" CreationDate="2010-11-13T21:59:40.213" Score="3" ViewCount="0" Body="&lt;p&gt;By definition, the hypothenuse is the longest side of a right triangle; so in &lt;em&gt;any&lt;/em&gt; case you could not possibly have a triangle in which the &quot;base&quot; is larger than the hypothenuse as you claim it &quot;should&quot; be here. In any case, &quot;base&quot; depends on orientation (so it is possible for the hypothenuse to &lt;em&gt;be&lt;/em&gt; the base). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your picture is misleading you somehow into thinking that the side opposite the right angle is the side of length $4$; it cannot be. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-13T21:59:40.213" CommentCount="2" />
  <row Id="10180" PostTypeId="1" CreationDate="2010-11-13T22:22:14.820" Score="1" ViewCount="79" Body="&lt;p&gt;I am reading some handwritten notes, and scribbled next to a restriction of scalars functor, are the words &quot;a pullback&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this might be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, consider a field $k$ with a map $\varphi:k\rightarrow A$ for $A$ an associative unital $k$-algebra. Then we consider the induced functor $\varphi_*:A-mod\rightarrow k-mod$ restriction of scalars.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\varphi_*$ the pullback of some diagram in the category of functors from $A$-mod to $k$-mod?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know this a fairly strange question, but I just cant see what the scribbler had in mind.&lt;/p&gt;&#xA;" OwnerUserId="16" LastEditorUserId="16" LastEditorDisplayName="" LastEditDate="2010-11-13T23:38:40.793" LastActivityDate="2010-11-14T00:12:49.563" Title="Is restriction of scalars a pullback?" Tags="&lt;category-theory&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10181" PostTypeId="1" AcceptedAnswerId="10691" CreationDate="2010-11-13T22:28:25.217" Score="2" ViewCount="157" Body="&lt;p&gt;I'd like to learn formal math. Preferably, though not necessarily, starting with predicate logic/first order logic rather than higher order logic. I am trying to find resources (papers, books etc.) for doing this, but I haven't found anything I really like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of resources for predicate and first order logic, but most do not approach the topics in a very formal way. For example, many text don't seem to try to define what they mean by &quot;variables&quot; or mention substitution as an important concept. Tries to explicitly describe as many of the rules of the game as it can. Many texts bring up &quot;truth tables&quot; without having formal rules for what you're allowed to do with those tables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have resources that fit these criteria?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: many of the answers are good and helpful, but I feel like I should add some clarifying remarks:&#xA;Many texts mention that you can view math as merely manipulation of symbols. I don't doubt that this can be done, but I would like to see it done. A resource that explains the process of producing proofs explicitly in terms of manipulating symbols rather than in terms of functions, statements etc (at least without first defining these terms) would be helpful. I'd like to be able to pretend I was a person who didn't know any math and was just acting as a human computer, producing proofs. I'd like a resource that explains producing proofs like I was such a computer (not necessarily ONLY like that).&lt;/p&gt;&#xA;" OwnerUserId="3480" LastEditorUserId="3480" LastEditorDisplayName="" LastEditDate="2010-11-18T04:40:19.263" LastActivityDate="2010-11-18T04:40:19.263" Title="Resources for learning formal math?" Tags="&lt;logic&gt;&lt;reference-request&gt;&lt;formal-systems&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="10182" PostTypeId="2" ParentId="10181" CreationDate="2010-11-13T22:44:56.813" Score="1" ViewCount="0" Body="&lt;p&gt;Aside from general wikipedia and searches from google I would advise going to either the public or collegiate library. When I wanted to learn more about that I got &quot;Introduction to Mathematical Structures and Proofs&quot; by Larry J. Gerstein. It was in a shelf of free books outside my college's library. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regard to the truth tables they should have. You will see they are generated from the ideas of Boolean algebra. In fact, &lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_tables&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Truth_tables&lt;/a&gt; explains it quite well. I believe the necessary part is to do problems, and for that all you need is to go through definitions and see their consequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for texts for important concepts you may wish to pick up a book on the philosophy of mathematics, though you can see simply about variables from &lt;a href=&quot;http://en.wikipedia.org/wiki/Variable_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Variable_%28mathematics%29&lt;/a&gt;. Some can be rather dense, I would suggest an introduction. I am reading &quot;The Philosophy of Mathematics&quot; by W.D. Hart. It is a collection of short essays pretty much on important philosophical issues and notions in mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun. &lt;/p&gt;&#xA;" OwnerUserId="3417" LastActivityDate="2010-11-13T22:44:56.813" />
  <row Id="10183" PostTypeId="1" CreationDate="2010-11-13T22:54:35.183" Score="1" ViewCount="117" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have problem calculating the following summation :&#xA;$$S = \sum_{j=1}^{k^2-1} \lfloor \sqrt{j}\rfloor.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand the mean of that summation it will be something like $$1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+\cdots$$&#xD;&#xA;and I suspect that the last summation number will be $(k-1)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I really can find the pattern of the equal simpler summation.   &lt;/p&gt;&#xA;" OwnerUserId="3382" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-13T23:04:36.577" LastActivityDate="2010-11-14T20:21:25.753" Title="Summation problem " Tags="&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="10184" PostTypeId="2" ParentId="10183" CreationDate="2010-11-13T23:00:48.240" Score="2" ViewCount="0" Body="&lt;p&gt;Try replacing it by a sum of the form $\sum_1^k k \cdot c_k$, where $c_k$ is the number of times that $\lfloor \sqrt{j} \rfloor = k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-13T23:00:48.240" />
  <row Id="10185" PostTypeId="2" ParentId="10183" CreationDate="2010-11-13T23:04:20.587" Score="7" ViewCount="0" Body="&lt;p&gt;Since the last value for $j$ is $k^2-1$, none of the terms of the sum are $k$; they are all between $1$ and $k-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many $1$'s will be in the sum? Well, we'll get $1$ when $j$ is any number between $1^2$ and $2^2-1$; then we'll get $2$ for each number between $2^2$ and $3^2-1$. Then we'll get $3$ for each number between $3^2$ and $4^2-1$. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $n\leq k-1$, how many times does it show up in the sum? It shows up exactly the number of times that there are integers between $n^2$ and $(n+1)^2-1$, inclusively. This is&#xA;$$(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n+1.$$&#xD;&#xA;So your sum has $2(1)+1 = 3$ ones; $2(2)+1 = 5$ twos; $2(3)+1=7$ threes; etc. Up to $k-1$, which appears exactly $2(k-1)+1 = 2k-1$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get that&#xA;$$S = \sum_{r=1}^{k-1} r(2r+1) = 2\left(\sum_{r=1}^{k-1}r^2\right) + \sum_{r=1}^{k-1}r.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-13T23:04:20.587" CommentCount="1" />
  <row Id="10186" PostTypeId="1" AcceptedAnswerId="10193" CreationDate="2010-11-13T23:05:46.813" Score="3" ViewCount="162" Body="&lt;p&gt;So I found this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=3}^{\infty} 1 - \tan( \pi/2^k)^4$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.missouristate.edu/lesreid/Advanced.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only ever done tests for convergence of infinite sums.&#xA;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinite_product&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; it shows a way to convert but&#xA;in this case an is less than one for all k. I can see that $ f(k) = 1 - \tan( \pi/2^k)^4 $ is a strictly increasing function and in fact it is increasing very quickly however it is bound [ $ 1 - \tan( \pi/2^3)^4$, 1) where the lower bound is quite nearly one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to mean it will converge to something slightly less than the lower bound of the function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that is converges. I want to know if there analytic method to provide a closed from of the solution such as with an alternating series. I figure I can simply do it computationally with MatLab and it would be very simple but I thought there must another way.&lt;/p&gt;&#xA;" OwnerUserId="3417" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:45:01.407" LastActivityDate="2010-12-17T01:45:01.407" Title="Finding a closed form for an infinite product" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10187" PostTypeId="2" ParentId="10177" CreationDate="2010-11-13T23:07:26.960" Score="5" ViewCount="0" Body="&lt;p&gt;Any compact, connected manifold $M$ of dimension $n$ embedded in $\mathbb{R}^{n+1}$ is orientable.&#xA;This follows from the fact that there exists a smooth function $f$ on $\mathbb{R}^{n+1}$ s.t. $M$ is the zero locus of $f$ and the derivative of $f$ does not vanish on $M$ (so the gradient of $f$ gives a smooth normal vector field).&#xA;This also proves Jordan-Brouwer's theorem ($M$ cuts $\mathbb{R}^{n+1}$ in two parts: $f&amp;gt;0$ and $f&amp;lt;0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of $f$ is a &quot;technical lemma&quot; ;)&#xA;First, using compactness, you can show that there is an $\epsilon &amp;gt; 0$ and a covering $\left( V_x \right)_{x \in M}$ of $M$ such that $B=\left\{ y \in \mathbb{R}^{n+1}\ |\ d(y,M) \lt \epsilon \right\}$ is the disjoint union of the $\left\{ x + t n_x \ |\ -\epsilon \lt t \lt \epsilon \right\}$ ($n_x$ being a vector of norm one orthogonal to $T_x M$), and so $B$ is locally $V_x \times ]-\epsilon,\epsilon[$.&#xA;From there you can deduce that $\mathbb{R^{n+1}}$ has an open covering $\left(A_i\right)_i$ with smooth functions $f_i$ on $A_i$ s.t. whenever $A_i \cap A_j \neq \emptyset$, $f_i = \pm f_j$ locally on $A_i \cap A_j$, and $\cup_i f_i^{-1}(0) = M$ has empty interior (for this you need to take a smooth  non-decreasing function $\lambda : \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f(x)=-1$ when $x \lt -\epsilon/2$, $f(x)=1$ when $x \gt \epsilon/2$, and $f$ is increasing inbetween. then for $(y,t) \in V_x \times ]-\epsilon,\epsilon[$, define $f_i(y,t)=\lambda(t)$, and take the $f_i$s to be $1$ outside the $\epsilon/2$-neighbourhood of $M$).&#xA;This, together with the simple connectedness of $\mathbb{R}^{n+1}$, gives a smooth function $f$ on $\mathbb{R}^{n+1}$ locally equal to the $\pm f_i$, and you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a long exercise to solve your technical problem ;)&#xA;I'm sorry I couldn't just give you a reference, but the only one I have (and from which I took the above sketch of proof) is a french book: Thèmes d'analyse pour l'agrégation, Calcul différentiel by Stéphane Gonnord and Nicolas Tosel (p. 100).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: actually they give a reference: Elon L. Lima, The Jordan-Brouwer separation theorem for smooth hypersurfaces, American Mathematical Monthly, Volume 95 Issue 1, Jan. 1988&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-11-13T23:22:58.433" LastActivityDate="2010-11-13T23:22:58.433" CommentCount="16" />
  <row Id="10188" PostTypeId="2" ParentId="10124" CreationDate="2010-11-13T23:10:45.117" Score="0" ViewCount="0" Body="&lt;p&gt;You can calculate the exact probability of drawing $t$ unique vertices using inclusion-exclusion. We can estimate this probability as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\binom{n}{t} \binom{\binom{t}{2}}{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but you're overcounting instances when less than $ t $ vertices actually appear; you can fix that using inclusion exclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, you can try to calculate the expectation directly.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can also try to work your way from Moron's formula. His formula gives the expectation when &lt;i&gt;k&lt;/i&gt; edges are drawn with replacement. We know that distribution of the number of distinct edges drawn, so we can express the expectation when &lt;i&gt;k&lt;/i&gt; edges are drawn with replacement using the expectations when $k_0$ unique edges are drawn for $k_0 \leq k$. We get linear equations and maybe we can solve them to find the probability when &lt;i&gt;k&lt;/i&gt; unique edges are drawn.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-13T23:10:45.117" />
  <row Id="10189" PostTypeId="2" ParentId="10186" CreationDate="2010-11-13T23:13:39.607" Score="3" ViewCount="0" Body="&lt;p&gt;Convert the product to a sum by taking logarithms. For $x$ small, $\tan x \approx x$ and $\log (1 - x) \approx -x$ and so after taking logs, your product is roughly like the convergent geometric series $\sum \pi 2^{-k}$. So it converges.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-13T23:13:39.607" />
  <row Id="10190" PostTypeId="2" ParentId="10183" CreationDate="2010-11-13T23:14:46.547" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Your displayed sum has $\rm\ (2^2 - 1^2)\ \ 1&amp;#39;s,\ \ (3^2-2^2)\ \ 2&amp;#39;s,\ \ (4^2-3^2)\ \ 3&amp;#39;s,\ \ldots $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-13T23:14:46.547" />
  <row Id="10191" PostTypeId="2" ParentId="10186" CreationDate="2010-11-13T23:15:06.460" Score="7" ViewCount="0" Body="&lt;p&gt;First of all I do not understand why you say the product converges to 1. In fact the product must me strictly less than the first term namely, $(1-\tan^4(\frac{\pi}{8}))$. You can however say this function converges since $\displaystyle \sum_{k=1} \tan^4(\frac{\pi}{2^k}) &amp;lt; \infty$. Hence the product converges absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: $\displaystyle \Pi_{k=1}^n (1+a_k) \leq \exp(\displaystyle \sum_{k=1}^n a_k)$. So convergence of $\displaystyle \sum_{k=1}^{\infty} a_k$ implies the convergence of the infinite product $\displaystyle \Pi_{k=1}^{\infty} (1+a_k)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find an approximate answer using Matlab and then make an educated guess about the limit and then try to prove it. However, the limit of the first $10^7$ terms give me $0.968946146259369$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to make an educated guess (something based on $\pi$,$e$ or the golden ratio). It will be interesting to see what this product converges to!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just for the sake of completion and the solution is due to $\textbf{Moron}$. Following the idea suggested by Moron, and since we know the product has to converge let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_n = \displaystyle \Pi_{k=1}^n (1-\tan^4(\frac{\pi}{2^{k+2}})) = \displaystyle \Pi_{k=1}^n \frac{\cos(\frac{\pi}{2^{k+1}})}{\cos^4(\frac{\pi}{2^{k+2}})} = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{1}{\displaystyle \Pi_{k=1}^n \cos^3(\frac{\pi}{2^{k+2}})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_n = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{8^n \times \sin^3(\frac{\pi}{2^{n+2}})}{\sin^3(\frac{\pi}{4})}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the limit as $n \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $p = \frac{\cos(\frac{\pi}{4})}{1} \times \frac{(\frac{\pi}{4})^3}{\sin^3(\frac{\pi}{4})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $p = 2 \times (\frac{\pi}{4})^3 = \frac{\pi^3}{32} \approx 0.968946146259369$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-14T01:35:46.243" LastActivityDate="2010-11-14T01:35:46.243" CommentCount="3" />
  <row Id="10192" PostTypeId="2" ParentId="10164" CreationDate="2010-11-13T23:37:11.247" Score="3" ViewCount="0" Body="&lt;p&gt;By change of variable $x=y+t$, we need to bound the expression $$(\frac{|y+t|}{1+|y|})^k |g(t)|.$$ Since $$\frac{|y+t|}{1+|y|}\le \frac{|y|+|t|}{1+|y|}\le 1+|t|,$$ and &#xA;$$\frac{(1+|t|)^k}{1+t^{2k}}$$ and $$(1+t^{2k})|g(t)|$$ are bounded, say by $B_k$ and $C_k$, we find that&#xA;$$(\frac{|y+t|}{1+|y|})^k |g(t)|\le (1+|t|)^k|g(t)|\le B_kC_k.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-13T23:37:11.247" CommentCount="3" />
  <row Id="10193" PostTypeId="2" ParentId="10186" CreationDate="2010-11-13T23:38:08.017" Score="9" ViewCount="0" Body="&lt;p&gt;Write $\displaystyle 1- \tan^4 \theta$ as $\dfrac{\cos^2 \theta - \sin^2 \theta}{\cos^4 \theta} = \dfrac{\cos 2\theta}{\cos^4 \theta}$ and make repeated use of the following trick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To evaluate $\displaystyle \prod_{k=1}^{n}\cos \dfrac{\theta}{2^k}$, multiply and divide by $\sin \dfrac{\theta}{2^n}$ and use $\displaystyle 2\sin \theta \cos \theta = \sin 2\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should give you a closed form formula for the product of first $\displaystyle n$ terms, which I believe can be evaluated easily as $\displaystyle n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on this you can refer: &lt;a href=&quot;http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vi%C3%A8te's_formula&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-13T23:50:11.557" LastActivityDate="2010-11-13T23:50:11.557" CommentCount="3" />
  <row Id="10194" PostTypeId="1" AcceptedAnswerId="10196" CreationDate="2010-11-14T00:04:29.133" Score="1" ViewCount="80" Body="&lt;p&gt;Suppose $64$ students are taking an exam. The exam has $16$ questions in multiple choice format. Students must fill an oval for every one of the $16$ questions. They can't revise their answers. The time taken for a students to answer a question is exponentially distributed with mean $8$ minutes. So $f(t) = \frac{1}{8}e^{-t/8}$. The times for answering questions are independent. Using the normal approximation calculate the probability that the average time taken by the students is more than $132$ minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want to find $P[T_1+T_2 + \cdots + T_{64} &amp;gt; 132(64)]$. What would be the distribution of a sum of exponential random variables? Would it be a gamma distribution? If we let $S = T_{1} + T_{2} + \cdots + T_{64}$, then $P[S &amp;gt; 132(64)] = 1-P[S \leq 132(64)]$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-14T00:25:47.663" LastActivityDate="2010-11-14T00:40:18.057" Title="Questions on a Test" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10195" PostTypeId="2" ParentId="10180" CreationDate="2010-11-14T00:12:49.563" Score="1" ViewCount="0" Body="&lt;p&gt;The sense in which &quot;pullback&quot; is being used here is the same sense it's being used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Pullback&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; Wikipedia article; that is, it's being used to refer to the process of &quot;pulling back&quot; a morphism $A \to \text{End}(R)$ (an $A$-module) along the morphism $k \to A$ to get a morphism $k \to \text{End}(R)$ (a $k$-module) by precomposition.  This is different from the categorical pullback.  I don't know why they're named the same.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-14T00:12:49.563" />
  <row Id="10196" PostTypeId="2" ParentId="10194" CreationDate="2010-11-14T00:24:11.923" Score="3" ViewCount="0" Body="&lt;p&gt;Yes. It will be a Gamma distribution. But here as the problem says you can approximate it by a normal distribution. If you look at the variable $T = \displaystyle \sum_{k=1}^{64} T_i$, it has mean $64 \times 16 \times 8$ and the variance is again $64 \times 16 \times 8^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that $T_i$ has mean $16 \times 8$ and has variance $16 \times 8^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence all you need now is to look at the random variable $\frac{T-64 \times 16 \times 8}{\sqrt{64 \times 16 \times 8^2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $P(T \geq 132 \times 64) = P(\frac{T-64 \times 16 \times 8}{32 \times 8} \geq \frac{132 \times 64 - 128 \times 64}{256}) \approx  P(Z \geq 1) = 0.158655$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Z$ is the standard normal random variable with zero mean and unit variance. Look up the table for this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, Kindly check my arithmetic calculations)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-14T00:40:18.057" LastActivityDate="2010-11-14T00:40:18.057" CommentCount="0" />
  <row Id="10197" PostTypeId="2" ParentId="10194" CreationDate="2010-11-14T00:26:41.887" Score="1" ViewCount="0" Body="&lt;p&gt;The sum of i.i.d exponential random variables follows the Erlang distribution (which is a special case of the Gamma distribution). For details, check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlang_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Erlang_distribution&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-14T00:26:41.887" />
  <row Id="10198" PostTypeId="1" CreationDate="2010-11-14T00:53:39.297" Score="1" ViewCount="29" Body="&lt;p&gt;Suppose $f(x,y)$ is bivariate normal. You are given $E(X)$, $E(Y)$, $\text{Var}(X)$ and $\text{Var}(Y)$. You are also given $\rho$ the correlation coefficient. How would one find $\text{Var}(Y|X=k)$? Would you use &lt;a href=&quot;http://math.stackexchange.com/questions/7040/conditional-and-total-variance&quot;&gt;the total law of variance&lt;/a&gt;? We know that $\rho = \frac{\text{Cov}(X,Y)}{\sigma_x \sigma_y}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-14T00:59:34.160" Title="Bivariate Normal Distribution" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="10199" PostTypeId="2" ParentId="10198" CreationDate="2010-11-14T00:59:34.160" Score="1" ViewCount="0" Body="&lt;p&gt;Check the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T00:59:34.160" />
  <row Id="10200" PostTypeId="2" ParentId="10164" CreationDate="2010-11-14T01:24:10.357" Score="1" ViewCount="0" Body="&lt;p&gt;Trying to be systematic here... your condition is equivalent to saying $|g(x)| \leq C_n{1 \over 1 + |x|^n}$ for all $n$. So if your inequality is going to be true, we'd expect for some $l$ depending on $k$ we would have the following for all $(x,y)$:&#xA;$${|x|^k \over 1 + |y|^k}{1 \over 1 + |x - y|^l} \leq C_k$$&#xD;&#xA;And if we can show this then your inequality follows. A natural place to start would be choosing $l = k$, in which case what is needed is&#xA;$$|x|^k \leq C_k (1 + |y|^k)(1 + |x - y|^k)$$&#xD;&#xA;Notice though that since $x = y + (x-y)$, $|x| \leq |y| + |x-y|$ and thus $|x|^k \leq A_k(|y|^k + |x - y|^k)$ by an easy argument. These two terms are part of the right hand side of the above, so that's all you need.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-14T01:24:10.357" CommentCount="1" />
  <row Id="10201" PostTypeId="1" AcceptedAnswerId="10206" CreationDate="2010-11-14T02:01:23.647" Score="0" ViewCount="77" Body="&lt;p&gt;How to prove that the power &lt;strong&gt;b ^ n&lt;/strong&gt; is an irrational number, where &lt;strong&gt;b&lt;/strong&gt; is an integer greater than 1 and &lt;strong&gt;n&lt;/strong&gt; is an irrational number?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-14T03:11:57.327" Title="b^n is irrational, where b&gt;1 is a integer and n is an irrational" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10202" PostTypeId="1" AcceptedAnswerId="10205" CreationDate="2010-11-14T02:31:37.860" Score="6" ViewCount="104" Body="&lt;p&gt;Suppose I have a 2-variable function $g(k,n)$ and I know that $g(k,n)=O(n^{f(k)})$, for fixed $k$ as $n \rightarrow \infty$, for some function $f=f(k)$.  Suppose I also know that $f(k)=O(\log k)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it legitimate to write $g(k,n)=O(n^{O(\log k)})$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-14T02:41:37.707" LastActivityDate="2010-11-24T22:21:22.187" Title="Is it legitimate to write nested big-Os in an asymptotic formula for a multivariable function?" Tags="&lt;asymptotics&gt;&lt;big-o&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="10203" PostTypeId="1" AcceptedAnswerId="10214" CreationDate="2010-11-14T02:37:28.513" Score="2" ViewCount="54" Body="&lt;p&gt;I'm dealing with tables that show connectivity within a network. Can anyone point me toward a technique for ordering the rows and columns so that nodes that are connected tend to be close to each other in the table?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I had a network like this (connections are directional), the structure isn't immediately obvious; it just looks random.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  1 2 3 4 5&#xA;1     x   x&#xA;2   x   &#xA;3     x  x&#xA;4   x  x &#xA;5        x &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But if you re-order the columns and rows this way, the structure is clear; you can see some   organisation in the network.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  2 4 5 3 1&#xA;2 x    &#xA;4 x x   &#xA;5   x   &#xA;3     x x &#xA;1     x x &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course, I'm dealing with larger tables, with ~1024 nodes. I could just code up something that tries to find the arrangement with the minimal average distance between connected nodes by some general-purpose method like gradient descent. But perhaps there's a technique that's particularly suited to this type of problem?&lt;/p&gt;&#xA;" OwnerUserId="3483" LastActivityDate="2010-11-14T05:29:03.673" Title="Technique for &quot;untangling&quot; directed networks?" Tags="&lt;analysis&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="10204" PostTypeId="2" ParentId="10202" CreationDate="2010-11-14T02:38:34.020" Score="2" ViewCount="0" Body="&lt;p&gt;It is legitimate provided you interpret it in the right sense. However, I do not understand what purpose it will serve though, since (obviously) you cannot write that it is equal to $\mathcal{O}(n^{log(k)})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading a blog by TerryTao and he has used this notation. So I assume, you can use it provided you give the right interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2010/11/20/the-guth-katz-bound-on-the-erdos-distance-problem/&quot; rel=&quot;nofollow&quot;&gt;http://terrytao.wordpress.com/2010/11/20/the-guth-katz-bound-on-the-erdos-distance-problem/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T22:21:22.187" LastActivityDate="2010-11-24T22:21:22.187" CommentCount="1" />
  <row Id="10205" PostTypeId="2" ParentId="10202" CreationDate="2010-11-14T02:50:20.440" Score="3" ViewCount="0" Body="&lt;p&gt;In this particular case, you can simplify and conclude that $g(k,n) = n^{O(\log k)}$, since you can put fold the first constant into the second (as long as $n,k \geq 1 + \epsilon$). This fails if you think of $n$ as tending to infinity, but $k$ could be arbitrarily close to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: My answer assumes that the constant in $g(k,n) = O(n^{f(k)})$ doesn't depend on &lt;i&gt;k&lt;/i&gt;, which is the usual convention in computer science. If it does, we'd write $g(k,n) = O_k(n^{f(k)})$ to emphasize this (or sometimes just emphasize it in words).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-14T03:17:33.953" LastActivityDate="2010-11-14T03:17:33.953" CommentCount="3" />
  <row Id="10206" PostTypeId="2" ParentId="10201" CreationDate="2010-11-14T03:11:57.327" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfond-Schneider theorem&lt;/a&gt; shows that if you assume that $n$ is algebraic, then $b^n$ is indeed irrational, in fact transcendental! You can even take $b$ to be algebraic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can ask whether there's an easy proof that $ b^n $ is irrational for &lt;i&gt;b&lt;/i&gt; integral and &lt;i&gt;n&lt;/i&gt; an &lt;i&gt;algebraic&lt;/i&gt; irrational.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T03:11:57.327" />
  <row Id="10207" PostTypeId="2" ParentId="10202" CreationDate="2010-11-14T03:12:55.517" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strike&gt;No (if you mean what I think you mean by that statement).&lt;/strike&gt;  If you have more than one variable in play it's always a good idea to keep track of which variables the implied constant for each big-O depends on.  &lt;strike&gt;In your example, the implied constant for the outer big-O depends on $k$, but the implied constant for the inner big-O doesn't, so it seems misleading to use notation which pretends they are the same.  And if the implied constant grows fast enough, the final statement you want is false (if an unadormed big-O means it doesn't depend on any of the variables which appear on the LHS).  Consider, as I said in my comment to Yuval Filmus, $g(k, n) = 2^k n^{\log k}, f(k) = \log k$.  In addition to $f(k) = O(\log k)$ you need to know that the implied constant doesn't depend on $n$.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  The above was nonsense.  Actually you are fine.  You have a constant $C_1$ such that $f(k) \le C_1 \log k$ and another constant $C_2$ such that $g(k) \le C_2 n^{f(k)}$, hence $g(k) \le C_2 n^{C_1 \log k}$, or as Yuval Filmus points out, $g(k) = n^{O(\log k)}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-14T12:23:28.557" LastActivityDate="2010-11-14T12:23:28.557" CommentCount="1" />
  <row Id="10208" PostTypeId="2" ParentId="10119" CreationDate="2010-11-14T03:41:15.440" Score="1" ViewCount="0" Body="&lt;p&gt;a^3 - b^3 = (a - b)(a^2 + ab + b^2). So x^3 - 8 = (x - 2)(x^2 + 2x + 4).&lt;br&gt;&#xA;For x^3 - 3x^2 + x - 3, factor by grouping the first two terms together and taking out the GCF and grouping the second two and doing the same. This should give you&lt;br&gt;&#xA;x^2(x - 3) + 1(x - 3). This gives you a greatest common factor of (x - 3) that you can now factor out.&lt;br&gt;&#xA;(x - 3)(x^2 + 1). And you are done.&lt;/p&gt;&#xA;" OwnerUserId="134" LastActivityDate="2010-11-14T03:41:15.440" />
  <row Id="10209" PostTypeId="1" CreationDate="2010-11-14T03:57:25.963" Score="3" ViewCount="165" Body="&lt;p&gt;I am going to a Computer Science Course in University next year. I heard that Discrete Mathematics is whats required for Comp Sci so, I am looking for resources/books that I can read to get started with. I picked up &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071615865&quot; rel=&quot;nofollow&quot;&gt;Schaum's Outline of Discrete Mathematics&lt;/a&gt;. I find that some things are alittle hard to understand still. I have also looked at some videos from &lt;a href=&quot;http://www.youtube.com/user/khanacademy?blend=1&amp;amp;ob=4&quot; rel=&quot;nofollow&quot;&gt;Khan Academy on YouTube&lt;/a&gt;, I find the videos easy to understand. Except that I don't really know where am I supposed to start 1st. I am on algebra now, but can someone tell me whats the sequence I should go from here, eg. topics/playlists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An you know, for maths, there must be alot of practice. So where can I find lots of practice questions? Schaum's Outline of Discrete Mathematics has quite little practice questions. &lt;/p&gt;&#xA;" OwnerUserId="2006" LastActivityDate="2010-11-18T03:07:33.170" Title="Resources/Books for Discrete Mathematics" Tags="&lt;discrete-mathematics&gt;&lt;books&gt;&lt;online-resources&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="10210" PostTypeId="2" ParentId="10209" CreationDate="2010-11-14T03:59:40.220" Score="0" ViewCount="0" Body="&lt;p&gt;&quot;Algorithm Design&quot; by Authors Kleinberg and Tardos is a good comprehensive one! There is a good amount of theory and also a good number of problems at the end of each chapter.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-14T03:59:40.220" />
  <row Id="10211" PostTypeId="2" ParentId="10209" CreationDate="2010-11-14T04:07:30.940" Score="3" ViewCount="0" Body="&lt;p&gt;I think Kenneth Rosen's book on Discrete Mathematics does a good job of being exhaustive. Although I have heard mixed reviews of the book from others, I myself, began studying discrete mathematics from this book and was very happy with it. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T04:07:30.940" />
  <row Id="10212" PostTypeId="1" AcceptedAnswerId="10215" CreationDate="2010-11-14T04:29:36.513" Score="8" ViewCount="117" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Abel's identity&lt;/a&gt; states that if $X(t)$ and $A(t)$ are $n\times n$ matrix-valued functions such that $X&amp;#39;(t)=A(t)X(t)$, then $\frac{d}{dt}(\det X(t)) = \mathrm{tr}\,A(t) \cdot \det X(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether there's a nice high-brow basis-independent way to see this. Given that you can state the problem without ever referring to matrix entries, I want prove it without ever referring to matrix entries. I'd expect the identity $\det e^A = e^{\mathrm{tr}(A)}$ to play a central role in the proof, but I have yet to come up with such an argument.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can prove this without too much trouble in a bare-hands way, but I don't see how to turn this into a basis-free proof. Suppose $X_i(t)$ is the matrix you get from $X(t)$ by taking the derivative of every entry in the $i$-th row. Then $\frac{d}{dt}(\det X(t)) = \sum_{i=1}^n \det(X_i(t))$. Using the relation $X&amp;#39;(t)=A(t)X(t)$ you can express the derivative of the $i$-th row of $X&amp;#39;(t)$ in terms of the entries of $A$ and $X$. When you do this, you find that $\det X_i(t)$ is simply $\det X(t)$ multiplied by the $(i,i)$-th entry of $A(t)$.&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2010-11-14T06:31:52.377" Title="Is there a basis-independent proof of Abel's identity?" Tags="&lt;differential-equations&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10213" PostTypeId="2" ParentId="10124" CreationDate="2010-11-14T05:11:31.547" Score="5" ViewCount="0" Body="&lt;p&gt;For choosing without replacement, here is an exact answer.  Assuming $n \geq 2$, so that there is at least one pair, and $1 \leq k \leq \binom{n}{2}$, so that you're choosing at least one pair but not more than the total number of pairs, the expected value is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n - \left(\frac{n^2 - 3n - 2k + 4}{n-1}\right) \frac{\binom{\binom{n}{2} - n + 1}{k-1}}{\binom{\binom{n}{2} - 1}{k-1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assume that we are choosing pairs in order.  Let $X_k$ be the number of distinct items from $S$ through $k$ pairs.  Let $Y_i$ be the number of items in the $i$th pair that did not appear in any of the previous pairs.  So $X_k = \sum_{i=1}^k Y_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $Y_i$ is either 0, 1, or 2.  Since there are $\binom{n}{2} - n + 1$ pairs that do not contain a given item and $\binom{n}{2} - 2n + 3$ pairs that do not contain either of two given items, we have&#xA;$$P(Y_i = 1) = \frac{\binom{\binom{n}{2} - n + 1}{i-1} + \binom{\binom{n}{2} - n + 1}{i-1} - 2 \binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}$$&#xD;&#xA;and&#xA;$$P(Y_i = 2) = \frac{\binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$&#xD;&#xA;Thus &#xA;$$E[Y_i] = 2\frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be proved by induction that &#xA;$$\sum_{i=0}^k \frac{\binom{M}{i}}{\binom{N}{i}} = \frac{(N+1)\binom{N}{k} - (M-k)\binom{M}{k}}{(N+1-M)\binom{N}{k}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &#xA;$$E[X_k] = \sum_{i=1}^k E[Y_i] = 2\sum_{i=1}^k \frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}} $$ &#xA;$$= 2\frac{(\frac{n(n-1)}{2}-1+1)\binom{\binom{n}{2}}{k-1} - (\frac{n(n-1)}{2} - n + 1 - k + 1)\binom{\binom{n}{2} - n + 1}{k-1}}{(\binom{n}{2} - 1+1-\binom{n}{2} + n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$&#xD;&#xA;$$= \frac{n(n-1)\binom{\binom{n}{2}}{k-1} - (n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$&#xD;&#xA;$$= n -\frac{(n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-14T16:14:21.260" LastActivityDate="2010-11-14T16:14:21.260" CommentCount="4" />
  <row Id="10214" PostTypeId="2" ParentId="10203" CreationDate="2010-11-14T05:29:03.673" Score="1" ViewCount="0" Body="&lt;p&gt;Rahul's link to the Wikipedia article on topological sorting led me to &lt;strong&gt;tsort&lt;/strong&gt;, which is a Unix command to do topological sorting. The networks I'm dealing with might have some cycles, but it seems that &lt;strong&gt;tsort&lt;/strong&gt; can deal with cycles gracefully, and report the cycles to stderr. It looks like that will do the job I need. Thank you, Rahul!&lt;/p&gt;&#xA;" OwnerUserId="3483" LastActivityDate="2010-11-14T05:29:03.673" />
  <row Id="10215" PostTypeId="2" ParentId="10212" CreationDate="2010-11-14T05:31:04.903" Score="5" ViewCount="0" Body="&lt;p&gt;Please forgive this argument that plays rather fast and loose with infinitesimals. If you ignore second-order terms, an infinitesimal change $\delta t$ in $t$ takes $X(t)$ to $\big(I + \delta t A(t)\big) X(t)$. Since the determinant distributes over multiplication, all you need to show is that $\det\big(I + \delta t A(t)\big)$ is $1 + \delta t \operatorname{tr}A$ for infinitesimal $\delta t $. More formally, you need $\frac{d}{d\tau} \det(I + \tau A) = \operatorname{tr}A$. Since $I + \tau A$ agrees with $e^{\tau A}$ to first order (according to the Taylor series), you can differentiate the identity $\det e^{\tau A} = e^{\operatorname{tr} \tau A}$ with respect to $\tau$ and get the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; For a high-brow view, I think it makes much more sense to cast this in a geometric light, as it has much to do with &lt;a href=&quot;http://mathoverflow.net/questions/13526/geometric-interpretation-of-trace/13530#13530&quot; rel=&quot;nofollow&quot;&gt;the geometric interpretation of the trace&lt;/a&gt;. Consider how the vector ODE $x&amp;#39;(t) = A(t) x(t)$ acts on Euclidean $n$-space. The identity $\det e^A = e^{\operatorname{tr}A}$ is a statement of the fact that for constant $A$, the volume of any region grows (or shrinks) exponentially at the rate $\operatorname{tr}A$ under this ODE. (This is fine for varying $A(t)$ too as long as we don't care about second-order effects.) Now think of $X(t) $ as defining a parallelepiped in $n$-space as it flows around under the action of the ODE. The volume of the parallelepiped is $\det X(t)$, and its rate of change is therefore $\operatorname{tr}A $ times that. I think this is the answer I should have given in the first place.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-14T06:31:52.377" LastActivityDate="2010-11-14T06:31:52.377" />
  <row Id="10216" PostTypeId="1" CreationDate="2010-11-14T05:39:10.997" Score="2" ViewCount="154" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Companion_matrix&quot; rel=&quot;nofollow&quot;&gt;The companion matrix&lt;/a&gt; of a monic polynomial in 1 variable over a field $F$ plays an important role in understanding the structure of finite dimensional $F[x]$-modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an important fact that the characteristic polynomial and the minimal polynomial of $C(f)$ are both equal to $f$. This can be seen quite easily by induction on the degree of $f$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a different proof of this fact? I would love to see a graph theoretic proof or a non inductive algebraic proof, but I would be happy with anything that makes it seem like more than a coincidence!&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2010-11-14T22:45:51.437" Title="The characteristic and minimal polynomial of a companion matrix" Tags="&lt;combinatorics&gt;&lt;algebra&gt;&lt;reference-request&gt;&lt;commutative-algebra&gt;" AnswerCount="3" />
  <row Id="10217" PostTypeId="2" ParentId="10216" CreationDate="2010-11-14T05:59:13.283" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose your matrix is over a field $\mathbb{F}$. Look at $G = F[x]/f$, where $f$ is your polynomial of degree $n$. Then $G$ is a vector space over $\mathbb{F}$, and $C(f)$ is the matrix (with respect to the basis $1,x,x^2,\ldots,x^{n-1}$) corresponding to the linear operator $g \mapsto x \cdot g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f = 0$ in $G$, also $fx^i = 0$ in $G$, and so $f$ is a polynomial of degree $n$ such that $f(C(f)) = 0$; thus it must be the characteristic polynomial of $C(f)$. Moreover, any polynomial $g$ of smaller degree does not reduce to $0$ in $G$, so in particular $g(C(f))$ applied to the vector $1$ does not equal the zero vector. So $f$ is the minimal polynomial of $C(f)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T05:59:13.283" CommentCount="2" />
  <row Id="10218" PostTypeId="1" CreationDate="2010-11-14T06:05:19.193" Score="1" ViewCount="57" Body="&lt;p&gt;Say I have an equation where the value is $m-4$ when $m \ge 5, n=3,4$; $n-4$ when $n \ge 5, m=3,4$; and $m+n-9$ when $m,n \ge 5$. How can I combine these all into one statement? How can I combine them with further ones, like $m-6$ where $m \ge 7, n \in [3,6]$, etc?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1562" LastEditorDisplayName="" LastEditDate="2010-11-14T06:11:49.940" LastActivityDate="2010-11-14T07:47:47.833" Title="How to combine cases" Tags="&lt;algebra&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="10219" PostTypeId="2" ParentId="10218" CreationDate="2010-11-14T06:25:44.917" Score="1" ViewCount="0" Body="&lt;p&gt;I think for the first statement the following formula should work with the given domain:&#xA;$f(m,n)=(m-4)H[m-5]+(n-4)H[n-5]+H[m-5]H[n-5]$&#xD;&#xA;where H is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow&quot;&gt;heaviside&lt;/a&gt; step function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the phrasing for the second part can be improved, judging from your comments above. From what I gather you can add terms such as $(m-6)H[m-7]$ for the later conditions.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T06:36:36.417" LastActivityDate="2010-11-14T06:36:36.417" />
  <row Id="10220" PostTypeId="2" ParentId="9265" CreationDate="2010-11-14T06:44:54.127" Score="1" ViewCount="0" Body="&lt;p&gt;Note that $\frac{d(f(x) \times y)}{dx} = \frac{d(f(x))}{dx} \times y + f(x) \times  \frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance, if $f(x) = x$, we get $\frac{d(x \times y)}{dx} = \frac{d(x)}{dx} \times y + x \times  \frac{dy}{dx} = 1 \times y + x \times \frac{dy}{dx} = y + x \times \frac{dy}{dx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $f(x) = c$, where $c$ is a constant, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d(c \times y)}{dx} = \frac{d(c)}{dx} \times y + c \times \frac{dy}{dx} = 0 \times y + c \times \frac{dy}{dx} = c \times \frac{dy}{dx}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-14T06:44:54.127" />
  <row Id="10221" PostTypeId="2" ParentId="10218" CreationDate="2010-11-14T06:48:27.003" Score="2" ViewCount="0" Body="&lt;p&gt;For your first question, it turns out that on the values you've specified, $f(m,n) = \max(m-5,0)+\max(n-5,0) + 1$. In general, if you're trying to get a handle on a function of many variables, it's a good idea to see if it can be separated into a combination of functions of fewer variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, if what you mean is that you want to add $m - 6$ to the output if certain conditions are satisfied, and leave it alone otherwise, you could define $g(m,n) = f(m,n) + h(m,n)$, where $h(m,n)$ is either $m-6$ or zero depending on etc etc. Or you could just write out all the cases of $g(m,n)$, which is the systematic but mind-numbing way.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-14T06:48:27.003" />
  <row Id="10223" PostTypeId="1" CreationDate="2010-11-14T07:38:57.790" Score="0" ViewCount="41" Body="&lt;p&gt;$\max(f(n), g(n)) = O(f(n) + g(n))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I prove this? Also I'd appreciate the markup being corrected, thanks.&lt;/p&gt;&#xA;" OwnerUserId="538" LastEditorUserId="538" LastEditorDisplayName="" LastEditDate="2010-11-14T07:46:41.360" LastActivityDate="2010-11-14T07:46:41.360" Title="Max of functions are in big O of the sum of their individual big Os" Tags="&lt;asymptotics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10224" PostTypeId="2" ParentId="10223" CreationDate="2010-11-14T07:43:54.527" Score="2" ViewCount="0" Body="&lt;p&gt;You can use $\max(f(n),g(n)) \leq f(n) + g(n)$, given that both functions are non-negative.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:43:54.527" CommentCount="2" />
  <row Id="10225" PostTypeId="2" ParentId="10218" CreationDate="2010-11-14T07:47:47.833" Score="0" ViewCount="0" Body="&lt;p&gt;$$ f(n,m) = \begin{cases} m-4 &amp;amp; m \geq 5,\, n=3,4, \\ n-4 &amp;amp; n \geq 5,\, m=3,4, \\ m+ n - 9 &amp;amp; m,n \geq 5. \end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:47:47.833" />
  <row Id="10226" PostTypeId="2" ParentId="10172" CreationDate="2010-11-14T07:55:38.470" Score="1" ViewCount="0" Body="&lt;p&gt;Have you tried &lt;a href=&quot;http://www.treewidth.com/&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;? They have a small collection of graphs and a library for computing their exact tree-width, which you can compare against your code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try constructing &quot;random&quot; graphs by following the definition, though in that case you only know an upper bound on the tree-width; probably every reasonable way of doing this will actually result in a graph of &quot;most probably&quot; the tree-width you're aiming at.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:55:38.470" CommentCount="1" />
  <row Id="10227" PostTypeId="2" ParentId="9265" CreationDate="2010-11-14T07:57:46.033" Score="0" ViewCount="0" Body="&lt;p&gt;If it's just $y$ alone then you don't need to use implicit differentiation, just differentiate the other side of the equation. If it's $cy$ then you can just divide both sides by $c$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T07:57:46.033" />
  <row Id="10228" PostTypeId="1" CreationDate="2010-11-14T08:33:33.490" Score="0" ViewCount="76" Body="&lt;p&gt;after traversing a wireless error filled channel packets arrive at a network node according to a Poisson process with rate λ. these packets could have corrupted bits due to the wireless channel which has a bit error probability &lt;em&gt;p&lt;/em&gt;. A packet is said to arrive at the node successfully if no bits are in error. the number of bits in a given packet is Poisson distributed with mean μ.Assume packets are not transmitted . find the rate at which successful packets arrive at the network node&lt;/p&gt;&#xA;" OwnerUserId="3490" LastEditorUserId="3490" LastEditorDisplayName="" LastEditDate="2010-11-14T08:52:30.610" LastActivityDate="2010-12-14T10:44:26.167" Title="What is the rate at which successful packets arrive?" Tags="&lt;probability&gt;&lt;homework&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10229" PostTypeId="1" CreationDate="2010-11-14T08:59:50.390" Score="1" ViewCount="73" Body="&lt;p&gt;Which are the parallel lines?&#xA;&lt;img src=&quot;http://i.imgur.com/IalyM.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prove that a and b are parallel but can't prove that c and d are.&#xA;The angle ,which is 135 degrees and the 45 one are alternate angles.They aren't equal ,so c &lt;em&gt;isn't&lt;/em&gt; parallel to d but the answer is that c is parallel to d.Am I right that they aren't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&#xA;They are a little inclined like these:&#xA;&lt;img src=&quot;http://i.imgur.com/q7buJ.jpg&quot; alt=&quot;http://www.math10.com/f/download/file.php?id=1169&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1795" LastEditorUserId="1795" LastEditorDisplayName="" LastEditDate="2010-11-14T10:10:42.993" LastActivityDate="2010-11-14T15:12:18.103" Title="Parallel straight lines" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="10230" PostTypeId="2" ParentId="10183" CreationDate="2010-11-14T09:25:15.643" Score="1" ViewCount="0" Body="&lt;p&gt;Let $\lfloor \sqrt{n} \rfloor = a$, then the following sum holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{0\le k &amp;lt; n} \lfloor \sqrt{k} \rfloor = na - \frac{a^3}{3} -&#xD;&#xA;    \frac{a^2}{2} - \frac{a}{6}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also enjoy the slightly more tricky sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{0\le k &amp;lt; n} \lfloor k^{1/3} \rfloor = nb - \frac{b^2}{4} - \frac{b^3}{2} - \frac{b^4}{4},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where this time $b =\lfloor n^{1/3} \rfloor$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to obtain these answers is to merely apply the summation identity that I &lt;a href=&quot;http://math.stackexchange.com/questions/8336/proving-sum-limits-k1n-bigl-lfloor-fracka-bigr-rfloor-bigl-lfloo/8368#8368&quot;&gt;mentioned here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-14T20:21:25.753" LastActivityDate="2010-11-14T20:21:25.753" />
  <row Id="10231" PostTypeId="1" AcceptedAnswerId="10235" CreationDate="2010-11-14T09:35:07.517" Score="2" ViewCount="141" Body="&lt;p&gt;How to solve this manually ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; In my module the answer is given as &lt;strong&gt;$0$&lt;/strong&gt; but when I used mathematica &lt;code&gt;N[Sin[10 Degree] + Sin[20 Degree] + Sin[30 Degree] - Sin[360 Degree],50]&lt;/code&gt; gives &lt;strong&gt;$1.0156683209925990818958162414 \cdots$&lt;/strong&gt; (truncated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess there is some mistake in the problem statement.&lt;/p&gt;&#xA;" OwnerUserId="3492" LastEditorUserId="3492" LastEditorDisplayName="" LastEditDate="2010-11-14T10:19:05.190" LastActivityDate="2010-11-16T19:58:25.793" Title="Find the value of $\sin 10^\circ + \sin 20^\circ + \sin 30^\circ -  \sin 360^\circ $" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10232" PostTypeId="1" CreationDate="2010-11-14T09:40:25.313" Score="2" ViewCount="30" Body="&lt;p&gt;It's best explained with an image of course:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/j2MpR.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L1, L2, h1 &amp;amp; h2 are all given, and I would like to calculate the length of L3. Is it possible, and how?&lt;/p&gt;&#xA;" OwnerUserId="3491" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-14T09:52:18.397" LastActivityDate="2010-11-14T09:58:22.093" Title="How to calculate the length of line in a trapezium, given some other lines." Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="10233" PostTypeId="1" CreationDate="2010-11-14T09:49:19.677" Score="15" ViewCount="271" Body="&lt;p&gt;I want to motivate the quadratic reciprocity theorem, which at first glance does not look too important to justify it being one of Gauss' favorites. So far I can think of two uses that are basic enough to be shown immediately when presenting the theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) With the QRT, it is immediate to give a simple, efficient algorithm (that can be done even by hand) for computing Legendre symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In Euler's proof of Fermat's claim on the conditions in which a prime $p$ is of the form $x^2+ny^2$ (for certain small values of $n$) the proof is reduced to finding the conditions under which $p$ divides $x^2+ny^2$ for some $x,y$, hence to the question under which conditions is $-n$ a quadratic residue modulo $p$, which leads immediately to the QTR (for example, for $n=3$, where we get that $p\equiv_3 1$). I really like this example since it begins with an &quot;historic&quot; problem and proceeds to &quot;discover&quot; the QTR through special cases (which is what Euler did in practice - see Cox's book on &quot;Primes of the form $x^2+ny^2$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am sure there are many more examples (and I'm especially curious as to how Gauss reached the theorem himself). I'd love to hear about them and receive references for further reading.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-18T18:27:13.693" Title="Uses of quadratic reciprocity theorem" Tags="&lt;number-theory&gt;&lt;quadratic-reciprocity&gt;&lt;famous-theorems&gt;&lt;motivation&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="2" />
  <row Id="10234" PostTypeId="1" AcceptedAnswerId="10239" CreationDate="2010-11-14T09:50:48.013" Score="3" ViewCount="73" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt; axioms demand that one has an inverse element. When it comes to multiplication zero has no inverse element because one cannot reconstruct the original number after it was multiplied by zero: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an irreversible loss of information!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore all numbers cannot form a group under multiplication as long as zero is included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to matrices some matrices have an inverse, some don't have one. &lt;a href=&quot;http://en.wikipedia.org/wiki/Singular_matrix&quot; rel=&quot;nofollow&quot;&gt;Singular matrices&lt;/a&gt; have &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow&quot;&gt;determinant&lt;/a&gt; zero and don't have one. When the column vectors are linear dependent the determinant is zero and the matrix is singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt;&#xA;Where does the loss of information happen here? How is 'more' information lost when a vector is multiplied by a singular matrix (compared to multiplying it by an invertible matrix)?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-14T10:06:51.683" Title="Loss of information when multiplying by singular matrix" Tags="&lt;linear-algebra&gt;&lt;group-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10235" PostTypeId="2" ParentId="10231" CreationDate="2010-11-14T09:52:28.927" Score="6" ViewCount="0" Body="&lt;p&gt;The last two terms have &lt;a href=&quot;http://en.wikipedia.org/wiki/Exact_trigonometric_constants&quot; rel=&quot;nofollow&quot;&gt;simple well-known values&lt;/a&gt;.&#xA;The first two are less appetizing; see &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometryAnglesPi18.html&quot; rel=&quot;nofollow&quot;&gt;here for 10 degrees&lt;/a&gt; and &lt;a href=&quot;http://mathworld.wolfram.com/TrigonometryAnglesPi9.html&quot; rel=&quot;nofollow&quot;&gt;here for 20 degrees&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-14T09:52:28.927" />
  <row Id="10236" PostTypeId="2" ParentId="10233" CreationDate="2010-11-14T09:57:19.160" Score="0" ViewCount="0" Body="&lt;p&gt;There is a very nice historical introduction at &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows how mathematicians started to think in the direction of quadratic reciprocity after seeing the work of Fermat, and later how Legendre &quot;almost&quot; proved it, etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of credit has to be attributed to Legendre. It is not Gauss alone who made contributions to arithmetic during that period. In fact, in the preface to his &quot;Disquisitiones&quot;, Gauss acknowledges the work of Legendre.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-14T09:57:19.160" />
  <row Id="10237" PostTypeId="2" ParentId="10232" CreationDate="2010-11-14T09:58:22.093" Score="1" ViewCount="0" Body="&lt;p&gt;Denote the angle of the triangle by $\alpha$. Express all segments of the hypotenuse using $\alpha$ and $h_2,L_1,L_2$, then write an equation for $\alpha$ using $h_1$. Solve for $\alpha$, and find $L_3$. &lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T09:58:22.093" />
  <row Id="10238" PostTypeId="2" ParentId="10234" CreationDate="2010-11-14T10:01:07.210" Score="6" ViewCount="0" Body="&lt;p&gt;The loss of information is measured by the nullity (co-rank) of the matrix. A matrix with nullity $k$ is similar to a block matrix with regular first block and $0_{k\times k}$ second block. So there's a basis where $k$ of the basis vectors are &quot;thrown away&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T10:01:07.210" />
  <row Id="10239" PostTypeId="2" ParentId="10234" CreationDate="2010-11-14T10:04:03.387" Score="7" ViewCount="0" Body="&lt;p&gt;Let $A$ and $x$ be given, and set $b=Ax$. If $A$ is invertible, $x$ can be uniquely recovered from $b$ as $x=A^{-1}b$ (no information is lost, provided that we know $A$). But if $A$ is singular, someone who knows $A$ and $b$ cannot say for sure what $x$ was, since there are infinitely many solutions of the linear system $Ax=b$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-14T10:04:03.387" CommentCount="2" />
  <row Id="10240" PostTypeId="2" ParentId="10228" CreationDate="2010-11-14T10:05:47.817" Score="0" ViewCount="0" Body="&lt;p&gt;Calculate the probability that a (single) random packet arrives successfully. Conclude the sought rate.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T10:05:47.817" CommentCount="3" />
  <row Id="10241" PostTypeId="2" ParentId="10234" CreationDate="2010-11-14T10:06:51.683" Score="3" ViewCount="0" Body="&lt;p&gt;When multiplying by an invertible matrix $A$, you might say that no information is lost because you can recover $B$ from $AB$ by multiplying by $A^{-1}$.  Another way to put it is that the map $B\mapsto AB$ is injective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is singular, then information is lost because the map $B\mapsto AB$ is no longer injective.  One explanation is that $A$ has nontrivial nullspace, and $AB=0$ for each matrix $B$ whose column space lies in the null space of $A$.  It follows that $AB=AC$ when the column space of $B-C$ lies in the null space of $A$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-14T10:06:51.683" CommentCount="3" />
  <row Id="10242" PostTypeId="2" ParentId="10233" CreationDate="2010-11-14T10:27:15.407" Score="8" ViewCount="0" Body="&lt;p&gt;This is a good question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own take on motivating quadratic reciprocity is recorded &lt;a href=&quot;http://math.uga.edu/~pete/4400qrlaw.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (these are lecture notes from an undergraduate course on introductory number theory).  If you look there, you will find that most of what I have said is an elaboration of the two points you bring up.  I think a crisp way of explaining what QR does for you is in the idea of the &quot;direct&quot; and &quot;inverse&quot; problems attached to the Legendre symbol $(\frac{n}{p})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, for the &lt;strong&gt;direct problem&lt;/strong&gt; we fix $p$ and ask which integers $n$ are squares modulo $p$.  This is clearly a finite problem.  On the other hand there is the &lt;strong&gt;inverse problem&lt;/strong&gt;: we fix an integer $n$ and ask for which primes $p$ we have that &#xA;$n$ is a square modulo $p$.  This is, &lt;em&gt;a priori&lt;/em&gt;, an infinite problem.  However, it is one of great relevance to classical number theory: e.g. all of the many proofs I have seen of Fermat's Two Squares theorem pass through the fact that $-1$ is a square modulo an odd prime $p$ iff $p \equiv 1 \pmod 4$.  More generally, if you look at the Diophantine equation $x^2 - n y^2 = p$, for $n$ a nonzero integer and $p$ a prime with $\operatorname{gcd}(p,n) = 1$, then reducing modulo $p$ gives the necessary condition &#xA;$(\frac{n}{p}) = 1$.  Quadratic reciprocity to the rescue!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, as you also say, quadratic reciprocity gives an efficient algorithm for answering whether a particular $n$ is a square modulo $p$, much faster than computing all $\frac{p-1}{2}$ squares modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, this is more than enough for students to appreciate the usefulness of Gauss' &lt;em&gt;aureum theorema&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-14T10:27:15.407" />
  <row Id="10243" PostTypeId="2" ParentId="10046" CreationDate="2010-11-14T10:33:39.517" Score="2" ViewCount="0" Body="&lt;p&gt;Just as additional information, the inverse of these binomial coefficients satisfy several wonderful identities. For example, on &lt;a href=&quot;http://mathworld.wolfram.com/CentralBinomialCoefficient.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt; I saw the following amazing identity (valid for $k\ge 1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 1}\frac{1}{n^k \binom{2n}{n}} = \frac{1}{2}{}_{k+1}F_k(\underbrace{1,\ldots,1}_{k+1}; \frac{3}{2}, \underbrace{2,\ldots,2}_{k-1}; \frac{1}{4}),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where ${}_{k+1}F_k$ is a generalized hypergeometric function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to derive the above result? I don't recall the answer immediately, but perhaps browsing &lt;a href=&quot;http://www.math.rutgers.edu/~zeilberg/&quot; rel=&quot;nofollow&quot;&gt;Zeilberger's website&lt;/a&gt; might be helpful.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-14T22:29:50.287" LastActivityDate="2010-11-14T22:29:50.287" CommentCount="4" />
  <row Id="10244" PostTypeId="2" ParentId="10216" CreationDate="2010-11-14T10:49:25.830" Score="2" ViewCount="0" Body="&lt;p&gt;This is essentially Yuval's answer expressed in a slightly different way.&#xA;Let your companion matrix be&#xA;$$C=\pmatrix{0&amp;amp;1&amp;amp;0&amp;amp;\cdots&amp;amp;0\\\\&#xD;&#xA;0&amp;amp;0&amp;amp;1&amp;amp;\cdots&amp;amp;0\\\\&#xD;&#xA;\vdots&amp;amp;\vdots&amp;amp;\vdots&amp;amp;\ddots&amp;amp;\vdots\\\\&#xD;&#xA;0&amp;amp;0&amp;amp;0&amp;amp;\cdots&amp;amp;1\\\\&#xD;&#xA;-a_0&amp;amp;-a_1&amp;amp;-a_2&amp;amp;\cdots&amp;amp;-a_{n-1}}.$$&#xD;&#xA;Then for the vector $v=(1\,\,0\,\,0\cdots 0)$,&#xA;$$v\sum_{j=0}^{n-1} b_j C^j=&#xD;&#xA;\pmatrix{b_0&amp;amp;b_1&amp;amp;b_2&amp;amp;\cdots&amp;amp;b_{n-1}}$$&#xD;&#xA;so that $g(C)\ne0$ for all nonzero polynomials $g$ of degree less than $n$.&#xA;So the minimal polynomial has degree $n$, and equals the&#xA;characteristic polynomial (via Cayley-Hamilton).&#xA;But $vC^n=(-a_0\,\, {-a_1}\,\, {-a_2}\cdots{-a_{n-1}})$&#xD;&#xA;and for $v(C^n+\sum_{j=0}^{n-1}b_j C^j)=0$ we need $a_j=b_j$. So the minimal&#xA;and characteristic polynomials both equal $f$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-14T10:49:25.830" CommentCount="2" />
  <row Id="10245" PostTypeId="2" ParentId="10155" CreationDate="2010-11-14T10:56:02.447" Score="2" ViewCount="0" Body="&lt;p&gt;This is a transcrition of my comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation of the picture is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the origine of the coordinate system $x,y,z$ is the centre of the piece &quot;eye&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) force $F_1$ (with a magnitude of 80 N) lies in the $x,z$ plane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) its direction is $(x,y,z)=(4,0,3)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) the angle between the force components $F_{1_{x}},F_{1_{z}}$ is $π/2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) the force component $F_{1_{z}}=0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) the side (of the right triangle) with length 5 is opposite the right angle formed by the sides with lengths 3 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangle lies in the  $x,z$ plane. Its vertices are $P(0,0,0)$, $Q(4,0,0)$ and $R(4,0,3)$. The  hypotenuse is $PR$ and the base $PQ$.  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-16T14:07:18.193" LastActivityDate="2010-11-16T14:07:18.193" />
  <row Id="10246" PostTypeId="1" AcceptedAnswerId="10247" CreationDate="2010-11-14T11:11:26.550" Score="1" ViewCount="90" Body="&lt;p&gt;A power type &lt;strong&gt;b^[(a^/(1/n)]&lt;/strong&gt; where &lt;strong&gt;b&lt;/strong&gt;, &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;n&lt;/strong&gt; are positive integers greater than 1 and a^(1/n) is irrational, always represents an irrational number?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-14T11:14:00.310" Title="b^[a^(1/n)]: is it an irrational number?" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="10247" PostTypeId="2" ParentId="10246" CreationDate="2010-11-14T11:14:00.310" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, it is in fact transcendental by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfond-Schneider theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-14T11:14:00.310" />
  <row Id="10248" PostTypeId="2" ParentId="10177" CreationDate="2010-11-14T11:45:15.483" Score="3" ViewCount="0" Body="&lt;p&gt;The normal bundle of a codimension 1 embedding of a compact closed manifold $M$ in $\mathbb R^{n+1}$ is indeed trivial. Otherwise, you could find a simple closed curve in $\mathbb R^{n+1}$ that intersects $M$ in a single point. This implies that both the curve and $M$ represent nontrivial mod $2$ homology classes in $\mathbb R^{n+1}$; this is because the intersection product is dual to the cup product and cannot be nonzero on trivial homology classes. However there are no nontrivial homology classes in $\mathbb R^{n+1}$ since it is contractible. So we get a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the normal bundle is seen to be trivial, one can use the ambient orientation of $\mathbb R^{n+1}$ to locally orient the manifold, since there is a well defined positive normal direction.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-11-14T11:45:15.483" CommentCount="6" />
  <row Id="10249" PostTypeId="2" ParentId="10209" CreationDate="2010-11-14T12:27:57.717" Score="3" ViewCount="0" Body="&lt;p&gt;If you are going to study computer science, you want to read &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Graham, Knuth, Patashnik: Concrete Mathematics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look into solving recursion, combinatorics and maybe a little stochastics. Getting the hand of infinite sequences and series is also advisable. Above stated book covers that all because it is made for computer scientists. You should have math courses covering all that, though.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-14T12:27:57.717" />
  <row Id="10250" PostTypeId="2" ParentId="10233" CreationDate="2010-11-14T12:31:42.837" Score="9" ViewCount="0" Body="&lt;p&gt;Quadratic reciprocity allows you to make precise certain intuitions about the primes.  More precisely, it tells you that for every finite set $p_1, p_2, ... p_n$ of primes and every function $f : \{ 1, 2, ... n \} \to \{ -1, 1 \}$ there exists an arithmetic progression such that any prime $q$ in that progression satisfies $\left( \frac{p_i}{q} \right) = f(i)$.  In other words, you can get primes to behave locally independently (with respect to being or not being a quadratic residue).  You can use this idea to give one proof that, for example, $\mathbb{Q}(\sqrt{2}, \sqrt{3}, ... \sqrt{p_n})$ has the degree over $\mathbb{Q}$ that you think it does; this is described &lt;a href=&quot;http://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-14T12:31:42.837" />
  <row Id="10251" PostTypeId="1" AcceptedAnswerId="10312" CreationDate="2010-11-14T15:10:15.830" Score="1" ViewCount="130" Body="&lt;p&gt;How the following conversion is justified ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos \biggl( \theta + \frac{2\pi}{3} \biggr) + \cos \biggl( \theta + \frac{4\pi}{3} \biggr) = \cos \biggl(  \frac{\pi}{2} - \theta \biggr) - \cos \biggl(  \frac{\pi}{3} +\theta \biggr) $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-14T23:30:25.673" Title="Explain this step " Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10252" PostTypeId="2" ParentId="10229" CreationDate="2010-11-14T15:12:18.103" Score="1" ViewCount="0" Body="&lt;p&gt;This is a problem with extra information. Part of solving problems is knowing what to ignore! You are thinking about 4 lines, but to show two lines are parallel, you only really need 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you don't need to use &lt;strong&gt;all&lt;/strong&gt; of the information given to solve this problem. In fact, you can &lt;em&gt;ignore line d&lt;/em&gt; and just focus on how line c interacts with the lines you want to prove are parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next look at the theorem about parallel lines and their &lt;a href=&quot;http://www.mathnstuff.com/math/spoken/here/2class/260/trans.htm&quot; rel=&quot;nofollow&quot;&gt;transverse&lt;/a&gt; again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you should have your proof.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-14T15:12:18.103" />
  <row Id="10253" PostTypeId="2" ParentId="10065" CreationDate="2010-11-14T15:17:41.997" Score="8" ViewCount="0" Body="&lt;p&gt;I completely forgot about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem&quot; rel=&quot;nofollow&quot;&gt;Stolz-Cesàro theorem&lt;/a&gt;, from which we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n\to \infty}=\frac{a_{n+1}-a_{n}}{n+1-n}=\lim_{n\to \infty}=\frac{\frac{1}{a_{n}}}{1}=\lim_{n\to \infty}=\frac{1}{a_{n}}=0$ (the same technique works for $\frac{(a_{n})^2}{n}$)&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-14T15:17:41.997" CommentCount="1" />
  <row Id="10255" PostTypeId="1" AcceptedAnswerId="10259" CreationDate="2010-11-14T15:36:44.623" Score="4" ViewCount="139" Body="&lt;p&gt;Is there a common name for curves, obtained from dragging a point along another curve, similar to how tractrix is obtained by dragging a point along a line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a parametric equation of such curve given the parametric equation of a curve along which the master goes?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-12-29T12:40:48.490" Title="Tractrix-like curves" Tags="&lt;geometry&gt;&lt;plane-curves&gt;&lt;analytic-geometry&gt;" AnswerCount="2" />
  <row Id="10256" PostTypeId="1" AcceptedAnswerId="10303" CreationDate="2010-11-14T15:47:09.413" Score="2" ViewCount="133" Body="&lt;p&gt;Let $A$ be a finite dimensional vector space, $\cdots \rightarrow A_{n+1}\rightarrow^{f_{n+1}} A_n \rightarrow^{f_{n}} A_{n-1}\rightarrow \cdots $ be an inverse system of finite dimensional vector spaces, and let $\phi_i:A\to A_i$ be linear maps inducing a map $\phi:A\to\lim\limits_{\leftarrow} A_i$. If $\phi$ is an injection (resp. surjection), does it follow that $\phi_i$ is an injection (resp. surjection) for some $i\gg 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that it does, since the dimensions $\dim ker \phi_i$ and $\dim coker \phi_i$ should be non-increasing sequences. &lt;/p&gt;&#xA;" OwnerUserId="3332" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T21:16:21.533" LastActivityDate="2010-11-15T00:38:44.463" Title="limits of finite dimensional vector spaces" Tags="&lt;linear-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10257" PostTypeId="1" AcceptedAnswerId="10320" CreationDate="2010-11-14T15:53:40.110" Score="6" ViewCount="211" Body="&lt;p&gt;I have changed the title (replaced &quot;well-behaved&quot; by &quot;càdlàg&quot;), since it seems that &quot;a well-behaved function&quot; might be interpreted as &quot;a function of bounded variation&quot; (rather than &quot;a càdlàg function&quot;, which I actually meant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f:[0,1] \to {\bf R}$ have a following property: $f$ is continuous except at the points $x_{k,n} = \frac{{2k - 1}}{{2^n }}$, $k=1,\ldots,2^{n-1}$, $n=1,2,3,\ldots$, where $\lim _{x \downarrow x_{k,n} } f(x) = f(x_{k,n} )$ but $f(x_{k,n} ) -&#xD;&#xA;\lim _{x \uparrow x_{k,n} } f(x) = a_n  &amp;gt; 0$. Can such a function exist if $a_n &amp;gt; 1/ \log(n)$ for all sufficiently large $n$?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-14T23:37:49.887" LastActivityDate="2010-11-15T01:25:57.267" Title="How much a càdlàg (i.e., right-continuous with left limits) function can jump?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="10258" PostTypeId="2" ParentId="10255" CreationDate="2010-11-14T16:05:50.593" Score="0" ViewCount="0" Body="&lt;p&gt;Some of them are called pursuit curves: &lt;a href=&quot;http://mathworld.wolfram.com/PursuitCurve.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PursuitCurve.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-11-14T16:05:50.593" CommentCount="1" />
  <row Id="10259" PostTypeId="2" ParentId="10255" CreationDate="2010-11-14T16:16:01.183" Score="4" ViewCount="0" Body="&lt;p&gt;Generalized tractrices (tractories) exist, see &lt;a href=&quot;http://www.2dcurves.com/derived/generalized%20tractrix.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://www-cabri.imag.fr/abracadabri/Courbes/Tract/Tract3.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://www.mathcurve.com/courbes2d/tractrice/tractoire.shtml&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. (the last two are in French, but with slightly more detail than the first one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://books.google.com/books?id=93rvAAAAMAAJ&amp;amp;pg=PA127&quot; rel=&quot;nofollow&quot;&gt;old book&lt;/a&gt; ought to be of interest as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://books.google.com/books?id=f9lAHgWV0vIC&amp;amp;pg=PA176&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; describes Euler's treatment of the problem of the tractory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, the problem of finding the generating curve, given the tractory, is a much easier problem (hint: use the tangent vector of a curve) than finding the tractory corresponding to a generating curve.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For those who have difficulty reading French, the third link I mentioned gives the prescription for generating the corresponding tractory from a generating curve with parametric equations $(f(t)\quad g(t))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parametric equations for the tractory of $(f(t)\quad g(t))$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}f(t)\\g(t)\end{pmatrix}-\frac{a}{\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}}\begin{pmatrix}\cos\;\alpha(t)&amp;amp;\sin\;\alpha(t)\\-\sin\;\alpha(t)&amp;amp;\cos\;\alpha(t)\end{pmatrix}\cdot\begin{pmatrix}f^{\prime}(t)\\g^{\prime}(t)\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the function $\alpha(t)$ satisfies the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d\alpha}{\mathrm dt}=\frac{f^{\prime}(t)g^{\prime\prime}(t)-g^{\prime}(t)f^{\prime\prime}(t)}{f^{\prime}(t)^2+g^{\prime}(t)^2}-\frac{\sin\alpha}{a}\sqrt{f^{\prime}(t)^2+g^{\prime}(t)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $a$ is the length of the segment running through the generating curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, here is an animation showing the curve $(3\cos\;t-2\cos^3 t+\cos 2t\quad 2\sin^3 t+\sin 2t)$ and its tractory with segment length 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/AeY1U.gif&quot; alt=&quot;nephroid tractory&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a less fancy version of the last bicycle animation in the third French link.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T12:40:48.490" LastActivityDate="2010-12-29T12:40:48.490" CommentCount="3" />
  <row Id="10260" PostTypeId="2" ParentId="10256" CreationDate="2010-11-14T16:45:38.703" Score="0" ViewCount="0" Body="&lt;p&gt;Like you said, we have $\ker\phi_{i+1}\subseteq \ker\phi_{i}$. Hence the decreasing sequence of kernels either stabilizes at some nontrivial subspace of $A$, in which case $\phi$ is not injective, or it decreases to $0$. But it has to decrease to $0$ after finitely many $i$, so after that, $\phi_i$ is injective.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-11-14T16:45:38.703" />
  <row Id="10261" PostTypeId="1" AcceptedAnswerId="10262" CreationDate="2010-11-14T17:02:56.153" Score="5" ViewCount="313" Body="&lt;p&gt;I feel like finding the inverse of $y=xe^x$ should have an easy answer but can't find it.&lt;/p&gt;&#xA;" OwnerUserId="1958" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T17:44:29.703" LastActivityDate="2010-12-01T17:44:29.703" Title="Inverse of $y=xe^x$" Tags="&lt;special-functions&gt;" AnswerCount="3" />
  <row Id="10262" PostTypeId="2" ParentId="10261" CreationDate="2010-11-14T17:07:45.197" Score="12" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lambert_W_function&lt;/a&gt;, see the section Applications.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-14T17:07:45.197" CommentCount="1" />
  <row Id="10263" PostTypeId="2" ParentId="9642" CreationDate="2010-11-14T17:33:41.537" Score="2" ViewCount="0" Body="&lt;p&gt;What you're looking for is described on pages 247-249 of Robert Vanderbei's &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.111.1824&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Linear Programming: Foundations and Extensions&lt;/a&gt; (2nd edition), although Vanderbei has some of his sign conventions opposite of yours.  Take a close look in particular at Figure 14.4 on page 249 (&quot;Adding a new node to accommodate an arc $(i,j)$ having an upper bound $u_{ij}$ on its flow capacity&quot;).  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-14T17:33:41.537" />
  <row Id="10264" PostTypeId="1" AcceptedAnswerId="10270" CreationDate="2010-11-14T17:35:43.267" Score="9" ViewCount="251" Body="&lt;p&gt;I'm just reviewing for my exam tomorow looking at old exams, unfortunately I don't have solutions.  Here is a question I found : determine if the series converges or diverges.  If it converges find it's limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum\limits_{n=1}^{\infty}\dfrac{\sin(n-\sqrt{n^2+n})}{n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've ruled down possible tests to the limit comparison test, but I feel like I've made a mistake somewhere.&lt;br&gt;&#xA;&lt;strong&gt;divergence test&lt;/strong&gt; - limit is 0 by the squeeze theorem&lt;br&gt;&#xA;&lt;strong&gt;integral test&lt;/strong&gt; - who knows how to solve this&lt;br&gt;&#xA;&lt;strong&gt;comparison test&lt;/strong&gt; - series is not positive&lt;br&gt;&#xA;&lt;strong&gt;ratio root tests&lt;/strong&gt; - on the absolute value of the series, this wouldn't work out&lt;br&gt;&#xA;&lt;strong&gt;alternating series test&lt;/strong&gt; - would not work, the series is not decreasing or alternating  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas what to compare this series here with or where my mistake is on my reasoning above?&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T05:42:57.737" LastActivityDate="2010-11-18T05:42:57.737" Title="Does the series $\sum\limits_{n=1}^{\infty}\frac{\sin(n-\sqrt{n^2+n})}{n}$ converge?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10270" PostTypeId="2" ParentId="10264" CreationDate="2010-11-14T17:59:45.080" Score="17" ViewCount="0" Body="&lt;p&gt;The key here is that $n - \sqrt{n^2 + n}$ converges to $-{1 \over 2}$ as $n$ goes to infinity:&#xA;$$n - \sqrt{n^2 + n}=  (n - \sqrt{n^2 + n}) \times  {n + \sqrt{n^2 + n} \over n + \sqrt{n^2 + n}}$$&#xD;&#xA;$$= {n^2 - (n^2 + n) \over n + \sqrt{n^2 + n}} = -{n \over  n + \sqrt{n^2 + n}}$$&#xD;&#xA;$$= -{1 \over 1 + \sqrt{1 + {1 \over n}}}$$&#xD;&#xA;Take limits as $n$ goes to infinity to get $-{1 \over 2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\sin(n - \sqrt{n^2 + n})$ converges to $\sin(-{1 \over 2})$, and the series diverges similarly to ${1 \over n}$, using the limit comparison test for example.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-14T18:05:30.553" LastActivityDate="2010-11-14T18:05:30.553" CommentCount="2" />
  <row Id="10271" PostTypeId="1" AcceptedAnswerId="10272" CreationDate="2010-11-14T18:03:30.463" Score="0" ViewCount="41" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1jymi.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a little basic, but my math is pretty rusty.  I know the point coordinates of the two blue dots.  I also know the distances from the middle blue dot to the edges of the square.  How can I calculate the coordinates of the red dot?&lt;/p&gt;&#xA;" OwnerUserId="3496" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-14T18:08:11.283" LastActivityDate="2010-11-14T18:10:11.780" Title="Calculate point on hypotenuse of right-angled triangle." Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="10272" PostTypeId="2" ParentId="10271" CreationDate="2010-11-14T18:10:11.780" Score="1" ViewCount="0" Body="&lt;p&gt;The two right angled triangles in the picture are &lt;a href=&quot;http://www.gcseguide.co.uk/similar_triangles.htm&quot; rel=&quot;nofollow&quot;&gt;similar&lt;/a&gt;. Hence the ratio of the corresponding sides are the same. You can use this to get the remaining coordinates.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T18:10:11.780" />
  <row Id="10273" PostTypeId="2" ParentId="10261" CreationDate="2010-11-14T18:20:34.727" Score="8" ViewCount="0" Body="&lt;p&gt;Inversion of $xe^{-x}$ is a problem with combinatorial structure, from which one can also invert $xe^x$ and derive most of the properties at the Wikipedia page.  At least for theoretical understanding it is easier to treat the inverse of $x\exp(-x)$ as the basic object and then translate it into &quot;Lambert W&quot; terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inverse of $f(x) = x \exp(-x)$ is the generating function for rooted trees $\Sigma n^{n-1} X^n/n!$.  That the series represents the inverse is equivalent to a simple combinatorial fact (&quot;removing the root and its edges from a tree, gives a collection of rooted trees&quot;).  The inverse of $xe^x$ is then $W(x)= -f^{-1}(-x)$, which is the same series with minus signs.  The radius of convergence $1/e$ follows from Stirling's approximation to $n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficients of $W(x)^r$ given in the Wikipedia article are, up to some signs, Abel polynomials.  This also follows from the combinatorial interpretation: $(f^{-1}(x))^r$ is the generating function whose $x^n$ coefficient is the number maps from a set of size $n$ to a set of $r$ elements with a rooted tree on each fiber of the map.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-14T18:57:19.797" LastActivityDate="2010-11-14T18:57:19.797" />
  <row Id="10274" PostTypeId="1" CreationDate="2010-11-14T18:30:13.967" Score="0" ViewCount="121" Body="&lt;p&gt;A ring R is a Boolean ring provided that a^2=a for every a in R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3498" LastActivityDate="2010-11-14T20:38:56.230" Title="How to show that every Boolean ring is commutative？" Tags="&lt;ring-theory&gt;" AnswerCount="5" />
  <row Id="10275" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:35:11.537" Score="2" ViewCount="0" Body="&lt;p&gt;Plug $a = x + y$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T18:35:11.537" />
  <row Id="10276" PostTypeId="2" ParentId="10261" CreationDate="2010-11-14T18:37:20.413" Score="-1" ViewCount="0" Body="&lt;p&gt;Asymptotically, $x \approx \log y - \log\log y$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T18:37:20.413" />
  <row Id="10277" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:42:39.553" Score="2" ViewCount="0" Body="&lt;p&gt;Every Boolean ring is of characteristic 2, since&#xA;$a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a$&#xD;&#xA;Now, for any $x,y$ in the ring&#xA;$x+y=(x+y)^2=x^2+xy+yx+y^2=x+y+xy+yx$&#xD;&#xA;So that $xy+yx=0$ and hence $xy=-yx$. But since the ring has characteristic 2, $yx=-yx=xy$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T18:42:39.553" />
  <row Id="10278" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:43:34.597" Score="0" ViewCount="0" Body="&lt;p&gt;As Yuval points out $(x+y)^{2} = x+y$ which implies $x^{2} + y^{2} + x \cdot y + y \cdot x = x+y$. Now from this you have $x \cdot y + y \cdot x =0$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-14T18:43:34.597" CommentCount="1" />
  <row Id="10279" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T18:53:49.673" Score="0" ViewCount="0" Body="&lt;p&gt;We want to show that $xy = yx$ for all $x,y \in R$. We know that $(x+y)^2 = x+y$. So $(x+y)^2 = (x+y)(x+y) = xx+xy+yx+yy = x+xy+yx+y = x+x^2y^2+y^2x^2+y$. This equals $x+(xy)+(yx) + y = x+y$ so that $xy = yx$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-14T18:53:49.673" />
  <row Id="10280" PostTypeId="1" AcceptedAnswerId="10309" CreationDate="2010-11-14T18:54:15.883" Score="0" ViewCount="153" Body="&lt;p&gt;Let $x \in \mathbb{R} \backslash \mathbb{Q}, x&amp;gt;0$ and $q \in \mathbb{N}, q&amp;gt;0$, prove that there is an $r \in \mathbb{N}, r&amp;gt;0$ with:&#xA;     $r \cdot x - \left\lfloor r \cdot x \right\rfloor &amp;lt; \frac{1}{q}$ or $1-(r \cdot x - \left\lfloor r \cdot x \right\rfloor) &amp;lt; \frac{1}{q}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given the hint to divide the sequence $a_r := r \cdot x - \left\lfloor r \cdot x \right\rfloor \in [0,1)$ into q intervals $[0,\frac{1}{q}),[\frac{1}{q},\frac{2}{q}),\ldots,[\frac{q-1}{q},1)$ and use the pigeonhole principle, but I cannot see how this would help to the problem.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-03T04:04:41.297" LastActivityDate="2011-01-03T04:04:41.297" Title="Proving that an equation is solvable, Floor function." Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10281" PostTypeId="2" ParentId="10280" CreationDate="2010-11-14T19:18:29.073" Score="0" ViewCount="0" Body="&lt;p&gt;Let's use the notation $\{x\} = x - \lfloor x \rfloor$. We can think of $\{\cdot\}$ as a function $\mathbb{R} \longrightarrow \mathbb{R}/\mathbb{Z}$, indeed a homomorphism of groups (with addition as the group operator). That means that $\{x + y\} = \{x\} + \{y\}$, or what matters more in your case, $\{x - y\} = \{x\} - \{y\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $\{r_1x\},\{r_2x\}$ are close, what does this say about $\{(r_1-r_2)x\}$?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T19:18:29.073" CommentCount="4" />
  <row Id="10282" PostTypeId="1" AcceptedAnswerId="10374" CreationDate="2010-11-14T19:22:31.127" Score="9" ViewCount="479" Body="&lt;p&gt;As a physics student, I've come across mathematical objects called &lt;strong&gt;tensors&lt;/strong&gt; in several different contexts. Perhaps confusingly, I've also been given both the mathematician's and physicist's definition, which I believe are slightly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently think of them in the following ways, but have a tough time reconciling the different views:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An extension/abstraction of scalars, vectors, and matrices in mathematics.&lt;/li&gt;&#xA;&lt;li&gt;A multi-dimensional array of elements.&lt;/li&gt;&#xA;&lt;li&gt;A mapping between vector spaces that represents a co-ordinate independent transformation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, I'm not even sure how correct these three definitions are. Is there a particularly relevant (rigorous, even) definition of tensors and their uses, that might be suitable for a mathematical physicist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct answers/explanations, as well as links to good introductory articles, would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-11-29T15:55:04.040" Title="An Introduction to Tensors" Tags="&lt;tensors&gt;" AnswerCount="5" CommentCount="11" FavoriteCount="5" />
  <row Id="10283" PostTypeId="1" AcceptedAnswerId="10285" CreationDate="2010-11-14T19:35:01.253" Score="0" ViewCount="51" Body="&lt;p&gt;$10 \%$ of people in a population suffer from high blood pressure. Calculate the probability that the twelfth person tested will be the second one with high blood pressure. So this is a negative binomial distribution. Now $P(X=k) = \binom{k+r-1}{k} (1-p)^{r}p^{k}$. So this is the probability of $k$ successes after $r$ failures. So $P(X=2) = \binom{11}{2}(0.9)^{10}(0.1)^2$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-14T20:15:08.267" Title="Success and Failure " Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10284" PostTypeId="1" CreationDate="2010-11-14T19:41:03.790" Score="0" ViewCount="107" Body="&lt;p&gt;Say we have a set of 12 numbers - (0,1,2,0,3,0,4,0,5,6,0).&#xA;We go thru this set one number at a time, and choose a number with a probability 1/3; we sample. What is the probability we end up with 3 0's out of the 4 numbers we will likely choose(on average)? &#xA;Is there a closed-form solution for this?&lt;/p&gt;&#xA;" OwnerUserId="3501" LastActivityDate="2010-11-14T21:18:32.603" Title="Probability Question" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="10" />
  <row Id="10285" PostTypeId="2" ParentId="10283" CreationDate="2010-11-14T19:45:26.750" Score="1" ViewCount="0" Body="&lt;p&gt;The 12th person tested must be one with high blood pressure. Thus only one of the first eleven can have high blood pressure. So the total probability is the probability that exactly one of the first eleven has high blood pressure multiplied by the probability that the 12th person has high blood pressure. Hence $$P = \bigg({11 \choose 1}(0.9)^{10}(0.1) \bigg) \cdot (0.1) = 11(0.9)^{10} (0.1)^2$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-14T19:45:26.750" />
  <row Id="10286" PostTypeId="2" ParentId="10282" CreationDate="2010-11-14T19:55:14.113" Score="1" ViewCount="0" Body="&lt;p&gt;The most general notion I know is that tensor product of modules. You can read about this here&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product_of_modules&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tensor_product_of_modules&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since vector spaces are modules, this definition specializes to vector spaces. The tensor product of elements in these vector spaces that one usually sees in engineering and physics texts (frequently matrices) is basically an element in the tensor product of the corresponding vector spaces.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-14T20:37:40.520" LastActivityDate="2010-11-14T20:37:40.520" CommentCount="4" />
  <row Id="10287" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T19:56:00.197" Score="3" ViewCount="0" Body="&lt;p&gt;If I understand the problem, you have a list of 12 objects, 5 of which are special (the 0s).  (I use the word &quot;list&quot; instead of &quot;set&quot; since the latter implies there is no repetition among the objects).  Each of the 12 objects will be independently considered, and &quot;chosen&quot; with probability 1/3.  You want to know the probability of 3 of the special objects being chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, the 7 non-special objects are irrelevant.  You just want to know the probability that exactly 3 out of 5 independent events occur, when each one occurs with probability 1/3.  This is the probability of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;binomial random variable&lt;/a&gt; with $n=5$ and $p=1/3$ taking on the value $3$.  The linked Wikipedia page has a formula for computing this.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-14T19:56:00.197" CommentCount="1" />
  <row Id="10288" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T19:56:30.130" Score="-1" ViewCount="0" Body="&lt;p&gt;The probability that you want is $(1/3)^4(2/3)^8 N$, where $N$ is the number of subsets of length $4$ containing exactly $3$ zeroes.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-14T19:56:30.130" CommentCount="2" />
  <row Id="10289" PostTypeId="2" ParentId="2981" CreationDate="2010-11-14T20:12:40.690" Score="0" ViewCount="0" Body="&lt;p&gt;There is no such thing as a &quot;countable closure of this process&quot;, as there is no a priori definition of the set of functions envisaged here, let alone a topology on this set. All we can speak of is the ${\it union}$ of the larger and larger sets of functions constructed in the process.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-14T20:12:40.690" CommentCount="1" />
  <row Id="10290" PostTypeId="2" ParentId="10283" CreationDate="2010-11-14T20:15:08.267" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, you can view this as a negative binomial distribution, but I think you are misunderstanding what the negative binomial represents. A negative binomial  with parameters $r$ and $p$  is the trial number which produces the $r$th success, where each success has probability $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you have a negative binomial random variable $X$ with parameters $r=2$ and $p=0.1$, and you should be looking for the probability that the 2nd success occurs on 12th trial, i.e $P\{X = 12\}$. I also think you have the formula for the probability mass function slightly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it the probability for the 12th trial to be a success is $0.1$. For the other 11 trials, we need to have exactly 1 out of 11 to be a success, so $(0.1)(0.9)^{10}$ for each case with $11 \choose 1$ cases. The formula should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=k)={k-1 \choose r-1}(1−p)^{k-r}p^r$ Use that and you should end up with what Brandon and kahen came up with.&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-14T20:15:08.267" />
  <row Id="10291" PostTypeId="2" ParentId="10282" CreationDate="2010-11-14T20:16:50.827" Score="8" ViewCount="0" Body="&lt;p&gt;In mathematics, tensors are one of the first objects encountered which cannot be fully understood without their accompanying universal mapping property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before talking about tensors, one needs to talk about the tensor product of vector spaces. You are probably already familiar with the direct sum of vector spaces. This is an addition operation on spaces. The tensor product provides a multiplication operation on vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key feature of the tensor product is that it replaces &lt;em&gt;bilinear maps&lt;/em&gt; on a cartesian product of vector spaces with &lt;em&gt;linear maps&lt;/em&gt; on the tensor product of the two spaces. In essence, if $V,W$ are vector spaces, there is a bijective correspondence between the set of bilinear maps on $V\times W$ (to &lt;em&gt;any&lt;/em&gt; target space) and the set of &lt;em&gt;linear&lt;/em&gt; maps on $V\otimes W$ (the tensor product of $V$ and $W$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be phrased in terms of a universal mapping property. Given vector spaces $V,W$, a tensor product $V\otimes W$ of $V$ and $W$ is a space together with a map $\otimes : V\times W \rightarrow V\otimes W$ such that for any vector space $X$ and any &lt;em&gt;bilinear&lt;/em&gt; map $f : V\times W \rightarrow X$ there exists a unique &lt;em&gt;linear&lt;/em&gt; map $\tilde{f} : V\otimes W \rightarrow X$ such that $f = \tilde{f}\circ \otimes$. In other words, every bilinear map on the cartesian product factors uniquely through the tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown using a basic argument that the tensor product is unique up to isomorphism, so you can speak of &quot;the&quot; tensor product of two spaces rather than &quot;a&quot; tensor product, as I did in the previous paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tensor is just an element of a tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must show that such a tensor product &lt;em&gt;exists&lt;/em&gt;. The standard construction is to take the free vector space over $V\times W$ and introduce various bilinearity relations. See my link at the bottom for an article that does this explicitly. In my experience, however, the key is to be able to use the above mapping property; the particular construction doesn't matter much in the long run. The map $\otimes : V\times W \rightarrow V\otimes W$ sends the pair $(v,w) \in V\times W$ to $v\otimes w \in V\otimes W$. The image of $\otimes$ is the space of so-called elementary tensors, but a general element of $V\otimes W$ is &lt;em&gt;not&lt;/em&gt; an elementary tensor but rather a linear combination of elementary tensors. (In fact, due to bilinearity, it is enough to say that a general tensor is a sum of elementary tensors with the coefficients all being 1.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most generic reason why tensors are useful is that the tensor product is a machine for replacing bilinear maps with linear ones. In much of mathematics and physics, one seeks to find linear approximations to things; tensors can be seen as one tool for this, although exactly how they accomplish it is less clear than many other tools in the same vein. Here are some more specific reasons why they are useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For finite-dimensional spaces $V,W$, the tensor product $V^*\otimes W$ is isomorphic to the space of homomorphisms $\text{Hom}(V,W)$. So in other words every linear map $V \rightarrow W$ has a tensor expansion, i.e., a representation as a tensor in $V^* \otimes W$. For instance, if $\{v_i\}$ is a basis of $V$ and $\{x_i\}$ is the dual basis of $V^*$, then $\sum x_i \otimes v_i \in V^* \otimes V$ is a tensor representation of the identity map on $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensor products tend to appear in a lot of unexpected places. For instance, in analyzing the linear representations of a finite group, once the irreducible representations are known it can be of benefit to construct also a &quot;tensor product table&quot; which decomposes the tensor products of all pairs of irreducible representations as direct sums of irreducible representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, one often talks about a rank $n$ tensor being an assembly of numbers which transform in a certain way under change of coordinates. What one is really describing here is all the different coordinate representations of an abstract tensor in a tensor power $V^{\otimes n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one takes the direct sum of all tensor powers of a vector space $V$, one obtains the tensor algebra over $V$. In other words, the tensor algebra is the construction $k\oplus V\oplus (V\otimes V) \oplus (V\otimes V\otimes V) \oplus \dots$, where $k$ is the base field.  The tensor algebra is naturally graded, and it admits several extremely useful quotient algebras, including the well-known exterior algebra of $V$. The exterior algebra provides the natural machinery for differential forms in differential geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of the exterior algebra in practice. Suppose one wishes to classify all nonabelian two-dimensional Lie algebras $\mathfrak{g}$. The Lie bracket $[\cdot,\cdot]$ is antisymmetric and bilinear, so the machinery of tensor products turns it into a &lt;em&gt;linear&lt;/em&gt; map $\bigwedge^2 V \rightarrow V$, where $V$ is the underlying vector space of the algebra. Now $\bigwedge^2 V$ is one-dimensional and since the algebra is nonabelian the Lie bracket is not everywhere zero; hence as a linear map the Lie bracket has a one-dimensional image. Then one can choose a basis $\{X,Y\}$ of $V$ such that $[X,Y] = X$, and we conclude that there is essentially only one nonabelian Lie algebra structure on a two-dimensional vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fantastic reference on tensor products of modules was written by Keith Conrad: &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-11-29T15:55:04.040" LastActivityDate="2010-11-29T15:55:04.040" CommentCount="2" />
  <row Id="10292" PostTypeId="1" AcceptedAnswerId="10294" CreationDate="2010-11-14T20:22:16.703" Score="1" ViewCount="82" Body="&lt;p&gt;Hi, &#xA;so studying for my midterm on Tuesday (intro to abstract algebra). The topics on the exam are Division Algorithm, Divisibility, Prime Numbers, FTA, Congruency, Congruent Classes and very brief introduction to rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading a few theorems about Congruency and have a couple of questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what a &quot;congruent class&quot; is. My notes say &quot;the congruence class of a modulo n&quot; is a set: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left\{ \text{all } b \in \mathbb{Z} | b \equiv a \pmod{n} \right\}  $ which is also saying&#xA;$ \left\{ \text{all } a + kn \in \mathbb{Z}  | k \in \mathbb{Z} \right\}  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;okay so got that. I just wrote it for some people who might need a refreshed (it is a 3rd undergrad course after all). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in my notes our professor has a following example:&#xA;$\left[ 60 \right]_{17} = \left[ 43\right]_{17}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;so the way I figured this out is that to check if they are equivalent, we subtract 60-43 and see if that is a multiple of n = 17. Is this how you can check if they are equal classes?&lt;/strong&gt;  If not, is there a better way to do so? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A certain theorem states: Let $n \in \mathbb{Z}_+; a, b \in \mathbb{Z}$ and $gcd(a,n) = d$ then $[a]x=[b]$ has exactly $d$ solutions. &lt;strong&gt;My question here is that is x a congruence class or a random integer? What is x and how do I solve for it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Is it true that if we are in $\mathbb{Z}_{12}$ then $[7]x=[11]$ can be rewritten as $ 7x \equiv 11 \pmod{12}$? If so, &lt;strong&gt;would finding the solution be similar to &lt;a href=&quot;http://math.stackexchange.com/questions/2991/not-understanding-simple-modulus-congruency&quot;&gt;solution in this question&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou. I am just very confused about congurency and stuff. I understand the theorems but I am hoping someone would give me an &quot;easy&quot; explanation of what is going on. I still don't know the difference between circle plus and regular plus except that circle plus has to satisfy certain axioms. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T21:58:07.597" LastActivityDate="2010-11-14T23:17:41.867" Title="Congruency and Congruent Classes" Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10293" PostTypeId="2" ParentId="10177" CreationDate="2010-11-14T20:23:22.630" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a nice solution I just thought of, which may in fact be logically equivalent to Jim's (if it's even correct!).  I welcome comments addressing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compactify $\mathbb{R}^{n+1}$ to $S^{n+1}$, so we consider $M \subset S^{n+1}$.  By Alexander duality, $$\tilde{H}_0(S^{n+1} \backslash M ; \mathbb{Z}/2) \cong \tilde{H}^{(n+1)-0-1}(M;\mathbb{Z}/2) = \tilde{H}^n(M;\mathbb{Z}/2)=\mathbb{Z}/2,$$ so $H_0(S^{n+1}\backslash M;\mathbb{Z}/2)=\mathbb{Z}/2 \oplus \mathbb{Z}/2$.  Hence $M$ separates $S^{n+1}$.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-14T20:23:22.630" CommentCount="2" />
  <row Id="10294" PostTypeId="2" ParentId="10292" CreationDate="2010-11-14T20:32:48.750" Score="1" ViewCount="0" Body="&lt;p&gt;There is a much more general definition of congruence classes, but I shall restrict to the one that is sufficient for your course. Given any integer $n$ the only possible remainders that we can get when we divide an integer $a$ by $n$ are $0,1,...,n-1$. &#xA;Each set of integers which leave the same remainder on division by $n$ form what is called as a congruence class modulo $n$. All such congruence classes are mutually disjoint since a number can leave only one remainder on division by $n$. Their union is the set of all integers. Each integer in any given congruence class is said to be a representative of the class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check whether two integers are in the same class, we check their difference and see if it is divisible by $n$ (since the remainders of these integers cancel out when we divide by $n$). So your approach to your first question is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, $x$ is indeed a congruence class, since otherwise the equation does not make sense. We can define operations on the congruence classes by the correponding operations on their representatives. Its easy to check these are well defined.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-14T20:32:48.750" CommentCount="3" />
  <row Id="10295" PostTypeId="2" ParentId="10274" CreationDate="2010-11-14T20:38:56.230" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ \ A = X+Y\ \ \Rightarrow\ \ X\ Y = - Y\ X\:.\ $ But $\rm -1 = 1\ $ via $\rm\ A = -1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-14T20:38:56.230" />
  <row Id="10296" PostTypeId="2" ParentId="10042" CreationDate="2010-11-14T20:40:19.490" Score="2" ViewCount="0" Body="&lt;p&gt;See the equivalent question on mathoverflow: &lt;a href=&quot;http://mathoverflow.net/questions/8829/what-manifolds-are-bounded-by-rpodd&quot; rel=&quot;nofollow&quot;&gt;What manifolds are bounded by RP^odd?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since it seems there is a good reason to have the answer recorded as such (and, borrowing from the suggestion &lt;a href=&quot;http://meta.math.stackexchange.com/questions/1148/what-should-one-do-when-ones-question-has-been-answered-in-the-comments&quot;&gt;here&lt;/a&gt;), I'm moving my comment here.  Since all I'm doing it linking to another place, I don't want to gain reputation for this, so I'm making it community wiki.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in an effort to personally gain &lt;i&gt;something &lt;/i&gt; from this, I'll provide a link to a similar &lt;a href=&quot;http://mathoverflow.net/questions/14698/what-manifold-has-mathbbhpodd-as-a-boundary&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt; I asked on MO which still hasn't been answered.  The question is: What manifold has $\mathbb{H}P^{odd}$ as a boundary?  Incidentally, the case of $\mathbb{C}P^{odd}$ is covered in my question.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-14T20:47:31.820" LastActivityDate="2010-11-14T20:47:31.820" CommunityOwnedDate="2010-11-14T20:40:19.490" CommentCount="2" />
  <row Id="10297" PostTypeId="1" AcceptedAnswerId="10305" CreationDate="2010-11-14T20:55:00.653" Score="1" ViewCount="109" Body="&lt;p&gt;Suppose Jill and Jane are individually painting separate rooms. Each of the rooms has $4$ walls. The time taken to paint each wall is exponentially distributed with mean $1$ hour. What is the probability that Jill will finish painting $3$ walls before Jane finishes $2$ walls? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to look somewhat similar to a negative binomial distribution. Is this the correct distribution to use?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-15T20:31:17.400" Title="Painting Walls and Probability" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10298" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T21:00:23.987" Score="2" ViewCount="0" Body="&lt;p&gt;If I understand the problem, you have a sequence of independent and identically distributed random variables $X_1,X_2,\ldots,X_n$ (in your example $n=11$, as noted by Yuval), such that ${\rm P}(X_i = j) = p_j$ (in your example $j \in \lbrace 0,1,\ldots,6 \rbrace $); you sample each $X_i$ with probability $p$ (so in average you pick $pn$ numbers; in your example $p=1/3$), and wants to find the probability $P(n,j,m)$ of obtaining $0 \leq m \leq n$ occurrences of the number $j$. Then, this probability can be computed, using the law of total probability, as follows:&#xA;$$&#xD;&#xA;P(n,j,m) = \sum\limits_{k = m}^n {{n \choose k}p^k (1 - p)^{n - k} {k \choose m} p_j^m (1 - p_j )^{k - m} }. &#xD;&#xA;$$&#xD;&#xA;The term ${n \choose k}p^k (1 - p)^{n - k}$ corresponds to sampling $k$ $X_i$'s out of $n$, and the term&#xA;${k \choose m} p_j^m (1 - p_j )^{k - m} $ corresponds to obtaining $m$ occurrences of the number $j$ given that $k (\geq m)$ $X_i$'s were sampled. The rest is left for you.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-14T21:00:23.987" />
  <row Id="10299" PostTypeId="2" ParentId="10284" CreationDate="2010-11-14T21:18:32.603" Score="3" ViewCount="0" Body="&lt;p&gt;In your original question it sounded like you wanted the conditional probability $P\{\text{we obtain 3 0&amp;#39;s}|\text{4 events are sampled}\}$ which is the number of size 4 subsets that contain 3 0's divided by the total number of size 4 subsets which would be, if $k$ is the number of 0's total, $$\frac{{k \choose 3}{12-k \choose 1}}{{12 \choose 4}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it sounds like what you really want is the probability that any single sample is either a high frequency event or a low frequency event. Say you have an event that occurs with a high frequency, but that each trial is independent, we can model this by giving it a probability, say $p_h=\frac{n}{100}$ where n is the average # that occur in any 100 trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say you sample each trial independently with probability $p_s$. Then the probability that any single sample is one of the high frequency events is the conditional probability $$P\{\text{event is the high frequency event}| \text{event is sampled}\} = p_h $$&#xD;&#xA;This is because, presumably, your sampling and your event occurrences are independent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also sounds like you would like the expected value of the number of high frequency events sampled given a certain number of total events, $n$. Consider any single event. In order for it to be a &quot;success&quot; it has to be both a high frequency event and be sampled. $$P\{\text{an event is high frequency and is sampled}\} = p_h p_s$$&#xD;&#xA;The probability to have $k$ high frequency events sampled when you have $n$ total events is a binomial distribution with parameter $p_hp_s$.&#xA;$$P\{k \text{ of our sampled events are high frequency events}\} = {n \choose k} (p_sp_h)^k(1-p_sp_h)^{n-k}$$&#xD;&#xA;The expected number of sampled events that are high frequency is then just $np_sp_h$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything works the same with a low frequency event, just use an appropriate $p_l$ instead of $p_h$.&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-14T21:18:32.603" />
  <row Id="10300" PostTypeId="2" ParentId="10282" CreationDate="2010-11-14T21:19:25.087" Score="12" ViewCount="0" Body="&lt;p&gt;Once you understand what a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_product&quot; rel=&quot;nofollow&quot;&gt;tensor product&lt;/a&gt; is and what a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dual_space&quot; rel=&quot;nofollow&quot;&gt;dual space&lt;/a&gt; is, then a tensor of type $(n, m)$ is an element of $V^{\ast \otimes m} \otimes V^{\otimes n}$ where $V$ is some vector space.  This is the same thing as a multilinear map $V^m \to V^{\otimes n}$ or, if you don't like the asymmetry, a multilinear map $V^{\ast n} \times V^{m} \to F$ (where $F$ is the underlying field).  Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A tensor of type $(0, 0)$ is a scalar.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(1, 0)$ is a vector.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(0, 1)$ is a covector.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(1, 1)$ is a linear transformation.&lt;/li&gt;&#xA;&lt;li&gt;A tensor of type $(0, 2)$ is a bilinear form, for example an inner product.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you pick a basis of $V$, you can write tensors in terms of the natural basis on $V^{\ast \otimes n} \otimes V^{\otimes m}$ coming from taking products of the basis on $V$ with the corresponding dual basis on $V^{\ast}$.  This is where the &quot;multidimensional array&quot; definition of a tensor comes from, since this is the natural generalization of writing a matrix as a square array (which is equivalent to writing an element of $V^{\ast} \otimes V$ in terms of the basis $e_i^{\ast} \otimes e_j$ where $\{ e_i \}$ is a basis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a physicist says &quot;tensor,&quot; sometimes they mean a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_field&quot; rel=&quot;nofollow&quot;&gt;tensor field&lt;/a&gt;.  This is a &quot;globalization&quot; of the above definition: it is a compatible set of choices, for each tangent space $V = T_p(M)$ of a smooth manifold $M$, of a tensor of type $(n, m)$ as defined above.  Note that $V^{\ast}$ is the cotangent space.  Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A tensor field of type $(0, 0)$ is a smooth function.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(1, 0)$ is a vector field.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(0, 1)$ is a differential $1$-form.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(1, 1)$ is a morphism of vector fields.&lt;/li&gt;&#xA;&lt;li&gt;A tensor field of type $(0, 2)$ which is symmetric and nondegenerate is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_tensor&quot; rel=&quot;nofollow&quot;&gt;metric tensor&lt;/a&gt;.  If it is also positive-definite, it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemannian_manifold&quot; rel=&quot;nofollow&quot;&gt;Riemannian metric&lt;/a&gt;.  If it has signature $(1, n-1)$, it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold#Lorentzian_manifold&quot; rel=&quot;nofollow&quot;&gt;Lorentzian metric&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-15T11:24:37.547" LastActivityDate="2010-11-15T11:24:37.547" CommentCount="8" />
  <row Id="10301" PostTypeId="2" ParentId="9791" CreationDate="2010-11-14T21:24:13.267" Score="4" ViewCount="0" Body="&lt;p&gt;You can find the first &lt;em&gt;n&lt;/em&gt; primes in time $n\log n/\log\log n$ with the Atkin-Bernstein sieve.  This is better than your naive $n\log^5n$ or so algorithm (or drop the exponent by 1 using pretesting with M-R), but still superlinear.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T18:07:32.193" LastActivityDate="2010-11-15T18:07:32.193" />
  <row Id="10302" PostTypeId="2" ParentId="10297" CreationDate="2010-11-14T21:32:00.060" Score="1" ViewCount="0" Body="&lt;p&gt;Let $T$ denote the period of time until Jill finishes painting $3$ walls. Then, $T$ has gamma density $f_T (t) = t^2 {\rm e}^{-t}/2$, $t &amp;gt; 0$. Viewing the number of walls painted by Jane up to time $t$ as a Poisson process (with unit rate), the question amounts to calculate&#xA;$$&#xD;&#xA;\int_0^\infty  {{\rm P}(N_t  &amp;lt; 2|T = t)f_T (t)\,{\rm d}t}&#xD;&#xA;$$&#xD;&#xA;(by virtue of the law of total probability, conditioning on $T$), where $\lbrace N_t : t \geq 0 \rbrace$ is a unit rate Poisson process. The required probability then follows to be $0.3125 = 5/16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete generalization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose Jill and Jane are individually painting separate rooms. Each of the rooms has $n \ge \max \lbrace m_1 , m_2 \rbrace$ walls. The time it takes Jill and Jane to paint each wall is exponentially distributed with mean $1/\lambda_1$ and $1/\lambda_2$ hour, respectively. What is the probability that Jill will finish painting $m_1$ walls before Jane finishes $m_2$ walls? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is&#xA;$$&#xD;&#xA;\bigg(\frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}\bigg)^{m_1 } \sum\limits_{k = 0}^{m_2  - 1} {{k + m_1  - 1 \choose k} \bigg(\frac{{\lambda _2 }}{{\lambda _1  + \lambda _2 }}\bigg)^k }.&#xD;&#xA;$$&#xD;&#xA;This can be proved easily either by using Mike's method, noting that ${\rm P}(X_1  &amp;lt; X_2 ) = \frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}$ for independent exponential rv's $X_1$ and $X_2$ with respective means $1/\lambda_1$ and $1/\lambda_2$, or by using my method, noting that the time $T$ it takes Jill to paint $m_1$ walls has gamma density $f_T (t) = \frac{{\lambda _1^{m_1 } {\rm e}^{ - \lambda _1 t} t^{m_1  - 1} }}{{(m_1  - 1)!}}$, $t &amp;gt; 0$. (Letting $\lambda_1 = \lambda_2 = 1$, $m_1=3$, and $m_2=2$, this formula gives $5/16$, the same as for the original problem.)&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-15T20:31:17.400" LastActivityDate="2010-11-15T20:31:17.400" />
  <row Id="10303" PostTypeId="2" ParentId="10256" CreationDate="2010-11-14T21:53:27.463" Score="3" ViewCount="0" Body="&lt;p&gt;If $\mathbf{v}\in A$ is in the kernel of $\phi_i$, then for all $j\leq i$ you have $\phi_j(\mathbf{v}) = f_{ij}\phi_j(\mathbf{v}) = f_{ij}(\mathbf{0}) = \mathbf{0}$ (where $f_{ij}\colon A_i\to A_j$ is the map obtained by composing the $f_k$). Therefore, $\mathrm{ker}(\phi_i)\subseteq \mathrm{ker}(\phi_j)$ for all $i\geq j$. Hence you have a descending sequence&#xA;\begin{equation*}&#xA;\cdots \subseteq \mathrm{ker}(\phi_{i+1})\subseteq \mathrm{ker}(\phi_i)\subseteq \cdots \subseteq \mathrm{ker}(\phi_0).&#xA;\end{equation*}&#xA;Since $\dim(A)\lt\infty$, the sequence must stabilize, so there exists $n_0$ such that for all $n\geq n_0$, $\mathrm{ker}(\phi_n)=\mathrm{ker}(\phi_{n_0})$. Let $\mathbf{N}$ be that subspace of $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{v}\in\mathbf{N}$, then $\phi_i(\mathbf{v})=\mathbf{0}$ for all $i$, hence $\phi(\mathbf{v})=\mathbf{0}$, so $\mathbf{N}\subseteq \mathrm{ker}(\phi)$. Conversely, anything in $\mathrm{ker}(\phi)$ lies in the intersection of all kernels of the $\phi_i$. Thus, $\mathbf{N}=\mathrm{ker}(\phi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, $\phi$ is one to one if and only if the $\phi_i$ are eventually one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For surjection, you are going to be out of luck. Consider the system with $A_i = \mathbf{F}^2$ for all $i$, $f_i\colon\mathbf{F}^2\to\mathbf{F}^2$ given by $f_i(a,b)=(a,0)$. I claim that the inverse limit of this system is $\mathbf{V}=\mathbf{F}$ with structure maps $\mathfrak{f}_i\colon\mathbf{V}\to\mathbf{F}^2$ given by $\mathfrak{f}(a) = (a,0)$. First, note that this is a consistent system of maps from $\mathbf{V}$ to the $A_i$. Next, suppose that $\mathbf{W}$ is any vector space and that you have maps $g_i\colon\mathbf{W}\to\mathbf{F}^2$ for all $i$, with the property that for all $i\geq j$, $g_j(\mathbf{w}) = f_{ij}(g_i(\mathbf{w}))$. If $g_i(\mathbf{w}) = (\mathbf{w}_{1i},\mathbf{w}_{2i})$, then this means that $g_j(\mathbf{w}) = (\mathbf{w}_{1i}, 0)$. In particular, for all $j$ we have that $\mathrm{Im}(g_j)$ is contained in the subspace $\{(x,0)\mid x\in\mathbf{F}\}$ of $A_j$, and since the $f_i$ are isomorphisms when restricted to those subspaces, then it follows that for each $\mathbf{w}\in\mathbf{W}$ there is an $a_{\mathbf{w}}\in\mathbf{F}$ such that $g_j(\mathbf{w}) = (a_{\mathbf{w}},0)$ for all $j$. Hence, we have a map $\mathfrak{g}\colon\mathbf{W}\to \mathbf{V}$ with $\mathfrak{f}_j\circ\mathfrak{g}=g_j$ for all $j$, namely $\mathfrak{g}(\mathbf{w}) = a_{\mathbf{w}}$, and the map is clearly unique. Thus, $(\mathbf{V},\mathfrak{f}_i)$ is the inverse limit of the system. Then $\mathbf{V}$ itself with the structure maps gives a counterexample: the induced map to the inverse limit is onto (in fact, it is the identity), but the structure maps are never onto, so it is false that $\phi_i$ is onto for sufficiently large $i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; However, if your structure maps $f_i$ are &lt;em&gt;eventually surjective&lt;/em&gt;, then your conclusion will follow. To see this, assume that $f_i$ is surjective for all $i\geq N$; I claim that the inverse limit maps $\mathfrak{f}_j\colon\lim\limits_{\leftarrow}A_i\to A_j$ are surjective for all $j\geq N$. Fix $j\geq N$, and let $\mathbf{v}\in A_j$. Let $\mathbf{v}_{j+1}\in A_{j+1}$ be any preimage of $\mathbf{v}$, and inductively define $\mathbf{v}_{j+k+1}$ to be any preimage of $\mathbf{v}_{j+k}$. Then define maps $g_i\colon\mathbf{F}\to A_i$ by letting $g_i(1) = \mathbf{v}_{j+m}$ if $i=j+m\geq N$, and $g_i(1)=f_{ji}(\mathbf{v})$ if $i\lt N$. This induces a map $\mathfrak{g}\colon\mathbf{F}\to\lim\limits_{\leftarrow}A_i$, and $g_i = \mathfrak{f}_i\circ\mathfrak{g}$. In particular, $\mathbf{v}\in\mathrm{Im}(\mathfrak{f}_j)$. Thus, $\mathfrak{f}_j$ is onto for all $j\geq N$. Then, if you have $\phi\colon A\to A_i$ as in your statement surjective, then $\phi_j = \mathfrak{f}_j\circ\phi\colon A\to A_j$ is a composition of surjective functions for alL $j\geq N$, so the $\phi_j$ are eventually surjective, as desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; To see how I came up with the example, notice that if you let $B_{i-1} = \mathrm{Im}(f_i)$, then you get an inverse system with the $B_i$ instead of the $A_i$; the inverse limit of this system is isomorphic to the inverse limit of the original system (try to prove this). This is because a consistent system of homomorphism $\phi_i\colon A\to A_i$ has to have $\mathrm{Im}(\phi_i)\subseteq B_i$. So what really matters is what is going on in the images of the structure maps $f_i$, hence surjectivity into the limit depends only on how your maps $f_i$ relate to the $B_i$, not how they relate to the $A_i$. That is why if the maps $f_i$ are eventually surjective you do get the implication.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T00:38:44.463" LastActivityDate="2010-11-15T00:38:44.463" />
  <row Id="10304" PostTypeId="1" AcceptedAnswerId="10306" CreationDate="2010-11-14T21:53:34.850" Score="2" ViewCount="73" Body="&lt;p&gt;Suppose we have an infinite sequence of intervals $[a_n,b_n]$, where for all n, $a_n&amp;lt;b_n$. Then is there some $x\in \mathbb{R}$ such that for each n there is some $k\in\mathbb{Z}$ such that $kx\in [a_n,b_n]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems unlikely, but I'm not sure where to begin with the proof. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastActivityDate="2010-11-15T13:44:39.623" Title="For a sequence of intervals can we find some x such that each interval contains a multiple of x?" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="10305" PostTypeId="2" ParentId="10297" CreationDate="2010-11-14T21:56:15.767" Score="3" ViewCount="0" Body="&lt;p&gt;Here's another way to get the answer of $\frac{5}{16}$.  The number of possible ways for Jill to finish three walls before Jane finishes two is small enough to just list.  They are (with $X$ denoting Jill finishing a wall and $Y$ denoting Jane finishing a wall)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;XXX     &#xA;YXXX    &#xA;XYXX    &#xA;XXYX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Because of the memorylessness of the exponential distribution, starting from any point in time the probability that the next wall will be finished by Jill is $\frac{1}{2}$.  Thus the four listed events have probabilities $\frac{1}{8}$, $\frac{1}{16}$, $\frac{1}{16}$, and $\frac{1}{16}$, respectively.  So the probability that Jill finishes three before Jane finishes two is $\frac{5}{16}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Update: This answer assumes that you're looking for the probability that Jill finishes &lt;em&gt;at least&lt;/em&gt; 3 walls before Jane finishes 2.  If you want the probability that Jill finishes &lt;em&gt;exactly&lt;/em&gt; 3 walls before Jane finishes 2, then Yuval Filmus is correct that you want $P(X = 3)$, where $X$ is a negative binomial random variable with success probability $p = \frac{1}{2}$ and number of failures $r = 2$.  Here, $P(X = 3) = \frac{1}{8}$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-15T05:12:23.920" LastActivityDate="2010-11-15T05:12:23.920" CommentCount="4" />
  <row Id="10306" PostTypeId="2" ParentId="10304" CreationDate="2010-11-14T22:05:15.383" Score="9" ViewCount="0" Body="&lt;p&gt;Sketch of a proof: Take $a_n = 2^{-n}$ and $b_n = 2^{-n} + 2^{-n-2}$. Suppose such $x$ exists, it must be non-zero since none of the intervals intersect the origin. WLOG assume it is positive. Then take $N$ sufficiently large such that $2^{-N + 1} &amp;lt; x$. Then there cannot be any integer $k$ such that $a_N &amp;lt; k x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; Since the OP asked about sequences of intervals bounded away from 0. Clearly there exists sequences such that the desired property is true: just take $[a_n,b_n]$ to be intervals containing $n$. But there can also be counter examples. Let $2&amp;gt; C_n &amp;gt; 1$ be a decreasing sequence of numbers, and $\delta_n &amp;lt; C_n - C_{n+1}$ a decreasing sequence of numbers tending to 0. $C_n$ will be prescribed at the end of the proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the intervals $a_n = C_n - \delta_n$, $b_n = C_n$. By construction the intervals do not intersect. Let $m$ be an integer. If there were to exist some $x$ such that there exists $k,k&amp;#39;\in \mathbb{Z}$ such that $kx \in [a_m,b_m]$ and $k&amp;#39;x \in [a_{m+1},b_{m+1}]$, this will require &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x \in [a_m / k, b_m/k] \cap [a_{m+1}/k&amp;#39;, b_{m+1}/k&amp;#39;] \subset [a_{m+1}/k, b_m/k] \cap [a_{m+1}/k&amp;#39;, b_m/k&amp;#39;] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are led to consider intersections of annuli with rescalings of itself. Now, if $\frac{b_m}{a_{m+1}} - 1 &amp;lt; 1 / N$, then for any $k,k&amp;#39; &amp;lt; N$, the intersection $[a_{m+1}/k, b_m/k] \cap [a_{m+1}/k&amp;#39;, b_m/k&amp;#39;] = \emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $C_n = \frac{n+1}{n}$. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b_m - a_{m+1} &amp;lt; b_m - b_{m+2} = \frac1m \frac2{m+2} &amp;lt; \frac1m b_{m+1} &amp;lt; \frac1m a_{m+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this means that for $x$ to have multiples belonging to both $[a_m,b_m]$ and $[a_{m+1},b_{m+1}]$, said multiplication factor needs to be at least $m$ (actually at least $m^2$, but $m$ suffices for the proof). But since $1 &amp;lt; a_m,b_m &amp;lt; 2$, this implies that $x &amp;lt; 2/m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $m\to \infty$, we see that $x$ must equal to 0, which is absurd. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-15T13:44:39.623" LastActivityDate="2010-11-15T13:44:39.623" CommentCount="2" />
  <row Id="10307" PostTypeId="2" ParentId="10292" CreationDate="2010-11-14T22:36:50.577" Score="1" ViewCount="0" Body="&lt;p&gt;Okay, let's start from the beginning. You define the congruence class of $a$ as:&#xA;\begin{equation*}&#xA;[a]_n = \{ b\in\mathbb{Z}\mid a\equiv a\pmod{n}\}.&#xA;\end{equation*}&#xA;Now, what does $a\equiv b\pmod{n}$ mean? It means that $n$ divides $a-b$, or equivalently, that $a$ and $b$ leave the same remainder when you divide them by $n$. Considering the last of the conditions, notice that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Since each number leaves the same remainder as itself when divided by $n$, $a\equiv a \pmod{n}$ for all $a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $a$ leaves the same remainder as $b$ when divided by $n$, then $b$ leaves the same remainder as $a$ when divided by $n$. So if $a\equiv b\pmod{n}$, then $b\equiv a \pmod{n}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $a$ leaves the same remainder as $b$, and $b$ leaves the same remainder as $c$, then $a$ and $c$ leave the same remainder as well. So if $a\equiv b\pmod{n}$ and $b\equiv c\pmod{n}$, then $a\equiv c\pmod{n}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That means that:&#xA;\begin{equation*}&#xA;[a]_{n} = [b]_n\Longleftrightarrow a\equiv b\pmod{n}.&#xA;\end{equation*}&#xA;Why? Well, suppose the congruence classes are the same. Since $a\in[a]_n=[b]_n$, that means that $b\equiv a \pmod{n}$ (since $a\in[b]_n$), so $a\equiv b\pmod {n}$. That proves that if the classes are equal, then $a\equiv b\pmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $a\equiv b\pmod{n}$. How do we prove that $[a]_{n}=[b]_{n}$? Since they are sets, the usual way is to show that each is contained in the other. If $c\in[a]_n$, then $a\equiv c\pmod{n}$ by definition; since we also have $b\equiv a\pmod{n}$ (since $a\equiv b\pmod{n}$ is our assumption), then $b\equiv c\pmod{n}$, so $c\in[b]_n$. Therefore, every that is in $[a]_{n}$ is also in $[b]_n$, so $[a]_n\subseteq [b]_n$. For the converse inclusion, if $c\in[b]_n$, then $b\equiv c\pmod{n}$, and since we also have $a\equiv b\pmod{n}$, then $a\equiv c \pmod{n}$ so $c\in[a]_n$. Therefore, $[b]_n\subseteq [a]_n$. Since we have the two inclusions, we conclude that $[a]_n=[b]_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have the following, which is perhaps more surprising:&#xA;\begin{equation*}&#xA;[a]_n\cap[b]_n\neq\emptyset\Longleftrightarrow [a]_n=[b]_n.&#xA;\end{equation*}&#xA;That is: the only way the class of $a$ and the class of $b$ have &lt;em&gt;anything&lt;/em&gt; in common is if they are identical. Why? Well, if they are identical they certainly have nonempty intersection, since the class of $a$ always includes at least $a$. And conversely, if $c\in[a]_n\cap[b]_n$, then $a\equiv c\pmod{n}$ and $b\equiv c\pmod{n}$, from which we conclude that $a\equiv b\pmod{n}$, so $[a]_n=[b]_n$ by what we just finished proving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; How do we check if $[a]_{17} = [b]_{17}$? Precisely the way you did it: the classes are equal if and only if $a\equiv b \pmod{17}$. How do you check if $a\equiv b\pmod{17}$? By checking to see if $17$ divides $a-b$. So how do you check in general whether $[a]_n=[b]_n$? You check to see if $a-b$ is a multiple of $n$. If it is, then the classes are equal. If $a-b$ is not a multiple of $n$, then they are not equal, and in fact they are disjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways: for example, if you could show that $[60]_{17}$ and $[43]_{17}$ have &lt;em&gt;any&lt;/em&gt; element in common, then you would be able to conclude that they are equal. &lt;em&gt;Sometimes&lt;/em&gt; it may be simpler to see that $[a]_n$ and $[b]_n$ have some element in common than to check if $n$ divides $a-b$; but the &lt;em&gt;standard&lt;/em&gt; way of checking is to see whether $n$ divides $a-b$, just as you did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; What you have is an equation in which $x$ is an unknown. You are really looking for all solutions to the congruence&#xA;\begin{equation*}&#xA;ax \equiv b \pmod {n}&#xA;\end{equation*}&#xA;That is, all integers $x$ that make the congruence true. For example, $x=3$ is a solution to&#xA;\begin{equation*}&#xA;2x \equiv 1 \pmod{5}&#xA;\end{equation*}&#xA;because $(2)(3)= 6\equiv 1 \pmod{5}$. In fact, because $3$ is a solution, so is $3+5k$ for any integer $k$; that is, &lt;em&gt;any&lt;/em&gt; element in $[3]_5$ is a solution, since $3$ is a solution. So the solutions will &lt;em&gt;actually&lt;/em&gt; be a collection of congruence classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if we replace $2$ in the equation with $7$, so that we are trying to solve $7x\equiv 1 \pmod{5}$, then anything which was a solution to $2x\equiv 1\pmod{5}$ is still a solution, and any solution to the first one is still a solution to the new congruence; why? because $2x\equiv 7x\pmod{5}$ for &lt;em&gt;all&lt;/em&gt; $x$, as $2x-7x = -5(x)$ is always a multiple of $5$. So we can replace $2$ with &lt;em&gt;any&lt;/em&gt; element of $[2]_5$ and not change the solutions. Likewise, we can replace $1$ with any element of $[1]_5$ and not change the solutions. So it's almost as if instead of trying to solve the single congruence&#xA;\begin{equation*}&#xA;2x \equiv 1 \pmod{5}&#xA;\end{equation*}&#xA;we are trying to solve the equation&#xA;\begin{equation*}&#xA;[2]_5 x = [1]_5&#xA;\end{equation*}&#xA;So that's why you have written $[a]x=[b]$. That really means the congruence $ax\equiv b \pmod{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you wrote, however, is incorrect. You are missing a clause: the congruence will have $d$ solutions if $d$ divides $b$. Otherwise, it's not going to have any. For an easy example, consider the congruence $2x\equiv 1 \pmod{4}$. Then $\gcd(2,4)=2$, but there are no solutions to the congruence, because $2x$ is always even, so $2x-1$ is always odd, so $4$ never divides $2x-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how do you solve a congruence $ax\equiv b\pmod{n}$ when $\gcd(a,n)$ divides $b$? let $d=\gcd(a,n)$. Then we can write $a=da&amp;#39;$, $b=db&amp;#39;$, and $n=dn&amp;#39;$. Then $ax\equiv b\pmod{n}$ if and only if $n$ divides $ax-b$. But $ax-b = d(a&amp;#39;x - b&amp;#39;)$, and $dn&amp;#39;$ divides $d(a&amp;#39;x-b&amp;#39;)$ if and only if $n&amp;#39;$ divides $a&amp;#39;x-b&amp;#39;$, if and only if $a&amp;#39;x\equiv b&amp;#39;\pmod{n&amp;#39;}$. So if we can solve a congruence when $\gcd(a,n)=1$, then we can solve any congruence where $\gcd(a,n)$ divides $b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you solve a system $ax \equiv b\pmod{n}$ when $\gcd(a,n)=1$ (which will &lt;em&gt;of course&lt;/em&gt; divide $n$)? Since $\gcd(a,n)=1$, then we can write $n$ as a linear combination of $a$ and $n$, $1=ar+ns$ for some integers $r$ and $s$ (for example, using the Euclidean algorithm). Multiplying through by $b$ you get $b=a(rb)+ n(sb)$. That means that $n(sb) = a(rb)-b$, so $a(rb)\equiv b \pmod{n}$, which means that $x=rb$ is a solution. If $y$ is any other solution, then $ay\equiv b\pmod{n}$, and $ax\equiv b\pmod{n}$, so $ax\equiv ay\pmod{n}$, hence $n$ divides $ax-ay=a(x-y)$, and since $\gcd(a,n)=1$, then $n$ divides $x-y$; so if $y$ is any other solution, then $x\equiv y\pmod{n}$. So the only &lt;em&gt;congruence class&lt;/em&gt; that is a solution is $[x]_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what about general congruences? Suppose you have $[a]_nx=[b]_n$ and $\gcd(a,n)=d$ divides $b$. Write $a=da&amp;#39;$, $b=db&amp;#39;$, and $n=dn&amp;#39;$. Notice that $\gcd(a&amp;#39;,n&amp;#39;)=1$. Find a solution $x_0$ to $[a&amp;#39;]_{n&amp;#39;}x=[b&amp;#39;]_{n&amp;#39;}$. Then $n&amp;#39;$ divides $a&amp;#39;x_0 -b&amp;#39;$, so $dn&amp;#39;=n$ divides $d(a&amp;#39;x_0-b&amp;#39;) = da&amp;#39;x_0-db&amp;#39; = ax_0-b$. So $x_0$ is also a solution to the original problem. If $y$ is any other solution, then as before we get that $n$ divides $a(x_0-y)$; so $dn&amp;#39;$ divides $da&amp;#39;(x_0-y)$, hence $n&amp;#39;$ divides $a&amp;#39;(x_0-y)$, and since $\gcd(a&amp;#39;,n&amp;#39;)=1$, then $n&amp;#39;$ divides $x_0 - y$; that is, $y\equiv x_0\pmod{n&amp;#39;}$, so $y=x_0+kn&amp;#39;$ for some $k$. Conversely, if $y = x_0+\ell n&amp;#39;$ for some $\ell$, then &#xA;\begin{equation*}&#xA;ay = a(x_0+\ell n') = ax_0 + \ell an' = ax_0 + \ell (da')n' = ax_0 + \ell a'(dn') = ax_0 + (\ell a')n,&#xA;\end{equation*} &#xA;which says that $ay \equiv ax_0 \pmod{n}$, so $y$ is also a solution. So what are the different solutions? Well, we have $x_0$, $x_0+n&amp;#39;$, $x_0+2n&amp;#39;$, $x_0+3n&amp;#39;,\ldots, x_0+(d-1)n&amp;#39;$. All of these are &lt;em&gt;not&lt;/em&gt; congruent to one another modulo $n$; the &quot;next&quot; one, however, will be $x_0+dn&amp;#39; = x_0+n$, which &lt;strong&gt;is&lt;/strong&gt; congruent to $x_0$ modulo $n$. So the only distinct &lt;em&gt;congruence classes&lt;/em&gt; that are solutions are $[x_0]_n$, $[x_0+n&amp;#39;]_n,\ldots,[x_0+(d-1)n&amp;#39;]_n$, giving you exactly $d$ distinct classes that are solutions to $[a]_nx=[b]_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this in practice, take $n= 77$, $a=21$, and $b=14$. We want to solve the system $[21]_{77} x = [9]_{77}$ Since $\gcd(77,21)=7$ divides $b$, the system has solutions; in fact, it has $7$ different congruence classes modulo $77$ as solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we write $21 = 7\cdot 7$, $14 = 7\cdot 2$, $77 = 7\cdot 11$. And we first solve the system $[3]_{11}x=[2]_{11}$. To do this, we write $1$ as a linear combination of $3$ and $11$, like so: $1 = 3(4) - 11(1)$. Then multiply by $2$ to get $2 = 3(8)-11(2)$. So $x_0=8$ is a solution (indeed, $3(8)=24\equiv 2\pmod{11}$). Now going back to the original congruence, we take $x_0$ and we add multiples of $11$ (why $11$? Because $n=77$ is $7\cdot 11$, and $7$ is the $d$ from before) until we have our $7$ different solutions. So the solutions are: $[8]_{77}$, $[19]_{77}$, $[30]_{77}$, $[41]_{77}$, $[52]_{77}$, $[63]_{77}$, and $[74]_{77}$. You can verify that they all are solutions to $[21]_{77}x = [14]_{77}$. For example, $21\cdot 41 = 861$, and $[861]_{77}=[14]_{77}$ because $861-14 = 847 = 77\cdot 11$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; Yes: that is exactly the meaning of $[a]_nx=[b]_n$; it means $ax\equiv b\pmod{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps, despite the length. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-14T23:17:41.867" LastActivityDate="2010-11-14T23:17:41.867" />
  <row Id="10308" PostTypeId="2" ParentId="10216" CreationDate="2010-11-14T22:45:51.437" Score="0" ViewCount="0" Body="&lt;p&gt;I have been thinking about the the problem a bit today. What Robin and Yuval have both shown is that if the Cayley hamilton theorem is true, then the characteristic and minimal polynomial of $C(f)$ are both equal to $f$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, assume that for all $f \in F[x]$, the the characteristic and minimal polynomial of $C(f)$ are both equal to $f$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be a finite dimensional $F$-vector space and $T : V \to V$ a linear transformation. we know from the classification theorem of modules over PIDs that there exists a basis $B$ of $V$ and $f_1, \dots, f_s \in F[x]$ such that $f_1 \mid \dots \mid f_s$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [T]_B = \begin{pmatrix} C(f_1) &amp;amp; &amp;amp; \\ &amp;amp; \ddots &amp;amp; \\ &amp;amp; &amp;amp; C(f_s)  \\ \end{pmatrix} := M$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the characteristic polynomial of $M$ is the product of the characteristic polynomials of all of the $C(f_i)$s and the minimal polynomial of $M$ is the least common multiple of the minimal polynomials of all the $C(f_i)$s. We see form the assumption that the characteristic polynomial of $T$ is $f_1 f_2 \dots f_s$ and the minimal polynomial of $T$ is $f_s$. This proves the Cayley hamilton theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the Cayley Hamilton theorem is equivalent to the fact &quot;for all $f \in F[x]$, the the characteristic and minimal polynomial of $C(f)$ are both equal to $f$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving the Cayley hamilton theorem without assuming knowledge of modules over PIDs or companion matrices is quite delicate (from what I remember from first year university).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to support the idea that in order prove to the Cayley hamilton theorem (or the fact about companion matrices), you need to at some point get your hands dirty (whether it be directly computing the minimal and characteristic polynomials of a companion matrix or going through a delicate proof of the Cayley hamilton theorem).&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2010-11-14T22:45:51.437" CommentCount="9" />
  <row Id="10309" PostTypeId="2" ParentId="10280" CreationDate="2010-11-14T23:09:06.213" Score="2" ViewCount="0" Body="&lt;p&gt;Write $\{ x\}$ for $x-\lfloor x\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prove that $\{ rx\}$ for $r=0,1,\cdots,q$ are all distinct.&#xA;(Here is where you use the condition that $x$ is irrational.)&lt;/li&gt;&#xA;&lt;li&gt;Using pigeonhole principle,  find  integers $r,s, 0\le r&amp;lt;s,$ such that&lt;br&gt;&#xA;$\{ sx \},\{ rx\}$ belong to the same subinterval.&lt;/li&gt;&#xA;&lt;li&gt;Then prove that $s-r$ is your solution.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-15T14:18:16.597" LastActivityDate="2010-11-15T14:18:16.597" />
  <row Id="10310" PostTypeId="2" ParentId="10046" CreationDate="2010-11-14T23:14:47.127" Score="1" ViewCount="0" Body="&lt;p&gt;Here is some more useful info related to your first sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that when summing $\sum_{n\ge 0} t_n$ we can use the idea of hypergeometrics, by computing the ratio $t_{n+1}/t_n$, and then reading-off from it the Hypergeometric function that solves the summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the ratio is&#xA;$$\frac{t_{n+1}}{t_n} = \frac{(n+1)(n+1)}{(n+1/2)4},$$ so the sum over all $t_n$ ranging from $n\ge 0$ instead of $n\ge 1$ as you ask, is nothing but the hypergeometric function&#xA;$${}_2F_1(1, 1; \frac{1}{2}; \frac{1}{4})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which may be verified using some of the ideas above to be equal to &#xA;$$\frac{2}{27} \left(18+\sqrt{3} \pi \right)$$&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-14T23:14:47.127" />
  <row Id="10312" PostTypeId="2" ParentId="10251" CreationDate="2010-11-14T23:30:25.673" Score="2" ViewCount="0" Body="&lt;p&gt;First, notice that &#xA;$$\cos(x+\pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) = \cos(x)(-1)-\sin(x)(0) = -\cos(x).$$&#xD;&#xA;So, rewriting, using the above, and using that $\cos(-x)=\cos x$ you have:&#xA;\begin{align*}&#xA;\cos\left(\theta+\frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right) &amp;amp;= \cos\left(\left(\theta-\frac{\pi}{3}\right)+\pi\right) + \cos\left(\left(\theta+\frac{\pi}{3}\right)+\pi\right)\\&#xA;&amp;amp;= -\cos\left(\theta-\frac{\pi}{3}\right) - \cos\left(\theta+\frac{\pi}{3}\right)\\&#xA;&amp;amp;= -\cos\left(\frac{\pi}{3}-\theta\right) - \cos\left(\frac{\pi}{3}+\theta\right).&#xA;\end{align*}&#xA;So: the equality as written cannot hold: you would need $\cos(\frac{\pi}{2}+\theta) = - \cos(\frac{\pi}{3}-\theta)$ for all $\theta$, and plugging in $\theta=0$ shows this cannot hold. I suspect your $\frac{\pi}{2}$ should have been a $\frac{\pi}{3}$, and that you are missing a minus sign before the first term. If this is not the case, then please edit the question to give the equality you are actually trying to prove. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-14T23:30:25.673" CommentCount="1" />
  <row Id="10314" PostTypeId="1" AcceptedAnswerId="10316" CreationDate="2010-11-14T23:31:54.750" Score="0" ViewCount="133" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ULjJd.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers are already given after the = sign of each question. But I dont know how they arrived at these answers. What does it mean to say f(A)=1 and so on? I cant find the connection..&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T00:39:20.513" LastActivityDate="2010-11-15T00:47:35.543" Title="Determining if two  graphs are isomorphic" Tags="&lt;graph-theory&gt;&lt;trees&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="10316" PostTypeId="2" ParentId="10314" CreationDate="2010-11-14T23:41:38.670" Score="3" ViewCount="0" Body="&lt;p&gt;A graph can be thought of as a pair $(V,E)$, where $V$ are the vertices and $E$ are the edges. An &lt;strong&gt;isomorphism&lt;/strong&gt; between two graphs, $G_1 = (V_1,E_1)$ and $G_2=(V_2,E_2)$ means a function $f\colon V_1\to V_2$ that is bijective between the vertices (one-to-one and onto, so every vertex in $G_1$ is mapped to a vertex of $G_2$, and each vertex in $G_2$ is the image of one and only one vertex in $G_1$) &lt;strong&gt;and&lt;/strong&gt; for vertices $v_1,v_2$ of $G_1$, $v_1$ is adjacent to $v_2$ in $G_1$ if and only if $f(v_1)$ is isomorphic to $f(v_2)$ in $G_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer they give is a description of an isomorphism between the second graph and the first: it is a function from the set of vertices of the second graph, $\{A,B,C,D,G,H,I,J\}$, and the vertices of the first graph, $\{1,2,3,4,5,6,7,8\}$, with the property that two vertices in the second graph $x$ and $y$, are adjacent if and only if $f(x)$ is adjacent to $f(y)$ in the second graph. For instance, $A$ and $I$ are adjacent, so $f(A)=1$ needs to be adjacent to $f(I)=4$ (which it is), and $A$ and $J$ are not adjacent, so $f(A)=1$ should &lt;strong&gt;not&lt;/strong&gt; be adjacent to $f(J)=7$ (which it is not).  You can check that the function described actually &lt;em&gt;is&lt;/em&gt; an isomorphism by checking that every pair of adjacent vertices are sent to adjacent vertices, and every pair of non-adjacent vertices are sent to non-adjacent vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... how did they come up with $f$ in the first place? Well, let's think about the graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second graph, each vertex is adjacent to exactly three vertices, so if there is going to be an isomorphism between the two graphs, then every vertex in the first graph needs to be adjacent to exactly three vertices. This is indeed true, which is a good start for a possible isomorphism (if it wasn't true, we would have &lt;em&gt;no&lt;/em&gt; hope of having an isomorphism, and if fact we would know that no isomorphism exists).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the second graph is &quot;bipartite&quot;: you can divide the vertices into two groups in such a way that all edges are between the two groups, with no edges between vertices in the same group. If we are going to find an isomorphism between this graph and the first, we need to be able to &lt;em&gt;also&lt;/em&gt; separate the vertices of the first graph into two groups of four, so that all edges are between the groups and no edge goes from one vertex in a group to another vertex in the &lt;em&gt;same&lt;/em&gt; group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we divide the vertices in the first graph into two sets of four so that all edges are between the sets? Well, let's see: if $1$ is in one set, then $2$, $4$, and $5$ should be on the other set because there can be no edges between two vertices in the same set (remember: we are trying to check if there is an isomorphism). Notice that if we do this, then there better not be any edges between $2$, $4$, and $5$, because we are going to put them in the same set (the set which is not the set with $1$ in it). Good, no edges between any of $2$, $4$, and $5$. And $6$ and $3$ need to be with $1$, because they are both adjacent to $2$, and so they cannot be in the same set with $2$; the only possibility left is the set with $1$ in it. That means that $1$, $3$, and $6$ need to be in the same set of vertices, which means there should be no edges between them (no edges between any two vertices in the same set, remember?). Good, no edges between any of $1$, $3$, and $6$. And $8$ needs to be with $1$, because it is adjacent to $4$. And so $7$ needs to be in with $2$, $4$, and $5$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we can find an isomorphism between the two graphs, then we can divide the vertices of the first graph into two sets, just like the ones in the second graph, and one set needs to have $1$, $3$, $6$, and $8$, and the other set needs to have $2$, $4$, $5$, and $7$. And all edges should be between the sets, not no edges between vertices in the same set. This all holds, so we are on the right track!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can we define an isomorphism based on this? There is a lot of symmetry, so it probably does not matter where we send $A$, just so long as we do things right. So let's send $A$ to $1$: $f(A)=1$. Because $G$, $H$, and $I$ are adjacent to $S$, we need $f(G)$, $f(H)$, and $f(I)$ to be adjacent to $f(A)=1$; so $f(G)$, $f(H)$, and $f(I)$ need to be $2$, $4$, and $5$ in some order. Say $f(G)=2$. Then $f(H)$ needs to be either $4$ or $5$. If we make it $5$, then since $2=f(G)$ and $5=f(H)$ have only two mutually adjacent vertex (namely $1$ and $6$), and $G$ and $H$ have only two mutually adjacent vertices (namely $A$ and $B$), and $A$ already maps to $1$, then we need $B$ to map to $6$. So so far we have $f(A)=1$, $f(B)=6$, $f(G)=2$, and $f(H)=5$. $B$ is adjacent to $J$, which is not adjacent to $A$, so $f(J)$ needs to be adjacent to $f(B)=6$, but not to $f(A)=1$; and it needs to be in the same group as $2$ (so it has to be one of $2$, $4$, $5$, or $7$); $5$ and $2$ are already taken, $4$ is adjacent to $1$, so we need to make $f(J)=7$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So so far we have $f(A)=1$, $f(B)=6$, $f(G)=2$, $f(H)=5$, and $f(J)=7$. What about $C$? $f(C)$ needs to be adjacent to $f(G)=2$ and to $f(J)=7$, but not to $f(H)=5$; the only possibility not yet taken is $f(C)=3$. Continuing this way you get the isomorphism they give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this is not the only possible isomorphism; other choices lead to other isomorphisms, of which there are many because the graphs are so symmetric. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T00:47:35.543" LastActivityDate="2010-11-15T00:47:35.543" CommentCount="3" />
  <row Id="10317" PostTypeId="1" CreationDate="2010-11-14T23:47:20.670" Score="1" ViewCount="54" Body="&lt;p&gt;Suppose $P(A) = 0.6$ and $P(B) = 0.7$. Find the smallest possible value of $P(A \cap B)$. Instead of deducing this from $P(A \cup B) = P(A)+P(B)-P(A \cap B)$, could we just use the fact that $P(A|B)P(B) = P(A \cap B)$ and $P(B|A)P(A) = P(A \cap B)$? I notice that $(0.6)(0.7)^2 = 0.294$ which is close to $0.3$, the smallest value of $P(A \cap B)$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-15T00:35:49.743" Title="Conditional Probability" Tags="&lt;probability&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10318" PostTypeId="1" AcceptedAnswerId="10321" CreationDate="2010-11-15T00:14:18.120" Score="1" ViewCount="155" Body="&lt;p&gt;Let X and Y be independent random variables,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  P(X = k) = P(Y = k) = p(1 - p)^{k-1} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you show that the pmf of $ Z = X + Y $, is negative binomial, and how do you find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P(X = Y) $?&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-11-15T05:51:13.513" Title="Sum of two independent geometric random variables" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10319" PostTypeId="1" AcceptedAnswerId="10323" CreationDate="2010-11-15T00:15:24.270" Score="2" ViewCount="84" Body="&lt;p&gt;Suppose $f$ is an unknown polynomial of degree $n$ (in one indeterminate) but the sequence $\{ f(k) \}_{k \in \mathbb{N}}$ is given. It is a nice exercise to show that one needs only the first $n+1$ terms of the sequence to determine the coefficients of $f$. That is, simply solve the matrix equation $A\mathbf{x} = b$, where $\mathbf{b} = (f(0), \dots, f(n))^{\top}$, $A$ is the Vandermonde matrix of $(i^{j})_{i,j = 0, \dots, n}$ and $\mathbf{x} = (c_{0}, \dots, c_{n})^{\top}$ (the unknown coefficients of $f$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is there a closed form expression for the coefficients of a finite degree polynomial $f$ in terms of the sequence $\{ f(k) \}_{k \in \mathbb{N}}$ that doesn't involve matrix inversion or differentiation or explicitly calculating the polynomial in question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) The Ehrhart polynomial counts the number of integer lattice points intersecting a dilate of a polytope and can be calculated by the residue of an associated complex rational function (see M. Beck's articles on the subject). Some of the coefficients of the Ehrhart polynomial can be related to an $n$-volume, a relative area and the euler characteristic of said polytope. However, computing coefficients of the Ehrhart polynomial is not a particularly easy task. Having simple formulas for them, say in terms of the residues above would be nice to have at one's disposal. A reasonable starting point is answering the question above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T02:57:13.520" LastActivityDate="2010-11-15T02:57:13.520" Title="Determining Coefficients of a Finite Degree Polynomial $f$ from the Sequence $\{f(k)\}_{k \in \mathbb{N}}$" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="10320" PostTypeId="2" ParentId="10257" CreationDate="2010-11-15T00:17:34.640" Score="5" ViewCount="0" Body="&lt;p&gt;If $f\colon[0,1]\to\mathbb{R}$ is cadlag (&lt;a href=&quot;http://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g&quot; rel=&quot;nofollow&quot;&gt;continu &amp;agrave; droite, limites &amp;agrave; gauche&lt;/a&gt;) then we can ask what the possible jumps are. That is, for what functions $g\colon(0,1]\to\mathbb{R}$ is there a cadlag function $f$ with $g(x)=\Delta f(x)\equiv f(x)-f(x-)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that $g$ occurs as the jumps of a cadlag function if and only if the set $\{x\in(0,1]\colon\vert g(x)\vert &amp;gt; \epsilon\}$ is finite for each $\epsilon &amp;gt; 0$. In particular, there is a cadlag function with jumps as you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the necessity: If $S=\{x\in(0,1]\colon\vert g(x)\vert &amp;gt; \epsilon\}$ was not finite, then it would contain a strictly increasing or strictly decreasing sequence $x_n$. Then, $\vert f(x_n)-f(y_n)\vert &amp;gt; \epsilon$ for some $y_n$ chosen arbitrarily close to $x_n$. Replacing $x_n$ by $y_n$ where necessary gives a strictly increasing or strictly decreasing sequence $x_n\in[0,1]$ such that $\vert f(x_{n+1})-f(x_n)\vert &amp;gt; \epsilon/2$. However, as f is cadlag and has left and right limits everywhere, this contradicts the requirement that $f(x_n)$ tends to a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can show sufficiency: Set $\epsilon_n=2^{-n}$ for $n\ge1$ and $\epsilon_0=\infty$. For each $n\ge1$, consider the finite set $S_n=\{x\in(0,1]\colon \epsilon_{n-1}\ge\vert g(x)\vert &amp;gt; \epsilon_n\}$. We can construct a function $f_n\colon[0,1]\to\mathbb{R}$ such that $\Delta f_n(x)=1_{\{x\in S_n\}}g(x)$ and $\vert f_n(x)\vert\le\epsilon_{n-1}$. The idea is to take $f_n(x)=g(x)$ and $f_n(x-)=0$ for points in $S_n$, and linearly interpolate between these.&#xA;More precisely, if $S_n$ is empty we set $f_n=0$. Otherwise,&#xA;$$&#xD;&#xA;f_n(x)= g(a)(b-x)/(b-a)&#xD;&#xA;$$&#xD;&#xA;for $a\le x &amp;lt; b$. Here, $a &amp;lt; b$ are consecutive points of $S_n$. Also set $f_n(x)=0$ for $x$ less than the minimum of $S_n$ and $f_n(x)=g(c)$ for $x\ge c=\max S_n$. This has the required jumps $\Delta f_n(x)=1_{\{x\in S_n\}}g(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally set $f(x)=\sum_{n=1}^\infty f_n(x)$. As $\vert f_n\vert \le 2^{1-n}$ for $n\ge 2$ this converges uniformly and,&#xA;$$&#xD;&#xA;\Delta f(x)=\sum_n\Delta f_n(x)=\sum_n1_{\{x\in S_n\}}g(x)=g(x).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-15T01:25:57.267" LastActivityDate="2010-11-15T01:25:57.267" CommentCount="2" />
  <row Id="10321" PostTypeId="2" ParentId="10318" CreationDate="2010-11-15T00:20:23.887" Score="2" ViewCount="0" Body="&lt;p&gt;$P(Z=r)=\Sigma_{k=1}^r P(X=k)P(Y=r-k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X=Y)=\Sigma_{k=1}^{\infty} P(X=k)P(Y=k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming $k$ is a positive integer. Otherwise, the summations need to be modified accordingly.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T00:20:23.887" CommentCount="2" />
  <row Id="10322" PostTypeId="1" AcceptedAnswerId="10338" CreationDate="2010-11-15T00:24:15.697" Score="2" ViewCount="81" Body="&lt;p&gt;in trying to find this limit, all the (x,y) paths i check approaching (0,0) produce the same result, namely, the function approaches 0. consequently, i came up with the simplest proof i could that the limit indeed exists and is zero. however wolframalpha tells me that the limit doesnt exist. assuming im wrong, can anyone shed light on where my proof is flawed, and possibly provide a path on which the function doesnt approach 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{(x,y) \rightarrow (0,0)}&#xA;\frac { x \cdot \sin(\frac{y}{\sqrt{x}}) }&#xA;{\sqrt{ x^2 + y^2 }}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;provided $x&amp;gt;0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets take $\varepsilon &amp;gt; 0$ and prove the existence of some $\delta &amp;gt; 0$ such that&lt;br&gt;&#xA;$\forall (x,y) \in B(0, \delta ) : |f(x,y)| &amp;lt; \varepsilon$  &#xA;(where f is our function above, and as stated $x&gt;0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking the x to the denominator, we get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac { \sin(\frac{y}{\sqrt{x}}) }&#xA;{\sqrt{ 1 + (\frac{y}{x})^2 }}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin is bounded between -1 and 1, so from here i intend from here to show that i can choose a relevant $\delta$ where either the denominator is large enough to make the expression smaller than our $\varepsilon$, and if not, the numerator is smaller than our $\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we find the relevant denominator bound by:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {1}&#xA;{\sqrt{ 1 + (\frac{y}{x})^2 }} &amp;lt; \varepsilon&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {y}{x} &gt; \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;clearly, if this inequality holds, then the absolute value of the function is indeed bound by our $\varepsilon$ as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so lets deal with the seconds case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {y}{x} \leq \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying through by $\sqrt {x}$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac {y}{\sqrt{x}}&#xA;\leq \sqrt{x} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\leq \sqrt{\delta} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the last inequality is because $x$ is obivously bound by the $\delta$ we choose&lt;br&gt;&#xA;it should be clear that the right hand side of the inequality is arbitrarily small, although we will formalise somewhat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;take $\delta &amp;lt; \varepsilon^4$:&#xA;\begin{equation}&#xA;\frac {y}{\sqrt{x}}&#xA;\leq \sqrt{\delta} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1}&#xA;&amp;lt; \sqrt{\varepsilon^4} \cdot \sqrt{\frac{1}{\varepsilon^2}}&#xA;= \varepsilon^2 \cdot \frac{1}{\varepsilon}&#xA;= \varepsilon&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;without too much formality, we say that for sufficiently small $\delta$, $\sin x \approx x$ and therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;{\sin{ \frac {y}{\sqrt{x}}}&#xA;\approx&#xA;\frac {y}{\sqrt{x}}&#xA;\leq \varepsilon}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of course, i was lazy and didnt add in the necessary stages to prove the approximation will hold in the inequality, although it does, because i can take the $\delta$ to be such that the difference between the sin and the value itself is less than $\frac{\varepsilon}{3}$ and also such that $\frac{y}{\sqrt {x}} &amp;lt; \frac{\varepsilon}{3}$ which means the sin is indeed less than $\varepsilon$ as required. this takes care of the second case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i didnt take much care with whether y is positive or negative, although the function is odd, so we can work with positives and then just prove the result for the negatives)&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2010-11-15T01:18:56.493" Title="multivariable limit" Tags="&lt;limit&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10323" PostTypeId="2" ParentId="10319" CreationDate="2010-11-15T00:32:13.097" Score="4" ViewCount="0" Body="&lt;p&gt;Yes; you can take &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite differences&lt;/a&gt;.  Let $\Delta f(x) = f(x+1) - f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt;  $$f(x) = \sum_{k=0}^{n} \Delta^k f(0) {x \choose k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where ${x \choose k}$ is the polynomial $\frac{x(x-1)...(x-(k-1))}{k!}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  Observe that $\Delta {x \choose k} = {x \choose k-1}$.  Take $k$ finite differences of both sides and set $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, this means you can work out what $f$ is by writing down a finite difference table.  This is really easy to do by hand, and then the top row of the table tells you what the coefficients $\Delta^k f(0)$ are above.  In linear algebra terms, what we are doing is using a basis with respect to which the matrix $A$ is upper-triangular, and this makes life much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a known explicit formula for the inverse of $A$ which is equivalent to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange interpolation formula&lt;/a&gt;.  This is sometimes useful for theoretical reasons (e.g. as a way to control the behavior of the interpolation polynomial or to deduce certain identities).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can extract a &quot;closed form&quot; for the coefficients of $f$ using either of the two methods above, although I think using finite differences is slightly nicer.  We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${x \choose k} = \frac{1}{k!} \sum_{i=0}^{k} s(k, i) x^i$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s(k, i)$ are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the first kind&lt;/a&gt;.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[x^i] f(x) = \sum_{k=i}^{n} \Delta^k f(0) \frac{s(k, i)}{k!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta^k f(0) = \sum_{j=0}^{k} (-1)^{k-j} {k \choose j} f(j).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $[x^i] f(x) = \sum_{j=0}^{n} a_{i,j} f(j)$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{i,j} = \sum_{k=j}^{n} (-1)^{k-j} {k \choose j} \frac{s(k, i)}{k!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether this sum can be simplified further.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-15T01:57:35.113" LastActivityDate="2010-11-15T01:57:35.113" CommentCount="0" />
  <row Id="10324" PostTypeId="2" ParentId="10317" CreationDate="2010-11-15T00:35:49.743" Score="1" ViewCount="0" Body="&lt;p&gt;From the equation in the second line&#xA;$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$&#xD;&#xA;we have&#xA;$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$&#xD;&#xA;In order to minimize $P(A \cap B)$, we should maximize $P(A \cup B)$ since both $P(A)$ and $P(B)$ are fixed. The maximum value for $P(A \cup B)$ is $1$, so we have the minimum value of $P(A \cap B)$ as&#xA;$$P(A \cap B) = 0.6 + 0.7 - 1 = 0.3$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-15T00:35:49.743" />
  <row Id="10326" PostTypeId="1" AcceptedAnswerId="10331" CreationDate="2010-11-15T00:42:31.290" Score="0" ViewCount="110" Body="&lt;p&gt;Let $X_{i} = 1$ if &lt;em&gt;ith&lt;/em&gt; person gets his or her own hat back and $0$ otherwise. Let $S_{n} = \sum_{i=1}^{n}X_{i}$, then $S_{n}$ is the total number of people who get their own hats back. Show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $E(X_{i}^{2}) = \frac{1}{n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) $E(X_{i}.X_{j}) = \frac{1}{n(n - 1)}$ for $i\neq j$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $E(S_{n}^{2}) = 2$ (using (a) and (b)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) $var(S_{n}) = 1$&lt;/p&gt;&#xA;" OwnerUserId="2976" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-15T20:07:35.533" LastActivityDate="2010-11-15T20:07:35.533" Title="Suppose that n people have their hats returned at random, find the following" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="10327" PostTypeId="1" AcceptedAnswerId="10344" CreationDate="2010-11-15T00:44:02.297" Score="7" ViewCount="236" Body="&lt;p&gt;Is it algebraic the number 0.2468101214 ...? (After point, the natural numbers are juxtaposed pairs).&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-15T00:49:44.323" LastActivityDate="2010-11-15T02:10:41.477" Title="0.246810121416...: Is it a algebraic number?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10328" PostTypeId="2" ParentId="10326" CreationDate="2010-11-15T00:55:58.623" Score="1" ViewCount="0" Body="&lt;p&gt;I'll give some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Notice that $X_i^2 = X_i$, since $X_i$ is $0$ or $1$.  Then, &#xA;$E X_i = 0 \cdot \mathrm{P}(X_i=0) + 1 \cdot \mathrm{P}(X_i = 1) = \mathrm{P}(X_i = 1)$.  What is this probability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  Again, $X_i X_j$ is $0$ or $1$.  Notice that $X_i X_j = 1$ if and only if person $i$ gets their hat AND person $j$ gets their hat.  What is the probability of this?  Let $A$ be the event person $i$ gets his hat and $B$ be the event person $j$ does.  Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{P}(X_i X_j = 1) = \mathrm{P}(A \cap B) = \mathrm{P}(A) \mathrm{P}(B|A) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are $\mathrm{P}(A)$ and $\mathrm{P}(B|A)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  After some algebra,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E S_n^2 = E(\sum_{i=1}^n X_i )^2 = E(\sum_{i=1}^n X_i^2 +  \sum_{i \neq j} X_i X_j)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distribute the $E$ through. Use 1) and 2). How many ways can $i \neq j$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4)  $\mathrm{Var} S_n = E S_n^2 - (E S_n)^2$  &lt;/p&gt;&#xA;" OwnerUserId="2515" LastActivityDate="2010-11-15T00:55:58.623" />
  <row Id="10329" PostTypeId="1" AcceptedAnswerId="10345" CreationDate="2010-11-15T00:56:46.890" Score="2" ViewCount="89" Body="&lt;p&gt;Let X be an infinite-dimensional Banach space, and T a compact operator from X to X. Why must 0 then be a spectral value for T? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is equivalent to saying that T is not bijective, but I am not sure how to show that injectivity implies the absence of surjectivity and the other way around (or if this is even the right way to approach the problem).&lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-15T02:16:51.390" Title="Invertibility of compact operators in infinite dimensions" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10331" PostTypeId="2" ParentId="10326" CreationDate="2010-11-15T01:01:26.243" Score="2" ViewCount="0" Body="&lt;p&gt;Please note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a) $X_{i}=1$ probability $\frac{1}{n}$ and 0, otherwise. Thus $X_{i}^{2}=1$ with probability $\frac{1}{n}$ and 0, otherwise, which &lt;em&gt;implies&lt;/em&gt; $E(X_{i}^{2})=\frac{1}{n}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;b) The probability that both $X_{i}$ and $X_{j}$ are 1 is $\frac{1}{n(n-1)}$ (Why?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;c) $E(S_{n}^{2}) = \sum\limits_{i} E[X_{i}^{2}] + \sum\limits_{i}\sum\limits_{j \neq i} E[X_{i}X_{j}= n \cdot \frac{1}{n} + n(n-1) \cdot \frac{1}{n(n-1)}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;d) $Var(S_{n}) = E[S_{n}^{2}] - (E[S_{n}])^{2}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-15T01:01:26.243" />
  <row Id="10332" PostTypeId="2" ParentId="10322" CreationDate="2010-11-15T01:02:00.650" Score="1" ViewCount="0" Body="&lt;p&gt;Wolframalpha is not infallible. Your proof looks OK to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One trick that sometimes helps for problems like this is to write $y = ax$ and then examine your expression. In this case you get ${1 \over \sqrt{1 + a^2}} \sin (a \sqrt{x})$. Then you can use that $|\sin(\theta)| \leq |\theta|$ for all $\theta$, to get that it is bounded in absolute value by ${a \over \sqrt{1 + a^2}} \sqrt{x}$. Since $\big|{a \over \sqrt{1 + a^2}}\big|$ is at most 1, your expression's absolute value is at most $\sqrt{x}$, which goes to zero as $(x,y)$ approaches the origin.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-15T01:15:15.853" LastActivityDate="2010-11-15T01:15:15.853" CommentCount="2" />
  <row Id="10333" PostTypeId="1" AcceptedAnswerId="10334" CreationDate="2010-11-15T01:02:34.673" Score="0" ViewCount="80" Body="&lt;p&gt;Does it make sense to say that there are more irrational numbers than rational?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-15T01:09:04.720" Title="Are there many more irrational numbers than rational?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10334" PostTypeId="2" ParentId="10333" CreationDate="2010-11-15T01:03:19.767" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, in the sense of cardinality. In the same sense that one can say that there are more real numbers than there are rationals (or integers; or natural numbers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational numbers are countable, but the irrationals are uncountable. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:09:04.720" LastActivityDate="2010-11-15T01:09:04.720" CommentCount="2" />
  <row Id="10335" PostTypeId="1" CreationDate="2010-11-15T01:12:24.920" Score="0" ViewCount="60" Body="&lt;p&gt;Hey guys, second post for today. I have a midterm on tuesday so I am just practicing some proofs from the textbook. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question states: &#xA;If $a \in \mathbb{Z}$, prove that $a^2$ is not congruent to 2 modulo 4 or 3 modulo 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our work:&#xA;It's an if then proof, so I am thinking we can do a contradiction proof. &#xA;Suppose $a^2 \equiv 2 \pmod{4}$ and $a^2 \equiv 3 \pmod{4}$. &#xA;Then you can write by definition &#xA;\begin{align*}&#xA;a^2 &amp;amp;= 4k + 2 \\&#xA; a^2 &amp;amp;= 4k + 3&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but hold on, another way to write 2 mod 4 is $[2]_{4}$ which means the infinite set of all the integers in the form of $2 + 4k$ and $[3]_{4}$ which means the infinite set of all the integers in the form of $3 + 4k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we one set is a set of even integers and one set is of odd integers. but our a^2 can't be in both right? &lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:18:32.807" LastActivityDate="2010-11-15T02:03:43.810" Title="Congruency proof" Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10336" PostTypeId="1" AcceptedAnswerId="10364" CreationDate="2010-11-15T01:14:54.623" Score="1" ViewCount="81" Body="&lt;p&gt;I figured out this order on $\omega^2$: elements of different columns are ordered by their column number; within column $m$, the order is $0 \gt 1 \gt 2 \gt \cdots \gt n$, where $n$ is the first such that $P(m,n)$ is true; then $0 \lt n+1 \lt n+2 \lt \cdots$. The column is an infinite descending sequence if there is no such $n$. This order is well-founded iff for all $m$ there is an $n$ such that $P(m,n)$ is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can reduce any $\Pi_{2}^{0}$ sentence to a computable ordering which is well-founded iff the sentence is true. I don't see how to do this for $\Sigma_{2}^{0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Wikipedia tells me Kleene's $O$ is $\Pi_{1}^{1}$ complete, so I'm still a long way away from what I should be able to do. Rather than being told what the full reduction is, if instead I could just get a hint for the $\Sigma_{2}^{0}$ case, then maybe I can pretend to myself that I figured &quot;most&quot; of it out on my own?&lt;/p&gt;&#xA;" OwnerUserId="3508" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-15T04:54:39.820" LastActivityDate="2010-11-15T04:54:39.820" Title="Hint on Kleene's O" Tags="&lt;algorithms&gt;&lt;logic&gt;&lt;computability&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10337" PostTypeId="1" AcceptedAnswerId="10513" CreationDate="2010-11-15T01:17:35.980" Score="3" ViewCount="254" Body="&lt;p&gt;The title is somewhat deceptive: I know that $|\mathbb{Q}|=|\mathbb{Z}|.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose I wanted to compare the sets, knowing that they are of the same cardinality but still wondering if there was some sense in which Q could be larger.  There are two usual ways of thinking about this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}\subset\mathbb{Q}$ but $\mathbb{Q}\not\subset\mathbb{Z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry, there's no \subsetneq symbol on math.stackexachange) and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'For any distinct integers m,n there are infinitely many rationals between m and n.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These methods both rely on the natural mapping between integers and rationals, to wit: $n\mapsto\frac n1$.  What if we didn't have this correspondence, or were working with an application where it wasn't sensible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we still have the usual order &amp;lt; on Q and Z, and the usual operations + and *, but we no longer have a reason to privilege the usual mapping over, say, $n\mapsto\frac2n$ or a diagonal mapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm much less concerned about the 'answer' part of the answers than I am of the 'question' part: the main point is 'how could this question be made precise?' rather than answering such question.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-11-16T04:08:33.300" Title="Are there many more rational numbers than integers?" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10338" PostTypeId="2" ParentId="10322" CreationDate="2010-11-15T01:18:56.493" Score="5" ViewCount="0" Body="&lt;p&gt;Your proof is correct, although somewhat long-winded. It's a good practice, once you have a proof, to go back and see if you can simplify it as much as possible, aiming to find the most direct path to the desired result. In the process, one often gains a better understanding of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you ended up using the fact that $\sin \theta \le \theta$ for all $\theta &amp;gt; 0$, let's see if we can use this from the outset. Since the function is odd with respect to $y$, it suffices to only consider $y \ge 0$ as you noticed. Then we have $$\frac{x\sin\frac{y}{\sqrt{x}}}{\sqrt{x^2+y^2}} \le \frac{y\sqrt{x}}{\sqrt{x^2+y^2}} \le \sqrt{x}$$&#xD;&#xA;because $y \le \sqrt{x^2+y^2}$. In other words, the function is positive and bounded above by $\sqrt{x}$. From here, you can quite easily get to where you want.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-15T01:18:56.493" />
  <row Id="10339" PostTypeId="2" ParentId="10335" CreationDate="2010-11-15T01:22:10.603" Score="0" ViewCount="0" Body="&lt;p&gt;You cannot assume &lt;em&gt;both&lt;/em&gt; that $a^2\equiv 2\pmod{4}$ *and* that $a^2\equiv 3\pmod{4}$. Your &quot;assumption&quot; should be &quot;either $a^2\equiv 2\pmod{4}$ *OR* $a^2\equiv 3\pmod{4}$.&quot; That is, you need to do it by cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This because your assumption for a proof by contradiction should be &lt;em&gt;exactly&lt;/em&gt; the negation of your conclusion, not something else.  Your conclusion here is &#xA;$$\neg\left( a^2\equiv 2\pmod{4} \text{ or } a^2\equiv 3\pmod{4}\right)$$&#xD;&#xA;so the negation is just the statement that &lt;strong&gt;either&lt;/strong&gt; $a^2$ is $2$ *&lt;em&gt;or&lt;/em&gt;* $3$ modulo $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, no; you're not doing it right here, because your &quot;assumption for contradiction&quot; is not the correct one.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:27:33.450" LastActivityDate="2010-11-15T01:27:33.450" CommentCount="4" />
  <row Id="10340" PostTypeId="2" ParentId="10335" CreationDate="2010-11-15T01:22:17.457" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: There are only 4 possible choices for $a$ modulo $4$. You can explicitly look at each one and calculate $a^2$ modulo $4$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-15T01:22:17.457" CommentCount="4" />
  <row Id="10341" PostTypeId="2" ParentId="10337" CreationDate="2010-11-15T01:40:36.537" Score="7" ViewCount="0" Body="&lt;p&gt;The two statements you give relate to two different issues; the first relates to the fact that there is a natural embedding of $\mathbb{Z}$ in $\mathbb{Q}$ arising from their constructions, and it makes sense to ask whether $\mathbb{Z}$ is a proper subset of $\mathbb{Q}$. This is certainly a sensible question whenever we have two sets $A$ and $B$ and $A$ is &lt;em&gt;given&lt;/em&gt; as a subset of $B$. For arbitrary sets it does not make sense, as your comments about &quot;other mappings&quot; show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement has to do with &lt;em&gt;order&lt;/em&gt;, but again with the &lt;em&gt;natural&lt;/em&gt; or &lt;em&gt;canonical&lt;/em&gt; ordering that both $\mathbb{Z}$ and $\mathbb{Q}$ inherit from the order in $\mathbb{N}$ (which is defined in terms of the successor function). The order of $\mathbb{Q}$ is dense, while the order of $\mathbb{Z}$ is not: $\mathbb{Z}$ is order-isomorphic to subsets of $\mathbb{Q}$, but not the other way around. If you &quot;order&quot; the linearly ordered sets by saying that $(P,\leq)$ is smaller than or equal to $(Q,\preceq)$ if and only if there is an order-preserving injection $P\to Q$, then you get equivalence classes of linearly ordered sets (much in the same way as with cardinals), and in &lt;em&gt;that&lt;/em&gt; way of ordering linearly ordered sets, $\mathbb{Z}$ with its natural ordering &lt;em&gt;is&lt;/em&gt; strictly smaller than $\mathbb{Q}$, also with its natural ordering. But if you think there is nothing special about the usual embedding $\mathbb{Z}\hookrightarrow \mathbb{Q}$, then presumably you also think there is nothing special about the natural ordering of $\mathbb{Z}$ or $\mathbb{Q}$, and then you can give them orderings that will make it come out &quot;the other way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can certainly make the question &quot;precise&quot; if you specify more information (&quot;as ordered sets with their usual ordering&quot;; &quot;as subsets of $\mathbb{R}$&quot;; etc). But that extra context needs to be explicit to make it make sense.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T01:46:07.263" LastActivityDate="2010-11-15T01:46:07.263" CommentCount="9" />
  <row Id="10342" PostTypeId="1" AcceptedAnswerId="10391" CreationDate="2010-11-15T02:02:04.803" Score="3" ViewCount="65" Body="&lt;p&gt;Consider a $G$-equivariant map $\pi:X\to Y$ for $G$ an affine algebraic group, such that $\pi$ is a good categorical quotient.  Is there any relationship between $H^*_G(X)$ and $H^*(Y)$?  Is there if $\pi$ is a good geometric quotient, or if the quotient space is smooth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A categorical quotient is an equivariant map $\pi:X\to Y$ that is constant on $G$-orbits.  It's good if the topology on $Y$ is induced by $X$ ($\pi$ is a surjective open submersion) and the map from the functions on any affine $V \subset Y$ to $G$-invariant functions on $\pi^{-1}(V)$ is an isomorphism.  It's a good geometric quotient if the $G$-orbits closed in $Y$.&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="3008" LastEditorDisplayName="" LastEditDate="2010-11-15T14:45:42.750" LastActivityDate="2010-11-15T14:45:42.750" Title="Quotient spaces and equivariant cohomology" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10343" PostTypeId="2" ParentId="10335" CreationDate="2010-11-15T02:03:43.810" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad (2\:n)^2\ =\ 4\:n^2,\quad (2\:n+1)^2\ =\ 4\:(n^2+n)+1\ \ $ so squares are $\rm\equiv 0,\: 1\ (mod\ 4)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T02:03:43.810" />
  <row Id="10344" PostTypeId="2" ParentId="10327" CreationDate="2010-11-15T02:10:41.477" Score="12" ViewCount="0" Body="&lt;p&gt;No, this number is transcendental. The proof by Mahler mentioned in a comment shows this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference to learn about basic transcendental number theory is the book &quot;Making transcendence transparent: an intuitive approach to classical transcendental number theory&quot;, by Edward Burger and Robert Tubbs, Springer-Verlag (2004).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chapter 1 of the book the proof of the transcendence of Mahler's constant $0.1234\dots$ is discussed. The idea is to show that the &quot;obvious&quot; rational approximations actually are very very close, to the point that they would contradict easy estimates (due to Liouville) for how quickly rational numbers can approximate irrational algebraic numbers. The Wikipedia entry on &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville_number&quot; rel=&quot;nofollow&quot;&gt;Liouville numbers&lt;/a&gt; discusses Liouville's approximation theorem and related results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha$ is algebraic of degree $d\ge 2$ then there is a constant $C$ such that for any rational $p/q$ with $q&amp;gt;0$, we have $$ \left|\alpha-\frac pq\right|&amp;gt;\frac{C}{q^d}. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there is a bit of work needed here. The estimates the book discusses together with a &lt;em&gt;strengthening&lt;/em&gt; of Liouville's theorem give the proof for Mahler's constant, and the same argument works for the number you are asking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem&quot; rel=&quot;nofollow&quot;&gt;strengthening&lt;/a&gt; we need is due to Klaus Roth in 1955, and he was awarded the Fields medal in 1958 for this result.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-15T02:10:41.477" />
  <row Id="10345" PostTypeId="2" ParentId="10329" CreationDate="2010-11-15T02:16:51.390" Score="1" ViewCount="0" Body="&lt;p&gt;You are correct that it is the same as saying that $T$ is not bijective, because it follows from the open mapping theorem that a bounded operator on a Banach space has a bounded inverse if it is bijective.  However, more straightforward answers for your question can be given without explicitly thinking in these terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show that if $T$ is compact and $S$ is bounded, then $ST$ is compact.  If $T$ were invertible, this would imply that $I=T^{-1}T$ is compact.  This in turn translates to saying that the closed unit ball of $X$ is compact.  One way to see that this is impossible in the infinite dimensional case is implicit in &lt;a href=&quot;http://math.stackexchange.com/questions/4815/fitting-an-infinite-collection-of-balls-in-an-infinite-dimensional-unit-ball&quot;&gt;this question&lt;/a&gt;.  There is an infinite sequence of points in the unit ball whose pairwise distances are bounded below, no subsequence of which is Cauchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compact operators on infinite dimensional spaces can be injective, but they can never be surjective.  The closed subspaces of the range of a compact operator are finite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what Qiaochu said in his comment above.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-15T02:16:51.390" />
  <row Id="10346" PostTypeId="1" AcceptedAnswerId="10605" CreationDate="2010-11-15T02:31:51.857" Score="7" ViewCount="178" Body="&lt;p&gt;Combining the answers given by me and Ralth to the probability question at &lt;a href=&quot;http://math.stackexchange.com/questions/10284/probability-question&quot;&gt;Probability Question&lt;/a&gt;, we get the following identity:&#xA;$$&#xD;&#xA;\sum\limits_{k = m}^n {{n \choose k}p^k (1 - p)^{n - k} {k \choose m} p_j^m (1 - p_j )^{k - m} } = {n \choose m} (p p_j)^m (1 - p p_j)^{n-m},&#xD;&#xA;$$&#xD;&#xA;where $m \geq 0$ and $n \geq m$ are arbitrary integers, and $p$ and $p_j$ are arbitrary probabilities (in Ralth's notation $m$ is $k$, $p$ is $p_s$, and $p_j$ is $p_h$). Can you prove this identity directly (i.e., in a non-probabilistic setting)? Whether you'll find this identity interesting or not, at least Ralth's answer may now gain its due recognition.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-15T07:34:18.140" LastActivityDate="2010-11-16T21:17:10.470" Title="Non-probabilistic proofs of a binomial coefficient identity from a probability question" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;algebra&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" />
  <row Id="10347" PostTypeId="1" AcceptedAnswerId="10357" CreationDate="2010-11-15T02:40:01.817" Score="0" ViewCount="136" Body="&lt;p&gt;How would I go about creating a polynomial $f$ such that&lt;br&gt;&#xA;\begin{align*}&#xA;f(x) &amp;amp;= A_1\\&lt;br&gt;&#xA;f(y) &amp;amp;= A_2\\&lt;br&gt;&#xA;f'(x) &amp;amp;= B_1\\&lt;br&gt;&#xA;f'(y) &amp;amp;= B_2&#xA;\end{align*}&lt;br&gt;&#xA;With $x$ not equal to $y$ and $A_1$, $A_2$, $B_1$, $B_2$ real numbers?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T04:08:40.783" LastActivityDate="2010-11-15T06:15:30.360" Title="Create a function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="21" />
  <row Id="10348" PostTypeId="1" AcceptedAnswerId="10354" CreationDate="2010-11-15T03:05:08.443" Score="2" ViewCount="81" Body="&lt;p&gt;I have a finite alphabet A and a set S of $n$ ordered pairs $(a, b)$ with $a,b\in A.$  I want to find out how many sequences $(a_1, b_1),\ldots,(a_n,b_n)$ there are where adjacent elements have either $a_{k-1}=a_k$ or $b_{k-1}=b_k.$ (Since S is a set, they can't have both.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how to go about finding this?  Formulas, generating functions, algorithms, or whatever you have.  In my particular application I have $|A|=10$ and $|S|=21$, but the question seems general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in my application $(a_1,b_1)$ and $(a_k,b_k)$ are &lt;em&gt;not&lt;/em&gt; adjacent. (They can have, but need not have, one member different.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For clarification, I'm &lt;em&gt;not&lt;/em&gt; allowing all elements from $A\times A$, just those in $S$.  No valid answer can ignore S.  For example, I might have S = {(1, 1), (3, 3)} in which case there are no valid orderings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those interested in using a matrix approach (not sure if this can work; the path needs to go through each pair exactly once), here's the adjacency matrix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;[0 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0]&#xA;[1 0 1 1 1 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0]&#xA;[1 1 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1]&#xA;[1 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0]&#xA;[0 1 0 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0]&#xA;[0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0]&#xA;[1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0]&#xA;[0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1]&#xA;[1 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 0 0]&#xA;[0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 0 0]&#xA;[0 0 1 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 1]&#xA;[0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 1 0 0]&#xA;[0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0]&#xA;[1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0]&#xA;[0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1]&#xA;[1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 1 0 0 0]&#xA;[0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0]&#xA;[0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0]&#xA;[0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0]&#xA;[0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0]&#xA;[0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0]&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Kudos to anyone who recognizes it (and thus my goal).&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T17:39:51.543" LastActivityDate="2010-11-15T20:02:06.317" Title="Sequences of pairs where adjacent elements differ in exactly one component" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;graph-theory&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="10349" PostTypeId="1" AcceptedAnswerId="10356" CreationDate="2010-11-15T03:13:38.293" Score="1" ViewCount="64" Body="&lt;p&gt;I have trouble understanding an argument in the proof of the Kummer-Dedekind theorem.&#xA;I am referring to a proof given in Peter Stevenhagen's notes.&#xA;&lt;a href=&quot;http://websites.math.leidenuniv.nl/algebra/ant.pdf&quot; rel=&quot;nofollow&quot;&gt;http://websites.math.leidenuniv.nl/algebra/ant.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is theorem 3.1 on page 27. For the proof of the second part, the first statement says that since $p_i$ contains $pR$, it is invertible. I am not sure what the author is invoking here. For every ideal contains a principal ideal, but that is not sufficient for invertibility. Also, the fact that $p_i$ contains $pR$ should follow from the first part of the theorem since $p_i=pR+g_i(\alpha)R$. So I don't see what fact is used in proving invertibility of $p_i$ based on its containment of $pR$. Any help would be very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T17:20:33.813" LastActivityDate="2010-11-15T17:20:33.813" Title="Invertibility of prime ideals in a number ring lying over prime numbers" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;commutative-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="10350" PostTypeId="2" ParentId="10346" CreationDate="2010-11-15T03:20:00.787" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the following two-step process: Start with $n$ elements. Pick each one with probability $p$. Among the $k$ elements chosen in the first phase, pick each one with probability $p_j$. The left-hand side describes the probability that the final set is of size $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also calculate this probability directly: every element is chosen to the final set with probability $pp_j$, whence the right-hand side.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-15T03:20:00.787" CommentCount="2" />
  <row Id="10351" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T03:24:48.350" Score="-1" ViewCount="0" Body="&lt;p&gt;Perhaps I'm misunderstanding your problem. But if you want to count how many sequences of pairs there are which where each two conscutive pairs have at least one element in common, the answer seems easy. Just count how many sequences there are that have not that property, and substract from the total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get, then, $a^{2n} - a^2 (a-1)^{2(n-1)} = a^2 [ a^{2(n-1)} - (a-1)^{2(n-1)}]$ (where $a = |A|$)&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-11-15T03:30:34.790" LastActivityDate="2010-11-15T03:30:34.790" CommentCount="4" />
  <row Id="10352" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T03:28:33.933" Score="0" ViewCount="0" Body="&lt;p&gt;EDIT: This answer assumes that $S = A \times A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a pair $(a,b)$, there are $2|A|-1$ pairs $(a&amp;#39;,b&amp;#39;)$ such that $a=a&amp;#39;$ or $b=b&amp;#39;$ (or both). Therefore if we don't care about the relation between first and last element, we get $|A|^2(2|A|-1)^{n-1}$ sequences.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-15T20:02:06.317" LastActivityDate="2010-11-15T20:02:06.317" CommentCount="1" />
  <row Id="10353" PostTypeId="1" AcceptedAnswerId="10386" CreationDate="2010-11-15T03:31:58.640" Score="1" ViewCount="85" Body="&lt;p&gt;Two vectors $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ are orthogonal if their scalar product is zero, i.e. $x_1 y_1 + x_2 y_2=0$. What surface does this equation describe in the four-dimensional $(x_1,x_2,y_1,y_2)$ space? How about the orthogonality condition for $3$-dimensional (or higher dimensional) vectors? What is the corresponding surface and how would one visualize it?&#xA;Thanks for the answers.&lt;/p&gt;&#xA;" OwnerUserId="3511" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T04:05:09.603" LastActivityDate="2010-11-15T22:35:45.837" Title="Surface described by orthogonality condition for vectors" Tags="&lt;vector-calculus&gt;" AnswerCount="2" />
  <row Id="10354" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T03:35:36.640" Score="4" ViewCount="0" Body="&lt;p&gt;Let $m = \lvert A\rvert$, and denote by $L_{m,m}$ what is variously called the $m\times m$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Rook%27s_graph&quot; rel=&quot;nofollow&quot;&gt;rook's graph&lt;/a&gt; or &lt;a href=&quot;http://mathworld.wolfram.com/LatticeGraph.html&quot; rel=&quot;nofollow&quot;&gt;lattice graph&lt;/a&gt;: it has $m\times m$ vertices arranged in a square, with an edge connecting two vertices if they are in the same row or column. There is a natural mapping between the vertices of $L_{m,m}$ and ordered pairs of elements from $A$, of which $S$ is a subset. Then a sequence of length $n$ in your problem is equivalent to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian path&lt;/a&gt; in the subgraph of $L_{m,m}$ induced by $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of such paths will certainly depend on the elements of $S$, and not just their cardinality. For example, let $m = 2$ and $n = \lvert S\rvert = 2$; if the two elements of $S$ are adjacent, then there is one Hamiltonian path, otherwise none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest algorithm I can think of is to simply generate all permutations of the $n$ vertices and check whether each one is a valid path. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path_problem&quot; rel=&quot;nofollow&quot;&gt;Finding a &lt;em&gt;single&lt;/em&gt; Hamiltonian path is NP-complete&lt;/a&gt;, so I'm not sure you can do much better. You could do a search for &quot;count number of hamiltonian paths&quot; and see what you get.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-15T18:25:46.680" LastActivityDate="2010-11-15T18:25:46.680" CommentCount="8" />
  <row Id="10355" PostTypeId="2" ParentId="10347" CreationDate="2010-11-15T03:38:07.267" Score="3" ViewCount="0" Body="&lt;p&gt;This would be a typical question given in the beginning of a linear algebra class. You have the information that $f(x)$ passes through the points $(X,A_1)$ and $(Y,A_2)$ and $f&amp;#39;(x)$ passes through the points $(X, B_1)$ and $(Y,B_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have a set of 4 linear equations for 4 unknowns. Try setting up a linear system and solving it.&#xA;Hint: The first condition could be written as&#xA;$$[X^3 \;\; X^2 \;\; X \;\;\; 1] \left[\begin{array}{c}a\\ b\\ c\\ d\\\end{array} \right] =A_1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you put together all 4 of your equations you end up with an augmented matrix which you can put in RREF form to find your answer&lt;/p&gt;&#xA;" OwnerUserId="2596" LastEditorUserId="2596" LastEditorDisplayName="" LastEditDate="2010-11-15T03:44:14.393" LastActivityDate="2010-11-15T03:44:14.393" CommentCount="1" />
  <row Id="10356" PostTypeId="2" ParentId="10349" CreationDate="2010-11-15T03:42:12.127" Score="1" ViewCount="0" Body="&lt;p&gt;Principal ideals are invertible, and so too are divisors of invertible ideals, so  $\mathfrak{p}\:|\:pR\ \Rightarrow\ \mathfrak{p}\ $ invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the ring in the theorem is not necessarily a a Dedekind domain so it need not be true that $\ I \supset J\ \Rightarrow\ I|J$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-15T03:50:20.213" LastActivityDate="2010-11-15T03:50:20.213" CommentCount="6" />
  <row Id="10357" PostTypeId="2" ParentId="10347" CreationDate="2010-11-15T04:16:13.943" Score="5" ViewCount="0" Body="&lt;p&gt;In general, there are many possible answers. But, suppose your polynomial looks like this:&#xA;$$f(X) = a_n X^n + \cdots + a_1 X + a_0$$&#xD;&#xA;with $a_0,\ldots,a_n$ real numbers. Then &#xA;$$f&amp;#39;(X) = na_n X^n + \cdots + a_1.$$&#xD;&#xA;Evaluating at $x$ and at $y$ will give you a bunch of equations that they need to satisfy:&#xA;\begin{align*}&#xA;a_n x^n + \cdots + a_1x + a_0 &amp;amp; = A_1\\&#xA;na_n x^{n-1} + \cdots + a_1 &amp;amp;= B_1\\&#xA;a_n y^n + \cdots  a_1y + a_0 &amp;amp;= A_2\\&#xA;na_n y^{n-1} + \cdots a_1 &amp;amp;= B_2.&#xA;\end{align*}&#xA;You cannot really to get an answer if the degree of the polynomial is $0$ or $1$: too many equations with too few unknowns. Degree $2$ would give you four equations with three unknowns, which may be too restrictive as well. So your best hope of having it be solvable and not too hard is for the degree (for $n$) to be $3$. Then you'll have a bunch of equations like so:&#xA;$$\begin{array}{rcrcrcrcl}&#xD;&#xA;a_3 x^3 &amp;amp;+&amp;amp; a_2 x^2 &amp;amp;+&amp;amp; a_1 x &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; A_1\\&#xD;&#xA;a_3 y^3 &amp;amp;+&amp;amp; a_2 y^2 &amp;amp;+&amp;amp; a_1 y &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; A_2\\&#xD;&#xA;3a_3 x^2 &amp;amp;+&amp;amp; 2a_2 x &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; B_1\\&#xD;&#xA;3a_3 y^2 &amp;amp;+&amp;amp; 2a_2 y &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; B_2.&#xD;&#xA;\end{array}$$&#xD;&#xA;So you go ahead and solve this system in one of the standard ways; for example, back substitution if you don't know the methods from linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; How to solve these if you don't know linear algebra? The less sophisticated method is probably to &quot;solve and backsubstitute&quot;. Using the last equation, you can write $a_1$ using $y$, $a_2$, $a_3$, and $B_2$. Substitute that value into each of the previous three equations, and then each of those equations only has three unknowns. Then use one of them to write $a_2$ in terms of the other two unknowns, and substitute that into the remaining two equations. Then use one of them to &quot;solve&quot; for one unknown in terms of the other, and substitute into the remaining one. Solve for the unknown, and then work backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose that $x=2$ and $y=3$, $A_1 = -2$, $A_2=1$, $B_1 = 0$, and $B_2 = 1$ (I just picked those at random). So the equations become:&#xA;$$\begin{array}{rcrcrcrcl}&#xD;&#xA;8a_3 &amp;amp;+&amp;amp; 4a_2  &amp;amp;+&amp;amp; 2a_1  &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; -2\\&#xD;&#xA;27a_3 &amp;amp;+&amp;amp; 9a_2  &amp;amp;+&amp;amp; 3a_1  &amp;amp;+&amp;amp; a_0 &amp;amp;=&amp;amp; 1\\&#xD;&#xA;12a_3  &amp;amp;+&amp;amp; 4a_2  &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; 0\\&#xD;&#xA;27a_3  &amp;amp;+&amp;amp; 6a_2  &amp;amp;+&amp;amp; a_1   &amp;amp; &amp;amp;     &amp;amp;=&amp;amp; 1.&#xD;&#xA;\end{array}$$&#xD;&#xA;From the last equation, you have that $a_1 = 1 - 27a_3 - 6a_2$. Plugging that into each of the other three equations you get:&#xA;$$\begin{array}{rcrcrcl}&#xD;&#xA;-36a_3 &amp;amp; - &amp;amp; 8a_2 &amp;amp; + &amp;amp; a_0 &amp;amp;=&amp;amp; -4\\&#xD;&#xA;-54a_3 &amp;amp; - &amp;amp; 9a_2 &amp;amp; + &amp;amp; a_0 &amp;amp; = &amp;amp; -2\\&#xD;&#xA;-15a_3 &amp;amp; - &amp;amp; 2a_2 &amp;amp;   &amp;amp;     &amp;amp; = &amp;amp; -1.&#xD;&#xA;\end{array}$$&#xD;&#xA;Then we can use the first equation to write $a_0 = 36a_3 + 8a_2 - 4$, and substituting into the second equation we get&#xA;$$\begin{array}{rcrcl}&#xD;&#xA;-18a_3 &amp;amp; - &amp;amp; a_2 &amp;amp; = &amp;amp; 2\\&#xD;&#xA;-15a_3 &amp;amp; - &amp;amp; 2a_2 &amp;amp; = &amp;amp; -1.&#xD;&#xA;\end{array}$$&#xD;&#xA;From the first equation you get $a_2 = -18a_3-2$, which plugged into the second equation gives $21a_3=-3$, or $a_3 = -\frac{1}{7}$. This gives you the value for $a_2$, since you know $a_2 = -18a_3-2$; once you have $a_2$ and $a_3$, you use $a_0 = 36a_3+8a_2-4$ to find $a_0$; once you have $a_0$, $a_2$, and $a_3$, you use $a1 = 1 - 27a_3 - 6a_2$ to find $a_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the back-substitution in any order; this is &lt;em&gt;not&lt;/em&gt; the best way of doing it, but it's certainly &lt;em&gt;a&lt;/em&gt; way that may have occurred to you if you have ever solved a system of two equations with two unknowns. It's called &quot;back-substitution&quot; because you use one equation to isolate a variable, then you &quot;substitute it&quot; into the equations that are &quot;at the bacK&quot;, the other equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to do this for more equations with more unknowns quickly becomes far too annoying, which is why there are better methods, such as Gaussian Elimination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; If, as your comments in the question suggest, you are supposed to come up with &lt;em&gt;generic&lt;/em&gt; formulas for arbitrary numbers $A_1, A_2,B_1,B_2$, $x$, and $y$, then this method is difficult, though you &lt;em&gt;can&lt;/em&gt; push it through if you are careful and are really comfortable with algebra. You have to remember that $x$, $y$, $A_1$, $A_2$, $B_1$, and $B_2$ are supposed to be &lt;em&gt;numbers&lt;/em&gt;, not variables, so if you can write $a_i$ in terms only of these, then you've &quot;solved&quot; $a_i$. Your final answer would express $a_0$, $a_1$, $a_2$, and $a_3$ in terms of $A_1$, $A_2$, $B_1$, $B_2$, $x$, and $y$. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 3:&lt;/strong&gt; Your comments &lt;em&gt;clearly&lt;/em&gt; say now that you are trying to come up with formulas that you can just plug $x$, $y$, $A_1$, $A_2$, $B_1$, and $B_2$ into to get your polynomial. &lt;em&gt;You are not going to be able to come up with them using the methods discussed so far.&lt;/em&gt; Not because it's impossible, but  because you are not exhibiting the kind of comfort with algebra that you would need for that. It would involve carrying around fairly ugly expressions around in a comfortable way, and it seems clear you cannot do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's try a slightly different tack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start slow. A simple way to get a polynomial that will give you $A_1$ when evaluated at $x$ and $A_2$ when evaluated at $y$ is to &quot;cheat&quot;: make it a sum of two polynomials, one zero at $x$ and the other zero at $y$, and then just adjust things so they come out right. So you can take, for example, something like&#xA;$$f(X) = K(X-x) + M(X-y).$$&#xD;&#xA;Then $f(x) = M(x-y)$ and $f(y)=K(y-x)$. So you get easy formulas for $M$ and $K$: since $M(x-y)=A_1$, then $M = A_1/(x-y)$ (which you can do because $x\neq y$); and since $K(y-x) = A_2$, then $K=A_2/(y-x)$. So we get&#xA;$$f(X) = \frac{A_2}{y-x}(X-x) + \frac{A_1}{x-y}(X-y).$$&#xD;&#xA;Now, this is not going to do it for the derivatives, though. So we are going to have to add things to this polynomial to get the derivatives to come out right. How can we add things so as not to mess up what we've already achieved (that the first two equations are satisfied)? Well, we can make sure we don't mess up what we already have if we make sure that whatever we add has value $0$ at both $x$ and $y$. So we'll try something like:&#xA;$$f(X) = \frac{A_2}{y-x}(X-x) + \frac{A_1}{x-y}(X-y) + (X-x)(X-y)g(X)$$&#xD;&#xA;for some $g(X)$. What is the derivative of such an $f(X)$? Remembering that $x$, $y$, $A_1$, and $A_2$ are &lt;em&gt;numbers&lt;/em&gt;, then applying the product rule we have:&#xA;$$f&amp;#39;(X) = \frac{A_2}{y-x} + \frac{A_1}{x-y} + (X-y)g(X) + (X-x)g(X) + (X-x)(X-y)g&amp;#39;(X).$$&#xD;&#xA;If you evaluate at $x$, we get&#xA;$$B_1 = f&amp;#39;(x) = \frac{A_2}{y-x}+\frac{A_1}{x-y}+(x-y)g(x).$$&#xD;&#xA;Everything is numbers that we'll know, except for $g(x)$. So that means that we need&#xA;$$g(x) = \frac{B_1}{x-y} - \frac{A_2}{(y-x)(x-y)} - \frac{A_1}{(x-y)^2}.$$&#xD;&#xA;And if we evaluate $f&amp;#39;(X)$ at $y$ we get&#xA;$$B_2 = \frac{A_2}{y-x} + \frac{A_1}{x-y} + (y-x)g(y).$$&#xD;&#xA;Again, everything except $g(y)$ are numbers we already are supposed to know. So we get:&#xA;$$g(y) = \frac{B_2}{y-x} - \frac{A_2}{(y-x)^2} - \frac{A_1}{(x-y)(y-x)}.$$&#xD;&#xA;Notice that the derivative of $g(X)$ turns out to not play any role at all after all! So we can make $g(X)$ as simple as we possibly can; say, a linear polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we achieve this? Well, we discussed above how to make a polynomial of degree $1$ that has specific values at two different points, right? So you can just use the same trick we used for $f(x)$ and $f(y)$, but with $g(x)$ and $g(y)$, using the (complicated) expressions we just found for $g(x)$ and for $g(y)$ instead of the $A_1$ and $A_2$ that we used for $f(X)$. So go to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; You can keep playing with this if you have still more points-and-values; it gets progresssively more difficult. You would need to start with polynomials of degree $n-1$ if you know the value at $n$ points; this leads you to Lagrange Interpolation polynomials. The advantage of the linear algebra method is that the complexity grows in a bit more manageable fashion than this method.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T06:15:30.360" LastActivityDate="2010-11-15T06:15:30.360" CommentCount="8" />
  <row Id="10358" PostTypeId="1" AcceptedAnswerId="10361" CreationDate="2010-11-15T04:17:30.847" Score="2" ViewCount="49" Body="&lt;p&gt;There are two equivalent definitions of a locally convex topological vector space. Note that since the vector space is topological, addition by a fixed vector x and multiplication by a fixed scalar r are both homeomophisms, which means it is enough to give a neighborhood basis at the vector 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first definition is that $0$ has a local base consisting of open sets that are &lt;em&gt;convex&lt;/em&gt; (if $x$ and $y$ are in $S$, then $\lambda x+(1-\lambda)y$ is also in $S$ for every $\lambda\in[0,1]$), &lt;em&gt;balanced&lt;/em&gt;, (if $x\in S$, then $r x\in S$ for every $r\in\mathbb{F}$ with $|r|=1$, where $\mathbb F$ is either the real or complex numbers), and &lt;em&gt;absorbing&lt;/em&gt; (for any vector $y$ there exists a $\lambda\in (0,\infty)$ such that $\lambda S$ contains $y$, i.e. $S$ absorbs $y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second definition is that $0$ has a local base given by the ''balls'' of radius $r$ for each $r$ of each semi-norm in some fixed collection of semi-norms (a semi-norm is just like a norm except that non-zero vectors can have a non-zero semi-norm). The two definitions are equivalent in one direction because &quot;balls&quot; of semi-norms are convex, balanced and absorbing, and in the other because we can define for any convex, balanced, absorbing set $S$ the semi-norm $f_S$ by assigning to $x$ the &quot;least&quot; (infimum) $\lambda$ for which $x$ is in $\lambda S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what are some examples of topological vector spaces for which $0$ does have a local base of convex open sets, but that no local base consists of convex sets that are also balanced and absorbing?&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-11-15T04:47:05.207" LastActivityDate="2010-11-15T04:47:05.207" Title="What is an example of a topological vector space which contains a non-absorbent or non-balanced open convex set around the origin?" Tags="&lt;functional-analysis&gt;" AnswerCount="2" />
  <row Id="10359" PostTypeId="2" ParentId="10358" CreationDate="2010-11-15T04:21:07.977" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; &lt;em&gt;As Matt notes in a comment below, the following does not really work!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a vector space $V$ of dimension at least $2$, pick a non-zero proper subspace $W\subseteq V$, and consider the unique linear topology on $V$ for which $\{W\}$ is a basis of neighborhoods of $0\in V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an Hausdorff example, let now $V$ be an infinite dimensional vector space, and consider a decreasing sequence $(W_n)_{n\geq1}$ of non-zero subspaces of $V$ such that $\bigcap_{n\geq1}W_n=0$. Now put on $V$ the unique linear topology which has $\{V_n\}_{n\geq1}$ as a basis of neighborhoods of $0\in V$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-11-15T04:39:28.797" LastActivityDate="2010-11-15T04:39:28.797" CommentCount="2" />
  <row Id="10360" PostTypeId="1" CreationDate="2010-11-15T04:30:00.103" Score="-2" ViewCount="151" Body="&lt;p&gt;If a student does badly in a professors course, but has to take another course with him.....would the professor be biased?&lt;/p&gt;&#xA;" OwnerUserId="3513" LastActivityDate="2010-11-15T05:40:02.717" ClosedDate="2010-11-15T05:53:51.370" Title="Courses and professors" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10361" PostTypeId="2" ParentId="10358" CreationDate="2010-11-15T04:39:53.527" Score="3" ViewCount="0" Body="&lt;p&gt;Unless I am misunderstanding the situation, there is no such example.  That is, if $V$ is a topological vector space (over $\mathbb R$ or $\mathbb C$) which has a basis of neighbourhoods of the origin consisting of convex sets,  then it also has a basis of neighbourhoods of the origin consisting of balanced convex sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any neighbourhood of $0$ is automatically absorbent (because of continuity of scalar multiplication).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reference is Robertson and Roberston, &lt;em&gt;Topological Vector Spaces&lt;/em&gt; (Cambridge Tracts in Mathematics and Mathematical Physics, 53), which is a kind of summary of Bourbaki's Topological Vector Spaces volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is not hard: one first uses continuity of scalar multiplication to show&#xA;that any neighbourhood of $0$ contains a balanced neighbourhood.  Now if $U$ is a convex n.h of $0$, and $W$ is a balanced n.h. of $0$ contained in $U$, then $V:= \cap_{|x| \geq 1} x U$ contains $W$, so is a n.h., is convex (being the intersection of convex sets), is balanced (by construction), and lies in $U$ (again by construction --- consider $x  = 1$).  Thus $V$ is a convex balanced n.h. of $0$ contained in $U$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-15T04:39:53.527" />
  <row Id="10362" PostTypeId="1" AcceptedAnswerId="10363" CreationDate="2010-11-15T04:42:46.957" Score="2" ViewCount="116" Body="&lt;p&gt;I'm going through an examples section (on improper integrals) but I got lost at this bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{t\to-\infty} \frac{t}{e^{-t}} = \lim_{t\to-\infty}\frac{1}{-e^{-t}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's a simple algebra trick but I don't see how the right hand side came to be. How did it become $1/-e^{-t}$?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-15T05:42:31.053" LastActivityDate="2010-11-15T17:19:52.387" Title="Help seeing algebra in calculus" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10363" PostTypeId="2" ParentId="10362" CreationDate="2010-11-15T04:45:05.340" Score="4" ViewCount="0" Body="&lt;p&gt;It is an application of &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule&quot; rel=&quot;nofollow&quot;&gt;l'Hôpital's rule&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-15T04:45:05.340" CommentCount="1" />
  <row Id="10364" PostTypeId="2" ParentId="10336" CreationDate="2010-11-15T04:53:17.637" Score="1" ViewCount="0" Body="&lt;p&gt;The &quot;usual proof&quot; I know for $\Sigma^0_2$ is the same as for general $\Pi^1_1$ formulas. For $\Sigma^0_2$ formulas in particular, this is one possible method: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take the negation of your original formula to get a $\Pi^0_2$ formula. Construct a partial order on a subset of $\omega$ such that the partial order has an infinite descending sequence if and only if the original formula is false.&lt;/li&gt;&#xA;&lt;li&gt;Find a way to turn this partial order into a linear order while keeping the relevant properties.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is intentionally just a hint. The usual method by which these things are proved is not particularly obvious because it relies on a clever relationship between computable linear orders and computable subtrees of $\omega^{&amp;lt;\omega}$ (this relationship goes by the name &quot;Kleene-Brouwer order&quot;). &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-15T04:53:17.637" CommentCount="1" />
  <row Id="10365" PostTypeId="2" ParentId="10362" CreationDate="2010-11-15T04:57:12.420" Score="1" ViewCount="0" Body="&lt;p&gt;The ratio of functions which you provide are an example of an indeterminate form at the limit point. L'Hospital's rule allows you to differentiate the numerator and denominator independently and then take the limit, giving the intended limit. For your example,&#xA;\begin{eqnarray}&#xA;\lim_{t \to -\infty} \frac{t}{e^{-t}} \stackrel{L.'H.}{=} \lim_{t \to -\infty} \frac{1}{-e^{-t}} = 0.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T17:19:52.387" LastActivityDate="2010-11-15T17:19:52.387" CommentCount="2" />
  <row Id="10367" PostTypeId="2" ParentId="10346" CreationDate="2010-11-15T05:22:55.737" Score="2" ViewCount="0" Body="&lt;p&gt;Writing the left-hand side as&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^{n - m} {{n \choose k + m} p^{k + m} (1 - p)^{n - (k + m)} {k + m \choose m} p_j^m (1 - p_j )^{k + m - m} } &#xD;&#xA;$$&#xD;&#xA;we find, after some algebra, that the original equation is equivalent to&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^l {{l \choose k}p^k (1 - p)^{l - k} (1 - p_j )^k }  = (1 - pp_j )^l ,&#xD;&#xA;$$&#xD;&#xA;where $l (=n-m) \geq 0$. Since the case $p=0$ is satisfied ($0^0=1$), we assume that $p = x/y$, with $0 &amp;lt; x \leq y$. Substituting this leads straightforwardly to the equivalent equation&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^l {{l \choose k}(x - x p_j)^k (y-x)^{l-k}  = (y - xp_j)^l}.&#xD;&#xA;$$&#xD;&#xA;By the binomial theorem, the left-hand side is equal to $[(x - xp_j ) + (y - x)]^l$, completing the proof.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-15T05:22:55.737" />
  <row Id="10370" PostTypeId="2" ParentId="10360" CreationDate="2010-11-15T05:40:02.717" Score="-1" ViewCount="0" Body="&lt;p&gt;Depending on the university/college, most of the grading is done by TAs, who're replaced each semester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if it's a large course, then the professor is not likely to remember your name anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, grading is usually done according to an &quot;objective&quot; scale, so everyone gets what they deserve.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-15T05:40:02.717" CommentCount="1" />
  <row Id="10371" PostTypeId="2" ParentId="10318" CreationDate="2010-11-15T05:51:13.513" Score="1" ViewCount="0" Body="&lt;p&gt;While Timothy Wagner's answer is correct, I thought you might like to see another way to answer your first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often the simplest way to prove that the sum of independent random variables has a particular distribution is to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function&quot; rel=&quot;nofollow&quot;&gt;moment-generating functions&lt;/a&gt;.  This is because 1) if $X$ and $Y$ are independent with mgf's $M_X(t)$ and $M_Y(t)$, then $M_{X+Y}(t) = M_X(t) M_Y(t)$ and 2) moment-generating functions (when they exist) characterize distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this to your problem, a geometric $(p)$ random variable has mgf $$\frac{pe^t}{1 - (1-p)e^t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$M_{X+Y}(t) = \left(\frac{pe^t}{1 - (1-p)e^t}\right)^{2}.$$&#xD;&#xA;Since this is the mgf of a negative binomial random variable, $X+Y$ must have a negative binomial distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are different conventions for defining negative binomial and geometric random variables, so depending on the convention used in a particular reference the mgf's there may be slightly different from the ones I give here.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-15T05:51:13.513" />
  <row Id="10372" PostTypeId="2" ParentId="9933" CreationDate="2010-11-15T06:41:49.017" Score="2" ViewCount="0" Body="&lt;p&gt;&#xA;&#xA;Informal justification of positive x negative = negative:&lt;br&gt;&#xA;&lt;br&gt;&#xA;Continue the pattern:&lt;br&gt;&#xA;2x3=6&lt;br&gt;&#xA;2x2=4&lt;br&gt;&#xA;2x1=2&lt;br&gt;&#xA;2x0=0&lt;br&gt;&#xA;2x-1= ? (Ans=-2)&lt;br&gt;&#xA;2x-2= ? (Ans=-4)&lt;br&gt;&#xA;2x-3= ? (Ans=-6)&lt;br&gt;&#xA;&lt;br&gt;&#xA;The number on the right-hand side keeps decreasing by 2.&lt;br&gt;&#xA;&lt;br&gt;&#xA;Informal justification of negative x negative = positive:&lt;br&gt;&#xA;&lt;br&gt;&#xA;Continue the pattern:&lt;br&gt;&#xA;2x-3=-6&lt;br&gt;&#xA;1x-3=-3&lt;br&gt;&#xA;0x-3=0&lt;br&gt;&#xA;-1x-3= ? (Ans=3)&lt;br&gt;&#xA;-2x-3= ? (Ans=6)&lt;br&gt;&#xA;-3x-3= ? (Ans=9)&lt;br&gt;&#xA;&lt;br&gt;&#xA;The number on the right-hand side keeps increasing by 3.&lt;br&gt;&#xA;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-15T06:41:49.017" />
  <row Id="10373" PostTypeId="1" CreationDate="2010-11-15T07:54:51.813" Score="2" ViewCount="190" Body="&lt;p&gt;How to describe all ring homomorphism from $Z \times Z \to Z$ ?&lt;/p&gt;&#xA;" OwnerUserId="3516" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-15T09:18:28.463" LastActivityDate="2010-11-15T15:00:49.867" Title="All homomorphisms from the ring $Z \times Z \to Z$" Tags="&lt;ring-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10374" PostTypeId="2" ParentId="10282" CreationDate="2010-11-15T08:19:43.697" Score="9" ViewCount="0" Body="&lt;p&gt;At least to me, it is helpful to think in terms of bases.&#xA;(I'll only be talking about tensor products of finite-dimensional vector&#xA;spaces here.)&#xA;This makes the universal mapping property that Zach Conn talks about&#xA;a bit less abstract (in fact, almost trivial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First recall that if $L:V \to W$ is a linear map, then $L$ is completely determined&#xA;by what it does to a basis $\{ e_i \}$ for $V$:&#xA;$$L(x)=L(\sum x_i e_i) = \sum x_i L(e_i).$$&#xD;&#xA;(The coefficients of $L(e_i)$ in a basis for $W$ gives the $i$th column&#xA;in the matrix for $L$ with respect to the given bases.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensors come into the picture when one studies &lt;em&gt;multilinear&lt;/em&gt; maps.&#xA;If $B:V \times U \to W$ is a bilinear map, then $B$&#xD;&#xA;is completely determined by the values $B(e_i,f_j)$ where&#xA;$\{ e_i \}$ is a basis for $V$&#xD;&#xA;and $\{ f_j \}$ is a basis for $U$:&#xA;$$B(x,y) = B(\sum x_i e_i,\sum y_j f_j) = \sum\sum x_i y_j B(e_i,f_j).$$&#xD;&#xA;For simplicity, consider the particular case when $W=\mathbf{R}$;&#xA;then the values $B(e_i,f_j)$&#xD;&#xA;make up a set of $N=nm$ real numbers (where $n$ and $m$ are the&#xA;dimensions of $V$ and $U$), and these numbers are all that we need to keep&#xA;track of in order to know everything about the bilinear map $B:V \times U \to \mathbf{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that in order to compute $B(x,y)$ we don't really need to know the&#xA;individual vectors $x$ and $y$, but rather the $N=mn$ numbers $\{ x_i y_j \}$.&#xA;Another pair of vectors $v$ and $u$ with $v_i u_j = x_i y_j$ for all $i$ and $j$&#xD;&#xA;will satisfy $B(v,u)=B(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the idea of splitting the computation of $B(x,y)$ into two stages.&#xA;Take an $N$-dimensional vector space $T$ (they're all isomorphic so it doesn't matter&#xA;which one we take) with a basis $(g_1,\dots,g_N)$.&#xA;Given $x=\sum x_i e_i$ and $y=\sum y_j f_j$,&#xA;first form the vector in $T$&#xD;&#xA;whose coordinates with respect to the basis $\{ g_k \}$ are given by the column vector&#xA;$$(x_1 y_1,\dots,x_1 y_m,x_2 y_m,\dots,x_2 y_m,\dots,x_n y_1,\dots,x_n y_m)^T.$$&#xD;&#xA;Then run this vector through the &lt;em&gt;linear&lt;/em&gt; map $\tilde{B}:T\to\mathbf{R}$ whose matrix&#xA;is the row vector&#xA;$$(B_{11},\dots,B_{1m},B_{21},\dots,B_{2m},\dots,B_{n1},\dots,B_{nm}).$$&#xD;&#xA;This gives, by construction, $\sum\sum B_{ij} x_i y_j=B(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll call the space $T$ the tensor product of the vector spaces $V$ and $U$&#xD;&#xA;and denote it by $T=V \otimes U$;&#xA;it is &quot;uniquely defined up to isomorphism&quot;.&#xA;The vector in $T$ that we formed from $x\in V$ and $y\in U$ in the first stage above&#xA;will be denoted $x \otimes y$;&#xA;it's a &quot;bilinear mixture&quot; of $x$ and $y$ which doesn't allow us to&#xA;reconstruct $x$ and $y$ individually,&#xA;but still contains exactly all the information needed&#xA;in order to compute $B(x,y)$ for any bilinear map $B$;&#xA;we have $B(x,y)=\tilde{B}(x \otimes y)$.&#xA;This is the &quot;universal property&quot;; any bilinear map $B$ from $V \times U$&#xD;&#xA;can be computed by taking a &quot;detour&quot; through $T$, and this detour&#xA;is unique, since the map $\tilde{B}$ is constructed uniquely from&#xA;the values $B(e_i,f_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To tidy this up, one would like to make sure that the definition is&#xA;basis-independent. One way is to check that everything transforms&#xA;properly under changes of bases. Another way is to do the construction&#xA;by forming a much bigger space and taking a quotient with respect to&#xA;suitable relations (without ever mentioning bases).&#xA;Then, by untangling definitions, one can for&#xA;example show that a bilinear map $B:V \times U \to \mathbf{R}$ can be&#xA;canonically identified with an element of the space $V^* \otimes U^*$,&#xA;and dually an element of $V \otimes U$ can be identified with a&#xA;bilinear map $V^* \otimes U^* \to \mathbf{R}$.&#xA;Yet other authors find this a convenient &lt;em&gt;starting&lt;/em&gt; point, so that they&#xA;instead &lt;em&gt;define&lt;/em&gt; $V \otimes U$ to be the space of bilinear maps $V^*&#xD;&#xA;\otimes U^* \to \mathbf{R}$.&#xA;So it's no wonder that one can become a little confused when trying&#xA;to compare different definitions...&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-11-15T09:22:31.033" LastActivityDate="2010-11-15T09:22:31.033" />
  <row Id="10375" PostTypeId="1" CreationDate="2010-11-15T09:09:06.967" Score="3" ViewCount="80" Body="&lt;p&gt;Specifically, if you have a topos with natural numbers object (assume choice if you want to), what familiar theorems don't hold? I've heard that showing the Dedekind reals are the same as the Cauchy reals is one. Where in the arguments is well-pointedness used? It seems hard to find examples of this.&lt;/p&gt;&#xA;" OwnerUserId="3518" LastEditorUserId="3518" LastEditorDisplayName="" LastEditDate="2010-11-15T20:40:36.003" LastActivityDate="2010-12-15T20:44:25.920" Title="Where is the well-pointedness assumption of ETCS used in everyday math?" Tags="&lt;logic&gt;&lt;category-theory&gt;&lt;topos-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10377" PostTypeId="1" AcceptedAnswerId="10381" CreationDate="2010-11-15T10:14:42.020" Score="3" ViewCount="99" Body="&lt;p&gt;For a sequence defined by a formula normally the usual limit rules allows&#xA;one to find its limit. But for a sequence defined by a recurrence, up to now&#xA;I have only seen some refined ad hoc methods, mostly in Problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Trick explains that &quot;&lt;a href=&quot;http://www.tricki.org/article/I_have_a_problem_about_the_convergence_of_a_sequence&quot; rel=&quot;nofollow&quot;&gt;There is one trick that is (...) first to prove&#xA;that a limit exists, and then to use the recurrence to determine what the&#xA;limit must be&lt;/a&gt;&quot; illustrated by the example of the recurrence $&#xD;&#xA;a_{n+1}=a_{n}/2+1/a_{n}$, $a_{0}=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Are there relatively general methods to find the limit of a&#xA;sequence defined by a recurrence?  &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-15T11:10:52.603" LastActivityDate="2010-11-15T11:10:52.603" Title="Methods to find the limit of a sequence defined by a recurrence" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;&lt;recursive&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10378" PostTypeId="1" CreationDate="2010-11-15T10:35:12.080" Score="0" ViewCount="47" Body="&lt;p&gt;I'm in a little bit of a pickle here, can someone tell me if it is actually possible to solve these? From the general rule of thumb, 4 variables, 4 equations should be able to right?&lt;/p&gt;&#xA;" OwnerUserId="3525" LastActivityDate="2010-11-15T11:10:37.977" Title="Is it possible to solve: xa = 4200, xb = 4410, ya=4000, yb = 4200 for x, y, a and b?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10379" PostTypeId="1" CreationDate="2010-11-15T10:37:38.807" Score="1" ViewCount="145" Body="&lt;p&gt;It is known that, in the universe of complex numbers, the only root of the equation x² - 2x + 1 = 0 is 1.  Could we say that the equation has two equal real roots? Or should we say that the equation has one real root with multiplicity 2?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-11-15T10:43:34.090" LastActivityDate="2010-11-15T22:49:37.140" Title="A root? Or two roots?" Tags="&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10381" PostTypeId="2" ParentId="10377" CreationDate="2010-11-15T10:47:06.160" Score="6" ViewCount="0" Body="&lt;p&gt;Not really. Even once you've shown that a limit exists, a limit of, say, $x_{n+1} = f(x_n)$ is the same as a fixed point of $f$, or a solution to $f(x) - x = 0$. Needless to say it does not make much sense to ask whether there are general methods to find solutions to equations. For example, one can write down solutions to differential equations as fixed points, and there are no general methods to solve differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense all methods to find limits of recurrences are &quot;ad hoc&quot; in the same way that all methods to solve differential equations or Diophantine equations are &quot;ad hoc.&quot;  We do what we can.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T10:47:06.160" />
  <row Id="10382" PostTypeId="2" ParentId="10378" CreationDate="2010-11-15T11:10:37.977" Score="1" ViewCount="0" Body="&lt;p&gt;The best you can do is $20x = 21y$ and $21a = 20b.$&#xD;&#xA;If you write your second pair of equations as&#xA;$$a \left( \frac{20}{21} x \right) = 4000,$$&#xD;&#xA;$$b \left( \frac{20}{21} x \right) = 4200$$&#xD;&#xA;we note that they are equivalent to your first pair&#xA;$$ax = 4200$$&#xD;&#xA;$$bx = 4410,$$&#xD;&#xA;simply by multipling up by the $20/21.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your general rule of thumb, note that a system of equations&#xA;can be &lt;a href=&quot;http://www.mathwords.com/i/inconsistent_system_of_equations.htm&quot; rel=&quot;nofollow&quot;&gt;inconsistent.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-15T11:10:37.977" CommentCount="3" />
  <row Id="10383" PostTypeId="1" AcceptedAnswerId="10388" CreationDate="2010-11-15T11:13:54.137" Score="1" ViewCount="156" Body="&lt;p&gt;Hey guys,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck with this problem I'm trying to solve from about an hour. Here's the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the remainder when (3^202)+137 is divided by 101?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 options -&gt; 36, 45, 56, 11&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the answer of the question with proper and possibly easiest method to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, waiting for reply. :)&lt;/p&gt;&#xA;" OwnerUserId="712" LastActivityDate="2010-11-18T00:31:54.393" Title="What will be the remainder?" Tags="&lt;arithmetic&gt;" AnswerCount="3" />
  <row Id="10384" PostTypeId="2" ParentId="10383" CreationDate="2010-11-15T11:20:44.413" Score="9" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Fermat's Little Theorem, which says that $a^{p-1} \equiv 1 \ (\text{mod} \ p)$ , where $p$ is a prime and $101$ is a prime. Remainder should be $45$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-15T11:31:04.090" LastActivityDate="2010-11-15T11:31:04.090" CommentCount="1" />
  <row Id="10385" PostTypeId="2" ParentId="10373" CreationDate="2010-11-15T11:52:58.200" Score="7" ViewCount="0" Body="&lt;p&gt;To sum up all the comments\answers so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Chandru wrote, if $f:\mathbb{Z}\times \mathbb{Z} \rightarrow \mathbb{Z}$ is a ring homomorphism, then it is also an abelian group homomorphism (for the addition operation), so $f([m,n]) = m \cdot f([1,0]) + n \cdot (f([0,1])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you also want the function to preserve multiplication, then you have other conditions.&#xA;For example, as Robin commented f preserves idempotent -$f([0,1])=f([0,1][0,1])=f([0,1])f([0,1])$ so $x=f([0,1])$ is an element of $\mathbb{Z}$ that satisfies $x^2=x$ - therefore it must be 1 or 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f also preserves zero divisors so $0=f([0,0])=f([0,1][1,0])=f([0,1])f([1,0])$ which means that at least one of them is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you are now left with three possibilities -&#xA;$f_1([m,n])=m,\; \; \;f_2([m,n])= n,\; \; \;f_3([m,n])=0$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-15T11:52:58.200" />
  <row Id="10386" PostTypeId="2" ParentId="10353" CreationDate="2010-11-15T11:53:58.760" Score="2" ViewCount="0" Body="&lt;p&gt;This is only a first approximation to your answer, but I think it's a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x$ and $y$ be vectors in $\mathbb{R}^n$ and assume $x\cdot y = 0$.  Now, I'm going to add an additional condition that neither $x$ nor $y$ is $0$.  To give away the punchline, it will turn out there's a nice description of &lt;i&gt;these&lt;/i&gt; points.  The remaining points (where either $x=0$ or $y=0$ or both) will be degenerate in a way, because then the dot product doesn't tell you anything.  It turns out these remaining points destroy the &quot;niceness&quot; of the others (at least, how I usually define nice, i.e., getting a smooth manifold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $X =\{(x,y)\in \mathbb{R}^{2n}|$ $x\neq 0$, $y\neq 0,$ and $x\cdot y = 0\}$.  The goal is to understand the shape of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to notice is that if $(x,y)\in X$, then so is $(\lambda x,y)$ and $(x,\mu y)$ for any $\lambda, \mu &amp;gt; 0$.  Further, if we set $Y =\{(x,y)\in X|\text{  } |x|=|y|=1 \}$, then it's clear that every point in $X$ is of the form $(\lambda x, \mu y)$ for a unique choice of $\lambda, \mu &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from this that $X = Y\times (0,\infty)\times (0,\infty)$.  That is, $X$ has the shape of $Y$ extended in 2 directions.  But what is the shape of $Y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment, let's take an apparent detour.  How do you describe a sphere in $\mathbb{R}^n$?  It's $S^n = \{ x\in\mathbb{R}^{n+1}| \text{  }|x|=1\}$.  (Note that the $n$ in $S^n$ is one less than the $n$ in $\mathbb{R}^{n+1}$)  What is a tangent vector on $S^n$ at a point $x\in S^n$?  More generally, what is the tangent plane through a point $x\in S^n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shouldn't be too hard to convince yourself by looking at this on the circle and usual sphere that the tangent plane/line through a point $x$ in $S^n$ is the plane with normal vector $x$ going through the point $x$ (so $x$ does double duty as both a point and a vector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying the normal vector is $x$ means that for any vector $y$ in the tangent plane, we must have $x\cdot y = 0$.  What if we take a unit vector $y$ in the tangent plane?  Well, we just find that $|x| = |y| = 1$ and $x\cdot y = 0$.  But these were exactly our conditions on belonging to $Y$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $Y$ is the so called Unit Tangent Bundle of the sphere - it's the collection of all $(p,v)$ where $p$ is a point on the sphere (i.e., $|p| = 1$, $v$ is a tangent vector to there sphere at $p$ (i.e., $p\cdot v = 0$) where $v$ has unit length (i.e., $|v| = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, in particular, you can think of these spaces as &quot;the collection of points on a sphere paired with a choice of unit length tangent vector&quot;.  It turns out this shape is fairly well understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when $n=1$, one gets the unit tangent bundle of a circle.  It is know that the tangent bundle of a circle is trivial - that is, is diffeomorphic to $S^1\times \mathbb{R}$.  From here it follows that the unit tangent bundle is diffeomorphic to $S^1\times S^1$.  This shape is called a Torus and is best physically described as a hollow donut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, in particular, almost answers your first question about what we get in $\mathbb{R}^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n =2$, one gets the unit tangent bundle of $S^2$.  It's known that this is diffeomorphic to $\mathbb{R}P^3$, which can almost be visualized.  For $n=3$, one gets the unit tangent bundle of $S^3$, which is known to be diffeomorphic to $S^3\times S^2$.  I don't know of a suitably nice description of any other unit tangent bundles except for $n=7$, one gets $S^7\times S^6$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-15T22:35:45.837" LastActivityDate="2010-11-15T22:35:45.837" CommentCount="1" />
  <row Id="10387" PostTypeId="1" AcceptedAnswerId="10394" CreationDate="2010-11-15T12:23:34.437" Score="2" ViewCount="45" Body="&lt;p&gt;I'm stuck with the dual of an exercise in Atiyah-MacDonald.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easily seen that tensoring over $R$ with $-\otimes M$ the sequence&#xA;$$&#xD;&#xA;0\to \mathfrak a\to R\to R/\mathfrak a\to 0&#xD;&#xA;$$&#xD;&#xA;one gets the isomorphism $R/\mathfrak a\otimes M\cong M/\mathfrak aM$.&#xA;What can I deduce from &quot;&lt;em&gt;Hom&lt;/em&gt;(-,M)ing&quot; the same sequence? I have, let's say&#xA;$$&#xD;&#xA;0\to \mathfrak a\xrightarrow{\alpha} R\xrightarrow{\beta} R/\mathfrak a\to 0&#xD;&#xA;$$&#xD;&#xA;and &#xA;$$&#xD;&#xA;0\to \hom(R/\mathfrak a,M) \xrightarrow{\beta^*} \hom(R,M)\cong M\xrightarrow{\alpha^*} \hom(\mathfrak a,M)&#xD;&#xA;$$&#xD;&#xA;but now what can I do? What is $\hom(R/\mathfrak a,M)$ like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, it's easy, but...&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-11-15T13:43:37.640" Title="How to &quot;Hom&quot; the canonical sequence in a ring" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="10388" PostTypeId="2" ParentId="10383" CreationDate="2010-11-15T12:24:02.227" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;I think Chandru1 is correct. The answer is 45.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did I get it?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) The remainder of 3^30 / 101 = 6&#xA;2) 6^6 = 46656.&#xA;3) 46656 * (The remainder of 3^22 / 101) = 2286144.&#xA;4) 2286144 + 137 = 2286281&#xA;5) The remainder of 2286281 / 101 = 45&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note in step 2 and 3 that 30 * 6 = 180. And 180 + 22 = 202&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler example: The reaminder when (2^10)+7 is divided by 6 (the answer is 5).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1) The reaminder of 2^4 / 6 = 4&#xA;2) 4^2 = 16&#xA;3) 16 * (The remainder of 2^2 / 6) = 64&#xA;4) 64 + 7 = 71&#xA;5) The remainder of 71 / 6 = 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3314" LastActivityDate="2010-11-15T12:24:02.227" CommentCount="1" />
  <row Id="10389" PostTypeId="1" CreationDate="2010-11-15T12:32:03.033" Score="2" ViewCount="194" Body="&lt;p&gt;I have to solve the following recurrence :$\displaystyle T(n)=2T(n/2)+T(n/3)+\theta(n^2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done the whole tree analyses and  now I have to prove that $\displaystyle T(n) \leq dn^{2}\log_{2}(n)$ but I can not find mathematical way to prove that, is there any trick with logarithms that I can use?&lt;/p&gt;&#xA;" OwnerUserId="3382" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-15T20:32:14.163" LastActivityDate="2010-11-15T23:44:30.360" Title="Recurrence trouble: $T(n)=2T(n/2)+T(n/3)+\theta(n^2)$" Tags="&lt;asymptotics&gt;&lt;recursive&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="10390" PostTypeId="2" ParentId="10282" CreationDate="2010-11-15T13:01:54.610" Score="3" ViewCount="0" Body="&lt;p&gt;Mathematicians and physicists use very different languages when they talk about tensors. Fortunately, they are talking about the same thing, but unfortunately, this is not obvious at all. Let me explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, I'm going to focus on &lt;strong&gt;covariant 2-tensors&lt;/strong&gt;, since this case already contains the main intuition. Also, I'm not going to talk about the distinction between covariant and contravariant, but I'll get all the indices right for future study.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Physicist's definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Definition: A covariant 2-tensor is a set of numbers $t_{ij}$ with two indices that transforms in a particular way under a change of coordinates… Wait, wait, coordinates in &lt;em&gt;what space&lt;/em&gt;? Physicists usually don't mention it, but they mean coordinates in a given &lt;em&gt;vector space&lt;/em&gt; $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, let $\{\vec e_i\}$ be a basis of the vector space $V$. Then, every &lt;strong&gt;vector&lt;/strong&gt; $\vec v$ can be expressed in terms of its &lt;em&gt;coordinates&lt;/em&gt; $v^i$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec v = \sum_i v^i \vec e_i .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are two objects: the vector $\vec v$ which I think of as &quot;solid&quot; or &quot;fundamental&quot;, and its coordinates $v^i$, which are &quot;ephemeral&quot;, since I have to choose a basis $\vec e_i$ before I can talk about them at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in a &lt;strong&gt;different basis&lt;/strong&gt; $\{\vec e&amp;#39;_i\}$ of our vector space, the coordinates of one and the same vector $\vec v$ are very different numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec v = \sum_i v^i \vec e_i = \sum_i v&amp;#39;^i \vec e&amp;#39;_i .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $v^i ≠ v&amp;#39;^i$. So, the vector is the fundamental thing. Its coordinates are useful for calculations, but they are ephemeral and heavily depend on the choice of basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when defining a &lt;strong&gt;covariant 2-tensor&lt;/strong&gt;, physicists do something very mysterious: they define a fundamental object (= the 2-tensor) not by describing it directly, but only by specifying how its ephemeral coordinates look like and change when switching to a different basis. Namely, a change of basis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec e&amp;#39;_i = \sum_j R_i^a \vec e_a $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will change the coordinates $t_{ij}$ of the tensor via&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t&amp;#39;_{ij} = \sum_{ab} R^a_i R^b_j t_{ab} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is not completely unintuitive, I don't know what is.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mathematician's definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mathematicians define tensors differently. Namely, the give a direct, fundamental description of what a 2-tensor is and only then ponder how it looks like in different coordinate systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the definition: a &lt;strong&gt;covariant 2-tensor $t$ is a bilinear map&lt;/strong&gt; $t : V\times V \to \mathbb{R}$. That's it. (Bilinear = linear in both arguments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a covariant 2-tensor $t$ is a thing that eats two vectors $\vec v$, $\vec w$ and returns a number $t(\vec v, \vec w) \in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what does this thing look like in coordinates? Choosing a basis $\lbrace \vec e_i \rbrace$, bilinearity allows us to write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t(\vec v, \vec w) = t(\sum_i v^i \vec e_i, \sum_j w^j \vec e_j) = \sum_{ij} v^iw^j t(\vec e_i,\vec e_j) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we simply call the numbers $t_{ij} = t(\vec e_i, \vec e_j)$ the &lt;strong&gt;coordinates of the tensor $t$&lt;/strong&gt; in the basis $\vec e_i$. You can calculate that these numbers will behave just like the physicists tell us when you change the basis to $\vec e_i&amp;#39;$. So, the physicist's tensor and the mathematician's tensor are one and the same thing.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Tensor product&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Actually, mathematicians do something more advanced, they define a so called &lt;strong&gt;tensor product&lt;/strong&gt; of vector spaces. The previous definition as a bilinear map is still correct, but mathematicians like to write this as &quot;$t\in V^*\otimes V^*$&quot; instead of &quot;$t: V\times V \to \mathbb{R}$ and $t$ bilinear&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for a first understanding of the physicist's vs the mathematician's definition, it is not necessary to understand the mathematical tensor product.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-11-15T13:01:54.610" CommentCount="1" />
  <row Id="10391" PostTypeId="2" ParentId="10342" CreationDate="2010-11-15T13:03:10.670" Score="2" ViewCount="0" Body="&lt;p&gt;I don't know what a good categorical quotient is, but if $X$ is a CW complex, and the $G$ action is free and cellular, then $H^*_G(X)\cong H^*(X/G)$. See Brown's &quot;Cohomology of Groups,&quot; p.173.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-11-15T13:03:10.670" CommentCount="0" />
  <row Id="10392" PostTypeId="1" AcceptedAnswerId="10420" CreationDate="2010-11-15T13:25:41.353" Score="5" ViewCount="175" Body="&lt;p&gt;Let's say I have a truth table like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X  Y   A&#xA;t  t  f&#xA;t  f  t&#xA;f  t  t&#xA;f  f  f&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, I have to find the formula for A. This case is rather easy because I can immediately see that this looks like the inverted values of X~B, thus ¬(X~B). Now, for a more complicated problem I do not know the method.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X  Y  Z  A&#xA;t  t  t  f&#xA;t  t  f  f&#xA;t  f  t  t&#xA;t  f  f  f&#xA;f  t  t  f&#xA;f  t  f  f&#xA;f  f  t  t&#xA;f  f  f  f&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is there an approach i'm missing here?&lt;/p&gt;&#xA;" OwnerUserId="3527" LastEditorUserId="3527" LastEditorDisplayName="" LastEditDate="2010-11-15T13:31:40.023" LastActivityDate="2010-11-23T18:21:44.007" Title="How to find the logical formula for a given truth table?" Tags="&lt;logic&gt;" AnswerCount="5" />
  <row Id="10393" PostTypeId="2" ParentId="10387" CreationDate="2010-11-15T13:27:28.597" Score="0" ViewCount="0" Body="&lt;p&gt;Asking «what is $M/\operatorname{im}\alpha^*$?» is &lt;em&gt;not&lt;/em&gt; the dual of the A-M exercise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want a dual statement, you should be looking for a description of $\hom(R/\mathfrak a,M)$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T13:27:28.597" CommentCount="1" />
  <row Id="10394" PostTypeId="2" ParentId="10387" CreationDate="2010-11-15T13:32:21.570" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: Any $R$-module homomorphism $f$ from $R/\mathfrak{a}$ to $M$ is determined once we know $f(1)$. But every such homomorphism is killed by $\mathfrak{a}$ since $\mathfrak{a}f(1)=f(\mathfrak{a})=f(0)=0$. So one can show that, $Hom_R(R/\mathfrak{a},M)\cong Ann_{\mathfrak{a}}(M)$ &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-15T13:43:37.640" LastActivityDate="2010-11-15T13:43:37.640" CommentCount="3" />
  <row Id="10395" PostTypeId="2" ParentId="10392" CreationDate="2010-11-15T13:32:58.527" Score="1" ViewCount="0" Body="&lt;p&gt;Solve first the problem of finding the formula for truth tables which have exactly one T on the rightmost table. Then go to the general case.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T13:32:58.527" />
  <row Id="10396" PostTypeId="2" ParentId="10392" CreationDate="2010-11-15T13:35:21.223" Score="9" ViewCount="0" Body="&lt;p&gt;I believe &lt;a href=&quot;http://en.wikipedia.org/wiki/Karnaugh_map&quot; rel=&quot;nofollow&quot;&gt;Karnaugh Maps&lt;/a&gt; are useful for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-15T13:35:21.223" CommentCount="1" />
  <row Id="10397" PostTypeId="2" ParentId="10389" CreationDate="2010-11-15T13:37:11.343" Score="3" ViewCount="0" Body="&lt;p&gt;Try applying the &lt;a href=&quot;http://en.wikipedia.org/wiki/Akra-Bazzi_theorem&quot; rel=&quot;nofollow&quot;&gt;Akra-Bazzi&lt;/a&gt; theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Akra-Bazzi, if $p$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2(1/2)^p + (1/3)^p = 1$, (notice that $\displaystyle 1 \lt p \lt 2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle T(x) = \theta(x^p(1 + \int_{1}^{x} x^{1-p})) = \theta(x^p(1+x^{2-p}))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\displaystyle 1 \lt p \lt 2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we must have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle T(x)  = \theta(x^2) = O(x^2 \log_{2} x)$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T14:26:07.273" LastActivityDate="2010-11-15T14:26:07.273" CommentCount="8" />
  <row Id="10398" PostTypeId="2" ParentId="10353" CreationDate="2010-11-15T13:47:27.977" Score="0" ViewCount="0" Body="&lt;p&gt;Cut your hypersurface $\{(x_1,x_2,x_3,x_4)\in\mathbb R^4:x_1x_2+x_3x_4=0\}$ with the hyperplanes $H_t=\{(x_1,x_2,x_3,x_4)\in\mathbb R^4:x_1=t\}$, for $t\in\mathbb R$, and visualize those intersections. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T13:47:27.977" />
  <row Id="10399" PostTypeId="1" AcceptedAnswerId="10407" CreationDate="2010-11-15T14:14:54.323" Score="1" ViewCount="47" Body="&lt;p&gt;I believe I've found all Amicable Pairs (n, m) such that n is in the closed interval [1, 34000000]. Here's my &lt;a href=&quot;http://tutoree7.pastebin.com/fg3mVNmF&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inspection I believe that for all n in the list, n is odd iff its last digit is 5. Can it be proved that this is true for all Amicable Pairs? &lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-15T14:54:10.467" Title="True?: Let (n, m) be an arbitrary Amicable Pair. Then n is odd iff its last digit is 5" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="10400" PostTypeId="1" AcceptedAnswerId="10406" CreationDate="2010-11-15T14:21:37.407" Score="0" ViewCount="99" Body="&lt;p&gt;Let $R$ be a commutative ring and $I_1, \cdots, I_n$ pairwise comaximal ideals in $R$, e.g. $I_i + I_j = R$ for $i \neq j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?&lt;/p&gt;&#xA;" OwnerUserId="3529" LastActivityDate="2010-11-15T18:13:58.270" Title="Comaximal ideals in a commutative ring" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10401" PostTypeId="2" ParentId="9712" CreationDate="2010-11-15T14:29:30.687" Score="1" ViewCount="0" Body="&lt;p&gt;I believe the answer given in my comments is correct. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-15T14:29:30.687" />
  <row Id="10403" PostTypeId="2" ParentId="10033" CreationDate="2010-11-15T14:35:42.240" Score="1" ViewCount="0" Body="&lt;p&gt;I have checked my derivation thoroughly, and I believe now that it is a typo.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-15T14:35:42.240" />
  <row Id="10404" PostTypeId="2" ParentId="10383" CreationDate="2010-11-15T14:37:57.587" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\  101\ $ is prime so a little Fermat $\rm\ \Rightarrow\ \ 3^{101}\ \equiv\ 3\ \ \Rightarrow\ \ 3^{202}\ \equiv\ \ldots\ (mod\ 101)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your comment reveals you are not familiar with modular arithmetic, here is an alternative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Fermat's little theorem $101$ divides $\: 3^{101}-3\: $ so it also divides $\rm\ (3^{101}-3)\ (3^{101}+3)\ =\ 3^{202}-9$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T00:31:54.393" LastActivityDate="2010-11-18T00:31:54.393" CommentCount="2" />
  <row Id="10405" PostTypeId="2" ParentId="10399" CreationDate="2010-11-15T14:38:59.363" Score="1" ViewCount="0" Body="&lt;p&gt;How about 87633 and 69615?  Found 87633 in &lt;a href=&quot;http://oeis.org/A063990&quot; rel=&quot;nofollow&quot;&gt;OEIS A063990&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-15T14:38:59.363" CommentCount="1" />
  <row Id="10406" PostTypeId="2" ParentId="10400" CreationDate="2010-11-15T14:40:00.533" Score="3" ViewCount="0" Body="&lt;p&gt;It is sufficient to prove this for the case two comaximal ideals, say $I,J$. Need to show, $I^m+J^n=R$ for any positive integers $m,n$. Now, $R=R^{m+n-1}=(I+J)^{m+n-1}\subseteq I^m+J^n$ since in the binomial expansion of $(I+J)^{m+n-1}$, in every term, either the power of $I$ is at least $m$ or the power of $J$ is at least $n$ by pigeonhole principle.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-15T14:57:05.397" LastActivityDate="2010-11-15T14:57:05.397" />
  <row Id="10407" PostTypeId="2" ParentId="10399" CreationDate="2010-11-15T14:54:10.467" Score="1" ViewCount="0" Body="&lt;p&gt;I think the first counterexample (n,m) with n &amp;lt; m where n is odd and not congruent to 5 mod 10 is (34765731, 36939357), which by misfortune is just outside your search region.&lt;/p&gt;&#xA;" OwnerUserId="3531" LastActivityDate="2010-11-15T14:54:10.467" CommentCount="3" />
  <row Id="10408" PostTypeId="1" AcceptedAnswerId="10412" CreationDate="2010-11-15T14:57:34.677" Score="3" ViewCount="117" Body="&lt;p&gt;I want to describe the polynomials with integer coefficients and the property that f'(x)|f(x) (the derivate divides the polynomial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know that f(x) divides g(x) if all of f(x)'s roots are roots of g(x). I also know that f'(x) has degree one less than f(x). Finally, f'(x) detects multiple roots. If a is a root of f(x) and f'(x), it is at least a double root. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these facts I've conjectured that f(x), if it has degree n, has 1 root with multiplicity n or 2 roots, one with multiplicity n-1. But I can't prove it and I fear that I am making a mistake thinking just about linear terms.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastActivityDate="2010-11-15T16:02:39.687" Title="Polynomials, derivatives and repeated roots" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="10409" PostTypeId="2" ParentId="10373" CreationDate="2010-11-15T15:00:49.867" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ hom's preserve orthogonal idempotents, so $\ \{\:(0,1), (1,0)\:\}\to \{0,1\}\ $ if nontrivial&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T15:00:49.867" />
  <row Id="10410" PostTypeId="2" ParentId="10167" CreationDate="2010-11-15T15:02:57.573" Score="1" ViewCount="0" Body="&lt;p&gt;Here is the way I hinted at the solution: proving first that a finitely generated torsion-free abelian group is free, then use the fact that free groups are projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument for the first part is that used by Lang in his &lt;em&gt;Algebra&lt;/em&gt;: suppose $A$ is a finitely generated torsion-free abelian group; we may assume it is nonzero. Pick $x_1,\ldots,x_n\in A$ that are maximal with respect to the property that if $\alpha_1x_1+\cdots +\alpha_nx_n=0$, then $\alpha_i=0$ for all $i$. Let $B$ be the subgroup generated by the $x_i$; note that $B$ is free abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y_1,\ldots,y_k$ be a generating set for $A$. For each $i$, &#xA;there exists an integer $d_i\neq 0$ and  integers $\beta_1,\ldots,\beta_n$, not all zero, such that $d_iy +\beta_1x_1 + \cdots +\beta_n x_n=0$; that is, $d_iy_i\in B$. Letting $d=\mathrm{lcm}(d_1,\ldots,d_k)$, we have that $dA$ is contained in $B$. Since $A$ is torsion free, the map $A\to B$ given by $x\mapsto dx$ is an injection, so $A$ is isomorphic to a subgroup of a free abelian group, hence $A$ is free abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $A$ is a finitely generated group, and $tA$ is its torsion group, we know that $A/tA$ is finitely generated and torsion-free, hence free. Let $x_1,\ldots,x_k$ be a basis for $A/tA$. For each $i$, let $a_i\in A$ be any element that maps to $x_i$ under the canonical projection $\pi\colon A\to A/tA$. Since $A/tA$ is free, we have a (unique) group homomorphism $\rho\colon A/tA\to A$ that maps $x_i$ to $a_i$. Since $\pi\circ\rho = \mathrm{id}_{A/tA}$, then $\rho$ must be one-to-one, hence $A/tA$ is isomorphic to its image under $\rho$ inside of $A$. Thus, $A$ has a subgroup that is isomorphic to $A/tA$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T17:36:14.650" LastActivityDate="2010-11-15T17:36:14.650" />
  <row Id="10411" PostTypeId="2" ParentId="10408" CreationDate="2010-11-15T15:04:39.160" Score="2" ViewCount="0" Body="&lt;p&gt;This is assuming you are talking about polynomials over $\mathbb{C}$ or $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $f$ has leading coefficient $1$ and is degree $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$nf(x) = f&amp;#39;(x)(x-a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(x)/nf(x) = 1/(x-a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log(f(x)) = n\log(x-a) + C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f(x) = K(x-a)^n$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T15:39:09.180" LastActivityDate="2010-11-15T15:39:09.180" CommentCount="4" />
  <row Id="10412" PostTypeId="2" ParentId="10408" CreationDate="2010-11-15T15:11:51.243" Score="8" ViewCount="0" Body="&lt;p&gt;I'll assume you want the divisibility to be in $\mathbb{Q}[x]$.  Write $\displaystyle f(x) = \prod_{i=1}^{k} (x - a_i)^{m_i}$ where the $a_i$ are distinct (in $\mathbb{C}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  $\displaystyle \frac{f&amp;#39;(x)}{f(x)} = \sum_{i=1}^{k} \frac{m_i}{x - a_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a direct consequence of the product rule and you do not need to use any calculus to prove it.  (That is, you don't need to take logarithms.  The derivative can be defined as a totally formal operation on polynomials and this identity holds in $\mathbb{Q}(x)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f&amp;#39;(x)$ divides $f(x)$, then $\frac{f&amp;#39;(x)}{f(x)} = \frac{n}{x - a}$ for some $a$.  Now, the functions $\frac{1}{x - a}$ are linearly independent (you also do not need to use any calculus to prove this) in $\mathbb{C}(x)$.  It follows that this is possible if and only if $f$ has $a$ as a repeated root of multiplicity $n$, so $a$ is the only root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $f$ has the form $c(px - q)^n$ for some integers $c, p, q$ with $a, p \neq 0$.  (I'm ignoring the degenerate cases.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-15T15:18:10.690" LastActivityDate="2010-11-15T15:18:10.690" CommentCount="3" />
  <row Id="10413" PostTypeId="1" AcceptedAnswerId="10434" CreationDate="2010-11-15T15:13:17.183" Score="1" ViewCount="67" Body="&lt;p&gt;Given a $1$ dimensional domain $R$ with fraction field $K$. Suppose $\mathfrak{m}$ is a maximal ideal of $R$. Then, I need to show that $R_m$ is a maximal subring of $K$ in the sense that if $x\in K-R_\mathfrak{m}$, then, $R_m[x]=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Is this true? How would you go about showing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)If this is true, can any of the hypothesis be relaxed (domain, $1$ dimensional,&quot;at a maximal ideal&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-15T15:20:33.493" LastActivityDate="2010-11-15T19:31:05.610" Title="Localization of a 1 dimensional domain at a maximal ideal is a maximal subring of the fraction field" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="10414" PostTypeId="1" CreationDate="2010-11-15T15:26:21.017" Score="16" ViewCount="396" Body="&lt;p&gt;Imagine we have a wizard that knows a few spells. Each spell has 3 attributes:&#xA;Damage, cooldown time, and a cast time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cooldown time: the amount of time (t) it takes before being able to cast that spell again. A spell goes on &quot;cooldown&quot; the moment it begins casting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast time: the amount of time (t) it takes to use a spell. While the wizard is casting something another spell cannot be cast and it cannot be canceled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&#xA;How would you maximize damage given different sets of spells?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to calculate the highest damage per cast time. But what about in situations where it is better to wait then to get &quot;stuck&quot; casting a low damage spell when a much higher one is available: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;1) 100 damage, 1 second cast time, 10 second cool down.&lt;/h1&gt;&#xA;&#xA;&lt;h1&gt;2) 10 damage, 4 second cast time, 0 second cool down.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So, in this case you would cast #1, #2, wait. Cast #1&lt;/p&gt;&#xA;" OwnerUserId="3534" LastEditorUserId="3534" LastEditorDisplayName="" LastEditDate="2010-11-15T16:52:12.767" LastActivityDate="2010-11-19T08:14:00.693" Title="Math Wizardry - Formula for selecting the best spell." Tags="&lt;optimization&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="2" />
  <row Id="10415" PostTypeId="2" ParentId="10379" CreationDate="2010-11-15T15:52:20.070" Score="2" ViewCount="0" Body="&lt;p&gt;&quot;Two equal roots&quot; and &quot;a root of multiplicity (at least) two&quot; mean the same thing. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T15:52:20.070" />
  <row Id="10416" PostTypeId="2" ParentId="10400" CreationDate="2010-11-15T15:55:43.607" Score="1" ViewCount="0" Body="&lt;p&gt;If you are familar  with ideal radicals then, as I mentioned on &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/b24d9f96f53a0b92&quot; rel=&quot;nofollow&quot;&gt;sci.math&lt;/a&gt;, the proof is a one-liner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm rad\ (I^m + \cdots\ + J^n) \ \supset\ I +\cdots+ J = 1\ \ \Rightarrow\ \ I^m +\cdots + J^n\ =\ 1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, and much more generally, it maybe viewed as an immediate consequence of the &lt;strong&gt;Freshman's Dream&lt;/strong&gt; binomial theorem $\rm\ (A + B)^n = A^n + B^n\ $. This theorem is true for both arithmetic of GCDs and (invertible) ideals simply because, &#xA;in both cases, multiplication is cancellative and addition is idempotent, &#xA;i.e. $\rm\ A + A = A\ $ for ideals and $\rm\ (A,A) = A\ $ for GCDs. Combining this with the &#xA;associative, commutative, distributive laws of addition and multiplication &#xA;we obtain an extremely elementary high-school-level proof of the Freshman's Dream: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad (A + B)^4 \ =\  A^4  +  A^3 B  +  A^2 B^2 + AB^3 + B^4 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\phantom{(A + B)^4 }\ =\  A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ {(A + B)^2 }\ =\ \  A^2 + B^2\ $ if $\rm\ A+B\ $ is cancellative, e.g. if $\rm A+B = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same proof works generally since, as above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad (A + B)^{2n}\ =\  A^n\ (A^n + \cdots+ B^n) + (A^n +\cdots+ B^n)\ B^n $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the GCD case  $\rm\ A+B\ := (A,B) = \gcd(A,B)\ $ for $\rm\:A,B\:$ in a GCD-domain, &#xA;i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$.  Hence &#xA;Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a domain, nonzero principal ideals are cancellable, so Dream is true &#xA;for ideals in a PID (e.g. $\mathbb Z\:$), or f.g. (finitely generated) ideals in a Bezout domain. &#xA;More generally, Dream also holds true in any Dedekind domain (e.g. &#xA;any number ring) since nonzero ideals are invertible hence cancellable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact this &quot;Freshman's Dream&quot; is true for all f.g. ideals in domain $\rm\:D\:$ &#xA;iff every nonzero f.g. ideal is invertible. Such domains are known as &#xA;Prufer domains. They're non-Noetherian generalizations of Dedekind &#xA;domains. Moreover they form an important class of domains because &#xA;they may also be equivalently characterized by a large number of &#xA;other important properties, e.g. they are precisely the domains &#xA;satisfying CRT (Chinese Remainder Theorem); $\ $ Gauss's Lemma: the &#xA;content ideal $\rm\ \ c(fg) = c(f)\ c(g)\:$;$\ $  nonzero f.g. ideals are cancellable; &#xA;$\ $ f.g. ideals satisfy contains $\Rightarrow$ divides; $\: $ etc. It's been estimated &#xA;that there are close to a hundred such characterizations known. &#xA;See my&lt;a href=&quot;http://groups.google.com/group/sci.math/msg/ba5ac69dddfc4a67&quot; rel=&quot;nofollow&quot;&gt; sci.math post&lt;/a&gt; for about thirty such characterizations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-15T18:13:58.270" LastActivityDate="2010-11-15T18:13:58.270" />
  <row Id="10417" PostTypeId="2" ParentId="10408" CreationDate="2010-11-15T16:02:39.687" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rolle%27s_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Rolle's theorem&quot;&gt;Rolle's theorem.&lt;/a&gt; Between two zeroes of function lies a zero of its derivative. If f has two distinct zeroes, there is a zero of f' which is not a zero of f. &lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-15T16:02:39.687" CommentCount="1" />
  <row Id="10418" PostTypeId="2" ParentId="10414" CreationDate="2010-11-15T16:47:53.417" Score="5" ViewCount="0" Body="&lt;p&gt;I don't have an answer, but I just want to point out that the greedy algorithm fails.  That is, if we choose to cast, at any point, the available spell which maximizes $\frac{ \text{damage} }{ \text{cast time} }$, we don't actually maximize our damage output because of cooldown.  Consider the pair of spells $(300, 5, 12)$ and $(50, 3, 0)$ (where the three numbers are damage, cast time in second times, and cooldown time in seconds): the greedy algorithm suggests the casting schedule&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cast spell 1 ($300$ damage, $5$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Repeat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which gives about $32.1$ damage per second.  However, the casting schedule&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cast spell 1 ($300$ damage, $5$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Cast spell 2 ($50$ damage, $3$ seconds),&lt;/li&gt;&#xA;&lt;li&gt;Wait $1$ second,&lt;/li&gt;&#xA;&lt;li&gt;Repeat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;gives about $33.3$ damage per second.  So a correct algorithm must take into account which spells are about to finish cooling down.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T16:47:53.417" />
  <row Id="10419" PostTypeId="1" AcceptedAnswerId="10425" CreationDate="2010-11-15T16:55:24.327" Score="0" ViewCount="60" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a function, implicitly defined, how would understand the following; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $y=y(x)$, find $y&amp;#39;(0)$ and $y&amp;#39;&amp;#39;(0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I implicitly derivate it the first time, I have an expression with $dy/dx$. Do I just fill in zero for $x$, remove the parts that gets canceled out? I'm not sure if the answer is a number or a function?&lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T17:37:40.703" LastActivityDate="2010-11-15T19:42:19.387" Title="Easy; understanding question about implicit diff?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10420" PostTypeId="2" ParentId="10392" CreationDate="2010-11-15T17:08:35.637" Score="4" ViewCount="0" Body="&lt;p&gt;Expanding on Mariano's answer: to get a formula for a truth table that has exactly one &lt;code&gt;t&lt;/code&gt; and the rest of the lines are &lt;code&gt;f&lt;/code&gt;, look at the line and write down the values of the variables in that line. For instance, if you wanted a formula for a truth table with three variables as in your second example which has a &lt;code&gt;t&lt;/code&gt; in the third line (corresponding to $X$ and $Z$ true, and $Y$ false) and &lt;code&gt;f&lt;/code&gt; in all the other ones, then since that line is &quot;$X$ is true, $Y$ is false, and $Z$ is true&quot;, then you use the formula $X \wedge (\neg Y)\wedge Z$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you have a formula with &lt;code&gt;t&lt;/code&gt;'s in &lt;em&gt;two&lt;/em&gt; rows and &lt;code&gt;f&lt;/code&gt;s everywhere else. Say, three variables, a &lt;code&gt;t&lt;/code&gt; in row three, a &lt;code&gt;t&lt;/code&gt; in row five, and &lt;code&gt;f&lt;/code&gt;'s everywhere else. A formula that has a &lt;code&gt;t&lt;/code&gt; in just row three is, as before, $X\wedge (\neg Y)\wedge Z$. A formula that has a &lt;code&gt;t&lt;/code&gt; in just row five is $(\neg X)\wedge (\neg Y)\wedge Z$. So a formula that as a &lt;code&gt;t&lt;/code&gt; in just row three &lt;em&gt;or&lt;/em&gt; row five is:&#xA;$$\Bigl( X\wedge (\neg Y)\wedge Z\Bigr) \vee \Bigl((\neg X)\wedge (\neg Y)\wedge Z\Bigr).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary number of &lt;code&gt;t&lt;/code&gt;'s, just take the disjunction of enough formulas, each of which corresponds to a table with just one &lt;code&gt;t&lt;/code&gt;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T17:08:35.637" CommentCount="2" />
  <row Id="10422" PostTypeId="1" AcceptedAnswerId="10458" CreationDate="2010-11-15T17:12:18.763" Score="4" ViewCount="168" Body="&lt;p&gt;Hi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a finite abelian group $G$. $H$ and $K$ subgroups of $G$ such that $G/H$ and $G/K$ are primary cyclic. $H$ and $K$ intersect trivially ($H\cap K=1$),  and $H\neq 1$,  $K\neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it then true that $HK=G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far: It's easy to show that $[G:HK] | ([G:H], [G:K])$. So if $G/H$ and $G/K$ are primary cyclic for different primes, then we're done. But I don't know what to do if they're not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T17:15:33.960" LastActivityDate="2010-11-16T03:45:32.357" Title="$G$ finite abelian. $G/H$, $G/K$ primary cyclic. $H\cap K=1$. $H\neq 1\neq K$. Is $HK=G$?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10423" PostTypeId="2" ParentId="10419" CreationDate="2010-11-15T17:15:02.810" Score="0" ViewCount="0" Body="&lt;p&gt;There are lot of resources on the web and i think you can figure out your problem by reading some sources. Here i give some files in which the concept of &lt;em&gt;implicit differentiation&lt;/em&gt; has been written nicely.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.mash.dept.shef.ac.uk/Resources/web-implicit.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mash.dept.shef.ac.uk/Resources/web-implicit.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=ximF06lmqPM&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=ximF06lmqPM&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Regarding your problem suppose assume you want to calculate the &lt;em&gt;derivative&lt;/em&gt; of the function $z=f(x,y)$, (say $z=x^{2}+y^{2}$) &lt;em&gt;where  $y$ is a function of $x$&lt;/em&gt;, then by chain rule you have $$\frac{dz}{dx} = 2x + 2y \cdot \frac{dy}{dx}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-15T17:15:02.810" CommentCount="2" />
  <row Id="10424" PostTypeId="2" ParentId="6314" CreationDate="2010-11-15T17:24:41.013" Score="0" ViewCount="0" Body="&lt;p&gt;The answer to the question as stated is no, as others have explained. However, if we relax the hypothesis from disjoint to non-overlapping, then the answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two intervals $I_1$ and $I_2$ are non-overlapping if $I_1^{\circ}\cap I_2^{\circ}=\emptyset$; that is, if their &lt;em&gt;interiors&lt;/em&gt; are disjoint. If the intervals are closed and non-overlapping, then they intersect at most in their boundaries. For example, in $\mathbb{R}$, the intervals $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right]$ are non-overlapping, but clearly not disjoint as they share the point $\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-15T17:24:41.013" />
  <row Id="10425" PostTypeId="2" ParentId="10419" CreationDate="2010-11-15T17:47:54.730" Score="2" ViewCount="0" Body="&lt;p&gt;If you have a formula for $\frac{dy}{dx}$ and you want to find the value of $y&amp;#39;(0)$, then &lt;em&gt;first&lt;/em&gt; you plug in $x=0$ to the implicit formula defining $y$, and evaluate to get the value of $y$. Then you plug in the value of $y$ you found and $x=0$ into the formula for $\frac{dy}{dx}$ to get the value of $y&amp;#39;(0)$.  Similarly for $y&amp;#39;&amp;#39;(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you are looking at the circle of radius $1$, $x^2+y^2=1$. Differentiating implicitly, we have:&#xA;\begin{align*}&#xA;\frac{d}{dx}(x^2+y^2) &amp;amp;= \frac{d}{dx}1\\&#xA;2x + 2y\frac{dy}{dx} &amp;amp;= 0\\&#xA;2y\frac{dy}{dx} &amp;amp;= -2x\\&#xA;\frac{dy}{dx} &amp;amp;= -\frac{x}{y}.&#xA;\end{align*}&#xA;To find $y&amp;#39;(0)$, we first plug in $x=0$ into $x^2+y^2=1$ to find the value of $y$ that corresponds to $x=0$; we get $y^2=1$, so $y=\pm 1$. Thus, we have two points, $(0,1)$ and $(0,-1)$. Then we plug in these values into the formula for $\frac{dy}{dx}$, and we get that in both cases we have $y&amp;#39;(0) = -\frac{0}{\pm 1}=0$. So $y&amp;#39;(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second derivative, we again take derivatives:&#xA;\begin{align*}&#xA;\frac{dy}{dx} &amp;amp;= -\frac{x}{y}\\&#xA;\frac{d}{dx}\left(\frac{dy}{dx}\right) &amp;amp;= -\frac{d}{dx}\left(\frac{x}{y}\right)\\&#xA;\frac{d^2y}{dx^2} &amp;amp;= -\left(\frac{y(x)' - x(y')}{y^2}\right)\\&#xA;\frac{d^2y}{dx^2} &amp;amp;= \frac{x\frac{dy}{dx} - y}{y^2}\\&#xA;\frac{d^2y}{dx^2} &amp;amp;= \frac{x\left(-\frac{x}{y}\right) - y}{y^2}.&#xA;\end{align*}&#xA;The last equality, because $\frac{dy}{dx}=-\frac{x}{y}$. So now we plug in $x=0$, $y=\pm 1$ into &lt;em&gt;this&lt;/em&gt; equation, and we get that $y&amp;#39;&amp;#39;(0) = -\frac{y}{y^2} = -\frac{1}{y} = -\frac{1}{\pm 1}=\mp 1$. That is, $y&amp;#39;&amp;#39;(0)=1$ on $(0,-1)$, and $y&amp;#39;&amp;#39;(0)=-1$ on $(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if your implicit curve has a single point on the vertical line $x=0$, then you only have one value of $y$ to contend with, so the process will be simpler (as far as evaluating). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;: Notice that you don't take the derivative of the function you get &lt;em&gt;after&lt;/em&gt; evaluating at $x=0$ to get the second derivative! You have to take the entire function and only &lt;em&gt;after&lt;/em&gt; differentiating &lt;em&gt;again&lt;/em&gt; do you once again plug in $x=0$ and the value of $y$ in order to find $y&amp;#39;&amp;#39;(0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; If all you want is to find the values of $y&amp;#39;&amp;#39;(0)$ and $y&amp;#39;(0)$, you can avoid some complications in the computation of $\frac{d^2y}{dx^2}$ by starting from before you solved for $y&amp;#39;$. In the example above, after we find that when $x=0$ we have $y=\pm 1$ and $y&amp;#39;(0)=0$, we can start from the second line in the computation of $y&amp;#39;$:&#xA;$$2x + 2y\frac{dy}{dx} = 0,$$&#xD;&#xA;and after simplifying by cancelling the $2$, we can take derivatives again: just be careful with the product rule:&#xA;\begin{align*}&#xA;x + 2yy' &amp;amp;= 0\\&#xA;\frac{d}{dx}(x+2yy') &amp;amp;= 0\\&#xA;1 + 2(y'y' + yy'') &amp;amp;= 0\\&#xA;2(y')^2 + yy'' &amp;amp;= -1.&#xA;\end{align*}&#xA;Now, plugging in $x=0$, $y&amp;#39;(0)=0$, and $y(0)=\pm 1$, we get $\pm 1y&amp;#39;&amp;#39;(0)=-1$, hence $y&amp;#39;&amp;#39;(0) = \mp 1$; i.e., $y&amp;#39;&amp;#39;(0)=1$ when $y=-1$ and $y&amp;#39;&amp;#39;(0)=-1$ when $y=1$. Then you don't have to worry about fractions until you start solving. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T19:42:19.387" LastActivityDate="2010-11-15T19:42:19.387" CommentCount="2" />
  <row Id="10427" PostTypeId="1" CreationDate="2010-11-15T18:26:12.963" Score="3" ViewCount="174" Body="&lt;p&gt;What is $x$ in closed form if $2x-\sin2x=\pi/2$, $x$ in the first quadrant?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="3538" LastEditorDisplayName="" LastEditDate="2010-11-15T19:21:46.793" LastActivityDate="2010-11-15T21:24:00.893" Title="Solving $2x - \sin 2x = \pi/2$ for $0 &lt; x &lt; \pi/2$" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10429" PostTypeId="1" AcceptedAnswerId="10430" CreationDate="2010-11-15T18:41:23.470" Score="4" ViewCount="196" Body="&lt;p&gt;All power sets contain an empty set, so does this make the power set itself an empty set? Or am I mistaken?&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T01:54:07.327" LastActivityDate="2010-11-20T01:54:07.327" Title="Is a power set null because it contains an empty set?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="10430" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T18:48:14.757" Score="6" ViewCount="0" Body="&lt;p&gt;The short answer to your question is &quot;A power set is &lt;em&gt;never&lt;/em&gt; null (or empty), even if it is the power set of the empty set.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A power set $\mathcal{P}(X)$ is the &lt;em&gt;set of subsets&lt;/em&gt; of a given set $X$. The elements of $\mathcal{P}(X)$ are themselves sets, trivial or not. Since the empty set $\emptyset$ -- the set containing no elements -- is a subset of every set $X$, then $\emptyset \in \mathcal{P}(X)$ as a &lt;em&gt;set&lt;/em&gt;. Thus, every power set is non-trivial and has at least one element, namely the empty set. For example, if $X =$ {$a, b,c$}, then $\mathcal{P}(X)$ has 8 elements: the empty set $\emptyset$, the singletons {$a$}, {$b$} and {$c$}, all pairs {$a,b$}, {$a,c$} and {$b,c$} and the entire set $X =$ {$a, b,c$}. In fact, if $X$ has $|X|$ elements, one can prove that $|\mathcal{P}(X)| = 2^{|X|}$, even in the case that $X$ is infinite. When $X$ is finite, the proof follows from the identity &#xA;\begin{eqnarray}&#xA;\sum_{k = 0}^{n} \binom{n}{k} = 2^{n},&#xA;\end{eqnarray}&#xA;which is the sum of all ways to choose $k$ objects from $n$ objects as $k$ increases from $0$ to $n$. I'll leave it as an exercise to show how this counts $|\mathcal{P}(X)|$ with $|X| = n$.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T20:57:33.710" LastActivityDate="2010-11-15T20:57:33.710" />
  <row Id="10431" PostTypeId="2" ParentId="10427" CreationDate="2010-11-15T18:49:01.140" Score="4" ViewCount="0" Body="&lt;p&gt;The solution is given by $$\displaystyle  x = \pi/4 + D/2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle D$ is the root of $\cos y = y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The root of $\displaystyle \cos y = y$ is nowadays known as the &lt;a href=&quot;http://mathworld.wolfram.com/DottieNumber.html&quot; rel=&quot;nofollow&quot;&gt;Dottie Number&lt;/a&gt; and apparently has no known &quot;closed form&quot; solution. If you consider this number to be part of your constants, then the above can be considered a closed form solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For a proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle y = \sin(2x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x  = \pi/2 + y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y = \sin 2x = \sin (\pi/2 + y) = \cos y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The root of $$\displaystyle y = \cos y$$ is $\displaystyle y = 0.739085\dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle \pi/2 &amp;gt; x \gt \pi/4$ (as $\displaystyle f(x) = 2x - \sin 2x$ is increasing in $\displaystyle [0,\pi/2]$), so if $\displaystyle x = \pi/4 + z$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sin(2x) = \sin(\pi/2 + 2z) = \cos 2z = 0.739085\dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus $\displaystyle z = \dfrac{0.739085\dots}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\displaystyle x \sim \pi/4 + \dfrac{0.739085}{2} \sim 1.154940730005\dots$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Also: &lt;a href=&quot;http://oeis.org/A003957&quot; rel=&quot;nofollow&quot;&gt;A003957&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-15T19:38:44.353" LastActivityDate="2010-11-15T19:38:44.353" CommentCount="1" />
  <row Id="10432" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T18:49:08.457" Score="13" ViewCount="0" Body="&lt;p&gt;You seem to be saying that if a set contains the empty set, it is itself empty. This not true. In fact the opposite is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is empty if and only it contains no elements. If a set contains the empty set, it has at least one element (the empty set) and is thus, by definition, not empty.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-15T18:49:08.457" />
  <row Id="10433" PostTypeId="2" ParentId="10348" CreationDate="2010-11-15T18:52:27.657" Score="2" ViewCount="0" Body="&lt;p&gt;Rubin [1] gives an algorithm for counting Hamiltonian paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica has tools for working with Hamiltonian paths, but the naive&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Length[HamiltonianCycles[G, All]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives an out-of-memory exception for this particular graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Frank Rubin, &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=321854&quot; rel=&quot;nofollow&quot;&gt;A search procedure for Hamilton paths and circuits&lt;/a&gt; (1974)&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-11-15T19:23:00.677" LastActivityDate="2010-11-15T19:23:00.677" />
  <row Id="10434" PostTypeId="2" ParentId="10413" CreationDate="2010-11-15T18:52:36.267" Score="1" ViewCount="0" Body="&lt;p&gt;No, I am afraid this maximality is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us always suppose that $R$ is a domain: else there is no fraction field (although there are substitutes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the result is certainly not true for higher dimensional rings. For example take&#xA;$R=\mathbb C[x,y] , m=(x,y) , K=\mathbb C(x,y)$ . &#xA;Then you have strict inclusions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb C[x,y]_m \subset \mathbb C[x,y]_m [1/x] \subset \mathbb C (x,y)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we see that the result in your case is not purely formal and we must use dimension one.  But unfortunately the result may not be true even then, which answers  question 1) negatively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed,  just observe   in the case $R=\mathbb C[[t^2,t^3]], \; m=(t^2,t^3), \; R=R_m$ the strict inclusions &#xA;$$R=\mathbb C[[t^2,t^3]] \subset \mathbb C [[t]]\subset \mathbb C ((t))$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To end on a cheerful note, your result is indeed true if $R_m$ is a discrete valuation ring.&#xA;Here is the proof. Take $a\in K\setminus R_m$ and consider $R_m\subset R_m [a]\subset K$ ( the first inclusion is strict). Since $a$ is not in $R_m$, we know that $1/a$ *is* in $R_m$: this is characteristic of valuation rings. Hence $1/a$ is &lt;em&gt;a fortiori&lt;/em&gt; in $R_m[a]$ so that $1\in R[a]$ and we have proved $R[a]=K$.  &lt;/p&gt;&#xA;" OwnerUserId="3217" LastEditorUserId="3217" LastEditorDisplayName="" LastEditDate="2010-11-15T19:31:05.610" LastActivityDate="2010-11-15T19:31:05.610" CommentCount="1" />
  <row Id="10435" PostTypeId="1" AcceptedAnswerId="10442" CreationDate="2010-11-15T18:57:47.547" Score="0" ViewCount="152" Body="&lt;p&gt;Let a statement P is &quot;X is true if and only if Y is true&quot;. What is the negation of P? I am little confused. It seems that digital equivalent of this statement is P = X and Y. Hence negation of P is (not X) or (not Y) i.e. Either X or Y is false. Am I right guys?   &lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2010-11-15T20:47:28.680" Title="Negation of if and only if?" Tags="&lt;logic&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="10436" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T18:59:41.827" Score="1" ViewCount="0" Body="&lt;p&gt;To add on the previous answers, I just want to clarify that as it was remarked before the power set of $A$, usually denoted by $\mathcal{P}(A)$ is the collection of all the subsets of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, if $A = \emptyset$ then it has only one subset - itself, as the empty set is a subset of every given set. Therefore, $\mathcal{P}(\emptyset) = \{\emptyset \}$, and again as remarked before - it means that it has one element, and more specifically that it is non empty.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-11-15T18:59:41.827" />
  <row Id="10437" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:04:30.670" Score="7" ViewCount="0" Body="&lt;p&gt;The digital equivalent is P = X XNOR Y, and thus the negation is (not P) = X XOR Y. In other words, P is false when X is true but Y is false, or when X is false but Y is true.&lt;/p&gt;&#xA;" OwnerUserId="759" LastActivityDate="2010-11-15T19:04:30.670" CommentCount="1" />
  <row Id="10439" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:04:54.610" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, but you have to be very precise here, because the negation of implication is &lt;em&gt;exclusive&lt;/em&gt; (not inclusive) OR. So the answer is &quot;Either $X$ or $Y$ is false, but not both&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you are confused, start with a truth table for implication and then negate it. Resulting table matches XOR (exclusive OR). &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-15T19:04:54.610" />
  <row Id="10440" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:07:36.467" Score="1" ViewCount="0" Body="&lt;p&gt;You are not right.  Let $X = (\ell$ is even) and $Y = (\ell$ is not odd).  Then clearly $X \Leftrightarrow Y$, but &quot;($\ell$ is not even) or ($\ell$ is odd)&quot; is strictly weaker; you want &quot;($\ell$ is not even) $\Leftrightarrow$ ($\ell$ is odd)&quot; to be true.&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-15T19:07:36.467" />
  <row Id="10441" PostTypeId="1" AcceptedAnswerId="10445" CreationDate="2010-11-15T19:09:12.910" Score="0" ViewCount="59" Body="&lt;p&gt;I'm working through an exercise in Hartschorne concerning direct limits of sheaves on noetherian topological spaces. It seems natural to use the the fact that every open subset of a noetherian topological space is the quasi-compact in its induced topology. I'm some trouble proving what I need when I only use this property. I'm guessing that I'll have to use noetherian-ness more directly, but I thought I would ask if anyone knew an example offhand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a non-notherian topological space which has this property?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="756" LastActivityDate="2010-11-16T23:00:22.443" Title="Are there non-noetherian topological spaces in which every open subset is quasi-compact?" Tags="&lt;topology&gt;&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="10442" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T19:15:00.453" Score="10" ViewCount="0" Body="&lt;p&gt;$X\leftrightarrow Y$ is the conjunction of $X\leftarrow Y$ and $X\rightarrow Y$. The negation of a conjunction is the disjunction of the negations; the negation of $P\rightarrow Q$ is $P\wedge \neg Q$. So we have:&#xA;\begin{align*}&#xA;\neg(X\leftrightarrow Y) &amp;amp;\Longleftrightarrow \neg\Bigl( (X\rightarrow Y)\wedge (Y\rightarrow X)\Bigr)\\&#xA;&amp;amp;\Longleftrightarrow \neg(X\rightarrow Y)\vee \neg(Y\rightarrow X)\\&#xA;&amp;amp;\Longleftrightarrow (X\wedge \neg Y) \vee (Y\wedge \neg X).&#xA;\end{align*}&#xA;So the negation of &quot;$X$ is true if and only if $Y$ is true&quot; is &quot;Either $X$ is true and $Y$ is false, or $X$ is false and $Y$ is true.&quot; &lt;em&gt;Added&lt;/em&gt;: as it happens, as noted by Rahul Narain in his comment, this is in turn equivalent to &quot;$X$ is true if and only if $Y$ is false&quot; (just compare the cases when they are each true). So you also get that&#xA;$$\neg(X\leftrightarrow Y) \Longleftrightarrow X\leftrightarrow \neg Y \Longleftrightarrow \neg X\leftrightarrow Y.$$&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T20:01:28.413" LastActivityDate="2010-11-15T20:01:28.413" CommentCount="6" />
  <row Id="10443" PostTypeId="1" AcceptedAnswerId="10448" CreationDate="2010-11-15T19:17:27.760" Score="7" ViewCount="271" Body="&lt;p&gt;In school and at university we never had eigenvalues nor differential equations, so these concepts were really giving me a hard time. Now I developed some intuition for both concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned that both are connected in some way, i.e. there is an eigenvector/eigenvalue approach  to solve differential equations. Unfortunately most of the text I find are way beyond my understanding and seem to require very extensive study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a possibility to provide me with an intuition of the connection anyway and perhaps give an easy example of how to use this approach to solve a differential equation?&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-15T21:17:38.093" Title="How are eigenvectors/eigenvalues and differential equations connected?" Tags="&lt;linear-algebra&gt;&lt;intuition&gt;&lt;differential-equations&gt;&lt;examples&gt;&lt;eigenvectors&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="10444" PostTypeId="2" ParentId="10441" CreationDate="2010-11-15T19:18:07.953" Score="1" ViewCount="0" Body="&lt;p&gt;No.  Noetherianness is equivalent to the condition that every open subset is compact, and is in fact equivalent to the condition that every subset is compact.  This is a nice exercise.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T19:18:07.953" />
  <row Id="10445" PostTypeId="2" ParentId="10441" CreationDate="2010-11-15T19:20:55.267" Score="2" ViewCount="0" Body="&lt;p&gt;If every open subspace of a space is quasi compact, then the space is Noetherian. &#xA;This is exercise 6.5 in Atiyah Macdonald. It's an easy proof. Let me know if you need more help with it.&#xA;Hint: If you consider an increasing sequence of open sets, its union is open and hence quasi compact. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-16T23:00:22.443" LastActivityDate="2010-11-16T23:00:22.443" CommentCount="1" />
  <row Id="10446" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T19:24:29.910" Score="2" ViewCount="0" Body="&lt;p&gt;A good reference is here: &lt;a href=&quot;http://www.sosmath.com/diffeq/system/linear/eigenvalue/eigenvalue.html&quot; rel=&quot;nofollow&quot;&gt;http://www.sosmath.com/diffeq/system/linear/eigenvalue/eigenvalue.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T19:24:29.910" />
  <row Id="10447" PostTypeId="2" ParentId="10427" CreationDate="2010-11-15T19:25:07.847" Score="1" ViewCount="0" Body="&lt;p&gt;To get a graphical impression of the solution have a look here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+2+x+-+sin+(2+x)+%3D+Pi/2+for+x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=solve+2+x+-+sin+(2+x)+%3D+Pi/2+for+x&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: That WA doesn't give a closed form is at least a hint that there might be none (very probable) &lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-15T19:25:07.847" CommentCount="4" />
  <row Id="10448" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T19:26:53.210" Score="5" ViewCount="0" Body="&lt;p&gt;You could think of eigenvectors as giving you the coordinate system in which your problem has the simplest form. It's analogous to working physical problems by turning your coordinate system the right way to make some term go away.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-15T19:26:53.210" />
  <row Id="10449" PostTypeId="2" ParentId="10379" CreationDate="2010-11-15T19:34:09.913" Score="0" ViewCount="0" Body="&lt;p&gt;If you want to be totally precise, you have to say that it has one real root with multiplicity 2. But &quot;two equal real roots&quot; is almost universally understood as a shortcut for saying that.&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-15T22:49:37.140" LastActivityDate="2010-11-15T22:49:37.140" />
  <row Id="10450" PostTypeId="1" AcceptedAnswerId="10451" CreationDate="2010-11-15T19:38:56.710" Score="3" ViewCount="55" Body="&lt;p&gt;I was reviewing quantum factorization and am slightly unclear on a classical detail of order-finding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a (suitably nice) periodic function $f$ with unknown period $r$ and a power of two $N &amp;gt; r^2$, the quantum subroutine yields (with bounded probability) a result of the form $\lfloor jr/N \rceil$ for $1 \le j \le r-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, $\left \lvert \lfloor jr/N \rceil/N - j/r \right \rvert \le 1/2N$, so since $N &amp;gt; r^2$ a theorem assures us that $j/r$ is a convergent in the continued fraction expansion of $\lfloor jr/N \rceil/N$. Now I see lots of hand-waving at this point in most discussions of the algorithm. If the length of the CFE is $poly(\log(N))$, then this hand-waving can be trivially dispensed with (perhaps at the cost of a slight inefficiency). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does the CFE of $M/N$ have length $poly(\log(N))$?&lt;/p&gt;&#xA;" OwnerUserId="241" LastActivityDate="2010-11-15T19:49:54.683" Title="What is the length of a continued fraction expansion of a rational number?" Tags="&lt;continued-fractions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10451" PostTypeId="2" ParentId="10450" CreationDate="2010-11-15T19:49:54.683" Score="4" ViewCount="0" Body="&lt;p&gt;The length of the CF of a rational number $\displaystyle \dfrac{a}{b}$ is closely related to the number of steps it takes in the Euclidean algorithm for finding gcd of $\displaystyle a,b$, which can, in the worst case, be shown to be proportional to the number of digits in $\displaystyle b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the claim is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://www.albany.edu/~hammond/gellmu/examples/confrac.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.albany.edu/~hammond/gellmu/examples/confrac.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm#Number_of_steps&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euclidean_algorithm#Number_of_steps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-15T19:49:54.683" />
  <row Id="10452" PostTypeId="1" CreationDate="2010-11-15T19:54:01.687" Score="1" ViewCount="156" Body="&lt;p&gt;If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then I would like to prove the following two identities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$$ \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$$ \frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots +  \frac{1}{a_n} \biggr)$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-15T20:18:20.577" LastActivityDate="2010-11-16T18:29:21.037" Title="Prove an identity involving terms in arithmetic progression." Tags="&lt;algebra-precalculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="17" />
  <row Id="10453" PostTypeId="1" AcceptedAnswerId="10455" CreationDate="2010-11-15T19:59:55.163" Score="2" ViewCount="63" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a newbie question, but I would be grateful for any reference that you could give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $f(x) \in \mathbb{A}$, where $x \in \mathbb{A}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a symbol to indicate the repeated application of $f$ unto its results, in a manner similar to $\sum$ and $\prod$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-15T20:04:11.983" Title="Newbie question about the re-application of a function on its result" Tags="&lt;analysis&gt;&lt;notation&gt;&lt;elementary&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="10454" PostTypeId="2" ParentId="10453" CreationDate="2010-11-15T20:03:26.117" Score="5" ViewCount="0" Body="&lt;p&gt;It is usually indicated by $f\circ f$ or if there are more than two repeated applications, simply $f^n(x)$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T20:03:26.117" CommentCount="3" />
  <row Id="10455" PostTypeId="2" ParentId="10453" CreationDate="2010-11-15T20:04:11.983" Score="5" ViewCount="0" Body="&lt;p&gt;The symbol $\circ$ is used for composition of functions: $g\circ h(x)$ means $g(h(x))$. For composing a function with itself, many people use $f^{\circ n}(x)$, defined inductively by&#xA;\begin{align*}&#xA;f^{\circ 1}(x) &amp;amp;= f(x),\\&#xA;f^{\circ(k+1)}(x) &amp;amp;= f\Bigl( f^{\circ k}(x)\Bigr).&#xA;\end{align*}&#xA;So $f^{\circ 2}(x) = f(f(x))$; $f^{\circ 3}(x) = f(f^{\circ 2}(x)) = f(f(f(x)))$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timothy Wagner also notes that $f^n(x)$ is often used; that is true, but it has the disadvantage of often being ambiguous. For example, $\sin^2(x)$ is (almost) always understood to mean $(\sin(x))^2$, and not $\sin(\sin(x))$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T20:04:11.983" />
  <row Id="10457" PostTypeId="2" ParentId="10452" CreationDate="2010-11-15T20:15:26.617" Score="5" ViewCount="0" Body="&lt;p&gt;This question is now old enough for some more complete answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For number 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^{n-1} \frac{1}{ \sqrt{a_k}+ \sqrt{a_{k+1}}} = &#xD;&#xA;\sum_{k=1}^{n-1} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$&#xD;&#xA;where $d$ is the common difference,&#xA;$$ = \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right)&#xD;&#xA;= \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})}$$&#xD;&#xA;$$= \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for number 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = 2 \sum_{k=1}^n \frac{1}{a_k} = &#xD;&#xA;\left( \frac{1}{a_1} + \frac{1}{a_n} \right) + &#xD;&#xA;\left( \frac{1}{a_2} + \frac{1}{a_{n-1}} \right) + \cdots +&#xD;&#xA;\left( \frac{1}{a_n} + \frac{1}{a_1} \right)$$&#xD;&#xA;$$ = \sum_{k=1}^n \frac{a_{n-k+1}+ a_k}{a_k a_{n-k+1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $ a_{n-k+1}+ a_k = 2a_1 + (n-1)d = a_1 + a_n$ and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = (a_1+a_n) \sum_{k=1}^n \frac{1}{a_k a_{n-k+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the result follows.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-16T10:55:17.520" LastActivityDate="2010-11-16T10:55:17.520" />
  <row Id="10458" PostTypeId="2" ParentId="10422" CreationDate="2010-11-15T20:16:58.440" Score="4" ViewCount="0" Body="&lt;p&gt;If $a\in G$ is of order $p^n$ for some prime $p$ and $a \notin H$ then $a+H$ is of order $p^d$ for some $d\leq n$ so $G/H$ must be a p-group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this argument we also know that $H$ must contain all elements that are not of order $p^m$ for some m. So if |G| has at least two different prime divisors, then $G/H$ and $G/K$ must be p-group and q-group for different primes (because $H\cap K= 1$). Therefore all the p-sylow groups are contained in at least one of H or K so HK=G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are now remained with the case where $|G|=p^n$. I tried to prove this case (for too much time), but it appears that this is not true. For example, take $G=\mathbb{Z}_2 \times \mathbb{Z}_4$ and let H be the subgroup $\{ (1,0), (0,0) \}$ and $K=\{ (1,2), (0,0) \}$, then $G/H\cong G/K \cong \mathbb{Z}_4 $ and $|HK|=4 &amp;lt; |G|=8$&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-15T20:16:58.440" CommentCount="2" />
  <row Id="10459" PostTypeId="2" ParentId="10389" CreationDate="2010-11-15T20:18:30.210" Score="0" ViewCount="0" Body="&lt;p&gt;For suitable constant $c$, $f(n) = (T(n)/n^2) + c$ satisfies a recurrence $f(n) = Af(n/2)+Bf(n/3)$ which has solutions (for continuous-valued $n$) of the form $n^\alpha$.  This should give a guess about the asymptotics for integer $n$ and one could probably incorporate lower order terms (fluctuations from the complex solutions with real part less than that of $\alpha$) by analyzing the zeros of $G(x)=1 - Ax^{\log 2} - Bx^{\log 3}$ or the generating function with real exponents (generalized power series expansion) for $1/G(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this type of exercise for the large $n$ asymptotics and translation to the continuous problem is in books such as (e.g.) Flajolet and Sedgewick's text on asymptotic combinatorics.  However, for this specific recurrence it might be possible to go beyond the standard asymptotics and handle integer-valued $n$ directly, because the operations of replacing $n$ by $P(n)=[n/2]$ and $Q(n)=[n/3]$ commute.  One could look at base 6 expansions of $n$, or classify integers according to which operations $P^a Q^b$ take them to $0$ and $1$, and sums over the $(a,b)$ would give an integer-valued expansion formula for $f(n)$, similar to the binomial theorem or the expansion of a rational generating function $1/(1-H(x))$ as a geometric series.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-15T23:44:30.360" LastActivityDate="2010-11-15T23:44:30.360" />
  <row Id="10461" PostTypeId="2" ParentId="10452" CreationDate="2010-11-15T20:24:20.257" Score="3" ViewCount="0" Body="&lt;p&gt;For the first one, use induction (or) note that $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$, where $d$ is the common difference between the successive terms. Now use the telescopic summation to cancel out the terms in the numerator and massage it to get the final expression on the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second one, try to write each term on the Left Hand Side as a difference of two terms and proceed.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-16T02:52:57.120" LastActivityDate="2010-11-16T02:52:57.120" CommentCount="1" />
  <row Id="10462" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T20:36:53.050" Score="3" ViewCount="0" Body="&lt;p&gt;The collection of all functions that are differentiable at least $n$ times (let's call it $\mathcal{F}_n$ just for the purposes of this post) form a vector space; and the operator $D$ defined by $D(f) = f&amp;#39;$ is linear ($D(f+g)=D(f)+D(g)$, and $D(\alpha f) = \alpha D(f)$ for all $f,g$ and scalars $\alpha$). So $D$ is a linear transformation from the vector space of functions that can be differentiated at least $n$ times, and functions that can be differentiated at least $n-1$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you have a &lt;em&gt;homogeneous&lt;/em&gt; linear differential equation: &#xA;$$ f_n(x) y^{(n)}(x) + \cdots + f_1(x)y&amp;#39;(x) + f_0(x)y(x) = 0.$$&#xD;&#xA;Then the collection of all function $y(x)$ that are solutions to this system form a subspace of $\mathcal{F}_n$: because if $y_1$ and $y_2$ are solutions, then so is $y_1+\alpha y_2$ for any scalar $\alpha$, and the zero function is certainly a solution. It is, in fact, the nullspace of a certain linear transformation, namely, the linear transformation $L$ given by&#xA;$$L(y) = f_n(x)y^{(n)}(x)+\cdots + f_1(x)y&amp;#39;(x) + f_0(x)y(x).$$&#xD;&#xA;So one can bring some linear algebra to bear on this problem; e.g., determining the dimension of the solution space, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in analogy to the case of systems of linear equations, consider a *&lt;em&gt;non&lt;/em&gt;*homogeneous linear differential equation:&#xA;$$f_n(x) y^{(n)}(x) + \cdots + f_1(x)y&amp;#39;(x) + f_0(x)y(x) = g(x).$$&#xD;&#xA;Suppose you could find a particular solution $y_p$ to this equation. Then, if $y_h$ is any solution to the corresponding homogeneous equation, then $y_p+y_h$ is a solution to the nonhomogeneous equation as well; and if $z_p$ is another solution to the nonhomogeneous equation, then $y_p-z_p$ is a solution to the homogeneous equation. So &lt;em&gt;every&lt;/em&gt; solution to the nonhomogeneous equation is of the form $y_p + y_h$, where $y_p$ is the particular solution to found, and $y_h$ is a solution to the associated homogeneous equation. This is exactly the same thing that happens with systems of linear equations (where the unknowns are numbers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases, you can show that any solutions to your equations will actually be infinitely differentiable (e.g., if your $f$s are infinitely differentiable). Then you can think of the linear transformation $L$ as a linear operator on the space of infinitely differentiable functions. Then eigenvectors and eigenvalues can come into play: eigenvectors and eigenvalues give you simpler ways of thinking about a linear transformation, so they give you simpler ways of thinking about &lt;em&gt;this&lt;/em&gt; particular linear transformation (which happens to correspond to solutions of a differential equation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, systems of linear differential equations very naturally lead to linear transformations where the eigenvectors and eigenvalues play a key role in helping you solve the system, because they &quot;de-couple&quot; the system, by allowing you to think of a complex system in which each of the variables affects the derivative of the others as a system in which you have some new variables that are completely independent of one another (or in the case of generalized eigenvectors, easily dependent on only some of the others). This makes the system easier to solve.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-15T20:36:53.050" />
  <row Id="10463" PostTypeId="2" ParentId="10427" CreationDate="2010-11-15T20:38:45.800" Score="2" ViewCount="0" Body="&lt;p&gt;Repeating the same thing as the other answers, but with a moderately more elegant one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = 2x - \frac{\pi}{2}$, then substituting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = sin(y+\frac{\pi}{2})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = cos(y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = D$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = \frac{y+\frac{\pi}{2}}{2} = \frac{D}{2} + \frac{\pi}{4}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditorDisplayName="" LastEditDate="2010-11-15T21:24:00.893" LastActivityDate="2010-11-15T21:24:00.893" />
  <row Id="10464" PostTypeId="2" ParentId="10375" CreationDate="2010-11-15T20:41:25.423" Score="0" ViewCount="0" Body="&lt;p&gt;Ah, I forgot about the Yoneda lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think it's impossible to show that a strictly increasing function into a partial order is monic, because the normal proof uses the fact that monic on global elements implies monic. In fact, this lemma is probably used a lot.&lt;/p&gt;&#xA;" OwnerUserId="3518" LastActivityDate="2010-11-15T20:41:25.423" />
  <row Id="10465" PostTypeId="2" ParentId="10435" CreationDate="2010-11-15T20:47:28.680" Score="0" ViewCount="0" Body="&lt;p&gt;The assertion $X \leftrightarrow Y$ can also be written as $X = Y$. So its negation is $X \neq Y$, which is the same as $X = \overline{Y}$ (since $X,Y \in \{0,1\}$), which is the same as $\overline{X} \leftrightarrow Y$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-15T20:47:28.680" />
  <row Id="10466" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T20:55:31.357" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose $\sum a_n y^{(n)} = 0$ is a homogeneous linear differential equation with constant coefficients.  Let $D$ be the linear operator defined on, say, the space $C^{\infty}(\mathbb{R})$ of smooth functions $\mathbb{R} \to \mathbb{R}$ which sends a function $y$ to its derivative $y&amp;#39;$.  Then the solutions to the above differential equation are the nullspace of the operator $\sum a_n D^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this nullspace is finite-dimensional, so we can use finite-dimensional linear algebra on it.  It is not just a vector space, though: it comes equipped with an action of $D$, which commutes with $\sum a_n D^n$, and which therefore splits the nullspace into eigenspaces of $D$.  In the simplest case, these eigenspaces will all be one-dimensional.  Now, the eigenvectors of $D$ are precisely the functions satisfying $Dy = \lambda y$ for some eigenvector $\lambda$, which are precisely the exponential functions $e^{\lambda y}$.  And it's not hard to see that such a function is a solution of the differential equation if and only if $\lambda$ is a root of the &lt;strong&gt;characteristic polynomial&lt;/strong&gt; $\sum a_n \lambda^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example.&lt;/em&gt;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_harmonic_motion&quot; rel=&quot;nofollow&quot;&gt;Simple harmonic motion&lt;/a&gt; is governed by the differential equation $m D^2 + k D^0 = 0$, which has characteristic polynomial $m \lambda^2 + k = 0$.  The roots of this polynomial are complex, but if we allow ourselves to work with complex numbers (formally, in the above situation we tensor with $\mathbb{C}$) we find that the eigenvalues are $\lambda = \pm i \sqrt{ \frac{k}{m} }$, so the set of solutions is all functions of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = A e^{ i \sqrt{ \frac{k}{m} } t } + B e^{-i \sqrt{ \frac{k}{m} } t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Euler's formula, if we restrict our solutions to be real we get the familiar periodic sine and cosine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the eigenspaces will not be one-dimensional and then the theory of Jordan normal form applies.  This occurs, for example, when finding the general form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Damping&quot; rel=&quot;nofollow&quot;&gt;damped harmonic motion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-15T20:55:31.357" />
  <row Id="10467" PostTypeId="1" AcceptedAnswerId="10471" CreationDate="2010-11-15T20:59:12.770" Score="1" ViewCount="56" Body="&lt;p&gt;In &lt;em&gt;Classical Fourier Analysis&lt;/em&gt; by Loukas Grafakos we have in Proposition 2.3.25 the following definition for $\mathcal{S}_\infty(\mathbf{R}^n)$, namely that these are all the Schwartz functions $\phi$ such that for all multi-indices $\alpha$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbf{R}^n} x^\alpha \phi(x) \, dx = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm trying to find non-trivial functions in this space. I know that the Fourier transform maps the Schwartz-functions to itself, so I note that the requirement is actually that the Fourier transform evaluated in $0$ of $x^\alpha \phi(x)$ is $0$. So, $x^\alpha \mapsto i^\alpha d/dx^\alpha$ in the Fourier domain, so we actually want a function $\phi$ such that (let's take $n = 1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left . \frac{d}{dx^\alpha} \widehat{\phi}(x) \right |_{x = 0} = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $n \geq 0$ An obvious candidate is $f(x) = \text{exp}(-1/x^2)$ for $x &amp;gt; 0$ and $0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I now compute the Fourier transform (or the inverse) of this function (with Maple) I get another function say $g$, but if I plot the real part of $g$ it is not smooth (it has a cusp in 0), how is this possible? Further the integral which I want to be zero is only zero for odd $n$, for even $n$ it is complex! What goes wrong? &lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-15T21:05:21.297" LastActivityDate="2010-11-15T21:08:42.213" Title="Fourier transform of a special Schwartz function." Tags="&lt;fourier-analysis&gt;&lt;harmonic-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10468" PostTypeId="1" CreationDate="2010-11-15T20:59:47.717" Score="0" ViewCount="153" Body="&lt;p&gt;Hi there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'm a CompSci university student and i want some help for my math questions.&#xA;Therefore i am looking for a good discussion board for &lt;strong&gt;math homework questions&lt;/strong&gt;.&#xA;I don't know any (good), because i come from an &lt;em&gt;non-native-english-speaking&lt;/em&gt; country.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any ideas ?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;note&lt;/strong&gt; : I know that i could post my questions here, but:&lt;br&gt;&#xA;  1. this is not a discussion board, where i can discuss &lt;strong&gt;about&lt;/strong&gt; my&#xA;  question&lt;br&gt;&#xA;  2. i won't annoy the experts with &quot;simple&quot; algebra question :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3549" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-15T22:51:08.417" LastActivityDate="2010-11-23T16:46:43.467" Title="Which discussion board is good for homework questions ?" Tags="&lt;homework&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="10469" PostTypeId="1" AcceptedAnswerId="10682" CreationDate="2010-11-15T21:00:39.633" Score="5" ViewCount="75" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading Hatcher's notes on spectral sequences and he mentions that steenrod squares commute with the coboundary operator for pairs (X,A) which would then explain why these operations commute with the transgression. It says it's because &#xA;that coboundary operator can be defined in terms of suspension and we know steenrod operations commute with suspension. Does anyone know the details of this reasoning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So...&#xA;Assuming the standard axioms of steenrod operations, how do we prove that it commutes with the coboundary operator for pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&lt;/p&gt;&#xA;" OwnerUserId="3550" LastActivityDate="2010-11-17T09:07:44.160" Title="Why steenrod commute with transgression" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10470" PostTypeId="2" ParentId="10468" CreationDate="2010-11-15T21:03:41.503" Score="2" ViewCount="0" Body="&lt;p&gt;Have you looked at this:&#xA;&lt;a href=&quot;http://www.artofproblemsolving.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/index.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-15T21:03:41.503" />
  <row Id="10471" PostTypeId="2" ParentId="10467" CreationDate="2010-11-15T21:08:42.213" Score="1" ViewCount="0" Body="&lt;p&gt;As $x\to\infty$, $\exp(-1/x^2)\to1$ so this isn't in the Schwartz&#xA;space. Why not multiply this by $\exp(-x)$ to get it decaying to&#xA;$0$ at infinity rapidly enough?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-15T21:08:42.213" CommentCount="1" />
  <row Id="10472" PostTypeId="2" ParentId="10443" CreationDate="2010-11-15T21:17:38.093" Score="3" ViewCount="0" Body="&lt;p&gt;If you have a linear system on matrix form, $dX/dt=AX$,&#xA;where $X(t)$ is a vector in $\mathbf{R}^n$ and $A$ is an&#xA;$n\times n$ constant real matrix,&#xA;then $X(t) = \exp(\lambda t) V$ is a solution to the system&#xA;if $V$ is an eigenvector of $A$ with eigenvalue $\lambda$.&#xA;(This works since $\exp(\lambda t)$ is an eigenfunction of&#xA;the differential operator $d/dt$ with eigenvalue $\lambda$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is diagonalizable, then the general&#xA;solution is a linear combination of terms like above:&#xA;$X(t)=\sum_{k=1}^n c_k \exp(\lambda_k t) V_k$.&#xA;(If the eigenvalues are complex, this is the general complex-valued solution;&#xA;one can perform some standard tricks to extract the general real-valued solution&#xA;from that.)&#xA;Nondiagonalizable matrices $A$ cause a bit more trouble, but&#xA;that can be handled too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this is related to what has been said in other answers already,&#xA;but with a system of this type one sees clearly that the eigenvectors play a role&#xA;too, not just the eigenvalues.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-15T21:17:38.093" />
  <row Id="10473" PostTypeId="2" ParentId="10468" CreationDate="2010-11-15T21:19:20.993" Score="0" ViewCount="0" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://en.allexperts.com/c/science/8.htm&quot; rel=&quot;nofollow&quot;&gt;http://en.allexperts.com/c/science/8.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-15T21:19:20.993" />
  <row Id="10474" PostTypeId="2" ParentId="10468" CreationDate="2010-11-15T21:25:27.040" Score="1" ViewCount="0" Body="&lt;p&gt;Here are some:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.mathhelpforum.com&quot; rel=&quot;nofollow&quot;&gt;http://www.mathhelpforum.com&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://mathforum.org/dr/math/&quot; rel=&quot;nofollow&quot;&gt;http://mathforum.org/dr/math/&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-15T21:25:27.040" />
  <row Id="10475" PostTypeId="1" AcceptedAnswerId="10481" CreationDate="2010-11-15T21:30:39.987" Score="2" ViewCount="79" Body="&lt;p&gt;While reading about Moufang loops in the book &quot;An introduction to Quasigroups and their Representations&quot; by Smith, I've encountered the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The set $ S^7 $ of nonzero octonions of norm 1 forms a Moufang loop under multiplication. Geometrically, this set is a 7-sphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I understand why this set indeed forms a Moufang loop, I'm not sure how it is viewed as a sphere, or what is the general connection between Moufang loops and this geometrical point of view. Could anyone elaborate?&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="766" LastEditorDisplayName="" LastEditDate="2010-11-19T10:11:36.787" LastActivityDate="2010-11-19T10:11:36.787" Title="Nonzero Octonions as a 7-sphere" Tags="&lt;abstract-algebra&gt;&lt;octonions&gt;&lt;loops&gt;&lt;quasigroups&gt;" AnswerCount="2" />
  <row Id="10476" PostTypeId="2" ParentId="6716" CreationDate="2010-11-15T21:31:26.457" Score="1" ViewCount="0" Body="&lt;p&gt;Perhaps consider a full subcategory of Set with just the functions you are allowed to create given ZF. Then find some function that in normal set theory requires AC. This should do the trick. Or use the topos of constructible sets and functions, same thing.&lt;/p&gt;&#xA;" OwnerUserId="3518" LastActivityDate="2010-11-15T21:31:26.457" CommentCount="1" />
  <row Id="10477" PostTypeId="1" CreationDate="2010-11-15T21:38:11.007" Score="1" ViewCount="201" Body="&lt;p&gt;Calculate N = 9 + 99 +999 + ... + 999 ... 99 (The last part consists of 999 digits equal to 9.)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-15T23:33:54.590" Title="N= 9 + 99 + 999 + ... 999...99" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="10478" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T21:45:00.623" Score="5" ViewCount="0" Body="&lt;p&gt;$\displaystyle N = \sum_{k=1}^{999} (10^k - 1) = \dfrac{10(10^{999}-1)}{9} - 999 $&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-15T21:45:00.623" />
  <row Id="10479" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T21:46:30.487" Score="2" ViewCount="0" Body="&lt;p&gt;$N = 9 + 99 + 999 + ... = $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(10 - 1) + (100 - 1) + (1000 - 1)+ ... =$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(10 + 100 + 1000 + ...) - 999 = $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$111...1110 - 999 = $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$111...10111$ &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-15T21:46:30.487" />
  <row Id="10480" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T21:49:41.060" Score="2" ViewCount="0" Body="&lt;p&gt;The answer is&#xA;\begin{eqnarray}&#xA;N = \sum_{k = 1}^{999} 10^{k} - 1 = \frac{10(10^{999}-1)}{10-1}  -999,&#xA;\end{eqnarray}&#xA;which consists of nine-hundred and ninety-nine $1$s and one $0$ in the thousands place.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-15T21:56:58.743" LastActivityDate="2010-11-15T21:56:58.743" />
  <row Id="10481" PostTypeId="2" ParentId="10475" CreationDate="2010-11-15T21:50:24.360" Score="5" ViewCount="0" Body="&lt;p&gt;The real  octonions are a &lt;a href=&quot;http://en.wikipedia.org/wiki/Normed_division_algebra&quot; rel=&quot;nofollow&quot;&gt;normed division algebra&lt;/a&gt;, and in such a thing you always have a &lt;em&gt;unit sphere&lt;/em&gt;. The unit sphere, because of multiplicativity of the norm, is always closed under the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of your question, though, refers to the &lt;em&gt;nonzero&lt;/em&gt; octonions, and that is surly not a sphere (it has the homotopy type of a sphere, and in fact the unit sphere is a strong deformation retract---but I guess that if you know what this term means then you also are aware of this fact!)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-15T21:50:24.360" />
  <row Id="10482" PostTypeId="2" ParentId="10379" CreationDate="2010-11-15T21:52:12.010" Score="1" ViewCount="0" Body="&lt;p&gt;Both are correct, as well as saying that the root is &lt;em&gt;degenerate&lt;/em&gt; (with multiplicity of two).&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-15T21:52:12.010" />
  <row Id="10483" PostTypeId="2" ParentId="10422" CreationDate="2010-11-15T21:53:32.470" Score="1" ViewCount="0" Body="&lt;p&gt;I think this is a counterexample (assuming I haven't messed up something; I spent a fair amount of time trying to prove the result is true, reducing and reducing and reducing, and then the distillation suggested this; I &lt;em&gt;really&lt;/em&gt; hope I didn't mess it up):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $G=C_4\times C_{16}$, with $C_4$ generated by $x$ and $C_{16}$ generated by $y$. Take $H=\langle (x,y^2)\rangle$ and $K=\langle(x,y^4)\rangle$. Then $G/H$ is generated by the images of $(x,1)$ and $(1,y)$. But $(x,1)H = (1,y^{-2})H=(1,y)^{-2}H$, so $G/H$ is generated by $(1,y)H$ and hence is cyclic (and since $G$ is a $p$-group, necessarily primary cyclic). Similarly, $(x,1)K = (1,y^{-4})K$, so $G/K$ is generated by $(1,y)K$ and hence is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(x,y^2)^r = (x,y^4)^s$ for some integers $r$ and $s$; then $r\equiv s\pmod{4}$ and $2r\equiv 4s\pmod{16}$. But if $2r\equiv 4s\pmod{16}$, then $r\equiv 2s\pmod{8}$, hence $s\equiv r\equiv 2s\pmod {4}$.  The only possibility is $s\equiv 0 \pmod{4}$, hence $x^s=1$. But we also now have $4s\equiv 0 \pmod{16}$, so $y^{4s}=1$. Therefore, $(x,y^4)^s = (x^s,y^{4s}) = (1,1)$. Therefore, $H\cap K=\{(1,1)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $H$ and $K$ are both contained in $\langle(x,1),(1,y^2)\rangle$, so $HK$ is a proper subgroup of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Some comments: your hypothesis imply in any case that $G$ is a product of two primary cyclic groups. To see this, note that if $G/H$ is $p$-primary, and $G/K$ is $q$-primary, then $H$ must contain all $p&amp;#39;$-parts of $G$, and $K$ must contain all $q&amp;#39;$-primary parts of $G$, so $H\cap K=1$ implies that there is no prime which is different to both $p$ and $q$. If $p\neq q$, then the $p$-part of $G$ is a direct sum of $p$-primary cyclic groups, the $q$-part a sum of $q$-primary cyclic groups. Say there are two $p$-primary cyclic groups; then $H$ must contain the entire $q$-part as a subgroup, and have a nontrivial intersection with the $p$-part; since $K$ contains the $p$-part, $H\cap K$ would be nontrivial. Symmetrically with $q$. Thus, if $p\neq q$, then $G=C_{p^a}\times C_{q^b}$, and $H=C_{q^b}$, $K=C_{p^a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p=q$, then $G$ is a $p$-group. Then $K \cong K/(H\cap K) = HK/H$, which is a subgroup of $G/H$, so $K$ is cyclic. Symmetrically, $H$ is cyclic. Thus, $G$ can be generated by $2$ elements, so the decomposition of $G$ into a direct factors of cyclic groups has at most two factors. But $G$ cannot be cyclic and and $p$-group, because then any two nontrivial subgroups have nontrivial intersection, contradicting the hypothesis. Thus, $G=C_{p^a}\times C_{p^b}$ in this case as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the conditions require that $G=C_{p^a}\times C_{q^b}$ with $p$ and $q$ primes, possibly equal, and $a$ and $b$ positive.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T03:45:32.357" LastActivityDate="2010-11-16T03:45:32.357" />
  <row Id="10484" PostTypeId="1" CreationDate="2010-11-15T22:03:10.377" Score="0" ViewCount="40" Body="&lt;p&gt;Let $R$ be a commutative semiring and let $I$ be an $R$-ideal.  Suppose $X \subseteq R$ satisfies the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$I \subseteq X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$a \in X\ \&amp;amp;\ b \in X \Leftrightarrow ab \in X$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$a+b \in X\ \&amp;amp;\ b \in I \Rightarrow a \in X$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Does such a subset have a standard name?  &lt;/p&gt;&#xA;" OwnerUserId="3551" LastEditorUserId="3551" LastEditorDisplayName="" LastEditDate="2010-11-15T22:24:55.203" LastActivityDate="2010-11-15T22:24:55.203" Title="Does this semiring-theoretic concept have a name?" Tags="&lt;abstract-algebra&gt;" CommentCount="4" />
  <row Id="10486" PostTypeId="2" ParentId="10475" CreationDate="2010-11-15T22:22:57.097" Score="2" ViewCount="0" Body="&lt;p&gt;Mariano's answer is of course right on the money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I myself happen to currently have my library's copy of Conway and Smith's book &lt;em&gt;Quaternions and Octonions&lt;/em&gt;, let me recommend this book to you as well.  In general, throughout the book they continually emphasize a geometric approach to the algebraic structures mentioned in the title, so it would be a good place to read up on the topic of your question.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-15T22:22:57.097" />
  <row Id="10487" PostTypeId="2" ParentId="10414" CreationDate="2010-11-15T22:24:24.813" Score="1" ViewCount="0" Body="&lt;p&gt;How about this algorithm?  At each moment your decision is which spell to cast next.  Having decided, if the cool down time for that one has expired, then go ahead and cast it.  If not, wait until it has expired, then cast it.  To make the decision, rank the spells in damage/time where time is from now to completion (including any cool down time left).  Then see if you can fit in one of the other spell without delaying the best one.  Taking Qiochu's example, at the start spell 1 is 300/5=60 dam/sec and spell 2 is 50/3=16.7dam/sec.  Having cast spell 1, it is now 300/17=17.6 dam/sec, still better than 2.  But we can get 2 done twice before 1 is available, so we should do that.  The question would be whether you can change the parameters so that you should use the weaker spell even though it will delay the strong one by just a little bit.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-15T22:24:24.813" CommentCount="1" />
  <row Id="10488" PostTypeId="2" ParentId="8814" CreationDate="2010-11-15T22:34:47.463" Score="2" ViewCount="0" Body="&lt;p&gt;Ah, this is one identity which comes into use for proving the &lt;em&gt;Euler's Partition Theorem&lt;/em&gt;. The identity is as follows: $$ (1+x)(1+x^{2})(1+x^{3}) \cdots  = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-15T22:34:47.463" CommunityOwnedDate="2010-11-15T22:34:47.463" />
  <row Id="10489" PostTypeId="2" ParentId="10429" CreationDate="2010-11-15T22:40:04.807" Score="1" ViewCount="0" Body="&lt;p&gt;An empty set contains nothing. A power set of some set $A$, $\mathscr{P}(A)$  always contains something, at least $A$ itself, thus $\mathscr{P}(A)$ is not empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I decided to add my two cents is to point out to the frequent source of confusion which is associating empty set with zero, and then associating zero with &quot;nothing&quot;. Then, as the (faulty) reasoning goes, the set that contains only the empty set contains nothing, and therefore it is the empty set. The major cause of such confusion is mostly due to notation: designating the empty set as $\emptyset$, which looks almost like $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My rule of thumb: if in doubt, replace $\emptyset$ with $\{ \, \}$ and then think of $\{ \, \}$ as a container that contains nothing. Empty container. Clearly, the empty set. But, say, we have some other set $A$ that contains only the empty set. Again, $A$ is a container, but now it contains another container that contains nothing: $\{ \ \{ \, \} \ \}$. Clearly, a set $A$ contains &lt;em&gt;something&lt;/em&gt;, so it's not empty. However writing the same thing as $\{ \ \emptyset \ \}$ may cause some confusion to those exploring the set theory: is that zero the set  $A$ contains? Looks like zero, anyway! But zero is nothing! So the set contains nothing. So it is the empty set. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-15T22:40:04.807" />
  <row Id="10490" PostTypeId="1" AcceptedAnswerId="10493" CreationDate="2010-11-15T22:40:49.673" Score="8" ViewCount="389" Body="&lt;p&gt;&lt;strong&gt;From WIKI:&lt;/strong&gt; In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The indeterminate forms include $0^{0},\frac{0}{0},(\infty - \infty),1^{\infty}, \ \text{etc}\cdots$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is can anyone give me a &lt;em&gt;nice explanation&lt;/em&gt; of why $1^{\infty}$ is considered to be an indeterminate form? Because, i don't see any justification of this fact. I am still perplexed.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-16T16:28:16.430" Title="Why is $1^{\infty}$ considered to be an indeterminate form" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="10491" PostTypeId="2" ParentId="10414" CreationDate="2010-11-15T22:41:18.587" Score="0" ViewCount="0" Body="&lt;p&gt;Let's rephrase to: what's the maximal damage that I can cause in, say, 100 seconds? The constraints then become:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can't spend more than 100 seconds casting&lt;/li&gt;&#xA;&lt;li&gt;Can't cast a single spell more than $\frac{100}{cast\ time + cool\ time}$ times&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let $n_i$ be the number of times I cast spell i. I want to maximize $\sum_i n_i damage_i$ subject to $\sum_i n_i cast_i &amp;lt; 100$ and each $n_i(cast_i + cool_i) &amp;lt; 100$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be rewritten to a linear programming problem as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first constraint is $[cast_0 \dots cast_m] [n_0 \dots n_m]^T&amp;lt;100$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;The other constraints are $[0 \dots (cast_i + cool_i) \dots 0] [n_0 \dots n_m]^T &amp;lt; 100$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;: My assumptions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You cannot cast two spells at the same time. Constraint #1 ensures that this is true.&lt;/li&gt;&#xA;&lt;li&gt;You can only cast a spell once every $cast + cool$ period. Constraint #2 ensures this is true. (Note that this does &lt;em&gt;not&lt;/em&gt; say you cannot cast a &lt;em&gt;different&lt;/em&gt; spell in this period; only that you cannot cast the &lt;em&gt;same&lt;/em&gt; spell in this period.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-11-16T01:36:56.903" LastActivityDate="2010-11-16T01:36:56.903" CommentCount="12" />
  <row Id="10492" PostTypeId="2" ParentId="10389" CreationDate="2010-11-15T22:44:33.120" Score="2" ViewCount="0" Body="&lt;p&gt;Do it by induction. First, let's set the constant in $\theta$ to 1, so we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n)=2T(n/2)+T(n/3)+n^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume $T(m) \leq k m^2$ for some constant $k$ for all $m &amp;lt; n$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n) = 2 T(n/2) + T(n/3) + n^2 \leq k n^2/2 + k n^2/9 + n^2 \leq (11 k /18 +1) n^2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if $k = 3$, say, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(n) \leq (11/6+1) n^2 \leq 3 n^2.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if our bound holds for all $m &amp;lt; n$, then our bound holds for $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this a real proof, you would need to take the case when $n$ is not a multiple of 6 into account, but the details will work out when you do them carefully.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-11-15T22:44:33.120" />
  <row Id="10493" PostTypeId="2" ParentId="10490" CreationDate="2010-11-15T22:44:54.653" Score="26" ViewCount="0" Body="&lt;p&gt;Forms are indeterminate because, depending on the specific expressions involved, they can evaluate to different quantities.  For example, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \to \infty} \left(1 + \frac{1}{\ln n}\right)^n = \infty$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To expand on this some (and this thought process can be applied to other indeterminate forms, too), one way to think about it is that there's a race going on between the expression that's trying to go to 1 and the expression that's trying to go to $\infty$.  If the expression that's going to 1 is in some sense faster, then the limit will evaluate to 1.  If the expression that's going to $\infty$ is in some sense faster, then the limit will evaluate to $\infty$.  If the two expressions are headed toward their respective values at essentially the same rate, then the two effects sort of cancel each other out and you get something strictly between 1 and $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other cases, too, like &#xA;$$\lim_{n \to \infty} \left(1 - \frac{1}{\ln n}\right)^n = 0,$$&#xD;&#xA;but this still has the expression going to $\infty$ &quot;winning.&quot;  Since $1 - \frac{1}{\ln n}$ is less than 1 (once $n &gt; 1$), the exponentiation forces the limit to 0 rather than $\infty$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-16T16:28:16.430" LastActivityDate="2010-11-16T16:28:16.430" CommentCount="3" />
  <row Id="10494" PostTypeId="1" AcceptedAnswerId="10523" CreationDate="2010-11-15T22:55:34.077" Score="0" ViewCount="91" Body="&lt;p&gt;so I dont know if this is really what the website is meant for so let me know.  I have a midterm on abstract algebra (intro) and I've been doing practice problems and I just need to know whether my answers are correct and if my way of doing them is right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does the following have a solution: $x^2 \equiv 1 \pmod{3}$&#xD;&#xA;Well we know that a solution exists if $\mathrm{gcd}(a, n) | b$ where $a = x^2$, $b = 1$ and $n = 3$.&#xA;we can rewrite the equation as $1 \equiv x^2 \pmod{3}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If $a \equiv 3 \pmod{4}$ prove that there are no integers c, d such that $a = c^2 + d^2$&#xD;&#xA;By definition we can write this as $a = 4n + 3, n \in \mathbb{Z}$. so does this prove it? since $4n + 3$ can never be in the form of $c^2 + d^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Show that no perfect square has 2, 3, 7, 8 as its last digit. The way we did this was we listed out all the classes from in $\mathbb{Z}_{10}$ which are $[0]_{10}..[9]_{10}$ and showed that $a^2 (0, 1, 4, 16, 25 ...)$ are never an element of the $[2], [3], [7], [8]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) If $a \in \mathbb{Z}$, prove that last digit of $a^4$ is $0, 1, 5, \text{or } 6$&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T03:55:56.670" LastActivityDate="2010-11-16T04:25:43.700" Title="Congruency proofs" Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10495" PostTypeId="1" AcceptedAnswerId="10503" CreationDate="2010-11-15T22:57:31.910" Score="7" ViewCount="104" Body="&lt;h3&gt;Background&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Recall that an integral scheme $X$ is a scheme which is both irreducible and reduced; equivalently, its ring of functions is an integral domain on every open subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any point $p$, there is a local ring $R_p$ at $p$, which is given by localizing the ring of functions $R_U$ on any affine neighborhood $U$ of $p$ at the prime corresponding to $p$.  This local ring then has a residue field $K_p/p$.  The &lt;strong&gt;characteristic&lt;/strong&gt; of $p$ is then the characteristic of the residue field at $p$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Its not too hard to come up with integral schemes where the characteristic of points jumps around.  The standard example of $Spec(\mathbb{Z})$ has a closed point of every positive characteristic, and the unique non-closed point (the generic point) has characteristic 0 (since the residue field is $\mathbb{Q}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we don't require connectedness, then the disjoint union of $Spec(\mathbb{Q})$ and $Spec(\mathbb{Z}/2)$ has closed points of order 0 and 2, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have been playing with examples, and I can't seem to come up with an example of an integral scheme that has &lt;em&gt;closed&lt;/em&gt; points of both kinds.  &lt;strong&gt;Can an integral scheme have closed points of both positive and zero characteristic?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be curious to see such an example, since there is a wide gap in my intuition between schemes whose closed points have positive characteristic (which are inherently arithmetic) and schemes whose closed points have characteristic zero (which are either geometric, or certain arithmetic localizations).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Algebraic Version of Question (No Schemes Needed)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I should mention, though my motivation and curiosity are geometric in origin, the problem has an algebraic version.  &lt;strong&gt;Is there an integral domain $R$, and two maximal ideals $m$ and $n$, such that $R/m$ has positive characteristic, and $R/n$ has zero characteristic?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-11-16T01:07:26.177" Title="Can an integral scheme have closed points of both positive and zero characteristic?" Tags="&lt;algebraic-geometry&gt;&lt;schemes&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="10497" PostTypeId="2" ParentId="10477" CreationDate="2010-11-15T23:33:54.590" Score="1" ViewCount="0" Body="&lt;p&gt;$\rm\phantom{\ 10\ S -} S\  =\ \ 9\cdots 9\ +\ \cdots\ +\ 999\ +\ 99\ +\ 9$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ 10\ S - S\ =\ \ 9\cdots 90\ +\ \cdots +\ (990-999)\ +\ (90-99)\ -\ 9  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \Rightarrow\ \  9\ S\ =\ \ \: 9\cdots90\  +\ 999\ (-9)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad \Rightarrow\quad\ \ S\ =\ \ 1\cdots10\ - \ 999$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T23:33:54.590" />
  <row Id="10498" PostTypeId="1" AcceptedAnswerId="10522" CreationDate="2010-11-15T23:34:49.423" Score="1" ViewCount="51" Body="&lt;p&gt;So some of you may have noticed my algebra exam is tomorrow!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: If $[a]_{n} = [1]_{n}$ prove that $\mathrm{gcd}(a, n) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I tried it one way. We know the above means the following: $a \equiv 1 \pmod{n}$. This is also written as $1 \equiv a \pmod{n}$&#xD;&#xA;Then the gcd of these are  as follows:&#xA;$$\mathrm{gcd}(a, n) := d  $$ and &#xA;$$\mathrm{gcd}(1, n) := d&amp;#39; $$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I thought $ d = d&amp;#39;$ which is NOT the case as pointed out by roommate. but lets assume it does so I finished the proof as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{gcd}(a,n) = \mathrm{gcd}(1, n)$$&#xD;&#xA;$$\mathrm{gcd}(a, n) = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but then obviously that's not the case (or is it). I then tried the following:&#xA;$$d = as + nt,  \quad s, t \in \mathbb{Z}$$ &#xA;$$d&amp;#39; = as&amp;#39; + nt&amp;#39;,  \quad s&amp;#39;, t&amp;#39; \in \mathbb{Z}$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I'm stuck again :(&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T03:48:11.560" LastActivityDate="2010-11-16T03:53:03.677" Title="relating GCD, congruency and congruent classes." Tags="&lt;abstract-algebra&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" />
  <row Id="10499" PostTypeId="2" ParentId="10494" CreationDate="2010-11-15T23:42:13.077" Score="0" ViewCount="0" Body="&lt;p&gt;For 1) you have just rewritten the equation backwards, so have not proved anything.  You can use the same approach as 3)-square each number mod 3 and see if any square to 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2) why do you know 4n+3 cannot be the sum of two squares?  You need to list the numbers mod 4 that are squares, then find that no pair of them sums to 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4) do it just like 3, but using the fourth power.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-15T23:42:13.077" CommentCount="2" />
  <row Id="10500" PostTypeId="2" ParentId="10498" CreationDate="2010-11-15T23:43:52.737" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ d\:|\:n,\:1+kn\ \Rightarrow\ d\:|\: 1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T23:43:52.737" CommentCount="2" />
  <row Id="10501" PostTypeId="2" ParentId="10490" CreationDate="2010-11-15T23:48:06.637" Score="0" ViewCount="0" Body="&lt;p&gt;Here is some intuitive explanation, suitable also for non-mathematicians. Suppose that an imaginary basketball player has a probability $p = 0.999$ of making a free throw. The probability that he makes $10000$ free throws in a row is very small, the probability that he makes $100$ is high, and the probability that he makes $1000$ is approximately $e^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-15T23:48:06.637" CommentCount="2" />
  <row Id="10502" PostTypeId="2" ParentId="10494" CreationDate="2010-11-15T23:55:47.340" Score="3" ViewCount="0" Body="&lt;p&gt;1) prime $\rm\: p\:|\:(x-1)(x+1)\ \Rightarrow\ p|(x-1)\ \ or\ \ p|(x+1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\rm\ even^2\equiv 0,\ odd^2\equiv 1\ (mod\ 4)\ $ so two squares can't have sum $\equiv 3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) By 3) $\rm\ a^2\not\equiv \{\pm2,\:\pm3\}\ \Rightarrow\ a^2 \equiv \{0,\:\pm1,\:\pm4,\:5\}\ \Rightarrow\ a^4\equiv \{0,1,6,5\}\ \:(mod\ 10)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-15T23:55:47.340" />
  <row Id="10503" PostTypeId="2" ParentId="10495" CreationDate="2010-11-16T00:24:17.553" Score="8" ViewCount="0" Body="&lt;p&gt;Yes, but you're not going to like it.  Let $R = \mathbb{Z}[x_1, x_2, ... ]$.  Let $q_1, q_2, ... $ be an enumeration of the rationals; then the map $R \to \mathbb{Q}$ which sends $x_i$ to $q_i$ is surjective, so its kernel is a maximal ideal $n$ such that $R/n$ has characteristic zero.  On the other hand, let $m = (p, x_1, x_2, ...)$.  Then $R/m \simeq \mathbb{F}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Okay, so here is a Noetherian example.  Let $R&amp;#39;$ denote the localization of $\mathbb{Z}$ at $p$ (since $\mathbb{Z}_p$ generally means something else) and let $R = R&amp;#39;[x]$.  Then the map $R \to \mathbb{Q}$ which sends $x$ to $\frac{1}{p}$ is surjective, and so is the map $R \to \mathbb{F}_p$ which sends $x$ to, say, $1$.  It seems interesting to try to picture $\text{Spec } R$; it might help to stare at &lt;a href=&quot;http://matrix.cmi.ua.ac.be/DATA2/mumforddrawing.jpg&quot; rel=&quot;nofollow&quot;&gt;Mumford's picture of $\text{Spec } \mathbb{Z}[x]$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-16T01:07:26.177" LastActivityDate="2010-11-16T01:07:26.177" CommentCount="7" />
  <row Id="10504" PostTypeId="1" AcceptedAnswerId="10508" CreationDate="2010-11-16T00:37:41.837" Score="0" ViewCount="76" Body="&lt;p&gt;If $X,Y$ and $Z$ have identical distributions and are independent, and can assume $0$ or $1$, what is $\text{Var}(XYZ)$? So this is either $0$ or $1$. It's $1$ with probability $1/8$ and $0$ with probability $7/8$. This means that $\text{Var}(XYZ) = 7/16$? Or is this not right? How does this use the definition of variance? We could note that $E(X) = E(X^2)$, etc....?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-16T00:47:12.873" Title="Variance and identical random variables" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10506" PostTypeId="2" ParentId="10504" CreationDate="2010-11-16T00:45:18.973" Score="2" ViewCount="0" Body="&lt;p&gt;You imply that $X, Y$, and $Z$ have equal chance of being 0 or 1.  In that case $E((XYZ)^2)=\frac{1}{8}$ and $(E(XYZ))^2=\frac{1}{64}$.  So the variance would be $\frac{7}{64}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T00:45:18.973" />
  <row Id="10507" PostTypeId="1" CreationDate="2010-11-16T00:46:01.780" Score="2" ViewCount="113" Body="&lt;p&gt;Given the natural numbers bbb...b (n digits equal to b) and bbbb...b(m digits equal to b), show that the greatest common divisor of these numbers is bbb ... b (M digits equal to b, where M is the greatest common divisor of n and m).&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-16T02:46:02.697" Title="gcd(bbb...b, bbb...b)" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10508" PostTypeId="2" ParentId="10504" CreationDate="2010-11-16T00:47:12.873" Score="3" ViewCount="0" Body="&lt;p&gt;$Var(X) = E[X^2] - E[X]^2$. Assuming each of the variables take 0 or 1 equally likely, as you noted, $E[XYZ] = E[X^2Y^2Z^2] = 1/8$. So $Var(X) = 1/8 - 1/64 = 7/64$.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-11-16T00:47:12.873" />
  <row Id="10509" PostTypeId="2" ParentId="10507" CreationDate="2010-11-16T00:49:13.140" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The Euclidean algorithm computing their GCD mimicks that for $\rm\: gcd(m,n)\:$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $\rm\ 1^{[n]}\: :=\ 1\cdots 1\ \ (n\ 1&amp;#39;s)\ =\ (10^n-1)/9\:.\ $ Then your gcd is $\rm\ b\ gcd(1^{[m]},\:1^{[n]}) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\rm\ 1^{[m]}-1^{[n]}\ =\  1^{[m-n]}\cdot 10^n\ \ \ $ [e.g. $\rm\ 11111 - 111\ =\ 11000\ =\ 1^{[2]}\cdot 10^3\ $]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\rm\ gcd(10,1^{[n]}) = 1\ $ since $\rm\ 9\cdot 1^{[n]}\ =\ 10^n-1\ $ and $\rm\ \rm\ gcd(10,\:10^n-1) = 1\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, choosing notation so that $\rm\:m\ge n\:,\ $ and applying the above observations we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \rm\quad\quad\:\ gcd(1^{[m]},1^{[n]})\ =\ gcd(1^{[m]}-1^{[n]},\:1^{[n]})\ =\ gcd(1^{[m-n]}\cdot 10^n,\:1^{[n]})\ =\  gcd(1^{[m-n]}\:,\:1^{[n]})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ \ gcd(1^{[m]},1^{[n]})\ =\ gcd(1^{[m-n]}\:,\:1^{[n]})\ $ just like $\rm\ gcd(m,n)\ =\ gcd(m-n,n)\: $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\ \ gcd(1^{[m]},1^{[n]})\ =\ 1^{[gcd(m,n)]}\ $ follows from the above by a simple induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a full proof and some further insight see my posts in this &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question&quot;&gt;prior question.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-16T02:46:02.697" LastActivityDate="2010-11-16T02:46:02.697" />
  <row Id="10510" PostTypeId="2" ParentId="10414" CreationDate="2010-11-16T01:07:03.400" Score="13" ViewCount="0" Body="&lt;p&gt;It is worth noting that, in extreme special cases, this problem degenerates to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knapsack_problem&quot; rel=&quot;nofollow&quot;&gt;Knapsack Problem&lt;/a&gt;, which is NP-complete to solve exactly.  To see this, imagine that there is one spell, henceforth called the Megaspell, which does a very, very large amount of damage, has zero casting time, and has some positive cooldown $n$.  If all the other spells do much less damage than the Megaspell, it will clearly be optimal to cast the Megaspell every $n$ seconds and then optimize the cooldown time with the weaker spells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume all the other spells also have cooldown $n$.  If one optimizes a given $n$-second downtime with these spells, then the same spell-sequence will also be possible in the next $n$-second downtime, and so we can assume the solution is $n$-periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem then reduces to optimizing the $n$ available seconds with the lesser spells, each of which may only be cast once.  If one replaces casting time with 'weight' and damage with 'value', one recovers the Knapsack Problem for maximum weight n.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-11-16T01:07:03.400" CommentCount="8" />
  <row Id="10512" PostTypeId="1" AcceptedAnswerId="10517" CreationDate="2010-11-16T01:22:39.143" Score="5" ViewCount="137" Body="&lt;p&gt;Let $G$ be a (finite?) group. By definition, the Eilenberg-MacLane space $K(G,1)$ is a CW complex such that $\pi_1(K(G,1)) = G$ while the higher homotopy groups are zero. One can consider the singular cohomology of $K(G,1)$, and it is a theorem that this is isomorphic to the group cohomologies $H^*(G, \mathbb{Z})$. According to one of my teachers, this can be proved by an explicit construction of $K(G, 1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it seems like there ought to be a categorical argument. $K(G, 1)$ is the object that represents the functor $X \to H^1(X, \mathbb{Z})$ in the category of pointed CW complexes, say, while the group cohomology consists of the universal $\delta$-functor that begins with $M \to M^G$ for $M$ a $G$-module. In particular, I would be interested in a deeper explanation of this &quot;coincidence&quot; that singular cohomology on this universal object happens to equal group cohomology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there one?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-02T03:49:56.953" Title="Why is the cohomology of a $K(G,1)$ group cohomology?" Tags="&lt;algebraic-topology&gt;&lt;group-cohomology&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="10513" PostTypeId="2" ParentId="10337" CreationDate="2010-11-16T01:37:28.773" Score="6" ViewCount="0" Body="&lt;p&gt;As I mentioned in the comments section of Arturo's post, $\mathbb{Q}$ is as large as possible among countable linearly ordered sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said another way, if $X$ is countable an linearly ordered, then there is an order preserving injection $X\rightarrow\mathbb{Q}$.  So, if one (partially) orders the set of all countable linearly ordered sets by $X\leq Y$ iff there is a 1-1 order preserving map from $X$ to $Y$, then $\mathbb{Q}$ is the largest element of this set.  (To be frank, I'm not positive this satisfies the antisymmetry condition of being a partial order, but whatever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is quite constructive.  The key properties of $\mathbb{Q}$ are that it has no first or last element, and inbetween any two elements of $\mathbb{Q}$ there is another one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these properties, the proof goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X is countable so list the elements $X=\{x_1,x_2,...\}$.  We will inductively define a function $f:X\rightarrow \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start the induction, let $f(x_1)$ be any element of $\mathbb{Q}$.  Next, assume $f$ has been defined on $\{x_1,...,x_n\}$ with $f$ injective and order preserving.  We now define $f(x_{n+1})$.  If $x_{n+1}$ is smaller than all of the previous $x_i$, define $f(x_{n+1}) = q$ where $q$ is any rational number smaller than all of $f(x_1),...,f(x_n)$.  Such a $q$ certainly exists since the set $\{f(x_1),...,f(x_n)\}$ is finite.  Likewise, if $x_{n+1}$ is bigger than all the previous $x_i$, choose $f(x_{n+1})$ bigger than all of the $f(x_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we assume $x_i &amp;lt; x_{n+1} &amp;lt; x_j$ for some previously some $x_i$ and $x_j$ in $\{x_1,...,x_n\}$.  Further, we choose $x_i$ from $\{x_1,...,x_n\}$ as large as possible but still less than that $x_{n+1}$, and likewise, choose $x_j$ as small as possible.  Then we define $f(x_{n+1})$ to be any rational number in between $f(x_i)$ and $f(x_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, $f$ is defined for all $x_i$, and this $f$ is the desired injection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With more care, one can show that if $X$ is dense in itself with no first or last element, then we can arrange for $f$ to be surjective, showing that these properties uniquely characterize $\mathbb{Q}$ up to isomorphism!   (The &quot;more care&quot; is to well order $\mathbb{Q}$ and at any stage where a choice is made, to choose the minimal rational number satisfies whatever it has to).&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-11-16T01:37:28.773" CommentCount="2" />
  <row Id="10514" PostTypeId="1" AcceptedAnswerId="10630" CreationDate="2010-11-16T01:38:36.183" Score="2" ViewCount="89" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $T$ be a distribution, $f$ be smooth. It is known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(f(x - \cdot) ) \in C ^ \infty$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for $\phi$ a test function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$( T \ast f ) (\phi) = T( f( - \cdot ) \ast \phi ) = T( \int f( x - \cdot ) \phi( \cdot ) )$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the convolution is still a distribution. However, with regard to the first result we expect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T( \int f( x - \cdot ) \phi( \cdot ) ) = \int T ( f( x - \cdot ) \phi( \cdot ) )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This intuitively makes sense, as $T$ is a distribution in the $x$-variable, whereas we integrate over the unnamed second variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I don't know a proof of the last equality, and I am not familiar enough with the topologies of the smooth functions, test functions and distribution, to use approximative arguments. Can you help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-11-16T19:25:59.247" LastActivityDate="2010-11-16T22:57:00.057" Title="When to exchange integration and application of a distribution" Tags="&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10515" PostTypeId="2" ParentId="10490" CreationDate="2010-11-16T01:40:01.297" Score="6" ViewCount="0" Body="&lt;p&gt;Look at the logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, consider $f(x)^{g(x)}$ as $x \to \infty$, where $\lim_{x \to \infty} g(x) = \infty$ and $\lim_{x \to \infty} f(x) = 1$. (This is something of form $1^\infty$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say $f(x) = e^{h(x)}$, so $h(x) = \log f(x)$. Then $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \log f(x) = \log \lim_{x \to \infty} f(x) = \log 1 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &#xA;$$\lim_{x \to \infty} f(x)^{g(x)} = \lim_{x \to \infty} \exp (g(x) \log f(x)) = \exp \lim_{x \to \infty} (g(x) \log f(x)) $$&#xD;&#xA;and since the limit of a product is the product of the limits, that's&#xA;$$ \exp [ (\lim_{x \to \infty} g(x)) \cdot (\lim_{x \to \infty} \log f(x))] $$&#xD;&#xA;or&#xA;$$ \exp [ (\lim_{x \to \infty} g(x)) \cdot (\lim_{x \to \infty} h(x)) ]. $$&#xD;&#xA;But the first limit is infinity, and the second is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the indeterminacy of $1^\infty$ follows directly from the indeterminancy of $\infty \cdot 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The indeterminacy of $\infty^0$ actually follows in the same way, by taking the factors in the other order.)&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2010-11-16T01:51:21.923" LastActivityDate="2010-11-16T01:51:21.923" CommentCount="3" />
  <row Id="10516" PostTypeId="1" CreationDate="2010-11-16T02:02:15.710" Score="5" ViewCount="122" Body="&lt;p&gt;Let $\bar{\mathcal{P}}$ denote the closed, convex polytope with vertices at the origin and the positive rational points $(b_{1}, \dots, 0), \dots, (0, \dots, b_{n})$. Define the Ehrhart quasi-polynomial $L_{\mathcal{P}}(t) = |t \bar{\mathcal{P}} \cap \mathbb{Z}^{n}|$, which has the form: $\sum c_{k}(t) \ t^{k}$ with periodic functions $c_{k}(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; When is the Ehrhart quasi-polynomial a polynomial, i.e., the functions $c_{k}(t)$ are constants? Does this phenomenon occur only when vertices are at integral lattice points (i.e., $b_{i}$ are positive integers)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Suppose I have an Ehrhart polynomial in an indeterminate $t$. What is the significance of the value of the function at rational (non-integral) $t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 3:&lt;/strong&gt; Suppose I'd like to count &lt;em&gt;positive solutions&lt;/em&gt; (instead of &lt;em&gt;non-negative solutions&lt;/em&gt;) of $\sum \frac{x_i}{b_i} \leq t$ with $t$ positive and fixed. Assuming that $b_{i}$ are positive integers, what is the corresponding &quot;Ehrhart-like&quot; polynomial in $t$ which enumerates the (positive) integral points in the $t$-dilate $t\bar{\mathcal{P}}$? Does it follow from a simple variable change in $t$ or $b_{i}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Update&lt;/em&gt;) Here is an example of what I'm trying to do. Suppose I'd like to calculate the non-negative integer solutions of &#xA;\begin{eqnarray}&#xA;21 x_{1} + 14 x_{2} + 6 x_{3} \leq 1&#xA;\end{eqnarray}&#xA;(corresponding to the number of positive integer solutions of $21 x_{1} + 14 x_{2} + 6 x_{3} \leq 42$). Equivalently, by dividing through by the product $6 \cdot 14 \cdot 21 = 1764$, we can consider &#xA;\begin{eqnarray}&#xA;\frac{x_{1}}{84} + \frac{x_2}{126} + \frac{x_3}{294} \leq \frac{1}{1764}.&#xA;\end{eqnarray}&#xA;Here, $\mathbf{b} = (84, 126,294)$, so the corresponding polytope is &lt;em&gt;integral&lt;/em&gt;. The Ehrhart polynomial for $t$-dilates is &#xA;\begin{eqnarray}&#xA;L_{\bar{\mathcal{P}}}(t) = 1 + 231 t + 18522 t^{2} + 518616 t^{3},&#xA;\end{eqnarray}&#xA;but setting $t = \frac{1}{1764}$ gives a meaningless answer. My initial impression is that along with violating the requirement that $t$ must be an integer, what I am actually calculating is the number of lattice points in the $t$-dilate of the polytope defined by $\frac{1}{1764}$ replaced with $1$. Is there an interpolation scheme to correctly calculate the number of non-negative solutions of the first equation by finding the values of $L_{\bar{\mathcal{P}}}(0)$ and $L_{\bar{\mathcal{P}}}(1)$? Thoughts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-16T21:13:05.680" LastActivityDate="2010-11-16T21:45:51.680" Title="Counting Lattice Points with Ehrhart Polynomials" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="10517" PostTypeId="2" ParentId="10512" CreationDate="2010-11-16T02:43:54.480" Score="7" ViewCount="0" Body="&lt;p&gt;Akhil, you're thinking of this the opposite of how I think group cohomology was discovered.  The concept of group cohomology originally centered around the questions about the (co)homology of $K(\pi,1)$-spaces, by people like Hopf (he called them aspherical rather than $K(\pi,1)$ spaces, and Hopf preferred homology to cohomology at that point).  I think the story went that Hopf observed his formula for $H_2$ of a $K(\pi,1)$, which was a description of $H_2$ entirely in terms of the fundamental group of the space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This motivated people to ask to what extent (co)homology is an invariant of the fundamental group of a $K(\pi,1)$-space. This was resolved by Eilenberg and Maclane.  Eilenberg and Maclane went the extra step to show that one can define cohomology of a group directly in terms of a group via what nowadays would be called a &quot;bar construction&quot; (ie skipping the construction of the associated $K(\pi,1)$-space).  Bar constructions exist topologically and algebraically and they all have a similar feel to them.  On the level of spaces, bar constructions are ways of constructing classifying spaces.  For groups they construct the cohomology groups of a group. The latter follows from the former -- if you're comfortable with the concept of the &quot;nerve of a category&quot;, this is how you construct an associated simplicial complex to a group (a group being a category with one object).  The simplicial (co)homology of this object is your group (co)homology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dieudonne's &quot;History of Algebraic and Differential Topology&quot; covers this in sections V.1.D and V.3.B.   I don't think that answers all your questions but it answers some. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-16T02:43:54.480" CommentCount="1" />
  <row Id="10518" PostTypeId="1" AcceptedAnswerId="10542" CreationDate="2010-11-16T03:09:07.210" Score="1" ViewCount="138" Body="&lt;p&gt;I'm approximating an integral with only exponentials.  i.e., it is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{-\pi}^\pi{\frac{\displaystyle\sum_{j=a}^b{c_j e^{i\cdot d_j \cdot t}}}{\displaystyle\sum_{k=a}^b{r_k e^{i \cdot s_k \cdot t}}} dt}$&#xD;&#xA;where $c_j, d_j, r_k, s_k \in \mathbb{Z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it's a fraction of exponentials over exponentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized that we can integrate the individual terms of the power series expansion to integrate the expression.  I also found that for the particular integrals that I'm working with, only the first $N$ terms (for some $N$) significantly contribute to the value of the integral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My idea&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can use feedback and recurrences to approximate this integral.  For example, the first two terms of the power series expansion of the function will be something like $c_0 + c_1x$.  If we attempt to integrate these terms, we should get $c_0x + \frac{c_1}{2}x$ evaluated at the two endpoints of the interval of integration.  But we can approximate this second function.  Suppose we take the original power series expansion ($c_0 + c_1x$), rewrite $x$ as $x/2$, and then multiply by $x$.  Bingo!  We have the integrated function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can make &quot;guesses&quot; at the integral using the rewriting technique above.  Without going into detail, I then have a way to determine the error of the guess.  So we can introduce a feedback system to get closer and closer to the actual integral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My Question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does a method like this already exist?  I've seen techniques for approximating integrals using intervals, but this technique attempts to do it using feedback, instead.  Has a similar technique been used?  I would like to see the method, if it exists.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-16T08:58:37.310" Title="Is this integration approximation method known/used?" Tags="&lt;calculus&gt;&lt;intuition&gt;&lt;integral&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10519" PostTypeId="1" AcceptedAnswerId="10532" CreationDate="2010-11-16T03:40:39.923" Score="0" ViewCount="129" Body="&lt;p&gt;We have points A, B, C. On 2d plane. How having points coordinates (x, y) calculate area of triangle formed by them?&lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T10:16:14.013" LastActivityDate="2010-11-16T10:16:14.013" Title="How to calculate area of triangle, having its points 2d coordinates?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10520" PostTypeId="2" ParentId="10518" CreationDate="2010-11-16T03:43:29.003" Score="2" ViewCount="0" Body="&lt;p&gt;You can read about Robmerg Integration in section 4.3 of &lt;a href=&quot;http://www.nrbook.com/c/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; or in &lt;a href=&quot;http://en.wikipedia.org/wiki/Romberg_integration&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and the references.  It is a more formal approach to what you are describing.  You talk about using feedback, but it is not clear where you get the feedback from.  If your rewriting is really term-by-term integration (see &lt;a href=&quot;http://math.stackexchange.com/questions/9622/using-derivatives-to-evaluate-an-integral-of-exponentials&quot;&gt;your previous question and my answer&lt;/a&gt;) you do have feedback from that.  The Romberg routine follows the changes in the integral as the order rises and makes a guess based on that.  As explained earlier in the book regarding fitting functions, there are risks in this-maybe the function goes wild in a small area.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T03:43:29.003" CommentCount="1" />
  <row Id="10521" PostTypeId="2" ParentId="10519" CreationDate="2010-11-16T03:45:36.587" Score="1" ViewCount="0" Body="&lt;p&gt;If by square you mean the area, &lt;a href=&quot;http://en.wikipedia.org/wiki/Heron%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Heron's formula&lt;/a&gt; is your friend.  Just calculate the side lengths and the semiperimeter.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T03:45:36.587" CommentCount="2" />
  <row Id="10522" PostTypeId="2" ParentId="10498" CreationDate="2010-11-16T03:53:03.677" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, the GCDs are the same, but your argument shows a lot of confusion, as Bill points out. First, when you claim that $d=d&amp;#39;$, the problem is that you are simply &lt;em&gt;asserting&lt;/em&gt; a result that is &lt;em&gt;stronger&lt;/em&gt; than what you are trying to prove. You are trying to prove the simpler case that if $a\equiv 1 \pmod{n}$ then $\gcd(a,n)=1$, and you are invoking the result that if $a\equiv b\pmod{n}$ then $\gcd(a,n)=\gcd(b,n)$. That's &lt;strong&gt;more&lt;/strong&gt; than what you want to prove, so you cannot simply &lt;em&gt;assume&lt;/em&gt; it's true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really, think about $[a]_n = [1]_n$. That means that $1 = a+kn$ for some integer $k$. That means that you can write $1$ as a linear combination of $a$ and $n$. That means that $\gcd(a,n)$ divides $1$. That means...&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-16T03:53:03.677" CommentCount="4" />
  <row Id="10523" PostTypeId="2" ParentId="10494" CreationDate="2010-11-16T04:03:10.653" Score="1" ViewCount="0" Body="&lt;p&gt;For (1), you confused the condition for a congruence of the form $ax\equiv b\pmod{n}$ to have a solution with the congruence you have here. This congruence is not a linear congruence, so the condition does not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, that does not even begin to work; it's completely different kinds of congruences (linear vs. quadratic). And you did not determine if it had solutions or not, you just said that solving $x^2\equiv 1 \pmod{3}$ is the same as solving $1\equiv x^2\pmod{3}$. If I asked you to figure out if there is an integer such that $x^2=4$, would saying &quot;that's the same as finding an integer such that $4=x^2$&quot; constitute a valid answer that solves the problem of &lt;em&gt;determining&lt;/em&gt; if there is or is not such an integer? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to solve &lt;em&gt;this&lt;/em&gt; one is to check: there are only three possible values for $x$ modulo $3$, namely $0$, $1$, or $2$. Do any of them give you a solution to the congruence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, you can use the ideas hinted at in Bill Dubuque's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2), you aren't proving anything; you are just repeating what you have been asked to prove. Proving it requires more than just repeating the statement with conviction. &lt;em&gt;Why&lt;/em&gt; is it that $4k+3$ can never be written in the form $a^2+b^2$? You never say, you just assert it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, think about what $a^2+b^2$ would be modulo $4$. What are the possible values of $a^2$ modulo $4$? Of $b^2$? So what are the possible values of $a^2+b^2$ modulo $4$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) is essentially correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (4), try the same approach as (3). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T04:25:43.700" LastActivityDate="2010-11-16T04:25:43.700" />
  <row Id="10524" PostTypeId="2" ParentId="10337" CreationDate="2010-11-16T04:08:33.300" Score="1" ViewCount="0" Body="&lt;p&gt;I think the intuitive point here is captured in the idea that between -n and n there are only finitely many integers and infinitely many rationals.  So over any finite range (and intuitively therefore over the entire range because it is uniform) the density of integers in the rationals is zero.  The effort is to carry over the idea of &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_density&quot; rel=&quot;nofollow&quot;&gt;natural density&lt;/a&gt; in the sense that there are fewer squares than naturals because the natural density of squares is zero. Natural density is a well defined operation, but may still not capture all the comparisons one would like to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, this is captured in a remark on a previous post that cardinality is a coarse way to compare sets.  Natural density is finer but demands an ordering.  It allows more distinctions to be made, most (all?) of which are in line with our intuition.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-16T04:08:33.300" CommentCount="2" />
  <row Id="10525" PostTypeId="2" ParentId="2384" CreationDate="2010-11-16T04:37:06.367" Score="0" ViewCount="0" Body="&lt;p&gt;I have developed some educational freeware to teach students how to write simple, though thoroughly rigorous mathematical proofs. The software verifies every line of proof as it is entered. It also provides some important insights into the nature of modern mathematics. One of the worked examples in the tutorial is to prove the set of all things cannot exist. (It would be too big, and would include at least one very troublesome object that would make the whole thing blow up!) I think your budding mathematicians would enjoy it. Download the software and tutorial at my website: &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-16T04:37:06.367" CommunityOwnedDate="2010-11-16T04:37:06.367" />
  <row Id="10526" PostTypeId="2" ParentId="10512" CreationDate="2010-11-16T04:50:45.147" Score="1" ViewCount="0" Body="&lt;p&gt;Well, you'll probably want a more conceptual proof, but one thing you can do is check they are computed by the same chain complex: for $K(G,1)$ take the simplicial construction of the classifying space $BG$ and compute its cohomology in the usual way for simplicial sets (using the dual to the complex of formal linear combinations of simplices); for the group cohomology take the free resolution of the trivial $\mathbb{Z}[G]$-module $\mathbb{Z}$ where the nth term is $\mathbb{Z}[G^n]$, which amounts to the same complex as before.&lt;/p&gt;&#xA;" OwnerUserId="1070" LastActivityDate="2010-11-16T04:50:45.147" />
  <row Id="10527" PostTypeId="1" AcceptedAnswerId="10528" CreationDate="2010-11-16T05:15:28.640" Score="3" ViewCount="78" Body="&lt;p&gt;Suppose $r_i$, $1 \le i \le n$, and $c_j$, $1 \le j \le m$, are nonnegative integers.  When does there exist an $n \times m$ matrix in $\text{Mat}_{n \times m} (\mathbb{Z}^+)$, i.e. nonnegative entries, such that $r_i$ is the sum of the entries in its $i$th row and $c_j$ is the sum of the entries in its $j$th columns?&lt;/p&gt;&#xA;" OwnerUserId="314" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-16T05:49:43.337" LastActivityDate="2010-11-16T05:49:43.337" Title="Constraints on sum of rows and columns of matrix" Tags="&lt;combinatorics&gt;&lt;algebra-precalculus&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10528" PostTypeId="2" ParentId="10527" CreationDate="2010-11-16T05:20:03.367" Score="5" ViewCount="0" Body="&lt;p&gt;There is a survey paper by Alexander Barvinok on this:&#xA;&lt;a href=&quot;http://www.math.lsa.umich.edu/~barvinok/linalg.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.lsa.umich.edu/~barvinok/linalg.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-16T05:20:03.367" CommentCount="2" />
  <row Id="10529" PostTypeId="1" CreationDate="2010-11-16T05:50:59.983" Score="0" ViewCount="76" Body="&lt;p&gt;what is the sigma field created by X(t,w) where t belongs to [0,1] and X(t,w) =1 if t=w and zero otherwise.&lt;/p&gt;&#xA;" OwnerUserId="3559" LastActivityDate="2010-11-16T11:31:34.993" Title="sigma algebra stochastic process" Tags="&lt;sigma-algebra&gt;" AnswerCount="1" />
  <row Id="10531" PostTypeId="2" ParentId="1730" CreationDate="2010-11-16T06:03:17.430" Score="0" ViewCount="0" Body="&lt;p&gt;Why not use a real &quot;function machine,&quot; which each of your students should have -- a scientific calculator? After all, most, if not all the functions in your course will be numerical examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a function of one variable, use any of the trigonometric, squaring, cubing, square roots, or log functions. Use square root, inverse trig and log functions for examples of restricted domains of definition. You don't have to explain what these functions mean at first, just emphasize that you put in one number, and get another number back. Have students make a table of values using the calculator to reinforce this fact. Be sure to include examples that leave values unchanged, e.g. sqrt(0)=0, sqrt(1)=1, sin(0)=0, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For functions of two variables, use the ordinary arithmetic operations: add, subtract, multiply, divide or exponent functions.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-16T06:03:17.430" />
  <row Id="10532" PostTypeId="2" ParentId="10519" CreationDate="2010-11-16T06:08:48.063" Score="3" ViewCount="0" Body="&lt;p&gt;To make Rahul's comment more explicit, the determinant formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac12 \begin{vmatrix}x_1&amp;amp;y_1&amp;amp;1\\x_2&amp;amp;y_2&amp;amp;1\\x_3&amp;amp;y_3&amp;amp;1\end{vmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $(x_i,y_i)$ are the coordinates of the corners, gives the (signed) area of the triangle. For a guaranteed positive result, the points are to be taken anticlockwise.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-16T06:08:48.063" CommentCount="2" />
  <row Id="10533" PostTypeId="1" CreationDate="2010-11-16T06:24:27.363" Score="2" ViewCount="123" Body="&lt;p&gt;Hello...in a game of contract bridge, partner and me together have 11 cards in a suite. The remaining 2 cards in the same suite can be distributed amongst our opponents as either 1-1 or 2-0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that it will be distributed as 1-1 and what is the probability it will be distributed as 2-0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the above is solved, how can you extend it if more cards are missing. That is, let us assume that partner and me have 8 cards between us. How do you calculate the probability for the distributions (5-0, 4-1, 3-2)?&lt;/p&gt;&#xA;" OwnerUserId="1791" LastActivityDate="2010-11-16T15:55:02.370" Title="Probability in a game of Contract Bridge" Tags="&lt;probability&gt;&lt;card-games&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10534" PostTypeId="2" ParentId="10533" CreationDate="2010-11-16T06:30:55.410" Score="1" ViewCount="0" Body="&lt;p&gt;The number of cards in suit a given opponent gets is distributed &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypergeometric_distribution&quot; rel=&quot;nofollow&quot;&gt;hypergeomtrically&lt;/a&gt;. In your particular case, $1-1$ is slightly more likely, with probability $13/25 = 1/2 + 1/50$. This is not surprising, since once you got one of the cards, the probability you'll get another drops, since then there's only one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The hypergeometric distribution with parameters $(N,m,n)$ concerns a situation in which the universe consists of $N$ cards, $m$ of which are special, and you are drawing $n$ of them. The probability that the number of special cards you draw is $k$ is exactly $$\frac{\binom{m}{k}\binom{N-m}{n-k}}{\binom{N}{n}},$$ since there are $\binom{m}{k}\binom{N-m}{n-k}$ choices for a hand with exactly $k$ special cards, and $\binom{N}{n}$ total hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the distribution of $1-1$, you plug in $k=1$. In order to find the distribution of $2-0$, you plug in $k=0$ and $k=2$, and add, since you don't care which player gets none and which gets both.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-16T15:55:02.370" LastActivityDate="2010-11-16T15:55:02.370" CommentCount="1" />
  <row Id="10535" PostTypeId="1" AcceptedAnswerId="10538" CreationDate="2010-11-16T06:39:47.603" Score="0" ViewCount="77" Body="&lt;p&gt;The area of a pyramid with a square base with side length $L$ and height $h$ is calculated as follows: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In $\mathbb{R}^3$ place the pyramid upright with one side flush with the z-axis so that the corners of the base are at $(0,0,\frac{1}{L})$, $(0,0,-\frac{1}{L})$, $(L,0,\frac{1}{L})$ and $(L,0,-\frac{1}{L})$.&#xA;&lt;img src=&quot;http://i.imgur.com/gNKcT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We will slice parallel to the yz-plane from $x=0$ up to $x=\frac{L}{2}$. (We can then double it to get the whole pyramid.) &lt;strong&gt;The slices are rectangles&lt;/strong&gt; that all have a base equal to L and height given by $f(x)=\frac{2hx}{L}$. The cross-sectional area is then, $A(x) = L\frac{2hx}{L}=2hx$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So, we have $2 \int_{0}^{\frac L2} (2hx) dx \neq \frac13 L^2 h$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What has gone wrong?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANS: The slices are NOT rectangles. DUH. Thanks, Rahul Narain. &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-11-16T07:00:59.963" LastActivityDate="2010-11-16T07:20:46.677" Title="How to explain why this method of slicing a solid to find volume fails?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10536" PostTypeId="1" AcceptedAnswerId="10537" CreationDate="2010-11-16T06:52:22.817" Score="1" ViewCount="57" Body="&lt;p&gt;Suppose that $\mathcal{B}$ is a base for a topology on a space $X$. Is there a nice way of thinking about how we can modify $\mathcal{B}$ (for instance, to simplify computations) without changing the topology it generates? It seems non-trivial to compute the topology generated by a base, but maybe some &quot;small enough&quot; changes to the base should be safe. The situation is delicate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have a fixed open set $U$, consider the family of sets $\mathcal{B&amp;#39;} = \{ B \in \mathcal{B} | B \subset U \text{ or } B \subset (X \setminus U) \}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This does not generate the same topology as $\mathcal{B}$, since $\mathcal{B&amp;#39;}$ won't, in general, be a base. For instance, if $X = \mathbb{R}$, and $U = (0,1)$, then $1$ is not contained in any element of $\mathcal{B&amp;#39;}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cover $X$ by open sets $U_i$, and set $\mathcal{B&amp;#39;} = \{ B \in \mathcal{B} | B \subset U_i \text{ for some } i \}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This, on the other hand, does sometimes work. For instance, consider $X = \text{Spec} A$ with the Zariski topology, and suppose that $X$ is covered by $U_i = \text{Spec} A_i$. A base for the Zariski topology on $X$ is given by sets of the form $D(f) = \{ \mathfrak{p} \subseteq A | f \not \in \mathfrak{p} \}$. We can restrict this base to only include those $D(f)$ that lie in some $U_i$, and we obtain the same topology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose my question is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\mathcal{B}$ is a base for a topology $\mathcal{T}$, are there some nice types of subbases of $\mathcal{B}$ (along the lines of the second example above) of $\mathcal{B}$ that will always generate $\mathcal{T}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this sort of thing comes up when checking that various properties of scheme morphisms are affine local, so I've also tagged this with [algebraic-geometry].&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-11-16T06:57:56.520" Title="When does a subbase of a base generate the same topology?" Tags="&lt;topology&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="10537" PostTypeId="2" ParentId="10536" CreationDate="2010-11-16T06:57:56.520" Score="2" ViewCount="0" Body="&lt;p&gt;Since a topology generated by a base consists of open sets that are union of basic open sets, you may drop, from a given base, any open set that is a union of open sets in the same base and get a smaller base.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-16T06:57:56.520" />
  <row Id="10538" PostTypeId="2" ParentId="10535" CreationDate="2010-11-16T07:20:46.677" Score="2" ViewCount="0" Body="&lt;p&gt;This is an elaboration of my comment, so the question can be marked as answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how you are placing your pyramid. &quot;One side flush with the z-axis&quot; is not enough information, as one may rotate the shape about the $z$-axis while keeping it flush. If the base lies on the $xz$-plane, its corners have $z$-coordinate $0$ or $L$, not $1/L$, and the slices parallel to the $yz$-plane will not be rectangles but trapeziums (a.k.a. trapeziods), as one can see by considering the slice at $L/2$ which is clearly a triangle (a degenerate trapezium) and not a rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slice at a given $x$-coordinate $x &amp;lt; L/2$ is a trapezium with height $2hx/L$, as you noticed, but with one parallel size of length $L$ and the other of length $L-2x$. The area of the trapezium is therefore $A = 2hx/L\cdot(L - x) = 2hx\cdot(1-x/L)$, and the volume of the pyramid is $V = 2\int_{x=0}^{L/2} A dx = hL^2/3$ as expected.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-16T07:20:46.677" />
  <row Id="10539" PostTypeId="1" CreationDate="2010-11-16T07:25:59.637" Score="2" ViewCount="97" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I am reading&#xA;Logic an Structure by Dirk van Dalen (2008).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am missing some basics I try to find related lectures on youtube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I frequently watch MIT, Stanford, and University of Nottinham lectures on youtube.&#xA;www.youtube.com/user/MIT&#xA;www.youtube.com/user/StanfordUniversity&#xA;www.youtube.com/user/NottmUniversity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howevery, I have a hard time to find lectures on topics like predicate logic, high order logic, and  intuitionistic logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone happen to have pointers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3377" LastActivityDate="2010-11-16T07:25:59.637" Title="logic lectures on youtube" Tags="&lt;soft-question&gt;&lt;logic&gt;&lt;higher-order-logic&gt;" CommentCount="2" />
  <row Id="10540" PostTypeId="2" ParentId="10516" CreationDate="2010-11-16T07:32:11.903" Score="5" ViewCount="0" Body="&lt;p&gt;For question 3, by the reciprocity law $(-1)^{n}P(-t)$&#xD;&#xA;counts positive integer solutions of $\sum x_i/b_i&amp;lt;t$&#xD;&#xA;where $P$ is the Ehrhart plynomial of the $n$-simplex&#xA;with vertices $(0,\ldots,0)$ and the $(0,\ldots,b_i,\ldots,0)$.&#xA;This is short of your target by the number of positive integer&#xA;solutions of $\sum x_i/b_i=t$. But that is $(-1)^{n-1}Q(-t)$&#xD;&#xA;where $Q$ is the Ehrhart polynomial of the $(n-1)$-simplex&#xA;with vertices the $(0,\ldots,b_i,\ldots,0)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-16T07:32:11.903" CommentCount="0" />
  <row Id="10541" PostTypeId="1" CreationDate="2010-11-16T08:30:44.107" Score="1" ViewCount="28" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;who of you in this forum uses R (http://www.r-project.org/) with the multicore, SNOW or CUDA packages, so for advanced calculations that need more power than a workstation CPU? On which hardware do you compute these scripts? At home/ at work or do you have data center access somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of these questions is the following: I am currently writing my M.Sc. thesis about R and High-Performance-Computing and need a strong knowledge about who actually uses R. I read that R had 1 million users in 2008, bu thats more or less the only user statistics I could find on this topic - so I hope for your answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely Heinrich&lt;/p&gt;&#xA;" OwnerUserId="3561" LastActivityDate="2010-11-16T08:30:44.107" Title="Who uses R with multicore, SNOW or CUDA package for resource intense computing?" Tags="&lt;statistics&gt;&lt;computer-science&gt;&lt;applications&gt;" CommentCount="2" />
  <row Id="10542" PostTypeId="2" ParentId="10518" CreationDate="2010-11-16T08:58:37.310" Score="2" ViewCount="0" Body="&lt;p&gt;Are you aware that integral of this type can be computed exactly by means of the residue theorem? The price to pay is that you have to find the roots of the equation $\sum_a^b r_k z^{s_k}=0$ in the unit disk. (You may assume all $s_k\ge0$.)&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-16T08:58:37.310" CommentCount="1" />
  <row Id="10543" PostTypeId="1" CreationDate="2010-11-16T09:11:51.597" Score="3" ViewCount="158" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard about &lt;a href=&quot;http://en.wikipedia.org/wiki/Yakov_Perelman#Books&quot; rel=&quot;nofollow&quot;&gt;Yakov Perelman&lt;/a&gt; and his books. I just finished reading his two volumes of &lt;strong&gt;Physics for Entertainment&lt;/strong&gt;. What a delightful read! What a splendid author. This is the exact book I've been searching for. I can use it to develop interest for science (math &amp;amp; physics) in my students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His math books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mathematics Can Be Fun&lt;/li&gt;&#xA;&lt;li&gt;Figures for Fun&lt;/li&gt;&#xA;&lt;li&gt;Arithmetic for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Geometry for Entertainment&lt;/li&gt;&#xA;&lt;li&gt;Lively Mathematics&lt;/li&gt;&#xA;&lt;li&gt;Fun with Maths &amp;amp; Physics&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;His physics books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Physics for Entertainment (1913)&lt;/li&gt;&#xA;&lt;li&gt;Physics Everywhere&lt;/li&gt;&#xA;&lt;li&gt;Mechanics for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Astronomy for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Tricks and Amusements&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I want to get all the above books. Because books from author like this cannot be disappointing. But unfortunately not all of them are available. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also read another amazing book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/069111966X&quot; rel=&quot;nofollow&quot;&gt;How to Solve It: A New Aspect of Mathematical Method by G.Poyla&lt;/a&gt;. This books actually teaches you &lt;em&gt;how to think&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the similar lines if you have any book suggestions (with very practical problems &amp;amp; case studies) for physics &amp;amp; Math (Please don't differentiate between math &amp;amp; physics here. If I can develop interest in one of the subject he'll gain interest in other.) please contribute.&lt;/p&gt;&#xA;" OwnerUserId="2264" LastActivityDate="2010-11-16T15:58:38.537" Title="Books that develop interest &amp; critical thinking among high school students." Tags="&lt;books&gt;&lt;teaching&gt;&lt;high-school&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="10544" PostTypeId="1" AcceptedAnswerId="10570" CreationDate="2010-11-16T09:22:58.160" Score="2" ViewCount="124" Body="&lt;p&gt;I have a set and its cardinality. How can I calculate the cardinality of its complementary set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Gotta apologize. I was talking about well-defined sets with a fairly low positive integer as cardinality. Qwirk comes pretty close when talking about subsets and a given superset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So see it this way. I have a certain set of numbers. Take some extra condition to get a subset, know its cardinality and now try to calculate the cardinality of a subset with the negated condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for being unclear.&lt;/p&gt;&#xA;" OwnerUserId="3016" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T14:34:08.780" LastActivityDate="2010-11-16T19:11:09.470" Title="Cardinality of complementary set" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10545" PostTypeId="1" AcceptedAnswerId="10550" CreationDate="2010-11-16T09:52:09.963" Score="3" ViewCount="194" Body="&lt;p&gt;If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle#By_relative_lengths_of_sides&quot; rel=&quot;nofollow&quot;&gt;type&lt;/a&gt; ? &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T20:55:06.180" LastActivityDate="2010-11-17T07:42:15.903" Title="What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="10546" PostTypeId="2" ParentId="10544" CreationDate="2010-11-16T10:36:00.583" Score="2" ViewCount="0" Body="&lt;p&gt;Possible slight rephrasing of the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a set $X$ and a subset $S$ then calculate the cardinality of the complement of $S$ with respect to $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, this can be uncountably infinite, countably infinite or even have finite cardinality!&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-16T10:36:00.583" CommentCount="3" />
  <row Id="10547" PostTypeId="1" AcceptedAnswerId="10567" CreationDate="2010-11-16T10:46:27.520" Score="2" ViewCount="146" Body="&lt;p&gt;This is extracted from my module :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a $\displaystyle\bigtriangleup ABC$,&#xA;$$\displaystyle \frac{\sin (A) + \sin(B)}{\sin(A) -\sin(B)} = \frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A,B,C$ are the angles and  $a,b,c$ are the opposite sides,I can't figure out how this step holds valid?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T18:38:12.960" LastActivityDate="2010-11-16T19:11:40.453" Title="Why is $\frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$ valid in a triangle?" Tags="&lt;trigonometry&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="10548" PostTypeId="2" ParentId="10529" CreationDate="2010-11-16T11:31:34.993" Score="1" ViewCount="0" Body="&lt;p&gt;$X_{t}$ is a stochastic process. Maybe you mean a natural filtration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sigma algebra generated by $X_{t}$ is $\sigma(X_{t})=\{\emptyset,[0;1],t,[0;t)\cup (t;1]\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural filtration $\mathcal{F}_{t}$ is the minimal sigma algebra which contains all $\sigma(X_{u})$ for $u\in [0;t]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\mathcal{F}_{t}$ consists of all countable subsets of $[0;t]$ and their complements.&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-11-16T11:31:34.993" CommentCount="3" />
  <row Id="10549" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:42:34.860" Score="3" ViewCount="0" Body="&lt;p&gt;The cosine formula states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 = b^2+ c^2 – 2bc \cos A$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos A = \frac{ \cot^2 (A/2) – 1}{ \cot^2(A/2)+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^2 = b^2 + c^2 -2bc \frac{ (b+c)^2 – a^2}{(b+c)^2 + a^2},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which reduces to $a^4 = (b^2 – c^2)^2.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so taking the square root $a^2 =b^2 – c ^2,$ where $b$ is the larger side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $b^2 = c^2+a^2$ and so we have a right-angled triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so it's the angle opposite the larger of $b$ and $c$ which is the right-angle.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-16T11:42:34.860" />
  <row Id="10550" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:43:08.213" Score="5" ViewCount="0" Body="&lt;p&gt;So by &lt;em&gt;Law of Sines&lt;/em&gt; we have $$ \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} =k (\text{say})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this your equation becomes, $$\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{k(\sin{B} + \sin{C})}{k \sin{A}} = \frac{\sin{B}+\sin{C}}{\sin{A}} = \frac{\cos\frac{A}{2} \cos\frac{B-C}{2}}{\sin\frac{A}{2} \cos\frac{A}{2}}$$ From this you get $$\frac{\pi}{2} - \Bigl(\frac{B+C}{2}\Bigr)= \pm \frac{B-C}{2}$$ and you can do it then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $B=\frac{\pi}{2}$ or $C =\frac{\pi}{2}$, which means the triangle is &lt;em&gt;right-angled&lt;/em&gt;. Hurray!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The keen idea when you see problems of this type is to use the &lt;em&gt;Sine rule&lt;/em&gt;. Do you think that this problem would have been difficult, if you had applied the sine rule as i did in the first step. Very often in &lt;em&gt;Mathematics&lt;/em&gt; starting the solution is the difficult part. Once you figure that out, then the solutions simply flows. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-17T07:42:15.903" LastActivityDate="2010-11-17T07:42:15.903" CommentCount="7" />
  <row Id="10551" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:43:13.433" Score="3" ViewCount="0" Body="&lt;p&gt;OK, first rewrite it as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos(A/2) = \frac{b+c}{a} \sin(A/2) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides by $2 \cos(A/2)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cos^2(A/2) = \frac{b+c}{a} \sin(A) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the rule $a/\sin(A)=b/\sin(B)=c/\sin(C)$ to transform this to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cos^2(A/2) = \sin(B)+\sin(C) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Simpson's rule on the second hand of the equation to sum the sines gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \cos^2(A/2) = 2\sin((B+C)/2)\cos((B-C)/2) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since in a triangle $A+B+C=\pi$ we have $ 2\sin((B+C)/2) = 2\sin((\pi-A)/2) = 2\cos(A/2) $ and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cos(A/2) = \cos((B-C)/2) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or $A+C=B$ (If $B&gt;C$). Combining this with $A+B+C=\pi$ this gives $B=\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-11-16T12:24:12.010" LastActivityDate="2010-11-16T12:24:12.010" />
  <row Id="10552" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T11:46:41.487" Score="1" ViewCount="0" Body="&lt;p&gt;Geometrically, this is not too hard to see. Presumably, you understand why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{k(\sin(A) + \sin(B))}{k(\sin(A) - \sin(B))}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let k be the length of the normal from C onto AB. What is then k sin(A) and k sin(B)?&lt;/p&gt;&#xA;" OwnerUserId="3536" LastActivityDate="2010-11-16T11:46:41.487" />
  <row Id="10553" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T11:53:28.710" Score="3" ViewCount="0" Body="&lt;p&gt;Begin with the Law of Cosines to compute $\cot^2\frac{A}{2}$ in terms of the lengths of the sides of the triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$&#xD;&#xA;$$\Rightarrow \begin{cases}&#xD;&#xA;\cos^2{\frac{A}{2}}=\frac{1+\cos{A}}{2}=\frac{2bc+b^2+c^2-a^2}{4bc} = \frac{(b+c)^2-a^2}{4bc} \\&#xD;&#xA;\sin^2{\frac{A}{2}}=\frac{1-\cos{A}}{2}=\frac{2bc-(b^2+c^2-a^2)}{4bc} = \frac{a^2-(b-c)^2}{4bc}&#xD;&#xA;\end{cases}$$&#xD;&#xA;$$\Rightarrow \cot^2\frac{A}{2}=\frac{\cos^2{\frac{A}{2}}}{\sin^2{\frac{A}{2}}}=\frac{(b+c)^2-a^2}{a^2-(b-c)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, work in the condition from the problem, and let Algebra bring you home ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\frac{(b+c)^2}{a^2} &amp;amp;=&amp;amp; \frac{(b+c)^2-a^2}{a^2-(b-c)^2} \\&#xD;&#xA;(b+c)^2\left(a^2-(b-c)^2\right)&amp;amp;=&amp;amp; a^2\left((b+c)^2-a^2\right) \\&#xD;&#xA;a^2(b+c)^2-(b+c)^2(b-c)^2 &amp;amp;=&amp;amp; a^2 (b+c)^2-a^4 \\&#xD;&#xA;a^4-(b+c)^2(b-c)^2 &amp;amp;=&amp;amp; 0 \\&#xD;&#xA;a^4-(b^2-c^2)^2 &amp;amp;=&amp;amp; 0 \\&#xD;&#xA;\left(a^2+(b^2-c^2)\right)\left(a^2-(b^2-c^2)\right) &amp;amp;=&amp;amp; 0 \\&#xD;&#xA;\end{eqnarray}$$&#xD;&#xA;$$\Rightarrow \hspace{0.25in} a^2+b^2=c^2 \hspace{0.25in}\text{or}\hspace{0.25in} a^2+c^2=b^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice little problem. I suspect there's a more-direct path to the solution, though. (&lt;strong&gt;Edit&lt;/strong&gt; I see a couple such paths had been posted when I was working on my answer! :)&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-11-16T12:22:50.560" LastActivityDate="2010-11-16T12:22:50.560" />
  <row Id="10554" PostTypeId="1" CreationDate="2010-11-16T12:19:46.390" Score="3" ViewCount="140" Body="&lt;p&gt;&lt;strike&gt;I claim that it is commonly believed that&lt;/strike&gt; Mathematical objects can be seen as genuinely static, with no &quot;Platonic&quot; time in which they do genuinely evolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless time has its place in mathematics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An endomorphism of a set (seen as a set of states of a system) into itself can be seen as evolution of the system in discrete time steps.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a function of a totally ordered set into a set (seen as above) the ordered set can be seen as &quot;time&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;as the time-like component in Minkowski space &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Questions (slightly modified after Qiaochu's comment and Vhailor's answer):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which other constructs do give you a &quot;time feeling&quot; or give rise to &quot;dynamic intuition&quot; &lt;strike&gt;admit a comparable straight-forward&#xA;  interpretation as &quot;time&quot;&lt;/strike&gt;?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The examples above are set-theoretical (&quot;concrete&quot;). Is there a more abstract modelling of &quot;time&quot;, maybe in category theory?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-16T14:06:29.940" LastActivityDate="2010-11-16T14:06:29.940" Title="Time in Mathematics" Tags="&lt;soft-question&gt;&lt;philosophy&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10555" PostTypeId="1" AcceptedAnswerId="10602" CreationDate="2010-11-16T13:01:53.193" Score="2" ViewCount="123" Body="&lt;p&gt;(1). I was wondering about what are the relation and differences between formal and informal logic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What topics does each of them have?&#xA;For example, topics such as Meaning and Definition, Syllogistic Logic, Inductive Logic, Probabilistic/Statistical Reasoning, Deductive Logic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is symbolic logic same as formal logic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) what is the relation between informal/formal logic and deductive/non-deductive reasoning/inference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;must formal logic be only for deduction, not for non-deductive reasoning/inference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) what are other usual/better ways to categorize various logic topics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T19:42:18.627" LastActivityDate="2010-11-16T19:42:18.627" Title="categorization of  logic" Tags="&lt;logic&gt;&lt;computer-science&gt;&lt;philosophy&gt;" AnswerCount="3" />
  <row Id="10556" PostTypeId="1" CreationDate="2010-11-16T13:19:19.807" Score="1" ViewCount="55" Body="&lt;p&gt;According to the wikipedia article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Levenberg_Marquardt&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Levenberg_Marquardt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(\boldsymbol\beta+\boldsymbol\delta) \approx \|\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta\|^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the derivative with respect to δ and setting the result to zero gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(J^{T}J)\boldsymbol \delta  = J^{T} [y - f(\boldsymbol \beta)])$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt to derive the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\|\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta\|^2&#xD;&#xA;= (\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta)^T(\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using product rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \|\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta\|^2}{\partial \boldsymbol\delta} = (-J^{T})(\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta) + (\mathbf{y} - \mathbf{f}(\boldsymbol\beta) - \mathbf{J}\boldsymbol\delta)^T(-J)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimensions of the left and right side don't match.  I believe there might be something wrong with my differentiation.  There seems to be a transpose missing, but I'm not sure what would cause a transpose in the differentiation operation.&lt;/p&gt;&#xA;" OwnerUserId="3572" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T13:23:56.733" LastActivityDate="2010-11-16T15:21:56.960" Title="How to derive the Levenberg–Marquardt algorithm with matrix calculus" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;&lt;optimization&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10557" PostTypeId="2" ParentId="10544" CreationDate="2010-11-16T13:21:43.477" Score="0" ViewCount="0" Body="&lt;p&gt;To elaborate on Qwirk's answer, look at $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{R} \setminus \mathbb{R}^+$ is uncountably infinite, even though $\mathbb{R}^+$ is uncountably infinite. On the other hand, let $T$ be the set of all real transcendental numbers. The set $\mathbb{R} \setminus T$ is exactly the set of all algebraic numbers, which we know to be countable. Finally, let $S = \mathbb{R} \setminus \{0, 1, \dots, n\}$ for some $n \in \mathbb{N}$. Taking the complement of $S$ in $\mathbb{R}$, you see that you can also get a finite set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in general, you can not say anything about the cardinality of the complement of a set. However, if the cardinality of a subset $S \subset X$ is strictly less than that of $X$, then the cardinality of $X - S$ must be the same as $X$. You should try to see this for yourself.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-16T13:21:43.477" CommentCount="1" />
  <row Id="10558" PostTypeId="2" ParentId="10533" CreationDate="2010-11-16T13:24:43.070" Score="2" ViewCount="0" Body="&lt;p&gt;You can use the technique of &lt;em&gt;Vacant Spaces&lt;/em&gt; to calculate probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During play, if you know that there is space for $\displaystyle x$ unknown cards with West and $\displaystyle y$ unknown cards with East, the probability that a specific card is with west is $\displaystyle x/(x+y)$. This assumes you ignore the carding of the opponents so far. This gives accurate results when computing the &lt;em&gt;apriori&lt;/em&gt; probabilities and gives reasonable values during the play of the hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, say you want to calculate the apriori probability of a 3-2 split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider a specific holding say (Q32 with West and 54 with East) and try to calculate the chances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chances that West gets the Q = 13/26. &lt;br/&gt;&#xA;Chances that West gets the 3 = 12/25.&lt;br/&gt;&#xA;Chances that West gets the 2 = 11/24. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chances that East gets the 5 = 13/23. &lt;br/&gt;&#xA;Chances that East gets the 4 = 12/22. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the chances of the specific split Q32-54 is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13/26 \times 12/25 \times 11/24 \times 13/23 \times 12/22$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 5 choose 3  + 5 choose 2 = 20 such splits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the probability of 3-2 split is 20 times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13/26 \times 12/25 \times 11/24 \times 13/23 \times 12/22$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 0.678\dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there is a 67.8% chance of a 3-2 break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good rule of thumb to remember is that if there are an odd number of cards outstanding, then apriori chances are that they will split as evenly as possible. So with 5 cards outstanding, 3-2 is split is more likely than 4-1 or 5-0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are an even number of cards outstanding, then they tend to break unevenly. i.e with 2 cards outstanding, 2-0 split is more likely than 1-1. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An article on Vacant Spaces: &lt;a href=&quot;http://www.clairebridge.com/textes/vacantplaces.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.clairebridge.com/textes/vacantplaces.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pavlicek's Suit Break Calculator: &lt;a href=&quot;http://www.rpbridge.net/xsb2.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.rpbridge.net/xsb2.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T13:34:31.530" LastActivityDate="2010-11-16T13:34:31.530" CommentCount="6" />
  <row Id="10559" PostTypeId="2" ParentId="10543" CreationDate="2010-11-16T13:26:02.737" Score="1" ViewCount="0" Body="&lt;p&gt;At the high school level, I like &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0935218084&quot; rel=&quot;nofollow&quot;&gt;Thinking Physics&lt;/a&gt;&lt;/em&gt; by Lewis Carroll Epstein very much. It's entertaining and illustrated by cartoons, and it delves deeply into the right and wrong answers to basic physics problems. An example: Can you float a battleship in a bathtub? &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-16T13:26:02.737" CommentCount="1" />
  <row Id="10560" PostTypeId="2" ParentId="460" CreationDate="2010-11-16T13:45:33.067" Score="0" ViewCount="0" Body="&lt;p&gt;Excerpts from Grillet’s Semigroups Chapter 1, Page 1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“There are 1160 distinct … semigroups of order 5; &#xA;15793 semigroups of order 6; 836,021 semigroups&#xA;of order 7…”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted, many of those semigroups are not that interesting.&#xA;However, transformation semigroups (not monoids unless &#xA;you count the identity map in) are interesting because most&#xA;transformations(functions) are not bijective. Thus, transformation&#xA;semigroups are more natural than permutation groups.&lt;/p&gt;&#xA;" OwnerUserId="3568" LastActivityDate="2010-11-16T13:45:33.067" CommunityOwnedDate="2010-11-16T13:45:33.067" />
  <row Id="10561" PostTypeId="2" ParentId="10554" CreationDate="2010-11-16T13:55:52.753" Score="2" ViewCount="0" Body="&lt;p&gt;I know of two other concepts that have a &quot;time&quot; feeling to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition for homotopy involves a parameter which can very intuitively be interpreted as time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lie group also has to me some feeling of time, since it adds on top of classical geometry the idea that isometries must be a sort of continuous motion through time, not just a &quot;teleportation&quot; between two states.&lt;/p&gt;&#xA;" OwnerUserId="887" LastActivityDate="2010-11-16T13:55:52.753" CommentCount="1" />
  <row Id="10562" PostTypeId="2" ParentId="10543" CreationDate="2010-11-16T13:56:02.583" Score="2" ViewCount="0" Body="&lt;p&gt;I would like to add that books that cultivate interest and critical thinking do not have to be primarily about math (or physics).  It is very important to offer young students a broad view of the culture of mathematics and science in general.  To this end I recommend Feynman's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393316041&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Surely You're Joking, Mr. Feynman&lt;/em&gt;&lt;/a&gt;, James Watson's &lt;a href=&quot;http://www.amazon.com/Double-Helix-James-D-Watson/dp/B000O8VDNG/ref=sr_1_2?s=books&amp;amp;ie=UTF8&amp;amp;qid=1289915557&amp;amp;sr=1-2&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Double Helix&lt;/em&gt;&lt;/a&gt;, and P. Yourgrau's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465092942&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;A World Without Time, The Forgotten Legacy of Godel and Einstein&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These books (and many others) offer a glimpse into the lives of some very prolific modern scientists.  I am certain that you will find others that can be described this way, and your students will find them on their own, once they know what to look for.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-11-16T13:56:02.583" />
  <row Id="10563" PostTypeId="1" AcceptedAnswerId="10569" CreationDate="2010-11-16T13:59:00.960" Score="3" ViewCount="92" Body="&lt;p&gt;Let $f$ be a distribution on $\mathbf{R}^n$ (in the Schwartz sense) such that&#xA;$$\frac{\partial f}{\partial x_i} = 0 \text{ for $i = 1, \ldots, n$.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how to prove that $f$ is a constant? I had this exercise in a class last year but I couldn't find how to do the induction step (for $n = 1$ it is clear of course).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-16T15:24:52.557" LastActivityDate="2010-11-16T16:24:02.230" Title="Derivatives distribution." Tags="&lt;fourier-analysis&gt;&lt;harmonic-analysis&gt;&lt;distribution-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10564" PostTypeId="1" AcceptedAnswerId="12760" CreationDate="2010-11-16T14:00:52.317" Score="1" ViewCount="44" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1D projective geometry,&#xA;I want to compute the 2x2 Homography matrix $H$ (in homogeneous coordinates), given 3 pairs of corresponding points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. I want to find H such that:&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;h_{11} &amp;amp;  h_{12}\\&#xD;&#xA;h_{21} &amp;amp; h_{22}\end{array}\right)\left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp; a &amp;amp; a+b\\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 1&#xD;&#xA;\end{array}\right)   = &#xD;&#xA;\left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp;a&amp;#39; &amp;amp;a&amp;#39;+b&amp;#39;\\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 1&#xD;&#xA;\end{array}\right).$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've got 6 equations here and only 3 unknowns. &#xA;(dof(H) = 4 elements less one for scaling = 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about 3 scaling factors that would add up to 6 unknowns, s.t. we would have a unique solution. But how exactly do I insert the scaling factors into the matrices and how can I compute H then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a clue?&lt;/p&gt;&#xA;" OwnerUserId="3573" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T15:09:28.383" LastActivityDate="2010-12-02T16:25:39.533" Title="How to compute a 2x2 Homography out of 3 corresponding points?" Tags="&lt;linear-algebra&gt;&lt;projective-geometry&gt;" AnswerCount="2" />
  <row Id="10565" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T14:08:08.977" Score="3" ViewCount="0" Body="&lt;p&gt;I'm not sure what $k$ is supposed to be, but another way to see that the first and third expressions are equal is to rewrite the first one as&#xA;$$\frac{\frac{\sin(A)}{\sin(B)}+1}{\frac{\sin(A)}{\sin(B)}-1}$$&#xD;&#xA;and then note that the law of sines tells you&#xA;$$\frac{\sin(A)}{\sin(B)} = \frac{a}{b}$$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-16T14:08:08.977" />
  <row Id="10567" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T14:38:14.857" Score="6" ViewCount="0" Body="&lt;p&gt;From the $\sin $ rule, you know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{\sin A}=\frac{b}{\sin B}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In a proportion such as&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{q}=\frac{r}{s},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;we have&lt;/em&gt; (&lt;em&gt;by a property of proportions, which you can verify algebraically&lt;/em&gt;): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p+r}{p-r}=\frac{q+s}{q-s}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $p=a$, $q=\sin A$, $r=b$, $s=\sin B$ gives you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a+b}{a-b}=\frac{\sin A+\sin B}{\sin A-\sin B}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: Algebraic verification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{p+r}{p-r}=\dfrac{q+s}{q-s}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow \left( p+r\right) \left( q-s\right) -\left( p-r\right)&#xD;&#xA;\left( q+s\right) =0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow 2qr-2ps=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow qr=ps$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow \dfrac{p}{q}=\dfrac{r}{s}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And clearly $\dfrac{q+s}{q-s}=\dfrac{k\left( q+s\right) }{k\left( q-s\right) }$, for $k\neq 0$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-16T19:11:40.453" LastActivityDate="2010-11-16T19:11:40.453" />
  <row Id="10568" PostTypeId="2" ParentId="10563" CreationDate="2010-11-16T14:40:22.763" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose that $(x_1,\ldots,x_n) \neq (y_1,\ldots,y_n)$. If we fix $x_2,\ldots,x_n$, then $f(x_1,x_2,\ldots,x_n)$ is constant with respect to $x_1$. Hence, $f(x_1,x_2,\ldots,x_n) = f(y_1,x_2,\ldots,x_n)$. Next, fix $y_1,x_3,\ldots,x_n$. Then, $f(y_1,x_2,\ldots,x_n)$ is constant with respect to $x_2$. Hence, $f(y_1,x_2,\ldots,x_n) = f(y_1,y_2,\ldots,x_n)$. Continue this way to conclude that $f(x_1,\ldots,x_n) = f(y_1,\ldots,y_n)$, that is, $f$ is constant.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-16T14:40:22.763" CommentCount="4" />
  <row Id="10569" PostTypeId="2" ParentId="10563" CreationDate="2010-11-16T14:50:21.667" Score="3" ViewCount="0" Body="&lt;p&gt;For $\partial f/\partial x_i$ to vanish, then $f$&#xD;&#xA;has to vanish on the space of test functions of the form&#xA;$\partial \phi_i/\partial x_i$. You effectively need to&#xA;show the subspace spanned by all of these (as $i$ varies)&#xA;is of codimension one in $C_c(\mathbb{R}^n)$. This subspace&#xA;has to be the space of all test functions integrating to zero.&#xA;To go from&#xA;$n-1$ to $n$, take a test function $\phi$ with integral $0$ and let&#xA;$\psi(x_1,\ldots,x_{n-1})=\int\phi(x_1,\ldots,x_n)dx_n$.&#xA;Then $\psi$ is a test function on $\mathbb{R}^{n-1}$&#xD;&#xA;with integral $0$ so is a sum of functions&#xA;$\partial \psi_i/\partial x_i$ where the $\psi_i$ are test&#xA;functions on $\mathbb{R}^{n-1}$. Extend to test functions&#xA;$\phi_i$ on $\mathbb{R}^n$ by letting $\phi_i(x_1,\ldots,x_n)&#xD;&#xA;=\psi_i(x_1,\ldots,x_{n-1})h(x_n)$ where $h$ is a fixed bump function&#xA;with integral $1$. Then $\psi-\sum\partial\phi_i/\partial x_i$&#xD;&#xA;has zero integral along all lines in the $x_n$-direction,&#xA;so is the $x_n$-partial derivative of a test function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;I should add that the above is effectively the proof that&#xA;$H_c^n(\mathbb{R}^n)=\mathbb{R}$ where $H_c^*$ denotes de Rham&#xA;cohomology with compact supports.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-16T16:24:02.230" LastActivityDate="2010-11-16T16:24:02.230" CommentCount="1" />
  <row Id="10570" PostTypeId="2" ParentId="10544" CreationDate="2010-11-16T15:01:26.880" Score="5" ViewCount="0" Body="&lt;p&gt;So, you have a given set $X$, and a specific subset $A$. You know the cardinality/size of $A$, and want to know the cardinality of $X\setminus A$, the (relative) complement of $A$ in $X$. Use $|X|$ to denote the size/cardinality of the set. I will assume the Axiom of Choice so that all sets have comparable cardinalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two cases, this is completely determined by the cardinalities of $A$ and of $X$. They are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $X$ is finite, of size $n$; then $A$ is finite, and the size of $X\setminus A$ is $|X|-|A|$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $X$ is infinite, and the cardinality of $A$ is &lt;em&gt;strictly smaller&lt;/em&gt; than the cardinality of $X$, then the cardinality of $X\setminus A$ equals the cardinality of $X$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Reason:&lt;/em&gt; If $\kappa$ and $\lambda$ are cardinals, and at least one of them is infinite, then $\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}$. Here, $\kappa+\lambda$ is the cardinality of the disjoint union of a set of cardinality $\kappa$ and a set of cardinality $\lambda$; $\kappa\lambda$ is the cardinality of the set $X\times Y$, where $|X|=\kappa$ and $|Y|=\lambda$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, $|\mathbb{R}\setminus\mathbb{Q}|=|\mathbb{R}|$, because $|\mathbb{Q}|=\aleph_0 \lt 2^{\aleph_0}=|\mathbb{R}|$. If $A$ is a finite subset of $|\mathbb{N}|$, then $|\mathbb{N}\setminus A|=\aleph_0 = |\mathbb{N}|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this is all you can say. If $X$ is infinite and $|A|=|X|$, then the cardinality of the complement $X\setminus A$ could be anything from $0$ and up to $|X|$. To get a set of size $n$ with $n$ finite, pick any subset $B$ of $|X|$ of size $n$, and take $A=X\setminus B$. To get a set with complement of cardinality $\kappa$ for any $\kappa\lt|X|$, biject $X$ with the &lt;em&gt;cardinal&lt;/em&gt; $|X|$, which has a subset of cardinality $\kappa$; take the complement. To get a set of cardinality $|X|$, use the fact that $|X|=|X\times X|$. Then, if $x\in X$ is a particular element, then the subset that corresponds under a given bijection to $X\times\{x\}$ has complement of size $|X|$. For specific sets $X$ it is possible to find specific examples. For $\mathbb{N}$, the even numbers have complement of size $|\mathbb{N}|$. In the real numbers, the unit interval has complement of size $|\mathbb{R}|$. And so on.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-16T19:11:09.470" LastActivityDate="2010-11-16T19:11:09.470" CommentCount="4" />
  <row Id="10571" PostTypeId="2" ParentId="10556" CreationDate="2010-11-16T15:21:56.960" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, I had a chance to see that.  But my issue is more related to matrix calculus than levenberg-marquardt.  To put it in simpler form, what's the difference between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial f(x)}{\partial x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial f(x)^T}{\partial x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where f(x) is a R^n -&gt; R^n vector function.&lt;/p&gt;&#xA;" OwnerUserId="3576" LastActivityDate="2010-11-16T15:21:56.960" CommentCount="3" />
  <row Id="10572" PostTypeId="1" AcceptedAnswerId="10591" CreationDate="2010-11-16T15:27:41.497" Score="5" ViewCount="62" Body="&lt;p&gt;Are there necessary and sufficient conditions for the existence of a circuit, or a disjoint set of circuits, that passes through each vertex once in a directed graph?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-11-16T17:17:53.670" Title="Existence of circuit passing through each vertex in a directed graph" Tags="&lt;graph-theory&gt;" AnswerCount="3" />
  <row Id="10574" PostTypeId="2" ParentId="10564" CreationDate="2010-11-16T15:39:10.187" Score="1" ViewCount="0" Body="&lt;p&gt;I believe the result should be in the form&#xA;$$\left(\begin{array}{ccc}&#xD;&#xA;0 &amp;amp; k_2a&amp;#39; &amp;amp; k_3(a&amp;#39;+b&amp;#39;)\\&#xD;&#xA;k_1 &amp;amp; k_2 &amp;amp; k_3&#xD;&#xA;\end{array}\right)$$&#xD;&#xA;with $k_1,k_2,k_3$ nonzero, to account for the homogeneous coordinates. Remember that the projective point $[x:y]$ is the same as the point $[kx:ky]$. So the middle answer doesn't have to look like $[a&amp;#39;:1]$, it can be any of the other representations of the point, hence $[k_2a&amp;#39;:k_2]$. That gives you three &quot;scaling&quot; factors. Is that what you were hoping for?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-16T15:39:10.187" />
  <row Id="10575" PostTypeId="2" ParentId="2694" CreationDate="2010-11-16T15:41:08.240" Score="-2" ViewCount="0" Body="&lt;p&gt;KCd,&#xA;I wouldn't agree with you about your assessment of a problem as important or unimportant--though in context and not in content. Whether or not something is important has been a question of the interests and attention of the particular researcher(s). If a large body of mathematical researchers find a mainstream topic interesting, then the questions arising from it become important to them. As a matter of fact, the importance of a mathematical problem is not REALISED until enormous effort from experts (and sometimes amateurs) have been expended to no avail. Fermat's last theorem was in the same state until the leading mathematiarans of the 19th century failed to settle it and some became &quot;ashamed&quot; that such a &quot;simple&quot; problem had not succumbed to the then prowess of the human mind. Until Kummer did a groundbreaking attack on FLT, which introduced &quot;new ideas&quot;, mathematicians wouldn't have elevated the status of FLT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a RULE OF THUMB, you must know that EVERY DIFFICULT QUESTION is IMPORTANT until PROVEN OTHERWISE.  For as long as its proof is undiscovered, its importance is shroude in a mystery that makes many to naively refer to it as UNIMPORTANT. What would you say, if someone comes with an ingeniously simple and &quot;isolated&quot; proof of FLT? Would that diminish the importance of Wile's proof? Perhaps yes, afterall then FLT could be settled without the need of advanced machinery from cohomology, commutative algebra, L-functions, p-adic, etc &lt;/p&gt;&#xA;" OwnerUserId="3579" LastActivityDate="2010-11-16T15:41:08.240" CommunityOwnedDate="2010-11-16T15:41:08.240" />
  <row Id="10576" PostTypeId="2" ParentId="10572" CreationDate="2010-11-16T15:48:09.817" Score="1" ViewCount="0" Body="&lt;p&gt;I don't think any are known. A quick search on the web gave &lt;a href=&quot;http://ajc.maths.uq.edu.au/pdf/23/ocr-ajc-v23-p115.pdf&quot; rel=&quot;nofollow&quot;&gt;a paper by Gutin and Yeo&lt;/a&gt;, which also directs to Chapter 1 of &lt;strong&gt;Digraphs - Theory, Algorithms, and Applications&lt;/strong&gt; by Bang-Jensen and Gutin (Springer Monographs, ISBN: 978-1-84800-997-4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if there were necessary and sufficient conditions, then taking any graph and replacing each edge by a pair of directed edges going back-and-forth would give you a characterization of Hamiltonian graphs, if I'm not very confused, which is not known.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-16T15:48:09.817" />
  <row Id="10577" PostTypeId="1" AcceptedAnswerId="10607" CreationDate="2010-11-16T15:50:50.080" Score="3" ViewCount="95" Body="&lt;p&gt;I've come up with the following variation of a logarithm, intended to work on values that can be 0, or can grow exponentially from zero in either positive or negative direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$myLog(x) =&#xD;&#xA;\begin{cases}&#xD;&#xA; \log(x+1), &amp;amp; \mbox{if }x \geq 0 \\&#xD;&#xA; -\log(-x+1), &amp;amp; \mbox{if }x \lt 0&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this formula or something similar to it? Has it been studied before? I'd like to read up on it if there is anything out there, but without knowing a name for it I haven't had much success with Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a graph of the function. It resembles arctangent, but it doesn't have asymptotes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/jWyYl.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="477" LastEditorUserId="477" LastEditorDisplayName="" LastEditDate="2010-11-16T16:02:55.680" LastActivityDate="2010-11-16T19:45:21.680" Title="Name for logarithm variation that works on non-positive values?" Tags="&lt;functions&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10578" PostTypeId="2" ParentId="10543" CreationDate="2010-11-16T15:58:38.537" Score="1" ViewCount="0" Body="&lt;p&gt;As a kid I really enjoyed &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0760723613&quot; rel=&quot;nofollow&quot;&gt;The Mathematical Tourist&lt;/a&gt; by Ivars Peterson (at the time, the 1988 edition).  The best part is that it gave just enough detail (for example, on how to compute a particular type of fractal) to inspire me to implement a computer program of my own to carry out the calculation or simulation.  In this way it really was a good guidebook for my own mathematical/computational explorations.&lt;/p&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2010-11-16T15:58:38.537" />
  <row Id="10579" PostTypeId="1" CreationDate="2010-11-16T16:02:22.027" Score="3" ViewCount="117" Body="&lt;p&gt;What is the difference between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\frac{\partial (\textbf{x}^{T}A)}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\frac{\partial (A\textbf{x})}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is a nxn matrix and x is a n sized column vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of a Jacobian on wikipedia (http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $A^{T}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $A$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is not correct because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textbf{x}^{T}A = (A^{T}\textbf{x})^{T}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;performing $\frac{\partial}{\partial \textbf{x}}$ on each side results&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^T=A$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not true.&#xA;Does $\frac{\partial}{\partial \textbf{x}}$  become $\frac{\partial}{\partial \textbf{x}^{T}}$ when moved inside the transpose on the right side?&lt;/p&gt;&#xA;" OwnerUserId="3580" LastActivityDate="2010-11-17T04:01:02.673" Title="First order differential rule for matrices" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10580" PostTypeId="2" ParentId="10563" CreationDate="2010-11-16T16:02:36.047" Score="2" ViewCount="0" Body="&lt;p&gt;If $f(x)$ is a smooth function, one gets that $f(x)$ is constant relatively easily as in Shai Covo's comment. For a general distribution, one can convolve with approximations to the identity: Let $\phi(x)$ be a fixed smooth function with compact support, and let $\phi_k(x) =  k^n\phi(kx)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall $f \ast \phi_k$ is a smooth function (the integral of $&amp;lt;f, \phi_y&amp;gt;$ with respect to $y$, where $\phi_y$ is the translate of $\phi$ by $y$) and is defined by $&amp;lt;f \ast \phi_k, s&amp;gt; = &amp;lt;f(x) , s \ast \phi_k(x)&amp;gt;$ for $C_c(R^n)$ functions $s(x)$. Since $s \ast \phi_k(x)$ converges to $s(x)$ uniformly, with the same property holding for derivatives of $s(x)$, $f \ast \phi_k$ converges to $f$ as distributions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But each $f \ast \phi_k(x)$ is a smooth function, and $&amp;lt;f \ast \phi_k, \partial_i s(x)&amp;gt;$ $=$  $&amp;lt;f(x), \partial_i s \ast \phi_k(x)&amp;gt;$ $=$  $&amp;lt;f(x), \partial_i (s \ast \phi_k(x))&amp;gt;$ $= 0$ by assumption. Hence by the smooth function case $f \ast \phi_k(x)$ is a constant. Thus $f(x)$ is the distributional limit of constants and must be constant itself.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-16T16:02:36.047" CommentCount="2" />
  <row Id="10582" PostTypeId="2" ParentId="10555" CreationDate="2010-11-16T16:33:20.230" Score="1" ViewCount="0" Body="&lt;p&gt;Unfortunately I'm not versed in philosophical logic, so let's start with some &quot;different categories of logic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Scope &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The first relevant category of logic is &lt;i&gt;scope&lt;/i&gt;, what is the logic talking about. &lt;b&gt;Propositional logic&lt;/b&gt; talks about nonparametric propositions. A typical rule (modus ponens) allows you to derive a proposition B given that you know A and that A implies B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Predicate logic&lt;/b&gt; talks about parametric propositions, for example &quot;X is a poster&quot;. We can derive &quot;There's X such that X is a poster&quot; from &quot;Tim is a poster&quot;, whereas in propositional logic &quot;Tim is a poster&quot; is monolithic and cannot be broken up to &quot;Tim&quot; and &quot;X is a poster&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicate logic is also known as first-order logic. If we quantify over predicates then we get &lt;b&gt;Second order logic&lt;/b&gt;. This allows us to use concepts such as &quot;There's a predicate P such that P(X) iff X=Tim&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scope is enlarged in a different direction by &lt;b&gt;Modal logic&lt;/b&gt;, where we can talk about events (propositions) eventually happening, happening until some other event happens, and so on.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Rules &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The rules of logic people are usually taught constitute &lt;b&gt;Classical logic&lt;/b&gt;. In classical logic you're allowed to use proof by contradiction. However, some mathematicians (well, logicians) dislike such rules since they're &lt;i&gt;non-constructive&lt;/i&gt; (this is like set-theorists who consider what happens when you're not allowed to use AC, the axiom of choice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By modifying (restricting) the rules, we arrive at &lt;b&gt;Intuitionistic logic&lt;/b&gt; (or &lt;b&gt;Constructive mathematics&lt;/b&gt;), in which every proposition that we prove has a constructive proof from the givens. In particular, you can't use a proof of the sort &quot;if X, then Y; if not X, then also Y&quot; unless you can decide whether X is in fact true. The resulting world is nice - for example all functions are continuous. On the other hand, not every theorem you can prove in classical logic is true in the constructive sense, although surprisingly many are (for example, the fundamental theorem of algebra is still provable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a different direction, in &lt;b&gt;Infinitary logic&lt;/b&gt; the proof is infinite and there are logical rules with infinitely many premises. For example, you could have a rule deducing &quot;P(n) for every natural n&quot; from the (countably) many propositions P(0), P(1), ... In general these logics are not well-behaved, but if you choose the parameters correctly, they are (look for $L_{\omega_1,\omega}$).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Goals &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Different areas of logic have different goals. For example, in &lt;b&gt;Structural Proof Theory&lt;/b&gt; one goal is to understand how simple a proof system can be made, and what are the consequences. Using this syntactic approach, you can show that certain statements are not provable in certain systems by converting them into a very simple form, bounding the resulting size of the simple proof, and show that it is too short to prove the given statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complementary direction is &lt;b&gt;Model Theory&lt;/b&gt;, which is all about what a given set of axioms describes. You can prove for example that a set of first-order axioms cannot characterize the size of the universe (if it's infinite). The model-theoretic approach to showing that some theorem is unprovable is to exhibit a model of the axioms in which the theorem isn't true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In computer science, people are interested in &lt;b&gt;Proof Complexity&lt;/b&gt;, which studies how long it takes to prove statements if you're only allowed to use basic means. The holy grail is proving a statement which is even stronger than $P \neq NP$. A different sub-field constructs systems in which every predicate proven to exist is efficiently computable.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Categorical logic &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I don't know anything about &lt;i&gt;categorical logic&lt;/i&gt;. Apparently there are some kinds of categories (in the sense of category theory) which describe &quot;universes of set theory&quot; along with their associated logic, which is often constructive. These things are called &lt;b&gt;Topoi&lt;/b&gt; (singular &lt;b&gt;Topos&lt;/b&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt; Philosophers' logic &lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Philosophers are interested in different aspects of logic. Unfortunately, I don't know much about these, so I'll only give some sporadic examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some interest in nice, succinct systems of logic whose axioms are &quot;as natural as possible&quot;. In mathematical logic, by and large people aren't interested in the exact form of the axioms but in what they describe and what can be proven about the system using them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other interest is in what the allowable rules of logic should be. I described classical logic and intuitionistic logic. There are many systems in between (and even systems weaker than the latter), which might be natural given some philosophical stance. Most mathematicians don't even bother with intuitionistic logic, although some logicians (and even people from other branches) are intrigued by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also some interest in making sense of paradoxes - given a statement which is paradoxical (for example, the Liar paradox), what should its truth value be? The mathematical solution is to avoid self-reference in some systematic way, since mathematicians are more interested (in this case) in the well-being of the system rather than these &quot;monstrous examples&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-16T16:33:20.230" />
  <row Id="10583" PostTypeId="2" ParentId="10572" CreationDate="2010-11-16T16:46:58.850" Score="2" ViewCount="0" Body="&lt;p&gt;A collection of disjoint cycles that cover all vertices of an (undirected) graph is known as a 2-factor. &lt;a href=&quot;http://www.is.kochi-u.ac.jp/~fujisawa/papers/08AFS.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; references some criterion of Tutte for the existence of these, which, paraphrasing the authors, in some sense completely solves the problem. It's Theorem 1 on page 2, which points to &quot;The factors of graphs&quot; by Tutte (1952).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-16T16:46:58.850" CommentCount="4" />
  <row Id="10584" PostTypeId="2" ParentId="10579" CreationDate="2010-11-16T16:49:03.767" Score="2" ViewCount="0" Body="&lt;p&gt;There is actually a hugh difference. For one, when you say $\frac{\partial (x^TA)}{\partial x}$, the numerator is a $1 \times n$ vector. Whereas in case of $\frac{\partial (Ax)}{\partial x}$, the numerator is a $n \times 1$ vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look up this [wiki][1] page for details on how matrix derivatives should be done and the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to look up &lt;a href=&quot;http://matrixcookbook.com/&quot; rel=&quot;nofollow&quot;&gt;http://matrixcookbook.com/&lt;/a&gt; which gives you a lot of matrix identities like these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_calculus&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Matrix_calculus&lt;/a&gt; page&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-16T16:49:03.767" />
  <row Id="10585" PostTypeId="2" ParentId="10547" CreationDate="2010-11-16T16:56:37.890" Score="4" ViewCount="0" Body="&lt;p&gt;The sine rule for a triangle goes as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$, where $R$ is the radius of the circumcircle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we get $\sin(A) = \frac{a}{2R}$ and $\sin(B) = \frac{b}{2R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem has $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)}$. Plug in the expressions for $\sin(A)$ and $\sin(B)$ from the above and we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{a}{2R}-\frac{b}{2R}} = \frac{\frac{a+b}{2R}}{\frac{a-b}{2R}} = \frac{a+b}{a-b}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{a+b}{a-b}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-16T16:56:37.890" />
  <row Id="10586" PostTypeId="2" ParentId="7195" CreationDate="2010-11-16T17:08:14.367" Score="0" ViewCount="0" Body="&lt;p&gt;Observe this formulas:&lt;br&gt;&#xA;$sin^2 u = \frac{1 - cos 2u}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cos^2 u = \frac{1 + cos 2u}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;br&gt;&#xA;$\int\frac{1}{1-\cos (2u)}\mathrm{d}u = - \frac{cot (u)}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using of this formulas and partial integration formula you can easily get last part &lt;a href=&quot;http://integrals.wolfram.com/index.jsp?expr=%281-m%2ASin%5Bx%5D%5E2%29%5E%281%2F2%29%2FSin%5Bx%5D%5E2&amp;amp;random=false&quot; rel=&quot;nofollow&quot; title=&quot;mathematica online integrator&quot;&gt;of the answer&lt;/a&gt;.&lt;br&gt;&#xA;May be this will help you.&lt;/p&gt;&#xA;" OwnerUserId="1035" LastActivityDate="2010-11-16T17:08:14.367" />
  <row Id="10587" PostTypeId="2" ParentId="1730" CreationDate="2010-11-16T17:08:44.147" Score="0" ViewCount="0" Body="&lt;p&gt;One way I heard a lecturer describe functions recently was that of the CD player analogy.  &lt;/p&gt;&#xA;" OwnerUserId="3581" LastActivityDate="2010-11-16T17:08:44.147" />
  <row Id="10588" PostTypeId="1" AcceptedAnswerId="10590" CreationDate="2010-11-16T17:13:37.003" Score="3" ViewCount="99" Body="&lt;p&gt;I need to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A \backslash B) = P(A) - P(A\cap B)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the axioms of probability but can't see where to start. Note: this is homework, so a hint only would be nice :)&lt;/p&gt;&#xA;" OwnerUserId="1973" LastEditorUserId="1973" LastEditorDisplayName="" LastEditDate="2010-11-16T17:41:01.567" LastActivityDate="2010-11-16T17:41:01.567" Title="Proofs using the Axioms of Probability" Tags="&lt;probability&gt;&lt;homework&gt;&lt;axioms&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10589" PostTypeId="2" ParentId="10588" CreationDate="2010-11-16T17:17:04.340" Score="2" ViewCount="0" Body="&lt;p&gt;Well,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B \subset A$, then $A = B \cup (A \setminus B)$ which is a disjoint union so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,&#xA;$$A \cup B = (A \setminus (A \cap B)) \cup (A \cap B) \cup (B \setminus (A \cap B))$$&#xD;&#xA;which is again a disjoint union. So...&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-16T17:17:04.340" />
  <row Id="10590" PostTypeId="2" ParentId="10588" CreationDate="2010-11-16T17:17:09.817" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: Think &quot;countable additivity&quot;. What can you say about $A\setminus B$ and $A\cap B$ if you are hoping to use this axiom.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-16T17:17:09.817" />
  <row Id="10591" PostTypeId="2" ParentId="10572" CreationDate="2010-11-16T17:17:53.670" Score="3" ViewCount="0" Body="&lt;p&gt;According to the classic text, &quot;Computers and Intractability, A Guide to the Theory of NP-Completeness&quot; by Garey and Johnson, the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Partition into Hamiltonian Subgraphs&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a directed graph $G=(V,A)$, can&#xA;  the vertices be partitioned into&#xA;  disjoint sets $V_1, V_2, \dots, V_k$&#xD;&#xA;  for some $k$, such that each $V_i$&#xD;&#xA;  contains at least three vertices and&#xA;  induces a subgraph of $G$ that&#xA;  contains a Hamiltonian Circuit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is NP-Complete, by a reduction from 3SAT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book also mentions that if we allow each $V_i$ to contain at least &lt;em&gt;two&lt;/em&gt; vertices, then this is solvable in polynomial time using Matching techniques.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-16T17:17:53.670" />
  <row Id="10592" PostTypeId="2" ParentId="10588" CreationDate="2010-11-16T17:24:15.400" Score="1" ViewCount="0" Body="&lt;p&gt;The second one as written is false.  I'm sure you meant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(A \cup B) = P(A) + P(B) - P(A \cap B)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jonas' answer is a good hint on how to apply the first part to prove the second part.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-16T17:24:15.400" />
  <row Id="10593" PostTypeId="1" CreationDate="2010-11-16T17:35:58.997" Score="1" ViewCount="200" Body="&lt;p&gt;A manifold M is simply connected if for every pair of 1-cubes c1,c2: [0,1] -&gt; M with&lt;br&gt;&#xA;c1(0) = c1(1) = c2(0) = c2(1) = t&lt;br&gt;&#xA;there is a 2-cube b such that&lt;br&gt;&#xA;1) b1,0 = c1 and b1,1 = c2&lt;br&gt;&#xA;2) for all p in [0,1] b2,0(p) = b2,1(p) = t  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if I'm given a manifold M,&lt;br&gt;&#xA;for example choose M = S^2 = {(x,y,z) in R^3 such that x^2 + y^2 + z^2 = 1}&lt;br&gt;&#xA;how do I prove that it is simply connected?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would you prove that a manifold is not simply connected?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="3333" LastEditorDisplayName="" LastEditDate="2010-11-17T02:12:01.550" LastActivityDate="2010-12-08T07:07:36.730" Title="How to prove a manifold is simply connected" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="10594" PostTypeId="1" CreationDate="2010-11-16T17:36:28.607" Score="1" ViewCount="22" Body="&lt;p&gt;Are there any good publications for introduction on Popov forms of polynomial matrices? I found a treatment in the book &lt;em&gt;Linear Systems&lt;/em&gt; by Thomas Kailath. But I'm wondering if there a better publication (in pedagogical sense).&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-11-16T17:36:28.607" Title="Learning material on Popov forms" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;polynomials&gt;" />
  <row Id="10595" PostTypeId="1" AcceptedAnswerId="10600" CreationDate="2010-11-16T18:01:38.653" Score="3" ViewCount="122" Body="&lt;p&gt;Given a countable family of semi-norms $p_i$, we can define a metric&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(f,g) = \sum \limits_{i=0}^{\infty} 2^{-i} \frac{ p_i(f-g) }{ 1 + p_i(f-g) }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have the locally convex topology induced by the semi-norms as above, as well as the topology induced by the metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the proof work to show their equality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am known to the proof of Rudin (Functional analysis), but utilizes a different metric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(f,g) = \max \limits_{i \in \mathbb N}  2^{-i} \frac{ p_i(f-g) }{ 1 + p_i(f-g) }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the proof for this metric is fairly easy - you can handle value of the sequence on its own - i do not see how a similar proof might work for the first metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One guess would be to show equivalence of both metrics, but I don't see even that, as on $l^1$, the sum-norm-topology is strictly finer than the max-norm-topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-11-16T19:08:59.770" LastActivityDate="2010-11-17T09:25:20.047" Title="Proof that Frechet-metric generates same topology as the semi-norms" Tags="&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10596" PostTypeId="2" ParentId="10555" CreationDate="2010-11-16T18:15:43.487" Score="2" ViewCount="0" Body="&lt;p&gt;Any formal system consists of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a list of symbols (a vocabulary or alphabet),&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;a list of of typographical rules to connect these symbols to write statements (a grammar or syntax), and&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;a list typographical rules for deriving new statements from other statements (axioms or rules of inference). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By typographical rules, I mean rules that tell you only how to manipulate the symbols without reference to any meaning or interpretation that these symbols might have. Consider for example a rule of inference in propositional logic that I call  the Join Rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If X and Y are true statements, then so is &quot;(X) &amp;amp; (Y)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I make no reference in the statement of this rule to any possible meaning of the &quot;&amp;amp;&quot; symbol. (You can probably guess that it is the AND-operator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play around with my own implementation of formal propositional logic using a freeware program I have developed, available at my website &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-16T18:15:43.487" />
  <row Id="10597" PostTypeId="2" ParentId="10452" CreationDate="2010-11-16T18:29:21.037" Score="1" ViewCount="0" Body="&lt;p&gt;Both identities can be proved quite easily with inductions. Let $d$ be the common difference, i.e. $d=a_{n+1}-a_n$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use induction. So we need to prove&#xA;$$ \frac{n-2}{\sqrt{a_1}+\sqrt{a_ {n-1}}}+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_ n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_ n}}.$$&#xD;&#xA;Rationalize the denominators and substitute $$a_{n-1}-a_1=(n-2)d, a_n-a_{n-1}=d, a_n-a_1=(n-1)d.$$ Upon cancellations, we find&#xA;$$\sqrt{a_{n-1} }-\sqrt{a_ 1}+\sqrt{a_n }-\sqrt{a_{n-1} }=\sqrt{a_ n}-\sqrt{a_ 1}$$ on the left and the same on the right.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use induction, assuming true for $m&amp;lt;n$. Canceling the sum $$ \frac{1}{a_1a_n}+\frac{1}{a_na_1}$$ from the left with the term $$\frac{2}{a_1+a_n}\left(\frac{1}{a_1}+\frac{1}{a_n}\right)$$ on the right and using the induction hypothesis on the remaining terms $$\frac{1}{a_2a_{n-1}}+\cdots +\frac{1}{a_{n-1}a_2},$$ we find that  we need to prove  $$\frac{2}{a_2+a_{n-1}}$$ on the left is equal to $$\frac{2}{a_1+a_{n}}$$ on the right. And this is true because $a_2+a_{n-1}=a_1+a_n=2a_1+(n-1)d$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-16T18:29:21.037" />
  <row Id="10598" PostTypeId="2" ParentId="10595" CreationDate="2010-11-16T18:37:48.123" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming that you forgot a $2^{-i}$ in the definition of $d$ we argue as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that a Hausdorff TVS is metrizable iff zero has a countable neighborhood basis. If this is true then the topology is even generated by a translation invariant metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another theorem states that a topology on a TVS is locally convex iff it is generated by a family of seminorms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we combine these and use that your metric is translation invariant, we obtain the result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is indirect, probably there is a direct proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Oh, I don't think the direct proof is so hard, I will see if I have time to add it. You should just observe that the neighborhood $\{x : p_n(x) \leq 2^{-k}\}$ contains the set $\{x : |x| \leq 2^{-m-k-1} \}$ and that the set $\{x : |x| \leq 2^{-k} \}$ contains the neighborhood $\{x : p_{k + 1}(x) \leq 2^{-k - 2} \}$. Where $|x| := d(x,0)$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-16T23:39:36.793" LastActivityDate="2010-11-16T23:39:36.793" CommentCount="3" />
  <row Id="10599" PostTypeId="1" CreationDate="2010-11-16T18:38:10.997" Score="1" ViewCount="42" Body="&lt;p&gt;Let $p:E\rightarrow B$ be a Serre fibration of path connected spaces with fiber $F$. Are the connecting homomorphisms $\partial:\pi_{n+1}(B)\rightarrow \pi_{n}(F)$ in the long exact sequence of $p$ induced by a &lt;strong&gt;continuous&lt;/strong&gt; map $\Omega B\rightarrow F$?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastActivityDate="2010-11-16T19:34:10.190" Title="Connecting Homomorphism in LES of fibration" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="10600" PostTypeId="2" ParentId="10595" CreationDate="2010-11-16T18:49:57.670" Score="3" ViewCount="0" Body="&lt;p&gt;To prove equivalence of the topologies you can use the fact that both topologies are characterized by their convergent sequences.  This is true in any first countable space.  Metric spaces are first countable, and the topology induced by a countable family of seminorms is first countable.  Here, since you have translation invariance, it is enough to check sequences converging to 0 (although this simplification is not essential).  That is, you can show that if $x_1,x_2,\ldots$ is a sequence in your vector space, then $d(x_n,0)\to 0$ as $n\to\infty$ if and only if for all $i$, $p_i(x_n)\to0$ as $n\to\infty$.  For the left to right direction, you can use the fact that $\frac{p_i}{1+p_i}\leq 2^id$.  For the other direction you can first bound the tail, then work with the remaining finitely many terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no actual need to consider sequences.  You basically want to show that the identity map is a homeomorphism, and this reduces to showing it is continuous at 0 in both directions, which in any case involves the same estimates you would make when working with sequences. (In other words, considering first countability a priori isn't necessary.  If you really want to, you could instead work with nets.)  Note that Jonas T's answer suggests a complementary approach (that was given around the same time), working directly with neighborhoods of $0$ instead of with sequences.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-17T09:25:20.047" LastActivityDate="2010-11-17T09:25:20.047" CommentCount="3" />
  <row Id="10601" PostTypeId="2" ParentId="1099" CreationDate="2010-11-16T19:08:05.717" Score="0" ViewCount="0" Body="&lt;p&gt;Barnsley's book FRACTALS EVERYWHERE has many examples of this.  I believe some edition of that book (long ago...) came with software that lets you make your own examples, too.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-11-16T19:08:05.717" />
  <row Id="10602" PostTypeId="2" ParentId="10555" CreationDate="2010-11-16T19:08:50.810" Score="2" ViewCount="0" Body="&lt;p&gt;Informal logic is, well, informal.  As such it is not unambiguously defined what the subject matter is, how different statements are related to each other, or (more ambitiously) what statements &quot;mean&quot;.   This type of activity is more popular in philosophy, linguistics and other areas that are not specifically mathematical.  Modern interest in informal reasoning is often about how to formalize it, as in AI, natural language processing or data mining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formalized logic is as described in the other answers.  Essentially, there are many types of formal logic, and each one is a game with its own well-defined rules (in principle, performable by a computer) for taking a set of statements -- informally thought of as &quot;premises&quot; -- and adding to it some additional statements, informally thought of as &quot;conclusions&quot;.  In spirit it is very close to computer programming and its study as a field resembles computer science. Some branches such as model theory or set theory are studied purely as mathematics in the sense that the explicitly combinatorial/linguistic aspect is in the background and the sentences are studied in a &quot;semantic&quot; approach that speaks in terms of the structures, objects and their properties (the material supposedly described by the logic) rather than the permissible syntactic operations in a formal logical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics is intermediate between these two extremes.  It uses logic(s) that could be called &lt;em&gt;potentially&lt;/em&gt; (or presumably, or in-principle) &lt;em&gt;formalizable&lt;/em&gt;.  Steps of the game are not codified precisely enough to be programmed on a computer, but they are standardized and sufficiently well-defined to correspond to known genuine moves in some actual formal systems, so there is a presumption that any correct non-formalized proof is robust enough to have many slightly different expansions into a very detailed formal proof in several of the formal systems suggested as a &quot;foundation&quot; for mathematics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned probabilistic or statistical reasoning.   This is a somewhat separate question because any deterministic setting can be enlarged a probabilistic one (e.g., instead of measuring attributes or their absence by $1$ or $0$, allow any numerical value between $1$ and $0$) and the logical rules modified accordingly in a known way.  Similarly, any setting that uses ordinary or probabilistic logic on objects thought of as &quot;data&quot; can be placed in the context of statistics, where one considers how the data might have been generated in addition to the data itself, and the rules for reasoning and speaking about this richer picture can again be updated in a routine way.   But these probabilistic/fuzzy/statistical enrichments are a kind of fixed recipe where the &quot;logic&quot; and &quot;probabilistic&quot; ingredients are cooked separately and mixed in an understood way.  You could discuss the result as a new form of logic but really it is a repackaging of what you started with, not a fundamentally different mode of reasoning.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T19:34:01.657" LastActivityDate="2010-11-16T19:34:01.657" />
  <row Id="10603" PostTypeId="1" AcceptedAnswerId="10674" CreationDate="2010-11-16T19:24:48.690" Score="9" ViewCount="181" Body="&lt;p&gt;What do the finite-dimensional continuous complex representations of the additive group $\mathbb{Q}$ with the usual topology look like?  With the discrete topology?  Which representations are indecomposable?  Irreducible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only ones I can think of are of the form $t \mapsto e^{tA}$ for some $A \in \mathcal{M}_n(\mathbb{C})$.  I would be willing to believe that these are the only ones in the first case, but I'm less sure in the second case.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-17T09:25:13.387" Title="Representation theory of the additive group of the rationals?" Tags="&lt;representation-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10604" PostTypeId="2" ParentId="10599" CreationDate="2010-11-16T19:28:30.387" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the path fibration $PB \to B$, assuming you've chosen a base-point in $B$.  This map is null homotopic.  A constant map $PB \to B$ lifts to unique base-point preserving map $PB \to E$ (constant the base-point in $F$).  So provided your fibration satisfies the homotopy lifting property for the special case of the path fibration over $B$, you get a map $PB \to E$ covering the path fibration $PB \to E$.  Restricting this to the loop space $\Omega B \subset PB$ gives a map $\Omega B \to F$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume it's known and pretty easy to check whether or not Serre fibrations have this property, but certainly general fibrations have them (say, taking the fibration definition in Hatcher's book). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were you interested in this technical detail on the distinction between fibrations and Serre fibrations, or were you more interested in the general question of when the connecting homomorphism is induced by an actual map of spaces? &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-16T19:34:10.190" LastActivityDate="2010-11-16T19:34:10.190" CommentCount="2" />
  <row Id="10605" PostTypeId="2" ParentId="10346" CreationDate="2010-11-16T19:34:54.637" Score="4" ViewCount="0" Body="&lt;p&gt;We can prove this using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;Multinomial Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have, using the Multinomial Theorem, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (a+b+c)^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} a^m b^{k-m} c^{n-k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $\displaystyle a = pp_j x$, $b = p(1-p_j)$, $c = 1-p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (pp_jx + p(1-p_j) + 1-p )^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} (pp_jx)^m (p(1-p_j))^{k-m} (1-p)^{n-k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (pp_jx + 1 - pp_j)^n = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose k}{k \choose m} p_j^m(1-p_j)^{k-m} p^{k}(1-p)^{n-k} x^m$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding the left hand side using binomial theorem and comparing the coefficients of $\displaystyle x^m$ gives the result.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T21:17:10.470" LastActivityDate="2010-11-16T21:17:10.470" />
  <row Id="10606" PostTypeId="1" AcceptedAnswerId="10621" CreationDate="2010-11-16T19:44:47.747" Score="1" ViewCount="110" Body="&lt;p&gt;I'm familiar with the notation $\mathrm{ord}_a(x)$, when $x$ is an integer.  However, I'm reading a book where this notation is used with $x$ rational.  I'm not sure how to interpret this?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T23:43:43.170" LastActivityDate="2010-11-16T23:43:43.170" Title="Notation: $\mathrm{ord}_a(x)$ for $x\in \mathbb{Q}$" Tags="&lt;number-theory&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="10607" PostTypeId="2" ParentId="10577" CreationDate="2010-11-16T19:45:21.680" Score="7" ViewCount="0" Body="&lt;p&gt;Given your requirements, here's how you could have invented the inverse hyperbolic sine, even if you had never heard of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want a monotonic function that grows logarithmically at both infinities. You try adding, subtracting, and multiplying logarithmic functions, but that doesn't work. However, since the function you want is monotonic, you can think about the &lt;em&gt;inverse&lt;/em&gt; of your function, which must grow &lt;em&gt;exponentially&lt;/em&gt; at both infinities. And adding together exponentials does do the job for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the inverse of the function you invented, which I'll call $f$. Its inverse is $$f^{-1}(y) = \begin{cases}e^y - 1 &amp;amp; \text{if } y \ge 0, \\ 1 - e^{-y} &amp;amp; \text{if } y \lt 0.\end{cases}$$ Since $e^{-y} \le 1$ when $e \ge 0$, and $e^y &amp;lt; 1$ when $y &amp;lt; 0$, the above inverse function is quite close to the much more nicely-behaved $$\begin{align}g(y) &amp;amp;= e^y - e^{-y} \\ &amp;amp;\equiv 2 \sinh y.\end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in some sense, $f(x)$ behaves like $\sinh^{-1} (x/2)$. As it turns out, they agree at $x = 0$ (although their derivatives differ by a factor of two), and as $x \to \infty$, they both approach $\ln x$ and the difference between them decreases as $1/x$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-16T19:45:21.680" CommentCount="1" />
  <row Id="10608" PostTypeId="2" ParentId="2694" CreationDate="2010-11-16T19:49:13.757" Score="9" ViewCount="0" Body="&lt;p&gt;Most of the answers so far have been along the general lines of 'Why hard problems are important', rather than 'Why the Collatz conjecture is important'; I will try to address the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the basic question being touched on is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In what ways does the prime factorization of $a$ affect the prime factorization of $a+1$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, one can always multiply out the prime factorization, add one, and then factor again, but this throws away the information of the prime factorization of $a$.  Note that this question is also meaningful in other UFDs, like $\mathbb{C}[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems very hard to come up with answers to this question that don't fall under the heading of 'immediate', such as distinct primes in each factorization.  This seems to be in part because a small change in the prime factorization for $a$ (multiplication by a prime, say) can have a huge change in the prime factorization for $a+1$ (totally distinct prime support perhaps).  Therefore, it is tempting to regard the act of adding 1 as an essentially-random shuffling of the prime factorization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most striking thing about the Collatz conjecture is that it seems to be making a deep statement about a subtle relation between the prime factorizations of $a$ and $a+1$.  Note that the Collatz iteration consists of three steps; two of which are 'small' in terms of the prime factorization, and the other of which is adding one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;multiplying by 3 has a small effect on the factorization.&lt;/li&gt;&#xA;&lt;li&gt;adding 1 has a (possibly) huge effect on the factorization.&lt;/li&gt;&#xA;&lt;li&gt;factoring out a power of 2 has a small effect on the factorization (in that it doesn't change the other prime powers in the factorization).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the Collatz conjecture seems to say that there is some sort of abstract quantity like 'energy' which is cannot be arbitrarily increased by adding 1.  That is, no matter where you start, and no matter where this weird prime-shuffling action of adding 1 takes you, eventually the act of pulling out 2s takes enough energy out of the system that you reach 1.  I think it is for reasons like this that mathematicians suspect that a solution of the Collatz conjecture will open new horizons and develop new and important techniques in number theory.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-11-16T19:49:13.757" CommunityOwnedDate="2010-11-16T19:49:13.757" CommentCount="1" />
  <row Id="10609" PostTypeId="2" ParentId="10603" CreationDate="2010-11-16T19:52:12.413" Score="5" ViewCount="0" Body="&lt;p&gt;In the usual topology, each continuous&#xA;representation of $(\mathbb{Q},{+})$ will extend to a continuous&#xA;representation of $(\mathbb{R},{+})$ so you'll only get the obvious ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the discrete topology there are non-obvious representations even&#xA;in dimension one. Consider the map&#xA;$\alpha:\mathbb{Q}\to \mathbb{Q}_p\to\mathbb{Q}_p/\mathbb{Z}_p&#xD;&#xA;\cong \mathbb{Z}[1/p]/\mathbb{Z}$. Then $t\mapsto\exp(2\pi i\alpha(t))$&#xD;&#xA;is a character of $\mathbb{Q}$, but not of the form $t\mapsto e^{at}$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-16T19:52:12.413" />
  <row Id="10610" PostTypeId="2" ParentId="10606" CreationDate="2010-11-16T19:53:06.580" Score="5" ViewCount="0" Body="&lt;p&gt;For any rational number $x$ and any prime integer $p$, we can write $x=p^r\frac{m}{n}$ uniquely (where $r$ is an integer thanks to Yuval for clarifying that), where $m,n$ are integers relatively prime to $p$. Then, we write $ord_p(x)=r$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-16T22:59:18.887" LastActivityDate="2010-11-16T22:59:18.887" CommentCount="2" />
  <row Id="10611" PostTypeId="2" ParentId="10231" CreationDate="2010-11-16T19:58:25.793" Score="0" ViewCount="0" Body="&lt;p&gt;No, the answer is definitely not zero. But, of course, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=0}^{36} \sin(n\times 10&amp;#176;) = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-16T19:58:25.793" />
  <row Id="10612" PostTypeId="2" ParentId="2694" CreationDate="2010-11-16T20:01:18.167" Score="5" ViewCount="0" Body="&lt;p&gt;I believe the $3x+1$ problem is considered a test case for ergodic theory, i.e., proving that certain probabilistic expectations true are true for orbits of a specific system.  There are papers by Sinai, Lagarias and others giving probabilistic models, similar to the asymptotic predictions made in questions about distribution of prime numbers, where the predictions are reliable but proving them in any particular case (twin primes, $n^2+1$, Goldbach, ...) is a centuries-old open problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is analogous to transcendence or irrationality proofs of specific numbers: proving that what is &quot;true with probability 1&quot; really holds in a particular instance is extremely hard and pushes the theory forward.  Other than that there are no problems outside the $3x+1$ conjecture that use the same iteration, so it is a sink and not a source in the applications-of-theory graph.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-16T20:06:52.173" LastActivityDate="2010-11-16T20:06:52.173" CommunityOwnedDate="2010-11-16T20:01:18.167" CommentCount="1" />
  <row Id="10613" PostTypeId="2" ParentId="10545" CreationDate="2010-11-16T20:15:29.453" Score="1" ViewCount="0" Body="&lt;p&gt;I tried a geometry proof for this and thought it was worth sharing it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to draw the figure. I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the triangle $ABC$. Let the angle bisector of $A$ meet $BC$ at $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By angle bisector theorem, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{AB}{AC} = \frac{DC}{DB}$. Hence, we get $\frac{b}{c} = \frac{DC}{DB}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add one to both sides, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{b+c}{c} = \frac{DC+DB}{DB} = \frac{a}{DB}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging we get $\frac{c}{DB} = \frac{b+c}{a} = \cot(\frac{A}{2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument (or using the fact that $\frac{c}{DB} = \frac{b}{DC}$) yields, $\frac{b}{DC} = \frac{b+c}{a} = \cot(\frac{A}{2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we are almost done. (Note that if $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$, the above is true.) However, we still need to prove that for no other possibility this is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the above. Draw a perpendicular to $AB$ at $B$ and similarly draw a perpendicular to $AC$ at $C$. Let these two will intersect $AD$ at $D&amp;#39;$ and $D&amp;#39;&amp;#39;$. Now note that $BD&amp;#39; = BD$ since $\frac{BD}{AB} = \cot(\frac{A}{2}) = \frac{BD&amp;#39;}{AB}$ (Since $D&amp;#39;BA$ is a right angled triangle). Similarly $CD&amp;#39;&amp;#39; = CD$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not possible unless $D&amp;#39; = D$ or $D&amp;#39;&amp;#39; = D$, which implies $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures. The argument I hope however gives you the idea behind the method.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-16T20:15:29.453" />
  <row Id="10614" PostTypeId="1" CreationDate="2010-11-16T20:55:06.290" Score="2" ViewCount="103" Body="&lt;p&gt;I figured that I can create a deterministic &quot;random&quot; numbers generator by utilizing a bit of &quot;magic&quot; that I picked up from some cryptography. However I seem to have missed a detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the function $\displaystyle f(n) = p^n \mod q$ should for $\displaystyle n$ in the range $\displaystyle [1,q-1]$ produce all the numbers in the same range, but in a scrambled order, however that is only true for some pairs of $\displaystyle p$ and $\displaystyle q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly do I find and verify such pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;I now know why I missed that detail, I'm think my computer will be done finding a few suitable number pairs by brute force long before I will be done understanding a more efficient approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'll run two of these generators in tandem, using their sum mod limit to generate a number in the field $[0;limit-1]$. For each pair I'll choose $q$ close to 2^32, $p$ close to 2^16, and assert that $p$ is prime, $q$ is prime, $(q-1)/2$ is prime and that $p$ is a primitive root of $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your input, but I'll leave the question open to anyone who can explain a computationally efficient method of asserting primitive rootness status of two numbers in a way that I understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;I have found the pairs 65539 (mod 4294967087) and 65537 (mod 4294965887), this gives the following JavaScript function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function seed(state1,state2){&#xA;    var mod1=4294967087&#xA;    var mul1=65539&#xA;    var mod2=4294965887&#xA;    var mul2=65537&#xA;    state1=state1%mod1&#xA;    state2=state2%mod2&#xA;    function random(limit){&#xA;        state1=(state1*mul1)%mod1&#xA;        state2=(state2*mul2)%mod2&#xA;        return (state1+state2)%limit&#xA;    }&#xA;    return random&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3584" LastEditorUserId="3584" LastEditorDisplayName="" LastEditDate="2010-11-17T16:11:06.517" LastActivityDate="2010-11-17T16:11:06.517" Title="Deterministic random numbers generator using $p^n \mod q$" Tags="&lt;prime-numbers&gt;&lt;random&gt;" AnswerCount="2" />
  <row Id="10615" PostTypeId="2" ParentId="10614" CreationDate="2010-11-16T21:02:13.357" Score="4" ViewCount="0" Body="&lt;p&gt;Assuming $\displaystyle q$ is a prime, you need $\displaystyle p$ to be a &lt;a href=&quot;http://mathworld.wolfram.com/PrimitiveRoot.html&quot; rel=&quot;nofollow&quot;&gt;Primitive Root&lt;/a&gt; of $\displaystyle q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that every prime $\displaystyle q$ has a primitive root. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact $\displaystyle q$ can also be a power of an odd prime or twice the power of an odd prime, see here: &lt;a href=&quot;http://www.math.uiuc.edu/~hildebr/453/notes/nt-notes5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/~hildebr/453/notes/nt-notes5.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It (according to the wiki link below) has been shown that the least primitive root of $\displaystyle q$, $\displaystyle g_q$ satisfies $\displaystyle g_q &amp;lt; C\sqrt[4]{q}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by assuming the generalized Riemann Hypothesis, in fact that $\displaystyle g_p = O(\log^6 q)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Order_of_magnitude_of_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;Magnitude of Primitive Roots&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given a $\displaystyle q$, you should be able to simply do a scan to find the least primitive root relatively quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in $\displaystyle p$ also being prime, see also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;Artin's Conjecture on Primitive Roots&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-16T21:14:57.413" LastActivityDate="2010-11-16T21:14:57.413" />
  <row Id="10616" PostTypeId="1" AcceptedAnswerId="10628" CreationDate="2010-11-16T21:03:07.290" Score="5" ViewCount="236" Body="&lt;p&gt;How can I prove that if G is an abelian group with elements a and b with orders m and n, respectively, then G contains an element whose order is the least common multiple of m and n?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an exercise from Hungerford's book, but it's not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when m and n are coprimes. I did this part. But I have no idea how to solve the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastActivityDate="2010-11-17T02:51:10.800" Title="Order of elements in abelian groups" Tags="&lt;group-theory&gt;" AnswerCount="4" />
  <row Id="10617" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T21:11:34.373" Score="2" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also one of the starred problems in &lt;em&gt;Herstein's: Topics in Algebra&lt;/em&gt; book. Take a look into page 3 of this file:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.csie.ntu.edu.tw/~b89053/herstein.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.csie.ntu.edu.tw/~b89053/herstein.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In case you aren't able to view the solution i type it down as i am not able to download this file, but able to view it in Google Docs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt; Let $a$ and $b$ be elements of order $m$ and $n$ respectively.  Write $\text{lcm} (m,n)= \prod\limits_{i=1}^{t}p_{i}^{\alpha_{i}}$, where the $p_{i}$'s are distinct primes, and $\alpha_{i} &amp;gt; 0$ for each $i$. For any $i \in [1,2, \cdots,t ]$, $p_{i}^{\alpha_{i}}$ divides, either $m$ or $n$. If say, $p_{i}^{\alpha_{i}} \mid  m$, then, $a^{m/p_{i}^{\alpha_{i}}}$ is of order $p_{i}^{\alpha_{i}}$. In this way we obtain elements $x_{1},\cdots,x_{t}$ of orders $p_{1}^{\alpha_{1}}, \cdots, p_{t}^{\alpha_{t}}$ respectively. Now consider elements in $G$ of the form $$x_{1}^{\beta_{1}}, \cdots, x_{t}^{\beta_{t}} \quad \cdots (3)$$ where $\beta_{i} \in \mathbb{N}$, and $i \in [1,2,\cdots,t]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our next goal, is to show that, the single elements, $(x_{1},\cdots,x_{t})$ generates all elements of the form $(3)$. For this purpose, we prove that for all $\beta_{i} \in \mathbb{N}$,  $i \in [1,2,\cdots t]$, the system $$ k = \beta_{1} \ (\text{mod} \ p_{1}^{\alpha_{1}})$$ $$ \cdot $$ $$\cdot $$ $$\cdot $$ $$k = \beta_{t} \ (\text{mod} \ p_{t}^{\alpha_{t}}$$ has a solution in $k$. This follows by setting $$ k =\sum\limits_{i=1}^{t} \beta_{i}c_{i} p_{1}^{\alpha_{1}} \cdots p_{i-1}^{\alpha_{i-1}}p_{i+1}^{\alpha_{i+1}} \cdots p_{t}^{\alpha_{t}}$$ where $c_{i}$ satisfies $$ c_{i} p_{1}^{\alpha_{1}} \cdots p_{i-1}^{\alpha_{i-1}}p_{i+1}^{\alpha_{i+1}} \cdots p_{t}^{\alpha_{t}} \equiv 1 \ (\text{mod} \ p_{i}^{\alpha_{i}})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to see that the elements of the form in Eq.(3) form a subgroup $H$ of $G$. Hence for each $i$ the order $p_{i}^{\alpha_{i}}$ of the subgroup generated by $x_{i}$ divides $|H|$ by &lt;em&gt;Lagrange's theorem.&lt;/em&gt;  This gives $|H| \geq \prod\limits_{i=1}^{t} p_{i}^{\alpha_{i}}$. Clearly $|H| \leq \prod\limits_{i=1}^{t} p_{i}^{\alpha_{i}}$ also holds. Since the element, $(x_{1}, \cdots, x_{t})$ generates exactly $H$ it has order $|H|$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-16T21:34:36.870" LastActivityDate="2010-11-16T21:34:36.870" CommentCount="1" />
  <row Id="10618" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T21:18:59.357" Score="3" ViewCount="0" Body="&lt;p&gt;Let $n = \prod_i p_i^{n_i}$ and $m = \prod_i p_i^{m_i}$, where the $p_i$s are the same. Thus $\mathrm{lcm}(n,m) = \prod_i p_i^{\max(n_i,m_i)}$. Let $N = \{ i : n_i \geq m_i \}$ and $M = \{ i : n_i &amp;lt; m_i \}$. Define $n&amp;#39; = \prod_{i \in N} p_i^{n_i}$ and $m&amp;#39; = \prod_{i \in M} p_i^{m_i}$. Now $a&amp;#39; = a^{n/n&amp;#39;}$ is of order $n&amp;#39;$, and $b&amp;#39; = b^{m/m&amp;#39;}$ is of order $m&amp;#39;$. Since $n&amp;#39;,m&amp;#39;$ are coprime, we know that there is an element of order $n&amp;#39;m&amp;#39; = \mathrm{lcm}(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can also prove it using the structure theorem for abelian groups, but that's an overkill.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-16T21:18:59.357" CommentCount="2" />
  <row Id="10619" PostTypeId="2" ParentId="10614" CreationDate="2010-11-16T21:32:54.020" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots&quot; rel=&quot;nofollow&quot;&gt;wiki article&lt;/a&gt; has an algorithm for testing primitive roots; it requires you to factor $q-1$.  But once you've done that, together with the GRH bound mentioned in Moron's answer you should at least be able to locate a primitive root in polynomial time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is a really inadvisable way of generating &quot;random&quot; numbers.  Given even two consecutive values of the sequence you can predict all of the others.  There are much better methods described, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudorandom_number_generator&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-16T21:32:54.020" CommentCount="1" />
  <row Id="10620" PostTypeId="2" ParentId="10516" CreationDate="2010-11-16T21:45:51.680" Score="2" ViewCount="0" Body="&lt;p&gt;(&lt;em&gt;Partial Answer to Question 1&lt;/em&gt;) No, this phenomenon isn't restricted to integral convex polytopes. There are examples of non-integral rational convex polytopes with Ehrhart quasi-polynomials that are polynomials. See this &lt;a href=&quot;http://adsabs.harvard.edu/abs/2003math......9329D&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-16T21:45:51.680" />
  <row Id="10621" PostTypeId="2" ParentId="10606" CreationDate="2010-11-16T21:48:23.970" Score="2" ViewCount="0" Body="&lt;p&gt;The notation is defined and explained in some detail &lt;a href=&quot;http://math.uga.edu/~pete/4400intro.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;: see pages 11-13.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-16T21:48:23.970" CommentCount="4" />
  <row Id="10622" PostTypeId="1" AcceptedAnswerId="10633" CreationDate="2010-11-16T21:48:39.773" Score="4" ViewCount="95" Body="&lt;p&gt;Given an open set $U \subset \mathbb R ^n $, there exists an exhaustion by compact sets, i.e. a sequence of compact sets $K_i$, s.t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\cup _{i=0}^{\infty} K_i = U$ and $\forall i \in \mathbb N : K_i \subset K_{i+1} ^{\circ}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can imagine that different exhaustions by compact sets 'propagate' at different speeds for the various parts $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us call an exhaustion $(K_i)_i$ 'stronger' than another exhaustions $(L_i)_i$, whenever we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall i \in \mathbb N \exists j \in \mathbb N : L_i \subset K_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call two exhaustions equivalent, if each one is stronger than the other - i.e. for each compact set in the first, we have a compact superset in the second, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Are all exhaustions of $U$ equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purpose: You encounter various settings, where you define a certain structure by such an exhaustion. If the above question has a positive answer, this would spare from wondering whether your construction is independent of such an exhaustion.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-17T10:24:04.733" Title="'Equivalent' Exhaustion by compact sets" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="10623" PostTypeId="1" AcceptedAnswerId="10632" CreationDate="2010-11-16T21:59:31.080" Score="1" ViewCount="51" Body="&lt;p&gt;I'm working on a program that displays a bunch of thumbnails on the bottom of the screen and I'm using the mouse's position to slide the images from side to side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of the screen is 0 on a number line. Every frame, the mouse's X position is either added or subtracted from the x position of each thumbnail. (So if the mouse is at -15, each thumbnail will move -15 pixels.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images should stop moving if they are too far from the center. &quot;Too far&quot; means that they are off the screen to a particular side, or if there are more images than fits on the screen, the image should stop sliding when there is enough room for the last image to appear on the opposite side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you could say that the x position of each thumbnail is a function of the mouse position, what number thumbnail it is and the the number of thumbnails. The screen width is 800 here (or -400 to 400)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the mathematical formula for the above scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT, a (perhaps) clearer explanation of what I want:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the images to scroll horizontally. If the mouse is on the right, they should move towards the left, and if the mouse is on the left, the images should scroll towards the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images should stop scrolling so that they never completely scroll offscreen. (If there are more images than the screen can show, allow images to scroll off just enough to show the other images.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of the scrolling depends on the distence of the mouse cursor from the center.&lt;/p&gt;&#xA;" OwnerUserId="2414" LastEditorUserId="2414" LastEditorDisplayName="" LastEditDate="2010-11-16T22:20:37.813" LastActivityDate="2010-11-16T22:52:50.920" Title="Formula for positioning each image in a row of images" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10626" PostTypeId="2" ParentId="10603" CreationDate="2010-11-16T22:15:36.697" Score="1" ViewCount="0" Body="&lt;p&gt;Taking Pontrjagin duals of the short exact sequence of discrete abelian groups &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \rightarrow \hat{\mathbb{Z}} \rightarrow \mathbb{Q}^{\vee} \rightarrow S^1 \rightarrow 0$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- here $\hat{\mathbb{Z}}$ is the profinite completion of $\mathbb{Z}$ -- so this classifies the one-dimensional &quot;unitary&quot; continuous representations of $\mathbb{Q}$.  I believe that the exact sequence above is split; of all places, this came up at a dinner party I attended last week.  (&lt;b&gt;Added&lt;/b&gt;: no, it is not split as a sequence of topological groups or even as a sequence of groups: see Matt E's comment below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B.: If you ask a number theorist (broadly construed) what the &quot;usual topology&quot; on $\mathbb{Q}$ is, she will say that it is the discrete topology.  This is the topology it inherits from the topology on the adele ring $A_{\mathbb{Q}}$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-17T09:25:13.387" LastActivityDate="2010-11-17T09:25:13.387" CommentCount="4" />
  <row Id="10627" PostTypeId="1" CreationDate="2010-11-16T22:28:08.083" Score="5" ViewCount="112" Body="&lt;p&gt;Functions from $\mathbb{C} \rightarrow \mathbb{C}$ are hard to visualize because of their 4-dimensional nature. One nice way of looking at them is by what's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Domain_coloring&quot; rel=&quot;nofollow&quot;&gt;domain coloring&lt;/a&gt;. An example from the wiki article is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we look at the graph of a real function ($\mathbb{R} \rightarrow \mathbb{R}$), we can get a feel for some of the function's properties: where its zeros are, if it's continuous, if it's differentiable, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what kinds of properties of complex functions can we &quot;see&quot; by looking at a domain coloring? In the example below, we can obviously see where the zeros are and where it blows up. I'm wondering, can we tell if a function is analytic? A rational function? Can we estimate an integral like we could by looking at a real graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/d/d3/Color_complex_plot.jpg/200px-Color_complex_plot.jpg&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z) = \frac{(z^2 − 1)(z − 2 − i)^2}{(z^2 + 2 + 2i)}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-11-16T23:26:31.397" Title="What can be gleaned from looking at a domain-colored graph of a complex function?" Tags="&lt;complex-analysis&gt;&lt;intuition&gt;&lt;visualization&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10628" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T22:29:12.933" Score="6" ViewCount="0" Body="&lt;p&gt;Below is an inductive proof excerpted from my post in this prior &lt;a href=&quot;http://math.stackexchange.com/questions/6852/maximum-order-of-integers-coprime-to-a-prime-p/6856#6856&quot;&gt;thread here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\ $ A finite abelian group $\rm\:G\:$ has an  lcm-closed  order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\ X,Y \in G\ \ \Rightarrow\ \ \exists\ Z \in G:\  o(Z) = lcm(o(X),o(Y))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \ $  By induction on  $\rm o(X)\ o(Y)\:.\ $ If it's $\:1\:$ then trivially $\rm\:Z = 1\:$. $\ $ Otherwise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;write  $\rm\ o(X)\ =\ AP,\: \ \ o(Y) = BP&amp;#39;,\ \ \ P&amp;#39;|P = p^m &amp;gt; 1\:,\ \ $  prime $\rm\: p\:$ coprime to $\rm\: A,B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then  $\rm\: o(X^P) = A,\ \ o(Y^{P&amp;#39;}) = B\:.\ $  By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so  $\rm\ \ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP&amp;#39;)\: =\: lcm(o(X),o(Y))\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ \ $ This is the element-wise form of what's known as &quot;Herstein's hardest problem&quot;, viz. 2.5.11, p. 41 in the first edition of Herstein's popular textbook &quot;Topics in Algebra&quot;. In the 2nd edition Herstein added  the following note (problem 2.5.26, p.48) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Don't be discouraged if you don't get this problem with what you know of group theory up to this stage. I don't know anybody, including myself, who has done it subject to the restriction of using material developed so far in the text.  But it is fun to try. I've had more correspondence about this problem than about any other point in the whole book.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-16T23:01:32.987" LastActivityDate="2010-11-16T23:01:32.987" CommentCount="6" />
  <row Id="10630" PostTypeId="2" ParentId="10514" CreationDate="2010-11-16T22:31:40.850" Score="2" ViewCount="0" Body="&lt;p&gt;I'll do it for when $f(x)$ has compact support; the general Schwartz function case is a modification of this. I use the fact that the Schwatrz functions are sequentially dense in the distributions, meaning that there is a sequence of Schwartz functions $g_n$ such that $\int g_n(x) \alpha(x)$ converges to $T(\alpha)$ as $n$ goes to infinity for any Schwartz function $\alpha(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace $T$ by $g_n(x)$, then your identity holds by Fubini's theorem. So one takes limits now as $n$ goes to infinity. The left hand side converges to $T \int f(x - \circ)\phi(\circ) d\circ$ immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the right-hand side, for each $x$ one has that $g_n(f(x - \circ) \phi(\circ))$ converges to $T(f(x - \circ) \phi(\circ))$. Since the $g_n(f(x - \circ) \phi(\circ))$ will be uniformly bounded (the construction of the $g_n$ is quite explicit and readily implies this) and since $f(x - \circ) \phi(\circ)$ is the zero distribution for $x$ outside a compact set, the dominated convergence theorem gives that the right hand side converges to $\int T (f(x - \circ) \phi(\circ)) dx$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-16T22:57:00.057" LastActivityDate="2010-11-16T22:57:00.057" />
  <row Id="10631" PostTypeId="2" ParentId="10616" CreationDate="2010-11-16T22:35:58.183" Score="5" ViewCount="0" Body="&lt;p&gt;Both responses you have received provide correct answers. In fact, you don't need to go so far as to obtain elements of relatively prime orders; all you need to worry about are primes with the property that they divide $nm$, and the largest power of $p$ that divides $n$ is equal to the largest power of $p$ that divides $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; Let $x$ and $y$ be commuting elements of a group $G$, and assume $x$ and $y$ are of finite orders $n$ and $m$, respectively. Suppose that for every prime $p$ that divides $nm$, the largest power of $p$ that divides $n$ is different from the largest power of $p$ that divides $m$. Then the order of $xy$ in $G$ is $\mathrm{lcm}(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $\ell=\mathrm{lcm}(n,m)$. It is easy to verify that $(xy)^{\ell}=1$, so we just need to show that $\ell$ is the smallest positive integer $k$ such that $(xy)^k=1$. Suppose that $(xy)^k = 1$, with $0\lt k\leq \ell$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(xy)^k = x^ky^k = 1$, then $x^k = y^{-k}$, and in particular $x^k$ has the same order as $y^k$. The order of $x^k$ is $n/\gcd(n,k)$, and the order of $y^k$ is $m/\gcd(n,k)$. Let $p$ be a prime that divides $\ell$, and let $a=\mathrm{ord}_p(n)$, $b=\mathrm{ord}_p(m)$, and $c=\mathrm{ord}_p(k)$. Assume $b\lt a$. If $c\lt a$, then $\mathrm{ord}_p(n/\gcd(n,k)) = a-c$, and $\mathrm{ord}_p(m/\gcd(m,k))=\max(0,b-c)\lt a-c$, which is impossible. Thus, $c=a$. A symmetric argument shows that if $a\lt b$, then $c=b$. That is, for all primes $p$ that divide $\ell$, we have $\mathrm{ord}_p(k) = \max(\mathrm{ord}_p(n),\mathrm{ord}_p(n)) = \mathrm{ord}_p(\ell)$. Hence $k=\ell$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now assume that $a$ is an element of order $n$ in an abelian group, and let $b$ be an element of order $m$. Let $p_1,\ldots,p_k$ be the primes that divide $nm$ and for which the $p$-order of $p_i$ in $m$ and in $n$ are equal. Then&#xA;$$\mathrm{lcm}(n,m) = \mathrm{lcm}\left(n, \frac{m}{p_1\cdots p_k}\right).$$&#xD;&#xA;Since $m/(p_1\cdots p_k)$ is the order of $b^{p_1\cdots p_k}$, then it follows from the lemma that $ab^{p_1\cdots p_k}$ has order $\mathrm{lcm}(n,m)$, as desired. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T02:51:10.800" LastActivityDate="2010-11-17T02:51:10.800" CommentCount="4" />
  <row Id="10632" PostTypeId="2" ParentId="10623" CreationDate="2010-11-16T22:52:50.920" Score="1" ViewCount="0" Body="&lt;p&gt;You're probably looking for something like a logit or an expit or even a sine function with the right parameters put in.&#xA;On a more practical level though, it might be easier to program in a piecewise-linear relationship b/w your mouse position and the strip movement based on where the strip is.&lt;/p&gt;&#xA;" OwnerUserId="3594" LastActivityDate="2010-11-16T22:52:50.920" />
  <row Id="10633" PostTypeId="2" ParentId="10622" CreationDate="2010-11-16T22:59:57.530" Score="5" ViewCount="0" Body="&lt;p&gt;Yes.  Let $\{L_i\}_{i=1}^\infty$, $\{K_i\}_{i=1}^\infty$ be two exhaustions.  It follows that $\bigcup_{i=1}^\infty K_i^\circ = U$, so for each $n$, $\{K_i^\circ\}_{i=1}^\infty$ is an open cover of $L_n$.  By compactness, it has a finite subcover $\{K_{i_1}^\circ, \dots, K_{i_l}^\circ\}$.  But if $m = \operatorname{max}(i_1, \dots, i_l)$, we have $K_{i_j}^\circ \subset K_{i_j + 1} \subset K_{m+1}$ for all $j=1,\dots, l$.  Thus $L_n \subset K_{m+1}$, and so $K_i$ is stronger than $L_i$.  By symmetry they are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-16T22:59:57.530" CommentCount="5" />
  <row Id="10634" PostTypeId="1" CreationDate="2010-11-16T23:14:18.827" Score="1" ViewCount="104" Body="&lt;p&gt;If we have a module M over a ring Z, and we consider the the canonical module homomorphism M→Q⊗M over Z, is it true that the kernel of this map is the torsion submodule of M? Why?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-11-17T01:23:59.550" Title="Torsion submodule equal to kernel of canonical map" Tags="&lt;modules&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10635" PostTypeId="2" ParentId="10627" CreationDate="2010-11-16T23:26:31.397" Score="2" ViewCount="0" Body="&lt;p&gt;Hans Lundmark's pages on &lt;a href=&quot;http://www.mai.liu.se/~halun/complex/domain_coloring-unicode.html&quot; rel=&quot;nofollow&quot;&gt;domain coloring&lt;/a&gt;, and &lt;a href=&quot;http://www.mai.liu.se/~halun/complex/elliptic/&quot; rel=&quot;nofollow&quot;&gt;his page on Elliptic functions&lt;/a&gt; is good places to start. (Picard's great theorems, Lucas's theorem, Weierstrass $\wp$ function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think we can see that a function is rational  -- just moving the roots in the function that you have displayed above by $\pi/100$ in some random direction would be sufficient to make it not a rational function, but the picture would more or less look the same.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-11-16T23:26:31.397" CommentCount="1" />
  <row Id="10636" PostTypeId="1" CreationDate="2010-11-16T23:48:38.097" Score="3" ViewCount="103" Body="&lt;p&gt;In the proof of Stone-Weierstrass theorem provided in Rudin's Principles of Mathematical Analysis, why do we only need to show that there exists a sequence of polynomials $P_n$ that converges uniformly to the continuous complex function on $[0,1]$?&lt;/p&gt;&#xA;" OwnerUserId="3498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-16T23:50:43.850" LastActivityDate="2010-11-17T00:00:42.733" Title="Stone-Weierstrass Theorem" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10637" PostTypeId="2" ParentId="10636" CreationDate="2010-11-17T00:00:42.733" Score="4" ViewCount="0" Body="&lt;p&gt;I think you are just asking why, &quot;We may assume, without loss of generality, that $[a, b] = [0,1]$.&quot; The reason is that if $f$ is continuous on $[a,b]$, then we can uniformly approximate $f(a+(b-a)x)$ on $[0,1]$ with a sequence of polynomials $(p_n)_{n=1}^\infty$, and then $(p_n((x-a)/(b-a)))_{n=1}^\infty$ will converge uniformly to $f$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-17T00:00:42.733" />
  <row Id="10639" PostTypeId="2" ParentId="10622" CreationDate="2010-11-17T00:17:23.330" Score="0" ViewCount="0" Body="&lt;p&gt;Ok, one version on my own, more technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be any compact set included in $U$. It remains to show $\exists i \in \mathbb N : V \subset K_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\forall i : V \cap K_i$ bounded again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, there is no index $i$ s.t. $V \subset K_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a nested sequence $V_i$ of non-empty bounded sets, $V = V_0, \forall i : V_i \cap K_i = \{\}$, .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_i \in V_i$ be a sequence of points, which exists by assumption. Then this sequence is bounded and w.l.o.g it is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence let $x_i \rightarrow x$. By assumption, $\exists k \in \mathbb N : \exists \epsilon &amp;gt; 0 : B_{\epsilon}(x) \subset {K_k}^{\circ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $x_i$ convergence, we skip a leading number of elements and w.l.o.g. $\forall i : x_i \in B_{\epsilon/2}(x) \subset B_{\epsilon}(x) \subset K_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reindexing the $V_i$ according to our choices of subsequences, we obtain a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, such there is an $i$ s.t. $V \subset K_i$. As $V$ has been any compact set, we are done.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-11-17T10:24:04.733" LastActivityDate="2010-11-17T10:24:04.733" />
  <row Id="10640" PostTypeId="1" CreationDate="2010-11-17T00:30:24.343" Score="0" ViewCount="86" Body="&lt;p&gt;So in algebraic geometry, the prime spectrum of the ring of polynomials over a field k is isomorphic to the set of k-homomorphisms from the ring of polynomials to k, and we do this by using the fact that the residue field (localise w.r.t. some prime ideal, then mod out the ideal generated by this in the localisation) is isomorphic to k, and hence any polynomial goes to an element of k, via the equivalence relations. This all makes sense as the ideal generated by a prime ideal of the polynomial ring in its localisation is clearly maximal, but I'm stuck on a very basic example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let C[x,y] be the polynomial ring over the complex numbers and (x - 1) be the prime ideal. What number does y go to? A C-algebra homomorphism is defined completely by its action on the generators of the polynomial ring (so just the n variables, and taking n = 2 we get x |--&gt; 1, but y |--&gt; ?), but I can't figure out how this works in this simple case. It should maybe be zero (otherwise how do you invert 1 + y?), but I don't know. Can anyone explain this?&lt;/p&gt;&#xA;" OwnerUserId="3597" LastActivityDate="2010-11-17T00:35:02.530" Title="The prime spectrum in multiple variables - a basic example/question" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10641" PostTypeId="2" ParentId="10634" CreationDate="2010-11-17T00:33:06.900" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: If $m\in M$ is a torsion element, it is killed by some integer, say $r$. Then, $1\otimes m=\frac{r}{r}\otimes m=\frac{1}{r}\otimes rm=\frac{1}{r}\otimes 0=0$ where the second equality follows from the bilinearity of the tensor product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working under the assumption that $Z$ is the ring of integers and $Q$ the ring of rationals. As Yuval points out, everything works the same way if you think of $Z$ as an arbitrary domain and $Q$ as its fraction field.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-17T01:23:59.550" LastActivityDate="2010-11-17T01:23:59.550" CommentCount="5" />
  <row Id="10642" PostTypeId="2" ParentId="10640" CreationDate="2010-11-17T00:35:02.530" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;em&gt;maximal&lt;/em&gt; ideals in a polynomial ring of an algebraically closed field $k$ is isomorphic to the set of $k$-homomorphisms from the polynomial ring into $k$. (This is the Nullstellensatz.) This is not true for arbitrary prime ideals, nor is it true for nonalgebraically closed fields. (Consider for instance the ideal $(X^2 +1) \subset \mathbb{R}[X]$; this morally wants to correspond to the points $\pm i$, but it doesn't, since those don't live in the reals. Nonetheless, it happens that the &lt;em&gt;scheme&lt;/em&gt; $\mathrm{Spec} \mathbb{R}[X]$ includes all prime ideals, so it includes a copy of $\pm i$ glued together, if you want to call it that.) In the terminology of algebraic geometry, one would say that the point you described is not &quot;rational over $\mathbb{C}$.&quot; Such points do exist, and make schemes of even finite type over an algebraically closed field different from varieties.&#xA;(Fortunately, it turns out that closed points alone determine many topological properties---the precise statement is that the closed points (or maximal ideals) are dense in any constructible sets.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-17T00:35:02.530" />
  <row Id="10643" PostTypeId="1" AcceptedAnswerId="10644" CreationDate="2010-11-17T00:37:51.050" Score="5" ViewCount="119" Body="&lt;p&gt;I am looking for the proof of the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $(a_n)$ be a sequence of real&#xA;  numbers.  Then there exists a function&#xA;  $f$ which is infinitely differentiable&#xA;  at 0, and  $$ \frac{d^nf}{dx^n}(0) =&#xD;&#xA; a_n, \ \ \text{for all } n.$$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would appreciate either a sketch of the proof or an online reference to it.  A general case is listed as &lt;a href=&quot;http://en.wikipedia.org/wiki/Borel%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Borel's lemma&lt;/a&gt; in Wikipedia, without proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part is when the power series $\sum_n \frac{a_n}{n!}x^n$ has a zero radius of convergence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks for the answers! &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-17T01:17:33.417" LastActivityDate="2010-11-17T01:17:33.417" Title="Every real sequence is the derivative sequence of some function" Tags="&lt;calculus&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="10644" PostTypeId="2" ParentId="10643" CreationDate="2010-11-17T00:44:30.800" Score="5" ViewCount="0" Body="&lt;p&gt;As you mentioned, this is a result of Borel. Proofs can be found in several sources, see for example the book on &quot;Complex variables&quot; by Berenstein and Gay (it may just be an exercise there, but at least there is a &quot;hint&quot;). The idea is to try a power series. Of course, this may not converge, so you use the kind of functions that come in constructions of smooth partitions of unity to help the convergence; the point is that you can ensure these functions decay to zero sufficiently fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; There is a proof in Wikipedia, actually. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-analytic_smooth_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-17T00:49:52.697" LastActivityDate="2010-11-17T00:49:52.697" />
  <row Id="10645" PostTypeId="2" ParentId="10643" CreationDate="2010-11-17T00:46:26.807" Score="4" ViewCount="0" Body="&lt;p&gt;This is a famous theorem of Borel. If you have Hormander's &quot;The analysis of partial differential operators I&quot; it's Theorem 1.2.6 there. He proves the result in any dimension. The basic idea is to write a sum of functions $\sum_n {a_n \over n!}\phi(m_n x)x^n$ where $\phi(x)$ is $1$ near $x = 0$ and is equal to zero outside of a small set containing $0$. If the $m_n$ are chosen carefully then the sum will have the desired properties. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-17T00:46:26.807" CommentCount="0" />
  <row Id="10646" PostTypeId="1" AcceptedAnswerId="10659" CreationDate="2010-11-17T00:59:01.017" Score="6" ViewCount="216" Body="&lt;p&gt;Hello, I've been trying to prove the following statement, but the converse has been giving me trouble. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A topological space $X$ is regular if and only if every closed subset $Z\subseteq X$ is the intersection of all open sets $U\subseteq X$ which contain it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the defintion of regular is that for any point $p\in X$ and any closed subset $Z\subset X$ not containing $p$, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. The space does not necessarily have to be a $T_1$ space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I showed the forward statement correctly. I suppose $X$ is regular, and clearly $Z\subseteq\cap\mathcal{U}$, where $\mathcal{U}$ is the family of all open sets containing $Z$. Then if I take $p\in\cap\mathcal{U}$, then $p$ is in every open subset containing $Z$. Towards a contradiction, I assume that $p$ is not in $Z$. Then since $X$ is regular, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. Thus $p\not\in F$, but $F$ is an open set containing $Z$, and thus must contain $p$, a contradiction. Thus $p\in Z$, and so $Z=\cap\mathcal{U}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The backwards step has stumped me for a while. I take any point $p$ and a closed set $Z$ which does not contain $p$. So $Z=\cap\mathcal{U}$, and since $p\not\in Z$, there exists an open set $F$ containing $Z$ which does not contain $p$. The only observation I've been able to make is that $F^c$ is a closed set containing $p$, and then $F^c=\cap\mathcal{W}$ where $\mathcal{W}$ is every open set containing $F^c$. Hence $p$ is in every open set containing $F^c$. Taking any open sets $E$ and $F$ such that $p\in E$ and $Z\subseteq F$, and I haven't been able to see a way that there exists two such open sets that are disjoint, to show the space is regular. Showing $F^c$ is an open set would work, but I'm not sure if that's even true. I was hoping someone could point out what I'm missing, thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-20T01:51:23.927" LastActivityDate="2010-11-20T01:51:23.927" Title="A space is regular if each closed set $Z$ is the intersection of all open sets containing $Z$?" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="10647" PostTypeId="1" CreationDate="2010-11-17T01:29:43.157" Score="1" ViewCount="131" Body="&lt;p&gt;I was reading Rudin's Principles of Mathematical Analysis, and I came across the definition 7.19, where it says that a sequence of functions $f_n(x)$ is pointwise bounded on E if there exists a finite-valued function $\phi$ defined on E such that &#xA;$$ |f_n(x)| &amp;lt; \phi(x) $$&#xD;&#xA;for x element of E, n = 1, 2 ,3 ... &#xA;While $f_n$ is uniformly bounded on E if there exists a number M s.t. &#xA;$|f_n(x)| &amp;lt; M$&#xD;&#xA;for x element of E, n = 1, 2 , 3 ... &#xA;But if we define the set U as the values of $\phi(x)$ from our first definition and define the sup of the set as R, then don't we get the second definition. Wouldn't that mean that pointwise bounded implies uniform bounded? &lt;/p&gt;&#xA;" OwnerUserId="2919" LastActivityDate="2010-11-17T06:21:45.373" Title="Why doesn't pointwise bounded imply uniform bounded?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10648" PostTypeId="1" AcceptedAnswerId="10650" CreationDate="2010-11-17T02:08:44.310" Score="27" ViewCount="873" Body="&lt;p&gt;When you draw a circle in a plane you can perfectly surround it with 6 other circles of the same radius. This works for any radius. What's the significance of 6? Why not some other number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an answer deeper than &quot;there are 6*60=360 degrees in a circle, so you can picture it&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-17T20:10:49.727" LastActivityDate="2010-11-18T05:59:44.103" Title="Why is a circle in a plane surrounded by 6 other circles." Tags="&lt;geometry&gt;&lt;differential-geometry&gt;&lt;discrete-geometry&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" />
  <row Id="10649" PostTypeId="1" AcceptedAnswerId="10652" CreationDate="2010-11-17T02:16:27.703" Score="1" ViewCount="88" Body="&lt;p&gt;I am given two random variables Y1 and Y2 and:&#xA;E(Y1) = 4&#xA;E(Y2) = -1&#xA;V(Y1) = 2&#xA;V(Y2) = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asked to find Cov(Y1,Y1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Cov(Y1,Y2) = E(Y1Y2)-E(Y1)E(Y2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if it is a typo, but it says Cov(Y1,Y1) NOT Cov(Y1,Y2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 2, but i'm not sure how to get E(Y1Y2) from the information given, or why it's asking for the Cov of the same variable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-11-17T02:42:43.850" Title="find cov(Y1,Y1) given expected value and variance" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;&lt;random-variable&gt;" AnswerCount="2" />
  <row Id="10650" PostTypeId="2" ParentId="10648" CreationDate="2010-11-17T02:23:25.467" Score="44" ViewCount="0" Body="&lt;p&gt;The short answer is &quot;because they don't work,&quot; but that's kind of a copout.  This is actually quite a deep question.  What you're referring to is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_packing&quot; rel=&quot;nofollow&quot;&gt;sphere packing&lt;/a&gt; in two dimensions, specifically the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kissing_number&quot; rel=&quot;nofollow&quot;&gt;kissing number&lt;/a&gt;, and sphere packing is actually quite a sophisticated and active field of mathematical research (in arbitrary dimensions).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one answer, which isn't complete but which tells you why $6$ is a meaningful number in two dimensions.  The packing you refer to is a special type of packing called a &lt;em&gt;lattice&lt;/em&gt; packing, which means it comes from an arrangement of regularly spaced points; in this case, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexagonal_lattice&quot; rel=&quot;nofollow&quot;&gt;hexagonal lattice&lt;/a&gt;.  The number $6$ appears here because the hexagonal lattice has $6$-fold symmetry.  So a natural question might be whether one can find lattices in two dimensions with, say $7$-fold or $8$-fold symmetry, since these might correspond to circle packings with more circles around a given circle.  (Intuitively, we expect more symmetric lattices to give rise to denser packings and to packings where each circle has more neighbors.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no: $6$-fold symmetry is the best you can do!  This is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallographic_restriction_theorem&quot; rel=&quot;nofollow&quot;&gt;crystallographic restriction theorem&lt;/a&gt;.  The generalization of the theorem to $n$ dimensions says this: it is possible for a lattice to have $d$-fold symmetry only if $\phi(d) \le n$, where $\phi$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_totient_function&quot; rel=&quot;nofollow&quot;&gt;Euler's totient function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization implies that you still cannot do better than $6$-fold symmetry in $3$ dimensions.  There are two natural lattice packings in $3$ dimensions, which &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexagonal_close_packing#Fcc_and_hcp_lattices&quot; rel=&quot;nofollow&quot;&gt;both occur&lt;/a&gt; in molecules and crystals in nature and which both have $6$-fold symmetry, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_conjecture&quot; rel=&quot;nofollow&quot;&gt;it turns out&lt;/a&gt; that these are the densest sphere packings in $3$ dimensions.  It also turns out that they give the correct kissing number in $3$ dimensions, which is $12$ (see the wiki article).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $4$ dimensions, the kissing number is $24$, and I believe the corresponding packing is a lattice packing coming from a lattice with $8$-fold symmetry, which is possible in $4$ dimensions.  In higher dimensions, only two other kissing numbers are known: $8$ dimensions, where the &lt;a href=&quot;http://en.wikipedia.org/wiki/E8_lattice&quot; rel=&quot;nofollow&quot;&gt;$E_8$ lattice&lt;/a&gt; gives kissing number $240$, and $24$ dimensions, where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Leech_lattice&quot; rel=&quot;nofollow&quot;&gt;Leech lattice&lt;/a&gt; gives kissing number $196560$!  These lattices are really mysterious objects and are related to a host of other mysterious objects in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great reference for this stuff, although it is a little dense, is Conway and Sloane's &lt;a href=&quot;http://www2.research.att.com/~njas/doc/splag.html&quot; rel=&quot;nofollow&quot;&gt;Sphere Packing, Lattices, and Groups&lt;/a&gt;.  &lt;strong&gt;Edit:&lt;/strong&gt;  And for a very accessible and engaging introduction to symmetry in the plane and in general, I highly recommend Conway, Burgiel, and Goodman-Strauss's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568812205&quot; rel=&quot;nofollow&quot;&gt;The Symmetries of Things&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T05:59:44.103" LastActivityDate="2010-11-18T05:59:44.103" CommentCount="6" />
  <row Id="10651" PostTypeId="2" ParentId="10646" CreationDate="2010-11-17T02:34:04.830" Score="5" ViewCount="0" Body="&lt;p&gt;Nate's example is a good one and actually solves the question as it is. But suppose you think that the Cromarty property, as Nate called it, is not good enough since 'almost all' spaces satisfy it. Then you try to substitute it for a stronger condition. You can require that every closed subset $Z \subseteq X$ is a countable intersection of open sets. That's exactly what a &lt;a href=&quot;http://en.wikipedia.org/wiki/G%CE%B4_space&quot; rel=&quot;nofollow&quot;&gt;$G_{\delta}$ space&lt;/a&gt; is. In this situation the example Nate gave still works, you just need to require that the space $X$ is countable. In fact, if $F$ is a closed subset of $X$ with the cofinite topology then $X-F$ is the countable union of the closed sets $\{x\}$, with $x \in X - F$. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \bigcap_{x \in X-F}  (X - \{x\}).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a countable set $X$ with the cofinite topology is a $G_{\delta}$ space. But then you conclude that your condition isn't strong enough. Now you try to impose some kind of separation. The question becomes: is there a Hausdorff $G_{\delta}$ space that isn't regular?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes. Let $K = \{ 1/n : n \in \mathbb{N} \}$ and consider the collection $\mathcal{B}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. It's easy to show that $\mathcal{B}$ is a basis for a topology on $\mathbb{R}$, which is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/K-topology&quot; rel=&quot;nofollow&quot;&gt;K-topology&lt;/a&gt;. When $\mathbb{R}$ is endowed with this topology it's denoted by $\mathbb{R}_K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The K-topology is finer than the standard topology on the real line. So it's a Hausdorff space (actually a &lt;a href=&quot;http://en.wikipedia.org/wiki/Urysohn_space&quot; rel=&quot;nofollow&quot;&gt;completely Hausdorff space&lt;/a&gt;). Also, $\mathbb{R}_K$ is second countable, since the colletion $\mathcal{B&amp;#39;}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$ and with $a,b \in \mathbb{Q}$ is a basis for the K-topology. $\mathbb{R}_K$ isn't regular, since the closed set and the point $0$ cannot be separated by disjoint open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's see why $\mathbb{R}_K$ is a $G_{\delta}$ space. It's easier to prove that it's an $F_{\sigma}$ space, that is, a space such that every open set can be written as a countable union of closed sets, which is equivalent to prove that it's a $G_{\delta}$ space. Since $\mathbb{R}_K$ is second countable, every open set can be written as a countable union of basis elements. If we prove that every basis element is an $F_{\sigma}$ set, then will follow that $\mathbb{R}_K$ is an $F_{\sigma}$ space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sets $(a,b)$ are $F_{\sigma}$, since $(a,b) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, b - \frac{1}{n}]$. Now consider $(a,b) - K$; we do not need to worry with the end point $b$, so suppose $b &amp;gt; 1$. If $a&amp;gt;0$, let $n_0$ be the largest natural such that $a &amp;lt; 1/n_0$. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (a,b)-K = (a, \frac{1}{n_0}) \cup \bigcup_{n=1}^{n_0-1} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the sets in the decomposition above are $F_{\sigma}$, thus $ (a,b)-K$ is $F_{\sigma}$. If $a=0$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (0,b)-K = \bigcup_{n=1}^{\infty} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $(0,b)-K$ is $F_{\sigma}$. If $a&amp;lt;0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (a,b)-K = (a,0] \cup ((0,b)-K) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, 0] \cup ((0,b)-K),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $(a,b)-K$ is an $F_{\sigma}$ and we conclude that $\mathbb{R}_K$ is an $F_{\sigma}$ space.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-17T23:53:44.933" LastActivityDate="2010-11-17T23:53:44.933" CommentCount="4" />
  <row Id="10652" PostTypeId="2" ParentId="10649" CreationDate="2010-11-17T02:35:22.097" Score="2" ViewCount="0" Body="&lt;p&gt;$cov(Y_1,Y_1)=var(Y_1)=2$ &#xA;This follows from the definition of covariance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cov(Y_1,Y_1)=E(Y_1\cdot Y_1)-E(Y_1)E(Y_1)=E(Y_1^2)-E(Y_1)^2$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-17T02:42:43.850" LastActivityDate="2010-11-17T02:42:43.850" CommentCount="2" />
  <row Id="10653" PostTypeId="2" ParentId="10649" CreationDate="2010-11-17T02:41:27.097" Score="2" ViewCount="0" Body="&lt;p&gt;Cov$(Y_1,Y_2) = \text{Cov}((Y_1-E[Y_1])(Y_2-E[Y_2])) = E[Y_1 Y_2]-E[Y_1]E[Y_2]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $Y_2 = Y_1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Cov$(Y_1,Y_2) = E[{Y_1}^2] -(E[Y_1])^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation should look familiar...the variance of $Y_1$&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-17T02:41:27.097" CommentCount="1" />
  <row Id="10654" PostTypeId="2" ParentId="7064" CreationDate="2010-11-17T02:47:50.947" Score="0" ViewCount="0" Body="&lt;p&gt;US notation  - which is taught in Australia&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-11-17T02:47:50.947" CommunityOwnedDate="2010-11-18T06:46:45.763" />
  <row Id="10655" PostTypeId="2" ParentId="10646" CreationDate="2010-11-17T03:40:42.593" Score="0" ViewCount="0" Body="&lt;p&gt;Isn't the trivial (indiscrete) topology a counterexample?  Perhaps there are more assumptions somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I had a different definition of &quot;regular&quot; in mind from the asker.  The question has been clarified, and the trivial topology is not a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-17T05:23:30.763" LastActivityDate="2010-11-17T05:23:30.763" CommentCount="5" />
  <row Id="10656" PostTypeId="2" ParentId="10579" CreationDate="2010-11-17T04:01:02.673" Score="0" ViewCount="0" Body="&lt;p&gt;Thank for the matrixcookbook, its a useful source.  I found the identity which confuses me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = (B + B^{T})\textbf{x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;via product rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = \frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}B\textbf{x} + \textbf{x}^{T}\frac{\partial B \textbf{x}}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I'm stuck because I'm not sure if $\frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}$ is identity $I$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this definition of the Jacobian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} \dfrac{\partial y_1}{\partial x_1} &amp;amp; \cdots &amp;amp; \dfrac{\partial y_1}{\partial x_n} \\ \vdots &amp;amp; \ddots &amp;amp; \vdots \\ \dfrac{\partial y_m}{\partial x_1} &amp;amp; \cdots &amp;amp; \dfrac{\partial y_m}{\partial x_n}  \end{bmatrix}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I write $\frac{\partial B \textbf{x}}{\partial \textbf{x}}$ component by component with the definition above, it becomes simply $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = B\textbf{x} + \textbf{x}^{T}B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not correct, as you stated the left side is a nx1 vector and the right side is a nx1 vector.  I'm not sure where my faulty assumption is.&lt;/p&gt;&#xA;" OwnerUserId="3572" LastActivityDate="2010-11-17T04:01:02.673" CommentCount="3" />
  <row Id="10657" PostTypeId="2" ParentId="10181" CreationDate="2010-11-17T04:15:22.113" Score="1" ViewCount="0" Body="&lt;p&gt;There is a good list on &lt;a href=&quot;http://hbpms.blogspot.com/2008/05/stage-4-foundations-and-discrete.html&quot; rel=&quot;nofollow&quot;&gt;this website.&lt;/a&gt; I can personally recommend the Walicki notes (free). It includes a good introduction to modal logic, so that's always nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the books, Geoffrey Hunter's Metalogic is very good and covers all bases.&lt;/p&gt;&#xA;" OwnerUserId="3564" LastActivityDate="2010-11-17T04:15:22.113" CommentCount="2" />
  <row Id="10658" PostTypeId="1" AcceptedAnswerId="10668" CreationDate="2010-11-17T04:26:05.377" Score="5" ViewCount="170" Body="&lt;p&gt;Suppose that there is a certain collected works of plays that is N symbols long in the following sense: a &quot;symbol&quot; is one of the 26 letters of the alphabet, a line break, period, space, or a colon; in other words there are 30 possible symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &quot;a monkey&quot; randomly &quot;types&quot; 1 of these 30 symbols at a rate of one per second, how long will it take M monkeys working at this rate, on average, for one of them to randomly write this specific N symbol long collected works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity let me state that I am assuming each monkey ceaselessly types random symbols at this rate, and unless a monkey immediately types the right things, the collected works will be preceded by gibberish.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-17T05:57:12.297" LastActivityDate="2010-11-17T06:25:56.077" Title="Monkeys and Typewriters" Tags="&lt;probability-theory&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10659" PostTypeId="2" ParentId="10646" CreationDate="2010-11-17T05:46:23.483" Score="5" ViewCount="0" Body="&lt;p&gt;Ok, let's try another counterexample.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, say a space with your property (any closed set is the intersection of its neighborhoods) is &quot;&lt;em&gt;Cromarty&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be an infinite set with the cofinite topology (the closed sets are all the finite sets and $X$ itself).  If $Z \subset X$ is closed (hence finite), let $\mathcal{U}$ be the collection of all open sets containing $Z$.  For any $x \notin Z$, $\{x\}^c$ is open (because it is cofinite) and contains $Z$, so $\{x\}^c \in \mathcal{U}$, and thus $x \notin \bigcap \mathcal{U}$; hence $Z^C \subset \bigcap \mathcal{U}$, so $\bigcap \mathcal{U}=Z$.  So $X$ is Cromarty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $X$ is not regular, since every pair of nonempty open sets has nonempty (in fact, infinite) intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To amplify, it looks like the Cromarty property is equivalent to being &lt;a href=&quot;http://en.wikipedia.org/wiki/R0_space&quot; rel=&quot;nofollow&quot;&gt;$R_0$&lt;/a&gt;, which means: if $x$ has a neighborhood not containing $y$, then $y$ has a neighborhood not containing $x$.  This is of course strictly weaker than being regular (as the cofinite topology shows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $X$ is $R_0$, and $Z$ is closed.  Suppose $x \notin Z$, $y \in Z$.  Then $Z^c$ is a neighborhood of $x$ not containing $y$, so $y$ has a neighborhood $U_y$ that does not contain $x$.  Now $U = \bigcup_{y \in Z} U_y$ is a neighborhood of $Z$ that does not contain $x$.  As above, it follows that $Z$ is the intersection of its neighborhoods, so $X$ is Cromarty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $X$ is Cromarty.  Suppose a point $x$ has a neighborhood $U$ not containing $y$.  $U^c$ is closed, hence the intersection of its neighborhoods, so $U^c$ has a neighborhood $V$ not containing $x$.  But $V$ is also a neighborhood of $y$, so $X$ is $R_0$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-17T13:58:16.870" LastActivityDate="2010-11-17T13:58:16.870" CommentCount="5" />
  <row Id="10660" PostTypeId="2" ParentId="10658" CreationDate="2010-11-17T05:48:08.830" Score="8" ViewCount="0" Body="&lt;p&gt;Unfortunately, an exact answer will depend on the specific sequence of $N$ symbols.  To see why, take a simple example in which you only have two possible symbols, A and B, with $N = 2$, and only one monkey.  Compare the expected times until the monkey types the sequence AA vs. the sequence BA.  Unless the monkey types AA at the very beginning (with probability $\frac{1}{4}$), the first time AA appears the BA sequence &lt;em&gt;must&lt;/em&gt; have already occurred.  So the latter sequence will have the shorter expected time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of these sorts of counterintuitive results about the likelihood of seeing a certain sequence before another certain sequence and about the expected time until a sequence is first seen.  For more information, see MathWorld's entry on &lt;a href=&quot;http://mathworld.wolfram.com/CoinTossing.html&quot; rel=&quot;nofollow&quot;&gt;Coin Tossing&lt;/a&gt;, or, for even more information, the article &quot;&lt;a href=&quot;http://bact.mathcircles.org/files/Summer2010/PenneyAnte.pdf&quot; rel=&quot;nofollow&quot;&gt;Penney Ante&lt;/a&gt;&quot; that appeared in the &lt;em&gt;UMAP Journal&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $N$ is large and the symbols are assumed to be evenly distributed in the target sequence, maybe you can avoid the problems with the two-symbol and unevenly distributed example I gave here.  Or, if you force each monkey to stop after typing exactly $N$ symbols, check to see if they have typed the target sequence, and then have them start over again from scratch if they have not, then you can definitely avoid these problems.  (For more on the latter, see this article &quot;&lt;a href=&quot;http://plus.maths.org/content/infinite-monkey-business&quot; rel=&quot;nofollow&quot;&gt;Infinite Monkey Business&lt;/a&gt;&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-17T05:48:08.830" />
  <row Id="10661" PostTypeId="1" AcceptedAnswerId="10667" CreationDate="2010-11-17T05:56:57.727" Score="3" ViewCount="84" Body="&lt;p&gt;How to find the value of&#xA;$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;manually ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T05:59:24.433" LastActivityDate="2010-11-17T06:24:06.243" Title="Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10662" PostTypeId="2" ParentId="9045" CreationDate="2010-11-17T06:10:04.373" Score="0" ViewCount="0" Body="&lt;p&gt;I suppose that your definition of unitary is that $A^*A=AA^*=I$. &lt;br/&gt;&#xA;Let $A$ be unitary and $x\in V$. Then $$\|Ax\|^2=\langle Ax,Ax\rangle=\langle x, A^*Ax\rangle =\langle x,Ix\rangle=\langle x, x\rangle=\|x\|^2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose $\|Ax\|=\|x\|$ for all $x\in V$. Then for all $x,y\in V$, one has&#xA;$$\|A(x-y)\|^2=\|Ax-Ay\|^2=\|Ax\|^2-2\langle Ax,Ay\rangle+\|Ay\|^2=\|x\|^2-2\langle Ax,Ay\rangle+\|y\|^2$$ and $$ \|x-y\|^2=\|x\|^2-2\langle x,y\rangle+\|y\|^2.$$ &#xA;Equating $\|A(x-y)\|^2$ and $\|x-y\|^2$ yields $\langle Ax,Ay\rangle=\langle x,y\rangle$. Thus, for all $x,y$,&#xA;$$\langle x,(A^*A-I)y\rangle=\langle x,A^*Ay\rangle-\langle x,y\rangle=\langle Ax,Ay\rangle-\langle x,y\rangle=0$$&#xD;&#xA;Then it follows from your cited previous question that $A^*A-I=0$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-17T06:17:16.627" LastActivityDate="2010-11-17T06:17:16.627" />
  <row Id="10663" PostTypeId="1" AcceptedAnswerId="10671" CreationDate="2010-11-17T06:14:31.963" Score="5" ViewCount="119" Body="&lt;p&gt;In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle#By_internal_angles&quot; rel=&quot;nofollow&quot;&gt;type&lt;/a&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-17T07:47:44.397" Title="What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $ ?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10664" PostTypeId="1" AcceptedAnswerId="10669" CreationDate="2010-11-17T06:18:46.987" Score="1" ViewCount="72" Body="&lt;p&gt;I tried solving this, and I am pretty sure I am integrating this correctly, however, my solution manual shows -1 in the equation when doing this and I do not know why.  The answer in the solution manual is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: Find the corresponding distribution function and use it to determine the probability that a random variable having the distribution function will take on a value between 0.4 and 1.6.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = x for 0 &amp;lt; x &amp;lt; 1&#xA;       2-x for 1 &amp;lt;= x &amp;lt; 2&#xA;       0 elsewhere&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so for F(0.4 &amp;lt; x &amp;lt; 1.6) I did after integrating:&lt;br&gt;&#xA;2(1.6) - [(1.6)^2 / 2] - [(0.4)^2 / 2] = 1.84&lt;br&gt;&#xA;however the correct answer is 0.84.  The solution manual has a -1 in their equation, but I do not know how they got it.&lt;/p&gt;&#xA;" OwnerUserId="3601" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-17T06:42:12.680" LastActivityDate="2010-11-17T06:42:12.680" Title="Calculating probability with a piecewise density function" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;&lt;integral&gt;&lt;random-variable&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10665" PostTypeId="2" ParentId="10647" CreationDate="2010-11-17T06:21:45.373" Score="1" ViewCount="0" Body="&lt;p&gt;The short answer is: there need not be a real number that is the supremum of the values of $\phi(x)$. You may have $\sup\{\phi(x)\mid x \in E\} = \infty$. If that is the case, you're out of luck.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-17T06:21:45.373" />
  <row Id="10667" PostTypeId="2" ParentId="10661" CreationDate="2010-11-17T06:24:06.243" Score="8" ViewCount="0" Body="&lt;p&gt;$$\displaystyle \sin(60^{\circ}-20^{\circ}) = \sin 40^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ}  - 2 \sin 20^{\circ} \cos 20^{\circ}  = \frac{1}{2} \sin 20^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply by $\displaystyle \frac{2}{\sin 20^{\circ}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ} = 1$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-17T06:24:06.243" CommentCount="5" />
  <row Id="10668" PostTypeId="2" ParentId="10658" CreationDate="2010-11-17T06:25:56.077" Score="4" ViewCount="0" Body="&lt;p&gt;We will estimate the probability for a &quot;generic&quot; string. The number of occurrences of the string in any given monkey's output is roughly distributed Poisson with $\lambda = 30^{-N}$. The time until the first event happens is thus roughly distributed exponentially with rate $\lambda = 30^{-N}$. The minimum of $M$ such processes is also distributed exponentially with rate $\lambda = M/30^N$. Thus the expected time is roughly $30^N/M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same estimate can be obtained if we calculate the expected number of appearances. The expected number of appearances in any given monkey's stream for the first $t+M-1$ characters is $t/30^N$. For $M$ monkeys, it is $tM/30^N$. This is $1$ for $t = 30^N/M$, and this gives a rough estimate for the actual expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, assuming that the string &quot;doesn't overlap with itself&quot;, we can get an exact expression for the expectation (depending only on $N$ and $M$) using Theorem 1.1 in &quot;String overlaps, pattern matching, and nontransitive games&quot; (Guibas and Odlyzko '81), which gives a generating function for the probability that a given monkey is not done after $t$ steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper also gives an expression for &quot;non-generic&quot; strings and for multiple strings, but the collected works are not going to overlap themselves; even if they do, it will probably have only a slight effect on the probabilities.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T06:25:56.077" />
  <row Id="10669" PostTypeId="2" ParentId="10664" CreationDate="2010-11-17T06:35:00.050" Score="2" ViewCount="0" Body="&lt;p&gt;Since $f(t)$ is defined piecewise, you have to be careful with the integral.  This is the source of your mistake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0 &amp;lt; x &amp;lt; 1$, &#xA;$$F(x) = \int_0^x f(t) dt = \int_0^x t dt = \left.\frac{1}{2}t^2\right|_0^x = \frac{1}{2}x^2,$$&#xD;&#xA;which you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $1 \leq x &amp;lt; 2$, then you have to break the integral that yields $F(x)$ up into pieces.  This is &#xA;$$F(x) = \int_0^x f(t) dt = \int_0^1 t dt + \int_1^x (2-t) dt = \left.\frac{1}{2}t^2\right|_0^1 + \left[2t - \frac{1}{2}t^2\right]_1^x $$&#xD;&#xA;$$= \frac{1}{2} + 2x - \frac{1}{2}x^2 - 2 + \frac{1}{2} = 2x - \frac{1}{2}x^2 - 1.$$&#xD;&#xA;Here's where the $-1$ comes in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a common mistake.  If it makes you feel any better, my probability students trip up over this all the time. :)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-17T06:35:00.050" CommentCount="1" />
  <row Id="10670" PostTypeId="2" ParentId="10664" CreationDate="2010-11-17T06:35:26.027" Score="1" ViewCount="0" Body="&lt;p&gt;First, probabilities are from $0$ to $1$, so you're certainly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you calculated the cdf incorrectly for the interval $1 \leq x &amp;lt; 2$. It should be $$F(x) = \int_0^1 t \, \mathrm{d}t + \int_1^x 2-t \, \mathrm{d}t.$$ You forgot the first part, and integrated the second part from $0$ instead of from $1$. Since $$\int_0^1 t \, \mathrm{d}t = 1/2$$ whereas $$\int_0^1 2-t \, \mathrm{d}t = 3/2,$$ you counter $1/2$ too little and $3/2$ too much, for a total gain of $+1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T06:35:26.027" />
  <row Id="10671" PostTypeId="2" ParentId="10663" CreationDate="2010-11-17T06:36:57.193" Score="7" ViewCount="0" Body="&lt;p&gt;This essentially means $\sin^2(A)+\sin^2(B)+\sin^2(C)=2$. (This follows from $\sin$ rule.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace $C = \pi - (A+B)$ to get $\sin^2(A+B) = \cos^2(A) + \cos^2(B)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand $\sin(A+B)$ and do the manipulations to get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\cos^2(A)\cos^2(B) = 2\sin(A)\sin(B)\cos(A)\cos(B)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means $\cos(A) = 0$ or $\cos(B) = 0$ or $\cos(A)\cos(B) = \sin(A)\sin(B) \Rightarrow \cos(A+B) = 0 \Rightarrow \cos(C) = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence either $A = \pi/2$ or $B = \pi/2$ or $C = \pi/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the triangle is a right-angled triangle.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-17T06:36:57.193" />
  <row Id="10672" PostTypeId="1" CreationDate="2010-11-17T06:54:15.730" Score="3" ViewCount="195" Body="&lt;p&gt;This is from the book Principles of Mathematical Analysis by Rudin,  number 4 of chapter 7. It says consider &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = \sum\limits_{n=1}^{\infty}{ 1/(1+ n^2 x)  } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks: &#xA;(1) For what values of x does the series converge absolutely. We got that the series converges when x $\not=$ 0 &amp;amp; x $\not= -1/k^2 $ when k is an integer since, there is a discontinuity when n reaches the value $k^2$ .&lt;br&gt;&#xA;However we don't understand how to do any of the following questions asked. Any hints would be greatly appreciated. We were told that this problem was suppose to be fairly hard for its position in the problem set (ie. 4th question in the Rudin book).&lt;br&gt;&#xA;(2) What interval does it converge uniformly?&lt;br&gt;&#xA;(3) On what intervals does it fail to converge uniformly ?&lt;br&gt;&#xA;(4) Is f continuous wherever the series converges?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;(5) Is f bounded?&lt;/strike&gt;&lt;/p&gt;  &#xA;" OwnerUserId="2919" LastEditorUserId="2919" LastEditorDisplayName="" LastEditDate="2010-11-17T22:08:00.610" LastActivityDate="2010-11-17T22:08:00.610" Title="Convergence of a particular Series Uniformly (Homework)" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10673" PostTypeId="2" ParentId="7064" CreationDate="2010-11-17T06:56:49.590" Score="0" ViewCount="0" Body="&lt;p&gt;In Israel I was taught (in the 90's) the US notation, with the divisor on the right (might have something to do with the fact that we write Hebrew right to left).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T06:56:49.590" CommunityOwnedDate="2010-11-18T06:46:45.763" />
  <row Id="10674" PostTypeId="2" ParentId="10603" CreationDate="2010-11-17T07:05:10.070" Score="10" ViewCount="0" Body="&lt;p&gt;One way to think of $\mathbb Q$ is the direct limit over positive integers $n$ of $\frac{1}{n} \mathbb Z$.  Thus giving a character of $\mathbb Q$ is the same&#xA;as giving an element in the projective limit of the character groups of&#xA;$\frac{1}{n}\mathbb Z$.  In particular, if we restrict to unitary characters,&#xA;we obtain the $\mathbb Q^{\vee}$ is the projective limit of circle groups $S^1$ under the $n$th power maps.  This object is (I think) called a solenoid; to number theorists it is better known as the adele class group $\mathbb A/\mathbb Q$.  (Here and throughout I am using the discrete topology; if one instead considers the&#xA;induced topology from $\mathbb R$, then, as Robin explains, one just gets &#xA;characters of $\mathbb R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact sequence $0 \to \hat{\mathbb Z} \to \mathbb Q^{\vee} \to S^1 \to 0$&#xD;&#xA;in Pete's answer arises from the map taking the solenoid to the base $S^1$; the fibres of this map are copies of $\hat{\mathbb Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we wanted not necessarily unitary characters, we would instead get&#xA;the projective limit of copies of $\mathbb C^{\times}$ under the $n$th power maps.  Since $\mathbb C^{\times} = \mathbb R_{&amp;gt; 0} \times S^1$, and since&#xA;$\mathbb R_{&amp;gt; 0}$ is uniquely divisible, this projective limit is simply &#xA;$\mathbb R_{&amp;gt; 0}$ times the solenoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a slightly tangential note, let me remark that&#xA;the relationship with the adeles is important (e.g. it is the first step in Tate's thesis):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the adeles are the (restricted) product of $\mathbb R$ and each $\mathbb Q_p$, and since these are all self-dual, it is easy to see that $\mathbb A$ is self-dual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One then has the exact sequence&#xA;$$0 \to \mathbb Q \to \mathbb A \to \mathbb A/\mathbb Q \to 0$$&#xD;&#xA;which is again self-dual (the duality swaps $\mathbb Q$ and the solenoid&#xA;$\mathbb A/\mathbb Q$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should compare this with the exact sequence&#xA;$$0 \to \mathbb Z \to \mathbb R \to \mathbb R/\mathbb Z = S^1 \to 0.$$&#xD;&#xA;This is again self-dual ($\mathbb R$ is self-dual,&#xA;and duality swaps the integers and the circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings out the important intuition that the adeles are to $\mathbb Q$ as&#xA;$\mathbb R$ is to $\mathbb Z$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-17T07:05:10.070" CommentCount="1" />
  <row Id="10675" PostTypeId="2" ParentId="10663" CreationDate="2010-11-17T07:14:26.483" Score="2" ViewCount="0" Body="&lt;p&gt;In general, in $\triangle{ABC}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sin^2 A + \sin^2 B + \sin^2 C \gt 2$ then $\triangle{ABC}$ is acute angled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sin^2 A + \sin^2 B + \sin^2 C  = 2$ then $\triangle{ABC}$ is right angled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sin^2 A + \sin^2 B + \sin^2 C \lt 2$ then $\triangle{ABC}$ is obtuse angled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $A \le B \lt \pi/2$ and $ A \le B \le C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if $k = \sin^2 A + \sin^2 B + \sin^2 C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3-2k = \cos 2A + \cos 2B + \cos (2A+2B)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3-2k = 2\cos(A+B)\cos(A-B) + 2\cos^2(A+B) -1 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4-2k = 4\cos(A+B)\cos A\cos B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $k &amp;gt; 2$, then $\cos(A+B) \lt 0$ hence acute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = 2$, then $\cos(A+B) = 0$ hence right triangled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k &amp;lt; 2$, then $\cos(A+B) \gt 0$, hence obtuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we can go further and show that the maximum possible value of $k$ is $k = \frac{9}{4}$ which corresponds to $\triangle{ABC}$ being equilateral, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4\sin^2 A + 4\sin^2 B + 4\sin^2 C = 9  + \delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2 - 2\cos2A) + (2-2\cos 2B) + 4(1- \cos^2 (A+B)) = 9 + \delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&#xA;$1 + 2\cos2A + 2\cos 2B + 4\cos^2(A+B) = -\delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin^2(A-B) + \cos^2(A-B) + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin^2(A-B) + (\cos (A-B) + 2\cos(A+B))^2 = -\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\delta \le 0$ and so $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $\delta = 0$ gives us $\sin(A-B) = 0$ and $\cos(A+B) = \frac{-1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $A=B=C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the max value of $\sin^2 A + \sin^2 B + \sin^2 C$ is $\frac{9}{4}$ and is achieved when $A=B=C$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-17T07:47:44.397" LastActivityDate="2010-11-17T07:47:44.397" />
  <row Id="10676" PostTypeId="2" ParentId="10648" CreationDate="2010-11-17T07:25:20.713" Score="14" ViewCount="0" Body="&lt;p&gt;The question is equivalent to asking why 6 equilateral triangles fit together exactly around a point, with no additional room left over.  The answer is &quot;because the Euclidean plane is flat&quot;, a condition equivalent to triangles having angle sum of 180 degrees (half the angle around a point), so that each vertex of a symmetrical triangle has 1/3 of half of a full rotation = 1/6.  That the six-circle arrangement exists for any radius is also a special feature of Euclidean geometry: scale invariance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For flat surfaces such as a cylinder (rolled up plane) or flat torus (as in the Asteroids video game) the perfect 6-circle configurations exist only for small enough radius of the circles.  These geometries are, in small regions, the same as the Euclidean plane but differ &quot;globally&quot;, e.g. there is a maximum distance between points of the torus.  So the magic number of 6 is really about local flatness (absence of curvature) and having this configuration for circles of all size is a global question about the space in which the circles live.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hyperbolic geometry there are tesselations of the plane by equilateral triangles with angle $180/n$ at each vertex, for any $n \geq 7$.  In the picture of the $n=7$ triangular tesselation at &lt;a href=&quot;http://www.plunk.org/~hatch/HyperbolicTesselations/3_7_trunc0_512x512.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.plunk.org/~hatch/HyperbolicTesselations/3_7_trunc0_512x512.gif&lt;/a&gt; (triangles in white with the dual tesselation by heptagons shown in blue) if at each triangle vertex you draw a circle inscribed in its heptagon, there will be 7 circles exactly surrounding each circle, with all circles of the same radius.  The same type of configuration exists for any $n$ and suitable radius of the circles.  So the flatness condition is necessary; the theorem is false in negatively curved two-dimensional geometry.  Six is also not the correct number for spherical geomety, and both spherical and hyperbolic geometry lack a radius-independent &quot;number of circles that fit around one circle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the geometry of surfaces, having 5-or-fewer as the local number of circles that can be fit around a single circle characterizes positive curvature, as in spherical geometry.  Having more-than-6 fit, or extra room when surrounding one circle by six, is a characterization of negative curvature, as in hyperbolic geometry. This is a statement about the local geometry of general 2-dimensional surfaces, and does not assume the surface has the same amount of curvature at all points, as would be the case for the spherical and hyperbolic analogues of Euclidean plane geometry, where there is a homogeneity assumption that &quot;geometry is the same at all points&quot;.  Having exactly 6 circles fit perfectly is a characterization of locally Euclidean (that is, zero curvature) geometry.  If you don't know what curvature is in a technical sense, for purposes of this discussion it is (for surfaces) a number that can be associated to any point on the surface, and curvature being zero in a region of surface is equivalent to the ability to make a distance-preserving planar map of that region.  Impossibility of doing this for surfaces of nonzero curvature, such as the sphere which is positively curved, was Gauss' &lt;em&gt;Theorema Egregium&lt;/em&gt; which ruled out perfect flat cartography of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flatness plays the same role in the higher dimensional, sphere-packing interpretation of the question that Qiaochu suggested.  The necessary packings exist only in a limited set of dimensions.  The reason is that exactly surrounding a sphere of radius $r$ by $k$ spheres of that radius means more than a kissing number of $k$, the optimal kissing configuration should also be &lt;em&gt;rigid&lt;/em&gt; (tight enough that the spheres cannot be moved).  Rigidity is false in three dimensions and is believed to be false in most other dimensions but in those cases where it is known or suspected to be true (i.e., dimensions $d = 2, 8, 24$ and possibly a few others) the existence of a configuration with the same set of equalities between various interpoint distances as in the optimal kissing arrangement (in flat Euclidean space $\mathbb{R}^d$)should force a homogeneous $d$-dimensional geometry to be flat.  This is because deforming the curvature of the space in the positive direction would reduce, and negative curvature would increase, the freedom to position such spheres around a central sphere.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-17T21:12:13.910" LastActivityDate="2010-11-17T21:12:13.910" />
  <row Id="10677" PostTypeId="1" AcceptedAnswerId="10684" CreationDate="2010-11-17T08:18:15.290" Score="1" ViewCount="79" Body="&lt;p&gt;Hello everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question about a transformation of a matrix. Lets say we have the following matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;M =&#xD;&#xA;\left[ {\begin{array}{cc}&#xD;&#xA; 4 &amp;amp; 3  \\&#xD;&#xA; 4 &amp;amp; 3  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I want to transform all numbers in the matrix with the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = (x - 4)^2 + 2x -4$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the correct notation for this?&lt;/p&gt;&#xA;" OwnerUserId="3605" LastActivityDate="2010-11-17T10:29:37.393" Title="Matrix transformation notation" Tags="&lt;matrices&gt;&lt;notation&gt;&lt;transformation&gt;" AnswerCount="2" />
  <row Id="10678" PostTypeId="2" ParentId="10672" CreationDate="2010-11-17T08:41:02.677" Score="3" ViewCount="0" Body="&lt;p&gt;For strictly positive $x$ you can already see that each term of the series, except the first one, is bounded by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1+n^2 x} &amp;lt; \frac{1}{(n-1)^2 x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series formed by the bounds converges, so by the Weierstrass M-test, you have uniform convergence for $x&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x&amp;lt;-1$ you can make the same story:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\frac{1}{1+n^2 x}\right| &amp;lt; \frac{1}{(n+1)^2 |x|}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series formed by the bounds converges, so by the Weierstrass M-test, you have uniform convergence for $x&amp;lt;-1$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part is of course what happens on the open interval $]-1,0[$, you have already excluded the endpoints as well as all the negative inverses of squares of integers within that interval. &lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-17T08:41:02.677" CommentCount="2" />
  <row Id="10680" PostTypeId="2" ParentId="10672" CreationDate="2010-11-17T09:02:06.350" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have seen a number of discontinuities at certain points.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you use the M-test between points of discontinuities? &#xA;1.a If you can then it is good.&#xA;1.b If you can not do that - why? Can you do estimates if you remove a part around these points?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you show that the convergence is NOT uniform near these points? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-17T09:02:06.350" />
  <row Id="10681" PostTypeId="2" ParentId="10677" CreationDate="2010-11-17T09:06:08.823" Score="1" ViewCount="0" Body="&lt;p&gt;Could use the &quot;Hadamard Product&quot;... I don't know whether there is a consistently accepted notation for such a product, however a few examples &lt;a href=&quot;http://planetmath.org/encyclopedia/HadamardProduct.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and on Wikipedia seem to use an open circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense, your expression would take the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(M-4O)\circ(M-4O)+2M-4O$ where $O= \left(  \begin{array} {cc} 1 &amp;amp;1 \\ 1&amp;amp;1 \end{array}  \right) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In MATLAB, use .* for element-wise multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably other possibilities.&lt;/p&gt;&#xA;" OwnerUserId="3564" LastActivityDate="2010-11-17T09:06:08.823" />
  <row Id="10682" PostTypeId="2" ParentId="10469" CreationDate="2010-11-17T09:07:44.160" Score="1" ViewCount="0" Body="&lt;p&gt;I realized that your question wasn't exactly about the Steenrod axioms themselves, but about the definition of the coboundary operator involving suspension.  In reduced homology, the boundary operator $\partial$ for the pair $(X,A)$ (where the inclusion $i:A\rightarrow X$ is a cofibration) can be defined to come from the &quot;topological boundary map&quot; $\partial^!$ followed by the inverse of the suspension isomorphism.  The former is itself a composition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial^! = \pi \circ \psi^{-1}: X/A \rightarrow Ci \rightarrow \Sigma A, $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Ci$ is the mapping cone of $i$, $\psi^{-1}$ is a homotopy inverse of the quotient $\psi: Ci \rightarrow Ci/CA=X/A$, and $\pi: Ci \rightarrow Ci/X=\Sigma A$.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial = (\Sigma_*)^{-1} \circ \partial^!_* : \tilde{H}_q(X/A) \rightarrow \tilde{H}_q(\Sigma A) \rightarrow \tilde{H}_{q-1}(A) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is true for any reduced homology theory.  See May's &quot;Concise Course&quot; for details, pp. 106-7.  I'm pretty sure that the situation for cohomology is very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: In this formulation, the coboundary operator is the composition of a map induced from an actual map on spaces and the (inverse of the (?)) suspension isomorphism.  Steenrod squares commute with both of these, so they commute with the coboundary operator.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-11-17T09:07:44.160" CommentCount="3" />
  <row Id="10683" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T09:28:27.033" Score="0" ViewCount="0" Body="&lt;p&gt;The most interesting applications requiring the services of Quadratic Reciprocity  are, in my opinion, related to the field of Mathematical Cryptography!!  As mentioned above, Quadratic Reciprocity is a particularly useful tool when you want to see if a number is a square mod p (p prime.)  One cryptosystem in particular that requires the help of Quadratic Reciprocity is the Goldwasser-Micali public key cryptosystem; this is the case because it poses the following question based on the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let p, q be (secret) primes and let N=pq be given.  For a given integer a, determine whether a is a square mod N, ie, determine if there exists an integer u satisfying u^2 = a mod N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it is especially easy for Bob, the receiver of the message who knows how to factor N, to solve this problem because a is a square mod pq iff (a/p)=1 and (a/q)=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all from the Springer book entitled &lt;em&gt;An Introduction to Mathematical Cryptography&lt;/em&gt; by Hoffstein, Pipher, and Silverman.  This is such a great book (it was the course text for a course on Cryptography I took last spring) and Id recommend it to anyone who is interested in learning more about the practical and totally was AWESOME applications of Algebraic Number Theory.  There is so much more to be said about Quadratic Reciprocity (mentioned in the end of the 3rd chapter in this book) so be sure to check it out!&lt;/p&gt;&#xA;" OwnerUserId="3608" LastActivityDate="2010-11-17T09:28:27.033" />
  <row Id="10684" PostTypeId="2" ParentId="10677" CreationDate="2010-11-17T10:29:37.393" Score="3" ViewCount="0" Body="&lt;p&gt;Maybe I don't understand your question, but: what's wrong with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;f(4) &amp;amp; f(3) \\\&#xD;&#xA;f(4) &amp;amp; f(3)&#xD;&#xA;\end{pmatrix}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if you have a matrix $(a^i_j)$, you could write the resulting matrix as $(f(a^i_j))$.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-17T10:29:37.393" CommentCount="1" />
  <row Id="10685" PostTypeId="1" CreationDate="2010-11-17T10:33:29.033" Score="3" ViewCount="90" Body="&lt;p&gt;At Wikipedia I found that &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Asymmetric_graph&quot; rel=&quot;nofollow&quot;&gt;according to a strengthened version of Frucht's theorem, there are infinitely many asymmetric cubic graphs&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there infinitely many asymmetric&#xA;  &lt;em&gt;planar&lt;/em&gt; cubic graphs, too?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If so, does it follow that there is an &lt;em&gt;infinite&lt;/em&gt; asymmetric planar cubic graph?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If so, how could this graph be characterized?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an homogeneous and isotropic (in the large) regular graph that could &quot;mimick&quot; a discretized plane (without distinguished directions as in a grid). So an infinite asymmetric 4-regular graph would be even better (reflecting dimension 2).&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-17T11:15:29.260" LastActivityDate="2010-11-18T05:12:56.353" Title="Asymmetric planar cubic graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" />
  <row Id="10686" PostTypeId="1" AcceptedAnswerId="10692" CreationDate="2010-11-17T11:12:08.157" Score="4" ViewCount="158" Body="&lt;p&gt;Reposting from stackoverflow :) - Told to go here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've got a issue with a math assignment, and I were hoping someone could at least point me to what I should be doing, because currently I'm just wasting my time, thats for sure. - Please don't solve it all, thats up for me to do so :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assignment sounds like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A swimming pool is circular with a 10-metre diameter. The depth is constant along east-west lines and increases linearly from 1meter at the south end to 2meters at the north end. - Find the volume of water in the pool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured out that its about using double integrals on polar coordinates, however from there I kinda stalled, first of all; (1) if the pool is circular and has a 10-metre diameter, am I certain that the diameter is 10metres everywhere on the pool? - or just on a single spot? - That is could the pool be oval? (2) How far goes a 'side' alike west, east, north and south, on such a circular pool? - if its circular when will I know when I'm in the east line, and what about the north-east? (3) Would it seem likely that I'm going to describe the pool as a polar function first, before even considering the use of Integrals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this might not be the correct fora, for math questions, however stackoverflow has always seemed like a nice helpfull community :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;//Skeen&lt;/p&gt;&#xA;" OwnerUserId="3611" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-17T16:38:34.637" LastActivityDate="2010-11-17T16:38:34.637" Title="Finding the volume of water in a swimming pool" Tags="&lt;homework&gt;&lt;integral&gt;&lt;polar-coordinates&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10687" PostTypeId="1" AcceptedAnswerId="10688" CreationDate="2010-11-17T11:16:06.377" Score="2" ViewCount="94" Body="&lt;p&gt;$G$ is a finite abelian group. $A \times B$ can be embedded in $G$.&#xA;Does this mean there exist $C$, $D$ such that $G=C \times D$ and $A$ can be embedded in $C$ and $B$ can be embedded in $D$?&lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2010-11-17T12:57:18.393" Title="$G$ finite abelian. $A \times B$ embedded in $G$. Is $G=C \times D$ such that $A$ embedded in $C$, $B$ embedded in $D$?" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10688" PostTypeId="2" ParentId="10687" CreationDate="2010-11-17T11:28:58.053" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, this reduces to the case of finite $p$-groups. Use the lemma&#xA;that if $G=Z_{p^{r_1}}\times Z_{p^{r_2}}\times\cdots\times Z_{p^{r_k}}$&#xD;&#xA;where $r_1\ge r_2\ge\cdots \ge r_k$&#xD;&#xA;then each subgroup of $G$ is isomorphic to&#xA;$Z_{p^{s_1}}\times Z_{p^{s_2}}\times\cdots\times Z_{p^{s_k}}$&#xD;&#xA;where each $s_i\le r_i$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T11:28:58.053" CommentCount="4" />
  <row Id="10689" PostTypeId="2" ParentId="10686" CreationDate="2010-11-17T11:51:43.363" Score="2" ViewCount="0" Body="&lt;p&gt;To address the question that stalled you: a circular pool isn't oval.  If the problem meant for you to consider an oval pool, it would have said &quot;an elliptical pool with (say) horizontal axis $a$ and vertical axis $b$,&quot; or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need polar coordinates for this, nor do you need a double integral.  Single-variable calculus will do the trick just fine.  For starters, orient the pool so that the circle which defines it is centered at the origin, and look down at the pool from overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the water in it as if it was a big block of jello.  Now imagine slicing that block of jello horizontally infinitely many times, so that you have a stack of jello slices, each of which is infinitely thin, and each of which has its own width and depth, and each of which is located at its own $y$ coordinate (that is to say, for each slice, the points that make up the slice all have the same $y$-coordinate).  Each slice is essentially a rectangular box with volume width times depth times &quot;thinness&quot;.  In the figure below, the blue lines represent the individual jello slices, with the thick one being the slice located at $y = -3$.&#xA;&lt;img src=&quot;http://i.imgur.com/1v00U.png&quot; alt=&quot;figure of pool&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the equation of the circle (which, since it's centered at the origin, is $x^2 + y^2 = r^2$, where $r$ is the radius of the circle), you can figure out the width of each jello slice.  Knowing that the depth increases linearly, you can also figure out the depth of each slice as a function of its $y$ coordinate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Actually, it's not important that the depth increase linearly, per se, just that it changes north-south (i.e. as a function of $y$) and not east-west (i.e. as a function of $x$).  If it changed east-west instead, you'd slice the jello vertically instead of horizontally.  If it changed in all directions, then you'd need a double integral.].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, once you know the width and the depth, you can compute the volume of each slice: $$V_{\mbox{slice}} = \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$ where $dy$ represents the infinite &quot;thinness&quot; of the slice.  Now it's just a question of adding them up, which is exactly the point of a definite integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total volume is:&#xA;$$V_{\mbox{total}} = \int_{-r}^{-r} V_{\mbox{slice}} = \int_{-r}^{r} \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave you to figure out the appropriate expressions (in terms of $y$) for the width and the depth of each slice.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2010-11-17T12:13:03.533" LastActivityDate="2010-11-17T12:13:03.533" />
  <row Id="10690" PostTypeId="2" ParentId="10687" CreationDate="2010-11-17T12:57:18.393" Score="2" ViewCount="0" Body="&lt;p&gt;My first reaction was that the answer is no, but I had misinterpreted the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different question is: If $A$ and $B$ are subgroups of the abelian group $G$ such that $A \times B$ embeds in $G$, is it true that there are subgroups $C$ and $D$ of $G$ with $A \subseteq C$, $B \subseteq D$ and $G \cong C \times D$.&lt;/p&gt;&#xA;" OwnerUserId="2604" LastActivityDate="2010-11-17T12:57:18.393" />
  <row Id="10691" PostTypeId="2" ParentId="10181" CreationDate="2010-11-17T12:57:26.453" Score="1" ViewCount="0" Body="&lt;p&gt;For an undergraduate-level book, none of these three can do you wrong:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mendelson, &lt;em&gt;Introduction to mathematical logic&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Enderton, &lt;em&gt;A mathematical introduction to logic&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Boolos, Burgess, and Jeffrey, &lt;em&gt;Computability and logic&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They certainly go into the details, and they will leave you in a position where you could go further. If you can only look at one, try Mendelson.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An older text, which is now reprinted by Dover, is Kleene's &lt;em&gt;Mathematical logic&lt;/em&gt;. It is also very thorough, and has the advantage of being inexpensive. (Be aware that Kleene's &lt;em&gt;Introduction to metamathematics&lt;/em&gt; is a completely different book, which is much more advanced.)&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-17T12:57:26.453" />
  <row Id="10692" PostTypeId="2" ParentId="10686" CreationDate="2010-11-17T12:58:50.337" Score="7" ViewCount="0" Body="&lt;p&gt;Imaging one more inverted pool of same size below our original pool. They two pools will exactly fit in and will together form a cylinder of diameter 10 metres and height 3 metres.&lt;/p&gt;&#xA;" OwnerUserId="3615" LastActivityDate="2010-11-17T12:58:50.337" CommentCount="1" />
  <row Id="10693" PostTypeId="1" CreationDate="2010-11-17T13:15:41.040" Score="3" ViewCount="177" Body="&lt;p&gt;Define a subprobability measure to be a measure on Borel sigma algebra of the real line $\mathbb{R}$  with the measure for the whole real line less or equal than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering about the definition of vague convergence of a sequence of subprobability measures $\{ \mu_n, n\geq 1 \}$ to another subprobablity measure $\mu$. The convergence can be defined in slightly two different ways as in &lt;a href=&quot;http://books.google.com/books?id=zu80w7wd40UC&amp;amp;printsec=frontcover&amp;amp;dq=Chung%27s+probability+theory&amp;amp;source=bl&amp;amp;ots=AnNrU86FDf&amp;amp;sig=EAzRmK6PIIdg0qLjvUiUHlFblCM&amp;amp;hl=en&amp;amp;ei=UNjjTOC0IoX6lwe_leWbDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCsQ6AEwAg&quot; rel=&quot;nofollow&quot;&gt;Chung's probability theory book&lt;/a&gt; p85 and p90:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a &amp;lt;b, \mu_n((a,b]) \rightarrow \mu((a,b])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a &amp;lt;b, \mu_n((a,b)) \rightarrow \mu((a,b))$ (the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to show these two definitions are equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side question: &#xA;is there a definition of vague congergence for general measures on more general sigma algebra with more general underlying space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much!&lt;/p&gt;&#xA;" OwnerUserId="3616" LastEditorUserId="3616" LastEditorDisplayName="" LastEditDate="2010-11-17T13:38:33.710" LastActivityDate="2010-11-22T13:08:27.933" Title="Vague convergence of measures" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="10694" PostTypeId="1" CreationDate="2010-11-17T13:30:21.210" Score="0" ViewCount="33" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I receive 800 data per day. I have to spot failures in thoses data.&#xA;I use average and std dev to detect site where there is a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some model for that kind of purpose ? Do you have books reference abou that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3617" LastActivityDate="2010-11-17T13:30:21.210" Title="Statistic failure detection" Tags="&lt;statistics&gt;" CommentCount="1" />
  <row Id="10695" PostTypeId="2" ParentId="10693" CreationDate="2010-11-17T13:41:57.257" Score="0" ViewCount="0" Body="&lt;p&gt;Try $(a,b] = \bigcap_{c \in D, c &amp;gt; b} (a,c)$ and $(a,b) = \bigcup_{c \in D, c &amp;lt; b} (a,c]$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T13:41:57.257" CommentCount="1" />
  <row Id="10696" PostTypeId="1" CreationDate="2010-11-17T14:03:12.540" Score="0" ViewCount="57" Body="&lt;p&gt;Is it possible to define distribution function for a measure on a general measurable space, which generalizes the definition of distribution function for a probability measure on the extended real line? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, how? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, how general/special can the measure and the measurable space be in order to define its distribution function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3616" LastActivityDate="2010-11-17T16:35:24.057" Title="Define distribution function on a general measure space." Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10697" PostTypeId="1" AcceptedAnswerId="10724" CreationDate="2010-11-17T15:03:22.310" Score="4" ViewCount="260" Body="&lt;p&gt;YARFMO (Yet another reposting from Mathoverflow) ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more you know about math the more you find conceptions previously thought correct to be false:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) math is not as exact as many believe - in many cases, as Gowers points out in his amazing book &quot;Mathematics. A very short intro&quot;, it is mainly about approximations, boundaries or even &quot;only&quot; existence. Example: The Riemann hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) math is not &quot;true&quot; (whatever that means) in its own right - it is only true within an axiomatic system (deductive approach). Example: The parallel axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) closed form solutions are not as closed as many think: &quot;closed-form&quot; is just shorthand for &quot;popular enough to be given a name and notation.&quot; Examples: $\pi, e, \mathrm{Si}(x), \mathrm{li}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have further common misconceptions about math (ideally also with examples)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the ones given above only scratch the surface because in the end the picture of math also changes historically the further we proceed on this never-ending endeavor of human mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As should be clear already, I am talking about the &quot;meta'-level, so please don't give examples like $(x+y)^2=x^2+y^2$ - this is already covered &lt;a href=&quot;http://mathoverflow.net/questions/23478&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/7864&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-17T15:20:32.130" LastActivityDate="2010-11-17T20:24:03.320" CommunityOwnedDate="2010-11-18T17:24:01.440" Title="Common misconceptions about math" Tags="&lt;soft-question&gt;&lt;big-list&gt;&lt;education&gt;&lt;philosophy&gt;&lt;meta-math&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="10698" PostTypeId="2" ParentId="10593" CreationDate="2010-11-17T15:34:15.453" Score="0" ViewCount="0" Body="&lt;p&gt;Proving that a manifold $M$ is not simply connected (if true) is more difficult than proving that $M$ is simply connected (if true), because the first requires an &quot;impossibility proof&quot; (like proving that you cannot trisect an angle with ruler and compass). For an impossibility proof you need a &quot;higher theory&quot;. E.g., in order to prove that an annulus $R$  in $\mathbb R^2$ is not simply connected you show that $R$ carries a closed vector field (namely $\nabla\arg $) which is not the gradient of a globally defined function. Such a thing would not exist on a simply connected domain.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-17T15:34:15.453" CommentCount="1" />
  <row Id="10699" PostTypeId="1" CreationDate="2010-11-17T15:36:41.443" Score="2" ViewCount="110" Body="&lt;p&gt;I was wondering what is wrong in the following proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proposition. Let $\{f_{n}\}$ be a sequence of integrable functions such that $f_{n}$ converges pointwise to a function $f$. Show that if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim \int |f_{n} - f| d\mu =0$ then $\int |f| d\mu = \lim \int |f_{n}| d\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I used the fact that $||f_{n}| - |f| | \leq |f_{n} - f|$ and integrating both sides and using the assumption that $\lim \int |f_{n} - f| d\mu =0$ I get the result. But I never used the fact that $f_{n}$ converges pointwise to $f$. Why do we need this assumption or what is wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3622" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T17:36:13.100" LastActivityDate="2010-11-17T18:45:30.613" Title="Doubt convergence of integrals" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10700" PostTypeId="1" CreationDate="2010-11-17T15:46:17.867" Score="0" ViewCount="36" Body="&lt;p&gt;We are suppose to compute the quaternion which performs 1/5 of the rotation of this quaternion:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[ 0.965  (0.149  -0.149   0.149)]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The answer provided is shown as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.imgur.com/FIRJY.jpg&quot; rel=&quot;nofollow&quot;&gt;Picture of the answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck at the second last row of the ans. How do you change from the '0.998 q0 + 0.202 q' to [0.998 0.03 -0.03 0.03]&lt;/p&gt;&#xA;" OwnerUserId="3625" LastActivityDate="2010-12-17T16:44:26.050" Title="Question regarding practical SLERP" Tags="&lt;quaternions&gt;" AnswerCount="1" />
  <row Id="10701" PostTypeId="1" CreationDate="2010-11-17T15:49:07.583" Score="1" ViewCount="73" Body="&lt;p&gt;Let $f: A \rightarrow B$ be a ring homomorphism between commutative rings with identity. Then there exists an induced map $f&amp;#39; : Spec(B) \rightarrow Spec(A)$. If $f&amp;#39;$ is surjective, then clearly every prime ideal of $A$ is a contracted ideal. Now my question is, is the converse true?&lt;/p&gt;&#xA;" OwnerUserId="3620" LastActivityDate="2010-11-17T15:59:19.440" Title="If every prime ideal is a contracted ideal, does it imply that the induced map between spectrums is surjective?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="10702" PostTypeId="1" CreationDate="2010-11-17T15:49:27.090" Score="3" ViewCount="91" Body="&lt;p&gt;If $\sin(A+B) + \sin(B+C) + \cos(C-A) = \frac{3}{2}$ show that, $$1.\sin A + \cos B + \sin C = 0$$ $$2. \cos A + \sin B +\cos C = 0$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-17T18:53:17.450" Title="Proving trignometrical identities:" Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="10703" PostTypeId="2" ParentId="10699" CreationDate="2010-11-17T15:51:12.987" Score="3" ViewCount="0" Body="&lt;p&gt;Nothing's wrong, your proof is fine.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-17T15:51:12.987" />
  <row Id="10704" PostTypeId="2" ParentId="10685" CreationDate="2010-11-17T15:54:31.930" Score="0" ViewCount="0" Body="&lt;p&gt;I am not sure if this gets at what you mean for the 4-valent case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the infinite graph 4-valent which arises from tiling the plane with 4-gons (squares). Pick one vertex and remove that vertex and the 4 edges attached to it. (We now have four 3-valent vertices.) Join pairs of these 3-valent vertices with two edges each having slope 1. We now have a 4-valent infinite graph with two new 3-gons and a single 6-gon.  We can get an asymmetric infinite graph be carrying out this process in various unsymmetrical locations of the original tiling. By starting with irregularly spaced lines (rectangles rather than squares) the two new added edges will have very different directions.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-17T15:54:31.930" />
  <row Id="10705" PostTypeId="2" ParentId="10701" CreationDate="2010-11-17T15:59:19.440" Score="1" ViewCount="0" Body="&lt;p&gt;Let $P$ be a prime ideal of $A$, and suppose that&#xA;$P=f^{-1}(I)$ where $I$ is some ideal of $B$.&#xA;Let $S=A-P$, a multiplicatively closed subset&#xA;of $A$, and $T=f(S)$. Then $T$ is a multiplicatively&#xA;closed subset of $B$ and is disjoint from $I$.&#xA;By Zorn's lemma, there is an ideal $Q$ of $B$&#xD;&#xA;maximal with respect to the conditions that $Q$&#xD;&#xA;contains $I$ and $Q$ is disjoint from $T$. By&#xA;a standard argument, $Q$ is prime. Moreover $f^{-1}(Q)=P$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T15:59:19.440" CommentCount="1" />
  <row Id="10706" PostTypeId="2" ParentId="10181" CreationDate="2010-11-17T16:03:58.397" Score="0" ViewCount="0" Body="&lt;p&gt;May I humbly suggest some educational freeware I have developed to teach formal logic and basic proof-writing skills. It is a downloadable, PC-based proof assistant (editor and logic-checker). Being a computer program, it is of necessity a purely formal system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Included with the download is a tutorial that can be used as an introduction to formal mathematics. It includes ten worked examples of formal proofs, plus exercises with hints and full solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For testimonials, and a free full-function download, visit my website: &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-17T16:03:58.397" CommentCount="2" />
  <row Id="10707" PostTypeId="2" ParentId="10700" CreationDate="2010-11-17T16:11:25.970" Score="0" ViewCount="0" Body="&lt;p&gt;Your answer omits what $q$ and $q_0$ are. I presume&#xA;this is just a real linear combination of quaternions;&#xA;in your peculiar notation:&#xA;$$t[a\ \ (b\ \ c\ \ d)]+t&amp;#39;[a&amp;#39;\ \ (b&amp;#39;\ \ c&amp;#39;\ \ d&amp;#39;)]&#xD;&#xA;=[ta+t&amp;#39;a&amp;#39;\ \ (tb+t&amp;#39;b&amp;#39;\ \ tc+t&amp;#39;c&amp;#39;\ \ td+t&amp;#39;d&amp;#39;)].$$&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T16:11:25.970" CommentCount="1" />
  <row Id="10708" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T16:20:47.243" Score="1" ViewCount="0" Body="&lt;p&gt;Davenport, in &quot;The Higher Arithmetic&quot; Section III.5 asserts that the law of quadratic reciprocity, in its original form as conjectured by Euler, was the following assertion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ be any natural number and $p,q$ any primes such that $p\equiv q$ mod $4a$ or $p\equiv -q$ mod $4a$. Then $a$ is a square mod $p$ if and only of $a$ is a square mod $q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Davenport proves that  the usual quadratic reciprocity formula for odd primes is equivalent to this assertion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests what to me is the most impressive of all applications of quadratic reciprocity: That the  prime divisors of the values of quadratic polynomials fall into residue classes. One is immediately led to speculate that, more generally, the prime divisors of the values of polynomials have an intelligible structure which is somehow related to residue classes. This is really most surprising, given that polynomials combine addition and multiplication arbitrarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, to follow up this idea it was necessary to generalize the idea of ``residue class'' to rings of algebraic integers, giving birth to a big chunk of modern number theory along the way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe to understand the importance of quadratic reciprocity, the main thing is not to accumulate a catalogue of specific applications, but to see quadratic reciprocity as a historical gateway to modern Number Theory. &lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2010-11-17T16:20:47.243" />
  <row Id="10709" PostTypeId="2" ParentId="10693" CreationDate="2010-11-17T16:22:38.790" Score="1" ViewCount="0" Body="&lt;p&gt;Here is the general notion of vague convergence, taken from Olav Kallenberg's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387953132&quot; rel=&quot;nofollow&quot;&gt;Foundations of Modern Probability (2nd edition)&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be a locally compact, second countable Hausdorff space equipped with its Borel $\sigma$-field ${\cal S}$. Let $\hat{\cal S}$ be the ring of relatively compact Borel sets, and ${\cal M}(S)$ the space of locally finite non-negative measures.&#xA;Locally finite means that $\mu(B)&amp;lt;\infty$ when $B\in \hat{\cal S}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vague topology on ${\cal M}(S)$ is the topology generated by the mappings &#xA;$\mu\mapsto \int f\ d\mu$ for every $f$ a non-negative continuous function with compact support.  &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-17T16:22:38.790" CommentCount="1" />
  <row Id="10710" PostTypeId="1" AcceptedAnswerId="10720" CreationDate="2010-11-17T16:22:50.537" Score="2" ViewCount="77" Body="&lt;p&gt;Given $1&amp;lt;q\leq2$ and $0\leq p\leq1$, let us consider the following function:&#xA;$$\phi\left(\alpha\right)=p\times\left|1-\alpha\right|^{q}+\left(1-p\right)\times\left|1+\alpha\right|^{q}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum over $\mathbb{R}$ I have found is:&#xA;$$\alpha\left(p\right)=\frac{p^{\frac{1}{q-1}}-\left(1-p\right)^{\frac{1}{q-1}}}{p^{\frac{1}{q-1}}+\left(1-p\right)^{\frac{1}{q-1}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to show that: $\exists c&amp;gt;0,\exists\gamma\in\left]0;1\right],\forall p\in\left[0;1\right],\left|1-2p\right|\leq c\left(1-\phi\left(\alpha\left(p\right)\right)\right)^{\gamma}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I asked my question, I was mistaken: there was $\alpha\left(p\right)$ instead of $\phi\left(\alpha\left(p\right)\right)$ in the right hand side above. Actually, you just have to show the following to conclude:&#xA;$$\forall p\in\left[0;1\right],\left(1-2p\right)^{2} = 1-4p\left(1-p\right)\leq 1-\phi\left(\alpha\left(p\right)\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is simple indeed! :)&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2380" LastEditorDisplayName="" LastEditDate="2010-11-18T19:01:52.787" LastActivityDate="2010-11-18T19:01:52.787" Title="Inequality based on the minimum of a function" Tags="&lt;homework&gt;&lt;functions&gt;&lt;optimization&gt;&lt;inequality&gt;&lt;proof&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="10711" PostTypeId="1" CreationDate="2010-11-17T16:23:54.433" Score="0" ViewCount="101" Body="&lt;p&gt;Prove that X is a Random Variable IFF sigma field generated by X is countably generated.&lt;/p&gt;&#xA;" OwnerUserId="3559" LastActivityDate="2010-11-17T16:45:21.867" Title="countably generated sigma algebra" Tags="&lt;random-variable&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10712" PostTypeId="2" ParentId="10696" CreationDate="2010-11-17T16:35:24.057" Score="2" ViewCount="0" Body="&lt;p&gt;There is a well-known definition of cumulative distribution function $F$ for measures on finite dimensional Euclidean space ${\mathbb R}^d$. A function $F$ is the distribution function of a measure on ${\mathbb R}^d$ if and only if $F$ is right continuous and has non-negative increments, where these notions are defined appropriately for multidimensional space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find details about this in Chapter 3 of Olav Kallenberg's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387953132&quot; rel=&quot;nofollow&quot;&gt;Foundations of Modern Probability (2nd edition)&lt;/a&gt;&lt;/em&gt;, for example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of useful extensions of the idea of distribution functions to other partially ordered spaces, but they may exist. &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-17T16:35:24.057" CommentCount="1" />
  <row Id="10713" PostTypeId="1" AcceptedAnswerId="10722" CreationDate="2010-11-17T16:45:18.310" Score="1" ViewCount="69" Body="&lt;p&gt;Find modules $M,N$ and a ring $A$ s.t. $M\otimes_A N\not \cong M\otimes_{\mathbb{Z}}N$.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-17T18:28:10.057" Title="Non-isomorphic tensorproducts" Tags="&lt;modules&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10714" PostTypeId="2" ParentId="10711" CreationDate="2010-11-17T16:45:21.867" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;If and only if&quot; doesn't make sense here, but you should be able to prove that for any map $X:\Omega\to {\mathbb R}$ the $\sigma$-field generated by $X$, that is, $X^{-1}({\cal B}(\mathbb R))$,  is countably generated. Hint: The $\sigma$-field of Borel sets of $\mathbb R$ is countably generated.    &lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-11-17T16:45:21.867" CommentCount="1" />
  <row Id="10715" PostTypeId="2" ParentId="10713" CreationDate="2010-11-17T16:52:08.380" Score="1" ViewCount="0" Body="&lt;p&gt;Let $A=\mathbb Z[X]$ the polynomial ring, let $M=\mathbb Z[X,X^{-1}]$ be the ring of Laurent polynomials views as an $A$ module, and let $N=A/(X)$. As $\mathbb Z$-modules, both $M$ and $N$ are free $\mathbb Z$-modules. You can easily check that $M\otimes_AN=0$ yet $M\otimes_\mathbb ZN\neq0$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-17T16:52:08.380" CommentCount="1" />
  <row Id="10716" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T17:08:07.363" Score="2" ViewCount="0" Body="&lt;p&gt;I guess this might as well be a separate answer.  You can use quadratic reciprocity to give elementary proofs of certain special cases of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlets_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;Dirichlet's theorem&lt;/a&gt;.  First, you should be aware of the following nice result and its &quot;Euclidean&quot; proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  Let $f(x) \in \mathbb{Z}[x]$ and let $P_f$ be the set of primes $p$ such that $p | f(n)$ for some $n$.  Then $P_f$ is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  If $f(0) = 0$ then this is obvious, so suppose otherwise.  Let $p_1, p_2, ... p_n$ be a finite set of primes in $P_f$.  Then for any $k$, $\frac{1}{f(0)} f(k f(0) p_1 ... p_n)$ must be divisible by a prime which is not one of the $p_i$, and choosing $k$ sufficiently large we can find a new prime $p_{n+1}$ in $P_f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate proof is given &lt;a href=&quot;http://qchu.wordpress.com/2009/09/02/some-remarks-on-the-infinitude-of-primes/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  Using this lemma and properties of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_polynomial#Applications&quot; rel=&quot;nofollow&quot;&gt;cyclotomic polynomials&lt;/a&gt;, you can prove that there exist infinitely many primes congruent to $1 \bmod n$ for any $n$ without any heavy machinery, so I will skip these cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using quadratic reciprocity, you can prove that the following arithmetic progressions also contain infinitely many primes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$11 \bmod 12$:  Let $f(x) = 3x^2 - 1$.  Then $p | f(n)$ implies $\left( \frac{3}{p} \right) = 1$.  However, we can modify the proof of the lemma to prove that infinitely many of the primes dividing $f$ must be congruent to $3 \bmod 4$.  To see this, let $p_1, .. p_n$ be finitely many primes with this property and consider $f(2 p_1 ... p_n) \equiv 3 \bmod 4$.  It follows that infinitely many primes $p$ satisfy $\left( \frac{3}{p} \right) = 1$ and $p \equiv 3 \bmod 4$, so by quadratic reciprocity $\left( \frac{p}{3} \right) = -1$, so $p \equiv 2 \bmod 3$.  Hence $p \equiv 11 \bmod 12$.  In particular, there are infinitely many primes congruent to $2 \bmod 3$ and infinitely many primes congruent to $3 \bmod 4$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$4 \bmod 5$:  Let $f(x) = x^2 - 5$.  Then $p | f(n)$ implies $\left( \frac{5}{p} \right) = 1$.  Again, we can modify the proof of the lemma to prove that infinitely many of these primes are not congruent to $1 \bmod 5$.  To see this, let $p_1, ... p_n$ be finitely many primes with this property, none of which are equal to $5$, and consider either $f(p_1 ... p_n)$ or $f(2 p_1 ... p_n)$, one of which is not congruent to $1 \bmod 5$ and which therefore has a prime factor which is not congruent to $1 \bmod 5$.  So infinitely many primes $p$ satisfy $\left( \frac{5}{p} \right) = 1$ and $p \not \equiv 1 \bmod 5$.  By quadratic reciprocity this gives $\left( \frac{p}{5} \right) = 1$, hence $p \equiv 4 \bmod 5$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$3 \bmod 8$:  Let $f(x) = x^2 + 2$.  Then $p | f(n)$ implies $\left( \frac{-2}{p} \right) = 1$.  Again, we can modify the proof of the lemma to prove that infinitely many of these primes are not congruent to $1 \bmod 8$.  To see this, let $p_1, ... p_n$ be finitely many primes with this property, all of which are odd, and consider $f(2p_1 ... p_n) \equiv 6 \bmod 8$.  So infinitely many primes $p$ satisfy $\left( \frac{-2}{p} \right) = 1$ and $p \not \equiv 1 \bmod 8$.  By quadratic reciprocity this gives $p \equiv 3 \bmod 8$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And so on.  For what progressions is it possible to give these kinds of proofs?  It turns out this is possible for the progression $a \bmod n$ if and only if $a^2 \equiv 1 \bmod n$.  For details, see Keith Conrad's &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf&quot; rel=&quot;nofollow&quot;&gt;Euclidean proofs of Dirichlet's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, quadratic reciprocity is the key to writing down the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Dedekind zeta functions&lt;/a&gt; of quadratic number fields explicitly, and trying to generalize this leads you into class field theory and so forth.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-17T17:13:40.803" LastActivityDate="2010-11-17T17:13:40.803" />
  <row Id="10717" PostTypeId="1" AcceptedAnswerId="10718" CreationDate="2010-11-17T17:16:02.907" Score="3" ViewCount="95" Body="&lt;p&gt;This is a problem in baby Rudin. Chapter 4, problem 15. I'm just looking at examples at the moment. It says that any continuous open map is monotone. Recall, an open map f: X -&gt; Y is such that any open set U in X will haves its image, f(U) open in Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;Let $$ f: \mathbb{R} \to [-1, 1] $$ take $$ x \mapsto sin(x) $$ and consider the open interval $$ (\frac{\pi}{3}, \frac{7\pi}{12}) $$ The image of this interval in [-1, 1] is $$ (\frac{\sqrt{3}}{2}, 1] $$ which is open in [-1, 1] (its complement is [-1, sqrt(3)/2], certainly closed). The problem is, the function is not monotonic on this interval. It is continuous, so it must not be an open map. I'm pretty sure I have the image right, so, where am I making a mistake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really hope I'm not making a dumb error!&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="2465" LastEditorDisplayName="" LastEditDate="2010-11-17T17:36:49.807" LastActivityDate="2010-11-17T18:13:07.397" Title="continuous open mappings are monotonic" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10718" PostTypeId="2" ParentId="10717" CreationDate="2010-11-17T17:27:47.293" Score="2" ViewCount="0" Body="&lt;p&gt;The mistake is misinterpreting for which codomain the statement holds.  Continuous open maps from $\mathbb{R}$ to $\mathbb{R}$ are monotone.  You have changed the codomain to the image with subspace topology, and this changes the meaning of open map.  Since the interval $\left(\frac{\sqrt{3}}{2},1\right]$ is not open in $\mathbb{R}$, your example demonstrates the fact that this non-monotone continuous map cannot be an open map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Athough in this case you have a map that is not open, note that if you are trying to show that a map &lt;em&gt;is&lt;/em&gt; open, it is not enough to give an example of an open set that is mapped to an open set.  You would have to prove that &lt;em&gt;every&lt;/em&gt; open set is mapped to an open set.  On the other hand, to show that a map is not open, you only need to demonstrate that there exists an open set that is not mapped to an open set, and this fact will come in handy if you approach this problem using contraposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side remark: You will at some point have to invoke completeness of $\mathbb{R}$ (or a theorem that uses completeness).  The function $f(x)=x^3-6x$ as a map from $\mathbb{Q}$ to $\mathbb{Q}$ is continuous and open but not monotone.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-17T18:13:07.397" LastActivityDate="2010-11-17T18:13:07.397" CommentCount="2" />
  <row Id="10719" PostTypeId="2" ParentId="10233" CreationDate="2010-11-17T18:06:37.897" Score="2" ViewCount="0" Body="&lt;p&gt;The quadratic reciprocity law in any of its forms shows that there is an un-obvious correlation between different primes.  The $(p,q)$ symbol constrains the $(q,p)$ symbol.   This is astonishing compared to other more &quot;linear&quot; theorems about congruences or unique factorization.  In its 20th-century reformulations quadratic reciprocity is seen as an avatar of other reciprocity laws in geometry (reciprocity for tame symbols) and even geometric topology (linking numbers where knots play the role of primes) and although these other theorems are in some ways easier to prove, the analogies between all of them are mysterious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you are not shocked by this theorem, you don't completely understand it.   Historically it was a hard-won, prize result and not an inevitable universal discovery like the Pythagorean formula or other theorems that were difficult in their time but found independently in many times and places.  Many cultures had knowledge of basic number theoretic facts but quadratic reciprocity is one of the first signs of number theory as a science.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T18:27:13.693" LastActivityDate="2010-11-18T18:27:13.693" CommentCount="2" />
  <row Id="10720" PostTypeId="2" ParentId="10710" CreationDate="2010-11-17T18:07:15.823" Score="2" ViewCount="0" Body="&lt;p&gt;If $\alpha$ is large (of either sign), $\phi$ increases without bound essentially as $\alpha ^q$.  Is $\alpha \in (-1,1)$ in which case we can delete the absolute value bars?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-17T18:07:15.823" CommentCount="3" />
  <row Id="10721" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T18:25:47.110" Score="8" ViewCount="0" Body="&lt;p&gt;Perhaps more meta than you wanted since it's not about how you perceive mathematics coming from mathematics, but rather how the rest of the world views it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's that whole misconception that if you're good at math you're good at arithmetic. You know... that thing where someone says something like &quot;Hey, you're good at math! Can you do this calculation (typically menial multiplication) for me in your head?&quot; Sure I &lt;em&gt;could&lt;/em&gt;, but that's because I can keep a lot of stuff in my head at once and make sure I do it correctly. It's not as if that says anything about mathematical ability.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-17T18:25:47.110" CommunityOwnedDate="2010-11-18T17:24:01.440" CommentCount="2" />
  <row Id="10722" PostTypeId="2" ParentId="10713" CreationDate="2010-11-17T18:28:10.057" Score="3" ViewCount="0" Body="&lt;p&gt;Well, $A\otimes_A A$ is certainly isomorphic to $A$. So all we need&#xA;is a ring for which $A\otimes_{\mathbb{Z}} A$ isn't isomorphic to $A$.&#xA;An obvious choice is a ring like the Gaussian integers $A=\mathbb{Z}[i]$&#xD;&#xA;which is free of rank $2$ as a $\mathbb{Z}$-module, for then&#xA;$A\otimes_{\mathbb{Z}} A$ is free of rank $4$ as a $\mathbb{Z}$-module.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-17T18:28:10.057" />
  <row Id="10723" PostTypeId="2" ParentId="10699" CreationDate="2010-11-17T18:45:30.613" Score="1" ViewCount="0" Body="&lt;p&gt;Zaricuse has already given the correct answer, but I want to add a few remarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you don't assume pointwise convergence, the hypothesis $\lim \int |f_{n} - f| d\mu =0$ implies that $(f_n)$ converges to $f$ in measure, which implies that there is a subsequence that converges to $f$ pointwise almost everywhere.  But as you have shown, this is not needed in the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the normed space $L^1(\mu)$ of absolutely $\mu$ integrable functions with norm $\|f\|=\int|f|d\mu$, what you have shown is that the norm is continuous as a function from $L^1(\mu)$ with the norm topology to $[0,\infty)$.  The same is true in any normed space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu mentioned in a comment, pointwise convergence (almost everywhere) allows you to prove the converse, which follows from a generalized form of the dominated convergence theorem.  &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-17T18:45:30.613" CommentCount="4" />
  <row Id="10724" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T18:47:56.960" Score="13" ViewCount="0" Body="&lt;p&gt;Here are a couple of misconceptions that people who don't know about mathematics commonly hold. Firstly, that it's immutable, fixed, unchanging; surely it's all been discovered? It's a belief that even a highly educated non-mathematician might hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point was illustrated to me not long ago when a friend, who speaks at least 10 languages fluently, asked why I found mathematics interesting. She believed that it was cold and lifeless and thus inherently less appealing than language, which she viewed as living and dynamic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to persuade her that mathematics grows and changes too. Areas such as Group Theory  were born from ideas used in Number Theory and Geometry. And old ways of doing things can be surpassed. The history of cryptography is a good example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversation also touched on a second misconception: that mathematicians simply follow set rules. I explained that imagination was very important in doing mathematics. And when I said that  there was great beauty in mathematics, I was faced with almost total incomprehension. Surely, mathematics was the very antithesis of beauty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're reading this I don't need to convince you that these are indeed misconceptions, but unfortunately I don't think I convinced my friend!&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-17T18:47:56.960" CommunityOwnedDate="2010-11-17T18:47:56.960" CommentCount="2" />
  <row Id="10725" PostTypeId="2" ParentId="10702" CreationDate="2010-11-17T18:53:17.450" Score="6" ViewCount="0" Body="&lt;p&gt;The problem seems to be missing some assumptions, as noted by Americo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, If $B = 0$ and $A=C$ are acute angles, such that $\sin A  = 1/4$ the we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(A+B) + \sin(B+C) + \cos(A-C) = 3/2$, but none of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin A + \cos B + \sin C$ or $\cos A + \sin B + \cos C$ are $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, this looks like a perfect problem for using complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B&amp;#39; = \pi/2 - B$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_1 = \cos A + i \sin A$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_2 = \cos B&amp;#39; + i \sin B&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z_3 = \cos C + i \sin C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given identity is $\cos (A-B&amp;#39;) + \cos (C - B&amp;#39;) + \cos (A-C) = 3/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{z_1}{z_2} +\frac{z_2}{z_1} +\frac{z_3}{z_2} +\frac{z_2}{z_3} +\frac{z_1}{z_3} +\frac{z_3}{z_1}  = 3$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two identities&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\sin A + \cos B + \sin C = 0$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\cos A + \sin B +\cos C = 0$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;are equivalent to showing that $z_1 + z_2 + z_3 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eq 1, says that the imaginary part of $z_1 + z_2 + z_3$ is $0$ and Eq 2 says that the real part is $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-17T18:53:17.450" CommentCount="5" />
  <row Id="10726" PostTypeId="1" AcceptedAnswerId="10769" CreationDate="2010-11-17T18:53:36.220" Score="3" ViewCount="91" Body="&lt;p&gt;Consider the following arithmetic function, $f(a,b) = (a+1, \lceil a- \frac{a}{b} \rceil , \lceil a- \frac{2a}{b} \rceil, \dots, \lceil \frac{a}{b} \rceil, 1)$, where all terms except the first and last are computed by the application of a ceiling function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any pair of positive integers $(a,b)$, the vector $f(a,b)$ consists of the terms of an integer partition of length $a+1$. Moreover, $f(a,b)$ and $f(b,a)$ represent dual partitions and have reflected Ferrers Diagrams. This can be proved by recalling the identity&#xA;\begin{eqnarray}&#xA;\sum_{j = 0}^{b-1} \left \lceil \tfrac{a(b-j)}{b} \right \rceil = \sum_{j = 0}^{a-1} \left \lceil \tfrac{b(a-j)}{a} \right \rceil.&#xA;\end{eqnarray}&#xA;(Going in the other direction, one can use the methods outlined here to give a neat combinatorial proof of this ceiling summation identity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, these integer partitions contain no $0$s, contain at least one $1$ and at most $a$ $1$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Are these partitions a special (read &lt;em&gt;named&lt;/em&gt;) subset of the standard partitions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 1&lt;/strong&gt;: Determine all pairs $(a,b)$ such that $f(a,b)$ are the terms of an integer partition of $k$ for a fixed $k &amp;gt; 1$. Determine a simple way to find $k$ from $a$ and $b$ (and not by simply adding the terms of the defining relation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 2&lt;/strong&gt;: Find other symmetries like $f(a,b) \leftrightarrow f(b,a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) For example, $f(2,3) = (4,2,1)$, $f(3,2) = (3,2,1,1)$, $f(1,5) = (6,1)$ and $f(5,1) = (2,1,1,1,1,1)$, which are &lt;em&gt;dual&lt;/em&gt; integer partitions of $7$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem 3&lt;/strong&gt;: Give an interpretation to the sequence {$f(a,b)$} in terms of $a$ and $b$ for fixed $a$ or $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) The sequence {$f(1,b)$} consists of terms of the form $(2,1, \dots, 1)$ with $b + 1$ $1$s, which are integer partitions of $b+2$; {$f(2,b)$} consists of terms of the form $(3, 1), (3,1,1), (3,2,1)$ or $(3, 2, \dots, 2, 1, \dots, 1)$ which are each integer partitions of those integers greater than 3 not divisible by $3$; {$f(3,b)$} consists of special integer partitions of integers congruent to {$1,2,5$} mod $6$. Similarly, {$f(4,b)$} consists of integer partitions of integers congruent to {$1, 2, 3$} mod $5$ or {$1,2,6,8$} mod $10$; {$f(5,b)$} consists of integer partitions of integers congruent to {$1, 2, 4$} mod $6$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some Mathematica code that might be useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f[a_, b_] := &#xA;  Flatten[{a + 1, Table[Ceiling[a - j a/b], {j, 1, b - 1}], 1}];&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = 2;&#xA;Table[Total@f[a, b], {b, 1, 100}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-17T22:30:56.940" LastActivityDate="2010-11-17T23:03:33.783" Title="A Vector Arithmetic Function, Special Integer Partitions and Residue Sequences" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="10727" PostTypeId="1" AcceptedAnswerId="10730" CreationDate="2010-11-17T19:01:17.643" Score="0" ViewCount="72" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is based on the diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Ko6HT.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taken from this &lt;a href=&quot;http://www.ies.co.jp/math/products/trig/applets/yogen_auto/yogen_auto.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;. I don't understand why the areas of A6 and A5 adds up to $2bc\cos(A)$.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T19:23:32.030" LastActivityDate="2010-11-17T19:24:50.437" Title="Law of Cosines Proof" Tags="&lt;trigonometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10728" PostTypeId="2" ParentId="10727" CreationDate="2010-11-17T19:18:19.383" Score="1" ViewCount="0" Body="&lt;p&gt;The other side of A6 has length $c \cos A$ and that of A5 length $b \cos A.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So their total area is $b \times c \cos A + c \times b \cos A = 2bc \cos A.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-17T19:18:19.383" />
  <row Id="10729" PostTypeId="2" ParentId="10727" CreationDate="2010-11-17T19:24:44.680" Score="0" ViewCount="0" Body="&lt;p&gt;First note that A5 = A6. (The visual explains it using the fact that parallelogram area between the parallel lines are the same). And note that A5 is nothing but the side of the square times the length along the side $AB$. The side of the square is $c$. The length along the side $AB$ is $b \cos(A)$. Hence $A5 = c b \cos(A)$. Hence $A5 + A6 = 2bc \cos(A)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-17T19:24:44.680" />
  <row Id="10730" PostTypeId="2" ParentId="10727" CreationDate="2010-11-17T19:24:50.437" Score="0" ViewCount="0" Body="&lt;p&gt;If you look at A5. It is a rectangle with side $b$ and, say, $x$.&#xA;Now $\cos A= x/c$, so $x=c\cos A$. Hence the area of A5 is $bx=bc\cos A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument tells you that the area of A6 is $cb\cos A$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-17T19:24:50.437" />
  <row Id="10731" PostTypeId="1" AcceptedAnswerId="10805" CreationDate="2010-11-17T19:34:56.667" Score="2" ViewCount="80" Body="&lt;p&gt;Let $K \subset R^n$ be compact, and let $C_0^{\infty}(K)$ be the space of smooth functions on with support in $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p \in [1,\infty), \alpha$ multiindex, let $|f|_{\alpha,p} = (\int_K |D^{\alpha}f|^p)^{\frac{1}{p}}$, and $|f|_{\alpha,\infty} = sup_K |D^{\alpha}f|$ - as usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p \in [1,\infty]$, let $\tau_p$ be the locally convex topology generated on $C_0^{\infty}(K)$ by the seminorms $(|f|_{\alpha,p})_{\alpha}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt; All the $\tau_p$ are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One proof of this might utilize the Sobolev theory (Kondrashow embedding theorem) which is too heavy as tool for my taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't know a more elementary proof. Furthermore I guess a beautiful proof of this might work similarly for other spaces like the Schwartz-space or the space of smooth functions on the whole of $R^n$. Do you have any good idea?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-18T19:54:53.957" Title="p-seminorms on smooth functions are equivalent" Tags="&lt;analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10733" PostTypeId="2" ParentId="10648" CreationDate="2010-11-17T19:48:49.283" Score="0" ViewCount="0" Body="&lt;p&gt;The centres of three circles of radius r just touching one another will form an equilateral triangle of side length 2r. Exactly 6 such non-overlapping equilateral triangles will be formed in this way as you keep adding more circles around the edge of the central circle. Why 6? These equilateral triangles will have a common vertex at the centre, each forming a 60 degree angle there. And as, you pointed out, &quot;there are 6*60=360 degrees in a circle.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-11-17T19:56:23.347" LastActivityDate="2010-11-17T19:56:23.347" CommentCount="7" />
  <row Id="10734" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T19:59:16.743" Score="1" ViewCount="0" Body="&lt;p&gt;I've noticed it's quite common among algebraists dealing with finite structures (like finite groups) to think of mathematics (limited to their area, of course) as the absolute God given truth, harmonious, non-contradictory, sort of Platonic ideal world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point of view can be possible accepted, as long we limit ourselves to finite mathematics, but what happens when we introduce the notion of infinity? The whole harmonious God given world goes out the window and the mathematics becomes what it really is: a kind of human activity which reflects both the power of our intelligence as well as its limitations. &lt;/p&gt;&#xA;" OwnerUserId="2450" LastActivityDate="2010-11-17T19:59:16.743" CommunityOwnedDate="2010-11-18T17:24:01.440" CommentCount="1" />
  <row Id="10735" PostTypeId="1" AcceptedAnswerId="10742" CreationDate="2010-11-17T20:11:16.077" Score="0" ViewCount="81" Body="&lt;p&gt;I have to do this for an upcoming exam, but cannot find anywhere (in the textbook or online) how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only really need to know a couple points to plot it... when x = 0, and then the earliest crossing of the x-axis.&lt;/p&gt;&#xA;" OwnerUserId="3325" LastActivityDate="2010-11-17T20:40:07.083" Title="How to sketch a sinc function by hand?" Tags="&lt;calculus&gt;&lt;fourier-analysis&gt;&lt;signal-processing&gt;" AnswerCount="3" />
  <row Id="10736" PostTypeId="2" ParentId="10735" CreationDate="2010-11-17T20:14:08.663" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-17T20:14:08.663" CommentCount="1" />
  <row Id="10738" PostTypeId="2" ParentId="10697" CreationDate="2010-11-17T20:24:03.320" Score="3" ViewCount="0" Body="&lt;p&gt;Following up on Kahen's comment, most people who only take math through High School think math is about calculation. For higher math, description is more correct than calculation, but that still doesn't capture the flavor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people think math is about numbers. Some is, some isn't.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-11-17T20:24:03.320" CommunityOwnedDate="2010-11-18T17:24:01.440" />
  <row Id="10740" PostTypeId="2" ParentId="10735" CreationDate="2010-11-17T20:32:40.027" Score="1" ViewCount="0" Body="&lt;p&gt;I think its helpful also to graph it with $\frac{1}{x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29%2C+1%2Fx&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=plot+sinc%28x%29%2C+1%2Fx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write the function as $f(x) = \frac{\sin(x)}{x}$ for $x \not=0$ and $1$ if $x=0$  you can see that it will be zero at exactly when $\sin(x)$ is zero (except at $0$ where it is $1$). &lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2010-11-17T20:32:40.027" />
  <row Id="10741" PostTypeId="1" CreationDate="2010-11-17T20:32:54.007" Score="12" ViewCount="1018" Body="&lt;p&gt;In answering a question I mentioned the Asteroids video game as an example -- at one time, the canonical example -- of a locally flat geometry that is globally different from the Euclidean plane.  It might be out of date in 2010.  This raises its own question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are there other real-life examples of geometries formed by identifications?  We know the cylinder and Moebius strip and there are probably some interesting equivalents of those.  Origami are coverings of a punctured torus and I heard that there are crocheted examples of complicated 2-d and 3-d objects.  &lt;em&gt;Are there simple, Asteroids-like conversational examples for flat surfaces formed as quotients?&lt;/em&gt;  Punctures and orbifold points and higher dimensional examples all would be interesting, but I am looking less for mathematically sophisticated than conversationally relevant examples, such as a famous game or gadget that makes a cellphone function as a torus instead of a rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit:  Pac-Man, board games such as Chutes and Ladders, or any game with magic portals that transport you between different locations, all illustrate identification of points or pieces of the space, but they lead to non-homogeneous geometries.  The nice thing about Asteroids was that it was clearly the whole uniform geometry of the torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit-2: the Flying Toasters screen-saver would have been an example of what I mean, except that video of it exists online and shows it to be a square window onto motion in the ordinary Euclidean plane.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T03:39:51.813" LastActivityDate="2010-11-18T05:10:31.957" Title="Quotient geometries known in popular culture, such as &quot;flat torus = Asteroids video game&quot;." Tags="&lt;geometry&gt;&lt;topology&gt;&lt;pedagogy&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="3" />
  <row Id="10742" PostTypeId="2" ParentId="10735" CreationDate="2010-11-17T20:40:07.083" Score="1" ViewCount="0" Body="&lt;p&gt;You know the value at zero, it goes to zero at $x = n \pi$ and in-between reaches the maximum and minimum at close to $x = (n+ \frac{1}{2}) \pi$ (when $n&amp;gt;0$ or $n&amp;lt;-1$) with the value at these points being $\frac{(-1)^n}{(n+ \frac{1}{2}) \pi}$. So knowing these you should be able to roughly sketch the function.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-17T20:40:07.083" />
  <row Id="10743" PostTypeId="1" AcceptedAnswerId="10751" CreationDate="2010-11-17T20:47:26.387" Score="1" ViewCount="88" Body="&lt;p&gt;If I have a value of 25%, and I want the value as it would be on the opposite side of the halfway point (75% in this case), what is a formula that can calculate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the appropriate tags for this so feel free to adjust them.  However, it is a programming-related question.&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T21:02:54.137" LastActivityDate="2010-11-17T21:02:54.137" Title="How to reverse percentage?" Tags="&lt;elementary&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="10744" PostTypeId="1" AcceptedAnswerId="10756" CreationDate="2010-11-17T20:47:52.190" Score="1" ViewCount="55" Body="&lt;p&gt;After asking a few questions earlier, I think I've been able to describe my main issue.  Could someone show me why this identity is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = (B + B^{T})\textbf{x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gone through things like Matrixcookbook which seem to just show this identity without any derivation of it.  I have tried to unsuccessfully derive the formula below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Via product rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = \frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}B\textbf{x} + \textbf{x}^{T}\frac{\partial B \textbf{x}}{\partial \textbf{x}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize product rule for scalars is not the same for matrices, but it still holds as long as order is preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I'm stuck because I'm not sure if $\frac{\partial \textbf{x}^{T}}{\partial \textbf{x}}$ is identity $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this definition of the Jacobian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} \dfrac{\partial y_1}{\partial x_1} &amp;amp;; \cdots &amp;amp;; \dfrac{\partial y_1}{\partial x_n} \\ \vdots &amp;amp;; \ddots &amp;amp;; \vdots \\ \dfrac{\partial y_m}{\partial x_1} &amp;amp;; \cdots &amp;amp;; \dfrac{\partial y_m}{\partial x_n}  \end{bmatrix}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I write $\frac{\partial B \textbf{x}}{\partial \textbf{x}}$ component by component with the definition above, it becomes simply $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \textbf{x}^{T} B \textbf{x}}{\partial \textbf{x}} = B\textbf{x} + \textbf{x}^{T}B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is not correct, the left side is a nx1 vector and the right side is a nx1 vector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where my faulty assumption is.  I've gone through various sources which don't explain the situation I showed above.  Am I missing some very simple point that those books assume?&lt;/p&gt;&#xA;" OwnerUserId="3580" LastActivityDate="2010-11-17T21:25:17.760" Title="Correct method to differentiate a first order vector/matrix" Tags="&lt;linear-algebra&gt;&lt;vector-calculus&gt;" AnswerCount="3" />
  <row Id="10745" PostTypeId="1" CreationDate="2010-11-17T20:48:21.847" Score="1" ViewCount="98" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: Calculate limit of $\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+3}} + \cdots + \frac{1}{\sqrt{n+n^2}}$ as $n$ approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution: Denote the above some as $X$, then we can bound it:&#xA;$$ \infty\longleftarrow\frac{1}{\sqrt{n+n^2}} \lt X \lt \frac{n^2}{\sqrt{n+1}} \lt \frac{n^2}{\sqrt{n}} = \sqrt{\frac{n^4}{n}}\longrightarrow \infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from the Squeeze Principle, $\lim X = \infty$. Am I doing the right thing?&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-17T21:41:43.823" LastActivityDate="2010-11-17T21:41:43.823" Title="Question about Squeeze Principle" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="10746" PostTypeId="2" ParentId="10741" CreationDate="2010-11-17T20:49:56.133" Score="3" ViewCount="0" Body="&lt;p&gt;In (some of) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Civilization_%28series%29&quot; rel=&quot;nofollow&quot;&gt;Civilization games&lt;/a&gt; you can decide whether you want to play on a (bounded) plane, or a cylinder, or even a torus, I think.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-11-17T20:49:56.133" />
  <row Id="10747" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T20:54:05.227" Score="0" ViewCount="0" Body="&lt;p&gt;The term $\frac{1}{\sqrt{n+n^2}}$ goes to zero as $n \rightarrow \infty$, so that side doesn't work.  You should think about how many terms you are adding.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-17T21:02:55.007" LastActivityDate="2010-11-17T21:02:55.007" CommentCount="2" />
  <row Id="10748" PostTypeId="2" ParentId="10743" CreationDate="2010-11-17T20:54:21.870" Score="2" ViewCount="0" Body="&lt;p&gt;Unless I completely missed the point of the question, the answer is rather trivially $100\% - p$.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-17T20:54:21.870" />
  <row Id="10749" PostTypeId="2" ParentId="5420" CreationDate="2010-11-17T20:57:10.527" Score="1" ViewCount="0" Body="&lt;p&gt;There is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_logarithm#Real_number&quot; rel=&quot;nofollow&quot;&gt;algorithm&lt;/a&gt; for computing $\log_2(x)$ that might suit you.&#xA;Combine that with the spigot algorithm for $e$, and you can get $\ln(x)$.&#xA;From there, you can use Newton-Raphson to get $\exp(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this roundabout way ends up doing any better than just recomputing.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-17T20:57:10.527" />
  <row Id="10750" PostTypeId="2" ParentId="10743" CreationDate="2010-11-17T20:59:14.890" Score="1" ViewCount="0" Body="&lt;p&gt;Never mind, it's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(100 - 25)&#xA;or&#xA;(100 - 34.04)&lt;/p&gt;&#xA;" OwnerUserId="3636" LastActivityDate="2010-11-17T20:59:14.890" CommentCount="2" />
  <row Id="10751" PostTypeId="2" ParentId="10743" CreationDate="2010-11-17T21:00:50.880" Score="3" ViewCount="0" Body="&lt;p&gt;It's unclear to me what it is you want. One possible interpretation is that you know how much 25\% is, and you want to know how much 75\% is. For this situation it is very easy: if $c$ is 25% of some unknown quantity $y$, and you want to know how much 75% of $y$ is, it is $3c$. For instance, if you know that $27$ is 25% of something, then $3(27)=81$ is 75% of that same something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross-multiplication#Rule_of_Three&quot; rel=&quot;nofollow&quot;&gt;Rule of Three&lt;/a&gt; will let you figure out how much $q$% is if you know what $p$% is. If you know that $c$ is $p$% of an unknown quantity, and you want to know how much $q$% of that same quantity will be, then you can set it up as a cross-multiplication problem:&#xA;$$\begin{array}{ccc}&#xD;&#xA;c &amp;amp; \text{---} &amp;amp; p\\&#xD;&#xA;x &amp;amp; \text{---} &amp;amp; q&#xD;&#xA;\end{array}$$&#xD;&#xA;so that $x=qc/p$. For example, if you know that $23$ is $17$% of some unknown quantity $x$, and you would like to know who much 77% of that same quantity would be, it is $23(77/17)  \approx 104.176$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after writing all of the above it strikes me that the question may be even more basic: you have a certain quantity $p$ between $0$ and $50$, which you interpret as a percentage, and you want to know what quantity $q$ is just as far from $50$ but &quot;on the other side&quot;. Since you want $q-50 = 50-p$, then you want $q=100-p$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-17T21:00:50.880" CommentCount="3" />
  <row Id="10752" PostTypeId="2" ParentId="10744" CreationDate="2010-11-17T21:03:39.837" Score="1" ViewCount="0" Body="&lt;p&gt;$\frac{\partial}{\partial x}$ is just the gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$xBx$ is scalar, namely $\sum_i x_i \sum_j B_{ij} x_j = \sum_i \sum_j B_{ij} x_i x_j$. This can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sum_i B_{ii} x_i^2 + \sum_{i&amp;lt;j} x_i x_j ( B_{ij} + B_{ji})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any index $i$, the derivative with respect to $i$ becomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 B_{ii} x_i + \sum_{j \neq i} = x_j ( B_{ij} + B_{ji})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for any $i$, the above term is just the i-th component of the vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(B + B^T)x$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-17T21:03:39.837" />
  <row Id="10753" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:07:14.717" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, you are doing the right thing, assuming you meant $n^2 / \sqrt{n+n^2} &amp;lt; X$ rather than $1 / \sqrt{n+n^2} &amp;lt; X$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-17T21:07:14.717" CommentCount="1" />
  <row Id="10754" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:13:43.063" Score="0" ViewCount="0" Body="&lt;p&gt;Each term is bounded from below by the last term of your sum. Therefore a lower bound for the sum is $n^2$ times the last term. Show that this lower bound diverges as $n$ goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you only really need a lower bound. Don't bother with an upper bound.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-17T21:13:43.063" />
  <row Id="10755" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:18:01.780" Score="2" ViewCount="0" Body="&lt;p&gt;In proving something diverges to $\infty$, you don't need to squeeze it both from below &lt;em&gt;and&lt;/em&gt; from above: it is enough to &quot;half-squeeze&quot; it from below by something that goes to $\infty$ (because the limit you want is getting &quot;pushed up&quot; to $\infty$). So here, you can just note that each summand is greater than or equal to $\frac{1}{\sqrt{n+n^2}}$, so the sum of $n^2$ of them is at least as large as $\frac{n^2}{\sqrt{n+n^2}}$, and since $\lim\limits_{n\to\infty}\frac{n^2}{\sqrt{n+n^2}}=\infty$ and $\frac{n^2}{n+n^2}\leq X(n)$ for each $n$ (notice that what you called $X$ actually &lt;em&gt;varies&lt;/em&gt; with each $n$, so you should not give it a name that makes it look like it is fixed), then $\lim\limits_{n\to\infty}X(n)=\infty$ as well.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-17T21:18:01.780" CommentCount="3" />
  <row Id="10756" PostTypeId="2" ParentId="10744" CreationDate="2010-11-17T21:25:00.523" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a somewhat simplified view of things to get you started (one can avoid&#xA;all of the un-necessary elementwise notation down there, but it seems that to&#xA;help you start, it might still prove to be useful):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $x \in \mathbb{R}^n$. Further assume the convention that $x$ is a&#xA;&lt;strong&gt;column vector&lt;/strong&gt; so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \begin{bmatrix} x_1\\\\ x_2\\\\ \vdots\\\\ x_n&#xD;&#xA;\end{bmatrix}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that we have some function $f: \mathbb{R}^n \to \mathbb{R}$ (i.e., a&#xA;&lt;strong&gt;scalar&lt;/strong&gt; valued function of the vector $x$). We &lt;em&gt;define&lt;/em&gt; the partial&#xA;derivatives of $f(x)$ as the vector&#xA;$$\frac{\partial f(x)}{\partial x} = &#xD;&#xA;\begin{bmatrix}&#xD;&#xA;  \frac{\partial f(x)}{\partial x_1}\\\\&#xD;&#xA;  \frac{\partial f(x)}{\partial x_2}\\\\&#xD;&#xA;  \vdots\\\\&#xD;&#xA;  \frac{\partial f(x)}{\partial x_n}&#xD;&#xA;\end{bmatrix}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us look at your function $f(x) = x^TBx$. To simplify things for our&#xA;brute-force attack, we rewrite this is:&#xA;$$f(x) = \sum_{ij} x_iB_{ij}x_j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us try to compute the partial derivative wrt the $p$-th coordinate of&#xA;$x$, i.e., $\partial f(x) / \partial x_p$. This is given by&#xA;$$\frac{\partial f(x)}{\partial x_p} = \frac{\partial}{\partial x_p}\sum_{ij}&#xD;&#xA;x_i B_{ij} x_j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \sum_{j} B_{pj}x_j + \sum_i x_iB_{ip}$$&#xD;&#xA;using the product rule for differentiation. But recognize that &#xA;$$\sum_i B_{ip}x_i = \sum_i B^T_{pi}x_i = (B^Tx)_p$$&#xD;&#xA;while&#xA;$$\sum_j B_{pj}x_j = (Bx)_p$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we conclude that $$\frac{\partial f(x)}{\partial x_p} = (Bx)_p +&#xD;&#xA;(B^Tx)_p$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collecting derivatives for $p=1,\ldots,n$ yields the desired identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can follow this approach to obtain derivatives for other functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point that might help you is this: If $f$ is a function as defined&#xA;above, i.e., $f$ is a function of a column-vector $x$, then $\partial f&#xD;&#xA;/ \partial x$ is a &lt;strong&gt;column-vector&lt;/strong&gt;, while $\partial f / \partial x^T$ is a&#xA;&lt;strong&gt;row-vector&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please have a look at the matrix-calculus book by Magnus that I have cited in&#xA;this answer:&lt;a href=&quot;http://math.stackexchange.com/questions/9891/reference-for-matrix-calculus/9988#9988&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-17T21:25:00.523" CommentCount="1" />
  <row Id="10757" PostTypeId="2" ParentId="10744" CreationDate="2010-11-17T21:25:17.760" Score="0" ViewCount="0" Body="&lt;p&gt;You're right about suspecting that $\frac{ \partial \mathbf x^T }{\partial \mathbf x} \neq I$ because the dimensions no longer match up. I think you might be wading into tensor territory with such operations.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-17T21:25:17.760" />
  <row Id="10758" PostTypeId="2" ParentId="10745" CreationDate="2010-11-17T21:25:23.190" Score="1" ViewCount="0" Body="&lt;p&gt;First  $\frac{1}{\sqrt{n+n^{2}}} \rightarrow 0 $ if $n\rightarrow \infty $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore, in the infinite case the squeeze principle used so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $a_{n}\leq b_{n}$ and $a_{n}\rightarrow \infty$ then $b_{n}\rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you sequence i do not understand, &lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-17T21:25:23.190" />
  <row Id="10759" PostTypeId="2" ParentId="10741" CreationDate="2010-11-17T21:58:25.237" Score="15" ViewCount="0" Body="&lt;p&gt;You could view a kaleidoscope as a picture of a pattern in the quotient of the plane by a suitable group action. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Z15gn.jpg&quot; alt=&quot;pic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, some Escher work is a picture of life inside hyperbolic surfaces -- quotients of the hyperbolic plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/rFcU4.jpg&quot; alt=&quot;Escher&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the more extreme end, $\mathbb R^2$ is a quotient of $\mathbb R$, so presumably you have a stricter question in mind than the one you've actually written, as you can get all kinds of things as quotients. $SO_3$ (rotations in 3-space) are a quotient of the unit quaternions.  This is used in computer graphics, among other things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the kaleidoscope, you're really viewing the quotient as an orbifold with its natural geometric structure, rather than just as a topological space. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-17T21:58:25.237" CommentCount="3" />
  <row Id="10760" PostTypeId="1" AcceptedAnswerId="10803" CreationDate="2010-11-17T21:59:39.963" Score="1" ViewCount="136" Body="&lt;p&gt;I was looking for alternate proofs of the theorem that &quot;a vector space $V$ of dimension greater than $1$ over an infinite field is not a union of fewer than $|\mathbf{F}|$ proper subspaces&quot; and possible generalizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple measure-theoretic proof over $\mathbb{R}$ is as follows: By countable additivity the sum of the measures of any collection of subspaces is zero since the measure of each subspace is zero, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to look at proofs over arbitrary infinite fields and would like to know if similar statements hold for say modules (finitely generated or otherwise) over infinite rings.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-17T22:35:45.317" LastActivityDate="2010-11-18T05:43:06.063" Title="A vector space over $R$ is not a countable union of proper subspaces" Tags="&lt;real-analysis&gt;&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="10761" PostTypeId="1" CreationDate="2010-11-17T22:07:15.927" Score="3" ViewCount="81" Body="&lt;p&gt;Let $\mu (r)&amp;gt;2$ be the &lt;a href=&quot;http://mathworld.wolfram.com/IrrationalityMeasure.html&quot; rel=&quot;nofollow&quot;&gt;irrationality measure&lt;/a&gt; of a transcendental number $r$, and consider the following set of points $P \in\mathbb{R}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=\{r\in \mathbb{R}: \mu(r)=Constant\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this set a fractal, and If so, then what is it's dimension?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-17T22:12:18.343" LastActivityDate="2010-11-17T22:12:18.343" Title="Sets of Constant Irrationality Measure" Tags="&lt;fractals&gt;&lt;transcendental-numbers&gt;" CommentCount="10" FavoriteCount="2" />
  <row Id="10762" PostTypeId="2" ParentId="10760" CreationDate="2010-11-17T22:08:19.860" Score="1" ViewCount="0" Body="&lt;p&gt;Have a look here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-17T22:08:19.860" CommunityOwnedDate="2010-11-17T22:08:19.860" CommentCount="1" />
  <row Id="10763" PostTypeId="1" AcceptedAnswerId="10786" CreationDate="2010-11-17T22:16:59.413" Score="1" ViewCount="74" Body="&lt;p&gt;A commutative ring $R$ with unity is Hermite if for all $x,y\in R$ there exists $t,u,v\in R$ such that $x=tu$, $y=tv$ and $(u,v)=(1)$. Is there a finite commutative ring with unity that is not Hermite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This characterisation is taken from theorem 3 of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Remarks About Elementary Divisor Rings, Leonard Gillman and Melvin Henriksen, Transactions of the American Mathematical Society, Vol. 82, No. 2 (Jul., 1956), pp. 362-365 &lt;/p&gt;&#xA;" OwnerUserId="3643" LastEditorUserId="3643" LastEditorDisplayName="" LastEditDate="2010-12-20T14:06:24.047" LastActivityDate="2010-12-20T14:06:24.047" Title="Is there a finite non-Hermite commutative ring with unity?" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10764" PostTypeId="1" AcceptedAnswerId="10765" CreationDate="2010-11-17T22:26:49.663" Score="1" ViewCount="80" Body="&lt;p&gt;Let $f(x)$  be a Lipschitz  real-valued function defined on a closed interval $I$. The derivative $f &amp;#39;(x)$ exists a.e. since $f$ is absolutely continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is $f &amp;#39;(x)$ necessarily square integrable, i.e. in $L^2(I)$ ?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-17T22:55:28.290" Title="Is the derivative of a Lipschitz function square integrable?" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="10765" PostTypeId="2" ParentId="10764" CreationDate="2010-11-17T22:29:03.633" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, because the derivative is bounded.  If $M$ is the Lipschitz constant, then $|f&amp;#39;(x)|\leq M$ everywhere it exists.  This follows from the fact that each of the difference quotients has this bound.  Lipschitz continuous functions on $I$ are precisely the indefinite integrals of bounded measurable functions on $I$, whereas AC functions are the indefinite integrals of integrable functions on $I$.  (Thus you find AC counterexamples to this by integrating integrable but not square integrable functions.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-17T22:55:28.290" LastActivityDate="2010-11-17T22:55:28.290" />
  <row Id="10766" PostTypeId="1" AcceptedAnswerId="10775" CreationDate="2010-11-17T22:33:16.037" Score="0" ViewCount="74" Body="&lt;p&gt;How does one show that any open interval in n-dimensional Euclidean space is connected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is homework, and I have been stuck on it for a few hours now. Seems like it should be easy, but I can't get a proof that I am comfortable with.&lt;/p&gt;&#xA;" OwnerUserId="3644" LastActivityDate="2010-11-17T23:28:55.937" Title="Any open interval in an n-dimensional Euclidean space is connected" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="10767" PostTypeId="2" ParentId="10741" CreationDate="2010-11-17T22:49:42.913" Score="8" ViewCount="0" Body="&lt;p&gt;Many role-playing games have toroidal worlds. Especially Dragon Quest and Final Fantasy games. It's a pity that few go through the trouble of implementing a true spheric topology. I don't mind other topologies if they fit the game world, but in RPGs it's mostly understood that the world should be Earth-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, there is also this marvelous site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.geometrygames.org/HyperbolicGames/&quot; rel=&quot;nofollow&quot;&gt;http://www.geometrygames.org/HyperbolicGames/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-17T22:49:42.913" />
  <row Id="10768" PostTypeId="1" CreationDate="2010-11-17T22:59:56.757" Score="0" ViewCount="73" Body="&lt;p&gt;Suppose $f$ is a real-valued function and $f(x) \geq 1$ for every real $x \in [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why we can always find a unique positive integer $n$ such that $2^{n} \leq f(x) &amp;lt; 2^{n+1}$ ?&lt;/p&gt;&#xA;" OwnerUserId="3645" LastActivityDate="2010-11-17T23:20:38.910" Title="Existence of integer" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10769" PostTypeId="2" ParentId="10726" CreationDate="2010-11-17T23:03:33.783" Score="3" ViewCount="0" Body="&lt;p&gt;Here's an answer to the second part of Problem 1: $$k = \frac{(a+1)(b+1) - \gcd(a,b) + 3}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to your definition,&#xA;$$k = 1 + (a+1) + \sum_{j=1}^{b-1} \left\lceil \frac{ja}{b} \right\rceil.$$ &#xA;Now, $\left\lceil \frac{ja}{b} \right\rceil = \left\lfloor \frac{ja + b - 1}{b} \right\rfloor.$  (See, for example, Exercise 3.12, p.96, in &lt;em&gt;Concrete Mathematics&lt;/em&gt;, 2nd edition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that $$\sum_{j=0}^{b-1} \left\lfloor \frac{aj + x}{b} \right\rfloor = d \left\lfloor \frac{x}{d} \right\rfloor + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$&#xD;&#xA;for any real number $x$, where $d = \gcd(a,b)$.  (See, for example, Eq. 3.32, p. 94, of &lt;em&gt;Concrete Mathematics&lt;/em&gt;, 2nd edition.)  We have $x = b-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$d \left\lfloor \frac{b-1}{d} \right\rfloor = d\frac{b-1}{d} - d \left\{\frac{b-1}{d}\right\} = b-1 - d \left\{\frac{b}{d} - \frac{1}{d}\right\} = b - 1  - d \frac{d-1}{d} = b-d,$$&#xD;&#xA;where $\{x\}$ denotes the fractional part of $x$.  (The second-to-last step holds because $d | b$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting everything together yields $$k = a + 2 + b - d + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$&#xD;&#xA;and then combining terms gets the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements in $f(a,b)$ are closely related to Spec$(\frac{a}{b}) = \{\lfloor \frac{a}{b}\rfloor, \lfloor\frac{2a}{b}\rfloor,  \lfloor\frac{3a}{b}\rfloor, \ldots \}$, although you want ceilings, and you want to stop when the multiplier is $b-1$.  You may find the discussion of Spec$(\alpha)$ in &lt;em&gt;Concrete Mathematics&lt;/em&gt; helpful.  (In fact, when I looked at your problem I thought, &quot;This reminds me of the discussion of spectra in &lt;em&gt;Concrete Math&lt;/em&gt;,&quot; and then using results in there it wasn't too difficult to find the expression for $k$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-17T23:03:33.783" />
  <row Id="10770" PostTypeId="2" ParentId="10768" CreationDate="2010-11-17T23:09:00.267" Score="0" ViewCount="0" Body="&lt;p&gt;You can't do it for $f(x)=3x+1$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-17T23:09:00.267" CommentCount="2" />
  <row Id="10771" PostTypeId="2" ParentId="10768" CreationDate="2010-11-17T23:20:38.910" Score="1" ViewCount="0" Body="&lt;p&gt;The intervals $[2^n,2^{n+1})$ cover the entire real half (existence) line $y\geq 1$ and no two of them intersect (uniqueness). So any point on the graph of the function must intersect one of these since $f(x)\geq 1$. You don't need any hypothesis on the domain of the function. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-17T23:20:38.910" />
  <row Id="10772" PostTypeId="1" AcceptedAnswerId="10777" CreationDate="2010-11-17T23:27:14.380" Score="1" ViewCount="60" Body="&lt;p&gt;I'm reading Number Theory I, by Kato, et al.  The authors claim that for $x$, $y\in \mathbb{Q}$ and $m$ a square free positive integer, if $2x$, $x^2-my^2 \in \mathbb{Z}$, then for any odd prime $p$ $\mathrm{ord}_p(m)+2\mathrm{ord}_p(y) \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get that $\mathrm{ord}_p(my^2)=\mathrm{ord}_p(m)+2\mathrm{ord}_p(y)$ and $\mathrm{ord}_p(x)\geq 0$.  I'm just not sure that I can say that $\mathrm{ord}_p(x^2-my^2)=\min\{\mathrm{ord}_p(x^2), \mathrm{ord}_p(my^2)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a bit of trouble seeing this.  Does anyone know what to do here?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T00:03:52.490" LastActivityDate="2010-11-18T00:10:42.903" Title="Proof for a fact involving $\mathrm{ord}_p(x)$ " Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10773" PostTypeId="2" ParentId="10766" CreationDate="2010-11-17T23:27:43.843" Score="0" ViewCount="0" Body="&lt;p&gt;Can you show that an interval is connected in $\mathbb{R}$?  Then can you say (a,b) is connected to (c,b) is connected to (c,d)?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-17T23:27:43.843" />
  <row Id="10774" PostTypeId="1" CreationDate="2010-11-17T23:27:53.927" Score="4" ViewCount="134" Body="&lt;p&gt;In Andy's answer to the question &quot;&lt;a href=&quot;http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform&quot; rel=&quot;nofollow&quot;&gt;What are fixed points of the Fourier Transform&lt;/a&gt;&quot; on Math Overflow, he shows that the Fourier Transform has eigenvalues {+1, +i, -1, -i} and that the projections of any function onto the corresponding four eigenspaces may be found through some simple linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to get a better feeling for these four eigenspaces of the fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I find some interesting members of each of these eigenspaces?&lt;/li&gt;&#xA;&lt;li&gt;How can I show that Hermite-Gaussians are in one (or more?) of the eigenspaces?&lt;/li&gt;&#xA;&lt;li&gt;How can one define usable projection operators onto these eigenspaces?&lt;/li&gt;&#xA;&lt;li&gt;The wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#Eigenfunctions&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; on the Fourier Transform mentions that Wiener defined the Fourier Transform via these projections.  What exactly was Wiener's approach?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2010-11-18T12:33:56.453" Title="How do I compute the eigenfunctions of the Fourier Transform?" Tags="&lt;linear-algebra&gt;&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="10775" PostTypeId="2" ParentId="10766" CreationDate="2010-11-17T23:28:55.937" Score="3" ViewCount="0" Body="&lt;p&gt;OK. This is a basic result which states that &lt;em&gt;Cartesian&lt;/em&gt; product of connected sets are connected. You may look here for the proof:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://planetmath.org/encyclopedia/ProofOfProductsOfConnectedSpaces.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/ProofOfProductsOfConnectedSpaces.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-17T23:28:55.937" />
  <row Id="10777" PostTypeId="2" ParentId="10772" CreationDate="2010-11-17T23:47:44.087" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ 2x\in \mathbb Z\ \Rightarrow\ ord_p(x)\ge 0\ \ $  so $\rm\ ord_p (x^2-my^2) \ge 0\ \Rightarrow\ ord_p(my^2) \ge 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall the &lt;strong&gt;principle of domination:&lt;/strong&gt;  $\rm\ ord_p\ Y &amp;lt; ord_p\ X\ \Rightarrow\ ord_p\:(X+Y) = ord_p\ Y\:$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T00:10:42.903" LastActivityDate="2010-11-18T00:10:42.903" CommentCount="4" />
  <row Id="10778" PostTypeId="1" CreationDate="2010-11-17T23:47:52.373" Score="0" ViewCount="60" Body="&lt;p&gt;Now, suppose the transformation(in 2d) I am working with has two&lt;br&gt;&#xA;separate functions&#xA;for x and y.&lt;br&gt;&#xA;That is, the transformation for x is of the form&lt;br&gt;&#xA;x'=x/(x+y)&lt;br&gt;&#xA;and the transformation of y is&lt;br&gt;&#xA;y'=y/(x+y)&lt;br&gt;&#xA;Each is an LFT(The schwarzian derivatives are 0) but is&lt;br&gt;&#xA;the transform as a whole still considered an LFT?&lt;/p&gt;&#xA;" OwnerUserId="3646" LastActivityDate="2010-11-17T23:47:52.373" Title="is this a linear fractional transformation(LFT)?" Tags="&lt;transformation&gt;" CommentCount="4" />
  <row Id="10779" PostTypeId="1" AcceptedAnswerId="10789" CreationDate="2010-11-18T00:04:35.267" Score="2" ViewCount="108" Body="&lt;p&gt;I want to prove that for non-negative random variables with distribution F:&#xA;$$E(X^{n}) = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following proof correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R.H.S = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx = \int_0^\infty n x^{n-1} (1-F(x)) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using integration by parts:&#xA;$$R.H.S = [x^{n}(1-F(x))]_0^\infty + \int_0^\infty x^{n} f(x) dx = 0 + \int_0^\infty x^{n} f(x) dx = E(X^{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not correct, then how to prove it?&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="953" LastEditorDisplayName="" LastEditDate="2010-11-18T00:20:13.273" LastActivityDate="2010-11-18T01:15:17.577" Title="Moments and non-negative random variables?" Tags="&lt;probability&gt;" AnswerCount="5" />
  <row Id="10780" PostTypeId="1" AcceptedAnswerId="10790" CreationDate="2010-11-18T00:11:40.313" Score="4" ViewCount="167" Body="&lt;p&gt;Let $f$ be the $2\pi$ periodic function which is the even extension of $$x^{1/n}, 0 \le x \le \pi,$$ where $n \ge 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a general theorem that implies that the Fourier series of $f$ converges to $f$, pointwise, uniformly or absolutely. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-18T13:35:32.977" LastActivityDate="2010-11-18T13:35:32.977" Title="Convergence of a Fourier series " Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="10781" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:13:16.117" Score="0" ViewCount="0" Body="&lt;p&gt;Only if $\lim_{x\rightarrow \infty} x^n(1-F(x)) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't look like that's true in general.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T00:13:16.117" CommentCount="0" />
  <row Id="10782" PostTypeId="1" AcceptedAnswerId="10785" CreationDate="2010-11-18T00:14:01.280" Score="4" ViewCount="111" Body="&lt;p&gt;Here is how I have worked it out so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log _2(x-4)+\log(x+2)=4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log _2((x-4)(x+2)) = 4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-4)(x+2)=2^4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x-4)(x+2)=16$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I proceed from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+2x-8 = 16$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+2x = 24$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(x+2) = 24$ Which I know is not the right answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2+2x-24 = 0$ Can't factor this&lt;/p&gt;&#xA;" OwnerUserId="3647" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T00:47:26.460" LastActivityDate="2010-11-18T00:47:26.460" Title="Solving $\log _2(x-4) + \log _2(x+2) = 4$" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10783" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:16:53.040" Score="0" ViewCount="0" Body="&lt;p&gt;As a matter of notation, you should write the expectation as $E(X^n)$ instead of $E(x^n)$. The expectation is not a function of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proof would work if $\lim_{x \to \infty} x^n (1-F(x)) = 0$. I'm not sure this is true for any distribution.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-18T00:16:53.040" CommentCount="1" />
  <row Id="10784" PostTypeId="2" ParentId="10782" CreationDate="2010-11-18T00:19:28.977" Score="1" ViewCount="0" Body="&lt;p&gt;After $(x-4)(x+2)=16$, you get $x^2-2x-24=0$ (the coefficient of $x$ is $-2$ not $2$). So $x=\frac{2\pm \sqrt{100}}{2}$ by the quadratic formula. So $x=6$ or $x=-4$  &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-18T00:28:27.017" LastActivityDate="2010-11-18T00:28:27.017" CommentCount="3" />
  <row Id="10785" PostTypeId="2" ParentId="10782" CreationDate="2010-11-18T00:19:38.557" Score="5" ViewCount="0" Body="&lt;p&gt;It is $x^2-2x-8 = 16$ my friend. So you get $x^2 - 2x -24 = 0$, which factors as $(x-6)(x+4) = 0$. Hence, $x=6$ or $x = -4$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-18T00:19:38.557" CommentCount="2" />
  <row Id="10786" PostTypeId="2" ParentId="10763" CreationDate="2010-11-18T00:19:54.493" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, $\mathbb{F}_2[x,y]/(x,y)^2$.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-18T00:19:54.493" CommentCount="2" />
  <row Id="10787" PostTypeId="2" ParentId="10782" CreationDate="2010-11-18T00:31:35.470" Score="0" ViewCount="0" Body="&lt;p&gt;The quadratic formula:&#xA;$$ x = \frac{ -b \pm \sqrt{b^2 - 4 a c} }{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-24) }}{2 \times 1} $$&#xD;&#xA;4 + 96 in the square root.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T00:31:35.470" />
  <row Id="10788" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:45:01.177" Score="3" ViewCount="0" Body="&lt;p&gt;A better way to prove it would be to use Fubini's theorem to change the order of integration. This also gives you a condition when the result you have is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $I = \displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact that $\displaystyle \int_x^{\infty} f(y)dy = 1 - F(x)$, we get $I = \displaystyle \int_0^{\infty}n x^{n-1} \int_x^{\infty} f(y) dy dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we first integrate with respect to $y$ (the inner integral) and $y$ goes from $x$ to $\infty$ and then integrate with respect to $x$ (the outer integral), $x$ goes from $0$ to $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change the order of integration. i.e. integrate with respect to $x$ first and then with respect to $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this can be done provided the integral $I &amp;lt; \infty$ (See Fubini's theorem). This is the condition svenkatr and trutheality get as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the order of integration, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \displaystyle \int_0^{\infty} \displaystyle \int_{0}^{y} nx^{n-1}f(y)dxdy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that now $x$ in the inner integral goes from $0$ to $y$ and $y$ goes from $0$ to $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the inner integral with respect to $x$ can performed easily and now we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \displaystyle \int_0^{\infty} y^{n}f(y)dy = E[X^n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have $\displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx = E[X^n]$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-18T00:50:25.087" LastActivityDate="2010-11-18T00:50:25.087" />
  <row Id="10789" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T00:53:05.457" Score="3" ViewCount="0" Body="&lt;p&gt;Here's another way.  (As the others point out, the statement is true if $E[X^n]$ actually exists.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y = X^n$.  $Y$ is non-negative if $X$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know &#xA;$$E[Y] = \int_0^{\infty} P(Y \geq t) dt,$$&#xD;&#xA;so&#xA;$$E[X^n] = \int_0^{\infty} P(X^n \geq t) dt.$$&#xD;&#xA;Then, perform the change of variables $t = x^n$.  This immediately yields &#xA;$$E[X^n] = \int_0^{\infty} n x^{n-1} P(X^n \geq x^n) dx = \int_0^{\infty} n x^{n-1} P(X \geq x) dx.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-18T00:53:05.457" />
  <row Id="10790" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T01:06:43.587" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps you can apply the one found here: &lt;a href=&quot;http://books.google.com/books?id=XqqNDQeLfAkC&amp;amp;pg=PA84&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=XqqNDQeLfAkC&amp;amp;pg=PA84&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snapshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ySQ0q.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T01:06:43.587" CommentCount="5" />
  <row Id="10791" PostTypeId="2" ParentId="9851" CreationDate="2010-11-18T01:09:28.220" Score="3" ViewCount="0" Body="&lt;p&gt;Do the &lt;a href=&quot;http://digitus.itk.ppke.hu/~matyi/optika/Phase_Diversity/AO82_PRComparison1.pdf&quot; rel=&quot;nofollow&quot;&gt;usual numerical phase retrieval algorithms&lt;/a&gt; not work?&lt;/p&gt;&#xA;" OwnerUserId="2191" LastActivityDate="2010-11-18T01:09:28.220" CommentCount="1" />
  <row Id="10792" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T01:13:18.283" Score="2" ViewCount="0" Body="&lt;p&gt;I looked it up on Wikipedia. Assuming that the articles are correct, it seems that your function satisfies some &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%B6lder_condition&quot; rel=&quot;nofollow&quot;&gt;Hölder condition&lt;/a&gt; and thus by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Pointwise_convergence&quot; rel=&quot;nofollow&quot;&gt;Dirichlet-Dini chriterion&lt;/a&gt; its Fourier series converges pointwise to &lt;em&gt;f&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also show that your function is Lipschitz and then you'd have &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Absolute_convergence&quot; rel=&quot;nofollow&quot;&gt;absolute convergence&lt;/a&gt; of its Fourier series.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-18T01:13:18.283" CommentCount="1" />
  <row Id="10793" PostTypeId="2" ParentId="10779" CreationDate="2010-11-18T01:15:17.577" Score="2" ViewCount="0" Body="&lt;p&gt;If ${\rm E}(X^n) &amp;lt; \infty$, then $\int_0^\infty  {x^n f(x){\rm d}x}  &amp;lt; \infty $, and in turn, $\int_M^\infty  {x^n f(x){\rm d}x}  \to 0$ as $M \to \infty$. Since $\int_M^\infty  {x^n f(x){\rm d}x} \geq \int_M^\infty  {M^n f(x){\rm d}x}  = M^n [1 - F(M)]$, we have that $M^n [1 - F(M)] \to 0$ as $M \to \infty$. Hence, your solution is correct (assuming that ${\rm E}(X^n) &amp;lt; \infty$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-18T01:15:17.577" CommentCount="1" />
  <row Id="10794" PostTypeId="2" ParentId="10209" CreationDate="2010-11-18T01:25:01.640" Score="1" ViewCount="0" Body="&lt;p&gt;Mathematics - A Discrete Introduction  by Edward R. Scheinerman is the best book for learning discrete math that I have come across. &#xA;Kenneth Rosen's book blows; sorry, but it's true - I had to suffer through it back in college. It wasn't fun. &#xA;Scheinerman's book made discrete math interesting again. I think it was the uncluttered presentation and clear explanations.&lt;/p&gt;&#xA;" OwnerUserId="3649" LastActivityDate="2010-11-18T01:25:01.640" />
  <row Id="10795" PostTypeId="2" ParentId="10078" CreationDate="2010-11-18T01:31:52.727" Score="1" ViewCount="0" Body="&lt;p&gt;As long as the sequence doesn't fall outside of the support of the function, it is a sequence that could be generated by the function (the all-heads sequence for a fair coin toss is a good example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can try to do is minimize the distance b/w the empirical CDF of the data and the CDF of your distribution.&#xA;You have to decide what distance metric to use (maximum absolute difference is one obvious choice) and once you have that it boils down to solving an optimization problem with each new data point added.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T01:31:52.727" />
  <row Id="10796" PostTypeId="1" AcceptedAnswerId="10798" CreationDate="2010-11-18T01:32:52.687" Score="0" ViewCount="45" Body="&lt;p&gt;The number of dropped connections per call follows a Poisson distribution. From four calls, the number of dropped connections is 2, 0 , 3, 1.&lt;br&gt;&#xA;Obtain maximum likelihood estimate that the next two calls will be completed without any accidental drops.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the maximum likelihood estimate of lamda is 1.5.&#xA;I think I am suppose to use this formula:&#xA;$$f(x|\lambda) = \frac{\lambda^x e^{-\lambda}}{x!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not know what x should be.  The answer should be 0.0498&lt;/p&gt;&#xA;" OwnerUserId="3601" LastEditorUserId="3601" LastEditorDisplayName="" LastEditDate="2010-11-18T01:39:23.010" LastActivityDate="2010-11-18T01:39:23.010" Title="Obtain maximum likelihood estimate" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10797" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T01:36:38.410" Score="2" ViewCount="0" Body="&lt;p&gt;You can actually do your example pretty explicitly. For $f(x) = |x|^{\alpha}$ you actually have that $\hat{f}(n) = c_{\alpha} n^{-1 - \alpha}  + O(n^{-2})$, so the cosine series is absolutely convergent. If a Fourier series of a continuous function converges, it has to converge to the original function; I refer you to a Fourier analysis text for this fact though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the above expression for $\hat{f}(n)$, recall that $\hat{f}(n)$ is defined as&#xA;$\hat{f}(n) = 2\int_0^1x^{\alpha} \cos(n\pi x)dx$. Integrating by parts, this is the same as ${2 \alpha \over \pi}n^{-1}\int_0^1x^{\alpha - 1} \sin(n\pi x)dx$. This in turn can be written as ${2 \alpha \over \pi}n^{-1} \int_0^\infty x^{\alpha - 1} \sin(n\pi x)dx - &#xD;&#xA;{2 \alpha \over \pi}n^{-1} \int_1^\infty x^{\alpha - 1} \sin(n\pi x)dx$. (These integrals are convergent improper integrals). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By changing variables $x$ to $nx$ the first integral becomes the main term $c_{\alpha}n^{-1 - \alpha}$. By doing one more integration by parts and then taking absolute values, the second term is bounded by $C n^{-2}$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-18T01:36:38.410" />
  <row Id="10798" PostTypeId="2" ParentId="10796" CreationDate="2010-11-18T01:36:54.530" Score="1" ViewCount="0" Body="&lt;p&gt;x is the number of dropped connections, right? So $x=0$ for no dropped connections.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-18T01:36:54.530" CommentCount="3" />
  <row Id="10800" PostTypeId="2" ParentId="10741" CreationDate="2010-11-18T02:22:38.327" Score="2" ViewCount="0" Body="&lt;p&gt;A running track is a good one-dimensional example.  We start with the real line and identify all points {x + n*400meter}, or equivalently we take a 400meter long line segment and glue the ends together.   Now, we can sit in the stands and watch an entire 800m, 1600m, or 3200m race, because we have a nice &quot;compact&quot; track!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you explain how tiling the real line with 400m segments leads to a running track, it may be easier to explain how two-dimensional tilings of the plane (or sphere or hyperbolic plane) can lead to quotient spaces.  Of course, we need to know how to glue the edges, and that point may be a little vague in a casual conversation-- but if Asteroids is not available, I think that tilings are one of the best ways to visualize quotient spaces.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-18T02:22:38.327" />
  <row Id="10801" PostTypeId="1" AcceptedAnswerId="10802" CreationDate="2010-11-18T02:29:52.660" Score="2" ViewCount="97" Body="&lt;p&gt;You are dealt five cards from 52 card deck that is thoroughly shuffled. What is the probability that two out of five cards are seven?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it:&#xA;$${4C2 . 12C3 . (4C1)^3} / 52C5$$&lt;/p&gt;&#xA;" OwnerUserId="953" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-11-18T05:43:05.843" LastActivityDate="2010-11-18T05:43:05.843" Title="Probability that two out of five cards are seven?" Tags="&lt;probability&gt;&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10802" PostTypeId="2" ParentId="10801" CreationDate="2010-11-18T02:33:00.333" Score="6" ViewCount="0" Body="&lt;p&gt;Kindly consult the Wikipedia page on Hypergeometric distribution. The correct formula is $$\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}.$$&#xD;&#xA;So $\binom{12}{3} \binom{4}{1}^3$ is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-18T02:33:00.333" />
  <row Id="10803" PostTypeId="2" ParentId="10760" CreationDate="2010-11-18T02:59:55.177" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a proof of the finite dimensional case over any field (finite or infinite):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $\mathbf{V}$ be a nonzero finite dimensional vector space over $\mathbf{F}$. If $\mathbf{V}$ is a union of $\kappa$ proper subspaces, then $\kappa\geq|\mathbf{F}|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Write $\mathbf{V}=\bigcup\limits_{k\in\kappa} W_{k}$, with $W_k$ a proper subspace of $\mathbf{V}$. By enlarging the $W_k$ as necessary, we may assume that $\dim(\mathbf{V}/W_i) = 1$ for all $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is by induction on $\dim(\mathbf{V})$. The result is trivially true if $\dim(\mathbf{V})=1$, because $\mathbf{V}$ is never the union of proper subspaces in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of $\dim(\mathbf{V})=2$, let $\{w_1\}$ be a basis for $W_1$, and let $v\notin W_1$. For each $\alpha\in \mathbf{F}$ there exists $j_{\alpha}\in\kappa$ such that $w_1+\alpha v\in W_{j_{\alpha}}$. Moreover, if $\alpha\neq\beta$, then $w_1+\alpha v$ and $w_1+\beta v$ are linearly independent, since $\{w_1,v\}$ are a basis for $\mathbf{V}$. Thus, no $W_k$ contains more than one $w_1+\alpha v$. This gives an injection from $\mathbf{F}$ to $\kappa$, proving that $\kappa\geq|\mathbf{F}|$, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the result holds for $n$-dimensional vector spaces, and let $\mathbf{V}$ be $(n+1)$-dimensional. Let $\{w_1,\ldots,w_n\}$ be a basis for $W_1$, and $v\notin W_1$. For each $\alpha\in\mathbf{F}$, consider the subspace $W_{\alpha}=\mathrm{span}(w_1+\alpha v,w_2,\ldots,w_n)$. If $W_{\alpha}$ is contained in some $W_k$, then $W_{\alpha}=W_k$ by dimension considerations; and if $W_{\alpha}=W_{\beta}$, then $\alpha=\beta$, for otherwise we would be able to find a nontrivial linear combination involving $w_1,\ldots,w_n,v$. So there is again an injection from the set&#xA;$$S=\{\alpha\in\mathbf{F} \mid W_{\alpha}=W_k\text{ for some }k\in \kappa\}$$&#xD;&#xA;to $\kappa$. If the set has cardinality $|\mathbf{F}|$ we are done. Otherwise, let $\alpha_0\in\mathbf{F}\setminus S$, and look at $W_{\alpha_0}$. For each $k\in \kappa$, $W_{\alpha_0}\cap W_{\kappa}\neq W_{\alpha_0}$; since&#xA;$$W_{\alpha_0} = W_{\alpha_0}\cap\mathbf{V} = W_{\alpha_0}\cap\left(\bigcup_{k\in\kappa}W_k\right) = \bigcup_{k\in\kappa}(W_{\alpha_0}\cap W_k),$$&#xD;&#xA;then by the induction hypothesis we have that $\kappa\geq|\mathbf{F}|$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;I suspect a similar argument can be made in the infinite dimensional case; certainly, we can construct the analogous set to $S$, and if $|S|=|\mathbf{F}|$ then we are done. But if not, then $\dim(W_{\alpha})=\dim(\mathbf{V})$, so we cannot really use a reduction argument. But I think there may be a way to tweak this.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Pete L. Clark has noted, the result does not hold in the infinite dimensional case.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T05:43:06.063" LastActivityDate="2010-11-18T05:43:06.063" CommentCount="1" />
  <row Id="10804" PostTypeId="2" ParentId="10209" CreationDate="2010-11-18T03:07:33.170" Score="1" ViewCount="0" Body="&lt;p&gt;It may be worth having a look at the following &lt;a href=&quot;http://hbpms.blogspot.com/2008/05/stage-1-introductory-discrete.html&quot; rel=&quot;nofollow&quot;&gt;blog&lt;/a&gt; for a selection of resources for introductory discrete mathematics resources, a lot of them free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking to buy a book, &lt;em&gt;Discrete Mathematics with Applications&lt;/em&gt; by Susanna S. Epp is particularly good. Plenty of exercises and examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really dislike Rosen's book. I don't find his examples instructive at all.&lt;/p&gt;&#xA;" OwnerUserId="3564" LastActivityDate="2010-11-18T03:07:33.170" CommentCount="1" />
  <row Id="10805" PostTypeId="2" ParentId="10731" CreationDate="2010-11-18T03:38:24.490" Score="1" ViewCount="0" Body="&lt;p&gt;Here's a remark added to original remarks below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned below, Hölder's inequality reduces this to showing that $\tau_1$ is finer than $\tau_\infty$.  This means showing that the $\tau_1$ to $\tau_\infty$ identity map  is continuous, while we already know that its inverse is continuous.  Hence if you prove that $\tau_1$ and $\tau_\infty$ are complete, this follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_mapping_theorem_%28functional_analysis%29#Generalizations&quot; rel=&quot;nofollow&quot;&gt;open mapping theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not sure this is worth posting, because I can only finish the easy case $n=1$, but here's a proof for that case from someone ignorant of Sobolev embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $1\leq p\leq q\leq\infty$, then because $K$ has finite measure, Hölder's inequality yields $|f|_{\alpha,p}\leq C|f|_{\alpha,q}$ for a constant $C$ depending only on $p$, $q$, and the measure of $K$.  Thus $\tau_q$ is finer than $\tau_p$, and the work is in showing that $\tau_1$ is finer than $\tau_\infty$.  Here, as Willie Wong pointed out, it is no longer possible to do so by directly comparing seminorms with the same multi-index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(f_k)$ is a sequence of functions such that $|f_k|_{\alpha,1}\to 0$ for all $\alpha$ as $k\to\infty$.  This implies that the same holds for the sequence $(D^{\alpha_0}f_k)$ for each fixed multi-index $\alpha_0$, so it is enough to check that $|f_k|_{0,\infty}\to0$, that is, $(f_k)$ converges uniformly to $0$.  Here's a way to see this when $n=1$.  Let $y$ be a point where all of the functions vanish, say $y=\inf K$ for definiteness.  For each $x\in K$, $|f_k(x)|=|f_k(x)-f_k(y)|=|\int_y^x f_k&amp;#39;(t)dt|\leq \int_K|f_k&amp;#39;(t)|dt=|f_k|_{1,1}$, which goes to $0$ independent of $x$.  Hence $(f_k)$ converges uniformly to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, with slightly more work the proof that $\tau_1$ is finer than $\tau_\infty$ for $n=1$ can be extended to the case of smooth functions with arbitrary support.  With the same assumptions on $(f_k)$, for each $x\in\mathbb{R}$, $|f_k(x)-f_k(0)|=|\int_0^x f_k&amp;#39;(t)dt|\leq \int|f_k&amp;#39;(t)|dt=|f_k|_{1,1}$, which goes to $0$ independent of $x$.  This means that there is a sequence $(\varepsilon_k)$ of positive numbers converging to $0$ such that each $f_k$ is within $\varepsilon_k$ of the constant function with value $f_k(0)$.  Since $|f_k|_{0,1}$ goes to $0$, the corresponding constants must converge to $0$, and this implies that $(f_k)$ converges uniformly to $0$.  As Willie Wong also pointed out, you don't get $\tau_q$ finer than $\tau_p$ for $q\gt p$ in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To extend this argument to $n\gt 1$, I would need bounds on $|f(x)-f(y)|$ in terms of volume integrals of sums of (a finite number of) partial derivatives of $f$, and this is where my ignorance keeps me from going further.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-18T19:54:53.957" LastActivityDate="2010-11-18T19:54:53.957" CommunityOwnedDate="2010-11-18T19:54:53.957" CommentCount="2" />
  <row Id="10806" PostTypeId="1" AcceptedAnswerId="10807" CreationDate="2010-11-18T04:14:49.887" Score="1" ViewCount="110" Body="&lt;p&gt;Hi,&#xA;I'm out of practice with algebra, and I'm having a mind blank on how to solve for $t$ in the following equation. It's for some collision detection if you're wondering.&#xA;$$\Bigl( \bigl(a-y-(qt)\bigr) \bigl(b-x-(rt)\bigr) \Bigr) - \Bigl( \bigl(c-x-(rt)\bigr) \bigl(d-y-(qt)\bigr) \Bigr) = z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really not seeing it at the moment. Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="3652" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-18T04:42:17.027" LastActivityDate="2010-11-18T04:42:17.027" Title="Solving a tricky equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10807" PostTypeId="2" ParentId="10806" CreationDate="2010-11-18T04:24:20.717" Score="5" ViewCount="0" Body="&lt;p&gt;You could denote the differences $(a-y)$, $(b-x)$, $(c-x)$, $(d-y)$ as a single symbol to simplify computation. That is, $(a-y)=A$, $(b-x)=B$, $(c-x)=C$, $(d-y)=D$. Then you have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(A-qt)(B-rt)-(C-rt)(D-qt)=z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(AB-qBt-rAt+qrt^2)-(CD-rDt-qCt+qrt^2)=z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the $qrt^2$ terms cancel upon distributing the negative, and then you have a linear equation for $t$, so $t$ will be easy to isolate, and then you can resubstitute the values for $A$, $B$, $C$, and $D$. I hope my algebra has been correct.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-18T04:24:20.717" CommentCount="2" />
  <row Id="10809" PostTypeId="2" ParentId="10741" CreationDate="2010-11-18T04:51:58.160" Score="4" ViewCount="0" Body="&lt;p&gt;The original Sonic the Hedgehog games (on the Sega Genesis and similar consoles) provide good examples. Many of the levels wrap around vertically, so that there is no bottom or top. They are equivalent to cylinders, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples from Sonic 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://info.sonicretro.org/File:Ic1map.PNG&quot; rel=&quot;nofollow&quot;&gt;http://info.sonicretro.org/File:Ic1map.PNG&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://info.sonicretro.org/File:Mg1map.PNG&quot; rel=&quot;nofollow&quot;&gt;http://info.sonicretro.org/File:Mg1map.PNG&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The designers used this to make the giant hill at the beginning of the ice cap level that is many times taller than the level itself. Also, the Sonic 3 Special stages are toroidal, just like Asteroids, though in the game they are presented with an apparent spherical curvature, which had me terribly lost and confused until I googled it. In hindsight, I should have realized that you can't cover a sphere with squares, but Sonic doesn't exactly give you a lot of time for thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://info.sonicretro.org/File:S%26KSS1.png&quot; rel=&quot;nofollow&quot;&gt;http://info.sonicretro.org/File:S%26KSS1.png&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=LpDbSlbhP5M&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=LpDbSlbhP5M&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="3653" LastEditorUserId="3653" LastEditorDisplayName="" LastEditDate="2010-11-18T05:10:31.957" LastActivityDate="2010-11-18T05:10:31.957" />
  <row Id="10810" PostTypeId="1" AcceptedAnswerId="10811" CreationDate="2010-11-18T04:57:07.050" Score="4" ViewCount="82" Body="&lt;p&gt;I'm trying to derive the integral form of the Bessel function by finding&#xA;the $k$th coefficient of the Laurent series expansion of the function&#xA;$f(z) =\exp [\lambda(z-\frac{1}{z})]$.&#xA;I managed to get it down to the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;J_k(\lambda) =\frac{1}{2\pi}\int_{0}^{2 \pi} e^{i(\lambda \sin\theta - k \theta)} d\theta&#xD;&#xA;=\frac{1}{2\pi}\int_{0}^{2 \pi} [\cos(\lambda \sin\theta - k \theta)&#xD;&#xA;+i\sin(\lambda \sin\theta - k \theta)]d\theta&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I need to show that this is equivalent to&#xA;$&#xD;&#xA;\frac{1}{2\pi}\int_{0}^{2 \pi} \cos(\lambda \sin\theta - k \theta)d\theta&#xD;&#xA;$.&#xA;In other words, I need to show that&#xA;$&#xD;&#xA;\int_{0}^{2 \pi}\sin(\lambda \sin\theta - k \theta)d\theta =0&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't figure out how to do this. I tried expanding using trig identities and then&#xA;writing sin and cos as Taylor Series and integrating term by term, but no luck. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-11-18T17:54:25.187" Title="Deriving Bessel's Integrals" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="10811" PostTypeId="2" ParentId="10810" CreationDate="2010-11-18T04:59:45.637" Score="1" ViewCount="0" Body="&lt;p&gt;Split the integral involving $\sin$ as $\int_0^{\pi}$ + $\int_{\pi}^{2\pi}$. Change variable in the second integral as $\phi = \theta - \pi$. Do some trigonometric manipulations like $\sin(\theta + n \pi) = (-1)^n \sin(\theta)$ and you will find that the second integral turns out to be negative of the first.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-18T17:54:25.187" LastActivityDate="2010-11-18T17:54:25.187" CommentCount="3" />
  <row Id="10812" PostTypeId="2" ParentId="10810" CreationDate="2010-11-18T05:01:17.787" Score="3" ViewCount="0" Body="&lt;p&gt;Use $\displaystyle \int_{a}^{b} f(x)\ dx = \int_{a}^{b} f(a+b-x) \ dx$ &lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T05:01:17.787" />
  <row Id="10813" PostTypeId="2" ParentId="10685" CreationDate="2010-11-18T05:12:56.353" Score="3" ViewCount="0" Body="&lt;p&gt;Unless I'm mistaken, you can add a sufficiently large &quot;cyclic ladder&quot; to the Frucht graph as in: &lt;img src=&quot;http://i.imgur.com/ZreVp.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would argue that, in an automorphism, the right-hand side would need to be mapped to itself.  Therefore, an automorphism of the above graph would imply an automorphism of the Frucht graph (and therefore must be trivial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite version could be constructed by appending two infinite ladders instead (actually, I'm pretty sure you could attach one infinite ladder).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-18T05:12:56.353" CommentCount="3" />
  <row Id="10814" PostTypeId="2" ParentId="10760" CreationDate="2010-11-18T05:22:39.007" Score="1" ViewCount="0" Body="&lt;p&gt;I wrote a &lt;a href=&quot;http://math.uga.edu/~pete/coveringnumbersv2.pdf&quot; rel=&quot;nofollow&quot;&gt;short note&lt;/a&gt; on exactly this problem.  It will appear in the Monthly one of these years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note in particular that your quoted statement is not necessarily true for infinite-dimensional vector spaces: an infinite-dimensional vector space over any field can be covered by $\aleph_0$-many proper subspaces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask also about modules over rings.  Yes, there has been work on that.  I believe that Apoorva Khare, a mathematician who independently proved most of the results in my note, also has some work on the case of modules over various rings.  If you google for &quot;covering numbers&quot;, you'll find that many, many papers on this topic have been written.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-18T05:22:39.007" CommentCount="2" />
  <row Id="10816" PostTypeId="2" ParentId="10780" CreationDate="2010-11-18T05:32:17.310" Score="2" ViewCount="0" Body="&lt;p&gt;I found the following theorems  from the book &quot;Introduction to classical real analysis&quot; by Karl R. Stromberg, 1981.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;(Zygmund) If $f$ satisfies a Holder (also called Lipschitz) condition of order $\alpha\gt 0$ and $f$ is of bounded variation on $[0,2\pi]$, then the Fourier series of $f$ converges absolutely (and hence uniformly).  p. 521.&lt;br/&gt;&#xA;This applies to the example in my  question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f$ is absolutely continuous, then the Fourier series of $f$ converges uniformly  but not necessarily absolutely.  p. 519 Exercise 6(d) and p.520 Exercise 7c.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Bernstein) If $f$ satisfies a Holder condition of order $\alpha\gt 1/2$ , then the  Fourier series of $f$ converges absolutely (and hence uniformly).  p.520 Exercise  8 (f)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Hille) For each $0&amp;lt;\alpha\le 1/2$, there exists a function that satisfies a Holder condition of order $\alpha$ whose Fourier series converges uniformly, but not absolutely.  p.520 Exercise  8 (f)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-18T05:32:17.310" />
  <row Id="10817" PostTypeId="1" AcceptedAnswerId="10887" CreationDate="2010-11-18T06:39:57.267" Score="2" ViewCount="197" Body="&lt;p&gt;On p.28 (&lt;a href=&quot;http://books.google.com/books?id=KvKdSkPwwmMC&amp;amp;lpg=PP1&amp;amp;dq=reid%20undergraduate%20algebraic%20geometry&amp;amp;pg=PA28#v=onepage&amp;amp;q=no%20rational%20parametrisation&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) Reid proves that the curve $y^2=x(x-1)(x-\lambda)$ for $\lambda\neq 0,1$ has no rational paramaterisation. At one point (on p.29), he has the equation $$r^2=ap(p-q)(p-\lambda q)$$ where $p$ and $q$ are polynomials over a field $k$, and $a$, $\lambda$ are constants in the field. He then says that &quot;by considering factorisation into primes, there exist nonzero constants $b,c,d\in k$ such that $$bp,c(p-q),d(p-\lambda q)$$ are all squares in $k[t]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; $p$ and $q$ are known to be coprime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like this should be obvious, but I don't follow his argument. Why do we know that those factors are squares (up to units)?&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-11-18T06:49:38.307" LastActivityDate="2010-11-18T20:08:28.730" Title="Proof from Reid's Undergraduate Algebraic Geometry: help with minor details" Tags="&lt;homework&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10818" PostTypeId="1" AcceptedAnswerId="10864" CreationDate="2010-11-18T07:38:45.867" Score="2" ViewCount="80" Body="&lt;p&gt;If $A$ is a positive definite matrix, then the solution to the minimization problem $(1/2)x^TAx - b^Tx$ is given by $A^{-1}b$.  I'm interested in the generalization of this to a Hilbert space.  What conditions on $A$ should be required for this to be true in a Hilbert space?  Also is there a reference that includes this? I searched Google books, but couldn't find it. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-18T21:05:36.203" LastActivityDate="2010-11-18T21:05:36.203" Title="Quadratic minimization in a Hilbert space" Tags="&lt;functional-analysis&gt;&lt;optimization&gt;&lt;hilbert-spaces&gt;&lt;calculus-of-variations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10819" PostTypeId="1" AcceptedAnswerId="10844" CreationDate="2010-11-18T09:21:26.283" Score="1" ViewCount="106" Body="&lt;p&gt;A rectangle can be cut into two connected pieces of equal area so that they can be rearranged to get a new rectangle with different side lengths than the original one. Show that this can be done in countable many ways. Can this be done in uncountable many ways?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-18T17:06:47.193" LastActivityDate="2010-11-18T17:06:47.193" Title="Splitting rectangle into two parts" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10820" PostTypeId="1" AcceptedAnswerId="10822" CreationDate="2010-11-18T09:36:20.920" Score="1" ViewCount="55" Body="&lt;p&gt;I have a unit sphere, on the surface of which are defined spherical caps. I typically characterize the caps by the unit vector $n$ from the center of the sphere to the top of the cap, and the angle $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: given a pair of spherical caps, how can I determine whether they intersect?&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-18T10:12:08.593" Title="How to test whether spherical caps intersect?" Tags="&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10821" PostTypeId="1" AcceptedAnswerId="10823" CreationDate="2010-11-18T09:50:18.043" Score="0" ViewCount="54" Body="&lt;p&gt;Today, I read these two new Logarithmic identity $\displaystyle$ $$a^{\log_a m} = m $$ $$\log_{a^q}{m^p} = \frac{p}{q} \log_a m$$ Both of them seems new to me,so even after solving some problems (directly) based on thesm I haven't fully understood how they holds good,Could anybody show me how to prove them ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T10:05:36.137" LastActivityDate="2010-11-18T10:28:26.197" Title="Proving Logarithmic identity" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="10822" PostTypeId="2" ParentId="10820" CreationDate="2010-11-18T10:12:08.593" Score="5" ViewCount="0" Body="&lt;p&gt;The angle between $n_1$ and $n_2$ is the analogue of the &quot;distance&quot; between the centres of the two caps. The caps intersect if this angle is less than the sum of $\theta_1$ and $\theta_2$, assuming the $\theta_i$ play the role of &quot;radius&quot; and not &quot;diameter&quot; in your notation.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-18T10:12:08.593" />
  <row Id="10823" PostTypeId="2" ParentId="10821" CreationDate="2010-11-18T10:21:16.420" Score="3" ViewCount="0" Body="&lt;p&gt;Just so this doesn't remain unanswered:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a statement of the fact that the functions $a^x$ and $\log_a\;x$ are inverses of each other; thus, $\log_a\;a^x=a^{\log_a\;x}=x$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Letting c be a positive real number not equal to one:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\log_{a^q}\;m^p=\frac{\log_c\;m^p}{\log_c\;a^q}=\frac{p\;\log_c\;m}{q\;\log_c\;a}=\frac{p}{q}\log_a\;m$&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T10:28:26.197" LastActivityDate="2010-11-18T10:28:26.197" />
  <row Id="10824" PostTypeId="1" CreationDate="2010-11-18T10:48:14.147" Score="0" ViewCount="105" Body="&lt;p&gt;The &lt;a href=&quot;http://www.math-help-ace.com/Logarithm-Tables.html&quot; rel=&quot;nofollow&quot;&gt;rule&lt;/a&gt;: &lt;strong&gt;&quot;Characteristic of a number($\lt$ 1) with n zeros after decimal and before a significant digit  = $\overline{( n + 1 )}$&quot;&lt;/strong&gt; This rule is mentioned in my module,Goggle lead me to this &lt;a href=&quot;http://www.math-help-ace.com/Logarithm-Tables.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I am trying to verify with some random numbers like &lt;code&gt;Log[10, 0.0004356]&lt;/code&gt; its giving me $-3.36091$ and &lt;code&gt;Log[10, 0.725]&lt;/code&gt; gives me $-0.139662$. Giving $-3$ and $-0$ as characteristic respectivley, however shouldn't the characteristic be $-4$ and $-1$ respectively,consistent to the rule ? what exactly I am missing here ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T15:51:20.280" LastActivityDate="2010-11-18T21:22:51.663" Title="Characteristic of Logarithms of positive numbers $\lt 1$" Tags="&lt;elementary&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10825" PostTypeId="1" CreationDate="2010-11-18T10:48:36.927" Score="1" ViewCount="62" Body="&lt;p&gt;Suppose that K is a finite extension over $\mathbb{Q}$. Let $p$ be a prime in $\mathbb{Q}$, let $p\mathcal{O}_K=\mathfrak{P}_1\ldots\mathfrak{P}_n$ be its prime decomposition in $\mathcal{O}_K$, then is it true that any prime ideal in $\mathcal{O}_K$ above $p$ is a factor in the decomposition of $p$?&lt;/p&gt;&#xA;" OwnerUserId="3661" LastActivityDate="2010-11-18T14:06:36.230" Title="prime decomposition in galois extensions" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="10826" PostTypeId="1" CreationDate="2010-11-18T10:49:13.427" Score="2" ViewCount="89" Body="&lt;p&gt;I have $n$ random elements from 0 to $k$. What is the probability that there is some element which appears more than $n/2$ times (or $n/2$ times).&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-19T05:58:08.070" LastActivityDate="2010-11-19T05:58:08.070" Title="Probability of having a majority element" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10827" PostTypeId="1" CreationDate="2010-11-18T11:10:13.843" Score="0" ViewCount="44" Body="&lt;p&gt;Suppose that q is a rational prime, and K is a p-extension over Q where p is prime, then q ramifies in K if and only if q is p or q is congruent to 1 mod p. I don't know why this is true, can someone convince me of this?&lt;/p&gt;&#xA;" OwnerUserId="3663" LastActivityDate="2010-11-19T12:49:47.727" Title="ramification in p-extensions" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="3" />
  <row Id="10828" PostTypeId="2" ParentId="10827" CreationDate="2010-11-18T11:21:59.320" Score="1" ViewCount="0" Body="&lt;p&gt;This can't possibly be true: by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;famous theorem of Dirichlet&lt;/a&gt;, there are infinitely many prime numbers $q$ such that $q \equiv 1 \pmod p$.  However, only finitely many &#xA;primes ramify in any finite extension $K/\mathbb{Q}$: those which divide the discriminant.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-18T11:49:00.020" LastActivityDate="2010-11-18T11:49:00.020" />
  <row Id="10829" PostTypeId="2" ParentId="10825" CreationDate="2010-11-18T11:31:05.190" Score="1" ViewCount="0" Body="&lt;p&gt;Yes.  I suppose that the definition of a prime ideal $\mathcal{P}$ being &quot;above $p$&quot; you are using is $\mathcal{P} \cap \mathbb{Z} = (p)$.  From this it is clear that if &#xA;$\mathcal{P}$ lies above $p$, then it contains $p\mathbb{Z}$, and -- since it is an ideal -- hence also $p \mathcal{O}_K$.  But recall that in a Dedekind domain &quot;to contain is to divide&quot;, hence $\mathcal{P}$ divides $p\mathcal{O}_K$ and appears in the (unique!) prime factorization of $\mathcal{O}_K$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this does not require that $K/\mathbb{Q}$ be Galois (as appears in the title but not the body of the question), and it has nothing to do with algebraic number fields &lt;em&gt;per se&lt;/em&gt;: in general, for a finite extension of Dedekind domains $S/R$, the prime ideals of $S$ lying over a given prime $\mathfrak{p}$ of $R$ are exactly the ones appearing in &#xA;the factorization of $\mathfrak{p}S$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-18T11:31:05.190" />
  <row Id="10830" PostTypeId="2" ParentId="10774" CreationDate="2010-11-18T11:40:31.950" Score="2" ViewCount="0" Body="&lt;p&gt;One method is to apply the &lt;a href=&quot;http://planetmath.org/encyclopedia/BargmannTransform.html&quot; rel=&quot;nofollow&quot;&gt;Bargmann transform:&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_f(z) = \int_{-\infty}^{\infty}f(x)exp(2 \pi x z - \pi x^2 - \frac{\pi}{2}z^2) dy$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the eigenvalue equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \lambda \int_{-\infty}^{\infty}f(y)exp(-2 \pi ix y) dy$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain the following relation for the Bargmann transformed eigenfunctions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_f(z) = \lambda B_f(-iz)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose solutions are the monomials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_f(z) = z^n$, corresponding to the eigenvalues $i^{n_{mod4}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the inverse Bargmann transform of the monomials are just the Hermite functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: The application of the Bargmann transform to the eigenvalue equation entails &#xA;a change of the integration order which is possible because the Hermite functions are bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projection operators onto the n-th subspace in the Bargamann &#xA;representation are given by the kernels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(v,\bar{z})=\frac{(v\bar{z})^n}{n!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which act on the Bargmann transformed functions according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pf(v) = \int_{\mathbb{C}}P(v,\bar{z}) f(z) exp(-z\bar{z})dzd\bar{z}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the time domain the projection kernels are just $P(t, \tau) = H_n(\tau) H_n(t)$ due to &#xA;the orthonormality of the Hermite funtions.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditorDisplayName="" LastEditDate="2010-11-18T12:33:56.453" LastActivityDate="2010-11-18T12:33:56.453" CommentCount="2" />
  <row Id="10831" PostTypeId="2" ParentId="10827" CreationDate="2010-11-18T11:41:07.283" Score="2" ViewCount="0" Body="&lt;p&gt;If $K/\mathbb{Q}$ is a degree $p$ extension of $\mathbb{Q}$&#xD;&#xA;then any particular prime can ramify: consider $K=\mathbb{Q}(q^{1/p})$.&#xA;Suppose that $K$ is a Galois and so cyclic degree $p$ extension.&#xA;Then only finitely many primes ramify in $\mathbb{Q}$ and they&#xA;can only have the form $q=p$ or $q\equiv1$ (mod $p$).&#xA;(I'm sure this is the result you are thinking of).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for $p$-power degree Galois extensions&#xA;(consider the ramification groups at their localizations).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-18T11:41:07.283" />
  <row Id="10832" PostTypeId="1" AcceptedAnswerId="10841" CreationDate="2010-11-18T11:42:19.020" Score="0" ViewCount="112" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given a simple function, very simple. f(x) = bla bla. With a constant and an X. The point was to find for which values of a(the constant) the function is not one-to-one. The domain of the function is limited from zero and upwards. Here is what I put down on the paper: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as a is negative, it will not be one-to-one. If a = -2. Then f(0) = 4 and f(4)=4. Thereby concluding that it's not one-to-one in it's domain as long as a &amp;lt; 0. And stating the point of one-to-one is that we shouldn't be able to reach the same output with different input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't get full score on this, what am I missing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: f(x) = (x+a)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I've added it! I not able to comment on my own post :( &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="2230" LastEditorDisplayName="" LastEditDate="2010-11-18T11:49:20.590" LastActivityDate="2010-11-18T15:47:20.520" Title="Calculus, given function. Say when it's not one-to-one? (What did i miss)" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10833" PostTypeId="2" ParentId="10832" CreationDate="2010-11-18T11:58:27.293" Score="0" ViewCount="0" Body="&lt;p&gt;Try to draw its graphic for $a=0$: maybe you'll recognise something familiar, won't you? Then, try to figure out what happens when you put some $a\neq 0$ (and you'll see that this doesn't change anything at all about injectivity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you could try to think about this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(x_1 +a )^2 = (x_2 + a)^2 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again: start with the case $a=0$, then see what happens for $a \neq 0$ (nothing new).&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-11-18T11:58:27.293" CommentCount="2" />
  <row Id="10834" PostTypeId="2" ParentId="10693" CreationDate="2010-11-18T12:29:41.023" Score="0" ViewCount="0" Body="&lt;p&gt;We will only prove here that definition (1) implies definition (2) (this is the easy part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the condition in definition (1) holds. That is, $\mu_n ((a,b]) \to \mu ((a,b])$ for any $a,b \in D$, $a &amp;lt; b$, where $D$ is a dense subset of $R$. Define $\tilde D = \lbrace x \in R:\mu (\lbrace x \rbrace ) = 0 \rbrace$. Then $\tilde D$ is dense in $R$, since its complement is at most countable. For any $a,b \in \tilde D$, $a &amp;lt; b$, and $a&amp;#39;,a&amp;#39;&amp;#39;,b&amp;#39;,b&amp;#39;&amp;#39; \in D$, $a&amp;#39; &amp;lt; b&amp;#39;$, $a&amp;#39;&amp;#39; &amp;lt; b&amp;#39;&amp;#39;$, such that $a&amp;#39;&amp;#39; &amp;lt; a &amp;lt; a&amp;#39;$ and $b&amp;#39; &amp;lt; b &amp;lt; b&amp;#39;&amp;#39;$, we have&#xA;$$&#xD;&#xA;\mu _n ((a&amp;#39;,b&amp;#39;]) \le \mu _n ((a,b)) \le \mu _n ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;]).&#xD;&#xA;$$&#xD;&#xA;By assumption, $\mu _n ((a&amp;#39;,b&amp;#39;]) \to \mu ((a&amp;#39;,b&amp;#39;])$ and $\mu _n ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;] \to \mu ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;])$ as $n \to \infty$. Then, since $a&amp;#39;,a&amp;#39;&amp;#39;$ and $b&amp;#39;,b&amp;#39;&amp;#39;$ can be chosen arbitrarily close to $a$ and $b$, respectively, we can conclude that &#xA;$$&#xD;&#xA;\mu ((a,b)) \le \mathop {\lim \inf }\limits_{n \to \infty } \mu _n ((a,b)) \le \mathop {\lim \sup }\limits_{n \to \infty } \mu _n ((a,b)) \le \mu ([a,b]).&#xD;&#xA;$$&#xD;&#xA;However, by assumption $\mu (\lbrace a \rbrace) = \mu (\lbrace b \rbrace) = 0$, and hence $\lim _{n \to \infty } \mu _n ((a,b)) = \mu ((a,b))$. Thus, the condition in definition (2) is satisfied (with $\tilde D$ playing the role of $D$ in that definition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is instructive to consider here the following simple example. As is customary, denote by $\delta_x$ the probability measure concentrated at $x$. Define $\mu_n = \delta_{-1/n}$, $n \geq 1$, and $\mu = \delta_0$. Then, for any $a,b \in R$, $a &amp;lt; b$, we have $\mu _n ((a,b]) \to \mu ((a,b])$, and hence, by definition (1), $\mu_n$ converges vaguely to $\mu$. On the other hand, $\mu _n ((-2,0)) = 1$ for any $n \geq 1$, whereas $\mu ((-2,0)) = 0$. However, if we define $\tilde D$ as above, then $\tilde D = R - \lbrace 0 \rbrace$, and for any $a,b \in \tilde D$, we do have $\mu _n ((a,b)) \to \mu ((a,b))$; hence, $\mu_n$ converges vaguely to $\mu$ also according to definition (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as for the OP's remark (concerning definition (2)) that &quot;the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?&quot;, it is clear that we are deliberately only given that $\mu_n ((a,b))$ converges -- this is what makes the converse implication (from definition (2) to (1)) a substantially more difficult problem.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-18T12:42:17.250" LastActivityDate="2010-11-18T12:42:17.250" CommentCount="3" />
  <row Id="10835" PostTypeId="1" CreationDate="2010-11-18T12:39:54.123" Score="3" ViewCount="85" Body="&lt;p&gt;Two numbers are equivalent if they have the same digits ex 2271 and 1227 are equivalent. How many equivalent classes are there between 1000 &amp;amp; 9999? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I fully understand how this works, is the answer to this question (9*10*10*10)/4!? First calculating the total numbers formed and divindg by 4! to get rid of the repetitions.&lt;/p&gt;&#xA;" OwnerUserId="3666" LastActivityDate="2010-11-18T17:36:00.353" Title="Combinations question" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" />
  <row Id="10836" PostTypeId="2" ParentId="10826" CreationDate="2010-11-18T12:51:33.573" Score="2" ViewCount="0" Body="&lt;p&gt;This is related to the problem of finding the integral of a normal distribution, in short this means that there is no simple formula that can solve your problem, you will have to do it as a sum of probabilities for a number of simpler cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of a specific element appearing exactly $m$ times in a group of $n$ elements chosen randomly from a range of $k$ elements is: $$f(m, n, k) = \frac{(1/k)^m\ ((k-1)/k)^{n-m}\ n!}{m!\ (n-m)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability of a specific element appearing $m$ or more times is: $$\sum_{p=roundup(n/2)}^n f(p, n, k)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of any element appearing $m$ or more times things get tricky, I can't think of any simple way of taking the correlation between the numbers of different elements into account, but for your specific question less will do. You can simply multiply the result of the previous formula by $k$, but for even values of $n$ you will have to account for the special case where half of the elements is one value and half is another, since that will have been counted twice, so you will have to subtract that probability: $$\frac{k\ (k-1)\ (1/k)^n\ n!}{2\ (n/2)!^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that my $k$ is different from your $k$, since there are $k+1$ numbers from $0$ to $k$ you will have to add 1 to your $k$ before using it in the formulas.&lt;/p&gt;&#xA;" OwnerUserId="3584" LastActivityDate="2010-11-18T12:51:33.573" />
  <row Id="10838" PostTypeId="2" ParentId="10832" CreationDate="2010-11-18T12:54:03.700" Score="1" ViewCount="0" Body="&lt;p&gt;The key here is that your domain is $[0,\infty)$. In your solution you gave a very loose reason why it is not one-to-one for $a &amp;lt; 0$, and did not say anything about the case where $a \geq 0$.&#xA;For $a &amp;lt; 0$, consider $f(0)$ and $f(-2a)$. Both are well-defined, as they are contained in the domain. Then&#xA;$$f(0) = (0 + a)^2 = a^2 = (-2a + a)^2 = f(-2a)$$&#xD;&#xA;Since $0 \neq -2a$ by assumption, $f$ is not one-to-one for $a &amp;lt; 0$.&#xA;To finish, you need to show that $f$ is one-to-one for $a \geq 0$. Hint: Derivatives should be useful here...&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-18T12:54:03.700" />
  <row Id="10839" PostTypeId="1" AcceptedAnswerId="10849" CreationDate="2010-11-18T13:36:42.663" Score="2" ViewCount="142" Body="&lt;p&gt;This question is about &lt;a href=&quot;http://projecteuler.net/index.php?section=problems&amp;amp;id=113&quot; rel=&quot;nofollow&quot;&gt;Project Euler 113&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Working from left-to-right if no digit&#xA;  is exceeded by the digit to its left&#xA;  it is called an increasing number; for&#xA;  example, 134468.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Similarly if no digit is exceeded by&#xA;  the digit to its right it is called a&#xA;  decreasing number; for example, 66420.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We shall call a positive integer that&#xA;  is neither increasing nor decreasing a&#xA;  &quot;bouncy&quot; number; for example, 155349.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As n increases, the proportion of&#xA;  bouncy numbers below n increases such&#xA;  that there are only 12951 numbers&#xA;  below one-million that are not bouncy&#xA;  and only 277032 non-bouncy numbers&#xA;  below 10^10.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How many numbers below a googol&#xA;  (10 pow 100) are not bouncy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is obviously not something you can brute force. I found a &lt;a href=&quot;http://freelancersunite.net/project_euler/project-euler-problem-113/&quot; rel=&quot;nofollow&quot;&gt;mathematical solution&lt;/a&gt; that does the trick in no time, the problem is, I don't understand it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point me in the right direction? Not having a math background, I'm not even sure where to begin! &lt;/p&gt;&#xA;" OwnerUserId="3668" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T19:56:51.413" LastActivityDate="2010-11-18T19:56:51.413" Title="Combinatorial counting" Tags="&lt;combinatorics&gt;&lt;project-euler&gt;" AnswerCount="2" />
  <row Id="10840" PostTypeId="2" ParentId="10825" CreationDate="2010-11-18T14:06:36.230" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ P \supset \mathfrak{P}_1\ldots\mathfrak{P}_n\ \Rightarrow\ P \supset \mathfrak P_i\ $ for some $\rm\:i\:.\ $ This is true for any prime ideal $\rm\ P\ $ in any ring. Further $\rm\ P = \mathfrak P_i\ $ when the $\ \mathfrak P_i\ $ are maximal, e.g. in one dimensional domains, i.e. domains where nonzero prime ideals are maximal.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-18T14:06:36.230" />
  <row Id="10841" PostTypeId="2" ParentId="10832" CreationDate="2010-11-18T14:12:08.827" Score="4" ViewCount="0" Body="&lt;p&gt;There are two things you are missing: you are missing the case in which $a\geq 0$; and you did not really justify your assertion when $a\lt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, the only case you actually established was the case $a=-2$. That leaves quite a few values of $a$ untouched!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one asserts that something happens &quot;for all numbers less than $0$&quot;, it is almost never good enough to check what happens in &lt;em&gt;one&lt;/em&gt; number less than $0$. That's called &quot;argument by example&quot;, and examples almost never establish general cases. It would be like trying to argue that &lt;em&gt;everyone&lt;/em&gt; in your class is going to fail by pointing to one student in the class who happens to be failing. It's certainly possible that all will fail, but just showing one person who will is not going to establish this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: first, you need to finish off the case $a\lt 0$: if $a$ is &lt;strong&gt;any&lt;/strong&gt; number less than $0$, why is $f(x)=(x+a)^2$ not one-to-one on $[0,\infty)$? Because, &lt;em&gt;no matter what $a$ is&lt;/em&gt;, you can always find at least two distinct points in $[0,\infty)$ on which the function takes the same value: $x=0$, and $x=-2a$ (which is positive because $a\lt 0$, and so both in the domain and different from $0$). Indeed, $f(0)=(0+a)^2 = a^2$, and $f(-2a) = (-2a+a)^2 = (-a)^2 = a^2=f(0)$. You'll note that this takes care of &lt;em&gt;all&lt;/em&gt; the values of $a$ that are negative. (&lt;em&gt;Note:&lt;/em&gt; These are not the only points you can use; for instance, $\frac{-a}{2} and $\frac{-3a}{2}$ will also have equal images. But to show that something is &lt;em&gt;not&lt;/em&gt; one-to-one it &lt;em&gt;does&lt;/em&gt; suffice to give a single example of two distinct points that map to the same thing, because the assertion &quot;is not one-to-one&quot; means &quot;there is &lt;strong&gt;at least one&lt;/strong&gt; pair $a,b$, with $a\neq b$ and $f(a)=f(b)$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we also need to deal with the case $a\geq 0$. What happens then? If $a\geq 0$, then $x+a\geq 0$ (since $x\geq 0$). If $f(x)=f(y)$, then we have:&#xA;\begin{align*}&#xA;f(x) &amp;amp;= f(y)\\&#xA;(x+a)^2 &amp;amp;= (y+a)^2\\&#xA; \sqrt{(x+a)^2} &amp;amp;= \sqrt{(y+a)^2}\\&#xA;|x+a|&amp;amp;=|y+a|&#xA;\end{align*}&#xA;(remember that $\sqrt{r^2}=|r|$). Now, because $x$, $y$, and $a$ are all nonnegative, so are $x+a$ and $y+a$, so they equal their absolute value: $|x+a|=x+a$ and $|y+a|=y+a$. So we get $x+a=y+a$, and so we deduce $x=y$. That is, if $a\geq 0$, then for $f(x)$ to equal $f(y)$ we &lt;em&gt;must&lt;/em&gt; have $x=y$. This means that in this case, $f$ is one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: you forgot to check $a\geq 0$. And you did not fully establish $a\lt 0$. It's not enough to just &quot;it is not one-to-one when $a\lt 0$&quot;, because this is only says what happens if $a\lt 0$; your answer is not presumed to be exhaustive (that is, we don't assume that you are telling us &lt;em&gt;all&lt;/em&gt; the values in which it is not one-to-one) because you are not saying it is exhaustive (to do so, you would have to say &quot;it is not one-to-one &lt;em&gt;if and only if&lt;/em&gt; $a\lt 0$&quot;, and even then you would have to discuss what happens when $a\geq 0$).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T15:47:20.520" LastActivityDate="2010-11-18T15:47:20.520" />
  <row Id="10842" PostTypeId="2" ParentId="10839" CreationDate="2010-11-18T14:23:23.930" Score="0" ViewCount="0" Body="&lt;p&gt;First, you are trying to count increasing and decreasing numbers.  They are related in that you only reverse the digits of one to get the other.  The only difference is that increasing numbers cannot start with 0, but decreasing ones can end with 0.  So let's just work on counting increasing ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many one digit increasing numbers are there? Nine.  How many one digit increasing numbers are there that end in 1? One.  So make a function f(x,n) where f(x,n) is the number of n digit increasing numbers that end in x, you should be able to make a recurrence for each one.  That is, (6,n)=some function of the f(x,n-1).  The increasing numbers under a googol are just the sum of f(x,100) over x.  Then you do the same for decreasing numbers but have to add f(0,n) to the mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't look at the program you cite to see if it works anything like this.  But this approach worked for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many counting problems that yield to techniques like this.  Instead of keeping a list of all the increasing numbers (which would grow too big), just keep track of how many there are of a given size and find a recurrence to get to the next size up.  The next three problems in the list can be attacked this way as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another approach.  Think of a series of 101 zeros.  Then all the increasing numbers can be constructed by putting up to nine &quot;step up&quot; marks between the digits, with one of them required to be between the first and second zeros (to avoid leading zeros).  You can even make it exactly nine steps if you allow some to be at the end.  I'll let somebody else make this more explicit&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-18T15:17:31.473" LastActivityDate="2010-11-18T15:17:31.473" />
  <row Id="10843" PostTypeId="2" ParentId="10835" CreationDate="2010-11-18T14:33:05.557" Score="0" ViewCount="0" Body="&lt;p&gt;One approach is to work by cases of the number of matching digits.  If no digits match and none is zero, there are in fact 4!=24 permutations.  So this accounts for 9*8*7*6/24 equivalence classes.  If no digits match and one is zero, the zero cannot be the first digit, so there are only 3*3*2*1=18 permutations and 1*9*8*7/18 equivalence classes.  If you have one pair of matching digits and no zero, there are how many permutations and how many equivalence classes?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-18T14:33:05.557" />
  <row Id="10844" PostTypeId="2" ParentId="10819" CreationDate="2010-11-18T15:13:40.633" Score="2" ViewCount="0" Body="&lt;p&gt;Let n &amp;ge; 1 be an integer.  Divide the width of an H-by-W rectangle into n+1 equal parts, and the height into n equal parts.  Separate the rectangle into two connected and congruent pieces along a path stepped as necessary from a side divided at the 1/(n+1) part of its length, down a distance of 1/n of its height, and so on alternating until the opposite side of the rectangle is reached.  The two parts may then be &quot;slid&quot; across one another to form a rectangle of height (n+1)/n H and width n/(n+1) W.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives a countable number of ways to rearrange the rectangle, at most one of which could be congruent to the original (i.e. when W = (n+1)/n H for some n, or conversely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe these are the only such dissections of a rectangle into two pieces, and if so there are not uncountably many ways to do it (though of course the roles of height and width can be exchanged).  The locations of the four original corners (right angles) are constrained so that they can only appear on the boundary of the rearranged rectangle, and the rearrangement works both ways.  The dissection must thus operate so as to separate the boundary of both rectangles into two parts (because an intact boundary of a rectangle cannot be encompassed by a different rectangle of equal area).  A bit of case analysis should then show that the two rectangles share a pair of opposite corners and that the path dividing the pieces must enter and exit at the &quot;new&quot; pair of corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, hm&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-18T15:13:40.633" CommentCount="1" />
  <row Id="10845" PostTypeId="1" AcceptedAnswerId="10847" CreationDate="2010-11-18T15:22:52.667" Score="1" ViewCount="66" Body="&lt;p&gt;Original equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \frac{(x+1)(x^2+2)(x^3+3)}{\sqrt{x^4+3}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I take ln on both sides, and than differentiate I get this:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{g(x)} = \frac{1}{x+1}+\frac{1}{x^2+2}+\frac{1}{x^3+3}-\frac{1}{2}* \frac{1}{x^4+3} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this good so far? I don't know what to do about the left side?&lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-18T15:27:05.847" Title="Calculus, differentiation, but first taking ln on both sides? " Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10846" PostTypeId="1" AcceptedAnswerId="10871" CreationDate="2010-11-18T15:26:29.500" Score="5" ViewCount="124" Body="&lt;p&gt;What is an example of a  &lt;strong&gt;continuous&lt;/strong&gt;, or even better, differentiable, $2\pi$ (or 1) periodic function whose Fourier series converges pointwise but not uniformly? (Such function cannot be of Holder class, or absolutely continuous.) &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-18T19:56:47.777" LastActivityDate="2010-11-18T19:56:47.777" Title="Pointwise but not uniform convergence of a Fourier series" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10847" PostTypeId="2" ParentId="10845" CreationDate="2010-11-18T15:27:05.847" Score="4" ViewCount="0" Body="&lt;p&gt;Close, but you forgot to apply the chain rule!&lt;/p&gt;&#xA;" OwnerUserId="1306" LastActivityDate="2010-11-18T15:27:05.847" CommentCount="4" />
  <row Id="10848" PostTypeId="1" AcceptedAnswerId="10854" CreationDate="2010-11-18T15:39:02.120" Score="5" ViewCount="132" Body="&lt;p&gt;Under the isomorphism of Hilbert spaces $L^2(S^1)\to\ell^2(\mathbb Z),\quad e^{2\pi i n t}\mapsto e_n$, smooth functions on the circle are mapped to rapidly decaying sequences (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series#Properties&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the converse also true? That is, does every rapidly decaying sequence in $\ell^2(\mathbb Z)$ represent a smooth function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: For certain arguments, it would be really nice to read off smoothness from the Fourier coefficients.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-18T15:47:47.700" LastActivityDate="2010-11-18T16:12:16.900" Title="Does rapid decay of Fourier coefficients imply smoothness?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="10849" PostTypeId="2" ParentId="10839" CreationDate="2010-11-18T15:49:53.917" Score="6" ViewCount="0" Body="&lt;p&gt;Suppose you wanted to find the number of decreasing numbers of exactly $\displaystyle k$ digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you knew that the digits were distinct, you chould choose k distinct digits and sort them, giving you 10 choose k  = $\displaystyle {10 \choose k}$ numbers. If the number was in base $n$, the corresponding number would be $\displaystyle {n \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given a non-decreasing number like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5543220&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can construct a &quot;sorted&quot; number with distinct digits in a &lt;em&gt;different&lt;/em&gt; base, by adding 0 to the rightmost digit, adding 1 to the digit to the left of it, 2 to the digit to the left of that, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5543220 becomes BA86430 in base 10 + 6 = 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A decreasing number with $\displaystyle k$ digits correponds in a 1-1 fashion to a sorted number of $k$ digits, in base $\displaystyle 10+k-1 = 9+k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we saw earlier the number of &quot;sorted&quot; numbers of $\displaystyle k$ digits in base $\displaystyle n$ is $\displaystyle {n \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of decreasing numbers is $\displaystyle {9+k \choose k}  = {9+k \choose 9}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total number of decreasing numbers $\displaystyle \lt 10^{100}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=1}^{99} {9+k \choose 9}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\displaystyle {n \choose r} - {n-1 \choose r} = {n-1 \choose r-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the above becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{k=1}^{99} {10+k \choose 10} - {9+k \choose 10}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a telescoping sum and becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  = {10+99 \choose 10} - {9+1 \choose 10} = {109 \choose 10} - 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do something similar for the increasing numbers.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T15:49:53.917" CommentCount="5" />
  <row Id="10850" PostTypeId="1" AcceptedAnswerId="10852" CreationDate="2010-11-18T15:49:56.607" Score="0" ViewCount="93" Body="&lt;p&gt;I am facing some trouble in solving this equation : $$\displaystyle3 \cdot x^{\log_5 2} + 2^{log_5 x} = 64$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give me some hints to proceed on this.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-18T16:38:16.850" Title="Solving an equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10851" PostTypeId="2" ParentId="10824" CreationDate="2010-11-18T15:54:12.273" Score="3" ViewCount="0" Body="&lt;p&gt;Read carefully: you take the characteristic as $-(n+1)$, but then you &lt;strong&gt;add a positive number&lt;/strong&gt; (the mantissa); so the actual number you will get will be $-(n+1)+c$, with $c\gt 0$, and so you will get a number $\ell$ that satisfies&#xA;$$-(n+1) \leq \ell \lt -n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your example $0.0004356$, you start with a characteristic of $-4$, but then you add a positive number to get a number strictly larger than $-4$, but less than $-3$. With $0.725$, you start with $-1$, then add a positive number to get a number strictly larger than $-1$ but less than $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Basically, what you are doing is taking your number and multiplying by a sufficiently large power of $10$ so that you &lt;em&gt;just&lt;/em&gt; get it to $1$ or over. If the first significant digit occurs in position $n+1$ after $n$ zeroes, then $10^{n+1}x$ will &quot;shift&quot; the decimal point $n+1$ spaces to the right, so you will get a number of the form $d_{n+1}.d_{n+2}d_{n+3}d_{n+4}\ldots$ (where your original number was $x = 0.0\ldots0d_{n+1}d_{n+2}d_{n+3}\ldots$, with $n$ zeroes after the decimal). You already know how to figure out these logarithms: they have characteristic $0$, and a certain mantissa $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you get then is that $\log_{10}(10^{n+1}x) = 0.m$. But by the properties of the logarithm, you know that &#xA;$$0.m = \log_{10}(10^{n+1}x) = \log_{10}(10^{n+1}) + \log_{10}(x) = (n+1) + \log_{10}(x).$$ &#xA;Since $\log_{10}(x)$ is what you &lt;em&gt;really&lt;/em&gt; want to know, that tells you that &#xA;$$\log_{10}(x) = -(n+1)+0.m$$&#xD;&#xA;So, we take $-(n+1)$ as the &quot;characteristic&quot;, and then add the &lt;em&gt;positive&lt;/em&gt; number $m$. So we get $-(n+1)\leq \log_{10}(x) \lt -n$.You get equality if and only if $m=0$, which happens if and only if $d_{n+2}=d_{n+3}=\cdots = 0$, which is exactly when $x=10^{-(n+1)}$. Otherwise, you get a number strictly between $-(n+1)$ and $-n$. But these numbers, when written in decimal, have $-n$ before the decimal, not $-(n+1)$. That's why you are getting $-3$ and $0$, and not $-4$ and $-1$, in your examples.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T21:22:51.663" LastActivityDate="2010-11-18T21:22:51.663" />
  <row Id="10852" PostTypeId="2" ParentId="10850" CreationDate="2010-11-18T16:03:38.923" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ x^{\: log_5 2}\ =\ 5^{\: log_5 x\ log_5 2}\ =\ 2^{\: log_5 x}\ \ $  (or take $\rm\:log_5\:$ of both sides if that's clearer to you)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the equation reduces to $\rm\ 2^{\: \log_5 x}\ =\ 2^4\ \Rightarrow\ log_5\: x\ =\ 4\ \Rightarrow\ x\ =\ \ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TIP&lt;/strong&gt; $\ $ Typically such exponential equations will be solvable in closed form only if the exponentials are all linearly dependent, so you should always check for that first.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T16:38:16.850" LastActivityDate="2010-11-18T16:38:16.850" CommentCount="5" />
  <row Id="10853" PostTypeId="2" ParentId="10826" CreationDate="2010-11-18T16:05:45.890" Score="1" ViewCount="0" Body="&lt;p&gt;If we interpret the $n$ random elements from $0$ to $k$ to mean that each different selection is equally likely, then this question does have a very nice closed form answer: see (1) below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, say we are selecting two objects from the set $ \lbrace 0,1 \rbrace $ then by each selection being equally likely we mean $ \lbrace 00 \rbrace, $ $ \lbrace 01 \rbrace $ and&#xA;$ \lbrace 11 \rbrace $ each have probability $1/3.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case where $k=1$ and $n=2,$ the probability that some object appears&#xA;$ &amp;gt; \lceil n/2 \rceil $ i.e. $2$ times, is $2/3.$ This is given by $(1)$ below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when $k=3$ and $n=3$ we're looking at the probability that some selection contains $3$ objects all the same. This can be seen to be $1/5$ by noting that if we select from&#xA;$ \lbrace 0,1,2,3 \rbrace $ then we have $4$ possibilities of the form&#xA;$ \lbrace 000 \rbrace $ (all the same), $12$ of type $ \lbrace 001 \rbrace $ (two the same and one of another) and $4$ of type $ \lbrace 012 \rbrace $ (all different).  As a check, $(1)$ gives the probability as $1/5.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the derivation of $(1).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways we can place $n$ identical objects into $k+1$ boxes is&#xA;${n+k \choose k}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we need to find the number of ways of placing the $n$ identical objects into the $k+1$ boxes where one box contains $ &amp;gt; \lceil n/2 \rceil $ objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are $ \lceil n/2 \rceil + i $ objects in one box,&#xA;$ 1 \le i \le \lfloor n/2 \rfloor,$ then the other $ \lfloor n/2 \rfloor - i $ objects can be distributed amongst the other $k$ boxes in&#xA;$$ { \lfloor n/2 \rfloor - i + k -1 \choose k – 1 }$$&#xD;&#xA;ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the total number of ways that one box can contain $ &amp;gt; \lceil n/2 \rceil $ objects is&#xA;$$ (k+1) \sum_{i=1}^{ \lfloor n/2 \rfloor}{ \lfloor n/2 \rfloor - i + k -1 \choose k – 1 }$$&#xD;&#xA;$$ = \frac {k+1}{k} (\lfloor n/2 \rfloor + k -1){ \lfloor n/2 \rfloor + k - 2 \choose k – 1 }.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the probability that one box contains $ &amp;gt; \lceil n/2 \rceil $ objects is given by&#xA;( $ k \ge 1, n \ge 1 $ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac {k+1}{k} \frac{ (\lfloor n/2 \rfloor + k -1){ \lfloor n/2 \rfloor + k - 2 \choose k – 1 } }&#xD;&#xA;{ {n+k \choose k} }. \qquad (1)$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-18T16:05:45.890" CommentCount="3" />
  <row Id="10854" PostTypeId="2" ParentId="10848" CreationDate="2010-11-18T16:06:52.880" Score="7" ViewCount="0" Body="&lt;p&gt;Yes. Smoothness is equivalent to the Fourier coefficients forming a sequence that decays rapidly (faster than any polynomial). To see the direction you asked about, note that if $\{c_n\}$ is a rapidly decaying sequence, then the sum $\sum c_n e^{in x}$ will converge uniformly as will every derivative. So the sum will represent a  smooth function. (The continuous analog of this is that the Fourier transform induces an isomorphism of the &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Schwartz_space&quot; rel=&quot;nofollow&quot;&gt;Schwartz space&lt;/a&gt; onto itself.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, one can use the Fourier coefficients to give an $L^2$ definition of differentiability. Morally, a function ought to be $C^k$ if its Fourier coefficients decay like $|n|^{-k}$. This is not quite accurate, but one can use this to define the &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Sobolev_space&quot; rel=&quot;nofollow&quot;&gt;Sobolev spaces&lt;/a&gt;&#xA;on the circle (more generally on the torus, or using the Fourier transform on euclidean space, and by an extension process on compact manifolds) that describe how many &quot;weak&quot; derivatives a function has. This sort of argument shows that $L^2$ differentiability is roughly comparable with normal differentiability, and using such Sobolev spaces turns out to be integral when one tries to prove explicit bounds about things such as &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Elliptic_regularity&quot; rel=&quot;nofollow&quot;&gt;elliptic regularity&lt;/a&gt;. There, working with Fourier coefficients (or transforms) and a multiplication operator is significantly easier than working in the original function space with the correspondingly more complicated differential operator. (I am using the fact that constant-coefficient differential operators correspond under the Fourier transform to multiplication by a polynomial.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-18T16:12:16.900" LastActivityDate="2010-11-18T16:12:16.900" />
  <row Id="10855" PostTypeId="2" ParentId="10850" CreationDate="2010-11-18T16:09:41.050" Score="0" ViewCount="0" Body="&lt;p&gt;Hint: 3y + y = 64&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further (basic) hint: $a^{\log _b c}  = (e^{\log a} )^{\log c/\log b} $&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-18T16:21:04.053" LastActivityDate="2010-11-18T16:21:04.053" CommentCount="1" />
  <row Id="10856" PostTypeId="1" AcceptedAnswerId="10870" CreationDate="2010-11-18T16:48:37.803" Score="3" ViewCount="80" Body="&lt;p&gt;I am given $n$ points in 2D.(Each of say approximately equal weight). I want to partition it into $m$ clusters ($m$ can be anything and it is input by the user) in such a way that the center of mass of each cluster is &quot;far&quot; from center of mass of all other clusters. What is a good heuristic approach (it should also be quick and easy to implement) for this? My current approach is to set up a binary tree at each step. What I am doing now is that the line I choose to separate cluster at each step which maximizes the moment of inertia of the set of points in the cluster I am splitting. Any suggestion welcome!&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T17:38:38.303" LastActivityDate="2010-11-18T18:43:58.177" Title="Partitioning a graph  (clustering of point sets in 2 dimensions)" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;&lt;computer-science&gt;" AnswerCount="3" />
  <row Id="10857" PostTypeId="2" ParentId="10856" CreationDate="2010-11-18T17:17:29.790" Score="4" ViewCount="0" Body="&lt;p&gt;You might find &lt;a href=&quot;http://en.wikipedia.org/wiki/K-means_clustering&quot; rel=&quot;nofollow&quot;&gt;K-Means Clustering&lt;/a&gt; (and related) helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the classic machine learning problems, you should find plenty of literature on this.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T17:17:29.790" />
  <row Id="10858" PostTypeId="2" ParentId="10856" CreationDate="2010-11-18T17:23:42.193" Score="5" ViewCount="0" Body="&lt;p&gt;The keyword is &quot;clustering&quot; as mentioned in Moron's answer.  Any problem of this type will be NP-complete.  In practice, K-means is not bad in its runtime or (depending very much on the application) its results.  Like the simplex algorithm for linear programming, it can take exponential time in the worst case, but its practical complexity is much lower.  The worst-case bound was proven only very recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, partitioning a &lt;em&gt;graph&lt;/em&gt; is a different problem.  Here you are partitioning a set of points and distances are used but not any graph structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(added:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the smoothed analysis.  K-means has polynomial runtime when averaged over (Gaussian) random perturbations of the input, which is not surprising considering the practical efficiency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0904.1113&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0904.1113&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-18T17:35:47.603" LastActivityDate="2010-11-18T17:35:47.603" CommentCount="1" />
  <row Id="10859" PostTypeId="1" CreationDate="2010-11-18T17:23:54.017" Score="3" ViewCount="83" Body="&lt;p&gt;In &quot;A Short Course on Spectral Theory&quot;, page 10, William Arveson asserts that the &quot;ax+b group&quot;, ie. the group generated by all dilations and translations of the real line, is isomorpic to the group of all (real) 2x2 matrices of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a     b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0     1/a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&gt;0, b real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very easy to check that the ax+b group is isomorphic to the group of all matrices of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a     b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0     1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a&gt;0, b real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these two matrix groups should be isomorphic. Is this correct? Can someone give me the isomorphism? I've tried for a while and can't seem to get it.&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2010-11-18T19:24:25.173" Title="matrix representation of ax+b group" Tags="&lt;group-theory&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10860" PostTypeId="2" ParentId="10835" CreationDate="2010-11-18T17:30:25.303" Score="2" ViewCount="0" Body="&lt;p&gt;Represent each equivalence class by the number which has it's digits sorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is uniquely determined by the number of 0s, number of 1s, ..., number of 9s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you need to find the total number of tuples $(n_0, n_1, \dots, n_9)$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=0}^{9} n_i = 4$ such that $n_i \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is given by $\displaystyle {13 \choose 4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this also counts $0000$, the answer for your problem is $\displaystyle {13 \choose 4}-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the number of disitnct $n$-tuples of non-negative integers which sum to $k$ is given by $\displaystyle {n+k-1 \choose k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-18T17:36:00.353" LastActivityDate="2010-11-18T17:36:00.353" />
  <row Id="10861" PostTypeId="1" AcceptedAnswerId="10881" CreationDate="2010-11-18T17:41:17.070" Score="0" ViewCount="58" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If set is closed under some operation,&#xA;  what does that tell us the about structure&#xA;  of the groupoid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a question from one of the previous math exams which was shown to me by one of my friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I can't find a good answer to it. As far as I know, the answer is nothing, because of the definition of groupoid we use. Here's what (among other things) my math textbook has to say about groupoids:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ordered pair (S,*), where arbitrary&#xA;  binary operation closed in  set S&#xA;  is marked with *, is called groupoid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="348" LastEditorUserId="348" LastEditorDisplayName="" LastEditDate="2010-11-18T17:55:09.073" LastActivityDate="2010-11-18T19:24:24.813" Title="If set is closed under some operation, what does that tell us about the  structure of the groupoid?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10862" PostTypeId="2" ParentId="10859" CreationDate="2010-11-18T17:49:04.240" Score="2" ViewCount="0" Body="&lt;p&gt;The isomorphism is given by mapping $\begin{pmatrix} a &amp;amp; b \\ 0 &amp;amp; a^{-1} \end{pmatrix}$ to $\begin{pmatrix} a^2 &amp;amp; a b \\ 0 &amp;amp; 1 \end{pmatrix}$.&#xA;The inverse is given by $\begin{pmatrix} a &amp;amp; b \\ 0 &amp;amp; 1 \end{pmatrix}&#xD;&#xA;\mapsto \begin{pmatrix} a^{1/2} &amp;amp; a^{-1/2} b \\ 0 &amp;amp; a^{-1/2}\end{pmatrix}.$&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-18T17:49:04.240" />
  <row Id="10863" PostTypeId="1" AcceptedAnswerId="10865" CreationDate="2010-11-18T17:50:41.790" Score="1" ViewCount="54" Body="&lt;p&gt;Find a general expression for the Laurent Series expansion of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(z) = (z-\alpha)^{-n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when $|z|&amp;gt;|\alpha|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried expanding the function using the binomial theorem, and I've tried&#xA;integrating with the general formula for Laurent Series coefficients &#xA;using Cauchy's Integral formula, but I can't get the solution,&#xA;which I know is supposed to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma_{j=n}^{\infty} \alpha^{j-n} \frac{(j-1)!}{(n-1)!(j-n)!}z^{-j}$&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-11-18T18:06:39.233" Title="Expanding Binomial using Laurent Series" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="1" />
  <row Id="10864" PostTypeId="2" ParentId="10818" CreationDate="2010-11-18T17:53:29.807" Score="2" ViewCount="0" Body="&lt;p&gt;What you need is to look at some texts on functional analysis, especially about &quot;Direct Methods&quot; in &quot;Calculus of Variations&quot;. I'll give a quick primer here to flesh out what Rahul Narain said in the comments, but it will be by no means complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let us review how the minimization problem is solved in finite dimensional spaces. We will consider a finite dimensional vector space $V$ with an inner product $\langle,\rangle$. Let $A$ be a self-adjoint linear operator $V\to V$, and $b$ some element of $V$. We consider the minimization problem with the non-linear functional $J(x) = \frac12 \langle x,Ax\rangle - \langle b,x\rangle$. There are two steps used in finding the minimizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1) &lt;strong&gt;Existence of a minimizer&lt;/strong&gt;.  First we need to show that $J(x)$ is bounded from below. This is where we first use that $A$ has only non-negative eigenvalues. If $A$ were to have negative eigenvalues, let $v$ be its eigenvector, and look at $J(kv)$ for $k$ a scalar. Letting $k\nearrow\infty$ we see that $J(kv)\searrow -\infty$ and there cannot be an minimum. Next, if $A$ were to have a non-empty kernel, such that there exists some vector $v$ with the property $Av = 0$ and $\langle b,v\rangle \neq 0$, then again taking $J(kv)$ and sending $k$ to either $\pm\infty$, we will have that the functional $J$ is not bounded below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $A$ must only have positive eigenvalues. Now, let $\lambda$ be the smallest of $A$'s eigenvalues. Then we have $\langle x,Ax\rangle &amp;gt; \lambda \|x\|^2$. Using Cauchy-Schwartz inequality, we then have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ J(x) &amp;gt; \frac12 \lambda \|x\|^2 - \|b\|\|x\| $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that for $\|x\| &amp;gt; 2\|b\| / \lambda$, $J(x) &amp;gt; 0$. And by continuity, $J(x)$ is bounded below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $j = \inf J &amp;lt; 0$. (The infimum is negative since by choosing $w$ a sufficiently small multiple of $b$, we see that $J(w) &amp;lt; 0$.) By the definition of infimum, there exists a sequence of vectors $(v_k)$ such that $J(v_k) \searrow j$. Since the infimum is less than 0, by our previous computations we know that the sequence $(v_k)$ must all satisfy $\|v_k\| \leq 2\|b\|/\lambda$. Now, this set is a bounded closed ball in a finite dimensional vector space. In particular, it is compact. Hence up to a subsequence $v_k$ must converge. Now using continuity of $J$, you have that the limit element $v_\infty$ of the sequence &lt;em&gt;attains&lt;/em&gt; the minimum. That is $J(v_\infty) = j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2) &lt;em&gt;Computing the minimizer&lt;/em&gt;. Here we just use the Euler-Lagrange equations. At the minimizer, the functional is stationary. That is $(\frac{\delta}{\delta x}J)(v_\infty) = 0$. Taking the variation of the functional $J$, we get, using again that $A$ is self-adjoint, that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\delta}{\delta x}J = 0 \iff Ax - b = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A$ only has positive eigenvalues, it is invertible, so we can solve $v_\infty = A^{-1}b$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Formally, we can try to do the same thing in infinite dimension. But there are a few caveats. First is that it is not enough that $A$ only has positive eigenvalues. You need that $A$ is &lt;em&gt;coercive&lt;/em&gt; in the following sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle x, Ax\rangle &amp;gt; \lambda \|x\|^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $\lambda$. In finite dimensions that $A$ is positive implies the existence of this $\lambda$. In infinite dimensions this is not true. Imagine the $\ell^2$ Hilbert space with standard inner product and the standard basis $(e_k)$. Let $Ae_k = \frac1k e_k$. Then $A$ has only positive eigenvalues. But by taking larger and larger $k$'s, we can make the eigenvalue approach 0 as closely as we want, and violate coercivity. Coercivity allows us to guarantee that the functional $J(x)$ has a bottom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is that a closed bounded set in a Hilbert space is not necessarily compact. So we cannot conclude immediately that a minimizing sequence converges in the norm topology. However, using Banach-Alaoglu theorem, the minimizing sequence must have a convergent subsequence in the weak topology of your Hilbert space. So it is sufficient to show that your functional $J(x)$ is what is called &quot;sequentially weakly lower-semicontinuous&quot; to get the desired conclusion. (That condition just guarantees that for any weakly converging sequence $w_k\to_w w_\infty$, $\lim\inf J(w_k) \geq J(w_\infty)$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our particular case, things can be made slightly easier. If we in addition assume that $A$ is a bounded linear operator. Then we can use $(x,y) = \langle x,Ay\rangle$ as an equivalent inner product on our &lt;em&gt;Hilbert space&lt;/em&gt;. Then $A^{-1}b$ exists by the Riesz representation theorem, and our functional can be written as $J(x) = \frac12 (x,x) - (A^{-1}b, x)$. Now we run through our previous arguments again. Now $w_k\to_w w_\infty$, and therefore $(A^{-1}b,w_k) \to (A^{-1}b,w_\infty)$ strongly by the definition of weak limits. And we also use the property of weak limits that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim\inf (w_k,w_k) \geq (w_\infty,w_\infty)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gurantees lower semi-continuity as desired. So now we've shown that the minimizer exists, and computing the variation we see that the minimizer must satisfy $w_\infty = A^{-1}b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to repeat, a sufficient condition is that $A$ is self-adjoint, bounded, &lt;em&gt;and&lt;/em&gt; coercive. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-18T17:53:29.807" CommentCount="1" />
  <row Id="10865" PostTypeId="2" ParentId="10863" CreationDate="2010-11-18T18:06:39.233" Score="2" ViewCount="0" Body="&lt;p&gt;$$(z-\alpha)^{-n} = \frac1{z^n} \left(1-\frac\alpha z\right)^{-n}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now expand the (...) term using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;generalized binomial theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-11-18T18:06:39.233" />
  <row Id="10866" PostTypeId="1" AcceptedAnswerId="10869" CreationDate="2010-11-18T18:36:36.790" Score="1" ViewCount="156" Body="&lt;p&gt;This actually triggered me in my mind from &lt;a href=&quot;http://math.stackexchange.com/questions/10850/solving-an-equation&quot;&gt;here&lt;/a&gt;. After some playing around I notice that the relation $a^{\log_b x} = x^{\log_b a}$ is true for any valid value of $a,b$ and $x$. I am very inquisitive to see how this holds ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-19T02:02:27.203" Title="How $a^{\log_b x} = x^{\log_b a}$ ?" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;&lt;logarithms&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="10867" PostTypeId="2" ParentId="803" CreationDate="2010-11-18T18:37:38.170" Score="1" ViewCount="0" Body="&lt;p&gt;I really like this one (image taken from &lt;a href=&quot;http://www.cut-the-knot.org/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;cut-the-knot&lt;/a&gt; #4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/NRfbG.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;along with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+b)^2 = 4\cdot\frac{1}{2}a b + c^2$$&#xD;&#xA;$$\Leftrightarrow c^2 = a^2 + b^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's so clear and easy ...&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-18T18:37:38.170" CommunityOwnedDate="2010-11-18T18:37:38.170" />
  <row Id="10868" PostTypeId="1" AcceptedAnswerId="10878" CreationDate="2010-11-18T18:41:21.983" Score="3" ViewCount="262" Body="&lt;p&gt;I am quite a baffled now, I am not getting by &lt;em&gt;how&lt;/em&gt; it can be written that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;&amp;amp;\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan 43^\circ \cdots \log_{10} \tan 50^\circ  \\&#xA;&amp;amp;= \log_{10} \tan 40^\circ + \log_{10} \tan 41^\circ + \log_{10} \tan 42^\circ + \log_{10} \tan 43^\circ + \cdots + \log_{10} \tan 50^\circ &#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it even valid ? If yes,how ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:48:09.500" LastActivityDate="2010-12-17T01:48:09.500" Title="Baffling identity: $\prod \log_{10} \tan = \sum \log_{10} \tan$" Tags="&lt;trigonometry&gt;&lt;elementary&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10869" PostTypeId="2" ParentId="10866" CreationDate="2010-11-18T18:41:21.910" Score="5" ViewCount="0" Body="&lt;p&gt;For any positive $r$ and any $s$, you have &#xA;$$r^s = b^{\log_b(r^s)} = b^{s\log_b(r)}.$$&#xD;&#xA;So, taking $r=a$ and $s=\log_b(x)$, we have:&#xA;\begin{align*}&#xA;a^{\log_b(x)} &amp;amp;= b^{\log_b(x)\log_b(a)}\\&#xA;&amp;amp;= b^{\log_b(a)\log_b(x)}\\&#xA;&amp;amp; = b^{\log_b(x^{\log_b(a)})}\\&#xA;&amp;amp;= x^{\log_b(a)}.&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-18T18:41:21.910" />
  <row Id="10870" PostTypeId="2" ParentId="10856" CreationDate="2010-11-18T18:43:58.177" Score="3" ViewCount="0" Body="&lt;p&gt;You might want to look at this survey paper on clustering methods:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cchen1.csie.ntust.edu.tw:8080/students/2011/Survey%20of%20clustering%20algorithms.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cchen1.csie.ntust.edu.tw:8080/students/2011/Survey%20of%20clustering%20algorithms.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-18T18:43:58.177" />
  <row Id="10871" PostTypeId="2" ParentId="10846" CreationDate="2010-11-18T18:44:03.263" Score="1" ViewCount="0" Body="&lt;p&gt;Consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{n,N}(z) = \sin(Nx) \sum_{k = 1}^n \frac{\sin(kx)}{k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_k \frac{1}{k^2} f_{2^{k^3}, 2^{k^3 - 1}}(z)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for $x = \pi / (4n)$ and $N = 2n$ we get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(\pi/4) \sum_1^n \frac{1}{k} &amp;gt; \frac{1}{\sqrt{2}} \log n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have for some $x$ that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|s_{n_k + 1} - s_{n_k}| \geq \frac{1}{\sqrt{2}} \frac{1}{k^2} \log n_k$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we cannot have uniform convergence. I believe this is due to Hugo Steinhaus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I didn't make a mistake, but it is along these lines, I can correct it if I made an error.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-18T18:44:03.263" CommentCount="6" />
  <row Id="10872" PostTypeId="2" ParentId="10866" CreationDate="2010-11-18T18:45:31.997" Score="3" ViewCount="0" Body="&lt;p&gt;Using log properties, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{\log_b(x)} = b^{\log_b\left(a^{\log_b(x)}\right)} = b^{\log_b(x)\log_b(a)} = \left(b^{\log_b(x)}\right)^{\log_b(a)} = x^{\log_b(a)}$&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-18T18:45:31.997" CommentCount="1" />
  <row Id="10873" PostTypeId="2" ParentId="803" CreationDate="2010-11-18T18:46:16.503" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.math.ntnu.no/~hanche/pythagoras/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ntnu.no/~hanche/pythagoras/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3679" LastActivityDate="2010-11-18T18:46:16.503" CommunityOwnedDate="2010-11-18T18:46:16.503" />
  <row Id="10874" PostTypeId="1" AcceptedAnswerId="10886" CreationDate="2010-11-18T18:49:47.623" Score="2" ViewCount="110" Body="&lt;p&gt;In algebra, we learn that if a function $ f(x) $ has a one-to-one mapping, then we can find the inverse function $ f^{-1}(x) $. The method that I have seen taught is the &quot;horizontal line test&quot;: if any horizontal line touches the graph of the function more than once, then it must not be one-to-one. This method is not exactly rigorous, since any function with a non-finite range can not be completely viewed on a graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an analytic method to determine if a function is one-to-one? Is this possible in elementary algebra or calculus?&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-11-18T20:18:22.383" Title="Analytic method for determining if a function is one-to-one" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10875" PostTypeId="2" ParentId="803" CreationDate="2010-11-18T18:51:15.880" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tgs7a.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2044" LastActivityDate="2010-11-18T18:51:15.880" CommunityOwnedDate="2010-11-18T18:51:15.880" />
  <row Id="10876" PostTypeId="2" ParentId="10874" CreationDate="2010-11-18T18:53:40.707" Score="0" ViewCount="0" Body="&lt;p&gt;A function is invertible if and only if it is bijective, that is surjective (onto) and injective (one-to-one), so your statement is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to determine that if a function is injective, you assume $f(x) = f(y)$ and derive $x = y$, alternatively you can assume $x \neq y$ and show that $f(x) \neq f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have a linear function $f$, then we see that we need to show that if $f(x) = f(y)$ we have that $f(x - y) = 0$, so in other words we need to show that $f(x) = 0$ implies $x = 0$. This is usually easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case you want to show a function is bijective (and hence invertible) you first show that it is injective and then that if $f: X \to Y$ that for every $y \in Y$ you can find an $x \in X$ such that $f(x) = y$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T19:28:21.570" LastActivityDate="2010-11-18T19:28:21.570" CommentCount="3" />
  <row Id="10877" PostTypeId="2" ParentId="10859" CreationDate="2010-11-18T18:55:47.087" Score="1" ViewCount="0" Body="&lt;p&gt;The isomorphism arises simply from factoring $\rm\ \ aa\ X + b\ \  =\ \ a\ (ax+b/a)\:,\ $ namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ \ (aa\ X + b)\ \: \circ\: \ (AA\ X+B)\quad =\quad a\ a\ (A\ A\ \ X\ \ +\ \ B)\ \ +\ \ b\quad\ \ \  =\quad\ \ \  aa\ \ AA\ X\ +\ aaB + b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\ a\bigg(aX+\frac{b}a\bigg)\circ \:A\:\bigg(AX+\frac{B}A\bigg)\ =\ aA\:\bigg(aX+\frac{b}{aA}\bigg)\circ\bigg(AX+\frac{B}A\bigg)\ =\ aA\bigg(aA\ X\ + \frac{aaB + b}{aA}\bigg)\ $&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T19:24:25.173" LastActivityDate="2010-11-18T19:24:25.173" />
  <row Id="10878" PostTypeId="2" ParentId="10868" CreationDate="2010-11-18T18:56:01.867" Score="9" ViewCount="0" Body="&lt;p&gt;Yes it is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_{10} \tan 40^{\circ} + \log_{10} \tan 50^{\circ} = \log_{10} \tan 40^{\circ} + \log_{10} \cot 40^{\circ} = \log_{10} 1 = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly combine 41 and 49, 42 and 48 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product on the left side is $0$ as $\log_{10} \tan 45^{\circ} = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T18:56:01.867" CommentCount="5" />
  <row Id="10879" PostTypeId="2" ParentId="10868" CreationDate="2010-11-18T18:57:40.183" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Observe that $$\log_{10} \tan{40} + \log_{10} \tan{50} = \log_{10} \sin{40} - \log_{10} \cos{40} + \log_{10} \sin{50} - \log_{10} \cos{50}$$ Now use the fact that $\sin(90 - x) =\cos{x}$ and see that they cancel out.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-18T18:57:40.183" />
  <row Id="10881" PostTypeId="2" ParentId="10861" CreationDate="2010-11-18T19:24:24.813" Score="3" ViewCount="0" Body="&lt;p&gt;Assuming &quot;the grupoid&quot; means the grupoid (using the definition you quote) made up of that specific set and that specific operation (as opposed to some other operation), then the answer is &quot;Nothing beyond the fact that it is a grupoid.&quot; Because any property you care to name that is a property of &lt;em&gt;some&lt;/em&gt; but &lt;strong&gt;not all&lt;/strong&gt; grupoids will arise from a particular example of a &quot;set that is closed under some operation,&quot; but will also fail to arise from another particular example. That is, absent more information, you can't say anything else. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-18T19:24:24.813" CommentCount="1" />
  <row Id="10882" PostTypeId="1" AcceptedAnswerId="10889" CreationDate="2010-11-18T19:29:31.913" Score="1" ViewCount="94" Body="&lt;p&gt;In MATLAB, when you plot something, there's a tool available which is called &quot;curve fitting&quot;. And if you have a set of data points and a linear correlation, this tool will easily come up with an equation on the form y = ax + c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do exactly that, but I need to do it programmatically. I know that I can use &lt;code&gt;polyfit&lt;/code&gt; to find the coefficient, and &lt;code&gt;polyval&lt;/code&gt; to evaluate as many data points as I want, but what I can't figure out is how to find the constant c. This is very important, since I need to be able to plot the line as well next to my data points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3680" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T23:16:44.157" LastActivityDate="2010-11-18T23:16:44.157" Title="Find equation of curve fit programmatically in Matlab?" Tags="&lt;statistics&gt;&lt;math-software&gt;&lt;matlab&gt;&lt;regression&gt;" AnswerCount="2" />
  <row Id="10883" PostTypeId="2" ParentId="10882" CreationDate="2010-11-18T19:32:53.673" Score="0" ViewCount="0" Body="&lt;p&gt;Use the equations &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_linear_regression&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-18T19:32:53.673" CommentCount="1" />
  <row Id="10884" PostTypeId="2" ParentId="10866" CreationDate="2010-11-18T19:45:03.643" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ log(A^{\log X})\ =\ log\ X\ \ log\ A\ =\ log(X^{\log A})\:,\ $ where $\rm\ log := log_b$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-18T19:45:03.643" />
  <row Id="10885" PostTypeId="1" AcceptedAnswerId="10901" CreationDate="2010-11-18T19:53:49.433" Score="8" ViewCount="217" Body="&lt;p&gt;Hi, do you have an example of a non separable locally compact connected metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3681" LastActivityDate="2010-11-19T13:44:22.690" Title="Non separable locally compact connected metric space" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10886" PostTypeId="2" ParentId="10874" CreationDate="2010-11-18T19:59:49.630" Score="8" ViewCount="0" Body="&lt;p&gt;First, let me answer your question; but please keep reading, because there's lots more to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes: there is an analytic way to see if a function is one-to-one. For this, you need the &quot;analytic definition&quot; of being one-to-one. The definition is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f\text{ is one-to-one if and only if for all }a,b\text{ if }a\neq b\text{ then } f(a)\neq f(b).$$&#xD;&#xA;Logically, this is equivalent to:&#xA;$$f\text{ is one-to-one if and only if for all }a,b\text{ if }f(a)=f(b)\text{ then }a=b.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this provides a way to check if the function is one-to-one: if you can find $a\neq b$ such that $f(a)=f(b)$, then $f$ is not one-to-one; and this is essentially the best way of doing it: &lt;strong&gt;exhibit&lt;/strong&gt; a pair of distinct numbers that map to the same thing. To &lt;em&gt;prove&lt;/em&gt; that $\sin(x)$ is not one-to-one, all I need to do is say: &quot;Look, $0$ and $\pi$ are different, but $\sin(0)=\sin(\pi)$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that a function &lt;em&gt;is&lt;/em&gt; one-to-one, you can do it in any of two (equivalent) ways: show that if $a$ and $b$ are &lt;em&gt;any&lt;/em&gt; numbers with the property that $f(a)=f(b)$, then it must be the case that $a=b$; or show that if $a\neq b$, then $f(a)$ must be different from $f(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, to show that $f(x)=x^3$ is one to one, we can note that if $f(a)=f(b)$, then $a^3=b^3$, and taking cubic roots we conclude that $a=\sqrt[3]{a^3} = \sqrt[3]{b^3} = b$. So if $f(a)=f(b)$, then $a=b$. QED. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we can argue that if $a\neq b$, then either $a\lt b$ or $a\gt b$. If $a\lt b$, then $a^3\lt b^3$ (you can prove this easily using the properties of real numbers and inequalities), so $f(a)\neq f(b)$. If $a\gt b$, then $f(a)=a^3\gt b^3=f(b)$, hence $f(a)\neq f(b)$. Either way, $a\neq b$ implies $f(a)\neq f(b)$, so $f$ is one-to-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the standard way of showing, analytically, that a function is one-to-one. How one establishes the implication will depend on the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are some things you may want to remember:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: to specify a function, we usually need to specify at least two things: the domain of the function, and the value of the function at any point in the domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More often, we are interested in functions between two specific sets. In that case, we actually need to specify &lt;em&gt;three&lt;/em&gt; things: the domain, the set in which the images will lie, and the value of the functions at any point of the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we say that we have a function $f\colon X\to Y$ between two sets, then we mean that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;em&gt;every&lt;/em&gt; $x\in X$, there is an element of $y\in Y$ such that $f(x)=y$; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $x\in X$, there is only one $y\in Y$ with $f(x)=y$ (unique value).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, we need &lt;strong&gt;every&lt;/strong&gt; element of $X$ to have a unique image. Different elements of $X$ may have the same image, but a single element should not have multiple images. We call $X$ the &lt;strong&gt;domain&lt;/strong&gt; of $f$, and we call $Y$ the &lt;strong&gt;codomain&lt;/strong&gt; of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus, however, we are usually a bit sloppy. We almost never mention either the domain or the codomain! Instead, we agree that if we will either specify the domain explicitly, or else we will mean &quot;the natural domain&quot;. And we almost never mention the codomain at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have a function $f\colon X\to Y$, then we say that a function $g\colon Y\to X$ &quot;going the other way&quot; is &quot;the inverse of $f$&quot; if and only if two things happen:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For every $x\in X$, $g(f(x)) = x$  ($g$ &quot;un-does&quot; what $f$ does); and&lt;/li&gt;&#xA;&lt;li&gt;For every $y\in Y$, $f(g(y)) = y$  ($f$ &quot;un-does&quot; what $g$ does).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, for this to actually work, we need two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Given an element $y\in Y$, there can be at most one $x\in X$ such that $y=f(x)$. This is the &quot;one-to-one&quot; condition. If you think in terms of calculus and the graph, it is precisely the &quot;horizontal line test&quot;: for each value of $y$ (each horizontal line), there is at most one point in the domain where $f$ takes value $y$ (it cuts the graph in at most one point).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Reason:&lt;/em&gt; If you had $x\neq x&amp;#39;$ but with $y=f(x)=f(x&amp;#39;)$, then you would need $g(y)=x$ so that $x=g(y)=g(f(x))$, but you would also need $g(y)=x&amp;#39;$ because $x&amp;#39;=g(y) = g(f(x&amp;#39;))$. But a function cannot have two different values at the same point, so this is an impossible situation for $g$. The only way to solve this quandry is for either $g$ not to exist, or for $x$ and $x&amp;#39;$ not to exist. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given any element $y\in Y$, there is &lt;em&gt;at least&lt;/em&gt; one $x\in X$ such that $y=f(x)$. This is the &quot;onto&quot; or &quot;surjective&quot; part people have mentioned.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Reason:&lt;/em&gt; Since we also want $f(g(y))=y$ for every $y\in Y$, we need there to be an element of $X$, namely $g(y)$, that maps to $y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So: if $f\colon X\to Y$ has an inverse, then it must be &lt;em&gt;both&lt;/em&gt; one-to-one, &lt;strong&gt;and&lt;/strong&gt; onto the set $Y$. This is &lt;em&gt;necessary.&lt;/em&gt; In fact, it is also sufficient, and one can show that if there is an inverse, then there is one and only one inverse, so we call it $f^{-1}$ instead of $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's the thing: if your function satisfies the &lt;em&gt;first&lt;/em&gt; condition (one-to-one), but &lt;em&gt;not&lt;/em&gt; the second, then you can &quot;cheat&quot;: instead of thinking of $f$ as a function from $X$ to $Y$, we let $Y&amp;#39;=\mathrm{Image}(f)$, and then look at the function $\mathfrak{f}\colon X\to Y&amp;#39;$, with $\mathfrak{f}(x)=f(x)$ for every $x$; the only thing we changed is what we want the set $Y$ to be. This is not really the same as $f$: $\mathfrak{f}$ one &lt;em&gt;is&lt;/em&gt; both one-to-one and onto, so $\mathfrak{f}$ *does* have an inverse, even though $f$ does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you think of the function $f\colon\mathbb{R}\to \mathbb{R}$ given by $f(x)=e^x$, then $f$ is one-to-one, but is not onto ($f(x)$ does not take negative values). So this function is not invertible. However, if we tweak the function and think of it instead as $\mathfrak{f}\colon\mathbb{R}\to (0,\infty)$, given by $\mathfrak{f}(x)=e^x$, then $\mathfrak{f}$ *is* onto, is also one-to-one, so $\mathfrak{f}$ *is* invertible. The inverse is a function $f^{-1}\colon (0,\infty)\to\mathbb{R}$, so the only valid inputs are positive numbers. (You may know who $f^{-1}$ is: it's the natural logarithm). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, notice that to check if a function has an inverse, you need to know &lt;strong&gt;both&lt;/strong&gt; what $X$ and what $Y$ is. But in Calculus we almost never mention $Y$. So what can we do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we agree that we will take $Y$ to be &quot;the image of $f$&quot;; that is, the collection&#xA;$$\{ f(x) \mid x\text{ is in the domain of }f\}.$$&#xD;&#xA;That means that we always &quot;automatically&quot; assume our functions are &quot;onto&quot;. We say this by saying that the function is &quot;onto its image&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that agreement, in order to figure out if $f$ has an inverse, we just need to know if $f$ is one-to-one, which is why your calculus book says things like &quot;a function has an inverse if and only if it is one to one, if and only if it passes the horizontal line test.&quot; They are refering exclusively to functions whose codomain is &lt;em&gt;always&lt;/em&gt; taken to be the image. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T20:18:22.383" LastActivityDate="2010-11-18T20:18:22.383" CommentCount="3" />
  <row Id="10887" PostTypeId="2" ParentId="10817" CreationDate="2010-11-18T20:08:28.730" Score="3" ViewCount="0" Body="&lt;p&gt;Because $p$ and $q$ are coprime, then $p$ and $p-q$ are coprime: any irreducible factor of both would divide $p-(p-q)=q$, so it would divide both $p$ and $q$. Likewise, since $\lambda\neq 0$, $p$ and $p-\lambda q$ are coprime (same reason). And since $\lambda\neq 1$, then $p-\lambda q$ and $p-q$ are coprime; any shared factor between them is a factor of $\lambda(p-q)-(p-\lambda q) = (\lambda-1)p$ (hence of $p$) and of $(p-q)-(p-\lambda q) = (\lambda-1)q$ (hence of $q$), but $p$ and $q$ have no common factors other than units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $p$, $(p-q)$, and $(p-\lambda q)$ are pairwise coprime as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, the argument is the same as in any UFD (e.g., the integers): you have $r^2 = xyz$, with $x$, $y$ and $z$ relatively prime; looking at each irreducible factor of $r$, and at each irreducible factor of $x$, of $y$, and of $z$, you get that each irreducible factor of $x$, of $y$, and of $z$ must appear to an even exponent, so each of $x$, $y$, and $z$ are squares.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-18T20:08:28.730" CommentCount="1" />
  <row Id="10888" PostTypeId="1" AcceptedAnswerId="10890" CreationDate="2010-11-18T20:30:36.250" Score="1" ViewCount="121" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All, I am having some trouble in checking whether a sequence of functions &lt;em&gt;converges pointwise or uniformly&lt;/em&gt;. The problem is, sometimes, my intuition is right and sometimes its wrong. Finding the &lt;em&gt;limit function&lt;/em&gt; is the most difficult task I would say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to understand some simple examples, such as a &lt;em&gt;pointwise convergence&lt;/em&gt; of $f_{n}(x)= \frac{x}{n}$ over $\mathbb{R}$. But when given complex problems, I am having trouble in guessing the limit function. For example, I had trouble in solving this.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Test the convergence of $f_{n}(x) = n \Bigl[ (x)^{1/n} - 1 \Bigr]$ over $X=[1,t]$ for $t&amp;gt;1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given such functions, how can I improve my ability of guessing the limit function, and then proceeding to check whether the convergence is uniform or not. One method which I found extremely useful is the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f_{n} \to f$ uniformly on $X$ iff $y_{n} =\sup \{|f_{n}(x)-f(x)| : x \in X \} \to 0$. But as I see with some examples, this result is not always successful. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T20:37:29.037" LastActivityDate="2010-11-18T20:56:56.860" Title="Common Problems while showing Uniform Convergence of functions" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10889" PostTypeId="2" ParentId="10882" CreationDate="2010-11-18T20:39:18.987" Score="3" ViewCount="0" Body="&lt;p&gt;I'm going to give two answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer 1: If you already know $a$, then $c = \bar{y} - a\bar{x}$, where $\bar{x}$ and $\bar{y}$ are the means of the $x$ and $y$ values, respectively.  (This is actually in that pile of formulas in the link provided by Yuval Filmus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer 2: At least in my version of Matlab, polyfit gives both $a$ and $c$.  For example, if you generate a set of data via (I'm modifying the example given in the Matlab help)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = (0: 0.1: 2.5)';&#xA;y = erf(x); &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then call polyfit with $n=1$ (for a linear curve fit, or linear regression)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p = polyfit(x,y,1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The output is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.3554    0.3191&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which means your linear equation is $y = 0.3554x + 0.3191$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T22:21:16.490" LastActivityDate="2010-11-18T22:21:16.490" CommentCount="1" />
  <row Id="10890" PostTypeId="2" ParentId="10888" CreationDate="2010-11-18T20:46:46.733" Score="7" ViewCount="0" Body="&lt;p&gt;Intuition is developed by doing examples and becoming familiar with these things, so in that respect you are heading in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as &quot;guessing the limit function&quot;, that's not what we want to do! We don't want to guess, we want to &lt;em&gt;figure it out&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly if there is going to be uniform convergence, then there &lt;em&gt;has&lt;/em&gt; to be pointwise convergence. So you can begin by considering the limit of $f_n(x)$ for fixed $x$; if it does not exist, you're done: no convergence, pointwise &lt;em&gt;or&lt;/em&gt; uniform. If it always exists, then the limit function &lt;em&gt;must&lt;/em&gt; be the function&#xA;$$f(x_0) = \lim_{n\to\infty}f_n(x_0)$$&#xD;&#xA;for each $x_0$. I hope that by &quot;guessing the function&quot; you actually mean &quot;recognize it as a function I may already know&quot;. This is not always possible in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the functions $f_n(x)=n\left((x)^{1/n}-1\right)$, you will first want to consider the convergence at a given point $a$ that lies in $[1,t]$. Does the limit exist for a fixed $a$? That is, does&#xA;$$\lim_{n\to\infty} n\left(a^{1/n}-1\right)$$&#xD;&#xA;exist? If so, you want to try to express it as a function of $a$. This is a problem in sequences (or limits). This is an $\infty\times 0$ indeterminate, so you can try using L'Hopital's Rule to find it:&#xA;\begin{align*}&#xA;\lim_{n\to\infty}n\left(a^{1/n}-1\right) &amp;amp;= \lim_{x\to\infty}\frac{a^{1/x}-1}{\frac{1}{x}} = \lim_{x\to\infty}\frac{e^{\ln(a)/x}-1}{\frac{1}{x}}\\&#xA;&amp;amp;= \lim_{x\to\infty}\frac{-\frac{1}{x^2}\ln(a)e^{\ln(a)/x}}{-\frac{1}{x^2}}\\&#xA;&amp;amp;= \lim_{x\to\infty}\ln(a)e^{\ln(a)/x} = \ln(a).&#xA;\end{align*}&#xA;So there &lt;em&gt;is&lt;/em&gt; a pointwise limit, and $\lim\limits_{n\to\infty}f_n(a) = \ln(a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells you that the function you are converging &lt;em&gt;pointwise&lt;/em&gt; to is $f(x)=\ln(x)$. Now that you have a target function, you can check to see if the convergence is uniform in any of the many ways we have of checking of the convergence is uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Mind you, just like we have ways of determining that a sequence converges without actually determining what it converges &lt;em&gt;to&lt;/em&gt; (for example, showing that the sequence is Cauchy), we also have ways of determining if a sequence of functions converges pointwise without actually determining &lt;em&gt;to what&lt;/em&gt;. &lt;strong&gt;And&lt;/strong&gt; we have ways of determining if a sequence of functions converges &lt;em&gt;uniformly&lt;/em&gt; to &lt;em&gt;something&lt;/em&gt; without having to first determine if it converges pointwise, or &lt;em&gt;what&lt;/em&gt; it converges to. You could show that the sequence is &quot;uniformly Cauchy&quot; or &quot;Cauchy in the sup norm&quot;: you define &#xA;$$||f_n-f_m||_{\infty} = \sup\{ |f_n(x)-f_m(x)|\}.$$&#xD;&#xA;We say the sequence $\{f_n\}$ is Cauchy in the sup norm if and only if for every $\epsilon\gt 0$ there exists $N\gt 0$ such that if $n,m\geq N$, then $||f_n-f_m||_{\infty}\lt \epsilon$. It is straightforward to show that if $\{f_n\}$ is Cauchy in this sense, then for each $a$ you have $\{f_n(a)\}$ is Cauchy (in the usual sense), so there must be pointwise convergence (to &lt;em&gt;something&lt;/em&gt;); and you can also prove that if the sequence $\{f_n\}$ is Cauchy in the sup norm, then the convergence &lt;em&gt;will&lt;/em&gt; be uniform (to whatever it is that it converges). This is to figuring it if $\{f_n\}$ converges uniformly to a given $f$ like showing a sequence $\{a_n\}$ is Cauchy is to figuring out if it converges to a given $L$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-18T20:56:56.860" LastActivityDate="2010-11-18T20:56:56.860" CommentCount="5" />
  <row Id="10891" PostTypeId="2" ParentId="10885" CreationDate="2010-11-18T21:05:53.103" Score="0" ViewCount="0" Body="&lt;p&gt;This isn't so much an answer (so please don't vote this up) so much as a pointer on some things that don't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever one cooks up it can't be an infinite-dimensional Banach space since those aren't locally compact. Uncountably many copies of the unit interval each horizontal line segment some fixed distance away from all others doesn't work either since it's not connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This slightly more clever construction is also no good: Consider the set $[0,1] \times [0,1]$ with the pseudometric $d&amp;#39;$ given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d&amp;#39;((x_1,y_1),(x_2,y_2)) = \begin{cases}&#xD;&#xA;\vert x_1 - x_2 \vert &amp;amp; \text{ if } y_1 = y_2 \lor x_1 = 0 \lor x_2 = 0  \\\\&#xD;&#xA;2                     &amp;amp; \text{ if } y_1 \neq y_2 \land x_1,x_2 \neq 0,&#xD;&#xA;\end{cases} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and let $X$ be its Kolmogorov quotient and $d$ its associated metric. It satisfies all the desired properties except local compactness - $\{0\}\times [0,1]$ doesn't have a compact neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in conjuring up an example seems to come from the fact that you want a nonseparable connected space where every point has a separable neighbourhood (compact metric implies 2nd-countable).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-18T21:05:53.103" CommentCount="1" />
  <row Id="10892" PostTypeId="1" CreationDate="2010-11-18T21:20:13.603" Score="1" ViewCount="144" Body="&lt;p&gt;How to prove the following theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the binomial numbers $C(n,p)$ and $C(n,q)$ are equal, with $p$ different from $q$, then $p+q = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($n$, $p$ and $q$ are natural numbers)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-18T22:19:50.610" LastActivityDate="2010-11-19T16:56:38.343" Title="$C(n,p) = C(n,q) =&gt;  p + q= n$" Tags="&lt;binomial-coefficients&gt;" AnswerCount="6" />
  <row Id="10893" PostTypeId="2" ParentId="10892" CreationDate="2010-11-18T21:21:33.990" Score="7" ViewCount="0" Body="&lt;p&gt;The binomial coefficients are &lt;a href=&quot;http://faculty.salisbury.edu/~despickler/pascgalois/classroom/NumberTheory/Fisher/Binomial.pdf&quot; rel=&quot;nofollow&quot;&gt;Unimodal&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-18T21:21:33.990" CommentCount="1" />
  <row Id="10894" PostTypeId="2" ParentId="7064" CreationDate="2010-11-18T21:22:15.223" Score="0" ViewCount="0" Body="&lt;p&gt;This is so interesting, especially the idea of language dictating direction!  I grew up in Argentina and attended a bilingual school...so I learned both European and US notation but am stronger in the US because we used it more often.  Now I'm teaching language and will do a lesson on the different &quot;formats&quot; for simple math.  Gracias a todos!&lt;/p&gt;&#xA;" OwnerUserId="3684" LastActivityDate="2010-11-18T21:22:15.223" CommunityOwnedDate="2010-11-18T21:22:15.223" CommentCount="1" />
  <row Id="10895" PostTypeId="2" ParentId="2949" CreationDate="2010-11-18T21:27:47.037" Score="1" ViewCount="0" Body="&lt;p&gt;Apart from the above sets of interesting and surprising results, this one should also be mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaussian quadrature rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $n$ point Gaussian quadrature rule, yields an exact result for polynomials of degree $2n − 1$ or less.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-18T21:27:47.037" CommunityOwnedDate="2010-11-18T21:27:47.037" CommentCount="1" />
  <row Id="10896" PostTypeId="1" AcceptedAnswerId="10915" CreationDate="2010-11-18T21:28:32.963" Score="2" ViewCount="94" Body="&lt;p&gt;Here's the probability (I think) that a particle in Brownian motion &#xA;(w/ standard deviation $\sqrt{t}$) will exceed $m$ between times $t_1$ and $t_2$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{2\sqrt{2\pi}}\int_{-\infty }^m \frac1{\sqrt{t_1}}e^{-\frac{x^2}{2t_1}}\left(1+\mathrm{erf}\left(\frac{m-x}{2\sqrt{t_2-t_1}}\right)\right)\mathrm{d}x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, in Mathematica (slightly different form): &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p[m_,t1_,t2_] := Integrate[ &#xA; PDF[NormalDistribution[0,Sqrt[t1]]][x]* &#xA; CDF[NormalDistribution[x, Sqrt[2]*Sqrt[t2-t1]]][m], &#xA;{x,-Infinity,m}, Assumptions -&amp;gt; { &#xA; t2 &amp;gt;= t1 &amp;gt;= 0, Element[m,Reals], Element[t2, Reals], Element[t1, Reals]} &#xA;] &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mathematica&lt;/em&gt; can numerically integrate this for specific values of &lt;code&gt;m&lt;/code&gt;, &#xA;&lt;code&gt;t1&lt;/code&gt;, and &lt;code&gt;t2&lt;/code&gt;, but it's not superfast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now want to find the partial derivatives of &lt;code&gt;p[]&lt;/code&gt; with respect to each &#xA;of its variables. Ideally in closed-form (&lt;em&gt;Mathematica&lt;/em&gt; can't find one), &#xA;but a good approximation if not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried power series and a few other techniques, but I've found &#xA;nothing good enough for a wide range of values for &lt;code&gt;m&lt;/code&gt;, &lt;code&gt;t1&lt;/code&gt;, and &lt;code&gt;t2&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this problem will calculate the &quot;Greeks&quot; for box options:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&quot;&gt;http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-11-19T16:19:59.153" LastActivityDate="2010-11-19T16:19:59.153" Title="Derivatives of Brownian motion or Box Options Greeks" Tags="&lt;calculus&gt;&lt;numerical-methods&gt;&lt;finance&gt;&lt;computer-algebra-systems&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10897" PostTypeId="1" AcceptedAnswerId="10899" CreationDate="2010-11-18T21:30:08.097" Score="1" ViewCount="92" Body="&lt;p&gt;I need to re-arrange the following equation to make $d$ the subject [i.e., solve for $d$]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \sqrt{\frac{d}{be}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square root part is really confusing me, any help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3686" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T00:15:15.650" LastActivityDate="2010-11-19T00:15:15.650" Title="Re-arranging an equation with a square root" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10898" PostTypeId="2" ParentId="10892" CreationDate="2010-11-18T21:35:14.540" Score="-3" ViewCount="0" Body="&lt;p&gt;Use the definition of binomial coefficients,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C(n,p) = n! / (p! * (n-p)!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C(n, n-p) = n! / ((n-p)! * p!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C(n,p) = C(n,n-p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let q = n - p, then we have C(n,p) = C(n,q), p != q (necessarily), and p + q = n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could be more rigorous, but this is the gist.&lt;/p&gt;&#xA;" OwnerUserId="3685" LastActivityDate="2010-11-18T21:35:14.540" CommentCount="1" />
  <row Id="10899" PostTypeId="2" ParentId="10897" CreationDate="2010-11-18T21:36:12.230" Score="2" ViewCount="0" Body="&lt;p&gt;Ok. Squaring both sides you obtain $$a^{2} = \frac{d}{be} \Longrightarrow d = (be) \times a^{2}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-18T21:56:47.377" LastActivityDate="2010-11-18T21:56:47.377" CommentCount="9" />
  <row Id="10900" PostTypeId="2" ParentId="10892" CreationDate="2010-11-18T21:42:31.100" Score="5" ViewCount="0" Body="&lt;p&gt;It is suffices to show that $q! (n-q)! = p! (n-p)!$ implies $p+q=n$ ($p \ne q$). Assume without loss of generality that $p &amp;gt; q$. Then, we need to show that $p!/q! = (n-q)!/(n-p)!$ implies $p+q=n$, or $(q+1) (q+2) \cdots p = (n-p+1) (n-p+2) \cdots (n-q)$ implies $p+q=n$. Since both sides of the last equality have identical number of terms ($p-q$), we must have $q+1 = n-p+1$, that is $p+q=n$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-18T21:57:59.487" LastActivityDate="2010-11-18T21:57:59.487" />
  <row Id="10901" PostTypeId="2" ParentId="10885" CreationDate="2010-11-18T21:48:21.550" Score="12" ViewCount="0" Body="&lt;p&gt;Well, no one will be able to provide an example, since it doesn't exist. In the appendix A to chapter $1$ of his book 'A Comprehensive Introduction to Differential Geometry, volume $1$' Spivak proves that a connected, locally compact, paracompact space is $\sigma$-compact. This last term means that you can write your space as a countable union of compact subspaces. It's easy to see that every $\sigma$-compact space is Lindelöf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you are considering a connected, locally compact metric space, so it is $\sigma$-compact and then Lindelöf. But, in metric spaces, it is equivalent to being separable.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-19T13:44:22.690" LastActivityDate="2010-11-19T13:44:22.690" CommentCount="1" />
  <row Id="10902" PostTypeId="2" ParentId="2949" CreationDate="2010-11-18T22:46:36.460" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Probabilistic_method&quot; rel=&quot;nofollow&quot;&gt;Erdös' Probabilistic Method&lt;/a&gt; because it is so elegant.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-18T22:46:36.460" CommunityOwnedDate="2010-11-18T22:46:36.460" CommentCount="1" />
  <row Id="10903" PostTypeId="1" CreationDate="2010-11-18T22:52:28.040" Score="3" ViewCount="57" Body="&lt;p&gt;For a university project, I am implementing an automated regression analysis tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have very little background in statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what books / articles / material would you suggest I could use to brush up on this topic, based on your experiences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3689" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-18T23:17:26.573" LastActivityDate="2010-11-19T03:40:53.907" Title="Recommended Reading on Regression Analysis?" Tags="&lt;statistics&gt;&lt;reference-request&gt;&lt;regression&gt;" AnswerCount="2" />
  <row Id="10904" PostTypeId="1" AcceptedAnswerId="10926" CreationDate="2010-11-18T22:56:04.143" Score="1" ViewCount="90" Body="&lt;p&gt;Question:&lt;br&gt;&#xA;Let $a$ be a quadratic residue to a prime modulus $p$. Prove $a^{(p-1)/2} \equiv 1 \pmod{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at a solution:&lt;br&gt;&#xA;\begin{align*}&#xA;&amp;amp;a\text{ is a quadratic residue}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow a\text{ is a residue class of $p$ which has even index $c$ relative to a primitive root $g$}\\ &#xA;&amp;amp;\Longrightarrow a \equiv g^c \pmod{p}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow a \equiv g^{2k} \pmod{p}\text{ where $2k=c$}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow g^{2kv} \equiv g^{c} \pmod{p}\text{ for some natural number $v$}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow 2kv \equiv c \pmod{p-1}\text{ (by a proof in class)}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow 2kv \equiv 2k \pmod{p-1}\\&#xA;&amp;amp;\Longrightarrow kv \equiv k \pmod{(p-1)/2}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow v \equiv k(k^{-1}) \pmod{(p-1)/2}\text{ since $\gcd(2k, p-1)$ is not equal to 1}\\&#xA;&amp;amp;\Longrightarrow k^{-1} \text{ (k inverse exists)}\\&lt;br&gt;&#xA;&amp;amp;\Longrightarrow v \equiv 1 \pmod{(p-1)/2}.&#xA;\end{align*}  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this implies that $g^{(p-1)/2} \equiv 1 \pmod{p}$, is this correct?&lt;br&gt;&#xA;Although what I was required to show was $a^{(p-1)/2} \equiv 1 \pmod{p}$, am I on the right track, how do I show this, I've spent quite some time on this and looked over all the proofs in my notes, I can't seem to find out how.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T04:38:39.400" LastActivityDate="2010-11-19T04:45:08.737" Title="Let a be a quadratic residue to a prime modulus p. Prove a^((p-1)/2) = 1 mod p" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10905" PostTypeId="1" AcceptedAnswerId="10908" CreationDate="2010-11-18T23:11:41.337" Score="0" ViewCount="41" Body="&lt;p&gt;My algebra is sooo rusty so this should be an easy one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area of a rectangle is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = W x H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you're given A (say 150) and a ratio of W:H (say 3:2) how can you calculate W and H?&lt;/p&gt;&#xA;" OwnerUserId="3690" LastActivityDate="2010-11-18T23:54:01.680" Title="Calculate width and Height of rectangle containing given area and conforming to given ratio." Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="10906" PostTypeId="2" ParentId="10903" CreationDate="2010-11-18T23:13:32.067" Score="4" ViewCount="0" Body="&lt;p&gt;Off the top of my head:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0072472278&quot; rel=&quot;nofollow&quot;&gt;Bevington &lt;em&gt;Data Reduction and Error Analysis for the Physical Sciences&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471170828&quot; rel=&quot;nofollow&quot;&gt;Draper, Smith &lt;em&gt;Applied Regression Analysis&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471746967&quot; rel=&quot;nofollow&quot;&gt;Chatterjee, Hadi &lt;em&gt;Regression Analysis by Example&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471691178&quot; rel=&quot;nofollow&quot;&gt;Belsley, Kuh, Welsch &lt;em&gt;Regression Diagnostics: Identifying Influential Data and Sources of Collinearity&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521791685&quot; rel=&quot;nofollow&quot;&gt;Monahan &lt;em&gt;Numerical Methods of Statistics&lt;/em&gt;&lt;/a&gt; (not entirely about regression, but the chapter on regression has pointers on implementation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there are other nice refs, but I'll let the others point them out.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-18T23:13:32.067" CommentCount="2" />
  <row Id="10907" PostTypeId="2" ParentId="10892" CreationDate="2010-11-18T23:15:33.880" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ In order to prove that $\rm\ \binom{n}{p}\ =\ \binom{n}{q}\ \:\Rightarrow\ \  \ p+q = n\ \ \ if\ \ \ p \ne q\ \ $ for $\rm\ \ p,q\in [0, n]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider how to prove that $\rm\ \ \ sin(p)\ =\ sin(q)\ \Rightarrow\ p+q = \pi\ \ \ if\ \ \ p \ne q\ \ $ for $\rm\ \ p,q\in [0,\pi]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MORAL&lt;/strong&gt; $\ $ A proper choice of $\rm \:sin\:$ can get you &quot;over the hump&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-18T23:40:38.043" LastActivityDate="2010-11-18T23:40:38.043" />
  <row Id="10908" PostTypeId="2" ParentId="10905" CreationDate="2010-11-18T23:30:02.880" Score="2" ViewCount="0" Body="&lt;p&gt;Right so  you have $W:H$ is $3:2$, which means $$150 = 3x \times 2x$$ So this is just a simple equation which you can solve for. That is $6x^{2}=150$ which says $x^{2} =25$ and hence $x =\sqrt{25}$ that is $x =5 $. So $W = 3 \times 5 =15 $ and $H= 2 \times 5 =10$ and you can see that they have ratio $3 : 2$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-18T23:49:34.907" LastActivityDate="2010-11-18T23:49:34.907" CommentCount="3" />
  <row Id="10909" PostTypeId="2" ParentId="10904" CreationDate="2010-11-18T23:32:14.123" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Put $\rm\ a = b^2\ $ and use Fermat's little theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ The converse is also true - it is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Euler's criterion&lt;/strong&gt;.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-19T03:50:13.880" LastActivityDate="2010-11-19T03:50:13.880" />
  <row Id="10910" PostTypeId="1" CreationDate="2010-11-18T23:51:54.423" Score="1" ViewCount="99" Body="&lt;p&gt;I'm writing a program to solve this proof, but I don't know how to go about solving it.&#xA;If anyone has some insight it would be great help.  Thanks&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For every odd integer $n$, $3 \leq n \leq 199$, there exists an integer $m \geq 0$, and a prime number $p$, such that $n = 2^{m} + p$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3691" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-18T23:54:46.883" LastActivityDate="2010-11-18T23:54:46.883" Title="Method to solving this proof with a java app" Tags="&lt;prime-numbers&gt;&lt;proof&gt;" CommentCount="6" />
  <row Id="10911" PostTypeId="2" ParentId="10905" CreationDate="2010-11-18T23:54:01.680" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\ \ \frac{W}H = \frac{3}2\ \Rightarrow\ 2\:W = 3\:H\:,\ $ so $\rm\ 300 = 2\:W\:H = 3\:H\:H\ \Rightarrow\ H = \ldots $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-18T23:54:01.680" CommentCount="2" />
  <row Id="10912" PostTypeId="2" ParentId="10885" CreationDate="2010-11-19T00:34:27.023" Score="3" ViewCount="0" Body="&lt;p&gt;Meanwhile, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_line_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;long line&lt;/a&gt; is an example of a space that is non-separable, locally compact, connected  and &lt;em&gt;locally&lt;/em&gt; metrizable, but not metrizable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long line is the space $[0,1)\times\omega_1$, consisting of $\omega_1$ many copies of the half-open unit interval, using the (inverse lexical) order topology, where $\omega_1$ is the first uncountable ordinal. These intervals fit together neatly, so that every initial segment of the order looks exactly like an interval in the real line. Thus, the long line is locally compact, locally metrizable and connected. But the long line is non-separable, since every countable subset of $\omega_1$ and hence of the long line is bounded. Lastly, it is not metrizable, since otherwise this would contradict Nuno's answer.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-11-19T00:43:03.973" LastActivityDate="2010-11-19T00:43:03.973" CommentCount="2" />
  <row Id="10913" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T00:38:33.103" Score="2" ViewCount="0" Body="&lt;p&gt;I was very surprised to learn about the Cantor set, and all of its amazing properties. The first one I learnt is that it is uncountable (I would never have told), and that it has measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was shown this example as a freshman undergraduate, for an example of a function that is Riemann-integrable but whose set of points of discontinuity is uncountable. (equivalently, that this set has measure zero). This came more as a shock to me, since I had already studied some basic integrals in high school, and we had defined the integral only for continuous functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, after learning topology and when learning measure theory, I was extremely shocked to see that this set can be modified to a residual set of measure zero! I think the existence of such sets and the disconnectednes of topology and measure still gives me the creeps...&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-19T00:38:33.103" CommunityOwnedDate="2010-11-19T00:38:33.103" />
  <row Id="10915" PostTypeId="2" ParentId="10896" CreationDate="2010-11-19T01:26:12.523" Score="4" ViewCount="0" Body="&lt;p&gt;In view of your previous &lt;a href=&quot;http://math.stackexchange.com/questions/9608/determining-distribution-of-maximum-of-dependent-normal-variables&quot;&gt;post&lt;/a&gt;,&#xA;the probability you are considering is actually&#xA;$$&#xD;&#xA;\bar P(m,t_1,t_2) = 1 - \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x},\;\; m \in {\bf R}.&#xD;&#xA;$$&#xD;&#xA;This is the complement of the probability considered in your previous post.&#xA;So, we may consider instead the partial derivatives of &#xA;$$&#xD;&#xA;P(m,t_1,t_2) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x}.&#xD;&#xA;$$&#xD;&#xA;Now, erf(x) has derivative ${\rm erf}&amp;#39;(x)= \frac{2}{{\sqrt \pi  }}e^{ - x^2 }$, $x &amp;gt; 0$. &#xA;Then, the partial derivative with respect to $t_2$ is obtained as follows:&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial t_2 }}P(m,t_1 ,t_2 ) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {\frac{\partial }{{\partial t_2 }}{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x},&#xD;&#xA;$$&#xD;&#xA;and explicitly,&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial t_2 }}P(m,t_1 ,t_2 ) = &#xD;&#xA;\frac{{ - 1}}{{\sqrt {8 \pi ^2 t_1 (t_2  - t_1 )^3 } }}\int_{ - \infty }^m {\exp \bigg[ - \frac{{(m - x)^2 }}{{4(t_2  - t_1 )}} - \frac{{x^2 }}{{2t_1 }}\bigg](m - x){\rm d}x}.&#xD;&#xA;$$&#xD;&#xA;(The probability that the maximum is smaller than $m$ decreases as $t_2$ increases, hence the minus sign in this equation.) To find the partial derivative with respect to $m$ corresponds to taking the derivative of a convolution, according to the rule $(f * g)&amp;#39; = f&amp;#39; * g$. Specifically, this says that&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial m }}P(m,t_1 ,t_2 ) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {\frac{\partial }{{\partial m }}{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2  - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x},&#xD;&#xA;$$&#xD;&#xA;and explicitly,&#xA;$$&#xD;&#xA;\frac{{\partial}}{{\partial m }}P(m,t_1 ,t_2 ) =&#xD;&#xA;\frac{{1}}{{\sqrt {2  \pi ^2 t_1 (t_2  - t_1 ) } }}\int_{ - \infty }^m {\exp \bigg[ - \frac{{(m - x)^2 }}{{4(t_2  - t_1 )}} - \frac{{x^2 }}{{2t_1 }}\bigg]{\rm d}x}.&#xD;&#xA;$$&#xD;&#xA;Maybe you can simplify further. The partial derivative with respect to $t_1$ is apparently a quite complicated expression.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-19T01:26:12.523" CommentCount="8" />
  <row Id="10916" PostTypeId="1" CreationDate="2010-11-19T01:42:05.760" Score="3" ViewCount="193" Body="&lt;p&gt;I learned of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Plate_trick&quot; rel=&quot;nofollow&quot;&gt;plate trick&lt;/a&gt; via Wikipedia, which states that this is a demonstration of the fact that SU(2) double-covers SO(3). It also offers a &lt;a href=&quot;http://gregegan.customer.netspace.net.au/APPLETS/21/21.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to an animation of the &quot;belt trick&quot; which is apparently equivalent to the plate trick. Since I've thought most about the belt version, I'll phrase my question in terms of the belt trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not clear on how the plate/belt trick relates to the double covering. Specifically, I am looking for a sort of translation of each step of the belt trick into the Lie group setting. For example, am I correct in interpreting the initial twisting of the belt as corresponding to the action of a point in SU(2)? Which point? Do I have the group right? &lt;/p&gt;&#xA;" OwnerUserId="3694" LastEditorUserId="3694" LastEditorDisplayName="" LastEditDate="2010-11-19T16:54:32.150" LastActivityDate="2010-11-19T16:54:32.150" Title="Can somebody explain the plate trick to me?" Tags="&lt;recreational-mathematics&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10917" PostTypeId="2" ParentId="10866" CreationDate="2010-11-19T02:02:27.203" Score="2" ViewCount="0" Body="&lt;p&gt;Otherwise :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the base of logarithm, we have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{b}a= \displaystyle\frac{\log_{x}a}{\log_{x}b}$$ and&#xA;$$\log_{x}b= \displaystyle\frac{\log_{b}b}{\log_{b}x} = \frac{1}{\log_{b}x} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By combine these two ecuations, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_{b}a= (\log_{b}x)(\log_{x}a) \Leftrightarrow (\log_{b}a)(\log_{x}x)= (\log_{b}x)(\log_{x}a) \Leftrightarrow \log_{x}x^{\log_{b}a} = \log_{x}a^{\log_{b}x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By last, canceling $\log_{x}$ on both sides, we have :&#xA;$$x^{\log_{b}a} = a^{\log_{b}x}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-19T02:02:27.203" />
  <row Id="10918" PostTypeId="2" ParentId="10916" CreationDate="2010-11-19T02:52:49.787" Score="6" ViewCount="0" Body="&lt;p&gt;A 360 degree rotation of the arm results in a very sore arm unless you unwind it or wind it a second time. Think about the rotation of the hand as being a path in the space of all rotations of 3-space. Or if you really want to rotate something, tape a belt to a soccer ball and tape the other end of the belt to the wall. Just rotate the ball a full rotation, the belt gets a twist. Now rotate again; the belt has two full twists. The twists in the belt can be undone by looping the ball back through the belt as in the animation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram in the lower right shows the paths in the space $SO(3)$ which is the $3$-ball with its antipodal points on its boundary identified. That is why the doubled path appears to be broken; it is passing through antipodal points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might play with a Mobius band and the pencil loop which maps twice around to gain more intuition. The point of thinking of the Mobius band is that it is embedded in the space of rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is really worth thinking this through while going to sleep or while riding the bus!&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-11-19T02:52:49.787" CommentCount="2" />
  <row Id="10919" PostTypeId="2" ParentId="10903" CreationDate="2010-11-19T03:40:53.907" Score="3" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Thisted, &lt;em&gt;Elements of Statistical&#xA;Computing&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;Rao, &lt;em&gt;Linear Statistical&#xA;Inference and Its Applications&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Chapter 3 and 4 of Hastie, Tibshirani and Friedman, &lt;em&gt;Elements of Statistical Learning&lt;/em&gt; (&lt;a href=&quot;http://www-stat.stanford.edu/~tibs/ElemStatLearn/&quot; rel=&quot;nofollow&quot;&gt;downloadable&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-11-19T03:40:53.907" CommentCount="1" />
  <row Id="10920" PostTypeId="1" AcceptedAnswerId="10925" CreationDate="2010-11-19T04:01:37.127" Score="0" ViewCount="92" Body="&lt;p&gt;Let a be a natural number.&lt;br&gt;&#xA;Suppose  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^4 \not\equiv 1 \pmod {29}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^7 \not\equiv 1 \pmod {29}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{14} \not\equiv 1 \pmod {29}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the order of a modulo 29.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Euler's Criterion, I have $a^{14} = a^{(29-1)/2} \not\equiv 1 \pmod {29}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about solving this problem, I have not seen this kind of problem before.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T04:12:07.643" LastActivityDate="2010-11-19T22:04:15.720" Title="Let a be a natural number, Suppose $a^4, a^7, a^{14} \not\equiv 1 \pmod{29}$, calculate the order of a modulo 29." Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="4" />
  <row Id="10921" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T04:03:54.773" Score="1" ViewCount="0" Body="&lt;p&gt;I remember a homework question in elementary school. Something like this: Billy and Jane's house is x blocks east and y blocks north of school. Billy walks home by walking east for x blocks and then north for y blocks. Jane decides to take a short cut: she walks alternately a block north and a block east. (There is a picture: Jane's route is a step-like hypotenuse.) Is Jane's route really a short cut?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it is exactly the same distance, but I found this really hard to digest. I knew that in the triangle the sum of the two sides would exceed the hypotenuse. &lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-11-19T04:03:54.773" CommunityOwnedDate="2010-11-19T04:03:54.773" CommentCount="4" />
  <row Id="10922" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:07:55.770" Score="3" ViewCount="0" Body="&lt;p&gt;First we assume $(a,29) = 1$. Else order doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use Fermat's little theorem (or) Euler's theorem $a^{\phi(n)} \equiv 1 \pmod n$, whenever $(a,n) = 1$. $\phi(29) = 28$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a^l \equiv 1 \pmod n$, then order of $a$ divides $l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So order has to divide $28$ but not $4$,$7$,$14$. So the order has to be $28$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T04:09:48.843" LastActivityDate="2010-11-19T04:09:48.843" CommentCount="1" />
  <row Id="10923" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:08:22.453" Score="2" ViewCount="0" Body="&lt;p&gt;What do you know about the multiplicative group modulo 29?  What are the possible orders of nonzero elements mod 29?  (It's not hard to answer if you know the structure of the group; it's also possible to compute directly, since there are only 28 elements to check.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the structure, you should be able to work out the answer.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-11-19T04:08:22.453" />
  <row Id="10924" PostTypeId="1" CreationDate="2010-11-19T04:12:27.157" Score="2" ViewCount="72" Body="&lt;p&gt;I'm not sure if this is the proper place to ask this, but I am looking for an English translation of Voronoy's paper on the generalized Pell's equation.  It might be his doctoral dissertation; I'm not sure.  I believe the title is &quot;On a generalization of the Algorithm of Continued Fractions&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastEditorUserId="2505" LastEditorDisplayName="" LastEditDate="2010-11-19T04:18:09.650" LastActivityDate="2010-11-19T06:56:01.820" Title="English translation of Voronoy paper on generalized Pell's equation." Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="10925" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:18:15.853" Score="3" ViewCount="0" Body="&lt;p&gt;Presumably, $\gcd(a,29)=1$; else, $a\equiv 0\pmod{29}$, and all your incongruences hold, and there is no order as Sivaram points out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Fermat's Little Theorem, you know that $a^{28}\equiv 1\pmod{29}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the &lt;em&gt;smallest&lt;/em&gt; positive integer $j$ such that $a^j\equiv 1\pmod{29}$? It is at most $28$. If we divide $28$ by $j$ with remainder, $28 = qj + r$, $0\leq r\lt j$, we have that&#xA;$$a^r = a^{28-qj} \equiv a^{28}(a^{j})^{-q} \equiv 1 \pmod{29}.$$&#xD;&#xA;Since $j$ is the &lt;em&gt;smallest&lt;/em&gt; positive integer such that $a^j\equiv 1 \pmod{29}$, but $r$ is nonnegative and smaller, you must have $r=0$. (This is a standard argument, shows up all the time; it is really the fact that the set of exponents $r$ for which $a^r\equiv 1 \pmod{29}$ form an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;ideal of the integers&lt;/a&gt;). So $j$ divides $28$. That is, the order must divide $28$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what can $j$ be? It must be positive, it must divide $28=2^2\times 7$. So it must be either $1$, $2$, $4$, $7$, $14$, or $28$. Given the information you have, what must $j$ be?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T22:04:15.720" LastActivityDate="2010-11-19T22:04:15.720" CommentCount="1" />
  <row Id="10926" PostTypeId="2" ParentId="10904" CreationDate="2010-11-19T04:45:08.737" Score="2" ViewCount="0" Body="&lt;p&gt;Bill has succintly told you how to prove the result. But you were also asking for comments on your proposed argument. I will address that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In line 5, where did the $v$ come from, and what is its role? Notice that you can take $v=1$ and what you write is true. So how is this giving you any information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In line 9, you are already using an inverse of $k$, even though you only assert its existence in the next line. You cannot do that: in order to use it, you must first show it exists, and you haven't done it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But assuming it does exist, and that your entire chain of argument holds, you'll notice that all you concluded was that $v\equiv 1\pmod{(p-1)/2}$. This is of course natural: you have $a=g^c=g^{2k}=g^{2kv}$; you can always take $v=1$ and that will work regardless of $k$, $g$, $a$... And you probably know now that it does not lead to a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you had not actually proven anything. You've only written $a$ as an even power of a primitive root, and &lt;em&gt;that's it&lt;/em&gt;. Lines 1 through 4 are correct; but from line 5 through the end, you are just spinning your wheels and not getting any closer to the result you want.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-19T04:45:08.737" CommentCount="3" />
  <row Id="10927" PostTypeId="1" AcceptedAnswerId="10930" CreationDate="2010-11-19T04:54:42.533" Score="11" ViewCount="262" Body="&lt;p&gt;This is motivated by looking at the calculus exams of some of my undergraduate students. A recurring mistake is assuming that the derivative of the product of functions is a product of derivatives and the derivative of the quotient of two functions is the quotient of their derivatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be an ill formed question, but do their exist a pair of functions such that these rules hold. Explicitly, do there exist $f(x),g(x)$ satisfying all of:&#xA;\begin{align*}&#xA;f&amp;amp;\neq 0;\\&#xA;\frac{d(fg)}{dx}&amp;amp;=\frac{df}{dx}\cdot \frac{dg}{dx}\\&#xA;\text{and }\frac{d(f/g)}{dx}&amp;amp;=\frac{df}{dx}/ \frac{dg}{dx}\text{ ?}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to clarify my question, looking at the responses so far. I am looking for functions satisfying all three conditions above simultaneously.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:29:34.687" LastActivityDate="2010-11-19T16:29:34.687" Title="When do the Freshman's dream product and quotient rules for differentiation hold?" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" />
  <row Id="10928" PostTypeId="2" ParentId="10920" CreationDate="2010-11-19T04:59:50.950" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\quad\quad D\:|\:28,\ \ \ D\:\nmid\:4,\:14\ \ \Rightarrow\ \ D = 28\:.\ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally $\rm\ \ \ D\ |\ N = P_1^{e_1}\:\cdots\ P_k^{e_k}\:,\ \ \ \forall\ i:\ D\:\nmid N/P_i\ \ \Rightarrow\ \ D = N\:,\ \ \ assuming\ all\  P_j\ are\ prime\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea is at the heart of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pocklington_primality_test&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Pocklington-Lehmer primality test.&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-19T06:49:28.677" LastActivityDate="2010-11-19T06:49:28.677" CommentCount="2" />
  <row Id="10930" PostTypeId="2" ParentId="10927" CreationDate="2010-11-19T05:22:19.897" Score="10" ViewCount="0" Body="&lt;p&gt;I don't think there is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;g&amp;#39;=f&amp;#39;g+g&amp;#39;f$ implies $\frac{f&amp;#39;}{f}=\frac{g&amp;#39;}{g&amp;#39;-g}$ where the denominators are nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{f&amp;#39;}{g&amp;#39;}=\frac{f&amp;#39;g-g&amp;#39;f}{g^2}$ implies $\frac{f&amp;#39;}{f}=\frac{(g&amp;#39;)^2}{g(g&amp;#39;-g)}$ where the denominators are nonzero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ and $g$ are nonzero and $g&amp;#39;-g$ is nonzero, this implies that $\frac{g&amp;#39;}{g&amp;#39;-g}=\frac{(g&amp;#39;)^2}{g(g&amp;#39;-g)}$, which implies $g&amp;#39;=g$.  Thus $g=g&amp;#39;$ if $f$ and $g$ are nonzero, so $g(x)=ce^x$.  But now from $f&amp;#39;g&amp;#39;=f&amp;#39;g+g&amp;#39;f$ we have $ce^xf(x)=0$, which implies $f=0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For solutions to just the product rule, the equations above suggest taking $g$ such that $\frac{g&amp;#39;}{g&amp;#39;-g}$ is integrable, and &#xA;$$f(x)=\exp\left(\int_a^x\frac{g&amp;#39;(t)}{g&amp;#39;(t)-g(t)}dt\right),$$ which in particular works for $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c$ (as seen in a now deleted answer).  Similarly, for just the quotient rule, one could try finding $g$ such that $\frac{(g&amp;#39;)^2}{g(g&amp;#39;-g)}$ is integrable, and then take &#xA;$$f(x)=\exp\left(\int_a^x\frac{g&amp;#39;(t)^2}{g(t)(g&amp;#39;(t)-g(t))}dt\right).$$  &#xA;For example, $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c^2$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-19T05:57:09.663" LastActivityDate="2010-11-19T05:57:09.663" CommentCount="4" />
  <row Id="10931" PostTypeId="2" ParentId="10927" CreationDate="2010-11-19T05:29:40.147" Score="6" ViewCount="0" Body="&lt;p&gt;The product rule equality gives $f&amp;#39;g+g&amp;#39;f = f&amp;#39;g&amp;#39;$, or $-fg&amp;#39; = f&amp;#39;g-f&amp;#39;g&amp;#39;$. From this, using the quotient rule equality we get&#xA;$$&#xD;&#xA;\frac{f&amp;#39;}{g&amp;#39;} = \frac{gf&amp;#39;-fg&amp;#39;}{g^2} = \frac{gf&amp;#39; +f&amp;#39;g - f&amp;#39;g&amp;#39;}{g^2}&#xD;&#xA;= f&amp;#39;\left(\frac{2g-g&amp;#39;}{g^2}\right).&#xD;&#xA;$$&#xD;&#xA;If $f$ is not constant, then $f&amp;#39;\neq 0$. So we get $\frac{1}{g&amp;#39;} = \frac{2g-g&amp;#39;}{g^2}$, or $g^2 - 2gg&amp;#39;+(g&amp;#39;)^2 = 0$, or $(g-g&amp;#39;)^2 = 0$; hence $g=g&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, from the product rule again we have $f&amp;#39;g&amp;#39;=f&amp;#39;g+fg&amp;#39;$, or $f&amp;#39;g = f&amp;#39;g+fg$, so $fg=0$. This requires $g=0$, which makes the quotient rule impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is constant, then the product rule equality requires $g$ constant, and then the quotient rule equality cannot hold. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-19T05:29:40.147" CommentCount="5" />
  <row Id="10932" PostTypeId="1" CreationDate="2010-11-19T05:38:14.587" Score="0" ViewCount="44" Body="&lt;p&gt;Is there any partial differential equation which doesn't have any classical as well as generalized fuction solution?&lt;/p&gt;&#xA;" OwnerUserId="3700" LastActivityDate="2010-11-19T05:38:14.587" Title="classical function as well as generalized function" Tags="&lt;functions&gt;" CommentCount="1" />
  <row Id="10933" PostTypeId="2" ParentId="10927" CreationDate="2010-11-19T05:45:16.930" Score="7" ViewCount="0" Body="&lt;p&gt;Another answer to your question from a different angle is this: They hold in a different kind of calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you swap subtraction for division and division for taking roots in the usual definition of the derivative, you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^*(x) = \lim_{\Delta x \to 0} \left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not too hard to prove that $(fg)^* = f^* g^*$ and $(f/g)^* = f^*/g^*$ under this definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculus that results from this definition of the derivative (and the corresponding definition of the integral) goes by various names, such as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_calculus&quot; rel=&quot;nofollow&quot;&gt;multiplicative calculus&lt;/a&gt;,&quot; &quot;non-Newtonian calculus,&quot; and &quot;product calculus.&quot;  Many of the standard results in the usual calculus (e.g., Fundamental Theorem, Mean Value Theorem, l'Hopital-type rules, Taylor's Theorem) can be redone for this multiplicative calculus.  The idea goes back at least to Vito Volterra in the late 1800s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Wikipedia page, &quot;Opinions differ as to the usefulness of the multiplicative calculi,&quot; but there are some applications.  For instance, the product derivative measures the multiplicative rate of change of some function and so is useful when you're interested in looking at, say, growth rates of stock prices.  The product integral has some corresponding applications; my favorite is that it can be used to calculate geometric means (in the same way that the usual integral can find arithmetic means).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplicative calculus also has a close relationship to the usual calculus.  For instance, the derivatives are related via &#xA;$$f^*(x) = \exp\left(\frac{d}{dx}\ln |f(x)|\right) = e^{f&amp;#39;(x)/f(x)}.$$&#xD;&#xA;The product integral and the usual integral have a similar relationship.  In fact, I think part of the reason some folks don't find this multiplicative calculus all that interesting is that product derivatives and integrals can so easily be expressed in terms of the usual ones.  The question, I suppose, is whether there is really anything new and different going on here.  (Note also that this relationship I just gave means that the product derivative is just $\exp$ of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_derivative&quot; rel=&quot;nofollow&quot;&gt;logarithmic derivative&lt;/a&gt; of $f$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, a few years ago I wrote a &lt;a href=&quot;http://math.pugetsound.edu/~mspivey/ProdCalc.pdf&quot; rel=&quot;nofollow&quot;&gt;survey paper&lt;/a&gt; and had a student do a &lt;a href=&quot;http://math.pugetsound.edu/~mspivey/ProdInt.pdf&quot; rel=&quot;nofollow&quot;&gt;summer research project&lt;/a&gt; on this before I realized how much was already out there and before the Wikipedia page appeared.  There are lots of other references and information on the Wikipedia page, so if you find this interesting you should check it out.  The page on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_integral&quot; rel=&quot;nofollow&quot;&gt;product integral&lt;/a&gt; is also a good source.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:28:56.747" LastActivityDate="2010-11-19T16:28:56.747" CommentCount="3" />
  <row Id="10934" PostTypeId="2" ParentId="653" CreationDate="2010-11-19T06:01:04.227" Score="2" ViewCount="0" Body="&lt;p&gt;Homotopy theory / algebraic topology was born out of applications rather than abstract nonsense considerations. So there's plenty of applications, as that's how the subject began. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the first topological proof would be the bridges of Konigsberg problem: &lt;a href=&quot;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where algebraic topology started getting off the ground was in the work of Poincare.  The Poincare-Hopf Index theorem: &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Hopf_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Hopf_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;was a landmark.  In its natural setting it was a relationship between Euler characteristic, tangent bundles and intersection theory.  But from the perspective of a differential equator it's a fundamental tool that allows you to determine whether or not differential equations have fixed points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applications have piled-up over the years.  Some of the more modern ones are listed in other people's responses.  The birth of topological dynamics in the mid 20-th century was of course a big one. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-19T06:01:04.227" />
  <row Id="10935" PostTypeId="2" ParentId="10924" CreationDate="2010-11-19T06:01:53.227" Score="3" ViewCount="0" Body="&lt;p&gt;If you seek a detailed exposition of Voronoi's algorithm then see below. Based on what they wrote, the authors may have an English translation of some of his work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/stable/2006105&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Calculation of the Regulator of a Pure Cubic Field&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By H. C. Williams, G. Cormack and E. Seah&lt;br&gt;&#xA;Mathematics of Computation, Vol. 34, No. 150 (Apr., 1980), pp. 567-6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abstract&lt;/strong&gt;. A description is given of a modified version of Voronoi's algorithm for obtaining the regulator of a pure cubic field $\mathbb Q(\sqrt[3]{D})$. This new algorithm has the advantage of executing relatively rapidly for large values of $D$. It also eliminates a computational problem which occurs in almost all algorithms for finding units in algebraic number fields. This is the problem of performing calculations involving algebraic irrationals by using only approximations of these numbers. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-19T06:56:01.820" LastActivityDate="2010-11-19T06:56:01.820" CommentCount="3" />
  <row Id="10936" PostTypeId="1" AcceptedAnswerId="10939" CreationDate="2010-11-19T06:11:29.823" Score="5" ViewCount="69" Body="&lt;p&gt;So, we are considering the subset&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \{(x, y) \in \mathbb{R^2} | (x \text{ and } y \in \mathbb{Q}) \text{ or }  (x \text{ and } y \notin \mathbb{Q})\} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And consider its complement $$ T = \mathbb{R^2} \backslash S $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set T is disconnected, actually I am fairly certain it is totally disconnected. I am just having problems showing that rigorously. I was trying to show it using straight lines but I don't think I was getting anywhere. I know that a totally disconnected set's only connected sets are the one point sets. I've been trying to show that given two arbitrary points, that a separation exists between them. It is more difficult since this is in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints at all would be a great help. Maybe I'm making mountains of molehills.&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T06:36:43.183" LastActivityDate="2010-11-19T06:41:37.377" Title="showing a set is totally disconnected" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10937" PostTypeId="1" AcceptedAnswerId="12083" CreationDate="2010-11-19T06:20:02.563" Score="0" ViewCount="59" Body="&lt;p&gt;Suppose you want to find $E(X)$ for a non-negative random variable $X$. So you can use $\int_{0}^{\infty} [1-F(x)] \ dx$. To get $1-F(x)$, all you have to do is integrate as follows: $\int_{x}^{\infty} f(t) \ dt$? In other words, instead of integrating from $t=0$ to $t= x$, we can just integrate from $t = x$ to $t = \infty$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-27T17:01:22.133" Title="Expected Value and Distribution Functions" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10939" PostTypeId="2" ParentId="10936" CreationDate="2010-11-19T06:32:59.553" Score="3" ViewCount="0" Body="&lt;p&gt;I had written out a full solution, but since this is homework, I've removed it and replaced it with this suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to prove that $T$ is totally disconnected is to show that whenever $p$ and $q$ are distinct points of $T$, then there are open sets $A$ and $B$ of $\mathbb{R}^2$ such that $T\subseteq A\cup B$, $(A\cap B)\cap T=\emptyset$, $p\in A$, and $q\in B$. This will show that there is a disconnection of $T$ in which $p$ and $q$ are in distinct components. In particular, $p$ and $q$ cannot be in the same connected component of $T$. If this holds for &lt;em&gt;all&lt;/em&gt; pairs of points $p$ and $q$, then the connected components of $T$ must be single points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, pick two distinct points $p$ and $q$ in $T$. Try to find a line that is completely contained in $S$ and which separates $p$ and $q$. One way to achieve a line completely contained in $S$ is to have it go through a rational point with rational slope. Then throw away the line to get your two sets $A$ and $B$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T06:41:37.377" LastActivityDate="2010-11-19T06:41:37.377" />
  <row Id="10940" PostTypeId="1" CreationDate="2010-11-19T06:33:59.070" Score="1" ViewCount="107" Body="&lt;p&gt;Hi,&#xA;I have some questions about Sasaki -and Sasaki-Einstein geometry.&#xA;I hope, you can answer me some of these questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Let $(M, g)$ a Riemannian manifold and $(C(M)=\mathbb{R}_{&amp;gt;0} \times M, \overline{g}=dr^{2}+r^{2}g)$ denote its metric &lt;/p&gt;&#xA;&#xA;&lt;p&gt;cone. In the paper &quot;Sasaki-Einstein Manifolds&quot; from Lames Sparks, I find that the Sasaki manifold $(M, g)$ is naturally isometrically embedded into the cone via the inclusion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;M=\lbrace r=1\rbrace=\lbrace1 \times M\rbrace \subset C(M).&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I see this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Let $\overline{\nabla}$ the Levi-Civita connection of $\overline{g}$. Is then $\overline{\nabla}_{\partial_{r}}\partial_{r}=0$&#xD;&#xA;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline{\nabla}_{\partial_{r}}\partial_{r}=\partial_{r}$ with &#xA;$\partial_{r}:=\frac{\partial}{\partial r}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) How can I show, that the Lie-derivative of $J$ in direction from $r \partial_{r}$ is equal to zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) If I define $\xi=r \partial_{r}$, how can I show that $\mathcal{L}_{\xi} \overline{g}=0$, i.e. $\xi$ is a Killing vectorfield. With $\mathcal{L}$ I mean the Lie-derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.) Let $\eta=d^{c} \log r=i(\partial-\overline{\partial}) \log r$, where $\partial$ and $\overline{\partial}$ the usual Dolbeault operators. How can I show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\eta(X)=\frac{1}{r^{2}} \overline{g}(J(r\partial_{r}), X)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Kähler 2-form on $C(M)$ is $\omega=\frac{1}{2}d(r^{2}\eta)=\frac{1}{2} i \partial \overline{\partial} r^{2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6.) We define $\Phi(X)$ as $\Phi(X)=\nabla_{X} \xi$. How can I show that $(\nabla_{X}\Phi) Y=g(\xi, Y)X-g(X, Y)\xi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.) &lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/9810/9810250v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/hep-th/pdf/9810/9810250v1.pdf&lt;/a&gt;, Proposition 1.1.2, page number 4. Can someone proof the equivalence (i) $\Leftrightarrow$ (ii)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8.) Let $M=S^{2m+1} \cong \mathbb{C}^{m+1}$. This is a Sasaki manifold. But what is $\xi$ oder $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9.) Analogous: Let $M=\mathbb{R}^{2m+1}$, it is also Sasaki, but what is $\xi$ or $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10.) I have read the following about 3-Sasaki-manifolds. The definition of 3-Sasaki-manifolds (cone is hyperkähler) implies &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Hol}(\overline{g}) \subset Sp(k) \subset SU(2k)$ and then that 3-Sasaki manifolds are also Sasaki-Einstein. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the last question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11.) Let $(M, g)$ a Sasaki-manifold of dimension $2m+1$. Then the following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $(M, g)$ is Sasaki with $\text{Ric}_{g}=2m \cdot g$.&#xA;ii) $(C(M), \overline{g})$ is Ricci flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the sketch of proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot and best regards&#xA;Ronald&lt;/p&gt;&#xA;" OwnerUserId="3702" LastEditorUserId="3702" LastEditorDisplayName="" LastEditDate="2010-11-19T06:41:37.577" LastActivityDate="2010-11-19T06:41:37.577" Title="11 questions about Sasaki -and Sasaki-Einstein geometry" Tags="&lt;differential-geometry&gt;" CommentCount="4" />
  <row Id="10941" PostTypeId="2" ParentId="1181" CreationDate="2010-11-19T07:03:26.147" Score="1" ViewCount="0" Body="&lt;p&gt;You can quickly reduce to the case of a transitive action, in which case we just want to explain why the total number of times that something gets fixed is exactly the size of the group.  But in this case everything is symmetric at all points in the (unique) orbit.  So to count all the times something gets fixed, we can just count how many times a particular x gets fixed, and multiply by the size of the orbit.  Now we've reduced to the fact that the size of an orbit is the index of the stabilizer.&lt;/p&gt;&#xA;" OwnerUserId="1392" LastEditorUserId="1392" LastEditorDisplayName="" LastEditDate="2010-11-19T07:22:51.490" LastActivityDate="2010-11-19T07:22:51.490" />
  <row Id="10942" PostTypeId="1" AcceptedAnswerId="10954" CreationDate="2010-11-19T07:48:07.137" Score="6" ViewCount="112" Body="&lt;p&gt;I heard about it sometime somewhere and want to read about it now, but I can't recall what the name is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with $a_1 = \ldots =a_n=1$. Choose a number between 1 and $n$ with probability $a_i/(a_1+ \ldots + a_n)$ to choose $i$. If $i_0$ is the number chosen, increase $a_i$ by 1 and now choose another number and so on indefinitely.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T16:14:56.903" LastActivityDate="2010-11-19T21:15:13.410" Title="What's the name of this stochastic process?" Tags="&lt;probability&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10943" PostTypeId="2" ParentId="10414" CreationDate="2010-11-19T08:14:00.693" Score="0" ViewCount="0" Body="&lt;p&gt;I was able to solve the problem with a computer algorithm but am still not certain how it can be done mathematically. It was pointed out by @Greg Muller that the knapsack problem is applicable but I just don't have the mathematical prowess to apply it. If someone could show how that can be done please do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will share my logic here, hopefully it is useful to someone out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to determine the spell with the most damage per cast time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This spell becomes the &quot;baseline&quot; spell since it will guarantee the highest damage per second. Meaning, you should always cast this spell if the following 2 conditions are met:&#xA;1) The baseline spell is available (not on cooldown).&#xA;2) You are not currently casting a spell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it then becomes a matter of filling in other spells while the baseline spell is on cooldown. Between (cast time) and (cooldown - cast time). However,  some overlapping can occur (rule 2 above is false).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then becomes a matter of recursing through all non-baseline spells to find all sequences of spells which do not violate the 2 rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For spells which DO overlap you must penalize them for potential damage the baseline spell could have done (up to its maximum damage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example, 2 spells &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;1: 300 damage, 3s cast time, 10s cooldown&lt;/h1&gt;&#xA;&#xA;&lt;h1&gt;2: 290 damage, 3s cast time,  3s cooldown&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The most damage comes from the sequence 1 - 2 - 2 - 2. Which causes an overlap of 2 seconds into a potential #1 cast. However, this is still beneficial since if you dont cast the third spell (ie. 1 - 2 - 2) you will do 880 damage with 1 second to spare. If you cast the extra #2 spell you will do 1170 - 2 second of #1 which is 200. So 970 damage is your relative damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algorithm is significantly faster than other algorithms which look for sequences that match a target goal: ie. time limit or damage.&lt;/p&gt;&#xA;" OwnerUserId="3534" LastActivityDate="2010-11-19T08:14:00.693" CommentCount="2" />
  <row Id="10944" PostTypeId="1" CreationDate="2010-11-19T09:22:02.837" Score="0" ViewCount="58" Body="&lt;p&gt;Is there a simple algorithm for efficient decomposition on this special kind of graph?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-11-19T09:22:02.837" Title="Tree decomposition for planar nearest neighbor graphs" Tags="&lt;graph-theory&gt;" CommentCount="0" />
  <row Id="10946" PostTypeId="1" CreationDate="2010-11-19T09:53:08.977" Score="1" ViewCount="50" Body="&lt;p&gt;I'm looking for an algorithm to solve this problem:&#xA;Given set of n points on 2D euclidean space, create a net of rectilinear edges, so that:&#xA;1. Every two points are connected with shortest edge.&#xA;2. Sum of all edges is minimal (minimized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's clearly a NP-hard problem and it's similar to Steiner tree problem but I can't use any of the common aproaches because of constraint 1. Moreover, in my case, every algorithm to solve that problem is feasible as long as it has polynomial complexity - the function of the objective is here the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any good ideas how to solve this? My current aproach is rather naive: connect every two points with an edge and then merge the edges that are overlapping.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2010-11-19T09:53:08.977" Title="Minimum rectilinear net" Tags="&lt;topology&gt;&lt;algorithms&gt;&lt;optimization&gt;" CommentCount="6" />
  <row Id="10947" PostTypeId="2" ParentId="7201" CreationDate="2010-11-19T09:59:39.423" Score="0" ViewCount="0" Body="&lt;p&gt;Perhaps what you're after is on page 31 in Wilf's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Proposition&lt;/strong&gt;. A formal power series $f=\sum_{n \geq 0} a_n x^n$ has a reciprocal if and only if $a_0 \neq 0$.  In that case the reciprocal is unique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-19T09:59:39.423" CommentCount="1" />
  <row Id="10948" PostTypeId="1" CreationDate="2010-11-19T10:00:34.853" Score="1" ViewCount="43" Body="&lt;p&gt;Suppose T is an elliptic Möbius transformation that fixes p and q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Prove that the pole and inverse pole lie on the perpendicular bisector of he line segment pq.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Show that T is the composition of two inversions about cline hat contain p and q.&lt;/p&gt;&#xA;" OwnerUserId="2888" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T10:27:42.693" LastActivityDate="2010-11-19T10:27:42.693" Title="Elliptic Möbius transformations" Tags="&lt;geometry&gt;&lt;homework&gt;&lt;complex-analysis&gt;" />
  <row Id="10949" PostTypeId="1" AcceptedAnswerId="11014" CreationDate="2010-11-19T10:06:51.457" Score="7" ViewCount="185" Body="&lt;p&gt;Is it known whether every finite abelian group is isomorphic to the ideal class group of the ring of integers in some number field? If so, is it still true if we consider only imaginary quadratic fields?&lt;/p&gt;&#xA;" OwnerUserId="546" LastEditorUserId="546" LastEditorDisplayName="" LastEditDate="2010-11-19T10:11:56.160" LastActivityDate="2010-11-20T12:30:20.480" Title="Finite abelian groups as class groups" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="10950" PostTypeId="2" ParentId="8001" CreationDate="2010-11-19T10:34:56.583" Score="2" ViewCount="0" Body="&lt;p&gt;I'm sorry I didn't see your question earlier.  The answer is in fact quite easy: Just permute the symbols with 3-&gt;6, 4-&gt;3 and 6-&gt;4 (or its inverse if you want to go in the other direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can permute the symbols arbitrarily in a Latin square and it will preserve row-completeness and column-completeness (or lack thereof).&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-19T10:50:15.237" LastActivityDate="2010-11-19T10:50:15.237" CommentCount="1" />
  <row Id="10951" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T10:36:33.417" Score="-1" ViewCount="0" Body="&lt;p&gt;It's not really a mathematical result, but after seeing all the nice entries here, I thought this lighter one would fit in well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Young man, in mathematics you don't understand things. You just get used to them.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I see all the examples here, this dictum by von Neumann comes to mind. I'm always remembered of how true it is.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-19T10:36:33.417" CommunityOwnedDate="2010-11-19T10:36:33.417" />
  <row Id="10952" PostTypeId="1" AcceptedAnswerId="10959" CreationDate="2010-11-19T10:43:52.690" Score="3" ViewCount="115" Body="&lt;p&gt;In Chapter 1 of &lt;em&gt;&lt;a href=&quot;http://www.springer.com/mathematics/algebra/book/978-3-642-03979-9&quot; rel=&quot;nofollow&quot;&gt;Polynomials&lt;/a&gt;&lt;/em&gt; by Victor Prasolov, Springer, 2001, the following theorem is proved. (p.3)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Theorem 1.1.4 (Ostrovsky). Let&#xA;  $f(x)=x^{n}-b_{1}x^{n-1}-\cdots -b_{n}$,&#xA;  where all the numbers $b_{i}$ are non-negative and at least one of them&#xA;  is nonzero. If the greatest common&#xA;  divisor of the indices of the positive&#xA;  coefficients $b_{i}$ is equal to&#xA;  $1$, then $f$ has a unique positive&#xA;  root $p$ and the absolute value of&#xA;  the other roots are $&amp;lt;$ p.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following is one of the Problems to Chapter 1 (p.41).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Problem 1.5 - Find the number of real&#xA;  roots of the following polynomials&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) ...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) $nx^{n}-x^{n-1}-\cdots -1$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: How to solve this Problem?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: $nx^{n}-x^{n-1}-\cdots -1=0$ $\Leftrightarrow x^{n}-\dfrac{1}{n}x^{n-1}-\cdots -\dfrac{1}{n}=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 2: &lt;a href=&quot;http://planetmath.org/encyclopedia/SturmsTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Sturm's Theorem.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T00:00:03.113" LastActivityDate="2010-11-20T00:00:03.113" Title="Problem from Victor Prasolov's Polynomials -- Finding the number of real roots of  $nx^{n}-x^{n-1}-\cdots -1$" Tags="&lt;polynomials&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10953" PostTypeId="2" ParentId="10134" CreationDate="2010-11-19T11:00:14.900" Score="1" ViewCount="0" Body="&lt;p&gt;For the sake of having less open questions and the fact that the final is probably already done I will show that if $f \in L^p(X,\mu)$, $0 &amp;lt; p &amp;lt; \infty$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|f\|_p^p = p \int_0^\infty x^{p - 1} \mu\{z : |f(z)| &amp;gt; x\} \, dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;p \int_0^\infty &amp;amp;x^{p - 1} \mu\{z : |f(z)| &gt; x\} \, dx\\&#xA;&amp;amp;= p \int_0^\infty \int_X 1_{{z : |f(z)| &gt; x}} \, d\mu(z) \, dx \text{ writing the measure as an integral.}\\&#xA;&amp;amp;= \int_X \int_0^{|f(z)|} p x^{p - 1} \, dx \, d\mu(z) \text{ Fubini,}\\&#xA;&amp;amp;= \int_X |f(x)|^p \, d\mu(x)&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you have $p = 1$, so&#xA;$$\int_{2}^\infty \mu \{x \in [0,1] : |f(x)| &amp;gt; t \} \, dt \leq \int_{2}^\infty \frac{1}{t \log^2 t} \, dt = \frac{1}{\log(2)}.$$&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-19T11:00:14.900" CommentCount="2" />
  <row Id="10954" PostTypeId="2" ParentId="10942" CreationDate="2010-11-19T11:54:28.483" Score="6" ViewCount="0" Body="&lt;p&gt;For this process and a substantial generalization, see &lt;a href=&quot;http://www.combinatorics.net/aoc/toc/v7n2/Chung/7_2_141.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&#xA;The case $p=0$ and $\gamma = 1$ corresponds to the process you described.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-19T11:54:28.483" CommentCount="1" />
  <row Id="10955" PostTypeId="1" AcceptedAnswerId="10976" CreationDate="2010-11-19T11:55:38.263" Score="1" ViewCount="74" Body="&lt;p&gt;I want to motivate the theory of $C_0$-semigroups to someone, and the following question was asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is an example of a non-separable linear PDE?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Preferably a simple homogeneous one.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-19T12:40:54.963" LastActivityDate="2010-11-19T16:22:59.473" Title="Non-separable linear PDE." Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10956" PostTypeId="2" ParentId="5595" CreationDate="2010-11-19T12:17:15.240" Score="0" ViewCount="0" Body="&lt;p&gt;@crasic, Can you outline the formal proof out here?&lt;/p&gt;&#xA;" OwnerUserId="2565" LastActivityDate="2010-11-19T12:17:15.240" CommentCount="2" />
  <row Id="10957" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T12:38:49.397" Score="1" ViewCount="0" Body="&lt;p&gt;I was very surprised when I discovered that $$0.\overline{99} = 1$$&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="2411" LastEditorDisplayName="" LastEditDate="2010-11-20T09:03:55.323" LastActivityDate="2010-11-20T09:03:55.323" CommunityOwnedDate="2010-11-19T12:38:49.397" />
  <row Id="10958" PostTypeId="2" ParentId="10827" CreationDate="2010-11-19T12:49:47.727" Score="0" ViewCount="0" Body="&lt;p&gt;For the case $K=\mathbb{Q}(q^{1/p})$ a $p$-extension over $\mathbb{Q}$ the discriminant of $K/\mathbb{Q}$ is $p^{3}q^{p-1}$ so both $p$ and $q$ ramify. I don't understand why $q$ should be congruent to 1 mod $p$.&lt;/p&gt;&#xA;" OwnerUserId="3708" LastActivityDate="2010-11-19T12:49:47.727" />
  <row Id="10959" PostTypeId="2" ParentId="10952" CreationDate="2010-11-19T13:00:23.703" Score="7" ViewCount="0" Body="&lt;p&gt;I think there is some value here in knowing how to do such problems &quot;by hand.&quot;  The proof in this case is quite simple.  If $|x| &amp;gt; 1$, then $|x^n| &amp;gt; |x^k|$ for $k &amp;lt; n$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n |x|^n &amp;gt; |x|^{n-1} + |x|^{n-2} + ... + |1| \ge |x^{n-1} + x^{n-2} + ... + 1|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the triangle inequality, so this polynomial $f(x)$ has no roots of absolute value greater than $1$.  It follows that any real roots lie in $[-1, 1]$.  By inspection $x = 1$ is a root and $x = 0, -1$ are not, so any remaining roots lie in $(0, 1)$ or $(-1, 0)$.  If $x \in (0, 1)$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{n-1} + x^{n-2} + ... + 1 &amp;gt; nx^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so there are no roots in $(0, 1)$.  To find any remaining roots in $(-1, 0)$, let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x) = f(x) (x - 1) = nx^n(x - 1) - x^n + 1 = nx^{n+1} - (n+1) x^n + 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $g&amp;#39;(x) = n(n+1) x^n - n(n+1) x^{n-1} = n(n+1) x^{n-1}(x - 1)$ has roots $x = 0, 1$, hence $g$ is monotonic on $(-1, 0)$, so to determine if there are roots on this interval it suffices to compute $g(-1)$ and $g(0)$.  We have $g(0) = 1$ and $g(-1) = -2n$ if $n$ is even and $g(-1) = 2n+2$ if $n$ is odd.  In the first case there is one real root in $(-1, 0)$ by the IVT and in the second case there are none.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-19T14:13:28.693" LastActivityDate="2010-11-19T14:13:28.693" CommentCount="2" />
  <row Id="10960" PostTypeId="1" CreationDate="2010-11-19T13:30:13.820" Score="2" ViewCount="57" Body="&lt;p&gt;Suppose $0.6$ of students submit assigned homework on time. In a typical class of $40$ students a professor has assigned homework. If a student submits homework on time, he gets $5$ points. Otherwise he gets $0$ points. Find the variance of the total number of points by all students. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is a binomial distribution. The variance is $40(0.6)(0.4)$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-19T13:36:46.593" Title="Binomial Distribution" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10961" PostTypeId="2" ParentId="10960" CreationDate="2010-11-19T13:36:46.593" Score="2" ViewCount="0" Body="&lt;p&gt;Close, but the variance has to depend upon the number of points given.  You have the variance in the number of students that turn in the work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-19T13:36:46.593" CommentCount="3" />
  <row Id="10963" PostTypeId="1" CreationDate="2010-11-19T14:16:32.193" Score="1" ViewCount="62" Body="&lt;p&gt;I have an interval $[2,t]$ containing some number of primes. I now want to divide this interval into two intervals $a=[2,m]$ and $b=]m,t]$ such that the number of primes in $a$ and $b$ is almost the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of the approximation $\pi(t)=\frac{t}{2\ln(t)}$, but can not seem to get any good and usable result out of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a computer implementation, so avoiding Lambert functions and likewise functions would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastActivityDate="2010-11-19T14:47:44.050" Title="Dividing an interval, such that the primes get divided even" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10964" PostTypeId="2" ParentId="10963" CreationDate="2010-11-19T14:41:12.310" Score="1" ViewCount="0" Body="&lt;p&gt;The approximation $\pi(x) \sim \frac{x}{\ln x}$ given by the prime number theorem tells you there are about $\frac{t}{\ln t}$ primes in $[2, t]$.  Now, the PNT also tells you that the $n^{th}$ prime $p_n$ behaves asymptotically like $p_n \sim n \ln n$.  We want $n \sim \frac{t}{2 \ln t}$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \sim \frac{t}{2 \ln t} \left( \ln t - \ln 2 - \ln \ln t \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this tells you is that for reasonably large values of $t$ (say a hundred digits) the approximation $m \sim \frac{t}{2 \ln t}$ should work quite well.  For small values of $t$ you are probably better off using the sieve of Erathosthenes to explicitly count primes.  You shouldn't really trust the other two terms anyway because the two approximations I gave above are only compatible up to ignoring them.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-19T14:47:44.050" LastActivityDate="2010-11-19T14:47:44.050" />
  <row Id="10965" PostTypeId="1" CreationDate="2010-11-19T14:45:53.723" Score="1" ViewCount="69" Body="&lt;p&gt;Hi,&#xA;i have a question to special cases of Sasaki-manifolds. Let $(M, g, \xi, \eta, \Phi)$ a Sasaki-manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In special case maybe $M=S^{2m+1} \cong \mathbb{C}^{m+1}$. This is a Sasaki manifold. But what is $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous: Let $M=\mathbb{R}^{2m+1}$, it is also Sasaki, but what is $\xi$ or $\Phi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and best regards&#xA;Ronald&lt;/p&gt;&#xA;" OwnerUserId="3702" LastActivityDate="2010-12-21T11:44:25.977" Title="Special types of Sasaki manifolds" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10967" PostTypeId="1" AcceptedAnswerId="10974" CreationDate="2010-11-19T15:29:02.330" Score="2" ViewCount="110" Body="&lt;p&gt;I recently read an article in which the author describes how to find some functions that obey to certain recursion relationships. If we want to find a function that satisfies, for example, &#xA;$f(x^a) = x \cdot f(x)$, then the author explains we can proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $ x &amp;gt; a &amp;gt; 1$ . Then $f(x) = f((x^{(1/a)})^a) = x^{1/a}f(x^{(1/a)})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= x^{1/a}f((x^{(1/a^2)})^a) = x^{1/a + 1/a^2}f(x^{(1/a^2)}) = x^{1/a + 1/a^2 + ... + 1/a^n}f(x^{1/a^n})$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the limit of $1/a^n$ is zero, when $n$ tends to infinity. The equality $ r + r^2 + r^3 + ... = r/(1-r)$ is also useful. When we finally set $f(1) = 1$, we may write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = x^{\frac{1/a}{1-1/a}} f(x^0) = x^{\frac1{a-1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the author and I wonder if a function can be found that satisifies the recurrence relationship $f(\log(x)) = x \cdot f(x)$ . For me, the main motivator for asking this question is plain curiosity. As always, pointers to relevant literature are very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: log(x) is the natural logarithm of x. &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-19T15:43:53.813" LastActivityDate="2010-11-19T17:23:15.133" Title="Find a function such that $f(\log(x)) = x \cdot f(x) $" Tags="&lt;functional-equations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10968" PostTypeId="1" AcceptedAnswerId="10971" CreationDate="2010-11-19T15:35:35.317" Score="1" ViewCount="64" Body="&lt;p&gt;$M$, $N$ are sets, $f: M \rightarrow N$ is a function, $R \subseteq M\times M$ is an equivalence relation on $M$, $\pi: M \rightarrow M/R$ is the canonical projection (i.e. $\pi(x)=[x]_R$), and $\ker f \subseteq M\times M$ is defined as $\{(x,y) \in M \times M: f(x)=f(y)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $R \subseteq \ker f \Leftrightarrow (\exists F: M/R \rightarrow N)(f=F\circ \pi)$ and that there is exactly one $F$, if both statements in the logical equivalence are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's see what I've come up with so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$\Rightarrow$&quot;: assume $R \subseteq \ker f$ and construct a function $F: M/R \rightarrow N$ which works like this: given a $c \in M/R$, choose an $x \in M$ with $c=[x]_R$ and let $F(c)=f(x)$. This works exactly because we know that $(\forall a,a&amp;#39; \in [x]_R)(f(a) = f(a&amp;#39;))$. Now let $x \in M$ and $y=\pi(x)=[x]_R$, then $F(y)=f(x)$ and $F(\pi(x))=(F\circ \pi)(x)=f(x)$ for all $x \in M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;$\Leftarrow$&quot;: assume $\exists F: M/R \rightarrow N$ with $f=F\circ \pi$: then $(\forall c \in \{[x]_R: x\in M\})(F(c)=f(x))$. Let $a \in M/R$ and $x,x&amp;#39; \in a$, then $\pi(x)=\pi(x&amp;#39;)$. We can then apply $F$ to both sides and see that $f(x)=F(\pi(x))=F(\pi(x&amp;#39;))=f(x&amp;#39;)$, i.e. $R\subseteq M\times M$ can be defined as $\{(x,y)\in M\times M: f(x)=f(y)\}$ and therefore $R\subseteq \ker f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I missed something important or is this correct? If so, how would I go about proving that there is exactly one F satisfying the given attributes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side-note: I'm very new to mathematics (and I'm no native speaker either), so any hints regarding style/language are very welcome.&lt;/p&gt;&#xA;" OwnerUserId="3546" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-11-19T17:14:44.307" LastActivityDate="2010-11-19T17:14:44.307" Title="Equivalence relation/kernel/projection: exactly one $F$ such that $f=F\circ \pi$" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;&lt;relations&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10969" PostTypeId="1" AcceptedAnswerId="10973" CreationDate="2010-11-19T15:36:01.133" Score="1" ViewCount="95" Body="&lt;p&gt;I was wondering if there are any general results to this problem. Explicitly, given an $n$-cube and $n$ distinct colors, in how many ways can we paint the faces of the cube. Here we consider two colorings the same if they can be obtained from one another by rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=3$ this number is $57$ by Burnside's lemma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any results for higher dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks for all the comments and sorry for the errors in my question. There is no intersection condition. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-19T17:13:47.887" LastActivityDate="2010-11-19T17:13:47.887" Title="n-coloring an n-cube" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="10970" PostTypeId="2" ParentId="10892" CreationDate="2010-11-19T15:46:18.353" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose that for a fixed $n$ we have the equality $\binom{n}{p} = \binom{n}{q}$ with $p \neq q$. Observe that&#xA;\begin{eqnarray}&#xA;0 = \frac{\pi}{2^{n-1}} \left[ \binom{n}{p} - \binom{n}{q} \right] = \int_{-\pi}^{\pi} \left( \cos (2p - n)x - \cos (2q - n)x \right) \ \cos^{n} x \ dx.&#xA;\end{eqnarray}&#xA;The cosine difference identity&#xA;\begin{eqnarray}&#xA;\cos A - \cos B = -2 \sin (\tfrac{A + B}{2}) \ \sin (\tfrac{A - B}{2})&#xA;\end{eqnarray}&#xA;implies &#xA;\begin{eqnarray}&#xA;\int_{-\pi}^{\pi} \left( \sin (p + q - n)x \  \sin (p- q)x \right) \ \cos^{n} x \ dx = 0 .&#xA;\end{eqnarray}&#xA;I'll leave it as an exercise to finish the proof and conclude the result.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-19T15:46:18.353" />
  <row Id="10971" PostTypeId="2" ParentId="10968" CreationDate="2010-11-19T15:46:25.957" Score="3" ViewCount="0" Body="&lt;p&gt;Your &quot;$\Leftarrow$&quot; proof is a muddled. First, $R$ is &lt;em&gt;not&lt;/em&gt; necessarily equal to $\mathrm{ker} f$, but that is what you are asserting is happening when you say &quot;$R$ can be defined as...&quot;.  Second, I don't see you showing that every element of $R$ is in $\mathrm{ker} f$, which is what you need to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that direction, you want to show that if $f=F\circ \pi$ for some function $F\colon M/R \to N$, then $R\subseteq \mathrm{ker} f$. Since you are trying to prove an inclusion, what you want to do is start by taking an element $(a,b)\in R$, and show that $(a,b)\in \mathrm{ker}f$. So you want to show that $f(a)=f(b)$. We know that $f(a) = F\circ\pi(a) = F(\pi(a)) = F([a]_R)$. And then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the function $F$ is unique, assume you have a function $G\colon M/R\to N$ with $f=G\circ\pi$. You want to show that $F=G$ (this will prove uniqueness). They have the same domain and they have the same codomain, so you only need to show that for every $[x]_R\in M/R$, you have $G([x]_{R})=F([x]_{R})$. Now, $G([x]_R) = G(\pi(x)) = f(x)$. What about $F([x]_{R})$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-19T15:46:25.957" CommentCount="1" />
  <row Id="10972" PostTypeId="1" CreationDate="2010-11-19T15:56:54.843" Score="5" ViewCount="328" Body="&lt;p&gt;Hello people, two integrals that got my attention because I really don't know how to solve them. They are a solution to the CDW equation below critical temperature of a 1D strongly correlated electron-phonon system. The second one  is used in the theory of superconductivity, while the first is a more complex variation in lower dimensions. I know the result for the second one, but without the whole calculus, it is meaningless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^b \frac{\tanh(c(x^2-b^2))}{x-b}\mathrm{d}x $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^b \frac{\tanh(x)}{x}\mathrm{d}x \approx \ln\frac{4e^\gamma b}{\pi} \text{as} \ b \to \infty$$&#xD;&#xA;where $\gamma = 0.57721...$ is Euler's constant&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-22T18:24:28.817" LastActivityDate="2010-11-23T00:41:44.187" Title="Integral with Tanh: $\int_{0}^{b} \tanh(x)/x \mathrm{d} x$" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="10973" PostTypeId="2" ParentId="10969" CreationDate="2010-11-19T16:06:15.277" Score="3" ViewCount="0" Body="&lt;p&gt;The answer is $1$ in all dimensions, if you mean what I think you mean by &quot;faces.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying graph you're trying to color is the graph whose vertices are the points $(\pm 1, 0, 0, ...), (0, \pm 1, 0, ...), ...$ where a vertex is adjacent to every other vertex except the opposite one (and note that by &quot;vertex&quot; here I mean &quot;face of an $n$-cube&quot;); in other words it is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tur%C3%A1n_graph&quot; rel=&quot;nofollow&quot;&gt;Turan graph&lt;/a&gt; $T_2(2n)$.  In particular the subgraph formed by $(1, 0, 0, ...), (0, 1, 0, ...), ...$ is a $K_n$, so we must use each of the $n$ colors exactly once to color each of these vertices.  But now it's not hard to see that, since a vertex and its opposite share the same neighbors, they must have the same color.  So the only valid colorings are those which use each of the $n$ colors exactly twice each, once for each pair of opposite vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now every permutation of the set of pairs of opposite vertices is realizable by a rotation, and it follows that the coloring is unique up to rotation.   &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-19T16:25:09.700" LastActivityDate="2010-11-19T16:25:09.700" CommentCount="3" />
  <row Id="10974" PostTypeId="2" ParentId="10967" CreationDate="2010-11-19T16:08:27.507" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Claim&lt;/strong&gt; given any function $g:(0,1]\to \mathbb{R}$, there exists a unique extension $f:\mathbb{R}\to\mathbb{R}$ satisfying your relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;. For $x &amp;gt; 1$, there exists a unique $n\in\mathbb{N}$ such that taken the $n$-fold logarithm of $x$ gives you a number in $(0,1]$. And for $x \leq 0$, $e^x \in (0,1]$. So by iteration the function is well-defined. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; Okay, let me do this explicitly. Fix your favourite function $g(x)$ on $(0,1]$. Mine happens to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_function&quot; rel=&quot;nofollow&quot;&gt;Cantor function&lt;/a&gt;. It doesn't matter at all for the construction what this function is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let your $f(x)$ be defined piecewise. For $x\in (0,1]$ define $f(x) = g(x)$. &#xA;For $x\leq 0$, define $f(x) = e^{x} g(e^{x})$. Notice that $e^x$ for $x\leq 0$ is a number in $(0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $1 &amp;lt; x \leq e$, define $f(x) = \frac1x g(\log x)$. For $e &amp;lt; x \leq e^e$, let $f(x) = \frac1x f(\log x) = \frac1{x\log x} g(\log \log x)$. For $e^e &amp;lt; x \leq e^{e^e}$, let $f(x) = \frac1x f(\log x) = \frac1{x \cdot \log x \cdot \log\log x} g(\log \log \log x)$. And so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your favourite function is $g(x) = 0$, then when you run this procedure you get Chandru's example where $f(x) = 0$ everywhere. If your favourite function is $g(x) = 1$, you have that $f(x) = e^{x}$ for $x \leq 0$, $f(x) = 1$ for $0 &amp;lt; x \leq 1$, $f(x) = \frac1x$ for $1 &amp;lt; x \leq e$, $f(x) = \frac{1}{x\log x}$ for $e &amp;lt; x \leq e^e$ and so on and so forth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For whatever function you choose to start with as $g(x)$ defined on $(0,1]$, you get one corresponding function $f(x)$ that solves your recurrence relation. Since there are uncountably many functions on $(0,1]$, there are also uncountably many possible solutions to your recurrence relation. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-11-19T17:23:15.133" LastActivityDate="2010-11-19T17:23:15.133" CommentCount="7" />
  <row Id="10975" PostTypeId="2" ParentId="9468" CreationDate="2010-11-19T16:12:56.710" Score="0" ViewCount="0" Body="&lt;p&gt;Not many people know that WA is also able of giving you all kinds of possible closed forms concerning fundamental constants, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=2.9299372410244&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=2.9299372410244&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also click &quot;more&quot;...&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-19T16:12:56.710" />
  <row Id="10976" PostTypeId="2" ParentId="10955" CreationDate="2010-11-19T16:22:59.473" Score="1" ViewCount="0" Body="&lt;p&gt;See my comments above on some general ideas. For specific examples, perhaps a good starting point would be L. P. Eisenhart, &quot;Enumeration of potentials for which one-particle Schrodinger equations are separable,&quot; Phys. Rev. 74, 87-89 (1948). So just take an arbitrary potential function $V(x)$ that is &lt;em&gt;not&lt;/em&gt; in Eisenhart's list, and consider the Schroedinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [-i \partial_t + \triangle + V(x)]\psi(t,x) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Linear, homogeneous, and aside from the trivial separability of the $t$ variable, satisfies your requirements. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-19T16:22:59.473" CommentCount="1" />
  <row Id="10978" PostTypeId="2" ParentId="10892" CreationDate="2010-11-19T16:56:38.343" Score="2" ViewCount="0" Body="&lt;p&gt;Mike Spivey correctly points out that unimodality does not suffice. Here's a quick way to fill the gap:  Consider the set $S = \{p,q,n-p,n-q\}$. Now $n\choose x$ assumes only one value for $x\in S$, by hypothesis and the fact that ${n\choose p}={n\choose n-p}$. But $p\neq q$ and $p+q\neq n$ together imply that $S$ has at least three distinct values.  This plus unimodality gives a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-11-19T16:56:38.343" CommentCount="2" />
  <row Id="10979" PostTypeId="2" ParentId="9505" CreationDate="2010-11-19T17:02:03.867" Score="2" ViewCount="0" Body="&lt;p&gt;Although this thing has already been answered, here a shorter proof&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $x^y = y^x $ is symmetric we first demand that $x&amp;gt;y$&#xD;&#xA;Then we proceed simply this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ x^y = y^x $       &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ x = y^{\frac x y } $         &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \frac x y = y^{\frac x y -1} $       &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \frac x y -1 = y^{\frac x y -1} - 1 $       &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we expand the rhs into its well-known exponential-series      &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ \frac x y -1 =  \ln(y)*(\frac x y -1) + \frac {((\ln(y)*(\frac x y -1))^2}{2!} + ... $       &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here by the definition x&gt;y the lhs is positive, so if $ \ln(y) $ &gt;=1 we had lhs $\lt$ rhs&#xA;Thus $ \ln(y) $ must be smaller than 1, and the only integer y&gt;1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x&amp;gt;4$ can be chosen.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ask for $x=4^{1+\delta}   ,\delta &amp;gt; 0 $ inserting the value 2 for y:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ 4^{(1+\delta)*2}=2^{4^{(1+\delta)}} $      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take log to base 2:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ (1+\delta)*4=4^{(1+\delta)} $      &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \delta =4^{\delta} - 1 $      &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ \delta = \ln(4)*\delta  + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $      &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ 0 = (\ln(4)-1)*\delta  + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because $ \ln(4)-1 &amp;gt;0 $ this can only be satisfied if $ \delta =0 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So indeed the only solutions, assuming x&gt;y, is $ (x,y) = (4,2)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-19T18:11:54.043" LastActivityDate="2010-11-19T18:11:54.043" />
  <row Id="10980" PostTypeId="1" AcceptedAnswerId="10986" CreationDate="2010-11-19T17:04:02.763" Score="4" ViewCount="153" Body="&lt;p&gt;Let $A$ be a commutative ring with identity. If $A$ has finite number of prime ideals $p_1,...p_n$ and moreover $\prod_{i=1}^n p_i^{k_i} = 0$ for some $k_i$. Are the prime ideals necessarily maximal? &lt;/p&gt;&#xA;" OwnerUserId="3620" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T17:49:26.193" LastActivityDate="2010-11-19T18:14:19.817" Title="Ring with finitely many prime ideals with an extra condition. Are they maximal?" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="10982" PostTypeId="1" AcceptedAnswerId="10983" CreationDate="2010-11-19T17:17:51.827" Score="2" ViewCount="83" Body="&lt;p&gt;The rank of a finitely-generated abelian group is the size of the maximal $\mathbb{Z}$-linearly-independent subset; loosely this means the number of distinct copies of $\mathbb{Z}$ contained as direct summands. I have come across a situation where a notion somewhat like rank may be useful for finite abelian groups (which obviously all have rank zero). I'm looking at graphs embedded on surfaces and I have a procedure for associated certain finite abelian groups to a graph embedding. I've found that the groups I'm getting this way are all quotients of $\mathbb{Z}^{2g}$, where $g$ is the genus of the surface. So for example, I can realize $(\mathbb{Z}/2\mathbb{Z})^2$ as a group coming from a graph embedded on a torus, but I can't get $(\mathbb{Z}/2\mathbb{Z})^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my question: for a finite abelian group $G$, define the &quot;finite rank&quot; of $G$ to be the minimal rank of a free abelian group that surjects onto $G$. Is this a common notion? Although there is the possibility this could be an argumentative question, is it a &lt;em&gt;useful&lt;/em&gt; notion (or conversely, can you make the case for it being &lt;em&gt;useless&lt;/em&gt;)? &lt;/p&gt;&#xA;" OwnerUserId="3694" LastEditorUserId="3694" LastEditorDisplayName="" LastEditDate="2010-11-19T17:32:55.283" LastActivityDate="2010-11-20T06:57:30.257" Title="Analogue of rank for finite abelian groups?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10983" PostTypeId="2" ParentId="10982" CreationDate="2010-11-19T17:26:24.630" Score="5" ViewCount="0" Body="&lt;p&gt;Unless I am mistaken, the &quot;finite rank&quot; is just the minimal size of a generating set for $G$. Indeed, if $X$ is a generating set for $G$, then the free abelian group on $X$ surjects onto $G$ by the map induced by the embedding of $X$ into $G$, so the &quot;finite rank&quot; is at most the smallest size of a generating set. And if a free abelian group $F$ of rank $k$ surjects onto $G$, then the image of a free generating set for $F$ maps to a generating set for $G$ (possibly not injectively), so $G$ has a generating set with at most $k$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number, in an arbitrary group, is usually written $d(G)$; I don't know of any special name for it, and none of my books seems to have any name attached to it. In $p$-groups, it is the index of the Frattini subgroup, $[G:\Phi(G)]$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T22:02:26.860" LastActivityDate="2010-11-19T22:02:26.860" CommentCount="2" />
  <row Id="10984" PostTypeId="2" ParentId="10972" CreationDate="2010-11-19T17:41:51.003" Score="3" ViewCount="0" Body="&lt;p&gt;I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle C = \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} = 0.81878\dots$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\displaystyle \log \frac{4e^{\gamma}}{\pi} = 0.81878\dots$, &lt;strike&gt;so your formula could be right! &lt;/strike&gt; It is in fact correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Derek's excellent &lt;a href=&quot;http://math.stackexchange.com/questions/10972/integral-with-tanh-int-0b-tanhx-x/11358#11358&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\displaystyle \dfrac{\tanh(x)}{x} = \dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $b &amp;gt; 1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(b) = \int_{0}^{b}\dfrac{\tanh(x)}{x} \ \text{dx} = \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx}  + \int_{1}^{b} (\dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)})\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(b) =  \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx}  + \log b - \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&#xA;$$\displaystyle \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} - \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsitute $t = \dfrac{1}{x}$ in the first integral, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{0}^{1}\dfrac{2}{t(e^{2/t}+1)}\ \text{dt}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(b) =  \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx}  + \log b + \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as $\displaystyle b \to \infty$, $\displaystyle \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} \to 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx \log b + \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle C = \int_{0}^{1}\left(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ \right)\text{dx} = 0.81878\dots$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T20:23:11.670" LastActivityDate="2010-11-22T20:23:11.670" CommentCount="6" />
  <row Id="10985" PostTypeId="1" CreationDate="2010-11-19T18:09:54.047" Score="4" ViewCount="51" Body="&lt;p&gt;Can I draw a net for a Steinmetz solid from two cylinders with a compass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, can we flatten the net? I often make a model using paper and a compass-- it looks about right... is it really a valid method of construction?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T12:31:00.400" LastActivityDate="2010-12-20T12:31:00.400" Title="Can I draw a net for a Steinmetz solid with a compass?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10986" PostTypeId="2" ParentId="10980" CreationDate="2010-11-19T18:14:19.817" Score="4" ViewCount="0" Body="&lt;p&gt;No, but the counterexample is trivial. Take any integral domain with finitely many prime ideals which is not a field. For example, the localization $\mathbb{Z}_{(p)}$ of the integers at a prime p. The zero ideal is non-maximal and prime so, trivially, $\prod_{i=1}^np_i=0$. Maybe this isn't exactly what you were meaning to ask?&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-11-19T18:14:19.817" CommentCount="6" />
  <row Id="10987" PostTypeId="1" AcceptedAnswerId="10988" CreationDate="2010-11-19T18:32:14.573" Score="0" ViewCount="97" Body="&lt;p&gt;Let $$ \theta(x) = \sum\limits_{p \leq x} \log{p} \quad \ ; \ \psi(x)=\sum\limits_{n=1}^{\infty} \theta(x^{1/n})$$ then how does one prove $$e^{\psi(x)}= \text{lcm}[ 1,2,\cdots , \lfloor{x\rfloor}]$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-19T18:49:45.387" Title="Showing $e^{\psi(x)}= \text{lcm}[ 1,2,\cdots , \lfloor{x\rfloor}]$" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="10988" PostTypeId="2" ParentId="10987" CreationDate="2010-11-19T18:43:54.660" Score="3" ViewCount="0" Body="&lt;p&gt;$e^{\psi(x)} = \exp({\displaystyle \sum_{n=1}^{\infty} \theta(x^{1/n})}) = \displaystyle \Pi_{n=1}^{\infty} \exp(\theta (x^{1/n})) = \displaystyle \Pi_{n=1}^{\infty} \exp(\displaystyle \sum_{p \leq x^{1/n}} \log(p))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = \Pi_{n=1}^{\infty} (\Pi_{p \leq x^{1/n}} p) = (\Pi_{p \leq x} p) \times (\Pi_{p \leq x^{1/2}} p) \times (\Pi_{p \leq x^{1/3}} p) ... = lcm[1,2,3,...,[x]]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$ To make the last equality a bit more clear, the highest power of $p$ dividing $lcm[1,2,3,...,[x]]$ is given by $[\log_p([x])]$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T18:49:45.387" LastActivityDate="2010-11-19T18:49:45.387" />
  <row Id="10989" PostTypeId="2" ParentId="10985" CreationDate="2010-11-19T18:55:42.467" Score="1" ViewCount="0" Body="&lt;p&gt;Imagine keeping one of the cylinders intact, but merely marking on it the curves that would have been the edges of the Steinmetz solid. If you unfold the cylinder, the curves must be periodic in the azimuthal direction, so they cannot be circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you solve $(x,y,z) = (\cos \theta, \sin \theta, z) = (x, \cos \phi, \sin \phi)$, you'll find that the curves are actually sinusoids, $z = \pm \cos \theta$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-19T18:55:42.467" />
  <row Id="10990" PostTypeId="1" CreationDate="2010-11-19T18:56:04.087" Score="1" ViewCount="177" Body="&lt;p&gt;Dear MSE-Community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was trying to find closed-form representations for odd zeta-values, I used &#xA;$$ \Gamma(z) = \frac{e^{-\gamma \cdot z}}{z} \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{\frac{z}{n}} $$ &#xA;and rearranged it to &#xA;$$ \frac{\Gamma(z)}{e^{-\gamma \cdot z}} = \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{z/n}. $$&#xD;&#xA;As we know that $$\prod_{n=1}^{\infty} e^{z/n} = e^{z + z/2 + z/3 + \cdots + z/n} = e^{\zeta(1) z},$$ &#xA;we can state that &#xA;$$\prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big) = \frac{e^{z(\zeta(1) - \gamma)}}{z\Gamma(z)}\qquad\text{(1)}$$&#xD;&#xA;I then stumbled upon the Wikipedia page of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramanujan_summation&quot; rel=&quot;nofollow&quot;&gt;Ramanujan Summation&lt;/a&gt; (see the bottom of the page), which I used to set $\zeta(1) = \gamma$ (which was, admittedly, a rather dangerous move. Remarkably, things went well eventually. Please don't stop reading). The $z^3$ -coefficient of both sides can now be obtained. Consider &#xA;\begin{align*}&#xA;(1-ax)(1-bx) &amp;amp;= 1 - (a+b)x + abx^2\\&#xA;&amp;amp;= 1-(a+b)x + (1/2)((a+b)^2-(a^2+b^2))&#xA;\end{align*}&#xA;and&#xA;\begin{align*}(1-ax)(1-bx)(1-cx) &amp;amp;= 1 - (a + b + c)x\\&#xA;&amp;amp;\qquad + (1/2)\Bigl((a + b + c)^2 - (a^2 + b^2 + c^2)\Bigr)x^2\\&#xA;&amp;amp;\qquad -(abc)x^3.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also set &#xA;\begin{align*}&#xA;(abc)x^3 &amp;amp;= (1/3)\Bigl((a^3 + b^3 + c^3) - (a + b + c)\Bigr)\\&#xA;&amp;amp;\qquad + (1/2)(a + b + c)^3 -(a + b + c)(a^2 + b^2 + c^2).&#xA;\end{align*}&#xA;It can be proved by induction that the x^3 term of $(1-ax)(1-bx)\cdots(1-nx)$ is equal to &#xA;\begin{align*}&#xA;(1/3)&amp;amp;\Bigl((a^3 + b^3 + c^3 +\cdots + n^3) - (a + b + c + \cdots + n^3))\\&#xA;&amp;amp;\qquad+ (1/2)(a + b + c + \cdots + n)^3 -(a + b + c + \cdots + n)(a^2 + b^2 + c^2 + \cdots + n^2).\qquad\text{(2)}&#xA;\end{align*}&#xA;On the right side of equation (1), the $z^3$-term can be found by looking at the z^3 term of the Taylor expansion series of $1/(z \Gamma(z))$, which is $(1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$. We then use (2) to obtain the equality&#xA;$$ (1/6)\gamma^3 - (1/2)\gamma \pi^2 - (1/6) \psi^{(2)}(1) = 1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$$ &#xA;to find that &#xA;$$\zeta(3) = - (1/2) \psi^{(2)} (1),$$ (3)&#xA;which is a true result that has been &#xA;known (&lt;em&gt;known&lt;/em&gt; should be a hyperlink but it isn't for some reason) for quite a long time. The important thing here is that I used $\zeta(1) = \gamma$, which isn't really true. Ramanujan assigned a summation value to the harmonic series (again, see Ramanujan Summation), and apparently it can be used to verify results and perhaps to prove other conjectures/solve problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is: Is this a legitimate way to prove (3) ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalizing this question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When and how are divergent series and their summation values used in mathematics? What are the 'rules' when dealing with summed divergent series and using them to (try to) find new results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-22T22:20:58.833" LastActivityDate="2010-11-22T22:20:58.833" Title="Uses of Divergent Series and their summation-values in mathematics ?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;big-list&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="2" />
  <row Id="10991" PostTypeId="2" ParentId="2949" CreationDate="2010-11-19T19:06:04.113" Score="1" ViewCount="0" Body="&lt;p&gt;Fintushel and Stern's construction of exotic K3's by surgery on torus fibered knots in S^3.  If the Alexander polynomial of the knot is not monic then the smooth structure doesn't admit a symplectic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/dg-ga/9612014.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/dg-ga/9612014.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also very beautifully explained in the last chapter of Scorpan's &quot;The Wild World of 4-manifolds.&quot; The entire construction is available in the Google books preview.&lt;/p&gt;&#xA;" OwnerUserId="3722" LastActivityDate="2010-11-19T19:06:04.113" CommunityOwnedDate="2010-11-19T19:06:04.113" />
  <row Id="10992" PostTypeId="1" CreationDate="2010-11-19T19:25:40.927" Score="1" ViewCount="172" Body="&lt;p&gt;Alright, I'm trying to figure out how to calculate a  critical value using t-distribution in Microsoft Excel... ex. a one-tailed area of 0.05 with 39 degrees of freedom: t=1.685&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the answer, but how do I get this? I've tried TDIST() TINV() and TTEST() but they all give me different answers. This web calculator: &lt;a href=&quot;http://www.danielsoper.com/statcalc/calc10.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.danielsoper.com/statcalc/calc10.aspx&lt;/a&gt; always gives me what I'm looking for but I cannot manage to get Excel to do the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-20T00:04:26.697" LastActivityDate="2010-11-20T00:09:10.717" Title="Finding critical value using t-distribution in Excel" Tags="&lt;statistics&gt;&lt;math-software&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10993" PostTypeId="1" CreationDate="2010-11-19T19:27:51.577" Score="4" ViewCount="254" Body="&lt;p&gt;Suppose $x \geq 0$, $y \geq 0$ and $0&amp;lt;p&amp;lt;1$. Why is the following inequality true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|x^{p}-y^{p}| \leq |x-y|^p$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3724" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T00:21:43.673" LastActivityDate="2010-11-20T05:32:44.807" Title="Please explain inequality $|x^{p}-y^{p}| \leq |x-y|^p$" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="10994" PostTypeId="2" ParentId="10992" CreationDate="2010-11-19T19:37:02.567" Score="0" ViewCount="0" Body="&lt;p&gt;Non-answer: Export the data to CSV and use software that is suited for statistical work, e.g. R.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-19T19:37:02.567" />
  <row Id="10995" PostTypeId="2" ParentId="5595" CreationDate="2010-11-19T19:40:17.960" Score="3" ViewCount="0" Body="&lt;p&gt;I don't really have the intuition for this, but I know the formal proof.  This is equivalent to showing that the probability that in a permutation of $[n]$ chosen uniformly at random, two elements chosen uniformly at random are in the same cycle is $1/2$.  By symmetry, it's enough to show that the probability that $1$ and $2$ are in the same cycle is $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to show this fact. For example: the probability that $1$ is in a cycle of length $k$ is $1/n$, for $1 \le k \le n$. This is true because the number of possible $k$-cycles containing $1$ is ${n-1 \choose k-1} (k-1)! = (n-1)!/(n-k)!$, and the number of ways to complete a permutation once a $k$-cycle is chosen is $(n-k)!$. So there are $(n-1)!$ permutations of $[n]$ in which $1$ is in a $k$-cycle.  Now the probability that $2$ is in the same cycle as $1$, given that $1$ is in a $k$-cycle, is $(k-1)/(n-1)$. So the probability that $2$ is in the same cycle as $1$ is&#xA;$$ \sum_{k=1}^n {k-1 \over n-1} {1 \over n} = {1 \over n(n-1)} \sum_{k=1}^n (k-1) = {1 \over n(n-1)} {n(n-1)\over 2} = 1/2. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_restaurant_process&quot; rel=&quot;nofollow&quot;&gt;Chinese restaurant process&lt;/a&gt; with $\alpha = 0, \theta = 1$ generates a uniform random permutation of $[n]$ at the $n$th step; $2$ is paired with $1$ at the second step with probability $1/2$. This is a bit more elegant but requires some understanding of the CRP.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-19T19:40:17.960" />
  <row Id="10996" PostTypeId="2" ParentId="9505" CreationDate="2010-11-19T19:49:10.333" Score="0" ViewCount="0" Body="&lt;p&gt;Say $x^y = y^x$, and $x &amp;gt; y &amp;gt; 0$. Taking logs, $y \log x = x \log y$; rearranging, $(\log x)/x = (\log y)/y$. Let $f(x) = (\log x)/x$; then this is $f(x) = f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $f^\prime(x) = (1-\log x)/x^2$, so $f$ is increasing for $x&amp;lt;e$ and decreasing for $x&amp;gt;e$. So if $x^y = y^x$ has a solution, then $x &amp;gt; e &amp;gt; y$.  So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x &amp;gt; 0 &amp;gt; y$, then $0 &amp;lt; x^y &amp;lt; 1$ and $y^x$ is an integer, so there are no such solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0 &amp;gt; x &amp;gt; y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{-y} = y^{-x}$. Then $(-x)^{-y} = (-y)^{-x}$ since $x$ and $y$ have the same parity. (The number of factors of $-1$ we introduced to each side differs by $x-y$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-19T19:49:10.333" />
  <row Id="10997" PostTypeId="2" ParentId="10993" CreationDate="2010-11-19T19:53:05.853" Score="10" ViewCount="0" Body="&lt;p&gt;Assume w.lo.g $x&amp;gt;y$. Then the statement becomes $x^p - y^p \leq (x-y)^p$. Since $y &amp;gt; 0$, divide through out by $y^p$. So we need to show $(\frac{x}{y})^p - 1 \leq (\frac{x}{y}-1)^p$ whenever $x &amp;gt; y &amp;gt;0$ and $0&amp;lt;p&amp;lt;1$. Let $t = \frac{x}{y}$. So we need to show $t^p - 1 \leq (t-1)^p$ whenever $t &amp;gt; 1$ and $0&amp;lt;p&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a calculus problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(t) = (t-1)^p - (t^p -1) $ where $0&amp;lt;p&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that the function $f(t)$ is increasing for $t \geq 1$ and when $0 &amp;lt; p &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $f(t) \geq f(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $f(1) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we get the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$ To show $f(t)$ is increasing for $t \geq 1$ and when $0 &amp;lt; p &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to show $0 &amp;lt; f&amp;#39;(t) = p (t-1)^{p-1} - pt^{p-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $p&amp;gt;0$, all we need to show is that $(t-1)^{p-1}&amp;gt;t^{p-1}$, $\forall t &amp;gt; 1$ and $0&amp;lt;p&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $0&amp;lt;p&amp;lt;1$, we need to show $t^{1-p} &amp;gt; (t-1)^{1-p}$ which is true since $p&amp;lt;1$ and $t&amp;gt;t-1&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T20:19:13.020" LastActivityDate="2010-11-19T20:19:13.020" CommentCount="2" />
  <row Id="10998" PostTypeId="1" AcceptedAnswerId="10999" CreationDate="2010-11-19T20:01:45.803" Score="9" ViewCount="160" Body="&lt;p&gt;A vector $v \in \mathbb{Z}^n$ is primitive if there does not exist some vector $v&amp;#39; \in \mathbb{Z}^n$ and some $k \in \mathbb{Z}$ such that $v = k v&amp;#39;$ and $k \geq 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a paper I'm writing right now, I'd like to know that a &quot;random&quot; vector in $\mathbb{Z}^n$ is primitive.  Let me make this precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\|\cdot\|_{1}$ be the $L^{1}$ norm on $\mathbb{Z}^n$, so $\|v\|_1 = \sum_{i=1}^n |v_i|$, where the $v_i$ are the components of $v$.  Define $\mathcal{V}_k$ to be the number of vectors $v$ in $\mathbb{Z}^n$ such that $\|v\|_1 \leq k$.  Define $\mathcal{P}_k$ to be the number of primitive vectors $v$ in $\mathbb{Z}^n$ such that $\|v\|_1 \leq k$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then want&#xA;$$\lim_{k \rightarrow \infty} \frac{\mathcal{P}_k}{\mathcal{V}_k} = 1.$$&#xD;&#xA;Assuming this is true, is there any nice estimate as to how fast it approaches $1$?&lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2010-11-20T13:25:22.573" Title="Probability that a vector in $\mathbb{Z}^n$ is primitive" Tags="&lt;linear-algebra&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10999" PostTypeId="2" ParentId="10998" CreationDate="2010-11-19T20:06:17.490" Score="13" ViewCount="0" Body="&lt;p&gt;In the case where $n = 2$, you're asking for the &quot;probability&quot; that two integers are relatively prime; this is well-known to be $6/\pi^2$, not 1. In the general-$n$ case, the probability that $n$ integers are relatively prime is $1/\zeta(n)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Coprime&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coprime&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-19T20:22:53.200" LastActivityDate="2010-11-19T20:22:53.200" CommentCount="4" />
  <row Id="11000" PostTypeId="1" AcceptedAnswerId="11006" CreationDate="2010-11-19T20:35:01.747" Score="4" ViewCount="112" Body="&lt;p&gt;In 'An Introduction to Probability Theory and Applications' by W. Feller I encountered this apparently innocuous problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A throw of twelve dice can result in&#xA;  $6^{12}$ different outcomes, to all of&#xA;  which we attribute equal&#xA;  probabilities. The event that each&#xA;  face appears twice can occur in as&#xA;  many ways as twelve dice can be&#xA;  arranged in six groups of two each.&#xA;  Hence the probability of the event is&#xA;  $\displaystyle \frac{12!}{2^{6}6^{12}}=0.003438$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning is that by doing that you're grouping two 1's, two 2's, ..., two 6's (each group using one partition in the multinomial) and the result is the number of different partitions that can be found with that particular characteristic. However, I had doubts about that answer. To understand better that problem, I did a simpler example with $4$ dice instead of $12$ (in this case, the event is the number of ways in which two faces appear twice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the same result I get as the probability $\displaystyle \frac{4!}{2^{2}6^{4}}=$ &lt;s&gt;0.46&lt;/s&gt; $0.0046$. Then, to see if that's true I ran a little simulation of this case in Mathematica: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dice = Table[Table[Random[Integer, {1, 6}], {i, 1, 4}], &#xA;{j, 1, 1000000}]; i=0;&#xA;f[{a_, b_, c_, d_}] := Which[a === b &amp;amp;&amp;amp; c === d &amp;amp;&amp;amp; a != c, &#xA;i++, a === c &amp;amp;&amp;amp; b === d &amp;amp;&amp;amp; a != b, &#xA;i++, a === d &amp;amp;&amp;amp; b === c &amp;amp;&amp;amp; a != b, i++;];&#xA;Map[f, dice, {1}];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After $1000000$ steps I got $69687$ cases in which two faces appear twice. This is equivalent to a probability of $0.069687$. Far smaller than what I expected based on the calculation above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this latter example is much more manageable than the one with twelve dice, I did the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With four dice we have the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Partition $r_{1}$ contains the first and second dice and partition $r_{2}$ contains the third and fourth dice.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{1}$ contains the second and fourth dice and partition $r_{2}$ contains the first and fourth third.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{1}$ contains the first and fourth dice and partition $r_{2}$ contains the second and third dice.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{2}$ contains the first and second dice and partition $r_{1}$ contains the third and fourth dice.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{2}$ contains the second and fourth dice and partition $r_{1}$ contains the first and fourth third.&lt;/li&gt;&#xA;&lt;li&gt;Partition $r_{2}$ contains the first and fourth dice and partition $r_{1}$ contains the second and third dice.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For each case, we can have $30$ outcomes in which two faces appear two times. For example, for the first case we have $1122, 1133, 1144, 1 155, 1166, 2211, 2233, 2244,2255, 2266,... 6611, 6622, 6633, 6644, 6655$. However, such outcomes are repeated twice (particularly, the outcomes of case 1 repeat the outcomes of case 4, etc). Therefore, the number of different outcomes which produce two faces appearing two times are $\frac{6}{2}6*5=90$. Since there are $6^{4}$ outcomes, we have a probability of $\frac{90}{6^{4}}=0.0694444$, which is the result that produces the simulation in Mathematica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it wrong the first reasoning? If so, is there a general approach to use the multinomial coefficient to solve this kind of problems. For instance, it appears that this only happens for $r_{1}=r_{2}=r_{k}$. Otherwise, there are not repeated outcomes.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-11-19T21:44:52.753" LastActivityDate="2010-11-19T22:12:39.000" Title="Probability for twelve dice" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="11001" PostTypeId="2" ParentId="10998" CreationDate="2010-11-19T20:43:34.930" Score="10" ViewCount="0" Body="&lt;p&gt;Further to Michael's answer, not only does $\mathcal{P}_k/\mathcal{V}_k\to1/\zeta(n)$, but we can calculate a bound for the rate of convergence. I'll also give an argument which is a little different from the one given in his Wikipedia reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that the set $\lbrace v\in\mathbb{R}^n\colon\Vert v\Vert_1\le k\rbrace$ has volume $ck^n$ (for a constant c depending only on the dimension n) and surface area proportional to $k^{n-1}$ gives&#xA;$$&#xD;&#xA;\mathcal{V}_{k}-1=ck^n + O(k^{n-1}).\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xD;&#xA;The '-1' on the left hand side is not relevant for large k as it can be absorbed into the &lt;em&gt;O&lt;/em&gt;(&lt;i&gt;k&lt;/i&gt;&lt;sup&gt;&lt;em&gt;n&lt;/em&gt;&amp;minus;1&lt;/sup&gt;) error term, and is just there so that (1) is also valid for small &lt;em&gt;k&lt;/em&gt;&amp;nbsp;&amp;lt;&amp;nbsp;1.&#xA;Noting that every nonzero $v\in\mathbb{Z}^n$ decomposes uniquely as $v=mv^\prime$ for integer $m\ge1$ and primitive $v^\prime\in\mathbb{Z}^n$ leads to the following relation between $\mathcal{P}_k$ and $\mathcal{V}_k$,&#xA;$$&#xD;&#xA;\mathcal{V}_k-1=\sum_{m=1}^\infty\mathcal{P}_{\frac{k}{m}}.&#xD;&#xA;$$&#xD;&#xA;This can be inverted via the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_function&quot; rel=&quot;nofollow&quot;&gt;M&amp;ouml;bius function&lt;/a&gt; &amp;mu;,&#xA;$$&#xD;&#xA;\mathcal{P}_k=\sum_{m=1}^\infty\mu(m)(\mathcal{V}_{\frac{k}{m}}-1).&#xD;&#xA;$$&#xD;&#xA;In dimension $n &amp;gt; 2$, substituting (1) into this expression gives&#xA;$$&#xD;&#xA;\mathcal{P}_k=\sum_{m=1}^\infty \mu(m)c k^n m^{-n} + O(k^{n-1}).\qquad\qquad{(2)}&#xD;&#xA;$$&#xD;&#xA;The $O(k^{n-1})$ comes from the sum $\sum_m (k/m)^{n-1}$ from the remainder term of (1) which, for $n &amp;gt; 2$, gives $k^{n-1}$ multiplied by a convergent sum.&#xA;Dividing through by $\mathcal{V}_k$,&#xA;$$&#xD;&#xA;\mathcal{P}_k/\mathcal{V}_k=\sum_{m=1}^\infty\mu(m)m^{-n}+O(1/k)=1/\zeta(n)+O(1/k).&#xD;&#xA;$$&#xD;&#xA;Edit: The case for $n=2$ is actually a little bit different, and we do not obtain such a good convergence rate. As the sum $\sum_m(k/m)^{n-1}$ does not converge, the error term in (2) does not apply. Instead, we can use $O(1_{\lbrace k\ge1\rbrace}k+1_{\lbrace k &amp;lt; 1\rbrace}k^2)$ for the error term in (1). This leads to an error of order $k\sum_{m\le k}m^{-1}+k^2\sum_{m &amp;gt; k}m^{-2}\sim k\log k$ in (2), giving&#xA;$$&#xD;&#xA;\mathcal{P}_k/\mathcal{V}_k=1/\zeta(2)+O(\log k/k).&#xD;&#xA;$$&#xD;&#xA;You can also look at the paper &lt;a href=&quot;http://dx.doi.org/10.1016/0022-314X%2872%2990038-8&quot; rel=&quot;nofollow&quot;&gt;On the probability that &lt;em&gt;k&lt;/em&gt; positive integers are relatively prime&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-20T13:25:22.573" LastActivityDate="2010-11-20T13:25:22.573" />
  <row Id="11002" PostTypeId="1" CreationDate="2010-11-19T21:14:35.510" Score="5" ViewCount="156" Body="&lt;p&gt;Can we determine if a binomial coefficient $C(n,p)$ is even or odd, without calculating its value? &#xA;($p\lt n$, $p$ and $n$ are positive integers)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T21:40:22.943" LastActivityDate="2010-11-20T05:25:14.973" Title="$C(n,p)$: even or odd?" Tags="&lt;binomial-coefficients&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11003" PostTypeId="2" ParentId="10942" CreationDate="2010-11-19T21:15:13.410" Score="2" ViewCount="0" Body="&lt;p&gt;Thanks for the pointer, Shai Covo. The name of this specific process is &lt;a href=&quot;http://en.wikipedia.org/wiki/Preferential_attachment&quot; rel=&quot;nofollow&quot;&gt;preferential attachment&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-11-19T21:15:13.410" />
  <row Id="11005" PostTypeId="2" ParentId="11002" CreationDate="2010-11-19T21:18:56.540" Score="7" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We only need to know how many powers of two appear in the prime factorization of C(n,p). The number of powers of two that appear in the prime factorization of $n!$ is $\lfloor n/2 \rfloor + \lfloor n/4 \rfloor + \lfloor n/8 \rfloor + \cdots$ -- count all the multiples of two, of four, of eight, and so on. This turns out to be $n$ minus the number of ones in the binary expansion of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this reduces to the following: C(n,p) is odd if and only if every bit of the binary expansion of $p$ is less than the corresponding bit of the binary expansion of $n$. This is due to Lucas.&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-19T21:18:56.540" />
  <row Id="11006" PostTypeId="2" ParentId="11000" CreationDate="2010-11-19T21:21:26.833" Score="6" ViewCount="0" Body="&lt;p&gt;There are two things going on here.  First, $\displaystyle \frac{4!}{2^{2}6^{4}}$ is about $0.0046$, not $0.46$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the problem described by Feller and your simplification to four dice are not the same problem.  Feller's problem requires that &lt;em&gt;each&lt;/em&gt; face appear twice.  Your four-dice problem requires that $some$ two faces appear twice.  There's only one way that Feller's event can be satisfied; namely, that each of 1, 2, 3, 4, 5, and 6 appear exactly twice.  There are many more ways that some two faces could appear twice.  For instance, you could have two 1's and two 2's, two 1's and two 3's, etc.  In fact, there are $\binom{6}{2} = 15$ different ways to choose the two faces that will appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that solves your problem.  Taking the correct result from Feller, $\displaystyle \frac{4!}{2^{2}6^{4}} \approx 0.0046$, and multiplying it by the $15$ different ways to choose the faces gives $\displaystyle \frac{90}{6^4} \approx 0.0694444$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to your question about the multinomial coefficient, $\displaystyle \binom{n}{k_1, k_2, \ldots, k_m}$ calculates the number of ways to partition a set of $n$ items into $m$ groups with $k_1$ items in the first group, $k_2$ items in the second group, and so forth.  So in Feller's problem $\displaystyle \binom{12}{2, 2, 2, 2, 2, 2}$ is calculating the number of ways to put two dice in the 1's group, two dice in the 2's group, and so forth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the four-dice problem you describe, you haven't specified a particular partition of dice into groups; there are several partitions that will satisfy &quot;two faces appearing twice.&quot;  So, in the four-dice problem, $\displaystyle \binom{4}{2, 2}$ counts the number of ways to put two dice in a 1's group and two dice in a 2's group.  It also counts the number of ways to put two dice in a 1's group and two dice in a 3's group, and it counts the number of ways to put two dice in a 1's group and two dice in a 4's group, and so forth.  So to calculate the total number of ways of obtaining &quot;two faces appearing twice&quot; you need to multiply $\displaystyle \binom{4}{2, 2}$ by the number of ways to pick two of the faces out of six possible faces, which is $\binom{6}{2}$.  Then you mutiply by $\frac{1}{6^4}$ to get the probability you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a general answer, the probability of throwing $n$ $d$-sided dice and having exactly $m$ faces appear $\frac{n}{m}$ times would be &#xA;$$\binom{n}{m} \frac{n!}{((\frac{n}{m})!)^m d^n} .$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-19T22:12:39.000" LastActivityDate="2010-11-19T22:12:39.000" CommentCount="12" />
  <row Id="11008" PostTypeId="2" ParentId="11000" CreationDate="2010-11-19T21:27:27.200" Score="2" ViewCount="0" Body="&lt;p&gt;The case of four dice is not so hard. You require XXYY in some order, where X != Y. So let the first throw be X. Then there are three possibilities: XXYY, XYXY, and XYYX. Each has probability 1/6 x 5/6 x 1/6 = 5/216. So we get 5/72, which is about 0.0694. This agrees very well with your simulation!&lt;br&gt;&#xA;The question is, where did you go wrong in your calculation? You calculated the probability of a &lt;em&gt;specific&lt;/em&gt; XXYY -- say, two ones and two fives. Now you have to multiply by the number of such (X,Y) pairs, which is 15. And 15 x 4! / (2^2 x 6^4) is 5/72.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-11-19T21:27:27.200" CommentCount="1" />
  <row Id="11009" PostTypeId="2" ParentId="11002" CreationDate="2010-11-19T21:30:04.267" Score="8" ViewCount="0" Body="&lt;p&gt;There is a beautiful result of Kummer that says that if $q$ is a prime, then the largest power of $q$ that divides $\binom{m+n}{n}$, $m$ and $n$ nonnegative, is the number of carries when $m$ and $n$ are added in base $q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, we get the Theorem of Lucas mentioned by Michael Lugo: we get a carry in the addition if and only if there is a $k$ such that the $k$th digit of $m+n$ in base $q$ is smaller than the $k$th digit in the base $q$ expansion of $n$, so $\binom{m+n}{n}$ is divisible by $q$ if and only if there is a $k$ such that the $k$th digit of the base $q$ expansion of $m+n$ is smaller than the $k$th digit of the base $q$ expansion of $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to determine if $\binom{n}{p}$ is odd, you just need to know whether there is at least one carry when adding $p$ and $n-p$ in base $2$; you get a carry exactly when the $k$th binary digit of both $n-p$ and $p$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just write $p$ and $n-p$ in binary, one digit at a time (from least significant to most significant digit), until you get that both are $1$; or write $p$ and $n$ one binary digit at a time, until you get a $1$ in $p$ and a $0$ in $n$. If this happens, $\binom{n}{p}$ is even; otherwise, it is odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrew Granville's paper &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/Binomial/toppage.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Arithmetic properties of binomial coefficients&lt;/em&gt;&lt;/a&gt; is chock full of interesting stuff about binomial coefficients and their divisibility. The proof of Kummer's Theorem can be found &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/paper/binomial/html/node3.html#SECTION00030000000000000000&quot; rel=&quot;nofollow&quot;&gt;there&lt;/a&gt;. There are also &lt;a href=&quot;http://www.cecm.sfu.ca/organics/papers/granville/&quot; rel=&quot;nofollow&quot;&gt;other formats available&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-20T05:25:14.973" LastActivityDate="2010-11-20T05:25:14.973" CommentCount="3" />
  <row Id="11010" PostTypeId="1" AcceptedAnswerId="11011" CreationDate="2010-11-19T21:42:36.717" Score="0" ViewCount="155" Body="&lt;p&gt;Which of the following mappings are linear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)$\varphi : \mathbb{R} \rightarrow \mathbb{R}^{2}, \varphi(x) = (1 - x, (1 - x)^{2}),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)$\varphi : \mathcal{F}(\mathbb{R},\mathbb{R}) \rightarrow \mathbb{R}, \varphi(f) = f(1),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)$\varphi : \mathbb{R}[t] \rightarrow \mathbb{R}[t], \varphi(f) = t \cdot f', $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d)Let $V$ be the $\mathbb{R}$-Vectorspace of converging sequences $(a_{n})$ of real numbers. Let $\varphi \rightarrow \mathbb{R}, \varphi((a_{n})) = a^{3}_{1} +$ lim $a_{n}.$&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2010-11-24T00:58:57.707" LastActivityDate="2010-11-24T17:38:46.187" Title="which of the following are linear mappings?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11011" PostTypeId="2" ParentId="11010" CreationDate="2010-11-19T21:45:50.647" Score="8" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; A mapping is said to be linear if $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verify whether the maps which you have given satisfy the above 2 conditions. If they satisfy then conclude that the map is linear. A simple example would be to consider $\varphi : \mathbb{R} \to \mathbb{R}$ defined by $\varphi(x)=cx$, note that $\varphi(x+y)=c(x+y)=cx+cy=\varphi(x)+\varphi(y)$ and $\varphi(ax)= a(cx)=a \varphi(x)$. Hence $\varphi$ is a linear map. With this example in mind, proceed for your question.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-19T21:51:08.537" LastActivityDate="2010-11-19T21:51:08.537" CommentCount="9" />
  <row Id="11012" PostTypeId="1" CreationDate="2010-11-19T21:48:43.227" Score="5" ViewCount="106" Body="&lt;p&gt;Assume $M$ is a (finitely generated) $A$-module such that $\wedge^n M$ is free of rank $1$ for some $n \geq 1$. Does it follow that $M$ is free of rank $n$? Or at least locally free of rank $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, is there a way of characterizing local free modules via exterior powers and tensor products? &lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-20T16:29:31.427" Title="characterizing free modules by exterior power" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="11013" PostTypeId="2" ParentId="10949" CreationDate="2010-11-19T21:51:41.947" Score="4" ViewCount="0" Body="&lt;p&gt;At least as of 1999, this was still an open question, according to the MathReview of a paper by Marc Perret (&lt;em&gt;On the ideal class group problem for global fields&lt;/em&gt;, J. Number Theory &lt;strong&gt;77&lt;/strong&gt; (1999), no. 1, pages 27-35; MR1695698 (2000d:11135), review by Bruno Anglès). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luther Claborn proved in 1966 that every abelian group is the ideal class group of &lt;em&gt;some&lt;/em&gt; Dedekind domain (&lt;em&gt;Every abelian group is a class group&lt;/em&gt;, Pacific J. Math. &lt;strong&gt;18&lt;/strong&gt; (1966), pages 219-222, MR0195889 (33 #4085)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gary Cornell (&lt;em&gt;Abhyankar's lemma and the class group&lt;/em&gt;. Number theory, Carbondale 1979 (Proc. Southern Illinois Conf., Southern Illinois Univ., Carbondale, Ill., 1979), pp. 82–88, Lecture Notes in Math., 751, Springer, Berlin, 1979, MR0564924 (82c:12007)) proved that every finite abelian group is a &lt;em&gt;subgroup&lt;/em&gt; of the ideal class group of a cyclotomic extension of $\mathbb{Q}$.  There have been some refinements, but I did not find anything in MathSciNet that would suggest the problem has been settled in the interim.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-19T22:01:13.447" LastActivityDate="2010-11-19T22:01:13.447" CommentCount="1" />
  <row Id="11014" PostTypeId="2" ParentId="10949" CreationDate="2010-11-19T22:33:38.823" Score="7" ViewCount="0" Body="&lt;p&gt;Virtually nothing is known about the question of which abelian groups can be the ideal class group of (the full ring of integers of) some number field.  So far as I know, it is a plausible conjecture that all finite abelian groups (up to isomorphism, of course) occur in this way.  Conjectures and heuristics in this vein have been made, but unfortunately for me I'm not so familiar with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for imaginary quadratic fields is different.  Here there is an absolute bound on the size of an integer $k$ such that the class group of an imaginary quadratic field can be isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$.  Conditionally on the Generalized Riemann Hypothesis, the largest such $k$ is $4$.  This has do to with &lt;strong&gt;idoneal numbers&lt;/strong&gt;, of which the following paper provides a very fine survey:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the truth is slightly stronger: let $H_D$ be the class group of the imaginary quadratic field $\mathbb{Q}(\sqrt{-D})$.  Then, as $D$ tends to negative infinity through squarefree numbers, the size of $2H_D$ (the image of multiplication by $2$) tends to infinity.  See for instance &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.0358v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.0358v2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some recent explicit bounds on this.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-20T12:30:20.480" LastActivityDate="2010-11-20T12:30:20.480" CommentCount="2" />
  <row Id="11015" PostTypeId="2" ParentId="10993" CreationDate="2010-11-19T22:50:52.363" Score="4" ViewCount="0" Body="&lt;p&gt;This follows from the more general inequality&#xA;\begin{equation}(1)\qquad\qquad|x+y|^p\le |x|^p+|y|^p\qquad\qquad\text{(for $x,y\in\mathbb{C}$ and $0\lt p\le1$)}\end{equation}&#xA;Indeed, if we replace $x$ by $x-y$ in (1) we get &#xA;$$|x|^p\le |x-y|^p+|y|^p$$&#xD;&#xA;which imply&#xA;$$|x|^p-|y|^p\le |x-y|^p$$&#xD;&#xA;To prove (1), first note&#xA;$$|x+y|^p\le(|x|+|y|)^p$$&#xD;&#xA;Hence it is sufficient to prove (1) for $x,y\ge0$ in which case we may apply Sivaram's argument in the previous answer.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-20T05:32:44.807" LastActivityDate="2010-11-20T05:32:44.807" CommentCount="5" />
  <row Id="11016" PostTypeId="1" AcceptedAnswerId="11017" CreationDate="2010-11-19T23:01:19.047" Score="4" ViewCount="98" Body="&lt;p&gt;I know it is unknown if there are infinitely many primes of the form $n^2+1$. Is it known if there is a positive integer $k$ such that $|\{n\in\mathbb{Z}:n^2+1 \text{ has at most k prime factors}\}|=\infty$?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-19T23:06:21.037" Title="Prime factors of $n^2+1$" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="11017" PostTypeId="2" ParentId="11016" CreationDate="2010-11-19T23:05:42.090" Score="8" ViewCount="0" Body="&lt;p&gt;Yes, Iwaniec, &quot;Almost-primes represented by quadratic polynomials&quot;, Inventiones Math., 47:171–188, 1978, proves that there exist inﬁnitely many &#xA;integers $n$ such that $n^2 + 1$ is either prime or the product of two &#xA;primes. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-19T23:05:42.090" CommentCount="3" />
  <row Id="11018" PostTypeId="2" ParentId="11016" CreationDate="2010-11-19T23:06:21.037" Score="1" ViewCount="0" Body="&lt;p&gt;Please see  Iwaniec, &quot;Almost-primes represented by quadratic polynomials&quot;, Inventiones Math&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-19T23:06:21.037" />
  <row Id="11019" PostTypeId="1" CreationDate="2010-11-19T23:08:19.103" Score="0" ViewCount="56" Body="&lt;p&gt;Let ${f_{n}}$ be a sequence in $L^{2}(X,\mu)$ such that $||f_{n}||_{2} \rightarrow 0$ as $n \rightarrow \infty$. How to show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \to \infty} \int_{X} |f_n(x)| \log(1+|f_{n}(x)|) d\mu = 0$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3724" LastActivityDate="2010-11-19T23:36:36.903" Title="Compute limit of integral" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="11020" PostTypeId="2" ParentId="11019" CreationDate="2010-11-19T23:12:10.163" Score="1" ViewCount="0" Body="&lt;p&gt;Use the fact that $\log (1+y) \leq y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \displaystyle \int_{X} |f_n(x)| |f_n(x)| d \mu = \displaystyle \int_{X} |f_n(x)|^2 d \mu = ||f_n||_2^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $0 \leq \displaystyle \lim_{n \rightarrow \infty} \displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \lim_{n \rightarrow \infty} ||f_n||_2^2 = 0$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-19T23:36:36.903" LastActivityDate="2010-11-19T23:36:36.903" CommentCount="3" />
  <row Id="11023" PostTypeId="1" CreationDate="2010-11-19T23:44:53.220" Score="0" ViewCount="64" Body="&lt;p&gt;If $f \in L(0,1)$ show that $x^{n} f(x) \in L(0,1)$ for $n \in \mathbb{N}$ and that $\displaystyle \int_{0}^{1} x^{n} f(x) dx \rightarrow 0$ as $n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following attempt correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $0&amp;lt;x&amp;lt;1$ then $0&amp;lt;x^{n}&amp;lt;1$ also $x^{n}$ is measurable (being cts) and $f$ is measurable because it is in $L(0,1)$ so $x^{n} f(x)$ is measurable. Now certainly $|x^{n}f(x)| \leq |f(x)|$ so this implies that $x^{n} f(x)$ is integrable so in $L(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to finish the second part can we simply say that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{n} f(x) \rightarrow 0$ as $n \rightarrow \infty$ pointwise so we may apply convergence theorem.&lt;/p&gt;&#xA;" OwnerUserId="3724" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-19T23:46:08.837" LastActivityDate="2010-11-19T23:49:49.557" Title="Elementary problem in integration theory" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11024" PostTypeId="2" ParentId="11023" CreationDate="2010-11-19T23:49:49.557" Score="1" ViewCount="0" Body="&lt;p&gt;I am assuming that by $L(0,1)$, you mean $\mathcal{L}^1 (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $x^nf(x)$ dominated by $f(x)$ in the interval $(0,1)$ and $f \in \mathcal{L}^1 (0,1)$. Hence, you can use dominated convergence theorem to swap the limit and the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And hence you will get the desired result since $\displaystyle \lim_{n \rightarrow \infty} x^n f(x) = 0$ almost surely.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-19T23:49:49.557" CommentCount="0" />
  <row Id="11025" PostTypeId="2" ParentId="10949" CreationDate="2010-11-19T23:59:46.147" Score="5" ViewCount="0" Body="&lt;p&gt;The smallest abelian group which is not the class group of an imaginary quadratic field is $(\mathbf{Z}/3 \mathbf{Z})^3$. There are six other groups of order&#xA;$&amp;lt; 100$ which do not occur in this way, of orders&#xA;$32$, $27$, $64$, $64$, $81$, and $81$ respectively.&#xA;The groups $(\mathbf{Z}/3 \mathbf{Z})^2$ and $(\mathbf{Z}/2 \mathbf{Z})^4$ occur as class groups of imaginary quadratic fields exactly once, for $D = -4027$ and $-5460$ respectively. &#xA;These results follow from the &quot;class number $100$&quot; problem, solved by Mark Watkins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you restrict to the $p$-part of the class group, then the answer (for general number fields) is positive. That is, for any abelian $p$-group $A$, there exists a number field $K$ with class group $C$ such that $C \otimes \mathbf{Z}_p = A$.&#xA;There is even a non-abelian analog of this. Namely, for any finite $p$-group $G$, there exists a number field $K$ such that the maximal Galois $p$-extension $L/K$ unramified everywhere has Galois group $G$. This is a very recent result of Ozaki.&lt;/p&gt;&#xA;" OwnerUserId="3728" LastActivityDate="2010-11-19T23:59:46.147" CommentCount="5" />
  <row Id="11026" PostTypeId="1" AcceptedAnswerId="11030" CreationDate="2010-11-20T00:00:57.793" Score="0" ViewCount="55" Body="&lt;p&gt;It is a well-known fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f\colon a \mapsto \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;0 &amp;amp; 1\\&#xD;&#xA;\end{array}\right), ~ b \mapsto \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; 0\\&#xD;&#xA;1 &amp;amp; 1\\&#xD;&#xA;\end{array}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a monomorphism from the free monoid generated by $a$ and $b$ to the matrix monoid $\mathbb{Z}^{2\times 2}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an efficient algorithm which computes the length of $f^{-1}(X)$ from $X \in \mathbb{Z}^{2\times 2}$ (that is, the number of symbols in corresponding word $x$, if $f^{-1}(X) = \{x\}$)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an efficient algorithm which computes $f^{-1}(X)$ from $X \in \mathbb{Z}^{2\times 2}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a standard reference for this problem?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3729" LastActivityDate="2010-11-20T13:41:53.387" Title="Inverse of a monoid homomorphism" Tags="&lt;algebra&gt;&lt;reference-request&gt;" AnswerCount="1" />
  <row Id="11027" PostTypeId="2" ParentId="10992" CreationDate="2010-11-20T00:03:53.703" Score="1" ViewCount="0" Body="&lt;p&gt;The right function is &lt;code&gt;TINV()&lt;/code&gt;, but the problem is that &lt;code&gt;TINV()&lt;/code&gt; assumes that you're inputting the two-tailed probability, not the one-tailed probability.  In other words, &lt;code&gt;TINV(p,d)&lt;/code&gt; outputs the value of $x$ such that $P(|X| \geq x) = p$, where $X$ has a $t$ distribution with $d$ degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if you want  &lt;code&gt;TINV()&lt;/code&gt; to give you the critical value associated with a one-tailed probability, you have to double the probability first to account for the other tail.  Thus, for example, &lt;code&gt;TINV(0.1,39)&lt;/code&gt; yields an output of &lt;code&gt;1.684875122&lt;/code&gt;, which is the critical value you want.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-20T00:09:10.717" LastActivityDate="2010-11-20T00:09:10.717" />
  <row Id="11028" PostTypeId="1" CreationDate="2010-11-20T00:07:04.263" Score="2" ViewCount="128" Body="&lt;p&gt;Stuck with this problem from Zgymund's book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f_{n} \rightarrow f$ almost everywhere and that $f_{n}, f \in L^{p}$ where $1&amp;lt;p&amp;lt;\infty$. Assume that $||f_{n}||_{p} \leq M &amp;lt; \infty$. Prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int f_{n}g \rightarrow \int fg$ as $n \rightarrow \infty$ for all $g \in L^{q}$ such that $\frac{1}{p} + \frac{1}{q} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right, so I estimate the difference of the integrals and using Hölder end up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\int f_{n} g - \int fg| \leq ||g||_{q} ||f_{n} - f||_{p}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I'm stuck because we are not assuming convergence in the seminorm but just pointwise convergence almost everywhere. How to proceed?&lt;/p&gt;&#xA;" OwnerUserId="3724" LastEditorUserId="3724" LastEditorDisplayName="" LastEditDate="2010-11-20T00:35:56.297" LastActivityDate="2010-11-21T02:47:11.463" Title="Convergence of integrals in L^p" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="11029" PostTypeId="1" AcceptedAnswerId="11032" CreationDate="2010-11-20T00:09:03.883" Score="3" ViewCount="28" Body="&lt;p&gt;For a 2D Normal distribution $N(0, \left[ \begin{array}{cc} 1 &amp;amp; -1/4 \\ -1/4 &amp;amp; 1 \end{array} \right])$, I am now trying to maximize its density function over $\{ x\geq 10, y \geq 10 \}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is that the optimal solution is $(10,10)$. My problem is that I am not sure which way to find the optimal solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastActivityDate="2010-11-20T00:26:57.090" Title="Maximize normal density function over a subset" Tags="&lt;probability&gt;&lt;optimization&gt;" AnswerCount="1" />
  <row Id="11030" PostTypeId="2" ParentId="11026" CreationDate="2010-11-20T00:09:46.863" Score="1" ViewCount="0" Body="&lt;p&gt;Yes; see the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith_normal_form&quot; rel=&quot;nofollow&quot;&gt;Smith normal form&lt;/a&gt;.  In two dimensions you are essentially using the Euclidean algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  The basic observation here is that if $M = \left[ \begin{array}{cc} x &amp;amp; y \\\ z &amp;amp; w \end{array} \right]$ is an integer matrix, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(a^{-1}) = \left[ \begin{array}{cc} x &amp;amp; y - z \\\ z &amp;amp; w - z \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1}) = \left[ \begin{array}{cc} x - y &amp;amp; y \\\ z - w &amp;amp; w \end{array} \right].$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in addition $\det M = 1$ and $x, y, z, w$ are non-negative integers, then exactly one of these operations will give a matrix with the same properties (this is equivalent to $f$ being a monomorphism).  So, as I said, essentially one is performing the Euclidean algorithm on the columns (or, if you multiply from the left instead, on the rows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, if $M = \left[ \begin{array}{cc} 8 &amp;amp; 5 \\\ 3 &amp;amp; 2 \end{array} \right]$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1}) = \left[ \begin{array}{cc} 3 &amp;amp; 5 \\\ 1 &amp;amp; 2 \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1} a^{-1}) = \left[ \begin{array}{cc} 3 &amp;amp; 2 \\\ 1 &amp;amp; 1 \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M f(b^{-1} a^{-1} b^{-1}) = \left[ \begin{array}{cc} 1 &amp;amp; 2 \\\ 0 &amp;amp; 1 \end{array} \right] = f(aa)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence $M = f(aabab)$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-20T13:41:53.387" LastActivityDate="2010-11-20T13:41:53.387" CommentCount="5" />
  <row Id="11031" PostTypeId="1" CreationDate="2010-11-20T00:22:32.650" Score="1" ViewCount="107" Body="&lt;p&gt;I am trying to do the following stochastic integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_0^T \mu(B_s) dB_s - \frac{\int_0^T (\mu(B_s))^2 ds}{2}&#xD;&#xA;$$&#xD;&#xA;where $ \{ B_t \}$ is a standard Brownian motion, and $\mu(x) = \frac{\beta}{2x} (x^2 - \frac{4\gamma}{\sigma^2} - \frac{1}{ \beta})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently only know how to integrate when the integrands are constant, i.e. $\mu$.is constant function. I was wondering how to integrate under more complicated case as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-23T02:27:51.583" LastActivityDate="2010-11-23T02:27:51.583" Title="Compute a stochastic integral" Tags="&lt;stochastic-processes&gt;&lt;stochastic-integrals&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11032" PostTypeId="2" ParentId="11029" CreationDate="2010-11-20T00:26:57.090" Score="4" ViewCount="0" Body="&lt;p&gt;The density of your distribution is&#xA;$$ \exp \left( {-1 \over 2(1-1/16)} \left[ x^2 + y^2 + {xy \over 2} \right] \right)$$&#xD;&#xA;multiplied by a normalizing constant. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Multivariate_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;; you have $\mu_x = \mu_y = 0, \sigma_x = \sigma_y = 1, \rho = -1/4$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your problem reduces to minimizing the quadratic $x^2 + y^2 + xy/2$ over $\{ x \ge 10, y \ge 10 \}$. This function is increasing in both $x$ and $y$ and so is minimized at $x = y = 10$. &lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-11-20T00:26:57.090" />
  <row Id="11033" PostTypeId="2" ParentId="11031" CreationDate="2010-11-20T00:31:19.877" Score="4" ViewCount="0" Body="&lt;p&gt;You can do this by different methods. One way I like is to use Ito's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider $B_t^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(B_t^2) = 2B_t dB_t + dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\int_{0}^{T} B_t dB_t = \frac{1}{2} \int_{0}^{T} (d(B_t^2) - dt) = \frac{1}{2} (B_T^2 - T)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider $B_t^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use Ito's formula,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(B_t^3) = 3B_t^2 dB_t + 3 B_t dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\int_{0}^{T} B_t^2 dB_t = \frac{1}{3} \int_{0}^{T} (d(B_t^3) - 3 B_t dt) = \frac{1}{3} (B_T^3 - 3 \int_{0}^{T} B_t dt)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use these to get your desired answer. (Plug in for $\mu(B_s)$ and then simplify to get integrals of the form $\int_{0}^{T} B_t^2 dB_t$ and $\int_{0}^{T} B_t dB_t$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-20T00:31:19.877" />
  <row Id="11034" PostTypeId="2" ParentId="11002" CreationDate="2010-11-20T00:33:04.650" Score="1" ViewCount="0" Body="&lt;p&gt;I already commented with one Wikipedia link above, but I just want to add that if you run the Pascal's triangle recurrence $\bmod 2$ what you get is a sequence of approximations to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sierpinski_triangle&quot; rel=&quot;nofollow&quot;&gt;Sierpinski triangle&lt;/a&gt;.  So the result that comes out of Lucas' theorem is an explicit nonrecursive algorithm for plotting the Sierpinski triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to take this opportunity to mention a paper which discusses generalizations of this problem, and which is both quite entertaining to read and perhaps has one of the strangest titles of any math paper: Granville's &lt;a href=&quot;http://www.gianpierobiancoli.it/wp-content/uploads/2009/10/beeb.pdf&quot; rel=&quot;nofollow&quot;&gt;Zaphod Beeblebrox's Brain and the Fifth-Ninth Row of Pascal's Triangle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-20T00:33:04.650" />
  <row Id="11035" PostTypeId="2" ParentId="11028" CreationDate="2010-11-20T00:56:09.103" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Egorov's theorem, convergence a.e. implies for every $\epsilon$ there exists $B$ with $|B| &amp;lt; \epsilon$ such that $f_n\to f$ uniformly on $X\setminus B$ (where $X$ is &quot;almost&quot; the whole space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split $\int (f_n - f)g$ in two pieces, one over $B$ and one over $X\setminus B$. On $X\setminus B$ uniform convergence implies the integral can be made as small as you want. Holder's inequality implies the integral on $B$ is controlled by $(M + \|f\|_p)\|g\|_{L^q(B)}$. Taking $B\searrow$ a measure zero set, then the integral on $B$ of $g$ goes to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finish by taking a diagonalizing sequence as usual. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-20T00:56:09.103" CommentCount="8" />
  <row Id="11036" PostTypeId="1" AcceptedAnswerId="11046" CreationDate="2010-11-20T00:59:16.760" Score="3" ViewCount="185" Body="&lt;p&gt;For every odd integer $n$, $3\leq n\leq 199$, there exists an integer $m\geq 0$, and a prime number $p$, such that $n=2^m+p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I created a software program in Java to solve this problem.  The statement is actually false.  The first odd number it fails at is 127.  I really appreciate everyone's input on this.&lt;/p&gt;&#xA;" OwnerUserId="3691" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T03:27:11.740" LastActivityDate="2010-11-21T03:27:11.740" Title="Can I solve this with induction?" Tags="&lt;discrete-mathematics&gt;&lt;proof-strategy&gt;&lt;induction&gt;&lt;proof&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="11038" PostTypeId="2" ParentId="11036" CreationDate="2010-11-20T01:32:04.000" Score="4" ViewCount="0" Body="&lt;p&gt;Induction can't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;.  As Arturo points out, I made too strong of a statement here.  Perhaps a more correct way to state it is &quot;Induction in the usual sense can't work&quot; or &quot;There is likely no natural inductive proof&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it DID work, then there wouldn't be an upper bound of of n = 199.  However, the case n = 251 provides a counterexample since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 1+ 2*125&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 2 + 3*83&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 4 + 13*19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 8 + 3*81&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 16 + 5*49&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 32 + 3*73&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 64 + 11*17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;251 = 128 + 3*41&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And all other powers of 2 are greater than 251.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I didn't check if this is the smallest counterexample or not)&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-20T05:04:12.220" LastActivityDate="2010-11-20T05:04:12.220" CommentCount="1" />
  <row Id="11039" PostTypeId="2" ParentId="11036" CreationDate="2010-11-20T01:40:20.127" Score="1" ViewCount="0" Body="&lt;p&gt;Induction is not valid, because it is used to show propositions for all integers positive or integers infinite. The condition $n\Rightarrow n+1$ involves integers infinites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the solution is prove each for find $m$ and $p$ in each case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$173= 2^{6}+109$$&#xD;&#xA;$$153=2 ^{6}+89$$&#xD;&#xA;$$117=2^{4}+101$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-20T01:50:35.410" LastActivityDate="2010-11-20T01:50:35.410" />
  <row Id="11040" PostTypeId="1" AcceptedAnswerId="11042" CreationDate="2010-11-20T02:23:30.837" Score="2" ViewCount="178" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taken a look at the following topics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3069/number-of-infinite-sets-with-different-cardinalities&quot;&gt;number of infinite sets with different cardinalities&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/1467/cardinality-of-all-cardinalities&quot;&gt;Cardinality of all cardinalities&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/8013/are-there-uncountably-infinite-orders-of-infinity&quot;&gt;Are there uncountably infinite orders of infinity?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/5378/types-of-infinity&quot;&gt;Types of infinity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still can't quite find/understand the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the easiest way to prove (if possible, without using ordinals etc. as my current math understanding of set theory counts only cardinals, and countable &amp;amp; uncountable sets) that the number of cardinalities that exists is not countable (that is, can't be put into bijection with N)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What exactly does it mean that the set of all cardinals is so big that it's not even a set, but a class? Where does contradiction that does not allow it to be a set arise? I have read Pete's notes at &lt;a href=&quot;http://math.uga.edu/~pete/settheorypart1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/settheorypart1.pdf&lt;/a&gt;, but am not quite sure how #20 leads up to that conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot! &lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T06:59:29.477" LastActivityDate="2010-12-27T06:59:29.477" Title="Cardinality of a set that consists of all existing cardinalities" Tags="&lt;set-theory&gt;&lt;infinity&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="11042" PostTypeId="2" ParentId="11040" CreationDate="2010-11-20T02:34:16.637" Score="6" ViewCount="0" Body="&lt;p&gt;There is no &quot;number of cardinalities&quot;. As you say, there are so many that they cannot form a set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that ${\mathcal A}$ is a set whose elements are sets, with the property that if $A,B\in{\mathcal A}$ and $A\ne B$, then $|A|\ne|B|$, i.e., $A,B$ have different cardinalities. Let $C=\bigcup{\mathcal A}$, i.e., $C=\bigcup_{A\in{\mathcal A}}A$. Clearly, $|C|\ge|A|$ for each $A\in{\mathcal A}$. Let $D={\mathcal P}(C)$ be the power set of $C$. Then $|D|&amp;gt;|C|$ so $|D|&amp;gt;|A|$ for any $A\in{\mathcal A}$. This proves that there cannot be a set of all cardinalities, because given any such set, we just found a new cardinality different from all the ones in the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if ${\mathcal A}$ is countable, this shows that the ''number'' of cardinalities is not countable. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is a small remark that may be worth making. The argument works just as well if we do not require that all sets in ${\mathcal A}$ have different cardinalities, but simply that for any prescribed cardinality we want to consider, there is at least one set in ${\mathcal A}$ of that size (but there may more than one). This is slightly more general, but there is also a technical advantage, namely, in this form, the argument does not depend in any version of the axiom of choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Finally: I just checked Pete's nice note that is linked to in the body of the question. His fact 20 there is essentially the argument I've shown here.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-20T04:32:35.753" LastActivityDate="2010-11-20T04:32:35.753" CommentCount="2" />
  <row Id="11043" PostTypeId="2" ParentId="2130" CreationDate="2010-11-20T02:51:11.153" Score="3" ViewCount="0" Body="&lt;p&gt;(Arriving late on the scene.)  One important thing about small categories which hasn't been mentioned yet is that they’re what you use to define small limits and colimits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to say, for instance, that the category $\mathbf{Top}$ of topological spaces is complete, in some sense.  But if you tried to say “&lt;em&gt;all&lt;/em&gt; (co)limits exist”, then you could take something like “the coproduct of $\mathrm{ob}(\mathbf{Top})$-many copies of the point”, which would get you into the territory of Russell’s paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;can&lt;/em&gt; show is statements like “the category of &lt;em&gt;small&lt;/em&gt; spaces has all &lt;em&gt;small&lt;/em&gt; (co)limits”.  (Or replace small with “$n$-small” for the $n$th Grothendieck universe, or with “$\lambda$-bounded” if you prefer set-theoretic language, and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this exemplifies an important thing about small categories: they’re the categories which play nicely in algebraic constructions with small objects of other kinds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another reason for wanting to talk about small categories is because we obviously want to discuss some kind of category of categories, $\mathbf{Cat}$.  But versions of Russell’s paradox tell us that this category has to be bigger than all the categories we put into it; so to form it as a category itself, we have to accept a size limitation of some kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, from another point of view (a different logical formalism, to be precise), if we want to really talk about the category of &lt;em&gt;all&lt;/em&gt; categories, we have to let ourselves talk about this as a proper class — in other words, step back to a perspective where we can see larger things than we could before; at which point, for want of a better term, we start calling the things we could see before “small”.&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-11-20T02:51:11.153" />
  <row Id="11044" PostTypeId="1" AcceptedAnswerId="11053" CreationDate="2010-11-20T03:14:23.077" Score="3" ViewCount="60" Body="&lt;p&gt;My PhD supervisor called the following result something along the lines: &quot;The theorem in the theory of Latin squares with the most incorrect proofs&quot; (then, to my embarrassment, proceeded to point out an error in my &quot;proof&quot; at the time).  I'll pose proving it as a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;:  A $k \times n$ *Latin rectangle* is a $k \times n$ matrix with symbols from $\{1,2,\ldots,n\}$ for which each symbol occurs exactly once in each row and at most once in each column.  For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 4 3&#xA;2 3 1 4&#xA;4 1 3 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is a $3 \times 4$ Latin rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;:  A Latin rectangle is called &lt;em&gt;reduced&lt;/em&gt; if the first row is $(1,2,\ldots,n)$ and the first column is $(1,2,\ldots,k)^T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above Latin rectangle is not reduced, but this one is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 2 3 4&#xA;2 3 4 1&#xA;3 4 1 2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let $L_{k,n}$ be the number of $k \times n$ Latin rectangles.  Let $R_{k,n}$ be the number of reduced $k \times n$ Latin rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;:  For all $1 \leq k \leq n$, \[L_{k,n}=\frac{n!(n-1)!}{(n-k)!}R_{k,n}.\]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;:  Prove it (and double-check your proof).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T04:06:19.677" Title="What is the number of Latin rectangles given the number of reduced Latin rectangles?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="11045" PostTypeId="2" ParentId="11040" CreationDate="2010-11-20T03:26:52.013" Score="1" ViewCount="0" Body="&lt;p&gt;If the OP or anyone else would like to suggest an addition or modification to what is written in my notes on this point, I would be glad to modify them accordingly.  (These notes exist for people to read and understand them and not for any other reason!)  To me, what I wrote looks essentially the same as Andres' answer above, but as I said, it doesn't matter much what I think -- there's obviously some room for improvement here.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-20T03:26:52.013" CommunityOwnedDate="2010-11-20T03:26:52.013" />
  <row Id="11046" PostTypeId="2" ParentId="11036" CreationDate="2010-11-20T03:27:58.347" Score="7" ViewCount="0" Body="&lt;p&gt;I sort of disagree with the statements saying that induction &quot;cannot work&quot;; but it is a subtle disagreement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can use &quot;induction&quot; to prove a statement for a restricted range of numbers. The most common type of such restricted ranges are statements that apply &quot;eventually&quot;: for every $n\geq N$ for some $N$. The two standard ways of doing so is to either do the &quot;base case&quot; with $n=N$ instead of $n=0$ (or $1$, depending on whether you are a set theorist or a number theorist&lt;sup&gt;1&lt;/sup&gt;); or to establish the equivalent proposition for $N+n$ (or $N-1+n$ if you are a number theorist) and do &quot;induction on $n$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one can take a statement such as the one you give and establish, &lt;em&gt;by mathematical induction&lt;/em&gt;, a proposition which is essentially what you want. It's just that it is almost always a wasteful exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see what I mean, let us take the statement you want to prove: for every odd integer $n$, $3\leq n\leq 199$, there is an integer $m$ and a prime number $p$ such that $n=2^m+p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the following statement P:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for all natural numbers $n$, either $n$ is even, or $n\lt 3$, or $n\gt 199$, or $n=2^m+p$ for some integer $m$ and some prime $p$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a statement about &lt;em&gt;all&lt;/em&gt; natural numbers, and as such could be proven by induction. What is more, the statement is &lt;strong&gt;equivalent&lt;/strong&gt; to the one you want to prove: if the statement above is true, then given an odd integer $n$, $3\leq n\leq 199$, the proposition will hold for $n$; but $n$ is not even, is not less than $3$, and is not greater than $199$, so therefore the clause &quot;$n=2^m+p$ for some integer $m$ and some prime $p$&quot; must hold for $n$, exactly what you want. And if the proposition you want to prove is true, then the statement above is true, since the other clauses of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Disjunction&quot; rel=&quot;nofollow&quot;&gt;disjunction&lt;/a&gt; will be trivially true for all &quot;other&quot; natural numbers. So in a sense, proving your statement is &quot;the same&quot; as proving the statement P I give above. So, in that sense, you &lt;em&gt;can&lt;/em&gt; prove your statement by induction, because you can prove the &lt;em&gt;equivalent&lt;/em&gt; statement P, which is something about all natural numbers, by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just that the induction argument would be very convoluted, and almost certainly a waste of time, because the inductive step would have to break up into cases depending on $k$: if $k\geq 199$, then the proof proceeds one way; if $k$ is odd, another; and so on, until you get down to showing what happens if $2\leq k\leq 198$, with $k$ even. At that point, your argument would almost certainly devolve into simply a direct proof of the fact, so that &quot;Induction&quot; is really a smokescreen that only confuses the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in a sense, &quot;yes, you could prove it by induction, but it is almost certainly a very, very bad idea to try to do so.&quot; The same is true for other statements in which $n$ is bounded above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Induction is particularly powerful when you are trying to prove something holds for &lt;em&gt;infinitely many&lt;/em&gt; natural numbers, because that is not something that we can verify by checking the natural numbers one at a time. When we are trying to show something holds for finitely many natural number, then induction is seldom the way to go.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; The standard way of defining the natural numbers within set theory is to construct them as the smallest &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_set_%28axiom_of_infinity%29&quot; rel=&quot;nofollow&quot;&gt;inductive set&lt;/a&gt;; then the set contains $\emptyset$, which is identified with $0$; or through &lt;a href=&quot;http://en.wikipedia.org/wiki/Peano_axioms&quot; rel=&quot;nofollow&quot;&gt;the Peano axioms&lt;/a&gt;, which usually include $0$ (though they need not do so). So for set theorists (and logicians), $\mathbb{N}$ contains $0$. But historically, $\mathbb{N}$ represented the counting numbers, and so &quot;began&quot; with $1$. So number theorists usually consider $\mathbb{N}$ to consist only of the positive integers.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-20T03:55:56.457" LastActivityDate="2010-11-20T03:55:56.457" CommentCount="7" />
  <row Id="11047" PostTypeId="1" CreationDate="2010-11-20T03:44:24.323" Score="0" ViewCount="68" Body="&lt;p&gt;How many committees of 5 can be chosen from 10 people if each committee includes Alice and&#xA;excludes Bob?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many committees of 5 can be chosen from 12 people if each committee&#xA;(i) includes Alice;&#xA;(ii) excludes Bob;&#xA;(iii) includes Alice and excludes Bob;&#xA;(iv) includes at least one of either Alice or Bob&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help answers this problem and include the explaination of how to get the answers. Thanks you. &lt;/p&gt;&#xA;" OwnerUserId="3732" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-20T04:02:44.077" LastActivityDate="2010-11-20T04:12:10.990" Title="Principle of counting problem" Tags="&lt;homework&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11048" PostTypeId="2" ParentId="11047" CreationDate="2010-11-20T03:51:05.027" Score="2" ViewCount="0" Body="&lt;p&gt;1) To form a committee of $5$ people which includes Alice, we need to choose just $4$ more members. Since it excludes Bob, we only have $8$ out of the $10$ people to choose from (Alice and Bob have been accounted for). So this can be done in ${8}\choose{4}$ ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Inclusion into a committee is to be regarded as one less person to choose and exclusion from a committee should be thought as one less person to choose from. The various cases are just a reapplication of these ideas once or more.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-20T04:12:10.990" LastActivityDate="2010-11-20T04:12:10.990" CommentCount="1" />
  <row Id="11049" PostTypeId="2" ParentId="11047" CreationDate="2010-11-20T04:00:24.797" Score="0" ViewCount="0" Body="&lt;p&gt;@Timothy Wagner: here is my suggestion.  Thanks for not doing the hardest one. There is a wide range in how people respond to homework questions, so don't take mine as gospel-make up your own mind.  But the differences between us are small.  It is in how big a hint to give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet this is homework.  Please tag as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)You have 10 people to choose from.  Alice is required and Bob is excluded.  So there are really only 8 to think about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)iii is the same (with 10-&gt;12) as the first.  i and ii and not so different&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-20T04:00:24.797" CommentCount="8" />
  <row Id="11050" PostTypeId="1" AcceptedAnswerId="11051" CreationDate="2010-11-20T04:46:13.610" Score="0" ViewCount="145" Body="&lt;p&gt;Let $\{f_n\}$ be a sequence of smooth functions which converges to a function $f$.  If the convergence is not uniform at a point $a$ the $f$ is discontinuous at $a$.&#xA;Is there any different type of convergence where if it happens at $a$ then $f$ is continuous at $a$ but is not differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\{f_n\}$ be a sequence of smooth functions which converges to a function $f$.If there  is a discontinuity in $f$ at some point $a$ then the convergence is nonuniform.Is there any different type of convergence needed for $f$ to be continuous at some point $a$ but not differentiable ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-20T18:37:22.323" LastActivityDate="2010-11-20T18:37:22.323" Title="Convergence and differentiability" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="23" FavoriteCount="1" />
  <row Id="11051" PostTypeId="2" ParentId="11050" CreationDate="2010-11-20T05:10:39.233" Score="1" ViewCount="0" Body="&lt;p&gt;P[a,b], the set of polynomials on [a,b] (obviously a subset of the smooth functions on [a,b]) is dense in C[a,b]. In addition the set of functions that are differentiable at (at least) a single point in [a,b] are of the first category in C[a,b]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense &quot;most&quot; convergent sequences of smooth functions converge to nowhere differentiable functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semi-related, but you might in interested in the fact that $C^k(\Omega)$, $\Omega \subset \mathbb{R}^n$ open, can be made into a Fréchet space for any $k = 1,2,\ldots, \infty$. See Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Fr%C3%A9chet_space#Examples&quot; rel=&quot;nofollow&quot;&gt;Fréchet spaces&lt;/a&gt;, the topology on $C^k(\Omega)$ being uniform convergence of $f_n$ and all of its derivatives up to order $k$ (q.v. &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiindex&quot; rel=&quot;nofollow&quot;&gt;multiindex&lt;/a&gt;) &lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-20T05:21:23.963" LastActivityDate="2010-11-20T05:21:23.963" CommentCount="1" />
  <row Id="11053" PostTypeId="2" ParentId="11044" CreationDate="2010-11-20T06:36:47.237" Score="1" ViewCount="0" Body="&lt;p&gt;Consider the following reduction algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move the unique column in which $1$ is the first element to the left (extract it and put it as the first column).&lt;/li&gt;&#xA;&lt;li&gt;Apply the unique injection that maps the $j$th element in the first column to $j$.&lt;/li&gt;&#xA;&lt;li&gt;Apply the unique permutation on the $n-1$ columns to the right of the first column which makes the first row $1\ldots n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Taking our steps in reverse to unscramble a reduced form, we have $(n-1)!$ choices in the third step, $(n-1)^{\underline{k-1}} = (n-1)!/(n-k)!$ choices in the second step, and $n$ choices in the first step, to a total of $$(n-1)!\frac{(n-1)!}{(n-k)!}n = \frac{n!(n-1)!}{(n-k)!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, we could define&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A_{k,n}$ as the number of Latin rectangles whose upper left value is $1$.&lt;/li&gt;&#xA;&lt;li&gt;$B_{k,n}$ as the number of Latin rectangles whose left-most column is reduced.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then we show the following:&#xA;$$\begin{align*} L_{k,n} &amp;amp;= n A_{k,n}, \\ A_{k,n} &amp;amp;= \frac{(n-1)!}{(n-k)!} B_{k,n}, \\ B_{k,n} &amp;amp;= (n-1)R_{k,n}. \end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-20T06:36:47.237" CommentCount="2" />
  <row Id="11054" PostTypeId="2" ParentId="8180" CreationDate="2010-11-20T06:56:50.723" Score="1" ViewCount="0" Body="&lt;p&gt;In my experience I haven't seen the gain go above one, but theoretically, it shouldn't go above one either, because then the product of the gain and innovation will lie outside the range of the predicted value and the measurement received. This will be incorrect, because the estimated value should be in between the prediction and the measurement received.&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-11-20T06:56:50.723" />
  <row Id="11056" PostTypeId="1" CreationDate="2010-11-20T07:02:49.970" Score="4" ViewCount="95" Body="&lt;p&gt;I was reading Martin Gardner's &lt;em&gt;Mathematical Games&lt;/em&gt; column on the Ulam spiral which appeared in the March 1964 issue of &lt;em&gt;Scientific American&lt;/em&gt;.  (The spiral actually featured on the cover of that issue.)  Gardner makes the following statement: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The grid on the cover suggests that throughout the entire number series expressions of this form are likely to vary markedly from those &quot;poor&quot; in primes to those that are &quot;rich,&quot; and that on the rich lines an unusual amount of clumping occurs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &quot;this form&quot; Gardner means the form $4x^2+bx+c$.  I'm curious - and a little bit skeptical - about his last statement concerning clumping.  I know that the existence of prime-rich and prime-poor polynomials is a longstanding conjecture, going back to Euler's discovery that polynomials such as $x^2-x+41$ generate unusually many primes, and that Hardy and Littlewood and also Bateman and Horn made concrete proposals as to what the density of primes in such polynomials ought to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether there is any evidence, either numerical or heuristic, that there should be a large amount of clumping in the primes of the form $x^2-x+41$.  Famously, the first 40 values of $x$ all give primes, but if one goes to higher values of $x$ are there more long clusters of primes than one would expect if the primes were randomly distributed?&lt;/p&gt;&#xA;" OwnerUserId="3736" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T07:14:24.540" LastActivityDate="2010-11-20T17:00:55.627" Title="Ulam spiral: Is there an &quot;unusual amount of clumping&quot; in prime-rich quadratic polynomials?" Tags="&lt;number-theory&gt;&lt;probability&gt;&lt;prime-numbers&gt;" AnswerCount="1" />
  <row Id="11057" PostTypeId="2" ParentId="10972" CreationDate="2010-11-20T07:03:19.470" Score="1" ViewCount="0" Body="&lt;p&gt;For $x$ large, $\tanh x$ is very close to $1$. Therefore for large $b$, $$\int_0^b \frac{\tanh x}{x} \, \mathrm{d}x \approx C + \int^b \frac{\mathrm{d}x}{x} = C&amp;#39; + \log b.$$ You can prove it rigorously and obtain a nice error bound if you wish. Your post indicates a specific value of $C&amp;#39;$, but for large $b$, any two &quot;close&quot; constants $C_1,C_2$ will satisfy $$\log b + C_1 \approx \log b + C_2,$$ so probably $\gamma + \log (4/\pi)$ has no significance other than being a number close to $C&amp;#39;$ and having a nice form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do the estimation rigorously, we will probably find out that $C&amp;#39;$ is well defined (i.e. the error in the first $\approx$ is $o(b)$), and then one can ask for its value. It probably has no nice closed form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In fact $\gamma + \log (4/\pi)$ is the correct constant, as shown in Derek's answer.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T00:41:44.187" LastActivityDate="2010-11-23T00:41:44.187" CommentCount="0" />
  <row Id="11058" PostTypeId="2" ParentId="6863" CreationDate="2010-11-20T07:03:25.303" Score="0" ViewCount="0" Body="&lt;p&gt;I'm not entirely sure of what parameters you're taking as an input, but as far as my experience goes with the H matrix, it is used to focus on a particular input, irrespective of the weightage you intend to give it (I've seen the H matrix in other forms too, but do not have the knowledge to comment on it). &#xA;What you seem to be looking for, is an algorithm to fuse the inputs of two sensors. The best algorithm available for it is IMM (interacting multiple model). It internally uses Kalman filters, and has steps for estimate fusion and mixing probabilities. A transition probability matrix will help you decide on emphasizing the positive properties and minimising the negative. The models correct each other, which plays a large role in getting a more accurate estimate.&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-11-20T07:03:25.303" />
  <row Id="11059" PostTypeId="1" CreationDate="2010-11-20T07:17:32.633" Score="0" ViewCount="23" Body="&lt;p&gt;In a Kalman filter taking into account NCT, the state matrix is [x; vx; y; vy; omega]. Where omega is the angular velocity.&#xA;What I don't understand is that when the F matrix for NCT is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;F=[ blah blah blah blah 0; &#xA;      blah blah blah blah 0; &#xA;      blah blah blah blah 0;&#xA;      blah blah blah blah 0;&#xA;      0    0    0    0    1; ];&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can the F matrix or any other matrix directly contribute to the omega value? How can the Kalman filter take into account the angular velocity of a particle if there is no contribution to omega, from any of the matrices in the filter? (the P matrix contributes a little, but that causes omega to become 1e-31).&#xA;Am I doing something wrong? Is there a correct way to use the NCT model?&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-12-20T08:44:26.027" Title="How does the NCT model take angular velocity into account? (Nearly coordinated turn)" Tags="&lt;matrices&gt;&lt;discrete-mathematics&gt;&lt;simulation&gt;" AnswerCount="1" />
  <row Id="11060" PostTypeId="2" ParentId="11059" CreationDate="2010-11-20T08:01:16.727" Score="0" ViewCount="0" Body="&lt;p&gt;The whole idea of a &quot;coordinated turn&quot; is that the turn is executed at constant angular velocity. The &quot;nearly&quot; is there to indicate that the only contribution to a change in omega is from noise.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-20T08:01:16.727" CommentCount="1" />
  <row Id="11061" PostTypeId="2" ParentId="10990" CreationDate="2010-11-20T08:03:27.953" Score="1" ViewCount="0" Body="&lt;p&gt;Since your last question looks much general: you might be interested in the nice introductory paper &lt;a href=&quot;http://www.math.nmsu.edu/~davidp/euler2k2.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Dances between discrete and continuous&quot;&lt;/a&gt;. Here D. Pengelley discusses some early motivations (and techniques how) to deal with divergent series (focused on L.Euler). I liked that paper much when I started reading about divergent summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thinking a second time about your qualified question I can't imagine you don't know this all (basic reference G.H.Hardy, &quot;divergent series&quot; et al.) and you are looking for a more modern historical overview over that subject - so my answer might be somehow misfocused...  &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-20T08:10:18.260" LastActivityDate="2010-11-20T08:10:18.260" CommentCount="2" />
  <row Id="11062" PostTypeId="1" AcceptedAnswerId="11067" CreationDate="2010-11-20T08:06:31.297" Score="2" ViewCount="124" Body="&lt;p&gt;How to find the domain of the function $\sqrt{ \log_{\frac{1}{2}} x}$ ?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-11-20T11:24:49.823" Title="Domain Problem: $\sqrt{ \log_{\frac{1}{2}} x}$" Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" />
  <row Id="11064" PostTypeId="2" ParentId="11012" CreationDate="2010-11-20T08:43:30.793" Score="3" ViewCount="0" Body="&lt;p&gt;This isn't a complete answer, but just an idea about your&#xA;first question. Let $P$ be a rank one projective module&#xA;over the commutative ring $A$, and $P^*=\mathrm{Hom}_A(P,A)$&#xD;&#xA;be its dual. Then for $M=P\oplus P^*$, $\bigwedge^2 M\cong&#xD;&#xA;P\otimes_A P^*\cong A$ is free. There must be $P$ for which $M$&#xD;&#xA;isn't free, but I can't think of any off the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is a Dedekind domain then $M$ is free. Taking $A=C^\infty(N)$&#xD;&#xA;where $N$ is a smooth manifold, then $P$ would correspond to&#xA;a line bundle on $N$. If $M$ is free then the direct sum of this&#xA;line bundle and its dual would be trivial. Surely there are manifolds&#xA;and line bundles for which this isn't true?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-20T08:43:30.793" CommentCount="2" />
  <row Id="11065" PostTypeId="1" CreationDate="2010-11-20T09:01:54.330" Score="0" ViewCount="38" Body="&lt;p&gt;I've got a Python script that grinds out Amicable Pairs by trying out each integer: i, i+1, i+2, ... . I've now found all APs with 1 &amp;lt;= n &amp;lt;= 65 million. At this point I'm finding that it takes 24 minutes per each additional million. To get into much larger n I'll need to find a way to filter out substantial numbers of n in advance. Have any been discovered?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My list of the first 202 APs: &lt;a href=&quot;http://tutoree7.pastebin.com/dpEc0RbZ&quot; rel=&quot;nofollow&quot;&gt;http://tutoree7.pastebin.com/dpEc0RbZ&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T11:13:29.880" LastActivityDate="2010-11-20T11:13:29.880" Title="Let (n, m), n &lt; m be an Amicable Pair. Looking for large sets of integers that cannot be n. " Tags="&lt;number-theory&gt;" CommentCount="2" />
  <row Id="11066" PostTypeId="1" CreationDate="2010-11-20T09:11:33.357" Score="1" ViewCount="166" Body="&lt;p&gt;Just today I had a bet with my friend over the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many winning configurations can you have in a nxn Tic-Tac-Toe game where players win if a they get n/2 in either a row or column, consecutively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a 4x4 game, players win if a they get 2 of their symbols in either a row or column, consecutively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet the figure to be &quot;2 * ( 2 * n ) * ( 3 ** ( n / 2 ) )&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I win?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to proceed if we were to count only draws? ( how many board configurations can there be so that they are always draws - i.e. no one wins )&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="3740" LastEditorDisplayName="" LastEditDate="2010-11-21T02:59:34.243" LastActivityDate="2010-11-21T05:55:12.600" Title="How many winning configurations can you have in a nxn Tic-Tac-Toe game where players win if a they get n/2 in either a row or column, consecutively. " Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11067" PostTypeId="2" ParentId="11062" CreationDate="2010-11-20T10:19:38.080" Score="6" ViewCount="0" Body="&lt;p&gt;Assuming the result is real, we must have $\log _{\frac{1}{2}}x\geq 0$.&#xA;Since $\log _{\frac{1}{2}}x\geq 0\Leftrightarrow 0&amp;lt;x\leq 1$, the domain is $%&#xD;&#xA;0&amp;lt;x\leq 1$, with $x\in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added 2: $\log _{\frac{1}{2}}x=\frac{\log x}{\log \frac{1}{2}}=\frac{\log x}{\log&#xD;&#xA;1-\log 2}=\frac{\log x}{0-\log 2}=-\frac{\log x}{\log 2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log _{\frac{1}{2}}x\geq 0\Leftrightarrow-\frac{\log x}{\log 2}\geq 0\Leftrightarrow\log x\le 0\Leftrightarrow 0&amp;lt;x\leq 1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: plot of $\log_{\frac{1}{2}}x$ (green) and $\sqrt{\log_{\frac{1}{2}}x}$ (blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8PsLd.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T11:24:49.823" LastActivityDate="2010-11-20T11:24:49.823" CommentCount="6" />
  <row Id="11068" PostTypeId="1" CreationDate="2010-11-20T10:24:42.697" Score="0" ViewCount="53" Body="&lt;p&gt;How to determine the width of a Gaussian membership function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mp1D5.jpg&quot; alt=&quot;definition&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3625" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T11:04:26.453" LastActivityDate="2010-12-20T16:44:25.987" Title="How to determine the width of a Gaussian membership function?" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11069" PostTypeId="2" ParentId="2949" CreationDate="2010-11-20T11:05:41.553" Score="1" ViewCount="0" Body="&lt;p&gt;The solution to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Hilbert's 10th problem&lt;/strong&gt;&lt;/a&gt;, i.e. the &lt;a href=&quot;http://en.wikipedia.org/wiki/MRDP_theorem&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;MRDP theorem&lt;/strong&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number theorists were trying to find a general method to solve Diophantine equations. Special cases of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diophantine_equation&quot; rel=&quot;nofollow&quot;&gt;Diophantine equations&lt;/a&gt; were/are studied extensively and the theorems are quite nice. Learning the fascinating fact that there is no general method (algorithm) to solve arbitrary Diophantine equations was surprising for me.&lt;/p&gt;&#xA;" OwnerUserId="468" LastActivityDate="2010-11-20T11:05:41.553" CommunityOwnedDate="2010-11-20T11:05:41.553" />
  <row Id="11070" PostTypeId="1" AcceptedAnswerId="11071" CreationDate="2010-11-20T11:11:05.943" Score="1" ViewCount="84" Body="&lt;p&gt;I've been having some rather morbid thoughts lately, so, naturally, I decided to share:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a deadly disease, which has a chance of killing me every day that it is left uncured. Also, suppose that there's this operation I can go through, but it is not without its risks: If it is successful, I won't have to worry about the disease killing me ever again. If it's a failure, I die there and then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, let $q : N \rightarrow [0,1]$ be the probability that the disease kills me at any given day $t \in N$. Also, let $p : N \rightarrow [0,1]$ be the probability the that the operation fails if I have it on day $t \in N$. Assume that both $p$ and $q$ are monotonically increasing. The question is: At what day $t_0$ do I have to go through the operation in order to maximise my expected lifespan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to solve this: It is obvious that at any day that $p(t) \leq q(t)$ you're better off having the operation because if you don't, you'd have the same chance of dying and no &#xA;chance of being cured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that can't possibly be the whole story. What if $p(t) &amp;gt; q(t)$ for every $t \in N$? You would still have to do the operation at some point, otherwise you'll eventually die because of the accumulated probability of all those days you're left uncured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that light, let $f(t) = \prod_{k=0}^{t}(1-q(k))$ be the probability that you survive $t$ days without going through the operation. Then, all you need to do is to solve the inequality $f(t_0) &amp;lt; f(t_0-1) \cdot (1-p(t_0))$. Is that correct? Or am I still missing something?&lt;/p&gt;&#xA;" OwnerUserId="3742" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-20T20:27:49.363" LastActivityDate="2010-11-20T20:27:49.363" Title="To operate or not to operate?" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="11071" PostTypeId="2" ParentId="11070" CreationDate="2010-11-20T11:34:51.220" Score="1" ViewCount="0" Body="&lt;p&gt;Elaborating on my comments above, the formula for the expected lifespan in the case of not taking the operation is&#xA;$$\sum_{t&amp;lt;T}t\bigl(f(t)-f(t-1)\bigr),$$&#xD;&#xA;while the expected lifespan in case of taking the operation is&#xA;$$\sum_{t&amp;lt;t_0}t\bigl(f(t)-f(t-1)\bigr)+T\cdot f(t_0).$$&#xD;&#xA;Here $T$ should be the age that you will die if you take the operation, survive it, and die a natural death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you just have to plug in and see which number is larger.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-20T12:38:02.247" LastActivityDate="2010-11-20T12:38:02.247" />
  <row Id="11072" PostTypeId="1" CreationDate="2010-11-20T11:40:24.690" Score="0" ViewCount="33" Body="&lt;p&gt;Goodmorning at all .&#xA;I'm a software development and i have a small question in mathematic. &#xA;I have a two variables function in a interval [(0,0),(6,6)]. I wont divide this interval in two equal subinterval . So i think [(0,0),(3,3)] and [(3,3),(6,6)]&#xA;It is correct ? &#xA;Which algorithm i should apply? &#xA;Thanks &lt;/p&gt;&#xA;" OwnerUserId="3743" LastActivityDate="2010-11-20T11:40:24.690" Title="divide an interval of function in two variables " Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;soft-question&gt;&lt;functions&gt;" CommentCount="1" />
  <row Id="11073" PostTypeId="1" AcceptedAnswerId="11115" CreationDate="2010-11-20T11:46:36.697" Score="1" ViewCount="116" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S^1$ the unit circle in $\mathbb{R}^2$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d: S^1\times S^1\to\mathbb{R}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d(\theta_1,\theta_2) = \left\{&#xD;&#xA;    \begin{array}{ll}&#xD;&#xA;        |\theta_1-\theta_2| &amp;amp; \mbox{if } |\theta_1-\theta_2|\le \pi \\&#xD;&#xA;        2\pi-|\theta_1-\theta_2| &amp;amp; \mbox{else}&#xD;&#xA;    \end{array}&#xD;&#xA;\right.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove that this function satisfies the triangle inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d(\theta_1,\theta_3)\le d(\theta_1,\theta_2)+d(\theta_2,\theta_3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three possible cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$|\theta_1-\theta_2|, |\theta_2-\theta_3|\le\pi$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$|\theta_1-\theta_2|\le\pi, |\theta_2-\theta_3|&amp;gt;\pi$ or $|\theta_2-\theta_3|\le\pi, |\theta_1-\theta_2|&amp;gt;\pi$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$|\theta_1-\theta_2|, |\theta_2-\theta_3|&amp;gt;\pi$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I proved the first two ones using the triangle inequality for the absolute value, but I'm stuck for the third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is equivalent to prove that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\theta_1-\theta_3|+|\theta_1-\theta_2|+|\theta_2-\theta_3|\le 4\pi\text{ if }|\theta_1-\theta_3|\le\pi$$&#xD;&#xA;and&#xA;$$|\theta_1-\theta_2|+|\theta_2-\theta_3|-|\theta_1-\theta_3|\le 2\pi\text{ if }|\theta_1-\theta_3|&amp;gt;\pi$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you give me a hint ? Or did I misdefine $d$ or any other error ?&lt;/p&gt;&#xA;" OwnerUserId="524" LastActivityDate="2010-11-21T13:40:09.720" Title="Unit circle metric" Tags="&lt;geometry&gt;&lt;metric-spaces&gt;&lt;circle&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11074" PostTypeId="2" ParentId="11073" CreationDate="2010-11-20T12:26:11.873" Score="2" ViewCount="0" Body="&lt;p&gt;Apologies: this should be a comment, but I am not &quot;reputable&quot; enough to leave comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you proved 1, then you should have no problems with 3. The key to the exercise is to understand this metric geometrically. Note that if you draw two radii on the unit circle, then there are two obvious angles between them, that you could measure. The two-case-rule just says &quot;always measure the smaller one of the two angles&quot;. In other words, the distance between two angles is always at most $\pi$, which should allow you to easily complete the proof.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-20T12:26:11.873" CommentCount="2" />
  <row Id="11075" PostTypeId="2" ParentId="9659" CreationDate="2010-11-20T12:55:19.353" Score="-1" ViewCount="0" Body="&lt;p&gt;It is clear that a null set is a subset of every set as it does no contains any element which is also an element of every set as per the definition of subset. Also by definition of propre subset null set should be the proper subset of every non-empty sets as no element of the null set is also an element of the non-empty set. Also there exist atleast one element in the non-empty set which is not an element of the null set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But so far why null set is called an improper subset of every non-empty sets?&lt;/p&gt;&#xA;" OwnerUserId="3745" LastActivityDate="2010-11-20T12:55:19.353" CommentCount="1" />
  <row Id="11076" PostTypeId="1" AcceptedAnswerId="11079" CreationDate="2010-11-20T13:06:42.590" Score="6" ViewCount="335" Body="&lt;p&gt;It is nature to understand the need for scalar(numbers), but why do we invent vectors? Who invent it and for what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As George Lowther pointed out the problem is too broad, I added the following questions as concrete supplements.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's easy for our human to understand the law of addition of scalar number, but why the addition of vector follow the parallelogram rule, why doesn't it follow some other laws?&#xA;link to physics: How do we know that the addition of forces follow the parallelogram law?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The length of 2-dimensional vector is the hypotenuse of the triangle whose is constructed from the two component of the vector and the length of 3-dimensional vector is also follows this way. But what about 4-dimensional vector?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do we define dot production law of vector like it is now? Is it because of its physical essence or its equivalence to the law of cosine?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-20T13:54:46.573" LastActivityDate="2010-12-20T10:52:40.393" Title="Why do we need vectors and who invented it?" Tags="&lt;vector-calculus&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="3" />
  <row Id="11077" PostTypeId="2" ParentId="11073" CreationDate="2010-11-20T13:20:40.893" Score="1" ViewCount="0" Body="&lt;p&gt;This is not a formal proof, but gives, I think, the geometric intuition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ObvOd.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-20T13:20:40.893" CommentCount="2" />
  <row Id="11079" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T13:37:46.463" Score="13" ViewCount="0" Body="&lt;p&gt;Vectors should be thought of, at a first approximation, as &quot;numbers with direction&quot;. For physical phenomena which carries a direction, such as velocity and displacement, vectors are immensely useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of a number with direction most likely dates to antiquity, as the making of maps and sign-posts already implicitly incorporates the notion. The modern representation of a vector/point in space with an ordered triplet of numbers is often attributed to the advent of &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_geometry&quot; rel=&quot;nofollow&quot;&gt;analytical geometry&lt;/a&gt; due to the philosopher &lt;a href=&quot;http://ualr.edu/lasmoller/descartes.html&quot; rel=&quot;nofollow&quot;&gt;Rene Descartes&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different notion of vectors also arose with the &quot;discovery&quot; of the complex numbers by Jerome Cardan: the imaginary numbers can be thought of as living on a different direction as the real scalars (so the complex numbers form a real vector space). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the past 400 years or so the notion of vector gradually evolved to become what we know today, with contributions from branches of mathematics that developed into modern analysis and algebra. A nice summary of that period of development is available &lt;a href=&quot;http://web.archive.org/web/20040126161844/http://www.nku.edu/~curtin/crowe_oresme.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. See Michael Crowe's &lt;a href=&quot;http://en.wikipedia.org/wiki/A_History_of_Vector_Analysis&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; for a fuller description of also the Greek contributions and the influences from the 16th century in this matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, vectors shouldn't be thought of as being &quot;invented&quot;, nor should it be attributed to one person alone.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-20T13:37:46.463" CommentCount="1" />
  <row Id="11080" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T13:48:20.963" Score="4" ViewCount="0" Body="&lt;p&gt;A case study of a particular application:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vectors are absolutely necessary for game development, for instance. How is a 3D model, which may be positioned at any point in space and rotated in any direction and is viewed by a camera pointing in an arbitrary direction, actually rendered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the model is stored in some local coordinate system. This is the data that actually fills the file from which the model is loaded. Although the exact contents will differ across different rendering systems, at the very least you will have the coordinates of the vertices relative to some arbitrarily chosen center point. These are viewed as vectors, and this viewpoint is crucial for what comes next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all the models are loaded by the rendering system, but they need to be placed into a global coordinate system. This includes putting them in their actual position in the world, rotating them, etc. This stage is accomplished by transforming each of the vectors in local coordinates according to a world transformation matrix--this matrix translates and rotates the model. [You may be wondering how translations are accomplished in this way. That's because most rendering systems store vertices in four-dimensional homogeneous coordinates, and the transformation matrices are $4\times 4$--it's a worthwhile exercise to see that this allows you to perform affine transformations on $\mathbb{R}^3$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next another matrix transformation is applied to all the vertex data to take into account the position and orientation of the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally yet another transformation is applied which projects all the data onto the viewing screen--this is not an orthogonal projection but a perspective projection, so that objects in the distance seem smaller than objects close to the camera.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-11-20T13:48:20.963" />
  <row Id="11081" PostTypeId="1" AcceptedAnswerId="11092" CreationDate="2010-11-20T13:55:12.470" Score="1" ViewCount="100" Body="&lt;p&gt;$\lim_{x\to\infty} \frac{e^x}{x^n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n is any natural number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using L'hopital doesn't make much sense to me. I did find this in the book: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In a struggle between a power and an exp, the exp wins.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I refer that line as an answer? If the fraction would have been flipped, then the limit would be zero. But in this case I the limit is actually $\infty$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2230" LastActivityDate="2010-11-20T16:17:50.117" Title="Calculus, find the limit, Exp vs Power?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;limit&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="11082" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T14:00:34.260" Score="7" ViewCount="0" Body="&lt;p&gt;Willie covered a lot of what I wanted to say; however I'd like to make a little historical digression: before we ever had the concept of a vector, there was the quaternion, William Rowan Hamilton's generalization of the usual complex numbers. They proved very convenient for physical applications, and thus the use of quaternions took off. In fact the electromagnetic equations of Maxwell were first couched in quaternion notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was Josiah Gibbs and independently Oliver Heaviside who looked at decomposing the quaternion into a scalar (real) part, and a vector (imaginary) part, and found that the manipulations in this new formulation were &quot;cleaner&quot;. Vector analysis took off, and quaternions became less prominent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say more, but a &lt;a href=&quot;http://books.google.com/books?id=y5-S5dmVqGIC&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; has already been written about this matter, so I refer you to it.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-20T14:00:34.260" CommentCount="3" />
  <row Id="11083" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T14:00:40.063" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; One way of looking at this would be: $$\frac{1}{x^{n}} \biggl[ \biggl(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}\biggr) + \frac{x^{n+1}}{(n+1)!} + \cdots \biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you understand why i put the brackets inside those terms.&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-20T14:00:40.063" CommentCount="1" />
  <row Id="11084" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T14:10:38.933" Score="1" ViewCount="0" Body="&lt;p&gt;Why isn't L'Hôpital a good solution? Just use induction.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-20T14:10:38.933" CommentCount="2" />
  <row Id="11085" PostTypeId="1" AcceptedAnswerId="11086" CreationDate="2010-11-20T14:22:42.477" Score="3" ViewCount="176" Body="&lt;p&gt;How do I calculate:&#xA;The height of a regular tetrahedron, side length 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-28T22:38:33.123" Title="Height of a tetrahedron" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11086" PostTypeId="2" ParentId="11085" CreationDate="2010-11-20T14:27:33.737" Score="7" ViewCount="0" Body="&lt;p&gt;The first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's a view of the geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Cj5bL.png&quot; alt=&quot;tetrahedron&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here's a view of the bottom face:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0QvQD.png&quot; alt=&quot;triangle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that the short sides of the isosceles triangle bisect the 60° angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30°, 30° and 120°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1=2\ell^2-2\ell^2\cos 120^{\circ}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\frac{1}{\sqrt{3}}$, as indicated &lt;a href=&quot;http://mathworld.wolfram.com/EquilateralTriangle.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h^2+\left(\frac{1}{\sqrt{3}}\right)^2=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $h$ in the above equation, we now find the height to be $\sqrt{\frac23}=\frac{\sqrt{6}}{3}$, as mentioned &lt;a href=&quot;http://mathworld.wolfram.com/Tetrahedron.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T22:00:33.340" LastActivityDate="2010-11-20T22:00:33.340" CommentCount="4" />
  <row Id="11087" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T14:51:09.867" Score="1" ViewCount="0" Body="&lt;p&gt;I believe the problem is tailor-made for repeated application of L'Hopital's Rule, but here are some thoughts ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could note that $e^{x} = (e^{x/n})^n$, and consider $\left( \lim \frac{e^{x/n}}{x}\right)^n$, so that you are comparing an exponential to a single power of $x$, which might be a bit less daunting for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more cleanly, and to make the numerator and denominator match better, define $y := \frac{x}{n}$. Then $$\frac{e^{x}}{x^n}=\frac{e^{ny}}{(ny)^n}=\frac{\left(e^{y}\right)^n}{n^n y^n}=\frac{1}{n^n}\frac{\left(e^{y}\right)^n}{y^{n}}=\frac{1}{n^n}\left(\frac{e^y}{y}\right)^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n$ is a constant, you can direct your limiting attention to $\frac{e^y}{y}$ (as $y \to \infty$, of course).&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-20T14:51:09.867" />
  <row Id="11088" PostTypeId="1" AcceptedAnswerId="11093" CreationDate="2010-11-20T15:07:58.707" Score="2" ViewCount="63" Body="&lt;p&gt;Let $f:\mathbb{R}\to \mathbb{C}$ be a Lebesgue measurable function. For each $z\in \mathbb{C}$ let $Arg (z)$ be the principal argument of $z$ (define it to be $0$ if $z=0$). Define $g(x)= \sqrt{|f(x)|}exp(\frac{1}{2} i Arg (f(x))$. Then $g(x)$ is measurable and $(g(x))^2=f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure the above statement is correct, but just wanted to confirm with you all. The only part of concern is, of course, whether the function $exp(\frac{1}{2} i Arg (f(x))$ is measurable. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-20T15:42:43.187" LastActivityDate="2010-11-20T17:40:20.023" Title="Square root of a complex measurable function" Tags="&lt;real-analysis&gt;&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="11089" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T15:12:12.557" Score="2" ViewCount="0" Body="&lt;p&gt;Use the fact that $e^x \ge \left( 1 + \frac{x}{n} \right)^n$ for any $n &amp;gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-20T15:12:12.557" />
  <row Id="11090" PostTypeId="2" ParentId="11085" CreationDate="2010-11-20T15:15:46.940" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\frac{2}{\sqrt 3}$, and since the edges of this tetrahedron have length $\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \sqrt{\frac{2}{3}}$.&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="3747" LastEditorDisplayName="" LastEditDate="2010-11-28T22:38:33.123" LastActivityDate="2010-11-28T22:38:33.123" CommentCount="1" />
  <row Id="11091" PostTypeId="1" CreationDate="2010-11-20T15:16:06.970" Score="5" ViewCount="398" Body="&lt;p&gt;I greatly enjoyed the Lights Out game described &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Lights_Out_%28game%29&amp;amp;oldid=395986075&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (I am sorry I had to link to an older page because some wikidiot keeps deleting most of the page). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its mathematical analysis is &lt;a href=&quot;http://www.jaapsch.net/puzzles/lomath.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (it's just linear algebra)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I just discovered a hexagonal version:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://cwynn.com/games/lightsoutmobile.htm&quot; rel=&quot;nofollow&quot;&gt;http://cwynn.com/games/lightsoutmobile.htm&lt;/a&gt;&lt;br&gt;&#xA;  (and hopefully soon) &lt;a href=&quot;https://sites.google.com/site/beyondlightsout/&quot; rel=&quot;nofollow&quot;&gt;https://sites.google.com/site/beyondlightsout/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there any mathematical results for this version of the game, before I dig in and start the analysis myself. I have the major references including  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Turning Lights Out with Linear Algebra&lt;/em&gt;, by M. Anderson and T. Feil (1998). &lt;em&gt;Math Magazine&lt;/em&gt;, vol. 71, no. 4, October 1998, pp. 300-303.  It's &lt;a href=&quot;http://www.math.ksu.edu/~dmaldona/math551/lights_out.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (thanks to J.M.).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Update: I've been playing with the iPhone T-Lights hexagonal versions. I can get most of them, except for the ones where the &quot;template&quot; is a Y shape. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2010-11-30T02:41:33.813" LastActivityDate="2010-11-30T02:41:33.813" Title="Lights out game on hexagonal grid" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="11092" PostTypeId="2" ParentId="11081" CreationDate="2010-11-20T15:21:37.597" Score="3" ViewCount="0" Body="&lt;p&gt;Repeated use of &lt;a href=&quot;http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule&quot; rel=&quot;nofollow&quot;&gt;L'Hôpital's rule&lt;/a&gt; ($n$ times):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{n}}=\underset{%&#xD;&#xA;x\rightarrow \infty }{\lim }\dfrac{e^{x}}{nx^{n-1}}=\underset{x\rightarrow&#xD;&#xA;\infty }{\lim }\dfrac{e^{x}}{n(n-1)x^{n-2}}=\cdots =$ $=\cdots=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{n(n-1)\cdots&#xD;&#xA;3\cdot x^{2}}=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{%&#xD;&#xA;n(n-1)\cdots 3\cdot 2x}=$ $=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{%&#xD;&#xA;n(n-1)\cdots 3\cdot 2\cdot 1}=\underset{x\rightarrow \infty }{\lim }\dfrac{%&#xD;&#xA;e^{x}}{n!}=\infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convince yourself: If you had $\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{10}}$ you would have to apply L'Hôpital's rule ten times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 2: Plot of $\dfrac{e^{x}}{x^{3}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nGHEu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T16:17:50.117" LastActivityDate="2010-11-20T16:17:50.117" />
  <row Id="11093" PostTypeId="2" ParentId="11088" CreationDate="2010-11-20T15:24:02.757" Score="2" ViewCount="0" Body="&lt;p&gt;Rewriting my answer to improve clarity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp({1 \over 2}i(Arg(f(x)))$ can be written as $h(g(f(x)))$, where $h(z) = e^{{1 \over 2} iz}$ and $g(z) = Arg(z)$. By Rudin thm 1.12d) for example, a measurable function followed by a Borel measurable function is measurable. So it suffices to show $g$ and $h$ are Borel measurable. For $h(z)$ this follows from the fact that $h$ is continuous. For $g(z)$, regardless of how you define $Arg(z)$,&#xA;the set $ \{z: a &amp;lt; Arg(z) &amp;lt; b\}$ is the union of at most two wedges which are Borel sets. So $\{z: g(z) \in (a,b) \}$ is a Borel set. This means $g(z)$ is Borel measurable as well.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-20T17:40:20.023" LastActivityDate="2010-11-20T17:40:20.023" CommentCount="10" />
  <row Id="11094" PostTypeId="1" AcceptedAnswerId="11100" CreationDate="2010-11-20T15:51:13.020" Score="1" ViewCount="45" Body="&lt;p&gt;I have two random variables $X$ and $Y$, both receiving values between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $E[X - Y] \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I get any inequality of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(X - Y \ge \delta) \le F(\delta,X,Y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F(\delta,X,Y)$ is a (reasonable) function of $\delta$ and $E[X-Y]$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markov inequality would be good here, for example, by setting $F(\delta,X,Y) = E[X-Y]/\delta$. However, Markov inequality would require $X-Y \ge 0$, and I do not necessarily have that.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2010-11-20T19:22:33.233" Title="can I get a bound on the probability of deviation, similar to Markov inequality?" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;inequality&gt;" AnswerCount="3" />
  <row Id="11095" PostTypeId="1" AcceptedAnswerId="11101" CreationDate="2010-11-20T16:02:50.690" Score="1" ViewCount="200" Body="&lt;p&gt;I've read about many attempts to find patterns in prime numbers.  First, is there a mathematical way to prove there is not a pattern to prime numbers?  Since there are ways to check if a number is prime or not, can these methods be combined into some large function that produces prime numbers?  And in case it isn't obvious, no, I never got passed pre-Cal.&lt;/p&gt;&#xA;" OwnerUserId="3748" LastActivityDate="2010-12-29T00:46:47.917" Title="Patterns in Prime numbers, and the null hypothesis" Tags="&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="11096" PostTypeId="2" ParentId="11076" CreationDate="2010-11-20T16:12:05.323" Score="8" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As with most anything in physics, the addition of forces follows the parallelogram law exactly because it agrees with every experiment mankind has ever made---in other words, ultimately it's an assumption we make based on what we see. But, it's perfectly intuitive that this is how forces should behave: say you're at the origin on the xy-plane and have a force pushing you in the positive x-direction and one pushing you in the positive y-direction, both of equal magnitude. Then it should physically and intuitively clear that the net result will be a force pushing you in the direction of (1,1). If the force in the x-direction was stronger, then the net result would still push you in some &quot;diagonal&quot; direction, but aligned more with the x-axis. This is exactly the parallelogram law.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Same exact thing. The problem is that we can't really picture what a right triangle in 4 dimensions looks like, but it's exactly the same idea.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Of course, there are many possible &quot;dot products&quot; that can be defined on vectors, and in many applications it is indeed useful to use one different from the standard one. But I would say the importance of the standard dot product indeed comes from its relation with the cosine of the angle between them: again if you try to think about this physically, this is a precise measure of how much the two vectors (or forces if you like) are working &quot;with&quot; or &quot;against&quot; each other. Note in particular that the sign of the dot product solely depends on whether or not the angle between your two vectors is less than, equal to, or greater than 90 degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1914" LastActivityDate="2010-11-20T16:12:05.323" />
  <row Id="11097" PostTypeId="2" ParentId="11095" CreationDate="2010-11-20T16:13:21.963" Score="3" ViewCount="0" Body="&lt;p&gt;It really depends on what you mean by patterns. Legendre showed that there is no rational algebraic function that outputs only primes. You can read more about prime generating functions here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The famous Green-Tao theorem states that there are arbitrarily large arithmetic progressions in primes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-20T16:13:21.963" />
  <row Id="11098" PostTypeId="2" ParentId="11068" CreationDate="2010-11-20T16:23:05.353" Score="0" ViewCount="0" Body="&lt;p&gt;You don't say what data you have.  If you have (x,y) pairs, you can just use a multidimensional fitting routine like &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; or Excel's goal seek function to get all the parameters.  If you have random samples it may be hard to distinguish the effects of m and s unless you have a lot of them.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-20T16:23:05.353" />
  <row Id="11099" PostTypeId="2" ParentId="11012" CreationDate="2010-11-20T16:29:31.427" Score="2" ViewCount="0" Body="&lt;p&gt;Following variation of Robin Chapman idea works in algebraic geometry.&#xA;Take $X$ elliptic curve minus point : it is affine variety. For general point $P \in X$ take $L=\mathcal O(P)$. Then $M=L\oplus L^*=\mathcal O(P) \oplus \mathcal O(-P)$ has properties:&#xA;A) $\Lambda ^2 (M)=\mathcal O$ is free of rank one.&#xA;B) $M$ is not free because it has no sections except zero (since  $\mathcal O(P)$&#xD;&#xA; and $\mathcal O(-P)$ have no sections except zero)&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-11-20T16:29:31.427" />
  <row Id="11100" PostTypeId="2" ParentId="11094" CreationDate="2010-11-20T16:31:43.097" Score="3" ViewCount="0" Body="&lt;p&gt;Since $0 \le X,Y \le 1$, you have $X-Y \ge -1$, so applying the Markov inequality to $X-Y+1$ gives&#xA;$$P(X-Y \ge \delta) = P(X-Y+1 \ge \delta+1) \le (E[X-Y]+1)/(\delta+1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without any restrictions on the range of $X,Y$, no such bound is possible.  For any numbers $\mu, \delta \in \mathbb{R}$, $p \in (0,1)$, there exists a random variable $Z$ (here $Z=X-Y$) with $P(Z \ge \delta) = p$ and $EZ = \mu$.  Thus $p$ cannot be bounded by any function of $\mu$ and $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $Z=a$ with probability $p$ and $Z=b$ with probability $1-p$.  Then we have $\mu = EZ = pa + (1-p)b$, so $b = (\mu-pa)/(1-p)$.  Since $b \to -\infty$ as $a \to \infty$, you can choose $a$ so large that $a &amp;gt; \delta$ and $b &amp;lt; \delta$, and then $P(Z \ge \delta) = p$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-20T16:39:54.057" LastActivityDate="2010-11-20T16:39:54.057" CommentCount="2" />
  <row Id="11101" PostTypeId="2" ParentId="11095" CreationDate="2010-11-20T16:49:59.673" Score="3" ViewCount="0" Body="&lt;p&gt;I like Dirichlet's Theorem, which states that for relatively prime $a,d \in \mathbb{Z}^+$, there are infinitely many primes in the progression $\{a + nd \mid n \in \mathbb{Z^+}\}$.  Further, the proportion of primes in any relatively prime residue class of $d$ is about $1/\phi(d)$, where $\phi$ is the Euler phi function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, the primes can be seen as somewhat evenly distributed.  You can see more below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-20T16:49:59.673" />
  <row Id="11102" PostTypeId="2" ParentId="11085" CreationDate="2010-11-20T16:51:17.207" Score="1" ViewCount="0" Body="&lt;p&gt;You can also use trig based on the dihedral angle between two faces of the tetrahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;|OA|^2 &amp;amp;=&amp;amp; |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\&#xD;&#xA;s^2 &amp;amp;=&amp;amp; \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\&#xD;&#xA;s^2 &amp;amp;=&amp;amp; \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\&#xD;&#xA;1 &amp;amp;=&amp;amp; \frac{3}{2} - \frac{3}{2} \cos{M} \\&#xD;&#xA;\frac{-1}{2} &amp;amp;=&amp;amp; - \frac{3}{2} \cos{M} \\&#xD;&#xA;\frac{1}{3} &amp;amp;=&amp;amp; \cos{M} \;\;\; (**)\\&#xD;&#xA;\Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&amp;amp;=&amp;amp; \sin{M}&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\cos\left({\text{angle between two sides of a regular triangle}}\right) &amp;amp;=&amp;amp; \frac{1}{2}\\&#xD;&#xA;\cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &amp;amp;=&amp;amp; \frac{1}{3}\\&#xD;&#xA;\cos\left({\text{angle between two facets of a regular n-simplex}}\right) &amp;amp;=&amp;amp; \frac{1}{n}&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Who would've suspected, upon first encountering it, that the &quot;$2$&quot; in &quot;$\cos{60^{\circ}}=\frac{1}{2}$&quot; was actually a reference to the dimension of the triangle?)&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-20T16:51:17.207" CommentCount="3" />
  <row Id="11103" PostTypeId="2" ParentId="11056" CreationDate="2010-11-20T17:00:55.627" Score="3" ViewCount="0" Body="&lt;p&gt;You will find of interest Fund and Williams, &quot;Quadratic polynomials which have a high density of prime values&quot;, Mathematics of Computation, 55:191, July 1990, 345-353.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2008810&quot; rel=&quot;nofollow&quot;&gt;Jstor.org link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-20T17:00:55.627" CommentCount="1" />
  <row Id="11104" PostTypeId="1" AcceptedAnswerId="11105" CreationDate="2010-11-20T17:04:40.953" Score="5" ViewCount="153" Body="&lt;p&gt;Hi,&#xA;Is $L^2(\mathbb{R})$ a Banach algebra, with convolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure the answer is no, because I think that&#xA;$f,g \in L^2(\mathbb{R})$ does not imply that $f*g \in L^2(\mathbb{R})$. However, I can't seem to find a counterexample, or a proof that this is not the case, using Fourier Transform for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give a hint?&#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3752" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-20T17:28:43.983" LastActivityDate="2010-11-21T19:46:11.597" Title="Is $L^2(\mathbb{R})$ with convolution a Banach Algebra?" Tags="&lt;functional-analysis&gt;&lt;fourier-analysis&gt;&lt;convolution&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="11105" PostTypeId="2" ParentId="11104" CreationDate="2010-11-20T17:19:33.153" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;.  The Fourier transform preserves norms on $L^2$, i.e. $\|f\|_2 = \|\hat f\|_2$.  Also $\hat{(f * g)} = \hat f  \hat g$.  So it is enough that you find a function which is in $L^2$ but whose square is not in $L^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit after the comment.&lt;/strong&gt; The proof of $\hat{(f * g)} = \hat f \hat g$ with $f,\ g \in L^2$ requires extending the Fourier transform to the space of tempered distributions.  But you can actually prove what you want with its $L^1$ version and a contradiction.  Assume that convolution is a Banach algebra, and take $f,\ g  \in L^1 \cap L^2$.  Then by assumption $\|f*g\|_2\leq \|f\|_2\|g\|_2$.  From this it follows by isometry that $\|\hat f \hat g\|_2 \leq \|\hat f\|_2\|\hat g\|_2$.  &lt;strike&gt;But since $L^1\cap L^2$ is dense in $L^2$, this must be true for all $f,\ g \in L^2$, for which you can find counter-examples.&lt;/strike&gt; In particular, taking $g = f$, one gets $\|\hat f ^2\|_2 \leq \|\hat f\|_2^2$ i.e.,  $\|\hat f \|_4 \leq \|\hat f\|_2$.  But a counter example can be produced for this by taking any function $f \in L^1\cap L^2$ and forming a sequence $f_n$ by scaling and normalizing it accordingly so that $ \|\hat f_n\|_2$ stays constant but  $\|\hat f_n\|_4$ blows up. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-21T19:46:11.597" LastActivityDate="2010-11-21T19:46:11.597" CommentCount="4" />
  <row Id="11106" PostTypeId="2" ParentId="11028" CreationDate="2010-11-20T17:20:07.770" Score="0" ViewCount="0" Body="&lt;p&gt;I'm not really convinced by Willie Wong's argument, although I don't doubt that it is correct, but I cannot completely fill in the details (especially the Egoroff part) so I made my own (inspired by his).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that the hypothesis that $f \in L^p$ is redundant by Fatou's lemma. Now note that we could as well consider $f_n \to 0$ weakly (the conclusion actually means weak-convergence) by linearity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that:&#xA;\begin{align*}&#xA;F_n &amp;amp;: L^q \to \mathbf{K}\\&#xA;g &amp;amp; \mapsto \int f_n g.&#xA;\end{align*}&#xA;defines a bounded operator linear operator on $L^q$ by Hölder with operatornorm smaller or equal to $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the stepfunctions are dense in $L^q$. So by linearity and continuity it is sufficient to consider indicatorfunctions $1_A$ with $A$ having finite measure. Now we have a finite measure space so we can use Egoroff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_A f_n \to 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now find a $B$ with $|B| \leq \epsilon$ such that $f_n$ converges uniformly to $0$ on $A \setminus B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for sufficiently large $n$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{A \setminus B} |f_n| \leq \epsilon |A \setminus B| \leq \epsilon |A|.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{B} |f_n| \leq |B|^{1/q} \|f_n\|_p \leq \epsilon^{1/q} \|f_n\|_p.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;KTHXBYE.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-20T17:20:07.770" />
  <row Id="11107" PostTypeId="2" ParentId="11104" CreationDate="2010-11-20T17:22:47.470" Score="4" ViewCount="0" Body="&lt;p&gt;You can take any $f,g\in L^2(\mathbb R)$ such that their pointwise product $fg$ does not belong to $L^2(\mathbb R)$ and use the fact that the Fourier transform of a product is a convolution (and that the Fourier transform is an isometry of $L^2(\mathbb R)$ onto itself). &lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-11-20T17:22:47.470" />
  <row Id="11108" PostTypeId="1" CreationDate="2010-11-20T17:25:56.213" Score="1" ViewCount="126" Body="&lt;p&gt;$g(x) =\frac{3-5x}{5-x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I am retarded for not being able to do this on my own. Algebraically you just have to solve for x. Even so, I can't do it. &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T21:45:19.563" LastActivityDate="2010-11-20T21:45:19.563" Title="Find the inverse of the following function" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11109" PostTypeId="2" ParentId="11108" CreationDate="2010-11-20T17:28:03.640" Score="5" ViewCount="0" Body="&lt;p&gt;Hint: $y=\frac{3-5x}{5-x}$ so $y(5-x)=3-5x$. Now collect terms involving $x$ on one side and those not involving $x$ on the other and solve for $x$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-20T17:28:03.640" CommentCount="3" />
  <row Id="11110" PostTypeId="2" ParentId="2949" CreationDate="2010-11-20T17:35:35.880" Score="4" ViewCount="0" Body="&lt;p&gt;In class of number theory, identities of Ramanujan(continued fractions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha, \beta &amp;gt;0$ with $\alpha\beta=\frac{1}{5}$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left\{ \left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}+ \displaystyle\frac{e^{-\frac{2\pi\alpha}{5}}}{1+\displaystyle\frac{e^{-2\pi\alpha}}{1+\displaystyle\frac{e^{-4\pi\alpha}}{1+...}}}\right\}\cdot\left\{ \left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}+ \displaystyle\frac{e^{-\frac{2\pi\beta}{5}}}{1+\displaystyle\frac{e^{-2\pi\beta}}{1+\displaystyle\frac{e^{-4\pi\beta}}{1+...}}}\right\} = 5\sqrt{5}\left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beautiful result!!!&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-20T17:35:35.880" CommunityOwnedDate="2010-11-20T17:35:35.880" CommentCount="1" />
  <row Id="11111" PostTypeId="2" ParentId="11108" CreationDate="2010-11-20T17:44:46.147" Score="4" ViewCount="0" Body="&lt;p&gt;Maybe you could start with a less complicated example and build up from there. Why don't you try finding the inverse function of $f(x) = x + 3$ using the idea that Timothy gave you of putting $y = x + 3$ and then solving for $x$. Then you can try $f(x) = 7x - 2$ for instance. Now you could also try $f(x) = \frac{3}{x}$ for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something that is sometimes very useful is to consider &quot;easier&quot; problems related to your original problem and trying to solve those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the point is to gain some experience on what the particular difficulty may be, and that the &quot;harder&quot; problem may combine some of the difficulties that you will have treated in the particular, less complicated examples that you considered. This advice is treated by Polya in his excellent book about problem solving.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-20T17:44:46.147" CommentCount="1" />
  <row Id="11112" PostTypeId="2" ParentId="11066" CreationDate="2010-11-20T17:48:37.673" Score="2" ViewCount="0" Body="&lt;p&gt;For the 4x4 case there are only two drawn configurations-a checkerboard.  As there are 12870 evenly filled boards, there are 12868 winning ones.  This is not the 144 you calculated.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-20T17:48:37.673" />
  <row Id="11113" PostTypeId="1" AcceptedAnswerId="11160" CreationDate="2010-11-20T17:55:08.553" Score="5" ViewCount="162" Body="&lt;p&gt;Suppose $R=k[[x_1,...,x_n]]$ is a formal power series ring over the field $k$, what can we say about the structure of $R/p$ if $p$ is a prime ideal of $R$ such that dim$(R/p)=1$. In particular, are these subrings of power series rings in one variable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: Given a subring of a power series ring in one variable which can be written as, $k[[x^{a_1},...,x^{a_n}]]$ where $a_1,...,a_n$ are distinct positive integers, we have a homomorphism $k[[x_1,...,x_n]]\to k[[x^{a_1},...,x^{a_n}]]$ which sends $x_i\to x^{a_i}$. The kernel of this homomorphism must be a prime ideal since $k[[x^{a_1},...,x^{a_n}]]$ is a domain. Moreover, the quotient modulo the kernel must be one dimensional since the image ring is one dimensional. I was wondering if we have a (partial) converse to this result. (I do realize the second paragraph would analogously work with polynomial rings, but it seems power series rings and much nicer (they are regular and local), so I was hoping for a nicer description of the quotient in the power series case).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-20T22:59:08.533" LastActivityDate="2010-11-22T03:31:02.417" Title="One dimensional quotients of formal power series rings" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="11114" PostTypeId="2" ParentId="1002" CreationDate="2010-11-20T18:11:51.367" Score="1" ViewCount="0" Body="&lt;p&gt;It took me quite a while to understand what exactly is meant by &lt;em&gt;Fourier transform&lt;/em&gt; since it can refer to various algorithms, operations and results. Though I'm quite new in this topic, I'll try to give a short but hopefully intuitive overview on what I came up with (feel free to correct me):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a function $f(t)$ that maps some time value $t$ to some value $f(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we'll try to approximate $f$ as the sum of simple harmonic oscillations, i.e. sine waves of certain frequencies $\omega$. Of course, there are some frequencies that fit well to $f$ and some that approximate it less well. Thus we need some value $\hat{f}(\omega)$ that tells us &lt;em&gt;how much&lt;/em&gt; of a given oscillation with frequency $\omega$ is present in the approximation of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example the red function from &lt;a href=&quot;http://de.wikipedia.org/w/index.php?title=Datei%3aSin%28x%29%2B0,13sin%283x%29.svg&amp;amp;filetimestamp=20100814211526&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/d9hST.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t) = \sin(t)+0.13\sin(3t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green oscillation with $\omega=1$ has the biggest impact on the result, so let's say $$\hat{f}(1)=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue sine wave ($\omega=3$) has at least some impact, but it's amplitude is much smaller. Thus we say $$\hat{f}(3)=0.13$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other frequencies may not be present in the approximation at all, thus we would write $$\hat{f}(\omega)  = 0$$ for these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we knew $\hat{f}(\omega)$ not only for some but all possible frequencies $\omega$, we could perfectly approximate our function $f$. And that's what the &lt;em&gt;continuous Fourier transform&lt;/em&gt; does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes some function $f(t)$ of time and returns some other function $\hat{f}(\omega) = \mathcal{F}(f)$, it's &lt;em&gt;Fourier transform&lt;/em&gt;, that describes how much of any given &lt;em&gt;frequency&lt;/em&gt; is present in $f$. It's just another representation of $f$, of equal information but with a completely different domain. Often though, problems can be solved much easier in this other representation (which is like finding the appropriate coordinate system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But given a Fourier transform, we can integrate over all frequencies, put together the weighted sine waves and get our $f$ again, which we call inverse Fourier transform $\mathcal{F}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now why should one want to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly, the Fourier transform has many nice mathematical properties (i.e. convolution is just multiplication). It's often much easier to work with the Fourier transforms than with the function itself. So we transform, have an easy job with filtering, transforming and manipulating sine waves and transform back after all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we want to do some noise reduction on a digital image. Rather than manipulating a function $\text{image} : \text{Pixel} \to \text{Brightness}$, we transform the whole thing and work with $\mathcal{F}(\text{image}) : \text{Frequency} \to \text{Amplitude}$. &#xA;Those party of high frequency that cause the noise can simply be cut off - $\mathcal{F}(\text{image})(\omega) = 0, \omega &amp;gt; ...Hz$. We transform back et voilà.&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-20T18:11:51.367" />
  <row Id="11115" PostTypeId="2" ParentId="11073" CreationDate="2010-11-20T18:14:46.610" Score="1" ViewCount="0" Body="&lt;p&gt;I would say that w.l.o.g. $\theta_1 \leq \theta_3$ and break it up into the cases of $\theta_1 \leq \theta_2 \leq \theta_3$, $\theta_2 \leq \theta_1 \leq \theta_3$, and $\theta_1 \leq \theta_3 \leq \theta_2$. That allows you to use more than just the triangle inequality on the real line.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-20T18:14:46.610" CommentCount="1" />
  <row Id="11116" PostTypeId="1" CreationDate="2010-11-20T18:17:56.747" Score="0" ViewCount="80" Body="&lt;p&gt;Hello,&lt;br&gt;&#xA;I have a probability space&#xA;$\omega = 2^{\{1,\ldots,n\}}$&#xD;&#xA;$\sigma$-algebra $2^\omega$&#xD;&#xA;and $P(\{s\})=(p^{|s|})*(1-p)^{(n-|s|)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that $n=2k$,$k$ natural number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find a random variable that will distribute like $\mathrm{Bin}(k,p^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me with this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;br&gt;&#xA;benny&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-21T05:10:38.163" LastActivityDate="2010-12-21T05:10:38.163" Title="random variable with binomial distribution" Tags="&lt;homework&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11117" PostTypeId="2" ParentId="8113" CreationDate="2010-11-20T18:20:02.643" Score="0" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;crasic has taken the words out of my mouth.  It's the same as the number of proper subsets of {2, 3, 4, 5, 6}, i.e. 2^5 - 1 = 31.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Only 63% like cheese, so there cannot be more than 63% who like both cheese and apples.  So B is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;37% don't like cheese, and 27% don't like apples (which I personally find hard to imagine).  The lower bound for x is found in the case where these two sets are disjoint - 64% either don't like cheese or don't like apples, leaving 36%.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3754" LastActivityDate="2010-11-20T18:20:02.643" />
  <row Id="11118" PostTypeId="1" AcceptedAnswerId="11143" CreationDate="2010-11-20T18:23:20.110" Score="1" ViewCount="118" Body="&lt;p&gt;(a)Is there a linear mapping $f : \mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with&#xA;$ f(1,1,1)=(3,2,7), f(0,2,1)=(2,1,-1), f(2,0,1)=(1,0,0)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)Is there a linear mapping $f :\mathbb{R}^{3} \rightarrow \mathbb{R}^3$ with &#xA;$ f(0,2,-1)=(3,-1,3), f(1,0,2)=(0,-2,1), f(-2,2,-5)=(3,3,1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substantiate your statements and if it exists, provide such a linear mapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) is it first important to determine whether $(1,1,1), (0,2,-1), (2,0,1)$ (called vectors, right??) are a basis for $\mathbb{R}^{3}$? in that case my understanding is a basis is linear independent vectors which span a vectorspace. Since $2(1,1,1)-(0,2,1)=(2,0,1)$ i would like to conclude that these vectors are linearly dependent $\Rightarrow$ do not form a basis for $\mathbb{R}^{3} \Rightarrow$ there is no linear map for these vectors. In case my reasoning is correct, I just wanted to double check is that i could have reached the linear dependence conclusion by turning the vectors into a matrix and seeing if that yielded a row of zeros or a free variable, right? (i happened to just look at it and recognize that this time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)actually after writing my (a) attempt i realize that (b) would also be linearly dependent, correct? i still haven't learned to create a matrix with latex yet, but in case someone is skeptical that it is linearly dependent i could show you my steps. however, first i'd prefer to know if i am even on the right track ;)&#xA;as a note i'm always happy to hear if i am using terminology improperly. thanks in advance for all your help!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2010-11-21T19:41:15.753" Title="Are there linear mappings for the following vectors, if so, what are they?" Tags="&lt;linear-algebra&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="11119" PostTypeId="2" ParentId="11095" CreationDate="2010-11-20T18:29:33.043" Score="4" ViewCount="0" Body="&lt;p&gt;I especially like this characterization of primes made by Don Zagier&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There are two facts about the&#xA;  distribution of prime numbers which I&#xA;  hope to convince you so overwhelmingly&#xA;  that they will be permanently engraved&#xA;  in your hearts.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The first is that despite their simple&#xA;  definition and role as the building&#xA;  blocks of the natural numbers, the&#xA;  prime numbers... grow like weeds among&#xA;  the natural numbers, seeming to obey&#xA;  no other law than that of chance, and&#xA;  nobody can predict where the next one&#xA;  will sprout.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The second fact is even more&#xA;  astonishing, for it states just the&#xA;  opposite: that the prime numbers&#xA;  exhibit stunning regularity, that&#xA;  there are laws governing their&#xA;  behaviour, and that they obey these&#xA;  laws with almost military precision.&quot;&#xA;  Don Zagier, Bonn University inaugural lecture&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is taken from &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/mrwatkin/isoc/quotes.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; site where you can find many more quotations from prominent mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you can find some formulas for primes.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-11-20T19:45:42.617" LastActivityDate="2010-11-20T19:45:42.617" />
  <row Id="11120" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:33:43.137" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: Suppose we have a linear map $f:\mathbb{R^3}\to \mathbb{R^3}$. Say, $f_1(x,y,z)=a_1x+b_1y+c_1z$, $f_2(x,y,z)=a_2x+b_2y+c_2z$, $f_3(x,y,z)=a_3x+b_3y+c_3z$ where $f_1,f_2,f_3$ are the $x,y,z$ components of $f$. Substituting the values $(1,1,1),(0,2,1),(2,0,1)$ in this map, we get the corresponding values of the output if such a function exists. Now this gives us $9$ equations in $9$ unknowns, $a_i,b_i,c_i$ with $i=1,2,3$. If these equations form a determined system, we have a unique solution and hence a map. If the system is overdetermined, i.e. one of the equations is a linear combination of the others, then we have several such maps. If the system is inconsistent, i.e. the equations cannot be satisfied simulatenously, then we have no solution and no such map. So it just boils down to solving these equations simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Adrian makes a useful remark, see his comment below.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-20T19:37:14.197" LastActivityDate="2010-11-20T19:37:14.197" CommentCount="5" />
  <row Id="11121" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:33:55.457" Score="0" ViewCount="0" Body="&lt;p&gt;A brute-force approach: try to write using different notation: e.g., say you are searching for a matrix $A$ that satisfies $AX=Y$, where the columns of $X$ are given by the different vectors that are arguments to your $f$, and the columns of $Y$ are the different vectors that you obtain on applying $f$. Then, see if you can &quot;solve&quot; to get $A=XY^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-11-20T18:33:55.457" />
  <row Id="11122" PostTypeId="1" CreationDate="2010-11-20T18:35:17.733" Score="1" ViewCount="106" Body="&lt;p&gt;Let $a$ and $b$ be real numbers and $p&amp;gt;0$. What is the best way to prove that $|a+b|^p&amp;lt;2^p(|a|^p+|b|^p)$? &lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2010-11-20T22:19:11.407" Title="Yet another inequality" Tags="&lt;real-analysis&gt;&lt;inequality&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="11123" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T18:40:58.060" Score="3" ViewCount="0" Body="&lt;p&gt;If $0&amp;lt;p\leq1$ you don't need any powers of $2$.  This came up in another &lt;a href=&quot;http://math.stackexchange.com/questions/10993/please-explain-inequality-xp-yp-leq-x-yp&quot;&gt;recent question&lt;/a&gt;.  If $1\leq p$ you can strengthen the inequality to $|a+b|^p\leq 2^{p-1}(|a|^p+|b|^p)$ by applying &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's inequality&lt;/a&gt; to the convex function $t\mapsto t^p$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-20T18:40:58.060" />
  <row Id="11124" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T18:41:24.887" Score="3" ViewCount="0" Body="&lt;p&gt;If you divide $2^p$ on both sides, then you get $(|a+b|/2)^p &amp;lt; |a|^p + |b|^p$. Can you see why that would be true? &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-11-20T18:41:24.887" CommentCount="1" />
  <row Id="11125" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:42:02.130" Score="1" ViewCount="0" Body="&lt;p&gt;If you define $f(x,y,z)=A\cdot \left(&#xD;&#xA;                        \begin{array}{}&#xD;&#xA;                         x \\&#xD;&#xA;                         y \\&#xD;&#xA;                         z \\&#xD;&#xA;                        \end{array}&#xD;&#xA;                      \right)$, where the matrix $A$ is given for  the 9 ecuations as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A\cdot \left(&#xD;&#xA;                        \begin{array}{}&#xD;&#xA;                         1 \\&#xD;&#xA;                         1 \\&#xD;&#xA;                         1 \\&#xD;&#xA;                        \end{array}&#xD;&#xA;                      \right) = \left(&#xD;&#xA;                        \begin{array}{}&#xD;&#xA;                         3 \\&#xD;&#xA;                         2 \\&#xD;&#xA;                         7 \\&#xD;&#xA;                        \end{array}&#xD;&#xA;                      \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find all coefficients of the matrix $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each relation as the previous we find 3 ecuation, in total 9 ecuations with 9 unknowns, then you problem is solves .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember this depends of that ecuations system can be solved !!!!&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-20T18:57:50.550" LastActivityDate="2010-11-20T18:57:50.550" CommentCount="1" />
  <row Id="11126" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:43:32.097" Score="0" ViewCount="0" Body="&lt;p&gt;For (a), your reasoning is incomplete. For example consider the problem of finding $f$ when&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(1,0,0) = (0,0,4)$, $f(2,0,0) = (0,0,8)$ and $f(3,0,0) = (0,0,12)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is clearly a linear function $f$ that satisfies this (infinitely many, in fact), even though the vectors $(1,0,0), (2,0,0), (3,0,0)$ are linearly dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you show that the &quot;output&quot; vectors are linearly independent, however, your justification would be good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timothy Wagner's answer is slightly incomplete: you have to solve the 3 equations in 3 unknowns problem 3 times: once for each output dimension.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastEditorUserId="3638" LastEditorDisplayName="" LastEditDate="2010-11-20T18:51:58.403" LastActivityDate="2010-11-20T18:51:58.403" />
  <row Id="11127" PostTypeId="2" ParentId="11116" CreationDate="2010-11-20T18:45:42.003" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: $2 = n/k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another hint: $p^2$ has the same exponent $2$ as $n/k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-20T19:29:47.240" LastActivityDate="2010-11-20T19:29:47.240" CommentCount="2" />
  <row Id="11128" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T18:51:38.593" Score="1" ViewCount="0" Body="&lt;p&gt;Given the linear dependence you observed in the input vectors in (a), you could check if the same dependence applies to the output vectors.  As a linear map requires that f(ax+b)=af(x)+f(b), if the output vectors do not satisfy the same dependence there is no map.  This would save you trying to solve the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you prove the input vectors independent, there is guaranteed to be a linear map.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T20:00:58.130" LastActivityDate="2010-11-20T20:00:58.130" CommentCount="0" />
  <row Id="11129" PostTypeId="1" CreationDate="2010-11-20T18:57:33.530" Score="1" ViewCount="113" Body="&lt;p&gt;The task: &#xA;$\lim_{x\to\infty} \sqrt{x^2+1} -x $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've multiplied with the conjugate expression ($\sqrt{x^2+1} +x$), then I get this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{x\to\infty} \frac{1}{\sqrt{x^2+1} +x} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct so far? And what would be the next step? &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="2230" LastEditorDisplayName="" LastEditDate="2010-11-20T19:11:25.593" LastActivityDate="2010-11-20T23:38:57.483" Title="Another limit task, I've multiplied by the conjugate, now what?" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="11130" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T18:59:31.047" Score="2" ViewCount="0" Body="&lt;p&gt;The multiplication of conjugates is not correct: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)=(x^2+1)-x^2=1$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-20T19:02:16.910" LastActivityDate="2010-11-20T19:02:16.910" CommentCount="2" />
  <row Id="11131" PostTypeId="2" ParentId="2949" CreationDate="2010-11-20T19:03:25.863" Score="1" ViewCount="0" Body="&lt;p&gt;How about the Anti-Calculus Proposition (Erdös): Suppose f is analytic throughout a closed disc and assumes its maximum modulus at the boundary point a. Then f´(a) is not equal to 0 unless f is constant. (Source Bak/Newman: Complex Analysis 2nd ed.)&lt;/p&gt;&#xA;" OwnerUserId="3757" LastActivityDate="2010-11-20T19:03:25.863" CommunityOwnedDate="2010-11-20T19:03:25.863" />
  <row Id="11132" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:03:26.753" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; What you have to do is this : $$ \sqrt{x^{2}+1} -x \times \frac{\sqrt{x^{2}+1} + x}{\sqrt{x^{2}+1} + x} = \frac{1}{\sqrt{x^{2}+1} +x}$$ then proceed to evaluate your limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now observe that $\lim_{x \to \infty} \frac{1}{x} = 0$. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-20T19:14:22.453" LastActivityDate="2010-11-20T19:14:22.453" />
  <row Id="11133" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:07:23.080" Score="1" ViewCount="0" Body="&lt;p&gt;$$\underset{x\rightarrow \infty }{\lim }\sqrt{x^{2}+1}-x=\underset{%&#xD;&#xA;x\rightarrow \infty }{\lim }\frac{\left( \sqrt{x^{2}+1}-x\right) \left( &#xD;&#xA;\sqrt{x^{2}+1}+x\right) }{\sqrt{x^{2}+1}+x}=\underset{x\rightarrow \infty }{%&#xD;&#xA;\lim }\frac{1}{\sqrt{x^{2}+1}+x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}=\infty$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_of_a_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia, limits involving infinity and properties&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{x\rightarrow \infty }{\lim }\frac{1}{\sqrt{x^{2}+1}+x}=\dfrac{\underset{x\rightarrow \infty }{\lim }1}{\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}}=\dfrac{1}{\infty}=0$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-20T21:01:06.323" LastActivityDate="2010-11-20T21:01:06.323" CommentCount="5" />
  <row Id="11135" PostTypeId="2" ParentId="11091" CreationDate="2010-11-20T19:15:09.670" Score="4" ViewCount="0" Body="&lt;p&gt;Yes there are linear algebra and graph theoretic results for this game, for any arrangement of switches and starting from all off position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that you can construct a simple graph out of the arrangement of lights, by taking the lights to be vertices, and making them adjacent in the graph if they are neighbours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Graph theoretic:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can found here: Problem 17 on &lt;a href=&quot;http://books.google.com/books?id=e99fXXYx9zcC&amp;amp;pg=PA42&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=e99fXXYx9zcC&amp;amp;pg=PA42&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition A:&lt;/strong&gt; Given a simple undirected graph $G(V,E)$, with vertex set $V$ and edge set $E$, we can partition $V$ into two sets $V_1$ and $V_2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall y \in V_1, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_1\}| = 0 \mod 2$ &lt;br/&gt;&#xA;$\forall y \in V_2, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_1\}| = 1 \mod 2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. any vertex in $V_1$ is incident to an &lt;em&gt;even&lt;/em&gt; number of vertices in $V_1$ and any vertex not in $V_1$ (i.e. in $V_2$) is incident to an &lt;em&gt;odd&lt;/em&gt; number of vertices in $V_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\circ$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So selecting the vertices in $V_1$ will turn on all the lights, if we start from the all off position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We tackle a closely related problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first show that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition B&lt;/strong&gt;: Given a simple undirected graph $G(V,E)$, there is a partition of the vertex set of $G$, $V = V_1 \cup V_2$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall y \in V_1, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_1\}| = 0 \mod 2$ &lt;br/&gt;&#xA;$\forall y \in V_2, \ \  |\{ x: \{x,y\} \in E \wedge x \in V_2\}| = 0 \mod 2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. Any vertex in $V_1$ is adjacent to an &lt;em&gt;even&lt;/em&gt; number of vertices in $V_1$ and any vertex in $V_2$ is adjacent to an &lt;em&gt;even&lt;/em&gt; number of vertices in $V_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\circ$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of Proposition B is by induction on the number of vertices in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base cases are easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $G$ has $n+1$ vertices. If all the vertices of $G$ have even degree, we are done, by taking $V_2 = \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $o$ is a vertex with odd degree and let $o_1, o_2, \cdots, o_k$ be the neighbours of $o$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Form a graph $G&amp;#39;$ as follows: If $o_i$ and $o_j$ are adjacent in $G$, they are not adjacent in $G&amp;#39;$ and vice versa. Also delete $o$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G&amp;#39;$ is a graph with $n$ vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply the induction hypothesis to $G&amp;#39;$. Say we get a partition $V&amp;#39; = X \cup Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $o$ must be incident to an even number of vertices in one of $X$ or $Y$. Say $X$. Add $o$ to $X$ to get a partition for $G$. We can easily show that this partition of $G$ is what we need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\square$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to apply this to our problem (Proposition A):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a new vertex $N$ to G and make it adjacent to every vertex of even degree in $G$ to get a new graph $G&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Proposition B to $G&amp;#39;$. Suppose the partition satisfying proposition B is $V = X \cup Y$. We can easily show that this is also the partition in $G$ (ignoring $N$) for proposition A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This not only proves the existence of a solution, but the induction proof actually gives you an O(n^2) time &lt;strong&gt;algorithm&lt;/strong&gt; to &lt;em&gt;find&lt;/em&gt; the solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear Algebra Proof due to Noga Alon:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over $\mathbb{F}_2$, let $A$ by any $n\times n$ symmetric matrix. We can show that if $d$ is the diagonal vector of $A$, then $d$ is in the space spanned by the rows of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the identity (valid over $\mathbb{F}_2$) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v^{T}Av = d^{T}v$, where $u^{T}$ is the transpose of $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $v$ is in the null space of $A$, then $d$ is orthogonal to $v$ and as a consequence, $d$ is in the row space of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider the matrix A to be $B+I$ where $B$ is the adjacency matrix of the corresponding graph we get, from the arrangement of the lights.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-29T03:41:55.760" LastActivityDate="2010-11-29T03:41:55.760" CommentCount="14" />
  <row Id="11136" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:15:11.783" Score="3" ViewCount="0" Body="&lt;p&gt;Remember that $a^{2}-b^{2}=(a-b)(a+b)$, put $a=\sqrt{x^{2}+1}$ and $b=x$ get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\sqrt{x^{2}+1}+x)(\sqrt{x^{2}+1}-x)=(\sqrt{x^{2}+1})^{2}-(x)^{2} = x^{2}+1-x^{2}= 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end...&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-20T22:01:56.897" LastActivityDate="2010-11-20T22:01:56.897" />
  <row Id="11137" PostTypeId="2" ParentId="11094" CreationDate="2010-11-20T19:19:14.873" Score="2" ViewCount="0" Body="&lt;p&gt;Nate's bound is optimal. Consider $X$ to be a fair coin toss and $Y = 1-X$. Then $EX=EY$ and $P(X-Y \geq 1) = 1/2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-20T19:19:14.873" />
  <row Id="11138" PostTypeId="2" ParentId="11094" CreationDate="2010-11-20T19:22:33.233" Score="1" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like the bound to go to 0 as E[X-Y] goes to 0 as well, and it won't happen with this bound. I should have specified it explicitly when I wrote &quot;reasonable&quot;, perhaps it was not clear. Any ideas about that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think such a bound is possible. Consider $X,Y$ i.i.d uniform. $E[X-Y] = 0$ yet $P(X-Y \geq \delta) &amp;gt; 0$ for all $\delta &amp;lt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-20T19:22:33.233" CommentCount="0" />
  <row Id="11139" PostTypeId="2" ParentId="11091" CreationDate="2010-11-20T19:28:44.260" Score="2" ViewCount="0" Body="&lt;p&gt;Here is another version of this game. You have a symmetric matrix $A$ over $GF(2)$ (i.e. $0,1$ with all computations done modulo $2$). Then you can show that the diagonal of $A$ is always in the range of $A$, and this is best possible in the sense that the range of $A$ could consist of only the zero vector and the diagonal of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was actually given this &quot;game&quot; as a riddle, so I wouldn't ruin it for you with an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;a href=&quot;http://www.cs.toronto.edu/~yuvalf/Range.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s my answer, which is probably very similar to Moron's proof. Of course Noga's proof is much neater, but it's not algorithmic (in any other respect it's much better).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-20T22:43:17.477" LastActivityDate="2010-11-20T22:43:17.477" CommentCount="0" />
  <row Id="11140" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T19:51:39.353" Score="2" ViewCount="0" Body="&lt;p&gt;We have $$\sqrt{1 + \epsilon} = 1 + \frac{\epsilon}{2} + O(\epsilon^2)$$ using the Taylor expansion. So $$\sqrt{x^2+1} = x\sqrt{1+x^{-2}} = x\left(1 + \frac{1}{2x^2} + O(x^{-4})\right) = x + \frac{1}{2x} + O(x^{-3}).$$ Hence $$\sqrt{x^2+1} - x = \frac{1}{2x} + O(x^{-3}) \longrightarrow 0.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-20T19:51:39.353" />
  <row Id="11141" PostTypeId="1" AcceptedAnswerId="11165" CreationDate="2010-11-20T20:14:14.713" Score="2" ViewCount="131" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question about solving a system of geometric equations. I really hope someone here can help me, it's been several months since I try to solve the problem but without success. As I am not an expert in maths, I count on your help. Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;br&gt;&#xA;I have 3 cercles C1, C2, C3. For example (&lt;a href=&quot;http://www.scribd.com/doc/43416793?secret_password=tz701s0bwws9k9jqzl4&quot; rel=&quot;nofollow&quot;&gt;refer to this figure&lt;/a&gt;)&lt;br&gt;&#xA;C1: $(x+1)^2+(y−4)^2=9$ &lt;br&gt;&#xA;C2: $(x+4)^2+y^2=25$ &lt;br&gt;&#xA;C3: $(x−2)^2+y^2=16$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N2 and N3 are two points:&lt;br&gt;&#xA;N2 is located at the intersection of C1 and C2, and is outside C3.&lt;br&gt;&#xA;N3 is located at the intersection of C1 and C3, and is outside C2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1, T2 and T3 are three points (defining the blue triangle in the figure):&lt;br&gt;&#xA;T1 is located at C1 (and inside C2 and C3).&lt;br&gt;&#xA;T2 is located at the intersection of C2 and the line that passes through T1 and N2.&lt;br&gt;&#xA;T3 is located at the intersection of C3 and the line that passes through T1 and N3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;Given &lt;strong&gt;any&lt;/strong&gt; three circles C1, C2, and C3, is there a point T1 (defined as above) such that distance(T1,T2)=α∗distance(T1,T3) (α is some constant, for example equal to 1) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried Maple, it gives me the solution, but not the way how to calculate it. Any idea ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: What I need is a proof of calculability.&#xA;For example, in a system of linear equation with $x$ variables, we know that $x-1$ equations are needed, otherwise there is an infinity of possible solutions. For my system, I know how to prove that if a solution exists, it is necessarily unique. But I don't know an algorithm that can calculate it because the equations are not linear.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3749" LastEditorUserId="3749" LastEditorDisplayName="" LastEditDate="2010-11-20T22:00:10.050" LastActivityDate="2010-11-21T01:43:17.597" Title="A system of geometric equations" Tags="&lt;geometry&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="11142" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T21:29:45.400" Score="3" ViewCount="0" Body="&lt;p&gt;Well,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;|a + b|^p \leq (|a| + |b|)^p &amp;amp;\leq 2^p \text{max}\{|a|^p, |b|^p\}\\&#xA;&amp;amp;\leq 2^{p - 1} (|a|^p + |b|^p - |a^p - b^p|)\\&#xA;&amp;amp;\leq 2^{p - 1} (|a|^p + |b|^p)&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-20T21:29:45.400" CommentCount="1" />
  <row Id="11143" PostTypeId="2" ParentId="11118" CreationDate="2010-11-20T21:46:58.903" Score="3" ViewCount="0" Body="&lt;p&gt;(i) You are sort of on the right track, but you are trying to (1) prove too much; and (2) drawing the wrong conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a bunch of linearly independent vectors, then you can &lt;em&gt;always&lt;/em&gt; find (at least one) linear transformation that maps them to whatever you want. This because we can always do it with a basis (decide where you want the basis vectors to go, and that gives you a linear transformation), and we can always take any collection of linearly independent vectors and complete it to a basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; your three vectors were linearly independent (hence a basis for $\mathbb{R}^3$, then you would know that there &lt;em&gt;is&lt;/em&gt; a linear transformation with the desired values, because you can always find a linear transformation that does whatever you want to a basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the three vectors are &lt;em&gt;not&lt;/em&gt; linearly independent, this does not preclude the possibility that there is a linear transformation with the desired properties! For instance, $(1,0,0)$, $(0,1,0)$, and $(1,1,0)$ do not form a basis for $\mathbb{R}^3$, but I can nonetheless find a linear transformation such that $f(1,0,0)=(2,1,-1)$, $f(0,1,0) = (0,1,7)$, and $f(1,1,0)=(2,2,6)$ (in fact, many); for instance, $f(x,y,z) = (2x,x+y,7y-x)$ will do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your argument is that you misusing statement. You know that &quot;If they are a basis, then a linear transformation exists&quot;. You are trying to use this by saying &quot;If they are not a basis, then there is no linear transformation&quot;, but that is false. What you &lt;em&gt;can&lt;/em&gt; conclude is that &quot;if there is no linear transformation, then they are not a basis&quot; (this is called the &quot;contrapositive&quot;). In general, if you know that &quot;If P then Q&quot;, then from &quot;P is false&quot; you cannot conclude &lt;em&gt;anything&lt;/em&gt;. Q may or may not hold. For instance, &quot;If I fall into the pool, I will get wet.&quot; Does that mean that if I don't fall into the pool, I will be dry? No; I could get caught in a rainshower. If I know I didn't fall into the pool, I just don't know whether I will be wet or dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this problem, thoguh, you have actually put your finger on exactly what you need to put your finger on: the fact that one of your vectors is a linear combination of the other two. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, that means that for &lt;strong&gt;any&lt;/strong&gt; linear transformation $T$, you &lt;em&gt;must&lt;/em&gt; have&#xA;$$T(2,0,1) = T\bigl(2(1,1,1)-(0,2,1)\bigr) = 2T(1,1,1) - T(0,2,1).$$&#xD;&#xA;So you would need $f(2,0,1)=(1,0,0)$ to be equal to $$2f(1,1,1)-f(0,2,1)=2(3,2,7)-(2,1,-1) = (6-2, 4-1, 14+1) = (4,3,15).$$&#xD;&#xA;Since $(1,0,0)\neq (4,3,15)$, there can be no linear transformation with the desired properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; The key here is that if you have vectors that satisfy some linear dependence, then any linear transformation must &lt;em&gt;also&lt;/em&gt; respect that linear dependence. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, then &lt;strong&gt;any&lt;/strong&gt; linear transformation $f$ will necessarily satisfy that $f(2,0,1) = 2f(1,1,1)-f(0,2,1)$. Since the values you have in (a) &lt;em&gt;don't&lt;/em&gt; satisfy this, then you know there is no such linear transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if they &lt;em&gt;do&lt;/em&gt; satisfy it, as in (b)? Here you have that $(0,2,-1)=2(1,0,2)+(-2,2,-5)$, so if $f$ is a linear transformation, then you must have&#xA;$$f(0,2,-1) = 2f(1,0,2) + f(-2,2,-5).$$&#xD;&#xA;Here, this &lt;em&gt;does&lt;/em&gt; happen, since &#xA;$$2f(1,0,2) + f(-2,2,-5) = 2(0,-2,1) + (3,3,1) = (3, -1, 3) = f(0,2,-1).$$&#xD;&#xA;So, since they satisfy the equations, does there exist a linear transformation that does this? Yes! In fact, there are &lt;em&gt;infinitely many&lt;/em&gt; different linear transformations that do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you find one? Well, this is where you can try doing systems of linear equations. You want to find a linear transformation&#xA;$$f(x,y,z) = (ax+by+cz, \ell x+my+nz, rx+sy+tz)$$&#xD;&#xA;such that if you plug in $x=0$, $y=2$, and $z=-1$, you get $(3,-1,3)$; and if you plug in $x=1$, $y=0$, $z=2$, you get $(0,-2,1)$. This gives you six equations in nine unknowns (the unknowns being $a$, $b$, $c$, $\ell$, $m$, $n$, $r$, $s$, and $t$), and it will have a solution. (The other equation you get when you plug in $x=-2$, $y=2$, and $z=-5$ is actually a &lt;em&gt;consequence&lt;/em&gt; of the six we already had; taht is what we checked to begin with). So you can try solving that system and getting one possible map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(When the three vectors you have are a basis, you will get nine equations in nine unknowns with a single possible solution, so that will be the linear transformation you want.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a way here to save yourself a lot of work: take your favorite two vectors from this set of three, and then find a fourth vector that, together with the first two, makes up a basis. Then define the linear transformation by picking an arbitrary value for the image of that one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, start with $(0,2,-1)$ and $(1,0,2)$. Notice that $(1,0,0)$ is not a linear combination of these two, so $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ makes a basis. Since we can always find a linear transformation that maps $(0,2,-1)$ to $(3,-1,3)$, $(1,0,2)$ to $(0,-2,1)$, and $(1,0,0)$ to anything (say $(0,0,0)$ because it makes things &lt;em&gt;easier&lt;/em&gt;), that one will do. Notice that, if $f(1,0,0) = (0,0,0)$, that tells you that $a=\ell=r=0$, so now we only have three more unknowns. In fact, if we can figure out what $f(0,1,0)$ must be, that will gives us automatically $b$, $m$, and $s$; and $f(0,0,1)$ will give us $c$, $n$, and $t$. Can we do that? Yes: because $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ are a basis: we have&#xA;$$(0,1,0) = \frac{1}{2}(0,2,-1) + \frac{1}{4}(1,0,2) -\frac{1}{4}(1,0,0).$$&#xD;&#xA;So&#xA;$$f(0,1,0) = \frac{1}{2}(3,-1,3) + \frac{1}{4}(0,-2,1) - \frac{1}{4}(0,0,0).$$&#xD;&#xA;Then to find the values of $c$, $n$, and $t$, we use the fact that&#xA;$$(0,0,1) = \frac{1}{2}(1,0,2) - \frac{1}{2}(1,0,0).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also works when you are trying to find your function $f$ explicitly when the three vectors you have &lt;em&gt;are&lt;/em&gt; a basis: try to write $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ as linear combinations of the three vectors (this amounts to solving three systems of three linear equations with three unknowns first; or to finding the inverse of a matrix; so it looks like the same amount of work, but because there are a lot of zeros it is often easier to solve these than the general one with $a$, $b,\ldots,t$). And then use the value that $f$ must have at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ to figure out the &quot;formula&quot; for $f$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T19:41:15.753" LastActivityDate="2010-11-21T19:41:15.753" CommentCount="7" />
  <row Id="11144" PostTypeId="2" ParentId="11104" CreationDate="2010-11-20T21:48:52.570" Score="2" ViewCount="0" Body="&lt;p&gt;As seen in the other answers $L^2(\mathbb{R})$ is not a Banach algebra. Instead we may consider a &lt;em&gt;weighted&lt;/em&gt; $L^2$-space, let $w$ be a positive continuous function defined on $\mathbb{R}$ and $L^2_w= L^2_w(\mathbb{R})$ be the space of all measurable functions $f$ such that&#xA;$fw\in L^2(\mathbb{R})$ - the norm is defined by&#xA;$$&#xD;&#xA;\|f\|_{L^2_w}=\|fw\|_{L^2}.&#xD;&#xA;$$&#xD;&#xA;It is an exercise to show that $L^2_w$ is a Banach space. If we defined convolution as in the non-weighted space, that is&#xA;$$f*g(x)=\int_\mathbb{R} f(y) g(x-y)dy\qquad\text{(note no weight here)} $$&#xD;&#xA;we will get a Banach algebra if $w^{-2}*w^{-2}\le w^{-2}$. Unfortunately, I do not know if this condition can be improved. See this thread for more details, and a proof of the last statement:  &lt;a href=&quot;http://math.stackexchange.com/questions/2952/when-is-the-weighted-space-ellp-mathbbz-omega-a-banach-algebra-p1&quot;&gt;When is the weighted space $\ell^p(\mathbb{Z},\omega)$ a Banach algebra ($p&gt;1$)?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-20T21:48:52.570" />
  <row Id="11145" PostTypeId="1" AcceptedAnswerId="11290" CreationDate="2010-11-20T22:03:37.527" Score="13" ViewCount="295" Body="&lt;p&gt;Let $\ell^n_p$ denote $\mathbb{R}^n$ with the $p$-norm; assume $n&amp;gt;1$.  Now it's well-known that for $p\ne q$, $\ell^n_p$ and $\ell^n_q$ are never isometric unless $n=2$ and $\{p,q\}=\{1,\infty\}$, but I realized recently I didn't actually know how to prove this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove most of it - the only case I still need to rule out is the case where $n=2$ and $q=p&amp;#39;$ (i.e. $\frac{1}{1-1/p}$).  Now of course, it is indeed possible that $q=p&amp;#39;$, because $p$ could be $2$ (so $p=q$) or $1$ or $\infty$ (the sole exceptional case).  But the problem then is showing that $p$ must actually be one of those three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore I can show that if there is such an isometry, then, after applying a translation and some signed permutation matrix, it must be $\lambda R$, where $\lambda=2^{1/p-1/2}=2^{1/2-1/p&amp;#39;}$ and $R$ is rotation by $\pi/4$.  So actually all we need to do is, given $p\notin \{1,2,\infty\}$, to find a point $v$ such that $||\lambda R v||_{p&amp;#39;}\ne ||v||_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, graphing things, we can see that in fact &lt;em&gt;any&lt;/em&gt; point that is not on the axes or the lines $y=\pm x$ will work; but I do not know how to prove it for even a single specified point, even, say, $(1,2)$, because the resulting equation for $p$ (assuming $p\ne 1,\infty$, of course) is just absolutely terrible, and attempting to, say, determine when each side is larger by differentiating seems to only make things worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum Nov 21&lt;/strong&gt;: I should perhaps point out - obviously the point can vary with $p$, but I have no idea how given $p$ one might pick a specific point that makes it easier to prove - especially since it is after all true for nearly any point!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how is this case handled? And is there by any chance a nicer approach that I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="2884" LastEditorUserId="2884" LastEditorDisplayName="" LastEditDate="2010-11-21T22:30:00.317" LastActivityDate="2010-11-23T10:38:59.083" Title="How to show $\ell^2_p$ not isometric to $\ell^2_{p'}$ unless $p\in\{1,2,\infty\}$?" Tags="&lt;linear-algebra&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="8" />
  <row Id="11147" PostTypeId="2" ParentId="11122" CreationDate="2010-11-20T22:19:11.407" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an alternative for the case $p\ge1$ we may also use Hölder's inequality&#xA;$$|a+b|=|1\cdot a +1\cdot b|\leq (1^q+ 1^q)^{1/q}(|a|^p+|b|^p)^{1/p}=2^{1/q}(|a|^p+|b|^p)^{1/p}$$&#xD;&#xA;where $1/q+1/p=1$, which reduces to&#xA;$$|a+b|^p\le 2^{p/q}(|a|^p+|b|^p)=2^{p(1-1/p)}(|a|^p+|b|^p)=2^{p-1}(|a|^p+|b|^p).$$&#xD;&#xA;For $0&amp;lt;p&amp;lt;1$, please see my answer in the thread &lt;a href=&quot;http://math.stackexchange.com/questions/10993/please-explain-inequality-xp-yp-leq-x-yp&quot;&gt;&quot;Please explain inequality $|x^p−y^p|\le|x−y|^p$&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-20T22:19:11.407" />
  <row Id="11148" PostTypeId="1" CreationDate="2010-11-20T22:34:39.103" Score="3" ViewCount="60" Body="&lt;p&gt;A &lt;em&gt;diagonal&lt;/em&gt; of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt; is a selection of n entries in which no two entries occur in the same row or column.  For example: the entries marked with an asterisk below form a diagonal.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1  2* 3  4&#xA;2  3  4  1*&#xA;3  4  1* 2&#xA;4* 1  2  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt;:  Every Latin square contains a diagonal in which no symbol appears thrice (or more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asterisked diagonal in the above example is a diagonal in which no symbol appears twice.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;: Prove the above theorem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is quite a fun problem to solve, but there is a trap.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-20T22:34:39.103" Title="Does every Latin square contain a diagonal in which no symbol appears thrice?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" CommentCount="3" />
  <row Id="11149" PostTypeId="2" ParentId="9286" CreationDate="2010-11-20T22:55:16.273" Score="3" ViewCount="0" Body="&lt;p&gt;I know this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f$ and $g$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x):=\left(\int_{0}^{x}e^{-t^{2}}dt\right)^{2} \ \  \ and  \  \ \  g(x):=\left(\int_{0}^{1}\frac{e^{-x^{2}(t^{2}+1)}}{t^{2}+1}dt\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $$f&amp;#39;(x)=2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g&amp;#39;(x)=\int_{0}^{1}\frac{\partial}{dx}\left[\frac{e^{-x^{2}(t^{2}+1)}}{t^{2}+1}\right]dt = -2xe^{-x^{2}}\int_{0}^{1}e^{-x^{2}t^{2}}dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting $t=tx$, get  $\displaystyle\int_{0}^{1}e^{-x^{2}t^{2}}dt= \frac{1}{x}\displaystyle\int_{0}^{x}e^{-t^{2}}dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we get; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g&amp;#39;(x)=-2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $f&amp;#39;(x)+g&amp;#39;(x)=0$ for all $x$, then $f(x)+g(x)$ is an constant function. Also $$f(0)+g(0)=\displaystyle\int_{0}^{1}\frac{1}{t^{2}+1}dt = \displaystyle\frac{\pi}{4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(x)+g(x)=\displaystyle\frac{\pi}{4}$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\lim_{x \to{+}\infty}{g(x)}=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$\displaystyle\frac{\pi}{4} = \lim_{x \to{+}\infty}{f(x)+g(x)}=\lim_{x \to{+}\infty}{f(x)}= \left(\int_{0}^{\infty}e^{-t^{2}}dt\right)^{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\infty}e^{-t^{2}}dt=\sqrt{\frac{\pi}{4}}= \frac{\sqrt{\pi}}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-22T01:21:02.573" LastActivityDate="2010-11-22T01:21:02.573" />
  <row Id="11150" PostTypeId="1" AcceptedAnswerId="11155" CreationDate="2010-11-20T23:02:43.643" Score="17" ViewCount="574" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone provide me with good explanation of why $0^0 = 1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My train of thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x &amp;gt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0^x = 0^{x-0} = 0^x/0^0$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0^0 = 0^x/0^x = ?$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0^0 * 0^x = 1 * 0^x$, so  $0^0 = 1$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$0^0 = 0^x/0^x = 0/0 = \text{undefined}$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I've read the &lt;a href=&quot;http://mathforum.org/dr.math/faq/faq.0.to.0.power.html&quot; rel=&quot;nofollow&quot;&gt;explanation&lt;/a&gt; on mathforum.org, but it isn't clear to me.&lt;/p&gt;&#xA;" OwnerUserId="3761" LastEditorUserId="3761" LastEditorDisplayName="" LastEditDate="2010-11-21T09:37:28.377" LastActivityDate="2010-11-22T11:49:17.377" Title="Zero to zero power" Tags="&lt;discrete-mathematics&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="11151" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:06:05.717" Score="9" ViewCount="0" Body="&lt;p&gt;$0^0$ is undefined. It is an Indeterminate form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at this post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form&quot;&gt;Why is $1^{\infty}$ considered to be an indeterminate form&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, $0^0$ has many possible interpretations and hence it is an indeterminate form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0^{+}} x^{x} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0^{+}} 0^{x} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0^{-}} 0^{x} = $ not defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{x \rightarrow 0} x^{0} = 1$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-22T11:49:17.377" LastActivityDate="2010-11-22T11:49:17.377" CommentCount="2" />
  <row Id="11152" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:08:18.603" Score="3" ViewCount="0" Body="&lt;p&gt;Some indeterminates forms $0^{0}, \displaystyle\frac{0}{0}, 1^{\infty}, \infty − \infty, \displaystyle\frac{\infty}{\infty}, 0 &amp;#215; \infty, $ and $\infty^{0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{ x \rightarrow 0+ }x^{0}=1$$ &#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{ x \rightarrow 0+ }0^{x}=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Indeterminate_form&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Indeterminate_form&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-20T23:43:10.867" LastActivityDate="2010-11-20T23:43:10.867" />
  <row Id="11153" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:10:07.073" Score="9" ViewCount="0" Body="&lt;p&gt;This is merely a definition, and can't be proved via standard algebra. However, two examples of places where it is convenient to assume this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The binomial formula: $(x+y)^n=\sum_{k=0}^n {n\choose k}x^ky^{n-k}$. When you set $y=0$ (or $x=0$) you'll get a term of $0^0$ in the sum, which should be equal to 1 for the formula to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If $A,B$ are finite sets, then the set of all functions from $B$ to $A$, denoted $A^B$, is of cardinality $|A|^{|B|}$. When both $A$ and $B$ are the empty sets, there is still one function from $B$ to $A$, namely the empty function (a function is a collection of pairs satisfying some conditions; an empty collection is a legal function if the domain $B$ is empty).&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-20T23:10:07.073" CommentCount="1" />
  <row Id="11155" PostTypeId="2" ParentId="11150" CreationDate="2010-11-20T23:14:47.343" Score="27" ViewCount="0" Body="&lt;p&gt;In Calculus and Analysis, there is no good answer as to what $0^0$ &quot;should&quot; be, so it is usually left undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus, algebra, we just don't want to give it any value, we just declare it undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote &quot;size&quot; (cardinalities), then the &quot;$A^B$&quot; is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there &lt;em&gt;is&lt;/em&gt; one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we &lt;em&gt;must&lt;/em&gt; define $0^0$ as $1$. So if we are talking about &lt;em&gt;cardinal exponentiation&lt;/em&gt;, then the only possible definition is $0^0=1$, and we define it that way, period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; the same holds in Discrete Mathematics, when we are mostly interested in &quot;counting&quot; things. In Dicrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is &lt;em&gt;really&lt;/em&gt; the same thing as &quot;maps from $\{1,2,\ldots,m\}$ to $\{1,2,\ldots,n\}$&quot; when interpreted appropriately, so it is again the same thing as in set theory). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly &lt;strong&gt;one&lt;/strong&gt; way of doing that: just sit and do nothing! So we  make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you &lt;em&gt;cannot&lt;/em&gt; do it, so the answer is that $0^1=0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &quot;train of thoughts&quot; don't really work: If $x\neq 0$, then $0^x$ means &quot;the number of ways to make $x$ choices from $0$ possibilities&quot;. This number is $0$. So for &lt;em&gt;any&lt;/em&gt; number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ &quot;should&quot; be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ &#xA;&lt;em&gt;is&lt;/em&gt; the number of ways of making no selections from no choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coda.&lt;/strong&gt; In the end, it is a matter of definition and utility. In Calculus, in analysis, in algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that &quot;useful and natural&quot; way &lt;em&gt;should&lt;/em&gt; be, so we define it that way.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T20:53:48.097" LastActivityDate="2010-11-21T20:53:48.097" CommentCount="15" />
  <row Id="11156" PostTypeId="1" AcceptedAnswerId="11157" CreationDate="2010-11-20T23:15:25.243" Score="6" ViewCount="116" Body="&lt;p&gt;I'm familiar with the &quot;standard&quot; proof using Galois theory that there is no general formula for solving an equation of fifth (or higher) degree using radicals (i.e. arithmetic and root-taking). However, now I'm wondering if other proofs of different nature were found (in particular ones relying on analysis rather than algebra). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sparked my interest was seeing a description of the solution of the 2nd, 3rd and 4th degree equations via something that looked like discrete Fourier transform.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-11-20T23:21:33.467" Title="A different proof of the insolubility of the quintic?" Tags="&lt;abstract-algebra&gt;&lt;proof&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11157" PostTypeId="2" ParentId="11156" CreationDate="2010-11-20T23:21:33.467" Score="4" ViewCount="0" Body="&lt;p&gt;This is the content of the Abel-Ruffini theorem (whose proof predates Galois')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological proof may be found in this paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tmna.ncu.pl/files/v16n2-02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tmna.ncu.pl/files/v16n2-02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-20T23:21:33.467" />
  <row Id="11159" PostTypeId="2" ParentId="11129" CreationDate="2010-11-20T23:38:57.483" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ \ $ Put $\rm\ f = x,\ g = 1\ $ in the following simple&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\displaystyle\ \ \ \sqrt{f^2+g}\ -\ f\ \to\ 0 \ \ \ if\ \ \ \frac{f}g \to\ \infty\ \ \ and\ \ \frac{1}g\ \to\ \infty\ \ or\ \ r\in \mathbb R$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\displaystyle\quad\quad\ \ \sqrt{f^2+g}\ -\ f\ \ =\ \frac{g}{\sqrt{f^2+g}\ +\ f}\ =\ \frac{1}{\sqrt{(\frac{f}g)^2+\frac{1}g}\ +\ \frac{f}g}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-20T23:38:57.483" />
  <row Id="11160" PostTypeId="2" ParentId="11113" CreationDate="2010-11-20T23:46:12.923" Score="2" ViewCount="0" Body="&lt;p&gt;The integral closure of a complete Noetherian local domain is also complete, local and Noetherian. So the integral closure of $R/p$ is a complete regular local ring containing a field (normal rings in dimension $1$ are regular by Serre criterion), and you win! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: add more details, by Tymothy's comments (now deleted): So let $S$ be the integral closure of $R/p$ in its quotient field. Then by definition, $S$ is integrally closed (normal), thus regular because $\dim S=1$. Clearly $R/p$ is a subring of $S$. By Cohen Structure theorem, $S$ is a power series ring with one variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact about integral closure in my first sentence is well-known. Check out  Theorem 4.3.4 of the book available online by Huneke-Swanson, or Eisenbud Chapter 13, or Matsumura somewhere (probably the finiteness of integral closure is true for excellent rings as well). &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditorDisplayName="" LastEditDate="2010-11-22T03:31:02.417" LastActivityDate="2010-11-22T03:31:02.417" CommentCount="3" />
  <row Id="11161" PostTypeId="1" AcceptedAnswerId="11164" CreationDate="2010-11-21T01:17:53.913" Score="2" ViewCount="198" Body="&lt;p&gt;Why does X divided by zero not equal X? After all, X is not being divided by anything.&lt;/p&gt;&#xA;" OwnerUserId="2414" LastActivityDate="2010-11-21T08:12:54.487" Title="Why does X divided by zero not equal X?" Tags="&lt;arithmetic&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11163" PostTypeId="2" ParentId="11161" CreationDate="2010-11-21T01:23:14.290" Score="3" ViewCount="0" Body="&lt;p&gt;I would say that if you're dividing by 1, then you're &quot;not really dividing&quot;, which is why $X/1=X$.  Dividing by $0$, however, is undefined.  Here's a way to see this based on basic properties of arithmetic.  Suppose we could define, say $3/0$ as a real number $r$.  Then multiplying both sides of the equation $3/0=r$ by $0$ yields $3=0\cdot r$.  But $0\cdot r=0$ for all real numbers $r$, so this is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally, dividing $X$ by $n$ tells you how big to make the pieces of pie if you have to serve them to $n$ people.  This means that $n$ pieces of size $X/n$ give a total amount of $X$.  But if $n=0$, then no matter the size of each piece, $0$ of them can never give a total amount of $X$.  (If $X=0$, then the reasoning becomes more intricate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This reminds me of another confusion involving $0$ and language.  I have seen a temptation to replace &quot;the solution is 0&quot; with &quot;there is no solution,&quot; and vice versa among precalculus or beginning calculus students.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this problem of contrasting dividing by $0$ and &quot;not dividing by anything&quot; can be further clarified by comparison with what happens when adding and subtracting.  When you add $0$, you're &quot;not adding anything&quot;, and sure enough $X+0=X$ holds for all $X$.  Similarly for subtraction: $X-0=X$, and here the interpretation &quot;not subtracting anything&quot; works out.  This works because $0$ is a &lt;em&gt;neutral element&lt;/em&gt; with respect to addition (by definition).  Changing operations to multiplication, $0$ is no longer the neutral element; $1$ is.  Multiplying by $1$ is what translates to &quot;don't change anything&quot;: $X\cdot 1=X$.  Thus &quot;undoing&quot; multiplication by $1$, naming dividing by $1$, also doesn't change anything.  But since $0$ no longer has this innocuous behavior with respect to multiplication, things similarly change for division.  It changes drastically, because multiplication by $0$ becomes something that &lt;em&gt;can't be undone&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-21T08:12:54.487" LastActivityDate="2010-11-21T08:12:54.487" CommentCount="2" />
  <row Id="11164" PostTypeId="2" ParentId="11161" CreationDate="2010-11-21T01:23:39.897" Score="6" ViewCount="0" Body="&lt;p&gt;Think of division as an &quot;inverse&quot; operation to multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let us look at $\frac{a}{b}$ means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x = \frac{a}{b}$. What it means is when $x$ is multiplied by $b$ it gives $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you want to make sense of $\frac{a}{0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $\frac{a}{0}$ is $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we want $y$ such that $y \times 0$ gives us $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know that $y \times 0 = 0$, $\forall y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a logical inconsistency here.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-21T01:23:39.897" />
  <row Id="11165" PostTypeId="2" ParentId="11141" CreationDate="2010-11-21T01:43:17.597" Score="4" ViewCount="0" Body="&lt;p&gt;Here's an approach that leverages the geometry of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K_i$ be the center of circle $C_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points $N_2$ and $N_3$ on $C_1$ can be considered &quot;known&quot;. Hence, the measure of the central angle $N_2 K_1 N_3$ (corresponding to the &quot;top&quot; arc $N_2 N_3$ in your diagram) is also known, and easily computable with dot products; call that $2\theta_1$. (Why the &quot;2&quot;? Read on...) By the Inscribed Angle Theorem, the angle $N_2 T_1 N_3$ --&lt;em&gt;no matter where $T_1$ is on &quot;bottom&quot; arc $N_2 N_3$&lt;/em&gt;-- is half the size of the central angle; that is $N_2 T_1 N_3$ has measure $\theta_1$. This must also be the measure of angle $T_2 T_1 T_3$, at the top of your blue triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the blue triangle's angle at $T_2$ have (unknown) measure $\theta_2$. Let $P_2$ be the (other) point at which line $T_2 T_3$ meets circle $C_2$. The angle $N_2 T_2 P_2$ has measure $\theta_2$ (because it's the &quot;same angle&quot; as $T_1 T_2 T_3$). By the Inscribed Angle Theorem again, the &lt;em&gt;central&lt;/em&gt; angle $N_2 K_2 P_2$ (measured along the &quot;outer&quot; arc $N_2 P_2$) must be twice as big: $2\theta_2$. Therefore, if we know the size of $\theta_2$, then we know how far around circle $C_2$ we have to travel from $N_2$ to get to $P_2$.  Likewise, with (unknown) $\theta_3$ the blue triangle's angle at $T_3$ and $P_3$ the (other) point where line $T_2 T_3$ meets $C_3$, we have that angle $N_3 K_3 P_3$ has measure $2\theta_3$, which tells us how to get to $P_3$ from $N_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we can find formulas for $\theta_2$ and $\theta_3$ in terms of $\theta_1$ and $a$, then we'll effectively know the locations of $P_2$ and $P_3$ relative to $N_2$ and $N_3$. The line joining the $P$s determines the points $T_2$ and $T_3$, which in turn give us the location of $T_1$ (as the intersection of lines $T_2 N_2$ and $T_3 N_3$ with circle $C_1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's get to it ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Law of Sines, and your assumption about the ratio of side lengths, tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sin{\theta_3}}{\sin{\theta_2}}=\frac{|T_1 T_2|}{|T_1 T_3|}=a$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a \sin{\theta_2} = \sin{\theta_3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\theta_3=\pi-\left(\theta_1+\theta_2\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a\sin{\theta_2} = \sin{\left(\pi-\left(\theta_1+\theta_2\right)\right)}=\sin{\left(\theta_1+\theta_2\right)}=\sin\theta_1 \cos\theta_2+\cos\theta_1 \sin\theta_2$$&#xD;&#xA;$$\sin{\theta_2}\left(a-\cos\theta_1\right)=\sin\theta_1 \cos{\theta_2}$$&#xD;&#xA;$$\sin^2{\theta_2}\left(a-\cos\theta_1\right)^2=\sin^2\theta_1 \cos^2{\theta_2}$$&#xD;&#xA;$$\left( 1 - \cos^2{\theta_2} \right)\left(a- \cos\theta_1\right)^2 = \sin^2\theta_1 \cos^2\theta_2$$&#xD;&#xA;$$\left(a-\cos\theta_1\right)^2 = \cos^2{\theta_2}\left(a^2 - 2 a \cos\theta_1+ \cos^2\theta_1+\sin^2\theta_1\right)= \cos^2{\theta_2}\left(1 +a^2 - 2a \cos\theta_1\right)$$&#xD;&#xA;$$\cos^2{\theta_2}=\frac{\left(a-\cos\theta_1\right)^2}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1+\cos\left(2\theta_2\right)}{2}=\frac{\left(a-\cos\theta_1\right)^2}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\left(2\theta_2\right)=\frac{a^2-1-2a\cos\theta_1+2\cos^2\theta_1}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, swapping $\theta_2 \leftrightarrow \theta_3$ and $a \leftrightarrow \frac{1}{a}$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\left(2\theta_3\right)=\frac{1-a^2-2a\cos\theta_1+2a^2\cos^2\theta_1}{1 + a^2 - 2 a \cos\theta_1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, use $N_2$ and $N_3$ and $2\theta_2$ and $2\theta_3$ to locate $P_2$ and $P_3$, then $T_2$ and $T_3$, and finally $T_1$. I doubt the formulas are particularly elegant, but each step in finding them should be relatively straightforward.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-11-21T01:43:17.597" CommentCount="1" />
  <row Id="11166" PostTypeId="2" ParentId="9286" CreationDate="2010-11-21T01:59:55.537" Score="3" ViewCount="0" Body="&lt;p&gt;It might be worth mentioning that one also can use spherical coordinates in 3-dimensions analogously to the polar coordinates Ross Millikan used above: If $I$ denotes $\int_{-\infty}^{\infty} e^{-x^2}dx$, then we have&#xA;$$I^3 =  \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-x^2 - y^2 - z^2}\,dx\,dy\,dz$$&#xD;&#xA;Switching to spherical coordinates this becomes&#xA;$$\int_0^{2\pi}\int_0^{\pi}\int_0^{\infty}\sin(\phi) \rho^2 e^{-\rho^2}\,d\rho\,d\phi\,d\theta$$&#xD;&#xA;Doing the theta and $\phi$ integrations this becomes &#xA;$$I^3 = 4\pi\int_0^{\infty}\rho^2 e^{-\rho^2}\,d\rho$$&#xD;&#xA;One can then integrate parts in this, differentiating $\rho$ and integrating $\rho e^{-\rho^2}$. This leads us to&#xA;$$I^3 = 2\pi \int_0^{\infty} e^{-\rho^2}\,d\rho$$&#xD;&#xA;Note the right-hand side is exactly $2\pi * {I \over 2} = \pi I$. Thus $I^3 = \pi I$ and thus $I = \sqrt{\pi}$ as needed. Obviously polar coordinates are faster. Just sayin'...&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-21T01:59:55.537" />
  <row Id="11167" PostTypeId="2" ParentId="11161" CreationDate="2010-11-21T02:08:33.080" Score="8" ViewCount="0" Body="&lt;p&gt;I just want to mention that in certain situations it is perfectly valid to think of something nonzero divided by something zero as &quot;infinity.&quot;  For example, the slope of the line between the points $(a, b)$ and $(c, d)$ is $\frac{d - b}{c - a}$, unless $c - a = 0$ and $d - b \neq 0$, where this expression stops making sense.  However, the corresponding line is vertical, and it is perfectly reasonable to think of its slope as being &quot;infinity.&quot;  More precisely, the space of slopes is a gadget called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_line&quot; rel=&quot;nofollow&quot;&gt;projective line&lt;/a&gt; $\mathbb{P}^1(\mathbb{R})$ and if all one cares about is the geometric structure of this space (that is, one is not too concerned with adding and multiplying) it is perfectly valid to think about &quot;infinity&quot; in the same way as any other element, since they're all taken to each other by projective transformations.  These ideas are very important in certain branches of mathematics, such as algebraic geometry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more concrete application is that the tangent function can be thought of as taking values in $\mathbb{P}^1(\mathbb{R})$ instead of in $\mathbb{R}$, and then it no longer has &quot;singularities&quot;; in fact, it defines a homeomorphism from the circle to $\mathbb{P}^1(\mathbb{R})$.  Doing the same thing over $\mathbb{C}$ allows you to think of meromorphic functions as functions to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere&quot; rel=&quot;nofollow&quot;&gt;Riemann sphere&lt;/a&gt; $\mathbb{P}^1(\mathbb{C})$, and this idea also has many fruitful applications.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T02:08:33.080" CommentCount="1" />
  <row Id="11168" PostTypeId="1" AcceptedAnswerId="11170" CreationDate="2010-11-21T02:24:36.227" Score="0" ViewCount="60" Body="&lt;p&gt;A homework problem states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An aquarium 5 ft long, 2ft wide, and 3ft deep is full of water. Part C: find the hydrostatic force on one end of the aquarium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions (I've completed part a and b. B asked for the force on the bottom which was 1875 lbs.:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What exactly does one end of the aquarium mean?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a standard to applying long and wide in mathematics problems? It may not matter in calculating area, but in other instances it might. For example, calculating the area of a vertical standing slice would either have to be 2dx or 5dx.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My solutions manual shows that part C requires setting up an integral with a range from 0 to 3. This means that the &quot;one end&quot; of the aquarium is the top side? How else could the range be 0 to 3?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's my drawing of the aquarium:&#xA;&lt;img src=&quot;http://i.imgur.com/IOgTo.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T03:28:32.623" LastActivityDate="2010-11-21T03:28:32.623" Title="Hydrostatic Force and Integration" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11169" PostTypeId="2" ParentId="11028" CreationDate="2010-11-21T02:47:11.463" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a proof which is not based on Egoroff's theorem. As Jonas T points out in an other answer, Fatou's Lemma imply&#xA;$$\int|f|^p=\int\lim |f_n|^p =\int\liminf |f_n|^p  \le \liminf\int |f_n|^p \le M^p$$&#xD;&#xA;and hence $f\in L^p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we may assume $f=0$ (consider the problem with $f_n$ replaced by $f_n-f$), and we must show $\int f_ng\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix a number $A\gt0$, and let $E_n=\{x: |f_n(x) g(x)|\le A|g(x)|^q\}$ where $q=p/(p-1)$ (the conjugate to $p$). By dominated convergence we have&#xA;$$\int_{E_n}f_ng\,dx\to0$$&#xD;&#xA;On the complement, $E_n^c =X-E_n$, we have $|g|^{q}\le A^{-1}|f_n g|$. Hence, by Hölder's inequality&#xA;$$\int_{E_n^c}|f_ng|\,dx\le\left(\int_{E_n^c} |f_n|^p\right)^{1/p}\left(\int_{E_n^c} |g|^q\right)^{1/q}\le M A^{-1/q}\left(\int_{E_n^c}|f_ng|\,dx\right)^{1/q}$$&#xD;&#xA;In other words&#xA;$$\left(\int_{E_n^c}|f_ng|\,dx\right)^{1-1/q}\le MA^{-1/q}$$&#xD;&#xA;that is &#xA;$$\limsup_n\int_{E_n^c}|f_ng|\,dx\le M^pA^{-p/q}.$$&#xD;&#xA;In total we have shown &#xA;$$\limsup_n\int|f_ng|\,dx\le M^pA^{-p/q}$$&#xD;&#xA;for all $A\gt0$, from which be conclude $$\lim_n\int|f_ng|=0.$$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-21T02:47:11.463" CommentCount="1" />
  <row Id="11170" PostTypeId="2" ParentId="11168" CreationDate="2010-11-21T02:50:38.443" Score="1" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Objects are assumed to be longer than they are wide (unless a front and back are explicitly specified).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You integrate over depth to get the force onto one side because at each horizontal cros-section the pressure is different (since it's a function of depth).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-11-21T02:50:38.443" />
  <row Id="11171" PostTypeId="1" CreationDate="2010-11-21T03:51:54.140" Score="1" ViewCount="60" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to do filter design using MATLAB, in particular there is one step where I form an expression &lt;code&gt;H&lt;/code&gt; and then use &lt;code&gt;[N D] = numden(H)&lt;/code&gt; to get the numerator and denominator for input into a &lt;code&gt;freqz(N, D)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems like the &lt;code&gt;N&lt;/code&gt; and &lt;code&gt;D&lt;/code&gt; do not reduce to lowest form (divide by common factors). How can I do that in MATLAB?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3766" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T05:05:31.563" LastActivityDate="2010-11-21T05:05:31.563" Title="MATLAB numden() does not simplify by lowest common factor." Tags="&lt;math-software&gt;&lt;matlab&gt;" CommentCount="2" />
  <row Id="11172" PostTypeId="2" ParentId="8903" CreationDate="2010-11-21T04:07:39.877" Score="1" ViewCount="0" Body="&lt;p&gt;Expand RHS, using distributive property and the property that $\langle u,cv\rangle=\bar{c}\langle u,v\rangle$. Simplify and cancel many terms to get the LHS.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-21T04:07:39.877" />
  <row Id="11173" PostTypeId="1" AcceptedAnswerId="11174" CreationDate="2010-11-21T04:19:34.987" Score="1" ViewCount="89" Body="&lt;p&gt;Show that the set of the R-homomorphism  $A\rightarrow B$  with factorization through an injective module is an subgroup of $Hom_{R}(A,B)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a question from  my last test in  abstract algebra, that I can't solve.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-21T15:51:24.193" LastActivityDate="2010-11-21T15:51:24.193" Title="Help  with $Hom_{R}(A,B)$" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11174" PostTypeId="2" ParentId="11173" CreationDate="2010-11-21T04:29:38.400" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: This just follows from the fact that a direct sum of two injective modules is injective. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-21T04:49:30.230" LastActivityDate="2010-11-21T04:49:30.230" CommentCount="1" />
  <row Id="11175" PostTypeId="1" CreationDate="2010-11-21T04:43:59.833" Score="2" ViewCount="83" Body="&lt;p&gt;Let $F \to E \to B$ be a fibration with $B$ simply connected (more generally, such that $\pi_1(B)$ acts trivially on the homology of $F$). Then there is a Serre spectral sequence $H_p(B, H_q(F)) \to H_{p+q}(E)$.  One can do the same for singular cohomology. However, for reasonable spaces (specifically, locally contractible spaces, e.g. CW complexes), singular cohomology is the same as sheaf cohomology of the constant sheaf $\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is another spectral sequence for sheaf cohomology: the Leray spectral sequence. Given spaces $X, Y$ and $f: X \to Y$, and a sheaf $\mathcal{F}$ on $X$, there is a spectral sequence $H^p(Y, R^q_f(\mathcal{F})) \to H^{p+q}(X, \mathcal{F})$.&#xA;The Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Leray_spectral_sequence&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; hints that the topological implications of this include in particular the Serre spectral sequence. I would be interested in this, because I like the machinery of the Grothendieck spectral sequence (from which the Leray spectral sequence easily follows), and would be curious if the Serre spectral sequence could be obtained as a corollary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-21T05:07:16.803" Title="Is the Serre spectral sequence a special case of the Leray spectral sequence?" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="11176" PostTypeId="2" ParentId="11175" CreationDate="2010-11-21T05:07:16.803" Score="2" ViewCount="0" Body="&lt;p&gt;Yes.  In fact, the result is basically obvious if you use Czech cohomology on the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Serre really had two key insights.  First, sheaf cohomology is a pain to compute, but if there is no fundamental group then for fiber bundles the Leray spectral sequence is really just using normal old-fashioned untwisted cohomology.  Second, you don't really need to work with fiber bundles -- all you need are Serre fibrations, and those are easy to construct.  In particular, you have the standard Serre fibration $\Omega X \rightarrow PX \rightarrow X$, where $\Omega X$ is the loop space of $X$ and $PX$ is the space of paths starting at the basepoint of $X$ and the map $PX \rightarrow X$ is &quot;evaluation at the endpoint&quot;.  Clearly $PX$ is contractible!  An amazing amount of milage can be had from this silly observation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Serre also really developed many of the key algebraic tricks one needs to work with spectral sequences.  For instance, he had the amazing idea that one can work modulo &quot;Serre classes&quot;, and thus ignore things like torsion.  It's like pretending to localize spaces long before Sullivan and Quillen realized you could do so for real!&lt;/p&gt;&#xA;" OwnerUserId="3768" LastActivityDate="2010-11-21T05:07:16.803" CommentCount="14" />
  <row Id="11177" PostTypeId="1" AcceptedAnswerId="11180" CreationDate="2010-11-21T05:13:35.040" Score="5" ViewCount="208" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking to expand my knowledge on set theory (which is pretty poor right now -- basic understanding of sets, power sets, and different (infinite) cardinalities). Are there any books that come to your mind that would be useful for an undergrad math student who hasn't taken a set theory course yet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for your suggestions!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T06:59:08.917" LastActivityDate="2010-12-27T12:43:26.587" Title="What are good books/other readings for elementary set theory?" Tags="&lt;reference-request&gt;&lt;elementary-set-theory&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="5" />
  <row Id="11178" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:25:09.780" Score="3" ViewCount="0" Body="&lt;p&gt;Enderton's book should be a gentle, easy read for an undergraduate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0122384407&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Elements-Set-Theory-Herbert-Enderton/dp/0122384407/ref=pd_sim_b_26&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-21T05:25:09.780" />
  <row Id="11179" PostTypeId="2" ParentId="7503" CreationDate="2010-11-21T05:26:28.373" Score="0" ViewCount="0" Body="&lt;p&gt;Let $\chi$ be the indicate function of $[-1,1]$. Then the Fourier transform of $\chi$ is $$g(\xi)=\frac{\sin(2\pi\xi)}{\pi\xi}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute the triple convolution $h(x)$ of $\chi$ and evaluate it at 0. That will be 3. From Fourier inverse formula one gets $$h(x)=\int_{-\infty}^\infty (g(\xi))^3e^{2\pi ix\xi}d\xi.$$ Evaluating at $x=0$ yields&#xA;$$ 3=\int_{-\infty}^\infty \frac{\sin^3(2\pi\xi)}{\pi^3\xi^3}d\xi,$$&#xD;&#xA;from which the result follows by substitution.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-21T05:26:28.373" />
  <row Id="11180" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:27:37.170" Score="8" ViewCount="0" Body="&lt;p&gt;I recommend Naive Set Theory by Halmos.  It's a friendly, thin and fun to read introduction to set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=x6cZBQ9qtgoC&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=x6cZBQ9qtgoC&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-11-21T05:27:37.170" />
  <row Id="11181" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:33:15.187" Score="2" ViewCount="0" Body="&lt;p&gt;Halmos's book mentioned above is very gentle and easy, and you should look there first.  Afterwards, when I was an undergraduate I remember learning a lot from Set Theory for the Working Mathematician by Krzysztof Ciesielski.  In particular, it has a long chapter showing how transfinite induction can be used to construct all sorts of odd subsets of $\mathbb{R}^n$ and what not.&lt;/p&gt;&#xA;" OwnerUserId="3768" LastActivityDate="2010-11-21T05:33:15.187" />
  <row Id="11182" PostTypeId="2" ParentId="11066" CreationDate="2010-11-21T05:33:54.157" Score="8" ViewCount="0" Body="&lt;p&gt;I believe the answer is 643888.  Here's how I arrived at this number:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let $C_k$ be the set of $4 \times 4$ $(0,1)$-matrices with $k$ 1's for which there do not exist two neighbouring 1's.&lt;/li&gt;&#xA;&lt;li&gt;Let $B_k$ be the set of $4 \times 4$ $(0,1)$-matrices with $k$ 1's for which there exists two neighbouring 1's, and for which deletion of some 1 implies membership in $C_{k-1}$.&lt;/li&gt;&#xA;&lt;li&gt;For any two (0,1)-matrices, $L$ and $M$, let $\chi(L,M)=1$ if $L$ and $M$ do not have a 1 in the same position and $0$ otherwise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We need the sets $B_k$ and $C_k$ to avoid multiple win situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number you seek is given by \[\sum_{k=0}^{n^2/2} \Big(\sum_{L \in B_k} \sum_{M \in C_{k-1}} \chi(L,M) + \sum_{L \in C_k} \sum_{M \in B_k} \chi(L,M)\Big).\]  The first term counts when player 1 wins, and the second term counts when player 2 wins.  I implemented this formula in GAP with the code below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function determines if a (0,1)-matrix has n/2 lines in a horizontal or vertical line.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740Win:=function(L,n)&#xA;  local m,i,j;&#xA;  if(n mod 2=1) then return fail; fi;&#xA;  m:=n/2;&#xA;  for i in [1..n] do for j in [1..m+1] do&#xA;    if(ForAll([0..m-1],k-&amp;gt;L[i][j+k]=1)) then return true; fi;&#xA;    if(ForAll([0..m-1],k-&amp;gt;L[j+k][i]=1)) then return true; fi;&#xA;  od; od;&#xA;  return false;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This function determines if a (0,1)-matrix has n/2 lines in a horizontal or vertical line, and that deletion of one of the 1's results in falsifying user3740Win(L,n).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740StrongWin:=function(L,n)&#xA;  local M,i,j;&#xA;  if not(user3740Win(L,n)) then return false; fi;&#xA;  M:=List([1..n],i-&amp;gt;List([1..n],j-&amp;gt;L[i][j]));&#xA;  for i in [1..n] do for j in [1..n] do&#xA;    M[i][j]:=0;&#xA;    if(user3740Win(M,n)=false) then return true; fi;&#xA;    M[i][j]:=L[i][j];&#xA;  od; od;&#xA;  return false;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This function generates the (0,1)-matrices with at most n^2/2 1's.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740MatricesStep1:=function(n)&#xA;  local A,B,L,M,i,j;&#xA;  A:=Filtered(Tuples([0,1],n^2),i-&amp;gt;Sum(i)&amp;lt;=n^2/2);&#xA;  B:=[];&#xA;  for L in A do&#xA;    M:=List([1..n],i-&amp;gt;List([1..n],j-&amp;gt;0));&#xA;    for i in [1..n] do for j in [1..n] do&#xA;      M[i][j]:=L[n*(i-1)+j];&#xA;    od; od;&#xA;    Append(B,[M]);&#xA;  od;&#xA;  return B;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Starting from A (the set of (0,1)-matrices with at most n^2/2 1's), I run through k=0..n^2/2.  B is $B_k$ above.  C1 is $C_{k-1}$ above.  C2 is $C_k$ above.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;user3740Count:=function(n)&#xA;  local A,B,C1,C2,S,i,j,k,L,M;&#xA;  A:=user3740MatricesStep1(n);&#xA;  S:=[];&#xA;  for k in [0..n^2/2] do&#xA;    B:=Filtered(A,L-&amp;gt;user3740StrongWin(L,n) and Sum(Sum(L))=k);&#xA;    C1:=Filtered(A,L-&amp;gt;(not user3740Win(L,n)) and Sum(Sum(L))=k-1);&#xA;    C2:=Filtered(A,L-&amp;gt;(not user3740Win(L,n)) and Sum(Sum(L))=k);&#xA;    Print(&quot;k=&quot;,k,&quot; &quot;,Size(B),&quot; &quot;,Size(C1),&quot; &quot;,Size(C2),&quot;\n&quot;);&#xA;    for L in B do for M in C1 do&#xA;      if(not 2 in Union(L+M)) then Append(S,[L+2*M]); fi;&#xA;    od; od;&#xA;    for L in C2 do for M in B do&#xA;      if(not 2 in Union(L+M)) then Append(S,[L+2*M]); fi;&#xA;    od; od;&#xA;    Print(&quot;count=&quot;,Size(S),&quot;\n&quot;);&#xA;  od;&#xA;  return S;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Just so the reader can independently confirm these calculations, here is the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; S:=user3740Count(2);;&#xA;k=0 0 0 1&#xA;count=0&#xA;k=1 4 1 0&#xA;count=4&#xA;k=2 0 0 0&#xA;count=4&#xA;gap&amp;gt; S:=user3740Count(4);;&#xA;k=0 0 0 1&#xA;count=0&#xA;k=1 0 1 16&#xA;count=0&#xA;k=2 24 16 96&#xA;count=2072&#xA;k=3 284 96 276&#xA;count=58808&#xA;k=4 1200 276 405&#xA;count=315836&#xA;k=5 2276 405 304&#xA;count=571892&#xA;k=6 2032 304 114&#xA;count=638352&#xA;k=7 848 114 20&#xA;count=643752&#xA;k=8 156 20 2&#xA;count=643888&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 0, 0, 0 ], [ 2, 0, 1, 2 ], [ 0, 1, 1, 0 ], [ 1, 2, 0, 0 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 2, 0, 2, 1 ], [ 1, 1, 0, 2 ], [ 0, 0, 1, 0 ], [ 1, 2, 0, 0 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 1, 1, 0 ], [ 1, 2, 0, 2 ], [ 0, 0, 0, 1 ], [ 2, 1, 0, 2 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 2, 1, 0 ], [ 0, 0, 0, 0 ], [ 1, 1, 1, 2 ], [ 2, 0, 0, 0 ] ]&#xA;gap&amp;gt; Random(S);&#xA;[ [ 0, 2, 0, 2 ], [ 1, 1, 2, 0 ], [ 1, 0, 0, 1 ], [ 0, 1, 2, 0 ] ]&#xA;gap&amp;gt; Size(Filtered(S,L-&amp;gt;Sum(Sum(L))=4));&#xA;336&#xA;gap&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The numbers are k (as in the above formula), |B|, |C1|, |C2|, then &quot;count&quot; is the running sum.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-21T05:55:12.600" LastActivityDate="2010-11-21T05:55:12.600" CommentCount="1" />
  <row Id="11183" PostTypeId="2" ParentId="8890" CreationDate="2010-11-21T05:37:04.460" Score="1" ViewCount="0" Body="&lt;p&gt;The first inequality follows from expanding $$(|z|-|w|)^2.$$ The second inequality follows from expanding&#xA;$$(z+w)(\bar{z}+\bar{w}).$$&#xD;&#xA;Note that $\bar{z}w+z\bar{w}=2\Re(\bar{z}w)$ and $\Re(c)\le |c|$. Then apply the first part.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-21T05:37:04.460" />
  <row Id="11184" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T05:41:37.247" Score="2" ViewCount="0" Body="&lt;p&gt;I recently bought the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821827316&quot; rel=&quot;nofollow&quot;&gt;Basic Set Theory&lt;/a&gt; by A. Shen and N.K. Vereshchagin and it has been a really nice read. It is very accessible and has a lot of exercises. It covers the basics and is very short, about a 100 pages or so. I would recommend it sincerely, although I'm not sure if it will be too basic for what you already know.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-21T05:41:37.247" CommentCount="2" />
  <row Id="11185" PostTypeId="1" AcceptedAnswerId="11186" CreationDate="2010-11-21T06:11:57.147" Score="0" ViewCount="90" Body="&lt;p&gt;Proof in progress.&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="3740" LastEditorDisplayName="" LastEditDate="2010-11-23T19:46:56.743" LastActivityDate="2010-11-23T19:47:28.787" Title="In a Lattice, x ∧ y ≤ x and x ∧ y ≤ y" Tags="&lt;order-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11186" PostTypeId="2" ParentId="11185" CreationDate="2010-11-21T06:30:45.377" Score="3" ViewCount="0" Body="&lt;p&gt;I do not see why you assert that $x\wedge y$ must be either $x$ or $y$. This is not true in an arbitrary lattice: for instance, in the lattice of subsets of a given set $X$ (with $\leq$ corresponding to inclusin, $\wedge$ corresponding to intersection, and $\vee$ corresponding to union) it is false that $x\cap y$ must be either $x$ or $y$. So you should really be feeling not comfortable about your assertions well before you actually do. But even if you got to that step, you are definitely in trouble: it does &lt;em&gt;not&lt;/em&gt; follow that $y\leq x$ &quot;as $L$ is a partially ordered set&quot;. That assertion has no justification whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your &quot;proof by contradiction&quot;, a proof by contradiction does not begin by negating your &quot;definition&quot; (as you claim earlier). A proof by contradiction begins by negating the &lt;strong&gt;proposed conclusion&lt;/strong&gt;. If you wanted to do this by contradiction, you would have to being by assuming that your conclusion, &quot;$x\wedge y \leq x$ and $x\wedge y \leq y$&quot; is false. So you would need to assume that &quot;$x\wedge y\not\leq x$ OR $x\wedge y \not\leq y$&quot; is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if you get to your third line, you have it exactly backwards: it does &lt;em&gt;not&lt;/em&gt; follow that $x\leq x\wedge y$ and $y\leq x\wedge y$. What makes you think that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... how do you prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you are defining the lattince operations in terms of the partial order. That is, you &quot;know&quot; what $\leq$ means, and then you define $\wedge$ and $\vee$ using $\leq$. The definition would be that $x\wedge y$ is the greatest lower bound of $\{x,y\}$ (which must exist if you call the poset a lattice) and $x\vee y$ is the least upper bound of $\{x,y\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, using that definition, $x\wedge y$ is the greatest lower bound of the set $\{x,y\}$. In particular, it is a lower bound of $\{x,y\}$. What is the definition of &quot;lower bound&quot;? What does that tell you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that it is the other way around: you &quot;know&quot; what $\wedge$ and $\vee$ mean, and you define a partial order in terms of them. If you define $\leq$ in terms of the lattice operations, then, explicitly, you &lt;em&gt;define&lt;/em&gt; the partial order associated to the lattice by saying that for $a,b\in L$, &#xA;$$a\leq b \Longleftrightarrow  a\wedge b=a,$$&#xD;&#xA;or else you define it by&#xA;$$a \leq b \Longleftrightarrow a\vee b = b.$$ &#xA;In fact, both definitions are equivalent, as can be shown using the properties of the two operations in a lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, what you want to show is that $(x\wedge y)\vee x = x$ and that $(x\wedge y)\vee y = y$. (Or if you defined it using $\wedge$, that $(x\wedge y)\wedge x = x\wedge y$ and that $(x\wedge y)\wedge y = x\wedge y$). These will follow directly from the properties that operations that make up a lattice must satisfy (commutativity, associativity, idempotency, and the two absorption laws). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T07:10:44.983" LastActivityDate="2010-11-21T07:10:44.983" CommentCount="7" />
  <row Id="11187" PostTypeId="2" ParentId="11185" CreationDate="2010-11-21T07:42:57.463" Score="0" ViewCount="0" Body="&lt;p&gt;Proof in progress.&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="3740" LastEditorDisplayName="" LastEditDate="2010-11-23T19:47:28.787" LastActivityDate="2010-11-23T19:47:28.787" CommentCount="6" />
  <row Id="11188" PostTypeId="1" CreationDate="2010-11-21T08:08:52.600" Score="-1" ViewCount="85" Body="&lt;p&gt;&lt;strong&gt;Added later:&lt;/strong&gt; Proof now in progress.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Let ( L, ≤ ) be a lattice, x, y, z ∈ L. Prove that $x ≤ y$ and $x ≤ z &amp;lt;=&amp;gt; x ≤ y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proceed to prove as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement can be split into two implications :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $x ≤ y$ and $x ≤ z =&amp;gt; x ≤ y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) $x ≤ y ∧ z =&amp;gt; x ≤ y$ and $x ≤ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) &lt;em&gt;Approach 1 : Proceeding by use of definitions:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x ≤ y$ and $x ≤ z =&amp;gt; x$ is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $w = y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since w is the GLB, by defn. x ≤ w, i.e. x ≤ y ∧ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Approach 2 : Proof by contradiction:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x ≤ y$ and $x ≤ z =&amp;gt; x$ not $≤ y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x ≤ y and x ≤ z =&gt; x is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $w = y ∧ z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since w is the GLB, by defn. x ≤ w, i.e. x ≤ y ∧ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a contradiction has been reached, the original assertion is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) &lt;em&gt;Approach 1 : Proceeding by use of definitions:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x ≤ y ∧ z =&amp;gt; x$ is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of LB, x ≤ for all w ∈ L, i.e. x ≤ y and x ≤ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Approach 2 : Proof by contradiction:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x ≤ y ∧ z =&amp;gt;$ not x ≤ y or not x ≤ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x ≤ y ∧ z =&amp;gt; x$ is a LB in L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of LB, x ≤ for all w ∈ L, i.e. x ≤ y and x ≤ z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a contradiction has been reached, the original assertion is true.&lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:31:50.677" LastActivityDate="2010-11-23T21:11:35.103" Title="In a Lattice, x ≤ y and x ≤ z &lt;=&gt; x ≤ y ∧ z" Tags="&lt;order-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="11189" PostTypeId="1" AcceptedAnswerId="11207" CreationDate="2010-11-21T08:42:39.337" Score="1" ViewCount="80" Body="&lt;p&gt;I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that&#xA;$$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$&#xD;&#xA;How this can be done? Is it possible to prove stronger inequalities like&#xA;$$10(a^6 + b^6 + c^6) + 41abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a),$$&#xD;&#xA;or even &#xA;$$a^6 + b^6 + c^6 + 50abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$&#xD;&#xA;without computer?&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-21T13:27:22.837" Title="Inequality of three variable polynomial" Tags="&lt;polynomials&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11190" PostTypeId="1" AcceptedAnswerId="11192" CreationDate="2010-11-21T09:30:23.737" Score="1" ViewCount="70" Body="&lt;p&gt;Let $r \in \mathbb{N}$. Observe the equivalence relation $\sim \subseteq \mathbb{Z} \times \mathbb{Z}$ defined as $x \sim y :\Leftrightarrow (\exists k \in \mathbb{Z})(y=x+kr)$. Show that an operation $&amp;#39;+&amp;#39;$ on $\mathbb{Z}/\sim$  given by $[x]_\sim + [y]_\sim := [x+y]_\sim$ is well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at a proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x\sim x&amp;#39;, y\sim y&amp;#39;$, then we know that $(\exists k_1 \in \mathbb{Z})(x=x&amp;#39;+k_1r)$ and $(\exists k_2 \in \mathbb{Z})(y=y&amp;#39;+k_2r)$. Adding $x$ and $y$, we see that $x+y=x&amp;#39;+y&amp;#39;+(k_1+k_2)r$. Because $\mathbb{Z}$ is closed under addition, $k_1+k_2 \in \mathbb{Z}$, and eventually $x+y\sim x&amp;#39;+y&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I'm not even sure whether I'm on the right track here. As far as I can see, I've shown that adding two elements of (different) equivalence classes yields an element, whose equivalence class is in $\mathbb{Z}/\sim$. Is that sufficient to show that $&amp;#39;+&amp;#39;$ is well-defined? &lt;/p&gt;&#xA;" OwnerUserId="3546" LastActivityDate="2010-11-21T10:20:22.543" Title="Equivalence relation: showing that an operation is well-defined" Tags="&lt;homework&gt;&lt;relations&gt;" AnswerCount="1" />
  <row Id="11192" PostTypeId="2" ParentId="11190" CreationDate="2010-11-21T10:07:19.903" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, you're on the right track.  &quot;Well-defined&quot; just means that it should be an actual operation that takes 2 equivalence classes and sends them to a (uniquely determined) equivalence class.  The way the operation is written makes use of specific representatives of the equivalence classes.  $[x+y]$ makes sense as an element of $\mathbb{Z}/\sim$, but you need to know that it is determined by the classes of $x$ and $y$, rather than depending on the choices of representatives.  To do so, you suppose that another arbitrary choice of representatives is made (your $x&amp;#39;$ and $y'$), and show that the sum $[x&amp;#39;+y&amp;#39;]$ is the same.  That is what your work shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The language in your second to last sentence isn't entirely accurate.  What you've shown is that adding 2 equivalence classes determines an equivalence class.  The key word is &quot;an&quot; (singular); you've shown that you couldn't get different equivalence classes by &quot;adding&quot; in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To contrast, suppose $r=3$ and we want to &quot;define&quot; an operation by $[x]^{[y]}=[x^y]$.  If we try to combine $[2]$ and $[1]$ using this &quot;operation&quot;, we get $[2]^{[1]}=[2^1]=[2]$.  But $[1]=[4]$, so we also get $[2]^{[1]}=[2]^{[4]}=[2^4]=[16]=[1]$.  Thus this &quot;operation&quot; is not well-defined.  (Of course, there are other problems, like defining powers of $0$ or negative powers, but the point here is the non-uniqueness, even if it did make sense.)  Better examples of non-well-defined &quot;operations&quot; arise when you study groups.  There you study equivalence classes called cosets of subgroups and try to combine them analogously to how equivalence classes mod $r$ are added in your problem.  You find that unless the subgroup satisfies a certain property called being &quot;normal&quot;, the &quot;operation&quot; is not well-defined.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-21T10:20:22.543" LastActivityDate="2010-11-21T10:20:22.543" CommentCount="1" />
  <row Id="11193" PostTypeId="1" AcceptedAnswerId="11291" CreationDate="2010-11-21T10:30:35.343" Score="2" ViewCount="151" Body="&lt;p&gt;What is the :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;most useful&lt;/li&gt;&#xA;&lt;li&gt;prettiest&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;way to visualise extra dimensions in shapes and charts?&lt;/p&gt;&#xA;" OwnerUserId="3746" LastEditorUserId="3746" LastEditorDisplayName="" LastEditDate="2010-11-21T16:47:20.277" LastActivityDate="2010-11-22T02:27:06.600" CommunityOwnedDate="2010-11-21T14:26:01.350" Title="Visualising extra dimensions" Tags="&lt;geometry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11194" PostTypeId="1" CreationDate="2010-11-21T10:35:12.407" Score="1" ViewCount="84" Body="&lt;p&gt;$$T(n)=1+2\sum_{i=1}^{n-1}T(i) , \quad n &amp;gt; 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(1)=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any hint or how to solve?&lt;/p&gt;&#xA;" OwnerUserId="3741" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T11:46:22.160" LastActivityDate="2010-11-21T20:16:20.090" Title="Solving the recurrence relation that contains summation of nth term." Tags="&lt;discrete-mathematics&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="11196" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T10:45:51.463" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: Consider $T(n) - T(n-1)$ in case the $1$ is outside of the sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\frac{T(n)}{n-1} - \frac{T(n-1)}{n-2}$ if you sum also over the $1$.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastEditorUserId="3330" LastEditorDisplayName="" LastEditDate="2010-11-21T10:58:21.110" LastActivityDate="2010-11-21T10:58:21.110" CommentCount="2" />
  <row Id="11197" PostTypeId="2" ParentId="10965" CreationDate="2010-11-21T10:55:44.663" Score="0" ViewCount="0" Body="&lt;p&gt;Have someone an idea???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and best regards&#xA;Ronald&lt;/p&gt;&#xA;" OwnerUserId="3775" LastActivityDate="2010-11-21T10:55:44.663" />
  <row Id="11198" PostTypeId="2" ParentId="11073" CreationDate="2010-11-21T11:21:10.870" Score="0" ViewCount="0" Body="&lt;p&gt;If the three points $z_1$, $z_2$, $z_3\in S^1$ can be covered by a semicircle $H$ then the restriction $d\restriction H$ acts on the $z_i$ like the usual metric on $\mathbb R$, if not, then the sum of any two distances between the $z_i$  is $&amp;gt;\pi$ and the third distance is $&amp;lt;\pi$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-11-21T13:40:09.720" LastActivityDate="2010-11-21T13:40:09.720" />
  <row Id="11199" PostTypeId="1" CreationDate="2010-11-21T11:21:40.247" Score="2" ViewCount="186" Body="&lt;p&gt;While experimenting with certain sums, I came to the following sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After rewriting the summand as&#xA;$$\frac{n-k+1}{(n+1) (n+2) (n-k+2)} = \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^2}$$&#xD;&#xA;and then feeding the sum to Mathematica, i obtained&#xA;$$S_n = \frac{7}{6(n+1)(n+2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But directly summing the first expression above does not yield this closed-form result, because this &quot;result&quot; is wrong! (see edits below) Where is the bug or what am i missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; There seems to be some bug / feature in Mathematica which is leading to the erroneous computation above; the correct answer is of course, as also observed in two answers below. If someone can explain the bug, then he / she should feel free to report it to Wolfram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To be specific, the answer above is using $7/6 = (1+2)-H_{1+2}$, instead of $n+2 - H_{n+2}$ as it should)&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-23T18:26:33.307" LastActivityDate="2010-11-23T19:47:54.120" Title="Bug in mathematica: computing the sum of the ratios $(n-k+1)/(n-k+2)$" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="13" FavoriteCount="1" />
  <row Id="11200" PostTypeId="2" ParentId="11199" CreationDate="2010-11-21T11:38:23.267" Score="2" ViewCount="0" Body="&lt;p&gt;You can write the sum as&#xA;$$\frac1{(n+1)(n+2)} \sum_{k=0}^{n} \frac{n+1-k}{n+2-k},$$&#xD;&#xA;so it's enough to compute the last part of it. This is just&#xA;$$\frac{n+1}{n+2} + \frac{n}{n+1} + \dots + \frac{n+1-n}{n+2-n} = 1 - \frac1{n+2} + 1 - \frac1{n+1} + \dots + 1 - \frac1{n+2-n}$$&#xD;&#xA;$$= n + 2 - H_{n+2},$$&#xD;&#xA;where $H_n = \sum_{k=1}^n \frac1{k}$ is the $n$:th Harmonic number.&#xA;Your sum is then&#xA;$$\frac{n + 2 - H_{n+2}}{(n+1)(n+2)}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3776" LastEditorUserId="3776" LastEditorDisplayName="" LastEditDate="2010-11-21T13:17:11.520" LastActivityDate="2010-11-21T13:17:11.520" CommentCount="1" />
  <row Id="11201" PostTypeId="1" AcceptedAnswerId="11202" CreationDate="2010-11-21T12:12:51.010" Score="1" ViewCount="42" Body="&lt;p&gt;I should prove or give a counterexample for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial (\bigcup_{i \in \mathbb{N}} A_i) \subset \bigcup_{i \in \mathbb{N}} (\partial A_i)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\partial$ is the boundary and (X,d) is a metric space with $A_{i} \subset X $ for each $i \in \mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot make up a counterexample for finite sets $A_{i}$ but I think there might be one for infinite ones, or am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-11-21T12:25:57.077" LastActivityDate="2010-11-21T12:28:33.090" Title="Boundary in metric spaces" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11202" PostTypeId="2" ParentId="11201" CreationDate="2010-11-21T12:27:39.490" Score="2" ViewCount="0" Body="&lt;p&gt;you can take the metric space $\mathbb{R}$ and the sets $A_i = \{\frac{1}{i}\}$.&#xA;the boundary of $\bigcup A_i = \{\frac{1}{i} \mid i\in \mathbb{N}\}=A$ is $A\cup \{0\}$ while $\partial A_i = A_i$ so $\bigcup \partial A_i = A $ &lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-21T12:27:39.490" />
  <row Id="11203" PostTypeId="2" ParentId="11201" CreationDate="2010-11-21T12:28:33.090" Score="1" ViewCount="0" Body="&lt;p&gt;Let $X = \{ 1/i , i \in \mathbb N \}$ considered as a metric sub-space of $\mathbb R$. Let $A_i = \{ 1/i\}$. The RHS is empty whereas LHS is the set $\{ 0\}$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T12:28:33.090" />
  <row Id="11204" PostTypeId="1" CreationDate="2010-11-21T12:33:45.863" Score="0" ViewCount="56" Body="&lt;p&gt;Let $\Omega \subset R^n$ be open, and let us equip $\mathcal E = C^\infty$ with the locally convex topology induced by the following family of semi-norms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $K \subset \Omega$ and $\alpha, \beta \in \mathbb N _ 0 ^n$, let $|f|_{K,\alpha,\beta} = sup_{x \in K} | \partial_x^{\alpha} f(x) |$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that compact sets are closed and bounded within that topology. But how does the opposite direction work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-11-21T15:19:42.647" Title="Why is $\mathcal E$ a Montel-space?" Tags="&lt;functional-analysis&gt;&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11205" PostTypeId="2" ParentId="11199" CreationDate="2010-11-21T12:36:49.420" Score="2" ViewCount="0" Body="&lt;p&gt;Here is the same answer with a variation on the calculations &#xA;$\begin{eqnarray*}&#xD;&#xA;S_{n} &amp;amp;=&amp;amp;\sum_{k=0}^{n}\dfrac{n-k+1}{(n+1)(n+2)(n-k+2)} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\dfrac{n-k+1}{n-k+2}\qquad (n+1)(n+2)%&#xD;&#xA;\text{ is independent of }k \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\frac{m-1}{m}\qquad \text{substitution }m=n-k+2 \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\left( 1-\dfrac{1}{m}\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{1}{(n+1)(n+2)}\left(\sum_{m=2}^{n+2}1\right)-\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}%&#xD;&#xA;\dfrac{1}{m} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{n+2-2+1}{(n+1)(n+2)}-\dfrac{1}{(n+1)(n+2)}\left( \sum_{i=1}^{n+2}%&#xD;&#xA;\dfrac{1}{i}\right) +\dfrac{1}{(n+1)(n+2)} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{n+1}{(n+1)(n+2)}-\dfrac{H_{n+2}}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)} \\&#xD;&#xA;&amp;amp;=&amp;amp;\dfrac{n+2-H_{n+2}}{(n+1)(n+2)},&#xD;&#xA;\end{eqnarray*}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_{n+2}=\displaystyle\sum_{i=1}^{n+2}\dfrac{1}{i}.$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-21T21:58:56.970" LastActivityDate="2010-11-21T21:58:56.970" CommentCount="1" />
  <row Id="11206" PostTypeId="1" AcceptedAnswerId="11254" CreationDate="2010-11-21T12:48:20.977" Score="0" ViewCount="100" Body="&lt;p&gt;How to find center of ellipse from two points(these are just points on the ellipse, not related to foci), and two radiuses (rx and ry, from standard definition of the ellipse x^2/rx^2 + y^2/ry^2=1) of that ellipse? (there will be two centers, I assume)&lt;/p&gt;&#xA;" OwnerUserId="3777" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T00:16:41.210" LastActivityDate="2010-11-22T00:16:41.210" Title="Calculating center of the ellipse" Tags="&lt;geometry&gt;&lt;conic-sections&gt;&lt;ellipse&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11207" PostTypeId="2" ParentId="11189" CreationDate="2010-11-21T13:09:54.293" Score="0" ViewCount="0" Body="&lt;p&gt;You want to apply AM-GM to some of the terms on the LHS to get some of the terms on the RHS.  A naive way to do this such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b^6 + c^6 + 2a^2 b^3 c \ge 4 a b^3 c^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;uses too many of the $11$ terms relative to the $40$ terms and only proves the weaker inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^6 + b^6 + c^6 + abc(ab^2 + bc^2 + ca^2) \ge 2abc(a^2 b + b^2 c + c^2 a)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after cyclic summation.  A slightly better combination of terms is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5a^6 + c^6 + 12a^2 b^3 c \ge 18 a^3 b^2 c$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this still uses too many of the $11$ terms relative to the $40$ terms and proves the inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^6 + b^6 + c^6 + 2abc(ab^2 + bc^2 + ca^2) \ge 3abc(a^2 b + b^2 c + c^2 a)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after cyclic summation.  You get a strong enough result by mixing some of the $40$ terms with each other to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4a^6 + 12a^2 b^3 c + 4a^3 b c^2 \ge 20 a^3 b^2 c$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which proves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^6 + b^6 + c^6 + 4abc(ab^2 + bc^2 + ca^2) \ge 5abc(a^2 b + b^2 c + c^2 a).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now multiply by $10$ and apply AM-GM to the remaining terms to conclude.  The question of whether one can do better than this using AM-GM is a geometric question about using convex linear combinations of the vectors $(6, 0, 0), (0, 6, 0), (0, 0, 6), (2, 3, 1), (3, 1, 2), (1, 2, 3)$ to obtain the vectors $(3, 2, 1), (2, 1, 3), (1, 3, 2)$ and one might be able to prove that the above result is optimal.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T13:09:54.293" />
  <row Id="11208" PostTypeId="1" CreationDate="2010-11-21T13:13:22.837" Score="0" ViewCount="119" Body="&lt;p&gt;By using the exponential formula of $y=ab^t$, we can find doubling time by $2^{t/T}= b^t$; another formula for the exponential is $y=ae^{kt}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be correct and also I am quite confused with how can we estimate the doubling time or halving time using graphs. I am quite confused; please help. Show me in both graphical form and using a formula. Giving any example would be fine as I am not quite a mathematician and can hardly make up a proper example, but please help give me the clear detail and idea. &lt;/p&gt;&#xA;" OwnerUserId="3732" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-22T23:15:31.377" LastActivityDate="2010-11-22T23:15:31.377" Title="Finding doubling time and halving time using graphs and exponential formula" Tags="&lt;algebra-precalculus&gt;" CommentCount="1" />
  <row Id="11209" PostTypeId="2" ParentId="11189" CreationDate="2010-11-21T13:27:22.837" Score="1" ViewCount="0" Body="&lt;p&gt;I don't have enough karma to comment, so I'll just drop this here:&#xA;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154225&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154225&lt;/a&gt;&#xA;discusses the problem.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-21T13:27:22.837" CommentCount="0" />
  <row Id="11210" PostTypeId="1" AcceptedAnswerId="11228" CreationDate="2010-11-21T13:34:30.273" Score="2" ViewCount="180" Body="&lt;p&gt;Investigating this sequence led me to the following conclusions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $a_{1}\geq 2$ then $a_{n}\geq 2$ and $a_{n}$ is decreasing, thus it converges and is bounded.&lt;/li&gt;&#xA;&lt;li&gt;if $-1\lt a_{1} \lt 2$ then $a_{n}$ is increasing, thus it converges and is bounded.&lt;/li&gt;&#xA;&lt;li&gt;if $a_{1}=-1$ or $a_{1}=2$ the sequence is constant and is bounded.&lt;/li&gt;&#xA;&lt;li&gt;if $a_{1}\lt -1$ then for some $k,~a_{k}\gt 0$ and then we're in one of the previous options.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But this seems overly complicated and very long to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of trying the following argument: define $f(x)=\frac{4x+2}{x+3}$. We see that $\displaystyle \lim_{x\to \infty}f(x)=4$ and $f(x)$ is continuous in the interval $[0, \infty)$ therefore  it's bounded there. Then we show that if $a_{1}&amp;lt;0$ then $\exists N,~\forall n\gt N,~a_{n}&amp;gt;0$ which gives us $a_{n+1}=f(a_{n})$ is bounded.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-11-22T10:31:59.510" Title="How can I show the sequence $a_{n+1}=\frac{4a_{n}+2}{a_{n}+3}$ is bounded for every $a_{1}\in \mathbb{R}$?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="11211" PostTypeId="2" ParentId="11150" CreationDate="2010-11-21T13:54:35.360" Score="5" ViewCount="0" Body="&lt;p&gt;$0^{0}$ is just one instance of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Empty_product&quot; rel=&quot;nofollow&quot;&gt;empty product&lt;/a&gt;, which means it is the multiplicative identity 1.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-11-21T13:54:35.360" />
  <row Id="11212" PostTypeId="1" CreationDate="2010-11-21T13:55:55.343" Score="1" ViewCount="98" Body="&lt;p&gt;I am a student taking engineering course and wish to learn more about Fourier Transforms. It seems very useful. Would highly appreciate it if anyone could advise me where to start.&lt;/p&gt;&#xA;" OwnerUserId="3778" LastActivityDate="2010-11-22T13:28:11.667" Title="Recommended books/links for Fourier Transform beginners?" Tags="&lt;fourier-analysis&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="11213" PostTypeId="2" ParentId="11212" CreationDate="2010-11-21T14:05:07.953" Score="1" ViewCount="0" Body="&lt;p&gt;I first learned Fourier techniques from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073039381&quot; rel=&quot;nofollow&quot;&gt;Bracewell's &lt;em&gt;The Fourier Transform and Its Applications&lt;/em&gt;&lt;/a&gt;, but I'm digging the non-conventional treatment in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3527404414&quot; rel=&quot;nofollow&quot;&gt;Meikle's &lt;em&gt;A New Twist to Fourier Transforms&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-21T14:05:07.953" CommentCount="5" />
  <row Id="11214" PostTypeId="1" CreationDate="2010-11-21T14:06:30.120" Score="2" ViewCount="80" Body="&lt;p&gt;My consideration might be total nonsense (as a high school student, I lack the mathematical knowledge to really check my idea), but I was just wondering whether one could find a continuous generalization of the Taylor series comparable to the continous Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like the Fourier transform decomposes a function $f(t)$ into a sum of trigonometric functions with a result $\hat{f}(\omega)=\mathcal{F}(f)(\omega)$ with the domain of &lt;em&gt;frequencies&lt;/em&gt;, could one define a Taylor transform $\mathcal{T}(f)$ that would operate on $n$th derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the trigonometric functions in Fourier transform, $f$ would be decomposed into a sum of polynomials of form $\frac{1}{n!}x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it a valid operation to generalize the Taylor series for continuous values of $n$ (i.e. using integrals instead of sums)? My naive approach would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t) = \intop_0^{\infty} \mathcal{T}(f)(n)\cdot \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact, the transform would define (or at least require) &lt;strong&gt;non-integer derivatives&lt;/strong&gt; of a function. Do these exist (like a non-integer &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_iteration&quot; rel=&quot;nofollow&quot;&gt;iterates&lt;/a&gt; of functions can be defined too)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;Assuming that $\frac{\mathrm{d}^ne^x}{\mathrm{d}x^n}(0) = 1$ for any $x\in \mathbb{R}$, I'd come up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^t = \intop_0^{\infty} \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$ as a continuous version of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $$\mathcal{T}(e^t)(n) = 1,\,n\in\mathbb{R}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a transform like I explained possible or - in case it is - even somehow useful?&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-21T14:50:31.750" Title="Is there a continuous Taylor/MacLaurin transform (like the Fourier transform)?" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="11215" PostTypeId="2" ParentId="11210" CreationDate="2010-11-21T14:07:09.790" Score="0" ViewCount="0" Body="&lt;p&gt;What about $a_1 = -3$? The function don't even exists there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, your argument is valid: if $a_n = f(n)$ and $\lim_{x\to \infty} f(x) = L$ then $\lim_{n\to \infty} a_n = L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You never care about the first terms of the sequence, because when you've convergence to the infinity, you can just pick a maximum value and set it as limit. For example, if your sequence converges to 4, you can set a bound as $\max (4, a_1, a_2, ..., a_k)$ for any k you like.&lt;/p&gt;&#xA;" OwnerUserId="3635" LastActivityDate="2010-11-21T14:07:09.790" CommentCount="1" />
  <row Id="11216" PostTypeId="2" ParentId="11212" CreationDate="2010-11-21T14:11:30.970" Score="1" ViewCount="0" Body="&lt;p&gt;For a general engineering perspective, Erwin Kreyszig's  book &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470458364&quot; rel=&quot;nofollow&quot;&gt;Advanced Engineering Mathematics&lt;/a&gt;&quot; would have some chapters on Fourier and other integral transforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more mathematical approach, but still with applications in mind, Sneddon's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486685225&quot; rel=&quot;nofollow&quot;&gt;Fourier Transforms&lt;/a&gt; is recommended. It has a lot of physics applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book of Taub and Schilling on &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0070629552&quot; rel=&quot;nofollow&quot;&gt;Principles of Communication Systems&lt;/a&gt; is very good from an electrical engineering point of view. I particularly liked this last book.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-11-21T14:19:20.913" LastActivityDate="2010-11-21T14:19:20.913" />
  <row Id="11217" PostTypeId="1" AcceptedAnswerId="11221" CreationDate="2010-11-21T14:13:32.950" Score="1" ViewCount="108" Body="&lt;p&gt;I am trying to simplify this expression by as usual the expansion way,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl(\frac{ 1+x^2}{1-x^2}\biggr)^2 = \frac{1}{1-y^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some steps I am getting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 - y^2 - 2x^2y^2 - x^4y^2 = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer suggested in my module is $x^2y = 2x - y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the answer to be correct I think what I should get is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 - y^2 - 4xy - x^4y^2 = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly I am doing wrong ? I tried to find an error in my solution, but unable to spot any(yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; For reference I am adding the other options mentioned the question (and now the question too):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $4\biggl[\frac{x^2}{1} + \frac{x^{6}}{3}+ \frac{x^{10}}{5} + \cdots \biggr] = y^2 + \frac{y^4}{2} + \frac{y^6}{3} + \cdots $, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2y = 2x+y \text{ or } x = 2y^2 - 1 \text{ or } x^2y = 2x + y^2$$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-21T14:42:05.480" LastActivityDate="2010-11-21T17:07:56.117" Title="Simplification : $\biggl(\frac{ 1+x^2}{1-x^2}\biggr)^2 = \frac{1}{1-y^2}$" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;elementary&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="11218" PostTypeId="1" AcceptedAnswerId="11219" CreationDate="2010-11-21T14:19:41.807" Score="0" ViewCount="78" Body="&lt;p&gt;How to prove :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln \biggl(\frac{x+1}{x-1}\biggr) = 2\biggl[\frac{1}{x} +  \frac{1}{3x^3} +  \frac{1}{5x^5} + \cdots \biggr]$$ where $|x| \gt 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to get how to proceed on this one ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-21T14:24:48.543" LastActivityDate="2010-11-22T01:19:06.663" Title="A proof  based on Logarithmic series:" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11219" PostTypeId="2" ParentId="11218" CreationDate="2010-11-21T14:30:39.450" Score="3" ViewCount="0" Body="&lt;p&gt;Note that $\frac{x+1}{x-1} = \frac{1 + 1/x }{1 - 1/x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the series for $\log (1+x)$ converges absolutely when $|x| &amp;lt;1$. Therefore it is permissible to rearrange the series in any way you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that $\log \left(\frac{1+1/x}{1-1/x}\right) = \log (1+1/x) - \log (1-1/x)$. Expand both logarithmic series and group the terms.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T14:30:39.450" CommentCount="2" />
  <row Id="11220" PostTypeId="1" AcceptedAnswerId="11225" CreationDate="2010-11-21T14:37:18.007" Score="2" ViewCount="76" Body="&lt;p&gt;Let $f_n,f \in L^p$ and $f_n \to f$ uniformly. Further let $g \in L^q$. Now, clearly $\int (f_n - f) g$ is defined. How do I show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int |f_n - f||g|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be made arbitrary small if I choose $n$ large enough? If we just naively apply Hölder we will not be able to make it small but just finite. Further, we cannot just replace $|f_n - f|$ by the bound &quot;$\epsilon$&quot; because that is larger than this. Does someone have a hint?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-21T15:05:14.313" Title="Uniform convergence and $L^p$-convergence." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11221" PostTypeId="2" ParentId="11217" CreationDate="2010-11-21T14:49:28.820" Score="4" ViewCount="0" Body="&lt;p&gt;My interpretation is that you want to know the relation between $y$ and $x$&#xD;&#xA;so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \dfrac{1+x^{2}}{1-x^{2}}\right) ^{2}=\dfrac{1}{1-y^{2}}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My detailed computation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\left( 1+x^{2}\right) ^{2}}{\left( 1-x^{2}\right) ^{2}}= \dfrac{x^{4}+2x^{2}+1}{x^{4}-2x^{2}+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( \dfrac{1+x^{2}}{1-x^{2}}\right) ^{2}=\dfrac{1}{1-y^{2}}\Leftrightarrow &#xD;&#xA;\dfrac{x^{4}+2x^{2}+1}{x^{4}-2x^{2}+1}=\dfrac{1}{1-y^{2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow \left( x^{4}+2x^{2}+1\right) \left( 1-y^{2}\right)&#xD;&#xA;=x^{4}-2x^{2}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left( x^{4}+2x^{2}+1\right) \left( 1-y^{2}\right)&#xD;&#xA;=2x^{2}-y^{2}+x^{4}-2x^{2}y^{2}-x^{4}y^{2}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2x^{2}-y^{2}+x^{4}-2x^{2}y^{2}-x^{4}y^{2}+1=x^{4}-2x^{2}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow 4x^{2}-y^{2}-2x^{2}y^{2}-x^{4}y^{2}=0\qquad\text{the same as in the question}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+2x^{2}+x^{4})y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+x^{2})^{2}y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+x^{2})y=\pm 2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the positive root, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y+x^{2}y=2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{2}y=2x-y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added: Or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow 4x^{2}-y^{2}-2x^{2}y^{2}-x^{4}y^{2}=0\qquad\text{the same as in the question}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+2x^{2}+x^{4})y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow (1+x^{2})^{2}y^{2}=4x^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the negative root gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+x^{2})y=-2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Leftrightarrow y+x^{2}y=-2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{2}y=-2x-y$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-21T15:30:51.493" LastActivityDate="2010-11-21T15:30:51.493" CommentCount="6" />
  <row Id="11222" PostTypeId="2" ParentId="11214" CreationDate="2010-11-21T14:50:31.750" Score="1" ViewCount="0" Body="&lt;p&gt;This is a very interesting question. I am acquainted only very little with these matters. Hopefully other people can fill you in better. There is indeed some analogy of the type you are seeking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fourier series of a periodic function captures all the information for a function(signal) over a fixed period. When you determine the Fourier coefficients, you compute suitable integrals over a fixed period and you capture all the frequency information for a certain period. Fourier transforms capture frequency domain information for non-periodic signals as well. If you look at the Fourier transform of a function, you would in general be at a loss to guess the function behavior at a particular point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, note that Taylor Series is giving information about a function specifically at a point. Taylor series captures very very &lt;em&gt;local&lt;/em&gt; information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems there is such a tradeoff always happening. If you want more information on the frequency domain, you have to sacrifice information on the time domain, and vice versa. On the one extreme we have the Taylor series and the other the Fourier theory. It seems there are a range of transforms called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wavelet_transform&quot; rel=&quot;nofollow&quot;&gt;wavelet transforms&lt;/a&gt; which are in between these two extremes and captures a bit of both frequency domain and time domain information without completely sacrificing one of these. It seems for such transforms for precisely formulating such a tradeoff, there is a certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainly_principle&quot; rel=&quot;nofollow&quot;&gt;uncertainty principle&lt;/a&gt; which is a precise theorem in Harmonic analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T14:50:31.750" CommentCount="1" />
  <row Id="11223" PostTypeId="1" CreationDate="2010-11-21T14:57:55.093" Score="2" ViewCount="170" Body="&lt;p&gt;There are 1000 light bulbs and 1000 tutors. All light bulbs are off. Tutor 1 goes flipping light bulb 1,2,3,4... tutor 2 then flips 2,4,6,8...tutor 3 then 3,6,9...etc until all 1000 tutors have done this.  What is the status of light bulb 25? 93? 576?  Is there a general solution to find out the status of any light bulb?  How many light bulbs are on after all 1000 tutors have gone by?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not homework but a UK university interview question. Please help.&lt;/p&gt;&#xA;" OwnerUserId="3780" LastActivityDate="2010-11-25T07:20:10.310" Title="A tedious puzzle (but not homework)" Tags="&lt;sequences-and-series&gt;&lt;algebra&gt;&lt;puzzle&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="11224" PostTypeId="2" ParentId="11223" CreationDate="2010-11-21T15:02:24.790" Score="5" ViewCount="0" Body="&lt;p&gt;Hint: Bulb number $N$ will be flipped by all tutors with number $m$ such that $m$ divides $N$ and no others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e.g. light bulb number $36$ will be flipped by tutors numbered $1,2,3,4,6,9,12,18,36$, so $9$ times. Since every bulb was initially off, an odd number of switches will turn it on, which is what will happen with number $36$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-21T15:26:29.957" LastActivityDate="2010-11-21T15:26:29.957" />
  <row Id="11225" PostTypeId="2" ParentId="11220" CreationDate="2010-11-21T15:05:14.313" Score="3" ViewCount="0" Body="&lt;p&gt;Outside of a sufficiently large compact set, $\|g\|_q &amp;lt;\epsilon$. Use Holder naively, and on that portion the integral is bounded above by $(\|f_n\|_p + \|f\|_p)\epsilon$. On the inside of the compact set $\|g\|_1 \leq C \|g\|_q$ where $C$ is given by the volume of the compact set. Then you can replace $|f_n - f|$ by $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this requires $\|f_n\|_p$ be uniformly bounded, which would be true if you additionally assume $f_n \to f$ in $L^p$. If you don't assume that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider over $\mathbb{R}$, take $f_n = \frac1n \chi_{[2^n,2^{n+1}]}$, where $\chi$ denotes the characteristic function. $f_n \to 0$ uniformly as a sequence of functions (i.e. $f_n\to f$ in $L^\infty$). For this sequence $\|f_n\|_p = \frac1n \cdot 2^{n/p} \nearrow \infty$ (unless $p = \infty$). (Note that since it is not a bounded sequence, you cannot assert that it has a weakly converging subsequence.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take $g$ be the function such that $g(x) = 2^{-n/(q-\delta)}$ if $2^n &amp;lt; |x| \leq 2^{n+1}$. By construction it is a function in $L^q$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $|f_n - f| |g| = \frac1n \cdot 2^{-n / (q - \delta)} \chi_{[2^n,2^{n+1}]}$. If you integrate it, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac1n \cdot 2^{n( 1 - \frac{1}{q - \delta}) }$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unless $q = 1$ and $p = \infty$, you can always choose $\delta$ sufficiently small such that the exponent in the integrated expression is positive, and provide a counterexample to what you want proved. (And of course, in the case where $p = \infty$, you can just replace $|f-f_n|$ by $\epsilon$ and be done with it.  &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-21T15:05:14.313" CommentCount="4" />
  <row Id="11226" PostTypeId="2" ParentId="11204" CreationDate="2010-11-21T15:11:55.077" Score="1" ViewCount="0" Body="&lt;p&gt;This is far too long to be a comment, so it's an answer instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think anyone usually gives $C^\infty(\Omega)$ another name. Now I presume you've already shown the space to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Barrelled_space&quot; rel=&quot;nofollow&quot;&gt;barreled&lt;/a&gt;, so we just need $X \subset C^\infty(\Omega)$ compact iff $X$ is closed and bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that your sufficient family of seminorms is too large. Because $\Omega$ is $\sigma$-compact, we can thin out the family by taking a countable family of compact sets $K_j$ that exhaust $\Omega$, i.e. $K_j \subsetneq K_{j+1}$ and $\cup_j K_j = \Omega$, so &#xA;$$ \Vert f \Vert_{i,j} = \max_{|\alpha|\leq i} \; \sup_{x \in K_j} |\partial^\alpha f(x)|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a countable family of seminorms that generate the same topology. From these we can define a metric on $C^\infty(\Omega)$ as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d(f,g) = \sum_{i=0}^\infty\; \sup_{j \in \mathbb{N}} \frac{\Vert f-g \Vert_{i,j}}{1 + \Vert f-g \Vert_{i,j}} 2^{-i}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to verify that this metric generates the same topology as the family of seminorms and that the metric is complete and translation-invariant. Thus $C^\infty(\Omega)$ is a Fréchet space and we know that the compact subsets of complete metric spaces are precisely those that are closed and totally bounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the strategy is clear. First show that $X$ bounded w.r.t. the seminorms implies $X$ bounded w.r.t. the metric. Then show that $X$ bounded w.r.t. the metric implies that $X$ is totally bounded.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-21T15:19:42.647" LastActivityDate="2010-11-21T15:19:42.647" CommentCount="1" />
  <row Id="11227" PostTypeId="1" AcceptedAnswerId="11289" CreationDate="2010-11-21T15:25:38.430" Score="2" ViewCount="123" Body="&lt;p&gt;Let $\mathbb{Z}_p$ be te ring of p-adic integers and let $T$,$T'$ be two free $\mathbb{Z}_p$-module with a continuous action of $G_{\mathbb{Q}_p}$ (the absolute Galois group of $\mathbb{Q}_p$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to see that if $T$ and $T&amp;#39;$ are of rank one, then there is a non zero $G_{\mathbb{Q}_p}$-equivariant map $T \to T&amp;#39; \otimes \mathbb{Q}_p / \mathbb{Z}_p$ if and only if $T \simeq T&amp;#39;$ mod $p$ (just write down explicitely the characters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true in a more general case, namely when $T&amp;#39;$ and $T&amp;#39;$ mod $p$ are irreducible and $T$ is any $\mathbb{Z}_p$-representation ?&lt;/p&gt;&#xA;" OwnerUserId="3781" LastActivityDate="2010-11-22T16:29:04.707" Title="torsion representation" Tags="&lt;number-theory&gt;&lt;representation-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11228" PostTypeId="2" ParentId="11210" CreationDate="2010-11-21T15:28:01.337" Score="2" ViewCount="0" Body="&lt;p&gt;You can formalize what you're doing by showing that $-1 \leq a_{n+1} &amp;lt; 4$ whenever $a_n \geq -1$. So once you get an $n$ such that $a_n \geq -1$ you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some $a_n &amp;lt; -3$, your recursion gives $a_{n+1} &amp;gt; -1$ and you're in the above situation. The only remaining option is that $-3 &amp;lt; a_n &amp;lt; -1$ for all $n$... but then the sequence is still bounded. So you're done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-21T15:54:16.287" LastActivityDate="2010-11-21T15:54:16.287" CommentCount="2" />
  <row Id="11229" PostTypeId="2" ParentId="11210" CreationDate="2010-11-21T15:32:09.963" Score="0" ViewCount="0" Body="&lt;p&gt;Your last argument is sound.  If you know that the terms are bounded over $[0,\infty)$, (or any interval, for that matter) that iteration doesn't take it out of that interval, and that all starting values eventually enter that interval, the terms are bounded.  As AkiRoss says, prepending terms cannot change the bounded nature, it only changes the bound.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-21T15:32:09.963" />
  <row Id="11230" PostTypeId="2" ParentId="154" CreationDate="2010-11-21T15:59:05.797" Score="1" ViewCount="0" Body="&lt;p&gt;It may help to think of negative numbers as something other than &quot;negative.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept that helped me was to think of -1 as the opposite of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the context was distance, for example, and I had a value of 1, then I went one unit in a direction.  If, on the other hand, I had a value of -1, then I'm traveling in the opposite direction of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In calculus and series work, it will become more clear that this is a terminology problem most people have.  So whether you're dealing with space-time or voltages or something else, the negative symbol is just a relative thing, not an absolute value.  In other words, don't think of -5 as negative 5 but as something opposite of something else... &lt;/p&gt;&#xA;" OwnerUserId="3783" LastActivityDate="2010-11-21T15:59:05.797" />
  <row Id="11231" PostTypeId="1" AcceptedAnswerId="11316" CreationDate="2010-11-21T16:03:13.343" Score="2" ViewCount="132" Body="&lt;p&gt;Let's say you have a single-lane road (single in this case meaning single lane in each direction, or what you could also call two-lane). Let's say you have a random number of vehicles on a given 10-mile stretch of road. The speed limit is 45 mph, but some people go faster and some go slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any math that describes the likelihood that faster vehicles will overtake slower vehicles at &quot;choke points&quot; (points where oncoming traffic makes passing impossible)? (Assume a straight, flat road where safe passing is always allowed.) EDIT: Choke points are points where oncoming cars make it impossible to pass without the overtaking vehicle having to slow down and wait for the oncoming car (or cars) to pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the answer has to be some kind of curve, because if there is only one vehicle on the road the probability is 0 that a choke point will be encountered, and if the road is completely filled with vehicles in both directions (or even one) the probability is 1 that a choke point will be encountered at any given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a stupid question. But it seems like there should be a way to express this mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2: Limiting the scope of the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Mike Spivey, and I hope this helps narrow the problem so that may be in some way answerable. If not, I'll just declare defeat and retire from the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that we have two cars going in one direction: a slow vehicle (Vehicle A) traveling at 30mph and a faster vehicle (Vehicle B) traveling at 60mph. At the start of the problem the A is somewhere between mile 2 and mile 4 on the 10-mile road, and B is at mile 0. Additionally, there will be four oncoming vehicles randomly distributed, and these vehicles are randomly traveling at 30 or 60mph. The overtaking vehicle requires .5 miles of &quot;clear&quot; space to pass safely if the oncoming vehicle is traveling at 60mph, and .25 if it is traveling at 30mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume further that the car lengths do not matter, but that a &quot;choke point&quot; (or bottleneck, if you prefer) is not reached until B gets within .05 miles of the rear bumper of A, at which point B has to pass A or decelerate. We will call the latter condition a &quot;deceleration event,&quot; (DE) and I would like to be able to calculate the probability that at least one DE will occur for Vehicle A along the 10-mile route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this clarification is still insufficient to render the problem solvable, but I appreciate anyone taking the time to consider it.&lt;/p&gt;&#xA;" OwnerUserId="3782" LastEditorUserId="3782" LastEditorDisplayName="" LastEditDate="2010-11-22T13:58:19.363" LastActivityDate="2010-11-22T13:58:19.363" Title="Probability in single-lane traffic flow: What are the odds of &quot;choke points&quot; being encountered?" Tags="&lt;probability&gt;&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11232" PostTypeId="1" AcceptedAnswerId="11281" CreationDate="2010-11-21T16:14:14.313" Score="13" ViewCount="166" Body="&lt;p&gt;Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be bijective map with following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $f|_{\mathbb{Q}^2}=id$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Image of any line under map $f$ is again a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right that $f=id$?&lt;/p&gt;&#xA;" OwnerUserId="3784" LastActivityDate="2010-11-21T21:23:42.827" Title="Almost identical map" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="11233" PostTypeId="1" CreationDate="2010-11-21T16:21:31.667" Score="0" ViewCount="117" Body="&lt;p&gt;For a 2D Gaussian distribution with&lt;br&gt;&#xA;$$ \mu = \begin{pmatrix} \mu_x \\ \mu_y \end{pmatrix}, \quad \Sigma = \begin{pmatrix} \sigma_x^2 &amp;amp; \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y &amp;amp; \sigma_y^2 \end{pmatrix},&#xD;&#xA;$$&#xD;&#xA;its probability density function is&#xA;$$&#xD;&#xA;    f(x,y) = \frac{1}{2 \pi \sigma_x \sigma_y \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_x)^2}{\sigma_x^2} + \frac{(y-\mu_y)^2}{\sigma_y^2} - \frac{2\rho(x-\mu_x)(y-\mu_y)}{\sigma_x \sigma_y} \right] \right), &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is also a similarly clean formula for 3D Gaussian distribution density? What is it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I ask is after taking the inverse of the covariance matrix, if the density has a clean form just as in 2D case?&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-21T17:10:38.580" LastActivityDate="2010-11-21T17:10:38.580" Title=" density of 3D Gaussian distribution" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="11235" PostTypeId="2" ParentId="11218" CreationDate="2010-11-21T16:36:14.017" Score="2" ViewCount="0" Body="&lt;p&gt;If $|x|&amp;gt;1$ then $\frac{1}{|x|}&amp;lt;1$, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(\frac{x+1}{x-1}\right)=\ln\left(\frac{x(1+\frac{1}{x})}{x(1-\frac{1}{x})}\right)=\ln\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)=\ln(1+t)-\ln(1-t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $t=\displaystyle\frac{1}{x}&amp;lt;1$, then $|t|&amp;lt;1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(1+t)=t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(1-t)=-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(1+t)-\ln(1-t)=\left(t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...\right)-\left(-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...\right)=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\left(t+\frac{t^{^3}}{3}+\frac{t^{5}}{5}...\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $t=\displaystyle\frac{1}{x}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left(\frac{x+1}{x-1}\right)=2\left(\frac{1}{x}+\frac{1}{3x^{^3}}+\frac{1}{5x^{^5}}+\frac{1}{7x^{^7}}...\right)$$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-22T01:19:06.663" LastActivityDate="2010-11-22T01:19:06.663" />
  <row Id="11236" PostTypeId="1" CreationDate="2010-11-21T16:49:02.163" Score="0" ViewCount="180" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim\limits_{x\rightarrow 0 } \frac{x}{ln x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was wrong, I got a big red wrong! Why doesn't L'Hopital work on this one? &#xA;The problem is that ln is not defined for 0. It needs to be rewritten?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks to everyone helping me out with my homework, due anxiety I'm not able attend the class workshops.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I did get the correct limit ($0$), but that was a coincidence.  &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T21:28:11.830" LastActivityDate="2010-11-21T21:28:11.830" Title="Another limit task, x over ln x, L'Hopital won't do it? " Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="11237" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T16:50:56.887" Score="1" ViewCount="0" Body="&lt;p&gt;I've looked at many set theory books and I think the best one for a beginner is one called Classic Set Theory by Goldrei.  It goes through loads of examples and is designed for self-study.&lt;/p&gt;&#xA;" OwnerUserId="3785" LastActivityDate="2010-11-21T16:50:56.887" />
  <row Id="11238" PostTypeId="2" ParentId="11223" CreationDate="2010-11-21T16:51:44.733" Score="1" ViewCount="0" Body="&lt;p&gt;You need to factorise each lightbulb's number. If it has an odd number of divisors then it is ON. If it has an even number of divisors it is OFF. I'm sorry I don't have a general n-th rule though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question, 25 is ON, 93 is OFF and 576 is ON.&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-21T16:51:44.733" CommentCount="1" />
  <row Id="11239" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T16:55:58.533" Score="6" ViewCount="0" Body="&lt;p&gt;L'Hopital's rule is best forgotten about. Of course as $x\to0^+$,&#xA;$\log x\to-\infty$ and so $1/\log x\to0$. &lt;em&gt;A fortiori&lt;/em&gt;&#xA;$x/\log x\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better problem is to find $\lim_{x\to0^+}x\log x$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-21T16:55:58.533" CommentCount="3" />
  <row Id="11240" PostTypeId="2" ParentId="11217" CreationDate="2010-11-21T17:00:56.377" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\ 0 \ \ = \ \ (y^2-1)\ (1+x^2)^2 + (1 - x^2)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\ = \ \ y^2\:(1+x^2)^2 - 4\:x^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad\quad\quad\ =\ \ (y\:(1+x^2)-2\:x)\ \ (y\:(1+x^2)+2\:x)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-21T17:07:56.117" LastActivityDate="2010-11-21T17:07:56.117" />
  <row Id="11241" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T17:02:58.813" Score="1" ViewCount="0" Body="&lt;p&gt;$\frac{0}{\infty}$  is not indeterminate form, is zero, then $\lim\limits_{ x \rightarrow 0^+ }\frac{x}{lnx}=0$ but is a right limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But L'Hôpital's rule applies..&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-21T21:26:44.843" LastActivityDate="2010-11-21T21:26:44.843" CommentCount="0" />
  <row Id="11242" PostTypeId="2" ParentId="11223" CreationDate="2010-11-21T17:03:22.567" Score="2" ViewCount="0" Body="&lt;p&gt;Each number that is a square will be ON.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-21T17:03:22.567" />
  <row Id="11243" PostTypeId="2" ParentId="11233" CreationDate="2010-11-21T17:05:18.977" Score="4" ViewCount="0" Body="&lt;p&gt;There is a standard, general formula for the density of the joint normal (or multivariate normal) distrubution of dimension $n$, provided that the ($n \times n$) covariance matrix $\Sigma$ is non-singular (see, e.g., &lt;a href=&quot;http://planetmath.org/encyclopedia/JointNormalDistribution.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Multivariate_normal_distribution&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;). In particular, you can apply for $n=3$. When the covariance matrix is singular, the distribution is expressed in terms of the characteristic function.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-21T17:05:18.977" CommentCount="14" />
  <row Id="11244" PostTypeId="1" AcceptedAnswerId="11258" CreationDate="2010-11-21T17:20:09.923" Score="2" ViewCount="73" Body="&lt;p&gt;Let $\hat{\mathbb{C}} = \mathbb{C} \cup \{ \infty \}$ denote the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere&quot; rel=&quot;nofollow&quot;&gt;Riemann Sphere&lt;/a&gt;. While reading the wikipedia article about it I found a passage that said that &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sphere#Rational_functions&quot; rel=&quot;nofollow&quot;&gt;every rational function on the complex plane can be extended to a continuous function on the Riemann Sphere&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular construction is as follows, let $R(z) = \frac{f(z)}{g(z)} \in \mathbb{C}(z)$ be a rational function. Lets assume for simplicity that $f(z)$ and $g(z)$ share no common factor. Then for any point $a \in \mathbb{C}$ such that $g(a) = 0$ but $f(a) \neq 0$ we define $R(a) = \infty$. Also we define $R(\infty) := \lim_{z \to \infty} R(z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to quote wikipedia, with this definitions $R(z)$ becomes a continuous function from the Riemann Sphere to itself. The problem for me is that the article doesn't add any details as to how one may go about showing that in fact $R(z)$ is continuous. So my question is exactly that, to make it simple, if I have for instance $R(z) = \frac{z-1}{z+1}$ or even $R(z) = \frac{1}{z}$, how do I show that it is a continuous function on $\hat{\mathbb{C}}$? I think I can build up from an easy example such as this (assumming this is easy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm a little bit confused about how to interpret this continuity, how should I see the continuity on the Riemann Sphere?, does it involve an argument with stereographic projection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added in the Riemann Surface tag just in case they are involved, which I'm not sure. Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-21T18:03:04.010" Title="Continuity of Rational Functions on the Riemann Sphere $\hat{\mathbb{C}}$" Tags="&lt;complex-analysis&gt;&lt;riemann-surface&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11245" PostTypeId="1" CreationDate="2010-11-21T17:26:49.300" Score="0" ViewCount="29" Body="&lt;p&gt;Let $S$ be a graded ring, which is generated by finitely elements in $S_1$. Let $M$ be a graded $S$-module and $\tilde{M}$ the associated sheaf of modules on $Proj(S)$. Is it possible to determine its global sections? There is a canonical morphism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_0 \to \Gamma(Proj(S),\tilde{M})$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it's neither injective nor surjective in general.&lt;/p&gt;&#xA;" OwnerUserId="1650" LastActivityDate="2010-11-21T17:26:49.300" Title="global sections of associated module on Proj(S)" Tags="&lt;algebraic-geometry&gt;" CommentCount="3" />
  <row Id="11246" PostTypeId="1" CreationDate="2010-11-21T17:28:23.137" Score="2" ViewCount="168" Body="&lt;p&gt;How can we prove the following trigonometric identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$&lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-21T22:17:17.057" LastActivityDate="2010-11-21T22:17:17.057" Title="How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="11247" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T17:34:09.720" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\quad$ Solve it the same way you would solve $\rm\ f(x)\ =\ 1 + 2 \int f(x)\ dx\ $ but instead of $\rm\ d/dx\ $ apply $\rm\ \Delta_n\: f(n) = f(n+1) - f(n)\ $ to eliminate (invert) the $\rm\: \sum = \Delta^{-1}\:,\ $ i.e. employ the the &lt;strong&gt;fundamental theorem of difference calculus&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-21T17:34:09.720" />
  <row Id="11249" PostTypeId="2" ParentId="11246" CreationDate="2010-11-21T17:41:59.247" Score="2" ViewCount="0" Body="&lt;p&gt;Hi--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the solution in this page:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://natto.2ch.net/math/kako/1002/10029/1002903143.html&quot; rel=&quot;nofollow&quot;&gt;http://natto.2ch.net/math/kako/1002/10029/1002903143.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Translation of the page into  &lt;em&gt;English.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = tan (3π/11) +4 sin (2π/11) &#xA;and t = 3π/11 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;　　　11t = 3π &#xA;　⇔ 6t = 3π-5t &#xA;　⇒ sin (6t) = sin (3π-5t) of the both sides took ← sin &#xA;　⇔ 2sin (3t) cos (3t) = sin5t double angle formula 2 ← &#xA;　2 ⇔ {3sint-4 (sint) ^ 3} {4 (cost) ^ 3-3cost} = 16 (sint) ^ 5-20 (sint) ^ 3 +5 (sint)-angle 3 ←, five double angle formula &#xA;　2 ⇔ {3-4 (sint) ^ 2} {4 (cost) ^ 3-3cost} = 16 (sint) ^ 4-20 (sint) ^ 2 +5 ← (sint) divided by 0 ≠ &#xA;　⇔ 32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = ← 0 (sint) ^ 2 = 1 - (cost) ^ 2, x = cost of using organized &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus x = cos (3π/11) is &#xA;32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = 0 solution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2π/11) = {1 - (9 / 11)} π = (π-3t) more &#xA;I = tan (3π/11) +4 sin (2π/11) &#xA;　= Tant +4 sin (π-3t) &#xA;　= Tant +4 sin3t &#xA;　= (Sint / cost) +4 {3sint-4 (sint) ^ 3} &#xA;　= (Sint / cost) {16 (cost) ^ 3-4 (cost) +1} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ^ 2 = (sint / cost) ^ 2 {16 (cost) ^ 3-4 (cost) +1} ^ 2 &#xA;　= {(1 - (cost) ^ 2) / (cost) ^ 2} {16 (cost) ^ 3-4 (cost) +1} ^ 2 &#xA;　= {(1-x ^ 2) (16x ^ 3-4x +1) ^ 2} / x ^ ← 2 x = cost &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecule {(1-x ^ 2) (16x ^ 3-4x +1) ^ 2} a &#xA;{32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1} is divided by &#xA;← 2 11x ^ quotient remainder omitted &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-21T17:41:59.247" />
  <row Id="11250" PostTypeId="1" CreationDate="2010-11-21T17:43:53.780" Score="1" ViewCount="71" Body="&lt;p&gt;I was wondering if f(x)=O(x^(c+a)) for all a&gt;0 then is it necessarily true that f(x)=O((x^c)logx)?&#xA;I suspect it's not true but want to know why.  (I know the converse is true.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is much appreciated,&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3785" LastActivityDate="2010-11-21T19:50:41.800" Title="If f is big O of x^c+a for all a&gt;0 then is f big O of (x^c)logx?" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11251" PostTypeId="2" ParentId="11193" CreationDate="2010-11-21T17:46:36.320" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose you want to visualize an $8$-dimensional space. One possible approach(in fact the only one I know) is to first visualize an $n$-dimensional space in the form of $\mathbb R^n$, and then let $n = 8$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-21T17:46:36.320" CommunityOwnedDate="2010-11-21T17:46:36.320" CommentCount="3" />
  <row Id="11252" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T17:48:27.080" Score="2" ViewCount="0" Body="&lt;p&gt;As others have mentioned, you don't need L'Hospital's rule for this simple example. However, in fact, there is a general form of the rule that does apply to this case, and it does come in handy in some circumstances - so it is worth knowing. For an exposition see e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. E. Taylor, $\ $ L'Hospital's Rule&lt;br&gt;&#xA;Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2307183&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2307183&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. M. Ostrowski, $\ $ Note on the Bernoulli-L'Hospital Rule&lt;br&gt;&#xA;Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2318210&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/2318210&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-21T17:48:27.080" CommentCount="4" />
  <row Id="11254" PostTypeId="2" ParentId="11206" CreationDate="2010-11-21T17:50:39.583" Score="0" ViewCount="0" Body="&lt;p&gt;You seem to be asking about an ellipse whose axes are aligned with the $x$ and $y$ directions. Note that an ellipse which is rotated by an arbitrary angle is still an ellipse. But you cannot uniquely determine such an ellipse from the information given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two points $(x_1, y_1)$ and $(x_2, y_2)$ pass through an ellipse centred at $(x_0, y_0)$ with &lt;em&gt;semi-axes&lt;/em&gt; along $x$ and $y$ of length $r_x$ and $r_y$ respectively, then the points $(x_1/r_x, y_1/r_y)$ and $(x_2/r_x, y_2/r_y)$ pass through a circle of unit radius with centre $(x_0/r_x, y_0/r_y)$. So if you can find the centre of this circle, you can find the centre of the corresponding ellipse.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-21T17:50:39.583" />
  <row Id="11256" PostTypeId="2" ParentId="11250" CreationDate="2010-11-21T17:53:52.177" Score="7" ViewCount="0" Body="&lt;p&gt;Consider $f(x) = x^c (\log x)^b$, where $b&amp;gt;1$. This function is $O(x^{(c+a)})$ but it is not $O(x^c \log x)$. &lt;/p&gt;&#xA;" OwnerUserId="2641" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2010-11-21T18:00:50.357" LastActivityDate="2010-11-21T18:00:50.357" CommentCount="1" />
  <row Id="11257" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T17:55:28.443" Score="1" ViewCount="0" Body="&lt;p&gt;Using a spreadsheet, I note that $T(n)=3^{(n-1)}$  This is easily verified by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(1)=1=3^0$.&lt;br&gt;&#xA;Then if it is true up to $n$, $$T(n+1)=1+2\sum_{i=0}^{n-1}3^i=1+2\frac{3^n-1}{3-1}$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-21T17:55:28.443" />
  <row Id="11258" PostTypeId="2" ParentId="11244" CreationDate="2010-11-21T17:56:59.753" Score="2" ViewCount="0" Body="&lt;p&gt;One way of doing this is to use the sequential definition of continuity... this works because of the way the Riemann sphere has a metric (distance function) defined on it, inherited from the regular Euclidean distance function in 3-d. In this viewpoint, the statement that $\lim_{z \rightarrow a} f(z) = \infty$ is equivalent to saying $\lim_{z \rightarrow a} |f(z)| = \infty$. In the case that $a$ itself is infinity, then $\lim_{z \rightarrow \infty} f(z) = \infty$ becomes $\lim_{|z| \rightarrow \infty } |f(z)| = \infty$. In other words, for every $N$ there's an $M$ such that if $|z| &amp;gt; M$ then $|f(z)| &amp;gt; N$. Similarly, for some finite $z_0$ the statement $\lim_{z \rightarrow \infty} f(z) = z_0$ means $\lim_{|z| \rightarrow \infty } f(z) = z_0$; for every $\epsilon &amp;gt; 0$ there's an $N$ such that $|z| &amp;gt; N$ implies $|f(z) - z_0| &amp;lt; \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way many statements like the ones you're trying to prove become pretty routine.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-21T18:03:04.010" LastActivityDate="2010-11-21T18:03:04.010" CommentCount="4" />
  <row Id="11259" PostTypeId="1" AcceptedAnswerId="11265" CreationDate="2010-11-21T17:59:01.543" Score="1" ViewCount="56" Body="&lt;p&gt;Hi there,&#xA;I have two questions and I hope it is alright to ask them both at the same time. I'm currently trying to solve an online-exercise about differential calculus and the Taylor series and I'm having some trouble there.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f: (a,b) \rightarrow \mathbb{R}$ be differentiable at the point $x_0$. Which of the following statements is true?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$f(x) = f(x_0) + O(x-x_0) \;\; (x \rightarrow x_0)$&#xD;&lt;/li&gt;&#xA;  &lt;li&gt;$f(x) = f(x_0) + f&amp;#39;(x_0)(x-x_0) + O((x-x_0)^2) \;\; (x \rightarrow x_0)$&#xD;&lt;/li&gt;&#xA;  &lt;li&gt;$f(x) = f(x_0) + f&amp;#39;(x_0)(x-x_0) + o((x-x_0)) \;\; (x \rightarrow x_0)$&#xD;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure the third statement is true, because that's basically our definition of the Taylor series. However, I'm not quite sure about the other two statements, though I believe either both or none of them is true. When proving the Fundamental Theorem of Algebra, we once &quot;replaced&quot; $o(r^k)$ by $O(r^{k+1})$ which already then I found hard to understand. Can one always do this? In that case, all three statements would be correct...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x) = e^{\frac{-1}{|x|}} \cdot \cos(x^{-1})$ for $x \neq 0$ and $0$ for $x = 0$. $f$ has a local maximum or minimum at $x=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, I am not sure at all. I plotted the function and came to the conclusion that it has a minimum at $x=0$. However, we usually have a minimum when $f&amp;#39;(x_0) = 0$ and $f^{(n)} (x_0) &amp;gt; 0$, where $n$ is an even number. This function is arbitrarily often differentiable and I believe at none of the derivatives $f^{(n)} (x_0) \neq 0$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be delighted if someone could help me out here.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-21T18:12:11.697" Title="Landau symbols and a question about the turning point of an arbitrarily often differentiable function" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11260" PostTypeId="2" ParentId="11250" CreationDate="2010-11-21T18:00:21.500" Score="0" ViewCount="0" Body="&lt;p&gt;Consider $f(x) = (\log x)^2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-21T18:00:21.500" />
  <row Id="11261" PostTypeId="1" AcceptedAnswerId="11266" CreationDate="2010-11-21T18:04:44.143" Score="2" ViewCount="136" Body="&lt;p&gt;What is your recommended symbolic computation program/software for free and commercial respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are its strength and weakness? For example, efficiency, comprehensiveness, etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-21T23:23:16.343" LastActivityDate="2010-11-21T23:23:16.343" CommunityOwnedDate="2010-11-22T07:36:53.763" Title="symbolic computation program/software" Tags="&lt;math-software&gt;&lt;symbolic-computation&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="11262" PostTypeId="2" ParentId="11259" CreationDate="2010-11-21T18:05:30.197" Score="1" ViewCount="0" Body="&lt;p&gt;For your second question, there are points arbitrarily close to zero, such as $\frac{1}{2k\pi}$ for k a natural number, where f(x)&gt;0 and also points $\frac{1}{(2k+1)\pi}$where f(x)&amp;lt;0.  So zero is neither a maximum nor a minimum.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-21T18:05:30.197" />
  <row Id="11263" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T18:06:51.427" Score="2" ViewCount="0" Body="&lt;p&gt;Both Mathematica and Maple are excellent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems they have added symbolic computation toolbox for scilab (which is free). Since you are linux user, you might want to explore it. I have never personally used it though (I have used scilab itself and it does a good job of being a lilliputian Matlab)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://packages.ubuntu.com/lucid/scilab-scimax-doc&quot; rel=&quot;nofollow&quot;&gt;http://packages.ubuntu.com/lucid/scilab-scimax-doc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-21T18:40:30.137" LastActivityDate="2010-11-21T18:40:30.137" CommunityOwnedDate="2010-11-22T07:36:53.763" CommentCount="5" />
  <row Id="11264" PostTypeId="2" ParentId="11259" CreationDate="2010-11-21T18:07:58.033" Score="1" ViewCount="0" Body="&lt;p&gt;For your first question, the third statement is the usual definition of the derivative. For the second statement, take the absolute value function and integrate it to get a function $f(x)$; check what happens at $x_0=0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-21T18:07:58.033" />
  <row Id="11265" PostTypeId="2" ParentId="11259" CreationDate="2010-11-21T18:12:11.697" Score="1" ViewCount="0" Body="&lt;p&gt;For your first question, the first statement is true; it follows from the third.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement is not always true; it essentially requires the function to be twice differentiable.  Try a function that is only once differentiable at $x_0$ and see what happens.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-21T18:12:11.697" CommentCount="3" />
  <row Id="11266" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T18:25:14.267" Score="6" ViewCount="0" Body="&lt;p&gt;I recommend  sageMath  &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.sagemath.org/&lt;/a&gt;, also Mathematica and Maple&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy to use  is Maxima  &lt;a href=&quot;http://maxima.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;http://maxima.sourceforge.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can  to work with sage here: &lt;a href=&quot;http://www.sagenb.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.sagenb.org/&lt;/a&gt; , not is necesary installing !!!&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-21T19:53:14.183" LastActivityDate="2010-11-21T19:53:14.183" CommunityOwnedDate="2010-11-22T07:36:53.763" />
  <row Id="11267" PostTypeId="1" CreationDate="2010-11-21T18:43:33.330" Score="1" ViewCount="128" Body="&lt;p&gt;John Von Neumann once said to Felix Smith, &quot;Young man, in mathematics you don't understand things. You just get used to them.&quot; This was a response to Smith's fear about the method of characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did he mean that with experience and practice, one obtains understanding? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-21T18:43:33.330" Title="What are some interpretations of Von Neumann's quote?" Tags="&lt;soft-question&gt;" CommentCount="6" />
  <row Id="11268" PostTypeId="1" AcceptedAnswerId="11274" CreationDate="2010-11-21T19:01:08.457" Score="4" ViewCount="77" Body="&lt;p&gt;Let $X$, $Y$ be pointed CW complexes, $Y$ connected and $f:X\to Y$ a mapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the assertion '$\Sigma f:\Sigma X\to\Sigma Y$ is a homotopy equivalence' imply that $f$ is a homotopy equivalence? '$\Sigma$' is the reduced suspension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, is it true with some additional hypotheses on $Y$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Addition:&lt;/em&gt;&#xA;Does the assertion '$\Omega f:\Omega X\to\Omega Y$ is a homotopy equivalence' imply that $f$ is a homotopy equivalence? '$\Omega$' is the loopspace.&lt;/p&gt;&#xA;" OwnerUserId="3789" LastEditorUserId="3789" LastEditorDisplayName="" LastEditDate="2010-11-21T20:32:13.053" LastActivityDate="2010-11-21T21:46:05.177" Title="Is a map a homotopy equivalence if its suspension is so?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11269" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T19:11:33.280" Score="2" ViewCount="0" Body="&lt;p&gt;Another very popular free program for symbolic computations is &lt;a href=&quot;http://pari.math.u-bordeaux.fr/&quot; rel=&quot;nofollow&quot;&gt;PARI/GP&lt;/a&gt;. It is aimed somehow at computations in number theory, but its functionality is not only restricted to this. You can read a little bit about it also in the &lt;a href=&quot;http://en.wikipedia.org/wiki/PARI/GP&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-21T19:11:33.280" CommunityOwnedDate="2010-11-22T07:36:53.763" />
  <row Id="11270" PostTypeId="2" ParentId="11177" CreationDate="2010-11-21T19:39:06.460" Score="3" ViewCount="0" Body="&lt;p&gt;I am going to go out on a limb and recommend a more elementary book than (I think) any of the ones others have mentioned.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that as a pure mathematician who is not a set theorist, all the set theory I have ever needed to know I learned from Irving Kaplansky's &lt;em&gt;Set Theory and Metric Spaces&lt;/em&gt;.  (And, you know, I also enjoyed the part about metric spaces).  Kaplansky spent most of his career at the University of Chicago.  Although he had left for MSRI by the time I got there in the mid 1990's, nevertheless his text was still used for the one undergraduate set theory course they offered there.  (Not that I actually took that course, but I digress...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact I think that if you work through this book carefully -- it's beautifully written and reads easily, but is not always as innocuous as it appears -- you will actually come out with more set theory than the average pure mathematician knows.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if you actually do need or want to know some more serious stuff: there's nothing about, say, cofinalities in there, let alone forcing and whatever else comes later on.  But maybe this answer will be appropriate for someone else, if not for you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: I suppose I might as well mention my own lecture notes, available online &lt;a href=&quot;http://www.math.uga.edu/~pete/expositions.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (scroll down to Set Theory).  I think it is fair to say that these are a digest version of Kaplansky's book, even though they were for the most part not written with that book in hand.  [However, last week David Speyer emailed me to kindly point out that I had completely screwed up (not his words!) one of the proofs.  He also suggested the correct fix, but I didn't feel sanguine about it until I went back to Kaplansky to see how he did it.]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description &lt;em&gt;All the set theory I have ever needed to know&lt;/em&gt; on the main page is not meant to be offensive to set theorists (and I hope it isn't) but rather an honest admission: here is the little bit of material that goes a very long way indeed.  Note especially the word &lt;em&gt;need&lt;/em&gt;: this is not to say that these 40 pages contain all the set theory I &lt;em&gt;want&lt;/em&gt; to know.  For instance, I own Cohen's book on forcing and the Continuum Hypothesis, and I would certainly like to know how that stuff goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Come to think of it: I would be highly amused and interested to read 40 pages of notes entitled &lt;em&gt;All the number theory I have ever needed to know&lt;/em&gt; written by one of the several eminent set theorists / logicians who frequent this site and MO.  What would make the cut?]&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-27T12:43:26.587" LastActivityDate="2010-12-27T12:43:26.587" />
  <row Id="11271" PostTypeId="2" ParentId="11261" CreationDate="2010-11-21T20:12:38.713" Score="1" ViewCount="0" Body="&lt;p&gt;I have been using Mathematica since fall 2007. I absolutely love it. It has good palettes included that help you typeset your equations more easily. It also has nice display options for giving presentations. However, if you are just wanting to typeset a paper, LaTeX is probably your best bet. TeXmaker is a very good free mac OS software for LaTeX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maple is another very popular computational software; however, I have not used it.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-21T20:12:38.713" CommunityOwnedDate="2010-11-22T07:36:53.763" />
  <row Id="11272" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T20:13:37.233" Score="2" ViewCount="0" Body="&lt;p&gt;You can also use generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $$\displaystyle f(x) = \sum_{n=1}^{\infty} T(n) x^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{2f(x)}{1-x} = \sum_{n=1}^{\infty} (2\sum_{k=1}^{n} T(k)) x^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{2f(x)}{1-x} = \sum_{n=1}^{\infty} (3T(n) - 1) x^n = 3f(x) - \frac{1}{1-x} + 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle f(x) = \frac{x}{1-3x} = \sum_{n=1}^{\infty} 3^{n-1} x^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle T(n) = 3^{n-1}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-21T20:13:37.233" />
  <row Id="11273" PostTypeId="2" ParentId="11194" CreationDate="2010-11-21T20:16:20.090" Score="0" ViewCount="0" Body="&lt;p&gt;Take a look at Kelley and Peterson's textbook [1]. They provide a very good discussion of difference equations in this text. I believe you can relate the information in this book to answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find further discussion regarding the answers posted here by Moron and Ross Millikan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you still cannot figure it out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-21T20:16:20.090" />
  <row Id="11274" PostTypeId="2" ParentId="11268" CreationDate="2010-11-21T20:18:47.157" Score="7" ViewCount="0" Body="&lt;p&gt;I believe the answer to your first question is no. Let $X$ be any connected acyclic CW-complex with non-trivial fundamental group, for example the space constructed as example 2.38 in Hatcher. Such a space has the property that $H_i(X) = 0$ for $i&amp;gt;0$ and $H_0(X) = \mathbb{Z}$, but $\pi_1(X) \neq 0$. (In particular $\pi_1(X)$ must be perfect). Consider the projection map $f: X \to pt$. By looking $\pi_1$, $f$ cannot be a homotopy equivalence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\Sigma f: \Sigma X \to \Sigma pt$ is a homotopy equivalence. To see this, note that suspension increases the connectivity, which implies that both spaces are simply connected. Hence the homology Whitehead theorem applies, which says that a map between simply connected CW-complexes is a homotopy equivalence if and only if it induces isomorphisms on all homology groups. Using the suspension axiom in homology, we see that all $H_i(\Sigma X)$ and $H_i(\Sigma pt)$ for all $i&amp;gt;0$ are zero and for $i=0$ are $\mathbb{Z}$. It is then easy to check that $\Sigma f$ is an isomorphism in all degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: in your addition, I think the answer is yes, if we replace homotopy equivalence by weak homotopy equivalence. The Whitehead theorem says that $f: X \to Y$ is homotopy equivalence if and only if induces an isomorphism on all $\pi_i$. Because $\Omega X$ and $\Omega Y$ have the homotopy type of CW-complexes, we can replace them by CW-complexes with the price of replacing homotopy equivalence with weak homotopy equivalence. Now note that $Map_+(S^n,\Omega X) \cong Map_+(S^{n+1},X)$ and similarly $Map_+(S^n,\Omega Y) \cong Map_+(S^{n+1},Y)$. Under this isomorphism $(\Omega f)_*$ corresponds to $f_*$.&lt;/p&gt;&#xA;" OwnerUserId="3790" LastEditorUserId="3790" LastEditorDisplayName="" LastEditDate="2010-11-21T21:46:05.177" LastActivityDate="2010-11-21T21:46:05.177" />
  <row Id="11275" PostTypeId="2" ParentId="11150" CreationDate="2010-11-21T20:23:48.617" Score="1" ViewCount="0" Body="&lt;p&gt;Take a look at WolframMathWorld's [1] discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if this gives you any clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Weisstein, Eric W. &quot;Indeterminate.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/Indeterminate.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Indeterminate.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-21T20:23:48.617" />
  <row Id="11276" PostTypeId="2" ParentId="11236" CreationDate="2010-11-21T20:27:33.590" Score="2" ViewCount="0" Body="&lt;p&gt;The limit, as written, does not exist because the function is not defined on an open neighborhood of the limit point: the function is not defined on $(-\infty,0)$, so you cannot take the limit as $x\to 0$. That would be a full answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the limit is taken only from the right, then the other answers you have received are correct: the limit is not an indeterminate form, it is equal to $0$. But you cannot get that by 'evaluating', because $\ln(x)$ is not defined at $0$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-21T20:27:33.590" CommentCount="1" />
  <row Id="11277" PostTypeId="1" AcceptedAnswerId="11278" CreationDate="2010-11-21T20:32:11.483" Score="2" ViewCount="76" Body="&lt;p&gt;Is there an easy way to see that the $n$-dimensional Fourier transform of $1/|x|^a$ is equal to $1/|x|^{n-a}$ (up to multiplicative constant) where $x$ is an $n$ dimensional vector (assuming that $ 0 &amp;lt; a &amp;lt; n $) ?&lt;/p&gt;&#xA;" OwnerUserId="3792" LastActivityDate="2010-11-21T21:45:12.383" Title="how to calculate fourier transform of a power of radial function" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="11278" PostTypeId="2" ParentId="11277" CreationDate="2010-11-21T20:45:50.943" Score="3" ViewCount="0" Body="&lt;p&gt;Yes... you use the way the definition of Fourier transform changes under rotations and dilations. Namely, recall that&#xA;$$\hat{f}(\xi) = \int_{R^n} f(x) e^{-2\pi i x \cdot \xi} \,dx$$&#xD;&#xA;So that if $f(x) = |x|^{-a}$ this becomes&#xA;$$\hat{f}(\xi) = \int_{R^n} |x|^{-a} e^{-2\pi i x \cdot \xi} \,dx$$&#xD;&#xA;Note this is an improper integral, so you have to be careful here with how one interprets the above. Next, note that if $R$ is a rotation, then $Rx \cdot \xi = x \cdot R^{-1}{\xi}$. So changing variables in the above, turning $x$ into $Rx$, and observing that $|x| = |Rx|$, leads to &#xA;$$\hat{f}(\xi) = \int_{R^n} |x|^{-a} e^{-2\pi i x \cdot R^{-1}\xi} \,dx$$&#xD;&#xA;$$ = \hat{f}(R^{-1}\xi)$$&#xD;&#xA;Since $R$ is any rotation this means that $\hat{f}$ is radial.&#xA;Next, change variables $x \rightarrow tx$ for some $t &amp;gt; 0$ in the above definition of $\hat{f}$. This time you get &#xA;$$\hat{f}(\xi) = \int_{R^n} |tx|^{-a} e^{-2\pi i tx \cdot \xi} t^n \,dx$$&#xD;&#xA;$$ = |t|^{n-a}\int_{R^n} |x|^{-a} e^{-2\pi i x \cdot t\xi} \,dx$$&#xD;&#xA;So $\hat{f}(\xi) = |t|^{n-a} \hat{f}(t\xi)$, or equivalently $\hat{f}(t\xi) = &#xD;&#xA;|t|^{a-n}\hat{f}(\xi)$. This means on each radial line, $\hat{f(t\xi)}$ is equal to $C|t|^{a-n}$. And since $\hat{f}(\xi)$ is a radial function, $C$ is independent of which radial line you're on. This gives what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-21T21:45:12.383" LastActivityDate="2010-11-21T21:45:12.383" CommentCount="3" />
  <row Id="11279" PostTypeId="2" ParentId="11212" CreationDate="2010-11-21T20:52:49.513" Score="0" ViewCount="0" Body="&lt;p&gt;If you have an interest in engineering optics or linear systems, try &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471292885&quot; rel=&quot;nofollow&quot;&gt;Gaskill's book&lt;/a&gt;.  It's good to start from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521709792&quot; rel=&quot;nofollow&quot;&gt;David Kammler's book&lt;/a&gt; is also a a good introduction.  The first part explains the connections between Fourier Series, Fourier Transform, DTFT, DFT very clearly which other books often gloss over.&lt;/p&gt;&#xA;" OwnerUserId="3792" LastActivityDate="2010-11-21T20:52:49.513" />
  <row Id="11280" PostTypeId="1" AcceptedAnswerId="11287" CreationDate="2010-11-21T20:54:45.377" Score="3" ViewCount="86" Body="&lt;p&gt;I'm reading Serre's &quot;Lie algebra and Lie groups&quot; now, and I found a description of Lie algebra of an algebraic matrix group, given for a free algebra (free as a module) $k&amp;#39;$ with a basis $\lbrace 1, \varepsilon\rbrace$, $\varepsilon^2 = 0$ as all $X \in M(n,k)$ such that $1 + \varepsilon X \in G(k&amp;#39;)$, where $G(k&amp;#39;)$ is a set of zeros of a number of polinomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k&amp;#39;$ struck me as being extremely close to how smooth infinitesimals are defined in &lt;a href=&quot;http://arxiv.org/abs/0805.3307&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0805.3307&lt;/a&gt; (a link I found on Wiki)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a grad student who recently switched from applied math, so I know very little about algebra, differential geometry etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please clarify how is such a free algebra related to smooth infinitesimals, and when were the latter invented, and how close is synthetic differential geometry to this construction? I currently don't know much about synthetic differential geometry, but I know it uses smooth infinitesimals to define differential, tangent space etc. differently from how they are usually defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if the questions are not clear enough :) I want to dive into the interesting subject, I just need a push in the right direction. It amazes me how the example fits the concept of a Lie algebra as being an 'infinitesimal Lie group' when you think of $\varepsilon$ as an infinitesimal, but I don't know what to make out of it, or even if I am right in my intuition.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-11-21T21:09:27.877" LastActivityDate="2010-11-21T22:05:46.853" Title="Synthetic differential geometry" Tags="&lt;abstract-algebra&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11281" PostTypeId="2" ParentId="11232" CreationDate="2010-11-21T21:23:42.827" Score="5" ViewCount="0" Body="&lt;p&gt;(Posting this as CW so Alex can accept the answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trutheality re-asked the question on MathOverflow, and it turns out the answer is given by what is known as the &quot;Fundamental Theorem of Affine Geometry&quot;. See &lt;a href=&quot;http://mathoverflow.net/questions/46854/continuity-in-terms-of-lines&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/46854/continuity-in-terms-of-lines&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-21T21:23:42.827" CommunityOwnedDate="2010-11-21T21:23:42.827" CommentCount="2" />
  <row Id="11282" PostTypeId="2" ParentId="11193" CreationDate="2010-11-21T21:33:55.047" Score="1" ViewCount="0" Body="&lt;p&gt;The most useful way to think of the $n$-cube is to imagine the set of configurations of $n$-points in the interval. For a more explicit recipe, wiggle your index finger; its set of configurations describes an interval. Wiggle two fingers: now you have a square. Wiggle $4$-fingers and pretend that the motions are independent (or ask an Indian percussionists to demonstrate); the set of configurations describes a $4$-cube. The configuration space method is useful if you want to decompose a (hyper)cube into simplicies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A beautiful rendition of an $n$-cube is obtained by projecting the unit coordinate vectors to the $n$-th roots of unity. As for &quot;how to&quot; see the MO link above.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-11-21T21:33:55.047" CommunityOwnedDate="2010-11-21T21:33:55.047" />
  <row Id="11283" PostTypeId="1" AcceptedAnswerId="11285" CreationDate="2010-11-21T21:35:01.053" Score="1" ViewCount="140" Body="&lt;p&gt;I am being more stupid than usual: embarrassingly so. Let $G$ be a group. Let $\phi:G \rightarrow G$ be an automorphism with inverse $\psi$. Is it true (and why is it true) that $\psi(x)=\phi(x^{-1})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I got myself in a gumption trap and have not thought my way out of it.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-11-22T00:25:22.773" Title="Inverse of an automorphism" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11285" PostTypeId="2" ParentId="11283" CreationDate="2010-11-21T21:44:39.000" Score="7" ViewCount="0" Body="&lt;p&gt;This is not true in general. For instance consider the case of the identity $I_{G}:G \rightarrow G $. It is its own inverse, nevertheless it isn't true that $I_{G}(x) = I_{G}(x^{-1})$ because that would imply that $x = x^{-1}$ for every $x \in G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another counterexample along the same lines. If $G$ is an abelian group then you have available the automorphism $\phi : G \rightarrow G$ given by $\phi (x) = x^{-1}$. Then in this case also $\phi$ is its own inverse and the same argument as before applies.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-21T22:05:30.567" LastActivityDate="2010-11-21T22:05:30.567" CommentCount="2" />
  <row Id="11286" PostTypeId="2" ParentId="11283" CreationDate="2010-11-21T21:48:51.587" Score="1" ViewCount="0" Body="&lt;p&gt;$\phi(x^{-1})=\phi(x)^{-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\phi(x)^{-1}= \phi^{-1}(x)$, not is necesarily true .&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-11-21T21:48:51.587" />
  <row Id="11287" PostTypeId="2" ParentId="11280" CreationDate="2010-11-21T22:05:46.853" Score="2" ViewCount="0" Body="&lt;p&gt;Your intuition is right. This $\varepsilon^2 = 0$ stuff is basically an algebraic way to extract differential information, but it's certainly not as sophisticated as the full edifice of synthetic differential geometry because for polynomials everything is trivial. The basic observation here is that $f(a + \varepsilon b) = f(a) + b f&amp;#39;(a) \varepsilon$ and from here everything else follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that if $f_1, ... f_k$ are a family of polynomials, then a $k$-homomorphism $k[x_1, ... x_n]/(f_1, ... f_k) \to k$ is the same thing as a $k$-point $(p_1, ... p_n)$ on the variety $f_1 = ... = f_k = 0$.  The appropriate generalization here is that a $k$-homomorphism $k[x_1, ... x_n]/(f_1, ... f_k) \to k[\varepsilon]/\varepsilon^2$ is the same thing as a &quot;generalized point&quot; $(p_1 + \varepsilon q_1, ... p_n + \varepsilon q_n)$ on the variety.  If you write down what it means for $f_1, ... f_k$ to vanish on this point, the condition you get is that $p = (p_1, ... p_n)$ must be a point in the ordinary sense and $(q_1, ... q_n)$ must be orthogonal to the gradients of all the $f_i$; but this is equivalent to being a &lt;strong&gt;tangent vector&lt;/strong&gt; at $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one way to define the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zariski_tangent_space&quot; rel=&quot;nofollow&quot;&gt;Zariski tangent space&lt;/a&gt; at a point of a variety, and from here the algebraic definition of the Lie algebra follows.  Again, since we only have to deal with polynomials, we don't need anything close to the full power of synthetic differential geometry, which is a much more recent development.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T22:05:46.853" />
  <row Id="11288" PostTypeId="2" ParentId="11246" CreationDate="2010-11-21T22:16:05.513" Score="8" ViewCount="0" Body="&lt;p&gt;This is a famous problem! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof, which I got from just googling, appears as a solution Problem 218 in the College Mathematics Journal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snapshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PYq30.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to find a couple of different proofs more and references here: &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.3755v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.3755v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-21T22:16:05.513" CommentCount="0" />
  <row Id="11289" PostTypeId="2" ParentId="11227" CreationDate="2010-11-21T22:30:29.227" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the exact sequence of trivial $G$-modules:&#xA;$$0 \rightarrow \mathbf{Z}/p \rightarrow&#xD;&#xA;\mathbf{Q}_p/\mathbf{Z}_p \rightarrow \mathbf{Q}_p/\mathbf{Z}_p \rightarrow 0,$$&#xD;&#xA;(where the second map is multiplication by $p$) and tensor with $T&amp;#39;$. Since $T&amp;#39;$ is free over $\mathbf{Z}_p$, it is flat, and so the sequence remains exact.&#xA;Thus $T&amp;#39;/p$ injects into $T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p$. In particular,&#xA;any non-trivial map $T/p \rightarrow T&amp;#39;/p$ extends to a non-trivial map&#xA;$$T \rightarrow T/p \rightarrow T&amp;#39;/p \rightarrow T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, for the &quot;if&quot; part, you only need the hypothesis that $T&amp;#39;$ is flat and there is a non-zero $G$-equivariant map from $T/p$ to $T&amp;#39;/p$ (for any $G$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, any map $T \rightarrow T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p$ factors through $T/p^n$ for some $n$, and hence as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T \rightarrow T/p^n \rightarrow (T&amp;#39; \otimes \mathbf{Q}_p/\mathbf{Z}_p)[p^n]&#xD;&#xA;= T&amp;#39;/p^n.$$&#xD;&#xA;If $T&amp;#39;/p$ is irreducible, then the composition factors of the RHS are all of the form $T&amp;#39;/p$, and hence there is a surjective map $T/p \rightarrow T&amp;#39;/p$, which must be an isomorphism if $T/p$ is also irreducible.&lt;/p&gt;&#xA;" OwnerUserId="3795" LastEditorUserId="3795" LastEditorDisplayName="" LastEditDate="2010-11-22T16:29:04.707" LastActivityDate="2010-11-22T16:29:04.707" CommentCount="2" />
  <row Id="11290" PostTypeId="2" ParentId="11145" CreationDate="2010-11-21T22:41:36.917" Score="1" ViewCount="0" Body="&lt;p&gt;Here's an idea, although I don't have time to try it: draw lines at angles of multiples of $\frac{\pi}{4}$ from the origin to both of the unit circles you're investigating and compute the curvature at the corresponding intersection points.  If you can show that the curvatures don't match up, the corresponding unit circles can't be related by a scaling and a rotation.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-21T22:41:36.917" CommentCount="4" />
  <row Id="11291" PostTypeId="2" ParentId="11193" CreationDate="2010-11-21T22:51:33.827" Score="0" ViewCount="0" Body="&lt;p&gt;Whenever I think about $S^{2n-1}$, I draw $n$ complex planes and a unit disk in each one.  $S^{2n-1}$ is the set of ordered $n$-tuples of points, one from each disk, so that the sum of the squares of their absolute values is one; you can rotate each &quot;coordinate&quot; in a circle in its own plane independently of the others, but if you move it &quot;outwards&quot; or &quot;inwards,&quot; the others have to change in magnitude as well.  From there, it's only a small leap to $\mathbb{CP}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For complex functions, which live in 4 dimensions, one idea is to graph either the real part or the magnitude as a function in 3-space, and then use color to represent the imaginary part or the argument.  Or you could just do several different graphs, &lt;a href=&quot;http://mathworld.wolfram.com/ExponentialFunction.html&quot; rel=&quot;nofollow&quot;&gt;like Wolfram MathWorld does&lt;/a&gt;.  This breaks down for things sitting in 4-space that aren't graphs of functions, because you could wind up with two points sharing x-, y-, and z-coordinates but with different &quot;colors.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe people who study knotted surfaces like to use &quot;movies,&quot; each &quot;frame&quot; of which is a 2-d cross section of the surface.  The advantage of this is that the Reidemeister moves for knots generalize to &quot;movie moves&quot; for knotted surfaces.  I don't know that much about this, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, my algebraic topology professor told me that he just thinks of CW-complexes in terms of their cell structure.  I asked him to draw $\mathbb{CP}^n$ and he just wrote $1,0,1,0,\dotsc$ in an upwards column, each number representing the number of cells in that dimension.  Obviously, you'd need to take attaching maps into account as well in the general case.  But this lets you just read off homology, etc.  Thinking like this is really why fields like algebraic topology exist: it's easier to visualize the invariants than the actual objects.  I've yet to talk to a geometer about this, though, and it would be interesting to hear how you think of, say, 3-manifolds when it's no longer &quot;up to homotopy equivalence.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is: depends on what you want to visualize!&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-11-21T22:51:33.827" CommunityOwnedDate="2010-11-21T22:51:33.827" />
  <row Id="11292" PostTypeId="2" ParentId="11085" CreationDate="2010-11-21T23:30:37.467" Score="-1" ViewCount="0" Body="&lt;p&gt;If you use a 60 degree coordinate system, the height of a regular tet is 1. It's volume is 1. It's area is 4.&lt;/p&gt;&#xA;" OwnerUserId="3797" LastActivityDate="2010-11-21T23:30:37.467" CommentCount="2" />
  <row Id="11293" PostTypeId="1" AcceptedAnswerId="11299" CreationDate="2010-11-21T23:42:20.600" Score="0" ViewCount="56" Body="&lt;p&gt;This isn't homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or dis-prove the statement: If $R$ is a regular language and both $L$ \ $R$ and $L \cup R$ are context-free, then $L$ is context-free.&lt;/p&gt;&#xA;" OwnerUserId="3051" LastActivityDate="2010-11-22T00:20:03.480" Title="context free languages closure properties" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="11294" PostTypeId="1" AcceptedAnswerId="11296" CreationDate="2010-11-21T23:45:02.283" Score="2" ViewCount="130" Body="&lt;p&gt;Let $f(x)= \displaystyle \sum \limits_{n=1}^\infty \frac{\sin(nx)}{n^3}.$ Show that $f(x)$ is differentiable and that the derivative $f&amp;#39;(x)$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In class we solved a similar problem, and I think we had to show that both $f(x)$ and $f&amp;#39;(x)$ converge uniformly, but I am not really sure &lt;em&gt;why&lt;/em&gt; that is what we have to show. I think we also showed that the partial sums are continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate it if someone could solve this problem and maybe explain a little the theory behind it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2467" LastActivityDate="2010-11-22T00:15:14.367" Title="Convergence of Series of Functions" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11295" PostTypeId="2" ParentId="11294" CreationDate="2010-11-21T23:53:47.643" Score="0" ViewCount="0" Body="&lt;p&gt;Well, to see that $f(x)$ *converges uniformly* please apply the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_M-test&quot; rel=&quot;nofollow&quot;&gt;Weierstrass M-Test&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, i am aware of the fact that term by term differentiation and integration is possible if the series is uniformly convergent. But i don't know much about this, i don't want to answer that, but i am sure someone here will help you out regarding this.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T00:00:39.197" LastActivityDate="2010-11-22T00:00:39.197" CommentCount="3" />
  <row Id="11296" PostTypeId="2" ParentId="11294" CreationDate="2010-11-22T00:02:32.597" Score="2" ViewCount="0" Body="&lt;p&gt;To show uniform convergence of the original series and the series of derivatives you can use the facts that $\sin$ and $\cos$ are bounded and that $\sum_{n=1}^\infty \frac{1}{n^3}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ converge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uniform limit of continuous functions is continuous.  To see that $f$ is differentiable with derivative equal to the series of derivatives, you can apply an integral convergence theorem.  E.g., uniform convergence allows you to pass limits through integrals, and (using the fundamental theorem of calculus) $f(x)=\sum_{n=1}^\infty \int_0^x \frac{\cos(nt)}{n^2}dt$.  You can use the fundamental theorem of calculus again to finish, also using the fact that the series of derivatives represents a continuous function by the first sentence of this paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm intentionally leaving some details for you to fill in.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-22T00:02:32.597" CommentCount="4" />
  <row Id="11298" PostTypeId="2" ParentId="11294" CreationDate="2010-11-22T00:15:14.367" Score="1" ViewCount="0" Body="&lt;p&gt;You had to show that both $(f_n)$ and $(f_n&amp;#39;)$ converge uniformly because you want to apply a theorem like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 10.7&lt;/strong&gt; (from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521497566&quot; rel=&quot;nofollow&quot;&gt;&quot;Real Analysis&quot;&lt;/a&gt; by N.L. Carothers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(f_n)$ is a sequence of real-valuled functions, each having a continuous derivative on $[a,b]$, and suppose that the sequence of derivatives $(f_n&amp;#39;)$ converges uniformly to a function $g$ on $[a,b]$. If $(f_n(x_0))$ converges at any point $x_0$ in $[a,b]$, then in fact, $(f_n)$ converges uniformly to a differentiable function $f$ on $[a,b]$. Moreover $f&amp;#39; = g$. That is, $(f_n&amp;#39;)$ converges uniformly to $f&amp;#39;$ on $[a,b]$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-22T00:15:14.367" />
  <row Id="11299" PostTypeId="2" ParentId="11293" CreationDate="2010-11-22T00:20:03.480" Score="2" ViewCount="0" Body="&lt;p&gt;Take $R = A^*$, where $A$ is your alphabet.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-22T00:20:03.480" />
  <row Id="11300" PostTypeId="2" ParentId="11283" CreationDate="2010-11-22T00:25:22.773" Score="1" ViewCount="0" Body="&lt;p&gt;After brushing off some cobwebs, I thought I might see for which groups the identity can hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Claim&lt;/em&gt;: Let $G$ be a finite group for which there exists $\phi \in \mathrm{Aut}(G)$ for which $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  Then $G$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_abelian_group#Automorphism_group&quot; rel=&quot;nofollow&quot;&gt;elementary abelian 2-group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;:  If $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$, then $\phi(\phi(x))=\phi^{-1}(\phi^{-1}(x^{-1}))$ and so on.  Hence $x=x^{-1}$ and $x^2=\mathrm{id}$ for all $x \in G$, and so $G$ must be an elementary abelian 2-group. $\square$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, a converse of the above is true, since each elementary abelian 2-group admits a the identity automorphism $\phi$, which satisfies the identity $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  But, as we will see, not all automorphisms of elementary abelian 2-groups satisfy the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Claim&lt;/em&gt;:  Let $G$ be a finite group for which each $\phi \in \mathrm{Aut}(G)$ satisfies $\phi(x)=\phi^{-1}(x^{-1})$ for all $x \in G$.  Then $G \cong \mathbb{Z}_2$ or $G$ is the trivial group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;:  By the previous claim, $G$ must be an elementary abelian 2-group.  Applying $\phi$ to both sides of the identity $\phi(x)=\phi^{-1}(x^{-1})$, we find that $\phi(\phi(x))=x^{-1}=x$ for all $x \in G$.  Hence $\phi^2$ is the identity automorphism for all $\phi \in \mathrm{Aut}(G)$.  Therefore, every automorphism of $G$ must not contain a 3-cycle (in fact t-cycle where $t \geq 3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elementary abelian 2-group of order 4 admits an automorphism that contains a 3-cycle.  Consequently, we can prove by induction that elementary abelian 2-group of order $2^a$ for all $a \geq 2$ each admit an automorphism that contains a 3-cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is true for the trivial group and $\mathbb{Z}_2$ by inspection. $\square$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-22T00:25:22.773" CommentCount="1" />
  <row Id="11301" PostTypeId="1" AcceptedAnswerId="11309" CreationDate="2010-11-22T00:49:45.567" Score="2" ViewCount="122" Body="&lt;p&gt;Gaussian integers are the set:&#xA;$$\mathbb{Z}[\imath] =\{a+b\imath : a,b \in\mathbb{Z} \}$$&#xD;&#xA;With norm: &#xA;$$\mathrm{N}(a+b\imath)=a^{2}+b^{2}.$$&#xD;&#xA;It satisfies $\mathrm{N}(\alpha\cdot \beta )=\mathrm{N}(\alpha)\cdot\mathrm{N}( \beta )$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units of $\mathbb{Z}[\imath]$ are  precisely: $1,-1,\imath,-\imath$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result: If $p=4k+1$ ( p : prime), there are $x&amp;lt;p$  such that $p/(x^{2}+1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Theorem$: If $p$ is a prime with&#xA;  $p=4k+1$, then $p$ is an sum of two&#xA;  squares.&#xA;  &lt;em&gt;Proof.&lt;/em&gt;&#xA;  If $p=4k+1$ there are $x&amp;lt;p$ such that $p/(x^{2}+1)$, then  $p/(x+\imath)(x-\imath)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that, if $p/(x+\imath)$, there are $(a+b\imath)$ such that:&#xA;$$(x+\imath)=p(a+b\imath)=pa+pb\imath.$$  &#xA;This implies that $x=pa$, this is imposible as $x\lt p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $p\nmid(x+\imath)$, also $p\nmid(x-\imath)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $p$ is not an Gaussian prime, so $p = \alpha \beta$ with $\mathrm{N}(\alpha)\gt 1$ and $\mathrm{N}(\beta)\gt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ \alpha=a+b\imath $ and $\beta=c+d\imath$, we get:&#xA;$$\mathrm{N}(p)=\mathrm{N}(\alpha \beta)=\mathrm{N}\alpha\cdot\mathrm{N}\beta,$$&#xD;&#xA;which implies:&#xA;$$p^{2}=(a^{2}+b^{2})(c^{2}+d^{2}).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the last equation is an integer, so $(a^{2}+b^{2})|p^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this last equation $(a^{2}+b^{2}),(c^{2}+d^{2})\neq 1,$ so therefore:&#xA;$$p=a^{2}+b^{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's beautiful!  Does anyone have other&#xA;  proofs  ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T02:36:58.293" LastActivityDate="2010-11-22T02:42:19.567" Title="Fermat's Christmas theorem on sums of two squares with Gaussian integers" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="11302" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T00:55:01.790" Score="0" ViewCount="0" Body="&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares#Lagrange.27s_proof_through_quadratic_forms&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; page has the other proofs which I know of. Other novel proofs are welcome as always.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-22T00:55:01.790" />
  <row Id="11303" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T01:09:03.303" Score="1" ViewCount="0" Body="&lt;p&gt;There is a proof by &lt;a href=&quot;http://eprints.maths.ox.ac.uk/677/&quot; rel=&quot;nofollow&quot;&gt;Roger Heath-Brown&lt;/a&gt; and another by &lt;a href=&quot;http://www.jstor.org/pss/2323282&quot; rel=&quot;nofollow&quot;&gt;John Ewell&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heath-Brown's proof was later adapted into a &quot;one-sentence&quot; proof by Zagier. Zagier's proof is available at the wikipedia link given in Sivaram's answer.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T01:09:03.303" CommentCount="1" />
  <row Id="11304" PostTypeId="1" AcceptedAnswerId="11318" CreationDate="2010-11-22T01:29:11.273" Score="7" ViewCount="138" Body="&lt;p&gt;Background: Let $k$ be an even integer. The Eisenstein series are defined by $$E_{k} =  1 - \frac{2k}{B_{k}}\sum_{n=1}^{\infty} \sigma_{k-1}(n)q^{n}$$ where &#xA;$$\sigma_{k-1}(n)= \sum\limits_{d \mid n,\;d \geq 1} d^{k-1}$$ and $B_{k}$ is the $k$-th Bernoulli number. If $k \gt 2$ then $E_{k}$ is a modular form of weight $k$. In the case when $k = 2$, $E_{k}$ is not a modular form. Let $P = E_{2}$. Thus $P = 1 - 24\sum_{n = 1}^{\infty} \sigma_{1}(n)q^{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\theta$ be the differential operator defined by $\theta = q \frac{d}{dq}$. If $f$ is a modular form of weight $k$, define $\Delta f = 12 \theta f - kPf$. (Note that the symbol $\circ$ means composition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &#xA;$$\alpha =  \left( \begin{array}{cc}&#xD;&#xA;a &amp;amp; b  \\&#xD;&#xA;c &amp;amp; d \\&#xD;&#xA;\end{array} \right) \in \textrm{SL}_{2}(\mathbf{R}).$$&#xD;&#xA;Define &#xA;$$\alpha \tau = \frac{a \tau + b}{c \tau + d}.$$&#xD;&#xA;Define $f \circ [a]_{k}(\tau) = f(\alpha \tau)(c \tau + d)^{-k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lang claims (as a lemma on page 161 of his &quot;Introduction to Modular Forms&quot;) that &#xA;$\Delta(f \circ [\alpha]_{k}) = (\Delta f) \circ [\alpha]_{k + 2}$. My question is why is this true?&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T03:03:16.653" LastActivityDate="2010-11-22T06:18:27.670" Title="why does a certain formula in Lang's book on modular forms hold? " Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;&lt;modular-forms&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11305" PostTypeId="1" AcceptedAnswerId="11308" CreationDate="2010-11-22T01:33:31.977" Score="3" ViewCount="60" Body="&lt;p&gt;Hello, I'm working through a little proof of this statement, and here is an excerpt of the problem I'm following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $p \equiv 1\pmod{4}$. Show that if $x^2-py^2=1$ the $x$ is odd and $y$ is even. Suppose that ${x_0}^2-p{y_0}^2=1$ with $y_0&amp;gt;0$, $y_0$ minimal. Show that $gcd(x_0+1,x_0-1)=2$. Deduce that one of two cases arises: Case 1: $x_0-1=2pu^2$, $x_0+1=2v^2$. Case 2: $x_0-1=2u^2$, $x_0+1=2pv^2$. Show that in Case 1, $v^2-pu^2=1$ with $|u|&amp;lt;y_0$ a contradiction to the minimality of $y_o$. show that in Case 2, $u^2-pv^2=-1$. Conclude that if $p\equiv 1\pmod{4}$ then the equation $x^2-py^2=-1$ has an integral solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I note that $0$ and $1$ are the only squares modulo $4$, thus $x^2$ and $y^2$ can only take values $0,1$ modulo $4$. Then mod 4, $x^2-py^2=1$ reduces to $x^2-y^2=1$, and this can only occur when $x^2\equiv 1$, $y^2\equiv 0$, and thus $x$ is odd, and $y$ is even. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since $x_0$ is then odd, $x_0-1$ and $x_0+1$ are both even, so their gcd is at least $2$. Also, any common divisor must divide $(x_0+1)-(x_0-1)=2$, and so the gcd is 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure why the two cases must arise, as I'm not sure what $u,v$ are. My guess is they are coprime factors of $y_0$? I'm thinking this because since $y_0$ is even, $y_0=2uv$ for some factors $u,v$ of $y_0$. Then $(x_0-1)(x_0+1)=p{y_0}^2=4pu^2v^2=(2v^2)(2pu^2)$? Is this the right train of thought? I'm not sure why it should follow that $(x_0+1),(x_0-1)$ must each be equal to one of $(2u^2),(2pv^2)$, and not some other arrangement. Is it because this is the only way to group the factors such that their products have gcd $2$? If so, I don't see why $u$ or $v$ would be coprime to $p$. Thanks for any explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, I believe I can show the remaining steps of the problem.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-22T01:50:42.360" LastActivityDate="2010-11-22T02:06:47.187" Title="Small question in showing if $p\equiv 1\pmod{4}$, then $x^2-py^2=-1$ has integral solution." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11306" PostTypeId="1" CreationDate="2010-11-22T01:48:42.800" Score="5" ViewCount="76" Body="&lt;h1&gt;Motivation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Suppose that $u \in \mathbb{R}^d$ is a unit-norm vector, $\|u\| = 1$, $a, b, c$ are some positive constants and $\xi \in [0,1]$ is another constant (usually chosen close to 1). I am interested in solving the following problem &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sup_{v \in \mathbb{R}^d}\ (\xi + (1-\xi)\|v\|_1^2)\left(\sqrt{&#xD;&#xA;\frac{a \langle u, v \rangle^2 + b}{\xi + (1-\xi)\|v\|_1^2}} - c&#xD;&#xA;\right)&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to $\|v\| = 1$. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;While any suggestions on how to find an optimal $v$ are welcome, I am specifically interested in the following question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find a vector $v$ that maximizes $\langle v, u \rangle$ and satisfies $\|v\|_2 = 1$ and $\|v\|_1 = x$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question (that may be an easier one): given a vector $v_1$ that maximizes $\langle v_1, u \rangle$ and satisfies $\|v_1\|_2 = 1$ and $\|v_1\|_1 = x$ and another vector $v_2$ that maximizes $\langle v_2, u \rangle$ and satisfies $\|v_2\|_2 = 1$ and $\|v_2\|_1 = x + \delta$, how to find the change $\langle v_1, u \rangle - \langle v_2, u \rangle$ as a function of $\delta$?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-22T06:14:24.600" Title="How to compute the change in the angle between two unit norm vectors as the $\ell_1$ norm of one vector changes?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;optimization&gt;&lt;euclidean-geometry&gt;" AnswerCount="0" CommentCount="4" />
  <row Id="11307" PostTypeId="1" CreationDate="2010-11-22T02:02:57.120" Score="2" ViewCount="152" Body="&lt;p&gt;If someone could give answers and explain, it would be greatly appreciated. Help required studying for a final.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One hundred tickets, numbered 1,2,3,…,100, are sold to 100 different people for a drawing.&#xA;Four different prizes are awarded, including a grand prize (a trip to Tahiti). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) How many ways are there to award the prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) How many ways are there to award the prizes if the person holding ticket 47 wins the grand prize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) How many ways are there to award the prizes if the person holding ticket 47 wins one of the prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D) How many ways are there to award the prizes if the person holding ticket 47 does not win a prize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E) How many ways are there to award the prizes if the people holding tickets 19 and 47 both win prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F) How many ways are there to award the prizes if the people holding tickets 19, 47, 73, and 97 all win prizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G) How many ways are there to award the prizes if none of the people holding tickets 19, 47, 73, and 97 wins a prize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H) How many ways are there to award the prizes if the grand prize winner is a person holding ticket 18, 47, 73, or 97?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I) How many ways are there to award the prizes if the people holding tickets 19 and 47 win prizes, but the people holding tickets 73 and 97 do not win prizes?&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="2778" LastEditorDisplayName="" LastEditDate="2010-11-22T07:28:25.833" LastActivityDate="2010-11-22T07:28:25.833" Title="How do I do Combinatorics / Counting?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="11308" PostTypeId="2" ParentId="11305" CreationDate="2010-11-22T02:06:47.187" Score="4" ViewCount="0" Body="&lt;p&gt;Firstly, since $(x_0+1)(x_0-1)=x_0^2-1 = py_0^2$, exactly one of $x_0+1$ and $x_0-1$ is divisible by $p$. If you take the common factor of 2 out of both of them and that factor of $p$ out of one of them, then you are left with two &lt;strong&gt;co-prime&lt;/strong&gt; integers whose product is a square, therefore each must be a square by unique factorisation. The co-primality condition is crucial for that step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $u$ and $v$ are simply new variables that are introduced to express the fact that either both $\frac{x_0+1}{2p}$ and $\frac{x_0-1}{2}$ or both $\frac{x_0+1}{2}$ and $\frac{x_0-1}{2p}$ are squares.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-22T02:06:47.187" CommentCount="2" />
  <row Id="11309" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T02:09:04.490" Score="1" ViewCount="0" Body="&lt;p&gt;This is best viewed from a slightly more general perspective as follows. In any $\rm UFD$, if $\rm\ a\ $ is not prime, i.e.  $\rm\ a\:|\:bc\ $ but $\rm\ a\nmid b,\ a\nmid c\ $ then $\rm\ \gcd(a,b)\ $ is a &lt;em&gt;proper&lt;/em&gt; factor of $\rm\:a\:$. Moreover this gcd can be computed  when a $\rm UFD$ has a constructive Euclidean algorithm, as does $\rm \mathbb Z[i]\:$. Therefore this yields a &lt;em&gt;constructive&lt;/em&gt; proof that nonprime nonunits are reducible in Euclidean domains (i.e. the nontrivial half of the &lt;a href=&quot;http://math.stackexchange.com/questions/4268/alternate-definition-of-prime-number/4399#4399&quot;&gt;equivalence&lt;/a&gt; of irreducible and prime elements in $\rm UFDs$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this above &lt;a href=&quot;http://math.stackexchange.com/questions/5877/efficiently-finding-two-squares-which-sum-to-a-prime/5895#5895&quot;&gt;we deduce&lt;/a&gt; that $\rm\ gcd(p,x-i)\ = a + bi $ is a proper factor of $\rm\:p\:$, so it must have norm a proper factor of $\rm\ N(p) = p^2\ $, i.e. it must have norm $\rm\:p\:$. Therefore $\rm\ p = a^2 + b^2\ $ as desired. This leads to an elegant efficient few-line Euclidean algorithm to compute a representation of a prime $\rm\ p = 4k+1\ $ as a sum of two squares - see my &lt;a href=&quot;http://math.stackexchange.com/questions/5877/efficiently-finding-two-squares-which-sum-to-a-prime/6159#6159&quot;&gt;post here&lt;/a&gt; for an implementation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-22T02:16:12.907" LastActivityDate="2010-11-22T02:16:12.907" />
  <row Id="11311" PostTypeId="2" ParentId="11193" CreationDate="2010-11-22T02:27:06.600" Score="0" ViewCount="0" Body="&lt;p&gt;If you want to visualize four dimensional space time, then imagine three dimensional space except each point has a clock; this is just three space where we associate a number with each point.  When the fourth dimension is curved, then time passes at different rates for these clocks.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-22T02:27:06.600" CommunityOwnedDate="2010-11-22T02:27:06.600" />
  <row Id="11312" PostTypeId="2" ParentId="11307" CreationDate="2010-11-22T02:33:02.213" Score="13" ViewCount="0" Body="&lt;p&gt;The basic rules of counting are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Sum rule:&lt;/em&gt; If one event can occur in $n$ different ways, and another, independent, event can occur in $m$ different ways, then the number of different ways in which either one or the other event can occur is $n+m$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Product rule:&lt;/em&gt; If one event can occur in $n$ different ways, and another, independent, event can occur in $m$ different ways, then the number of different ways in which &lt;em&gt;both&lt;/em&gt; evens can occur is $nm$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When making choices, you have two basic kinds: if the order in which you make the choices matters, they are called &lt;strong&gt;permutations.&lt;/strong&gt; (For example, choosing the Cesar salad as an appetizer and the shrimp cocktail as the main course is different from choosing the shrimp cocktail as an appetizer and the Cesar salad as a main course). If the order in which you make the choices does not matter, you have &lt;strong&gt;combinations.&lt;/strong&gt; (For example, when choosing teams, it doesn't matter if Bill is chosen first to join team A and Clara is chosen second also for team A, or if Clara is chosen first to join team A and Bill is chosen second also for team A; all that matters is that both Bill and Clara are in team A). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these, you get some basic formulas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices, with $m$ options for each, allowing repetitions but where the order of the choice matters (permutations with repetitions), is $m^n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices, with $m$ possibilities, where the order matters but with repetitions not allowed (permutations without repetitions) is $m(m-1)(m-2)\cdots(m-n+1) = \frac{m!}{(m-n)!}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices out of $m$ options, if the order does not matter and repetitions are not allowed, is $\binom{m}{n} = \frac{n!}{n!(m-n)!}$ (called &quot;$m$-choose-$n$&quot;, because you are choosing $n$ out of $m$). These are &quot;combinations&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The number of ways in which you can make $n$ choices, with $m$ options, if the order does not matter and repetitions &lt;em&gt;are&lt;/em&gt; allowed is $\binom{m+n-1}{n}$. (Combinations with repetitions).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You probably knew all that, but still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. You need to choose 4 tickets, out of 100 possibilities; you are not allowed to choose the same ticket twice (no repetitions). Since the prizes are all different, the order in which you make the choices &lt;em&gt;matters&lt;/em&gt; (draw for the 4th prize, then the 3rd prize, then the 2nd prize, then the Grand Prize; or in whichever order you want). So, which formula above applies, and what is the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. If ticket 47 will win the Grand Prize, you still need to assign the remaining three prizes. They have to be awarded to people &lt;em&gt;other&lt;/em&gt; than ticket 47, no repetitions, but order still matters. How many options do you have, and how many choices to do you need to make? Order matters, no repetitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. This is similar to the above, but &lt;em&gt;this&lt;/em&gt; time, ticket 47 may win the Grand Prize, the second, the third, or the fourth place prize. Count each of the four outcomes separately, then apply the &lt;em&gt;Sum rule&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. Well, if you are going to exclude ticket 47, you have to choose from among the remaining 99 tickets. Order still matters, repetitions are not allowed. So the only difference is the number of options you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E. Well, you have two prizes awarded, and two more to award. First award the other two prizes (i.e., count how many ways you can do that). Then decide which prizes go to tickets 19 and 47: you have four prizes you can award, so you need to choose two prizes (to give to 19 and 47); order matters (first prize chosen goes to 19, second to 47), no repetitions. Count &lt;em&gt;that&lt;/em&gt;. In total, you need to (i) pick to other winners; pick them in order, so first winner gets the top prize not awarded to 19 or 47, second gets the lower prize not awarded); and (ii) pick which prizes to give to 19 and to 47. You need &lt;em&gt;both&lt;/em&gt; things to happen, so you should then use the &lt;em&gt;Product rule.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F. Now you know who the four winners are; you just need to figure out which prizes they get. Count how many ways you can distribute the prizes among the four winners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G. Similar to D, but now you have to exclude four people instead of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H. First decide who wins the Grand Prize. Then pick the other three winners. Then use the &lt;em&gt;Product rule.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. This is a combination of the ideas from E and D. Use the same method as in E to figure out how many ways there are to award prizes to 19 and 47; then figure out how many ways there are to assign the remaining two prizes to the remaining people if you exclude 73 and 97. Then combine the two answers using an appropriate rule mentioned above. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T06:21:04.260" LastActivityDate="2010-11-22T06:21:04.260" CommentCount="2" />
  <row Id="11313" PostTypeId="2" ParentId="11307" CreationDate="2010-11-22T02:35:24.787" Score="0" ViewCount="0" Body="&lt;p&gt;Try to solve the questions one award at a time, for each award asking yourself how many people could receive this award? Note that this approach may lead to branching, question C for instance you would get 100*99*98 for the first 3 awards, but the number of persons who may receive the final award is either 1 or 97 depending on whether ticket 47 has already been awarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the questions may be simpler to answer if you initially consider what awards a specific person or persons may have, and thereafter how the rest may be distributed.&lt;/p&gt;&#xA;" OwnerUserId="3584" LastActivityDate="2010-11-22T02:35:24.787" />
  <row Id="11314" PostTypeId="2" ParentId="11301" CreationDate="2010-11-22T02:42:19.567" Score="0" ViewCount="0" Body="&lt;p&gt;There is a nice proof using &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowskis_theorem&quot; rel=&quot;nofollow&quot;&gt;Minkowski's theorem&lt;/a&gt; which also proves the four-square theorem.  It is Proof #2 &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/blog.php?u=14052&amp;amp;b=10589&amp;amp;&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-22T02:42:19.567" />
  <row Id="11315" PostTypeId="1" CreationDate="2010-11-22T03:52:14.470" Score="6" ViewCount="149" Body="&lt;p&gt;Suppose you are given a question that goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider three planets that revolve&#xA;  around a star on separate circular&#xA;  orbits sharing the same orbital plane.&#xA;  The first planet, A, takes 385 days to&#xA;  complete its orbit. The second planet,&#xA;  B, takes 814 days, while the third&#xA;  planet, C, takes 1672 days. Suppose,&#xA;  at one moment in time, the three&#xA;  planets and their star are in one&#xA;  straight line among each other (i.e.&#xA;  they are collinear). Calculate how&#xA;  many days the star and planets will be&#xA;  collinear again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like a simple question. Obvious enough, there are at least two trivial solutions for this kind of question. The first is the linear product of all the figures: 385 x 814 x 1672 = 523988080 days, while the second is their Least Common Multiple (LCM): LCM(385, 814, 1672) = 2165240 days. But surprisingly, it just takes $\frac{2165240}{39}$ days for the planets to be collinear again, albeit not at the same location. The question does not restrict the answer to be an integer, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My &quot;Orbital Collinearity Theorem&quot; :P&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By trial and error, I found out that the number of days for the planets to become collinear again can be calculated by the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{LCM(d_1, d_2, ... , d_n)}{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where LCM(…) denotes Least Common Multiple and M is the largest integer for which the following modular congruence holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d_1}{G} \equiv \frac{d_2}{G} \equiv ... \equiv \frac{d_n}{G} \pmod{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where G is the Greatest Common Divisor of the numbers of days the planets take to complete their respective orbits ($d_1$, $d_2$, ... , $d_n$). In other words, M is the largest integer that can give the same remainder when it divides each of $\frac{d_1}{G}$, $\frac{d_2}{G}$, ... , $\frac{d_n}{G}$. If there is no such M that is larger than 1 (since all integers will give remainder of 0 when divided by 1), the answer will be the LCM, which is very much expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 385, 814 and 1672, the value of M is 39:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$GCD(385, 814, 1672) = 11$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{385}{11}=35$ and $35 \bmod 39 = 35$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{814}{11}=74$ and $74 \bmod 39 = 35$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1672}{11}=152$ and $152 \bmod 39 = 35$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don’t really have the proof/working since it is trial and error, but so far after testing with many samples the formula seems to hold for each case. So let’s check if after $\frac{2165240}{39}$ days, the three planets are collinear or not:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet A: if it takes 385 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{385}=144 \frac{3080}{15015} = 144 \frac{8}{39}$ cycles&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet B: if it takes 814 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{814}=68 \frac{6512}{31746} = 68 \frac{8}{39}$ cycles&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planet C: if it takes 1672 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{1672}=33 \frac{13376}{65208} = 33 \frac{8}{39}$ cycles&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It should be obvious that the numbers of cycles that the three planets have gone have the same fractional part, which is $\frac{8}{39}$. Meaning that they are all $\frac{8}{39}$ cycle into their current cycle (34th, 69th and 145th). Hence they all form the same angle from their last collinear position, thus they are collinear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second sample: minute and hour hands&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The minute and hour hands overlap at&#xA;  12 o’clock. At what time will the&#xA;  hands overlap again?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By considering the hands as planets revolving around the clock center, we can use the formula to solve it. First consider that minute hand takes 60 minutes to complete a cycle, while hour hand takes 720 minutes. LCM(60,720) = 720 and GCD(60, 720) = 60. That will give us M = 11. Thus it will take $\frac{720}{11} = 65.454545454545454545454545454545$ minutes for them to overlap again. And translating that into time gives us 1:05 and 27.27272727272 seconds. Correct. Moreover, this sample is an example of why we must divide the numbers by their GCD to get the M, a step that was not significant with the three planets question; because if we didn’t, M would be 60, giving us 12 minute as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, my real question is: Is there official name and proof for this formula/theorem?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p/s: forgive the inappropriate tags since I don't have enough points to add new tags and there are no suitable tags among the existing ones.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I've worked out the proof half-way but I still can't connect it to the last piece. I'm appending the half-proof here with hope that any of you guys can help me complete it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L = LCM(d_1, d_2, ... , d_n)$. Then it's a given that L is the least number of days for the planets to be collinear again at the same position (i.e. all of them have revolves $c_1, c_2, ... , c_n$ integral number of cycles around the star). Proof by contradiction: suppose there exists another number $K &amp;lt; L$ for which the planets are collinear at the same position, then L must be a common multiple of $d_1, d_2, ... , d_n$, but $L$ is the least of their common multiples, hence contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, we still have a possibility that L could be the M-th time the planets are collinear regardless of position. If such $M &amp;gt; 1$ exists, then the number of cycles that the planets have undergone after L days, $c_1, c_2, ... , c_n$, should have the same remainder if divided by M. Therefore, the following modular congruence must hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{L}{d_1} \equiv \frac{L}{d_2} \equiv ... \equiv \frac{L}{d_n} \pmod{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where I am now. I'm trying to connect the above congruence to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d_1}{G} \equiv \frac{d_2}{G} \equiv ... \equiv \frac{d_n}{G} \pmod{M}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's trivial to do so with $d_1$ and $d_2$ only because $LCM(d_1,d_2)GCD(d_1,d_2) = d_1d_2$ but the relationship between LCM and GCD becomes less simplistic for more than two numbers. The simplest I could find is $\displaystyle a_1 ... a_n = GCD(a_1, a_2, ... a_n) LCM \left( \frac{a_1 ... a_n}{a_1}, \frac{a_1 ... a_n}{a_2}, ... \frac{a_1 ... a_n}{a_n} \right)$ (&lt;a href=&quot;http://math.stackexchange.com/questions/1579/n-ary-version-of-gcda-blcma-b-ab/1581#1581&quot;&gt;from another question in Math SE&lt;/a&gt;) but it cannot be used directly in the formula because the GCD and LCM terms are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="3803" LastEditorUserId="3803" LastEditorDisplayName="" LastEditDate="2010-11-23T01:31:53.753" LastActivityDate="2010-11-23T01:31:53.753" Title="Is there formal name and proof for this formula/theorem (for now I'm calling it Orbital Collinearity Theorem)?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="11316" PostTypeId="2" ParentId="11231" CreationDate="2010-11-22T04:06:05.287" Score="5" ViewCount="0" Body="&lt;p&gt;Your question is an interesting one, but I think the reason you haven't received an answer to it yet is that it's ill-constrained.  That's not a criticism but a technical term.  Basically it means that the problem isn't well-defined enough for there to be a clear, distinct answer.  (This is what the second part of Rahul Narain's comment is getting at.)  To borrow from the Math Overflow FAQ: &quot;The site works best for well-defined questions: math questions that actually have a specific answer.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, your question would be great in my mathematical modeling course.  I would pose it and then ask the students what assumptions would need to be made in order to get a well-defined mathematical model, or what aspects of the problem would be parameters to the model that we could vary and then see how the solution changes (such as those Rahul Narain mentions), or what existing mathematical tools we could use to help understand traffic behavior (such as queuing theory, as Yuval Filmus mentions).  An answer to the question would then require some student taking this on as an extended assignment.  He or she would have to make and justify assumptions, create a model (or two or three, which might require some programming), testing it (or them), refine the model(s), vary parameters, and then interpret the output from the model(s).  There probably wouldn't be a single answer but a range of answers along the lines of &quot;Based on these assumptions our model says this.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've actually just described the process of mathematical modeling.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect of your question is that traffic problems are hard.  They frequently show up as problems in the &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/&quot; rel=&quot;nofollow&quot;&gt;Mathematical Contest in Modeling&lt;/a&gt; (see, for example, the &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/contests/2009/problems/&quot; rel=&quot;nofollow&quot;&gt;2009&lt;/a&gt; and &lt;a href=&quot;http://www.comap.com/undergraduate/contests/mcm/contests/2005/problems/&quot; rel=&quot;nofollow&quot;&gt;2005&lt;/a&gt; contests).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traffic problems are an active area of research, too, because there's a lot we don't understand about the way traffic behaves.  There's not even a consensus about the best way to go about modeling traffic.  &lt;a href=&quot;http://math.mit.edu/projects/traffic/&quot; rel=&quot;nofollow&quot;&gt;Fluid flow models&lt;/a&gt; seem to be the most popular, but some people use &lt;a href=&quot;http://www.rci.rutgers.edu/~gursoy/papers/ifacconf-Stochastic%20Traffic%20Flow%20Models-v2.pdf&quot; rel=&quot;nofollow&quot;&gt;queuing&lt;/a&gt;, and I've also seen &lt;a href=&quot;http://link.aip.org/link/?JCNDDM/3/021206/1&quot; rel=&quot;nofollow&quot;&gt;discrete dynamical system&lt;/a&gt; and even &lt;a href=&quot;http://sjsu.rudyrucker.com/~han.jiang/paper/&quot; rel=&quot;nofollow&quot;&gt;cellular automata&lt;/a&gt; models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in learning more about research on traffic, you could also check out &lt;a href=&quot;http://www.physorg.com/news117283969.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; (&quot;Traffic jam mystery solved by mathematicians&quot;) or the book &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?hl=en&amp;amp;lr=&amp;amp;id=U-Aj1nvo3sgC&amp;amp;oi=fnd&amp;amp;pg=PA1&amp;amp;dq=mathematical+theories+of+traffic+flow&amp;amp;ots=etyrVdoQko&amp;amp;sig=xiApBvgh8ijvc8HP036Ezg3HZrQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mathematical Theories of Traffic Flow&lt;/a&gt;&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-22T04:06:05.287" CommentCount="4" />
  <row Id="11317" PostTypeId="2" ParentId="11315" CreationDate="2010-11-22T04:32:23.403" Score="1" ViewCount="0" Body="&lt;p&gt;Here's how you'd start working on it. Suppose $t$ is a time in which planets $i$ and $j$ are collinear. Then $t/d_i - t/d_j$ is some integer $n$. Thus $$t = \frac{nd_id_j}{d_i-d_j}.$$ Therefore you're looking for the smallest $t$ in $$\bigcap_{i &amp;lt; j} \frac{d_id_j}{d_i-d_j} \mathbb{Z}_+.$$ We can take the common denominator and find the answer (up to sign) $$\frac{LCM(\{d_id_j\prod_{(l,k)\neq (i,j)} (d_k-d_l) : i&amp;lt;j\})}{\prod_{k &amp;lt; l} (d_k-d_l)}.$$ Of course, that form is not as explicit as yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is the explicit result (up to sign) for $n=3$, where the orbits are denoted $x,y,z$: $$\frac{LCM(xy(z-x)(z-y),xz(y-x)(y-z),yz(x-y)(x-z))}{(x-y)(x-z)(y-z)}.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-22T08:40:32.490" LastActivityDate="2010-11-22T08:40:32.490" CommentCount="4" />
  <row Id="11318" PostTypeId="2" ParentId="11304" CreationDate="2010-11-22T04:56:32.030" Score="7" ViewCount="0" Body="&lt;p&gt;Note to begin with that the symbol $\circ$ does &lt;em&gt;not&lt;/em&gt; mean composition; rather, it means the operation given by the formula further on in your question, namely:&#xA;$$f\circ[\alpha]_k(\tau) = (c\tau + d)^k f(\alpha \tau).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to say that $f$ is a modular form of weight $k$ is to say that&#xA;$$f\circ[\alpha]_k(\tau) = f(\tau),$$&#xD;&#xA;for all $\alpha \in SL_2(\mathbb Z),$&#xD;&#xA;or equivalently,&#xA;that $$f(\alpha \tau) = (c\tau + d)^k f(\tau).$$&#xD;&#xA;Differentiating both sides with respect to $\tau,$ &#xA;we find that&#xA;$$f&amp;#39;(\alpha \tau) (c\tau + d)^{-2} = (c\tau + d)^k f&amp;#39;(\tau) + c k (c\tau + d)^{k-1}f(\tau).$$&#xD;&#xA;Since $\theta = \dfrac{1}{2\pi i} \dfrac{d}{d\tau},$&#xD;&#xA;we may rewrite this as&#xA;$$(c\tau + d)^{-(k+2)}(\theta f)(\alpha\tau) = (\theta f)(\tau)  + \dfrac{c k}{2 \pi i(c\tau + d)} f(\tau),$$&#xD;&#xA;or equivalently,&#xA;that $$(\theta f)\circ [\alpha]_{k+2}(\tau) = (\theta f)(\tau) &#xD;&#xA;+ \dfrac{c k}{c \tau + d} f(\tau).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &#xA;$$(k P f)\circ[\alpha]_{k+2}(\tau) = k (P\circ[\alpha]_2)(f\circ [\alpha]_k)(\tau),$$&#xD;&#xA;and so the formula you ask about is equivalent to the formula&#xA;$$P\circ [\alpha]_2(\tau) = P(\tau) + \dfrac{12 c}{2 \pi i(c\tau + d)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a standard formula for $P$ (which shows that although $P$ is not&#xA;a modular form of weight 2, it is not so far off).&#xA;One way to obtain it is to consider the weight 12 cuspform&#xA;$\Delta = q\prod_{n = 1}^{\infty}(1 - q^n)^{24}.$&#xD;&#xA;(Sorry for the conflict in notation, but $\Delta$ is the standard notation here.&#xA;The operator that you have labelled $\Delta$ is often denoted $\delta$, perhaps &#xA;to avoid precisely this notational conflict.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computation (using the product formula for $\Delta$) shows that $\theta(\log \Delta) = P.$  &#xA;Now writing down the modularity equation $\Delta(\alpha \tau) =&#xD;&#xA;(c\tau + d)^{12} \Delta$ and applying $\theta$ to the log of each side,&#xA;we find that indeed $P$ satisfies the required functional equation,&#xA;and thus the formula you asked about holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, note that one can also reverse this last argument: suppose that we know that the formula&#xA;in the question is true.  Applying it to the modular form&#xA;$\Delta$, we find that $\theta \Delta - P \Delta$ is a cuspform of weight $14$, but there are no non-zero such cuspforms. Thus we find that&#xA;$\theta\Delta = P\Delta,$ or equivalently,&#xA;that $\theta(\log \Delta) = P.$  If you look in Serre's &lt;em&gt;Course in arithmetic&lt;/em&gt;,&#xA;you will see that this is how he proves the product formula for $\Delta$:&#xA;he verifies the functional equation for $P$ directly (in the special case&#xA;when $\alpha = \begin{pmatrix}0 &amp;amp; 1 \\ - 1 &amp;amp; 0\end{pmatrix}$; but since&#xA;the functional equation is obvious for $\begin{pmatrix} 1 &amp;amp; 1 \\ 0 &amp;amp; 1 \end{pmatrix},$ and since these two matrices together generate &#xA;$PSL_2({\mathbb Z})$, verifying it for this single choice of $\alpha$ is enough), and then&#xA;essentially applies the argument I just gave to deduce the product formula for &#xA;$\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, I highly recommend Serre's book; as an introduction to modular forms it is both shorter and clearer than Lang's.  If you then want to go further with the theory, there are other, better sources.  (For example, after&#xA;reading Serre's book you could try directly reading Atkin and Lehner's paper on newforms.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-22T06:18:27.670" LastActivityDate="2010-11-22T06:18:27.670" CommentCount="1" />
  <row Id="11319" PostTypeId="1" AcceptedAnswerId="11324" CreationDate="2010-11-22T05:26:02.240" Score="2" ViewCount="102" Body="&lt;p&gt;The question :&#xA;Prove that :$$ \text{ if } y = 2x^2 - 1,\text{ then } \biggl[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5}+ \cdots \biggr]$$  is equal to $$\frac{1}{2} \biggl[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \cdots \biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I have modified the question, in the actual question (from my paper) there were four other options given,my approach was to reduce the first expression to $\frac{1}{2} \ln \biggl( \frac{y+1}{y-1} \biggr) $  and then trying to check for each options to find the match, now is there any other approach for this one ? Since it took me sometime for checking each options :( and the desired answer is at the last option!&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-11-22T05:46:32.290" Title="Any tricky method to solve this one ? " Tags="&lt;algebra-precalculus&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11320" PostTypeId="1" AcceptedAnswerId="11323" CreationDate="2010-11-22T05:35:39.357" Score="4" ViewCount="212" Body="&lt;p&gt;This question might turn out to be really trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ is a one-to-one function from the interval $[0,1]$ to $\mathbb{R}$. Is it necessary that $\exists q \in \mathbb{Q}$ such that $f(x) = q$ for some $x \in [0,1]$ i.e. is it necessary that the image of $f$ contains a rational number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this question when I was browsing through some website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is false. But I am unable to come up with a counter example.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-22T06:40:09.337" LastActivityDate="2010-11-22T06:40:09.337" Title="One-to-one function from the interval $[0,1]$ to $\mathbb{R}\setminus\mathbb{Q}$?" Tags="&lt;real-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="11321" PostTypeId="1" AcceptedAnswerId="11326" CreationDate="2010-11-22T05:38:38.167" Score="2" ViewCount="86" Body="&lt;p&gt;I'm interested because I want to show that $x^2-34y^2\equiv -1\pmod{m}$ has solutions for all integers $m$. I started by using the following reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $3\nmid m$, then $gcd(m,3)=1$. Then there exists a multiplicative inverse $\bar{3}$ modulo $m$. I note that $5^2-34=-(3^2)$, and thus $\bar{3}^2(5^2-34)\equiv (\bar{3}\cdot 5)^2-34(\bar{3}^2)\equiv -(\bar{3})^2(3^2) \equiv -1\pmod{4}$. And thus $(\bar{3}\cdot 5, \bar{3})$ is a solution modulo $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $5\nmid m$, then $(m,5)=1$. Then since $3^2-34=-(5^2)$, then I also have $\bar{5}^2(3^2-34)\equiv (\bar{5}\cdot 3)^2-34(\bar{5}^2)\equiv -(\bar{5})^2(5^2)\equiv -1\pmod{m}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for any $m$ not divisible by $3$ or $5$, there exists a solution. Then for $m$ such that $3|m$ and $5|m$, then $m$ has prime factorization $m=3^a5^b{p_1}^{q_1}\cdots {p_r}^{q_r}$. This would give the system of congruences&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2-34y^2 \equiv -1 \pmod{3^a},&#xD;&#xA;x^2-34y^2 \equiv -1 \pmod{5^b},&#xD;&#xA;x^2-34y^2 \equiv -1 \pmod{{p_i}^{q_i}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $5\nmid 3^a$, $3\nmid 5^b$, and $3\nmid {p_i}^{q_i}$ and $5\nmid {p_i}^{q_i}$, so each of the congruences has a solution. Does the Chinese Remainder Theorem then imply that there is a solution modulo $m$? I know that it holds for polynomials in one variable $x$, and that the number of solutions is the product of the number of solutions for each prime power modulus. Would the same result hold now that there are two variables in the polynomial? I haven't found any proofs to support or contradict the result. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-22T05:55:31.483" LastActivityDate="2010-11-22T06:04:20.537" Title="How does the Chinese Remainder Theorem behave with polynomials with multiple variables?" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11322" PostTypeId="2" ParentId="11320" CreationDate="2010-11-22T05:40:11.580" Score="2" ViewCount="0" Body="&lt;p&gt;It is obviously false since $[0,1]$ and $\mathbb{R}\setminus\mathbb{Q}$ have the same cardinality, so there exists a function $f: [0,1] \to \mathbb{R}\setminus\mathbb{Q}$ that is bijective. Do codomain expansion and you get an injective map from the unit interval to $\mathbb{R}$ that has no rationals in its range.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-22T05:40:11.580" CommentCount="1" />
  <row Id="11323" PostTypeId="2" ParentId="11320" CreationDate="2010-11-22T05:43:54.097" Score="9" ViewCount="0" Body="&lt;p&gt;If you want an explicit counterexample, define $f:[0,1]\to\mathbb{R}$ by $f(x)=x$ if $x$ is irrational, $f(x)=\sqrt2+x$ if $x$ is rational.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-22T05:56:41.603" LastActivityDate="2010-11-22T05:56:41.603" CommentCount="2" />
  <row Id="11324" PostTypeId="2" ParentId="11319" CreationDate="2010-11-22T05:46:32.290" Score="3" ViewCount="0" Body="&lt;p&gt;One you simplify to $\frac{1}{2}\log(\frac{y+1}{y-1})$, plug in $y = 2x^2 -1$ and simplify it to get $-\frac{1}{2} \log(1-\frac{1}{x^2})$. Now expand to get the desired answer.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-22T05:46:32.290" />
  <row Id="11326" PostTypeId="2" ParentId="11321" CreationDate="2010-11-22T05:57:58.547" Score="3" ViewCount="0" Body="&lt;p&gt;The Chinese Remainder Theorem says that if the modulii $m_1,\ldots,m_k$ are pairwise relatively prime, and $a_1,\ldots,a_k$ are arbitrary, then there is a solution $x$ to the congruence system&#xA;\begin{align*}&#xA;x &amp;amp;\equiv a_1 &amp;amp;\pmod{m_1}\\&#xA;x &amp;amp;\equiv a_2 &amp;amp;\pmod{m_2}\\&#xA;&amp;amp;\vdots\\&#xA;x &amp;amp;\equiv a_k &amp;amp;\pmod{m_k}&#xA;\end{align*}&#xA;and the solution is unique modulo $m_1\cdots m_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each $i$ you find a value $r_i$ and a value $s_i$ such that $r_i^2 - 34s_i^2 \equiv -1 \pmod{p_i^{q_i}}$, and values $r$ and $s$ for the congruence modulo $3^a$, and values $r&amp;#39;$ and $s&amp;#39;$ for the congruence modulo $5^b$. Apply the Chinese Remainder Theorem to the $r_i$, $r$, and $r&amp;#39;$ (with appropriate modulii) to get a single value of $x$ that is congruent to what you want for each congruence. Do the same with the $s_i$, $s$, and $s&amp;#39;$ to get a value for $y$.  The single value of $x$ and $y$ is the solution you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that you are not applying the Chinese Remainder Theorem to a 2-variable polynomial (or to a one-variable polynomial other than $p(x)=x$, for that matter): rather, you are finding the values you want for each variable for each prime power, and then you are using the Chinese Remainder Theorem to find a single value that will have the appropriate remainders for each of the variables separately.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T06:04:20.537" LastActivityDate="2010-11-22T06:04:20.537" CommentCount="2" />
  <row Id="11329" PostTypeId="1" AcceptedAnswerId="11333" CreationDate="2010-11-22T07:45:18.040" Score="2" ViewCount="62" Body="&lt;p&gt;I've got a multiplicative-with-noise model  $F(x,y)=S(x)*R(y)*D(x,y)+N$, where $S(x)$ and $R(y)$ are unknown functions, $D(x,y)$ is a distance function, that is, a function that depends only on $|x-y|$ and decreases quickly when distance increases. $N$ is an uncorrelated &quot;small random noise&quot; function. All of the functions except noise are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have $F(x,y)$ sampled for almost any not-very-distant pair of discrete $(x,y)$, that is, for $(x,y): |x-y| \leq D_{max}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to decompose $F(x,y)$ to obtain the &quot;form&quot; of $S(x), R(y), D(x,y)$. How can I do it? My only idea is to calculate averages $S_{est}(x_i)=average(F(x,y), x=x_i)$, $R_{est}(y_j)=average(F(x,y),y=y_j)$ etc, assume $N=0$, then produce an estimated $F_{est}=S_{est}(x)*R_{est}(y)*D_{est}(x,y)$. I dont know what to do next.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2010-11-22T08:38:18.530" Title="How to decompose a result of a multiplication?" Tags="&lt;approximation&gt;" AnswerCount="1" />
  <row Id="11330" PostTypeId="1" AcceptedAnswerId="11334" CreationDate="2010-11-22T07:59:53.717" Score="6" ViewCount="151" Body="&lt;p&gt;Is there a proof in the spirit of Euclid to prove Dirichlet's theorem on primes in arithmetic progression? (By the spirit of Euclid, I mean assuming finite number of primes we try to construct another number which has a prime factor which falls in the same equivalence class as the other primes but the number is not divisible by any of the primes we considered in the initial list.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of the proof using $L$ functions but I am curious to know if Euclid's &quot;simple&quot; idea can be extended to all other cases as well. I tried googling but am unable to find a proof other than the ones relying on $L$ functions. If it is not possible, to what cases can Euclid's idea be extended to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other proofs are welcome as well.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-22T09:27:41.657" Title=" Dirichlet's theorem on primes in arithmetic progression" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11331" PostTypeId="2" ParentId="9807" CreationDate="2010-11-22T08:11:11.890" Score="2" ViewCount="0" Body="&lt;p&gt;I got the same answer as the book: 20,077,200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's the equation: 27 * 26 * 26 * 11 * 10 * 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, since the license plate can have 2 or 3 letters and 2 or 3 numbers, you can consider the first letter and first number to have extra one choice: the 'null' letter/number!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first letter can be A to Z or null: 27 choices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second and third letters can be A to Z: 26 choices each&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number can be 0 to 9 or null: 11 choices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second and third letters can be 0 to 9: 10 choices each&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply them together to get 20077200.&lt;/p&gt;&#xA;" OwnerUserId="3803" LastActivityDate="2010-11-22T08:11:11.890" />
  <row Id="11332" PostTypeId="2" ParentId="11330" CreationDate="2010-11-22T08:30:28.940" Score="0" ViewCount="0" Body="&lt;p&gt;There are similar proofs for primes of the forms $4n+1$ and $4n+3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p_1,\ldots,p_k$ are all primes of the form $4n+1$. Consider $$P = (p_1 \cdots p_k)^2 + 2^2.$$ Since $P$ is a sum of two squares but not a square (or twice a square) itself, it must have a prime factor $q$ of the type $4n+1$ (by Fermat's theorem on numbers representable as sums of two squares). Since $P \mod p_i = 4 \mod p_i \neq 0$, $q$ is a &lt;i&gt;new&lt;/i&gt; prime of the form $4n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p_1,\ldots,p_k$ are all primes of the form $4n+3$. Consider $P = p_1 \cdots p_k + C$, where $C$ is chosen to be either $2$ or $4$ so that $P$ itself is of the form $4n+3$. Since primes of the form $4n+1$ never multiply to a number of the form $4n+3$, $P$ must have a prime number $q$ of the type $4n+3$. As before, $q$ must be new.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-22T08:30:28.940" CommentCount="1" />
  <row Id="11333" PostTypeId="2" ParentId="11329" CreationDate="2010-11-22T08:38:18.530" Score="3" ViewCount="0" Body="&lt;p&gt;The standard way would be to minimize an error function, perhaps least-min-squares $$\mathbb{E} [F(x,y) - S(x)R(y)D(|x-y|)]^2,$$ where $S,R,D$ are all variables. You can also add a regularization term reflecting your belief that $D(d)$ is small for large $d$, e.g. something like $\sum c_d D(d)^2$ for some chosen weights $c_d \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you have a clear goal, you can try to find an explicit solution by finding the minimum analytically. Alternatively, use any standard iterative minimizing technique in the literature, for example gradient descent, to solve the problem in practice.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-22T08:38:18.530" CommentCount="4" />
  <row Id="11334" PostTypeId="2" ParentId="11330" CreationDate="2010-11-22T08:43:21.320" Score="9" ViewCount="0" Body="&lt;p&gt;As Jonas says, Keith Conrad's paper will tell you that the answer is essentially no.  He defines a certain notion of &quot;Euclidean&quot; proof coming from writing down a polynomial with certain properties and gives two classic results which say that these proofs exist for primes in arithmetic progression $a \bmod n$ if and only if $a^2 \equiv 1 \bmod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first case these proofs can't handle is primes congruent to $2 \bmod 5$ (equivalently, primes congruent to $3 \bmod 5$).  The basic problem is that there is no way to force a positive integer to have factors congruent to $2 \bmod 5$ and not congruent to $3 \bmod 5$ (or the other way around) solely by controlling its residue class $\bmod 5$.  In more sophisticated language, $2$ and $3$ lie in all the same subgroups of $(\mathbb{Z}/5\mathbb{Z})^{\ast}$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-22T08:43:21.320" CommentCount="1" />
  <row Id="11335" PostTypeId="1" CreationDate="2010-11-22T08:46:26.950" Score="1" ViewCount="55" Body="&lt;p&gt;Suppose that a in the plane a given conic curve is compelled to pass through two fixed points of that plane.&#xA;What are the curves covered by a fixed point of the conic, its center (for an ellipse), its focus, etc ?&#xA;(I apologize for the bad english ...)  &lt;/p&gt;&#xA;" OwnerUserId="3808" LastActivityDate="2010-11-22T08:46:26.950" Title="Curve of a fixed point of a conic compelled to pass through 2 points" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;" CommentCount="4" />
  <row Id="11337" PostTypeId="2" ParentId="11330" CreationDate="2010-11-22T09:27:41.657" Score="6" ViewCount="0" Body="&lt;p&gt;A couple more references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ram Murty and Nithum Thain,&#xA;&lt;a href=&quot;http://www.mast.queensu.ca/~murty/murty-thain2.pdf&quot; rel=&quot;nofollow&quot;&gt;Prime numbers in certain arithmetic progressions&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul Pollack, &#xA;&lt;a href=&quot;http://www.math.uiuc.edu/~pppollac/impossibility.pdf&quot; rel=&quot;nofollow&quot;&gt;Hypothesis H and an impossiblity theorem of Ram Murty&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-22T09:27:41.657" />
  <row Id="11338" PostTypeId="2" ParentId="11210" CreationDate="2010-11-22T10:31:59.510" Score="3" ViewCount="0" Body="&lt;p&gt;Write $a_n={2+b_n \over 1-b_n}$. Then $b_{n+1}={3\over5} b_n$. This implies $\lim_{n\to\infty} b_n=0$ and $\lim_{n\to\infty} a_n=2$, unless $b_n=1$ for some $n\ge0$. This is the case if $b_0=(5/3)^k$ for some $k\ge0$. In addition there is the fixed point $a_0=a_1=\ldots=-1$ which is not covered by this &quot;Ansatz&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-22T10:31:59.510" CommentCount="2" />
  <row Id="11339" PostTypeId="1" CreationDate="2010-11-22T10:44:39.157" Score="0" ViewCount="97" Body="&lt;p&gt;Suppose $T_1, T_2 : \ell^2(Z_N)-&amp;gt; \ell^2(Z_N)$ are translation invariant linear transformation. How to prove that the composition $T_2 \circ T_1$ is translation invariant?&#xA;Recall the translation operator $R_k$ defined by &#xA;$(R_k z)(n)=z(n-k) $&#xD;&#xA;Also recall definition:&#xA;Let $T : \ell^2(Z_N)-&amp;gt; \ell^2(Z_N)$ be a linear transformation. $T$ is translation invariant if &#xA;$T(R_k z)=R_k T(z)$&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T15:28:30.147" LastActivityDate="2010-12-22T18:44:26.177" Title="How to prove that the composition $T_2 \circ T_1$ is translation invariant?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11340" PostTypeId="1" CreationDate="2010-11-22T11:35:01.250" Score="2" ViewCount="98" Body="&lt;p&gt;I ask this in a partly recreational, and partly research-related spirit, and I realize my problem might be ill-posed, so any suggestions for clarification might go a long way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Succinctly, my problem can be stated as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the period of a &quot;black box&quot;&#xA;  pseudorandom number generator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but I suppose the following formulation might be more attractive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most programs for playing MP3 files (and MP3-playing devices as well) possess a &quot;shuffle&quot; feature that allows the player to play a list of songs in essentially &quot;random&quot; order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run a music player set to shuffle long enough, you will note that sometimes the time interval between the instances your favorite song in the playlist can be short, or can be long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a number of reasons, you can't take a look at what pseudorandom number generator your music player is using; you are however interested in how frequently your favorite song in the playlist would be played.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one now asks,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you determine how often your favorite song in the playlist would be played based on the time intervals between the different instances your song is played?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How long should your playlist be? Obviously, you can't conclude much if you only have two songs in your playlist...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How long should you leave your music player playing to answer question number 1?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I should probably make the drastic assumption that all the songs in the playlist have the same length, e.g. 3 minutes or so.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T11:41:58.763" LastActivityDate="2010-11-23T01:06:46.327" Title="Determining the period of a music player's &quot;shuffle&quot; feature." Tags="&lt;random&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="11341" PostTypeId="1" CreationDate="2010-11-22T11:41:21.037" Score="0" ViewCount="115" Body="&lt;p&gt;How to prove that convolution is associative, that is &#xA;$(x*y)*z = x*(y*z)$, for $x,y,z \in l^2(Z_N)$. Either directly from the definition of convolution( Definition of convolution: For $z,w \in l^2(Z_N)$, the convolution $z*w \in l^2(Z_N)$ is the vector with components &#xA;$z*w(m) = \sum_{n=0}^{N-1} z(m-n)w(n)$ ) or using following lemma:Suppose $z,w \in l^2(Z_N)$. Then for each m, $(z*w)\hat{}(m) = \hat{z(m)}\hat{w(m)}$( where ^ is discrete fourier transform $\sum_{n=0}^{N-1} z(n)e^{-2 \pi i m n/N}$) and also Fourier inversion formula: $z(n) = 1/N \sum_{m=0}^{N-1} \hat{z(m)} e^{2 \pi i m n/N}$, for $n=0,1,\dots, N-1$.&#xA;In my solution I have used lemma above and Fourier inversion formula as following:&#xA;\begin{align*}&#xA;(x*y)*z(n) &amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} ((x*y)*z)\widehat{(m)}e^{2 \pi i m n/N}\\ &#xA;&amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} (x*y)\widehat{(m)}*\widehat{z(m)}e^{2 \pi i m n/N}\\&#xA;&amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} \widehat{x(m)}\widehat{y(m)}\widehat{z(m)}e^{2 \pi i m n/N} \\&#xA;&amp;amp;= &#xA;\frac{1}{N} \sum_{m=0}^{N-1} \widehat{x(m)}(y*z)\widehat{(m)}e^{2 \pi i m n/N}\\&#xA;&amp;amp;= \frac{1}{N} \sum_{m=0}^{N-1} (x*(y*z))\widehat{(m)}e^{2 \pi i m n/N}\\&#xA;&amp;amp;= x*(y*z).&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T15:33:10.703" LastActivityDate="2010-11-22T15:33:10.703" Title="How to prove that convolution is associative?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11342" PostTypeId="1" CreationDate="2010-11-22T12:15:12.087" Score="0" ViewCount="68" Body="&lt;p&gt;Suppose $b_1, b_2 \in l^2(Z_N)$. How to prove that the composition $T_{b_2} \circ T_{b_1} $ of the convolution operators $T_{b_2}$ and $T_{b_1}$ is the convolution operator $T_b$ with $b=b_2 \cdot b_1$? Hint use previous question where I asked: How to prove that convolution is associative? and also &#xA;$b_2 \cdot b_1(m) = \sum_{n=0}^{N-1} b_2(m-n)b_1(n)$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T15:27:17.357" LastActivityDate="2010-12-23T14:44:26.140" Title="How to prove that the composition $T_{b_2} \circ T_{b_1} $ of the convolution operators $T_{b_2}$ and $T_{b_1}$ is the convolution operator $T_b$ ?" Tags="&lt;linear-algebra&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11343" PostTypeId="1" CreationDate="2010-11-22T12:36:51.950" Score="1" ViewCount="138" Body="&lt;p&gt;$$\sum_{i=1}^n \frac1{4i-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I have to integrate the function but from what to find  lower and upper bound. &lt;/p&gt;&#xA;" OwnerUserId="3741" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-22T16:30:19.150" LastActivityDate="2010-11-23T01:18:05.417" Title="how to find bounds of the series $\sum_{i=1}^n \frac1{4i-1}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="11344" PostTypeId="2" ParentId="11343" CreationDate="2010-11-22T13:01:01.287" Score="1" ViewCount="0" Body="&lt;p&gt;$\sum_{i=1}^n \frac{1}{4i - 1} = \frac{1}{3} + \sum_{i=2}^n \frac{1}{4i-1}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply the usual technique for obtaining a lower bound on the sum on the right.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-22T13:01:01.287" CommentCount="2" />
  <row Id="11345" PostTypeId="2" ParentId="10693" CreationDate="2010-11-22T13:08:27.933" Score="0" ViewCount="0" Body="&lt;p&gt;In my previous answer, I have shown that definition (1) implies definition (2). Now I prove the converse implication.&#xA;Suppose that the condition in definition (2) holds. That is, $\mu_n ((a,b))$ converges for any $a,b \in D$, $a &amp;lt; b$, where $D$ is a dense subset of $\mathbb{R}$. We are now going to specify the limit. For any $a,b \in D$, $a &amp;lt; b$, define&#xA;$$&#xD;&#xA;\mu ((a,b)) = \mathop {\lim }\limits_{n \to \infty } \mu _n ((a,b)).&#xD;&#xA;$$&#xD;&#xA;Since $\mu_n$ is a sequence of subprobability measures (s.p.m.), $\mu ((a,b)) \in [0,1]$.&#xA;Next, fix any $a,b \in {\mathbb R}$, $a &amp;lt; b$. If $a_m$ is a decreasing sequence converging to $a$, and $b_m$ is an increasing sequence converging to $b$, $a_m,b_m \in D$, then, by the last definition, $\mu ((a_m,b_m))$ is a monotone increasing sequence, bounded from above by $1$. Hence, this sequence converges, and we can define&#xA;$$&#xD;&#xA;\mu ((a,b)) = \mathop {\lim }\limits_{m \to \infty } \mu ((a_m ,b_m )) \in [0,1].&#xD;&#xA;$$&#xD;&#xA;This actually determines a s.p.m. $\mu$ on $\mathbb{R}$. Now, define $\tilde D = \lbrace x \in {\mathbb R}:\mu (\lbrace x \rbrace ) = 0 \rbrace$ and proceed similarly to my previous answer, based on &#xA;$$&#xD;&#xA;\mu _n ((a&amp;#39;,b&amp;#39;)) \le \mu _n ((a,b]) \le \mu _n ((a&amp;#39;&amp;#39;,b&amp;#39;&amp;#39;)),&#xD;&#xA;$$&#xD;&#xA;to conclude that for any $a &amp;lt; b$ in (the dense set) $\tilde D$, $\mu _n ((a,b]) \to \mu ((a,b])$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-22T13:08:27.933" />
  <row Id="11346" PostTypeId="1" CreationDate="2010-11-22T13:15:22.943" Score="5" ViewCount="195" Body="&lt;p&gt;Given two vectors $u, v \in \mathbb{R}^d$ represented the spherical coordinates is there a simple formula to compute the angle between the two vectors? Without loss of generality, we can assume that the vectors $u$ and $v$ have unit norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure that the following notation is standard, but I assume that the vector $u$ is defined with $\rho = 1$ and the angular components $\theta_1, \ldots, \theta_{d-1}$. We can obtain the Euclidean components as follows: $$u_{x_1} = \cos \theta_1$$ $$u_{x_2} = \sin \theta_1 \cos \theta_2$$ $$\ldots$$ $$u_{x_{d-1}} = \sin \theta_1 \ldots \sin \theta_{d-2} \cos \theta_{d-1}$$ $$u_{x_d} = \sin \theta_1 \ldots \sin \theta_{d-2} \sin \theta_{d-1}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to find the angle is to represent the two vectors in the Euclidean coordinates and compute arccos of the dot product. Is there a simpler way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in comments, is there a generalization of the Haversine formula?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastEditorUserId="2384" LastEditorDisplayName="" LastEditDate="2010-11-23T10:02:27.780" LastActivityDate="2010-11-23T10:02:27.780" Title="How to compute the angle between two vectors expressed in the spherical coordinates?" Tags="&lt;geometry&gt;" CommentCount="12" FavoriteCount="1" />
  <row Id="11347" PostTypeId="2" ParentId="11212" CreationDate="2010-11-22T13:28:11.667" Score="0" ViewCount="0" Body="&lt;p&gt;This link is really, really good!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourier Transform in one day&#xA;&lt;a href=&quot;http://www.dspdimension.com/admin/dft-a-pied/&quot; rel=&quot;nofollow&quot;&gt;http://www.dspdimension.com/admin/dft-a-pied/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3783" LastActivityDate="2010-11-22T13:28:11.667" />
  <row Id="11348" PostTypeId="1" CreationDate="2010-11-22T13:29:14.397" Score="0" ViewCount="34" Body="&lt;p&gt;If you have a non-causal function (i.e. the Hodrick-Prescott filter) and you want to turn this into a causal filter that resembles it as closely as possible, how would you do this? Train a  Multilayer feedforward network? Try to find a FIR filter with .. coefficients? Fuzzy systems? A RBF network?&lt;/p&gt;&#xA;" OwnerUserId="3813" LastActivityDate="2010-11-22T13:29:14.397" Title="turn non-causal spline into causal filter" Tags="&lt;statistics&gt;" />
  <row Id="11349" PostTypeId="1" CreationDate="2010-11-22T14:04:24.587" Score="0" ViewCount="67" Body="&lt;p&gt;I was recently studying some literature on chaos theory and non-linear equations . where in various ciphers were created using non- linear equations like Lorenz equation . Are the data generated from these equations truly random , is it true that the cipher created using non -linear equations are much stronger than AES and other ciphers used currently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I am very naive in maths and I apologies , if this sounds ridiculous. &lt;/p&gt;&#xA;" OwnerUserId="3814" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-22T14:49:08.100" LastActivityDate="2010-11-22T17:52:06.023" Title="Are the values generated by non-linear equations truly random ?" Tags="&lt;random&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11350" PostTypeId="2" ParentId="11341" CreationDate="2010-11-22T14:16:44.083" Score="3" ViewCount="0" Body="&lt;p&gt;This (very standard) computation is done in more generality on Proposition 46, p. 41 of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/integral.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/integral.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP is asking for the special case in which the algebra $R$ is $\mathbb{C}$ and the semigroup $M$ is the (group) $(\mathbb{Z}/N\mathbb{Z},+)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course use of the language of algebras and semigroups is overkill for this particular special case.  But if you look at the three lines of computation in the middle of the page, you see that the point is just to expand each of $(x*y)*z$ and $x*(y*z)$ into a triple sum and then notice that these two triple sums are identical.  In other words, &lt;em&gt;just do it&lt;/em&gt;: it sounds harder than it actually is.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-22T14:37:33.027" LastActivityDate="2010-11-22T14:37:33.027" CommentCount="3" />
  <row Id="11351" PostTypeId="2" ParentId="11349" CreationDate="2010-11-22T14:17:08.323" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;Random&quot; is usually a poorly defined term, so it is impossible to give a clear answer.  Many times what is called random behavior in a dynamical system is more appropriately called chaotic.  If you plot the output values (however you take them) you notice that certain values are much more common than others, just from one value you can't predict the next.  Usually for random numbers the first of many requirements is that all numbers are equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you speak of the strength of an encryption algorithm, what you are usually asking is &quot;given certain data (read about cryptology attacks to see what sorts of data are considered), can you learn something about the key or about the plaintext that was encrypted.  Encryption algorithms are (should be) carefully designed on this point.  Generally speaking you need to scramble the bits in well-defined ways.  Dynamical systems are not designed with this in mind and may well have vulnerabilities when used for encryption.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-22T14:17:08.323" CommentCount="2" />
  <row Id="11352" PostTypeId="1" AcceptedAnswerId="11354" CreationDate="2010-11-22T14:36:58.423" Score="1" ViewCount="96" Body="&lt;p&gt;I'm looking for a program or utility to help construct DFA/NFA/PDA etc. and the like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason I'm interested in this, is because I want to be able to move the various states around without breaking their component connections, or having to redraw all the connections to their new place (otherwise I'd be doing this by hand! :P )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if there's a better SE site for this.&lt;/p&gt;&#xA;" OwnerUserId="1760" LastActivityDate="2010-11-22T16:58:34.243" Title="Program for working with DFA/NFA/PDA ?" Tags="&lt;computer-science&gt;" AnswerCount="5" CommentCount="9" />
  <row Id="11353" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T15:01:00.460" Score="4" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to construct and test and play with your automata, you might find &lt;a href=&quot;http://www.cs.duke.edu/csed/jflap/&quot; rel=&quot;nofollow&quot;&gt;JFlap&lt;/a&gt; usefull.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to produce images and are familiar with LaTeX then TikZ is great (&lt;a href=&quot;http://www.texample.net/tikz/examples/feature/automata-and-petri-nets/&quot; rel=&quot;nofollow&quot;&gt;TikZ automata examples&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to produce images maybe a vector drawing tool like &lt;a href=&quot;http://inkscape.org/download/&quot; rel=&quot;nofollow&quot; title=&quot;Inkscape&quot;&gt;Inkscape&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="571" LastActivityDate="2010-11-22T15:01:00.460" CommentCount="1" />
  <row Id="11354" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T15:06:30.580" Score="1" ViewCount="0" Body="&lt;p&gt;JFlap is a pretty useful tool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also used Grapher which may be downloaded from here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ii.pw.edu.pl/~ratajcza/csm/grapher/grapher.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ii.pw.edu.pl/~ratajcza/csm/grapher/grapher.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-22T15:06:30.580" CommentCount="1" />
  <row Id="11355" PostTypeId="2" ParentId="11341" CreationDate="2010-11-22T15:08:27.850" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;  $\ $ Put $\rm\ \hat f = \sum_{i=0}^{N-1} \:f(i)\: X^i\ $ and note $\rm\ \hat f \:(\hat g\: \hat h) \equiv (\hat f\: \hat g)\:\hat h\ \ (mod\:\ X^N - 1)\ $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-22T15:08:27.850" />
  <row Id="11356" PostTypeId="1" AcceptedAnswerId="11364" CreationDate="2010-11-22T15:16:36.017" Score="0" ViewCount="43" Body="&lt;p&gt;I want to show that if $X$ and $Y$ are the two loss variables such that $X\leq Y$, then $\text{VaR}_\delta(X)\leq\text{VaR}_\delta(Y)$.  &lt;/p&gt;&#xA;" OwnerUserId="626" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-11-22T15:56:58.743" LastActivityDate="2010-11-22T17:10:27.743" Title="proof that value at risk VaR is monotonic" Tags="&lt;probability&gt;&lt;finance&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11357" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T15:44:24.527" Score="0" ViewCount="0" Body="&lt;p&gt;Have a look into the &lt;a href=&quot;http://en.wikipedia.org/wiki/DOT_language&quot; rel=&quot;nofollow&quot;&gt;dot file format&lt;/a&gt;. It is a quasi-standard for graph descriptions. Once you have written down your graph, you can generate an SVG from it using e.g. &lt;a href=&quot;http://www.graphviz.org/&quot; rel=&quot;nofollow&quot;&gt;graphviz&lt;/a&gt; and edit the result in &lt;a href=&quot;http://inkscape.org/&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-22T15:44:24.527" />
  <row Id="11358" PostTypeId="2" ParentId="10972" CreationDate="2010-11-22T16:02:56.867" Score="6" ViewCount="0" Body="&lt;p&gt;The constant $C$ given in Moron's answer, as suspected, is exactly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma + \log \frac{4}{\pi},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, together with Moron's answer, nicely rounds off the second part of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ \text{sech} x = 2(e^{-x} – e^{-3x} + e^{-5x} + \cdots ) \quad (1)$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 \frac{\tanh x}{x} dx =&#xD;&#xA;2\int_0^1 \frac{\sinh x}{x}(e^{-x} – e^{-3x} + e^{-5x} + \cdots ) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\int_0^1 \frac{\sinh x}{x} e^{-x} dx = - Ei(-2) + \gamma + \log 2$$&#xD;&#xA;$$2\int_0^1 \frac{\sinh x}{x} e^{-3x} dx = - Ei(-4) + Ei(-2) + \log 2$$&#xD;&#xA;$$2\int_0^1 \frac{\sinh x}{x} e^{-5x} dx = - Ei(-6) + Ei(-4) + \log (3/2)$$&#xD;&#xA;$$2\int_0^1 \frac{\sinh x}{x} e^{-7x} dx = - Ei(-8) + Ei(-6) + \log (4/3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so on, where $Ei(x)$ is the &lt;a href=&quot;http://mathworld.wolfram.com/ExponentialIntegral.html&quot; rel=&quot;nofollow&quot;&gt;exponential integral.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, interchanging the order of summation, summing and using &lt;a href=&quot;http://en.wikipedia.org/wiki/Wallis_product&quot; rel=&quot;nofollow&quot;&gt;Wallis's product&lt;/a&gt;&#xA;we obtain&#xA;$$\int_0^1 \frac{\tanh x}{x} dx = \gamma + \log \frac{4}{\pi}&#xD;&#xA;-2Ei(-2)+2Ei(-4)-2Ei(-6) + \cdots. \quad (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $(1)$ for $\text{sech} x$ we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 \frac{2}{x(e^{2/x}+1) } dx = 2 \int_1^\infty \frac{1}{x(e^{2x}+1)} dx$$&#xD;&#xA;$$=  \int_1^\infty \frac{\text{sech} x}{x} e^{-x} dx&#xD;&#xA;= 2  \int_1^\infty \frac{e^{-2x}}{x} - \frac{e^{-4x}}{x} + \frac{e^{-6x}}{x} - \cdots dx $$&#xD;&#xA;$$=  -2Ei(-2)+2Ei(-4)-2Ei(-6) + \cdots.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the result follows from $(2).$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-22T18:06:27.977" LastActivityDate="2010-11-22T18:06:27.977" CommentCount="1" />
  <row Id="11359" PostTypeId="1" CreationDate="2010-11-22T16:36:20.750" Score="5" ViewCount="80" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that for any locally compact Hausdorff space X, we can define a Hausdorff one-point compactification containing X.&#xA;In the case of the (differentiable) manifold $\mathbb R^n$ this one-point compactification turns out to be (homeomorphic to) $\mathbb S^n$, which is again a (differentiable) manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the picture look like in the general case for compactifications of an arbitrary manifold $M$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the one-point compactification of $M$ is not a manifold in general (e.g. $\mathbb R^n - 0$); is it possible to view every manifold as an open (dense?) subset of a compact manifold by taking some other kind of compactification?&#xA;In the differentiable case? In the $C^0$-case?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I had thought along the following lines at first:&#xA;By the Whitney embedding theorem, every manifold $M$ can be thought of as a closed submanifold of $\mathbb R^n$ for some $n$. And by embedding $\mathbb R^n$ into $\mathbb S^n$, we can think of $M$ as an embedded submanifold of a compact manifold. &#xA;But I guess taking the closure of $M$ in $\mathbb S^n$ will not in general leave us with a manifold anymore (?), so this does not answer my question...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Has this been looked into?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S.L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2010-11-22T16:36:20.750" Title="Compactification of Manifolds" Tags="&lt;topology&gt;&lt;differential-topology&gt;" CommentCount="2" />
  <row Id="11360" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T16:53:24.157" Score="1" ViewCount="0" Body="&lt;p&gt;If I were attempting to draw a DFA or NFA, I would probably use &lt;a href=&quot;http://www.graphviz.org/&quot; rel=&quot;nofollow&quot;&gt;Graphviz&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-11-22T16:53:24.157" />
  <row Id="11361" PostTypeId="2" ParentId="11352" CreationDate="2010-11-22T16:58:34.243" Score="1" ViewCount="0" Body="&lt;p&gt;If you already have access to Mathematica, you can also use it to make images of DFAs.  The recent versions &lt;a href=&quot;http://reference.wolfram.com/mathematica/tutorial/GraphDrawingIntroduction.html&quot; rel=&quot;nofollow&quot;&gt;have the ability to draw graphs&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-11-22T16:58:34.243" CommentCount="2" />
  <row Id="11362" PostTypeId="1" AcceptedAnswerId="11427" CreationDate="2010-11-22T17:02:10.780" Score="1" ViewCount="54" Body="&lt;p&gt;I've come across a problem with the size of coefficients of a series expansion.  Here's an example of an expression that I expand into a power series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{1-e^{i3n\cdot t}}{1-e^{i\cdot n\cdot t}}$ where $n \in \mathbb{N}$ and $t \in \mathbb{R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I'd like to work with small coefficient values, but the values increase exponentially with $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One question that seems natural to ask in this case is, &quot;Can we find another series with smaller coefficients?&quot;.  Let me explain a little.  I'm fiddling with an integration technique for specific forms of integrals, and I use the series expansions to help with the integration.  I can fairly easily work with more terms, but it's hard to work with larger values of coefficients.  This may seem a bit contrary to many mathematicians' experiences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm getting at is that I'd like to find alternatives to power series expansions.  In other words, is there literature on series expansions other than power series expansions?  I've seen series like Dirchlet series and whatnot, but I'm specifically interested in finding series expansions for functions of exponentials.  And, of course, I'm hoping that I can find ones whose terms are smaller than the corresponding power series expansions. &lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-22T17:16:39.560" LastActivityDate="2010-11-23T01:47:35.287" Title="References to Information on Alternative Series Representations" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="11363" PostTypeId="1" AcceptedAnswerId="11372" CreationDate="2010-11-22T17:08:13.750" Score="21" ViewCount="265" Body="&lt;p&gt;Let $\mu\left(n\right)$ be the Mobius function. Let $\phi\left(n\right)$ be Euler's totient function. Let $\sigma\left(n\right)$ be the sum of divisors and $\tau\left(n\right)$ be the number of divisors functions. I am curious to know whether or not the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu\left(n\right)=a$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi\left(n\right)=b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma\left(n\right)=c$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau\left(n\right)=d$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has at most one solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I remember a number theory assignment I had where we were given particular values for each of these functions and asked to recover the original number. I can't for the life of me remember how (or if) I managed to solve this problem. I tried to work out a general proof, but couldn't. I also wrote a loop in maple to check for counterexamples, but haven't found any yet. I feel like this is something I should know, but probably have forgotten the relevant facts to approaching this problem.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-11-22T18:45:54.480" Title="Do the Mobius function, totient function, sum of divisors and number of divisors uniquely specify a number?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="11364" PostTypeId="2" ParentId="11356" CreationDate="2010-11-22T17:10:27.743" Score="2" ViewCount="0" Body="&lt;p&gt;From the link given by whuber, we have &#xA;$$&#xD;&#xA;{\rm VaR}_\alpha  (L) = \inf \lbrace l \in \mathbb{R}:{\rm P}(L &amp;gt; l) \le 1 - \alpha \rbrace. &#xD;&#xA;$$&#xD;&#xA;If $X$ and $Y$ are random variables such that $X \leq Y$, then obviously ${\rm P}(X &amp;gt; l) \leq {\rm P}(Y &amp;gt; l)$ for any $l \in \mathbb{R}$.&#xA;So, if $l$ belongs to the set $\lbrace l \in \mathbb{R}:{\rm P}(Y &amp;gt; l) \le 1 - \alpha \rbrace$, then $l$ belongs also to the set $\lbrace l \in \mathbb{R}:{\rm P}(X &amp;gt; l) \le 1 - \alpha \rbrace$. Hence, in particular,&#xA;$$&#xD;&#xA;\inf \lbrace l \in \mathbb{R}:{\rm P}(X &amp;gt; l) \le 1 - \alpha \rbrace \leq  &#xD;&#xA;\inf \lbrace l \in \mathbb{R}:{\rm P}(Y &amp;gt; l) \le 1 - \alpha \rbrace,&#xD;&#xA;$$&#xD;&#xA;that is ${\rm VaR}_\alpha  (X) \leq {\rm VaR}_\alpha  (Y)$. Of course, nothing changes if the $\leq$ in $X \leq Y$ refers to stochastic dominance.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-22T17:10:27.743" CommentCount="1" />
  <row Id="11365" PostTypeId="1" CreationDate="2010-11-22T17:21:12.550" Score="0" ViewCount="46" Body="&lt;p&gt;I'm searching for a statistical method to determine if a player is cheating in an online game. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a number  of positive points and a number of negative points per player (score) and given $n$ players ($n &amp;lt;= 64$). Each player can cheat independent of each other. I used standard deviation, which works great on a single player but fails miserably as soon as several players are cheating at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any useful algorithm or formula for this problem?&lt;/p&gt;&#xA;" OwnerUserId="3234" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T18:02:13.687" LastActivityDate="2010-11-22T21:55:22.927" Title="Method to reliably determine abnormal statistical values" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11366" PostTypeId="2" ParentId="11349" CreationDate="2010-11-22T17:52:06.023" Score="0" ViewCount="0" Body="&lt;p&gt;The random (chaotic) result from x&amp;lt;-4x(1-x) (which is truly chaotic if x starts as almost any irrational number between 0 and 1) actually ends up reflecting the final bit of rounding error when using any precision floating point representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still pretty random, but not worth using for any cryptographic uses.&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2010-11-22T17:52:06.023" />
  <row Id="11367" PostTypeId="2" ParentId="11339" CreationDate="2010-11-22T18:14:24.647" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Some advices:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, try to break down the problem into pieces you are familiar with and understand - if you feel uncomfortable with some part, then you should study that again and again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you attend lectures, try to study the chapter before the lecture.   &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-22T18:14:24.647" />
  <row Id="11368" PostTypeId="2" ParentId="11363" CreationDate="2010-11-22T18:21:37.903" Score="7" ViewCount="0" Body="&lt;p&gt;First, congratulations on asking a number theory question which is quite unlike any I have seen before.  It's certainly more fun to read about questions like this rather than computing $a^b$ modulo $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the assignment you got: one can certainly find particular values of $a$, $b$, $c$ and $d$ such that there is exactly one $n$ solving the equations.  For instance, if $p$ is a prime number then taking $a = -1$, $b = p-1$, $c = p+1$, $d = 2$ has the unique &#xA;solution $n = p$: $\tau(n) = 2$ forces $n$ to be prime and thus $\varphi(n) = n-1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am a number theorist, I don't have an expert opinion on the general question.  (I'm not really &quot;the right kind&quot; of number theorist to answer this question.  You might ask Carl Pomerance, for instance.)  I suppose the more reasonable guess is that there are values of $a,b,c,d$ which yield more than one solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the first thing to do is what you're doing: search for counterexamples by computer.  When you get tired of doing that, let us know how far you looked (and, of course, whether you've found any).  The second thing to do is to come up with a heuristic on how often one might expect multiple solutions to be found...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-22T18:21:37.903" CommentCount="4" />
  <row Id="11369" PostTypeId="2" ParentId="8854" CreationDate="2010-11-22T18:21:39.680" Score="1" ViewCount="0" Body="&lt;p&gt;I would be extremely surprised if there existed a translation or a LaTeX version. I have a searchable pdf of it though. Annoying Megaupload link: &lt;a href=&quot;http://www.megaupload.com/?d=F6GBYUBU&quot; rel=&quot;nofollow&quot;&gt;http://www.megaupload.com/?d=F6GBYUBU&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-11-22T18:21:39.680" />
  <row Id="11370" PostTypeId="2" ParentId="6568" CreationDate="2010-11-22T18:27:26.227" Score="1" ViewCount="0" Body="&lt;p&gt;We could also write set difference in terms of intersection and complement as : $A \cap B = A - (A - B)$. (I do also prefer $A - B$ to $A \setminus B$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expression comes handy when you are dealing with cardinality.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-22T18:34:27.377" LastActivityDate="2010-11-22T18:34:27.377" />
  <row Id="11371" PostTypeId="1" AcceptedAnswerId="11397" CreationDate="2010-11-22T18:30:19.813" Score="0" ViewCount="68" Body="&lt;p&gt;Let $s = (\frac{1}{\sqrt{d}}, \ldots, \frac{1}{\sqrt{d}})$ and $u \in \mathbb{R}^d$ be two distinct unit norm vectors in the first orthant. Consider moving along the great circle defined by $s$ and $u$ (in the direction from $s$ to $u$) until the first intersection of the great circle and one of the following hyperplanes $\{x_1 = 0, x_2 = 0, \ldots, x_d = 0\}$ is reached. Let $v$ denote that intersection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to explicitly give an equation for the point $v$ or at least the angle between $u$ and $v$?&lt;/p&gt;&#xA;" OwnerUserId="2384" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T19:12:31.710" LastActivityDate="2010-11-22T22:33:35.407" Title="How to find the intersection between the great circle and a hyperplane?" Tags="&lt;geometry&gt;&lt;linear-algebra&gt;&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11372" PostTypeId="2" ParentId="11363" CreationDate="2010-11-22T18:31:14.103" Score="15" ViewCount="0" Body="&lt;p&gt;The answer is No. The smallest counterexamples I could find are {1836, 1824), {5236, 4960}, {5742, 5112}, {6764, 6368}, {9180, 9120} and {9724, 9184}. I think those are all the pairs in which both numbers are less than 10,000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, both $n=1836$ and $n=1824$ satisfy $\mu(n)=0$, $\varphi(n)=576$, $\sigma(n)=5040$ and $\tau(n)=24$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: here's the code of the program I used in GAP.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;vec := function(n)&#xA; return [MoebiusMu(n), Phi(n), Sigma(n), Tau(n)];&#xA;end;&#xA;&#xA;dic:=NewDictionary([1,2,3,4], true);&#xA;&#xA;for i in [2..10000] do&#xA; v:=vec(i);&#xA; if (LookupDictionary(dic, v) &amp;lt;&amp;gt; fail) then Print(i,&quot; &amp;lt;=&amp;gt; &quot;, LookupDictionary(dic, v), &quot;\n&quot;); fi;&#xA; AddDictionary(dic, v, i);&#xA;od;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2010-11-22T18:45:54.480" LastActivityDate="2010-11-22T18:45:54.480" CommentCount="3" />
  <row Id="11373" PostTypeId="1" AcceptedAnswerId="11375" CreationDate="2010-11-22T18:56:18.640" Score="2" ViewCount="106" Body="&lt;p&gt;I just want to make sure that I understand these two objects correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a ring. A subset $R&amp;#39;\subseteq R$ is a subring provided $R&amp;#39;$ is an additive subgroup that is closed with respect to multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset $I\subseteq R$ is an ideal provided $I$ is an additive subgroup such that $rI\subseteq I$ and $Ir\subseteq I$ for all $r\in R$ (where $rI$ and $Ir$ are defined in the &quot;obvious&quot; way). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in my notes, for the definition of a subring, I didn't specify what I meant by &quot;with respect to multiplication.&quot; Looking back now, I assume I meant left multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then, if this is correct, an ideal is a subring that is closed with respect to both left and right multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-25T07:35:13.650" Title="Relationship between subrings and ideals..." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="11374" PostTypeId="1" AcceptedAnswerId="11380" CreationDate="2010-11-22T18:57:41.380" Score="2" ViewCount="59" Body="&lt;p&gt;Suppose you have two options for making a payment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Pay $90 \%$ of the purchase price two months after the date of the sale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) Deduct $X \%$ off the purchase price and pay cash on the date of sale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine $X$ so that a customer is indifferent between the two options when valuing them using an effective annual interest rate of $8 \%$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be the purchase price. So option (B) is: $P \left(1- \frac{X}{100} \right)$. Now option (A) would be $0.9P(1.08)^{1/6}$ since we want to know what $0.9P$ would be worth in $2$ months? Then set $P \left(1- \frac{X}{100} \right) = 0.9P(1.08)^{1/6}$ to find $X$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-22T21:13:21.997" Title="Payments and Interest Rates" Tags="&lt;finance&gt;" AnswerCount="1" />
  <row Id="11375" PostTypeId="2" ParentId="11373" CreationDate="2010-11-22T19:03:37.757" Score="5" ViewCount="0" Body="&lt;p&gt;A left-ideal is an abelian subgroup (under addition) of the ring which is closed under left-multiplication by elements of the ring, and not just elements in the ideal. In the commutative case we identify left and right multiplication and then an ideal is a subring of the ring. However, that is not enough. Since the subring requires closure under multiplication of elements contained in itself while a left or right ideal requires closure under corresponding multiplication by elements of the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the polynomial ring $R=k[x]$ in one variable over a field $k$. Then $R$ is a commutative ring with identity. $k$ is a subring of $R$ being a field. However, any ideal that contains any element of $k$ is actually the whole ring (since if it contains an element of $k$, it contains $1$ by multiplying it by it's inverse and hence contain every element of $R$). So being an ideal is a stronger requirement.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-22T19:46:19.123" LastActivityDate="2010-11-22T19:46:19.123" CommentCount="2" />
  <row Id="11376" PostTypeId="2" ParentId="3006" CreationDate="2010-11-22T19:12:28.373" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;A particle on a trigonometric hyperbola has acceleration described by the same hyperbola.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, differentiation is idempotent on the hyperbolic cosine. Since $e^{x} = (\cosh x)^{\prime} + \cosh x$, the claim follows immediately.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-22T19:27:45.473" LastActivityDate="2010-11-22T19:27:45.473" CommentCount="1" />
  <row Id="11377" PostTypeId="2" ParentId="11373" CreationDate="2010-11-22T19:15:46.010" Score="2" ViewCount="0" Body="&lt;p&gt;In the case of the multiplication operation in a (sub)ring the phrase 'with respect to multiplication' doesn't refer to any 'left multiplication': In a ring there is just one multiplication which may be or not be commutative. Being closed with respect to multiplication means that if $a$ and $b$ belong to $R&amp;#39;$, then also $ab$&amp;#160;(and $ba$ by exchanging the roles of $a$ and $b$) belong to $R&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the non-commutative situation we often distinguish the products $ab$ and $ba$ of two elements $a$ and $b$&amp;#160;by saying that we 'multiply $b$ on left/right by $a$'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideal is more than a subring in the sense that we require that products stay in the ideal even when multiplying by elements that don't belong to the ideal.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-22T19:15:46.010" CommentCount="1" />
  <row Id="11378" PostTypeId="2" ParentId="11373" CreationDate="2010-11-22T19:27:31.020" Score="5" ViewCount="0" Body="&lt;p&gt;Remember from groups and vector spaces the notions of &quot;sub-group&quot; and of &quot;sub-space&quot;: you must have a &#xA;&lt;strong&gt;subset&lt;/strong&gt; which is also &quot;closed&quot; with respect to the operations. That is, if you take elements &lt;em&gt;from the subset&lt;/em&gt;, and you subject them to the operations you have, you should end up with results that are also in your subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With groups, that means that if a subset $S$ of a group $G$ is going to be a group, then it must be the case that whenever $s,t\in S$, then $st\in S$ (&quot;closed under multiplication&quot;; if you take two things in $S$ and multiply them, you don't need to go &quot;outside of $S$&quot; to find the answer, it is all already inside $S$; no need to go outside); and if $s\in S$, then you need $s^{-1}\in S$ (&quot;closed under inverses&quot;); and you need $e\in S$ (closed under the operation that gives the identity element).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With vector spaces, if a subset $W$ of a vector space $V$ is going to be a subspace, you need that $\mathbf{0}\in W$ (closed under the operation that gives the identity element); that if $w_1,w_1\in W$, then $w_1+w_2\in W$ (closed under vector addition); and that if $w\in W$ and $\alpha\in F$, then $\alpha w\in W$ (closed under scalar multiplication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for rings: if you have a ring $R$, then in order for a subset $S$ to be a &lt;strong&gt;subring&lt;/strong&gt; you need it to be (i) closed under addition; (ii) closed under additive inverses; (iii) contain the identity element of the sum; and (iv) closed under multiplication (if $a,b\in S$, then $ab\in S$). Saying &quot;is a subgroup&quot; is the same as saying (i), (ii), and (iii). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideals, however, are a bit more than simply subrings. They play exactly the analogous role to rings as normal subgroups play to groups. You may recall that a normal subgroup $N$ of a group $G$ is a &quot;subgroup-with-something-extra&quot;: not only is it &quot;closed&quot; under the three usual operations of the group (identity element, addition, and inverses), it must also be &quot;closed&quot; under a bunch of other operations (conjugation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly with ideals: not only must they be &quot;closed&quot; under the four usual operations of the ring (additive identity element, addition, additive inverses, and multiplication), it must also be &quot;closed&quot; under a bunch of other operations (that are like the scalar multiplications in the vector space case): left multiplication by any element of $R$, and right multiplication by any element of $R$. Just like &quot;normal subgroup&quot; is &quot;subgroup with something extra&quot;, likewise ideals are &quot;subrings with something extra.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with saying &quot;an ideal is a subring that is closed under left and right multiplication&quot; is that saying &quot;closed under left and right multiplication&quot; &lt;em&gt;only&lt;/em&gt; says that if $a,b\in I$, then $ab\in I$; you really want to add the coda of &quot;...left and right multiplication &lt;strong&gt;by elements of the ring.&lt;/strong&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples abound. Take $R=\mathbb{R}[x]$, the polynomials with coefficients in the real numbers, and consider $\mathbb{Z}[x]$, the polynomials with integer coefficients. The latter is a subggroup of $R$, and if $p(x),q(x)\in\mathbb{Z}[x]$, then the product $p(x)q(x)$ is also in $\mathbb{Z}[x]$. So $\mathbb{Z}[x]$ is a subring of $R$. It is not, however, an &lt;em&gt;ideal&lt;/em&gt; of $R$, because $x\in\mathbb{Z}[x]$, and $\pi\in R$, but their product $\pi x\notin\mathbb{Z}[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or take $S=\mathbb{R}[x^2]$, the polynomials with real coefficients in which only even powers of $x$ occur. This is a subring of $R$ but not an ideal, because $x^2 \in S$, $x\in R$, but $x(x^2)\notin S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $S=\{p(x)\in R\mid p(0)=0\}$, the collection of all polynomials with constant term equal to $0$, is both a subring and an ideal: if $p(x)\in S$ and $q(x)\in R$, then $q(x)p(x)\in S$ since $q(0)p(0)=0$ (and $p(x)q(x)=q(x)p(x)$). That is, $q(x)S\subseteq S$ and $Sq(x)\subseteq S$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-22T19:27:31.020" />
  <row Id="11379" PostTypeId="1" CreationDate="2010-11-22T19:58:21.487" Score="0" ViewCount="73" Body="&lt;p&gt;How to prove that a power series in more than one variable is invertible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this still true if the variables are noncommutative?&lt;/p&gt;&#xA;" OwnerUserId="3817" LastActivityDate="2010-11-22T21:06:02.253" Title="Inverting a power series " Tags="&lt;power-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11380" PostTypeId="2" ParentId="11374" CreationDate="2010-11-22T20:02:10.430" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Edited.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not quite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it works: let $P$ be the price. Under option (A), you take the money $P$ you would have paid, put it in the bank generating interest, and at the end of the two months you pay $0.9P$, keeping the difference between what you have at the end of the two months and the price you pay. So under option (A), at the end of two months you have both the item plus $P(1.08)^{1/6}-0.9P$ in cash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under option (B), you pay $(1- X/100)P$ now, keeping $(\frac{X}{100})P$ of the price, which you can then put in the bank for two months generating interest. So under option (B), at the end of the two months, you have both the item and $\left(\frac{X}{100}P\right)(1.08)^{1/6}$ in cash. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trying to find the value of $X$ that will make both options equal. (Note that the value $P$ does not really matter, because it will cancel out.) So the quantities you need to set equal to one another are not exactly what you have, but something slightly different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt;&#xA;Equivalently, as you note in the comments, we may instead want to compare the present value of option (A) with option (B) (rather than comparing how much money you will end up with if you set aside $P$ under options (A) and (B), two months down the line). The present value of option (A) is the amount of money $Y$ you would need to put in the bank at $8\%$ interest so that at the end of two months you have $0.9P$ to pay for the item; that is, $Y(1.08)^{1/6} = 0.9P$, or $Y = \frac{0.9P}{(1.08)^{1/6}}$. So you want to find $x$ for which the current expenditure $(1-x)P$ equals the current value of option (A), $(1-x)P = \frac{0.9P}{(1.08)^{1/6}}$. Again, $P$ drops out, and we get the formula you have in the comments,&#xA;$$ 1 - x = \frac{0.9}{(1.08)^{1/6}}.$$&#xD;&#xA;This gives $x = \frac{X}{100} = 1 - \frac{0.9}{(1.08)^{1/6}}$, or a little over $11$% discount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you proceed as I described originally, and fix the mistake I originally made, you get exactly the same formula and the same answer. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-22T21:13:21.997" LastActivityDate="2010-11-22T21:13:21.997" CommentCount="4" />
  <row Id="11381" PostTypeId="2" ParentId="11379" CreationDate="2010-11-22T20:06:43.407" Score="2" ViewCount="0" Body="&lt;p&gt;(I am answering with respect to the commutative case.) This is not completely true. A power series with a nonzero constant term over a field is invertible. The power series ring in finitely many variables $k[[x_1,...,x_n]]$ over a field is a local ring and the ideal generated by all the variables is the unique maximal ideal. By an elementary result on local rings, every element outside the unique maximal ideal is invertible. So if a power series has a nonzero constant term, it is not in the unique maximal ideal and hence invertible. If the constant term is zero, it is in this maximal ideal and hence not invertible (this ideal is proper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: You only need the ring of coefficients to be local (and not necessarily a field) for the formal power series ring to be local. In this case the maximal ideal is $m+(x_1,...,x_n)$ where $m$ is the maximal ideal of the coefficient ring $R$. So in this case a power series is invertible iff the constant term is outside the maximal ideal of the coefficient ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if this is true in the noncommutative case.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-22T21:06:02.253" LastActivityDate="2010-11-22T21:06:02.253" CommentCount="2" />
  <row Id="11382" PostTypeId="1" CreationDate="2010-11-22T20:23:14.117" Score="26" ViewCount="1986" Body="&lt;p&gt;I'm hoping that this isn't such a basic question that it gets completely laughed off the site, but why would I want to multiply two polynomials together? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I flipped through some algebra books and have googled around a bit, and whenever they introduce polynomial multiplication they just say 'Suppose you have two polynomials you wish to multiply', or sometimes it's just as simple as 'find the product'. I even looked for some example story problems, hoping that might let me in on the secret, but no dice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that a polynomial is a basically a set of numbers (or, if you'd rather, a mapping of one set of numbers to another), or, in another way of thinking about it, two polynomials are functions, and the product of the two functions is a new function that lets you apply the function once, provided you were planning on applying the original functions to the number and then multiplying the result together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elementary multiplication can be described as 'add X to itself Y times', where Y is a nice integer number of times. When Y is not a whole number, it doesn't seem to make as much sense. &lt;/p&gt;&#xA;" OwnerUserId="3818" LastActivityDate="2010-11-24T04:17:12.877" Title="Why would I want to multiply two polynomials?" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;" AnswerCount="13" CommentCount="13" FavoriteCount="6" />
  <row Id="11383" PostTypeId="2" ParentId="11379" CreationDate="2010-11-22T20:32:00.143" Score="0" ViewCount="0" Body="&lt;p&gt;Timothy Wagner's answer extends to the case of power series over a field with in noncommuting variables with nonzero constant terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the power series with zero constant terms is both a maximal left ideal and a maximal right ideal. The same argument as in the commutative case gives left and right inverses, and by associativity both must be equal.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T20:32:00.143" />
  <row Id="11384" PostTypeId="1" CreationDate="2010-11-22T20:42:15.287" Score="1" ViewCount="73" Body="&lt;p&gt;What is the order of the set of distinct (up to similarity) nxn matrices over $\mathbb{R}$ with determinant equal to some non-zero scalar... say 6? (eg. countable, uncountable etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of matrices over $\mathbb{R}$ is uncountable. So is the order of the set consisting of classes of matrices with the same determinant. In each of those classes we have further subsets, the equivalence classes formed by grouping similar matrices. Each equivalence class of similar matrices represents one linear transformation expressed in terms of all possible bases of $\mathbb{R}^n$, so the size of each equivalence class is also uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm not certian about is the size of the set of &quot;different&quot; transformations that have the same determinant. Is it uncountable too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I put it in a correspondence with to show this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if this is poorly worded-- let me know if there is a better way to ask this question.&lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-22T22:07:04.793" Title="What is the order of the set of distinct (up to similarity) nxn matrices over R?" Tags="&lt;linear-algebra&gt;&lt;elementary-set-theory&gt;&lt;cardinals&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11385" PostTypeId="2" ParentId="11384" CreationDate="2010-11-22T20:47:06.030" Score="3" ViewCount="0" Body="&lt;p&gt;It has the same cardinality as $\mathbb{R}$ if $n\gt 1$.  Consider diagonal matrices with entries $(x,6/x,1,1,\ldots,1)$, $x\neq0$.  Two such are similar only in a situation where $\{x,6/x\}=\{y,6/y\}$, hence the conclusion on cardinality.  (Of course $6$ is not special.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-22T20:47:06.030" CommentCount="3" />
  <row Id="11386" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T20:47:27.863" Score="1" ViewCount="0" Body="&lt;p&gt;If you were working with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_ring&quot; rel=&quot;nofollow&quot;&gt;polynomial ring&lt;/a&gt;, you'd definitely want to be able to multiply polynomials together. This can allow you to use power series to solve recurrences with generating functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polynomial rings are also used in some cryptographic systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a few years, you'll take calculus, and then you'll need to multiply polynomials in order to take derivatives of rational functions, to give one example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebra you learn now will be very useful to your mathematics education later on. You want to get this stuff down cold now, so that later on you can focus on more complicated subjects.&lt;/p&gt;&#xA;" OwnerUserId="2994" LastActivityDate="2010-11-22T20:47:27.863" CommentCount="4" />
  <row Id="11388" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T20:53:56.683" Score="28" ViewCount="0" Body="&lt;p&gt;You have two questions, the explicit one about why you would want to multiply polynomials, and an implicit one in your final paragraph about what multiplication by a non-integer might mean or why we would care to multiply by a non-integer in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address the last one first: once you have multiplication by integers, multiplication by fractions will very quickly rear its head. What does multiplying by &quot;one and a half&quot; mean, if multiplying by 2 means &quot;add to itself&quot;, etc? Well, imagine you have a chocolate bar, those that are made up of smaller squares. You can imagine breaking the bar in half, and then figuring out what three times that half will be; that will be multiplying by &quot;three halves&quot; (a.k.a one and a half). You are really multiplying by an integer, after suitably modifying $X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you need to multiply $X$ by a fraction, $\frac{p}{q}$, imagine dividing $X$ into $q$ equal parts, and then multiplying such a $q$th part of $X$ by $p$ in the sense you have above. That is the same as &quot;multiplying by $\frac{p}{q}$&quot;. So multiplying by a fraction &lt;em&gt;is&lt;/em&gt; like &quot;abbreviated addition&quot;: it means &quot;break up into $q$ equal parts, and then add a $q$th part repeatedly $p$ times.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least, multiplying by fractions makes just as much &quot;natural sense&quot; as multiplying by integers does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why bother with numbers other than fractions then? Well, in one sense you don't have to: you can try to stick to fractions and nothing more complicated than that, and you can go very far. But as the Greeks discovered a long time ago, you also run into very big walls very quickly. For instance, if you draw a square which is $1$ foot long on each side, and you try to measure how long its diagonal is (say, for construction purposes), then it turns out that the diagonal is &lt;em&gt;not&lt;/em&gt; a number that can be expressed as a fraction; it is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Irrational_number&quot; rel=&quot;nofollow&quot;&gt;irrational number&lt;/a&gt;. So very soon you end up having to consider numbers that are &lt;em&gt;not&lt;/em&gt; fractions, and if they are lying around sooner or later you are going to have to multiply them to compute stuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you end up having to find some way of multiplying irrationals as well, even though they no longer seem to fit with that same &quot;natural&quot; meaning they had back when we started with integers. One solution is that every irrational can be approximated by a suitable sequence of fractions (think about computing the decimals one at a time; every time you stop, what you have so far as a rational; for example, $\sqrt{2} = 1.4142\ldots$, and you get that $1.4 = \frac{14}{10}$, $1.41=\frac{141}{100}$, $1.414=\frac{1414}{1000}$, etc.) We know what it means to multiply $X$ by each of those fractions in a sensible way, so we say that multiplying $X$ by $\sqrt{2}$ is the number you get by doing the successive multiplications, just like $\sqrt{2}$ is the number you get by doing the successive fractional approximations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This no longer makes sense as &quot;abbreviated addition&quot;, but it turns out that it is very, very necessary and very, very useful, in order to make sense of things and be able to compute things that we need to be able to compute (areas, productivity, interest, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for multiplying polynomials...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One answer: multiplying functions lets you construct more complicated functions out of simpler ones. Or more to the point, it lets you express more complicated functions in terms of simpler ones. This is particularly important if you want to perform complex computations, as then you my be able to &quot;get away&quot; with performing much simpler computations and then multiplying the results, rather than do the really complicated expression instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, say you have a single polynomial like $p(x) = x^2-7x+10$. If you realize that $p(x)$ is the result of multiplying the simpler polynomial $x-2$ by the (also simpler) polynomial $x-5$, then whenever you need to evaluate $p(x)$ at a number, say $17$, instead of having to square $17$, then multiply it by $7$, subtract &lt;em&gt;that&lt;/em&gt; form the square you computed, and then adding $10$ (three multiplications and two additions/subtractions), you can instead take $17$, subtract $2$ to get $15$; then take $17$, subtract $5$ to get $12$; and then multiply $15$ by $12$ (one multiplication and two additions/subtractions), because $x^2-7x+10 = (x-2)(x-5)$, so $(17)^2 - 7(17) + 10 = (17-2)(17-5)$. Much simpler to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another: it is usually very hard to find a value $x$ for which the result of doing some complex series of operations will be a desired quantity, $d$. For example, you want to know how much money to put in the bank so that, at the end of five months at a particular interest rate, you will have exactly the amount of money you need to buy that new wide-screen TV. This involves solving equations. Many natural equations can be written down in the form $p(x)=c$ where $p(x)$ is a polynomial expression in the unknown quantity $x$, and $c$ is the desired value. Solving such equations can be dificult in general. If you don't know the quadratic formula, then figuring out the values of $x$ for which the polynomial above $x^2-7x+10$ is equal to zero can be pretty difficult. Or think about something like $x^4 + x^3 - 120x^2 - 121x = 121$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, figuring out when a product is equal to $0$ is very easy, because the only way for a product to be zero is if one of the two factors is equal to zero. So if you take the equation above and you write it as $x^4+x^3-120x^2-121x-121 =0$, then you are trying to find when a certain polynomial is equal to $0$. If you can write $q(x)=x^4+x^3-120x^2-121x-121$ as a &lt;em&gt;product&lt;/em&gt;, $q(x) = p(x)r(x)$, then you have that $q(x)=0$ if and only if either $p(x)=0$ or $r(x)=0$. With some luck, $p$ and $r$ will be &quot;easier&quot; than $q$, so you can solve them. (In the above case, $q(x) = (x^2-121)(x^2+x+1)=(x-11)(x+11)(x^2+x+1)$, so the only way you can get $q(x)=0$ is if $x=11$ or $x=-11$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is one way to figure out the quadratic formula (did you ever wonder where it came from?). Why are the solutions to $ax^2 + bx+c = 0$ given by $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$? You can factor out $a$ and get $a(x^2 + Bx + C) = 0$, with $B=\frac{b}{a}$ and $C=\frac{c}{a}$. For this to be zero, you need $x^2+Bx+C=0$. Now, imagine you could write it as a product, &#xA;$$x^2 + Bx+C = (x-r_1)(x-r_2).$$&#xD;&#xA;What would $r_1$ and $r_2$ be? If you know how to multiply polynomials, you get that $(x-r_1)(x-r_2)=x^2 - (r_1+r_2)x + r_1r_2$, so you need $r_1r_2=C$ and $r_1+r_2 = -B$. Squaring the latter you get $(r_1+r_2)^2 = B^2$; but&#xA;$(r_1+r_2)^2 = r_1^2 +2r_1r_2 + r_2^2$. On the other hand, &#xA;$$(r_1-r_2)^2 = r_1^2 - 2r_1r_2 + r_2^2 = (r_1^2+2r_1r_2+r_2^2) - 4r_1r_2 = B^2 - 4C.$$&#xD;&#xA;So $(r_1-r_2)^2 =B^2-4C$. Taking square roots, you have that $r_1-r_2 = \pm \sqrt{B^2-4C}$. And you already know that $r_1+r_2 = -B$. Adding them you get&#xA;$$2r_1 = -B\pm\sqrt{B^2-4C}\qquad\text{or}\qquad r_1 = \frac{-B\pm\sqrt{B^2-4C}}{2}$$&#xD;&#xA;and taking the difference between $r_1+r_2 = -B$ and $r_1-r_2 = \pm\sqrt{B^2-4C}$ you get&#xA;$$2r_2 = -B\mp\sqrt{B^2-4C}\qquad\text{or}\qquad r_2 = \frac{-B\mp\sqrt{B^2-4C}}{2}.$$&#xD;&#xA;So you get that $r_1 = \frac{-B+\sqrt{B^2-4C}}{2}$ and $r_2 = \frac{-B-\sqrt{B^2-4C}}{2}$, and plugging in $B=\frac{b}{a}$ and $C=\frac{c}{a}$ gives the usual quadratic formula. No way to find it without knowing how to multiply polynomials!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to Calculus (&lt;em&gt;added:&lt;/em&gt; I'm assuming you will &quot;get to Calculus&quot; because you tagged the question as being (algebra-precalculus), so presumably you are taking a course labeled as 'precalculus'; but this may not be the case. If you are not going to &quot;get to Calculus&quot;, then this paragraph will not tell you anything useful), you will find that there is a particular operation (differentiation, taking derivatives) which is &lt;em&gt;very&lt;/em&gt; useful and &lt;em&gt;very&lt;/em&gt; important. It tells you how fast a certain quantity is changing, and it can be used to find all sorts of useful things, like what production level will maximize profit in a factory, how big a dose of medicine and how often you should give to a patient based on how fast they metabolize it, and many other applications. Computing derivatives from first principles with an arbitrary function is pretty labor-intensive; but by reconizing a function as being &quot;made up&quot; (through sums, products, quotients, and compositions) of other, simpler, functions, makes it a very straighforward and easy job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in order to be able to recognize that a function is a product of two other functions, you first need to know how to multiply two functions together. Polynomials are one case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another situation occurs when the polynomials are measuring different things, and their product is somehow meaningful; maybe one polynomial gives you the length and the other polynomial gives you the width of a certain figure? Their product will be the area, which may be something you need to compute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And more generally, you can think of polynomials as &quot;abbreviations&quot; for more complicated operations that you are doing with numbers, just like you are thinking of multiplication as &quot;abbreviated addition&quot;. In that case, multiplying the two polynomials represents another complicated operation that you need to express in terms of the two simpler ones (addition and multiplication). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T04:17:12.877" LastActivityDate="2010-11-24T04:17:12.877" CommentCount="4" />
  <row Id="11389" PostTypeId="1" CreationDate="2010-11-22T20:57:48.267" Score="0" ViewCount="43" Body="&lt;p&gt;Let ( L, ≤ ) be a lattice, x, y, z ∈ L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to understand why  ( x ∧ z ) ∧ ( y ∧ z ) = z ∧ ( y ∧ x ) is mentioned as fact in  Y.N. Singh's &quot;Mathematical Foundation of Computer Science&quot; - Pg 157&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I prove this somehow, or is it an axiom?&lt;/p&gt;&#xA;" OwnerUserId="3740" LastActivityDate="2010-11-22T21:37:10.783" Title="In a lattice, ( x ∧ z ) ∧ ( y ∧ z ) = z ∧ ( y ∧ x )" Tags="&lt;order-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11390" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T21:08:40.630" Score="1" ViewCount="0" Body="&lt;p&gt;When you take calculus, you will need to factor a polynomial p as a product of two polynomials a and b.  If you know how polynomial multiplication works, then finding factorizations is easier.  Learn how to multiply now so that you can factor easily later. :)&lt;/p&gt;&#xA;" OwnerUserId="3331" LastActivityDate="2010-11-22T21:08:40.630" CommentCount="4" />
  <row Id="11391" PostTypeId="1" AcceptedAnswerId="11392" CreationDate="2010-11-22T21:08:54.667" Score="0" ViewCount="81" Body="&lt;p&gt;It is well known that $\text{Ext}(\mathbb Z_p,\mathbb Z)$ is the trivial group, because $\mathbb Z_p$ is projective; this seems to be in contradiction with the Exercise 1.1 in Hilton - Stammbach, pag. 88: the two sequences&#xA;$$&#xD;&#xA;0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon}\mathbb Z_3 \to 0&#xD;&#xA;$$&#xD;&#xA;and&#xA;$$&#xD;&#xA;0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon&amp;#39;}\mathbb Z_3 \to 0&#xD;&#xA;$$&#xD;&#xA;where $\mu$ is the multiplication by 3 in $\mathbb Z$ and $\epsilon(n)=n\pmod{3}$, $\epsilon&amp;#39;(n)=n+1\pmod{3}$ are not equivalent, and then represent different elements in the set of extensions of $\mathbb Z_3$ and $\mathbb Z$... Where am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-11-22T21:33:40.643" Title="Computing $\text{Ext}(\mathbb Z_p,\mathbb Z)$" Tags="&lt;commutative-algebra&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11392" PostTypeId="2" ParentId="11391" CreationDate="2010-11-22T21:18:16.347" Score="2" ViewCount="0" Body="&lt;p&gt;Well, $\mathbb Z_p=\mathbb Z/p\mathbb Z$ certainly isn't projective as a&#xA;$\mathbb Z$-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&#xA;Now $\mathrm{Ext}(\mathbb Z/p\mathbb Z,\mathbb Z)$ is cyclic of&#xA;order $p$. The zero element corresponds to the split extension of&#xA;$\mathbb Z$ by $\mathbb Z_p$. The other elements correspond to extensions&#xA;$$0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\mathbb Z\to0.\qquad\qquad\qquad(*)$$&#xD;&#xA;In this exact sequence the surjection can be taken to map $1\in \mathbb Z$&#xD;&#xA;to any given nonzero $a\in\mathbb Z/p\mathbb Z$. Thus there are&#xA;$p-1$ non-isomorphic extensions looking like $(*)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-22T21:33:40.643" LastActivityDate="2010-11-22T21:33:40.643" CommentCount="1" />
  <row Id="11393" PostTypeId="1" CreationDate="2010-11-22T21:21:12.863" Score="2" ViewCount="177" Body="&lt;p&gt;Suppose I can calculate the extraordinary cohomology encoded in &lt;em&gt;topological&lt;/em&gt; $K$-groups of a topological space $X$ with CW structure. What information does this give me about $C^{*}$-algebras associated with $X$? What is the algebraic analogue of topological suspension or the algebraic version of Bott Periodicity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bigger Question&lt;/strong&gt;: More generally, what is the &lt;em&gt;deep&lt;/em&gt; connection between topological $K$-theory and algebraic $K$-theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;: I've computed the reduced topological $K$-groups of a wedge sum of $m$ $n$-spheres. They are given by&#xA;\begin{eqnarray}&#xA;\tilde{K}^{p}_{\text{top}} \left(\bigvee^{m} S^{n} \right) \cong&lt;br&gt;&#xA;\mathbb{Z}^{m} \quad \text{if} \quad p + n \equiv 0 \mod 2 &#xA;\end{eqnarray} &#xA;and trivial otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is the positive integer $m$ the dimension of an algebra associated with the wedge sum? If so, is there an intuitive description of this algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-22T21:32:55.537" LastActivityDate="2010-11-22T23:05:01.803" Title="Topological vs. Algebraic $K$-Theory" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;&lt;algebraic-k-theory&gt;&lt;topological-k-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="11394" PostTypeId="2" ParentId="11389" CreationDate="2010-11-22T21:37:10.783" Score="2" ViewCount="0" Body="&lt;p&gt;There are at least two ways to do this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As Robin Chapman suggests, show that each side of the equation is just the infimum of the set $\{x,y,z\}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove that $\wedge$ is an associative and commutative operation, and then use this to reduce the left side to $x \wedge y \wedge (z \wedge z)$ which is $x \wedge y \wedge z$ because $z \wedge z = z$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-11-22T21:37:10.783" CommentCount="2" />
  <row Id="11395" PostTypeId="1" CreationDate="2010-11-22T21:50:01.963" Score="0" ViewCount="46" Body="&lt;p&gt;The upper-half plane $\mathbb H$ carries a hyperbolic metric and the geodesics are semicircles with base on the real line. We consider oriented geodesics. Let $x \in \mathbb H$ and let $v$ be a unit tangent vector at $v$. How to prove the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists a unique geodesic $\gamma$ on $\mathbb H$ such that $\gamma^\prime(0) = v$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3822" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-23T00:46:15.253" LastActivityDate="2010-12-23T00:46:15.253" Title="Geodesic on half-plane determined by tangent vector" Tags="&lt;differential-geometry&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11396" PostTypeId="2" ParentId="11365" CreationDate="2010-11-22T21:55:22.927" Score="3" ViewCount="0" Body="&lt;p&gt;Your question needs some more information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is their score generated (what kind of game is it)? What should your non-cheating data look like? &#xA;How do people cheat? How will their score be different (in a statistical sense) when they are not cheating?&#xA;Do you know roughly the proportion that are cheating? Or is that something you also want to find out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also look at outlier detection algorithms: wikipedia looks useful on this topic (&lt;a href=&quot;http://en.wikipedia.org/wiki/Outlier#Identifying_outliers&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;). Using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Q-Q_plot&quot; rel=&quot;nofollow&quot;&gt;Q-Q plot&lt;/a&gt; on your data may also be useful if your non-cheating data should be approximately Normally distributed; points that are significantly above the line might be cheaters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also try posting this on the stats.stackexchange.com since this is really more of a statistics question.&lt;/p&gt;&#xA;" OwnerUserId="3480" LastActivityDate="2010-11-22T21:55:22.927" />
  <row Id="11397" PostTypeId="2" ParentId="11371" CreationDate="2010-11-22T21:58:51.227" Score="1" ViewCount="0" Body="&lt;p&gt;$s$ and $u$ span a 2-dimensional plane, and the great circle that contains them is precisely this plane intersected with the unit sphere.  Let $p_i$ be the point of intersection of the great circle and the hyperplane $\{x_i = 0\}$.  Then you know $p_i$ is of the form $as + bu$, and you can solve the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$as_i + bu_i = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lVert as + bu \rVert = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $s_i$ and $u_i$ indicate the $i^{th}$ components of $s$ and $u$, respectively.  This gives you the (typically) two points where the the great circle crosses the hyperplane, and by finding the $i$ whose solution has all nonnegative coordinates, you should get your answer.&lt;/p&gt;&#xA;" OwnerUserId="3823" LastEditorUserId="3823" LastEditorDisplayName="" LastEditDate="2010-11-22T22:33:35.407" LastActivityDate="2010-11-22T22:33:35.407" />
  <row Id="11398" PostTypeId="1" CreationDate="2010-11-22T22:05:20.470" Score="14" ViewCount="117" Body="&lt;p&gt;This question is a follow up to &lt;a href=&quot;http://math.stackexchange.com/questions/11363/do-the-mobius-function-totient-function-sum-of-divisors-and-number-of-divisors&quot;&gt;this&lt;/a&gt; excellent mathematics stackexchange question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mu(n)$ be the Möbius function, $\phi(n)$ Euler's totient function, $\sigma(n)$ the sum of divisors function and $\tau(n)$ the number of divisors function. Define the set $S_N,$ for a natural number $N,$ by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_N = \lbrace (m,n) \in \mathbb{N} \times \mathbb{N} \mid m \ne n, \,&#xD;&#xA;\mu(m)=\mu(n), \, \phi(m)=\phi(n),$$&#xD;&#xA;$$\sigma(m)=\sigma(n), \, \tau(m)=\tau(n) \textrm{ and }&#xD;&#xA;\text{max} \lbrace m,n \rbrace \le N \rbrace .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How large is the set $ S_N $ ? &lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-22T22:05:20.470" Title="The order of the number of integer pairs satisfying certain arithmetical function relationships." Tags="&lt;number-theory&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="11400" PostTypeId="2" ParentId="872" CreationDate="2010-11-22T22:08:07.980" Score="0" ViewCount="0" Body="&lt;p&gt;The basic concept of the Hodge dual has been around since Grassmann -- it is essentially Grassmann's complement. The treatments of the Hodge dual you can find on the web are usually sketchy and unmotivated. My notes on vector spaces give a more thorough treatment than most of what you can find on the web; just google &quot;Vector Spaces, Vector Algebras, and Vector Geometries&quot;. --Richard     &lt;/p&gt;&#xA;" OwnerUserId="3824" LastActivityDate="2010-11-22T22:08:07.980" />
  <row Id="11401" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T22:10:21.410" Score="9" ViewCount="0" Body="&lt;p&gt;Suppose you want to make error-correcting codes (to do things like communicating with spacecraft, manufacturing CD's, and designing computer memory systems). Encoding and decoding BCH codes involves multiplying polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, nowadays the best error correcting codes are turbo codes or LPDC codes, which aren't based on polynomials. But for nearly 40 years many digital communications systems were based on multiplying polynomials.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-11-22T22:10:21.410" CommentCount="2" />
  <row Id="11402" PostTypeId="1" AcceptedAnswerId="11404" CreationDate="2010-11-22T22:22:24.053" Score="2" ViewCount="111" Body="&lt;p&gt;In a few examples i noted that the existence of k-regular graph on n vertices is :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;True , for k or n even.&lt;/li&gt;&#xA;&lt;li&gt;False , for k and n odd . But we can find a graph with n-1 vertices with&#xA;degree k and one vertex with degree  k-1. There doesn't exists a k-regular graph for &#xA;k and n odd  because k=deg(G) = 2*|E(G)| / |V(G)|&#xA;|E(G)| = k*n/2 , and |E(G)|= m is not a natural number if n and k is odd.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any proof idea ??&lt;/p&gt;&#xA;" OwnerUserId="3825" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T00:54:25.427" LastActivityDate="2010-11-23T00:54:25.427" Title="Existence of k-regular graph " Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11403" PostTypeId="2" ParentId="11343" CreationDate="2010-11-22T22:25:03.013" Score="2" ViewCount="0" Body="&lt;p&gt;It sounds like you may be asking for a representation of $$\sum_{i=1}^n \frac{1}{4i-1}$$ as a Riemann sum estimate of two different integrals, one of which underestimates the true value of the integral (giving you an upper bound on your sum), and the other of which overestimates the true value of the integral (giving you a lower bound on your sum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right-hand Riemann sum approximation of $$\int_0^n \frac{1}{4x-1} dx$$ with $\Delta x = 1$ is $\sum_{i=1}^n \frac{1}{4i-1}$ and will underestimate the integral.  However, the integrand is undefined at $x = 1/4$, and so you're better off looking at $$\frac{1}{3} + \sum_{i=2}^n \frac{1}{4i-1}$$ as an underestimate of $$\frac{1}{3} + \int_1^n \frac{1}{4x-1} dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the left-hand Riemann sum approximation of $$\int_0^n \frac{1}{4(x+1)-1} dx$$ with $\Delta x = 1$ is $\sum_{i=1}^n \frac{1}{4i-1}$ and will overestimate the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing the Riemann sum approximations and the graphs of the functions will help to see this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^n \frac{1}{4x-3} dx &amp;lt; \sum_{i=1}^n \frac{1}{4i-1} \leq \frac{1}{3} + \int_1^n \frac{1}{4x-1} dx,$$&#xD;&#xA;with equality holding in the second case only when $n=1$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-22T22:48:04.687" LastActivityDate="2010-11-22T22:48:04.687" />
  <row Id="11404" PostTypeId="2" ParentId="11402" CreationDate="2010-11-22T22:30:28.427" Score="2" ViewCount="0" Body="&lt;p&gt;There is a theorem (&lt;a href=&quot;http://mathworld.wolfram.com/GraphicSequence.html&quot; rel=&quot;nofollow&quot;&gt;Erdos-Gallai&lt;/a&gt;) on degree sequences:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle d_i$ is a degree sequence of some graph if and only if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{m} d_i \leq m(m-1) + \sum_{i=m+1}^{n} \min \{d_i, m\} \ \  \text{for} \ \ m \in \{1,2, \dots, n\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{n} d_i$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be easy to verify for the case when $\displaystyle d_i = k \ \  \forall i \in \{1,2, \dots, n\}$ (cumbersome verification at the end of the answer).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The case $\displaystyle k=n-1$, we trivially know the existence of a regular graph ($\displaystyle K_n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\displaystyle k \lt n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for $\displaystyle 1 \le m \le k$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m(m-1) + (k-m)m + (n-k)k -mk = nk - k^2 - m \ge nk-k^2 - k = k(n-k-1) &amp;gt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle  m \gt k$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m(m-1) + (n-m)k - mk = m^2 - m(2k+1) + nk = (m-(2k+1)/2)^2 +nk - ((2k+1)/2)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $k \lt n-1$ we have that $\displaystyle 4nk \gt 4k^2 + 4k$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $4nk \ge 4k^2 + 4k + 1 = (2k+1)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &#xA;$\displaystyle m(m-1) + (n-m)k - mk \ge 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $\displaystyle kn$ is even, there exists a $k$-regular graph on n vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other part I leave to you.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-22T23:11:52.450" LastActivityDate="2010-11-22T23:11:52.450" CommentCount="2" />
  <row Id="11405" PostTypeId="2" ParentId="227" CreationDate="2010-11-22T22:32:04.050" Score="0" ViewCount="0" Body="&lt;p&gt;Of interest to anyone who finds this page, a new stack exchange Economics site is nascent and needs your commitment if you can lend it: &lt;a href=&quot;http://area51.stackexchange.com/proposals/1618/economics?referrer=iym-fVnycBaB379vI_74Bg2&quot;&gt;http://area51.stackexchange.com/proposals/1618/economics?referrer=iym-fVnycBaB379vI_74Bg2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-22T22:32:04.050" CommunityOwnedDate="2010-11-22T22:32:04.050" />
  <row Id="11406" PostTypeId="2" ParentId="11395" CreationDate="2010-11-22T22:48:35.610" Score="1" ViewCount="0" Body="&lt;p&gt;Here is an attempt at solution without getting into the intricacies of Riemannian geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a geodesic for the upper-half plane have the center $(a,0)$ and radius $r$. Thus it has the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x-a)^2 + y^2 = r^2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifying a unit tangent vector at a point is a choice of a direction there, which might be given by a directed line passing through that point. Your requirement translates to the assertion that the slope of the circle at this point agrees with that of the directed line corresponding to the unit vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two unknowns in this situation, viz., $a$ and $r$. Partial differentiation of the circle equation to get the $x$ and $y$ components of slope would give two linear equations, for which there would be a unique solution.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T22:48:35.610" />
  <row Id="11408" PostTypeId="2" ParentId="8854" CreationDate="2010-11-22T22:53:10.733" Score="1" ViewCount="0" Body="&lt;p&gt;There is no translation in my knowledge. But there is a later book by Katz and Mazur on &quot;Arithmetic Moduli of Elliptic Curves&quot;, which is in English and which ought to a good amount of the things in the paper of Deligne and Rapoport.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-22T22:53:10.733" CommentCount="1" />
  <row Id="11409" PostTypeId="2" ParentId="11393" CreationDate="2010-11-22T23:05:01.803" Score="3" ViewCount="0" Body="&lt;p&gt;I cannot fully answer your questions, but hopefully I can help orient you.  I know a little about C*-algebra K-theory, and less about topological K-theory, so take this with a grain of salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $K_0$ group of a (complex) C*-algebra $A$ is defined as the Grothendieck group of a commutative monoid of equivalence classes of projections in the matrix algebras over $A$.  The $K_1$ group of $A$ can be defined as a group of equivalence classes of unitary elements of matrix algebras over $A$.  (The subscripts are used instead of superscripts because the corresponding functors are covariant rather than contravariant.)  To define higher $K_n$ groups, &quot;suspensions&quot; are used.  In this setting, the suspension $SA$ of $A$ is defined to be $C_0(0,1)\otimes A\cong C_0((0,1),A)$, which can be thought of as the C*-algebra of continuous $A$-valued functions on the unit interval that vanish at $0$ and $1$.  When $n$ is greater than $1$, $K_n(A)$ is defined to be $K_{n-1}(SA)$.  It turns out that this is consistent with the $n=1$ case, i.e., $K_1(A)$ is isomorphic to $K_0(SA)$.  Bott periodicity says that $K_{n+2}(A)$ is isomorphic to $K_n(A)$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the commutative case, there is a locally compact Hausdorff space $X$ such that $A$ is isomorphic to $C_0(X)$, the algebra of complex-valued continuous functions on $X$ vanishing at infinity.  The homeomorphism class of $X$ is determined by $A$, and the isomorphism class of $A$ is determined by $X$.  One can consider either the operator K-theory of $C_0(X)$ or the topological K-theory of $X$.  It turns out that the groups are isomorphic: $K_n(C_0(X))\cong K^n(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The K-groups don't tell you anything literally about the dimension of the C*-algebra.  For example, the infinite dimensional C*-algebra $C(S^2)$ of continuous complex-valued functions on the $2$-sphere has K-groups $K_0(C(S^2))=\mathbb{Z}^2$ and $K_1(C(S^2))=0$.  These are the same as the K-groups of the algebra $\mathbb{C}\oplus\mathbb{C}$ (which is the algebra of continuous complex-valued functions on a $2$-point discrete space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a terminological aside, usually a distinction is made between &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_K-theory&quot; rel=&quot;nofollow&quot;&gt;algebraic K-theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Operator_K-theory&quot; rel=&quot;nofollow&quot;&gt;K-theory of operator algebras&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some references.  For topological K-theory, see &lt;a href=&quot;http://books.google.com/books?id=RfArvjhOG44C&amp;amp;dq=atiyah+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Atiyah&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=fJzlLkHBYTYC&amp;amp;dq=karoubi+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Karoubi&lt;/a&gt;.  For operator K-theory, in increasing order of difficulty and assumed background, see &lt;a href=&quot;http://books.google.com/books?id=EVOjQgAACAAJ&amp;amp;dq=k-theory%20wegge-olsen&amp;amp;source=gbs_book_other_versions&quot; rel=&quot;nofollow&quot;&gt;Wegge-Olsen&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=SMiB8VIB5UIC&amp;amp;dq=rordam+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Rørdam et al.&lt;/a&gt;, and &lt;a href=&quot;http://books.google.com/books?id=214a1Wri63QC&amp;amp;dq=blackadar+k-theory&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Blackadar&lt;/a&gt;.  Blackadar's book is a wonderful reference and the most comprehensive, while the other two serve as good introductions.  (Rørdam et al. is the one I've studied the most.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some insight on the relationship between the two theories can be gained in light of &lt;a href=&quot;http://en.wikipedia.org/wiki/Swan%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Swan's theorem&lt;/a&gt; on the equivalence between finitely generated projective modules over $C(X)$ and vector bundles over $X$.  Roughly speaking, since projective modules are direct summands of free modules, they correspond to idempotents in the endomorphism rings of free modules, and in the C*-algebra case this corresponds to the projections in matrix algebras mentioned above.  Each of the operator K-theory books mentioned above includes at least some material on the relationship to topological K-theory.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-22T23:05:01.803" CommentCount="1" />
  <row Id="11410" PostTypeId="2" ParentId="11382" CreationDate="2010-11-22T23:14:42.253" Score="5" ViewCount="0" Body="&lt;p&gt;When solving real-world problems with simple algebra, it is not uncommon to have polynomials scattered throughout the equation. Given that 1/(3x+4)=2x-1, which isn't an extraordinarily complex equation, the first thing you might want to do is multiply both sides by 3x+4, which means you will need to be able to deal with (2x-1)*(3x+4).&lt;/p&gt;&#xA;" OwnerUserId="1995" LastActivityDate="2010-11-22T23:14:42.253" />
  <row Id="11411" PostTypeId="1" AcceptedAnswerId="11413" CreationDate="2010-11-22T23:25:46.047" Score="4" ViewCount="165" Body="&lt;p&gt;I was re-reading &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486661695&quot; rel=&quot;nofollow&quot;&gt;this old book&lt;/a&gt; of mine; and I noticed that in defining the rules of differential forms, it &quot;makes sense&quot; that we have the rule $dx \wedge dx=0$ because if $dx$ is infinitesimal, then to first order approximations we can ignore powers of $dx$.  Similarly, the definition for the exterior derivative $d$, of a differential form $\omega=Adx+Bdy+Cdz$, $d\omega=\frac{dA}{dx}dx + \frac{dB}{dy}dy + \frac{dC}{dz}dz $ &quot;makes sense&quot; because it feels like we are just multiplying the top and bottom by the differentials $dx,dy,$ and $dz$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is practically a miracle that by introducing the simple anti-symmetrical commutation relations for differential forms, and applying very elementary operations, we can arrive at all the results of vector calculus such as gradient and cross product, among a large amount of other well known results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular book, the authors motivate the anti-symmetry condition by properties of determinants and Jacobian's for change of variables in integration.  But I was wondering if there are other ways to think about why differential forms should commute anti-symmetrically which might provide some more intuition on just why this &quot;miracle&quot; works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-23T00:17:41.593" Title="Intuition behind $dx \wedge dy=-dy \wedge dx$." Tags="&lt;differential-forms&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11412" PostTypeId="1" AcceptedAnswerId="11423" CreationDate="2010-11-23T00:10:38.773" Score="8" ViewCount="207" Body="&lt;p&gt;Let $B$ be a Stone space (compact, Hausdorff, and totally disconnected).  Then I am basically certain (because of Stone's representation theorem) that if $a, b \in B$ are two distinct points in $B$, then $B$ can be written as a disjoint union $U \cup V$ of open sets where $a \in U, b \in V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can't seem to prove this directly.  The proof should be fairly straightforward, so I am sure I'm missing something obvious.  (As an exercise for myself, I'm trying to prove Stone's representation theorem, and I need this as a lemma.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-23T16:25:46.667" Title="Any two points in a Stone space can be disconnected by clopen sets" Tags="&lt;topology&gt;&lt;general-topology&gt;&lt;boolean-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11413" PostTypeId="2" ParentId="11411" CreationDate="2010-11-23T00:11:44.077" Score="5" ViewCount="0" Body="&lt;p&gt;One way of looking at the antisymmetric relation is a consequence of $dx∧dx=0$ (which feels intuitive to you). Applied to $(dx+dy)∧(dx+dy)=0$, we get $(dx∧dx)+(dx∧dy)+(dy∧dx)+(dy∧dy)=0$. So, $(dx∧dy)+(dy∧dx)=0$, so $(dx∧dy)=-(dy∧dx)$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T00:11:44.077" CommentCount="2" />
  <row Id="11414" PostTypeId="2" ParentId="11411" CreationDate="2010-11-23T00:17:41.593" Score="4" ViewCount="0" Body="&lt;p&gt;I like the motivation given by Jack Lee's book Introduction to Smooth Manifolds.  Roughly, we want to capture volume by the exterior algebra: say $\omega$ is a tensor that we want to apply to $n$ vectors to get the $n$-dimensional volume of the parallelogram they form. In the case $n=2$ for example, we should have $\omega(X,X) = 0$ since we get a line and not a 2-d region (so 0 area).  Now by linearity, $\omega(X,X) = 0$ forces $X$ to be alternating (as in Timothy's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algebra of forms is the algebra of alternating tensors.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-11-23T00:17:41.593" />
  <row Id="11415" PostTypeId="1" AcceptedAnswerId="11416" CreationDate="2010-11-23T00:23:01.947" Score="1" ViewCount="106" Body="&lt;p&gt;The chance to throw a 6 with one die is 1/6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And 6 times 1/6 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I throw with 6 dice, the chance to throw at least 1 six should be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I throw 6 dice, I sometimes don't throw any 6 at all..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come?&lt;/p&gt;&#xA;" OwnerUserId="3535" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T01:05:14.870" LastActivityDate="2010-11-23T01:05:14.870" Title="chance on throwing a six with 6 dice" Tags="&lt;probability&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="11416" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:29:31.463" Score="6" ViewCount="0" Body="&lt;p&gt;With $6$ dice there are $6^6$ possible outcomes. Of these, $5^6$ don't lead to six on any of the dice. So the number of outcomes that lead to at least one six are $6^6-5^6$, so the probability of at least one six is $\frac{6^6-5^6}{6^6}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T00:29:31.463" CommentCount="2" />
  <row Id="11418" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:43:14.107" Score="3" ViewCount="0" Body="&lt;p&gt;The &lt;i&gt;expected&lt;/i&gt; number of dice showing $6$ is $1$ when throwing six dice. The &lt;i&gt;probability&lt;/i&gt; to see at least one $6$ is $1 - (5/6)^6$, as explained in Timothy's answer.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T00:43:14.107" />
  <row Id="11419" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:44:02.953" Score="1" ViewCount="0" Body="&lt;p&gt;If you flip a coin, there's a 50-50 chance you get heads. If you flip two coins there's a $3/4$ chance you get at least one head, since there's a $1/2 \times 1/2 = 1/4 $ chance they're both tails. So similarly, there's a $5/6$ chance each die is not $6$. The chance they're all not $6$ is $(5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) = (5/6)^6$. So the chance that at least one is a six is $1 - (5/6)^6$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-23T00:44:02.953" />
  <row Id="11420" PostTypeId="2" ParentId="11415" CreationDate="2010-11-23T00:51:37.973" Score="3" ViewCount="0" Body="&lt;p&gt;The probability of the union of independent events (unlike disjoint events) is not the sum of the individual probabilities. If $E_i$ denotes the event &quot;$6$ is obtained on the $i$th throw&quot;, then $E_i$ are independent events, and it does not hold ${\rm P}(E_1  \cup  \cdots  \cup E_6 ) = {\rm P}(E_1 ) +  \cdots  + {\rm P}(E_6 )$. The left-hand side probability can be found as follows.&#xA;The complement of the event $E_1  \cup  \cdots  \cup E_6$ is $E_1^c  \cap  \cdots  \cap E_6^c $, where $E_i^c$ is the event &quot;$6$ is not obtained on the $i$th throw&quot;. Hence, &#xA;$$&#xD;&#xA;{\rm P}(E_1  \cup  \cdots  \cup E_6 ) = 1 - {\rm P}(E_1^c  \cap  \cdots  \cap E_6^c ).&#xD;&#xA;$$&#xD;&#xA;Now, the probability of the intersection of independent events is the product of their individual probabilities.&#xA;So, ${\rm P}(E_1^c  \cap  \cdots  \cap E_6^c ) = {\rm P}(E_1^c) \cdots {\rm P}(E_6^c) = (5/6)^6$.&#xA;Hence, ${\rm P}(E_1  \cup  \cdots  \cup E_6 ) = 1 - (5/6)^6$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T00:51:37.973" CommentCount="2" />
  <row Id="11421" PostTypeId="2" ParentId="11402" CreationDate="2010-11-23T00:54:03.867" Score="1" ViewCount="0" Body="&lt;p&gt;At the outset, you should assume that $k &amp;lt; n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n = 2m$ is even, construct a graph with vertex set $$ \{ X_i : i \in \mathbb{Z}_m \} \cup \{ Y_i : i \in \mathbb{Z}_m \}, $$ where $\mathbb{Z}_m$ is the integers modulo $m$. Connect $X_i$ to $Y_j$ if $j-i \in \{1,\ldots,k\}$, where subtraction is done modulo $m$. Each $X_i$ is connected to $Y_{i+1},\ldots,Y_{i+k}$, and each $Y_j$ is connected to $Y_{j-1},\ldots,Y_{j-k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = 2l$ is even, construct a graph with vertex set $$\{X_i : i \in \mathbb{Z}_n\}.$$ Connect $X_i$ and $X_j$ if $i \neq j$ and $i-j \in \{-l,\ldots,l\}$. This relation is symmetric (since $j-i = -(i-j)$), and every $X_i$ is connected to $X_{i-l},\ldots,X_{i+l}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mentioned, when both $n,k$ are odd, we have $2e = nk$ where $e$ is the number of edges (this formula is obtained by counting all endpoints of all edges), which is a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T00:54:03.867" CommentCount="4" />
  <row Id="11422" PostTypeId="2" ParentId="8814" CreationDate="2010-11-23T00:59:40.413" Score="2" ViewCount="0" Body="&lt;p&gt;Let $f$ be a symbol with the property that $f^n = n!$. Consider $d_n$, the number of ways of putting $n$ letters in $n$ envelopes so that no letter gets to the right person (aka derangements). Many people initially think that $d_n = (n-1)! = f^{n-1}$ (the first object has $n-1$ legal locations, the second $n-2$, ...). The correct answer isn't that different actually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_n = (f-1)^n$. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-11-23T00:59:40.413" CommunityOwnedDate="2010-11-23T00:59:40.413" />
  <row Id="11423" PostTypeId="2" ParentId="11412" CreationDate="2010-11-23T01:00:31.443" Score="7" ViewCount="0" Body="&lt;p&gt;Before proving the fact you want, we'll need the notion of quasicomponents and some basic propositions about it. In these terms, you are just asking if distinct points are in distinct quasicomponents. Let $X$ be a topological space. Given $x,y \in X$ , define $x \sim y$ if $X$ cannot be written as a disjoint union of open sets $U$ and $V$ containing $x$ and $y$, respectively. It is straightforward to verify that $\sim$ is an equivalence relation. The equivalence classes are called the quasicomponents of $X$. It's easy to prove that the quasicomponent of a point $x$ is the intersection of all closed-open subsets of $X$ containing $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every topological space, the component $C$ of a point $x$ is contained in the quasicomponent $Q$ of the point $x$. In fact, if $F$ is a closed-open set containing $x$, then $X = F \ \cup (X-F \ )$ is a separation of $X$. Since $C \cap F \ne \emptyset$, it follows that $C \subseteq F$. Thus we have $C \subseteq Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll prove now, based on Engelking's proof in the book 'General Topology', that in every compact Hausdorff space, components and quasicomponents coincide. Let $C$ and $Q$ as above. We just need to prove that the quasicomponent $Q$ is connected. Then will follow that $ Q = C$. Suppose that $Q = X_1 \cup X_2$, where $X_1, X_2$ are two disjoint closed subsets of the space $Q$. We have that $X_1$ and $X_2$ are closed in $X$, since $Q$ is closed in $X$. By normality of compact Hausdorff spaces, there exist disjoint open subsets $U,V$ of $X$ containing $X_1, X_2$, respectively. Hence, we have $Q \subseteq U \cup V$ and, by compactness, there exist closed-open sets $F_1, \ldots, F_k$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q \subseteq \bigcap_{i=1}^k F_i \subseteq U \cup V.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \bigcap_{i=1}^k F_i$ is clearly closed-open. Since $ \overline{U \cap F} \subseteq \overline{U} \cap F = \overline{U} \cap (U \cup V) \cap F = U \cap F$, the intersection $U \cap F$ is also closed-open. As $x \in U \cap F$, we have $Q \subseteq U \cap F$ and $X_2 \subseteq Q \subseteq U \cap F \subseteq U$. It follows that $X_2 \subseteq U \cap V = \emptyset$, which shows that the set $Q$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's simple to prove your statement. Let $B$ be a Stone Space. Since $B$ is compact Hausdorff, quasicomponents coincide with components. $B$ is totally disconnected, so the quasicomponent of a point $a \in B$ is $\{a \}$. If $b \in B$ is a different point, then $b$ isn't in the quasicomponent of $a$. Thus there exist disjoint open sets $U,V$ containing $a,b$, respectively, such that $B = U \cup V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; This was motivated by Pete's answer. Actually, to prove that a locally compact Hausdorff totally disconnected space $X$ is zero-dimensional, we just need what I've proved for compact Hausdorff spaces. Indeed, let $x \in X$ and $U$ an open set containing $x$. Since $X$ is regular, there is an open set $V$ containing $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$. Also, $\overline{V}$ is a totally disconnected space. Using the fact that quasicomponents and components coincide in compact Hausdorff spaces, we have that the quasicomponent of $x$ in $\overline{V}$ is $\{x\}$. Now, the compactness of $\overline{V}$ guarantees that there are closed-open sets $F_1, \ldots, F_k$ such that $x \in F_1 \cap \ldots \cap F_k \subseteq V$. Let $F$ be $\bigcap_{i=1}^k F_i$. Then $F \ $ is an open set in $V$ and since $V \ $ is open in $X$, $F \ $ is open in $X$. Also, $F \ $ is compact, since each $F_i$ is a closed subset of the compact space $\overline{V}$, and then $F \ $ is closed in $X$. We conclude that for each open set $U$ containing $x$, there is a closed-open set $F \ $ such that $x \in F \subseteq U$. Thus $X$ is zero-dimensional. &lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T16:25:46.667" LastActivityDate="2010-11-23T16:25:46.667" CommentCount="4" />
  <row Id="11424" PostTypeId="2" ParentId="11340" CreationDate="2010-11-23T01:06:46.327" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a practical procedure, which assumes that the Shuffler is cyclic. Construct some hash function $H$ from $k$-tuples of songs to $n$ bits (the parameters $k,n$ will be decided upon later). Store the value of $H$ on the first $k$ songs, and look for a matching value of $H$ on a future $k$-tuple. If $k$ is large enough, then the expected number of occurrences of this $k$-tuple in the entire cycle is $&amp;lt;1$, and so its likely that a repeat means that you've reached the end of the cycle. If $n$ is big enough (larger than $\sqrt{N}$, where $N$ is the period), then it's probable that matching hashes means matching $k$-tuples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to increase your certainty, you can store a much larger prefix, and then use it whenever the hash value matches. In fact, you can probably use several hash functions this way to optimize the procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't assume that the Shuffler is cyclic, but rather that from some point on it will cycle, you can store all the hashes in a hash table (along with their time) and wait for a hash that appears twice. If you're allowed to read the input twice, you can use any of the Pollard rho algorithms, as hardmath commented.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T01:06:46.327" CommentCount="1" />
  <row Id="11425" PostTypeId="2" ParentId="11343" CreationDate="2010-11-23T01:18:05.417" Score="0" ViewCount="0" Body="&lt;p&gt;You can write $$\frac{1}{4i-1} = \frac{1}{4i} + \frac{1}{4i(4i-1)}.$$ Therefore $$\sum_{i=1}^n \frac{1}{4i-1} = \frac{1}{4} \sum_{i=1}^n \frac{1}{i} + \sum_{i=1}^n \frac{1}{4i(4i-1)}.$$ The second term is bounded (by comparison with the convergent infinite series $\sum_{i=1}^\infty 1/i^2$), and the first term is familiar.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T01:18:05.417" />
  <row Id="11426" PostTypeId="1" CreationDate="2010-11-23T01:43:55.140" Score="0" ViewCount="168" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Show that the term ZZ where Z is λz.λx. x(z z x) satisfies the requirement for fixed point combinators that ZZM =β M(ZZM).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3829" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T04:43:05.273" LastActivityDate="2011-01-08T02:39:00.537" Title="Proving a combinator is a fixed point" Tags="&lt;lambda-calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11427" PostTypeId="2" ParentId="11362" CreationDate="2010-11-23T01:47:35.287" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not entirely sure how helpful this suggestion (this was supposed to be a comment, but it got very long) will ultimately be, but you might be interested in knowing that there exists a generalization of the usual Taylor series, called the (Lagrange-)Bürmann series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, given a function $f(x)$ to be expanded, a &quot;basis function&quot; $g(x)$, an expansion point $a$, and the assumption that $g^{\prime}(x)$ is nonzero over some interval containing $a$, such that $g(x)-g(a)$ is monotonic as $x$ increases over said interval, the Bürmann series of $f(x)$ with basis function $g(x)$ and expansion point $x=a$ reads as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=f(a)+\sum_{k=1}^{\infty} \frac{\beta_k(a)}{k!}(g(x)-g(a))^k$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $\beta_k(x)$ satisfy the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta_0(x)=f(x),\qquad \beta_k(x)=\frac1{g^{\prime}(x)}\frac{\mathrm d}{\mathrm dx}\beta_{k-1}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The series may or may not converge, of course, so the analysis of convergence (which &lt;em&gt;highly&lt;/em&gt; depends on the nature of your $f(x)$ and $g(x)$) is still your burden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not terribly popular, probably because it's difficult to choose $g(x)$ such that your expansion of $f(x)$ is meaningful, but it's worth a shot.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-23T01:47:35.287" CommentCount="4" />
  <row Id="11428" PostTypeId="1" AcceptedAnswerId="11432" CreationDate="2010-11-23T02:01:30.070" Score="3" ViewCount="161" Body="&lt;p&gt;How to solve  $y&amp;#39;&amp;#39;-\displaystyle\frac{y&amp;#39;}{x}=4x^{2}y$ ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the solution of this equation is: $y = e^{x^{2}}$, but I cannot resolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I thought that $z=y&amp;#39;$ could be, but was not.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-23T04:05:58.730" LastActivityDate="2010-11-23T04:05:58.730" Title="How to solve this differential equation?  $y''-\frac{y'}{x}=4x^{2}y$ " Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11429" PostTypeId="1" CreationDate="2010-11-23T02:16:19.473" Score="0" ViewCount="76" Body="&lt;p&gt;A loan for $50,000$ has level payments to be made at the end of each year for 10 years at an annual rate of $9 \%$. (a) Find the balance at the end of $3$ years. (b) Find the interest paid in the third payment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think this involves using amortization. That is, one pays the interest due first and then the principal afterwards. In other words, one covers the &quot;creeping costs&quot; first, and then pays the full amount. So this is a immediate annuity. So the interest due would be $50000(1.09)^{10}$?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-26T21:56:21.903" Title="Loan Repayment and Annuities" Tags="&lt;finance&gt;" AnswerCount="3" />
  <row Id="11431" PostTypeId="1" AcceptedAnswerId="11442" CreationDate="2010-11-23T02:17:14.730" Score="2" ViewCount="146" Body="&lt;p&gt;Today I was thinking about composition of functions.  It has nice properties, its always associative, there is an identity, and if we restrict to bijective functions then we have an inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I thought about commutativity.  My first intuition was that bijective self maps of a space should commute but then I saw some counter-examples. The symmetric group is only abelian if n&amp;lt;=2 so clearly there needs to be more restrictions on functions than bijectivity for them to commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only examples I could think of were boring things like multiplying by a constant or maximal tori of groups like O(n) (maybe less boring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: In a euclidean space, what are (edit) some nice characterizations of sets of functions that commute? What about in a more general space? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: Is this notion of commutativity important anywhere in analysis?&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2010-11-23T04:07:44.667" LastActivityDate="2010-11-23T04:07:44.667" Title="When functions commute under composition" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11432" PostTypeId="2" ParentId="11428" CreationDate="2010-11-23T02:17:31.750" Score="6" ViewCount="0" Body="&lt;p&gt;Note that we have $xy&amp;#39;&amp;#39; - y&amp;#39; = 4x^3y$. The LHS in some sense is dimensional consistent and looks something similar to a quotient rule provided we divide by $x^2$. So dividing by $x^2$ and doing algebraic manipulations we get $(\frac{1}{x}y&amp;#39;)&amp;#39; = 4xy$. Now this looks familiar to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting, we get $(\frac{y&amp;#39;}{2x})&amp;#39; = 2xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\frac{y&amp;#39;}{2x} = z(x)$. Plug this in and simplify to get $z&amp;#39; = \frac{y&amp;#39;}{z}y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Replace $2x$ by $\frac{y'}{z}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $z^2 = y^2 + c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have now converted a second order differential equation in terms of first order differential equation, viz,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2x}\frac{dy}{dx} = \pm \sqrt{y^2 + c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You could plug this in and check that this satisfies the second order differential equation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need other conditions (boundary/ initial conditions) to completely solve the problem i.e. to eliminate $c$ and other constant which will arise after solving the first order differential equation to get $y(x) = \exp(x^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that taking $c =0 $ we get a simple ode and the solution to which is $y(x) = y(0) \exp(\pm x^2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first order ODE can be solved as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{CASE 1:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c &amp;gt; 0$, then let $c = a^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{\sqrt{y^2 + a^2}} = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a \tan(\theta)$, we get $dy = a \sec^2(\theta) d\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the ode now becomes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) d\theta = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute for $\theta$ in terms of $y$ to get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{y}{a} \pm \sqrt{1+(\frac{y}{a})^2} = K \exp(\pm x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{CASE 2:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c &amp;gt; 0$, then let $c = -a^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{\sqrt{y^2 - a^2}} = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a \sec(\theta)$, we get $dy = a \sec(\theta) \tan(\theta) d\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the ode now becomes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) d\theta = \pm d(x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute for $\theta$ in terms of $y$ to get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{y}{a} \pm \sqrt{(\frac{y}{a})^2 - 1} = K \exp(\pm x^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{CASE 3:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $c = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation, we have now is $\frac{dy}{dx} = \pm 2xy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving, we get $y(x) = K \exp(\pm x^2)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T02:58:11.263" LastActivityDate="2010-11-23T02:58:11.263" CommentCount="1" />
  <row Id="11433" PostTypeId="2" ParentId="11429" CreationDate="2010-11-23T02:31:31.470" Score="0" ViewCount="0" Body="&lt;p&gt;You need to determine the payment so you come out even at the end of 10 years.  There should be a formula in your text for that.  If I recall correctly, it looks like payment = interest * principal/(1-(1+i)^-t), but I don't know what symbols you use for the terms.  Then you are right, as each payment is made you deduct the interest that has accrued in the past year, then apply the rest to the principal.  The interest due in the third year will be somewhat less than 9% of 50000 because the principal will have been reduced by the first two year's payments.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T02:31:31.470" />
  <row Id="11434" PostTypeId="1" CreationDate="2010-11-23T02:37:24.847" Score="0" ViewCount="135" Body="&lt;p&gt;Hello, I need help studying for an exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give a recursive form (including bases) for the following functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;f(n) = 5 + (-1)^n&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;f(n) = n(n+3)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(0) = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S(1) = 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derive the closed form for this recursion and prove it true by induction.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2010-11-27T18:02:32.883" Title="How do I do Recursion?" Tags="&lt;discrete-mathematics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="11435" PostTypeId="1" CreationDate="2010-11-23T02:55:16.287" Score="4" ViewCount="123" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What is the correct notion of a &lt;em&gt;product&lt;/em&gt; of integral (or rational) polytopes which induces a factorization of its Ehrhart (quasi-)polynomial into two primitive Ehrhart (quasi-)polynomials corresponding to its constituent polytopes, viz., $L_{P \times Q}(t) = L_{P}(t) L_{Q}(t)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Motivation&lt;/em&gt;) Given two closed integral polytopes $P$ and $Q$ each with vertices at $\{ \mathbf{0} , b_{1} \mathbf{e}_{1}, \dots, b_{n} \mathbf{e}_{n} \}$ and $\{ \mathbf{0} , d_{1} \mathbf{e}_{1}, \dots, d_{m} \mathbf{e}_{m} \}$, respectively, where $n, b_{i}, m, d_{j} \in \mathbb{N}$, define the integral polytope $R$ with vertices at $\{ \mathbf{0}, b_{1} \mathbf{e}_{1}, \dots, b_{n} \mathbf{e}_{n}, d_{1} \mathbf{e}_{n+1}, \dots, d_{m} \mathbf{e}_{n+m} \}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above construction cannot be the sought after product $P \times Q$. Suppose $P$ and $Q$ are defined by $b_{1} = b_{2} = d_{1} = d_{2} = 2$. It is easy to show that $L_{P}(1) = L_{Q}(1) = 6$. Define $R$ as above with vertices of $P$ and $Q$. It is true that $L_{R}(1) = 15 \neq 6^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What is $R$ in terms of $P$ and $Q$? Is it special in some way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-23T04:41:18.210" LastActivityDate="2010-11-23T16:55:25.710" Title="Product of Polytopes" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" />
  <row Id="11436" PostTypeId="2" ParentId="11428" CreationDate="2010-11-23T02:57:05.413" Score="4" ViewCount="0" Body="&lt;p&gt;$\rm\quad\quad\quad \ y&amp;#39;\ =\ f\ y $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\Rightarrow\ y&amp;#39;&amp;#39;\ =\ f\:&amp;#39;\ y + f^{\:2}\ y $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle\Rightarrow\ y&amp;#39;&amp;#39;\ = \frac{f\:&amp;#39;}f\ y&amp;#39; + f^{\:2}\ y $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\rm\quad\displaystyle y&amp;#39;&amp;#39;\ = \ \frac{1}x\ y&amp;#39; + 4x^2\ y\ \ \Rightarrow\ \ f\ =\ \pm 2\:x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\rm\ y&amp;#39;\ =\ \pm 2\:x\ y\ \ \Rightarrow\ \ y\ =\ c\ e^{\pm x^2}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T03:02:41.553" LastActivityDate="2010-11-23T03:02:41.553" CommentCount="6" />
  <row Id="11438" PostTypeId="1" AcceptedAnswerId="11445" CreationDate="2010-11-23T03:04:25.443" Score="2" ViewCount="94" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to compute the Kolmogorov distance between two univariate gaussian distributions $\mathcal{N}(0,n)$ and $\mathcal{N}(0,2n)$ for large $n$.  I have a feeling this should be simple but whatever I have tried so far doesn't work.  Could anyone give me some hints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Kolmogorov distance between two distributions $P$ and $Q$ I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle max_t | \Pr[P&amp;gt;t] - \Pr[Q&amp;gt;t] | $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="3831" LastActivityDate="2010-11-23T03:38:16.203" Title="Kolmogorov distance between univariate gaussians" Tags="&lt;real-analysis&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11439" PostTypeId="2" ParentId="11434" CreationDate="2010-11-23T03:22:31.793" Score="3" ViewCount="0" Body="&lt;p&gt;Recursion refers to a method where you figure out the &quot;next&quot; value in terms of previous values. For example, to solve the &lt;a href=&quot;http://math.stackexchange.com/questions/2650/how-to-prove-the-optimal-towers-of-hanoi-strategy&quot;&gt;Towers of Hanoi&lt;/a&gt; puzzle, you can explain the solution recursively: to move $n$ disks, first move the top $n-1$ disks to the auxiliary rod, then move the bottom disk to the target rod, then move the other $n-1$ disks to the target rod. The &quot;move the top $n-1$ disks&quot; parts of the instruction are recursive, because you need to know how to solve the $n-1$ case. And you solve that case by solving the $n-2$ case. And you solve &lt;em&gt;that&lt;/em&gt; case by solving the $n-3$ case. And you ... etc. Eventually you need to touch bottom, which you do at the $n=1$ case, the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's think about $f(n)=5+(-1)^n$. Can you express $f(n+1)$ in terms of $f(n)$? If so, then this expression plus the value of $f(0)$ will give you the recursive forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To guide you, let me do a different example. Consider $f(n) = 2^{n}-1$. How do we express it recursively? First, I try to see if I can express $f(n+1)$ in terms of $f(n)$:&#xA;$$f(n+1) = 2^{n+1}-1 = 2(2^n) - 1 = 2\Bigl((2^n - 1)+1\Bigr) -1 = 2(f(n)+1)-1 = 2f(n)+1.$$&#xD;&#xA;(Why did I do that? Because I need to &quot;find&quot; the expression $2^n-1$ somehow &lt;em&gt;in&lt;/em&gt; the expression $2^{n+1}-1$, so that I can replace it with $f(n)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, together with the fact that $f(0)=2^0-1 = 0$ gives me the recursive form:&#xA;\begin{align*}&#xA;f(0) &amp;amp;= 0;\\&#xA;f(n+1) &amp;amp;= 2f(n) + 1.&#xA;\end{align*}&#xA;Try something similar with your two $f(n)$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your &quot;lastly&quot;, I have no idea what you have. Giving just the first two values does not tell you what a &quot;closed form&quot; for $S$ will be (here, &quot;closed form&quot; is the opposite process: if you know the recursive formula, you want to find a formula of the form $S(n) = $blah, where blah depends only on $n$ and not on previous values of $S$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T03:22:31.793" />
  <row Id="11440" PostTypeId="2" ParentId="11434" CreationDate="2010-11-23T03:23:38.097" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ (-1)^{n+1}/(-1)^n = -1,\ $ so substituting $\rm\ (-1)^n\ =\ f(n)-5\ \Rightarrow\ \ldots$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, note that $\rm\ f(n+1) - f(n)\ $ reduces the degree of polynomial $\rm\ f(n)\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-23T03:23:38.097" />
  <row Id="11441" PostTypeId="1" CreationDate="2010-11-23T03:30:13.690" Score="1" ViewCount="68" Body="&lt;p&gt;I'd like to know about the best numerical methods for approximating an integral.  Unfortunately, I want to know about a fairly general case, so I cannot give a lot of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I have an integral that is full of exponential functions and little else.  I therefore assume as given, although possibly incorrectly, that the power series expansion of the integrand can be integrated to give an exact value to the integral.  In other words, we can take the power series expansion of the integrand (the function inside the integral), and integrate it to obtain the same value as integrating the integral itself gives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then &quot;define&quot; the value $n$, saying that we need at least $n$ terms of the power series expansion to integrate to the wanted degree of accuracy/precision.  I also &quot;define&quot; $m$, such that by numerical methods, the integral must be evaluated at at least $m$ points to obtain the same degree of accuracy/precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hoping I've stated the integral and defined some restraints well, I have some curiosities.  I'm wondering how well numerical methods can hope to achieve these requirements.  In other words, how many calculations, and how much memory, etc. is needed to approximate this integral?  For example, if we know that we have to evaluate the integral at $m$ points, does it take $O(m)$ calculations to evaluate this integral, or perhaps $O(m \log(m))$?  Similarly, if we know that the power series expansion requires $n$ terms to get the job done, how much computing power do we need?  $O(n)$?  $O(n \log(n))$?  Some other value?  I know that this is an extremely general case, but I'm curious as to how much computing requirements various methods take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking because I want to compare a method that I'm experimenting with.  I doubt that I can do as well as the best methods, but I would like to know what the best method is so I can eventually use it.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-23T03:30:13.690" Title="Numerical Methods:  Approximating an Integral with Exponentials" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;intuition&gt;&lt;numerical-methods&gt;&lt;numerical-optimization&gt;" CommentCount="4" />
  <row Id="11442" PostTypeId="2" ParentId="11431" CreationDate="2010-11-23T03:32:23.577" Score="8" ViewCount="0" Body="&lt;p&gt;A &lt;a href=&quot;http://www.ams.org/journals/tran/1923-025-03/S0002-9947-1923-1501252-3/S0002-9947-1923-1501252-3.pdf&quot; rel=&quot;nofollow&quot;&gt;classic result of Ritt&lt;/a&gt; shows that polynomials that commute under composition must be, up to a linear homeomorphism, either both powers of $x$, both iterates of the same polynomial, or both  Chebychev polynomials. Actually Ritt proved a more general rational function case - follow the link. His work was motivated by work of Julia and Fatou's work on Julia sets of rational functions, e.g. see &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.47.8482&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for a modern presentation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T03:45:15.153" LastActivityDate="2010-11-23T03:45:15.153" CommentCount="3" />
  <row Id="11443" PostTypeId="1" CreationDate="2010-11-23T03:36:51.807" Score="9" ViewCount="166" Body="&lt;p&gt;Let $n$ be a positive integer. Suppose we have an equilateral polygon in the Euclidean plane with the property that all angles except possibly two consecutive ones are an integral multiple of $\pi/n$, then all angles are an integral multiple of $\pi/n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is #28 on page 61 in these notes restated here for convenience:&#xA;&lt;a href=&quot;http://websites.math.leidenuniv.nl/algebra/ant.pdf&quot; rel=&quot;nofollow&quot;&gt;http://websites.math.leidenuniv.nl/algebra/ant.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen a number-theoretic proof of this. I was wondering if there are any geometric (or at least non number-theoretic) proofs of this result.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-12T01:02:54.250" LastActivityDate="2010-12-12T01:02:54.250" Title="Equilateral polygon in a plane" Tags="&lt;geometry&gt;&lt;algebraic-number-theory&gt;" CommentCount="3" />
  <row Id="11444" PostTypeId="2" ParentId="11431" CreationDate="2010-11-23T03:37:47.033" Score="4" ViewCount="0" Body="&lt;p&gt;According to Wikipedia, a set of diagonalizable matrices commute if and only if they are simultaneously diagonalizable. There is a far-reaching generalization, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelfand_representation#Statement_of_the_commutative_Gelfand-Naimark_theorem&quot; rel=&quot;nofollow&quot;&gt;Gelfand representation theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gelfand representation theorem for commutative $C^*$ algebras represents every commutative $C^*$ algebra as an algebra of functions with pointwise multiplication; the domain of the latter algebra is the spectrum of the former algebra.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T03:37:47.033" CommentCount="2" />
  <row Id="11445" PostTypeId="2" ParentId="11438" CreationDate="2010-11-23T03:38:16.203" Score="2" ViewCount="0" Body="&lt;p&gt;Because the Gaussians have the same mean and Gaussians are symmetric about the mean, the expression $\Pr[P&amp;gt;t] - \Pr[Q&amp;gt;t]$ should have a single maximum and a single minimum whose absolute values are the same.  So you could just solve $\max_t (\Pr[P&amp;gt;t] - \Pr[Q&amp;gt;t])$.  To do that, I would express $\Pr[P&amp;gt;t]$ as &#xA;$$\int_t^{\infty} \frac{1}{\sqrt{2\pi n}} e^{-x^2/(2n)} dx,$$&#xD;&#xA;and $\Pr[Q&amp;gt;t]$ similarly.  Then take the derivative of the differences with respect to $t$ via the Fundamental Theorem of Calculus, set it to $0$, and solve for $t$.  Since you don't want a full answer, I'll stop there. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-23T03:38:16.203" CommentCount="3" />
  <row Id="11446" PostTypeId="2" ParentId="11426" CreationDate="2010-11-23T03:42:29.170" Score="0" ViewCount="0" Body="&lt;p&gt;Hint: Expand ZZM using the definition of Z &lt;i&gt;once&lt;/i&gt; (for the outer Z).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T03:42:29.170" />
  <row Id="11447" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T04:16:16.673" Score="2" ViewCount="0" Body="&lt;p&gt;In terms of simply the math, it is essentially a skill you need to learn. Mathematics becomes (a long with many other sciences) more &lt;em&gt;abstract&lt;/em&gt; as you go up in rank, so to speak. You can think of multiplying and adding polynomials as the next logical step after learning how to multiply decimals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a natural progression, that is both historic and customary to progress from the most concrete to the more abstract. In grade school, the first mathematical concept you learned is numbers, shortly thereafter you were introduced to how you could &lt;em&gt;combine&lt;/em&gt; two numbers to form a third one (namely through addition and multiplication). After being introduced to natural numbers, the concept of combining two numbers was expanded to something called fractions, which also introduced the concept of division (the opposite of multiplication, if you will). After getting comfortable with fractions, you were introduced to decimals and &lt;em&gt;irrational&lt;/em&gt; numbers, and the rules for combining and manipulating those numbers. At this point, mathematical education turns to more abstract concepts, the idea of &lt;em&gt;functions&lt;/em&gt; and &lt;em&gt;variables&lt;/em&gt;. You spend plenty of time (depending on how far you go in math, maybe the rest of your mathematical career) studying functions, how they behave, how to manipulate them. It turns out that polynomials, are some of the simplest functions you can think of, so it is only natural that you start with them. Following the trend as before, after understanding what functions (specifically polynomials) &lt;em&gt;are&lt;/em&gt;, a natural next step is considering how you can &lt;em&gt;combine&lt;/em&gt; those polynomials to make new polynomials. Later on in math, you will be introduced with even more ways of combining and manipulating functions to create new functions. This is where you are right now, you are learning to manipulate and combined some (if you think about it) incredibly abstract concept, you are learning to manipulate an object that took thousands of years of human thought to finalize. Of course, in reality this is just a small stepping stone to the greener pastures ahead, assuming you go on in mathematics you will encounter some of the most mind boggling abstract concepts (some of the negatively voted answers to this question will give you just a small hint of whats out there). You will find that the very simple concepts of addition and multiplication (and their respective inverses) will go a long way in the study of mathematics, but they are stripped of all their cuddly original meaning, to become the most beefed up behemoths of abstract mathematical thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, enough with the theoretical &quot;its the way it should be&quot;. Lets move on to practicalities. Applications of numbers, are easy enough to see, addition,multiplication, subtraction, and division of numbers is an everyday thing. Considering how to apply polynomials to your everyday life (especially why you would ever think of combining polynomials) requires a little greater stretch of the mind (or a little more mathematics :D ). Without appealing to calculus (which is certainly an easy cop-out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polynomials, Describe a large variety of every day processes and events. The most common being the quadratic polynomial (this is due to a very deep physical fact, but I promised to avoid calculus). The free fall trajectory of an object in a gravitational field is described (neglecting friction) by a quadratic polynomial. Even in simple physical problems you can see the multiplication of monomials and quadratics. For example, solving for when a shot cannon hits the ground is merely the root of the polynomial, conversely, if you know where the canon hit the ground you can multiply to monomials $(x-r)$ where $r$ is the root, to get the trajectory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we deviate a little from the math you have been exposed to, there are more abstract polynomials called &lt;a href=&quot;http://en.wikipedia.org/wiki/Laguerre_polynomials&quot; rel=&quot;nofollow&quot;&gt;Laguerre Polynomials&lt;/a&gt; that are present in the solution to the hydrogen atom. So if you (for some reason) wanted to calculate the shape of a 3D orbital of the hydrogen atom, you would need to multiply a lot of polynomials. Related polynomials play important roles in other problems in quantum mechanics and classic physics (Legendre Polynomials for example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this clears some of it up, unfortunately in a large sense it is really a stepping stone, into the world of higher mathematics. They are important on their own, but for reasons that... also become much more obvious later on. &lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-11-23T04:16:16.673" CommentCount="3" />
  <row Id="11448" PostTypeId="1" AcceptedAnswerId="11449" CreationDate="2010-11-23T04:22:22.423" Score="4" ViewCount="185" Body="&lt;p&gt;Let &lt;em&gt;n&lt;/em&gt; and &lt;em&gt;k&lt;/em&gt; be integers with 1 $\leq$ k $\leq$ n. Show that $\sum_{k=1}^n {n \choose k}{n \choose k-1} = \frac {2n+2 \choose n+1} {2} - {2n \choose n} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told this is supposed to use a combinatorial proof and while I'm not that comfortable with that, many similar proofs use a mathematical proof that shows equality. Any guidance would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-23T15:40:32.217" LastActivityDate="2010-12-25T02:48:04.800" Title="Combinatorial Proof of Binomial Coefficient Summation" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" />
  <row Id="11449" PostTypeId="2" ParentId="11448" CreationDate="2010-11-23T04:30:10.307" Score="8" ViewCount="0" Body="&lt;p&gt;The Combinatorial Proof goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider three sets of boxes $A$,$B$ and $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Box $A$ contains $n$ balls, Box $B$ contains $n$ balls, and Box $C$ contains $2$ balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to choose $n+1$ balls out of these three boxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways is $C(2n+2,n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be split as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You choose $2$ balls from box $C$ and the remaining $n-1$ balls from boxes $A$ and $B$ or you choose $1$ ball from box $C$ and the remaining $n$ balls from boxes $A$ and $B$ or you choose all the $n+1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Case 1:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you choose $2$ balls from box $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one way of choosing these $2$ balls from the box $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need the number of ways of choosing $n-1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if you choose $k$ balls from box $A$, you need to choose $n-k-1$ balls from box $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways of choosing $k$ balls from box $A$ and $n-k-1$ balls from box $B$ is given by $C(n,k)C(n,n-k-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now summing over all possible values of $k$ from $0$ to $n-1$ we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total number of ways is $\displaystyle \sum_{k=0}^{n-1} C(n,k)C(n,n-k-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $C(n,r) = C(n,n-r)$. (For a combinatorial argument of this see the bottom of the post).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, Total number of ways for the current case is $1 \times \displaystyle \sum_{k=0}^{n-1} C(n,k)C(n,k+1) = \displaystyle \sum_{k=1}^{n} C(n,k-1)C(n,k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Case 2:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you choose $1$ ball from the box $C$. You need to now choose the remaining $n$ balls from boxes $A$ and $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways of choosing $1$ ball from box $C$ is given by $C(2,1) = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $2n$ balls in total in boxes $A$ and $B$ and hence the total number of ways of choosing $n$ balls from boxes $A$ and $B$ is $C(2n,n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the total number of ways for the current case is $2 \times C(2n,n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Case 3:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you choose all the $n+1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways of choosing no balls from box $C$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now need to count the number of ways of choosing $n+1$ balls from boxes $A$ and $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to Case 1 and we get the total number of ways for this case is $1 \times \displaystyle \sum_{k=1}^{n} C(n,k)C(n,n+1-k) = \displaystyle \sum_{k=1}^{n} C(n,k)C(n,k-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: Here $k$ goes from $1$ to $n$ since we need to choose atleast $1$ from one of the boxes and we can choose atmost $n$ from each box).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also Note: You could directly state that Case 1 and Case 3 should be the same since $C(2n,n-1)=C(2n,n+1)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine all the three cases to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \times \displaystyle \sum_{k=1}^{n} C(n,k)C(n,k-1) + 2 \times C(2n,n) = C(2n+2,n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now do the desired algebra to get your result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{APPENDIX:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combinatorial argument of $C(n,r) = C(n,n-r)$ is that the number of ways of selecting $r$ out of $n$ things is the same as the number of ways of discarding $n-r$ out of $n$ things.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-25T02:48:04.800" LastActivityDate="2010-12-25T02:48:04.800" CommentCount="6" />
  <row Id="11450" PostTypeId="2" ParentId="11448" CreationDate="2010-11-23T04:32:02.817" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: Use $\displaystyle {n \choose k} = {n \choose n-k}$ and try to count something which equals $\displaystyle {n \choose n-k}{n \choose k-1}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-23T04:32:02.817" />
  <row Id="11451" PostTypeId="1" CreationDate="2010-11-23T04:33:50.287" Score="2" ViewCount="136" Body="&lt;p&gt;Hello, I need help studying for an exam. How do I find &quot;Big Oh,&quot; &quot;Big Omega,&quot; and &quot;Big Theta&quot;?&#xA;How do I combine this with Induction?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prove that $f(x)$ is $O(x^n)$, where&#xA;$$f(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_2 x^2 + a_1 x + a_0$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find &quot;Big O&quot;: (Usually log in Computer Science is assumed to be base 2)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(2n+\log(n!)+3)(n^2+3\log n)(2^n+2)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$(3n+1)\log(n^5+7)+n!\log(n+\log n)$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each function, circle the sets that $f(n)$, $g(n)$ and $h(n)$ belong to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(n) = n\log(n!) + 100$: $O(g(n))$, $\Theta(g(n))$, $\Omega(g(n))$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$g(n) = (n \log n)^2 + n^2$:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$O(n^3)$, $\Theta(n^3)$, $\Omega(n^3)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n^2)$, $\Theta(n^2)$, $\Omega(n^2)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n \log n)$, $\Theta(n \log n)$, $\Omega(n \log n)$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;$h(n) = n(1000 + 2^n) + n!$:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$O(2^n)$, $\Theta(2^n)$, $\Omega(2^n)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n!)$, $\Theta(n!)$, $\Omega(n!)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$O(n^n)$, $\Theta(n^n)$, $\Omega(n^n)$&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T04:57:16.560" LastActivityDate="2010-11-23T05:56:40.463" Title="How do I find &quot;Big Oh,&quot; &quot;Big Omega,&quot; and &quot;Big Theta&quot;?" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11452" PostTypeId="1" AcceptedAnswerId="11462" CreationDate="2010-11-23T04:36:04.340" Score="0" ViewCount="62" Body="&lt;p&gt;I have a formula which is G-force = velocity*omega/9.8. Omega is the angular velocity.&#xA;I've seen on the internet that G force is actually acceleration/9.8. I'm confused as to which formula is correct. For simulating the motion of particle taking a turn, would omega simply be velocity divided by radius of turn? Assuming Cartesian coordinates.&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-11-23T05:40:21.483" Title="Calculation of G force" Tags="&lt;computer-science&gt;&lt;physics&gt;&lt;simulation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11453" PostTypeId="1" CreationDate="2010-11-23T04:37:59.397" Score="4" ViewCount="137" Body="&lt;p&gt;I found this puzzle posted on a student website I frequent, but no one including the poster nor I could solve it. So I'm posting this puzzle here with hope that some of math whiz in here could illuminate on how to approach it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is an alien on the surface of a&#xA;  spherical planet. The alien can run at a top&#xA;  speed of $u$. An alien hunter is hunting&#xA;  the alien in his spaceship which can&#xA;  fly at a top speed of $v$. Once the&#xA;  hunter sees the alien, he fires, and&#xA;  the alien dies. Show that the alien&#xA;  will always die if $v &amp;gt; 10u$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that there's no information given about the planet radius nor the altitude at which the spaceship hovers so I'm suspecting that the solution somehow ignores them or they get canceled out somewhere in the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p/s: again I repeat that I do not know of the solution myself&lt;/p&gt;&#xA;" OwnerUserId="3803" LastActivityDate="2010-11-23T05:28:01.250" Title="Alien vs Alien Hunter Puzzle" Tags="&lt;geometry&gt;&lt;puzzle&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11454" PostTypeId="1" CreationDate="2010-11-23T04:40:57.877" Score="0" ViewCount="118" Body="&lt;p&gt;Help would be appreciated. The notes are poor on the subject, and im clueless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verify the following equalities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) SIII=βI, where S is λxyz.(xz)(yz) and I is λx.x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) twice (twice) f x= β f(f(f(f x))), where twice is λfx.f(f x)&lt;/p&gt;&#xA;" OwnerUserId="3829" LastActivityDate="2010-12-23T06:44:25.983" Title="lambda calculus, equalities" Tags="&lt;lambda-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11455" PostTypeId="2" ParentId="11453" CreationDate="2010-11-23T04:49:11.133" Score="2" ViewCount="0" Body="&lt;p&gt;There's a similar puzzle &lt;a href=&quot;http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/challenges/May2001.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (with solution).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T04:49:11.133" CommentCount="2" />
  <row Id="11456" PostTypeId="2" ParentId="11454" CreationDate="2010-11-23T04:58:34.577" Score="0" ViewCount="0" Body="&lt;p&gt;Try substituting the arguments in the definitions of the combinators.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T06:15:58.457" LastActivityDate="2010-11-23T06:15:58.457" CommentCount="4" />
  <row Id="11457" PostTypeId="2" ParentId="11412" CreationDate="2010-11-23T05:04:11.173" Score="5" ViewCount="0" Body="&lt;p&gt;The point of this answer is mostly to provide some terminology with which to express the distinction here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space is &lt;strong&gt;totally disconnected&lt;/strong&gt; if the only nonempty connected subsets are the singleton sets.  Equivalently, the connected components are all singleton sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space is &lt;strong&gt;zero-dimensional&lt;/strong&gt; if it admits a base of clopen [i.e., both open and closed] sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A topological space is &lt;strong&gt;separated&lt;/strong&gt; (more traditionally: $T_1$) if all the singleton sets are closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First easy observation: A totally disconnected space is separated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proof: Otherwise the closure of a point would give a larger connected component.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second easy observation: In a separated zero-dimensional space $X$, for any two distinct points $x_1$, $x_2$, there exists a separation $X = U_1 \coprod U_2$ with $x_i \in U_i$. &#xA;(In the terminology of Nuno's answer -- which is relatively standard if not well-known -- the conclusion is that the quasi-components are singleton sets.) In particular, $X$ is totally disconnected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Proof: Let $x_1$, $x_2$ be distinct points in $X$.  Since $X$ is separated, $N = X \setminus \{x_1\}$ is open.  By definition of a base, there exists a clopen set $U_1$ with $x \in U_1 \subset N$ and then $U_1$, $U_2 = X \setminus U_1$ is the desired separation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, we have the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nontrivial result: A locally compact Hausdorff space is zero-dimensional iff it is totally disconnected.  For instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-dimensional_space&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt; gives a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Nuno's answer (which should be the accepted one, IMO) gives a proof of this in the compact case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a cleaner approach is to &quot;redefine&quot; a Stone space to be a space which is compact, Hausdorff and zero-dimensional.  Then when you are given a space that someone says is a Stone space, the first thing you'll do is look for a base of clopen sets, which tends to be a good idea anyway.  For instance, given any profinite space (i.e., an inverse limit of finite discrete spaces, yet another term for the same class of spaces!) it is very easy to do this.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-23T05:04:11.173" CommentCount="3" />
  <row Id="11458" PostTypeId="1" AcceptedAnswerId="11459" CreationDate="2010-11-23T05:08:23.137" Score="1" ViewCount="109" Body="&lt;p&gt;How to proceed in this problem ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-23T05:14:45.273" LastActivityDate="2010-11-23T06:27:53.110" Title="If $ x = \log_{12} 27 \text {,then what is the  the value of } \log_6 16 $ ?" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11459" PostTypeId="2" ParentId="11458" CreationDate="2010-11-23T05:15:09.143" Score="5" ViewCount="0" Body="&lt;p&gt;$x = \log_{12}(27) = 3 \log_{12}(3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{x} = \frac{1}{3} \frac{1}{\log_{12}(3)} = \frac{1}{3} \log_{3}(12) = \frac{1}{3} \log_{3}(3 \times 4) = \frac{1}{3} (1 + 2 \log_{3}(2))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying, we get $\log_{3}(2) = \frac{3}{2x} - \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $y = \log_{6}(16) = 4 \log_{6}(2) = 4 \frac{1}{\log_{2}(6)} = \frac{4}{1 + \log_{2}(3)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now make use of the fact that $\log_{3}(2) = \frac{1}{\log_{2}(3)}$ to get $y$ in terms of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get $y = 4 (\frac{3-x}{3+x})$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T06:25:40.647" LastActivityDate="2010-11-23T06:25:40.647" CommentCount="1" />
  <row Id="11460" PostTypeId="1" CreationDate="2010-11-23T05:22:43.470" Score="0" ViewCount="33" Body="&lt;p&gt;What does it mean full 3D per-point anisotropic covariance estimates of a matrix?&lt;/p&gt;&#xA;" OwnerUserId="3836" LastActivityDate="2010-11-23T05:22:43.470" Title="Anisotropic covariance estimates" Tags="&lt;linear-algebra&gt;" CommentCount="1" />
  <row Id="11461" PostTypeId="2" ParentId="11453" CreationDate="2010-11-23T05:28:01.250" Score="3" ViewCount="0" Body="&lt;p&gt;Let us measure distance in planet radius and time such that the alien can make one radian per unit time on the surface.  If I fly along the surface at altitude h, I can see a swath of width $2 \arccos (\frac{1}{1+h})$ and can fly around the equator in $\frac{2 \pi (1+h)}{10}$, so as long as $2 \arccos (\frac{1}{1+h})-\frac{2 \pi (1+h)}{10} \gt 0$ I can make a spiral starting at one pole and push the alien in front of me down to the other pole.  At h of about 0.55, this only requires my speed to be about 5.6 instead of 10.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T05:28:01.250" CommentCount="5" />
  <row Id="11462" PostTypeId="2" ParentId="11452" CreationDate="2010-11-23T05:40:21.483" Score="0" ViewCount="0" Body="&lt;p&gt;For circular motion, the two expressions are equivalent.  Yes, in circular motion omega is velocity/radius.  Acceleration in circular motion is velocity^2/radius.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T05:40:21.483" />
  <row Id="11463" PostTypeId="2" ParentId="11458" CreationDate="2010-11-23T05:45:31.477" Score="1" ViewCount="0" Body="&lt;p&gt;Put $\rm\ \ell\: n\ =\ \log_6 n\:.\ \rm\ 27\ =\ 12^x\ \Rightarrow\ \ell\: 27\ =\ x\:(\ell\:6 + \ell\: 2)\ = x\:(1 + \ell\:2)\:.\ $ Now solve this for $\rm\:\ell\:2\:$ and plug the result into $\rm \ell\:16\ =\ 4\ \ell\: 2\ =\ \ldots$ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T06:27:53.110" LastActivityDate="2010-11-23T06:27:53.110" CommentCount="6" />
  <row Id="11464" PostTypeId="1" AcceptedAnswerId="11549" CreationDate="2010-11-23T05:52:25.493" Score="3" ViewCount="182" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$&#xD;&#xA;how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3838" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-23T05:58:44.127" LastActivityDate="2010-11-23T18:42:09.403" Title="How to compute the following formula" Tags="&lt;discrete-mathematics&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="11465" PostTypeId="1" CreationDate="2010-11-23T05:56:22.870" Score="7" ViewCount="83" Body="&lt;p&gt;I am an undergraduate learning about gauge theory and I have been tasked with working through the two examples given on pages 65 and 66 of &quot;&lt;a href=&quot;http://www.jstor.org/pss/1971013&quot; rel=&quot;nofollow&quot;&gt;Characteristic forms and geometric invariants&lt;/a&gt;&quot; by Chern and Simon. I will recount the examples and my progress at a solution. For ease here is the relevant text:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Example 1.&lt;/em&gt; Let $M = \mathbb{R}P^3 = SO(3)$ together with the standard metric of constant curvature 1. Let $E_1, E_2, E_3$ be an orthonormal basis of left invariant fields on $M$, oriented positively. Then it is easily seen that $\nabla_{E_1}E_2 = E_3, \nabla_{E_1}E_3 = - E_2, \text{ and } \nabla_{E_2}E_3 = E_1$. Let $\chi : M \rightarrow F(M)$ be the cross-section determined by this frame.&#xA;  $$\Phi(SO(3)) = \frac{1}{2}.$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Example 2.&lt;/em&gt; Again let $M = SO(3)$, but this time with left invariant metric $g_{\lambda}$, with respect to which $\lambda E_1, E_2, E_3$ is an orthonormal frame. Direct calculation shows&#xA;  $$\Phi(SO(3),g_{\lambda}) = \frac{2\lambda^2 - 1}{2\lambda^4}.$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For each of these examples I am expected to calculate&lt;br&gt;&#xA;$$\Phi(M) = \int_{\chi(M)} \frac{1}{2} TP_1(\theta)$$&#xD;&#xA;which lies in $\mathbb{R}/\mathbb{Z}$. Previously in the paper they give an explicit formulation of $TP_1(\theta)$ in terms of the &quot;component&quot; forms of the connection $\theta$ and its curvature $\Omega$,&#xA;$$TP_1(\theta) = \frac{1}{4\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} + \theta_{12}\wedge\Omega_{12} + \theta_{13}\wedge\Omega_{13} + \theta_{23}\wedge\Omega_{23}\right).$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have verified this formula for myself given the information in the paper. Using the structural equation $\Omega = d\theta + \theta\wedge\theta$ I am able to reduce the expression for $TP_1(\theta)$ to&#xA;$$TP_1(\theta) = \frac{-1}{2\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe I have assumed anything about the structure of $M$ during that reduction so I believe it should hold for both examples. I continue by claiming that since $E_1, E_2, E_3 \in so(3)$, the Lie algebra of $SO(3)$ I should be able to compute $\theta$ by considering&#xA;$$\nabla_{E_i}E_j := (\nabla E_j)(E_i) = \sum_k E_k \otimes \theta^{k}{}_{ij}(E_i)$$&#xD;&#xA;and comparing it with the given derivatives.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example one this yielded for me $\theta_{12} = E^3, \theta_{13} = -E^2, \theta_{23} = E^1$ where $E^i$ are the 1-forms dual to the basis $E_i$. Then I think that $\chi^*$ should act trivially on $TP_1(\theta)$ as it is a horizontal form in $\Lambda^*(T^*F(M))$. Therefore I find that $\chi^*(TP_1(\theta)) = \frac{1}{2\pi^2}\omega$, where $\omega$ is the volume form of $M$, and when integrated this yields the correct answer of $\frac{1}{2}$ for the first example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my approach fails completely for the second example. I assume that the set $\lambda E_1, E_2, E_3$ obeys the same derivate relationships as given in the first example, but this does not seem to give me enough factors of $\lambda$. I suspect that I am not handling the computation of the $\theta_{ij}$ forms or the application of $\chi^*$ correctly, however I am uncertain what my exact issue is. Is there a fundamental flaw in my understanding? I am hoping someone with more experience can point me in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-11-23T05:56:22.870" Title="Computing the Chern-Simons invariant of SO(3)" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" CommentCount="1" />
  <row Id="11466" PostTypeId="2" ParentId="11451" CreationDate="2010-11-23T05:56:40.463" Score="2" ViewCount="0" Body="&lt;p&gt;For the first part, since $x^k \leq x^n$ for $k \leq n$ (whenever $x \geq 1$), the polynomial is at most $$(|a_n| + \cdots + |a_0|)x^n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, the following facts are useful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\log n! = O(n\log n)$.&lt;/li&gt;&#xA;&lt;li&gt;If $g(n) = o(f(n))$ then $f(n) + g(n) = O(f(n))$. For example, $2^n + 2 = O(2^n)$. The notation $g(n) = o(f(n))$ means that $\lim_{n\rightarrow\infty} \frac{g(n)}{f(n)} = 0$, i.e. $g$ grows much slower than $f$.&lt;/li&gt;&#xA;&lt;li&gt;If $f_1(n) = O(g_1(n))$ and $f_2(n)) = O(g_2(n))$ then $f_1(n)f_2(n) = O(g_1(n)g_2(n))$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the third part, the following facts are useful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$n! = \Theta(\sqrt{n}(n/e)^n)$ (Stirling's approximation).&lt;/li&gt;&#xA;&lt;li&gt;$\log n = O(n^{\epsilon})$ for every $\epsilon &amp;gt; 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T05:56:40.463" />
  <row Id="11467" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T06:06:14.420" Score="9" ViewCount="0" Body="&lt;p&gt;There are at least five ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use plain induction.&lt;/li&gt;&#xA;&lt;li&gt;Differentiate the identity for the sum of a geometric series.&lt;/li&gt;&#xA;&lt;li&gt;Find some combinatorial interpretation for both sides and provide a bijection.&lt;/li&gt;&#xA;&lt;li&gt;Write as a sum of geometric series, sum each individual series, and sum the resulting geometric series.&lt;/li&gt;&#xA;&lt;li&gt;Divide both sides by $2^d$ and slightly modify the sum so that it has some probabilistic interpretation. Use known properties of the relevant random variable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T06:06:14.420" />
  <row Id="11468" PostTypeId="1" CreationDate="2010-11-23T06:07:37.113" Score="3" ViewCount="157" Body="&lt;p&gt;Need help studying for an exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practice Question:&#xA;If there are 200 students in the library, how many ways are there for them to be split among the floors of the library if there are 6 floors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: The students can not be told apart (they are indistinguishable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer must bein terms of P(n,r), C(n,r), powers, or combinations of these. The answers do not have to be calculated.&lt;/p&gt;&#xA;" OwnerUserId="3800" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-21T01:42:54.787" LastActivityDate="2010-12-21T03:00:26.217" Title="If there are 200 students in the library, how many ways are there for them to be split among the floors of the library if there are 6 floors?" Tags="&lt;combinatorics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11470" PostTypeId="2" ParentId="11468" CreationDate="2010-11-23T06:12:03.060" Score="3" ViewCount="0" Body="&lt;p&gt;Since the students are indistinguishable, we can number them from $1$ to $200$, and can assume that students $a_i$ to $a_{i+1}-1$ get assigned to floor $i$, where $a_1 = 1$ and $a_7 = 201$. Since $a_1,a_7$ are fixed, we have to choose $a_2,\ldots,a_6$. Note that $1 \leq a_2 \leq \cdots \leq a_6 \leq 200$, and this is the only restriction on them.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T06:12:03.060" />
  <row Id="11471" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T06:12:49.290" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\displaystyle\rm\ \ \ r\ x^r\ =\ x \frac{d}{dx} (x^r)\:.\ $ Apply this to $\rm\ \sum_{r=1}^d x^r\ $ then put $\rm\ x = 2\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T06:28:40.533" LastActivityDate="2010-11-23T06:28:40.533" />
  <row Id="11472" PostTypeId="2" ParentId="11145" CreationDate="2010-11-23T06:15:24.477" Score="6" ViewCount="0" Body="&lt;p&gt;You seem to have solved this on your own with &quot;bare hands&quot; but I wanted to point out some connections that this problem has with slightly more general concepts from Banach space theory.  It is a little sad that nobody else has pointed these things out yet, but not surprising; Banach space theory does not seem to be very popular on the mathematical Internet.  (I suspect if you had posted this problem on Mathoverflow, a few people would vote to close it as &quot;too specialized&quot; out of their own ignorance, and then gone off to write answers to much simpler, much more specific problems in algebraic geometry.)  In what follows I write $\ell^p$ for what you have written $\ell^n_p$; suppressing the dependence on $n$ since it can clearly be determined from the dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to distinguish these spaces is with the modulus of convexity.  For $X$ a Banach space, the modulus of convexity of $X$ is the function $\delta_X: (0,2] \to [0,1]$ given by&#xA;$$&#xD;&#xA;\delta_X(\epsilon) = \inf \{1 - \frac{1}{2} \|x+y\|: x, y \in X, \|x\|=\|y\|=1, \|x-y\| \geq \epsilon\}.&#xD;&#xA;$$&#xD;&#xA;(There is not total agreement on this; you will see the data encoded slightly differently in the literature.  For example what I have just written is what some people would call $\delta_X(\epsilon/2)$, and they would define it only for $\epsilon \in (0,1]$.  Keep your eyes open.)  It is evident from the definition that the modulus of convexity is invariant under isometric isomorphism, that is, if $X$ and $Y$ are Banach spaces and there is an isometric isomorphism $f: X \to Y$, then $\delta_X = \delta_Y$ as functions on $(0,2]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the modulus of convexity for $\ell^p$, go and prove Hanner's inequality for $\ell^p$.  For $p \in [1,2]$ this is the statement&#xA;$$&#xD;&#xA;\|x+y\|_p^p + \|x-y\|_p^p \geq (\|x\|_p + \|y\|_p)^p + | \|x\|_p - \|y\|_p|^p, \qquad x, y \in \ell^p,&#xD;&#xA;$$&#xD;&#xA;and for $p \in (2, \infty)$ the inequality reverses.  There are various nice proofs of Hanner's inequality in the literature--- Hanner's original proof is pretty straightforward, although maybe not the simplest.  My advice would be to search Math Reviews for a proof that is to your liking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of work, that I omit, shows that Hanner's inequality completely determines the modulus of convexity for $\ell^p$, in that you can write down an actual formula for it.  And once you have done that, you see that different $p$s give you different moduli of convexity.  (For $p \geq 2$ it is something like $1 - (1 - (\epsilon/2)^p)^{1/p}$; for $p \in [1,2)$ the formula is more simply written with $\delta$ as an &quot;implicit&quot; function of $\epsilon$.)  The details are in Hanner's original paper, &quot;On the uniform convexity of $L^p$ and $\ell^p$&quot;, if you don't want to supply them yourself.  (To recover $p$ from the formula in the $[2, \infty)$ case, you can look at the value at $\epsilon = 1$, for example, and use calculus to show that this is strictly decreasing in $p$.  Or you could look at the limit of the second $\epsilon$-derivative as $\epsilon$ approaches $0$; I think it's a constant multiple of $(p-1)$.  In any case this distinguishes all the $\ell^p$ with $p \in [2, \infty)$ from one another, and doing the calculation for $p \in [1,2)$ you can finish the job in a similar way.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read up on the modulus of convexity you see it has a very geometric interpretation (it quantifies in some sense &quot;how convex&quot; the unit ball of $X$ is) and contains many of the same ideas you were probably getting at with an ad hoc approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is also solved in Banach's textbook (Theory of linear operators), I think, in another way.  I don't have the book on me at the moment but the idea is that any sequence $(x_k)_{k=1}^{\infty}$ in $\ell^p$ that converges to $0$ has a subsequence $(x_{k_j})_{j=1}^{\infty}$ with the property that the function $n \mapsto \|\sum_{j=1}^n x_{k_j}\|$ is $O(n^{1/p})$, and if you think about this long enough, this can be used to distinguish the $\ell^p$ from one another.  There is a more modern or high tech approach that abstracts this idea to an invariant defined more generally in Banach spaces (e.g. you can distinguish the $\ell^p$ using the indices of &quot;Rademacher type&quot; and &quot;Rademacher cotype&quot;).&lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2010-11-23T06:15:24.477" CommentCount="5" />
  <row Id="11473" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T06:37:00.943" Score="5" ViewCount="0" Body="&lt;p&gt;Perhaps a sixth way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle S = \sum_{r=1}^{d} r\cdot 2^r$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2S = \sum_{r=1}^{d} r\cdot 2^{r+1} = \sum_{r=2}^{d+1} (r-1)2^{r}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2S -S = d\cdot 2^{d+1} - \sum_{r=1}^{d} 2^r = d\cdot 2^{d+1} - 2^{d+1} +2  = (d-1)2^{d+1} + 2$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-23T06:37:00.943" CommentCount="1" />
  <row Id="11474" PostTypeId="1" AcceptedAnswerId="12178" CreationDate="2010-11-23T06:44:49.313" Score="1" ViewCount="139" Body="&lt;p&gt;I recall reading somewhere that the supremum of a family of topologies on a set is simply the topology generated by the union of all topologies in the family. My question is, what does &quot;generated&quot; mean in this case? For instance, I know when dealing with filter bases, the generated filter is the set of all subsets containing some set in the filter base. But for topologies, how do we know what subsets to include in the generated topology? I would like to be able to visualize what kind of subsets are contained in this supremum. Thanks for any insight.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-28T04:42:50.983" Title="What exactly is the topology generated by the union of a family of topologies?" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="11475" PostTypeId="2" ParentId="11468" CreationDate="2010-11-23T06:53:21.640" Score="13" ViewCount="0" Body="&lt;p&gt;Note that if they are distinguishable then the number of ways is given by $6^{200}$ since each of the 200 students have $6$ choices of floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we are given that the students are indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we are essentially interested in solving $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 200$, where $a_i$ denotes the number of students in the $i^{th}$ floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constraints are $0 \leq a_i \leq 200$, $\forall i \in \{1,2,3,4,5,6\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will in fact look at a general version of this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to find the total number of natural number solutions for the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{i=1}^{n} a_i = N$, where $a_i \in \mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $N$ sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| | | | | | | | ... | | |$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to do partition these $N$ sticks into $n$ parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done if we draw $n-1$ long vertical lines in between these $N$ sticks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of gaps between these $N$ sticks is $N-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total number of ways of drawing these $n-1$ long vertical lines in between these $N$ sticks is $C(N-1,n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of natural number solutions for $\displaystyle \sum_{i=1}^{n} a_i = N$ is $C(N-1,n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are interested in the number of non-negative integer solutions, all we need to do is replace $a_i = b_i - 1$ and count the number of natural number solutions for the resulting equation in $b_i$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $\displaystyle \sum_{i=1}^{n} (b_i - 1) = N$ i.e. $\displaystyle \sum_{i=1}^{n} b_i = N + n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of non-negative integer solutions to $\displaystyle \sum_{i=1}^{n} a_i = N$ is given by $C(N+n-1,n-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for the current problem assuming that some of the floors can be empty, the answer is $C(200+5,5) = C(205,5) = 2872408791$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T18:41:02.233" LastActivityDate="2010-11-23T18:41:02.233" CommentCount="1" />
  <row Id="11476" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T07:13:28.263" Score="2" ViewCount="0" Body="&lt;p&gt;Denote the solution by $\rm\:S(d)\:,$ and put $\rm\ s(d) = S(d)/2 - 1   = (d-1)\:2^d\:.\ $ It is the unique solution of the recurrence $\rm\ s(2) = 4,\ \ s(d+1)/s(d) = 2d/(d-1)\ \:$ i.e. $\rm\ (d-1)\ s(d+1)\ =\ 2d\ s(d)\:.\: $ It suffices to verify that $\rm\ sum/2 - 1$ satisfies the same recurrence - a simple calculation (probably essentially the same calculation that is in Moron's answer, but I haven't checked that).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T17:53:58.640" LastActivityDate="2010-11-23T17:53:58.640" />
  <row Id="11477" PostTypeId="1" AcceptedAnswerId="11479" CreationDate="2010-11-23T07:15:38.417" Score="6" ViewCount="243" Body="&lt;p&gt;Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-23T17:21:23.490" LastActivityDate="2010-12-24T22:01:27.810" Title="Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$" Tags="&lt;discrete-mathematics&gt;&lt;fibonacci-number&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="1" />
  <row Id="11478" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:19:19.893" Score="1" ViewCount="0" Body="&lt;p&gt;One simple way is to use the formula for the $n^{th}$ Fibonacci number, viz, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_n = \frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$ where $\phi$ is the golden ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi = \frac{1 + \sqrt{5}}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or another equally simple way is to use induction on $n$ and then on $m$ or just using induction on one of these might turn out to suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be interesting to see a direct proof...&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-23T07:19:19.893" CommentCount="5" />
  <row Id="11479" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:28:31.730" Score="1" ViewCount="0" Body="&lt;p&gt;Fix $m \in \mathbb{N}$. We shall use induction on $n$. For $n=1$, the &lt;strong&gt;RHS&lt;/strong&gt; of the equation becomes $$F_{m-1}F_{1} + F_{m}F_{2} = F_{m-1} + F_{m}$$, which is equal to $F_{m+1}$. When $n=2$, the equation is also true.( I hope you can prove this!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume, that the result is true for $k=3,4, \cdots , n$. We want to show that the result is true for $k=n+1$. $$ \text{For} \ k=n-1 \ \text{we have} \quad F_{m+n-1} = F_{m-1}F_{n-1} + F_{m}f_{n}$$ and $$ \text{For} \ k = n \ \text{we have} \quad F_{m+n}=F_{m-1}F_{n} + F_{m}F_{n+1}$$ Adding both the sides you will get $$F_{m+n-1} + F_{m+n} = F_{m+n-1} = F_{m-1}F_{n+1} + F_{m}F_{n+2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, i reversed the notations. This is for proving $$F_{m+n} = F_{m-1}F_{n} + F_{m}F_{n+1}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-23T07:28:31.730" CommentCount="1" />
  <row Id="11480" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:30:49.987" Score="1" ViewCount="0" Body="&lt;p&gt;These can almost always be solved by induction. (In general sequences that are defined by recursion go often well together with induction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this formula we can do the following: Induct on $m$. If $m=1$, then&#xA;$F_{n+m} = F_{n}F_{m-1} + F_{n-1}F_m = F_{n}F_{0} + F_{n-1}F_1 = F_n + F_{n-1} = F_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose then that the result holds for $m$ and smaller. Now&#xA;$$F_{n+m+1} = F_{n+m} + F_{n+m-1} = F_{n}F_{m-1} + F_{n-1}F_m + F_{n}F_{m-2} + F_{n-1}F_{m-1}$$&#xD;&#xA;$$= F_{n}(F_{m-1} + F_{m-2}) + F_{n-1}(F_m + F_{m-1}) = F_{n}F_{m} + F_{n-1}F_{m+1}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-23T07:30:49.987" />
  <row Id="11481" PostTypeId="2" ParentId="11435" CreationDate="2010-11-23T07:32:21.323" Score="3" ViewCount="0" Body="&lt;p&gt;Isn't this obvious (or have I missed something?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appropriate product of rational polytopes $P\subseteq\mathbb R^m$&#xD;&#xA;and $Q\subseteq\mathbb R^n$ is surely their Cartesian product&#xA;$P\times Q\subseteq\mathbb R^{m+n}$. For a positive&#xA;integer $t$, the integer points in $t(P\times Q)$ are those of the form&#xA;$(a,b)$ where $a\in tP$ and $b\in tQ$, so there are $L_P(t)L_Q(t)$&#xD;&#xA;of them.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-23T07:32:21.323" CommentCount="2" />
  <row Id="11482" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:35:51.993" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ If you put the Fibonacci recurrence into matrix form then the result is obvious, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M^n\ :=\ \left(\begin{array}{ccc} &#xD;&#xA;1 &amp;amp; 1 \\\&#xD;&#xA;1 &amp;amp; 0 \end{array}\right)^n\ =\ \left(\begin{array}{ccc}&#xD;&#xA;F_{n+1} &amp;amp; F_n \\\&#xD;&#xA;F_n &amp;amp; F_{n-1} \end{array}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compute $\ M^{n+m}\ =\ M^n \ M^m$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T07:46:22.153" LastActivityDate="2010-11-23T07:46:22.153" CommentCount="3" />
  <row Id="11483" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:37:09.553" Score="5" ViewCount="0" Body="&lt;p&gt;If pressed, I'd nuke with Binet's formula myself, but here's another&#xA;approach. By an easy induction, if&#xA;$$A=\pmatrix{0&amp;amp;1\\\\ 1&amp;amp;1}$$&#xD;&#xA;then&#xA;$$A^n=\pmatrix{F_{n-1}&amp;amp;F_n\\\\ F_n&amp;amp;F_{n+1}}.$$&#xD;&#xA;Comparing the top right entry in the equation $A^m A^n=A^{m+n}$&#xD;&#xA;gives&#xA;$$F_{m-1}F_n+F_m F_{n+1}=F_{m+n}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-23T07:37:09.553" />
  <row Id="11484" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T07:44:40.313" Score="3" ViewCount="0" Body="&lt;p&gt;Try proving this statement:&#xA;Claim: If $f(n) = f(n-1)+f(n-2)$, then $f(n) =  F_n f_1 + F_{n-1} f_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &quot;fix&quot; $m$ and think of $F_{n+m}$ as a linear recurrence in $n$ with initial conditions $F_{m+1}$ and $F_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then your claim will follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, there is a short and clean proof of Claim, but you should know it uses the fact $F_{-1} = 1$.(Something you can easily verify if you did not already know).&lt;/p&gt;&#xA;" OwnerUserId="899" LastActivityDate="2010-11-23T07:44:40.313" />
  <row Id="11485" PostTypeId="2" ParentId="11145" CreationDate="2010-11-23T07:55:49.753" Score="0" ViewCount="0" Body="&lt;p&gt;An elementary approach involving the third mixed partial derivatives shows that such an isometry in the form given in the question is only possible with $p=q=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $1&amp;lt; p,\ q &amp;lt; \infty$ and $1/p + 1/q = 1$. Assume also that  there exist an isometry of the given form, i.e., a normalized rotation of $\pi/4$ between $(\mathbb R^2,\|\ \|_p)$ and $(\mathbb R^2,\|\ \|_q)$.   Then the corresponding matrix would be $A = \lambda&amp;#39;\left(\matrix{1 &amp;amp; 1\\1 &amp;amp; -1}\right)$ for some scalar $\lambda&amp;#39;$. It follows that &#xA;$$|x|^p + |y|^p = \lambda &amp;#39;&amp;#39;(|x + y|^q + |x - y|^q)^{p/q}$$  &#xA;for some constant $\lambda&amp;#39;&amp;#39;$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compute&lt;sup&gt;1&lt;/sup&gt; the mixed third derivative $\frac {\partial^3}{\partial x \partial y^2}$ of both sides at the point $(1,0)$, after simplifications we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = (-2+q) (-1+q)+\left(-1+\frac{p}{q}\right) (-1+q) q.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plus the assumptions on $p$ and $q$ gives us that the unique solution is $p=2$ and $q =2$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; with the help of Mathematica: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;D[((x + y)^q + (x - y)^q)^(p/q), x, y, y] /. {x -&amp;gt; 1, y -&amp;gt; 0}&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-23T10:38:59.083" LastActivityDate="2010-11-23T10:38:59.083" />
  <row Id="11486" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T07:59:39.253" Score="4" ViewCount="0" Body="&lt;p&gt;This is an arithmetic-geometric progression. I remember doing these&#xA;at school, I think even before O-levels. The typical A-G progression&#xA;has the form&#xA;$$ab,ar(b+d),ar^2(b+2d),\ldots,a r^n(b+nd)\ldots$$&#xD;&#xA;(it's the pointwise product of an arithmetic progression and a geometric&#xA;progression). To obtain the sum of the first $N$ terms, one uses the same&#xA;trick as for summing $N$ terms of a GP. Let&#xA;$$S=\sum_{k=0}^{N-1} ar^k(b+kd).$$&#xD;&#xA;Then&#xA;$$rS=\sum_{k=0}^{N-1} ar^{k+1}(b+kd)=\sum_{k=1}^N ar^k(b+kd-d).$$&#xD;&#xA;Hence&#xA;$$(1-r)S=ab-ar^N(b+(N-1)d)+\sum_{k=1}^{N-1}adr^k.$$&#xD;&#xA;The last sum is that of a GP, which one already knows about...&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-23T07:59:39.253" CommentCount="4" />
  <row Id="11487" PostTypeId="1" CreationDate="2010-11-23T08:33:57.043" Score="1" ViewCount="119" Body="&lt;p&gt;Y= X+N where X,N are normally distributed with zero mean &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i know that Y is also normally distributed with zero mean and sum of variances &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i don't know how to get f(X|y) in order to calculate E(X|y)&lt;/p&gt;&#xA;" OwnerUserId="3490" LastActivityDate="2010-11-23T21:08:32.720" Title="Y= X+N what is E(X|Y=y)" Tags="&lt;probability&gt;&lt;expectation&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="11488" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T09:01:20.477" Score="1" ViewCount="0" Body="&lt;p&gt;\begin{equation}&#xA;p(x|y) = \frac{p(y|x)p(x)}{p(y)} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know $p(y)$ and $p(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;p(y|x) = \frac{1}{\sqrt{2 \pi \sigma_N^2}} e^{\frac{-(y-x)^2}{2\sigma_N^2}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these expressions, you can easily get both $p(x|y)$ and $E(X|Y=y)$.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-23T09:01:20.477" />
  <row Id="11489" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T09:07:59.393" Score="4" ViewCount="0" Body="&lt;p&gt;It's possible to write down the set of matches between two strings as the result of multiplying two polynomials constructed from the strings. This turns out to be very useful: indeed, fast algorithms for string matching exploit the fast Fourier transform method used to multiply polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this works, think of a binary string as representing coefficients of a polynomial. Then matches between the strings correspond to nonzero coefficients of the product of the polynomial from one string and the reverse of the polynomial from the second. Here, the &quot;multiplication&quot; is over the AND-OR field, rather than the usual integer or GF(2) field. &lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2010-11-23T09:07:59.393" CommentCount="2" />
  <row Id="11490" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T09:11:11.540" Score="1" ViewCount="0" Body="&lt;p&gt;You can find ${\rm E}(X|Y=y)$ immediately as follows.&#xA;If $X \sim {\rm N}(0,\sigma_1^2)$ and $N \sim {\rm N}(0,\sigma_2^2)$&#xD;&#xA;(and $X$ and $N$ are independent), then&#xA;$$&#xD;&#xA;{\rm E}(X|X + N) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}(X + N),&#xD;&#xA;$$&#xD;&#xA;so, in particular,&#xA;$$&#xD;&#xA;{\rm E}(X|X + N = y) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}y.&#xD;&#xA;$$&#xD;&#xA;I will (probably) justify this (i.e., the first formula) later on today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that, in some respect, the normal distribution does not play a special role in this result.&#xA;For example, if $X$ and $N$ are independent Poisson random variables with means $\lambda_1$ and $\lambda_2$, respectively, then&#xA;$$&#xD;&#xA;{\rm E}(X|X + N) = \frac{{\lambda _1 }}{{\lambda_1  + \lambda_2 }}(X+N).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is common to the normal distribution and the Poisson distribution in this context is that both are infinitely divisible. More details later on.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-23T15:06:55.690" LastActivityDate="2010-11-23T15:06:55.690" CommentCount="2" />
  <row Id="11491" PostTypeId="2" ParentId="11342" CreationDate="2010-11-23T09:18:10.753" Score="1" ViewCount="0" Body="&lt;p&gt;$T_{b_2} \circ T_{b_1} = T_{b_2}(T_{b_1}(z))= T_{b_2}(b_1*z)=b_2*(b_1*z)=(b_2*b_1)*z = b*z$.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T14:43:51.860" LastActivityDate="2010-11-23T14:43:51.860" />
  <row Id="11492" PostTypeId="1" CreationDate="2010-11-23T09:43:15.927" Score="3" ViewCount="129" Body="&lt;p&gt;I would like to show that the quantity: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is negative on the surface:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some sufficiently large value of $C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was not able to massage the first quantity any more in order to make it look like the second. I also considered a change of coordinates, but had no luck.&#xA;$\sigma, b, r$  are positive parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a step in exercise 9.2.2 from Strogatz Nonlinear Dynamics and Chaos.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-12-23T21:23:13.623" Title="Create a trapping region for Lorenz Attractor" Tags="&lt;geometry&gt;&lt;dynamical-systems&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11493" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T09:57:25.633" Score="1" ViewCount="0" Body="&lt;p&gt;Multiplication of polynomials has a very nice combinatorial interpetation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a simple example: $1+x=x^0+x^1$. If you perform an experiment once, that experiment can either succeed or fail. You can fail the experiment in only one way in one performance, or you can succeed in one way. This is precisely encoded in the coefficients of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look now at the product: $(1+x)\times(1+x)=1\times x^0+2\times x^1+1\times x^2$. The coefficients now encode in how many ways you can fail twice, how many ways you can succeed once and fail once and how many ways you can succeed twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I guess you get the trick now, $(1+x)^n$ will have encoded in its $k$th coefficient in how many ways you can succeed $k$ times but fail $n-k$ times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalized to all kinds of polynomials.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-23T09:57:25.633" />
  <row Id="11494" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T10:24:48.220" Score="1" ViewCount="0" Body="&lt;p&gt;Apart from the mundane reasons, e.g. &quot;you need it to do certain computations&quot; there are also some more technical reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Stone-Weierstrass theorem: &lt;em&gt;Let $X$ be a compact metric space and let $A$ be a $\mathbb{C}$-sub&lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_over_a_field&quot; rel=&quot;nofollow&quot;&gt;algebra&lt;/a&gt; of $C(X,\mathbb{C})$. If $A$ separates points in $X$ and vanishes at no point of $X$ and is self-conjugate, then $A$ is dense in $C(X,\mathbb{C})$.&lt;/em&gt; &lt;br/&gt;&#xA;The polynomial functions on $[a,b]$ with complex coefficients are the prototypical example of such a subalgebra. Of course verifying that it is an algebra and has the stated properties is an application of multiplication of polynomials.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Orthogonal polynomials. These are somewhat important in the theory of special functions. For example in &lt;a href=&quot;http://books.google.com/books?id=nMm13WXpLt8C&amp;amp;pg=PA240&quot; rel=&quot;nofollow&quot;&gt;[Andrews, Askey and Roy]&lt;/a&gt; there is an entire chapter devoted to families of orthogonal polynomials.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T10:35:22.103" LastActivityDate="2010-11-23T10:35:22.103" CommentCount="2" />
  <row Id="11495" PostTypeId="1" CreationDate="2010-11-23T10:37:34.663" Score="0" ViewCount="211" Body="&lt;p&gt;Hi to all, &#xA;I'm in Italian guy almost in my 30. I have a regular job as a programmer. When in school I've never been much interested nor good in math and even at the university I've been studying languages, hence something more on the literary side that on the scientific one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When in high school I've had mostly literary classes, so, again, my current knowledge of algebra, geometry and arithmetic is quite basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always been interested in programming, though, up to making it my profession. With my growing interest in math, I'm tinkering with the idea of even getting a Computer Science or Computer Engineering degree at the university. Both rely heavily on maths and I would have to study it by myself, as I cannot afford following classes (as said before, I have a full-time job), starting from the basis that I don't have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before attending to it, I want to start learning math by myself, to gain the necessary base knowledge I don't currently have. Do you think it is possible to successfully study mathematics with such a short initial knowledge and not having my full day to do it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any general advice or book / site to get me started with the &quot;preparation&quot; study? What are the basic areas I should concentrate on to be prepared to the first year math exams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="3847" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T10:40:50.583" LastActivityDate="2010-11-25T14:18:04.083" CommunityOwnedDate="2010-11-25T12:19:27.163" Title="Preparing for first year CS" Tags="&lt;soft-question&gt;&lt;self-learning&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="11496" PostTypeId="2" ParentId="11495" CreationDate="2010-11-23T10:55:51.707" Score="1" ViewCount="0" Body="&lt;p&gt;Your first course with programming in it will probably use some functional programming language such as an ML variant, Scheme, Haskell, a LISP, etc. As I understand it, universities start with that for two reasons, (i) it works well for teaching (this is debatable, though), and (ii) whatever misconceptions and terrible habits the neophytes bring to the table from unguided tinkering with e.g. Visual Basic will be almost &lt;em&gt;entirely useless&lt;/em&gt;, thus ensuring that you also have a good chance of beating some bad habits out of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first heavy math class in a CS education (other than the typically required first course in mathematics) is probably algorithms. Curiously the math requirements for an algorithms course are rather low in the &quot;skills&quot; department, but high in the &quot;mathematical maturity&quot; department. That first course in math will teach you the method of induction and - equally important - the ability to reason mathematically.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-23T10:55:51.707" CommunityOwnedDate="2010-11-25T12:19:27.163" CommentCount="1" />
  <row Id="11497" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T10:58:55.453" Score="2" ViewCount="0" Body="&lt;p&gt;I think a better metaphor for multiplication comes from physics.  If your speed s is constant over some time t, then your distance traveled is just s times t.  Whatever fraction of time you travel, you will expect to have traveled some distance.  So even fractional multiplication makes some sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the regular world, it's very common to use a polynomials (often linear ones) to approximate things that are changing.  For instance, if you are a grocer and you are trying to predict how many tomatoes you will sell tomorrow, it's not a bad approach to guess it will be the same as yesterday.  It's also not a bad approach to guess it might be the average of the last two days.  Or, you might come up with a more complicated formula involving the last 50 days.  All of these attempts to guess will have an error.  This is the point of studying these things, to do a better job and make smaller errors.  (For reasons, I can't go into, the squares and cubes of values can sometimes help make approximations more accurate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that any quantity is a potential polynomial.  You might approximate your customers per day as a polynomial of the number of customers of the last two days.  You might approximate the value of sales per customer as another polynomial.  Well, if you wanted to know your potential income, customers times sales, you might multiply them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use polynomials all the time.  For instance, when I was helping a friend lose weight, I approximated their weight loss behavior using a polynomial.  It was fairly accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, keep with polynomials, they were worth it in the long run!&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-11-23T10:58:55.453" />
  <row Id="11498" PostTypeId="1" CreationDate="2010-11-23T11:06:48.587" Score="0" ViewCount="148" Body="&lt;p&gt;I'm really embarrassed to ask this - my apologies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a piece of software I'm writing, I am using the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y = X+\frac{X}{10}N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all fine - but how do I calculate X when I know Y and N? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it's really basic, but for some reason I get stuck half-way through moving the equation around. &lt;/p&gt;&#xA;" OwnerUserId="3848" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T11:25:49.097" LastActivityDate="2010-11-23T15:18:45.993" Title="Basic algebra gone out the window" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="11499" PostTypeId="1" AcceptedAnswerId="11512" CreationDate="2010-11-23T11:26:45.163" Score="1" ViewCount="213" Body="&lt;h2&gt;Edit:&lt;/h2&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answers I seem to be getting here are that this is not possible, but I have come across a formulae for subdividing an icosahedron (which I dont have to hand, but goes something like this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V = 10 * u² + 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E = 30 * u²&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 20 * u² &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where U = the frequency of subdivision&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I may have it the wrong way around but you get the picture) this allows us to calculate the amount of vertecies, edges and faces for each subdivision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this forumlea (or the correct one anyway) you can incrediment the frequency untill the number of faces reach 1000. It would appear in theory this will allow me to imitate a sphere with 1000 congruent polygon's (faces) by constructing a subdivided icosahedron because using V-E+F=2 I get 502-1500+1000=2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone confirm if the faces will wrap around to make a full icosahedron without any missing faces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Know the answer? reply and you will most likley get the vote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;complete icosahedron with 20,40,80,120,160,200,240,280,320...1000 faces = true?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hi, Does anyone know if it possible to reprisent a sphere with 1000 equal surfaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say equal surfaces, I am talking in terms of area. However it would also be preferable if the shape of these surfaces match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I am looking at subdivided icosahedron's, all surfaces are equal triangles in terms of shape and area, but to produce a nice sphere it would seem the number of surfaces would need to be 20,80,320 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I have come across zonohedron's - Does this allow any amount of surface? I can see that the surfaces are different shapes (but i should hope they have the same area)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this may sound abit confusing, but any feedback will be much appreachiated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Chris&lt;/p&gt;&#xA;" OwnerUserId="3849" LastEditorUserId="3849" LastEditorDisplayName="" LastEditDate="2010-12-03T14:14:24.483" LastActivityDate="2010-12-03T19:41:07.227" Title="Possible to imitate a sphere with 1000 congruent polygons?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="11500" PostTypeId="2" ParentId="5713" CreationDate="2010-11-23T12:00:56.430" Score="0" ViewCount="0" Body="&lt;p&gt;I think the question requires that the language must be finite. So, it's quite hard.&lt;/p&gt;&#xA;" OwnerUserId="3851" LastActivityDate="2010-11-23T12:00:56.430" />
  <row Id="11501" PostTypeId="1" CreationDate="2010-11-23T12:09:32.897" Score="0" ViewCount="103" Body="&lt;p&gt;I'm trying to teach myself mathematics. I only started last year with a previous knowledge of basic arithmetic. I often make a detour into regions featuring problems that are a bit too deep for me. I'm often too curious to wait until I was able to catch up on the necessary bedding. I hope you'll excuse me :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=f%28x%29+%3D+f%28x-1%29%2ab&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha says&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = f(x-1)*b -&amp;gt; f(x) = c_1 b^(x-1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is how I would compute the original recursive algorithm(?):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(x) = f(x-1)*b&#xA;&#xA;f(4) = f(4-1)*b&#xA;     = f(f(3-1)*b)*b&#xA;     = f(f(f(2-1)*b)*b)*b&#xA;     = f(f(f(f(1-1)*b)*b)*b)*b&#xA;     = ...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;how do I compute the recurrence relation f(x) = c_1 b^(x-1)&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would compute the Fibonacci numbers like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Fn = Fn-1 + Fn-2&#xA;F0 = 0&#xA;F1 = 1&#xA;&#xA;F5 = F5-1 + F5-2&#xA;   = F4 + F3&#xA;   = (F3 + F2) + (F2 + F1)&#xA;   = ((F2 + F1) + (F1 + F0)) + ((F1 + F0) + F1)&#xA;   = (((F1 + F0) + F1) + (F1 + F0)) + ((F1 + F0) + F1)&#xA;   = 5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm also not sure about the difference between a recursive algorithm and its recurrence relation. I'd be grateful if someone could enlighten me or simply link me up. &lt;/p&gt;&#xA;" OwnerUserId="3850" LastActivityDate="2010-11-23T16:13:04.837" Title="f(x) = f(x-1)*b -&gt; f(x) = c_1 b^(x-1)" Tags="&lt;algorithms&gt;&lt;recursive&gt;&lt;wolframalpha&gt;" AnswerCount="2" />
  <row Id="11502" PostTypeId="1" AcceptedAnswerId="11506" CreationDate="2010-11-23T12:15:30.550" Score="3" ViewCount="97" Body="&lt;p&gt;Is there any &quot;algorithm&quot; or steps to follow to get a formula from a table of values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X                Result&#xA;1                3&#xA;2                5&#xA;3                7&#xA;4                9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'd like to obtain:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Result = 2X+1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe using excel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not going to be always a polynomial and it may have several parameters (I think 2).&lt;/p&gt;&#xA;" OwnerUserId="3314" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T14:30:15.317" LastActivityDate="2010-11-23T14:30:15.317" Title="Find formula from values" Tags="&lt;functions&gt;&lt;approximation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="11503" PostTypeId="2" ParentId="11501" CreationDate="2010-11-23T12:19:24.133" Score="2" ViewCount="0" Body="&lt;p&gt;I believe that your way to compute $f(n)$ is not correct, as you have outlined in the question. Observe that $f(4) = b * f(3) = b^2 * f(2)$ and not $f(4) = b * f(f(2)*b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you have $$f(n) = b * f(n-1) = b^2 * f(n-2) = \ldots = b^{n-1} * f(1).$$&#xD;&#xA;Say that your initial condition states that $f(1) = c_1$, you can conclude that $f(n) = c_1 * b^{n-1}$.&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-23T12:19:24.133" />
  <row Id="11504" PostTypeId="2" ParentId="11502" CreationDate="2010-11-23T12:24:55.393" Score="1" ViewCount="0" Body="&lt;p&gt;One way to obtain the desired relationship is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Regression_analysis&quot; rel=&quot;nofollow&quot;&gt;regression analysis&lt;/a&gt;. In the case that you know what form of the relationship you are expecting, say you know that $f(x) = a * x + b$ it is very simple to find the parameters $a$ and $b$ (see for &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_regression&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;). In the case that you don't know what form your formula may take, it is common to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonparametric_regression&quot; rel=&quot;nofollow&quot;&gt;nonparametric regression&lt;/a&gt; techniques.&lt;/p&gt;&#xA;" OwnerUserId="2384" LastActivityDate="2010-11-23T12:24:55.393" />
  <row Id="11505" PostTypeId="2" ParentId="11502" CreationDate="2010-11-23T12:38:40.513" Score="2" ViewCount="0" Body="&lt;p&gt;An easier method (assuming that the values were generated by a polynomial) would be to note that successive divided differences $\frac{y_{i+1}-y_i}{x_{i+1}-x_i}$ are constant and equal to 2; thus your function is of the form $y=2x+c$. The constant $c$ is then determined by replacing both $x$ and $y$ with appropriate values, and then solving for $c$.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-23T12:38:40.513" CommentCount="1" />
  <row Id="11506" PostTypeId="2" ParentId="11502" CreationDate="2010-11-23T13:07:28.313" Score="3" ViewCount="0" Body="&lt;p&gt;The best tool for doing this is that very, very impressive piece of software:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ccsl.mae.cornell.edu/eureqa&quot; rel=&quot;nofollow&quot;&gt;http://ccsl.mae.cornell.edu/eureqa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For your abovementioned very easy example, even WA will find the right formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=3,+5,+7,+9&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=3,+5,+7,+9&lt;/a&gt;,...&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-23T13:07:28.313" CommentCount="4" />
  <row Id="11508" PostTypeId="2" ParentId="10392" CreationDate="2010-11-23T13:51:31.893" Score="0" ViewCount="0" Body="&lt;p&gt;You can also use &lt;a href=&quot;http://en.wikipedia.org/wiki/Disjunctive_normal_form&quot; rel=&quot;nofollow&quot;&gt;disjunctive normal form&lt;/a&gt; or conjunctive Normal Form(no link because of spam prevention mechanism)&#xA;for finding the right formula. So for the first answe&lt;/p&gt;&#xA;" OwnerUserId="3844" LastActivityDate="2010-11-23T13:51:31.893" />
  <row Id="11509" PostTypeId="1" AcceptedAnswerId="11516" CreationDate="2010-11-23T13:54:10.520" Score="2" ViewCount="117" Body="&lt;p&gt;Hi all, I've got a question concerning how to proof the Pythagoras theorem using the following assumption:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ is perpendicular to $y$ (if and only if) $||x+y||^2 = ||x||^2 + ||y||^2$,&#xA;where $x$ and $y$ are vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a basic understanding of linear algebra however I'm a beginner with this. The question provides hints how to prove the above mentioned equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely that I should use the properties of the dot product and the definition of the norm of a vector.&#xA;Those being symmetry, scaling and distributivity as the dot product properties and the norm of a vector being the squared root of the dot product between the same vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about using the fact that if a vector is perpendicular to another vector the dot product between those vectors should be 0. But that is not provided as a hint so I'm not sure. I know the under lying thought behind it is the cosine rule for vectors, that being:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\cdot y = ||x||\,||y|| \cos(\theta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the angle between the two vectors is perpendicular you should use $\cos(\pi/2)$ which is $0$ and $||x||\cdot 0 = 0$ and $||y||\cdot 0 = 0$ with the vectors not necessarily being $0$. Thus $x\cdot y = 0$.&#xA;How would I apply this to the equation I first mentioned to prove the Pythagoras theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few more thoughts on how I could prove this but I'm not sure if they're correct. &#xA;I hope someone could point me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, Rope.&lt;/p&gt;&#xA;" OwnerUserId="3854" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-23T15:40:26.223" LastActivityDate="2010-12-05T23:07:00.273" Title="How to prove the Pythagoras theorem using vectors" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11511" PostTypeId="2" ParentId="11509" CreationDate="2010-11-23T14:05:44.590" Score="0" ViewCount="0" Body="&lt;p&gt;If I understood the question correctly, you have to prove that&#xA;$$\|x+y\|^2 = \|x\|^2 + \|y\|^2$$&#xD;&#xA;if and only if $x$ and $y$ are perpendicular. That is, as you noted, $x \cdot y = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: You can use the following formulas:&#xA;$$\|x\|^2 = x \cdot x$$&#xD;&#xA;$$(x+y) \cdot z = x \cdot z + y \cdot z$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-23T14:05:44.590" CommentCount="4" />
  <row Id="11512" PostTypeId="2" ParentId="11499" CreationDate="2010-11-23T14:11:11.433" Score="3" ViewCount="0" Body="&lt;p&gt;If you want the faces congruent, it will not work.  If you try to  use congruent polygons you cannot satisfy Euler's formula.  This says if you use pentagons and hexagons there must be 12 pentagons.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Truncated_icosahedron&quot; rel=&quot;nofollow&quot;&gt;truncated icosahedron&lt;/a&gt;, which has 12 pentagonal and 20 hexagonal faces is quite close to a sphere and there are only two kinds of faces.  I thnk you could subdivide the hexagons into 7 and pentagons into one pentagon and five hexagons to get 200 hexagons and 12 pentagons, but I don't think the hexagons are all the same shape any more.  Another subdivision has 1460 hexagons and 12 pentagons.  I think there are more in between, but a quick web search didn't turn up any good references.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T14:11:11.433" CommentCount="2" />
  <row Id="11513" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T14:16:56.950" Score="1" ViewCount="0" Body="&lt;p&gt;Have a look at Kelley and Peterson's textbook[1]. I think this book can give you a little more understanding of where everything comes from. It talks about difference calculus, which is the discrete analogue of continuous calculus. In difference calculus, we have the delta and sum operators where we have the derivative and integral respectively in continuous calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at it and let me know if it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T14:16:56.950" />
  <row Id="11514" PostTypeId="2" ParentId="11492" CreationDate="2010-11-23T14:18:50.737" Score="0" ViewCount="0" Body="&lt;p&gt;Just simplifying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right) &amp;lt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2} &amp;gt; 0 ... \sigma &amp;gt; 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+bz^{2}-2brz+br^{2}-br^{2} &amp;gt; 0 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+bz^{2}-2brz &amp;gt; 0 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+bz^{2} &amp;gt; 2brz $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; z&amp;lt;=0$ or $z&amp;gt;=2r$ for any $C$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ rx^{2}+y^{2} &amp;gt; 2brz - bz^{2}, 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,&#xA;$rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}+(\sigma-1) y^{2}+\sigma\left(z-2r\right)^{2}=C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; rx^{2}+y^{2}= C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the original expression is negative when $z$ is outside $(0,2r)$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 2brz - bz^{2} &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; bz(2r - z) &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; 0 &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}+bz(z - 2r), 0&amp;lt;z&amp;lt;2r $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt; 0 &amp;lt; C -(\sigma-1) y^{2}-\sigma\left(z(1-b/2)-2r\right)^{2}+(bz/2)^{2}, 0&amp;lt;z&amp;lt;2r $ ... completing the square&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="820" LastEditorDisplayName="" LastEditDate="2010-11-23T14:33:11.407" LastActivityDate="2010-11-23T14:33:11.407" />
  <row Id="11515" PostTypeId="1" AcceptedAnswerId="11532" CreationDate="2010-11-23T14:22:48.797" Score="2" ViewCount="134" Body="&lt;p&gt;Let's say that we have three points: $p = (x_p,y_p)$, $q = (x_q,y_q)$ and $a = (x_a,y_a)$. How can i find point $b$ which is reflection of $a$ across a line drawn through $p$ and $q$? I know it's simple to calculate, when we have $p$, $q$ etc. But I want to do this in my program, and I'm not sure, how to compute this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I've found solution by myself (but answers in this topic really helped me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, that we have a line $Ax+By+C=0$, and $A^2+B^2 \not= 0$. $M (a,b)$ reflection across the line is point: $M&amp;#39; (\frac{aB^2-aA^2-2bAB-2AC}{A^2+B^2}, \frac{bA^2-bB^2-2aAB-2BC}{A^2+B^2})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, we don't have line, but only 2 points. How we can find $A,B,C$? It's simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say, that $p=(p_x,p_y)$ and $q = (q_x,q_y)$. Line equation is: $(y-p_y)(q_x-p_x) - (q_y-p_y)(x-p_x) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some calculations we have: $y(q_x-p_x) - x(q_y-p_y) - (p_y(q_x-p_x)+p_x(p_y-q_y)) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&#xA;$A = p_y-q_y$, $B = q_x-p_x$ and $C = -p_y(q_x-p_x)-p_x(p_y-q_y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all.&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorUserId="3856" LastEditorDisplayName="" LastEditDate="2010-11-23T21:42:03.613" LastActivityDate="2010-11-23T21:42:03.613" Title="Point reflection across a line" Tags="&lt;reflection&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11516" PostTypeId="2" ParentId="11509" CreationDate="2010-11-23T14:28:38.993" Score="2" ViewCount="0" Body="&lt;p&gt;$x,y$ are perpendicular if and only if $x\cdot y=0$. Now, $||x+y||^2=(x+y)\cdot (x+y)=(x\cdot x)+(x\cdot y)+(y\cdot x)+(y\cdot y)$. The middle two terms are zero if and only if $x,y$ are perpendiculat. So, $||x+y||^2=(x\cdot x)+(y\cdot y)=||x||^2+||y||^2$ if and only if $x,y$ are perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T14:28:38.993" CommentCount="1" />
  <row Id="11517" PostTypeId="2" ParentId="11515" CreationDate="2010-11-23T14:31:50.813" Score="2" ViewCount="0" Body="&lt;p&gt;You find the equation of the line through p and q, one version of which is $$\frac{y-y_p}{x-x_p}=\frac{y_q-y_p}{x_q-x_p}$$  The slope is the RHS.  So the slope of the line through a and perpendicular to the line through p and q is $$\frac{-(x_q-x_p)}{y_q-y_p}$$  and the line is $$y-y_a=\frac{-(x_q-x_p)}{y_q-y_p}(x-x_a)$$  Calculate the distance from a to the line through p and q and extend that far again.  I think there is a formula for this, but don't know it, so I would have to solve the two equations for x and y to get the intersection.  Note that there is a divide by zero concern.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T14:31:50.813" CommentCount="5" />
  <row Id="11518" PostTypeId="1" AcceptedAnswerId="11530" CreationDate="2010-11-23T14:42:02.587" Score="2" ViewCount="112" Body="&lt;p&gt;How to prove that $\mathcal{O}_{\sqrt[3]{3}}$ is an euclidean domain? I heard that one should prove the following but why it is enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $ a,b,c\in\mathbb{R}$, prove that there are $ x,y,z\in\mathbb{R}$ such that $ x-a,y-b,z-c\in\mathbb{Z}$ and that&#xA;$$-1\leq x^3+3y^3+9z^3-9xyz\leq 1.$$&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-23T16:47:21.970" Title="Euclidean domain" Tags="&lt;inequality&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11519" PostTypeId="2" ParentId="11474" CreationDate="2010-11-23T14:44:17.990" Score="2" ViewCount="0" Body="&lt;p&gt;The topology generated by any family of subsets $F$ of a set $X$ is the intersection of all topologies on $X$ containing $F$; $F$ then forms a &lt;a href=&quot;http://en.wikipedia.org/wiki/Subbase&quot; rel=&quot;nofollow&quot;&gt;subbase&lt;/a&gt; for the resulting topology.  In practice this means that you take all finite intersections of the elements of $F$ to get a &lt;a href=&quot;http://en.wikipedia.org/wiki/Base_(topology)&quot; rel=&quot;nofollow&quot;&gt;base&lt;/a&gt; $F&amp;#39;$, then take arbitrary unions of the elements of $F&amp;#39;$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-23T14:44:17.990" />
  <row Id="11520" PostTypeId="2" ParentId="11499" CreationDate="2010-11-23T14:46:28.063" Score="3" ViewCount="0" Body="&lt;p&gt;Draw 1000 meridians enclosing an angle of 0.36 degrees between two successive ones. Or, if you prefer 1000 congruent triangles: Draw an equator and 500 meridians enclosing an angle of 0.72 degrees between two successive ones.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-12-03T19:41:07.227" LastActivityDate="2010-12-03T19:41:07.227" CommentCount="1" />
  <row Id="11521" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T14:53:25.453" Score="1" ViewCount="0" Body="&lt;p&gt;For some later purposes, I give here the complete solution based on svenkatr's approach (thus also verifying my previous answer). In what follows, I use  a different (somewhat more natural) notation, so don't get confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X$ and $Y$ are independent ${\rm N}(0,\sigma_1^2)$ and ${\rm N}(0,\sigma_2^2)$ variables, respectively.&#xA;We can find ${\rm E}(X|X+Y=z)$ as follows.&#xA;$$&#xD;&#xA;{\rm E}(X|X + Y = z) = \int_{ - \infty }^\infty  {xf_{X|X + Y} (x|z)\,{\rm d}x} = \int_{ - \infty }^\infty  {x\frac{{f_X (x)f_{X + Y|X} (z|x)}}{{f_{X + Y} (z)}}\, {\rm d}x}&#xD;&#xA;$$&#xD;&#xA;(the notation should be clear from the context). Noting that $f_{X+Y|X}(\cdot|x)$ is the ${\rm N}(x,\sigma_2^2)$ density function, some algebra shows that the right-hand side integral is equal to&#xA;$$&#xD;&#xA;\int_{ - \infty }^\infty  {x\sqrt {\frac{{\sigma _1^2  + \sigma _2^2 }}{{\sigma _1^2 \sigma _2^2}}} \frac{1}{{\sqrt {2\pi } }}\exp \bigg\{  - \frac{{[x - z \sigma _1^2 /(\sigma _1^2  + \sigma _2^2 )]^2 }}{{2[\sigma _1^2 \sigma _2^2 /(\sigma _1^2  + \sigma _2^2 )]}}} \bigg \} \,{\rm d}x.&#xD;&#xA;$$&#xD;&#xA;Now, from $\int_{ - \infty }^\infty  {x\frac{1}{{\sqrt {2\pi \sigma ^2 } }}{\rm e}^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x}  = \mu $ (expectation of a ${\rm N}(\mu,\sigma^2)$ random variable), we can find that&#xA;$$&#xD;&#xA;{\rm E}(X|X + Y = z) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}z.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T14:53:25.453" />
  <row Id="11522" PostTypeId="2" ParentId="11498" CreationDate="2010-11-23T15:18:45.993" Score="1" ViewCount="0" Body="&lt;p&gt;If $Y=X+\frac{X}{10}N=X\left(1+\frac{N}{10}\right)$, then $X=\frac{Y}{1+\frac{N}{10}}$ .&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-23T15:18:45.993" CommunityOwnedDate="2010-11-23T15:18:45.993" />
  <row Id="11523" PostTypeId="1" CreationDate="2010-11-23T15:25:17.580" Score="1" ViewCount="42" Body="&lt;p&gt;Hello friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be a smooth projective variety with canonical class K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a be defined to be the maximum dimension of the image of X under the rational map induced by the linear system |nK| as n ranges over all positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let b be defined to be the smallest positive integer c such that the sequence (dim(H^0 (X, nK))/n^c) is bounded (as n gets large).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second definition I guess doesnt include the case with kodaira dimension smaller than 0 (ie -1 or -infinity depending on how you like to define it), so we're only talking about non-negative kodaira dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will someone please tell me why a=b?&lt;/p&gt;&#xA;" OwnerUserId="3860" LastActivityDate="2010-11-23T15:25:17.580" Title="Different definitions of Kodaira dimension" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="11524" PostTypeId="1" AcceptedAnswerId="11526" CreationDate="2010-11-23T15:28:43.410" Score="4" ViewCount="96" Body="&lt;p&gt;I'm asking here because no textbook or website that I know of gives a definition of the above mentioned term.&#xA;Since there's no obvious way (that I can think of) to define a normal subloop, I don't see how the definition of a simple group can be modified for the case of loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - &lt;strong&gt;what is the definition of a simple loop?&lt;/strong&gt; (And what is the motivation behind the definition?)&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-11-23T15:55:52.660" Title="What is a simple loop?" Tags="&lt;abstract-algebra&gt;&lt;definition&gt;&lt;loops&gt;&lt;quasigroups&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11525" PostTypeId="2" ParentId="11509" CreationDate="2010-11-23T15:32:46.947" Score="2" ViewCount="0" Body="&lt;p&gt;The definition of $||x||$ for vectors is:&#xA;$$||x|| = \sqrt{x\cdot x}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you have that&#xA;\begin{align*}&#xA;||x+y||^2 &amp;amp;= (x+y)\cdot(x+y)  &amp;amp;\text{(by definition)}\\&#xA;&amp;amp;= x\cdot x + x\cdot y + y\cdot x + y\cdot y&#xA;&amp;amp;\text{(by distributivity)}\\&#xA;&amp;amp;= x\cdot x + y\cdot y + 2(x\cdot y) &amp;amp;\text{(by symmetry)}\\&#xA;&amp;amp;= ||x||^2 + ||y||^2 + 2(x\cdot y) &amp;amp;\text{(by definition)}\\&#xA;&amp;amp;= ||x||^2 + ||y||^2 + 2||x||\,||y||\cos(\theta),&#xA;\end{align*}&#xA;where $\theta$ is the angle between $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $||x+y||^2 = ||x||^2 + ||y||^2$ if and only if $2||x||\,||y||\cos(\theta)=0$. One possibility is $||x||=0$; another is $||y||=0$; and the final one is $\cos(\theta)=0$. So the equality holds if and only if one of the following happens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$||x||=0$;&lt;/li&gt;&#xA;&lt;li&gt;$||y||=0$; or&lt;/li&gt;&#xA;&lt;li&gt;$\cos(\theta)=0$ where $\theta$ is the angle between $x$ and $y$, $x\neq 0$, $y\neq 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T23:07:00.273" LastActivityDate="2010-12-05T23:07:00.273" CommentCount="2" />
  <row Id="11526" PostTypeId="2" ParentId="11524" CreationDate="2010-11-23T15:55:52.660" Score="6" ViewCount="0" Body="&lt;p&gt;A subloop $H$ of a loop $L$ is a normal subloop if $H$ is the kernel of a homomorphism from $L$ onto some loop $K$, and $L$ is a simple loop if it has no nontrivial normal subloops; that is, the only homomorphic images of $L$ have order $1$ or are isomorphic to $L$. (This is Qiaochu's answer above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the definition out of the review  (MR0977475 (90b:20057)) of&#xA;Bannai, Eiichi; Song, Sung-Yell, &lt;em&gt;The character tables of Paige's simple Moufang loops and their relationship to the character tables of ${\rm PSL}(2,q)$&lt;/em&gt;.&#xA;Proc. London Math. Soc. (3) &lt;strong&gt;58&lt;/strong&gt; (1989), no. 2, 209–236,  &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T15:55:52.660" />
  <row Id="11527" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T16:01:59.470" Score="6" ViewCount="0" Body="&lt;p&gt;With Fibonacci numbers usually there are &lt;em&gt;multiple&lt;/em&gt; ways of proving identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way (which is one of my favourites) to prove your identity is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A person climbs up $\displaystyle n$ steps, by taking either one step, or two steps at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways the person can climb up all the $\displaystyle n$ steps is $\displaystyle F_{n+1}$ (Why?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider climbing $\displaystyle m+n-1$ steps and split into the cases when the person lands on step $\displaystyle n$ and the cases when the person lands on step $\displaystyle n-1$ and takes two steps at that point (and so does not land on step $\displaystyle n$ in those cases). These two cases cover all possibilities, and so we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle F_{m+n} = F_{n+1}F_{m} + F_{n}F_{m-1}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-23T16:01:59.470" CommentCount="5" />
  <row Id="11528" PostTypeId="1" AcceptedAnswerId="11545" CreationDate="2010-11-23T16:12:44.790" Score="1" ViewCount="120" Body="&lt;p&gt;What is the next term of the following sequences ? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ 3,8,13,24,41, \cdots $ &#xA;       &lt;strong&gt;OPTIONS :&lt;/strong&gt; $\{ 70,75,80,85 \}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$ 4,23,60,121, \cdots $  &lt;strong&gt;OPTIONS :&lt;/strong&gt; $\{212,221,241,242\}$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This two question is taken from a test paper of mine,I am no able to figure out the correct-answer.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-23T18:13:52.983" LastActivityDate="2010-11-23T18:46:16.560" Title="Find the missing term :" Tags="&lt;elementary&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="11529" PostTypeId="2" ParentId="11501" CreationDate="2010-11-23T16:13:04.837" Score="-1" ViewCount="0" Body="&lt;p&gt;A &lt;i&gt;recurrence relation&lt;/i&gt; defines a function depending on its values on points of lower &quot;rank&quot;. Since you want to be able to calculate the function, you also need some bases cases that are just given (like axioms). For example, the recurrence relation defining the Fibonacci numbers defines $F_n$ in terms of $F_{n-1},F_{n-2}$; the base cases $F_1=1,F_0=0$ are not defined in terms of previous elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can have more complicated definitions of &quot;rank&quot;. Here are two examples. One can define the binomial coefficients using Pascal's identity $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$ with appropriate bases cases. Written out, that would be $$C(n,k) = C(n-1,k-1) + C(n-1,k), \quad C(0,k) = 1_{k=0}.$$ So in that cases $(n-1,\cdot)$ is of lower rank than $(n,\cdot)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the determinant - it's a function on $n \times n$ matrices (arrays of numbers) which is defined as the single value in case $n=1$, and using some formula involving $(n-1)\times(n-1)$ matrices otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for the definition to work, the notation of rank must be a &lt;i&gt;well ordering&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A recursive algorithm is a procedure which calls itself. For example, you can use the following pseudocode to calculate the Fibonacci numbers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if ((n == 0) || (n==1)) return n;&#xA;else return F(n-1) + F(n-2);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Every recurrence relation can be computed this way. In this particular case, the algorithm for calculating $F_n$ runs in exponential time (in other words, it's very inefficient), and there are better ways to calculate $F_n$; for example, you could calculate the values iteratively:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F[0] = 0; F[1] = 1;&#xA;for (int i = 2; i &amp;lt;= n; i++) F[i] = F[i-1] + F[i-2];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are even better algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, recursive algorithms have other uses, for example the best algorithms for sorting a list of numbers are recursive and use the &quot;divide and conquer&quot; paradigm.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T16:13:04.837" />
  <row Id="11530" PostTypeId="2" ParentId="11518" CreationDate="2010-11-23T16:16:49.640" Score="2" ViewCount="0" Body="&lt;p&gt;Note that the ring of integers of $\mathbb{Q}(\sqrt[3]{3})$ is $\mathbb{Z}[\sqrt[3]{3}]$ with basis $1,\sqrt[3]{3},\sqrt[3]{9}$ over $\mathbb{Z}$. You want to show that the norm, defined by&#xA;\begin{eqnarray*}&#xA;N(a+b\sqrt[3]{3} + c\sqrt[3]{9}) &amp;amp; = &amp;amp; (a+b\sqrt[3]{3} + c\sqrt[3]{9})(a+\zeta_3b\sqrt[3]{3} + \zeta_3^2c\sqrt[3]{9})(a+\zeta_3^2b\sqrt[3]{3} + \zeta_3c\sqrt[3]{9})\\&#xA;&amp;amp; = &amp;amp; a^3 + 3b^3 + 9c^3 - 9abc&#xA;\end{eqnarray*}&#xA;gives in fact a Euclidean norm upon taking absolute values, where $\zeta_3$ is a fixed primitive cube root of unity. Now, the first (and maybe the main) thing you might wonder about is where I pulled this norm out. For that you have to know a little bit of Galois theory. The basic idea is that from the perspective of $\mathbb{Q}$, the elements $\sqrt[3]{3}$ and $\zeta_3\sqrt[3]{3}$ are indistinguishable: they are both just &lt;strong&gt;some&lt;/strong&gt; roots of the polynomial $x^3-3$ and can be thought of as mirror images of each other. Essentially, the factors that I am multiplying are like &quot;mirror images&quot; of my given element of $\mathbb{Z}[\sqrt[3]{3}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the strategy is exactly the same as in the case of some quadratic rings like $\mathbb{Z}[\sqrt{2}]$: given $\alpha = u+v\sqrt[3]{3} + w\sqrt[3]{9}$ and&#xA;$\beta = f+g\sqrt[3]{3} + h\sqrt[3]{9}\in \mathbb{Z}[\sqrt[3]{3}]$, you want to show that there exist $p$ and $r$ in that ring such that&#xA;\begin{eqnarray*}\alpha = p\beta + r,&#xA;\end{eqnarray*}&#xA;where either $r=0$ or $N(r)&amp;lt;N(\beta)$. Dividing by $\beta$, this becomes equivalent to showing that given any $\alpha/\beta=a+b\sqrt[3]{3} + c\sqrt[3]{9}$ in the field of fraction $\mathbb{Q}(\sqrt[3]{3})$ of $\mathbb{Z}[\sqrt[3]{3}]$, there exists $r/\beta\in \mathbb{Q}(\sqrt[3]{3})$ such that the difference is in the ring of integers itself (since the difference is $p$) and $|N(r/\beta)|&amp;lt;1$. That is the statement you posted.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-23T16:16:49.640" CommentCount="1" />
  <row Id="11531" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T16:16:59.267" Score="1" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;3+8+2=13,8+13+3=24,13+24+4=41,24+41+5=70&lt;/li&gt;&#xA;&lt;li&gt;I don't know&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;PS: you should use the tag &quot;homework&quot;&lt;/p&gt;&#xA;" OwnerUserId="3631" LastActivityDate="2010-11-23T16:16:59.267" CommentCount="3" />
  <row Id="11532" PostTypeId="2" ParentId="11515" CreationDate="2010-11-23T16:18:22.383" Score="5" ViewCount="0" Body="&lt;p&gt;It is useful to think about this by using vectors. Suppose that your points are $p,q,a \in \mathbb{R^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $x = p + (q-p)t$ is a point on the line $pq$. We wish to find the projection of $a$&amp;#160;on the line. To do this we require&#xA;$\|x - a\|^2 = \|x\|^2 + \|a\|^2 - 2 (x \cdot a)$ to be minimal. That is we have to minimize&#xA;$$\|p + (q-p)t\|^2 + \|a\|^2 - 2 (p + (q-p) t) \cdot a)$$&#xD;&#xA;w.r.t $t$. To do this we write this as&#xA;$$\|p\|^2 + \|q-p\|^2 t^2 + 2 t p \cdot (q-p) + \|a\|^2 - 2(p \cdot a) - 2 t (q-p) \cdot a.$$&#xD;&#xA;This is a quadratic in $t$ with minimum at the tip of the parabola:&#xA;$$t = \frac{-2 p \cdot (q-p) + 2 (q-p) \cdot a}{2\|q-p\|^2} = \frac{(q-p) \cdot (a-p)}{\|q-p\|^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the projection is given by&#xA;$$x = p + (q-p) \frac{(q-p) \cdot (a-p)}{\|q-p\|^2}$$ and the reflection is then just $x + (x-a) = 2x-a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method doesn't have problems with infinite slopes.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-23T16:18:22.383" CommentCount="5" />
  <row Id="11533" PostTypeId="2" ParentId="11435" CreationDate="2010-11-23T16:21:22.647" Score="2" ViewCount="0" Body="&lt;p&gt;The (Motivation) description appears to be similar to a direct sum (a.k.a. tegum product) as defined in Section 1.2 of this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.reed.edu/~davidp/homepage/seniors/mcmahan.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.reed.edu/~davidp/homepage/seniors/mcmahan.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that for a direct sum, 0 is in the relative interior of both of the factor polytopes, rather than a common vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same article answers your question about what the nm vertices of a Cartesian product are.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-11-23T16:55:25.710" LastActivityDate="2010-11-23T16:55:25.710" />
  <row Id="11534" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T16:21:28.437" Score="4" ViewCount="0" Body="&lt;p&gt;For part (2), try adding $4$ to the numbers.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-23T16:21:28.437" CommentCount="2" />
  <row Id="11535" PostTypeId="2" ParentId="11518" CreationDate="2010-11-23T16:40:32.590" Score="2" ViewCount="0" Body="&lt;p&gt;As Alex mentioned, the proof that you have in mind amounts to showing that your cubic field is Euclidean with respect to the norm. In fact it is known that there are only three pure cubic fields $\rm\ Q(\sqrt[3] m)\ $ that are norm Euclidean, viz. $\rm\ \mathbb Q(\sqrt[3] 2),\ \mathbb Q(\sqrt[3] 3),\ \mathbb Q(\sqrt[3] {10})\:.\ $ You can find the proofs in Cioffari: &lt;a href=&quot;http://www.ams.org/journals/mcom/1979-33-145/S0025-5718-1979-0514835-5/S0025-5718-1979-0514835-5.pdf&quot; rel=&quot;nofollow&quot;&gt;The Euclidean Condition in Pure Cubic and Complex Quartic Fields.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-23T16:47:21.970" LastActivityDate="2010-11-23T16:47:21.970" CommentCount="3" />
  <row Id="11536" PostTypeId="2" ParentId="10468" CreationDate="2010-11-23T16:46:43.467" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://tutorteddy.com/site/&quot; rel=&quot;nofollow&quot;&gt;http://tutorteddy.com/site/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3864" LastActivityDate="2010-11-23T16:46:43.467" />
  <row Id="11537" PostTypeId="1" AcceptedAnswerId="11546" CreationDate="2010-11-23T17:02:38.837" Score="2" ViewCount="81" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Discontinuous_linear_map#Axiom_of_choice&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, you need the Axiom of Choice to find a discontinuous map between two Banach spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this procedure also apply for Banach algebras yielding a discontinuous multiplicative linear map?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, is there some obstruction, ensuring that every algebraic homomorphism between two Banach algebras is continuous? (I know that this is true for $*$-homomorphisms between C*-algebras.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-23T21:04:26.220" LastActivityDate="2010-11-23T21:04:26.220" Title="Is there an algebraic homomorphism between two Banach algebras which is not continuous?" Tags="&lt;set-theory&gt;&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="11538" PostTypeId="1" AcceptedAnswerId="11540" CreationDate="2010-11-23T17:04:54.977" Score="5" ViewCount="142" Body="&lt;p&gt;I understand the formal definition of uniform continuity of a function,&#xA;and how it is different from standard continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is there an intuitive way to classify a function&#xA;on $\mathbb{R}$ as being&#xA;uniformly continuous, just like there is for regular continuity?&#xA;I mean, that for a &quot;nice&quot; function $f:\mathbb{R} \to \mathbb{R}$,&#xA;it is usually easy to tell if it is continuous on an interval by&#xA;looking at or thinking of the graph of the function on the interval,&#xA;or on all of $\mathbb{R}$. Can I do the same for uniform continuity?&#xA;Can I tell that a function is uniformly continuous just by what it&#xA;looks like? Ideally, this intuition would fit with the Heine-Cantor theorem for&#xA;compact sets on $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-11-23T21:04:02.507" Title="Intuition for uniform continuity of a function on $\mathbb{R}$" Tags="&lt;real-analysis&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="11539" PostTypeId="1" AcceptedAnswerId="11544" CreationDate="2010-11-23T17:06:00.450" Score="0" ViewCount="124" Body="&lt;p&gt;Consider a function f:[0,1] -&gt; R  with f(0)=0&lt;br&gt;&#xA;having a finite derivative at each x in (0,1)&lt;br&gt;&#xA;Prove that if f' is an increasing function (at least on the interval (0,1))&lt;br&gt;&#xA;then h(x)=f(x)/x is also increasing&lt;/p&gt;&#xA;" OwnerUserId="3333" LastActivityDate="2010-11-23T17:42:57.830" Title="Prove an increasing derivative implies an increasing function" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11540" PostTypeId="2" ParentId="11538" CreationDate="2010-11-23T17:14:16.323" Score="9" ViewCount="0" Body="&lt;p&gt;I like to think of the following fact: a $C^1$ function on $\mathbb{R}$ with bounded derivative is uniformly continuous.  So in order for a function not to be uniformly continuous, there have to be places where its graph is &quot;arbitrarily steep&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the theorem that any continuous function on a compact set is uniformly continuous, this can only happen if you make the function steeper and steeper as you go off to $\infty$, either unboundedly (like $x^2$) or boundedly (like $\sin(x^2)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As pointed out, the converse is false: a function with unbounded derivative can still be uniformly continuous.  $\sin(x^4)/(1+x^2)$ is such an example, I believe.  So this may not be great intuition after all.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-11-23T21:04:02.507" LastActivityDate="2010-11-23T21:04:02.507" CommentCount="6" />
  <row Id="11541" PostTypeId="2" ParentId="11538" CreationDate="2010-11-23T17:14:43.090" Score="6" ViewCount="0" Body="&lt;p&gt;One possible intuition for uniform continuity is that a function shouldn't oscillate arbitrarily wildly. If the function is differentiable e.g. on an open interval, then the derivative should be bounded in absolute value. That's something that you can see by looking at the graph. For example if you consider the function $sin(1/x)$ on the open interval $(0,1)$, then just by looking at it, you will realise that it's not uniformly continuous, since it's slope becomes arbitrarily steep as you approach $0$ from the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way this intuition ties in with the Heine-Cantor theorem is that on compact sets, bounded functions attain their extrema. If your function is differentiable, then this is in particular true for its derivative, so the slope of the function cannot get arbitrarily steep on a compact interval.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-23T17:58:47.390" LastActivityDate="2010-11-23T17:58:47.390" CommentCount="2" />
  <row Id="11542" PostTypeId="1" AcceptedAnswerId="11572" CreationDate="2010-11-23T17:15:28.323" Score="5" ViewCount="154" Body="&lt;p&gt;I'm trying to understand a proof in &quot;Combinatorial Group Theory&quot; by Lyndon &amp;amp; Schupp, page 26, Lemma 4.7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F$ is a free group with $X$ as basis. $x \in X$. $\alpha \in IA(F)$ - meaning it's an automorphism of $F$ which induces the identity automorphism on $F/[F,F]$. $k \in F_n$ - a term in the descending central series for F. $x\alpha$ and $xk$ are conjugates.&#xA;In addition: for some $N\geq1$, $\alpha^N \in JA(F)$ - the group of inner automorphisms of F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Elementary Fact&quot; used in the proof: $\alpha \in IA(F)$ and $k \in F_n$ implies $k \alpha = k$ (modulo $F_{n+1}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right after mentioning this &quot;elementary fact&quot; they say it implies: $x\alpha^N = xk^N$ (modulo $F_{n+1}$). Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: A little clarification: $x\alpha$ is the image of $x$ under the automorphism $\alpha$, following the notation in this book. $xk$ is just $x$ times $k$, multiplication in the group $F$.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:44:43.387" LastActivityDate="2010-11-28T21:36:56.230" Title="Combinatorial Group Theory / Lyndon &amp; Schupp, Lemma 4.7. Automorphisms of free groups." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="11543" PostTypeId="2" ParentId="11538" CreationDate="2010-11-23T17:20:06.753" Score="3" ViewCount="0" Body="&lt;p&gt;This is slightly tricky because your intuitive picture for &quot;continous&quot; probably more closely matches &quot;uniformly continuous.&quot;  That is if you can &lt;em&gt;actually&lt;/em&gt; draw a graph without lifting your pencil then it's uniformly continuous.  In order to get something continuous but not uniformly continuous you have to do something that you can't actually draw like going off to infinity on an open interval or oscillating really wildly (as explained in the other two answers).&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-11-23T17:20:06.753" />
  <row Id="11544" PostTypeId="2" ParentId="11539" CreationDate="2010-11-23T17:23:06.327" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: show that $h&amp;#39;(x)$ is positive on $(0,1)$, using the mean value theorem applied to $f(x)-f(0)$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T17:23:06.327" CommentCount="3" />
  <row Id="11545" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T17:32:16.657" Score="3" ViewCount="0" Body="&lt;p&gt;For the second one, take a look at the Online Encyclopedia of Integer Sequences[1]. Sequence #A030653 is what you are wanting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can look at [1] for more extensive information including sources, but I will tell a bit about it here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a(n)$ be the n-th number of the sequence for $n \in \{1,2,3...\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that $a(n) = n^{3} + 3n^{2} + 3n - 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(1) = 1 + 3(1) + 3(1) - 3 = 4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(2) = 8 + 3(4) + 3(2) -3 = 8 + 12 + 6 - 3 = 23$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next number in the sequence, we need to calculate $a(5)$. Using the formula above, we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(5) = 125 + 3(25) + 3(5) - 3 = 125 + 75 + 15 -3 = 212$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 212 is the next number in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://oeis.org/A030653&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A030653&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T17:32:16.657" CommentCount="2" />
  <row Id="11546" PostTypeId="2" ParentId="11537" CreationDate="2010-11-23T17:42:54.460" Score="6" ViewCount="0" Body="&lt;p&gt;Rasmus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There are&quot; examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is &lt;em&gt;independent&lt;/em&gt; of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, &quot;A discontinuous homomorphism from $C(X)$ without CH&quot;, J. London Math. Soc. (2) 48 (1993), no. 2, 299-315:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $X$ is an infinite compact Hausdorff space and let $C(X)$ be the algebra of continuous real-valued functions on $X$. Then $C(X)$ is a commutative Banach algebra relative to the supnorm: ${}\|f\|= \sup\{f(p)\mid p\in X\}$. A well-known question of I. Kaplansky posed around 1947 asks whether every algebra homomorphism of $C(X)$ into a Banach algebra $B$ is necessarily continuous. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There is a discontinuous homomorphism of $C(X)$ if and only if there is a discontinuous homomorphism of $C(X, {\mathbb C})$, the $C^*$-algebra of continuous complex-valued functions on $X$. We prefer to deal with the real case; some of the references adopt the complex view. The question is now known to be independent of the axioms of set theory, ZFC. H. G. Dales &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin%27s_axiom&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; and J. Esterle [3] independently constructed discontinuous homomorphisms of $C(X)$ for any infinite space $X$ assuming the &#xA;  Continuum Hypothesis, CH. About the same time R. Solovay [4] proved that it is relatively consistent with ZFC that every homomorphism of $C(X)$ for any space X is necessarily continuous. Solovay's result was improved [5] fairly soon thereafter, to obtain the relative consistency with ZFC + &lt;a href=&quot;http://en.wikipedia.org/wiki/Martin%27s_axiom&quot; rel=&quot;nofollow&quot;&gt;Martin's Axiom&lt;/a&gt; (ZFC + MA) that every homomorphism of $C(X)$ for any space $X$ is continuous. We refer the reader to &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_forcing_axiom&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt; for an exposition of the latter result concerning MA, historical points and related results. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After these results several questions remained. This paper is concerned with the question of whether the existence of a discontinuous homomorphism of $C(X)$ is possible given the failure of the Continuum Hypothesis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The references listed above are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;H. G. DALES, &quot;A discontinuous homomorphism from $C(X)$&quot; Amer. J. Math. 101 (1979) 647-734. &lt;/li&gt;&#xA;&lt;li&gt;H. G. DALES and H. WOODIN, An introduction to independence for analysts, London Mathematical Society Lecture Notes 115 (University Press, Cambridge, 1987). &lt;/li&gt;&#xA;&lt;li&gt;J. R. ESTERLE, &quot;Injection de semi-groupes divisibles dans des algebres de convolution et construction d'homomorphismes discontinus de $C(K)$&quot; Proc. London Math. Soc. (3) 36 (1978) 59-85. &lt;/li&gt;&#xA;&lt;li&gt;R. SOLOVAY, personal communication, unpublished. &lt;/li&gt;&#xA;&lt;li&gt;H. WOODIN, Set theory and discontinuous homomorphisms from Banach algebras, Ph.D. Thesis, University of California at Berkeley, 1984. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I believe Solovay's original construction is simply superseded by Hugh's result, as described in his thesis. In any case, reference 2 is particularly good to learn about this fascinating topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly after, in Stevo Todorcevic, Partition Problems in Topology (Contemporary Mathematics), American Mathematical Society (January 1989), it is shown that automatic continuity of homomorphisms between Banach algebras is a theorem of ZFC + PFA. PFA is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_forcing_axiom&quot; rel=&quot;nofollow&quot;&gt;proper forcing axiom&lt;/a&gt;, a strengthening of Martin's axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a much more recent reference: H. Garth Dales, Banach Algebras and Automatic Continuity (London Mathematical Society Monographs New Series), Oxford University Press, USA (May 17, 2001).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-23T17:42:54.460" />
  <row Id="11547" PostTypeId="2" ParentId="11539" CreationDate="2010-11-23T17:42:57.830" Score="0" ViewCount="0" Body="&lt;p&gt;This seems to be a straightforward application of Cauchy's Mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-11-23T17:42:57.830" />
  <row Id="11548" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T17:48:34.737" Score="1" ViewCount="0" Body="&lt;p&gt;To add to Gerry's answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first sequence, you can use a recurrence relation to find the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(n+2) = y(n+1) + y(n) + n - 1$, $y(1) = 3$, and $y(2) = 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(3) = y(2) + y(1) + 3 - 1 = 8 + 3 + 3 - 1 = 13$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(4) = y(3) + y(2) + 4 - 1 = 13 + 8 + 4 - 1 = 24$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are trying to find the next number in the sequence which is $y(6)$. So we have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(6) = y(5) + y(4) + 6 - 1 = 41 + 24 + 6 - 1 = 70$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, 70 should be the next number in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I get a chance to find a formula for the sequence that does not involve recursion, I will post it here.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T17:48:34.737" />
  <row Id="11549" PostTypeId="2" ParentId="11464" CreationDate="2010-11-23T18:13:11.567" Score="1" ViewCount="0" Body="&lt;p&gt;As noted above, observe that&#xA;\begin{eqnarray}&#xA;\sum_{r = 1}^{d} x^{r} = \frac{x(x^{d} - 1)}{x - 1}.&#xA;\end{eqnarray}&#xA;Differentiating both sides and multiplying by $x$, we find&#xA;\begin{eqnarray}&#xA;\sum_{r = 1}^{d} r x^{r} = \frac{dx^{d + 2} - x^{d+1}(d+1) + x}{(x - 1)^{2}}.&#xA;\end{eqnarray}&#xA;Substituting $x = 2$, &#xA;\begin{eqnarray}&#xA;\sum_{r = 1}^{d} r 2^{r} =  d2^{d + 2} - (d+1) 2^{d+1} + 2 = (d - 1) 2^{d+1} + 2.&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-23T18:42:09.403" LastActivityDate="2010-11-23T18:42:09.403" CommentCount="1" />
  <row Id="11550" PostTypeId="2" ParentId="10392" CreationDate="2010-11-23T18:21:44.007" Score="1" ViewCount="0" Body="&lt;p&gt;Pick out the rows where a &lt;em&gt;t&lt;/em&gt; appears in the rightmost column, and write down a disjunctive normal form.  In your example, there are only two rows with a &lt;em&gt;t&lt;/em&gt; and your expression will have two terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (X \cdot \bar{Y} \cdot Z)  + (\bar{X} \cdot \bar{Y} \cdot Z) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a logical formula for your truth table. You can stop there, or you can use the laws of &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_algebra_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;boolean algebra&lt;/a&gt; to get a simpler expression. In this case, you can use the distributive law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (X \cdot \bar{Y} \cdot Z)  + (\bar{X} \cdot \bar{Y} \cdot Z) = (X + \bar{X})\cdot (\bar{Y} \cdot Z) = 1 \cdot (\bar{Y} \cdot Z) = \bar{Y} \cdot Z$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-11-23T18:21:44.007" />
  <row Id="11551" PostTypeId="1" CreationDate="2010-11-23T18:24:11.590" Score="10" ViewCount="343" Body="&lt;p&gt;Why in a Hausdorff sequentially compact space the size of the closure of a countable subset is less or equal than $c$ ? I can see why this is true when the space if first countable but we are not assuming so.&lt;/p&gt;&#xA;" OwnerUserId="3867" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-23T22:57:26.453" LastActivityDate="2010-12-27T16:44:28.200" Title="Size of the closure of a set" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" />
  <row Id="11552" PostTypeId="1" AcceptedAnswerId="11558" CreationDate="2010-11-23T18:34:15.410" Score="1" ViewCount="152" Body="&lt;p&gt;$f\colon (a,b) \to \mathbb{R}$ is continuous, with finite derivatives everywhere in $(a,b)$, except maybe at $c$.&lt;br&gt;&#xA;If &#xA;$$\lim_{x\to c}f&amp;#39;(x) = B,$$&#xD;&#xA;show that $f&amp;#39;(c)$ exists and equals $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to start on this. I've tried using the definitions of continuity and f' but it isn't working out  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I started out with &#xA;$$\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f&amp;#39;(c)$$&#xD;&#xA;is equivalent to saying&lt;br&gt;&#xA;given $\epsilon\gt 0$ there exists $\delta\gt 0$ such that&lt;br&gt;&#xA;$$|x-c| \lt \delta \Longrightarrow \left|\frac{f(x)-f(c)}{x-c} - f&amp;#39;(c)\right| \lt\epsilon.$$  &#xA;I was trying to prove $f&amp;#39;(c)$ must exist since $f$ is continuous but I feel like I'm assuming what I'm trying to prove. &lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T19:32:50.280" LastActivityDate="2010-11-24T00:42:04.950" Title="Prove a derivative exists" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11553" PostTypeId="2" ParentId="11552" CreationDate="2010-11-23T18:41:57.230" Score="0" ViewCount="0" Body="&lt;p&gt;So what can go wrong?  We can certainly define a function g(x)=something nice except when x=c, something different at c.  Then $\lim_{x\rightarrow c} g(x) \neq g(c)$ Can you rule out this behavior in $f&amp;#39;(x)$?  Or maybe $f&amp;#39;(x)$ has a step at $c$.  Why can't it?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T18:41:57.230" CommentCount="1" />
  <row Id="11554" PostTypeId="2" ParentId="11528" CreationDate="2010-11-23T18:46:16.560" Score="1" ViewCount="0" Body="&lt;p&gt;I am adding lots of answers here, sorry. I think since they are giving different information, they are best put separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have not had experience with this, you may be interested in how people derive formulas for these problems. Difference calculus is a nice method to use. Kelley and Peterson's textbook [1] is a very good resource for this (I tend to recommend this on a lot of different questions on here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can take a look at the notes I took in my difference equations course for this type of problem [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any additional questions, I'd be happy to try to help you. Feel free to email me, my contact information is on my website ( &lt;a href=&quot;http://www.tylerclark12.com/blog/?page_id=11&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/?page_id=11&lt;/a&gt; ). Good luck with your exam! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T18:46:16.560" />
  <row Id="11555" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T18:48:31.940" Score="5" ViewCount="0" Body="&lt;p&gt;While some of the other answers look good, there are two particular related uses for polynomial multiplication which I didn't see while skimming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is related to probabilities. We can use a polynomial to represent the possible outcomes of a random process (so long as there are finitely many), where there is a variable for each possible outcome, and the coefficient of that variable is the probability of it occurring. (We'll just go with linear polynomials for now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, a toss of a fair coin might be represented as 1/2 H + 1/2 T. An unfair coin might be 2/3 H + 1/3 T. Someone's plays in the game of rock/paper/scissors might be summarised as 3/5 rock + 1/5 paper + 1/5 scissors. (Dude likes rock.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what would it mean to multiply two such polynomials? Well, supposing that the random events are independent of each other (that is, the outcome of one doesn't affect the probabilities involved in the other), the product of two such polynomials tells us the likelihood of various pairs of outcomes. If we flip that unbalanced coin, and have that guy throw rock/paper/scissors, the combined possibilities are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2/3 H + 1/3 T) (3/5 rock + 1/5 paper + 1/5 scissors) = 2/5 H rock + 1/5 T rock + 2/15 H paper + 1/15 T paper + 2/15 H scissors + 1/15 T scissors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can read off that, for instance, the combined probability of getting a head on the coin and having the guy throw scissors is the coefficient of H scissors, which is 2/15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's a decent way to organise information about probabilities in a way that lets us combine multiple independent random processes. Raising a polynomial to the power of n in this setting gives us the probabilities when the process is repeated n times, assuming again that the outcome of the previous trials doesn't affect the next one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let's flip the unbalanced coin three times in a row:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2/3 H + 1/3 T)^3 = 8/27 H^3 + 4/9 H^2 T + 2/9 H T^2 + 1/27 T^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that the probability that we get two heads and one tail is 4/9, and so on. (If we didn't allow H and T to commute with each other, we would get the probability of each possible sequence.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that it's quite effective to use polynomials in keeping track of proportions when forming mixtures of things, which are then mixed together (say, in chemistry), using substitution of one variable representing a mixture with the polynomial representing its contents. Multiplication doesn't immediately have such a useful interpretation there though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing we can do is to use the coefficients of polynomials to count the number of things of a given type. If we have a (finite) bunch of things with integer &quot;sizes&quot; or &quot;weights&quot;, and we want to record the number of things with a given weight, we can use the coefficient of x^n in a polynomial to record the number of objects with weight n. The product of two such polynomials will then count the number of ways to form a pair of objects with total weight n, for each n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we want to know the number of ways to put change for n cents into a vending machine using 5 coins, for each possible n. (And we'll assume that the coins have values {1, 5, 10, 25, 100, 200}). We can use the polynomial c(x) = (x^1 + x^5 + x^10 + x^25 + x^100 + x^200) to represent our set of coins. Sequences of 5 coins are then represented by (c(x))^5. This is a polynomial with a fairly large number of terms (representing the fact that there are lots of ways to choose a sequence of 5 coins in order, specifically, 6^5 = 7776, and moreso the fact that there are many different possible sums that those coins could have).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This polynomial is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^5 + 5 x^9 + 10 x^13 + 5 x^14 + 10 x^17 + 20 x^18 + 5 x^21 + &#xA; 30 x^22 + 10 x^23 + x^25 + 20 x^26 + 30 x^27 + 5 x^29 + 5 x^30 + &#xA; 30 x^31 + 10 x^32 + 20 x^33 + 10 x^35 + 20 x^36 + 30 x^37 + &#xA; 20 x^38 + 10 x^40 + 25 x^41 + 60 x^42 + 10 x^45 + 60 x^46 + &#xA; 30 x^47 + 21 x^50 + 60 x^51 + 10 x^53 + 30 x^55 + 20 x^56 + &#xA; 30 x^57 + 20 x^60 + 30 x^61 + 30 x^62 + 15 x^65 + 60 x^66 + &#xA; 30 x^70 + 30 x^71 + 30 x^75 + 10 x^77 + 10 x^80 + 20 x^81 + &#xA; 10 x^85 + 20 x^86 + 20 x^90 + 10 x^95 + 5 x^101 + 5 x^104 + &#xA; 5 x^105 + 20 x^108 + 5 x^110 + 30 x^112 + 20 x^113 + 20 x^116 + &#xA; 60 x^117 + 5 x^120 + 60 x^121 + 30 x^122 + 21 x^125 + 60 x^126 + &#xA; 20 x^128 + 30 x^130 + 20 x^131 + 60 x^132 + 20 x^135 + 60 x^136 + &#xA; 60 x^137 + 25 x^140 + 120 x^141 + 60 x^145 + 60 x^146 + 60 x^150 + &#xA; 30 x^152 + 20 x^155 + 60 x^156 + 30 x^160 + 60 x^161 + 60 x^165 + &#xA; 30 x^170 + 20 x^176 + 20 x^180 + 20 x^185 + 5 x^200 + 10 x^203 + &#xA; 5 x^204 + 30 x^207 + 20 x^208 + 30 x^211 + 60 x^212 + 20 x^213 + &#xA; 10 x^215 + 80 x^216 + 60 x^217 + 35 x^220 + 90 x^221 + 30 x^222 + &#xA; 50 x^225 + 60 x^226 + 30 x^227 + 20 x^228 + 40 x^230 + 80 x^231 + &#xA; 60 x^232 + 50 x^235 + 120 x^236 + 60 x^237 + 85 x^240 + 120 x^241 + &#xA; 90 x^245 + 60 x^246 + 60 x^250 + 30 x^251 + 30 x^252 + 50 x^255 + &#xA; 60 x^256 + 60 x^260 + 60 x^261 + 60 x^265 + 30 x^270 + 10 x^275 + &#xA; 20 x^276 + 20 x^280 + 20 x^285 + 5 x^300 + 10 x^302 + 20 x^303 + &#xA; 20 x^306 + 60 x^307 + 10 x^310 + 80 x^311 + 60 x^312 + 40 x^315 + &#xA; 120 x^316 + 70 x^320 + 60 x^321 + 60 x^325 + 20 x^326 + 60 x^327 + &#xA; 40 x^330 + 120 x^331 + 80 x^335 + 120 x^336 + 120 x^340 + 60 x^345 + &#xA; 10 x^350 + 60 x^351 + 60 x^355 + 60 x^360 + 20 x^375 + 5 x^401 + &#xA; 30 x^402 + 10 x^403 + 5 x^405 + 60 x^406 + 30 x^407 + 35 x^410 + &#xA; 90 x^411 + 30 x^412 + 70 x^415 + 60 x^416 + 60 x^420 + 30 x^421 + &#xA; 35 x^425 + 60 x^426 + 30 x^427 + 70 x^430 + 60 x^431 + 90 x^435 + &#xA; 60 x^436 + 60 x^440 + 30 x^445 + 30 x^450 + 30 x^451 + 30 x^455 + &#xA; 30 x^460 + 10 x^475 + x^500 + 20 x^501 + 30 x^502 + 20 x^505 + &#xA; 60 x^506 + 50 x^510 + 60 x^511 + 60 x^515 + 30 x^520 + 20 x^525 + &#xA; 60 x^526 + 60 x^530 + 60 x^535 + 30 x^550 + 5 x^600 + 30 x^601 + &#xA; 10 x^602 + 30 x^605 + 20 x^606 + 40 x^610 + 20 x^611 + 20 x^615 + &#xA; 10 x^620 + 30 x^625 + 20 x^626 + 20 x^630 + 20 x^635 + 10 x^650 + &#xA; 10 x^700 + 20 x^701 + 20 x^705 + 20 x^710 + 20 x^725 + 10 x^800 + &#xA; 5 x^801 + 5 x^805 + 5 x^810 + 5 x^825 + 5 x^900 + x^1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just used my computer to calculate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us, for instance, that there are 120 distinct ways to put coins in a vending machine to make $1.41, since the coefficient of x^141 is 120. There is of course, only one way to make change for 10 dollars (which is to put in all 2 dollar coins), so the coefficient of x^1000 is 1. Note also that if we were to evaluate this polynomial at x = 1, we would get the total number of possible sequences, which is 7776. It's also maybe interesting to see which terms are missing from this polynomial (have a coefficient of 0), telling us that there's no way to make change for that amount using 5 coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, both of these approaches generalise to power series (which are like polynomials, but which have infinite sequences of terms), and the effectiveness of the approach really becomes more apparent in that more general setting. This technique is called generating series (or the somewhat unfortunate, but perhaps more common name &quot;generating functions&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I think that gives some impression of some of the variety of things that one can do with polynomials. (I'll just leave that pun there for the algebraic geometers. ;)&lt;/p&gt;&#xA;" OwnerUserId="3869" LastEditorUserId="3869" LastEditorDisplayName="" LastEditDate="2010-11-23T19:05:30.990" LastActivityDate="2010-11-23T19:05:30.990" />
  <row Id="11556" PostTypeId="2" ParentId="11551" CreationDate="2010-11-23T18:49:41.970" Score="0" ViewCount="0" Body="&lt;p&gt;It seems like you don't need all the conditions, so maybe I am missing something.  There are only continuum many sequences of points from a countable set.  So the limit points are at most continuum many.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T18:49:41.970" CommentCount="5" />
  <row Id="11557" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T18:52:26.927" Score="5" ViewCount="0" Body="&lt;p&gt;This is really several questions.  In increasing order of depth:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why multiply polynomials?  Because they are functions, and multiplication is a natural or desirable operation on functions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why multiply functions?   Because the values of the functions are integers, or rational numbers, or real numbers, or objects of some other more complicated kind that can be multiplied (such as matrices or rotations).  If multiplication is natural for these numbers or number-like objects, it is natural for functions whose values are such objects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why multiply integers, rational numbers, real numbers, etc?  Here is where explanation becomes difficult.  The natural operation is &lt;em&gt;not&lt;/em&gt; multiplication but the more structured operation of &quot;tensor multiplication&quot; that retains information about the factors that are multiplied.  One can represent an Area as a product of type Length x Length that remembers its two-dimensionality, instead of a one-dimensional numerical value of that product that has forgotten its origins.  In the same way, for integer multiplication it is most direct to multiply 5 and 6 by drawing 5x6 as a rectangular 2-dimensional array of dots instead of the &quot;numerical evaluation&quot; of that array as a one-dimensional string of 30 dots. The latter is less natural in that it requires a method of enumerating the dots in the grid and there is no preferred order in which to count them.  This is also reflected in the ability to naturally multiply 5 gadgets by 6 gizmos, without there being any given ordering of the gadgets or gizmos.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Polynomials, especially polynomials in several variables ($x +3x^2y + 2 z^{10} x$ and such), have more inner structure than numbers and thus can reflect -- in fact, they can be defined and derived from -- the tensor structure of multiplication.   So it is not that polynomials exist and we &lt;em&gt;might&lt;/em&gt; want to multiply them; it is that multiplication of numbers (or of finite sets) naturally involves more than just numerical information and polynomials are an enhancement of numbers that more directly embody that information &lt;em&gt;so that&lt;/em&gt; the multiplication can be performed in a way that retains more of the inner structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hint of this is that integers in base 10 are values of polynomials at $x = 10$ (and those polynomials can be thought of as a &quot;liberation&quot; or &quot;upgrading&quot; of the numbers).  Multiplication of integers can sometimes be seen to replicate the patterns in the coefficients when the polynomials are multiplied for general $x$, e.g., compare powers of $102$ and $x^2 + 0x + 2$.  Later there are things like generating functions and convolutions that directly exploit polynomials as carriers of information that is enriched compared to using numbers alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is glossing over some technical questions about commutativity. Also, areas should be sums of LxL products, and one should explain the role of sums as well as products.  But these are details that do not affect the main point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In applications, multiplication represents interaction or correlation between different effects.  Situations where there are several independent processes isolated from each other and contributing to some outcome lead to sums  of functions of the different variables, such as $f(x) + g(y) + h(z)$, the terms in the sum accounting for the separate effects.  Situations where different parts of the mechanism can interact with each other in producing the outcome, or where there is correlation between different effects (e.g., one intensifies or suppresses the other), almost always involve multiplication when expressed mathematically.   For example, if you count the number of handshakes in a group of $n$ people the answer will be of order $n^2$, and if you count the number of possible 3-person interactions in this group it will be of order $n^3$.  Nonlinearities reflect the ability to organize the people into pairs, triples etc, and this again reflects the fact that the natural multiplication of two finite sets $A$ and $B$ is the set of ordered pairs $(a,b)$ of elements, one from each set, and the same for triples and higher numbers of sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat faddish term for related ideas is (de- or re-)&lt;em&gt;categorification&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-23T23:55:32.883" LastActivityDate="2010-11-23T23:55:32.883" />
  <row Id="11558" PostTypeId="2" ParentId="11552" CreationDate="2010-11-23T19:38:26.030" Score="1" ViewCount="0" Body="&lt;p&gt;You are right that you are kind of assuming what you want, because you cannot even write down $f&amp;#39;(c)$ until you prove it exists; this can be fixed if you replace $f&amp;#39;(c)$ by a specific limit, and the specific limit you want is probably $B$. But even so ran into a bit of an alley, so let me help you out.&#xA;Here is the beginning of half a proof, following the Hint of Robin Chapman:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first the limit as $x\to c^+$. For any $h\gt 0$, the function $f(x)$ is continuous on $[c,c+h]$, and is differentiable on $(c,c+h)$, so by the Mean Value Theorem there exists a point $d_h$ (which depends on $h$) such that &#xA;$$\frac{f(c+h)-f(c)}{h} = f&amp;#39;(d_h).$$&#xD;&#xA;Therefore, we have that&#xA;\begin{align*}&#xA;\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c} &amp;amp;= \lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}\&#xA;&amp;amp;=\lim_{h\to 0^+}f'(d_h).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've managed to turn the (one-sided) limit that will tell us whether $f$ is differentiable at $c$ into a limit about $f&amp;#39;(x)$. Luckily, we have information about limits of $f&amp;#39;(x)$. So perhaps we can leverage that information into an actual value for this limit? If so, perhaps we can try doing the same thing for the limit from the left (as $x\to c^-$), so that the two together shows us that &#xA;$$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$&#xD;&#xA;exists; and will tell us what the value of $f&amp;#39;(c)$ must be in that case.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T00:42:04.950" LastActivityDate="2010-11-24T00:42:04.950" />
  <row Id="11559" PostTypeId="2" ParentId="11382" CreationDate="2010-11-23T19:38:55.593" Score="5" ViewCount="0" Body="&lt;p&gt;Why not?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="3757" LastActivityDate="2010-11-23T19:38:55.593" CommentCount="2" />
  <row Id="11560" PostTypeId="1" AcceptedAnswerId="11564" CreationDate="2010-11-23T19:45:37.543" Score="1" ViewCount="242" Body="&lt;p&gt;I have 2 matrices and have been trying to multiply them but to no avail. Then I found this &lt;a href=&quot;http://easycalculation.com/matrix/matrix-multiplication.php&quot; rel=&quot;nofollow&quot;&gt;online&lt;/a&gt; site and trying feeding it the values but yet no success.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;- R' . T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is what i would like to do but R is a 3x3 (3 rows, 3 cols) matrix and T is a 1 x 3 matrix (1 row, 3 cols).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how to go ahead with this?&lt;/p&gt;&#xA;" OwnerUserId="3866" LastActivityDate="2010-11-23T20:06:23.857" Title="How to multiply a 3x3 matrix with a 1x3 matrix?" Tags="&lt;matrices&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11561" PostTypeId="2" ParentId="11199" CreationDate="2010-11-23T19:47:54.120" Score="1" ViewCount="0" Body="&lt;p&gt;Mathematica apparently does not have a nice way of looking at the sum as it is given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you calculate this sum as a definite integral, for both the equation above and the one you derived, ie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n=p} \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where p is the number you enter to make it a definite sum, you get the same calculation as when using the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially used the FactorialPower[n,k] command and then I thought that maybe the domain for k was causing issues, so I rewrote the falling factorials using the gamma function to see if that would help. If you are not familiar with how to do this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n^{\underline{k}} = \frac{ \Gamma (n+1)}{ \Gamma (n-k+1)}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way I write the falling factorial, I still come up with the equation you are getting when I use an indefinite sum, which is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_n = \frac{7}{6(n+1)(n+2)}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, this is in fact not correct. I am not sure why it works for the definite sum, but not the indefinite sum. I will keep thinking about this and let you know what I come up with, if I come up with anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find an answer, please let us know here. Sorry I cannot be of more help.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T19:47:54.120" CommentCount="1" />
  <row Id="11563" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T19:49:51.883" Score="2" ViewCount="0" Body="&lt;p&gt;The usual matrix multiplication is only defined for multiplying an $m\times n$ matrix with an $n\times R$ matrix. So the number of columns of the first matrix must be equal to the number of rows of the second for matrix multiplication to be defined. This is not satisfied by the matrices you have. So you cannot multiply them in that order. You can, however, multiply $T$ with $R$ i.e. the product $TR$ is defined but not $RT$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T19:49:51.883" CommentCount="2" />
  <row Id="11564" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T19:50:56.560" Score="3" ViewCount="0" Body="&lt;p&gt;You can only multiply an $n\times m$ matrix by a $p\times q$ matrix, in that order, if $m=p$. That is, only if the number of columns of the first factor is equal to the number of rows of the second factor. So you &lt;em&gt;cannot&lt;/em&gt; multiply a $3\times 3$ matrix by a $1\times 3$ matrix; on the other hand, you &lt;em&gt;can&lt;/em&gt; multiply a $1\times 3$ matrix by a $3\times 3$ matrix (so you can do $TR$); or you can multiply the $3\times 3$ matrix by the &lt;em&gt;transpose&lt;/em&gt; of $T$, which would be $3\times 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; try extending $T$ into a $3\times 3$ by adding two rows of zeros (that is what the site you linked to does), but then you are not really multiplying $R$ by $T$. All you are doing is multiplying the first column of $R$ by the first entry of $T$, the second column of $R$ by the second entry of $T$, and the third column of $R$ by the third entry of $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real question is: &lt;em&gt;where&lt;/em&gt; did the matrices come from and &lt;em&gt;why&lt;/em&gt; are you trying to multiply them in that order? Perhaps it turns out you shouldn't be doing that product at all...&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T19:50:56.560" CommentCount="9" />
  <row Id="11565" PostTypeId="1" AcceptedAnswerId="11571" CreationDate="2010-11-23T19:58:31.467" Score="1" ViewCount="49" Body="&lt;p&gt;This is initially a funny question, because I've found this on old notes but I do not find/recover my own derivation... But then the question is more general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1:     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = - \frac {2x^2+3x}{(x+1)^2}  $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expressed this by a powerseries $ f_1(x) = -3x + 4x^2 - 5x^3 + 6x^4 - \ldots $ &#xA;and stated without the derivation that this is also&#xA;$  f_2(x) = \frac {-2}{1} -\frac {-1}{x} + 0 - \frac  {1} {x^3} + \frac {2}{x^4} - \ldots + \ldots $ &lt;br&gt;&#xA;and - well: hell, - don't see it now how I did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was interesting to me was, that after looking for the fixpoints $ x_0=0, x_{1,2} =-2 $ the range of convergence in the expression by $f_1$ is obviously $ |x|&amp;lt;1 $ limited to the unit-interval but in that by $f_2$ it is infinity and $ |x|&amp;gt;1 $ .      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2:&lt;br&gt;&#xA;I would like to be able to translate also other powerseries into an $f_2$-type-expression. (I remember to have read a remark of &lt;em&gt;&quot;expanding a powerseries at infinity&quot;&lt;/em&gt; but have never seen an explanation of this - so this might be irrelevant for this case?) So: what is the technique to do this given a function in terms of a usual powerseries, for instance for the geometric series $ g(x)=1+x+x^2+ \ldots $ or some series $ h(x) = K + a*x + b*x^2 + c*x^3 + \ldots $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit: minus-sign in f(x) was missing, one numerator in f2 was wrong]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-23T21:12:30.750" LastActivityDate="2010-11-23T21:12:30.750" Title="conversion of a powerseries $-3x+4x^2-5x^3+\ldots $ into $ -2+\frac 1 x - 0 - \frac  1 {x^3} + \ldots $" Tags="&lt;functions&gt;&lt;functional-equations&gt;&lt;taylor-expansion&gt;" AnswerCount="1" />
  <row Id="11566" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T19:59:26.810" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose you have two matrices A and B. Let A be a 3x3 matrix and B be a 1x3 matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above, you can only multiply matrix C (mxn) by matrix D (pxq) if n=p. That is, the number of columns in C = the number of rows in D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that for your problem, we have the following information,&#xA;$m=n=q=3$ and $p=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that $ n \neq p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insist on multiplying these two matrices together, you could always take the transpose of matrix B, denoted $B^{T}$. (For information about transposed matrices, see [1].)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B^{T}$ would be a 3x1 matrix. So you could in fact do $AB^{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on matrix multiplication in general, see [2]. I hope this is helpful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Weisstein, Eric W. &quot;Transpose.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/Transpose.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Transpose.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Weisstein, Eric W. &quot;Matrix Multiplication.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/MatrixMultiplication.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/MatrixMultiplication.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T19:59:26.810" />
  <row Id="11567" PostTypeId="1" CreationDate="2010-11-23T20:03:55.613" Score="4" ViewCount="108" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question&quot;&gt;Number theory proving question?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since b, x, y, z, ... integers greater than 1, how can we prove that &#xA;gcd (b ^ x - 1, b ^ y - 1, b ^ z - 1 ,...)= [b ^ gcd (x, y, z, .. .)] - 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-24T04:42:38.897" LastActivityDate="2010-11-24T04:42:38.897" ClosedDate="2010-11-25T20:07:55.940" Title="$\gcd(b^x - 1, b^y - 1, b^ z- 1,...) = b^{\gcd(x, y, z,...)} -1$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="11568" PostTypeId="2" ParentId="11560" CreationDate="2010-11-23T20:06:23.857" Score="0" ViewCount="0" Body="&lt;p&gt;From what I can see, perhaps you have the R and Xw matrix switched around. If you had Xw*R, then I think it would be okay. Did you derive this formula or did you find it somewhere?&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T20:06:23.857" CommentCount="1" />
  <row Id="11570" PostTypeId="2" ParentId="11567" CreationDate="2010-11-23T20:07:58.660" Score="7" ViewCount="0" Body="&lt;p&gt;It suffices to prove it for two terms, that is, $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n,m)} - 1$.  The basic idea is that we can use the Euclidean algorithm on the exponents, as follows: if $n &amp;gt; m$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gcd(a^n - 1, a^m - 1) = \gcd(a^n - 1, a^n - a^{n-m}) = \gcd(a^{n-m} - 1, a^m - 1).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can keep subtracting one exponent from the other until we get $\gcd(n, m)$ as desired.  Another way to look at this computation is to write $d = \gcd(a^n - 1, a^m - 1)$ and note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^n \equiv 1 \bmod d, a^m \equiv 1 \bmod d \Rightarrow a^{nx+my} \equiv 1 \bmod d$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which it readily follows, as before, that $a^{\gcd(n,m)} \equiv 1 \bmod d$, so $d$ dividess $a^{\gcd(n,m)} - 1$.  On the other hand, $a^{\gcd(n, m)} - 1$ also divides $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really nice about this result is that it holds both for particular values of $a$ and also for $a$ as a &lt;strong&gt;variable&lt;/strong&gt;, e.g. in a polynomial ring with indeterminate $a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can readily deduce several seemingly nontrivial results from this; for example, the sequence defined by $a_0 = 2, a_n = 2^{a_{n-1}} - 1$ is a sequence of pairwise relatively prime integers, from which it follows that there are infinitely many primes.  By working only slightly harder you can deduce that in fact there are infinitely many primes congruent to $1 \bmod p$ for any prime $p$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-23T20:07:58.660" />
  <row Id="11571" PostTypeId="2" ParentId="11565" CreationDate="2010-11-23T20:11:41.853" Score="3" ViewCount="0" Body="&lt;p&gt;Divide the numerator and denominator of $f(x)$ by $x^2$ and set $y=1/x$ then expand for $y$ and you have your expansion at infinity.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-23T20:11:41.853" CommentCount="2" />
  <row Id="11572" PostTypeId="2" ParentId="11542" CreationDate="2010-11-23T20:14:59.527" Score="3" ViewCount="0" Body="&lt;p&gt;For the first assertion, I will work with $F_n$ being the $n$th term of the lower central series; but the argument should be about the same for any central series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We show that if $\alpha$ is the identity on $F/[F,F]$, then it induces the identity on $F_n/F_{n+1}$ for all $n$. The proof is by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assertion for $n=1$ is the original assumption. Assume now that the result holds for $k$, so that if $w\in F_k$, then $w\alpha\equiv w\pmod{F_{k+1}}$. We want to show that the result holds for $k+1$. The generators of $F_{k+1}$ are contained in the set $[F,F_{k+1}]$, so let $u\in F$ and $w\in F_k$; if we prove the result for $[u,w]$, then it will follow for $F_{k+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use the following commutator identity: in any group, for all $x,y,z,t$, we have:&#xA;$$[xy,zt] = [x,t]^y[y,t][x,z]^{yt}[y,z]^t,$$&#xD;&#xA;where $a^b = b^{-1}ab$ is conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $u\alpha\equiv u\pmod{F_2}$, then $u\alpha=uc$ with $c\in F_2$. And since $w\alpha\equiv w\pmod{F_{k+1}}$, then $w\alpha = wd$ with $d\in F_{k+1}$. Therefore,&#xA;$$[u,w]\alpha = [u\alpha,w\alpha]= [uc,wd]&#xD;&#xA;= [u,d]^{c}[c,d][u,w]^{cd}[c,w]^{d} \equiv [u,w]\pmod{F_{k+2}},$$&#xD;&#xA;because $[u,d]\in[F_1,F_{k+1}]$, $[c,d]\in[F_2,F_{k+1}]$, $[c,w]\in[F_2,F_k]$, and $[u,w]^{cw}=[u,w][[u,w],cw]\equiv [u,w]\pmod_{F_{k+2}}$. &#xA;Therefore, $\alpha$ acts as the identity on $F_{k+1}/F_{k+2}$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Alternatively, the basic commutators of weight $k$ give a basis for the free abelian group $F_{k}/F_{k+1}$, and it is an easy matter to check that if $\alpha$ induces the identity on $F/F_{2}$, then it maps a basic commutator of weight $c$ to the product of itself times commutators of weight greater than $c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still working on the second assertion, but for instance, for $n=1$, we have that $x\alpha= xk$ for some $k\in F_2$, hence $x\alpha\equiv xk\pmod{F_3}$, and since $k\alpha\equiv k\pmod{F_3}$, you get that $x\alpha^2 \equiv x\alpha k \equiv x k^2\pmod{F_3}$, and so inductively $x\alpha^N \equiv x k^{N}\pmod{F_3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; I double-checked the book on my way out of the office and to the gym, and I noticed that the book did not have &quot; $x\alpha^N\equiv xk^N\pmod{F_{n+1}}$, but only $x\alpha^N\equiv xk^N$. I can show that under the given assumptions, $x\alpha^N$ is &lt;em&gt;conjugate&lt;/em&gt;, modulo $F_{n+1}$, to $xk^N$, and that should be enough for the purposes of the argument, since the next step of the proof is to note that since $\alpha^N\in JA(F)$, then this means that $uxu^{-1}\equiv xk^N\pmod{F_{n+1}}$ for some $u$, and this will also hold if all you have is that $x\alpha^N$ is conjugate to $xk^N$ modulo $F_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show this weaker claim, I proceed by induction on $m$. We know that $x\alpha$ is conjugate to $xk$, hence $x\alpha \equiv (xk)^u\pmod{F_{n+1}}$ for some $u$. Assume that $x\alpha^m\equiv (xk^m)^v\pmod{F_{n+1}}$ for some $v\in F$, where $a^b = b^{-1}ab$ is just conjugation. Applying $\alpha$ on both sides, we have $x\alpha^{m+1}$ on the left hand side, and &#xA;\begin{align*}&#xA;(xk^m)^v\alpha &amp;amp;= ((x\alpha)(k^m\alpha))^{v\alpha}\\&#xA;&amp;amp; \equiv \bigl((xk)^u k^m\bigr)^{v\alpha}\pmod{F_{n+1}}\\&#xA;&amp;amp;\equiv \bigl( (xk^{m+1})^u\bigr)^{v\alpha}\pmod{F_{n+1}} &amp;amp;\text{(as $k$ is central in $F_/F_{n+1}$)}\\&#xA;&amp;amp;\equiv (xk^{m+1})^{u(v\alpha)}\pmod{F_{n+1}}.&#xA;\end{align*}&#xA;Therefore, $x\alpha^{m+1}\equiv (xk^{m+1})^w\pmod{F_{n+1}}$ for some $w\in F$; that is, $x\alpha^{m+1}$ is conjugate, modulo $F_{n+1}$, to $xk^{m+1}$. In particular, $x\alpha^N$ is conjugate, modulo $F_{n+1}$ to $xk^{N}$, and since $x$ is conjugate to $x\alpha^N$, then $x$ is conjugate to $xk^N$ modulo $F_{n+1}$. Now I think the argument given in the book can be continued from this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;hr/&gt;&#xA;So, continuing on with the proof of the proposition. We know that $x\alpha^N$ is conjugate to $xk^N$ modulo $F_{n+1}$, and that $\alpha^N$ is equal to conjugation by some element. So there exists $u$ such that $x\alpha^N = u^{-1}xu = x[x,u]$ is equal, modulo $F_{n+1}$, to $xk^N$, and therefore $[x,u]=k^N \pmod{F_{n+1}}$ by cancelling $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $F_n/F_{n+1}$ is a finitely generated free abelian group generated by the basic commutators; $k\in F_n$, so we can write $k$ as a product of basic commutators, and the same is true for $[x,u]$. Because $[x,u]$ is an $N$th power, commutator calculus tells us that in fact $u$ must be an $N$th power, so that in fact we can find a $v$ such that $[x,v]\equiv k \pmod{F_{n+1}}$ (essentially, every commutator that shows up is an $N$th power, so dividing the exponent by $N$ we can construct an appropriate $v$). So now we have $[x,v]\equiv k\pmod{F_{n+1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: $x\alpha$ is conjugate to $xk$, so $v^{-1}x\alpha v$ is conjugate to $vxkv^{-1}$ modulo $F_{n+1}$. But $vxkv^{-1} \equiv vxv^{-1}k\pmod{F_{n+1}}$ (since $k$ is central in $F/F_{n+1}$), so &#xA;$$vxkv^{-1} \equiv vxv^{-1}k \equiv x[x,v^{-1}]k \equiv x[x,v]^{-1}k \equiv xk^{-1}k\pmod{F_{n+1}}$$&#xD;&#xA;(the next to last congruence by using commutator identities, since we are in the lowest term of the lower central series of $F/F_{n+1}$). In summary, $x\alpha$ is conjugate to $x$ modulo $F_{n+1}$. That means that there exists some $k&amp;#39;\in F_{n+1}$ and some $w\in F$ such that $w^{-1}(x\alpha)w\equiv x \pmod{F_{n+1}}$. But this means that $w^{-1}(x\alpha)w =xk&amp;#39;$ for some $k&amp;#39;\in F_{n+1}$, which is precisely what we wanted to prove to complete the inductive step.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T21:36:56.230" LastActivityDate="2010-11-28T21:36:56.230" CommentCount="1" />
  <row Id="11573" PostTypeId="1" AcceptedAnswerId="11575" CreationDate="2010-11-23T20:19:33.050" Score="0" ViewCount="73" Body="&lt;p&gt;Let $f\colon A\to B$, and $g\colon B\to C$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove: If $g\circ f$ is onto and $g$ is 1-1, then $f$ is onto.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Here is all I can think of:&#xA;$g\circ f\colon A\to C$, and for all $x\in B$ there is a unique $y\in C$.   &lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:25:58.047" LastActivityDate="2010-11-23T20:35:49.577" Title="Prove: If $g\circ f$ is onto and $g$ is 1-1, then $f$ is onto" Tags="&lt;functions&gt;&lt;proof&gt;" AnswerCount="2" />
  <row Id="11574" PostTypeId="1" CreationDate="2010-11-23T20:23:15.320" Score="0" ViewCount="45" Body="&lt;p&gt;This is just a simple doubt I wanted to clear. I'm calculating the Hamming and Euclidean distance between two columns in a matrix, where each column is storing time information in seconds of when a particular event occurred. When I calculate the Hamming and Euclidean distance between the two columns, I get back a certain value. What is the units of measure for that value? Is it in seconds? &lt;/p&gt;&#xA;" OwnerUserId="2747" LastActivityDate="2010-11-23T20:29:26.867" Title="Units of Hamming, and Euclidean distance" Tags="&lt;statistics&gt;&lt;discrete-mathematics&gt;&lt;recreational-mathematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11575" PostTypeId="2" ParentId="11573" CreationDate="2010-11-23T20:28:17.560" Score="3" ViewCount="0" Body="&lt;p&gt;Given an element of $B$ we need to find an element in $A$ which maps to it under $f$. Given $b\in B$, there is an element $c\in C$ such that $g(b)=c$. Since $g\circ f$ is onto, there exists an element $a\in A$ such that $g(f(a))=c$. Since $g$ is $1-1$ $f(a)=b$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T20:28:17.560" />
  <row Id="11576" PostTypeId="2" ParentId="11573" CreationDate="2010-11-23T20:29:09.400" Score="2" ViewCount="0" Body="&lt;p&gt;Since $g\circ f$ is onto, then for every $c\in C$ there exists (at least one) $a\in A$ such that $g\circ f(a) = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, since $f(a)=b\in B$, and $g(b)=c$, then $g$ is onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $g$ is onto, and $g$ is also 1-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $g$ is ......&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you fill in the blank, use that to get that $f$ is onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; This is not the most direct way of doing it; Timothy Wagner's direct argument is more streamlined. I did it this way, because through long use,  &quot;if a $g\circ f$ is onto, then $g$ is onto&quot; and &quot;if $g\circ f$ is 1-1 then $f$ is 1-1&quot; are so ingrained that they immediately jump out, just like if I see &quot;$2\times 7$&quot; I immediately think &quot;$14$&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:35:49.577" LastActivityDate="2010-11-23T20:35:49.577" CommentCount="1" />
  <row Id="11577" PostTypeId="2" ParentId="11574" CreationDate="2010-11-23T20:29:26.867" Score="2" ViewCount="0" Body="&lt;p&gt;The Euclidean distance has the same units as the positions are measured in.  The Hamming distance has no units.  Think of changing units, say from meters to centimeters.  The numeric value of the Euclidean distance will multiply by 100, but the Hamming distance will not change.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T20:29:26.867" />
  <row Id="11578" PostTypeId="1" CreationDate="2010-11-23T20:34:19.207" Score="0" ViewCount="46" Body="&lt;p&gt;Let's say I have an &quot;S&quot; shaped curve made up of 50 points.  I am only given the data points, and there is no simple or obvious equation for each point on the curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given only the data points, how would I be able to calculate the slope of the curve at a particular data point?&lt;/p&gt;&#xA;" OwnerUserId="3874" LastActivityDate="2010-11-23T21:03:18.360" Title="Calculating The Slope Of A Line At A Specific Point" Tags="&lt;calculus&gt;" AnswerCount="3" />
  <row Id="11579" PostTypeId="1" AcceptedAnswerId="11583" CreationDate="2010-11-23T20:35:05.960" Score="1" ViewCount="54" Body="&lt;p&gt;Suppose $f\colon [a,b] \to \mathbb{R}$ is continuous and has a finite derivative $f&amp;#39;$ everywhere on $(a,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(a)=f(b)=0$ prove that for every $y\in\mathbb{R}$ there is some $c$ in $(a,b)$ with $f&amp;#39;(c) = yf(c)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were given a hint to apply Rolle's theorem to $h(x)f(x)$ for an $h$ that depends on $y$ but I don't understand how to apply Rolle's theorem here. &lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-23T20:50:23.867" LastActivityDate="2010-11-23T20:50:23.867" Title="For every $y\in \mathbb{R}$ there is some $c$ in $(a,b)$ with $f'(c) = yf(c)$" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11580" PostTypeId="2" ParentId="11578" CreationDate="2010-11-23T20:39:49.293" Score="1" ViewCount="0" Body="&lt;p&gt;Estimate the tangent by taking the secants you can compute; you the tangent will lie somewhere between the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: you cannot compute the slope exactly, but you can estimate it from the information you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have immediately preceding and immediately succeeding points, $t_{i-1}$ and $t_{i+1}$, as well as value $f(t_{i-1})$, $f(t_i)$, and $f(t_{i+1})$. The secant between $(t_{i-1},f(t_{i-1}))$, and $(t_i,f(t_i))$ has a certain slope $L_1$; the secant between $(t_i,f(t_i))$ and $(t_{i+1},f(t_{i+1}))$ has a certain slope $L_2$. Unless the graph is very nasty, the tangent at $(t_i,f(t_i))$ should have a slope that is somewhere between $L_1$ and $L_2$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T20:39:49.293" />
  <row Id="11581" PostTypeId="2" ParentId="11578" CreationDate="2010-11-23T20:40:23.367" Score="0" ViewCount="0" Body="&lt;p&gt;The simplest is just to take one of the neighboring points and find the slope of the line through those two.  This approach is sensitive to measurement error in the points, particularly if they are measured at x values that are close together.  Another approach is to fit some functional form (like an arctangent or logistic function is your data is S shaped) through the data points.  This is smoothing, so will be less sensitive to measurement error (particularly if it has fewer parameters than you have measurements), but is sensitive to modeling error.  There is a discussion in chapters 3 and 15 of &lt;a href=&quot;http://www.nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-23T20:40:23.367" />
  <row Id="11582" PostTypeId="2" ParentId="11578" CreationDate="2010-11-23T20:42:18.177" Score="0" ViewCount="0" Body="&lt;p&gt;Cubic Splines! You fit cubic spline through the points, and then take slope from the cubic spline definition. Note that if the data have any noise, you need to smooth it first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;links&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathworld.wolfram.com/CubicSpline.html&quot; rel=&quot;nofollow&quot;&gt;Cubic Splines - Wolfram&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spline_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Splines - Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nrbook.com/a/bookf90pdf.php&quot; rel=&quot;nofollow&quot;&gt;Interpolation - Numerical Recipies Online&lt;/a&gt; - Open charter B3&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once the spline is defined, for each segment with curve coordinates ($x(\alpha)$,$y(\alpha)$) the slope is $$s=\frac{\mathrm{d}y/\mathrm{d}\alpha}{\mathrm{d}x/\mathrm{d}\alpha}$$ evaluated at the desired location parameter $\alpha=0\ldots1$.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="3301" LastEditorDisplayName="" LastEditDate="2010-11-23T21:03:18.360" LastActivityDate="2010-11-23T21:03:18.360" />
  <row Id="11583" PostTypeId="2" ParentId="11579" CreationDate="2010-11-23T20:43:14.997" Score="2" ViewCount="0" Body="&lt;p&gt;$\textbf{HINT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $y \in \mathbb{R}$, consider the function $g_y(x)$ as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_y(x) = e^{-yx} f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply Rolle's to $g_y(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In general, if you find equations where the derivative is proportional to the function as in this case where $f&amp;#39;(c) \propto f(c)$, the detective in you must suspect immediately that the exponential function must be there somewhere in the picture)&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-23T20:50:01.293" LastActivityDate="2010-11-23T20:50:01.293" />
  <row Id="11584" PostTypeId="2" ParentId="11579" CreationDate="2010-11-23T20:49:45.753" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose you have a function $h(x)$ which is differentiable, and you consider the function $\mathcal{F}(x) = h(x)f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ and $f$ are both continuous on $[a,b]$, so is $\mathcal{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h$ and $f$ are both differentiable on $(a,b)$, so is $\mathcal{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f(a)=f(b)=0$, then $\mathcal{F}(a)=h(a)f(a)=0$ and $\mathcal{F}(b)=h(b)f(b)=0$, regardless of what $h(x)$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\mathcal{F}$ will &lt;em&gt;also&lt;/em&gt; satisfy Rolle's Theorem, and so you know that there exists $c\in (a,b)$ such that $\mathcal{F}&amp;#39;(c)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is $\mathcal{F}&amp;#39;(c)$? Well,&#xA;$$\mathcal{F}&amp;#39;(x) = h&amp;#39;(x)f(x) + h(x)f&amp;#39;(x).$$&#xD;&#xA;So $\mathcal{F}&amp;#39;(c) = h&amp;#39;(c)f(c) + h(c)f&amp;#39;(c)$. You want $f&amp;#39;(c)=yf(c)$, so you want&#xA;$$ 0 = h&amp;#39;(c)f(c) + h(c)f&amp;#39;(c) = h&amp;#39;(c)f(c) + h(c)yf(c) = f(c)\Bigl(h&amp;#39;(c)+yh(c)\Bigr).$$&#xD;&#xA;So either $f(c)=0$, or else $h&amp;#39;(c)+yh(c)=0$. You don't want $f(c)=0$ unless $y=0$, so in general you just want $h&amp;#39;(c)+yh(c)=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about trying to find some function $h(x)$ such that $h&amp;#39;(x) = -yh(x)$ for all $x$? Then use that for $\mathcal{F}$. Obviously, your choice of $h$ will depend on $y$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-23T20:49:45.753" CommentCount="2" />
  <row Id="11585" PostTypeId="1" AcceptedAnswerId="11589" CreationDate="2010-11-23T21:07:21.840" Score="3" ViewCount="120" Body="&lt;p&gt;How  to show :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $A \subseteq \mathbb{I}$ is not&#xA;  countable, then $A$ is not closed set&#xA;  in $\mathbb{R}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that if $A$ is closed, then $A^{c}$ is open thus: $A^{c}$ is an union of open intervals, but...&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-23T21:11:13.493" LastActivityDate="2010-11-23T22:13:49.767" Title="Question about irrational numbers" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="11586" PostTypeId="2" ParentId="11487" CreationDate="2010-11-23T21:08:32.720" Score="0" ViewCount="0" Body="&lt;p&gt;What follows is too long to be a comment. It is also very useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first that $X_i$, $i=1,2$, are independent ${\rm N}(0,\sigma_i^2)$ variables. Their characteristic functions are ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\sigma _i^2 ( - z^2 /2)}$. Suppose now that $X_i$, $i=1,2$, are independent Poisson variables, with means $\lambda_i$. Their characteristic functions are ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\lambda _i ({\rm e}^{{\rm i}z}  - 1)}$. For the normal distribution case&#xA;${\rm E}(X_1 | X_1 + X_2) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}(X_1  + X_2 )$, and for the Poisson case&#xA;${\rm E}(X_1 | X_1 + X_2) = \frac{{\lambda_1 }}{{\lambda_1  + \lambda_2 }}(X_1  + X_2 )$ (the later result is easy to prove). Finally, consider independent rv's $X_i$, $i=1,2$, having finite expectations and characteristic functions of the form ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{c _i \psi(z)}$, $c_i &amp;gt; 0$. What would you expect ${\rm E}(X_1 | X_1 + X_2)$ to be, based on the previous observations? Though very informal, this is very useful, since the conclusion can be applied  to a vast number of rv's, e.g. gamma rv's (specifically, the conclusion applies to all integrable infinitely divisible rv's).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-23T21:08:32.720" />
  <row Id="11587" PostTypeId="1" AcceptedAnswerId="11593" CreationDate="2010-11-23T21:10:14.273" Score="2" ViewCount="93" Body="&lt;p&gt;I am trying to upper bound $\sum_{l=1}^\infty (1 - n! 2^{-ln} {2^l \choose n})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to use the simple ${2^l \choose n} \ge (\frac{2^l}{n})^n$, but then $l$ disappears and it diverges. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastActivityDate="2010-11-23T22:55:56.877" Title="Upper bound binomial sum" Tags="&lt;inequality&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="11588" PostTypeId="2" ParentId="11188" CreationDate="2010-11-23T21:11:35.103" Score="1" ViewCount="0" Body="&lt;p&gt;With the identity, $ x \leq y \Leftrightarrow x = x \wedge y$, the implication is straightforward to prove.  That is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;x \leq y \text{ and } x \leq z &amp;amp; \Leftrightarrow x = x \wedge y \text{ and } x = x \wedge z \\&#xD;&#xA;&amp;amp; \Leftrightarrow x \wedge x = (x\wedge y) \wedge (x\wedge z)\\&#xD;&#xA;&amp;amp; \Leftrightarrow x = x \wedge (y \wedge z)\\&#xD;&#xA;&amp;amp; \Leftrightarrow x \leq y \wedge z. &#xD;&#xA;\end{align}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3877" LastActivityDate="2010-11-23T21:11:35.103" />
  <row Id="11589" PostTypeId="2" ParentId="11585" CreationDate="2010-11-23T21:23:53.143" Score="4" ViewCount="0" Body="&lt;p&gt;The statement is not true.  There exist perfect&lt;sup&gt;1&lt;/sup&gt; (hence closed) subsets of $\mathbb R$ that doesn't intersect $\mathbb Q$, and &lt;a href=&quot;http://pirate.shu.edu/~wachsmut/ira/topo/proofs/pfctuncb.html&quot; rel=&quot;nofollow&quot;&gt;perfect sets are not countable&lt;/a&gt;.  See: &lt;a href=&quot;http://math.stackexchange.com/questions/1064/perfect-set-without-rationals&quot;&gt;Perfect set without rationals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;  A set $S$ is perfect if it is closed and every point of $S$ is an accumulation point of $S$. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-23T21:39:37.690" LastActivityDate="2010-11-23T21:39:37.690" />
  <row Id="11590" PostTypeId="1" AcceptedAnswerId="11591" CreationDate="2010-11-23T21:38:59.967" Score="1" ViewCount="64" Body="&lt;p&gt;Given $c \in \mathbb{R}$ and $m,n \in \mathbb{N}$, is it possible to combine the inequations&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n &amp;lt; m$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\frac{8}{c} &amp;lt; m$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;into one inequation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I argued that due $n &amp;lt; m$ and $\frac{8}{c} &amp;lt; m$, there has to be one $n$ between $m$ and $\frac{8}{c}$, being $n = \left\lceil \frac{8}{c} \right\rceil$, so that $m &amp;gt; n &amp;gt; \frac{8}{c}$, but I'm not sure if that is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall c \in R^{+} \exists n \in \mathbb{N}: \forall m \in \mathbb{N} \land m &amp;gt; n: |f(m)|&amp;lt; c*|g(m)) \implies f \in \mathcal{o}(g)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this is about the landau-notation of $f(x)$ raises slower than $g(x)$, the task is to show that a specific case ($f(x) = 8x$ and $g(x) = x^2$) complies to this definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I already solved it until I encounter both inequations together and want to show that there is an $n$ for all cases.&lt;/p&gt;&#xA;" OwnerUserId="3880" LastEditorUserId="3880" LastEditorDisplayName="" LastEditDate="2010-11-23T21:55:08.433" LastActivityDate="2010-11-23T22:41:28.817" Title="Combining two inequations" Tags="&lt;homework&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11591" PostTypeId="2" ParentId="11590" CreationDate="2010-11-23T21:47:00.620" Score="1" ViewCount="0" Body="&lt;p&gt;Nope. What if $m=2$, $c=7$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's really no way of finding out the what the relationship b/w $n$ and $8/c$ is without additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the task of finding an $n$ that satisfies certain properties is very different from the task of determining whether $n$ satisfies certain properties. (It's the difference between &quot;there exists&quot; and &quot;for all&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit RE: Your edits:&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you correctly, what you actually need is to find an $n$ such that $n&amp;lt;m \Rightarrow \frac8c&amp;lt;m$. Then $n = \lfloor \frac8c \rfloor$ should satisfy that.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastEditorUserId="3638" LastEditorDisplayName="" LastEditDate="2010-11-23T22:32:24.797" LastActivityDate="2010-11-23T22:32:24.797" />
  <row Id="11592" PostTypeId="1" AcceptedAnswerId="11989" CreationDate="2010-11-23T21:52:36.407" Score="1" ViewCount="125" Body="&lt;p&gt;Let $a,b,c,x,y,z\in\mathbb{R}$. Prove that $$ \left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\geq(c-a)^2+(c-b)^2$$&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-11-26T20:28:41.563" Title="Second degree polynomial inequality " Tags="&lt;polynomials&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="11593" PostTypeId="2" ParentId="11587" CreationDate="2010-11-23T22:03:26.360" Score="3" ViewCount="0" Body="&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{n!}{2^{kn}} { 2^k \choose n } =&#xD;&#xA;\left( 1 - \frac{1}{2^k} \right) \left( 1 - \frac{2}{2^k} \right) \cdots&#xD;&#xA;\left( 1 - \frac{n-1}{2^k} \right) \ge&#xD;&#xA;1 - \frac{n(n-1)}{2^{k+1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n(n-1)}{2^{k+1}} \ge 1 - \frac{n!}{2^{kn}} { 2^k \choose n }.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-23T22:03:26.360" CommentCount="1" />
  <row Id="11594" PostTypeId="2" ParentId="11585" CreationDate="2010-11-23T22:09:52.533" Score="4" ViewCount="0" Body="&lt;p&gt;As another counterexample, enumerate the rationals as $\{q_k\}_{k=1}^\infty$, and let $U = \bigcup_{k=1}^\infty (q_k - 2^{-(k+2)}, q_k + 2^{-(k+2)})$.  $U$ is open and contains all the rationals, so its complement $U^c$ is closed and contains none of them.  But $U$ is a union of intervals of lengths $2^{-(k+1)}$, so its Lebesgue measure (i.e. length) is at most $\sum_{k=1}^\infty 2^{-(k+1)} = 1/2$.  So $U^c$ has measure at least $1/2$ and in particular must be uncountable.  You can modify this construction to make the  measure of $U^c$ as close to $1$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-23T22:09:52.533" />
  <row Id="11595" PostTypeId="2" ParentId="11585" CreationDate="2010-11-23T22:13:49.767" Score="1" ViewCount="0" Body="&lt;p&gt;As AgCl pointed out (and François G. Dorais, in the answer AgCl linked to), the statement is false. (And Nate also pointed out, as I was writing my answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would even say that the point is that ${\mathbb R}\setminus{\mathbb Q}$ is &lt;em&gt;homeomorphic&lt;/em&gt; to the set ${\mathbb N}^{\mathbb N}$ (with the product topology of discrete ${\mathbb N}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, obviously, ${\mathbb N}^{\mathbb N}$ contains lots of perfect subsets, such as $\{1,2\}^{\mathbb N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[A really pretty example of such a homeomorphism in the first chapter of the book &quot;Descriptive Set Theory and Forcing: how to prove theorems about Borel sets the hard way&quot; by Arnold Miller (which you can download from his webpage). The usual construction (which is also pretty) is to note that a sequence $(a_1,a_2,\dots)$ can be identified with the continued fraction of an irrational in $(0,1)$, and a bit of thought shows that this map is indeed a homeomorphism.]&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-23T22:13:49.767" />
  <row Id="11596" PostTypeId="1" CreationDate="2010-11-23T22:23:29.743" Score="0" ViewCount="37" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the univariate central limit theorem, the Berry-Esseen theorem gives a quantitative bound on the rate of convergence of distributions to the Normal distribution under Kolmogorov distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are similar statements known for the multivariate version of the central limit theorem, that use some standard distance measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem#Multidimensional_central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Central_limit_theorem#Multidimensional_central_limit_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="3831" LastActivityDate="2010-11-23T22:23:29.743" Title="Quantitative bounds for multivariate central limit theorem" Tags="&lt;real-analysis&gt;&lt;probability-theory&gt;&lt;reference-request&gt;" CommentCount="1" />
  <row Id="11597" PostTypeId="2" ParentId="11590" CreationDate="2010-11-23T22:33:20.130" Score="0" ViewCount="0" Body="&lt;p&gt;So, you need to show that for every $c&amp;gt;0 \in \mathbb{R}$ there exists $n \in \mathbb{R}^+$ such that $|f(x)|&amp;lt;c|g(x)|$ for all $x&amp;gt;n$. Given a $c&amp;gt;0$, let $n=\frac{8}{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then...&#xA;$x&amp;gt;n \implies |f(x)|&amp;lt;c|g(x)|$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2596" LastEditorUserId="2596" LastEditorDisplayName="" LastEditDate="2010-11-23T22:41:28.817" LastActivityDate="2010-11-23T22:41:28.817" />
  <row Id="11598" PostTypeId="1" AcceptedAnswerId="11607" CreationDate="2010-11-23T22:46:53.097" Score="0" ViewCount="67" Body="&lt;p&gt;I've been playing around with this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\int_{-\pi}^\pi{\displaystyle\frac{1-e^{3it}}{1-e^{it}}dt}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it seems to me that we can (possibly) split the integral into four seperate integrals; one for each quadrant of the complex plane.  Doing this, we can determine the complex part if we know the real part, and vice versa.  So I therefore ponder that it might be possible to simplify the exponential expressions into expressions of variables.  In other words, I wonder if we can rewrite the $e^{it}$s as $u$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm extremely interested in the potential of a variable substitution, but maybe I'm missing the obvious.  I can't tell you exactly what I'm looking for, except for information concerning the possibility of this method.  What is known?  If there is a known way, I'd really enjoy seeing it.  It would make many series expansions that I'm thinking about much easier, I'm almost certain.  If no methods are known, I'd appreciate it very much if someone could take the time to elaborate on all of the reasons why this method is not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very interested in going into detail on this matter, because it seems to decide issues concerning a great amount of potential.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-11-23T23:35:41.940" Title="Complex integrals and the possibility of avoiding parametric equations" Tags="&lt;complex-analysis&gt;&lt;intuition&gt;&lt;complex-variables&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11599" PostTypeId="2" ParentId="11477" CreationDate="2010-11-23T22:54:08.843" Score="4" ViewCount="0" Body="&lt;p&gt;There are several good answers already, but I thought I would add the following derivation because it is one of the few uses I know for the sum property of permanents; namely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$, $B$, and $C$ are matrices with identical entries except that one row (column) of $C$, say the $k^{th}$, is the sum of the $k^{th}$ rows (columns) of $A$ and $B$, then $\text{ per } A + \text{ per } B = \text{per } C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the matrices $\begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_0 &amp;amp; F_1 \end{bmatrix}$ and $\begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_1 &amp;amp; F_2 \end{bmatrix}$.  Since $F_0 = 0$ and $F_1 = F_2 = 1$, they have permanents $F_n$ and $F_n + F_{n-1} = F_{n+1}$, respectively.  Applying the Fibonacci recurrence and the permanent sum property,  we have $\text{ per } \begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_2 &amp;amp; F_3 \end{bmatrix} = F_{n+2}$.  By continuing to construct new matrices whose second rows are the sums of the second rows of the previous two matrices, this process continues until we have $F_n F_{m+1} + F_{n-1}F_m = \text{ per} \begin{bmatrix} F_n &amp;amp; F_{n-1} \\ F_m &amp;amp; F_{m+1} \end{bmatrix} = F_{n+m}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on this approach (but with determinants), see this paper I wrote a few years ago: &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/FibDetermPageProofs.pdf&quot; rel=&quot;nofollow&quot;&gt;Fibonacci Identities via the Determinant Sum Property&lt;/a&gt;,&quot; &lt;em&gt;The College Mathematics Journal&lt;/em&gt;, 37 (4): 286-289, 2006. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-24T22:01:27.810" LastActivityDate="2010-12-24T22:01:27.810" CommentCount="1" />
  <row Id="11600" PostTypeId="2" ParentId="11587" CreationDate="2010-11-23T22:55:56.877" Score="0" ViewCount="0" Body="&lt;p&gt;I was working on this and noticed an interesting thing. Do not know if it is obvious. I wanted to try and improve the bound Derek has (which is a quadratic $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reformulate the problem as Derek has (since for this we need $2^k \geq n$, so we forget the original problem and think the problem given is as follows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. Let $S(n) = \displaystyle \sum_{k=1}^{\infty} (1 - (1-\frac{1}{2^k})(1-\frac{2}{2^k}) \cdots (1-\frac{n-1}{2^k}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(1) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(2) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(3) = \frac{7}{3} \approx 2.3333$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(4) = \frac{67}{21} \approx 3.1904$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(5) = \frac{407}{105} \approx 3.8762$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(6) = \frac{4789}{1085} \approx 4.4138$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(7) = \frac{5289}{1085} \approx 4.8747$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(8) = \frac{726093}{137795} \approx 5.2694$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(9) = \frac{118399669}{21082635} \approx 5.61598$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the growth is slow as $n$ increases. My question is if this converges to some limit or it diverges. I have been working on it for the past couple of hours but am unable to come to a conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is what is $\displaystyle \lim_{n \rightarrow \infty} S(n)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it converges, can we find the limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it diverges, how fast does it diverge?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-23T22:55:56.877" CommentCount="2" />
  <row Id="11601" PostTypeId="1" CreationDate="2010-11-23T23:00:49.960" Score="3" ViewCount="181" Body="&lt;p&gt;From its definition a combination $(^n_k)$, is the number of distinct subsets of size &lt;em&gt;k&lt;/em&gt; from a set of &lt;em&gt;n&lt;/em&gt; elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly an integer, however I was curious as to why the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n!}{k!(n-k)!}$ always evaluates to an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I figured:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.&lt;/p&gt;&#xA;" OwnerUserId="3881" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-23T23:03:28.677" LastActivityDate="2010-11-24T15:35:06.263" Title="Proof that a Combination is an integer" Tags="&lt;binomial-coefficients&gt;&lt;proof&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11602" PostTypeId="1" AcceptedAnswerId="11662" CreationDate="2010-11-23T23:02:25.277" Score="5" ViewCount="242" Body="&lt;p&gt;This is a follow up of &lt;a href=&quot;http://math.stackexchange.com/questions/11587/upper-bound-binomial-sum&quot;&gt;Upper bound binomial sum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working on the problem in the above thread and noticed an interesting thing. I wanted to try and improve the bound Derek had (which was a quadratic in $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we reformulate the problem as Derek has (since for this we need $2^k \geq n$, so we forget the original problem and think the problem given is as follows)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S(n) = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(1) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(2) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(3) = \frac{7}{3} \approx 2.3333$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(4) = \frac{67}{21} \approx 3.1904$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(5) = \frac{407}{105} \approx 3.8762$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(6) = \frac{4789}{1085} \approx 4.4138$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(7) = \frac{5289}{1085} \approx 4.8747$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(8) = \frac{726093}{137795} \approx 5.2694$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(9) = \frac{118399669}{21082635} \approx 5.61598$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(10) = \frac{9120486643}{1539032355} \approx 5.92612$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(11) = \frac{105065594573}{16929355905} \approx 6.20612$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(12) = \frac{31986101239583}{4950627362505} \approx 6.4610$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the growth is slow as $n$ increases. My question is if this converges to some limit or it diverges. I have been working on it for the past couple of hours but am unable to come to a conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is what is $\displaystyle \lim_{n \rightarrow \infty} S(n)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it converges, can we find the limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it diverges, how fast does it diverge?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\textbf{EDIT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So Mike has proved that $S(n)$ in fact diverges.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The conjecture is now that $\lim_{n \rightarrow \infty} (2 \log_{2}(n) - S(n)) = \alpha$ where $\alpha \approx 0.667$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &lt;a href=&quot;http://math.stackexchange.com/questions/11726/limit-of-sn-displaystyle-sum-k1-infty-left1-prod-j1n-1-le&quot;&gt;Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$ - Part II&lt;/a&gt; for further discussions.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T22:55:14.717" LastActivityDate="2010-11-24T22:55:14.717" Title="Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="11603" PostTypeId="2" ParentId="11601" CreationDate="2010-11-23T23:08:13.087" Score="5" ViewCount="0" Body="&lt;p&gt;As Jonas mentioned, it counts something so it has to be a natural number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to notice that product of $m$ consecutive natural numbers is divisible by $m!$.(Prove this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we write $n! = n(n-1)(n-2) \cdots (k+1) \times (k!)$, we find that $k!$ divides $k!$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(n-1)(n-2) \cdots (k+1)$ is a product of $(n-k)$ consecutive natural numbers and hence $(n-k)!$ divides it.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-23T23:08:13.087" CommentCount="3" />
  <row Id="11604" PostTypeId="2" ParentId="11601" CreationDate="2010-11-23T23:14:00.720" Score="5" ViewCount="0" Body="&lt;p&gt;Well, one noncombinatorial way is to induct on $n$ using Pascal's triangle; that is, using the fact that ${n \choose k} = {n-1 \choose k - 1} + {n-1 \choose k}$ (easy to verify directly) and that each ${n - 1 \choose 0}$ is just $1$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-23T23:14:00.720" />
  <row Id="11605" PostTypeId="2" ParentId="11602" CreationDate="2010-11-23T23:21:28.080" Score="5" ViewCount="0" Body="&lt;p&gt;If $(x-1)(x-2)\cdots(x-(n-1)) = \sum_{j=0}^{n-1} a_j x^j$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As Mike points out, the $a_j$ are nothing but &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind#Definition&quot; rel=&quot;nofollow&quot;&gt;Stirling Numbers of the first kind&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the $k^{th}$ term is $$\displaystyle - \sum_{j=0}^{n-2} \frac{a_j}{2^{(n-1-j)k}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by setting $\displaystyle x=2^k$ in the above polynomial and dividing by an appropriate power of $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the limit is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n) = \displaystyle - \sum_{j=0}^{n-2} \frac{a_j}{2^{n-1-j}-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for $\displaystyle n=5$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x-1)(x-2)(x-3)(x-4) = x^4 - 10x^3 + 35x^2 - 50x + 24$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle S(5) = 10/1 - 35/3 + 50/7 - 24/15 = 407/105$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-23T23:47:28.730" LastActivityDate="2010-11-23T23:47:28.730" CommentCount="6" />
  <row Id="11606" PostTypeId="2" ParentId="11601" CreationDate="2010-11-23T23:25:25.263" Score="7" ViewCount="0" Body="&lt;p&gt;See my &lt;a href=&quot;http://math.stackexchange.com/questions/2158/division-of-factorials/2192#2192&quot;&gt;post here&lt;/a&gt; for a simple purely arithmetical proof that every binomial coefficient is an integer. The proof shows how to rewrite any binomial coefficient fraction as a product of fractions whose denominators are all coprime to any given prime $\rm\:p\:$. This implies that no primes divide the denominator (when written in lowest terms), so the fraction is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key property that lies at the heart of this proof is that, among all products of $\rm\: n\:$ consecutive integers, $\rm\ n!\ $ has the least possible power of $\rm\:p\:$ dividing it - for all primes $\rm\:p\:$. Thus $\rm\ n!\ $ divides every product of $\rm\:n\:$ consecutive integers, since it has a smaller power of every prime divisor. Therefore&#xA;$$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{n\:(n-1)\quad\quad\:\cdots\:\quad\quad 1\quad\quad}\ \in\ \mathbb Z$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T15:35:06.263" LastActivityDate="2010-11-24T15:35:06.263" />
  <row Id="11607" PostTypeId="2" ParentId="11598" CreationDate="2010-11-23T23:27:17.077" Score="5" ViewCount="0" Body="&lt;p&gt;I don't really see why you don't want to just divide and then integrate the three terms... but if you insist on not using parameterizations: you can rewrite your integral as&#xA;$$-i\int_{-\pi}^{\pi} {1 - e^{3it} \over 1 - e^{it}} ie^{it} {1 \over e^{it}}\,dt$$&#xD;&#xA;Then you do a complex variables version of the $u$-substitution you are trying. You let $z = e^{it}$ and $dz = i e^{it} dt$. Your integral becomes &#xA;$$-i\int_{|z| = 1} {1 - z^3 \over z(1 - z)}\,dz$$&#xD;&#xA; Dividing $(1 - z)$ into $(1 - z^3)$ turns this into&#xA;$$-i \int_{|z| = 1} ({1 \over z} + 1 + z) \,dz$$&#xD;&#xA;Using the residue theorem this immediately evaluates into $-i 2\pi i = 2\pi$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-23T23:35:41.940" LastActivityDate="2010-11-23T23:35:41.940" CommentCount="5" />
  <row Id="11608" PostTypeId="1" AcceptedAnswerId="11615" CreationDate="2010-11-23T23:29:46.517" Score="1" ViewCount="75" Body="&lt;p&gt;I am currently a mathematics graduate student at Western Kentucky University in Bowling Green, KY. I am looking for some kind of summer opportunity to participate in during summer 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any suggestions of good opportunities or a good list of opportunities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am more interested in pure mathematics than I am in applied mathematics. My favorite areas are set theory, complex variables/analysis, topology, difference equations, difference(discrete) calculus, and time scale calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really looking to broaden my horizons though with new and interesting topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, I am going to speak with my advisor and professors, but I would like to get insight from others as well.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-23T23:54:01.360" Title="Mathematics Graduate Student Summer Opportunities" Tags="&lt;education&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="11609" PostTypeId="1" AcceptedAnswerId="11623" CreationDate="2010-11-23T23:40:13.187" Score="1" ViewCount="152" Body="&lt;p&gt;Question: Let $v = (1,1,0,1) \in \mathbb{R}^{4}$ Let $$V:= \{f: \mathbb{R}^{4} \rightarrow \mathbb{R}^{2} \mbox{ linear } | f(v) = 0\}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Show that $V$ forms a vectorspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Determine dim$(V)$. (Tip: Describe the elements in $V$ by extending $v$ to a basis of $\mathbb{R}^{4}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt: for (a): ok i know that a vectorspace is defined through those 8 axioms, since it says &quot;linear&quot; i know that $f(v+u) = f(v)+f(u)$ and $\lambda f(v) = f(\lambda v)$, however i don't see how to proceed. what more should i be inferring from the given information?&#xA;for (b) if i understand correctly, i can extend $v=(1,1,0,1)$ to a basis of $\mathbb{R}^{4}$ with $u=(0,1,0,0), w=(0,0,1,0), x=(0,0,0,1)$ correct? of course i wouldn't know what to do with that.. in the problems i've encountered so far it seems that dim$(V)$ has always been the amount of elements in a basis, however i'm doubting that it's so simple as dim$(V)=4$ just because dim$(\mathbb{R}^{4})=4$... i notice that $f(v)=0$, perhaps i could do something with the sum of the kernel and the image? as always, any help is greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2010-11-24T21:27:35.580" Title="Show that this is a vector space and determine the dimension" Tags="&lt;linear-algebra&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11610" PostTypeId="1" CreationDate="2010-11-23T23:41:35.090" Score="0" ViewCount="57" Body="&lt;p&gt;Dear friends,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of any curve that goes from A to B (other than the line segment AB) is greater than the length of that segment. This statement may be a theorem? Or, necessarily, is an axiom?&#xA;It seems to me that any attempt to build a theorem eventually turning into begging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paulo Argolo&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T02:47:35.123" LastActivityDate="2010-11-24T03:03:10.427" Title="Lower curve from A to B:  axiom or theorem?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11611" PostTypeId="1" CreationDate="2010-11-23T23:43:40.127" Score="1" ViewCount="214" Body="&lt;p&gt;I was trying to wrap my brain around this but could not think of anything. I am very poor at maths but trying to learn of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Exact question is&#xA;If x^2+x-6 is a composite function f(g(x)); then figure out f(x) and g(x) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me information looks incomplete to figure out such details.&lt;/p&gt;&#xA;" OwnerUserId="3882" LastEditorUserId="3882" LastEditorDisplayName="" LastEditDate="2010-11-23T23:55:33.267" LastActivityDate="2010-12-25T09:44:26.000" Title="How to decompose function like  x^2+x-6 ?" Tags="&lt;polynomials&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="11612" PostTypeId="2" ParentId="11610" CreationDate="2010-11-23T23:44:46.427" Score="0" ViewCount="0" Body="&lt;p&gt;This is a (simple) theorem.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Arc_length#Definition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arc_length#Definition&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3883" LastActivityDate="2010-11-23T23:44:46.427" CommentCount="4" />
  <row Id="11615" PostTypeId="2" ParentId="11608" CreationDate="2010-11-23T23:54:01.360" Score="3" ViewCount="0" Body="&lt;p&gt;Robert Krasny from University of Michigan, has an excellent resource page&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.lsa.umich.edu/~krasny/fellowships_etc.html&quot; rel=&quot;nofollow&quot;&gt;http://www.math.lsa.umich.edu/~krasny/fellowships_etc.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a good list here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phds.org/graduate-school/preparing/summer-internships-2/&quot; rel=&quot;nofollow&quot;&gt;http://www.phds.org/graduate-school/preparing/summer-internships-2/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-23T23:54:01.360" CommentCount="1" />
  <row Id="11616" PostTypeId="2" ParentId="11611" CreationDate="2010-11-23T23:56:43.687" Score="0" ViewCount="0" Body="&lt;p&gt;We always have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x+a)&lt;em&gt;(x+b) = (x&lt;/em&gt;(x+b))+(a*(x+b)) = (x*x)+(x*b)+(a*x)+(a*b) = (x^2)+(b*x)+(a*x)+(a*b)&#xA;(x^2)+(b*x)+(a*x)+(a*b) = (x^2)+(a*x)+(b*x)+(a*b) = (x^2)+((a+b)*x)+(a*b)&#xA;(x+a)*(x+b) = (x^2)+((a+b)*x)+(a*b)&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Converting your equation to that form gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x^2+x-6 = (x^2)+(1*x)+(-6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we want&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(x^2)+(1*x)+(-6) = (x^2)+((a+b)*x)+(a*b)&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;So, we're looking for numbers whose sum is 1 and whose product is -6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a+b = 1  and  a*b = -6&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;That should be enough for you to figure it out, but in case it's not,&#xA;&lt;br&gt;&#xA;if you just want the answer you can put&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;n vf rdhny gb artngvir gjb naq o vf rdhny gb cbfvgvir guerr&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into rot13.com&lt;/p&gt;&#xA;" OwnerUserId="3883" LastActivityDate="2010-11-23T23:56:43.687" CommentCount="2" />
  <row Id="11618" PostTypeId="1" CreationDate="2010-11-24T01:20:51.570" Score="1" ViewCount="151" Body="&lt;p&gt;Prove the following: $\displaystyle a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one could use induction on $n$? Could one also use trichotomy or some type of combinatorial argument? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-24T04:38:48.420" LastActivityDate="2010-11-24T15:29:24.757" Title="Algebraic Identity $a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="11619" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T01:22:32.263" Score="2" ViewCount="0" Body="&lt;p&gt;Can we build a combinatorial argument along these lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if we have say $n$ students to be allotted in $a$ rooms with $b$ of the $a$ room being non air-conditioned. (Assume the students are distinguishable so we can order the student as $1,2,3,...,n$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of ways is $a^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose all students are allotted to the $b$ rooms, the number of ways is $b^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first $n-1$ students are allotted to the $b$ rooms, and the final dude in some other room, the number of possible ways is $b^{n-1} \times (a-b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if the first $n-2$ students are allotted to the $b$ rooms, and the remaining two students are now left. If the $(n-1)^{th}$ student chooses from the $b$ rooms then we are back to the earlier case. So the $(n-1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the $n^{th}$ student can choose from any of the $a$ rooms. The number of possible ways is $b^{n-2} \times (a-b) \times a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if the first $n-k$ students are allotted to the $b$ rooms, and the remaining $k$ students are now left. If the $(n-k+1)^{th}$ student chooses from the $b$ rooms then we are back to the previous case. So the $(n-k+1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the students from $(n-k+2)$ to $n$ can choose from any of the $a$ rooms. The number of possible ways is $b^{n-k} \times (a-b) \times a^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the total number of ways is $b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the counting must add up and hence we get $a^n = b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use geometric series to conclude the result as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand side is $(a-b) b^{n-1} \displaystyle \sum_{k=0}^{n-1} (\frac{a}{b})^k = (a-b) b^{n-1} \frac{((\frac{a}{b})^n - 1)}{(\frac{a}{b})-1} = (a-b) \frac{a^n - b^n}{a-b} = a^n - b^n$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T01:50:08.750" LastActivityDate="2010-11-24T01:50:08.750" CommentCount="2" />
  <row Id="11620" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T01:24:38.537" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, you could use induction on n.  I don't see an easy trichotomy or combinatorial argument.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-24T01:24:38.537" />
  <row Id="11621" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T01:36:10.593" Score="7" ViewCount="0" Body="&lt;p&gt;I have no idea what you mean by &quot;use trichotomy,&quot; but here is the combinatorial argument.  $a^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ and $b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... b \}$.  Assume $a &amp;gt; b$.  Then $a^n - b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ such that at least one letter is greater than $b$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given such a word, suppose the last letter greater than $b$ occurs at position $k+1$.  Then there are $a - b$ choices for this letter, $a^k$ choices for the letters before this letter, and $b^{n-k-1}$ choices for the letters after this letter.  Thus there are $(a - b) a^k b^{n-k-1}$ such words, and summing over all $k$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^k b^{n-k-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-24T01:36:10.593" CommentCount="3" />
  <row Id="11622" PostTypeId="1" AcceptedAnswerId="11625" CreationDate="2010-11-24T01:48:44.867" Score="2" ViewCount="42" Body="&lt;p&gt;This is a follow up to a question I asked earlier. I had previously asked that if a space was regular if and only if any closed set $Z$ was the intersection of all open sets containing $Z$, but the backwards implication turned out to be untrue. I'm now curious to see if this new bi-implication is fully true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My definition of regular here is that for any $p\in X$ and any closed set $Q$ not containing $p$, there exist open sets $U,V$ such that $p\in U$, $Q\subseteq V$, and $U\cap V=\emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume $X$ is regular. I think it is clear that $$Z\subseteq\bigcap_{\text{closed}\ N\in\mathscr{N}_Z}N$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other containment, I suppose that $p\not\in Z$. Then since $X$ is regular, there exist open sets $U,V$ such that $p\in U$, $Z\subseteq V$, and $U\cap V=\emptyset$. Then $U\subseteq V^c\subseteq Z^c$, and thus $U^c\supseteq V\supseteq Z$. So $U^c$ is a closed neighborhood of $Z$ which does not contain $p$. Hence $p\not\in\cap\mathcal{N}$, where $\mathcal{N}$ is the family of closed neighborhood of $Z$, and the equality follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel I have a hole in my reasoning to show the other implication. I suppose now that $Z=\cap\mathcal{N}$, and take some $p\not\in Z$, and so $p\not\in\cap\mathcal{N}$. Anyway, $p\in Z^c$, which is an open neighborhood of $p$. I then take some closed neighborhood $Q$ of $p$ such that $Q\subseteq Z^c$. I then let $U={Q}^{\circ}$, and $V=Q^c$. So both are are open and disjoint. Also, $p\in U$, and $V=Q^c\supseteq (Z^c)^c=Z$, so $X$ is regular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't think I'm able to make the assumption that there is some closed neighborhood of $p$ contained in the open neighborhood $Z^c$. Is there some way to use the fact that $p\not\in\cap\mathcal{N}$ to show this is indeed true? Thanks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In case it is not entirely clear, my definition of a neighborhood $N$ of a set $Z$ is that $Z$ is contained in the interior of $N$, so $N$ is a neighborhood of each point $Z$. Then my definition of a neighborhood $M$ of a point $p$ is that $M$ contains an open set containing $p$. $\mathscr{N}_Z$ is my notation for the set of all neighborhoods of $Z$, and $\mathcal{N}$ is a shortcut way of writing the set of $\text{closed}\ N\in\mathscr{N}_Z$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-24T02:08:36.840" LastActivityDate="2010-11-24T02:09:45.777" Title="A topological space is regular if and only if any closed set $Z$ is the intersection of its closed neighborhoods?" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11623" PostTypeId="2" ParentId="11609" CreationDate="2010-11-24T02:04:00.557" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Re-rewritten&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems, from your comments, that you aren't even very comfortable with the idea of the set $V$ as a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, it is better to start from a bit further back, and consider the set of &lt;strong&gt;all&lt;/strong&gt; linear transformations from $\mathbb{R}^4\to\mathbb{R}^2$, which I'm going to call $\mathbf{V}$; that is,&#xA;$$\mathbf{V}=\left\{ T\colon\mathbb{R}^4\to\mathbb{R}^2\mid \text{ $T$ is linear}\right\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to make $\mathbf{V}$ into a vector space. So the &lt;em&gt;elements&lt;/em&gt; of $\mathbf{V}$ are going to be our &quot;vectors&quot;; the fact that these vectors are actually functions when they are at home is immaterial: when they are at their &quot;job&quot; in $\mathbf{V}$, they are &quot;vectors&quot;. In order for this to be a vector space, we need to have a way of adding these vectors, and of multiplying them by scalars so that the operations satisfy the axioms of a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are we going to add these vectors? Remember from Calculus that if we have two functions $f\colon \mathbb{R}\to\mathbb{R}$ and $g\colon\mathbb{R}\to\mathbb{R}$, we can make a new function called the &quot;sum of $f$ and $g$&quot;. This function is called &quot;$f+g$&quot;. In order to tell you what function &quot;$f+g$&quot; is, I need to tell you how to evaluate it at any real number $r$. So here's the rule: To find the value of $f+g$ at $r$, first find the value of $f$ at $r$, then find the value of $g$ at $r$, and then add the two results. That is, $$(f+g)(r) = f(r)+g(r).$$ &#xA;Now, there are two &lt;code&gt;+&lt;/code&gt; symbols in that equation, but they mean different things: the first &lt;code&gt;+&lt;/code&gt; sign is the symbol that represents the sum of two functions; the second &lt;code&gt;+&lt;/code&gt; is the usual sum sign that represents the sum of two real numbers (notice that both $f(r)$ and $g(r)$ are real numbers). So, for example, if $f(x) = 2x-1$ and $g(x)=x^3$, then to find the value of $(f+g)(7)$, we first find $f(7)$ (which is $13$), then we find the value of $g(7)$ (which is $343$), and then we add them ($13+343=356$, so $(f+g)(7) = 356$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to do the same thing with the elements of $\mathbf{V}$. The elements of $\mathbf{V}$ are functions from $\mathbb{R}^4$ to $\mathbb{R}^2$. As it happens, we know how to &quot;add&quot; two elements of $\mathbb{R}^2$ (because this is a vector space). That means that we can play the same game with functions $T\colon\mathbb{R}^4\to\mathbb{R}^2$ and $U\colon \mathbb{R}^4\to\mathbb{R}^2$ that we played in calculus: we are going to define a new function, which I'm going to call $T\oplus U$ (I'm using a different symbol so we are clear that this is a &lt;em&gt;new&lt;/em&gt; kind of addition which I am defining), which is going to be a function from $\mathbb{R}^4$ to $\mathbb{R}^2$. In order to tell you what this function is, I need to tell you how you evaluate it at any element of $\mathbb{R}^4$. Okay, here goes: given any $\mathbf{v}\in\mathbb{R}^4$, first find out what $T(\mathbf{v})$ is; then find out what $U(\mathbf{v})$ is. Both of these are elements of $\mathbb{R}^2$, so we can add them. The result of adding them is what $(T\oplus U)(\mathbf{v})$ is. That is, if &lt;code&gt;+&lt;/code&gt; represents the sum of vectors in $\mathbb{R}^2$, then&#xA;$$ (T\oplus U)(\mathbf{v}) = T(\mathbf{v}) + U(\mathbf{v}).$$&#xD;&#xA;(I put the parentheses around $T\oplus U$ so that we are clear that it is a single entity with a long name which we are evaluating at $\mathbf{v}$; that way we don't think that we are doing $U(\mathbf{v})$ first, and then $\oplus$-ing something called $T$ to &lt;em&gt;that&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this makes sense for &lt;em&gt;any&lt;/em&gt; two functions from $\mathbb{R}^4$ to $\mathbb{R}^2$, whether they are linear or not. For instance, if $T\colon\mathbb{R}^4\to\mathbb{R}^2$ is the function $T(a,b,c,d) = (a+b,cd)$, and $U\colon\mathbb{R}^4\to\mathbb{R}^2$ is the function $U(a,b,c,d) = (a,d)$, then the value of $T\oplus U$ at $(1,2,3,4)$ is the result of adding $T(1,2,3,4)=(3,12)$ and $U(1,2,3,4)=(1,4)$; so $(T\oplus U)(1,2,3,4)=(3,12)+(1,4) = (4,16)$. If $T$ and $U$ are &lt;em&gt;any&lt;/em&gt; functions from $\mathbb{R}^4$ to $\mathbb{R}^2$, then $T\oplus U$ is &lt;em&gt;also&lt;/em&gt; a function from $\mathbb{R}^4$ to $\mathbb{R}^2$. (As in Calculus, though, if $T$ and $U$ happen to have &quot;nice&quot; properties, then there is a good chance that these properties will be inherited by $T\oplus U$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of operation is called &lt;strong&gt;pointwise addition&lt;/strong&gt;: whenever you have functions from the same domain to the same codomain, and you know how to &quot;add&quot; elements of the codomain, you can define a &quot;sum of functions&quot; $f+g$ by saying &quot;first figure out the value of $f$, then figure out the value of $g$, and then add the results together.&quot; It's &quot;pointwise&quot; because the way we compute the sum is by first figuring out the values at &lt;em&gt;each point&lt;/em&gt;, and then adding those values. Notice that the domain $\mathbb{R}^4$ is not playing any role here; you could replace it with any set and still be able to define the sum of functions, because the sum is taking place in the &lt;em&gt;image&lt;/em&gt;, not in the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can play the same game with scalar multiplication. If $T\colon\mathbb{R}^4\to\mathbb{R}^2$ is &lt;em&gt;any&lt;/em&gt; function, and $\alpha$ is any real number, then we can remember that we know how to &quot;multiply&quot; any element of $\mathbb{R}^2$ by $\alpha$; so we can define a &quot;pointwise multiplication by $\alpha$&quot; as well. I'm going to define a new function, called $\alpha\odot T\colon\mathbb{R}^4\to\mathbb{R}^2$. The rule for evaluating $\alpha\odot T$ is: to find the value at $\mathbf{v}$, first evaluate $T$ at $\mathbf{v}$, and then multiply the result by $\alpha$. That is,&#xA;$$(\alpha\odot T)(\mathbf{v}) = \alpha\cdot T(\mathbf{v}).$$&#xD;&#xA;Again, I put the parenthesis around $\alpha\odot T$ to remind us that it is a single entity with a long name. The $\cdot$ on the right hand side is the usual scalar multiplication of $\mathbb{R}^2$. For instance, going back to my $T\colon\mathbb{R}^4\to\mathbb{R}^2$ from the previous paragraph, $T(a,b,c,d)=(a+b,cd)$, if $\alpha=\frac{3}{2}$ then $\frac{3}{2}\odot T$ at $(1,2,3,4)$ is computed by first doing $T(1,2,3,4)=(3,12)$, and then doing $\frac{3}{2}(3,12) = (\frac{9}{2},18)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this definition makes sense for &lt;em&gt;any&lt;/em&gt; function into $\mathbb{R}^2$, whether linear or not, whether from $\mathbb{R}^4$ or not. This one is called &lt;strong&gt;pointwise (scalar) multiplication&lt;/strong&gt;, because again the &quot;multiplication&quot; is happening after we figure out the value &quot;at each point&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now let's go back and look at the set $\mathbf{V}$,&#xA;$$\mathbf{V}= \bigl\{ T\colon\mathbb{R}^4\to\mathbb{R}^2\ \bigm|\ \text{$T$ is linear}\bigr\}.$$&#xA;This is a set of functions, and if we take two functions $T$ and $U$ from $\mathbf{V}$ then we can &quot;add&quot; them using $\oplus$ to get a function $T\oplus U\colon\mathbb{R}^4\to\mathbb{R}^2$, and if we take $\alpha\in\mathbb{R}$ and $T\in\mathbf{V}$, we can &quot;scalar multiply them&quot; and get a function $\alpha\odot T\colon\mathbb{R}^4\to\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that if $T$ and $U$ are in $\mathbf{V}$ (that is, if in addition to just being functions they are also &lt;em&gt;linear&lt;/em&gt; functions), then $T\oplus U$ is also going to be in $\mathbf{V}$ (it's already a function from $\mathbb{R}^4$ to $\mathbb{R}^2$, so we will just need to check if it is linear). And that if $T\in\mathbf{V}$ and $\alpha\in\mathbb{R}$, then $\alpha\odot T$ is &lt;em&gt;also&lt;/em&gt; going to be in $\mathbf{V}$. That is, $\oplus$ will give me a way of adding elements of $\mathbf{V}$ and get elements of $\mathbf{V}$, and $\odot$ will give me a way to multiply elements of $\mathbf{V}$ by scalars to get elements of $\mathbf{V}$. (This is the &lt;em&gt;first&lt;/em&gt; requirement for a set to be a vector space: it must have some kind of &quot;vector addition&quot; and some kind of &quot;scalar multiplication&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose that you start by taking $T$ and $U$ in $\mathbf{V}$. In order for $T\oplus U$ to also belong to the club $\mathbf{V}$, $T\oplus U$ must satisfy the membership requirements: it must be a function, it must map $\mathbb{R}^4$ to $\mathbb{R}^2$, and it must be linear. We already know it is a function that maps $\mathbb{R}^4$ to $\mathbb{R}^2$, so we just need to check that it is linear. Remember that each of $T$ and $U$ are themselves linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one check that &lt;em&gt;a&lt;/em&gt; function $\mathfrak{f}\colon\mathbb{R}^4\to\mathbb{R}^2$ is linear? We check that if $u,w\in\mathbb{R}^4$ are any vectors, then $\mathfrak{f}(u+w) = \mathfrak{f}(u)+\mathfrak{f}(w)$ (the first &lt;code&gt;+&lt;/code&gt; is the sum in $\mathbb{R}^4$; the second is the sum in $\mathbb{R}^2$; the equality will take place in $\mathbb{R}^2$); and we check that for any $u\in\mathbb{R}^4$ and any real number $\alpha$, $\mathfrak{f}(\alpha u) = \alpha\mathfrak{f}(u)$ (first scalar multiplication is in $\mathbb{R}^4$; the second is in $\mathbb{R}^2$; and the equality is taking place in $\mathbb{R}^2$). So, we do this for our function $T\oplus U$: let $u,w\in\mathbb{R}^4$. Then:&#xA;\begin{align*}&#xA;(T\oplus U)(u+w) &amp;amp;= T(u+w) + U(u+w) &amp;amp;\text{(by definition of $(T\oplus U)$)}\\&#xA;&amp;amp;= \bigl(T(u) + T(w)\bigr) + \bigl(U(u) + U(w)\bigr) &amp;amp;\text{(because $T$ and $U$ are each linear)}\\&#xA;&amp;amp;= \bigl(T(u) + U(u)\bigr) + \bigl(T(w) + U(w)\bigr) &amp;amp;\text{(because addition in $\mathbb{R}^2$}\\&#xA;&amp;amp;&amp;amp;\text{ is associative and commutative)}\\&#xA;&amp;amp;= (T\oplus U)(u) + (T\oplus U)(w). &amp;amp;\text{(by definition of $(T\oplus U)$)}&#xA;\end{align*}&#xA;And if $u\in\mathbb{R}^4$ and $\alpha\in\mathbb{R}$, then&#xA;\begin{align*}&#xA;(T\oplus U)(\alpha u) &amp;amp;= T(\alpha u) + U(\alpha u) &amp;amp;\text{(by definition of $(T\oplus U)$)}\\&#xA;&amp;amp;= \alpha T(u) + \alpha U(u) &amp;amp;\text{(because $T$ and $U$ are each linear)}\\&#xA;&amp;amp;= \alpha(T(u) + U(u)) &amp;amp;\text{(because scalar multiplication distributes over sum in $\mathbb{R}^2$)}\\&#xA;&amp;amp;=\alpha(T\oplus U)(u).&amp;amp;\text{(by definition of $T\oplus U$)}&#xA;\end{align*}&#xA;So, $T\oplus U$ is indeed a linear function from $\mathbb{R}^4$ to $\mathbb{R}^2$, and so by rights belongs to $\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, to check that if $T\in\mathbf{V}$ and $\gamma\in\mathbb{R}$ then $\gamma \odot T$ is linear (I'm using $\gamma$ so we don't confuse it with the $\alpha$ that we need to check linearity) let $u,w\in\mathbb{R}^4$; then:&#xA;\begin{align*}&#xA;(\gamma\odot T)(u+w) &amp;amp;= \gamma T(u+w) &amp;amp;\text{(by definition of $\gamma\odot T$)}\\&#xA;&amp;amp;= \gamma\bigl(T(u) + T(w)\bigr) &amp;amp;\text{(because $T$ is linear)}\\&#xA;&amp;amp;= \gamma T(u) + \gamma T(w) &amp;amp;\text{(because scalar multiplication distributes in $\mathbb{R}^2$)}\\&#xA;&amp;amp;= (\gamma\odot T)(u) + (\gamma\odot T)(w). &amp;amp;\text{(by definition of $\gamma\odot T$)}&#xA;\end{align*}&#xA;So $(\gamma\odot T)(u+w) = (\gamma\odot T)(u) + (\gamma\odot T)(w)$. And, if $u\in \mathbb{R}^4$ and $\alpha\in\mathbb{R}$, then:&#xA;\begin{align*}&#xA;(\gamma\odot T)(\alpha u) &amp;amp;= \gamma T(\alpha u) &amp;amp;\text{(by definition of $\gamma\odot T$)}\\&#xA;&amp;amp;= \gamma\left(\alpha T(u)\right) &amp;amp;\text{(because $T$ is linear)}\\&#xA;&amp;amp;= (\gamma \alpha)T(u) &amp;amp;\text{(scalar multiplication is associative in $\mathbb{R}^2$)}\\&#xA;&amp;amp;= (\alpha\gamma)T(u) &amp;amp;\text{(real multiplication is commutative)}\\&#xA;&amp;amp;= \alpha\left(\gamma T(u)\right) &amp;amp;\text{(scalar multiplication is associative in $\mathbb{R}^2$)}\\&#xA;&amp;amp; = \alpha (\gamma\odot T)(u). &amp;amp;\text{(by definition of $\gamma\odot T$)}&#xA;\end{align*}&#xA;So $(\gamma\odot T)(\alpha u) = \alpha(\gamma\odot T)(u)$. Together with the additivity, this tells us that $\gamma\odot T$ is linear, and therefore is, by all rights, a member of $\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... now we have a set $\mathbf{V}$, and operations $\oplus$ and $\odot$ on $\mathbf{V}$. To check that this makes it into a vector space, we need to check the axioms. That is, we need to check that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $T$ and $U$ are in $\mathbf{V}$, then $T\oplus U = U\oplus T$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But what does it mean for $T\oplus U$ to equal $U\oplus T$? Well, they are both functions, so we need to check that they have the same domain and the exact same value at every point of the domain; that is, we need to check that for every $u\in\mathbb{R}^4$, $(T\oplus U)(u) = (U\oplus T)(u)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $S$, $T$, and $U$ are in $\mathbf{V}$, then $(S\oplus T)\oplus U = S\oplus(T\oplus U)$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That is, the two functions, $(S\oplus T)\oplus U$ and $S\oplus(T\oplus U)$ have the same domain, and for every $u$ in the domain they have the same value at $u$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a function $O$ in $\mathbf{V}$ such that for all functions $T\in \mathbf{V}$, $O\oplus T = T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $T\in\mathbf{V}$ there is a function $U\in\mathbf{V}$ such that $T\oplus U = O$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $T\in\mathbf{V}$, $1\odot T = T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each pair of real numbers $\alpha$ and $\beta$ and every $T\in\mathbf{V}$, $\alpha\odot(\beta \odot T) = (\alpha\beta)\odot T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each $\alpha\in\mathbb{R}$ and every $T$ and $U$ in $\mathbf{V}$, $\alpha\odot(T\oplus U) = (\alpha\odot T) \oplus (\alpha\odot U)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For every $\alpha$ and $\beta$ in $\mathbb{R}$ and every $T\in\mathbf{V}$, $(\alpha+\beta)\odot T = (\alpha\odot T)\oplus (\beta\odot T)$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Again, both $(\alpha+\beta)\odot T$ and $(\alpha\odot T)\oplus(\beta\odot T)$ are functions, so to check if they are equal you need to check that they have the same domain and the same value at each point of the domain.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If all of these conditions are satisfied, then by all rights we must acknowledge that $\mathbf{V}$, with the operations $\oplus$ for &quot;vector addition&quot; and $\odot$ as &quot;scalar multiplication&quot; &lt;strong&gt;is&lt;/strong&gt; a vector space. The &quot;vectors&quot; of this vector space are linear transformations when they are relaxing at home, but at their job in $\mathbf{V}$ they are &quot;vectors&quot; (because the elements of a vector space are called &quot;vectors&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you have several different meanings of &quot;vector&quot; at play: there's the &quot;vectors&quot; that are elements of $\mathbf{V}$ (who, in reality, are linear maps from $\mathbb{R}^4$ to $\mathbb{R}^2$). You also have the &quot;vectors&quot; that are elements of $\mathbb{R}^4$ (at home, they are $4$-tuples of real numbers); and you have the elements of $\mathbb{R}^2$ (who are &quot;really&quot; ordered pairs of real numbers). You need to keep them straight, but it is not hard to do that if you just keep your wits about you and simpy &lt;em&gt;refuse&lt;/em&gt; to get confused.  &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;So, hopefully now we know that $\mathbf{V}$ actually &lt;em&gt;is&lt;/em&gt; a vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our set $V$ is a &lt;em&gt;subset&lt;/em&gt; of $\mathbf{V}$.  In order to show that $V$ is itself a vector space (with the same operations that we have for $\mathbf{V}$, we are in the situation of checking if something is a &lt;em&gt;subspace&lt;/em&gt;: because $\oplus$ is associative and commutative for &lt;em&gt;all&lt;/em&gt; functions in $\mathbf{V}$, then it is also so in $V$; similarly with axioms 5 through 8 for $\odot$ and $\oplus$. Really, the only sticking points for $V$ to be a vector space are four things: what I call the &quot;hidden axioms&quot; of a vector space, and axioms 3 and 4. Explicitly:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We know that if we take $T$ and $U$ in $\mathbf{V}$, then $T\oplus U$ is also in $\mathbf{V}$. But for $V$ to be a vector space, we need to make sure that if we take $T$ and $U$ in $V$, then $T\oplus U$ is not just going to be in $\mathbf{V}$, but actually in $V$ (we know the sum is in the same &quot;state&quot;, but we actually  need to make sure it lives in the same &lt;em&gt;city&lt;/em&gt;). This is one of the &quot;hidden axioms&quot; of vector spaces, because it is not listed in the usual list; it is implicit when we talk about addition of vectors being an &quot;operation&quot;. So we need to check that if $T$ and $U$ are in $V$ (and not merely in $\mathbf{V}$), then $T\oplus U$ is &lt;em&gt;also&lt;/em&gt; in $V$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Likewise, we know that if $T$ is in $\mathbf{V}$ and $\alpha\in\mathbb{R}$, then $\alpha\odot T$ is going to be in $\mathbf{V}$. So if we start with $T\in V$, then certainly $\alpha\odot T$ is going to be in $\mathbf{V}$, but we actually need it to be in the more exclusive neighborhood of $V$. So we need to check &lt;em&gt;that&lt;/em&gt;: if $T\in V$ and $\alpha\in\mathbb{R}$, we need to make sure that $\alpha\odot T$ is &lt;em&gt;also&lt;/em&gt; in $V$ (and not merely in $\mathbf{V}$, which we know is the case).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Axiom 3 is not immediate either: we know that there is a vector $O$ in $\mathbf{V}$ such that for every $T\in\mathbf{V}$, $O\oplus T = T$. In particular, if we take a $T\in V$, then we wil also have $O\oplus T=T$. But the problem is that for $V$ to be a vector space, we need this special vector $O$ to actually be in $V$, and not somewhere outside of $V$. So we need to check that this happens.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Axiom 4 is also not immediate: if $T\in V$, then since $T$ is also in $\mathbf{V}$ we know there is a $U$ in $\mathbf{V}$ such that $T\oplus U = O$. But for $V$ to be a vector space, we need to make sure that $U$ is &lt;em&gt;also&lt;/em&gt; in $V$, and not just in $\mathbf{V}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But if we can check these four new items, then we will have that $V$ is also a vector space. (In fact, there are shortcuts, which you may or may not have seen if you have already studied &quot;subspaces&quot;; you are really trying to show that $V$ is a subspace of $\mathbf{V}$, so if you know the shortcuts, you can use them here, keeping in mind that &quot;vectors&quot; means &quot;elements of $V$&quot; (or $\mathbf{V}$), etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how do we check that if $T$ and $U$ are in $V$, then $T\oplus U$ is in $V$? Well, we know $T\oplus U$ is linear, and maps $\mathbb{R}^4$ to $\mathbb{R}^2$ (because we know that $T\oplus U$ is in $\mathbf{V}$). The only extra membership requirement to be in $V$ is that you &lt;em&gt;must&lt;/em&gt; have that $(T\oplus U)(v) = 0$ (that is, that $(T\oplus U)(1,1,0,1) = (0,0)$).  Since we are assuming that $T$ and $U$ are &lt;em&gt;each&lt;/em&gt; in $V$, that means that $T(v)=0$ and that $U(v)=0$. But that means that&#xA;\begin{align*}&#xA;(T\oplus U)(v) &amp;amp;= T(v) + U(v) &amp;amp;\text{(by definition of $T\oplus U$)}\\&#xA;&amp;amp;= 0 + 0 &amp;amp;\text{(since $T$ and $U$ are each in $V$)}\\&#xA;&amp;amp;= 0. &amp;amp;\text{(well... duh)}&#xA;\end{align*}&#xA;So... guess what? $T\oplus U$ satisfies all membership requirements for belonging to $V$, so it is in $V$! Great!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we need to check the rest of the conditions similarly.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;What about (b)? There are a couple of ways; following the suggestion, remember that the elements of $V$ are &lt;em&gt;really&lt;/em&gt; linear transformations. To &quot;know&quot; what a linear transformation is, it is enough to know what it does to a basis, any basis. So, let's start by taking the vector $v$, and extending $\{v\}$ to a basis for $\mathbb{R}^4$; say $\beta=[v, v_2, v_3, v_4]$. In order for me to give you an element $f$ of $\mathbf{V}$, it is enough for me to tell you what $f(v)$, $f(v_2)$, $f(v_3)$, and $f(v_4)$ are. From that information, and knowing that it is a linear map, you can figure out who $f$ is. And, what is more, any values I give you will give you an element of $\mathbf{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what do I need to give you in order to describe an element of $V$? I still need to tell you what $f(v)$, $f(v_2)$, $f(v_3)$ and $f(v_4)$ are, but as it happens, in order for $f$ to be in $V$ we need $f(v)=0$. So I just need to tell you what happens to $v_2$, $v_3$, and $v_4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be tempted to think this means the dimension of $V$ is $3$, because you have three degrees of freedom, but in fact this is not the case: each choice of $f(v_2)$, $f(v_3)$ and $f(v_4)$ is a &lt;em&gt;vector&lt;/em&gt; in $\mathbb{R}^2$; so you actually have to specify &lt;em&gt;two&lt;/em&gt; things for &lt;em&gt;each&lt;/em&gt; of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should suggest what the answer &lt;em&gt;should&lt;/em&gt; be; you still have to actually prove it, though.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T21:27:35.580" LastActivityDate="2010-11-24T21:27:35.580" CommentCount="20" />
  <row Id="11624" PostTypeId="1" CreationDate="2010-11-24T02:08:54.810" Score="6" ViewCount="125" Body="&lt;p&gt;I'm trying to classify (at least as fully as possible) the values of $t$ in $(0,\pi/2)$ for which the following equation has a solution for some natural number, $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any results that will help me?&lt;/p&gt;&#xA;" OwnerUserId="3884" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T02:35:37.193" LastActivityDate="2010-11-24T04:38:51.990" Title="When is $\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$?" Tags="&lt;number-theory&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11625" PostTypeId="2" ParentId="11622" CreationDate="2010-11-24T02:09:45.777" Score="5" ViewCount="0" Body="&lt;p&gt;I agree that you don't know that there is a closed neighborhood of $p$ contained in $Z^c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this perhaps will work: Since $p$ is not in the intersection $\cap \mathcal{N}$, there is a closed set $V$ with $Z\subseteq V^{\circ}$ and with $p\notin V$. Since $V$ is closed, then $V^c$ is an open set containing $p$; and the interior of $V$ is an open set containing $Z$. Of course, $V^c$ is disjoint from $V^{\circ}$, so don't $V^{\circ}$ and $V^c$ give you open sets, disjoint, one containing $Z$ and the other containing the point $p\notin Z$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-24T02:09:45.777" CommentCount="3" />
  <row Id="11626" PostTypeId="1" CreationDate="2010-11-24T02:38:16.783" Score="2" ViewCount="232" Body="&lt;p&gt;I have a few weeks off from school soon, and I was hoping to self-study a bit of algebra. I don't think this question has been asked on here before, but does anyone have any suggestions for algebra books which are particularly suited to self-study? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is some background and motivation, if it's helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently a junior in high school, but I have some familiarity with groups, rings, polynomials and fields, as I went through Fraleigh's book. I liked the writing style of that book, but I felt many of the problems were not helpful. I would like to get better at algebra with some other book, particularly one that's aimed at someone who already has a little familiarity with the objects, but is by no means completely comfortable with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my longer-term goals is to learn a bit of algebraic number theory, as right now I've only been studying elementary number theory. That's why I feel it would be good to learn more algebra first. (I'd also be interested in suggestions of other good subjects to learn prior to tackling algebraic number theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was considering using one of Lang's books, since they seem to be pretty universal, but I've heard rumors that they are very terse and not good for learning by oneself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions would be very appreciated, thank you!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-24T04:33:54.587" CommunityOwnedDate="2010-11-24T06:01:08.860" Title="Good books for self-studying algebra?" Tags="&lt;big-list&gt;&lt;books&gt;&lt;self-learning&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="11627" PostTypeId="2" ParentId="11626" CreationDate="2010-11-24T02:52:10.173" Score="1" ViewCount="0" Body="&lt;p&gt;For Algebra you can look at these books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Topics in Algebra&lt;/em&gt; by I.N. Herstein&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Abstract Algebra&lt;/em&gt; by Dummit and Foote&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Algebra&lt;/em&gt; by Michael Artin&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Algebra&lt;/em&gt; by T.Hungerford (Springer)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Lectures in Abstract Algebra&lt;/em&gt; by N.Jacobson (Has 3 volumes!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Algebra&lt;/em&gt; by Anthony Knapp. (2 Volumes.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My feeling of &lt;em&gt;Herstein&lt;/em&gt; is it has lot of problems which are challenging. For theory part i would like to use &lt;em&gt;Dummit and Foote&lt;/em&gt;. Artin's Algebra is very well written and contains a lot of Linear Algebra. Anthony Knapps treatment of Algebra is very comprehensive, and contains a lot of Algebra. Since your aim is to read &lt;em&gt;Algebraic Number Theory&lt;/em&gt; you might want to learn some &lt;em&gt;Galois theory&lt;/em&gt; also for which there many good books like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Lectures in Galois theory&lt;/em&gt; by Emil Artin&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Field theory and its Classical problems&lt;/em&gt; by Charles Hadlock.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Galois theory&lt;/em&gt; by J.Rotmain (Springer.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-24T02:59:32.057" LastActivityDate="2010-11-24T02:59:32.057" CommunityOwnedDate="2010-11-24T06:01:08.860" CommentCount="2" />
  <row Id="11628" PostTypeId="2" ParentId="11611" CreationDate="2010-11-24T02:58:33.177" Score="0" ViewCount="0" Body="&lt;p&gt;The 'simplest' possibility is f(x) := x-3 and g(x) := x^2+x-3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the more natural of the 2 pairs of integer polynomials of degree at most 2 that minimize the sum of squares of their coefficients while composing to give x^2+x-6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and apparently it's not this easy to get back the account I created on another computer.&lt;/p&gt;&#xA;" OwnerUserId="3888" LastActivityDate="2010-11-24T02:58:33.177" CommentCount="0" />
  <row Id="11629" PostTypeId="2" ParentId="11610" CreationDate="2010-11-24T03:03:10.427" Score="1" ViewCount="0" Body="&lt;p&gt;It's a consequence of the so-called triangle inequality.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T03:03:10.427" />
  <row Id="11630" PostTypeId="2" ParentId="11624" CreationDate="2010-11-24T03:11:22.560" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT: This solution is wrong! The error is highlighted below (thanks for Zaricuse and Rahul).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $t$, define a function $$f(n) = \frac{\sin(nt)}{n}.$$&#xD;&#xA;Its derivative is $$f&amp;#39;(n) = \frac{\cos(nt)nt - \sin(nt)}{n^2}.$$&#xD;&#xA;If $f(n) = f(n+2)$ then the derivative must have a zero somewhere in between, say $f&amp;#39;(m) = 0$. That implies $$\tan(mt) = mt.$$ However, for $x \in (0,\pi)$ we have $\tan(x) &amp;gt; x$ (consider the Taylor expansion of $\tan$, for example). &lt;b&gt;The unfortunate error is that although $t &amp;lt; \pi$, $x = mt$ is not necessarily less than $\pi$.&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-24T03:58:33.910" LastActivityDate="2010-11-24T03:58:33.910" CommentCount="3" />
  <row Id="11631" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T03:15:20.887" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof by induction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=1$ is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supose valid by n, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{n+1}-b^{n+1}=a(a^{n})+b(b^{n})$$, using the hipotesis :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(a^{n})+b(b^{n})=a\left[b^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + b\left[a^{n}-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ab^{n}+(b-a)a\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + \left[a^{n}b-(b-a)b\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}\right] + \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+(b-a)b^{n}-(b-a)b^{n}\right] +$$&#xD;&#xA;$$ \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}-(b-a)a^{n}+(b-a)a^{n}\right]=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+b^{n}]+b^{n+1}\right] +\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}+a^{n}]-a^{n+1}\right] +$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}+b^{n+1}\right] +\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}-a^{n+1}\right] =$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] =$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&#xA;$$a^{n+1}-b^{n+1}=-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] $$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2(a^{n+1}-b^{n+1})=2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^{n+1}-b^{n+1}=\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $n+1$ is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete the proof&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-24T03:59:39.820" LastActivityDate="2010-11-24T03:59:39.820" CommentCount="1" />
  <row Id="11632" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T03:27:39.387" Score="3" ViewCount="0" Body="&lt;p&gt;Someone should mention the &quot;polynomial multiplication&quot; or &quot;telescoping&quot; proof, which may be viewed as a variant of the &quot;geometric series&quot; method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*} (a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k} &amp;amp;= \sum_{k=0}^{n-1} a^{k+1} b^{n-1-k} - \sum_{k=0}^{n-1} a^k b^{n-k} \\ &amp;amp;= \sum_{k=1}^n a^k b^{n-k} - \sum_{k=0}^{n-1} a^k b^{n-k} = a^n - b^n. \end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T03:27:39.387" />
  <row Id="11633" PostTypeId="2" ParentId="11624" CreationDate="2010-11-24T03:32:53.173" Score="3" ViewCount="0" Body="&lt;p&gt;I don't know if there is any closed form for the solutions, but by replacing $\sin\big((n+2)t\big)$ with $\sin nt \cos 2t + \cos nt \sin 2t$ and rearranging, your problem is equivalent to finding the solutions of $$\tan nt = \frac{n \sin 2t}{n + 2 - n \cos 2t}.$$ For any $n &amp;gt; 0$, the RHS is bounded, while the LHS crosses from $-\infty$ to $\infty$ exactly $2n$ times over a $2 \pi$ interval. Some experimentation suggests that there are indeed exactly $2n$ solutions; possibly you could bound the slopes of the two expressions to make this rigorous. This would also let you find bounds for the locations of the solutions, and easily find them numerically through a bisection procedure.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-24T04:03:06.170" LastActivityDate="2010-11-24T04:03:06.170" CommentCount="5" />
  <row Id="11634" PostTypeId="1" AcceptedAnswerId="11649" CreationDate="2010-11-24T03:38:12.787" Score="1" ViewCount="125" Body="&lt;p&gt;Assume $f$ has a finite derivative and $|f&amp;#39;(x)| \leq y &amp;lt; 1$ for all $x \in (a,b)$ &lt;br&gt;&#xA;$f$ is continuous and $a \leq f(x) \leq b$ for all $x \in [a,b]$. Prove $f$ has a unique fixed point in $[a,b]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have for every c in (a,b) |f'(c)| ≤ y&#xA;=&gt; lim x-&gt;c |f(x) - f(c)|/|x-c| ≤ y&#xA;=&gt; lim x-&gt;c |f(x) - f(c)| ≤ y lim x-&gt;c |x-c|&lt;br&gt;&#xA;Would that be the definition of a contractive map in R?&lt;br&gt;&#xA;Therefore by Banach Fixed Point Theorem, f has a unique fixed point.&lt;br&gt;&#xA;Can I prove Banach's theorem using the mean value theorem?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-11-24T03:48:00.407" LastActivityDate="2010-11-24T04:37:01.133" Title="Prove Fixed Point Theorem using the Mean Value Theorem" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11635" PostTypeId="2" ParentId="11626" CreationDate="2010-11-24T03:39:27.930" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568814542&quot; rel=&quot;nofollow&quot;&gt;A Survey of Modern Algebra&lt;/a&gt; by Birkhoff and Maclane.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-24T03:39:27.930" CommunityOwnedDate="2010-11-24T06:01:08.860" CommentCount="1" />
  <row Id="11636" PostTypeId="2" ParentId="11567" CreationDate="2010-11-24T03:39:59.053" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ mod\ d:\ \ b^M,\:b^N\equiv 1\iff ord(b)\:|\:M,N\iff ord(b)\:|\:gcd(M,N)\iff b^{gcd(M,N)}\equiv 1\ $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ The conceptual structure at the heart of this simple proof is the ubiquitous &lt;strong&gt;order ideal.&lt;/strong&gt; See my &lt;a href=&quot;http://math.stackexchange.com/questions/4713/if-the-order-divides-a-prime-p-then-the-order-is-p-or-1/4715#4715&quot;&gt;post here&lt;/a&gt; for more on this and the more familiar additive form of a &lt;strong&gt;denominator ideal.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally $\rm\ gcd(f(m), f(n))\ =\ f(gcd(m,n))\ \ \ if\ \ \ f(n)\ \equiv\ f(n-m)\ \ (mod\ f(m)),\ \ \ f(0)\ =\ 0\:.\ $ See my &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question/7561#7561&quot;&gt;post here&lt;/a&gt; for a simple inductive proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there is a q-analog: the result also holds true for polynomials $\rm\ \ f(n)\ =\ (x^n-1)/(x-1)\:,\ $ and $\rm\ x\to 1\ $ yields the integer case (Bezout identity) - see my &lt;a href=&quot;http://math.stackexchange.com/questions/7473/number-theory-proving-question/7479#7479&quot;&gt;post here&lt;/a&gt; for a simple proof.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-24T03:39:59.053" />
  <row Id="11639" PostTypeId="1" AcceptedAnswerId="11641" CreationDate="2010-11-24T03:48:54.650" Score="1" ViewCount="97" Body="&lt;p&gt;Is there a fundamental relationship between circuits and cycles in an arbitrary graph? Are they mutually independent objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-24T15:55:15.157" Title="Eulerian Circuits and Hamiltonian Cycles" Tags="&lt;graph-theory&gt;" AnswerCount="3" />
  <row Id="11640" PostTypeId="1" CreationDate="2010-11-24T03:50:07.473" Score="2" ViewCount="77" Body="&lt;p&gt;As is well known, if $x,y$ are natural numbers then their GCD $(x,y)$ has a representation $(x,y) = ax + by$ where $a,b$ are natural. Now let's prove that if $(x,y)=1$ then $(xy,x+y) = 1$. Starting from $ax+by=1$, $$1 = (ax+by)^2 = a^2x^2 + b^2y^2 + 2abxy = (a^2x+b^2y)(x+y) - (a-b)^2xy.$$ The reverse implication is trivial in this method. The proof can be generalized to show that if $(x,y,z)=1$ then $(xyz,xy+xz+yz,x+y+z)=1$ and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;Can this idea be molded into a reasonably strong formal proof system for some fragment of number theory?&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're competing with divisibility proofs, which we'd like to forbid somehow. For example, the original statement can be proven as follows. Suppose $p$ is a prime dividing $(xy,x+y)$. Since $p|xy$ and $p$ is prime, either $p|x$ or $p|y$. From $p|x+y$ we get that $p|(x,y)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T04:13:52.040" Title="A proof system for GCD?" Tags="&lt;number-theory&gt;&lt;logic&gt;" AnswerCount="1" />
  <row Id="11641" PostTypeId="2" ParentId="11639" CreationDate="2010-11-24T03:53:40.013" Score="3" ViewCount="0" Body="&lt;p&gt;A cycle with two chords has a Hamiltonian cycle but no Eulerian circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lemniscate ($\infty$, two cycles pasted at a vertex) has an Eulerian circuit but no Hamiltonian cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cycle has both a Hamiltonian cycle and an Eulerian circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A star with at least 3 edges has neither a Hamiltonian cycle nor an Eulerian circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia describes the graphs which have Eulerian circuits; Hamiltonian cycles are much more complicated, and in particular it is very probable that there's no simple characterization of graphs that have them (unless P=NP).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-24T04:09:21.113" LastActivityDate="2010-11-24T04:09:21.113" CommentCount="5" />
  <row Id="11642" PostTypeId="2" ParentId="11634" CreationDate="2010-11-24T03:55:46.253" Score="3" ViewCount="0" Body="&lt;p&gt;To show that there is some fixed point, consider $g(x) = f(x)-x$. Then $g(a)\geq 0$ and $g(b) \leq 0$. To show that it is unique, use the mean-value theorem for the two purported fixed points.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T03:55:46.253" CommentCount="4" />
  <row Id="11643" PostTypeId="2" ParentId="11639" CreationDate="2010-11-24T04:08:14.900" Score="0" ViewCount="0" Body="&lt;p&gt;By definition a &lt;em&gt;path graph&lt;/em&gt; cannot have an Eulerian circuit or a Hamiltonian cycle. A &lt;em&gt;loop&lt;/em&gt; graph (consisting of one edge and one vertex) has both an Eulerian circuit and a Hamiltonian cycle. As above, there are examples where a graph might have one but not the other. The answer to your question is that there is &lt;em&gt;no&lt;/em&gt; fundamental relationship between the two.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-24T04:46:12.090" LastActivityDate="2010-11-24T04:46:12.090" />
  <row Id="11644" PostTypeId="2" ParentId="11640" CreationDate="2010-11-24T04:08:33.573" Score="4" ViewCount="0" Body="&lt;p&gt;First, work with ideals, which provides much more power and transparency. For example, in terms of ideals your first proof is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad (x,y) \ \supset\ (xy,\ x+y)\ \supset\ (x,y)^2\ \ $ therefore $\rm\ \ (xy,\ x+y)\ =\ 1\ \iff\ (x,y)\ =\ 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how employing ideals has eliminated obfuscatory information such as the extraneous coefficients $\ a,\:b\ $ in the original proof. That done, the innate structure becomes much clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there are various generalizations of the Grobner basis algorithm over Euclidean domains. Whether or not they will suffice for your application is hard to say without knowing further details. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T04:13:52.040" LastActivityDate="2010-11-24T04:13:52.040" CommentCount="2" />
  <row Id="11645" PostTypeId="1" AcceptedAnswerId="11652" CreationDate="2010-11-24T04:10:55.083" Score="1" ViewCount="102" Body="&lt;p&gt;I was wondering if we have a finite group $G$ with subgroups $H$ and $K$.  With $X:=G/H$.&#xA;Is there a simple way of finding $X^K$ and when will $X^K=\emptyset$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also does this have any baring on when $G/H≅ G/K$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the first question I've ever asked here so apologies for errors etc.)&lt;/p&gt;&#xA;" OwnerUserId="3890" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T04:33:19.707" LastActivityDate="2010-11-25T01:14:06.170" Title="Fixed points of finite group." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="11646" PostTypeId="2" ParentId="11624" CreationDate="2010-11-24T04:20:09.017" Score="2" ViewCount="0" Body="&lt;p&gt;EDIT: See J.M.'s comment to Rahul's answer: it's better to use the &quot;other&quot; Chebyshev polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square your identity: $$(n+2)^2 \sin^2 nt  = n^2 \sin^2 (n+2)t.$$ Convert from sines to cosines: $$(n+2)^2 + n^2 \cos^2 (n+2)t = n^2 + (n+2)^2 \cos^2 nt.$$ Now let $x = \cos t$ to get $$(n+2)^2 + n^2 T_{n+2}(x)^2 = n^2 + (n+2)^2 T_n(x)^2,$$ where the $T_m$ are the Chebyshev polynomials. We get a polynomial equation of degree $2(n+2)$ in $x$ which you can solve to find all solutions (you will need to check that the signs of the sines match).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-24T04:38:51.990" LastActivityDate="2010-11-24T04:38:51.990" CommentCount="1" />
  <row Id="11648" PostTypeId="2" ParentId="11626" CreationDate="2010-11-24T04:33:54.587" Score="0" ViewCount="0" Body="&lt;p&gt;I learned algebra (before taking Artin's course) by self-studying Nathan Jacobson's &lt;em&gt;Basic Algebra I and II&lt;/em&gt;. This is a beautiful algebra textbook. However, it may help to supplement it with other presentations since it is a bit too concise in some places. A good choice for a supplement is Karlheinz Spindler's &lt;em&gt;Abstract algebra with applications in two volumes&lt;/em&gt;. It has many interesting well-chosen motivational examples.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-24T04:33:54.587" CommunityOwnedDate="2010-11-24T06:01:08.860" CommentCount="1" />
  <row Id="11649" PostTypeId="2" ParentId="11634" CreationDate="2010-11-24T04:37:01.133" Score="2" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim_{x\to c} |f(x) - f(c)|\leq y\ \lim_{x\to c} |x-c|$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Would that be the definition of a contractive map in R?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, this is just the statement that $f$ is continuous at $c$, because the right-hand side is $0$.  The fixed point theorem will apply, but to show that $f$ is contractive you will want to use the mean value theorem.  Suppose that $a\leq z\lt x\leq b$.  By the mean value theorem, there is a $c$ in $(z,x)$ such that $f&amp;#39;(c)=\frac{f(x)-f(z)}{x-z}$.  Apply absolute values, rearrange, and use the hypothesis on the derivative to conclude that $f$ is contractive.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-24T04:37:01.133" CommentCount="1" />
  <row Id="11650" PostTypeId="1" AcceptedAnswerId="11667" CreationDate="2010-11-24T04:42:09.430" Score="9" ViewCount="229" Body="&lt;p&gt;Quote from Don Zagier (Mathematicians: An Outer View of the Inner World):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;    I like explicit, hands-on formulas. To me they have a beauty of their own. They can be deep or not. As an example, imagine you have a series of numbers such that if you add 1 to any number you will get the product of its left and right neighbors. Then this series will repeat itself at every fifth step! For instance, if you start with 3, 4 then the sequence continues: 3, 4, 5/3, 2/3, 1, 3, 4, 5/3, etc. The difference between a mathematician and a nonmathematician is not just being able to discover something like this, but to care about it and to be curious about why it's true, what it means, and what other things in mathematics it might be connected with. In this particular case, the statement itself turns out to be connected with a myriad of deep topics in advanced mathematics: hyperbolic geometry, algebraic K-theory, the Schrodinger equation of quantum mechanics, and certain models of quantum field theory. I find this kind of connection between very elementary and very deep mathematics overwhelmingly beautiful.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recursion described here is a(n+1)=(1+a(n))/a(n-1) with a(0)=1, a(1)=3.&#xA;It's fairly straightforward to show that for every positive a(0),a(1) the sequence repeats after 5 steps (and a(0)=a(1)=golden ratio is a fixed point). One can also think of this recursion as the planar map f(x,y)=((1+x)/y,x), and every point in the plane is a period-5 point of this map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the connection with the deep topics listed in the quote? &lt;/p&gt;&#xA;" OwnerUserId="3891" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-24T17:37:13.890" LastActivityDate="2010-12-21T10:16:22.013" Title="What is the connection of the sequence 3, 4, 5/3, 2/3, 1 with deep topics?" Tags="&lt;sequences-and-series&gt;&lt;hyperbolic-geometry&gt;&lt;algebraic-k-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="11651" PostTypeId="1" AcceptedAnswerId="11655" CreationDate="2010-11-24T04:56:57.320" Score="1" ViewCount="63" Body="&lt;p&gt;What would be an example of a matrix that would not have a $LDL^T$ decomposition? This is a trivial case but I was thinking of a zero matrix which would result in L being an identity matrix and D would be a zero matrix. When you compute $LDL^T$ it still gives you the original zero matrix so I don't believe this is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a singular matrix?&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="1880" LastEditorDisplayName="" LastEditDate="2010-11-24T05:03:09.293" LastActivityDate="2010-11-24T05:20:43.807" Title="Symmetric Matrix without $LDL^T$ Decomposition" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="11652" PostTypeId="2" ParentId="11645" CreationDate="2010-11-24T05:00:03.433" Score="4" ViewCount="0" Body="&lt;p&gt;I'm assuming that $G/H$ represents the left cosets of $H$ in $G$, that is the sets of the form $gH$ with $g\in G$, with the action of $G$ on $X$ being given by $x\cdot(gH) = xgH$ (left multiplication), and that $X^K$ is the set of fixed points of $K$; that is, all $gH\in X$ such that for all $k\in K$, $k(gH)=gH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $kgH = gH$, then $g^{-1}kg\in H$. Thus, $gH$ is fixed by all $k\in K$ if and only if $g^{-1}Kg\subseteq H$. So&#xA;$$X^K = \{ gH\in G/H \mid g^{-1}Kg\subseteq H\}.$$&#xD;&#xA;Note that this is well defined: if $gH = g&amp;#39;H$, then $g&amp;#39;^{-1}g\in H$, so if $g^{-1}Kg\subseteq H$, then &#xA;$$g&amp;#39;^{-1}Kg&amp;#39; = (g&amp;#39;^{-1}g)(g^{-1}Kg)(g&amp;#39;^{-1}g)^{-1}\subseteq g&amp;#39;^{-1}gH(g&amp;#39;^{-1}g)^{-1}=H.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives an easy way to find $X^K$ (by looking at the conjugates of $K$) and to figure out if $X^K=\emptyset$ (if no conjugate of $K$ is contained in $H$; e.g., if the order of $K$ is larger than the order of $H$; if $H$ is normal and $K$ is not contained in $H$; many other criteria).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when are the $G$-sets $G/H$ and $G/K$ isomorphic? First, you must have $[G:H]=[G:K]$. Also, there must be a bijection $\psi\colon G/H\to G/K$ such that for all $x,g\in G$, $\psi(gxH) = g\psi(xH)$. In particular, for all $h\in H$ we must have $\psi(H) = \psi(hH) = h\psi(H)$. If $\psi(H) = xK$, then that means that $HxK = xK$, or that $x^{-1}Hx\subseteq K$. In particular, we need a conjugate of $H$ to be contained in $K$. Since $G$ is finite and $H$ and $K$ must have the same order, that means that $H$ is a conjugate of $K$. If $x^{-1}Hx=K$, then for any coset $gH$ we must have $\psi(gH) = g\psi(H) =gxK$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if $H$ is a conjugate of $K$, $x^{-1}Hx = K$, then let $\psi\colon G/H\to G/K$ be the map that sends the coset $gH$ to the coset $gxK$. First, I claim this is well defined: if $yH=zH$, we need to show that $yxK = zxK$. Since $z^{-1}y\in H$, then $x^{-1}z^{-1}yx\in x^{-1}Hx = K$, so $yxK=zxK$, as claimed. If $gH\in G/H$ and $a\in G$, then $\psi(a\cdot gH) = \psi(agH) = agxK = a\cdot gxK = a\psi(gH)$, so $\psi$ is a $G$-set homomorphism. The map $G/K\to G/H$ given by $gK\mapsto gx^{-1}K$ is also a $G$-set homomorphism (as $xKx^{-1}=H$), and is the inverse of the previous map, so $\psi$ is an isomorphism. Thus, the $G$-set $G/H$ is isomorphic to the $G$-set $G/K$ if and only if $H$ and $K$ are conjugate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; It is interesting to note that the definition of $\psi$ above &lt;em&gt;does&lt;/em&gt; depend on the choice of $x$; if $x$ and $y$ are two distinct elements such that $xHx^{-1}=yHy^{-1}=K$, then the maps $\psi(gH) = gxK$ and $\phi(gH) = gyK$ are the same map if and only if $y^{-1}x\in K$, that is, if and only if $y$ and $x$ are in the same coset of $K$. So you get one map for each coset of $K$ that intersects the set of elements that conjugate $H$ to $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that this does not give you any light into the situation in which $H$ and $K$ are normal, and you are interested in knowing whether the quotient groups $G/H$ and $G/K$ are isomorphic &lt;em&gt;as groups&lt;/em&gt;: obviously, if $H$ and $K$ are normal, then they are conjugate if and only if they are equal, but it is trivial to find examples of two distinct normal sugroups of a finite group $G$ that give you isomorphic quotients. For example, take any two distinct nontrivial proper subgrups of the Klein $4$-group.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T01:14:06.170" LastActivityDate="2010-11-25T01:14:06.170" CommentCount="2" />
  <row Id="11653" PostTypeId="2" ParentId="11650" CreationDate="2010-11-24T05:04:38.730" Score="1" ViewCount="0" Body="&lt;p&gt;Google &quot;cluster algebra&quot;, e.g. see this paper &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0602/0602408v5.pdf&quot; rel=&quot;nofollow&quot;&gt;Combinatorial interpretations for rank-two cluster algebras of affine type.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-24T05:04:38.730" CommentCount="9" />
  <row Id="11654" PostTypeId="1" AcceptedAnswerId="11716" CreationDate="2010-11-24T05:17:50.057" Score="1" ViewCount="149" Body="&lt;p&gt;Let $f(x,y)$, with $0\le x,y\le 1$, be integrable in $y$ for each $x$. Further suppose $\partial f(x,y)/\partial x$ is a bounded function of $(x,y)$. Show that $\partial f(x,y)/\partial x$ is a measurable function of $y$ for each $x$ and&#xA;$$\frac{d}{dx}\int_0^1{f(x,y)\,dy} = \int_0^1{\frac{\partial}{\partial x}f(x,y)\,dy}\,.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I had some work to really show. At this point all I've done is written down each side of the desired equation in terms of limits, and I know ultimately this problem is actually about passing limits inside integrals... But we've done almost no examples of this sort of thing (we literally just finished the limiting theorems today: Fatou, MCT, DCT, etc), and I'm not sure where to begin.&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-26T04:27:09.650" Title="Passing the derivative inside the integral" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11655" PostTypeId="2" ParentId="11651" CreationDate="2010-11-24T05:20:43.807" Score="5" ViewCount="0" Body="&lt;p&gt;So that this does not remain unanswered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}0&amp;amp;1\\1&amp;amp;0\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the simplest matrix that does not possess an $\mathbf L\mathbf D\mathbf L^T$ decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, any symmetric matrix whose leading submatrix (the submatrix formed by the first few rows and columns) is singular will not possess such a decomposition.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-24T05:20:43.807" CommentCount="1" />
  <row Id="11656" PostTypeId="2" ParentId="11654" CreationDate="2010-11-24T05:46:34.500" Score="2" ViewCount="0" Body="&lt;p&gt;You can write the limits of difference quotients as limits of sequences of difference quotients.  The measurability of the partial derivative follows from the fact that a pointwise limit of a sequence of measurable functions is measurable.  The left-hand side of the equation you want to prove is a limit of integrals of difference quotients, and the right-hand side is an integral of limits of difference quotients, so you want to justify passing a limit through an integral.  You can do this because the sequence of difference quotients is uniformly bounded, by the mean value theorem and the hypothesis that $\frac{\partial f}{\partial x}$ is bounded. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-24T05:46:34.500" />
  <row Id="11657" PostTypeId="1" AcceptedAnswerId="11660" CreationDate="2010-11-24T05:58:32.797" Score="0" ViewCount="99" Body="&lt;p&gt;For&lt;br&gt;&#xA;$$f(x,y) = \begin{cases}&#xD;&#xA;\frac{xy}{x^2+y^2} &amp;amp;\text{if }(x,y)\neq (0,0);\\&#xD;&#xA;0 &amp;amp;\text{if }(x,y)=(0,0).\end{cases}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to prove that $\frac{\partial f}{\partial x}$, the partial derivative with respect to $x$, exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the partial derivative freehand I get&lt;br&gt;&#xA;$$\frac{y(x^2+y^2)-xy(2x)}{(x^2+y^2)^2}=  &#xD;&#xA;\frac{y(y^2-x^2)}{(x^2+y^2)^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But using the limits I get&#xA;$$\lim_{x\to c}\frac{f(x,y)-f(c,y)}{x-c} = y(y^2-c^2).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what I'm doing wrong...&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T06:08:30.007" LastActivityDate="2010-11-24T06:29:27.237" Title="Partial Derivative proof" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11658" PostTypeId="2" ParentId="11657" CreationDate="2010-11-24T06:12:34.177" Score="0" ViewCount="0" Body="&lt;p&gt;You have an error in the calculation of the limit (see Arturo's remark about the denominator). When you correct it, you'll get the expression you calculated &lt;i&gt;freehand&lt;/i&gt;, with $c$ substituted for $x$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-24T06:12:34.177" />
  <row Id="11659" PostTypeId="1" AcceptedAnswerId="11664" CreationDate="2010-11-24T06:15:14.050" Score="1" ViewCount="33" Body="&lt;p&gt;How many elements are in the set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{(x,y,z) \mid x+y+z = k \text{ and } x,y,z \in \{0,\dotsc,n\}\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that when $k \le n$, it's $\binom{k+2}{2}$.&#xA;And, by symmetry, the value for $k$ is the same as that for $3n-k$.&#xA;So, I've figured out the two pyramids, and I'm just missing the octahedron in the middle.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-11-24T06:41:14.063" Title="Number of nodes on diagonal planes of a cube" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="11660" PostTypeId="2" ParentId="11657" CreationDate="2010-11-24T06:16:37.003" Score="1" ViewCount="0" Body="&lt;p&gt;I have no idea what you did with the limit. For $y\neq 0$ or $c\neq 0$, we have:&#xA;\begin{align*}&#xA;\lim{x\to c}\frac{f(x,y)-f(c,y)}{x-c} &amp;amp;= \lim_{x\to c}\frac{\frac{xy}{x^2+y^2} - \frac{cy}{c^2+y^2}}{x-c}\\&#xA;&amp;amp;= \lim_{x\to c}\frac{xy(c^2+y^2)-cy(x^2+y^2)}{(c-y)(x^2+y^2)(c^2+y^2)}\\&#xA;&amp;amp;= \lim_{x\to c}\frac{(x-c)(y^3-xyc)}{(x-c)(x^2+y^2)(c^2+y^2)}\\&#xA;&amp;amp;= \lim{x\to c}\frac{y^3-xyc}{(x^2+y^2)(c^2+y^2)}\\&#xA;&amp;amp;= \frac{y(y^2-c^2)}{(c^2+y^2)^2}.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how did you get what you claim you got?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added.&lt;/em&gt; For the partial at $(x,y)=(0,0)$, you have:&#xA;$$\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x-0} = \lim_{x\to 0}\frac{0 - 0}{x} = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the limits exist, the partial derivative exists at each point.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T06:29:27.237" LastActivityDate="2010-11-24T06:29:27.237" CommentCount="2" />
  <row Id="11661" PostTypeId="1" AcceptedAnswerId="11663" CreationDate="2010-11-24T06:30:08.983" Score="0" ViewCount="84" Body="&lt;p&gt;How can I prove that $a^x$ &amp;lt; $a^{x + \Delta x}$, where $\Delta x &amp;gt; 0$, $a$ is a constant, and $a &amp;gt; 1$ (in my case, $a=2$)?&#xA;I don't want to use a graph, of course.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-11-24T06:35:54.327" LastActivityDate="2010-11-24T06:38:40.500" Title="Proof that $a^x$ &lt; $a^{x + \Delta x}$" Tags="&lt;proof&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="11662" PostTypeId="2" ParentId="11602" CreationDate="2010-11-24T06:36:42.197" Score="5" ViewCount="0" Body="&lt;p&gt;$S(n)$ diverges at a rate at least as large as $\log_2 n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n &amp;gt; 2^k$.  Then, for some $1 \leq j \leq n-1$, $j = 2^k$.  Thus&#xA;$$\prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$$S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right) &amp;gt; \sum_{k=1}^{\log_2 (n-1)} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right) = \sum_{k=1}^{\log_2 (n-1)} 1 = \lfloor \log_2 (n-1) \rfloor.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;As far as an upper bound, the following graph is of $2 \log_2 n - S(n)$ for $n \leq 300$.  I conjecture that there exists some $\alpha \approx \frac{2}{3}$ such that $S(n) \leq 2 \log_2 n - \alpha$ for $n \geq 2$ and that $\lim_{n \to \infty} (2 \log_2 n - S(n)) = \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2OH3g.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More numerical evidence: The value of $2 \log_2 n - S(n)$, is, for $n = 1000$, $2000$, and $3000$, respectively, $0.667734$, $0.667494$, and $0.667413$.)&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-24T06:56:50.807" LastActivityDate="2010-11-24T06:56:50.807" CommentCount="9" />
  <row Id="11663" PostTypeId="2" ParentId="11661" CreationDate="2010-11-24T06:38:40.500" Score="2" ViewCount="0" Body="&lt;p&gt;It all depends on what you know. Do you know that if $a\gt 1$ and $b\gt 0$ then $a^b\gt 1$? If so, $2^{x+\Delta x}-2^x = 2^x(2^{\Delta x}-1)$ is a product of positive numbers.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-24T06:38:40.500" />
  <row Id="11664" PostTypeId="2" ParentId="11659" CreationDate="2010-11-24T06:41:14.063" Score="3" ViewCount="0" Body="&lt;p&gt;When $n &amp;lt; k &amp;lt; 2n$, the number of points for which $x,y,z \in \mathbb N$ is $\binom{k+2}{2}$. Of these, $\binom{k-n+1}{2}$ have $x &amp;gt; n$ (as you can see by letting $x = x'+n+1$), and similarly for $y$ and $z$. These are the only exceptions, because when $k &amp;lt; 2n$, more than one of the variables cannot exceed $n$. So the number of points with $x,y,z \le n$ is $\binom{k+2}{2} - 3\binom{k-n+1}{2}$. This looks somewhat asymmetrical, but it's actually equal to $(k-n)(2n-k) + \binom{n+2}{2}$. I like writing it in this form because then it's obvious that it satisfies the &quot;boundary conditions&quot;.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-24T06:41:14.063" />
  <row Id="11665" PostTypeId="1" AcceptedAnswerId="12982" CreationDate="2010-11-24T07:02:41.940" Score="2" ViewCount="185" Body="&lt;p&gt;Let $x$ be a positive rational number of the form $\displaystyle x = \sum\limits_{k=1}^{n}  \frac{a_k}{k!}$ where each $a_k$ is a nonnegative integer with $a_k \leq k-1$ for $k \geq 2$ and $a_n &amp;gt;0$. Prove that $a_1 = [x]$, $a_k = [k!x]-k[(k-1)!x]$ for $k = 2, \dots, n$, and that $n$ is the smallest integer such that $n!x$ is an integer. Conversely, show that every positive rational number can be expressed in this form in one and only one way. Note that $[x]$ is the greatest integer function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think there are two parts to this: (i) an inductive proof and (ii) a proof by contradiction. Would this be the correct &quot;high level&quot; approach to this problem?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-04T15:34:05.293" Title="Rational Numbers and Uniqueness" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11666" PostTypeId="2" ParentId="5627" CreationDate="2010-11-24T07:19:41.617" Score="1" ViewCount="0" Body="&lt;p&gt;For a treatment of these basic facts about what happens to injectivity and surjectivity under composition, see Section 2.5 of &lt;a href=&quot;http://www.math.uga.edu/~pete/3200relationsfunctions.pdf&quot; rel=&quot;nofollow&quot;&gt;these lecture notes&lt;/a&gt; from a &quot;transition to upper level mathematics&quot; course that I taught twice at UGA.  Note that elsewhere in the course there were exercises asking students to prove that certain other statements which are &lt;em&gt;not&lt;/em&gt; listed here are not always true: e.g. if $g \circ f$ is surjective, then $g$ must be surjective (as is in the notes) but $f$ need not be.  Similarly, if $g \circ f$ is injective, then $f$ must be injective (...) but $g$ need not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given how fundamental and ubiquitous these results are, it's surprising that they are not always covered in standard texts.  (I guess the student is supposed to figure them out for herself, but in my experience most actual undergraduates do much better if they are exposed to these concepts in the context of an actual formal course.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-24T07:19:41.617" />
  <row Id="11667" PostTypeId="2" ParentId="11650" CreationDate="2010-11-24T07:37:05.133" Score="6" ViewCount="0" Body="&lt;p&gt;Hyperbolic geometry / algebraic K-theory must mean that $(1+x)/y$ and its iterates appear in the functional equations of one of the dilogarithm functions (Abel's relation or similar).  There is a 5-term functional equation with some cyclic symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Models of QFT may refer to the &quot;pentagon relation&quot; from integrable models.  This again reflects the order 5 property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[added:]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does appear to be what Zagier was pointing to.  Back in the 1990's,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADE functional dilogarithm identities and integrable models&#xA;F. Gliozzi,  R. Tateo&#xA;&lt;a href=&quot;http://xxx.lanl.gov/abs/hep-th/9411203&quot; rel=&quot;nofollow&quot;&gt;http://xxx.lanl.gov/abs/hep-th/9411203&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;formulated a functional equation generalizing some identities for the Rogers dilogarithm evaluated at roots of unity.  The latter identities were known to correspond to torsion in $K$-theory of fields (see papers of Frenkel and Szenes on dilogarithm identities, also early and mid-90's, on arxiv, including proofs of Gliozzi-Tateo formulae and some comments on $K_3(\mathbb{C})$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 4-5 of the paper it is pointed out that the 5-fold periodicity comes from an inspection of the cyclic symmetry of the Abel 5-term functional equation of the Rogers dilogarithm.  If the terms are numbered correctly then $(1+x)/y$ is the generator of this symmetry.  The periodicity of this transformation was probably observed about 200 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cluster algebras appeared later and provided the technology to prove (among many other things) more general periodicity conjectures from integrable models of QFT, going back to earlier papers of Zamolodchikov.  However, the connection of this one period-5 transformation to the dilogarithm is probably easiest to see in the Gliozzi/Tateo paper above.  K-theory and hyperbolic geometry have no known direct connection to cluster algebras, though the latter are related to triangulations of polygons and this is sometimes connected to hyperbolic 2- and 3-dimensional geometry.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-24T17:33:48.167" LastActivityDate="2010-11-24T17:33:48.167" />
  <row Id="11668" PostTypeId="1" CreationDate="2010-11-24T08:39:02.637" Score="1" ViewCount="137" Body="&lt;p&gt;As far as I am concerned, probability &lt;strong&gt;distribution&lt;/strong&gt; function is for discrete random variables while probability &lt;strong&gt;density&lt;/strong&gt; function is for continuous random variables. To find the probability value of continuous random variable, we have to take the total area under the function which differ from discrete random variable, where we can take the probability value directly from the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's my confusion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say Z=X+Y. X is discrete and can take in value of -1 and 1. Y is Gaussian random variable. So,Z is Gaussian random variable with probability &lt;strong&gt;density&lt;/strong&gt; function of p(Z)=p(Z|X=1)+p(Z|X=-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when we try to find p(X=1|Z), it is equals to [p(Z|X=1)p(X=1)]/p(Z).&#xA;My questions, how can we time p(Z|X=1) and p(X=1) since former is probability &lt;strong&gt;density&lt;/strong&gt; function while the latter is probability &lt;strong&gt;distribution&lt;/strong&gt; function? What's more, p(Z) is also a probability &lt;strong&gt;density&lt;/strong&gt; function. In the end, what is p(X=1|Z), a probability &lt;strong&gt;density&lt;/strong&gt; or &lt;strong&gt;distribution&lt;/strong&gt; function?&lt;/p&gt;&#xA;" OwnerUserId="3778" LastActivityDate="2010-11-28T06:14:09.763" Title="Confusion between probability distribution function and probability density function" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11669" PostTypeId="1" AcceptedAnswerId="11671" CreationDate="2010-11-24T09:24:05.267" Score="101" ViewCount="24305" Body="&lt;p&gt;The division of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatic_scale&quot; rel=&quot;nofollow&quot;&gt;chromatic scale&lt;/a&gt; in 7 natural notes (white keys in a piano) and 5 &lt;a href=&quot;http://en.wikipedia.org/wiki/Accidental_%28music%29&quot; rel=&quot;nofollow&quot;&gt;accidental&lt;/a&gt; ones (black) seems a bit arbitrary to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, adjacent notes in a piano (including white or black) are always separated by a semitone. Why the distinction, then? Why not just have scales with 12 notes? (apparently there's a musical scale called &lt;a href=&quot;http://en.wikipedia.org/wiki/Swara&quot; rel=&quot;nofollow&quot;&gt;Swara&lt;/a&gt; that does just that)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've asked several musician friends, but they lack the math preparation for giving me a valid answer. &quot;Notes are like that because they are like that&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need some mathematician with musical knowledge (or a musician with mathematical knowledge) to help me out with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, is there any difference between white and black notes, or do we make the distinction just for historical reasons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I tried to tag this as 'music' but apparently I don't have enough reputation for creating new tags - I had to go with signal-processing. If someone with more reputation could re-tag this question, I'd be grateful)&lt;/p&gt;&#xA;" OwnerUserId="3893" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T09:57:20.713" LastActivityDate="2010-11-27T22:28:27.287" Title="Mathematical difference between white and black notes in a piano?" Tags="&lt;music-theory&gt;" AnswerCount="21" CommentCount="11" FavoriteCount="53" />
  <row Id="11670" PostTypeId="2" ParentId="11668" CreationDate="2010-11-24T09:25:37.827" Score="1" ViewCount="0" Body="&lt;p&gt;You have assumed $X$ to be discrete, so $P(X=1|Z)$ had better be discrete or something really weird is going on. Just because you know $Z$ doesn't still change that $X$ still &quot;lives on&quot; a countable set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will all make a lot more sense after you've had measure theory and you see that it's all really the same thing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the continuous case, the probability that you get an outcome in $B \subset \mathbb{R}$ is $\int_B p(x) \; \mathrm{d}m(x)$, where $p$ is your density and $m$ is the Lebesgue measure. &quot;Of course&quot; this can be rewritten into $\mu(B)$ for some probability measure $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the discrete case, the probability that you get an element in $A \subset \mathbb{N}$ is exactly $\sum_{a \in \mathbb{N}} p(a)$, but this can be rewritten as an integral with respect to the counting measure $c$: $\int_\mathbb{N} p(a) \; \mathrm{d}c(a)$. And this too can be rewritten into a probability measure: $\nu(A)$, where $\nu = \sum_{a \in \mathbb{N}} p(a) \delta_a$, $p$ being your distribution and $\delta_i$ the Dirac measure at $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is &quot;just&quot; integrals with respect to some measure. And finally a word of warning: one shouldn't forget that the statement: &quot;Probability theory is just measure theory in the special case of total measure one&quot; is a gross oversimplification.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-24T09:25:37.827" CommentCount="4" />
  <row Id="11671" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T09:49:53.367" Score="121" ViewCount="0" Body="&lt;p&gt;The first thing you have to understand is that notes are not uniquely defined.  Everything depends on what tuning you use.  I'll assume we're talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Equal_temperament&quot; rel=&quot;nofollow&quot;&gt;equal temperament&lt;/a&gt; here.  In equal temperament, a half-step is the same as a frequency ratio of $\sqrt[12]{2}$; that way, twelve half-steps makes up an octave.  Why twelve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, what we want out of our musical frequencies are nice ratios of small lintegers.  For example, a perfect fifth is supposed to correspond to a frequency ratio of $3 : 2$, or $1.5 : 1$, but in equal temperament it doesn't; instead, it corresponds to a ratio of $2^{ \frac{7}{12} } : 1 \approx 1.498 : 1$.  As you can see, this is not a fifth; however, it is quite close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a perfect fourth is supposed to correspond to a frequency ratio of $4 : 3$, or $1.333... : 1$, but in equal temperament it corresponds to a ratio of $2^{ \frac{5}{12} } : 1 \approx 1.335 : 1$.  Again, this is not a perfect fourth, but is quite close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on.  What's going on here is a massively convenient mathematical coincidence: several of the powers of $\sqrt[12]{2}$ happen to be good approximations to ratios of small integers, and there are enough of these to play Western music.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how this coincidence works.  You get the white keys from $C$ using (part of) the circle of fifths.  Start with $C$ and go up a fifth to get $G$, then $D$, then $A$, then $E$, then $B$.  Then go down a fifth to get $F$.  These are the &quot;neighbors&quot; of $C$ in the circle of fifths.  You get the black keys from here using the rest of the circle of fifths.  After you've gone up a &quot;perfect&quot; perfect fifth twelve times, you get a frequency ratio of $3^{12} : 2^{12} \approx 129.7 : 1$.  This happens to be rather close to $2^7 : 1$, or seven octaves!  And if we replace $3 : 2$ by $2^{ \frac{7}{12} } : 1$, then we get exactly seven octaves.  In other words, the reason you can afford to identify these intervals is because $3^{12}$ happens to be rather close to $2^{19}$.  Said another way, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_2 3 \approx \frac{19}{12}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;happens to be a good rational approximation, and this is the main basis of equal temperament.  (The other main coincidence here is that $\log_2 \frac{5}{4} \approx \frac{4}{12}$; this is what allows us to squeeze major thirds into equal temperament as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fundamental fact of music that $\log_2 3$ is irrational, so it is impossible for any kind of equal temperament to have &quot;perfect&quot; perfect fifths regardless of how many notes you use.  However, you can write down good rational approximations by looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fraction&lt;/a&gt; of $\log_2 3$ and writing down convergents, and these will correspond to equal-tempered scales with more notes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can use other types of temperament, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Well_temperament&quot; rel=&quot;nofollow&quot;&gt;well temperament&lt;/a&gt;; if you stick to $12$ notes (which not everybody does!), you will be forced to make some intervals sound better and some intervals sound worse.  In particular, if you don't use equal temperament then &lt;strong&gt;different keys sound different.&lt;/strong&gt;  This is a major reason many Western composers composed in different keys; during their time, this actually made a difference.  As a result when you're playing certain sufficiently old pieces you aren't actually playing them as they were intended to be heard - you're using the wrong tuning.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  I suppose it is also good to say something about why we care about frequency ratios which are ratios of small integers.  This has to do with the physics of sound, and I'm not particularly knowledgeable here, but this is my understanding of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably know that sound is a wave.  More precisely, sound is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Longitudinal_wave&quot; rel=&quot;nofollow&quot;&gt;longitudinal wave&lt;/a&gt; carried by air molecules.  You might think that there is a simple equation for the sound created by a single note, perhaps $\sin 2\pi f t$ if the corresponding tone has frequency $f$.  Actually this only occurs for tones which are produced electronically; any tone you produce in nature carries with it &lt;a href=&quot;http://en.wikipedia.org/wiki/Overtone&quot; rel=&quot;nofollow&quot;&gt;overtones&lt;/a&gt; and has a Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum \left( a_n \sin 2 \pi n f t + b_n \cos 2 \pi n f t \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficients $a_n, b_n$ determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Timbre&quot; rel=&quot;nofollow&quot;&gt;timbre&lt;/a&gt; of the sound; this is why different instruments sound different even when they play the same notes, and has to do with the physics of vibration, which I don't understand too well.  So any tone which you hear at frequency $f$ almost certainly also has components at frequency $2f, 3f, 4f, ...$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you play two notes of frequencies $f, f&amp;#39;$ together, then the resulting sound corresponds to what you get when you add their Fourier series.  Now it's not hard to see that if $\frac{f}{f&amp;#39;}$ is a ratio of small integers, then many (but not all) of the overtones will match in frequency with each other; the result sounds a more complex note with certain overtones.  Otherwise, you get dissonance as you hear both types of overtones simultaneously and their frequencies will be similar, but not similar enough.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  You should probably check out &lt;a href=&quot;http://www.maths.abdn.ac.uk/~bensondj/html/maths-music.html&quot; rel=&quot;nofollow&quot;&gt;the book Rahul Narain recommended&lt;/a&gt; in the comments for the full story.  There was a lot I didn't know, and I'm only in the introduction!&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-26T04:07:28.117" LastActivityDate="2010-11-26T04:07:28.117" CommentCount="13" />
  <row Id="11672" PostTypeId="2" ParentId="11668" CreationDate="2010-11-24T10:04:37.307" Score="1" ViewCount="0" Body="&lt;p&gt;Let $X$ be random variable over $\mathbb{Z}$ with probability weights $p_k = Pr[X=k]$. Many discrete scenario can be expressed in these terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability distribution function is by virtue of definition $F(x) = Pr[X \leq x] = \sum\limits_{k=-\infty}^{\lfloor x \rfloor}Pr[X=k] = \sum\limits_{k=-\infty}^{\lfloor x \rfloor} p_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A density function corresponds to the weights $p_k$ while the definition of distribution function remains the same in the continuous case, only that $Pr[X \leq x]$ has to be calculated by an integral instead of a sum. See kahen's answer for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, you should not exchange distribution function and density. They do different things.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-11-24T10:04:37.307" CommentCount="4" />
  <row Id="11673" PostTypeId="1" CreationDate="2010-11-24T10:08:28.270" Score="0" ViewCount="73" Body="&lt;p&gt;The following result must be well known, but it is instructive to give it here in view of the recent posts &lt;a href=&quot;http://math.stackexchange.com/questions/11487/y-xn-what-is-exyy&quot;&gt;Y= X+N what is E(X|Y=y)&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/47168/ex-1-x-1-x-2-where-x-i-are-integrable-independent-infinitely-divisib/47204#47204&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/47168/ex-1-x-1-x-2-where-x-i-are-integrable-independent-infinitely-divisib/47204#47204&lt;/a&gt;. This is also an interesting exercise in its own right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X_1$ and $X_2$ are independent ${\rm Gamma}(c_i,\lambda)$ rv's, meaning that $X_i$, $i=1,2$, has density function $f_{X_i } (x) = \lambda ^{c_i } {\rm e}^{ - \lambda x} x^{c_i  - 1} /\Gamma (c_i )$, $x &amp;gt; 0$ ($c_i$ are positive constants, $\Gamma$ is the gamma function). Show that&#xA;$$&#xD;&#xA;{\rm E}(X_1 | X_1 + X_2 = z) = \frac{{c_1 }}{{c_1  + c_2 }} z.&#xD;&#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-27T04:39:49.707" Title="$E(X_1 | X_1 + X_2)$ for independent gamma random variables" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11674" PostTypeId="1" CreationDate="2010-11-24T10:31:14.427" Score="4" ViewCount="64" Body="&lt;p&gt;For Riemann zeta function and $L$-functions of number fields, the width of critical strip is $1$. For $L$-functions of modular forms of weight $k$, the width of the critical strip is $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there a variation in the width of the critical strip for various $L$-functions? Is there a conceptual explanation or an underlying heuristics? &lt;/p&gt;&#xA;" OwnerUserId="3894" LastActivityDate="2010-11-24T11:00:24.293" Title="Why is width of critical strip what it is?" Tags="&lt;analytic-number-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="11675" PostTypeId="1" CreationDate="2010-11-24T10:42:43.470" Score="3" ViewCount="105" Body="&lt;p&gt;For proving the quadratic reciprocity, Gauss sums are very useful. However this seems an ad-hoc construction. Is this useful in a wider context? What are some other uses for Gauss sums?&lt;/p&gt;&#xA;" OwnerUserId="3895" LastActivityDate="2010-11-24T17:03:56.313" Title="Applications of Gauss sums" Tags="&lt;number-theory&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11676" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T10:45:47.540" Score="3" ViewCount="0" Body="&lt;p&gt;I'm no musician, but as far as I know, audio waves are felt only then &quot;round/sound&quot;, iff they repeat faster than a specific frequency. That frequency is probably that of our brain waves: being awake and in a non-meditating state, that is faster than like 18 or more Hz; neither can you shiver faster nor hear lower frequencies than your brain waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Audio waves have a length of $$\mathrm{lcm}\{m,n\}&amp;#183;2\pi$$ if they are of the shape $$a_1&amp;#183;\sin(m&amp;#183;2\pi&amp;#183;t+s_1)+a_2&amp;#183;\sin(n&amp;#183;2\pi&amp;#183;t+s_2)$$. The notes double their frequency each octave; therefore they have a logarithmic scale and not a linear. Good violin and harp players can play all fitting (&quot;sound&quot; sounding) combination of frequencies, but instruments with keys lack the variety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Qiaochu Yuan did answer faster than me, while I was on the phone. Seems to be more complete than I could have answered. I have nothing to add.)&lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T03:20:39.270" LastActivityDate="2010-11-26T03:20:39.270" />
  <row Id="11677" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T10:55:36.000" Score="5" ViewCount="0" Body="&lt;p&gt;Not just &lt;strong&gt;quadratic&lt;/strong&gt; reciprocity, one can use them to prove higher&#xA;reciprocity laws: see Ireland and Rosen's &lt;em&gt;A Classical Introduction&#xA;to Modern Number Theory&lt;/em&gt;. They also turn up in the functional&#xA;equation for Dirichlet L-functions (and are massively generalized in the&#xA;topic of root numbers).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T11:15:31.403" LastActivityDate="2010-11-24T11:15:31.403" />
  <row Id="11678" PostTypeId="2" ParentId="11674" CreationDate="2010-11-24T11:00:24.293" Score="5" ViewCount="0" Body="&lt;p&gt;Roughly speaking the critical strip is where the $L$-function&#xA;is hard to compute/understand. Its right edge is the boundary&#xA;of the region where the Dirichlet series/Euler product converges&#xA;nicely (absolutely, locally uniformly etc). Its left edge&#xA;is the image of the right edge under the functional equation&#xA;(to compute it to the left of the critical strip you can use&#xA;the functional equation to reduce it to comuting a nice series/product).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-24T11:00:24.293" />
  <row Id="11679" PostTypeId="1" CreationDate="2010-11-24T11:11:59.440" Score="0" ViewCount="48" Body="&lt;p&gt;The context is the definition of Hecke Größencharakter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hecke_character&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hecke_character&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is supposed to generalize the Dirichlet $L$-series for number fields. Dirichlet characters are characters of the multiplicative groups of $\mathbb Z/p\mathbb Z$. An appropriate generalization would be instead to consider characters of the multiplicative group of $cal O_K/\cal P$ where $\cal P$ is a prime ideal in the ring of integers of a number field $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Hecke Größencharakter goes to more trouble than this. It brings in ideles and such for a more complicated generalization. Why is this necessary?&lt;/p&gt;&#xA;" OwnerUserId="3895" LastActivityDate="2010-11-24T11:19:51.820" Title="Motivation for Hecke characters" Tags="&lt;analytic-number-theory&gt;" AnswerCount="1" />
  <row Id="11680" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T11:17:28.050" Score="2" ViewCount="0" Body="&lt;p&gt;It can be used to prove a certain theorem called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stickelberger_theorem&quot; rel=&quot;nofollow&quot;&gt;Stickelberger's theorem&lt;/a&gt;, the likes of which can in turn be used for proving things like &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_conjecture&quot; rel=&quot;nofollow&quot;&gt;Catalan's Conjecture&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-24T11:17:28.050" />
  <row Id="11681" PostTypeId="2" ParentId="11679" CreationDate="2010-11-24T11:19:51.820" Score="1" ViewCount="0" Body="&lt;p&gt;One reason for the greater generality is that Hecke&#xA;characters do more than describe Abelian extensions of number fields&#xA;(essentially Dirichlet characters describe Abelian extenions&#xA;of $\mathbb Q$). For instance the L-function of an elliptic&#xA;curve with complex multiplication is the L-function of a Hecke&#xA;character of infinite order.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-24T11:19:51.820" />
  <row Id="11682" PostTypeId="1" CreationDate="2010-11-24T11:28:18.193" Score="0" ViewCount="57" Body="&lt;p&gt;Let X be a topological space and A be an abelian group. Give A the discrete topology. For any open set U of X, Let $\cal A(U)$ be the group of all continuous aps of U into A. Thus with the usual restriction maps we obtain a sheaf $\cal A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why for every connected open set U, $\cal A(U)\cong A$ for all U?&#xA; And, what is the sheafifcation of the presheaf $U \mapsto A$?&lt;/p&gt;&#xA;" OwnerUserId="3897" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T17:22:55.067" LastActivityDate="2010-11-24T17:22:55.067" Title="the sheafification of a constant presheaf" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="11683" PostTypeId="1" CreationDate="2010-11-24T11:46:31.063" Score="0" ViewCount="93" Body="&lt;p&gt;I need to solve this equation. Please help me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{-\mathrm d^2u}{\mathrm dt^2}+\dfrac{u}{LC}=\dfrac{V}{LC},\qquad L,C,V,I=\text{constant}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $u(0)=0$, $\left. \dfrac{\mathrm du}{\mathrm dt}\right\vert_{t=0}=\dfrac{I}{C}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3898" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-24T16:08:24.397" LastActivityDate="2010-11-24T16:08:24.397" Title="Solution for the equation $\frac{-\mathrm d^2u}{\mathrm dt^2}+\frac{u}{LC}=\frac{V}{LC}$" Tags="&lt;differential-equations&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11684" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T11:47:18.227" Score="1" ViewCount="0" Body="&lt;p&gt;You can apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Ruffini%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Ruffini's rule&lt;/a&gt;. Here is a copy from my Algebra text book (Compêndio de Álgebra, VI, by Sebastião e Silva and Silva Paulo) where the following formula is obtained:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1}).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HtIOy.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translation: The Ruffini's rule can be used to find the quotient of $x^n-a^n$ by $x-a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Figure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $n$ is a natural number, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1})$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-24T15:29:24.757" LastActivityDate="2010-11-24T15:29:24.757" CommentCount="3" />
  <row Id="11685" PostTypeId="2" ParentId="4619" CreationDate="2010-11-24T12:01:14.003" Score="0" ViewCount="0" Body="&lt;p&gt;An example of a decidable, NP-Hard but not NP-Complete problem is the Towers of Hanoi problem. Intuitively, the T.of H. problem is not in NP because the exploration of all possible configurations cannot be done in polynomial time, on a non-deterministic machine.&lt;/p&gt;&#xA;" OwnerUserId="3899" LastActivityDate="2010-11-24T12:01:14.003" CommentCount="1" />
  <row Id="11686" PostTypeId="1" AcceptedAnswerId="11691" CreationDate="2010-11-24T12:57:27.540" Score="2" ViewCount="232" Body="&lt;p&gt;Is there a rational number (with denominator not greater than 200) between 15/106 and 16/113? &lt;/p&gt;&#xA;" OwnerUserId="3901" LastActivityDate="2011-01-08T12:14:04.437" Title="Is there a rational number (with denominator not greater than 200) between 15/106 and 16/113? " Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;divisibility&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="11687" PostTypeId="2" ParentId="11682" CreationDate="2010-11-24T13:09:44.737" Score="2" ViewCount="0" Body="&lt;p&gt;Some hints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your first question: $A$ has the discrete topology, so elements of $A$ are open and closed. Then, what properties does the preimage of a single element under a continuous map $U \to A$ have? Recall the definition of connectedness. What does this imply for continuous maps $U \to A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, write out explicitly the definition of the sheafification functor for your example and see that the result you will get is exactly the sheaf you described in the first part of your question.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2010-11-24T13:09:44.737" />
  <row Id="11688" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T13:44:16.563" Score="6" ViewCount="0" Body="&lt;p&gt;A brute-force search in &lt;em&gt;Mathematica&lt;/em&gt;, using&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Select[Flatten[Table[m/n, {n, 2, 200}, {m, 1, n - 1}]], (15/106 &amp;lt; # &amp;lt; 16/113) &amp;amp;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;indicates that there is no rational number satisfying your requirement.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-24T13:44:16.563" CommentCount="5" />
  <row Id="11689" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T13:44:53.840" Score="5" ViewCount="0" Body="&lt;p&gt;Just to let you see that other tunings are possible and thus other keyboards:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.kylegann.com/tuning.html&quot; rel=&quot;nofollow&quot;&gt;http://www.kylegann.com/tuning.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-24T13:44:53.840" />
  <row Id="11690" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T13:45:22.360" Score="10" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $Google &quot;farey sequence&quot; and &quot;mediant&quot; and / or see my many sci.math posts on these topics, e.g. see &lt;a href=&quot;http://groups.google.com/groups?selm=y8zpsy7ixpf.fsf%40nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; which explains the lucid lattice geometric viewpoint, and see &lt;a href=&quot;http://groups.google.com/groups?selm=y8zwv4xtxj3.fsf%40nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; on how to apply this to perform Farey mediant binary searching to find the simplest fraction(s) between two fractions. See also my &lt;a href=&quot;http://groups.google.com/groups/search?q=group%3a%2Amath%2A+author%3Adubuque+%28farey+OR+mediant%29&quot; rel=&quot;nofollow&quot;&gt;other posts&lt;/a&gt; on these topics, which includes the fascinating connection with &lt;a href=&quot;http://en.wikipedia.org/wiki/Pick%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Pick's theorem&lt;/a&gt; on the area of lattice polygons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ \ $ If someone here is handy with graphing programs and/or Java you'd be doing a great service if you could draw/animate the mediant lattice construction depicted in ascii art in my first post. Or perhaps someone has done this elsewhere? That yields a &quot;proof without words&quot; that succinctly captures the essence of the matter.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T15:05:04.850" LastActivityDate="2010-11-24T15:05:04.850" CommentCount="1" />
  <row Id="11691" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T13:47:18.153" Score="11" ViewCount="0" Body="&lt;p&gt;The sandwich ${15\over 106}&amp;lt;{p\over q}&amp;lt; {16\over 113}$ implies $15 q&amp;lt;106p$ and $16q&amp;gt;113p$, so necessarily $q&amp;gt;7p$. Write $q=7p+ q_1$ with $q_1&amp;gt;0$. It follows that $15q_1&amp;lt;p$ and $16 q_1&amp;gt;p$, so necessarily $p= 15 q_1+p_1$ with $0&amp;lt;p_1&amp;lt;q_1$. The optimal choice is $p_1=1$, $q_1=2$ from which we get $p=31$, $q=219&amp;gt;200$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-24T13:47:18.153" CommentCount="2" />
  <row Id="11692" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T13:53:01.273" Score="5" ViewCount="0" Body="&lt;p&gt;Gauss sums are not an ad-hoc construction!  I know two ways to motivate the definition, one of which requires that you know a little Galois theory and the other which is totally mysterious to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the Galois-theoretic explanation.  Let $\zeta_p$ be a primitive $p^{th}$ root of unity, for $p$ prime.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotomic_field&quot; rel=&quot;nofollow&quot;&gt;cyclotomic field&lt;/a&gt; $\mathbb{Q}(\zeta_p)$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_extension&quot; rel=&quot;nofollow&quot;&gt;Galois&lt;/a&gt;, so one can define its &lt;a href=&quot;http://en.wikipedia.org/wiki/Galois_group&quot; rel=&quot;nofollow&quot;&gt;Galois group&lt;/a&gt;, the group of all field automorphisms which preserve $\mathbb{Q}$.  Such an automorphism is determined by what it does to $\zeta_p$, and it must send $\zeta_p$ to another primitive $p^{th}$ root of unity.  It follows that the Galois group $G = \text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\times}$, which is cyclic of order $p-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $p$ is odd.  As a cyclic group of even order, $G$ has a unique subgroup $H$ of index two given precisely by the multiplicative group of quadratic residues $\bmod p$, so by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of Galois theory&lt;/a&gt; the fixed field $\mathbb{Q}(\zeta_p)^H$ is the unique quadratic subextension of $\mathbb{Q}(\zeta_p)$.  And it's not hard to see that this unique quadratic subextension must be generated by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{\sigma \in H} \sigma(\zeta_p) = \sum_{a \text{ is a QR}} \zeta_p^a = \frac{1}{2} \left( \sum_{a=1}^{p-1} \zeta_p^{a^2} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you will of course recognize as a Gauss sum!  So the Gauss sum generates a quadratic subextension, and any of various methods will tell you that this subextension is precisely $\mathbb{Q}(\sqrt{p^{\ast}})$ where $p^{\ast} = (-1)^{ \frac{p-1}{2} } p$.  (This does not actually require any computation: if you know enough algebraic number theory, it follows from a consideration of which primes ramify in cyclotomic extensions.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The totally mysterious explanation is that Gauss sums naturally appear when you start thinking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;discrete Fourier transform&lt;/a&gt;.  For example, the trace of the DFT matrix is a Gauss sum.  But more mysteriously, Gauss sums are eigenfunctions of the DFT in a certain sense.  (I sketch how this works &lt;a href=&quot;http://qchu.wordpress.com/2010/01/11/some-quadratic-reciprocity/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)  There is a sort of mysterious connection here to the Gaussian distribution, which is an eigenfunction of the continuous Fourier transform; see &lt;a href=&quot;http://mathoverflow.net/questions/11053/whats-the-relationship-between-gauss-sums-and-the-normal-distribution&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;.  Again, I don't know what to make of this.  There is a book by Berg called &lt;a href=&quot;http://books.google.com/books?id=fxVZybOHF8EC&amp;amp;pg=PR17&amp;amp;lpg=PR17&amp;amp;dq=fourier-theoretic+proof+of+quadratic+reciprocity&amp;amp;source=bl&amp;amp;ots=o4aiY9ferP&amp;amp;sig=Wj7JvvAdomtfCwy-UAFfgOKYPWs&amp;amp;hl=en&amp;amp;ei=ARntTKCBF6GShAf_6eTNDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBIQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;The Fourier-analytic proof of quadratic reciprocity&lt;/a&gt; and it may or may not be about this construction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-24T13:53:01.273" />
  <row Id="11693" PostTypeId="1" CreationDate="2010-11-24T13:57:56.773" Score="9" ViewCount="104" Body="&lt;p&gt;The Manin-Drinfeld theorem asserts that for a modular curve $X_0(N)$ and Jacobian $J_0(N)$ with the former being embdedded in the latter under the map that takes $i\infty$ to $0$, the cusps are torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof Manin-Drinfeld theorem seems to use Hecke operators and thus seems to be valid only for congruence subgroups of $PSL_2(\mathbb Z)$. Is there possibly a finite-index subgroup of $PSL_2(\mathbb Z)$ such that the statement of Manin-Drinfeld theorem is still true for the quotient? &lt;/p&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2010-11-27T06:58:21.037" Title="Generality of the Manin-Drinfel'd theorem" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="11694" PostTypeId="2" ParentId="11669" CreationDate="2010-11-24T14:02:58.830" Score="2" ViewCount="0" Body="&lt;p&gt;If you only had a repetitive series of keys on your piano it would be a bit difficult to visually get some reference points. I think this is the main reason wh&lt;/p&gt;&#xA;" OwnerUserId="3904" LastActivityDate="2010-11-24T14:02:58.830" />
  <row Id="11695" PostTypeId="2" ParentId="11618" CreationDate="2010-11-24T14:05:00.467" Score="-1" ViewCount="0" Body="&lt;p&gt;If you know the sum of a geometric sequence, then set $x=a/b$ and conclude&#xA;$&#xD;&#xA;x^n-1 = (x-1)(1+x+x^2+\cdots+x^{n-1})&#xD;&#xA;$. Now multiply by $b^n$. (If $b=0$, then the identity is obvious.)&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-11-24T14:05:00.467" CommentCount="1" />
  <row Id="11696" PostTypeId="1" CreationDate="2010-11-24T14:15:01.887" Score="0" ViewCount="119" Body="&lt;p&gt;The Coordinates of two points are A(-2,6) and B(9,3).  Find the coordinates of the point C on the x-axis such that AC = BC&lt;/p&gt;&#xA;" OwnerUserId="3905" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-24T14:20:22.367" LastActivityDate="2010-11-24T15:52:54.900" Title="Find the coordinates of a point equidistant from two given points" Tags="&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11697" PostTypeId="2" ParentId="11673" CreationDate="2010-11-24T14:34:07.723" Score="0" ViewCount="0" Body="&lt;p&gt;First of all, we note that $X_1 + X_2 \sim {\rm Gamma}(c_1 + c_2 , \lambda)$.&#xA;Given that $X_1 + X_2 = z$, $X_1$ cannot exceed $z$. We can thus find ${\rm E}(X_1|X_1+X_2=z)$ as follows.&#xA;$$&#xD;&#xA;{\rm E}(X_1|X_1 + X_2 = z) = \int_{0}^z  {xf_{X_1|X_1 + X_2} (x|z)\,{\rm d}x} = \int_{0}^z  {x\frac{{f_{X_1} (x)f_{X_1 + X_2|X_1} (z|x)}}{{f_{X_1 + X_2} (z)}}\, {\rm d}x}. &#xD;&#xA;$$&#xD;&#xA;Noting that $f_{X_1 + X_2|X_1} (z|x) = f_{X_2} (z-x)$, it follows after some algebra and a change of variable that the right-hand side integral is equal to&#xA;$$&#xD;&#xA;z\frac{{\Gamma (c_1  + c_2 )}}{{\Gamma (c_1 )\Gamma (c_2 )}}\int_0^1 {x^{c_1 } (1 - x)^{c_2  - 1} \,{\rm d}x}  = z\frac{{\Gamma (c_1  + c_2 )}}{{\Gamma (c_1 )\Gamma (c_2 )}}{\rm B}(c_1  + 1,c_2 ),&#xD;&#xA;$$&#xD;&#xA;where ${\rm B}(a,b) = \int_0^1 {t^{a - 1} (1 - t)^{b - 1} \,{\rm d}t}$ is the beta function. Finally, from the alternative form ${\rm B}(a,b) = \frac{{\Gamma (a)\Gamma (b)}}{{\Gamma (a + b)}}$ followed by $\Gamma (p+1) = p \Gamma (p)$, we obtain&#xA;$$&#xD;&#xA;{\rm E}(X_1|X_1 + X_2 = z) = \frac{{c_1 }}{{c_1  + c_2 }} z.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-24T14:34:07.723" />
  <row Id="11698" PostTypeId="1" CreationDate="2010-11-24T14:44:30.650" Score="0" ViewCount="51" Body="&lt;p&gt;If I have an arc (which comes as part of an ellipse), can I represent it (or at least closely approximate) by cubic Bezier curve? And if yes, how can I calculate control points for that Bezier curve?&lt;/p&gt;&#xA;" OwnerUserId="3777" LastEditorUserId="3777" LastEditorDisplayName="" LastEditDate="2010-11-24T15:26:48.757" LastActivityDate="2010-11-24T15:26:48.757" Title="How elliptic arc can be represented by cubic Bezier curve?" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;&lt;elliptic-curves&gt;&lt;ellipse&gt;" CommentCount="3" />
  <row Id="11699" PostTypeId="2" ParentId="11683" CreationDate="2010-11-24T14:45:22.737" Score="0" ViewCount="0" Body="&lt;p&gt;Hint:  What functions do you know whose second derivative is proportional to the function?  There should be a two-dimensional solution space to the homogeneous equation (where the RHS is set to zero)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-24T14:45:22.737" />
  <row Id="11700" PostTypeId="2" ParentId="11683" CreationDate="2010-11-24T14:49:52.833" Score="2" ViewCount="0" Body="&lt;p&gt;Here are some hints. It's a standard form of 2nd order ordinary differential equation. You first find a &quot;particular solution&quot; $u_p$ to your equation (not including your conditions on $u(0)$ and ${du \over dt}(0)$). In this case it's just a constant solution. Then you find the general solution to the homogeneous equation $-{d^2 u \over dt} + {1 \over LC} u = 0$. This will be of the form $a u_1(t) + b u_2(t)$ for the right two functions, which will be certain exponentials here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then any solution to $-{d^2 u \over dt} + {1 \over LC} u = {V \over LC}$ will be of the form $a u_1(t) + b u_2(t) + u_p(t)$. Plug in the conditions on $u(0)$ and ${du \over dt}(0)$ to figure which $a$ and $b$ correspond to your situation. Voila, you're done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-24T14:49:52.833" />
  <row Id="11701" PostTypeId="2" ParentId="7570" CreationDate="2010-11-24T14:50:17.427" Score="0" ViewCount="0" Body="&lt;p&gt;If they commute with each other we can interchange the limits........&lt;/p&gt;&#xA;" OwnerUserId="3908" LastActivityDate="2010-11-24T14:50:17.427" />
  <row Id="11702" PostTypeId="2" ParentId="7570" CreationDate="2010-11-24T14:55:56.357" Score="1" ViewCount="0" Body="&lt;p&gt;If the total sum converges absolutely, you can reorder the terms however you want.  Can you prove that?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-24T14:55:56.357" />
  <row Id="11703" PostTypeId="1" AcceptedAnswerId="11704" CreationDate="2010-11-24T14:56:28.337" Score="0" ViewCount="244" Body="&lt;p&gt;Proof to the formula &#xA;$$1+x+x^2+x^3+\cdots+x^n = \frac{x^n-1}{x-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="3907" LastEditorUserId="3" LastEditorDisplayName="" LastEditDate="2010-12-14T15:40:49.413" LastActivityDate="2010-12-14T15:58:54.113" Title="proof of the formula $1+x+x^2+x^3+...+x^n =\frac{x^{n+1}-1}{x-1}$" Tags="&lt;algebra&gt;&lt;proof&gt;" AnswerCount="7" CommentCount="3" />
  <row Id="11704" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T15:02:54.680" Score="4" ViewCount="0" Body="&lt;p&gt;Let $S=1+x+x^2+...+x^n$. Then, $xS=x+x^2+...+x^{n+1}=1+x+x^2+...+x^n+(x^{n+1}-1)=S+x^{n+1}-1$. So, $xS-S=x^{n+1}-1$. So, $S=\frac{x^{n+1}-1}{x-1}$. (The exponent of the $x$ in the numerator of the RHS should be $n+1$ not $n$).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-24T15:02:54.680" />
  <row Id="11705" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T15:10:23.360" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ The sum $\rm\:S\:$ is &quot;almost&quot; preserved by a shift symmetry $\rm\ S \to x\:S$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examine the discrepancy $\rm\ x\:S - S\:.\ \ $ It's just the finite case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s%5Fparadox%5Fof%5Fthe%5FGrand%5FHotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's infinite Hotel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T15:15:42.807" LastActivityDate="2010-11-24T15:15:42.807" />
  <row Id="11706" PostTypeId="1" CreationDate="2010-11-24T15:29:47.680" Score="2" ViewCount="163" Body="&lt;p&gt;In English (I think this is universal anyway) we use the 1s, 10s, and 100s in a cycle. One, one thousand, one million; twenty two, twenty two thousand, twenty two million; one hundred and forty six, one hundred and forty six thousand, one hundred and forty six million, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who decided to cycle every (3x+1)th digit?&lt;/p&gt;&#xA;" OwnerUserId="3911" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-11-24T15:53:01.650" LastActivityDate="2010-11-24T17:37:15.740" Title="Why do our number start over at million, billion, etc." Tags="&lt;soft-question&gt;&lt;math-history&gt;&lt;counting&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="11707" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T15:31:06.573" Score="1" ViewCount="0" Body="&lt;p&gt;It equals (x^(n+1)-1)/(x-1), not what you wrote.&lt;/p&gt;&#xA;" OwnerUserId="3913" LastActivityDate="2010-11-24T15:31:06.573" CommentCount="1" />
  <row Id="11708" PostTypeId="2" ParentId="11696" CreationDate="2010-11-24T15:52:54.900" Score="3" ViewCount="0" Body="&lt;p&gt;A general approach: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two points $A$ and $B$, the line that is perpendicular to the segment $AB$ and goes through its midpoint is &lt;em&gt;precisely&lt;/em&gt; the set of points that are equidistant from $A$ and $B$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-24T15:52:54.900" CommentCount="3" />
  <row Id="11709" PostTypeId="2" ParentId="11686" CreationDate="2010-11-24T15:54:17.200" Score="5" ViewCount="0" Body="&lt;p&gt;Any rational number between 15/106 and 16/113 is closer to their mean than either of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this mean 3391/(2*106*113) has denominator larger than 200.  However if we examine the best approximations to this mean by rational numbers of increasing denominators, we will find if there are any before the denominator 200 is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say a rational number $p/q$ is a &lt;strong&gt;best rational approximation&lt;/strong&gt; to real number $x$ iff no rational number with denominator of smaller absolute value is closer to x.  Without loss of generality we restrict discussion to the case p,q,x &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of such best rational approximations can be derived from &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt; the convergents of its continued fraction expansion&lt;/a&gt;.  The convergents are themselves entries in this sequence, and the only other entries are &quot;interpolants&quot; between the convergents in the following sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $p_1/q_1$ and $p_2/q_2$ are a consecutive pair of convergents to $x$ (in the terminology of continued fractions).  The next convergent $p_3/q_3$ will by definition have the form $(p_1 + d*p_2)/(q_1 + d*q_2)$ where $d$ is a positive integer unless already $x$ = $p_2/q_2$.  For integers $a = 1,...,d-1$ the &quot;semiconvergents&quot; $(p_1 + a*p_2)/(q_1 + a*q_2)$ are also best rational approximations, and having denominators between $q_2$ and $q_3$, they furnish the entries in the sequence of best rational approximations between $p_2/q_2$ and $p_3/q_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We thus begin by &lt;a href=&quot;http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfCALC.html&quot; rel=&quot;nofollow&quot;&gt; forming the convergents&lt;/a&gt; to 3391/(2*106*113) until denominators exceed 200:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0/1&lt;br&gt;&#xA;1/7&lt;br&gt;&#xA;15/106&lt;br&gt;&#xA;31/219&lt;br&gt;&#xA;...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the only semiconvergent between the last two convergents is exactly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16/113 = (1+15)/(7+106)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves there is no rational number between 15/106 and 16/113 with denominator less than the next entry in the sequence 31/219, whose denominator exceeds 200.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-24T15:54:17.200" CommentCount="1" />
  <row Id="11710" PostTypeId="2" ParentId="11639" CreationDate="2010-11-24T15:55:15.157" Score="1" ViewCount="0" Body="&lt;p&gt;Even for planar 3-connected graphs, which are the vertex-edge graphs that arise from convex 3-dimensional polyhedra, one can have all four possibilities: non-HC and non-Eulerian graphs, non-HC and Eulerian graphs; HC and Non-Eulerian graphs; HC and and Euler graphs, where HC means has a Hamiltonian circuit, and Eulerian means has an Eulerian circuit. (It is well known that a connected graph has an Eulerian circuit if and only if it is even-valent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for planar and 3-connected graphs it is of considerable interest to look at operations on the graphs that preserve the property of having an HC or Eulerian circuit or special kinds of Eulerian circuits. For example if G is plane, 3-connected and has an HC when does the dual of G have an HC? When does the medial graph for G have an HC, etc.? A famous problem which remains open is known as Barnette's conjecture: Does every 3-valent, bipartite, plane, 3-connected graph have an HC? It is also unkown if all the planar 3-connected 3-valent graphs with exactly 12 pentagons and some number of hexagonal faces have HC's.  (There graphs are known as the fullerenes.)&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-24T15:55:15.157" />
  <row Id="11711" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T16:20:53.090" Score="2" ViewCount="0" Body="&lt;p&gt;Observe that&#xA;\begin{eqnarray}&#xA;x^{n+1} - 1 = x^{n+1} + (x^{n} - x^{n}) + \cdots + (x - x) - 1 = (x^{n} + x^{n-1} + \cdots + x + 1)(x - 1).&#xA;\end{eqnarray}&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-11-24T16:20:53.090" />
  <row Id="11714" PostTypeId="1" AcceptedAnswerId="11727" CreationDate="2010-11-24T16:59:56.897" Score="2" ViewCount="99" Body="&lt;p&gt;Factoring a rational function as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(x-a)(x-b)\cdots(x-z)}{(x-A)(x-B)\cdots(x-Z)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly displays the x-intercepts (roots) and vertical asymptotes (poles) of a rational function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, dividing the denominator D into the numerator N to get N = DQ+R and writing N/D as Q + R/D clearly displays the end-behavior (horizontal or even diagonal asymptotes and so forth).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Is there some similar factorization that displays the local minimum and maximums of the real-valued real-input rational function?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, if the local extrema occur at a,b,…,z then the expression for the function has the a,b,…,z explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only example I can think of is &quot;completing the square&quot; exhibits the vertex of a parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; After a few hours of work, I've also managed to easily express a cubic in terms of its two vertices (I had no idea cubics were so symmetrical).  I'm hoping similar ideas will work for any polynomial.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-11-25T03:50:53.090" LastActivityDate="2010-11-25T03:50:53.090" Title="Exhibiting local extrema of rational functions" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11715" PostTypeId="2" ParentId="11675" CreationDate="2010-11-24T17:03:56.313" Score="2" ViewCount="0" Body="&lt;p&gt;They are also used to describe something called the Talbot Effect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phy.bris.ac.uk/people/berry_mv/research.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phy.bris.ac.uk/people/berry_mv/research.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;look at #8 in the list. I attended a seminar by Mike Berry about 12 years ago where he claimed that the Talbot Effect was a physical manifestation of Gauss Sums.&lt;/p&gt;&#xA;" OwnerUserId="3919" LastActivityDate="2010-11-24T17:03:56.313" />
  <row Id="11716" PostTypeId="2" ParentId="11654" CreationDate="2010-11-24T17:14:51.317" Score="2" ViewCount="0" Body="&lt;p&gt;Write $\frac{d}{dx}\int_0^1 f(x,y)dy$ as $$\lim_{h\to 0}\int_0^1 \frac{f(x+h,y)-f(x,y)}{h}dy.$$(You can do that because integration is linear.) If $$|\frac{\partial f}{\partial x}|\le M,$$ then by mean valued theorem&#xA;$$|\frac{f(x+h,y)-f(x,y)}{h}|\le M.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply Bounded Convergence Theorem (a weaker form of Lebesgue convergence theorem in this case.)&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-26T04:27:09.650" LastActivityDate="2010-11-26T04:27:09.650" CommentCount="3" />
  <row Id="11717" PostTypeId="2" ParentId="11010" CreationDate="2010-11-24T17:20:15.197" Score="1" ViewCount="0" Body="&lt;p&gt;To amplify Chandru1's response, it is important to note the difference in how we check that a map is linear and how we check that a map is &lt;em&gt;not&lt;/em&gt; linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that a map $\varphi$ is linear, we need to show that &lt;strong&gt;for every&lt;/strong&gt; $\mathbf{x}$ and &lt;strong&gt;every&lt;/strong&gt; $\mathbf{y}$ $\varphi(\mathbf{x}+\mathbf{y})=\varphi(\mathbf{x})+\varphi(\mathbf{y})$; and that &lt;strong&gt;for every&lt;/strong&gt; scalar $\alpha$ and &lt;strong&gt;every&lt;/strong&gt; $\mathbf{x}$, we have $\varphi(\alpha\mathbf{x})=\alpha\varphi(\mathbf{x})$. This is often done at the level of &quot;general formulas&quot;, checking that both expressions in each case lead to &lt;em&gt;identical&lt;/em&gt; formulas. Thus, for example, to check that you function (b) is a linear transformation, we verify that both $\varphi(f+g)$ and $\varphi(f)+\varphi(g)$ lead to the same expression:&#xA;$$\varphi(f+g) = (f+g)(1) = f(1)+g(1);\qquad \varphi(f)+\varphi(g) = f(1)+g(1);$$&#xD;&#xA;and likewise with $\varphi(af)$ and $a\varphi(f)$:&#xA;$$\varphi(af) = (af)(1) = a\cdot f(1);\qquad a\varphi(f) = a\cdot f(1).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, what you write in your comment for (a) is &lt;em&gt;not&lt;/em&gt; a good proof that the map $\varphi(x) = \bigl(1-x, (1-x)^2\bigr)$ is not linear. When you write something like:&#xA;$$ \varphi(x+y) = \bigl( 1-x-y, (1-x-y)^2\bigr) \neq (2 - x -y, (1-x)^2+(1-y)^2\bigr) = \varphi(x)+\varphi(y)$$&#xD;&#xA;you are actually asserting that the two expressions as &lt;strong&gt;never equal&lt;/strong&gt;. But in many instances, students will write this even when the expressions may sometimes be equal, yet think this constitutes a proof that the map is not linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;em&gt;not&lt;/em&gt; the case. You must either explain why the two expressions &lt;em&gt;actually&lt;/em&gt; are never equal (this happens in this case, because $1-x-y$ is never equal to $2-x-y$), or you must exhibit specific values of $x$ and $y$ for which the two expressions evaluate to different values. It is &lt;em&gt;not&lt;/em&gt; enough for the formulas to &quot;look different&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that the function is not linear, you must show that there exist &lt;strong&gt;specific&lt;/strong&gt; $x$ and $y$ such that $\varphi(x+y)$ is not equal to $\varphi(x)+\varphi(y)$, or that there exist a &lt;strong&gt;specific&lt;/strong&gt; $\alpha$ and a &lt;strong&gt;specific&lt;/strong&gt; $x$ such that $\varphi(\alpha x)\neq \alpha\varphi(x)$. That is, the verification that the equality does not always hold is &lt;em&gt;not&lt;/em&gt; done at the level of &quot;the formulas don't look the same&quot;, but rather by saying: &lt;em&gt;&quot;here are specific values, and, look, they don't match&quot;&lt;/em&gt;. So to show that (a) is &lt;em&gt;not&lt;/em&gt; linear, rather than working out the expressions and seeing that they don't &lt;em&gt;seem&lt;/em&gt; to be identical (just because two things don't &lt;em&gt;look&lt;/em&gt; identical doesn't mean they don't always take the same value; &quot;$\sin^2(x)+\cos^2(x)$&quot; looks very different from &quot;$1$&quot;, but they always take the same value), you should find &lt;em&gt;specific&lt;/em&gt; values of $x$ and $y$ for which the answers don't come out the same. For instance, you could show that if $x=y=1$, then&#xA;$$\varphi(1+1) = \varphi(2) = \bigl( 1-2, (1-2)^2\bigr) = (-1,1),$$&#xD;&#xA;but &#xA;$$\varphi(1)+\varphi(1) = \bigl( 1-1, (1-1)^2\bigr) + \bigl(1-1, (1-1)^2\bigr) = (0,0)+(0,0)=(0,0),$$&#xD;&#xA;and since $(-1,1)\neq (0,0)$, then $\varphi(1+1)\neq\varphi(1)+\varphi(1)$. &lt;em&gt;This&lt;/em&gt; shows the map is not linear, not the fact that the formulas for $\varphi(x+y)$ and for $\varphi(x)+\varphi(y)$ don't seem to be identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it this way: to show that not every student in your class has red hair, it is not enough to argue that it would be highly unlikely that everyone in a big class would have such an uncommon color of hair, that it never happens, that it doesn't look that way when one takes a quick look, etc. The &lt;em&gt;sure&lt;/em&gt; way of showing that not every student in your class has red hair is to stand someone up and &lt;em&gt;show&lt;/em&gt; that he/she does not have red hair, and then you're absolutely done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for (d).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T17:38:46.187" LastActivityDate="2010-11-24T17:38:46.187" CommentCount="2" />
  <row Id="11718" PostTypeId="2" ParentId="11706" CreationDate="2010-11-24T17:25:37.780" Score="3" ViewCount="0" Body="&lt;p&gt;This is not universal. In Spanish, for example, you need a thousand thousands to get a million. And &quot;billion&quot; is not, as in the United States, a thousand million; no, &quot;billion&quot; means &quot;a million millions&quot;; then trillion means &quot;a billion billions&quot;; quatrillions are &quot;a trillion trillion&quot;, etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T17:35:01.840" LastActivityDate="2010-11-24T17:35:01.840" CommunityOwnedDate="2010-11-24T17:25:43.877" CommentCount="1" />
  <row Id="11719" PostTypeId="1" CreationDate="2010-11-24T17:26:11.820" Score="1" ViewCount="32" Body="&lt;p&gt;For my understanding of modular forms, I have &quot;properly&quot; read the last chapter of Serre's book on arithmetic. I have not &quot;properly&quot; read the setup of Hecke operators for congruence subgroups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was intending to read the paper of Atkin and Lehner for filling this lacuna, as well as for understanding the proof of Atkin-Lehnher theorem. For some reason I could not relate very well with the exposition in the book of Diamond &amp;amp; Shurman or the book of Shimura.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be most grateful if somebody can point out some good reference for the Atkin-Lehner theorem other than these.&lt;/p&gt;&#xA;" OwnerUserId="3921" LastActivityDate="2010-11-24T17:26:11.820" Title="Reference for Atkin-Lehner theorem" Tags="&lt;number-theory&gt;&lt;modular-forms&gt;" CommentCount="1" />
  <row Id="11720" PostTypeId="1" CreationDate="2010-11-24T17:30:40.493" Score="2" ViewCount="92" Body="&lt;p&gt;What is a good reference for the story of congruences such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \tau(n) \equiv \sigma(11)(n)\ mod\ 691$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with a conceptual explanation with connections to \'etale cohomology, etc?&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-24T17:35:26.283" LastActivityDate="2010-11-24T18:26:24.980" Title="Ramanujan congruences and etale cohomology" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;modular-forms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11721" PostTypeId="2" ParentId="11706" CreationDate="2010-11-24T17:37:15.740" Score="3" ViewCount="0" Body="&lt;p&gt;Warning: I am not a linguist, so my answer is to be taken as an educated guess, no more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is essentially historical and the development of this system is told at  &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_and_short_scales&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Long_and_short_scales&lt;/a&gt; . As said above, it is not universal, even in other base-10 languages : East Asia and India have different systems. Even in western scientific tradition, Archimedes used a myriad (10⁴) based system for big numbers and French use the long scale, a 10⁶ based system. The choice of a system grouping the digits by 3 is therefore an historical accident, and we could as well have ended with a different grouping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the need of very large number seems to only exists for some technical use (science, accounting, religion) and do not emerge naturally in the language evolution. Therefore, the systems for big numbers are constructed by the specialist in question and do not blend well with the other numeral. In French, for example, the word &quot;mille&quot; (thousand) is grammatically a number (a cardinal numeral) but &quot;million&quot; is a simple noun. Hence the &quot;need&quot; for a &quot;restart&quot;. I guess it's similar for other languages.&lt;/p&gt;&#xA;" OwnerUserId="3918" LastActivityDate="2010-11-24T17:37:15.740" />
  <row Id="11722" PostTypeId="2" ParentId="11720" CreationDate="2010-11-24T17:43:05.237" Score="2" ViewCount="0" Body="&lt;p&gt;The paper of Swinnerton-Dyer in Lecture Notes in Math 350, as well as Serre's Bourbaki talk on the same topic. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-24T17:43:05.237" />
  <row Id="11723" PostTypeId="2" ParentId="11703" CreationDate="2010-11-24T17:50:23.837" Score="0" ViewCount="0" Body="&lt;p&gt;For a more mechanical proof, you could use induction. The proof then boils down to finding a common denominator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x^{n+1}-1}{x-1} + x^{n+1} = \frac{x^{n+1}-1+(x-1)x^{n+1}}{x-1} = \frac{x^{n+2}-1}{x-1}$&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-11-24T17:50:23.837" />
  <row Id="11724" PostTypeId="1" CreationDate="2010-11-24T17:50:43.870" Score="6" ViewCount="196" Body="&lt;p&gt;What is the compelling need for introducing a theory of $p$-adic integration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the existing theories of $p$-adic integration use some kind of analogues of Lebesgue measures? That is, do we put a Lebesgue measure on $p$-adic spaces, and just integrate real or complex valued functions on $p$-adic spaces, or is something more possible like integrating $p$-adic valued functions on $p$-adic spaces? What is the machinery used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, does the integration on spaces like $\mathbb C_p$ give something more than the usual integration in real analysis? I mean, the integration of complex valued functions of complex variables, or more precisely holomorphic functions, is much a much more interesting topic than measure theory. Is a similar analogue true in $p$-adic cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also seen mentioned that Grothendieck's cohomology theories like etale cohomology, crystalline cohomology etc., fit into such $p$-adic integration theories. What could possibly be the connection?&lt;/p&gt;&#xA;" OwnerUserId="3924" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-25T18:43:01.407" LastActivityDate="2010-11-26T05:33:52.650" Title="Theories of $p$-adic integration" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;arithmetic-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11726" PostTypeId="1" AcceptedAnswerId="12020" CreationDate="2010-11-24T17:56:34.630" Score="4" ViewCount="292" Body="&lt;p&gt;This is a follow up of &lt;a href=&quot;http://math.stackexchange.com/questions/11602/limit-of-sn-displaystyle-sum-k1-infty-left1-prod-j1n-1-le&quot;&gt;Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details can be found in the above thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S(n) = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mike has proved that $S(n)$ in fact diverges at-least faster than $\log_2(\lfloor n-1 \rfloor)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now based on what Mike has worked this conjectures arises:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \rightarrow \infty} (2 \log_{2}(n) - S(n)) = \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can $\alpha$ be expressed in terms of other familiar constants. $\frac{\pi \gamma}{e}$ seems to be a close guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical evidence seem to suggest they are true.  For example, we have the following graph of $2 \log_2 n - S(n)$ for $n \leq 300$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/E5byB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More numerical evidence: The value of $2 \log_2 n - S(n)$, is, for $n = 1000$, $2000$, and $3000$, respectively, $0.667734$, $0.667494$, and $0.667413$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative expression for $S(n)$ was worked out by Moron in the previously-mentioned question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n) = - \sum_{k=1}^{n-1} \frac{s(n,k)}{2^{n-k}-1},$$ &#xA;where $s(n,k)$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling number of the first kind&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-27T05:10:59.820" LastActivityDate="2010-11-27T21:14:25.560" Title="Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$ - Part II" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;combinatorics&gt;&lt;sequences-and-series&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="11727" PostTypeId="2" ParentId="11714" CreationDate="2010-11-24T18:08:55.120" Score="3" ViewCount="0" Body="&lt;p&gt;I don't think there is a nice answer for the general case of your question. However, there is nice answer to a closely related question - see the review below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR1752251 (2001c:11035) 11D25 (11D41 11G05 11G30)&lt;br&gt;&#xA;Buchholz, Ralph H.; MacDougall, James A.(5-NEWC)&lt;br&gt;&#xA;&lt;strong&gt;When Newton met Diophantus: a study of rational-derived polynomials&lt;br&gt;&#xA;and their extension to quadratic fields.&lt;/strong&gt;&lt;br&gt;&#xA;J. Number Theory 81 (2000), no. 2, 210–233.&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1006/jnth.1999.2473&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1006/jnth.1999.2473&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an interesting paper, which surveys the problem of determining the set $D(n)$ of all &quot;$k$-derived'' univariate polynomials of degree $n$ (where a polynomial $f \in k[x]$ is $k$-derived if $f$ and each of its successive derivatives has all roots in the ground field $k$). Define two polynomials $f_1,f_2\in k[x]$ to be equivalent if $f_1(x)=r f_2(s x+t)$ for $r,s,t\in k$, $r,s \neq 0$. Then up to equivalence, the following is known about $\mathbb Q$-derived polynomials: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(1)=\{x\};\quad D(2)=\{x^2,x(x-1)\};$$&#xD;&#xA;$$D(3)=\{x^3\}\cup\bigg\{x(x-1)(x-a)\ :\ a=\frac{w(w-2)}{w^2-1},w\in \mathbb Q\bigg\};$$&#xD;&#xA;$$ D(4)\supseteq \{x^4\}\cup\bigg\{x^2(x-1)(x-a)\ :\ a=\frac{9(2w+z-12)(w+z)}{(z-w-18)(8w+z)},&#xD;&#xA; (w,z)\in E(\mathbb Q), E\colon z^2=w(w-6)(w+18)\bigg\};$$&#xD;&#xA;$$ D(n)\supseteq \{x^n, x^{n-1}(x-1)\}\ {\rm for}\ n\geq 5.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors prove that determining $D(n)$ in general devolves into two conjectures: (1) that no quartic with four distinct roots is $\mathbb Q$-derived; (2) that no quintic of type $x^3(x-a)(x-b)$, $a\neq b,\ a,b\neq0$, is $\mathbb Q$-derived. The first conjecture can be solved by determining all rational points on a hyperelliptic surface of degree 10. The second conjecture can be solved by determining all rational points on a curve of genus 2 (E. V. Flynn [&quot;On $\mathbb Q$-derived polynomials'', Preprint; per revr.] has now proved this second conjecture). The authors also discuss briefly the situation of $K$-derived polynomials for quadratic extensions $K$ of $\mathbb Q$; there is, for example, the quartic $y^2=x^2(x-1)(x-\frac{37-20\sqrt{3}}{13})$ which is a ${\mathbb Q}(\sqrt{3})$-derived polynomial.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reviewed by Andrew Bremner&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-24T18:08:55.120" CommentCount="2" />
  <row Id="11728" PostTypeId="1" CreationDate="2010-11-24T18:12:55.553" Score="0" ViewCount="44" Body="&lt;p&gt;An annuity immediate has quarterly payments of $1000$ for $25$ years at a rate of $6 \%$ converted quarterly. Find its present value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective interest rate $i$ is $0.06/4$. So let $v = \frac{1}{1+i}$. Then wouldn't the present value be $1000 \left(\frac{1-v^{25}}{i} \right)$? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-11-24T19:25:27.380" Title="Annuity immediate" Tags="&lt;finance&gt;" AnswerCount="3" />
  <row Id="11729" PostTypeId="2" ParentId="11720" CreationDate="2010-11-24T18:16:12.040" Score="0" ViewCount="0" Body="&lt;p&gt;The paper of Deligne, Formes modulaires et représentations $l$-adiques, available at &lt;a href=&quot;http://www.numdam.org/numdam-bin/fitem?id=SB_1968-1969__11__139_0&quot; rel=&quot;nofollow&quot;&gt;numdam&lt;/a&gt;, has these things.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-11-24T18:26:24.980" LastActivityDate="2010-11-24T18:26:24.980" />
  <row Id="11731" PostTypeId="1" CreationDate="2010-11-24T18:24:52.840" Score="2" ViewCount="49" Body="&lt;p&gt;In various descriptions of the moduli space of elliptic curves with level structures, such as the description of $X_0(N)$ being defined over $Spec\ \mathbb{Z}[1/N]$, the primitive $N^{th}$ roots of unity pop up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is somewhere a description available as to how this connection arises? I mean, somewhere manageable, without wading through the heavy volumes of Deligne &amp;amp; Rapoport or Katz &amp;amp; Mazur?&lt;/p&gt;&#xA;" OwnerUserId="3927" LastActivityDate="2010-11-24T18:24:52.840" Title="Primitive nth roots of unity and moduli of elliptic curves" Tags="&lt;arithmetic-geometry&gt;" />
  <row Id="11732" PostTypeId="2" ParentId="11728" CreationDate="2010-11-24T18:32:19.490" Score="1" ViewCount="0" Body="&lt;p&gt;For starters, there are 100 periods, not 25.  If you just reduce each payment to present value, what is the value of the nth payment?  Can you then sum this from 1 to 100?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-24T19:25:27.380" LastActivityDate="2010-11-24T19:25:27.380" />
  <row Id="11733" PostTypeId="2" ParentId="11728" CreationDate="2010-11-24T18:36:55.193" Score="0" ViewCount="0" Body="&lt;p&gt;The present value of an income stream with $n$ equally spaced payments, with an interest of $r$ at each payment, each payment of $C$, is obtained by computing the present value of each payment and adding it up. Here, you have $r=0.06/4 = 0.015$, as you note, so you get&#xA;$$1000 + \frac{1000}{1+r} + \frac{1000}{(1+r)^2} + \cdots + \frac{1000}{(1+r)^{100}},$$&#xD;&#xA;since there are 100 payments (one payment per quarter, 25 years, beginning now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just&#xA;$$1000\left(1+\frac{1}{1+r} + \left(\frac{1}{1+r}\right)^2 + \cdots + \left(\frac{1}{1+r}\right)^{100}\right)$$&#xD;&#xA;which is&#xA;$$1000\left( \frac{1 - \frac{1}{(1+r)^{101}}}{1-\frac{1}{1+r}}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only get 1000 payments (no payment at the end of the last quarter 25 years from now) then the sum only goes up to exponent 99; adjust accordingly.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-24T19:23:21.733" LastActivityDate="2010-11-24T19:23:21.733" />
  <row Id="11735" PostTypeId="1" AcceptedAnswerId="11738" CreationDate="2010-11-24T18:56:11.987" Score="7" ViewCount="156" Body="&lt;p&gt;Hi there,&#xA;I want to find the volume of a torus with a given thickness and a given radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let r be the radius of a circle with its midpoint at $M(0|b)$ ($b \geq r$). Now I want to rotate this circle about the x-axis, that is to say about a circular path which has the length $2 \pi \cdot |b|$. So I thought I'd simply integrate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \int\limits_0^{2 \pi \cdot |b|} \pi r^2 dz = 2 \pi \cdot r^2 \cdot |b|$, which turns out to be the correct result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't find it trivial that the volume of this torus is the same as the volume of a cylinder with the corresponding height. I read the article on Wikipedia about the torus and it said that this was due to Cavalieri's theorem, which to my mind doesn't really have a lot to do with the torus vs. the cylinder...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some easy way to prove that a torus has the same volume as a cylinder with the height equal to the torus' perimeter?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-24T20:37:55.747" Title="How to prove that a torus has the same volume as a cylinder (with the height equal to the torus' perimeter)" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="11736" PostTypeId="2" ParentId="11728" CreationDate="2010-11-24T19:19:42.583" Score="1" ViewCount="0" Body="&lt;p&gt;If the first payment is received at the end of the first quarter, then the present value is $$1000\left(\frac{1-v^{100}}{i}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first payment is received at the beginning of the first quarter (this seems to be the case, since you stated &quot;immediate&quot;), then the present value is $$1000\left(\frac{1+i-v^{99}}{i}\right),$$&#xD;&#xA;assuming a total of 100 payments are made.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-24T19:19:42.583" />
  <row Id="11737" PostTypeId="2" ParentId="11735" CreationDate="2010-11-24T19:34:49.757" Score="-1" ViewCount="0" Body="&lt;p&gt;Perhaps I don't understand the question, but if you have a cylinder, you can imagine bending it so that the top and bottom are connected. Then you have a torus.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-11-24T19:34:49.757" CommentCount="3" />
  <row Id="11738" PostTypeId="2" ParentId="11735" CreationDate="2010-11-24T19:47:09.163" Score="12" ViewCount="0" Body="&lt;p&gt;(i) Slice the torus into a million near-disks.&lt;br&gt;&#xA;(ii) Rotate every second disk through 180 degrees.&lt;br&gt;&#xA;(iii) Stick them all together again.&lt;br&gt;&#xA;You get a near-cylinder whose height is nearly 2πb. Now let a million tend to infinity.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-11-24T19:47:09.163" CommentCount="1" />
  <row Id="11739" PostTypeId="1" AcceptedAnswerId="11741" CreationDate="2010-11-24T20:02:47.903" Score="3" ViewCount="124" Body="&lt;p&gt;If $\ln(1-x+x^2)  = a_1x+a_2x^2 + \cdots \text{ then } a_3+a_6+a_9+a_{12} + \cdots = $ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach is to write $1-x+x^2 = \frac{1+x^3}{1+x}$ then expanding the respective logarithms,I got a series (of coefficient) which is nothing but $\frac{2}{3}\ln 2$.But this approach took some time for me (I can't solved it during the test but after the test I solved it)... Any other quick method?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-24T20:31:48.770" LastActivityDate="2010-11-24T23:24:26.027" Title="Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$" Tags="&lt;calculus&gt;&lt;logarithms&gt;&lt;power-series&gt;" AnswerCount="3" />
  <row Id="11740" PostTypeId="2" ParentId="11726" CreationDate="2010-11-24T20:07:33.100" Score="-4" ViewCount="0" Body="&lt;p&gt;We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(n) - \beta \log(n) = \displaystyle S(n) - \sum_{k=1}^{\infty} \beta \frac{x^k}{k} = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \frac{x^k}{k}\right)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x = \frac{n-1}{n}$. If in (i) we can guarantee the interchange of the limit and the summation we can conclude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \lim_{n \rightarrow \infty} S(n) - \beta \log(n) = \lim_{n \rightarrow \infty} \left[  \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \frac{x^k}{k}\right) \right] = \sum_{k=1}^{\infty}  \left(1 - \lim_{n \rightarrow \infty}\prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \lim_{n \rightarrow \infty}\frac{x^k}{k}\right)= \sum_{k=1}^{\infty} 1 - \frac{\beta}{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am, however, still thinking about whether the interchange is legitimate, I think it should be... perhaps by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Dominated convergence theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Possible Completion of Proof of Divergence:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We can show that the interchange of limits holds by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Dominated convergence theorem&lt;/a&gt; if $\displaystyle \sum_{k=1}^{\infty} \left|1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right) - \beta \frac{x^k}{k}\right| &amp;lt; \infty$ does converge absolutely. But then the interchange of limits shown above shows that (i) diverges. &lt;/p&gt;&#xA;" OwnerUserId="1077" LastEditorUserId="1077" LastEditorDisplayName="" LastEditDate="2010-11-24T21:06:41.467" LastActivityDate="2010-11-24T21:06:41.467" CommentCount="4" />
  <row Id="11741" PostTypeId="2" ParentId="11739" CreationDate="2010-11-24T20:13:20.440" Score="11" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $f(x) = \sum a_n x^n$ be a power series.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum a_{3n} x^{3n} = \frac{f(x) + f(\omega x) + f(\omega^2 x)}{3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega = e^{ \frac{2\pi i}{3} }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;  Ignoring convergence it suffices to prove this for a single term, and then it boils down to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1 + \omega^n + \omega^{2n}}{3} = \begin{cases} 1 \text{ if } 3 | n \\\ 0 \text{ otherwise} \end{cases}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;discrete Fourier transform&lt;/a&gt;.  Applying the lemma, we readily obtain that the desired sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ln 1 + \ln (1 - \omega + \omega^2) + \ln (1 - \omega^2 + \omega)}{3} = \frac{1}{3} \ln (2 \omega^2 \cdot 2 \omega) = \frac{2}{3} \ln 2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we use the fact that $1 + \omega + \omega^2 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-24T20:13:20.440" CommentCount="8" />
  <row Id="11744" PostTypeId="2" ParentId="11735" CreationDate="2010-11-24T20:37:55.747" Score="2" ViewCount="0" Body="&lt;p&gt;To use Cavalieri's theorem, lay the torus and cylinder on a table and slice them with planes parallel to the table. Then it suffices to show that the torus slices (annuli) have the same area as the cylinder slices (rectangles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At height $h$ (measured from the centre of the torus or cylinder), the annulus has inner and outer radii $$R_\pm = |b| \pm \sqrt{r^2-h^2},$$ so its area is $$A_1 = \pi R_+^2 - \pi R_-^2 = \pi(R_+ + R_-)(R_+ - R_-) = \pi \cdot 2|b| \cdot 2\sqrt{r^2-h^2}.$$ The rectangle has width $2\sqrt{r^2-h^2}$ and length $2\pi |b|$, so its area is $$A_2 = 2\sqrt{r^2-h^2} \cdot 2\pi|b|.$$ The areas are equal, so we're done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer Pappus's centroid theorem though...&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2010-11-24T20:37:55.747" />
  <row Id="11746" PostTypeId="2" ParentId="11739" CreationDate="2010-11-24T21:23:20.390" Score="1" ViewCount="0" Body="&lt;p&gt;Generally one can employ evaluations at roots of unity to take arbitrary such &lt;strong&gt;multisections&lt;/strong&gt; of power series. See my &lt;a href=&quot;http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint/3514#3514&quot;&gt;post here&lt;/a&gt; and the links there for discussion and examples. Generally, if we let $\rm\:w\:$ denote a primitive $\rm n$'th root of unity, then the $\rm m$'th $\rm n$-section of a power series selects the linear progression of $\rm\: m + k\:n\:$ indexed terms, viz. if $\rm\ f\:(X)\ =\ a_0 + a_1 X + a_2 X^2 +\: \cdots\ $ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad a_{m}\ X^m + a_{m+n}\ X^{m+n} + a_{m+2n}\ X^{m+2n}\ +\: \cdots $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\displaystyle\ =\ \frac{1}n\: (\:f\:(X) + f\:(Xw)\ w^{-m} + f\:(X\:w^2)\ w^{-2m} +\:\cdots\: \:+\ f\:(X\:w^{n-1})\ w^{(1-n)\:m}\:) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Riordan's classic textbook &quot;Combinatorial Identities&quot; for many applications of  multisections.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T23:24:26.027" LastActivityDate="2010-11-24T23:24:26.027" />
  <row Id="11747" PostTypeId="1" AcceptedAnswerId="11752" CreationDate="2010-11-24T21:37:08.870" Score="1" ViewCount="273" Body="&lt;p&gt;How can I prove that dividing an irrational number by an irrational number (besides himself) &lt;strong&gt;can&lt;/strong&gt; result in an integer?&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-02T04:29:41.227" LastActivityDate="2011-01-02T04:29:41.227" Title="Proof that dividing irrational number by an irrational number can result in an integer?" Tags="&lt;real-number-computation&gt;&lt;irrational-numbers&gt;" AnswerCount="9" CommentCount="10" />
  <row Id="11748" PostTypeId="2" ParentId="3510" CreationDate="2010-11-24T21:40:12.020" Score="0" ViewCount="0" Body="&lt;p&gt;An easy way out to answer the question is when we try to extend the definition of exponential to complex plane in a &quot;nice&quot; way (read &quot;nice&quot; as holomorphic) we then end up with this definition. And as you are probably aware there is only at most one such extension. So if we want to define our exponential in the complex plane such that the exponential function is holomorphic, and it matches with the conventional exponential function on the real line, we end up with $e^{it} = \cos(t) + i \sin(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other answers I think are circular in some sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, how do we know that we can expand $e^{i x}$ as a power series. All we know is that we can expand $e^{x}$ as a power series when $x$ is real. We do not know apriori that the expansion can be carried forward if $x$ happens to be complex.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T21:54:47.170" LastActivityDate="2010-11-24T21:54:47.170" CommentCount="1" />
  <row Id="11749" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:40:40.650" Score="5" ViewCount="0" Body="&lt;p&gt;By giving an example. For instance, $2\sqrt{2}$ divided by $\sqrt{2}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note well:&lt;/em&gt; To prove that something &lt;strong&gt;can&lt;/strong&gt; happen, all you need to do is exhibit a single instance of it happening. To prove that something &lt;strong&gt;must&lt;/strong&gt; happen you need to show that it happens always. So to show that dividing an irrational number by an irrational number other than itself &lt;strong&gt;can&lt;/strong&gt; be an integer, all you need to do is cook up a single example of it happening. If you wanted to prove that, say, a rational divided by rational other than zero &lt;strong&gt;must&lt;/strong&gt; be a rational, an example would not suffice; you would need to show that for &lt;em&gt;any&lt;/em&gt; rational $q$ and &lt;em&gt;any&lt;/em&gt; rational $r\neq 0$, you would have $\frac{q}{r}$ a rational. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-24T21:40:40.650" />
  <row Id="11750" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:03.377" Score="4" ViewCount="0" Body="&lt;p&gt;A trivial example would be $2\pi/\pi=2$. In fact every example would be of this form (one irrational would be an integer multiple of the other), for if $a,b$ are distinct irrational numbers such that $\frac{a}{b}=c$ where $c$ is an integer, then, $a=bc$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-24T21:47:07.847" LastActivityDate="2010-11-24T21:47:07.847" />
  <row Id="11751" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:35.700" Score="0" ViewCount="0" Body="&lt;p&gt;$\frac{nr}{r}$, where $n \in \mathbb{Z} \backslash \{0\}$ and $r \in \mathbb{R} \backslash \mathbb{Q}$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-24T21:42:35.700" />
  <row Id="11752" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:36.153" Score="4" ViewCount="0" Body="&lt;p&gt;An example would suffice to prove this claim. If you don't want to divide an irrational number by itself, how about $2\sqrt{2}/\sqrt{2}$. You should be able to prove that if a number is irrational, then twice that number is also irrational.&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-11-24T21:42:36.153" />
  <row Id="11753" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:42:45.263" Score="1" ViewCount="0" Body="&lt;p&gt;@Arturo Magidin: Did you read the &quot;besides himself&quot; part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Alon: How about $\frac{2 \pi}{\pi}$ or something similar?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-24T21:42:45.263" CommentCount="1" />
  <row Id="11754" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:44:35.793" Score="2" ViewCount="0" Body="&lt;p&gt;Start by assuming that you have a case where it &lt;em&gt;does&lt;/em&gt; result in an integer, and see what that says about the irrational numbers.  From there you should be able to give examples.  That is, if $x/y = n$ for some integer $n$, what does that say about the relationship between $x$ and $y$?  Choosing a specific integer $n$ and a specific $y$ should lead you to an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(While writing this, many answers were posted.  I'm adding this anyway to give a naive approach to use if you don't already see how it will come out.)&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-24T21:44:35.793" />
  <row Id="11755" PostTypeId="2" ParentId="11739" CreationDate="2010-11-24T21:51:11.717" Score="2" ViewCount="0" Body="&lt;p&gt;Qiaochu Yuan gives an excellent exposition of the general method. Sometimes in specific cases you can get the answer via some hands-on calculations as well, and expanding the logs as you mentioned can actually be done pretty quickly: using that $1 - x + x^2 = {1 + x^3 \over 1 + x}$ one gets that $\ln(1 - x + x^2) = \ln(1 + x^3) - \ln(1 + x)$. You now can use the fact that the sum of $a_3 + a_6 + ...$ is obtained by subtracting the corresponding sums for $\ln(1 + x^3)$ and $\ln(1 + x)$. Since the power series for $\ln(1 + x^3)$ only contains powers of $x^3$, the contribution to $a_3 + a_6 + ...$ coming from that term is what you get from plugging in $x = 1$, namely $\ln(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power series of $\ln(1 + x)$ may be written as $-\sum_{n &amp;gt; 0} {(-x)^n \over n}$.&#xA;Taking every third term gives $-\sum_{n &amp;gt; 0}  {(-x)^{3n} \over 3n} = -{1 \over 3}\sum_{n &amp;gt; 0} {(-x^3)^n \over n}$. Note this series is again the series of $\ln(1 + x)$, but applied to $x^3$ in place of $x$. Plugging in $x = 1$ gives ${1 \over 3}\ln(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer you want is $\ln(2) - {1 \over 3}\ln(2) = {2 \over 3}\ln(2)$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-24T22:03:06.470" LastActivityDate="2010-11-24T22:03:06.470" />
  <row Id="11756" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T21:51:13.367" Score="1" ViewCount="0" Body="&lt;p&gt;If x is an irrational then x/-x works and is the canonical example that satisfies the OP's rules&lt;/p&gt;&#xA;" OwnerUserId="3878" LastActivityDate="2010-11-24T21:51:13.367" />
  <row Id="11757" PostTypeId="1" CreationDate="2010-11-24T22:31:20.513" Score="3" ViewCount="132" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The series for $\log (1+x)$ is convergent(by estimating radius of convergence) only with $|x| &amp;lt;1$. Then how is it still true for $x = 1$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is it still apparently true when $x$ is a roots of unity other than $-1$? Such as in the answers to this question: &lt;a href=&quot;http://math.stackexchange.com/questions/11739&quot;&gt;Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3939" LastActivityDate="2010-11-24T22:53:04.547" Title="Convergence for log 2" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11758" PostTypeId="2" ParentId="11757" CreationDate="2010-11-24T22:36:12.127" Score="1" ViewCount="0" Body="&lt;p&gt;Look up alternating series test or generalized alternating series test (Also known as Dirichlet's test as George pointed out).&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-24T22:53:04.547" LastActivityDate="2010-11-24T22:53:04.547" CommentCount="2" />
  <row Id="11759" PostTypeId="2" ParentId="11757" CreationDate="2010-11-24T22:38:21.207" Score="13" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Abel's theorem&lt;/a&gt; is the key to why you can say that the limit as you approach the boundary comes out to the same as evaluating the series at the boundary.  Sivaram's answer addresses why the series converges at those boundary points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly, Abel's theorem implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{r\nearrow 1}\log(1+zr)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}z^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for each complex number $z$ with $|z|=1$ such that the series converges.  By continuity of the logarithm, this will be $\log(1+z)$ in such cases.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-24T22:51:44.813" LastActivityDate="2010-11-24T22:51:44.813" />
  <row Id="11760" PostTypeId="2" ParentId="11747" CreationDate="2010-11-24T22:46:54.420" Score="2" ViewCount="0" Body="&lt;p&gt;The basic idea works much more generally: it amounts to how a subgroup acts on its complement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm If\:\ \ a,b \in A \subset B\ \ \: then\ \ ab \in  A \iff b \in A\ \ $ if  $\rm\ \ a^{-1}\in A,\ $ and $\rm\ A,\:B\ $ are monoids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\ \ b\not\in A\ \ \Rightarrow\ \ ab\not\in A\ \ and\ \ (ab)/b = a\in A\ $ as you desire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case is simply $\rm\ \ A = \mathbb Q,\ \ B = \mathbb R,\ \ a\in \mathbb Z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key property above is essentially this &quot;complementary&quot; view of a subgroup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\ \ $ A nonempty subset $\rm\:A\:$ of abelian group $\rm\:B\:$ &#xA;comprises a subgroup $\rm\iff\ A\ \bar A\ = \bar A\ $ where  $\rm\: \bar A\:$ is the complement of $\rm\:A\:$ in $\rm\:B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instances of this are ubiquitous in concrete number systems, e.g. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;                                 transcendental &#xA;     algebraic * nonalgebraic  =  nonalgebraic  if  nonzero &#xA;      rational * irrrational   =   irrational   if  nonzero &#xA;          real *   nonreal     =    nonreal     if  nonzero &#xA;&#xA;         even  +     odd       =      odd          additive example&#xA;       integer + noninteger    =   noninteger&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-24T23:16:32.743" LastActivityDate="2010-11-24T23:16:32.743" />
  <row Id="11761" PostTypeId="1" CreationDate="2010-11-24T23:29:25.130" Score="3" ViewCount="145" Body="&lt;p&gt;From the wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Motive_(algebraic_geometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Motive_(algebraic_geometry&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it appears that the category of Mixed motives $MM(k)$ over a field $k$ is still conjectural; but there is a good derived category $DMM(k)$ already constructed for this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good reference for this construction, and why this derived category is the &quot;suitable&quot; one.&lt;/p&gt;&#xA;" OwnerUserId="3942" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-25T10:24:07.307" LastActivityDate="2010-11-26T05:50:07.007" Title="Status of mixed motives" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;homological-algebra&gt;" AnswerCount="2" />
  <row Id="11762" PostTypeId="2" ParentId="5231" CreationDate="2010-11-24T23:35:50.863" Score="0" ViewCount="0" Body="&lt;p&gt;At the moment, I went with a somewhat simple solution that allows me to draw a plane/face with 4 vertices with arbitrary rotations, in 2D:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how it works:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PVector[] unRotateVerts(PVector[] verts,PVector n){&#xA;  //get the angle between the face4 normal and Y&#xA;  angle = PVector.angleBetween(n,y);//acos(n.dot(y));&#xA;  //get the axis of rotation, by getting the perpendicular&#xA;  axis = n.cross(y);&#xA;  axis.normalize();&#xA;  //clone vertices&#xA;  int vl = verts.length;&#xA;  PVector[] clone = new PVector[vl];&#xA;  for(int i = 0; i&amp;lt;vl;i++) clone[i] = verts[i].get();&#xA;  //make a rotation matrix and rotate it&#xA;  PMatrix3D rMat = new PMatrix3D();&#xA;  rMat.rotate(angle,axis.x,axis.y,axis.z);&#xA;  //multiply each vertex with the rotation matrix&#xA;  PVector[] dst = new PVector[vl];&#xA;  for(int i = 0; i&amp;lt;vl;i++) {&#xA;    dst[i] = new PVector();&#xA;    rMat.mult(clone[i],dst[i]);&#xA;  }&#xA;  return dst;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This works perfectly for aligning with XZ, but rotations on the Z axis are still there. Any hints on how to remove that would be handy.&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2010-11-24T23:35:50.863" />
  <row Id="11763" PostTypeId="1" CreationDate="2010-11-24T23:50:56.167" Score="4" ViewCount="139" Body="&lt;p&gt;It is a bit easier to understand the homology $H_1(X, \mathbb Z)$ for various compact surfaces in analogy with handles and so on. There seems to be a nice intuitive picture with handles, holes, etc to think of the first homology group, and similar heuristics for higher homology groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But almost all axiomatic treatment of homology groups uses instead the relative homology. But it is not so intuitively clear how to visualize the relative homology groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some intuitive crutches for dealing with these relative homology groups, particularly for surfaces?&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2010-12-27T17:28:46.127" Title="Meaning of relative homology" Tags="&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11764" PostTypeId="1" CreationDate="2010-11-24T23:55:37.283" Score="0" ViewCount="77" Body="&lt;p&gt;Sheaf cohomology is a well-studied topic with a lot of references available. For example Hartshorne's book. But for a certain paper I am reading now, I have to understand sheaf homology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone tell me some references? This is just sheaf homology in the topological sense; not in algebraic geometry.&lt;/p&gt;&#xA;" OwnerUserId="3944" LastActivityDate="2010-11-25T20:42:07.207" Title="References for sheaf homology" Tags="&lt;algebraic-topology&gt;" AnswerCount="2" />
  <row Id="11765" PostTypeId="1" CreationDate="2010-11-25T00:04:45.040" Score="2" ViewCount="39" Body="&lt;p&gt;Consider the power series ring in noncommutative associative formal variables $X, Y$ over the field $\mathbb C$. Let $P$ be a power series. Define the exponential $e^P$ and logarithm $\log (1+P)$ using the usual series expansions. The exponential maps from a neighborhood of $0$ to a neighborhood of $1$ and the logarithm goes the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be such a power series with zero constant term. How to show that the logarithm and exponential maps defined above are inverses of each other?&lt;/p&gt;&#xA;" OwnerUserId="3945" LastActivityDate="2010-11-25T00:17:01.087" Title="Exponential and Logarithm in noncommutative setting" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11766" PostTypeId="2" ParentId="11765" CreationDate="2010-11-25T00:17:01.087" Score="1" ViewCount="0" Body="&lt;p&gt;Powers of $P$ commute so you need not worry about the non-commutative underlying variables. Expand $e^{\log(1+P)}$ in powers of $P$ and calculate the power of $P^t$ for each $t$; you'll get $1+P$. That's something like Moebius inversion. You can also do it the opposite way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you don't even need to calculate it, since we already know the answer by substitution for the corresponding &quot;real&quot; power series, which do converge in some neighborhood, and the answers must match.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-25T00:17:01.087" />
  <row Id="11767" PostTypeId="2" ParentId="11747" CreationDate="2010-11-25T00:24:36.527" Score="1" ViewCount="0" Body="&lt;p&gt;An example from Euler together with the proof of the irrationality of $\pi^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi^2}{\zeta(2)}=6,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(2)=\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-25T00:24:36.527" />
  <row Id="11768" PostTypeId="1" CreationDate="2010-11-25T00:27:07.227" Score="0" ViewCount="74" Body="&lt;p&gt;I was wondering about the generalization of the concept of convolution from the familiar one on real spaces and how many properties still remain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convolution on Lebesgue-integrable real-valued functions defined on a group equipped with measure, does each of the following properties holds? If not, what extra requirements needed on the group, measure and/or functions in order that it can become true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Commutativity: $f * g = g * f \,$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Associativity: $f * (g * h) = (f * g) * h \,$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Linearity (This, I am more sure, holds, but I might be wrong):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distributivity:    $f * (g + h) = (f * g) + (f * h) \,$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associativity with scalar multiplication: $  a (f * g) = (a f) * g = f * (a g) \, $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Time invariant: $f(t+\tau)*g(t)=h(t+\tau), \forall t, \tau$ in the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) the reverse of (3) and (4), i.e.: Any linear and time-invariant operation on the set of Lebesgue-integrable real-valued functions must be the convolution with some Lebesgue-integrable real-valued function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition for the convolution on a group with measure is defined according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution#Convolutions_on_groups&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If G is a suitable group endowed with&#xA;  a measure λ, and if f and g are real&#xA;  or complex valued integrable functions&#xA;  on G, then we can define their&#xA;  convolution by&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$(f * g)(x) = \int_G f(y) g(-y+x)\,d\lambda(y). \,$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-25T04:36:46.480" LastActivityDate="2010-11-25T10:02:20.810" Title="Convolution on group with measure" Tags="&lt;integral&gt;&lt;convolution&gt;&lt;group&gt;&lt;measure&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11769" PostTypeId="1" CreationDate="2010-11-25T00:44:33.407" Score="7" ViewCount="85" Body="&lt;p&gt;Let $M$ be a smooth connected manifold. Is it always possible to find a connected dense open subset $U$ of $M$ which is diffeomorphic to an open subset of &lt;strong&gt;R&lt;/strong&gt;$^n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we don't require $U$ to be connected, the answer is yes: it is enough to construct a countable collection of disjoint open &quot;affines&quot; whose union is dense, and this is not terribly difficult.&lt;/p&gt;&#xA;" OwnerUserId="546" LastEditorUserId="546" LastEditorDisplayName="" LastEditDate="2010-11-25T01:16:20.663" LastActivityDate="2010-11-25T02:19:45.387" Title="Do all manifolds have a densely defined chart?" Tags="&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11770" PostTypeId="1" AcceptedAnswerId="11772" CreationDate="2010-11-25T00:56:44.640" Score="0" ViewCount="60" Body="&lt;p&gt;I'm trying to solve a differential equation problem that has to do with barometric pressure at a certain altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate at which barometric pressure decreases with altitude is proportional to the pressure at that altitude. If the altitude in feet, then the constant of proportionality is 3.7 x 10&lt;sup&gt;5&lt;/sup&gt;. The barometric pressure at sea level is 29.92 inches of mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I calculate the barometric pressure, in inches of mercury, at any altitude using the information above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential equation that I came up with looks like this:&lt;br /&gt;&#xA;Let $P=$ Barometric Pressure&lt;br /&gt;&#xA;Let $a=$ Altitude&lt;br /&gt;&#xA;$\dfrac{dP}{da}=-k(29.92-P)$ &lt;br /&gt;&#xA;But when I solve it, it doesn't look like it's the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="3946" LastActivityDate="2010-11-25T01:02:45.637" Title="Differential Equations: Calculating Barometric Pressure at an Altitude" Tags="&lt;calculus&gt;&lt;differential-equations&gt;" AnswerCount="1" />
  <row Id="11771" PostTypeId="1" AcceptedAnswerId="11774" CreationDate="2010-11-25T01:01:01.740" Score="3" ViewCount="89" Body="&lt;p&gt;I have a question about the set of functions from a set to another set. I am wondering about the degenerate cases. Suppose $X^Y$ denotes the set of functions from a set $Y$ to a set $X$, why is $|Y^{\emptyset}|=1$ but $|\emptyset^Y|=0$ where $Y\neq \emptyset$?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T01:03:23.953" LastActivityDate="2010-11-25T01:23:23.637" Title="Why is $|Y^{\emptyset}|=1$ but $|\emptyset^Y|=0$ where $Y\neq \emptyset$" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="11772" PostTypeId="2" ParentId="11770" CreationDate="2010-11-25T01:02:45.637" Score="2" ViewCount="0" Body="&lt;p&gt;Pressure decreases exponentially with altitude and thats what you will get provided you fix your signs in the equation. Your equation will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{da} = -kP$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is some positive constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this to get $P(a) = P(0) e^{-ka}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and at $a = 0$, $P(0) = 29.92$. Make sure to keep track of all your units.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-25T01:02:45.637" CommentCount="1" />
  <row Id="11773" PostTypeId="2" ParentId="11771" CreationDate="2010-11-25T01:05:21.173" Score="4" ViewCount="0" Body="&lt;p&gt;Because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Empty_function&quot; rel=&quot;nofollow&quot;&gt;empty function&lt;/a&gt; is the unique function from the empty set to an arbitrary set $Y$, while if $Y\neq\emptyset$, then there exists $y\in Y$, but there's no place in $\emptyset$ for a function to send $y$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-25T01:05:21.173" CommentCount="2" />
  <row Id="11774" PostTypeId="2" ParentId="11771" CreationDate="2010-11-25T01:09:10.903" Score="6" ViewCount="0" Body="&lt;p&gt;The definition of $A^B$ is &quot;the set of all functions with domain $B$ and codomain $A$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function $f$ from $B$ to $A$ is a set of ordered pairs such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $(x,y)\in f$, then $x\in B$ and $y\in A$.&lt;/li&gt;&#xA;&lt;li&gt;For every $b\in B$ there exists $a\in A$ such that $(b,a)\in f$.&lt;/li&gt;&#xA;&lt;li&gt;If $(b,a)$ and $(b,a&amp;#39;)$ are in $f$, then $a=a&amp;#39;$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, what happens if $B=\emptyset$? Well, then there can be no pair in $f$, because you cannot have $x\in B$. But notice that in that case, 2 is satisfied &quot;by vacuity&quot; (if it were false, you would be able to exhibit a $b\in\emptyset$ for which there is no $a\in A$ with $(b,a)\in f$; but there &lt;em&gt;are&lt;/em&gt; no $b\in\emptyset$, so you cannot make such an exhibition; the statement is true because the premise, &quot;$b\in\emptyset$&quot;, can never hold). Likewise 3 holds by vacuity. So it turns out that if we take $f=\emptyset$, then $f$ satisfies 1, 2, and 3, and therefore it is by all rights a &quot;function from $\emptyset$ to $A$&quot;.  But this is the only possible function from $\emptyset$ to $A$, because only the empty set works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, if $A=\emptyset$, but $B\neq\emptyset$, then no set $f$ can satisfy both 1 and 2, so no set can be a function from $B$ to $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that $Y^{\emptyset}$ always contains exactly one element, namely the &quot;empty function&quot;, $\emptyset$. But if $Y\neq\emptyset$, then $\emptyset^Y$ contains &lt;em&gt;no&lt;/em&gt; elements; that is, it is empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, since $Y^{\emptyset}$ has exactly one element, $|Y^{\emptyset}|=1$ regardless of what $Y$ is. But if $Y\neq\emptyset$, then $\emptyset^{Y}$ is empty, so $|\emptyset^{Y}| = 0$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T01:15:45.263" LastActivityDate="2010-11-25T01:15:45.263" CommentCount="2" />
  <row Id="11776" PostTypeId="2" ParentId="11769" CreationDate="2010-11-25T01:41:19.397" Score="3" ViewCount="0" Body="&lt;p&gt;If the manifold is compact, the answer is yes.  The geometric idea is to put a small balloon in the manifold, and to inflate it.  Since the manifold is compact, you eventually fill the manifold, so you're left with the interior of the balloon (an open ball) with identifications on the boundary, so it's dense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make this precise by putting a Riemann metric on the manifold and using the Hopf-Rinow Theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation between my answer and Jason's comment would be to take a handle decomposition of the manifold and to consider a maximal tree in the dual 1-skeleton as giving a &quot;tree-like ball&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: responding to Qiaochu's comment, I believe the answer is affirmative for connected, non-compact manifolds as well.  The technique of the proof has to be adapted some.  Step 1: construct a proper Morse function on $M$, meaning a Morse function $f : M \to [0,\infty)$ such that it has only one local minimum -- which is the global minimum, $0$, and demand that $f^{-1}[0,c]$ is connected and compact for all $c\geq 0$.  Step 2: you construct the charts on $f^{-1}[c_i,c_{i+1}]$ where $c_1&amp;lt;c_2&amp;lt;\cdots$ are an increasing sequence $\lim_{n\to\infty} c_n =\infty$ which are regular values of $f$.  Step 3: paste the charts together along little arcs connecting them, pairwise. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-11-25T02:19:45.387" LastActivityDate="2010-11-25T02:19:45.387" CommentCount="1" />
  <row Id="11777" PostTypeId="2" ParentId="11769" CreationDate="2010-11-25T02:06:48.710" Score="1" ViewCount="0" Body="&lt;p&gt;Depending on what you consider a manifold, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Long_line_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;long line&lt;/a&gt; may be a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for a non-connected manifold, surely the answer is no?  Take your favorite smooth manifold, and take a disjoint union of more than $\mathfrak{c}$ copies of it.  Again, unless your definition of &quot;manifold&quot; rules this out (by assuming separability, etc).&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-25T02:06:48.710" CommentCount="2" />
  <row Id="11778" PostTypeId="1" CreationDate="2010-11-25T02:54:13.690" Score="0" ViewCount="108" Body="&lt;p&gt;I hope that someone's up for the challenge; I'm attempting to solve this via computer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int_{-\pi}^\pi{\displaystyle \frac{e^{i\cdot a\cdot t}(e^{i\cdot b\cdot t}-1)(e^{i\cdot c \cdot t}-1)}{(e^{i\cdot t}-1)(e^{i\cdot d \cdot t}+1)(e^{i\cdot f \cdot t}-1)} \dots dt}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know if it's possible to break this up into simpler subproblems.  You can use just about anything to do this, but there is one restriction.  In computer science terms, I want this to be in $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to explain.  I don't want the work I have to do grow exponentially.  For instance, if I break up the integral into two integrals, I don't want to double the amount of work I have to do.  I don't want to dramatically increase the amount of work I have to do by breaking things apart, I want to keep things fairly fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;  I'd prefer to see integration techniques used for this.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-25T03:23:14.230" LastActivityDate="2010-11-25T03:23:14.230" Title="Insidious exponential integral" Tags="&lt;intuition&gt;&lt;integral&gt;" CommentCount="9" />
  <row Id="11779" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T02:55:59.880" Score="2" ViewCount="0" Body="&lt;p&gt;I agree with the theory that the distinction between the notes is used for visual aid and reference points. In addition to that, it was meant to be treated as a vertically rising instrument as if you were to go up a ladder of sorts and those accidentals ( in the case of C, the black notes) are the grips to reach to the next level. As we would refer to them as leading notes. There is also another reason why there are that many notes. Almost all scales are a variation of the major scale or aeolian mode. This scale is designed to have a certain number of tones and semitones to give it the feel of a major scale. If there were too many tones or too many semitones it would not be the same because it will produce too much dissonance or invariably consonance. That is why there is a standard tuning for pianos i.e. A440. If the interval in vibration were to be changed it would not be the same because if the vibrations aren't in sync, the resonation will be totally off. That is why there can be only soo many notes on a piano and make sense to the human ear. Other tunings are possible but the same effect is made that the intervals are kept in a strict way to keep harmony. So, getting back to your question Mathematically, yes there is a reason for that specific order of white keys and black keys. Most of its relation deals with the mode theory of 12 notes and the circle of fifths where if you were to expand the notes on the piano it will form a perfect circle in diminished chords of C as the cardinal points. If you were to go in fifths in a clockwise direction the circle will be C g d A e b F#/Gb d#/db Ab Eb Bb F where when compressed into one octave it turns out with 7 natural notes and 5 accidentals&lt;/p&gt;&#xA;" OwnerUserId="3954" LastActivityDate="2010-11-25T02:55:59.880" />
  <row Id="11780" PostTypeId="1" CreationDate="2010-11-25T03:06:04.110" Score="5" ViewCount="55" Body="&lt;p&gt;I have the following problem. There is a binary relation $\succeq$ defined on a separable metric space $X$, and $\succeq$ is total, transitive, and closed (as a subset of $X\times X$, in the product metric). There exists a set $Y\subseteq X$ that bounds the elements of $X$ in order (that is, for each $x \in X$ one can find $y_{1},y_{2} \in Y$ such that $y_{1}\succeq x\succeq y_{2}$). The set $X$ is itself convex, and the order $\succeq$ is affine when restricted to $Y$ (you may interpret the restriction of $\succeq$ to $Y$ as a convex subset of $Y \times Y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence class $[x]$ the element $x$ belongs to intersects the set $Y$. Informally, I want to construct a measurable selection from that intersection: a (Borel) measurable function that maps $x$ to an element $y$ of $Y$ such that $y\in [x]$. When the set $X$ is compact I can construct it easily. In general, for $X$ separable, I am not able to find one. I came across some measurable selection theorems, but they all work for Polish spaces. I also found a theorem in Kechris's book (Classical Descriptive Set Theory) that could be useful, but it also works for Polish spaces only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My question is&lt;/em&gt;: Where can I find (measurable) selection theorems from equivalence classes? Is there any special theory for it (or counterexamples)? (Sorry for the vague question, but I am not quite acquainted with the literature.)&lt;/p&gt;&#xA;" OwnerUserId="3959" LastActivityDate="2010-11-25T03:06:04.110" Title="Measurable selection from equivalence classes" Tags="&lt;reference-request&gt;" CommentCount="0" />
  <row Id="11781" PostTypeId="2" ParentId="11495" CreationDate="2010-11-25T03:31:23.667" Score="0" ViewCount="0" Body="&lt;p&gt;Get a solid foundation in problem solving. Many CS educations are actually not that math heavy but often rely on what is generally know as a &quot;certain mathematical maturity&quot;. Luckily quite a few good books exists about this subject, but none more important (and accssible/enjoyable) than: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/How-Solve-Aspect-Mathematical-ebook/dp/B001R23FWA/&quot; rel=&quot;nofollow&quot;&gt;How to solve it&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stay clear of the graphics/optimization (and image processing) curses then you might get a graduate degree without doing any Linear Algebra, however you should properly pick up a basic book on discrete mathematics. Plus if you plan to do any programming lanugage theory you should read a book about mathematical proofs (In which case you should pick &lt;a href=&quot;http://www.amazon.com/How-Prove-Structured-Approach-ebook/dp/B000VHZRU0/&quot; rel=&quot;nofollow&quot;&gt;How to prove it&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And last, the only way to learn math is to do math, and in the context of CS you can get lots of good training by starting on &lt;a href=&quot;http://projecteuler.net/&quot; rel=&quot;nofollow&quot;&gt;project Euler&lt;/a&gt;. Finally top it off with reading a good algorithm book like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1848000693&quot; rel=&quot;nofollow&quot;&gt;The Algorithm Design Manual&lt;/a&gt; by&#xA;Steven S. Skiena.&lt;/p&gt;&#xA;" OwnerUserId="3493" LastActivityDate="2010-11-25T03:31:23.667" CommunityOwnedDate="2010-11-25T12:19:27.163" />
  <row Id="11782" PostTypeId="1" AcceptedAnswerId="11783" CreationDate="2010-11-25T03:35:12.920" Score="0" ViewCount="55" Body="&lt;p&gt;Let $\{r_1, r_2, r_3,\ldots \}$ be an enumeration of the rationals in $[0,1]$, and let $(a_k)$ be a sequence such that $\sum_{k=1}^{\infty}{a_k}$ converges absolutely. Set $f_k(x) = a_k\,|x-r_k|^{-1/2}$. I'd like to show that $$\sum_{k=1}^{\infty}{\int_{[0,1]}{|f_k(x)|\,dx}}&amp;lt;\infty\,.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've rewritten the sum as $$\sum{|a_k|\,\int_0^1{\dfrac{1}{|x-r_k|^{1/2}}\,dx}}\,.$$ For each $k$, I broke up the integral into two integrals to account for the absolute value. After integrating, I got $2(\sqrt{r_k}+\sqrt{1-r_k})$. (If anyone is willing to double check that... that'd be awesome.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now how do I show that $$2\sum{|a_k|\,(\sqrt{r_k}+\sqrt{1-r_k})}$$ converges? That's where I'm stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: You know, I typed all this stuff up, and then as soon as I submitted, I realized that $\sqrt{r_k}$ and $\sqrt{1-r_k}$ are bounded by $1$. So $\sum{|a_k|\,(\sqrt{r_k}+\sqrt{1-r_k})}\le \sum{2|a_k|}$, and we know that $\sum{|a_k|}$ is convergent. It's a simple comparison. (Someone please tell me I'm correct, haha.)&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-11-25T03:52:11.700" Title="Show a series of integrals is convergent" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11783" PostTypeId="2" ParentId="11782" CreationDate="2010-11-25T03:40:13.207" Score="1" ViewCount="0" Body="&lt;p&gt;$\textbf{HINT:}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\sqrt{r_k} + \sqrt{1-r_k} \leq \sqrt{2}$, $\forall r_k \in \mathbb{Q} \cap [0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove the above by Arithmetic mean - Geometric Mean Inequality&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T03:52:11.700" LastActivityDate="2010-11-25T03:52:11.700" />
  <row Id="11784" PostTypeId="1" AcceptedAnswerId="11800" CreationDate="2010-11-25T03:45:49.537" Score="2" ViewCount="145" Body="&lt;p&gt;So I am studying for my calculus midterm and I had the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f_{x}(a, b)$ and $f_{y}(a,b)$ both exist then $f$ is differentiable at $(a,b)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answered true, but the answer is false. Why is that? I haven't done real analysis yet (dropped it this term) so I won't know what your talking about if its too advanced. If you attempt to use epsilons and deltas then please try to explain it as simple as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T20:00:07.407" LastActivityDate="2010-11-25T23:14:23.170" Title="Differentiable function - Easy Calc." Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="11785" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T03:50:14.870" Score="1" ViewCount="0" Body="&lt;p&gt;Think about when can you say a function is differentiable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that $f_x$ and $f_y$ gives you the partial derivatives in just 2 directions namely $X$ axis and $Y$ axis respectively.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-25T03:50:14.870" />
  <row Id="11786" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T03:52:24.927" Score="4" ViewCount="0" Body="&lt;p&gt;$f=\frac{xy}{x^2+y^2}$ when $(x,y)\neq (0,0)$ and $f(0,0)=0$ is a standard counterexample for this. $f_x(0,0)=f_y(0,0)=0$, but $f$ is not differentiable (or even continuous at the origin).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-25T03:52:24.927" />
  <row Id="11787" PostTypeId="1" AcceptedAnswerId="11794" CreationDate="2010-11-25T04:00:45.723" Score="4" ViewCount="96" Body="&lt;p&gt;Why is it that cubics are separable over fields that are not of characteristic $2$ or $3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the starting point for some a discussion of the Galois group of a cubic, but I seem to be stuck right off the bat. &lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:19:48.467" LastActivityDate="2010-11-25T07:39:52.907" Title="Why are cubics separable over fields that are not characteristic 2 or 3" Tags="&lt;abstract-algebra&gt;&lt;polynomials&gt;&lt;elementary&gt;&lt;galois-theory&gt;" AnswerCount="2" />
  <row Id="11788" PostTypeId="2" ParentId="11495" CreationDate="2010-11-25T04:01:30.817" Score="0" ViewCount="0" Body="&lt;p&gt;I'm currently in CS, so i could give you some titles of the books i have used over the years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discrete mathematics with applications by Susanna S.Epp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus by Larson, Hostetler, Edwards ISBN:  0-618-14918-X I like this book the best for calc. I have like 5 calc books so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also You will probably be learning some sort of assembly language. I can't remember if i even had a book for that, but we use MIPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had to learn java for first year programming language, which almost any book from amazon should do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For C, once you learn java you should pick up C pretty quickly. I really don't even read this book. The class i am taking now is Haskell, ruby and prolog in one semester. So, haskell would be a nice little language to learn. Its pretty cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i would definitely learn Calc and discrete math or statistics. discrete math is somewhat similar to stats so it would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if a place to start right now is Calc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also i recommend CLEPing some of your general lower division course. This will save you money. Any REA book or even &quot;5 steps to A 5&quot; by mcgraw hill. While it will say AP what subject you take i found it useful for passing. I studied probably 2 weeks for one class i have never taken in my life and passed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also plenty of resources out there like MIT online and Stanford online. They have free courses online.&lt;/p&gt;&#xA;" OwnerUserId="3961" LastActivityDate="2010-11-25T04:01:30.817" CommunityOwnedDate="2010-11-25T12:19:27.163" />
  <row Id="11789" PostTypeId="2" ParentId="11044" CreationDate="2010-11-25T04:06:19.677" Score="0" ViewCount="0" Body="&lt;p&gt;So I'll give the proof I'm used to.  It relies on what (I've been led to believe) Brendan McKay coined the &quot;switching principle&quot;:  Suppose you have a bipartite multi-graph, with parts $A$ and $B$.  Suppose every vertex in $A$ has degree $c$ and every vertex in $B$ has degree $d$.  Then $Ac=Bd$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say a Latin rectangle is &lt;em&gt;normalised&lt;/em&gt; if the first row is $(1,2,\ldots,n)$.  Let $K_{k,n}$ be the number of $k \times n$ normalised Latin rectangles.  By permuting the columns of a $k \times n$ Latin rectangle, we generate $n!$ distinct Latin rectangles of which exactly one is normalised.  Hence $L_{k,n}=n! K_{k,n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now prove the identity $K_{k,n}=\frac{(n-1)!}{(n-k)!} R_{k,n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take A to be the set of all &lt;em&gt;normalised&lt;/em&gt; $k \times n$ Latin rectangles and take B to be the set of all &lt;em&gt;reduced&lt;/em&gt; $k \times n$ Latin rectangles.  [Actually, B is a subset of A, so this is strictly speaking a bit suspect, but this can be overcome this problem by labelling (I won't do this here, as it's a bit messy).]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G$ be the group of permutations of $\{1,2,\ldots,n\}$ that fix 1.  Given any reduced Latin rectangle $L \in B$ and $\alpha \in G$ we can construct a normalised Latin rectangle by: starting from $L$, permute the columns by $\alpha$, then permute the symbols by $\alpha$.  Call this Latin square $L_\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We construct the bipartite multigraph by: for every $L \in B$ add an edge from $L$ to $L_\alpha$ for all $\alpha \in G$.  Since $L_\alpha$ might equal $L_\beta$, the graph might not be simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every vertex in $B$ has degree $(n-1)!$.  So it is sufficient to show that every vertex in $A$ has degree $(n-k)!$.  There is an edge between $L=(l_{ij}) \in A$ and some vertex in $B$ for each permutation $\alpha \in G$ satisfying $\alpha(i)=l_{i1}$ for all $1 \leq i \leq k$.  There are $(n-k)!$ such permutations, and hence every vertex in $A$ has degree $(n-k)!$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T04:06:19.677" />
  <row Id="11790" PostTypeId="1" CreationDate="2010-11-25T04:10:10.633" Score="0" ViewCount="33" Body="&lt;p&gt;Can elliptic arc (defined as part of an ellipse, with extent not bigger than 90 degrees) be represented by quadratic Bezier curve?&lt;/p&gt;&#xA;" OwnerUserId="3777" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T04:33:59.260" LastActivityDate="2010-11-25T04:33:59.260" Title="Can elliptic arc be represented by quadratic Bezier curve?" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" CommentCount="4" />
  <row Id="11792" PostTypeId="1" CreationDate="2010-11-25T04:26:48.053" Score="3" ViewCount="121" Body="&lt;p&gt;I went through this thread on &lt;a href=&quot;http://math.stackexchange.com/questions/329/best-ever-book-on-number-theory&quot;&gt;this question&lt;/a&gt; asking about a text on number theory, but my question is slightly different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a short and sweet book on elementary number theory. I don't need it to be a definitive treatise on number theory. I just need to have book which covers enough material to advance to algebraic number theory. I have looked at the books by Niven-Zuckerman-Montgomery and the Jones' and the one by Burton. All of them were very readable and I would like to read at least one of them at some point. For now, I would just like a book (say 100 pages long) which can get me acquainted with most of the material needed for algebraic number theory. Basically there is a course on algebraic number theory that I wish to take which begins in a month's time and I have not done enough elementary number theory. &lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:47:23.330" LastActivityDate="2010-11-25T09:47:12.820" Title="Best introductory text on elementary number theory that is concise" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11794" PostTypeId="2" ParentId="11787" CreationDate="2010-11-25T04:32:33.070" Score="4" ViewCount="0" Body="&lt;p&gt;A polynomial has multiple roots if and only if it is not relatively prime to its formal derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $f(x) = x^n + a_{n-1}x^{n-1}+\cdots+a_0$ is an irreducible polynomial of degree $n\gt 0$ over a field $\mathbf{F}$. Then $f&amp;#39;(x) = nx^{n-1}+(n-1)a_{n-1}x^{n-1}+\cdots + a_1$. Because $f(x)$ does not have any proper factors in $\mathbf{F}[x]$, then the only possible common factors of $f(x)$ and $f&amp;#39;(x)$ are, up to units, $1$ and $f(x)$. But if $f&amp;#39;(x)\neq 0$, then it is of degree strictly smaller than $f(x)$, so it cannot be divisible by $f(x)$. So &lt;em&gt;either&lt;/em&gt; $f&amp;#39;(x)=0$, or else $\gcd(f,f&amp;#39;)=1$. But the only way in which we can have $f&amp;#39;(x)=0$ is if $n=0$ in $\mathbf{F}$; that is, the characteristic of $\mathbf{F}$ divides $n$. (In fact, more is true: $f(x)$ must be expresssible as $g(x^p)$, where $g(x)$ is a polynomial and $p$ is the characteristic of $\mathbf{F}$, but that is not needed here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a polynomial is separable if and only if its &lt;em&gt;irreducible factors&lt;/em&gt; are separable. Take a cubic polynomial $f(x) = x^3 + ax^2+bx+c$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f(x)$ is irreducible, then the only way it can fail to be separable is if $\mathrm{char}(\mathbf{F})|3$, that is, if the characteristic of $\mathbf{F}$ is $3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f(x)$ is the product of an irreducible quadratic and a linear polynomial, $f(x) = \ell(x)q(x)$, then $\ell(x)$ is certainly separable, so the only way for $f(x)$ to fail to be separable is for $q(x)$ to not be separable; this is a monic polynomial of degree $2$, so for $q(x)$ to &lt;em&gt;not&lt;/em&gt; be separable we need $\mathrm{char}(\mathbf{F})$ to divide $2$; that is, $\mathbf{F}$ to be of characteristic $2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f(x)$ is a product of three linear polynomials (possibly repeated), then $f(x)$ is separable, because each irreducible factor is linear and hence separable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, a &lt;em&gt;necessary&lt;/em&gt; condition for a cubic polynomial to &lt;strong&gt;not&lt;/strong&gt; be separable is for the characteristic of the field of definition to be either $2$ or $3$. In consequence, if $f(x)$ is a cubic over a field of characteristic not equal to $2$ or to $3$, then it &lt;em&gt;must&lt;/em&gt; be separable.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-25T04:32:33.070" CommentCount="2" />
  <row Id="11795" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T04:45:25.623" Score="3" ViewCount="0" Body="&lt;p&gt;Need the partial derivatives exist and are $continuous$ in a open  ball around $(a, b)$. This is sufficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;You can check : &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_derivative&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Partial_derivative&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-25T13:22:23.633" LastActivityDate="2010-11-25T13:22:23.633" />
  <row Id="11796" PostTypeId="2" ParentId="7090" CreationDate="2010-11-25T04:45:47.753" Score="4" ViewCount="0" Body="&lt;p&gt;A complete residue system modulo $n$ is a set of $n$ integers containing a representative element from each congruence class modulo $n$ (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for more details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commonly used complete residue system is $\{0,1,\ldots,n-1\}$ where it is easy to see that any two elements are incongruent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim states that if gcd(a,n)=1, then $\{0,a,2a,\ldots,(n-1)a\}$ is a complete residue system modulo $n$.  Since there are exactly $n$ elements in this set, in order to prove the claim, it is sufficient to show that $ka \not\equiv ra \pmod n$ whenever $0 \leq k&amp;lt;r \leq n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: $ka \equiv ra \pmod n$ implies $k \equiv r \pmod n$.  While true, I think it should be remarked that we may &quot;divide&quot; by $a$ only because it is comprime to $n$.  In fact, it's not really division, it's multiplication by an integer $b$ such that $ab \equiv 1 \pmod n$ (which can be shown to exist using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt;).  So $ka \equiv ra \pmod n$ implies $kab \equiv rab \pmod n$ implies $k \equiv r \pmod n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: $k \equiv r \pmod n$ implies $n$ divides $k-r$.  There's a typo in the OP's statement.  This is by the definition of congruent modulo n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: $n$ divides $k-r$ implies $k=r$ because $0 \leq k&amp;lt;r \leq n-1$.  There's some information left out here in the OP's statement.  In order to have $n$ dividing $k-r$ while both $k$ and $r$ are between $0$ and $n-1$, we must have $k=r$ (note that 0 is divisible by n).&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T04:45:47.753" />
  <row Id="11797" PostTypeId="2" ParentId="11792" CreationDate="2010-11-25T04:46:26.843" Score="3" ViewCount="0" Body="&lt;p&gt;The thing about algebraic number theory is that, although the problems you will be looking at come from &quot;classical&quot; number theory, the techniques owe a lot more to abstract algebra than to classical number theory. You would be better served making sure that you have a solid understanding of commutative rings, field extensions, Galois Theory, and the basics of modules over commutative rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that respect, the book by Ireland and Rosen that was mentioned in that thread, &lt;em&gt;A Classical Introduction to Modern Number Theory&lt;/em&gt;, may be a good one, since it is leading you in &lt;em&gt;that&lt;/em&gt; direction in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I took Algebraic Number Theory (which I absolutely loved; it was a graduate two-course sequence), I had not had any number theory beyond the very elementary congruence stuff; I had never seen anything like Quadratic Reciprocity, or Fermat's Christmas Theorem, or anything like that. I did not feel hobbled by this lack. But I certainly had to lean heavily on the commutative algebra and Galois Theory I knew! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was only &lt;em&gt;later&lt;/em&gt; that I sat in on an upper-division number theory course (based on the Ireland-Rosen book mentioned above), and that I read through Gauss's &lt;em&gt;Disquisitiones&lt;/em&gt; (on my own) and other number theory books. Then again, I'm not a number theorist, so perhaps I am in fact deeply handicapped by this but I don't know it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T04:52:17.443" LastActivityDate="2010-11-25T04:52:17.443" CommentCount="3" />
  <row Id="11798" PostTypeId="2" ParentId="6575" CreationDate="2010-11-25T05:02:56.693" Score="1" ViewCount="0" Body="&lt;p&gt;There's a lot of similar things that are called &quot;trees&quot; in graph theory.  I interpret this question to mean full binary trees (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_tree&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;) -- that is, each non-leaf node has a &quot;left&quot; and a &quot;right&quot; descendant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;:  How can you modify a full binary tree with n+1 nodes to a full binary tree with n nodes?  Well, you could try deleting a leaf-node, but then you will end up with a non-leaf node with only one descendant (which isn't a full binary tree).  So this isn't going to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can you ensure that a full binary tree (with fewer nodes) is constructed?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T05:02:56.693" />
  <row Id="11799" PostTypeId="2" ParentId="6575" CreationDate="2010-11-25T05:16:57.633" Score="0" ViewCount="0" Body="&lt;p&gt;How do you define a leaf?  The idea of structural induction would be to show that whenever you add a new node to an existing tree, the difference between leaves and nodes without right siblings stays constant.  Without the leaf definition I can't help further.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-25T05:16:57.633" />
  <row Id="11800" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T05:24:19.337" Score="3" ViewCount="0" Body="&lt;p&gt;Others have given you some of the standard examples that you can verify are indeed examples. Let me try to give you some intuition of &lt;em&gt;why&lt;/em&gt; the partials are not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, a function $f\colon\mathbb{R}^2\to\mathbb{R}$ is differentiable at a point $(x_0,y_0)$ if you can find a plane that is tangent to the graph of $f$ at the point $(x_0,y_0,f(x_0,y_0))$. Here, &quot;tangent&quot; is interpreted the same way that a line is said to be tangent to a curve when dealing with $g\colon\mathbb{R}\to\mathbb{R}$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the one-variable case, a line $y=mx+b$ is tangent to the graph of $y=g(x)$ at the point $c$ if and only $mc+b=g(c)$ (they go through the same point) &lt;em&gt;and&lt;/em&gt; the approximation given by the tangent as $x$ approaches $c$ goes to $0$ &#xA;&lt;em&gt;faster&lt;/em&gt; than $x$ approaches $c$; that is, &#xA;$$\lim_{x\to c}\frac{g(x)-(mx+b)}{x-c}=0.$$&#xD;&#xA;&lt;em&gt;Note:&lt;/em&gt; To see that this certainly will happen for the tangent, remember that the tangent will have $m=g&amp;#39;(c)$ and $b=g(c) - g&amp;#39;(c)c$, so the expression in the limit becomes&#xA;$$\frac{g(x) - (g&amp;#39;(c)x + g(c)-g&amp;#39;(c)c)}{x-c} = \frac{g(x) - g(c) - g&amp;#39;(c)(x-c)}{x-c} = \frac{g(x)-g(c)}{x-c} - g&amp;#39;(c).$$&#xD;&#xA;But in fact, one can &lt;em&gt;define&lt;/em&gt; the tangent as the (unique) line with this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do the same thing with two variables, but with planes instead of lines. Differentiability at $(x_0,y_0)$ for $f(x,y)$ means that there is a plane $z=ax+by+c$, with $ax_0 + by_0 + c = f(x_0,y_0)$, and such that &#xA;$$\frac{f(x,y) - (ax + by + c)}{||(x,y)-(x_0,y_0)||}$$&#xD;&#xA;goes to $0$ as $(x,y)\to (x_0,y_0)$: the plane approaches the graph, in &lt;em&gt;every&lt;/em&gt; direction, faster than the arguments are approaching the point. (Here, $||\cdot||$ is the usual norm in $\mathbb{R}^2$: $||(a,b)-(c,d)||=\sqrt{(a-c)^2+(b-d)^2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this has to happen along &lt;em&gt;all&lt;/em&gt; directions of approach to $(x_0,y_0)$; just like with limits of two variables, approaching along vertical lines alone or along horizontal lines alone is not enough (even approaching along &lt;em&gt;lines&lt;/em&gt; is not enough; it really has to be along &lt;em&gt;all&lt;/em&gt; directions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, the partial derivative with respect to $x$ only tell you that if you &quot;slice&quot; the graph with a plane parallel to the XZ-plane and you look at the &quot;shadow&quot; on that plane (which will look just like the graph of a one-variable function), then you can approach the bit of graph with an appropriate line; that is, you can approach the graph of $z=f(x,y)$ (which is a three-dimensional thing) with a &lt;em&gt;line&lt;/em&gt; parallel to the XZ plane; and the partial derivative with respect to $y$ tells you that you can approach the graph with a &lt;em&gt;line&lt;/em&gt; parallel to the YZ-plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, &lt;em&gt;if&lt;/em&gt; you can approach the graph with a plane appropriately, then you can also approach it with lines (just make sure the lines are contained in the plane). But being able to approach it in the two coordinate directions alone is not enough to make sure you can actually get a whole plane close to the graph in &lt;em&gt;all&lt;/em&gt; directions. That is why knowing that the partial derivatives exist does not tell you that the function is differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nice thing, as mentioned by Bryan Yocks, is that if if the partial derivatives, viewed as functions of two variables, not only exist but are also &lt;em&gt;continuous&lt;/em&gt;, then you can leverage this extra information into showing you can find a plane. Intuitively, what happens here is that the lines do give you enough information because you know that points close to $(x_0,y_0)$ will have similar lines that are very close to the ones you have, and so you can picture in your head perhaps that as you draw all these lines, &quot;locally&quot; it's going to look very much like a plane because the change in either of the two directions is very small. But if all you know is that the partial derivatives exist, this is not enough for the reasons described above. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T05:46:08.157" LastActivityDate="2010-11-25T05:46:08.157" />
  <row Id="11801" PostTypeId="2" ParentId="11787" CreationDate="2010-11-25T05:28:06.573" Score="4" ViewCount="0" Body="&lt;p&gt;You can actually do this &quot;by hand&quot; too. Our goal is to show that a monic irreducible polynomial $f(x)$ of degree $2$ or $3$ over $F$ splits into distinct linear factors in an algebraic closure $E$, as long as the characteristic of the $F$ is not $2$ or $3$. Suppose the irreducible $f(x)$ does not split into linear factors in $E$; we will get a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if the polynomial is of the form $(x - a)^3 = x^3 - 3ax^2 + 3a^2x + a^3$, then the $x^2$ coefficient $-3a$ must be in $F$. So $a$ is in $F$ too, using that $F$ is not of characteristic $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining case is that the polynomial is of the form $(x - a)^2(x - b)$.&#xA;For this we use the fact that for any polynomial $p(x)$, ${p&amp;#39;(x) \over p(x)} = \sum {1 \over x - r}$, where the sum is taken over the roots of $p(x)$. Applying this to $f(x)$ gives that ${f&amp;#39;(x) \over f(x)} = {2 \over (x - a)} + {1 \over x - b}$.  Note that $f(x) = x^3 - (b + 2a)x^2 + ...$, so that $b + 2a \in F$. Hence ${b + 2a \over 3} \in F$ too. Substituting this value of $x$ into ${f&amp;#39;(x) \over f(x)}$ gives gives ${9 \over 2(b - a)}$, which must be in $F$. Thus $b - a\in F$; note we use the characteristic is not $2$ or $3$ here. So $b = {2 \over 3}(b - a) + {1 \over 3}(b + 2a)$ is in $F$ as well, again contradicting the irreducibility of $f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-25T07:39:52.907" LastActivityDate="2010-11-25T07:39:52.907" />
  <row Id="11802" PostTypeId="1" AcceptedAnswerId="11805" CreationDate="2010-11-25T05:34:07.760" Score="0" ViewCount="118" Body="&lt;p&gt;I've never had to do this before, so I'm not really sure how to do it. These problems also don't even really relate to what the subject of the book is as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: $f(px+(1-p)y)\le pf(x)+(1-p)f(y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)\left\{ \begin{array}{cc}&#xD;&#xA;x &amp;amp; x\le1\\&#xD;&#xA;2-x &amp;amp; x&amp;gt;1\end{array}\right\} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $f(x)$ is convex on $-\infty &amp;lt; x \le 1$ and convex on $1 &amp;lt; x &amp;lt; \infty$ but not convex on $-\infty &amp;lt; x &amp;lt; \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem using the same convex definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f(x)$ is defined on the interval I. If $x_{1}&amp;lt;x_{2}&amp;lt;x_{3}$ are in I and $f(x_{1})&amp;lt;f(x_{2})$ and $f(x_{3})&amp;lt;f(x_{2})$, then show that $f(x)$ is not convex on I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Im not sure how to use the definition. Thats kind of where i am stuck. I'm sure it's easy, but like i said i've never seen this before, the book has nothing to do with this, it's just randomly thrown in here. There are no examples for this either.&lt;/p&gt;&#xA;" OwnerUserId="3961" LastEditorUserId="3961" LastEditorDisplayName="" LastEditDate="2010-11-26T22:22:25.193" LastActivityDate="2010-11-26T22:22:25.193" Title="show that function is not convex" Tags="&lt;convex-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11803" PostTypeId="2" ParentId="11802" CreationDate="2010-11-25T05:39:42.907" Score="0" ViewCount="0" Body="&lt;p&gt;You just need to apply the definition for these two problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first problem,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two points $x,y$ such that $x,y \leq 1$. Argue why the definition of convexity is true in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider two points $x,y$ such that $x,y &amp;gt; 1$. Argue why the definition of convexity is true in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $x &amp;lt; 1$ and $y &amp;gt; 1$ and argue why the definition is violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second problem, your statement should read &quot;If $x_1 &amp;lt; x_2 &amp;lt; x_3$ are in $I$ and $\ldots$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the same idea as before (argue from the definition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x_2$ can be written as a weighted linear combination of $x_1$ and $x_3$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T05:45:13.480" LastActivityDate="2010-11-25T05:45:13.480" />
  <row Id="11804" PostTypeId="2" ParentId="11792" CreationDate="2010-11-25T05:56:55.420" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ktCZ2MvgN3MC&amp;amp;printsec=frontcover&amp;amp;dq=sierpinski&amp;amp;hl=en&amp;amp;ei=e_rtTK2HNcSAlAeC1rWtDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCMQ6AEwAA#v=onepage&amp;amp;q=sierpinski&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Elementary Theory of Numbers&lt;/a&gt; by Sierpinski.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-25T05:56:55.420" CommentCount="3" />
  <row Id="11805" PostTypeId="2" ParentId="11802" CreationDate="2010-11-25T05:57:21.693" Score="2" ViewCount="0" Body="&lt;p&gt;If you look at the definition geometrically (on $\mathbb{R}$) it states that a function is convex if each point of the graph $(x, f(x))$ is below any line that adjoins graph points $(a, f(a))$ and $(b, f(b))$ where $a&amp;lt;x&amp;lt;b$. In other words, for any ordered triple $a&amp;lt;x&amp;lt;b$ &#xA;the point $(x,f(x))$ is below the lint through $(a,f(a))$ and $(b,f(b))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about the graph of your function - look at $x&amp;lt;1$... look at $x&amp;gt;1$... look at $x=1$...&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-25T05:57:21.693" />
  <row Id="11806" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T06:10:02.810" Score="3" ViewCount="0" Body="&lt;p&gt;Check out &lt;a href=&quot;http://arxiv.org/abs/0711.1873&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which is about the regular 12-gon and music theory.  It will help you answer this question, as well as many others that are similar to it.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-25T06:10:02.810" />
  <row Id="11807" PostTypeId="1" AcceptedAnswerId="11816" CreationDate="2010-11-25T06:12:32.357" Score="1" ViewCount="69" Body="&lt;p&gt;I was trying to solve this congruence equation: $2ax+b\equiv 0 \pmod 5$. Here, $b$ is any integer and $a$ is an integer not divisible by $5$. Does a solution always exist? I don't understand why they have set the condition above on $a$. As should be clear from my other question on elementary number theory, I am not very good at it, and I have no idea where to begin when solving congruences. Any help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T06:32:29.103" LastActivityDate="2010-11-25T07:32:58.683" Title="Solution of a congruence $2ax+b\equiv 0 \pmod 5$" Tags="&lt;elementary-number-theory&gt;" AnswerCount="3" />
  <row Id="11808" PostTypeId="2" ParentId="11807" CreationDate="2010-11-25T06:25:02.020" Score="1" ViewCount="0" Body="&lt;p&gt;If $\ a\equiv 0\ $ then $\ b\equiv 0\:,\ $ so it wouldn't be solvable for $b\not\equiv 0$. Conversely, if $\ a\not\equiv 0 \ $ then $\ 1/a \equiv a^3\ $ since $\ a^4\equiv 1 \ $ by Fermat's little theorem, so you can cancel the leading coefficient and solve for $x$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-25T06:25:02.020" />
  <row Id="11809" PostTypeId="1" AcceptedAnswerId="11813" CreationDate="2010-11-25T06:25:19.633" Score="2" ViewCount="107" Body="&lt;p&gt;Question: Let $U$ denote a subset of $\mathbb{R}$. We say that $U$ is an open set when for every $x \in U$ there is an $\epsilon &amp;gt; 0$ for which $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is contained in $U$. Show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) For $a,b \in \mathbb{R}$ the interval $(a,b)$ is open, but the interval $[a,b]$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) If $U,W \subset \mathbb{R}$ are open, so are $U \cup W$ and $U \cap W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cup_{i \in I}U_{i}$ is also open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iv) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cap_{i \in I}U_{i}$ is not necessarily open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&#xA;I think that with (i) i can conceptually understand everything:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x \in (a,b): a&amp;lt; x - \frac{|a-x|}{2} &amp;lt; x &amp;lt; x + \frac{|b-x|}{2} &amp;lt; b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;gt; \frac{|b-a|}{2}$ let $\epsilon = \frac{|b-x|}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;lt; \frac{|b-a|}{2}$ let $\epsilon = \frac{|a-x|}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \forall x \in (a,b) \exists \epsilon &amp;gt; 0 : U_{\epsilon}(x) = (x - \epsilon, x +\epsilon) \Rightarrow$ the interval (a,b) is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;max$[a,b] = b \Rightarrow \forall x \in [a,b]: b \geq x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\epsilon &amp;gt; 0, b \ngeq b+ \epsilon \Rightarrow$ there exists an $x$ for which  $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is not contained in $U \Rightarrow$ the interval is not open&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With (ii) and (iii) i think i understand, but i am not sure how to start stating that they are true mathematically. to be sure,  $\cup_{i \in I}U_{i}$ denotes several unions in a row, correct? i'm even reading a proof which goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $v \in U \cap W$, there exists an open ball $B$ of radius $r$ centered at $v$ contained in $U$, and there exists an open ball $B&amp;#39;$ of radius $r&amp;#39;$ centered at $v$ contained in $W$. Let $\delta =$ min$(r, r')$. Then the open ball of radius $\delta$ centered at $v$ is contained in $U \cap W$, which is therefore open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately i'm having trouble interpreting this and am wondering if there is another way to put it or approach it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With (iv) i am not even sure intuitively why it should be true. i can't think of a situation where it wouldn't be true, i guess it has to do with whether the sets are finite or not, but i don't see how this would make a difference (especially given that openness is apparently preserved with infinite unions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can help provide suggestions, corrections, hints, etc i appreciate it, thanks!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-11-25T14:25:45.493" LastActivityDate="2010-11-25T19:31:35.010" Title="Proving statements related to open sets" Tags="&lt;analysis&gt;&lt;general-topology&gt;" AnswerCount="1" />
  <row Id="11810" PostTypeId="1" CreationDate="2010-11-25T06:27:07.237" Score="8" ViewCount="155" Body="&lt;p&gt;This question arises from a Fourier transform class I took about a year back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poisson summation formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{n= - \infty}^{\infty} f(n) = \displaystyle \sum_{k= - \infty}^{\infty} \hat{f}(k)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{f}$ is the Fourier Transform of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting since this is true for all $f(x)$ for which we can define Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a nice (probably physical) interpretation for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if this characterizes some property which is invariant, some sort of conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if we consider Parseval's theorem, one interpretation of it is that the total energy across all time is the total energy across all of its frequency components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from a mathematical standpoint what does this mean? Is this a manifestation of some property of integers?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T08:41:44.670" LastActivityDate="2010-11-25T11:34:33.910" Title="Interpretation of Poisson Summation Formula" Tags="&lt;number-theory&gt;&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11811" PostTypeId="2" ParentId="11807" CreationDate="2010-11-25T06:41:52.683" Score="0" ViewCount="0" Body="&lt;p&gt;Solving the congruence in this particular example means, we need to get solution for $x$ as $x \equiv m (\bmod 5)$, where $m$ depends on $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when $5|a$, we have $2ax \equiv 0 (\bmod 5)$. Hence, for arbitrary integer $b$, $2ax + b \equiv b (\bmod 5)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if we want a solution to exist for $2ax + b \equiv 0 (\bmod 5)$, then this forces $b$ to be $0 (\bmod 5)$, which is not necessarily true since $b$ is an arbitrary integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are getting started on Number Theory, then you could argue by cases. This will be a good exercise when you try to generalize this to other numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $a \equiv 1 (\bmod 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $ax \equiv x (\bmod 5)$, $2ax \equiv 2x (\bmod 5)$ and hence $2ax + b \equiv 2x+b (\bmod 5)$. And now we want this to be $0 (\bmod 5)$. So we need $2x + b \equiv 0 (\bmod 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the equation by $3$ and we get $6x + 3b \equiv 0 (\bmod 5)$. Now observe that $5x \equiv 0 (\bmod 5)$. Hence, we get $x + 3b \equiv 0 (\bmod 5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason you multiply by $3$ is $3 \times 2 \equiv 1 (\bmod 5)$. So $3$ is the inverse of $2$, when you are working in $(\bmod 5)$ setting. Also note that this gives you the reason why you couldn't do when $5|a$, since $5$ doesn't have an inverse in the $(\bmod 5)$ setting.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is $x \equiv -3b (\bmod 5)$, when $a \equiv 1 (\bmod 5)$.Proceed via the same idea for other cases as well viz $a \equiv 2 (\bmod 5)$, $a \equiv 3 (\bmod 5)$, $a \equiv 4 (\bmod 5)$.You could then generalize this once you observe what is happening. Note that here since $5$ is a prime all the other elements will have inverses. Think what will happen when $5$ is replaced by '$n$' where '$n$' is a composite number.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-25T07:13:16.883" LastActivityDate="2010-11-25T07:13:16.883" />
  <row Id="11813" PostTypeId="2" ParentId="11809" CreationDate="2010-11-25T06:57:37.330" Score="4" ViewCount="0" Body="&lt;p&gt;Intuitively, a set $U$ is open if every point in the set can be completely surrounded by points that are all in the set; here, &quot;surrounded&quot; means that an entire little interval centered at $x$ is completely contained in $U$, which means, in a sense, that $x$ is &lt;em&gt;really&lt;/em&gt; inside $U$. A set is open if every point in the set is &lt;em&gt;really&lt;/em&gt; &quot;inside&quot; the set. Compare with $[a,b]$ (which you show is not open), where one might say that although $b$ &#xA;&lt;em&gt;is&lt;/em&gt; in the set, it is &quot;barely&quot; in the set: move just a bit to the right, and you fall outside $[a,b]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, on to your work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea for the first part of (i) is okay, but the way you describe $\varepsilon$ is not. For example, say $x=1000$, $a=999.9$, and $b=1000.5$. Then $|b-a|/2 = .3$, and $x\gt |b-a|/2$, so you take $\varepsilon = 0.3$. But $(x-0.3,x+0.3)$ is not contained in $(a,b)$, because $x-0.3 = 999.7\lt a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;really&lt;/em&gt; want to say is that $\varepsilon=\min(|x-b|/2, |x-a|/2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, you've got it right, but there is no need to say &quot;there exists an $x$ ...&quot;, just say that for every $\varepsilon\gt 0$, $(b-\varepsilon,b+\varepsilon)$ is not contained in $[a,b]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (ii); not quite: $\cup_{i\in I} U_i$ is not &quot;several unions in a row&quot;, because the set may not something that you can do &quot;one after another&quot;; maybe there is one for every real number! No, $\cup_{i\in I}U_i$ just means &quot;the union of all the $U_i$&quot;; this is the collection of all points that are in at least one of the $U_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you mean $U\cap W$ when you said $V\cap V$. But you've go the idea right for the intersection. Conceptually: you know that you can find an open interval around $x$ that &quot;works&quot; for $U$, and one that works for $W$; to find one that works for &lt;em&gt;both&lt;/em&gt;, just take the smaller one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (iv), it's &lt;em&gt;not&lt;/em&gt; true: you are being asked to show it is not true. Can you think of a way to get a closed interval as an intersection of open ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Okay, here's a hint for (iv): for every positive integer $n$, $\frac{1}{n}$ is stricly larger than $0$; but the infinimum of $\{ \frac{1}{n}\mid n\in\mathbb{Z}_{\gt 0}\}$ is &lt;em&gt;not&lt;/em&gt; strictly larger than $0$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T19:31:35.010" LastActivityDate="2010-11-25T19:31:35.010" CommentCount="2" />
  <row Id="11814" PostTypeId="2" ParentId="11223" CreationDate="2010-11-25T07:20:10.310" Score="0" ViewCount="0" Body="&lt;p&gt;You simply need to look at the number of factors that a given number has.  This equals the number of tutors that have &quot;changed&quot; that light bulb.  For instance, 7 only has one and itself as a factor, so only 2 tutors affect bulb 7; tutors 1 and 7 (the bulb goes from off, to on then off where it remains).  169 on the other hand, has three factors; 1,13, and 169, so this bulb is affected by the tutors numbered 1,13 and 169 (from off to on to off back to on).  &lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T07:20:10.310" />
  <row Id="11815" PostTypeId="2" ParentId="10073" CreationDate="2010-11-25T07:27:23.627" Score="0" ViewCount="0" Body="&lt;p&gt;I'm Not too sure how to answer the first part of your question.  For the second part, a complex variable function can be thought of as a real variable function whose domain and range are both subsets of R^2.  This is usually different then those functions you study in multivariable calculus: functions from R^n to R.  One noticible difference is that for functions of a single complex variable, there is a total derivaitve which exists if the cauchy riemaan differential equations hold.  Real functions on R^2 would have two partial derivatives instead.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T07:27:23.627" />
  <row Id="11816" PostTypeId="2" ParentId="11807" CreationDate="2010-11-25T07:32:58.683" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps you've come across the following theorem in your studies?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $gcd(a,m)=1$, then there is an $x$ such that $ax\equiv 1\pmod{m}$, and any two such $x$ are congruent $\pmod{m}$. If $gcd(a,m)&amp;gt;1$, then there is no such $x$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Essentially, this means an integer $a$ has a multiplicative inverse modulo $m$ if and only if it $a$ is relatively prime to the modulus $m$. This nice thing about primes is that if an integer $a$ is not divisible by a prime $p$, then it is relatively prime to that prime, as the only other common factor can be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since $5\nmid a$, you know $gcd(a,5)=1$, and so there is some $x$, let's call it $\bar{a}$, such that $\bar{a}a\equiv 1\pmod{5}$. This allows you to solve for $x$ in this linear congruence. Note &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2ax+b\equiv 0\Rightarrow 2ax\equiv -b\Rightarrow \bar{2}2ax\equiv -\bar{2}b\Rightarrow \bar{a}ax\equiv -\bar{a}\bar{2}b\Rightarrow x\equiv -\bar{2}\bar{a}b\equiv -3\bar{a}b\pmod{5}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $gcd(2,5)=1$, you also know that $2$ has a multiplicative inverse modulo $5$, namely $3$. This is the same reason why there is such a condition on $a$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-25T07:32:58.683" CommentCount="1" />
  <row Id="11817" PostTypeId="2" ParentId="11373" CreationDate="2010-11-25T07:35:13.650" Score="1" ViewCount="0" Body="&lt;p&gt;In noncommutative rings, the terms &quot;left ideal&quot; and &quot;right ideal&quot; are used indicate if you are multiplying from the left or right.  The term &quot;ideal&quot; usualy means that it is a left and right ideal at the same time.  In commutative ring theory, which is probably what you are studying, this notion of left and right is meaningless since ab and ba are always the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, an ideal is a very special type of subring, with the added property that if a is in the ideal and r is in the ring, r times a is in the ideal (even if r is not in the ideal).  Subrings don't have this extra property in general.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T07:35:13.650" />
  <row Id="11818" PostTypeId="1" AcceptedAnswerId="11819" CreationDate="2010-11-25T08:03:48.497" Score="2" ViewCount="101" Body="&lt;p&gt;In the situation of having a high entropy random number generator, that generates numbers in the range of 0 and 2,147,000,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i have a list of 1,000,000 integer values, what are the chances that a random number will already be on my list?&lt;/p&gt;&#xA;" OwnerUserId="3472" LastActivityDate="2010-11-25T17:32:29.240" Title="Probability of Random number repeating" Tags="&lt;probability&gt;&lt;random&gt;" AnswerCount="2" />
  <row Id="11819" PostTypeId="2" ParentId="11818" CreationDate="2010-11-25T08:10:49.267" Score="3" ViewCount="0" Body="&lt;p&gt;One minus the probability that they are all different:&#xA;$$1 - \left( 1- \frac{1}{n} \right)&#xD;&#xA;\left( 1- \frac{2}{n} \right)&#xD;&#xA;\cdots \left( 1- \frac{k-1}{n} \right),$$&#xD;&#xA;where $n=2,147,000,001$ (since you include $0$) and $k=1,000,000.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Birthday_problem&quot; rel=&quot;nofollow&quot;&gt;birthday problem&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-25T08:10:49.267" CommentCount="0" />
  <row Id="11820" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T09:21:21.250" Score="23" ViewCount="0" Body="&lt;p&gt;The first answer is great, so I'll try to approach the question from another angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there are several different scales, and different cultures use different ones. It depends on the mathematics of the instruments as much as on cultural factors. Our scale has a very long history that can be traced to the ancient Greeks and Pythagoras in particular. They noticed (by hearing) that stringed instruments could produce different notes by adjusting the length of the string, and that some combinations sounded better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Greeks had a lot of interest in mathemathics, and it seemed &quot;right&quot; for them to search for &quot;perfect&quot; combinations - perfect meaning that they should be expressed in terms of fractions of small integer numbers. They noticed that if you double or halve the string length, you get the same note (the concept of an octave); other fractions, such as 2/3, 3/4, also produced &quot;harmonic&quot; combinations. That's also the reason why some combinations sound better, as it can be explained by physics. When you combine several sine waves, you hear several different notes that are the result of the interference betwen the original waves. Some combinations sound better while others produce what we call &quot;disonance&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in theory, you can start from an arbitrary frequency (or note) and build a scale of &quot;harmonic&quot; notes using these ratios (I'm using quotes because the term harmonic has a very specific meaning in music, and I'm talking in broad and imprecise terms). The major and minor scales of Western music can be approximately derived from this scheme. Both scales (major and minor) have 7 notes. The white keys in the piano correspond to the major scale, starting from the C note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you get the C note and use the &quot;perfect&quot; fractions, you'll get the &quot;true&quot; C major scale. And that's where the fun begins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take any note in the C major scale, you can treat that note as the start of another scale. Take for instance the fifth of C (it's the G), and build a new major scale, now starting from G instead of C. You'll get another seven notes. Some of them are also on teh scale of C; others are very close, but not exactly equal; and some fall in the middle of the notes in the scale of C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you repeat this exercise with all notes, you'll end up building 12 different scales. The problem is that the interval is not regular, and there are some imprecisions. You need to retune the instrument if you want to have the perfect scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of &quot;chromatic&quot; scale (with 12 notes, equaly spaced) was invented to solve this &quot;problem&quot;. The chromatic scale is a mathematical approximation, that is close enough for MOST people (but not all). People with &quot;perfect&quot; ear can listen the imperfections. In the chromatic scale, notes are evenly space using the twelfth root of two. It's a geometric progression, that matches with good precision all possible major and minor scales. The invention of the chromatic scale allows players to play music in arbitrary scales without retuning the instrument - you only need to ajust the scale by &quot;offsetting&quot; a fixed number of positions, or semitones, from the base one of the original scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, that's just convention, and a bit of luck. The white keys are an &quot;historical accident&quot;, being the keys of the major scale of C. The other ones are needed to allow for transposition. Also bear in mind that (1) the keys need to have a minimum width to allow for a single finger, and (2) if you didn't have the black keys, the octave would be too wide for &quot;normal&quot; hands to play. So the scheme with a few intermediate keys is needed anyway, and the chromatic scale that we use is at least as good (or better) as any other possible scale.&lt;/p&gt;&#xA;" OwnerUserId="3981" LastActivityDate="2010-11-25T09:21:21.250" CommentCount="1" />
  <row Id="11821" PostTypeId="2" ParentId="3510" CreationDate="2010-11-25T09:26:35.567" Score="-1" ViewCount="0" Body="&lt;p&gt;Ok, here is &quot;an engineer's proof&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not simply differentiate the Euler formula $e^{ix}=\cos x+i\sin x$ twice by $x$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i^2e^{ix}=-cos x-i\sin x$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $i^2=-1$, the proof is ended.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Sorry, i am not true mathematician :)}&lt;/p&gt;&#xA;" OwnerUserId="2323" LastActivityDate="2010-11-25T09:26:35.567" />
  <row Id="11822" PostTypeId="2" ParentId="11611" CreationDate="2010-11-25T09:27:29.497" Score="0" ViewCount="0" Body="&lt;p&gt;The context of this question suggests that $f=f(x)$ and $g=g(x)$ are going to be polynomials (with either real, rational or integer coefficients).  If this is the case then...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;: While there's going to be an infinite number of functions $f,g$ for which $f(g(x))=x^2+x-6$, there will only be two situations when this can occur (consider the degrees of $f$ and $g$, and how that effects $f(g(x))$).  These can be classified by setting $f$ and $g$ as arbitrary polynomials of the appropriate degrees, composing them, then equating coefficients with $f(g(x))=x^2+x-6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ and $g$ are not restricted to polynomials, there's going to be all sorts of functions $f$ and $g$ for which $f(g(x))=x^2+x-6$.  For example: Let $A=\left(\begin{matrix}-2 &amp;amp; 1 \\ 4 &amp;amp; 1 \end{matrix}\right)$.  Then the characteristic polynomial $\det(A-xI)=(-2-x)(1-x)-4=x^2+x-6$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T09:27:29.497" />
  <row Id="11823" PostTypeId="2" ParentId="11792" CreationDate="2010-11-25T09:47:12.820" Score="0" ViewCount="0" Body="&lt;p&gt;The book of Harold Davenport, &lt;strong&gt;The Higher Arithmetic: An introduction to the theory of numbers&lt;/strong&gt;,  fits your bill. &lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T09:47:12.820" />
  <row Id="11824" PostTypeId="2" ParentId="11810" CreationDate="2010-11-25T09:55:26.433" Score="1" ViewCount="0" Body="&lt;p&gt;If you want analogies from electrical engineering, one possible way to understand the consequences of the Poisson summation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;nofollow&quot;&gt;Nyquist sampling theorem&lt;/a&gt;. It says that periodic sampling of a signal is enough to capture all the signals whose frequency is less than or equal to half the sampling rate.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T09:55:26.433" CommentCount="2" />
  <row Id="11825" PostTypeId="1" CreationDate="2010-11-25T09:56:46.607" Score="0" ViewCount="77" Body="&lt;p&gt;Let $\{e_0, e_1, \dots, e_{N-1} \}$ be the Euclidian basis for $l^2(Z_N)$, and let $\{F_0, F_1, \dots, F_{N-1} \}$ be the fourier basis &#xA;i. How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$ for all k? Notice that $\hat{e}_m$ is very nearly ( up to a reflection and a normalization ) an element of the Fourier basis.&#xA;ii. How to show that $\hat{F}_m = e_m$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First some definitions $F_m=N^{-1/2} E_m = \frac{1}{N}e^{2 \pi i m n/N}$ and&#xA;$\hat{z}(m) = \sum_{n=0}^{N-1} z(n) e^{-2 \pi i m n/N}$&#xD;&#xA;i.How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$?&#xA;(my) Solution:&#xA;$\hat{e}_m(k) = \sum_{n=0}^{N-1} e_m(k) e^{-2 \pi i m k/N} = e^{-2 \pi i m k/N}$&#xD;&#xA;ii.  How to show that $\hat{F}_m = e_m$? &#xA;(my solution:&#xA;$\hat{F}_m = \sum_{n=0}^{N-1} \frac{1}{N}e^{2 \pi i m n/N} e^{-2 \pi i m n/N} = \sum_{n=0}^{N-1} \frac{1}{N} = ? $&#xD;&#xA;Any comments on my solutions.&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-11-25T10:43:38.750" LastActivityDate="2010-11-25T10:43:38.750" Title="How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$ for all k?" Tags="&lt;homework&gt;&lt;algebra&gt;&lt;linear&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="11826" PostTypeId="2" ParentId="11768" CreationDate="2010-11-25T10:02:20.810" Score="1" ViewCount="0" Body="&lt;p&gt;The proofs of all these assertions in the case when your group is $\mathbb R$, can be found in any elementary book defining convolution. These same proofs will work in the case of arbitrary groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take the case of associativity. A proof for the case of real numbers is available at &lt;a href=&quot;http://planetmath.org/encyclopedia/AssociativityOfConvolution.html&quot; rel=&quot;nofollow&quot;&gt;planetmath&lt;/a&gt;. The same proof technique carries over to the general case. All you need additionally is to formally state two theorems: 1. Fubini's theorem on integration wrt product measures, and 2. translation invariance of the measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here note that you always have to use a Haar measure.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T10:02:20.810" />
  <row Id="11828" PostTypeId="2" ParentId="11693" CreationDate="2010-11-25T10:07:54.377" Score="2" ViewCount="0" Body="&lt;p&gt;I like your question! I wish I could give a full answer, but I'm not familiar enough with the Manin-Drinfeld theorem.  After I finish a couple weeks of work, I hope I can return and think about it in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess the answer is yes, and I have some candidates you could try.  The Fermat curves can be identified with compactified quotients $\Gamma\backslash\mathcal{H}$ for finite index subgroups $\Gamma$ in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modular function $\lambda(z) = \frac{\theta_2^4(q)}{\theta_3^4(q)}=16 q^{1/2} -128q +704q^{3/2} + \ldots$ &#xA;is a Hauptmodul of the genus 0 congruence subgroup $\Gamma(2)$.  (I.e., it is invariant under the action &#xA;of $\Gamma(2)$ by fractional linear transformation, and parametrizes the genus 0 modular surface $X_{\Gamma(2)} = \Gamma(2) \backslash \mathcal{H} \cup \{0,1,i_\infty\}$.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define Hauptmoduln $x:= \sqrt[n]{\lambda}$ and $y:= \sqrt[n]{1-\lambda}$, which determine finite index  genus 0 subgroups $\Gamma_x$ and $\Gamma_y$. Then $\Gamma:= \Gamma_x \cap \Gamma_y$ is subgroup with compactified quotient $X_{\Gamma}: x^n +  y^n = 1$, and $\Gamma$ is a noncongruence subgroup for $n \neq 1,2,4,8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction is from a survey paper by Ling Long.  We can use essentially the same idea to obtain quotient curves of the Fermat curves: If we set $x:= -\sqrt[5]{\lambda}$ and $y:= \sqrt{1-\lambda}$, we obtain a noncongruence subgroup $\Gamma$ with model $y^2=x^5+1$, which is a nice genus 2 curve, and its Jacobian has CM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can verify what happens in this case, I would be interested to hear about it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Added later:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just did a search and found a relevant paper: &quot;The Manin—Drinfeld theorem and Ramanujan sums&quot; by V. Kumar Murty and Dinakar Ramakrishnan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They give the same construction of the Fermat curves as I gave above, attributing it to Fricke and Klein.  They also cite a result of Rohrlich which establishes that the cusps of $\Gamma$ map to torsion points of the Jacobian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is yes; the noncongruence groups $\Gamma$ corresponding to the Fermat curves also satisfy the statement of the Manin-Drinfeld theorem.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-11-27T06:58:21.037" LastActivityDate="2010-11-27T06:58:21.037" CommentCount="1" />
  <row Id="11829" PostTypeId="2" ParentId="11761" CreationDate="2010-11-25T10:16:59.820" Score="6" ViewCount="0" Body="&lt;p&gt;The Wikipedia references are a reasonable starting point in the sense that some of them contain Voevodsky's construction(s) and related expositions.  Beyond that the literature on motives is enormous and difficult and relies on many ideas that are not apparent from the papers themselves -- not to mention unpublished letters and oral tradition found in a few major mathematical centers.  The state of documentation is far better in the internet era but this is still a highly technical field that is difficult to penetrate by casual reading.  In one respect the main breakthrough made by Voevodsky is that he created a theory with a framework that can be learned (e.g., by topologists) knowing scheme theory and a bounded number of other things, whereas before that motives were considered a pinnacle of inaccessible mystery, or a foundation-less heuristic used informally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what is &quot;suitable&quot;, Voevodsky proved the universality of motivic cohomology as defined in his approach; in the 2-category of homology theories satisfying some axioms, he constructs an initial object (similar to characterizing singular homology as the universal model of Eilenberg-Steenrod axioms).  Also, he builds a motivic homotopy theory with analogies to the topological theory, and not &quot;only&quot;(!!) the motivic homology and cohomology.  There are other respects in which his approach is better motivated (so to speak) or conceptual compared to other approaches.  (Grothendieck's approach is also conceptual but assumes Standard Conjectures and the work on mixed motives was not published.)   So although nobody knows the ultimate construction of mixed motives, Voevodsky's approach is the leading contender at present.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-25T18:34:26.653" LastActivityDate="2010-11-25T18:34:26.653" />
  <row Id="11830" PostTypeId="2" ParentId="11724" CreationDate="2010-11-25T10:50:29.990" Score="2" ViewCount="0" Body="&lt;p&gt;I must say I do not know anything much about it; but the following is the general idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the case of cohomology theories for smooth projective varieties over $\mathbb C$, there are various canonical isomorphisms between de Rham cohomology, Betti cohomology, etale cohomology, etc.. For instance the isomorphism between de Rham cohomology and Betti cohomology involves integrating differential forms over homology classes. Actually the situation is a bit more intricate; the de Rham cohomology decomposes into various other spaces via the Hodge decomposition. This whole stuff is all done over the complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $p$-adic case, an analogous theory is built, called the $p$-adic Hodge theory. Some details are available at the &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_Hodge_theory&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;. There is an isomorphism between algebraic de Rham cohomology and $p$-adic etale cohomology, as per the Hodge-Tate conjecture. There is another conjectural functor from the algebraic de Rham cohomology to $p$-adic etale cohomology, which is called Grothendieck's mysterious functor. A theory for this was constructed by Jean-Marc Fontaine and it was applied in various situations such as the study of $p$-adic Galois representations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theories of $p$-adic integration are useful in connection with such investigations. Pierre Colmez has a theory and so does Robert Coleman. Coleman's homepage has a &lt;a href=&quot;http://math.berkeley.edu/~coleman/fontaine.html&quot; rel=&quot;nofollow&quot;&gt;course&lt;/a&gt; on this material. Also see Perrin-Riou's Asterisque volume [3]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3]: Fonctions L p-adiques des représentations p-adiques, Perrin-Riou, B.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T10:50:29.990" />
  <row Id="11831" PostTypeId="1" CreationDate="2010-11-25T11:23:56.250" Score="0" ViewCount="105" Body="&lt;p&gt;Need some help on following induction problem:&#xA;$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}+...+\frac{1}{n(n+1)} = \frac{n}{n+1}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3989" LastActivityDate="2010-11-25T14:02:49.233" Title="proof by induction: n/(n+1)" Tags="&lt;proof&gt;&lt;induction&gt;" AnswerCount="3" />
  <row Id="11832" PostTypeId="2" ParentId="11831" CreationDate="2010-11-25T11:31:38.847" Score="3" ViewCount="0" Body="&lt;p&gt;Hint:&#xA;$$\frac{1}{k(k+1)}= \frac{1}{k} - \frac{1}{k+1}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n}{n+1} = 1 - \frac{1}{n+1}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-25T11:31:38.847" />
  <row Id="11833" PostTypeId="2" ParentId="11810" CreationDate="2010-11-25T11:34:33.910" Score="0" ViewCount="0" Body="&lt;p&gt;The formula is not about the integers in the sense that it doesn't involve their multiplicative structure; rather, it's about how the integers sit inside the reals as a discrete subgroup.  You can see this in the way the formula generalizes to $\mathbb{Z}^n$ sitting inside $\mathbb{R}^n$ or more generally in how it generalizes to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Selberg_trace_formula&quot; rel=&quot;nofollow&quot;&gt;Selberg trace formula&lt;/a&gt;.  The trace formula makes it clear that Poisson summation is a representation-theoretic fact; indeed one of its special cases is Frobenius reciprocity.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-25T11:34:33.910" />
  <row Id="11834" PostTypeId="1" AcceptedAnswerId="12298" CreationDate="2010-11-25T11:34:48.933" Score="2" ViewCount="250" Body="&lt;p&gt;For a common basis, I'll state basic definitions of a category and the relation type I'm thinking of. They're here for quick clarity, not precision, so feel free to revise for an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category:&lt;br&gt;&#xA;A collection of objects, a collection of arrows each with a source and target among said objects, identity arrows, and a composition operator satisfying an associative law over arrows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflexive relation on AxA:&lt;br&gt;&#xA;A collection of objects, a labelled multi-set of pairs of said objects satisfying the reflexive law, and the standard relation combinator operation (which satisfies the associative law).&lt;br&gt;&#xA;Note that labelled multi-set relations are commonly expressed and utilized in discrete math as multi-graphs. I collapse graphs and relations here for unified axiomatic treatment. This is not to treat, e.g., paths over graphs as different from relation composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What, exactly, separates these two? They appear essentially identical in definition and meaning to me. Relations immediately appear more general, given that categories only analogize to a certain restricted class of relations.&lt;br&gt;&#xA;I note some notational difference, such as how categories name each arrow; I don't see how that would change mathematical power. Similarly with the explicit treatment of composition. While such contrivances can come in handy for certain proofs or explorations, I don't see how it justifies treating the two as separate branches rather than syntactic shims on identical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDITS: fixed associativity statement, extended relations to multi-set representation with graph analogy]&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="3976" LastEditorDisplayName="" LastEditDate="2010-11-29T00:03:05.563" LastActivityDate="2010-11-29T07:51:23.107" Title="What is the difference between Categories and Relations?" Tags="&lt;category-theory&gt;&lt;relations&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="11835" PostTypeId="2" ParentId="11831" CreationDate="2010-11-25T11:36:13.693" Score="2" ViewCount="0" Body="&lt;p&gt;$\textbf{HINT:}$ Actually this is the answer in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all that you will need when you do induction.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-25T11:36:13.693" />
  <row Id="11836" PostTypeId="1" AcceptedAnswerId="11841" CreationDate="2010-11-25T11:46:10.217" Score="2" ViewCount="224" Body="&lt;p&gt;Is the essential difference between one-variable calculus and multi-variable calculus exterior differential or something else?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-11-25T21:03:45.707" Title="What is the essential difference between one-variable calculus and multi-variable calculus?" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11838" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T12:22:24.783" Score="2" ViewCount="0" Body="&lt;p&gt;It seems to me that an important difference is that while in one-variable calculus one only deals with one derivative, in multi-variable calculus there are infinitely many derivatives, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Directional_derivative&quot; rel=&quot;nofollow&quot;&gt;directional derivatives&lt;/a&gt;, a particular case of which are the partial derivatives. There are still the total derivatives for functions whose variables depend on another variable. Though all these derivatives are a generalization of the derivative of a single variable function.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-25T20:42:59.967" LastActivityDate="2010-11-25T20:42:59.967" />
  <row Id="11839" PostTypeId="2" ParentId="18" CreationDate="2010-11-25T12:34:05.153" Score="0" ViewCount="0" Body="&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  a = transverse radius = semi-major axis (for ellipse/oblate spheroid);&#xA;  b = conjugate radius = semi-minor axis (&quot;     &quot;     &quot;);&#xA; oe = angular eccentricity = arccos(b/a);&#xA;  f = flattening = (a-b)/a = 1 - b/a) = 1 - cos(oe) = 2*sin(.5*oe)^2;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then b = a*cos(oe) = a*(1-f).&lt;/p&gt;&#xA;" OwnerUserId="3996" LastActivityDate="2010-11-25T12:34:05.153" />
  <row Id="11841" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T13:12:14.430" Score="3" ViewCount="0" Body="&lt;p&gt;There's essentially no difference in the &lt;em&gt;concept&lt;/em&gt; of the derivative. It is still the best linear approximation to your function at each point. I.e. $f$ differentiable at $x_0$ still means that&#xA;$$ f(x_0+h) = f(x_0) + f&amp;#39;(x_0)x_0 + \varphi(h), $$&#xD;&#xA;for $h$ in some nbhd. of $0$ and $\varphi$ a continuous function that tends to $0$ at $0$ faster than linear. Except now $f&amp;#39;(x_0)$ is an $m \times n$ matrix called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix&quot; rel=&quot;nofollow&quot;&gt;Jacobian matrix&lt;/a&gt; and much more complicated to work with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first new thing we get when we go to higher dimensions is the notion of directional derivative, i.e. &quot;how much does $f$ change in the direction of $v$?&quot; We actually already have that on $\mathbb{R}$, it's just that there is only one direction (actually two, but it just comes down to a sign change), so it's never really looked at that way. The notion of directional derivative is exactly what you want when you want to generalize further to smooth manifolds, except you have to be a bit clever since you don't have an ambient space in which to have tangent vectors and instead use &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivation_%28abstract_algebra%29&quot; rel=&quot;nofollow&quot;&gt;derivations&lt;/a&gt; (of e.g. smooth functions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrals in multiple variables are much more complicated that the usual Riemann integral. Even when the functions are continuous. As demonstrated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fubini%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Fubini's theorem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tangentially calculus-related (really it's more analysis, but those are related anyway): The fact that $\mathbb{R}$ has an ordering allows one to define things like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral&quot; rel=&quot;nofollow&quot;&gt;Henstock–Kurzweil integral&lt;/a&gt;. Such an extension of the Lebesgue integral is (AFAIK) not possible in $\mathbb{R}^n$ for $n &amp;gt; 1$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-25T13:12:14.430" />
  <row Id="11842" PostTypeId="2" ParentId="11831" CreationDate="2010-11-25T14:02:49.233" Score="7" ViewCount="0" Body="&lt;p&gt;Every question of the form: prove by induction that&#xA;$$\sum_{k=1}^n f(k)=g(n)$$&#xD;&#xA;can be done by verifying two facts about the functions&#xA;$f$ and $g$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(1)=g(1)$&#xD;&#xA;and&lt;/li&gt;&#xA;&lt;li&gt;$g(n+1)-g(n)=f(n+1)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-25T14:02:49.233" />
  <row Id="11843" PostTypeId="1" AcceptedAnswerId="12753" CreationDate="2010-11-25T14:05:24.377" Score="4" ViewCount="202" Body="&lt;p&gt;Hi all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets imagine that we have some experiments. Each experiment may result in one of the outcomes: A, B, C. So we have probability distributions for each experiment $P_A, P_B, P_C$ which is context-dependent, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Context_1 \Rightarrow \{P_A^1, P_B^1, P_C^1\},$ experimental outcome is A&lt;/li&gt;&#xA;&lt;li&gt;$Context_2 \Rightarrow \{P_A^2, P_B^2, P_C^2\},$ experimental outcome is B&lt;/li&gt;&#xA;&lt;li&gt;$Context_3 \Rightarrow \{P_A^3, P_B^3, P_C^3\},$ experimental outcome is A&lt;/li&gt;&#xA;&lt;li&gt;$Context_4 \Rightarrow \{P_A^4, P_B^4, P_C^4\},$ experimental outcome is C&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These probabilities are calculated by some function $F:Context\rightarrow \{P_A, P_B, P_C\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to estimate an absolute trust rate of this function. In other words, I want to be able to say &quot;we can trust this function on 86%&quot; like we do when we deal with Pearson's chi-square test.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="4003" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T15:01:38.453" LastActivityDate="2010-12-02T13:46:47.453" Title="Chi-square analog for context-dependent distributions" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="11844" PostTypeId="2" ParentId="11495" CreationDate="2010-11-25T14:18:04.083" Score="0" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you think it is possible to&#xA;  successfully study mathematics with&#xA;  such a short initial knowledge and not&#xA;  having my full day to do it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That depends on your university course and how faster you can learn and get skills, but you will like to devote at least 4-5 hours day, and will want to do as much exercises as you can.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When in high school I've had mostly&#xA;  literary classes, so, again, my&#xA;  current knowledge of algebra, geometry&#xA;  and arithmetic is quite basic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should start with that basic stuff, they are the foundations for everything that comes next and doing that way will prevent you to go back very often while you are learning the college's math. Once you get in touch with Algebra, Calculus and Discrete Math you will see how there is sinergy about programming and doing math that makes the hours worthy besides&#xA;the CS degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to give you an advice from my personal experience being a math student but like CS more ;) : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;half efforts don't produce half results, you must do as more exercises as you can practicing&#xA;both theory and plumping work (when you get to lineral algebra you might remember my words),&#xA;don't let that the amount of work to-do grow bigger every week, be sure to understand the theorems (that is, dont left the desk every time you get stuck, happens to everyone). After you dive in for a couple of weeks you will start noticing the difference... you will become more analytic and insightful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any general advice or book / site to&#xA;  get me started with the &quot;preparation&quot;&#xA;  study? What are the basic areas I&#xA;  should concentrate on to be prepared&#xA;  to the first year math exams?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consult your courses specifications, they are many books out there, first read the ones your professor recommends... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vi auguro buona fortuna (i wish you good luck right?)&lt;/p&gt;&#xA;" OwnerUserId="4005" LastActivityDate="2010-11-25T14:18:04.083" CommunityOwnedDate="2010-11-25T14:18:04.083" />
  <row Id="11845" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T14:25:00.390" Score="9" ViewCount="0" Body="&lt;p&gt;Note also that many cultures use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pentatonic_scale&quot; rel=&quot;nofollow&quot;&gt;pentatonic scale&lt;/a&gt;. This would correspond to playing only the notes CDEGA. As explained in Qiaochu's answer, we want notes that are in small rational intervals, and particularly notes that are in small rational intervals from the tonic. Exactly which set of notes is chosen varies from culture to culture, with Western music using the 7 white keys, but many other cultures only using the 5 pentatonics.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-11-25T14:25:00.390" CommentCount="1" />
  <row Id="11846" PostTypeId="1" CreationDate="2010-11-25T14:39:18.680" Score="1" ViewCount="109" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am learning about discrete probability distributions and found 2 definitions for Geometric Distributions from &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_distribution&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&#xA;1. The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ...}&#xA;2. The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... }&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like a subtle difference, but i'm having trouble wrapping my head around when i would use which? Any insights appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4007" LastActivityDate="2010-11-27T05:28:44.690" Title="Which geometric distribution to use?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="11847" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T14:42:02.847" Score="1" ViewCount="0" Body="&lt;p&gt;The question is too wide - the answer depends on who you ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essential in analysis is to look at limits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In one real variable there are two ways/directions to narrow a point. &#xA;In two variables there are infinitely lines through a point, hence there are infinitely many directions to narrow a point and moreover you might get different results on different other kind of curves. (I think this is what Américo Tavares meant by infinitely many derivatives - directional derivatives). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Integration over a 2-dimensional set might be done in at least three ways - a double integral or two iterated integrations -  which if the conditions of Fubini-Tonelli's theorem is fulfilled has the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The fundamental theorem of calculus is a bit harder - Stoke's theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Partial differential equations are, generally speaking, harder than ordinary differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Also, several variables makes one variable easier, as an example you will learn an easy way to calculate &#xA;$$\int_{-\infty}^\infty e^{-x^2}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) ...&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-25T14:42:02.847" />
  <row Id="11848" PostTypeId="2" ParentId="11834" CreationDate="2010-11-25T14:50:32.157" Score="11" ViewCount="0" Body="&lt;p&gt;Qiaochu is right: categories are much more general. Let me provide a concrete example which, while in a sense trivial, may illustrate just how much more you could get from a category than you can from a reflexive and &lt;em&gt;transitive&lt;/em&gt; relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the positive integers $\mathbb P$. The usual total order $\leqslant$ is a perfectly good reflexive, transitive relation on this set. We can think of this as a category which consists of the identity map $\mathrm{id}_n$ on each integer &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;isin;&amp;nbsp;$\mathbb P$, and the composition of the successor maps $\sigma_n: n \to n+1$. The result is essentially an &lt;em&gt;almost&lt;/em&gt; acyclic directed graph (all circuits are stationary walks on a single vertex).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let me describe to you a different category on the positive integers: the category of injective linear maps on finite-dimensional real vector spaces &amp;mdash; which is to say, the category whose objects are the positive integers, and whose arrows are &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; matrices over $\mathbb R$, with rank &lt;em&gt;n&lt;/em&gt;, for &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;ge;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;An aside&lt;/em&gt;&lt;/strong&gt;. &quot;What!?&quot; you may exclaim; &quot;How can you talk about matrices having domains and codomains consisting of positive integers?&quot; Well, the shape of each matrix defines what sorts of other matrices it can be composed with, and any action of a matrix &lt;em&gt;M&lt;/em&gt; on a vector &lt;strong&gt;v&lt;/strong&gt;&amp;nbsp;&amp;isin;&amp;nbsp;$\mathbb R^n$ could be viewed as composing the &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; matrix &lt;em&gt;M&lt;/em&gt; with the &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;1 matrix &lt;strong&gt;v&lt;/strong&gt; to obtain an &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;times;&amp;nbsp;1 matrix &lt;em&gt;M&lt;/em&gt;&lt;strong&gt;v&lt;/strong&gt;. The integers stand in for the dimension of the vector space $\mathbb R^n$ or $\mathbb R^m$ which are the &quot;real&quot; domain and codomain of the matrix; and because we can substitute the linear action of operators on $\mathbb R^n$ with the effect of composing two linear operators (one of which happens to have a domain of $\mathbb R^1$), we may ignore the structure inside of the vector spaces and refer to them just by their dimensions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As with the directed graph generated by the total order &amp;le;, there exist arrows &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; for any &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;ge;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; (by definition). All of the maps from &lt;em&gt;n&lt;/em&gt; to &lt;em&gt;m&lt;/em&gt; for &lt;em&gt;m&lt;/em&gt;&amp;nbsp;&amp;gt;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;+1 can be obtained by composing maps &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;+1 and &lt;em&gt;n&lt;/em&gt;+1&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;m&lt;/em&gt;; and furthermore, all maps &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt;+1 can be obtained by composing &lt;em&gt;any single designated&lt;/em&gt; map $\sigma_n: n \to n+1$ with a suitable map $\tau: n \to n$. So this category has a number of things in common with the one defined by the total order &amp;le;. But there are important differences, owing in part to the fact that there are infinitely maps &lt;em&gt;n&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;n&lt;/em&gt; for each &lt;em&gt;n&lt;/em&gt;. Because of this, this second category has much more structure than a binary relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, it becomes meaningful and interesting to talk about &lt;em&gt;coproducts&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;Definition.&lt;/em&gt;&lt;/strong&gt; Let &lt;em&gt;A,B&lt;/em&gt; be two objects. A &lt;em&gt;coproduct&lt;/em&gt; of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is an object (&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;), together with two maps &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;A&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;) and &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;B&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;(&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;) such that, for any other object &lt;em&gt;X&lt;/em&gt; and maps &lt;em&gt;f&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;A&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;X&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt;&amp;nbsp;:&amp;nbsp;&lt;em&gt;B&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;X&lt;/em&gt;, there exists a unique map (&lt;em&gt;f&lt;/em&gt;&amp;nbsp;|&amp;nbsp;&lt;em&gt;g&lt;/em&gt;)&amp;nbsp;:&amp;nbsp;(&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;)&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;X&lt;/em&gt; such that &lt;em&gt;f&lt;/em&gt;&amp;nbsp;=&amp;nbsp;(&lt;em&gt;f&lt;/em&gt;&amp;nbsp;|&amp;nbsp;&lt;em&gt;g&lt;/em&gt;)&amp;nbsp;$\circ$&amp;nbsp;&lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt;&amp;nbsp;=&amp;nbsp;(&lt;em&gt;f&lt;/em&gt;&amp;nbsp;|&amp;nbsp;&lt;em&gt;g&lt;/em&gt;)&amp;nbsp;$\circ$&amp;nbsp;&lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt;&amp;nbsp;. That is, in the following diagram, any two directed walks between a common pair of points represent the same map:&lt;br&gt;&lt;br&gt;&#xA;  $\begin{matrix} \qquad\! A \!\!&amp;amp;\!\! \xrightarrow{\textstyle \;i_A\;} \!\!&amp;amp;\!\! (A \amalg B) \!\!&amp;amp;\!\! \xleftarrow{\textstyle \;i_B\;} \!\!&amp;amp;\!\! B \!\qquad \\&#xD;&#xA;\mathrm{id_A}\Bigg\updownarrow &amp;amp; &amp;amp; \Bigg\downarrow(f|g)\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&amp;amp; &amp;amp;\Bigg\updownarrow\mathrm{id_B} \\&#xD;&#xA;\qquad\! A \!\!&amp;amp;\!\! \xrightarrow[\textstyle \quad f \quad] \!\!\!\!\!\!\!\!\!\!&amp;amp;\!\! X \!\!&amp;amp;\!\!\!\!\!\!\!\!\!\! \xleftarrow[\textstyle \quad g \quad] \!\!&amp;amp;\!\! B\!\qquad \end{matrix}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the case of the partial order &amp;le;, the coproduct of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is just &amp;thinsp;&lt;b&gt;max&lt;/b&gt;&amp;thinsp;{&amp;thinsp;&lt;em&gt;A, B&lt;/em&gt;&amp;thinsp;}&amp;thinsp;, which you should verify for yourself.&#xA;&#xA;&lt;li&gt;In the category of matrices we described above, the coproduct of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is not the maximum of &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt;, but is instead the integer &lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt;, corresponding to splitting a vector of dimension &lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt; into a block of size &lt;em&gt;A&lt;/em&gt; and a block of size &lt;em&gt;B&lt;/em&gt;.&lt;br&gt;&lt;br&gt;&#xA;&lt;ul&gt;&lt;li&gt; The &quot;inclusion&quot; map $i_A$ would correspond to mapping a vector in &lt;b&gt;v&lt;/b&gt;&amp;nbsp;&amp;isin;&amp;nbsp;$\mathbb R^A$ into the first &lt;em&gt;A&lt;/em&gt; coefficients of a vector in $\mathbb R^{A+B}$, with the others set to zero;&#xA;&lt;li&gt; The &quot;inclusion&quot; map $i_B$ would be similar, mapping vectors in $\mathbb R^B$ into the final &lt;em&gt;B&lt;/em&gt; coefficients.&#xA;&lt;li&gt; For functions &lt;em&gt;f&lt;/em&gt;:&amp;nbsp;&lt;em&gt;A&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt; and &lt;em&gt;g&lt;/em&gt;:&amp;nbsp;&lt;em&gt;B&lt;/em&gt;&amp;nbsp;&amp;rarr;&amp;nbsp;&lt;em&gt;A&lt;/em&gt;+&lt;em&gt;B&lt;/em&gt;, the map (&lt;em&gt;f&lt;/em&gt;&amp;thinsp;|&amp;thinsp;&lt;em&gt;g&lt;/em&gt;) corresponds to performing the map &lt;em&gt;f&lt;/em&gt; to the first block and &lt;em&gt;g&lt;/em&gt; to the second block of the vector.&lt;/ul&gt;&#xA;&lt;br&gt;Of course, there's more than one way to decompose $\mathbb R^{A+B}$ into blocks; we could have chosen a different definition for the maps &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt;, by having &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt; map vectors in $\mathbb R^B$ into the first &lt;em&gt;B&lt;/em&gt; coefficients, and &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; map vectors into the final &lt;em&gt;A&lt;/em&gt; coefficients. This way of embedding $\mathbb R^A$ and $\mathbb R^B$ into the same space at the same time is equivalent to the one we describe above; in fact we can show that there is a unique isomorphism between these two ways of constructing (&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;). So, the coproduct (&lt;em&gt;A&lt;/em&gt;&amp;nbsp;$\amalg$&amp;nbsp;&lt;em&gt;B&lt;/em&gt;) is dependent on the choice of maps &lt;em&gt;i&lt;sub&gt;A&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;i&lt;sub&gt;B&lt;/sub&gt;&lt;/em&gt; &amp;mdash; but not in any way which is really important. &#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Even though the first category generated by &amp;le; is what you get when you forget the matrix structure of the second category &amp;mdash; by replacing each infinite collection of arrows in each case with a single arrow between the same points &amp;mdash; that second category of injective linear operators is substantially different from the simple one described by &amp;le;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just scratching the surface of one fingernail of category theory; but hopefully it convinces you that categories and reflexive, transitive binary relations are not trivially equivalent.&lt;/p&gt;&#xA;" OwnerUserId="439" LastEditorUserId="439" LastEditorDisplayName="" LastEditDate="2010-11-25T21:00:27.800" LastActivityDate="2010-11-25T21:00:27.800" CommentCount="5" />
  <row Id="11849" PostTypeId="1" CreationDate="2010-11-25T15:23:39.890" Score="2" ViewCount="93" Body="&lt;p&gt;I am looking for a proof that this inequality:&#xA;$$(W_1+W_2+\cdots+W_n)^a \leq  W_1^a +\cdots +  W_n^a$$ &#xA;is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a power function $f(W)=W^a$ where $a$ is a real number, constant but usually $0\lt a\lt 1$. The $W_i$ are real positive numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I want to check if $\left(\sum W_i\right)^a \leq \sum(W_i^a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example for $n=2$ : $(x+y)^a \leq x^a + y^a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4011" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T23:36:50.523" LastActivityDate="2010-11-25T23:36:50.523" Title="$(W_1+W_2+\cdots+W_n)^a \leq  W_1^a +\cdots +  W_n^a$  for $n$ integer, $n\geq 2$, $W\gt 0$ and $a$ constant, real, $0\lt a\lt 1$" Tags="&lt;inequality&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11850" PostTypeId="2" ParentId="11849" CreationDate="2010-11-25T15:48:29.343" Score="1" ViewCount="0" Body="&lt;p&gt;To show that $(w_1+w_2)^a\leq w_1^a+w_2^a$ whenever $w_1$ and $w_2$ are positive, we can divide by $w_1^a$ on both sides and rewrite $\frac{w_2}{w_1}$ as $t$ to get the equivalent statement $(1+t)^a\leq 1+ t^a$ when $t\geq 0$.  Notice that the two sides agree when $t$ goes to zero, and the derivative of the right-hand side with respect to $t$ is always bigger than that of the left-hand side (because $0\lt a\lt 1$), so the inequality follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of general $n$ follows by induction, using the $n=2$ case for the inductive step. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-25T15:53:40.880" LastActivityDate="2010-11-25T15:53:40.880" />
  <row Id="11851" PostTypeId="2" ParentId="11849" CreationDate="2010-11-25T16:04:55.173" Score="4" ViewCount="0" Body="&lt;p&gt;This is equivalent to proving $(\frac{W_1}{W_1+W_2+...+W_n})^a+(\frac{W_2}{W_1+W_2+...+W_n})^a+...+(\frac{W_n}{W_1+W_2+...+W_n})^a\geq 1$. Write $a_i=\frac{W_1}{W_1+W_2+...+W_n}$. Then, $a_1+a_2+...+a_n=1$, $0&amp;lt; a_1,a_2,...,a_n,a &amp;lt; 1$. So, $a_i^a &amp;gt; a_i$. So, $a_1^a+...+a_n^a&amp;gt; a_1+a_2+...+a_n=1$. QED. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think the inequality should be strict since $a&amp;lt;1$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-25T21:41:55.580" LastActivityDate="2010-11-25T21:41:55.580" CommentCount="4" />
  <row Id="11852" PostTypeId="1" CreationDate="2010-11-25T16:37:51.623" Score="0" ViewCount="82" Body="&lt;p&gt;Let f be a permutation on n letters. I want to count the number of k-long increasing subsequences quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I use divide and conquer to split the problem into finding increasing subsequences in pairs of numbers. The resulting 'tree' has height logn. I'm wondering how I can combine the subproblems in an efficient way. I thought about prefix sums and using mergesort, but I can't see how to effectively implement them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea to save time when combining two subproblems is to compare elements at the end of your increasing subsequences in your first subproblem with the values in your second subproblem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to get to O(knlogn) time but this doesn't do it.&lt;/p&gt;&#xA;" OwnerUserId="4015" LastActivityDate="2010-11-25T16:46:35.393" Title="computing the number of increasing k-long subsequences" Tags="&lt;algorithms&gt;&lt;permutations&gt;" AnswerCount="0" CommentCount="2" />
  <row Id="11854" PostTypeId="2" ParentId="3048" CreationDate="2010-11-25T17:19:43.410" Score="1" ViewCount="0" Body="&lt;p&gt;It seems not to have yet been mentioned here that there is no universal agreement on the meaning of countable.  &quot;Countably infinite&quot; is unambiguous, but some authors use &quot;countable&quot; to mean countably infinite, while many (perhaps most) use countable to mean finite or countably infinite, as the other answers indicate.  When authors use countable to refer only to sets in bijection with $\mathbb{N}$, they often end up using the phrase &quot;at most countable&quot;.  This is seen for example in Rudin's analysis texts.   Springer's online encyclopedia also &lt;a href=&quot;http://eom.springer.de/C/c026710.htm&quot; rel=&quot;nofollow&quot;&gt;defines&lt;/a&gt; countable to mean countably infinite. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-25T17:19:43.410" />
  <row Id="11855" PostTypeId="2" ParentId="11818" CreationDate="2010-11-25T17:32:29.240" Score="0" ViewCount="0" Body="&lt;p&gt;If the list is drawn with replacement, the chance that a given number (the new draw) is not in it is $$(1-\frac{1}{2147000001})^{1000000}$$  The birthday problem only comes up if you ask for the chance of a collision anywhere in the 1000001 numbers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-25T17:32:29.240" />
  <row Id="11856" PostTypeId="1" AcceptedAnswerId="12092" CreationDate="2010-11-25T17:40:15.383" Score="3" ViewCount="83" Body="&lt;p&gt;I think the cone (or what is also called the &quot;half cone&quot;) is a differential manifold but not a smooth manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me understand this the nuts and bolts way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How explicitly can I write down the differential structure on the cone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what is the natural metric on it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Is that what is called the ``Sasakian Metric&quot;? If yes then can one kindly explain that or give an expository reference on this?)&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-11-27T18:13:57.443" Title="Differential and Riemannian structure on the cone" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="11857" PostTypeId="2" ParentId="11849" CreationDate="2010-11-25T17:48:03.253" Score="0" ViewCount="0" Body="&lt;p&gt;See my answer to &lt;a href=&quot;http://math.stackexchange.com/questions/10993/please-explain-inequality-xp-yp-leq-x-yp&quot;&gt;Please explain inequality $|x^p-y^p|\le |x-y|^p$&lt;/a&gt; -&#xA;the inequality you ask for follows by induction.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-25T17:48:03.253" />
  <row Id="11858" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T17:57:36.350" Score="3" ViewCount="0" Body="&lt;p&gt;Start at F and go up a fifth (to C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xEMSY.jpg&quot; alt=&quot;piano&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In a keyboard with 12-key octaves, that's 7 steps.) Repeat that process (through the circle of fifths). You'll hit all the white keys and then all the black keys  -- F, C, G, D, A, E, B, F#, C#, G#, D#, A# -- note, these keys are usually represented with flats). So it turns out if you're splitting tones on 3 : 2 (fifth) or 4 : 3 (fourth), the least common multiple is twelve. In practice, the 3 : 2 is similar enough that it gives a sort of 'secure' or content feeling. The 4 : 3 gives a slightly edgier feeling but one which is somewhat counterposed perfectly against this secure feeling. So a fourth + a fifth will give you an octave. So why we want all twelve keys is that we're saying that we want the fifth (dominant) and the fourth (subdominant) to come together and make a whole. This is sort of mirrored with the fact that the first 7/12 of the circle of fifths form the basis and the second 5/12 are 'overlayed' on top (with black keys).&lt;/p&gt;&#xA;" OwnerUserId="4022" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T03:16:40.100" LastActivityDate="2010-11-26T03:16:40.100" />
  <row Id="11859" PostTypeId="1" AcceptedAnswerId="11860" CreationDate="2010-11-25T18:00:35.497" Score="3" ViewCount="125" Body="&lt;p&gt;I have seen this claim on Wikipedia and other places. Which branch of mathematics does this result come from?&lt;/p&gt;&#xA;" OwnerUserId="4025" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T23:06:21.997" LastActivityDate="2010-11-25T23:06:21.997" Title="Why can't erf be expressed in terms of elementary functions?" Tags="&lt;special-functions&gt;" AnswerCount="3" />
  <row Id="11860" PostTypeId="2" ParentId="11859" CreationDate="2010-11-25T18:04:38.467" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_Galois_theory&quot; rel=&quot;nofollow&quot;&gt;Differential Galois theory.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-11-25T18:04:38.467" />
  <row Id="11861" PostTypeId="2" ParentId="11859" CreationDate="2010-11-25T18:06:45.210" Score="1" ViewCount="0" Body="&lt;p&gt;I am not aware of any easy proof of this. Here is a proof that $\int e^{x^2}$ cannot be expressed in terms of elementary functions:&#xA;&lt;a href=&quot;http://math.hunter.cuny.edu/ksda/papers/rick_09_02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.hunter.cuny.edu/ksda/papers/rick_09_02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-25T18:06:45.210" />
  <row Id="11862" PostTypeId="2" ParentId="11859" CreationDate="2010-11-25T18:07:15.880" Score="2" ViewCount="0" Body="&lt;p&gt;This is a consequence of the Liouville-Ritt theory of integration in finite terms. You can find a brief sketch and references in my &lt;a href=&quot;http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral/5723#5723&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-25T18:07:15.880" />
  <row Id="11863" PostTypeId="2" ParentId="11856" CreationDate="2010-11-25T18:10:35.103" Score="1" ViewCount="0" Body="&lt;p&gt;The differentiable structure of the cone can be viewed as the same as the one of $S^1\times\mathbb R$, or $\mathbb{R}^2 - \{0\}$ . That is, you construct a homeomorphism between $\mathbb{R}^2 - \{0\}$ and the cone (as subsets of $\mathbb{R}^3$) and say that this is a diffeomorphism, in the most intuitive way: $(x,y)\mapsto (x,y,\sqrt{x^2 + y^2})$. With this imposition, the differentiable structure of $\mathbb{R}^2 - \{0\}$ is carried to the cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &quot;natural&quot; metric is the one induced from $\mathbb{R}^3$: $\tilde g = i^*g$, where $i$ is the inclusion map from the cone to $\mathbb{R}^3$ and $g$ is the Euclidean metric of $\mathbb{R}^3$. &lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-11-25T18:10:35.103" CommentCount="2" />
  <row Id="11864" PostTypeId="1" AcceptedAnswerId="11893" CreationDate="2010-11-25T18:30:18.393" Score="3" ViewCount="136" Body="&lt;p&gt;Let $A$ be the infinitesimal generator of a $C_0$-semigroup $(S(t))_{t \geq 0}$. Now, for every $x_0 \in X$ the map $t \mapsto S(t) x_0$ is a mild solution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\label{eq:1}&#xA;\dot{x} = Ax, \quad x(0) = x_0. \qquad \text{(*)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a continuous function $x: [0, \infty) \to X$ is called a mild solution of $\text{(*)}$ if $\int_0^t x(s) \, ds \in D(A)$ where $D(A)$ is the domain of $A$, $x(0) = x_0$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t) - x(0) = A \int_0^t x(\tau) \, d\tau \text{ for all $t \geq 0$}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have a proof of this but it uses Hille's theorem, but it is quite involved (needs a few tricks) and Hille's theorem is not elementary, does someone know an elementary proof of the uniqueness?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-04T23:51:13.470" LastActivityDate="2010-12-04T23:51:13.470" Title="Uniqueness mild solution of $\dot{x} = A x$." Tags="&lt;differential-equations&gt;&lt;semigroup&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11865" PostTypeId="1" AcceptedAnswerId="11900" CreationDate="2010-11-25T18:33:11.130" Score="3" ViewCount="164" Body="&lt;p&gt;&lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/Biographies/Khayyam.html&quot; rel=&quot;nofollow&quot;&gt;Omar Khayyam&lt;/a&gt; is known for his significant progress in solving cubic polynomial equations. For example, his biography on &lt;a href=&quot;http:///www-history.mcs.st-andrews.ac.uk&quot; rel=&quot;nofollow&quot;&gt;www-history.mcs.st-andrews.ac.uk&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(...) This problem in turn led Khayyam to&#xA;  solve the cubic equation x^3 + 200x =&#xA;  20x^2 + 2000 and he found a positive&#xA;  root of this cubic by considering the&#xA;  intersection of a rectangular&#xA;  hyperbola and a circle. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(...) Indeed Khayyam did produce such a work, the Treatise &#xA;  on Demonstration of Problems of&#xA;  Algebra  which contained a complete&#xA;  classification of cubic equations with&#xA;  geometric solutions found by means of&#xA;  intersecting conic sections.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I still can't see the big picture of those days.  I'm possibly omitting something about the idea of geometric solutions of algebraic equations, but why were they trying hard to find intersections of conic sections, and building large classification schemes for it?. If the idea was to get a numerical value out of these constructions by measuring lengths on paper,  they could just have prepared a careful template for the function $y = x^3$, and then solved all the cubic equations by intersecting it with a parabola, like in the figure below for the mentioned equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate answers that would clarify my confusion.  Was it that they did not conceive $y=x^3$ as a curve, if they were interested in getting a numerical value? Or was it a conceptual challenge to show that all cubic equations can be represented as an intersection of two conic sections? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/E4U2q.png&quot; alt=&quot;plot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-25T18:51:20.547" LastActivityDate="2010-11-26T03:58:11.750" Title="Khayyam's work on cubic equations" Tags="&lt;math-history&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11866" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T18:43:35.760" Score="3" ViewCount="0" Body="&lt;p&gt;Other responses do a good job of explaining the 12-note chromatic scale. From those 12 tones, if one starts to build a series of tones starting on a single note and going up the circle of fifths, there are two natural stopping points where you have a complete-sounding scale that spans the octaves and has relatively equal spacing between the notes with no gaps: five notes, which gives whole-step and minor-third intervals; and seven notes, which gives whole-step and half-step intervals. These two scales (pentatonic and heptatonic) correspond to the spacing of the black keys and white keys on the keyboard.  They are mirror images of each other around the circle of fifths. So the two colors of notes are not &quot;different,&quot; but rather a natural division into two symmetrical sclaes built from going opposite directions around the circle of fifths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the standard tuning system C is &quot;privileged&quot; because it is (essentially) the note where we start building the circle of fifths to create these two scales.&lt;/p&gt;&#xA;" OwnerUserId="4030" LastActivityDate="2010-11-25T18:43:35.760" />
  <row Id="11867" PostTypeId="2" ParentId="6647" CreationDate="2010-11-25T18:48:17.030" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a good writeup about implementing Stirling's approximation, along with a reference implementation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://threebrothers.org/brendan/blog/stirlings-approximation-formula-clojure/&quot; rel=&quot;nofollow&quot;&gt;http://threebrothers.org/brendan/blog/stirlings-approximation-formula-clojure/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4032" LastActivityDate="2010-11-25T18:48:17.030" />
  <row Id="11868" PostTypeId="2" ParentId="11865" CreationDate="2010-11-25T18:57:25.520" Score="5" ViewCount="0" Body="&lt;p&gt;I don't think they had the idea of y=x^3 as a curve. For one thing there was no analytic geometry until Descartes. With coordinates some problems become much simpler. &lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-11-25T18:57:25.520" CommentCount="4" />
  <row Id="11869" PostTypeId="2" ParentId="6546" CreationDate="2010-11-25T19:24:01.080" Score="3" ViewCount="0" Body="&lt;p&gt;Have you tried taking the semi-direct product of two subgroups (one needs to be a normal subgroup).  The order of G is the product of the orders of the subgroups.  One subgroup needs to be normal.  If the other is not, then the semidirect product is usually (if not always) non-abelian&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-25T19:24:01.080" />
  <row Id="11870" PostTypeId="2" ParentId="11764" CreationDate="2010-11-25T19:28:34.587" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a reference -first result in google:&#xA;&lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.ijm/1256060422&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.ijm/1256060422&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author (who now works in a bank) needs his construction of sheaf homology to prove a criterion for the vanishing of certain integrals of differential forms, though the paper treats general topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overarching goal seems interesting, as the names of Langlands, Igusa, Denef are mentioned, but is beyond me, unhelpfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway that looks like very nice mathematics. There are actually 2 constructions of sheaf homology. Both rely on triangulations, the first proceeds defining modules of sections on simplices and restriction maps to subsimplices and extending by linearity to complexes. The second, used to prove Poincaré duality/isomorphism with (usual) sheaf cohomology, proceeds defining sheaves &quot;on simplicial complexes&quot; and taking the homology of the chain complex of global sections of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2 constructions are isomorphic, and I guess they could be isomorphic to a more direct construction which only uses open covers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the paper you are referring to?&lt;/p&gt;&#xA;" OwnerUserId="3407" LastActivityDate="2010-11-25T19:28:34.587" />
  <row Id="11871" PostTypeId="1" AcceptedAnswerId="11877" CreationDate="2010-11-25T19:39:48.103" Score="3" ViewCount="125" Body="&lt;p&gt;Given $N$ random iid variables, $X_1, \ldots, X_N$, with a uniform probability distribution on $[0, 1)$ what is the distribution of $\displaystyle \max_{i = 1 \ldots N}(X_i)$?&lt;/p&gt;&#xA;" OwnerUserId="958" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T01:29:11.853" LastActivityDate="2010-11-26T01:29:11.853" Title="Probability distribution of  the maximum of random variables" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11872" PostTypeId="1" CreationDate="2010-11-25T19:43:38.947" Score="0" ViewCount="191" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number&quot;&gt;$a^{1/2}$ is either an integer or an irrational number.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Will every $n^{th}$ root of $2$ be an irrational number? If yes, how can I prove that?&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T01:08:29.040" LastActivityDate="2010-11-26T01:08:29.040" ClosedDate="2010-11-26T20:05:27.770" Title="Is $n^{th}$ root of 2 an irrational number?" Tags="&lt;proof&gt;&lt;roots&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11873" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T19:46:22.057" Score="12" ViewCount="0" Body="&lt;p&gt;The math of frequency relationships here is sound (pun intended) but they don't help explain the white vs black key piano layout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the historical imperitive thay led to this layout for &quot;Western Music&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the major triad: root + third + fifth notes of the &quot;diatonic scale&quot;.&#xA;They follow the harmonic series:&#xA;1 - root&#xA;2 - octave (doubling of root frequency)&#xA;3 - fifth (triple of the root - 3:2 relationship to the octave)&#xA;4 - double octave (4x)&#xA;5 - 10th (double octave of a third)&#xA;6 - octave fifth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are notes a static length of tubing can produce by blowing into it: the bugle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combinations of these notes create frequencies that make choirs sound heaven-ly.&#xA;The frequencies align and blend into pure complex vibraations that are the sum and&#xA;the differencies (harmonic overtones) of these relationships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choirs can tune themselves dynamically to create these frequency alignments that&#xA;are percieved as being perfectly consonant. Upbeat western music focus on the 3 major&#xA;chords found in the diatonic scale:&#xA;1+3+5 root major chord - white keys C - E - G&#xA;4+6+1 4th chord - white keys F - A - C&#xA;5+7+2 5th chord - whote keys G - B - D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basics of western folk music are the 1 - 4 - 5 sequences of chords.&#xA;Learn C, F and G on a guitar and you can play the bulk of the classic Country &#xA;song book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the notes of these chords into a scale and you get that row of 7 white keys:&#xA;C - D - E - F - G - A - B (repeat until you can't hear it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, they western scale is based upon frequency relationships that make combinations&#xA;of notes &quot;ring&quot; in consonance in it's purest form... like the Gregorian Chants of&#xA;the Roman Church.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a basic &quot;western keyboard&quot; could be made from just these 7 notes repeated across the frequency spectrum. Look at the layout of a Greek Lyre (a harp) and that's what you&#xA;will find. A sequence following the diatonic scale which sounds pleasant if you just strum across the strings due to the tuning of even multiples (adjusted by octaves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK... now adding the black keys is a compromise of tuning specific notes so that you can build these 1+3+5 chords from any starting point and thus play a song adjusted up or down to any starting point. The piano will never achieve that sonic mathematical glimpse into the &quot;music of the spheres&quot; that the self-adjusting choir can to make a chord mathematically perfect in alignment but it's the &quot;keyboard&quot; for the modern composer... the effective &quot;musical qwerty&quot; that a composer or a pianist begins to visualize chord &quot;shapes&quot; as hand positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a lot of practice a pianist can pre-visualize sound in terms of finger and hand movements much like a solid touch typist starts to set words and sentences as a sequence of movements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of the black keys was called a &quot;Well Tempered&quot; tuning and Bach was one of the first composers to create whole bodies of compositions that worked through the Major and Minor keys of the 12 scales that you noticed intially when inspecting the keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look into other musical cultures you will find a different approaches to standardizing sound relationships that do not focus on the 1 - 4 - 5 chords.&#xA;This music to a culturally trained western ear is less predicatable in nature and&#xA;that lack of predictablility can make the music frustrating or exciting... music &quot;speaks&quot; to us in terms of pure sensory inputs that can move, excite, bore or confuse us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the piano keyboard is designed to be the perfect delivery system for an individual to produce the range of complexity that western music has achieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modern keyboard synthesizers are now able to produce the full range of the western orchestra in terms of &quot;instruments&quot; and I'm hoping someone create one that micro-adjusts notes based upon the surrounding context... shifting a note up or down slightly from the &quot;well tempered&quot; compromise to the pitch that makes a chord &quot;ring&quot; and&#xA;produce the upper harmonic overtones that make a great orchestra truly &quot;heavenly&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it's already been done.&lt;/p&gt;&#xA;" OwnerUserId="4040" LastActivityDate="2010-11-25T19:46:22.057" CommentCount="2" />
  <row Id="11874" PostTypeId="1" CreationDate="2010-11-25T19:48:04.957" Score="4" ViewCount="185" Body="&lt;p&gt;What does this expression mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f\in C^2[a,b]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I don't know what $C$ means.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-26T05:17:06.330" LastActivityDate="2010-11-26T05:17:06.330" Title="What does &quot;$f\in C^2[a,b]$&quot; mean?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11875" PostTypeId="2" ParentId="11872" CreationDate="2010-11-25T19:50:16.513" Score="7" ViewCount="0" Body="&lt;p&gt;Yes. In fact, for every integer $k$ and every $n\gt 1$, the $n$th root of $k$ is either an integer or irrational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to prove it is to use exactly the same idea as for proving the square root of $2$ is irrational: suppose $\sqrt[n]{k} =\frac{p}{q}$, with $p$ and $q$ integers, relatively prime. Then $q^nk = p^n$. Now think about the prime factorizations: every prime that divides $q$ must divide $p$, but $p$ and $q$ are relatively prime, so $q=1$. That means that you must have $k=p^n$ with $p$ an integer. That is, the only way for the $n$th root of $k$ to be a rational is if $k$ is an $n$th power of an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can use the Rational Root Test: an $n$th root of $k$ is a root of the polynomial $x^n - k$. But a rational root of a polynomial with integer coefficients that is written in lowest term $\frac{p}{q}$ must have denominator $q$ that divides the leading coefficient and numerator $q$ that divides the constant coefficient. So any rational root of $x^n-k$ must be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting this back to your question, since $2$ is not an $n$th power of an integer for any $n\gt 0$, $\sqrt[n]{2}$ is not a rational for any integer $n\gt 0$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T20:00:33.820" LastActivityDate="2010-11-25T20:00:33.820" />
  <row Id="11876" PostTypeId="2" ParentId="11874" CreationDate="2010-11-25T19:51:54.963" Score="5" ViewCount="0" Body="&lt;p&gt;$f : [a,b] \rightarrow \mathbb{R}$ is a function that is twice differentiable with each derivative continuous.  That is, $f&amp;#39;$ and $f&amp;#39;&amp;#39;$ both exist and are both continuous.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-11-26T01:19:25.570" LastActivityDate="2010-11-26T01:19:25.570" CommentCount="5" />
  <row Id="11877" PostTypeId="2" ParentId="11871" CreationDate="2010-11-25T19:51:58.687" Score="4" ViewCount="0" Body="&lt;p&gt;Hint: $\max \lbrace X_1,...,X_N \rbrace \leq x$ if and only if $X_i \leq x$ for all $i=1,...,N$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-25T19:51:58.687" />
  <row Id="11878" PostTypeId="1" AcceptedAnswerId="11901" CreationDate="2010-11-25T19:53:03.613" Score="2" ViewCount="115" Body="&lt;p&gt;Dear MSE-community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we consider the Taylor Series expansion of $f(x)=b^x$ for some $b \in \mathbb{R}$, we see that $$b^x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}x^n.$$ We can substitute $x$ for $b^x$ to find that $$ b^{b^{x}} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{xn}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say that we want to find a function that accurately describes how we should raise $b$ to $b$'th power $x$ times. We say that $b^{(b)}_x$ says that $b$ should be raised to itself $x$ times, for some $x\in\mathbb{R}$. We can generalize the previous examples, and write $$b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-1}.\qquad \text{(1)}$$ (Please note that $b^{(b)n}_{x}$ is &lt;em&gt;not&lt;/em&gt; equal to $b^{(bn)}_{x}$.) Using this equation we can also state that $$b^{(b)}_{x-1} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2}.\qquad\text{(2)}$$&#xD;&#xA;When we substitute (2) in (1), we find that $$ b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}\Big( 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2} \Big).$$&#xD;&#xA;From now on, we write $ \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!} = A$, because then the next equations look more clear. &#xA;We could go on substituting in this manner $*$, until we arive at $b^{(b)}_{(x-(x-1))}=b$. We then see that the equality $$ b^{(b)}_x = 1 + \sum A \Big( 1 + \sum A \Big(\cdots \Big(1 + \sum_{n=1}^{\infty} \frac{(\log(b^b))^n}{n!}\Big)\Big)\Big) $$&#xD;&#xA;holds when we iterate $f(x) = 1 +  \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}$ an $x-2$ amount of times, for real $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Is this possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; If so, how is this done? How do you describe the formula that precisely defines how $f(x)$ looks like after being iterated $x-2$ times? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 3:&lt;/strong&gt; If such a formula is found, would it imply that a nice way of finding the fourth hyper operator (or &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetration&quot; rel=&quot;nofollow&quot;&gt;tetration&lt;/a&gt;&quot;) is found for real numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$*$EDIT: When we proceed in this manner when $x$ is not an integer, we will not find $x=1$. We should be able to iterate the function a real amount of times to find $x=1$.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-26T06:26:46.490" LastActivityDate="2010-11-26T06:26:46.490" Title="Could real iterates of the Taylor Series expansion of $b^x$ help to find a way to define tetration?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11879" PostTypeId="2" ParentId="11874" CreationDate="2010-11-25T19:55:45.337" Score="3" ViewCount="0" Body="&lt;p&gt;$f:[a,b]\to \mathbb{R}$ or $\to \mathbb{C}$. And both the first  and second derivatives of $f$ are continuous.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-25T19:55:45.337" />
  <row Id="11880" PostTypeId="2" ParentId="8040" CreationDate="2010-11-25T20:03:39.290" Score="0" ViewCount="0" Body="&lt;p&gt;It works because the Leibniz's notation for derivatives dx/dt means two things that are both true:&#xA; (1) a name for the derivative x'(t);&#xA; (2) the fraction or quotient of differentials: numerator dx, denominator dt (arbitrary number); so dx=x'(t)dt.&lt;/p&gt;&#xA;" OwnerUserId="4042" LastActivityDate="2010-11-25T20:03:39.290" />
  <row Id="11881" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T20:12:28.107" Score="1" ViewCount="0" Body="&lt;p&gt;If you really want to know all about it then you should read 'On the sensation of tone' by Helmholz.&#xA;&lt;a href=&quot;http://store.doverpublications.com/0486607534.html&quot; rel=&quot;nofollow&quot;&gt;http://store.doverpublications.com/0486607534.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4045" LastActivityDate="2010-11-25T20:12:28.107" CommentCount="1" />
  <row Id="11882" PostTypeId="1" AcceptedAnswerId="11884" CreationDate="2010-11-25T20:12:29.370" Score="3" ViewCount="146" Body="&lt;p&gt;Prove that $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ using an epsilon delta proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have most of the work done.  I choose $\delta = min{\frac{1}{2}, y}$,&lt;br&gt;&#xA;$f(x)$ factors out to $\frac{|x-3||x-2|}{|x+1|}$  &#xA;But $|x-3| \lt \frac{3}{2}$ for $\delta = \frac{1}{2}$ and also $|x+1| &amp;gt; 5/2$ (I'll spare you the details).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to choose my y here. If I take $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ &amp;lt; $(3/5) \delta$   How do I choose my epsilon here (replace y with this) to satisfy this properly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T20:13:57.857" LastActivityDate="2010-11-26T04:57:02.610" Title="$\epsilon$-$\delta$ limit proof, $\lim_{x \to 2} \frac{x^{2}-2x+9}{x+1}$" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;real&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11884" PostTypeId="2" ParentId="11882" CreationDate="2010-11-25T20:26:34.630" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;First:&lt;/strong&gt; You &lt;strong&gt;don't&lt;/strong&gt; choose $\epsilon$! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an $\epsilon$-$\delta$ proof, you begin by taking an &lt;em&gt;arbitrary&lt;/em&gt; $\epsilon$; you then need to show that it is possible to choose a $\delta$ that &quot;works&quot; for &lt;em&gt;that&lt;/em&gt; $\epsilon$. So you are going at things &lt;em&gt;exactly&lt;/em&gt; backwards, by first trying to work with $\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second:&lt;/strong&gt; You cannot &quot;prove $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}$&quot; using $\epsilon$-$\delta$ proofs. What you &lt;em&gt;can&lt;/em&gt; try to prove is that the limit is equal to &lt;em&gt;something&lt;/em&gt; (in this case, to $3$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third:&lt;/strong&gt; $f(x)$ does not factor as $\frac{|x-2||x-3|}{|x+1|}$; what factors that way is $|f(x)-3|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So. Let's start from scratch, shall we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $\epsilon$-$\delta$ proof that $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}=3$ would work as follows: given some arbitrary $\epsilon\gt 0$, we must find a $\delta\gt 0$ (which may depend on $\epsilon$), with the property that if $0\lt |x-2|\lt \delta$, then we will have $\left|\frac{x^2-2x+9}{x+1} - 3\right|\lt \epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $\epsilon\gt 0$ be given. You want to make sure that&#xA;$$\left|\frac{x^2-2x+9}{x+1} - 3\right| = \left|\frac{x^2-2x+9-3x-3}{x+1}\right| = \left|\frac{x^2-5x+6}{x+1}\right| = \left|\frac{(x-3)(x-2)}{x+1}\right|$$&#xD;&#xA;is small, by making sure that $|x-2|$ is small (that is, that $0\lt |x-2|\lt\delta$ for some chosen $\delta$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, you also want to make sure that $\delta\lt \frac{|x+1|}{|x-3|}\epsilon$, because then we will have:&#xA;$$\left|\frac{x^2-x+0}{x+1}-3\right| = \frac{|x-3||x-2|}{|x+1|} \lt \delta\frac{|x-3|}{|x+1|} \lt \frac{|x+1|\epsilon}{|x-3|}\frac{|x-3|}{|x+1|} = \epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make sure that $\frac{1}{|x+1|}$ is no larger than some constant by making sure that $x$ is close enough to $2$. You can do the same thing for $|x-3|$. So &lt;strong&gt;then&lt;/strong&gt; you can pick a $\delta$ that is simultaneously small enough to ensure that $\frac{1}{|x+1|}$ is smaller than some $C$, that $|x-3|$ is smaller than some $D$, &lt;em&gt;and&lt;/em&gt; that $|x-2|$ is smaller than $CD\epsilon$. Try that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T04:57:02.610" LastActivityDate="2010-11-26T04:57:02.610" CommentCount="5" />
  <row Id="11885" PostTypeId="2" ParentId="11872" CreationDate="2010-11-25T20:26:36.580" Score="2" ViewCount="0" Body="&lt;p&gt;If $\rm\ \sqrt[n] c = a/b,\ \ gcd(a,b) = 1\ $ then $\rm\ c\ b^n = a^n\ \Rightarrow \ b\:|\: a^n\:.\: $ But $\rm\ gcd(a,b) = 1\ \Rightarrow\ gcd(a^n,b) = 1\ $ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's Lemma&lt;/a&gt;. Thus $\rm b = 1\ $ hence $\rm\:\ c \  =\ a^n\:.\ $ In particular $\rm\ c = 2\ \Rightarrow\ n = 1\:.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possible variations on such irrationality proofs, e.g. using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;Rational Root Test&lt;/a&gt; or directly using Unique Factorization of integers, or using the principality of &lt;a href=&quot;http://math.stackexchange.com/questions/4713/if-the-order-divides-a-prime-p-then-the-order-is-p-or-1/4715#4715&quot;&gt;denominator ideals&lt;/a&gt;. Perhaps the most elegant is to employ Dedekind's notion of the &lt;a href=&quot;http://math.stackexchange.com/questions/2883/show-that-3p2q2-implies-3p-and-3q/2941#2941&quot;&gt;&lt;strong&gt;conductor  ideal&lt;/strong&gt;&lt;/a&gt; - which yields a one-line proof that a  $\rm PID$ is integrally closed, i.e. satisfies the monic case of the rational root test.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-25T20:42:04.907" LastActivityDate="2010-11-25T20:42:04.907" />
  <row Id="11886" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T20:36:27.457" Score="4" ViewCount="0" Body="&lt;p&gt;To add to this thread, you can understand why certain notes sound good/bad together by looking at trig sum/product formulas, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cos(a) + cos(b) = 2 * cos(a - b) * cos(a + b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that when you add two tones/frequencies 'a' and 'b', it is equivalent to taking one wave of frequency 'a + b' and modulating its amplitude with another of frequency 'a - b'. The frequency 'a + b' will be a faster vibration, and the frequency 'a - b' will be a slower vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the two original frequencies are close (e.g. A = 440Hz and A# = 466Hz), the 'a - b' component will be heard as an unpleasant low frequency beating (here, 26Hz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the two original frequencies are integer ratios of each other (e.g. 3/2, 4/3) as in chords, then the resulting 'a + b' and 'a - b' frequencies will also be integer ratios of each other. The resulting wave will be simple and sounds harmonious. This is why integer ratios of notes are so important in music.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps to plot sums of sines graphically to see this in action.&lt;/p&gt;&#xA;" OwnerUserId="4048" LastActivityDate="2010-11-25T20:36:27.457" CommentCount="1" />
  <row Id="11887" PostTypeId="2" ParentId="11882" CreationDate="2010-11-25T20:36:47.203" Score="0" ViewCount="0" Body="&lt;p&gt;I'm going to go out on a limb and guess that you're trying to show the limit is 3 and that $f(x) = {x^2 - 2x + 9 \over x + 1} - 3$. I suggest trying to translate what you've done into the fact that  $|{x^2 - 2x + 9 \over x + 1} - 3| &amp;lt; {3 \over 5}|x - 2|$ whenever $|x - 2| &amp;lt; {1 \over 2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you choose any $\epsilon &amp;lt; {1 \over 2}$, then you have that $|{x^2 - 2x + 9 \over x + 1} - 3| &amp;lt; {3 \over 5}\epsilon$ whenever $|x - 2| &amp;lt; \epsilon$. So, given $\epsilon$, the natural choice for $\delta$ is ${3 \over 5}\epsilon$. (you got the $\delta$ and $\epsilon$ reversed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now verify that the &quot;for every $\epsilon$ there is a $\delta$&quot; definition is satisfied in this way. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-25T20:36:47.203" />
  <row Id="11888" PostTypeId="2" ParentId="11764" CreationDate="2010-11-25T20:42:07.207" Score="1" ViewCount="0" Body="&lt;p&gt;See Glen  Bredon's book &lt;em&gt;Sheaf Theory&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-25T20:42:07.207" />
  <row Id="11889" PostTypeId="2" ParentId="11836" CreationDate="2010-11-25T21:03:45.707" Score="2" ViewCount="0" Body="&lt;p&gt;The topology of $R^n$ is much more complicated than the topology of $R$. For example, a simply connected subset of $R$ is just an interval. A simply connected subset in $R^n$ can be very complicated.  This matters because simple connectivity enters as a hypothesis in several theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary of an interval in $R$ is simply a pair of points (or 1 or 0 points if the interval is unbounded).  But the boundary of an open set in $R^n$ can be a manifold or something more complicated.  Since the fundamental theorem of calculus requires integrating over a boundary, the theorem is more complicated in several variables. Integration over the boundary of an interval in $R$ is simply evaluating a function at two points.  Integration over a manifold is more subtle and requires a large amount of machinery to do formally.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-25T21:03:45.707" />
  <row Id="11890" PostTypeId="2" ParentId="11669" CreationDate="2010-11-25T21:25:24.103" Score="3" ViewCount="0" Body="&lt;p&gt;The &quot;circle of fifths&quot; is a by product of the preference for diatonic scales.&#xA;If you layout the chromatic (12-tone) scale without the white and raised black arrangement you'd use the same logic to describe a &quot;circle of sevenths&quot; (counting&#xA;upo semitones from C to G).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the arrangement makes solid sense when applied to the human hand. We need to be able to span interesting distances with the &quot;octave&quot; interval be very useful for most pianists&#xA;as a basic required for anyone older that 10-12. Some pianists can span 10ths with relative ease but they are in the minority. The piano music of Rachmaninoff is riddled with these massive but musically sonorus intervals. The are the major third expanded to&#xA;the pure natural interval (10 keys apart) of the &quot;bugle&quot; overtone series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can reach the 10th's on the white to white instances but the black to white (Bb to D for example) are beyond me. And doing them quickly and accurately is the mark of true mastery of the instrument... it's like being able to dunk: genetics help and no amount of effort can help a small handed pianist.&lt;/p&gt;&#xA;" OwnerUserId="4040" LastActivityDate="2010-11-25T21:25:24.103" />
  <row Id="11892" PostTypeId="1" CreationDate="2010-11-25T21:55:07.347" Score="3" ViewCount="116" Body="&lt;p&gt;Suppose that $f:U\subset\mathbb{C}\to\mathbb{C}$, where $U$ is a region in the complex plane, is a holomorphic function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c\in\mathbb{R}$ is a regular value for $\text{Re}(f(z))$ then it follows from implicit function theorem that &#xA;$\text{Re}(f(z))^{-1}(c)$ is at least locally a differentiable curve in the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt; Question: &lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- If $c$ is a regular value is any connected component of $\text{Re}(f(z))^{-1}(c)$ a global differentiable curve ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-If $c$ is not a regular value and $\text{Re}(f(z))^{-1}(c)$ have at least one cluster point is this set locally a curve ?&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2010-11-28T06:50:18.003" Title="What Re(f(z))=c is if f is a holomorphic function ?" Tags="&lt;complex-analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="11893" PostTypeId="2" ParentId="11864" CreationDate="2010-11-25T22:14:50.810" Score="1" ViewCount="0" Body="&lt;p&gt;You just need some simple properties of the Banach space-valued integral, namely linearity (the integral commutes with continuous linear operators) substitution for translations and the fundamental theorem of calculus, which says: $\frac{1}{h}\int_0 ^h f(t) d t \to f(0)$ (in the Banach space) for $h \to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after writing out all definitions (and using linearity and the semigroup properties), your job is to show that for any $t&amp;gt;0$ we have $$\frac{1}{h}\int_0 ^t (S(\tau+h) x_0 - S(\tau)x_0) d \tau \to S(t)x_0 - x_0$$ for $h\to 0$ from above. We can assume $t&amp;gt;h$. Then split the first integral into $\int_0 ^{t-h}+\int_{t-h} ^t$ and the second one into $\int_0 ^h+ \int_h ^t$. By translation the integrals $\int_0 ^{t-h}$ and $\int_h ^t$ cancel, and you just have to apply the fundamental theorem of calculus.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-11-25T22:14:50.810" CommentCount="6" />
  <row Id="11894" PostTypeId="2" ParentId="4170" CreationDate="2010-11-25T22:44:52.867" Score="0" ViewCount="0" Body="&lt;p&gt;For my work in this area, I refer to :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Richard Epstein &quot;Classical Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;li&gt;Wolfgang Rautenberg &quot;A Concise Introduction to Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;li&gt;Jon Barwise &quot;Handbook of Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;li&gt;Jean Heijenoort &quot;From Frege to Godel&quot;&lt;/li&gt;&#xA;&lt;li&gt;We Li &quot;Mathematical Logic&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rautenberg has a lot of examples, exercise, but is very heavy going (at least for me).  Epstein is fairly recent and very well laid out. Barwise is the most comprehensive for when you need to deep dive.&lt;/p&gt;&#xA;" OwnerUserId="4057" LastActivityDate="2010-11-25T22:44:52.867" CommunityOwnedDate="2010-11-25T22:44:52.867" />
  <row Id="11895" PostTypeId="2" ParentId="4170" CreationDate="2010-11-25T23:03:32.873" Score="0" ViewCount="0" Body="&lt;p&gt;Yuri Manin, A course in mathematical logic.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-25T23:03:32.873" CommunityOwnedDate="2010-11-25T23:03:32.873" />
  <row Id="11897" PostTypeId="1" CreationDate="2010-11-25T23:07:42.280" Score="15" ViewCount="363" Body="&lt;p&gt;Many years ago a lecturer of mine had a photocopy of a page from a book containing a math problem ( I think it was a simple quadradic equation ) that was stated/solved in Cuneiform, Arabic, Latin scripts and Finally in modern math notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have contacted my lecturer but he has no idea where it was from, nor I have been able to find it using google books searches etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know where to find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-25T23:11:59.090" LastActivityDate="2011-01-10T07:44:15.123" Title="Image of a math problem that was stated in Cuneiform, Arabic, Latin and Finally in modern math notation" Tags="&lt;reference-request&gt;&lt;math-history&gt;" CommentCount="5" FavoriteCount="3" />
  <row Id="11898" PostTypeId="2" ParentId="11784" CreationDate="2010-11-25T23:14:23.170" Score="0" ViewCount="0" Body="&lt;p&gt;An even easier example than Timothy's  is $f(x,y)=1$ when $x=0$ or $y=0$, and $f(x,y)=0$ for all other points. Then $f_x(0,0)=0, f_y(0,0)=0$ but $f$ is not differentiable (or even continuous at the origin).&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-25T23:14:23.170" />
  <row Id="11899" PostTypeId="1" AcceptedAnswerId="11913" CreationDate="2010-11-25T23:17:48.850" Score="5" ViewCount="171" Body="&lt;p&gt;Hello alltogether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At university, we are currently introduced in various methods of proving that a series converges. For example the ComparisonTest, the RatioTest or the RootTest. However we aren't told of how to calculate the point of convergence. The only series of which we know the point of convergence is the geometric series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this to be a non trivial task in general, but there have to be at least some methods to calculate the point of convergence for simple series, haven't there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could point me in the direction of some of these methods I would very much appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3188" LastEditorUserId="3188" LastEditorDisplayName="" LastEditDate="2010-11-25T23:26:47.783" LastActivityDate="2010-11-26T19:08:24.190" Title="Convergence of Series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="11900" PostTypeId="2" ParentId="11865" CreationDate="2010-11-25T23:27:02.570" Score="4" ViewCount="0" Body="&lt;p&gt;There's a brief note in &lt;a href=&quot;http://books.google.com/books?id=LLEZQC74gVcC&amp;amp;pg=PA157&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; on how Khayyam bumped into having to solve a cubic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll only make the note that you should remember the context of the time: there was no concept of negative, much less complex, solutions. Corresponding to our current Cartesian system, Khayyam only looked at intersections in the first quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another note should be made that the curves of the time were constructed with geometric tools (straightedge, compass, and a bunch of other contraptions), and $y=x^3$ isn't really a sort of curve that easily lends itself to such a construction (but is now easily constructed thanks to our current knowledge of coordinate geometry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=EIdtVPeD7GcC&amp;amp;pg=PA107&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a more explicit mention of the hyperbola-circle intersection problem Khayyam studied and was mentioned in the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ucQXDt2WRJgC&amp;amp;pg=PA89&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a (more or less) complete table of all the intersection cases Khayyam studied. (The book has an appendix containing a (translated) section of Khayyam's work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=V_B5SIGTnnoC&amp;amp;pg=PA92&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is yet another reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll keep updating this answer as I comb through more books; watch this space! As an aside, it's funny that my attempts to look for answers to &lt;a href=&quot;http://math.stackexchange.com/questions/11897&quot;&gt;this question&lt;/a&gt; are leading me to references for this question!)&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T03:58:11.750" LastActivityDate="2010-11-26T03:58:11.750" CommentCount="2" />
  <row Id="11901" PostTypeId="2" ParentId="11878" CreationDate="2010-11-25T23:31:46.023" Score="2" ViewCount="0" Body="&lt;p&gt;The process you describe can be better expressed in terms of the Bell-/Carleman-matrix (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleman_matrix&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; )  associated to function $f(x) = b^x$ . I got myself used to the following notation:&lt;br&gt;&#xA;let $V(x)$ denote a rowvector of infinite dimension of consecutive powers of an argument x $V(x)=[1,x,x^2,x^3, \dots]$ then let &lt;em&gt;B&lt;/em&gt; denote the infinite matrix which performs $ V(x) * B = V(b^x) $ . The columns of &lt;em&gt;B&lt;/em&gt; contain the coefficients for the powerseries for the consecutive powers of $(b^x)^0, (b^x)^1 , (b^x)^2, \ldots $. &lt;em&gt;B&lt;/em&gt; is then the (transposed) Carleman-matrix.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your formula and iteration constructs can simply be expressed by the notion of powers of &lt;em&gt;B&lt;/em&gt;; the h-fold nested infinite sums (your &lt;em&gt;A&lt;/em&gt;) are captured by the formal matrixproducts $ V(x)*B*B*\ldots*B = V(x)*B^h = V(b^(b^(b^\ldots (b^x)))  $ . Unfortunately the fractional part of iteration must then be expressed by a fractional power of &lt;em&gt;B&lt;/em&gt; --- which is not trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is only one problem. We have already the problem of powers of &lt;em&gt;B&lt;/em&gt; (or nested summation of your example): the convergence using matrices of finite size gets lost after few iterations and it is a special art to find closed forms of arbitrarily precise computable expressions for the powerseries only for third or fourth iteration. I've tried this different ways and could not get significant improvement of the convergence behaviour as long as I used that nesting which you describe above for more than some exotic bases b near 1.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are ways to express the coefficients of the powerseries in finite expressions/sums including &lt;em&gt;exp/log&lt;/em&gt; which we can assume to be available always in arbitrary precision. This can be found by triangular decomposition of &lt;em&gt;B&lt;/em&gt; and a modified way to arrive at the required powers. (See for a not very explanative technical description &lt;a href=&quot;http://go.helms-net.de/math/tetdocs/TTetrationExactEntries_short.htm&quot; rel=&quot;nofollow&quot;&gt;exact entries&lt;/a&gt;, remark: there might be better descriptions...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all this covers only the part of integer height (integer number of iterations). And for this we do not really need that formal powerseries/matrixpowers approximated by finitely truncated matrices: we have precise exponentiation for each base at hand. The crux is the part of the fractional height/iteration count. Here there are some different approaches available; one of it is the approach to compute fractional powers of &lt;em&gt;B&lt;/em&gt; by diagonalization to get the coefficients for a formal powerseries which represents the fractional iteration of $b^x$ However - to have series with real coefficients we must restrict ourselves to bases b between $exp(-exp(1))\ldots exp(exp(-1)) $ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-25T23:40:34.727" LastActivityDate="2010-11-25T23:40:34.727" CommentCount="1" />
  <row Id="11902" PostTypeId="2" ParentId="11899" CreationDate="2010-11-25T23:38:27.723" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;there have to be at least some methods to calculate the point of convergence for simple series, haven't there?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are sure a given sequence of numbers converges, you have to just keep iterating everything manually, unless you can think of a clever trick.  I suggest you study how Euler proved that $\zeta(2)=\frac{\pi^2}{6}$, that will be very instructive for you because it will demonstrate how clever you have to sometimes be in order to solve this type of problem.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-25T23:38:27.723" />
  <row Id="11903" PostTypeId="1" CreationDate="2010-11-25T23:42:05.523" Score="2" ViewCount="124" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/11849/w1w2-wna-w1a-wna-for-n-integer-2-w0-and-aconstant-real-0&quot;&gt;a previous post&lt;/a&gt;, the following inequality has been proven&#xA;$${\left( {\sum\limits_{i = 1}^n {{W_i}} } \right)^a} \le \sum\limits_{i = 1}^n {{W_i}^a}$$&#xD;&#xA;where $W_i\gt 0$, $0\lt a\lt 1$.&#xA;I guess it is more correct to say that this is always greater, and it is valid for $a\gt 0$ not just $0\lt a \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to see if one can generalize it to something like&#xA;$$f\left( \sum\limits_{i = 1}^n W_i  \right) \le \sum\limits_{i = 1}^n {f({W_i})}$$ &#xA;where ${W_i}\gt 0$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under what circumstances and functions can this be true ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been proven for the power function $f(x)=kx^a$ where $a,k\gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other cases? Do you think there is any basic inequality to prove this?&lt;/p&gt;&#xA;" OwnerUserId="4011" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-25T23:59:29.310" LastActivityDate="2010-11-26T11:41:21.750" Title="$f\left(\sum X_i\right) \leq \sum  f(X_i)$,  where $X_i\gt 0$; for what functions is this true ?" Tags="&lt;functions&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="11905" PostTypeId="2" ParentId="11899" CreationDate="2010-11-25T23:52:20.970" Score="3" ViewCount="0" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right after the geometric series, the easiest method I know of to calculate the sum of a series is the property of derivation term-by-term of convergent power series. I don't know if you've already covered it in the lessons, so I'll try to give you an idea of how it works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that $\sum x^n=\frac{1}{1-x}$ if $x&amp;lt;1$; now, you will see that you can derive a series like the one before with respect to $x$, and you can do it term-by-term so that you obtain (by differentiating both sides of the previous identity): $\frac{d}{dx}\sum x^n=\sum \frac{d}{dx}x^n=\sum nx^{n-1}=\frac{1}{(1-x)^2}$.&lt;/p&gt;&#xA;" OwnerUserId="4054" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T00:04:00.927" LastActivityDate="2010-11-26T00:04:00.927" CommentCount="1" />
  <row Id="11906" PostTypeId="2" ParentId="11903" CreationDate="2010-11-26T00:00:08.793" Score="2" ViewCount="0" Body="&lt;p&gt;Many inequalities similar to this one come from &lt;a href=&quot;http://en.wikipedia.org/wiki/Jensen%27s_inequality&quot; rel=&quot;nofollow&quot;&gt;Jensen's Inequality&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4054" LastEditorUserId="4054" LastEditorDisplayName="" LastEditDate="2010-11-26T11:41:21.750" LastActivityDate="2010-11-26T11:41:21.750" CommentCount="5" />
  <row Id="11907" PostTypeId="2" ParentId="11899" CreationDate="2010-11-26T00:02:26.013" Score="5" ViewCount="0" Body="&lt;p&gt;There are some convergent series whose exact sum can be evaluated. For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{n}{(n+1)!}=\dfrac{(n+1)-1}{(n+1)!}=\dfrac{1}{n!}-\dfrac{1}{(n+1)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the series &lt;a href=&quot;http://en.wikipedia.org/wiki/Telescoping_series&quot; rel=&quot;nofollow&quot;&gt;telescopes&lt;/a&gt;. Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=1-\underset{n\rightarrow\infty}{\lim}\dfrac{1}{(n+1)!}=1-0=1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in general it is a difficult task to find the exact sum, as you said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Unlike $\displaystyle\sum_{n=1}^{\infty }\frac{1}{n^{2}}$ whose sum equals $\dfrac{\pi ^{2}}{6}$, the series $\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{3}}$, although convergent, nobody knows a closed form in terms of other mathematical constants. This sum is therefore a mathematical constant in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 2: Another example that uses integration and differentiation of the geometric series $$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x},\qquad\left\vert x\right\vert &amp;lt;1$$  is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1%&#xD;&#xA;}{e-1}\right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added 3: From&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x}\qquad\left\vert x\right\vert &amp;lt;1,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get for $\left\vert x\right\vert &amp;lt;1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \sum_{n=1}^{\infty }x^{n}dx=\sum_{n=1}^{\infty }\frac{1}{n+1}%&#xD;&#xA;x^{n+1}=\int \frac{x}{1-x}dx=-x-\log \left\vert 1-x\right\vert .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{1}{n+1}x^{n}=-1-\frac{1}{x}\log \left\vert&#xD;&#xA;1-x\right\vert $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we diferentiate, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n-1}=\frac{1}{x^{2}}\log \left\vert&#xD;&#xA;1-x\right\vert -\frac{1}{x^{2}-x},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or equivalently&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n}=\frac{1}{x}\log \left\vert&#xD;&#xA;1-x\right\vert -\frac{1}{x-1}\qquad\left\vert x\right\vert &amp;lt;1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally for $x=e^{-1}$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1%&#xD;&#xA;}{e-1}\right) .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-26T19:08:24.190" LastActivityDate="2010-11-26T19:08:24.190" CommentCount="3" />
  <row Id="11908" PostTypeId="1" AcceptedAnswerId="11910" CreationDate="2010-11-26T00:07:50.230" Score="3" ViewCount="66" Body="&lt;p&gt;I am working through an economics &lt;a href=&quot;http://faculty.wcas.northwestern.edu/~lchrist/papers/aeragency.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; and I need to take the derivative of the following function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h\left(\overline{\omega}\right) = \int^{\infty}_{\overline{\omega}} \omega \Phi \left(d\omega\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though I don't understand it well, I can do the derivative for the case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g\left(\overline{\omega}\right) = \int^{\overline{\omega}}_{0} \omega \Phi \left(d\omega\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the derivative is simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g&amp;#39;\left(\overline{\omega}\right) = \overline{\omega} \phi \left(\overline{\omega}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for $h\left(\overline{\omega}\right)$ where the upper bound is $\infty$ I really have no idea of what to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me? Any explanation or even a pointer to where I can learn those things would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="85" LastActivityDate="2010-11-26T00:27:34.907" Title="Derivative of an expectation (using integrals)" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;expectation&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="11909" PostTypeId="2" ParentId="11908" CreationDate="2010-11-26T00:17:09.643" Score="3" ViewCount="0" Body="&lt;p&gt;Since $$h(\overline{\omega})=-\int_{\infty}^{\overline{\omega}}\omega\Phi(d\omega),$$&#xD;&#xA;$h&amp;#39;(\overline{\omega})=-\overline{\omega}\phi(\overline{\omega})$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T00:17:09.643" CommentCount="1" />
  <row Id="11910" PostTypeId="2" ParentId="11908" CreationDate="2010-11-26T00:17:43.187" Score="4" ViewCount="0" Body="&lt;p&gt;If the integral is sufficiently nice by which I mean it doesn't blow up to infinity, we can write $h(\bar{\omega}) = \int_{l}^{\infty} \omega d(\Phi(w)) - \int_{l}^{\bar{\omega}} \omega d(\Phi(w))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $l$ is some constant number. It could be $0$ or $\infty$ based on your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first integral is just a constant, and the second integral is similar to $g(\bar{\omega})$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T00:27:34.907" LastActivityDate="2010-11-26T00:27:34.907" CommentCount="10" />
  <row Id="11911" PostTypeId="2" ParentId="11908" CreationDate="2010-11-26T00:18:29.647" Score="4" ViewCount="0" Body="&lt;p&gt;Note that $h(\bar \omega ) + g(\bar \omega )$ is constant, and use the result for $g(\bar \omega )$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-26T00:18:29.647" />
  <row Id="11912" PostTypeId="2" ParentId="11903" CreationDate="2010-11-26T00:27:33.893" Score="3" ViewCount="0" Body="&lt;p&gt;The condition $$f\left( {\sum\limits_{i = 1}^n {{W_i}} } \right) \le \sum\limits_{i = 1}^n {f({W_i})} $$ where ${W_i}$&gt;0 is equivalent to $$f\left( {\sum\limits_{i = 1}^2 {{W_i}} } \right) \le \sum\limits_{i = 1}^2 {f({W_i})} $$ where ${W_i}$&gt;0. These are subadditive functions. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Subadditivity&quot; rel=&quot;nofollow&quot;&gt; Subadditivity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T00:27:33.893" CommentCount="1" />
  <row Id="11913" PostTypeId="2" ParentId="11899" CreationDate="2010-11-26T01:03:30.250" Score="6" ViewCount="0" Body="&lt;p&gt;The short answer is that there are no general methods and we use whatever's available.  The first thing you can do is learn the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;Taylor series&lt;/a&gt; of common functions; then evaluation of the Taylor series at a point where the series converges to the function corresponds to evaluation of the function.  For example, $e^x = \sum \frac{x^n}{n!}$, hence $e = \sum \frac{1}{n!}$.  Less trivially, $\ln (1 + x) = \sum \frac{(-1)^{n-1} x^n}{n}$, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \pm ....$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more Taylor series you know, the more series you can evaluate &quot;explicitly&quot; in some sense.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, life gets difficult.  Mastering the use of Taylor series is already highly nontrivial - especially recognizing when the method is applicable - so worry about learning how to do that properly first.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-26T01:03:30.250" CommentCount="2" />
  <row Id="11916" PostTypeId="1" AcceptedAnswerId="11919" CreationDate="2010-11-26T02:45:05.160" Score="1" ViewCount="87" Body="&lt;p&gt;Let $$p_n(x)e^{-x^2}$$ be the $n$th derivative of $$e^{-x^2}.$$&#xD;&#xA;Find a formula for $p_n(x)$. We have $p_1(x)=-2x, p_2(x)=4x^2-2$, etc. But what is the general formula for $p_n$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T02:57:47.390" Title="Higher derivatives of  an exponential function" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;classical-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11917" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T02:48:20.080" Score="9" ViewCount="0" Body="&lt;p&gt;The math in this thread is awesome, but I'm not sure it addresses the original question about the &quot;difference between white and black notes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other responses in this thread provide enough math to understand that each octave can be more-or-less naturally divided into twelve semitones. The Western music tradition further evolved to be based around what's called the &quot;diatonic scale&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A musical scale is a sequence of pitches within one octave; scales can be defined by the number of semitones between each successive note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Whole Tone Scale consists entirely of whole tones; it has six distinct pitches, each of which is two semitones higher than the last. So you might represent it with the string '222222' — that is, take a note, then the note 2 semitones higher, then the note 2 semitones higher, etc., until the last &quot;2&quot; takes you to the note an octave above where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Diatonic Scale that Western music is based around could likewise be represented by the string '2212221'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with a C on a keyboard and go up, you'll see that the white keys conform to that pattern of semitones. That, generally, is why the black keys are in that particular pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can start a scale on any pitch, not just C. That's why the &quot;same&quot; diatonic scale in a different key will involve a unique set of sharps and flats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Diatonic Scale can also be represented by '2212221' shifted to the left or right any number of times. For example, '2122212', '1222122', etc. are also Diatonic; these are called the &quot;modes&quot; of the Diatonic scale. Each Diatonic mode can be played on only the white keys of the piano by starting on a different pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2212221 is called the Ionian mode (this is also generically called the Major scale), and can be played on the white keys starting with C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2122212 is the Dorian mode and can be played on the white keys starting with D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1222122 is the Phrygian mode, starting on E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2221221 is the Lydian mode, starting on F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2212212 is the Mixolydian mode, starting on G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2122122 is the Aeolian mode (the Minor scale), starting on A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1221222 is the (awesome) Locrian mode, starting on B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each mode has its own unique &quot;sound&quot;, which (in my opinion, at least) derives precisely from the different placement of the semitones within each scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course there are scrillions of non-Diatonic scales that have nothing whatsoever to do with how the modern keyboard came to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to add a shorter, less implicit answer: The white keys alone can be used to play the set of diatonic scales listed above; the black keys are &quot;different&quot; because they are the remaining chromatic pitches not used in that set of diatonic scales.&lt;/p&gt;&#xA;" OwnerUserId="4067" LastEditorUserId="4067" LastEditorDisplayName="" LastEditDate="2010-11-27T22:21:51.630" LastActivityDate="2010-11-27T22:21:51.630" CommentCount="2" />
  <row Id="11918" PostTypeId="2" ParentId="11916" CreationDate="2010-11-26T02:49:28.713" Score="1" ViewCount="0" Body="&lt;p&gt;$p_n$ satisfies the recurrence, $p_{n+1}(x) = p_n&amp;#39;(x) - 2xp_n(x)$ with $p_0(x) = 1$. &lt;s&gt; This looks to me like it might give you some Tchebyshev polynomial. Something like $p_n(x) = 2 (-1)^n T_n(x)$ &lt;/s&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;I was wrong with the Tchebyshev polynomial thing. Sorry.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T02:54:44.400" LastActivityDate="2010-11-26T02:54:44.400" />
  <row Id="11919" PostTypeId="2" ParentId="11916" CreationDate="2010-11-26T02:57:47.390" Score="7" ViewCount="0" Body="&lt;p&gt;That depends on what you mean by &quot;general formula.&quot;  These are (up to some normalization) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hermite_polynomials&quot; rel=&quot;nofollow&quot;&gt;Hermite polynomials&lt;/a&gt;.  They satisfy a nice recurrence and have a nice generating function.  You could torture some kind of general formula out of the generating function but I really don't see the point.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-26T02:57:47.390" CommentCount="4" />
  <row Id="11920" PostTypeId="1" AcceptedAnswerId="11921" CreationDate="2010-11-26T03:06:35.913" Score="3" ViewCount="102" Body="&lt;p&gt;I am reminded of this question that appeared in a regional Physics Olympiad I had appeared. I was wondering if there is a &quot;mathematical&quot; way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start from a point $A$ at midnight along a specified path down a mountain and reach a specified point $B$ exactly 24 hours later. At this point you reverse your direction of travel and travel along the same path to point $A$ and reach $A$ exactly 24 hours later. At any point of time during these two days, your velocity can be positive, negative or zero (and of course less than $c$). Prove that there exists at least one point along the path where you were at the same time on day 1 and day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &quot;seems&quot; like an application of intermediate value theorem for some appropriately defined function, but I am not sure (though I will tag it as calculus for the moment). Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-26T04:00:22.167" Title="Traveling up and down a mountain" Tags="&lt;calculus&gt;" AnswerCount="1" />
  <row Id="11921" PostTypeId="2" ParentId="11920" CreationDate="2010-11-26T03:10:45.550" Score="9" ViewCount="0" Body="&lt;p&gt;This a classic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;layman&quot; proof is to consider you and your ghost from the previous day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you turn back, let your ghost of the previous day start from point $A$. At some point, you and ghost must meet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In stricter mathematical terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two continuous functions, which measure the distance from point $A$ with time (in days say). Let the distance from $A$ to $B$ be 1km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have that $f(0) = 0, f(1) = 1, g(0) = 1, g(1) = 0$. $f$ is the function for day 1 and $g$ is the function for day 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $h(x) = f(x) - g(x)$ is continuous with $h(0) &amp;lt; 0$ and $h(1) &amp;gt; 0$ and so must be $0$ at some point in between, using the intermediate value theorem.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-26T04:00:22.167" LastActivityDate="2010-11-26T04:00:22.167" CommentCount="2" />
  <row Id="11922" PostTypeId="1" CreationDate="2010-11-26T03:52:09.197" Score="1" ViewCount="91" Body="&lt;p&gt;Let $A$ and $B$ be finite-dimensional Hermitian matrices, and let $v$ $:=$ $max${$||A||,||B||$}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that $||(e^{-iAt/m}e^{-iBt/m})^m -e^{-i(A+B)t}||\leq\epsilon$ provided $m = \Omega(v^2t^2/\epsilon)$.&lt;/p&gt;&#xA;" OwnerUserId="4073" LastEditorUserId="4073" LastEditorDisplayName="" LastEditDate="2010-11-26T04:15:59.430" LastActivityDate="2010-11-26T07:42:41.853" Title="Lie Product Formula" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11923" PostTypeId="1" AcceptedAnswerId="11926" CreationDate="2010-11-26T04:15:14.580" Score="7" ViewCount="248" Body="&lt;p&gt;Can anyone recommend a good self-contained reference for completion of rationals to get reals using Cauchy sequences?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-28T03:08:15.393" Title="Completion of rational numbers via Cauchy sequences" Tags="&lt;real-analysis&gt;&lt;reference-request&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="11924" PostTypeId="1" AcceptedAnswerId="11927" CreationDate="2010-11-26T04:37:25.723" Score="2" ViewCount="80" Body="&lt;p&gt;I made the following observation following a suggestion from a friend. This question is about the Microsoft calculator program implemented on Windows. Since I don't have access to any other calculator at the moment, I cannot confirm whether this is a general feature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you enter an integer in the calculator program, say $N$, hit the minus button followed by the same integer again, you get zero as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you input $N^2$, then hit square root button, you get $N$, then subtract $N$, you get a very small number, but not zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been my observation for several integers that I have tried. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $N=5$, you get $1.232\times 10^{-24}$ in the final step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I understand computers use finite precision (and hence the square root function must be working under finite precision). However, I don't understand why this shows up in the final step only? Is it only because this is a small number in comparison to any integer, but when you get $0$, you can no longer make that approximation?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-26T04:52:48.210" LastActivityDate="2010-11-26T05:53:23.320" Title="Rounding trouble in a calculator's square root function" Tags="&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11925" PostTypeId="1" AcceptedAnswerId="11928" CreationDate="2010-11-26T04:43:11.550" Score="2" ViewCount="110" Body="&lt;p&gt;Let $f:A \rightarrow B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complements indicated below are taken within $A$ or $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to prove that&#xA;$f(S)^c \subseteq f(S^c)$, $\forall S \subseteq A$, if and only if $f$ is surjective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I need to prove $f(A) = B$ right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove this?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T05:40:50.090" LastActivityDate="2010-11-26T06:00:15.420" Title="$f(S)^c \subseteq f(S^c)$, $\forall S\subseteq A$, if and only if $f\colon A\to B$ is surjective. " Tags="&lt;elementary-set-theory&gt;&lt;proof&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="1" />
  <row Id="11926" PostTypeId="2" ParentId="11923" CreationDate="2010-11-26T04:50:49.867" Score="4" ViewCount="0" Body="&lt;p&gt;Here is a link that came up in a Google search.  You can also search for &quot;construction of real numbers&quot;.  Since this is standard material, you can find it in many books on analysis.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.mit.edu/~tkemp/18.024/R.pdf&quot; rel=&quot;nofollow&quot;&gt;math.mit.edu/~tkemp/18.024/R.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-26T06:54:27.670" LastActivityDate="2010-11-26T06:54:27.670" CommentCount="4" />
  <row Id="11927" PostTypeId="2" ParentId="11924" CreationDate="2010-11-26T04:52:33.207" Score="7" ViewCount="0" Body="&lt;p&gt;You'll find your answer and much more in the articles below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://docs.sun.com/source/806-3568/ncg_goldberg.html&quot; rel=&quot;nofollow&quot;&gt;What Every Computer Scientist Should Know About Floating-Point Arithmetic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see this article on &lt;a href=&quot;http://www.rskey.org/~mwsebastian/miscprj/forensics.htm&quot; rel=&quot;nofollow&quot;&gt;Calculator Forensics&lt;/a&gt;, esp. &quot;Hints and Tricks For Obtaining Guard Digits&quot;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-26T04:52:33.207" CommentCount="1" />
  <row Id="11928" PostTypeId="2" ParentId="11925" CreationDate="2010-11-26T05:04:40.917" Score="3" ViewCount="0" Body="&lt;p&gt;You want to prove &lt;em&gt;two&lt;/em&gt; things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; the complement of the image is always contained in the image of the complement, &lt;em&gt;then&lt;/em&gt; $f(A)=B$; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; $f(A)=B$, then for every subset $S$ of $A$ you have that the complement of the image is contained in the image of the complement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, &quot;I have to prove that $f(A)=B$&quot; is only true for half of the problem at issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how you prove 1 above, well, since the containment holds for &lt;em&gt;all&lt;/em&gt; subsets $S$ of $A$, perhaps you can pick a &lt;em&gt;particular&lt;/em&gt; subset $S$ of $A$ in some clever way that will tell you that $f(A)$ must be equal to $B$. (&lt;strong&gt;HINT:&lt;/strong&gt; $f(A)=B$ if and only if $f(A)^c$ is... )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2, you have to &lt;strong&gt;assume&lt;/strong&gt; that $f(A)=B$ (that $f$ is surjective). Then you need to take an &lt;em&gt;arbitrary&lt;/em&gt; subset $S$ of $A$. Then to test whether $f(S)^c\subseteq f(S^c)$ holds, you need to take an arbitrary $b\in f(S)^c$, and show that it must lie in $f(S^c)$; that is, that there exists some $a\notin S$ such that $b=f(a)$. Now go to it.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T05:32:17.450" LastActivityDate="2010-11-26T05:32:17.450" />
  <row Id="11929" PostTypeId="2" ParentId="11925" CreationDate="2010-11-26T05:15:49.450" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ (\Rightarrow)\ $ Put $\rm S\ =\ \ldots\ \ (\Leftarrow)\ $ Consider $\rm\  {\overline {f(S)}} \cap (f(S) \cup f(\overline S))$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-26T05:15:49.450" CommentCount="3" />
  <row Id="11930" PostTypeId="2" ParentId="11724" CreationDate="2010-11-26T05:33:52.650" Score="2" ViewCount="0" Body="&lt;p&gt;I would normally take $p$-adic integration to mean &quot;integration of $p$-adic valued functions&quot; or &quot;integration of differential forms with some kind of $p$-adic valued functions as coefficients&quot;, where the integration is also taking place over some kind of $p$-adic space or manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for wanting such theories are various.  One reason is indicated in George S.'s answer: there are &lt;em&gt;known&lt;/em&gt; analogues of classical Hodge theory, known as $p$-adic Hodge theory, whose proofs however are not analytic, but rather proceed via arithmetic geometry.  One would like to have more analytic ways of thinking about them, and this is one goal of Robert Coleman's theory.  (In a recent volume of Asterisque, namely vol. 331, Coleman and Iovita have an article, &lt;a href=&quot;http://math.berkeley.edu/~coleman/Hidden/HiddenStr.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Hidden structures on semistable curves&quot;&lt;/a&gt;, related to this problem.)  (Note also that $p$-adic Hodge theory relates $p$-adic etale cohomology to crystalline cohomology, which gives on answer to your question of how $p$-adic integration might be related to those topics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is that many integral formulas (involving usual archimedean integrals) appear in the theory of classical $L$-functions attached&#xA;to automorphic forms, and one would like, at least in certain contexts, to be able to write down $p$-adic analogues so as to construct $p$-adic $L$-functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what machinery is used: in the theory of $p$-adic $L$-functions and related contexts in Iwasawa theory, often nothing more is used than basic computations with Riemann sums.  In the material related to $p$-adic Hodge theory, much more substantial theoretical foundations are used: tools from arithemtic geometry, rigid analysis, possibly Berkovic spaces, and related topics.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-26T05:33:52.650" />
  <row Id="11931" PostTypeId="2" ParentId="11925" CreationDate="2010-11-26T05:39:11.903" Score="1" ViewCount="0" Body="&lt;p&gt;You need to prove more than that $f(A) = B$. You need to show that if $f$ is surjective, then $(f(S))^c \subset f(S^c)$ for every possible subset $S$ of $A$, and that if $f$ is not surjective, then there is some $S$ for which $(f(S))^c$ is not a subset of  $f(S^c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Largish hint for the first part: If $f$ is surjective then $f(S) \cup f(S^c)$ is all of $B$, regardless of what $S$ is. Use this to prove the contrapositive of the first part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint for the second part: Pick a special subset $S$ of $A$ and show it works for that $S$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-26T06:00:15.420" LastActivityDate="2010-11-26T06:00:15.420" CommentCount="6" />
  <row Id="11932" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T05:42:04.033" Score="7" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mathematically, is there any difference between white and black notes, or do we make the distinction just for historical reasons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no mathematical difference between the white and black notes. Adjacent notes on modern piano keyboards are typically tuned 1/12 of an octave apart. Quiaochu explains this most completely, but what it boils down to is that there is no difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We haven't always and don't today always use equal temperament on keyboard instruments but even then the difference between white and black notes would be arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction is for historical and cultural reasons. There is a cool picture here showing Nicholas Farber's Organ (1361), which used an 8 + 4 layout rather than the modern 7 + 5 layout we see today. &lt;a href=&quot;http://en.wikipedia.org/wiki/Musical_keyboard#Size_and_historical_variation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Musical_keyboard#Size_and_historical_variation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples of instruments in use today that use a chromatic keyboard with no differentiation between the &quot;white&quot; and the &quot;black&quot; notes. See the Bayan and the Bandoneon accordion type instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the New England Conservatory in the classroom where they teach a class on quarter tones, they keep two pianos tuned a quarter tone off from each other. In that case, a full 24-note chromatic quarter tone octave must be played alternating notes on the two pianos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only the beginning of this particular rabbit hole.&lt;/p&gt;&#xA;" OwnerUserId="4076" LastActivityDate="2010-11-26T05:42:04.033" CommentCount="1" />
  <row Id="11933" PostTypeId="2" ParentId="11761" CreationDate="2010-11-26T05:50:07.007" Score="5" ViewCount="0" Body="&lt;p&gt;As a supplement to T..'s nice answer, let me write something that is more of a cultural remark than a direct answer to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people learnt about the yoga of mixed motives by reading &lt;a href=&quot;http://www.math.ias.edu/files/deligne/GaloisGroups.pdf&quot; rel=&quot;nofollow&quot;&gt;Deligne's article&lt;/a&gt;&#xA;on the thrice punctured sphere.  (This was certainly the case for me; and a hat-tip to &lt;a href=&quot;http://quomodocumque.wordpress.com/2009/07/24/le-groupe-fondamental-de-la-droite-projective-moins-trois-points-is-now-online/&quot; rel=&quot;nofollow&quot;&gt;Quomodocumque&lt;/a&gt; for providing the link.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article helps explain what one should expect from the category of mixed motives, and thus helps one recognize the (or perhaps a) correct construction when one sees it. &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-11-26T05:50:07.007" CommentCount="0" />
  <row Id="11934" PostTypeId="2" ParentId="11924" CreationDate="2010-11-26T05:53:23.320" Score="0" ViewCount="0" Body="&lt;p&gt;J.M., since when is 24 decimal digits accuracy considered a bad approximation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original poster, they could show sqrt(5)**2 as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.000000000000000000000001232&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since the display probably has fewer display digits than this, they chop it to &quot;5&quot; as that is the closest representation in the allowed number of digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I just tried this in &quot;calc&quot; in windows 7 and the 5-sqrt(5)**2 is 1.7599e-37&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want exact math, you will need to use a symbolic math package.&lt;/p&gt;&#xA;" OwnerUserId="4080" LastActivityDate="2010-11-26T05:53:23.320" CommentCount="2" />
  <row Id="11935" PostTypeId="2" ParentId="11923" CreationDate="2010-11-26T06:03:21.543" Score="2" ViewCount="0" Body="&lt;p&gt;Pages &lt;a href=&quot;http://books.google.com/books?id=3-nrPB7BQKMC&amp;amp;lpg=PP1&amp;amp;pg=PA142#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;142&lt;/a&gt;-154 of Stoll's &lt;em&gt;Set theory and logic&lt;/em&gt; looks like a good self-contained treatment.  The chapter starts on &lt;a href=&quot;http://books.google.com/books?id=3-nrPB7BQKMC&amp;amp;lpg=PP1&amp;amp;pg=PA130#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 130&lt;/a&gt; and is called &quot;The extension of the natural numbers to the real numbers.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-26T06:03:21.543" />
  <row Id="11936" PostTypeId="1" CreationDate="2010-11-26T06:16:29.437" Score="1" ViewCount="67" Body="&lt;p&gt;Let Q= {q_1,...,q_n} be a set of permutations of some finite set X.  Let &#xA;Q' be {q in Q^* | q=1 in the symmetric group on X}.  Is Q' a regular language?&lt;/p&gt;&#xA;" OwnerUserId="4081" LastActivityDate="2010-11-27T04:22:51.837" Title="regular languages and permutations" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="11937" PostTypeId="2" ParentId="11936" CreationDate="2010-11-26T06:20:37.763" Score="3" ViewCount="0" Body="&lt;p&gt;Construct a graph $\Gamma$ having as vertices the elements of the symmetric group $S(X)$, and for each $g\in S(X)$ and each $i\in\{1,\dots,n\}$ an arrow from the vertex $g$ to the vertex $q_ig$ labeled $q_i$. Upgrade $\Gamma$ into a deterministic automatic so that the initial state is the identity element $e$ of $S(X)$, and the only accepted state is $e$ itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the language accepted by the automaton $\Gamma$?&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-26T06:20:37.763" />
  <row Id="11938" PostTypeId="2" ParentId="11846" CreationDate="2010-11-26T06:25:36.740" Score="1" ViewCount="0" Body="&lt;p&gt;When you would use which really depends on the random variable you're interested in.  If you're interested in the number of trials needed to obtain the first success, use the first kind of geometric distribution.  If you're interested in the number of successes you are able to achieve before the first failure occurs then use the second kind of geometric distribution.  (Inverting the interpretation of the second version like this also requires you to redefine the success probability as $1-p$ and the failure probability as $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Wikipedia page says, &quot;Which of these one calls 'the' geometric distribution is a matter of convention and convenience.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the two different versions confusing myself.  When I read somewhere that $X$ has a &quot;geometric distribution,&quot; the writer isn't always careful to specify which one is meant, and then I have to spend time figuring out which one it is.  This confusion also spreads to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_binomial_distribution&quot; rel=&quot;nofollow&quot;&gt;negative binomial distribution&lt;/a&gt;, which is a generalization of the geometric distribution.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;exponential distribution&lt;/a&gt;, which is a continuous version of the geometric distribution, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution&quot; rel=&quot;nofollow&quot;&gt;gamma distribution&lt;/a&gt; (a generalization of the exponential), have more than one definition, too.  It's too bad that these weren't standardized with one definition for each, but part of the reason they weren't is that the different versions are useful in different scenarios. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to Josh Guffin's comment: Yes, in many contexts it is easy to figure out which geometric distribution a writer is referring to.  However, in some it is not.   &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page on moment-generating functions&lt;/a&gt; is a classic example.  The table there gives the mgf's for the negative binomial (and thus geometric), exponential, and gamma distributions, but it doesn't specify which convention for each one is being used.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-27T05:28:44.690" LastActivityDate="2010-11-27T05:28:44.690" CommentCount="1" />
  <row Id="11939" PostTypeId="1" AcceptedAnswerId="11959" CreationDate="2010-11-26T06:51:20.807" Score="2" ViewCount="117" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a 2x2 linear system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;u&lt;/strong&gt; = &lt;strong&gt;b&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution will still be the same even after we interchange the rows in A and B. I know this to be true because algebraically, we will get the same set of equations before and after the row interchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the vectors in columns of A and B are different. So how can the system still have the same solution as before the row interchange?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2010-11-26T10:03:08.690" Title="Elementary Row Operations - Interchange a Matrix's rows" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11940" PostTypeId="2" ParentId="9934" CreationDate="2010-11-26T07:01:58.430" Score="3" ViewCount="0" Body="&lt;p&gt;Just thought, I would make this obvious remark, extending Arturo's answer: since, even in the infinite case, any subgroup of finite index in $G$ must be normal for $G$ to satisfy the requirement, it follows that for any $H$ of finite index in $G$, any subgroup of the quotient $G/H$ will also correspond to a subgroup of $G$ of finite index. In particular, for any $H$ of finite index, $H$ must be normal and the quotient $G/H$ must be cyclic by Arturo's argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A class of such groups considerably extending that of cyclic groups are the pro-cyclic group, i.e. inverse limits of cyclic ones. Examples include $\mathbb{Z}_p$ or any product $\prod_p \mathbb{Z}_p$ over distinct primes $p$. In particular, $\hat{\mathbb{Z}}$ is another example. In fact, Arturo's argument shows that any pro-finite group satisfying the above condition must be pro-cyclic.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T07:01:58.430" CommentCount="3" />
  <row Id="11941" PostTypeId="1" AcceptedAnswerId="11950" CreationDate="2010-11-26T07:02:12.060" Score="3" ViewCount="182" Body="&lt;p&gt;I was assisting a TA for an introductory calculus class with the following limit,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I came to simple solution which involved evaluating the &quot;reciprocal&quot; limit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{z \rightarrow 0} \left(\frac{1}{1+\frac{z}{5}}\right)^{8/z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by using the Taylor expansion of $\log(1+z)$ around $z=0$. However, the TA claims that the students have not learned about series expansions so that would not be a valid solution for the course. I tried applying L'Hopital's rule, which I was told the class did cover, but I was unsuccessful. As a note I will mention that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \rightarrow \infty} \left(\frac{20x}{20x+4}\right)^{8x} = e^{-8/5}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas for a solution to this problem using only knowledge from a first quarter (or semester) calculus course which hasn't covered series expansions?&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-12-08T02:58:40.887" Title="Evaluation of the limit, $\lim_{x\rightarrow\infty} \left(\frac{20x}{20x+4}\right)^{8x}$, using only elementary methods." Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11942" PostTypeId="2" ParentId="11939" CreationDate="2010-11-26T07:02:28.527" Score="3" ViewCount="0" Body="&lt;p&gt;Rearranging the rows of $A$ and $b$ corresponds to premultiplying them with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Permutation_matrix&quot; rel=&quot;nofollow&quot;&gt;permutation matrix&lt;/a&gt; $P$. It's not hard to see that the new system $PAu = Pb$ has the same solution as the original system. This holds for linear systems of all sizes, not just $2\times 2$ ones.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-26T07:02:28.527" />
  <row Id="11944" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T07:08:32.163" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: Consider taking logarithms before applying L'Hospital's rule.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-26T07:08:32.163" />
  <row Id="11945" PostTypeId="1" AcceptedAnswerId="11969" CreationDate="2010-11-26T07:08:40.347" Score="8" ViewCount="443" Body="&lt;p&gt;Where does this sequence converge?&#xA;$\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,...&lt;/p&gt;&#xA;" OwnerUserId="4082" LastEditorUserId="4082" LastEditorDisplayName="" LastEditDate="2010-11-26T17:59:40.423" LastActivityDate="2010-11-27T00:03:34.820" Title="Limit of $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$" Tags="&lt;real-analysis&gt;" AnswerCount="3" />
  <row Id="11946" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T07:09:13.630" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\lim_{x \to \infty} \biggl(\frac{20x}{20x+4}\biggr)^{8x} &amp;amp;=\lim_{x \to \infty} \biggl( 1 - \frac{4}{20x+4}\biggr)^{8x} \\ &amp;amp;=\lim_{x \to \infty} \biggl(1 - \frac{4}{20x+4}\biggr)^{\frac{20x+4}{4} \cdot \frac{8x}{20x+4} \cdot 4}&amp;amp;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the fact that $$\lim_{n \to \infty} \biggl(1 + \frac{1}{n}\biggr)^{n} =e$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt; After simplyfing you should get this. $$e^{\displaystyle \lim_{x \to \infty} - \frac{8x}{20x+4} \cdot 4} = e^{-8/5}$$  &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-12-08T02:58:40.887" LastActivityDate="2010-12-08T02:58:40.887" />
  <row Id="11947" PostTypeId="2" ParentId="11945" CreationDate="2010-11-26T07:11:20.163" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Suppose the sequence converges to $x$. Then solve for $x = \sqrt{7+x}$&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-11-26T07:11:20.163" />
  <row Id="11948" PostTypeId="1" CreationDate="2010-11-26T07:12:40.850" Score="1" ViewCount="140" Body="&lt;p&gt;I have read a result on computing the highest power of a prime that divides $n!$. I was wondering if there are any results on how to compute the highest power of a prime dividing $f(x)$, where $f$ is a function on the real numbers that outputs integer values. &lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-29T19:30:19.347" Title="The highest power of a prime that divides $f(x)$" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11949" PostTypeId="2" ParentId="11939" CreationDate="2010-11-26T07:14:08.910" Score="4" ViewCount="0" Body="&lt;p&gt;Perhaps I don't fully understand your question. So suppose you have a $2\times 2$ linear system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}a &amp;amp; b\\c &amp;amp; d \end{bmatrix}\begin{bmatrix}u_1\\ u_2\end{bmatrix}=\begin{bmatrix} e\\ f\end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you the augmented matrix&#xA;$$\begin{bmatrix}a &amp;amp; b &amp;amp; e\\ c &amp;amp; d &amp;amp; f\end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the set of equations &#xA;$$au_1+bu_2=e\quad\text{and}\quad cu_1+du_2=f.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you interchange the rows, you get the augmented matrix&#xA;$$\begin{bmatrix}c &amp;amp; d &amp;amp; f\\ a &amp;amp; b &amp;amp; e\end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which indeed has different columns. But this gives the equations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$cu_1+du_2=f\quad\text{and}\quad au_1+bu_2=e$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;just the other way around as before. The entries of $\textbf{u}$ haven't been switched, if that was your concern.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-26T07:14:08.910" />
  <row Id="11950" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T07:16:19.770" Score="4" ViewCount="0" Body="&lt;p&gt;If they know the definition of $e$ as&#xA;$$\lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n,$$&#xD;&#xA;then set $\alpha=\lim_{x\rightarrow\infty}\left(\frac{20x}{20x+4}\right)^{8x}$&#xD;&#xA;and note that&#xA;$$1/\alpha^5 = \lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{40x}=&#xD;&#xA;\left(\lim_{x\rightarrow\infty}\left(1 + \frac{1}{5x}\right)^{5x}\right)^8 = e^8.$$&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T07:16:19.770" CommentCount="2" />
  <row Id="11952" PostTypeId="2" ParentId="11945" CreationDate="2010-11-26T07:18:02.840" Score="14" ViewCount="0" Body="&lt;p&gt;Hint: It morally converges to $\sqrt{7+\sqrt{7+\sqrt{7+.....}}}$. Call this limit $l$. Then, since the nested series of square roots extends indefinitely, $l=\sqrt{7+l}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-26T07:18:02.840" CommentCount="4" />
  <row Id="11953" PostTypeId="1" CreationDate="2010-11-26T07:24:23.833" Score="1" ViewCount="91" Body="&lt;p&gt;I have the following constraint functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{i_{min}} \leq y_{i+1}-y_{i} \leq g_{i_{max}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{i_{max}}-y_{i} \geq h_{i}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{i_{min}} \leq \Biggl[\frac{(y_{i+1}-y_{i})^{3} (h_{i}+a)^{4}}{(h_{i}+b)^{4}}\Biggr] \leq v_{i_{max}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{i_{min}} \leq \Biggl[\frac{(y_{i+1}-y_{i})^{3} (h_{i}+a)^{10}}{(h_{i}+b)^{4}}\Biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i=1,2,3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables for minimization are $y_i$ and $h_i$. &lt;strong&gt;Note that $y_i$ are continuous and $h_i$ are discrete.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All $g_{i_{min}}, g_{i_{max}}, Q_{i_{min}},v_{i_{min}},v_{i_{max}}, y_{i_{max}},a,b$ are constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;two&lt;/em&gt; objectives functions to minimize against:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{1}=\sum_{i=1}^3 h_i  \quad\quad \text{(1)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{2}=-\sum_{i=1}^4 y_i  \quad\quad \text{(2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that $(1)$ should always be minimized first, and in the case of a draw, minimize $(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how to tackle this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a second thought, I think for my application, I can combine the two objective functions into a single function. &lt;strong&gt;So my question how is is there anyway to minimize a set of nonlinear constraint equations, which some of the variables are discrete, and some are continuous?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-11-30T15:34:41.643" LastActivityDate="2010-12-30T15:44:30.017" Title="Optimizing Nonlinear Constraint Equations with Discrete Variables and Multiple Objective Functions" Tags="&lt;nonlinear-optimization&gt;&lt;integer-programming&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11955" PostTypeId="2" ParentId="11922" CreationDate="2010-11-26T07:42:41.853" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Consider the product of power series $\exp(A/m)\exp(B/m) = I + \frac{A}{m} + \frac{B}{m} + O\left(\frac{1}{m^2}\right)$. Then use this to estimate the logarithm for sufficiently large $m$, then exponentiate and raise to the $m$-th power.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from the proof in section 2.4 of &quot;Lie Groups, Lie Algebras, and Representations&quot; by Brian C. Hall.&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-11-26T07:42:41.853" CommentCount="1" />
  <row Id="11956" PostTypeId="2" ParentId="11941" CreationDate="2010-11-26T08:15:46.117" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Always start by simplifying&#xA;$$\left(\frac{20x}{20x+4}\right)^{8x}=\left(\frac{20x+4}{20x}\right)^{-8x}=\left(1+\frac{1}{5x}\right)^{-8x}$$&#xD;&#xA;perhaps a substitution helps... &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-26T08:15:46.117" />
  <row Id="11957" PostTypeId="2" ParentId="11948" CreationDate="2010-11-26T08:26:57.417" Score="4" ViewCount="0" Body="&lt;p&gt;Finding divisors of combinatorial numbers can range from trivial to extremely difficult.  I'm going to explain a technique that I used numerous times to find divisors of combinatorial numbers, particularly the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin squares&lt;/a&gt; and Latin rectangles. (check out my website for numerous papers including this tactic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a set $C$ and want to find $|C| \pmod \mu$ for some $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;: Identify a group $G$ such that (a) $G$ acts on $C$ and (b) $|G|$ divisible by $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B=\{x \in C:$ there exist a non-trivial $g \in G$ such that $gx=x\}$, that is, $B$ is the set of objects in $C$ that admit non-trivial automorphisms in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $G$ acts on $C \setminus B$, and moreover, the orbit of any $x \in C \setminus B$ is of cardinality $|G|$ (by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_action&quot; rel=&quot;nofollow&quot;&gt;Orbit-Stabiliser Theorem&lt;/a&gt;, when all the stabilisers are trivial).  We can therefore deduce $|G|$ divides $|C \setminus B|$, and hence $|C| \equiv |B| \pmod {|G|}$ (and hence $|C| \equiv |B| \pmod \mu$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, $x \in C$ will not admit a non-trivial automorphism in $G$.  This means, that if we want to compute $|C| \pmod \mu$, we can instead find $|B|$, which already results in a significant computational saving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, however, this approach is still inappropriate (e.g. $|B|$ could be massive).  But not all is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;: Identify a set $A$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$B \subseteq A \subseteq C$,&lt;/li&gt;&#xA;&lt;li&gt;$G$ acts on $A$,&lt;/li&gt;&#xA;&lt;li&gt;members of $A$ have some property (such as a certain substructure) that can be used to find a divisor $\nu$ of $|A|$ (in the case of Latin squares, the substructure is a subsquare, which we can replace by any other subsquare, inducing an equivalence relation on $A$, thus giving a divisor of $|A|$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the same argument as above implies $|C| \equiv |A| \pmod {|G|}$ and so $\gcd(|A|,|G|)$ divides $|C|$.  Since we know $\nu$ divides $|A|$ and $\mu$ divides $|G|$, we can deduce that $\gcd(\mu,\nu)$ divides $|C|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I depict this concept below.  The area shaded gray represents $C$, and is partitioned into orbits by $G$.  The blue orbits have cardinality $|G|$ and the red orbits have cardinality strictly less than $|G|$.  We &quot;break a bit off&quot; and call it $A$.  If we can find a divisor of $|A|$, then we can use it to find a divisor of $|C|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/3zg3S.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-26T08:26:57.417" />
  <row Id="11958" PostTypeId="1" CreationDate="2010-11-26T09:15:52.073" Score="4" ViewCount="91" Body="&lt;p&gt;I start from a simple example. I through a (possibly unfair) coin 20 times. I got &quot;eagle&quot; and &quot;tail&quot; 15 and 5 times, respectively. Now I need to estimate probabilities for eagle and tail. For eagle I get 15/20 = 0.75 and for the tail I get 0.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I know that the probability that I got are not accurate. It is still possible that I have a fair coin (with probability for eagle equal to 0.5) and I get more eagles in my experiment just by chance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to estimate probabilities of probabilities. In other words I want to know how likely it is that probability for eagle is 0.5 (or 0.3, or 0.9). I can solve this problem, but I would like to know if there is a name for this problem. I an also interested in the generalization of the problem for the case of more than two events (not just &quot;eagle&quot; and &quot;tail&quot;).&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2010-11-26T17:26:53.003" Title="What is the standard method to estimate probabilities of events based on observations?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11959" PostTypeId="2" ParentId="11939" CreationDate="2010-11-26T09:51:42.353" Score="3" ViewCount="0" Body="&lt;p&gt;Let us consider a $2 \times 2$ example. We will then extend this higher dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$A = \begin{bmatrix}A_{11} &amp;amp; A_{12}\\A_{21} &amp;amp; A_{22} \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b = \begin{bmatrix}b_1 \\b_2 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you now want to solve $Ax_1 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_1$ is given by $A^{-1}b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you swap the two rows of $A$ and $b$. Call them $\tilde{A}$ and $\tilde{b}$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{A} = \begin{bmatrix}A_{21} &amp;amp; A_{22}\\A_{11} &amp;amp; A_{12} \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{b} = \begin{bmatrix}b_2 \\b_1 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do we relate $\tilde{A}$ and $A$ and similarly $\tilde{b}$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation is given by a Permutation matrix $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{A} = P A$ and $\tilde{b} = P b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix $P$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{P} = \begin{bmatrix}0 &amp;amp; 1\\1 &amp;amp; 0 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check that $\tilde{A} = P A$ and $\tilde{b} = P b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we look at solving the system $\tilde{A}x_2 = \tilde{b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute for $\tilde{A}$ and $\tilde{b}$ in terms of $A$ and $b$ respectively to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PAx_2 = Pb$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the important thing to note is that $P^2 = I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be verified algebraically or by a simple argument by seeing that $P^2$ swaps and swaps again which reverts back to the original giving $I$ or the other way of looking is $P^2$ is $P$ applied to $P$ which swaps the two rows of $P$ giving back the identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from $P^2 = I$, we get $P^{-1} = P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $PAx_2 = Pb$ and premultiplying by $P^{-1}$ gives $Ax_2 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $Ax_1 = b$ and $Ax_2 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we assume $A$ is invertible this gives us a unique solution and hence we get $x_1 = x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the other way to look at is to write $x_2 = \tilde{A}^{-1} \tilde b = (PA)^{-1}Pb = A^{-1} P^{-1} P b = A^{-1} I b = A^{-1} b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to observe in the above step is that the matrix $P$ is invertible and hence the matrix $(PA)$ is also invertible (since $A$ is assumed to be invertible and that $(PA)^{-1} = A^{-1}P^{-1}$ and matrix multiplication is associative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument with permutation matrix holds true for a $n \times n$ system as well.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T10:03:08.690" LastActivityDate="2010-11-26T10:03:08.690" />
  <row Id="11960" PostTypeId="2" ParentId="11958" CreationDate="2010-11-26T11:17:48.863" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Maximum_likelihood&quot; rel=&quot;nofollow&quot;&gt;maximum likelihood estimate&lt;/a&gt; is a typical method for estimating the probability distribution given some outcomes.  To describe it simply, you assume that your probability distribution you are sampling from is actually one from some set $\Omega$, and you are trying to find which one fits best the data (i.e. which is most likely to be correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you could reasonably assume that the probability distribution $\theta_\pi$ for which &quot;heads&quot; has probability $\pi$ and &quot;eagle&quot; has probability $1-\pi$.  So the question is:  Which $\pi \in [0,1]$ is most likely to by the actual distribution, given the observed outcomes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A description of how to answer the question is given on the linked Wikipedia page under &quot;Discrete distribution, finite parameter space&quot;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-26T11:17:48.863" />
  <row Id="11961" PostTypeId="1" CreationDate="2010-11-26T12:16:55.353" Score="0" ViewCount="73" Body="&lt;p&gt;Why are consistently ordered matrices important?  Are there any applications where they are turn up?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastActivityDate="2010-11-26T12:16:55.353" Title="Consistently ordered matrices." Tags="&lt;matrices&gt;" CommentCount="3" />
  <row Id="11962" PostTypeId="1" AcceptedAnswerId="12318" CreationDate="2010-11-26T12:38:38.577" Score="1" ViewCount="64" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question is not off-topic, but I'm currently discovering the existence of mixing processes, which attempt to express the dependence between random variables using miscellaneous coefficients ($\alpha$, $\beta$, $\phi$ and $\rho$ are the ones I've met so far).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any references you would advise that survey them, perhaps along with more or less concrete examples of situations where these various coefficients can be used?&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-11-29T09:12:29.183" Title="Mixing processes" Tags="&lt;probability-theory&gt;&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="11963" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T12:42:33.087" Score="1" ViewCount="0" Body="&lt;p&gt;One day I shall do a serious study of this!  there is truth in all these answers, the white notes give us our do-re-mi (major scale) starting on C, this scale has a mixture of tones and semitones, and dictate where the black notes should go and how many we need.  The re-tuning to equal temperament is a fudge, and if you were to analyse a tuned keyboard, not all semitones are equally spaced.  Other intervals are also compromised, so a major third in one key may have notes further apart than one in a different key.  Composers have long been aware of this, and aware that the key they select for a composition can make a significant difference to the &quot;mood&quot; (that is after you have selected major or minor).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical Indian Music uses a system of scales (ragas).  There are several hundred of these, and they will be fitted to specific moods, times of day, types of occasion etc. These are not random variations from any Western scale, and have nothing to do with the keyboards we commonly use.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our keyboard system is just for keyboards - a string instrument may not play exactly the same pitch as a piano for a given note (unless it is an open string), because they will tend to use something closer to the original pythagorean scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS I am a working musician with a bachelors degree in maths!&lt;/p&gt;&#xA;" OwnerUserId="4094" LastActivityDate="2010-11-26T12:42:33.087" />
  <row Id="11964" PostTypeId="1" CreationDate="2010-11-26T13:02:17.070" Score="1" ViewCount="167" Body="&lt;p&gt;Came across this math problem while programming something:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given positive $y$, $a$, and $b$, find an integer $x$, $a\lt x\lt b$, so that $y\bmod x$ (the remainder of $y$ when divided by $x$) is positive and less than or equal to $100$,&#xA;$$0 \lt y\bmod x \leq 100.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know sometimes it's not solvable depending on $y$ and the range. I'm not really sure where to begin because I wasn't very good at modular algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't want to have to try every possible number for $x$ between $a$ and $b$ because the range I'm using it for has over 25 million in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Sorry the question wasn't very clear. $y$, $a$, and $b$ are given. And I'm trying to find an $x$ that satisfies those conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: I've realize $y \bmod x$ need to be less than or equal to $100$ and also over $0$.&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="4093" LastEditorDisplayName="" LastEditDate="2010-11-27T07:31:17.580" LastActivityDate="2010-11-29T10:45:22.450" Title="Solving for x in $0 &lt; y \bmod x \le 100$ where x is in the range (a, b)" Tags="&lt;inequality&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11965" PostTypeId="1" AcceptedAnswerId="11966" CreationDate="2010-11-26T15:40:46.563" Score="2" ViewCount="70" Body="&lt;p&gt;This is my first post here. Apologies if there are any misformats or mis-tagging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if the post doesn't belong here, sorry, and I will delete it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I derived this formula from a probability calculation. How would I go about finding the maximum value for the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{T}{S}\frac{S!}{(S-D)!}\frac{(T-D)!}{T!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T&amp;gt;S&amp;gt;D$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and T, S, and D are all integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4102" LastEditorUserId="4102" LastEditorDisplayName="" LastEditDate="2010-11-26T15:48:45.963" LastActivityDate="2010-11-26T15:50:55.353" Title="Maximum value of probability" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11966" PostTypeId="2" ParentId="11965" CreationDate="2010-11-26T15:50:55.353" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: We can rewrite this as (by multiplying and dividing by $(D-1)!$) as $\frac{{S-1}\choose {D-1}}{{T-1}\choose {D-1}}$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-26T15:50:55.353" CommentCount="1" />
  <row Id="11967" PostTypeId="2" ParentId="11726" CreationDate="2010-11-26T16:05:30.503" Score="4" ViewCount="0" Body="&lt;p&gt;Here's an argument that pushes the lower bound for $S(n)$ closer to a factor of $2$ times&#xA;$\log_2 n.$ More precisely, we show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S(n) \ge \left( 2 - \frac{1}{e} \right) \left( \lfloor \log_2 n \rfloor – 1 \right). $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ a= \lfloor \log_2 n \rfloor $ and write&#xA;$ f(n,x) = 1 - \prod_{j=1}^{n-1} ( 1 – jx),$&#xD;&#xA;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S(n) = \sum_{k=1}^\infty \left( 1 - \prod_{j=1}^{n-1} \left( 1 - \frac{j}{2^k} \right) \right)$$&#xD;&#xA;$$ = \sum_{k=1}^\infty f(n, \frac{1}{2^k} ) = &#xD;&#xA;a-1 + \sum_{k=a}^\infty f(n, \frac{1}{2^k} ),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since for each $k \le a-1$ $\exists$ $j=2^k$ in $\prod_{j=1}^{n-1} ( 1 - j/2^k ),$ and thus this product is $0.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So neglecting terms with $k \ge 2a-1,$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(n) \ge a-1 + \sum_{k=a}^{2a-2} f(n, \frac{1}{2^k} ). \qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the AM-GM for $x \le 1/(m-1)$ we have for $m \ge 2$ and $2/m(m-1) \le x \le 1/(m-1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1-x)(1-2x) \cdots (1-(m-1)x) \le&#xD;&#xA;\left( 1 - \frac{mx}{2} \right)^{m-1} \le&#xD;&#xA;\left( 1 - \frac{1}{m-1} \right)^{m-1} &amp;lt; \frac{1}{e}. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $k=a, a+1,\ldots, 2a-2$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(n, \frac{1}{2^k} ) \ge 1 - \frac{1}{e}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substituting this into $(1)$ the result follows.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-11-26T16:05:30.503" CommentCount="1" />
  <row Id="11968" PostTypeId="1" AcceptedAnswerId="11972" CreationDate="2010-11-26T16:18:56.453" Score="4" ViewCount="80" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Can a finite group G have a cyclic Sylow 3-subgroup of order 9, such that the intersection of the Sylow 3-subgroups of G has order exactly 3, without having non-identity normal subgroups of order coprime to 3?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In classifying the finite groups with cyclic Sylow 3-subgroups of order 9, it seems reasonable to split into cases based on the size of the 3-core (the intersection of the Sylow 3-subgroups) mod the 3'-core (the largest normal subgroup of order coprime to 3).  3-Cores of sizes 1 and 9 are easy to handle, but size 3 is devolving into a number of cases none of which seem to work, but for no systematic reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some systematic reason it cannot occur, or is there an example I've overlooked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The original left out the important condition on normal subgroups of order coprime to 3, without which the classification is intractable.  Alex's answer shows how to use such normal subgroups to get fairly arbitrary behavior (for any type of Sylow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; If the perfect residuum X of the group is nontrivial, it contains Ω(P) of the cyclic Sylow p-subgroup P, and so O&lt;sub&gt;p&lt;/sub&gt;(G) ≤ O&lt;sub&gt;p&lt;/sub&gt;(X).  In this case, O&lt;sub&gt;p&lt;/sub&gt;(X) = 1.  Hence the group is solvable, and Fit(G) = O&lt;sub&gt;p&lt;/sub&gt;(G), so G/Fit(G) ≤ Aut(p) = p−1 has the wrong order.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-11-26T17:53:13.760" LastActivityDate="2010-11-26T17:53:13.760" Title="Cyclic sylow subgroup of order 9 with 3-core of order 3" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11969" PostTypeId="2" ParentId="11945" CreationDate="2010-11-26T16:21:11.883" Score="17" ViewCount="0" Body="&lt;p&gt;For a proof of convergence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the sequence as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_{0} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_{n+1} =\sqrt{7 + x_n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\displaystyle x_n \geq 0 \ \ \forall n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle x^2 - x - 7 = (x-a)(x-b)$ where $\displaystyle a \lt 0$ and $\displaystyle b \gt 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \lt b$ &lt;br/&gt;&#xA;ii) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \gt x_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_n \lt b = b^2 - 7$ and so $x_n +7  \lt b^2$ and thus by taking square roots $x_{n+1} \lt b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a proof of ii) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x_{n+1})^2 - (x_n)^2 = -(x^2_n - x_n -7) = -(x_n-a)(x_n-b) \gt 0$ if $x_n \lt b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \{x_{n}\}$ is monotonically increasing and bounded above and so is convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By setting $L = \sqrt{7+L}$, we can easily see that the limit is $\displaystyle b = \dfrac{1 + \sqrt{29}}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In fact, we can show that the convergence is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{b^2 - (7+x_n)}{(b+\sqrt{7+x_n})(b-x_n)} = \dfrac{1}{b + x_{n+1}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \lim_{n\to \infty}  \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{1}{2b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also show something a bit stronger:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle t_n = b - x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The we have shown above that $\displaystyle t_n \gt 0$ and $\displaystyle t_n \lt b^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle b - t_{n+1} = \sqrt{7 + b - t_n} = \sqrt{b^2 - t_n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $\displaystyle b$ throughout we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 - \dfrac{t_{n+1}}{b} = \sqrt{1 - \dfrac{t_n}{b^2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\displaystyle 1 - \dfrac{x}{2} \gt \sqrt{1-x} \gt 1 - x \ \ 0 \lt x \lt 1$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1 - \dfrac{t_n}{2b^2} \geq 1 - \dfrac{t_{n+1}}{b} \geq 1 - \dfrac{t_n}{b^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  \dfrac{t_n}{2b}  \leq t_{n+1} \leq \dfrac{t_n}{b}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us that $\displaystyle  b - \dfrac{b}{b^n} \leq x_n \leq b - \dfrac{b}{(2b)^n}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-27T00:03:34.820" LastActivityDate="2010-11-27T00:03:34.820" />
  <row Id="11970" PostTypeId="2" ParentId="11964" CreationDate="2010-11-26T16:28:43.027" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure I understood your question. Let's suppose your question is the following : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $100=c&amp;lt;a&amp;lt;b&amp;lt;y$ be 4 given integers, how to find $x$ such that&#xA;$$&#xD;&#xA;a\le x \le b \text{ and }&#xD;&#xA;y \mod x \le c&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a tentative algorithm, which is probably not optimal, but (slightly) better that trying all the possible values for $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, try the first possibility $x=a$ and perform an euclidean division to find $q$ and $r$ such that $y=qa+r$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $0&amp;lt;r\le c$, $x=a$ and you're done&lt;/li&gt;&#xA;&lt;li&gt;otherwise, divide $r$ by $q$ to find $k,l$ such that $r=kq+l$.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if $0&amp;lt;l\le c$, $x=a+k$ is your solution if it is in $[a,b]$. If not, try $b$. If $b$ is not a solution either, then tere is no solution.&lt;/li&gt;&#xA;&lt;li&gt;otherwise, increase $a$ by $k+1$ and start again from the beginning.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, you stop repeating the above as soon as $a$ becomes bigger than $b$, which would mean that ther is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to take into account the $0 &amp;lt; y \mod c$ condition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think this condition change the search much, at least if you use the above algorithm. Hoxever, if you have a $z$ such that $y \mod z =0$, you can write $y=mz$, which might be helpful. For example, if $m\le c$, $x=z+1$ will be a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to correct for the $a+k&amp;gt;b$ case&lt;/strong&gt;&#xA;  &lt;strong&gt;Edit&lt;/strong&gt;&#xA;To be more clear, the algorithm above find $k,q,l$ such that $y=(a+k)q+l$, and tries to get $l$ as small as possible (without trying too much)&lt;/p&gt;&#xA;" OwnerUserId="3918" LastEditorUserId="3918" LastEditorDisplayName="" LastEditDate="2010-11-29T10:45:22.450" LastActivityDate="2010-11-29T10:45:22.450" CommentCount="10" />
  <row Id="11971" PostTypeId="1" CreationDate="2010-11-26T16:40:07.103" Score="4" ViewCount="141" Body="&lt;p&gt;I am learning group theory, and while learning automorphisms, I came across conjugation as an example in many textbooks. Though the definition itself, (and when considering the case of abelian groups), it seems pretty innocent, I have to admit that I have no intuition about whats happening, or why even such a map is important. Did anyone feel this way when you learnt it? Any examples/references that you have found useful when you learnt it?&lt;/p&gt;&#xA;" OwnerUserId="1441" LastActivityDate="2010-11-26T17:18:41.670" Title="Intuition behind conjugation in group theory" Tags="&lt;group-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="11972" PostTypeId="2" ParentId="11968" CreationDate="2010-11-26T16:52:22.483" Score="3" ViewCount="0" Body="&lt;p&gt;Such groups are easy to construct: let $C_9$ act on something through a $C_3$-quotient. E.g. construct the semi-direct product $C_7\rtimes C_9$ such that $C_9$ acts on $C_7$ through the unique automorphism of order 3. Then, the Sylow 3-subgroups are not normal, so there is more than one of them, but there is a unique $C_3$, which is the centre of $G$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T16:52:22.483" CommentCount="4" />
  <row Id="11973" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T16:57:07.053" Score="0" ViewCount="0" Body="&lt;p&gt;My intuition goes back to permutations.  Namely, conjugating everything by some group element can be identified with some permutation of the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the importance of conjugation, one easy answer is that it gives insight into the commutative portions of the group.  If $x$ is closed under conjugation by  $g$, that is $x = gxg^{-1}$, then we have $xg = gx$, so the elements commute.  The set of elements that are closed under conjugation by all group elements forms the &lt;em&gt;center subgroup&lt;/em&gt;, while any subgroup that is closed under conjugation by all group elements is a &lt;em&gt;normal subgroup&lt;/em&gt;.  These are fundamental to understanding the structure of a given group.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-11-26T16:57:07.053" />
  <row Id="11974" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T16:59:04.650" Score="2" ViewCount="0" Body="&lt;p&gt;There are several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Conjugation is a way of measuring, how commutative or otherwise your group is. Namely $aba^{-1} = b\Leftrightarrow ab=ba$. Since abelian groups are much easier to understand than arbitrary ones, this measure is important.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Conjugation is more or less the only known way of obtaining automorphisms of any given group without knowing anything further about the group (of course, these automorphisms can all turn out to be trivial if your group is abelian).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have to bear in mind that normal subgroups and the action of conjugation were first introduced by Galois, because for a subgroup to be normal was exactly the criterion that he needed for an automorphism of a field to give an automorphism of a subfield upon restriction. This comment might be slightly over your head if you are only learning groups at the moment, but bear in mind that historically, the notion of normal subgroups arose in the context of interpreting groups as symmetries of field extensions, when Galois developed a theory to decide whether a given polynomial was solvable by using elementary operations of arithmetic and roots.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-26T16:59:04.650" />
  <row Id="11975" PostTypeId="2" ParentId="11669" CreationDate="2010-11-26T17:07:45.710" Score="1" ViewCount="0" Body="&lt;p&gt;A somewhat grapical representation of what Carlos Ribeiro was talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If you take any note in the C major scale, you can treat that note as the start of another scale. Take for instance the fifth of C (it's the G), and build a new major scale, now starting from G instead of C. You'll get another seven notes. Some of them are also on teh scale of C; others are very close, but not exactly equal; and some fall in the middle of the notes in the scale of C. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the semi-tone interval EF and BC on the C scale.&#xA;When trying to reproduce the same scale starting at D, we run into a problem.&#xA;Alphabetically, the third note should be a F, but F is a semitone too low for that spot. In order to maintain the same sounding scale, we need to introduce a NEW note, called F#.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C - D - E F - G - A - B C  (C scale)&lt;/li&gt;&#xA;&lt;li&gt;D - E - F#G - A - B - C#D  (D scale)&lt;/li&gt;&#xA;&lt;li&gt;E - F#-G#A - B - C#- D#E  (E scale)&lt;/li&gt;&#xA;&lt;li&gt;F - G - AA#- C - D - E F  (F scale)  &lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;br&gt;&#xA;Note that in actual writing of music the A-A# would be written as A-Bb so that the 'A' line of the staff wouldn't be ambiguous.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4109" LastActivityDate="2010-11-26T17:07:45.710" />
  <row Id="11976" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T17:12:52.693" Score="8" ViewCount="0" Body="&lt;p&gt;One way to think about conjugation is as a generalization of changing coordinates when rewriting matrices or, from a physical point of view, as the process of &quot;seeing a group from a different perspective.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's make this precise.  Let $G$ be a group and let $X$ be a set on which $G$ acts faithfully.  For example, if $G$ is the Euclidean group of isometries of the plane, $X$ could be... well, the plane.  Let's say we have a reasonably concrete description of what a particular element $g \in G$ does to $X$.  For example, if $G, X$ are as above, then perhaps $g$ is a rotation counterclockwise by $\theta$ centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that description, what does the element $hgh^{-1}$ do?  Well, this is actually quite simple: in the description of $g$, replace all the elements $x \in X$ that occur by the corresponding elements $hx \in X$.  For example, if $h$ is a translation by a vector $v$, then $hgh^{-1}$ is a rotation counterclockwise by $\theta$ centered at $v$ instead of the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what conjugation does in terms of group actions (and it is always a good idea to think of groups in terms of their actions) is it corresponds to a &quot;change of coordinates&quot; on the underlying set $X$.  This is a basic reason conjugation and group theory are important in understanding Newtonian mechanics (where $G$ is the Galilean group) or relativity (where $G$ is the Lorentz group) or really much of modern physics in general: in physics groups are always studied because of their actions and we always want our concepts and equations to be invariant under these actions.  (For example, mass and charge are invariant concepts.  The mass or charge of an object doesn't change when you rotate or translate it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a very intuitive definition of a normal subgroup: it's a subgroup that &quot;looks the same from every perspective.&quot;  For example, the subgroup of translations in the Euclidean group is always normal because the description &quot;$g$ is a translation&quot; is the same from every perspective (that is, it's invariant under conjugation).  It's a good exercise to look at some groups you're familiar with and see if you can identify which subgroups are normal based on this principle.  (This principle underlies the importance of normal subgroups in Galois theory as well.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-26T17:18:41.670" LastActivityDate="2010-11-26T17:18:41.670" CommentCount="4" />
  <row Id="11977" PostTypeId="2" ParentId="11971" CreationDate="2010-11-26T17:13:43.613" Score="3" ViewCount="0" Body="&lt;p&gt;One example of conjugation you probably already know from linear algebra is &lt;strong&gt;change of base&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear transformation A that takes (x,y) to (3x−y,2x) is reasonably nice and can be represented by the matrix $$A = \begin{pmatrix}3&amp;amp;2\\-1&amp;amp;0\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the basis u=2x−y, v=x−y, Au = 2Ax − Ay = 2(3x−y) − 2x = 4x − 2y = 2u and Av = Ax−Ay = (3x−y) - 2x = x−y = v.  So in this basis, the linear transformation is incredibly nice and can be represented by the matrix $$\tilde A = \begin{pmatrix}2&amp;amp;0\\0&amp;amp;1\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the matrix $$B = \begin{pmatrix} 2&amp;amp;1\\-1&amp;amp;-1\end{pmatrix}$$&#xD;&#xA;then $B^{-1} A B = \tilde A$ expresses the matrix in the new basis defined by B, (u,v) = (2x-1y,1x-1y).  Note that Bx = u, and By = v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense this is what conjugation does in general: it changes the basis on which your group acts.  If you conjugate by g, it changes the way G acts by converting the old basis to the a new basis already acted on by g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example to work through to see this conjugation in a new context is with &lt;strong&gt;permutations&lt;/strong&gt;.  If a permutation A takes 1,2,3 to 3,2,1 and the permutation B takes 1,2,3 to 1,3,2, then the conjugated element $B^{-1} A B$ takes 1,3,2 to 2,3,1.  In other words, instead of acting on 1,2,3 it acts on 1,3,2.  In cycle notation, you can write this as A=(1,3) and B=(2,3) and then A^B = (1,2).  The &quot;3&quot; in (1,3) has been replaced by &quot;2&quot;, since B replaces 3s by 2s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to think of groups as acting on something.  Either let them be matrices moving vectors around, or permutations moving arbitrary things around.  Conjugation just lets you &lt;strong&gt;relabel&lt;/strong&gt; the things you are moving around.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-26T17:13:43.613" />
  <row Id="11978" PostTypeId="2" ParentId="11958" CreationDate="2010-11-26T17:26:53.003" Score="1" ViewCount="0" Body="&lt;p&gt;We want to find $p,0\le p\le 1,$ such that $${20\choose 15}p^{15}(1-p)^5$$ is maximum.&#xA;By simple calculus, this $p$ turns out to be 15/20, as expected.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T17:26:53.003" />
  <row Id="11979" PostTypeId="1" CreationDate="2010-11-26T17:30:09.023" Score="-2" ViewCount="224" Body="&lt;p&gt;How many moves can a horse make on a chessboard?&lt;/p&gt;&#xA;" OwnerUserId="2779" LastEditorUserId="2779" LastEditorDisplayName="" LastEditDate="2010-11-26T17:50:38.587" LastActivityDate="2010-11-27T00:12:39.823" Title="How many moves can a horse make  on a chessboard?" Tags="&lt;homework&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="0" />
  <row Id="11980" PostTypeId="1" AcceptedAnswerId="11984" CreationDate="2010-11-26T17:39:46.090" Score="2" ViewCount="91" Body="&lt;p&gt;In other words, if X = Y in distribution, is it true that EX = EY?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this must be true, but I've tried to prove it a few times and I always get stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any hints or reference.&lt;/p&gt;&#xA;" OwnerUserId="4111" LastActivityDate="2010-11-26T23:10:20.193" Title="Does equality in distribution imply equality of expected value?" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="11981" PostTypeId="2" ParentId="11979" CreationDate="2010-11-26T17:42:11.340" Score="-1" ViewCount="0" Body="&lt;p&gt;That piece can move in an L-shape. So potential, it can move in 1 of 4 directions (unless it's restricted by one edges of the board).&lt;/p&gt;&#xA;" OwnerUserId="4007" LastActivityDate="2010-11-26T17:42:11.340" CommentCount="2" />
  <row Id="11983" PostTypeId="1" CreationDate="2010-11-26T17:44:47.037" Score="3" ViewCount="137" Body="&lt;p&gt;How would one go about showing how many solutions the following congruence has?&#xA;$$x^2 + y^2 \equiv 23 \pmod{93}.$$&lt;/p&gt;&#xA;" OwnerUserId="4112" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T19:16:00.140" LastActivityDate="2010-11-26T22:22:32.653" Title="Quadratic Congruence and Sum of Two Squares" Tags="&lt;number-theory&gt;&lt;number&gt;&lt;quadratic-reciprocity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="11984" PostTypeId="2" ParentId="11980" CreationDate="2010-11-26T17:45:13.637" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expectation value depends only on the measure $P^X$ on $\mathbb R$ induced by the random variable $X$, and $X$ and $Y$ are, by definition, equal in distribution if and only if $P^X=P^Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find this here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Expectation_value&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Expectation_value&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-26T23:10:20.193" LastActivityDate="2010-11-26T23:10:20.193" CommentCount="1" />
  <row Id="11985" PostTypeId="2" ParentId="11983" CreationDate="2010-11-26T18:16:01.507" Score="0" ViewCount="0" Body="&lt;p&gt;This is a start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would like to know for which y it holds that 23 - y^2 is a square modulo 93. As 93 is 31*3, you can find these y by finding the squares of the form 23 - y^2 modulo 3 and modulo 31. Can you go on from there? &lt;/p&gt;&#xA;" OwnerUserId="4113" LastActivityDate="2010-11-26T18:16:01.507" CommentCount="3" />
  <row Id="11986" PostTypeId="1" CreationDate="2010-11-26T19:17:07.363" Score="2" ViewCount="148" Body="&lt;p&gt;My homework question:  Show that all zeros of $p(z)=z^4 + 6z + 3$ lie in the circle of radius $2$ centered at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know $p(z)$ has a zero-count of $4$ by using the  Fundamental Theorem of Algebra. Then using the Local Representation Theorem the $\int \frac{n}{z+a} = 4(2 \pi i)$. I am assuming $a=0$ since we are centered at the origin. I apologize for my lack of math-type.  What does $= 8 \pi i$ mean?  Am I going around the unit circle $4$ times?  Or is it even relevant to my final answer.  Which I am assuming is finding the coordinates to the $4$ singularities. I have always looked for my singularities in the values that make the denominator zero, but in this question my denominator is $z$.  $z=0$ doesn't seem right.  So the question is, am I suppose to factor the polynomial $z^4 + 6z + 3$ to find the zeros?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4115" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-26T20:39:59.130" LastActivityDate="2010-11-26T20:39:59.130" Title="Guidance on a Complex Analysis question" Tags="&lt;homework&gt;&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11987" PostTypeId="2" ParentId="11986" CreationDate="2010-11-26T19:25:15.167" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; This kind of questions are usually handled using &lt;a href=&quot;http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Rouche's Theorem&lt;/a&gt;. I suggest you look it up in the wikipedia article, where you can see an example of its usage. Also here's an &lt;a href=&quot;http://books.google.co.cr/books?id=wDsrQTPUaSsC&amp;amp;pg=PA77&amp;amp;dq=rouche%27s+theorem&amp;amp;hl=es&amp;amp;ei=YAvwTNmTOIX7lwej4JH9DA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCoQ6AEwAQ#v=onepage&amp;amp;q=rouche%27s%20theorem&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is choosing wisely another function $f(z)$ with which to compare in the inequality in Rouche's theorem and such that you can easily decide how many zeroes does $f(z)$ have inside the region you are considering, which in your case is the circle $|z| &amp;lt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About your other question, you don't need to factor the polynomial in order to answer this.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-26T20:06:27.580" LastActivityDate="2010-11-26T20:06:27.580" />
  <row Id="11988" PostTypeId="2" ParentId="11983" CreationDate="2010-11-26T19:26:11.877" Score="4" ViewCount="0" Body="&lt;p&gt;If you are working modulo a prime $p$, then you can obtain the result using the Legendre symbol or by simple trial and error if it is small. For instance, working modulo $3$, as CJost suggests, you are trying to solve $x^2 + y^2 \equiv 23 \equiv 2 \pmod{3}$. then you want to solve $x^2 \equiv 2-y^2\pmod{3}$. Since $y^2$ can only be congruent to $0$ or to $1$ modulo $3$, you only have solutions when $x^2\equiv y^2\equiv 1\pmod{3}$. That gives you four solutions modulo $3$: $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to do something similar for $31$: for each possibly value of $y^2$, decide if $23-y^2$ is a square modulo $31$; if it is, that gives you two possible values for $x$ and two possible values for $y$ (unless one of the two needs to be $0$ modulo $23$). But remember that you are working &lt;em&gt;modulo&lt;/em&gt; $31$; for example, with $y=5$, there is no problem with the fact that $23-y^2 = -2$, because you are trying to solve $x^2\equiv -2\pmod{31}$, and this is the same as $x^2\equiv 29 \pmod{31}$; the negative number is not a problem. And of course, to decide if $x^2\equiv 29\pmod{31}$ has a solution, you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_symbol&quot; rel=&quot;nofollow&quot;&gt;Legendre symbol&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_reciprocity&quot; rel=&quot;nofollow&quot;&gt;quadratic reciprocity&lt;/a&gt; to compute it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; As it happens, $23$ is &lt;em&gt;not&lt;/em&gt; a square modulo $31$, so any solutions to $x^2+y^2\equiv 23\pmod{31}$ will necessarily have both $x$ and $y$ not congruent to zero modulo $31$; so you don't need to worry about that when you are counting solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you find a solution $x=a$ and $y=b$ modulo $31$. How do you get solutions modulo $93$? You take one solution modulo $3$, say $(1,1)$, and the solution modulo $31$, say $(a,b)$, and you use the Chinese Remainder Theorem: you can find an integer $x$ such that $x\equiv 1 \pmod{3}$ and $x\equiv a\pmod{31}$; then find an integer $y$ such that $y\equiv 1\pmod{3}$ and $y\equiv b\pmod{31}$. Then the pair $(x,y)$ will be a solution modulo $93$ that corresponds to the solution $(1,1)$ modulo $3$ and the solution $(a,b)$ modulo $31$. So each choice of a solution modulo $3$ and a solution modulo $31$ will give you a solution modulo $93$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T22:22:32.653" LastActivityDate="2010-11-26T22:22:32.653" />
  <row Id="11989" PostTypeId="2" ParentId="11592" CreationDate="2010-11-26T20:28:41.563" Score="2" ViewCount="0" Body="&lt;p&gt;With help from Maple, I got &#xA;$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2-(c-a)^2-(c-b)^2$$&#xD;&#xA;equal to&#xA;$$\frac{(c(x^3+y^3+z^3)+(a-c)(x^2y+y^2z+z^2x)+(b-c)(x^2z+y^2x+z^2y)-3(a+b-c)xyz)^2}{(x-y)^2(y-z)^2(x-z)^2}$$ which of course is $\ge 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with no help from a computer algebra, how would one prove this? &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T20:28:41.563" />
  <row Id="11991" PostTypeId="1" AcceptedAnswerId="11993" CreationDate="2010-11-26T20:57:09.110" Score="1" ViewCount="106" Body="&lt;p&gt;Are &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_theory&quot; rel=&quot;nofollow&quot;&gt;order&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_relation&quot; rel=&quot;nofollow&quot;&gt;binary relation&lt;/a&gt; the same thing, or order is just some special binary relation? If it is the latter, what kinds of properties distinguish an order from a general binary relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sorry for the confusion already caused. What I thought was that a partial order is just some special order, and there are other orders that are not partial orders, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Preorder&quot; rel=&quot;nofollow&quot;&gt;preorder&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Directed_set&quot; rel=&quot;nofollow&quot;&gt;the order on a directed set&lt;/a&gt;. So I was wondering how an order was defined then, and I saw there was no definition for an order in the Wikipedia page for order theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe  I was not lazy or could not read, what I understood was different from yours, and I did not realize that when I first posted my question. So sorry about that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, if possible, I would like to know if there is a concept that can include partial order, preorder and the &quot;order&quot; on a directed set, but more special than a binary relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-11-26T22:33:24.433" LastActivityDate="2010-11-26T22:33:24.433" Title="order and binary relation" Tags="&lt;elementary-set-theory&gt;&lt;order-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="11992" PostTypeId="2" ParentId="11429" CreationDate="2010-11-26T21:00:44.720" Score="0" ViewCount="0" Body="&lt;p&gt;I am not sure if this is a homework. So I will give you the answers and you figure out yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $i=0.09, v=(1+i)^{-1}=0.917431192$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer for (a): $$50000\frac{1-v^7}{1-v^{10}}$$&#xD;&#xA;Answer for (b): $$50000\times 0.09\times\frac{1-v^8}{1-v^{10}}$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-26T21:00:44.720" />
  <row Id="11993" PostTypeId="2" ParentId="11991" CreationDate="2010-11-26T21:01:56.217" Score="5" ViewCount="0" Body="&lt;p&gt;Orders are a special kind of binary relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A binary relation $R$ on a set $A$ is just a collection of ordered pairs $R\subseteq A\times A$, with absolutely no conditions on $R$. So long as it is a subset, it is a binary relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (partial) order on $A$, by contrast, is a subset $P\subseteq A\times A$ which satisfies three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The relation is &lt;strong&gt;reflexive&lt;/strong&gt;: For every $a\in A$, $(a,a)\in P$.&lt;/li&gt;&#xA;&lt;li&gt;The relation is &lt;strong&gt;anti-symmetric&lt;/strong&gt;: for every $a,b\in A$, if $(a,b)\in P$ and $(b,a)\in P$, then $a=b$.&lt;/li&gt;&#xA;&lt;li&gt;The relation is &lt;strong&gt;transitive&lt;/strong&gt;: for every $a,b,c\in A$, if $(a,b)\in P$ and $(b,c)\in P$, then $(a,c)\in P$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The partial order is a &lt;strong&gt;total&lt;/strong&gt; or linear order if, in addition, for every $a,b\in A$, either $(a,b)\in P$ or $(b,a)\in P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an order on $A$ is a special kind of binary relation, because it must satisfy the three properties above, whereas a &quot;general binary relation&quot; doesn't have to satisfy anything other than being a collection of ordered pairs of elements of the set.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Okay, addressing your additions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most basic notion is &quot;partial order&quot;. Every other kind of order is a &quot;partial order with extra conditions&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one exception is a &lt;strong&gt;pre-order&lt;/strong&gt;, because, as the name indicates, a pre-order is something which is not yet an order (again, &quot;order&quot; is just a short way of saying &quot;partial order&quot;), but may become one (when it grows up, so to speak).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pre-orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;pre-order&lt;/strong&gt; is a binary relation which is reflexive and transitive, but is not necessarily anti-symmetric. An example of this is the relation &quot;divides&quot; in the integers: define $a\preceq b$, with $a$ and $b$ integers, if and only if $a$ divides $b$. This is reflexive (every integer divides itself), and transitive (if $a$ divides $b$ and $b$ divides $c$, then $a$ divides $c$), but it is not anti-symmetric, because if $a$ divides $b$ and $b$ divides $a$, the best you can conclude is that $|a|=|b|$; for instance, $2$ and $-2$ are &lt;em&gt;different&lt;/em&gt;, but $2\preceq -2$ and $-2\preceq 2$. So this relation is &lt;em&gt;not&lt;/em&gt; a partial order because it lacks anti-symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to solve this lack is to use the relation to define something which &lt;em&gt;is&lt;/em&gt; a partial order on a closely related set. We define an &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation&quot; rel=&quot;nofollow&quot;&gt;equivalence relation&lt;/a&gt; on the integers by saying $a\sim b$ if and only if $a\preceq b$ and $b\preceq a$ (in this case, if and only if $|a|=|b|$). Then we can consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_relation#Quotient_set&quot; rel=&quot;nofollow&quot;&gt;quotient set&lt;/a&gt;, and define the relation $\leq$ on the quotient set by $[a]\leq [b]$ if and only if $a\preceq b$ (here, $[a]$ is the equivalence class of $a$ under $\sim$, etc). Then one can show that this is a partial order on the quotient set; it is patently closely related to the original pre-order $\preceq$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why relations that are reflexive and transitive are called &lt;strong&gt;pre-orders&lt;/strong&gt;: you only need one further step to get to a partial order, though not usually on the set you started with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other kinds of orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic notion, as I said above, is &quot;partial order&quot;. There are lots of extra conditions one can put on a partially ordered set to make it a &quot;nicer&quot; kind of ordered set. These extra conditions define special kinds of orders. A random sampling off the top of my head (I'm using $\leq$ to denote whatever the partial order relation in question is; $a\leq b$ means $(a,b)$ is in the relation that defines the order):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Total&lt;/strong&gt; or &lt;strong&gt;linear orders&lt;/strong&gt;. In a partial order, it is possible for two elements to not be comparable at all, that is, you have neither $a\leq b$ nor $b\leq a$. For example, if you take a set $X$, you can define a partial order on $\mathcal{P}(X)$, the power set of $X$ (the set of all subsets of $X$) by saying that $A\leq B$ if and only if $A\subseteq B$ (in fact, this is in a sense the most basic kind of &quot;partial order&quot;). If $X$ has more than one element, though, then you have elements of $\mathcal{P}(X)$ that cannot be compared: take $x,y\in X$, $x\neq y$; then $\{x\}\not\leq \{y\}$ and $\{y\}\not\leq\{x\}$. This in contrast to the order relations we are used to in the natural numbers, real numbers, etc., where any two elements can be compared: one is bigger than the other. So we add a condition: the partially ordered set $A$ is &lt;strong&gt;totally ordered&lt;/strong&gt; or &lt;strong&gt;linearly ordered&lt;/strong&gt; if the partial order &lt;em&gt;also&lt;/em&gt; satisfies the condition that for every $a,b\in A$, either $a\leq b$ or $b\leq a$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Well order.&lt;/strong&gt; This is a very strong generalization of total orders. In a total order, any subset with finitely many elements has a smallest element. A partially ordered set $A$ is said to be &lt;strong&gt;well-ordered&lt;/strong&gt; if &lt;em&gt;any&lt;/em&gt; nonempty subset $B$ of $A$ has a smallest element. For instance, the natural numbers with their usual order is well-ordered, but the real numbers with their usual order are not (the positive reals have no least element). A well-order must be a total order, but the converse is not true in general.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Lattice order.&lt;/strong&gt; A lattice order is a partial order where, even though it is not true that any two elements are comparable, it is nevertheless the case that any two elements have a &lt;em&gt;least upper bound&lt;/em&gt; (given $a,b\in A$, there exists a $c\in A$ such that $a\leq c$, $b\leq c$, and for every $d\in A$, if $a\leq d$ and $b\leq d$, then $c\leq d$), and a &lt;em&gt;greatest lower bound&lt;/em&gt; (the dual concept; just reverse all the inequalities in the definition above). The typical example is going back to the partially ordered set of subsets of a given $X$ under inclusion; the least upper bound of $A$ and $B$ is $A\cup B$, and the greatest lower bound is $A\cap B$. You can weaken the definition to require only least upper bounds (&lt;em&gt;upper semi-lattice&lt;/em&gt;) or just greatest upper bounds (&lt;em&gt;lower semi-lattices&lt;/em&gt;). Or you can strengthen it to require least upper bound and/or greatest lower bounds for any nonempty subset (&lt;em&gt;complete lattices&lt;/em&gt;); you can also require the existence of a smallest element and/or a greatest elements, or the existence of &quot;complements&quot; (if $0$ is the smallest element and $1$ is the greatest element, then a complement of $a\in A$ is an element $b\in A$ such that the least upper bound of $\{a,b\}$ is $1$ and the greatest lower bound of $\{a,b\}$ is $0$; in the case of $\mathcal{P}(X)$, the complement is the usual set-complement relative to $X$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Directed sets.&lt;/strong&gt; This is a kind of compromise between partially ordered sets and linear sets, in one direction: a partially ordered set $A$ is &lt;strong&gt;directed&lt;/strong&gt; if and only if for every $a,b\in A$ there exists a $c\in A$ such that $a\leq c$ and $b\leq c$ (any two elements have a common upper bound). the dual notion, where any two elements have a common lower bound, is called an &lt;strong&gt;inversely directed sets.&lt;/strong&gt; They play an important role in many areas of mathematics to construct special kinds of limits; when the index set is a directed system, you get something called a &lt;em&gt;directed limit&lt;/em&gt;; when the index set is an inversely directed set, you get something called an &quot;inverse limit&quot;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/P-adic_number&quot; rel=&quot;nofollow&quot;&gt;$p$-adic integers&lt;/a&gt; are an example of an inverse limit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; These are just a few of the ones I know, and they are likely a very limited sample of the ones that people who actually work with orders know. I don't do research anywhere near these topics, I just use orders all the time, like most mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strict orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, there is a &quot;sister&quot; notion to partial orders, called &quot;strict orders&quot;. A binary relation $\prec$ is a &lt;em&gt;strict order&lt;/em&gt; if and only if it is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Anti-reflexive: for every $a\in A$, $a\not\prec a$.&lt;/li&gt;&#xA;&lt;li&gt;Asymmetric: for all $a,b\in A$, if $a\prec b$, then $b\not\prec a$. &lt;/li&gt;&#xA;&lt;li&gt;Transitive: for every $a,b,c\in A$, if $a\prec b$ and $b\prec c$, then $a\prec c$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, strict and partial orders are actually closely connected. If you have a partial order $\preceq$, then you can define a strict order $\prec$ by &#xA;$$a\prec b \Longleftrightarrow a\preceq b\text{ and }a\neq b.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if you have a strict order $\prec$, then you can use it to define a partial order $\preceq$ by&#xA;$$a\preceq b \Longleftrightarrow a\prec b\text{ or } a=b.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can take either notion as your &quot;original&quot; notion, and define the other one in terms of it. These days, it is more common to take &quot;partial order&quot; as the original, and define strict order in terms of it. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-26T22:00:31.167" LastActivityDate="2010-11-26T22:00:31.167" CommentCount="13" />
  <row Id="11994" PostTypeId="2" ParentId="250" CreationDate="2010-11-26T21:43:39.813" Score="1" ViewCount="0" Body="&lt;p&gt;The Fold-and-Cut Theorem is pretty unintuitive. &lt;a href=&quot;http://erikdemaine.org/foldcut/&quot; rel=&quot;nofollow&quot;&gt;http://erikdemaine.org/foldcut/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-11-26T21:43:39.813" CommunityOwnedDate="2010-11-26T21:43:39.813" />
  <row Id="11995" PostTypeId="1" AcceptedAnswerId="11996" CreationDate="2010-11-26T22:48:19.783" Score="2" ViewCount="160" Body="&lt;p&gt;Hi, first question!&#xA;Equations it's on Latex Format, I'm still trying to understand how MathJax works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^{+\infty } \frac{1}{x \sqrt{x}} \, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that does not converge.&#xA;And if the integral was a indefinite one, also I'm almost sure the result is  $\ln(\sqrt x)$ using substitution.&#xA;If I am correct, why wolfram alpha says that the result is $- 2/ \sqrt x$ instead of $\ln(\sqrt x)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathway.com/&quot; rel=&quot;nofollow&quot;&gt;MathWay&lt;/a&gt; shows the correct results for both definite and indefinite integrals.&#xA;I'm new to any mathematical soft like wolfram alfa /mathematica, and my idea is to test my pen and paper results with a software that checks my results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints will be greatly appreciated.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-26T23:25:09.550" LastActivityDate="2010-11-27T00:58:05.090" Title="Comparing integral resolutions using Wolfram Alpha / Mathematica" Tags="&lt;calculus&gt;&lt;math-software&gt;&lt;mathematica&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11996" PostTypeId="2" ParentId="11995" CreationDate="2010-11-26T22:56:22.283" Score="2" ViewCount="0" Body="&lt;p&gt;The indefinite integral is $$\int \frac{1}{x\sqrt{x}} \, dx = \int \frac{1}{x^{^{3/2}}}\, dx = \frac{-2}{\sqrt{x}} + C$$ I suppose you made a mistake when integrating it by making a wrong substitution $u = \sqrt{x}$. Probably that's how you got the $\ln{\sqrt{x}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the convergence issue, you're right, the integral is divergent as you can see by taking the corresponding limit at $0$ of the indefinite integral $\frac{-2}{\sqrt{x}}$.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-26T23:03:15.240" LastActivityDate="2010-11-26T23:03:15.240" CommentCount="3" />
  <row Id="11997" PostTypeId="1" AcceptedAnswerId="11999" CreationDate="2010-11-26T23:20:10.337" Score="0" ViewCount="70" Body="&lt;p&gt;Given: $f(px+(1-p)y)\le pf(x)+(1-p)f(y)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $f(x)$ is defined on the interval I. If $x_{1}&amp;lt;x_{2}&amp;lt;x_{3}$ are in I and $f(x_{1})&amp;lt;f(x_{2})$ and $f(x_{3})&amp;lt;f(x_{2})$, then show that $f(x)$ is not convex on I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure where to begin on this one&lt;/p&gt;&#xA;" OwnerUserId="3961" LastActivityDate="2010-11-26T23:37:40.517" Title="convex with generic function" Tags="&lt;convex-analysis&gt;" AnswerCount="2" />
  <row Id="11999" PostTypeId="2" ParentId="11997" CreationDate="2010-11-26T23:27:54.853" Score="1" ViewCount="0" Body="&lt;p&gt;First express $x_2$ as a weighted linear combination of $x_1$ and $x_3$ i.e. $x_2 = tx_1 + (1-t)x_3$ such that $0 &amp;lt; t &amp;lt;1$. This is always possible since $x_2$ lies on the line segment connecting $x_1$ and $x_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $f(x)$ is convex in the interval $[x_1,x_2]$, then $f(x_2) = f(tx_1 + (1-t)x_3) &amp;lt; tf(x_1) + (1-t)f(x_3)$. But we are given $f(x_1) &amp;lt; f(x_2)$ and $f(x_3) &amp;lt; f(x_2)$. And hence $tf(x_1) + (1-t)f(x_3) &amp;lt; t f(x_2) + (1-t) f(x_2) = f(x_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we get $tf(x_1) + (1-t)f(x_3) &amp;lt; f(x_2)$ which contradicts the fact that $f$ is convex on the interval.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-26T23:27:54.853" />
  <row Id="12000" PostTypeId="2" ParentId="11997" CreationDate="2010-11-26T23:29:45.743" Score="0" ViewCount="0" Body="&lt;p&gt;What you say is &quot;given&quot; is actually the definition of convex, which you want to show does &lt;em&gt;not&lt;/em&gt; hold for any $f$ with the property given subsequently.  The convexity property says that the function value at a point between $x_1$ and $x_3$ has to be less than or equal to a certain weighted average of the values $f(x_1)$ and $f(x_3)$.  This is impossible if the in-between value is larger than the endpoint values.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-26T23:29:45.743" />
  <row Id="12001" PostTypeId="2" ParentId="11979" CreationDate="2010-11-26T23:44:19.787" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps you want to know the total number of moves that a single knight could theoretically make, where (since this is a graph theory homework question) I'll offer the following hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint&lt;/em&gt;:  You could construct a graph (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knight%27s_graph&quot; rel=&quot;nofollow&quot;&gt;knight's graph&lt;/a&gt;) where edges represent the moves of the knight.  The total number of moves can be deduced from a graph-theoretical property of this graph.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-27T00:12:39.823" LastActivityDate="2010-11-27T00:12:39.823" CommentCount="1" />
  <row Id="12002" PostTypeId="1" CreationDate="2010-11-27T00:17:31.380" Score="1" ViewCount="86" Body="&lt;p&gt;Given a problem, we are asked to use a statistical method to come up with a conclusion. e.g. XYZ drug decreases heart diseases. (null hypothesis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After searching on internet, I got a list of these tests:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis of variance (ANOVA),&#xA;Chi-square test,&#xA;Correlation,&#xA;Factor analysis,&#xA;Mann–Whitney U,&#xA;Mean square weighted deviation (MSWD),&#xA;Pearson product-moment correlation coefficient,&#xA;Regression analysis,&#xA;Spearman's rank correlation coefficient,&#xA;Student's t-test,&#xA;Time series analysis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume all methods above can be applied. So I pick up student's t-test as my choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I know whether it is a maximum likelihood estimation which is unbiased, consistent and efficient? What I think is that we should know the Mean and Variance of the estimate to conclude. But it is nowhere in the notes I found online. =/&lt;/p&gt;&#xA;" OwnerUserId="3778" LastEditorUserId="2641" LastEditorDisplayName="" LastEditDate="2010-11-27T05:19:17.803" LastActivityDate="2010-11-27T05:19:17.803" Title="Maximum likelihood estimation, " Tags="&lt;probability-theory&gt;&lt;estimation-theory&gt;" CommentCount="7" />
  <row Id="12003" PostTypeId="1" AcceptedAnswerId="12004" CreationDate="2010-11-27T00:23:49.990" Score="4" ViewCount="307" Body="&lt;p&gt;This is probably a bit of a boring and ridiculously noob question, so apologies in advance, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is everything in maths essentially built on adding, subtracting, multiplying and dividing?&lt;/p&gt;&#xA;" OwnerUserId="3229" LastEditorUserId="3229" LastEditorDisplayName="" LastEditDate="2010-12-01T22:34:49.730" LastActivityDate="2010-12-02T02:16:21.710" Title="Is everything in mathematics built on add, subtract, multiply and divide?" Tags="&lt;soft-question&gt;&lt;elementary&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="12004" PostTypeId="2" ParentId="12003" CreationDate="2010-11-27T00:38:48.287" Score="22" ViewCount="0" Body="&lt;p&gt;Not really. For one thing, before you can &quot;add&quot;, &quot;subtract&quot;, &quot;multiply&quot;, and &quot;divide&quot;, you have to have &lt;strong&gt;something&lt;/strong&gt; that you are &quot;adding&quot;, &quot;subtracting&quot;, &quot;multiplying&quot;, or &quot;dividing.&quot; So that &quot;something&quot; has to precede these operations. And for another, there are many notions that cannot really be expressed in these terms. To take a familiar one if you've taken Calculus, the idea of &quot;limit&quot; is not one that can be expressed in terms of adding, subtracting, multiplying, or dividing, but rather is expressed in terms of &quot;approaching&quot;, or of being &quot;arbitrarily close to&quot;.  (This is one reason why Newton and Leibnitz are so rightly hailed as revolutionaries: the ideas of Calculus were very much unlike what mathematics had been up to that point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one has to start &lt;em&gt;somewhere&lt;/em&gt;. Historically, mathematics started from different starting points in different cultures; counting was certainly one of the firsts things that comes up. But some cultures then realized that they need some more general ideas that went beyond simply counting (with its addition, multiplication, etc). Greeks settled on geometry as their underlying structure, and in fact translated everything about numbers into statements about geometry: lines, points, circles, squares, etc. Other cultures did not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days we usually take the route of &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiomatic_theory&quot; rel=&quot;nofollow&quot;&gt;axiomatic theories&lt;/a&gt;, where we have some &quot;primitive notions&quot; that are not defined, and some rules about how we can talk about them and their properties, and then we try to build everything on top of &lt;em&gt;that&lt;/em&gt;. One &lt;em&gt;can&lt;/em&gt; start from the natural numbers (the &quot;counting numbers&quot;) using, for instance, &lt;a href=&quot;http://en.wikipedia.org/wiki/Peano_axioms&quot; rel=&quot;nofollow&quot;&gt;the Peano axioms&lt;/a&gt;, and start building from there. Then you are actually starting &lt;em&gt;before&lt;/em&gt; you get to &quot;addition&quot;, &quot;multiplication&quot;, &quot;subtraction&quot;, and &quot;division&quot;, with only the notion of &quot;number&quot;, &quot;next one&quot;, and induction. Or we can have other starting points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;em&gt;today&lt;/em&gt;, there is usually one of two starting points: Set Theory, and Category Theory. Both presume that we agree on the basics of logic (the meaning of things like &quot;and&quot;, &quot;or&quot;, &quot;if... then...&quot;, and &quot;not&quot;, the meaning of &quot;is equal to&quot;, and things like that); and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rules_of_inference&quot; rel=&quot;nofollow&quot;&gt;rules of inference&lt;/a&gt;, which are rules that let you go from some statements others; for instance, the rule that says that if you know that A implies B, and you also know that A is true, then you know that B is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Set Theory, you start with the idea of &quot;set&quot; and &quot;element of&quot;; these ideas are not actually defined, you just give some axioms that describe things you can say about them. From these axioms one can &lt;em&gt;define&lt;/em&gt; numbers, addition, multiplication, etc. And also build &lt;em&gt;other&lt;/em&gt; notions that are orthogonal to natural numbers and their usual operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Category Theory, you start with the idea of &quot;objects&quot; and &quot;maps&quot; between objects (again, not defined; you only get some axioms that tell you how you can talk about them); then you can use these notions to construct everything else in terms of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, rather, in modern mathematics everything is either &quot;sets all the way down&quot;, or everything is &quot;categories all the way down&quot;. If there is one common set of presuppositions, it's just &lt;a href=&quot;http://en.wikipedia.org/wiki/First_order_logic&quot; rel=&quot;nofollow&quot;&gt;basic logic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T00:44:05.333" LastActivityDate="2010-11-27T00:44:05.333" />
  <row Id="12005" PostTypeId="2" ParentId="11995" CreationDate="2010-11-27T00:58:05.090" Score="2" ViewCount="0" Body="&lt;p&gt;If you are at the point of computing integrals, then you are probably already comfortable with computing derivatives.  The way to check whether a potential antiderivative is correct is to take its derivative.  If you take the derivative of $f(x)=\ln(\sqrt x)$ (either by first simplifying to $\frac{1}{2}\ln(x)$ or using the chain rule) you get $f&amp;#39;(x)=\frac{1}{2x}$. If you take the derivative of $g(x)=-\frac{2}{\sqrt x}=-2x^{-1/2}$, you get $g&amp;#39;(x)=x^{-3/2}=\frac{1}{\sqrt{x}^3}=\frac{1}{x\sqrt{x}}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-27T00:58:05.090" />
  <row Id="12006" PostTypeId="1" CreationDate="2010-11-27T01:32:19.110" Score="1" ViewCount="111" Body="&lt;p&gt;I remember reading in RPF's biographical book, &quot;Surely you're joking Mr. Feynman&quot; that he used to have timed contest about writing down the biggest number using standard symbols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the time restriction, I was wondering what if we restrict the number of characters, to say $N=10$. What is the largest number we can write using $N$ characters using numbers and standard functions known to most mathematicians? Of course I think the real difficulty would be in proving the said number is the largest. What is the largest number you can think of with $N=10$?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-27T05:09:27.390" Title="Largest number that can be written using N characters" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12007" PostTypeId="1" AcceptedAnswerId="12029" CreationDate="2010-11-27T01:35:20.923" Score="8" ViewCount="121" Body="&lt;p&gt;I'm trying to reconstruct a problem I saw in the Monthly, years ago. Perhaps it'll look familiar to someone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the integer lattice in the plane, we grow a tree in the following natural way: Initially the tree is just the origin. At each step, we find the set of lattice points that are neighbors (distance 1) to precisely one vertex of our tree, and add them (simultaneously) to the tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus on day 0 the tree is $\{(0,0\}$; on day 1 it contains $\{(0,0), (1,0), (-1,0),(0,1),(0,-1)\}$; on day 2 it contains those vertices along with $(2,0),(-2,0),(0,2)$ and $(0,-2)$ (note that $(1,1)$ is not added because it has two neighbors already in the tree), and on day 3 we add 12 new vertices. It looks like a pretty familiar fractal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing I'm not sure of is what exactly was asked of that tree... Possible candidates include its asymptotic density, some sort of simple formula to determine which lattice points ultimately make it into the tree and the # of vertices added on day $n$. There are lots of interesting questions and I'm happy to try and solve them but I prefer to work on the ones that were actually posed! &lt;/p&gt;&#xA;" OwnerUserId="308" LastEditorUserId="308" LastEditorDisplayName="" LastEditDate="2010-11-28T15:59:47.607" LastActivityDate="2010-11-28T23:54:13.190" Title="Reconstructing a Monthly problem: tree growth on the 2D integer lattice" Tags="&lt;number-theory&gt;&lt;reference-request&gt;&lt;discrete-geometry&gt;&lt;fractals&gt;&lt;fun-problem&gt;" AnswerCount="1" />
  <row Id="12008" PostTypeId="1" CreationDate="2010-11-27T01:40:47.643" Score="1" ViewCount="93" Body="&lt;p&gt;I'm trying to understand the proof for Lemma 4.8 in Lyndon &amp;amp; Schupp Combinatorial Group Theory, page 26, proposition 4.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very end of this proof, we have: $N \unlhd F$ where $F$ is a finitely generated free group. $[F:N]=p$, $p$ prime. $K \unlhd N$, $K \unlhd F $. $u,v \in N$ such that their images in $N/K$ are not conjugate.&#xA;It is then claimed that the images of $u,v$ in $F/K$ are not conjugate as well. Why is that? &lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2010-11-28T22:38:23.330" Title="Simplified Version: $N \unlhd F$, $a,b \in N$. $[F:N]=p$, prime. $a \sim b$ in $F$ (conjugates). Are $a \sim b$ in $N$? (more inside)" Tags="&lt;group-theory&gt;&lt;group&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12009" PostTypeId="2" ParentId="11923" CreationDate="2010-11-27T02:40:20.727" Score="16" ViewCount="0" Body="&lt;h3&gt;Update 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;should&lt;/em&gt; be the final edit. Lots of typos have been corrected and the presentation in the section on existence has been greatly improved. It should be very close to a form that can be used as a basis for a project for students.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Even more typos corrected. Added example of non-archimedean ordered field. Fleshed out a few bits more. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update 3&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Fixed up the proof sketch in the last exercise. Also I keep finding these small annoying errors... I'm sorry that I keep up bumping this thread, I just hope that you don't mind too much.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Introduction&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'll do one better: I'll write you one (mostly because I actually have already written one - consider this sort of a Cliff's notes version with most of the details left as exercises). Recall that a first course in analysis typically asserts that there exists a field with an order and the least upper bound property, and you then develop the theory of sequences and series from there. Thus our goal should be to prove existence and uniqueness (up to some isomorphism notion) of such an object.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Uniqueness&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 1.&lt;/strong&gt; An &lt;em&gt;ordered field&lt;/em&gt; is a field, $K$ along with a total order $\leq$ on $K$ such that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\forall x,y,z \in K : x \leq y \implies x+z \leq y+z$ (Transitivity).&lt;/li&gt;&#xA;&lt;li&gt;$\forall x,y \in K : 0 \leq x \land 0 \leq y \implies 0 \leq xy$ (Positivity of product).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Moreover, define the &lt;em&gt;absolute value&lt;/em&gt; function $\vert \cdot \vert: K \to K$ by &#xA;$$\vert x \vert = \begin{cases} &#xD;&#xA;    x  &amp;amp; \text{if } 0 \leq x \\\\ &#xD;&#xA;    -x &amp;amp; \text{otherwise.}&#xD;&#xA;\end{cases} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 1.&lt;/strong&gt; There is an alternate definition of an ordered field: $K$ is said to be ordered if there is a subset $P$ of $K$ - called the &lt;em&gt;positive elements of $K$&lt;/em&gt; - such that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$K$ is the disjoint union of $P$, $-P$ and $\lbrace 0 \rbrace$.&lt;/li&gt;&#xA;&lt;li&gt;$x,y \in P \implies x+y \in P$ and $xy \in P$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Make formal the colloquial statement &quot;The two definitions of ordered field are equivalent&quot; and prove said equivalence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we could just as well require that the inequality in Definition 1 be strict since $x \lt y \lor x = y \iff x \leq y$ allows us to pass between strict and non-strict orderings as we please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 2.&lt;/strong&gt; Let $K$ and $L$ be ordered fields. An order-preserving ($f$ is order-preserving means that $x\leq  y \implies f(x)\leq f(y)$) ringhomomorphism is called an &lt;em&gt;embedding&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that any embedding must be injective since the kernel of a ringhomomorphism is and ideal and $0 \lt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 3.&lt;/strong&gt; If $K$ is an ordered field, then $K$ has a subfield isomorphic to $\mathbb{Q}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 2.&lt;/strong&gt; Prove Proposition 3 by showing that $\varphi: \mathbb{N} \to K$ defined by &#xA;$$ \varphi(n) = \underbrace{1+\dotsb+1}_{n \text{ } 1\text{s}} $$&#xD;&#xA;extends to an embedding of $\mathbb{Q}$ in $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 4.&lt;/strong&gt; An ordered field $K$ is called an &lt;em&gt;archimedean field&lt;/em&gt; if for all nonzero $x$ in $K$ there exists a natural number $n$ such that $n|x| \gt 1$. Or equivalently: there exists a natural number $n$ such that $n \gt |x|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 5.&lt;/strong&gt; Let $K$ be an ordered field. If $K$ has the least upper bound property, then $K$ is archimedean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 3.&lt;/strong&gt; Show the contrapositive of Proposition 5 by considering the set&#xA;$$ I = \lbrace x \in K \;\vert\; 0 \lt x \land \forall n \in \mathbb{N} : nx &amp;lt; 1 \rbrace. $$&#xD;&#xA;Show that $I$ is nonempty, $1$ is an upper bound for $I$ and $I$ has no least upper bound [Hint: assume that $y$ is the least upper bound and split in to cases (i) for $y \in I$ look at $2y$, and (ii) for $y \notin I$ look at $y/2$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a bit of an abuse of language, the elements of $I$ can be said to be &quot;infinitely small&quot;. For an example of an ordered field which is not archimedean, consider $\mathbb{Q}(X)$, the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_function&quot; rel=&quot;nofollow&quot;&gt;rational functions&lt;/a&gt; in one variable over $\mathbb{Q}$ (see [&lt;a href=&quot;http://en.wikipedia.org/wiki/Ordered_field#Examples_of_ordered_fields&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;] for how the order is defined) - $0 \lt \frac{1}{X} \lt q$ for all positive rationals $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's time to consider the order topology on ordered and archimedean fields, i.e. the topology generated by the open intervals $(a,b)$ defined in the obvious way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 4.&lt;/strong&gt; Show that the open intervals in an ordered field $K$ are a basis for a topology on $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 6.&lt;/strong&gt; $\mathbb{Q}$ is dense in any archimedean field $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sketch of proof.&lt;/em&gt; Since the open intervals are a basis, we just need to verify that any open interval contains a rational. Define the ceiling function $\lceil \cdot \rceil : K \to \mathbb{Z}$ by &#xA;$$ \lceil x \rceil = \min \lbrace n \in \mathbb{Z} \;\vert\; n \geq x\rbrace, $$&#xD;&#xA;which is well-defined since $K$ is archimedean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $x \in K$ and a $K \owns \varepsilon \gt 0$ be given. Since $K$ is archimedean we can find $n \in \mathbb{N}$ such that $n\varepsilon &amp;gt; 1$. Now we have:&#xA;$$ 0 \leq \left\lvert \frac{\lceil nx \rceil}{n} - x \right\rvert &#xD;&#xA;     \leq \left\lvert \frac{\lceil nx \rceil - nx}{n} \right\rvert&#xD;&#xA;     \leq \frac{1}{n} \lt \varepsilon&#xD;&#xA;$$&#xD;&#xA;Hence $\frac{\lceil nx \rceil}{n} \in (x-\varepsilon,x+\varepsilon)$ from which that result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary 7.&lt;/strong&gt; $\mathbb{Q}$ is densely ordered in any archimedean field, i.e. if $K$ is archimedean, $x,y \in K$, and $x \lt y$, then there exists a $q \in \mathbb{Q}$ such that $x \lt q \lt y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 8.&lt;/strong&gt; Any archimedean field is 2nd-countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 5.&lt;/strong&gt; Prove Proposition 8 [Hint: Consider the set of intervals with rational endpoints. Show that it is a basis for a topology. Argue that it is coarser than the standard topology, and then show that any basis set for the standard topology (i.e. an interval) contains an interval with rational endpoints]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 9.&lt;/strong&gt; Any archimedean field $K$ is a topological field, i.e. $K \times K \owns (x,y) \mapsto x+y \in K$, $K^\times \times K^\times \owns (x,y) \mapsto xy \in K^\times$ and $K^\times \owns x \mapsto x^{-1} \in K^\times$ are continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 6.&lt;/strong&gt; Prove Proposition 9 [Hint: Consider sequences $x_n$ and $y_n$ that converge to $x$ and $y$ respectively, then use that convergent sequences are bounded].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that proposition 9 in fact holds for ordered fields too. You just have to consider nets instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 10.&lt;/strong&gt; Let $G$ be a topological group. A sequence $(x_n)$ in $G$ is said to be a &lt;em&gt;Cauchy sequence&lt;/em&gt; if for every neighbourhood $U$ of $e$ exists a $N$ such that $m,n \geq N \implies x_nx_m^{-1} \in U$. We shall denote the set of Cauchy sequences in $G$ by $G^\mathbb{N}_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this definition too can be used to define Cauchy nets in topological groups, but we shall not need that. Observe that the intervals with rational endpoints gives us a countable basis at $0$ in any archimedean field, so we see that for archimedean fields, the definition boils down to this:&#xA;$$ \forall \varepsilon \in \mathbb{Q}^+ \exists N : n,m \geq N \implies |x_n - x_m| &amp;lt; \varepsilon. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 11.&lt;/strong&gt; Let $K$ be an archimedean field and $(x_n)$ a sequence in $K$. The following hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $(x_n)$ is convergent, then $(x_n)$ is a Cauchy sequence, and any subsequence of $(x_n)$ converges to the same limit.&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is a Cauchy sequence, then it is bounded.&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is bounded, then $(x_n)$ has a Cauchy subsequence (&quot;pre-Bolzano-Weierstraß&quot;)&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is a Cauchy sequence then any subsequence is a Cauchy sequence. Moreover if $(x_n)$ has a convergent subsequence, then $(x_n)$ converges to the same limit.&lt;/li&gt;&#xA;&lt;li&gt;If $(x_n)$ is a Cauchy sequence and $(x_{k_n})$ is a subsequence, then $\lim_{n \to \infty} |x_n - x_{k_n}| = 0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 7.&lt;/strong&gt; Prove Proposition 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 12.&lt;/strong&gt; Let $K$ be an archimedean field. TFAE:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$K$ has the least upper bound property.&lt;/li&gt;&#xA;&lt;li&gt;Any Cauchy sequence in $K$ is convergent (&quot;$K$ is complete&quot;).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 8.&lt;/strong&gt; Flesh out the following sketch of a proof of Proposition 12: &lt;br/&gt;&#xA;$1 \implies 2$: Let $(x_n)$ be a Cauchy sequence in $K$. If $(x_n)$ is eventually constant, then it is obviously convergent, so assume that it is not so, and consider the set &#xA;$$ X = \lbrace x \in K \;\vert\; \exists N \in \mathbb{N} : n \geq N \implies x \lt x_n \rbrace. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ is nonempty and has an upper bound (why?) and thus $\sup X$ exists, and we claim that $x_n \to \sup X$. Given an $\varepsilon &amp;gt; 0$ find $N$ such that $|x_n - x_m| &amp;lt; \frac{\varepsilon}{2}$, and consider the interval $(x_N-\frac{\varepsilon}{2}, x_N + \frac{\varepsilon}{2})$. There are only finitely many elements of the sequence outside of this interval, so $x_N - \frac{\varepsilon}{2} \in X$. Moreover, $x + \frac{\varepsilon}{2}$ is an upper bound for $X$. The triangle inequality (note: prove the triangle inequality for archimedean fields) now gives &#xA;$$ |x_n - \sup X| \leq |x_n - x_N| + |x_N - \sup X| \lt \varepsilon. $$&#xD;&#xA;$2 \implies 1$: Let $A \subset K$ be nonempty and bounded above by $y_0 \in K$. Let $x_0$ be some non-upper bound for $A$ e.g. $a-1$ for some $a \in A$, and recursively define a pair of sequences $(x_n)$ and $(y_n)$ over $\mathbb{N}_0$ by&#xA;$$ \begin{align}&#xD;&#xA;y_{n+1} &amp;amp; = \begin{cases}&#xD;&#xA;    \tfrac{y_n + x_n}{2} &#xD;&#xA;        &amp;amp; \text{if } \tfrac{y_n+x_n}{2} \text{ is an upper bound for } A, \\\\&#xD;&#xA;    y_n &amp;amp; \text{otherwise}&#xD;&#xA;\end{cases} \\\\&#xD;&#xA;x_{n+1} &amp;amp; = \begin{cases}&#xD;&#xA;    \tfrac{y_n+x_n}{2}&#xD;&#xA;        &amp;amp; \text{if } \tfrac{y_n+x_n}{2} \text{ is a non-upper bound for } A, \\\\&#xD;&#xA;    x_n &amp;amp; \text{otherwise.}&#xD;&#xA;\end{cases}&#xD;&#xA;\end{align} $$&#xD;&#xA;By induction we have that $(x_n)$ is increasing, $(y_n)$ is decreasing, and  for all $i,j$: $x_i \leq y_j$. Likewise by induction we get that for all $i$: $0 \leq y_i - x_i \leq 2^{-i}(y_0 - x_0)$. It is now easy to show that $(y_n)$ is a Cauchy sequence. Assume that $n \geq m$ and we have:&#xA;$$ \begin{align}&#xD;&#xA;y_m-y_n &amp;amp; = y_m - y_{m+1} + y_{m+1} - \dotsb - y_{n-1} + y_{n-1} - y_n \\\\&#xD;&#xA;        &amp;amp; \leq (2^{-m-1} + 2^{-m-2} + 2^{-n})(y_0 - x_0) \\\\&#xD;&#xA;        &amp;amp; \leq 2^{-m}(y_0 - x_0)&#xD;&#xA;\end{align} $$&#xD;&#xA;from which it is easy to conclude that $(y_n)$ is Cauchy. Similarly $(x_n)$ is shown to be Cauchy. Thus the two sequences are convergent, and their limits must be the same since $y_n - x_n \leq 2^{-n}(y_0 - x_0)$ -- denote the limit by $s$. Finally observe that for all $a \in A$, $a \leq y_n$, so $a \leq s$ (why?) i.e. $s$ is an upper bound for $A$. Moreover if $u$ is an upper bound for $A$ then $x_n \leq u$ for all $n$, so $s \leq u$ (why?). Hence $s$ is the least upper bound for $A$. QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 12.&lt;/strong&gt; Let $K$ and $L$ be archimedean fields. If $f: K \to L$ and $g: L \to K$ are embeddings, then $K$ and $L$ are isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $g \circ f = \phi: K \to K$. By definition $\phi$ is an embedding. Assume that $\phi$ is not the identity. Then there exists an $x$ for which $\phi(x) \neq x$. Find a $q \in \mathbb{Q}$ between $x$ and $\phi(x)$ and observe that $\phi$ has to be the identity on $\mathbb{Q}$, but it can't preserve the ordering of $x$ and $q$. This gives a contradiction, and thus $\phi$ must be the identity. The proof that $f \circ g$ is the identity is similar. QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 13.&lt;/strong&gt; Let $G$ and $H$ be topological groups and let $G&amp;#39;$ be a dense subgroup of $G$. If $\phi: G \to H$ is continuous and $\phi|_{G&amp;#39;}$ is a homomorphism, then $\phi$ is a homomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $x,y \in G$ and assume that $\phi(xy) \neq \phi(x)\phi(y)$. Since $G&amp;#39;$ is dense there exists nets $(x_\alpha)$ and $(y_\alpha)$ in $G&amp;#39;$ converging to $x$ and $y$ respectively. Since $\phi$ is a homomorphism on $G&amp;#39;$ we have for all $\alpha$ that $\phi(x_\alpha y_\alpha) = \phi(x_\alpha)\phi(y_\alpha)$, and since $\phi$ is continuous and $\cdot$ is continuous we have that $\phi(xy) = \phi(x)\phi(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark.&lt;/em&gt; This is not a vacuous use of reductio ad absurdum, since a direct proof would have to work with arbitrary nets converging to $x$ and $y$. Of course this lemma immediately generalizes to also apply to rings in an obvious manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 14.&lt;/strong&gt; (Embedding theorem for archimedean fields). Let $K$ and $L$ be archimedean fields where $L$ is complete. Then there exists an embedding $\phi: K \to L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Define $\phi: K \to L$ by $\phi(x) = \sup \lbrace q \in \mathbb{Q} \;\vert\; q \lt x \rbrace$. It is easy to show that $\phi$ is order-preserving, so the only hard part is showing that it's a homomorphism. We wish to use the preceding lemma, so first we note that it's obvious that $\phi$'s restriction to $\mathbb{Q}$ is the identity, so the only hard part is verifying that it's continuous. Consider a $k \in K$ and let $V$ be a neighbourhood of $\phi(k)$. Find rationals $p$ and $q$ such that $\phi(k) \in (p,q) \subset V$. We claim that $\phi((p,q)) \subset (p,q)$. Let $k&amp;#39; \in (p,q) \subset K$ and find $q&amp;#39; \in \mathbb{Q}$ such that $p \lt k&amp;#39; \lt q&amp;#39; \lt q$. It's easy to see that $p \lt \phi(k&amp;#39;) \leq q&amp;#39; \lt q$, so $\phi(k&amp;#39;) \in (p,q)$, and we conclude that $\phi$ is continuous. QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 15.&lt;/strong&gt; (Uniqueness of complete ordered field) There exists up to order-preserving isomorphism one and only one complete ordered field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $K$ and $L$ be complete ordered fields. Theorem 14 gives embeddings $\phi: K \to L$ and $\varphi: L \to K$. Proposition 12 gives that $K$ and $L$ are isomorphic. QED.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Existence&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Much more has been written on this subject than uniqueness, so I'll write a lot less here. Recall that $\mathbb{Q}$ under addition is a topological group, so we use the previously established notation for its set of Cauchy sequences: $\mathbb{Q}^\mathbb{N}_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 16.&lt;/strong&gt; $\mathbb{Q}^\mathbb{N}_c$ is a commutative unital ring under coordinatewise addition and multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 8.&lt;/strong&gt; Prove Proposition 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition 17.&lt;/strong&gt; The set of rational Cauchy sequences that converge to zero, i.e.&#xA;$$ \mathfrak{z} = \left\lbrace (z_n) \in \mathbb{Q}^\mathbb{N}_c \;\bigg\vert\; z_n \to 0 \right\rbrace,$$&#xD;&#xA;is a maximal ideal in $\mathbb{Q}^\mathbb{N}_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 9.&lt;/strong&gt; Prove Proposition 17. [Hint: Consider an ideal $\mathfrak{a} \supset \mathfrak{z}$ with a $(q_n) \in \mathfrak{a}\setminus\mathfrak{z}$. Show that there exists an $N$ such that $n \geq N \implies q_n \neq 0$ and consider the sequence $(z_n) = (1-q_1,\dotsc,1-q_N,0,0,\dotsc)$. Argue that $(q_n + z_n) \in \mathfrak{a}$ and conclude that $\mathfrak{a} = \mathbb{Q}^\mathbb{N}_c$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\mathfrak{z}$ is a maximal ideal and hence the quotient, which we shall denote by $R$, must be a field. From here on out the rest is sketchy as there are a lot of details to check and it's really rather boring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We adopt the convention that we write sequences with &lt;strong&gt;bold&lt;/strong&gt;, e.g. $(x_n) = {\bf x}$ and use the notation $[{\bf x}]$ for the equivalence class containing ${\bf x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\lt$ on $R$ by $[{\bf p}] \lt [{\bf q}] \iff \exists (p_n) \in [{\bf p}], (q_n) \in [{\bf q}], \alpha \in \mathbb{Q}^+ : p_n \lt q_n + \alpha$ eventually. In particular note that the existence of the positive $\alpha$ cannot be dropped; e.g. look at $(1/n)$, $(0)$ and $(-1/n)$. Without the requirement that such a positive $\alpha$ exists, you'd thus have $0 \lt 0$, which is clearly rubbish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 10.&lt;/strong&gt; Verify that if $(p_n) \in [{\bf p}]$ and $(q_n) \in [{\bf q}]$ are witnesses that $[{\bf p}] \lt [{\bf q}]$, then &lt;em&gt;any&lt;/em&gt; pair of representatives will do just as well, i.e. if $(p&amp;#39;_n) \in [{\bf p}]$ and $(q&amp;#39;_n) \in [{\bf q}]$, then $(p&amp;#39;_n)$ and $(q&amp;#39;_n)$ are witnesses as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 11.&lt;/strong&gt; Show that $\lt$ is a strict total order, i.e. $\lt$ is irreflexive, transitive and total [Hint (for totality): You need to show that exactly one of $[{\bf x}] \lt [{\bf y}]$, $[{\bf x}] = [{\bf y}]$ and $[{\bf y}] \lt [{\bf x}]$ holds, so first show that at least one holds by assuming that none of them holds and derive a contradiction, and then show that at most one holds likewise by contradiction and using that $[{\bf x}] \nless [{\bf y}] \iff [{\bf y}] \leq [{\bf x}]$. You'll probably need to pass to &quot;nicer&quot; subsequences many times, so Proposition 11(5) will come in handy].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And sit back and reap the reward with the following easy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 12.&lt;/strong&gt; Show that $R$ is an ordered field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 13.&lt;/strong&gt; Show that $R$ is archimedean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 14.&lt;/strong&gt; Fill out the details of the following sketch of a proof that any Cauchy sequence in $R$ is convergent: &lt;br/&gt;&#xA;&lt;em&gt;Proof.&lt;/em&gt; Starting with an arbitrary Cauchy sequence $([{\bf q}]^i)$ of equivalence classes, Proposition 11(4) allows us to assume WLOG that for all $i,j$ with $j \geq i$ that $\left\lvert[{\bf q}]^i - [{\bf q}]^j]\right\rvert \lt 4^{-i}$. Note that convergence means showing some inequalities hold, so we only need to show those for &lt;em&gt;some&lt;/em&gt; representatives, so we are free to pick nice ones. &#xA;Pick representatives, $(q^i_n)$, such that $|q^i_n - q^i_m| \lt 4^{-n}$.&#xA;Take ${\bf x} = (q^1_1,q^2_2,\dotsc)$. This is obviously a Cauchy sequence (why?), and we claim that $[{\bf q}]^i \to [{\bf x}]$. Convince yourself that this is the same as this: &#xA;$$\forall \varepsilon \in \mathbb{Q}^+ \exists K,N : k \geq K, n \geq N \implies \left\lvert q^k_n - q^n_n \right\rvert \lt \varepsilon. $$&#xD;&#xA;Note that the requirement of the existence of the positive $\alpha$ dropped out since the above has to hold for every $\varepsilon &amp;gt; 0$. So let $\varepsilon \gt 0$ be given. We use the standard &quot;trick&quot; of inserting a clever element into the inequality: $|q^k_n - q^n_n| \leq |q^k_n - q^k_k| + |q^k_k - q^n_n|$ and use that the both the sequence and the representatives have that &quot;$4^{-m}$&quot;-property to conclude that if we choose $N,K$ such that $N=K$ and $2^{-N} \lt \varepsilon$ then RHS is eventually smaller than $\varepsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CONGRATULATIONS! You're done and you now have a fully certified driver's license for the real numbers!&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-11-28T03:08:15.393" LastActivityDate="2010-11-28T03:08:15.393" CommentCount="8" />
  <row Id="12010" PostTypeId="2" ParentId="12008" CreationDate="2010-11-27T02:46:56.587" Score="2" ViewCount="0" Body="&lt;p&gt;I don't have the book handy, but in your setup, a,b could easily be non-conjugate in N; similarly u,v could easily be conjugate in F/K.  Take F/K to be nonabelian of order 6, N/K to be index 2 and order 3.  None of its distinct elements are conjugate in N/K, but its non-identity elements are all conjugate in F/K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, something you haven't mentioned so far is critical for their proof.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-27T02:46:56.587" CommentCount="0" />
  <row Id="12011" PostTypeId="2" ParentId="12008" CreationDate="2010-11-27T02:47:12.480" Score="2" ViewCount="0" Body="&lt;p&gt;As you have stated it, I think the result is not true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a counterexample: for odd prime $p$, take the nonabelian group of order $p^3$ and exponent $p$, &#xA;$$G = \langle a,b\,\bigm|\, a^p = b^p = [a,b]^p = [a,b,a]=[a,b,b]=1\rangle.$$&#xD;&#xA;Let $F(x,y,z)$ be the free group on three generators. Consider $\varphi\colon F\to G$ given by $x\mapsto a$, $y\mapsto b$, and $z\mapsto [a,b]$. Let $K$ be the kernel of $\varphi$, and let $N$ be the preimage under $\varphi$ of the subgroup $\langle b,[a,b]\rangle$. Then $N\triangleleft F$, $K\triangleleft N$, $K\triangleleft F$, and $[F:N] = [G:\langle b,[a,b]\rangle] = p$. Note that since $N/K$ is abelian, if $u,v\in N$ are conjugate in $N/K$, then they are equivalent modulo $K$. Now consider $y$ and $yz$ in $N$; viewing $N/K$ as being $\langle b,[a,b]\rangle$, you have the elements $b$ and $b[a,b]$, which are not equal and hence not conjugate in $N/K$. However, $y^x = y[x,y]$. But $y[x,y] \equiv yz \pmod{K}$, since $z$ maps to $[a,b]$, and so does $[x,y]$. So $y$ and $yz$ &#xA;&lt;em&gt;are&lt;/em&gt; conjugate in $F/K$, even though they are not conjugate in $N/K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can cook up a similar example with the quaternion group or the dihedral group with $p=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either the lemma is incorrect, or else your simplification has eradicated some crucial bit of information.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-27T02:47:12.480" />
  <row Id="12013" PostTypeId="2" ParentId="275" CreationDate="2010-11-27T03:14:15.540" Score="2" ViewCount="0" Body="&lt;p&gt;I think Mathematics : A very short Introduction by Timothy Gowers is a very interesting read. As such, it's not something you &quot;learn&quot; from, but it makes for a very short and sweet introduction to someone who is just curious about mathematics. (Also its also a great read for people who actually do work in mathematics, because that's where I lift my examples from, when I explain to my family and friends what I am doing!) ;)&lt;/p&gt;&#xA;" OwnerUserId="1441" LastActivityDate="2010-11-27T03:14:15.540" CommunityOwnedDate="2010-11-27T03:14:15.540" />
  <row Id="12015" PostTypeId="2" ParentId="275" CreationDate="2010-11-27T03:45:13.880" Score="1" ViewCount="0" Body="&lt;p&gt;I'm surprised nobody has mentioned &lt;strong&gt;Mathematics and the Imagination&lt;/strong&gt; by Edward Kasner and James Newman; originally published in 1940, still &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486417034&quot; rel=&quot;nofollow&quot;&gt;available&lt;/a&gt; in an edition from Dover. Among other things, the book that introduced the world to the name &quot;googol&quot; for $10^{100}$. It's a classic, and I've never heard anything bad about it. The book is meant for non-mathematicians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I second the recommendation of Martin Gardner's columns as a follow-up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more recent addition to this genre is &lt;strong&gt;The Calculus Diaries: How math can help you lose weight, win in Vegas, and survive zombie apocalypse&lt;/strong&gt; by Jennifer Ouellette; it was reviewed favorably in NPR's &quot;Science Friday&quot;; written by a non-mathematician who never got through Calculus in school, also for non-mathematicians. I've heard some minor criticisms of the style, but otherwise generally positive reviews.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-27T03:45:13.880" CommunityOwnedDate="2010-11-27T03:45:13.880" />
  <row Id="12016" PostTypeId="1" CreationDate="2010-11-27T03:48:58.057" Score="2" ViewCount="69" Body="&lt;p&gt;I have just begun delving into p-adic number theory. I was wondering, given a poynomial $f(x)$ with integer coefficients, what does it mean when we say, $f(x)$ has a root in $\mathbb{Z}_2$, for instance.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-27T03:48:58.057" Title="What does it mean for a polynomial with integer coefficients to have a root in p-adic integers?" Tags="&lt;number-theory&gt;" CommentCount="2" />
  <row Id="12017" PostTypeId="1" CreationDate="2010-11-27T03:50:45.647" Score="2" ViewCount="34" Body="&lt;p&gt;The group $S_n \times S_n \times S_n$ acts on the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin squares&lt;/a&gt; $L$ of order $n$, with $\theta:=(\alpha,\beta,\gamma)$, starting from $L$, permuting the rows by $\alpha$, permuting the columns by $\beta$ and permuting the symbols by $\gamma$.  Each $\theta \in S_n \times S_n \times S_n$ is called an &lt;em&gt;isotopism&lt;/em&gt;.  In some instances $\theta(L)=L$, whence we call $\theta$ an &lt;em&gt;autotopism&lt;/em&gt; of $L$.  The set of autotopisms of a Latin square forms a group under composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Side note: we reserve the name automorphism for autotopisms of the form $(\alpha,\alpha,\alpha)$ to correspond with the notions in group theory and quasigroup theory]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Brendan McKay's &lt;a href=&quot;http://cs.anu.edu.au/~bdm/data/latin.html&quot; rel=&quot;nofollow&quot;&gt;data&lt;/a&gt;, we can deduce that no Latin square of order 7 has an autotopism group of order 7.  [There are autotopisms of order 7, but no autotopism &lt;em&gt;group&lt;/em&gt; has order 7.]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's a clever proof of this observation that doesn't look through all the non-isotopic Latin squares?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alexander Hulpke, Petteri Kaski, Patric R. J. Östergård, &lt;a href=&quot;http://arxiv.org/abs/0909.3402&quot; rel=&quot;nofollow&quot;&gt;The number of Latin squares of order 11&lt;/a&gt;, claims that there is a Latin square of order 11 that admits an autotopism group of order $11$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T03:50:45.647" Title="Why is there no Latin square of order 7 with an autotopism group of order 7?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" CommentCount="2" />
  <row Id="12018" PostTypeId="2" ParentId="11673" CreationDate="2010-11-27T04:22:37.850" Score="2" ViewCount="0" Body="&lt;p&gt;another way to get this result is to first recall that $S := X_1+X_2$ and $R := X_2/X_1$ are independent. [this is well-known and can be proved by the usual method of derived distributions for two random variables. the mapping $(X_1,X_2) \to (R,S)$ is 1-1 and easily inverted. this is a standard example or exercise in many intro probability and math stat books - such as  &lt;a href=&quot;http://books.google.com/books?id=L95tVBSOgpUC&amp;amp;pg=PA377&amp;amp;dq=functions+of+two+gamma+variables&amp;amp;hl=en&amp;amp;ei=PXvwTJDpEIT68Abu3N2HDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=35&amp;amp;ved=0CNYBEOgBMCI#v=onepage&amp;amp;q=functions%20of%20two%20gamma%20variables&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, noting that $\frac{X_1}{X_1+X_2} = \frac{1}{1+R}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {\mathrm E}\{ X_1 | S\} = {\mathrm E}\{ \frac{S}{1+R} | S\} = S\kern1pt {\mathrm E}\{ \frac{1}{1+R} | S\} = S\kern1pt{\mathrm E}\frac{1}{1+R}. \kern10pt (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the last equality in (1) follows from independence. also, [on repeating the argument - or just taking unconditional expectations in (1)],&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_1 = {\mathrm E}X_1 = {\mathrm E}\frac{S}{1+R} = {\mathrm E}\frac{1}{1+R}\kern2pt {\mathrm E}S = {\mathrm E}\frac{1}{1+R}(c_1+c_2),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kern84pt {\mathrm E}\frac{1}{1+R} = \frac{c_1}{c_1+c_2}, \kern84pt (2)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be plugged into (1) to get the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[btw, (2) says that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\mathrm E}\frac{X_1}{X_1+X_2} = \frac{{\mathrm E}X_1}{{\mathrm E}(X_1+X_2)},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a relation which, in general, is a fond wish of students in an intro prob or stats course - and perhaps the instructors too.]&lt;/p&gt;&#xA;" OwnerUserId="1547" LastEditorUserId="1547" LastEditorDisplayName="" LastEditDate="2010-11-27T04:39:49.707" LastActivityDate="2010-11-27T04:39:49.707" />
  <row Id="12019" PostTypeId="2" ParentId="11936" CreationDate="2010-11-27T04:22:51.837" Score="0" ViewCount="0" Body="&lt;p&gt;Thanks Mariano!&lt;/p&gt;&#xA;" OwnerUserId="4127" LastActivityDate="2010-11-27T04:22:51.837" />
  <row Id="12020" PostTypeId="2" ParentId="11726" CreationDate="2010-11-27T05:05:05.037" Score="4" ViewCount="0" Body="&lt;p&gt;Here's a proof.  The value of $\alpha$ is the value of the infinite sum $$\sum_{m= -\infty}^{\infty} (e^{-2^{-m-1}} - [m \geq 1]),$$  where $[m \geq 1]$ is 1 if $m \geq 1$ and 0 otherwise.  Mathematica gives this value (to 6 decimal places) as $0.667253$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full argument in all its rigor is too long to post on this site, so I'm only going to give an extended outline.  There are a couple of strange claims in here, but bear with me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&lt;strong&gt;Part 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my &lt;a href=&quot;http://math.stackexchange.com/questions/11602/limit-of-sn-sum-k1-infty-left1-prod-j1n-1-left1-fracj/11662#11662&quot;&gt;previous post&lt;/a&gt; we know that $1 - \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) = 1$ when $k \leq \lfloor \log_2 (n-1) \rfloor$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p = 2 \log_2 n - \lfloor 2 \log_2 n \rfloor$.    Thus  $2 \log_2 n - S(n)$ is  $$2 \log_2 n - \lfloor 2 \log_2 n \rfloor + \sum_{k= \lfloor \log_2 (n-1) \rfloor +1}^{\lfloor 2 \log_2 n \rfloor} \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) + \sum_{k= \lfloor 2 \log_2 n \rfloor + 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) - 1\right)$$&#xD;&#xA;$$= p + \sum_{k= \lfloor \log_2 (n-1) \rfloor +1}^{2 \log_2 n - p} \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) + \sum_{k= 2 \log_2 n - p + 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) - 1\right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the expression $$\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right)$$ is very close to 0 when $k &amp;lt; 2 \log_2 n$ and very close to 1 when $k &amp;gt; 2 \log_2 n$.  So most of the contribution to $2 \log_2 n - S(n)$ occurs when $k$ is close to $2 \log_2 n$.  The next step, then, is to reindex with $m = k - \lfloor 2\log_2 n \rfloor$.  Now we basically have &#xA;$$2 \log_2 n - S(n) = p + \sum_{m= - \log_2 n}^0 \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) + \sum_{m= 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) - 1\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&lt;strong&gt;Part 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we need a good approximation to $\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right)$.  It turns out that $e^{-2^{-m+p-1}}$ is an excellent approximation (which surprises me some - despite the fact that I have verified it numerically - as it is independent of $n$).  To see this, rewrite $\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right)$ as $$\exp \left( \sum_{j=1}^{n-1} \ln\left(1 - \frac{j}{2^{m-p} n^2}\right)\right)$$ and then expand the $\log$ expression with the Maclaurin series for $\ln (1+x)$.  The first term in the expansion dominates when $m$ is positive or constant, and we get&#xA;$$\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) = \exp \left(\sum_{j=1}^{n-1} \left(- \frac{j}{2^{m-p} n^2} \right) + O\left(\frac{j^2}{4^{m-p} n^4}\right) \right)$$&#xD;&#xA;$$=\exp \left( - \frac{1}{2^{m-p+1}} + O\left(\frac{1}{2^m n}\right) \right) = \exp \left( - \frac{1}{2^{m-p+1}}\right) + O\left(\frac{1}{2^m n}\right)$$&#xD;&#xA;Thus $$\sum_{m= 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) - 1\right) = \sum_{m= 1}^{\infty} \left(\exp \left( - \frac{1}{2^{m-p+1}} \right) - 1\right) + O\left(\frac{1}{n}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $m$ is negative and not constant, things are a little trickier, as the higher-order Maclaurin series terms make $\exp \left( - \frac{1}{2^{m-p+1} } \right)$ not a good relative approximation to the product.  However, all the terms in the Maclaurin series are negative, so truncating after the first term does yield an upper bound.  In addition, $\exp \left( - \frac{1}{2^{m-p+1}} \right)$ goes to zero extremely fast as $m \to -\infty$.  (For example, if $m = - \log_2 (\log n)$ (which heads to $-\infty$ very slowly), we still have $\exp \left( - \frac{1}{2^{m-p+1}} \right) = \frac{1}{n^{2^{p-1}}}$.)  Thus $\exp \left( - \frac{1}{2^{m-p+1}} \right)$ is still an excellent absolute approximation for the product when $m$ is negative.  Since there are only $\log_2 n$ negative terms in the sum we are considering, we have &#xA;$$\sum_{m= - \log_2 n}^0 \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) = \sum_{m = - \log_2 n}^0 \exp \left( - \frac{1}{2^{m-p+1}} \right) + E(n),$$&#xD;&#xA;where $E(n) \to 0$ as $n \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&lt;strong&gt;Part 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $$f(p) = \sum_{m= - \infty}^0  e^{-2^{-m+p-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m+p-1}} - 1).$$&#xD;&#xA;The function $f$ is linear in $p$!  (Despite having verified this numerically and despite the argument below, this still seems weird to me!)  The slope is $-1$.  To see this, differentiate to get $$f&amp;#39;(p) = - \ln 2 \sum_{m= - \infty}^{\infty} 2^{-m+p-1} e^{-2^{-m+p-1}}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin summation formula&lt;/a&gt;.  Because we have the product of two exponentials, $f^{(k)}(p)$ will have the expression $2^{-m+p-1} e^{-2^{-m+p-1}}$ in every term.  As $m \to \infty$, $2^{-m+p-1} \to 0$ and $e^{-2^{-m+p-1}} \to 1$.  As $m \to -\infty$, $2^{-m+p-1} \to \infty$ and $e^{-2^{-m+p-1}} \to 0$, but the latter expression dominates.  Thus  $f^{(k)}(p) \to 0$ as $m \to \infty$ and as $m \to -\infty$.  Thus the Euler-Maclaurin formula says that&lt;br&gt;&#xA;$$f&amp;#39;(p) = - \ln 2 \int_{-\infty}^{\infty} 2^{-m+p-1} e^{-2^{-m+p-1}} dp = \left. e^{-2^{-m+p-1}}\right|_{-\infty}^{\infty} = -1.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $$\lim_{n \to \infty} \left(2 \log_2 n - S(n)\right) = p + \sum_{m= - \infty}^0  e^{-2^{-m+p-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m+p-1}} - 1) $$&#xD;&#xA;$$= p - p + \sum_{m= - \infty}^0  e^{-2^{-m-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m-1}} - 1) = \sum_{m= - \infty}^0  e^{-2^{-m-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m-1}} - 1).$$&#xD;&#xA;Again, this value is approximately $0.667253$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-27T21:14:25.560" LastActivityDate="2010-11-27T21:14:25.560" CommentCount="8" />
  <row Id="12021" PostTypeId="2" ParentId="12006" CreationDate="2010-11-27T05:09:27.390" Score="2" ViewCount="0" Body="&lt;p&gt;Well, just for fun I thought I'd try to answer the question with the operations restricted to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;nullary operations $\{0,1,2,\ldots,9\}$ and&lt;/li&gt;&#xA;&lt;li&gt;binary operation $*$ (i.e. multiplication).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It turns out that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;9999999999&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;is the greatest in this case.  The closest runner-up is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;99999*9999 = 999890001&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can make some simplifying assumptions:  (a) all 10 slots are to be used (otherwise, we just add in another 9 at the end) and (b) any of $0,1,2\ldots,8$ is better off being replaced by a $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This exercise didn't seem all that satisfying really, there's so many sequences that result in syntax errors.  E.g.:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*999*99999&#xA;999999999*&#xA;999999**99&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In my opinion, it gets even less satisfying if you keep going.  E.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is 1/0 allowed?&lt;/li&gt;&#xA;&lt;li&gt;Does exponentation require a character?  E.g. &lt;code&gt;2^6&lt;/code&gt; vs. $2^6$.&lt;/li&gt;&#xA;&lt;li&gt;Do you require suitable bracketing?  When?  &lt;code&gt;1/2^{-6}&lt;/code&gt; vs. &lt;code&gt;1/2^-6&lt;/code&gt;.  What about &lt;code&gt;9^9^9&lt;/code&gt;?&lt;/li&gt;&#xA;&lt;li&gt;How many characters does $\log$ require?&lt;/li&gt;&#xA;&lt;li&gt;The result varies on human factors -- e.g. what number system is used (base 10?), how many characters it takes to write a function, etc.&lt;/li&gt;&#xA;&lt;li&gt;And, of course, there's the inevitable question of self-referential sequences (which might be difficult to arrange for 10 characters).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T05:09:27.390" CommentCount="1" />
  <row Id="12022" PostTypeId="1" AcceptedAnswerId="12024" CreationDate="2010-11-27T05:09:36.953" Score="0" ViewCount="86" Body="&lt;p&gt;The problem is &quot;Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$, where L is a smooth, simple closed, and postively oriented curve that does not pass through the orgin&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution:&lt;br&gt;&#xA;Let $$P(x,y) = \frac{-y}{x^2 + y^2}, Q(x,y) = \frac{x}{x^2 + y^2}$$&#xD;&#xA;Get $$\frac{\partial{P}}{\partial{y}} = \frac{y^2 - x^2}{(x^2 + y^2)^2} = \frac{\partial{Q}}{\partial{x}}$$&#xD;&#xA;Acorrding to the Green formula: &#xA;$$\oint_{L} \frac{xdy - ydx}{x^2 + y^2} = \iint (\frac{\partial{Q}}{\partial{x}} - \frac{\partial{P}}{\partial{y}})dxdy = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with my solution?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-27T05:50:47.697" LastActivityDate="2010-11-27T06:05:29.370" Title="Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$" Tags="&lt;calculus&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="12023" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T05:12:47.750" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: What does the hypothesis of &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Green's theorem&lt;/a&gt; say about the requirements on functions $P$ and $Q$?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-27T05:12:47.750" />
  <row Id="12024" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T05:16:45.993" Score="0" ViewCount="0" Body="&lt;p&gt;That is valid as long as $L$ is a simple closed curve and the origin is not &quot;inside&quot; $L$, i.e., in the bounded region determined by $L$.  For example, this would work if $L$ is a unit circle centered at $(2,0)$, but not if it is the unit circle centered at the origin, and not if it is not a simple closed curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As Timothy Wagner already pointed out before I finished writing, you need to check the hypotheses on the theorem you invoked to see when it works.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $L$ is closed, your integral should be $2\pi$ times the number of times $L$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Winding_number#Differential_geometry&quot; rel=&quot;nofollow&quot;&gt;winds&lt;/a&gt; around the origin.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-27T05:25:47.780" LastActivityDate="2010-11-27T05:25:47.780" CommentCount="10" />
  <row Id="12025" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T05:21:03.107" Score="0" ViewCount="0" Body="&lt;p&gt;The answer should be $2\pi$ if the origin lies &quot;inside&quot; $L$, otherwise it should be 0.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-27T06:01:41.497" LastActivityDate="2010-11-27T06:01:41.497" CommentCount="2" />
  <row Id="12026" PostTypeId="1" AcceptedAnswerId="12033" CreationDate="2010-11-27T06:00:20.420" Score="0" ViewCount="92" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/12022/calculating-oint-l-fracxdy-ydxx2-y2&quot;&gt;this&lt;/a&gt; problem, I know that the hypothsis of Green theorem must ensure that the simple closed curve is smooth, but what is smooth? Could you give a definition and an intuitive explaination?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-11-27T06:55:08.457" Title="What does smooth curve mean?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="12027" PostTypeId="2" ParentId="12026" CreationDate="2010-11-27T06:04:52.053" Score="-1" ViewCount="0" Body="&lt;p&gt;Smooth means differentiable (at least once). In other words, smooth means continuous and without &quot;corners&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: changed &quot;edges&quot; to &quot;corners&quot; after J.M.&lt;/p&gt;&#xA;" OwnerUserId="3891" LastEditorUserId="3891" LastEditorDisplayName="" LastEditDate="2010-11-27T06:12:40.937" LastActivityDate="2010-11-27T06:12:40.937" CommentCount="3" />
  <row Id="12028" PostTypeId="2" ParentId="12022" CreationDate="2010-11-27T06:05:29.370" Score="0" ViewCount="0" Body="&lt;p&gt;One way of understanding why Green's theorem doesn't necessarily work in this context is to view Green's theorem as two separate equations added together: $\oint_{L} P dx = \int\int-{\partial P \over \partial y} dy dx$ and $\oint_{L} Q dy = \int\int{\partial Q \over \partial x} dx dy$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, Green's theorem holds because you can prove each of these equations by integrating by parts. In the first equation you integrate by parts in $y$ for a fixed $x$ and then integrate the result in $x$. In the second, you integrate by parts in $x$ for a fixed $y$ and then integrate the result in $y$. Since in your case both $-{\partial P \over \partial y}$ and  ${\partial Q \over \partial x}$ blow up at the origin, if the origin is contained in the interior of your curve these integrations by parts won't work... and you'll get the wrong answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The curve is assumed to be simple and closed in the above.)&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T06:05:29.370" />
  <row Id="12029" PostTypeId="2" ParentId="12007" CreationDate="2010-11-27T06:06:41.177" Score="8" ViewCount="0" Body="&lt;p&gt;I think I got it. It seems to be problem 10360, originally published in 1994 (volume 101, issue 1, page 76), proposed by Richard Stanley. I'm looking at &lt;a href=&quot;http://www.jstor.org/pss/2589004&quot; rel=&quot;nofollow&quot;&gt;the JSTOR page of the solution&lt;/a&gt; (by &lt;a href=&quot;http://math.stackexchange.com/users/226/robin-chapman&quot;&gt;Robin Chapman&lt;/a&gt;), which appeared in October 1998, volume 105, no 8, pages 769-771). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the statement of the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L$ be the &lt;em&gt;integer lattice&lt;/em&gt; in $\mathbb{R}^d$, i.e., the $L$ is the set of points $(x_1,x_2,\ldots,x_d)$ with all $x_j\in\mathbb{Z}$. Consider $L$ as a graph by declaring two lattice points to be adjacent if the distance between them is $1$. Define a sequence $S_0$, $S_1,\ldots$ of subsets of $L$ inductively as follows:&#xA;\begin{align*}&#xA;S_0 &amp;amp;= \Bigl\{ (0,0,\ldots,0)\Bigr\}\\&#xA;S_{n} &amp;amp;= \Bigl\{ P\in L-\mathop{\cup}\limits_{0\leq k\lt n}S_k\ \Bigm|\  \text{$P$ is adjacent to exactly one element of $\mathop{\cup}\limits_{0\leq k\lt n}S_k$}\Bigr\}.&#xA;\end{align*}&#xA;Let $S$ be the subgraph of $L$ whose vertices are $\cup S_n$. Thus, $P\in S$ is adjacent to $P&amp;#39;\in S$ if the distance between $P$ and $P&amp;#39;$ is $1$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a simple condition for a point of $L$ to belong to $S$.&lt;/li&gt;&#xA;&lt;li&gt;For $P\in S$, find a simple rule to determine $i$ such that $P\in S_i$.&lt;/li&gt;&#xA;&lt;li&gt;How many elements are in $S_i$?&lt;/li&gt;&#xA;&lt;li&gt;How many $P\in S_i$ are adjacent to no points in $S_{i+1}$?&lt;/li&gt;&#xA;&lt;li&gt;Show that $S$ is a tree.&lt;/li&gt;&#xA;&lt;li&gt;Investigate the (vertex) density of $S$ in $L$, and compare it to the largest density of a subset of $L$ for which the induced subgraph is a tree.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T23:54:13.190" LastActivityDate="2010-11-28T23:54:13.190" CommentCount="3" />
  <row Id="12030" PostTypeId="1" AcceptedAnswerId="12037" CreationDate="2010-11-27T06:10:58.757" Score="1" ViewCount="77" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/12022/calculating-oint-l-fracxdy-ydxx2-y2&quot;&gt;The orginal problem&lt;/a&gt; is &quot;Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$, where L is a smooth, simple closed, and postively oriented curve that does not pass through the orgin&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I modify the hypothesis and allow non-simple closed curve? I mean is there something like green formula that allows us calculuating &quot;non-simple closed curve integral&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;br&gt;&#xA;It seems to me that this integral should also resemble the simple closed one. Because common interior line integral should be canceled only remaining the outter curve. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-27T07:54:22.260" LastActivityDate="2010-11-27T09:14:57.473" Title="Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$ PartII " Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="4" />
  <row Id="12032" PostTypeId="1" AcceptedAnswerId="12036" CreationDate="2010-11-27T06:16:00.823" Score="2" ViewCount="173" Body="&lt;p&gt;Was trying to calculate $$\int_{0}^{\infty}e^{-x}\ln x dx=-\gamma$$ and I found this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to analyze $$\int\frac{e^{-x}}{x}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $u=\displaystyle\frac{1}{x} \Rightarrow du = \displaystyle\frac{-1}{x^{2}} dx  $,  and $dv=e^{-x} \Rightarrow v=-e^{-x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{e^{-x}}{x}dx = \displaystyle\frac{1}{x}\cdot-e^{-x}-\int-e^{-x}\cdot\displaystyle\frac{-1}{x^{2}} dx = -\displaystyle\frac{e^{-x}}{x}-\int \displaystyle\frac{e^{-x}}{x^{2}} dx$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating from same form get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{e^{-x}}{x}dx = -\displaystyle\frac{e^{-x}}{x} + \displaystyle\frac{e^{-x}}{x^{2}} + 2\int\frac{e^{-x}}{x^{3}}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These calculations are correct?, and more is  valid say : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{e^{-x}}{x}dx = \displaystyle\sum\limits_{n=0}^\infty (-1)^{n+1}n!\frac{e^{-x}}{x^{n+1}}\ ?$$  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\bf{EDIT}$: This serie  helps me to&#xA;  calculate it ? :&#xA;  $$\int_{0}^{\infty}e^{-x}\ln xdx=-\gamma$$ I don't know  how to turn&#xA;  this serie in something armonic. If&#xA;  not is this way to calculate that this&#xA;  integral converges to $-\gamma$, which&#xA;  is the form ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-28T20:20:08.303" LastActivityDate="2010-11-28T20:20:08.303" Title="These calculations are correct ? About  $\int\frac{e^{-x}}{x}dx$ $ \bf{[EDITED]}$:" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12033" PostTypeId="2" ParentId="12026" CreationDate="2010-11-27T06:17:36.133" Score="1" ViewCount="0" Body="&lt;p&gt;There are many ways you can characterize the smoothness of a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, people use the notation $C^{(n)}(\Omega)$ where $n \in \mathbb{N} \cup \{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we say $f(x) \in C^{(n)}(\Omega)$, we mean that $f(x)$ has $n$ derivatives in the entire domain ($\Omega$ denotes the domain of the function) and the $n^{th}$ derivative of $f(x)$ is continuous i.e. $f^{n}(x)$ is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also by convention, if $f(x)$ is just continuous, then we say $f(x) \in C^{(0)}(\Omega)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $f(x) \in C^{(\infty)}$ if the function is differentiable any number of times. For instance, $e^{x} \in C^{(\infty)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example to illustrate is to consider the following function $f: \mathbb{R} \rightarrow \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = 0, x \leq 0$$ $$f(x) = x^2, x&amp;gt;0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is in $C^{(1)}(\mathbb{R})$ but not in $C^{(2)}(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when the domain of the function is the largest set over which the function definition makes sense, then we omit $\Omega$ and write that $f \in C^{(n)}$ the domain being understood as the largest set over which the function definition makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note the containment $C^{(n)} \subseteq C^{(m)}$ whenever $n&amp;gt;m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of Green's theorem when we apply the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint_c (Ldx + Mdy) = \int \int_D (\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}) dx dy$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we need $L,M \in C^1{(\Omega)}$ where $\Omega$ is a domain containing the curve and the interior of the curve viz $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple closed curve $C$ should be piecewise smooth or more generally the curve $C$ should be in $C^{(0)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that the curve $C$ is piecewise smooth curve when the two conditions below are satisfied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $C \in C^{(0)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The domain over which the curve is defined can be partitioned into disjoint subsets such that the curve is in $C^{(\infty)}$ over each of these subsets&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-27T06:55:08.457" LastActivityDate="2010-11-27T06:55:08.457" CommentCount="8" />
  <row Id="12034" PostTypeId="1" AcceptedAnswerId="12041" CreationDate="2010-11-27T06:20:08.473" Score="0" ViewCount="100" Body="&lt;p&gt;What are the conditions for existence of fourier series expansion of a function $f\colon\mathbb{R}\to\mathbb{R}$&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T06:22:13.937" LastActivityDate="2010-11-27T07:11:42.837" Title="What are the conditions for existence of fourier series expansion of a function $f\colon\mathbb{R}\to\mathbb{R}$" Tags="&lt;fourier-series&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="12035" PostTypeId="2" ParentId="12026" CreationDate="2010-11-27T06:20:11.717" Score="0" ViewCount="0" Body="&lt;p&gt;Consider the following curve in the plane, $(x(t),y(t))$, this curve is called smooth if the functions $x(t)$ and $y(t)$ are smooth, which simply means that for all $N$, the derivatives $\frac{d^Nx}{dt^N}$ and $\frac{d^Ny}{dt^N}$ exist.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-11-27T06:20:11.717" />
  <row Id="12036" PostTypeId="2" ParentId="12032" CreationDate="2010-11-27T06:24:13.777" Score="2" ViewCount="0" Body="&lt;p&gt;The series diverges, but converges to your integral asymptotically: If you add up the first $n$ terms the ratio of the error to the $n$th term goes to zero as $x$ goes to infinity&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T06:24:13.777" CommentCount="1" />
  <row Id="12037" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T06:25:30.037" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Suppose $L$ intersects with it self at finitely many points. Then, if we consider a parametrization of $L$, we deduce $L=L_1\cup L_2\cup\cdots L_n$ where $L_k$ is simple. &#xA;Now consider&#xA;$$ \oint_L f = \sum_k\oint_{L_k} f$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-27T06:25:30.037" />
  <row Id="12038" PostTypeId="1" CreationDate="2010-11-27T06:44:35.637" Score="0" ViewCount="42" Body="&lt;p&gt;For the simpler flexagons, the notion of a face is obvious. However, there are flexagons where it's possible to mix faces. In those cases, how is a face defined?&lt;/p&gt;&#xA;" OwnerUserId="4131" LastActivityDate="2010-11-27T06:44:35.637" Title="What is a face of a flexagon?" Tags="&lt;topology&gt;" />
  <row Id="12039" PostTypeId="2" ParentId="11434" CreationDate="2010-11-27T06:49:18.007" Score="0" ViewCount="0" Body="&lt;p&gt;Finding a recurrence for the first equation should be obvious (just look at the first few values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second function $f(n) = n(n+3)$, here's an approach that seems natural to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to express $f(n)$ in terms of $f(n-1)$, $f(n-2)$, etc.  So start with $f(n)=a f(n-1) + b f(n-2)$ and equate coefficients.  That is, we have \[n^2+3n=a(n^2+n-2)+b(n^2-n-2)\] so&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a+b=1$,&lt;/li&gt;&#xA;&lt;li&gt;$a-b=3$,&lt;/li&gt;&#xA;&lt;li&gt;$-2a-2b=0$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We observe that this system of linear equations has no solutions (the first and third equations are inconsistent).  So, we now consider a deeper recurrence: Let $f(n)=a f(n-1) + b f(n-2) + c f(n-3)$, now solve the resultant system of linear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: in both cases, remember to describe sufficiently many initial values.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T06:49:18.007" />
  <row Id="12040" PostTypeId="2" ParentId="8814" CreationDate="2010-11-27T06:58:45.720" Score="1" ViewCount="0" Body="&lt;p&gt;\[\sqrt{n}^{\log n}=n^{\log \sqrt{n}}\]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T06:58:45.720" CommunityOwnedDate="2010-11-27T06:58:45.720" CommentCount="2" />
  <row Id="12041" PostTypeId="2" ParentId="12034" CreationDate="2010-11-27T07:11:42.837" Score="6" ViewCount="0" Body="&lt;p&gt;If $f\in L^1_\text{loc}(\mathbb{R})$, then on an interval $I=(a,b)$ we can define &#xA;$$\hat{f}(n)=\frac{1}{b-a}\int_a^b f(x)e^{-2\pi inx/(b-a)}dx.$$&#xD;&#xA;However, in order for the formal Fourier series $$S[f](x)=\sum_{-\infty}^{\infty} \hat{f}(n)e^{2\pi inx/(b-a)}$$ to converge we need more conditions on $f$. &#xA;Kolmogorov proved in 1925 that there is there is $f\in L^1(0,2\pi)$ such that $S[f]$ diverges almost everywhere. In 1966 Carleson proved that $S[f]$ converges almost everywhere provided $f\in L^2(0,2\pi)$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-27T07:11:42.837" CommentCount="6" />
  <row Id="12042" PostTypeId="1" AcceptedAnswerId="12055" CreationDate="2010-11-27T07:30:50.853" Score="4" ViewCount="157" Body="&lt;p&gt;Let $L(n)=\lfloor(1+\sqrt{5})^n\rfloor$. What kind of a linear recurrence is satisfied by $L(n)$? I have no idea how to go about this, because of the presence of the greatest integer function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to retag it as I kept getting an error on every tag I thought was appropriate.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-27T07:36:08.510" LastActivityDate="2010-11-27T11:33:53.923" Title="Recurrence relation satisfied by $\lfloor(1+\sqrt{5})^n\rfloor$" Tags="&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12043" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T07:42:01.740" Score="1" ViewCount="0" Body="&lt;p&gt;You should parameterize your curve using polar coordinates:&#xA;write the typical point $(x,y)=(r(t)\cos\theta(t),r(t)\sin\theta(t))$&#xD;&#xA;where $r$ and $\theta$ are functions of $t$ on an interval $[a,b]$.&#xA;Also note that as the curve &quot;closes up&quot; then $r(b)=r(a)$&#xD;&#xA;and $\theta(b)=\theta(a)+2m\pi$ for some integer $m$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-27T07:42:01.740" />
  <row Id="12044" PostTypeId="2" ParentId="12042" CreationDate="2010-11-27T07:46:54.087" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a possible approach. If $x=1+\sqrt 5$, notice that $x^2=4+2x$, and you could use this to simplify higher powers: $(a+bx)x=4b+(a+2b)x$ leads to recurrence relations for $a_n$ and $b_n$ in the expression $x^n=a_n+b_nx$.  If you determine simple expressions for $a_n$ and $b_n$ and thus $x^n$, you can then find $[x^n]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to study the closely related &lt;a href=&quot;http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio&quot; rel=&quot;nofollow&quot;&gt;closed-form expression for the Fibonacci sequence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: In the original version I made a mistake in the equation for $x^2$.]&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-27T08:04:19.437" LastActivityDate="2010-11-27T08:04:19.437" />
  <row Id="12045" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T07:50:00.240" Score="0" ViewCount="0" Body="&lt;p&gt;First you need to prove that $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$ is the same as calculating the integral over a circle of radius '$r$' such that the circle of radius '$r$' lies completely within the domain enclosed by the curve $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you prove that the problem now becomes simple. All you need is to evaluate the integral over a circle. This can be done by parameterizing the circle as $x = r \cos(\theta)$ and $y = r \sin(\theta)$. Note that $dx = - r \sin(\theta)d\theta$ and $dy = r cos(\theta)d\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $x^2 + y^2 = r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral then becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{0}^{2 \pi} \frac{r\cos(\theta) \times r \cos(\theta) d\theta - r\sin(\theta) \times -r \sin(\theta) d\theta}{r^2} = \int_{0}^{2 \pi} d\theta =2\pi$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $$\oint_{L} \frac{xdy - ydx}{x^2 + y^2} = 2\pi$$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-11-27T07:50:00.240" />
  <row Id="12046" PostTypeId="2" ParentId="11953" CreationDate="2010-11-27T07:56:11.597" Score="1" ViewCount="0" Body="&lt;p&gt;This is not a complete answer. I'm assuming that you also have non-negativity constraints on all variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After looking at your question more carefully, I rearranged it to the following constraints:&#xA;$$\begin{align}&#xD;&#xA;y_{i+1} - y_i &amp;amp; \in [g_{i_{\min}}, g_{i_{\max}}], \\&#xD;&#xA;y_i &amp;amp; \le y_{i_{\max}} - h_i, \\&#xD;&#xA;y_{i+1} - y_i &amp;amp; \in \left(1 + \frac{c}{h_i+b}\right)^{4/3}[w_{i_{\min}}, w_{i_{\max}}], \\&#xD;&#xA;y_{i+1} - y_i &amp;amp; \ge R_{i_{\min}} \left(1 + \frac{c}{h_i+b}\right)^{4/3}\frac{1}{(h_i+a)^2},\end{align}$$&#xD;&#xA;where $c = b - a$, $w_{i_{*}} = v_{i_{*}}^{1/3}$, and $R_{i_{\min}} = Q_{i_{\min}}^{1/3}$. Hopefully the abuse of notation in the third equation is understandable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all the $y_i$ are on the left side of the constraints and the $h_i$ are on the right. This form should be easier to reason about. For example, I was initially concerned that the optimal $f_1$ might always be $0$. Note that all the right-hand sides get larger as the $h_i$ get smaller, so a solution with all $h_i = 0$ would only fail to be feasible if one of the lower bounds for $y_{i+1} - y_i$ became too large, i.e. incompatible with one of the upper bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since all the constraints except one involve only the &lt;em&gt;differences&lt;/em&gt; between the $y_i$, you can decrease all $y_i$ by the same amount and still have a feasible solution, as long as none of them fall below zero. So when you get an optimal solution for $f_1$, you can subtract $\min\{y_j\}$ from $y_i$ and get the solution that minimizes $f_2$ for fixed $h_i$. That's not the same thing as only fixing $\sum h_i$, but it's something...&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-27T07:56:11.597" CommentCount="2" />
  <row Id="12047" PostTypeId="1" AcceptedAnswerId="12056" CreationDate="2010-11-27T08:05:39.530" Score="3" ViewCount="347" Body="&lt;p&gt;How to prove the following trigonometric identities ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $\displaystyle \tan (\alpha) \cdot \tan(\beta) = 1 \text{ then } \alpha + \beta = \frac{\pi}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to prove it by using the the formula for $\tan(\alpha + \beta)$ but ain't it valid only when $\alpha + \beta \neq \frac{\pi}{2}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this one I tried substituting them with the sides of the triangle, but not successful to the final result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not my homework, I am trying to learn maths almost on my own, so ...&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-27T13:27:04.257" LastActivityDate="2010-12-01T15:06:28.400" Title="Proving basic/standard trigonometric identities" Tags="&lt;trigonometry&gt;&lt;elementary&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="12048" PostTypeId="2" ParentId="12047" CreationDate="2010-11-27T08:26:25.987" Score="1" ViewCount="0" Body="&lt;p&gt;1) tan(a)tan(b) = 1 is equivilant to sin(a)sin(b) = Cos(a)cos(b) **using the fact that tan is sine divided by cosine.  Next use &#xA;                    cos(a+b) = cos(a)cos(b) -sin(a)sin(b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies      sin(a)sin(b)  = cos(a)cos(b) -cos(a+b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this on the left hand side of ** to get &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                              cos(a+b) = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which implies a+b = Pi/2 or -Pi/2&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-27T08:26:25.987" CommentCount="5" />
  <row Id="12049" PostTypeId="2" ParentId="11669" CreationDate="2010-11-27T08:37:15.947" Score="-1" ViewCount="0" Body="&lt;p&gt;ALL comments: tl;dr.  But I read quite a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get you all straight...you are all attempting to figure out an algorithm that corresponds correctly with soundwaves and specifically with music?  If ANYONE finds a way to measure that, that would be HUGE.&lt;/p&gt;&#xA;" OwnerUserId="4132" LastActivityDate="2010-11-27T08:37:15.947" CommentCount="1" />
  <row Id="12050" PostTypeId="2" ParentId="12030" CreationDate="2010-11-27T09:14:57.473" Score="0" ViewCount="0" Body="&lt;p&gt;In the punctuated plane $\dot{\mathbb R}^2$ we have the arg-&quot;function&quot; taking values in $\mathbb R/(2\pi\mathbb Z)$. Its gradient is a well defined vector field, namely $\nabla$arg$=(-{y\over x^2+y^2}, {x\over x^2+y^2})$. It follows that  for any nice curve $L$, closed or not, simple or not (but not going through the origin), your integral sums up the ongoing argument increases/decreases along $L$. If $L$ is closed then these arg-increases/decreases sum up to $2\pi$ times the winding number of $L$ around the origin.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-11-27T09:14:57.473" />
  <row Id="12051" PostTypeId="2" ParentId="12047" CreationDate="2010-11-27T10:29:19.570" Score="2" ViewCount="0" Body="&lt;p&gt;For (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}= \frac{1 + \sin \theta}{\cos \theta}$$ $$ = \frac{(1 + \sin \theta)\cdot (1 - \sin \theta)}{\cos \theta \cdot(1 - \sin \theta)}\text{ [Multiplying both sides by } (1 - \sin \theta)\text{]} $$&#xD;&#xA;$$ = \frac{\cos^2 \theta}{ \cos \theta - \sin \theta \cdot \cos \theta} = \frac{1}{\sec \theta - \tan \theta} \text{ (Q.E.D) }$$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-27T10:42:26.570" LastActivityDate="2010-11-27T10:42:26.570" />
  <row Id="12052" PostTypeId="1" AcceptedAnswerId="12053" CreationDate="2010-11-27T10:38:02.127" Score="5" ViewCount="97" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the following series: $\sum_{n=1}^{\infty} (\sqrt[3]{n+1} - \sqrt[3]{n-1})^{\alpha }$ &#xA;where $\alpha \in \mathbb{R}$. Does the series converge or diverge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempts to solve the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\lim_{n\to\infty} (\sqrt[3]{n+1} - \sqrt[3]{n-1})^{\alpha } = 0$ - not helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Used the formula $a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ - not helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The ration test is not helpful either.&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2010-11-27T10:57:16.480" Title="Series converge or diverge" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12053" PostTypeId="2" ParentId="12052" CreationDate="2010-11-27T10:51:22.867" Score="5" ViewCount="0" Body="&lt;p&gt;The formula $a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ actually can be quite helpful.  Note that it implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{n+1}-\sqrt[3]{n-1}=\frac{2}{(n+1)^{2/3}+(n^2-1)^{1/3}+(n-1)^{2/3}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sufficiently large $n$ ($n\gt2$ will do for sure, and you can get tighter bounds by going further), that denominator is squeezed between $\frac{3}{2}n^{2/3}$ and $6n^{2/3}$, and therefore the convergence question is the same as for the series $\sum_{n=1}^\infty n^{-2\alpha/3}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-27T10:57:16.480" LastActivityDate="2010-11-27T10:57:16.480" CommentCount="2" />
  <row Id="12054" PostTypeId="1" AcceptedAnswerId="12061" CreationDate="2010-11-27T11:13:22.527" Score="5" ViewCount="194" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Lagrange's remainder, I have to prove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log(1+x) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n}, \; \forall |x| &amp;lt; 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not quite sure how to do this. I started with the Taylor series for $x_0 = 0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_0) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} \cdot x^n + r_n$, where $r_n$ is the remainder. Then, I used induction to prove that the n-th derivative of $\log(1+x)$ can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{(n)} = (-1)^{n+1} \cdot \frac{(n-1)!}{(1+x)^n}, \forall n \in \mathbb{N}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plugged this formula into the Taylor series for $\log(1+x)$ and ended up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x_0) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n} + r_n$, which already looked quite promising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the formula which I have to prove doesn't have that remainder $r_n$, I tried to show that $\lim_{n \to \infty} r_n = 0$, using Lagrange's remainder formula (for $x_0 = 0$ and $|x| &amp;lt; 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I basically showed that the formula was valid for $x \to x_0 = 0$. I also showed that the radius of convergence of this power series is $r = 1$, that is to say the power series converges $\forall |x| &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is bugging me, is the fact, that to my opinion, the formula is only valid for $x \to 0$. I mean sure, the radius of convergence is 1, but does this actually tell me that the formula is valid within $(-1,1)$? I've never done something like this before, thus the insecurity. I'd be delighted, if someone could help me out and tell me, whether the things I've shown are already sufficient or whether I still need to prove something.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-10T17:00:14.863" Title="How to use the Lagrange's remainder to prove that log(1+x) = sum(...)?" Tags="&lt;analysis&gt;&lt;taylor-expansion&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12055" PostTypeId="2" ParentId="12042" CreationDate="2010-11-27T11:33:53.923" Score="3" ViewCount="0" Body="&lt;p&gt;Do you have any reason to suspect $L(n)$ should satisfy a linear recurrence?  Here's a way you can prove that it doesn't satisfy a linear recurrence of depth 2 (which can be generalised to any depth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 1&lt;/em&gt;: Compute $L(n)$ for small $n$:  3, 10, 33, 109, 354, 1148, 3716, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 2&lt;/em&gt;: Assume $b L(n-2)+a L(n-1)=L(n)$ for some $a,b$.  Using the data from Step 1 we obtain the system of linear equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$3b+10a=33$,&lt;/li&gt;&#xA;&lt;li&gt;$10b+33a=109$,&lt;/li&gt;&#xA;&lt;li&gt;$33b+109a=354$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, we can keep going forever adding equations $109b+354a=1148$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Step 3&lt;/em&gt;: Solve the system of linear equations (or get your computer to do it for you (using e.g. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=3b%2B10a%3D33,+10b%2B33a%3D109,+33b%2B109a%3D354&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt;)).  In this case there are no solutions, so $L(n)$ does not satisfy a linear recurrence of depth 2.  If you feel that the starting point shouldn't be $L(1)$, you can use the same argument starting later in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I coded things correctly, I have checked that $L(n)$ doesn't satisfy a linear recurrence of depth 10 (or less).  [It's probably a good idea to check this yourself if you end up relying on this result.]  I also attempted to find a linear recurrence with polynomial coefficients for $L(n)$ to a limited extent (see &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt; about Sister Celine's Technique for more info).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you're allowed to use an auxiliary function, then let $s(1)=1+\sqrt{5}$ and for $n \geq 2$ let $s(n)=(1+\sqrt{5}) \cdot s(n-1)$.  Then $L(n)=\lfloor s(n) \rfloor$ for all $n \geq 1$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-27T11:33:53.923" CommentCount="2" />
  <row Id="12056" PostTypeId="2" ParentId="12047" CreationDate="2010-11-27T11:35:04.217" Score="4" ViewCount="0" Body="&lt;p&gt;Added (and corrected twice):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \alpha \tan \beta =1\Leftrightarrow \dfrac{\sin \alpha }{\cos \alpha }%&#xD;&#xA;\dfrac{\sin \beta }{\cos \beta }=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $\cos\alpha\cos\beta\ne 0$, gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin \alpha \sin \beta -\cos \alpha\cos \beta =0 \Leftrightarrow \cos (\alpha +\beta )=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha +\beta =\dfrac{\pi }{2}+n\pi,\qquad (\ast)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to which we still have to add the condition written above ($\cos\alpha\cos\beta\ne 0$), which means the constraint &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha,\beta\ne\dfrac{\pi}{2}+n\pi,\qquad (\ast\ast)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In the original equation $\tan \alpha \tan \beta =1$, neither $\alpha$ nor $\beta$ can be zero.  The combined conditions $(\ast)$ and $(\ast\ast)$ assures that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The identity $$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\qquad \theta \neq (2n+1)\dfrac{\pi}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is equivalent to $$\sin ^{2}\theta +\cos ^{2}\theta =1.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta \neq (2n+1)\dfrac{\pi }{2}\Leftrightarrow \sin \theta \neq \pm 1 \Leftrightarrow \dfrac{\pm1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }\neq 0\Leftrightarrow \pm\sec \theta -\tan \theta \neq 0,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\Leftrightarrow \left( \sec \theta +\tan \theta \right) \left( \sec \theta&#xD;&#xA;-\tan \theta \right) =1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Leftrightarrow \sec ^{2}\theta -\tan ^{2}\theta =1\Leftrightarrow \dfrac{1}{\cos ^{2}\theta }-\dfrac{\sin ^{2}\theta }{\cos^{2}\theta }=1\Leftrightarrow 1-\sin ^{2}\theta=\cos ^{2}\theta$$ $$\Leftrightarrow \sin ^{2}\theta +\cos ^{2}\theta =1.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T15:06:28.400" LastActivityDate="2010-12-01T15:06:28.400" CommentCount="5" />
  <row Id="12057" PostTypeId="1" AcceptedAnswerId="12059" CreationDate="2010-11-27T11:45:24.857" Score="0" ViewCount="138" Body="&lt;p&gt;Let $p,q$ belong to $\mathbb{N}$ and are relatively prime to each other. If $\alpha,\beta$ belong to $\mathbb{N}$, are also relatively prime to each other,then are $(p\beta+q\alpha)$ and $q\beta$ always relatively prime ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-27T12:29:03.593" LastActivityDate="2010-12-28T02:15:51.310" Title="a basic question on relatively prime numbers" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="12058" PostTypeId="1" CreationDate="2010-11-27T11:49:37.903" Score="1" ViewCount="41" Body="&lt;p&gt;Hi everyone,&#xA;I have a problem of commutative algebra. Take $A$ an excellent integrally closed noetherian ring with field of quotients $K$, $I\subset A$ a radical ideal such that $A$ is $I$-adically complete. Consider $G=Spec(A)\times \mathbb{G}_m^r$ and $Y\subset G(K)$ a subgroup isomorphic to $\mathbb{Z}^r$. Let $X$ the set of character of $\mathbb{G}_m^r$ and consider an homomorphism $f: Y\rightarrow X$ such that $X_{f(y)}(z)=X_{f(z)}(y)$ and $X_{f(y)}(y)\in I$ for every $y,z\in Y$ where $X_{a}$ is the element of $\Gamma(G,O_G)$ corresponding to $a\in X$.&#xA;The question is the following: why there exists only a finite set of minimal primes $p\subset A$ such that $ord_p X_{f(y)}(z)\neq 0$ for all $y,z\in Y$?&lt;/p&gt;&#xA;" OwnerUserId="4136" LastActivityDate="2010-11-27T11:49:37.903" Title="existence of finitely many non zero valuation" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="12059" PostTypeId="2" ParentId="12057" CreationDate="2010-11-27T11:51:38.610" Score="7" ViewCount="0" Body="&lt;p&gt;The answer is clearly no: take $p=\alpha$, $q=\beta$. Then $p\beta+q\alpha=2pq$ and $q\beta=pq$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however $\beta$ is co-prime to $q$, then the answer is equally clearly yes, since no divisor of $q$ divides $p\beta$ and no divisor of $\beta$ divides $q\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-27T11:51:38.610" CommentCount="0" />
  <row Id="12060" PostTypeId="2" ParentId="12057" CreationDate="2010-11-27T11:55:00.463" Score="2" ViewCount="0" Body="&lt;p&gt;No, e.g. $p=3,q=5,\alpha=3,\beta=5$ is a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T11:55:00.463" />
  <row Id="12061" PostTypeId="2" ParentId="12054" CreationDate="2010-11-27T11:56:16.257" Score="6" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$f(x_0) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n} + r_n$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That should say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\sum_{n=1}^k (-1)^{n+1} \cdot \frac{x^n}{n} + r_k(x),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_k$ is the error term of the $k^\text{th}$ partial sum.  You want to use estimates to show that the error term goes to $0$ as $k$ goes to $\infty$, which will justify convergence of the series to $f(x)=\log(1+x)$.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: I've struck through part of my answer that relied on a wrong estimate of the derivatives, as pointed out by Robert Pollack.  With the missing $k!$ term, the estimate only works on $[-\frac{1}{2},1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: To make this answer a little more useful, I decided to look up a correct method.  Spivak in his book &lt;em&gt;Calculus&lt;/em&gt; (3rd Edition, page 423) uses the formula &#xA;$$\frac{1}{1+t}=1-t+t^2-\cdots+(-1)^{n-1}t^{n-1}+\frac{(-1)^nt^n}{1+t}$$&#xD;&#xA;in order to write the remainder as $r_n(x)=(-1)^n\int_0^x\frac{t^n}{1+t}dt$.  The estimate $\int_0^x\frac{t^n}{t+1}dt\leq\int_0^xt^ndt=\frac{x^{n+1}}{n+1}$ holds when $x\geq0$, and the harder estimate &#xA;$\left|\int_0^x\frac{t^n}{1+t}\right|\leq\frac{|x|^{n+1}}{(1+x)(n+1)}$, when $-1\lt x\leq0$,&#xA;is given as Problem 11 on page 430.  Combining these, you can show that the sequence of remainders converges uniformly to $0$ on $[-r,1]$ for each $r\in(0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Lagrange's form of the error term can be used to do this.  The estimates, which follow from Taylor's theorem, are also found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_of_the_remainder&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.  In this case, if $0\lt r\lt 1$, then $|f^{k+1}(x)|\leq \frac{1}{(1-r)^{k+1}}$ whenever $x\geq-r$, so you have the estimate $|r_k(x)|\leq \frac{r^{k+1}}{(1-r)^{k+1}}\frac{1}{(k+1)!}$ for all $x$ in $(-r,r)$, which you can show goes to $0$ (because (k+1)! grows faster than the exponential function $\left(\frac{r}{(1-r)}\right)^{k+1}$), thus showing that the series converges uniformly to $\log(1+x)$ on $(-r,r)$.  Since $r$ was arbitrary, this shows that the series converges on $(-1,1)$, and the convergence is uniform on compact subintervals.&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-10T04:41:03.077" LastActivityDate="2010-12-10T04:41:03.077" CommunityOwnedDate="2010-12-10T04:41:03.077" CommentCount="6" />
  <row Id="12062" PostTypeId="1" CreationDate="2010-11-27T12:24:07.040" Score="1" ViewCount="131" Body="&lt;p&gt;With help from Maple, I got &#xA;$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2-(c-a)^2-(c-b)^2$$&#xD;&#xA;equal to&#xA;$$\frac{(c(x^3+y^3+z^3)+(a-c)(x^2y+y^2z+z^2x)+(b-c)(x^2z+y^2x+z^2y)-3(a+b-c)xyz)^2}{(x-y)^2(y-z)^2(x-z)^2}$$ which of course is $\ge 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with no help from a computer algebra, how would one prove:$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\ge (c-a)^2+(c-b)^2 ?$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T17:20:52.033" Title="How to prove this inequality without use of computers?" Tags="&lt;inequality&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="12063" PostTypeId="1" CreationDate="2010-11-27T12:41:52.823" Score="1" ViewCount="72" Body="&lt;p&gt;Recently, I consider a problem: Let $E/F$ is a Galois extension of number field, denote the idele group of $E$ (resp $F$) by $I_E$ (resp $I_F$). There is a homomorphism induced by norm map $N_{E/F}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N: I_E/E^{\ast} \to I_F/F^{\ast}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a finite abel extension $K/E$ ,from Global Class Field Theory,there is an isomorphism called Artin map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_K/E: I_E/E^{\ast} \text{Nm}(I_K) \to \text{Gal}(L/K)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a subset $U$ of $I_E/E^{\ast}$ s.t $A_K/E(U) = 1$, where $1$ is the identity element of $\text{Gal}(K/E)$, So i really want know is there a special relationship between $U$ and the set $\text{Ker}(I_E/E{\ast} \to I_F/F^{\ast})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4138" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-27T12:51:42.113" LastActivityDate="2010-11-27T12:51:42.113" Title="A subtle relationship from class field theory" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" CommentCount="1" />
  <row Id="12064" PostTypeId="2" ParentId="9274" CreationDate="2010-11-27T13:01:31.827" Score="0" ViewCount="0" Body="&lt;p&gt;(deleted as duplicate)&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-11-28T09:19:44.817" LastActivityDate="2010-11-28T09:19:44.817" CommentCount="1" />
  <row Id="12065" PostTypeId="1" CreationDate="2010-11-27T14:08:20.037" Score="3" ViewCount="201" Body="&lt;p&gt;How can we prove that the product of n consecutive integers is divisible by n factorial?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note: In this &lt;a href=&quot;http://math.stackexchange.com/questions/12067/the-product-of-n-consecutive-integers-is-divisible-by-n-without-using-the-prope&quot;&gt;subsequent question&lt;/a&gt; and the comments here the OP has clarified that he seeks a proof that &quot;does not use the properties of binomial coefficients&quot;. Please post answers in said newer thread so that this incorrectly-posed question may be closed as a duplicate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T00:12:14.287" LastActivityDate="2010-11-28T00:12:14.287" Title="The product of n consecutive integers is divisible by n factorial" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="12" />
  <row Id="12066" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T14:18:45.863" Score="4" ViewCount="0" Body="&lt;p&gt;This is almost immediate from the fact that the binomial coefficient $$\binom{k+n}{k}$$ is an integer. Just write the product $(k+1) \cdots (k+n)$ accordingly and you'll have your answer.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-11-27T14:18:45.863" />
  <row Id="12067" PostTypeId="1" CreationDate="2010-11-27T14:33:38.090" Score="0" ViewCount="174" Body="&lt;p&gt;How can we prove, without using the properties of binomial coefficients, the product of n consecutive integers is divisible by n factorial?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastActivityDate="2010-11-27T20:59:43.410" Title="The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="17" />
  <row Id="12068" PostTypeId="1" AcceptedAnswerId="12080" CreationDate="2010-11-27T14:49:08.863" Score="2" ViewCount="131" Body="&lt;p&gt;If $x_i$ has exponentially distributed $(i=1,...,n)$ with parameter $\lambda$, what is below expectation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E\left\{ \left(\sum_{i=1}^n {x_i} \right)^2 \right\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in terms of $n$ and $\lambda$ and possibly other constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; a copy of this question is also asked on &lt;a href=&quot;http://stats.stackexchange.com/q/4959/2148&quot;&gt;http://stats.stackexchange.com/q/4959/2148&lt;/a&gt; that is still unanswered.&lt;/p&gt;&#xA;" OwnerUserId="4051" LastEditorUserId="4051" LastEditorDisplayName="" LastEditDate="2010-11-27T14:55:23.383" LastActivityDate="2010-11-27T16:57:14.777" Title="expectation of $E\left\{ \left(\sum_{i=1}^n {x_i} \right)^2 \right\}$" Tags="&lt;statistics&gt;&lt;expectation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12069" PostTypeId="1" AcceptedAnswerId="12071" CreationDate="2010-11-27T14:58:54.910" Score="1" ViewCount="76" Body="&lt;p&gt;For example, if $f(x) = \sin x$ and $x$ is uniformly distributed on $[0, \pi]$, how is the equation found that satisfies the probability distribution function of $f(x)$? I imagine the distribution function will be greater when the derivative of $f(x)$ is closer to zero, but this is just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize if this question is vague or not advanced enough, but I can't find the answer anywhere.&lt;/p&gt;&#xA;" OwnerUserId="227" LastActivityDate="2010-12-29T18:03:44.410" Title="Given a function $f(x)$ where $x$ is uniformly distributed between $a$ and $b$, how do I find the probability density function of $f$?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="2" />
  <row Id="12070" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T15:13:55.950" Score="0" ViewCount="0" Body="&lt;p&gt;The identity below shows that the problem is equivalent to the fact that binomial coefficients are integral - for which various proofs are known, e.g. using their recursion, or their well-known combinatorial  interpretation, or their minimality in terms of prime divisors - see this &lt;a href=&quot;http://math.stackexchange.com/questions/11601/proof-that-a-combination-is-an-integer&quot;&gt;prior question&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{n\:(n-1)\quad\quad\:\cdots\:\quad\quad 1\quad\quad}$$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-27T15:13:55.950" />
  <row Id="12071" PostTypeId="2" ParentId="12069" CreationDate="2010-11-27T15:30:32.713" Score="2" ViewCount="0" Body="&lt;p&gt;Note that $\sin(x)$ increases from $x = 0$ to $x = {\pi \over 2}$, then decreases from ${\pi \over 2}$ to $\pi$, in a way symmetric about ${\pi \over 2}$. So for a given $0 \leq \alpha \leq 1$, the $x \in [0,\pi]$ for which $\sin(x) \leq \alpha$ consists of two segments, $[0,\beta]$ and $[\pi - \beta, \pi]$, where $\beta$ is the number for which $\sin(\beta) = \alpha$. In other words $\beta = \arcsin(\alpha)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x$ is uniformly distributed on $[0,\pi]$, the probability $x$ is in $[0,\beta]$ is ${\beta \over \pi}$, and the probability $x$ is in $[\pi - \beta, \pi]$ is also ${\beta \over \pi}$. So the chance that $x$ is in one of these two segments is $2{\beta \over \pi}$. This means the probability $\sin(x) \leq \alpha$ is $2{\beta \over \pi}$, or ${2 \over \pi} \arcsin(\alpha)$. Thus this gives the distribution function of $\sin(x)$. The density function is obtained by differentiating with respect to $\alpha$; the result is ${2 \over \pi \sqrt{1 - \alpha^2}}$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T15:30:32.713" CommentCount="1" />
  <row Id="12072" PostTypeId="2" ParentId="12067" CreationDate="2010-11-27T15:33:11.357" Score="0" ViewCount="0" Body="&lt;p&gt;Since this statement is equivalent to the fact that binomial coefficients are integral it is a bit tricky to make precise the notion of a proof that does not &quot;use properties of binomial coefficients&quot;. I will interpret this proviso to mean that the proof is not essentially combinatorial, i.e. it does not employ properties that are immediate from the combinatorial interpretation of binomial coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my &lt;a href=&quot;http://math.stackexchange.com/questions/2158/division-of-factorials/2192#2192&quot;&gt;post here&lt;/a&gt; you'll find a purely arithmetical proof that $\rm\: n!\: $ divides the product of $\rm\:n\:$ consecutive integers. The proof shows how to rewrite the associated fraction as a product of fractions whose denominators are all coprime to any given prime $\rm\:p\:$. This implies that no primes divide the denominator (when written in lowest terms), hence the fraction is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key property that lies at the heart of this proof is that, among all products of $\rm\: n\:$ consecutive integers, $\rm\ n!\ $ has the least possible power of $\rm\:p\:$ dividing it - for all primes $\rm\:p\:$. Thus $\rm\ n!\ $ divides every product of $\rm\:n\:$ consecutive integers, since it has a smaller power of every prime divisor. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-27T15:45:52.353" LastActivityDate="2010-11-27T15:45:52.353" CommentCount="1" />
  <row Id="12073" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T15:37:57.733" Score="2" ViewCount="0" Body="&lt;p&gt;You might be interested in this blog post of Timothy Gowers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gowers.wordpress.com/2010/09/18/are-these-the-same-proof/&quot; rel=&quot;nofollow&quot;&gt;http://gowers.wordpress.com/2010/09/18/are-these-the-same-proof/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3736" LastActivityDate="2010-11-27T15:37:57.733" CommunityOwnedDate="2010-11-27T15:37:57.733" CommentCount="1" />
  <row Id="12074" PostTypeId="2" ParentId="11669" CreationDate="2010-11-27T16:03:00.670" Score="1" ViewCount="0" Body="&lt;p&gt;I think in a tiny tiny nutshell... the reason for the 12-division is because a very practical solution for Western music, and the layout of black/white &quot;evolved&quot; into this form because it lacked re-engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no particularly &quot;mathemagical&quot; thing about it. In other words... square two: it's an arbitrary choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for something that uses 12-divided octave as a practical solution and is engineered for facility, check out the layout of the Russian Bayan (accordion). It's pretty awesome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for something that is engineered for facility but does not divide the octave into 12 parts, your common fretless string instruments are good examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, all I've said has been mentioned above. Just beware of the overtly &quot;mathemagical&quot; ones, they don't say much about the music but rather put it in a fancy straightjacket.&lt;/p&gt;&#xA;" OwnerUserId="4142" LastActivityDate="2010-11-27T16:03:00.670" />
  <row Id="12075" PostTypeId="2" ParentId="11551" CreationDate="2010-11-27T16:08:52.553" Score="0" ViewCount="0" Body="&lt;p&gt;I can't find any reference for this, so please check carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is true that the sequential closure of a set is equal to the union of the closures of its countable subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would follow from this, that the closure of a countable set is the same as its sequential closure, regardless of whether the containing space is first countable.&lt;/p&gt;&#xA;" OwnerUserId="3457" LastActivityDate="2010-11-27T16:08:52.553" CommentCount="4" />
  <row Id="12076" PostTypeId="2" ParentId="12067" CreationDate="2010-11-27T16:10:22.777" Score="3" ViewCount="0" Body="&lt;p&gt;There is an alternate proof at the following link which uses the result about the highest power of a prime dividing a factorial (scroll down for the proof)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://2000clicks.com/MathHelp/BasicFactorialConsecutiveIntegerProducts.aspx&quot; rel=&quot;nofollow&quot;&gt;http://2000clicks.com/MathHelp/BasicFactorialConsecutiveIntegerProducts.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-27T16:10:22.777" CommentCount="1" />
  <row Id="12077" PostTypeId="1" AcceptedAnswerId="12079" CreationDate="2010-11-27T16:27:24.993" Score="1" ViewCount="58" Body="&lt;p&gt;Let $(X,d)$ be a metric space and $S \subset X$. Show that $d_S(x):=\text{inf}\{d(x,s): s \in S\}=0 \Leftrightarrow x \in \overline S .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: $\overline S$ is the closure of S. Maybe you can use that a closed set is also closed for sequences in the set? I think the difficult part is when S is open, otherwise its trivial as the closure would be equal to S.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-11-27T16:39:37.637" Title="Closure of a subset in a metric space" Tags="&lt;topology&gt;&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12078" PostTypeId="2" ParentId="12077" CreationDate="2010-11-27T16:36:58.123" Score="1" ViewCount="0" Body="&lt;p&gt;If $x \in \overline{S}$, then there is a sequence $(s_n)$ in $S$ converging to $x$ (e.g. take $s_n$ to be some element in $S \cap B(x;\tfrac{1}{n})$). Then $d_S(x) \leq \inf\lbrace d(x,s_n) \;\vert\; n \in \mathbb{N}\rbrace = 0$ (why?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other directtion, we prove the contrapositive: If $x \notin \overline{S}$, then there is some ball around $x$ disjoint from $S$. This gives that the infimum in $d_S(x)$ must be greater than $0$ (why?).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-11-27T16:36:58.123" CommentCount="2" />
  <row Id="12079" PostTypeId="2" ParentId="12077" CreationDate="2010-11-27T16:39:37.637" Score="4" ViewCount="0" Body="&lt;p&gt;If $x\notin \overline{S}$, then there exists $r&amp;gt;0$ such that $B_r(x)\cap S=\phi$. It follows that $d(x,s)\ge r&amp;gt;0$ for all $s\in S$ and hence $d_S(x)&amp;gt;0$. &#xA;And these steps can be reversed, i.e. the steps above are actually &quot;if and only if&quot;. So you get the proof.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T16:39:37.637" CommentCount="2" />
  <row Id="12080" PostTypeId="2" ParentId="12068" CreationDate="2010-11-27T16:43:33.123" Score="4" ViewCount="0" Body="&lt;p&gt;In what follows, $x_j$ are assumed to be independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[x_0^2] = \int_0^{\infty} \lambda t^2 e^{-\lambda t}  dt = \frac{2}{\lambda^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[x_0 x_1] = \int_0^{\infty} \int_0^{\infty} \lambda^2 t_0 t_1 e^{-\lambda t_0} e^{-\lambda t_1 } dt_0 dt_1 = \frac{1}{\lambda^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E[ (\sum_{j=0}^{n-1} x_j)^2 ] = n E[ x_0^2] + n(n-1) E[x_0 x_1] = \frac{ n^2 + n }{ \lambda^2 } $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4143" LastEditorUserId="4143" LastEditorDisplayName="" LastEditDate="2010-11-27T16:57:14.777" LastActivityDate="2010-11-27T16:57:14.777" CommentCount="3" />
  <row Id="12081" PostTypeId="1" CreationDate="2010-11-27T16:59:19.983" Score="1" ViewCount="67" Body="&lt;p&gt;In my website, users can either &quot;like&quot; or &quot;dislike&quot; a posted comment. I want to put a link to sort comments by liking such that the most liked ones becomes on top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I cannot just sort by the number of likes only. I have to subtract the dislikes. But what if the difference (likes - dislikes) are the same. e.g. 10 - 8 = 2 and 5 - 3 = 2. I think in this case, the comment of 10 likes and 8 dislikes has to come before the 5 likes and 3 dislikes comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there an equation that you feed it the number of likes and number of dislikes and then it gives you a meaningful rating number that I can sort with?&lt;/p&gt;&#xA;" OwnerUserId="4145" LastActivityDate="2010-11-27T23:50:19.010" Title="Equation to calculate best liked comments" Tags="&lt;sorting&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12082" PostTypeId="1" CreationDate="2010-11-27T17:00:27.580" Score="5" ViewCount="145" Body="&lt;p&gt;If there is a field $F$ that is a field reduction of the real numbers, that is $F(a)=\mathbb{R}$ for some $a$, let's also denote this $F=\mathbb{R}(\setminus a)$, then given $x \in \mathbb{R}$ is there a general method to determine whether $x$ is in $F$ or $x$ is in $\mathbb{R}\setminus F$ ?&lt;/p&gt;&#xA;" OwnerUserId="4144" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:20:59.577" LastActivityDate="2010-11-28T04:20:59.577" Title="Field reductions" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12083" PostTypeId="2" ParentId="10937" CreationDate="2010-11-27T17:01:22.133" Score="1" ViewCount="0" Body="&lt;p&gt;I guess your question is: Why is $E(X)=\int_0^\infty 1-F(x) dx$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is as follows: By definition $$E(X)=\int_0^\infty tf(t) dt.$$&#xD;&#xA;Now \begin{eqnarray*} &amp;amp; &amp;amp; \int_0^\infty tf(t) dt\&#xA;&amp;amp;=&amp;amp;\int_0^\infty f(t)\int_0^t 1 dx dt\&#xA;&amp;amp;=&amp;amp;\int_0^\infty\int_0^t f(t) dx dt\&#xA;&amp;amp;=&amp;amp;\int_0^\infty\int_x^\infty f(t) dt dx \&#xA;&amp;amp;=&amp;amp;\int_0^\infty 1-F(x) dx.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-27T17:01:22.133" />
  <row Id="12084" PostTypeId="1" AcceptedAnswerId="12086" CreationDate="2010-11-27T17:17:57.340" Score="0" ViewCount="76" Body="&lt;p&gt;The empty set is a member of $P({a,b}) \times P({p,q})$. True or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct was false, since the empty set is a member of each power set individually, but when multiplied together, you get {0,0}, which I'm not sure represents the empty set. But my counter argument is that the empty set is a member of the power set of anything, right?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T21:30:59.580" LastActivityDate="2010-11-27T21:30:59.580" Title="Empty set and power sets" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="12085" PostTypeId="2" ParentId="12062" CreationDate="2010-11-27T17:20:52.033" Score="3" ViewCount="0" Body="&lt;p&gt;This problem may be amenable to the &quot;Purkiss Principle&quot;. I'll leave it as an exercise for you to determine if it can be applied here. But even if not you should read the following beautiful article on it by Wm. Waterhouse &lt;a href=&quot;http://mathdl.maa.org/mathDL/22/?pa=content&amp;amp;sa=viewDocument&amp;amp;nodeId=2941&quot; rel=&quot;nofollow&quot;&gt;Do Symmetric Problems Have Symmetric Solutions?&lt;/a&gt; I recall thinking that this was one of the most beautiful Monthly articles that I ever read as an undergraduate. Apparently others felt similarly since it won a prestigious Lester R. Ford award for expository excellence.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-11-27T17:20:52.033" CommentCount="1" />
  <row Id="12086" PostTypeId="2" ParentId="12084" CreationDate="2010-11-27T17:21:16.683" Score="7" ViewCount="0" Body="&lt;p&gt;Your first instinct was right: $P(a,b) \times P(p,q)$ contains $(\emptyset, \emptyset)$, but not $\emptyset$. $(\emptyset, \emptyset)$ is not the empty set, so $P(a,b) \times P(p,q)$ does not contain the empty set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, every powerset contains the empty set, but $P(a,b) \times P(p,q)$ is not a powerset, it's the cartesian product of two powersets.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-27T17:21:16.683" />
  <row Id="12087" PostTypeId="2" ParentId="12082" CreationDate="2010-11-27T17:37:59.340" Score="7" ViewCount="0" Body="&lt;p&gt;In fact there are no nontrivial &quot;field reductions&quot; of $\mathbb{R}$: if $\mathbb{R} = F(a)$, then $a \in F$ and $F = \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1: $a$ is algebraic over $F$, hence $F(a) = F[a]$, so $d = [F[a]:F]$ is finite.  Then $[\mathbb{C}:F] = 2d$, i.e., &quot;the&quot; algebraic closure of $F$ has finite degree over $F$.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Theorem: Let $K$ be a field and $\overline{K}$ any algebraic closure.  If $[\overline{K}:K] = d$ is finite, then $d = 1$ or $d = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: This is essentially the Grand Artin-Schreier Theorem (see e.g. Section 12.5 of &lt;a href=&quot;http://math.uga.edu/~pete/FieldTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/FieldTheory.pdf&lt;/a&gt; for a proof of that.)  Namely, Artin-Schreier says that if $\overline{K}/K$ is a finite &lt;strong&gt;Galois&lt;/strong&gt; extension, then the degree is either $1$ or $2$.  Certainly $\overline{K}/K$ is normal.  And for the purposes of this question we are in characteristic zero, so everything is separable.  Therefore $\overline{K}/K$ is Galois.  (But here is a proof of the separability in the general case: if $\overline{K}/K$ is not separable, then there is a nontrivial subextension $L$ such that $[\overline{K}:L] = p^n$ and $\overline{K} = L(a^{p^{-n}})$.  But this is impossible: the polynomial $t^p - a$ is irreducible over $L$ iff all of the polynomials $t^{p^n} - a$ are irreducible over $L$.  So $\overline{K}/K$ is separable.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thus we must have $d = 1$, i.e., $F = \mathbb{R}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: $a$ is transcendental over $F$.  But then the field $F(a)$ is isomorphic to the rational function field $F(t)$.  Such a field cannot be isomorphic to $\mathbb{R}$, because it admits finite extensions of degree $n$ for all $n \in \mathbb{Z}^+$, e.g. &#xA;$F(t^{\frac{1}{n}})$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-27T18:03:03.437" LastActivityDate="2010-11-27T18:03:03.437" CommentCount="18" />
  <row Id="12088" PostTypeId="1" AcceptedAnswerId="12089" CreationDate="2010-11-27T17:52:30.617" Score="2" ViewCount="92" Body="&lt;p&gt;No subset of the real numbers is countable. True or false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_number#Uses&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; for real numbers, I'm not really clear on the answer to this. They use the words computable and countable, and I'm not sure of the difference. Also, I dont think I fully understand the term countable. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;definition&lt;/a&gt; states that a set is countable if it has the same cardinality as some subset of the natural numbers, but I'm not sure what that means. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T21:30:29.440" LastActivityDate="2010-11-27T21:30:29.440" Title="Real numbers and countability" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12089" PostTypeId="2" ParentId="12088" CreationDate="2010-11-27T17:54:52.020" Score="4" ViewCount="0" Body="&lt;p&gt;A set is said to be countable if there is bijection between a  subset of natural numbers (or even integers) and that set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Countable_set&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Countable_set&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cantor's diagonalization argument shows that the set of reals is uncountable. So, there are uncountably many subsets of real numbers that are countable. For example the set consisting of any single real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computability has nothing to do with counting the number of elements in a set. This is related to the idea of computable functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Computable_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Computable_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A real number is computable if there is finite algorithm that can yield it's digits to any desired level of accuracy.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-27T18:29:27.190" LastActivityDate="2010-11-27T18:29:27.190" CommentCount="4" />
  <row Id="12090" PostTypeId="2" ParentId="11434" CreationDate="2010-11-27T18:02:32.883" Score="0" ViewCount="0" Body="&lt;p&gt;I'd look at it this way:&lt;br&gt;&#xA;$ f(n)  -f(n-1) = (n^2 + 3n) - (n^2+n-2) = +2n + 2 $ &lt;br&gt;&#xA;$ f(n-1)-f(n-2) = (n^2 + n-2) - (n^2-n-2)= +2n  $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by the first differences the square terms vanish and by the second order differences also the linear terms vanish, so only a constant term remains if second order difference is used, so we can leave this as    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (f(n)-f(n-1)) - (f(n-1)-f(n-2)) = 1*f(n) - 2*f(n-1) + 1*f(n-2)  = 2 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally get by this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(n) = 2*f(n-1)-f(n-2) + 2 $ &lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2010-11-27T18:02:32.883" />
  <row Id="12091" PostTypeId="1" CreationDate="2010-11-27T18:11:17.953" Score="2" ViewCount="153" Body="&lt;p&gt;well i was wondering if, there s a way in which one could compare these two theories on the ground that which one will provide more accurate answers to a given problem. &#xA;Fuzzy logic always focuses on analysing between 0 and 1 and logic only dolls down to 0 and 1 , so does that imply that fuzzy logic provides us with more insight into a given problem or creates more parameters than necessary and compounds the problem ?&lt;/p&gt;&#xA;" OwnerUserId="3814" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-20T23:37:32.260" LastActivityDate="2010-12-20T23:37:32.260" Title="Fuzzy Logic vs Logic " Tags="&lt;logic&gt;&lt;fuzzy-logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12092" PostTypeId="2" ParentId="11856" CreationDate="2010-11-27T18:13:57.443" Score="1" ViewCount="0" Body="&lt;p&gt;There seems to be fair bit of confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of, there are different notions of &quot;cone&quot;. The standard one refers to a cone on a topological space (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cone_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;wiki article&lt;/a&gt;), and is defined to be  $CX = (X \times [0,1])/(X \times \{0\})$. This can, in particular, be applied to manifolds. What one gets is a topological space which, in general, is not a manifold, but can, in some special cases, be given a structure of a smooth manifold (see for example the discussion &lt;a href=&quot;http://math.stackexchange.com/questions/7175/topology-of-a-cone-of-rp2&quot;&gt;here&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if one removes the &quot;cone point&quot; $X\times \{0\}$, what remains is just the product  $X \times (0,1]$, which, as any product of a manifold and a manifold with boundary, is a manifold with boundary. The boundary is $X\times \{1\}$, which can be identified with $X$. This is the kind of cone one talks about in &lt;a href=&quot;http://en.wikipedia.org/wiki/Sasakian_manifold&quot; rel=&quot;nofollow&quot;&gt;Sasakian geometry&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a completely separate issue, but the distinction between smooth and C^1 manifolds is largely nonexistent, due to a theorem of &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_structure#Existence_and_uniqueness_theorems&quot; rel=&quot;nofollow&quot;&gt;Whitney&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Sasaki manifolds, those are very special types of Riemannian manifolds. As the wiki referenced above explains, one way to define when a given Riemannian manifold $X$ is Sasaki, is to use the Riemannian metric on $X$ to put a metric  on the cone, and see if the resulting metric makes the cone &lt;a href=&quot;http://en.wikipedia.org/wiki/K%C3%A4hler_manifold&quot; rel=&quot;nofollow&quot;&gt;Kahler&lt;/a&gt;. (The article uses $X \times (0, \infty)$, which is diffeomorphic to the interior of the manifold we have, so this amounts to the same thing.) Usually it will not (Kahler metrics are quite special), but sometimes it will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is done this way because Kahler manifolds are more well known - they appear as projective or affine complex varieties. The meaning of this construction is that Sasaki manifolds are boundaries of (non-compact) Kahler manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with these things sorted out, one can actually ask how does one put the Riemannian metric on the cone $X \times (0, \infty)$. The formula is in the wiki article - it calls it the &quot;cone metric&quot;. If it is unclear what it means, please ask - here as a comment, or elsewhere.  &lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-27T18:13:57.443" CommentCount="6" />
  <row Id="12093" PostTypeId="1" AcceptedAnswerId="12094" CreationDate="2010-11-27T18:33:02.860" Score="2" ViewCount="66" Body="&lt;p&gt;Imagine a two-dimensional grid consisting of 20 points along the x-axis and&#xA;10 points along the y-axis. Suppose the origin (0,0) is in the bottom-left corner and the&#xA;point (20,10) is the top-right corner. A path on the grid consists of a series of moves in&#xA;which each move is either one unit to the right or one unit up. Diagonal moves are not&#xA;allowed. How many diﬀerent ways are there to construct a path starting at (0,0) and&#xA;ending at (20,10)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little stuck on this one. I feel I'm headed towards the right direction, but I'm not sure if I'm doing this right. For every move, there are 2 possible choices. If we want to get to (20,10), then there are 200 points from the origin to this point. And I think order matters here, so we would use permutations, so I come up with 200 P 2, which is 39,800.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-11-27T18:35:26.780" LastActivityDate="2010-11-27T23:16:53.997" Title="Counting number of moves on a grid" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12094" PostTypeId="2" ParentId="12093" CreationDate="2010-11-27T18:36:29.087" Score="5" ViewCount="0" Body="&lt;p&gt;You have 30 moves to make, 10 of which must be up and 20 of which must be right.  How many ways to pick which are the 10?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-27T18:36:29.087" CommentCount="6" />
  <row Id="12095" PostTypeId="2" ParentId="12081" CreationDate="2010-11-27T18:49:49.670" Score="1" ViewCount="0" Body="&lt;p&gt;You have to define what criterion will say that a pair $(L_1,D_1)$ for comment 1 is better than $(L_2,D_2)$ for comment 2.  One way is just to subtract, so $(L_1,D_1) \geq (L_2,D_2)$ if $L_1-D_1 \geq L_2-D_2$.  But on stackexchange there are many more upvotes than downvotes, so maybe you want $(L_1,D_1) \geq (L_2,D_2)$ if $L_1-10*D_1 \geq L_2-10*D_2$ or some such.  Maybe you want to compare on $\frac{L-D}{L+D}$.  There are many choices, and you need to consider your audience and their behavior to select one.  Any such function will map (L,D) to some number, which you can then sort.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-27T18:49:49.670" />
  <row Id="12096" PostTypeId="2" ParentId="12093" CreationDate="2010-11-27T19:06:30.277" Score="3" ViewCount="0" Body="&lt;p&gt;There are two ways of doing this. One is Ross Millikan's: you will make ten &quot;up&quot; moves, and 20 &quot;right&quot; moves; the only question is which order you make them in. Imagine placing the &quot;right&quot; moves on a row; now you need to decide where to do the &quot;up&quot; moves: you do so by inserting them &quot;in between&quot; (or before, or after) the &quot;right&quot; moves. So you need to choose ten places to put &quot;up&quot; moves: there are 21 locations for them (nineteeen in between the &quot;right&quot; moves, one before all of them, one after), and you are allowed to choose the same location more than once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a combinations-with-repetitions: the formula is $\binom{n+r-1}{r}$, where you have $n$ possibilities, and must make $r$ choices with repetitions allowed. In this case, $n=21$, $r=10$, so you get $\binom{30}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another way of doing it, which is more graphical. I'll do it with a 4 by 3 array so you see how it works. You have this array:&#xA;$$\begin{array}{cccc}&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Now, you start at the bottom left, so there is only one way to get there; we put a $1$ next to it.&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Then, you can go either up or right; there is only one way to get to those points (via the first move); we put a $1$ next to them:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot\;1 &amp;amp; \cdot &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Now: to get to $(1,1)$, you can either get to it from $(1,0)$ or from $(0,1)$; since there is only one way to get to each of those, there are two ways to get to $(1,1)$. On the other hand, only one way to get to $(2,0)$ or to $(0,2)$:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot\;1 &amp;amp; \cdot &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;2 &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Next: to get to $(1,2)$, you can arrive either from $(0,2)$ (one way of being there), or from $(1,1)$ (two ways of getting &lt;em&gt;there&lt;/em&gt;); so in total, three ways. Likewise, you have three ways to get to $(2,1)$, because you can either go up from $(2,0)$, and there is only one way to do all of that, or you can go right from $(1,1)$ (and there are two ways of doing that, corresponding to the two ways there are to get &lt;em&gt;to&lt;/em&gt; $(1,1)$; so we have:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot \;1 &amp;amp;\cdot\;3 &amp;amp; \cdot &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;2 &amp;amp; \cdot\;3 &amp;amp; \cdot\\&#xD;&#xA;\cdot\;1&amp;amp; \cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot&#xD;&#xA;\end{array}$$&#xD;&#xA;Continuing this way, we get:&#xA;$$\begin{array}{llll}&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;3 &amp;amp; \cdot\;6 &amp;amp; \cdot\;10\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;2 &amp;amp; \cdot\;3 &amp;amp; \cdot\;4\\&#xD;&#xA;\cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot\;1 &amp;amp; \cdot\;1&#xD;&#xA;\end{array}$$&#xD;&#xA;So there are $10$ ways to get to the top right corner in the 4 by 3 case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may even recognize that these numbers are just Pascal's triangle lying on its side! Well, there is a combinatorial formula for the entries of Pascal's triangle: the $r$th entry in the $m$th row corresponds to the coefficient of $a^{m-r}b^{r-1}$ in the binomial expansion of $(a+b)^{m-1}$, so it equals $\binom{m-1}{r-1}$. To figure out the entry that corresponds to the top right corner, note that you go &quot;down&quot; one row for each position on the $x$-axis, and another one for each step up. So here we have gone to the 4th row on the horizontal steps, and then to the 6th row on the by going up twice. Each step up is a move &quot;right&quot; on the row. So with a 4 by 3, we are in row $4+(3-1)=6$, and we are in position $3$ of that row. According to the formula above, that corresponds to $\binom{4+2-1}{3-1}=\binom{5}{2}$. This corresponds to the need to make $3$ moves right and two moves up, so we need choose where to place the two up moves among the three right moves; there are four places to put them in (before the three, after the three, or in the two spaces in between), so the formula I gave above gives this answer as well.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:16:53.997" LastActivityDate="2010-11-27T23:16:53.997" CommentCount="1" />
  <row Id="12097" PostTypeId="1" AcceptedAnswerId="12272" CreationDate="2010-11-27T19:08:33.473" Score="5" ViewCount="207" Body="&lt;p&gt;Original post on Mathoverflow &lt;a href=&quot;http://mathoverflow.net/questions/47487/probability-of-the-maximum-levy-stable-random-variable-in-a-list-being-greater&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a list of identical and independently distributed &lt;a href=&quot;http://en.wikipedia.org/wiki/Stable_distribution&quot; rel=&quot;nofollow&quot;&gt;Levy Stable&lt;/a&gt; random variables, $(X_0, X_1, \dots, X_{n-1})$, what is the is the probability that the maximum exceeds the sum of the rest?  i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = \text{Max}(X_0, X_1, \dots, X_{n-1}) $$&#xD;&#xA;$$ \text{Pr}( M &amp;gt; \sum_{j=0}^{n-1} X_j - M ) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, in &lt;a href=&quot;http://academic2.american.edu/~jpnolan/stable/chap1.pdf&quot; rel=&quot;nofollow&quot;&gt;Nolan&lt;/a&gt;'s notation, $X_j \in S(\alpha, \beta=1, \gamma, \delta=0 ; 0)$, where $\alpha$ is the critical exponent, $\beta$ is the skew, $\gamma$ is the scale parameter and $\delta$ is the shift.  For simplicity, I have taken the skew parameter, $\beta$, to be 1 (maximally skewed to the right) and $\delta=0$ so everything has its mode centered in an interval near 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From numerical simulations, it appears that for the region of $0 &amp;lt; \alpha &amp;lt; 1$, the probability converges to a constant, irregardless of $n$ or $\gamma$.  Below is a plot of this region for $n=500$, $0&amp;lt; \alpha &amp;lt; 1$, where each point represents the result of 10,000 random draws.  The graph looks exactly the same for $n=100, 200, 300$ and $400$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qX5W7.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $1 &amp;lt; \alpha &amp;lt; 2$ it appears to go as $O(1/n^{\alpha - 1})$ (maybe?) irregardless of $n$ or $\gamma$.  Below is a plot of the probability for $\alpha \in (1.125, 1.3125)$ as a function of $n$.  Note that it is a log-log plot and I have provided the graphs $1/x^{.125}$ and $1/x^{.3125}$ for reference.  It's hard to tell from the graph unless you line them up, but the fit for each is a bit off, and it appears as if the (log-log) slope of the actual data is steeper than my guess for each.  Each point represents 10,000 iterations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/niTeg.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\alpha=1$ it's not clear (to me) what's going on, but it appears to be a decreasing function dependent on $n$ and $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried making a heuristic argument to the in the form of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Pr}( M &amp;gt; \sum_{j=0}^{n-1} X_j - M) \le n \text{Pr}( X_0 - \sum_{j=1}^{n-1} X_j &amp;gt; 0 )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using formula's provided by Nolan (pg. 27) for the parameters of the implicit r.v. $ U = X_0 - \sum_{j=1}^{n-1} X_j$ combined with the tail approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Pr}( X &amp;gt; x ) \sim \gamma^{\alpha} c_{\alpha} ( 1 + \beta ) x^{-\alpha} $$&#xD;&#xA;$$ c_{\alpha} = \sin( \pi \alpha / 2) \Gamma(\alpha) / \pi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this leaves me nervous and a bit unsatisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for comparison, if $X_j$ were taken to be uniform r.v.'s on the unit interval, this function would decrease exponentially quickly.  I imagine similar results hold were the $X_j$'s Gaussian, though any clarification on that point would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting closed form solutions for this is probably out of the question, as there isn't even a closed form solution for the pdf of Levy-Stable random variables, but getting bounds on what the probability is would be helpful.  I would appreciate any help with regards to how to analyze these types of questions in general such as general methods or references to other work in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this problem is elementary, I would greatly appreciate any reference to a textbook, tutorial or paper that would help me solve problems of this sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;:  George Lowther and Shai Covo have answered this question below.  I just wanted to give a few more pictures that compare their answers to some of the numerical experiments that I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the probability of the maximum element being larger than the rest for a list size of $n=100$ as a function of $\alpha$, $\alpha \in (0,1)$.  Each point represents 10,000 simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/khCj3.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are two graphs for two values of $\alpha \in \{1.53125, 1.875\}$.  Both have the function $ (2/\pi) \sin(\pi \alpha / 2) \Gamma(\alpha) n (( \tan(\pi \alpha/2) (n^{1/\alpha} - n))^{-\alpha} $ with different prescalars in front of them to get them to line up ( $1/4$ and $1/37$, respectively) superimposed for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/6Bt49.png&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/7OBsH.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As George Lowther correctly pointed out, for the relatively small $n$ being considered here, the effect of the extra $n^{1/\alpha}$ term (when $1 &amp;lt; \alpha &amp;lt; 2$) is non-negligible and this is why my original reference plots did not line up with the results of the simulations.  Once the full approximation is put in, the fit is much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I get around to it, I will try and post some more pictures for the case when $\alpha=1$ as a function of $n$ and $\gamma$.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastEditorUserId="4143" LastEditorDisplayName="" LastEditDate="2010-12-01T07:36:23.203" LastActivityDate="2010-12-01T07:36:23.203" Title="Probability of the maximum (Levy Stable) random variable in a list being greater than the sum of the rest?" Tags="&lt;probability&gt;&lt;analysis&gt;&lt;reference-request&gt;" AnswerCount="4" CommentCount="13" FavoriteCount="1" />
  <row Id="12098" PostTypeId="1" AcceptedAnswerId="12105" CreationDate="2010-11-27T19:09:33.090" Score="2" ViewCount="189" Body="&lt;p&gt;It's not really typical math question,today while studying graphs I suddenly got inquisitive about whether there exiat any function that could possible to draw a heart-shaped graph.Out of sheer curiousity I clicked on Google, which took me to this &lt;a href=&quot;http://mathworld.wolfram.com/HeartCurve.html&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The page seems informative, and I am glad to learn certain new things! Now I am interested in drawing them by my own by using mathematica.So my question is it possible to draw them in mathematica if yes please show me how.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T01:17:21.810" LastActivityDate="2010-11-28T03:15:14.563" Title="Drawing heart in mathematica" Tags="&lt;geometry&gt;&lt;plane-curves&gt;&lt;mathematica&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="12099" PostTypeId="2" ParentId="12098" CreationDate="2010-11-27T19:14:38.520" Score="1" ViewCount="0" Body="&lt;p&gt;This is really about plotting polar plots, parametric plots and implicitly defined functions in Mathematica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the info on how to draw polar plots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/PolarPlot.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PolarPlot.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parametric plots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/ParametricPlot.html&quot; rel=&quot;nofollow&quot;&gt;http://reference.wolfram.com/mathematica/ref/ParametricPlot.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This provides info on implicit plots&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://grosz.math.txstate.edu/~dhaz/prob_sets/LTs09cal1lab8.pdf&quot; rel=&quot;nofollow&quot;&gt;http://grosz.math.txstate.edu/~dhaz/prob_sets/LTs09cal1lab8.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-27T19:30:45.427" LastActivityDate="2010-11-27T19:30:45.427" CommentCount="2" />
  <row Id="12100" PostTypeId="2" ParentId="12088" CreationDate="2010-11-27T19:19:12.403" Score="2" ViewCount="0" Body="&lt;p&gt;It is false, because you can always pick a countable subset of your choice (e.g. {1, 2}). A set is countable if it can be put into one-to-one correspondence with the set of Natural numbers or some subset of Natural numbers. Since Natural numbers are contained in Real numbers, you can always pick a subset (like the one mentioned above) that would be countable.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-11-27T19:37:02.873" LastActivityDate="2010-11-27T19:37:02.873" />
  <row Id="12101" PostTypeId="2" ParentId="11964" CreationDate="2010-11-27T19:25:40.980" Score="0" ViewCount="0" Body="&lt;p&gt;Your question is equivalent to asking whether any element of {y-1, y-2, $\ldots$, y-100} has a divisor in the range (a,b).  So if you can factor all the 100 numbers, you can check that.  Unfortunately, factoring is pretty hard and then you have the NP complete knapsack problem to solve in looking through all the factors for one that is in range.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-27T19:25:40.980" />
  <row Id="12102" PostTypeId="1" AcceptedAnswerId="12103" CreationDate="2010-11-27T19:40:09.830" Score="1" ViewCount="86" Body="&lt;p&gt;Suppose a graph with 12 vertices is colored with exactly 5 colors. By the pigeonhole principle, each color appears on at least two vertices. True or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is false, but I assumed it to be true. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-27T19:45:10.213" Title="Pigeonhole principle question" Tags="&lt;number-theory&gt;&lt;counting&gt;&lt;pigeonhole-principle&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="12103" PostTypeId="2" ParentId="12102" CreationDate="2010-11-27T19:45:10.213" Score="2" ViewCount="0" Body="&lt;p&gt;The Pigeonhole Principle implies that there is &lt;em&gt;at least one&lt;/em&gt; color which appears on at least $2$ vertices, not that &lt;em&gt;each&lt;/em&gt; color appears.  Is it possible to color $12$ vertices with five colors in such a way that one of the colors is used only on a single vertex?&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-27T19:45:10.213" />
  <row Id="12105" PostTypeId="2" ParentId="12098" CreationDate="2010-11-27T20:00:55.967" Score="6" ViewCount="0" Body="&lt;p&gt;For the fifth function in the link you &lt;a href=&quot;http://mathworld.wolfram.com/HeartCurve.html&quot; rel=&quot;nofollow&quot;&gt;mentioned&lt;/a&gt; (which I thought it was the most similar to a heart):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PolarPlot[(Sin[t]Sqrt[Abs[Cos[t]]])/(Sin[t]+7/5)-2Sin[t]+2, {t, 0, 10}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Similarly, using W|A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/pZEwC.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2010-11-27T20:00:55.967" CommentCount="2" />
  <row Id="12106" PostTypeId="1" AcceptedAnswerId="12108" CreationDate="2010-11-27T20:02:46.873" Score="2" ViewCount="69" Body="&lt;p&gt;My TI-83 calculator doesnt allow me to do this, but using Windows calculator, I can compute the factorial of say 5.8. What does this mean and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-27T20:05:56.630" Title="Factorial of a non-integer number" Tags="&lt;number-theory&gt;&lt;factorial&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12107" PostTypeId="1" AcceptedAnswerId="12115" CreationDate="2010-11-27T20:04:11.850" Score="6" ViewCount="240" Body="&lt;p&gt;This is probably simple, but I'm solving a practice problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \to \infty}\frac{1}{n}\left( \cos{\frac{\pi}{n}} + \cos{\frac{2\pi}{n}} + \ldots +\cos{\frac{n\pi}{n}} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognize this as the Riemann sum from 0 to $\pi$ on $\cos{x}$, i.e. I think its the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^\pi{ \cos{x}dx }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is 0, but the book I'm using says it should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{1}{\pi}\int_0^\pi{ \cos{x}dx }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still 0 anyway, but where did the $\frac{1}{\pi}$ in front come from?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:28:55.760" LastActivityDate="2010-11-28T14:40:10.017" Title="Is $\lim\limits_{n \to \infty}\frac{1}{n}\left( \cos{\frac{\pi}{n}} + \cos{\frac{2\pi}{n}} + \ldots + \cos{\frac{n\pi}{n}} \right)$ a Riemann sum?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="12108" PostTypeId="2" ParentId="12106" CreationDate="2010-11-27T20:05:56.630" Score="10" ViewCount="0" Body="&lt;p&gt;Perhaps you're looking for the Gamma function. It restricts to the usual factorial on natural numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gamma_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-27T20:05:56.630" CommentCount="1" />
  <row Id="12110" PostTypeId="1" AcceptedAnswerId="12111" CreationDate="2010-11-27T20:24:55.517" Score="4" ViewCount="122" Body="&lt;p&gt;I accidentally typed &lt;code&gt;4 ** 5&lt;/code&gt; into Google and received &lt;code&gt;1,024&lt;/code&gt; instead of &lt;code&gt;20&lt;/code&gt;. But &lt;code&gt;4^5&lt;/code&gt; is also &lt;code&gt;1024&lt;/code&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the double multiplication sign also mean the power sign? Why two symbols?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2010-11-28T03:26:01.440" Title="What is the double multiplication sign for?" Tags="&lt;powers&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12111" PostTypeId="2" ParentId="12110" CreationDate="2010-11-27T20:25:21.527" Score="8" ViewCount="0" Body="&lt;p&gt;Exponentiation in some software. So a**b means a^b.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-27T20:25:21.527" />
  <row Id="12112" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T20:25:46.763" Score="1" ViewCount="0" Body="&lt;p&gt;The sum is also the real part of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{n}\left(e^{i\frac{\pi}{n}}+e^{i\frac{2\pi}{n}}+\ldots+e^{i\frac{n\pi}{n}}\right) \; .$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-27T20:25:46.763" CommentCount="2" />
  <row Id="12113" PostTypeId="1" AcceptedAnswerId="12117" CreationDate="2010-11-27T20:36:19.150" Score="1" ViewCount="50" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/UkWn5.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer for this turns out to be only irreflexive. However, how is this not transitive? The definition I have for transitive states &quot;whenever there is a&#xA;path from x to y then there must be a direct arrow from x to y&quot;. So for the above graph, if there exists a path from one point to another, then there should be a direct arrow. Well that appears to be the case for the above image, no? Theres a path from C to B, and there is a direct arrow. There is no path from C to E, so no arrow needed. No path from B to C, B to D, so no arrows needed here. There is a path from Z to A and there is also a direct arrow, and same for A to Z. Am I misunderstanding this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Transitive_relation#Properties_of_transitivity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article for transitive relation&lt;/a&gt; says:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a transitive relation the&#xA;  following are equivalent:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Irreflexivity  &lt;/li&gt;&#xA;  &lt;li&gt;Asymmetry   &lt;/li&gt;&#xA;  &lt;li&gt;Being a strict partial order&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this mean that if a relation is transitive, then it is also irreflexive? If thats the case, then shouldn't the above graph then be transitive?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-27T20:49:07.193" Title="Transitive and reflexive graph" Tags="&lt;graph-theory&gt;&lt;relations&gt;" AnswerCount="1" />
  <row Id="12114" PostTypeId="2" ParentId="12067" CreationDate="2010-11-27T20:36:59.297" Score="3" ViewCount="0" Body="&lt;p&gt;You could argue by induction on $n$. The result is obvious if $n=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inductive step, assume that $(n-1)!$ divides any product of $n-1$ consecutive integers. Now consider products of $n$ consecutive integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your product is $P(k)=(k+1)(k+2)\dots(k+n)$. First, you may assume $k\ge0$: If one of the factors $k+j$ is 0, then $P(k)=0$ and obviously $n!$ divides it. If $k+n=-t-1&amp;lt;0$, this is the same as $$(-1)^n(t+1)(t+2)\dots(t+n)=(-1)^nP(t),$$ and $t\ge0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now argue by induction on $k$. The result clearly holds if $k=0$, since $P(0)=n!$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $n!|P(k)$. Then $P(k+1)=(k+2)(k+3)\dots(k+n+1)$. Split the last factor as $(k+1)+n$. We have $$P(k+1)=[(k+2)\dots(k+n)](k+1)+[(k+2)\dots(k+n)]n.$$ The first summand is $P(k)$, which is divisible by $n!$ by the inductive assumption on $k$. The second summand is $n$ times a product of $n-1$ consecutive integers, thus $n$ times a multiple of $(n-1)!$, by the inductive assumption on $n$. Clearly, $n$ times a multiple of $(n-1)!$ is a multiple of $n!$, and the sum of two multiples of $n!$ is a multiple of $n!$, so $n!|P(k+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are done by induction.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-27T20:59:43.410" LastActivityDate="2010-11-27T20:59:43.410" />
  <row Id="12115" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T20:39:24.593" Score="6" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sum#Definition&quot; rel=&quot;nofollow&quot;&gt;Riemann sum&lt;/a&gt; is given by $$S=\sum_{i=1}^{n}f(y_{i})(x_{i}-x_{i-1})$$ where $x_{i-1}\leq y_{i}\leq x_{i}$. If you choose $f(y_{i})=\cos(y_{i})$, $y_{i}=\frac{i \pi}{n}$ and $x_{i}=\frac{i \pi}{n}$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\pi} \cos(x)dx=\lim_{n\rightarrow{\infty}}\sum_{i=1}^{n}\cos\left(\frac{i \pi}{n}\right)(x_{i}-x_{i-1})$$ which is just the sum you have with an extra factor $\pi$. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\pi}\int_{0}^{\pi} \cos(x)dx=\lim_{n\to\infty}\frac{1}{n}\left( \cos{\frac{\pi}{n}} + \cos{\frac{2\pi}{n}} + \ldots \cos{\frac{n\pi}{n}} \right)$$&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:26:22.783" LastActivityDate="2010-11-27T23:26:22.783" CommentCount="7" />
  <row Id="12116" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T20:43:09.740" Score="2" ViewCount="0" Body="&lt;p&gt;Assume that $x_{k-1}\leq c_{k}\leq x_{k}$, $x_{0}=0,x_{n}=\pi $, and that the interval of integration is divided into $n$ sub-intervals of equal width. Under these circumstances $x_{k}-x_{k-1}=\dfrac{\pi }{n}$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\cos \frac{k\pi }{n}&amp;amp;=&amp;amp;\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\cos \left( c_{k}\right) \frac{%&#xD;&#xA;x_{k}-x_{k-1}}{\pi }\\&#xD;&#xA;&amp;amp;=&amp;amp;\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{1}{\pi }\cos \left(&#xD;&#xA;c_{k}\right)\right) (x_{k}-x_{k-1})\\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\frac{1}{\pi }\cos x\ dx\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum $\displaystyle\sum_{k=1}^{n}\left(\dfrac{1}{\pi }\cos \left(&#xD;&#xA;c_{k}\right)\right) (x_{k}-x_{k-1})$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_sum#Definition&quot; rel=&quot;nofollow&quot;&gt;Riemann Sum&lt;/a&gt; of the function $f(x)=\dfrac{1}{\pi }\cos x$ in the interval $[0,\pi]$, but &lt;em&gt;not&lt;/em&gt; of the function $\cos x$.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-28T14:40:10.017" LastActivityDate="2010-11-28T14:40:10.017" />
  <row Id="12117" PostTypeId="2" ParentId="12113" CreationDate="2010-11-27T20:43:19.557" Score="2" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, how is this not transitive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is an edge from Z to A and one from A to Z, but no edge from Z to itself or from A to itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean that if a relation is transitive, then it is also irreflexive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it means that if a transitive relation is irreflexive, it is also asymmetric and a strict partial order (and if it is asymmetric, it is also irreflexive etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the graph was transitive, its irreflexivity would imply that it's also antisymmetric, but since the graph is not transitive, this does not apply.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditorDisplayName="" LastEditDate="2010-11-27T20:49:07.193" LastActivityDate="2010-11-27T20:49:07.193" CommentCount="3" />
  <row Id="12118" PostTypeId="1" CreationDate="2010-11-27T20:55:38.597" Score="3" ViewCount="80" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;grassmann algebra&lt;/a&gt; $G$ is the algebra over a field $\mathbb{F}$ generated by the variables $e_i$ such that $e_i^2=0$ and $e_i e_j = -e_j e_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for some references on algebras $G\otimes A$ (In this case the grassmann algebra is over $\mathbb{C}$) where A is the group algebra $\mathbb {C} \mathbb{Z}_n$ generated by $U_t$ and $e_i U_t = \zeta U_t e_i$ where $\zeta$ is a $n$-th primitive root of unity. Or in other words, I want to attach another variable $t$ to the grassman algebra such that $t^n=1$ and $e_i t = \zeta t e_i $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask this, is that I came across algebras that have the same polynomial identities as the ones above, so I want to see what can I say about them from only the identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think now that these algebras are Matrix algebras with coefficients in $\mathbb{Q}(\zeta)G$ such that each diagonal is multiplied by a (single) element from G. something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\left(\begin{array}{cccc}&#xD;&#xA;g_{0} &amp;amp; g_{1} &amp;amp; g_{2} &amp;amp; g_{3}\\&#xD;&#xA;g_{3} &amp;amp; g_{0} &amp;amp; g_{1} &amp;amp; g_{2}\\&#xD;&#xA;g_{2} &amp;amp; g_{3} &amp;amp; g_{0} &amp;amp; g_{1}\\&#xD;&#xA;g_{1} &amp;amp; g_{2} &amp;amp; g_{3} &amp;amp; g_{0}\end{array}\right)&#xD;&#xA;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g_0, g_2 \in G_0$ (even) and $g_1, g_3 \in G_1$ (odd). The $t$ variable is the diagonal matrix with coefficients $1,i,-1,-i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-11-28T06:43:32.953" LastActivityDate="2010-11-28T06:43:32.953" Title="Grassmann Algebras" Tags="&lt;algebra&gt;&lt;exterior-algebra&gt;" CommentCount="2" />
  <row Id="12119" PostTypeId="2" ParentId="12097" CreationDate="2010-11-27T20:57:04.353" Score="0" ViewCount="0" Body="&lt;p&gt;So $M=\mbox{Max}(X_0,\ldots,X_{n-1})$ then you know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(M\geq x)=1-F(x)^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $F(x)$ the Levy-stable cumulative distribution. Now, you also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(M\geq \sum X_i) = \int p(M\geq \sum X_i | \sum X_i = x) p(x) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $p(x)$ is the density function for Levy-stable distributions. (Since by definition Levy-stable distributions are stable under summation of random Levy-stable variables.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems to me you have in principle all the ingredients to try some asymptotics, even if you don't have an explicit formula for all the distributions. Maybe you can try with some for which the formula exists, like Cauchy and normal distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the sloppiness, just trying to sketch an approach for now.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-11-27T21:12:39.637" LastActivityDate="2010-11-27T21:12:39.637" />
  <row Id="12120" PostTypeId="1" AcceptedAnswerId="12127" CreationDate="2010-11-27T21:07:29.847" Score="0" ViewCount="54" Body="&lt;p&gt;There's one basilar math thing that keeps bugging me: the fact that a really simple 2D geometrical figure (like a circle) might not be a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what the definition of a function is. A circle is not a function (of one variable) because it would associate two values of the codomain to a single value of the domain. But this doesn't help my intuitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds terribly weird that a given curve (say a sinusoidal) is a function only unless you rotate it by 45 or more degrees...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is there any simple way (a concept similar to that of how most people imagine a function: a curve on a graph) to represent 2D geometrical figures like a circle (or a rotated sinusoidal, or whatever)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only one I can think of is using a function of two or more variables, but this sounds pretty dirty to me: why should I use a function in three dimensions just to see its shadow on two dimensions?&lt;br&gt;&#xA;Besides if I think of the function as a real object (in our real, 3D space), I cannot help thinking that it's not a 2D circle, it's a 3D weird object which can be seen as a circle when rotated in a particular way (just like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_stairs&quot; rel=&quot;nofollow&quot;&gt;Penrose stairs&lt;/a&gt; look &lt;em&gt;possible&lt;/em&gt; when rotated in a special way).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2010-11-27T21:42:29.897" Title="From geometrical figures to function" Tags="&lt;geometry&gt;&lt;functions&gt;&lt;plane-curves&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12121" PostTypeId="2" ParentId="12065" CreationDate="2010-11-27T21:16:00.180" Score="0" ViewCount="0" Body="&lt;p&gt;Let us prove that $m^{(k)}=m(m+1)...(m+k-1)$ is divided by $k!$ for all integer $m$.&#xA;Induction by $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k=1$: Every integer $m$ is divided by $1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k\to k+1$: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;induction by $m$: $m=0$: $0^{k+1}=0$&#xD;&#xA;is divided by $(k+1)!$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m\to m+1$: $(m+1)^{(k+1)}=((m+1)^{(k+1)}-m^{(k+1)})+m^{(k+1)}=$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=(m+1)(m+2)...(m+k+1)-m(m+1)...(m+k)+m^{(k+1)}=$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(k+1)(m+1)...(m+k)+m^{(k+1)}=(k+1)(m+1)^{(k)}+m^{(k+1)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and first term is divided by $(k+1)\cdot k!=(k+1)!$ because of induction by k and the second term is divided by $(k+1)!$ because of induction by $m$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the same works for $m\to m-1$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; Oops, essentially the same proof found in the thread mentioned in this &lt;a href=&quot;http://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial/12073#12073&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="425" LastEditorDisplayName="" LastEditDate="2010-11-27T21:21:26.043" LastActivityDate="2010-11-27T21:21:26.043" CommunityOwnedDate="2010-11-27T21:21:26.043" />
  <row Id="12122" PostTypeId="1" AcceptedAnswerId="12124" CreationDate="2010-11-27T21:16:24.347" Score="1" ViewCount="77" Body="&lt;p&gt;Question: Let $V$ be a $K$-Vectorspace and $f: V \rightarrow V$ be linear. It holds that $f \circ f = f$. Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt: So i guess that the $\oplus$ denotes a direct sum which means i have to show that (i) $V = \mbox{ker}(f) + \mbox{im}(f)$ and (ii) $\mbox{ker}(f) \cap \mbox{im}(f) = \{0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to do (ii) first: Let $v \in \mbox{im}(f) \cap \mbox{ker}(f)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow \exists u: f(u)=v \wedge f(v) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(can i put a &quot;Rightarrow&quot; here?) $(f \circ f)(u)=f(f(u)) = f(v) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (i) i am having difficulty with an approach to showing that $V = \mbox{ker}(f) + \mbox{im}(f)$. Should I even be trying to do this in the first place? if so, any advice as to how?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2010-11-27T21:30:47.993" Title="Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$" Tags="&lt;linear-algebra&gt;" AnswerCount="3" />
  <row Id="12123" PostTypeId="2" ParentId="12122" CreationDate="2010-11-27T21:25:55.747" Score="0" ViewCount="0" Body="&lt;p&gt;$v = (v - f(v)) + f(v)$&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T21:25:55.747" />
  <row Id="12124" PostTypeId="2" ParentId="12122" CreationDate="2010-11-27T21:28:30.790" Score="3" ViewCount="0" Body="&lt;p&gt;Let's see (ii) first: (Yes, your use of the implication symbol $\Rightarrow$ is appropriate.) You want to use that $f\circ f=f$, which you do not mention explicitly. The point is: In (ii) you need to show that the $v$ you are considering is $0$. You have: $f\circ f=f$, so $v=f(u)=(f\circ f)(u)$ etc, as you did. This gives $v=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i), let $v$ be any element of $V$. What can we say about $v-f(v)$? Let's call it $w$. We have $f(w)=f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$. Here, I used that $f$ is linear and (again) that $f\circ f=f$. This shows that $w\in{\rm ker}(f)$. So, $v=(v-f(v))+f(v)=w+f(v)$ is the sum of an element of the kernel (namely, $w$) and one of the image (namely, $f(v)$), and that is precisely what you needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A meta-question here is how would one think of considering $v-f(v)$. Ok, let's try to work backwards. Suppose we already know that $v=a+b$ with $a$ in the kernel and $b$ in the range. Then $f(v)=f(a)+f(b)=f(b)$, since $a$ is in the kernel. Also, $f(v)=f(f(v))$, so it is reasonable, as a first approach, to let $b=f(v)$, and see if we can directly prove that $a=v-b$ is in the kernel. And this is precisely what we did.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T21:28:30.790" CommentCount="2" />
  <row Id="12125" PostTypeId="2" ParentId="12122" CreationDate="2010-11-27T21:30:47.993" Score="0" ViewCount="0" Body="&lt;p&gt;Hi.  For 1) let v be an arbitrary element in the intersection of ker f and image f.  This second statement, as you have pointed out, means that v=f(u) for some u, and the first statement, means that f(v) = 0.  So 0=f(v)=f(f(u)) which is f(u) (or v) by assumption.  This shows v=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2) I believe you can do this with nothing more than set theory.  Just prove V is a subset of ker(f) + im(f) and then prove the reverse inclusion (this second inclusion is trivial since ker(f) and im(f) are clearly both subsets of V)&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-27T21:30:47.993" />
  <row Id="12126" PostTypeId="1" AcceptedAnswerId="12151" CreationDate="2010-11-27T21:40:04.383" Score="10" ViewCount="163" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lennart_Carleson&quot; rel=&quot;nofollow&quot;&gt;Lennart Carleson&lt;/a&gt; proved Luzin's conjecture that &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;the Fourier series of each $f\in L^2(0,2\pi)$ converges almost everywhere&lt;/a&gt;. Also, &lt;a href=&quot;http://en.wikipedia.org/wiki/Richard_Allen_Hunt&quot; rel=&quot;nofollow&quot;&gt;Richard Hunt&lt;/a&gt; extended the result to &lt;a href=&quot;http://en.wikipedia.org/wiki/Carleson%27s_theorem#CITEREFHunt1968&quot; rel=&quot;nofollow&quot;&gt;$L^p$ ($p&gt;1$)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some time ago I tried to read Carleson's paper, but I would say it is fairly hard to assimilate. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an easier proof? Can someone point out of the core or give an outline?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What did Hunt do? Can someone give an outline of that proof?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-28T00:52:20.577" Title="Accessible proof of Carleson's $L^2$ theorem" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" />
  <row Id="12127" PostTypeId="2" ParentId="12120" CreationDate="2010-11-27T21:42:29.897" Score="2" ViewCount="0" Body="&lt;p&gt;Some possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you might also want to switch to &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinates&quot; rel=&quot;nofollow&quot;&gt;polar coordinates&lt;/a&gt;, in which points are defined through angle $\varphi$ and radius $r$ rather than $x$/$y$ coordinates. For example, in a polar coordinate system, a circle (let's take the unit circle) $K$ actually &lt;strong&gt;is&lt;/strong&gt; &lt;em&gt;a function&lt;/em&gt; (now of form $r(\varphi)$ rather than $y(x)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the radius is constant, we end up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K: r(\varphi) = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most figures still aren't functions in polar coordinates, so we might have to take a more general approach: A curve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curve is a function that produces &lt;em&gt;coordinates&lt;/em&gt; rather than a single value from some parameter, i.e. $\mathbb{R} \to \mathbb{R}^2$ in our case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be our unit circle again - now parameterized by some angle $\varphi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(\varphi) = (\cos \varphi, \sin \varphi)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least, this should work for most figures. The most general form though is simply an equation that the coordinates have to satisfy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K: x^2+y^2 = 1$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-27T21:42:29.897" />
  <row Id="12128" PostTypeId="1" CreationDate="2010-11-27T21:44:52.540" Score="2" ViewCount="73" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the series: $\sum_{n=0}^{\infty}(-1)^{n}(\sqrt[n]{n} - 1)^{n}$. Does the series converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt to solution (might be incorrect):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\sqrt[n]{n} - 1)^{n}&amp;gt; (1+\frac{1}{n})^{n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+\frac{1}{n})^{n} \to e \Rightarrow (\sqrt[n]{n} - 1)^{n}$ lower-bounded by $e$. Based on Leibniz Criterion the sequence $\{A_n\}$ (in our case, $(\sqrt[n]{n} - 1)^{n}$) is monotone decreasing, but its limit is not $0$ at infinite $\Rightarrow$ series diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it enough to say that since the sequence is lower-bounded, the limit of it at infinite is not $0$, or should I actually calculate the limit of the sequence? &lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2010-11-27T23:32:58.540" Title="Leibniz criterion for alternating series" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12129" PostTypeId="2" ParentId="12107" CreationDate="2010-11-27T21:49:36.213" Score="3" ViewCount="0" Body="&lt;p&gt;$$\sum_{k=0}^n \cos\left(\frac{k\pi}{n}\right) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expression equals to $-1/n$, and the limit is trivially equal to zero.&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-27T21:49:36.213" />
  <row Id="12130" PostTypeId="2" ParentId="12128" CreationDate="2010-11-27T22:06:42.410" Score="2" ViewCount="0" Body="&lt;p&gt;Your lower bound for $(n^{1/n} -1)^n$ is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: To show that&#xA;$(n^{1/n} -1)^n \rightarrow 0$ write $n^{1/n} = 1 + r_n$ and use the binomial theorem&#xA;to show that for $n \ge 2$ we have $r_n &amp;lt; \sqrt{2/(n-1)}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2: When you expand $(1+r_n)^n$ the important term in the expansion&#xA;is the one in ${r_n}^2 .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way you can show that the series converges absolutely, you don't really need the Leibniz criterion although they will do the job. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-11-27T22:24:09.040" LastActivityDate="2010-11-27T22:24:09.040" />
  <row Id="12131" PostTypeId="2" ParentId="12128" CreationDate="2010-11-27T22:15:01.340" Score="1" ViewCount="0" Body="&lt;p&gt;Write your series as $\sum_{n = 1}^\infty (-1)^n a_n$ where $a_n = (n^{1/n} - 1)^n$. Note that $(a_n)$ is monotonically decreasing to zero and all terms are positive, hence by the alternating series test (or Leibniz' test) the series converges.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-11-27T22:15:01.340" />
  <row Id="12132" PostTypeId="1" AcceptedAnswerId="12148" CreationDate="2010-11-27T22:15:37.003" Score="4" ViewCount="97" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets consider a 3x3 matrix, A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse of this matrix can be found by finding the adjoint of the matrix and dividing it by its determinant. Can someone please explain why this process work and how it can be visualized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the determinant can be thought of as the volume spanned by the column vectors and that the adjoint of matrix A is the transpose of A's cofactors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a 3x3 matrix, A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cofactors of this matrix can be found by the cross product of each of its column vectors. So what does the cofactor matrix tell us? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2010-11-28T00:13:09.543" Title="Finding the inverse of a Matrix using determinants" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12134" PostTypeId="1" AcceptedAnswerId="12136" CreationDate="2010-11-27T22:18:15.423" Score="2" ViewCount="84" Body="&lt;p&gt;I've recently read a article 'Theories of computational complexity'. In the article it states that with Rice's theorem one can prove that, for example, {$i∈[\mathbb N:\phi_i$ is a constant function}, is not recursive, but I can't see how. I already know the theorem, but I don't really get how I can use it. Thank you for your answer!&lt;/p&gt;&#xA;" OwnerUserId="4149" LastActivityDate="2010-11-27T22:35:40.373" Title="Rice’s theorem explanation" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="12135" PostTypeId="2" ParentId="11669" CreationDate="2010-11-27T22:28:27.287" Score="7" ViewCount="0" Body="&lt;p&gt;The answers given are pretty good from a musical, mathematical, and socialogical / historical reason.  But they miss the fundamental reason why there are 12 notes in a western scale (or 5 notes in an eastern pentatonic, etc.), and why it's those particular 12 notes (or 5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qiaochu almost nailed it by pointing out that we like notes which are simple integer ratios.  But why?  The fundamental reason stems from the physics of common early instruments -- flutes (including the human voice) and plucked strings -- and from the physics of the tympanum in the ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu noted, sound is not composed of a single sine wave frequency but rather a sum of many sine waves.  The &quot;note&quot; we hear is the frequency of the primary (loudest) wave coming from these instruments.  But frequencies exist in that wave as well, albeit largely masked by the primary.  These are known informally as harmonics or overtones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first several harmonics of flutes and plucked strings are similar and very straightforward:  If the primary is normalized to frequency 1, then the second loudest harmonic is typically 1/2 (an octave above), the third is usually 1/3 (an octave and a fifth above), the fourth is usually 1/4 (two octaves), the fifth is usually 1/5 (two octaves and a major third), and the sixth is usually 1/6 (two octaves and a fifth).  If the primary note is C1, these translate roughly into C2, G3, C4, E4, and G4.  If the harmonics continued in this way -- and they don't always -- various other notes appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matters because if you want to play TWO instruments together, you'd like their harmonics to coincide even if they're playing different notes.  Otherwise the excess of harmonics sounds bad to the ear.  In the worst case, very close but not entirely overlapping harmonics create &quot;beats&quot; -- seeming alternating loud and soft periods of time -- which are irritating to listen to and tough on the ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get harmonics to coincide in multiple instruments or even successive notes, you have to pick notes for them to play where their harmonics have a strong overlap.  For example, this is also why the major fourth is useful even though it doesn't often appear early.  It's because if one instrument is playing C, if the other instrument is playing major fourth but &lt;em&gt;lower&lt;/em&gt; by an octave, they'll overlap nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe these note selections (guaranteeing harmonics in harmony, so to speak) influenced the evolution of scale choices -- especially the pentatonic, that is, the black notes), and the division of the octave into 12 pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One early instrument which is totally out of whack from this is the bell.  Bells and gongs can be tuned to have a variety of harmonics, but the most common ones -- foundry bells -- have a very loud, unusual third harmonic: minor third or E flat.  It is so loud and incongruous that they sound terrible, even disturbing, when played along with strings, flutes, voices, etc.  In fact, entire musical pieces have to be written specially for carillons (large multibell instruments) in order to guarantee proper overlap of harmonics.  Generally this means that the entire piece has to be written in fully diminished chords.  Major chords sound among the worst because of the clash between the major third in the chord and the minor third coming from the root's loud third harmonic.&lt;/p&gt;&#xA;" OwnerUserId="4151" LastActivityDate="2010-11-27T22:28:27.287" CommentCount="1" />
  <row Id="12136" PostTypeId="2" ParentId="12134" CreationDate="2010-11-27T22:35:40.373" Score="3" ViewCount="0" Body="&lt;p&gt;Rice's theorem says that for any subset $F$ of the class $T$ of partial computable functions, the set $\{i\mid \phi_i\in F\}$ is recursive iff $F=\emptyset$ or $F=T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F$ be the set of partial computable functions that are constant on their domain. $F$ is certainly non-empty, since the function that always takes the value 0 is certainly computable. $F$ is also not the whole class $T$, since the function $\phi$ with $\phi(0)=0$, $\phi(n)=1$ for $n&amp;gt;0$ is certainly computable and not constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the set you are interested in is not recursive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your confusion lies in how to translate the usual statement of Rice's theorem (in terms of decision problems) into the version I wrote, simply remember that formally a decision problem is a set $D\subseteq{\mathbb N}$, and the problem is decidable iff $D$ is recursive. Of course, this is usually stated by saying that there is a Turing machine that, given $n$, decides whether $n\in D$ or not. But this is certainly the same as saying that the machine computes the characteristic function of $D$, which is (the definition of) being recursive.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T22:35:40.373" CommentCount="1" />
  <row Id="12137" PostTypeId="1" AcceptedAnswerId="12138" CreationDate="2010-11-27T22:49:27.500" Score="1" ViewCount="52" Body="&lt;p&gt;State in words the negation of the following sentence: For every martian M, if M is green, then M is tall and ticklish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the right answer to this, give or take a few words, but this is a question of form more than anything. After converting this statement to symbols and negating everything, I come up with: $\exists M(P \wedge (\neg \text{Tall } \vee \neg\text{ Ticklish})$ and so in word format that would be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exists a martian such that it is&#xA;  green and not tall or not ticklish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However the really correct answer is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a martian M such that M is&#xA;  green but M is not tall or M is not&#xA;  ticklish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The difference between these two is a 'but' and an 'and'. Does this mean anything mathematically? Is my version correct?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:51:59.617" LastActivityDate="2010-11-27T23:51:59.617" Title="Negating a statement" Tags="&lt;propositional-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12138" PostTypeId="2" ParentId="12137" CreationDate="2010-11-27T22:52:07.410" Score="4" ViewCount="0" Body="&lt;p&gt;Your answer is &quot;really correct&quot;. It is perhaps more usual in English to use &quot;but&quot; rather than &quot;and&quot; in a sentence as the one in your example. Formally (i.e., mathematically), there is no difference.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T22:52:07.410" CommentCount="1" />
  <row Id="12139" PostTypeId="1" AcceptedAnswerId="12143" CreationDate="2010-11-27T23:29:20.897" Score="4" ViewCount="96" Body="&lt;p&gt;Consider a non-empty set A containing n objects. How many relations on A are both symmetric and reflexive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this is $2^p$ where $p=$ $n \choose 2$. However, I dont understand why this is so. Can anyone explain this?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-27T23:54:00.673" LastActivityDate="2010-11-28T03:42:11.043" Title="Number of relations that are both symmetric and reflexive" Tags="&lt;combinatorics&gt;&lt;discrete-mathematics&gt;&lt;relations&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12140" PostTypeId="2" ParentId="12128" CreationDate="2010-11-27T23:32:58.540" Score="2" ViewCount="0" Body="&lt;p&gt;The alternatingness of the series is something of a red herring as the series converges absolutely. By the root test, to show this it suffices to show that $\lim_{n \rightarrow \infty} |n^{1 \over n} - 1| = 0$. In other words, it suffices to show that $\lim_{n \rightarrow \infty} n^{1 \over n} = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few ways to show this limit is in fact $1$. One way is to note that $\ln (n^{1 \over n}) = {\ln(n) \over n}$, and the latter is seen to go to zero as $n$ goes to infinity using L'Hopital's rule. Since the natural log of the $n$th term of the sequence goes to zero, the $n$th term of the sequence goes to $e^0 = 1$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-27T23:32:58.540" />
  <row Id="12142" PostTypeId="2" ParentId="12139" CreationDate="2010-11-27T23:45:04.277" Score="5" ViewCount="0" Body="&lt;p&gt;Being reflexive means that $(x,x)\in R$ for all $x\in A$. Being symmetric means that $(x,y)\in R$ implies that $(y,x)\in R$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Begin by listing $A$ as $A=\{a_1,\dots,a_n\}$. Then let $B$ be the set $$\{(a_i,a_j)\mid 1\le i&amp;lt;j\le n\}.$$ Note that if $x\ne y$ are elements of $A$, then either $(x,y)\in B$ or $(y,x)\in B$ but not both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be any subset of $B$. Let $$R_S=S\cup\{(y,x)\mid (x,y)\in S\}\cup\{(x,x)\mid x\in A\}.$$ Then $R_S$ is a symmetric and reflexive relation on $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are $2^{|B|}$ subsets of $B$, and that if $S\ne S&amp;#39;$ are subsets of $B$, then $R_S\ne R_{S&amp;#39;}$. Also, note that $|B|=\binom{n}2$. (If the last equality is not clear, note that $$B=\{(a_1,a_j)\mid j&amp;gt;1\}\cup\{(a_2,a_j)\mid j&amp;gt;2\}\cup\dots$$ so $|B|=(n-1)+(n-2)+\dots+1$, and it is well-known that the last sum equals $n(n-1)/2=\binom n2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the number of symmetric, reflexive relations on $A$ is at least $2^p$ with $p=\binom n2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the equality, it is enough to check that any such relation $R$ is $R_S$ for some $S\subseteq B$. But, given $R$, let $S=\{(a_i,a_j)\in R\mid i&amp;lt;j\}$. This is a subset of $B$, and it is easy to check that $R=R_S$. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-27T23:45:04.277" />
  <row Id="12143" PostTypeId="2" ParentId="12139" CreationDate="2010-11-27T23:45:20.950" Score="8" ViewCount="0" Body="&lt;p&gt;To be reflexive, it must include all pairs $(a,a)$ with $a\in A$. To be symmetric, whenever it includes a pair $(a,b)$, it must include the pair $(b,a)$. So it amounts to choosing which $2$-element subsets from $A$ will correspond to associated pairs.  If you pick a subset $\{a,b\}$ with two elements, it corresponds to adding both $(a,b)$ and $(b,a)$ to your relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many $2$-element subsets does $A$ have? Since $A$ has $n$ elements, it has exactly $\binom{n}{2}$ subsets of size $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you want to pick a collection of subsets of $2$-elements. There are $\binom{n}{2}$ of them, and you can either pick or not pick each of them. So you have $2^{\binom{n}{2}}$ ways of picking the pairs of distinct elements that will be related. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-27T23:45:20.950" />
  <row Id="12144" PostTypeId="2" ParentId="12081" CreationDate="2010-11-27T23:50:19.010" Score="0" ViewCount="0" Body="&lt;p&gt;You can try $$f(L,D) = L - D + \frac{1}{D+2}.$$ This would sort first according to $L - D$ (ascending) and then according to $D$ (descending).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-27T23:50:19.010" CommentCount="2" />
  <row Id="12145" PostTypeId="2" ParentId="12139" CreationDate="2010-11-27T23:52:59.200" Score="3" ViewCount="0" Body="&lt;p&gt;Hi!&#xA;Maybe you can see it like this: a relation $R$ on $A$ is a subset of $A\times A$, and it is symmetric if and only if $(x,y)\in R \implies (y,x)\in R$, moreover, if the relation is reflexive, then $(x,x)\in R$ for all $x\in A$. Then you can determine uniquely such a relation by saying which subsets of two distinct elements of $A$ &quot;belong&quot; to $R$, in the sense that $\{x,y\}\in R \iff (x,y),(y,x)\in R$. Now, you know that the number of subsets with two distinct elements of $A$ is $\binom{n}{2}$, and the number of subset of a set with $p$ elements is $2^p$.&#xA;I'm sorry if i was too obscure. &lt;/p&gt;&#xA;" OwnerUserId="4054" LastActivityDate="2010-11-27T23:52:59.200" CommentCount="1" />
  <row Id="12147" PostTypeId="2" ParentId="12139" CreationDate="2010-11-28T00:09:09.953" Score="1" ViewCount="0" Body="&lt;p&gt;There is only one way to make the relation reflexive -- all ordered pairs $(x,x), x\in A$ must be in the relation. So the number of reflexive symmetric relations on $A$ is the same as the number of ways of adding symmetric pairs $(a,b),(b,a)$, where $a\neq b$ into the relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S$ be a subset of $2^A$ consisting of subsets of 2 elements. Then $S$ gives rise to exactly one reflexive symmetric relation on $A$. For example, if $A=\lbrace 1,2,3,4\rbrace$, then an example of $S$ is $\lbrace \lbrace 1,2\rbrace, \lbrace 1,4\rbrace, \lbrace 2,3\rbrace\rbrace$.   The relation induced by $S$  is $$\lbrace (1,2), (2,1), (1,4), (4,1), (2,3), (3,2)\rbrace$$ plus all $(x,x), x\in A$. &#xA;Conversely, every reflexive symmetric relation on $A$ arises in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are $p={n\choose 2}$ subsets of 2 elements, there are $2^p$  such $S$'s. The answer to your question is therefore $2^p$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-28T03:42:11.043" LastActivityDate="2010-11-28T03:42:11.043" CommentCount="2" />
  <row Id="12148" PostTypeId="2" ParentId="12132" CreationDate="2010-11-28T00:13:09.543" Score="1" ViewCount="0" Body="&lt;p&gt;There is some explanation in the Geometric Interpretation &lt;a href=&quot;http://en.wikipedia.org/wiki/Cramer%27s_rule&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but I don't really like it too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's easier to think this way: The cross product of $v$ and $w$ is a vector $x$ orthogonal to the two given ones, whose length is twice the area of the triangle. Hence the volume of the tetrahedron on $u, v, w$ is one sixth of the the dot product of $x$ with the remaining vector $u$ (one should be careful with signs, but it works out). The dot product of $x$ with $v$ or $w$ is zero. Similar things are true about the cross product of $u$ and $w$ dotted with $v$ (and $u$ and $w$) and the cross product of $u$ and $v$ dotted with $w$ (and $u$ and $v$).  What all of this implies is that the matrix of cross product vectors (transposed) multiplied with the matrix of $u$, $v$, $w$ is diagonal, with all diagonal entries equal to six times the volume of the tetrahedron, i.e. the determinant of the $u$, $v$, $w$ matrix. The result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher dimensional versions are fairly similar, with volumes of $n$ and $n-1$ simplices.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-28T00:13:09.543" />
  <row Id="12149" PostTypeId="1" CreationDate="2010-11-28T00:22:26.130" Score="4" ViewCount="62" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem&quot; rel=&quot;nofollow&quot;&gt;Freyd-Mitchell's embedding theorem&lt;/a&gt; states that: if A is a small abelian category, then there exists a ring R and a full, faithful and exact functor F: A → R-Mod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quite the theorem and has several useful applications (it allows one to do diagram chasing in abstract abelian categories, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been asked to state and prove the theorem in class (a homological algebra course). However, by reading the texts I was recommended, I'm about to give in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Freyd's Abelian Categories says that the text, excepting the exercises, tries to be a geodesic leading to the theorem. If you take out the exercises, probably the text is 120 pages long. Impossible to do in 2:30 hours. To give you an idea, the course I'm taking is based in Rotman's &quot;An Introduction to Homological Algebra&quot; which works in R-Mod...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mitchell's Theory of Categories is very hard to read, and also to prove the theorem you have tons of definitions and propositions and lemmas to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weibel's An Introduction to Homological Algebra redirects me to Swan, The Theory of Sheaves, a book which is unavailable in my university's library. I've leafed through Swan's Algebraic K-Theory: the theorem is proved, but it is also long, hard and painful to read, and assumes a lot of knowledge I don't have (I had never seen a weakly effaceable functor, or a Serre subcategory; and it certainly is not well known to me that the category of additive functors from a small abelian category to the category of abelian groups is well-powered, right complete, and has injective envelopes!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm starting to believe it's an impossible task. But maybe there are more modern proofs which require less heavy machinery and technicalities?&lt;/p&gt;&#xA;" OwnerUserId="2614" LastActivityDate="2010-11-28T00:22:26.130" Title="Freyd-Mitchell's embedding theorem" Tags="&lt;reference-request&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="12150" PostTypeId="1" CreationDate="2010-11-28T00:37:21.787" Score="2" ViewCount="89" Body="&lt;p&gt;On my mid-term exam tonight, I had a problem which went something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which algebraic structure is $(M,*)$, where $M= \left \{ \begin{bmatrix}&#xD;&#xA;1 &amp;amp; 0\\ &#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;\end{bmatrix} \right., \begin{bmatrix}&#xD;&#xA;0 &amp;amp; 1\\ &#xD;&#xA;-1 &amp;amp;0 &#xD;&#xA;\end{bmatrix},\begin{bmatrix}&#xD;&#xA;-1 &amp;amp; 0\\ &#xD;&#xA;0 &amp;amp; -1&#xD;&#xA;\end{bmatrix},\begin{bmatrix}&#xD;&#xA;0 &amp;amp;-1 \\ &#xD;&#xA;1 &amp;amp; 0&#xD;&#xA;\end{bmatrix}\left.  \right \}  $   and * is matrix multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be some smart way of doing it, but I couldn't figure it out. In the end I made a table with all elements of M and relationship between them and then just read structure's properties. To me this just feels like excessive application of brute force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was trying to prove that operation is closed, I tried with something like $x=\begin{bmatrix}&#xD;&#xA;a &amp;amp; b\\ &#xD;&#xA;c &amp;amp; d&#xD;&#xA;\end{bmatrix}, y= \begin{bmatrix}&#xD;&#xA;p &amp;amp; q\\ &#xD;&#xA; s&amp;amp; t&#xD;&#xA;\end{bmatrix}$ and multiply that and then prove that result is part of the set, but I run out of ideas on how to prove that result is part of the set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had basically same problem for inverse element and commutativity too.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-11-28T01:39:12.027" Title="Looking for a smart way to find out what algebraic structure this is..." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12151" PostTypeId="2" ParentId="12126" CreationDate="2010-11-28T00:52:20.577" Score="13" ViewCount="0" Body="&lt;p&gt;A few years after Carleson's proof Fefferman came up with a shorter proof of the $L^2$ and $L^p$ results. Later, in the context of their work in multilinear harmonic analysis Lacey and Thiele came up with a quite short and easy to understand proof for the $L^2$ theorem which is to some extent a descendant of Fefferman's proof. It's only 10 pages long and can be found in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lacey, Michael; Thiele, Christoph (2000), &quot;A proof of boundedness of the Carleson operator&quot;, Mathematical Research Letters 7 (4): 361–370.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, this paper has an amusing Mathscinet review which begins with &quot;This is one of the greatest papers written in Fourier analysis.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also wrote an expository article describing this and a number of related results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lacey, Michael T. (2004), &quot;Carleson's theorem: proof, complements, variations&quot;, Publicacions Matemàtiques 48 (2): 251–307&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also put an expanded version of this last paper on the arxiv &lt;a href=&quot;http://arxiv.org/abs/math/0307008v4&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math/0307008v4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-28T00:52:20.577" CommentCount="4" />
  <row Id="12152" PostTypeId="2" ParentId="12150" CreationDate="2010-11-28T01:08:31.190" Score="2" ViewCount="0" Body="&lt;p&gt;Max has got the right idea in the comments.  If you look at what these matrices do to points in the plane the answer should be clear.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-28T01:08:31.190" />
  <row Id="12154" PostTypeId="1" AcceptedAnswerId="12155" CreationDate="2010-11-28T01:23:45.460" Score="2" ViewCount="40" Body="&lt;p&gt;If any edge from a given spanning tree T0 is contained in some minimum spanning tree T*, does this imply that T0 is also a minimum spanning tree ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I'm trying to draw on paper some graphs to prove that it doesn't. Please correct me if it does, or help me find an example if it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. smb please add MST or spanning-trees tag.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-28T01:32:00.020" LastActivityDate="2010-11-28T01:34:09.270" Title="Fast question about minimum spanning trees" Tags="&lt;graph-theory&gt;&lt;trees&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12155" PostTypeId="2" ParentId="12154" CreationDate="2010-11-28T01:34:09.270" Score="4" ViewCount="0" Body="&lt;p&gt;Consider a triangle with edge weights $2,2,1$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T01:34:09.270" CommentCount="1" />
  <row Id="12156" PostTypeId="2" ParentId="12150" CreationDate="2010-11-28T01:39:12.027" Score="0" ViewCount="0" Body="&lt;p&gt;Another option is to take one of the matrices and apply it several times. Say you take the second matrix and apply it to $(x,y)$. Magically, you will immediately recognize the other matrices (in order), so you've got a cyclic group of order $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Max and Qiaochu have commented, there's a geometric interpretation (think what geometric transformation keeps the origin and has order $4$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T01:39:12.027" CommentCount="2" />
  <row Id="12157" PostTypeId="2" ParentId="12098" CreationDate="2010-11-28T02:18:04.533" Score="3" ViewCount="0" Body="&lt;p&gt;You can plot the &lt;a href=&quot;http://www.wolframalpha.com/entities/surfaces/taubin%27s_heart_surface/cr/rn/yc/&quot; rel=&quot;nofollow&quot;&gt;Taubin's heart surface&lt;/a&gt; using ContourPlot3D&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ContourPlot3D[(2 x^2 + y^2 + z^2 - 1)^3 - (1/10) x^2 z^3 - y^2 z^3 == 0, {x, -1.5, 1.5}, {y, -1.5, 1.5}, {z, -1.5, 1.5}, Mesh -&amp;gt; None, ContourStyle -&amp;gt; Directive[Red, Opacity[0.8]]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would post a image but it won't let me because I don't have reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lXM95.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1938" LastEditorUserId="1938" LastEditorDisplayName="" LastEditDate="2010-11-28T03:15:14.563" LastActivityDate="2010-11-28T03:15:14.563" CommentCount="1" />
  <row Id="12158" PostTypeId="2" ParentId="12098" CreationDate="2010-11-28T02:49:36.433" Score="2" ViewCount="0" Body="&lt;p&gt;Consider the map $T \colon \mathbb R^2 \rightarrow \mathbb R^2, \ (x,y) \mapsto (x, y+ \sqrt{|x|})$.  With a little examination, you can see that this will define a warping on the plane that will map the unit circle to a heart shaped curve: &#xA;&lt;img src=&quot;http://i.imgur.com/qpKuu.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you know that a parametrization for the circle is $(\cos(t),\ \sin(t)),\  t\in [-\pi,\pi]$, then the parametrization for its heart-shaped image would be $(\cos(t),\ \sin(t) + \sqrt{|\cos(t)|}),\  t\in [-\pi,\pi]$. You can plot the curve with the following Mathematica code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ParametricPlot[{Cos[t], Sin[t] + Sqrt[Abs[Cos[t]]]}, {t, -Pi, Pi}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-11-28T02:49:36.433" />
  <row Id="12159" PostTypeId="2" ParentId="12107" CreationDate="2010-11-28T03:21:44.597" Score="0" ViewCount="0" Body="&lt;p&gt;The key to this last assertion is the simple fact that $\cos(\pi - x) = -cos(x)$.  Said symmetry can be observed directly from the definition of the cosine function via the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T03:21:44.597" />
  <row Id="12160" PostTypeId="1" AcceptedAnswerId="12270" CreationDate="2010-11-28T03:25:57.893" Score="8" ViewCount="193" Body="&lt;p&gt;I was wondering if the following properties of the Legendre polynomials are true in general. They hold for the first ten or fifteen polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Are the roots always simple (ie. multiplicity 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Except for low-degree cases, the roots can't be calculated exactly,only approximated.(unlike Chebyshev polynomials) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Are roots of the entire family of Legendre Polynomials dense in the interval [0,1] (ie, its not possible to find a subinterval, no matter how small, that doesn't contain at least one root of one polynomial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows of an article/text that proves any of the above, please let me know.  The definition of these polynomials can be found on Wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_polynomials&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Legendre_polynomials&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T13:13:05.537" LastActivityDate="2011-01-08T09:44:03.503" Title="Roots of Legendre Polynomial" Tags="&lt;real-analysis&gt;&lt;polynomials&gt;&lt;special-functions&gt;&lt;roots&gt;" AnswerCount="5" CommentCount="12" />
  <row Id="12161" PostTypeId="2" ParentId="12110" CreationDate="2010-11-28T03:26:01.440" Score="1" ViewCount="0" Body="&lt;p&gt;Caveat: The symbol ^ in C family languages is bitwise xor.  This is true in Python, too.  The ** symbol is also seen in Python.  &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T03:26:01.440" />
  <row Id="12162" PostTypeId="1" CreationDate="2010-11-28T03:39:58.390" Score="2" ViewCount="43" Body="&lt;p&gt;In an attempt to model the cassette tape spooling action, I'm hoping to solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required feed speed 5cm/second.&#xA;Two Spindles with starting radius of &#xA;S1: 0.8cm (empty spool) &#xA;S2: 1.8cm (full spool)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spooled tape thickness is unknown but can be assumed to be around 15micrometer if needed.&#xA;The time to complete the spooling from one spool to the other spool is required to be 45 minutes.&lt;/p&gt;&#xA;" OwnerUserId="4153" LastActivityDate="2010-11-28T05:13:57.377" Title="How do I calculate the instantaneous angular speed of a spindle given a spooled radius and required feed speed?" Tags="&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12163" PostTypeId="1" CreationDate="2010-11-28T03:42:31.793" Score="4" ViewCount="140" Body="&lt;p&gt;How can we find $\displaystyle \lim_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$ without using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) L'Hôpital's rule, (b) the limit $\displaystyle \lim_{h \to 0}\frac{e^h-1}{h} = 1$, and (c) the fact that $\displaystyle \frac{d}{dx}\left(e^x\right) = e^x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the conditions is that with this limit I'm trying to prove (c), and I've done so with (b) and I gather it would be circular to use (a). So that's that. Also, I would appreciate if you could share one or more ways of proving that the derivative of $e^x$ is $e^x$. Thanks a lot for your time.  &lt;/p&gt;&#xA;" OwnerUserId="3207" LastActivityDate="2010-11-28T10:17:29.837" Title="Calculate $\displaystyle \lim_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="12164" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T03:47:04.190" Score="0" ViewCount="0" Body="&lt;p&gt;Say $y = b(1+\epsilon)$. Then $\ln y - \ln b \approx \epsilon$ whereas $y - b = b\epsilon$. So it all boils down to showing $\ln (1+\epsilon) \approx \epsilon$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T03:47:04.190" CommentCount="1" />
  <row Id="12165" PostTypeId="2" ParentId="11474" CreationDate="2010-11-28T03:49:10.973" Score="0" ViewCount="0" Body="&lt;p&gt;You can also do this.  The intersection of any family of topologies on a set is itself a topology.  The discrete topology (all subsets open) is a topology for any set.  So, take the set of all topologies containing the family of topologies (remember: the discrete one is here) and you will get a topology.  It will be the smallest topology containing all topologies in the family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is integral to the notion of weak topology.  The first answer has a second correct characterization of this topology.  See the book {\it Topology} by Wilansky.  &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T03:49:10.973" CommentCount="1" />
  <row Id="12166" PostTypeId="1" CreationDate="2010-11-28T03:50:11.767" Score="2" ViewCount="66" Body="&lt;p&gt;&quot;When you draw a circle in a plane of radius 1 you can perfectly surround it with 6 other circles of the same radius.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT when you draw a circle in a plane of radius 1 and try to perfectly surround the central circle with 7 circles you have to change the radius of the surround circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can i find the radius of the surround circles if i want too use more that 6 cirles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ex :&#xA;7 circles = radius 0.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 circles = radius 0.2&lt;/p&gt;&#xA;" OwnerUserId="4155" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-28T04:33:18.257" LastActivityDate="2010-11-28T04:33:18.257" Title="Numbers of circles around a circle" Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12167" PostTypeId="1" AcceptedAnswerId="12168" CreationDate="2010-11-28T03:59:04.013" Score="0" ViewCount="129" Body="&lt;p&gt;The set of rationals $\mathbb{Q}$ has the same cardinality as&#xA;the set of integers $\mathbb{Z}$. True or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a question on an old exam for our class. The correct answer is true. However, I did some additional reading and came across Cantor's transfinite numbers. In the book I'm reading, it says that &quot;there are more real numbers (which include rational and irrational numbers) than there are integers&quot;. So can it also be said that there are more rational numbers than integers? And so can we say that the above statement is false?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:00:34.500" LastActivityDate="2010-12-22T21:21:46.343" Title="The set of rationals has the same cardinality as the set of integers" Tags="&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="12168" PostTypeId="2" ParentId="12167" CreationDate="2010-11-28T04:02:11.240" Score="2" ViewCount="0" Body="&lt;p&gt;No. The statement is still true.&#xA;The cardinality of the natural number set is the same as the cardinality of the rational number set. In fact, this cardinality is the first transfinite number denoted by $\aleph_0$ i.e. $|\mathbb{N}| = |\mathbb{Q}| = \aleph_0$. By first I mean the &quot;smallest&quot; infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinality of the set of real numbers is typically denoted by $\mathfrak{c}$. We have $\mathfrak{c}  &amp;gt; \aleph_0$, since we can set up a bijection from $\mathbb{R}$ to the power set of the natural numbers and by Cantor's theorem, for any set $X$, we have $|X| &amp;lt; |2^{X}|$. So we have $|\mathbb{R}| = |2^{\mathbb{N}}| &amp;gt; |\mathbb{N}|$. So what this essentially says is that &quot;there are more real numbers (which include rational and irrational numbers) than there are integers&quot; in some sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Continuum_Hypothesis&quot; rel=&quot;nofollow&quot;&gt;continuum hypothesis&lt;/a&gt; states that &quot;there is no set whose cardinality is strictly between that of the natural numbers and that of the real numbers&quot; which essentially means real numbers form the second &quot;smallest&quot; infinity.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-28T05:11:34.897" LastActivityDate="2010-11-28T05:11:34.897" CommentCount="7" />
  <row Id="12169" PostTypeId="2" ParentId="9703" CreationDate="2010-11-28T04:03:09.460" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a very prosaic answer. If you go into a grocery store with an order, you expect the cash register to read 0 before any items are rung up.  This is because $0 + x = x$ for any number $x$.  It is the fact that zero is the neutral element of addition that makes it a valid start for the register; starting there will not cheat you on your order or give you any unfair advantage.  To wit: an empty sum is 0.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, suppose we are multiplying numbers in a multiplying machine.  What is neutral there?  We know $1*x = x$ for any number $x$.  Since an empty product, like an empty sum should be neutral and not affect the product of the numbers coming in after it, the multiplying machine should be set at 1 before it starts work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $2^0$ is an empty product so $2^0 = 1$.  Said identity should be true for any base.&lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T04:03:09.460" />
  <row Id="12170" PostTypeId="2" ParentId="12167" CreationDate="2010-11-28T04:05:02.967" Score="6" ViewCount="0" Body="&lt;p&gt;You can say whatever you want. However, if you want to be &lt;em&gt;correct&lt;/em&gt;, and you are talking about cardinalities, then you cannot say that there are more rationals than integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we &lt;em&gt;can&lt;/em&gt; say that there are more reals than rational numbers is because they actually have &lt;em&gt;greater cardinality&lt;/em&gt;, not just because the set of reals properly contains the natural numbers. While we can find a bijection between $\mathbb{N}$ and $\mathbb{Q}$, &lt;em&gt;no&lt;/em&gt; function from $\mathbb{N}$ to $\mathbb{R}$ is surjective (and, naturally, no function from $\mathbb{Q}$ to $\mathbb{R}$ can be surjective either). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:11:10.290" LastActivityDate="2010-11-28T04:11:10.290" />
  <row Id="12171" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T04:07:35.833" Score="4" ViewCount="0" Body="&lt;p&gt;If $e^x=\sum_{n\geq0}x^n/n!$, then you can show easily that $e^{x+y}=e^xe^y$. Then $$\frac{e^{x+h}-e^x}{h} = \frac{e^h-1}{h}e^x,$$ and to conclude that $\frac{d}{dx}e^x=e^x$ we need only then show that $$\frac{e^h-1}{h}\to1$$ if $h\to0$. Using your definition, that is easy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-11-28T04:07:35.833" CommentCount="6" />
  <row Id="12172" PostTypeId="1" AcceptedAnswerId="12206" CreationDate="2010-11-28T04:08:17.420" Score="1" ViewCount="125" Body="&lt;p&gt;Does there exist a nonzero linear transformation $T$ such that&#xA;$$\alpha^t T \alpha = 0\text{ for all $\alpha\in\mathbb{R}^n$,}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where ${\alpha}^{t}$ denotes the transpose of the matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! =D&lt;/p&gt;&#xA;" OwnerUserId="647" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T04:12:59.900" LastActivityDate="2010-11-28T13:06:50.410" Title="Does there exist a nonzero linear transformation $T$ such that ${\alpha}^{T} T \alpha = 0$ for all $\alpha \in \mathbb{R}^n$?" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="12173" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T04:13:32.813" Score="2" ViewCount="0" Body="&lt;p&gt;Sure. If $n=2$, let $T$ be a $90^\circ$ rotation. A suitable modification of this works for all $n\ge 2$: Simply rotate the first two coordinates $90^\circ$, and map the rest to 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n=1$, there is an $r\in{\mathbb R}$ such that $T\alpha=r\alpha$ for all $\alpha$, and $\alpha^T T\alpha$ is simply $r\alpha^2$, so if $n=1$, $T$ must be 0.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-28T04:13:32.813" />
  <row Id="12174" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T04:15:38.770" Score="2" ViewCount="0" Body="&lt;p&gt;If $n$ is even then the block matrix  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \begin{pmatrix} 0 &amp;amp; I \\ -I &amp;amp; 0 \end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;works. For odd $n$ the matrix $T$ may be adjusted by simply including $\mathbb{R}^n$ in $\mathbb{R}^{n-1}$ by sending one component to zero. Of course this fails for $n=1$, and as noted in other answers $T$ must be zero for $n=1$.&lt;/p&gt;&#xA;" OwnerUserId="2887" LastEditorUserId="2887" LastEditorDisplayName="" LastEditDate="2010-11-28T04:19:29.717" LastActivityDate="2010-11-28T04:19:29.717" />
  <row Id="12175" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T04:15:46.870" Score="4" ViewCount="0" Body="&lt;p&gt;Presumably, your vectors are column vectors and $T$ is $n\times n$, so that $\alpha^t T\alpha$ is a $1\times 1$ matrix (i.e., a scalar). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying $v^t$ by $w$ amounts to taking the dot product of $v$ with $w$. Since $v\cdot w=0$ if and only if $v$ and $w$ are orthogonal, the question is equivalent to asking if you can find a linear transformation that sends every vector to a vector orthogonal to itself, and does not send all vectors to zero. In two dimensions, this is trivial. If you can do it in two dimensions, you can do it in any larger number (think about it). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-28T04:15:46.870" />
  <row Id="12176" PostTypeId="2" ParentId="12166" CreationDate="2010-11-28T04:32:42.077" Score="7" ViewCount="0" Body="&lt;p&gt;Imagine there are $n \geq 3$ circles surrounding your unit circle.  Then the situation would look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/M9dm8.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whence $\cos(x)=\frac{r}{r+1}$.  The angle $x$ is half of the &lt;a href=&quot;http://www.mathsisfun.com/geometry/interior-angles-polygons.html&quot; rel=&quot;nofollow&quot;&gt;interior angle&lt;/a&gt; of the corresponding regular polygon, so $x=\frac{n-2}{2n} \cdot 180^{\circ}$.  You can then solve for $r=\frac{\cos(x)}{1-\cos(x)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;when $n=4$ we have $r=\frac{\cos(45^{\circ})}{1-\cos(45^{\circ})}=1+\sqrt{2}=2.41421...$.&lt;/li&gt;&#xA;&lt;li&gt;when $n=6$ we have $r=\frac{\cos(60^{\circ})}{1-\cos(60^{\circ})}=1$.&lt;/li&gt;&#xA;&lt;li&gt;when $n=8$ we have $r=\frac{\cos(67.5^{\circ})}{1-\cos(67.5^{\circ})}=0.619914...$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-28T04:32:42.077" />
  <row Id="12177" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T04:38:45.723" Score="5" ViewCount="0" Body="&lt;p&gt;Let us look at the limit in the title and the definition of $\log x$  (please excuse me for using $\log$ notation instead of $\ln$) as in your comment,&#xA;$$\log x =\int_1^x \frac{1}{t}dt$$&#xD;&#xA;By the fundamental theorem of calculus we have &#xA;$$\frac{d}{dx}\log x = \frac{1}{x}$$&#xD;&#xA;on the other hand &#xA;$$f&amp;#39;(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$$&#xD;&#xA;hence &#xA;$$\lim_{y\to x}\frac{y-x}{\log y-\log x}=\lim_{y\to x}\frac{1}{\frac{\log y-\log x}{y-x}}=\frac{1}{1/x}=x$$&#xD;&#xA;where in the last step we used the quotient rule &#xA;$$\lim_{y\to a}\,g(y)=A\, \text{ and }\,\lim_{y\to a}\,h(y)=B\ne0\text{ implies }\lim_{y\to a}\,\frac{g(y)}{h(y)}=\frac{A}{B}$$&#xD;&#xA;with $g(y)=1$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-28T04:38:45.723" />
  <row Id="12178" PostTypeId="2" ParentId="11474" CreationDate="2010-11-28T04:42:50.983" Score="1" ViewCount="0" Body="&lt;p&gt;The following short piece of lecture notes is almost entirely devoted to an answer of your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/TopSection4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/TopSection4.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, for reasons that now escape me, I decided to break the file here, so the important remark that the statements &quot;$\tau$ is the topology generated by the family of sets $\mathcal{F}$&quot; and &quot;The family of sets $\mathcal{F}$ is a subbase for the topology $\tau$&quot; are equivalent does not appear until the beginning of the next little piece of notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/TopSection5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/TopSection5.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-28T04:42:50.983" CommentCount="2" />
  <row Id="12179" PostTypeId="1" CreationDate="2010-11-28T04:46:09.903" Score="1" ViewCount="133" Body="&lt;p&gt;How does one diagonalize a large sparse symmetric matrix to get the eigenvalues and the eigenvectors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the matrix could be very large (though it is sparse), at most 2500*2500. Is there a good algorithm to do that and most importantly, I can implement it into my own code. Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="2419" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-28T14:36:05.683" LastActivityDate="2010-11-28T14:36:05.683" Title="how to diagonalize a large sparse symmetric matrix, to get the eigenvalues and eigenvectors " Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12180" PostTypeId="1" AcceptedAnswerId="12184" CreationDate="2010-11-28T05:02:05.237" Score="0" ViewCount="158" Body="&lt;p&gt;A function $f(x)$ is defined $\forall x \in \mathbb{Q}$ and $ x \in (0,1)$ as $f(x) = q $ where $x = (p/q)$ in simplest form. Find $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)$    where $ a \in \mathbb{Q}$ and $ a \in (0,1)$ and $ \tau \in (\alpha,\beta)$ where   $ 0&amp;lt; \alpha &amp;lt; \beta &amp;lt;2 $ and $ \alpha,\beta \in \mathbb{Q} $. $\sup$ and $\min$ are taken over only those $x$ where $f(x)$ is defined. Is the supremum infinity ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supremum may or may not depend on $\alpha$ and $\beta$...this has not been ruled out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will try something !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\tau = l/m$ where $l,m \in \mathbb{N}$ in simplest form and $\tau \in \mathbb{Q}\cap(0,2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is minimum possible value of $q$ where $\tau-a = p/q $ written in simplest form&#xA;such that $a \in \mathbb{Q} \cap (0,1)$ and $(\tau-a) \in (0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Yuval Filmus suggested in his answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one way to choose $a$ is that it has same denominator (in simplest form) as that of $\tau$ and numerator $n$ such that $l-n$ is relatively prime to $m$, (If such a choice is possible under given conditions), then $f(a)+f(\tau-a)$ becomes $2m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question is whether such a choice leading to $f(a)+f(\tau-a) = 2m$ is the minimum possible value ? If not I request you to give a counter example.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is, then the supremum (over $\tau$ varying in any arbitrary open interval contained in $(0,2)$) is $\infty$ as we can increase $m$ arbitrarily by varying $\tau$ in any arbitrary open interval $(\alpha,\beta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;small change in Question...instead of&lt;/strong&gt; $0&amp;lt;\alpha&amp;lt;\beta&amp;lt;2$ we have $-1&amp;lt;\alpha&amp;lt;\beta&amp;lt;1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;this change is not made with any intention of bringing out a certain solution.Also its influence on the solution is hoped to be very small.Inconvenience is regretted.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-28T17:58:43.037" LastActivityDate="2010-11-28T21:26:39.943" Title="A question on Rational numbers in simplest form." Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12181" PostTypeId="1" CreationDate="2010-11-28T05:02:21.190" Score="1" ViewCount="99" Body="&lt;p&gt;I propose the following restriction on universal instantiation: UI may not be used to introduce new variables. The variable specified should be an &quot;old&quot; variable, i.e. it must already have been introduced by either an active premise or by existential instantiation (E-elimination). (An &lt;em&gt;active&lt;/em&gt; premise is one that has not been closed-off or discharged by what I call a &lt;em&gt;conclusion&lt;/em&gt; statement.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary benefit of this restriction is to eliminate the need to consider dependencies among variables that are created with existential instantiation. Such considerations never seem to come up in &quot;real mathematics.&quot; The restriction I propose seems to be implicit even in the most rigorous mathematical proofs, if not in standard formal logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this restriction in place, dependencies among variables need not be considered because the only variables on which you could then make universal generalizations are those introduced in a premise that has subsequently been closed-off or discharged by what I call a conclusion statement. The conclusion statement and subsequent statements that are derived from it and also refer to these variables can always be universally generalized regardless of the presence of any other variables that were introduced by E-elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have implemented this restriction in my proof checker/editor (available free at &lt;a href=&quot;http://www.dcproof.com&quot; rel=&quot;nofollow&quot;&gt;http://www.dcproof.com&lt;/a&gt;). Play around with it. Try to &quot;break it&quot; if you can. Work through the brief tutorial to learn the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another benefit, from a pedagogical perspective, has been the elimination from my program of several what must have been confusing warning messages about dependencies among variables. It also simplifies the requirements for universal generalizations (A-introduction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following is a summary of the rules of inference I use in my system that are relevant to the handling of free variables:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Free variables may be introduced only by means of a premise (assumption) or by existential specification (E-elimination).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Existential specification (E-elimination) allows an unused free variable to be specified for any active, existentially quantified statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Universal specification (A-elimination) allows any free variable introduced by an active premise or any algebraic expression in such free variables to be specified for any active, universally quantified statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Existential generalization (E-introduction) may be applied to any free variable or any algebraic expression in one or more free-variables that is found in any active statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Universal generalizations (A-introduction) may be applied to any free variable that (a) is found in an active statement, and (b) is &lt;em&gt;not&lt;/em&gt; referred to by any active premise, and (c) was &lt;em&gt;not&lt;/em&gt; introduced by existential specification (E-elimination). (There are no considerations given for any dependencies among variables.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Free variables introduced after a premise statement and before the corresponding conclusion statement may &lt;em&gt;not&lt;/em&gt; appear in that conclusion statement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In my program, I use color-coding for free variables. Green indicates that a Universal Generalization is possible. Red indicates that a Universal Generalization is &lt;em&gt;not&lt;/em&gt; possible. When first introduced, all free variables are red. A free variable that was introduced by a premise is changed to green when that premise is closed-off (or discharged) by a conclusion statement. It is effectively reintroduced, changing back to red if it is referred to in subsequent premise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your comments would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-29T17:44:51.867" LastActivityDate="2010-11-29T17:44:51.867" Title="Proposed Restriction on Universal Instantiation (Natural Deduction)" Tags="&lt;logic&gt;&lt;proof-theory&gt;" AnswerCount="1" />
  <row Id="12182" PostTypeId="2" ParentId="12181" CreationDate="2010-11-28T05:11:26.617" Score="1" ViewCount="0" Body="&lt;p&gt;You can always add some tautology as an assumption (say $x \rightarrow x$) and then later discharge it, so it seems like your restriction is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &quot;Structural Proof Theory&quot;, a book by von Plato and Negri, which discusses such issues; it also sponsors a proof editor PESCA by Aarne Ranta.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T05:11:26.617" CommentCount="5" />
  <row Id="12183" PostTypeId="2" ParentId="12162" CreationDate="2010-11-28T05:13:57.377" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose that the feed spindle has outer radius $r_f(t)$ and the intake spindle has outer radius $r_i(t)$, similarly let the spindle's angular velocity be given by $\omega_f(t)$ and $\omega_i(t)$. Note that both $r_f$ and $r_i$ lie in the interval $[r_0, R_0]$, where $r_0= .8$ cm and $R_0 = 1.8$ cm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only constraint is that the tape must pass between the two spindles at a constant rate of $s = 5$ cm/s. For the tape to remain taught under this condition the tangential velocity of each spindle must be $s$. Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s = r_f(t)\cdot\omega_f(t) = r_i(t)\cdot\omega_i(t).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega(t) = \frac{s}{r(t)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for either spool. So now we simply need to find a description of the outer radius of the spindle. Consider that the tape has thickness $a \ll r_0$ and width $w$, thus the volume of tape transfered per unit time is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dV}{dt} = \pm a s w$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $V = \pi w r(t)^2$ then we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dV}{dt} = 2\pi r(t) \dot{r}(t) w$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then by equating these expressions and solving the IVP for the feeding spindle yields,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_f(t) = \sqrt{R_0^2 - \frac{as}{\pi} t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the IVP for the intake spindle yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_i(t) = \sqrt{r_0^2 + \frac{as}{\pi} t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From those expressions it is easy to see that the proper speeds to drive the spindles are approximately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_f(t) = \left(\sqrt{\left(\frac{R_0}{s}\right)^2 - \frac{a}{s\pi} t}\right)^{-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_i(t) = \left(\sqrt{\left(\frac{r_0}{s}\right)^2 + \frac{a}{s\pi} t}\right)^{-1}.$$&lt;/p&gt;&#xA;" OwnerUserId="2887" LastActivityDate="2010-11-28T05:13:57.377" CommentCount="3" />
  <row Id="12184" PostTypeId="2" ParentId="12180" CreationDate="2010-11-28T05:15:03.290" Score="2" ViewCount="0" Body="&lt;p&gt;EDIT: This is a consolidated solution which previously appeared in parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1. For each natural $M$ and real $\alpha &amp;lt; \beta$ there's a rational number in $(\alpha,\beta)$ whose reduced form has a denominator larger than $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Choose some power of two $2^n &amp;gt; \max\left(M, \frac{2}{\beta-\alpha}\right)$. Since $2^{-n} &amp;lt; \frac{\beta-\alpha}{2}$, there are at least two fractions of the form $x/2^n$ in $(\alpha,\beta)$. One of them has odd $x$ and so is in reduced form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2. If $p_1/q_1 + p_2/q_2 = P/Q$, all rationals in reduced form, then $(q_1+q_2)^2 \geq 4Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Taking a common denominator, $\frac{p_1 q_2 + p_2 q_1}{q_1q_2} = \frac{P}{Q}$ and so $Q|q_1q_2$ (since $P/Q$ is reduced). In particular $Q \leq q_1q_2 \leq \left(\frac{q_1+q_2}{2}\right)^2$, where the last inequality is the arithmetic-geometric mean inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem. For every real $M$ and reals $\alpha &amp;lt; \beta$ there is a rational $r \in (\alpha,\beta)$ such that $\min_{a \in \mathbb{Q}} f(a) + f(r-a) &amp;gt; M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Using lemma 1, take a rational $r \in (\alpha,\beta)$ with denominator $Q &amp;gt; M^2/4$. Using lemma 2, we conclude that for every rational $a$, $f(a) + f(r-a) \geq 2\sqrt{Q} = M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary. $$\sup_r \min_a f(a) + f(r-a) = \infty.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-28T21:26:39.943" LastActivityDate="2010-11-28T21:26:39.943" CommentCount="15" />
  <row Id="12185" PostTypeId="2" ParentId="9934" CreationDate="2010-11-28T05:31:50.503" Score="2" ViewCount="0" Body="&lt;p&gt;Let $G$ be a group.  The &lt;em&gt;canonical residually finite quotient&lt;/em&gt; of $G$ is $R(G)=G/K$ where $K$ is the intersection of all the finite-index subgroups of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt;  &lt;em&gt;If $G$ is finitely generated (update)&lt;/em&gt; then $G$ has at most one subgroup of each index if and only if $R(G)$ is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; First, note that $R(G)$ is residually finite.  If every finite quotient of $R(G)$ is cyclic then $R(G)$ is residually cyclic, and it follows that $R(G)$ is abelian.  So $R(G)$ has a non-cyclic finite quotient unless $R(G)$ is cyclic.  Therefore, if $R(G)$ is not cyclic then $R(G)$, and hence $G$, has a finite non-cyclic quotient, and hence, by Artuto's answer, has a two distinct finite-index subgroups of the same index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $R(G)$ is cyclic.  Every finite-index subgroup of $G$ contains $K$, so the quotient map $G\to R(G)$ maps finite-index subgroups to finite-index subgroups bijectively and preserves the index.  Therefore, if $R(G)$ is cyclic then $G$ has at most one subgroup of each index.  &lt;em&gt;QED&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that it is an open question whether or not there is an algorithm to determine whether a fp group has a proper finite-index subgroup, ie whether or not $R(G)$ is non-trivial.  So it may be open whether or not it is possible to determine if $R(G)$ is cyclic, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Earlier, I forgot to mention that I had implicitly assumed that $G$ is finitely generated.  This assumption is clearly necessary; otherwise the additive group of the rationals is a counterexample.  If $G$ is not finitely generated, then the same argument shows that if $G$ has at most one subgroup of each finite index then $R(G)$ is &lt;em&gt;residually&lt;/em&gt; cyclic.  But it's not clear to me that the converse of this statement is true.   So I'll finish with a question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $G$ is residually cyclic, does $G$ have at most one subgroup of each finite index?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-11-28T15:01:03.837" LastActivityDate="2010-11-28T15:01:03.837" CommentCount="1" />
  <row Id="12186" PostTypeId="1" CreationDate="2010-11-28T05:32:49.567" Score="1" ViewCount="89" Body="&lt;p&gt;I posted this on &lt;a href=&quot;http://gamedev.stackexchange.com/q/6009/3637&quot;&gt;gamedev.stackexchange.com&lt;/a&gt; earlier&lt;s&gt;, but I didn't get any answers... and only a few views&lt;/s&gt;.&#xA;EDIT: I've got two answers there (as of now), and I'm investigating them now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find out the distance traveled by a Bezier Curve?&#xA;For example, the distance traveled by a Linear Bezier Curve is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$distance = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what of a Quadratic, Cubic, Nth-Degree Bezier Curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to figure out the &quot;resolution&quot; ($resolution = 1-t$) beforehand, so I don't have to waste time checking if the next point is touching the previous point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\mathbf{B}(t) = \sum_{i=0}^n {n\choose i}(1-t)^{n-i}t^i\mathbf{P}_i$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-11-28T21:35:12.987" LastActivityDate="2010-11-28T21:35:12.987" Title="Arc Length of Bezier Curves" Tags="&lt;bezier-curve&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="12188" PostTypeId="2" ParentId="11668" CreationDate="2010-11-28T06:14:09.763" Score="2" ViewCount="0" Body="&lt;p&gt;the real complication here is the &lt;em&gt;joint&lt;/em&gt; probability diatribution of $X$ and $Y$. in introductory courses, one learns about the joint pdf of two continuous random variables, or the joint pmf of two discrete random variables. the mixed case you mention, where one variable is continuous and the other is discrete is not commonly mentioned in introductory texts - altho some manage to sneak it in without calling attention to just what they are dealing with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a common example is this: suppose $X$ has the binomial distribution bin($n,u$), where $u$ is the success probability. each day a new $X$ is observed. however, the value of $u$ changes from day to day in a random way: it is selected each time at random uniformly from the interval (0,1). because the success probability $u$ is also random, we should denote this random quantity [before it is selected] by $U$. [in bayesian contexts, this uniform distribution for $U$ is referred to as a 'prior pdf' for $U$. however, unlike the bayesian context, which focuses on $U$, our focus here is on $X$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have here a mixed pair of random variables $(X,U)$, where one is discrete and one is continuous [as in the OP's question.] here the [mixed] joint probability distribution of $(X,U)$ is specified by two distributions: the [marginal] pdf of $U$: $f_U(u) = 1$ for $0 &amp;lt; u &amp;lt; 1$ [and is $0$ elsewhere] -  and what is really the conditional distribution of $X$ given $U$, for which the conditional pmf is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{X|U}(x|u) = P(X=x|U=u) = P(bin(n,u) = x) = {n\choose x} u^x(1-u)^{n-x},\quad x = 0,1,\cdots, n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it turns out that the joint [mixed] pmf-pdf of $(X,U)$ can be written as the product of the marginal pdf of $U$ and the conditional pmf of $X|U$, analogous to what one does in the case where both variables are discrete, or both are continuous. [the justification for this involves what kahen alludes to in his answer - but it goes beyond the scope of this reply.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can write the mixed joint pmf-pdf of $(X,U)$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_{X,U}(x,u) = f_U(u)f_{X|U}(x|u) = {n\choose x} u^x(1-u)^{n-x},\quad x = 0,1,\cdots, n,\quad 0&amp;lt;u&amp;lt;1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[here $f_{X,U}(x,u)$ is understood to vanish for other values of $(x,u)$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this mixed pmf-pdf can be treated similarly to the way one manipulates a joint pmf or pdf [to get marginal and conditional distributions, for example] - except that one integrates out $u$ to get the marginal pmf of $X$ and sums out $x$ to get the marginal pdf of $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \int_0^1 f_{X,U}(x,u)du  =\int_0^1 {n\choose x} u^x(1-u)^{n-x}du = \frac{1}{n+1},\quad x = 0,1,\dots,n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus $X$ is marginally uniformly distributed on the set of outcomes $\{ 0,1,\dots, n\} $.  [this means that over a large number of days, all of the possible values of $X$ (from $0$ to $n$) will turn up about equally often. given the way $U$ varies from day to day, this seems plausible.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large \mathbf [$ the evaluation of the integral is straightforward on noting that it is a beta function integral, so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^1 u^x(1-u)^{n-x}du = Beta(x+1, n-x+1) = \frac{x!(n-x)!}{(n+1)!}. {\mathbf ]}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one can similarly get the conditional pdf of $U|X$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{U|X}(u|x) = \frac{f_{X,U}(x,u)}{f_X(x)} = (n+1){n\choose x}u^x(1-u)^{n-x},\quad 0&amp;lt;u&amp;lt;1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this means that looking only at the days when $X=x$, conditionally, $U$ has a beta distribution with parameters $(x+1, n-x+1)$. this result is what interests bayesians in their contexts - altho their scenario would usually involve considering the value of $X$ only for day 1, and their interpretation of $f_{U|X}(u|x)$ isn different from that given here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the bottom line is that in working with a mixed pmf-pdf, each variable retains its discrete or continuous nature whether one considers marginal, joint or conditional distributions, and the joint mixed pmf-pdf  can be manipulated as in the pure [= unmixed] cases, remembering to sum or integrate as is appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this reply does not address the specific problem posed in the OP, but it can be handled by the same methods. [only the imputs are a bit different: there one starts with [the joint pmf-pdf of] $(X,Y)$ and proceeds to $(X,Z)$. starting from the mixed pmf-pdf of $(X,Y)$, it is straightforward to get the conditional pmf of $Z|X$ and then the joint mixed pmf-pdf of $(X,Z)$.] &lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-11-28T06:14:09.763" />
  <row Id="12189" PostTypeId="1" AcceptedAnswerId="12191" CreationDate="2010-11-28T06:43:58.010" Score="2" ViewCount="49" Body="&lt;p&gt;As a homework assignment, I've been given a particular subring of $M_2(\mathbb{Q})$, and asked to list all the ideals... For reference, $S$ = { set of matrices in $M_2(\mathbb{Q})$ with bottom left entry $0$ } is the subring in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really see how to get a handle on this. Is there any sort of algorithmic way of doing this, or...? Where do I start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I posted this over at Ask an Algebraist originally: &lt;a href=&quot;http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2257&quot; rel=&quot;nofollow&quot;&gt;http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2257&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T12:12:45.637" LastActivityDate="2010-11-28T12:12:45.637" Title="Ideals of subrings of $M_2(\mathbb{Q})$" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="12190" PostTypeId="2" ParentId="11892" CreationDate="2010-11-28T06:50:18.003" Score="1" ViewCount="0" Body="&lt;p&gt;(1) To use your notation, if $c \in \mathbb{R}$ is a regular value, then the level set $Re(f(z))^{-1}(c)$ will be a 1-dimensional embedded submanifold of $U \subset \mathbb{R}^2$.  Therefore, every connected component will be a connected 1-manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm not sure what exactly you mean by &quot;global differentiable curve.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean &quot;something of the form $f(t) = (x(t), y(t))$,&quot; then it follows from the comments in &lt;a href=&quot;http://math.stackexchange.com/questions/4699/which-smooth-1-manifolds-can-be-represented-by-a-single-smooth-parametrization&quot;&gt;this question&lt;/a&gt; that yes, every connected component can be put in that form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean &quot;something of the form $y = f(x)$ or $x = f(y)$&quot; then the answer is (I think) no.  For example, consider $f(z) = \log z$ on $U = R^2 - \{x \geq 0, y = 0\}$, i.e. the plane with the non-negative x-axis deleted.  Then $Re(f(z)) = \log(\sqrt{x^2+y^2})$, so the level set $Re(f(z))^{-1}(1)$ is the circle $x^2 + y^2 = e^2$ minus the point $(e,0)$.  So, it doesn't seem like you'd be able to represent this curve by a single function $x = f(y)$ or $y = f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-11-28T06:50:18.003" CommentCount="3" />
  <row Id="12191" PostTypeId="2" ParentId="12189" CreationDate="2010-11-28T06:50:24.407" Score="7" ViewCount="0" Body="&lt;p&gt;Start by looking at ideals generated by one element - choose a matrix and see what happens when you multiply it from the left and right, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every ideal contains an ideal generated by one element, so now you need to see if you can extend some of the ideals you got to larger ideals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing is that ideals in your ring are also vector spaces over the Rationals, because the ring &quot;contains&quot; $\mathbb{Q}$ as the scalar matrices, so you know that once you have an ideal of dimension 2 it is maximal and cannot be extended (since the entire vector space is of dimension 3).&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-11-28T06:50:24.407" />
  <row Id="12192" PostTypeId="1" CreationDate="2010-11-28T06:57:48.860" Score="3" ViewCount="49" Body="&lt;p&gt;I would appreciate some help with factoring a trivariate polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polynomial in question is&#xA;$p(x,y,z)=a_1 x^7+a_2 x^5*y+a_3 x^3*y^2+a_4 x*y^3+a_5 x^4*z+a_6 x^2*y*z+a_7 y^2*z+a_8 x*z^2$, where the coefficients are integers. I would like to assign values to these coefficients such that $p$ can be factored.&#xA;For example, one possible factorization might be&#xA;$p(x,y,z)=(x^3+b_1 x*y+b_2 z)(b_3 x^4+b_4 x^2*y+b_5 x*z+b_6 y^2)$.&#xA;One can expand this factorization and equate the coefficients of the expansion with those of the above equation and hope that the system of equations can be solved for integers b's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one find other possible factorizations? Is there an easy way?&lt;/p&gt;&#xA;" OwnerUserId="3891" LastActivityDate="2010-11-28T06:57:48.860" Title="Factoring a trivariate polynomial" Tags="&lt;algebra&gt;&lt;diophantine-equations&gt;" />
  <row Id="12193" PostTypeId="1" AcceptedAnswerId="12267" CreationDate="2010-11-28T07:40:27.607" Score="2" ViewCount="180" Body="&lt;p&gt;This is a follow-up to the question &#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/12082/field-reductions&quot;&gt;on field reductions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT: Originally this question used the notation $\{\mathbb{R}(\setminus a)\}$, but now uses $\mathbb{R}(\setminus a)$ instead.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't a unique largest subfield of $\mathbb{R}$ not containing $a$,&#xA;so let $\mathbb{R}(\setminus a)$ denote the set of maximal subfields of $\mathbb{R}$ not containing $a$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The question &quot;What is the cardinality of $\mathbb{R}(\setminus a)$ ?&quot; has now been asked on &lt;a href=&quot;http://mathoverflow.net/questions/47810/set-of-maximal-subfields-not-containing-particular-elements&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the previous question if $F\in \mathbb{R}(\setminus a)$ then $F$ is uncountable, but what is the size of $\mathbb{R}\setminus F$ the complement of $F$ in $\mathbb{R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $F \in \mathbb{R}(\setminus a)$ should be arrived at by removing elements of $\mathbb{R}$, but if you start with $\mathbb{R}$ and remove $a$ and then remove more elements until you get a field then surely you arrive at a single field, so where do the multiple maximal fields come from ? - It must be because you get different fields depending on the order in which elements are removed from $\mathbb{R}$, in which case what is the most natural order to remove elements to get a definition of a single canonical subfield $\mathbb{R}(\setminus a)$ of $\mathbb{R}$ that doesn't contain $a$, that we could call 'the' field reduction of $\mathbb{R}$ by $a$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Given Arturo Magidin's comments and answer, if we abandon the idea of a distinguished element in the abstract/in general, then since the notation $\{\mathbb{R}(\setminus a)\}$ looks like a set containing a single element called $\mathbb{R}(\setminus a)$, it makes more sense to drop the brackets and just write $\mathbb{R}(\setminus a)$ on the understanding that this is a set of field reductions i.e. a set of subfields not a single subfield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Pete L. Clark's answer to &lt;a href=&quot;http://math.stackexchange.com/questions/12082/field-reductions&quot;&gt;the previous question&lt;/a&gt;, if $F(a)=\mathbb{R}$ then $F=\mathbb{R}$, so let's define the notion of a super-extension of $F$ by $a$, $F(a..)=R$ being the sequence of field extensions to get to $\mathbb{R}$ from $F\in\mathbb{R}(\setminus a)$. More generally, let $@K(\setminus a)$ be an element of $K(\setminus a)$ then $@K(\setminus a)(a..)=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also reduce $K$ by a set $A$, with $K(\setminus A)$ being the set of maximal subfields that don't contain any elements from $A$, and again write the super-extension $@K(\setminus A)(A..)=K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb{A}&amp;#39;$ be the set of non-rational algebraic numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the cardinality of $\mathbb{R}(\setminus \mathbb{A}&amp;#39;)$ ?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="4157" LastEditorDisplayName="" LastEditDate="2010-11-30T17:01:02.337" LastActivityDate="2010-11-30T17:01:02.337" Title="Field reductions. part two" Tags="&lt;abstract-algebra&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="12194" PostTypeId="1" CreationDate="2010-11-28T08:30:24.823" Score="5" ViewCount="391" Body="&lt;p&gt;Ibid the topic title.&lt;/p&gt;&#xA;" OwnerUserId="4158" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-29T02:22:57.623" LastActivityDate="2010-11-29T02:22:57.623" Title="I know that &quot;x exp(x) = e&quot; means x = 1, but how do you solve for it?" Tags="&lt;calculus&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="12195" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T08:42:52.680" Score="17" ViewCount="0" Body="&lt;p&gt;If you want to prove the proposition that if $x e^x = e$ then $x = 1$, one way to do it is to just show that (1) $x = 1$ is indeed a solution of $x e^x = e$ by plugging the value in, and that (2) $x = 1$ is the &lt;em&gt;only&lt;/em&gt; solution, for which you can argue that $x e^x$ is an increasing function on $x \ge 0$, and the equation clearly has no solution for $x &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that makes it sounds like you have to know the answer beforehand before you can solve for it, well, that's not far from the truth. In general, if the right-hand side is an arbitrary real number $y$ instead of $e$, then you cannot solve $x e^x = y$ for $x$ in terms of elementary functions. One in fact &lt;em&gt;defines&lt;/em&gt; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambert_W_function&quot; rel=&quot;nofollow&quot;&gt;Lambert W function&lt;/a&gt; as precisely the solution of this equation.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-28T09:07:58.680" LastActivityDate="2010-11-28T09:07:58.680" CommentCount="3" />
  <row Id="12196" PostTypeId="2" ParentId="11091" CreationDate="2010-11-28T09:20:19.337" Score="2" ViewCount="0" Body="&lt;p&gt;I &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Lights_Out_%28game%29&amp;amp;oldid=399287365&quot; rel=&quot;nofollow&quot;&gt;tried&lt;/a&gt; (and &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Lights_Out_%28game%29&amp;amp;oldid=399483159&quot; rel=&quot;nofollow&quot;&gt;tried again&lt;/a&gt;) to improve the Wikipedia article but the edits were promptly reverted as if well-known solutions of Lights Out are &quot;original research&quot;, despite providing several citations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any Stack Exchange members are Wikipedians who can assist in restoring the article, it would be greatly appreciated. Also, I would direct this request more appropriately if I could, but don't see a way to contact John Smith directly. Feel free to delete it.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="1174" LastEditorDisplayName="" LastEditDate="2010-11-29T08:10:25.650" LastActivityDate="2010-11-29T08:10:25.650" CommentCount="2" />
  <row Id="12197" PostTypeId="1" AcceptedAnswerId="12226" CreationDate="2010-11-28T09:29:22.530" Score="5" ViewCount="102" Body="&lt;p&gt;My friend claims that one could reduce the number of states of a given turning  machine by somehow blowing up the tape alphabet. He does not have any algorithm though. He only has the intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I say it's not possible. Else one could arbitrarily keep decreasing the states via the same algorithm and arrive at some constant sized machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is right?&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2010-11-28T17:53:33.493" Title="Can we reduce the number of states of a Turing Machine?" Tags="&lt;algorithms&gt;&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12198" PostTypeId="2" ParentId="12163" CreationDate="2010-11-28T10:17:29.837" Score="3" ViewCount="0" Body="&lt;p&gt;We can convert your $\lim $ into the inverse of the derivative of $f(y)=\ln y&#xD;&#xA;$, evaluated at $y=b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{y\rightarrow b}{\lim }\dfrac{y-b}{\ln y-\ln b}=\dfrac{1}{\underset{&#xD;&#xA;y\rightarrow b}{\lim }\dfrac{\ln y-\ln b}{y-b}}=\dfrac{1}{\dfrac{d}{dy}\left.&#xD;&#xA;\ln y\right\vert _{y=b}}=\dfrac{1}{\left. \dfrac{1}{y}\right\vert _{y=b}}=%&#xD;&#xA;\dfrac{1}{\dfrac{1}{b}}=b.$$&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-11-28T10:17:29.837" />
  <row Id="12199" PostTypeId="2" ParentId="12197" CreationDate="2010-11-28T10:21:09.927" Score="0" ViewCount="0" Body="&lt;p&gt;I think this in the same vein as creating a compression algorithm that will compress any given file, i.e. than we can compress the output again and again, until we reach a single bit that will represent all possible files. Yet, compression algorithms do exist, and they do compress &lt;strong&gt;some&lt;/strong&gt; files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, even it the number of states of a given Turing machine is reducible, it does not mean that all Turing machines are reducible, since that would mean that all Turing machines are just different interpretations of one and the same one-state machine.&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditorDisplayName="" LastEditDate="2010-11-28T10:41:52.147" LastActivityDate="2010-11-28T10:41:52.147" CommentCount="2" />
  <row Id="12201" PostTypeId="2" ParentId="12179" CreationDate="2010-11-28T12:05:01.273" Score="3" ViewCount="0" Body="&lt;p&gt;$2500 \times 2500$ is a &lt;strong&gt;small&lt;/strong&gt; matrix by current standards. The standard eig command of matlab should be able to handle this size with ease. Iterative sparse matrix eigensolvers like those implemented in ARPACK, or SLEPc will become more preferable if the matrix is much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you want to implement an eigensolver into your own code, just use the LAPACK library that comes with very well developed routines for such purpose. Matlab also ultimately invokes LAPACK routines for doing most of its numerical linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semi-related note: the matrix need not be explicitly available for the large sparse solvers, because they usually just depend on being able to compute $A*x$ and $A&amp;#39;*x$.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="1709" LastEditorDisplayName="" LastEditDate="2010-11-28T12:15:30.517" LastActivityDate="2010-11-28T12:15:30.517" CommentCount="4" />
  <row Id="12202" PostTypeId="1" AcceptedAnswerId="12207" CreationDate="2010-11-28T12:10:29.180" Score="1" ViewCount="89" Body="&lt;p&gt;Define $A$ as a nonempty set, $\mathcal{B}:=\{f: A \rightarrow \mathbb{R}: f(A) \text{is bounded} \} ,d_\infty:=\text{sup}\{|f(x)-g(x)|:x \in A\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which $A$ is $\overline {B_1(0)} \subset \mathcal{B}$ compact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: ${B_1(0)}$ is the unit sphere. I already proved that $(\mathcal{B},d_\infty)$ is a metric space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts: I investigated a bit on this and found some proofs that the unit sphere in a banach space is compact when the banach space is finite dimensional. But I don't want to use this here, I don't even know if $(\mathcal{B},d_\infty)$ is a banach space. All proofs we did who depend on the dimension of a metric space is that the intersections of compact subsets is compact and that the finite union of compact subsets is compact. Maybe we can use this?&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-11-28T13:55:34.680" LastActivityDate="2010-11-28T13:55:34.680" Title="Unit sphere compactness in a metric space" Tags="&lt;functions&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12203" PostTypeId="1" AcceptedAnswerId="12210" CreationDate="2010-11-28T12:40:48.570" Score="1" ViewCount="90" Body="&lt;p&gt;Hello!&#xA;I want to ask, how can I simplify this formula ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ e^{-i0.5t}+e^{i0.5t} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it can be simplify to $\cos(0.5t)$, but I don't know how :/&lt;/p&gt;&#xA;" OwnerUserId="4164" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-28T15:27:20.403" LastActivityDate="2010-11-28T15:27:20.403" Title="Simplify formula $e^{-i0.5t}+e^{i0.5t}$" Tags="&lt;complex-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12204" PostTypeId="2" ParentId="12203" CreationDate="2010-11-28T12:47:20.613" Score="3" ViewCount="0" Body="&lt;p&gt;Use $e^{ia}  = \cos a + i\sin a$, and find that the answer is slightly different from what you wrote.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-28T12:47:20.613" />
  <row Id="12205" PostTypeId="1" CreationDate="2010-11-28T12:53:14.977" Score="5" ViewCount="127" Body="&lt;p&gt;I wonder if someone could shed some light in the following question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x,y)$ denote the greatest common divisor of $x$ and $y$, and let $x_1,y_1,x_2,y_2$ be integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is  the following statement true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(x_1,y_1)=(x_2,y_2)=(x_1^2+y_1^2,x_2^2+y_2^2)=1$, then $$(x_1x_2\pm y_1y_2, x_1y_2\mp x_2y_1)=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what further hypotheses are necessary to guarantee the claim?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;Guillermo&lt;/p&gt;&#xA;" OwnerUserId="4165" LastActivityDate="2010-11-29T11:11:11.917" Title="About the greatest common divisor of certain expressions..." Tags="&lt;number-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="12206" PostTypeId="2" ParentId="12172" CreationDate="2010-11-28T13:06:50.410" Score="10" ViewCount="0" Body="&lt;p&gt;The property holds if and only if $T$ is skew-symmetric (meaning that $T^t=-T$). (The other answers so far are special cases of this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Let $x=\alpha^t T \alpha$. Since $x$ is a $1\times 1$ matrix, it equals its own transpose. Thus $2x=x+x^t=\alpha^t T \alpha+(\alpha^t T \alpha)^t =\alpha^t (T+T^t)\alpha$, which is a quadratic form in $\alpha$ that corresponds to the symmetric matrix $T+T^t$; thus it is zero for all $\alpha$ iff $T+T^t=0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-28T13:06:50.410" />
  <row Id="12207" PostTypeId="2" ParentId="12202" CreationDate="2010-11-28T13:50:09.807" Score="0" ViewCount="0" Body="&lt;p&gt;The result you mention has a converse: the closed unit ball in a Banach space is compact &lt;strong&gt;if and only if&lt;/strong&gt; the Banach space is finite-dimensional.  That should suggest to you what is true here.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-28T13:50:09.807" CommentCount="2" />
  <row Id="12209" PostTypeId="2" ParentId="12160" CreationDate="2010-11-28T14:18:33.437" Score="3" ViewCount="0" Body="&lt;p&gt;I'll answer question 1 only for now, but I might edit this to address the others later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should note that corresponding to any set of orthogonal polynomials, there exists a symmetric tridiagonal matrix, called a &lt;strong&gt;Jacobi matrix&lt;/strong&gt;, whose characteristic polynomial is the monic (leading coefficient is 1) version of the set of orthogonal polynomials considered. To use the Legendre polynomials as an explicit example, we first note that the monic Legendre polynomials satisfy the following two-term recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{P}_{n+1}(x)=x \hat{P}_n(x)-\frac{n^2}{4 n^2-1}\hat{P}_{n-1}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{P}_n(x)=\frac{(n!)^2 2^n}{(2n)!}P_n(x)$ is the monic Legendre polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can derive an explicit expression for the corresponding Jacobi matrix (here I give the 5-by-5 case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}0&amp;amp;\frac{1}{\sqrt{3}}&amp;amp;0&amp;amp;0&amp;amp;0\\\frac{1}{\sqrt{3}}&amp;amp;0&amp;amp;\frac{2}{\sqrt{15}}&amp;amp;0&amp;amp;0\\0&amp;amp;\frac{2}{\sqrt{15}}&amp;amp;0&amp;amp;\frac{3}{\sqrt{35}}&amp;amp;0\\0&amp;amp;0&amp;amp;\frac{3}{\sqrt{35}}&amp;amp;0&amp;amp;\frac{4}{\sqrt{63}}\\0&amp;amp;0&amp;amp;0&amp;amp;\frac{4}{\sqrt{63}}&amp;amp;0\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the general pattern is that you have $\frac{n}{\sqrt{4 n^2-1}}$ in the $(n,n+1)$ and $(n+1,n)$ positions, and 0 elsewhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now note that $\frac{n}{\sqrt{4 n^2-1}}$ can never be 0, and then use the fact that if a symmetric tridiagonal matrix has no zeroes in its sub- or superdiagonal, then all its eigenvalues have multiplicity 1. (A proof of this fact can be found in Beresford Parlett's &lt;a href=&quot;http://books.google.com/books?id=mWin91cLndsC&amp;amp;pg=PA134&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Symmetric Eigenvalue Problem&lt;/em&gt;&lt;/a&gt;.) Thus, all the roots of the Legendre polynomial are simple roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more conventional proof of this fact is in page 27 of Theodore Chihara's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0677041500&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;An Introduction to Orthogonal Polynomials&lt;/em&gt;&lt;/a&gt;. Briefly, the argument is that $P_n(x)$ changes sign at least once within $[-1,1]$ (and thus has at least one zero of odd multiplicity within the support interval) since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-1}^1 P_n(u)\mathrm du=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_n(x)\prod_{j=1}^k(x-\xi_j)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $\xi_j$ are the distinct zeroes of odd multiplicity within $[-1,1]$, should be greater than or equal to zero within $[-1,1]$, and thus its integral over $[-1,1]$ should be greater than zero. However, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-1}^1 P_n(u) u^k\mathrm du=0\qquad\text{if}\qquad k &amp;lt; n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have a contradiction, and thus all the roots of the Legendre polynomial are simple (and within the support interval $[-1,1]$).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-28T14:18:33.437" CommentCount="1" />
  <row Id="12210" PostTypeId="2" ParentId="12203" CreationDate="2010-11-28T14:27:32.040" Score="1" ViewCount="0" Body="&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{ix}=\cos x+i\sin x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $x=-0.5t$ , we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{-0.5t}=e^{i\left( -0.5t\right) }=\cos \left( -0.5t\right) +i\sin \left(&#xD;&#xA;-0.5t\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for $x=0.5t$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{0.5t}=\cos \left( 0.5t\right) +i\sin \left( 0.5t\right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\cos$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_function#Even_functions&quot; rel=&quot;nofollow&quot;&gt;even function&lt;/a&gt; and $\sin$ an &lt;a href=&quot;http://en.wikipedia.org/wiki/Even_function#Even_functions&quot; rel=&quot;nofollow&quot;&gt;odd function&lt;/a&gt;, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;e^{-0.5t}+e^{0.5t} &amp;amp;=&amp;amp;\cos \left( -0.5t\right) +i\sin \left( -0.5t\right)&#xD;&#xA;+\cos \left( 0.5t\right) +i\sin \left( 0.5t\right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \cos \left( -0.5t\right) +\cos \left( 0.5t\right) \right) +i\left(&#xD;&#xA;\sin \left( -0.5t\right) +\sin \left( 0.5t\right) \right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \cos \left( 0.5t\right) +\cos \left( 0.5t\right) \right) +i\left(&#xD;&#xA;-\sin \left( 0.5t\right) +\sin \left( 0.5t\right) \right)  \\&#xD;&#xA;&amp;amp;=&amp;amp;2\cos \left( 0.5t\right) +i\cdot 0 \\&#xD;&#xA;&amp;amp;=&amp;amp;2\cos \left( 0.5t\right) &#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-28T14:34:24.890" LastActivityDate="2010-11-28T14:34:24.890" CommentCount="2" />
  <row Id="12211" PostTypeId="2" ParentId="12180" CreationDate="2010-11-28T15:10:24.373" Score="1" ViewCount="0" Body="&lt;p&gt;For a given $\tau=\frac{1}{q}, q$ prime it seems that $a$ should be (perhaps among other choices) $\frac{\tau}{2}$ with $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)=2q$.  Can you prove this?  If so, then the sup over $\tau$ is infinity as you can find $\tau$ with arbitrarily large denominator.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-28T15:29:24.163" LastActivityDate="2010-11-28T15:29:24.163" CommentCount="2" />
  <row Id="12212" PostTypeId="2" ParentId="8224" CreationDate="2010-11-28T15:22:04.847" Score="0" ViewCount="0" Body="&lt;p&gt;We have (using ab for those who favor I and II, n for those who favor none, etc)&#xA;a+b+c+2ab+2ac+2bc+3abc=?&#xA;a+b+c+2ab+2ab+2ac=?&#xA;a+b+c+ab+ac+bc+abc=?&#xA;a+b+c+ab+ac+bc=?&#xA;and so?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-28T15:22:04.847" />
  <row Id="12213" PostTypeId="1" AcceptedAnswerId="12214" CreationDate="2010-11-28T15:32:13.077" Score="5" ViewCount="126" Body="&lt;p&gt;Let C be a collection of subsets of $\mathbb{N}$ such that $A,B\in C \Rightarrow A \not\subseteq B$. Can C be uncountable?&lt;/p&gt;&#xA;" OwnerUserId="2934" LastActivityDate="2010-11-28T16:22:32.067" Title="Can a collection of subsets of N such that no one set contains another  be uncountable?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="12214" PostTypeId="2" ParentId="12213" CreationDate="2010-11-28T15:45:08.570" Score="8" ViewCount="0" Body="&lt;p&gt;Yes. Consider the collections of subsets which contain exactly one of $2i$ and $2i+1$, for all $i$.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-11-28T15:45:08.570" CommentCount="2" />
  <row Id="12215" PostTypeId="2" ParentId="8224" CreationDate="2010-11-28T16:16:04.023" Score="1" ViewCount="0" Body="&lt;p&gt;The simplest way is probably to draw Venn diagrams and stuff, but here's how you do it using the inclusion-exclusion principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any three sets $A,B,C$, we have $$|A \cup B \cup C| = S_1 - S_2 + S_3,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S_1 = |A|+|B|+|C|$, $S_2 = |A \cap B| + |B \cap C| + |C \cap A|$, and $S_3 = |A \cap B \cap C|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $A,B,C$ denote the obvious sets, you are given the values of $|A \cup B \cup C|$, $S_1$ and $S_3$ so you can solve for $S_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are asked to find $x = |(A \cap B) \cup (B \cap C) \cup (C \cap A)|$, which we can expand with the same formula (using the fact that $(A \cap B) \cap (B \cap C) = A \cap B \cap C$ etc.) as $$x = |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| + 3|A \cap B \cap C| = S_2 - 2S_3.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do this right, and assuming I've done it right, the answer should be 17%.&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2010-11-28T16:16:04.023" />
  <row Id="12216" PostTypeId="2" ParentId="12205" CreationDate="2010-11-28T16:16:43.400" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \: $ Interpret it in Gaussian integers $\rm\: \mathbb Z[i]\:.\: $ Define $\rm\ a + b\ i\ $ to be primitive if it has no rational prime divisor, i.e. $\rm\ (a,b) = 1\:.\:$ Then the statement says that the product of two primitive Gaussian integers is primitive if they have coprime norms. Put $\rm\ \ \alpha = x_1+y_1\ i,\ $ $\rm\ \beta = x_2+y_2\ i\:.\ $ Suppose $\rm\ p\:|\:\alpha\beta\ $ for a prime $\rm\: p\in \mathbb Z\:.\:$ If $\rm \: p\:$ is prime in $\rm\: \mathbb Z[i]\: $ then $\rm\ p\:|\:\alpha\ $ or $\rm\ p\:|\:\beta\ $ contra $\ldots\ $  Else $\rm\: p = \pi\pi&amp;#39;\: $ for a prime $\rm\: \pi \in \mathbb Z[i]\:$ hence $\rm\ \pi\pi&amp;#39;|\:\alpha\beta\ \Rightarrow\ \ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you are not familiar with Gaussian integer arithmetic then you may employ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity&quot; rel=&quot;nofollow&quot;&gt;Brahmagupta–Fibonacci identity&lt;/a&gt; for composition of squares. It is a consequence of the fact that the norm is &lt;em&gt;multiplicative:&lt;/em&gt; $\rm\ \ \ N(\alpha)\ N(\beta)\ =\ N(\alpha\beta)\ $ i.e. $\rm\ \alpha\alpha&amp;#39;\beta\beta&amp;#39; = (\alpha\beta) (\alpha\beta)&amp;#39;\: $ which, rationally, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm (x_1^2 + y_1^2)\ (x_2^2 + y_2^2)\ =\ (x_1x_2\pm y_1y_2)^2 + ( x_1y_2\mp x_2y_1)^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper $\pm$ signs are from $\rm\ N(\alpha)\ N(\beta&amp;#39;)\ = \ N(\alpha\beta&amp;#39;)\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T17:43:56.817" LastActivityDate="2010-11-28T17:43:56.817" CommentCount="1" />
  <row Id="12217" PostTypeId="2" ParentId="12213" CreationDate="2010-11-28T16:22:32.067" Score="4" ViewCount="0" Body="&lt;p&gt;Here is another one, perhaps due to Donald J Newman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each real number $x$ consider an (infinite) sequence of rationals $R_x = \{r_{x1}, r_{x2}, \cdots \}$ which converges to $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x \neq y$, $R_x$ and $R_y$ have at most finite intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the set $S = \{R_x : x \in \mathbb{R}\}$ is an uncountable collection of infinite subsets of a countable set, none of which is a subset of other. In fact, any two members of the collection have only a finite intersection.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-28T16:22:32.067" />
  <row Id="12218" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T16:27:39.797" Score="10" ViewCount="0" Body="&lt;p&gt;Another way is thinking of this as asking where do the graphs of the functions given by $f(x) = e^{x-1}$ and $g(x) = \frac{1}{x}$ intersect. This is because $$ xe^{x} = e \Leftrightarrow e^{x-1} = \frac{1}{x} \quad \text{for $x \neq 0$}$$ Then the answer you have would correspond to the only intersection point $(1, 1)$. So if you have no idea of where to start you could just draw the corresponding graphs and that could give you an idea of what is happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/upxAx.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the graphs maybe would suggest to you the following line of reasoning. If $0 &amp;lt; x &amp;lt; 1$ then $$ \frac{1}{x} &amp;gt; 1 \quad \text{but} \quad e^{x-1} &amp;lt; e^{0} = 1$$ so their graphs can't meet in the interval $]0, 1[$. Then if $x &amp;gt; 1$ we have $$\frac{1}{x} &amp;lt; 1 \quad \text{but} \quad e^{x-1} &amp;gt; e^{0} = 1$$ so again there's no intersection point in the interval $]1, \infty[$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rules out all the possible solutions apart from $x = 1$ which you already knew and which could be guessed by looking at the graphs.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-11-28T18:21:15.210" LastActivityDate="2010-11-28T18:21:15.210" CommentCount="4" />
  <row Id="12219" PostTypeId="1" AcceptedAnswerId="12228" CreationDate="2010-11-28T16:33:02.907" Score="4" ViewCount="73" Body="&lt;p&gt;Let $T$ be a bounded linear operator on $L^2(\mathbb R)$. So, let us now assume that $T$ commutes with the translations $\tau_x$. How do I now show that $T$ is given by a convolution with respect to a distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I know I can probably find the proof somewhere in one of Stein's books, but I would like to prove it myself without knowing what it should be but I'm struggling a bit. So I would like some hints. Especially I would like a method of deriving the result without knowing what it should be. If that is not possible, an intuitive argument why it should be true would be nice as well.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-11-28T16:40:04.340" LastActivityDate="2010-11-28T19:01:19.687" Title="Operators commuting with translations." Tags="&lt;harmonic-analysis&gt;&lt;distribution-theory&gt;" AnswerCount="2" />
  <row Id="12220" PostTypeId="1" AcceptedAnswerId="12756" CreationDate="2010-11-28T16:52:56.750" Score="2" ViewCount="68" Body="&lt;p&gt;Hi all, I have the following situation:&#xA;I have a hermitian Matrix $A$ that satisfies some symmetries which I can express via &#xA;$AS = SA$ for a unitary matrix $S$. Now I am interested in the eigenvectors of $A$, but I want that these eigenvectors also respect my symmetries ( compare to Bloch waves in physics where the eigenvectors are chosen to reflect the translational invariance of the lattice ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A$ has degenerate (repeated) eigenvalues, the standard numerical techniques will return some arbitrary (yet orthonormal) eigenvectors spanning the eigenspace. I, however, want to obtain unique results and thus want to make use of the symmetries. How can I do this numerically? I know that commuting matrices can be diagonalized simultaneously - in theory. But I don't know how to do it practically. Would I have to diagonalize one of them, apply the unitary transform thus obtained to the other one, arriving at a block diagonal form where I then have to diagonalize each block separately? Or is there something more elegant I can do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The matrix is dense, but quite small (12x12 to 18x18).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetry would be something like translation symmetry:&#xA;$\begin{pmatrix} 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; -1 \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\ 0  &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \end{pmatrix}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each entry is a 3x3-block in the case of a 12x12 matrix, which looks something like&#xA;$\begin{pmatrix} a &amp;amp; b &amp;amp; 0 &amp;amp; -b\\ b&amp;amp;a&amp;amp;b&amp;amp;0\\ 0&amp;amp;b&amp;amp;a&amp;amp;b\\ -b&amp;amp;0&amp;amp;b&amp;amp;a\end{pmatrix}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4171" LastEditorUserId="4171" LastEditorDisplayName="" LastEditDate="2010-11-28T17:09:58.957" LastActivityDate="2010-12-02T15:45:40.987" Title="Symmetrize eigenvectors of degenerate (repeated) eigenvalue" Tags="&lt;linear-algebra&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12221" PostTypeId="2" ParentId="11843" CreationDate="2010-11-28T17:05:27.957" Score="2" ViewCount="0" Body="&lt;p&gt;Your function is really &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F: Context \mapsto \mathbb{P}(X|Context)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in other words you just want to determine the probability of a particular outcome, conditional on a certain context. The question is, what are your data? Do you know about the different contexts? Or do you only dispose of the outcomes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, since $$\mathbb{P}(X|Context)=\frac{\mathbb{P}(X \cap Context)}{\mathbb{P}(Context)} \; ,$$ I think any test you'd want to devise for this function would boil down to the Chi-squared test anyway.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-28T17:05:27.957" CommunityOwnedDate="2010-11-28T17:05:27.957" />
  <row Id="12222" PostTypeId="2" ParentId="12219" CreationDate="2010-11-28T17:09:14.560" Score="2" ViewCount="0" Body="&lt;p&gt;Maybe you should look at $F \circ T \circ F^{-1}$ where $F$ denotes Fourier transform, so that $\tau_x$ becomes multiplication by $t \mapsto e^{itx}$ and convolution becomes multiplication by something as well.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-11-28T17:09:14.560" />
  <row Id="12223" PostTypeId="1" CreationDate="2010-11-28T17:11:35.790" Score="4" ViewCount="118" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Logical_connective#Order_of_precedence&quot; rel=&quot;nofollow&quot;&gt;the precedence of logical connectives&lt;/a&gt;, operator $\rightarrow$ gets higher precedence than $\leftrightarrow$ operator. But what about associativity of $\rightarrow$ operator? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not locate any book/webpage that mentions about associativity of logical operators in discrete mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please also cite the reference (book/reliable webpage) that you use to answer my question (as I'm planning to add this to wikipedia page about 'logical connectives').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I got this question when I saw this problem:&#xA;Check if following compound  proposition is tautology or not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{p} \leftrightarrow (\mathrm{q} \wedge \mathrm{r}) \rightarrow \neg\mathrm{r} \rightarrow \neg\mathrm{p}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4170" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T05:44:58.277" LastActivityDate="2010-11-29T05:44:58.277" Title="Associativity of logical connectives" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="12225" PostTypeId="1" AcceptedAnswerId="12241" CreationDate="2010-11-28T17:42:28.737" Score="1" ViewCount="84" Body="&lt;p&gt;For a given matrix $A$ the question &quot;is there a vector (or are there vectors)&#xA;whose direction remain the same when it is multiplied with the matrix&quot;&#xA;leads to the well-known eigenvalue problem $A\mathbf{v}=\lambda\mathbf{v}$,&#xA;where $\mathbf{v}$ is an eigenvector of the matrix. Since a vector is fully&#xA;described by its direction and amplitude, one can also ask the other obvious question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given matrix $A$ is there a vector (or are there vectors) whose&#xA;amplitude remains the same when it is multiplied with the matrix? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would&#xA;amount to finding &quot;eigenrotations&quot; that satisfy&#xA;$\left\Vert A\mathbf{r}\right\Vert =\left\Vert \mathbf{r}\right\Vert .$&#xD;&#xA;The eigenrotations are thus characterized by&#xA;$\mathbf{r}^{T}\left(  A^{T}A-I\right)  \mathbf{r}=0,$&#xD;&#xA;and under some conditions on the coefficients of $A$ there exist such vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: In what contexts do &quot;eigenrotations&quot; come up? (other than a mathexchange question)&lt;/p&gt;&#xA;" OwnerUserId="3891" LastActivityDate="2010-11-28T19:17:54.067" Title="&quot;Eigenrotations&quot; of a matrix" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12226" PostTypeId="2" ParentId="12197" CreationDate="2010-11-28T17:53:33.493" Score="4" ViewCount="0" Body="&lt;p&gt;Yes it is possible to reduce the number of states a Turing machine uses to decide a problem $X$ by increasing the number of symbols.&#xA;However it gets very tricky when you are close to the &lt;em&gt;minimum number&lt;/em&gt; of states/symbols needed to solve  $X$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a nice survey paper about the efforts to find the minimum number of states and symbols Turing machines need for universality. &#xA;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1498068&quot; rel=&quot;nofollow&quot;&gt;http://portal.acm.org/citation.cfm?id=1498068&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The complexity of small universal Turing machines: A survey&quot;&#xA;Damien Woods and Turlough Neary,&#xA;Theoretical Computer Science archive&#xA;Volume 410 Issue 4-5, February, 2009&lt;/p&gt;&#xA;" OwnerUserId="571" LastActivityDate="2010-11-28T17:53:33.493" />
  <row Id="12227" PostTypeId="1" AcceptedAnswerId="12233" CreationDate="2010-11-28T18:11:01.550" Score="1" ViewCount="102" Body="&lt;p&gt;I'm trying to wrap my head around this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \equiv b^{-1}\pmod{y}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x \equiv c^{-1} \pmod{y}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know all values but $x$ and $y$. Is it possible to solve this problem for $x$?&#xA;Can anyone point out a good resource on this kind of problems?&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T19:09:56.787" LastActivityDate="2010-11-29T00:19:09.960" Title="Understanding congruence equations" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="12228" PostTypeId="2" ParentId="12219" CreationDate="2010-11-28T18:14:37.663" Score="1" ViewCount="0" Body="&lt;p&gt;Some suggestions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the delta function is not in $L^2$, heuristically what does your condition on $T$ say when applied to the delta function? Now, given the answer to the previous question and the fact that every function is an average of translated delta functions, what does one expect $Tf$ to be for an arbitrary $L^2$ function $f(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do this on the Fourier transform side.. i.e. consider $G = F \circ T \circ F^{-1}$ as Plop suggested and then look at how $G$ behaves on a given $\delta(x - a)$. Then again use the idea that an $L^2$ function is an average of translated delta functions.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-28T18:14:37.663" CommentCount="5" />
  <row Id="12229" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T18:17:24.903" Score="2" ViewCount="0" Body="&lt;p&gt;Assuming that we are only looking for real solutions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$xe^x = e.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $e^x$ is always positive and so no solution for this equation is possible for $x&amp;lt;0$ or $x=0$.  As x increases, both x and $e^x$ are strictly increasing.  Therefore, there is only one value for which $xe^x = e$.  We already know that solution to be $x=1$.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1004" LastEditorDisplayName="" LastEditDate="2010-11-29T00:33:55.103" LastActivityDate="2010-11-29T00:33:55.103" />
  <row Id="12233" PostTypeId="2" ParentId="12227" CreationDate="2010-11-28T18:36:38.487" Score="3" ViewCount="0" Body="&lt;p&gt;If $b,c$  have an inverse in $\pmod y$ then is valid :$$ ab\equiv 1 \pmod y $$ and $$ xc\equiv 1 \pmod y $$&#xD;&#xA;&#xD;&#xA;From this you get:  $$ ab-1 =  k_{1}y  $$  $$ xc-1 =  k_{2}y  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For some  $k_{1}$ and $k_{2}$  integers )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can find $x$ and $y$ from this ecuations system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Edit:$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $y=5$, the elements with inverse are: $2,3$, also $b=2$ and $b^{-1}=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Know that (with $a=8$ ):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 8 \equiv 3 \pmod 5 $, is say , $ a \equiv b^{-1} \pmod y $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 16 =8 \cdot 2\equiv 1 \pmod 5 $, is say , $ ab \equiv 1 \pmod y $ &lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T00:19:09.960" LastActivityDate="2010-11-29T00:19:09.960" CommentCount="4" />
  <row Id="12234" PostTypeId="2" ParentId="12227" CreationDate="2010-11-28T18:39:25.890" Score="0" ViewCount="0" Body="&lt;p&gt;$\rm x\ =\ (c^{-1}\ mod\ y)\ +\ y\ \mathbb Z\ \ $ where $\rm\ \ y\:|\: ab-1,\ \ gcd(y,bc)=1$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T18:45:17.410" LastActivityDate="2010-11-28T18:45:17.410" />
  <row Id="12235" PostTypeId="2" ParentId="12223" CreationDate="2010-11-28T18:39:43.923" Score="2" ViewCount="0" Body="&lt;p&gt;Presumably $p \Rightarrow q \Rightarrow r$ means $(p \Rightarrow q) \Rightarrow r$, since the other option $p \Rightarrow (q \Rightarrow r)$ is equivalent to $(p \land q) \Rightarrow r$. That's similar to the laws of powering $p^{q^r} = p^{\left(q^r\right)}$ since $\left(p^q\right)^r = p^{qr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best practice is just &lt;i&gt;not&lt;/i&gt; to put the reader into a state of ambiguity. Some mathematical writers dislike punctuation, but that's not an excuse for the rest of us to take advantage of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\land$ and $\lor$ are associative, there's no ambiguity there, and it makes sense not to put parentheses. Sometimes one even thinks of an expression like $p \land q \land r$ as a conjunction $\land(p,q,r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter convention is the case in proof complexity when one is measuring the depth of a formula. If a first-order &lt;i&gt;quantified&lt;/i&gt; formula is translated to propositional logic, then $\forall$ will correspond to a conjunction, and $\exists$ to a disjunction; it makes sense not to penalize the formula for the size of the underlying universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even more daring &quot;convention&quot; is having infinite conjunctions/disjunctions - this wouldn't make sense at all for $\Rightarrow$. So your textbook authors are at fault (even though having the $\Leftrightarrow$ bind weakly is a reasonable convention if not overused).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T18:39:43.923" CommentCount="4" />
  <row Id="12237" PostTypeId="1" AcceptedAnswerId="12249" CreationDate="2010-11-28T18:46:26.933" Score="1" ViewCount="55" Body="&lt;p&gt;For a given group of data, I have both a mean and variance (already calculated). Looking over the raw spread of data, and its numerous outliers, I expect that variance will be high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is -- now that I have my variance calculated, what do I need to compare it to to determine if the variance is significant (i.e., indicative of highly skewed data)?&lt;/p&gt;&#xA;" OwnerUserId="1760" LastActivityDate="2010-11-28T21:02:08.393" Title="How do I determine if my variance is significant?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12238" PostTypeId="1" CreationDate="2010-11-28T18:49:35.293" Score="1" ViewCount="65" Body="&lt;p&gt;I have skimmed this video that I found on mathoverflow:&#xA;&lt;a href=&quot;http://tube.sfu-kras.ru/video/407?playlist=397&quot; rel=&quot;nofollow&quot;&gt;http://tube.sfu-kras.ru/video/407?playlist=397&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At about 15:05 the lecturer wrote down an equality&#xA;$\sum F(m_1, \ldots, m_m)z^{m_1}\ldots z^{m_m} = \frac{1}{det(zI - A)}$, where F is given in terms of coefficients of the $m$-by-$m$ matrix $A$. The definition of $F$ doesn't yet make much sense to me, but that's beyond the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how is this equality possible? the right hand side obviously has poles—the eigenvalues of $A$, but the left hand side is just a polinomial, it can't have poles! Is there something I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: well, it was actually&#xA;$\sum F(m_1, \ldots, m_m) z_1^{m_1}\ldots z_m^{m_m} = \frac{1}{det(z - A)}$.&#xA;Here I have abused the notation slightly: the z means actually the diagonalization of the corresponding vector. However, when you take $z_1 = \ldots = z_m$, the question still stands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The domain is real, as I understand from the context of the lecture :), so the use of the term pole isn't really correct, I just forgot the right word in English :)&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-11-28T19:33:11.947" LastActivityDate="2010-11-28T19:33:11.947" Title="A relation between permanents and determinants" Tags="&lt;real-analysis&gt;&lt;algebra&gt;" CommentCount="8" />
  <row Id="12239" PostTypeId="1" CreationDate="2010-11-28T18:51:44.143" Score="3" ViewCount="68" Body="&lt;p&gt;I know this is a trivial problem but am stuck with this thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \in \mathbb{R}^{m \times m}$ is a symmetric positive definite matrix and $C \in \mathbb{R}^{m \times p}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I show that the matrix $$B = \begin{bmatrix} A &amp;amp; C \\ C^T &amp;amp; 0 \end{bmatrix}$$ is indefinite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can show that there exists a vector $v$ such that $v^TBv &amp;gt; 0$.&#xA;(This is trivial. For instance, $v = [x , 0]^T$ where $x \in \mathbb{R}^{1 \times m}$ and $0 \in \mathbb{R}^{1 \times p}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do I find a vector such that $v^TBv &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not need the answer. A clue/hint is welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume that $p \leq m$ and the matrix $C$ is full rank that is to say that all the constraints are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-28T22:54:36.977" LastActivityDate="2010-11-29T07:25:28.463" Title="Indefiniteness of KKT system" Tags="&lt;linear-algebra&gt;&lt;homework&gt;&lt;optimization&gt;" AnswerCount="2" />
  <row Id="12240" PostTypeId="1" AcceptedAnswerId="12242" CreationDate="2010-11-28T18:59:17.207" Score="3" ViewCount="66" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ and $Q$ be complex polynomials with $deg(P) &amp;lt; deg(Q)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Q(z) = (z-z_1)^{k_1} (z-z_2)^{k_2}...(z-z_m)^{k_m}, z_i \in \mathbb{C} \text{ and } k_i \in \mathbb{N}$ be a complete decomposition. Then, I have to prove that there exists a unique exposition of the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P(z)}{Q(z)} = \sum\limits_{i=1}^m \sum\limits_{j=1}^{k_i} \frac{a_{ij}}{(z-z_i)^j}, a_{ij} \in \mathbb{C}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with the proof of the uniqueness of this exposition and succeeded. However, I don't feel like I am doing well now, while trying to prove its existence. Basically, I want to do two inductions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I want to do an induction over $deg(Q)$ with $deg(P) = 0$, then I want to go on with an induction over $deg(P)$ with an arbitrary $deg(Q)$. The start of the first induction is easy, but I can't get to an end at the induction step. I have something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P}{Q} = \frac{P}{(z-z*)Q&amp;#39;} = \frac{1}{z-z*} \frac{P}{Q&amp;#39;}$, where $Q&amp;#39;$ is a polynomial with $deg(Q&amp;#39;) = n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I do next? I know that a unique exposition exists for the latter fraction and if $z* = z_i$ for some $i$, it's proven I guess. But what if this $z*$ is a completely new complex number?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-11-28T19:55:03.160" Title="Stuck at the proof of the existence of the partial fraction expansion" Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="12241" PostTypeId="2" ParentId="12225" CreationDate="2010-11-28T19:17:54.067" Score="3" ViewCount="0" Body="&lt;p&gt;Allow me to generalize the question a bit. It's a very natural operation to multiply a matrix by a scalar. Most properties of matrices, including the eigenvectors and eigenvalues, are either not affected by scalar multiplication or are multiplied by the same scalar. But $\lVert Ax \rVert = \lVert x \rVert$ does not remain true for the same vector $x$. So it's more natural to look at all the sets $\lVert Ax \rVert = c \lVert x \rVert$ as a function of a scalar $c$. These are just the level sets of $f(x) := \lVert Ax \rVert / \lVert x \rVert$. Of course, unlike the eigendecomposition, where you get a discrete set of (at most) $n$ vectors out of it, you will generically have infinitely many solutions for any $c$. Since all nonzero scalar multiples of $x$ have the same value of $f(x)$, we can without loss of generality restrict to the surface of the unit sphere $\lVert x \rVert = 1$. Then we are talking about solutions to the equations $\lVert x \rVert = 1$, $\lVert Ax \rVert = c$. Clearly, the solutions form a $(n-2)$-dimensional manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the only context in which I have seen such a construction. I think it's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Poinsot%27s_ellipsoid&quot; rel=&quot;nofollow&quot;&gt;Poinsot's ellipsoid&lt;/a&gt;. Consider a rigid body with moment of inertia $I$ tossed into the air with some angular velocity $\omega$. Since no external forces are acting on the body, its angular momentum $L = I\omega$ is conserved, and so is its rotational kinetic energy $T = \frac{1}{2}\omega^T I\omega$. Take a reference frame attached to the body. Then $I$ becomes a constant matrix. $L$ is no longer constant, but its magnitude still must be. So we get two equations: $\lVert L \rVert = \mathrm{const}$, and $\frac{1}{2} L^T I^{-1} L = \mathrm{const}$, which has almost the same form as your equations. This blog post called &quot;&lt;a href=&quot;http://gmunu.wordpress.com/2008/06/29/on-tossing-stuff-into-the-air/&quot; rel=&quot;nofollow&quot;&gt;On tossing stuff into the air&lt;/a&gt;&quot; has some nice pictures, as well as discussion about how this tells you why things wobble if you spin them about the intermediate moment of inertia, and what happens when you toss a bag of peas into the air.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-28T19:17:54.067" />
  <row Id="12242" PostTypeId="2" ParentId="12240" CreationDate="2010-11-28T19:27:48.717" Score="5" ViewCount="0" Body="&lt;p&gt;By the inductive hypothesis, $\frac{P}{Q&amp;#39;}$ has a partial fraction decomposition.  Show that by multiplying this decomposition by $\frac{1}{z - z^{\ast}}$ you can write down a partial fraction decomposition for $\frac{P}{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-28T19:27:48.717" />
  <row Id="12243" PostTypeId="2" ParentId="2923" CreationDate="2010-11-28T19:38:16.860" Score="0" ViewCount="0" Body="&lt;p&gt;Hello, I don't have Billingsley's book here, but I guess what he says is that uniform integrability and convergence in distribution imply convergence &lt;strong&gt;of the means&lt;/strong&gt; (and not &lt;strong&gt;in mean&lt;/strong&gt;), i.e. E[X_n] converges to E[X]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple (though not truely elementary) way to see it is to use Skorokhod's theorem stating that convergence in distribution is equivalent to almost-sure convergence of copies of the random variables in some abstract probability state. Then we just apply the usual Vitali theorem about uniform integrability for these copies. It gives convergence in mean of the copies of X_n to the copy of X, and therefore convergence of the mean of (the copy of) X_n to that of (the copy of) X. I put parentheses since the mean only depends on the law and hence is the same for X_n or a copy of it. This is the result. &lt;/p&gt;&#xA;" OwnerUserId="4178" LastActivityDate="2010-11-28T19:38:16.860" />
  <row Id="12244" PostTypeId="2" ParentId="12240" CreationDate="2010-11-28T19:49:58.297" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad\displaystyle 1 = gcd(F,G)\ \ \Rightarrow\ \ 1 = AF+BG\ \ \Rightarrow\ \ \frac{1}{FG}\ = \ \frac{A}G + \frac{B}F$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-28T19:55:03.160" LastActivityDate="2010-11-28T19:55:03.160" />
  <row Id="12245" PostTypeId="2" ParentId="12194" CreationDate="2010-11-28T20:02:37.150" Score="4" ViewCount="0" Body="&lt;p&gt;With logarithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly  $x&amp;gt;0$, Then :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$xe^{x}=e \Rightarrow \ln (xe^{x})=\ln e \Rightarrow \ln x + x = 1 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;gt;1$ then $ \ln x &amp;gt;0$, thus $\ln x +x &amp;gt; 1+0 =1$  (contradiction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x&amp;lt;1$ then $ \ln x &amp;lt; 0$, thus $\ln x +x &amp;lt; 1+0 =1$  (contradiction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $x=1$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T02:02:20.783" LastActivityDate="2010-11-29T02:02:20.783" />
  <row Id="12246" PostTypeId="2" ParentId="10092" CreationDate="2010-11-28T20:09:30.240" Score="1" ViewCount="0" Body="&lt;p&gt;Here is something even more spectacular. Let $\epsilon &amp;gt;0$.  Let $z_n$ be an enumeration of all points in $R^d$ with all rational coordinates. We know this sequence's range is dense in $R^d$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About each $z_n$ choose an open ball of small radius $B_n$ whose volume is less than $\epsilon/2^n$.  Now write&#xA;$$U_\epsilon = \bigcup_{n=1}^\infty B_n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a union of open balls and is therefore open.  Its total Lebesgue measure (volume) is less than $\epsilon$. It is an open dense subset of $R^d$.  An open dense subset of $R^d$, can be very small indeed.  &lt;/p&gt;&#xA;" OwnerUserId="4154" LastActivityDate="2010-11-28T20:09:30.240" />
  <row Id="12247" PostTypeId="1" AcceptedAnswerId="12248" CreationDate="2010-11-28T20:13:22.163" Score="2" ViewCount="85" Body="&lt;p&gt;If a cyclic subgroup has a proper subgroup of infinite order, how many finite subgroups does it have?  Explain.  Any help appreciated!  Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4179" LastActivityDate="2010-11-28T20:38:40.020" Title="Abstract Algebra - Cyclic Subgroups with a Proper Subgroup of Infinite Order" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12248" PostTypeId="2" ParentId="12247" CreationDate="2010-11-28T20:38:40.020" Score="5" ViewCount="0" Body="&lt;p&gt;It's easy to see that a cyclic infinite group is isomorphic to the set of integers $\mathbb{Z}$: being cyclic, it is generated my one element, $x$ say, and being infinite, no power of $x$ is equal to the identity. Consequently it is formed by the integer powers of $x$, and not two of them are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question consequently is that an infinite cyclic group cannot have a nontrivial finite subgroup.&lt;/p&gt;&#xA;" OwnerUserId="2137" LastActivityDate="2010-11-28T20:38:40.020" />
  <row Id="12249" PostTypeId="2" ParentId="12237" CreationDate="2010-11-28T20:49:21.240" Score="0" ViewCount="0" Body="&lt;p&gt;Taken from &lt;a href=&quot;http://forumserver.twoplustwo.com/15/poker-theory/how-does-standard-deviation-relate-poker-867847/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; post of mine from a poker forum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You need to somehow compute the standard deviation for your distribution, and then count how many standard deviations away from the expected value your actual result is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we are flipping coins, and each time it lands heads you give me a dollar, and each time it's tails I give you a dollar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let's keep track of how much money I have made or lost. On average I expect to make nothing, but in practice I will most likely lose or win a finite amount. In fact if we flip the coin $N$ times, then I am most likely to win or lose somewhere around the $\sqrt{N}$ dollars. Why is this true? It's actually easy to show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call my profit/loss $M$. Then $M$ starts at zero, and the average change in $M$ after one or more flips is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the clever trick: let's consider the average change in $M^2$. Well after one flip it's quite clear that this will be 1, since +1 and -1 squared are both equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that after one flip the average profit/loss is zero, but in reality I either gained or lost a dollar. $M^2$ however is 1, so you can see it sort of measures the fluctuations in my profit/loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call $M(N)$ my profit/loss after N flips. Then what is $M(N+1)^2$? Since I can either gain or lose 1 dollar, we have either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(N+1) = M(N) + 1$ or $M(N+1) = M(N) - 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets square these two expressions and take the average, which I write as $&amp;lt;M(N+1)^2 &amp;gt;$. If you do this, which is quite easy, you will find that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;M(N+1)^2&amp;gt;$ = $&amp;lt;M(N)&amp;gt;^2 + 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it, this implies that after $N$ flips, the average of $M^2$ will be $N$; therefore the square root of the average of $M^2$ is $\sqrt{N}$, which is just a fancy way of saying the SD for flipping a coin N times is $\sqrt{N}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This shows that the SD for flipping a coin (1:1 odds) is $\sqrt{N}$, and it's simple to modify the above argument to find the SD for weighted coins.  Therefore, besides telling us how good/bad you are running, the magnitude of the SD also tells you, in poker, how well on average you are getting your money in.  In other words, SD can be used to measure what average chance you have to win the pot when you put all your chips in with cards left to come.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-28T21:02:08.393" LastActivityDate="2010-11-28T21:02:08.393" />
  <row Id="12250" PostTypeId="1" AcceptedAnswerId="12265" CreationDate="2010-11-28T21:07:49.127" Score="0" ViewCount="92" Body="&lt;p&gt;Euler's formula for connected planar graphs (i.e. a single connected component) states that $v-e+f=2$. State the generalization of Euler's formula for planar graphs with $k$ connected components (where $k\geq1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is $v-e+f=1+k$, but I'm not understanding the reasoning behind it. Anyone care to share some insight?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-28T23:33:57.980" Title="Euler's formula for connected planar graphs" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12251" PostTypeId="2" ParentId="12250" CreationDate="2010-11-28T21:33:25.040" Score="2" ViewCount="0" Body="&lt;p&gt;The usual proof of Euler's formula works by first triangulating the graphs, then removing triangles one by one until you reach a single triangle; all these respect the Euler characteristic $v-e+f$. The proof is completed by calculating $v-e+f$ for a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the same here - the first phase is the same, and in the end you will be left with $k$ triangles. Then all you have to do is calculate $v-e+f$ for $k$ unconnected triangles. Since $v=e$ for any collection of triangles, the result is the same as the number of faces, which is $1+k$: one face per triangle, and an extra outer face.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-28T21:33:25.040" />
  <row Id="12252" PostTypeId="2" ParentId="8875" CreationDate="2010-11-28T21:53:51.233" Score="1" ViewCount="0" Body="&lt;p&gt;First, one has $$e^{ir}-1=O(|r|) $$ for all $r\in \mathbb{R}.$ This means that there exist constant (independent of $r$) $A&amp;gt;0$ such that $$|e^{ir}-1|\le A|r| \mbox{ for all } r\in \mathbb{R} $$ (In fact, you can take $A=1$, as $|e^{ir}-1|$ is the length of the chord from 1 to $e^{ir}$ and $r$ is the corresponding arc length on the unit circle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us prove your question in the case $X $ is contiuous. (The discrete case can be proved similarly.) Suppose $f(x)$ is the probability density function of $X$.  Then&#xA;$$ \phi(t)=E(e^{itX})=\int_{-\infty}^\infty f(x)e^{itx} dx$$ and $\phi(0)=\int_{-\infty}^\infty f(x) dx =1$. So by linearity of integral&#xA;$$\phi&amp;#39;(0)=\lim_{t\to 0}\frac{\phi(t)-\phi(0)}{t}=\lim_{t\to 0}\int_{-\infty}^\infty f(x)\frac{e^{itx}-1}{t} dx.$$ Since&#xA;$$\left|f(x)\frac{e^{itx}-1}{t}\right|\le f(x) A \frac{|tx|}{|t|}=A f(x)|x|,$$ and $$\lim_{t\to 0}\frac{e^{itx}-1}{t}=\frac{d}{dt} e^{itx}|_{t=0}=ix e^0=ix,$$&#xD;&#xA;it follows from DCT that if $E(|X|)&amp;lt;\infty$, then $$\phi&amp;#39;(0)= \int_{-\infty}^\infty f(x)ix dx=i E(X).$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-28T22:46:44.740" LastActivityDate="2010-11-28T22:46:44.740" />
  <row Id="12253" PostTypeId="2" ParentId="12223" CreationDate="2010-11-28T22:03:41.640" Score="3" ViewCount="0" Body="&lt;p&gt;Some logical operators are associative: both $\wedge$ and $\vee$ are associative, as a simple check of truth tables verifies. Likewise, the biconditional $\leftrightarrow$ is associative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the implication $\rightarrow$ is &lt;em&gt;not&lt;/em&gt; associative. Compare $(p\rightarrow q)\rightarrow r$ and $p\rightarrow(q\rightarrow r)$. If $p$ and $r$ are true and $q$ is false, then $(p\rightarrow q)$ is false, so $(p\rightarrow q)\rightarrow r$ is true. But under the same conditions $q\rightarrow r$ is true, and therefore $p\rightarrow(q\rightarrow r)$ is true. So $(p\rightarrow q)\rightarrow r$ and $p\rightarrow(q\rightarrow r)$ cannot be equivalent; that is, $\rightarrow$ is not associative.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-28T22:03:41.640" CommentCount="4" />
  <row Id="12254" PostTypeId="2" ParentId="12097" CreationDate="2010-11-28T22:06:46.883" Score="0" ViewCount="0" Body="&lt;p&gt;Some rough idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I consider only the special case where the $X_i$ are positive rv's such that, by self-similarity, $X_1 + \cdots + X_{n-1}$ is equal in distribution to $(n-1)^{1/\alpha}X_1$ where $\alpha$ is the associated index. Then, it seems useful to start (similarly to what you did above) with&#xA;$$&#xD;&#xA;P = n{\rm P}(X_0  &amp;gt; X_1  +  \cdots  + X_{n-1}),&#xD;&#xA;$$&#xD;&#xA;where $P$ is the probability in the title. Then, recalling the self-similarity argument, a straightforward application of the law of total probability (conditioning on $X_0$) yields&#xA;$$&#xD;&#xA;P = n\int F (s/(n - 1)^{1/\alpha } )f(s){\rm d}s,&#xD;&#xA;$$&#xD;&#xA;where $F$ and $f$ are the common distribution and density functions of the $X_i$, respectively. See if you can continue from here.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-28T23:08:02.063" LastActivityDate="2010-11-28T23:08:02.063" CommentCount="2" />
  <row Id="12255" PostTypeId="1" AcceptedAnswerId="12257" CreationDate="2010-11-28T22:08:43.243" Score="4" ViewCount="86" Body="&lt;p&gt;I'm trying to help my son study 8th grade math (in Texas).  He's having trouble with the &quot;Nth rule&quot;.  Is anyone familiar with this terminology?  I think that it may be related to the Nth term in a series, but I also think that it has a more specific meaning.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastActivityDate="2010-11-28T22:23:02.583" Title="Terminology - Nth rule" Tags="&lt;sequences-and-series&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12256" PostTypeId="1" AcceptedAnswerId="12260" CreationDate="2010-11-28T22:14:58.773" Score="2" ViewCount="90" Body="&lt;p&gt;What's the largest regular tetrahedron (side length x) you can fit inside a sphere with a radius of one?&lt;/p&gt;&#xA;" OwnerUserId="3746" LastActivityDate="2010-11-29T00:42:21.850" Title="Tetrahedron inside a sphere" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="12257" PostTypeId="2" ParentId="12255" CreationDate="2010-11-28T22:15:32.140" Score="9" ViewCount="0" Body="&lt;p&gt;A quick google search suggests that &quot;the $n$th rule&quot; means finding the formula for the $n$th term of a given sequence of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given $1$, $3$, $5$, $7,\ldots$, the expected &quot;nth rule&quot; is probably &quot;$2n-1$&quot;, since the 1st term is $2(1)-1$, the 2nd term is $2(2)-1$, the 3rd term is $2(3)-1$, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What follows is a bit of a diatribe completely irrelevant to your question:&lt;/em&gt; It is an utterly silly request, as noted long ago by Carl Linderholm in his book, &lt;strong&gt;Mathematics made difficult&lt;/strong&gt;. Any answer that gives the right values at the given points is valid; and if you pick your favorite number for the &quot;next&quot; number, you can find a formula that will give &lt;em&gt;those&lt;/em&gt; using Lagrange interpolation; for example, if you have the sequence $1$, $2$, $4$, $8$, $16,\ldots$, then the &quot;next term&quot; is $31$, either through Lagrange interpolation, or by counting the number of sectors that a circle is divided into if you place $n$ points on its circumference and connect them with straight lines. Both show much deeper mathematical understanding than &quot;$2^{n-1}$&quot;. The problem is not really  &quot;What is the formula for the $n$th term of this sequence?&quot; the question is &quot;of all possible formulas that would give these values for $n=1,2,\ldots$, which is the one that the person who is grading this test is thinking of?&quot;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-28T22:23:02.583" LastActivityDate="2010-11-28T22:23:02.583" CommentCount="1" />
  <row Id="12258" PostTypeId="2" ParentId="12255" CreationDate="2010-11-28T22:16:23.460" Score="4" ViewCount="0" Body="&lt;p&gt;What they mean is this: Suppose I give you a sequence like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,4,6,8,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I ask you: How can I obtain the n-th term in this sequence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you'll say: It's just $2n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for sequences of the form $an + b$, it is quite easy to obtain the rule: Just look how the element increases in each step, this gives you $a$. Then see what the first element is and choose $b$ so that it's obtained for $n = 1$.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-11-28T22:16:23.460" />
  <row Id="12259" PostTypeId="2" ParentId="12239" CreationDate="2010-11-28T22:28:49.753" Score="1" ViewCount="0" Body="&lt;p&gt;Ok. So I figured it out. But I think my argument is really convoluted. I would appreciate if someone could come up with a simpler argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without loss of generality, we can assume that the matrix $C$ is of full-rank $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let $B = \begin{bmatrix} A &amp;amp; C \\ C^T &amp;amp; 0 \end{bmatrix}$, where $A \in \mathbb{R}^{m \times m}$ and $C \in \mathbb{R}^{m \times p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first observation is that all the eigenvalues of this matrix have to be real. This is so since the matrix $B$ is symmetric. Also, the matrix $B$ has an eigenvalue decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $\exists x$ such that $x^TBx &amp;gt; 0$. For instance, $x = [x_1 , 0]^T$ where $x_1 \in \mathbb{R}^{1 \times m}$ and $x_1 \neq 0$ and $0 \in \mathbb{R}^{1 \times p}$. Then $x^TBx =x_1^TAx_1 &amp;gt; 0$ since $A$ is positive definite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we consider $x=[0,x_2]^T$ where $x_2 \in \mathbb{R}^{1 \times p}$ and $x_2 \neq 0$ and $0 \in \mathbb{R}^{1 \times m}$, then $x^TBx = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $B$ is not positive definite. And since $B$ is symmetric, there is an eigenvalue decomposition for $B$. And hence $\exists \lambda$ such that $\lambda \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need to prove that $\exists \lambda$ such that $\lambda &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we recognize that the matrix $B$ is full-rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen from the fact that the null space of the matrix $B$ is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $ z = \begin{bmatrix} x \\ y \end{bmatrix}$ such that $Bz = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $Ax + Cy = 0$ and $C^Tx = 0$. So we have $x = -A^{-1}Cy$ and hence we get $C^TA^{-1}Cy = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix $C^TA^{-1}C$ is positive definite. This is because $C$ is full rank and is a skinny matrix. So whenever $x \neq 0$, $x^* = Cx \neq 0$ and hence $x^TC^TA^{-1}Cx = {x^*}^TA^{-1}x^* &amp;gt; 0$ (Since $A^{-1}$ is also positive definite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $y=0$ is the only possible solution which implies $x=0$ and hence $z=0$ is the only vector in the null space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the matrix $B$ is of full-rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, none of the eigenvalues are zero i.e. $\forall \lambda$, $\lambda \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from previous argument, we have that $\exists \lambda$ such that $\lambda \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we can conclude that $\exists \lambda$ such that $\lambda &amp;lt; 0$. So if I choose the (or) a eigenvector $v$ corresponding to this $\lambda$, then I have $v^TBv = - \lambda ||v||_2^2 &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the matrix $B$ is Indefinite.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-11-28T22:59:06.410" LastActivityDate="2010-11-28T22:59:06.410" CommentCount="4" />
  <row Id="12260" PostTypeId="2" ParentId="12256" CreationDate="2010-11-28T22:36:20.223" Score="5" ViewCount="0" Body="&lt;p&gt;We are clearly looking for the regular tetrahedron inscribed in a sphere of radius 1 (i.e. with all its vertices lying on the sphere). The neat trick with regular tetrahedra is to inscribe them in a cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a picture of the two regular tetrahedra you can find in a cube: &lt;a href=&quot;http://en.wikipedia.org/wiki/File:CubeAndStel.gif&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:CubeAndStel.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube inscribed in a unit sphere has side length $\frac{2}{\sqrt 3}$, so the regular tetrahedron has side length $x = \sqrt{2} \frac{2}{\sqrt 3} = \sqrt{\frac{8}{3}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example of this idea: &lt;a href=&quot;http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron/11090#11090&quot;&gt;Height of a tetrahedron&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2010-11-28T22:36:20.223" />
  <row Id="12262" PostTypeId="1" CreationDate="2010-11-28T23:04:34.433" Score="1" ViewCount="66" Body="&lt;p&gt;At Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Erlangen_program#Abstract_returns_from_the_Erlangen_program&quot; rel=&quot;nofollow&quot;&gt;Erlangen program&lt;/a&gt; I read that &quot;quite often, it appears there are two or more distinct geometries with isomorphic automorphism groups&quot;. Some examples are given.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But what are examples of geometries nevertheless &lt;em&gt;being&lt;/em&gt;&#xA;  uniquely defined (up to isomorphism) &#xA;  by the isomorphism class of their&#xA;  automorphism group?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Are there examples of non-Euclidean&#xA;  geometries with an automorphism group&#xA;  isomorphic to the Euclidean group?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-28T23:47:03.413" LastActivityDate="2010-11-28T23:47:03.413" Title="Euclidean geometry and the Euclidean group" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;lie-groups&gt;" CommentCount="3" />
  <row Id="12263" PostTypeId="1" AcceptedAnswerId="12333" CreationDate="2010-11-28T23:23:38.753" Score="0" ViewCount="64" Body="&lt;p&gt;I'm attempting to understand how B-Spline derivatives work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, i've generated some data in MATLAB to plot sin(x) over 0 - 2*pi. I then use the splinetool to fit an order 2 spline to the data which it fits perfectly. However, when I look at the 1st derivative of this sine wave data it's max is 0.017 and it's min is -0.017.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why the derivative of this data doesn't give exactly cos(x) with a max and min of 1, -1. Any insights please?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2010-11-29T11:08:09.980" Title="B-Spline Derivatives" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12264" PostTypeId="2" ParentId="12256" CreationDate="2010-11-28T23:25:37.967" Score="2" ViewCount="0" Body="&lt;p&gt;You are looking at the regular tetrahedron inscribed in a sphere of radius 1. Denote the center of the sphere by O, and the vertices by A, B, C and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact: In the regular tetrahedron, the altitude from A is cut by O in 3:1 ratio (Note: In an equilateral triangle the analogous ratio is 2:1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:  The four vectors from O to the vertices sum to zero (the sum vector is invariant under rotations preserving the tetrahedron, so must be zero). Pick a vertex A. The projections of the other three vectors OB, OC and OD onto the line OA are all equal and, as their sum must cancel OA, each of these projections is equal to 1/3 of OA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the base of the altitude by H. Then we just showed OH=1/3, AH=4/3. Now by Pythagoras theorem, $HB^2=BO^2-OH^2=1^2-(1/3)^2=8/9$. Finally, by Pythagoras again, $AB^2=AH^2+HB^2=(4/3)^2+8/9=24/9$, so $AB=\sqrt(8/3)$.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-11-29T00:42:21.850" LastActivityDate="2010-11-29T00:42:21.850" />
  <row Id="12265" PostTypeId="2" ParentId="12250" CreationDate="2010-11-28T23:33:57.980" Score="3" ViewCount="0" Body="&lt;p&gt;For many approaches to Euler's formula look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ics.uci.edu/~eppstein/junkyard/euler/&quot; rel=&quot;nofollow&quot;&gt;http://www.ics.uci.edu/~eppstein/junkyard/euler/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-11-28T23:33:57.980" />
  <row Id="12266" PostTypeId="1" CreationDate="2010-11-28T23:38:29.083" Score="1" ViewCount="32" Body="&lt;p&gt;Hecke operators are things that act on modular forms and give rise to a lot of interesting arithmetical results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hecke_operator&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hecke_operator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand on the wikpedia page for Hecke algebras, which should naively be an algebra of Hecke operators, the term seems to acquire very different meanings, such as deformations of group algebra of a Coxeter group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the connection between the two?&lt;/p&gt;&#xA;" OwnerUserId="4183" LastActivityDate="2010-11-28T23:38:29.083" Title="Connection between Hecke operators and Hecke algebras " Tags="&lt;modular-forms&gt;&lt;representation-theory&gt;" CommentCount="2" />
  <row Id="12267" PostTypeId="2" ParentId="12193" CreationDate="2010-11-28T23:45:33.107" Score="2" ViewCount="0" Body="&lt;p&gt;I'll place something as an answer in the interest of not having questions with no answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what the cardinality of $\{\mathbb{R}(\setminus a)\}$ is, but I would be surprised if it is not uncountable. I'll see if I can think about it some in the next couple of days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your final paragraph, the different fields come not from the order in which you would remove elements, but rather on the &lt;em&gt;choices&lt;/em&gt; you make as to what to remove at each (of infinitely many) steps. As I noted in the comments, think of the similar process whereby we look for subgroups of a given group that are maximal among those that do not contain a given element. If we start with the Klein $4$-group, $C_2\times C_2$, and remove the element $a=(1,1)$, then we get two different choices of what other element to remove in order to get a group: we can remove &lt;em&gt;either&lt;/em&gt; $(1,0)$ &#xA;&lt;em&gt;or&lt;/em&gt; &#xA;$(0,1)$; each of the choices will lead to a different subgroup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, consider for instance the case where $a=\sqrt{2}$; then as noted in your previous question, a maximal subfield of $\mathbb{R}$ that does not contain $a$ may contain either $\sqrt{3}$ or $\sqrt{6}$, but not both, and you get one field that is maximal and &lt;em&gt;does&lt;/em&gt; contain $\sqrt{3}$ (simple application of Zorn's Lemma), and one that is maxiimal and &lt;em&gt;does&lt;/em&gt; contain $\sqrt{6}$ (ditto); so with your process you can either remove $\sqrt{3}$ at some point, or remove $\sqrt{6}$ at some point, and these choices lead to different maximal subfields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very word &quot;choices&quot; suggests how the Axiom of Choice is playing a role in this &quot;top-down&quot; construction of elements of $\{\mathbb{R}(\setminus a)\}$ (in the &quot;bottoms-up&quot; approach, it plays a role via Zorn's Lemma). In order to be able to obtain something one could call &lt;strong&gt;the&lt;/strong&gt; &quot;field reduction&quot;, we would need some procedure or algorithm that determines &lt;em&gt;which&lt;/em&gt; choice we make at any stage. One way would be by endowing $\mathbb{R}$ with a well-ordering, and then at each step we can consider the set of all elements we can remove; if it is empty, we are done; if it is not empty, we remove the least element of the set. But, of course, well-ordering $\mathbb{R}$ is a bit of a problem in and of itself: there is no consistently agreed-upon well-ordering of $\mathbb{R}$ (and in some cases, not even an explicit one). We could try to pick a particular model in which there are explicit well-orderings (as mentioned by Adres Caicedo in the comments, such models exist), &lt;em&gt;and&lt;/em&gt; a particular well-ordering in there, and &lt;em&gt;then&lt;/em&gt; define &quot;the&quot; field reduction, but that seems to be as much of a problem in and of itself: which model, and which order, and why? Perhaps there are other mechanisms whereby one could try to establish a distinguished element in $\{\mathbb{R}(\setminus a)\}$ (or perhaps, a distinguished element for specific instances of $a$), but it seems to me to be a hopeless task in the abstract/in general. &#xA;&lt;hr/&gt;&#xA;&lt;em&gt;Added and edited:&lt;/em&gt; Recall that any field extension $K/k$ can be broken up as $K/F/k$, where $K/F$ is algebraic and $F/k$ is a purely transcendental extension. In particular, if $T$ is a transcendence basis for $K$ over $k$, then $k(T)$ will be a maximal subfield of $K$ that contains $k$ and does not contain any elements of $K$ that are algebraic over $k$ but not in $k$. So for any field extension $K/k$, if we let&#xA;\begin{equation*}&#xA;\mathbb{A}_{K/k} =\bigl\{ a \in K\ \bigm|\ \mbox{$a$ is algebraic over $k$, $a\notin k$}\bigr\},&#xA;\end{equation*}&#xA;then &#xA;$$K(\setminus \mathbb{A}_{K/k}) = \{ k(T)\ |\ \mbox{$T$ is a transcendence basis for $K$ over $k$}\}.$$&#xA;If $K$ is algebraic over $k$, then you just get $k$, of course; if $K$ is purely transcendental over $k$, then $\mathbb{A}_{K/k}$ is empty, so you get $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had originally written that the set would always be singleton, but this is not the case. If $T$ is any transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$, you can modify an element of $T$ by multiplying it by a non-rational algebraic number and you get a slightly different subfield. For example, fix $t\in T$ and replace $t$ with $\sqrt{2}t$ to obtain a new set $T&amp;#39;$. This is still a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$; but whereas $\sqrt{2}t\in \mathbb{Q}(T&amp;#39;)$, it does not lie in $\mathbb{Q}(T)$. If it did, then we would immediately get $\sqrt{2}\in\mathbb{Q}(T)$, contradicting the fact that $\mathbb{Q}(T)$ is purely transcendental over $\mathbb{Q}$. You can do the same thing taking arbitrary subsets of $T$ and multiplying them through by a nonrational algebraic number. Since $|T|=\mathfrak{c}$, this gives at least $2^{\mathfrak{c}}$ distinct elements in $\mathbb{R}(\setminus\mathbb{A}&amp;#39;)$, and since that is the number of &lt;em&gt;subsets&lt;/em&gt; of $\mathbb{R}$, it follows that the cardinality of $\mathbb{R}(\setminus\mathbb{A}&amp;#39;)$ is $2^\mathfrak{c}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T18:16:40.417" LastActivityDate="2010-11-29T18:16:40.417" CommentCount="9" />
  <row Id="12268" PostTypeId="1" CreationDate="2010-11-29T00:03:20.383" Score="1" ViewCount="82" Body="&lt;p&gt;Problem:&#xA;Let $\omega\in\Omega^r(M^n)$ suppose that $\int_\sum \omega = 0$ for every oriented smooth manifold $\sum \subseteq M^n$ that is diffeomorphic to $S^r$. Show that $d\omega = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;Assume $d\omega \neq 0$. Then there exists $v_1, \ldots, v_{r+1}\in T_pM$ such that $d\omega_p(v_1, \ldots, v_{r+1}) \neq 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D^{r+1}\subseteq \mathbb{R}^{r+1}$ a smooth submanifold of $\mathbb{R}^{n}$ with boundary $S^r$. Let $(h,U)$ be a chart around $p$ such that $D^{r+1}$ (with &lt;em&gt;some&lt;/em&gt; radius) is mapped to $N = h^{-1}(D^{r+1})$ around $p$. Then $N$ is a smooth submanifold of $M^n$ with boundary equal to $\partial N =  h^{-1}(S^r)$ (diffeomorphic to $S^r$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of the integral and Stokes' theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{\partial N} \omega = \int_N d\omega = \int_{D^{r+1}}(h^{-1})^* (d\omega)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\alpha = (h^{-1})^*(d\omega)$. Then $\alpha = f(x)dx_1\wedge\ldots\wedge dx_{r+1}$(topform in $\mathbb{R}^{r+1}$). Since $f(x) \neq 0$, it has to be different from zero on a small domain. Assume that $f(x) &amp;gt; 0$. Then &#xA;$\int_{D^{r+1}}\alpha = \int_{D^{r+1}}f(x)d\mu_{r+1} &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- I feel my idea is correct, but I'm not fully sure this is a full good proof. Could this have been done easier? I'm grateful for any feedback.&lt;/p&gt;&#xA;" OwnerUserId="2900" LastActivityDate="2010-11-29T00:03:20.383" Title="Diffeomorphisms and Stokes' theorem" Tags="&lt;differential-topology&gt;" CommentCount="3" />
  <row Id="12269" PostTypeId="1" AcceptedAnswerId="12693" CreationDate="2010-11-29T00:03:37.653" Score="6" ViewCount="174" Body="&lt;p&gt;Recall that a &lt;strong&gt;model category&lt;/strong&gt; is a complete and cocomplete category with classes of morphisms called &lt;em&gt;cofibrations, fibrations,&lt;/em&gt; and &lt;em&gt;weak equivalences.&lt;/em&gt; These are closed under composition and satisfy certain axioms, such as lifting properties. Furthermore, according to Mark Hovey (but this apparently varies in the literature), any morphism must admit a functorial factorization into the composite of a cofibration and a trivial fibration (i.e. a fibration which is a weak equivalence), as well as a composite of a trivial cofibration and a fibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard examples of model categories  are simplicial sets and chain complexes. Yet the words &quot;fibration&quot; and &quot;cofibration&quot; suggest not category theory but the topological homotopy lifting property. A morphism of topological spaces $f: X \to Y$ is called a (Hurewicz) fibration if whenever $p: T \times [0,1]  \to Y$ is a map and $\widetilde{p}: T \to X$ is a map lifting $p|_{T \times \{0\}}$, $\widetilde{p}$ can be extended to a lifting of $p$. Cofibrations can be defined by an analogous homotopy extension property, and for Hausdorff spaces this is equivalent to being a deformation retract of a suitable neighborhood (defined as $\{x: u(x)&amp;lt;1\}$ where $u$ is a suitable continuous function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, although this is not explicitly stated anywhere I see, I take it that topological spaces with the usual notions of fibrations and cofibrations (with weak equivalences the homotopy equivalences) do indeed form a model category. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Am I right in assuming this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axioms are slightly tricky to check in general. Topological spaces admit limits and colimits. I believe that the lifting properties of cofibrations with respect to trivial fibrations (or trivial cofibrations with respect to fibrations) follows directly from the definition of a cofibration (namely, if you can extend something up to homotopy, you can do it exactly for a cofibration). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm getting slightly confused on the functorial factorizations. Let $f: X \to Y$ be a morphism. We need a &lt;em&gt;canonical&lt;/em&gt; way of factoring this (actually, two canonical ways). &lt;del&gt;One way is to use the inclusion of $X$ in the mapping cylinder $M_f$ (which is a cofibration because $(M_f, X)$ is checked to be an NDR-pair).&#xA;Moreover, $M_f \to Y$ is a homotopy equivalence. &#xA;I don't see why this is a fibration though (it's true that the fibers have the same homotopy type at least).&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correction: As Aaron observes in the comments, there are easy counterexamples for when the map from the mapping cylinder is not a fibration. This means that another approach is needed to construct the functorial factorization of a map into a cofibration and a trivial fibration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone clarify?&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-29T02:59:24.847" LastActivityDate="2010-12-02T03:18:40.370" Title="Why is Top a model category?" Tags="&lt;algebraic-topology&gt;&lt;category-theory&gt;&lt;homotopy-theory&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="12270" PostTypeId="2" ParentId="12160" CreationDate="2010-11-29T00:12:26.997" Score="3" ViewCount="0" Body="&lt;p&gt;To resolve the second question, note first that the Legendre polynomials are odd functions for odd order (0 then is one root of the polynomial), and even functions for even order. Thus, with regards to solubility in terms of radicals, you should be able to derive (possibly complicated!) radical expressions at least up until $P_9(x)$. To use that as an example, note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_9(\sqrt{x})}{\sqrt{x}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a quartic; thus, one can use the quartic formula to derive explicit expressions for its roots, and then you can easily derive the roots of $P_9(x)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{10}(x)$ is where your trouble starts. If we take a look at the polynomial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{10}(\sqrt{x})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have a quintic to contend with. I'll skip the relatively tedious details, but you can verify that its Galois group is not a solvable group, and thus the solution cannot be expressed in terms of radicals (you can use theta or hypergeometric functions, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, not much hope in the symbolic front. In the &lt;em&gt;numeric&lt;/em&gt; front, things are much easier. The slickest way of getting accurate values of the roots of the Legendre polynomial is to use the Jacobi matrix in my previous answer. Since there exist stable and efficient algorithms (e.g. QR algorithm or divide-and-conquer) for the symmetric eigenproblem (in &lt;a href=&quot;http://www.netlib.org/lapack/&quot; rel=&quot;nofollow&quot;&gt;LAPACK&lt;/a&gt;, for instance), and things can be set such that only eigenvalues are returned, you have a good way of generating good approximate values of Legendre polynomial roots. (In the context of Gaussian quadrature, where the roots of orthogonal polynomials play a pivotal role, the scheme is referred to as the &lt;a href=&quot;http://www.ams.org/journals/mcom/1969-23-106/S0025-5718-69-99647-1/S0025-5718-69-99647-1.pdf&quot; rel=&quot;nofollow&quot;&gt;Golub-Welsch algorithm&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, as I mentioned in the comments, there exist asymptotic approximations for the roots, which can then be subsequently polished with a few applications of Newton-Raphson. One such asymptotic approximation is due to Francesco Tricomi. Letting $\xi_{n,k}$ be the $k$-th root of $P_n(x)$, ordered in decreasing order, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\xi_{n,k}\approx\left(1-\frac1{8n^2}+\frac1{8n^3}\right)\cos\left(\pi\frac{4k-1}{4n+2}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $O(n^{-4})$ and further terms are omitted. Other asymptotic approximations due to Luigi Gatteschi use roots of Bessel functions, but I won't say more about those.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-11-29T00:12:26.997" />
  <row Id="12271" PostTypeId="2" ParentId="7096" CreationDate="2010-11-29T01:28:25.750" Score="0" ViewCount="0" Body="&lt;p&gt;I cannot answer your question but if we take $E$ to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Topologist%27s_sine_curve&quot; rel=&quot;nofollow&quot;&gt;topologist's sine curve&lt;/a&gt; and its projection to $B=[0,1]$ we have a topological (i.e. continuous) family of points (compact subspaces, analogs of the compact complex submanifolds) which however is not proper as a map of topological spaces, since $E=\pi^{-1}(B)$ is not compact while $B$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may obtain similar nonproper examples of topological families with compact fibers modifying the fibers or the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the base is a manifold but not the total space/family, and I do not see immediately such counterexamples with families of compact topological manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually I do not understand how Ehresmann's theorem is used in your reasonings, in particular whether you obtain that $\pi$ is proper from it.&lt;/p&gt;&#xA;" OwnerUserId="3407" LastActivityDate="2010-11-29T01:28:25.750" CommentCount="1" />
  <row Id="12272" PostTypeId="2" ParentId="12097" CreationDate="2010-11-29T01:57:26.827" Score="3" ViewCount="0" Body="&lt;p&gt;What you are asking for is a special case of the joint distribution of the maximum and sum of a sequence of random variables. The fact that the variables are stable simplifies this and allows us to compute the joint distribution in the limit as n goes to infinity. This limit can be described in terms of the joint distribution of the value of a stable &lt;a href=&quot;http://en.wikipedia.org/wiki/L%C3%A9vy_process&quot; rel=&quot;nofollow&quot;&gt;L&amp;eacute;vy process&lt;/a&gt; at time $t=1$ and its maximum jump size over $t\le1$. L&amp;eacute;vy processes are right continuous with left limits (&lt;a href=&quot;http://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g&quot; rel=&quot;nofollow&quot;&gt;cadlag&lt;/a&gt;) and have stationary independent increments. Stable L&amp;eacute;vy processes have increments with stationary distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X_i$ each have the $\alpha$-stable distribution denoted by $S(\alpha,\beta,\gamma,\delta)=S(\alpha,\beta,\gamma,\delta;0)$ in the text you link, then the sum $S=\sum_{i=0}^{n-1}X_i$ has the $S(\alpha,\beta,n^{1/\alpha}\gamma,\delta_n)$ distribution where&#xA;$$&#xD;&#xA;\delta_n=\begin{cases}&#xD;&#xA;n\delta + \gamma\beta\tan\frac{\pi\alpha}{2}(n^{1/\alpha}-n),&amp;amp;\textrm{if }\alpha\not=1,\\&#xD;&#xA;n\delta +\gamma\beta\frac{2}{\pi}n\log n,&amp;amp;\textrm{if }\alpha=1.&#xD;&#xA;\end{cases}\qquad\qquad{\rm(1)}&#xD;&#xA;$$&#xD;&#xA;This is equation (1.9) on page 20 of the linked notes. It will be helpful to rescale the sum to remove the dependence on n. The random variable $Y=n^{-1/\alpha}(S-\delta_n)$ has the $S(\alpha,\beta,\gamma,0)$ distribution (by Proposition 1.16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to relate this to L&amp;eacute;vy processes. Let $Z_t$ be the L&amp;eacute;vy process with $Z_1\sim Y$. We can take $X_i=n^{1/\alpha}(Z_{(i+1)/n}-Z_{i/n})+\delta_n/n$, which will be independent with the required $S(\alpha,\beta,\gamma,\delta)$ distribution. In fact, in this case we have $Z_1=Y$ and $S=n^{1/\alpha}Z_1+\delta_n$.&#xA;Writing $M=\max_iX_i$ and $\Delta Z^*_t$ for the maximum of $\Delta Z_s=Z(s)-Z(s-)$ over $s\le t$ gives $n^{-1/\alpha}(M-\delta_n/n)\to\Delta Z^*_1$ as n goes to infinity.&#xA;We can then compute the probability &lt;em&gt;P&lt;/em&gt; that you are asking for to leading order as &lt;em&gt;n&lt;/em&gt; becomes large,&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;P&amp;amp;\equiv\mathbb{P}\left(M &amp;gt; S-M\right)\\&#xD;&#xA;&amp;amp;=\mathbb{P}\left(2n^{-1/\alpha}(M-\delta_n/n) &amp;gt; n^{-1/\alpha}(S-2\delta_n/n)\right)\\&#xD;&#xA;&amp;amp;\sim\mathbb{P}\left(2\Delta Z^*_1 &amp;gt; Z_1+n^{-1/\alpha}\delta_n(1-2/n)\right)&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;So,&#xA;$$&#xD;&#xA;P\sim\mathbb{P}\left(2\Delta Z^*_1 - Z_1 &amp;gt; n^{-1/\alpha}\delta_n\right)&#xD;&#xA;$$&#xD;&#xA;and the leading term in the asymptotics for &lt;em&gt;P&lt;/em&gt; are determined by the distribution of $2\Delta Z^*_1-Z_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\alpha &amp;lt; 1$ it can be seen from (1) that $n^{-1/\alpha}\delta_n\to\gamma\beta\tan\frac{\pi\alpha}{2}$ as &lt;em&gt;n&lt;/em&gt; goes to infinity, so we get a finite and positive limit&#xA;$$&#xD;&#xA;P\to\mathbb{P}\left(2\Delta Z^*_1-Z_1 &amp;gt; \gamma\beta\tan\frac{\pi\alpha}{2}\right)&#xD;&#xA;$$&#xD;&#xA;as $n\to\infty$. The tangent has only appeared in the expression here because of the particular parameterization that we are using. It is a bit simpler if we rewrite this in terms of $\tilde Z_t=Z_t+\gamma\beta t\tan\frac{\pi\alpha}{2}$. Then, for the $\beta=1$ case you are using, $\tilde Z_t$ has support $[0,\infty)$ (Lemma 1.10 in the linked text). This means that $\tilde Z$ is a &lt;em&gt;stable subordinator&lt;/em&gt; and the required probability is&#xA;$$&#xD;&#xA;P\to\mathbb{P}\left(2\Delta\tilde Z^*_1 &amp;gt; \tilde Z_1\right)=\frac{\sin(\pi\alpha)}{\pi\alpha}.&#xD;&#xA;$$&#xD;&#xA;[I've been a bit sneaky here, and substituted in the expression that Shai Covo gave in &lt;a href=&quot;http://math.stackexchange.com/questions/12097/probability-of-the-maximum-levy-stable-random-variable-in-a-list-being-greater/12347#12347&quot;&gt;his answer&lt;/a&gt; for $\mathbb{P}(2\Delta\tilde Z^*_1 &amp;gt; \tilde Z_1)$ and simplified a bit using &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Reflection_formula&amp;amp;oldid=358060311&quot; rel=&quot;nofollow&quot;&gt;Euler's reflection formula&lt;/a&gt;.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $\alpha &amp;gt; 1$ then (1) gives $n^{-1/\alpha}\delta_n\sim n^{1-1/\alpha}(\delta-\gamma\beta\tan\frac{\pi\alpha}{2})$ and, if $\alpha=1$, then $n^{-1/\alpha}\delta_n\sim\gamma\beta\frac{2}{\pi}\log n$. In this case, P tends to zero at a rate dependent on the tail of the distribution function of $2\Delta Z^*_1-Z_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to use the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=L%C3%A9vy_process&amp;amp;oldid=396531725#L.C3.A9vy.E2.80.93Khintchine_representation&quot; rel=&quot;nofollow&quot;&gt;L&amp;eacute;vy-Khintchine representation&lt;/a&gt; to compute the tail of the distribution function of $2\Delta Z^*_1-Z_1$. The jumps of a L&amp;eacute;vy process are described by a L&amp;eacute;vy measure $\nu$ on the real numbers, so that the jumps of size belonging to a set $A$ occur according to a &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Poisson_process&amp;amp;oldid=391609118&quot; rel=&quot;nofollow&quot;&gt;Poisson process&lt;/a&gt; of rate $\nu(A)$. An $\alpha$-stable process has L&amp;eacute;vy measure $d\nu(x)=c\vert x\vert^{-\alpha-1}\,dx$, where $c$ only depends on the sign of $x$. We can back out the value of $c$ by calculating the characteristic function of $Z_1$ using the L&amp;eacute;vy-Khintchine formula (essentially the Fourier transform of $\nu$, see &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=L%C3%A9vy_process&amp;amp;oldid=396531725#L.C3.A9vy.E2.80.93Khintchine_representation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, which uses W for the L&amp;eacute;vy measure) and comparing with expression (1.4) from your text. Going through this computation gives&#xA;$$&#xD;&#xA;d\nu(x)=\pi^{-1}\alpha\gamma^\alpha\sin\frac{\pi\alpha}{2}\Gamma(\alpha)(1+\beta{\rm sign}(x))\vert x\vert^{-\alpha-1}\,dx.&#xD;&#xA;$$&#xD;&#xA;The number of jumps of size greater than $x$ is Poisson distributed of rate $\nu((x,\infty))$ giving,&#xA;$$&#xD;&#xA;\begin{align}\mathbb{P}(\Delta Z^*_1&amp;gt; x) &amp;amp;= 1-e^{-\nu((x,\infty))}\\&#xD;&#xA;&amp;amp;\sim\nu((x,\infty))\\&#xD;&#xA;&amp;amp;=\pi^{-1}\gamma^\alpha\sin\frac{\pi\alpha}{2}\Gamma(\alpha)(1+\beta)x^{-\alpha}.&#xD;&#xA;\end{align}\qquad{\rm(2)}&#xD;&#xA;$$&#xD;&#xA;Comparing with your linked text (Theorem 1.12), this is exactly the same as the tail of $Z_1$! That is, $\mathbb{P}(\Delta Z^*_1 &amp;gt; x)\sim\mathbb{P}(Z_1 &amp;gt; x)$. This means that the tail of the distribution is entirely due to the occasional big jump of the L&amp;eacute;vy process. This does actually correspond to experience. I have plotted Cauchy process sample paths before, and you do get some paths with a single big jump drowning out all other behaviour. We can also see why this should be the case. The density of the jump distribution is sub-exponential, proportional to $\vert x\vert^{-\alpha-1}$. So one big jump of size $x$ has a much bigger probability than &lt;em&gt;n&lt;/em&gt; small jumps of size $x/n$. It also suggests that, in your calculation of $X_i$, the samples contributing to the probability come mainly from the case where all of them are reasonably close to the mean except for one extreme (positive or negative) $X_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from the argument above, the leading asymptotic of the probability &lt;em&gt;P&lt;/em&gt; comes from the case where $\Delta Z_1^*$ is very large compared to $Z_1-\Delta Z_1^*$, or when the largest negative jump $\Delta(-Z)^*_1$ is large compared to $Z_1+\Delta(-Z)^*_1$. This gives&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\mathbb{P}(2\Delta Z^*_1-Z_1 &amp;gt; x)&#xD;&#xA;&amp;amp;\sim\mathbb{P}(\Delta Z^*_1 &amp;gt; x)&#xD;&#xA;+\mathbb{P}(\Delta (-Z)^*_1 &amp;gt; x)\\&#xD;&#xA;&amp;amp;\sim 2\pi^{-1}\gamma^{\alpha}\sin\frac{\pi\alpha}{2}\Gamma(\alpha)x^{-\alpha},&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;where I substituted in (2). Plugging in the asymptotic form for $x=n^{-1/\alpha}\delta_n$ calculated above gives&#xA;$$&#xD;&#xA;P\sim2\pi^{-1}\gamma^{\alpha}\sin\frac{\pi\alpha}{2}\Gamma(\alpha)\left(\delta-\gamma\beta\tan\frac{\pi\alpha}{2}\right)^{-\alpha}n^{1-\alpha}&#xD;&#xA;$$&#xD;&#xA;for $\alpha &amp;gt; 1$. This confirms your $O(1/n^{\alpha-1})$ guess from the numerical simulation! Also, using $x=n^{-1/\alpha}\delta_n\sim\gamma\beta\frac{2}{\pi}\log n$ for $\alpha=1$ gives &#xA;$$&#xD;&#xA;P\sim\frac{1}{\beta\log n}.&#xD;&#xA;$$&#xD;&#xA;Hopefully this is all correct, but I can't rule out making an arithmetic slip above. &lt;strong&gt;Edit&lt;/strong&gt;: The limit above for $\alpha &amp;gt; 1$ does not look very good. It appears to be slightly above the plot in the original post. However, even for n=1000, it should not be very close to the asymptotic limit described. Replacing the asymptotic limit for $\delta_n$ by its exact form effective replaces the $n^{1-\alpha}$ term for &lt;em&gt;P&lt;/em&gt; by $(n^{1-1/\alpha}-1)^{-\alpha}$. For n=1000 this approximately doubles the calculated number so, the more accurate value for $\delta_n$ moves you further away from the plot. Something has gone wrong somewhere...&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-11-30T01:04:58.173" LastActivityDate="2010-11-30T01:04:58.173" CommentCount="20" />
  <row Id="12273" PostTypeId="1" AcceptedAnswerId="12277" CreationDate="2010-11-29T02:01:34.797" Score="5" ViewCount="117" Body="&lt;p&gt;This is some sort of standard puzzle, which many do it using trial and error or brute-force method. The question goes like this, given the numbers, 11,13,31,33,42,44,46 what is choices of numbers (a number can be chosen more than once) so that sum adds up to 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formula like way or approach to this other than brute force only?&lt;/p&gt;&#xA;" OwnerUserId="4186" LastEditorUserId="4186" LastEditorDisplayName="" LastEditDate="2010-11-29T05:06:15.197" LastActivityDate="2010-11-29T17:53:34.207" Title="Minimum multi-subset sum to a target." Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12274" PostTypeId="1" CreationDate="2010-11-29T02:05:25.773" Score="2" ViewCount="35" Body="&lt;p&gt;Are there any interesting examples of second order stationary processes on ${\mathcal R}^2$ or ${\mathcal R}^3$ that are not isotropic? The book I am looking at has no examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update : I asked this &lt;a href=&quot;http://mathoverflow.net/questions/47684/stationary-non-isotropic-spatial-stochastic-processes&quot; rel=&quot;nofollow&quot;&gt;question&lt;/a&gt; in mathoverflow, apparently such examples are not easy to come by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Processes with &lt;a href=&quot;http://books.google.com/books?id=ZvoibTTS9QwC&amp;amp;pg=PA113&amp;amp;dq=anisotropic+variograms&amp;amp;hl=en&amp;amp;ei=fvbzTMm-I8WqlAeP4aX0DA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCcQ6AEwAA#v=onepage&amp;amp;q=anisotropic%2520variograms&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;anisotropic variograms&lt;/a&gt; are examples of non-isotropic stationary processes.&lt;/p&gt;&#xA;" OwnerUserId="404" LastEditorUserId="404" LastEditorDisplayName="" LastEditDate="2010-11-29T18:56:46.977" LastActivityDate="2010-11-29T18:56:46.977" Title="stationary non-isotropic spatial stochastic processes" Tags="&lt;probability-theory&gt;&lt;statistics&gt;" />
  <row Id="12275" PostTypeId="1" AcceptedAnswerId="12280" CreationDate="2010-11-29T02:06:10.020" Score="11" ViewCount="128" Body="&lt;p&gt;Could you prove or disprove the following statement? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let &#xA;  $f\colon[0,1]^2\rightarrow \mathbb R$ &#xA;  be a continuous function.  Then&#xA;  there are continuous functions &#xA;  $g,\ h\colon [0,1]\rightarrow \mathbb R$ and&#xA;  $\Phi\colon \mathbb R \to \mathbb R$&#xD;&#xA;  such that  $$ f(x,y) = \Phi(g(x) + h(y)).$$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(This problem popped up in my mind while I was thinking about &lt;a href=&quot;http://mathoverflow.net/questions/47557/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; related one on MO.  I couldn't find an easy proof or a disproof. This version is much weaker than the one asked at MO, since $g$ and $h$ do depend on $f$ here.)&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-11-29T04:47:34.353" Title="Writing  $f(x,y)$ as $\Phi(g(x) + h(y))$" Tags="&lt;real-analysis&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="12276" PostTypeId="1" AcceptedAnswerId="12286" CreationDate="2010-11-29T02:07:17.977" Score="17" ViewCount="244" Body="&lt;p&gt;Green's Theorem has the form: &#xA;$$\oint P(x,y)dx = - \iint \frac{\partial P}{\partial x}dxdy , \oint Q(x,y)dy = \iint \frac{\partial Q}{\partial y}dxdy $$&#xD;&#xA;The Cauchy-Riemann equations have the following form:(Assuming $z = P(x,y) + iQ(x,y)$)&#xA;$$\frac{\partial P}{\partial x} = \frac{\partial Q}{\partial y}, \frac{\partial P}{\partial y} = - \frac{\partial Q}{\partial x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any connection between this two equations?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T02:58:50.557" LastActivityDate="2010-11-29T03:40:57.813" Title="Is there any connection between Green's Theorem and the Cauchy-Riemann equations?" Tags="&lt;calculus&gt;&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12277" PostTypeId="2" ParentId="12273" CreationDate="2010-11-29T02:17:33.820" Score="5" ViewCount="0" Body="&lt;p&gt;SUBSET-SUM is NP-complete, so in general it's difficult to tell even if there is a choice of numbers that adds up to a given target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Moron has raised the valid point that numbers can be reused.&#xA;We therefore replace each number $x$ with a sequence $x,2x,4x,\ldots,2^kx,\ldots$, stopping before reaching a number larger than our target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's carefully analyze the blowup in size. Suppose the original numbers $x_i$ had size $l_i$, and the target $T$ had size $L$. We replace each number $x_i$ with $L$ numbers of size at most $L$, and so the new size is $$\sum_i L^2 + L \leq L^2 (\sum_i x_i + L),$$ so the blowup is at most cubic, and in particular polynomial. &lt;i&gt;So the problem is still NP-hard (it is trivially in NP):&lt;/i&gt; no, since the reduction is in the wrong direction... (thanks, Moron, for noticing)&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-29T16:41:42.380" LastActivityDate="2010-11-29T16:41:42.380" CommentCount="5" />
  <row Id="12278" PostTypeId="2" ParentId="12276" CreationDate="2010-11-29T02:17:39.937" Score="6" ViewCount="0" Body="&lt;p&gt;If you consider $P(x,y)$ and $Q(x,y)$ as  the part real and imaginary of an analitic function you can get the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $P(x,y)$ and $Q(x,y$) verify the Cauchy-Riemann equations, thus :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial P}{\partial x} = \frac{\partial Q}{\partial y} ,  \frac{\partial P}{\partial x} = -\frac{\partial Q}{\partial y} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending of oriented get that the integral is Zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This satisfies the Cauchy's integral theorem that an analytic function  on a closed curve is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see it &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, where the proof of Cauchy's integral theorem  uses Green's Theorem .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little deeper you can see, Complex Analysis by Lars Ahlfors, section 4.6 page 144.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T03:26:53.690" LastActivityDate="2010-11-29T03:26:53.690" />
  <row Id="12279" PostTypeId="2" ParentId="12275" CreationDate="2010-11-29T02:57:21.143" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose $f$ is such that $f(x,0)$ and $f(0,y)$ are monotonic in $x$ and $y$ respectively over $[0,1]$. Since $f(x,0) = \Phi(g(x) + h(0))$, it must be the case that $g$ is monotonic over $[0,1]$. Similarly, $h$ is monotonic over $[0,1]$. So $g(x) + h(y)$ is the sum of monotonic functions. It's not hard to show that for any $(x,y)$ in the interior of $[0,1]^2$, there must be a point on the boundary with the same value of $g(x) + h(y)$, and therefore the same value of $\Phi(g(x) + h(y))$. Since it's easy to construct a continuous $f$ which satisfies the first assumption yet takes a larger range of values in the interior than it does on the boundary (for any $f$ that doesn't, try adding $M x(1-x)y(1-y)$ for large enough $M$), this contradicts the conjecture.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-29T03:15:50.177" LastActivityDate="2010-11-29T03:15:50.177" CommentCount="2" />
  <row Id="12280" PostTypeId="2" ParentId="12275" CreationDate="2010-11-29T03:05:07.337" Score="8" ViewCount="0" Body="&lt;p&gt;The statement is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the counterexample simpler, I will take $f:[-1,1]^2 \rightarrow \mathbb{R}$.  Let &lt;em&gt;f(x,y)=xy&lt;/em&gt;.  So we have &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;---0+++&#xA;---0+++&#xA;0000000&#xA;+++0---&#xA;+++0---&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Consider the image of &lt;em&gt;xy=0&lt;/em&gt; in &lt;em&gt;g(x)+h(y)&lt;/em&gt;.  Since it's a connected, compact set, the image is connected and compact, so an interval [m,n], with $\Phi([m,n])=0$.  We may assume that $\Phi(n+\epsilon)&amp;gt;0$ and $\Phi(m-\epsilon)&amp;lt;0$.  Let &lt;em&gt;A=g(1)&lt;/em&gt;, &lt;em&gt;a=g(-1)&lt;/em&gt;, &lt;em&gt;B=h(1)&lt;/em&gt;, &lt;em&gt;b=h(-1)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $A+B&amp;gt;n, \; a+b&amp;gt;n, \;  A+b&amp;lt;m, \; a+B&amp;lt;m$, which leads to a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="1461" LastEditorDisplayName="" LastEditDate="2010-11-29T04:47:34.353" LastActivityDate="2010-11-29T04:47:34.353" CommentCount="2" />
  <row Id="12281" PostTypeId="1" AcceptedAnswerId="12282" CreationDate="2010-11-29T03:13:36.400" Score="3" ViewCount="56" Body="&lt;p&gt;In propositional logic, for example:&#xA;$$\neg p \vee q.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p$ is true at the outset, does that mean it must be considered false when comparing with q in the disjunction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I am unsure about tags for this question.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T03:23:09.577" LastActivityDate="2010-11-29T05:03:01.170" Title="Are standalone propositions affected by negation operators?" Tags="&lt;logic&gt;&lt;propositional-calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12282" PostTypeId="2" ParentId="12281" CreationDate="2010-11-29T03:17:14.653" Score="4" ViewCount="0" Body="&lt;p&gt;If $p$ is true, then $\neg p$ is false. To evaluate $\neg p \vee q$, you must evaluate $\neg p$ and you must evaluate $q$. If either $\neg p$ is true or $q$ is true, then $\neg p\vee q$ is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you really need to figure out $(\neg p)\vee q$, performing first the operation inside the parentheses, then the disjunction. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-29T03:17:14.653" />
  <row Id="12283" PostTypeId="2" ParentId="12281" CreationDate="2010-11-29T03:17:16.960" Score="4" ViewCount="0" Body="&lt;p&gt;If $p$ is true, then $\lnot p \lor q \Leftrightarrow q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, $p$ and $\lnot p$ have the opposite value: if one is true then the other is false, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of $p$ as some proposition, say &quot;today is Sunday&quot;. Then $\lnot p$ stands for &quot;today is &lt;i&gt;not&lt;/i&gt; Sunday&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T03:17:16.960" />
  <row Id="12284" PostTypeId="2" ParentId="12273" CreationDate="2010-11-29T03:21:59.810" Score="4" ViewCount="0" Body="&lt;p&gt;Assuming that the numbers are positive integers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is closely related to the &lt;a href=&quot;http://mathworld.wolfram.com/CoinProblem.html&quot; rel=&quot;nofollow&quot;&gt;Frobenius Coin Problem&lt;/a&gt; which says that there is a maximum number $\displaystyle F$ (called the Frobenius number) which is &lt;em&gt;not&lt;/em&gt; representable. It is NP-Hard to find out the Frobenius number when there are at least $\displaystyle 3$ numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a formula like approach to determine if such a representation is possible or not, you can use generating functions, which can be used to give a pseudo polynomial time algorithm, polynomial in size $\displaystyle W = n_1 + n_2 + \cdots + n_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the numbers are $\displaystyle n_1, n_2, \dots, n_k$ and you need to see if they can be summed to $\displaystyle S$ then the number of ways it can be done is the coefficient of $\displaystyle x^S$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1+x^{n_1} + x^{2n_1} + x^{3n_1} + \cdots )(1+ x^{n_2} + x^{2n_2} +  x^{3n_2} + \cdots ) \cdots (1 + x^{n_k} + x^{2n_k} + x^{3n_k} + \cdots )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle  = \dfrac{1}{(1-x^{n_1})(1-x^{n_2}) \cdots (1-x^{n_k})}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using partial fractions this can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{m} \dfrac{C_j}{c_j - x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle C_j$ and $\displaystyle c_j$ are appropriate complex numbers and $\displaystyle m \le n_1 + n_2 + \cdots + n_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of $\displaystyle x^S$ is thus given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{m} \dfrac{C_j}{c_j^{S+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you need to check is zero or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this might require quite precise floating point operations and does not actually tell you what numbers to choose.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-11-29T17:53:34.207" LastActivityDate="2010-11-29T17:53:34.207" CommentCount="0" />
  <row Id="12285" PostTypeId="1" CreationDate="2010-11-29T03:24:32.057" Score="9" ViewCount="143" Body="&lt;p&gt;Why is the following true?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $(X,d)$ is a compact metric space and $f: X \rightarrow X$ is non-expansive (i.e $d(f(x),f(y)) \leq d(x,y)$) and surjective then $f$ is an isometry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4188" LastEditorUserId="4188" LastEditorDisplayName="" LastEditDate="2010-11-29T03:36:07.740" LastActivityDate="2010-11-29T11:20:42.060" Title="Isometry in compact metric spaces" Tags="&lt;metric-spaces&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" />
  <row Id="12286" PostTypeId="2" ParentId="12276" CreationDate="2010-11-29T03:40:57.813" Score="13" ViewCount="0" Body="&lt;p&gt;Absolutely yes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Section 7 of &lt;a href=&quot;http://math.uga.edu/~pete/handouteight.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes on Green's Theorem&lt;/a&gt;, I explain how Green's Theorem plus the Cauchy-Riemann equations immediately yields the Cauchy Integral Theorem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably most serious students of mathematics realize this on their own at some point, but it is surprising how few standard texts make this connection.  In (especially American) undergraduate texts on Subject X, there is a distressing trend to politely ignore the existence of Subject Y, even when any student of Subject X will almost surely have already have studied / be concurrently studying / soon be studying Subject Y.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-29T03:40:57.813" CommentCount="2" />
  <row Id="12287" PostTypeId="1" AcceptedAnswerId="12290" CreationDate="2010-11-29T04:23:53.613" Score="3" ViewCount="153" Body="&lt;p&gt;It is my question when I was in Senior High School. Up to now, I have no idea about the correct explanation. I just accept it by faith :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here  is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HuwRD.gif&quot; alt=&quot;tangent line diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the symbol $\Delta x \to 0$ does not mean $\Delta x =0$, how can the points $A$ and $B$ get overlapped which in turn, how can the line joining $A$ and $B$ become $C$, the tangent line to the curve $y=f(x)$ at point $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="3901" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T04:55:35.967" LastActivityDate="2010-11-29T15:37:20.493" Title="Approaching to zero, but not equal to zero, then why do the points get overlapped? " Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;functions&gt;&lt;limit&gt;&lt;differential&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="12288" PostTypeId="2" ParentId="12287" CreationDate="2010-11-29T04:34:59.907" Score="7" ViewCount="0" Body="&lt;p&gt;$A$ and $B$ do not literally become overlapped, and the line $C$ is not obtained literally as a line joining points $A$ and $B$ on the curve.  The diagram is designed to motivate the plausibility of the fact that as $B$ gets &quot;sufficiently close&quot; (but not equal) to $A$, the line connecting $A$ and $B$ gets &quot;arbitrarily close&quot; to the tangent line $C$ to the curve at the point $A$.  This is consistent with the idea that $\Delta x$ &quot;approaches $0$&quot;, but is never equal to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, the slope of the line $C$ is the derivative of $f$ at $x_0$, $f&amp;#39;(x_0)$, defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(x_0)=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More informally again, this &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_of_a_function&quot; rel=&quot;nofollow&quot;&gt;limit&lt;/a&gt; means that you can get &quot;arbitrarily close&quot; to $f&amp;#39;(x_0)$ by taking $\Delta x$ &quot;sufficiently small&quot; (but not $0$) in the difference quotient $\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$, which is the slope of the line connecting $A$ to $B$ in the diagram.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-29T05:25:53.573" LastActivityDate="2010-11-29T05:25:53.573" CommentCount="4" />
  <row Id="12289" PostTypeId="2" ParentId="12285" CreationDate="2010-11-29T04:37:03.290" Score="5" ViewCount="0" Body="&lt;p&gt;In fact more is true. For a compact, metric space, $f$ locally non-expansive $\implies$ $f$ local isometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is theorem 4.2 in the paper linked below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.pjm/1102970464&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.pjm/1102970464&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-29T04:37:03.290" CommentCount="1" />
  <row Id="12290" PostTypeId="2" ParentId="12287" CreationDate="2010-11-29T04:51:14.567" Score="13" ViewCount="0" Body="&lt;p&gt;Basically, what you have put your finger on is the original difficulty with &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinitesimals&quot; rel=&quot;nofollow&quot;&gt;infinitesimals&lt;/a&gt;, and the basis on which Bishop Berkeley made some of his famous objections (mind you, Berkeley didn't really care about the logical foundations of Calculus; he was engaged in a theological debate at the time&lt;sup&gt;1&lt;/sup&gt;): if infinitesimals are not zero, then what you have is not a tangent but a secant, because the line touches the curve at two distinct points; but if the two points are the same, then they don't define a tangent because you cannot determine a line with a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that when we talk about limits, we are talking about what the quantities are &lt;em&gt;approaching&lt;/em&gt;, not what the quantities &lt;em&gt;are&lt;/em&gt;. The point $B$ never &quot;gets overlapped&quot; with $A$, it just &lt;em&gt;approaches&lt;/em&gt; $A$; the line between $A$ and $B$ never &quot;becomes&quot; the tangent (which in your diagram is $C$), but its slope &lt;em&gt;approaches&lt;/em&gt; the slop of $c$. These &quot;approaches&quot; have a very precise meaning (made formal by Weierstrass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, what would it mean to say that the values of a function $g(x)$ &quot;approach&quot; the value $L$ as $x$ approaches $a$? It means that you can make all the values of $g(x)$ be arbitrarily close to $L$ provided that $x$ is &quot;close enough&quot; to $a$. If you specify a narrow horizontal band around  the value $y=L$, then by specifying a narrow vertical band around the value $a$ we can guarantee that the graph of $y=g(x)$ for values of $x$ inside the narrow vertical band will necessarily lie inside the horizontal line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is sensible is that if the values of $g(x)$ approach some other number $M$ as well, then by specifying a band around $L$ which is smaller than the distance from $M$ to $L$ you will always run into problems: the graph of $y=g(x)$ will always end up with parts outside this band, because $M$ is outside the band and you are also approaching $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, we say it like this: the limit of $g(x)$ as $x$ approaches $a$ is $L$, which is written:&#xA;$$\lim_{x\to a}g(x) = L$$&#xD;&#xA;if and only if for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that if $0\lt |x-a|\lt \delta$, then $|f(x)-L|\lt \epsilon$; $\epsilon$ is how wide the horizontal band around $L$ is, $\delta$ is how thin you need to make the vertical band around $a$ needs to be to make sure the graph is completely inside the rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the picture suggests is that as $\Delta x$ approaches $0$, the slope of the line joining $A$ and $B$  &lt;em&gt;should&lt;/em&gt; approach the slope of the tangent at $a$; however, the line joining $A$ and $B$ never actually &quot;becomes&quot; the tangent $C$, and the point $B$ never actually &quot;becomes&quot; the point $A$. They are just approaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, does the line joining $A$ and $B$ have a slope that &lt;em&gt;really&lt;/em&gt; approaches the slope of the tangent? The key is that we can characterize the tangent algebraically: the tangent is the unique line that affords the best approximation to $y=f(x)$ near the point $x_0$, where by &quot;best approximation&quot; we mean one in which the relative error approaches zero. That is: if we take a line $mx+b$ that goes through $(x_0,f(x_0))$, then we can ask for the &quot;relative error&quot; in approximating using $mx+b$ instead of $f(x)$:&#xA;$$\frac{f(x) - (mx+b)}{x-x_0}.$$&#xD;&#xA;The tangent is the unique line through $(x_0,f(x_0))$ for which the relative error approaches $0$ as $x$ approaches $x_0$ (if such a line exists at all). Since $y=mx+b$ goes through $(x_0,f(x_0))$ if and only if $f(x_0) = mx_0 + b$, then $b= f(x_0)-mx_0$; so we have that the relative error will be&#xA;$$\frac{f(x) - mx - f(x_0)+mx_0}{x-x_0} = \frac{f(x)-f(x_0) - m(x-x_0)}{x-x_0} = \frac{f(x)-f(x_0)}{x-x_0} - m.$$&#xD;&#xA;So if we want the relative error to approach $0$, then we need &#xA;$\frac{f(x)-f(x_0)}{x-x_0}$&#xD;&#xA;to approach $m$; that is, the slope of the tangent &lt;em&gt;must&lt;/em&gt; be whatever quantity the numbers&#xA;$$\frac{f(x)-f(x_0)}{x-x_0}$$&#xD;&#xA;are approaching as $x$ approaches $x_0$ (if they are approaching any particular number). Setting $\Delta x = x-x_0$, so that $x = x_0+\Delta x$, the above fraction is the same as&#xA;$$\frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}.$$&#xD;&#xA;And $x$ approaches $x_0$ if and only if $\Delta x$ approaches $0$. So the slope of the tangent needs to be whatever quantity this fraction approaches as $\Delta x$ approaches $0$; that is, the slope of the tangent at $x_0$ is $m$ if and only if &#xA;$$ \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x} = m.$$&#xD;&#xA;Turning this assertion into a picture gives precisely the picture you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;sup&gt;1&lt;/sup&gt; In case anyone is interested: many people viewed Newtonian mechanics and its consequent &quot;clockwork universe&quot; as a direct attack on the Christian notion of a deity that was directly involved in and modifying his creation, and also on the notion of free will. Some deists were in fact using the new physics as evidence in favor of the deist God, who created the universe, set it motion, but does not actively participate in it, in contrast to the theistic deity. By attacking the mathematical foundation of the new physics, Berkeley was defending the notions of free will and of the active deity. If you read Augustus de Morgan's &lt;em&gt;A Budget of Paradoxes&lt;/em&gt;, he reviews many pamphlets and booklets written during those years which attack Newton and Calculus because they view the latter as an attack on religion. The &quot;morality&quot; arguments raised against Calculus are eerily similar (when not downright &lt;em&gt;identical&lt;/em&gt;) to those raised against Darwin and the Theory of Evolution in later days. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T15:37:20.493" LastActivityDate="2010-11-29T15:37:20.493" CommentCount="5" />
  <row Id="12291" PostTypeId="1" AcceptedAnswerId="12535" CreationDate="2010-11-29T04:58:08.367" Score="5" ViewCount="79" Body="&lt;p&gt;Given a Noetherian, local ring $(R,m)$, can we always find a system of parameters whose images in the cotangent space $m/m^2$ are linearly independent. We can do this in the regular case, by just choosing a basis for the cotangent space and looking at their preimages in $m$ under the canonical map. By Nakayama's lemma these generate $m$ and hence form a system of parameters. Can we do this for any general Noetherian, local ring?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-30T21:29:09.977" Title="System of parameters which have linear independent images in the cotangent space" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12292" PostTypeId="2" ParentId="12281" CreationDate="2010-11-29T05:03:01.170" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not sure I understand your question, but this may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truth Table for ~p v q:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~ p v q&lt;br&gt;&#xA;F T T T&lt;br&gt;&#xA;F T F F&lt;br&gt;&#xA;T F T T&lt;br&gt;&#xA;T F T F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If p is true, and ~p v q is true (first line only), then q is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that ~p v q is logically equivalent to p =&gt; q. &lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-29T05:03:01.170" />
  <row Id="12293" PostTypeId="2" ParentId="12287" CreationDate="2010-11-29T05:22:33.170" Score="0" ViewCount="0" Body="&lt;p&gt;You have already got good answers. However, I would just want to add a suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Take a look at the diagram and picture other points $B_1,B_2,\ldots$ closer and closer to $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think of the lines $L_1,L_2,\ldots$ joining $A$ with and the points $B_1,B_2,\ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see how the lines approach the tangent line $C$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-29T05:22:33.170" />
  <row Id="12294" PostTypeId="2" ParentId="12223" CreationDate="2010-11-29T05:31:29.347" Score="1" ViewCount="0" Body="&lt;p&gt;Counter example: Suppose p, q, r are all false. Then (p=&gt;q)=&gt;r is false, and p=&gt;(q=&gt;r) is true.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-29T05:31:29.347" CommentCount="3" />
  <row Id="12295" PostTypeId="1" AcceptedAnswerId="12297" CreationDate="2010-11-29T05:48:06.060" Score="1" ViewCount="43" Body="&lt;p&gt;I encountered a problem which is asks for the first negative term in the expansion of $(1+x)^{\frac{27}{5}}$, where $x$ is always positive,the solution suggested in my module is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \text{ For the first negative term: } n-r+1 \lt 0 \text{ or, } r \gt 6\frac{2}{5}$, since $n = \frac{27}{5}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the 8th term is the first negative term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my problem is that I am not getting the rationale behind this approach.Also,if you are aware of any other method for the same problem you may like to post too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:Inquisitive to verify the same using Mathematica,I tried &lt;code&gt;Expand[(1 + x)^(27/5)]&lt;/code&gt; but I guess this is not the correct function.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-29T05:53:25.327" LastActivityDate="2010-11-29T06:07:27.920" Title="First non-negative term in an Expansion" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;" AnswerCount="1" />
  <row Id="12296" PostTypeId="1" CreationDate="2010-11-29T05:48:19.867" Score="0" ViewCount="60" Body="&lt;p&gt;Let H be a subgroup of G and V the trivial representation of C[H]. Prove that we have an isomorphism of C[G]-modules between $\mathrm{Ind}_{H}^{G}(V)$ and C[G/H].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="4190" LastActivityDate="2010-11-29T06:23:52.073" Title="Induced representation" Tags="&lt;representation-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12297" PostTypeId="2" ParentId="12295" CreationDate="2010-11-29T05:52:47.003" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: Use generalized binomial theorem, when does a negative factor appear in the binomial coefficient. (I am asssuming $x$ is positive here and $n,r$ have their usual meanings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${n\choose r}=\frac{(n)(n-1)...(n-r+1)}{r!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $r,x$ are positive here. So the only way we can get a negative term is having a negative coefficient and hence only by having a negative numerator of the coefficient. Since the numerator is a product of real numbers it first becomes negative when the last factor is negative (since the first factor is positive and the value of the factors keeps decreasing by one successively). So this happens when $n-r+1&amp;lt;0$. Set $n=\frac{27}{5}$, so $\frac{27}{5}-r+1&amp;lt;0\implies \frac{32}{5}-r&amp;lt;0 \implies r&amp;gt;\frac{32}{5}=6\frac{2}{5}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-29T06:07:27.920" LastActivityDate="2010-11-29T06:07:27.920" CommentCount="5" />
  <row Id="12298" PostTypeId="2" ParentId="11834" CreationDate="2010-11-29T06:11:21.257" Score="6" ViewCount="0" Body="&lt;p&gt;A directed multigraph can be thought of as a one-dimensional simplicial complex: the vertices are points, and the edges are 1-simplices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a category, we can form its nerve, and the 1-skeleton will essentially be a multi-graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observation of the OP is that these two constructions are roughly inverse to each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between simplicial sets and categories is an important one,&#xA;and is at the basis of the theory of higher categories and its relationship to homotopy theory.  Thus the OP's observation and question shouldn't be dismissed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I'm not sure that I agree with the sentiment of the OP's final non-parenthetical sentence:&#xA;&quot;I don't see how it justifies treating the two as separate branches rather than syntactic shims on identical concepts.&quot;   Certainly (multi-)graph theorists and (higher-)category/homotopy theorists both make itensive studies of 1-dimensional&#xA;simplicial sets,&#xA;but the questions they ask and the constructions they are concerned with are typically extremely different.   If they do find common ground, that's great, but they needn't be obliged to seek it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most working mathematicians using categories are, furthermore, not thinking in a homotopical/simplicial way at all about the structures involved, but are just taking advantage of the very useful language and range of concepts that category theory provides.  These are concepts that (as far as I know) are not used at all by those studying multi-graphs and related structures, and so I don't see any reason that people should be obliged or even encouraged to translate (what to them are) familiar category-theoretic notions into a different terminological framework of relations/multi-graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added: Ryan Budney's comment above makes essentially the same point in a pithier fashion.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-11-29T07:51:23.107" LastActivityDate="2010-11-29T07:51:23.107" CommentCount="1" />
  <row Id="12299" PostTypeId="1" AcceptedAnswerId="12303" CreationDate="2010-11-29T06:20:14.720" Score="2" ViewCount="81" Body="&lt;p&gt;Hi, my Calculus II class is nearing the end of the quarter and we've just started differential equations to get ready for Calculus III. In my homework, I came upon these problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the general solution to the differential equation &#xA;$$\frac{dy}{dt} = t^3 + 2t^2 - 8t.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The teacher just said to integrate. So I did. Then in question 8a it gives the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dt} = y^3 + 2y^2 - 8y.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and asks &quot;Why can't we find a solution like we did to the previous problem? My guess was: &quot;In 7 we were integrating with respect to t. Since this equation is the highest order derivative, we can't solve it like # 7&quot;. Although, I have no confidence in that answer and I'm not sure it makes total sense even to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, part 8B. asks: Show that the constant function $y(t) = 0$ is a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a problem like this before, except that it wasn't a constant function. This problem seems like a question that asks: &quot;show that every member of the family of functions $y = (\ln x + C)/x$ is a solution to the differential equation (some diff. equation)&quot; except it seems a little bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints on how I can solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="3514" LastEditorDisplayName="" LastEditDate="2010-11-29T06:31:55.743" LastActivityDate="2010-11-29T06:52:08.630" Title="Help solving a differential equation" Tags="&lt;homework&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="12300" PostTypeId="2" ParentId="12296" CreationDate="2010-11-29T06:23:52.073" Score="1" ViewCount="0" Body="&lt;p&gt;For instance, one of the standard descriptions of the induced module $\operatorname{Ind}_H^G(V)$ is as a direct sum $\bigoplus_{s \in S} s V$ of copies of $V$ indexed by a set $S$ of coset representatives of $G/H$ with a certain natural $G$-action.  See for instance Lemma 2.1 &lt;a href=&quot;http://math.berkeley.edu/~serganov/math252/notes4.pdf&quot; rel=&quot;nofollow&quot;&gt;in these notes&lt;/a&gt;.  (I have no particular connection to these course notes; they just came up as one of the first hits upon googling &quot;induced module&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are trying to prove follows almost immediately from this description of the induced module.  Indeed, in the aforementioned lecture notes, your exercise appears as Example 1 on the following page (with no further explanation).  If you can't see how to deduce this, you should probably book up (and possibly ask further questions) on the definition of an induced module.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-29T06:23:52.073" />
  <row Id="12301" PostTypeId="2" ParentId="12299" CreationDate="2010-11-29T06:30:20.560" Score="3" ViewCount="0" Body="&lt;p&gt;The difference is that the main variable is $y$ and you separate the variables to integrate and obtain the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation is very easy, so :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dt}=t^3 + 2t^2 - 8t \Rightarrow dy = (t^3 + 2t^2 - 8t) dt \Rightarrow $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \int dy = \int (t^3 + 2t^2 - 8t) dt $$   ( which is the left side ? )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other.&#xA;$$\frac{dy}{dt}=y^3 + 2y^2 - 8y \Rightarrow  \frac{1}{y^3 + 2y^2 - 8y}dy =  dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then proceed integrating.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T06:52:08.630" LastActivityDate="2010-11-29T06:52:08.630" />
  <row Id="12302" PostTypeId="1" CreationDate="2010-11-29T06:30:21.690" Score="3" ViewCount="125" Body="&lt;p&gt;Find the value of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots  $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T06:35:56.060" LastActivityDate="2010-11-29T07:17:31.593" Title="Suggest a tricky method for this problem" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" />
  <row Id="12303" PostTypeId="2" ParentId="12299" CreationDate="2010-11-29T06:34:44.813" Score="3" ViewCount="0" Body="&lt;p&gt;Parts of problems that say &quot;show that [explicitly given function] is a solution&quot; can be solved by simply &quot;plugging in&quot;.  E.g., if you're supposed to show that $y(t)=\frac{1}{1-t}$ is a solution to the differential equation $dy/dt=y^2$, then you should compute $dy/dt=\frac{1}{(1-t)^2}$, compute $y^2=\frac{1}{(1-t)^2}$, and &quot;plug them in&quot; to check that the differential equation is satisfied for this choice of $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct that the difference in the second problem is that you no longer have the derivative with respect to $t$ expressed as a function of $t$.  When the equation involves both a function $y$ and its derivative (or derivatives), it can generally be more difficult to solve.  In this case, a hint I'll give you is that $dy/dt=1/(dt/dy)$ (e.g., see &lt;a href=&quot;http://en.wikipedia.org/wiki/Derivative_rule_for_inverses&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). You can first turn things around and think of $t$ as a function of $y$.  You can find $t(y)$ using similar methods to the first problem (but with a more complicated function), and then find the inverse function to get $y(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat more general method of solving differential equations that applies in this case is &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;separation of variables&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-29T06:34:44.813" />
  <row Id="12304" PostTypeId="2" ParentId="12299" CreationDate="2010-11-29T06:34:50.583" Score="1" ViewCount="0" Body="&lt;p&gt;In the first question, you are given the derivative in terms of the variable. But in the second question, you are given an expression for the derivative that involves &lt;em&gt;the function&lt;/em&gt;. For instance, it would be one thing if you were told $\frac{dy}{dx} = x$ (which would mean that $y = \frac{1}{2}x^2 + C$), and a completely different thing if you are told $\frac{dy}{dx}=y$ (this tells you that the function $y$ is equal to its derivative; which means that $y=Ae^x$ for some constant $A$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, we &lt;em&gt;can&lt;/em&gt; solve the second differential equation; but we don't solve it by simple integration, just like we don't solve the equation $y&amp;#39; = y$ by integrating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Showing that $y(t)=0$ is a solution to the second equation is just a question of plugging in $0$ for $y$ and verifying that you get a solution by verifying that the resulting equality is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You never said what question 8a was, through; were you asked to &lt;em&gt;solve&lt;/em&gt; the differential equation, or just to say what is the difference between it and the previous one?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-29T06:34:50.583" />
  <row Id="12305" PostTypeId="2" ParentId="12302" CreationDate="2010-11-29T06:50:49.783" Score="6" ViewCount="0" Body="&lt;p&gt;Hint: What is the generalized binomial expansion of $(1-2(\frac{1}{10})^2)^{-\frac{1}{2}}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-29T06:50:49.783" CommentCount="3" />
  <row Id="12306" PostTypeId="2" ParentId="12302" CreationDate="2010-11-29T06:50:51.537" Score="5" ViewCount="0" Body="&lt;p&gt;Note that we can rewrite the series as &#xA;$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n}$$&#xD;&#xA;which is exactly&#xA;$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot x^{2n}$$&#xD;&#xA;evaluated at $x = \frac{1}{10}$. It is easy to see that this sum has radius of convergence $\frac{1}{\sqrt{2}}$, and so converges absolutely for $x = \frac{1}{10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that this sum is exactly the Taylor series for $\frac{1}{\sqrt{1-2x^2}}$ centered at $x = 0$. Thus&#xA;$$ \displaystyle\sum\limits_{n=0}^\infty \frac{(2n)!}{2^n(n!)^2} \cdot \bigg(\frac{1}{10}\bigg)^{2n} = \frac{10}{7\sqrt{2}}$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-11-29T06:50:51.537" />
  <row Id="12307" PostTypeId="1" CreationDate="2010-11-29T06:56:47.140" Score="2" ViewCount="277" Body="&lt;p&gt;I know from an online calculator &lt;a href=&quot;http://www.numberempire.com/derivatives.php&quot; rel=&quot;nofollow&quot;&gt;http://www.numberempire.com/derivatives.php&lt;/a&gt; that &#xA;$\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$.  How do you calculate this step by step?&lt;/p&gt;&#xA;" OwnerUserId="4192" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-29T11:09:38.797" LastActivityDate="2010-11-29T18:12:27.953" Title="How to calculate  $\displaystyle\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$" Tags="&lt;calculus&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="12308" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T07:03:24.933" Score="7" ViewCount="0" Body="&lt;p&gt;This is more or less by definition of $e$, depending on which definition you use. Do you know the definition $e=\lim_{n\rightarrow\infty} (1+\frac{1}{n})^n$? Then set $x=2n$ and you are done.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T07:03:24.933" CommentCount="3" />
  <row Id="12309" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T07:04:26.867" Score="2" ViewCount="0" Body="&lt;p&gt;Rewriting the function whose limit is taken $$\left(1 + \frac{2}{x}\right)^x,$$ we immediately recognize the definition of $e^2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T07:04:26.867" />
  <row Id="12310" PostTypeId="1" CreationDate="2010-11-29T07:12:34.383" Score="1" ViewCount="41" Body="&lt;p&gt;Three planes &lt;em&gt;A&lt;/em&gt;, &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt; intersect at point P.  The dihedral angle between &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; is $\theta$ and the dihedral angle between &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt; is $\psi$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Solve for the dihedral angle between by &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planes &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; intersect plane &lt;em&gt;C&lt;/em&gt; in two lines that form angle QPR.  That is, point Q lies in both &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt;, and point R lies in both &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;C&lt;/em&gt;. Solve for the angle of QPR in terms of $\theta$ and $\psi$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does this fall under trigonometry? If not, what?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4158" LastActivityDate="2010-11-29T09:04:12.357" Title="Solve for the third dihedral angle in an intersection of three planes." Tags="&lt;trigonometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12311" PostTypeId="2" ParentId="12302" CreationDate="2010-11-29T07:17:31.593" Score="3" ViewCount="0" Body="&lt;p&gt;This can also be done using Wallis' product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance see a similar problem: &lt;a href=&quot;http://math.stackexchange.com/questions/3763/summing-the-series-frac12n1-frac12-cdot-frac12n3-cdots/3766#3766&quot;&gt;Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/3502/summing-the-series-1k-frac2k2k1-a2k1&quot;&gt;Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \sin^{2n} x \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} =  \dfrac{1}{2^n} \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n} \dfrac{1}{10^{2n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your sum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle  \sum_{n=0}^{\infty} \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} =  \dfrac{2}{\pi} \int_{0}^{\pi/2} \sum_{n=0}^{\infty} \left(\dfrac{2\sin^2 x}{100}\right)^n  \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle = \dfrac{2}{\pi}\int_{0}^{\pi/2} \dfrac{100}{100 - 2\sin^2x} \ \text{dx} = \dfrac{2}{\pi} \dfrac{5\pi}{7\sqrt{2}} = \dfrac{10}{7\sqrt{2}}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-29T07:17:31.593" CommentCount="2" />
  <row Id="12312" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T07:17:38.487" Score="1" ViewCount="0" Body="&lt;p&gt;Find a step-by-step derivation here (and not only for this special problem):&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=limit+((x%2B2)/x&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=limit+((x%2B2)/x&lt;/a&gt;)^x+x-%3Eoo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(if the link doesn't work copy and paste the whole line)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...or go to &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com&lt;/a&gt; directly and type:&lt;br&gt;&#xA;&quot;limit ((x+2)/x)^x x-&gt;oo&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on &quot;Show steps&quot; - Done! ;-)&lt;/p&gt;&#xA;" OwnerUserId="346" LastActivityDate="2010-11-29T07:17:38.487" />
  <row Id="12313" PostTypeId="2" ParentId="12239" CreationDate="2010-11-29T07:25:28.463" Score="2" ViewCount="0" Body="&lt;p&gt;Hint for a less convoluted argument: Unless $C$ is zero (in which case the statement is false), there are vectors $x$ and $y$ such that $x^T C y$ is a nonzero $1\times 1$ matrix. Now consider $v=[x,ky]^T$ for large $k$ (positive or negative, depending on the sign of $x^T C y$).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-29T07:25:28.463" />
  <row Id="12314" PostTypeId="1" AcceptedAnswerId="12315" CreationDate="2010-11-29T07:53:59.113" Score="4" ViewCount="111" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give me an example of a (smooth) $n$-manifold $(n=2, 3)$ which cannot be embedded (or immersed) in $\mathbb R^4$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. L.&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2010-11-29T09:19:58.180" Title="Examples of  manifolds that cannot be embedded in $\mathbb R^4$" Tags="&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="2" />
  <row Id="12315" PostTypeId="2" ParentId="12314" CreationDate="2010-11-29T08:02:02.410" Score="11" ViewCount="0" Body="&lt;p&gt;All 2-manifolds embed in $\mathbb R^4$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $3$-manifolds, $\mathbb RP^3$ doesn't embed in $\mathbb R^4$.  An interesting aspect for 3-manifolds in $\mathbb R^4$ is the subject fractures into tame topological embeddings and smooth/PL embeddings.  For example, Poincare Dodecahedral space has a tame topological embedding in $\mathbb R^4$ but not a smooth or PL-embedding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more examples, check out this pre-print: &lt;a href=&quot;http://front.math.ucdavis.edu/0810.2346&quot; rel=&quot;nofollow&quot;&gt;http://front.math.ucdavis.edu/0810.2346&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-11-29T08:02:02.410" CommentCount="1" />
  <row Id="12316" PostTypeId="1" AcceptedAnswerId="12365" CreationDate="2010-11-29T08:47:53.323" Score="1" ViewCount="97" Body="&lt;p&gt;In Linear Algebra, when computing an inner product $&amp;lt;x,y&amp;gt; = y^*Wx$, what is the name of the matrix W?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't have a name, where can I find a practical explanation of how to construct it for a particular problem or space?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a text on the subject that explains this simply? A note on my background, I am approaching this from the perspective of an engineering student and not that of a mathematician; I don't have an understanding of the finer points of topology or differential geometry (not yet at least :) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this is the same matrix used in vector calculus to perform a change of variables. As in, we perform our change of variables, then multiply the new expression by $\frac{det(J(W))}{det(J(V))}$ where $W$ is this magic matrix in the new space, $V$ is this magic matrix the old space, and J(X) is the Jacobian operator.  Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Also, is this related to one of the the matrices that comes up in Singular Value Decomposition (namely the diagonal matrix containing the singular values)?&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="4194" LastEditorUserId="4194" LastEditorDisplayName="" LastEditDate="2010-11-29T18:18:01.140" LastActivityDate="2010-11-29T18:18:01.140" Title="What is the name of the matrix used to weight an inner product?" Tags="&lt;linear-algebra&gt;&lt;differential-geometry&gt;&lt;inner-product-space&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12317" PostTypeId="2" ParentId="12310" CreationDate="2010-11-29T09:04:12.357" Score="1" ViewCount="0" Body="&lt;p&gt;Question 1: The third angle is not uniquely determined by $\theta$ and $\psi$. If $a$, $b$ and $c$ are the unit normal vectors to the planes $A$, $B$ and $C$, then the problem is equivalent to determining $a \cdot c$ from $a \cdot b$ and $c \cdot b$, which cannot be done. (Think of $b$ being fixed; then giving values of $a \cdot b$ and $c \cdot b$ constrains $a$ and $c$ to lie on certain cones around $b$, but the positions of $a$ and $c$ on these cones are arbitrary.) &lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-29T09:04:12.357" />
  <row Id="12318" PostTypeId="2" ParentId="11962" CreationDate="2010-11-29T09:12:29.183" Score="0" ViewCount="0" Body="&lt;p&gt;I found the following references by Richard C. Bradley:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.emis.de/journals/PS/images/ps-2005-37baf3.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Basic Properties of Strong Mixing Conditions. A Survey and Some Open Questions&quot;&lt;/a&gt;;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.kendrickpress.com/MixingConditions.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Introduction to Strong Mixing Conditions, Volumes 1-3&quot;&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've only been able to have a look at the first one; not very intuitive (to me, at least), but still a great overview.&lt;/p&gt;&#xA;" OwnerUserId="689" LastActivityDate="2010-11-29T09:12:29.183" />
  <row Id="12319" PostTypeId="2" ParentId="12314" CreationDate="2010-11-29T09:19:58.180" Score="5" ViewCount="0" Body="&lt;p&gt;No compact nonorientable $(n-1)$-manifold embeds in $\mathbb R^n$:&#xA;this follows from the Alexander duality theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immersibility is a harder problem....&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-11-29T09:19:58.180" />
  <row Id="12320" PostTypeId="1" AcceptedAnswerId="12321" CreationDate="2010-11-29T09:30:07.420" Score="3" ViewCount="161" Body="&lt;p&gt;I am looking for a correct proof of this statement: If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $G/Z(G)$ is isomorphic to $\operatorname{Inn}(G)$ and is cyclic, and then for every $a$ and $b$ in $G$ the inner isomorphisms $\gamma_a$ and $\gamma_b$ satisfy $\gamma_a \gamma_b = \gamma_{ab} = \gamma_{ba} = \gamma_b \gamma_a$, and therefore for every $a,b \in G$, $ab = ba$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that proof complete, or am I missing something? Thanks a lot for the help.&lt;/p&gt;&#xA;" OwnerUserId="4195" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-29T21:38:36.243" LastActivityDate="2010-11-29T21:38:36.243" Title="Proof that  if group $G/Z(G)$ is cyclic, then $G$ is commutative." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12321" PostTypeId="2" ParentId="12320" CreationDate="2010-11-29T09:44:29.367" Score="3" ViewCount="0" Body="&lt;p&gt;A problem with your attempt is that you seem to want to use only the fact that the inner automorphism group is abelian, but this does not suffice.  There are nonabelian groups $G$ such that $G/Z(G)$ is abelian, like the group of symmetries of the square.  Thus, it does not follow from $\gamma_{ab}=\gamma_{ba}$ that $ab=ba$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to use the hypothesis that $G/Z(G)$ has a single generator.  Robin Chapman pointed out to you &lt;a href=&quot;http://mathoverflow.net/questions/47650/proof-that-if-group-g-zg-is-cyclic-then-g-is-commutative-closed&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; what you can conclude from this, so I might as well quote him: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each element of $G$ has the form $a^nz$ where $a$ is fixed and $z\in Z(G)$...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-29T09:44:29.367" CommunityOwnedDate="2010-11-29T09:44:29.367" CommentCount="1" />
  <row Id="12322" PostTypeId="1" AcceptedAnswerId="12324" CreationDate="2010-11-29T10:04:31.390" Score="4" ViewCount="142" Body="&lt;p&gt;How to prove DeMorgan's Law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A - (B \cup C) = (A - B) \cap (A - C)$$&#xD;&#xA;$$A - (B \cap C) = (A - B) \cup (A - C)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;br&gt;&#xA;Here is what I have tried so far:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the first equation, assuming $x \in A - (B \cup C)$ then $x \in A$ and $x \not\in B$ and $x \not\in C$, while the right hand means ($x \in A$ and $x \not\in B$) or ($x \in A $ and $x \not\in C$) which is the same as $x \in A$ and $x \not\in B$ and $x \not\in C$. So the two set is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I do not know whether this is sufficient for a proof. Am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T15:55:21.217" LastActivityDate="2010-11-29T16:02:17.670" Title="How to prove DeMorgan's law?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12323" PostTypeId="1" AcceptedAnswerId="12357" CreationDate="2010-11-29T10:08:56.793" Score="3" ViewCount="209" Body="&lt;p&gt;A lot of textbooks said it was hard for human to accept zero when it was first introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could it be? It seems to me as natural as postive integer which represent there is no elements at all.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2010-11-30T19:25:41.630" Title="What's the hard part of zero?" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="12324" PostTypeId="2" ParentId="12322" CreationDate="2010-11-29T10:10:30.407" Score="5" ViewCount="0" Body="&lt;p&gt;I'd &quot;reduce&quot; it to logic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\in A-(B\cup C) \iff x\in A \text{ and not } (x\in B \text{ or } x\in C)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus by DeMorgan in logic (which can be proved with truth table)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\in A \text{ and } x\notin B \text{ and } x\notin C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $(x \in A  \text{ and } x \notin B) \text{ and } (x \in A  \text{ and } x \notin C)$. Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\in (A- B)\cap (A- C)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="52" LastEditorDisplayName="" LastEditDate="2010-11-29T16:02:17.670" LastActivityDate="2010-11-29T16:02:17.670" />
  <row Id="12325" PostTypeId="2" ParentId="12322" CreationDate="2010-11-29T10:12:26.040" Score="3" ViewCount="0" Body="&lt;p&gt;Maybe you're familiar with the fact that if $A\subseteq B$ and $B\subseteq A$, then $A=B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're viewing $A,B,C$ as sets, you can prove these by showing the set on the left of $=$ is a subset of the set on the right of $=$, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose $x\in A-(B\cup C)$. So $x\in A$, but $x\not\in B\cup C$. In particular, $x\not\in B$, and $x\not\in C$. It follows that $x\in A-B$ and $x\in A-C$, that is, $x\in (A-B)\cap (A-C)$. Thus $A-(B\cup C)\subseteq (A-B)\cap (A-C)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try using a similar method to show the reverse containment, and you'll get your result. This works for the second law also.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-29T10:12:26.040" />
  <row Id="12326" PostTypeId="2" ParentId="12323" CreationDate="2010-11-29T10:18:55.953" Score="0" ViewCount="0" Body="&lt;p&gt;You can't count out zero 'things'; I would imagine that that was the big issue.  Same goes for the negative side of $\mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="4194" LastActivityDate="2010-11-29T10:18:55.953" CommentCount="5" />
  <row Id="12327" PostTypeId="2" ParentId="12323" CreationDate="2010-11-29T10:27:42.093" Score="9" ViewCount="0" Body="&lt;p&gt;Of course it seems natural to you; you grew up in the modern world, where everyone accepts zero.  More importantly, people now accept the abstract concept of numbers and are capable of divorcing them from the things that they represent.  This is a sophisticated point of view.  From a more naive point of view, a number is a property of a collection of objects: when I say there are $2$ apples somewhere, that is a property of a collection of apples.  If there are no apples, then what is there to &quot;hold&quot; the corresponding property?  So instead of saying there are $0$ apples, people said there are no apples.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In computer science terms, &lt;code&gt;apples.num()&lt;/code&gt; is not defined if there is no &lt;code&gt;apples&lt;/code&gt; variable!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to understand that &quot;human&quot; in your statement means (from my understanding) &quot;Western mathematicians.&quot;  Indian mathematicians had no trouble with zero.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-29T10:27:42.093" CommentCount="5" />
  <row Id="12328" PostTypeId="1" AcceptedAnswerId="12367" CreationDate="2010-11-29T10:45:34.587" Score="2" ViewCount="151" Body="&lt;p&gt;I stumbled across this paragraph in a paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hence, user b cannot decrypt C&#xA;  directly. But using e and d , user b&#xA;  can quickly factor N.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is  it possible to speedup the prime factorization when knowing e (public key) and d (private key)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RSA provides us with these equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = pq$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi = (p-1)(q-1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$gcd(e, \phi) = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$de = 1\pmod{\phi}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to determine $p$ and $q$ an attacker has to factor n which is not feasible. However the paper stated that it is easy to reconstruct $p$ and $q$ when a person knows both (his) private and public keys.&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-29T12:39:06.533" LastActivityDate="2010-11-29T17:23:39.080" Title="RSA: Fast factorization of N if d and e are know" Tags="&lt;number-theory&gt;&lt;cryptography&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12329" PostTypeId="1" AcceptedAnswerId="12343" CreationDate="2010-11-29T10:45:56.917" Score="4" ViewCount="102" Body="&lt;p&gt;How possible methods for finding groups of given order we have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sylow theorems - and next? We know something about center of group, about possible orders of these elements and subgroups. But what is most effective way, to find all non-isomorphic groups of given order? Is there any other strong &quot;weapon&quot; for this (like Sylow theorems)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I took groups order 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutative group is only $\mathbb{Z}_{15}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About non-commutative we know, they have 1 Syllow 3-subgroup and 1 Syllow 5-subgroup. And next, that $M_3 \cup M_5 = G$ (if $M_3$, resp. $M_5$ is Syllow 3-subgroup, resp. Syllow 5-subgroup).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_3$ and $M_5$ are single generated, than we can choose elements $f$ and $g$ satysfaing $&amp;lt;f&amp;gt;=M_3$ and $&amp;lt;g&amp;gt;=M_5$. Because there are characterictis, we get these equations: $f^{-1}gf=g^m$, where $m \in \{1,2,3,4\}$ and $g^{-1}fg=f^n$, where $n \in \{1,2\}$. $m$ and $n$ must supplying $|g|=|g^m|$ and $|f|=|f^n|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is end of my way, i don't know how continue... Can anyone help? (Sorry for bad English)&lt;/p&gt;&#xA;" OwnerUserId="4195" LastActivityDate="2010-11-29T13:16:34.893" Title="How to find finite groups" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12330" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T11:00:10.287" Score="5" ViewCount="0" Body="&lt;p&gt;Observe that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \dfrac{x+2}{x}\right) ^{x}=\left( \left( 1+\dfrac{1}{\frac{x}{2}}%&#xD;&#xA;\right) ^{x/2}\right) ^{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, by the definition of $e$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e=\lim_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\lim_{x\rightarrow \infty }\left( \dfrac{x+2}{x}\right) ^{x}&#xD;&#xA;&amp;amp;=&amp;amp;\lim_{x\rightarrow \infty }\left( \left( 1+\frac{1}{x/2}\right) ^{x/2}\right) ^{2} \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \lim_{x\rightarrow \infty }\left( 1+\frac{1}{x/2}\right) ^{%&#xD;&#xA;x/2}\right) ^{2} \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right)&#xD;&#xA;^{2} \\&#xD;&#xA;&amp;amp;=&amp;amp;e^{2}&#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For instance, by the same argument, for $k\in\mathbb{N}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x\rightarrow \infty }\left( \dfrac{x+k}{x}\right) ^{x}=\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right)&#xD;&#xA;^{k}=e^{k}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-29T18:12:27.953" LastActivityDate="2010-11-29T18:12:27.953" />
  <row Id="12331" PostTypeId="1" AcceptedAnswerId="12372" CreationDate="2010-11-29T11:03:17.620" Score="1" ViewCount="115" Body="&lt;p&gt;Here is what my textbook told me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the formula for surface $\Sigma$ is $$F(x,y,z) = 0$$ &#xA;Suppose $X_0 = (x_0, y_0, z_0)$ is a point on the surface $\Sigma$ and we assuming F(x,y,z) is differentiable and $$\mathbf{J}F(X_0) = (\frac{\partial F(X_0)}{\partial x}, \frac{\partial F(X_0)}{\partial y}, \frac{\partial F(X_0)}{\partial z}) \neq 0$$ &#xA;Draw a line $\Gamma$ in the surface $\Sigma$ passing through the point $X_0$, assuming the equations for $\Sigma$ is $$x = x(t), y = y(t), z = z(t)$$  $t = t_0$ correspond to the point $X_0$ and $x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0)$ does not all vanish. Because of the line $\Gamma$ is on the surface $\Sigma$, so $$F(x(t), y(t), z(t)) = 0$$ So $$ \frac{dF}{dt}\mid_{t=t_0} = {F_x}&amp;#39;(X_0)x&amp;#39;(t_0) + {F_y}&amp;#39;(X_0)y&amp;#39;(t_0) + {F_z}&amp;#39;(X_0)z&amp;#39;(t_0) = 0 $$ So $$ ({F_x}&amp;#39;(X_0), {F_y}&amp;#39;(X_0), {F_z}&amp;#39;(X_0))\cdot(x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0)) = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the vector $\mathbf{T} = (x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0))$ is the tangent vector for the line $\Gamma$ on the point $X_0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why the vector $\mathbf{T}$ is the tangent vector for line $\Gamma$ at point $X_0$?&lt;/li&gt;&#xA;&lt;li&gt;Why the $\mathbf{J}F(X_0)$ should not equal to zero? What if it is zero?&lt;/li&gt;&#xA;&lt;li&gt;Why $x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0)$ should not all vanish? What if all vanish?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-11-29T11:14:31.350" LastActivityDate="2010-11-29T18:26:05.883" Title="Help understanding the  definition of tangent vector or tangent plane?" Tags="&lt;calculus&gt;&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12332" PostTypeId="1" CreationDate="2010-11-29T11:03:39.490" Score="2" ViewCount="90" Body="&lt;p&gt;For $n &amp;gt; 2$, is it possible to embed $\underbrace{S^1 \times \cdots \times S^1}_{n\text{ times}}$ into $\mathbb R^{n+1}$?&lt;/p&gt;&#xA;" OwnerUserId="4199" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-11-29T11:16:24.500" LastActivityDate="2010-11-29T18:14:35.043" Title="Embedding torus in Euclidean space" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12333" PostTypeId="2" ParentId="12263" CreationDate="2010-11-29T11:08:09.980" Score="1" ViewCount="0" Body="&lt;p&gt;A B-spline of order 2 is a continuous piecewise-linear function that interpolates between the assigned data points (&quot;knots&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the slopes of such a function vary between +0.017 and -0.017 on the interval $[0,2\pi]$, then by the Mean Value Theorem there cannot be more than $0.034\pi$ or about 0.1 between the minimum and maximum function values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either there are no data points near the minimum and maximum of $sin(x)$ on this interval or else some error has been made in computation (perhaps a mistake in scaling the x-axis that stretches everything out).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-11-29T11:08:09.980" CommentCount="1" />
  <row Id="12334" PostTypeId="2" ParentId="12205" CreationDate="2010-11-29T11:11:11.917" Score="0" ViewCount="0" Body="&lt;p&gt;@Bill: Thanks, I see that $\pi\pi&amp;#39;|\alpha\beta$ implies that $p^2|N(\alpha)N(\beta)$. Since $(N(\alpha),N(\beta))=1$ we have that $p^2$ divides $N(\alpha)$ or $N(\beta)$, say $p^2|N(\alpha)$. Thus, $p$ must divide $\alpha$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="4200" LastActivityDate="2010-11-29T11:11:11.917" />
  <row Id="12335" PostTypeId="2" ParentId="12285" CreationDate="2010-11-29T11:20:42.060" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the map $f\times f: X\times X \mapsto X\times X$. It is continuous (because f is)  and surjective (because f is). Consider the orbit of $(x,y)$ under $f$, i.e. the sequence $p_n=(f^n(x), f^n(y))$. Because $X\times X$  is compact, it has a convergent subsequence, hence for any $\epsilon$ (sorry, I hoped to do no epsilons, but can't), there is an $N_0, k&amp;gt;0$, such that $f^{N_0}(x)$ is $\epsilon$ close to $f^{k+N_0}(x)$ and $f^{N_0}(y)$ is epsilon close to $f^{k+N}(y)$, hence $d(f^{k+N_0}(x), f^{k+N_0}(y))$ is $2\epsilon$ close to $d(f^{N_0}(x), f^{N_0}(y))$. Then because $f$ does not expand distances, induction shows the same is true for all $N \geq N_0$. But we want more. We want this to be true for some fixed $N$ (and varying $k$) for all $(x,y)$. That is achievable by compactness as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we showed (replacing $\epsilon$ by $\epsilon/2$) we have $d(f^{k+N_0}(x), f^{k+N_0}(y))$ is $\epsilon$ close to $d(f^{N_0}(x), f^{N_0}(y))$. Because $f$ does not expand distances (and triangle inequality), this means that for any $(v,w)$ with $d(v,x)&amp;lt;\epsilon/4$ and $d(w,y)&amp;lt;\epsilon/4$ the we have $d(f^{k+N_0}(v), f^{k+N_0}(w))$ is $2\epsilon$ close to $d(f^{N_0}(v), f^{N_0}(w))$. Now cover $X$ with $\epsilon/4$ balls and pick a finite subcover. For any pair of centers $(x_i,y_i)$ of the balls we have some $N_0$ which gives $\epsilon$-closeness, and hence gives $2\epsilon$ closeness on the ball, hence the maximum $M$ of these $N_0$'s gives $2\epsilon$ closeness everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can prove $f$ is an isometry. To get contradiction, assume you have $(x,y)$ with $d(f(x), f(y))&amp;lt; d(x,y)$, hence for some $\epsilon$ also $d(f(x), f(y))&amp;lt; d(x,y)-2\epsilon$. Then for any $k&amp;gt;0$, we have $d(f^k(x), f^k(y))&amp;lt; d(x,y)-2\epsilon$.  Finally we use surjectivity. Take $(a,b)$ with $f^M(a)=x, f^M(b)=y$. Then $d(f^{M+k}(a), f^{M+k} (b))= d(f^k(x), f^k(y))$ is $2\epsilon$ close to $d(f^M (a), f^M(b))=d(x,y)$ for some $k$, contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-29T11:20:42.060" />
  <row Id="12336" PostTypeId="1" CreationDate="2010-11-29T11:26:59.793" Score="0" ViewCount="47" Body="&lt;p&gt;Hi all,&#xA;I have a uniform grid of square cells and a point on that grid.  I want to calculate the intersection of cells on the grid, with a circular area centered on the point, with a radius R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of an efficient way of doing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4201" LastActivityDate="2010-11-29T11:26:59.793" Title="Simple method for detecting grid intersection with circle" Tags="&lt;geometry&gt;" CommentCount="1" />
  <row Id="12337" PostTypeId="1" CreationDate="2010-11-29T11:36:17.353" Score="1" ViewCount="69" Body="&lt;p&gt;Follow-up to &lt;a href=&quot;http://math.stackexchange.com/questions/12193/field-reductions-part-two&quot;&gt;Field reductions. part two&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a field $K$, an element $a \in K$, and $K(\setminus a)$ the set of field reductions (maximal subfields of $K$ not containing $a$) are there any interesting equivalence classes on $K(\setminus a)$ or binary operations, or distance functions, or ...? Real numbers are represented by infinite decimals. How would you represent elements of $\mathbb{R}(\setminus a)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2010/11/27/nonstandard-analysis-as-a-completion-of-standard-analysis/&quot; rel=&quot;nofollow&quot;&gt;latest blog post&lt;/a&gt; is about the concept of extending and completing things. What about going the other way and reducing things ? So moving on from fields to general spaces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a space, i.e. a set with some added structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $a \in X$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X(\setminus a)$ be the set of maximal subspaces of $X$ not containing $a$, called the $a$-reduced spaces of $X$ or space reductions of $X$ by $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What examples are there in mathematics that have been previously studied that could be described as a maximal subspace not containing $a$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $@X(\setminus a)$ denote an element of $X(\setminus a)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a subspace $S$ of $X$, let $S(a..)$ denote the extension of $S$ to $X$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so then $@X(\setminus a)(a..)=X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some category-theoretic framework that can be used here ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morphisms from $X(\setminus x_1)$ to $X(\setminus x_2)$ or $Y(\setminus y_1)$ or $Z(\setminus z_1)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you extend a field by adding a new element and then also adding all its arithmetic combinations with the existing elements, then this is like adding an idea to a body of knowledge or piece of information to your list and then combining it in all possible ways with all existing items to get an extended list, like in Sudoku or Logic problems or ... , so information is like a universal algebra with lots of operations of different arities, where not only the elements can be added to, but also the operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove a piece of information from a list and remove all the information that depended on it then there seems to be only one way of doing that so the analogy between information reduction and field reduction breaks down, but the analogy can be repaired by removing not only what depends on it, but also what leads to it, and there would be a choice because removing one thing that leads to it would mean not needing to remove another thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_mathematics&quot; rel=&quot;nofollow&quot;&gt;Reverse mathemtaics&lt;/a&gt;: take a set of axioms of a theory and adjoin all true statements, now take a statement that you are interested in finding the necessary axioms for, remove that statement and anything that follows from it or leads to it, then you'd get maximal subtheories for which that statement was independent of. If any of the axioms are not in a subtheory then the statement depended on that axiom (Probably. Correct me if I'm wrong.) &lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="4157" LastEditorDisplayName="" LastEditDate="2010-11-29T14:19:17.853" LastActivityDate="2010-11-29T14:19:17.853" Title="Field reductions. part three" Tags="&lt;abstract-algebra&gt;&lt;category-theory&gt;&lt;field-theory&gt;" CommentCount="1" />
  <row Id="12338" PostTypeId="2" ParentId="12332" CreationDate="2010-11-29T11:40:08.027" Score="2" ViewCount="0" Body="&lt;p&gt;Let $e_0, \ldots, e_n$ be the standard basis of $\mathbb {R}^{n+1}$. Take $\epsilon$ small. Consider the vector $v_1$ of length 1 in the span of  $e_0, e_1$. Then the  vector $v_2$ of length $\epsilon$ in the span of  $v_1, e_2$, and in general the vector $v_i$  of length $\epsilon^{i-1}$ in the span of  $v_{i-1}, e_i$. Now consider the vector $w=v_1+\ldots v_n$  For small $\epsilon$ the set of $w$'s is a torus embedded $\mathbb {R}^{n+1}$ (any $\epsilon &amp;lt; 1$ will do, actually).&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-11-29T18:14:35.043" LastActivityDate="2010-11-29T18:14:35.043" />
  <row Id="12340" PostTypeId="1" AcceptedAnswerId="12341" CreationDate="2010-11-29T11:59:15.350" Score="7" ViewCount="217" Body="&lt;p&gt;What is the limit of the series 1/(2n)! for n in [0, infinity) ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've ground out the sum of the 1st 1000 terms to 1000 digits using Python,&#xA;(see &lt;a href=&quot;http://tutoree7.pastebin.com/pv0i1XQb&quot; rel=&quot;nofollow&quot;&gt;http://tutoree7.pastebin.com/pv0i1XQb&lt;/a&gt; ), but how would a mathematician calculate the limit? And what is it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, this isn't homework. I'm 73. Just curious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-29T15:10:38.763" LastActivityDate="2010-11-29T15:10:38.763" Title="What is limit of $\sum_{n=0}^{\infty}\frac{1}{(2n)!} $ ?" Tags="&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12341" PostTypeId="2" ParentId="12340" CreationDate="2010-11-29T12:12:36.387" Score="17" ViewCount="0" Body="&lt;p&gt;It's half the sum of $e^1=\sum 1/n!$ and $e^{-1}=\sum (-1)^{n}/n!$ (or $\cosh 1$, in other words).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-11-29T12:12:36.387" CommentCount="13" />
  <row Id="12343" PostTypeId="2" ParentId="12329" CreationDate="2010-11-29T13:04:44.323" Score="4" ViewCount="0" Body="&lt;p&gt;If you don't like semi-direct products, then you will not get very far in your endeavour. I recommend that you play around with them until you start liking them. The reason is that the single most powerful result in trying to classify all finite groups of given order is the so-called Schur-Zassenhaus theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Schur-Zassenhaus Theorem&lt;/strong&gt;: Let $N$ be a normal subgroup of a finite group $G$ and suppose that the order of $N$ is co-prime to its index. Then $G$ is a semidirect product $G=N\rtimes H$ for a subgroup $H$ of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this useful? I will show you, how to finish your project for $n=15$ essentially in one line: as you have noted, there is one Sylow 5-subgroup in $G$, so it's normal. Since its index is 3, Schur-Zassenhaus applies and tells us that $G=C_5\rtimes C_3$. But the automorphism group of $C_5$ is cyclic of order 4, so there are no non-trivial maps from $C_3$ to the automorphism group of $C_5$ and $G$ must in fact be a direct product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: classify all finite groups of order 20 using this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. One more thing: don't ever take unions of subgroups! What you wanted to say is that $G=\langle M_3,M_5\rangle$, i.e. is generated by (all possible products of elements of) the subgroups.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T13:04:44.323" />
  <row Id="12344" PostTypeId="2" ParentId="12307" CreationDate="2010-11-29T13:05:54.377" Score="2" ViewCount="0" Body="&lt;p&gt;For ease rewrite as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\rightarrow \infty}\left(1+\frac{2}{x}\right)^x.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we compute &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty}\ln\left(\left(1+\frac{2}{x}\right)^x\right).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using laws of logarithms we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty}x\ln\left(1+\frac{2}{x}\right)  =\lim_{n\to\infty}\frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now in a position to apply L'Hopital's Rule.  Taking derivatives gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty} \frac{\left(\frac{1}{1+\frac{2}{x}}\right)\left(-\frac{2}{x^2}\right)}{-\frac{1}{x^2}}=\lim_{n\to\infty}\frac{2}{1+\frac{2}{x}}=2.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your limit is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{n\to\infty}e^{\ln\left(\left(1+\frac{2}{x}\right)^x\right)}=e^2&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-11-29T13:05:54.377" />
  <row Id="12345" PostTypeId="2" ParentId="12329" CreationDate="2010-11-29T13:16:34.893" Score="2" ViewCount="0" Body="&lt;p&gt;Here's how to finish without assuming the Sylow 5-subgroup is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume by way of contradiction:) If $n=2$, so $g^{-1} f g = ff$, then: $$g^{-2} f g^2 = g^{-1}( g^{-1} f g ) g = g^{-1}( ff )g = g^{-1}( fg g^{-1}f)g = (g^{-1}fg) (g^{-1}fg) = (ff) (ff) = f^4.$$ Every time you apply $g^{-1}( * )g$, you double the number of $f$, so $g^{-i} f g^i = f^{(2^i)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use $g^5 = 1$.  If you apply $g^{-1}( * )g$ five times, you get $g^{-5} f g^5 = 1 f 1 = f$ on the one hand, but $f^{(2^5)} = f^{32} = f^2$ on the other, using the fact that 2 ≡ 32 mod 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can $f = f^2$?  It does not, since $f$ has order 3.  Assuming $n=2$ gives a contradiction, so $n=1$, and $g^{-1} f g = f$.  Multiply on the left by $g$ to get $f g = g f$.  The group must be abelian, since all of its generators commute.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now try this with &quot;5&quot; replaced by &quot;2&quot; and you will still get the Sylow 3-subgroup is normal, but the Sylow 2-subgroup need not be normal.  The same argument will get you that both n=1 (cyclic group of order six), and n=2 (dihedral group of order six) are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, you'll check that $g^{-2} f g^2 = f^{(2^2)} = f^4 = f$ just as it should be.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Congratulations, you have just analyzed some semi-direct products using automorphisms!  With cyclic Sylow subgroups it makes a lot of sense.  We didn't even need to call &quot;n&quot; an automorphism, nor did we need to call M&lt;sub&gt;3&lt;/sub&gt;∪M&lt;sub&gt;5&lt;/sub&gt; a semi-direct product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the subgroups are not Sylow subgroups, then it is harder to check (1) that they are normal or (2) that M∩N = 1, but there are more general techniques for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the subgroups are not cyclic, then you cannot just use one &quot;n&quot; for the Sylow 3-subgroup.  You get a matrix of &quot;m&quot;s and &quot;n&quot;s.  Not just any matrix will do, but the ones that do work are called automorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the general case of semi-direct products is similar to what you are now doing.  Don't give up.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-29T13:16:34.893" />
  <row Id="12347" PostTypeId="2" ParentId="12097" CreationDate="2010-11-29T13:53:37.097" Score="2" ViewCount="0" Body="&lt;p&gt;My previous answer was useless. The new answer below completes George's answer for the  $0 &amp;lt; \alpha &amp;lt; 1$ case, illustrated in the first graph above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this case, George provided the asymptotic approximation $P_n \sim {\rm P}(2 \Delta Z^*_1 - Z_1 &amp;gt; 0)$ as $n \to \infty$, where $P_n$ is the probability the OP asked for, $Z = \lbrace Z_t : t \geq 0 \rbrace$ is a strictly stable L\'evy process of index $\alpha$, and $\Delta Z^*_1$ is the largest jump of $Z$ in the time interval $[0,1]$. It is not easy to obtain &#xA;${\rm P}(2 \Delta Z^*_1 - Z_1 &amp;gt; 0)$ in closed form, but indeed possible, and in fact the following more general result is known:&#xA;$$&#xD;&#xA;{\rm P}\bigg(\frac{{\Delta Z_1^* }}{{Z_1 }} &amp;gt; \frac{1}{{1 + y}}\bigg) = \frac{{y^\alpha  }}{{\Gamma (1 - \alpha )\Gamma (1 + \alpha )}}, \;\; y \in [0,1],&#xD;&#xA;$$&#xD;&#xA;where $\Gamma$ is the gamma function. Letting $y = 1$ thus gives $P_n \sim 1/[\Gamma(1 - \alpha) \Gamma(1 + \alpha)]$. &#xA;This asymptotic expression agrees very well with the graph above, as I confirmed by checking various values of $\alpha \in (0,1)$. (For example, $\alpha = 1/2$ gives $P_n \sim 2/\pi \approx 0.6366$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A generalization. Instead of just considering the probability ${\rm P}(M &amp;gt; \sum\nolimits_{i = 0}^{n - 1} {X_i }  - M)$, we can consider the probability ${\rm P}(M &amp;gt; y^{-1}[\sum\nolimits_{i = 0}^{n - 1} {X_i }  - M])$, $y \in (0,1]$. In view of George's answer, this should correspond to the formula given above for ${\rm P}(\Delta Z_1^* / Z_1 &amp;gt; 1/(1+y))$.&#xA;This can be further generalized, in various directions, based on known results from the literature (three useful references in this context are indicated somewhere in the comments above/below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As George observed, the term $\Gamma(1 - \alpha) \Gamma(1 + \alpha)$ can be simplified to $(\pi\alpha) / \sin(\pi\alpha)$. The former expression corresponds to $[\Gamma(1 - \alpha)]^k \Gamma(1 + k \alpha)$, which is incorporated in the explicit formula for ${\rm P}(\Delta Z_1^* / Z_1 &amp;gt; 1/(1+y))$, $y &amp;gt; 0$, of the form&#xA;$\sum\nolimits_{k = 1}^{\left\lceil y \right\rceil } {a_k \varphi _k (y)} $. The function $\varphi _k (y)$ is some $(k-1)$-dimensional integral, hence that formula is computationally very expensive already for moderate values of $y$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-29T21:19:24.743" LastActivityDate="2010-11-29T21:19:24.743" />
  <row Id="12348" PostTypeId="1" AcceptedAnswerId="12351" CreationDate="2010-11-29T14:07:57.407" Score="2" ViewCount="180" Body="&lt;p&gt;folks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am new to this forum and not a math expert. so please bear with me if am asking silly questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the question is &quot;probability of getting 50 heads from tossing a coin 100 times&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the answer for this is, i guess, (100 C 50)*(2 ^-100) [[ fixed typo ]]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so all am trying to get is easier way to calculate 100 C 50. or another approach to the parent problem only&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks all, appreciate that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;raj&lt;/p&gt;&#xA;" OwnerUserId="4205" LastEditorUserId="4205" LastEditorDisplayName="" LastEditDate="2010-11-30T07:49:13.857" LastActivityDate="2010-11-30T12:15:03.573" Title="probability of getting 50 heads from tossing a coin 100 times" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12349" PostTypeId="2" ParentId="12331" CreationDate="2010-11-29T14:14:09.347" Score="3" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The coordinates $x(t)$ and so on tell you how to trace out a curve, as $t$ changes. The vector &lt;strong&gt;T&lt;/strong&gt; describes the change in the $x$, $y$ and $z$-coordinates as $t$ changes infinitesimally. Surely, that's exactly what the tangent vector to a curve should do. More algebraically, the tangent to the curve at $X_0$ is defined as&#xA;$$\lim_{\delta\rightarrow 0} \frac{(x(t_0),y(t_0)z(t_0)) - (x(t_0+\delta),y(t_0+\delta)z(t_0+\delta))}{\delta},$$&#xD;&#xA;whenever the limit is defined.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As J.M. wrote, in that case you don't have a well-defined tangent. Again, algebraically, your calculation has shown that &lt;strong&gt;J&lt;/strong&gt;$F(X_0)$ is normal to tangents to all curves on the surface passing through $X_0$, so it &lt;em&gt;should&lt;/em&gt; be the normal vector to the surface at that point. But if the surface has a singularity, then there is no well-defined normal vector.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As remarked above, the derivatives $x&amp;#39;(t_0)$ and so on describe, how quickly the coordinates change as you trace out your curve. In pictorial terms, you can imagine walking down the curve as the parameter $t$ slowly increases. If all these derivatives vanish, it means that as $t$ approaches $t_0$, your walk is slowing down to a grinding halt. That means that either you have chosen a bad parametrisation, i.e. you are staying in one place for some range of $t$, which you might as well leave out then, or you are approaching a singularity, which is bad news if you are computing tangents.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T14:14:09.347" />
  <row Id="12350" PostTypeId="2" ParentId="12348" CreationDate="2010-11-29T14:16:05.217" Score="4" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://www.johndcook.com/blog/2008/04/24/how-to-calculate-binomial-probabilities/&quot; rel=&quot;nofollow&quot;&gt;How to calculate binomial probabilities&lt;/a&gt; for how to calculate these probabilities while avoiding numerical overflow or underflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I believe you meant to type 2^-100 instead of 2^-10 in your question.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-11-29T14:16:05.217" CommentCount="1" />
  <row Id="12351" PostTypeId="2" ParentId="12348" CreationDate="2010-11-29T14:31:55.533" Score="3" ViewCount="0" Body="&lt;p&gt;The coefficient ${2n \choose n}$ for $n$ large can be approximated well (using Stirling's formula) by ${2n \choose n} \approx 4^n / \sqrt{n \pi} $. In particular, letting $n=50$ gives ${100 \choose 50} \approx 4^{50}/ \sqrt {50 \pi}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-29T14:31:55.533" />
  <row Id="12352" PostTypeId="1" AcceptedAnswerId="12355" CreationDate="2010-11-29T14:35:05.700" Score="0" ViewCount="71" Body="&lt;p&gt;how to calculate the sides and hypotenuse length of the &lt;strong&gt;right triangle&lt;/strong&gt; if I know&#xA;the bigger side = 60 &#xA;one angle = 60 second angle = 30 (3rd = 90)&lt;/p&gt;&#xA;" OwnerUserId="699" LastEditorUserId="699" LastEditorDisplayName="" LastEditDate="2010-11-29T14:47:22.153" LastActivityDate="2010-11-29T15:38:50.647" Title="calculate sides of the right triangle if I know 1 side and all the angles" Tags="&lt;geometry&gt;" AnswerCount="4" />
  <row Id="12353" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T14:40:57.320" Score="3" ViewCount="0" Body="&lt;p&gt;You could use the identities &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$sin(\theta)=\frac{oposite-side}{hypotenuse} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$cos(\theta)=\frac{adyacent-side}{hypotenuse} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(60)=\frac{60}{hypotenuse} \Rightarrow hypotenuse = \frac{120}{\sin(60)}=\frac{120}{\sqrt{3}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(60)=\frac{adyacent-side}{hypotenuse} \Rightarrow adyacent-side = \cos(60)\cdot hypotenuse =\frac{60}{\sqrt{3}} $$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-29T14:53:27.953" LastActivityDate="2010-11-29T14:53:27.953" CommentCount="2" />
  <row Id="12354" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T14:41:23.213" Score="1" ViewCount="0" Body="&lt;p&gt;tan(60) = side1/side2 = 60/side2 =&gt; side2 = 60/tan(60) = 60/sqrt(3) = 20sqrt(3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(60) = side1/hypotenuse =&gt; hypotenuse = 60/sin(60)=120/sqrt(3) = 40sqrt(3)&lt;/p&gt;&#xA;" OwnerUserId="1030" LastActivityDate="2010-11-29T14:41:23.213" CommentCount="1" />
  <row Id="12355" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T14:41:45.183" Score="2" ViewCount="0" Body="&lt;p&gt;Some hints: what is the third angle? Draw a picture. Can you spot some symmetry?&#xA;If not, then decode the following hint with &lt;a href=&quot;http://www.rot13.com/index.php&quot; rel=&quot;nofollow&quot;&gt;http://www.rot13.com/index.php&lt;/a&gt;.&#xA;But please, spend some time trying first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ersyrpg gur gevnatyr va gur evtug natyr. Jung vf fb fcrpvny va gur erfhygvat gevnatyr? Gel znxvat hfr bs gur rkgen flzzrgel.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-29T14:41:45.183" CommentCount="4" />
  <row Id="12356" PostTypeId="1" AcceptedAnswerId="12359" CreationDate="2010-11-29T14:48:41.560" Score="1" ViewCount="87" Body="&lt;p&gt;What are the necessary and sufficient conditions on a vector field $F$ for the divergence $\nabla\cdot F$ to exist at a given point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Divergence&quot; rel=&quot;nofollow&quot;&gt;Divergence&lt;/a&gt; in the second line under the heading &quot;&lt;strong&gt;Application in Cartesian coordinates&lt;/strong&gt;&quot;, why is it assumed that $\vec{F}$ to be a continuously differentiable vector field ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally one would expect each component of $F$ to be differentiable at a given point $\vec{a}$ no matter through which continuous contour you traverse the point $\vec{a}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it that each component of $F$ to be differentiable and the derivative being continuous at a given point $\vec{a}$ no matter through which continuous contour you traverse the point $\vec{a}$. &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-11-29T17:06:39.950" LastActivityDate="2010-11-29T17:06:39.950" Title="Conditions for existence of divergence of a vector field." Tags="&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="12357" PostTypeId="2" ParentId="12323" CreationDate="2010-11-29T15:27:21.570" Score="1" ViewCount="0" Body="&lt;p&gt;The problem was that math started out of necessity.  Shepherds needed to count how many sheep they own, parents needed to count how many children they had, etc.  For most simple mathematics, zero is simply a place-holder.  Who needs to count zero sheep?  Even the most basic forms of counting--using your fingers--who has zero fingers?  Zero is also pretty closely tied to the concept of negative numbers as well, which are a very non-natural inclusion for mathematics that simply evolve out of necessity.  What shepherd in his right mind has -1 sheep.  Wrapping your head around the idea of zero is easy once the idea is conveyed, but somewhat non-intuitive to arbitrarily include.  Even worse, once you do include zero, how do you deal with all the problems that come along with it such as division by zero.  A great history on the story of how we came to understand zero can be found in the book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0140296476&quot; rel=&quot;nofollow&quot;&gt;Zero: The Biography of a Dangerous Idea&lt;/a&gt; by Charles Seife&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vXuTw.jpg&quot; alt=&quot;Book Cover&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="2470" LastEditorDisplayName="" LastEditDate="2010-11-30T19:25:41.630" LastActivityDate="2010-11-30T19:25:41.630" />
  <row Id="12358" PostTypeId="2" ParentId="12352" CreationDate="2010-11-29T15:38:50.647" Score="2" ViewCount="0" Body="&lt;p&gt;Well, you don't even need to assume that the triangle is a right triangle. Simply use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_sines&quot; rel=&quot;nofollow&quot;&gt;Law of sines&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-29T15:38:50.647" />
  <row Id="12359" PostTypeId="2" ParentId="12356" CreationDate="2010-11-29T15:48:54.700" Score="1" ViewCount="0" Body="&lt;p&gt;Perhaps the point is that existence of partial derivatives is not enough for a function of several variables to be &lt;strong&gt;differentiable&lt;/strong&gt; in the multivariable sense, or even necessarily continuous: see &lt;a href=&quot;http://mathoverflow.net/questions/19303/good-example-of-a-non-continuous-function-all-of-whose-partial-derivatives-exist&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt; for a standard example of a function of two variables all of whose partial (and even all directional) derivatives exist but is not even continuous at the origin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;b&gt;Edit&lt;/b&gt;: the following paragraph was added after reading Qiaochu's comment.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible for all the partial derivatives in a given coordinate system to exist -- e.g. $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ -- but for directional derivatives in other directions not to exist.  (For instance, let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be the function which is $0$ at $(x,y,z)$ if at least two of the variables are zero or if all of $x$,$y$,$z$ have rational coordinates and which is $1$ at all other points.)  Here the divergence will be defined with respect to the standard coordinate system but not with respect to a different coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, in general a function which has a continuous derivative (or partial derivatives) is much better behaved than a merely differentiable function.  Inserting the hypothesis that a function is $C^1$ -- unless you really need to be considering weaker hypotheses than that for a specific application -- is generally a prudent practice.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-29T16:03:25.687" LastActivityDate="2010-11-29T16:03:25.687" CommentCount="6" />
  <row Id="12360" PostTypeId="2" ParentId="12160" CreationDate="2010-11-29T16:07:17.367" Score="1" ViewCount="0" Body="&lt;p&gt;The Abramowitz–Stegun Handbook of Mathematical Functions claims on page 787 that all the roots are simple: &lt;a href=&quot;http://convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150&amp;amp;Page=787&quot; rel=&quot;nofollow&quot;&gt;http://convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150&amp;amp;Page=787&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2307" LastActivityDate="2010-11-29T16:07:17.367" />
  <row Id="12361" PostTypeId="2" ParentId="12160" CreationDate="2010-11-29T16:35:11.370" Score="4" ViewCount="0" Body="&lt;p&gt;I think there is a simpler proof that the roots are simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Legendre polynomial $\displaystyle P_n(x)$ satisfies the differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1-x^2) y&amp;#39;&amp;#39; -2x y&amp;#39; + n(n+1) y = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, we scale the polynomials so that $\displaystyle P_n(1) = 1$, so if $\displaystyle \alpha$ is a root, then $\displaystyle \alpha \neq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\displaystyle \alpha$ is a root of multiplicity $\displaystyle \gt 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we must have that $P_n(\alpha) = P_n&amp;#39;(\alpha) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above equation implies that $P_n&amp;#39;&amp;#39;(\alpha) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, by differentiating the above equation we can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1-x^2) y^{(k+2)&amp;#39;} - f_k(x) y^{(k+1)&amp;#39;} + g_k(x) y^{k&amp;#39;} = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle y^{l&amp;#39;}$ is the $\displaystyle l^{th}$ derivate of $\displaystyle y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle \alpha \neq 1$ we see that $\displaystyle P_n^{k&amp;#39;}(\alpha) = 0 \ \ \forall k$ and thus the only root of $\displaystyle P_n(x)$ is $\displaystyle \alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is easy to show that $\displaystyle P_n(x) \neq C(x-\alpha)^n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-11-29T16:35:11.370" />
  <row Id="12362" PostTypeId="1" CreationDate="2010-11-29T16:42:23.533" Score="0" ViewCount="260" Body="&lt;p&gt;What is the square root of $-4$ (negative 4)?&lt;/p&gt;&#xA;" OwnerUserId="4211" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T17:10:37.350" LastActivityDate="2010-12-11T04:54:58.200" Title="what is the square root of $-4$?" Tags="&lt;complex-numbers&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="12363" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T16:43:19.227" Score="1" ViewCount="0" Body="&lt;p&gt;Plus or minus 2i&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-29T16:43:19.227" />
  <row Id="12364" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T16:44:07.667" Score="2" ViewCount="0" Body="&lt;p&gt;-4 has no real square roots, since $x^2 \geq 0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-29T16:44:07.667" CommentCount="4" />
  <row Id="12365" PostTypeId="2" ParentId="12316" CreationDate="2010-11-29T16:44:46.227" Score="1" ViewCount="0" Body="&lt;p&gt;I think you will need to give more details on what is motivating your search for $W$ before you'll get a satisfactory answer. I will make a guess that you have a vector space that has a preferred basis and you are seeking an inner product which renders this basis orthonormal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of a name for $W$; it is the matrix of the inner product. It should be (symmetric) positive definite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have $d$ linearly independent vectors $\{x_i\}$ in $\mathbb{R}^d$ that are a convenient basis for our work, except that they aren't orthonormal. We may as well assume they are normalized with respect to the usual inner product on $\mathbb{R}^d$. Make a matrix $X$ whose columns are these vectors. Then $X$ is the matrix of a change of basis. Specifically, $X^{-1}y=x$ gives us $x$ as the representation of $y$ in the basis $\{x_i\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $x^Tx$. That is the inner product we want. It is the inner product which renders our basis orthonormal. Since $x^Tx = y^TX^{-T}X^{-1}y$, we can set $W = X^{-T}X^{-1}$ and we have the desired inner product without having to explicitly perform a change of basis. (Here $X^{-T}$ is the inverse of the transpose of $X$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One context where I have dealt with this is when $X$ was the complete basis of eigenvectors of a linear operator, and I wanted an inner product for which the operator was self-adjoint.&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2010-11-29T16:44:46.227" CommentCount="3" />
  <row Id="12366" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T16:45:52.063" Score="4" ViewCount="0" Body="&lt;p&gt;no it's not -2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square root is a number $x$ such that its square is the original number. Depending on the number system you are looking at there might be no square root, for example in the integers or reals. Over the complex numbers there is a square root which other posters have given as 2i or -2i&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2010-12-11T04:30:17.237" LastActivityDate="2010-12-11T04:30:17.237" />
  <row Id="12367" PostTypeId="2" ParentId="12328" CreationDate="2010-11-29T17:17:20.133" Score="4" ViewCount="0" Body="&lt;p&gt;It is not difficult to prove that one can factor $\rm N$ in polynomial time given any multiple of $\rm \phi(N)$ (here $\rm \:de-1\:$).$\ $ See, for example, Gary Miller: &lt;a href=&quot;http://www.cs.cmu.edu/~glmiller/Publications/Papers/Mi76.pdf&quot; rel=&quot;nofollow&quot;&gt;Riemann's hypothesis and tests for primality. 1976&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ This fact was well-known to the discoverers of RSA. Indeed they mention it explicitly in section IX of the &lt;a href=&quot;http://people.csail.mit.edu/rivest/Rsapaper.pdf&quot; rel=&quot;nofollow&quot;&gt;original 1978 paper on RSA&lt;/a&gt;, which is quite readable.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-11-29T17:23:39.080" LastActivityDate="2010-11-29T17:23:39.080" />
  <row Id="12368" PostTypeId="1" AcceptedAnswerId="12371" CreationDate="2010-11-29T17:19:56.447" Score="0" ViewCount="94" Body="&lt;p&gt;Given x(t) = f(t) * g(t), what is the Fourier transform of x(t).  If possible, please explain your answer.&lt;/p&gt;&#xA;" OwnerUserId="4212" LastActivityDate="2010-11-29T17:55:23.220" Title="What is the Fourier transform of the product of two functions?" Tags="&lt;fourier-analysis&gt;&lt;transformation&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="12369" PostTypeId="1" AcceptedAnswerId="12406" CreationDate="2010-11-29T17:23:10.080" Score="2" ViewCount="99" Body="&lt;p&gt;I have a couple questions that I'd like answered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  How do I evaluate this integral?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)  Is it possible to do a substitution?  What about $u=e^{i \cdot t}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3)  What if 5 and 3 are replaced by other constant integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4)  What should be the most effective method to handle question (3), using the fewest calculations (like in numerical methods)?  In other words, what is the most effecient way to handle the general case?  (Note that it doesn't have to be a numerical method; I just want something fast!)&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-29T17:34:56.600" LastActivityDate="2010-11-29T23:21:38.443" Title="Help With $\int_{-\pi}^\pi{\cos{(y e^{i\cdot 5\cdot t} + z e^{i\cdot 3\cdot t})}dt}$" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12370" PostTypeId="2" ParentId="12368" CreationDate="2010-11-29T17:41:52.147" Score="1" ViewCount="0" Body="&lt;p&gt;Transforms such as Fourier transform or Laplace transform, takes a product of two functions to the convolution of the integral transforms, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;Convolution Theorem&lt;/a&gt;, and is available with proof at &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-11-29T17:41:52.147" />
  <row Id="12371" PostTypeId="2" ParentId="12368" CreationDate="2010-11-29T17:43:13.563" Score="1" ViewCount="0" Body="&lt;p&gt;Fourier transform of a product is the convolution of the corresponding transforms. For details on conditions on the functions refer links below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fourier_transform&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fourier_transform#Convolution_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-29T17:55:23.220" LastActivityDate="2010-11-29T17:55:23.220" CommentCount="2" />
  <row Id="12372" PostTypeId="2" ParentId="12331" CreationDate="2010-11-29T18:26:05.883" Score="3" ViewCount="0" Body="&lt;p&gt;This is intended as a complement to Alex Bartel's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  Consider the 2D case given by $x = x(t)$, $y = y(t)$.  Here is a graph of a parametrized function with $(x(t_0),y(t_0))$ (in red) and $(x(t_0 + \delta), y(t_0 + \delta))$ (in yellow) marked on it.  The vector $(x(t_0 + \delta), y(t_0 + \delta)) - (x(t_0),y(t_0))$ (from the numerator of the derivative definition) is turquoise, sits almost on top of the graph of the function, and thus is very close to being tangent to the graph of the function at $(x(t_0), y(t_0))$.  Dividing by $\delta$ only changes the length of that vector, not its direction.  Now imagine letting $\delta \to 0$ and watching the yellow and turquoise vectors change.  Graphically, the turquoise vector is getting closer and closer to being a tangent vector at $t_0$.  In the limit, it should actually be a tangent vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/l8N24.png&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) (from another perspective)  If $(x(t), y(t), z(t))$ describes the position of some object moving through space, then $(x&amp;#39;(t), y&amp;#39;(t), z&amp;#39;(t))$ is its velocity.  Since the direction of the velocity vector gives the direction the object is moving at an instant in time, $(x&amp;#39;(t_0), y&amp;#39;(t_0), z&amp;#39;(t_0))$ will be tangent to the graph of the position at $t_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Here's an example in the 2D case where $x&amp;#39;(t)$ and $y&amp;#39;(t)$ both vanish.  The parametrized function is $x(t) = 1+t^3$ and $y(t) = t^2$, with $x&amp;#39;(0) = y&amp;#39;(0) = 0$.  Graphically, we have a cusp, and so there is no well-defined tangent vector.  This is not the only situation that can occur (see Alex's answer and J.M.'s comment), but it does show why an assumption like this is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2u7Wm.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-29T18:26:05.883" CommentCount="3" />
  <row Id="12373" PostTypeId="2" ParentId="12362" CreationDate="2010-11-29T18:49:53.527" Score="4" ViewCount="0" Body="&lt;p&gt;There is no solution to this in the real numbers. The square of any number is greater or equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In complex numbers however, we have &lt;em&gt;defined&lt;/em&gt; something called $i$ with $i \cdot i = -1$ or equivalently $i = \sqrt{-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 = -4 = 4\cdot(-1)$$&#xD;&#xA;$$x = \pm \sqrt{4}\cdot\sqrt{-1} = \pm2i$$&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-11-29T18:49:53.527" />
  <row Id="12374" PostTypeId="1" CreationDate="2010-11-29T19:09:07.240" Score="2" ViewCount="88" Body="&lt;p&gt;I have a system that is modeled by the following differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;db/dt = j(t)&lt;em&gt;ha + k&lt;/em&gt;(Ta-b(t))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where db/dt is the rate of temperature change, j(t) is an input, ha, k, Ta are all constants, and b(t) is an output.  Note that this is newtonian cooling with a heating input, j(t)*ha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the transfer function is the laplace domain, B(s)/J(s).&#xA;Taking the laplacian of the equation of interest, assuming all IC's are 0, yields:&#xA;s*B(s)+k*B(s)-(k*Ta)/s = ha*J(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't figure out is the term k*Ta/s is not a function of J(s) or B(s) so I can not get a transfer function of purely B(s)/J(s).  Does anyone know how to solve this equation or a better way to find the transfer function relating the input to the output?&lt;/p&gt;&#xA;" OwnerUserId="4217" LastActivityDate="2010-11-30T21:01:14.707" Title="Transfer Function, Newtonian Cooling" Tags="&lt;differential-equations&gt;&lt;control-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12376" PostTypeId="1" CreationDate="2010-11-29T19:20:56.743" Score="0" ViewCount="46" Body="&lt;p&gt;I have data for points of a 3D grid. The points of the grid are generated from three nonorthogonal vectors: each grid point has coordinates $\mathbf{q}_{ijk} = i \mathbf{a} + j \mathbf{b} + k \mathbf{c}$, where $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are nonorthogonal, noncoplanar vectors. (In cristallography, it's called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Triclinic&quot; rel=&quot;nofollow&quot;&gt;triclinic system&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: how would you adapt the marching tetrahedra (or marching cubes, if easier) to this case? Has this been treated already somewhere, is there software for that (I haven't found any).&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-11-29T22:43:41.683" Title="Marching tetrahedra (or cubes) on a non-orthogonal 3D grid" Tags="&lt;algorithms&gt;&lt;marching-cubes&gt;&lt;marching-tetrahedra&gt;" AnswerCount="1" />
  <row Id="12377" PostTypeId="1" CreationDate="2010-11-29T19:20:58.407" Score="0" ViewCount="73" Body="&lt;p&gt;Suppose I is an index set (not necessarily countable). Say we have a family of subsets $\{C_{i}: i \in I\}$ and suppose that this family is a cover of a space $X$, that is, for each $x$ in $X$ we have $x \in C_{j}$ for some $j \in I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we say the following? for each $x \in X$ there exists some natural number $n$ such that $x \in C_{i_{n}}$ where $i \in I$ ? or do we require countability of $I$ in order to guarantee this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me state the question more precisely. I'm trying to understand the proof of the following fact: Let $X$ be a $T_{1}$ space with locally-finite basis. Then $X$ must be discrete. Proof of the author: say $B$ is a locally finite basis. Now pick $x \in B$ then by definition there is some element $B_{1}$ of $B$ such that $x \in B_{1}$. Using recursion and the assumption of $T_{1}$ we can find $B_{n}$ in B such that $x$ in all $B_{n}$ and for all n we have the inclusion $B_{n+1} \subset B_{n}$. Thus $B$ cannot be locally finite, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't the above proof assuming I is countable? Can you please ellaborate more?&lt;/p&gt;&#xA;" OwnerUserId="4219" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T20:18:34.903" LastActivityDate="2010-11-29T20:42:42.567" Title="Doubt (index sets)" Tags="&lt;topology&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12379" PostTypeId="2" ParentId="12377" CreationDate="2010-11-29T19:29:13.577" Score="1" ViewCount="0" Body="&lt;p&gt;We cannot say it. It depends on the specific family you consider. This is possible even if $I$ is uncountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say $X$ has size $\aleph_1$, the first uncountable cardinal. Let $I=X$, and suppose $i\in C_i$ and $X\setminus C_i$ is &lt;em&gt;countable&lt;/em&gt; for each $i\in X$. A natural example is to let $X=\omega_1$, the first uncountable ordinal, and $C_i=[i,\infty)$ with the usual order of ordinals. Then Only countably many $C_i$ are needed to cover $X$, so we can pick for each $x$ an $i_n$ as you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $I=X$ and $C_i=\{i\}$ for each $i$, then of course all $C_i$ are needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we can easily put a topology in each case so the $C_i$ are open sets, so that additional condition would not change the answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(So, to &lt;em&gt;guarantee&lt;/em&gt; that it is always possible to pick such $i_n$, we need $I$ to be countable, or some topological condition if we want the $C_i$ to be open, such as $X$ being Lindelöf, which is a weakening of being compact.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-29T19:29:13.577" CommentCount="2" />
  <row Id="12380" PostTypeId="2" ParentId="11948" CreationDate="2010-11-29T19:30:19.347" Score="0" ViewCount="0" Body="&lt;p&gt;This brought to mind a conjecture I made a couple of years ago which involved a generalisation of the factorial function.&#xA;The integer n! is the product of x taken over all integers from 1 to n and as you know Legendre (and also I think de Polignac) proved the highest power of a prime p diving n! is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(floor(n/p)) + (floor(n/p^2)) + (floor(n/p^3)) + . . ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the arithmetical function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(n) := Product {1&amp;lt;=x&amp;lt;=n,1&amp;lt;=y&amp;lt;=n} GCD(x,y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here GCD(x,y) is the greatest common divisor of x and y. Clearly, we can consider this function as a generalisation of n! = Product {1&amp;lt;=x&amp;lt;=n} GCD(x). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence F(n) begins [1,1,2,6,96,480,414720,..], which I submitted as A092287 in the OEIS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There I conjectured that the highest power of a prime p dividing F(n) is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(floor(n/p))^2 + (floor(n/p^2))^2 + (floor(n/p^3))^2 + . . ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So F(n) is, in some sense, the next step up from n!.&#xA;Continuing further in this manner, consider products of the GCD of triples of integers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G(n) := Product {1&amp;lt;=x&amp;lt;=n,1&amp;lt;=y&amp;lt;=n,1&amp;lt;=z&amp;lt;=n} GCD(x,y,z).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See A129454 in the OEIS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculation suggests that the highest power of a prime p diving G(n) is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(floor(n/p))^3 + (floor(n/p^2))^3 + (floor(n/p^3))^3 + . . ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may have an infinite sequence n!, F(n), G(n), ... of arithmetical functions answering your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also use the least common multiple function LCM instead of the GCD function in the above to get similar though more complicated conjectural factorisation results. See A090494 in Sloane for an example. &#xA;Finally, take a look at A092143 and A129365 for yet more conjectural factorisations which have a similar feel to Legendre's result.&lt;/p&gt;&#xA;" OwnerUserId="4221" LastActivityDate="2010-11-29T19:30:19.347" />
  <row Id="12381" PostTypeId="1" CreationDate="2010-11-29T19:37:40.500" Score="3" ViewCount="85" Body="&lt;p&gt;Dear All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that $f(x)=f^{1}(x)$ and that $f(f(x))=f^{2}(x)$. Moreover, $f^{n}(x)$ is the n-th iterate of $f(x)$, for $n \in \mathbb{N}$. I'm curious about extending iteration to larger number sets. For $n \in \mathbb{R}$, there's the concept &lt;a href=&quot;http://en.wikipedia.org/wiki/Flow_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;flow&lt;/a&gt; (I think?). I don't understand the Wikipedia-article on this subject very well, though. I was hoping for some nice, concrete examples of iterated functions extended to the real or even complex numbers with which I might understand things better. If we take $f(x) = x^2 +3$, for example, what would $f^{\sqrt(2)}(x)$ be? Or, even more ambitiously, say that $g(x)=e^x$ How do we find $g^{\pi^2 + 3i}(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Editorial to the moderators: perhaps this should be CW?&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-30T13:35:07.610" LastActivityDate="2010-11-30T13:35:07.610" CommunityOwnedDate="2010-12-17T11:21:07.453" Title="Explicit examples of functions with flow?" Tags="&lt;big-list&gt;&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12382" PostTypeId="1" AcceptedAnswerId="12398" CreationDate="2010-11-29T19:43:46.313" Score="2" ViewCount="82" Body="&lt;p&gt;Given a topological space (with nice enough conditions, maybe Hausdorff, compactly generated, or CW complex, I'm not sure) $X$ and a subspace $A\subset X$, is it true that $H^n(X,A)\cong H^n(X\backslash A)$ whenever there is an open set containing $A$ which can be retracted to $A$?  Can this be shown easily using the Mayer-Vietoris Sequence, or something else similar to that? (I'm basically assuming simplicial or singular cohomology here, with integer coefficients.)&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-11-29T21:22:23.360" Title="Relative Cohomology Isomorphic to Cohomology of Quotient" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12383" PostTypeId="1" AcceptedAnswerId="12385" CreationDate="2010-11-29T19:46:45.943" Score="5" ViewCount="108" Body="&lt;p&gt;Question: (a) Let $f: V \rightarrow W$ with $ V,W \simeq \mathbb{R}^{3}$ given by: $$f(x_1, x_2, x_3) = (x_1 - x_3, 2x_1 -5x_2 -x_3, x_2 + x_3).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the matrix of $f$ relative to the basis $\{(0,2,1),(-1,1,1),(2,-1,1)\}$ of $V$ and $\{(-1,-1,0),(1,-1,2),(0,2,0)\}$ of $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Let $n \in \mathbb{N}$ and $U_n$ the vector space of real polynomials of degree $\leq n$. The linear map $f: U_n \rightarrow U_n$ is given by $f(p) = p&amp;#39;$. Determine the matrix of $f$ relative to the basis $\{1,t,t^{2},...,t^{n}\}$ of $U_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far: &#xA;(a): First relative to the bases of $W$ I found the coordinates of an arbitrary vector: $\left( \begin{array}{r} a \\ b \\ c \end{array} \right) = x \left( \begin{array}{r} -1 \\ -1 \\ 0 \end{array} \right) + y \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) + z \left( \begin{array}{c} 0 \\ 2 \\ 0 \end{array} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{l} a = -x + y \\ b = - x - y + 2z \\ c = 2y \end{array}$ or $\begin{array}{l} x = -a + \frac{1}{2}c \\ z = -\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c \\ y = \frac{1}{2}c \end{array}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I believe I have the linear combinations of the given basis in $W$ for an arbitrary vector, so next I take the vectors from $V$ and send them to $W$ using the given function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{l} f(v_1) = f(0,2,1) = (-1,-11,3) = (1 + \frac{3}{2})w_1 + \frac{3}{2}w_2 + (\frac{1}{2} - \frac{11}{2} + \frac{3}{2})w_3 \\ f(v_2) = f(-1,1,1) = (-2,-8,2) = (2+1)w_1 + w_2 + (1 - 4 +1)w_3 \\ f(v_3) = f(2,-1,1) = (1,8,0) = w_1 + (-\frac{1}{2} + 4)w_3 \end{array}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $\left( \begin{array}{rrc} \frac{5}{2} &amp;amp; 3 &amp;amp; 1 \\ \frac{3}{2} &amp;amp; 1 &amp;amp; 0 \\ -\frac{7}{2} &amp;amp; -2 &amp;amp; \frac{7}{2}\end{array} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was I taking the correct steps? I didn't really do anything differently based on the fact that $V,W$ were isometric... Is there a particular significance or interpretation for the resulting matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b): Not really sure here... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(p) = p&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would it make sense to write something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(1,t,t^{2},\dots, t^{n}) = (0,1,2t, \dots, nt^{n-1})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if a basis for $(1,t,t^{2},\dots, t^{n})$ would be $A = \left( \begin{array}{ccccc} 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; \cdots &amp;amp; 0 \\ \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\0 &amp;amp; \cdots &amp;amp; \cdots &amp;amp; 0 &amp;amp; 1 \end{array} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could i write: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A&amp;#39; =  \left( \begin{array}{ccccc} 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; \cdots &amp;amp; 0 \\ \vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \end{array} \right)$?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T20:14:02.773" LastActivityDate="2010-11-29T21:01:20.560" Title="Determine the matrix relative to a given basis" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12384" PostTypeId="1" AcceptedAnswerId="12388" CreationDate="2010-11-29T20:01:46.603" Score="-6" ViewCount="263" Body="&lt;p&gt;&lt;strong&gt;This question has been edited in view of comments received.Hope it is better now.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of all points on the circumference of a circle and the set of all points inside a circle. what is difference in the measure of these two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is &lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;Bertrand&lt;/a&gt; paradox related to measure theory ? Can bertrand paradox be resolved with the notion of a measure of a set. If so please suggest some good references/books for measure theory. &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-12-09T08:42:59.643" LastActivityDate="2010-12-09T08:42:59.643" Title="A soft Question Suggestions for good introductory books" Tags="&lt;measure-theory&gt;&lt;paradox&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="12385" PostTypeId="2" ParentId="12383" CreationDate="2010-11-29T20:02:26.830" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, your solution to (a) is correct assuming your computations for how to express the elements is correct. Though you don't need to find how to write an &lt;em&gt;arbitrary&lt;/em&gt; vector in terms of the basis of $W$, you only need to find out how to write $f(0,2,1)$, $f(-1,1,1)$, and $f(2,-1,1)$ (the images of the vectors in the given basis for $V$) in terms of the given basis of $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, you never need to use the fact that $V$ and $W$ are isomorphic (we don't consider a metric here, so &quot;isometric&quot; is not appropriate here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer for (b), on the other hand, is not well done (or correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis for your vector space is $\{1,t,\ldots,t^n\}$. To find the matrix representation of $f$ relative to this basis, you need to find the image of each basis vector in the domain, and express it in terms of the basis vectors of the range. But remember: the &lt;em&gt;vectors&lt;/em&gt; in $U_n$ are not tuples, they are &lt;em&gt;polynomials&lt;/em&gt;. So $(1,t,\ldots,t^n)$ is not an element of $U_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you would have to find each of $f(1)$, $f(t)$, $f(t^2),\ldots, f(t^n)$, and then express them in terms of the basis of the range, which happens to again be $\{1,t,t^2,\ldots,t^n\}$. So for instance, $f(t^2) = 2t$, so&#xA;$$f(t^2) = 0\cdot 1 + 2\cdot t + 0\cdot t^2 + \cdots + 0\cdot t^n.$$&#xD;&#xA;That means that the third column of the matrix (the one corresponding to the third basis vector, which is $t^2$) will be the transpose of $(0,2,0,\ldots,0)$, that is&#xA;$$\left(\begin{array}{c} 0\\2\\0\\ \vdots \\ 0\end{array}\right).$$&#xD;&#xA;Etc. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T21:01:20.560" LastActivityDate="2010-11-29T21:01:20.560" CommentCount="2" />
  <row Id="12386" PostTypeId="2" ParentId="12384" CreationDate="2010-11-29T20:04:47.433" Score="2" ViewCount="0" Body="&lt;p&gt;Here's one:&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387952241&quot; rel=&quot;nofollow&quot;&gt; Beginning Functional Analysis by Karen Saxe &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2010-11-29T20:04:47.433" />
  <row Id="12387" PostTypeId="2" ParentId="12381" CreationDate="2010-11-29T20:07:39.473" Score="4" ViewCount="0" Body="&lt;p&gt;I thought the wikipedia article is pretty straightforward. Define a function $\phi: \mathbb{R}^2 \to \mathbb{R}: (x,t) \mapsto \phi(x,t)$. Now, you want the second parameter $t$ to be interpreted as the number of times you have applied the function to $x$, in a way. To formalize this, you introduce the following rule on $\phi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(\phi(x,t),s)=\phi(x,t+s)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this for all $x,t$ and $s$. In particular, you see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi(\phi(x,t),t)=\phi(x,2t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and more generally if we define $\phi_t:\mathbb{R}\to\mathbb{R}:x\mapsto\phi(x,t)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_t^n(x)=\phi(x,nt)=\phi_{nt}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is exactly the behaviour you would like to have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining a flow $\phi(x,t)$ starting from the condition that $\phi(x,1)=f(x)$ is not an easy task however and can not be done for any arbitrary function I think. I remember another post related to the question. I think it was a question about the Vieta product.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-11-29T20:07:39.473" CommunityOwnedDate="2010-12-17T11:21:07.453" CommentCount="3" />
  <row Id="12388" PostTypeId="2" ParentId="12384" CreationDate="2010-11-29T20:08:55.213" Score="4" ViewCount="0" Body="&lt;p&gt;Lang's book is a fairly comprehensive treatise on Differential Geometry which will introduce you to Riemannian manifold and other notions you mention in first paragraph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=VfxGB5nYv1MC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=VfxGB5nYv1MC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure theory can be picked up from Paul Halmos' book&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387900888&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/dp/0387900888&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Following Rajesh D's comment below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I liked Munkres book on point set Topology. It also gives a flavor of Algebraic Topology towards the end. Though I have heard mixed opinions from others about this. But it can be read with almost zero background in any sort of math. It builds up from basic set theory (this was incidentally the first mathematics text I ever read).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131816292&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/dp/0131816292&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Janich book on basic topology is wonderful and reads like a novel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387908927&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/dp/0387908927&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-30T01:36:29.813" LastActivityDate="2010-11-30T01:36:29.813" CommentCount="6" />
  <row Id="12389" PostTypeId="2" ParentId="12381" CreationDate="2010-11-29T20:11:55.010" Score="2" ViewCount="0" Body="&lt;p&gt;Max, besides the existence of a lot of literature about this subject I've written two very basic/introductive articles about it. Perhaps the better is the shorter (and newer) one, I deal with the iteration of $ f(x) = 1/(1+x) $ The idea stems originally from a consideration of continued fractions in a general form $ x; 1/(1+x); 1/(1+1/(1+x));....$ . On the other hand, it is simple and the iteration of the function can be approached from different viewpoints, so the approaches are crosschecked mutually.&#xA;See : &lt;a href=&quot;http://go.helms-net.de/math/tetdocs/FracIterAltGeom.htm&quot; rel=&quot;nofollow&quot;&gt;fractional functional iteration of f(x)=1/(1+x) &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;additional comment: the difficulties of defining a continuous iteration of $f(x) = e^x$ compared with that of a polynomial $f(x)=x^2 + 3$ are so eminent that putting them into the same question requires a much too wide area for discussion in a MO-thread[Edit: Oh sorry, we're not in MO, I was jumping between tabs in my browser, sorry]. However in the last years a lot of material, including scans of historical articles, was put online and can be read. Possibly I can provide useful links if you ask more specific questions. Also you might browse the tetration-forum at &lt;a href=&quot;http://math.eretrandre.org/tetrationforum/index.php&quot; rel=&quot;nofollow&quot;&gt;Tetration-forum&lt;/a&gt; There is also an online reference-database. &lt;a href=&quot;http://math.eretrandre.org/tetrationforum/showthread.php?tid=365&quot; rel=&quot;nofollow&quot;&gt;see tet-forum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Update] i constructed a function f(x,h) where f(x,0)=x, f(x,1)=x^2+3 and f(x,h) the h'th iterate using fixpoint-shift and matrix-diagonalization. Unfortunately this gives complex powerseries for fractional iterates. The construction gives the following complex plot for x=1,h=0..2 . note that for real, noninteger heights h the curve extends into the complex plane! So I think, there should be a better solution...    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/g8FDe.png&quot; alt=&quot;provisorial construction of f(x,h)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-11-29T23:06:47.277" LastActivityDate="2010-11-29T23:06:47.277" CommunityOwnedDate="2010-12-17T11:21:07.453" CommentCount="3" />
  <row Id="12390" PostTypeId="2" ParentId="12383" CreationDate="2010-11-29T20:18:10.467" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a suggestion for seeing the bigger picture in your first question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The key idea below is that the column vectors of a matrix $M$ are the images of the standard ordered basis when the matrix $M$ is viewed as a transformation&#xA;$M: \mathbb{R}^n \to \mathbb{R}^n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is easy to find a matrix of f with respect to the standard ordered basis of $\mathbb{R}^3$: it is given by a matrix with columns $\{1, 2, 0\}$, $\{0, -5, 1\}$, and $\{-1, -1, 1\}$.  Call this matrix $C$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The matrix which transforms the given ordered basis of $V$ to the standard basis of $\mathbb{R}^3$ is given by writing the basis vectors as columns of this matrix. Similarly, this can be done for the basis of $W$.  Let's call these matrices $A$ and $B$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You desire a matrix which represents $f$.  Interpreting the matrices above as transformations, consider the following diagram:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}&#xD;&#xA;V &amp;amp; \to &amp;amp; \mathbb{R}^3\\&#xD;&#xA;\downarrow &amp;amp; \qquad &amp;amp; \downarrow\\&#xD;&#xA;W &amp;amp; \to &amp;amp; \mathbb{R}^3&#xD;&#xA;\end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the horizontal maps are $A$ and $B$ and the right vertical map is $C$.&#xA;Your desired matrix is the left vertical arrow.  It is now clear that it is given by the matrix $B^{-1} C A$.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-29T20:26:18.437" LastActivityDate="2010-11-29T20:26:18.437" CommentCount="1" />
  <row Id="12391" PostTypeId="2" ParentId="12381" CreationDate="2010-11-29T20:19:22.710" Score="2" ViewCount="0" Body="&lt;p&gt;Since you ask for explicit examples, the solution to any time-invariant ordinary differential equation as a function of time and the initial value of the variable is a flow. For example, $dx/dt = \lambda x$ with initial value $x(0) = x_0$ has the solution $f(t,x_0) = e^{\lambda t} x_0$, which in your notation can be interpreted as $f^t(x) = e^{\lambda t} x$. This is a somewhat boring example; the interesting ones happen in more than one dimension. Rotation, for example: let $f^t(x)$ be the rotation of $x$ about a fixed axis by an angle $\omega t$. I'll leave it to you to think of more complicated ODEs.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-11-29T20:19:22.710" CommunityOwnedDate="2010-12-17T11:21:07.453" CommentCount="0" />
  <row Id="12392" PostTypeId="2" ParentId="12377" CreationDate="2010-11-29T20:42:42.567" Score="1" ViewCount="0" Body="&lt;p&gt;You are misunderstanding the author's argument.  Suppose $x \in B_1$.  By local finiteness, $B_1$ only contains finitely many elements of $X$, say $n$.  If $y \in B_1$ is any element other than $x$, then by the $T_1$ assumption $B_1 - \{ y \} = B_2$ is also an open set containing $x$.  Continuing in this way we obtain a sequence $B_1 \supset B_2 \supset ... $ of strictly decreasing open sets containing $x$ which stabilizes after at most $n$ steps; this gives an open set $B_n = \{ x \}$, hence the topology is discrete.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-29T20:42:42.567" />
  <row Id="12393" PostTypeId="1" CreationDate="2010-11-29T20:45:09.880" Score="1" ViewCount="63" Body="&lt;p&gt;I'm looking for a graphic that shows the isomorphism theorems illustrated using a finite vector space. (that is a finite dimensional vector space over a finite field.) Something that shows nicely how the cosets &quot;collapse&quot; in to single elements or how cosets at different &quot;levels&quot; are matched up to form the isomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried searching but I keep finding arrow diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, suggestions for good examples that don't take too long to draw would be appreciated.   &lt;/p&gt;&#xA;" OwnerUserId="2563" LastActivityDate="2010-11-29T20:45:09.880" Title="Pictorial representations of the isomorphism theorems" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;" CommentCount="1" />
  <row Id="12394" PostTypeId="1" AcceptedAnswerId="12409" CreationDate="2010-11-29T20:46:47.580" Score="2" ViewCount="141" Body="&lt;p&gt;I have a radius, &lt;code&gt;R&lt;/code&gt;, for an aircraft traveling at velocity, &lt;code&gt;V&lt;/code&gt;. If we start at point, &lt;code&gt;(X,Y)&lt;/code&gt;, what is the position of the point at time, &lt;code&gt;t&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;The aircraft is at point &lt;code&gt;(0,0)&lt;/code&gt; and traveling at &lt;code&gt;250&lt;/code&gt; knots and initiates a turn with a bank angle, phi, of &lt;code&gt;5&lt;/code&gt; degrees. Assume that the aircraft can instantaneously rotate to the five degree bank. The equation for the turn radius, R where g is the acceleration due to gravity (9.81) is:&#xA;\begin{equation}&#xA;\text{R} = \frac{V^2}{\text{g} \tan{\phi}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this example, &lt;code&gt;R = 10.4&lt;/code&gt; nautical miles. Where is the aircraft at &lt;code&gt;t = 2&lt;/code&gt; if the aircraft is traveling at a heading of &lt;code&gt;90&lt;/code&gt; degrees (straight along the y axis)?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-30T14:22:07.523" LastActivityDate="2010-11-30T14:22:07.523" Title="Given a radius and velocity calculate position of an aircraft banking to make a turn" Tags="&lt;geometry&gt;&lt;homework&gt;&lt;physics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="12395" PostTypeId="2" ParentId="12394" CreationDate="2010-11-29T20:51:01.497" Score="1" ViewCount="0" Body="&lt;p&gt;I don't know what you mean by R.  If you start at the origin, $v_x=v\cos(\theta)$ and $v_y=v\sin(\theta)$, $x=v_xt$ and $y=v_yt$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-29T20:51:01.497" CommentCount="4" />
  <row Id="12396" PostTypeId="2" ParentId="1901" CreationDate="2010-11-29T20:51:56.813" Score="1" ViewCount="0" Body="&lt;p&gt;The following (famous) question is perhaps in the spirit of the OP's question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ and $N$ are closed aspherical $n$-manifolds with isomorphic fundamental groups, are $M$ and $N$ homeomorphic?  (This is known as the Borel Conjecture.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scott Carter's example illustrates the need for the assumption that the manifolds are closed.  But the question above suggests that perhaps the fundamental group is sufficient to characterize the topological type of a closed aspherical manifold.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-11-29T20:51:56.813" />
  <row Id="12397" PostTypeId="1" AcceptedAnswerId="16108" CreationDate="2010-11-29T21:07:00.270" Score="1" ViewCount="69" Body="&lt;p&gt;For a given radius, theta, and position, can you calculate a Bezier curve?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2011-01-02T09:44:32.633" Title="Given a radius, calculate Bezier curve" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12398" PostTypeId="2" ParentId="12382" CreationDate="2010-11-29T21:22:23.360" Score="4" ViewCount="0" Body="&lt;p&gt;It should say reduced cohomology of $X/A$, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it is always true that the reduced cohomology of $X \cup CA$ (for $CA$ the cone) is isomorphic to the relative cohomology. To see this,  one removes a small contractible piece near the top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact sequence shows that $\widetilde{H}^*(X \cup CA) \simeq \widetilde{H}^*(X \cup CA, C A)$ because the cone is contractible. Then, excise the top of the cone. The pair then becomes homotopy equivalent to the pair $(X, A)$. So this proves the claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next fact that one needs to show is that $X/A$ has the same cohomology as $X \cup CA$. This is true if the inclusion $A \to X$ is a cofibration. In general, it is a fact that if $B \subset Y$ is a cofibration, and $B$ is contractible, then $Y$ and $Y/B$  have the same homotopy type. (Apply this with $Y = X \cup &#xD;&#xA;CA, B = CA$ to see that $X \cup CA$ and $X/A$ have the same homotopy type under cofibration conditions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, one takes a contracting homotopy $B \times I \to B$ of the identity to a point, and extends it to a homotopy $\phi: Y \times I \to Y$ by the cofibration property. This sends $B \times \{1\}$ into a point. So $\phi(., 1)$ is a map $Y/B \to Y$. The composite of $\phi(.,1)$ and $Y \to Y/B$ is homotopic to $1_{Y/B}$ by $\phi$ (since $\phi(.,t)$ always sends $B$ into itself).&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-29T21:22:23.360" CommentCount="1" />
  <row Id="12399" PostTypeId="2" ParentId="12394" CreationDate="2010-11-29T22:18:39.520" Score="2" ViewCount="0" Body="&lt;p&gt;There is still not enough information to answer the question.  The initial speed is specified, but not the direction.  Given the direction of travel, you can find the center of the circle.  Then (as in my last comment) the angular velocity is V/R and you can apply the same trig functions.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-29T22:18:39.520" CommentCount="3" />
  <row Id="12400" PostTypeId="1" CreationDate="2010-11-29T22:23:35.473" Score="4" ViewCount="68" Body="&lt;p&gt;Let $F_n$ be an $n$-generator free group with a free basis $x_1,\ldots,x_n.$ Is it true that the stabilizer of $x_1$ in $\mathrm{Aut}(F_n)$ is generated by all left and right Nielsen moves $\lambda_{ij}$ and $\rho_{ij}$ such that $i \ne 1$ and by the element of order two $\epsilon_n$ such that $\epsilon_n(x_n)=x_n^{-1}$ while other elements of the basis remain fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $i \ne j$ and $1 \le i,j \le n.$ The left Nielsen move $\lambda_{ij}$ takes $x_i$ to $x_j x_i$ and the right Nielsen move $\rho_{ij}$ takes $x_i$ to $x_i x_j;$ both $\lambda_{ij}$ and $\rho_{ij}$ fix all $x_k$ with $k \ne i.$ &lt;/p&gt;&#xA;" OwnerUserId="4223" LastEditorUserId="4223" LastEditorDisplayName="" LastEditDate="2010-11-29T23:29:20.220" LastActivityDate="2010-11-29T23:29:20.220" Title="Stab(x_1) in Aut( F(x_1,...x_n ) )" Tags="&lt;group-theory&gt;" CommentCount="3" />
  <row Id="12401" PostTypeId="2" ParentId="12376" CreationDate="2010-11-29T22:43:41.683" Score="1" ViewCount="0" Body="&lt;p&gt;Nothing in Marching Cubes (or Marching Tetrahedra, AFAIK) actually requires that the grid be cubical or even orthogonal; the algorithms are inherently topological (or arguably, combinatorial) in nature.  The heart of Marching Cubes is really just a table matching the $2^8$ combinations of positive/negative function values at the cube corners to a set of topological triangulations that are consistent with the 'simplest' isosurface of a function taking on those values (e.g., when only one vertex is positive we have a single triangle, with vertices along the three edges connecting the positive vertex to its neighbors), and those triangulations are invariant under linear transformations - which is all your skewed grid represents.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-11-29T22:43:41.683" CommentCount="1" />
  <row Id="12402" PostTypeId="1" CreationDate="2010-11-29T22:46:59.990" Score="3" ViewCount="75" Body="&lt;p&gt;Problem: The series $\sum_{n=1}^{\infty} a_n$ diverge and positive. What can be said on the series $\sum_{n=1}^{\infty} \frac{a_n}{1+n^{2}a_n}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, I tried separating it into two cases: $a_n \to \infty$, $a_n \to A$ (where $A$ is some const value). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tried the ratio test and got nowhere with that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think that the right approach is to check each case separately, but my hunch tells me that there is a workaround.&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2010-11-29T22:52:19.520" Title="Question about Infinite Series" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12403" PostTypeId="2" ParentId="4362" CreationDate="2010-11-29T22:48:46.400" Score="3" ViewCount="0" Body="&lt;p&gt;Noam Elkies gives an example &lt;a href=&quot;http://www.math.harvard.edu/~elkies/Misc/sol8.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2010-11-29T22:48:46.400" />
  <row Id="12404" PostTypeId="2" ParentId="12402" CreationDate="2010-11-29T22:52:07.147" Score="7" ViewCount="0" Body="&lt;p&gt;Hint: Rewrite as $$\sum_{n=1}^{\infty} \frac{1}{\frac{1}{a_n} + n^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-11-29T22:52:07.147" CommentCount="3" />
  <row Id="12405" PostTypeId="2" ParentId="12402" CreationDate="2010-11-29T22:52:19.520" Score="2" ViewCount="0" Body="&lt;p&gt;Divide the numerator and denominator by $a_n$, and compare what you get with $1/n^2$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-11-29T22:52:19.520" />
  <row Id="12406" PostTypeId="2" ParentId="12369" CreationDate="2010-11-29T23:11:25.843" Score="3" ViewCount="0" Body="&lt;p&gt;This is another contour integral. Replace $y$ by $a$ and $z$ by $b$ to simplify notation. Then your integral is&#xA;$$-i \int_{-\pi}^{\pi}{\cos(ae^{5it} + be^{3it}) \over e^{it}} ie^{it}\,dt$$&#xD;&#xA;$$= -i \int_{|z| = 1}{ \cos(az^5 + bz^3) \over z}\,dz$$&#xD;&#xA;By the residue theorem this integral is $2\pi$ times the residue of $\cos(az^5 + bz^3)$ at $z = 0$. Since $\cos(az^5 + bz^3) = 1$ when $z = 0$, this residue is $1$. So your overall integral is $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it doesn't matter what $a$ or $b$ are, and that $3$ and $5$ can be replaced by any nonnegative integers.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-11-29T23:21:38.443" LastActivityDate="2010-11-29T23:21:38.443" CommentCount="6" />
  <row Id="12407" PostTypeId="1" CreationDate="2010-11-29T23:34:12.947" Score="0" ViewCount="56" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/bBDqg.png&quot; alt=&quot;N=2762&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N=2762&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been exploring a data set that seems to give rise to this kind of qqplot rather frequently. Would you say this is one population with a different than normal population? Or are two populations confounding the normal distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It used matplotlib and scipy.stats:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      (x,y), (slope, inter, cor) = stats.probplot(data, dist='norm')&#xA;&#xA;      osmf = x.take([0, -1])  # endpoints&#xA;      osrf = slope * osmf + inter&#xA;      self.ax.plot(x, y, ',', osmf, osrf, '-', **self.kwargs)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4225" LastActivityDate="2010-11-29T23:34:12.947" Title="Does this qqplot suggest a distribution?" Tags="&lt;statistics&gt;" CommentCount="2" />
  <row Id="12408" PostTypeId="1" AcceptedAnswerId="12411" CreationDate="2010-11-29T23:40:13.343" Score="2" ViewCount="54" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Regular_sequence_%28algebra%29&quot; rel=&quot;nofollow&quot;&gt;regular sequence&lt;/a&gt; is an $n$-fold collection $\{r_1, \cdots, r_n\} \subset R$ of elements of a ring $R$ such that for any $2 \leq i \leq n$, $r_i$ is not a zero divisor of the quotient ring&#xA;$$ \frac R {\langle r_1, r_2, \cdots, r_{i-1} \rangle}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the order of the $r_i$'s matter? That is, is any permutation of a regular sequence regular?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-11-29T23:59:15.177" Title="Is a regular sequence ordered?" Tags="&lt;commutative-algebra&gt;" AnswerCount="2" />
  <row Id="12409" PostTypeId="2" ParentId="12394" CreationDate="2010-11-29T23:42:45.413" Score="4" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Bank_angle#Banked_turn_in_aeronautics&quot; rel=&quot;nofollow&quot;&gt;banked turn in  aeronautics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a motion along a circular path centered at $(0,0)$ (see figure below) with constant angular velocity $\omega$ the position of a point $P(x,y)$ (in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;Cartesian coordinates&lt;/a&gt; and $P(r,\varphi)$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;cylindrical coordinates&lt;/a&gt;) starting at $(r,0)$ at $t=0$ is given by the law of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_motion&quot; rel=&quot;nofollow&quot;&gt;uniform circular motion&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos(\omega t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=r\sin(\omega t),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r$ is the radius of the circumference and $\omega=\dfrac{d\varphi}{dt}$. The magnitude of the tangential velocity is $v=\omega r$ and the radial velocity is zero. Thus $\omega=\dfrac{v}{r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos\left(\frac{vt}{r}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=r\sin\left(\frac{vt}{r}\right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case $r=R$, $v=V$ (it is assumed that the aircraft mantains the speed $V$) and the centre of the circumference is $(X,Y)=(-R,0)$. If you make the change of coordinates $X=x-R$ and $Y=y$, you get the following $X,Y$ coordinates as a function of time $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=R\left(\cos\left(\frac{V}{R}t\right)-1\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=R\sin\left(\frac{V}{R}t\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this $XY$-coordinate system the motion starts at $(X,Y)=(0,0)$ ($t=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed that the direction of the turn is to the left. If it is to the right, then one has a symmetric motion with respect to the $y-$axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=R\left(-\cos\left(\frac{V}{R}t\right)+1\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=R\sin\left(\frac{V}{R}t\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/S7zj4.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical result is obtained for $t=2$ and the other given data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: For a different starting position $\left( X_{0},Y_{0}\right) $ at $t=0$ we have to incorporate these values in the motion equations. Integrating $%&#xD;&#xA;\omega =\dfrac{d\varphi }{dt}$, we obtain $\varphi =\omega t+C$, where $C$ is&#xA;a constant. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos \left( \omega t+C\right) ,y=r\sin \left( \omega t+C\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{0}=r\cos \left( C\right) ,y_{0}=r\sin \left( C\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{y_{0}}{x_{0}}=\dfrac{\sin \left( C\right) }{\cos \left( C\right) }%&#xD;&#xA;=\tan \left( C\right) ,C=\arctan \left( \dfrac{y_{0}}{x_{0}}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right)&#xD;&#xA;,y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=x-r=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right)&#xD;&#xA;\right) -r,Y=y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}%&#xD;&#xA;\right) \right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_{0}=x_{0}-r,Y_{0}=y_{0},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+r}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=R,\omega =\dfrac{v}{r}=\dfrac{V}{R},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+R}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we obtain the equations of the motion of the aircraft in the $X,Y$-plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=x-R=R\left( \cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) -1\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\left( X_{0},Y_{0}\right) $ is the position of the aircraft at $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the direction of the turn is to the right, then the motion is symmetric with respect to the $y-$axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=-x+R=R\left( -\cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) +1\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: I changed the notation; in the question the position of the aircraft&#xA;at $t=0$ is $\left( X,Y\right) $.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The velocity vector is $\overrightarrow{v}=v\overrightarrow{e}_{\varphi }$ and the acceleration vector $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-30T12:49:11.563" LastActivityDate="2010-11-30T12:49:11.563" CommentCount="2" />
  <row Id="12410" PostTypeId="1" AcceptedAnswerId="12420" CreationDate="2010-11-29T23:43:27.350" Score="4" ViewCount="125" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a particle starting at the the origin and moving along the positive real line at a constant speed of 1.  Suppose there is a counter which clicks at random time intervals following the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;exponential distribution&lt;/a&gt; with parameter $\lambda$ and whenever the counter clicks, the position $x &amp;gt; 0$ of the particle at that time instantaneously changes to the position $x/2$.  We wish to calculate the expected average speed of the particle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really have any idea of how to go about solving this.  Here are a couple of related problems which seem even more difficult to me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Modify the puzzle so that when the counter clicks, the particle moves from $x$ to a point chosen uniformly at random from $[0,x]$.&lt;/li&gt;&#xA;&lt;li&gt;The particle starts moving as above but whenever the counter clicks, its speed increases by 1 (the initial speed was 1).  What is the expected time when the particle hits the position 1?  What is the expected speed when the particle hits the position 1?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is not a homework problem.  Any solutions, hints, thoughts will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="3831" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-29T23:53:19.333" LastActivityDate="2010-12-01T06:14:38.493" Title="Particle moving at constant speed with Poisson setbacks" Tags="&lt;probability&gt;&lt;puzzle&gt;&lt;expectation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="12411" PostTypeId="2" ParentId="12408" CreationDate="2010-11-29T23:44:55.880" Score="4" ViewCount="0" Body="&lt;p&gt;Not in general. A standard example is $R=k[x,y,z]$, where $k$ is a field. Then, $x,y(1-x),z(1-x)$ is regular but $y(1-x),z(1-x),x$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the bright side, if $R$ is Noetherian, local then every permutation of a regular sequence is regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact more is true. We can extend this notion to modules over rings analogously. Then, if $M$ is a finitely generated module over a Noetherian, local ring, then every permutation of a regular sequence is regular.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-29T23:59:15.177" LastActivityDate="2010-11-29T23:59:15.177" CommentCount="2" />
  <row Id="12412" PostTypeId="2" ParentId="12408" CreationDate="2010-11-29T23:45:50.257" Score="2" ViewCount="0" Body="&lt;p&gt;In general, no. If $R$ is a local noetherian ring, then yes, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counterexample is if $R = k[x,y,z]/(x-1)z$ over a field $k$, and the elements $x, (x-1)y$. This is a regular sequence even though $(x-1)y$ is a zerodivisor (so the swapped sequence is not regular).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true under local hypotheses by the Krull intersection theorem; see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://amathew.wordpress.com/2010/10/30/the-notion-of-a-regular-sequence/&quot; rel=&quot;nofollow&quot;&gt;http://amathew.wordpress.com/2010/10/30/the-notion-of-a-regular-sequence/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-11-29T23:45:50.257" />
  <row Id="12413" PostTypeId="1" AcceptedAnswerId="12415" CreationDate="2010-11-29T23:59:40.840" Score="3" ViewCount="64" Body="&lt;p&gt;Currently taking a logic class and trying to understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two set A and B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sets are empty sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is set A a subset of the complement of set B?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the context is the universal set.&lt;/p&gt;&#xA;" OwnerUserId="4227" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-11-30T00:09:57.003" LastActivityDate="2010-11-30T00:09:57.003" Title="Is there and empty set in the complement of an empty set?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12414" PostTypeId="2" ParentId="12413" CreationDate="2010-11-30T00:06:03.700" Score="1" ViewCount="0" Body="&lt;p&gt;The empty set is a subset of any other set.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-30T00:06:03.700" />
  <row Id="12415" PostTypeId="2" ParentId="12413" CreationDate="2010-11-30T00:09:35.100" Score="7" ViewCount="0" Body="&lt;p&gt;The answer is yes. But there are several comments that need to be made: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is &lt;em&gt;only one&lt;/em&gt; empty set. So it is better to say that $A=B=\emptyset$ rather than saying that &quot;both $A$ and $B$ are empty sets&quot;, as the latter erroneously suggests that there is more than one. This is because two sets are equal precisely if they have the same elements. So any two empty sets are equal, since they have precisely the same elements (namely, none).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I assume by &quot;context&quot; you mean that the complement of $B$ is computed with respect to the &quot;universal set.&quot; In the standard system of set theory, there can be no &quot;universal set&quot;, as assuming its existence leads to problems (Russell paradox). [Though, yes, it is usual to talk of a &quot;universal set&quot; as a way to delineate what objects we are interested in.]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason why $A$ is contained in the complement of $B$ is that $A$ (being the empty set) is a subset of &lt;em&gt;any&lt;/em&gt; set. This is because we define &quot;$A$ is contained in $C$&quot; to mean that any element of $A$ is also an element of $C$. Now, since nothing is an element of $A$, this condition is satisfied in this case (one typically says that it is satisfied &lt;em&gt;vacuously&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-30T00:09:35.100" />
  <row Id="12416" PostTypeId="1" AcceptedAnswerId="12417" CreationDate="2010-11-30T00:13:04.817" Score="4" ViewCount="74" Body="&lt;p&gt;This is a follow up to a &lt;a href=&quot;http://math.stackexchange.com/questions/12042/recurrence-relation-satisfied-by-lfloor1-sqrt5n-rfloor&quot;&gt;question&lt;/a&gt; I had asked earlier about a linear recurrence relationship satsified by $\lfloor(1+\sqrt{5})^n\rfloor$. I messed up there, and I actually meant to ask about $L(n)=\lfloor(1+\sqrt{3})^n\rfloor$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following Douglas' suggestion I have determined that the values (at least the first 1000) satisfy the following recurrence: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(2n+5)=8L(2n+3)-4L(2n+1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how do I prove something like this. I can prove the recurrence for the values inside the floor function, but floor function in general does not commute with addition and multiplication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly, it's easy to show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+\sqrt{3})^{2n+5}=8(1+\sqrt{3})^{2n+3}-4(1+\sqrt{3})^{2n+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I am not sure how to prove the recurrence from here.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-11-30T00:18:19.133" Title="Recurrence relation satisfied by $\lfloor(1+\sqrt{3})^n\rfloor$ " Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" />
  <row Id="12417" PostTypeId="2" ParentId="12416" CreationDate="2010-11-30T00:18:19.133" Score="8" ViewCount="0" Body="&lt;p&gt;Prove the same relation for $1-\sqrt 3$, and check that the resulting powers are all smaller than 1, all negative, and when added to the same powers of $1+\sqrt 3$ you end up with an integer: For all $k=1,2,\dots$, $$(1+\sqrt 3)^k+(1-\sqrt 3)^k $$ is an integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways of checking this fact. For example, by induction. Or by using the Binomial theorem.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-30T00:18:19.133" CommentCount="4" />
  <row Id="12418" PostTypeId="1" CreationDate="2010-11-30T00:23:22.993" Score="0" ViewCount="71" Body="&lt;p&gt;$$\sum_{n=1}^{\infty}\frac{1}{2^n3^{(n-1)}}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4228" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-30T00:30:15.337" LastActivityDate="2010-11-30T00:38:20.147" Title="whether the following series converge or diverge" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12419" PostTypeId="2" ParentId="12418" CreationDate="2010-11-30T00:29:03.877" Score="4" ViewCount="0" Body="&lt;p&gt;Your series can be written as $$\sum_{n=1}^\infty \frac {3}{6^n}.$$&#xD;&#xA;It is a geometric series with $a=3/6=1/2, r=1/6$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-30T00:29:03.877" />
  <row Id="12420" PostTypeId="2" ParentId="12410" CreationDate="2010-11-30T00:31:24.137" Score="3" ViewCount="0" Body="&lt;p&gt;Sketch of solution: Let $X_t$ denote the location of the particle at time $t$, and set $f(t) = {\rm E}(X_t)$. Consider the time $t + \Delta t$, $\Delta t \approx 0+$. With probability of about $1 - \lambda \Delta t$ the particle continues to position $X_t + \Delta t$, whereas with probability of about &#xA;$\lambda \Delta t$ it moves to position of about $X_t/2$. This leads straightforwardly to an elementary differential equation in terms of $f(t)$, from which you obtain $f(t)$. The expected average speed is then $f(t)/t$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-30T00:48:41.180" LastActivityDate="2010-11-30T00:48:41.180" CommentCount="7" />
  <row Id="12421" PostTypeId="2" ParentId="12418" CreationDate="2010-11-30T00:38:20.147" Score="1" ViewCount="0" Body="&lt;p&gt;You could use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_test&quot; rel=&quot;nofollow&quot;&gt;comparison test&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\sum_{n=1}^\infty \frac{1}{2^n 3^{n-1}}&amp;lt;\sum_{n=1}^\infty \frac{1}{2^n}=1.\]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-30T00:38:20.147" />
  <row Id="12422" PostTypeId="2" ParentId="12374" CreationDate="2010-11-30T00:42:29.967" Score="1" ViewCount="0" Body="&lt;p&gt;When you take the Laplace transform and solve for B(s), you should consider &lt;em&gt;both&lt;/em&gt; nonhomogeneous terms in the input:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;B(s) = \frac{1}{s+k} \left (h_a J(s) + \frac{kT_a}{s} \right ).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function is just first factor,  $H(s) = \displaystyle \frac{1}{s+k}$, and the solution $b(t)$ is given by the convolution of $h(t) = e^{-kt}$ with the forcing term $h_a j(t) + k T_a$.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-11-30T00:42:29.967" />
  <row Id="12423" PostTypeId="1" CreationDate="2010-11-30T00:44:48.953" Score="5" ViewCount="75" Body="&lt;p&gt;I am in the process of proving that if a space curve (in R^3) has infinite length and the curvature tends towards 0 as the natural parameter s tends to infinity, the curve must be unbounded - i.e. not contained in any sphere of finite radius. This seems correct intuitively, but I have no guarantee it is correct, unless I am missing something obvious. One way to prove my hunch, I have deduced, is to use a lemma that any curve contained in the open unit ball with curvature always less than one must have a finite upper bound on its length (possibly 2π, but it could be greater for all I know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might one go about proving such an upper bound exists, or if it exists? It might also be nice to know what the bound specifically is, too. I've thought it might be possible to pose this as a variational problem - maximizing length - and then reducing it into a simpler problem, but that appears to be hellishly complicated. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-11-30T03:14:06.033" Title="Supremum length of space curves contained in the open unit ball having always less than unity curvature" Tags="&lt;differential-geometry&gt;&lt;calculus-of-variations&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12424" PostTypeId="2" ParentId="12091" CreationDate="2010-11-30T00:51:32.490" Score="2" ViewCount="0" Body="&lt;p&gt;This is the type of questions to which the answer should be obvious: It depends on what you want to do. If you are interested in practical things like decisions about some outcomes of an experiment or even electrical circuits, then fuzzy logic may be the choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it is a generalization of two-valued boolean logic, which is just a small (and mathematically not that interesting) part of logic. For many applications, you do not even explicitly use truth values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;logic vs. fuzzy logic&quot; is the wrong question.&lt;/p&gt;&#xA;" OwnerUserId="1858" LastActivityDate="2010-11-30T00:51:32.490" />
  <row Id="12425" PostTypeId="1" CreationDate="2010-11-30T01:09:08.080" Score="1" ViewCount="76" Body="&lt;p&gt;{An} is a sequence of events which may NOT be independent. Also A is any event such that P(A) greater than zero. Again the question says nothing about whether {An} is an independent sequence and also it says nothing about whether A or{An} are independent (we assume no independence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we are asked to prove P[An i.o.]=1 iff ∑ P(A ∩ An) = ∞. (Sum goes fron n=1 to infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I have so far. I think if we assume that ∑ P(A ∩ An) = ∞,  That means &#xA;P[(A ∩ An) i.o.]=1, that is the    lim sup (A ∩ An) =1. If I am not wrong,lim sup (A ∩ An) is a subset of  lim sup (An).Therefore P(lim sup (An)) must be greater than or equal to &#xA;P (lim sup (A ∩ An)) and we have proved that P[An i.o.]= P( lim sup An ) = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume P[An i.o.]= P( lim sup An ) = 1, how can we prove this implies ∑ P(A ∩ An) = ∞ ? Once again please remember we cannot use independence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can someone enlighten me on this one please :)&lt;/p&gt;&#xA;" OwnerUserId="4230" LastEditorUserId="4230" LastEditorDisplayName="" LastEditDate="2010-11-30T01:22:28.637" LastActivityDate="2010-11-30T01:29:02.453" Title="Problem Related To Borel Cantelli Lemma and independence" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12426" PostTypeId="1" CreationDate="2010-11-30T01:13:44.897" Score="1" ViewCount="159" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an ESL learner and most of the time I am able to understand English but In my courses, we don't study mathematics, and I think that it is a mistake or wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-What are really the best sources (book or books) to study and review completely high school mathematics?&#xA;-titles&#xA;-authors&#xA;-edition&#xA;-publishers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-What are really the best sources (book or books) to study and review completely college mathematics?&#xA;-titles&#xA;-authors&#xA;-edition&#xA;-publishers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-What are really the best sources (book or books) to study and review completely university mathematics; the basic courses before to study a specialty in a university for example. I ignore how matematics courses work in an American, Canadian or English university.&#xA;-titles&#xA;-authors&#xA;-edition&#xA;-publishers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-What is or are the best calculator or calculators? I mean software applications used to calculate or the usual electronic device used in university courses? I know DreamCalc but are there others? I know what is spreadsheet but is there a better thing that I can use? I know of the IrisPen to develop solutions to math problems using a computer but is there a better thing to use? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sincerely thank you to develop extensively comments and suggest really good books, products, software applications and os on.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-11-30T01:26:19.480" LastActivityDate="2010-12-01T01:39:58.107" Title="I would like to buy references in order to review and study mathematics." Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12427" PostTypeId="2" ParentId="12425" CreationDate="2010-11-30T01:29:02.453" Score="1" ViewCount="0" Body="&lt;p&gt;This is not correct as stated.  Suppose $B$ is an event with probability $1/2$, and let $A$ and all the $A_n$ be equal to $B$.  Then $P(A_n \text{i.o.}) = P(B) = 1/2$ but $\sum_{n=1}^\infty P(A \cap A_n) = \sum_{n=1}^\infty \frac{1}{2} = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sentence beginning &quot;I think if we assume&quot; is not true in general.  It is the so-called second Borel-Cantelli lemma, which requires independence.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-11-30T01:29:02.453" CommentCount="2" />
  <row Id="12428" PostTypeId="1" CreationDate="2010-11-30T02:02:10.550" Score="0" ViewCount="102" Body="&lt;p&gt;Prove that any isomorphism between two cyclic groups always maps every generator to a generator.&lt;/p&gt;&#xA;" OwnerUserId="4179" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T04:22:59.433" LastActivityDate="2010-11-30T04:22:59.433" Title="Isomorphism, Cyclic Groups, and Generators" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12429" PostTypeId="2" ParentId="12428" CreationDate="2010-11-30T02:05:18.543" Score="1" ViewCount="0" Body="&lt;p&gt;Step 1: Show that any homomorphism from a cyclic group is determined by its image on a generator of the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Show that the image of the homomorphism is generated by the image of that generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: What can you say about the kernel of such a homomorphism? What powers of the generator does it send to the identity?&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-30T02:05:18.543" />
  <row Id="12430" PostTypeId="2" ParentId="12428" CreationDate="2010-11-30T02:26:20.640" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose you have two cyclic groups $G$ and $G&amp;#39;$ and some isomorphism $\phi\colon G\to G&amp;#39;$. Since $\phi$ is surjective, for any $g&amp;#39;\in G&amp;#39;$ there exists some $g\in G$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g&amp;#39;=\phi(g).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try expressing $g$ and/or $g&amp;#39;$ in terms of the generators, and see what happens from there. The fact that $\phi$ is a homomorphism is particularly useful here.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-30T02:26:20.640" />
  <row Id="12431" PostTypeId="1" AcceptedAnswerId="12503" CreationDate="2010-11-30T02:37:58.180" Score="0" ViewCount="154" Body="&lt;p&gt;I have deleted the content of the original post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following exercise is inspired by answers to the questions &lt;a href=&quot;http://math.stackexchange.com/questions/11487/y-xn-what-is-exyy&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/11673/ex-1-x-1-x-2-for-independent-gamma-random-variables&quot;&gt;this&lt;/a&gt;, but no knowledge of probability theory is required. The solution is straightforward assuming familiarity with the relevant theorems/properties. The proof is supposed to be informal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\hat \varphi (\omega)$ denote the Fourier transform of a function $\varphi$, and $*$ convolution.&#xA;Express $\hat g(\omega)$ in terms of $\hat f(\omega)$ and a constant $a \in (0,1)$, so that&#xA;$$&#xD;&#xA;(f_2  * g)(t) = at(f * g)(t),&#xD;&#xA;$$ &#xA;where $f_2 (x) = xf(x)$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-30T14:14:50.950" LastActivityDate="2010-12-01T09:08:45.633" Title="A convolution identity (revision of the question &quot;Is this convolution identity known?&quot;)" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;convolution&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12433" PostTypeId="1" AcceptedAnswerId="12450" CreationDate="2010-11-30T02:58:10.820" Score="6" ViewCount="256" Body="&lt;p&gt;Hi,&#xA;Can someone please guide me to a way by which I can solve the following problem.&#xA;There is a die and 2 players. Rolling stops as soon as some exceeds 100(not including 100 itself). Hence you have the following choices: 101, 102, 103, 104, 105, 106. &#xA;Which should I choose given first choice. &#xA;I'm thinking Markov chains, but is there a simpler way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I wrote dice instead of die. There is just one die being rolled&lt;/p&gt;&#xA;" OwnerUserId="2988" LastEditorUserId="2988" LastEditorDisplayName="" LastEditDate="2010-11-30T03:49:15.467" LastActivityDate="2010-12-01T01:26:14.597" Title="Probability of dice sum just greater than 100" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="12434" PostTypeId="1" AcceptedAnswerId="12441" CreationDate="2010-11-30T03:04:30.347" Score="1" ViewCount="92" Body="&lt;p&gt;If $(x_n)$ and $(y_n)$ are positive real sequences such that $(y_n)$ is bounded and $\lim\frac{(x_n)}{(y_n)} = 0$, prove that $\lim(x_n) = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since $(y_n)$ is bounded, there exists a real number $M$ such that for all n natural numbers, $(y_n) &amp;lt; |M|$  &#xA;So by the limit definition :&lt;br&gt;&#xA;$\lim|\frac{(x_n)}{(y_n)}-0| = \lim\frac{(x_n)}{(y_n)} &amp;lt; \epsilon$&#xD;&#xA;So $|(x_n)| &amp;lt; \epsilon(y_n)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to find a natural number $N$ such that for all $n &amp;gt; N$, $|(x_n)| &amp;lt; \epsilon$ is this correct which would prove this statement.  However I'm having trouble picking something that will work here?  Am I on the right track? I have to be very careful also about anything I use to quote the theorem since this is an introduction to real analysis class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T04:55:13.807" LastActivityDate="2010-11-30T04:55:13.807" Title="$(y_n)$ bounded sequence and $\lim\frac{(x_n)}{(y_n)}=0$ show $\lim(x_n) = 0$" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" />
  <row Id="12435" PostTypeId="2" ParentId="12423" CreationDate="2010-11-30T03:04:47.087" Score="1" ViewCount="0" Body="&lt;p&gt;Here is something along these lines, coming from estimates using elementary calculus, but it is much weaker than the lemma you want to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\gamma$ be a regular $C^2$ curve in $\mathbb{R}^3$ parametrized with respect to arclength $s$ (I will assume for simplicity that $s$ starts at $0$).  If the curvature $\|\gamma&amp;#39;&amp;#39;(s)\|$ is always less than $K$ and $\gamma$ is contained in a ball of radius $R\leq\frac{1}{4K}$, then the length of $\gamma$ is less than $\frac{1}{K}(1-\sqrt{1-4KR})$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, write $\gamma(s)=\gamma(0)+s\gamma&amp;#39;(0)+\int_0^s(s-t)\gamma&amp;#39;&amp;#39;(t)dt$ (derived as in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem#Integral_version&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;).  Moving $s\gamma&amp;#39;(0)$ to the left and $\gamma(s)$ to the right and applying the triangle inequality, &#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;s &amp;amp;\leq \|\gamma(s)-\gamma(0)\|+\|\int_0^s(s-t)\gamma&amp;#39;&amp;#39;(t)dt\| \\&#xD;&#xA; &amp;amp;\leq 2R +\int_0^s\|(s-t)\gamma&amp;#39;&amp;#39;(t)\|dt \\&#xD;&#xA; &amp;amp;\lt 2R +K\int_0^s(s-t)dt \\&#xD;&#xA; &amp;amp;=2R+\frac{K}{2}s^2.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this inequality to always hold, $s$ must remain smaller than the smallest root of $\frac{K}{2}x^2-x+2R$.  Thus,  $s\lt\frac{1}{K}(1-\sqrt{1-4KR})$ as claimed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could still apply this to prove the original result as stated, but hopefully someone comes along with a stronger result.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-11-30T03:14:06.033" LastActivityDate="2010-11-30T03:14:06.033" CommentCount="2" />
  <row Id="12437" PostTypeId="2" ParentId="12433" CreationDate="2010-11-30T03:24:06.690" Score="2" ViewCount="0" Body="&lt;p&gt;Maybe I didn't understand the question, but it seems clear, considering $\lbrace 95 + j + k:j = 0, \ldots ,5;k = 1, \ldots 6 \rbrace$, that you should choose $101$. Here, $95+j$ corresponds to the sum just before exceeding $100$ for the first time, and $95+j+k$ to the sum when exceeding $100$ for the first time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate a little on my argument above. First, you can easily show by induction on $j$ that if the current sum is $100 - j$, $j=0,\ldots,4$, then there is no strictly better choice than $101$ (the base case $j=0$ is trivial). Then consider the case where the current sum is $95$, to conclude that $101$ is strictly the best choice.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-11-30T08:30:07.790" LastActivityDate="2010-11-30T08:30:07.790" CommentCount="8" />
  <row Id="12438" PostTypeId="2" ParentId="12434" CreationDate="2010-11-30T03:26:38.350" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|y_{n}|\leq M$ then $1 \leq \frac{M}{|y_{n}|} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$| x_{n}| \leq |x_{n}| \cdot \frac{M}{|y_{n}|} = |\frac{x_{n}}{y_{n}}| \cdot M &amp;lt; \varepsilon_{0} \cdot M = \frac{\varepsilon}{M} \cdot M = \varepsilon$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $|\frac{x_{n}}{y_{n}}| &amp;lt; \varepsilon_{0}$ *&lt;em&gt;for hypotesis&lt;/em&gt;* and choosing $\varepsilon_{0} = \frac{\varepsilon}{M}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-30T04:13:11.950" LastActivityDate="2010-11-30T04:13:11.950" CommentCount="8" />
  <row Id="12439" PostTypeId="1" AcceptedAnswerId="12452" CreationDate="2010-11-30T03:41:14.323" Score="5" ViewCount="97" Body="&lt;p&gt;Let $H=\left\{ z\in\mathbb{C}\mid\Im\left(z\right)&amp;gt;0\right\}$  be the upper-half Poincare plane. Let $GL\left(2,\mathbb{R}\right)$ be the general linear group, $Z\left(GL\left(2,\mathbb{R}\right)\right)$ be the center of the general linear group and $O\left(2,\mathbb{R}\right)$ be the orthogonal subgroup of $GL\left(2,\mathbb{R}\right)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean to say $H=GL\left(2,\mathbb{R}\right)/\left(Z\left(GL\left(2,\mathbb{R}\right)\right)\cdot O\left(2,\mathbb{R}\right)\right)$? The left-hand side is a metric space and the right hand side is a set of cosets of $GL\left(2,\mathbb{R}\right)$. So I'm confused about what it means to write that they are equal or to say &quot;the upper half plane is...&quot; It seems like this would be the group of orientation preserving isometries of H, but I still find the terminology confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to figure out what this could possibly mean, but my searches on the internet have not been fruitful. I've also looked at 2 sources on standard modular groups but they make no mention of this fact. An explanation or reference would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I am reading a paper titled &quot;On Modular Functions in characteristic p&quot; by Wen-Ch'ing Winnie Li which can be found at &lt;a href=&quot;http://www.jstor.org/stable/1997973&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/stable/1997973&lt;/a&gt;. The claim appears on page 3 of the pdf (page 232 of the journal). It is also stated on the wikipedia page: &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T04:24:22.840" LastActivityDate="2010-11-30T05:26:58.993" Title="What does $H=GL(2,\mathbb{R})/(Z(GL(2,\mathbb{R}))\cdot O(2,\mathbb{R}))$ mean?" Tags="&lt;geometry&gt;&lt;group-theory&gt;&lt;reference-request&gt;" AnswerCount="2" />
  <row Id="12440" PostTypeId="2" ParentId="12439" CreationDate="2010-11-30T04:03:59.797" Score="1" ViewCount="0" Body="&lt;p&gt;H is not a group, but rather a space with a group acting on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If K ≤ G is a subgroup, then G/K is not usually a group, but rather a set of cosets on which G acts.  G/K is only a group if K is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your situation, the subgroup is not normal, so you only get a group action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One keyword form the wikipedia article is &quot;isotropy subgroup&quot;.  If G acts transitively on a space H, then the set K of elements of G that fix some particular point of H forms a subgroup (the &quot;isotropy subgroup&quot;), and the action of G on H is isomorphic to the action of G on the cosets in G/K.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-11-30T04:03:59.797" CommentCount="5" />
  <row Id="12441" PostTypeId="2" ParentId="12434" CreationDate="2010-11-30T04:07:25.533" Score="4" ViewCount="0" Body="&lt;p&gt;I would approach this by contradiction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sequence does not converge to 0, then there is a subsequence that is bounded away from 0. So we may fix $a&amp;gt;0$ such that $|x_n|&amp;gt;a$ for infinitely many values of $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sequence of $y_n$ is bounded, say by $M$, we have $|y_n|&amp;lt;M$ for all $n$, so if $n$ is one of the indices in our subsequence, we have $|x_n/y_n|&amp;gt;a/M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This inequality holds for infinitely many $n$, and we have found a subsequence of $(x_n/y_n)$ that does not converge to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense? &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-11-30T04:07:25.533" CommentCount="5" />
  <row Id="12442" PostTypeId="1" CreationDate="2010-11-30T04:08:33.717" Score="8" ViewCount="155" Body="&lt;p&gt;I am trying to get a grasp on what a representation is, and a professor gave me a simple example of representing the group Z_12 as the twelve roots of unity, or corresponding 2x2 matrices. Now I am wondering how GL(1,C) and GL(2,R) are related, since the elements of both groups are automorphisms of the complex numbers. GL(C), the group of automorphisms of C, is (to my understanding) isomorphic to both GL(1,C) and GL(2,R) since the complex numbers are a two-dimensional vector space over R. But it doesn't seem like these two groups are isomorphic to each other. &lt;/p&gt;&#xA;" OwnerUserId="4238" LastActivityDate="2010-11-30T22:29:31.530" Title="How is GL(1,C) related to GL(2,R)?" Tags="&lt;group-theory&gt;&lt;lie-groups&gt;&lt;representation-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="12444" PostTypeId="1" CreationDate="2010-11-30T04:16:18.093" Score="0" ViewCount="37" Body="&lt;p&gt;I posted this question in MO (Mathoverflow) and didn't get an answer, so I try my luck here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $C$ be a nonempty bounded closed convex subset, say the unit ball, of $\ell^2(\mathbb{N})$. Let $T: C\to 2^C$ be a map such that $T(x)$ is nonempty closed for each $x$, and that $$D(Tx,Ty)\le \|x-y\|$$ for all $x,y\in C$, where $D$ is the Hausdorff metric defined by $$D(A,B)=\inf\lbrace r&amp;gt;0: N_r(A)\supset B, N_r(B)\supset A\rbrace,$$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the $r$-neighborhood of $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether $T$ has a fixed point, i.e. a point $x\in C$ such that $x\in Tx$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes if $T$ is a contraction, i.e. replacing the $\le \|x-y\|$ by $\le \lambda \|x-y\|$ for some $0\le\lambda&amp;lt;1$ in the definition; or if $Tx$ is compact for each $x$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-11-30T04:16:18.093" Title="Multi-valued nonexpansive maps" Tags="&lt;geometry&gt;&lt;analysis&gt;&lt;functional-analysis&gt;&lt;metric&gt;" />
  <row Id="12445" PostTypeId="2" ParentId="12442" CreationDate="2010-11-30T04:22:46.940" Score="7" ViewCount="0" Body="&lt;p&gt;No, they're not isomorphic. $GL(1,\mathbb{C})$ injects into $GL(2, \mathbb{R})$ but the latter consists of the set of &lt;em&gt;real&lt;/em&gt; automorphisms of $\mathbb{R}^2$, while the former consists of the set of all &lt;em&gt;complex&lt;/em&gt; automorphisms of $\mathbb{C}$. For instance, an element of $GL(1, \mathbb{C})$ is the composite of a scaling (dilation) by a positive real number and a rotation, so is $\mathbb{R}_{&amp;gt;0} \times SO(2) = \mathbb{R}_{&amp;gt;0} \times S^1$. However, $GL(2, \mathbb{R})$ includes more complicated things (e.g. reflections, things which even if scaled do not become orthogonal transformations).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-30T12:52:26.117" LastActivityDate="2010-11-30T12:52:26.117" CommentCount="6" />
  <row Id="12446" PostTypeId="2" ParentId="12442" CreationDate="2010-11-30T04:23:20.933" Score="4" ViewCount="0" Body="&lt;p&gt;$GL(1,\mathbb{C})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a complex vector space.  Each such automorphism is given by multiplication by a nonzero complex number. On the other hand, $GL(2,\mathbb{R})$ can be thought of as the group of automorphisms of $\mathbb{C}$ as a &lt;em&gt;real&lt;/em&gt; vector space, and hence it is larger.  In terms of matrices, $GL(1,\mathbb{C})$ corresponds to all $2$-by-$2$ real matrices of the form &#xA;$\begin{bmatrix}&#xD;&#xA;a &amp;amp; -b \\&#xD;&#xA;b &amp;amp; a&#xD;&#xA;\end{bmatrix}$ with $a^2+b^2\gt0$, which is a proper subgroup of the group $GL(2,\mathbb{R})$ of all $2$-by-$2$ real invertible matrices.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-11-30T04:23:20.933" CommentCount="3" />
  <row Id="12447" PostTypeId="1" AcceptedAnswerId="12448" CreationDate="2010-11-30T04:30:35.970" Score="4" ViewCount="41" Body="&lt;p&gt;Hello, I started reading about uniform spaces in Bourbaki, and the closure operation has been defined as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{A}=\bigcap_{V\in\mathcal{U}}V(A)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mathcal{U}$ is the uniform structure of the space, and $V(A)$ is the set of all left-relatives of $A$. One thing I don't understand is why $\bar{A}=\bar{\bar{A}}$, as is necessary for a closure operation. In particular, I don't see why $\bar{\bar{A}}\subseteq\bar{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After playing around with it for a while, I have this. Take some $s\in\bar{\bar{A}}$. So $s\in V(\bar{A})$ for all $V\in\mathcal{U}$. So taking any $V$, there is some $r\in\bar{A}$ such that $(s,r)\in V$. But $r\in V(A)$, so $(r,t)\in V$ for some $t\in A$. Is there some why to show that $(s,t)\in V$, to conclude that $s\in V(A)$? At best I can see that $(s,t)\in V\circ V$. &lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-30T04:43:45.740" Title="Closure Operation on a Uniform Topology" Tags="&lt;topology&gt;" AnswerCount="1" />
  <row Id="12448" PostTypeId="2" ParentId="12447" CreationDate="2010-11-30T04:36:49.727" Score="3" ViewCount="0" Body="&lt;p&gt;For any $V\in {\cal U}$ choose $U\in {\cal U}$ such that $U\circ U\subset V$. Then you proceed with your proof above with $U$ (instead of $V$).&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-30T04:43:45.740" LastActivityDate="2010-11-30T04:43:45.740" CommentCount="1" />
  <row Id="12449" PostTypeId="2" ParentId="12426" CreationDate="2010-11-30T04:38:00.173" Score="1" ViewCount="0" Body="&lt;p&gt;Hi&#xA;1) &quot;High School&quot; Math really varies from region to region, but if you aim is the review for University, you should get any book on analytic geometry, or &quot;college&quot; algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,3) Calculus and Linear algebra make up the bulk of first year University Math.  Any old edition of &quot;James Stewart Calculus&quot; should suit you fine.  I think David Lay's &quot;Linear algebra and it's applications&quot; is also pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Most University math is done without a calculator, but some students purchase the computer software &quot;Maple&quot; as an aid.  Maple is easier to learn than other programs like Mathamatica, and you shouldn't need anything more advanced than this for intro college/university&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-11-30T04:38:00.173" CommentCount="1" />
  <row Id="12450" PostTypeId="2" ParentId="12433" CreationDate="2010-11-30T04:40:46.490" Score="5" ViewCount="0" Body="&lt;p&gt;My attempt: a combination of my comment and Shai's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A partition means a partition of n using only 1,2,3,4,5,6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 101: Number of partitions of 101.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 102: partitions of 96 + partitions of 97... + partitions of 100. Since we can have 96+6, 97+5, ...100+2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 106: Partitions of 100. Since we can get to 106 only by adding to 100 then rolling 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A partition of n will be the nth coefficient x in $\prod_{k=0}^6 \frac{1}{1-x^k}$ using the geometric series. But an easier observation is that $a_n = a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+a_{n-5}+a_{n-6}$. Using this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 101 is  $a_{101}=a_{100}+a_{99}+a_{98}+a_{97}+a_{96}+a_{95}$&#xD;&#xA;Number of ways to get to 102 is $a_{100}+a_{99}+a_{98}+a_{97}+a_{96}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of ways to get to 106 is $a_{100}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a_n &amp;gt; 0$ we see that 101 will occur the most frequently.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-11-30T04:40:46.490" />
  <row Id="12451" PostTypeId="1" CreationDate="2010-11-30T05:25:30.740" Score="3" ViewCount="254" Body="&lt;p&gt;Can someone please share references (websites or books) where I can find problems related with Fubini's theorem and applications of Radon-Nikodym theorem? I have googled yes and don't find many problems. What are the &quot;typical&quot; problems (if there are any) related with these topics? [Yes, exam is coming soon so I don't know what to expect and don't have access to midterms from previous years].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2011-01-06T17:32:20.957" Title="Typical applications of Fubini's theorem and Radon-Nikodym" Tags="&lt;measure-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="12452" PostTypeId="2" ParentId="12439" CreationDate="2010-11-30T05:26:58.993" Score="5" ViewCount="0" Body="&lt;p&gt;My comments were getting too long, so I will post this as an answer: the isomorphism referred to is an isomorphism of $G$-sets. This works in great generality: let $H$ be a set, let $G$ act on $H$ transitively. Pick an arbitrary point $z\in H$ and let $K=\text{Stab}_G(z)$ be the point stabiliser in $G$. Note that $K$ is not normal in general, since the group $gKg^{-1}$ stabilises the point $g(z)$ (my action is on the left). So, $K$ is normal if and only if it acts trivially on $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the set of cosets $G/K$ is always a $G$-set, i.e. a set with an action of $G$:&#xA;$$g: hK\mapsto (gh)K$$&#xD;&#xA;for all $g\in G$ and $hK\in G/K$. This is the usual coset action. Now, check that the map&#xA;$$\phi:G/K\rightarrow H,\;gK\mapsto g(z)$$&#xD;&#xA;is a bijection of $G$-sets, i.e. a bijection of sets that respects the $G$-action.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-11-30T05:26:58.993" CommentCount="2" />
  <row Id="12453" PostTypeId="1" AcceptedAnswerId="12474" CreationDate="2010-11-30T06:38:39.760" Score="10" ViewCount="148" Body="&lt;p&gt;I heard that using some relatively basic differential geometry, you can show that the only spheres which are Lie groups are $S^0$, $S^1$, and $S^3$.  My friend who told me this thought that it involved de Rham cohomology, but I don't really know anything about the cohomology of Lie groups so this doesn't help me much.  Presumably there are some pretty strict conditions we can get from talking about invariant differential forms -- if you can tell me anything about this it will be a much-appreciated bonus :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A necessary condition for a manifold to be a Lie group is that is must be parallelizable, since any Lie group is parallelized (?) by the left-invariant vector fields generated by a basis of the Lie algebra.  Which happens to mean, by some pretty fancy tricks, that the only spheres that even have a chance are the ones listed above plus $S^7$.  The usual parallelization of this last one comes from viewing it as the set of unit octonions, which don't form a group since their multiplication isn't associative; of course this doesn't immediately preclude $S^7$ from admitting the structure of a Lie group.  Whatever.  I'd like to avoid having to appeal to this whole parallelizability business, if possible.)&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-11-30T08:45:21.293" LastActivityDate="2010-11-30T14:02:13.187" Title="Is there an easy way to show which spheres can be Lie groups?" Tags="&lt;differential-geometry&gt;&lt;differential-topology&gt;&lt;lie-groups&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="12454" PostTypeId="1" AcceptedAnswerId="12455" CreationDate="2010-11-30T06:53:01.767" Score="1" ViewCount="78" Body="&lt;p&gt;On a 2D plane, how to construct a graph with 7 vertexes and 21 edges? I tried various combination but couldn't seem to draw that kind of contrived graph on a paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my understanding is that &lt;a href=&quot;http://mathoverflow.net/questions/30759/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-bases-with-each-an/30767#30767&quot; rel=&quot;nofollow&quot;&gt;it is possible&lt;/a&gt;. So anyone can help me with it?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-11-30T20:23:48.337" Title="Construct a Graph with 7 Vertex, 21 Edges on 2D Plane" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="12455" PostTypeId="2" ParentId="12454" CreationDate="2010-11-30T07:20:01.227" Score="5" ViewCount="0" Body="&lt;p&gt;My guess is you're after $K_7$ drawn on the torus; see &lt;a href=&quot;http://www.amotlpaa.org/math/k7torus.html&quot; rel=&quot;nofollow&quot;&gt;http://www.amotlpaa.org/math/k7torus.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer you question as to why $K_5$ is not planar:  If a complete graph is planar, then for every $K_3$ subgraph, either every vertex (that's not part of the $K_3$) is inside the $K_3$ or outside the $K_3$ (otherwise there's a crossing edge from outside to inside the $K_3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we attempt to draw $K_5$ vertex-by-vertex, then we first draw a triangle $K_3$.  We can place the 4-th vertex either inside or outside of the triangle.  In either case, the drawing obtained will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vSihQ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, wherever you put the 5-th vertex, you will form some $K_3$ subgraph for which the other two vertices are not both inside and not both outside.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-30T07:42:24.270" LastActivityDate="2010-11-30T07:42:24.270" CommentCount="2" />
  <row Id="12456" PostTypeId="2" ParentId="12451" CreationDate="2010-11-30T07:25:49.127" Score="1" ViewCount="0" Body="&lt;p&gt;Radon-Nikodym is used to prove the existence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conditional_expectation#Formal_definition&quot; rel=&quot;nofollow&quot;&gt;conditional expectation&lt;/a&gt; in probability theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fubini's theorem is, among other things, a very useful device to compute integrals over product spaces.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-30T08:05:58.640" LastActivityDate="2010-11-30T08:05:58.640" CommentCount="3" />
  <row Id="12457" PostTypeId="2" ParentId="12433" CreationDate="2010-11-30T07:43:40.307" Score="1" ViewCount="0" Body="&lt;p&gt;101 is the most probable stop state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operation &quot;replace the final toss of the dice by the one that gives a sum of 101&quot; is a probability-conserving transformation on the set of possible games.  It does not cover the whole set (i.e. the image of the transformation is a subset) of possible games stopping at 101, such as the ones with a transition from 95 to 101 by a roll of 6.  This shows that all the terminal states higher than 101 have a smaller probability than that of finishing at 101. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-30T22:26:17.523" LastActivityDate="2010-11-30T22:26:17.523" />
  <row Id="12458" PostTypeId="1" CreationDate="2010-11-30T07:50:47.630" Score="-2" ViewCount="82" Body="&lt;p&gt;Suppose $f(n)$ is $2n$.&#xA;Suppose $g_i(n)$ is $n+i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we have a recurrence given by $a_{2n} = c_0 a_n + c_1 a_{n+1} + \dots c_m a_{n+m}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) What is known about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) What type of recurrence is this?  (My guess is that this is still a linear recurrence, since it is based linearly on parts of the series.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Where are there references and information on recurrences such as this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) What, if anything, remains unknown about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we simplify this to $a_{2n} = a_n + n$.  Then what is known?&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-11-30T08:22:32.270" LastActivityDate="2010-11-30T09:36:29.490" Title="Concerning $a_{f(n)} = \displaystyle\sum_{i=0}^m{c_i a_{g_i(n)}}$" Tags="&lt;sequences-and-series&gt;&lt;intuition&gt;&lt;recursive&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12459" PostTypeId="1" AcceptedAnswerId="12491" CreationDate="2010-11-30T08:01:01.063" Score="3" ViewCount="52" Body="&lt;p&gt;I have a network, with $V$ vertex, $E$ edges and $C$ cycles, which has &quot;current&quot; flowing in and out from the network, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1Micr1JiaNxGw3FTF-vJOR3h94IjNPjq1mQqEBS7QkGY&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each edge has a current value associated to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations governs the current flow in the network are the so-called &quot;current conservation theorem&quot;, which means that all the current flowing into a node is equal to the current flowing outside from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the above network, such equations can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{in}=e1+e2    \quad\quad \text{(at v1)}$$  &#xA;$$e3+e13=e7+e5+e6   \quad\quad \text{(at v4)}$$&#xD;&#xA;$$e8+e10=e11+Q_{out2} \quad\quad \text{(at v8)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at every vertex. the node location and the value of $Q_{outi}$ are given beforehand, but only the location of $Q_{in}$ are given, the value of $Q_{in}$ is not given. $e_i$ is not known as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing to note is that from the above sets of equation, one can deduce that, for any kinds of graph system, all the incoming external sources &lt;strong&gt;must&lt;/strong&gt; equal to the outgoing external sources, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{in}=Q_{out1}+Q_{out2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way one can deduce the value of $Q_{in}$ quite easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to note is that in a $V$ vertexes network with $E$ edges, the total number independent equations one can form from the above requirements are $V-1$. This is because if you manipulate the above equations for any kinds of graph, you will always find that there are only $V-1$ independent equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the number of current value that one can arbitrary set is $E-(V-1)$. Recall that $C=E-V+1$, one can conclude that &lt;em&gt;the number of cycles corresponds to the number of indeterminate variables&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words this is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Indeterminate_system&quot; rel=&quot;nofollow&quot;&gt;indeterminate system&lt;/a&gt; with $E$ variables but $V-1$ independent equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, out of all the $e1, e2, e3$ and so on, how to choose which $e_{i}$ for arbitrary value assignment ( so that the system will be reduced to a &lt;em&gt;unique system&lt;/em&gt;, with $V-1$ variables and $V-1$ independent equations)? My hunch is that I will use all the edges that are not inside the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spanning_tree&quot; rel=&quot;nofollow&quot;&gt;spanning tree&lt;/a&gt; of the graph and set their current value arbitrarily. But I can't prove ( or disprove) this. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-11-30T15:14:09.423" LastActivityDate="2010-12-02T10:36:30.170" Title="How to Set the Redundant Current Value in a Network?" Tags="&lt;linear-algebra&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12460" PostTypeId="1" AcceptedAnswerId="12461" CreationDate="2010-11-30T08:10:41.627" Score="3" ViewCount="84" Body="&lt;p&gt;Is Knopp's &lt;em&gt;Theory and Application of Infinite Series&lt;/em&gt; out of date? It's looks terrific to me, but the Dover edition I bought new maybe a year ago: &lt;a href=&quot;http://preview.tinyurl.com/2eprqps&quot; rel=&quot;nofollow&quot;&gt;http://preview.tinyurl.com/2eprqps&lt;/a&gt; seems to be the same as an edition published in 1951 and may go back as far as 1921. 60 or 90 years is a lot of math years. How about it? Does my book leave out some important developments? Is it old-fashioned in some other ways?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen this question: &lt;a href=&quot;http://math.stackexchange.com/questions/6845/what-is-the-current-state-of-the-art-in-methods-of-summing-exotic-series&quot;&gt;what is the current state of the art in methods of summing &quot;exotic&quot; series?&lt;/a&gt; but it doesn't have a full answer yet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-11-30T08:14:56.833" Title="Is Knopp's &quot;Theory and Application of Infinite Series&quot; out of date? " Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="12461" PostTypeId="2" ParentId="12460" CreationDate="2010-11-30T08:14:56.833" Score="0" ViewCount="0" Body="&lt;p&gt;Konrad Knopp's books are nice to read and the results still holds. I remember some strange &quot;hunt&quot; for a &quot;convergence border&quot; in that book, which is a bit absurd to me..&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-11-30T08:14:56.833" CommentCount="1" />
  <row Id="12462" PostTypeId="1" AcceptedAnswerId="12497" CreationDate="2010-11-30T08:25:43.217" Score="6" ViewCount="263" Body="&lt;p&gt;I've never actually done a delta-epsilon proof, so I thought I'd try my hand at one. I decided to try it out for $f(x)=1/x$. If I understand correctly from the wikipedia article, I want to show for any $\epsilon&amp;gt;0$, there exists a $\delta&amp;gt;0$ such that if $|x-c|&amp;lt;\delta$, then $|f(x)-f(c)|&amp;lt;\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I noticed that I want something like $|f(x)-f(c)|=|1/x-1/c|=\frac{|x-c|}{|xc|}&amp;lt;\epsilon$. So $|x-c|&amp;lt;|xc|\epsilon$, which looks similar to the fact that I want $|x-c|&amp;lt;\delta$. However, I've also heard that one is never supposed to let $\delta$ depend on $x$. Is this the right direction? How would I use this information to find a corresponding $\delta$ for each $\epsilon$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-11-30T20:23:11.010" Title="Is $f(x)=1/x$ continuous on $(0,\infty)$?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="12463" PostTypeId="1" AcceptedAnswerId="12467" CreationDate="2010-11-30T08:30:39.873" Score="3" ViewCount="46" Body="&lt;p&gt;Let $Z$ be a random variable with finite mean and consider the function $F: R \rightarrow R$ defined by $$ F(x) = E[\max(x+Z,0)]$$ How to prove that $F(x)$ is a continuous function of $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not homework. A textbook I am reading states that $F(x)$ is continuous without providing any explanation, and I can't seem to fill the gap. I would be happy with a hint. &lt;/p&gt;&#xA;" OwnerUserId="4244" LastEditorUserId="4244" LastEditorDisplayName="" LastEditDate="2010-11-30T08:50:19.473" LastActivityDate="2010-11-30T09:50:57.533" Title="Continuity of $E[\max(x+Z,0)]$ as a function of $x$" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="12464" PostTypeId="2" ParentId="12462" CreationDate="2010-11-30T08:45:37.620" Score="2" ViewCount="0" Body="&lt;p&gt;Hi, your proof looks ok.  Delta can depend on epsilon and on c in the definition of &quot;continuity&quot;.  A function is called uniformly continuous if you can proof that given epsion, the required value of delta depends on epsilon but NOT c.  1/x is NOT uniformly continuous on (0,1) which is why you can't get around the fact that delta depends on c.  1/x would be uniformly continous on other intervals however (intervals where f' is bounded).&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T15:04:15.263" LastActivityDate="2010-11-30T15:04:15.263" CommentCount="5" />
  <row Id="12465" PostTypeId="1" AcceptedAnswerId="12471" CreationDate="2010-11-30T08:46:25.597" Score="0" ViewCount="35" Body="&lt;p&gt;This is a slightly different question than the one asked &lt;a href=&quot;http://math.stackexchange.com/questions/1340/in-a-graph-is-it-always-possible-to-construct-a-set-of-cycle-basis-with-each-an&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for non-planar graph there are edges in the graph which has more than 2 cycles sharing on it. But for &lt;a href=&quot;http://en.wikipedia.org/wiki/Planar_graph&quot; rel=&quot;nofollow&quot;&gt;planar graph&lt;/a&gt;, can we prove that &lt;em&gt;it is always possible to find a complete set of cycle basis that each edge is shared by at most 2 cycles?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-11-30T11:26:52.927" Title="For a Planar Graph, is it always possible to construct a set of cycle basis, with each and every edge Is shared by at most 2 cycle bases?" Tags="&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="12466" PostTypeId="2" ParentId="12462" CreationDate="2010-11-30T08:51:19.753" Score="2" ViewCount="0" Body="&lt;p&gt;The statement in your first paragraph would be the definition of uniform continuity (I interpret what you wrote as &quot;for all $\epsilon$, there exists a $\delta$ such that for all $x$...&quot;). For a related question, see &lt;a href=&quot;http://math.stackexchange.com/questions/11538/intuition-for-uniform-continuity-of-a-function-on-mathbbr&quot;&gt;Intuition for uniform continuity of a function on $\mathbb{R}$&lt;/a&gt;. Uniform continuity implies continuity, but is strictly stronger, and in fact, the function at hand is continuous on $(0,\infty)$, but not uniformly continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement of continuity would be that for all $\epsilon$ and all $x$ in the range, there exists a $\delta$ such that... So $\delta$ depends on $\epsilon$ *and* on $x$. Intuitively: the greater the slope at $x$, the closer you will have to get to $x$ for the function values to be close. &lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-30T08:56:26.097" LastActivityDate="2010-11-30T08:56:26.097" CommentCount="6" />
  <row Id="12467" PostTypeId="2" ParentId="12463" CreationDate="2010-11-30T08:53:23.590" Score="3" ViewCount="0" Body="&lt;p&gt;If $\lvert x-y\rvert\leq\delta$, then $\lvert (x+Z)-(y+Z)\rvert\leq\delta$, and also $$\lvert \mathrm{max}(x+Z,0)-\mathrm{max}(y+Z,0)\rvert\leq\delta,$$&#xD;&#xA;right? (Notice that $\mathrm{max}(\bullet+Z,0)$ is 1-Lipschitz.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we also have $\lvert F(x)-F(y)\rvert\leq\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Remark: In order for $\mathrm{max}(x+Z,0)$ to be well-defined, we should assume that $Z$ is bounded, I guess.&lt;/s&gt; (According to the comments this is not neccessary.)&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-11-30T09:50:57.533" LastActivityDate="2010-11-30T09:50:57.533" CommentCount="3" />
  <row Id="12468" PostTypeId="2" ParentId="12458" CreationDate="2010-11-30T09:17:09.390" Score="0" ViewCount="0" Body="&lt;p&gt;To ensure that the equations define a nonzero, finite dimensional space of sequences, some condition on the functions $f$ and $g_i$ is needed.  If they are too fast-growing the sequence will be underconstrained and if they are slow-growing it will be overdetermined.  To avoid excessive generality such as a trivial encoding of the $3x+1$ problem (or, I think, the halting problem) as a statement about a recurrence of this type, the functions should be of an easy to analyze kind, such as polynomial or linear.  In that case asymptotic analysis should be possible but the details will depend on the individual recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit: I see the question has changed to include specific functions $f(n)=2n$ and $g_i=n+i$.  In that case it should be true that $a_n$ has relatively smooth asymptotics, such as $n^u (\log n)^b$, once it is well-defined, and this could be substituted in the recurrence to get the approximation $a_{2n} = Ca_n$ where $C = \Sigma c_i$, so that $u$ can be calculated and the procedure &quot;bootstrapped&quot; to get more precise information.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-11-30T09:36:29.490" LastActivityDate="2010-11-30T09:36:29.490" CommentCount="2" />
  <row Id="12469" PostTypeId="1" CreationDate="2010-11-30T10:28:15.360" Score="0" ViewCount="44" Body="&lt;p&gt;Question: (a) Determine each of the absolute conditions of the linear map $f: \mathbb{R} \rightarrow \mathbb{R}:$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$f(x) = |x|$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$f(x) = \frac{\pi}{2}$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(b) Provide both an example of a well conditioned and an ill conditioned function evaluation (both the function $f$ and the argument $x$). Do not use any of the examples from (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt: so although this seems really simple I'm unfortunately having trouble with exactly what to do. The given definition of absolute condition $\kappa_{abs}$ is the smallest number for which: $|f(x_0) - f(x)| \leq \kappa_{abs}|x_0 - x| + o(|x_0 - x|)$. However in the one example I have it seems that the last term with $o$ is ignored (I assume it is too small...). I was thinking that perhaps for (a) 2. I could write something like: $|f(x_0) - f(x)| = |\frac{\tilde{\pi}}{2} - \frac{\pi}{2}|= \frac{1}{2}|\tilde{\pi}-\pi| \Rightarrow \kappa_{abs} = \frac{1}{2}$ ? With 1. (a) I am don't know which sort of answer would be expected...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) I am guessing it must be measured in terms of relative condition, right? Since it appears that the rule is condition(not specified which) greater than 1 is ill conditioned. I know that subtraction of almost equal numbers does not do well. I thought a way to guarantee that the function would always subtract two such numbers would to be to let $f(x) = x - \frac{99x}{100}$ using the formula for relative condition of subtraction($\kappa = \frac{|x|+|y|}{|x-y|}$) I would have $$\frac{|x| + |\frac{99x}{100}|}{|x| - |\frac{99x}{100}|} = 199 &amp;gt; 1$$ $\Rightarrow$ This function is ill-conditioned ? For the well conditioned function I have the theorem: for $x,y &amp;gt; 0$ $$\frac{|(x+y) - (\tilde{x} + \tilde{y})|}{|x+y|} \leq 1\epsilon \Rightarrow \kappa = 1$$ but I don't understand how to use that here since it seems self explanatory that a positive number divided by a larger positive number is less than 1. Can I simply substitue any positive number for $y$ and let my function be $f(x) = x^{2} +10$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As if it weren't obvious I am very lost with this and any help, tips, advice, etc is as usual greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2010-11-30T12:32:21.680" LastActivityDate="2010-11-30T12:32:21.680" Title="Determine absolute conditions of mappings, provide examples of ill/well conditioned functions" Tags="&lt;numerical-methods&gt;" CommentCount="5" />
  <row Id="12470" PostTypeId="2" ParentId="12348" CreationDate="2010-11-30T10:52:22.457" Score="0" ViewCount="0" Body="&lt;p&gt;The numbers in question here, of course, can be computed exactly.  For example, using &lt;a href=&quot;http://gmplib.org/&quot; rel=&quot;nofollow&quot;&gt;bignum&lt;/a&gt; or &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt; (or even &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; -- the exactly link won't work on here, but I'm sure you can type in &quot;(100 choose 50)*2^(-100)&quot; yourself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my home computer, in GAP, it took less than a millisecond.  To make it more interesting, I also computed ${100000 \choose 50000} \cdot 2^{-100000}$, which took a bit more than 17 seconds.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; Binomial(100,50)/2^100;&#xA;12611418068195524166851562157/158456325028528675187087900672&#xA;gap&amp;gt; time;&#xA;0&#xA;gap&amp;gt; Binomial(100000,50000)/2^100000;&#xA;&amp;lt;&amp;lt;an integer too large to be printed&amp;gt;&amp;gt;/&amp;lt;&amp;lt;an integer too large to be printed&amp;gt;&amp;gt;&#xA;gap&amp;gt; time;&#xA;17266&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, provided the coin has probability 1/2, the probability will always have a terminating decimal expansion (since binomial coefficients are integers, and 2 divides 10).  Here it is in this case:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.0795892373871787614981270502421704614029315404247333213573478705171737601631321012973785400390625&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-30T12:15:03.573" LastActivityDate="2010-11-30T12:15:03.573" CommentCount="1" />
  <row Id="12471" PostTypeId="2" ParentId="12465" CreationDate="2010-11-30T11:26:52.927" Score="1" ViewCount="0" Body="&lt;p&gt;This is of course true, with the basis cycles being boundaries of the bounded faces (after killing off all pairs {edge e, reverse of edge e}). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you looking for a rigorous proof? It should follow from the argument establishing the fact that there are such things as faces etc (see for example chapter 4 &lt;a href=&quot;http://diestel-graph-theory.com/basic.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). How much of relevant topology are you willing to assume? &lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-11-30T11:26:52.927" CommentCount="4" />
  <row Id="12472" PostTypeId="1" CreationDate="2010-11-30T11:47:54.887" Score="0" ViewCount="162" Body="&lt;p&gt;Can this expression be simplified?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{i=0}^n (k^i)/i!  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;This expression I have got as the number of possible ways to select n objects or less from k different infinite objects (you can select as many as you can from any type)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it must be equal to the number of possible ways to select n objects from k+1 different infinite objects, where the extra +1 is a dummy object, selecting one of this type means letting an empty space in the final selection (i.e. selecting n-1 not n)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so this must be equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (k+1)^n/n! $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why that is not right??&lt;/p&gt;&#xA;" OwnerUserId="896" LastEditorUserId="896" LastEditorDisplayName="" LastEditDate="2010-11-30T16:33:09.223" LastActivityDate="2010-11-30T16:33:09.223" Title="Can this be simplified?" Tags="&lt;combinatorics&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="12474" PostTypeId="2" ParentId="12453" CreationDate="2010-11-30T12:11:06.713" Score="12" ViewCount="0" Body="&lt;p&gt;Here is the sketch of the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a compact connected Lie group G.  Let's break into 2 cases - either $G$ is abelian or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ is abelian, then one can easily show the Lie algebra is abelian, i.e., $[x,y]=0$ for any $x$ and $y$ in $\mathfrak{g}$.  Since $\mathbb{R}^n$ is simply connected and has the same Lie algebra as $G$, it must be the universal cover of $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $G$ is a sphere, it's $S^1$, since all the others are simply connected, and hence are their own universal covers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we move onto the case where $G$ is nonabelian.  For $x,y,$ and $z$ in the lie algebra, consider the map $t(x,y,z) = \langle [x,y], z\rangle$.  This map is clearly multilinear.  It obviously changes sign if we swap $x$ and $y$.  What's a bit more surprising is that it changes sign if we swap $y$ and $z$ or $x$ and $z$.  Said another way, $t$ is a 3 form!  I believe $t$ is called the Cartan 3-form.  Since $G$ is nonabelian, there are some $x$ and $y$ with $[x,y]\neq 0$.  Then $t(x,y,[x,y]) = ||[x,y]||^2 \neq 0$ so $t$ is not the 0 form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, use left translation on $G$ to move $t$ around: define $t$ at the point $g\in G$ to be $L_{g^{-1}}^*t$, where $L_{g^{-1}}:G\rightarrow G$ is given by $L_{g^{-1}}(h) = g^{-1}h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differential 3-form is automatically left invariant from the way you've defiend it everywhere.  It takes a bit more work (but is not too hard) to show that it's also right invariant as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next one argues that a biinvariant form is automatically closed.  This means $t$ defines an element in the 3rd de Rham cohomology of $G$.  It must be nonzero, for if $ds = t$, then we may assume wlog that $s$ is biinvariant in which case $ds = 0 = t$, but $t$ is not $0$ as we argued above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for a nonabelian Lie group, $H^3_{deRham}(G)\neq 0$.  But this is isomporphic to singluar homology.  Hence, for a sphere to have a nonabelian Lie group stucture, it must satisfy $H^3(S^n)\neq 0$.  This tells you $n=3$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-30T14:02:13.187" LastActivityDate="2010-11-30T14:02:13.187" CommentCount="15" />
  <row Id="12476" PostTypeId="1" AcceptedAnswerId="12482" CreationDate="2010-11-30T12:57:59.400" Score="3" ViewCount="58" Body="&lt;p&gt;This is just a completely random question of the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be the set of bijections $\phi:\mathbb{N} \cup \{0\} \rightarrow \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can imagine any $\phi \in \Omega$ to be equivalent to a wacky &quot;staircase&quot;, where for $x \in \mathbb{R}^+$ we have $x \mapsto \phi(\lfloor x \rfloor)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we walk along this staircase, the biggest step we need to take is of size $\sup_{n \geq 0} |\phi(n+1)-\phi(n)|$.  Suppose we want to make the steps as small as possible.  How small can we make them?  That is, what is \[\inf_{\phi \in \Omega} \sup_{n \geq 0} |\phi(n+1)-\phi(n)|\ \ ?\]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-30T14:01:49.270" Title="Which bijection from the natural numbers to the rationals has the smallest steps?" Tags="&lt;real-analysis&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12477" PostTypeId="1" AcceptedAnswerId="12483" CreationDate="2010-11-30T13:12:27.553" Score="5" ViewCount="126" Body="&lt;p&gt;Many introductory books on vector spaces mention that the scalars need not be reals, and might even have sections discussing complex vector spaces or vector spaces over the integers mod 2.  I have never seen any such book mention that all of the theory goes through as well if one restricts the scalars to be just rational numbers.  Perhaps this is because there is a dearth of interesting problems about such vector spaces accessible at this level that couldn't simply be discussed in the context of real scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there is an interesting introductory-level problem or topic about vector spaces that would be most naturally conducted by allowing rational number scalars.  Does anyone know of such, perhaps one with a number-theoretic aspect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By introductory:  I envision a first course on linear algebra, including non-math majors.  They would be seeing vector spaces (and that level of abstraction) for the first time.  Perhaps they would be seeing matrix multiplication for the first time.  Usually, in my experience, such courses primarily use the real numbers as scalars.)&lt;/p&gt;&#xA;" OwnerUserId="4253" LastEditorUserId="4253" LastEditorDisplayName="" LastEditDate="2010-11-30T15:24:26.533" LastActivityDate="2010-11-30T16:43:27.057" Title="Does anyone know an interesting introductory topic involving vector spaces over the rationals" Tags="&lt;linear-algebra&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="12478" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T13:26:22.097" Score="4" ViewCount="0" Body="&lt;p&gt;An abelian group is torsionfree and divisible iff it has the (unique) structure of a vector space over $\mathbb{Q}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this observation the classification of vector spaces over a field in terms of their dimension gives rise to a classification of torsionfree divisible abelian groups.  In particular, since if $V$ is a vector space over a field $K$ with $\operatorname{dim} V&amp;gt; \# K$ we have $\operatorname{dim} V = \# V$, we find that any two uncountable torsionfree divisible abelian groups of the same cardinality are isomorphic.  In particular $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as abelian groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;b&gt;Added&lt;/b&gt;: In model-theoretic terms, we have established that the elementary theory of torsionfree divisible abelian groups is &lt;strong&gt;uncountably categorical&lt;/strong&gt;.  Since it has no finite models, by Vaught's Test this theory is complete.  See &lt;a href=&quot;http://www.math.uga.edu/~pete/modeltheory2010Chapter4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~pete/modeltheory2010Chapter4.pdf&lt;/a&gt;, especially Theorem 6.  (This is one of the most inexpensive examples of a complete theory that I know.)  Now I feel honorbound to advise you &lt;em&gt;not&lt;/em&gt; to mention this in a linear algebra class!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether one would or could want to introduce any of this in a linear algebra class is highly dubious (a graduate-level linear algebra class, perhaps).  However, I can't help but recall that Paul Halmos was very fond of the conclusion in the last paragraph and apparently liked to include as a problem on his exams: &quot;Can $\mathbb{R}$ be given the &#xA;structure of a vector space over $\mathbb{C}$?&quot;  He describes this in his autobiography. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am a number theorist, no number-theoretic applications are springing to mind.  A little voice in my head is saying &quot;Hilbert 90&quot;, but that's probably not a direction you want to go in either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may well know this already, but there are lots of cool combinatorial problems which -- in a highly non-obvious way! -- can be solved using linear algebra over finite fields, especially $\mathbb{F}_2$.  I highly recommend L. Babai's manuscript &quot;Linear algebra methods in combinatorics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-30T13:39:45.843" LastActivityDate="2010-11-30T13:39:45.843" CommentCount="1" />
  <row Id="12479" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T13:32:47.147" Score="2" ViewCount="0" Body="&lt;p&gt;The question is kind of general, but here is an attempt at an answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In the general introductory books on vector spaces, one usually tries to develop as much of the theory as possible without specifying the field of scalars at all. This has a good reason: the main advantage in talking about finite dimensional vector spaces rather than tuples of numbers is that one gets rid of baggage, as it were, and leaves only the essential parts of a theory. The same reasoning suggests to not specify the field. This way, the relevant structure becomes visible and is not obstructed by unnecessary details. There is no need to specify the field, if you want to define vectors, linear maps, bases, matrices, change of basis, determinants etc. In the usual order of teaching these topics, the first place when it becomes necessary to specify the field is when one starts talking about inner products, because to define non-degeneracy, one needs a notion of positivity. To sum up point one: the reason that books rarely say &quot;by the way, this works also for the rationals&quot; is that most books state and prove as much as possible by starting &quot;Let $V$ be a vector space over a field $F$...&quot; They don't feel that they have to remind the reader that $\mathbb{Q}$ is one such field. Indeed, there are soooo many others out there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are however areas neighbouring linear algebra where the field of scalars becomes important and in many of them, $\mathbb{Q}$ is one of the fields that receive special attention. One example is: classical representation theory is usually first developed over $\mathbb{C}$. But when all the necessary theory is in place, one starts wondering about fields of definition. Questions like &quot;which representations are defined over $\mathbb{Q}$&quot; and &quot;how many irreducible rational representations does a finite group have&quot; are very natural and very non-trivial.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since your thoughts are already going in that direction, I will also mention, that some theories such as number theory are even more interested in free $\mathbb{Z}$-modules than in vector spaces. Those are even more special (and difficult) than rational vector spaces. Such beasts arise naturally in algebraic number theory. For example Galois groups of finite extensions of number fields naturally act on many free $\mathbb{Z}$-modules, such as rings of integers, units therein (modulo torsion) and many more. Determining the structure of these actions is a very active area of research, but you will have difficulty introducing that in an early undergraduate course. Sometimes, determining the integral structure is so difficult that one is forced to tensor with $\mathbb{Q}$ and turn these modules into rational vector spaces, which brings us back to where you started from.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So to summarise all this: the very basic theory of linear algebra tries to be oblivious to the field itself. The more advanced theory is very much interested in the field of definition. $\mathbb{C}$ tends to be the easiest to understand, then comes $\mathbb{R}$, the rationals are much more difficult than these two. So I guess, the reason that you don't see $\mathbb{Q}$ often in introductory undergraduate books is that, as soon as the field actually matters, $\mathbb{Q}$ is a difficult one to work with.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-11-30T13:50:42.007" LastActivityDate="2010-11-30T13:50:42.007" CommentCount="6" />
  <row Id="12480" PostTypeId="2" ParentId="12476" CreationDate="2010-11-30T13:58:12.100" Score="4" ViewCount="0" Body="&lt;p&gt;If $\phi$ is a bijection, so is $r \phi$ for any $r \in \mathbb{Q}$, so if the supremum is ever finite then the infimum you ask for is zero.  And it's not hard to describe a bijection with bounded step size:  number the rationals in $[-1, 1]$ with denominator at most $2$ from left to right, then weave back and number the rationals in $[-2, 1]$ with denominator at most $4$ that you haven't already, then weave back and number the rationals in $[-2, 2]$ with denominator at most $8$ that you haven't already... and so forth.  The step size here, I think, is always bounded by $\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-30T13:58:12.100" />
  <row Id="12481" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T14:01:19.037" Score="3" ViewCount="0" Body="&lt;p&gt;One nontrivial example of vector spaces over $\mathbb{Q}$ being used to solve a problem is the construction of the Dehn invariant in the resolution of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_third_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's third problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  After reading your comments on Alex Bartel's answer, I guess this is not really &quot;introductory level&quot; as there are two nontrivial $\mathbb{Q}$-vector spaces involved and one has to tensor them... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think if you want to impress on your students that stuff like the rank-nullity theorem goes through over an arbitrary field, you should be looking for combinatorial applications over $\mathbb{F}_2$.  If your students can digest the definitions, these are very elegant examples of just how general the theory is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, much of the theory covered in a standard textbook - eigenvalues, eigenvectors - does &lt;em&gt;not&lt;/em&gt; go through over an arbitrary field.  Perhaps this will all just be unnecessarily confusing.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-11-30T14:15:36.087" LastActivityDate="2010-11-30T14:15:36.087" CommentCount="4" />
  <row Id="12482" PostTypeId="2" ParentId="12476" CreationDate="2010-11-30T14:01:49.270" Score="3" ViewCount="0" Body="&lt;p&gt;I would claim the limit is zero.  First, here is a version with the step under 2.  Partition $\mathbb{R}^+$ into sets $(m,m+1]$ for $m\in\mathbb{N}$.  There are countably many rationals in each interval, so order them in type $\omega$.  Now let $\phi(0)$ be the first item in (0,1], $\phi(1)$ the first item in (1,2], $\phi(2)$ the second item in (0,1], $\phi(3)$ the second item in (1,2], $\phi(4)$ the first item in (2,3], etc..  We step up through the intervals, going one higher each run, then back down to (0,1], and start up again.  No step is longer than twice an interval.  Now we can shrink the intervals as small as we wish.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-11-30T14:01:49.270" CommentCount="3" />
  <row Id="12483" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T14:09:28.543" Score="4" ViewCount="0" Body="&lt;p&gt;One standard example of using $\mathbb{Q}$-vector spaces is to construct discontinuous additive functions $f: \mathbb{R} \to \mathbb{R}$. The approach is to choose a $\mathbb{Q}$-basis for $\mathbb{R}$. For each $x \in \mathbb{R}$, define $f(x)$ to be the sum of the coefficients in this basis. This is linear but discontinuous (it maps into $\mathbb{Q}$).&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-11-30T14:59:16.517" LastActivityDate="2010-11-30T14:59:16.517" CommentCount="2" />
  <row Id="12484" PostTypeId="2" ParentId="12472" CreationDate="2010-11-30T14:30:47.830" Score="1" ViewCount="0" Body="&lt;p&gt;In response to the edited question, it is not right.  If we just let n=2, $1+k+\frac{k^2}{2}\neq \frac{(k+1)^2}{2}$.  The difference is $\frac{1}{2}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-11-30T15:13:14.800" LastActivityDate="2010-11-30T15:13:14.800" CommentCount="1" />
  <row Id="12485" PostTypeId="1" AcceptedAnswerId="12490" CreationDate="2010-11-30T14:31:25.957" Score="4" ViewCount="120" Body="&lt;p&gt;I've run across the recurrence $a_{n+1} = (a_n)^2 + 1$ in the past.  Unfortunately, the referrence escapes me.  However, my impression was that recurrences involving the product of previous terms (such as $a_{n+1} = (a_n)(a_{n-1})$) are difficult to solve.  I'm wondering what is known for this very general problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)  Is there a known way to solve recurrences involving the product of previous terms?  Or, what is known about these?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2)  What are these recurrences called?  Do they have a general name (such as non-linear recurrences)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3)  Where can I find more literature on the subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4)  Who are some experts that have dealt with this?&lt;/p&gt;&#xA;" OwnerUserId="2626" LastActivityDate="2010-12-16T01:24:43.107" Title="What's known about recurrences involving $(a_n)^2$?" Tags="&lt;sequences-and-series&gt;&lt;recursive&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="12486" PostTypeId="1" AcceptedAnswerId="12496" CreationDate="2010-11-30T14:31:53.077" Score="5" ViewCount="81" Body="&lt;p&gt;A general problem for the Calculus of Variations asks us to minimize the value of a functional $A[f]$, where $f$ is usually a differentiable function defined on $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if, however, the domain of $A$ is not actually all differentiable functions. Suppose there is a constraint equation on $f$, such as (for example):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L[f] = \int_{-1}^1 \sqrt{1 + f&amp;#39;(x)^2} dx = \pi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we want to minimize over functions satisfying the above and the property that $f(-1)=f(1)=0$ the functional&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A[f] = \int_{-1}^1 f(x) dx $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of problem seems to me to be very similar to the problem in multivariate calculus of minimizing a function $f(x)$ with respect to a constraint equation $g(x) = 0$. In this case we are trying to minimize a functional $A[f]$ with respect to a functional constraint equation $L[f] = \pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former, one can use Lagrange multipliers to reduce the problem to that of solving a system of equations. Is there such a technique for the variational version?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-30T15:25:50.587" LastActivityDate="2010-11-30T15:25:50.587" Title="Calculus of Variations and Lagrange Multipliers" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;calculus-of-variations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12487" PostTypeId="1" CreationDate="2010-11-30T14:34:47.747" Score="1" ViewCount="73" Body="&lt;p&gt;There are presumably three ways of characterizing the abstract Euclidean space $E^n$ that are quite different in spirit:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;axiomatically (with axioms concerning dimension)&lt;/li&gt;&#xA;&lt;li&gt;by the abstract Euclidean group $E(n)$ (as its symmetry group, determining $E^n$ uniquely)&lt;/li&gt;&#xA;&lt;li&gt;by presupposing a metric and requiring &lt;em&gt;that the space is a maximal one with respect to the property&lt;/em&gt; that the $(n+1)$-dimensional Cayley-Menger determinant vanishes for all $(n+2)$-tuples of points &lt;em&gt;and does not vanish for all $k$-tuples of points &quot;in general position&quot; for $k &amp;lt; n+2$&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[Having tried to &quot;rescue&quot; 3 by adding &lt;em&gt;conditions in italics&lt;/em&gt;, due to Robin's comment.]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Is it correct, actually, that $E^n$ is uniquely&#xA;  determined via 2 and 3?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; What are other ways of&#xA;  characterizing $E^n$ that are quite different in spirit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditorDisplayName="" LastEditDate="2010-11-30T15:23:43.617" LastActivityDate="2010-11-30T15:23:43.617" Title="Characterizations of Euclidean space" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" CommentCount="1" />
  <row Id="12488" PostTypeId="1" AcceptedAnswerId="13069" CreationDate="2010-11-30T14:37:48.863" Score="4" ViewCount="132" Body="&lt;p&gt;In a planar graph $G$, one can easily find all the cycle basis by first finding the spanning tree ( &lt;em&gt;any&lt;/em&gt; spanning tree would do), and then use the remaining edge to complete cycles. Given Vertex $V$, edge $E$, there are $C=E-V+1$ number of cycles, and there are $C$ number of edges that are inside the graph, but not inside the spanning tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;a href=&quot;http://math.stackexchange.com/questions/12465/for-a-planar-graph-is-it-always-possible-to-construct-a-set-of-cycle-basis-with&quot;&gt;there always exists a set of cycle basis such that each and every edge inside the $G$ is shared by at most 2 cycles&lt;/a&gt;. My question is, is there any algorithm that allows me to find such a set of cycle basis? The above procedure I outlined only guarantees to find a set of cycle basis, but doesn't guarantee that all the edges in the cycle basis is shared by at most two cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: Coordinates for each vertex are &lt;em&gt;not&lt;/em&gt; known, even though we do know that the graph &lt;em&gt;must&lt;/em&gt; be planar.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;del&gt;Note: All the edges are of the same weight.&lt;/del&gt; (deleted for not relevant)&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-01T12:24:58.440" LastActivityDate="2010-12-05T07:10:16.833" Title="For a Planar Graph, Find the Algorithm that Constructs A Cycle Basis, with each Edge Shared by At Most 2 Cycles " Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="12489" PostTypeId="1" AcceptedAnswerId="12494" CreationDate="2010-11-30T14:44:04.150" Score="2" ViewCount="90" Body="&lt;p&gt;(Pun definitely intended.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear MSE-Community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to choose one artist that has made interesting works of art not only because of their beauty but also because of their connections to Mathematics, I would choose Escher. His works have always intrigued me. Some of his paintings are mind-boggling when looked at for a long time, but I have the feeling that it can be described accurately with mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare some drawings. In the first and second drawing ((1),(2)), the artist chooses to depict some simple geometrical objects in one- and two point perspective respectively. Although the objects seem to float in space, both the images look &quot;all right&quot; to me. These drawing are, however, though not 'Eschers', illusory too. The brain somehow creates 3D space from 2D space, but that's more of a biological issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Point Perspective&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ilLt5.gif&quot; alt=&quot;alt text&quot;&gt;  (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two Point Perspective&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7HxIg.gif&quot; alt=&quot;alt text&quot;&gt;  (2)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sub-question 1: How does one describe these seemingly &quot;sound&quot; drawings mathematically? How do (1) and (2) compare to one another?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, lets get to to part I find most interesting, Escher's etchings, prints and lithographs. When I look at the following pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- M.C. Escher 1960 lithograph Ascending and Descending&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ViM3c.jpg&quot; alt=&quot;alt text&quot;&gt;  (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- M.C. Escher 1953 Relativity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FeUTO.gif&quot; alt=&quot;alt text&quot;&gt;  (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognize that these are two different types of paradoxes because Escher plays with perspective in two different ways. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sub-question 2: How could the difference between these and other visual paradoxes of artists (mostly Escher, but I guess there are a lot more artists that mimic and extend his style) be formalized with the aid of mathematics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Muller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm sorry these images are all of different sizes and some are too large. I'm a bit in a hurry so I didn't make them equally large. &lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-11-30T15:06:25.473" LastActivityDate="2010-11-30T15:09:25.460" Title="Can we put the mathematics of paradoxes in visual art into perspective?" Tags="&lt;visualization&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="1" />
  <row Id="12490" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T14:47:34.863" Score="11" ViewCount="0" Body="&lt;p&gt;$a_{n+1} = a_n a_{n-1}$ happens to be quite easy to solve: if $b_n = \log a_n$ then $b_n$ satisfies the Fibonacci recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only recurrences of the form $a_{n+1} = a_n^2 + c$ that have anything like a closed form solution occur when $c = 0$ (obvious) and when $c = -2$ (where the solution is given by $a_n = 2 \cos 2^n \theta$ where $\theta$ is determined by $a_0$).  For all other values of $c$ you get very complicated behavior, and the situation for more general nonlinear recurrences is even worse.  People study questions like these in &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_system&quot; rel=&quot;nofollow&quot;&gt;dynamical systems&lt;/a&gt; and related areas; one keyword you might use to start your exploration is &lt;a href=&quot;http://en.wikipedia.org/wiki/Julia_set&quot; rel=&quot;nofollow&quot;&gt;Julia set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like many other classes of sufficiently general problems in mathematics, the problem is hopelessly complicated in general and people study interesting special cases.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-30T14:47:34.863" />
  <row Id="12491" PostTypeId="2" ParentId="12459" CreationDate="2010-11-30T14:53:22.897" Score="2" ViewCount="0" Body="&lt;p&gt;Update: Upon a bit of reflection I decided that the original answer is unnecesarily long. What is really being computed is the homology of the graph. Let $C(V)$ be the vector space (over real numbers, since the flows are real) freely generated by the vertices, and $C(E)$ the vector space freely generated by the edges. We have the map $\delta: C(E) \mapsto C(v)$ which assigns to a flow (thought of as a weighted set of edges) the tuple of net flows through vertices. The question is what is it's kernel, which is of course the $H_1$ of the graph an is in fact exactly the subspace of cycles in $C(E)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See  the last paragraph for interpretation of what follows in this language.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you can certainly take any spanning tree $T$. The point is that any edge in the complement of $T$ can be completed to a cycle using edges in $T$ in a unique way, and the resulting cycles form a basis of the cycle space. This is your ambiguity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general statement is: Flows on the graph with prescribed sinks and sources are in bijection with the linear maps on the cycle space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument is as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need two facts:&#xA;Fact 1) If the flow around any cycle is zero, then the flow actually comes from a potential - that is a function $f$ on vertices, such that flow on edge e is the difference of the value of $f$ on its head and its tail. This $f$ is uniquely determined by the flow up to an overall scalar (assuming the graph is connected; start at any vertex and any value of $f$ for it and fololow edges assigning the unique possible values for $f$ as you go; the flow around cycle being zero gurantees that you never assign contradictory values). We will show that this potential is also uniquely determined, up to overall scalar, by the net flow in each vertex, and as a consequence the corresponding flow on edges is uniquely defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact 2) If the net flow at each vertex is zero (no sources or sinks) then the flow is uniquely determined by a function on the cycle space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of Fact 1. Consider the map $Fl$ sending a potential $f$ to the set of net flows at each vertex. $Fl$ is linear, and constant potentials are in the kernel. I claim that this is the whole kernel. Indeed, if there is a vertex with a maximal potential, it will have a negative net flow from it, unless all its neighbours have the same potential. As the graph is connected, the only potentials with no flow at any vertex are constant ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Fl$ is a map between vector spaces of the same dimension. As the kernel of $Fl$ is one dimensional, so is the cokernel.  Note that $Fl$ lands in the subset of all tuples of net flows with the zero sum (in the summ of all net flows each edge is counted twice, once with plus and once with minus). Hence $Fl$ hits any net flow tuple with total flow zero, and any such tuple of net flows comes from unique potential, up to an overall scalar, as wanted. End of proof of Fact 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof of Fact 2: Given any linear function $G$ on the cycle space, we can define the flow $E=F(G)$ on each edge $e$ to be the sum of $G(c)$ for cycles $c$ with $e$ in $c$. This of course lands in the flows with zero net flow at each vertex (this is clear by considering what happens to a dual function to any cycle $c$ and the fact that the assignment $F$ is linear). Observe that there is a basis $c_k$ of the cycle space and edges $e_k$ with $e_k$ in $c_k$ and no $c_l$ for $l \neq k$ (the edges in the complement of the spanning tree will do), so $F$ is injective. Finally, dimension count shows $F$ is an isomorphism. End of proof of Fact 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we fix a collection of sources and sinks, this determines a unique flow $E$ which is zero on cycles by Fact 1. The difference of any flow $E&amp;#39;$ with these sinks and cycles and $E$ has zero net flow on each vertex, and hence corresponds to a linear function on the cycle space by Fact 2. The end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of original answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the long argument above does is as folows: One can view flows as cochains instead of chains. That is, since $C(E)$ comes with a basis, it has a prefered isomorphism $I_e$ to $\operatorname{Hom}(C(E), \mathbb{R})$. Similarly there is a prefered isomorphism $I_v$ from $C(v)$ to $\operatorname{Hom}(C(V), \mathbb{R})$. The map $\delta$ has the dual $\delta ^*:\operatorname{Hom}(C(V), \mathbb{R}) \mapsto \operatorname{Hom}(C(E), \mathbb{R})$ sending a potential (thought of as a map from vertices to $\mathbb{R}$) to its flow (there should be a (non-commutative) diagram here). Fact 1 above is that the composition $\delta \cdot I_c \cdot \delta^*$ is an isomorphism from $\operatorname{Hom}(C(V), \mathbb{R})/\operatorname{constants}$ to $\operatorname{Im} \delta$. Fact 2 is that the kernel of $\delta$ is the cycle space (or its dual). All of this presumably in some standard book. &lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-12-02T10:36:30.170" LastActivityDate="2010-12-02T10:36:30.170" CommentCount="7" />
  <row Id="12492" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T14:57:34.720" Score="2" ViewCount="0" Body="&lt;p&gt;If we have a sequence $a_{n+1} = f(a_n, ..., a_{n-k})$ where $f$ is a quadratic polynomial in $k$ variables, then we can use this recurrence relation to write down a &quot;simplified&quot; recurrence relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{n+1} = g(a_{n-k})$ where $g$ is a polynomial of even degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a new sequence $b_j = a_{(k+1) \cdot j}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $b_{j+1} = a_{kj + k + j + 1} = g(a_{(k+1)j}) = g(b_j)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that if $a_n$ converges, then so does $b_n$ and $\lim a_n = \lim b_n$. Denote $\lim b_n$ by $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $g$ is continuous,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{n \rightarrow \infty} b_n = \lim_{n \rightarrow \infty} g(b_{n+1}) = g( \lim_{n \rightarrow \infty} b_n )$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $g(b) = b$, so the limit of the sequence $a_n$ must be one of the fixed points of $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the fixed points of $g(x)$, one must find the roots of the polynomial $g(x) - x$ (of which there are only finitely many), and to solve this problem one can use a variety of methods, like for example Newton's method.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2010-11-30T15:21:54.147" LastActivityDate="2010-11-30T15:21:54.147" CommentCount="2" />
  <row Id="12493" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T14:59:23.063" Score="2" ViewCount="0" Body="&lt;p&gt;Sometime ago I considered the sequence of fractions defined by $a_1 = 1/2$, $a_{n+1} = (a_n^2 + 1)/2$, in some probabilistic context; see &lt;a href=&quot;http://oeis.org/A167424&quot; rel=&quot;nofollow&quot;&gt;OEIS seqence A167424 &lt;/a&gt;. This sequence is closely related to the one defined by $a_1 = 1/2$, $a_{n+1}=a_n-a_n^2$, see &lt;a href=&quot;http://oeis.org/A076628&quot; rel=&quot;nofollow&quot;&gt;OEIS sequence A076628&lt;/a&gt;, whose asymptotics as $n \to \infty$ was analyzed &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/99/somos&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-16T01:24:43.107" LastActivityDate="2010-12-16T01:24:43.107" />
  <row Id="12494" PostTypeId="2" ParentId="12489" CreationDate="2010-11-30T15:09:25.460" Score="3" ViewCount="0" Body="&lt;p&gt;Question 1: this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Projective_geometry&quot; rel=&quot;nofollow&quot;&gt;projective geometry&lt;/a&gt;.  In this context the horizon line is known as the line at infinity, and the drawings illustrate that &quot;parallel lines&quot; in the projective plane meet at the line at infinity while non-parallel lines meet at a finite point and intersect the line at infinity in different points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: I don't know that you have to do anything sophisticated to describe what is going on in Ascending and Descending.  The drawing suggests the existence of four heights $h_1, h_2, h_3, h_4$ (the heights of the corners) such that $h_1 &amp;gt; h_2 &amp;gt; h_3 &amp;gt; h_4 &amp;gt; h_1$, and this is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is paradoxical about Relativity?  As far as I can tell, that room is buildable.  Maybe I don't understand what you are trying to ask here.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-30T15:09:25.460" CommentCount="2" />
  <row Id="12496" PostTypeId="2" ParentId="12486" CreationDate="2010-11-30T15:18:39.097" Score="5" ViewCount="0" Body="&lt;p&gt;Yes. For the general theorem, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multipliers_on_Banach_spaces&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt;. In the particular case which you consider, you can actually consider the minimization problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{-1}^1 f(x) + \lambda (\sqrt{ 1 + f&amp;#39;(x)^2} - \pi) dx $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leads to the Euler-Lagrange equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \left( \frac{f&amp;#39;}{\sqrt{1 + (f&amp;#39;)^2}}\right)&amp;#39; = \lambda^{-1} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leads to a somewhat unappealing looking Lagrangian with $f&amp;#39;&amp;#39;$ in the denominator when you plug $\lambda$ back in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is, of course, not completely unexpected, as your functional $A$ is not bounded below on the set $L[f] = \pi$. To see this, it suffice to note that $L[f + c] = L[f]$ by definition, while $A[f + c] = A[f] + c$. So for a decreasing sequence of $c\searrow -\infty$, $A[f + c]$ decreases without bound.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-30T15:18:39.097" CommentCount="2" />
  <row Id="12497" PostTypeId="2" ParentId="12462" CreationDate="2010-11-30T15:21:00.873" Score="10" ViewCount="0" Body="&lt;p&gt;There have been some confusing comments regarding dependence on $x$ or on $c$, so let me try to put it all together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that $\delta$ should not depend on $x$. However, when one is proving that $f(x)$ is continuous &lt;em&gt;at $c$&lt;/em&gt;, then $\delta$ is allowed to depend on both $\epsilon$  &lt;em&gt;and&lt;/em&gt; $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the definition: $f(x)$ is continuous &lt;em&gt;at $c$&lt;/em&gt; if and only for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that for all $x$, if $|x-c|\lt \delta$, then $|f(x)-f(c)|\lt \epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the existence of $\delta$ is mentions &lt;em&gt;before&lt;/em&gt; $x$ ever comes into the picture? That's an indication that $\delta$ cannot depend on $x$. On the other hand, both $f(x)$, $\epsilon$, and $c$ occur &lt;em&gt;before&lt;/em&gt; $\delta$, which means that, absent any indication to the contrary, $\delta$ is allowed to depend on $f(x)$ (obviously), on $\epsilon$, &lt;em&gt;and&lt;/em&gt; on $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here, you cannot pick $\delta=|xc|\epsilon$, because that would make $\delta$ depend on $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to get around it is to get rid of the dependence on $x$. The key here is that since we are trying to make sure everything works if $x$ is &quot;close enough&quot; to $c$, then we will also have that $|x|$ will be very close to $|c|$. So we should be able to to control that division by $x$ in the expression $\frac{|x-c|}{|xc|}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How? Well, if a particular $\delta_0$ works, then any smaller one will work as well. So we can always shrink $\delta$ a bit more if needed. Wo the first thing we can note is that we can always require that $\delta$ be smaller than both $1$ and than $\frac{c}{2}$; that is, we will require $\delta\lt\min\{1,\frac{c}{2}\}$. Why $1$? Because then I know that $c-1\lt x \lt c+1$; if $c-1\gt 0$, then this means that $\frac{1}{c+1}\lt \frac{1}{x} \lt \frac{1}{c-1}$, so we can &quot;control&quot; the value of $\frac{1}{x}$. Why less than $\frac{c}{2}$? Just in case $c-1\lt 0$. So let $\mu=\min\{1,\frac{c}{2}\}$. Then we can conclude that $\frac{1}{x}\lt \frac{1}{c-\mu}$. (We could get away with simply putting $\delta\lt\frac{c}{2}$; restricting it to less than $1$ is a common practice, though, which is why I put it here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by requiring that $\delta\lt\min\{1,\frac{c}{2}\}$, we guarantee that $\frac{1}{|x|}\lt \frac{1}{c-\mu}$ (remember that we are working on $(0,\infty)$). What do we gain by this? Well, look:&#xA;$$|f(x)-f(c)| = \left|\frac{1}{x}-\frac{1}{c}\right| = \left|\frac{c-x}{xc}\right| = |x-c|\frac{1}{c}\cdot\frac{1}{x} \leq |x-c|\frac{1}{c(c-\mu)}.$$&#xD;&#xA;So if we &lt;em&gt;also&lt;/em&gt; ask that $\delta\lt c(c-\mu)\epsilon$, then we have:&#xA;$$|f(x)-f(c)|\leq |x-c|\frac{1}{c(c-\mu)} \lt \frac{\delta}{c(c-\mu)} \lt \frac{c(c-\mu)\epsilon}{c(c-\mu)} = \epsilon$$&#xD;&#xA;which is what we want!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, what do we need? We need to make sure that $\delta$ is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Less than $1$;&lt;/li&gt;&#xA;&lt;li&gt;Less than $\frac{c}{2}$; (both of these to ensure $\frac{1}{x}\lt\frac{1}{c-\mu}$);&lt;/li&gt;&#xA;&lt;li&gt;Less than $c(c-\mu)\epsilon$, where $\mu=\min\{1,\frac{c}{2}\}$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, for instance, we can just let $\delta = \frac{1}{2}\min(1,\frac{c}{2},c(c-\mu)\epsilon)$, where $\mu=\min\{1,\frac{c}{2}\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you can let your $\delta$ depend only on $f(x)$ and on $\epsilon$ but not $c$, then we say $f(x)$ is &lt;em&gt;uniformly continuous&lt;/em&gt;. This is a stronger condition than continuity, and often very useful. $\frac{1}{x}$ is not uniformly continuous on $(0,\infty)$, however (though it is on $[a,\infty)$ for any $a\gt 0$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T20:23:11.010" LastActivityDate="2010-11-30T20:23:11.010" CommentCount="4" />
  <row Id="12498" PostTypeId="2" ParentId="12485" CreationDate="2010-11-30T15:21:05.617" Score="2" ViewCount="0" Body="&lt;p&gt;You may want to take a look at Kelley and Peterson's textbook [1]. The problem you are looking at here is a recurrence relation as you have stated. You also hear them called difference equations. If you are familiar with differential equations, these difference equations are the discrete analog of differential equations [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kelley and Peterson go very in depth with solving different types of difference equations. They give a nice overview of several different methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you! If you have any other questions, feel free to email me or contact me through my blog &lt;a href=&quot;http://www.tylerclark12.com/blog&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Weisstein, Eric W. &quot;Recurrence Equation.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/RecurrenceEquation.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/RecurrenceEquation.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-30T15:21:05.617" />
  <row Id="12500" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T15:30:10.837" Score="0" ViewCount="0" Body="&lt;p&gt;I just thought of one that is at about the level I envisioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let V be a two-dimensional Q-subspace of the Q-vector space R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)  Show that if V is closed under multiplication of real numbers, and if the nonzero vectors in V are closed under division of real numbers, then V has a basis of the form {1, \alpha} where alpha is a root of a quadratic polynomial whose coefficients are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)  Show that under the same hypotheses as in part (a), V has a basis of the form {1, \sqrt{N}} where N is a positive integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)  Show that if V has a basis of the form {1,\sqrt{N}} where N is a positive integer, then V is closed under multiplication of real numbers and the nonzero vectors of V are closed under division of real numbers.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2010-11-30T15:30:10.837" />
  <row Id="12501" PostTypeId="1" AcceptedAnswerId="12517" CreationDate="2010-11-30T16:18:42.870" Score="-1" ViewCount="106" Body="&lt;p&gt;For a constant, &lt;code&gt;N&lt;/code&gt;, what value of &lt;code&gt;x&lt;/code&gt; will maximize the cosine (or any trig) function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;1 = \cos{(Nx)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for the exact form, not the approximation because,&#xA;\begin{equation}&#xA;\frac{\arccos{(1)}}{N} = x = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1+%3D+cos%2819.013+%2a+x%29&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha.com&lt;/a&gt; states that if &lt;code&gt;N = 19.013&lt;/code&gt;, then,&#xA;\begin{equation}&#xA;x = \frac{2000 \pi n}{19013} , n \text{ } \varepsilon \text{ } \text{set of integers}&#xA;\end{equation}&#xA;How was that solution calculated?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2010-11-30T18:55:06.283" Title="Calculate exact solution to maximize cosine" Tags="&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="12503" PostTypeId="2" ParentId="12431" CreationDate="2010-11-30T16:28:40.380" Score="2" ViewCount="0" Body="&lt;p&gt;I may be messing up my usage of Fourier transforms and convolutions with distributions, but I think this can be solved with a few identities (namely how transforms work on products and convolutions, and how convolutions work with derivatives and Dirac deltas). Given the starting point $(xf) * g = a t (f * g)$, take the (unitary, ordinary frequency) transform to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\hat{x}*\hat{f} ) \hat{g} = a\hat{x} * (\hat{f} \hat{g}) \implies (\delta&amp;#39;*\hat{f} ) \hat{g} = a\delta&amp;#39; * (\hat{f} \hat{g}) \implies \hat{f}&amp;#39; \hat{g} = a (\hat{f}\hat{g})&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you foil and rewrite you get the simple differential equation $a \hat{g} &amp;#39; + (a-1) (\hat{f}&amp;#39; / \hat{f}) \hat{g} =0 $, which is easy to solve and comes out with $\hat{g} = \hat{f}^{1/a - 1}$ (I hope I did that all correctly!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT - thanks to below comment.]&lt;/p&gt;&#xA;" OwnerUserId="3122" LastEditorUserId="3122" LastEditorDisplayName="" LastEditDate="2010-11-30T16:45:30.553" LastActivityDate="2010-11-30T16:45:30.553" CommentCount="2" />
  <row Id="12504" PostTypeId="2" ParentId="275" CreationDate="2010-11-30T16:39:01.727" Score="0" ViewCount="0" Body="&lt;p&gt;I rather enjoyed Professor Stewart's book [1]. Take a look at it; I hope you enjoy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have blogged about a selection from his book, you can view it at [2]. This is just one of the many different mathematical concepts covered in the book. It is more of a fun book than lots of theory. It will get you to think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Stewart, I. (2009). Professor Stewart’s Hoard of Mathematical Treasures. New York: Basic Books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.tylerclark12.com/blog/?p=159&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/?p=159&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-11-30T16:39:01.727" CommunityOwnedDate="2010-11-30T16:39:01.727" />
  <row Id="12505" PostTypeId="2" ParentId="12477" CreationDate="2010-11-30T16:43:27.057" Score="3" ViewCount="0" Body="&lt;p&gt;Continuing Akhil's answer, let's prove a theorem of Dehn: if a rectangle is tiled by squares, the the ratio of the lengths of its sides is rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose to the contrary that the sides of the rectangle $x,y$ are &lt;i&gt;not&lt;/i&gt; rationally dependent. Then we can find some linear homomorphism $f\colon \mathbb{R} \rightarrow \mathbb{Q}$ such that $f(x) = 1$ and $f(y) = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the $f$-area $A(R)$ of a rectangle $R$ with edge lengths $h,v$ to be $f(h)f(v)$. If a rectangle $R$ is tiled by rectangles $R_i$ &lt;i&gt;forming a grid&lt;/i&gt;, then from linearity it immediately follows that $$A(R) = \sum_i A(R_i).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the big rectangle $R$ and the squares $S_i$. Take your tiling and extend all the lines to form a grid inside the rectangle. Denote the grid rectangles by $G_j$. Then $$A(R) = \sum_j A(G_j) = \sum_i A(S_i).$$ Since a square has both sides equal, $A(S_i) \geq 0$. On the other hand, by construction $A(R) &amp;lt; 0$. This contradiction shows that the two sides of the big rectangle are, in fact, rationally dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of taking a linear mapping from $\mathbb{R}$ to $\mathbb{Q}$ we could take a linear mapping from a smaller, finite dimensional domain by only considering the lengths in the grid $G_j$ - then everything becomes beginner's linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-11-30T16:43:27.057" />
  <row Id="12506" PostTypeId="2" ParentId="275" CreationDate="2010-11-30T17:06:10.613" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=Oac_PgAACAAJ&amp;amp;dq=how+to+solve+it&amp;amp;hl=en&amp;amp;ei=0S71TPb-BIX7lweo3viUBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CD4Q6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;How to solve it&lt;/a&gt; by G. Polya.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=NNQEAAAAYAAJ&amp;amp;printsec=frontcover&amp;amp;dq=value+of+science&amp;amp;hl=en&amp;amp;ei=LS_1TO6XG8SAlAfYwvCbBg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCsQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Value of Science&lt;/a&gt; by H. Poincare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=T43Cmu_EaZAC&amp;amp;printsec=frontcover&amp;amp;dq=symmetry&amp;amp;hl=en&amp;amp;ei=-i_1TO6qKcGqlAf90sDQBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Symmetry&lt;/a&gt; by H. Weyl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=R6E0AAAAMAAJ&amp;amp;printsec=frontcover&amp;amp;dq=flatland&amp;amp;hl=en&amp;amp;ei=VjD1TKnfAoOdlgeuh-jiBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CC0Q6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Flatland&lt;/a&gt; by E. Abbott&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=vslEPgAACAAJ&amp;amp;dq=chaos&amp;amp;hl=en&amp;amp;ei=hi_1TIyJBIOClAevkJXTBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCwQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;Chaos: Making of a New Science&lt;/a&gt; by J. Gleick&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-11-30T17:13:39.373" LastActivityDate="2010-11-30T17:13:39.373" CommunityOwnedDate="2010-11-30T17:06:10.613" CommentCount="1" />
  <row Id="12507" PostTypeId="1" AcceptedAnswerId="12510" CreationDate="2010-11-30T17:18:30.947" Score="8" ViewCount="127" Body="&lt;p&gt;I've been wondering about the following question:&#xA;Suppose that $P$ is a polynomial of degree $n$&amp;#160;with complex coefficients. Assume that $a_0, a_1, \dots, a_n \in \mathbb{C}$ are distinct. Are the polynomials $$P(x + a_0), P(x + a_1), \dots, P(x + a_n)$$ linearly independent?&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-30T19:27:52.670" Title="Are translations of a polynomial linearly independent?" Tags="&lt;linear-algebra&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12508" PostTypeId="2" ParentId="275" CreationDate="2010-11-30T17:30:51.137" Score="0" ViewCount="0" Body="&lt;p&gt;Here's what you're looking for. It's an amazing back, and you'll be blown away by all the stuff mentioned in here:&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1402757964&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Math-Book-Pythagoras-Milestones-Mathematics/dp/1402757964&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-11-30T17:30:51.137" CommunityOwnedDate="2010-11-30T17:30:51.137" />
  <row Id="12510" PostTypeId="2" ParentId="12507" CreationDate="2010-11-30T17:33:50.993" Score="6" ViewCount="0" Body="&lt;p&gt;Consider the matrix of coefficients, where each row corresponds to an evaluation at a point, and each column to a power of $x$. Any linear dependency among the columns amounts to some degree~$n$ polynomial in the point, which is supposed to have $n+1$ roots. Since this can't happen, the polynomials are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example (almost Vandermonde) - $P(x) = x^2$: $$\begin{pmatrix} 1 &amp;amp; 2a &amp;amp; a^2 \\ 1 &amp;amp; 2b &amp;amp; b^2 \\ 1 &amp;amp; 2c &amp;amp; c^2 \end{pmatrix}$$ A linear dependency of the columns is a quadratic polynomial having three different roots $a,b,c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to clarify what happens in the general case. Here's the matrix corresponding to $P(x) = x^2 + \alpha x + \beta$: $$\begin{pmatrix} 1 &amp;amp; 2a + \alpha &amp;amp; a^2 + \alpha a + \beta \\ 1 &amp;amp; 2b + \alpha &amp;amp; b^2 + \alpha b + \beta \\ 1 &amp;amp; 2c + \alpha &amp;amp; c^2 + \alpha c + \beta \end{pmatrix}$$&#xD;&#xA;Since the degree of the polynomial in column $t$ is $t$ (if we start counting from zero), it's easy to see that the any non-zero combination of columns results in a non-zero polynomial (consider the rightmost non-zero column).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-11-30T18:27:05.197" LastActivityDate="2010-11-30T18:27:05.197" CommentCount="2" />
  <row Id="12511" PostTypeId="2" ParentId="12501" CreationDate="2010-11-30T17:53:34.637" Score="0" ViewCount="0" Body="&lt;p&gt;We know that   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;1 = \cos{(2\pi)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for a given &lt;code&gt;N&lt;/code&gt; to &quot;maximize&quot; cosine:&lt;br&gt;&#xA;\begin{aligned}&#xA;2\pi = Nx \&#xA;\end{aligned}&#xA;\begin{aligned}&#xA;x = \frac{2\pi}{N}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above equation:&#xA;\begin{aligned}&#xA;x &amp;amp;= \frac{2\pi}{19.013} \&#xA;\end{aligned}&#xA;\begin{aligned}&#xA;x &amp;amp;\approx 0.33046 \&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="4222" LastEditorDisplayName="" LastEditDate="2010-11-30T18:02:03.180" LastActivityDate="2010-11-30T18:02:03.180" CommentCount="1" />
  <row Id="12512" PostTypeId="1" CreationDate="2010-11-30T17:56:21.013" Score="3" ViewCount="164" Body="&lt;p&gt;Is it even worth the theorem below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every positive integer $n$, there is a real number $r$, &#xA;and $\frac{1}{12n+1} \lt r \lt\frac{1}{12n}$, such that:&#xA;$$ n! = \sqrt{2n\pi}\left(\frac{n}{e}\right)^n e^r.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this statement on some sites, but got no further details.&#xA;I think the statement refers to an exact value of $n!$, not an approximation.&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T18:27:00.250" LastActivityDate="2010-11-30T18:44:19.250" Title="$n! =\sqrt{2n\pi}\left(\frac{n}{e}\right)^n e^r$?" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12513" PostTypeId="2" ParentId="12512" CreationDate="2010-11-30T17:59:17.463" Score="13" ViewCount="0" Body="&lt;p&gt;It is the strirling's aproximation  &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation&quot; rel=&quot;nofollow&quot;&gt;See article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n!\approx \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error estimates are very interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n!=\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n} \cdot e^{r_{n}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\frac{1}{12n+1}&amp;lt;r_{n}&amp;lt;\frac{1}{12n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the strirling's aproximation with remainder &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This aproximation is very useful, for example in the calculation of limits of sequences&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\sqrt[n]{n!}}{n} \approx \frac{\sqrt[n]{\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}}}{n} = \frac{\left(\frac{n}{e}\right)\sqrt[n]{\sqrt{2\pi n}}}{n} = \frac{\sqrt[2n]{2\pi n}}{e} \longrightarrow \frac{1}{e}, ( n\longrightarrow \infty ) $$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact $\sqrt[n]{n} \longrightarrow 1,  ( n\longrightarrow \infty )$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-11-30T18:44:19.250" LastActivityDate="2010-11-30T18:44:19.250" />
  <row Id="12514" PostTypeId="2" ParentId="12512" CreationDate="2010-11-30T17:59:28.430" Score="2" ViewCount="0" Body="&lt;p&gt;I think you are referring to the Stirling's approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stirlings_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stirlings_approximation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-30T17:59:28.430" />
  <row Id="12515" PostTypeId="2" ParentId="12507" CreationDate="2010-11-30T18:14:23.440" Score="0" ViewCount="0" Body="&lt;p&gt;If you write $P(x) = \sum_{j=0}^n p_j x^j$ and set $ \sum_{i=0}^n c_i P(x+a_i) = 0$, the full expression is written $\sum_{i=0}^n \sum_{k=0}^n \sum_{j=0}^k c_i p_k {k \choose j} a_i^{k-j} x^j$. If we find the coefficient of each power $x_j$ as a matrix times the vector $(c_0, ..., c_n)$, and then setting the system equal to zero, we can deduce the determinant of the matrix is zero. The matrix is $A_{ij} = \sum_{k=j}^n p_k {k \choose j} a_i^{k-j}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting the determinant equal to zero gives a polynomial in all $a_i$. I don't know if this polynomial is simpler than the matrix hints at, but it is certainly possible to find coefficients $a_i$ such that the polynomials are linearly independent, though for &quot;most&quot; values it won't be the case, as the solution set must be of at least one dimension less than $C^{n+1}$ (seen by fixing all $a_1, ..., a_n$ arbitrarily and thereby determining a finite set of possible $a_0$).&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-11-30T18:14:23.440" CommentCount="2" />
  <row Id="12516" PostTypeId="2" ParentId="12501" CreationDate="2010-11-30T18:52:23.657" Score="0" ViewCount="0" Body="&lt;p&gt;$x=0$ is always a solution, since $\cos(N\times0)=\cos0=1$ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the set of all solutions, use the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\theta=1 \Leftrightarrow \theta=2k\pi \text{ with } k\in\mathbb{Z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $\theta$ by $Nx$ above gives you the set of solutions :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1=\cos{Nx}\Leftrightarrow x\in \left\{\frac{2k\pi}N\right\}_{k\in\mathbb{Z}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the solution given by Wolfram Alpha for $N=19.013=\frac{19013}{1000}$.&lt;/p&gt;&#xA;" OwnerUserId="3918" LastActivityDate="2010-11-30T18:52:23.657" />
  <row Id="12517" PostTypeId="2" ParentId="12501" CreationDate="2010-11-30T18:55:06.283" Score="1" ViewCount="0" Body="&lt;p&gt;Cosine takes it's maximum when the argument is $2k\pi$, where $k$ is any integer. Therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}2k\pi = Nx \\\\&#xD;&#xA;&#xD;&#xA;x=\frac{2k\pi}{N}\end{aligned}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for $N=19.013$&#xD;&#xA;$$x= \frac{2k\pi}{19.013} = \frac{2000k\pi}{19013}$$.&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2010-11-30T18:55:06.283" />
  <row Id="12518" PostTypeId="2" ParentId="12323" CreationDate="2010-11-30T19:13:18.450" Score="1" ViewCount="0" Body="&lt;p&gt;When did 'none' come to be a number? My guess is that it was when merchants or scribes first codified symbolic addition and subtraction to calculate quantities of food and supplies on hand. Intuitively, they must have known that 'none' would be the result of subtracting 100 bushels of wheat from same. And that adding 100 bushels to 'none' would give you a measurable quantity. Philosophers may have quibbled about whether 'none' was a real and true number, but I'm sure it was all too real to the merchants and scribes of antiquity.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-11-30T19:13:18.450" />
  <row Id="12519" PostTypeId="2" ParentId="12507" CreationDate="2010-11-30T19:27:52.670" Score="0" ViewCount="0" Body="&lt;p&gt;Here a fancy proof by induction that uses differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case $n=0$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A polynomial of degree $0$ has the form $P(x)=c_0$ with $c_0\neq 0$ which is clearly linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case $n=k+1$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that there is a linear dependence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_0 P(x+a_0) + \lambda_1 P(x+a_1) + \dots + \lambda_n P(x+a_n) = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this relation holds for all $x$, we can differentiate this relation. After differentiating $n$ times, we obtain a relation for the $n$-th derivative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_0 P^{(n)}(x+a_0) + \lambda_1 P^{(n)}(x+a_1) + \dots + \lambda_n P^{(n)}(x+a_n) = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since $P$ has degree $n$, we know that $P^{(n)}(x+a_i)$ is equal to a constant $c\neq 0$, which implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_0 + \lambda_1 + \dots + \lambda_n = 0 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By expressing $\lambda_0$ in terms of the other coefficients, we obtain a relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_1 (P(x+a_1) - P(x+a_0)) + \dots + \lambda_n (P(x+a_n) - P(x+a_0)) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of polynomials of degree $n-1$. By induction, we conclude that all the $\lambda_i$ must vanish.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-11-30T19:27:52.670" CommentCount="1" />
  <row Id="12520" PostTypeId="1" AcceptedAnswerId="12529" CreationDate="2010-11-30T19:28:21.500" Score="1" ViewCount="169" Body="&lt;p&gt;Can we call the &lt;a href=&quot;http://mathworld.wolfram.com/FloorFunction.html&quot; rel=&quot;nofollow&quot;&gt;greatest integer function&lt;/a&gt; as a periodic function with no fundamental period or is it just non-periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain your answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To my understanding, if we consider $f(x) = [x]$ now, $f(3) = [3] = 3$ and $f(3+0.5) = [3.5] = 3$ So, can't we say that it is periodic (&lt;strong&gt;A constant function is periodic with no fundamental period&lt;/strong&gt;) ? But the problem is to derive the fundamental period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; After checking out some aswer I am quite inquisitive to know is it really necessary to have a fundamental period to call a function periodic? However,If you go by my book it is not.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-11-30T19:48:38.933" LastActivityDate="2010-12-01T16:47:46.237" Title="Doubt on greatest integer function" Tags="&lt;functions&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="12521" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:30:34.430" Score="1" ViewCount="0" Body="&lt;p&gt;A periodic functions repeats it's output values after regular intervals. That is there exists $T$ such that $f(x+T)=f(x)$ for all $x$. The greatest integer function does not satisfy this equation for any $T$. So it is not periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Periodic_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental period is just the smallest (positive) such $T$ (if one exists). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sawtooth function which measures the deviation from the greatest integer function (also called floor function) is periodic with period 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sawtooth_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sawtooth_function&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-30T19:52:12.423" LastActivityDate="2010-11-30T19:52:12.423" />
  <row Id="12522" PostTypeId="1" AcceptedAnswerId="12527" CreationDate="2010-11-30T19:35:38.783" Score="3" ViewCount="77" Body="&lt;p&gt;What is the value of n $\in \mathbb{Z} $ for which the function $\displaystyle f(x) = \frac{\sin nx} { \sin \biggl( \frac{x}{n} \biggr) } \text { has } 4\pi $ as period?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also could it be possible to solve this if we need $x\pi$ as period ?I am interested in learning the general approach for this particular type of the problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-11-30T19:54:53.840" LastActivityDate="2010-12-02T12:03:03.493" Title="Period of a function" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12523" PostTypeId="2" ParentId="7096" CreationDate="2010-11-30T19:36:19.993" Score="1" ViewCount="0" Body="&lt;p&gt;I think the following argument answers your question. It reminds me a lot of the exercises in the basic topology course I took during my undergrad. Fun times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\pi$ is a submersion, so for any $x$ in $E$ there exists a neighborhood of the form $U \times V$, such that $\pi$ identifies with the projection $U \times V \to V$. Suppose we only know that $E_t := \pi^{-1}(t)$ is compact for any $t$ in $B$, and let's show that $\pi$ is proper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K \subset B$ be compact and set $L = \pi^{-1}(K)$. Let $(\mathcal U_\alpha)$ be an covering of $L$ by open subsets of $E$. We will show that there exists a finite covering of $E$ by open subsets of $(\mathcal U_\alpha)$, and thus by sets of the original covering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $t \in K$. Find neighborhoods $\mathcal U_1, \ldots \mathcal U_{n(t)}$ that cover $E_t$. For any point $x \in E_t$, find neighborhoods $U_{t,x} \times V_{t,x} \subset \mathcal U_{j(x)}$ for some $j(x)$ (the $\mathcal U_{j(x)}$ being one of the above), such that the map $\pi$ identifies with the projection $U_{t,x} \times V_{t,x} \to V_{t,x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compactness of $E_t$ gives finitely many such $U_{t,x_\nu} \times V_{t,x_\nu}$ which cover $E_t$ and such that there is a $V_t \subset B$ contained in the image of any $V_{t,x_\nu}$ by $\pi$ (the intersection of the finitely many such $V$ contains $t$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: $K$ is compact, so it is covered by finitely many of those $V_t$. The corresponding finite collection $(U_{t,x_\nu} \times V_{t,x_\nu})$ covers $L$, and consists of subsets of elements of $(\mathcal U_\alpha)$. Thus finitely many elements of $(\mathcal U_\alpha)$ cover $L$.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-11-30T19:36:19.993" />
  <row Id="12524" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:37:28.103" Score="1" ViewCount="0" Body="&lt;p&gt;I am inclined to say that the greatest integer function (floor function) is not periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathworld [1] tells us that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function f(x) is said to be periodic (or, when emphasizing the presence of a single period instead of multiple periods, singly periodic) with period p if $f(x)=f(x+np)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the floor function, we can see that $\lfloor x+n \rfloor = \lfloor x \rfloor + n$ for $ x \not \in \mathbb{Z}$ and $n \in \mathbb{Z}$. This would give us reason to believe that we cannot find a period $p$ such that $f(x)=f(x+np)$ is satisfied for the greatest integer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Weisstein, Eric W. &quot;Periodic Function.&quot; From MathWorld--A Wolfram Web Resource. &lt;a href=&quot;http://mathworld.wolfram.com/PeriodicFunction.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/PeriodicFunction.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I know what you are saying about the constant function now though. I suppose that if you restrict the domain of the function, then you could say that the greatest integer function is periodic. Suppose that we just want to look at the greatest integer function on the interval [0,1). Then we can say that the definition of periodicity is satisfied. However on the entire domain of the greatest integer function, we do not have this property.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastEditorUserId="3791" LastEditorDisplayName="" LastEditDate="2010-11-30T19:49:38.823" LastActivityDate="2010-11-30T19:49:38.823" CommentCount="6" />
  <row Id="12525" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:40:43.717" Score="0" ViewCount="0" Body="&lt;p&gt;Even if you choose your period $T$ arbitrarily small, you can't satisfy $f(x+T) = f(x)$ for the arguments that are closer to any integer than T, so yo can't say it's a function with no fundamental period, it is non-periodic.&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2010-11-30T19:40:43.717" />
  <row Id="12526" PostTypeId="1" AcceptedAnswerId="14774" CreationDate="2010-11-30T19:47:04.550" Score="7" ViewCount="174" Body="&lt;p&gt;Let $B$ be the unit ball of $\ell^2(\mathbb{N})$, i.e. $B=\lbrace x\in \ell^2(\mathbb{N}): \|x\|\le 1\rbrace.$ For each $x=(x_1,x_2,\cdots)\in B$, let &#xA;$$f(x)=(1-\|x\|,x_1,x_2,\cdots).$$ Define $T:B\to 2^B$ by $$T(x)=B(f(x),r(x))\cap B, \mbox{ where }r(x)=\frac{1}{2}(\|x-f(x)\|).$$&#xD;&#xA;Is it true that $$D(Tx,Ty)\le \|x-y\| \mbox{ for all }x,y\in B?$$&#xD;&#xA;Here $B(y,r)$ denotes the closed ball with radius $r$ centered at $y$, and $D$ is the Hausdorff metric defined by $$D(A,B)=\inf\lbrace r&amp;gt;0: N_r(A)\supset B, &#xD;&#xA;N_r(B)\supset A\rbrace,$$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the &#xA;$r$-neighborhood of $S$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-11-30T20:25:59.820" LastActivityDate="2010-12-18T17:38:16.397" Title="A map on the unit ball " Tags="&lt;geometry&gt;&lt;functional-analysis&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" />
  <row Id="12527" PostTypeId="2" ParentId="12522" CreationDate="2010-11-30T19:49:01.723" Score="4" ViewCount="0" Body="&lt;p&gt;With $n=2$, $\sin(2x)$ has period $\pi$ and $\sin(x/2)$ has period $4\pi$ so their ratio must have a period of $4\pi$ since the latter period is an integral multiple of the former.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T12:03:03.493" LastActivityDate="2010-12-02T12:03:03.493" CommentCount="9" />
  <row Id="12528" PostTypeId="2" ParentId="12522" CreationDate="2010-11-30T19:53:28.817" Score="5" ViewCount="0" Body="&lt;p&gt;You want $$\frac{\sin n(x + 4\pi)}{\sin \frac{x + 4\pi}{n}} = \frac{\sin nx}{\sin \frac{x}{n}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent with&#xA;$$\sin \frac{x + 4\pi}{n} = \sin \frac{x}{n}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\frac{x}{n} = \frac{x + 4\pi}{n} + 2k\pi$ or $\frac{x}{n} = \pi - \frac{x + 4\pi}{n} + 2k\pi$ for some $k \in \mathbb{Z}$. In the first case $x = x + 4 \pi + 2k \pi n$ and thus $n = \frac{4}{2k}$. In the second case $x = \pi n - x - 4\pi + 2k\pi n$ and thus $n = \frac{2x + 4\pi}{2k\pi + \pi}$, which is impossible since this should hold for every $x \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $n = \pm 1$ or $\pm 2$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-11-30T19:53:28.817" />
  <row Id="12529" PostTypeId="2" ParentId="12520" CreationDate="2010-11-30T19:54:23.467" Score="10" ViewCount="0" Body="&lt;p&gt;A function $f(x)$ is &quot;periodic with period $c$&quot;, with $c\gt 0$, if $f(x+c) = f(x)$ for all values of $x$. In particular, $f(x+mc) = f(x)$ for all integers $m$, by using induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A function is periodic if it is periodic with period $c$ for some $c\neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every periodic function has lots of periods: if $c$ is a period, then so is $mc$ for any positive integer $m$, at the very least; it may have others. So we may be interested in knowing what is the &quot;quickest&quot; that the function starts repeating. For instance, $\tan x$ certainly repeats every $2\pi$, since $\tan(x+2\pi) = \tan(x)$ for all $x$ (either both are defined and equal, or both are undefined); and by the comment above, it will repeat every $2m\pi$ for any integer $m$. But it actually repeats &lt;em&gt;more often&lt;/em&gt; than every $2\pi$: we know that $\tan(x+\pi) = \tan(x)$ for all $x$. So altough it is true that $\tan(x)$ is periodic with period $2\pi$, it is &lt;em&gt;also&lt;/em&gt; true that $\tan(x)$ is periodic with period $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we would like to have something more than just a &quot;period&quot;; we would like to be able to talk about the &lt;em&gt;smallest&lt;/em&gt; possible period. That's what the &quot;fundamental period&quot; is trying to capture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that a periodic function $f(x)$ has &quot;fundamental period&quot; $k$ if and only if $k$ is the smallest positive number such that $f(x)$ is periodic with period $k$. That means that we require:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f$ to be periodic with period $k$: so $f(x) = f(x+k)$ for all $x$;&lt;/li&gt;&#xA;&lt;li&gt;$k$ to be positive; and&lt;/li&gt;&#xA;&lt;li&gt;if $c$ is a positive constant such that $f(x+c) = f(x)$ for all $x$, then $k\leq c$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, in general, we would take the set of all positive periods of the function $f$, and look for its minimum (smallest element). The one possible problem is that not every set of positive real numbers has a smallest element, so it may be possible, at least in principle, that a function is periodic but has no fundamental period, because it &lt;em&gt;does&lt;/em&gt; have periods, but it doesn't have one that is the &lt;em&gt;smallest&lt;/em&gt; period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant functions, $f(x) = a$ for all $x$, are an example of this pathology: they are periodic, because they satisfy the defintion of being periodic. In order to show that they satisfy the definition, we need to show that there is at least one number $c\gt 0$ such that $f(x)=f(x+c)$ for all $x$. Well, we can take $c=1$, because $f(x) = a = f(x+1)$ for all $x$. So the constant function is certainly periodic. But in fact, for &lt;em&gt;every&lt;/em&gt; positive number $c$, $f(x+c) = a =f(x)$. That means that if you take &lt;em&gt;any&lt;/em&gt; $c\gt 0$, then you can rightly say that &quot;$f(x)$ is periodic with period $c$&quot;. So if we look at the set of all periods,&#xA;$$P=\{ c \in \mathbb{R}\mid c\gt 0\text{ and $f(x)$ is periodic of period $c$}\}$$&#xA;then $P=(0,\infty)$. Since the &quot;fundamental period&quot; is supposed to be the &lt;em&gt;smallest&lt;/em&gt; element of $P$, but $P$ does &lt;em&gt;not&lt;/em&gt; have a smallest element, then the constant function does &lt;em&gt;not&lt;/em&gt; have a fundamental period. But it &lt;em&gt;is&lt;/em&gt; periodic (we just saw it satisfies the definition).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the greatest integer function, $f$ is not periodic at all. Again, &quot;$f$ is periodic&quot; means &quot;there exists $c\gt 0$ such that $f(x+c)=f(x)$ for all $x$&quot;. So the negation of &quot;$f$ is periodic&quot; is &quot;for every $c\gt 0$, there exists at least one $x$ such that $f(x)\neq f(x+c)$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that the greatest integer function $f(x)=\lfloor x\rfloor$ is &lt;strong&gt;not&lt;/strong&gt; periodic, we need to show that given any $c\gt 0$ (think of it as a candidate for a period), there is at least one point $x$ for which $f(x)\neq f(x+c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $c\gt 0$ be given. I'm going to come up with some point $x$ (which will depend on $c$) with the property that $\lfloor x\rfloor\neq \lfloor x+c\rfloor$. The idea is simply to take a value of $x$ that is before some integer, but once you add $c$ to it you go &lt;em&gt;above&lt;/em&gt; that integer. Then the greatest integer function will give two different values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest example to take is $x=-\frac{c}{2}$; this number is negative, so $f(-\frac{c}{2}) = \lfloor -\frac{c}{2}\rfloor\lt 0$; on the other hand, $x+c = -\frac{c}{2}+c = \frac{c}{2}$ is positive, so $f(x+c) = \lfloor \frac{c}{2}\rfloor \geq 0$. Since $f(x)$ is strictly less than $0$ and $f(x+c)$ is at least $0$, $f(x)\neq f(x+c)$. So $f(x)$ is not periodic with period $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $c$ was an &lt;em&gt;arbitrary&lt;/em&gt; positive number. That means that we cannot say &quot;$f(x)$ is periodic with period $c$&quot; for &lt;em&gt;any&lt;/em&gt; positive number, and that means, by definition, that $f(x) = \lfloor x\rfloor$ is &lt;strong&gt;not&lt;/strong&gt; periodic.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T20:28:01.800" LastActivityDate="2010-11-30T20:28:01.800" CommentCount="5" />
  <row Id="12530" PostTypeId="2" ParentId="12454" CreationDate="2010-11-30T19:59:04.300" Score="0" ViewCount="0" Body="&lt;p&gt;Although the software below does not allow one to check for planarity you might find it useful in seeing drawings of graphs (not typically drawn in the plane) with a small number of vertices and specified number of edges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gfredericks.com/sandbox/graphs/browse&quot; rel=&quot;nofollow&quot;&gt;http://www.gfredericks.com/sandbox/graphs/browse&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2010-11-30T20:23:48.337" LastActivityDate="2010-11-30T20:23:48.337" />
  <row Id="12531" PostTypeId="1" AcceptedAnswerId="12532" CreationDate="2010-11-30T20:32:25.087" Score="8" ViewCount="87" Body="&lt;p&gt;Question: Let $V \subset M(n,n,\mathbb{R})$ be the set of all symmetric, real $(n \times n)$ matrices, that is $a_{ij} = a_{ji}$ for all $i,j$. Show that $V$ is a subspace of $M(n,n,\mathbb{R})$ and calculate dim$(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&#xA;First part: To show that $V$ is a subspace I need to show:&#xA; (a) $ 0 \in V$ and &#xA; (b) $\forall A,B \in V:&#xD;&#xA; 	(i) A + B \in V&#xD;&#xA; 	(ii) \lambda A \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (a) I would say: Let $a_{ij} \in 0$(this should represent a zero matrix, is that how to write it?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{ij} = 0 = a_{ji} \Rightarrow 0 \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) I am actually confused since I would first think: both a $(2 \times 2)$ matrix and a $(3 \times 3)$ matrix belong to $V$ but addition of matrices of different size is undefined $\Rightarrow$ $V$ is not closed under addition $\Rightarrow$ $V$ is not a subspace of $M(n,n,\mathbb{R})$... what am I missing here? (To start I don't really understand the notation $M(n,n,\mathbb{R})$... what exactly does the $\mathbb{R}$ represent there?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding my confusion I would still try to show (b), but my mathematical notation is still lacking competence... Is the following somewhat clear? Would anyone ever use &quot;$\in$&quot; to denote &quot;is an element of matrix&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)Let $a_{ij},a_{ji} \in A$ and $b_{ij}, b_{ji} \in B$. Let $A,B \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow a_{ij} = a_{ji}, b_{ij} = b_{ji}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A + B = C \Rightarrow c_{ij} = (a_{ij}+b_{ij}) = (a_{ji} + b_{ij}) = (a_{ij} + b_{ji}) = c_{ji} = (a_{ji} + b_{ji})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow C \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Let $A\in V, \lambda \in \mathbb{R}$. Let $a_{ij},a_{ji} \in A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow a_{ij} = a_{ji}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda \cdot A = A&amp;#39;$ with $\lambda a_{ij} = \lambda a_{ji} \Rightarrow A&amp;#39; \in V$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second part: I feel that I understand the answer... For an $(n \times n)$ matrix, the diagonal length $ = n$ and these are the elements which have no counterpart and are not critical to the symmetry. When these elements are subtracted from the total$(n^{2})$, half of the remainder can be independently selected and the other half will follow as a result. Therefore I think it makes sense to write that dim$(V) = n + \frac{n^{2}-n}{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? If so, given the context of the exercise, how could I make my answer more acceptable?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T11:01:04.557" LastActivityDate="2010-12-01T11:01:04.557" Title="Show that the set of all symmetric, real matrices is a subspace, determine the dimension" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12532" PostTypeId="2" ParentId="12531" CreationDate="2010-11-30T20:44:48.927" Score="7" ViewCount="0" Body="&lt;p&gt;To write that the matrix is the zero matrix, you should write &quot;let $a_{ij}=0$ for all $i$ and $j$&quot;, not &quot;$a_{ij}\in 0$&quot;. (Nothing is an element of $0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b): No, notice that the $n$ is fixed. You are only considering matrices that are symmetric of a &lt;em&gt;fixed size&lt;/em&gt;. If $n=2$, then you only consider $2\times 2$ matrices; if $n=3$, then you only consider $3\times 3$ matrices. You never consider &lt;em&gt;both&lt;/em&gt; $2\times 2$ and $3\times 3$ matrices at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(n,n,\mathbb{R})$ means:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Matrices  (that's the &quot;$M$&quot;);&lt;/li&gt;&#xA;&lt;li&gt;with $n$ rows (that's the first $n$);&lt;/li&gt;&#xA;&lt;li&gt;with $n$ columns (that's the second $n$);&lt;/li&gt;&#xA;&lt;li&gt;and each entry is a real number (that is the $\mathbb{R}$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So $M(2,3,\mathbb{Z})$ would mean &quot;matrices with 2 rows and 3 columns each, and each entry is an integer.&quot; $M(7,2,\mathbb{C})$ means &quot;matrices with 7 rows and 2 columns, and every entry is a complex number. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way you want to say (b) is: Let $A=(a_{ij})$ and $B=(b_{ij})$ (that is, let's call the entries of $A$ &quot;$a_{ij}$&quot;, and let's call the entries of $B$ &quot;$b_{ij}$&quot;). Because $A$ is symmetric, we know that for all $i$ and $j$, $a_{ij}=a_{ji}$; and since $B$ is symmetric we know that for all $i$ and $j$, $b_{ij}=b_{ji}$. Now, let $C=A+B$. If we call the $(i,j)$th entry of $C$ &quot;$c_{ij}$&quot;, then you want to show that $c_{ij}=c_{ji}$ for all $i$ and $j$. How do you do that? You use the fact that you can express $c_{ij}$ in terms of the entries of $A$ and of $B$, and that $A$ and $B$ are symmetric, pretty much how you did; it's just a matter of writing it clearly. Same with the scalar multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument for the second part is essentially correct. To make it water tight, you should produce a basis of the appropriate size. Each element would be used to determine a particular entry as you describe them. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-11-30T20:44:48.927" CommentCount="0" />
  <row Id="12533" PostTypeId="2" ParentId="12374" CreationDate="2010-11-30T21:01:14.707" Score="0" ViewCount="0" Body="&lt;p&gt;Thanks cch.  Another approach a coworker recommended is to create a new time dependent term that includes all the constants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c(t) = (ha/k)*j(t)+Ta&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that ha, k, and Ta are all constant so c(t) and j(t) have no dynamics between themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the differential equation simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;db/dt = k*(c(t)-b(t))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the laplacian becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sB(s) + kB(s) = kC(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the transfer function becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B(s)/C(s) = k/(s+k)&lt;/p&gt;&#xA;" OwnerUserId="4262" LastActivityDate="2010-11-30T21:01:14.707" />
  <row Id="12534" PostTypeId="1" CreationDate="2010-11-30T21:21:20.580" Score="2" ViewCount="99" Body="&lt;p&gt;Given an angle $\theta$ (in radians), write $\theta = 2\pi t + \phi$, where $\phi$ is between $0$ and $2\pi$, $0\leq \phi\lt 2\pi$, and $t$ is an integer. Call $k=2\pi t$ the number of &quot;full circles&quot; or &quot;full turns&quot; corresponding to the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta = \frac{13\pi}{4}$, &#xA;am I right that k=$2\pi$?&lt;/p&gt;&#xA;" OwnerUserId="4263" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-11-30T22:21:10.897" LastActivityDate="2010-11-30T22:21:10.897" Title="Basic trigonometry" Tags="&lt;homework&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12535" PostTypeId="2" ParentId="12291" CreationDate="2010-11-30T21:29:09.977" Score="2" ViewCount="0" Body="&lt;p&gt;By induction let's just worry about picking one parameter element $x$. We need $x$ to be in $m$, but  outside all the minimal primes of $R$ and $m^2$. This smells exactly like &lt;a href=&quot;http://commalg.wiki-site.com/index.php/Prime_avoidance_lemma&quot; rel=&quot;nofollow&quot;&gt;Prime Avoidance&lt;/a&gt; (note that they don't have to be all primes)! Now replace $R$ by $R/(x)$ and repeat.&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-11-30T21:29:09.977" />
  <row Id="12536" PostTypeId="1" CreationDate="2010-11-30T21:47:11.727" Score="1" ViewCount="186" Body="&lt;p&gt;I heard that they are difficult. Is this true? Are they like the qualifying exams in grad school? For example, is the probability exam and the financial math exam comparable to qualifying exams (e.g. needs months and months of study)?&lt;/p&gt;&#xA;" OwnerUserId="4264" LastActivityDate="2010-12-01T01:09:52.810" Title="Are the actuarial exams hard?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="12537" PostTypeId="1" CreationDate="2010-11-30T21:48:54.800" Score="2" ViewCount="62" Body="&lt;p&gt;Let C be a category of sets, which has objects all sets and arrows all functions, with usual identity functions and the usual composition of functions. For any set S, the assignment s-s for all s in S describes the identity function. If S is a subset of Y, the assignment also describes the inclusion function from S to Y. These functions are different unless S is equal to Y. My question is why these functions are different since they have the same domain and assignment.&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-11-30T22:10:06.230" Title="Category of sets" Tags="&lt;category-theory&gt;" AnswerCount="2" />
  <row Id="12538" PostTypeId="2" ParentId="12537" CreationDate="2010-11-30T21:52:09.857" Score="3" ViewCount="0" Body="&lt;p&gt;In Categories, we have two operations called &lt;em&gt;Domain&lt;/em&gt; and &lt;em&gt;Codomain&lt;/em&gt; from the collection of all arrows to the collection of all objects. Each arrow $f$ belongs to the collection $C(\mathrm{dom}f,\mathrm{codom} f)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the Category of Sets, we actually are interested not just in the domain and the assignment of each function (which is what the arrows are), but also in the codomain. That is, you want to think of a function in the category $\mathcal{S}et$ as a triple, consisting of the domain, the codomain, and the actual set of pairs that make up the function (the assignment), $(A,B,f)$ for a function $f\colon A\to B$. Then the operator &lt;em&gt;Domain&lt;/em&gt; will just give you the first component, $A$; the operator &lt;em&gt;Codomain&lt;/em&gt; will give you the second component $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, two functions are equal if and only if they have the same domain, the same assignment, &lt;em&gt;and the same codomain&lt;/em&gt;. The codomain is important because it is an important attribute of arrows in categories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the inclusion function from $S$ to $Y$ has domain $S$, codomain $Y$, and rule $s\mapsto s$. The identity function from $S$ to itself has domain $S$, codomain $S$, and rule $s\mapsto s$. So the first function corresponds to&#xA;$$\Bigl(S, Y, \{(s,s)\in S\times S\}\Bigr)$$&#xD;&#xA;while the second function is&#xA;$$\Bigl(S,S,\{(s,s)\in S\times S\}\Bigr).$$&#xD;&#xA;If the two arrows are equal, then the value of &lt;em&gt;Domain&lt;/em&gt; and of &lt;em&gt;Codomain&lt;/em&gt; must be the same in both; that is, we must have $S=Y$. And of course, if $S=Y$, then they are the same function.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-11-30T22:10:06.230" LastActivityDate="2010-11-30T22:10:06.230" />
  <row Id="12539" PostTypeId="2" ParentId="12537" CreationDate="2010-11-30T21:53:00.773" Score="1" ViewCount="0" Body="&lt;p&gt;The two functions have different codomains. Any map $S\to S$ lies in $Mor(S,S)$ while any map $S\to Y$ lies in $Mor(S,Y)$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-11-30T21:53:00.773" />
  <row Id="12540" PostTypeId="2" ParentId="12534" CreationDate="2010-11-30T21:53:28.863" Score="3" ViewCount="0" Body="&lt;p&gt;I'm guessing that the homework problem is something like &quot;find an angle between 0 and $2\pi$ that's equal to $13\pi/4$&quot;.  That is, for what $k$ such that $k$ is a multiple of $2\pi$ do we have &#xA;$$ {13\pi\over 4} = k + {5\pi\over 4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is indeed $k=2\pi$.  The first equation in the original question should just have &quot;$k$&quot; in place of &quot;($2\pi$)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-11-30T21:53:28.863" />
  <row Id="12541" PostTypeId="2" ParentId="12536" CreationDate="2010-11-30T22:20:53.110" Score="1" ViewCount="0" Body="&lt;p&gt;If you are referring to the actuarial exams in the United States, obviously different people have different reactions, but they are designed to be hard in the sense that they are designed so that students will study for months to pass them. The pass rate for the first exam (exam P) is about 40 percent, and the pass rate only slowly increases with the higher exams. Most actuaries fail exams along the way, and it's not uncommon for an actuary to have one exam in particular he/she gets stuck on and has to retake several times. And some give up altogether. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-11-30T22:20:53.110" />
  <row Id="12542" PostTypeId="2" ParentId="12536" CreationDate="2010-11-30T22:27:09.490" Score="1" ViewCount="0" Body="&lt;p&gt;Personal experience: I took the first two of the &lt;a href=&quot;http://www.soa.org/&quot; rel=&quot;nofollow&quot;&gt;SOA&lt;/a&gt;/ASA/CAS exams (which is, if you are in the United States, the same as what you called the Exam P-Probability and Exam FM-Financial Mathematics) when I was finishing high school and starting college. Compared to graduate school (in mathematics) qualifying exams, they are much, much easier. (Though I heard that the latter exams in the CAS or ASA requirement series get quite a bit harder.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability exam, modulo some terminology/jargon issue, is actually not beyond the grasp of high-school level pre-calculus. You just need to have some experience dealing with probability and statistics. I think it is more of a test on problem solving skills than anything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The financial mathematics one has quite a bit more specialist knowledge involved. I remember having to spend several months studying for it just because there were many concepts which were new to me. At the time I managed to pass it by mentally associating the various words with the mathematical concepts/formulae they involved; in hindsight it was definitely the wrong way to study. Anyway, the study material from the SOA is actually quite comprehensive for the first two exams. If you spend a few months going through them you shouldn't have too much trouble.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-11-30T22:27:09.490" CommentCount="4" />
  <row Id="12543" PostTypeId="2" ParentId="12442" CreationDate="2010-11-30T22:29:31.530" Score="0" ViewCount="0" Body="&lt;p&gt;I'd like to supplement the above answers with two important examples.  Consider a &lt;em&gt;hyperbolic&lt;/em&gt; transformation $H(x, y) = (2x, y/2)$ and a &lt;em&gt;shear&lt;/em&gt; $S(x,y) = (x+y, y)$, both elements of $GL(2,R)$.  If you draw how these act on points of $R^2$ you will see that these distort angles.  On the other hand, elements of $GL(1,C)$ always preserve angles (and thus so do holomorphic maps, away from critical points).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the extra algebraic structure preserved by $GL(1,C)$ translates to the geometric property of preserving angles and &quot;handedness&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-11-30T22:29:31.530" />
  <row Id="12544" PostTypeId="1" CreationDate="2010-11-30T22:35:36.933" Score="6" ViewCount="224" Body="&lt;p&gt;Can n! be a perfect square when n is an integer greater than 1?&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-12-04T06:36:18.557" LastActivityDate="2010-12-04T06:36:18.557" Title="Can n! be a perfect square when n is an integer greater than 1?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="12545" PostTypeId="2" ParentId="12544" CreationDate="2010-11-30T22:37:22.353" Score="4" ViewCount="0" Body="&lt;p&gt;There is a prime between n/2 and n, if I am not mistaken. &lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2010-11-30T22:37:22.353" />
  <row Id="12546" PostTypeId="2" ParentId="12544" CreationDate="2010-11-30T22:56:09.077" Score="10" ViewCount="0" Body="&lt;p&gt;Assume, $n\geq 4$. By Bertrand's postulate there is a prime, let's call it $p$ such that $\frac{n}{2}&amp;lt;p&amp;lt;n$ . Suppose, $p^2$ divides $n$. Then, there should be another number $m$ such that $p&amp;lt;m\leq n$ such that  $p$ divides $m$. So, $\frac{m}{p}\geq 2$, then, $m\geq 2p &amp;gt; n$. This is a contradiction.  So, $p$ divides $n!$ but $p^2$ does not. So $n!$ is not a perfect square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bertrand_postulate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bertrand_postulate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves two more cases. We check directly, $2!=2$ and $3!=6$ are not perfect squares.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-11-30T23:09:44.190" LastActivityDate="2010-11-30T23:09:44.190" CommentCount="2" />
  <row Id="12547" PostTypeId="1" AcceptedAnswerId="12569" CreationDate="2010-11-30T22:57:19.120" Score="13" ViewCount="254" Body="&lt;p&gt;Please excuse the naive question.  I have had two classes now in which this theorem was taught and proven, but I have only ever seen a single (indirect?) application involving the quantum harmonic oscillator.  Even if this is not the strongest spectral theorem, it still seems useful enough that there should be many nice examples illustrating its utility.  So... what are some of those examples?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I couldn't readily find any nice examples looking through a few functional analysis textbooks, either.  Maybe I have the wrong books.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T03:35:44.840" Title="What is the spectral theorem for compact self-adjoint operators on a Hilbert space actually for?" Tags="&lt;big-list&gt;&lt;functional-analysis&gt;&lt;examples&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="12548" PostTypeId="1" AcceptedAnswerId="12549" CreationDate="2010-11-30T23:11:41.863" Score="1" ViewCount="71" Body="&lt;p&gt;I'm not quite sure where such a concept fits. Suppose $X$ is a topological space. I know then that the diagonal $\Delta=\{(x,x)\ | x\in X\}$, so $\Delta\subseteq X\times X$. What then would a neighborhood of the diagonal be comprised of? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this I mean, suppose $E\in\mathscr{N}_\Delta$, where $\mathscr{N}_\Delta$ is the filter of all neighborhoods of the diagonal. Is $E\in\mathscr{P}(X)$, or is $E\in\mathscr{P}(X\times X)$? What is the criterion for a subset to be in a neighborhood of $\Delta$? My guess is that $E\in\mathscr{P}(X\times X)$, and if this is correct, we would say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x,x&amp;#39;)\in E\quad\text{if and only if...}?$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-11-30T23:19:01.417" LastActivityDate="2010-11-30T23:23:19.850" Title="What exactly is a neighborhood of the diagonal on a topological space?" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12549" PostTypeId="2" ParentId="12548" CreationDate="2010-11-30T23:18:17.513" Score="4" ViewCount="0" Body="&lt;p&gt;A neighborhood of the diagonal $A$ is a subset of $X\times X$ containing an open set $E$ in $X\times X$ with the product topology such that $\Delta\subset E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of all neighborhoods of the diagonal is a subset of $\mathcal{P}(X\times X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in checking &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_space&quot; rel=&quot;nofollow&quot;&gt;uniform spaces&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="2614" LastEditorDisplayName="" LastEditDate="2010-11-30T23:23:19.850" LastActivityDate="2010-11-30T23:23:19.850" CommentCount="1" />
  <row Id="12550" PostTypeId="1" CreationDate="2010-12-01T00:08:39.127" Score="2" ViewCount="118" Body="&lt;p&gt;If topology is called rubber-sheet geometry, would it be accurate to describe the &quot;cut and shuffle&quot; topic of &quot;piecewise isometries&quot; as broken glass geometry ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isometry sounds more geometrical than topological. What would be a good name for &quot;piecewise topology&quot; ?  If you take a topological space and break it up and put it back together in a random way, not necessarily in one piece, then what could you say about the topology of the resulting jumble ?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastActivityDate="2010-12-01T01:41:34.937" Title="Broken glass geometry" Tags="&lt;geometry&gt;&lt;topology&gt;&lt;terminology&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12551" PostTypeId="2" ParentId="12547" CreationDate="2010-12-01T00:14:38.257" Score="9" ViewCount="0" Body="&lt;p&gt;Maybe a definition of functional calculus? Using quantum mechanics notation, if $A$ is self-adjoint and compact, then $A = \sum \lambda_k |k\rangle\langle k|$, which means that for $f:\mathbb{R}\to\mathbb{R}$ we can define $f(A) = \sum f(\lambda_k) |k\rangle\langle k|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows us to give a simple demonstration of Stone's theorem for such operators: that $\exp itA$ is a strongly continuous one-parameter unitary group on your Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also gives very simple motivation for the construction of Green's functions and resolvent operators.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Besides the usual quantum harmonic oscillator, a similar construction can be used to give the decomposition of $L^2$ via eigenfunctions of the Laplacian on a compact manifold. This naturally leads to the Hodge decomposition, which I'm told is generally considered to be somewhat useful :-) &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That on a compact manifold, the inverse of the Laplacian is a compact operator means that, discarding the harmonic functions, the Laplacian has a lowest eigenvalue. This fact (and that self-adjointness allows it to be diagonalized) allows you to define the Zeta-function &lt;a href=&quot;http://en.wikipedia.org/wiki/Functional_determinant&quot; rel=&quot;nofollow&quot;&gt;determinant&lt;/a&gt; of the Laplacian using some analytic continuation tricks. On 2-dimensional closed surfaces, this is an interesting invariant with nice geometric properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that in the case of non-compact domains, the inverse Laplacian is no longer a compact operator, and the Laplacian has a continuous spectrum. So the summation in the zeta-function determinant no longer makes any sense...)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-01T00:14:38.257" />
  <row Id="12552" PostTypeId="2" ParentId="12550" CreationDate="2010-12-01T00:41:31.763" Score="2" ViewCount="0" Body="&lt;p&gt;If the pieces are of given shapes, then this is essentially a jigsaw puzzle, perhaps in a higher dimension than 2.  A mathematical term for problems of this type is &quot;dissection&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pieces don't have a specific form and don't have to be put back together &quot;in one piece&quot;, I don't think much can be proved about the &quot;resulting jumble&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox&quot; rel=&quot;nofollow&quot;&gt; Banach-Tarski paradox&lt;/a&gt; gives an example of what is possible, at least if the Axiom of Choice is allowed.  This is a result that proves a ball (solid sphere) can be partitioned into a finite number of subsets that can be recombined to form two balls, each congruent to the original.  (The subsets are not measurable.  It is not a true paradox, but it is so called because it shows the Axiom of Choice has implications that defy common intuition.)&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-01T00:41:31.763" CommentCount="1" />
  <row Id="12553" PostTypeId="2" ParentId="12433" CreationDate="2010-12-01T01:03:33.780" Score="1" ViewCount="0" Body="&lt;p&gt;It's a nice problem. The chance of hitting $101$ first is surprisingly large. &#xA; Let $a(x)$ be the chance of hitting $101$ first, starting at $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We solve recursively by setting $a(106)=a(105)=a(104)=a(103)=a(102)=0$ and $a(101)=1$. Then, for $x$ from $100$ to $1$, put $a(x)={1\over 6}\sum_{j=1}^6 a(x+j)$. By the renewal theorem, $a(x)$ converges to $1/\mu$, where $\mu=(1+2+3+4+5+6)/6=7/2$. &#xA;The value $a(1)$ is very close to this limit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the whole hitting distribution is approximately $[6/21,5/21,4/21,3/21,2/21,1/21]$.&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-12-01T01:26:14.597" LastActivityDate="2010-12-01T01:26:14.597" CommunityOwnedDate="2010-12-01T01:03:33.780" CommentCount="1" />
  <row Id="12554" PostTypeId="2" ParentId="12536" CreationDate="2010-12-01T01:09:52.810" Score="2" ViewCount="0" Body="&lt;p&gt;I have taken the US SOA exams up through M and taken a qualifying exam at a US graduate school.  They are very different kinds of exams and so are a different kind of hard.  The mathematics involved in an actuarial exam isn't particularly advanced, but there is a lot of terminology to learn and concepts that you wouldn't spend much time on in math degree.  You shouldn't consider them trivial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the later exams is they get closer to quals in character.  They are several hour written exams and they serve a similar purpose (making sure you have the required knowledge of the discipline).  In particular, there are a set of exams in the US called the EA exams that require memorization and application of a large chunk of IRS regulations.  I'm glad I decided to go to grad school before having to take those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nice thing about actuarial exams (and I think this is relatively universal) is that once you get a job you will get paid study time and financial rewards for passing them.   &lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2010-12-01T01:09:52.810" />
  <row Id="12555" PostTypeId="1" AcceptedAnswerId="12558" CreationDate="2010-12-01T01:13:36.377" Score="20" ViewCount="469" Body="&lt;p&gt;Sorry if this is a really naive question, but in my reading of a lot of textbooks and articles, there is a lot of mention of how many decimals we know of a certain number today, such as $\pi$ or $e$. An excerpt from my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1748, Leonard Euler used the sum of the infinite series of $e$ (mentioned in the book in a section about Taylor Series) to find the value of $e$ to 23 digits. In 2003, Shigeru Kondo, again using the series, computed $e$ to 50 billion decimals places&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is &lt;em&gt;why&lt;/em&gt; does it matter how many decimals we know? Isn't this just a huge waste of time? What could we ever do with so many decimal places? And, if $e$ can be represented as a sum of infinite series of $1/n!$, can't we just plug that into a computer that just loops the same equation but increasing $n$ every iteration, and find as many decimals of $e$ as we like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Once again, I realize this may be an ignorant/naive question, but I've always been curious about this)&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T02:56:16.027" LastActivityDate="2010-12-01T03:51:45.757" Title="$e$ to 50 billion decimal places" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;taylor-expansion&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="1" />
  <row Id="12556" PostTypeId="2" ParentId="12555" CreationDate="2010-12-01T01:23:32.527" Score="6" ViewCount="0" Body="&lt;p&gt;The answer to your last question is no, for a certain value of &quot;no.&quot;  The problem with your idea is that as $n$ gets bigger it gets harder to calculate $n!$, so it gets harder to tell what the digits of $\frac{1}{n!}$ are.  In other words, if you actually tried to carry out your plan, you would quickly find that it is computationally infeasible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead one has to resort to a smarter algorithm.  Thus being able to compute constants to high precision is a measure of how smart our algorithms are.  If your algorithm is smarter than mine, the most concrete way to prove that is to use it to compute more digits in a reasonable amount of time than I can.  So while it's probably true that nobody is actually going to use these digits for anything (except possibly to verify certain conjectures), the fact that we know them is a measure of our knowledge, both about algorithms and about $e$.  (It is also a measure of how good our hardware is, but I'm trying to emphasize the math here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is an &lt;a href=&quot;http://www.numberworld.org/misc_runs/pi-5t/announce_en.html&quot; rel=&quot;nofollow&quot;&gt;announcement from Kondo&lt;/a&gt; about a more recent version of this computation (for $\pi$) and here are, among other things, his &lt;a href=&quot;http://www.numberworld.org/misc_runs/pi-5t/details.html&quot; rel=&quot;nofollow&quot;&gt;reasons for doing it&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T01:23:32.527" />
  <row Id="12558" PostTypeId="2" ParentId="12555" CreationDate="2010-12-01T01:26:32.460" Score="20" ViewCount="0" Body="&lt;p&gt;There are a couple of reasons why people compute $\pi$ and $e$ to so many digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is simply that it is a way to test hardware. The values are known (or can be known), and you want to see if your hardware is fast and accurate with these computations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is that there are actually a lot of questions about the decimal expansions of $\pi$ and $e$ (and other numbers) for which we simply don't know the answer yet. For example: is $\pi$  &lt;em&gt;normal in base 10&lt;/em&gt;? That is, will a particular sequence of digits occur in its decimal expansion in about the frequency you expect? More precisely, given a specific sequence of $n$ digits, will&#xA;$$\frac{\text{number of times that the specific sequence occurs in the first $m$ digits of $\pi$}}{m}$$&#xA;approach $1/10^n$ as $m\to\infty$ (there are $10^n$ different sequences of $n$ digits, so this is what you would expect if the sequence was completely random)? While the answer cannot be known simply by computing digits of $\pi$, knowing more and more digits helps us see whether it seems at least plausible or not, at least early on. We can also perform other tests of randomness to see whether the digits of $\pi$ (or $e$) seem to pass it or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question about $e$, yes, in principle we can do that. But the process gets more and more complex as $n$ gets larger. Normally, a computer only knows how to represent numbers that are no larger than a certain quantity (depending on the number of bytes it uses to represent numbers), so you need to &quot;explain&quot; to your computer how to handle big numbers like $n!$ for large $n$. The storage space needed to perform the computations also gets larger and larger and larger. And that idea only works if you start from a known point. So while theoretically we could compute $e$ to as many digits as we want, in practice if we want the computations to finish sometime before the Sun runs out of hydrogen we can't really go that far. There are other algorithms known to compute decimals of $\pi$ or $e$ that are faster, or that don't require you to know the previous digits to figure out the next one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that leads to yet another thing that people who are computing so many digits of $\pi$ and $e$ may be doing: testing algorithms for large number computations, for floating point accuracy, or for parallel computing (we are not very good at parallel computing, and trying to figure out how to do it effectively is a Very Big Deal; coming up with ways to do computations such as &quot;compute $\pi$ to the $n$th millionth digit&quot; are ways to test ideas for doing parallel computing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leads to a third interest in the computations: coming up with mathematical ideas that can zero in on desired digits of the decimal expansion quickly; not because we are particularly interested in the digits of $\pi$,  necessarily, but because we often &lt;em&gt;are&lt;/em&gt; interested in particular digits of &lt;em&gt;other&lt;/em&gt; numbers. $\pi$ and $e$ are just very convenient benchmarks to test things with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did Euler compute the numbers? Because he wanted to show off some of the uses for Taylor series; up to that time, the methods for approximating the value of $\pi$ were much more onerous. Ludolph van Ceulen famously used the methods of Archimedes (inscribing and circumscribing polygons in circles) to find the value of $\pi$ to 16 decimals (and after many more years of effort, to 25 places); this was in the late 16th century, before it was known that $\pi$ was transcendental (or even before the notion of being transcendental was very clear), so trying to compute $\pi$ precisely actually had a point. He was so proud of the accomplishment (given all the hard work it had entailed) that he had the value of $\pi$ he computed put into his tombstone. Euler showed that Taylor series could be used to obtain the same results with a lot less effort and with more precision.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T03:51:45.757" LastActivityDate="2010-12-01T03:51:45.757" CommentCount="5" />
  <row Id="12559" PostTypeId="2" ParentId="12426" CreationDate="2010-12-01T01:39:58.107" Score="0" ViewCount="0" Body="&lt;p&gt;At first when I had the idea to ask this question, I tought about books such as Schaum's Outline of College Mathematics and other books of the same kind for high school and university mathematics courses; then I found out this Website and I had the idea to demand to &quot;math experts.&quot; I knew Mathematica, this great math software application, but I am also happy that you talked about Maple that also really seems to be a great tool too. Thank you to tell me that Maple is less complex. At least, it seems that they developed portals and resources for students. It &quot;sounds&quot; great. In practice it is sometimes &quot;another game.&quot; In my opinion, I suppose that someone who buys the IrisPen to develop math problems on a computer and on paper, and that is also enable to use an application such as Maple 14 Academic or another version is really in a good position to be good at solving math problems and tasks of the like.      &lt;/p&gt;&#xA;" OwnerUserId="4269" LastActivityDate="2010-12-01T01:39:58.107" />
  <row Id="12560" PostTypeId="2" ParentId="12550" CreationDate="2010-12-01T01:41:34.937" Score="3" ViewCount="0" Body="&lt;p&gt;You might be interested in reading about &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_third_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Third Problem&lt;/a&gt;. Hilbert asked the question: Given two polytopes of the same volumes, can we cut one up into polytopal pieces and reassemble it to form the other. The answer is NO in dimensions $\geq 3$; there are many additional subtle obstacles. &lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-01T01:41:34.937" />
  <row Id="12561" PostTypeId="1" CreationDate="2010-12-01T02:01:46.293" Score="2" ViewCount="65" Body="&lt;p&gt;Theorem: If a function f is analytis throughout a simply connected domain D, then for every closed contour C lying in D, the integral is o along the contour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a theorem from the text I am using without complete proof. The problem is that when the closed contour has an infinite number of self-intersection points. Who can give me a rigorous proof or a reference about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2010-12-01T03:07:47.390" Title="Simply connected domains" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12562" PostTypeId="2" ParentId="12555" CreationDate="2010-12-01T02:08:46.537" Score="13" ViewCount="0" Body="&lt;p&gt;I'd just like to give two quotes from the book &lt;a href=&quot;http://www.siam.org/books/100digitchallenge/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The SIAM 100-Digit Challenge: A Study in High-Accuracy Numerical Computing&lt;/em&gt;&lt;/a&gt; that might help explain motivation. Here is the one from their chapter on computing constants to 10,000 digits:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While such an exercise might seem frivolous, the fact is we learned a&#xA;  lot from the continual refinement of our algorithms to work efficiently at&#xA;  ultrahigh precision. The reward is a deeper understanding of the theory,&#xA;  and often a better algorithm for low-precision cases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and here is something from the foreword written by David Bailey, one of the pioneers of &lt;a href=&quot;http://www.experimentalmath.info/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;experimental mathematics&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some may question why anyone would  care about such prodigious precision,&#xA;  when in the “real” physical world, hardly any quantities are known to an&#xA;  accuracy beyond about 12 decimal digits. For instance, a value of π&#xA;  correct to 20 decimal digits would suffice to calculate the circumference&#xA;  of a circle around the sun at the orbit of the earth to within the width&#xA;  of an atom. So why should anyone care about finding any answers to 10,000&#xA;  digit accuracy?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In fact, recent work in experimental mathematics has provided an important&#xA;  venue where numerical results are needed to very high numerical&#xA;  precision, in some cases to thousands of decimal digit accuracy. In&#xA;  particular, precision of this scale is often required when applying integer&#xA;  relation algorithms to discover new mathematical identities. An integer&#xA;  relation algorithm is an algorithm that, given $n$ real numbers&#xA;  ($x_i,\quad 1\leq i\leq n$), in the form of high-precision floating-point&#xA;  numerical values, produces $n$ integers, not all zero, such that&#xA;  $a_1x_1+a_2x_2+\cdots+a_n x_n=0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The best known example of this sort is the discovery in 1995 of a new formula&#xA;  for π:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\pi=\sum_{k=0}^{\infty}\frac1{16^k}\left(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac1{8k+5}-\frac1{8k+6}\right)$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This formula was found by a computer program implementing the PSLQ integer&#xA;  relation algorithm, using (in this case) a numerical precision of&#xA;  approximately 200 digits. This computation also required, as an input&#xA;  real vector, more than 25 mathematical constants, each computed to 200-digit&#xA;  accuracy. The mathematical significance of this particular formula is that it &#xA;  permits one to directly calculate binary or hexadecimal digits of &#xA;  π beginning at any arbitrary position, using an algorithm that is very simple,&#xA;  requires almost no memory, and does not require multiple-precision arithmetic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-01T02:08:46.537" />
  <row Id="12563" PostTypeId="1" CreationDate="2010-12-01T02:13:14.107" Score="5" ViewCount="88" Body="&lt;p&gt;I want to solve equations like this (mod $2^n$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcrcrcr} 3x&amp;amp;+&amp;amp;4y&amp;amp;+&amp;amp;13z&amp;amp;=&amp;amp;3&amp;amp;\pmod{16} \\ x&amp;amp;+&amp;amp;5y&amp;amp;+&amp;amp;3z&amp;amp;=&amp;amp;5&amp;amp;\pmod{16} \\ 4x&amp;amp;+&amp;amp;7y&amp;amp;+&amp;amp;11z&amp;amp;=&amp;amp;12&amp;amp;\pmod{16}\end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are working over a ring and not a field, Gaussian elimination doesn't work. So how can I still solve these types of equations?&lt;/p&gt;&#xA;" OwnerUserId="3163" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T17:02:29.487" LastActivityDate="2010-12-01T17:02:29.487" Title="Solving systems of linear equations over a finite ring" Tags="&lt;number-theory&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12564" PostTypeId="2" ParentId="12563" CreationDate="2010-12-01T02:23:23.043" Score="2" ViewCount="0" Body="&lt;p&gt;You basically do Gaussian elimination as usual, although you can get stuck if at some point all coefficients of a variable are, say, even. This just means that you'll have two solutions to that variable. In general, you'll pick the row in which the coefficient has the least power of $2$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T02:23:23.043" />
  <row Id="12565" PostTypeId="2" ParentId="12563" CreationDate="2010-12-01T02:27:34.803" Score="6" ViewCount="0" Body="&lt;p&gt;You can still use Gaussian elimination as long as you don't &quot;divide&quot; by things that are not relatively prime to the modulus. In this case, you can &quot;divide&quot; by any odd number, and perform all the usual computations. In this case, you can perform Gaussian pretty well:&#xA;\begin{align*}&#xA;\left(\begin{array}{ccc|c}&#xA;3 &amp;amp; 4 &amp;amp; 13 &amp;amp; 3\\&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;4 &amp;amp; 7 &amp;amp; 11 &amp;amp; 12&#xA;\end{array}\right) &amp;amp;\rightarrow&#xA;\left(\begin{array}{ccc|c}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;3 &amp;amp; 4 &amp;amp; 13 &amp;amp; 3\\&#xA;4 &amp;amp; 8 &amp;amp; 11 &amp;amp; 12&#xA;\end{array}\right) &amp;amp;&amp;amp; \rightarrow&#xA;\left(\begin{array}{ccr|c}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;0 &amp;amp; 5 &amp;amp; 4 &amp;amp; 4\\&#xA;0 &amp;amp; 4 &amp;amp; -1 &amp;amp; 8&#xA;\end{array}\right)\\&#xA;&amp;amp;\rightarrow&#xA;\left(\begin{array}{ccr|r}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;0 &amp;amp; 1 &amp;amp; 5 &amp;amp; -4\\&#xA;0 &amp;amp; 4 &amp;amp; -1 &amp;amp; 8&#xA;\end{array}\right)&#xA;&amp;amp;&amp;amp;\rightarrow&#xA;\left(\begin{array}{ccr|r}&#xA;1 &amp;amp; 5 &amp;amp; 3 &amp;amp; 5\\&#xA;0 &amp;amp; 1 &amp;amp; 5 &amp;amp; -4\\&#xA;0 &amp;amp; 0 &amp;amp; 11 &amp;amp; 8&#xA;\end{array}\right).&#xA;\end{align*}&#xA;So here you get that $11z\equiv 8 \pmod{16}$. Since $11^{-1} \equiv 3\pmod{16}$, this means $z \equiv 24 \equiv 8\pmod{16}$. Then you can backsubstitute and solve. (Assuming I didn't make any mistakes with my modular arithmetic, anyway...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are unlucky enough to get a congruence in which all the coefficients are even, then you can divide through by $2$ and get a congruence modulo $8$ (instead of $16$); that will lead to two solutions modulo $16$ (if you get $x\equiv 4 \pmod{8}$, that means $x\equiv 4 \pmod{16}$ or $x\equiv 8+4=12\pmod{16}$, for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, so long as you are careful, you can &lt;em&gt;certainly&lt;/em&gt; do Gaussian elimination. You can even do it over more general rings, through in that case you have to other restrictions on what you can or cannot conclude.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-01T02:27:34.803" />
  <row Id="12566" PostTypeId="1" AcceptedAnswerId="12588" CreationDate="2010-12-01T02:39:10.637" Score="2" ViewCount="246" Body="&lt;p&gt;How to prove this: $$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;em&gt;Apostol's&lt;/em&gt; number theory text i know that $$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$$ But how can i use this to prove my claim. &lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-12-02T16:46:37.057" Title="Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$" Tags="&lt;analytic-number-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="12567" PostTypeId="2" ParentId="12561" CreationDate="2010-12-01T03:07:47.390" Score="1" ViewCount="0" Body="&lt;p&gt;You can find a proof in &lt;a href=&quot;http://books.google.com/books?id=9LtfZr1snG0C&amp;amp;lpg=PP1&amp;amp;dq=conway%20complex&amp;amp;pg=PA87#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Section IV.6&lt;/a&gt; of J.B. Conway's &lt;em&gt;Functions of one complex variable&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-01T03:07:47.390" CommunityOwnedDate="2010-12-01T03:07:47.390" />
  <row Id="12568" PostTypeId="1" AcceptedAnswerId="12581" CreationDate="2010-12-01T03:13:53.077" Score="3" ViewCount="73" Body="&lt;p&gt;Suppose I take $n$ sets $\{S_i\}$ where each $S_i$ is some subset of integers. Let $P$ indicate a set of integers obtainable by drawing exactly one factor from each $\{S_i\}$ and multiplying them together. What conditions on these sets do I need to ensure that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$P$ forms a contiguous interval&lt;/li&gt;&#xA;&lt;li&gt;Each element of $P$ has a unique expansion of this form&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is probably a well known problem, so any keyword pointers are welcome&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-12-01T07:57:14.490" LastActivityDate="2010-12-01T07:57:14.490" Title="Generalizing decimal expansions" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12569" PostTypeId="2" ParentId="12547" CreationDate="2010-12-01T03:35:44.840" Score="7" ViewCount="0" Body="&lt;p&gt;What is the spectral theorem for compact operators good for?  Here are some examples.  (I am ignoring the self-adjoint aspects, since they don't really play&#xA;a role in the theorem.  And it is valid for more general spaces than Hilbert spaces too, so I will also ignore that part, in the sense that I won't pay too much attention to whether my examples deal with Hilbert spaces on the nose, rather than some variant.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Proving the Peter--Weyl theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Proving the Hodge decomposition for cohomology of manifolds (using the&#xA;fact that the inverse to the Laplacian is compact); Willie noted this example in his answer too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Proving the finiteness of cohomology of coherent sheaves on compact complex analytic manifolds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In its $p$-adic version, the theory of compact operators is basic to the theory of $p$-adic automorphic forms: e.g. in the construction of so-called eigenvarieties parameterizing $p$-adic families of automorphic Hecke eigenforms of finite slope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is also a basic tool in more classical problems, such as the theory of integral equations.  (It is in this context that the theory was first developed; see Dieudonne's book on the history of functional analysis for a very nice account of the historical development of the theory.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-01T03:35:44.840" CommentCount="8" />
  <row Id="12570" PostTypeId="1" CreationDate="2010-12-01T04:06:38.983" Score="1" ViewCount="46" Body="&lt;p&gt;Just curious. You pay electronically, and you get a string of numbers. My question, what is the likelihood for a person to type in the same serial number to say, top up their mobile credit? What is the range of probability electronic payment companies are using?Then, does electronic payment in China have longer serial number?&lt;/p&gt;&#xA;" OwnerUserId="3778" LastActivityDate="2010-12-01T04:35:37.950" Title="What probability methodology is electronic payment using?" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12571" PostTypeId="1" CreationDate="2010-12-01T04:10:02.817" Score="4" ViewCount="88" Body="&lt;p&gt;If $K$ is a field complete with respect to a non-archimedean absolute value, then the $n$th term test (checking whether the $n$th term of a series goes to zero) is sufficient to check convergence of a series in $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if we have any results that guarantee, under additional hypothesis, whether a series in $\mathbb{R}$ converges by just making sure that the $n$th term goes to zero. An example would be the alternating series test. Are there others?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T05:17:25.667" LastActivityDate="2010-12-01T05:17:25.667" Title="When is the $n$th term sufficient to guarantee convergence of a series" Tags="&lt;real-analysis&gt;" CommentCount="2" />
  <row Id="12572" PostTypeId="1" AcceptedAnswerId="12579" CreationDate="2010-12-01T04:13:39.087" Score="3" ViewCount="111" Body="&lt;p&gt;Every Fourier transform formula that I know of has a corresponding Fourier series analog, except the multiplication formula&#xA;$$\int_{-\infty}^\infty f(x)\hat{g}(x) dx=\int_{-\infty}^\infty \hat{f}(y)g(y) dy.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether a Fourier series analog for this formula exists.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-01T04:22:05.970" LastActivityDate="2010-12-01T16:27:58.560" Title="Fourier series analog of a formula in Fourier transform" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="3" />
  <row Id="12573" PostTypeId="1" CreationDate="2010-12-01T04:28:04.810" Score="3" ViewCount="73" Body="&lt;p&gt;Let $f$ be Lebesgue measurable in $[0,1]$ and assume $f$ takes finitely many values. Assuming $f(x) - f(y)$ is Lebesgue measurable in $[0,1] \times [0,1]$ show that $f$ is integrable over $[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stuck for a while with this one. (Not homework, just practice)&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2010-12-01T04:56:34.397" Title="Integration over the unit square" Tags="&lt;measure-theory&gt;" AnswerCount="1" />
  <row Id="12574" PostTypeId="2" ParentId="12570" CreationDate="2010-12-01T04:35:37.950" Score="1" ViewCount="0" Body="&lt;p&gt;Credit card transactions generally involve an input of much more information that just the credit card number (usually there is name, address, security code, card type as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even assuming just the card number is required, assuming it is 16 digits (15 with AMEX), since each digit can be anything from $0-9$, there are $10$ possibilities for each place, and hence $10^{16}$ possible card numbers. The current population of the world is of the order of $10^{10}$. Even if everyone in the world owns a card issued by the same company, the chances of someone just &quot;guessing&quot; someone else's number are pretty slim (of the order of $10^{-6}$). Of course as I said above there are more security measures in place and most merchants generally don't allow multiple attempts to enter card info beyond a certain threshold. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-01T04:35:37.950" />
  <row Id="12575" PostTypeId="2" ParentId="12573" CreationDate="2010-12-01T04:49:27.463" Score="4" ViewCount="0" Body="&lt;p&gt;The fact that $f(x)-f(y)$ is Lebesgue measurable implies that it is defined almost everywhere on the square. For this to be possible, $f$ cannot be $+\infty$ or $-\infty$ on a set $S$ of measure greater than 0 (otherwise $f(x)-f(y)$ will be undefined on the set $S\times S$ of positive measure). Since $f$ also takes on only finitely values, it is integrable.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-01T04:56:34.397" LastActivityDate="2010-12-01T04:56:34.397" CommentCount="0" />
  <row Id="12576" PostTypeId="1" CreationDate="2010-12-01T05:16:16.157" Score="2" ViewCount="151" Body="&lt;p&gt;Hello everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to ask these questions here, feeling that this place place is without doubt is better than anywhere else, feeling hesitant and shy to ask such question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For several reasons, the environment which I live in doesn't encourage the learning of natural sciences, it permits it but doesn't put efforts in teaching it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result I missed a great deal of exciting things as I went to college which show me the several weak points I developed through the years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being taught by other people who came from different locations, environments cultures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could see the huge contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feeling the pain of such experience, I decided to relearn everything form scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beginning from basics of math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I'm studying in college Computer Information Systems, I have some college level Math courses (Calc 1, Calc 2, Discrete Math, Linear algebra and some other courses like QA). And I am delaying them to prepare myself to accomplish basic courses like algebra1, algebra2, Geometry, College algebra,Trignometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I think about it. I found that I don't have the time (though I enjoy them+ feeling pain that I missed learning them properly) to do like all these algebra courses in a short time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it ok just to take a quick review algebra 1,2 using (cliffs notes) then study this college algebra book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0618825150&quot; rel=&quot;nofollow&quot;&gt; College Algebra and Trigonometry&lt;/a&gt;  or should I finish algebra 1 then jump into this book, as I find algebra one some how very easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually want to shorten the time to focus on geometry, so I guess learning geometry prepares me for trignometry and learning algebra 1 (at least) prepares me for college algebra which leads to higher mathematics. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="4272" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T05:21:13.573" LastActivityDate="2010-12-01T05:46:19.343" Title="about math in general." Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12577" PostTypeId="1" AcceptedAnswerId="12578" CreationDate="2010-12-01T05:36:04.520" Score="1" ViewCount="72" Body="&lt;p&gt;Let $\rho : S_n \to \mathrm{GL}(\mathbb{C}^n)$, where $\rho(\sigma)(x_1, \ldots, x_n) = (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$.  How can you prove that $W = \{ (x_1, \ldots, x_n) : x_1 + \cdots + x_n = 0 \} \subset \mathbb{C}^n$ is an irreducible representation?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-12-01T10:24:39.140" Title="Representation is irreducible" Tags="&lt;representation-theory&gt;" AnswerCount="2" />
  <row Id="12578" PostTypeId="2" ParentId="12577" CreationDate="2010-12-01T05:41:21.107" Score="4" ViewCount="0" Body="&lt;p&gt;Use the trace formula for $\rho$ to get that $\rho$ can be decomposed into two irreducible representations.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T05:41:21.107" />
  <row Id="12579" PostTypeId="2" ParentId="12572" CreationDate="2010-12-01T05:44:06.397" Score="8" ViewCount="0" Body="&lt;p&gt;Your formula is another way of writing &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#The_Plancherel_theorem_and_Parseval.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parseval's theorem&lt;/a&gt;, namely that Fourier transform preserves inner products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty} f \overline{g} = \int_{-\infty}^{\infty} \hat{f}\overline{\hat{g}} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You should combine this with the identities $\hat{\hat{g}}(x) = g(-x)$&#xD;&#xA;and $\overline{\hat{g}}(y) = \hat{\overline{g}}(-y).$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a Fourier series version of Parseval's theorem: namely&#xA;if $f$ (with period 1) has Fourier series $a_n$ and $g$ (with period 1)&#xA;has Fourier series $b_n$, then&#xA;$$\int_0^1 f \overline{g} = \sum_{n = -\infty}^{\infty} a_n \overline{b}_n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't see how to rewrite this in a way analogous to your formula,&#xA;just because (unlike in the Fourier transform context, where the funtion $f$&#xD;&#xA;and the Fourier transform $\hat{g}$ are both functions on the real line) a periodic function and its Fourier series are different beasts, and so it's not clear (to me) how one could mix them together in an integral or something similar.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-01T05:51:55.020" LastActivityDate="2010-12-01T05:51:55.020" />
  <row Id="12580" PostTypeId="2" ParentId="12576" CreationDate="2010-12-01T05:46:19.343" Score="2" ViewCount="0" Body="&lt;p&gt;You will need some geometry for calculus.  You should be able to teach yourself everything you need to know in terms of geometry within 1 month max, I would say if you work hard 2-3 weeks is a realistic timeframe.&#xA;You really need a solid foundation in algebra if you need to take calc 1 and calc 2 though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend taking some sort of algebra class but not more than one.  Alternatively, if you are disciplined, you can self-teach yourself, but this is not quite as good in my opinion.  I'm in my 2nd year of math/computer science joint program and my math classes are making my computer science courses a breeze.  In my (biased) opinion a solid foundation in math will REALLY help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-12-01T05:46:19.343" CommentCount="1" />
  <row Id="12581" PostTypeId="2" ParentId="12568" CreationDate="2010-12-01T05:48:58.480" Score="1" ViewCount="0" Body="&lt;p&gt;Let's assume that all the integers are in fact at least $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\max(P) = \prod_i s_i$. If $\max(P) \neq \min(P)$ then $\max(P)-1$ must be representable. Now $(s_1 - 1) \prod_{i&amp;gt;1} s_i = \prod_i s_i - \prod_{i&amp;gt;1} s_i$, so there must be an index $I$ such that $S_i = \{1\}$ for $i \neq I$. $S_I$ itself is of course of the form $\{1,\ldots,\max(P)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing, we have found two solutions: $|S_i|=1$ for all $i$ (the sets are otherwise arbitrary), and $S_i = \{1\}$ apart for one set which is an interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sets are infinite (i.e. $\max(P) = \infty$) then for each prime $p$ take the set $S_p = \{p^k : k \geq 0\}$. There are other solutions: you can decompose the set of primes arbitrarily as the disjoint union $\bigcup P_i$ and then take as $S_i$ all numbers whose prime factors are in $P_i$. All these use the convention that you only select finitely many numbers which are not $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open question: are these the only solutions?&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T05:48:58.480" CommentCount="0" />
  <row Id="12582" PostTypeId="2" ParentId="12566" CreationDate="2010-12-01T05:56:28.040" Score="1" ViewCount="0" Body="&lt;p&gt;You can try to break up the sum according to some exponential scale in $p$, and then use the previous estimate to sum up every step of the ladder, multiplying the result by $\sqrt{p_{\min}}$ where $p_{\min}$ is the lowest prime in your ladder (you should use a smarter factor since you'll be summing each $1/p$ several times, and you want the sum of the relevant factors to be $\sqrt{p_{\min}}$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T05:56:28.040" />
  <row Id="12583" PostTypeId="2" ParentId="12410" CreationDate="2010-12-01T06:14:38.493" Score="3" ViewCount="0" Body="&lt;p&gt;Just for the record, I like Shai Covo's answer better.  But the OP asked me to post my solution as well, so here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_t$ be the position of the object at time $t$.  Given $N$ clicks in $[0,t]$, let $\tau_1, \tau_2, \ldots \tau_N$ be the times of those clicks.  Let $T_i$ be the $i$th interarrival time, so that $T_1 = \tau_1$, $T_{N+1} = t - \tau_N$, and $T_i = \tau_i - \tau_{i-1}$ otherwise.  Thus $t = \sum_{i=1}^{N+1} T_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By properties of the exponential distribution, $E[T_i|N] = E[T_j|N]$ for all $i, j$.  Thus $$t = \sum_{i=1}^{N+1} E[T_i|N] = E[T_i|N] (N+1) \Rightarrow E[T_i|N] = \frac{t}{N+1}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N=0$, then $X_t = T_1$.  If $N = 1$, $X_t = \frac{1}{2}T_1 + T_2$.  If $N = 2$, $X_t = \frac{1}{4}T_1 + \frac{1}{2}T_2 + T_3$, and, in general, $X_t = \sum_{i=0}^N \frac{T_{N+1-i}}{2^i} $.  Thus &#xA;$$E[X_t|N] = \sum_{i=0}^N \frac{E[T_{N+1-i}|N]}{2^i}  = \frac{t}{N+1}\left(2 - \frac{1}{2^N}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $E[X_t] = E[E[X_t|N]]$, we just have to calculate $E\left[\frac{1}{N+1}\right]$ and $E\left[\frac{1}{(N+1)2^N}\right]$.  Since $N$ is Poisson$(\lambda t)$, we have&#xA;$$E\left[\frac{1}{N+1}\right] = \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} e^{-\lambda t}}{(n+1) n!} = e^{-\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n} }{(n+1)!} = \frac{e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^{n+1} }{(n+1)!} $$&#xD;&#xA;$$= \frac{e^{-\lambda t}}{\lambda t} \left(\sum_{n=0}^{\infty} \frac{(\lambda t)^n }{n!} - 1 \right) = \frac{e^{-\lambda t}}{\lambda t} \left(e^{\lambda t} - 1 \right) = \frac{1}{\lambda t} - \frac{e^{-\lambda t}}{\lambda t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, &#xA;$$E\left[\frac{1}{(N+1)2^N}\right] = \frac{2e^{-\lambda t}}{\lambda t} \sum_{n=0}^{\infty} \frac{(\frac{\lambda t}{2})^{n+1} }{(n+1)!} = \frac{2e^{-\lambda t}}{\lambda t} \left(e^{\lambda t/2} - 1 \right) = \frac{2e^{-\lambda t/2}}{\lambda t} - \frac{2e^{-\lambda t}}{\lambda t}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[X_t] = E\left[\frac{2t}{N+1}\right] - E\left[\frac{t}{(N+1)2^N}\right] = \frac{2}{\lambda}\left(1 - e^{-\lambda t/2}\right),$$&#xD;&#xA;which is exactly what you get if you solve the differential equation in Shai Covo's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected average speed (velocity, actually, since the average speed is technically 1) is &#xA;$$\frac{2}{\lambda t}\left(1 - e^{-\lambda t/2}\right).$$&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-01T06:14:38.493" />
  <row Id="12584" PostTypeId="1" AcceptedAnswerId="12585" CreationDate="2010-12-01T06:24:32.777" Score="5" ViewCount="81" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Doubly_stochastic_matrix&quot; rel=&quot;nofollow&quot;&gt;doubly-stochastic matrix&lt;/a&gt; is an $n\times n$ matrix $P$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{i=1}^n{p_{ij}}=1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_{j=1}^n{p_{ij}}=1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_{ij}\ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone suggest an algorithm for generating these matrices uniform randomly?&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-12-01T21:03:49.730" Title="How do I Generate Doubly-Stochastic Matrices Uniform Randomly?" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12585" PostTypeId="2" ParentId="12584" CreationDate="2010-12-01T06:36:25.930" Score="5" ViewCount="0" Body="&lt;p&gt;What we want is to generate a bistochastic matrix according to the Haar measure, which is the unique distribution which is invariant to multiplication by bistochastic matrices from both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard algorithm is to take some iid matrix (each entry is chosen iid from some distribution over non-negative numbers) and then repeatedly make it row-stochastic and column-stochastic - this is like projecting the matrix to the linear subspace of stochastic matrices. The process converges pretty fast, and we get some random bistochastic matrix, even though not according to the wanted distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chaos.if.uj.edu.pl/~karol/pdf/CSBZ09.pdf&quot; rel=&quot;nofollow&quot;&gt;A recent paper&lt;/a&gt; discusses these issues in the context of estimating the measure of all bistochastic matrices (an open problem).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T06:36:25.930" CommentCount="1" />
  <row Id="12586" PostTypeId="2" ParentId="12584" CreationDate="2010-12-01T06:36:55.790" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose you want a random rational point in the Birkhoff polytope (set of double stochastic matrices), with denominators dividing $N$.  This is equivalent to a random contingency table with all row and column sums equal to $N$.  A reversible Markov chain exists on the set of tables and is rapidly mixing.  This can be used for approximate sampling by following the chain for a large number of steps.  The limit distribution of the chain is close to uniform because away from the boundary of the polytope, all vertices of the chain have the same degree.  Exact sampling from the limit distribution of the chain is possible, but because the limit distribution is not quite uniform this is not the same as obtaining an exact sample from the uniform distribution on contingency tables.  The latter is a hard unsolved problem as far as I know. &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-01T21:03:49.730" LastActivityDate="2010-12-01T21:03:49.730" CommentCount="2" />
  <row Id="12587" PostTypeId="1" CreationDate="2010-12-01T06:47:37.353" Score="9" ViewCount="147" Body="&lt;p&gt;I have a homework problem in my textbook that has stumped me so far. There is a similar one to it that has not been assigned and has an answer in the back of the textbook. It reads: How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 609638400, but no matter what I try I cannot reach that number. I have tried doing 2(8!5!/3!)*(8!/5!) since each woman must be paired with a man in order to prevent two women getting near each other, but of course it's the wrong answer. What am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="3071" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-01T07:11:15.033" LastActivityDate="2010-12-01T07:20:40.270" Title="Homework related: Permutations and Combinations question." Tags="&lt;homework&gt;&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" />
  <row Id="12588" PostTypeId="2" ParentId="12566" CreationDate="2010-12-01T07:01:53.993" Score="10" ViewCount="0" Body="&lt;p&gt;We can use your second estimate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + R(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle R(x) = \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to show the asymptotic estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \mathcal{\theta}\left(\frac{\sqrt{x}}{\log x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact something stronger is known, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle R(x) = \mathcal{O}\left(\frac{1}{\log^2{x}}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I believe this gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log ^2x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, we use &lt;a href=&quot;http://books.google.com/books?id=iJx1lP9ZcIkC&amp;amp;pg=PA25&quot; rel=&quot;nofollow&quot;&gt;Abel's Identity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\displaystyle A(x) = \sum\limits_{n \leq x} \frac{a(n)}{n}$, where $\displaystyle a(n) = 1$ iff $\displaystyle n$ is prime and $\displaystyle 0$ otherwise. The above estimate tells us that $\displaystyle A(x) = \log\log x + A + R(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $\displaystyle f(t) = \sqrt{t}$ and using Abel's Identity we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} + C = A(x) \sqrt{x} - \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t = \int_{3}^{x} \frac{\log \log t + A + R(t)}{2\sqrt{t}} \ \text{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle  = C^{\prime} + A\sqrt{x} + \sqrt{x} \log \log x - \mathrm{Ei}\left(\frac{\log x}{2}\right) + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int \frac{\log \log x }{\sqrt{x}} = 2\sqrt{x} \log \log x - 2\mathrm{Ei}\left(\frac{\log x}{2}\right) + K$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \int \frac{R(x)}{\sqrt{x}} = \mathcal{O}\left(\int \frac1{\log^2 x \sqrt{x}}\right) = \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle \mathrm{Ei}(x)$ is the exponential integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All gotten using Wolfram Alpha).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle \mathrm{Ei}(x) = \frac{e^x}{x} + \mathcal{O}\left(\frac{e^x}{x^2}\right)$ (again using Wolfram Alpha), we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} = C + R(x) \sqrt{x} + \frac{2\sqrt{x}}{\log{x}} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hopefully, I got the computations right).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I will leave my earlier answer here (which is more relevant to the question as asked)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An elementary lower bound&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one elementary way (which does not use your second estimate or the prime number theorem) which proves something slightly stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the fact the each integer is uniquely expressible as the product of a square and a square free integer, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \prod_{p \le n} \left(1 + \dfrac1{\sqrt{p}}\right) \sum_{k=1}^{n} \dfrac1{k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle e^x \gt 1 + x$, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \left(\prod_{p \le n} e^{\dfrac1{\sqrt{p}}}\right)\left(\sum_{k=1}^{n} \dfrac1{k}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the estimate that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{j=1}^{n} \dfrac{1}{\sqrt{j}} \ge 2\sqrt{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{k=1}^{n} \dfrac1{j} \le \log (n+2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2\sqrt{n-1} \leq \log(n+2) \prod_{p \le n} e^{\dfrac1{\sqrt{p}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking logs, we see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{1}{2}\log (n-1) - \log \log(n+2)  + \log 2 \le \sum_{p \le n} \dfrac1{\sqrt{p}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle \log \log (n+2) - \log \log n \to 0 \ \text{as} \ n \to \infty$, $\displaystyle \log (n) - \log (n-1) \to 0 \ \text{as} \ n \to \infty$ and $\displaystyle \log 2 \gt 0$, you just need to verify you inequality for finite number of $\displaystyle n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T16:46:37.057" LastActivityDate="2010-12-02T16:46:37.057" CommentCount="4" />
  <row Id="12590" PostTypeId="2" ParentId="12587" CreationDate="2010-12-01T07:04:51.107" Score="17" ViewCount="0" Body="&lt;p&gt;For this type of problem first consider the position of the men, then the position of the women.In general first consider the the group with more people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many possible ways are there to arrange eight men in a row? It will be $ ^8P_8 = 8! = 40320$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as &lt;code&gt;no two women stand next to each other&lt;/code&gt;,we can imagine the situation as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;                   * M * M * M * M * M * M * M * M *&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence, we need to find how many ways we can arrange $5$ women in the $9$ possible (as shown above) places,this is actually $ ^9P_5 = 9*8*7*6*5 = 15120$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now applying the fundamental law of counting (precisely &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_product&quot; rel=&quot;nofollow&quot;&gt;product rule&lt;/a&gt;), total number of possible arrangements satisfying both constraints is: $15120 * 40320 = 609638400$ which is your required/desired answer. &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-01T07:20:40.270" LastActivityDate="2010-12-01T07:20:40.270" CommentCount="3" />
  <row Id="12591" PostTypeId="1" CreationDate="2010-12-01T07:10:41.960" Score="1" ViewCount="93" Body="&lt;p&gt;Suppose $\frac{dx}{dt}=ax+b$ and then assume that $a=c+g$ where $g$ is a Wiener process. &lt;/p&gt;&#xA;" OwnerUserId="4274" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-02T01:04:38.047" LastActivityDate="2010-12-02T01:04:38.047" Title="How can I obtain from a differential equation a stochastic version?" Tags="&lt;differential-equations&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12592" PostTypeId="2" ParentId="12566" CreationDate="2010-12-01T07:28:01.853" Score="9" ViewCount="0" Body="&lt;p&gt;I wasn't sure that you wanted a proof that used that fact from Apostol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One easy method &lt;em&gt;not&lt;/em&gt; using the result you quote from Apostol is as follows:&#xA;$$&#xD;&#xA;\sum_{p\le x} \frac{1}{\sqrt{p}} &amp;gt; \sum_{p \le x} \frac{1}{\sqrt{x}} = \frac{\pi(x)}{\sqrt{x}} &amp;gt; c \frac{\sqrt{x}}{\log{x}}&#xD;&#xA;$$&#xD;&#xA;where $c$ is a constant you can get from Chebyshev or &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.ijm/1255631807&quot; rel=&quot;nofollow&quot;&gt;Rosser and Schoenfeld&lt;/a&gt;, and maybe do a little computation to take care of small $x$, and you've got it.  This is a much better lower bound than what you are trying to prove.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-12-01T07:28:01.853" CommentCount="1" />
  <row Id="12593" PostTypeId="1" CreationDate="2010-12-01T07:52:34.953" Score="1" ViewCount="94" Body="&lt;p&gt;Is there a &quot;SETI at Home&quot; style initiative for calculating constants with distributed computing?&lt;/p&gt;&#xA;" OwnerUserId="4277" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T08:16:06.370" LastActivityDate="2010-12-31T09:44:30.140" Title="Distributed software for calculating $\pi$ or $e$" Tags="&lt;math-software&gt;&lt;pi&gt;" AnswerCount="1" />
  <row Id="12594" PostTypeId="1" AcceptedAnswerId="12595" CreationDate="2010-12-01T07:57:13.950" Score="2" ViewCount="71" Body="&lt;p&gt;For what values of $x = (x_0 x_1 ... x_n)$ the following inequality holds for all positive values of the vector $a = (a_0 a_1 ... a_n)$: $a x^T \ge 0$?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="1041" LastEditorDisplayName="" LastEditDate="2010-12-01T08:56:35.547" LastActivityDate="2010-12-01T08:56:35.547" Title="Simple-looking vector inequality ..." Tags="&lt;inequality&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12595" PostTypeId="2" ParentId="12594" CreationDate="2010-12-01T08:08:59.437" Score="4" ViewCount="0" Body="&lt;p&gt;The condition is that $x_i\geq 0$ for all $i$.  It is clear that $ax^T\geq0$ if this holds.  On the other hand, if $x_i\lt0$ for some $i$, then choose $a$ such that $a_i=1$ and $|a_jx_j|\lt\frac{|x_i|}{n}$ for all $j\neq i$.  Then $ax^T=x_i+(a_0x_0+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_nx_n)&#xD;&#xA;\lt x_i +n\frac{|x_i|}{n}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if it was assumed that $a$ is fixed beforehand, then your inequality would describe a &lt;a href=&quot;http://en.wikipedia.org/wiki/Half-space&quot; rel=&quot;nofollow&quot;&gt;half-space&lt;/a&gt; whose boundary is the hyperplane described by $ax^T=0$, which would include those vectors $x$ such that $x_i\geq0$ for all $i$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-01T08:14:20.887" LastActivityDate="2010-12-01T08:14:20.887" CommentCount="1" />
  <row Id="12596" PostTypeId="1" AcceptedAnswerId="12606" CreationDate="2010-12-01T08:15:22.203" Score="2" ViewCount="185" Body="&lt;p&gt;While learning &lt;em&gt;topology&lt;/em&gt; one learns about compact set. The standard definition is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A set $X$ is said to be compact &lt;em&gt;if open cover has a finite subcover.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since $[0,1]$ is compact, if we take a open cover for this we should be able to get a finite subcover. I know, that $(0,1)$ is not compact, so there must exists some open cover for $(0,1)$ which doesn't admit any finite subcover. But how does one prove this fact?&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-12-01T12:03:50.010" Title="Finite sub cover for $(0,1)$." Tags="&lt;analysis&gt;&lt;topology&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12598" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T08:20:33.367" Score="15" ViewCount="0" Body="&lt;p&gt;Isn't this clear from $(0,1) = (0,1/2) \cup (0,2/3) \cup (0,3/4) \cup  \cdots$?&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-01T08:20:33.367" />
  <row Id="12599" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T08:26:08.650" Score="3" ViewCount="0" Body="&lt;p&gt;Let $A$ be a non-closed subset of a metric space, and let $x\in\overline A\setminus A$.  Then $A$ is covered by open sets given by the complements of closed balls of positive radius centered at $x$, but not by any finite subcollection.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-01T08:26:08.650" />
  <row Id="12600" PostTypeId="1" AcceptedAnswerId="12602" CreationDate="2010-12-01T08:43:07.687" Score="3" ViewCount="53" Body="&lt;p&gt;I have a plane defined through a point P and 2 3d-vectors Xdir and yDir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to convert coordinates of points on this plane between local 2d-parametric and world 3d coordinate systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the conversion from 2d Parametric to 3d is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C(u, v) = P + u*XDir + v*YDir &lt;/p&gt;&#xA;&#xA;&lt;p&gt;however i have been unable to find a way for the inverse case &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C'(x,y,z) which should give me the parameters u and v for any point x,y,z in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this conversion work?&lt;/p&gt;&#xA;" OwnerUserId="1357" LastActivityDate="2010-12-01T09:25:08.900" Title="Given a plane and a 3d-Vertex, what are U and V?" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="12601" PostTypeId="1" CreationDate="2010-12-01T08:46:19.227" Score="1" ViewCount="47" Body="&lt;p&gt;I am reading by myself this book &lt;a href=&quot;http://tinyurl.com/37z4bbt&quot; rel=&quot;nofollow&quot;&gt;http://tinyurl.com/37z4bbt&lt;/a&gt;. But to be honest, I have several problems to fully understand some part of the text. Maybe because I have not yet solid knowledge or I still need to learn more about the subject, hence asked a little help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be exact, is this part of the construction concerning to Algebraic Blow-Up that I do not understand like trying...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quote part of the building in which I do not understand how to deal with some concepts&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the canonical map from $\mathbb{C}^2$ to the projective line $\mathbb{CP}^1$ that associates with each point $(x,y)$ different from the origin, the line {$(tx, ty) : t \in \mathbb{C}$} passing through this point. The graph of this map is a complex 2-dimensional surface in the complex 3-dimensional manifold (the Cartesian product) $\mathbb{C}^2$ × $\mathbb{CP}^1$, &lt;strong&gt;which is not closed&lt;/strong&gt;. To obtain the closure, one has to add the exceptional curve $E={0} \times \mathbb{CP^{1}} \subset \mathbb{C^{2}} \times \mathbb{CP^{1}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then, I do not understand is how to handle this graph. If, as a &lt;strong&gt;topological space&lt;/strong&gt; or as &lt;strong&gt;complex manifold&lt;/strong&gt;. Well, in the sense of complex manifold, say it is not closed, it means that the manifold is not compact without boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really a little confused&lt;/p&gt;&#xA;" OwnerUserId="4278" LastActivityDate="2010-12-01T08:46:19.227" Title="Blow-up. Graph of canonical map: is a complex manifold no closed or a topological space no closed" Tags="&lt;topology&gt;&lt;differential-equations&gt;&lt;algebraic-topology&gt;&lt;analytic&gt;" CommentCount="1" />
  <row Id="12602" PostTypeId="2" ParentId="12600" CreationDate="2010-12-01T08:57:52.457" Score="3" ViewCount="0" Body="&lt;p&gt;The inverse of the 3-by-3 matrix $A$ whose columns are given by $X$, $Y$, and the cross product $X\times Y$ changes basis from $\{X,Y,X\times Y\}$ to the standard basis.  Given $(x,y,z)^T=P+uX+vY$ in the plane, subtracting $P$ and multiplying $A^{-1}$ by the resulting column vector gives the column vector $(u,v,0)^T$.  So one way to write the map would be $\pi(A^{-1}((x,y,z)^T-P))$, where $\pi:\mathbb{R}^3\to\mathbb{R}^2$ is projection onto the first two coordinates.  To write this more explicitly, you could use the matrix $\left(\begin{matrix}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0&#xD;&#xA;\end{matrix}\right)&#xD;&#xA;$&#xD;&#xA;of $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product isn't really necessary, it just seemed like the most straightforward way to complete $\{X,Y\}$ to a basis for $\mathbb{R}^3$.  Any vector $Z$ not in the span of $\{X,Y\}$ would do.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-01T09:25:08.900" LastActivityDate="2010-12-01T09:25:08.900" />
  <row Id="12603" PostTypeId="2" ParentId="12431" CreationDate="2010-12-01T09:08:45.633" Score="0" ViewCount="0" Body="&lt;p&gt;My answer is similar to chroma's answer, only that I get &#xA;$$&#xD;&#xA;(xf) * g = at(f * g) \Rightarrow \hat f&amp;#39;\hat g = a(\hat f \hat g)&amp;#39; &#xD;&#xA;$$&#xD;&#xA;straight from the following standard property (combined, of course, with the convolution theorem):&#xA;if $\psi = x \varphi$, then $\hat \psi = i \hat \varphi&amp;#39;$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-01T09:08:45.633" />
  <row Id="12604" PostTypeId="1" CreationDate="2010-12-01T09:30:33.900" Score="3" ViewCount="70" Body="&lt;p&gt;Hi I have a problem with quadratic splines, I am supposed to find $&#xD;&#xA;S_1 $ and $S_2$ that interpolates the following points $S(-1)=0$, $S(0)=1$, $S(1)=2$, and at the same time we want to find $S$ such that  $\int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x$  is minimal. The answer is on the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_1(x)=a_1 x^2 +b_1x +c_1$ on $[-1,0]$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_2(x)=a_2 x^2 +b_2x +c_2$ on $[0,1]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the data points and find that $a_1=-a_2$, $b_1=b_2$ and $c_1=c_2=1$, but I have no idea how to use minimize $\int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x$&#xD;&#xA;, can I divide it up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min \int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x=\min (\int_{-1}^0 \! (S_1(x))^2 \, \mathrm{d}x + \int_0^1 \! (S_2(x))^2 \, \mathrm{d}x )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I should get an expression and probably set the derivative to zero but I just don't know how to attack the minimizing integral since the function has two parts. Help greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4281" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T11:52:46.080" LastActivityDate="2010-12-02T11:58:20.910" Title="Quadratic splines, minimizing integral" Tags="&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="1" />
  <row Id="12605" PostTypeId="2" ParentId="12593" CreationDate="2010-12-01T09:41:25.023" Score="1" ViewCount="0" Body="&lt;p&gt;There was &lt;a href=&quot;http://oldweb.cecm.sfu.ca/projects/pihex/index.html&quot; rel=&quot;nofollow&quot; title=&quot;Pi Hex&quot;&gt;PiHex&lt;/a&gt; which has now ended. I am not aware of any others - someone may correct me&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-12-01T09:41:25.023" />
  <row Id="12606" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T09:43:37.913" Score="5" ViewCount="0" Body="&lt;p&gt;To prove that a set $X$ isn't compact is generally easy. You only need to give an example of a cover, that doesn't admit a finite subcover. To expand on Shai's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal U := \{(0,n/(n+1): n \in \mathbb N\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and observe that if you take any $U_1,U_2,\dots U_n \in \mathcal U$, then &#xA;there exists some k such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(0,k/(k+1))=\bigcup_{i=1}^n U_1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't obvious, note that for any two elements in $\mathcal U$ that one is contained the other. So if you take a finite number of them, then one of them will contain the rest. Furthermore since $(0,k/(k+1))$ does not contain $(0,1)$,&#xA;$\mathcal U$ does not have a finite subcover, so X is not compact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you should also show that $\mathcal U$ is a cover; I'll leave it as an exercise to the reader.&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2010-12-01T09:43:37.913" />
  <row Id="12607" PostTypeId="2" ParentId="12604" CreationDate="2010-12-01T10:03:43.013" Score="7" ViewCount="0" Body="&lt;p&gt;Here's a hint: since you need your piecewise expression to be a &lt;em&gt;spline&lt;/em&gt;, one thing you should have is the imposition of a derivative condition at the point $x=0$; namely, $S_1^{\prime}(0)=S_2^{\prime}(0)$. (This is apart from the obvious equations you get from substituting appropriate values of $x$ and $y$ into the expressions for $S_1$ and $S_2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far, you should have five equations in six unknowns; you can eliminate some unknowns, and be able to express some of them in terms of just one of the unknowns, such that both $S_1$ and $S_2$ depend on a single parameter. You had the right idea to split up the integral into integrals over $[-1,0]$ and $[0,1]$ ; if you do that, you should end up with a quadratic in a single variable, whose minimum is trivial to find.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Shorter version of the hint: you should be able to express $a_1$ (and thus $a_2$ as well) in terms of either $b_1$ or $b_2$. Picking $b_1$ as the variable of interest, and substituting into your integral, you should have an expression of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(b_1)^2+v\cdot b_1+w$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u$, $v$, and $w$ are certain constants. If $u$ is positive, then $-\frac{v}{2u}$ is the minimum value of $b_1$ you seek. Substitute into all your other relations, and you should now have the desired spline.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T11:58:20.910" LastActivityDate="2010-12-02T11:58:20.910" CommentCount="1" />
  <row Id="12608" PostTypeId="1" AcceptedAnswerId="12640" CreationDate="2010-12-01T10:22:34.880" Score="0" ViewCount="57" Body="&lt;p&gt;Let $S$ be a collection of disjoint sets. Let a binary operation '*' defined $\forall x,y$ each belonging to two different sets or a same set in $S$ with the property that $z=x*y$ belongs to some set in $S$ . Let $\forall A_1,A_2 \in S$ a  binary operation $ \otimes$ is defined between $A_1,A_2$ as $B=A_1 \otimes A_2$ as the collection of all $z=(x*y)$ where $x\in A_1,y\in A_2$. Now the set $S$ is closed under the binary operation $\otimes$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to study the properties of such a set $S$ under the operation $\otimes$.My question is what is the subject in mathematics which deals with such a situation ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T10:45:40.170" LastActivityDate="2010-12-01T17:07:43.413" Title="a reference request to study properties of a set which is closed under a binary operation." Tags="&lt;reference-request&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" />
  <row Id="12609" PostTypeId="2" ParentId="12577" CreationDate="2010-12-01T10:24:39.140" Score="3" ViewCount="0" Body="&lt;p&gt;The first time I was assigned this exercise I was explicitly told not to use character theory.  There is a straightforward direct proof (showing directly that the orbit of any nonzero vector is all of $W$) and I implore you to try to find it.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T10:24:39.140" CommentCount="2" />
  <row Id="12610" PostTypeId="1" AcceptedAnswerId="12612" CreationDate="2010-12-01T11:18:05.400" Score="1" ViewCount="70" Body="&lt;p&gt;In a 2-d plane, let $A$ be the set of all points inside a circle $C$ and $P$ be the set of all points on the perimeter of the circle $C$. Is there any generalized measure in set theory which distinguishes between two such sets ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-01T11:20:14.460" LastActivityDate="2010-12-01T14:22:23.760" Title="a question about a measurable property of a set" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="12611" PostTypeId="2" ParentId="12520" CreationDate="2010-12-01T11:45:26.763" Score="2" ViewCount="0" Body="&lt;p&gt;Though the floor function is not periodic, it is &lt;em&gt;quasiperiodic&lt;/em&gt;; that is, it is a function that satisfies the functional equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x+\alpha)=f(x)+ux+v$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific case of the floor function, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lfloor x+1\rfloor=\lfloor x\rfloor+1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so 1 can be thought of as the floor function's quasiperiod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a rather trivial example of a quasiperiodic function, though.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-01T11:45:26.763" />
  <row Id="12612" PostTypeId="2" ParentId="12610" CreationDate="2010-12-01T11:50:21.670" Score="5" ViewCount="0" Body="&lt;p&gt;Yes there is. It's called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hausdorff_measure&quot; rel=&quot;nofollow&quot;&gt;Hausdorff measure&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-01T11:50:21.670" />
  <row Id="12613" PostTypeId="2" ParentId="12596" CreationDate="2010-12-01T12:03:50.010" Score="1" ViewCount="0" Body="&lt;p&gt;Alternatively, use the fact that every compact subspace of a Hausdorff space is closed. Since (0,1) is not closed, it cannot be compact. (this is theorem 26.3 in Munkres)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-01T12:03:50.010" />
  <row Id="12614" PostTypeId="1" CreationDate="2010-12-01T12:34:49.837" Score="0" ViewCount="81" Body="&lt;p&gt;Suppose a &lt;a href=&quot;http://en.wikipedia.org/wiki/Parabolic_mirror&quot; rel=&quot;nofollow&quot;&gt;parabolic mirror&lt;/a&gt; of a telescope is large enough (or equivalently that the camera &lt;a href=&quot;http://en.wikipedia.org/wiki/Shutter_speed&quot; rel=&quot;nofollow&quot;&gt;exposure time&lt;/a&gt; is short enough) that you need to take into account the transit time of the light signal from hitting the mirror to reaching the focal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What shape should the mirror be to have the same effect as a parabolic mirror that doesn't take into account light transit time ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added after comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to view surface details on an exoplanet with an array of space mirrors spanning millions or even billions of km, also redirecting photons towards a detector at a focal point. The property to preserve is photons from an event on the exoplanet arriving at the focal point at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the transit time from hitting the mirror to reaching the focus is long enough then given that objects aren't ideal or at infinity, then any discrepancy at all in arrival times is magnified by the huge transit-time so if the object being viewed changes fast it can change significantly during the discrepancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In talking about shutter speed I was thinking about viewing an object whose properties change very quickly (say clouds moving on a giant exoplanet). If there is a discrepancy in the time it takes for photons leaving the object to reach the focal point and the object changes during that time then the image would be blurred (temporally aberrated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross Millikan points out distances to the focus point should be the same, so I was misunderstanding things, however looking at Wikipedia exact focus only exists for point sources otherwise there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Coma_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;coma aberration&lt;/a&gt;. The article also says there is something called bestform or aplantic lenses to minimise this, but you can't apply a lense to an array of mirrors spanning billions of km, so the original question stands but with a changed understanding.&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="4157" LastEditorDisplayName="" LastEditDate="2010-12-01T15:50:52.493" LastActivityDate="2010-12-01T15:50:52.493" Title="Temporal aberration. Shape of telescope mirror large enough that light transit time is significant" Tags="&lt;geometry&gt;&lt;physics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="12615" PostTypeId="1" AcceptedAnswerId="12616" CreationDate="2010-12-01T12:42:11.133" Score="3" ViewCount="148" Body="&lt;p&gt;I know that there was a guy that could get 100 decimal digits of $\pi$ before computers were able to get thousands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the guy do that?&lt;/p&gt;&#xA;" OwnerUserId="3314" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T16:37:55.730" LastActivityDate="2010-12-01T16:37:55.730" Title="Get $\pi$ decimals manually" Tags="&lt;numerical-methods&gt;&lt;pi&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12616" PostTypeId="2" ParentId="12615" CreationDate="2010-12-01T12:48:11.177" Score="5" ViewCount="0" Body="&lt;p&gt;According to Wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Machin&quot; rel=&quot;nofollow&quot;&gt;John Machin&lt;/a&gt; combined the formula &#xA;$$\frac{\pi}{4}=4\cot^{-1}5-\cot^{-1}239$$&#xD;&#xA;with the Taylor series expansion for the inverse tangent in order to compute $\pi$ to 100 decimal places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A previous record was due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Abraham_Sharp&quot; rel=&quot;nofollow&quot;&gt;Abraham Sharp&lt;/a&gt; who used an arcsine series to find 72 decimal digits. &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-01T13:02:06.933" LastActivityDate="2010-12-01T13:02:06.933" />
  <row Id="12617" PostTypeId="2" ParentId="12572" CreationDate="2010-12-01T13:00:13.157" Score="1" ViewCount="0" Body="&lt;p&gt;Given an orthogonal function system $\phi_{n}(x)$ ($n=1,2,\dots$) with norm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\vert \left\vert \phi _{n}\right\vert \right\vert =\sqrt{\left( \phi_{n}\cdot \overline{\phi }_{n}\right) }=\displaystyle\sqrt{\displaystyle\int_{a}^{b}\phi _{n}\left( x\right) \,\overline{\phi }_{n}\left( x\right) \;dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and two functions $f(x)$ e $g(x)$ represented by the Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)\sim\displaystyle\sum_{n\ge 1} c_{n}\phi_{n}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(x)\sim\displaystyle\sum_{n\ge 1} d_{n}\phi_{n}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with coefficients $c_n,d_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n=\dfrac{\left ( f\cdot \overline{\phi }_{n}\right)}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}=\dfrac{\displaystyle\int_{a}^{b}f\left( x\right)\overline{\phi }_{n}\left( x\right) \;dx}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_n=\dfrac{\left ( f\cdot \overline{\phi }_{n}\right)}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}=\dfrac{\displaystyle\int_{a}^{b}g\left( x\right)\overline{\phi }_{n}\left( x\right) \;dx}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}\left\vert f(x)-\displaystyle\sum_{n=1}^{\infty}c_{n}\phi_{n}(x)\right\vert ^{2}dx=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}\left\vert g(x)-\displaystyle\sum_{n=1}^{\infty}d_{n}\phi_{n}(x)\right\vert ^{2}dx=0,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the following equality holds &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}f(x)\overline{g}_{n}(x)\; dx=\displaystyle\sum_{n\ge 1} c_{n}\overline{d}_{n}||\phi_n||^2,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a particular case of which, for $g(x)=f(x)$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\int_{a}^{b}|f(x)|^2\; dx=\displaystyle\sum_{n\ge 1} |c_{n}|^2||\phi_n||^{2}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T15:24:57.847" LastActivityDate="2010-12-01T15:24:57.847" />
  <row Id="12618" PostTypeId="2" ParentId="12615" CreationDate="2010-12-01T13:09:51.073" Score="3" ViewCount="0" Body="&lt;p&gt;This page on the &lt;a href=&quot;http://www-history.mcs.st-and.ac.uk/history/HistTopics/Pi_chronology.html&quot; rel=&quot;nofollow&quot;&gt;chronology of pi&lt;/a&gt; contains many useful notes on how the pre-computer era calculations of $\pi$ were performed.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-01T13:09:51.073" />
  <row Id="12619" PostTypeId="1" CreationDate="2010-12-01T13:14:59.247" Score="1" ViewCount="82" Body="&lt;p&gt;Hello!&#xA;I am looking for a website that contains a collection of functions, more than Matlab contains, to download I use from Matlab. &#xA;The extension should be *.m or the code should appear.&#xA;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="4255" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T16:41:17.973" LastActivityDate="2010-12-01T16:41:17.973" Title="Matlab: Website with a lot of functions to download" Tags="&lt;functions&gt;&lt;math-software&gt;&lt;matlab&gt;" CommentCount="2" />
  <row Id="12620" PostTypeId="1" CreationDate="2010-12-01T13:21:00.520" Score="2" ViewCount="53" Body="&lt;p&gt;The &lt;a href=&quot;http://stackoverflow.com/questions/4324490/gaussian-distribution-hash-tables&quot;&gt;original question&lt;/a&gt; I posted on StackOverflow. I think it's more mathematically inclined so I posted it here again. In terms of math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a class of students. Each students has a score between 0 and 300 (gaussian, with some known average and standard deviation). I need to find classes of marks (like 0-100, 101-125, 126-140, 141-160, 161-175, 176-200, 201-300) such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;The number of classes is minimum&lt;/li&gt;&lt;li&gt;The number of students in each class is minimum&lt;/li&gt;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do I go about doing this? Is this a standard problem? Also, is it possible to prove that only this set of classes will have the above property.&lt;/p&gt;&#xA;" OwnerUserId="4285" LastActivityDate="2010-12-01T14:07:28.767" Title="Gaussian Distribution + Hash Tables" Tags="&lt;probability&gt;&lt;normal-distribution&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12621" PostTypeId="1" AcceptedAnswerId="12624" CreationDate="2010-12-01T13:25:39.363" Score="2" ViewCount="46" Body="&lt;p&gt;Let us assume that we are given a closed, orientable 2D manifold embedded in $R^3$, and lets call it $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is clear that in a coordinate neighborhood $(U, \phi)$ it is possible to at each point define a unit normal vector $N_p$ by looking at the tangent space $T_pM$ as a subspace of $T_pR^3$ and considering the 1D vector space $T_pM^\perp$. However, there are always two choices of $N_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, an orientation should should define $N_p$ uniquely, but I am not sure how to use that $M$ is oriented to do this. If we have an orientation 2-form $\Omega$ on $M$, how can we use this to figure out which of the two possible $N_p$'s is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since we have $N_p$ defined at each point in $U$, we have a vector field on $U$. I was not sure how to show that this is a smooth vector field, but I believe that once you've done this, you can somehow use a partition of unity to extend $N$ to the entire manifold. But, I think that there is some subtlety even to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, for a non-orientable manifold such as a mobius strip, it seems to me that you can define locally a smooth unit-normal vector field, but when you try to extend it to the whole manifold it is no longer a continuous vector field, so somehow the orientation must play a role in the extension, and I am not sure how.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-01T14:45:04.013" Title="Unit Normal Field on a 2D manifold embedded in R^3" Tags="&lt;calculus&gt;&lt;topology&gt;&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="12622" PostTypeId="2" ParentId="12620" CreationDate="2010-12-01T13:47:15.827" Score="1" ViewCount="0" Body="&lt;p&gt;You question is a bit underdefined, but in general what you're looking for is index (order?) statistics. For example, if the number of classes is to be $2$, you need to cut them at the median. If it's $3$, you cut them at the $1/3$ and $2/3$ order statistics, and so on. The classes will be entirely evenly-spread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former paragraph assumed the classes can depend on the data. If the classes are not supposed to depend only on the distribution, then you just need to find the index statistics for the distribution. For example, the median of a Gaussian is equal to its mean, and for the other points you can consult a table (there is no closed form). You can then estimate how evenly spread the scores are going to be, e.g. the spread in the L2 spread is given by a $\chi^2$ distribution, whose mean you can look up in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T13:47:15.827" />
  <row Id="12623" PostTypeId="2" ParentId="12620" CreationDate="2010-12-01T13:47:23.690" Score="1" ViewCount="0" Body="&lt;p&gt;To have the minimum number of classes, take a class from 0 to 300.  To have the minimum number of students per class, take each different score to be a class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously, you generally need to define the number of classes you want, and whether you want all the classes to be about a given size.  You might want this when giving grades, say 50% A's, 40% B's and 10% C's.  For other applications you are just trying to group &quot;things that are close enough together that we can treat them the same.&quot;  If the distribution is a true Gaussian, with no noise, there is no nice answer.  But with a reasonably small sample, you can just plot a histogram, look for scores that are relatively rare, and make the cut there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more than one dimension it gets harder.  You might look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Cluster_analysis&quot; rel=&quot;nofollow&quot;&gt;cluster analysis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-01T13:47:23.690" />
  <row Id="12624" PostTypeId="2" ParentId="12621" CreationDate="2010-12-01T14:04:51.550" Score="2" ViewCount="0" Body="&lt;p&gt;There is a standard unit volume form $\omega = dx\wedge dy\wedge dz$&#xD;&#xA;on $\wedge R^3$. Contracting with repect to a unit vector $v$ gives&#xA;a 2-form $\omega&amp;#39;$. Pulling back to your surface if $v$ is normal&#xA;to $M$ at $p$ then $\omega&amp;#39;$ pulls back to a multiple of $\Omega$&#xD;&#xA;at $p$. If this is a positive multiple then $v$ and $\Omega$ are&#xA;compatibly oriented.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-01T14:45:04.013" LastActivityDate="2010-12-01T14:45:04.013" CommentCount="0" />
  <row Id="12625" PostTypeId="2" ParentId="12620" CreationDate="2010-12-01T14:07:28.767" Score="2" ViewCount="0" Body="&lt;p&gt;Statisticians would usually refer to your &quot;classes&quot; as &quot;bins&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your requirements for &quot;minimum number of bind&quot; and &quot;minimum size of the bins&quot; are in opposition, and you would have to clarify how the conflict should be resolved.  For instance, if one arrangement has 10 bins with 5 marks each, and another has 5 bins with 10 marks each, which is better?  Or, one arrangement has 10 bins with 10 marks each, and another has 10 bins total, 9 with 9 marks each, and the last one with 19 marks.  You have to specify a rule for how to identify which of any two arrangements is better.  Only then can you ask how to achieve the best one, and whether it is unique.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-12-01T14:07:28.767" CommentCount="1" />
  <row Id="12626" PostTypeId="1" AcceptedAnswerId="12643" CreationDate="2010-12-01T14:12:55.940" Score="2" ViewCount="94" Body="&lt;p&gt;Let $L,K$ be number fields and $L|K$ a galois extension.&#xA;Let $(0)\neq Q$ a prime ideal in $\mathcal O_L$ (=ring of integers in $L$) and $P=Q \cap \mathcal O_K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_Q $ denotes the decomposition field of $Q$ and $T_Q$ denotes the inertia field of $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $Q&amp;#39; :=Q\cap Z_Q$ and $Q&amp;#39;&amp;#39; :=Q\cap T_Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one prove, that $e(Q|Q&amp;#39;&amp;#39;)=e(Q|P)$ and $f(Q&amp;#39;&amp;#39;|Q&amp;#39;)=f(Q|P)$, if $e$ denotes the ramification index and $f$ the inertia degree?&lt;/p&gt;&#xA;" OwnerUserId="4287" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T17:09:36.673" LastActivityDate="2010-12-01T22:02:45.747" Title="Ramification index and inertia degree" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="12627" PostTypeId="2" ParentId="12614" CreationDate="2010-12-01T14:18:38.163" Score="1" ViewCount="0" Body="&lt;p&gt;As Willie Wong points out, this is not an issue for a parabolic mirror.  In fact, another way to think about optical focus is that it makes constructive interference at the image point.  This means that all light rays coming from the subject take the same travel time to reach the focus or else the interference wouldn't be constructive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that shutter speed doesn't matter unless you are thinking about exposing different parts of the sensor at different times (like the old 35mm curtain cameras) and the subject is moving.  Focus is focus for any shutter speed.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-01T14:18:38.163" CommentCount="6" />
  <row Id="12628" PostTypeId="2" ParentId="12610" CreationDate="2010-12-01T14:22:23.760" Score="0" ViewCount="0" Body="&lt;p&gt;The area of A is positive, while the area of P is zero.  Did you mean to distinguish between the interior of the circle and the closure of that set (i. e. the union of the perimeter and the interior)?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-01T14:22:23.760" />
  <row Id="12629" PostTypeId="1" AcceptedAnswerId="12632" CreationDate="2010-12-01T15:22:00.353" Score="10" ViewCount="244" Body="&lt;p&gt;Can you partition an infinite set, into an infinite number of infinite sets?&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-01T16:21:38.633" LastActivityDate="2011-01-09T06:16:48.997" Title="Partitioning an infinite set" Tags="&lt;elementary-set-theory&gt;&lt;infinity&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="12630" PostTypeId="2" ParentId="12629" CreationDate="2010-12-01T15:27:18.297" Score="7" ViewCount="0" Body="&lt;p&gt;Certainly.  Look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor pairing function&lt;/a&gt; that shows a correspondence between the naturals and pairs of naturals.  You have $f:\mathbb{N}\mapsto \mathbb{N\times N}$ which  is a byjection.  All sets of the form (x,y) for a given x are infinite.  So the sets of n that go to (x,y) for a given x are infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, this is too complicated.  $(x,y) \in \mathbb{N \times N}$ is infinite.  The sets of the form $(x,1)$ are infinite, as are $(x,2)$, as are $\ldots$.  So we have divided an infinite set into an infinite number of infinite sets.  The pairing function just projects this onto $\mathbb{N}$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-01T15:37:21.730" LastActivityDate="2010-12-01T15:37:21.730" />
  <row Id="12631" PostTypeId="2" ParentId="12615" CreationDate="2010-12-01T15:33:12.300" Score="1" ViewCount="0" Body="&lt;p&gt;See also &lt;a href=&quot;http://www.springer.com/mathematics/history+of+mathematics/book/978-0-387-20571-7&quot; rel=&quot;nofollow&quot;&gt;Pi: A Source Book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-01T15:33:12.300" />
  <row Id="12632" PostTypeId="2" ParentId="12629" CreationDate="2010-12-01T15:33:25.900" Score="14" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that it's enough to do it for countably infinite sets.  For if $X$ is uncountable and $Y$ is a countably infinite subset, then $Z = X-Y$ is also infinite.  Then, if we divide $Y$ into infintely many infinite sets, we've divided $X$ into infinitely many infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's consider a countable set $Y$.  Of course, we may as well label the elements of $Y$ as $0,1,2,...$.  In short, I'm going to break $\mathbb{N}$ into infinitely many infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that there are infinitely many prime numbers.  Further, if $p$ and $q$ are prime numbers, and if $p^a = q^b$, then we must have $p=q$ and $a=b$.  This follows from unique factorization of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for each prime $p$, consider the set $Y_p = \{ p^a$ with $a&amp;gt;0\in\mathbb{N}\}$.  Then, e.g., $Y_2 = \{2,4,8,16,32,64,128,...\}$ and $Y_3 =\{3,9,27,81,243,...\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, each $Y_p$ is infinite.  Further, if $Y_p$ and $Y_q$ have anything in common, then by what we said before, we must have that $p=q$.  In other words, for different primes $p$ and $q$, the sets $Y_p$ and $Y_q$ have no overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are infinitely many primes, there are infinitely many $Y_p$ with no overlap.  Finally, let $Z$ be all the naturals which are NOT powers of a prime number.  $Z$ is also infinite since, for example, it contains $2*3, 2*3^2, 2*3^3, 2*3^4,...$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we've dividied $\mathbb{N}$ into infinitely many infinite sets:  $Z$ and each of the $Y_p$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-12-01T15:33:25.900" />
  <row Id="12633" PostTypeId="2" ParentId="12629" CreationDate="2010-12-01T15:54:49.077" Score="6" ViewCount="0" Body="&lt;p&gt;For another example with a somewhat different flavour, chop the real line up at integers, dividing it into countable union of unit intervals, each of which has uncountably many points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{R} = \bigcup_{n \in \mathbb{Z}} [n,n+1)$$&lt;/p&gt;&#xA;" OwnerUserId="2439" LastActivityDate="2010-12-01T15:54:49.077" />
  <row Id="12635" PostTypeId="1" AcceptedAnswerId="12658" CreationDate="2010-12-01T15:59:04.203" Score="5" ViewCount="81" Body="&lt;p&gt;Update: Steve points out in comments that many direct products of residually cyclic groups will be counterexamples.  Indeed, I should have spotted this: every finitely generated abelian group is residually cyclic!  I should have asked about &lt;strong&gt;fully&lt;/strong&gt; residually cyclic groups, ie groups G such that for any finite subset $X\subseteq G\smallsetminus 1$, there is a homomorphism from G to a cyclic group that doesn't kill any elements of $X$.  I suspect the question is now quite easy, though I haven't had a chance to think about it yet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This question is motivated by &lt;a href=&quot;http://math.stackexchange.com/questions/9934/groups-having-at-most-one-subgroup-of-any-given-finite-index&quot;&gt;this recent question&lt;/a&gt;, which asked for a characterisation of groups with at most one subgroup of each finite index.  Arturo Magidin's answer showed that every such finite group is cyclic, but the questioner, Louis Burkill, added in comments that he is really interested in the infinite case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my answer to that question, I argue that a &lt;em&gt;finitely generated&lt;/em&gt; group has at most one subgroup of each finite index if and only if its canonical residually finite quotient, R(G), is cyclic. The `finitely generated' hypothesis is necessary - otherwise the additive group of the rationals provides a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, one can reduce from the case of G to R(G) as before (this gets rid of pathological examples like infinite simple groups), and the same argument shows that if R(G) has at most one subgroup of each finite index then R(G) is &lt;em&gt;fully&lt;/em&gt; residually cyclic, in particular abelian.  But the converse is not clear to me.  Hence the question in the title, which I'll reiterate here, with some extra hypotheses that should indicate where the difficulty lies.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If G is an infinitely generated, &lt;strong&gt;fully&lt;/strong&gt; residually cyclic (in particular, abelian) group, must G have at most one subgroup of each finite index?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="2287" LastEditorDisplayName="" LastEditDate="2010-12-01T18:34:22.190" LastActivityDate="2010-12-01T23:28:26.750" Title="If G is fully residually cyclic, does G have at most one subgroup of each finite index?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12636" PostTypeId="2" ParentId="12572" CreationDate="2010-12-01T16:27:58.560" Score="1" ViewCount="0" Body="&lt;p&gt;There is such a formula, but $f$ and $g$ live on different domains: Let $f: t\mapsto f(t)$ be periodic with period $1$ and $g: k\mapsto g(k)$ be a function defined on $\mathbb Z$. Both these functions have Fourier transforms defined by&#xA;$$\hat f(k)=\int_0^1 f(t) e^{-2\pi i k t} dt, \ \ \ \hat g(t)=\sum_{k=-\infty}^\infty g(k)e^{-2\pi i k t}.$$ Put $h(t):= \overline{\hat g(t)}$; this is a function of period 1. Then $$h(t)= \sum_{k=-\infty}^\infty \overline{g(k)}e^{2\pi i k t},$$ so the Fourier coefficients of $h$ are given by $\hat h(k)=\overline{g(k)}$.  Therefore, by Parseval's theorem for Fourier series one has $$\int_0^1 f(t)\hat g(t)dt = \int_0^1 f(t)\overline{h(t)}dt=\sum_{k=-\infty}^\infty \hat f(k)\overline{\hat h(k)}=\sum_{k=-\infty}^\infty \hat f(k) g(k).$$&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-01T16:27:58.560" CommentCount="0" />
  <row Id="12637" PostTypeId="1" CreationDate="2010-12-01T16:29:31.807" Score="4" ViewCount="124" Body="&lt;p&gt;How do we show the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a topological space and let $x \in X$. Show that if $x$ has a countable neighborhood basis in $X$ then $x$ has a countable neighborhood basis in $\beta X$. Here $\beta X$ denotes the Stone-Cech compactification of $X$.&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2010-12-01T16:29:31.807" Title="Stone-Cech compactification problem" Tags="&lt;topology&gt;" CommentCount="2" />
  <row Id="12638" PostTypeId="2" ParentId="12520" CreationDate="2010-12-01T16:47:46.237" Score="1" ViewCount="0" Body="&lt;p&gt;The answers so far do not catch the fact that the floor function has indeed some built in periodicity: The function $f(x):=\lfloor x\rfloor - x$ is in fact periodic and so can be developped into a Fourier series. &quot;For all practical purposes&quot; one has&#xA;$$\lfloor x\rfloor =x-{1\over2} +{1\over\pi}\sum_{k=1}^\infty{\sin(2\pi k t)\over k},$$&#xD;&#xA;but here the right hand side for $t\in\mathbb Z$ does not produce the correct value.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-01T16:47:46.237" />
  <row Id="12639" PostTypeId="1" AcceptedAnswerId="12649" CreationDate="2010-12-01T17:05:13.137" Score="4" ViewCount="115" Body="&lt;p&gt;My understanding of integration-by-parts is a little shaky.  In particular, I'm not totally certain that I understand how to properly calculate the limits of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the formula I have is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{v_1}^{v_2}{u dv} = (u_2 v_2 - u_1 v_1) - \int_{u_1}^{u_2}{v du}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to see how to calculate $u_1$ and $u_2$, preferably in a complete example (that solves a definite integral.)  I'm really interested in an example where the limits of integration change; i.e. $u_1$ and $u_2$ are different than $v_1$ and $v_2$, if possible.&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="2626" LastEditorDisplayName="" LastEditDate="2010-12-01T17:44:36.893" LastActivityDate="2010-12-01T18:54:32.033" Title="Understanding Limits of Integration in Integration-by-Parts" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12640" PostTypeId="2" ParentId="12608" CreationDate="2010-12-01T17:07:43.413" Score="1" ViewCount="0" Body="&lt;p&gt;Sets with a binary operation that is not assumed to be associative are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Magma_%28algebra%29&quot; rel=&quot;nofollow&quot;&gt;Magmas&lt;/a&gt; (they are sometimes also called &quot;quasigroups&quot;, especially in older literature, but that term has a different meaning in category Theory). Your set $S$ with operation $\otimes$ is a magma. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the operation happens to be associative, then you have &lt;a href=&quot;http://en.wikipedia.org/wiki/Semigroup&quot; rel=&quot;nofollow&quot;&gt;semigroup&lt;/a&gt;, though that will depend, in this case, on the binary operation *. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-01T17:07:43.413" CommentCount="1" />
  <row Id="12641" PostTypeId="1" AcceptedAnswerId="12650" CreationDate="2010-12-01T17:26:03.100" Score="1" ViewCount="120" Body="&lt;p&gt;Can you show me how to prove this::&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$66^{11 k+66} + 11^{101 k+55}\equiv \left\{\begin{array}{ll}&#xA;22 \pmod{165}&amp;amp; \text{if $k$ is odd,}\\&#xA;77 \pmod{165} &amp;amp; \text{if $k$ is even.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-01T17:55:23.933" LastActivityDate="2010-12-01T18:17:27.710" Title="hard problem of modular arithmetic" Tags="&lt;number-theory&gt;&lt;proof&gt;" AnswerCount="3" />
  <row Id="12642" PostTypeId="2" ParentId="12639" CreationDate="2010-12-01T17:31:22.747" Score="0" ViewCount="0" Body="&lt;p&gt;Okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_1^2 \ln x \, dx = [x \ln x]_{x = 1}^2 - \int_1^2 1 \, dx = 2 \ln 2 - 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prototype example. Where $u = \ln x$ and $v = x$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-01T17:31:22.747" CommentCount="3" />
  <row Id="12643" PostTypeId="2" ParentId="12626" CreationDate="2010-12-01T17:42:30.580" Score="1" ViewCount="0" Body="&lt;p&gt;This is essentially the argument in Daniel Marcus's &lt;em&gt;Number Fields&lt;/em&gt; (which I very highly recommend, especially for its exercises), Theorem 28 on page 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $G=\mathrm{Gal}(L/K)$ be the Galois group. The decomposition group of $Q$ is&#xA;$$D = \{ \sigma\in G\mid \sigma Q = Q\}$$&#xD;&#xA;and the inertia group of $Q$ is&#xA;$$E = \{\sigma\in G\mid \sigma(a)\equiv a\pmod{Q}\text{ for all $a\in \mathcal{O}_L$}\}.$$&#xA;Then $Z_Q$ is defined to be the fixed field of $D$, and $T_Q$ is the fixed field of $E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, let $[L:K]=n = efr$, where $e$ is the ramification degree and $f$ is the inertia degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I claim that $[Z_Q:K]=r$. Indeed, by the Fundamental theorem of Galois Theory, $[Z_Q:K] = [G:D]$. If $\tau\in G$, then every element of the coset $\tau D$ maps $Q$ to $\tau Q$; moreover, if $\tau Q=\rho Q$, then $\rho^{-1}\tau \in T$. So we have a one-to-one correspondence between the cosets of $D$ in $G$, and the primes over $P$ of the form $\tau Q$ with $\tau\in G$.  Since $L$ is Galois over $K$, the action of $G$ is transitive on the primes lying above $P$, and there are $r$ of them. So the index $[G:D]$ equals $r$, hence $[Z_Q:K]=r$, as claimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we show that $e(Q&amp;#39;|Q)=f(Q&amp;#39;|Q)=1$. Notice that $Q$ is the only prime of $\mathcal{O}_L$ that lies over $Q&amp;#39;$: because $\mathrm{Gal}(L/Z_Q)=D$ by the Fundamental Theorem of Galois Theory, and $D$ acts transitively on the primes of $\mathcal{O}_L$ lying over $Q&amp;#39;$; but every element of $D$ maps $Q$ to itself, so $Q$ is the only prime lying over $Q&amp;#39;$. Since $[L:Z_Q]=e(Q|Q&amp;#39;)f(Q|Q&amp;#39;)r(Q|Q&amp;#39;) = e(Q|Q&amp;#39;)f(Q|Q&amp;#39;)$, and since $erf=[L:K]=[L:Z_Q][Z_Q:K]=[L:Z_Q]r$, then $[L:Z_Q]=ef$. And since $e(Q|Q&amp;#39;)\leq e$ and $f(Q|Q&amp;#39;)\leq f$, it follows that we must have $e(Q|Q&amp;#39;)=e$, $f(Q|Q&amp;#39;)=f$. And since $e=e(Q|P) = e(Q|Q&amp;#39;)e(Q&amp;#39;|P)$, and $f=f(Q|P) = f(Q|Q&amp;#39;)f(Q&amp;#39;|P)$, then $e(Q&amp;#39;|P)=f(Q&amp;#39;|P)=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, since $1 = e(Q&amp;#39;|P) = e(Q&amp;#39;|Q&amp;#39;&amp;#39;)e(Q&amp;#39;&amp;#39;|P)$, then $e(Q&amp;#39;|Q&amp;#39;&amp;#39;)=1$. Thus, $e(Q|Q&amp;#39;&amp;#39;) = e(Q|Q&amp;#39;)e(Q&amp;#39;|Q&amp;#39;&amp;#39;) = e = e(Q|P)$, which proves the first equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we show that $f(Q|Q&amp;#39;&amp;#39;) = 1$, or equivalently that $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;$ is equal to $\mathcal{O}_L/Q$. For this it suffices to show the corresponding Galois group is trivial.  If we can establish this, then we will have &#xA;$f = f(Q|Q&amp;#39;) = f(Q|Q&amp;#39;&amp;#39;)f(Q&amp;#39;&amp;#39;|Q&amp;#39;) = f(Q&amp;#39;&amp;#39;|Q&amp;#39;)$, which will give the second equality you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $\mathcal{O}_L/Q$ is the trivial extension of $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;$, it is enough to show that for every $\overline{a}\in\mathcal{O}_L/Q$, the polynomial $(x-\overline{a})^m$ lies in $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;[x]$ for some positive $m$. If this is the case, then every element of the Galois group must send $\overline{a}$ to itself. Pick any preimage $a\in \mathcal{O}_L$ of $\overline{a}$. Then the polynomial&#xA;$$\prod_{\sigma\in E}(x - \sigma(a))$$&#xD;&#xA;has coefficients in the fixed field of $E$, that is, in $T_Q$; reduce modulo $Q&amp;#39;&amp;#39;$ to get a polynomial with coefficients in $\mathcal{O}_{T_Q}/Q&amp;#39;&amp;#39;$; since $\sigma(a)\equiv a \pmod{Q}$ for all $\sigma\in E$, the reduced polynomial is of the form $(x-\overline{a})^m$, with $m=|E|$. This proves that every element of $\mathcal{O}_L/Q$ is fixed by every element of the Galois group, so the extension is trivial, hence the inertia degree is $1$. This proves the second equality, as outlined above.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T22:02:45.747" LastActivityDate="2010-12-01T22:02:45.747" CommentCount="2" />
  <row Id="12644" PostTypeId="1" AcceptedAnswerId="13587" CreationDate="2010-12-01T17:45:53.383" Score="6" ViewCount="175" Body="&lt;p&gt;A student came to me showing a question from his exam in basic group theory, in which they are asked to prove that $\mathbb{C}^*$ modulo the subgroup of roots of unity is isomorphic to $\mathbb{R}^+$ (in both cases we mean the multiplicative groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this seems to me to be a simple error in the question. I believe they meant to ask to prove that $\mathbb{C}^*$ modulo &lt;strong&gt;all&lt;/strong&gt; the elements of absolute value 1 is isomorphic to $\mathbb{R}^+$ which is very easy to prove (take the homomorphism mapping $z$ to $|z|$ and use the first homomorphism theorem). However, I couldn't prove the claim about the roots of unity is wrong; is there an easy way to show this?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastActivityDate="2010-12-09T00:35:28.517" Title="Is $\mathbb{C}^*$ modulo the roots of unity isomorphic to $\mathbb{R}^+$?" Tags="&lt;group-theory&gt;&lt;complex-numbers&gt;&lt;counter-examples&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12645" PostTypeId="1" AcceptedAnswerId="12654" CreationDate="2010-12-01T17:46:56.190" Score="1" ViewCount="76" Body="&lt;p&gt;Hello everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently trying to do some Fourier transformations, or at least trying to understand them. The only thing I'm worried about is the complex part of the function. All I have is some basic, self thought, understanding about complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I'm concerned, a complex number exists of a real part and a complex part. In the following DFT it seems it has no real part. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/EUcLT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can simplify it to the following, since the others are just modifiers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{-i}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for some reason, im not getting the good value's. Perhaps i'm doing something wrong, but im not quite sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just realised this may be a bit vague, but my questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does the function F(k,l) return a complex number, with no real part.&lt;/li&gt;&#xA;&lt;li&gt;Is there a special rule about $e^{-i}$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, I'm sorry for asking these basic questions, this is all way beyond the math I've learned at school xD. But I understand most of the DFT now. Only there's one thing I don't understand and that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What does the $F(k,l)$ part of the function define. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3605" LastEditorUserId="3605" LastEditorDisplayName="" LastEditDate="2010-12-03T08:02:36.057" LastActivityDate="2010-12-03T08:02:36.057" Title="Getting the value of a Fourier Transform, problem with the complex part" Tags="&lt;fourier-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12646" PostTypeId="1" AcceptedAnswerId="12681" CreationDate="2010-12-01T17:46:57.720" Score="3" ViewCount="198" Body="&lt;p&gt;Has anyone ever come across a good cheat sheet for a list of definitions for the various algebraic structures out there, i.e. groups, fields, rings etc. Every time I come across the name of some structure, I have to look it up on wikipedia just to be sure I'm thinking of the right one, figured it would be cool to print out a cheat sheet and hang it on a wall nearby. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_structure&quot; rel=&quot;nofollow&quot;&gt;algebraic structure&lt;/a&gt; article on wikipedia is almost what I want, however it's a bit cryptic and lacks some structures, e.g. that of a vector space or a module.&lt;/p&gt;&#xA;" OwnerUserId="4291" LastActivityDate="2010-12-02T04:59:23.770" Title="Algebraic structure cheat sheet anyone ?" Tags="&lt;abstract-algebra&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="12647" PostTypeId="1" AcceptedAnswerId="12653" CreationDate="2010-12-01T17:51:31.473" Score="1" ViewCount="73" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose, $f(x,y)$ is a bounded continuous function on $\mathbb{R}^2$. Consider&#xA;$$\lim_{ y \rightarrow y&amp;#39; }\&amp;gt; \sup_{x \in \mathbb{R}} f(x,y).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In how far can you switch suprema and limits, or which possible term comes closest to switching these two? Ideally, there would be something like&#xA;$$\sup_{x \in \mathbb{R}}\&amp;gt; \lim_{ y \rightarrow y&amp;#39; } f(x,y).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T18:03:30.413" LastActivityDate="2010-12-01T19:37:22.027" Title="Switching Limits and Suprema" Tags="&lt;analysis&gt;" AnswerCount="1" />
  <row Id="12648" PostTypeId="1" AcceptedAnswerId="12699" CreationDate="2010-12-01T17:53:21.500" Score="5" ViewCount="323" Body="&lt;p&gt;I have an idea for solving the knapsack problem, but it looks too good to be true. I would like someone to explain potential problems with this approach. I'll give an example: I want to find a subset of {2,7,11} which has elements that sum to 13. Here is the algorithm for solving this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In binary notation 2=0010, 7=0111, 11=1011, 13=1101. Suppose that 2x+7y+11z=13, where x,y,z are {0,1}. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then y+z=1 (mod 2), since the last bits of 7, 11, and 13 are 1 and the last bit of 2 is 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And (y+z-1)+2*(x+y+z)=0 (mod 4), since the second to last bits of 2, 7, and 11 are 1 and the second to last bit of 13 is 0. (The y+z-1 is the carry bit from before.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And ((y+z-1)+2*(x+y+z)-0)+4y=4 (mod 8), since the third to last bits of 7 and 13 are 1 and the third to last bit of 2 and 11 are 0. (The (y+z-1)+2*(x+y+z)-0 is the carry bit from before.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally ((y+z-1)+2*(x+y+z)-0)+4y-4)+8z=8 (mod 16), since the first bits of 11 and 13 are 1 and the first bit of 2 and 7 are 0. (The ((y+z-1)+2*(x+y+z)-0)+4y-4 is the carry bit from before.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have four linear equations over a finite ring, if we change each one to mod 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8y+8z=8 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8x+12y+12z=4 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4x+14y+6z=10 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x+7y+11z=13 (mod 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, a solution is x=1, y=0, z=1, as one would expect, since the set {2,11} has elements that sum to 13. We could have gotten this through Gaussian elimination (according to answers to my last question on this site).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By request, here is the general algorithm that I just wrote (which can be induced from the above example): We want to solve a_1*x_1 + ... a_n*x_n = b, where each a_j and b is an integer and each x_j is in {0,1}. Putting each a_j and b in binary, we have a_j=(a_{mj} ... a_{1j}) and b=(b_m ... b_1), for suitable m. For instance, a_j=5 in binary is (a_{4j} ... a_{1j})=(0 1 0 1) - highest bit on the left side, lowest on the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the simple algorithm for constructing a new matrix A=(a_{ij}) from this original (a_{ij}) and a new b=(b_i) from this original b:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;`Let b_1 = 2^{m-1}*b_1;&#xA;For j=1 to n &#xA; Let a_{1j} = 2^{m-1}*a_{1j};&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For i=2 to m {&#xA; Let b_i = 2^{m-1}*b_i + b_{i-1}/2;&#xA; For j=1 to n&#xA;  Let a_{ij} = 2^{m-1}*a_{ij} + a_{(i-1),j}/2;&#xA;}`&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next solve the matrix equation Ax=b (mod 2^m), for the new A=(a_{ij}) and b=(b_i). If there is a solution to the knapsack problem, there should be a solution for x in {0,1} in Ax=b (mod 2^m) and vice versa (since the last row is just the original a_j and b). I'm curious to see if it works for large m and n. Anyone want to give it a shot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with this approach? It shouldn't work, but I don't know why.&lt;/p&gt;&#xA;" OwnerUserId="3163" LastEditorUserId="3163" LastEditorDisplayName="" LastEditDate="2010-12-02T02:10:03.117" LastActivityDate="2010-12-02T04:04:41.053" Title="knapsack algorithm that looks too good to be true" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="3" CommentCount="13" FavoriteCount="2" />
  <row Id="12649" PostTypeId="2" ParentId="12639" CreationDate="2010-12-01T17:53:22.890" Score="7" ViewCount="0" Body="&lt;p&gt;A more precise notation is this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx=\left(&#xD;&#xA;u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right) -\int_{x_{1}}^{x_{2}}u^{\prime&#xD;&#xA;}(x)v(x)dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is derived from the derivative rule for the product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(u(x)v(x))^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(x)v^{\prime }(x)=(u(x)v(x))^{\prime }-u^{\prime }(x)v(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx&#xD;&#xA;&amp;amp;=&amp;amp;\int_{x_{1}}^{x_{2}}(u(x)v(x))^{\prime }dx-\int_{x_{1}}^{x_{2}}u^{\prime&#xD;&#xA;}(x)v(x)dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left. (u(x)v(x))\right\vert&#xD;&#xA;_{x=x_{1}}^{x=x_{2}}-\int_{x_{1}}^{x_{2}}u(x)v(x)dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}u^{\prime }(x)v(x)dx.&#xD;&#xA;\end{eqnarray*}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write $dv=v^{\prime }(x)dx$ and $du=u^{\prime }(x)dx$, you get your&#xA;formula but with $u,v$ as a function of $x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{v_{1}(x)}^{v_{2}(x)}u(x)dv=\left(&#xD;&#xA;u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)&#xD;&#xA;-\int_{u_{1}(x)}^{u_{2}(x)}v(x)du$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Assume you want to evaluate $\int_{x_{1}}^{x_{2}}\log&#xD;&#xA;xdx=\int_{x_{1}}^{x_{2}}1\cdot \log xdx$. You can choose $v^{\prime }(x)=1$&#xD;&#xA;and $u(x)=\log x$. Then $v(x)=x$ (omitting the constant of integration) and &#xA;$u^{\prime }(x)=\frac{1}{x}$. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\int_{x_{1}}^{x_{2}}\log xdx &amp;amp;=&amp;amp;\int_{x_{1}}^{x_{2}}1\cdot \log xdx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}\frac{1}{x}\cdot xdx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(&#xD;&#xA;x_{2}-x_{1}\right) &#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The same example with your formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u=\log x,v=x,dv=dx,v=x,du=\frac{1}{x}dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{2}=\log x_{2},u_{1}=\log x_{1},v_{2}=x_{2},v_{1}=x_{1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\int_{v_{1}}^{v_{2}}udv &amp;amp;=&amp;amp;\left( u_{2}v_{2}-u_{1}v_{2}\right)&#xD;&#xA;-\int_{u_{1}}^{u_{2}}vdu \\&#xD;&#xA;\int_{x_{1}}^{x_{2}}\log xdx &amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot&#xD;&#xA;x_{1}\right) -\int_{\log x_{1}}^{\log x_{2}}xdu \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)&#xD;&#xA;-\int_{x_{1}}^{x_{2}}x\cdot \frac{1}{x}dx \\&#xD;&#xA;&amp;amp;=&amp;amp;\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(&#xD;&#xA;x_{2}-x_{1}\right). &#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The limits of integration, although different in terms of $u(x),v(x)$, when expressed in terms of the &lt;em&gt;same&lt;/em&gt; variable $x$ of functions $u(x),v(x)$ are the &lt;em&gt;same&lt;/em&gt; in both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a strategy on how to chose the $u$ and $v$ terms see &lt;a href=&quot;http://math.stackexchange.com/questions/10146/intuition-behind-the-ilate-rule&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-01T18:54:32.033" LastActivityDate="2010-12-01T18:54:32.033" />
  <row Id="12650" PostTypeId="2" ParentId="12641" CreationDate="2010-12-01T17:53:28.990" Score="2" ViewCount="0" Body="&lt;p&gt;It's  $\rm \equiv 0\ (mod\ 11)\:,\: $ and $\rm\: mod\ 15\ $ it's $\rm\ 6^n \equiv 6\ $ plus $\rm\ (-4)^n \equiv -4,\:1,-4\:, 1\:, \ldots\ $ equals $\rm\ 2, 7, 2, 7\:\ldots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence it's $\rm\ 22,77,22,77\:\ldots\ (mod\ 11*15)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-01T18:17:27.710" LastActivityDate="2010-12-01T18:17:27.710" />
  <row Id="12651" PostTypeId="2" ParentId="12641" CreationDate="2010-12-01T17:54:13.747" Score="7" ViewCount="0" Body="&lt;p&gt;First, $66^2 \equiv 66 \pmod{165}$. Therefore $66^{11k+66} \equiv 66 \pmod{165}$. Second, $11^3 \equiv 11 \pmod{165}$. Therefore for even $n &amp;gt; 0$, $11^n \equiv 121 \pmod{165}$, whereas for odd $n &amp;gt; 0$, $11^n \equiv 11 \pmod{165}$. If $n=101k+55$ then $n$ is odd iff $k$ is even. Thus for odd $k$ we get $66+121 \equiv 22 \pmod{165}$, and for even $k$ we get $66+11 \equiv 77 \pmod{165}$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T17:54:13.747" CommentCount="1" />
  <row Id="12652" PostTypeId="2" ParentId="12641" CreationDate="2010-12-01T17:54:38.723" Score="1" ViewCount="0" Body="&lt;p&gt;Since $165=3\times 5\times 11$, work modulo each of $3$, $5$, and $11$ separately and then combine them using the Chinese Remainder Theorem.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T18:00:15.907" LastActivityDate="2010-12-01T18:00:15.907" />
  <row Id="12653" PostTypeId="2" ParentId="12647" CreationDate="2010-12-01T18:09:55.803" Score="3" ViewCount="0" Body="&lt;p&gt;You cannot in general. To take a very simple example, consider $f(x,y) = \sin(xy)$, and $y&amp;#39;=0$. Then we have:&#xA;$$\lim_{y\to 0}\&amp;gt;\sup_{x\in\mathbb{R}}\sin(xy) = 1.$$&#xD;&#xA;(Remember that if we take the limit as $y\to 0$, then we do not consider $y=0$, so $y\neq 0$ in $\sin(xy)$). &#xA;But&#xA;$$\sup_{x\in\mathbb{R}}\&amp;gt;\lim_{y\to 0}\sin(xy) = \sup_{x\in\mathbb{R}}\ 0 = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the limit of $f(x,y)$ as $y\to y&amp;#39;$ exists for every $x$, and the limit of the suprema exist, then you get one inequality:&#xA;$$\sup_{x\in\mathbb{R}}\&amp;gt;\lim_{y\to y&amp;#39;}f(x,y) \leq \lim_{y\to y&amp;#39;}\&amp;gt;\sup_{x\in\mathbb{R}} f(x,y).$$&#xD;&#xA;To see this, note that for each fixed $x_0\in\mathbb{R}$, $f(x_0,y) \leq \sup\limits_{x\in\mathbb{R}}f(x,y)$, so taking limits we have&#xA;$$\lim_{y\to y&amp;#39;} f(x_0,y) \leq \lim_{y\to y&amp;#39;}\&amp;gt;\sup_{x\in\mathbb{R}}f(x,y).$$&#xD;&#xA;Since this holds for each $x_0$, the supremum also satisfies the inequality, so&#xA;$$\sup_{x\in\mathbb{R}}\&amp;gt;\lim_{y\to y&amp;#39;}f(x,y) \leq \lim_{y\to y&amp;#39;}\&amp;gt;\sup_{x\in\mathbb{R}}f(x,y).$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T19:37:22.027" LastActivityDate="2010-12-01T19:37:22.027" CommentCount="10" />
  <row Id="12654" PostTypeId="2" ParentId="12645" CreationDate="2010-12-01T18:20:56.897" Score="4" ViewCount="0" Body="&lt;p&gt;Your sum has both real and imaginary parts. Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{ix}= \cos(x) + i \sin(x) \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $f(a,b)$ in your formula a real number? Then the transform has real and imaginary components in general. (For certain choices of $f(a,b)$, you can have purely real numbers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean $e^{-i}$, then use the formula I provided you and you should be able to compute it.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-01T18:20:56.897" CommentCount="8" />
  <row Id="12655" PostTypeId="1" AcceptedAnswerId="12660" CreationDate="2010-12-01T19:13:38.683" Score="1" ViewCount="104" Body="&lt;p&gt;Just a quick question about the geometry of Hilbert spaces from an intuitive standpoint.  Maybe just assuming we're working with $L^2$ would simplify the situation. Basically, in something like $\mathbb{R}^2$ we have the situation that $\cos(\theta)=\frac{\langle a,b\rangle}{\vert a\vert\cdot\vert b\vert}$, and the idea of an angle between vectors is very meaningful, geometrically.  We can easily extend this idea to $\mathbb{R}^n$ because when we talk about the angle between two vectors, we mean we are choosing the plane that both of them lie in, and picking the vector in there.  But what does this really mean in a Hilbert space like $L^2$? I have a good intuition about functions, and about geometry (topology) separately, but not really the &quot;geometry of functions&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there may be no visualization of this in $L^2$, and I'm not asking for one, but is there any sense to doing geometry (i.e. actual polygons, things like that) in a space like $L^2$?  Also, what sort of applications do ideas like this have in functional analysis?  Are we ever interested in ideas like &quot;planes&quot; of functions, polygons, surfaces, solids, etc.? What do we really mean by angles, projections, normal vectors?  And do these sorts of things ever have any sort of interesting relationships?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm primarily asking for an intuitive idea here.  It's easy to just do the math, prove theorems about inner products, norms, etc. Maybe geometry gives some clues or intuitive ideas when doing functional analysis?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-02T04:23:52.583" Title="Soft Question Hilbert Space Geometry " Tags="&lt;soft-question&gt;&lt;functional-analysis&gt;&lt;hilbert-spaces&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="12656" PostTypeId="1" CreationDate="2010-12-01T19:16:14.860" Score="1" ViewCount="65" Body="&lt;p&gt;First let me start by saying, before you spend too much time on this, that it is still an open question, so i am just trying to gather some ideas and views on the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\Omega ,$ $\mathcal{A},$ $&#xD;&#xA;\mathbb{P)}$ be a probability space where $(B_{t})_{t\geq 0}$ is a &lt;em&gt;brownian motion&lt;/em&gt; starting at 0. Let us define the &lt;em&gt;local times&lt;/em&gt; at 0 : $(L_{t})_{t\geq 0}$ and its inverse $\tau _{s}=\inf \{t:L_{t}&amp;gt;s\}$. A simple consequence of the tanaka formula and the Levy characterisation of the &lt;em&gt;brownian motion&lt;/em&gt; is that :&#xA;\begin{align*}&#xA;$\tilde{B}_{t}=\left| B_{t}\right| -L_{t}=\int_{0}^{t}sgn(B_{s})dB_{s}$&#xD;&#xA;\end{align*}&#xA;is a &lt;em&gt;brownian motion&lt;/em&gt;.&lt;br&gt;&#xA;In other words, the &lt;strong&gt;Levy transform&lt;/strong&gt; : &#xA;\begin{align*}&#xA;T:\mathbb{W}&amp;amp; \longrightarrow \mathbb{W} \\&#xA;B&amp;amp; \longmapsto \tilde{B}&#xA;\end{align*}&#xA;defined almost everywhere on the Wiener space $\mathbb{W}=\mathcal{C}&#xD;&#xA;\mathbb{(R}_{+},\mathbb{R}\mathbb{)}$ leave the Wiener measure $\mu$ invariant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{F}_{t}=\sigma (B_{s},s\leq t)$ being the natural filtration of the brownian motion $B$, with the eventual addition of the &lt;em&gt;negligeables&lt;/em&gt; $\sigma$-field (generated by the element of probability 0) $\mathcal{N}$, $\mathcal{F}_{t}^{(1)}=\sigma (\tilde{B}_{s},s\leq t)$ and similarly $\mathcal{F}_{t}^{(n)}$ the natural filtration of $T^{n}(B)$, then one can wonder if the asymptotic filtration is trivial i.e. if : &#xA;\begin{equation*}&#xA;\forall t\geq 0\text{  }\bigcap_{n\geq 0}\mathcal{F}_{t}^{(n)}=\mathcal{%&#xA;N}&#xA;\end{equation*}&#xA;In a &lt;em&gt;Ergodic Theory&lt;/em&gt; context it means that the transformation is &lt;em&gt;exact&lt;/em&gt; implying that it is &lt;em&gt;ergodic&lt;/em&gt;. There is an indication that it is as the discrete case is. But the many attempts in the continuous case have been unsucessfull. Although we are making steps in the right direction, for example that orbits are dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see book by Revuz&amp;amp;Yor, articles, by Dubins, Smorodinsky, Malriq,... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2905" LastEditorUserId="2905" LastEditorDisplayName="" LastEditDate="2010-12-02T21:13:37.223" LastActivityDate="2010-12-02T21:13:37.223" Title="Is the Levy Transform Ergodic?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;stochastic-processes&gt;" CommentCount="4" />
  <row Id="12657" PostTypeId="1" AcceptedAnswerId="12666" CreationDate="2010-12-01T19:18:21.753" Score="8" ViewCount="119" Body="&lt;p&gt;This is a problem that a professor proposed for the highschool mathematical olympiad in Costa Rica that we haven't been able to solve. Therefore it cannot be asked since we don't have a solution yet in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal{F}_{k}$ for a fixed $k \in \mathbb{N}$   be the family of subsets $A_i \subset \mathbb{N}$ that satisfy the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The cardinality of $A_i$ is $k$ for every  index $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For every $A_i$ it holds that given any two different subsets of two elements $ \{ x, y \} \neq \{ z, w \}$, the absolute value of the differences between the elements of each subset are different&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |x - y| \neq |z - w|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we define a function $f: \mathcal{F}_k \rightarrow \mathbb{N}$ given by $$f(A_i) =  \max{A_i}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to find the minimum of the image of $f$, that is to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min{f(\mathcal{F}_k)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, we know that for $k = 4$ the answer is $\min{f(\mathcal{F}_4)} = 7$ and for $k = 3 $ the answer is $\min{f(\mathcal{F}_3)} = 4$ but we don't have a general pattern for the solution, basically these were done by &quot;brute force&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would very much appreciate any help with this problem. Thanks a lot in advance.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-01T20:54:27.800" LastActivityDate="2010-12-02T12:08:20.180" Title="Finding the least element of the greatest elements of certain subsets of the natural numbers" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="12658" PostTypeId="2" ParentId="12635" CreationDate="2010-12-01T19:32:34.033" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;The odds are against us.  Even &quot;fully&quot; is not enough.&lt;/strong&gt;&#xA;With the &quot;fully&quot; hypothesis, probably it is easy:  If G has two subgroups H,K of finite index n, then their intersection has finite index.  Take X to be a set of (non-identity) coset reps of G/H∩K, then there is a quotient where no element non-identity element of G/H∩K is sent to the identity &lt;s&gt;(and WLOG, G/H∩K is that quotient)&lt;/s&gt;, but since G is fully residually cyclic, that quotient is cyclic, and &lt;s&gt;the two subgroups must be identical by the lattice homomorphism theorem&lt;/s&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;In other words, you seem to have exactly answered your own question!&lt;/s&gt; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least K4 is no longer a counterexample.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-01T23:28:26.750" LastActivityDate="2010-12-01T23:28:26.750" CommentCount="6" />
  <row Id="12659" PostTypeId="2" ParentId="5819" CreationDate="2010-12-01T19:35:26.107" Score="1" ViewCount="0" Body="&lt;p&gt;As an addition: the separating closed disjoint sets part is often called normality (a space is normal if it satisfies this), so $T_4$ is normal &lt;strong&gt;plus&lt;/strong&gt; $T_1$, and similarly for regular: $T_3$ is regular &lt;strong&gt;and&lt;/strong&gt; $T_1$. Spaces that have no disjoint non-empty closed sets (besides the trivial topology, we have examples like $\mathbf{N}$ with the topology generated by the sets of the form $U(n) = \{ k : k \ge n \}$ e.g.) trivially satisfy normality. $T_4$ is to avoid these pathologies: the extra $T_1$ ensures that at least all finite sets are closed, so we have some &quot;relevant&quot; closed sets to apply normality to...&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-01T19:35:26.107" />
  <row Id="12660" PostTypeId="2" ParentId="12655" CreationDate="2010-12-01T19:45:09.070" Score="2" ViewCount="0" Body="&lt;p&gt;I think in infinite-dimensional spaces one should interpret the angle as a measure of how correlated two functions are.  The intuition manifests itself most clearly when the functions are random variables of mean zero; then their inner product is precisely their covariance, which is zero when they are independent (that is, uncorrelated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly one of the most important thing about Hilbert spaces is that there is a good notion of projection, and things one would expect to intuitively be true about projection (for example, that the projection of a vector onto a subspace is the part of the subspace closest to the vector) are actually true because it turns out the inner product makes everything work.  So in that sense, one can sometimes reliably use finite-dimensional intuition in the infinite-dimensional case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps one can also draw intuition from quantum mechanics, where the inner product is a &quot;probability amplitude.&quot;  I don't know enough about quantum mechanics to really elaborate on this point, though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T19:45:09.070" CommentCount="1" />
  <row Id="12661" PostTypeId="1" AcceptedAnswerId="12672" CreationDate="2010-12-01T20:10:21.987" Score="0" ViewCount="70" Body="&lt;p&gt;When ever I find definition of the quantum plane it says $A_q^2 = C&amp;lt;x,y&amp;gt;/I$, where $I = C&amp;lt;xy-qyx&amp;gt;$. What I want to know is whether they mean the unital free algebra or just the free algebra. In brief, is the quantum plane unital?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, when people write $A_q^N$, they mean the free (unital) algebra with $N$ generaterators, where every generator just commutes with every other generator? &lt;/p&gt;&#xA;" OwnerUserId="2099" LastActivityDate="2010-12-01T21:37:22.393" Title="Definition of the Quantum Plane" Tags="&lt;algebra&gt;&lt;quantum-groups&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12662" PostTypeId="1" AcceptedAnswerId="12663" CreationDate="2010-12-01T20:10:37.217" Score="4" ViewCount="120" Body="&lt;p&gt;Prove or disprove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\rho : \mathbb{N} \rightarrow \mathbb{N}$ injective. Let $(a_{n})_{n \in \mathbb{N}}$ be a sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ absolutely converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my understanding is that a series converges if the infinite sum of the series is the limit? It converges absolutely if $\displaystyle\sum\limits_{n=1}^\infty |a_{n}|$ converges. The way I am reading the $a_{\rho(n)}$ 's is that they denote some sort of permutation or rearrangement of the original series. Aside from that though, I am absolutely lost when it comes to approaching this problem... In general am I looking for an $\epsilon &amp;gt;0$ which is greater than $a_{\rho(1)}+ \dots +a_{\rho(n)}$ ? If so, how do I begin trying to find it?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T11:06:02.013" LastActivityDate="2010-12-02T19:48:05.987" Title="Proving or disproving statements related to convergent and absolutely convergent series" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="12663" PostTypeId="2" ParentId="12662" CreationDate="2010-12-01T20:30:19.370" Score="7" ViewCount="0" Body="&lt;p&gt;For (i):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\sum a_n$ converges to $a \in \mathbb{R}$. So if $\epsilon &amp;gt; 0$ there exists N such that if $n, l &amp;gt; N$ and $s_n = a_1 + \ldots + a_n$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|a - s_n| &amp;lt; \epsilon \text{    and    } \sum_{k = N + 1}^l |a_n| &amp;lt; \epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $M \in \mathbb N$ such that the terms $a_1, \ldots, a_N$ are contained as sum elements in $t_m = a_{\rho(1)} + \ldots + a_{\rho(M)}$. So now we have that for $m \geq M$ that $t_m - s_n$ is a sum of finitely many terms $a_l$ with $l &amp;gt; N$. So, for some $l &amp;gt; N$ we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|t_m - s_n| \leq \sum_{k = N + 1}^l |a_n| &amp;lt; \epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for $m \geq M$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|t_m - a| \leq |t_m - s_n| + |s_n - a| &amp;lt; \epsilon + \epsilon = 2\epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rearrangement converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) is not true. Let $\sum_n (-1)^n/n$ then take the $\rho$ which maps the integers injectively to the even integers.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-01T20:30:19.370" />
  <row Id="12664" PostTypeId="1" AcceptedAnswerId="12665" CreationDate="2010-12-01T20:31:42.973" Score="6" ViewCount="117" Body="&lt;p&gt;Let $G$ be a group and $G&amp;#39;$ its commutator subgroup. Let $\pi: G\to G/G&amp;#39;$ be the natural projection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1: $G/G&amp;#39;$ is the largest Abelian quotient of $G$ in the sense that if $H\unlhd G$ and $G/H$ is Abelian, then $G&amp;#39;\le H$. Conversely, if $G&amp;#39;\le H$, then $H\unlhd G$ and $G/H$ is Abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 2: If $\varphi:G\to A$ is any homomorphism of $G$ into an Abelian group $A$, then $\varphi$ factors through $G&amp;#39;$; i.e., $G&amp;#39;\le \ker{\varphi}$ and there is a homomorphism $\hat{\varphi}:G/G&amp;#39;\to A$ such that $\varphi(g) = (\hat{\varphi}\circ \pi)(g)$. (That is, we have a fancy commutative diagram.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from Dummit and Foote, p.169, Proposition 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of (1) is very straightforward. However, the authors claim that (1) is a restatement of (2) in terms of homomorphisms. Can anyone explain this? Because it is not clear to me. Also, if I wanted to prove (2) outright, what should the map $\hat{\varphi}$ be? My first thought was defining it as $\hat{\varphi}(aG&amp;#39;)= \varphi(a)$, but I don't think this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-12-01T22:00:09.133" Title="These two group theory statements are &quot;the same&quot;?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12665" PostTypeId="2" ParentId="12664" CreationDate="2010-12-01T20:41:53.207" Score="5" ViewCount="0" Body="&lt;p&gt;Your idea for (2) exactly works.  Any two representatives $a$, $b$ with $aG&amp;#39; = bG&amp;#39;$ will be related by an element $g&amp;#39; \in G&amp;#39;$ with $ag&amp;#39; = b$.  Then, because $G&amp;#39;$ is a subset of the kernel of $\varphi$, $$\varphi(b) = \varphi(ag&amp;#39;) = \varphi(a)\varphi(g&amp;#39;) = \varphi(a).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This really is a restatement of the first isomorphism theorem for groups to a &quot;first homomorphism theorem for groups&quot;, so to speak.  The isomorphism theorem says that map $\theta: G \to A$ with kernel $K$ will factor through the quotient map $G \to G/K$ --- but really for any subgroup $N$ of $K$ with $N$ normal in $G$ will give a map $G \to G / N$ by the same recipe described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific case, when the target is abelian, the kernel necessarily contains the commutator subgroup $G&amp;#39;$ of $G$, and so you use the &quot;first homomorphism theorem&quot; to get your map $\hat\varphi: G / G&amp;#39; \to A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the first isomorphism theorem can be weakened in this way means that, whereas the first isomorphism theorem alone relates surjective maps off $G$ to normal subgroups of $G$, the weakening says that the inclusions $N_1 \subseteq N_2$ of normal subgroups of $G$ is also reflected on the level of homomorphisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="2574" LastActivityDate="2010-12-01T20:41:53.207" CommentCount="1" />
  <row Id="12666" PostTypeId="2" ParentId="12657" CreationDate="2010-12-01T20:43:53.347" Score="8" ViewCount="0" Body="&lt;p&gt;Here are some more values, starting from $\mathcal{F}_1$: $$1,2,4,7,12,18,26,35,45,56,73,86,107,128,152,178,200,217,247,284,334,357,373,426,481,493,\ldots$$&#xD;&#xA;You're trying to find the shortest &lt;a href=&quot;http://en.wikipedia.org/wiki/Golomb_ruler&quot; rel=&quot;nofollow&quot;&gt;Golomb ruler&lt;/a&gt; of given size. I believe that the exact optimal values are unknown in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is &lt;a href=&quot;http://oeis.org/A003022&quot; rel=&quot;nofollow&quot;&gt;A003022&lt;/a&gt; (in which all numbers are $1$ less, e.g. $(0,)1,3,6,\ldots$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T12:08:20.180" LastActivityDate="2010-12-02T12:08:20.180" CommentCount="2" />
  <row Id="12667" PostTypeId="2" ParentId="12664" CreationDate="2010-12-01T20:44:21.803" Score="6" ViewCount="0" Body="&lt;p&gt;By the Homomorphism Theorem, any homomorphism $f\colon G\to K$ factors through $G/\mathrm{ker}f$, meaning that there is a map $\hat{f}\colon G/\mathrm{ker}f \to K$ such that $ f = \hat{f}\pi$. The map is indeed $\hat{f}(g\,\mathrm{ker}f) = f(g)$. This applies to the specific case given in Statement 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; In fact, the full Statement 1 is not equivalent to Statement 2, in the sense that if you replace $G&amp;#39;$ with an arbitrary subgroup $M$ of $G$ in both statements, then Statement 1 characterizes $G&amp;#39;$, but Statement 2 does not. That is, if you have&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Statement 1': If $H\triangleleft G$ and $G/H$ is abelian, then $M\subseteq H$; and if $M\subseteq H$, then $H\triangleleft G$ and $G/H$ is abelian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Statement 2': If $\varphi\colon G\to A$ is any homomorphism of $G$ into an abelian group $A$, then $\varphi$ factors through $M$; that is, $M\subseteq \ker\varphi$ and there is a homomorphism $\hat{\varphi}\colon G/M\to A$ such that $\varphi(g) = \hat{\varphi}\circ\pi(g)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only subgroup $M$ of $G$ that satisfies Statement 1' is $M=G&amp;#39;$. However, any subgroup of $G&amp;#39;$ that is normal in $G$ will satisfy Statement 2'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Statement 2 is equivalent to the first clause of Statement 1, namely that if $H\triangleleft G$ and $G/H$ is abelian, then $G&amp;#39;\subseteq H$, plus the implicit assertion that $G&amp;#39;$ itself is normal in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the first clause of Statement 1 plus the fact that $G&amp;#39;\triangleleft G$, if $\varphi\colon G\to A$ is a homomorphism, then by the Homomorphism Theorem, letting $H=\ker\varphi$, then $G/H$ is (isomorphic to) a subgroup of $A$, hence abelian, so we must have $G&amp;#39;\subseteq H = \mathrm{ker}\varphi$; this is Statement 2 (with the final clause of 2 given by the homomorphism theorem as above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming Statement 2, (which implicitly asserts that $G&amp;#39;$ is normal) suppose that $H$ is a normal subgroup of $G$ such that $G/H$ is abelian. Then considering $\pi\colon G\to G/H$ and applying 2, you conclude that $G&amp;#39;\subseteq \mathrm{ker}\pi = H$. And normality of $G&amp;#39;$ follows from the statement of 2, which requires it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, they aren't quite equivalent, because Statement 1 has &lt;em&gt;another&lt;/em&gt; clause, namely the &quot;Conversely...&quot; clause, which is not a consequence of assuming Statement 2. But the first part of Statement 1 (plus &quot;$G'\triangleleft G$&quot;) &lt;em&gt;is&lt;/em&gt; equivalent to Statement 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added earlier:&lt;/em&gt; To see that the two are not quite equivalent as stated, let me give you an example of a subgroup $M$ of $G$ that satisfies Statement 2' but not Statement 1': consider the case of $G=S_4$; then $G&amp;#39; = A_4$. Now let $M = \{ 1, (12)(34), (13)(24), (14)(23)\}$. Then $M\triangleleft G$, and the statement in 2 holds for $M$: given any homomorphism $f\colon G\to A$ with $A$ abelian, the map $f$ factors through $G/M$ and there exists a homomorphism $\hat{f}\colon G/M\to A$ such that $f=\hat{f}\pi$. However, $M$ is not the commutator subgroup of $G$.  What is missing in Statement 2 for it to be a true equivalent of Statement 1 is some statement that corresponds to the assertion that $G/G&amp;#39;$ is &lt;em&gt;itself&lt;/em&gt; abelian, which is what follows from the &quot;Conversely...&quot; clause in Statement 1. One way to do it is to simply state that $G/G&amp;#39;$ is itself abelian. Another is to consider the intersection of all kernels of all homomorphisms into abelian groups, and say that $G&amp;#39;$ must be equal to that intersection.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-01T22:00:09.133" LastActivityDate="2010-12-01T22:00:09.133" CommentCount="2" />
  <row Id="12668" PostTypeId="2" ParentId="12648" CreationDate="2010-12-01T20:48:28.740" Score="1" ViewCount="0" Body="&lt;p&gt;When you solve your equations mod 16, you get x, y, and z mod 16. Not mod 2. So all you have is integers x, y, and z between 0 and 15 inclusive such that 2x+7y+11z=13.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-01T20:48:28.740" />
  <row Id="12669" PostTypeId="2" ParentId="12662" CreationDate="2010-12-01T21:28:05.090" Score="6" ViewCount="0" Body="&lt;p&gt;I just want to give an alternative proof for (i).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that $\rho : \mathbb{N} \rightarrow \mathbb{N}$ is injective means that for every $n \in \mathbb{N}$, $\rho(n) \in \mathbb{N}$, and if $n \neq m$, then $\rho(n) \neq \rho(m)$. Define $S^*_n = |a_1| + |a_2| + \cdots + |a_n|$, and $\tilde S^*_n = |a_{\rho(1)}| + |a_{\rho(2)}| + \cdots + |a_{\rho(n)}|$. Obviously, since $\rho$ is injective, $\tilde S^*_n \leq |a_1| + |a_2| + |a_3| + \cdots$. But the right-hand side exists as a finite nonnegative number (since $\displaystyle\sum\nolimits_{n=1}^\infty a_{n}$ absolutely converges, or, equivalently, $S^*_n$ converges). So, $\tilde S^*_n$ is a monotone increasing sequence, bounded from above by $\sum\nolimits_{n = 1}^\infty  {|a_n |}$. Hence, $\tilde S^*_n$ converges to a finite nonnegative number $\tilde S^* \leq \sum\nolimits_{n = 1}^\infty  {|a_n |}$. That is, $\displaystyle\sum\nolimits_{n=1}^\infty a_{\rho(n)}$ converges absolutely to $\tilde S^*$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-01T21:28:05.090" />
  <row Id="12670" PostTypeId="1" AcceptedAnswerId="12673" CreationDate="2010-12-01T21:35:26.410" Score="2" ViewCount="86" Body="&lt;p&gt;Show that: $\sum _{k=n}^{\infty } \frac{1}{k!}\leq \frac{2}{n!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am clueless here, I tried to multiply both sides with $n!$ but it doesn't make things better. I know that the left one converges against $e$ for $n=0$, but I better don't want to use its numerical value.&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-12-01T21:42:59.960" Title="Series inequality" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12671" PostTypeId="2" ParentId="12670" CreationDate="2010-12-01T21:36:07.697" Score="4" ViewCount="0" Body="&lt;p&gt;Estimate it using a geometric series.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-01T21:36:07.697" CommentCount="2" />
  <row Id="12672" PostTypeId="2" ParentId="12661" CreationDate="2010-12-01T21:37:22.393" Score="0" ViewCount="0" Body="&lt;p&gt;Kassel, p. 3 ($k$ is the ground field):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An algebra is a ring together with a ring map $\eta_A : k \to A$ whose image is contained in the center of $A$. ... A morphism of algebras or an algebra morphism_ is a ring map $f : A \to B$ such that $f \circ \eta_A = \eta_B.$ (1.1)  As a consequence of (1.1), $f$ preserves the &lt;strong&gt;units&lt;/strong&gt;, i.e., we have $f(1) = 1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is also a definition of free algebras on p. 7 where he explicitly states that they have a unit.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-01T21:37:22.393" CommentCount="4" />
  <row Id="12673" PostTypeId="2" ParentId="12670" CreationDate="2010-12-01T21:42:59.960" Score="5" ViewCount="0" Body="&lt;p&gt;Hint:&#xA;$$\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots $$&#xD;&#xA;$$= \frac{1}{n!} \left( \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \cdots \right) $$&#xD;&#xA;$$ &amp;lt; \frac{1}{n!} \times \text{some geometric series}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-01T21:42:59.960" CommentCount="2" />
  <row Id="12674" PostTypeId="1" CreationDate="2010-12-01T21:58:29.103" Score="0" ViewCount="74" Body="&lt;p&gt;Say I have a numeric sequence A and a set of sequences B that vary with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that there is a relationship between one or more of the B sequences and sequence A,  that changes in Bn are largely or wholly caused by changes in sequence A.  However there is an unknown time delay between changes in A and their effect on each of the B sequences (they are each out of phase by varying amounts)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a means of finding the most closely correlating B to A regardless of the time delay.  What options are available to me? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;** EDIT **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of the problem here is that I have millions of B sequences to test, and there are approx 2 million data points within the lag window that I would like to test over.  Working out a correllation for each B for each possible lag scenario is just going to be too computationally expensive (especially as in reality there will be a more dynamic relationship than just lag between A and B, so I will be looking to test variations of relationships as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I am looking for is a means of taking the lag out of calculation.&lt;/p&gt;&#xA;" OwnerUserId="4301" LastEditorUserId="4301" LastEditorDisplayName="" LastEditDate="2010-12-02T15:14:05.743" LastActivityDate="2011-01-01T15:44:30.517" Title="Correlation between out of phase signals" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12675" PostTypeId="1" CreationDate="2010-12-01T22:06:59.783" Score="3" ViewCount="56" Body="&lt;p&gt;&lt;strong&gt;The context:&lt;/strong&gt;&#xA;I'm working on geocoded documents. This is, documents have  latitude and longitude attributes, as well as some other geo attributes such as an address. Right now, I'm performing text search operations on the indexed docs and showing their geolocation on a map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem:&lt;/strong&gt;&#xA;Most of search criterias include a location expressed by a name. But this name may appear in the document in fields other than those that express location. This usally gives irrelevan results on the map. &#xA;In other cases, a doc within a resultset may be wrongly geocoded.&#xA;I both cases, the consequence is that the map appears with some relevant results geographically grouped and a few irrelevant results scattered far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The required solution:&lt;/strong&gt;&#xA;I'm stuck trying to find an algorithm that processes the latitude/longitude of each doc in the results to determine which points are grouped and discard those that are not grouped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2011-01-02T17:44:32.993" Title="Algorithm for discarding geographically disgregate points" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12676" PostTypeId="1" AcceptedAnswerId="12682" CreationDate="2010-12-01T22:13:15.593" Score="2" ViewCount="74" Body="&lt;p&gt;What exactly is &quot;Approximation Theory&quot;? If I read the wikipedia-article I doesn't get much clearer. Why are &quot;pure&quot; mathematicians interested in it? I see a lot of people that do harmonic analysis also do approximation theory.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-01T22:27:58.087" LastActivityDate="2010-12-01T23:47:25.097" Title="Approximation Theory." Tags="&lt;reference-request&gt;&lt;approximation-theory&gt;" AnswerCount="1" />
  <row Id="12678" PostTypeId="2" ParentId="12591" CreationDate="2010-12-01T22:22:35.347" Score="1" ViewCount="0" Body="&lt;p&gt;I think that 'Stochastic Differential Equations' by Bernt Oksendal is a good book about SDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According your equation, lets rewrite it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dx = ax\,dt +b\,dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and add noise&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dx = (cx+b)\,dt + xg\,dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $g$ is a Wiener process, then $g\,dt$ is a Brownian motion, so your SDE equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dX_t(\omega) = (cX_t(\omega)+b)\,dt + X_t(\omega)dB_t(\omega)$$&lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2010-12-01T22:22:35.347" CommentCount="4" />
  <row Id="12679" PostTypeId="2" ParentId="12674" CreationDate="2010-12-01T22:30:01.530" Score="0" ViewCount="0" Body="&lt;p&gt;Will depend a bit on what kind of sequences you have, but assuming you are talking about discrete sequences, let's say $A=(A(t))_{t=-\infty}^{\infty}=(\ldots,A(-2),A(-1),A(0),A(1),A(2),\ldots)$ and the same for a sequence $B$. If your sequence doesn't run indefinitely, just put past values and future values equal to zero from a certain time on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can look at the following quantity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C(A,B;t_0,T,\tau)=\sum_{t=t_0}^{t_0+T} [A(t) - \bar{A}(t_0,T)][B(t-\tau)-\bar{B}(t_0-\tau,T)] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{A}(t_0,T) = \frac{1}{T+1} \sum_{t=t_0}^{t_0+T} A(t) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a moving average over a time window $T+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity $C(A,B;t_0,T,\tau)$ then measures correlation between signals $A$ and $B$. You can normalize it by dividing with the square roots of the autocorrelations of $A$ and $B$, that is $C(A,A;t_0,T,0)$ and $C(B,B;t_0-\tau,T,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me take a periodic sequence as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A= (\ldots,0,1,2,0,1,2,0,1,2,0,1,2,\ldots)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the pattern $(0,1,2)$ repeats itself indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time average over a window of 6 time units is $\bar{A}=(0+1+2+0+1+2)/6=1$ and this regardless of where I chose to start the sum. (I chose 6 on purpose, this independence need not be generally true.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The autocorrelation of A with itself is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C(A,A;0,6,\tau)=\sum_{t=0}^{6} [A(t) - 1][A(t-\tau)-1] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\tau=0$, this sum will be $$C(A,A;0,6,0)= [0-1]^2 + [1-1]^2 + [2-1]^2 + [0-1]^2 + [1-1]^2 + [2-1]^2 = 4 \; .$$ In fact, if $\tau$ is any multiple of 3, you should get the same response, since the function is periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\tau=1$, this sum will be $$C(A,A;0,6,1)= 2\left([0-1][2-1] + [1-1][0-1] + [2-1][1-1]\right) = -2 \; .$$ Which in absolute value is smaller than the previous result, meaning that the correlation is less strong, plus since the sign is negative, it can be interpreted as being closer to anti-correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's how you'll be able to read patterns from these correlation functions.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-01T23:53:39.087" LastActivityDate="2010-12-01T23:53:39.087" CommentCount="4" />
  <row Id="12680" PostTypeId="1" AcceptedAnswerId="12707" CreationDate="2010-12-01T22:40:52.470" Score="1" ViewCount="92" Body="&lt;p&gt;$S$ is a collection of disjoint sets. $(S,\cdot)$ is a commutative monoid and $(S,*)$ is a commutative semigroup. The identity element $e$ of $(S,\cdot)$ is the zero element of $(S,*)$.&#xA;The monoid $(S,\cdot)$ does not have a zero element. The binary operator $&amp;#39;*&amp;#39;$ is distributive over the binary operator $&amp;#39;\cdot&amp;#39;$. What can be said about such an algebraic structure ? What could be its usefulness ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a third binary operator $\otimes$ with which $(S,\otimes)$ is a commutative semigroup and the operator $&amp;#39;\otimes&amp;#39;$ is distributive over $&amp;#39;\cdot&amp;#39;$. $(S,\otimes)$ does not have a zero (absorption) element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation behind the question :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had something and wanted to check where it fits in a formalism and it happened to turn out like this. I have a naive question, why should i even bother about such a formalism like semigroup,semiring etc., how is it useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I thought it is OK/appropriate to edit a question to add to ask more on the same subject. Please let me know if it is not OK.I can make it as another question.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2010-12-02T16:52:20.410" LastActivityDate="2010-12-02T16:52:20.410" Title="On a set which is a monoid and a semigroup under two different binary operators with some properties." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12681" PostTypeId="2" ParentId="12646" CreationDate="2010-12-01T23:06:24.683" Score="4" ViewCount="0" Body="&lt;p&gt;This might not be as comprehensive as you wish, but here is one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.glump.net/content/abstract_algebra_cheat/index.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.glump.net/content/abstract_algebra_cheat/index.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few others here if you wish&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math-blog.com/2008/09/20/13-useful-math-cheat-sheets/&quot; rel=&quot;nofollow&quot;&gt;http://math-blog.com/2008/09/20/13-useful-math-cheat-sheets/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still reckon there is no substitute to doing it on your own. You can cater it to your needs and typeset it as beautifully as you want in tex&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-01T23:06:24.683" CommentCount="1" />
  <row Id="12682" PostTypeId="2" ParentId="12676" CreationDate="2010-12-01T23:47:25.097" Score="5" ViewCount="0" Body="&lt;p&gt;Approximation theory includes many subject areas of analysis, but the common idea is how well a target in a topological space (often a metric space) can be approximated by the points of a narrower subspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples will illustrate the breadth of this topic.  Given a real number x of a certain kind (e.g. algebraic), what are the rational numbers of bounded denominator that best approximate x, using the least absolute value of their difference as the objective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a real function $f$ on $[0,1]$ of a certain kind (e.g. twice continuously differentiable), what are the polynomials of bounded degree that best approximate $f$?  There are a variety of objectives that might be used, such as minimizing the square integral of the difference or minimizing the maximum difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems involving function approximation can be extended to higher dimensions, and the scope of approximating candidates can be varied endlessly (splines, trigonometric series, rational functions, etc.) and subject to many different restrictions (monotonicity, analyticity, symmetry, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a strong sense all of analysis uses approximation theory.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-01T23:47:25.097" CommentCount="2" />
  <row Id="12683" PostTypeId="1" CreationDate="2010-12-01T23:53:59.913" Score="0" ViewCount="44" Body="&lt;p&gt;I'm currently looking at the equation for a multiclass classifier of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{i=1}^{N_I} \sum\limits_{k=1,\atop k \neq y_i}^{N_K} L(1+ \mathbf{w_k}\cdot\mathbf{x_i}-\mathbf{w_{y_i}}\cdot\mathbf{x_i})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;br/&gt;&#xA;$X$ is a $N_K \times N_F$ data matrix&lt;br/&gt;&#xA;$y$ is a vector of class labels  &lt;br/&gt;&#xA;$W$ is an $N_K \times N_I$ matrix where each corresponds to the weights for the hyperplane splitting one class from the rest &lt;br/&gt;&#xA;$L$ is some arbitrary loss function returning a real number eg square loss, absolute, hinge etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been told that an answer for the above equation looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{w_k} = \sum\limits_{i=1 \atop y_i \neq k}^{N_I} \ L&amp;#39;(1+ \mathbf{w_k}\cdot\mathbf{x_i}-\mathbf{w_{y_i}}\cdot\mathbf{x_i})\cdot\mathbf{x_i}&#xD;&#xA;&#xD;&#xA;-&#xD;&#xA;\sum\limits_{i=1 \atop y_i =k}^{N_I} \sum\limits_{l=1,\atop l \neq k}^{N_K} L&amp;#39;(1+ \mathbf{w_k}\cdot\mathbf{x_i}-\mathbf{w_{y_i}}\cdot\mathbf{x_i})\cdot\mathbf{x_i}$$&#xD;&#xA;Where $L&amp;#39;$ is the derivative of the loss function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer doesn't make sense to me. Where does the second summation ($\sum\limits_{i=1 \atop y_i =k}^{N_I} \sum\limits_{l=1,\atop l \neq k}^{N_K} )$ come from? After all, in the initial equation, when $i$ is such that $y_i=k$, we skip over it. How then does this case have a large impact on the partial derivative? When I've attempted to do this myself, I've gotten the first part of the derivative but not the second.&lt;/p&gt;&#xA;" OwnerUserId="4286" LastEditorUserId="4286" LastEditorDisplayName="" LastEditDate="2010-12-03T01:04:17.290" LastActivityDate="2010-12-03T01:04:17.290" Title="Partial derivative of a multiclass linear classifier" Tags="&lt;calculus&gt;" CommentCount="2" />
  <row Id="12684" PostTypeId="1" CreationDate="2010-12-02T00:21:21.883" Score="0" ViewCount="68" Body="&lt;p&gt;To construct a multivariate B-spline, we simply take the Kronecker tensor product between the univariate basis functions constructed for each individual dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is how do you construct a B-spline derivative for a multivariate function? Is it the same method of using the tensor product, but with the univariate B-spline derivatives? &lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T00:45:18.820" LastActivityDate="2010-12-02T00:45:18.820" Title="Multivariate B-Spline Derivatives" Tags="&lt;approximation&gt;" CommentCount="5" />
  <row Id="12685" PostTypeId="2" ParentId="12646" CreationDate="2010-12-02T01:11:17.840" Score="2" ViewCount="0" Body="&lt;p&gt;What might be an even better question is How can I remember all of the different common algebraic structures? How do I keep track of all of the &quot;different&quot; definitions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use quotes because most of the definitions are very similar.&#xA;It is important that we understand homomorphisms : whenever someone says homomorphism they mean it preserves all of the structure that it can! (unit, product, sum, everything in sight)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly two rules that you always have whenever you could have them (for the most part) :&#xA;1) associative law : whenever there is an operation it is associative, this is a rule that only involves one operation. Again, almost every algebraic structure has every binary operation being associative... except for non-associative algebras.&#xA;2) distributive law : whenever there are two operations there is the distributive law. I am pretty sure this one is always there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most basic :&#xA;Monoids (some people call them different things, and there are more basic notions, but i dont think they are actively studied in algebra. In fact, I dont think monoids are actively studied, this definition is handy though for later purposes) : You have one operation and a unit identity element for that operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutative monoid : like above but the operation is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups : a monoid with inverses FOR EVERYTHING!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abelian Group : a commutative monoid with inverses FOR EVERYTHING!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point on everything is an abelian group plus some extra stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two operations :&#xA;these come in one of two flavors: another operation, or the action of something external on the guy you are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;another operation (usually we think of one of the opeerations as additive and the other as multiplicative, this is common and don't let it scare or worry you) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rings with unit : these are monoid objects in the category of abelian groups... whatever that means. The right way to think about that last sentence is that there is some structure sitting on top of the abelian group structure, and they play nicely together (the distributive law! multiplication by a fixed element is a homomorphism of abelian groups). (after we understand the rules for rings with units we can think about rings without units... if we even want to worry about them now. Some people take ring to mean ring with unit, it really depends on the book)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutative Ring with unit : like a commutative monoid in the category of abelian groups right? yep... whatever that means. Mostly just look at the above. Looking a tthe commutative versions of things is usually an easy transition to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Division algebra : This is like a group object in the category of abelian groups except there can't be an inverse for zero, or rather the additive identity. So $R$ is a division algebra if $x \in R\setminus 0$ has an inverse for all $a \in R$. Note here $R$ must have a multiplicative identity. Essentially $R$ is a ring and $R\setminus 0$ is a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Field: Like a division algebra except commutative, so $R$ is a ring and $R\setminus 0$ is an abelian group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;External operation : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these are Modules in one form or another. Always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Module: an abelian group with an action of a ring on it so that the ring action satisfies some distributivity condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vector space: A module over a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideal: A submodule of $R$. Note that if $R$ is a ring than we can think of it as acting on itself via multiplication, this is the sense in which an ideal is a submodule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later I may come back and add examples if you like.&#xA;Let me know if I left something out that you want me to add. I strongly think the above intuitive definition is far better than memorizing the equations one needs. If you learn something about commutative diagrams then you oly need the diagrams, and there are only really 2 or 3 of those. Then you will see it really is always the same concept.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T01:16:03.197" LastActivityDate="2010-12-02T01:16:03.197" CommentCount="1" />
  <row Id="12686" PostTypeId="2" ParentId="12269" CreationDate="2010-12-02T01:52:29.923" Score="2" ViewCount="0" Body="&lt;p&gt;As Aaron observes in the comments, this is in Strom's &quot;The homotopy category is a homotopy category.&quot;&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-02T01:52:29.923" CommunityOwnedDate="2010-12-02T01:52:29.923" />
  <row Id="12687" PostTypeId="3" CreationDate="2010-12-02T02:12:13.200" Score="0" ViewCount="0" Body="&lt;p&gt;I have been an active member of math.SE since the public beta, and before that I was an active member of MathOverflow; I have participated in meta on both sites.  While MathOverflow is a uniquely valuable resource to the research mathematics community, math.SE is a valuable resource to the wider community of those interested in mathematics, and for that reason I am committed to seeing it continue to grow and to run smoothly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I check math.SE frequently, so it would be relatively easy for me to carry out janitorial duties such as closing spam questions.  I am particularly interested in merging tags; they are quite cluttered at the moment.  I won't take a particularly active role in closing questions other than those which obviously don't belong.  I just want to make it easier for interesting questions to be found and answered.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-02T02:12:13.200" LastActivityDate="2010-12-02T02:12:13.200" CommunityOwnedDate="2010-12-02T02:12:13.200" CommentCount="6" />
  <row Id="12688" PostTypeId="2" ParentId="12003" CreationDate="2010-12-02T02:16:21.710" Score="1" ViewCount="0" Body="&lt;p&gt;This is an interesting rhetorical question. On the one hand, many theoretical mathematical concepts transcend arithmetic. However, computer algebra systems (such as Mathematica) are able to express answers to those sorts of questions using a CPU that performs a limited range of simple computations. So the answer is a matter of opinion. Or there is no answer?&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2010-12-02T02:16:21.710" />
  <row Id="12689" PostTypeId="1" AcceptedAnswerId="12717" CreationDate="2010-12-02T02:32:32.990" Score="4" ViewCount="78" Body="&lt;p&gt;A little help here. Exercise 21, Ch. 2 from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471257087&quot; rel=&quot;nofollow&quot;&gt;Feller's book&lt;/a&gt; reads &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a town a $n+1$ inhabitants, a person tells a rumor to a second person, who in turn repeats it to a third person, etc. At each step, the recipient of the rumor is chosen at random from the $n$ people available. Find the probability that the rumor will be told $r$ times without: a) returning to the originator, b) being repeated to any person. Do the same problem when at each step the rumor is told by one person to a gathering of $N$ randomly chosen people. (The first question is the special case N=1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I already did a) and b) for the first description of the problem and a) for the case when the rumor is spreading through a gathering of $N$ people, however, my solution for b) in this second case is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reasoned in the following way: In a first instance, $n$ people to receive the rumor, however, it's needed to spread such rumor through a group of $N$ people, therefore, there are $\displaystyle n \choose N$ ways to choose those gatherings. Once one of these people is chosen, he/she can choose from another gathering of $N$ people, taking care of not choosing someone who already know the rumor, which is, there are $\displaystyle n-1 \choose N$, and so on, until we reach the $r$ step in this process. Therefore, the probability I get is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\displaystyle {n \choose N} {n-1 \choose N} {n-2 \choose N} ... {n-r+1 \choose N}}{\displaystyle {n \choose N}^{r}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the book, the solution must be $\displaystyle \frac{(n)_{Nr}}{(n_{N})^{r}}$ (which is not equivalent to the first expression).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will appreciate any help.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-02T03:51:20.727" LastActivityDate="2010-12-02T05:04:38.650" Title="Probability on spreading of rumors" Tags="&lt;probability&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12690" PostTypeId="1" CreationDate="2010-12-02T02:50:17.483" Score="3" ViewCount="73" Body="&lt;p&gt;I understand Lie groups are defined by the structure constants associated with the lie brackets, which are treated as commutators in quantum mechanics, but i dont know of a math theory related to group theory to define or use an anti commutator. If Lie groups theory uses the commutator, what theory uses the anti commutator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finite groups (not Lie groups, which are continuous), can be specified by structure equations analogous to a Lie bracket, but more general, of which commutation or anti commutation relations are just one of infinite possibilities. Is there such variety of possible structure equations in continuous groups too?&lt;/p&gt;&#xA;" OwnerUserId="4307" LastActivityDate="2010-12-02T23:06:19.323" Title="What is the formal mathematical theory behind the concept of the anti commutator used to quantize fermions?" Tags="&lt;lie-groups&gt;" AnswerCount="2" />
  <row Id="12691" PostTypeId="2" ParentId="12488" CreationDate="2010-12-02T02:51:53.617" Score="3" ViewCount="0" Body="&lt;p&gt;I agree with all the commenters who say that you should just find a planar embedding. However, I happened to stumble across a description that might make you happy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $G$ be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Connectivity_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;three-connected&lt;/a&gt;&#xA;  planar graph and let $C$ be a cycle.&#xA;  Let $G/C$ be the graph formed by&#xA;  contracting $C$ down to a point. Then&#xA;  $C$ is a face of the planar graph if&#xA;  and only if $G/C$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Connectivity_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;two-connected&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, this lemma is useful in proving that a three-connected graph can only be planar in one way, a result of Whitney.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But testing this for every cycle is much less efficient than just finding the planar embedding.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-02T02:51:53.617" />
  <row Id="12692" PostTypeId="2" ParentId="12690" CreationDate="2010-12-02T03:07:29.580" Score="5" ViewCount="0" Body="&lt;p&gt;If Lie algebras are (in light of the Poincare-Birkhoff-Witt theorem) a complete axiomatization of the antisymmetric multiplication $AB-BA$ in an associative algebra, Jordan algebras are an almost-complete axiomatization of the symmetric multiplication $(AB+BA)/2$. ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Jordan_algebra&lt;/a&gt; ).  They are the closest thing known to an intrinsic structure related to anticommutators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike Lie algebras, there are exceptional Jordan algebras that do not come from the multiplication in associative algebras, and Jordan algebras are not known to be infinitesimal objects for groups or any other structure.  So they do not arise as often outside of (or in) quantum mechanics and the initial hopes for a Lie-like theory have not been realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean not the anti-commutator but the supercommutator, $AB \pm BA$ with the sign depending on parity of $A$ and $B$ or their consituents (e.g., negative for bosons and positive for fermions) there is a theory of Lie superalgebras and Lie supergroups. There are new phenomena in the classification, such as continuous families of nonisomorphic simple objects.  The theory can be phrased as the &quot;Lie group theory in the category of super-vector spaces&quot;, which in turn is a special case of Lie theory (or group theory, or algebraic geometry) in a tensor category.  So in principle there are many possible theories of commutator-like objects, but I don't know if any have been found to be interesting besides the usual theory, its super-version, and analogues in characteristic $p$.  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-02T19:20:50.783" LastActivityDate="2010-12-02T19:20:50.783" />
  <row Id="12693" PostTypeId="2" ParentId="12269" CreationDate="2010-12-02T03:18:40.370" Score="4" ViewCount="0" Body="&lt;p&gt;This is in many places. I think that the whole point of model categories is that people saw you could get a lot of mileage out of the adjunction between Top and sSet (geometric realization and singular simplices). Just how good is the combinatorial model for topological spaces? The answer is that it is a Quillen equivalence (whatever that means). But for that sentence to exist we need to talk about model categories. I would say that Top is one of the most basic examples, before simplicial sets and chain complexes. For one thing the model structure on sSet requires use of the model structure on Top (the standard definition of a weak equivalence of simplicial sets is that its geometric realization is a weak equivalence in Top). The model structure on chain complexes sort of came out of homological algebra, and i don't know how early it was realized that this model category stuff shed light on the homological algebra that was going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When thinking about this stuff I find the model theoretic approach helps me understand what is going on in top. Also, the weak equivalences are the maps that induce isos in homotopy for all choices of basepoints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps Peter May has a write up of a verification of the model category axioms for Top under Misc notes on his website. (There is tons of great stuff on there)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pps Dwyer and Spalinski is a great resource as well &lt;a href=&quot;http://www.nd.edu/~wgd/&quot; rel=&quot;nofollow&quot;&gt;http://www.nd.edu/~wgd/&lt;/a&gt; number 75&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-02T03:18:40.370" CommentCount="2" />
  <row Id="12694" PostTypeId="1" CreationDate="2010-12-02T03:27:32.737" Score="1" ViewCount="131" Body="&lt;p&gt;So i've been stuck on this problem for about an hour. I can't figure out how to do it, and help would be absolutely amazing. This is what's given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 + 1 = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x (x + 0 = x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall x \forall y [x + (y+1)= (x+y)+1]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[0 + 1 = 1 + 0 \wedge \forall x(x + 1 = 1 + x \rightarrow (x+1)+1 = 1+(x+1)))] \rightarrow  \forall x (x+1=1+x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with these premises, I need to get $\forall x (x+1=1+x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help would be great. I keep getting closeish, only to find I'm actually doing it totally wrong.&lt;/p&gt;&#xA;" OwnerUserId="4309" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T04:11:21.460" LastActivityDate="2010-12-02T17:29:35.757" Title="Formal logic to prove x+1 = 1 +x" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12695" PostTypeId="2" ParentId="10512" CreationDate="2010-12-02T03:32:04.433" Score="0" ViewCount="0" Body="&lt;p&gt;This is really a comment on ryan'sanswer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to disagree with Ryan. Group cohomology was in its early stages before Eilenberg and Maclane came along. There are awful and ugly formulations of just $H^1$ and $H^2$ that lead me to believe that they must have been formulated before E&amp;amp;M did their work. I am thinking of factor sets and cocycle conditions that people wrote down a long time ago. I think this is a matter of history though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just checked wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_cohomology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Group_cohomology&lt;/a&gt; the history section near the end. It seems to agree with what I said above.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-02T03:49:56.953" LastActivityDate="2010-12-02T03:49:56.953" />
  <row Id="12696" PostTypeId="2" ParentId="12655" CreationDate="2010-12-02T03:43:26.097" Score="0" ViewCount="0" Body="&lt;p&gt;Let us say we are dealing with real Hilbert space $X=L^2[0,1]$. Any two functions $f,g\in X$ are contained in the plane $P$ spanned by $f,g$. This plane is isometrically isomorphic to the Euclidean space $\mathbb{R}^2$; this means that $P$ is spanned by two orthonormal vectors $e_1, e_2$ (any two orthonormal vectors), and the linear map $T$ that maps $e_1$ to $(1,0)$ and $e_2$ to $(0,1)$  preserves distance and inner product (and hence also angle). The angle between $f$ and $g$ is precisely the angle between $Tf$ and $Tg$ in $\mathbb{R}^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f, g$ are linearly independent, $e_1,e_2$ can be obtained from them by Gram-Schmidt process, i.e.&#xA;$$ e_1= \frac{f}{\|f\|},  e_2=\frac{g-(e_1,g)e_1}{\|g-(e_1,g)e_1\|}.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-02T04:23:52.583" LastActivityDate="2010-12-02T04:23:52.583" />
  <row Id="12697" PostTypeId="2" ParentId="12648" CreationDate="2010-12-02T03:54:24.893" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not sure how one sees that finding solutions to the linear system over {0,1} can be accomplished in polynomial time.  Reducing the matrix to row-echelon form is not sufficient for finding such solutions.  Suppose that the reduced system has a parametrized solution set, as happens in your example.  Is there any way, short of plugging in 0 and 1 for the free parameters in all possible ways and then checking the values of the remaining variables, of finding {0,1} solutions?  Maybe, but I don't think this is immediately clear.  It seems possible that there might be many free parameters, and that there would then be a combinatorial explosion in the number of cases that needed to be checked in order to find {0,1} solutions.&lt;/p&gt;&#xA;" OwnerUserId="3736" LastActivityDate="2010-12-02T03:54:24.893" />
  <row Id="12698" PostTypeId="1" AcceptedAnswerId="12701" CreationDate="2010-12-02T04:01:40.807" Score="5" ViewCount="149" Body="&lt;p&gt;I want to why $\mathbb{Z}[x]/(1-x,p)$ is isomorphic to $\mathbb{Z_{p}}$ where p is a prime integer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I have so far, but I am unsure if I am correct.   Every $f\in \mathbb{Z}[x]$ can be written as $(1-x)q+ r$ where $q\in \mathbb{Z}[x]$ and $r$ is in $\mathbb{Z}$.   Does It follows that there are $p$ cosets of $(1-x,p)$  ( namely 0+(1-x,p), 1+(1-x,p),2+(1-x,p)  etc ..) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would imply $\mathbb{Z}[x]/(1-x,p)$ is isomorphic to $\mathbb{Z_{p}}$&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-02T17:49:30.520" Title="Why is $\mathbb{Z}[x]/(1-x,p)$ isomorphic to $\mathbb{Z_{p}}$ where p is a prime integer." Tags="&lt;abstract-algebra&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="12699" PostTypeId="2" ParentId="12648" CreationDate="2010-12-02T04:04:41.053" Score="7" ViewCount="0" Body="&lt;p&gt;The problem is that the equations generated are all linearly dependent on the original equation, so finding solutions of the matrix is just as hard as finding solutions to just the original equation. It works in the given example only because the matrix is so small that trial-and-error makes a solution apparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use your example, let $S = \{2,7,11\}$ and $N = 13$. We want to find a subset of $S$ whose elements sum to $K$. Looking at your generated solutions modulo $2^n$, notice that&#xA;$$2x + 7y + 11z \equiv 13 \pmod{16} \, \Rightarrow \, 4x + 14y + 6z \equiv 10 \pmod{16}$$&#xD;&#xA;$$4x + 14y + 6z \equiv 10 \pmod{16} \, \Rightarrow \, 8x + 12y + 12z \equiv 4 \pmod{16}$$&#xD;&#xA;$$8x + 8y + 12z \equiv 4 \pmod{16} \, \Rightarrow \, 8y + 8z \equiv 8 \pmod{16}$$&#xD;&#xA;Since the reverse implications are not true in general, &quot;simplifying&quot; the equation in this manner does no good, as solving a simpler equation has no effect than to rule out possibilities (and as the problem scales it becomes infeasible to keep track of all remaining possibilities).&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-12-02T04:04:41.053" CommentCount="1" />
  <row Id="12700" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:07:37.480" Score="0" ViewCount="0" Body="&lt;p&gt;Remember that finite fields are isomorphic if and only if they have the same number of elements.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-02T04:07:37.480" CommentCount="1" />
  <row Id="12701" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:08:41.387" Score="5" ViewCount="0" Body="&lt;p&gt;Basically, yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit easier if you consider it in steps, though. First, consider $\mathbb{Z}[x]/(p)$, which is isomorphic to $\mathbb{Z}_p[x]$; then consider $\mathbb{Z}_p[x]/(1-x)$, which is isomorphic to $\mathbb{Z}_p$ under the map &quot;evaluate at 1&quot; (or using the division algorithm, like you do above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can do it the other way. First consider the quotient $\mathbb{Z}[x]/(x-1)$, which is isomorphic to $\mathbb{Z}$ under evaluation at $x=1$; then mod out by $(p)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(What you are doing if you go this route is using the homomorphism theorems; the ideal $(p + (x-1))$ of $\mathbb{Z}[x]/(x-1)$ corresponds to the ideal $(p,x-1)$ of $\mathbb{Z}[x]$, and $(R/I)/(J/I)\cong R/J$, where $I$ and $J$ are ideals of $R$ with $I\subseteq J$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T04:16:30.017" LastActivityDate="2010-12-02T04:16:30.017" />
  <row Id="12702" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:10:45.877" Score="2" ViewCount="0" Body="&lt;p&gt;Define a map $\mathbb{Z}[x]\to \mathbb{Z}_p$ that sends $f(x)$ to the equivalence class of $f(1)$ modulo $p$. This is a homomorphism of rings. Since, every $f(x)$ can be written as $(x-1)q+r$ as you did, this map is surjective (just look at the images of the constant functions $0,1,...,p-1$), and the kernel is precisely $(x-1,p)$. This gives the required isomorphism by the first isomorphism theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting cosets only gives you an abstract isomorphism, while this gives you a canonical one.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T04:10:45.877" />
  <row Id="12703" PostTypeId="2" ParentId="12694" CreationDate="2010-12-02T04:12:53.720" Score="5" ViewCount="0" Body="&lt;p&gt;The last axiom is basically your induction statement: it says that if you can prove that $x+1 = 1+x$ for $x=0$, and you can prove that if it holds for $x$ then it holds for $x+1$, then you can conclude that the statement $x+1=1+x$ holds for all $x$. So you want to use your first and second axioms, as well as your third axiom, to show that the premises of the last axiom hold. That will give the conclusion you want.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T04:12:53.720" />
  <row Id="12704" PostTypeId="1" AcceptedAnswerId="12712" CreationDate="2010-12-02T04:30:10.350" Score="3" ViewCount="62" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume I have linear equation system in $n$ variables and the solution is $x_1=x_2=\cdots = x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What properties must the matrix of the system have, or how do I prove that the system has such a solution?&lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T06:47:33.773" LastActivityDate="2010-12-30T06:47:33.773" Title="matrix properties if all the solutions are the same" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" />
  <row Id="12705" PostTypeId="1" AcceptedAnswerId="12710" CreationDate="2010-12-02T04:32:06.397" Score="3" ViewCount="60" Body="&lt;p&gt;What does $\mathbb{Z}[\omega]$ usually mean when $\omega$ is a primitive root of unity?&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-02T04:39:41.703" Title="Definition of $\mathbb{Z}[\omega]$ where $\omega$ is a primitive root of unity." Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12706" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:32:23.993" Score="2" ViewCount="0" Body="&lt;p&gt;Just to add something to the nice answers you already have. In fact this is a special case of a more general proposition. It can be shown that all the maximal ideals $\mathfrak{m}$ of the polynomial ring $\mathbb{Z}[x]$ have the form $\mathfrak{m} = (g(x), p)$ where $p \in \mathbb{Z}$ is a prime number and $g(x) \in \mathbb{Z}[x]$ satisfies that its reduction modulo $p$ is an irreducible element $\bar{g}(x)$ in the polynomial ring $\mathbb{Z}_{p}[x]$. As a consequence of this you get that the quotient ring $ \mathbb{Z}[x] / \mathfrak{m} $, which in fact is a field since $\mathfrak{m}$ is maximal, is a finite algebraic extension field of $\mathbb{Z}_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find this theorem for instance as &lt;strong&gt;proposition 1.5&lt;/strong&gt; in page 22 of Miles Reid's &lt;a href=&quot;http://books.google.co.cr/books?id=mUL1us0mRrAC&amp;amp;pg=PA22&amp;amp;dq=undergraduate+commutative+algebra+Worked+examples&amp;amp;hl=es&amp;amp;ei=6B_3TN6YIYKB8gbjnszjBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCoQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Undergraduate Commutative Algebra&lt;/a&gt;. Another place where you may want to look is &lt;strong&gt;Section II.4.3&lt;/strong&gt; in page 84 of Eisenbud and Harris' &lt;a href=&quot;http://books.google.co.cr/books?id=BpphspzsasEC&amp;amp;printsec=frontcover&amp;amp;dq=The+Geometry+of+Schemes&amp;amp;hl=es&amp;amp;ei=5az3TPLGEYG0lQf9nIyhAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Geometry of Schemes&lt;/a&gt;, where this is discussed more generally in the context of the prime spectrum $\text{Spec} \, \mathbb{Z}[x]$ and you can see a nice picture of it and how its &quot;points&quot; correspond to the points in $\text{Spec} \,  \mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-02T14:29:23.610" LastActivityDate="2010-12-02T14:29:23.610" />
  <row Id="12707" PostTypeId="2" ParentId="12680" CreationDate="2010-12-02T04:33:59.763" Score="1" ViewCount="0" Body="&lt;p&gt;I think what you are describing is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiring&quot; rel=&quot;nofollow&quot;&gt;semirings without identity&lt;/a&gt;: a semiring would be aset $R$ with two operations, $+$ and $\cdot$, such that $(R,+)$ is a commutative monoid, $(R,\cdot)$ is a commutative monoid, $\cdot$ distributes over $+$, and the neutral element relative to $+$ is a zero element relative to $\cdot$. The only difference between this and your structure is the existence of an identity for the operation *. Some authors allow semirings to not have a multiplicative identity, which would be exactly what you have.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T04:33:59.763" />
  <row Id="12708" PostTypeId="1" CreationDate="2010-12-02T04:34:21.510" Score="2" ViewCount="53" Body="&lt;p&gt;The idea of a read-only right-moving Turing Machine already exists [1].&#xA;Is there a relaxed version of this where an input-reading head only moves &#xA;right to read inputs as needed? There may be other heads doing the computation&#xA;on work tapes; the movement of these heads is unrestricted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the right-moving head could simply copy the contents to another unrestricted TM tape and start that; I'm trying to study TM configurations where the TM works as much as possible before requesting another input character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this machine is Turing complete; correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References with your answer would be helpful. Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://en.wikipedia.org/wiki/Read-only_right_moving_Turing_Machines&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Read-only_right_moving_Turing_Machines&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1884" LastEditorUserId="1884" LastEditorDisplayName="" LastEditDate="2010-12-02T05:16:04.220" LastActivityDate="2010-12-02T05:16:04.220" Title="Right-moving Turing Machines" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="12709" PostTypeId="2" ParentId="12694" CreationDate="2010-12-02T04:36:31.470" Score="1" ViewCount="0" Body="&lt;p&gt;Step 1: Prove 0+1=1+0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply premise #2 for x=1, and substitute into premise #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Prove Ax [x+1=1+x -&gt; x+1+1=1+(x+1)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose k+1=1+k (premise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply premise #3 for x=1, y=k. Substitute from last premise. Apply symmetry of = to get k+1+1=1+(k+1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalize. Apply premise #4 to get required result.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-12-02T17:29:35.757" LastActivityDate="2010-12-02T17:29:35.757" />
  <row Id="12710" PostTypeId="2" ParentId="12705" CreationDate="2010-12-02T04:36:42.240" Score="4" ViewCount="0" Body="&lt;p&gt;$\mathbb{Z}[\omega]$ is the ring generated by $\mathbb{Z}$ and $\omega$ (inside, say $\mathbb{C}$). You can think of this as a ring whose elements are polynomials in $\omega$ with coefficients in $\mathbb{Z}$. Since, $\omega^n=1$ for the appropriate $n$, you need to only consider polynomials of degree less than $n$. Then the usual addition and multiplication of polynomials give you the ring structure.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T04:36:42.240" CommentCount="1" />
  <row Id="12711" PostTypeId="2" ParentId="12705" CreationDate="2010-12-02T04:39:41.703" Score="1" ViewCount="0" Body="&lt;p&gt;When $\omega$ is any algebraic element of $\mathbb{C}$ satisfying a monic polynomial with integral coefficients, $\mathbb{Z}[\omega]$ denotes the ring $\{a_0+a_1\omega+\ldots+a_{n-1}\omega^{n-1}: a_i\in \mathbb{Z}, n=\text{deg }\omega\}$, where the degree of $\omega$ is the degree of the minimal polynomial of $\omega$ over $\mathbb{Q}$. When $\omega$ is a primitive $n$-th root of unity, this ring happens to be the ring of integers of the cyclotomic field $\mathbb{Q}(\mu_n)$, but for general algebraic $\omega$, $\mathbb{Z}[\omega]$ is a proper finite index subring of the ring of integers of $\mathbb{Q}[\omega]$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-02T04:39:41.703" CommentCount="2" />
  <row Id="12712" PostTypeId="2" ParentId="12704" CreationDate="2010-12-02T04:49:24.830" Score="6" ViewCount="0" Body="&lt;p&gt;I assume that you have an $n\times n$ linear system $Ax=0$, whose solutions are precisely $x_1=x_2=\cdots=x_n$. &#xA;This means that the null space is the linear span of the vector $e=(1,1,\cdots,1)$. So the row space is the orthogonal complement of $e$. That means the sum of each row of $A$ must be 0, and that $n-1$ of these rows must be linearly independent.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T04:49:24.830" />
  <row Id="12713" PostTypeId="2" ParentId="12698" CreationDate="2010-12-02T04:51:39.257" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\quad S\ :=\ R[x]/(x-a,b)\ \cong\ R/b\ \ $ for $\rm\ a,b\in R\ $ any ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ \ $ In $\rm\:S\ $ we have $\rm\ x = a\ $ hence $\rm\ f(x) = f(a)\:.\ $ Therefore the natural  map of $\rm R $ into $\rm S$ is onto, with kernel $\rm K \supset b\:R\:.\ $  If $\rm\ c\in K\ $ then $\rm\ c = (x-a)\ f(x) + b\ g(x)\ \Rightarrow\ c\in b\:R\ \ $  via $\ $ eval $\: $ at $\rm\ x = a\:.$ Therefore $\rm\ \  \ K = b\:R\ \ $ so $\rm\: \ S \cong R/K =  R/b\:.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-02T17:49:30.520" LastActivityDate="2010-12-02T17:49:30.520" />
  <row Id="12714" PostTypeId="2" ParentId="12708" CreationDate="2010-12-02T04:53:48.047" Score="2" ViewCount="0" Body="&lt;p&gt;You could copy the input to a work tape and go on from there.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T04:53:48.047" />
  <row Id="12715" PostTypeId="1" AcceptedAnswerId="12730" CreationDate="2010-12-02T04:55:20.510" Score="1" ViewCount="48" Body="&lt;p&gt;Suppose I have a function $f \in L^2(R^n)$ with compact support, and a mollifier $\phi(x) \in C^\infty(R^n)$ with compact support s.t. $\int \phi(x) dx = \int \phi_\epsilon(x) = 1$ and $lim_{\epsilon \to 0} \phi_\epsilon(x) = \delta(x)$ where $\phi_{\epsilon}(x) = \epsilon^{-n} \phi(x/\epsilon)$ and $\delta(x)$ denotes the dirac delta. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_\epsilon = f * \phi_\epsilon$, where $*$ denotes convolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, is it possible to pass the limit of the integral of $f_\epsilon$ inside and say that  $\lim_{\epsilon \to 0} \int f * \phi_\epsilon = \int \lim_{\epsilon \to 0} f *\phi_\epsilon$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if it is not, what type of stronger condition would I need to be able to do that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about using the Dominated Convergence Theorem, but I wasn't sure how to come up with a dominating function that works for every $\epsilon$.   &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-12-02T06:10:10.757" Title="Integrating mollifiers and their limits" Tags="&lt;real-analysis&gt;" AnswerCount="1" />
  <row Id="12716" PostTypeId="2" ParentId="12646" CreationDate="2010-12-02T04:59:23.770" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a different suggestion on how to remember all the different algebraic structures, one that doesn't require memorization of the list of axioms: think of a &quot;canonical&quot; example for each type, an example which doesn't fit into any of the stronger structures.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T04:59:23.770" />
  <row Id="12717" PostTypeId="2" ParentId="12689" CreationDate="2010-12-02T05:04:38.650" Score="4" ViewCount="0" Body="&lt;p&gt;Liberalkid is right. Using his suggestion, you get $$\frac{\binom{n}{N}\binom{n-N}{N}\cdots\binom{n-(r-1)N}{N}}{\binom{n}{N}^r} = \frac{n_N (n-N)_N \cdots (n-(r-1)N)_N}{(n_N)^r} = \frac{n_{Nr}}{(n_N)^r}.$$ In the first step you cancel $N!$ from each side $r$ times.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T05:04:38.650" CommentCount="3" />
  <row Id="12718" PostTypeId="2" ParentId="12675" CreationDate="2010-12-02T05:09:43.920" Score="0" ViewCount="0" Body="&lt;p&gt;You can look, for each point, at the maximum distance among the $k$ nearest neighbors, and throw the point away if the maximum is too big (perhaps, relative to some average score on the specific map). You can determine $k$ by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, these points are known as &lt;i&gt;outliers&lt;/i&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T05:09:43.920" />
  <row Id="12719" PostTypeId="1" CreationDate="2010-12-02T05:15:48.240" Score="1" ViewCount="52" Body="&lt;p&gt;Earlier I had posted question about Borel cantelli Lemma. Thanks to Nate and Shai for your kind response. Please see my profile for the earlier questions. As you can see I do not understand the lemma well. Here are some confusions I have. I hope someone will be kind enough to help me understand the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that P ([An i.o.]∩ A) = P [( A ∩ An) i.o.]. If so, how do we prove this? or is it trivial? And assuming that we know that P[An i.o.] =1 , can we establish that the sum {P(A ∩ An)} diverges (meaning = infinity) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="4230" LastActivityDate="2010-12-02T06:09:28.087" Title="Borell Cantelli lemma question part 2 " Tags="&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="12720" PostTypeId="1" AcceptedAnswerId="12721" CreationDate="2010-12-02T05:40:02.223" Score="1" ViewCount="111" Body="&lt;p&gt;I like to understand how things work. It's the only way I'll remember the rule. If anyone has a way of breaking this down into logical steps to get from one to the other... it would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X^n\cdot X^m = X^{n+m}$ ?&lt;/p&gt;&#xA;" OwnerUserId="4313" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T05:42:25.683" LastActivityDate="2010-12-02T05:49:46.937" Title="why is $X^n\cdot X^m = X^{n+m}$" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12721" PostTypeId="2" ParentId="12720" CreationDate="2010-12-02T05:43:19.613" Score="6" ViewCount="0" Body="&lt;p&gt;$X^n$ is $X$ multiplied by itself $n$ times. $X^m$ is $X$ multiplied by itself $m$ times. When you multiply these together, you'll get $X$ multiplied by itself $n+m$ times, which is $X^{n+m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example with $n=2$ and $m=3$: $$X^2 \cdot X^3 = (X \cdot X) \cdot (X \cdot X \cdot X) = X^5.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T05:43:19.613" CommentCount="1" />
  <row Id="12722" PostTypeId="2" ParentId="12720" CreationDate="2010-12-02T05:49:46.937" Score="6" ViewCount="0" Body="&lt;p&gt;If $n$ and $m$ are positive integers, then remember that $X^n$ is shorthand for &#xA;$$ \underbrace{X\cdot X\cdots X}_{\text{$n$ factors}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $X^n\cdot X^m$ is:&#xA;$$X^n \cdot X^m = \underbrace{(X\cdot X\cdots X)}_{\text{$n$ factors}} \cdot \underbrace{(X\cdot X \cdots X)}_{\text{$m$ factors}} = \underbrace{X \cdot X\cdots X}_{\text{$n+m$ factors}} = X^{n+m}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For negative integers, remember that if $n\gt 0$, then $X^{-n} = \frac{1}{X^n}$. So we have that &#xA;$$X^{-n}\cdot X^{-m} = \frac{1}{X^n}\cdot \frac{1}{X^m} = \frac{1}{X^n\cdot X^m} = \frac{1}{X^{n+m}} = X^{-(n+m)} = X^{-n-m}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, using the other exponent rules, you can also get that it works for exponents that are fractions; and then once you define arbitrary exponents, for any exponent.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T05:49:46.937" />
  <row Id="12723" PostTypeId="1" AcceptedAnswerId="12727" CreationDate="2010-12-02T05:53:21.107" Score="2" ViewCount="60" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/xLuHb.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part b, I'm getting a little stuck. So I'm trying to show that if $(x,y,r)R(a,b,s)$ and $(a,b,s)R(x,y,r)$, then $(x,y,r)=(a,b,s)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $(x,y,r)R(a,b,s)$ implies $\sqrt{(x-a)^2+(y-b)^2} \leq s-r$&#xD;&#xA;and $(a,b,s)R(x,y,r)$ implies $\sqrt{(a-x)^2+(b-y)^2} \leq r-s$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some algebra, here's what I get:&#xA;For the first one, &#xA;$x^2+y^2-s^2-2xa-2yb+2sr+b^2+a^2 \leq r^2$&#xD;&#xA;and for the second one, &#xA;$x^2+y^2-s^2-2ax-2by+2sr+b^2+a^2 \leq r^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so they come out to be the same, but in an example from the notes, I think I should have came up with something like: $a \leq c$ and $c \leq a$, so $c=a$, and I didnt get anything of that form, so I'm wondering if my steps above are also correct. Have I proven anything, or if not, what have I done/assumed wrong?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T05:56:03.483" LastActivityDate="2010-12-02T06:34:56.087" Title="Proving a relation is antisymmetric" Tags="&lt;relations&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12724" PostTypeId="2" ParentId="12723" CreationDate="2010-12-02T05:56:42.343" Score="3" ViewCount="0" Body="&lt;p&gt;Your two square roots are the same, so  $r=s$ follows from the antisymmetry of $\leq$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-02T05:56:42.343" />
  <row Id="12725" PostTypeId="2" ParentId="12723" CreationDate="2010-12-02T05:58:41.727" Score="2" ViewCount="0" Body="&lt;p&gt;The relation is saying that the first circle is completely included inside the second circle. If two circles are completely included in one another, they must be the same circle.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-02T06:10:08.557" LastActivityDate="2010-12-02T06:10:08.557" CommentCount="5" />
  <row Id="12726" PostTypeId="2" ParentId="7966" CreationDate="2010-12-02T06:06:41.990" Score="1" ViewCount="0" Body="&lt;p&gt;If you know a bit of algebraic topology Hatcher's book mentioned above is great. I thought it was a little slow going, so instead I prefer Mosher and Tangora, which is now a Dover book!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are some great exercises that really made me feel comfortable with Spectral sequences. The first Two SS I would look at are Bockstein and Serre. First look at the description/construction/definition of $sq^1$ in MT. It IS the first differential in the mod two bockstein SS. so now you can compute integral cohomology from mod two cohomology once you know how $sq^1$ acts (this also shed light on $sq^1$ for me). To do these computations you should adjoin some indeterminate and use adams grading. then $d_1$ goes up one and to the left one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can use the serre SS and various simple fibre sequences to compute the cohomology of all the complex and real projective spaces WITH their ring structure! (first compute it for the infinite cases then restrict). This is all spelled out in MT (maybe not the part about restricting to get the finite dimensional projective spaces). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then compute the cohomology of $\Omega S^n$ for every n, with ring structure!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then prove the Thom Isomorphism Thm with a relative form of the Serre SS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when you are looking at the bockstein SS draw one of boardman's unrolled exact couples and look at what is going on. And dont use their construction of the steenrod operations, there are better places imo.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-02T06:06:41.990" CommentCount="2" />
  <row Id="12727" PostTypeId="2" ParentId="12723" CreationDate="2010-12-02T06:07:50.847" Score="3" ViewCount="0" Body="&lt;p&gt;Part (a) would give you the intuition. But the first thing to notice here (even before you notice that the square roots have the same value) is that since the square root is always nonnegative, the first condition gives you that $s-r\geq 0$ and the second gives you that $r-s\geq 0$. Those two together tell you &lt;em&gt;what&lt;/em&gt; about $s$ and $r$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have that, you know something about the square root; and since the quantity inside is a sum of squares, the only way that &lt;em&gt;something&lt;/em&gt; can happen is if both squares are -&lt;strong&gt;fill in the blank&lt;/strong&gt;-. Which tells you something about $x$, $y$, $a$, and $b$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T06:34:56.087" LastActivityDate="2010-12-02T06:34:56.087" CommentCount="11" />
  <row Id="12728" PostTypeId="1" CreationDate="2010-12-02T06:08:11.007" Score="7" ViewCount="137" Body="&lt;p&gt;Hello all this is my first math question, I hope this is the right place to post.&#xA;Anyway, this is basic stuff but I'm a bit confused here, here's the problem's legend:&#xA;&quot;Find the equation of the plane that passes through the intersection line ($l$) of the planes $2x-y-z=2, x+y-3z+4=0$ and has a distance of 2 from the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, So here's what I've done, the family of planes that pases through $l$ is $$2x-y-z-2+k(x+y-3z+4)=0$$ We work the equation so it looks more like a planes general form: $$Ax+By+Cz+D=0$$ giving us this:$$(2+k)x+(k-1)y+(-3k-1)z +2(2k-1)=$$&#xA;where$$A=(2+k)$$ $$B=(k-1)$$ $$C=(-1-3k)$$ $$D=2(2k-1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem states that it's distance to the origin is 2 and the normal form of a plane($x\cos{\alpha}+y\cos{\beta}+z\cos{\gamma}-p=0$) where $p$ is the distance of the plane to the origin. We know that to transform a planes equation from the general form to the normal form we must divide each term by $$r=\pm \sqrt{A^2+B^2+C^2}$$&#xD;&#xA;So $p=\frac{D}{r}$ and we have this:$$2=\frac{2(2k-1)}{\pm \sqrt{(2+k)^2+(k-1)^2+(-1-3k)^2}}$$&#xA;And here is where my doubt is, what sign should $r$ have positive or negative, the following theorem says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;if $D \neq 0$, $r$ is opposite sign of $D$&lt;/li&gt;&#xA;&lt;li&gt;if $D=0$ and $C \neq 0$ $r$ is equal sign of $C$&lt;/li&gt;&#xA;&lt;li&gt;if $D=0$ and $C=0$ and $B \neq 0$ $r$ is equal sign of $B$&lt;/li&gt;&#xA;&lt;li&gt;if $D=0$ and $C=0$ and $B=0$ and $A \neq 0$ $r$ is equal sign of $A$&lt;/li&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we don't know the value of k and $D(k)=2(2k-1)$ and to have $D$ we must find k but k is determined by the sign of D(what a mess hehe) how do we chose $r$'s sign?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T10:57:55.980" LastActivityDate="2010-12-02T16:07:19.603" Title="Analytic Geometry Doubt" Tags="&lt;geometry&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12729" PostTypeId="2" ParentId="12719" CreationDate="2010-12-02T06:09:28.087" Score="2" ViewCount="0" Body="&lt;p&gt;The events &quot;$A_n$ for infinitely many $n$; and $A$&quot; and &quot;$A_n \land A$ for infinitely many $n$&quot; are one and the same since $\land$ distributes over $\lor$: $$ A \land \bigwedge_{n \geq 0} \bigvee_{m \geq n} A_m \Leftrightarrow \bigwedge_{n \geq 0} A \land \bigvee_{m \geq n} A_m \Leftrightarrow \bigwedge_{n \geq 0} \bigvee_{m \geq n} A \land A_m.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, if $A$ never happens then $P(A \land A_n) = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T06:09:28.087" />
  <row Id="12730" PostTypeId="2" ParentId="12715" CreationDate="2010-12-02T06:10:10.757" Score="3" ViewCount="0" Body="&lt;p&gt;Note that &#xA;$$\int f \ast \phi_{\epsilon}(x)\,dx= \int(\int f(x - y)\phi_{\epsilon}(y)\,dy)\,dx$$&#xD;&#xA;So by applying Fubini's theorem and integrating with respect to $y$ first, $\int f \ast \phi_{\epsilon}(x)\,dx = \int f \int \phi_{\epsilon} = \int f$. So to show your equality, it suffices to show that $\lim_{\epsilon \rightarrow 0} f \ast \phi_{\epsilon}(x) = f(x)$ for almost all $x$. It's actually true at all Lebesgue points of $f(x)$ which is a standard fact about these mollifiers. So this gives you what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way: Since your functions are compactly supported you can use that $\int |f -  f \ast \phi_{\epsilon}| \leq C (\int |f -  f \ast \phi_{\epsilon}|^2)^{1 \over 2}$. (This follows by Schwartz's inequality for example). Then you can use Plancherel's theorem on the right hand side and get that&#xA;$$\int |f -  f \ast \phi_{\epsilon}|^2 = \int |\hat{f} - \hat{\phi_{\epsilon}}\hat{f}|^2$$&#xD;&#xA;Since $\phi_{\epsilon}$ are mollifiers, $\hat{\phi_{\epsilon}}(x) = \hat{\phi}(\epsilon x)$ converges to 1 for all $x$ as $\epsilon$ goes to zero. So $\hat{f} - \hat{\phi_{\epsilon}}\hat{f}$ converges to zero. Since $|\hat{f} - \hat{\phi_{\epsilon}}\hat{f}|^2$ has the dominating function $|2\hat{f}|^2$, one can now apply the dominated convergence theorem to the right-hand side above and get that zero is the limit. As a result, the left-hand side must also converge to zero. So since $\int |f -  f \ast \phi_{\epsilon}| \leq C (\int |f -  f \ast \phi_{\epsilon}|^2)^{1 \over 2}$, one also has that $\int |f -  f \ast \phi_{\epsilon}|$ goes to zero. So the same is true for $|\int f - \int f \ast \phi_{\epsilon}| = |\int (f - f \ast \phi_{\epsilon})| \leq \int |f -  f \ast \phi_{\epsilon}|$, which is what you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-02T06:10:10.757" CommentCount="1" />
  <row Id="12731" PostTypeId="1" CreationDate="2010-12-02T06:10:55.977" Score="10" ViewCount="456" Body="&lt;p&gt;Given a number of 3in squares and 2in squares, how many of each are needed to get a total area of 35 in^2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through quick trial and error (the method they wanted I believe) you find that you need 3 3in squares and 2 2in squares, but I got to thinking on how to solve this exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 2 unknowns and the following info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4x + 9y = 35&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x &gt;= 0, y &gt;= 0, x and y are both integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also follows then that x &amp;lt;= 8 and y &amp;lt;= 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to use the inequalities or the integer only info to form a direct 2nd equation in order to solve the system of equations.  How would you do this without trial and error?&lt;/p&gt;&#xA;" OwnerUserId="4315" LastActivityDate="2010-12-02T07:07:45.390" Title="My Daughter's 4th grade math question got me thinking" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12732" PostTypeId="2" ParentId="12728" CreationDate="2010-12-02T06:23:04.383" Score="1" ViewCount="0" Body="&lt;p&gt;Correct me if I'm wrong here, but if two planes are not co-planar (and these are not), then the intersection between the plane will form a single line.  That line will be whatever distance it happens to be from the origin.  I'm assuming that by distance from origin, they mean closest point from origin to the line.  But anyways, that distance from the line seems just extra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, I'm probably missing the point here.  All I can think of when seeing this problem is forming normals for the 2 planes and forming the cross product of those vectors to get the intersecting vector.  Been doing computer graphics for too long :)&lt;/p&gt;&#xA;" OwnerUserId="4315" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-02T16:07:19.603" LastActivityDate="2010-12-02T16:07:19.603" CommentCount="1" />
  <row Id="12733" PostTypeId="2" ParentId="12731" CreationDate="2010-12-02T06:29:25.640" Score="10" ViewCount="0" Body="&lt;p&gt;A quick way to see the answer is to convert both sides of the equation mod 4. So the left hand side is y  (mod 4) (b/c 4=0, 9=1 mod 4), and the right hand side is 3 (mod 4). So y=3 mod 4. Since $y\le 3$ as you observed, the only solution (if there is any ) is $y=3$. Then you check that $4x+27=35$ and hence $x=2$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T06:29:25.640" CommentCount="4" />
  <row Id="12734" PostTypeId="1" AcceptedAnswerId="12737" CreationDate="2010-12-02T06:37:40.700" Score="2" ViewCount="126" Body="&lt;p&gt;I have read &lt;a href=&quot;http://dx.doi.org/10.1016%2F0020-0190%2879%2990002-4&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; and the wikipedia summary &lt;a href=&quot;http://en.wikipedia.org/wiki/2-satisfiability&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and even &lt;a href=&quot;http://docs.google.com/viewer?a=v&amp;amp;q=cache%3aDwWbM7Q3o1EJ%3awww.soe.ucsc.edu/classes/cmps132/Winter05/hw/hw8sols.pdf+2sat+in+p&amp;amp;hl=en&amp;amp;gl=us&amp;amp;pid=bl&amp;amp;srcid=ADGEESjTJ-300yEMXJh2u4othjtyh57czm-IkIBugnUlIZiWBNFLo82JfweSmUHniqQYVtU23zWxRQuuwXVM0XIwoAOPmuZpRqutv8YXt4vkB-RnJT777wC-4nuT1zSlquia3m0lZ7P5&amp;amp;sig=AHIEtbRZa564hezVFCgcggCp2rDREeEiGg&quot; rel=&quot;nofollow&quot;&gt;this homework problem&lt;/a&gt;, but I do not understand how 2SAT is in P. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my concern: the idea is that we start by setting a random variable. Then we set any implied variables. Repeat, and if we eventually reach a contradiction, then the problem has no solution. For example, that homework solutions thing says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Begin with an arbitrary vertex α such that (α, ¬α) is not an edge in the graph. [Fill out implied variables]. Repeating this process for every unassinged literal in I creates a&#xA;  satisfying assignment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not think this is always works. Take $(x\vee y)\wedge(\bar{x}\vee\bar{x})$. Set x = true. Now the second clause doesn't work, and we've reached a contradiction. But this formula does have a solution: $\bar{x}\wedge y$. Yet if we look at the implication for the second clause, it is $x\implies x$, so it's not as though we didn't set an implied variable. There is no rule $x\implies\bar{x}$ either, so it doesn't contradict that constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=5&amp;amp;ved=0CDAQFjAE&amp;amp;url=http%3A%2F%2Fwww.cs.tau.ac.il%2F~safra%2FComplexity%2F2SAT.ppt&amp;amp;rct=j&amp;amp;q=2sat%20in%20p&amp;amp;ei=zy73TJedO8P38AaqsemxBw&amp;amp;usg=AFQjCNGvmrsStd0iVzF5g8FEEvQkhIJRrw&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; says to &quot;rewind&quot; once if you reach a contradiction, but I don't see why it's not possible to reach contradictions any number of times even if the proposition is satisfiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-02T14:01:12.943" LastActivityDate="2010-12-02T14:01:12.943" Title="Help understanding the proof that 2SAT is in P" Tags="&lt;algorithms&gt;&lt;computer-science&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12735" PostTypeId="3" CreationDate="2010-12-02T06:45:55.713" Score="0" ViewCount="0" Body="&lt;p&gt;I've been a user of math.SE and MO since the inception of both, as well as the&#xA;meta forums on each. I would take a relatively light approach to moderation:&#xA;that is, let the community as a whole close questions (except for obvious spam&#xA;or duplicates), and allow generally free discussions on meta. On the other hand, I'd be certainly be interested in &lt;em&gt;discussing&lt;/em&gt; policies (about, say, homework) with the community. I would of course be willing to intervene when discussions degenerate into flamewars, but on the whole, I would aim to facilitate a more open community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it a useful and worthwhile project to create an active mathematical Q/A&#xA;at a more general level than MathOverflow and would be happy to serve if elected. Outside of here, I've participated in various aspects of the internet mathematical community (e.g. blogs, MathOverflow, and for a brief while nLab); it appears to hold promise to me, as I have found involvement in it to be productive and enlightening. &lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-02T06:57:02.543" LastActivityDate="2010-12-02T06:57:02.543" CommunityOwnedDate="2010-12-02T06:45:55.713" CommentCount="2" />
  <row Id="12736" PostTypeId="2" ParentId="12728" CreationDate="2010-12-02T06:48:02.467" Score="1" ViewCount="0" Body="&lt;p&gt;First of all, your equation should be $$(k+1)x+(k-1)y-\cdots.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the distance formula is  $$2=\frac{|2(2k-1)|}{\cdots}$$ (no $\pm$). Square both sides (note that square of $|\text{anything}|$ is $\text{anything}^2$, then solve a quadratic equation. You will get two $k$'s. Substitute back to see which one is correct. In this case, it turns out that both $k$'s are correct.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-02T16:00:01.310" LastActivityDate="2010-12-02T16:00:01.310" CommentCount="2" />
  <row Id="12737" PostTypeId="2" ParentId="12734" CreationDate="2010-12-02T06:50:50.143" Score="3" ViewCount="0" Body="&lt;p&gt;The graph in the paragraph you're quoting is the transitive closure of the original graph. Your algorithm can be adapted as follows: choose an arbitrary literal $\alpha$ and make all the necessary assignments; if you reach $\lnot \alpha$ (or reach both $\beta$ and $\lnot \beta$ for some other variable $\beta$), then $\alpha$ must be false, so try again from $\lnot \alpha$; if you reach $\alpha$ (or reach both $\beta$ and $\lnot \beta$ for some other variable $\beta$), then the instance is unsatisfiable. Continue until you set the values of all variables.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T06:50:50.143" CommentCount="5" />
  <row Id="12738" PostTypeId="2" ParentId="154" CreationDate="2010-12-02T06:55:08.583" Score="1" ViewCount="0" Body="&lt;p&gt;I'm of the idea that complex numbres exist in the same way real number exists, we can be ware of real number every time because they're involved in our daily operations. Lets talk about &lt;strong&gt;phasors&lt;/strong&gt;, they aren't more that convenient ways to express &lt;em&gt;senoidal equations&lt;/em&gt;, based on the &lt;strong&gt;Eulers Identity&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e^jT = Cos(T) + jSin(T) &amp;lt;- Eulers identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this is easir to express and solve &lt;em&gt;differential equations&lt;/em&gt; that involves senoidal functions in terms of exponential notation, then they become the prefered manner for &lt;em&gt;analysis of electrical circuits of alternating current (CA)&lt;/em&gt;. Some quantities sucha as impedance (Z=V/I) where V,I are phasors, have no sense without the complex numbers. In this application, physically the complex part is associated with &lt;em&gt;reactive elements&lt;/em&gt; (inductors and capacitors) while the real part is associated with &lt;em&gt;resistive elements&lt;/em&gt; (resistors). Also other kinds of transformation such as Laplace transformation (for solving linear differential equations) are based on complex numbers. I have had some &lt;a href=&quot;http://www.monografias.com/trabajos-pdf/principios-fisicos-matematicos-analisis-sistemas/principios-fisicos-matematicos-analisis-sistemas.pdf&quot; rel=&quot;nofollow&quot;&gt;work&lt;/a&gt; about these kind of &lt;em&gt;notations and transformations&lt;/em&gt; and their application in Electrical Enginnering.&lt;/p&gt;&#xA;" OwnerUserId="4314" LastActivityDate="2010-12-02T06:55:08.583" />
  <row Id="12739" PostTypeId="2" ParentId="12731" CreationDate="2010-12-02T07:00:42.280" Score="7" ViewCount="0" Body="&lt;p&gt;There is an algorithmic way to solve this which works when you have two types of squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $\displaystyle \text{gcd}(a,b) = 1$, then for any integer $c$ the linear diophantine equation $\displaystyle ax + by = c$ has an infinite number of solution, with integer $\displaystyle x,y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact if $\displaystyle x_0, y_0$ are such that $\displaystyle  a x_0 - b y_0 = 1$, then all the solutions of $\displaystyle ax + by = c$ are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x = -tb + cx_0$, $\displaystyle y = ta - cy_0$, where $\displaystyle t$ is an arbitrary integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x_0 , y_0$ can be found using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you also need $\displaystyle x \ge 0$ and $\displaystyle y \ge 0$ you must pick a $\displaystyle t$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c x_0 \ge tb$ and $ta \ge cy_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no such $\displaystyle t$, then you do not have a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $\displaystyle a= 9, b= 4$, we need a solution of $\displaystyle ax + by = 35$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily see that $\displaystyle x_0 = 1, y_0 = 2$ gives us $\displaystyle a x_0 - by_0 = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we need to find a $\displaystyle t$ such that $ 35 \ge t\times 4$ and $ t\times 9 \ge 35\times 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 35/4 \ge t \ge 35\times 2/9$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  8.75 \ge t \ge 7.77\dots$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $t = 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us $\displaystyle x = cx_0 - tb  = 3$, $\displaystyle y = ta- cy_0 = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I have swapped your x and y).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T07:07:45.390" LastActivityDate="2010-12-02T07:07:45.390" />
  <row Id="12741" PostTypeId="1" AcceptedAnswerId="12742" CreationDate="2010-12-02T09:38:43.603" Score="5" ViewCount="60" Body="&lt;p&gt;How to find the least positive root of the equation $\cos 3x + \sin 5x = 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach so far is to represent $\sin 5x$ as $\cos \biggl(\frac{\pi}{2} - 5x\biggr)$ then the whole equation reduces to $$2\cos \biggl(\frac{\pi}{4} - x\biggr)\cdot \cos \biggl(\frac{\pi}{4} - 4x\biggr) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl(\frac{\pi}{4} - x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl(\frac{\pi}{4} - 4x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there can be infinitely many solutions for this, what I am not getting how to compute the minimum among them? And what about if I am asked to find the maximum?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-12-02T10:10:34.063" LastActivityDate="2010-12-02T10:17:59.270" Title="Finding the least positive root" Tags="&lt;trigonometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12742" PostTypeId="2" ParentId="12741" CreationDate="2010-12-02T09:51:55.717" Score="4" ViewCount="0" Body="&lt;p&gt;You've already done the difficult part! Now just find all the solutions of $\cos(\pi/4-x)=0$ and $\cos(\pi/4-4x)=0$, and pick the smallest positive one.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-02T09:51:55.717" CommentCount="9" />
  <row Id="12743" PostTypeId="2" ParentId="12741" CreationDate="2010-12-02T10:00:39.527" Score="0" ViewCount="0" Body="&lt;p&gt;EDIT: So apparently you want a full solution... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A product of two real numbers is $0$ precisely when a factor is $0$, so you must have that &#xA;$$ \begin{align}&#xD;&#xA;&amp;amp; \cos\left(\tfrac{\pi}{4} - x\right) = 0 \qquad\;\; (1) \text{, or} \\\\&#xD;&#xA;&amp;amp; \cos\left(\tfrac{\pi}{4} - 4x\right) = 0 \qquad (2).&#xD;&#xA;\end{align} $$&#xD;&#xA;Now we use that $\cos a = 0 \iff a \in \tfrac{\pi}{2} + \pi\mathbb{Z}$:&#xA;$$ \begin{align}&#xD;&#xA;\cos \left(\tfrac{\pi}{4} - x\right) = 0 &#xD;&#xA;  &amp;amp; \iff \tfrac{\pi}{4} - x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\&#xD;&#xA;  &amp;amp; \iff x \in \tfrac{3\pi}{4} + \pi \mathbb{Z},&#xD;&#xA;\end{align} $$&#xD;&#xA;the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{4}$.&#xA;$$ \begin{align}&#xD;&#xA;\cos \left(\tfrac{\pi}{4} - 4x\right) = 0 &#xD;&#xA;  &amp;amp; \iff \tfrac{\pi}{4} - 4x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\&#xD;&#xA;  &amp;amp; \iff 4x \in \tfrac{3\pi}{4} + \pi \mathbb{Z} \\\\&#xD;&#xA;  &amp;amp; \iff x \in \tfrac{3\pi}{16} + \tfrac{\pi}{4}\mathbb{Z},&#xD;&#xA;\end{align} $$&#xD;&#xA;the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{16}$. The problem is now reduced to picking the smallest of &lt;em&gt;two&lt;/em&gt; real numbers.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-02T10:17:59.270" LastActivityDate="2010-12-02T10:17:59.270" CommentCount="3" />
  <row Id="12744" PostTypeId="3" CreationDate="2010-12-02T11:09:03.700" Score="0" ViewCount="0" Body="&lt;p&gt;&#xA;    On Stack Exchange, we believe the core moderators should come &lt;i&gt;from&lt;/i&gt; the community, and be elected by the community&#xA;    itself through popular vote. We hold regular elections to determine who these community moderators will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;    See &lt;a href=&quot;http://blog.stackoverflow.com/2009/05/a-theory-of-moderation/&quot; rel=&quot;nofollow&quot;&gt;a theory of moderation&lt;/a&gt; for the typical roles and abilities of a moderator. Once elected, moderators may hold the position as long as they wish, unless they become inactive or exhibit gross misbehavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;    Community moderators are accorded the highest level of privilege and trust on our community, and should themselves be exemplars of&#xA;    positive behavior within the community. Our general criteria for moderators is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;    &lt;li&gt;patient and fair&lt;/li&gt;&#xA;    &lt;li&gt;leads by example&lt;/li&gt;&#xA;    &lt;li&gt;shows respect for their fellow community members in their actions and words&lt;/li&gt;&#xA;    &lt;li&gt;open to some light but firm moderation to keep the community on track and resolve (hopefully) uncommon disputes and exceptions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&#xA;    Every election has three phases:&#xA;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;    &lt;li&gt;Nomination&lt;/li&gt;&#xA;    &lt;li&gt;Primary&lt;/li&gt;&#xA;    &lt;li&gt;Election&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;&#xA;Depending on the number of nominees that enter, and the number of moderator positions to be filled, in some circumstances the election may skip the Primary phase and proceed directly to the Election phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;    Please participate in the moderator elections by voting, and perhaps even by nominating yourself to be a community moderator!&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You may pose questions for the nominees over the &lt;a href=&quot;http://meta.math.stackexchange.com/questions/1254/list-of-questions-for-candidates-in-2010-moderator-election&quot;&gt;meta thread&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="153" LastEditorDisplayName="" LastEditDate="2010-12-09T00:16:21.587" LastActivityDate="2010-12-09T00:16:21.587" />
  <row Id="12745" PostTypeId="1" AcceptedAnswerId="12747" CreationDate="2010-12-02T11:18:02.913" Score="2" ViewCount="123" Body="&lt;p&gt;I'm reading a paper on fluid dynamics and it references a unit vector between two particles i and j. I'm not clear what it means by a unit vector in this instance. How do I calculate the unit vector between the two particles?&lt;/p&gt;&#xA;" OwnerUserId="4318" LastActivityDate="2010-12-02T12:28:48.240" Title="How do you calculate the unit vector between two points?" Tags="&lt;geometry&gt;&lt;vector-space&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="12746" PostTypeId="2" ParentId="12745" CreationDate="2010-12-02T11:24:50.750" Score="3" ViewCount="0" Body="&lt;p&gt;If particle $i$'s position is described by a position vector $\vec{r}_i$ and particle $j$'s position is described by a position vector $\vec{r}_j$, then you can define the position of $j$ relative to $i$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{r}_{ji}= \vec{r}_j-\vec{r}_i$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you divide this vector by its length: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\vec{r}_{ji}}{\|\vec{r}_{ji}\|}=\frac{\vec{r}_j-\vec{r}_i}{\|\vec{r}_j-\vec{r}_i\|}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get a vector with unit length and aligned along the direction of the line through particles $i$ and $j$, pointing towards $j$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-02T11:24:50.750" CommentCount="1" />
  <row Id="12747" PostTypeId="2" ParentId="12745" CreationDate="2010-12-02T11:26:37.180" Score="3" ViewCount="0" Body="&lt;p&gt;Two particles i, j are located in some reference frame at vectorial positions $\vec{r}_i$ and $\vec{r}_j$. Therefore the vector which starts at the position of i and ends at j, is just the difference $\vec{r}_j-\vec{r}_i$; its modulus $||\vec{r}_j-\vec{r}_i||$ is the distance between the particles so one can construct the unit vector in that direction (from i to j) by just&#xA;$$\vec{u}_{ij}=\frac{1}{||\vec{r}_j-\vec{r}_i||}(\vec{r}_j-\vec{r}_i)$$&#xD;&#xA;Indeed this is a unit vector for its a multiple of the original with unit modulus since $||\vec{u}_{ij}||=\left|\frac{1}{||\vec{r}_j-\vec{r}_i||}\right|\cdot||\vec{r}_j-\vec{r}_i||=1$ using the property $||\lambda\cdot\vec{v}||=|\lambda|\cdot ||\vec{v}||$.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditorDisplayName="" LastEditDate="2010-12-02T12:28:48.240" LastActivityDate="2010-12-02T12:28:48.240" />
  <row Id="12748" PostTypeId="2" ParentId="8349" CreationDate="2010-12-02T12:18:02.770" Score="4" ViewCount="0" Body="&lt;p&gt;If someone gives you a random large number, the last thing you want to do is perform a deterministic primality test -- it's very likely to be composite.  Firstly, you should perform:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;trial division and&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat_primality_test&quot; rel=&quot;nofollow&quot;&gt;Fermat's test&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These can be performed using PFGW (from &lt;a href=&quot;http://tech.groups.yahoo.com/group/primeform/&quot; rel=&quot;nofollow&quot;&gt;http://tech.groups.yahoo.com/group/primeform/&lt;/a&gt;).  You might even want to perform a stronger pseudo-primality test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once these tests have been passed, then you should be somewhat confident you have a prime.  If you're happy with a probable prime, then you can stop.  Otherwise, you'll need a proof.  [Curiously, if sufficiently many probabilistic primality tests have been performed, the probability that the number is composite can be less than the probability of a random error occurring in a &quot;proof&quot; of its primality.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your number has no particularly special form you can exploit, then your best hope is &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve_primality_proving&quot; rel=&quot;nofollow&quot;&gt;elliptic curve primality proving&lt;/a&gt; (see &lt;a href=&quot;http://primes.utm.edu/top20/page.php?id=27&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for ECPP records).  You could use &lt;a href=&quot;http://www.ellipsa.eu/&quot; rel=&quot;nofollow&quot;&gt;Primo&lt;/a&gt; or &lt;a href=&quot;http://www.alpertron.com.ar/ECM.HTM&quot; rel=&quot;nofollow&quot;&gt;Alpertron&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if your candidate prime happens to have a special form, then there are various deterministic primality proofs you could use.  For example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test&quot; rel=&quot;nofollow&quot;&gt;Lucas-Lehmer test&lt;/a&gt; (cf. &lt;a href=&quot;http://www.mersenne.org/&quot; rel=&quot;nofollow&quot;&gt;GIMPS&lt;/a&gt;), Proth's test (http://en.wikipedia.org/wiki/Proth's_theorem), &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer%E2%80%93Riesel_test&quot; rel=&quot;nofollow&quot;&gt;Lucas-Lehmer-Riesel test&lt;/a&gt;.  In fact, these tests are great if you're the type of person who aspires to finding a &lt;a href=&quot;http://primes.utm.edu/primes/home.php&quot; rel=&quot;nofollow&quot;&gt;top-5000 prime&lt;/a&gt; (something I used to do: see &lt;a href=&quot;http://primes.utm.edu/bios/page.php?id=776&quot; rel=&quot;nofollow&quot;&gt;my profile&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, before using these tests, you should perform trial division.  If you're looking for primes that can be proved by the latter two tests, you will probably find &lt;a href=&quot;http://primes.utm.edu/programs/NewPGen/&quot; rel=&quot;nofollow&quot;&gt;NewPGen&lt;/a&gt; invaluable.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-02T12:18:02.770" CommentCount="2" />
  <row Id="12749" PostTypeId="1" AcceptedAnswerId="12751" CreationDate="2010-12-02T13:16:39.020" Score="7" ViewCount="165" Body="&lt;p&gt;I can't use any of the convergence tests I learned because I have no information on $f(x)$, in particular I don't know if it's continuous or positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I could think of was that if $\displaystyle \int_{1}^{\infty}f(x)dx$ was absolutely convergent, then $|f(x)\sin x| \leq |f(x)|$ would imply by the comparison test that $\displaystyle \int_{1}^{\infty}f(x)\sin xdx$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I want to find a counter-example I have to pick $f(x)$ so that $\displaystyle \int_{1}^{\infty}f(x)dx$ conditionally converges, but I can't think of one.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T14:36:19.560" LastActivityDate="2010-12-02T14:36:19.560" Title="if $\int_1^{\infty}f(x)\mathrm dx$ converges, must $\int_1^{\infty}f(x)\sin x\mathrm dx$ converge?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="12750" PostTypeId="1" CreationDate="2010-12-02T13:19:42.263" Score="5" ViewCount="81" Body="&lt;p&gt;Can we alter &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_axioms&quot; rel=&quot;nofollow&quot;&gt;Hilbert's axioms&lt;/a&gt; to have $\mathbb{Q}^3$ as a unique model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical axioms seem to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_axioms#IV._Congruence&quot; rel=&quot;nofollow&quot;&gt;congruence axioms&lt;/a&gt; IV.1 and IV.4, and presumably the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_axioms#V._Continuity&quot; rel=&quot;nofollow&quot;&gt;line completeness axiom&lt;/a&gt; V.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how are they to be modified? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IV.1 might be replaced by requiring that there are counter-examples (irrationality of $\sqrt{2}$) and appropriately relaxing &quot;congruent&quot; to &quot;almost congruent&quot; (= &quot;arbitrarily close to congruence&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about line completeness then, since it might be possible to add irrational points to $\mathbb{Q}^3$ such that the modified axioms still hold?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastActivityDate="2010-12-02T13:19:42.263" Title="Can we alter Hilbert's axioms to have $\mathbb{Q}^3$ as a unique model?" Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;" CommentCount="3" />
  <row Id="12751" PostTypeId="2" ParentId="12749" CreationDate="2010-12-02T13:24:24.397" Score="15" ViewCount="0" Body="&lt;p&gt;Consider $f(x)=\sin(x) / x$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-02T13:24:24.397" CommentCount="5" />
  <row Id="12752" PostTypeId="1" AcceptedAnswerId="12754" CreationDate="2010-12-02T13:40:19.077" Score="7" ViewCount="135" Body="&lt;p&gt;It is proved in &lt;em&gt;Advanced Calculus&lt;/em&gt; by Angus Taylor, § 20.8, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log n!=\log \left( \left( \frac{n}{e}\right) ^{n}\sqrt{2\pi n}\right)&#xD;&#xA;+r_{n},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{n}=\sum_{k=1}^{\infty }S_{k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{k}=\sum_{p=n+1}^{\infty }\frac{k}{2(k+1)(k+2)p^{k+1}}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula for $r_{n}$ provides a method for finding the estimate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{12\left( n+1\right) }&amp;lt;r_{n}&amp;lt;\frac{1}{12\left( n-1\right) }.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{k\left( n+1\right) ^{k}}=\sum_{p=n+1}^{\infty }\int_{p}^{p+1}\frac{%&#xD;&#xA;1}{x^{k+1}}\mathrm dx&amp;lt;\sum_{p=n+1}^{\infty }\frac{1}{p^{k+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{p=n+1}^{\infty }\frac{1}{p^{k+1}}&amp;lt;\sum_{p=n+1}^{\infty }\int_{p-1}^{p}%&#xD;&#xA;\frac{1}{x^{k+1}}\mathrm dx=\frac{1}{kn^{k}},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2(k+1)(k+2)\left( n+1\right) ^{k}}&amp;lt;S_{k}&amp;lt;\frac{1}{2(k+1)(k+2)n^{k}}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{12\left( n+1\right) }&amp;lt;\sum_{k=1}^{\infty }\frac{1}{%&#xD;&#xA;2(k+1)(k+2)\left( n+1\right) ^{k}}&amp;lt;r_{n}&amp;lt;\sum_{k=1}^{\infty }\frac{1}{%&#xD;&#xA;2(k+1)(k+2)n^{k}}&amp;lt;\frac{1}{12\left( n-1\right) }.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;:The &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling%27s_approximation#Speed_of_convergence_and_error_estimates&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; pointed to in the comment gives an approximation based on the Euler-MacLaurin formula in terms of the Bernoulli numbers and states that it can also be obtained by repeated integration by parts. The Stirling series that appears in the same article is different from the one above. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; (edited): Are there better estimates of $r_{n}$ based on different methods?&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-04T17:53:48.850" LastActivityDate="2010-12-04T17:53:48.850" Title="Estimates for Stirling's formula remainder" Tags="&lt;calculus&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12753" PostTypeId="2" ParentId="11843" CreationDate="2010-12-02T13:46:47.453" Score="1" ViewCount="0" Body="&lt;p&gt;If your predictions are repetitive I think you can do this.  For example, if you always predict 50% chance of one choice, 30% for a second and 20% for the third (even if they are permuted on a particular play) you can do a standard chi-square test.  Did you in fact get 50% right on the one you claim?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your predictions vary continuously, you can still pick the highest probability and compute the observed number of times it is right compared to the prediction and the variance using the variance on each prediction is p(1-p).  Then you can compare with the null hypothesis of 1/3, 1/3, 1/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to understand the statement &quot;we can trust this function on 86%&quot;.  I think the chi-square test can say &quot;it is 86% probably that my predictions are better than the null hypothesis&quot;, but that doesn't tell you that the predictions are right.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-02T13:46:47.453" />
  <row Id="12754" PostTypeId="2" ParentId="12752" CreationDate="2010-12-02T14:40:17.450" Score="7" ViewCount="0" Body="&lt;p&gt;Here's a paper &lt;a href=&quot;http://biblio.ugent.be/input/download?func=downloadFile&amp;amp;fileOId=451038&quot; rel=&quot;nofollow&quot;&gt;Stirling's Series Made Easy&lt;/a&gt; by Chris Impens that states in Corollary 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{12n} - \frac{1}{360n^3} &amp;lt; r_n &amp;lt; \frac{1}{12n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and (the interesting part) that the constant $1/360$ cannot be improved. I confess to not having&#xA;read the paper closely.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-02T14:40:17.450" />
  <row Id="12755" PostTypeId="1" AcceptedAnswerId="12758" CreationDate="2010-12-02T15:41:43.920" Score="2" ViewCount="115" Body="&lt;p&gt;Set $S$ is a collection of disjoint sets each having cardinality of that of $\mathbb{R}$.The cardinality of set $S$ is also that of $\mathbb{R}$. $F$ is the set which is union of all the elements of $S$. Is the cardinality of set $F$ is equal to that of $\mathbb{R}$ ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-12-02T18:35:48.717" Title="on cardinality of a set." Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="12756" PostTypeId="2" ParentId="12220" CreationDate="2010-12-02T15:45:40.987" Score="0" ViewCount="0" Body="&lt;p&gt;I think I solved it: If matrices $A$ and $S$ commute, i.e. $AS = SA$, then I eigenspaces of one matrix are invariant under the other matrix, because if $Ax = \lambda x$, then &#xA;$A Sx = SAx = S\lambda x$, so if $x$ is eigenvector of $A$ with eigenvalue $\lambda$, then so is $S\lambda x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in mind, I choose one of the matrices and diagonalize it, say I choose $S$. Then I get&#xA;$S = T D T^\top$ with $T$ being unitary and $D$ being diagonal. If I then use $T$ to transform $A$, I get &#xA;$A&amp;#39; = T A T^\top$ is &lt;em&gt;block diagonal&lt;/em&gt;, because of the invariance mentioned earlier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can then diagonalize each block of $A$ on its own, because that way, the resulting eigenvectors of $A$ will remain eigenvectors of $S$. That way, I have simultaneously diagonalized $A$. The nice thing is that this can help make the resulting eigenvectors unique, because now eigenvectors are characterized not only by their $A$-matrix eigenvalue but also by their $S$-matrix eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't suffice, I could search for more symmetries that all simultaneously commute with each other. An example from quantum mechanics is the hydrogen atom, where one find that the Hamiltonian commutes with the modulus of the angular momentum and its z-component, which ultimately gives rise to the three quantum numbers n, l and m.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-12-02T15:45:40.987" />
  <row Id="12757" PostTypeId="2" ParentId="12755" CreationDate="2010-12-02T15:48:53.870" Score="6" ViewCount="0" Body="&lt;p&gt;You are asking whether $\mathbb{R} \times \mathbb{R}$ has the same cardinality as $\mathbb{R}$.  The answer is yes, thanks to the fact that $|\mathbb{R}| = 2^{\aleph_0}$ and $2^{\aleph_0} \times 2^{\aleph_0} = 2^{\aleph_0 + \aleph_0} = 2^{\aleph_0}$.  More explicitly, $\mathbb{R}$ has the same cardinality as the set $S = \{ 0, 1 \}^{\mathbb{N}}$ of binary sequences, and there is an obvious bijection $S \times S \to S$ given by &quot;interweaving&quot; sequences: that is, sending&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a_1, a_2, a_3, ...) \times (b_1, b_2, b_3, ...) \to (a_1, b_1, a_2, b_2, a_3, b_3, ...).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the statement that $A \times A$ has the same cardinality as $A$, for $A$ infinite, is true for all &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;alephs&lt;/a&gt;.  I think I have been told that whether this statement is true for all infinite sets is equivalent to AC.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-02T15:48:53.870" CommentCount="3" />
  <row Id="12758" PostTypeId="2" ParentId="12755" CreationDate="2010-12-02T15:49:22.963" Score="10" ViewCount="0" Body="&lt;p&gt;Yes. The cardinality of $F$ is $\sum\limits_{x\in S}|x| = \sum\limits_{x\in S}\mathfrak{c} = \mathfrak{c}\mathfrak{c}=\mathfrak{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cardinal arithmetic, if $\kappa$ and $\lambda$ are nonzero cardinals, and at least one is infinite, you have&#xA;$$\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T18:35:48.717" LastActivityDate="2010-12-02T18:35:48.717" CommentCount="2" />
  <row Id="12759" PostTypeId="1" AcceptedAnswerId="12761" CreationDate="2010-12-02T16:17:04.223" Score="0" ViewCount="65" Body="&lt;p&gt;Assume that Y has a beta distribution with parameters a and b. Find the density function of U = 1 - Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do then when they give the density function of Y, but i'm confused here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-02T16:46:09.100" Title="density function of U when Y has a beta distribution" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12760" PostTypeId="2" ParentId="10564" CreationDate="2010-12-02T16:23:32.170" Score="0" ViewCount="0" Body="&lt;p&gt;Your answer is mathematically correct, however I figured out another way to solve this equation, which leads to a simpler result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I applied the technique for 2D projective geometry, which is described &lt;a href=&quot;http://books.google.de/books?id=si3R3Pfa98QC&amp;amp;pg=PA35&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; to the 1D case and it works out fine.&lt;/p&gt;&#xA;" OwnerUserId="3573" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-02T16:25:39.533" LastActivityDate="2010-12-02T16:25:39.533" />
  <row Id="12761" PostTypeId="2" ParentId="12759" CreationDate="2010-12-02T16:46:09.100" Score="3" ViewCount="0" Body="&lt;p&gt;First, for the density function of a beta$(\alpha,\beta)$ random variable, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Beta_distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $Y$ is beta distributed, then it takes values in $(0,1)$. Hence, $U = 1 - Y$ also takes values in $(0,1)$.&#xA;In order to find the density of $U$, it is useful to find first its distribution function, and then differentiate it.&#xA;The distribution function of $U$ at $x \in (0,1)$ is the tail distribution function of $Y$ at $1-x$. By taking complement, you get the distribution function of $U$ expressed in terms of that of $Y$. Differentiating it, you get the density of $U$ expressed in terms of that of $Y$. You should find out that $U$ and $Y$ are very closely related.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-02T16:46:09.100" CommentCount="3" />
  <row Id="12762" PostTypeId="3" CreationDate="2010-12-02T16:57:08.190" Score="0" ViewCount="0" Body="&lt;p&gt;I am an active member of both math.stackexchange.com and&#xA;mathoverflow.net. So far mathoverflow has been a valuable&#xA;source of research-level questions and answers, and I see&#xA;math.se becoming the same at a more general level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see m.se as a site for well-focused mathematical&#xA;questions and I would take a robust line against spammers,&#xA;trolls, flamers and so on. So far the site has attracted few&#xA;of these individuals, but with increasing success, the danger&#xA;is that more would come. As a former regular on sci.math, over the years I&#xA;watched the newsgroup degenerate into a bear pit&#xA;filled with flame wars and crankery. I wish to maintain the high&#xA;signal-to-noise ratio presently here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were I elected, I would regard my appointment as for a period&#xA;of not more than one year, and I would resign or seek re-election&#xA;before or at the end of this time.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-02T16:57:08.190" LastActivityDate="2010-12-02T16:57:08.190" CommunityOwnedDate="2010-12-02T16:57:08.190" CommentCount="1" />
  <row Id="12763" PostTypeId="1" AcceptedAnswerId="12794" CreationDate="2010-12-02T17:01:18.817" Score="3" ViewCount="136" Body="&lt;p&gt;In the following picture, the author of the &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201128195&quot; rel=&quot;nofollow&quot;&gt;Field and Wave Electromagnetics&lt;/a&gt;&lt;/em&gt; shows the geometrical meaning of the direction of the gradient. That is, only by following the direction of the normal vector to the curve at that pointer could the rate of change be the maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/zpSIu.jpg&quot; alt=&quot;alt text&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the geometrical interpretation of the &lt;strong&gt;magnitude&lt;/strong&gt; of gradient &lt;strong&gt;generally&lt;/strong&gt; or maybe is there a &lt;strong&gt;geometrical interpreation&lt;/strong&gt; of the magnitude of graident?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-12-02T17:25:09.053" LastActivityDate="2010-12-03T16:58:11.923" Title="What's the geometrical interpretation of the magnitude of gradient generally?" Tags="&lt;calculus&gt;&lt;vector-analysis&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="12764" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T17:06:33.880" Score="2" ViewCount="0" Body="&lt;p&gt;If you look at the set of points satisfying f(x) = c, the gradient of f is normal to the surface and points in the direction of greatest increase of f. The magnitude of the gradient is proportional to the rate of increase.&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="247" LastEditorDisplayName="" LastEditDate="2010-12-02T20:22:23.850" LastActivityDate="2010-12-02T20:22:23.850" CommentCount="4" />
  <row Id="12765" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T17:13:03.340" Score="1" ViewCount="0" Body="&lt;p&gt;If you're standing on a point on a hill, the gradient gives the direction of the steepest ascent from that location immediately outward. The magnitude of the gradient gives how steep that ascent is, by elementary slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work out the algebra, the rate of change in a function $f$ in the direction of the unit vector $\eta$ is $\nabla f \cdot \eta$ (This is simply the chain rule.). This quantity reaches its maximum when $\eta$ shares the same direction as $\nabla f$ (to maximize the cosine of the angle that's part of the dot product formula). Thus the magnitude of $\nabla f$ gives the rate of change in that direction, as well.&lt;/p&gt;&#xA;" OwnerUserId="3122" LastEditorUserId="3122" LastEditorDisplayName="" LastEditDate="2010-12-02T17:18:32.663" LastActivityDate="2010-12-02T17:18:32.663" CommentCount="1" />
  <row Id="12766" PostTypeId="1" AcceptedAnswerId="12797" CreationDate="2010-12-02T17:20:03.920" Score="1" ViewCount="73" Body="&lt;p&gt;I have another question about correctness of my proof. We know, that $N$ and $G/N$ are p-groups, than $G$ is p-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $|N|=p^m$ and thus $|G|=p^m q$. Suppose that $(p^m,q)=1$, than in $G$ exists element of order $k$, such that $k | q$ and $(k,p^m)=1$ - suppose it's $b \in G$. Element $bN \in G/N$ is also of order $k$, and thus $k | \left| G/N \right|$ which contradicts with prerequisite, that $G/N$ is p-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's correct? It's here simply way to prove that? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4195" LastActivityDate="2010-12-02T19:40:50.273" Title="Proof that if group $N$ and $G/N$ are p-groups, than $G$ is p-group." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12767" PostTypeId="1" AcceptedAnswerId="12774" CreationDate="2010-12-02T17:27:22.437" Score="3" ViewCount="64" Body="&lt;p&gt;My question is similar to &lt;a href=&quot;http://math.stackexchange.com/questions/9627/cohomology-of-a-tensor-product-of-sheaves&quot;&gt;this&lt;/a&gt;, but not identical.  I believe the following to be true, but I'd like a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given (quasicoherent?) sheaves of $\mathcal O_X$ modules $E$ and $F$ on a projective variety $X$, &#xA;$$&#xD;&#xA;H^n(X, E\otimes F) \cong \bigoplus_{p+q=n} H^p(X, E) \otimes H^q(X, F).&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this true, and what is a good reference or counterexample? If true, is quasicoherent necessary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-12-02T18:57:28.790" Title="Cohomological decomposition of tensor sheaves?" Tags="&lt;algebraic-geometry&gt;&lt;reference-request&gt;&lt;quasicoherent-sheaves&gt;" AnswerCount="2" />
  <row Id="12768" PostTypeId="1" AcceptedAnswerId="12950" CreationDate="2010-12-02T17:29:26.363" Score="5" ViewCount="216" Body="&lt;p&gt;I'm reading &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/059680279X&quot; rel=&quot;nofollow&quot;&gt;High Performance JavaScript&lt;/a&gt;, and I think the graphs in one chapter are just plain wrong. &lt;a href=&quot;http://books.google.com/books?id=ED6ph4WEIoQC&amp;amp;pg=PA41&quot; rel=&quot;nofollow&quot;&gt;Here is one on Google Books&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The y axis is &quot;Times faster&quot;, and it runs from -1.5 to +4.0. Now, I would have thought that &quot;1 times faster&quot; means &quot;no faster&quot;, &quot;2 times faster&quot; means &quot;twice as fast&quot;, and &quot;0.5 times faster&quot; means &quot;half as fast&quot;/&quot;twice as slow&quot;. Have they just got completely confused in that graph, or is it me?&#xA;&lt;img src=&quot;http://i.imgur.com/q48tl.jpg&quot; alt=&quot;screenshot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4331" LastEditorUserId="4331" LastEditorDisplayName="" LastEditDate="2010-12-02T17:39:23.700" LastActivityDate="2010-12-03T22:01:03.900" Title="What does -1.13 times faster mean?" Tags="&lt;statistics&gt;" AnswerCount="6" CommentCount="9" />
  <row Id="12769" PostTypeId="1" AcceptedAnswerId="12773" CreationDate="2010-12-02T17:30:53.573" Score="1" ViewCount="62" Body="&lt;p&gt;Let $a_n$ be a sequence in $\mathbb{R}$, $\phi: \mathbb{N} \rightarrow \mathbb{N}$ a monotonically increasing function with $\phi (0)=0$. Show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum _{n=0}^{\infty } a_n$ converges $\Rightarrow  \sum _{i=0}^{\infty } {(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My notes: Can you use that ${(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ has to be a cauchy sequence and can be as small as you want? Certainly it will be smaller than $a_n$ for all $n$ bigger than some $N_e$&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-12-02T17:47:08.323" Title="Convergence of Sequences" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="12770" PostTypeId="1" AcceptedAnswerId="12783" CreationDate="2010-12-02T17:31:50.120" Score="2" ViewCount="93" Body="&lt;p&gt;I'm given the following exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\iint\limits_D \exp(x^{2}+y^{2})dA$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I dont even know where to start, any chance someone could give me a hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D is a half circle, given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9\le x^{2}+y^{2}\le 16, y\ge 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheres A, I don't know what it is, its not given in the exercise. - Thats partly whats bugging me.&lt;/p&gt;&#xA;" OwnerUserId="4332" LastEditorUserId="4332" LastEditorDisplayName="" LastEditDate="2010-12-02T17:43:25.630" LastActivityDate="2010-12-02T19:59:23.273" Title="Multiple integral Issue" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="12771" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T17:36:21.893" Score="1" ViewCount="0" Body="&lt;p&gt;I would say $-1.13$ faster means 1.13 slower. For example $-1.13$ faster than 160 mph is $-180.8$ mph, i.e. 180.8 going in the opposite direction.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T17:36:21.893" CommentCount="5" />
  <row Id="12772" PostTypeId="2" ParentId="12769" CreationDate="2010-12-02T17:44:59.620" Score="2" ViewCount="0" Body="&lt;p&gt;When you take a fixed N you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum _{i=0} ^{N} \sum _{j=\phi(i)} ^{\phi(i+1)-1} a_j = \sum_{j=0}^{j=\phi(N+1)-1} a_j$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so basically the new sequence (meaning the partial summations) is a subsequence of the original one.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2010-12-02T17:44:59.620" CommentCount="2" />
  <row Id="12773" PostTypeId="2" ParentId="12769" CreationDate="2010-12-02T17:47:08.323" Score="1" ViewCount="0" Body="&lt;p&gt;I suppose you meant $\phi$ is strictly increasing, otherwise the inner sum could be an empty sum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $b_i=\sum_{j=\phi(i)}^{\phi(i+1)-1} a_j$. For any $\epsilon &amp;gt;0$, you can find $N$ such that $$|a_j+\cdots +a_{j+k}|&amp;lt;\epsilon$$ for all $j\ge N, k&amp;gt;0$. Let $I$ be such that $\phi(I)\ge N$. Then for all $i\ge I$ you have &#xA;$$ |b_i+\cdots +b_{i+k}|&amp;lt;\epsilon $$&#xD;&#xA;for all $k&amp;gt;0$. This proves that your series is Cauchy and hence converges.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-02T17:47:08.323" />
  <row Id="12774" PostTypeId="2" ParentId="12767" CreationDate="2010-12-02T17:53:09.903" Score="5" ViewCount="0" Body="&lt;p&gt;No, it is false, consider e.g. $E=F=\mathcal{O}_X$ on a curve $X$ of genus one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here is something that is true though: let $p$ and $q$ be the two projection maps $X \times X \to X$. Then $H^\bullet(X\times X,p^\ast E \otimes q^\ast F) \cong H^\bullet(X,E) \otimes H^\bullet(X,F)$. There are lots of basic variants of this theorem and all of them are called the Kunneth formula.)&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-12-02T17:53:09.903" CommentCount="4" />
  <row Id="12775" PostTypeId="1" AcceptedAnswerId="12778" CreationDate="2010-12-02T17:56:08.750" Score="3" ViewCount="113" Body="&lt;p&gt;I'm looking for an example of a non-commutative ring, $R$, with identity s.t $R$ does not contain a non-trival 2 sided ideal and $R$ is not a division ring&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-12-02T21:00:30.620" LastActivityDate="2010-12-02T21:00:30.620" Title="Example of a non-commutative rings with identity that do not contain non-trival ideals and are not division rings" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12776" PostTypeId="1" AcceptedAnswerId="12782" CreationDate="2010-12-02T17:59:21.677" Score="1" ViewCount="87" Body="&lt;p&gt;-5 = 2x + 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to &quot;re-start&quot; my mathematical education.  I am needing it more and more at work.  A co-worker asked me to break that down (he couldn't remember either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where should I start?  What books?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4333" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-02T18:07:04.320" LastActivityDate="2010-12-02T18:07:04.320" Title="20 years since I was in high school.  How do I break this down?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="12777" PostTypeId="2" ParentId="12770" CreationDate="2010-12-02T17:59:43.527" Score="0" ViewCount="0" Body="&lt;p&gt;The A in $dA$ refers to the infinitesimal thing you're integrating over. It just means that instead of a line, you integrate over a surface. A for Area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a hint to start: Since you are dealing with circles, polar coordinates are a natural choice. I am sure these coordinates have been discussed in your lecture before. Note that $x^2 + y^2 = r^2$, where $r$ is the distance of the point from the coordinate origin. Also note that $dA = r dr d\phi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you can characterize the area $D$ by simple intervals for your polar coordinates: r will go from $3$ to $4$ and $\phi$ will go from $0$ to $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-12-02T17:59:43.527" CommentCount="2" />
  <row Id="12778" PostTypeId="2" ParentId="12775" CreationDate="2010-12-02T18:00:12.253" Score="8" ViewCount="0" Body="&lt;p&gt;If &quot;ideal&quot; means &quot;two-sided ideal,&quot; then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_algebra&quot; rel=&quot;nofollow&quot;&gt;Weyl algebra&lt;/a&gt; $\mathbb{C} \langle x, y \rangle / (xy - yx - 1)$ is an example.  So is $\mathcal{M}_n(R)$ for any division ring $R$ and $n \ge 2$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-02T18:00:12.253" />
  <row Id="12779" PostTypeId="2" ParentId="12775" CreationDate="2010-12-02T18:00:44.797" Score="7" ViewCount="0" Body="&lt;p&gt;If you mean two-sided ideals, you are looking for simple rings (that are not division rings):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_ring&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Simple_ring&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. as in the wikipedia article, is the ring of matrices (of a certain size) over a field. Clearly this is not a division ring since not every matrix is invertible (matrices with zero determinants are not invertible).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-02T19:25:27.297" LastActivityDate="2010-12-02T19:25:27.297" />
  <row Id="12780" PostTypeId="2" ParentId="12776" CreationDate="2010-12-02T18:05:36.787" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: In general for such one variable problems, where you need to find the value of the variable from a given single equation, you collect terms in the variable on one side of the equation and terms not containing the variable on the other side. Then divide the equation by the coefficient of the variable ($x$ in this case) to solve for it.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T18:05:36.787" />
  <row Id="12781" PostTypeId="2" ParentId="12776" CreationDate="2010-12-02T18:06:20.140" Score="0" ViewCount="0" Body="&lt;p&gt;This is elementary algebra, there are Schaum's books on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you solve for $x$ like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-5 - 5 = 2x + 5 - 5$ (substract 5 from each side), this gives $-10 = 2x$. Now divide both sides by $2$ to get $-10/2 = 2x/2$ so $x = -5$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-02T18:06:20.140" />
  <row Id="12782" PostTypeId="2" ParentId="12776" CreationDate="2010-12-02T18:06:42.183" Score="11" ViewCount="0" Body="&lt;p&gt;This is an equation, because it says &quot;Left hand side = Right hand side&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think ow an equation as a scale in perfect balance. Now you have to rearrange everything, without destroying the balance, so that the $x$ will appear at one side alone. Because then you know what $x$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you take away the same amount from &lt;em&gt;both&lt;/em&gt; sides, the scale will remain in balance. So, let us take away $5$ from both sides, and we end up with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-10 = 2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the scale is in balance now, it will remain in balance when we make each side half as big, because if two things are equal, then their halves are equal as well. This leaves us with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-5 = x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale is in balance, on the left side we have a $-5$ and on the right side we have $x$. Therefore, $x$ has to be $-5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to re-start your education: The problem above is elementary algebra, so I'd pick up with an algebra book that starts from the beginning. Something like &quot;Math for Dummies&quot; might be a nice idea as they are fun to read.&lt;/p&gt;&#xA;" OwnerUserId="4171" LastActivityDate="2010-12-02T18:06:42.183" CommentCount="1" />
  <row Id="12783" PostTypeId="2" ParentId="12770" CreationDate="2010-12-02T18:09:52.880" Score="5" ViewCount="0" Body="&lt;p&gt;The notation $\underset{D}{\displaystyle\int \int }f(x,y)dA$ has the same meaning as $\underset{D}{\displaystyle\int \int }f(x,y)dxdy$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To evaluate an integral in a different coordinate system one has to find the absolute value of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;Jacobian determinant&lt;/a&gt; of the transformation. Using polar coordinates, we can transform the initial integral &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underset{D}{\int \int }e^{x^{2}+y^{2}}dA=\underset{D}{\int \int }e^{x^{2}+y^{2}}dxdy$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) d\theta ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the conversion factor $r$ is the Jacobian determinant (see Example 3 on the Wikipedia article), and observe that $r^{2}=x^{2}+y^{2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;\underset{D}{\int \int }e^{x^{2}+y^{2}}dA &amp;amp;=&amp;amp;\underset{D}{\int \int }%&#xD;&#xA;e^{x^{2}+y^{2}}dxdy \\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) d\theta  \\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\left[ \frac{1}{2}e^{r^{2}}\right] _{3}^{4}d\theta  \\&#xD;&#xA;&amp;amp;=&amp;amp;\int_{0}^{\pi }\frac{1}{2}(e^{16}-e^{9})d\theta  \\&#xD;&#xA;&amp;amp;=&amp;amp;\frac{\pi }{2}(e^{16}-e^{9}).&#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-02T19:59:23.273" LastActivityDate="2010-12-02T19:59:23.273" />
  <row Id="12785" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T18:21:32.430" Score="0" ViewCount="0" Body="&lt;p&gt;I suppose Skilldrick is right, and $-x$ times faster is $x$ times slower or $1/x$ times faster. For example, if A takes 2 seconds and B is $2$ times faster, B takes 1 second; if C is $-2$ times faster, it takes 4 seconds. And A is $0$ times faster than itself.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-02T18:21:32.430" />
  <row Id="12786" PostTypeId="2" ParentId="12770" CreationDate="2010-12-02T18:31:31.963" Score="0" ViewCount="0" Body="&lt;p&gt;Some confusion is coming from not understanding what the $dA$ represents, so here's a link on &lt;a href=&quot;http://keep2.sjfc.edu/faculty/kgreen/vector/Block3/jacob/node4.html&quot; rel=&quot;nofollow&quot;&gt;the area element in polar coordinates&lt;/a&gt; that hopefully will help clear up this point.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-02T18:31:31.963" />
  <row Id="12787" PostTypeId="1" CreationDate="2010-12-02T18:37:21.577" Score="4" ViewCount="149" Body="&lt;p&gt;My son is taking algebra and I'm a little rusty. Not using a calculator or the internet, how would you find the roots of 2x&lt;sup&gt;4&lt;/sup&gt; + 3x&lt;sup&gt;3&lt;/sup&gt; - 11x&lt;sup&gt;2&lt;/sup&gt; - 9x + 15 = 0. Please list step by step. Thanks, Brian&lt;/p&gt;&#xA;" OwnerUserId="4338" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T02:39:10.197" LastActivityDate="2010-12-03T02:39:10.197" Title="Finding roots of fourth degree polynomial" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;" AnswerCount="2" />
  <row Id="12788" PostTypeId="2" ParentId="9092" CreationDate="2010-12-02T18:41:42.373" Score="0" ViewCount="0" Body="&lt;p&gt;First, what's $\lim_{a \to 0^+} \int_1^\infty ae^{-ax} \: dx$?  Second, can you show that your integral is not so far from $\int_1^\infty ae^{-ax} \: dx$?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2010-12-02T18:41:42.373" />
  <row Id="12789" PostTypeId="2" ParentId="12787" CreationDate="2010-12-02T18:43:51.280" Score="7" ViewCount="0" Body="&lt;p&gt;You can try first finding the rational roots using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rational_root_theorem&quot; rel=&quot;nofollow&quot;&gt;rational root theorem&lt;/a&gt; in combination with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Factor_theorem&quot; rel=&quot;nofollow&quot;&gt;factor theorem&lt;/a&gt; in order to reduce the degree of the polynomial until you get to a quadratic, which can be solved by means of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula&quot; rel=&quot;nofollow&quot;&gt;quadratic formula&lt;/a&gt; or by &lt;a href=&quot;http://en.wikipedia.org/wiki/Completing_the_square&quot; rel=&quot;nofollow&quot;&gt;completing the square&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to complement a little bit on Moron's answer, the first solution he found $x = 1$ can be guessed by using the rational root theorem since the theorem tells you to look for rational solutions only in the set of fractions $\frac{\pm a}{b}$ where $a, b \in \mathbb{Z}$ are integers such that $a$ divides $15$ and $b$ divides $2$. Thus you list all divisors of $15$, which are $\pm 1, \pm 3, \pm 5, \pm 15$ and the divisors of $2$ are $\pm 1, \pm 2$. then your list of possible rational roots would be $\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{5}{1}, \pm \frac{5}{2}, \pm \frac{15}{1}, \pm \frac{15}{2}$, and you have to start proving to see if any of those is a root of your polynomial.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-02T18:58:34.990" LastActivityDate="2010-12-02T18:58:34.990" />
  <row Id="12790" PostTypeId="2" ParentId="12787" CreationDate="2010-12-02T18:47:35.440" Score="9" ViewCount="0" Body="&lt;p&gt;Guessing one root sometimes opens up the whole equation for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First notice that $\displaystyle x=1$ gives 0. So $\displaystyle x-1$ is a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, rewrite as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x^4 - 2x^3 + 5x^3 - 5x^2 -6x^2 + 6x - 15x + 15$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to try and get $x-1$ as a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 2x^3(x-1) + 5x^2(x-1) - 6x(x-1) - 15(x-1) = (x-1)(2x^3 + 5x^2 - 6x - 15)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now notice that $\displaystyle 2x^3 - 6x = 2x(x^2-3)$ and $5x^2 - 15 = 5(x^2 - 3)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (x-1)(2x^3 + 5x^2 - 6x - 15) = (x-1)(2x(x^2 - 3) + 5(x^2 - 3)) = (x-1)(x^2-3)(2x+5)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the roots are $\displaystyle 1, \pm\sqrt{3}, -\frac{5}{2}$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-02T18:47:35.440" CommentCount="2" />
  <row Id="12791" PostTypeId="2" ParentId="12767" CreationDate="2010-12-02T18:57:28.790" Score="2" ViewCount="0" Body="&lt;p&gt;It is false.  Take $E = \mathcal O_{\mathbb P^n}(2)$ and $F= \mathcal O_{\mathbb P^n}(-2)$.  Then,&#xA;$$&#xD;&#xA;H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}) = &#xD;&#xA;H^0(\mathbb P^n, E\otimes F) \ne &#xD;&#xA;H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(2)) \otimes&#xD;&#xA;H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(-2)).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2010-12-02T18:57:28.790" />
  <row Id="12794" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T19:16:23.437" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a geometric way of thinking that might be helpful: Consider a family of level surfaces $f(x,y,z)=C$ for some evenly spaced values of $C$ (where the spacing should be fairly small). These level surfaces will lie closely stacked in space near points where $|\nabla f|$ is large, and farther apart near points where $|\nabla f|$ is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The two-dimensional counterpart is curves of constant elevation on a map; they are densely packed where the slope of the terrain is steep.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-02T19:16:23.437" />
  <row Id="12795" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T19:24:37.093" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose browser $A$ finishes loading a given page in five seconds, while browser $B$ completes in half the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times faster is $B$ than $A$? Well, twice, of course. But what about the inverse? How many &quot;times faster&quot; is $B$ than $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to answer this is to assume that &quot;$-x$ times faster than&quot; just means &quot;$1/x$ times the speed of&quot;. The intuitive problem with this is determining how much faster something is moving compared to something moving slower than it. It's just a really awkward way of putting it, but basically we can look at the number of &quot;times faster&quot; as a negative &lt;em&gt;or&lt;/em&gt; a fractional value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're both more or less valid interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we ask the question, how many times slower than $B$ is $A$, we get the original answer -- it's twice as slow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how many times faster is it? It depends on which interpretation we're using. It's correct to say that it is -2 times &quot;as fast&quot; since it is twice as slow; but also (and more intuitively) correct to say the speed of $A$ is 0.5 times the speed of $B$.&lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="4300" LastEditorDisplayName="" LastEditDate="2010-12-02T23:20:47.490" LastActivityDate="2010-12-02T23:20:47.490" />
  <row Id="12796" PostTypeId="1" CreationDate="2010-12-02T19:31:09.090" Score="1" ViewCount="47" Body="&lt;p&gt;What is the probability of getting 3 heads in a row? Would it be 1/8?&lt;/p&gt;&#xA;" OwnerUserId="4339" LastActivityDate="2010-12-02T19:36:45.130" Title="probability of heads" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12797" PostTypeId="2" ParentId="12766" CreationDate="2010-12-02T19:33:05.680" Score="3" ViewCount="0" Body="&lt;p&gt;There is a slight mistake in your argument: a priori, you don't know if the order $bN$ in $G/N$ is exactly $k$, but you do know that it is of order &lt;em&gt;dividing&lt;/em&gt; $k$, hence relatively prime to $p$. In order to show that the order cannot be $1$, you need to use the fact that $N$ itself is a $p$-group (did you notice that you never used it?) so that $g\notin N$. Once you know that, &lt;em&gt;then&lt;/em&gt; your contradiction would follow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to a simpler way, it depends on your definition of $p$-group! If to you a $p$-group is a group whose order is a power of $p$, then the very simplest way is simply to remember that for any group $G$ and any subgroups $H\subseteq K\subseteq G$, you have $[G:H]=[G:K][K:H]$ (cardinal multiplication in the case of infinite indices if necessary). So $|G|=[G:1]=[G:N][N:1] = |G/N||N|$ is a power of $p$, hence $G$ is a $p$-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are meanings of $p$-group: some authors define $G$ to be a $p$-group if for every $g\in G$ there exists $k\gt 0$ such that $g^{p^k}=1$ (that is, the order of every element is a power of $p$). Lagrange's and Cauchy's Theorems tell you that for finite groups, the two definitions coincide. But  there are infinite groups that satisfy the second meaning (but obviously not the first); the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pr%C3%BCfer_group&quot; rel=&quot;nofollow&quot;&gt;Prüfer group for $p$&lt;/a&gt; for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we prove the statement under this definition, for possibly infinite groups (it's also true)? Suppose $N$ is a $p$-group and normal in $G$, and $G/N$ is a $p$-group. Let $g\in G$. We want to show that the order of $g$ is some power of $p$. Look at $gN$ in $G/N$: since $G/N$ is a $p$-group by hypothesis, then there exists $k\gt 0$ such that $(gN)^{p^k} = eN$; that is, $g^{p^k}N = eN$. That means that $g^{p^k}\in N$. Now use the fact that $N$ is itself a $p$-group (under this definition) to deduce that the order of $g$ must be a power of $p$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T19:40:50.273" LastActivityDate="2010-12-02T19:40:50.273" />
  <row Id="12798" PostTypeId="2" ParentId="12796" CreationDate="2010-12-02T19:36:05.940" Score="4" ViewCount="0" Body="&lt;p&gt;If you are assuming 3 coin tosses of a fair coin, then there are $2\times 2\times 2=8$ outcomes. Three consective heads can occur only one way. So, yes, the probability is $\frac{1}{8}$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-02T19:36:05.940" CommentCount="3" />
  <row Id="12799" PostTypeId="2" ParentId="12796" CreationDate="2010-12-02T19:36:45.130" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming a fair coin, how many sequences of three throws are there? Are they equally likely? How many of those are &quot;3 heads in a row&quot;? &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-02T19:36:45.130" />
  <row Id="12800" PostTypeId="1" AcceptedAnswerId="12807" CreationDate="2010-12-02T19:48:03.197" Score="5" ViewCount="167" Body="&lt;p&gt;What are some motivating consequences of the axiom of choice (or its omission)? I know that weak forms of choice are sometimes required for interesting results like Banach-Tarski; what are some important consequences of a strong formulation of the axiom of choice? &lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="4300" LastEditorDisplayName="" LastEditDate="2010-12-02T19:54:35.620" LastActivityDate="2010-12-02T21:29:50.693" Title="Motivating implications of the axiom of choice?" Tags="&lt;set-theory&gt;&lt;teaching&gt;&lt;axioms&gt;&lt;axiom-of-choice&gt;" AnswerCount="9" CommentCount="1" FavoriteCount="1" />
  <row Id="12801" PostTypeId="2" ParentId="12662" CreationDate="2010-12-02T19:48:05.987" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a third alternative to statement (i).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A=\sum_{n=1}^\infty |a_n|$ and consider the sequence of partial sums $S_N=\sum_{n=1}^N |a_{\rho(n)}|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Since $\rho$ is injective $S_N\le A$ for all $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Note that $S_N\le S_{N+1}$ for all $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By (1) $S_N$ has a finite least upper bound (that is $\sup S_N&amp;lt;\infty$), and by (2) the sequence grows to that bound.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-02T19:48:05.987" />
  <row Id="12802" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:49:26.703" Score="5" ViewCount="0" Body="&lt;p&gt;Axiom of Choice $\iff$ A non-empty product of non-empty sets is non-empty.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-02T19:49:26.703" CommentCount="4" />
  <row Id="12803" PostTypeId="1" AcceptedAnswerId="12812" CreationDate="2010-12-02T19:52:16.873" Score="1" ViewCount="86" Body="&lt;p&gt;According to Wikipedia, a (non-commutative) ring R is local if and only if there do not exist two proper (principal) (left) ideals I1, I2 such that R = I1 + I2. It is easy to see that local rings have this property, but I'm having trouble showing the other way: if there do not exist two proper co-maximal (principal) (left) ideals then the ring is local.&lt;/p&gt;&#xA;" OwnerUserId="4341" LastEditorUserId="4341" LastEditorDisplayName="" LastEditDate="2010-12-02T21:43:39.270" LastActivityDate="2010-12-02T21:43:39.270" Title="Local rings characterization" Tags="&lt;abstract-algebra&gt;" AnswerCount="1" />
  <row Id="12804" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:53:42.660" Score="2" ViewCount="0" Body="&lt;p&gt;A consequence that motivates me is: If $I$ is some non-empty class and $X_i$ is a non-empty class $i\in I$, then $$\prod_{i\in I}X_i\ne\emptyset$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other consequence that motivates me is: If $I$ is some non-empty set and $X_i$ is a compact topological space $i\in I$, then $\prod_{i\in I}X_i\ne\emptyset$ is compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course these are known equivalences of the Axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-02T19:53:42.660" CommentCount="1" />
  <row Id="12805" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:54:16.073" Score="4" ViewCount="0" Body="&lt;p&gt;The Axiom of choice is required to prove for example that every ring has a maximal ideal (or a minimal prime), or the existence of an algebraic closure.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-02T19:54:16.073" CommentCount="1" />
  <row Id="12806" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:54:38.803" Score="2" ViewCount="0" Body="&lt;p&gt;To construct left- and right inverses of functions where the domain is not well-ordered we need the axiom of choice. Also, we can always get a well-ordering (this is equivalent to the axiom of choice) so that way we also could obtain this result (then you take for the many inverses the smallest one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is important in functional analysis, for example Hahn-Banach depends on it. The axiom of choice implies Hahn-Banach but not the other way around.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-02T19:54:38.803" />
  <row Id="12807" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T19:55:09.400" Score="6" ViewCount="0" Body="&lt;p&gt;Each of the following is equivalent to the Axiom of Choice:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Every vector space (over any field) has a basis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every surjection has a right inverse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Zorn's Lemma.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first is &lt;em&gt;extremely&lt;/em&gt; important and useful. The third is used all the time, especially in algebra, also very important and useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could write an entire book on important consequences (and equivalents) of the Axiom of Choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, any publisher worth his salt would reject it, since both have already been written:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Rubin, Herman; Rubin, Jean E.&#xA;&lt;strong&gt;Equivalents of the axiom of choice.&lt;/strong&gt; North-Holland Publishing Co., Amsterdam 1963 xxiii+134 pp.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rubin, Herman; Rubin, Jean E.&#xA;&lt;strong&gt;Equivalents of the axiom of choice. II.&lt;/strong&gt;&#xA;Studies in Logic and the Foundations of Mathematics, 116. North-Holland Publishing Co., Amsterdam, 1985. xxviii+322 pp. ISBN: 0-444-87708-8 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Howard, Paul; Rubin, Jean E.&#xA;&lt;strong&gt;Consequences of the axiom of choice.&lt;/strong&gt; Mathematical Surveys and Monographs, 59. American Mathematical Society, Providence, RI, 1998. viii+432 pp. ISBN: 0-8218-0977-6 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These books are probably not the best place to start, though; the first book is okay, listing some of the most important equivalents as they were known before Cohen's work, but at least the last is pretty difficult to slough through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a good introduction to the Axiom of Choice and some idea of its uses, Horst Herrlich's &lt;strong&gt;Axiom of Choice&lt;/strong&gt;, Lecture Notes in Mathematics v. 1876, Springer-Verlag (2006) ISBN: 3-540-30989-6 is pretty good, discussing some of the bad things that happen if you don't have AC, some of the bad things that happen if you &lt;em&gt;do&lt;/em&gt; have AC, and some alternative axioms that contradict AC but lead to very nice theorems. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T20:17:44.290" LastActivityDate="2010-12-02T20:17:44.290" CommentCount="3" />
  <row Id="12808" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T20:01:36.133" Score="2" ViewCount="0" Body="&lt;p&gt;There Lebesgue measure is non-trivial - there are non-measurable sets in $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-02T20:01:36.133" CommentCount="2" />
  <row Id="12809" PostTypeId="1" AcceptedAnswerId="12814" CreationDate="2010-12-02T20:02:26.253" Score="2" ViewCount="146" Body="&lt;p&gt;This argument appeared in a proof in a paper (I am paraphrasing), and I am not sure why it is true. It should be rather simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\Omega,P)$ be some sample space. Let $X$ be a random variable between $0$ and $1$. Let $U \subseteq \Omega$ and let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U^\prime = ${$ \omega \colon X(\omega) \ge E[X \mid U] - 1/m$}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $P(U^\prime) \ge P(U)/m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't see exactly why this is true.&lt;/p&gt;&#xA;" OwnerUserId="4342" LastActivityDate="2010-12-05T05:08:55.283" Title="a simple argument?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12810" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T20:03:52.293" Score="1" ViewCount="0" Body="&lt;p&gt;Adding a 2nd answer in order to keep things to one example per answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was some time ago a question on MO about whether &lt;a href=&quot;http://mathoverflow.net/questions/45928/does-arzela-ascoli-require-choice&quot; rel=&quot;nofollow&quot;&gt;Arzelà-Ascoli requires choice&lt;/a&gt;, and the answer turns out to be that it indeed does require choice - though not the full strength of AC.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-02T20:03:52.293" />
  <row Id="12811" PostTypeId="2" ParentId="12763" CreationDate="2010-12-02T20:10:23.287" Score="0" ViewCount="0" Body="&lt;p&gt;If you are at a point ${\bf p}$ in the domain of $f$ then the rate of change of $f$ when starting from ${\bf p}$ to nearby points depends on the direction you take. The gradient $\nabla f({\bf p})$ is a vector attached at the point ${\bf p}$; it points into the direction of maximal positive rate of change of $f$. The actual rate of change in this direction is $|\nabla f({\bf p})|$,  and in the opposite direction it is $-|\nabla f({\bf p})|$. For an arbitrary unit vector ${\bf e}$ attached at ${\bf p}$ the $directional\ derivative$ of $f$ in direction ${\bf e}$ is given by the scalar product $\nabla f({\bf p})\bullet {\bf e}$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-02T20:10:23.287" />
  <row Id="12812" PostTypeId="2" ParentId="12803" CreationDate="2010-12-02T20:16:31.983" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If $\rm\ R \ne 0\ $ is nonlocal then there exists an $\rm\: r\:$ such that both $\rm\: r\:$ and $\rm\: 1-r\:$ are nonunits. Choose maximal ideals containing them.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-02T20:16:31.983" CommentCount="5" />
  <row Id="12813" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T20:18:41.263" Score="3" ViewCount="0" Body="&lt;p&gt;The Hahn-Banach theorem and Tychonoff's theorem are two major ways in which the axiom of choice gets used in analysis (I am echoing other answers here because these are really important).  In addition to &lt;a href=&quot;http://en.wikipedia.org/wiki/Zorns_lemma&quot; rel=&quot;nofollow&quot;&gt;Zorn's lemma&lt;/a&gt;, the third &quot;classic&quot; theorem known to be equivalent to AC is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-ordering_theorem&quot; rel=&quot;nofollow&quot;&gt;well-ordering theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a somewhat surprising one I was recently made aware of: you cannot prove that a countable union of countable sets is countable without some form of countable choice.  The problem is that in order for any of the standard proofs to go through, you need to choose, for each countable set in the union, a bijection of it with $\mathbb{N}$, and there is no way to do this canonically in general.  (Fortunately, for most countable sets we encounter in practice we &lt;em&gt;can&lt;/em&gt; define explicit bijections, so there's usually no problem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A weak form of choice, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem&quot; rel=&quot;nofollow&quot;&gt;Boolean prime ideal theorem&lt;/a&gt; (or equivalently the ultrafilter lemma), also has many useful consequences.  For example, it is equivalent (I think) to Tychonoff's theorem for compact Hausdorff spaces.  It is equivalent to &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_completeness_theorem&quot; rel=&quot;nofollow&quot;&gt;Gödel's completeness theorem&lt;/a&gt;, as well as to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Compactness_theorem&quot; rel=&quot;nofollow&quot;&gt;compactness theorem&lt;/a&gt;.  It allows you to construct &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraproduct&quot; rel=&quot;nofollow&quot;&gt;ultraproducts&lt;/a&gt;, which has a whole genre of uses such as, for example, defining &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-standard_analysis&quot; rel=&quot;nofollow&quot;&gt;nonstandard models&lt;/a&gt; of $\mathbb{R}$ or more exotic things.  Terence Tao has written several good blog posts, including &lt;a href=&quot;http://terrytao.wordpress.com/2007/06/25/ultrafilters-nonstandard-analysis-and-epsilon-management/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;, on these and related subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in browsing &lt;a href=&quot;http://consequences.emich.edu/CONSEQ.HTM&quot; rel=&quot;nofollow&quot;&gt;http://consequences.emich.edu/CONSEQ.HTM&lt;/a&gt; for more consequences of AC.  There are really quite a lot.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-02T20:24:27.280" LastActivityDate="2010-12-02T20:24:27.280" CommentCount="1" />
  <row Id="12814" PostTypeId="2" ParentId="12809" CreationDate="2010-12-02T20:21:52.057" Score="2" ViewCount="0" Body="&lt;p&gt;We can assume that $U = \Omega$. So you want to show $$\Pr[X \geq E[X] - \alpha] \geq \alpha,$$ where $\alpha = 1/m$. If not, then $$E[X] &amp;lt; \alpha + (1-\alpha)(E[X] - \alpha) = (1-\alpha) E[X] + \alpha^2,$$ so that $\alpha E[X] &amp;lt; \alpha^2$ and $E[X] &amp;lt; \alpha$. But in that case, the inequality trivially holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Some more details. Suppose that $\Pr[X &amp;gt; E[X] - \alpha] = p &amp;lt; \alpha$. Given $p$, the distribution maximizing $E[X]$ under this constraint is $$\Pr[X=E[X]-\alpha]=1-p,\quad \Pr[X=1]=p.$$ Its expectation is $$p + (1-p)(E[X] - \alpha)= E[X] - \alpha + p(1 - E[X] + \alpha).$$ Since $p &amp;lt; \alpha$ and $1 - E[X] + \alpha &amp;gt; 0$, we get that $$E[X] &amp;lt; E[X] - \alpha + \alpha(1 - E[X] + \alpha).$$ Rearranging, $$\alpha (1 + E[X]) &amp;lt; \alpha (1 + \alpha)$$ and so $E[X] &amp;lt; \alpha$, in which case the inequality trivially holds.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-03T05:12:09.257" LastActivityDate="2010-12-03T05:12:09.257" CommentCount="5" />
  <row Id="12815" PostTypeId="1" AcceptedAnswerId="12819" CreationDate="2010-12-02T20:43:23.167" Score="7" ViewCount="221" Body="&lt;p&gt;Starting with the famous Basel problem, Euler evaluated the Riemann zeta function for all even positive integers and the result is a compact expression involving Bernoulli numbers. However, the evaluation of the zeta function at odd positive integers (in terms of getting a closed form sum) is still open. There has been some progress in the form of Apery's theorem and other results such as &quot;infinitely many of $\zeta(2n+1)$ are irrational&quot; or &quot;at least one of $\zeta(5),\zeta(7),\zeta(9)$ or $\zeta(11)$ is irrational&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question(s): Is there a high level understanding for this disparity between even and odd integers? Is it a case of there being a simple expression for $\zeta(3)$ that is out there waiting for an ingenious attack like Euler did with $\zeta(2)$? Or is the belief that such a closed form summation is unlikely? Where do the many many proofs powerful enough to evaluate $\zeta(2n)$ stumble when it comes to evaluating $\zeta(2n+1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: The Basel problem and Euler's solution are my all-time favorites for the sheer surprise factor and ingenuity of proof (what do $\pi$ and $\frac{sin(x)}{x}$ have to do with $\zeta(2)$??). However, I currently lack the more advanced analytical tools to appreciate the deeper results of this area. I have wondered for a while about the questions above and Internet search hasn't helped much. I would greatly appreciate any answers/references. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2011-01-04T13:40:59.233" Title="Riemann zeta function at odd positive integers" Tags="&lt;analytic-number-theory&gt;&lt;riemann-zeta&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="12816" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T20:43:25.520" Score="1" ViewCount="0" Body="&lt;p&gt;There are many abuses like this in the world.  I have seen things on sale that were 400% OFF!!!  Would they give me 3x the original price if I took it?  Maybe -1.13 times faster means that you send them 113% as many bits as you received in the base case.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-02T20:43:25.520" />
  <row Id="12817" PostTypeId="1" CreationDate="2010-12-02T20:49:30.647" Score="2" ViewCount="153" Body="&lt;p&gt;Given the following formula:&#xA;$$\bigg[\forall x P(x,x) \wedge \forall x \forall y \forall z\bigg( P(x,y) \wedge  P(y,z) \Rightarrow P(x,z)\bigg) \wedge \forall x \forall y (P(x,y) \vee P(y,x)) \bigg] \Longrightarrow \exists y \forall x P(y,x)&#xD;&#xA;$$ &#xA;It is known that this formula is always true if x,y,z are in any domain with finite size, but not always true in infinite case!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the first part can be proved by induction! If the domain contains only one point x then formula is true. Suppose its true for a domain of size k, how to prove the k+1 case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part I need to find a counterexample for such P in some infinite domain, say the integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?!&lt;/p&gt;&#xA;" OwnerUserId="4343" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-07T07:56:37.257" LastActivityDate="2010-12-07T07:56:37.257" Title="Model Theory-logic" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12818" PostTypeId="2" ParentId="12817" CreationDate="2010-12-02T20:52:50.333" Score="4" ViewCount="0" Body="&lt;p&gt;The premises tell you that $P$ is reflexive and transitive, as well as total; this means that it is a total pre-order on the domain. The conclusion is that the pre-order has a smallest element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you go about doing it by induction? Well, suppose you know the result holds in any domain with fewer than $k$ elements, and your domain has $k$ elements. Pick $x_0\in D$ ($D$ the domain), and let $D&amp;#39;=D\setminus\{x_0\}$, and let $P&amp;#39;=P\cap D&amp;#39;\times D&amp;#39;$. If you can prove that $P&amp;#39;$ satisfies the premises of the formula relative to $D&amp;#39;$ (given that $P$ satisfies them relative to $D$), then you'll be able to conclude that there is a $y\in D&amp;#39;$ such that $(y,x)\in P&amp;#39;$ for every $x\in D&amp;#39;$. Now, how can $y$ fail to be an element with the right property if you now consider it in $D$? And if that happens, can you find an element that &lt;em&gt;will&lt;/em&gt; work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to a counterexample, given my first paragraph, does something suggest itself?&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Okay, you are asking for some more help on getting the induction going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n$ be the number of elements in the model $D$. We want to prove the formula holds for $D$. We do so by induction on $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base case is $n=1$. To establish the proposition, we are allowed to assume the premise holds: that for all $x\in D$ we have $P(x,x)$, that for all $x$, $y$, and $z$ in $D$, if $P(x,y)$ and $P(y,z)$ both hold, then $P(x,z)$ holds; and that for all $x$ and $y$ in $D$, either $P(x,y)$ holds or $P(y,x)$ holds. We want to show that there exists a $y\in D$ such that for all $x\in D$, $P(y,x)$ holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $D = \{x_1\}$ has a single element, what can $y$ be? It must be $x_1$. So you need to show that for every $x$ in $D$, $P(x_1,x)$ holds. Surely this is easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so now assume that you know the proposition is true in &lt;em&gt;any&lt;/em&gt; domain with $k$ elements, and assume that $D=\{x_1,x_2,\ldots,x_k,x_{k+1}\}$ has $k+1$ elements. We are also allowed to assume that $P(x,x)$ holds for all $x\in D$, that for all $x,y,z$ in $D$, if $P(x,y)$ and $P(y,z)$ holds, then $P(x,z)$ holds; and that for all $x$ and $y$ in $D$, either $P(x,y)$ holds or $P(y,x)$ holds. We want to show that there is a $y\in D$ such that for all $x\in D$, $P(y,x)$ holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $D&amp;#39;=D-\{x_{k+1}\}$, and let $P&amp;#39;$ be a binary relation on $D&amp;#39;$ defined by &#xA;$$P'(x,y)\text{ in $D'$} \Leftrightarrow P(x,y) \text{in $D$.}$$&#xA;Verify that $P&amp;#39;$ satisfies the premise of the proposition in $D&amp;#39;$: for all $x\in D&amp;#39;$, $P&amp;#39;(x,x)$ holds; if $P&amp;#39;(x,y)$ and $P&amp;#39;(y,z)$ hold, then $P&amp;#39;(x,z)$ hold; and for all $x$ and $y$ in $D&amp;#39;$, either $P&amp;#39;(x,y)$ holds or $P&amp;#39;(y,x)$ holds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the induction hypothesis, the formula is &lt;em&gt;true&lt;/em&gt; in $D&amp;#39;$; so if the &lt;em&gt;premise&lt;/em&gt; is also true in $D&amp;#39;$, then the conclusion is true in $D&amp;#39;$. So there is a $y_0\in D&amp;#39;$ such that for all $x\in D&amp;#39;$ you have $P&amp;#39;(y_0,x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, go back to $D$; $y_0$ is in $D$, and for every $x\in D$, if $x\in D&amp;#39;$ (that is, if $x\neq x_{k+1}$) then $P(y_0,x)$ holds. Now, what happens to $x_{k+1}$? Well, by our premises, either $P(y_0,x_{k+1})$ holds, or else $P(x_{k+1},y_0)$ holds because $y_0$ and $x_{k+1}$ are both in $D$. Think about what each of those two things would tell you in terms of trying to find a $y\in D$ such that $P(y,x)$ holds for all $x$ in $D$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T04:56:10.807" LastActivityDate="2010-12-03T04:56:10.807" CommentCount="4" />
  <row Id="12819" PostTypeId="2" ParentId="12815" CreationDate="2010-12-02T21:00:02.087" Score="12" ViewCount="0" Body="&lt;p&gt;The zeta function is defined as a sum over the positive integers, but as far as actually evaluating it, it turns out to be more natural to think of it as a sum over all nonzero integers; thus we should really be thinking about $\sum_{n \neq 0} \frac{1}{n^k}$.  For $k$ even this is just $2 \zeta(k)$ and there are various ways to evaluate this more symmetric sum, e.g. by writing down a meromorphic function with the right poles, or a Fourier series with the right coefficients, etc.  But for $k$ odd this is equal to zero, since terms cancel with their negatives!  Written in this way, the zeta function at even integers reveals its alter ego as an &lt;a href=&quot;http://en.wikipedia.org/wiki/Eisenstein_series&quot; rel=&quot;nofollow&quot;&gt;Eisenstein series&lt;/a&gt; in one dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cancellation phenomenon occurs in Euler's classic &quot;proof,&quot; since the infinite product for $\sin z$ that he uses has zeroes at all integer multiples of $\pi$, not just the positive ones.  It also occurs in the general proof that proceeds by considering the generating function $\frac{z}{e^z - 1}$ for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli_number&quot; rel=&quot;nofollow&quot;&gt;Bernoulli numbers&lt;/a&gt;.  As you might know, the closed form of $\zeta(2k)$ involves Bernoulli numbers, and again $\frac{z}{e^z - 1}$ has poles at $2 \pi i n$ for all nonzero integers $n$, not just the positive ones.  I describe how this works in slightly more detail &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n1-infty-frac1n2/8373#8373&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about the difference between the even and odd cases is that one can think of the even cases as $L^2$ norms of appropriate Fourier series; this is precisely how a standard proof of the evaluation of $\zeta(2)$ works.  But for the odd cases we don't get an $L^2$ norm; instead we get a mysterious inner product.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-02T21:05:33.360" LastActivityDate="2010-12-02T21:05:33.360" CommentCount="2" />
  <row Id="12820" PostTypeId="1" AcceptedAnswerId="12861" CreationDate="2010-12-02T21:08:26.180" Score="0" ViewCount="77" Body="&lt;p&gt;Motivated by my ignorance &lt;a href=&quot;http://math.stackexchange.com/questions/12767/cohomological-decomposition-of-tensor-sheaves/12774#12774&quot;&gt;here&lt;/a&gt;, if $X$ is a projective toric variety, is &#xA;$$H^m(X, \mathcal O_X) \cong &#xD;&#xA;\begin{cases}&#xD;&#xA;0 &amp;amp; m &amp;gt; 0 \\&#xD;&#xA;\mathbb C &amp;amp; m = 1&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&#xA;as for $\mathbb P^n$?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="3008" LastEditorDisplayName="" LastEditDate="2010-12-02T23:10:42.287" LastActivityDate="2010-12-03T03:31:10.913" Title="Cohomology of $\mathcal O_X$ for toric varieties" Tags="&lt;algebraic-geometry&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12821" PostTypeId="1" CreationDate="2010-12-02T21:11:20.800" Score="1" ViewCount="75" Body="&lt;p&gt;If $F_{1}$ is a subfield of $F_{2}$, is $GL(N,F_{1})$ contained in $GL(N,F_{2})$?   I am particular interested in the case of when $F_{1} =\mathbb{Q}(\omega)$ and $F_{2} = \mathbb{C}$.   Here $\omega$ is a complex root of unity.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T04:57:45.000" LastActivityDate="2010-12-03T04:57:45.000" Title="If $F_{1}$ is a subfield of $F_{2}$, is $GL(N,F_{1})$ contained in $GL(N,F_{2})$?" Tags="&lt;abstract-algebra&gt;" CommentCount="9" />
  <row Id="12822" PostTypeId="1" AcceptedAnswerId="12827" CreationDate="2010-12-02T21:11:34.823" Score="2" ViewCount="155" Body="&lt;p&gt;I am trying to better understand how unique factorization of algebraic integers in an algebraic number ring implies that the class number of that number ring is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking for some examples of this to get me started.  &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-02T23:22:17.380" Title="Examples of Class Group" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="12823" PostTypeId="1" AcceptedAnswerId="12830" CreationDate="2010-12-02T21:19:13.233" Score="0" ViewCount="92" Body="&lt;p&gt;Hi. I need to re-arrange the following equation (see link JPG below please, sorry I'm not familiar with how to write an equation on here and as I'm a newbie it won't let me post images) to solve for $d$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = 4\pi C_x d^2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3686" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T21:37:44.287" LastActivityDate="2010-12-02T22:32:09.323" Title="Re-arranging a complex equation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12824" PostTypeId="1" AcceptedAnswerId="12835" CreationDate="2010-12-02T21:22:21.943" Score="6" ViewCount="214" Body="&lt;p&gt;So one way to define a group presentation is to say, well, let's generate the free group with some number of generators, and then quotient by saying certain elements (relators) cancel just as $aa^{-1}$ does (and other things, you have to take the normal subgroup generated by the elements, but the basic idea works, I think). That is, we have for some words $R_1, R_2, \ldots$ the relations $R_1 = R_2 = \ldots = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example is the cyclic group of order n: one presentation is $\langle g \mid g^n \rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can and do we do this for rings? I'm imagining you would do something like take the &quot;free ring&quot; consisting of all sums of products of generators, and perhaps instead of having all the relators equal to the trivial group they would equal the zero ideal, because then if you wanted some word $R$ to equal the multiplicative identity you would just say $R - 1 = 0$. (In particular, the comments on the vaguely related question that got me thinking about this is &lt;a href=&quot;http://math.stackexchange.com/questions/12698/why-is-mathbbzx-1-x-p-isomorphic-to-mathbbz-p-where-p-is-a-prime&quot;&gt;here&lt;/a&gt;: could you formulate a rigorous proof by &quot;dividing by the relations&quot;?)&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-12-30T23:47:04.843" Title="Do people study &quot;ring presentations&quot;? Is this a dumb question?" Tags="&lt;reference-request&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="12825" PostTypeId="1" CreationDate="2010-12-02T21:25:50.507" Score="6" ViewCount="123" Body="&lt;p&gt;My research has brought me to the following, very general problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a fixed, but arbitrary, natural number, $\displaystyle v$, consider the following family of polynomials: The $\displaystyle (n-1)^{th}$ derivative of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (1-x^2)^{v+n} \ \ \forall n \in \mathbb{N} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to prove (or disprove) that the roots of this entire family of polynomials forms a dense subset of the interval $\displaystyle [0,1]$ for any value of $\displaystyle v$ (I am not interested in roots outside the interval $\displaystyle [0,1]$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, given any subinterval, $\displaystyle [a,b]$,no mater how small, at least one of these polynomials has at least one root in the interval $\displaystyle [a,b]$ (for any fixed value of $\displaystyle v$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize my question is very general and will happily accept any partial solutions.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-02T22:23:01.593" LastActivityDate="2011-01-08T09:38:31.403" Title="density of roots of a family of polynomials: $(1-x^2)^{v+n}$" Tags="&lt;real-analysis&gt;&lt;polynomials&gt;&lt;special-functions&gt;&lt;roots&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="12826" PostTypeId="2" ParentId="12800" CreationDate="2010-12-02T21:29:50.693" Score="1" ViewCount="0" Body="&lt;p&gt;I think that one of the most important implications of the axiom of choice is actually the equivalence of continuity between the Cauchy definition of $\epsilon$-$\delta$ and the Heine definition using sequences.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-02T21:29:50.693" />
  <row Id="12827" PostTypeId="2" ParentId="12822" CreationDate="2010-12-02T21:30:20.677" Score="4" ViewCount="0" Body="&lt;p&gt;I'm not sure what you mean by giving you examples... the abstract argument seems simple enough to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every principal ideal that is generated by a prime element is a prime ideal: suppose $ab\in (p)$; then $p|ab$, and since $p$ is prime, $p|a$ or $p|b$, so $a\in (p)$ or $b\in (p)$. Conversely, a nonzero principal ideal is prime if and only if the generator is a prime element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\mathfrak{M}$ is a maximal ideal of the number ring; let $a\in\mathfrak{M}$. Then $a = p_1\cdots p_k$ for some primes $p_i$ by unique factorization into primes, hence $(a) = (p_1\cdots p_k) = (p_1)(p_2)\cdots(p_k)\subseteq \mathfrak{M}$. Since $\mathfrak{M}$ is maximal, it is prime, so there exists $i$ such that $(p_i)\subseteq \mathfrak{M}$. But since $p_i$ is a prime element, then $(p_i)$ is a prime ideal, and in a number ring every nonzero prime ideal is maximal; hence $\mathfrak{M}=(p_i)$. Therefore, every maximal ideal is principal; in particular, every nonzero prime ideal is principal, since nonzero prime ideals are maximal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\mathfrak{I}$ be any ideal; if $\mathfrak{I}=(0)$, then it is principal. Otherwise, $\mathfrak{I}$ is a product of prime ideals by the Fundamental Theorem of Dedekind Domains, so $\mathfrak{I}=(p_1)\cdots (p_k)$ with $p_i$ a prime element (since primes are maximal, which are principal, and therefore generated by prime elements); but $(p_1)\cdots(p_k) = (p_1\cdots p_k)$ is principal, so $\mathfrak{I}$ is principal. Thus, if you have unique factorization, then every ideal is principal, and therefore the class number is $1$: given any two nonzero ideals $\mathfrak{I}$ and $\mathfrak{J}$, we can find elements $a$ and $b$ such that $\mathfrak{I}=(a)$ and $\mathfrak{J}=(b)$, and therefore we have that $b\mathfrak{I}=a\mathfrak{J}$, hence $\mathfrak{I}\sim \mathfrak{J}$ in the ideal class group. Since this holds for any two nonzero ideals, the ideal class group consists of exactly one class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;The key really is that you have unique factorization into prime ideals and that nonzero primes are maximal; without that, you would be out of luck. One problem with the simplest examples ($\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{-2}]$, for instance) is that are also Euclidean domains, so that's really how you get the unique factorization (as a further consequence, a prior one already being that it is a PID). I think the smallest imaginary quadratic which is a PID but not Euclidean is $\mathbb{Z}[(1+\sqrt{-19})/2]$, and I'm not sure I would want to try to work out explicitly the argument above with specific elements of that ring.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-02T21:35:27.703" LastActivityDate="2010-12-02T21:35:27.703" CommentCount="1" />
  <row Id="12829" PostTypeId="2" ParentId="12822" CreationDate="2010-12-02T21:55:54.123" Score="3" ViewCount="0" Body="&lt;p&gt;$\rm PID\:$s are &lt;em&gt;precisely&lt;/em&gt; the $\rm UFD\:$s having dimension $\le 1\ $ (i.e. every nonzero prime ideal is maximal). Therefore $\rm UFD $ Dedekind domains are $\rm PID\:,$ so they have trivial class group. Here's the key result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$ &lt;br/&gt;&#xA;$1)\ $ prime ideals are maximal if nonzero&lt;br&gt;&#xA;$2)\ $ prime ideals are principal&lt;br&gt;&#xA;$3)\ $ maximal ideals are principal&lt;br&gt;&#xA;$4)\ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$ &lt;br/&gt;&#xA;$5)\ $ $\rm D$ is Bezout&lt;br&gt;&#xA;$6)\ $ $\rm D$ is a $\rm PID$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$ &lt;br/&gt;&#xA;$2\Rightarrow 3)$ $\ \: $ Clear.&lt;br&gt;&#xA;$3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$ &lt;br/&gt;&#xA;$4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$ &lt;br/&gt;&#xA;$5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors&lt;br&gt;&#xA;$6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-02T22:44:15.793" LastActivityDate="2010-12-02T22:44:15.793" CommentCount="0" />
  <row Id="12830" PostTypeId="2" ParentId="12823" CreationDate="2010-12-02T22:32:09.323" Score="1" ViewCount="0" Body="&lt;p&gt;As Arturo said&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=4 \pi C_x d^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies d^2 = \frac{L}{4 \pi C_x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies d= \pm\sqrt{\frac{L}{4 \pi C_x}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation looks like it is physical, so the negative square root may not be relevant&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-12-02T22:32:09.323" />
  <row Id="12831" PostTypeId="2" ParentId="12822" CreationDate="2010-12-02T22:54:29.837" Score="2" ViewCount="0" Body="&lt;p&gt;This is actually a frequently used (e.g. in the theory of divisors) fact of commutative algebra. A noetherian domain $A$ is a UFD if and only if every prime ideal of height one (by Krull's principal ideal theorem, this is the same thing as being minimal over a nonzero element) is principal (which corresponds, with a little work, to the statement that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weil_divisor&quot; rel=&quot;nofollow&quot;&gt;Weil divisor class group&lt;/a&gt; of a normal ring is trivial iff it is factorial). In the case of a Dedekind domain (of which a ring of integers in a number field is a paradigm example), this means that every prime ideal is principal (as $(0)$ obviously is).&#xA;In a Dedekind domain, every ideal is a product of prime ideals. So if every prime ideal is principal, so is every ideal. This is the statement that the class number is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this answers your question by reducing it to another result. You can find the proof and discussion of this result as Theorem 18.6 in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://people.fas.harvard.edu/~amathew/CAnotes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.fas.harvard.edu/~amathew/CAnotes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-02T23:22:17.380" LastActivityDate="2010-12-02T23:22:17.380" CommentCount="10" />
  <row Id="12832" PostTypeId="2" ParentId="12690" CreationDate="2010-12-02T23:00:54.493" Score="0" ViewCount="0" Body="&lt;p&gt;The algebraic structure corresponding most naturally to the anticommutator is that of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clifford_algebra&quot; rel=&quot;nofollow&quot;&gt;Clifford algebras&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can construct a Clifford algebra by defining $n$ generating elements $\mathbf{e}_j$ with $j \in \{1,\ldots,n\}$ such that they satisfy the conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{e}_i \mathbf{e}_j + \mathbf{e}_j \mathbf{e}_i = 2 \delta_{ij} \; ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta_{ij}$ is the Kronecker delta. From these elements you can form products of elements of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{e}_{i_1}\mathbf{e}_{i_2}\ldots\mathbf{e}_{i_p} \; ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $p$ varying from $0$ to $n$. There's a total of $2^n$ products that can be formed and from these, taking linear combination you can generate the rest of the algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check that Pauli matrices and quaternions form Clifford algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you take a Clifford algebra with $n$ generating elements, and construct the elements $M_{ij}=\frac{1}{2}\mathbf{e}_{i}\mathbf{e}_{j}$, you can show that they form the basis of the Lie algebra $so(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another construct related to fermions, and which is important in the path integral representation of fermions, are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grassmann_number&quot; rel=&quot;nofollow&quot;&gt;Grassman numbers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-02T23:06:19.323" LastActivityDate="2010-12-02T23:06:19.323" CommentCount="3" />
  <row Id="12833" PostTypeId="1" CreationDate="2010-12-02T23:06:40.527" Score="5" ViewCount="93" Body="&lt;p&gt;If $X$ is a locally compact and metrizable space such that its Alexandroff compactification is &lt;em&gt;not&lt;/em&gt; first countable. Does this imply that no other compactification of $X$ can be first countable? Why?&lt;/p&gt;&#xA;" OwnerUserId="4348" LastActivityDate="2010-12-02T23:06:40.527" Title="Alexandroff compactification question" Tags="&lt;topology&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="12834" PostTypeId="2" ParentId="7516" CreationDate="2010-12-02T23:07:35.427" Score="0" ViewCount="0" Body="&lt;p&gt;An excellent book is also &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444520775&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Lectures-Curry-Howard-Isomorphism-Foundations-Mathematics/dp/0444520775&lt;/a&gt;.  This book is more focused on the $\lambda$-Calculus but it has some really excellent sections on general intuitionist logic, including Heyting Algebras, Hilbert Proofs, the (Gentzen's) sequent calculus and things like that.  &lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-02T23:07:35.427" />
  <row Id="12835" PostTypeId="2" ParentId="12824" CreationDate="2010-12-02T23:16:05.387" Score="10" ViewCount="0" Body="&lt;p&gt;You certainly can: the constructions of the tensor algebra, symmetric algebra, and exterior algebra can all be performed using generators and relations (though are perhaps usually thought of conceptually via  a universal property). The operation of localization is also an example of this: one formally adjoins &quot;denominators&quot; satisfying the requisite relations (localization of $A$ at a multiplicative set $S$ can be thought of adjoining variables $x_s, s \in S$ with the relations $s x_s = 1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of a ring having a &quot;finite presentation&quot; over another ring $R$ (i.e. be  the quotient of a polynomial ring $R$ by a finitely generated ideal--that is, by finitely many generators and relations) is of importance in algebraic geometry, as many properties proved in, say, the finite type case over noetherian schemes extend here (e.g. I believe Chevalley's theorem of the constructibility of the image of a constructible set under a morphism of schemes is true under finite presentation hypotheses). I am not sufficiently qualified to say much more on this topic, but you might find BCnrd's answer &lt;a href=&quot;http://mathoverflow.net/questions/1788/does-finitely-presented-mean-always-finitely-presented-answered-yes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; enlightening.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-02T23:16:05.387" CommentCount="1" />
  <row Id="12836" PostTypeId="2" ParentId="12768" CreationDate="2010-12-02T23:28:16.910" Score="0" ViewCount="0" Body="&lt;p&gt;Strictly speaking, $-1.13$ times faster would mean that the program is doing things in reverse order. And Opera would completely stand still. Since that can't be what they really mean, it must be an abuse of language as has already been pointed out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be though that the original scale was logarithmic and that the person then just interpreted the numbers incorrectly? Which would mean that the graph is correct, but the person who added the 'times faster' doesn't get the concept of logarithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $-1.13$ should be $a^{-1.13}$ in some suitable base, probably base $10$, $2$ or $e$?&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-02T23:28:16.910" CommentCount="2" />
  <row Id="12837" PostTypeId="1" AcceptedAnswerId="12840" CreationDate="2010-12-02T23:42:49.027" Score="6" ViewCount="146" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are ten people in a room:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Person 1 knows 9 people. Person 2&#xA;  knows 8 people. etc. Person 9 knows 1&#xA;  person.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How many people does person 10 know?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the answer and what is the principle behind this question?&lt;/p&gt;&#xA;" OwnerUserId="4350" LastEditorUserId="4350" LastEditorDisplayName="" LastEditDate="2010-12-02T23:51:36.443" LastActivityDate="2010-12-03T00:01:17.413" Title="Ten people in a room" Tags="&lt;logic&gt;&lt;puzzle&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12838" PostTypeId="1" CreationDate="2010-12-02T23:50:53.137" Score="2" ViewCount="63" Body="&lt;p&gt;(Assume a unit speed parametrization)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that an admissible curve $c:(a,b) \to \mathbb R^3$ (hence $c^{\prime} \times c^{\prime\prime}$ is never zero) with a zero torsion is contained in a plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why this should be true, if there is no torsion element to the curve then $c$ can't twist into the third dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T = c^{\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N = c^{\prime\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B = \mathbf T \times \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T^{\prime} = \kappa \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B^{\prime} = -\tau\mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the torsion $\tau$ is zero, $\mathbf B^{\prime} = \mathbf 0$, therefore $\mathbf B$ is a constant not equal to zero. I'm not sure where to go from here. If the normal vector of $\mathbf T$ and $\mathbf N$ is a constant, does that mean no matter what point you are at on the curve $\mathbf B$ points in the same direction?&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T00:33:05.113" LastActivityDate="2010-12-03T02:46:16.250" Title="Admissible curve contained in a plane" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12839" PostTypeId="1" AcceptedAnswerId="12841" CreationDate="2010-12-02T23:57:41.700" Score="2" ViewCount="116" Body="&lt;p&gt;The following is a multiplication problem in traditional base ten notation. Each letter represents a different digit. What digit does each letter represent? How did you get your foot in the door? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/cXC4Y.png&quot; alt=&quot;i hate this course&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T05:14:15.310" LastActivityDate="2010-12-03T05:14:15.310" Title="Algorithm (cryptarithm) question, yes it is homework, but I can't get it" Tags="&lt;homework&gt;&lt;arithmetic&gt;&lt;cryptarithm&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12840" PostTypeId="2" ParentId="12837" CreationDate="2010-12-03T00:01:17.413" Score="5" ViewCount="0" Body="&lt;p&gt;You can think of this as a problem in &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_theory&quot; rel=&quot;nofollow&quot;&gt;graph theory&lt;/a&gt;.  You have a graph $G$ on ten vertices (the ten people in the room ${ 1, 2, 3, ... 10 }$) where two people are connected by an edge if and only if they know each other.  The &lt;strong&gt;degree&lt;/strong&gt; or &lt;strong&gt;valence&lt;/strong&gt; of a vertex $v \in G$, denoted $d_v$, is the number of edges connected to it (the number of people someone knows).  You are given that $d_1 = 9, d_2 = 8, ... d_9 = 1$ and you want to know $d_{10}$ (so the problem gives you information here: it tells you that $d_{10}$ is uniquely determined by these conditions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one way to solve it, in steps.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Person $1$ must know everyone else; that is, he must be connected to all of the other vertices (including person $10$).  In particular, he must be the only person that person $9$ knows.  &lt;/li&gt;&#xA;&lt;li&gt;Person $2$ must know everyone else except one person, and since person $9$ only knows person $1$, person $2$ must know everyone else except person $9$ (including person $10$).  In particular, he must be the only other person that person $8$ knows.&lt;/li&gt;&#xA;&lt;li&gt;Person $3$ must know everyone else except two people, and since persons $8$ and $9$ have all their friends accounted for, person $3$ must know everyone else except persons $8$ and $9$ (including person $10$).  In particular, he must be the only other person that person $7$ knows.&lt;/li&gt;&#xA;&lt;li&gt;... and so forth.  By &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_induction&quot; rel=&quot;nofollow&quot;&gt;induction&lt;/a&gt; we conclude that person $10$ knows persons $1, 2, 3, 4, 5$ but not persons $6, 7, 8, 9$.  So $d_{10} = 5$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-03T00:01:17.413" CommentCount="2" />
  <row Id="12841" PostTypeId="2" ParentId="12839" CreationDate="2010-12-03T00:03:01.467" Score="2" ViewCount="0" Body="&lt;p&gt;For the first equation $X=1,Y=0$ works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be the only solution since, $XY\cdot YX= XY$ implies $XY=0$ or if $XY\neq 0$ then, $YX=1$. If $XY=0$ then $X=Y=0$ (remember $X,Y$ are digits here), so they are not distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, $Y=0$, so the only way to get a third digit in the final solution, would be to have $X+Z=1+Z$ be a two digit sum. This can only happen if $Z=9$, so $V=0$ and $W=1$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-03T00:15:39.067" LastActivityDate="2010-12-03T00:15:39.067" CommentCount="3" />
  <row Id="12842" PostTypeId="1" AcceptedAnswerId="12843" CreationDate="2010-12-03T00:03:31.640" Score="4" ViewCount="123" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A newspaper launches a competition. It said that readers should submit one number between 1 and 1000. A £2000 prize would be awarded to the person that got the closest to 2/3 of the mean of all the entries. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What would a smart entry be? What is the underlying principle behind this type of question. I know it involves game theory but I do not fully understand this type of problem yet. Can anyone draw the game theory tables involved / explain the Nash Equilibria?&lt;/p&gt;&#xA;" OwnerUserId="4350" LastActivityDate="2010-12-03T00:34:24.143" Title="Newspaper competition" Tags="&lt;puzzle&gt;&lt;game-theory&gt;&lt;nash-equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12843" PostTypeId="2" ParentId="12842" CreationDate="2010-12-03T00:17:20.233" Score="12" ViewCount="0" Body="&lt;p&gt;The only Nash equilibrium is that everybody submits $1$.  To see this, consider any other set of strategies for all the players.  Suppose each player $i, 1 \le i \le n$ submits numbers with an expected value of $p_i$ and that at least one of the $p_i$ is greater than $1$.  Then the expected value of the mean of the entries is $\frac{p_1 + ... + p_n}{n}$, and there must be at least one player $j$ for which this mean is less than or equal to their number.  But then the expected value of $\frac{2}{3}$ of the mean must be strictly less than $p_j$, so player $j$ has an incentive to switch to a lower number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a funny example.  Although the Nash equilibrium suggests that playing $1$ should be an optimal strategy, in practice you will always lose when you do this.  The issue, I think, is that it only makes sense to adopt the Nash equilibrium strategy if you expect everybody to be playing optimally, but in practice most people will not think this game through enough to actually find the Nash equilibrium strategy.  I've played this game a few times, and what seems to be the case in practice is that to most players you can associate a number $n$ such that the number they pick is $\left( \frac{2}{3} \right)^n$ times the maximum.  I like to think of $n$ as being a very rough measure of how far the players think ahead, since one can think of $n$ as being the number of times they recurse through the following argument:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since the maximum you can play is $1000$, the maximum the average could be is $\frac{2}{3} \cdot 1000$, so I should never guess more than that.&lt;/li&gt;&#xA;&lt;li&gt;But since nobody should guess more than $\frac{2}{3} \cdot 1000$, the maximum the average could be is $\left( \frac{2}{3} \right)^2 1000$, so I should never guess more than that.&lt;/li&gt;&#xA;&lt;li&gt;But...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so forth.  If the people you're playing with are smart enough, $n$ will actually measure something much more complicated: each player will be trying to guess how far everyone else will think ahead, and then guessing how far everyone else will guess how far everyone else thinks ahead, and then... this phenomenon, in my opinion, is much more interesting than Nash equilibria; I think it is related to the idea of &lt;a href=&quot;http://en.wikipedia.org/wiki/Common_knowledge_(logic)&quot; rel=&quot;nofollow&quot;&gt;common knowledge&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The last time I played this game, the possible numbers were $0$ through $100$.  I and one other person submitted $100$ just to throw off the results.  The actual average ended up being something like $19$.  So here $n \approx 9.7$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here's a cute observation.  In general the existence of Nash equilibria can be proven using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nash_equilibrium#Alternate_proof_using_the_Brouwer_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Brouwer fixed point theorem&lt;/a&gt;.  The second argument I gave above establishes the existence (and uniqueness) of the Nash equilibrium using essentially the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach fixed point theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-03T00:34:24.143" LastActivityDate="2010-12-03T00:34:24.143" CommentCount="3" />
  <row Id="12845" PostTypeId="2" ParentId="12842" CreationDate="2010-12-03T00:26:04.687" Score="2" ViewCount="0" Body="&lt;p&gt;When solving problems like this the first thing to do is (almost) always to try and reduce the problem to a simpler problem; in this case, for example, pretending the number of allowed numbers is 10, and not 1000, is a good first thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also simplify the problem by allowing everyone to know the current mean value. That way you'll see that everybody has an incentive to submit lower and lower numbers - resulting in Qiaochu's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(as a &quot;by the way&quot;, I'll recommend G. Polya's book &quot;How to solve it&quot;, which talks about general problem solving techniques.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This problem reminds me of a &quot;paradox&quot; I read about yesterday:&#xA;&quot;A prisoner is awating death sentence, and is told he will be executed withing this week (it's Monday now), but on a day he's not expecting it. The prisoner reasons that then he can't be executed on Sunday (because if he lives then, that must be the day), and the same reasoning tells us that he will not be executed on Saturday either, and so on. The prisoner concludes that he wont be executed after all. (without expecting it, he is executed on Monday)&quot;&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-03T00:26:04.687" CommentCount="1" />
  <row Id="12846" PostTypeId="1" AcceptedAnswerId="12847" CreationDate="2010-12-03T01:20:36.677" Score="2" ViewCount="123" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a \equiv 12^{772} \pmod{71}$, when $0 \leq a &amp;lt; 71$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I am having troubles getting it started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{772}{71}$ is $10$ with a remainder of $62$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I do this question?&lt;/p&gt;&#xA;" OwnerUserId="4352" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T01:34:47.233" LastActivityDate="2010-12-03T18:47:18.977" Title="Using Fermat's little theorem" Tags="&lt;number-theory&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12847" PostTypeId="2" ParentId="12846" CreationDate="2010-12-03T01:24:50.010" Score="3" ViewCount="0" Body="&lt;p&gt;$12^{772}=12^{71\cdot 10+62}\equiv 12^{10+62}=12^{71+1}\equiv 12\cdot 12=144\equiv 2$ mod $71$. So $a=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12^{772}=12^{70\cdot 11+2}\equiv 12^2=144\equiv 2$ mod $71$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one uses the form $a^p\equiv a$ mod $p$ and the second uses the form $a^{p-1}\equiv 1$ mod $p$ if $gcd(a,p)=1$ where $p$ is prime.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-03T03:28:40.430" LastActivityDate="2010-12-03T03:28:40.430" />
  <row Id="12848" PostTypeId="1" AcceptedAnswerId="12857" CreationDate="2010-12-03T01:29:35.727" Score="4" ViewCount="157" Body="&lt;p&gt;Is there an infinite set of strings whose Kolmogorov complexities are computable?&lt;/p&gt;&#xA;" OwnerUserId="4353" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-03T02:58:30.117" LastActivityDate="2010-12-03T02:58:30.117" Title="Is there an infinite set of strings whose Kolmogorov complexities are computable?" Tags="&lt;computability&gt;&lt;kolmogorov-complexity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12849" PostTypeId="2" ParentId="12839" CreationDate="2010-12-03T01:50:00.883" Score="3" ViewCount="0" Body="&lt;p&gt;The image changes the problem and turns this into a classic 'long multiplication' cryptarithm problem.  For starters, we know (for the same reasons given above) that X=1, because multiplying X by XY gives XY again.  The second line tells us that Y*XY = YZ; in particular, since Y*X = Y, we know that Y*Y can't introduce a 'carry' digit to pollute the 1-by-2 multiplication.  This means that Y is either 2 or 3, leaving Z to be either 4 or 9 respectively (i.e., '2*12 = 24' or '3*13 = 39' - notice how 4*14 = 56, and the carry digit I mentioned before turns the tens digit into a 5, one more than the Y that it needs to be).  From here it's just trial and error, performing the rest of the calculation and checking to see whether either one gives duplicate digits.  For XY=12, we get XY*YX = 12*21 = 252 - but notice that this would be written as 'YVY', since we're told that each letter corresponds to a unique digit; we couldn't have W and Y both be 2.  However, with XY=13, we get XY*YX = 13*31 = 403, so ultimately we get the solution: X=1, Y=3, Z=9, W=4, V=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ETA: The multiplication can be confusing if you haven't seen it before - this is the old-style 'long multiplication' way of doing products, which I seem to recall isn't taught as much any more.  Each line between the two horizontal rules corresponds to the result of multiplying one of the lower digits by the entire upper number, and then those lines are all added together to produce the sum beneath the lower horizontal rule.)&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-12-03T01:50:00.883" />
  <row Id="12850" PostTypeId="1" AcceptedAnswerId="12854" CreationDate="2010-12-03T01:53:56.307" Score="3" ViewCount="56" Body="&lt;p&gt;Assume $f:(a,b) \to \mathbb R^3$ is an admissible unit speed curve (hence $f^{\prime} \times f^{\prime\prime}$ is never zero) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ lies on the sphere with center $a$ and radius $r$ prove that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f = a - (1/\kappa) \mathbf N - (1/\kappa)^{\prime} (1/\tau) \mathbf B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $|f^{\prime}| = 1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T = f^{\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N = f^{\prime\prime}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B = \mathbf T \times \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf T^{\prime} = \kappa \mathbf N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf B^{\prime} = -\tau\mathbf N$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been using the hint that since $f$ lies on the sphere then  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f-a) \cdot (f-a) = r^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And trying to differentiate it (three times) to get what I want but I'm not getting very far. I start with  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f-a) \cdot (f-a) = r^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\mathbf T \cdot (f-a) = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 + 2\kappa\mathbf N \cdot (f-a) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf N \cdot \mathbf T + (-\kappa\mathbf T + \tau\mathbf B + 2(\kappa^{\prime}/\kappa)\mathbf N) \cdot (f-a) = 0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I'm not sure where to go.&lt;/p&gt;&#xA;" OwnerUserId="3333" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T01:57:57.877" LastActivityDate="2010-12-03T02:33:01.223" Title="Prove using an admissible unit speed curve and a Frenet frame" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12851" PostTypeId="2" ParentId="12846" CreationDate="2010-12-03T02:00:38.827" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ X^N \equiv X\ \ \Rightarrow\ \ X^{A + B\ N + C\ N^2 + \cdots} \equiv\ X^{A + B + C + \cdots}\ (mod\ N)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ This is nothing but &quot;casting out orders&quot; - a cyclic group analog of &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting_out_nines&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;casting out nines&lt;/strong&gt;.&lt;/a&gt; Namely, $\rm\ X^{N-1} = 1\ \Rightarrow\ X^N = X\ $ so we can map $\rm\ N\to 1$ when calculating integral powers of $\rm\:X\:$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\rm\:X\:$ has order dividing $\:9\:,\:$ then it amounts precisely to casting out nines from its exponents.&#xA;For example, $\rm\ mod\ 27:\ (-2)^9 \equiv 1\ \Rightarrow\ (-2)^{721} \equiv -2\ \ $ since $\rm\ \ 721\equiv 7+2+1\equiv 1\ \:(mod\ 9)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T18:47:18.977" LastActivityDate="2010-12-03T18:47:18.977" />
  <row Id="12852" PostTypeId="1" AcceptedAnswerId="12853" CreationDate="2010-12-03T02:20:36.133" Score="1" ViewCount="75" Body="&lt;p&gt;Hi, I'm not very expert in math, but I have a curiosity that could help me many times ..&#xA;Is there any way to find a function, even just similar, from a set of values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;example:&#xA;I have some values (x, y):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{(200, 11), (72, 5), (36,3), (28,3), (18,2), (12,2), ...}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;what is the function that can return these values and all others who follow this trend on a graph, more easily and better possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you, hello!&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T02:28:18.387" LastActivityDate="2010-12-03T02:31:43.187" Title="Find a function from values" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12853" PostTypeId="2" ParentId="12852" CreationDate="2010-12-03T02:25:37.267" Score="3" ViewCount="0" Body="&lt;p&gt;Assuming those values came from a polynomial, what you want to do is &lt;em&gt;polynomial interpolation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, since you have six points, the degree of the polynomial is at most five, since six coefficients are needed to uniquely determine a quintic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Alpha is able to determine interpolating polynomials from given data, e.g. &lt;a href=&quot;http://tinyurl.com/35kldba&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T02:31:43.187" LastActivityDate="2010-12-03T02:31:43.187" CommentCount="9" />
  <row Id="12854" PostTypeId="2" ParentId="12850" CreationDate="2010-12-03T02:33:01.223" Score="4" ViewCount="0" Body="&lt;p&gt;For each $t\in(a,b)$, $\{T(t),N(t),B(t)\}$ forms an orthonormal basis for $\mathbb{R}^3$, so you can express $f(t)-a$ in terms of this basis as $f(t)-a=c_1(t)T(t)+c_2(t)N(t)+c_3(t)B(t)$, and the coefficients are obtained by taking the dot products with the corresponding vectors.  Thus the result of differentiating once shows that $c_1(t)\equiv 0$.  Solving for $(f-a)\cdot N$ after differentiating twice shows that $c_2(t)=-1/\kappa(t)$.  After differentiating three times, notice that $N$ and $T$ are orthogonal, and use what you already know from the previous steps to solve for $(f-a)\cdot B$ and obtain $c_3(t)$.   You will want to use the fact that $(1/\kappa)&amp;#39;=-\kappa&amp;#39;/\kappa^2$ to simplify the final coefficient to the given form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that it is not correct that $f&amp;#39;&amp;#39;=N$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-03T02:33:01.223" />
  <row Id="12855" PostTypeId="2" ParentId="12838" CreationDate="2010-12-03T02:46:16.250" Score="1" ViewCount="0" Body="&lt;p&gt;Show that if $t_0$ is in $(a,b)$, then for all $t\in(a,b)$, $c(t)-c(t_0)$ lies in the plane through the origin orthogonal to (the constant vector) $B$ by showing that the function $f(t)=(c(t)-c(t_0))\cdot B$ is identically zero. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-03T02:46:16.250" />
  <row Id="12856" PostTypeId="1" CreationDate="2010-12-03T02:53:26.077" Score="5" ViewCount="177" Body="&lt;p&gt;I was just reading a proof of the dimension theorem in Steven Roman's Advanced Linear Algebra. In addressing the cases of infinite bases, Roman proceeds to show that if $\mathcal{B}$ and $\mathcal{C}$ are bases of a space $V$, then $|\mathcal{B}|\leq |\mathcal{C}|$, working up to an application of the BSC-Theorem. Anyway, he uses the string&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\mathcal{B}|\leq\aleph_0|\mathcal{C}|=|\mathcal{C}|.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if it's an elementary question, but why does the equality follow? Here $\mathcal{C}$ is any infinite basis. Is it definition? I tried looking up multiplication of ordinals, but didn't find anything useful. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditorDisplayName="" LastEditDate="2010-12-03T02:58:49.797" LastActivityDate="2010-12-03T08:43:45.570" Title="Multiplying Cardinal Numbers" Tags="&lt;linear-algebra&gt;&lt;set-theory&gt;" AnswerCount="3" />
  <row Id="12857" PostTypeId="2" ParentId="12848" CreationDate="2010-12-03T02:54:37.263" Score="8" ViewCount="0" Body="&lt;p&gt;I think you are asking this: is there an infinite r.e. set of pairs $(\sigma,n)$ where $\sigma \in 2^{&amp;lt;\omega}$ is a string of Kolmogorov complexity $n$. The answer to that is no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a contradiction, assume such a list is r.e. - then there arbitrarily long strings in it, and thus strings of arbitrarily high Kolmogorov complexity. Define a function $P$ that takes input $\tau \in 2^{&amp;lt;\omega}$ and does the following. First, it effectively enumerates that list until it finds a pair $(\sigma, n)$ where $n &amp;gt; |\tau|$. Then it prints out $\sigma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumptions we have made ensure that $P$ is a total computable function.  Therefore, applying Kleene's recursion theorem to $P$ gives a program $e_0$ that, when run with no input, computes $P(e_0)$.  Thus the output of program $e_0$, run with no input, is a string of Kolmogorov complexity larger than $|e_0|$, which is impossible. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-03T02:54:37.263" CommentCount="3" />
  <row Id="12859" PostTypeId="2" ParentId="12856" CreationDate="2010-12-03T03:25:17.113" Score="5" ViewCount="0" Body="&lt;p&gt;No, it's not a definition. A theorem of Tarski says that if $A$ is an infinite set, then $|A| = |A \times A|$. An acessible proof of this can be found in Hungerford's &quot;Algebra&quot;. So using it and the Schröder-Bernstein theorem, it's easy to prove that if $A$ is an infinite set and $B$ is a non-empty set such that $|B| \le |A|$, then $|A||B| = |A|$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-03T03:38:06.880" LastActivityDate="2010-12-03T03:38:06.880" CommentCount="2" />
  <row Id="12860" PostTypeId="2" ParentId="12856" CreationDate="2010-12-03T03:25:54.303" Score="5" ViewCount="0" Body="&lt;p&gt;By hypothesis $\mathcal C$ is infinite so $\aleph_0|\mathcal{C}|= \max(\aleph_0,|\mathcal C|)  = |\mathcal{C}|$ by basic &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number&quot; rel=&quot;nofollow&quot;&gt;cardinal arithmetic.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-03T03:25:54.303" CommentCount="1" />
  <row Id="12861" PostTypeId="2" ParentId="12820" CreationDate="2010-12-03T03:31:10.913" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, this is true, at least for varieties over the complex numbers $\mathbb{C}$.  Indeed, a toric variety over an algebraically closed field is rational (i.e., birational to projective space).  In characteristic zero, rational connectedness is a birational invariant, so toric varieties are rationally connected.  Finally, any rationally connected variety is $\mathcal{O}$-acyclic, which is the name for the conclusion that you want.  See e.g. &lt;a href=&quot;http://www.aimath.org/WWN/qptsurface2/articles/html/14a/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for this last implication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion might well hold more generally; I am not an expert in these matters.  You may want to ask your question on MathOverflow if you are not satisfied with this answer.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-03T03:31:10.913" />
  <row Id="12862" PostTypeId="1" AcceptedAnswerId="12866" CreationDate="2010-12-03T04:53:31.777" Score="5" ViewCount="139" Body="&lt;p&gt;Suppose I have a function f that is infinitely differentiable on some interval I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I construct a Taylor series P for it, using some point a in I, does f(x) = P(x) for all x in I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused as to whether Taylor series approximate (or equal - I'm not about that either) functions on one point, or on an interval.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-03T06:19:46.927" LastActivityDate="2010-12-03T08:31:15.613" Title="On what interval does a Taylor series approximate (or equal?) its function?" Tags="&lt;calculus&gt;&lt;elementary&gt;&lt;taylor-expansion&gt;" AnswerCount="4" />
  <row Id="12863" PostTypeId="1" AcceptedAnswerId="12868" CreationDate="2010-12-03T04:54:44.397" Score="9" ViewCount="159" Body="&lt;p&gt;On an infinite grid of ideal one-ohm resistors, what's the equivalant resistance between two nodes a knights move away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgs.xkcd.com/comics/nerd_sniping.png&quot; alt=&quot;I first saw this problem on the Google Labs Aptitude Test.  A professor and I filled a blackboard without getting anywhere.  Have fun.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;(please fix the tags, I didn't really know where to put it)&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4360" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-03T05:13:24.073" LastActivityDate="2010-12-03T05:13:24.073" Title="What is the resistance between two points a knights move away on a infinite grid of 1-ohm resistors" Tags="&lt;physics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="12864" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T05:01:56.893" Score="1" ViewCount="0" Body="&lt;p&gt;The (truncated) Taylor expansion agrees with the function only at the point of expansion; otherwise, the approximation gradually becomes worse the further you get away from your point of expansion. How bad it gets depends on the behavior of the coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you're asking if a Taylor expansion and the function where it came from are the same: in most cases, yes, but there are exceptions like $\exp\left(-\frac1{x^2}\right)$ ...&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-03T05:01:56.893" CommentCount="4" />
  <row Id="12865" PostTypeId="2" ParentId="12815" CreationDate="2010-12-03T05:03:07.680" Score="5" ViewCount="0" Body="&lt;p&gt;One expects that the numbers $\zeta(2n+1)$ are algebraically independent of one another, and of $\pi$, and so one should think of them as ``new'' numbers; you can't expect any closed form expression in terms of powers of $\pi$, say.  Unfortunately, this conjecture seems very much out of reach at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explanation of why people believe this conjecture, one can see for example&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/30659/establishing-zeta3-as-a-definite-integral-and-its-computation/30698#30698&quot; rel=&quot;nofollow&quot;&gt;this answer&lt;/a&gt; to a related mathoverflow question.  &lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-03T05:03:07.680" CommentCount="1" />
  <row Id="12866" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T05:03:40.717" Score="11" ViewCount="0" Body="&lt;p&gt;Let $f$ be an infinitely differentiable function, and let $T(x) = \sum_{n=0}^{\infty} a_n x^n$ be its Taylor series (say at $x = 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The Taylor series of $f$ need not converge at any point other than $x = 0$.  Indeed, by a famous Theorem of Borel, for any sequence $\{a_n\}_{n=0}^{\infty}$ of real numbers whatsoever, there exists an infinitely differentiable function with Taylor series equal to $\sum_{n=0}^{\infty} a_n x_n$.  If the $a_n$ grow too rapidly -- e.g. $a_n = n!$ -- then there will be convergence only at $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Even if the Taylor series has a positive radius of convergence $R$, there is no guarantee that $f(x)$ and $T(x)$ must be equal on $(-R,R)$.  A function with this property is said to be &lt;strong&gt;analytic&lt;/strong&gt; at $x = 0$.  (It should be mentioned that most of the familiar functions one encounters in calculus are analytic.)  Perhaps the simplest example of a non-analytic function is $f(x) = e^{\frac{-1}{x^2}}$ (and $f(0) = 0$), for which $f$ and all of its derivatives vanish at $x = 0$, so $T(x) = 0$.  But $f(x)$ is clearly positive at any $x \neq 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Added&lt;/b&gt;: To show that a function is equal to its Taylor series in some interval $I$, one has to show, for each $x$ in $I$, that the remainder function $R_n(x) = f(x) - T_n(x) = f(x) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^n$ approaches $0$ as $n$ approaches infinity.  The most useful basic tool for this is Taylor's Formula for the Remainder.  The issue comes down to having a good understanding of the growth of the derivatives at $0$ as $n$ increases.  For instance, if there exists a fixed $M$ such that &#xA;$|f^{(n)}(a)| \leq M$ for all $a \in I$, then Taylor's Formula immediately implies that &#xA;$R_n(x) \rightarrow 0$ for all $x \in I$.  This is the case for instance for $f(x) = \cos x, \sin x, e^x$ (in the last case we need to assume that the interval $I$ is bounded, although we can take it to be bounded as large as we want, so the eventual conclusion will be that $e^x$ is equal to its Taylor series on the entire real line).  In general this can be a hard problem, for instance because the conclusion need not be true!  Books on advanced calculus / elementary real analysis will have some worked examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;More Added&lt;/b&gt;: Gosh, for a few minutes there I forgot about the thousand pages or so of lecture notes I have online!  In particular see &lt;a href=&quot;http://math.uga.edu/~pete/243functions2.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for some further details on applying Taylor's Formula with Remainder.  However this discussion is at a somewhat higher level than strictly necessary (it came from a second semester undergraduate real analysis course).  As it happens, starting in January I'll be teaching a sophomore-level course on sequences and series, so in the fullness of time I might have more detailed notes.  Anyway, I'm sure that a little googling will find plenty of better notes on this topic...(In fact if you flip towards the back of a good calculus book, you'll certainly find this material there.)&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-03T05:50:12.883" LastActivityDate="2010-12-03T05:50:12.883" CommentCount="4" />
  <row Id="12867" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T05:06:08.683" Score="2" ViewCount="0" Body="&lt;p&gt;No, in fact the Taylor series may not even converge to the function. Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \left\{&#xD;&#xA;&#xD;&#xA; \begin{array}{ll}&#xD;&#xA;  f(x)=e^{-1/x^2}  &amp;amp; \mbox{if } x \ne 0 \\&#xD;&#xA;  f(x)=0 &amp;amp; \mbox{if } x = 0&#xD;&#xA; \end{array}&#xD;&#xA;\right.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="2219" LastEditorDisplayName="" LastEditDate="2010-12-03T05:07:32.177" LastActivityDate="2010-12-03T05:07:32.177" />
  <row Id="12868" PostTypeId="2" ParentId="12863" CreationDate="2010-12-03T05:07:26.817" Score="8" ViewCount="0" Body="&lt;p&gt;After Google asked this in their aptitude test, this has become a famous problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a nice discussion and a more general solution here: &lt;a href=&quot;http://www.mathpages.com/home/kmath668/kmath668.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mathpages.com/home/kmath668/kmath668.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer for your case is $\displaystyle \dfrac{4}{\pi} - \dfrac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T05:07:26.817" CommentCount="2" />
  <row Id="12869" PostTypeId="2" ParentId="12809" CreationDate="2010-12-03T05:18:43.110" Score="0" ViewCount="0" Body="&lt;p&gt;looks like chebyshev inequality.&lt;/p&gt;&#xA;" OwnerUserId="4339" LastActivityDate="2010-12-03T05:18:43.110" CommentCount="3" />
  <row Id="12870" PostTypeId="1" CreationDate="2010-12-03T05:34:39.887" Score="0" ViewCount="69" Body="&lt;p&gt;I have been interested in learning formal math formally enough so that I could write a proof assistant using some simple parsing tools, and explain to someone else with little math knowledge how to write one. Naturally, it would be a very clunky and tedious assistant, but this is the standard I have set for myself for &quot;really understanding&quot; a formal system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to do this with a typed calculus, but you work with what you can get. The most formal explanations out there seem to be for the untyped lambda calculus. However, I haven't found a good formal explanation of how one detects/avoids problems with bound/unbound variables and problems with the same variable being bound in overlapping scopes. It seems like these kinds of problems should not just be avoided but somehow simply the result of an incorrect usage of the formal rules for constructing expressions. Does anyone know of a good way to understand these problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked a related question &lt;a href=&quot;http://math.stackexchange.com/questions/10181/resources-for-learning-formal-math&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3480" LastActivityDate="2010-12-03T21:00:18.173" Title="formal rules for avoiding bound/unbound variable problems in lambda calculus " Tags="&lt;logic&gt;&lt;reference-request&gt;&lt;formal-systems&gt;&lt;type-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12871" PostTypeId="1" CreationDate="2010-12-03T06:06:27.330" Score="3" ViewCount="104" Body="&lt;p&gt;The usual statement of the Lebesgue Decomposition Theorem says that given two $\sigma$-finite measures $\mu$ and $\nu$ on a measure space, we can decompose $\nu = \nu_1 + \nu_2$, where $\nu_1$ is absolutely continuous with respect to $\mu$ and $\nu_2$ and $\mu$ are mutually singular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Lebesgue Decomposition Theorem&quot;&gt;link text&lt;/a&gt;) says that there is a &quot;refinement&quot; of this result, where the singular part $\nu_2$ can be further decomposed into a discrete measure and a singular continuous measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand what a discrete measure is, but what exactly is the definition of a singular continuous measure? I was also wondering if anyone knew of a reference for this refined result, since I haven't been able to find it anywhere. &lt;/p&gt;&#xA;" OwnerUserId="1736" LastActivityDate="2010-12-03T08:45:52.027" Title="Lebesgue Decomposition Theorem" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12872" PostTypeId="1" AcceptedAnswerId="12880" CreationDate="2010-12-03T06:12:19.150" Score="13" ViewCount="385" Body="&lt;p&gt;I understand that $\pi$ and e are transcendental and that these are not simple facts. I mean, I have been told that these results are deep and difficult, and I am happy to believe them. I am curious what types of techniques are used and just how difficult of a problem it is. Would this result be a reasonable capstone to any course? (any course that isn't essentially &quot;how to prove...&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another part of this is the following observation: as time passes deep results become easier to understand or rather assimilate into ones body of knowledge and some problems are just hard. I am wondering if people feel like this result is something that a grad student could spend some leisure time and understand, or if it really is something only graspable by &quot;experts&quot; (meaning people in the appropriate field and not a general mathematical audience). How specialized are the techniques used for the problem at hand? Have they been used to prove different results? are the techniques drastically different for e and $\pi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this isn't too soft of a question. I was talking with my roommate, also a math grad student, and it came up. I said that it was a classically difficult result, but then wondered if that was right, so here I am.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I don't want a proof or a sketch of one, but maybe a heuristic as to why new techniques were needed or explaining the troubles one has when being naive or using early methods to attack the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Matt E pointed out I should also ask: what are the old techniques?&#xA;Also, did I even tag the question correctly? These seem to be the areas I would put them in, but I don't know the anything about this stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above it also isnt clear that I am wondering if the proof of this result is gotten by some clever new trick or by lots of hard had work that people couldn't/didn't do before? essentially, is it all elbow grease, or some clever new machinery or something completely different?&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-03T06:38:13.070" LastActivityDate="2010-12-23T04:50:52.643" Title="How hard is the proof of $\pi$ or e being transcendental?" Tags="&lt;analysis&gt;&lt;soft-question&gt;&lt;math-history&gt;&lt;analytic-number-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="4" />
  <row Id="12875" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T06:30:54.970" Score="6" ViewCount="0" Body="&lt;p&gt;As mentioned above, it can be easily proven that e is transcendental. One can then combine it with $e^{iπ}+1=0$ (Euler's equation). Lindemann (I think) proved that if $e^x+1=0$, then $x$ has to be transcendental. And since $i$π is transcendental, ($i$ is obviously algebraic), π has to be transcendental, too. So, this way the proof seems rather easy, but there are much more complicated ways to prove it (as far as I know), which I will leave for others to elaborate on.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-23T04:50:52.643" LastActivityDate="2010-12-23T04:50:52.643" CommentCount="10" />
  <row Id="12876" PostTypeId="1" AcceptedAnswerId="12877" CreationDate="2010-12-03T06:51:43.297" Score="2" ViewCount="74" Body="&lt;p&gt;Does anybody know where I can find a general form, in terms of n, of the sum $\left(\sum\limits_{i=1}^n a_i\right)^n$.  What I mean is, there appears to be some sort of pattern, if you look at $\left(\sum\limits_{i=1}^2 a_i\right)^2$ you get $a_1^2+2a_1a_2+a_2^2$ and if you let $n=3$ then you get $a_1^3+a_2^3+a_3^3+3a_1^2(a_2+a_3)+3a_2^2(a_1+a_3)+3a_3^2(a_2+a_3)+6a_1a_2a_3$.  I'm working with something right now where I'd like to subtract out the cubed terms and have a general formula left over for the remaining terms which I can write simply in terms of $n$.  Sorry about the lack of clarity, I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="3595" LastEditorDisplayName="" LastEditDate="2010-12-03T07:03:19.853" LastActivityDate="2010-12-03T07:03:19.853" Title="General form for $(\sum\limits_{i=1}^n a_i)^n$" Tags="&lt;sequences-and-series&gt;&lt;reference-request&gt;&lt;arithmetic&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12877" PostTypeId="2" ParentId="12876" CreationDate="2010-12-03T06:54:57.637" Score="6" ViewCount="0" Body="&lt;p&gt;It isn't really clear what you are asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this will help: &lt;a href=&quot;http://en.wikipedia.org/wiki/Multinomial_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Multinomial_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T06:54:57.637" CommentCount="2" />
  <row Id="12878" PostTypeId="2" ParentId="12870" CreationDate="2010-12-03T06:55:47.583" Score="1" ViewCount="0" Body="&lt;p&gt;A great reference for $\lambda$-calculus and formal logic in general, which treats the subject highly rigorously is Sorensen and Urzyczyn's &quot;Lectures on the Curry-Howard Isomorphism.&quot;  I'm doing a little reading on the subject myself, but not enough to really answer your question.&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-03T06:55:47.583" />
  <row Id="12879" PostTypeId="1" AcceptedAnswerId="12883" CreationDate="2010-12-03T07:06:16.723" Score="1" ViewCount="115" Body="&lt;p&gt;Hi, this step in the proof is confusing me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_1^\infty {\frac{4^{n}}{3^{n-1}}}\qquad \longrightarrow \qquad\sum_1^\infty 4\left(\frac43\right)^{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please explain how/why this happened?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cheers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gregg&lt;/p&gt;&#xA;" OwnerUserId="4362" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T14:10:27.563" LastActivityDate="2010-12-03T14:10:27.563" Title="sequence /series convergence of 2^2n 3^(1-n)" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12880" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T07:14:07.567" Score="13" ViewCount="0" Body="&lt;p&gt;This is an answer elaborating on my comment above, and largely addressing the&#xA;edits to the original question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there were any &quot;old techniques&quot; before Hermite proved&#xA;transcendence of $e$ in the early 1870s.  As far as I know, the only transcendental numbers known before then were Liouville's interesting but&#xA;somewhat artificial examples from early that century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subject of transcendence, and the related subject of Diophantine approximation (i.e. approximating irrational numbers, especially irrational algebraic numbers, by rationals), is relatively new.  Liouville proved the first results showing that it is not so easy to approximate an irrational algebraic number by rational ones, and used this to construct his transcendental numbers, which he could recognize as being transcendental because they are too well approximated by rational numbers.  (&quot;Too well&quot; and &quot;not so easy&quot; here refer to the following problem: if you try to approximate $\alpha$ by the rational number $p/q$, can you get within a distance of $O(q^{-n})$ for some given $n$ as you let the denominator $q$ get arbitrarily large. (The larger $n$ is, the smaller $q^{-n}$ is, and so the better the rate of approximation.) Liouville showed, using the pigeon hole principal more-or-less, that if $\alpha$ is algebraic of degree $d$ then you can't do better than $n = d$.)&#xA;But this left open the problem of showing that various given numbers (like $e$ and $\pi$) are transcendental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like, here is one way to think of the problem: if you want to show that $\alpha$ is transcendental, then&#xA;you want to show that $f(\alpha) \neq 0$ for any non-zero polynomial $f$&#xD;&#xA;with rational coefficients.   The difficulty is that there will be lots of polynomials with real coefficients that have $\alpha$ as a root, and any one of them can be approximated as closely as you like by an $f$ with rational coefficients, so we &lt;em&gt;can&lt;/em&gt; find (lots of!) $f$ with rational coefficients such that $f(\alpha)$ is as close to zero as we like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to find some way to pin down the difference between $f(\alpha)$ being zero and $f(\alpha)$ being very close to zero.  This is not so easy!&#xA;(For example, computationally, you can't tell the difference between $0$ and any real number that is smaller than your computational accuracy can recognize.)&#xA;And now one sees why Diophantine approximation ideas of the type mentioned above&#xA;are relevant.  They are related to quantifying how close we can make $f(\alpha)$ to zero while bounding the denominators of the rational numbers involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not coincidence that bounding the denominators is relevant: morally this is an attempt to pass from working over $\mathbb Q$ to working over $\mathbb Z$.&#xA;Why do we want to do this?  Well, as I already noted, it's pretty hard to tell the difference between $\mathbb Q$ and $\mathbb R$, since the former is dense in the latter, but we &lt;em&gt;can&lt;/em&gt; tell the difference between $\mathbb Z$ and $\mathbb R$, since the former is discrete: a non-zero integer is some definite positive distance (i.e. at least 1) away from $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding remarks are somewhat philosophical, and they reflect my (limited) experience of thinking about these kinds of questions.  If you look at the proofs in the link above, it may not be obvious that they are relevant, but I believe that they in fact do have some relevance: e.g. you will see that the arguments reduce to considering integer rather than rational polynomials, and that growth considerations play a key role.  Another thing you will see is that certain auxiliary polynomials enter the proof, and a key fact about them is that they have a high order of vanishing at their zeroes.  The appearance of auxiliary polynomials, often with a high order of vanishing, is ubiquitous in this theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more (somewhat cultural) remark:&#xA;Roth's theorem, for which he got the Fields medal, is the ultimate strengthening of Liouville's theorem: he shows that if $\alpha$ is irrational algebraic, then one can't do better than $O(q^{-2})$ in the problem of Diophantine approximation discussed above.  The proof involves (among other things) constructions with auxiliary polynomials.  So my impression is that Liouville, Hermite, and Lindemann (and there are probably other names that should be here) invented a new subject, namely Diophantine approximation and transcendence theory, whose modern methods are an outgrowth of the methods that they introduced.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  Reading the first few pages of Baker's book &lt;a href=&quot;http://books.google.com/books?id=SmsCqiQMvvgC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;Transcendental Number Theory&lt;/a&gt;&#xA;(Google Books link) might help.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T11:21:37.447" LastActivityDate="2010-12-03T11:21:37.447" CommentCount="2" />
  <row Id="12881" PostTypeId="1" AcceptedAnswerId="12884" CreationDate="2010-12-03T07:26:04.077" Score="3" ViewCount="61" Body="&lt;p&gt;I'm trying to figure out whether my hardware function for mapping operating system pages to DDR memory controllers is &lt;a href=&quot;http://en.wikipedia.org/wiki/Injective_function&quot; rel=&quot;nofollow&quot;&gt;injective&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x) = a x \pmod{b},\text{ where $x$ in $[0,b)$, $a$ and $b$ are co-prime}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hunch is that it is injective and I think it has to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chinese_remainder_theorem&quot; rel=&quot;nofollow&quot;&gt;Chinese Remainder Theorem&lt;/a&gt;, though it may be some other theorem.  Also, is &quot;$a,b$ are co-prime&quot; sufficient?  Or does $a$ also need to be greater than $b$?&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T15:53:46.107" LastActivityDate="2010-12-03T15:53:46.107" Title="When $a$ and $b$ are co-prime, is $F(x) = a  x \mod b$,   $x$ in $[0,b)$ , $a$ and $b$ co-prime, an injective function?" Tags="&lt;prime-numbers&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="12882" PostTypeId="2" ParentId="12881" CreationDate="2010-12-03T07:32:24.750" Score="3" ViewCount="0" Body="&lt;p&gt;It is injective, your hunch is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\displaystyle ax = ay \mod b$ then $\displaystyle a(x-y) = 0 \mod b$. Since $\displaystyle a$ and $\displaystyle b$ are co-prime, $\displaystyle x-y= 0 \mod b$ and hence $\displaystyle x = y$, as $\displaystyle x,y \in [0,b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not matter whether $\displaystyle a \gt b$ or not.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T07:32:24.750" />
  <row Id="12883" PostTypeId="2" ParentId="12879" CreationDate="2010-12-03T07:38:59.017" Score="6" ViewCount="0" Body="&lt;p&gt;The $n^\text{th}$ term was rewritten by pulling out a factor of $4$ from the numerator.  Maybe seeing a couple of extra steps will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{4^n}{3^{n-1}}=\frac{4\cdot4^{n-1}}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-03T07:38:59.017" CommunityOwnedDate="2010-12-03T07:38:59.017" CommentCount="4" />
  <row Id="12884" PostTypeId="2" ParentId="12881" CreationDate="2010-12-03T07:40:28.533" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, $\rm\ gcd(a,b)=1,\ \ b\:|\:ax\ \Rightarrow\ b\:|\:x\ $ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclid%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Euclid's Lemma&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said functionally: $\rm\ ax\equiv 0\ \Rightarrow\ x\equiv 0\ \ (mod\ b)\ $ so $\rm\ x\to ax\ $ is injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More intuitively, by &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout's identity&lt;/a&gt;, $\rm\ gcd(a,b)=1\ \Rightarrow\ a^{-1}\ $ exists $\rm\ (mod\ b)\:.\: $ Therefore one can multiply $\rm \ ax\equiv ay\ $ by $\rm\:a^{-1}\:$ to deduce $\rm\ x\equiv y\:$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T07:56:37.950" LastActivityDate="2010-12-03T07:56:37.950" />
  <row Id="12885" PostTypeId="1" AcceptedAnswerId="12886" CreationDate="2010-12-03T07:43:12.430" Score="3" ViewCount="88" Body="&lt;p&gt;I am not getting this theorem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&lt;strong&gt;Angle between chord AB and tangent at A is the same as subtended by segment AB at any point on the circumference.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to prove this theorem?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-03T12:59:55.380" Title="Angle between chord and tangent" Tags="&lt;geometry&gt;" AnswerCount="3" />
  <row Id="12886" PostTypeId="2" ParentId="12885" CreationDate="2010-12-03T07:50:38.160" Score="4" ViewCount="0" Body="&lt;p&gt;Use the fact that if $O$ is the center of circle, and $C$ is foot of the perpendicular from $O$ to $AB$, then the angle subtended at any point on the circumference is half $\angle{AOB}$ which is $\angle{AOC}$ and try showing that the angle between chord $AB$ and tangent at $A$ is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Eowxw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, you do have to be careful here, there are two different angles subtended by points on the circumference, corresponding to the two different arcs which $AB$ forms. In fact those two angles add up to $180^{\circ}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-03T08:03:33.340" LastActivityDate="2010-12-03T08:03:33.340" CommentCount="4" />
  <row Id="12887" PostTypeId="2" ParentId="12885" CreationDate="2010-12-03T07:54:33.553" Score="2" ViewCount="0" Body="&lt;p&gt;It is well known that the angle subtended by segment $AB$ at any point on the circumference is constant on either of the arcs $AB$. Since the sum of those angles is $\pi$, it suffices to prove this for the point $C$&amp;#160;that is opposite to point $A$. Then it is well known that the angle $ABC$ is a right angle and also the segment $CA$ meets the tangent at $A$ in a right angle. Therefore $CAB = \frac{\pi}{2} - BCA$ and your angle is then just $\frac{\pi}{2} - CAB = BCA$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-03T07:54:33.553" />
  <row Id="12888" PostTypeId="1" CreationDate="2010-12-03T08:13:24.933" Score="0" ViewCount="62" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/12848/is-there-an-infinite-set-of-strings-whose-kolmogorov-complexities-are-computable&quot;&gt;Is there an infinite set of strings whose Kolmogorov complexities are computable?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I &lt;a href=&quot;http://math.stackexchange.com/questions/12848/&quot;&gt;asked a similar question&lt;/a&gt; but it turned out to be ambiguous. I am attempting to clarify what I meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an infinite, recursively enumerable set of pairs $(\sigma,n)$ where $\sigma \in S$ is a string of Kolmogorov complexity $n$ where $S$ is some infinite set of strings each over the same finite alphabet for some finite alphabet?&lt;/p&gt;&#xA;" OwnerUserId="4353" LastActivityDate="2010-12-03T08:13:24.933" ClosedDate="2010-12-17T11:23:13.160" Title="Is Kolmogorov complexity recursively enumerable for some infinite set of strings?" Tags="&lt;computability&gt;&lt;kolmogorov-complexity&gt;" CommentCount="6" />
  <row Id="12889" PostTypeId="2" ParentId="12862" CreationDate="2010-12-03T08:31:15.613" Score="2" ViewCount="0" Body="&lt;p&gt;Since you speak about intervals (on the real line), perhaps it should also be mentioned that the &quot;natural habitat&quot; for power series is really the complex plane; computing a power series involves only +, -, *, /, and limits, which are well defined operations on complex numbers. And for so-called &quot;complex analytic&quot; (or &quot;holomorphic&quot;) functions, which includes most functions that you encounter in calculus, it is a fact that the Taylor series at any point in the complex plane is convergent (equal to the function) in a circle around that point. The size of this circle is such that it exactly reaches out to the nearest singularity of the function. (Circles in the complex plane are the counterparts to intervals on the real line in this context.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example is $f(x)=1/(1+x^2)$. If you just look at the graph of this function (for $x$ real) it looks perfectly nice, and there seems to be no reason why the Taylor series at $x=0$ only manages to converge to the function in the interval $(-1,1)$. But if you think of $x$ as a complex variable, it's clear that there are singularities (division by zero) at the points $x=\pm i$, which lie at distance one from the origin, and that explains why the Taylor series converges inside the circle with radius one. (Note that the intersection of this disk with the real axis is just the interval $(-1,1)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing goes for $g(x)=\arctan x$. It's derivative is $f(x)$ above, and where the derivative has singularitites, the function has too. So the Taylor series for $g(x)$ at $x=0$ is also convergent inside the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-03T08:31:15.613" CommentCount="1" />
  <row Id="12890" PostTypeId="2" ParentId="12856" CreationDate="2010-12-03T08:43:45.570" Score="3" ViewCount="0" Body="&lt;p&gt;Seeing how the essential question was answered, I want to stress something else in your post which needs to be pointed out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that cardinals (namely, &lt;a href=&quot;http://en.wikipedia.org/wiki/Aleph_number&quot; rel=&quot;nofollow&quot;&gt;Aleph numbers&lt;/a&gt;) are usually treated as ordinals, however the multiplications and addition of cardinals and ordinals are very different, and most of all - exponentiation is different as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ordinals $\alpha$ and $\beta$ we define the sum to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\alpha + 0 = \alpha$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\alpha + (\beta + 1) = (\alpha + \beta) + 1$ (where $+1$ is the successor ordinal)&lt;/li&gt;&#xA;&lt;li&gt;$\alpha + \beta$ for a limit ordinal $\beta$ is the limit of $\alpha+\gamma$ for $\gamma&amp;lt;\beta$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One can notice that it is usually non-commutative as $2+\omega = \sup\{2+n\colon n&amp;lt;\omega\} = \omega \neq \omega+2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_arithmetic#Multiplication&quot; rel=&quot;nofollow&quot;&gt;ordinal multiplication&lt;/a&gt; is defined in a similar way, as well &lt;a href=&quot;http://en.wikipedia.org/wiki/Ordinal_arithmetic#Exponentiation&quot; rel=&quot;nofollow&quot;&gt;exponentiation&lt;/a&gt;. (Namely a simple rule for zero, and successor and a limit for limits) and an interesting result is that $\omega^\omega$ is &lt;strong&gt;countable&lt;/strong&gt; when dealing with ordinal exponentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, if $\lambda$ and $\mu$ are cardinals then $\lambda + \mu = \mu + \lambda = \lambda \cdot \mu = \mu \cdot \lambda = \max \{\lambda, \mu\}$, and the exponentiation is defined as $\lambda^\mu = |\{f | f\colon\mu\to\lambda\}|$ - that is the cardinality of the collection of functions from $\mu$ into $\lambda$.&#xA;For further information and definitions you can see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number#Cardinal_arithmetic&quot; rel=&quot;nofollow&quot;&gt;this wikipedia link&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So once you were dealing with the cardinality of the basis you were looking for cardinal arithmetics and not ordinal arithmetics. Which are two different things.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-03T08:43:45.570" />
  <row Id="12891" PostTypeId="2" ParentId="12871" CreationDate="2010-12-03T08:45:52.027" Score="3" ViewCount="0" Body="&lt;p&gt;That decomposition is commonly encountered in probability theory. In a more general setting, suppose that $\rho$ is a $\sigma$-finite measure on $\mathcal{B}(\mathbb{R}^n)$. Then there is a unique decomposition $\rho = \rho_{ac} + \rho_d + \rho_{cs}$, such that: 1) $\rho_{ac}$ is absolutely continuous, that is, $\rho_{ac}$ is zero on sets of Lebesgue measure zero; 2) $\rho_d$ is discrete, that is, $\rho_d$ is zero on the complement of some countable set $C$;  3) $\rho_{cs}$ is continuous singular, that is, $\rho_{cs}$ is zero at every point $x \in \mathbb{R}^n$ (= continuous measure), and is zero on the complement of some set $B$ of Lebesgue measure zero (= singular measure).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-03T08:45:52.027" CommentCount="1" />
  <row Id="12892" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T10:02:46.547" Score="4" ViewCount="0" Body="&lt;p&gt;For the proofs, you would of course need to know the necessary notions from mathematical analysis, covered in a good course, for instance the equivalent of what is in W. Rudin's principles of mathematical analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are comfortable with the analytic notions, then proving the transcendence of $e$ is quite easy. It can be done using the series expansion for $e$. This is done in the introductory algebra book of I. N. Herstein, &quot;Topics in algebra&quot;. The transcendence of $pi$ is harder. But it is still within the grasp of a beginning graduate student, though the proof is usually not included in the curriculum. It is given for instance in the algebra book of N. Jacobson, &quot;Basic Algebra&quot;, Vol 1.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof sketch of the Lindemann-Weierstrass theorem is in fact given in &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to your question of how difficult the theorems are, the answer is that they are accessible to beginning graduate students, and undergraduates who are willing to put in the extra labor.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-03T10:02:46.547" CommentCount="2" />
  <row Id="12893" PostTypeId="2" ParentId="12872" CreationDate="2010-12-03T10:38:51.663" Score="6" ViewCount="0" Body="&lt;p&gt;This paper by Ivan Niven,&#xA;The transcendence of π. &#xA;&lt;em&gt;Amer. Math. Monthly&lt;/em&gt; 46, (1939). 469–471, is short and readable. His book &lt;em&gt;Irrational numbers&lt;/em&gt; contains this result and much more and is highly recommended.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-03T10:38:51.663" CommentCount="2" />
  <row Id="12894" PostTypeId="1" AcceptedAnswerId="12901" CreationDate="2010-12-03T11:21:39.693" Score="2" ViewCount="59" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in functional analysis, you encounter the terms 'adjoint' and 'formal adjoint'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does 'formal' in that case mean? It Sounds like a hint that 'formal adjoints' lack a certain property to make them a 'true' adjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have nowhere found a definition, and would be eager to know.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-12-03T12:39:51.327" Title="Distinction between 'adjoint' and 'formal adjoint'" Tags="&lt;terminology&gt;&lt;functional-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="12895" PostTypeId="1" AcceptedAnswerId="12899" CreationDate="2010-12-03T11:26:27.880" Score="9" ViewCount="197" Body="&lt;p&gt;And what about the other direction? If $(u_{n})$ is increasing, what about $(a_{n})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm &lt;em&gt;guessing&lt;/em&gt; the former is true since we know that if $(a_{n})$ converges, $(u_{n})$ converges to the same limit. That tells me that around infinity the sequences behave roughly the same. I tried proof by induction but got stuck.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-12-03T18:56:08.390" Title="If $(a_{n})$ is increasing, is $u_{n}=\frac{a_{1}+\cdots+a_{n}}{n}$ increasing as well?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="12896" PostTypeId="1" CreationDate="2010-12-03T11:51:18.423" Score="1" ViewCount="96" Body="&lt;p&gt;Guys is my logic correct? If Earth travels around the Sun in 29.78 km/s and I have to calculate how long will it take to travel 1km I did following:&lt;br&gt;&#xA;1s / 29.78 = 0.03358 (to 5 d.p)&lt;br&gt;&#xA;I think it's ok but I would like to get confirmation on this.&lt;br&gt;&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4366" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T12:01:16.990" LastActivityDate="2010-12-03T12:01:16.990" Title="Earth and Sun - fast question" Tags="&lt;arithmetic&gt;" CommentCount="4" />
  <row Id="12897" PostTypeId="1" AcceptedAnswerId="12898" CreationDate="2010-12-03T11:52:56.943" Score="8" ViewCount="269" Body="&lt;p&gt;How do i show that $f_{1}(x)=1$, $f_{2}(x)=e^{x}$ and $f_{3}(x)=\sin{x}$ are linearly independent, as elements of the vector space, of continuous functions $\mathcal{C}[0,1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for showing these elements are linearly independent, one needs to show that if $$ a_{1} \cdot 1 + a_{2} \cdot e^{x} + a_{3} \cdot \sin{x}=0$$ then from this we should conclude that $a_{1}=a_{2}=a_{3}=0$. But i am not being able to deduce this. &lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-12-03T14:37:52.323" LastActivityDate="2010-12-04T21:19:58.240" Title="Showing $1,e^{x}$ and $\sin{x}$ are linearly independent in $\mathcal{C}[0,1]$" Tags="&lt;linear-algebra&gt;&lt;analysis&gt;&lt;vector-space&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="12898" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T12:01:10.767" Score="14" ViewCount="0" Body="&lt;p&gt;If you differentiate your equation with respect to $x$ four times, you get that $a_2 e^x+a_3 \sin x = 0$, from which it follows that $a_1=0$. Now, letting $x=0$, you see that $a_2 e^x=0$, so $a_2=0$, and it follows readily that $a_3=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With problems like this, always remember that you are dealing with functions. That means that you can treat them as functions as well: you can differentiate them and set their values equal to something.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-03T12:01:10.767" CommentCount="1" />
  <row Id="12899" PostTypeId="2" ParentId="12895" CreationDate="2010-12-03T12:06:34.003" Score="16" ViewCount="0" Body="&lt;p&gt;For the first question, we have for increasing $a_n$:&#xA;\begin{align}&#xA;u_{n+1} - u_n &#xA;    &amp;amp;= \frac{a_1+\dotsb+a_{n+1}}{n+1} - \frac{a_1+\dotsb+a_n}{n} \\&#xA;    &amp;amp;= \frac{n(a_1+\dotsb+a_{n+1})}{n(n+1)} - \frac{(n+1)(a_1+\dotsb+a_n)}{n(n+1)} \\&#xA;    &amp;amp;= \frac{na_{n+1} - \sum_{k=1}^n a_k}{n(n+1)} \\&#xA;    &amp;amp;= \frac{a_{n+1}}{n+1} - \frac{u_n}{n+1} \\&#xA;    &amp;amp;\geq \frac{a_{n+1}}{n+1} - \frac{a_n}{n+1} \gt 0,&#xA;\end{align}&#xA;where we have used that $u_n \leq a_n$ (the mean is at most the largest of the elements we're averaging over).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding whether $u_n$ increasing $\implies a_n$ increasing, this seems to be false. For example we can take $a_1 = 0$, $a_2 = 1$, and for $n&amp;gt;2$: $a_n = \frac{1}{2}(u_{n-1}+a_{n-1})$. For $n \geq 2$ we have $a_{n+1} \lt a_n$ and $u_{n+1} \gt u_n$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-03T12:37:26.057" LastActivityDate="2010-12-03T12:37:26.057" CommentCount="3" />
  <row Id="12900" PostTypeId="2" ParentId="12894" CreationDate="2010-12-03T12:18:05.543" Score="1" ViewCount="0" Body="&lt;p&gt;Is it this thing you're looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_operator#Formal_adjoint_in_one_variable&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Differential_operator#Formal_adjoint_in_one_variable&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-03T12:18:05.543" />
  <row Id="12901" PostTypeId="2" ParentId="12894" CreationDate="2010-12-03T12:32:17.310" Score="7" ViewCount="0" Body="&lt;p&gt;Are you talking about differential operators on functions over a domain in $\mathbb{R}^d$? (This is the context in which the phrase &quot;formal adjoint&quot; usually comes up.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that working with, say, smooth functions with compact support, we have the integration by parts formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int Du\cdot v ~dx + \int u\cdot Dv~ dx = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a linear partial differential operator is defined as $P = \sum A_\alpha D^\alpha$ where $\alpha$ are multi-indices, you can write $P&amp;#39;$ as a linear partial differential operator $P&amp;#39;\phi = \sum (-1)^{|\alpha|} D^\alpha(A_\alpha \phi)$ and generalize the integration by parts formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int Pu \cdot v~dx = \int u \cdot P&amp;#39;v~ dx $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which looks, in &lt;em&gt;form&lt;/em&gt;, suspiciously like the adjoint with respect to the $L^2$ inner product. That is, writing $\langle,\rangle$ for the $L^2$ inner product of real valued functions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle Pu,v\rangle = \langle u, P&amp;#39;v\rangle $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that we call this a formal adjoint is because, technically, to take an adjoint (in the Hilbert space sense, there is also a different notion for Banach spaces) of an operator, you need to specify which Hilbert space you are working over. In the case of the formal adjoint, it is left unspecified: indeed, the formula only really hold for sufficiently smooth function decaying sufficiently fast at infinity, and not in general for arbitrary functions $u,v\in L^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general for differential operators, the operator itself will not be bounded on an $L^2$ Hilbert space, and so the operator is only densely defined on your Hilbert space. Therefore the adjoint can only be defined on another subset of the Hilbert space, the domain of the adjoint. (In the most general cases, the domain of the adjoint can be a much, much smaller set [even finite dimensional], so does not make much sense as an operator on the original Hilbert space. For differential operators, the adjoint is still densely defined using the density of $C^\infty_0$ in $L^2$.) (Note that also if the spactial domain has a boundary, the integration by parts formula picks up a boundary term in general, so you pick up a further problem with the notion of adjoints, related to the fact that $C^\infty_0(\Omega)$ is not dense in the Sobolev space $W^{1,2}(\Omega)$ when $\Omega$ has boundary.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the word &quot;formal&quot; is, I think, not mentioned explicitly, a lot of the problems that can arise when you deal with unbounded operators are discussed in chapter 8 of Reed-Simon, &quot;Methods of mathematical physics&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-03T12:39:51.327" LastActivityDate="2010-12-03T12:39:51.327" />
  <row Id="12902" PostTypeId="2" ParentId="12885" CreationDate="2010-12-03T12:54:24.840" Score="2" ViewCount="0" Body="&lt;p&gt;As a diagram of the other common method of proof of the Alternate Segment Theorem has not yet been posted, here it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9XvQV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AB is the diameter and so $ \angle BCA $ is a right angle. Hence $y = \pi/2 - \angle BAC = x,$&#xD;&#xA;and we just note that the angle subtended by $AC$ is constant in this segment.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-03T12:59:55.380" LastActivityDate="2010-12-03T12:59:55.380" />
  <row Id="12904" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T13:11:58.180" Score="6" ViewCount="0" Body="&lt;p&gt;You may calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wronskian&quot; rel=&quot;nofollow&quot;&gt;Wronskian&lt;/a&gt; of $f_1(x)$, $f_2(x)$, $f_3(x)$ and check that it doesn't vanish on $[0,1]$. This implies that the functions are linearly independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A caveat.&lt;/strong&gt; If the Wronskian of a set of functions is equal to zero identically, it doesn't&#xA;follow in general that the functions are linearly dependent.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-03T13:11:58.180" />
  <row Id="12905" PostTypeId="1" CreationDate="2010-12-03T13:25:56.560" Score="0" ViewCount="26" Body="&lt;p&gt;So I've got just the update step of the k-means algorithm running for a single cluster in a program.  I'm really only calculating a single centroid given a set number of observations.  So what does this make the running time?  Is it O(n)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aqT6I.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what if I add a single computation to the end of this update step.  Would this be added to the running time or would it be omitted as its not the leading order?  Resources to learn this would be cool too...&lt;/p&gt;&#xA;" OwnerUserId="3105" LastActivityDate="2010-12-03T13:25:56.560" Title="What is the time complexity of the update step of the K-Means Algorithm?" Tags="&lt;algorithms&gt;" CommentCount="1" />
  <row Id="12906" PostTypeId="1" AcceptedAnswerId="12907" CreationDate="2010-12-03T13:44:56.300" Score="28" ViewCount="1051" Body="&lt;p&gt;What is wrong with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/GU8wd.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.bestwtf.com/2010/11/explaining-why-pi-is-4.html&quot; rel=&quot;nofollow&quot;&gt;SOURCE&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-12-03T19:18:13.137" Title="Is value of $\pi$ = 4 ?" Tags="&lt;pi&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="14" />
  <row Id="12907" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T13:56:11.760" Score="25" ViewCount="0" Body="&lt;p&gt;This question is usually posed as the length of the diagonal of a unit square.  You start going from one corner to the opposite one following the perimeter and observe the length is 2, then take shorter and shorter stairsteps and the length is 2 but your path approaches the diagonal.  So $\sqrt{2}=2$  In both cases, you are approaching the area but not the path length.  You can make this more rigorous by breaking into increments and following the proof of the Riemann sum.  The difference in area between the two curves goes nicely to zero, but the difference in arc length stays constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  making the square more explicit.  Imagine dividing the diagonal into n segments and a stairstep approximation.  Each triangle is $(\frac{1}{n},\frac{1}{n},\frac{\sqrt{2}}{n})$.  So the area between the stairsteps and the diagonal is $n \frac{1}{n^2}$ which converges to 0.  The path length is $n \frac{2}{n}$, which converges even more nicely to 2.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-03T19:06:58.983" LastActivityDate="2010-12-03T19:06:58.983" CommentCount="7" />
  <row Id="12909" PostTypeId="1" AcceptedAnswerId="12911" CreationDate="2010-12-03T14:01:25.873" Score="4" ViewCount="100" Body="&lt;p&gt;I'm interested in the potential of such a technique.  I got the idea from Moron's answer to &lt;a href=&quot;http://math.stackexchange.com/questions/9402/9409#9409&quot;&gt;this question&lt;/a&gt;, which uses the technique of differentiation under the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'd like to consider this integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\pi}^\pi \cos{(y(1-e^{i\cdot n \cdot t}))}\mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to differentiate with respect to y.  This will give the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\pi}^\pi -(1-e^{i\cdot n \cdot t})(\sin{(y(1-e^{i\cdot n \cdot t}))}\mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...If I'm correct.  Anyways, I'm interested in obtaining the results to this second integral, using this technique.  So I'm wondering if solving the first integral can help give results for the second integral.  I'm thinking of setting $y=1$ in the second integral.  This should eliminate $y$ from the result, and give me the integral involving $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, I'm not sure I can use the technique of differentiation under the integral.  I want to know how I can apply this technique to the integrals above.  Any pointers are appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for what values of $y$ is this valid?&lt;/p&gt;&#xA;" OwnerUserId="2626" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T14:05:41.890" LastActivityDate="2010-12-03T14:14:49.530" Title="Will moving differentiation from inside, to outside an integral, change the result?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="12910" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T14:13:12.053" Score="3" ViewCount="0" Body="&lt;p&gt;Following up on Fredrik Meyer's post, you can just pick points to evaluate the functions at and show that the resulting equations require $a_1, a_2$, and $a_3$ are 0.  So at 0 you have $a_1+a_2=0$.  At 1 you have $a_1+a_2\exp (1) +a_3\sin (1)=0$.  At $\frac{\pi}{4}$ you have $a_1+a_2\exp(\frac{\pi}{4}) + a_3\frac{\sqrt{2}}{2}=0$.  These will show that $a_1, a_2$, and $a_3$ are 0.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-03T14:13:12.053" />
  <row Id="12911" PostTypeId="2" ParentId="12909" CreationDate="2010-12-03T14:14:49.530" Score="6" ViewCount="0" Body="&lt;p&gt;Wikipedia doesn't seem to have a precise statement of this theorem.  Here's a very general statement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Let $U$ be an open subset of $\mathbb{R}$ and let $E$ be a measure space (which you can freely take to be any open subset of $\mathbb{R}^n$ if you want).  Let $f : U \times E \to \mathbb{R}$ have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x \mapsto f(t, x)$ is integrable for all $t$,&lt;/li&gt;&#xA;&lt;li&gt;$t \mapsto f(t, x)$ is differentiable for all $x$,&lt;/li&gt;&#xA;&lt;li&gt;for some integrable function $g$, for all $x \in E$, and for all $t \in U$, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\left| \frac{\partial f}{\partial t}(t, x) \right| \le g(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the function $x \mapsto \frac{\partial f}{\partial t}(t, x)$ is integrable for all $t$.  Moreover, the function $F : U \to \mathbb{R}$ defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(t) = \int_E f(t, x) \mu(dx)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is differentiable, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F&amp;#39;(t) = \int_E \frac{\partial f}{\partial t}(t, x) \mu(dx).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the only condition that isn't easily satisfiable is the third one.  It is satisfied in this case, so you're fine (for all $y$).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-03T14:14:49.530" CommentCount="2" />
  <row Id="12912" PostTypeId="1" AcceptedAnswerId="12915" CreationDate="2010-12-03T14:20:02.480" Score="3" ViewCount="98" Body="&lt;p&gt;I run a website that generates &lt;a href=&quot;http://www.degreedays.net/&quot; rel=&quot;nofollow&quot;&gt;degree days&lt;/a&gt;, a specialist form of weather data used for calculations relating to building energy consumption.  Without going into unnecessary detail, temperature is a function of time, and degree days are essentially the integral of that function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment our site calculates the data using an approximation method based on daily average, maximum, and minimum temperatures.  But I'm working on improving that method by using finer-grained temperature measurements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These finer-grained temperature measurements are taken throughout each day, and the recording interval can be anything from 1 minute to several hours - it depends on the weather station making the recordings.  For most weather stations, the readings are fairly regular, but they're not completely regular.  A weather station might typically record the temperature every half hour, but there will often be extra readings or missing readings, or readings taken at less regular intervals for certain periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I've been using the trapezoidal method to numerically integrate the function of temperature against time.  It's working pretty well, but I'm wondering if I might be able to improve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mathematician, and my understanding of numerical integration is only very basic.  I understand that Simpson's 1/3 rule and Simpson's 3/8 rule typically work better than the trapezoidal rule when numerically integrating mathematical functions.  But real-world temperature readings don't follow an exact mathematical function.  Also I understand that Simpson's rules require equal intervals, which my temperature readings don't consistently have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it might make sense to use Simpson's rules to integrate stretches of temperature readings that have 2 or more consecutive time intervals of equal length, and use the trapezoidal rule for stretches of irregular readings.  But then I see &lt;a href=&quot;http://www.emis.de/journals/JIPAM/images/031_02_JIPAM/031_02.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (a paper that I don't pretend to understand properly) that the trapezoidal rule can often work better than Simpson's rule for various classes of &quot;rougher&quot; functions.  I would guess that outside air-temperature variation would be classed as pretty rough - the temperature jumps up and down throughout the day for all sorts of reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could probably come up with some way to estimate the effectiveness of various methods, but it's tricky because there's no &quot;right answer&quot; to compare figures against.  So I'm trying to figure out what method would make most sense from a theoretical standpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think the trapezoidal rule is likely to be the best approach for me?  Or are there other approaches that might make more sense?&lt;/p&gt;&#xA;" OwnerUserId="4367" LastEditorUserId="4367" LastEditorDisplayName="" LastEditDate="2010-12-03T14:28:14.690" LastActivityDate="2010-12-13T15:39:03.940" Title="Numerical Integration of Irregular Temperature Readings" Tags="&lt;integral&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="12913" PostTypeId="1" AcceptedAnswerId="12918" CreationDate="2010-12-03T14:38:43.360" Score="1" ViewCount="60" Body="&lt;p&gt;If I have matrices $\mathbf A$ and $\mathbf B$ then what is $\mathbf x$ to minimize $\|\mathbf A-\mathbf x\mathbf B\|_F$?&lt;/p&gt;&#xA;" OwnerUserId="4370" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T14:46:33.037" LastActivityDate="2010-12-03T15:35:06.470" Title="Minimize Frobenius norm" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="12914" PostTypeId="1" CreationDate="2010-12-03T14:39:38.547" Score="3" ViewCount="82" Body="&lt;p&gt;Let $n$ be composite. I'm trying to figure out if the set $H$ of $a$ such that&#xA;1) $a$ is relatively prime to $n$ and &#xA;2) the Miller-Rabin test fails to show compositeness of $n$ with $a$&#xD;&#xA;is a subgroup of the multiplicative group mod $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct is no: I am trying to show that the set is not closed. But then again we won't get any factors of $n$ by multiplication because of condition 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I've run some experiments and it seems that the number of strong liars relatively prime to n divides the order of the multiplicative group. I wanted to get a contradiction with Lagrange's theorem.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastEditorUserId="2298" LastEditorDisplayName="" LastEditDate="2010-12-03T18:57:51.663" LastActivityDate="2010-12-03T18:57:51.663" Title="Miller-Rabin Primality Testing failure and a subgroup" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" CommentCount="3" />
  <row Id="12915" PostTypeId="2" ParentId="12912" CreationDate="2010-12-03T14:44:29.250" Score="4" ViewCount="0" Body="&lt;p&gt;Since your temperature readings aren't equispaced, you can't directly apply Simpson's rule; the approach equivalent to this is to construct the parabola that interpolates three consecutive points (i.e., across two panels), and integrate that. The problem with this approach, of course, is that you need to have an odd number of data points (even number of panels) to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the trapezoidal rule, of course, across each panel, but a probably better idea might be to construct a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_Hermite_spline&quot; rel=&quot;nofollow&quot;&gt;cubic Hermite interpolant&lt;/a&gt; for each panel, and then integrate that. An obvious problem is that four conditions are needed to uniquely determine a cubic for each panel (two points and two derivative values), but estimates of derivative values can be constructed from the data such that the piecewise interpolant is &lt;em&gt;locally monotonic&lt;/em&gt;; briefly, an piecewise interpolant is locally monotonic if there are no spurious inflection points or extrema within a panel. One approach to estimating derivative values for monotonic interpolation, due to &lt;a href=&quot;http://dx.doi.org/10.1137/0717021&quot; rel=&quot;nofollow&quot;&gt;Fritsch and Carlson&lt;/a&gt;, is implemented in FORTRAN as &lt;a href=&quot;http://www.netlib.org/slatec/pchip/&quot; rel=&quot;nofollow&quot;&gt;the &lt;code&gt;pchip&lt;/code&gt; package&lt;/a&gt;, and in MATLAB as &lt;a href=&quot;http://www.mathworks.com/help/techdoc/ref/pchip.html&quot; rel=&quot;nofollow&quot;&gt;the function &lt;code&gt;pchip&lt;/code&gt;&lt;/a&gt;. A more modern approach, and one which may be better is some situations, is due to &lt;a href=&quot;http://adsabs.harvard.edu/full/1990A%26A...239..443S&quot; rel=&quot;nofollow&quot;&gt;Steffen&lt;/a&gt;. You may have to experiment which of the trapezoidal rule, Fritsch-Carlson, or Steffen would be best for integrating your data.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me detail the way one uses cubic Hermite interpolation for integrating data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each panel is bounded by two points, $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$. The trapezoidal rule consists of constructing the line joining these two points (&lt;em&gt;linear interpolation&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_i(x)=\frac{x_{i+1}-x}{x_{i+1}-x_i}y_i+\frac{x-x_i}{x_{i+1}-x_i}y_{i+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and integrating that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_i}^{x_{i+1}}f_i(x)\mathrm dx=\frac{x_{i+1}-x_i}{2}(y_i+y_{i+1})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating with cubic Hermite interpolation can be considered as a further &quot;improvement&quot; of the trapezoidal rule; briefly, in addition to points, one has derivative values $y_i^{\prime}$ and $y_{i+1}^{\prime}$, from which one constructs the cubic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_i(x)=y_i+y_i^{\prime}(x-x_i)+c_i(x-x_i)^2+d_i(x-x_i)^3$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle c_i=\frac{3\frac{y_{i+1}-y_i}{x_{i+1}-x_i}-2y_i^{\prime}-y_{i+1}^{\prime}}{x_{i+1}-x_i}$ and $\displaystyle d_i=\frac{y_i^{\prime}+y_{i+1}^{\prime}-2\frac{y_{i+1}-y_i}{x_{i+1}-x_i}}{(x_{i+1}-x_i)^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating $g_i(x)$ might look slightly complicated, however, there is a nice expression for the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_i}^{x_{i+1}}g_i(x)\mathrm dx=\frac{x_{i+1}-x_i}{6}\left(y_i+4g_i\left(\frac{x_{i+1}+x_i}{2}\right)+y_{i+1}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose verification I'll leave up to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll just add the note that in the case of equispaced data, integration with a piecewise cubic Hermite interpolant is equivalent to integration with the trapezoidal rule plus corrections at the beginning and end.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-13T15:39:03.940" LastActivityDate="2010-12-13T15:39:03.940" CommentCount="4" />
  <row Id="12916" PostTypeId="2" ParentId="12913" CreationDate="2010-12-03T15:28:11.300" Score="3" ViewCount="0" Body="&lt;p&gt;The problem is equivalent to minimizing $\|A - xB\|_F^2$, which is in turn:&#xA;$\min_{x} \sum_{i, j}(A_{ij}-xB_{ij})^2$&#xD;&#xA;This is a simple quadratic in $x$, whose minimizer is obtained by setting the derivative to 0.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-12-03T15:28:11.300" CommentCount="0" />
  <row Id="12917" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T15:32:35.210" Score="15" ViewCount="0" Body="&lt;p&gt;This problem illustrates the fact that two functions can be very close: $|f(x)-g(x)|&amp;lt;\epsilon$&#xD;&#xA;for all $x\in [0,1]$, but their derivatives can still be far apart, $|f&amp;#39;(x)-g&amp;#39;(x)|&amp;gt;c$ for some&#xA;constant $c&amp;gt;0$. &#xA;In our case, let $x=a(t),y=b(t),0\le t\le 1$ and $x=c(t),y=d(t), 0\le t\le 1$ be the &#xA;parametrizations of the two curves. By smoothing the corners, we may assume that both&#xA;are smooth. $$ \|(a(t),b(t))\|\approx \|(c(t),d(t))\|$$ does not imply&#xA;$$  \|(a&amp;#39;(t),b&amp;#39;(t))\|\approx \|(c&amp;#39;(t),d&amp;#39;(t))\|$$&#xD;&#xA;Therefore $\int_0^1 \|(a&amp;#39;(t),b&amp;#39;(t))\| dt$ need not be close to $\int_0^1 \|(c&amp;#39;(t),d&amp;#39;(t))\| dt.$&#xD;&#xA;Here $\|(x,y)\|$ denotes $\sqrt{x^2+y^2}$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-03T19:18:13.137" LastActivityDate="2010-12-03T19:18:13.137" CommentCount="3" />
  <row Id="12918" PostTypeId="2" ParentId="12913" CreationDate="2010-12-03T15:35:06.470" Score="3" ViewCount="0" Body="&lt;p&gt;The Frobenius norm is the regular vector Euclidean norm when $A$ and $B$ are viewed as $n^2$-dimensional vectors. So the $xB$ minimizing $||A - xB||_F$ would be the vector projection of $A$ on $B$, given by $(A \cdot {B \over ||B||_F}) B$, where $\cdot$ is the $n^2$ dimensional dot product. Hence $x$ is given by $A \cdot {B \over ||B||_F}$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-03T15:35:06.470" />
  <row Id="12919" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T15:40:10.990" Score="11" ViewCount="0" Body="&lt;p&gt;The pithy expression for this &quot;paradox&quot; is as follows: let $x_n(t)$ be a sequence of parameterized curves which converges uniformly to a limit curve $x(t)$.  Then it need not be the case that the arclengths of $x_n(t)$ approach the arclength of $x(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;b&gt;Added after seeing TCL's answer&lt;/b&gt;: it is also true that uniform convergence of a sequence of functions does not imply convergence of their derivatives.  See Section 3 &lt;a href=&quot;http://math.uga.edu/~pete/243functions1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for some discussion of this.  As TCL points out, since arclength elements are computed using derivatives, the observation about derivatives may be in some sense more fundamental.  In other words, I think I like TCL's answer better than mine.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross Millikan points out, this is more familiarly shown by approximating the hypotenuse of a right triangle by a staircase pattern of horizontal and vertical line segments.  I still remember being a senior in high school and having a friend (whom I had had no prior mathematical interactions with) show this to me.  I definitely remember thinking that it was not paradoxical but certainly surprising.  (And I have mathematically respected this person ever since, even though I haven't seen her since I was a teenager.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-03T15:48:09.107" LastActivityDate="2010-12-03T15:48:09.107" />
  <row Id="12920" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T15:57:41.907" Score="11" ViewCount="0" Body="&lt;p&gt;Hilarious! Of course, the circumference is &lt;em&gt;not&lt;/em&gt; approximated by the sum of lengths of the lines constructed as shown, but by the sum of the hypotenuses of each of the right-angle triangles formed around the edge of the circle (forming a polygon with vertices on the circle).&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-12-03T18:05:55.073" LastActivityDate="2010-12-03T18:05:55.073" />
  <row Id="12921" PostTypeId="1" CreationDate="2010-12-03T16:00:11.567" Score="-1" ViewCount="117" Body="&lt;p&gt;What is an R-trivial language? What is an R-trivial monoid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context: Formal languages. Afaik, R-trivial languages is a subset of the starfree languages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mostly have background in formal languages and automata theory but not so much with the syntactic monoid characterization. So it would be nice to give a basic definition with maybe a small example of such a language.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(In order to support multiple QA-sites because I don't want to have any QA-site stay behind and to have that question also represented there, I have also posted this question on these other sites: &lt;a href=&quot;http://stackoverflow.com/questions/4346608/formal-languages-what-does-r-trivial-mean&quot;&gt;stackoverflow.com&lt;/a&gt;, &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3503/formal-languages-what-does-r-trivial-mean&quot;&gt;cstheory.stackexchange.com&lt;/a&gt;, &lt;a href=&quot;http://mathoverflow.net/questions/48181/formal-languages-what-does-r-trivial-mean&quot; rel=&quot;nofollow&quot;&gt;mathoverflow.net&lt;/a&gt;. In general I am against cross-posting but in this case, as they all have the same goal to be a complete reference of questions in the specific area, having the question cross posted is the best thing you can do.)&lt;/p&gt;&#xA;" OwnerUserId="4371" LastEditorUserId="4371" LastEditorDisplayName="" LastEditDate="2010-12-03T20:09:38.077" LastActivityDate="2010-12-04T17:05:22.223" Title="formal languages: what does R-trivial mean?" Tags="&lt;monoid&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12922" PostTypeId="1" CreationDate="2010-12-03T16:18:53.813" Score="3" ViewCount="97" Body="&lt;p&gt;I've been attempting to find out more research and resources on generating random math problems based on performance and calculated difficulty. There's a few places that I'm just now stumbling upon, but I figured I'd ask the community what they recommend reading up on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.faqs.org/patents/app/20090017427&quot; rel=&quot;nofollow&quot;&gt;http://www.faqs.org/patents/app/20090017427&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an article here on an application based on research at MIT&#xA;&lt;a href=&quot;http://www.mathscore.com/math/mathProblems.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mathscore.com/math/mathProblems.html&lt;/a&gt;&#xA;I'm wondering where specifically that research resides or what I can learn from it and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to first understand how probability, priority queues and anything else plays a role (if at all) in generating varying math problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any resources and recommended reading would be most useful if you can provide that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="4372" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-04T05:12:51.430" LastActivityDate="2010-12-04T05:12:51.430" Title="A method for generating random math problems based on performance" Tags="&lt;probability-theory&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12924" PostTypeId="2" ParentId="12897" CreationDate="2010-12-03T16:23:05.820" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ y(0) = a+b,\ \ y&amp;#39;(0) = b+c,\ \ y&amp;#39;&amp;#39;(0) = b\ $ and they must all be $\:0\:$ if $\rm\ y(x) = 0 $&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-03T17:53:54.223" LastActivityDate="2010-12-03T17:53:54.223" />
  <row Id="12925" PostTypeId="1" AcceptedAnswerId="12980" CreationDate="2010-12-03T16:38:15.683" Score="2" ViewCount="159" Body="&lt;p&gt;There are 2 improper points: $0, 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that $\displaystyle \int_{\frac{1}{2}}^{1}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}\mathrm dx$ exists for $\alpha &amp;lt; 1$ (by using the limit comparison test, with $g(x)=\frac{1}{(1-x)^\alpha}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm having trouble picking $g(x)$ for the other interval. If $\alpha \lt 0$ then $\ln^\alpha(1+x) \underset{x\to 0}{\to} \infty$, and everything I try gives me $\frac{f(x)}{g(x)}\to 0$ or $\frac{f(x)}{g(x)}\to \infty$ which doesn't give me all possible solutions.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T11:48:51.367" LastActivityDate="2010-12-04T11:48:51.367" Title="For which values of $\alpha, \beta$, $\int_0^1 \frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}\mathrm dx$ exists?" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;improper-integrals&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="12926" PostTypeId="2" ParentId="12906" CreationDate="2010-12-03T16:38:51.283" Score="8" ViewCount="0" Body="&lt;p&gt;What is wrong with this? Fundamentally, that you have jumped in without a definition of the length of a arc.&lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-12-03T16:38:51.283" />
  <row Id="12927" PostTypeId="1" AcceptedAnswerId="12933" CreationDate="2010-12-03T16:43:54.613" Score="2" ViewCount="101" Body="&lt;p&gt;The problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Three poles standing at the points A,B and C subtend angles $\alpha$, $\beta$ and $\gamma$ respectively,at the circumcentre of the triangle ABC.If the heights of these poles are in arithmetic progression; then show that $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,what I could not understand is subtending of the angle part,precisely &lt;strong&gt;how a point subtend angle at another point?&lt;/strong&gt; So,what I am looking for a proper explanation of the problem statement with a figure,since it's troubling me from sometime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:I am not looking for the solution (as of now) or any hints regarding the solution,just a clear explanation will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My solution using &lt;a href=&quot;http://math.stackexchange.com/users/1102/moron&quot;&gt;Moron&lt;/a&gt;'s &lt;a href=&quot;http://math.stackexchange.com/questions/12927/help-to-understanding-the-problem/12933#12933&quot;&gt;interpretation&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a$ $b$ and $c$ are the length of three sides of the poles and $O$ be the circumcentre then, $$ \cot \alpha = \frac{OA}{a}$$ $$ \cot \beta = \frac{OB}{b}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \cot \gamma = \frac{OC}{c}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As O is the circumcentre,$OA = OB = OC = k $(say) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, $a$ $b$ and $c$ are in arithmetic progression,hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \cdot \frac{k}{\cot \alpha} = \frac{k}{\cot \beta} + \frac{k}{\cot \gamma}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canceling $k$ from both sides,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \cdot \frac{1}{\cot \alpha} = \frac{1}{\cot \beta} + \frac{1}{\cot \gamma}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression. (&lt;strong&gt;QED&lt;/strong&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-04T07:37:28.083" LastActivityDate="2010-12-04T07:37:28.083" Title="Help to understanding the problem" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="2" />
  <row Id="12928" PostTypeId="2" ParentId="12927" CreationDate="2010-12-03T16:47:35.803" Score="0" ViewCount="0" Body="&lt;p&gt;I would read it that the poles have a diameter&gt;0.  The pole at A has diameter so the angle seen from the circumcenter is $\alpha$.  See if that works.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-03T16:47:35.803" CommentCount="1" />
  <row Id="12929" PostTypeId="1" CreationDate="2010-12-03T16:52:44.297" Score="7" ViewCount="127" Body="&lt;p&gt;Let $\displaystyle AD$, $\displaystyle BE$, $\displaystyle CF$ be 3 cevians concurrent at $\displaystyle P$ inside the triangle $\displaystyle ABC$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \dfrac{AD}{AP} + \dfrac{BE}{BP} + \dfrac{CF}{CP} \ge \dfrac{9}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-04T07:35:51.297" LastActivityDate="2010-12-04T20:23:30.670" Title="An inequality on Cevians" Tags="&lt;geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12930" PostTypeId="2" ParentId="12675" CreationDate="2010-12-03T16:56:17.963" Score="0" ViewCount="0" Body="&lt;p&gt;It sounds like you are marking all the pizza joints in an area on Google maps, or something similar. More like a programming problem than a math problem. I have two notions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, look under the covers of a Hash Table for ideas. Not my field, but I believe they have ways to group a set in one pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, second, use a statistical sampling approach. Find the median and quartiles of the lat/long of a subset and make your decisions based on that. A little research plus some experimentation will give an idea about how big the subset has to be. &lt;/p&gt;&#xA;" OwnerUserId="1312" LastActivityDate="2010-12-03T16:56:17.963" />
  <row Id="12931" PostTypeId="1" CreationDate="2010-12-03T17:07:01.760" Score="2" ViewCount="63" Body="&lt;p&gt;One can write out an integral whose solution gives the number of solutions to the NP-Complete Number Partition Problem and I'm wondering if anyone has an suggestions or ideas on who to solve or approximate this integral analytically or numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $a_0, a_1, \dots, a_{n-1} \in \mathbb{Z}$, the Number Partition Problem asks for a partition of the $a_k$'s such that $\sum_{k=0}^{n-1} \sigma_k a_k = 0$, where $\sigma_k \in  (  -1,1 ) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following Integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I(a_0, a_1, \dots, a_{n-1}) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \prod_{k=0}^{n-1} (e^{ 2 \pi i a_k \theta } + e^{ -2 \pi i a_k \theta } ) d\theta $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(\dots)$ will count the number of solutions for a given instance of $(a_0, a_1, \dots, a_{n-1})$.  Solving this integral is worse than NP-Hard (it's #P) so asking for a general solution is out of the question.  But can one do any sort of approximation, either analytically or numerically?  If you choose the $a_k$'s with some distribution, say uniform on some interval, can you exploit that randomness to help you approximate this integral?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note:  This has been studied by Borgs et all for the &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.89.9577&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;NPP Phase Transition&lt;/a&gt; and that's where I first saw this integral representation of the Number Partition Problem, but their analysis relies on approximating the family of instances given a uniform distribution on the $a_k$'s rather than trying to solve a particular instance, as I'm trying to do above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can write out an integral whose solution gives the number of solutions to the NP-Complete Number Partition Problem and I'm wondering if anyone has an suggestions or ideas on who to solve or approximate this integral analytically or numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $a_0, a_1, \dots, a_{n-1} \in \mathbb{Z}$, the Number Partition Problem asks for a partition of the $a_k$'s such that $\sum_{k=0}^{n-1} \sigma_k a_k = 0$, where $\sigma_k \in  (  -1,1 ) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following Integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I(a_0, a_1, \dots, a_{n-1}) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \prod_{k=0}^{n-1} (e^{ 2 \pi i a_k \theta } + e^{ -2 \pi i a_k \theta } ) d\theta $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(\dots)$ will count the number of solutions for a given instance of $(a_0, a_1, \dots, a_{n-1})$.  Solving this integral is worse than NP-Hard (it's #P) so asking for a general solution is out of the question.  But can one do any sort of approximation, either analytically or numerically?  If you choose the $a_k$'s with some distribution, say uniform on some interval, can you exploit that randomness to help you approximate this integral?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been studied by Borgs et all for the &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.89.9577&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;NPP Phase Transition&lt;/a&gt; and that's where I first saw this integral representation of the Number Partition Problem, but their analysis relies on approximating the family of instances given a uniform distribution on the $a_k$'s rather than trying to solve a particular instance, as I'm trying to do above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;: This is a crosspost, and I have asked this on &lt;a href=&quot;http://mathoverflow.net/questions/38253/approximating-an-integral-representation-of-the-number-partition-problem&quot; rel=&quot;nofollow&quot;&gt;mathoverflow&lt;/a&gt;.  I have posted it here in hopes of getting more answers. &lt;/p&gt;&#xA;" OwnerUserId="4143" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T13:14:40.793" LastActivityDate="2010-12-04T13:14:40.793" Title="Approximating an integral representation of the Number Partition Problem" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;integral&gt;&lt;numerical-methods&gt;" CommentCount="4" />
  <row Id="12932" PostTypeId="1" AcceptedAnswerId="12934" CreationDate="2010-12-03T17:16:01.230" Score="1" ViewCount="41" Body="&lt;p&gt;This is a pretty simple question, I'm afraid I have been out of math-land for too long!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two currencies, the U.S. dollar and the rupee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$1 us dollar = .015 rupees&lt;/code&gt; (just an example, probably no where near the actual conversion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have $100,000$ rupees, how do I figure out the new value of my rupees if the value raises to be equal to the U.S. dollar ($1 U.S. dollar = 1 rupee)&lt;/p&gt;&#xA;" OwnerUserId="4374" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-03T18:00:58.833" LastActivityDate="2010-12-03T18:00:58.833" Title="(really) Simple percentage calculation with currency conversion" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12933" PostTypeId="2" ParentId="12927" CreationDate="2010-12-03T17:24:51.063" Score="1" ViewCount="0" Body="&lt;p&gt;My guess is the poles are of (different) heights $h_A$, $h_B$, $h_C$ and the angle is from the foot of pole to the circumcentre of $\triangle ABC$ to the top of the pole.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-03T17:24:51.063" CommentCount="11" />
  <row Id="12934" PostTypeId="2" ParentId="12932" CreationDate="2010-12-03T17:30:47.873" Score="2" ViewCount="0" Body="&lt;p&gt;Multiply the number of rupees by the value in dollars of one rupee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the problem is that your conversion &quot;goes the other way&quot; (that is, you have something like 1 US dollar = 0.75 rupees, then &quot;solve&quot; for how many dollars equals one rupee first:&#xA;\begin{align*}&#xA;1\text{ dollars} &amp;amp;= 0.75\text{ rupees}\\&#xA;\frac{1}{0.75} \text{ dollars} &amp;amp;= \frac{0.75}{0.75} \text{ rupees}\\&#xA;\frac{4}{3}\text{ dollars} &amp;amp;= 1\text{ rupees.}&#xA;\end{align*}&#xA;and then you can just multiply the number of rupees by $\frac{4}{3}$ to get the number of dollars.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-03T17:30:47.873" CommentCount="1" />
  <row Id="12937" PostTypeId="1" AcceptedAnswerId="12946" CreationDate="2010-12-03T18:47:31.257" Score="1" ViewCount="48" Body="&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R(t) = u + ct - \sum_{k=1}^{X(t)}Z_{k}, t\geq 0 $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be a risk process, where $u&amp;gt; 0$ is the initial capital of the insurance company and $c&amp;gt; 0$ is a premium rate. We know that the number of claims received by the insurance company is in accordance with a homogenous Poisson Process $X(t), t\geq 0$, with parameter (intensity) $\lambda &amp;gt; 0$. Assume that the amounts $Z_k, k = 1, 2, ...$ of successive claims are independent random variables with common probability density function $f_{Z_k}(z), z \geq 0$. Suppose that random variables $Z_{k}, k = 1, 2, ...$ are independent of $X(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would first like to be able to compute the ranges of $q, q \geq 0$ such that the function $f(z)=q\alpha \exp (-\alpha z) + (1 - q)\beta \exp (-\beta z); \beta &amp;gt; \alpha &amp;gt; 0,$ may be taken as $f_{Z_k}(z), z \geq 0$ in the risk process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I would like to be able to compute $E[R(t)]$, and $Var[R(t)]$ in terms of the parameters $u, c, \lambda, \alpha, \beta$ and $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help? I'm pretty stuck with this problem.&lt;/p&gt;&#xA;" OwnerUserId="2976" LastActivityDate="2010-12-03T21:30:51.537" Title="Risk process in Insurance" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;applications&gt;" AnswerCount="1" />
  <row Id="12938" PostTypeId="1" CreationDate="2010-12-03T18:51:18.770" Score="0" ViewCount="121" Body="&lt;p&gt;I have a question concerning Gödel's second incompleteness theorem.&lt;br&gt;&#xA;Let $\sharp \psi$ be the Gödel number of a sentence $\psi\in L_A$. $\mathcal{N}$ is the standard model of Peano arithmetic,$\mathcal{P}$.&#xA;Using Gödel's second incompleteness theorem,one can show that $\mathcal{N}\models \psi\rightarrow\exists bPf_\mathcal{P}(\underline{\sharp\psi},b)$ is not true. ($Pf_\mathcal{P}(a, b)$ means that $b$ is the Gödel number of proof from $\mathcal{P}$ of the $L_A$ formula with Gödel number $a$.)  My question is how can I understand $Con_\mathcal{P}$, in this case and more generally?&#xA;Thank you for your answer.&lt;/p&gt;&#xA;" OwnerUserId="4377" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T13:11:07.507" LastActivityDate="2010-12-04T13:11:07.507" Title="Gödel's second incompleteness theorem" Tags="&lt;logic&gt;" CommentCount="1" />
  <row Id="12939" PostTypeId="2" ParentId="12895" CreationDate="2010-12-03T18:56:08.390" Score="2" ViewCount="0" Body="&lt;p&gt;Here is a &quot;proof by physics&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average of $a_1, \dots, a_n$ is the center of mass of $n$ point masses, each with unit mass, having positions (along the $x$-axis) given by their values, i.e. $a_i$ has mass 1 and position $x = a_i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $(n+1)$ unit point masses, $a_1 \leq \dots \leq a_{n+1}$, let $u_n$ be the center of mass of the first $n$ points.  The center of mass $u_{n+1}$ can be computed by replacing the first $n$ points by a mass of $n$ with position $u_{n}$.  Since $a_{n+1} \geq a_n \geq u_n$, the addition of $a_{n+1}$ to the system shifts the center of mass to the right.  Hence, $u_{n+1} \geq u_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $n$ points of unit mass are distributed evenly along the $x$-axis, it is easy to see that the addition of a unit mass at a position to the right of $u_n$ but to the left of $a_n$ will shift the center of the mass to the right (even though the position of the additional mass is to the left of $a_n$).  Hence, the converse is false.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-03T18:56:08.390" />
  <row Id="12941" PostTypeId="1" AcceptedAnswerId="12942" CreationDate="2010-12-03T19:25:22.180" Score="4" ViewCount="104" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have two empty bags and 100 bread&#xA;  rings. The goal is to put all your&#xA;  rings into the bags in such a way that one&#xA;  of the bags will contain exactly&#xA;  twice as many rings then the other.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You can't eat, break and lose the rings.&#xA;  All of them should go into the bags.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me know if my explanation is not clear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I know the answer but I'd like to share it with you. IMHO, it is pretty nice.&lt;/p&gt;&#xA;" OwnerUserId="3761" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-03T19:33:04.053" LastActivityDate="2010-12-03T19:33:04.053" Title="Nice puzzle: 100 bread rings and two bags" Tags="&lt;puzzle&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12942" PostTypeId="2" ParentId="12941" CreationDate="2010-12-03T19:30:35.103" Score="7" ViewCount="0" Body="&lt;p&gt;Put half in one bag, and put half in the other bag. Then put one bag inside the other. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-03T19:30:35.103" CommentCount="4" />
  <row Id="12944" PostTypeId="1" AcceptedAnswerId="12988" CreationDate="2010-12-03T20:54:35.600" Score="2" ViewCount="202" Body="&lt;p&gt;what is the product of delta function with itself ? what is the dot product with itself ?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T15:15:40.067" LastActivityDate="2010-12-04T22:00:27.020" Title="what is product of delta function with itself ?" Tags="&lt;distribution-theory&gt;" AnswerCount="2" CommentCount="15" />
  <row Id="12945" PostTypeId="2" ParentId="12870" CreationDate="2010-12-03T21:00:18.173" Score="0" ViewCount="0" Body="&lt;p&gt;There are a number of ways of dealing with free/bound variables. The simplest (and least user friendly) is &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn_indices&quot; rel=&quot;nofollow&quot;&gt;de Bruijn Indices&lt;/a&gt;. In this system, a bound variable is named based on how &quot;far&quot; it is from the binder. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda x. \lambda y.~ x~y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda . \lambda .~ 1~0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in de Bruijn notation. This makes substitution of closed terms straight-forward. Free variables are represented by indices &quot;beyond&quot; the number of binders. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda x. x~a$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda. 0~1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is only one binder, $1$ is free (note that this is indexed from 0). For the full details, look at the wikipedia page linked above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great resource for this and other basic questions about lambda calculi is Benjamin Pierce's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0262162091&quot; rel=&quot;nofollow&quot;&gt;Types and Programming Languages&lt;/a&gt;. Very well written and readable. It's the closest we have to a &quot;standard&quot; Programming Languages book.&lt;/p&gt;&#xA;" OwnerUserId="4388" LastActivityDate="2010-12-03T21:00:18.173" />
  <row Id="12946" PostTypeId="2" ParentId="12937" CreationDate="2010-12-03T21:30:51.537" Score="1" ViewCount="0" Body="&lt;p&gt;I'll start with some suggestions that should work.  If you're still stuck after trying them, just comment under my answer, and I'll give more detail.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For the range of $q$, it sounds like you want to find the values of $q$ that make $f(z)$ a legitimate probability density function.  If so, you want to find the range of $q$ values for which $f(z) \geq 0$ and $\int_0^{\infty} f(z) dz = 1$.  That should be doable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $E[R(t)]$, use the fact that $E[R(t)] = E[E[R(t)|X(t)]]$.  Knowing the distribution of $X(t)$ should make this doable as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $Var[R(t)]$, try $Var[R(t)] = E[Var[R(t)|X(t)]] + Var[E[R(t)|X(t)]]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-03T21:30:51.537" />
  <row Id="12947" PostTypeId="1" CreationDate="2010-12-03T21:33:11.577" Score="0" ViewCount="55" Body="&lt;p&gt;Hello,&#xA;where can I find a proof that truncated octahedron tiles Euclidean 3D space?&lt;/p&gt;&#xA;" OwnerUserId="4389" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T23:59:12.943" LastActivityDate="2010-12-03T23:59:12.943" Title="Truncated octahedron tiles 3D space. Proof?" Tags="&lt;geometry&gt;&lt;tiling&gt;" CommentCount="1" />
  <row Id="12949" PostTypeId="1" AcceptedAnswerId="12956" CreationDate="2010-12-03T21:39:57.937" Score="0" ViewCount="139" Body="&lt;p&gt;Let $$F(x)=\sum_{n=1}^\infty \frac{(nx)}{n^2},$$ where $(x)=x$ for $x\in (-1/2,1/2]$ and $(x)$ is continued to $\mathbb{R}$ by periodicity, that is $(x+1)=(x)$. Prove that $F$ is discontinuous whenever $x=m/2n$, where $m,n\in \mathbb{Z}$ with $m$ odd and $n\neq 0$. This does not seem obvious to me.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-04T04:11:59.337" Title="An integrable function with many discontinuities" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12950" PostTypeId="2" ParentId="12768" CreationDate="2010-12-03T22:01:03.900" Score="4" ViewCount="0" Body="&lt;p&gt;sorry for the confusion, but you got it absolutely right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 times faster means two times as fast :) &#xA;A takes 100ms&#xA;B takes 200ms&#xA;so A is two times faster than B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular graph is probably the most confusing taken out of the flow of the previous ones in the chapter. This one shows an optimization that most often you shouldn't bother. Especially when there are cases of stuff being 100 times faster. And especially when it's not consistent among browsers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point was not to worry about micro-optimizations (unless it's critical to you and you've done all else). So 20% faster (or 1.2 times faster) is probably not worth it most of the time when there are other optimization that will make something 10 times faster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&#xA;Stoyan&lt;/p&gt;&#xA;" OwnerUserId="4390" LastActivityDate="2010-12-03T22:01:03.900" CommentCount="1" />
  <row Id="12951" PostTypeId="1" CreationDate="2010-12-03T22:09:42.717" Score="1" ViewCount="76" Body="&lt;p&gt;Any idea how to proof that when 3D space is tiled with truncated octahedra, all vertices can be colored black and white such that no 2 vertices sharing the same color are adjacent?&lt;/p&gt;&#xA;" OwnerUserId="4389" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T00:36:19.090" LastActivityDate="2010-12-04T00:36:19.090" Title="Truncated octahedron is bipartite. Proof?" Tags="&lt;geometry&gt;&lt;tiling&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12952" PostTypeId="1" CreationDate="2010-12-03T22:12:54.927" Score="0" ViewCount="43" Body="&lt;p&gt;I need to know what are these beasts, but I can't get my hands on Gray &amp;amp; Hervella paper :) Is there some other paper that describes them and is available for free on the Internet?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-03T23:44:28.627" LastActivityDate="2010-12-03T23:44:28.627" Title="Gray-Hervella classes" Tags="&lt;geometry&gt;&lt;reference-request&gt;" CommentCount="3" />
  <row Id="12953" PostTypeId="2" ParentId="7090" CreationDate="2010-12-03T22:45:09.197" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ a\:$ invertible $\rm\:(mod\ n)\ \ \Rightarrow\ \ a\ \{0,1,2,\ldots,n-1\}\ \equiv\ \{0,1,2,\ldots, n-1\}\ \ (mod\ n)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the map $\rm\ x \to a\:x\ $ is a bijection $\rm\ (mod\ n)\ $ so it simply permutes the congruence classes.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-03T22:45:09.197" />
  <row Id="12954" PostTypeId="2" ParentId="12951" CreationDate="2010-12-03T22:55:56.250" Score="1" ViewCount="0" Body="&lt;p&gt;The graph of a truncated octahedron is bipartite but there are space fillers with regular polygons as faces which are not bipartite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Space-FillingPolyhedron.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/Space-FillingPolyhedron.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-03T22:55:56.250" CommentCount="3" />
  <row Id="12956" PostTypeId="2" ParentId="12949" CreationDate="2010-12-03T23:41:34.770" Score="1" ViewCount="0" Body="&lt;p&gt;Note that $\lim_{x \rightarrow a}(x) = a$ except when $a$ is an odd multiple of ${1 \over 2}$, in which case $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 1$. Another way of saying this is that $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 0$ unless $a$ is an odd multiple of ${1 \over 2}$ in which case it is 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the term $f_n(x) = {(nx) \over n^2}$, $\lim_{x \rightarrow a^-}f_n(x) - \lim_{x \rightarrow a^+}f_n(x) = 0$ unless $x = {m \over 2n}$ for some odd integer $m$, in which case it is ${1 \over n^2}$. So for your arbitrary ${m \over 2n}$ with $m$ odd, it will incur a jump of ${1 \over n^2}$ due to $f_n(x)$, and other terms can only increase or keep the same the size of the jump. For any $\epsilon &amp;gt; 0$, if enough terms have been added, the absolute value of the sum of the remaining terms will be at most $\epsilon$, so that no matter how these remaining terms add up there will still be a jump of at least ${1 \over n^2} - \epsilon$ at ${m \over 2n}$. Letting $\epsilon \rightarrow 0$ this means the infinite sum has a jump of at least  ${1 \over n^2}$ at ${m \over 2n}$. (By jump I now mean the $\liminf$ from the left minus the $\limsup$ from the right.) In particular it's discontinuous there.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-04T01:43:42.233" LastActivityDate="2010-12-04T01:43:42.233" CommentCount="2" />
  <row Id="12957" PostTypeId="1" CreationDate="2010-12-04T00:13:39.370" Score="0" ViewCount="66" Body="&lt;p&gt;How do you find out how much a put contract would cost?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The premium for a 99.00 Eurodollar futures put option is 0.55 and the&#xA;  spot price is 98.51&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I have the intrinsic value which is &lt;code&gt;$1225&lt;/code&gt;, but I am not sure how to compute the time value and how much that put contract would cost. Still trying to understand options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3961" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T00:29:28.070" LastActivityDate="2010-12-04T07:27:52.770" Title="Stock/Futures put options " Tags="&lt;finance&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12958" PostTypeId="1" CreationDate="2010-12-04T00:17:33.350" Score="3" ViewCount="81" Body="&lt;p&gt;Both answers to this question seem equally reasonable to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is positive, I have no idea what the construction of such a space would look like....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is negative, I assume one would try to subdivide the cells somehow... but I don't really know how that would go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this came up because I was trying to think of an example of a CW-complex that wasn't homeomorphic to a Delta-complex... and figured the easiest way to make such a thing would be to make one that is not homotopic. This, of course, doesn't seem much easier to build, but at least easier to prove once you're done building.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-12-04T00:17:33.350" Title="Is every CW complex homotopic to a Delta-Complex?" Tags="&lt;topology&gt;&lt;algebraic-topology&gt;" CommentCount="2" />
  <row Id="12959" PostTypeId="1" AcceptedAnswerId="13005" CreationDate="2010-12-04T00:21:22.370" Score="6" ViewCount="115" Body="&lt;p&gt;Let $\{B(t), t \in \mathbb{R} \}$ be a two sided brownian motion defined as&#xA;$$&#xD;&#xA;B(t) = \begin{cases} B_1(t),\quad t &amp;gt;0 \\&#xD;&#xA;0, \quad t = 0 \\&#xD;&#xA;B_2(-t), \quad t &amp;lt; 0 \end{cases}&#xD;&#xA;$$&#xD;&#xA;where $B_1$ and $B_2$ are independent standard Brownian motions on $\mathbb{R}^+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $x_0 &amp;gt; 0$ and let $x_k = B(x_{k-1})$ for $k=1,2,\dots$.  What can we say about $\lim_{k\rightarrow \infty} x_k$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it converges, then I imagine the limit would have to be to $0$ a.s., since the limit $y$ would satisfy $B(y) = y$ a.s..  But I don't know how to show this sequence converges.  Any ideas? (this isn't homework, just a problem my friend and I thought up)&lt;/p&gt;&#xA;" OwnerUserId="2515" LastEditorUserId="2515" LastEditorDisplayName="" LastEditDate="2010-12-04T19:19:00.593" LastActivityDate="2010-12-04T19:19:00.593" Title="Question about Brownian motion" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12960" PostTypeId="1" AcceptedAnswerId="13017" CreationDate="2010-12-04T01:53:24.480" Score="12" ViewCount="100" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ZlY7gZacogsC&amp;amp;pg=PA172&quot; rel=&quot;nofollow&quot;&gt;Exercise 7.11 of  Kurzweil–Stellmacher&lt;/a&gt; asks me to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that the centralizer of an involution is nonsolvable inside any perfect group with generalized quaternion Sylow 2-subgroups of order at least 16.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to oblige, but I'm having trouble getting much of a handle on the centralizer itself, rather than just its action.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The center T=⟨t⟩ of a quaternion group is weakly closed, and so the normalizer X=N&lt;sub&gt;G&lt;/sub&gt;(T)=C&lt;sub&gt;G&lt;/sub&gt;(t) of T in that perfect group G controls G-fusion within any Sylow 2-subgroup P that happens to contain T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only fusion possible on P in a perfect group is the &quot;full&quot; fusion where all (both classes of) quaternion subgroups of order 8 are acted on by their full automorphism group (and otherwise no fusion beyond that induced by P itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know X contains at least two copies of SL(2,3) and X/Z contains at least two copies of S4.  Since G is perfect, P = [P,X] ≤ [X,X].  I think the copies of SL(2,3) pairwise intersect in a particular cyclic subgroup of P of order 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: If Q is a quaternion subgroup of P of order 8, then Y=N&lt;sub&gt;X&lt;/sub&gt;(Q) contains SL(2,3) and Y/Z contains S4.  [Y,Q] ≥ Q.  P is generated by its [T:Q] quaternion subgroups of order 8, so also by its [Y,Q]s.  The intersection of any pair of distinct Qs is the order 4 cyclic subgroup of the (unique) maximal subgroup that happens to be cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is a special configuration that has to fit lots of solvable groups inside, but I don't see any particular reason the result could not be solvable.  The closest thing to nonsolvable I get is that the derived length of N has to be at least 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get roughly the same results for a quaternion Sylow of order 8, except one only has the single SL(2,3).  I have not found a perfect G with solvable X in this case, but of course I can if I only require the same fusion pattern, since SL(2,3) and SL(2,5) have the same 2-fusion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More:&lt;/strong&gt; With a little reminder from reading the work of Bender's students, I can now do this using Brauer–Suzuki or Z*, but this does not seem much like a homework solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth seems to be X⋅O(G)=G, and every non-trivial quotient of a perfect group is perfect, so the homomorphic image G/O(G) of X is perfect.  Since G has Sylow 2-subgroups, G/O(G) is non-identity, so X has a non-identity perfect quotient, so X is nonsolvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to show X⋅O(G)=G without using tools significantly beyond chapter 7.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Smaller question:&lt;/strong&gt; I vaguely recall Suzuki proved that if G is a perfect group with no non-identity normal subgroups of odd order and with quaternion Sylow 2-subgroups, then each coset of C&lt;sub&gt;G&lt;/sub&gt;(t) contains exactly one conjugate of t, where t is an involution.  In other words, if the product of two involutions centralizes t, then that product is the identity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Can someone prove this with ideas from K–S up to chapter 7?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is hard for me to check since in reality such a group has exactly one involution t and G = C&lt;sub&gt;G&lt;/sub&gt;(t), but that comes later in my solution.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-04T18:22:12.760" LastActivityDate="2010-12-05T12:16:24.170" Title="Involution centralizer of perfect group with quaternion Sylow 2-subgroups" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="12961" PostTypeId="1" AcceptedAnswerId="12971" CreationDate="2010-12-04T04:06:03.290" Score="3" ViewCount="83" Body="&lt;p&gt;Got stuck on this one:&#xA;Show that there is a constant $C&amp;gt;0$ such that $$\left|\int_0^x \frac{\sin (N+1/2)t}{\sin t/2} dt \right|\le C$$ for all&#xA;$x\in [-\pi,\pi]$ and integer $N\ge 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought this should follow from $\int_0^\infty \frac{\sin t}{t}dt&amp;lt;\infty$, but somehow&#xA;don't get the connection.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-04T04:53:43.513" LastActivityDate="2010-12-04T05:58:23.280" Title="Boundedness of an integral" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12962" PostTypeId="2" ParentId="12949" CreationDate="2010-12-04T04:11:59.337" Score="1" ViewCount="0" Body="&lt;p&gt;A direct computation shows that &#xA;$$F(x+)=F(x)-\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)-\frac{\pi^2}{16n^2},$$&#xD;&#xA;$$F(x-)=F(x)+\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)+\frac{\pi^2}{16n^2},$$ &#xA;when $x=m/2n$, with $m$ and $2n$ being relatively prime. (The infinite sum is easily derived  from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Basel_problem&quot; rel=&quot;nofollow&quot;&gt;Euler result&lt;/a&gt;, of course.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The function $F(x)$ was suggested by Riemann in his thesis as an example of a (Riemann)-integrable function which has a point of discontinuity in every interval (see &#xA;&lt;a href=&quot;http://books.google.com/books?id=QdrCXs4L4v4C&amp;amp;printsec=frontcover&amp;amp;dq=Gesammelte+Mathematische+Werke+und+++Wissenschaftlicher+Nachlass&amp;amp;cd=2#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;&quot;Gesammelte Mathematische Werke und Wissenschaftlicher Nachlass&quot;&lt;/em&gt;&lt;/a&gt;, Springer -Verlag 1990, pp. 259-296).&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-04T04:11:59.337" CommentCount="2" />
  <row Id="12963" PostTypeId="2" ParentId="12961" CreationDate="2010-12-04T04:13:12.267" Score="0" ViewCount="0" Body="&lt;p&gt;If you expand the numerator you get a sum of two terms.  One doesn't depend on t, one is proportional to cot(t/2)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-04T04:13:12.267" CommentCount="4" />
  <row Id="12965" PostTypeId="1" AcceptedAnswerId="12972" CreationDate="2010-12-04T04:35:53.060" Score="1" ViewCount="96" Body="&lt;p&gt;I am trying to see why $\frac{\sin(nx)}{1+\vert x\vert^2}$ has no convergent subsequence in $L^1(\mathbb{R})$.  This is an &quot;optional&quot; homework problem from my disorganized analysis professor which is worth no points, and which I wouldn't do anyway, because I'm terrible at analysis and have no clue how to deal with this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems clear though that I want to be looking at the integral $\int\limits_\mathbb{R} \frac{\vert\sin(n_kx)-\sin(n_jx)\vert}{1+\vert(x)\vert^2}$ and finding some kind of constant, maybe in terms of when $n_k\neq n_j$ based on vague memories of undergrad PDE.  Any hints?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-04T06:05:19.033" Title="Convergence of $\frac{\sin(nx)}{1+\vert x\vert^2}$ " Tags="&lt;real-analysis&gt;&lt;functional-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12967" PostTypeId="2" ParentId="12922" CreationDate="2010-12-04T05:07:49.920" Score="3" ViewCount="0" Body="&lt;p&gt;The only stated relation to MIT at the second link is &quot;developed by MIT graduates&quot;. That or equivalent words appear in a few places on the web site, but I see no claim that the program or patents are related to research conducted at any university. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a page on adaptive testing: &lt;a href=&quot;http://en.wikipedia.org/wiki/Computerized_adaptive_testing&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Computerized_adaptive_testing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideas are similar to the statistical protocols used in medical trials, that allow for early stopping in case of unexpectedly good or bad results.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-04T05:07:49.920" CommentCount="1" />
  <row Id="12968" PostTypeId="1" AcceptedAnswerId="12969" CreationDate="2010-12-04T05:22:56.020" Score="13" ViewCount="302" Body="&lt;p&gt;I was wondering if there are examples of results in mathematics that were first proven using axiom of choice and later someone found a proof of the result without using the axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T06:12:11.653" LastActivityDate="2010-12-04T18:40:50.923" Title="Axiom of choice - to use or not to use" Tags="&lt;reference-request&gt;&lt;axiom-of-choice&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" />
  <row Id="12969" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T05:33:25.243" Score="14" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Cantor-Schroder-Berstein&lt;/a&gt; theorem was &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem#Original_proof&quot; rel=&quot;nofollow&quot;&gt;first proved&lt;/a&gt; by Cantor using the axiom of choice.  Proofs without using it came later. &lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2010-12-04T05:33:25.243" CommentCount="3" />
  <row Id="12970" PostTypeId="2" ParentId="12961" CreationDate="2010-12-04T05:46:30.377" Score="2" ViewCount="0" Body="&lt;p&gt;Consider: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin (Nt + \frac12 t) = \frac1{2i} \left[ \exp (i Nt + \frac{i}2 t) - \exp (-Nt - \frac{i}2 t)\right] = \sin \frac{t}2  \sum_{m = -N}^{N} \exp (imt) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your integrand reduces to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 + 2\sum_{m = 1}^N \cos(mt) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\int_0^x \cos(mt) dt = \frac{\sin (mx)}{m}$. To estimate $\sum_1^N \frac{\sin(mx)}{m}$ you can now use your fact that $\int_1^\infty \frac{\sin t}{t} dt &amp;lt; \infty$.  &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-04T05:46:30.377" CommentCount="2" />
  <row Id="12971" PostTypeId="2" ParentId="12961" CreationDate="2010-12-04T05:58:23.280" Score="3" ViewCount="0" Body="&lt;p&gt;This is a classic integral from summing Fourier series. I saw the following argument in Katznelson's book I believe. By symmetry we can assume $x &amp;gt; 0$. We divide the integral into $0$ to ${1 \over N}$ and ${1 \over N}$ to $x$ parts. (The second term will be $0$ if $x$ is small enough). Since $|\sin((N + {1 \over 2})t)| \leq (N + {1 \over 2})t$ and $\sin({t \over 2}) \geq C&amp;#39;t$ the integrand is bounded by $CN$ and the first term is at most $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second term, you can integrate by parts, integrating the $\sin((N + {1 \over 2})t)$ and differentiating the ${1 \over \sin({t \over 2})}$. You get a factor of ${1 \over N + {1 \over 2}}$ from the integration, and the resulting integrand is now bounded by $C{1 \over Nt^2}$. Taking absolute values and integrating from ${1 \over N}$ to $x$ again gives a bound of $C$. The left endpoint term in this integration by parts is bounded by $C{1 \over Nt}$ at $t = {1 \over N}$, so once again you just get a $C$. The right-endpoint term is even smaller. Thus you're done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-04T05:58:23.280" />
  <row Id="12972" PostTypeId="2" ParentId="12965" CreationDate="2010-12-04T06:05:19.033" Score="6" ViewCount="0" Body="&lt;p&gt;This is homework-like so I'll just give some hints. Show that if ${\sin(n_kx) \over 1 + x^2}$ converges in $L^1$ to some $f(x)$, then $\int_R{\sin(n_kx) \over 1 + x^2} g(x)$ converges to $\int_R f(x)g(x)$ for any bounded $g(x)$. Show this contradicts the Riemann-Lebesgue lemma in some way. Note you have to consider the case where $f(x) = 0$ too.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-04T06:05:19.033" CommentCount="1" />
  <row Id="12973" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T06:47:56.037" Score="6" ViewCount="0" Body="&lt;p&gt;This  article is an example of sorts where Doyle and Conway prove without using AC that for any two sets $A,B$ such that there is a bijection between the sets $3\times A$ and $3\times B$, where $3=\{0,1,2\}$, there is a bijection between $A$ and $B$. &lt;a href=&quot;http://www.math.dartmouth.edu/~doyle/docs/three/three.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.dartmouth.edu/~doyle/docs/three/three.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be relatively easy to prove using well ordering principle which is equivalent to axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-04T06:47:56.037" />
  <row Id="12974" PostTypeId="1" AcceptedAnswerId="12977" CreationDate="2010-12-04T06:50:37.313" Score="0" ViewCount="41" Body="&lt;p&gt;So, I'm playing a game where there is a certain percent value that changes over time.  I look at it occasionally, it seems to be approximately linear in time.  However, the change is slow enough that I want something better than simple linear regression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) := ax+b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) := \mathrm{round}(f(x))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given pairs $\{(x_1,g(x_1)),\dots,(x_n,g(x_n))\}$, estimate $a$ and $b$.&lt;/p&gt;&#xA;" OwnerUserId="3888" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T09:02:29.130" LastActivityDate="2010-12-04T09:02:29.130" Title="Estimating the equation of a line from its rounded values" Tags="&lt;approximation&gt;&lt;regression&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12975" PostTypeId="2" ParentId="12929" CreationDate="2010-12-04T07:06:42.020" Score="6" ViewCount="0" Body="&lt;p&gt;The inequality is true!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ffgYO.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that (see proof at the end of the answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF}  = 2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $\displaystyle 1 - \frac{PD}{AD} = \frac{AP}{AD}$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the inequality (easily shown using $\text{AM} \ge \text{GM}$) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle (\frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF})(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP} \ge \frac{9}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the equality occurs only when $\displaystyle \frac{AP}{AD} = \frac{BP}{BE} = \frac{CP}{CF} = \frac{2}{3}$, which implies that $\displaystyle P$ is the centroid.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us try showing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the figure (repeated from above for convenience).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ffgYO.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, if you are worried about acute triangle vs obtuse etc, a simple affine transformation will do to transform the triangle into an equilateral triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be the foot of perpendicular from A to BC and Y be the foot of the perpendicular from P to BC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \triangle AXD$ and $\triangle PYD$ are similar and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{PY}{AX} = \frac{PD}{AD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle \frac{PY}{AX} = \frac{|\triangle PBC|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle |\triangle MNO|$ is the area of $\displaystyle \triangle MNO$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle \frac{PD}{AD} = \frac{|\triangle PBC|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{PE}{BE} = \frac{|\triangle PAC|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{PF}{CF} = \frac{|\triangle PAB|}{|\triangle ABC|}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = \frac{|\triangle PAB| + |\triangle PAC| + |\triangle PBC|}{|\triangle ABC|} = \frac{|\triangle ABC|}{|\triangle ABC|} = 1$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-04T20:23:30.670" LastActivityDate="2010-12-04T20:23:30.670" CommentCount="3" />
  <row Id="12976" PostTypeId="2" ParentId="12957" CreationDate="2010-12-04T07:27:52.770" Score="1" ViewCount="0" Body="&lt;p&gt;Skipping your example, for a &lt;em&gt;European&lt;/em&gt; put option on a future contract you can apply the formula known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_model&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Black -76&lt;/em&gt;&lt;/a&gt;. For &lt;em&gt;American&lt;/em&gt; or &lt;em&gt;Bermudan&lt;/em&gt; options on futures this Black -76 is also adequate because if you exercise early than you would stand with the future which is delivered at expiry anyway - of course the difference is that it might be easier to sell the future..   &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-04T07:27:52.770" CommentCount="2" />
  <row Id="12977" PostTypeId="2" ParentId="12974" CreationDate="2010-12-04T07:58:04.087" Score="1" ViewCount="0" Body="&lt;p&gt;Given pairs $(x_i,y_i)$, you can try to find $a,b$ that minimize the $L^\infty$ distance from $y_i$ to $ax_i+b$ using LP, which is $\max_i |y_i-ax_i-b|$; if $y_i$ indeed depends linearly on $x_i$ up to rounding, then there's a choice of $a,b$ with $L^\infty$ distance of less than $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be a simpler way to solve this particular optimization problem, but an LP would work, especially with your small dataset. But if your model is wrong, you might get garbage because of the unforgiving $L^\infty$ metric. Minimizing the $L^1$ metric can also be formulated as an LP, and minimizing the $L^2$ metric is the well-known least-min-squares, for which there is an explicit solution.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-04T07:58:04.087" CommentCount="1" />
  <row Id="12980" PostTypeId="2" ParentId="12925" CreationDate="2010-12-04T09:49:44.480" Score="0" ViewCount="0" Body="&lt;p&gt;We can pick $g(x)=\frac{1}{x^(\beta-\alpha)}$ and then $\frac{f(x)}{g(x)}\to 1$ as $x\to 0$. By the LCT we get $\int_{0}^{\frac{1}{2}}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}$ exists iif $\beta-\alpha \lt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall $\int_{0}^{1}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}$ exists iif $\beta-1 \lt \alpha \lt 1$.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2010-12-04T09:49:44.480" />
  <row Id="12982" PostTypeId="2" ParentId="11665" CreationDate="2010-12-04T10:38:21.023" Score="3" ViewCount="0" Body="&lt;p&gt;Since $a_k \le k-1$ for $k \ge 2$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lfloor x \rfloor = \left\lfloor \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor \le a_1&#xD;&#xA;+ \left\lfloor \sum_{k=2}^n \frac{k-1}{k!} \right\rfloor = a_1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the latter term is $0$ since $ \sum_{k=2}^n \frac{k-1}{k!} = \sum_{k=2}^n \lbrace \frac{1}{(k-1)!} - \frac{1}{k!} \rbrace&#xD;&#xA;= 1 – 1/n! &amp;lt; 1.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show $ a_m = \lfloor m! x \rfloor – m \lfloor (m-1)! x \rfloor $ for $m=2,3,\ldots,n$ we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lfloor m! x \rfloor = \left\lfloor m! \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor = &#xD;&#xA;A_m + \left\lfloor m! \sum_{k=m+1}^n \frac{a_k}{k!} \right\rfloor, $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A_m$ is an integer given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_m = m! \frac{a_1}{1!} + m! \frac{a_2}{2!} + \cdots + m! \frac{a_m}{m!}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m! \sum_{k=m+1}^n \frac{a_k}{k!} \le m! \sum_{k=m+1}^n \frac{k-1}{k!}&#xD;&#xA;= m! \sum_{k=m+1}^n \left\lbrace \frac{1}{(k-1)!} - \frac{1}{k!} \right\rbrace $$&#xD;&#xA;$$ = m! \left( \frac{1}{m!} - \frac{1}{n!} \right) = 1 - \frac{m!}{n!} &amp;lt; 1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\lfloor m! \sum_{k=m+1}^n \frac{a_k}{k!} \right\rfloor = 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for $ m &amp;gt; 1 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m \lfloor (m-1)! x \rfloor = m \left\lfloor (m-1)! \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor $$&#xD;&#xA;$$=m \left\lbrace  (m-1)! \frac{a_1}{1!} + (m-1)! \frac{a_2}{2!} + \cdots + (m-1)! &#xD;&#xA;\frac{a_{m -1}}{(m-1)!} \right\rbrace + m \left\lfloor \sum_{k=m}^n \frac{a_k}{k!} \right\rfloor $$&#xD;&#xA;$$= \left( A_m - a_m \right) + m \left\lfloor \sum_{k=m}^n \frac{a_k}{k!} \right\rfloor&#xD;&#xA;=  A_m – a_m . $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k=m}^n \frac{a_k}{k!}  \le \sum_{k=m}^n \frac{k-1}{k!}&#xD;&#xA;= \frac{1}{(m-1)!} - \frac{1}{n!} &amp;lt; 1 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which proves that for $ m &amp;gt; 1 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lfloor m! x \rfloor -  m \lfloor (m-1)! x \rfloor = A_m – (A_m – a_m) = a_m . \qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that $n$ is the smallest integer such that $n! x $ is an integer, suppose that&#xA;$ (n-1)! x $  is an integer. Then by $(1)$ we have for $n &amp;gt; 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_{n-1} = (n-1)! x – n! x &amp;lt; 0 $$&#xD;&#xA;contradicting the fact that $a_k \ge 0 .$ And so $ (n-1)! x $ cannot be an integer, and if $ m! x $ is an integer for any $ m &amp;lt; n $ we have $(n-1)(n-2) \cdots m! x = (n-1)! x $ is an integer, another contradiction. Hence $ m! x $ cannot be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that every positive rational number $x$ can be expressed in this form, let $ x = p/q, $ where $ gcd(p,q)=1 \textrm{ and } p,q \in \mathbb{N} $ and let $ n $ be the smallest integer such that $ n! p/q $ is an integer. Define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align*}&#xD;&#xA;a_1 &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor&#xD;&#xA;\\  a_m &amp;amp;= \left\lfloor m! \frac{p}{q} \right\rfloor -  m \left\lfloor (m-1)!  \frac{p}{q} \right\rfloor&#xD;&#xA;\quad \textrm{ for } m &amp;gt; 1. \quad (2)&#xD;&#xA;\end{align*} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\lfloor (n-1)!  \frac{p}{q} \right\rfloor &amp;lt; (n-1)!  \frac{p}{q} , $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $ (n-1)! p/q $ is not an integer, and so $a_n &amp;gt; 0 .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since $ (m-1)! p/q $ is not an integer, for $ m \le n $ we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (m-1)!  \frac{p}{q} = N_m + r_m \quad \textrm{ where } 0 &amp;lt; r_m &amp;lt; 1 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $N_m$ is an integer. Hence for $m=2,3,\ldots,n$ from $(2)$ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_m = \lfloor mN_m + mr_m \rfloor - mN_m = \lfloor m r_m \rfloor \le m-1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the $a_m$ are non-negative. Now assume that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{p}{q} = \sum_{k=1}^n \frac{a_k}{k!} \quad (3)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then as the $a_k$ satisfy the conditions that each is a non-negative integer with $a_k \le k-1 $ for $k=2,3\ldots,n$ they are uniquely determined by $(1)$ and $a_1 = \lfloor p/q \rfloor .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It only remains to prove $(3).$ To show this we note that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align*} &#xD;&#xA;\sum_{k=1}^n \frac{a_k}{k!} &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor +  \sum_{k=2}^n \frac{a_k}{k!}&#xD;&#xA;\\ &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor +&#xD;&#xA;\sum_{k=2}^n \left\lbrace&#xD;&#xA;\frac{ \left\lfloor k! \frac{p}{q} \right\rfloor -  k \left\lfloor (k-1)!  \frac{p}{q} \right\rfloor }{k!}&#xD;&#xA;\right\rbrace&#xD;&#xA;\\ &amp;amp;= \left\lfloor \frac{p}{q} \right\rfloor + &#xD;&#xA;\sum_{k=2}^n \frac{ \left\lfloor k! \frac{p}{q} \right\rfloor }{k!} - &#xD;&#xA;\sum_{k=2}^n \frac{ \left\lfloor (k-1)! \frac{p}{q} \right\rfloor }{(k-1)!}&#xD;&#xA;\\ &amp;amp;= \frac {n! p/q}{n!} = \frac{p}{q},&#xD;&#xA;\end{align*} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $n! p/q$ is an integer and all the terms cancel, except the last. This completes the proof.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-04T15:34:05.293" LastActivityDate="2010-12-04T15:34:05.293" CommentCount="2" />
  <row Id="12983" PostTypeId="1" CreationDate="2010-12-04T11:43:06.430" Score="6" ViewCount="116" Body="&lt;p&gt;I'm studying statistics, and would like to better understand the Central Limit Theorem. The proof I found on Wikipedia requires some previous knowledge I do not currently possess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a quick intuitive explanation you can give as to why this theorem is correct?&lt;/p&gt;&#xA;" OwnerUserId="4401" LastActivityDate="2010-12-04T20:21:32.580" Title="Intuition about the Central Limit Theorem" Tags="&lt;statistics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="12984" PostTypeId="2" ParentId="8004" CreationDate="2010-12-04T12:09:59.680" Score="3" ViewCount="0" Body="&lt;p&gt;Your proof of (2) is incorrect. The quantifiers are wrong. &#xA;You have shown that for any $N$ (neighbourhood of $x$, in the remainder): $N$ intersects $E$ or $N$ intersects $F$.&#xA;But you have to show (for any $N$: $N$ intersects $E$) or (for any $N$, $N$ intersects $F$), i.e. that you intersect the &lt;strong&gt;same&lt;/strong&gt; set for any $N$. &#xA;So you have to argue by contradiction: suppose $x \in \overline{E \cup F}$, but $x \notin \overline{E}$ &lt;strong&gt;and&lt;/strong&gt; $x \notin \overline{F}$. The latter 2 imply that there is a neighbourhood $N_1$ of $x$ that misses $E$ and a neighbourhood $N_2$ of $x$ that misses $F$. But then $N_1 \cap N_2$ is a neighbourhood of $x$ as well and this misses $E \cup F$, contradicting that $x \in \overline{E \cup F}$. Done.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-04T12:09:59.680" CommentCount="1" />
  <row Id="12985" PostTypeId="2" ParentId="12983" CreationDate="2010-12-04T13:03:27.457" Score="6" ViewCount="0" Body="&lt;p&gt;I don't think you should expect any short, snappy answers because I think this is a very deep question.  Here is a guess at a conceptual explanation, which I can't quite flesh out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our starting point is something called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_maximum_entropy&quot; rel=&quot;nofollow&quot;&gt;principle of maximum entropy&lt;/a&gt;, which says that in any situation where you're trying to assign a probability distribution to some events, you should choose the distribution with maximum entropy which is consistent with your knowledge.  For example, if you don't know anything and there are $n$ events, then the maximum entropy distribution is the uniform one where each event occurs with probability $\frac{1}{n}$.  There are lots more examples in &lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/analysis/entropypost.pdf&quot; rel=&quot;nofollow&quot;&gt;this expository paper&lt;/a&gt; by Keith Conrad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a bunch of independent identically distributed random variables $X_i$ with mean $\mu$ and variance $\sigma^2$.  You know exactly what the mean of $\frac{X_1 + ... + X_n}{n}$ is; it's $\mu$ by linearity of expectation.  Variance is also linear, at least on independent variables (this is a probabilistic form of the Pythagorean theorem), hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Var}(X_1 + ... + X_n) = \text{Var}(X_1) + ... + \text{Var}(X_n) = n \sigma^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since variance scales quadratically, the variance of $\frac{X_1 + ... + X_n}{n}$ is actually $\frac{\sigma^2}{n}$; in other words, it goes to zero!  This is a simple way to convince yourself of the (weak) &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;law of large numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can convince ourselves that (under the assumptions of finite mean and variance) the average of a bunch of iid random variables tends to its mean.  If we want to study &lt;em&gt;how&lt;/em&gt; it tends to its mean, we need to instead consider $\frac{(X_1 - \mu) + ... + (X_n - \mu)}{\sqrt{n}}$, which has mean $0$ and variance $\sigma^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we suspected, for one reason or another, that this tended to some fixed limiting distribution in terms of $\sigma^2$ alone.  We might be led to this conclusion by seeing this behavior for several particular distributions, for example.  Given that, it follows that we don't know anything about this limiting distribution except its mean and variance.  So we should choose the distribution of maximum entropy with a fixed mean and variance.  And this is &lt;strong&gt;precisely&lt;/strong&gt; the corresponding normal distribution!  Intuitively, each iid random variable is like a particle moving randomly, and adding up the contributions of all of the random particles adds &quot;heat,&quot; or &quot;entropy,&quot; to your system.  (I think this is why the normal distribution shows up in the description of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_kernel&quot; rel=&quot;nofollow&quot;&gt;heat kernel&lt;/a&gt;, but don't quote me on this.)  In information-theoretic terms, the more iid random variables you sum, the less information you have about the result.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-04T14:20:18.843" LastActivityDate="2010-12-04T14:20:18.843" />
  <row Id="12986" PostTypeId="1" AcceptedAnswerId="12987" CreationDate="2010-12-04T14:11:34.690" Score="6" ViewCount="168" Body="&lt;p&gt;Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges. To make it more challenging, produce examples where $a_n$ and $b_n$ are positive and decreasing.&lt;/p&gt;&#xA;" OwnerUserId="4403" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-04T14:19:51.253" LastActivityDate="2010-12-04T14:43:48.013" Title="find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12987" PostTypeId="2" ParentId="12986" CreationDate="2010-12-04T14:29:45.943" Score="12" ViewCount="0" Body="&lt;p&gt;Here is a hint for the first part: you can make min$(a_n,b_n)=0$ for all $n$ (it doesn't get more convergent than that), while the two series $\sum a_n$ and $\sum b_n$ are very far from converging. Once you solve this one, let me know and I will give you a hint for the challenging part. It will build on ideas from the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Nice solutions to the first part! Now for the second: of course the idea will have to be the same: the sequences will have to somehow alternate in which one is lower at any given point. But think about it: if they alternate at each step, like before, then we will have that $\sum_n a_{2n}$ converges and also $a_{2n+1} &amp;lt; a_{2n}$, so we will have that $\sum_n a_n &amp;lt; 2\sum_n a_{2n}$ will converge. That's not good. If you think about this issue for a while, you will realise that the intervals at which one sequence dives below the other one must get longer and longer for the two to diverge. Now, can you make your previous idea work?&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-04T14:43:48.013" LastActivityDate="2010-12-04T14:43:48.013" CommentCount="4" />
  <row Id="12988" PostTypeId="2" ParentId="12944" CreationDate="2010-12-04T14:34:34.800" Score="6" ViewCount="0" Body="&lt;p&gt;A distribution is actually a linear functional on the space of compactly supported infinitely differentiable functions (the so called &quot;test functions&quot;). A function $f$ is compactly supported if $\overline{\{x : f(x) \neq 0\}}$ is compact (the overline denotes the closure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\delta$-distribution is a linear functional such that for all $\phi \in C_c^\infty(\mathbb{R}^n)$ we have that $\langle \delta, \phi \rangle = \phi(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to compute the product of distributions the problem is that you don't have a property which you would really like to have, that is associativity. So for distributions $\alpha$, $\beta$ and $\gamma$ we usually have that $(\alpha \cdot \beta) \cdot \gamma \neq \alpha \cdot (\beta \cdot \gamma)$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Distribution_%28mathematics%29%23Problem_of_multiplication&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives an example. However this does not really turn out to be a problem in applications. What we do have is convolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we want to do convolution we prefer a smaller class of distributions (for example because on the smaller class the Fourier transform of a distribution in this class is again a distribution in this class). This actually has a rougher class of test-functions, as test functions here we take the Schwartz functions, that are the smooth functions of which the function itself and all its derivatives are rapidly decreasing. $f$ is said to be rapidly decreasing if there are constants $M_n$ such that $|f(x)| \leq M_N |x|^{-N}$ as $x \to \infty$ for $N = 1,2,3,\ldots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin defining the convolution we first define what the convolution of Schwartz-function is with a tempered distribution. Let $f$ be our tempered distribution, then we can show that the following definition actually makes sense:&#xA;$$\langle \phi * f, \psi \rangle := \langle f, \tilde{\phi} * \psi \rangle$$&#xD;&#xA;where $\tilde{\phi}(x) = \phi(-x)$. Note that the RHS is well-defined. Convolution is a nice thing, we can see that if we start with a tempered distribution and convolute it with a test function, the result will be smooth.&#xA;Now, $L_1 * L_2$ is the unique distribution $L$ with the property that $L * \phi = L_1 * (L_2 * \phi)$. We can show that this is commutative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine, now note that $\delta * \phi(x) = \phi(x - y)|_{y = 0} = \phi(x)$. So we see that $\delta * \delta = \delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want me to comment on the dot product of distributions, you first would have to explain what you mean with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far for this short digression on distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Okay, you want to compute $\delta^2$. Let $\phi_n$ be an approximation to the identity and let it converge to $\delta$ in the sense of distributions, but $\phi_n^2$ does not converge at all since the integral against a test function that does not vanish at the origin blows up as $n \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-04T15:53:08.383" LastActivityDate="2010-12-04T15:53:08.383" />
  <row Id="12989" PostTypeId="1" AcceptedAnswerId="12990" CreationDate="2010-12-04T14:48:57.280" Score="5" ViewCount="78" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering how much a smooth function may be non-analytic, because in proofs, whilst there non-analytic smooth functions, it would suffice if a smooth function were analytic on only a &quot;small set&quot;. More exactly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U \in \mathbb R^n$ be open, $C^\infty = C^\infty(U,\mathbb R)$. A smooth function in $C^\infty$ is analytic in $a \in U$, iff there exists $\epsilon &amp;gt; 0$, s.t. the function is equal to its own Taylor series in $B_\epsilon(a)$. There exist smooth functions that are non-analytic, i.e. there exists $f \in C^\infty, b \in U, \epsilon &amp;gt; 0$ s.t. the function is not its taylor series at $x$ in $B_\epsilon (b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be the union of all $\epsilon$-Balls in $U$ where $f$ is analytic. By definition, $A$ is open. It's complement $C = A^c$is the closed set of points where $f$ is non-analytic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $C$ have an interior?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-12-04T15:09:18.907" Title="A smooth function's domain of being non-analytic" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12990" PostTypeId="2" ParentId="12989" CreationDate="2010-12-04T15:09:18.907" Score="10" ViewCount="0" Body="&lt;p&gt;Yes, that can happen. The canonical example (as far as I know) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fabius_function&quot; rel=&quot;nofollow&quot;&gt;Fabius function&lt;/a&gt; which is smooth and &lt;em&gt;nowhere&lt;/em&gt; analytic.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-04T15:09:18.907" />
  <row Id="12991" PostTypeId="1" CreationDate="2010-12-04T15:28:27.717" Score="1" ViewCount="108" Body="&lt;p&gt;Is Robinson Arithmetic complete in the sense of Gödels completeness theorem? And is Robinson Arithmetic incomplete in the sense of Gödels first incompleteness theorem?&#xA;If RA is both it would be a good example to explain the different notions of &quot;completeness&quot; used in the two theorems mentioned. &lt;/p&gt;&#xA;" OwnerUserId="4405" LastActivityDate="2010-12-05T11:33:39.327" Title="Is Robinson Arithmetic complete and not-complete?" Tags="&lt;logic&gt;&lt;incompleteness&gt;" AnswerCount="1" />
  <row Id="12992" PostTypeId="1" AcceptedAnswerId="12993" CreationDate="2010-12-04T15:52:01.147" Score="1" ViewCount="141" Body="&lt;p&gt;If I have 16 different symbols, howmany of these symbols can I combine to create no more than 16.777.216 unique combiations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg. I have all symbols of the Hexadecimal language (16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howmany combinations can I create with X amount of symbols to not have more than 16.777.216 unique combinations maximum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of each symbol is of importance.&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-04T15:54:18.193" LastActivityDate="2010-12-04T15:55:20.860" Title="If I have 16 different symbols, howmany of these symbols can I combine to create no more than 16.777.216 unique combiations?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12993" PostTypeId="2" ParentId="12992" CreationDate="2010-12-04T15:55:20.860" Score="2" ViewCount="0" Body="&lt;p&gt;Six, because $16^6 = 16.777.216$.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-12-04T15:55:20.860" CommentCount="6" />
  <row Id="12994" PostTypeId="2" ParentId="12991" CreationDate="2010-12-04T15:58:53.587" Score="3" ViewCount="0" Body="&lt;p&gt;Gödel's completeness theorem isn't about theories but about the logic used in the language of a theory, namely about the connection between the syntax and semantics of that logic. The logic used to describe Robinson's arithmetic (first order logic) is complete and at the same time RA is not complete as a theory. The same applies to first order Peano Arithmetic. The logic used in the language of PA is complete but PA is not a complete theory. Same thing applies to ZFC. Generally, every incomplete theory described in first order logic has that property.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastEditorUserId="1772" LastEditorDisplayName="" LastEditDate="2010-12-05T11:33:39.327" LastActivityDate="2010-12-05T11:33:39.327" CommentCount="8" />
  <row Id="12996" PostTypeId="1" CreationDate="2010-12-04T16:33:14.373" Score="3" ViewCount="99" Body="&lt;p&gt;Let $R$ a root system and $\Delta$ be a simple system of roots of a Lie algebra $\mathfrak g$, $\Delta&amp;#39;\subset \Delta$ and $R(\Delta&amp;#39;)=R\cap \mathbb Z(\Delta&amp;#39;)$. Define &#xA;$$p(\Delta&amp;#39;)=\mathfrak h \bigoplus_{\alpha \in R(\Delta&amp;#39;)} \mathfrak g_{\alpha}  \bigoplus_{\alpha \in R^+ \setminus R^+(\Delta&amp;#39;)}\mathfrak g_{\alpha}$$ the parabolic subalgebra associated to $\Delta&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\alpha$ is a simple root in $R^+(\Delta)\setminus R^+ (\Delta&amp;#39;)$, then $\beta(h_\alpha)=0$ for all $\beta$ in $R(\Delta&amp;#39;)$???&lt;/p&gt;&#xA;" OwnerUserId="2764" LastEditorUserId="2764" LastEditorDisplayName="" LastEditDate="2010-12-04T19:42:11.300" LastActivityDate="2010-12-28T19:35:11.343" Title="Parabolic subalgebra" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="12997" PostTypeId="2" ParentId="12921" CreationDate="2010-12-04T17:05:22.223" Score="0" ViewCount="0" Body="&lt;p&gt;The best answer has been given &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3503/formal-languages-what-does-r-trivial-mean/3515#3515&quot;&gt;here&lt;/a&gt; by &lt;a href=&quot;http://cstheory.stackexchange.com/users/508/michael-blondin&quot;&gt;Michael Blondin&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A semigroup $S$ is $R\text{-trivial}$ iff $a \: R \: b \Rightarrow a = b$ for all $a, b \in S$ where $R$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_relations#The_L.2C_R.2C_and_J_relations&quot; rel=&quot;nofollow&quot;&gt;Green's relation&lt;/a&gt; $a \: R \: b \Leftrightarrow aS^1 = bS^1$. The set of $R\text{-trivial}$ monoids forms a variety which can be ultimately defined by the equations $(xy)^n x = (xy)^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A language is $R\text{-trivial}$ if its &lt;a href=&quot;http://en.wikipedia.org/wiki/Syntactic_monoid&quot; rel=&quot;nofollow&quot;&gt;syntactic monoid&lt;/a&gt; is $R\text{-trivial}$. This variety of languages is alternatively defined as the set of all languages that can be written as a disjoint union  of languages of the form $A_0^* a_1 A_1^* a_2 \ldots a_n A_n^*$ where $n \geq 0$, $a_1, \ldots, a_n \in A$, $A_i \subseteq A \setminus \{a_{i+1}\}$ for $0 \leq i \leq n-1$. Another definition given in &lt;a href=&quot;http://www.liafa.jussieu.fr/~jep/Resumes/Varietes.html&quot; rel=&quot;nofollow&quot;&gt;[Pin]&lt;/a&gt;, which I'm not familiar with, uses the so-called &lt;em&gt;automates extensifs&lt;/em&gt; (&quot;extensive automata&quot;). You can find more results about those languages in &lt;a href=&quot;http://www.liafa.jussieu.fr/~jep/Resumes/Varietes.html&quot; rel=&quot;nofollow&quot;&gt;[Pin]&lt;/a&gt;. There is an English version of this book, I've never read it but I'm pretty sure that you can find the same content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of completeness, here is an example of a $R\text{-trivial}$ language: $\{b\}^* a \{a,c\}^* b \{a\}^* b \{a,b,c\}^* \cup \{d\}^* \cup abcd$. You can build other examples with the previous definitions. Note that all of the properties of varieties of languages hold for $R\text{-trivial}$ languages, therefore they are closed under union, intersection and complementation. It should help to build more complicated languages.&lt;/p&gt;&#xA;" OwnerUserId="4371" LastActivityDate="2010-12-04T17:05:22.223" CommentCount="2" />
  <row Id="12998" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T17:16:06.933" Score="1" ViewCount="0" Body="&lt;p&gt;Things come full circle:  Constructive proofs of the Hilbert Basis theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.andrew.cmu.edu/user/avigad/Students/hertz.pdf&quot; rel=&quot;nofollow&quot;&gt;Click with all of your justice&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-04T17:16:06.933" CommentCount="2" />
  <row Id="12999" PostTypeId="1" AcceptedAnswerId="13002" CreationDate="2010-12-04T17:23:18.677" Score="2" ViewCount="59" Body="&lt;p&gt;When you multiply two power series' that have difference intervals of convergence, what is the product's interval of convergence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex.  f(x) converges for |x|&amp;lt;6, g(x) converges for |x|&amp;lt;4.  What does f(x)g(x) converge for?&lt;/p&gt;&#xA;" OwnerUserId="3691" LastActivityDate="2010-12-04T18:08:14.847" Title="Multiplication of Power Series and their convergences" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="2" />
  <row Id="13000" PostTypeId="1" CreationDate="2010-12-04T17:25:53.403" Score="2" ViewCount="49" Body="&lt;p&gt;The 4-associahedron is the 4-dimensional version of Mac Lane's pentagon diagram.  If you look at &lt;a href=&quot;http://math.ucr.edu/home/baez/trimble/tetracategories.html&quot; rel=&quot;nofollow&quot;&gt;Trimble's notes on tetracategories&lt;/a&gt;, you can see the obvious difficulty in working with such a diagram (using it to verify things like coherence etc).  Are there any techniques and/or approaches to working with  such unwieldy higher-dimensional diagrams?  &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-04T17:25:53.403" Title="Practical approaches to working with nonplanar commutative diagrams?" Tags="&lt;category-theory&gt;" />
  <row Id="13001" PostTypeId="2" ParentId="12999" CreationDate="2010-12-04T17:30:59.047" Score="0" ViewCount="0" Body="&lt;p&gt;The power series for a function $f:\mathbb{C}\to\mathbb{C}$ converges for all $z$ within the largest open disk such that $f$ is analytic within the disk (i.e. no poles, discontinuities, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;$f(z)=\frac{1}{1+z^2}$. The function is analytic within $|z| &amp;lt; 1$, but has poles at $\pm i$. Its power series converges within the same disk. Multiplying with for example $g(z)=e^z$ (everywhere analytic), yields $h(z)=f(z)g(z)$ with radius of convergence $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;(actually, if $R,S$ is the radius of convergence for $f,g$ respectively, then the radius of convergence for $fg$ is $\min(R,S)$.)&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I lived in a complex world, see Pete L. Clark's comment below.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="4284" LastEditorDisplayName="" LastEditDate="2010-12-04T18:08:14.847" LastActivityDate="2010-12-04T18:08:14.847" CommentCount="4" />
  <row Id="13002" PostTypeId="2" ParentId="12999" CreationDate="2010-12-04T17:31:33.167" Score="4" ViewCount="0" Body="&lt;p&gt;Assuming both power series have the same center, the radius of convergence of the product is at least the minimum of the radii of the convergence of the two individual series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Proposition 2 in &lt;a href=&quot;http://math.stackexchange.com/users/299/pete-l-clark&quot;&gt;Pete Clark'&lt;/a&gt;s notes here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/243series7.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/243series7.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-04T17:31:33.167" CommentCount="1" />
  <row Id="13003" PostTypeId="1" AcceptedAnswerId="13089" CreationDate="2010-12-04T17:48:13.253" Score="3" ViewCount="88" Body="&lt;p&gt;Consider a parallelogram which is neither a rhombus nor a rectangle. It is well-known that such shapes do not have an axis of symmetry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a simple proof for this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I prefer a proof that I can give to a kid (~12 year old), but more involved proofs will do as well.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2010-12-05T09:55:37.517" Title="Parallelograms &amp; Axis of Symmetry" Tags="&lt;geometry&gt;&lt;symmetric-groups&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="13004" PostTypeId="1" AcceptedAnswerId="13007" CreationDate="2010-12-04T17:50:41.633" Score="3" ViewCount="81" Body="&lt;p&gt;Are there any advanced topics--preferably at the research frontier--which can nevertheless be explained accurately, if not efficiently, using very down-to-earth ideas which would be accessible to most college students, not necessarily even math majors? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a prototypical example of what I mean, I offer the book &lt;em&gt;QED&lt;/em&gt; by Feynman, which explains the ideas of quantum electrodynamics without even using the word &quot;complex number.&quot; In mathematics, the book &lt;em&gt;Naive Lie Theory&lt;/em&gt; by John Stillwell comes close to what I'm looking for. &lt;em&gt;Visual Complex Analysis&lt;/em&gt; by Tristan Needham is a pseudo-example, the main issue being that its subject matter is very standard undergraduate material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, any such explanation cannot be a fully faithful representation of the original work, but ideally it would still be possible to convey the essence of the methods without introducing any real inaccuracies.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-04T18:19:05.367" Title="Examples of advanced results and ideas explained in a down-to-earth way" Tags="&lt;explanation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13005" PostTypeId="2" ParentId="12959" CreationDate="2010-12-04T17:52:30.063" Score="3" ViewCount="0" Body="&lt;p&gt;For any fixed $x_0$, there is positive probability that the sequence starting from $x_0$ fails to converge.   To see why, take, say, $x_0 = 15$.  There is positive probability that $50 &amp;lt; B_t &amp;lt; 60$ for $10 &amp;lt; t &amp;lt; 15$ and $10 &amp;lt; B_t &amp;lt; 20$ for $50 &amp;lt; t &amp;lt; 60$.  On this event the sequence oscillates between the intervals $(10,20)$ and $(50,60)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, almost surely there exist infinitely many $t$ in any interval $(0,\epsilon)$ with $B_t = t$, so even when the sequence does converge the limit need not be 0.  For a proof, note that $P(B_t &amp;gt; t) = P(N &amp;gt; \sqrt{t}) \ge 1/4$ for sufficiently small $t$ (here $N$ is a standard normal random variable).  So for any sequence $t_n$ decreasing to $0$, we have $P(B_{t_n} &amp;gt; t_n \text{ i.o.}) \ge 1/4$; by the Blumenthal 0-1 law, $P(B_{t_n} &amp;gt; t_n \text{ i.o.}) =1$.  However, we also have $P(B_{t_n} &amp;lt; t_n) \ge P(B_{t_n} &amp;lt; 0) = 1/2$ so by a similar argument $P(B_{t_n} &amp;lt; t_n \text{ i.o.}) =1$.  The result follows by continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could ask some other questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For a fixed $x_0$, what is the probability that the sequence starting from $x_0$ converges?  My guess is $0$ but I don't see a proof offhand.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the (random) set $C$ of $x$ such that the sequence starting from $x$ converges.  What is the Lebesgue measure of $C$?  My guess is that $m(C) = 0$ a.s. but again no proof.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Another interesting fact is that almost surely, for every starting point $x_0$, the sequence $x_k$ is bounded, and hence has a convergent subsequence.  Let $M_r = \sup_{t \in [-r,r]} |B_t|$.  By the strong law of large numbers, $B_t/t \to 0$ a.s. as $t \to \pm \infty$, and it follows that $M_r / r \to 0$ a.s. as $r \to \infty$. In particular, a.s. there exists $r &amp;gt; x_0$ with $M_r &amp;lt; r$, and then $|x_k| \le M_r$ for all $k \ge 1$.&lt;/p&gt;&#xA;" OwnerUserId="822" LastEditorUserId="822" LastEditorDisplayName="" LastEditDate="2010-12-04T18:17:45.897" LastActivityDate="2010-12-04T18:17:45.897" CommentCount="1" />
  <row Id="13007" PostTypeId="2" ParentId="13004" CreationDate="2010-12-04T18:19:05.367" Score="5" ViewCount="0" Body="&lt;p&gt;I can think of a couple of books that may be close to what you claim to be looking for. One is &lt;a href=&quot;http://books.google.co.cr/books?id=h7oUr88S7JQC&amp;amp;printsec=frontcover&amp;amp;dq=fearless+symmetry&amp;amp;hl=es&amp;amp;ei=_4H6TLKHFoL88AaU7sWCCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Fearless Symmetry&lt;/a&gt; by Avner Ash and Robert Gross. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They try to &quot;explain&quot; (without entering in the details of course) the general ideas behind Wiles' proof of Fermat's Last Theorem. So they begin in chapter 1 with an intuitive explanation of the purpose and aim of the idea of a representation and follow their way through elliptic curves, reciprocity laws and even Galois Representations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is certainly a book that &quot;explains&quot; very advanced stuff, but that is aimed at a non-specialist audience. It is supposed to be a popular math book, but I'm not sure how accurate that is since it is not that easy to digest if you don't have at least some undergraduate mathematics under your belt. At least that's how I see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other book I have in mind is &lt;a href=&quot;http://books.google.co.cr/books?id=T_o0CnMZecMC&amp;amp;printsec=frontcover&amp;amp;dq=visual+group+theory&amp;amp;hl=es&amp;amp;ei=LIH6TM_bGYL58AaUy7zJCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Visual Group Theory&lt;/a&gt;, but this is closer to the spirit of Visual Complex Analysis, I suppose, and as in that case it explains standard undergraduate mathematics.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-04T18:19:05.367" CommentCount="1" />
  <row Id="13008" PostTypeId="2" ParentId="12968" CreationDate="2010-12-04T18:26:21.143" Score="7" ViewCount="0" Body="&lt;p&gt;The issue is complicated by the fact that many (most?) &quot;mainstream mathematicians&quot; are not really keeping track of whether they are using some form of AC in their arguments.  Moreover, they often do not write their proofs in such a way which makes clear either whether they are using AC or whether this use of AC is really necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example from my own experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem (Luther Claborn): For every abelian group $A$, there exists a Dedekind domain $R$ such that the ideal class group of $R$ is isomorphic to $A$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of three proofs of this theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Claborn's original 1966 proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/claborn66.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/claborn66.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) C. Leedham-Green's 1972 proof: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/leedhamgreen.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/leedhamgreen.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) My 2008 proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.uga.edu/~pete/ellipticded.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.uga.edu/~pete/ellipticded.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these, I am sorry to say that the only one I am truly familiar with is my own.  My proof uses transfinite induction.  To be quite honest, I was pretty tickled by this: it was maybe the fourth time in my life that I had written down a transfinite induction argument and the first time that it was for any serious purpose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I soon heard from the eminent mathematician Bjorn Poonen (then at Berkeley) that he had been thinking about this very result because a student of Eisenbud's had recently given a seminar talk on Leedham-Green's proof (which I would describe as &quot;clever and complicated&quot;).  The proof I had given was much more up his own alley, and he immediately found a way to recast / improve it.  Among other things, his recasting completely avoided use of the Axiom of Choice.  (But I hasten to add that there was more to it than that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We speculated briefly on the possibility that this modified argument was the &lt;em&gt;first&lt;/em&gt; proof of Claborn's theorem to avoid AC.  I asked whether Leedham-Green's proof used AC and he told me that indeed it was also a transfinite induction argument.  So then I looked back at Claborn's paper and asked my student about it.  We were able to point to some part of Claborn's argument (which, if you look at the paper, you'll see is rather terse: I wish the paper were more than four pages long!) which we thought used AC.  But I don't remember what that was, and I am only moderately confident that we were right: if you told me otherwise, I wouldn't be too surprised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I am afraid to have to tell you that I didn't include in my paper any hint that Poonen's modification would give an AC-less proof (and possibly the first such proof!) &#xA;of Claborn's theorem.  I felt that Poonen's argument was nice enough so that it deserved further exploration in its own right: I wanted to use it -- and still do -- to give a complete characterization of all abelian groups which can be the group of rational points on some elliptic curve over some field.  So in essence I have been keeping to myself the fact that this theorem of Claborn (which lots of people know about) can be proved independently of AC.  Sorry!&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-04T18:40:50.923" LastActivityDate="2010-12-04T18:40:50.923" />
  <row Id="13009" PostTypeId="1" AcceptedAnswerId="13011" CreationDate="2010-12-04T18:28:19.487" Score="3" ViewCount="66" Body="&lt;p&gt;What order is the field: $\mathbb{F}_p(\zeta_{p-1}^{1/n})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{F}_p^{\times}$ is size $p-1$ and cyclic (with a generator we shall call $\zeta_{p-1}$). Naively, it seems that by taking the $n^{th}$ root of $\zeta_{p-1}$ we get a group of units of size $C_{n(p-1)}$. This, of course, doesn't make sense. What should be true is that $C_{n(p-1)}$ is some subgroup of the group of units of the $\mathbb{F}_{p^r}$ we get.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What is the order of the field $\mathbb{F}_p[x]/(m_{\zeta_{p-1}^{1/n}})$ where $m_{\zeta_{p-1}^{1/n}}$ is the minimal polynomial of $\zeta_{p-1}^{1/n}$ (which, unless I'm mistaken, is the same as the splitting field of $x^n-\zeta_{p-1}$)? Does it depend on whether $p|n$ or not?&lt;/p&gt;&#xA;" OwnerUserId="4410" LastEditorUserId="4410" LastEditorDisplayName="" LastEditDate="2010-12-04T18:41:16.477" LastActivityDate="2010-12-04T18:45:57.403" Title="What field is $\mathbb{F}_p(\zeta_{p-1}^{1/n})$" Tags="&lt;finite-fields&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13010" PostTypeId="1" AcceptedAnswerId="13012" CreationDate="2010-12-04T18:33:24.827" Score="2" ViewCount="137" Body="&lt;p&gt;So I read the answers in &lt;a href=&quot;http://math.stackexchange.com/questions/713/how-to-visualize-a-rank-2-tensor&quot;&gt;this&lt;/a&gt; question, and don't feel that much closer to an answer about how tensors as multi-linear maps and tensors as &quot;multi-dimensional&quot; matrices are truly related.  For instance, it seems that a (1,1)-tensor should be able to be realized as a 2 by 2 matrix, as something that looks like $(x_1,x_2)\otimes(y_1,y_2)$ and as a multi-linear map from $\mathbb{R}^2\times(\mathbb{R}^2)^\ast$ to $\mathbb{R}$.  Is there a formal isomorphism between these three ideas?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motivation for asking this is that I'm trying to work with deRham cohomology without just doing the mechanics of it.  I'd like to actually know what's going on, and so I'm trying to see &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How is $\mathrm{Alt}^p(\mathbb{R}^n)$ actually the same as $\wedge^p(\mathbb{R}^n)$ where the former is the space of alternating p-linear maps and the latter is the algebraically defined quotient of $\mathbb{R}^n\otimes\ldots\otimes\mathbb{R}^n$, the tensor product of $p$ copies of $\mathbb{R}^n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How is $\Omega^p(\mathbb{R}^n)$ actually the dual of p-vector fields (if in fact it is...)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-04T19:37:59.113" Title="Tensors as matrices vs. Tensors as multi-linear maps" Tags="&lt;soft-question&gt;&lt;algebra&gt;&lt;differential-geometry&gt;&lt;cohomology&gt;&lt;tensors&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13011" PostTypeId="2" ParentId="13009" CreationDate="2010-12-04T18:45:57.403" Score="1" ViewCount="0" Body="&lt;p&gt;Every nonzero element of a finite field is a root of unity. Roots of unity behave very nicely, since the other roots of unity of the same order are just powers of the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance i is a primitive 4th root of unity, and the other one, -i, is just a power i,-1,-i,1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots of unity form these nice cyclic subgroups of the group of units of a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want one primitive k'th root of unity, you get all of them for free, so we are just interested in the smallest finite field of characteristic p with a primitive k'th root of unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you get into trouble if p divides k: a finite field of characteristic p has size p^m, so its group of units has order p^m-1, and so k has to divide p^m-1.  Such numbers are coprime to p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, you'll want to assume n is coprime to p, otherwise you won't get very good roots.  For instance, when p=2, the 8th roots of unity are exactly {1}; none of them are primitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups of units of finite fields are very well behaved: they are all cyclic.  So to find a primitive k'th root of unity, you just want to find the smallest m such that k divides p^m-1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The field obtained by adjoining a primitive k=n(p-1)st root of unity is the field of size p&lt;sup&gt;m&lt;/sup&gt;, where m is the order of p in the group of units of the ring &lt;strong&gt;Z&lt;/strong&gt; / k &lt;strong&gt;Z&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that k has to be coprime to p in order for a primitive k'th root of unity to exist.  Just divide k by p until it is coprime if you want to adjoin the &quot;closest to primitive&quot; root.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-04T18:45:57.403" />
  <row Id="13012" PostTypeId="2" ParentId="13010" CreationDate="2010-12-04T18:49:16.793" Score="5" ViewCount="0" Body="&lt;p&gt;0)  Let $V$ be a (finite-dimensional) real vector space.  A $(1, 1)$-tensor over $V$ is any of the following equivalent objects:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A linear transformation $V \to V$,&lt;/li&gt;&#xA;&lt;li&gt;An element of $V^{\ast} \otimes V$,&lt;/li&gt;&#xA;&lt;li&gt;A linear map $V \otimes V^{\ast} \to \mathbb{R}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The isomorphism between these three pictures comes from the definition of the dual and the universal property of the tensor product.  A linear transformation is an object such that, when you feed it an element of $V$, it spits out an element of $V$.  So if you feed it an element of $V$ and an element of $V^{\ast}$, then by the dual pairing (tensor contraction) you get an element of $\mathbb{R}$.  Explicitly, if $f : V \to V$ is a linear transformation, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle f(v), w^{\ast} \rangle : V \times V^{\ast} \to \mathbb{R}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is bilinear (where $\langle \cdot, \cdot \rangle$ denotes the dual pairing), and every bilinear map can be uniquely written this way.  Then one uses the universal property of the tensor product.  That gives you the identification with the third picture.  The identification with the second picture comes from the fact that dual distributes over tensor product (which again comes down to tensor contraction) and the fact that $V^{\ast \ast} \simeq V$.  Alternately, again by tensor contraction, there is a natural bilinear map $V \times (V^{\ast} \otimes V) \to V$ which identifies an element of $V^{\ast} \otimes V$ with a linear transformation $V \to V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get the &quot;explicit&quot; version of all these pictures by taking a basis $e_1, ... e_n$ of $V$ and the corresponding dual basis $e_1^{\ast}, ... e_n^{\ast}$ of $V^{\ast}$.  These define bases of every space of $(m, n)$-tensor (just multiply).  In particular the space of $(1, 1)$-tensors has basis $e_i^{\ast} \otimes e_j, 1 \le i, j \le n$.  Writing a $(1, 1)$-tensor in this basis corresponds exactly to writing a linear transformation as a square matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  It's not.  If by $\text{Alt}^p(V)$ you mean the space of alternating $p$-linear maps $V^p \to \mathbb{R}$, this is naturally isomorphic to $\Lambda^p (V^{\ast})$ where $V^{\ast}$ is the dual space to $V$ (the space of linear maps $V \to \mathbb{R}$).  You can't identify $V$ with its dual without, say, an inner product.  To see the isomorphism, $(V^{\ast})^{\otimes p}$ can naturally be identified with the space of all $p$-linear maps $V^p \to \mathbb{R}$, more or less by definition of the dual space and tensor product, and taking the subspace of alternating maps in one picture corresponds to taking the quotient in the other.  You can make this really explicit by writing down a basis for both spaces if you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  This all comes down to tensor contraction.  Locally a $p$-form is the same thing as an alternating $p$-linear map $V^p \to \mathbb{R}$ where $V$ is the tangent space, and by the universal property of the exterior power this is the same thing as a linear map $\Lambda^p V \to \mathbb{R}$.  And a $p$-vector is the same as an element of $\Lambda^p V$.  So the natural dual pairing (tensor contraction) comes into play.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-04T19:08:55.333" LastActivityDate="2010-12-04T19:08:55.333" />
  <row Id="13013" PostTypeId="1" AcceptedAnswerId="13014" CreationDate="2010-12-04T18:49:34.687" Score="1" ViewCount="70" Body="&lt;p&gt;So, I guess divide it into six sections representing the difference between the numbers, so the first is $0\rightarrow n$, the second through fifth are $1\rightarrow n$ and the sixth is $0\rightarrow n$ again, and that becomes a generating function of $((1+x+x^2+\cdots+x^n)^2)((x^2+\cdots+x^n)^4)$. Does this make sense, and what coefficient represents what value of $n$?&lt;/p&gt;&#xA;" OwnerUserId="4411" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-04T19:08:53.213" LastActivityDate="2010-12-07T14:57:15.787" Title="Generating function for modeling the number of ways to select five integers from 1 to n where no two are consecutive." Tags="&lt;combinatorics&gt;&lt;generating-functions&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13014" PostTypeId="2" ParentId="13013" CreationDate="2010-12-04T18:58:28.560" Score="0" ViewCount="0" Body="&lt;p&gt;Let $f_{k,n}$ be the number of ways to select $k$ integers in $\{ 1, 2, ... n \}$ such that no two are consecutive.  There are two ways to do this: either the largest one is not $n$, and then you choose $k$ integers in $\{ 1, 2, ... n-1 \}$, or the largest one is $n$, and then you choose $k-1$ integers in $\{ 1, 2, ... n-2 \}$.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{k,n} = f_{k,n-1} + f_{k-1, n-2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generating function you want is $\sum_{n \ge 0} f_{5,n} x^n$, but I think it will be easier to find the bivariate generating function $\sum_{n,k \ge 0} f_{k, n} y^k x^n$ first.  Do you know how to do this from a recursion?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-04T18:58:28.560" CommentCount="3" />
  <row Id="13015" PostTypeId="2" ParentId="356" CreationDate="2010-12-04T19:02:05.233" Score="0" ViewCount="0" Body="&lt;p&gt;The Mythical Man-Month more for software engineering but absolutely one of the best books ever written about programing IMO. &lt;/p&gt;&#xA;" OwnerUserId="4412" LastActivityDate="2010-12-04T19:02:05.233" CommunityOwnedDate="2010-12-04T19:02:05.233" />
  <row Id="13016" PostTypeId="2" ParentId="12983" CreationDate="2010-12-04T19:04:58.737" Score="1" ViewCount="0" Body="&lt;p&gt;There's an almost-formal argument using cumulants. Given a random variable $X$, define its moment generating function $$M(X) = E[e^{tX}].$$ It's called the moment generating function since opening the Taylor series of the exponential, we get $$M(X) = 1 + E[X]t + \frac{1}{2}E[X]^2t^2 + \cdots.$$ The moment generating function is useful because of its relation to convolution of two &lt;i&gt;independent&lt;/i&gt; random variables: $$M(X+Y) = E[e^{t(X+Y)}] = E[e^{tX}e^{tY}] = E[e^{tX}]E[e^{tY}] = M(X)M(Y).$$ One proof of the CLT takes the route of the mgf, but we would like to replace the multiplication by a addition since we only really know how to handle sums. So we define the cumulant generating function $$K(X) = \log M(X).$$ We can calculate the first few coefficients (which are called cumulant) by substituting into the (formal) power series of $\log(1+x) = x - x^2/2 + \cdots$: $$K(X) = \log (1+E[X]t+E[X^2]t^2/2 + \cdots) = E[X]t + E[X^2]t^2/2 - (E[X]^2t^2 + E[X]E[X^2]t^3 + E[X^2]t^4/4)/2 + \cdots = E[X]t + V[X]t^2/2 + \cdots.$$ Also, if $X$ and $Y$ are independent then $$K[X+Y] = K[X]+K[Y].$$ Now suppose $X_1,\ldots,X_n$ are iid variables distributed like $X$ with zero expectation. Then $$K[X_1+\cdots+X_n] = nK[X] = \frac{1}{2}nV[X]t^2 + \frac{1}{6}nK_3(X)t^3 + \cdots,$$ where $K_m(X)$ are just the (normalized) coefficients of the cgf, i.e. the cumulants (they are normalized by $1/m!$). If we scale this sum down by $\sqrt{n}$, then the second cumulant becomes $V[X]$ (i.e. the variance is the same), but the rest of the cumulants $K_m$ for $m \geq 3$ get multiplied by $n^{1-m/2} \rightarrow 0$, so in the limit they disappear, and the cumulant of the limit is just $$K\left[\frac{X_1+\cdots+X_n}{\sqrt{n}}\right] = \frac{1}{2}V[X]t^2.$$ Therefore there is one 'domain of attraction' for distributions, which must be the normal distribution with zero mean and variance $V[X]$; it can be calculated directly from this representation. The same idea can be used to analyze the case where the variables are independent but not identically distributed. The main step missing to make this proof formal is reasoning about the limit distribution from the limit cgf; this is the Levy continuity lemma, which shows that the 'inverse Fourier transform' is continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Had we taken the route of mgf's, we would have had to use the identity $(1+1/n)^n \rightarrow e^n$ somewhere, but otherwise the argument would be much the same.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-04T19:04:58.737" />
  <row Id="13017" PostTypeId="2" ParentId="12960" CreationDate="2010-12-04T19:06:04.510" Score="5" ViewCount="0" Body="&lt;p&gt;Let $P$ be generated by $x$ of order $2^n &amp;gt; 4$ and $y$ of order $4$, and let $z = x^{2^{n-2}}$ be a power of $x$ of order 4. Then the two classes of $Q_8$ in $P$ are represented by $\langle z,y \rangle$ and $\langle z, xy \rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you say, you know that both of these subgroups are normalized in $X$ by subgroups isomorphic to ${\rm SL}(2,3)$. That means that $z$ is conjugate in $X$ to both $y$ and to $xy$. I don't think that's possible in a solvable group $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $X$ is solvable, so it has a chief series with elementary abelian factors. So $z \in M \setminus N$ for one of these factors $M/N$. If $z$ is conjugate to both $y$ and $xy$, then we have $y, xy \in M \setminus N$. But then $x \in M$ and so $x^2 \in N$ and hence $z \in N$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED: For your smaller question if, for conjugates $u$ and $v$ of $t$, $x=uv$ is nontrivial and centralizes $t$, then $x$ has odd order and both $t$ and $u$ lie in the normalizer $N$ of $\langle x \rangle$ and hence are conjugate in $N$, but that is impossible because $t$ centralizes and $u$ inverts $x$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastEditorUserId="2820" LastEditorDisplayName="" LastEditDate="2010-12-05T12:16:24.170" LastActivityDate="2010-12-05T12:16:24.170" CommentCount="2" />
  <row Id="13019" PostTypeId="1" CreationDate="2010-12-04T19:08:13.823" Score="2" ViewCount="129" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $\Gamma(y) = \int_0^{\infty}{e^{-x}x^{y-1}\,dx}$ is finite for $y&amp;gt;0$ both as an improper Riemann integral and as a Lebesgue integral. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Show $\Gamma&amp;#39;(y) = \int_0^{\infty}{e^{-x}x^{y-1}\ln{x}\,dx}$ for $y&amp;gt;0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For one: I've tried simply integrating it as an improper Riemann integral, but you always end up with another integral of the &quot;same type&quot; (which is how you eventually show $\Gamma(y+1)=y\Gamma(y)$ ). How do I get around this? As for the Lebesgue integral, I think it'd be easiest to compare the integrand to a larger function whose integral converges, but I haven't come up with a good candidate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two: Fix $y_0&amp;gt;0$. Write $$\Gamma&amp;#39;(y) = \lim_{y\to y_0}{\int_0^{\infty}{\frac{e^{-x}x^{y-1}-e^{-x}x^{y_0-1}}{y-y_0}\,dx}}\,.$$ By the MVT, there exists $\eta$ between $y$ and $y_0$ such that the above limit is equal to $$\lim_{y\to y_0}{\int_0^{\infty}{e^{-x}x^{\eta-1}\ln{x}\,dx}}\,.$$ But now I'm not sure what to do. This is similar to a previous question I posted; for that problem we knew the derivative of the original integrand was bounded, so we applied the bounded convergence theorem. Would it be enough to prove that the derivative of my integrand is bounded, and apply BCT? &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-12-04T19:08:13.823" Title="Some questions about the gamma function" Tags="&lt;real-analysis&gt;&lt;homework&gt;" CommentCount="3" />
  <row Id="13020" PostTypeId="2" ParentId="13010" CreationDate="2010-12-04T19:37:59.113" Score="0" ViewCount="0" Body="&lt;p&gt;This interpretation makes judicious use of duality.  Recall that $Hom_R(A\otimes B, C)\cong Hom_R(A,Hom_R(B,C))$ (the proof of this fact is extremely simple.  Please give it a try.).  With finite dimensional vector spaces, we have a canonical isomorphism $V\cong V^{**}=Hom_R(Hom_R(V,R),R)$.  Then to give a map $A\to B$, this is the same as giving a map $A\to B^{**}$ by composition, and by the earlier note about the hom-tensor adjunction, this is identical to giving a map $A\otimes B^*\to R$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to your two questions, they fall out from these elementary observations.  I urge you to try to work them out.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-04T19:37:59.113" CommentCount="1" />
  <row Id="13022" PostTypeId="2" ParentId="13003" CreationDate="2010-12-04T20:02:32.490" Score="4" ViewCount="0" Body="&lt;p&gt;Perhaps this will be convincing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The axis of symmetry intersects one of the sides at point which is not the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, by considering the perpendicular to the axis at the point of intersection of the axis and the side, we see that the axis has to be perpendicular to the side and intersects it at the midpoint. In which case it is not an axis of symmetry as it won't intersect the opposite side at the midpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The axis of symmetry intersects a side at a vertex. Then by 1) it also intersects the opposite side at the vertex, and hence has to be a diagonal, in which case, it is not an axis of symmetry either.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-04T20:02:32.490" CommentCount="1" />
  <row Id="13023" PostTypeId="2" ParentId="12983" CreationDate="2010-12-04T20:21:32.580" Score="1" ViewCount="0" Body="&lt;p&gt;Working out a few simple examples might help. This would indeed show you that the theorem works in special cases. Thus it would go a long way towards convincing oneself of the validity of the central limit theorem. The central limit theorem first appeared in the work of Abraham de Moivre, in which he proved that the normal distribution to approximate the distribution of the number of heads resulting from many tosses of a fair coin. Later Laplace showed that the same for the binomial distribution, approximating it with the normal distribution. I suggest that you work out these two simpler cases to get a feeling of how the approximation happens. All the necessary background for doing this yourself is available in the book of Hoel, Port and Stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find the theorem hard to understand, it might make you feel better to hear that probabilists needed a long time to properly formulate and understand the theorem. It was only done in the twentieth century by Lyapunov.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are oriented towards practical applications, then getting used to some topics of your choice, for instance noise analysis in communication theory, might help you in convincing yourself of the truth of the central limit theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to understand the central limit theorem would of course be to take a course in probability theory. An introductory course usually ends with a proof of this theorem. And if you take a course, you would see other interesting theorems such as the weak and strong laws of large numbers, and this would put the central limit theorem in better perspective. Even after all this, you might still need to contemplate a bit to really absorb the theorem. The proof I have seen is using some &quot;characteristic functions&quot; and a sort of &quot;Fourier transform&quot;. I have to regretfully confess that I didn't fully understand it when I took the course. I never had to take up probability theory later; but if the occasion arises, I intend to go all properly through the proof and understand the machinery.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-04T20:21:32.580" />
  <row Id="13024" PostTypeId="2" ParentId="13003" CreationDate="2010-12-04T20:22:43.750" Score="1" ViewCount="0" Body="&lt;p&gt;Let's prove the converse: if a parallelogram $ABCD$ has a symmetry axis, it's either a rectangle or a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the image of a vertex is a vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some of the vertices (say $A$) is mapped to itself, than there must be another vertex mapped to itself (because otherwise the rest of vertices must be split into pairs by mapping). This another vertex must be $C$, because if it's adjacent to $A$, than the symmetry axis is one of the edges, which is impossible. So, $AC$ is the axis, $B$ maps to $D$, thus $BD$ is perpendicular to $AC$, therefore the parallelogram $ABCD$ is a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case when every vertex is mapped to some other vertex. If $A$ maps to $C$, than $B$ must map to $D$, so the axis should be a line perpendicular to both $AC$ and $BD$, which is impossible since $AC$ and $BD$ are parallel. So, $A$ maps to an adjacent vertex (let it be $B$), $C$ maps to $D$, so the axis connects the midpoints of $AB$ and $CD$ and is perpendicular to them. This implies that $AB$ is perpendicular to $CD$.&lt;/p&gt;&#xA;" OwnerUserId="2977" LastEditorUserId="2977" LastEditorDisplayName="" LastEditDate="2010-12-04T20:28:15.070" LastActivityDate="2010-12-04T20:28:15.070" />
  <row Id="13025" PostTypeId="2" ParentId="12897" CreationDate="2010-12-04T20:30:45.060" Score="2" ViewCount="0" Body="&lt;p&gt;You have $a + b \sum_{k\geq 0} \frac{x^k}{k!} + c \sum_{k\geq 0} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the coefficients must be 0 above, so, for the constant term we have $a+b = 0$ and comparing coefficients of $x$ we have $b+c=0$, comparing coefficient of $x^2$ we have $b=0$, from which it follows $a=b=c=0.$&lt;/p&gt;&#xA;" OwnerUserId="404" LastActivityDate="2010-12-04T20:30:45.060" />
  <row Id="13026" PostTypeId="1" AcceptedAnswerId="13048" CreationDate="2010-12-04T20:56:59.197" Score="1" ViewCount="101" Body="&lt;p&gt;OK, so I was told in &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3516/are-there-problems-that-cant-be-expressed-as-languages&quot;&gt;CSTheory&lt;/a&gt; that I should be asking here. So my question is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've taken my first course on Language Theory and we saw the &lt;a href=&quot;http://www.cs.lmu.edu/~ray/notes/languagetheory/&quot; rel=&quot;nofollow&quot;&gt;&quot;standard&quot;&lt;/a&gt; classification of languages. We saw that many problems can be expressed as a membership question of a corresponding language. I'm wondering if there are any problems for which no language can be specified.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1965" LastActivityDate="2010-12-04T23:39:38.327" Title="Are there problems that can't be expressed as languages?" Tags="&lt;computability&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13027" PostTypeId="2" ParentId="12897" CreationDate="2010-12-04T21:04:35.023" Score="0" ViewCount="0" Body="&lt;p&gt;You need to show that there are no $\alpha$, $\beta$ such that $\alpha e^x + \beta \sin x = \mathrm{const} \neq 0$ on $[0, 1]$. To do that, observe that $e^x$ is strictly convex and $\sin x$ is strictly concave on $[0, 1]$ and both are strictly increasing. It is easy to show that convexness and concaveness are invariant in respect to multiplication by a positive number, and they swap under multiplication by a negative number, so if $\alpha &amp;gt; 0$ and $\beta &amp;gt; 0$ or both are less than zero, then the linear combination is not constant (otherwise the summands would be either both strictly convex or both strictly concave). If $\alpha &amp;gt; 0$ and $\beta &amp;lt; 0$, then one summand is strictly increasing while the other is strictly decreasing, which raises the same problem. The last case to check is when either $\alpha$ or $\beta$ is zero, but it is then obvious.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-04T21:19:58.240" LastActivityDate="2010-12-04T21:19:58.240" />
  <row Id="13028" PostTypeId="1" CreationDate="2010-12-04T21:05:17.920" Score="2" ViewCount="121" Body="&lt;p&gt;Let T1 and T2 be two theories having the same set of symbols. Assume that any interpretation of T1 is a model of T1 if and only if it is not a model of T2. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1 and T2 are finitely axiomatizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i.e. there are finite sets of sentences A1 and A2 such that, for any sentence S:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T1 proves S if and only if A1 proves S, and T2 proves S if and only if A2 proves S).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/The proof will be by contradiction; assume T1 or T2 are not finitely axiomatizable, then .....?/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any one have any idea of how to prove this argument!&lt;/p&gt;&#xA;" OwnerUserId="4413" LastActivityDate="2010-12-05T12:26:08.600" Title="Finitely axiomatizable theories" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13029" PostTypeId="1" CreationDate="2010-12-04T21:22:55.387" Score="2" ViewCount="125" Body="&lt;p&gt;How to show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9 $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4415" LastActivityDate="2010-12-05T08:22:48.990" Title="AM-GM application" Tags="&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="13031" PostTypeId="2" ParentId="13029" CreationDate="2010-12-04T21:31:10.213" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; The function   $f(x,y,z)=(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ has the absolute minimum in $(1,1,1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial f}{\partial x}=\frac{1}{y}+\frac{1}{z}-\frac{y+z}{x^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial f}{\partial y}=\frac{1}{x}+\frac{1}{z}-\frac{x+z}{y^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial f}{\partial x}=\frac{1}{x}+\frac{1}{y}-\frac{x+y}{z^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that you must resolve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{&#xD;&#xA;    \begin{array}{ll}&#xD;&#xA;      \displaystyle\frac{\partial f}{\partial x}=0  \\&#xD;&#xA;      \displaystyle\frac{\partial f}{\partial y}=0  \\&#xD;&#xA;      \displaystyle\frac{\partial f}{\partial z}=0 &#xD;&#xA;    \end{array}&#xD;&#xA;  \right.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution of this system is: (1,1,1) (critical point).&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-04T21:54:40.327" LastActivityDate="2010-12-04T21:54:40.327" CommentCount="4" />
  <row Id="13033" PostTypeId="2" ParentId="13029" CreationDate="2010-12-04T21:41:48.563" Score="1" ViewCount="0" Body="&lt;p&gt;One way, although not very thoughtfull, is to simply expand the brackets.  You will get a sum of 9 terms on the LHS.  3 of these terms will be 1's and the other 6 will be fractions involving the three &quot;a&quot; terms.  Get a common denominator for the pair of fractions a/b and b/a (three such pairs) then you can use the following to show each of these three fractions is at least 2, producing a total of at least 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, you can only use this inequality if the numbers are non-negative.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2010-12-04T21:49:12.443" LastActivityDate="2010-12-04T21:49:12.443" />
  <row Id="13034" PostTypeId="2" ParentId="13029" CreationDate="2010-12-04T21:44:52.820" Score="6" ViewCount="0" Body="&lt;p&gt;Hint: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality&quot; rel=&quot;nofollow&quot;&gt;Cauchy-Schwarz&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint2: Two applications of the AM-GM will also solve it: one is $(x+y+z)/3 \ge \cdots $ and the other is ... (try to fill in the details).&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-04T22:08:04.297" LastActivityDate="2010-12-04T22:08:04.297" />
  <row Id="13035" PostTypeId="1" AcceptedAnswerId="13041" CreationDate="2010-12-04T21:46:39.317" Score="4" ViewCount="81" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to calculate the surface area of the solid of revolution which is produced from rotating $f: (-1,1) \rightarrow \mathbb{R}$, $f(x) = 1-x^2$ about the $x$-axis. I do know there is a formula:&#xA;$$S=2 \pi \int_{a}^b f(x)   \sqrt{1+f&amp;#39;(x)^2}\, \mathrm dx$$&#xD;&#xA;Which will work very well. However, I am not very comfortable with the integral&#xA;$$\int_{-1}^1 (1-x^2)\sqrt{1+4x^2}\, \mathrm dx$$&#xD;&#xA;which I would have to calculate in order to get to the surface area (I tried to substitute $x=\frac{1}{2} \sinh(u)$, but it did not work out too well). Thus, I had the idea to apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem&quot; rel=&quot;nofollow&quot;&gt;Pappus' centroid theorem&lt;/a&gt;. I first found the centroid of the area between the parabola and the x-axis to be at $y=\frac{2}{5}$, hence the surface area of the solid of revolution would be:&#xA;$$S = 2 \pi \frac{2}{5} \int_{-1}^1 \sqrt{1+4x^2}\, \mathrm dx$$&#xD;&#xA;But this leads me to a different result than I should get (I calculated the value of the first integral with the help of wolframalpha, it's about ~11...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I do wrong? My best guess is that I misunderstood Pappus' centroid theorem, but what's the mistake? How can I fix it?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T00:38:17.257" LastActivityDate="2010-12-05T00:38:17.257" Title="Calculate the surface area of a solid of revolution" Tags="&lt;calculus&gt;" AnswerCount="3" />
  <row Id="13036" PostTypeId="2" ParentId="12944" CreationDate="2010-12-04T22:00:27.020" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a heuristic to suggest that it will be difficult to define the square of the delta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fourier transform has the property that it takes the convolution of two functions to the product of their Fourier transforms, and vice versa, ie, it takes the product of two functions to their convolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the Fourier transform of the delta function is the constant function($=1$). Now suppose that $\delta^2$ exists. Then its Fourier transform would be the convolution of two constant functions. Such a convolution would shoot to infinity at every point. Even the theory of distributions can't handle such kind of stuff. So how would you imagine the inverse Fourier transform of this? How would it make sense?&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-04T22:00:27.020" />
  <row Id="13037" PostTypeId="2" ParentId="13026" CreationDate="2010-12-04T22:14:17.260" Score="1" ViewCount="0" Body="&lt;p&gt;Let $P$ be a property of objects of type $X$, i.e. of $x \in X$. Then $P$ can be expressed as membership in $$\{ x \in X : P(x) \}.$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-04T22:14:17.260" />
  <row Id="13038" PostTypeId="2" ParentId="13035" CreationDate="2010-12-04T22:17:48.007" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not too sure, but if I recall correctly, one way you can make this a little easier is to note that $g(x)=(1-x^2)\sqrt{1+4x^2}$ is even $g(-x) = g(x)$ and its range is from $-a$ to $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can do $2\int_0^a g(x)\, dx$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw: Can anyone show me a link on how to type out formulas so that they appear in mathematical format?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-04T22:24:56.963" LastActivityDate="2010-12-04T22:24:56.963" CommentCount="2" />
  <row Id="13039" PostTypeId="2" ParentId="13035" CreationDate="2010-12-04T22:24:46.670" Score="4" ViewCount="0" Body="&lt;p&gt;If you want to try the original integral again, you can use the trig substitution x = 0.5 tan(t).  This will give you an integral involving tan and sec which can be worked out.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-04T22:24:46.670" CommentCount="1" />
  <row Id="13041" PostTypeId="2" ParentId="13035" CreationDate="2010-12-04T22:37:10.217" Score="5" ViewCount="0" Body="&lt;p&gt;You did misinterpret Pappus' Theorem.  You used the geometric centroid of the &lt;em&gt;region&lt;/em&gt; between $1-x^2$ and the $x$-axis, whereas Pappus' Theorem wants you to use the geometric centroid of the &lt;em&gt;curve&lt;/em&gt; $1-x^2$.  The geometric centroid of the latter is not $\frac{2}{5}$ but (by definition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\int_{-1}^1 (1-x^2) \sqrt{1+4x^2} dx}{\int_{-1}^1 \sqrt{1+4x^2} dx} \approx 0.59002.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, multiplying this by $2\pi$ times the arc length just gives you the integral you started with.  So it doesn't appear that Pappus' Theorem is an easier route to take.  You could also try switching to an integral in $dy$, but I doubt that will be any better.  I would try Joe's suggestion on your first integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on finding geometric centroids of curves, see &lt;a href=&quot;http://www.maplesoft.com/applications/view.aspx?SID=32536&amp;amp;view=html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-04T22:42:16.967" LastActivityDate="2010-12-04T22:42:16.967" />
  <row Id="13043" PostTypeId="1" AcceptedAnswerId="13184" CreationDate="2010-12-04T22:52:00.983" Score="1" ViewCount="93" Body="&lt;p&gt;X|N,P ~ Binomial(N, P)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N ~ Poisson(11)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P ~ Beta(2,3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the moment generating function for X?&lt;/p&gt;&#xA;" OwnerUserId="4416" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-06T00:04:43.037" LastActivityDate="2010-12-06T05:04:58.393" Title="binomial-Poisson/beta hierarchy" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="13044" PostTypeId="1" AcceptedAnswerId="13046" CreationDate="2010-12-04T22:54:43.247" Score="9" ViewCount="129" Body="&lt;p&gt;I'm going over a chapter by chapter review for my calculus final and discovered this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\int_{\sqrt{x}}^{x^3}\sqrt{t}\sin{t}\;\mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They split it up so that it became:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int_1^{\sqrt{x}}\sqrt{t}\sin{t}\;\mathrm dt + \int_1^{x^3}\sqrt{t}\sin{t}\;\mathrm dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why 1? How was that determined?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T00:34:57.230" LastActivityDate="2010-12-05T01:52:42.163" Title="Ranges and the Fundamental Theorem of Calculus 1" Tags="&lt;calculus&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13045" PostTypeId="2" ParentId="13044" CreationDate="2010-12-04T23:07:19.263" Score="5" ViewCount="0" Body="&lt;p&gt;You can take any lower limit, it doesn't need to be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_b^c f(x) dx = \int_a^c f(x) dx - \int_a^b f(x) dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just think of it as the area under the curve.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-04T23:07:19.263" />
  <row Id="13046" PostTypeId="2" ParentId="13044" CreationDate="2010-12-04T23:18:38.440" Score="13" ViewCount="0" Body="&lt;p&gt;It doesn't have to be $1$; all it needs to be is in the domain of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on is that you have the Fundamental Theorem of Calculus, which tells you that if $f(x)$ is continuous on an interval containing $a$, then the function $F(x)$ defined on that interval by:&#xA;$$F(x) = \int_a^x f(t)\,dt$$&#xD;&#xA;is differentiable, and in fact $F&amp;#39;(x) = f(x)$ for all $x$ in the interval. But this requires (i) the lower limit to be constant; and (ii) the upper limit to be the variable with respect to which you are taking the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first question is: what does one do if you have the variable in the &lt;em&gt;lower&lt;/em&gt; limit instead of the upper limit? Well, that only requires you to use the property of the integral that says&#xA;$$\int_a^b f(t)\,dt = -\int_b^a f(t)\,dt,$$&#xD;&#xA;and the fact that $(-G(x))&amp;#39; = -G&amp;#39;(x)$ for any $G$. So if &#xA;$$F_2(x) = \int_x^a f(t)\,dt$$&#xD;&#xA;then&#xA;$$F_2&amp;#39;(x) = \frac{d}{dx}\int_x^a f(t)\,dt = \frac{d}{dx}\left(-\int_a^x f(t)\,dt\right) = -\frac{d}{dx}\int_a^x f(t)\,dt = -f(x),$$&#xD;&#xA;with the last equality holding by the FTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, what if the upper limit is not $x$, but a function of $x$, say&#xA;$$F_3(x) = \int_a^{g(x)} f(t)\,dt\ ?$$&#xD;&#xA;Then we use the Chain Rule:&#xA;$$\frac{dF_3}{dx} = \frac{dF_3}{dg}\frac{dg}{dx} = &#xD;&#xA;\left(\frac{d}{dg}\int_a^{g(x)}f(t)\,dt\right)\frac{dg}{dx} = f(g(x))g&amp;#39;(x),$$&#xD;&#xA;with the last equality again by the FTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the lower limit is the function and the upper limit the constant? We combine the two &quot;tricks&quot; above to get the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, what if both upper and lower limit are functions? Then we use the property of the integrals mentioned by Huy:&#xA;$$\int_a^b f(t)\,dt = \int_a^cf(t)\,dt + \int_c^b f(t)\,dt$$&#xD;&#xA;for any $c$ such that $f(x)$ is defined and continuous on an interval containing $a$, $c$, and $b$. In the case at hand, they pick $1$ simply because it's an easy point; you can use any point on $[0,\infty)$ (cannot be a negative point because you have $\sqrt{t}$ in the integrand). Pick your favorite $a$ on $[0,\infty)$, and you have&#xA;$$F(X) = \int_{\sqrt{x}}^{x^2}f(t)\,dt = \int_{\sqrt{x}}^a f(t)\,dt + \int_a^{x^2}f(t)\,dt = -\int_a^{\sqrt{x}}f(t)\,dt + \int_a^{x^2}f(t)\,dt$$&#xD;&#xA;and each of the integrals on the right is an integral we know how to do using the FTC. So if we let the first integral be some $G(X)$ and the second be $H(X)$, then $F(X)=G(X)+H(X)$, so $F&amp;#39;(X) = G&amp;#39;(X)+H&amp;#39;(X)$, and we can find $G&amp;#39;(X)$ and $H&amp;#39;(X)$ using the methods outlined above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They did it with $a=1$, but you could do it with $a=\pi$, or $a=4$, or any nonnegative $a$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T01:52:42.163" LastActivityDate="2010-12-05T01:52:42.163" CommentCount="1" />
  <row Id="13047" PostTypeId="2" ParentId="13028" CreationDate="2010-12-04T23:24:24.127" Score="4" ViewCount="0" Body="&lt;p&gt;There's a result using compactness that says that if two sets of sentences, $T_1,T_2$ have no common model then there exists a sentence $\phi$ such that $T_1\models\phi$ and $T_2\models\lnot\phi$. To see this use the contrapositive: If for every sentence $\phi$ we have $T_1\models\phi$ has as a consequence that $T_2\cup\{\phi\}$ is satisfiable, then for every sentence $\phi$ that is a conjunction of sentences of $T_1$ we have $T_2\cup\{\phi\}$ is satisfiable and through compactness $T_1$ and $T_2$ share a model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\phi$ be such that $T_1\models\phi$ and $T_2\models\lnot\phi$. Then $\phi$ is an axiomatization of $T_1$ and $\lnot\phi$ is an axiomatization of $T_2$. This is because, if a model satisfies $\phi$ then it can't satisfy $T_2$ and therefore satisfies $T_1$ and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the answer I came up with, but maybe there's a more direct way to it.&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2010-12-04T23:24:24.127" CommentCount="1" />
  <row Id="13048" PostTypeId="2" ParentId="13026" CreationDate="2010-12-04T23:39:38.327" Score="3" ViewCount="0" Body="&lt;p&gt;There are many types of &quot;problems&quot;.  Not all types of problems correspond to languages, but some do.  Here are some relevant examples.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Decision_problem&quot; rel=&quot;nofollow&quot;&gt;decision problem&lt;/a&gt;&lt;/em&gt; is the problem of deciding membership in some set. Usually in CS the set is a set of natural numbers or strings. Any such problem can be represented as a language in the most general sense. In the notes you linked, they only consider languages on sets of strings over a finite alphabet; these languages correspond to a certain type of decision problem, which is related to the usual reducibilities such as Turing reducibility and P-time reducibility. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Function_problem&quot; rel=&quot;nofollow&quot;&gt;function problem&lt;/a&gt;&lt;/em&gt; is a problem that asks you to be able to compute a specific function. For example, given an integer, produce an explicit prime factorization. These problems do not directly correspond to languages, but each function problem can be converted into the related decision problem of the graph of the function. When the underlying set is countable, this conversion preserves Turing degree.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;em&gt;mass problem&lt;/em&gt; is a more general type of problem that is represented by a subset of $\omega^\omega$. The &quot;problem&quot; is to produce some element of the set. These problems correspond to Medvedev reducibility and Muchnik reducibility.  They do not correspond to decision problems for sets of strings on finite alphabets. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-04T23:39:38.327" CommentCount="1" />
  <row Id="13049" PostTypeId="2" ParentId="13003" CreationDate="2010-12-05T01:29:32.397" Score="1" ViewCount="0" Body="&lt;p&gt;I would start by pointing out that parallelogram has two points farther apart than any other pair-two of the vertices.  Any symmetry must leave them in place or exchange them.  To leave them in place, the axis has to go through them.  To exchange them, the axis has to be the perpendicular bisector of the segment between them.  By inspection, neither is a symmetry.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T01:29:32.397" />
  <row Id="13050" PostTypeId="1" AcceptedAnswerId="13052" CreationDate="2010-12-05T02:16:37.627" Score="6" ViewCount="405" Body="&lt;p&gt;I was reading the following article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrafinitism&quot; rel=&quot;nofollow&quot;&gt;Ultrafinitism&lt;/a&gt;, and it mentions that one of the reasons ultrafinitists believe that &lt;strong&gt;N&lt;/strong&gt; is not infinite is because the floor of $e^{e^{e^{79}}}$ is not computable. I was wondering if that's the case because of technological limitations, or whether there is another reason we cannot find a floor of this number.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-05T12:53:08.707" Title="$e^{e^{e^{79}}}$ and ultrafinitism" Tags="&lt;finitism&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" />
  <row Id="13052" PostTypeId="2" ParentId="13050" CreationDate="2010-12-05T02:37:36.223" Score="16" ViewCount="0" Body="&lt;p&gt;In the formal meaning of &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Computable_number&quot; rel=&quot;nofollow&quot;&gt;computable&lt;/a&gt;&quot; the floor of that number is indeed computable. This is to say that a patient immortal human with access to unlimited paper and pencil could, in principle, work out the answer. (Here I assume, for technical reasons, that the number in question is not an integer - I assume someone who knows enough number theory will be able to cite a result that implies this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article linked makes the weaker claim that the value has not yet been calculated, which seems likely to me. The issue they are concerned about is that humans are not immortal and that our supply of paper is very limited. If the number of decimal digits in the value is too large, it would be impossible to actually represent it in any physical way within our universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I think it is more accurate to say that ultrafinitists don't accept that the set of all natural numbers is a coherent entity - not that they think it is finite. However, as the article you linked alludes, it is very difficult to find a coherent but non-arbitrary way to say what natural numbers are without accepting that there are an infinite number of them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Addendum&lt;/em&gt;&lt;/strong&gt; Here is why I am worried whether $e^{e^{e^{79}}}$ is an integer. It's certainly correct that no matter what, the floor of that number is an integer and is therefore computable. That part of my argument is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $e^{e^{e^{79}}}$ is not an integer, then I can tell you a specific algorithm to use to compute it. Namely, compute better and better upper and lower bounds until they fall strictly between two consecutive integers (which they must, since their limit is not an integer) and then pick the smaller of those two integers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $e^{e^{e^{79}}}$ is an integer, then that algorithm won't work, because it will never stop. But if we knew that $e^{e^{e^{79}}}$ was an integer then we could take better and better upper and lower approximations until they straddle a single integer, and then pick that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the reason that I am interested whether the number is an integer is that, beyond merely knowing that the floor is an integer, I'd like to know which algorithm could be used to compute it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I don't think that the point of the example was to pick a number that is not known to be integer or known to not be an integer. The point of the example should be to pick a number which is simply too large to represent physically.  I was hoping that someone would have a quick answer that confirms $e^{e^{e^{79}}}$ is not an integer, so I could edit my response with that info. But the non-integer property seems more difficult than I thought. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-05T12:53:08.707" LastActivityDate="2010-12-05T12:53:08.707" CommentCount="6" />
  <row Id="13053" PostTypeId="1" AcceptedAnswerId="14407" CreationDate="2010-12-05T02:46:48.630" Score="3" ViewCount="242" Body="&lt;p&gt;The following is a plot comparing &lt;code&gt;Exp[Derivative[1,0][Zeta][0,x]+1/2Log[2 Pi]]&lt;/code&gt; and &lt;code&gt;Gamma[x]&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Adlxh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory the blue and the red graphs should be identical, giving the Gamma function. But in reality we have the blue plot (the plot of &lt;code&gt;Exp[Derivative[1,0][Zeta][0,x]+1/2Log[2 Pi]]&lt;/code&gt;) representing the absolute value of the Gamma function instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: is this my fault (bad formula); and if so, how then do I correct it? Or it is a &lt;em&gt;Mathematica&lt;/em&gt; bug?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:22:56.877" LastActivityDate="2010-12-15T11:34:42.273" Title="Is it problem of Mathematica or my own?" Tags="&lt;math-software&gt;&lt;special-functions&gt;&lt;gamma-function&gt;&lt;riemann-zeta&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="13054" PostTypeId="1" CreationDate="2010-12-05T02:47:36.093" Score="26" ViewCount="836" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/13050/eee79-and-ultrafinitism/&quot;&gt;this question&lt;/a&gt;, I needed to assume in my answer that $e^{e^{e^{79}}}$ is not an integer. Is there some standard result in number theory that applies to situations like this? &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-23T05:54:38.943" Title="How to show $e^{e^{e^{79}}}$ is not an integer" Tags="&lt;number-theory&gt;&lt;transcendence-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="10" />
  <row Id="13055" PostTypeId="1" CreationDate="2010-12-05T03:08:06.427" Score="5" ViewCount="175" Body="&lt;p&gt;In some math job advertisments, it shows &quot;the deadeline is...&quot;, but to surprised some schools review applications before that, does that mean we submit application as earlier as possible could be in a better position? (because I normally do that when date is close to the deadline). (I just want to make sure I understand things correctly). Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-12-07T07:10:53.883" Title="deadline in math jobs application" Tags="&lt;soft-question&gt;" AnswerCount="2" />
  <row Id="13056" PostTypeId="2" ParentId="13055" CreationDate="2010-12-05T03:21:50.113" Score="10" ViewCount="0" Body="&lt;p&gt;I attended a panel talk about this recently and the uniform opinion of the panelists was that one should submit job applications as soon as possible regardless of the deadline. Some schools start reviewing applications much earlier than the deadline (this includes my university). Of course, as long as you send in your application before the deadline, it will be looked at. The reason the panelists mentioned for early applications is that review committees tend to swamped just before and after the deadline. So if you get in your application early, there is more chance that you will get a more through look and if you are a candidate who might otherwise be on the fringe, this might work in your favor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a useful article in the AMS: (page 1024 answers your question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/notices/200609/fea-lewis.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/notices/200609/fea-lewis.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quoting the relevant part for convenience:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When Do I Apply for Jobs?&#xA;  Most deadlines are between December 1 and January&#xA;  15. For nonpostdoctoral positions, it is very&#xA;  helpful to send in your application by mid-&#xA;  December even if the deadline is much later, especially&#xA;  if you will be at the Joint Mathematics&#xA;  Meetings in January. For schools that conduct interviews&#xA;  at the JMM, you are better off if you have&#xA;  already applied and the schools have had the&#xA;  chance to look over your application. In mid-December,&#xA;  you may want to email the schools and remind&#xA;  them that you will be at the JMM and would&#xA;  be happy to meet with them there. If you are giving&#xA;  a talk, this also lets you send them the title, time,&#xA;  and location if you didn’t know those earlier. A&#xA;  short polite email, especially if you mention&#xA;  specifics about the individual school, may lead directly&#xA;  to an interview and is unlikely to bother anyone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-05T04:39:54.773" LastActivityDate="2010-12-05T04:39:54.773" CommentCount="1" />
  <row Id="13058" PostTypeId="1" AcceptedAnswerId="13081" CreationDate="2010-12-05T04:13:49.767" Score="2" ViewCount="63" Body="&lt;p&gt;Let $$D_N(x)=\frac{\sin [(N+(1/2))t]}{\sin (t/2)}$$ be the Dirichlet kernel. Let $x(N)$ be the number in $0&amp;lt;x&amp;lt;\pi/N$ such that $D_N(x)=1$. Is $$\left|\int_{x(N)}^{\pi/N} D_N(t)\mathrm dt \right|=O\left(\frac1{N}\right)$$ true?&#xA;This question arises from my attempt to give a rigorous proof of Gibbs phenomenon. In fact, I only need that the limit is 0 as $N\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T05:36:58.337" LastActivityDate="2010-12-05T09:06:44.097" Title="A limit related to the Gibbs phenomenon" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="1" />
  <row Id="13060" PostTypeId="2" ParentId="12809" CreationDate="2010-12-05T05:08:55.283" Score="0" ViewCount="0" Body="&lt;p&gt;We want to show that &#xA;$$&#xD;&#xA;{\rm P}[X \ge {\rm E}(X|U) - 1/m] \ge {\rm P}(U)/m.&#xD;&#xA;$$&#xD;&#xA;This is trivially not true if $0 &amp;lt; m &amp;lt; 1$, for if $U = \Omega$, the right-hand side is greater than $1$. So, assume that $m \geq 1$, and ${\rm P}(U)&amp;gt;0$. Further, if ${\rm E}(X|U) - 1/m \leq 0$, then the inequality is trivial; hence we assume that&#xA;${\rm E}(X|U) - 1/m &amp;gt; 0$. Let $P_{X|U}$ denote the conditional distribution of $X$ given $U$, so that&#xA;$$&#xD;&#xA;P_{X|U} {(A)} = {\rm P}(X \in A | U) = \frac{{{\rm P}([X \in A],U)}}{{{\rm P}(U)}}.&#xD;&#xA;$$&#xD;&#xA;Then,&#xA;$$&#xD;&#xA;{\rm E}(X|U) = \int_{[0,1]} {xP_{X|U} ({\rm d}x)}. &#xD;&#xA;$$&#xD;&#xA;Next, decompose $[0,1]$ into the union of the intervals $I_1 = \lbrace 0 \leq x &amp;lt; {\rm E}(X|U) - 1/m \rbrace$ and $I_2 = \lbrace {\rm E}(X|U) - 1/m \leq x \leq 1 \rbrace$. On the one hand,&#xA;$$&#xD;&#xA;\int_{I_1 } {xP_{X|U} ({\rm d}x)}  \le  [{\rm E}(X|U) - 1/m]P_{X|U} (I_1 ) \le {\rm E}(X|U) - 1/m,&#xD;&#xA;$$ &#xA;and on the other hand, &#xA;$$&#xD;&#xA;\int_{I_2 } {xP_{X|U} ({\rm d}x)} \leq  \int_{I_2 } {P_{X|U} ({\rm d}x)} = P_{X|U} {(I_2)} \leq \frac{{{\rm P}(X \in I_2 )}}{{{\rm P}(U)}}.&#xD;&#xA;$$&#xD;&#xA;Combining it all, we get&#xA;$$&#xD;&#xA;{\rm E}(X|U) \leq {\rm E}(X|U) - 1/m + {\rm P}(X \in I_2)/{\rm P}(U).&#xD;&#xA;$$&#xD;&#xA;That is ${\rm P}(X \in I_2) \geq {\rm P}(U)/m$, which is exactly what we want.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-05T05:08:55.283" />
  <row Id="13062" PostTypeId="1" AcceptedAnswerId="13066" CreationDate="2010-12-05T06:29:28.217" Score="2" ViewCount="168" Body="&lt;p&gt;Let $G$ be a finitely generated group and $H$ a subgroup of $G$. If the index of $H$ in $G$ is finite, show that $H$ is also finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T09:08:17.637" LastActivityDate="2010-12-06T18:05:35.107" Title="subgroups of finitely generated groups with a finite index." Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13063" PostTypeId="1" CreationDate="2010-12-05T06:47:33.047" Score="1" ViewCount="25" Body="&lt;p&gt;I know a little bit of probability; but I do not know anything from dynamical systems. I want to understand the notion of Lyapunov exponent for a certain sequence of random variables arising out of a dynamical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the wikipedia article on Lyapunov exponent is not giving any suitable references. I would be most grateful if someone can give some helpful pointers.&lt;/p&gt;&#xA;" OwnerUserId="4427" LastActivityDate="2010-12-05T07:59:01.827" Title="Lyapunov exponent reference" Tags="&lt;reference-request&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13065" PostTypeId="1" AcceptedAnswerId="13070" CreationDate="2010-12-05T06:56:27.863" Score="1" ViewCount="34" Body="&lt;p&gt;I don't understand an equation I am reading in my notes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, $||$ is a nonarchimedean absolute value on a field $K$ complete wrt this absolute value. Suppose, $|a_0|&amp;gt;|a_i|$ for all $i&amp;gt;0$. Then, $|a_0+a_1+a_2+...|=|a_0|$. I understand that $|a_0+a_1+a_2+...|\leq max\{|a_i|:i\geq 0\}=|a_0|$ by ultrametric inequality. I am not sure how to get the reverse inequality.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="3947" LastEditorDisplayName="" LastEditDate="2010-12-05T07:17:06.090" LastActivityDate="2010-12-05T07:17:06.090" Title="A question regarding non archimedean absolute values" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="13066" PostTypeId="2" ParentId="13062" CreationDate="2010-12-05T06:58:59.920" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Suppose $G$ has generators $g_1, \ldots, g_n$. We can assume that the inverse of each generator is a generator. Now let $Ht_1, \ldots, Ht_m$ be all right cosets, with $t_1 = 1$. For all $i,j$, there is $h_{ij} \in H$ with $t_i g_j = h_{ij} t_{{k}_{ij}}$, for some $t_{{k}_{ij}}$. It's not hard to prove that $H$ is generated by all the $h_{ij}$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T07:31:51.330" LastActivityDate="2010-12-05T07:31:51.330" CommentCount="4" />
  <row Id="13067" PostTypeId="1" CreationDate="2010-12-05T07:02:17.123" Score="4" ViewCount="169" Body="&lt;p&gt;When I read some issues here I see from time to time incorrect references to the field &lt;em&gt;special functions&lt;/em&gt;, it might e.g. be a discussion around Dirac's $\delta$-function which is tagged (special-functions) or a discussion around some function reminding of the Weierstrass no where differentiable continuous function. These examples make me think - what would classify a special function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vague bad definition could be &quot;A function is a special function if it has some resemblance to some Hypergeometric function&quot; or &quot;A function is a special function if it fits into the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bateman_manuscript_project&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Bateman manuscript project&lt;/em&gt;&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;Gamma function&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Zeta function&lt;/a&gt; are definitively special functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have worked on &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_function&quot; rel=&quot;nofollow&quot;&gt;Legendre functions&lt;/a&gt; $P_\lambda$ and $Q_\lambda$ of the first and second kind , which I would call special functions, but not individually however (except perhaps $P_{-1/2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not say that elementary functions (such as trigonometric functions and the exponential function) are special functions - but I am not totally convinced about this..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not agree with &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_functions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, it says: &#xA;&quot;Special functions are particular mathematical functions which have more or less established names and notations due to their importance in mathematical analysis, functional analysis, physics, or other applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no general formal definition, but the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_mathematical_functions&quot; rel=&quot;nofollow&quot;&gt;list of mathematical functions&lt;/a&gt; contains functions which are commonly accepted as special. In particular, elementary functions are also considered as special functions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, looking at the Wikipedia list (linked above) the indicator function, step functions, the absolute value function and the sign are special functions -- this sounds very wrong to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is a special function and what should be under the (special-function) tag?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-06T06:19:27.813" Title="What is a special function?" Tags="&lt;special-functions&gt;&lt;definition&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="13068" PostTypeId="1" AcceptedAnswerId="13085" CreationDate="2010-12-05T07:07:58.413" Score="2" ViewCount="114" Body="&lt;p&gt;Evaluate the integral as a power series:&lt;br&gt;&#xA;$\int(x^{11}(\tan^{-1}(x^2))\mathrm dx)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have been using Abel's theorem to do this (and the fact that the function is differentiable and integrable on it's interval of convergence in this case)&#xA;From what I can tell, an approach to this question would be to find a closed form similar to $\frac1{1-x}$ by taking the derivative of $\tan^{-1}(x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so let $f(x) = \int(x^{11}(\tan^{-1}(x^2))\mathrm dx)$&lt;br&gt;&#xA;then let $t = x^2, g(t) = f&amp;#39;(x^2)$ &#xA;then divide by $x^{11/2}$,&lt;br&gt;&#xA;$\frac{g(t)}{t^{11/2}} = \tan^{-1}(t)$&lt;br&gt;&#xA;then take the derivative of both sides so&lt;br&gt;&#xA;derivative of $\frac{g(t)}{t^{11/2}} = \frac1{t^2+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am stuck here relating this to $\frac1{1-x}$.  Maybe I am looking at this the wrong way.  Any help would be appreciated. Thanks&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="2533" LastEditorDisplayName="" LastEditDate="2010-12-06T07:38:07.793" LastActivityDate="2010-12-06T07:38:07.793" Title="Evaluate the integral as a power series $\int(x^{11}(\tan^{-1}(x^2))\mathrm dx)$" Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13069" PostTypeId="2" ParentId="12488" CreationDate="2010-12-05T07:10:16.833" Score="2" ViewCount="0" Body="&lt;p&gt;To suppliment Moron's comment: yes, using plane embedding algorithm will do. One of such algorithm is &lt;a href=&quot;http://jgaa.info/accepted/2004/BoyerMyrvold2004.8.3.pdf&quot; rel=&quot;nofollow&quot;&gt;Boyer and Myrvold&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-12-05T07:10:16.833" />
  <row Id="13070" PostTypeId="2" ParentId="13065" CreationDate="2010-12-05T07:14:33.407" Score="2" ViewCount="0" Body="&lt;p&gt;Let $a = \sum a_i$. You can apply the argument to $a_0 = a- (a_1 + \dots + a_n)$. We have then that &#xA;$$ |a_0| \leq \max(|a|, |a_1 + \dots a_n|) = |a|$$&#xD;&#xA;because $|a_0|$ is bigger than $|a_1 + \dots + a_n|$, and consequently the second term cannot contribute to the maximum. This implies that $|a| \geq |a_0|$, which is the reverse inequality you wanted.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-05T07:14:33.407" CommentCount="3" />
  <row Id="13072" PostTypeId="1" CreationDate="2010-12-05T07:52:18.373" Score="0" ViewCount="32" Body="&lt;p&gt;Let $R$ be a unitary ring, $\mathfrak J$ an ideal of $R$ such that ($\mathfrak J+G=\{a+g\mid a\in\mathfrak J,\ g\in G\}$, and $R^\times$ is obviously the group of unity of $R$.) $\mathfrak J+G\le R^\times$ for any $G\le R^\times$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $G$ is said to be $\mathfrak J-$ &lt;em&gt;seminal&lt;/em&gt; iff $\mathfrak J+G=R^\times$. Let now $\Sigma_R(\mathfrak J)$ be the set of all $\mathfrak J-$seminal subgroups of $R^\times$, and let $\sigma_R(\mathfrak J)$ denote the intersection of all elements in $\Sigma_R(\mathfrak J)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set $\sigma_R(\mathfrak J)$ is the &lt;em&gt;seed&lt;/em&gt; of $R$ &lt;em&gt;rooted&lt;/em&gt; at the ideal $\mathfrak J$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the set &#xA;$$&#xD;&#xA;N_Z(R)=\mathcal Z(R)\cap \mathcal N(R)&#xD;&#xA;$$ &#xA;intersection of the center of $(R,\cdot)$ with the set of all nilpotent elements (which coincides with $\bigcap \text{Spec}(R)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $N_Z(R)-$seminal ideals of $R$ are said to be its &lt;em&gt;nilseminal&lt;/em&gt; ideals.&#xA; The seed of $R$ rooted at $N_Z(R)$ is said to be the &lt;em&gt;nilseed&lt;/em&gt; of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring $R$ is said to be &lt;em&gt;nilseminal&lt;/em&gt; iff its nilseed is an element of $\Sigma_R(N_Z(R))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the key step is to show that in fact &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$N_Z(R)+G\le R^\times$ for any $G\le R^\times$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other questions (those marked with ``true'' have been solved) which involve more general rings are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true A) If $n$ is a squarefree integer, then $\mathbb Z/n\mathbb Z$ is nilseminal;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;true B) If $n=p^\alpha$ for some $\alpha\in\mathbb N$, then $\mathbb Z/n\mathbb Z$ is nilseminal;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $R$ and $S$ be isomorphic rings via $\phi\colon R\to S$. Prove or disprove that for any ideal $\mathfrak J$ of $R$, a subgroup $G\le R^\times$ is $\mathfrak J-$seminal iff $\phi(G)$ is $\phi(\mathfrak J)-$seminal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a ring $R$ with nilpotent elements of any positive order be nilseminal? Can a nilseminal ring have nilpotents of any positive order?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $S$ be a unitary subring of a ring $R$. Prove or disprove that $R$ is nilseminal only if $S$ is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prove or disprove that ``being nilseminal'' is a local property, i.e. $R$ is nilseminal iff $R_\mathfrak p$ is nilseminal for any $\mathfrak p\in \text{Spec}(R)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-12-07T11:41:44.787" LastActivityDate="2010-12-07T11:41:44.787" Title="Properties of (nil)seminal rings" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" CommentCount="1" />
  <row Id="13073" PostTypeId="2" ParentId="13063" CreationDate="2010-12-05T07:59:01.827" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=FIYHiBLWCJMC&amp;amp;printsec=frontcover&amp;amp;dq=nonlinear+dynamics&amp;amp;hl=en&amp;amp;ei=2kX7TKeKFISClAeVs5GlDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;sqi=2&amp;amp;ved=0CCoQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Non Linear Dynamics and Chaos&lt;/a&gt; by Strogatz, page 322.  I read this a long time ago, and the whole book is worth reading in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastActivityDate="2010-12-05T07:59:01.827" />
  <row Id="13074" PostTypeId="2" ParentId="13068" CreationDate="2010-12-05T07:59:49.040" Score="2" ViewCount="0" Body="&lt;p&gt;$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)}$.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-05T07:59:49.040" CommentCount="1" />
  <row Id="13075" PostTypeId="1" AcceptedAnswerId="13079" CreationDate="2010-12-05T08:12:02.450" Score="1" ViewCount="37" Body="&lt;p&gt;Hi i am getting confused with this diagram &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://img35.imageshack.us/img35/8175/picture1654x310.jpg&quot; rel=&quot;nofollow&quot;&gt;http://img35.imageshack.us/img35/8175/picture1654x310.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bold line is a vector. I haven't worked much with vectors, but it looks similar to a linear graph (polynomial degree 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would i be correct in saying these are the differences: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working with linear graphs (polynomial degree 1) we deal with multiple x values and multiple y values. But vectors are different to linear graphs because we only deal with the length of the line (and direction) and the length of the projected lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that a linear graph (polynomial degree 1) represents multiple times. But a vector only represents an instant in time (but you can have multiple vectors for each time).&lt;/p&gt;&#xA;" OwnerUserId="4429" LastActivityDate="2010-12-05T09:03:51.810" Title="What are the differences between linear graphs (polynomial degree 1) and vectors " Tags="&lt;vector-analysis&gt;" AnswerCount="2" />
  <row Id="13076" PostTypeId="2" ParentId="13075" CreationDate="2010-12-05T08:18:35.780" Score="1" ViewCount="0" Body="&lt;p&gt;The vector you are referring here is a Euclidean vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_vector&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euclidean_vector&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A degree 1 polynomial is a straight line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Line_(geometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Line_(geometry&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-05T08:18:35.780" />
  <row Id="13077" PostTypeId="2" ParentId="13029" CreationDate="2010-12-05T08:22:48.990" Score="1" ViewCount="0" Body="&lt;p&gt;One more way is to note that it's just the AM-HM:&#xA;$$\frac{a_1 + a_2 + a_3}{3} \ge \frac{3}{\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it's useful to remember the power mean inequality: Suppose that $\alpha \ge \beta$. Then&#xA;$$\left(\frac{a_1^\alpha + \dots + a_n^\alpha}{n}\right)^{1/\alpha} \ge \left(\frac{a_1^\beta + \dots + a_n^\beta}{n}\right)^{1/\beta}.$$ The inequality works also in the case that $\alpha$&amp;#160;or $\beta$&amp;#160;is zero, when we take that to mean the geometric mean.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-05T08:22:48.990" CommentCount="2" />
  <row Id="13079" PostTypeId="2" ParentId="13075" CreationDate="2010-12-05T08:32:35.367" Score="0" ViewCount="0" Body="&lt;p&gt;The graphs of degree one polynomial functions are the same as non-vertical lines. Vectors are not lines; as you indicate, they consist of direction and (finite) length (and, depending on context, a base point).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a close relationship between the two.  A line is determined by taking a point in the plane and then choosing the direction to travel from that point.  This direction can be specified by a vector.  So instead of thinking of a line as the graph of a degree one polynomial function you could think of it as the set of all points $p+tv$, where $p$ is point in the plane, $v$ is a nonzero vector, and $t$ ranges over the real numbers.  The parameter $t$ tells you how much you travel in the $v$ direction (technically, divided by the length of $v$), with the understanding that when $t$ is negative you are going $|t|$ much in the opposite direction.  This is actually more general than thinking of graphs of functions, because you could also describe vertical lines by choosing $v$ to point straight up or down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: A vector and a starting point determine a line, and a line determines a direction vector up to multiplication by a nonzero real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be concrete, consider $f(x)=2x+3$.  The point $p=(0,3)$ is on the graph, and the slope $2$ determines the direction.  This direction can be described by the vector $v$ with $x$ component $1$ and $y$ component $2$.  We can write $v=(1,2)$ to summarize this information. Here going $t$ in the $v$ direction from $p$ takes you to the point $p+tv=(0,3)+(t,2t)=(t,2t+3)$, which is the same as the point $(t,f(t))$ on the graph of $f$.  Notice that the slope of the line is the ratio of the $y$ component to the $x$ component of the vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, say we take the point $p=(3,-1)$ and the vector $v=(-3,6)$.  The line through $p$ in the $v$ direction consists of all of the points $p+tv=(3,-1)+(-3t,6t)=(3-3t,6t-1)$ as $t$ ranges over the real numbers.  To determine the corresponding function, just find the slope-intercept form of the equation of the line through the point $(3,-1)$ with slope $\frac{6}{-3}=-2$, namely $y=-2x+5$, so that $f(x)=-2x+5$ is the desired function.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-05T09:03:51.810" LastActivityDate="2010-12-05T09:03:51.810" />
  <row Id="13080" PostTypeId="1" AcceptedAnswerId="13101" CreationDate="2010-12-05T08:44:20.740" Score="1" ViewCount="84" Body="&lt;p&gt;I have a sequence of  points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(x1, y1), (x2, y2), (x3, y3)....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would like to know is it exponential sequence and how to get expression of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/0PHR0.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4430" LastEditorUserId="4430" LastEditorDisplayName="" LastEditDate="2010-12-05T09:26:11.593" LastActivityDate="2010-12-05T12:07:10.690" Title="How to detect exponent" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="13081" PostTypeId="2" ParentId="13058" CreationDate="2010-12-05T09:06:44.097" Score="2" ViewCount="0" Body="&lt;p&gt;Looks true to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x(N) = \dfrac{\pi}{N+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\displaystyle D_n(x) = 1 + 2\sum_{k=1}^{n} \cos(kx)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\displaystyle |\sin x - \sin y| \leq |x-y|$ (easily seen using Mean Value Theorem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \left|\int_{\pi/(N+1)}^{\pi/N} D_N(t) \ \text{dt}\right| = \left|\int_{\pi/(N+1)}^{\pi/N} 1 + 2\sum_{k=1}^{N} \cos(kx)  \ \text{dt}\right|$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle = \frac{\pi}{N(N+1)} + 2 \sum_{k=1}^{N} \mathcal{O}(\frac{1}{N^2})  = \mathcal{O}(\frac{1}{N})$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-05T09:06:44.097" CommentCount="1" />
  <row Id="13085" PostTypeId="2" ParentId="13068" CreationDate="2010-12-05T09:15:16.733" Score="6" ViewCount="0" Body="&lt;p&gt;Recall that&#xA;$$\tan^{-1}x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}.$$&#xD;&#xA;Use this to find a power series expansion for the integrand,&#xA;and integrate it termwise.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-05T09:15:16.733" CommentCount="2" />
  <row Id="13087" PostTypeId="2" ParentId="13080" CreationDate="2010-12-05T09:24:38.310" Score="4" ViewCount="0" Body="&lt;p&gt;Set $Y=\ln y$. If the points $(x_k,y_k)$ lie on the curve $y=a e^{bx}$, then the points $(x_k,Y_k)$ lie on the straight line $Y=(\ln a) + bx$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-05T09:24:38.310" CommentCount="2" />
  <row Id="13088" PostTypeId="1" CreationDate="2010-12-05T09:36:08.783" Score="2" ViewCount="58" Body="&lt;p&gt;Hi, I need some help with my homework in probability.&#xA;I need to prove that if&#xA;$X(n) =$ the number of different coupons that the collector has in time $n$&#xD;&#xA;then $X(n)$ represents a Markov chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proved that &#xA;$$P(X(m+1)=j)= \left\{\begin{array}{ll}&#xA;\frac{X(m)}{n}&amp;amp;\text{if $j=X(m)$}\\&#xA;1-\frac{X(m)}{n}&amp;amp;\text{if $j=X(m)+1$}\\&#xA;0 &amp;amp;\text{otherwise}&#xA;\end{array}\right.$$&#xA;Now I need to show from there that $P(X(m+1=j)|X(m)=Km,\ldots ,X(0)=K0)=P(X(m+1)=j|X(m)=Km)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for the help.&#xA;benny.&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-05T21:59:31.023" LastActivityDate="2010-12-05T21:59:31.023" Title="coupon collector and Markov chains" Tags="&lt;homework&gt;&lt;probability-theory&gt;" CommentCount="1" />
  <row Id="13089" PostTypeId="2" ParentId="13003" CreationDate="2010-12-05T09:43:15.270" Score="2" ViewCount="0" Body="&lt;p&gt;If a symmetry pairs two vertices, then the angles at those vertices must be congruent. A symmetry that pairs &lt;em&gt;adjacent&lt;/em&gt; vertices of a parallelogram requires that the figure be a rectangle: adjacent angles are supplementary; if they're also congruent, then they must be right angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having ruled out rectangles, your proposed symmetry must pair each vertex (a) with its non-adjacent counterpart, or (b) with itself. In (a), the axis would be the perpendicular bisector of the diagonal between the paired vertices; in (b) the axis would contain the vertex. If both (a) and (b) occur, then we have a vertex (from (b)) on the perpendicular bisector of the diagonal joining two non-adjacent others (from (a)); this implies that we have two congruent adjacent edges, which in turn implies that the figure is a rhombus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid both rectangles and rhombi, we must have that the axis provides only-(a) symmetry, or only-(b) symmetry. An axis of only-(a) symmetry must be the perpendicular bisector of both diagonals, which (as perpendiculars of that axis, and &lt;em&gt;lying in the same plane of that axis&lt;/em&gt;) must therefore lie on the same line: the vertices of the figure are collinear. An axis of only-(b) symmetry contains all the vertices, making them collinear (but in a different arrangement than required by (a)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you disallow &quot;degenerate&quot; parallelograms with all vertices collinear, then you have your impossibility argument. However, I prefer to recognize degenerate figures as legitimate, and I try to avoid stigmatizing them whenever possible. (After all, they come in handy as intermediate steps in transforming one figure smoothly into another. Besides, it's not like the &quot;Parallelogram Law of Vector Addition&quot; becomes invalid when the vectors involved are linearly dependent.) So, I would go so far as to observe that an axis of only-(a) symmetry implies the existence of a (separate) axis of only-(b) symmetry, and vice versa; that is, we've determined a third class of parallelograms --along with rectangles and rhombi-- that have &lt;em&gt;two&lt;/em&gt; axes of symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Consideration of parallelograms whose vertices coincide is left to the reader.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: An only-(a) axial symmetry can also be realized with an axis perpendicular to the plane of the parallelogram (which need not be degenerate), but you've ruled this out.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-12-05T09:55:37.517" LastActivityDate="2010-12-05T09:55:37.517" />
  <row Id="13090" PostTypeId="1" CreationDate="2010-12-05T09:43:55.817" Score="2" ViewCount="109" Body="&lt;p&gt;I need help this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;prove that every finite ring has the Invariant Dimension Property(IDP).( Assume $1_R \neq 0_R$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I know I should do.  Let $X$ and $Y$ be two sets such that the free module with basis $X$ is isomorphic to the free module with basis $Y$. Next, I have to show that $|X| = |Y|$. This is where I'm stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4433" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T09:48:26.370" LastActivityDate="2010-12-05T10:20:07.750" Title="Invariant dimension property" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13091" PostTypeId="1" CreationDate="2010-12-05T09:47:10.480" Score="4" ViewCount="71" Body="&lt;p&gt;Hello all, this is me again, I have another problem which I haven't been able to solve, the &#xA;legend goes like this:&lt;br&gt;&#xA;Find the equation of the plane that contains the points $P_1(1,0,-1)$, $P_2(2,0,2)$ and forms a 60&lt;sup&gt;0&lt;/sup&gt; ($\theta$) angle with the other plane $2x-2y+z+6=0$ (Two Solutions).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we already know two points, P&lt;sub&gt;1&lt;/sub&gt;(1,0,-1) and P&lt;sub&gt;2&lt;/sub&gt;(2,0,2) they both satisfy a planes equation:&lt;br&gt;&#xA;$$Ax+By+Cz+D=0$$&#xD;&#xA;So, we substitute $x,y,z$ respectively for each point, giving use these two equations&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$A(1)+B(0)+C(-1)+D=0$&lt;/li&gt; &#xA;&lt;li&gt;$A(2)+B(0)+C(2)+D=0$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Simplifying we get:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; $A-C+D=0$&lt;/li&gt; &#xA;&lt;li&gt; $2A+2C+D=0$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, according to the following theorem the angle that is formed by the normals of to planes is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos{\theta} = \pm \frac{AA&amp;#39;+BB&amp;#39;+CC&amp;#39;}{\sqrt{A^2+B^2+C^2} \sqrt{(A&amp;#39;)^2+(B&amp;#39;)^2+(C&amp;#39;)^2}}$$ Where $[A,B,C]$ Are the director numbers of the normal in the first plane and $[A&amp;#39;,B&amp;#39;,C&amp;#39;]$ Are the director numbers of the normal in the second plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given those and taking the known angle of 60&lt;sup&gt;o&lt;/sup&gt;, we substitute the director numbers of $2x-2y+z+6-0:[A&amp;#39;=2,B&amp;#39;=-2,C&amp;#39;=1]$ and the angle in the last formula, giving us:&#xA;$$\cos{60^\circ} = \pm \frac{A(2)-B(-2)+C(1)}{\sqrt{A^2+B^2+C^2} \sqrt{(2)^2+(-2)^2+(1)^2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying, we get(positive because $\theta$ is acute):&#xA;$$\frac{1}{2}=\frac{2A-2B+C}{3 \sqrt{A^2+B^2+C^2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another simplification finally for:&#xA;$$3(A^2+B^2+C^2)=(4A-4B+2C)^2$$&#xD;&#xA;Now, we have 3 equations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; $A-C+D=0$&lt;/li&gt; &#xA;&lt;li&gt; $2A+2C+D=0$&lt;/li&gt;&#xA;&lt;li&gt;$3(A^2+B^2+C^2)=(4A-4B+2C)^2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If we take 1) and 2) and write them as $D(A,C)$ we get:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt; &#xA;&lt;li&gt;$D=C-A$ &#xA;&lt;li&gt;$D=-2A-2C$&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From which we can obtain A in terms of C&#xA;$$-2A-2C=C-A$$&#xD;&#xA;$$A=-3C$$&#xD;&#xA;Now we substitute back in 1):&#xA;$$(-3C)-C+D=0$$&#xD;&#xA;$$C=\frac{1}{4}D$$&#xD;&#xA;Since I want every director number in terms of D we substitute again but in A since A was in function of C.&#xA;$$A=-3C$$&#xD;&#xA;$$A=-3(\frac{1}{4}D)$$&#xD;&#xA;$$A=\frac{-3}{4}D$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have found $A(D)$, $C(D)$ we have yet to find $B(D)$ I'll explain why I think and want everything in terms of D, if we take the general form of a planes equation $Ax+By+Cz+D=0$ and substitute $A$ and $C$ with our known values we get:&#xA;$$(\frac{-3}{4}D)x+By+(\frac{1}{4}D)z+D=0$$&#xD;&#xA;if we divide by $D \neq 0$ and then multiply by 4 to eliminate fractions:&#xA;$$\frac{-3}{4}x+\frac{B}{D}y+\frac{1}{4}z+1=0$$&#xD;&#xA;$$-3x+\frac{4B}{D}y+z+4=0$$&#xD;&#xA;From here it's clear that if we find B in terms of D and B is linear we have the planes equation, so, in 3) $3(A^2+B^2+C^2)=(4A-4B+2C)^2$ we substitute A and C:&#xA;$$3((\frac{-3}{4}D)^2+B^2+(\frac{1}{4}D)^2)=(4(\frac{-3}{4})-4B+2(\frac{1}{4}D))^2$$&#xD;&#xA;$$\vdots$$&#xD;&#xA;$$104B^2+80DB+35D^2=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is were I get stuck, I get imaginary parts, since I have to use the quadratic formula, so any help is really appreciated, the problem states that there are two solutions, but this is what I have thinked of if anyone can supply any solution I'll be very gratefull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards.. Tristian&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditorDisplayName="" LastEditDate="2010-12-07T07:34:55.500" LastActivityDate="2010-12-07T08:35:25.010" Title="Analytic Geometry | Two Planes and a Angle | Two Solutions" Tags="&lt;geometry&gt;&lt;homework&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13092" PostTypeId="2" ParentId="13090" CreationDate="2010-12-05T09:55:12.417" Score="1" ViewCount="0" Body="&lt;p&gt;Just count the number of elements. Since $R$ is finite, you can say exactly how many elements a free modulo of rank $k$ has. For finite $k$, that will do it. In the infinite case, you'll need to do a bit of cardinal work.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-05T09:55:12.417" />
  <row Id="13094" PostTypeId="1" AcceptedAnswerId="13095" CreationDate="2010-12-05T09:58:12.893" Score="7" ViewCount="393" Body="&lt;p&gt;There is a formula given in my module :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle\sqrt[n]{a^n} = a \text{ if n is odd } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \displaystyle\sqrt[n]{a^n} = |a| \text{ if n is even } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand the differences between them,kindly explain with an example.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-12-06T00:09:02.257" Title="Significance of $\displaystyle\sqrt[n]{a^n} $?!" Tags="&lt;elementary&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13095" PostTypeId="2" ParentId="13094" CreationDate="2010-12-05T10:01:07.087" Score="14" ViewCount="0" Body="&lt;p&gt;$\sqrt{(-1)^2} =\sqrt{1} = 1 = |-1|$ because in order to make the square root an unambiguous operation, we agree that &lt;strong&gt;the&lt;/strong&gt; square root of a nonnegative number $x$ is always &lt;strong&gt;the&lt;/strong&gt; (unique) nonnegative number $r$ such that $r^2=x$.  But with cubic roots there is no problem: $\sqrt[3]{-1}=-1$, because &lt;em&gt;every&lt;/em&gt; real number has a &lt;em&gt;unique&lt;/em&gt; cubic root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true with 4th, 6th, 8th, 10th, etc. powers, since $a^n = (-a)^n$, and the 4th, 6th, 8th, 10th, etc roots are defined to be the unique nonnegative real number that &quot;works&quot;, so that they are unambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, there are two numbers which when squared will given you the value $2^2$: both $2$ and $-2$. There are two number that when taken to the fourth power will give you $(-6)^4$: both $-6$ and $6$. And so on. Generally, both $a$ and $-a$ will, when raised to an even $n$th power, give the same answer: $a^n = (-a)^n$. And we agree that a square root (fourth root, sixth root, etc) will always be the nonnegative answer, so the $n$th root of $a^n$ will be $|a|$ when $n$ is even. (Don't let the big $-$ in &quot;$-a$&quot; fool you; that does not mean that $-a$ is negative, it just means the additive inverse of whatever $a$ is; if $a$ is positive, then $-a$ is negative, but if $a$ is negative, say $a=-3$, then $-a$ is positive, $-a = -(-3) = 3$. Repeat after me: the proper way to pronounce &quot;-a&quot; is not &quot;&lt;em&gt;negative a&lt;/em&gt;&quot;, the proper pronunciation is &quot;&lt;em&gt;minus a&lt;/em&gt;&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if $n$ is odd, then every number has a unique $n$th root. In particular, the only number that when cubed gives $2^3$ is $2$; the only number which, when raised to the fifth power, gives $(-6)^5$, is $-6$. There is no longer the problem that both $6$ and $-6$ are possible answers, so we can simply say that the cubic root of $(-2)^3$ is $-2$, the fifth root of $7^5$ is $7$, etc.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T00:09:02.257" LastActivityDate="2010-12-06T00:09:02.257" CommentCount="8" />
  <row Id="13096" PostTypeId="2" ParentId="13091" CreationDate="2010-12-05T10:04:02.950" Score="3" ViewCount="0" Body="&lt;p&gt;The normal vector to the sought plane is orthogonal to the&#xA;line $P_1 P_2$ and at angle $\pi/3$ to the normal to your given plane.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-05T10:04:02.950" CommentCount="6" />
  <row Id="13097" PostTypeId="1" CreationDate="2010-12-05T10:19:38.933" Score="1" ViewCount="72" Body="&lt;p&gt;I have two sets $M$ and $H$. $M$ is an arbitrary string of length $k$ and $H$ is an  string of length $p$. Both are constructed from a charset of length $r$. And $p&amp;lt;k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hash function $f(m)=h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $r^p$ hashes are possible and that $r^k-r^p$ collisions occur for the complete set $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can I predict the overall probability for finding any element of $M$ that correctly maps to a given element of $h$?&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T16:52:04.937" LastActivityDate="2010-12-05T16:53:46.627" Title="Understanding birthday attack probabilities" Tags="&lt;probability&gt;&lt;cryptography&gt;" AnswerCount="1" />
  <row Id="13098" PostTypeId="2" ParentId="13090" CreationDate="2010-12-05T10:20:07.750" Score="1" ViewCount="0" Body="&lt;p&gt;As $R$ is finite, it is Artinian, so the modules $R^m$ and $R^n$&#xD;&#xA;have finite length. You can now use the Jordan-Holder theorem&#xA;to deduce that $R^m\cong R^n$ implies $m=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can just count :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-05T10:20:07.750" />
  <row Id="13099" PostTypeId="2" ParentId="2611" CreationDate="2010-12-05T10:28:01.937" Score="4" ViewCount="0" Body="&lt;p&gt;It should intuitively feel like it's not (theoretically) possible -- the cost to play and the average win are determined by a bookie (or a casino, etc.), whereas the probability of winning is determined by the game itself.  Given two games, one with a probability of winning $p_1$ and the other with a probability of winning $p_2$, the bookie can adjust the returns on bets so as the cost of play and average win remain constant.  For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider a tossing coin game, the player bets 1 dollar, if the coin is &quot;heads&quot; it will return 2 dollars and if &quot;tales&quot; then there is no return.  So the expected win is +1-1=0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now consider a die rolling game, where the player bets 1 dollar, if the die rolls 6, then 6 dollars are returned, otherwise there is no return.  Here the expected win is +5-1-1-1-1-1=0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In both cases the cost to play is 1 dollar and in both cases the expected win is 0, but the probability of winning is different (1/2 vs. 1/6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, you might be able to infer an approximate probability of winning based on past experiences of the bookie, familiarity with the game being played, etc.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-05T10:28:01.937" />
  <row Id="13100" PostTypeId="2" ParentId="13062" CreationDate="2010-12-05T11:57:38.973" Score="1" ViewCount="0" Body="&lt;p&gt;Well, the standard argument is as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &#xA;$$&#xD;&#xA;g \mapsto [g] \qquad (g \in G)&#xD;&#xA;$$&#xD;&#xA;be a function which is constant on all right cosets of $H,$ and we require&#xA;$$&#xD;&#xA;[e]=e.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that&#xA;$$&#xD;&#xA;u [u]^{-1} \in H, \quad [[u]]=[u], \quad [[u]v]=[uv] \qquad (u,v \in G)  \qquad \qquad (*)&#xD;&#xA;$$&#xD;&#xA;Now let&#xA;$$&#xD;&#xA;S = \{ [g] : g \in G\}&#xD;&#xA;$$&#xD;&#xA;and $Y=Y^{-1}$ be a symmetric generating set of $G.$ Then the set&#xA;$$&#xD;&#xA;\{ s y [sy]^{-1} : s \in S, y \in Y\}&#xD;&#xA;$$&#xD;&#xA;is a generating set of $H$ (a &lt;i&gt;finite&lt;/i&gt; one, if both $S$ and $Y$ are finite $\iff$ the index of $H$ in $G$ is finite and $G$ is finitely generated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For suppose that a product $y_1 \ldots y_r$ is in $H$ where $y_k \in Y$ ($k=1,\ldots,r$). Let, for example's sake, $r=3.$ Then&#xA;$$&#xD;&#xA;y_1 y_2 y_3 = y_1 [y_1]^{-1} \cdot [y_1] y_2 [[y_1] y_2]^{-1} \cdot [[y_1] y_2] y_3&#xD;&#xA;[[[y_1] y_2] y_3]^{-1} \qquad \qquad (**)&#xD;&#xA;$$&#xD;&#xA;where in the right hand side we have a product of elements of $H$ by (*), since&#xA;$$&#xD;&#xA;[[[y_1] y_2] y_3]=[y_1 y_2 y_3]=e;&#xD;&#xA;$$&#xD;&#xA;the same $(*)$ also simplifies the right hand side of $(**)$ as&#xA;$$&#xD;&#xA;y_1 [y_1]^{-1} \cdot [y_1] y_2 [y_1 y_2]^{-1} \cdot [y_1 y_2] y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3 [y_1 y_2 y_3]^{-1}=y_1 y_2 y_3&#xD;&#xA;$$ &#xA;Now the induction step in general must be easy. &lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditorDisplayName="" LastEditDate="2010-12-06T18:05:35.107" LastActivityDate="2010-12-06T18:05:35.107" CommentCount="1" />
  <row Id="13101" PostTypeId="2" ParentId="13080" CreationDate="2010-12-05T12:07:10.690" Score="4" ViewCount="0" Body="&lt;p&gt;I suppose it's time to be putting my dough where my pie hole is. This'll be a tad long, and I hope your collective eyes don't glaze over too much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I'd approach the problem of fitting $y=ab^x$ to a given set of points contaminated with error is to do a &lt;em&gt;weighted&lt;/em&gt; linear regression, similar to Hans's proposal, but with a tiny wrinkle that is required for reasons similar to what I mentioned in this &lt;a href=&quot;http://math.stackexchange.com/questions/3625/3636#3636&quot;&gt;answer&lt;/a&gt; to a related question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our linearization here takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln\;y=\ln\;a+x\ln\;b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the supplied $y_i$ are in fact $\hat{y}_i\pm\sigma_i$ where the $\hat{y}_i$ are the &quot;true values&quot; obscured by the measurement errors $\sigma_i$ , any application of linearization should take into account that the errors as well as the &quot;true values&quot; are being transformed (in this case, by a logarithm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now use the fact that if the $y_i$ are transformed by a function $f(y)$, the $\sigma_i$ are transformed according to $f&amp;#39;(y_i)\sigma_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, the objective function that needs to be minimized for a weighted linear fit is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(a,b)=\sum_i{y_i^2\;(\ln\;y_i-\ln\;a-x_i\ln\;b)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the recipe for getting the required $a$ and $b$ now reads as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=\sum_i y_i^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{x}=\frac{\displaystyle \sum_i y_i^2 x_i}{m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\sum_i y_i^2 (x_i-\bar{x})^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=\exp\left(\frac{\displaystyle \sum_i y_i^2\ln\;y_i (x_i-\bar{x})}{t}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\exp\left(\frac{\displaystyle \sum_i y_i^2\ln\;y_i}{m}-b\bar{x}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that to do a nonlinear fit, we don't even need a provisional value for the linear parameter $a$; we can set things up so that the nonlinear expression we have to minimize only involves $b$. We can then use the $b$ returned by the above formula as a seed, have a secant or Newton-Raphson iteration polish that provisional $b$, and use the polished $b$ to get a proper $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skipping the details (which can be adapted from the derivation I gave in the answer I linked to), the nonlinear equation that has to be solved for $b$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\sum_i y_i b^{x_i}\right)\left(\sum_i x_i b^{2x_i-1}\right)-\left(\sum_i x_i y_i b^{x_i-1}\right)\left(\sum_i b^{2x_i}\right)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\left(\sum_i y_i b^{x_i}\right) / \left(\sum_i b^{2x_i}\right)$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-05T12:07:10.690" CommentCount="4" />
  <row Id="13102" PostTypeId="2" ParentId="13028" CreationDate="2010-12-05T12:26:08.600" Score="4" ViewCount="0" Body="&lt;p&gt;The union $T_1\cup T_2$ has no models, and so by the Compactness theorem there is a finite subtheory with no models. This amounts to finite $A_1\subset T_1$ and $A_2\subset T_2$ such that $A_1\cup A_2$ has no models. Any model $M$ of $A_1$ is therefore not a model of $A_2$ and so $M$ is not a model of $T_2$ and hence by your assumption it is a model of $T_1$. So $A_1\vdash T_1$ and similarly $A_2\vdash T_2$, so both theories are finitely axiomatizable. &lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-12-05T12:26:08.600" />
  <row Id="13103" PostTypeId="2" ParentId="13094" CreationDate="2010-12-05T12:27:56.003" Score="-3" ViewCount="0" Body="&lt;p&gt;How could one say square roots are always nonnegative? What is the definition of square root? if by square root of $x$ we mean a number $ a$ s.t. $ a^2 = x $ then there are 2 numbers that satisfy the criteria, for example square roots of 4 are 2 and -2 because both $ 2^2 = -2^2 = 4  $ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the formula is confusing is that there is no order of operation is defined, Why &#xA;$\sqrt{(-1)^2} =\sqrt{1} = 1$ ? Why not $\sqrt{(-1)^2} = {(-1)}^{2 ( ^1/_2 )} = {-1}^{( ^1/_1 )} =-1 $ instead? since bot 1 and -1 squared are equal to 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That formula is taught only at elementary level. and at proper mathematical level is meaningless because every square root has two roots, one negative and the other positive, and a root of order n has exactly n roots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you are confused with that equation is that it is a meaningless and has no place in mathematics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to get the marks at elementary mathematics level accept what the teacher is preaching here, but look up the fundamental theorem of algebra and complex numbers and know (for yourself) that equation is as meaningless as it gets.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-05T12:27:56.003" CommentCount="27" />
  <row Id="13104" PostTypeId="1" AcceptedAnswerId="13105" CreationDate="2010-12-05T13:11:52.473" Score="2" ViewCount="69" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;If the $5$-th term of the G.P is its middle term and its value is 2,then find the product of all terms of this GP?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $a_5$ is the middle term, so the GP contains 9 terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_9 = a\cdot ar\cdot ar^2 \cdot ar^3 \cdot 2\cdot ar^5 \cdots ar^8$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leaves us $P_9 = 2\cdot (ar^4)^8$ ... I am not getting how to proceed further? Is this approach even correct?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-05T13:19:34.233" Title="Product of a geometric series from it's middle term" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13105" PostTypeId="2" ParentId="13104" CreationDate="2010-12-05T13:19:34.233" Score="4" ViewCount="0" Body="&lt;p&gt;The fourth term is $2/r$ and the sixth term is $2r$.  What happens when you multiply these?  What about the third and seventh terms?&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-05T13:19:34.233" CommentCount="2" />
  <row Id="13106" PostTypeId="2" ParentId="13067" CreationDate="2010-12-05T14:07:13.197" Score="3" ViewCount="0" Body="&lt;p&gt;The term &quot;special function&quot; has a historical connection with solutions of ordinary differential equations, often second order differential equations that arise from a separation of variables treatment of second order linear partial differential equations with constant coefficients.  This connection is mentioned briefly &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_functions&quot; rel=&quot;nofollow&quot;&gt; in the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.sagemath.org/doc/reference/sage/functions/special.html&quot; rel=&quot;nofollow&quot;&gt; Sage documentation of special functions&lt;/a&gt;, for example, lists the ODE satisfied by the function together with remarks about boundary conditions.  Stephen Wolfram gave a nice historical review in his talk &lt;a href=&quot;http://www.stephenwolfram.com/publications/recent/specialfunctions/&quot; rel=&quot;nofollow&quot;&gt; The History and Future of Special Functions&lt;/a&gt;, in which he says, &quot;Most continuous special functions are in effect defined implicitly, usually from differential equations.&quot;  Naturally he is not shy about what he perceives as Mathematica's contribution to this history, developing automated searches for accurate numerical methods that proceed directly from the underlying definition by differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is not alone in exploring the automatic generation of algorithms for special functions using symbolic computation.  Focusing on special functions that are the solutions of linear ordinary differential equations with polynomial coefficients, &lt;a href=&quot;http://algo.inria.fr/meunier/articles/MeSa03.pdf&quot; rel=&quot;nofollow&quot;&gt; Meunier and Salvy (2003)&lt;/a&gt; describe the design of their website &lt;a href=&quot;http://algo.inria.fr/esf&quot; rel=&quot;nofollow&quot;&gt; The Encyclopedia of Special Functions (ESF)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elementary transcendental functions fit easily into this scheme as special special functions, since they solve differential equations too.  For example, the exponential function solves $y&amp;#39; = y$ and sine and cosine, with different boundary conditions, solve the constant coefficient ODE $y&amp;#39;&amp;#39; = -y$.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-05T14:07:13.197" CommentCount="1" />
  <row Id="13107" PostTypeId="1" AcceptedAnswerId="13108" CreationDate="2010-12-05T14:09:01.353" Score="3" ViewCount="68" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Prove that if the system of equations $x+2ay + az = 0$,$x+3by+bz = 0$ ,$x+4cy+cz = 0$ has a non-zero solution then a,b,c are in harmonic progression.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a suitable approach for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-05T14:54:35.277" Title="Prove in harmonic progression" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13108" PostTypeId="2" ParentId="13107" CreationDate="2010-12-05T14:29:41.690" Score="4" ViewCount="0" Body="&lt;p&gt;Write these equations in determinants form and use the following operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$C_{2} \to C_{2} -2C_{3}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$R_{3} \to R_{3} - R_{2}$ and $R_{2} \to R_{2}-R_{1}$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When these two operations are done, then the resulting determinant should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{lll} 1 &amp;amp; 0 &amp;amp; a \\ 0 &amp;amp; b &amp;amp; b-a \\ 0 &amp;amp; 2c-b &amp;amp; c-b \end{array}\right| = b(c-b) - (b-a)(2c-b) =0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which will give $$ \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-12-05T14:29:41.690" />
  <row Id="13109" PostTypeId="1" CreationDate="2010-12-05T14:35:14.010" Score="0" ViewCount="172" Body="&lt;p&gt;I have some troubles proving soundness for $\forall$ Introduction. Let $T_1,...,T_n\vdash Q$. $\forall$ Intro gives that $T_1,...,T_n\vdash \forall x Q[x/t]$. Here, t is a constant symbol that does not occur in ${T_1,...,T_n}$, x is a variable which does not occur in Q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand it intuitively, but have some problems with tarskis' truth definitions. Thank you for your answer.&lt;/p&gt;&#xA;" OwnerUserId="4439" LastActivityDate="2010-12-09T08:22:30.440" Title="Soundness for $\forall$Introduction" Tags="&lt;logic&gt;" AnswerCount="2" />
  <row Id="13110" PostTypeId="2" ParentId="13107" CreationDate="2010-12-05T14:54:35.277" Score="2" ViewCount="0" Body="&lt;p&gt;OK, Chandru1 solved it, so I'm gonna explain the way I would have approached it. (Which is actually the same approach up to ordering my equations differently.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/ A homogeneous system of equations has a non-zero solution if the determinant is zero. Therefore look at this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2a &amp;amp; a \\&#xD;&#xA;1 &amp;amp; 3b &amp;amp; b \\&#xD;&#xA;1 &amp;amp; 4c &amp;amp; c &#xD;&#xA;\end{array}\right| = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ Since we want to check for a harmonic progression, we need to consider the inverses of $a,b$ and $c$. If one of these coefficients is zero but not the others, there can't be a harmonic progression, but you can also show that in these cases, the system of equations can't have a non-zero solution, so they are ruled out anyway. If two or more of these coefficients are zero, the system is underdetermined and there is an infinity of solutions. So, it seems that there is a small loophole here in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming that all coefficients are different from zero, we can rewrite the determinant as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{ccc}&#xD;&#xA;1/a &amp;amp; 2 &amp;amp; 1 \\&#xD;&#xA;1/b &amp;amp; 3 &amp;amp; 1 \\&#xD;&#xA;1/c &amp;amp; 4 &amp;amp; 1 &#xD;&#xA;\end{array}\right| = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting the second row from the first and third row, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\begin{array}{ccc}&#xD;&#xA;1/a-1/b &amp;amp; -1 &amp;amp; 0 \\&#xD;&#xA;1/b &amp;amp; 3 &amp;amp; 1 \\&#xD;&#xA;1/c-1/b &amp;amp; 1 &amp;amp; 0 &#xD;&#xA;\end{array}\right| = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this can be developed with respect to the third column to get as an end result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-05T14:54:35.277" />
  <row Id="13111" PostTypeId="1" CreationDate="2010-12-05T16:21:32.023" Score="3" ViewCount="222" Body="&lt;p&gt;So Ive been trying to prove that there's no simple group of order 300. This is what I did and I was wondering if it was enough. Let |$G$|=2$^2$ . 3 . 5$^2$. Suppose $G$ is simple. Then there would be 6 Sylow 5-groups, one of which will have an index of 6. But then |$G$|=300 does not divide 6! which leads to a contradiction. So the number of Sylow 5-groups is 1 and $\exists$ a proper normal Sylow 5-subgroup in $G$. Hence $G$ is not simple.  &lt;/p&gt;&#xA;" OwnerUserId="4426" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-05T16:51:21.117" LastActivityDate="2010-12-09T09:15:49.477" Title="No simple group of order 300" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="13112" PostTypeId="2" ParentId="13097" CreationDate="2010-12-05T16:43:36.843" Score="1" ViewCount="0" Body="&lt;p&gt;If the hash function is well constructed, the elements of $M$ will be equally distributed across the hash values.  So each element of $M$ will have $r^{-p}$ chance of mapping to a given element of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this model, I'm not sure how you are counting collisions.  Each element of $H$ will receive $r^{(k-p)}$ of the strings in $M$.  If your count is to go through all the strings in $M$ and count one for each that hashes to an already used value, you are correct.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T16:53:46.627" LastActivityDate="2010-12-05T16:53:46.627" />
  <row Id="13113" PostTypeId="1" CreationDate="2010-12-05T16:57:46.950" Score="0" ViewCount="44" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a column stochastic matrix A(n-by-n real, non-negative matrix) and want to solve the following equation in C++: Ax=x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that I need to find out an eigenvector, x, where eigenvalue would have to be 1(right?) but I couldn't figure it out in C++. So far I have checked out some math-libs such as Seldon, CPPScaLapack, Eigen... Among them, Eigen seems a good option but I couldn't understand how to utilize any of them to solve the equation above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me some suggestions/code snippets or ideas to solve the equation? Any help is highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4442" LastActivityDate="2010-12-05T16:57:46.950" Title="How to calculate the eigenvector of a column stochastic matrix in C++" Tags="&lt;linear-algebra&gt;&lt;eigenvectors&gt;" CommentCount="1" />
  <row Id="13114" PostTypeId="1" CreationDate="2010-12-05T17:16:29.470" Score="1" ViewCount="112" Body="&lt;p&gt;Let $\mathbf G$ be a given $m \times n$ matrix, $\mathbf y$ a given $m \times 1$ column vector and $\mathbf x$ an unknown $n \times 1$ column vector such that $\mathbf x \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How do you find $\displaystyle \min_{\mathbf x} (\mathbf y - \mathbf G\mathbf x)^T(\mathbf y - \mathbf G\mathbf x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Whether there is a solution of that problem similar to the simple pseudo-inverse $x=(G^TG)^{−1}G^Ty$ used for solving the least-squares problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) And what if we have $n \times n$ matrix $\mathbf C$ and the constraint $\mathbf C\mathbf x \ge 0$ instead of $\mathbf x \ge 0$? &lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="1041" LastEditorDisplayName="" LastEditDate="2010-12-08T10:38:29.843" LastActivityDate="2010-12-08T10:38:29.843" Title="Finding $\min_{\mathbf x} (\mathbf y - \mathbf G\mathbf x)^T(\mathbf y - \mathbf G\mathbf x)$" Tags="&lt;linear-algebra&gt;&lt;optimization&gt;&lt;vector-space&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="13115" PostTypeId="1" CreationDate="2010-12-05T17:17:37.503" Score="0" ViewCount="26" Body="&lt;p&gt;Consider a discrete or a continuous time dynamical system. Show that the complement of a forward invariant set is backward invariant, and vice versa. &lt;/p&gt;&#xA;" OwnerUserId="4443" LastActivityDate="2010-12-05T17:17:37.503" Title="Complement of forward invariant set" Tags="&lt;dynamical-systems&gt;" CommentCount="1" />
  <row Id="13116" PostTypeId="1" AcceptedAnswerId="13123" CreationDate="2010-12-05T17:18:56.507" Score="5" ViewCount="76" Body="&lt;p&gt;Let $p(n)$ be a polynomial of degree $a$. Start of with plunging in arguments from zero and go up one integer at the time. Go on until you have come at an integer argument $n$ of which $p(n)$'s value is not prime and count the number of distinct primes your polynomial has generated. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: what is the maximum number of distinct primes a polynomial of degree $a$ can generate by the process described above? Furthermore, what is the general form of such a polynomial $p(n)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question was inspired by &lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Please note that your polynomial does not need to generate consecutive primes, only primes at consecutive positive integer arguments.]&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-05T17:48:22.160" LastActivityDate="2010-12-05T18:42:09.023" Title="What is the maximum number of primes generated consecutively generated by a polynomial of degree $a$ ?" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;&lt;polynomials&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13118" PostTypeId="2" ParentId="13114" CreationDate="2010-12-05T17:37:16.823" Score="2" ViewCount="0" Body="&lt;p&gt;You can start by writing the expression as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;F(x) = (y^T - x^T G^T)(y - Gx) = y^T y - y^T Gx - x^TG^Ty + x^TG^TGx &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then differentiate wrt $x$ and set the expression to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;F'(x) = -2G^Ty + 2G^TG x = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get $x = (G^TG)^{-1}G^T y$.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-12-05T17:37:16.823" CommentCount="2" />
  <row Id="13119" PostTypeId="1" CreationDate="2010-12-05T17:39:47.893" Score="1" ViewCount="76" Body="&lt;p&gt;Is there a way to verify whether a non-empty intersection exists between two simple polygons (not necessarily convex) using the Chazelle's simplicity test ?&lt;/p&gt;&#xA;" OwnerUserId="4444" LastActivityDate="2011-01-04T19:44:35.003" Title="Using Chazelle's simplicity test to verify simple polygons intersection" Tags="&lt;computational-geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13120" PostTypeId="1" CreationDate="2010-12-05T17:40:28.237" Score="2" ViewCount="103" Body="&lt;p&gt;Louis de Branges  had a paper on his &lt;a href=&quot;http://www.math.purdue.edu/~branges/site//Papers&quot; rel=&quot;nofollow&quot;&gt;homepage&lt;/a&gt; claiming a solution for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariant_subspace_problem&quot; rel=&quot;nofollow&quot;&gt;Invariant Subspace Problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see that paper anymore, though he still has a &quot;proof&quot; of  Riemann Hypothesis on his website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know whether he actually admitted his proof of the Invariant Subspace Problem was wrong? Or someone pointed that out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His proposed proof of the RH has been on his website for many years now, and yet no one has pointed out an error.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:11:44.603" LastActivityDate="2010-12-06T08:12:08.410" Title="Invariant Subspace Problem" Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;functional-analysis&gt;&lt;open-problem&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" />
  <row Id="13121" PostTypeId="2" ParentId="13116" CreationDate="2010-12-05T17:50:02.830" Score="3" ViewCount="0" Body="&lt;p&gt;Here is result by Rabinowitsch for quadratic polynomials.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$n^2+n+A$ is prime for $n=0,1,2,...,A-2$ if and only if $d=1-4A$ is squarefree and the class number of $\mathbb{Q}[\sqrt{d}]$ is $1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See this article for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8911.pdf&quot; rel=&quot;nofollow&quot;&gt;http://matwbn.icm.edu.pl/ksiazki/aa/aa89/aa8911.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also here is a list of imaginary quadratic fields with class number $1$&#xD;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_number_fields_with_class_number_one#Imaginary_quadratic_fields&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_number_fields_with_class_number_one#Imaginary_quadratic_fields&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other articles about prime generating (quadratic) polynomials that you can google.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-05T17:50:02.830" CommentCount="3" />
  <row Id="13122" PostTypeId="1" CreationDate="2010-12-05T17:57:55.257" Score="1" ViewCount="115" Body="&lt;p&gt;For any positive integer $N$, the binomial$(N!,p)$ distribution has the following property: for any $1 \leq n \leq N$,&#xA;there exist i.i.d. random variables $X_1,\ldots,X_n$ such that $X_1 + \cdots + X_n \sim {\rm binomial}(N!,p)$ (specifically, we take $X_1,\ldots,X_n$ to be i.i.d. binomial$(N!/n,p)$ rv's). It may be interesting to consider the following question: given $N \geq 3$, arbitrary but fixed, is there a continuous bounded distribution $\mu = \mu_N$ having the same property? (I stress: continuous and bounded.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Well, it turns out this is a really trivial problem, but worth remembering...; see my answer below.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T15:33:14.530" LastActivityDate="2010-12-06T15:33:54.187" Title="A continuous analogue of the binomial distribution" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13123" PostTypeId="2" ParentId="13116" CreationDate="2010-12-05T18:00:02.710" Score="6" ViewCount="0" Body="&lt;p&gt;The Green-Tao Theorem states that there are arbitrarily long arithmetic progressions of primes; that is, sequences of primes of the form&#xA;$$ b , b+a, b+2a, b+3a,... ,b+na $$&#xD;&#xA;Since such a progression will be the first $n$ values of the polynomial $ax+b$, this implies that even for degree 1, there is no upper bound to how many primes in a row a polynomial can generate.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-12-05T18:00:02.710" CommentCount="6" />
  <row Id="13124" PostTypeId="1" AcceptedAnswerId="13130" CreationDate="2010-12-05T18:19:28.223" Score="3" ViewCount="60" Body="&lt;p&gt;I have heard that given two sheaves $A$ and $B$ on a variety, one can identify elements of $Ext^d(A,B)$ with complexes of sheaves&#xA;$$0\to B \to C_1 \to \cdots \to C_d \to A \to 0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I see that this is true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have obtained an element of $Ext^n$ by some other method, can I explicitly construct the $C_j$ sheaves and the differentials?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am sure this is well-known, so I'm marking it also as &quot;reference-request&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2010-12-05T18:55:40.853" Title="Ext &amp; Complexes" Tags="&lt;reference-request&gt;&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13125" PostTypeId="1" AcceptedAnswerId="13133" CreationDate="2010-12-05T18:31:22.570" Score="6" ViewCount="93" Body="&lt;p&gt;On $\mathbb{R}^n$ and $p\ge 1$ the $p$-norm is defined as $$\|x\|_p=\left ( \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$&#xD;&#xA;and there is the $\infty$-norm which is $\|x\|_\infty=\max _j |x_j|$. It's called the $\infty$ norm because it is the limit of $\|\cdot\|_p$ for $p\to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can use the definition above for $p&amp;lt;1$ as well and define a $p$-&quot;norm&quot; for these $p$. The triangle inequality is not satisfied, but I will use the term &quot;norm&quot; nonetheless. For $p\to 0$ the limit of $\|x\|_p$ is obviously $\infty$ if there are at least two nonzero entries in $x$, but if we use the following modified definition&#xA;$$\|x\|_p=\left ( \frac{1}{n} \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$&#xD;&#xA;then this should have a limit for $p\to 0$, which should be called 0-norm. What is this limit?&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-12-05T19:05:01.900" Title="What is the 0-norm?" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13127" PostTypeId="2" ParentId="13116" CreationDate="2010-12-05T18:42:09.023" Score="1" ViewCount="0" Body="&lt;p&gt;Here is a related fact which might also be of interest.  There exists a polynomial in 26 variables with the property that, if you plug in integers for all 26 variables, and the output is a positive number (it will always be an integer), then the output is prime.  The polynomial can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_for_primes&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Formula_for_primes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, every prime occurs as an output of this polynomial.  This is a very indirect way to generate primes, because there is no easy way to know if a given input will give a positive output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is apparently another polynomial in 10 variables that does this, but Wikipedia doesn't explicitly write it.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-12-05T18:42:09.023" />
  <row Id="13128" PostTypeId="1" CreationDate="2010-12-05T18:45:15.730" Score="0" ViewCount="37" Body="&lt;p&gt;Okay, I am in no way a mathematics student but I happened to be writing a program on a map with coordinates of locations displayed on the map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that the points could be scattered on the map with an indefinite boundary and I need to find a boundary (rectangle) around these points and thus I need to get a formula to form an algorithm so that the program will find the optimum boundary itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is understandable and thanks in anticipation of any response.&lt;/p&gt;&#xA;" OwnerUserId="4446" LastActivityDate="2010-12-05T23:45:08.293" Title="optimum rectangle around a group of coordinates (longitude and latitude)" Tags="&lt;polar-coordinates&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13129" PostTypeId="1" AcceptedAnswerId="13167" CreationDate="2010-12-05T18:47:36.463" Score="3" ViewCount="105" Body="&lt;p&gt;Let $K$ be a field and let $K(x)$ be the field of rational functions in $x$ whose coefficients are in $K$. Let $\theta(x)$ $\in$ Aut($K(x))$ such that $\theta|_K = id_K$.  Show that $\theta(x)$ =$\frac{ax+b}{cx+d}$, with $ad$ $\neq$ $bc$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\theta(x)$ = $\frac{f}{g}$, $f$, $g$ $\in $K$[$x$] $, with gcd($f, $g)=1. let $h$ $\in$ $K(x)$. Then  $h(\frac{f}{g})$ = $x$ and $\frac{f(x)}{g(x)}$ $\neq$ $\frac{f(y)}{g(y)}$ if $x\neq y$ . Suppose deg $f$ &gt; 1 or deg $g$ &gt; 1, then the equation $g(x)b=f(x)$, $b\in K$ ,  will have more than one solution for $x$. Hence $f$ and $g$ have degrees at most 1 and thus $\theta(x)$ =$\frac{ax+b}{cx+d}$ with inverse function $\theta(x)$ =$\frac{dx-b}{a-cx}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is if Ive approached the problem in the right manner. Please any input will be very much appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastActivityDate="2010-12-06T16:32:44.347" Title="Automorphism of the Field of rational functions." Tags="&lt;homework&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="13130" PostTypeId="2" ParentId="13124" CreationDate="2010-12-05T18:48:54.713" Score="3" ViewCount="0" Body="&lt;p&gt;For modules, Weibel discusses this in &quot;Introduction to homological algebra.&quot; Section 3.4 deals with d=1 case and in Vista 3.4.6 is about the general case. He gives no proof for d&gt;1 and refers to Bourbaki &quot;Algebre homologique&quot; 7.5 and Maclane &quot;Homology&quot; pp82-87.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="2633" LastEditorDisplayName="" LastEditDate="2010-12-05T18:55:40.853" LastActivityDate="2010-12-05T18:55:40.853" CommentCount="2" />
  <row Id="13131" PostTypeId="1" CreationDate="2010-12-05T18:58:45.990" Score="5" ViewCount="98" Body="&lt;p&gt;Prove that for any finite sequence of decimal digits, there exists an $n$ such that the decimal expansion of $2^n$ begins with these digits.&lt;/p&gt;&#xA;" OwnerUserId="4448" LastActivityDate="2010-12-05T19:06:57.797" Title="Starting digits of 2^n" Tags="&lt;homework&gt;&lt;dynamical-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13132" PostTypeId="2" ParentId="13122" CreationDate="2010-12-05T19:04:21.770" Score="0" ViewCount="0" Body="&lt;p&gt;OK, I think I've got one for you, but I have to admit it's a bit of a cheat. If you replace independence in your requirements by &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_probability&quot; rel=&quot;nofollow&quot;&gt;free independence&lt;/a&gt; then there exists a continuous bounded and infinitely divisible distribution, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_semicircle_distribution&quot; rel=&quot;nofollow&quot;&gt;Wigner semi-circle distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the only example I can think of for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also give this &lt;a href=&quot;http://books.google.be/books?id=NUyG98QZEl8C&amp;amp;printsec=frontcover&amp;amp;dq=free+probability+denes+petz&amp;amp;source=bl&amp;amp;ots=YAwO-jcpk7&amp;amp;sig=KRweScXQiDuUJCAhcFNxuxkFJSg&amp;amp;hl=nl&amp;amp;ei=7eL7TM2sJcqZ8QPX7bWHDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CDMQ6AEwAw#v=onepage&amp;amp;q=free%20probability%20denes%20petz&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; for details.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-05T19:04:21.770" CommentCount="2" />
  <row Id="13133" PostTypeId="2" ParentId="13125" CreationDate="2010-12-05T19:05:01.900" Score="8" ViewCount="0" Body="&lt;p&gt;When $p$ is small, $$x^p = \exp(p \log x) \approx 1 + p\log x.$$ Therefore $$\frac{1}{n} \sum_{j=1}^n x_j^p \approx 1 + p\frac{1}{n} \sum_{j=1}^n \log x_j = 1 + p \log \sqrt[n]{\prod_{i=1}^n x_j}.$$ On the other hand, we have $$(1+py)^{1/p} \longrightarrow \exp(y),$$ and so we easily get that the norm approaches the geometric mean, as Raskolnikov commented.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T19:05:01.900" />
  <row Id="13134" PostTypeId="2" ParentId="13131" CreationDate="2010-12-05T19:06:57.797" Score="8" ViewCount="0" Body="&lt;p&gt;Take $\log_{10} (2^n) = n \log_{10} 2$, note that $\log_{10} 2$ is irrational, and use the equidistribution theorem (or prove what you want directly using the pigeonhole principle).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T19:06:57.797" CommentCount="1" />
  <row Id="13135" PostTypeId="1" AcceptedAnswerId="13136" CreationDate="2010-12-05T19:07:54.323" Score="1" ViewCount="101" Body="&lt;p&gt;Let $f\colon A\to A$; prove that if $f\circ f = \mathrm{id}_A$ then $f$ is a bijection.&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:02:04.847" LastActivityDate="2010-12-06T00:00:22.393" Title="Let $f\colon A\to  A$;  prove that if $f\circ f = \mathrm{id}_A$ then $f$ is a bijection" Tags="&lt;elementary-set-theory&gt;&lt;proof&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13136" PostTypeId="2" ParentId="13135" CreationDate="2010-12-05T19:13:45.250" Score="3" ViewCount="0" Body="&lt;p&gt;Hints: Injectivity: Suppose $f(x)=f(y)$, what can you say about $f(f(x))$ and $f(f(y))$.&#xA;Surjectivity: Given $a\in A$, there exists $b\in A$ such that $f(b)=a$ since, $f(f(a))=id(a)=a$ so b=?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-05T19:13:45.250" CommentCount="7" />
  <row Id="13137" PostTypeId="1" CreationDate="2010-12-05T19:24:07.973" Score="6" ViewCount="136" Body="&lt;p&gt;If $A$ is a subset of $\mathbb{R}$ with Lebesgue measure strictly greater than $0$, does it follow then that are there $a$ and $b$ such that the measure of $[a,b]\cap A$ is $b-a$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2934" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:03:41.930" LastActivityDate="2010-12-17T02:27:25.063" Title="If A is a subset of R with Lebesgue measure &gt; 0 then are there a,b such that the measure of $[a,b]\cap A$ is b-a? " Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13138" PostTypeId="2" ParentId="13128" CreationDate="2010-12-05T19:26:54.550" Score="2" ViewCount="0" Body="&lt;p&gt;If the sides of the rectangle must be aligned with the edges of the map, you can just take the maximum and minimum coordinates in each direction.  If you allow arbitrary orientation of the rectangle you have a one dimensional minimization problem.  You can compute the coordinates of each point as a function of rotation angle, again take the maximum and minimum in each dimension, calculate the perimeter, and minimize over the rotation angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not devoted to a rectangle, maybe you would prefer the &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_hull&quot; rel=&quot;nofollow&quot;&gt;convex hull&lt;/a&gt; of your points.  There are efficient algorithms to find it and it doesn't need the minimization.  It also has a smaller perimeter yet.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T23:45:08.293" LastActivityDate="2010-12-05T23:45:08.293" />
  <row Id="13139" PostTypeId="1" CreationDate="2010-12-05T19:27:51.217" Score="1" ViewCount="82" Body="&lt;p&gt;Hi, I have a school assignment to do, but I have no idea, where to start. I hope you can help. Here it is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a finite sequence $A = (a_1, a_2,\ldots, a_n)$, length of $A$ is $n$, elements of $A$ are natural numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S(i,j)$ be partial sum of this sequence from $a_i$ to $a_j$, where $1\leq i\leq j\leq n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally written as $S(i,j) = a_i+a_{i+1}+\cdots+a_j$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can call sequence $A$ &quot;nice&quot;, when all partial sums $S(i,j)$ are different from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally written - For every $1\leq i\leq j\leq n$ and $1\leq k\leq l\leq n$ it holds that &quot;$S(i,j)=S(k,l)$ implies $i=k$ and $j=l$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. $(1,3,2)$ is &quot;nice&quot; and $(1,4,2,3)$ is not &quot;nice&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have following tasks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prove that for every natural $n$ a &quot;nice&quot; sequence exists.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the smallest possible value of the greatest element of a &quot;nice&quot; sequence $A$ of length $n$? Or in other words - for every $n$ find the smallest natural number $p$ for which a &quot;nice&quot; sequence $A$ of length $n$, in which no element is larger than $p$, exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: search for upper and lower bounds on $p$ in the form of appropriate functions of $n$.&lt;/p&gt;&#xA;" OwnerUserId="4447" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-05T20:30:21.133" LastActivityDate="2010-12-05T20:30:21.133" Title="Series partial sums" Tags="&lt;sequences-and-series&gt;" CommentCount="2" />
  <row Id="13140" PostTypeId="2" ParentId="13119" CreationDate="2010-12-05T19:31:01.860" Score="2" ViewCount="0" Body="&lt;p&gt;Independent of Chazelle, a discussion of the case of intersecting convex polygons can be found at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iro.umontreal.ca/~plante/compGeom/algorithm.html&quot; rel=&quot;nofollow&quot;&gt;http://www.iro.umontreal.ca/~plante/compGeom/algorithm.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/q744325k7350m237/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/q744325k7350m237/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-05T19:31:01.860" CommentCount="1" />
  <row Id="13141" PostTypeId="1" AcceptedAnswerId="13144" CreationDate="2010-12-05T20:08:56.147" Score="1" ViewCount="39" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the sum of the first $p$ terms of an arithmetic progression is $q$ and the sum of the first $q$ terms is $p$, then find the sum of $p+q$ terms.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the problem we can write (considering $a$ is the first term and $d$ is the common difference):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p}{2}\cdot \biggl[2a + (p-1)d \biggr] = q \qquad \cdots (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{q}{2}\cdot \biggl[2a + (q-1)d \biggr] = p \qquad \cdots (2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in my module it is given that from these we can write $\displaystyle d= \frac{-2(p+q)}{pq}$; I am not getting how we can get that value of $d$ ?!&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-05T20:25:11.777" LastActivityDate="2010-12-05T20:25:11.777" Title="A small doubt on calculation" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" />
  <row Id="13142" PostTypeId="1" AcceptedAnswerId="13143" CreationDate="2010-12-05T20:15:38.940" Score="0" ViewCount="56" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a_1,a_2,a_3 \cdots a_n$ are in HP then find the value of $ a_1 \cdot a_2 +  a_2 \cdot a_3 +   a_3 \cdot a_4 + \cdots +  a_{n-1} \cdot a_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My initial approach,using the property of HP that $ \frac{1}{a_1},  \frac{1}{a_2}, \frac{1}{a_3},\cdots ,\frac{1}{a_n}$ is in AP, I am getting this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a_1-a_2}{a_1 \cdot a_2} = \frac{a_2-a_3}{a_2 \cdot a_3} = \cdots = \frac{a_{n-1}-a_n}{a_{n-1} \cdot a_n}= d$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to proceed next?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-05T20:34:29.087" LastActivityDate="2010-12-05T20:34:29.087" Title="Suggest a tricky method" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13143" PostTypeId="2" ParentId="13142" CreationDate="2010-12-05T20:20:57.893" Score="1" ViewCount="0" Body="&lt;p&gt;Write $a_i a_{i+1}$ as a difference of two fractions to get a telescopic sum.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T20:20:57.893" />
  <row Id="13144" PostTypeId="2" ParentId="13141" CreationDate="2010-12-05T20:23:32.937" Score="3" ViewCount="0" Body="&lt;p&gt;$q \times (1) - p \times (2)$ will give you $d$ and then you can work out $a.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-05T20:23:32.937" />
  <row Id="13145" PostTypeId="2" ParentId="13141" CreationDate="2010-12-05T20:24:04.617" Score="0" ViewCount="0" Body="&lt;p&gt;You get two equations in two unknowns $a,d$; solve them to find $d$. Since we only want to find $d$, we can equate the coefficient of $a$ by multiplying the first equation by $q$, the second by $p$, and subtracting to get a linear equation for $d$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-05T20:24:04.617" />
  <row Id="13146" PostTypeId="1" CreationDate="2010-12-05T20:47:09.183" Score="1" ViewCount="81" Body="&lt;p&gt;Define $h\colon \mathbb{Z}/\sim ~16 \to \mathbb{Z}/\sim 24$ by $h([a]16) = [3a]24$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Prove $h$ is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Compute $h(a)$ where $a = \{[0]16, [3]16, [6]16\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Compute $h^{-1}([10]24)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. $h$ is well defined since each for all $a$ $|h([a]16)|\equiv 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. $\{[0]24, [9]24, [18]24\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. $\emptyset$&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:04:51.303" LastActivityDate="2011-01-04T22:44:35.277" Title="Define $h\colon \mathbb{ Z}/\sim 16 \to \mathbb{Z}/\sim 24$ by $h([a]16) = [3a]24$" Tags="&lt;elementary-set-theory&gt;&lt;proof&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13147" PostTypeId="2" ParentId="13146" CreationDate="2010-12-05T21:04:42.233" Score="0" ViewCount="0" Body="&lt;p&gt;Your answers to (b) and (c) are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (a), you need to prove that if a=b mod 16, i.e. $a-b=16n $ for some integer $n$, then $3a=3b$ mod 24, i.e. $3a-3b=24m$ for some integer $m$. Can you do that?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-05T21:04:42.233" />
  <row Id="13148" PostTypeId="1" CreationDate="2010-12-05T21:07:44.233" Score="0" ViewCount="110" Body="&lt;p&gt;How do you show that the convex hull of a given set of points S, always has the minimum perimeter ? By perimeter i mean the length of the boundary of the hull&lt;/p&gt;&#xA;" OwnerUserId="4444" LastActivityDate="2010-12-06T00:01:07.710" Title="Convex hull has the smallest perimeter" Tags="&lt;algorithms&gt;&lt;convex-analysis&gt;&lt;computational-geometry&gt;&lt;convex-sets&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13149" PostTypeId="1" CreationDate="2010-12-05T21:21:43.067" Score="2" ViewCount="36" Body="&lt;p&gt;The rough question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How do the finite subgroups of SL(2,R/I) vary as I varies, where R is a nice local ring?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SL(2,A) = { [a,b;c,d] : ad-bc=1, a,b,c,d in A } is the group of a 2&amp;times;2 matrices with entries from A and determinant 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two cases seem basic and approachable but already very interesting:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R the 2-adic integers, and R/I = &lt;strong&gt;Z&lt;/strong&gt;/2&lt;sup&gt;n&lt;/sup&gt;&lt;strong&gt;Z&lt;/strong&gt; with n varying, and&lt;/li&gt;&#xA;&lt;li&gt;R the power series over the field with 2 elements, and R/I = &lt;strong&gt;Z&lt;/strong&gt;[x]/(2,x&lt;sup&gt;n&lt;/sup&gt;) with n varying.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In each case G = SL(2,R/I) is finite and solvable, with G/O&lt;sub&gt;2&lt;/sub&gt;(G) ≅ SL(2,2) ≅ S&lt;sub&gt;3&lt;/sub&gt; being a nice familiar group.  In other words, nothing radically strange can happen in any one particular group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having quite a bit of trouble understanding how the subgroups vary with n.  Call the first sequence of groups G(2,n) and the second sequence of groups G(x,n).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which groups G(2,n) and G(x,n) contain Q&lt;sub&gt;8&lt;/sub&gt;, a quaternion group of order 8?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Q8 seems to be contained in G(x,2n+1) and not contained in G(x,2n) or G(x,1) or any G(2,n).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which dihedral groups are contained in each G(2,n) and G(x,n)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The G(2,n+1) seem to be bereft of dihedrals, but G(2,1) has one, and G(x,n+1) seem to have the full crew.  Why does the &quot;2&quot; destroy dihedrals while the &quot;x&quot; invites more? On the abelian side:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the 2-rank (maximum n such that G contains the direct product of n cyclic groups of order 2) of each G(2,n) and G(x,n)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again the pattern seems a bit wobbly, especially if you count how many subgroups of each rank there are (these do not form monotonic sequences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answer to the bold question should be able to address the strange (conjectured) answers to the smaller questions, but I'm also just fine with some ad hoc answers to the small ones as these groups are currently a complete mystery.  I think larger primes (or residue fields) would help with the very restricted nature of the subgroups available for testing, but at the beginning p=2 seems to help.  More independent nilpotents has not struck me as counter-intuitive yet.  Also in my limited experiments, &lt;strong&gt;GL&lt;/strong&gt; seems to be a lot less strange, but feel free to share observations about it as well.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-05T21:29:04.137" LastActivityDate="2010-12-05T21:29:04.137" Title="Subgroups of SL(2,R/I) for R local as I varies" Tags="&lt;group-theory&gt;" CommentCount="1" />
  <row Id="13150" PostTypeId="1" AcceptedAnswerId="13165" CreationDate="2010-12-05T21:24:32.117" Score="1" ViewCount="98" Body="&lt;p&gt;Hi, how can i extract rotation, scale values from 2d transformation matrix? i mean a have a 2d transformation&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;matrix = [1, 0, 0, 1, 0, 0]&#xA;&#xA;matrix.rotate(45 / 180 * PI)&#xA;matrix.scale(3, 4)&#xA;matrix.translate(50, 100)&#xA;matrix.rotate(30 / 180 * PI)&#xA;matrix.scale(-2, 4)&#xA;now my matrix have values [a, b, c, d, tx, ty]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;lets forget about the processes above and imagine that we have only the values a, b, c, d, tx, ty&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can i find total rotation and scale values via a, b, c, d, tx, ty&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry for my english&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks your advance&lt;/p&gt;&#xA;" OwnerUserId="4452" LastActivityDate="2010-12-07T11:02:02.000" Title="extracting rotation, scale values from 2d transformation matrix" Tags="&lt;matrices&gt;&lt;transformation&gt;" AnswerCount="2" />
  <row Id="13151" PostTypeId="1" AcceptedAnswerId="13173" CreationDate="2010-12-05T21:29:10.530" Score="2" ViewCount="78" Body="&lt;p&gt;Question: Calculate the rank of the following matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \left( \begin{array}{cc} 1 &amp;amp; n \\ n &amp;amp; 1 \end{array} \right), n \in \mathbb{Z}$ and $B = \left( \begin{array}{ccc} 1 &amp;amp; x &amp;amp; x^{2} \\ 1 &amp;amp; y &amp;amp; y^{2} \\ 1 &amp;amp; z &amp;amp; z^{2} \end{array} \right)$,  $x,y,z \in \mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the way I understand rank($A$), is the number of pivots in an echelon form of $A$. To put $A$ into echelon form I would subtract $n$ times the first row from the second row: $A \sim \left( \begin{array}{cc} 1 &amp;amp; n \\ n &amp;amp; 1 \end{array} \right) \sim \left( \begin{array}{cc} 1 &amp;amp; n \\ 0 &amp;amp; 1 - n^{2} \end{array} \right) \Rightarrow $rank$(A) = 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $B$ I would have done pretty much the same thing, subtracting row 1 from both row 2 and row 3: $B \sim \left( \begin{array}{ccc} 1 &amp;amp; x &amp;amp; x^{2} \\ 1 &amp;amp; y &amp;amp; y^{2} \\ 1 &amp;amp; z &amp;amp; z^{2} \end{array} \right) \sim \left( \begin{array}{ccc} 1 &amp;amp; x &amp;amp; x^{2} \\ 0 &amp;amp; y - x &amp;amp; y^{2} - x^{2} \\ 0 &amp;amp; z - x &amp;amp; z^{2} - x^{2} \end{array} \right)$ (at this point I could multiply row 2 by $-(\frac{z-x}{y-x})$ and add it to row 3 which ends up being a long polynomial....) However, with both parts, I am pretty confident that it is not so simple and that I am missing the point of this exercise. Could somebody please help point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T22:54:09.620" LastActivityDate="2010-12-05T23:36:11.253" Title="Calculate the rank of the following matrices" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13152" PostTypeId="1" AcceptedAnswerId="13158" CreationDate="2010-12-05T21:33:05.617" Score="2" ViewCount="40" Body="&lt;p&gt;Prove that a vector perpendicular to the line $Ax+By+C=0$ in the $xy$-plane is $&amp;lt;A, B&amp;gt;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So intuitively this is clear, since we can always find a vector normal to a line by looking at the coefficients of $x,y,z$ So in this case $&amp;lt;A, B, 0&amp;gt;$ or $&amp;lt;A, B&amp;gt;$ in 2d space is clearly a normal vector of the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using this approach but got stuck :  take two points on the line, find a vector parallel to the line, take the dot product with $&amp;lt;A, B&amp;gt;$ and show this is $= 0$&lt;br&gt;&#xA;So&lt;br&gt;&#xA;Let $P_1 = Ax_1 + By_1 = -C$    be a point on the line and&lt;br&gt;&#xA;Let $P_2 = Ax_2 + By_2 = -C$    be another point on the line.&lt;br&gt;&#xA;(let this denote the vector) Then $P_1P_2 = &amp;lt;Ax_2 - Ax_1, By_2 - By_1&amp;gt;$   is parallel to the line.&lt;br&gt;&#xA;So  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;A, B&amp;gt; \cdot P_1P_2 = A^2x_2-A^2x_1+B^2y_n-B^2y_1$&lt;br&gt;&#xA;                     $= A^2x_2 + B^2y_n - (A^2x_1+B^2y_1)$&lt;br&gt;&#xA;But I wasn't able to get this to equal 0 using the fact that $Ax + By = -C$&lt;br&gt;&#xA;Any help is appreciated!&lt;br&gt;&#xA;Thanks.  &lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-12-05T22:44:05.747" Title="Prove that a vector perpendicular to the line $Ax+By+C=0$ in the $xy$-plane is $&lt;A, B&gt;$" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13153" PostTypeId="1" AcceptedAnswerId="13161" CreationDate="2010-12-05T21:44:23.017" Score="1" ViewCount="101" Body="&lt;p&gt;The problem was to find $\int \tan^5(x)\, \sec^7(x)\, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution the book got was different from mine, but I'm sure I did the right steps. Since the solutions are completely different, I have a feeling they might be the same thing, but I'm not sure how I would know this. Here's what the book got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sec^{11}(x)/11-2\sec^9(x)/9+\sec^7(x)/7+C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I got:&#xA;$\tan^6(x)/6 + \tan^{11}(x)/11 + C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So my answer is wrong then, so let me list my steps see if you can catch my mistake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \tan^5(x)\sec^7(x)\, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \tan^4x \tan x \sec^5x \sec^2x\, dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \tan^4 x\tan x \ (1+\tan^5x)\,\sec^2x\,dx$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u=\tan x \ du=\sec^2x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int u^4(u)(1+u^5)\,du$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int u^5(1+u^5)\,du$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int u^5+u^{10}\,du$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\tan^6(x)/6 + \tan^{11}(x)/11 + C$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T23:47:24.810" LastActivityDate="2010-12-06T21:45:42.803" Title="Trigonometric integral question" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13154" PostTypeId="1" AcceptedAnswerId="13204" CreationDate="2010-12-05T21:53:38.613" Score="0" ViewCount="39" Body="&lt;p&gt;Hello,&#xA;how do I found the resolvent of this formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A \vee C) \wedge ( \neg A \vee B) \wedge C$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it as easy as take every clause with the same statements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A \vee C)$ and $( \neg A \vee B)$ and put out the contradiction (in this case $A$ and $\neg A$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you recommend me some source for resolution method? except &lt;a href=&quot;http://en.wikipedia.org/wiki/Resolution_%28logic%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, because there is too easy examples :-)&lt;/p&gt;&#xA;" OwnerUserId="4449" LastActivityDate="2010-12-06T05:22:35.313" Title="The resolution method" Tags="&lt;soft-question&gt;&lt;logic&gt;&lt;proof&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13155" PostTypeId="1" AcceptedAnswerId="13164" CreationDate="2010-12-05T21:53:43.930" Score="9" ViewCount="210" Body="&lt;p&gt;In most undergraduate differential geometry courses -- I am thinking of do Carmo's &quot;Differential Geometry of Curves and Surfaces&quot; -- the topic of study is curves and surfaces in $\mathbb{R}^3$.  However, the definition of &quot;curve&quot; and &quot;surface&quot; are usually presented in very different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curve is defined simply as a differentiable map $\gamma\colon I \to \mathbb{R}^3$, where $I \subset \mathbb{R}$ is an interval.  Of course, some authors prefer to define a curve as the image of such a map, and others require piecewise-differentiability, but the general concept is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, surfaces are essentially defined as 2-manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in graduate courses on manifolds -- I am thinking of John Lee's &quot;Introduction to Smooth Manifolds&quot; -- one talks about curves $\gamma\colon I \to M$ in a manifold, and can do line integrals over such curves, but talks separately about embedded/immersed 1-dimensional submanifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, then, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why make (parametrized) curves the object of study rather than 1-manifolds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Earlier, I asked a &lt;a href=&quot;http://math.stackexchange.com/questions/4699/which-smooth-1-manifolds-can-be-represented-by-a-single-smooth-parametrization&quot;&gt;question&lt;/a&gt; that was perhaps meant to hint at this one, though I didn't say so explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I would simply like to say &quot;curves are 1-manifolds and surfaces are 2-manifolds,&quot; and am looking for reasons why this is correct/incorrect or at least a good/bad idea.  (So, yes, I'm looking for a standard definition of &quot;curve.&quot;)&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-06T13:34:16.697" Title="Why study &quot;curves&quot; instead of 1-manifolds?" Tags="&lt;soft-question&gt;&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13156" PostTypeId="1" CreationDate="2010-12-05T21:55:55.790" Score="0" ViewCount="95" Body="&lt;p&gt;How is a complex number a group under addition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is a complex number a group under multiplication(without zero)&lt;/p&gt;&#xA;" OwnerUserId="4453" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-06T00:54:41.647" LastActivityDate="2010-12-06T00:54:41.647" Title="How does the set of complex numbers (resp., nonzero complex numbers) form a group under addition (resp., multiplication)?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13157" PostTypeId="1" AcceptedAnswerId="13160" CreationDate="2010-12-05T21:56:17.250" Score="1" ViewCount="87" Body="&lt;p&gt;A friend and I were doing some math together when we had to integrate 2(100 + t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just multiplied it out and integrated 200 + 2t, which should be 200t + t^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He did u-substitution and got (100 + t)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I take the derivative of both of ours, I seem to get the same thing. But (100 + t)^2 = 10,000 + 200t + t^2, which is not equal to just 200t + t^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on here?&lt;/p&gt;&#xA;" OwnerUserId="3143" LastActivityDate="2010-12-05T22:39:46.440" Title="Integrating 2(100+t)" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13158" PostTypeId="2" ParentId="13152" CreationDate="2010-12-05T21:56:32.920" Score="1" ViewCount="0" Body="&lt;p&gt;The only problem you have is that your expression for the vector $\overrightarrow{P_1P_2}$ is slightly off. As you say, let $P_1 = (x_1,y_1)$ and $P_2 = (x_2,y_2)$ be points on the line. Then what are the coordinates of the vector $\overrightarrow{P_1P_2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you calculate that, then notice that the coordinates of $P_1$ satisfy the equation $Ax_1 + By_1 = -C$, and similarly for $P_2$. Then calculate the scalar product of $(A,B)$ with the vector $\overrightarrow{P_1P_2}$, and interpret what comes out.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-12-05T21:56:32.920" CommentCount="1" />
  <row Id="13159" PostTypeId="2" ParentId="13152" CreationDate="2010-12-05T22:00:28.517" Score="1" ViewCount="0" Body="&lt;p&gt;vector is equals $&amp;lt;x_2 - x_1, y_2 - y_1&amp;gt;$, not $&amp;lt;Ax_2 - Ax_1, By_2 - By_1&amp;gt;$ then $&amp;lt;A,B&amp;gt; \cdot P_1P_2 = A\cdot x_2-A\cdot x_1+B\cdot y_2-B\cdot y_1$&lt;/p&gt;&#xA;" OwnerUserId="4454" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T22:44:05.747" LastActivityDate="2010-12-05T22:44:05.747" />
  <row Id="13160" PostTypeId="2" ParentId="13157" CreationDate="2010-12-05T22:02:07.500" Score="4" ViewCount="0" Body="&lt;p&gt;The difference is just the constant of integration.  Both your solutions are correct, as is any one of the form C+200t+t^2 for any constant C.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T22:02:07.500" />
  <row Id="13161" PostTypeId="2" ParentId="13153" CreationDate="2010-12-05T22:02:24.687" Score="3" ViewCount="0" Body="&lt;p&gt;If they differ by a constant then their difference should be constant. You can check that this is not so by setting $x=0$ and $x=\pi.$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-05T22:02:24.687" CommentCount="7" />
  <row Id="13162" PostTypeId="2" ParentId="13157" CreationDate="2010-12-05T22:03:18.187" Score="1" ViewCount="0" Body="&lt;p&gt;When you integrate indefinitely, you must add an arbitrary constant. So in essence, your answers are the same, up to addition of a constant (10,000 in this case). In other words, his constant will be 10,000 less than yours.&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-12-05T22:03:18.187" />
  <row Id="13163" PostTypeId="2" ParentId="13156" CreationDate="2010-12-05T22:04:26.297" Score="5" ViewCount="0" Body="&lt;p&gt;A complex number is not a group under addition.  The set of all complex numbers is a group under addition.  Just look at the definition of a group and see that you can verify the axioms.  Similarly for the set of complex numbers without zero and multiplication.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T22:04:26.297" />
  <row Id="13164" PostTypeId="2" ParentId="13155" CreationDate="2010-12-05T22:17:03.660" Score="10" ViewCount="0" Body="&lt;p&gt;Essentially because the connected 1-manifolds are (&quot;up to...&quot;) $(0,1)$ and $S^1$, so the notion of curve captures all of the possibilities of 1-manifolds sitting in higher-dimensional manifolds. In other words, the situation for 1-dimensional manifolds is so simple that it really makes no sense to use the full machinery of embeddings and immersions to talk about them other than just checking that the definitions of embedding and immersion are compatible with the definition of curve.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-05T22:17:03.660" />
  <row Id="13165" PostTypeId="2" ParentId="13150" CreationDate="2010-12-05T22:18:44.373" Score="2" ViewCount="0" Body="&lt;p&gt;Esentially you need to solve the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[\begin{array}{ccc}&#xD;&#xA;\mathrm{a} &amp;amp; \mathrm{b} &amp;amp; \mathrm{ty}\\&#xD;&#xA;\mathrm{c} &amp;amp; \mathrm{d} &amp;amp; \mathrm{ty}\end{array}\right]=\left[\begin{array}{ccc}&#xD;&#xA;s_{x}\cos\psi &amp;amp; -s_{x}\sin\psi &amp;amp; x_{c}\\&#xD;&#xA;s_{y}\sin\psi &amp;amp; s_{y}\cos\psi &amp;amp; y_{c}\end{array}\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s_x$, $s_y$ are the scalings, $x_c$, $y_c$ is the translation and $\psi$ is the rotation angle. The results I get are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{c}=\mathrm{tx}$$&#xD;&#xA;  $$y_{c}=\mathrm{ty}$$&#xD;&#xA;  $$s_{x}=\mathrm{sign(a)\,}\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$$&#xD;&#xA;  $$s_{y}=\mathrm{sign(d)\,}\sqrt{\mathrm{c}^{2}+\mathrm{d}^{2}}$$&#xD;&#xA;  $$\tan\psi=-\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{d}}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="3301" LastEditorDisplayName="" LastEditDate="2010-12-05T22:30:43.847" LastActivityDate="2010-12-05T22:30:43.847" CommentCount="3" />
  <row Id="13166" PostTypeId="1" AcceptedAnswerId="13185" CreationDate="2010-12-05T22:29:10.250" Score="2" ViewCount="248" Body="&lt;p&gt;Sometimes when you want to solve an equation you can just use algebra and rearrange it then you are done. But sometimes no amount of algebra can ever prove the equation, and then you need an idea, here is what I mean by the idea:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;parity (or modular arithmetic for higher numbers)&lt;/li&gt;&#xA;&lt;li&gt;complex integers.&lt;/li&gt;&#xA;&lt;li&gt;the irrational number&lt;/li&gt;&#xA;&lt;li&gt;a special function&lt;/li&gt;&#xA;&lt;li&gt;a strange curve&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A lot of equations can be proved impossible if the left hand side is odd and the right hand side is even, sometimes I had a pair of equations that turned out to be easy if you read it as a real and imaginary equation, I read to solve Pell's equation you need a fraction close to the square roots of $d$ and to solve $ax + by = 1$ you have use the greatest common divisor function. To solve Fermat's Last theorem you need elliptic curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions, Where can I find more examples of this (an equation and then what it is really - which lets you prove it)? and how do you find out what is behind the equation aka. what is the idea (because I only see the + - * ^ and variables..)?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-06T00:27:05.840" LastActivityDate="2010-12-07T01:40:06.353" Title="What are Diophantine equations REALLY?" Tags="&lt;number-theory&gt;&lt;soft-question&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13167" PostTypeId="2" ParentId="13129" CreationDate="2010-12-05T22:36:59.673" Score="1" ViewCount="0" Body="&lt;p&gt;It is unclear to me on what grounds you claim that $g(x)b=f(x)$ will have more than one solution for $x$. Even assuming that $K$ were algebraically closed, how do you know that you don't have one solution which is a multiple solution? For a very simple example, consider the case of $K=\mathbf{F}_2$, the field with two elements, and take $g(x) = x^2+x+1$ and $f(x)=x$. Then $g(x)0 = f(x)$ has one solution ($x=0$) and $g(x)1 = f(x)$ has one solution ($x=1$), so it does not follow merely from the degrees of $f$ and $g$: there is more work to be done. I don't know for sure if you can push it through, but you are correct that the first step is showing that the degrees have to be at most one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know two proofs: a direct way and a clever way. The direct way is to start as you do, with $\theta(x)=\frac{f(x)}{g(x)}$. Then consider an $\alpha = p(x)/q(x)$. If you write $f(x) = a_nx^n+\cdots +a_0$ and $g(x)=b_nx^n+\cdots+b_0$, with $a_n$ and $b_n$ not both zero, and you write $p(x) = c_mx^m+\cdots+c_0$, $q(x) = d_mx^m+\cdots+d_0$, with $c_m$ and $d_m$ both nonzero, then you can think about the degrees of the numerator and denominator of $\theta(\alpha)$. Try to think about what you would need for the degrees to drop after applying $\theta$ in order to conclude information about what can map &lt;strong&gt;to&lt;/strong&gt; $x$ under $\theta$. From there what you want will follow. This proof basically amounts to working things through in the obvious manner until you get to where you want to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second proof I know is the &quot;clever proof&quot;. Again, take $\theta(x)=\frac{f(x)}{g(x)}$ as you do. Note that $\theta(x)\notin K$. Show that $\theta(x)$ is transcendental over $K$ by showing that $x$ is algebraic over $K(\theta(x))$, and find its irreducible polynomial. Then use the fact that $K(\theta(x))$ is isomorphic to $K(x)$ to deduce that the irreducible polynomial must be of degree $1$. The irreducible polynomial turns out to be closely related to $f$ and $g$, of course. (I also like the fact that Gauss's Lemma comes into play to show that the obvious polynomial satisfied by $\theta(x)$ is in fact irreducible in the field you are looking at).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T16:32:44.347" LastActivityDate="2010-12-06T16:32:44.347" CommentCount="0" />
  <row Id="13168" PostTypeId="2" ParentId="13157" CreationDate="2010-12-05T22:39:46.440" Score="1" ViewCount="0" Body="&lt;p&gt;Failure to recognize that an indefinite integral is only defined up to a constant can lead to strange results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\int \frac{1}{x} dx &amp;amp; = \frac{1}{x} x - \int x \frac{(-1)}{x^2} dx \\ &#xD;&#xA;\int \frac{1}{x} dx &amp;amp; = 1 + \int \frac{1}{x} dx \end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, subtracting the integral from both sides, we obtain $0=1$. As troll face would say:&quot;Math will go bankrupt!&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-05T22:39:46.440" />
  <row Id="13169" PostTypeId="1" AcceptedAnswerId="13175" CreationDate="2010-12-05T22:41:48.630" Score="4" ViewCount="163" Body="&lt;p&gt;Let $X = \textrm{Spec} \mathbf{C}[[x,y]]/(y^{2} - x^{3} - x^{2})$. I would like to describe X set-theoretically. My questions are: Can one explicitly say what the elements in X are? Is it possible to interpret them geometrically? And is X irreducible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not really sure where to get started. Any help would be much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="4457" LastActivityDate="2010-12-05T23:37:16.537" Title="What is Spec $\mathbf{C}[[x,y]]/(y^{2} - x^{3} - x^{2})$?" Tags="&lt;homework&gt;&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13170" PostTypeId="2" ParentId="13122" CreationDate="2010-12-05T22:49:00.013" Score="1" ViewCount="0" Body="&lt;p&gt;EDIT: NEW ANSWER&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is convenient to express the solution in terms of convolutions: for any distribution $\mu$, positive integer $N$, and $1 \leq n \leq N$, we have&#xA;$$&#xD;&#xA;\mu ^{*N!}  = \big(\mu ^{*(N!/n)} \big)^{*n}.&#xD;&#xA;$$ &#xA;So simple, yet instructive...&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T15:33:54.187" LastActivityDate="2010-12-06T15:33:54.187" CommentCount="3" />
  <row Id="13171" PostTypeId="2" ParentId="13166" CreationDate="2010-12-05T22:52:37.047" Score="2" ViewCount="0" Body="&lt;p&gt;It sounds like what you mean is how can I find the motivation for studying a Diophantine equation by looking at the equation.  In the case of Fermat's Last Theorem, it is quite a natural equation to study.  I would disagree that it is really about elliptic curves, but rather say that it is about the natural numbers.  It so happens that elliptic curves gave the tools to solve the problem.  Similarly for Pell's equation, you can just wonder about solutions to $x^2-2y^2=1$, find several by hand, and find the recurrence that generates them without knowing about continued fractions and the expansion of $\sqrt{2}$.  Knowing about continued fractions and rings where you add a square root of some element suggest new areas of study, which has been quite productive.  But I don't know how to discern that from the equation itself.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T22:52:37.047" CommentCount="7" />
  <row Id="13172" PostTypeId="2" ParentId="13151" CreationDate="2010-12-05T22:56:33.190" Score="1" ViewCount="0" Body="&lt;p&gt;For your first matrix, the rank could be 1 if $n=1$ or $n=-1$ (because there would only be one pivot column).  For you second example, the rank could be 1,2,or 3 depending on x,y, and z.  For instance, if $x=y=z$ there are only non-zero entries in the first row of the reduced matrix.  You may want to look at the invertible matrix theorem to help you with this second example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.sunysb.edu/~yan2000/ams210_f2005/TheInvertibleMatrixTheorem.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.sunysb.edu/~yan2000/ams210_f2005/TheInvertibleMatrixTheorem.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, a square matrix has &quot;full rank&quot; iff it is invertible.  This makes your first question trivial.  For the second one, think about the values of $x,y,z$ that make the matrix singular, then classify these as rank 1 or 2.  Any combination of $x,y,z$ making the matrix invertible implies the resulting matrix has rank 3.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-05T23:36:11.253" LastActivityDate="2010-12-05T23:36:11.253" />
  <row Id="13173" PostTypeId="2" ParentId="13151" CreationDate="2010-12-05T23:01:27.443" Score="4" ViewCount="0" Body="&lt;p&gt;You seem to be assuming that because &quot;$1-n^2$&quot; doesn't look like $0$, then it cannot be zero. That is a common, but often fatal, mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that $n$ stands for &lt;em&gt;some&lt;/em&gt; integer. Once you get to&#xA;$$A = \left(\begin{array}{cc}&#xD;&#xA;1 &amp;amp; n\\&#xD;&#xA;0 &amp;amp; 1-n^2&#xD;&#xA;\end{array}\right),$$&#xD;&#xA;you cannot just jump to saying there are two pivots: your next step would be to divide the second row by $1-n^2$ to make the second pivot, but whenever you divide by something, that little voice in your head should be whispering in your ear: &quot;Wait! Are you sure you are not dividing by zero?&quot; (remember, if you divide by zero, &lt;em&gt;the universe explodes!&lt;/em&gt;). And the thing is, you &lt;strong&gt;aren't&lt;/strong&gt; sure you are not dividing by zero. It depends on what $n$ is! So, your answer should be that it will be rank $2$  &lt;em&gt;if&lt;/em&gt; $1-n^2\neq 0$, and rank $1$  &lt;em&gt;if&lt;/em&gt; $1-n^2 = 0$. But you don't want the person who is grading/reading to have to figure out &lt;em&gt;when&lt;/em&gt; that will happen. You want them to be able to glance at the original matrix, and then be able to immediately say (correctly) &quot;Rank is 1&quot; or &quot;Rank is 2&quot;. So you should express the conditions in terms of $n$ alone, not in terms of some computation involving $n$. So your final answer should be something like &quot;$\mathrm{rank}(A)=2$ if $n=\text{someting}$, and $\mathrm{rank}(A)=1$ if $n=\text{something else}$.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens with the second matrix: in order to be able to multiply by $-(\frac{z-x}{y-x})$, that little voice in your head will whisper &quot;Wait! are you sure you are not dividing by zero?&quot;, which leads you to consider what happens when $y-x=0$. But more: even if you are sure that $y-x\neq 0$, that meddlesome little voice should be whispering &quot;Wait! Are you sure you are not &lt;em&gt;multiplying&lt;/em&gt; the row by zero?&quot; (because, remember, multiplying a row by zero is &lt;em&gt;not&lt;/em&gt; an elementary row operation). (And be careful: if you don't pay attention to that voice, it's going to start yelling instead of whispering...) So that means that you &lt;em&gt;also&lt;/em&gt; need to worry about what happens when $z-x=0$. The answer on the rank of $B$, then, will depend on how $x$, $y$, and $z$ relate, and so your solution should reflect that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-05T23:16:28.437" LastActivityDate="2010-12-05T23:16:28.437" CommentCount="2" />
  <row Id="13174" PostTypeId="1" AcceptedAnswerId="13197" CreationDate="2010-12-05T23:23:32.597" Score="7" ViewCount="157" Body="&lt;p&gt;I have the following system of equations where the $m$'s are known but $a, b, c, x, y, z$ are unknown. How does one go about solving this system? All the usual linear algebra tricks I know don't apply and I don't want to do it through tedious substitutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;a + b + c &amp;amp;= m_0 \\&#xA;ax + by + cz &amp;amp;= m_1 \\&#xA;ax^2 + by^2 + cz^2 &amp;amp;= m_2 \\&#xA;ax^3 + by^3 + cz^3 &amp;amp;= m_3 \\&#xA;ax^4 + by^4 + cz^4 &amp;amp;= m_4 \\&#xA;ax^5 + by^5 + cz^5  &amp;amp;= m_5&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="1427" LastEditorUserId="1427" LastEditorDisplayName="" LastEditDate="2010-12-06T02:58:13.257" LastActivityDate="2010-12-06T15:17:48.337" Title="Solving a peculiar system of equations" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="13175" PostTypeId="2" ParentId="13169" CreationDate="2010-12-05T23:37:16.537" Score="13" ViewCount="0" Body="&lt;p&gt;Geometrically you are considering germ of singular node at the origin, that is at the singularity. The node is  $  N = \textrm{Spec} ( \mathbf{C}[x,y]/(y^{2} - x^{3} - x^{2})) $  and is irreducible affine curve, because polynomial $y^{2} - x^{3} - x^{2}$ is irreducible in polynomial ring  $\mathbf{C}[x,y]$. However your germ $X$ is reducible: this is because polynomial $y^{2} - x^{3} - x^{2}$ become reducible in formal powers series ring : $\; y^{2} - x^{3} - x^{2}=(y-x\sqrt{1+x})(y+x\sqrt{1+x})\in \mathbf{C}[[x,y]]$&#xD;&#xA; where $\sqrt{1+x}= 1+\frac{1}{2}x+...$ can be developed by Newton binomial in &#xA;$\mathbf{C}[[x,y]]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in scheme sense $X$ contains three points:  origin (= singularity )and two generic points of two irreducible components of $N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very intersting because node remains irreducible in every neighbourhood of origin in affine plane  $\textrm{Spec} ( \mathbf{C}[x,y])$ but by going to formal series you obligate curve to split in two components. So intuition should be that going to formal series is strong form of localization.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2010-12-05T23:37:16.537" CommentCount="1" />
  <row Id="13176" PostTypeId="1" AcceptedAnswerId="13179" CreationDate="2010-12-05T23:42:22.073" Score="2" ViewCount="107" Body="&lt;p&gt;Given the line $x+2=\frac{y+4}{2}=\frac{z-5}{-2}$&lt;br&gt;&#xA;I want to find the closest point on this line to $(1,1,1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the details here don't matter but in general how is this done?  We need a vector perpendicular to the line but that also reaches $(1,1,1)$.&lt;/p&gt;&#xA;" OwnerUserId="2533" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-06T00:00:13.110" LastActivityDate="2010-12-06T00:19:47.157" Title="How to find a point on a line closest to another given point?" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="13177" PostTypeId="2" ParentId="13153" CreationDate="2010-12-05T23:44:58.857" Score="0" ViewCount="0" Body="&lt;p&gt;Instead of working with $\tan $ ,$\sec $ , use their $\sin \text {and} \cos $ definitions in their place, the problem will be reduced to reverse application of product rule. way easier than to play around with than  $\tan $ and $\sec $ .&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-05T23:44:58.857" />
  <row Id="13178" PostTypeId="2" ParentId="13176" CreationDate="2010-12-05T23:49:12.113" Score="2" ViewCount="0" Body="&lt;p&gt;Using multi-variable calculus you would write a formula for distance from the point (1,1,1) to the arbitrary point (x,y,z) (this second points needs to satisfy your equation since it lies on the line).  You would then apply the multi-variable optimization procedure:  Find the critical points of this functions (those points where both partial derivatives equal zero or those points where either partial derivative is undefined) and examine these to find the minimum value for distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do want to use linear algebra, a vector from the point (x,y,z) (on the  line) to the point (1,1,1) is $&amp;lt;1-x,1-y,1-z&amp;gt;$  Now take a directional vector from your line and compute the dot product of this with $&amp;lt;1-x,1-y,1-z&amp;gt;$.  As you said, this dot product should be zero so solve for x,y, and z.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-05T23:49:12.113" CommentCount="1" />
  <row Id="13179" PostTypeId="2" ParentId="13176" CreationDate="2010-12-05T23:59:36.993" Score="6" ViewCount="0" Body="&lt;p&gt;One point on the line is (-2,-4,5) and another is (0,0,1) (found by setting the equation to 0, then to 2).  So the general point is (-2t, -4t,1+4t).  You want the dot product of the vector from a point on the line to (1,1,1) and a vector along the line to be zero.  So&#xA;$$(-2,-4,4)\cdot(-2t-1,-4t-1,4t)=0$$ &#xA; $$4t+2+16t+4+16t=0$$&#xD;&#xA; $$t=\frac{-1}{6}$$&#xD;&#xA;And the point is $$(\frac{1}{3},\frac{2}{3},\frac{1}{3})$$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-05T23:59:36.993" />
  <row Id="13180" PostTypeId="2" ParentId="13135" CreationDate="2010-12-06T00:00:22.393" Score="2" ViewCount="0" Body="&lt;p&gt;This is a job for the &lt;strong&gt;Green and Brown Fact&lt;/strong&gt;, namely Theorem 8 on page 14 of &lt;a href=&quot;http://www.math.uga.edu/~pete/3200relationsfunctions.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt;.  (The name comes from the colors of the chalk I used to draw a box around it that day in class.)  If you have a composition of functions $g \circ f$, this tells you that if the composite function is injective, $f$ is injective, and if the composite function is surjective, $g$ is surjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asked by the OP fits so nicely into this framework that I feel embarrassed not to have discussed it either time I taught the course.  Next time I will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ooh, another good exercise: what if $f \circ f \circ f = \operatorname{Id}_A$, and so forth.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-06T00:00:22.393" />
  <row Id="13181" PostTypeId="2" ParentId="13148" CreationDate="2010-12-06T00:01:07.710" Score="4" ViewCount="0" Body="&lt;p&gt;Take an arbitrary simple polygon $\pi$ including every point of $S$, and take some segment $PQ$ of the convex hull. Extend it into a line $\ell$. Form a new polygon $\pi&amp;#39;$ by locating the &quot;left&quot;most and &quot;right&quot;most intersection points $L,R$ of $\ell$ with $\pi$, adding new vertices to $\pi$ at $L,R$, and replacing the part of $\pi$ between $L$ and $R$ by the single line segment $LR$. The resulting polygon $\pi&amp;#39;$ is a simple polygon including all points of $S$, and the triangle inequality implies that its perimeter is smaller or equal to the perimeter of $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-06T00:01:07.710" CommentCount="1" />
  <row Id="13182" PostTypeId="2" ParentId="13176" CreationDate="2010-12-06T00:01:28.890" Score="1" ViewCount="0" Body="&lt;p&gt;First express $y$ and $z$ in terms of $x$: $y=y(x)$, $z=z(x)$. You want to find the minimum of $f(x) = (x-1)^2 + (y(x)-1)^2 + (z(x)-1)^2$. Noting that $f(x)$ is a parabola, you should find easily the (single) $x$ that minimizes $f(x)$ (global minimum). Then, you get $y(x)$ and $z(x)$ by substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Explicitly, $y(x)=2x$ and $z(x)=1-2x$. This leads to $f(x)=9x^2-6x+2$, whose minimum is obtained at $x=1/3$. The solution is then $(x,y,z)=(1/3,2/3,1/3)$, as already given in two previous answers.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T00:19:47.157" LastActivityDate="2010-12-06T00:19:47.157" />
  <row Id="13183" PostTypeId="2" ParentId="13176" CreationDate="2010-12-06T00:03:07.703" Score="1" ViewCount="0" Body="&lt;p&gt;One method is to parametrize the line. So&#xA;$$\vec{r}(t) = (t-2)\hat{\imath} + (2t-4)\hat{\jmath} + (-2t+5)\hat{k}$$&#xD;&#xA;We can then define the real-valued function $d(t)$ representing the distance from the line to $(1,1,1)$ for any value of $t$. Hence&#xA;$$d(t) = \sqrt{(t-3)^2 + (2t-5)^2 + (-2t+4)^2} = \sqrt{9t^2 - 42t + 40}$$&#xD;&#xA;To find the minimum, it's now only a single variable optimization problem. So by standard procedure, look at $d&amp;#39;(t)$&#xD;&#xA;$$d&amp;#39;(t) = \frac{9t - 21}{\sqrt{9t^2 - 42t + 40}}$$&#xD;&#xA;The critical point occurs at $t = \frac{7}{3}$, so the closest point is $\vec{r}(\frac{2}{3}) = (\frac{1}{3},\frac{2}{3},\frac{1}{3})$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-06T00:08:13.737" LastActivityDate="2010-12-06T00:08:13.737" />
  <row Id="13184" PostTypeId="2" ParentId="13043" CreationDate="2010-12-06T00:04:27.217" Score="0" ViewCount="0" Body="&lt;p&gt;You've got the right start by conditioning on $N$ and $P$ to get $E[e^{tX}] = E[(Pe^t + 1-P)^N]$.  Now, evaluate the second expression via&#xA;$$E[(Pe^t + 1-P)^N] = E[E[(Pe^t + 1-P)^N|P]].$$&#xD;&#xA;Since $N$ is Poisson, you will be able to get a closed-form answer for the inner expectation.  The only remaining variable will then be $P$.  But since $P$ is beta with small integer parameters the outer expectation will involve an integral of a low-degree polynomial times an exponential.  That can be evaluated via integration by parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Update with more details, after comments below.  This problem is more complicated than I had realized at first!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the calculations described above, we have, where $\lambda = 11$,&#xA;$$E[(Pe^t + 1-P)^N|P] = e^{\lambda P(e^t-1)}.$$&#xD;&#xA;Then $E[e^{\lambda P(e^t-1)}]$ must be done piecewise, depending on the value of $t$.  If $t = 0$, we have&#xA;$$E[e^{\lambda P(e^t-1)}] = E[1] = 1.$$&#xD;&#xA;Otherwise, $E[e^{\lambda P(e^t-1)}]$ is a much more complicated expression with, as the OP indicates, $(e^t-1)^4$ in the denominator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[e^{\lambda P(e^t-1)}] = \frac{12 \left(6 - 6 e^{\lambda (e^t-1)} + 2 \lambda (2 + e^{\lambda (e^t-1)}) (e^t-1) + &#xD;&#xA;   \lambda^2 (e^t-1)^2\right)}{\lambda^4 (e^t-1)^4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the limit of this latter expression as $t \to 0$ is $1$, $E[e^{\lambda P(e^t-1)}]$ is still continuous at $0$, despite the fact that it has to be calculated piecewise.  It also turns out to be differentiable at $t = 0$ as well, and so one can still find moments with it.  For example, to find $E[X]$ we differentiate the expression for $E[e^{\lambda P(e^t-1)}]$, $t \neq 0$, once with respect to $t$ and then take the limit as $t \to 0$.  This gives $E[X] = 4.4$, which is the same result one would obtain by calculating $E[X]$ directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, a similar situation occurs with the continuous uniform distribution on $[a,b]$.  Its mgf is, for $t \neq 0$, $$\frac{e^{tb}-e^{ta}}{t(b-a)}.$$  The moments can be found by successively differentiating this expression and then taking the limit as $t \to 0$ rather than by substituting $0$ for $t$.  See, for example, MathWorld's article on the &lt;a href=&quot;http://mathworld.wolfram.com/UniformDistribution.html&quot; rel=&quot;nofollow&quot;&gt;uniform distribution&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-06T05:04:58.393" LastActivityDate="2010-12-06T05:04:58.393" CommentCount="4" />
  <row Id="13185" PostTypeId="2" ParentId="13166" CreationDate="2010-12-06T00:40:46.630" Score="6" ViewCount="0" Body="&lt;p&gt;I interpret the question as &quot;how do I know how to attack a Diophantine problem&quot;. The question is in fact not easy and it is this skill that algebraic number theorists hone. Moreover, the methods that you listed are where the story was about 300 years ago. Today, there are more sophisticated and more general techniques available. If you are more interested in the motivation, you should have a look at &lt;a href=&quot;http://mathoverflow.net/questions/42406/why-certain-diophantine-equations-are-interesting-and-others-are-not&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt;, which somewhat goes in your direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I start listing the modern methods, let me directly address your questions &quot;what is behind a given equation&quot; and &quot;how do I know what to try?&quot;, starting with the latter: you don't. After a while you develop some intuition, but still, the basic way to find an approach that works is to try them all out in the order of decreasing likelihood of success. It is this likelihood that you can estimate better and better, as you become more proficient, but you will never know for sure, until you try an approach. As for the question &quot;what is behind a Diophantine equation&quot;, there is no good way of making sense of this question at present. Some people will view the equation as describing a geometric object (see last paragraph of this post), some people will look at it from a &quot;modular angle&quot; (see penultimate paragraph). But at the end of the day, when you are interested in integral or rational solutions, the equation is just that: a Diophantine equation. If you must categorise equations, then to categorise them according to the geometry and topology of the set of complex solutions is probably the most sensible thing to do (see last paragraph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two or three rather broad themes in modern research, where modern means everything from the last 150-odd years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a classical method that you have hinted at but that has much more potential is that often, to understand integer solutions, you are forced to work in a bigger ring. You have listed square roots of integers, but the technique is more general than that. To master it, you need to learn some classical algebraic number theory, as it was developed at the end of the 19th - beginning of the 20th century and that's also where I would recommend you to start reading. Have a look at an introductory book into algebraic number theory, such as &lt;a href=&quot;http://books.google.co.uk/books?id=PIibasv45boC&amp;amp;dq=algebraic+number+theory+books&amp;amp;printsec=frontcover&amp;amp;source=in&amp;amp;hl=en&amp;amp;ei=riz8TKfEEZCosAPB69X3DQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=12&amp;amp;sqi=2&amp;amp;ved=0CGgQ6AEwCw#v=onepage&amp;amp;q=algebraic%20number%20theory%20books&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the book by Ian Stewart&lt;/a&gt;, which I personally quite like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another broad theme is the one successfully used by Ribet, Frey, Wiles and several others along the way to prove Fermat's Last Theorem. It is nowadays subsumed under the mysterious bracketing term &quot;modularity&quot;. To start understanding, what this is about, you first need to learn about modular forms and elliptic curves. The basic idea is that the Shimura-Tanyiama-Weil conjecture, which was the actual result Wiles proved, relates two seemingly unrelated objects: rational elliptic curves and modular forms. This is extremely useful, because modular forms are extremely well-behaved. The &quot;modularity&quot;-idea of solving Diophantine equations then is to construct an elliptic curve out of a putative solution to your given equation that has such strange properties that it cannot possibly be modular. That would then contradict Wiles's theorem, so there cannot be such a solution. The places to start reading about elliptic curves and modular forms are (after you have completely read an introductory book on algebraic number theory and done all the exercises) &lt;a href=&quot;http://books.google.com/books?id=Z90CA_EUCCkC&amp;amp;lpg=PR1&amp;amp;ots=3K9dkyWh06&amp;amp;dq=Silverman%20elliptic%20curves&amp;amp;pg=PR1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Silverman&lt;/a&gt; - the classic on elliptic curves, and maybe &lt;a href=&quot;http://www.springer.com/mathematics/numbers/book/978-0-387-23229-4&quot; rel=&quot;nofollow&quot;&gt;the book by Diamond and Schurman&lt;/a&gt; for modular forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a very broad theme is that often, the geometry or the topology of the complex solutions of the equation controls its arithmetic (i.e. Diophantine) behaviour. It is difficult to point to one place where to learn about this, but elliptic curves are definitely the right point to start. I think, once you have read a book on algebraic number theory and one on elliptic curves, you should just come back here and ask this question again with your new background.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-07T01:40:06.353" LastActivityDate="2010-12-07T01:40:06.353" CommentCount="2" />
  <row Id="13186" PostTypeId="1" AcceptedAnswerId="13189" CreationDate="2010-12-06T00:51:34.697" Score="5" ViewCount="179" Body="&lt;p&gt;I was working through a textbook on topology and I came across a problem I couldn't solve.&lt;/p&gt;&#xA;&#xA; &lt;p&gt;1) It is known that &lt;b&gt;if a space is T1, it is countably compact if and only if every &lt;i&gt;countable&lt;/i&gt; open cover has a &lt;i&gt;finite&lt;/i&gt; subcover.&lt;/b&gt; (See below for the definition of countably compact, which might be different than the conventional definition.)&lt;/p&gt; &#xA;&#xA;&lt;p&gt;2) It is also true that &lt;b&gt;if a space is T1, it is countably compact if and only if every &lt;i&gt;infinite&lt;/i&gt; open cover has a &lt;i&gt;proper&lt;/i&gt; subcover.&lt;/b&gt; (why?)&lt;/p&gt; &#xA;&#xA;&lt;p&gt;Intuitively, both properties seem to talk about how open covers can be removed of unnecessary elements and still work as a cover, under conditions where points are sufficiently close together. However, I cannot figure out a proof for the second statement. Because the cover may contain uncountably many sets, it is very hard to deal with.&lt;/p&gt; &#xA;&#xA;&lt;p&gt;This question appears in &quot;Elements of Point Set Topology&quot; by John D. Baum as exercise 3.33. The question and related hint can be viewed &lt;a href=&quot;http://books.google.com/books?id=hlC6nK0tFBEC&amp;printsec=frontcover&amp;dq=elements+of+point+set+topology&amp;hl=en&amp;ei=KkH8TOrdGYKclgeUpIiZBQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCwQ6AEwAA#v=onepage&amp;q&amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Terminology used in this text:&lt;/b&gt;&lt;/p&gt; &#xA;&#xA;&lt;p&gt;A &lt;b&gt;T1 space&lt;/b&gt; is a topological space such that, if x is an element of the space, the set {x} is closed.&lt;/p&gt; &#xA;&#xA;&lt;p&gt;A &lt;b&gt;countably compact space&lt;/b&gt; is a space such that every infinite subset of the space has a limit point in the space. I am under the impression that other texts refer to this property as &lt;b&gt;limit point compactness&lt;/b&gt;.&lt;/p&gt; &#xA;&#xA;&lt;p&gt;An &lt;b&gt;infinite open cover&lt;/b&gt; is a collection of infinitely many open sets which cover the space.&lt;/p&gt; &#xA;" OwnerUserId="4460" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T02:13:54.223" LastActivityDate="2010-12-06T23:10:01.890" Title="A challenging question about T1 spaces and countable compactness" Tags="&lt;general-topology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13187" PostTypeId="1" CreationDate="2010-12-06T01:43:10.033" Score="0" ViewCount="89" Body="&lt;p&gt;On a practice final exam (for a Computer Security course), I am given the following equation to solve, but I have no idea how to solve it:&#xA;$$324x \mod{121} = 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any direction?&lt;/p&gt;&#xA;" OwnerUserId="4461" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:03:06.080" LastActivityDate="2010-12-06T06:50:43.517" Title="Newbie to modular arithmetic needs help" Tags="&lt;algebra&gt;&lt;modular-arithmetic&gt;" AnswerCount="2" />
  <row Id="13188" PostTypeId="2" ParentId="13187" CreationDate="2010-12-06T01:48:18.857" Score="3" ViewCount="0" Body="&lt;p&gt;You need the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt; which solves the problem of finding modular inverses.  &lt;a href=&quot;http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity&quot; rel=&quot;nofollow&quot;&gt;Bezout's identity&lt;/a&gt; guarantees that there is a solution.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T01:48:18.857" />
  <row Id="13189" PostTypeId="2" ParentId="13186" CreationDate="2010-12-06T02:17:48.520" Score="7" ViewCount="0" Body="&lt;p&gt;We shall show that if $X$ is a $T_1$ space, then it is countably compact if and only if every infinite open cover has a proper subcover. The key idea is to proceed by contrapositive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\Rightarrow$ Needs $T_1$) Suppose that $ \ \mathcal{U} \ $ is an infinite open cover of $X$ with no proper subcover. Then for each $U \in \mathcal{U}$, there is a point $p_U \in U$ that doesn't belong to any other member of $ \ \mathcal{U} \ $. The set $A = \{p_U : U \in \mathcal{U} \}$ is infinite and doesn't have a limit point. Indeed, if $x \in X$, then there is $U \in \mathcal{U}$ containing $x$ and $ U \cap A = \{ p_U \}$. If $x = p_U$, then it isn't a limit point of $A$. If $x \ne p_U$, then, using $T_1$, $U- \{ p_U \}$ is open and $(U- \{ p_U \}) \cap A =\emptyset$. So $x$ isn't a limit point of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;George Lowther gave an example showing that the $T_1$ hypothesis is fundamental. Since there are many comments, I'll reproduce it here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Consider the example of the real numbers where the open sets are unions of intervals $[n,a)$ for integer $n$ and real $a&amp;gt;n$. This is $T_0$ but not $T_1$. It is also countably compact, but the infinite cover $\mathcal{U}=\{[n,n+1)\colon n\in\mathbb{Z}\}$ has no proper subcovers. So, $T_1$ is needed.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A direct approach to prove this implication was suggested by Carl Mummert. Let $ \ \mathcal{U} \ $ be an infinite open cover of $X$ and $ \ \mathcal{U}_0 $ be a countably infinite subset of $ \ \mathcal{U} $. Now consider $ \ \mathcal{U}_1 = \mathcal{U} - \mathcal{U}_0$ and let $V \ $ be the union of all sets in $ \ \mathcal{U}_1$. Then $ \ \mathcal{U}_0 \cup \{ V \ \}$ is a countable open cover of $X$ and follows from $(1)$ that it has a finite subcover $U_1, \ldots, U_n$. Now the set consisting of all $U_i$, with possible exception of $V$, adjoined with the sets $ U \in \mathcal{U}_1$ is a proper subcover of $ \ \mathcal{U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\Leftarrow$ Doesn't need $T_1$) Now, suppose that $X$ isn't countably compact. Then there is an infinite set $A$ with no limit points. Thus $A$ is closed. By the definition of limit point, for each $x \in A$, there is an open set $U_x$ containing $x$ such that $ U_x \cap A = \{ x \} $. Consider $ \ \mathcal{U} \ $ the set of all $U_x$ plus, if necessary, $X-A$. Then $ \ \mathcal{U} \ $ is an infinite cover of $X$ with no proper subcovers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Thanks Carl Mummert, George Lowther and Mark for your efforts to clarify and improve this answer.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-06T23:10:01.890" LastActivityDate="2010-12-06T23:10:01.890" CommentCount="40" />
  <row Id="13190" PostTypeId="2" ParentId="13187" CreationDate="2010-12-06T02:22:43.773" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ One can easily solve it mod $11$ then lift the solution to one mod $11^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ mod\ 11:\ \ \ \ x = 1/(324) = 1/(-6) = -2\ \ $  so $\rm\ x = -2 + 11\ n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ \ mod\ 121:\ \ \: 1 = 324\ x = (-6 + 11\cdot 30)\ (-2 + 11\ n)\ \ =\ 55\ n - 43\ \ \Rightarrow\ \ 55\ n = 44$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by $11$ we deduce $\rm\ 5\ n = 4\ \:(mod\ 11)\ $ so $\rm\ n =\ \ldots\ \: $ The method generalizes, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad a\ x\ =\ 1 + b\ m\ \ \Rightarrow\ \ a\ x\ (2-a\ x)\ =\ (1 + b\ m)\ (1-b\ m) = 1 - (b\ m)^2\equiv 1\:\ (mod\ m^2) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $\rm\ a^{-1}\equiv x\:\ (mod\ m)\ $ lifts to $\rm\ a^{-1}\equiv x\ (2-a\ x)\ \ (mod\ m^2)\:.\:$&#xD;&#xA;This method of lifting solutions is a special case of a very powerful generalization of Newton's method known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hensel%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Hensel's Lemma.&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-06T06:50:43.517" LastActivityDate="2010-12-06T06:50:43.517" />
  <row Id="13191" PostTypeId="1" AcceptedAnswerId="13192" CreationDate="2010-12-06T02:38:08.510" Score="0" ViewCount="77" Body="&lt;p&gt;How many residue classes satisfy the congruence $x^3 \equiv 3 \pmod{21}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what this question is asking me to do.&lt;br&gt;&#xA;Can someone simplify the question for me, thanks.&lt;/p&gt;&#xA;" OwnerUserId="2588" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:28:26.797" LastActivityDate="2010-12-06T04:27:49.063" Title="How many residue classes satisfy the congruence $x^3 \equiv 3 \pmod{21}$?" Tags="&lt;modular-arithmetic&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="13192" PostTypeId="2" ParentId="13191" CreationDate="2010-12-06T02:48:31.450" Score="7" ViewCount="0" Body="&lt;p&gt;The question is asking you to check which of the 21 residue classes of integers modulo $21$ (to wit, the class of $0$, the class of $1$, the class of $2$, etc) are solutions to $x^3\equiv 3\pmod{21}$. If nothing else occurs to you, you can certainly plug and chug and figure out which ones are solutions and which ones are not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:07:26.107" LastActivityDate="2010-12-06T03:07:26.107" />
  <row Id="13193" PostTypeId="2" ParentId="13191" CreationDate="2010-12-06T02:49:30.527" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ mod\ 7:\ \ x^3 = 3\ \Rightarrow\ x^6 = \ldots\ $ contra a well-known &quot;little&quot; theorem.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-06T03:03:37.717" LastActivityDate="2010-12-06T03:03:37.717" />
  <row Id="13194" PostTypeId="2" ParentId="13191" CreationDate="2010-12-06T02:59:13.417" Score="3" ViewCount="0" Body="&lt;p&gt;Just a little something to add to Arturo's answer. Note that any solution to $x^3\equiv 3 \pmod{21}$ will also be a solution to $x^3\equiv 3\pmod{3}$ and $x^3\equiv 3\pmod{7}$. So instead of checking all $21$ congruence classes, you can begin by checking the congruence classes modulo the prime powers of $21$. If one happens to have no solution, then modulo $21$ there should be no solution either. It should save you some time, as there are fewer classes to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an application of the following theorem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x)$ be a fixed polynomial with integral coefficients, and for any positive interger $m$, let $N(m)$ denote the number of solutions of the congruence $f(x)\equiv 0\pmod{m}$. If $m=m_1m_2$, where $gcd(m_1,m_2)=1$, then $N(m)=N(m_1)N(m_2)$. If $m=\prod p^\alpha$ is the canonical factorization of $m$, then $N(m)=\prod N(p^\alpha)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-06T04:27:49.063" LastActivityDate="2010-12-06T04:27:49.063" CommentCount="7" />
  <row Id="13195" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T03:22:30.797" Score="4" ViewCount="0" Body="&lt;p&gt;Look at the first three equations as a system of linear equations determining $a$, $b$ and $c$ in terms of $x$, $y$, $z$, $m_0$, $m_1$, $m_2$; the matrix of coefficients is a van der Monde matrix for the triple $x$, $y$, $z$, so it not hard to use Cramer's rule to solve it when no two of them are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now replace the values you got for $a$, $b$ and $c$ in the second group of three equations. Magically, after simplifying and canceling things common to numerators and denominators, it is a system of &lt;em&gt;polynomial&lt;/em&gt; equations in $x$, $y$ and $z$. Moreover, the three left hand sides are symmetric polynomials in $x$, $y$ and $z$, so after expressing those left hand sides in terms of the elementary symmetric polynomials on $x$, $y$ and $z$, you end up with a not too difficult polynomial system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have the values of $x+y+z$, $xy+xz+yz$ and $xyz$. Solving a cubic equation now gives you $x$, $y$ and $z$ and, from the first part, also $a$, $b$ and $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; Following these steps is easily doable in &lt;em&gt;Mathematica&lt;/em&gt; and you get an answer. The answer is simply too huge, and I doubt it can be massaged into something sensible---and I hence doubt there is a sensible way to solve the system of equations by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mma&lt;/em&gt;, on the other hand, did not solve the system as it is originally given in the time I gave it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB2:&lt;/strong&gt; For comparison, the system $$\left\{\begin{aligned}a + b = m_0\\ax+by=m_1\\ax^2+by^2=m_2\\ax^3+by^3=m_3\end{aligned}\right.$$ has as solution, according to &lt;em&gt;Mma&lt;/em&gt; (and this one it does compute it by itself):&#xA;$$\begin{aligned}&#xD;&#xA; &amp;amp;a= \frac{-2 m_1^3+3 m_0 m_2 m_1+m_0 \left(\sqrt{4 m_3 m_1^3-3&#xD;&#xA;   m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}-m_0 m_3\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2&#xD;&#xA;   m_1^2-6 m_0 m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0&#xD;&#xA;   m_3^2\right)}} \\&#xD;&#xA; &amp;amp;b = \frac{2 m_1^3-3 m_0 m_2 m_1+m_0 \left(m_0 m_3+\sqrt{4 m_3&#xD;&#xA;   m_1^3-3 m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6 m_0&#xD;&#xA;   m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0 m_3^2\right)}} \\&#xD;&#xA; &amp;amp;x = \frac{-m_1 m_2+m_0 m_3+\sqrt{\left(m_1 m_2-m_0&#xD;&#xA;   m_3\right){}^2+4 \left(m_1^2-m_0 m_2\right) \left(m_1&#xD;&#xA;   m_3-m_2^2\right)}}{2 \left(m_0 m_2-m_1^2\right)} \\&#xD;&#xA; &amp;amp;y = \frac{m_1 m_2-m_0 m_3+\sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6 m_0&#xD;&#xA;   m_2 m_3 m_1+4 m_0 m_2^3+m_0^2 m_3^2}}{2 m_1^2-2 m_0 m_2}&#xD;&#xA;\end{aligned}&#xD;&#xA;$$ and&#xA;$$\begin{aligned}&#xD;&#xA; &amp;amp;a=\frac{2 m_1^3-3 m_0 m_2 m_1+m_0 \left(m_0 m_3+\sqrt{4 m_3&#xD;&#xA;   m_1^3-3 m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6 m_0&#xD;&#xA;   m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0 m_3^2\right)}} \\&#xD;&#xA; &amp;amp;b=\frac{-2 m_1^3+3 m_0 m_2 m_1+m_0 \left(\sqrt{4 m_3 m_1^3-3&#xD;&#xA;   m_2^2 m_1^2-6 m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2&#xD;&#xA;   m_3^2}-m_0 m_3\right)}{2 \sqrt{4 m_3 m_1^3-3 m_2^2&#xD;&#xA;   m_1^2-6 m_0 m_2 m_3 m_1+m_0 \left(4 m_2^3+m_0&#xD;&#xA;   m_3^2\right)}} \\&#xD;&#xA; &amp;amp;x=-\frac{m_1 m_2-m_0 m_3+\sqrt{\left(m_1 m_2-m_0&#xD;&#xA;   m_3\right){}^2+4 \left(m_1^2-m_0 m_2\right) \left(m_1&#xD;&#xA;   m_3-m_2^2\right)}}{2 \left(m_0 m_2-m_1^2\right)} \\&#xD;&#xA; &amp;amp;y=-\frac{-m_1 m_2+m_0 m_3+\sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6&#xD;&#xA;   m_0 m_2 m_3 m_1+4 m_0 m_2^3+m_0^2 m_3^2}}{2&#xD;&#xA;   \left(m_1^2-m_0 m_2\right)}&#xD;&#xA;\end{aligned}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-06T03:43:36.293" LastActivityDate="2010-12-06T03:43:36.293" />
  <row Id="13196" PostTypeId="1" AcceptedAnswerId="13199" CreationDate="2010-12-06T03:47:10.210" Score="2" ViewCount="119" Body="&lt;p&gt;Given that $f\colon [-1,1] \to \mathbb{R}$ is a continuous function such that $ \int_{-1}^{1} f(t) \ dt =1$, how do I evaluate the limit of this integral: $$\lim_{n \to \infty} \int_{-1}^{1} f(t) \cos^{2}{nt} \,dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did was to write $\cos^{2}{nt} = \frac{1+\cos{2nt}}{2}$ and substitute it in the integral so that I can make use of the given hypothesis of $\int_{-1}^{1} f(t) \ dt =1$. So the integral becomes, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int_{-1}^{1} f(t)\cos^{2}{nt} \ dt = \int_{-1}^{1} f(t) \biggl[\frac{1+\cos{2nt}}{2}\biggr] \ dt  &amp;amp;  \\  \hspace{3cm} =&#xA;\frac{1}{2}\int_{-1}^{1}f(t) \ dt + \int_{-1}^{1} \frac{f(t)\cos{2nt}}{2} \ dt \end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't really know how I can evaluate the second integral and also I can't realize as to why that &lt;em&gt;integral condition&lt;/em&gt; on $f$ has been assumed. Moreover without assuming that condition on $f$ is it possible to evaluate this integral? If yes, then what would the answer be?&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T03:59:09.770" LastActivityDate="2010-12-06T06:26:10.663" Title="Evaluating the integral $\mathop{\lim}\limits_{n \to \infty} \int_{-1}^{1} f(t)\cos^{2}(nt) \ dt $" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" />
  <row Id="13197" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T03:54:11.680" Score="6" ViewCount="0" Body="&lt;p&gt;Let $\displaystyle x,y,z$ be roots of $\displaystyle t^3 + pt^2 + qt + r = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle x^3 + px^2 + qx + r = 0$ &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;   -----   (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle y^3 + py^2 + qy + r = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; -----  (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle z^3 + pz^2 + qz + r = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the (1) by $\displaystyle a$, (2) by $\displaystyle b$ and (3) by $\displaystyle c$ and adding gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m_3 + p m_2 + q m_1 + rm_0 = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the (1) by $\displaystyle ax$, (2) by $\displaystyle by$ and (3) by $\displaystyle cz$ and adding gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m_4 + pm_3 + qm_2 + rm_1 = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the (1) by $\displaystyle ax^2$, (2) by $\displaystyle by^2$ and (3) by $\displaystyle cz^2$ and adding gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle m_5 + pm_4 + qm_3 + rm_2 = 0$&amp;nbsp;&amp;nbsp;&amp;nbsp; ----- (6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (4), (5), (6) is a set of 3 linear equations in 3 variables ($\displaystyle p,q,r$) and can be solved easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This give us the cubic which $\displaystyle x,y,z$ satisfy (because we can find out that $\displaystyle p,q,r$ are) which can be solved using &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method&quot; rel=&quot;nofollow&quot;&gt;Cardano's Method&lt;/a&gt;, or more simply by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Trigonometric_.28and_hyperbolic.29_method&quot; rel=&quot;nofollow&quot;&gt;Trigonometric and Hyperbolic Method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know $\displaystyle x,y,z$ you can solve for $\displaystyle a,b,c$, as those become just linear equations.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-06T07:23:11.153" LastActivityDate="2010-12-06T07:23:11.153" CommentCount="2" />
  <row Id="13198" PostTypeId="1" AcceptedAnswerId="13200" CreationDate="2010-12-06T04:01:47.673" Score="2" ViewCount="99" Body="&lt;p&gt;Hello all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably one of the more embarrassing moments of my life, but I will soldier on. Having not dealt with inequalities for quite some time, I was rather befuddled by a question from a friend of mine regarding solving for two variables given a pair of inequalities. The problem goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $x + y \lt 11$ and $x \gt 6$, what is the lower bound of $x - y$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure she told me the answer was $4$, but I've exhausted my bag of tricks trying to get there. If these were equalities, it would be fairly easy, of course, but would someone be so kind as to help nudge me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot in advance.&lt;/p&gt;&#xA;" OwnerUserId="1720" LastEditorUserId="1720" LastEditorDisplayName="" LastEditDate="2010-12-06T06:44:31.230" LastActivityDate="2010-12-06T06:44:31.230" Title="Solving for two unknowns in an inequality" Tags="&lt;algebra&gt;&lt;inequality&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="13199" PostTypeId="2" ParentId="13196" CreationDate="2010-12-06T04:04:39.920" Score="9" ViewCount="0" Body="&lt;p&gt;Define $g(t):[-\pi,\pi]\to \mathbb{R}$ such that $g(t)=f(t)$ for $t\in [-1,1]$ and $g(t)=0$ elsewhere. Then $g$ is integrable and by Lebesgue-Riemann lemma  $$\lim_{n\to\infty}\int_{-\pi}^\pi  g(t)\cos 2nt dt=0.$$ But this is just the same as $$\lim_{n\to\infty}\int_{-1}^1  f(t)\cos 2nt dt=0.$$ &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-06T04:04:39.920" />
  <row Id="13200" PostTypeId="2" ParentId="13198" CreationDate="2010-12-06T04:15:00.920" Score="7" ViewCount="0" Body="&lt;p&gt;Are $x$ and $y$ required to be &lt;em&gt;integers&lt;/em&gt;? Because if they are not, then the answer cannot be $4$: take $x=6.5$ and $y=4$. Then $x+y = 10.5\lt 11$, but $x-y = 2.5 \lt 4$. In fact, taking $x=6+\epsilon$ and $y=5-(\epsilon+\delta)$ with $\epsilon\gt 0$ and $\delta\gt 0$ arbitrary but positive, and this gives you $x+y = 11-\delta\lt 11$, but $x-y = 1+2\epsilon+\delta$, so we can make $x-y$ as close to $1$ as we like (without it ever being equal to $1$); on the other hand, since $x\gt 6$ and $x+y\lt 11$, then $y\lt 11-x \lt 11-6 = 5$, so $x-y \gt 6-5 = 1$. Thus, the difference is always greater than $1$; but since we can make it as close to $1$ as we want, the best possible lower bound for $x-y$ is $1$ (without it actually being achieved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $x$ and $y$ are required to be integers, then the smallest value of $x-y$ will occur when $x$ is as small as possible and $y$ as large as it possibly can be given that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; To see why, first suppose that $x+y\lt 11$ with $x\gt 6$, but $y$ is not as large as it can be; that is, that $x+(y+1)$ is &lt;em&gt;also&lt;/em&gt; less than $11$. Setting $y&amp;#39;=y+1$, you have that $x-y&amp;#39; = x-y-1 \lt x-y$, so you get a smaller value with $y&amp;#39;$. That means that in order to find the smallest possible value of $x-y$ for a given $x$, you may assume $y$ is as large as it possibly can be. Once you have that, you can check that larger values of $x$ will give you &lt;em&gt;larger&lt;/em&gt; differences, so you also want $x$ to be as small as it possibly can be.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T04:37:30.830" LastActivityDate="2010-12-06T04:37:30.830" CommentCount="1" />
  <row Id="13201" PostTypeId="2" ParentId="13198" CreationDate="2010-12-06T04:15:23.557" Score="2" ViewCount="0" Body="&lt;p&gt;The answer is $x-y&amp;gt;1$. What you can do is to graph the set $A$: $x+y&amp;lt;11, x&amp;gt;6$ and note that $x-y=1$ passes through the intersection of the two lines $x+y=11$ and $x=6$. The set $A$ is to the right of $x-y=1$, which is $x-y&amp;gt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x,y$ are integers, then the smallest integer $k$ such that the set $x-y\ge k$ includes all the points in $A$ is 4 ( $(7,3)$ lies on the line $x-y=4$). This is clear from the graph.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-06T06:00:18.477" LastActivityDate="2010-12-06T06:00:18.477" CommentCount="2" />
  <row Id="13202" PostTypeId="1" AcceptedAnswerId="13203" CreationDate="2010-12-06T05:08:25.753" Score="0" ViewCount="30" Body="&lt;p&gt;Alright, so I'm still trying to make my players movements relative to the camera. I got a good answer &lt;a href=&quot;http://stackoverflow.com/questions/4321382?tab=votes#tab-top&quot;&gt;here&lt;/a&gt; But he ends it with: $s(Jx \widehat{\mathbf{x}} + Jy \widehat{\mathbf{y}}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can I turn that equation into an $x$ and $y$?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T05:10:37.467" LastActivityDate="2010-12-06T05:21:49.807" Title="Splitting this into a x and y point?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13203" PostTypeId="2" ParentId="13202" CreationDate="2010-12-06T05:21:49.807" Score="1" ViewCount="0" Body="&lt;p&gt;If you read the answer you reference, $\widehat{\mathbf{x}}$ is a unit vector perpendicular to the vector from the camera to the player and $\widehat{\mathbf{y}}$ is a unit vector pointing towards the player from the camera.  Each has x and y coordinates in your coordinate plane.  $Jx$ is the x coordinate of the joystick position with a range depending on how you read it out.  Maybe it goes from 0 to 10 degrees.  Similarly for $Jy$.  So you form a new vector by multiplying and adding, then scale it by a constant $s$ that represents how fast the player should move for a given motion of the joystick.  This gives you a vector that you add to the current player position to get the new player position.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T05:21:49.807" CommentCount="2" />
  <row Id="13204" PostTypeId="2" ParentId="13154" CreationDate="2010-12-06T05:22:35.313" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, you're going to find clauses with contradictory literals and apply the resolution rule to those. So you're going to resolve the first two clauses in your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a couple of pages you can look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ai.mit.edu/courses/6.825/fall02/pdf/6.825-lecture-07.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ai.mit.edu/courses/6.825/fall02/pdf/6.825-lecture-07.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://logic.stanford.edu/classes/cs157/2005/notes/chap05.pdf&quot; rel=&quot;nofollow&quot;&gt;http://logic.stanford.edu/classes/cs157/2005/notes/chap05.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1884" LastActivityDate="2010-12-06T05:22:35.313" CommentCount="1" />
  <row Id="13205" PostTypeId="1" CreationDate="2010-12-06T05:24:11.497" Score="2" ViewCount="221" Body="&lt;p&gt;I would like to know why $\mathbb{C}[x,y]$ is not isomorphic to $\mathbb{C}[x] \otimes _{\mathbb{Z}} \mathbb{C}[y]$ as rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! 1&lt;/p&gt;&#xA;" OwnerUserId="4465" LastEditorUserId="4465" LastEditorDisplayName="" LastEditDate="2010-12-06T06:03:04.490" LastActivityDate="2010-12-06T13:56:26.000" Title="Why is $\mathbb{C}[x,y]$ not isomorphic to $\mathbb{C}[x] \otimes _{\mathbb{Z}} \mathbb{C}[y]$ as rings?" Tags="&lt;abstract-algebra&gt;&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="13206" PostTypeId="2" ParentId="13205" CreationDate="2010-12-06T05:35:24.417" Score="5" ViewCount="0" Body="&lt;p&gt;Because $\mathbb C \otimes_{\mathbb Z} \mathbb C$ is not isomorphic to&#xA;$\mathbb C$.  (There is a natural map from the first to the second, with an&#xA;absolutely enormous kernel.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: In light of the discussion below, it might help to add the remark that&#xA;this kernel is full of non-trivial zero-divisors (see e.g. Robin's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of this is that $\mathbb C \otimes_{\mathbb Z} \mathbb C = \mathbb C \otimes_{\mathbb Q} \mathbb C$ contains $\overline{\mathbb Q}\otimes_{\mathbb Q} \overline{\mathbb Q}$, and this latter ring already is full of zero divisors.  (Robin's answer gives one explicit example.  More generally,&#xA;$\overline{\mathbb Q}$ is the union of all finite Galois extensions $L$ of $\mathbb Q$ lying in $\mathbb C$, and for any such $L$, the tensor product&#xA;$L \otimes_{\mathbb Q} L$ is isomorphic to a product of $[L:\mathbb Q]$ copies of $L$.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-06T13:56:26.000" LastActivityDate="2010-12-06T13:56:26.000" CommentCount="5" />
  <row Id="13207" PostTypeId="1" CreationDate="2010-12-06T05:37:50.593" Score="1" ViewCount="117" Body="&lt;p&gt;Find the complete integral of partial differential equation&lt;br&gt;&#xA;&lt;code&gt;z^2 = pqxy&lt;/code&gt; ?&lt;br&gt;&#xA;I have solved this equation till auxiliary equation:   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dp/(-pqy+2pz) = dq/(-pqx+2qz) = dz/(2pqxy) = dx/(qxy) = dy/(pxy)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I have unable to find value of p and q.&lt;br&gt;&#xA;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p = ∂z/∂x&#xA;q = ∂z/∂y&#xA;r = ∂²z/∂x²  = ∂p/∂x&#xA;s = ∂²z/∂x∂y = ∂p/∂y or ∂q/∂x&#xA;t = ∂²z/∂y²  = ∂q/∂y&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4463" LastEditorUserId="4463" LastEditorDisplayName="" LastEditDate="2010-12-06T15:50:43.167" LastActivityDate="2011-01-05T20:44:39.410" Title="Charpit's Method" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13208" PostTypeId="2" ParentId="13205" CreationDate="2010-12-06T05:54:49.707" Score="4" ViewCount="0" Body="&lt;p&gt;Note that the two are not even isomorphic as $\mathbb{C}$-vector spaces, since the LHS has countable dimension, while for the RHS $\{1\otimes z|\;\bar{z}\in \mathbb{C}/\mathbb{Z}\}$ is an uncountable set that is linearly independent over $\mathbb{C}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: another way to see that the rings are not isomorphic, which is just an addition to Matt's answer, is e.g. to note that there is a different number of elements of finite multiplicative order $n$ for any $n&amp;gt;2$. E.g. on the LHS, an element must clearly be a scalar to be of finite multiplicative order, so the only elements of order 4 are $i$ and $-i$. On the other hand, on the RHS, $i\otimes 1$, $-i\otimes 1$, $1\otimes i$ and $-1\otimes i = 1\otimes -i$ are all distinct elements of order 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that two elements of the tensor product are really distinct (this applies to any tensor products), you have to construct a $\mathbb{Z}$-bilinear map from $\mathbb{C}[x]\times\mathbb{C}[y]$ to some other $\mathbb{Z}$-module that takes the two corresponding elements (in this case ($i$,1) and (1,$i$), say) to different images. To understand, why that is a necessary and sufficient condition for them to be distinct in the tensor product, go back to the definition of tensor product via a universal property.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-06T07:35:49.417" LastActivityDate="2010-12-06T07:35:49.417" CommentCount="6" />
  <row Id="13209" PostTypeId="2" ParentId="13109" CreationDate="2010-12-06T05:55:11.647" Score="1" ViewCount="0" Body="&lt;p&gt;Note that $x$ cannot occur free in any of the $T_{i}$ in order to apply the Generalization Theorem (namely: $\forall$ Introduction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick look at my old copy of [1] revealed the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Enderton proves the Generalization Theorem purely syntactically by induction on the logical axioms, formulas in which $x$ does not occur, and formulas derivable by modus ponens. (In Enderton's system, modus ponens is the only actual rule of inference; all the rest is lumped into lots of logical axioms.)&lt;/li&gt;&#xA;&lt;li&gt;He then proves Soundness entirely separately from the Generalization Theorem with a similar inductive argument with a lemma added to assure the validity of the logical axioms which surprisingly occupies the bulk of the proof.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.math.ucla.edu/~hbe/amil/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucla.edu/~hbe/amil/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1884" LastEditorUserId="1884" LastEditorDisplayName="" LastEditDate="2010-12-06T06:10:37.450" LastActivityDate="2010-12-06T06:10:37.450" />
  <row Id="13210" PostTypeId="2" ParentId="13062" CreationDate="2010-12-06T06:07:30.803" Score="4" ViewCount="0" Body="&lt;p&gt;Here's the topological argument.  The fact that $G$ is finitely generated means that $G=\pi_1(K)$ for $K$ a CW-complex with finite 1-skeleton.  Let $\widehat{K}$ be the covering space corresponding to $H$.  Then $H=\pi_1(\widehat{K})$, and $\widehat{K}$ also has finite 1-skeleton, so $H$ is finitely generated.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-12-06T06:07:30.803" CommentCount="2" />
  <row Id="13211" PostTypeId="2" ParentId="13067" CreationDate="2010-12-06T06:19:27.813" Score="0" ViewCount="0" Body="&lt;p&gt;Maybe it would better to say that &quot;&lt;em&gt;Special functions&lt;/em&gt; does not refer to any particular function, it is merely a mathematical field devoted to the study of some particular explicit function or class of functions in mathematical analysis.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-06T06:19:27.813" />
  <row Id="13212" PostTypeId="2" ParentId="13196" CreationDate="2010-12-06T06:26:10.663" Score="1" ViewCount="0" Body="&lt;p&gt;\begin{equation}&#xA;\int_{-1}^1 f(t) \cos^2 nt dt = 1 - \int_{-1}^1 f(t) \sin^2 nt dt &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use $\sin^2 nt = \cos^2(nt + \pi/2)$ and some manipulation to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{n \to \infty} \int_{-1}^1 f(t) \cos^2 nt dt = \lim_{n \to \infty} \int_{-1}^1 f(t) \sin^2 nt dt&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{n \to \infty} \int_{-1}^1 f(t) \cos^2 nt dt = \frac{1}{2}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-12-06T06:26:10.663" />
  <row Id="13213" PostTypeId="1" AcceptedAnswerId="13214" CreationDate="2010-12-06T07:07:14.383" Score="3" ViewCount="58" Body="&lt;p&gt;&lt;strong&gt;If $f(x)$ is a function satisfying $ \displaystyle f(x+y) = f(x) \cdot f(y)  \text { for all } x,y \in \mathbb{N} \text{ such that } f(1) = 3 \text { and } $ $ \sum_{x=1}^{n} f(x) = 120 $, then find the value of $n$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to approach this one?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-06T07:12:31.667" Title="Function Problem" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" />
  <row Id="13214" PostTypeId="2" ParentId="13213" CreationDate="2010-12-06T07:12:31.667" Score="7" ViewCount="0" Body="&lt;p&gt;Hint: Knowing the value of $f(1)$, you can get the value of $f(2)=f(1+1)$ from the defining equation. Knowing these two, you can get value of $f(3)=f(1+2)$ and so on.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-06T07:12:31.667" CommentCount="2" />
  <row Id="13215" PostTypeId="1" CreationDate="2010-12-06T07:59:47.883" Score="1" ViewCount="92" Body="&lt;p&gt;Let $(X,M,\mu)$ be a measure space. Why are the following statements equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) There exists a sequence of pairwise disjoint measurable sets $\{A_{n}\}$ in $M$ , each of finite measure, such that $\mu(A) = \sum_{n=1}^{\infty} \mu(A \cap A_{n})$ for every measurable set $A \in M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $\mu$ is a countable sum of pairwise mutually singular finite measures.&lt;/p&gt;&#xA;" OwnerUserId="4467" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T19:28:52.773" LastActivityDate="2010-12-07T19:28:52.773" Title="Equivalence of two properties of a measure space" Tags="&lt;measure-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13217" PostTypeId="2" ParentId="13215" CreationDate="2010-12-06T08:23:09.347" Score="1" ViewCount="0" Body="&lt;p&gt;(ii) $\implies$ (i)&#xA;Suppose $\mu = \sum_{n=1}^\infty \mu_n$, where $\mu_i \perp \mu_j$ for all $i \neq j$.  By definition of &quot;mutually singular,&quot; there exist pairwise disjoint $A_n \in M$ such that $\mu_n(A) = \mu_n(A \cap A_n)$ for all $A \in M$.  Therefore, $\mu(A_k) = \sum_{n=1}^\infty \mu_n(A_k) = \mu_k(A_k) &amp;lt; \infty$, so that each $A_n$ has finite $\mu$-measure.  Moreover, $\mu(A) = \sum_{n=1}^\infty \mu_n(A_n) = \sum_{n=1}^\infty \mu_n(A \cap A_n) = \sum_{n=1}^\infty \mu(A \cap A_n)$, where in the last step we used the fact that $\mu(A \cap A_n) = \mu_n(A \cap A_n)$ for all $n \in N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $\implies$ (ii)&#xA;Let $\mu_n$ be a measure concentrated on $A_n \in M$, normalized so that $\mu_n(A_n) = 1$.  Then $\mu(A) = \sum_{n=1}^\infty \mu(A\cap A_n) = \sum_{n=1}^\infty \mu_n(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT following Nate's comment: We can define $\mu_n(A) = \mu(A \cap A_n)$.)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-06T18:42:38.047" LastActivityDate="2010-12-06T18:42:38.047" CommentCount="2" />
  <row Id="13218" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T08:29:30.680" Score="2" ViewCount="0" Body="&lt;p&gt;@Moron's solution in insightful, but I'll present the (tedious, but thoughtlessly mechanical) methodology I tend to use to approach such problems: &lt;a href=&quot;http://mathworld.wolfram.com/Resultant.html&quot; rel=&quot;nofollow&quot;&gt;resultants&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resultant of two polynomials is a polynomial with one variable eliminated. Therefore, you can proceed to solve a polynomial system in roughly the same way you would with a linear system (in theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, convert your equations to polynomials, $p_i$, so that the &lt;em&gt;roots&lt;/em&gt; of these polynomials are the solutions to your equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;p_0 &amp;amp;=&amp;amp; a+b+c-m_0\\&#xD;&#xA;p_1 &amp;amp;=&amp;amp; a x + b y + c z - m_1\\&#xD;&#xA;p_2 &amp;amp;=&amp;amp; a x^2 + b y^2 + c z^2 - m_2\\&#xD;&#xA;p_3 &amp;amp;=&amp;amp; a x^3 + b y^3 + c z^3 - m_3\\&#xD;&#xA;p_4 &amp;amp;=&amp;amp; a x^4 + b y^4 + c z^4 - m_4\\&#xD;&#xA;p_5 &amp;amp;=&amp;amp; a x^5 + b y^5 + c z^5 - m_5&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick your favorite one (say, $p_0$), and use it to compute a bunch of resultants (using &lt;em&gt;Mathematica&lt;/em&gt;'s ${\tt Resultant[]}$ function) that eliminate one variable (say, $a$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;r_0 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_1, a {\tt ]} = -m_1 - b x - c x + m_0 x + b y + c z\\&#xD;&#xA;r_1 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_2, a {\tt ]} = -m_2 - b x^2 - c x^2 + m_0 x^2 + b y^2 + c z^2 \\&#xD;&#xA;r_2 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_3, a {\tt ]} = -m_3 - b x^3 - c x^3 + m_0 x^3 + b y^3 + c z^3\\&#xD;&#xA;r_3 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_4, a {\tt ]} = -m_4 - b x^4 - c x^4 + m_0 x^4 + b y^4 + c z^4\\&#xD;&#xA;r_4 &amp;amp;=&amp;amp; {\tt Resultant[} p_0, p_5, a {\tt ]} = -m_5 - b x^5 - c x^5 + m_0 x^5 + b y^5 + c z^5\\&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, the above amounts to simply solving $p_0 = 0$ for $a$ and substituting into the other $p_i$, since the system is linear in $a$. The same will be true when eliminating $b$ and $c$, so using resultants here is a little bit of overkill, but the process is mechanical enough to be about as easy to (not-)think about as, say, Gaussian elimination.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we use, say, $r_0$ to compute resultants that eliminate $b$, and so forth. For your special system, the polynomials at each step happen to have common factors that give rise to possible solutions; you might need to consider those separately, but they tend to represent &quot;special case&quot; scenarios. In the following, I base the next resultant step on the uncommon factors (indicated with upper-case variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;s_i &amp;amp;=&amp;amp; {\tt Resultant[} r_0, r_i, b {\tt ]} = S_i \cdot (x-y), \hspace{0.2in}i = 1, 2, 3, 4\\&#xD;&#xA;t_i &amp;amp;=&amp;amp; {\tt Resultant[} S_0, S_i, c {\tt ]} = T_i \cdot (y-z)(z-x), \hspace{0.2in}i = 1, 2, 3 \\&#xD;&#xA;u_i &amp;amp;=&amp;amp; {\tt Resultant[} T_0, T_i, x {\tt ]} = U_i \cdot (m_2-m_1 y - m_1 z + m_0 y z ), \hspace{0.2in}i = 1, 2 \\&#xD;&#xA;v_1 &amp;amp;=&amp;amp; {\tt Resultant[} U_0, U_1, y {\tt ]}&#xD;&#xA;\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final polynomial, $v_1$, has just one variable, $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}v_1 &amp;amp;=&amp;amp; \left(m_2^2 - m_1 m_3 - m_1 m_2 z + m_0 m_3 z + m_1^2 z^2 - m_0 m_2 z^2\right)^4 \\ &#xD;&#xA;&amp;amp;&amp;amp; \left(-m_3^3 + 2 m_2 m_3 m_4 - m_1 m_4^2 - m_2^2 m_5 + m_1 m_3 m_5 \right.\\&#xD;&#xA;&amp;amp;&amp;amp;\left. + z\left(m_2 m_3^2 - m_2^2 m_4 - m_1 m_3 m_4 + m_0 m_4^2 + m_1 m_2 m_5 - m_0 m_3 m_5 \right) \right. \\&#xD;&#xA;&amp;amp;&amp;amp;\left. +z^2\left(- m_2^2 m_3 + m_1 m_3^2 + m_1 m_2 m_4 - m_0 m_3 m_4 - m_1^2 m_5 + m_0 m_2 m_5\right) \right. \\&#xD;&#xA;&amp;amp;&amp;amp;\left. + z^3\left( m_2^3 - 2 m_1 m_2 m_3 + m_0 m_3^2 + &#xD;&#xA;   m_1^2 m_4 - m_0 m_2 m_4\right)\right)^2\end{eqnarray}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first factor is another &quot;special case&quot;, so that $z$, in general, should be a root of the cubic polynomial in the second factor. To get $y$, and $x$, and the rest, you could back-substitute into polynomials from earlier in the resultant chain (just as you would with a linear system), or you could just compute resultant chains that eliminate every variable except $y$, then every variable except $x$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that (as one might expect) your particular system is such that the &quot;final&quot; polynomials for $x$, $y$, and $z$ are equivalent; that is, $x$, $y$, and $z$ are roots of the cubic in the second factor of $v_1$ above. (This is, in fact, @Moron's cubic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll note that resultants in general are computationally expensive (especially in symbolic form). Very often, the process can bog down completely in &lt;em&gt;Mathematica&lt;/em&gt;, even for fairly modest systems, and/or the final polynomial can be of enormous degree. Sometimes, a bit of finesse in the order of elimination helps. My first attempt at finding (a chain of resultants that eliminated everything but) $a$ got boggy, but eliminating in the order $b$, $c$, $z$, $y$, $x$ gave me a cubic equation (and some &quot;special case&quot; factors) ... which is &lt;em&gt;huge&lt;/em&gt;, so I won't present it here. A more-practical approach here might be to find $x$, $y$, and $z$ from the cubic polynomial, substitute these values back into three of the $p_i$, and solve the linear system for $a$, $b$, and $c$.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-12-06T08:29:30.680" />
  <row Id="13219" PostTypeId="1" AcceptedAnswerId="13227" CreationDate="2010-12-06T08:47:45.277" Score="3" ViewCount="59" Body="&lt;p&gt;From Statistical Inference Second Edition (George Casella, Roger L. Berger)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;My telephone rings 12 times each week, the calls being randomly distributed among the 7 days. What is the probability that I get a least one call each day?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is .2285, but I don't know how they got it. My reasoning was as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 12 calls and thus 13 places to put &quot;day dividers&quot; to produce possible distributions of calls. There should be 6 day dividers for the week. One possibility:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;_1|2_3|4_5_6|7|8||9_10_11_12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(1 on Monday, 2 on Tuesday, 3 on Wednesday, 1 on Thursday, 1 on Friday, 0 on Saturday, 4 on Sunday)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;code&gt;13^6&lt;/code&gt; possible distributions (using this method). In order to satisfy 1 call/day, dividers can't be at the beginning or end, nor can they be on top of one another (signifying a day with no calls). This means there are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;11*10*9*8*7*6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Valid distributions and a probability of: 0.069. Where am I going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. This isn't homework as I'm not in school.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I don't think each &quot;distribution&quot; is equally likely. That's probably my error. But I still don't know how to get to the correct answer :)&lt;/p&gt;&#xA;" OwnerUserId="4468" LastEditorUserId="4468" LastEditorDisplayName="" LastEditDate="2010-12-06T08:53:31.400" LastActivityDate="2010-12-06T11:51:31.283" Title="Statistical Inference Question" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13220" PostTypeId="1" AcceptedAnswerId="13222" CreationDate="2010-12-06T09:03:55.007" Score="1" ViewCount="72" Body="&lt;p&gt;I have a vector A and a point P. My problem is how to find out if point is on the left or right side of vector looking from the point of origin of a vector in direction of it.&lt;/p&gt;&#xA;" OwnerUserId="3563" LastActivityDate="2010-12-06T12:01:45.943" Title="On which side of vector the point is" Tags="&lt;euclidean-geometry&gt;&lt;vector-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13222" PostTypeId="2" ParentId="13220" CreationDate="2010-12-06T09:58:44.330" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/44096/detecting-whether-directed-cycle-is-clockwise-or-counterclockwise/44098#44098&quot; rel=&quot;nofollow&quot;&gt;This answer&lt;/a&gt; on MO should do it for you.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-06T09:58:44.330" />
  <row Id="13224" PostTypeId="2" ParentId="13111" CreationDate="2010-12-06T11:05:53.543" Score="3" ViewCount="0" Body="&lt;p&gt;OK, the following results lead to a solution to this and similar problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem. Let $G$ be a finite simple group with a subgroup $H$ of index $n&amp;gt;1$. Then $n \ge 5$, and $|G|$ divides $n!/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Let $\phi: G \rightarrow S_n$ be the permutation representation of $G$ acting by (left or right depending on whether you use left or right group actions) multiplication on the set of (left or right) cosets of $H$ in $G$. Then  $G/{\rm Ker}(\phi) \cong {\rm im}(\phi) \le S_n$. Since $n&amp;gt;1$ and ${\rm im}(\phi)$ is transitive, $|{\rm im}(\phi)| &amp;gt; 1$ and so $G$ simple implies ${\rm Ker}(\phi) = 1$, and hence $G \cong {\rm im}(\phi)$. Now $S_n$ is solvable for $n &amp;lt; 5$, so we must have $n \ge 5$. Furthermore, we must have ${\rm im}(\phi) \le A_n$, since otherwise ${\rm im}(\phi) \cap A_n$ would be a normal subgroup of ${\rm im}(\phi)$ of index 2, and so $G$ would not be simple.&#xA;Hence $|G|$ divides $|A_n| = n!/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary. Let $G$ be a finite simple group and $n = |{\rm Syl}_p(G)|$ for some prime $p$ dividing $|G|$. Then  $n \ge 5$ and $|G|$ divides $n!/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Let $P \in {\rm Syl}_p(G)$. We cannot have $n=1$ because then  $P$ would be normal in $G$. Now apply the theorem to the subgroup $N_G(P)$ of index $n$ in $G$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-06T11:05:53.543" CommentCount="1" />
  <row Id="13225" PostTypeId="2" ParentId="13205" CreationDate="2010-12-06T11:17:46.260" Score="6" ViewCount="0" Body="&lt;p&gt;To show that $1\otimes 1+i\otimes i$ is nonzero&#xA;in $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$ note that it maps&#xA;to $1\otimes 1+i\otimes i$ in&#xA;$\mathbb{C}[X]\otimes_{\mathbb{R}}\mathbb{C}[X]$.&#xA;This is a tensor product over a field, so a basis as an&#xA;$\mathbb{R}$-vector space is got by tensoring togther bases&#xA;on each side. Now as the set of elements of the forms $X^n$&#xD;&#xA;and $iX^n$ are bases of $\mathbb{C}[X]$ over $\mathbb{R}$&#xD;&#xA;then $1\otimes 1$ and $i\otimes i$ are linearly independent&#xA;over $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As&#xA;$$(1\otimes 1+i\otimes i)(1\otimes 1-i\otimes i)=0$$&#xD;&#xA;then $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$&#xD;&#xA;isn't an integral domain, unlike $\mathbb{C}[X]\otimes_{\mathbb{C}}\mathbb{C}[X]$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-06T11:17:46.260" />
  <row Id="13226" PostTypeId="2" ParentId="13220" CreationDate="2010-12-06T11:33:13.707" Score="0" ViewCount="0" Body="&lt;p&gt;Let $B$ be the vector from the end point of $A$ to $P$, then observe the sign of the coefficient of $A \wedge B$ in a chosen basis. The exact sign is up to $O(2)/SO(2)$, but you only need to distinguish between the two cases, so it's ok :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD: Now that I think about it, letting $B$ be the vector from the starting point of $A$ to $P$ is even neater.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-06T12:01:45.943" LastActivityDate="2010-12-06T12:01:45.943" />
  <row Id="13227" PostTypeId="2" ParentId="13219" CreationDate="2010-12-06T11:51:31.283" Score="2" ViewCount="0" Body="&lt;p&gt;Here is an easy approach to solve the problem. Consider the tuples $(n_1,n_2,\ldots,n_7)$ such that $n_i$ are positive integers satisfying $n_1 + n_2 + \cdots + n_7 = 12$. Further, assume for the moment that $n_1 \geq n_2 \geq \cdots \geq n_7$.&#xA;You should find that there are only seven such tuples. Take, for example, the tuple $(4,3,1,1,1,1,1)$. The corresponding probability is, according to the multinomial distribution with parameters $n=12$ and $p_1 = p_2 = \cdots = p_7 = 1/7$, &#xA;$$&#xD;&#xA;\frac{{12!}}{{4!3!1!1!1!1!1!}}\bigg(\frac{1}{7}\bigg)^{4}\bigg(\frac{1}{7}\bigg)^{3}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1} = &#xD;&#xA;\frac{{12!}}{{4!3!}}\bigg(\frac{1}{7}\bigg)^{12}.&#xD;&#xA;$$&#xD;&#xA;Now, if we don't assume that $n_1 \geq n_2 \geq \cdots \geq n_7$, then you should multiply the above probability by ${7 \choose 2}2$. Doing the same for the other tuples, you obtain seven probabilities. Their sum is the probability you are looking for. I have done the calculation and the result is $\approx 0.228452440447$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-06T11:51:31.283" CommentCount="1" />
  <row Id="13228" PostTypeId="1" CreationDate="2010-12-06T12:30:24.297" Score="0" ViewCount="23" Body="&lt;p&gt;Had asked this in the Physics forum, and they told me to ask it here.&#xA;I've used a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter&quot; rel=&quot;nofollow&quot;&gt;Kalman filter&lt;/a&gt; with a constant velocity motion model, where the $X$ matrix is [$x$ $v_x$ $y$ $v_y$]'.&#xA;But for the Nearly Coordinated Turn model, the $X$ matrix is [$x$ $v_x$ $y$ $v_y$ $\omega$]'. &lt;em&gt;($\omega$ is angular velocity)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried an NCT model from &lt;a href=&quot;http://books.google.co.in/books?id=xz9nQ4wdXG4C&amp;amp;printsec=frontcover&amp;amp;dq=estimation+with+applications+to+tracking+and+navigation&amp;amp;source=bl&amp;amp;ots=Vn_Vt3d1u0&amp;amp;sig=MWNdwWa8V53k2wN-8bg8PO8wtxo&amp;amp;hl=en&amp;amp;ei=m3j8TJWqOcjJrAe5qqz-Bw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBwQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; book, which says that the omega value has to be estimated with a separate equation containing white noise acceleration. I want to know how this could be possible, as filtering is done to eliminate noise as much as possible. Then how can adding noise to the angular velocity help, because wouldn't that make the predictions randomly turn left and right even if the target is turning left constantly? (If I'm asking the question in the wrong forum, kindly redirect me. I'll delete this question)&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2010-12-06T12:30:24.297" Title="How is a Kalman filter with a Nearly Coordinated Turn model meant to work?" Tags="&lt;algorithms&gt;&lt;taylor-expansion&gt;&lt;matlab&gt;" FavoriteCount="1" />
  <row Id="13229" PostTypeId="1" CreationDate="2010-12-06T12:30:37.850" Score="0" ViewCount="47" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am computing the integral of some discrete data, in the form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t  m^3/s&#xA;0  0.75116&#xA;30 0.55621&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, the results for the smallest possible step (120) are not good at all:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Trapezoids rule&#xA;S: 0.111159 (step 120)&#xA;S': 0.111177 (step 60)&#xA;S'': 0.111188 (step 30)&#xA;------------------&#xA;Convergence Quotient ~= 1.73291 (should be 4&#xA;Error ~= 3.44458e-006 (meaningless for CQ != 4)&#xA;&#xA;&#xA;Simpson's method&#xA;------------------&#xA;h: 120&#xA;------------------&#xA;S: 0.111003 (step 120)&#xA;S': 0.111183 (step 60)&#xA;S'': 0.111191 (step 30)&#xA;------------------&#xA;Convergence quotient ~= 23.0815 (should be 16)&#xA;Error ~= 5.20611e-007 (meaningless for CQ != 16)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Samples range from 0 to 86400 in increments of 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? Floating point imprecisions? Something else I am missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4470" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:04:00.683" LastActivityDate="2010-12-07T15:04:00.683" Title="Trapezoid/Simpson's method for discrete data gives unreliable results" Tags="&lt;integral&gt;&lt;numerical-methods&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="13231" PostTypeId="1" AcceptedAnswerId="13232" CreationDate="2010-12-06T12:40:53.987" Score="0" ViewCount="110" Body="&lt;p&gt;What does D = max[0; M(x)] mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M(x) is a function.&lt;/p&gt;&#xA;" OwnerUserId="3532" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T00:04:59.307" LastActivityDate="2010-12-07T00:04:59.307" Title="What does max[] mean?" Tags="&lt;notation&gt;&lt;elementary&gt;" AnswerCount="2" />
  <row Id="13232" PostTypeId="2" ParentId="13231" CreationDate="2010-12-06T12:52:36.967" Score="3" ViewCount="0" Body="&lt;p&gt;Taking the maximal number amongst the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\max\{x_1,x_2\} = \cases{x_1, \text{if }x_1 &amp;gt; x_2\\x_2, \text{otherwise}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can define like that the maximum of any finitely many elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the parameters are an infinite set of values, then it is implied that one of them is maximal (namely that there is a greatest one, unlike the set $\{-\frac{1}{n} | n\in\mathbb{N}\}$ where there is no greatest element)&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-06T12:52:36.967" CommentCount="1" />
  <row Id="13233" PostTypeId="1" AcceptedAnswerId="13239" CreationDate="2010-12-06T13:23:19.617" Score="1" ViewCount="44" Body="&lt;p&gt;Can somebody prove that when we add 2 bit integers and if there is an overflow then the result always will be lesser than the 2 operands used??&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-06T14:01:52.920" LastActivityDate="2010-12-06T15:31:47.027" Title="detecting 2 n bits integer addition overflow" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13234" PostTypeId="2" ParentId="13155" CreationDate="2010-12-06T13:34:16.697" Score="1" ViewCount="0" Body="&lt;p&gt;Also, in agreement to what kahen said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a regular parametrization, you can always find a metric that is constant along the curve leading to zero intrinsic curvature. This is also a reason why it might be somehow not very enlightening to discuss curves as one dimensional manifolds in its own right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curves, on the other hand, can have extrinsic curvature which can for some body moving along a trajectory be interpreted as the acting force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greets&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2010-12-06T13:34:16.697" />
  <row Id="13235" PostTypeId="2" ParentId="13231" CreationDate="2010-12-06T13:35:33.057" Score="4" ViewCount="0" Body="&lt;p&gt;In this case, I can also think of $D$ as a function in $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(x) = max(0, M(x)) = \begin{cases} M(x), &amp;amp;M(x) &amp;gt;0 \\ 0, &amp;amp;else\end{cases}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is essentially $M$ &quot;cut off&quot; at $0$.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-12-06T13:35:33.057" CommentCount="3" />
  <row Id="13236" PostTypeId="2" ParentId="13109" CreationDate="2010-12-06T13:38:39.067" Score="0" ViewCount="0" Body="&lt;p&gt;Hi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I understand your query correctly that you want to universally abstract from constant $t$, to a variable $x$. Actually I would write $Q[x/t]$ instead of $Q[t/x]$, since you are replacing $t$ by $x$, right? Also I rather prefer writing $Q[c/x]$ instead $Q[t/x]$ to indicate dealing with a constant c and not with a term t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it can be shown that your rule is a combination of a lemma and another rule. If you can show soundness of the lemma and the other rule you are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lemma is the lifting lemma which holds for a great deal of logics. It can be formulated for variables and it can be formulated for constants. The version for &#xA;constants reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \vdash Q$            c/t does not cause name clash&lt;br&gt;&#xA;---------------- (Lifting)&lt;br&gt;&#xA;$T[c/t] \vdash Q[c/t]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above lemma can be proved via induction on the proof. The other rule is&#xA;now abstract of a variable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \vdash Q$            x not in T&lt;br&gt;&#xA;---------------  (Right Forall)&lt;br&gt;&#xA;$T \vdash \forall x Q$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how can we combine the two to arrive at your result? We simply need to observe&#xA;that when c does not occur in T, then T[c/t]=T. Therefore we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \vdash Q$        c not in T&lt;br&gt;&#xA;-----------  (Lifting)  c/x does not cause name clash&lt;br&gt;&#xA;$T \vdash Q[c/x]$&lt;br&gt;&#xA;-------------------- (Right Forall) x not in T&lt;br&gt;&#xA;$T \vdash \forall x Q[c/x]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see an additional condition sneaks in. We do not only have to have c not&#xA;in $T$, but also $x$ not in $T$ and $c/x$ does not cause a name clash. I am not &#xA;sure whether the later two conditions correspond to your condition that $x$ may not &#xA;occur in $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should $x$ not occur in $Q$ mean not occur free or not occur bound? Normally non &#xA;occurence refers to free occurence. If it occurs bound we might observe a name &#xA;clash and generate unsoundness. Need to find a counter example still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-09T08:22:30.440" LastActivityDate="2010-12-09T08:22:30.440" />
  <row Id="13237" PostTypeId="1" AcceptedAnswerId="13240" CreationDate="2010-12-06T13:40:40.000" Score="0" ViewCount="58" Body="&lt;p&gt;Hey!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mathematical function such that;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f(3, 5)   = 3&#xA;f(10, 2)  = 2&#xA;f(14, 15) = 14&#xA;f(9, 9)   = 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It would be even more cool if there's a function that takes three (3) parameters, but that one could be solved by using recursive functionality;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f( f(3, 5), 4) = 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4477" LastActivityDate="2010-12-07T15:31:56.650" Title="Arithmetic function to return lowest in-parameter" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13239" PostTypeId="2" ParentId="13233" CreationDate="2010-12-06T13:45:08.993" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose $a$ and $b$&amp;#160;are the numbers you are adding. Suppose that $0 \le a,b &amp;lt; 2^n$. Then we must have that $a+b - 2^n &amp;lt; a$ and $a+b - 2^n &amp;lt; b$. Hence if overflow occurs the result is smaller than $a$ or $b$.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-06T13:45:08.993" />
  <row Id="13240" PostTypeId="2" ParentId="13237" CreationDate="2010-12-06T13:46:36.730" Score="6" ViewCount="0" Body="&lt;p&gt;$$f(x,y)=\frac{x+y-|x-y|}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Oscar gave a nice interpretation of the above formula in &lt;a href=&quot;http://math.stackexchange.com/questions/13253&quot;&gt;his follow-up question&lt;/a&gt;, but I'll give a dumb derivation here for completeness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making use of &lt;a href=&quot;http://mathworld.wolfram.com/IversonBracket.html&quot; rel=&quot;nofollow&quot;&gt;Iversonian brackets&lt;/a&gt;, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(x,y)=x[y \geq x]+y[y &amp;lt; x]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $[\neg p]=1-[p]$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(x,y)=x[y \geq x]+y(1-[y \geq x])=y-(y-x)[y-x \geq 0]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{u+|u|}{2}=u[u \geq 0]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(x,y)=y+\frac{x-y-|x-y|}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to the desired expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extension to more than two arguments is no longer as compact, though, since one now has to contend with products of Iversonian brackets ($[p \land q]=[p]\cdot[q]$).&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:31:56.650" LastActivityDate="2010-12-07T15:31:56.650" CommentCount="3" />
  <row Id="13241" PostTypeId="1" AcceptedAnswerId="13243" CreationDate="2010-12-06T13:47:32.250" Score="0" ViewCount="111" Body="&lt;p&gt;We know that the set of continuous functions and the set of smooth functions in $\mathcal{L}^1$ or $\mathcal{L}^2$ are dense in $\mathcal{L}^1$ and $\mathcal{L}^2$. Is the set of $\mathcal{C}^k$ functions dense in $\mathcal{L}^1$ or $\mathcal{L}^2$ for any $k\in\mathbb{N}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the set $\mathcal{C}^k\setminus \mathcal{C}^{\infty}$ in $\mathcal{L}^1$ or $\mathcal{L}^2$ dense in $\mathcal{L}^1$ or $\mathcal{L}^2$. How would one prove this one way or the other?&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T15:18:47.637" LastActivityDate="2010-12-06T15:18:47.637" Title="Are the $\mathcal{C}^k$  functions dense in either $\mathcal{L}^2$ or $\mathcal{L}^1$ ?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13242" PostTypeId="1" CreationDate="2010-12-06T14:03:19.453" Score="0" ViewCount="28" Body="&lt;p&gt;I recently made a model to solve a conflict problem in sports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model is based on Bayesian Games: a complete definition can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayesian_game&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that I am now looking to compute the equilibria of the system, but I don't know any algorithm to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which language do you think I should use, I was a priori going to use R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody has any experience with this kind of problem?&lt;/p&gt;&#xA;" OwnerUserId="3354" LastActivityDate="2010-12-06T14:03:19.453" Title="Computing Bayesian Nash Equilibrium" Tags="&lt;game-theory&gt;" />
  <row Id="13243" PostTypeId="2" ParentId="13241" CreationDate="2010-12-06T14:23:55.523" Score="0" ViewCount="0" Body="&lt;p&gt;Here is the idea behind the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Step functions are dense in $L^p$ ($p=1,2$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Given a step function $T$ and $\varepsilon&amp;gt;0$ there is a $g\in C^k$ &#xA;$$\|T-g\|_{L^p}\lt\varepsilon$$&#xD;&#xA;($p=1,2$). &lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-06T14:23:55.523" CommentCount="6" />
  <row Id="13244" PostTypeId="1" CreationDate="2010-12-06T14:29:56.613" Score="3" ViewCount="107" Body="&lt;p&gt;Let $G$ and $H$ be two possibly directed, non necessarily simple, vertex-labelled graphs with respective adjacency matrices $A_G$ and $A_H$ and $V(G)=V(H)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the name of the graph $M$ with adjacency matrix $A_M=A_HA_G$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In the unlikely event that I am the first to think about using that operation, which symbols should I NOT use to denote it in order to avoid confusion with &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_product&quot; rel=&quot;nofollow&quot;&gt;other graph products&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; I have now asked this question on &lt;a href=&quot;http://mathoverflow.net/questions/49233/name-of-an-operation-on-graphs&quot; rel=&quot;nofollow&quot;&gt;MathOverflow&lt;/a&gt; as well.&lt;/p&gt;&#xA;" OwnerUserId="689" LastEditorUserId="689" LastEditorDisplayName="" LastEditDate="2010-12-13T14:18:04.337" LastActivityDate="2010-12-13T14:18:04.337" Title="Name of an operation on graphs" Tags="&lt;reference-request&gt;&lt;graph-theory&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="13245" PostTypeId="1" AcceptedAnswerId="13385" CreationDate="2010-12-06T14:32:13.693" Score="2" ViewCount="234" Body="&lt;p&gt;I'm having trouble understanding this proof:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lemma:&lt;/strong&gt; Let $\lambda$ be the Lebesgue measure restricted to the Borel sets of $[0,1]$. Then there exists a decreasing sequence $(E_n)_{n\ge0}$ of subsets of $[0,1]$ such that each of them is $\lambda$-thick and $\bigcap E_n = \emptyset$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; Let $E_n = [0,1]\cap\{f\ge n\}$, where $f$ is a noncontinuous solution $f:\mathbb{R}\longrightarrow\mathbb{R}$ of $f(x+y) = f(x) + f(y)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A continuous solution of that functional equation is a line through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ~ be the equivalence relation on $\mathbb{R}-\{0\}$ such that $x$ ~ $y$ $\leftrightarrow$ $\exists q\in\mathbb{Q}$ s.t. $x = qy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ be a set of representatives from each equivalence class. Let $g:A\longrightarrow\mathbb{R}$ be a non-constant &quot;assignment of slopes&quot;. Let $f:\mathbb{R} - \{0\}\longrightarrow\mathbb{R}$ s.t. $x \mapsto x\cdot g(y)$ where $x$~$y\in A$, let $f(0) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it roughly, a noncontinuous solution is a stack of lines through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;a) Is this description of a noncontinuous solution correct?&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No it's not! As TCL helped me realize, this construction does satisfy the desired functional equation only for x and y in the same equivalence class. I need to require something more from g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Why does the proof work? I fail to see why are the $E_n$ &lt;strike&gt;measurable, or why are they &lt;/strike&gt;$\lambda$-thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't need to be measurable! In fact, the concept of thickness is introduced to get around measurability issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's construct $f$ this way: let H be an Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, and $a\in H$. Let $g(x) = x$ for $x\in$ H$- {a}$ and $g(a) = 0$. Let's extend this function to $f:\mathbb{R}\longrightarrow\mathbb{R}$ by linearity using the Hamel basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this $f$ we have that $E_0 = [0,1]$, $E_1 = \{1\}$, which clearly isn't $\lambda$-thick. &lt;strike&gt;Where's the flaw?&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flaw is that $E_1 \ne \{1\}$ because... Hamel bases don't work that way :)&#xA;See the comments for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I give up. I will ask my professor about this.&lt;/p&gt;&#xA;" OwnerUserId="4471" LastEditorUserId="4471" LastEditorDisplayName="" LastEditDate="2010-12-09T09:41:07.940" LastActivityDate="2010-12-10T15:17:26.020" Title="Noncontinuous solutions of f(x+y) = f(x) + f(y), their use in a measure theory proof." Tags="&lt;homework&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="13246" PostTypeId="2" ParentId="13111" CreationDate="2010-12-06T14:46:02.463" Score="0" ViewCount="0" Body="&lt;p&gt;Assume $G$ is simple.  Then the existence of 6 Sylow 5-groups implies $G$ embeds in $S_{6}$ (let $G$ act on the Sylow 5-subgroups by conjugation and use the assumption that $G$ is simple).   But 300 does not divide 6 factorial.   So $G$ is not simple.  &lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-06T14:46:02.463" CommentCount="1" />
  <row Id="13247" PostTypeId="2" ParentId="13241" CreationDate="2010-12-06T14:49:10.050" Score="1" ViewCount="0" Body="&lt;p&gt;One basic way to see that smooth functions are dense in any reasonable function space is to use &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nascent_delta_function#nascent_delta_function&quot; rel=&quot;nofollow&quot;&gt;approximations to the identity&lt;/a&gt;.&quot; The point is that convolving with the delta function is the identity operator. However, the delta function can be approximated by smooth functions (take a smooth function supported in a small ball of total integral one). If $\phi$ is such a smooth function (which approximates the $\delta$ function), and $f$ is any function, then $\phi \ast f$ is probably close to $f$ in the metric you are interested in (this is always true in $L^p$, at least). But $\phi \ast f$ is smooth since to differentiate this, just differentiate $\phi$. (It is a nice property of convolution that convolving anything with a smooth function gives a smooth function---&lt;em&gt;if&lt;/em&gt; the integrals all converge nicely.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-06T14:49:10.050" CommentCount="1" />
  <row Id="13248" PostTypeId="2" ParentId="13174" CreationDate="2010-12-06T15:04:40.140" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Exploit the innate symmetry of the problem - here power sums as symmetric functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power sum $\rm\ \ \Sigma_n = a\ x^n + b\ y^n + c\ z^n\ \ $ satisfies the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \Sigma_{n+3} - (x+y+z)\ \Sigma_{n+2} + (xy+yz+zx)\ \Sigma_{n+1} - xyz\ \Sigma_n\ =\ 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\rm\ n = 0,1,2\ $ the above yields a linear system for $\rm\: \ x+y+z,\ \ xy+yz+zx,\ \ xyz\ ,\:$ which are the coefficients of the cubic whose roots are $\rm\ x, y, z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ As I mentioned in an &lt;a href=&quot;http://math.stackexchange.com/questions/6965/system-of-equations/6990#6990&quot;&gt;answer&lt;/a&gt; to a similar question, readers familiar with the  theory of &lt;strong&gt;Lucas-Lehmer sequences&lt;/strong&gt; may recognize this recursion as one of many useful identities satisfied by sums of powers.  Such identities arise in many diverse contexts. Here is an interesting example: Capdegelle's work on &lt;strong&gt;FLT&lt;/strong&gt; (Fermat's Last Theorem) employed the power sum recursion $\rm\ F_{n+3} + S_2\: F_{n+2} + S_1\: F_{n+1} + S_0\: F_n = 0\ $ where $\rm\ F_n = X^n + Y^n - Z^n\ $ is the Fermat polynomial and the polynomials $\rm\ S_k\ $ are the elementary symmetric polynomials $\rm\ S_0 = -XYZ,$ $\rm\ S_1 = XY + XZ + YZ,\ $ $\rm S_2 = -(X+Y+Z)\:.\ $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-06T15:17:48.337" LastActivityDate="2010-12-06T15:17:48.337" />
  <row Id="13249" PostTypeId="2" ParentId="13233" CreationDate="2010-12-06T15:31:47.027" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ a+b-c\ &amp;lt;\ a,b\ \iff\ a,b\ &amp;lt;\ c$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-06T15:31:47.027" />
  <row Id="13250" PostTypeId="1" CreationDate="2010-12-06T15:56:00.750" Score="4" ViewCount="114" Body="&lt;p&gt;We know that $$e^{i \pi} = -1 .$$ We can transform all of the components of this identity into (generalized) continued fractions. When we start of with $\pi$, we see that $$  \Big(3+ \mathbf{K}_{i=1}^{\infty} \frac{(\sqrt{a_{i-1}}+2)^2}{6}\Big)^{e \cdot i} = -1 $$ When we take the $\frac{1}{i}$'th power, we see that &#xA;$$ \Big(3 + \mathbf{K}_{i=1}^{\infty} \frac{(\sqrt{a_{i-1}}+2)^2}{6}\Big)^{e} = (-1)^{-i} ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which $a_1=1^2,a_2=3^2$. We could re-state this equality by writing $e, -1$ and the root the latter as &#xA;continued fractions, but I think the question is quite clear without doing so (also: other &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction#Pi&quot; rel=&quot;nofollow&quot;&gt;GCF's of pi&lt;/a&gt;). What information about (generalized) continued fractions and calculating with them can we extract from Euler's identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation: I was hoping that, if we could find a good explanation for this, we would have more insight in how continued fractions behave under arithmetic operations. With that insight, perhaps we could find ways to combine the continued fractions of certain constants and finding the values of some constants, like Apery's Constant and Catalan's, of which we &lt;em&gt;do&lt;/em&gt; know the continued fraction representations but few more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-06T16:55:06.563" LastActivityDate="2010-12-06T16:55:06.563" Title="What can Euler's identity teach us about (generalised) continued fractions?" Tags="&lt;number-theory&gt;&lt;continued-fractions&gt;" CommentCount="8" FavoriteCount="1" />
  <row Id="13251" PostTypeId="1" AcceptedAnswerId="13252" CreationDate="2010-12-06T16:21:33.083" Score="2" ViewCount="29" Body="&lt;p&gt;I'm currently reading Artificial Intelligence, by Russel &amp;amp; Norvig. They state that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) A sentence is valid if it is true in all models&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) The deduction theorem: &quot;For any sentences $\alpha$ and $\beta$, $\alpha \models \beta$ if and only if ($\alpha \implies \beta$) is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My mental blockage consists of the fact that implication is False if $\alpha = T$ and $\beta = F$ -- thus resulting in one model that is not true. But, in order for ($\alpha \implies \beta$) to be valid, it has to be true for &lt;em&gt;all&lt;/em&gt; models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="4478" LastActivityDate="2010-12-06T16:24:58.600" Title="The deduction theorem according to AIMA" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="13252" PostTypeId="2" ParentId="13251" CreationDate="2010-12-06T16:24:58.600" Score="2" ViewCount="0" Body="&lt;p&gt;In order for $\alpha\Rightarrow\beta$ to be valid, it must hold in all models; for $\alpha\Rightarrow\beta$ to not be valid, there must be a model where it is false. If there is a model where it is false, then there is a model in which $\alpha$ is true but $\beta$ is false, which means that $\alpha\models\beta$ does not hold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember: you are proving an implication. You are trying to prove that &lt;strong&gt;if&lt;/strong&gt; $\alpha\models\beta$, &lt;strong&gt;then&lt;/strong&gt; $\alpha\Longrightarrow \beta$ is valid. You are not trying to prove that $\alpha\Longrightarrow\beta$ is valid in all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, you also need to prove the converse: &lt;strong&gt;if&lt;/strong&gt; $\alpha\Longrightarrow\beta$ is valid, so that it holds in all models, &lt;strong&gt;then&lt;/strong&gt; you need to show that $\alpha\models\beta$ holds). &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-06T16:24:58.600" CommentCount="5" />
  <row Id="13253" PostTypeId="1" CreationDate="2010-12-06T16:35:46.163" Score="11" ViewCount="239" Body="&lt;p&gt;As we saw &lt;a href=&quot;http://math.stackexchange.com/questions/13237/13240#13240&quot;&gt;here&lt;/a&gt;, the minimum of two quantities can be written using elementary functions and the absolute value function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's even a nice intuitive explanation to go along with this: If we go to the point half way between two numbers, then going down by half their difference will take us to the smaller one. So my question is: &quot;Is there a similar formula for three numbers?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $\min(a,\min(b,c))$ will work, but this gives us the expression:&#xA;$$\frac{a+\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)}{2} - \frac{\left|a-\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)\right|}{2},$$&#xD;&#xA;which isn't intuitively the minimum of three numbers, and isn't even symmetrical in the variables, even though its output is. Is there some nicer way of expressing this function?&lt;/p&gt;&#xA;" OwnerUserId="1149" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:12:43.280" LastActivityDate="2010-12-07T15:12:43.280" Title="Nice expression for minimum of three variables?" Tags="&lt;functions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="13254" PostTypeId="2" ParentId="13253" CreationDate="2010-12-06T16:46:41.853" Score="15" ViewCount="0" Body="&lt;p&gt;This probably isn't what you were thinking of, but if $a_1, ... a_n$ are non-negative, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\min(a_1, ... a_n) = \lim_{k \to -\infty} \sqrt[k]{a_1^k + ... + a_n^k}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max(a_1, ... a_n) = \lim_{k \to \infty} \sqrt[k]{a_1^k + ... + a_n^k}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several applications of these identities (at least the second one), e.g. to functional analysis.  They are also related to the way in which tropical arithmetic arises as a &quot;limit&quot; of ordinary arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-06T16:46:41.853" />
  <row Id="13255" PostTypeId="2" ParentId="13244" CreationDate="2010-12-06T16:48:22.057" Score="1" ViewCount="0" Body="&lt;p&gt;I believe this is known as &lt;i&gt;Graph Squaring&lt;/i&gt;. See also &lt;a href=&quot;http://mathworld.wolfram.com/GraphPower.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-06T16:48:22.057" CommentCount="1" />
  <row Id="13256" PostTypeId="2" ParentId="13253" CreationDate="2010-12-06T16:54:22.103" Score="10" ViewCount="0" Body="&lt;p&gt;If you want a symmetric expression, you can take $\frac13 (\min(a,\min(b,c))+\min(b,\min(a,c))+\min(c,\min(a,b)))$. But if you rewrite it using that absolute-value trick, I still don't think it gives something that's &quot;intuitively the minimum&quot;, I'm afraid...&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-06T16:54:22.103" />
  <row Id="13257" PostTypeId="1" CreationDate="2010-12-06T17:07:08.530" Score="3" ViewCount="104" Body="&lt;p&gt;Hi all. I am stuck in trying to solve the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two points $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$, to determine the parameters $a$ and $b$ in the equation:&lt;br&gt;&#xA;$$y=\frac{e^{a+bx} - e^{a}}{1+e^{a+bx}}.$$&lt;br&gt;&#xA;In other words, I have two unknowns and two set of points, so there's for sure a solution, but how can the following system of equation be rewritten explicitly on $a$ and $b$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\begin{matrix}&#xD;&#xA;y_{1}&amp;amp;=\frac{e^{a+bx_{1}} - e^{a}}{1+e^{a+bx_{1}}}\\ &#xD;&#xA;y_{2}&amp;amp;=\frac{e^{a+bx_{2}} - e^{a}}{1+e^{a+bx_{2}}}&#xD;&#xA;\end{matrix}\right.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I implement an equation solver instead? In which case, do you have any suggestion on the method?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much, I hope it's not a too stupid question :)&lt;/p&gt;&#xA;" OwnerUserId="2916" LastEditorUserId="2916" LastEditorDisplayName="" LastEditDate="2010-12-06T19:33:51.187" LastActivityDate="2010-12-06T20:07:48.427" Title="Solving a &quot;logistic-like&quot; function with two unknowns, two data points." Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13258" PostTypeId="2" ParentId="13244" CreationDate="2010-12-06T17:09:53.357" Score="1" ViewCount="0" Body="&lt;p&gt;As described in Brualdi and Cvetkovic's &lt;a href=&quot;http://books.google.com/books?id=pwx6t8QfZU8C&amp;amp;printsec=frontcover&amp;amp;dq=combinatorial+approach+to+matrix+theory&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Approach to Matrix Theory&lt;/a&gt;,  rather than associating a directed graph in the usual way to a matrix, it's more natural to associate the &lt;strong&gt;König digraph&lt;/strong&gt; for the purposes of thinking about matrix multiplication.  For an $m \times n$ non-negative integer matrix $A$, the Konig digraph is a bipartite digraph with vertex classes $X$ (with $n$ elements) and $Y$ (with $m$ elements) such that the number of edges from $x \in X$ to $y \in Y$ is $A_{yx}$.  Composition of matrices then corresponds to &quot;plugging&quot; one digraph into another, and whatever purpose you want to multiply adjacency matrices for, I think it will be more natural (or at least equivalent) phrased in this language instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've described is a category whose objects are the non-negative integers $[n], n \ge 0$ and where a morphism from $[n]$ to $[m]$ is a collection of arrows from an $n$-element set to an $m$-element.  This category is a certain refinement of the category of finite sets and &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_relation&quot; rel=&quot;nofollow&quot;&gt;relations&lt;/a&gt; and is a very natural setting for certain combinatorial arguments (as well as for, as the book says, a combinatorial approach to matrix theory).  I don't know if it has a name, though.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-06T17:09:53.357" />
  <row Id="13259" PostTypeId="1" CreationDate="2010-12-06T17:17:11.367" Score="1" ViewCount="73" Body="&lt;p&gt;Hi, I hope you can help me with this excersise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets have finite sequence of characters 0 and 1. We are doing following algorithm:&#xA;&lt;br&gt;If the sequence starts with 0, we add 00 to the end.&#xA;&lt;br&gt;If the sequence starts with 1, we add 1101 to the end.&#xA;&lt;br&gt;Then we remove first three characters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Prove, that if the sequence repeats in cycles, the cycles are of even length.&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;br&gt;&#xA;We are changing the length of the sequence, doing &lt;i&gt;n&lt;/i&gt; steps, by &lt;i&gt;n&lt;/i&gt;. To get to the starting state of the loop, the length has to be changed doing &lt;i&gt;n&lt;/i&gt; steps, by -&lt;i&gt;n&lt;/i&gt;. The total number of steps is &lt;i&gt;2n&lt;/i&gt;, which is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume, that if &lt;i&gt;n = n+1&lt;/i&gt;, the number of steps is still even.&lt;br&gt;&#xA;We are changing the length of the sequence, doing &lt;i&gt;n+1&lt;/i&gt; steps, by &lt;i&gt;n+1&lt;/i&gt;. To get to the starting state of the loop, the length has to be changed doing &lt;i&gt;n+1&lt;/i&gt; steps, by -&lt;i&gt;(n+1)&lt;/i&gt;. The total number of steps is &lt;i&gt;2n + 2&lt;/i&gt;, which is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to create a 2 step loop, we need a sequence starting which some time comes to state &lt;br&gt;0xx1xx or &lt;br&gt;1xx0x. &lt;br&gt;Both these constructions will generate the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a 6 step loop, we need &lt;br&gt;1xx1xx1xx0xx0 or &lt;br&gt;1xx1xx0xx0xx0x or &lt;br&gt;1xx0xx0xx0xx1xx or &lt;br&gt;0xx0xx0xx1xx1xx1 or &lt;br&gt;0xx0xx1xx1xx1xx or &lt;br&gt;0xx1xx1xx1xx0x. &lt;br&gt;They generate the other one in that order.&lt;/p&gt;&#xA;" OwnerUserId="4447" LastEditorUserId="4447" LastEditorDisplayName="" LastEditDate="2010-12-06T20:55:05.813" LastActivityDate="2010-12-06T20:55:05.813" Title="Cycling sequence of 0 and 1 - is cycle length even?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13260" PostTypeId="1" AcceptedAnswerId="13271" CreationDate="2010-12-06T17:41:30.083" Score="5" ViewCount="114" Body="&lt;p&gt;I'm interested in methods of computing the homology groups and Euler characteristics of the classical Lie groups ($GL(n,\mathbb{R}), SL(n,\mathbb{R})$, etc.). (But I'd be interested in techniques which are more generally applicable to arbitrary Lie groups as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the current state of my knowledge on this issue with some explicit questions interspersed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A compact Lie group of positive dimension admits a nowhere vanishing vector field and hence has Euler characteristic zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I remember reading in the past that a connected Lie group deformation retracts onto its maximal compact subgroup, which by the foregoing has Euler characteristic zero. Given this, we can say that a connected Lie group is either contractible or has Euler characteristic zero. However, my knowledge of Lie theory is (very) weak at best, so I am not fully comfortable accepting this fact. Is there a simple proof? (I feel that there should be but I'm just missing it.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Euler characteristic respects fiber bundle structures: if $F \rightarrow E \rightarrow B$ is a fiber bundle with total space $E$, base $B$, and fiber $F$, then $\chi(E) = \chi(F)\cdot\chi(B)$. Some suitable conditions are needed; I think $B$ path-connected suffices, but I'm not 100% certain. I can think of one explicit example of this: Identify $S^3$ with $SU(2)$ and implement the Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$. The subgroup $U(1)$ is realized as $S^1$ and the quotient $SU(2)/U(1)$ is realized as $S^2$. So we have a fibration $SU(2) \rightarrow SU(2)/U(1)$ with fiber $U(1)$; hence $\chi(SU(2)) = \chi(SU(2)/U(1)) \cdot \chi(U(1)) = 0$. But this is already known since $SU(2)$ is compact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The most direct approach to computing both the homology and the Euler characteristics would possibly be to find an explicit cellular decomposition of these groups. Again, my knowledge of basic Lie theory is failing me here: I don't see how to find such decompositions of the matrix groups.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surely there are other methods of computing the homology which would exploit the Lie group structure, but I don't know of any.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-06T18:58:10.187" Title="Homology and Euler characteristics of the classical Lie groups" Tags="&lt;algebraic-topology&gt;&lt;lie-groups&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13261" PostTypeId="1" CreationDate="2010-12-06T17:47:01.107" Score="0" ViewCount="60" Body="&lt;p&gt;I'm doing a raytracing exercise. I have a vector representing the normal of a surface at an intersection point, and a vector of the ray to the surface. How can I determine what the reflection will be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the below image, I have &lt;code&gt;d&lt;/code&gt; and &lt;code&gt;n&lt;/code&gt;. How can I get &lt;code&gt;r&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://imgur.com/9BHM2.png&quot; alt=&quot;Vector d is the ray; n is the normal; t is the refraction; r is the reflection&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2755" LastActivityDate="2010-12-06T23:44:55.710" Title="How to get a reflection vector?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13262" PostTypeId="2" ParentId="13244" CreationDate="2010-12-06T17:51:15.623" Score="1" ViewCount="0" Body="&lt;p&gt;Maybe it would be helpful... from position that $M$ is a hypergraph, value of matrix element $A_M(i,j)$ seems to be equal to the number of edges from vertex i to vertex j with length 2 (minus edges from j to i). But I'm not sure, if $A_M$ is adjacency matrix for hypergraph $M$... Especially after googling &lt;a href=&quot;http://books.google.ru/books?id=DWZc9Ti_vBwC&amp;amp;pg=PA312&amp;amp;lpg=PA312&amp;amp;dq=hypergraph+adjacency+matrix+definition&amp;amp;source=bl&amp;amp;ots=2R8o1vC0Z3&amp;amp;sig=h-2ubx2ytHCYCatkQ8YlG12mWtA&amp;amp;hl=ru&amp;amp;ei=8RX9TN_eH4eXOp_nwdQK&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;sqi=2&amp;amp;ved=0CDcQ6AEwAw#v=onepage&amp;amp;q=hypergraph%20adjacency%20matrix%20definition&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&quot;hypergraph adjacency matrices&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4454" LastActivityDate="2010-12-06T17:51:15.623" />
  <row Id="13263" PostTypeId="2" ParentId="13261" CreationDate="2010-12-06T17:54:33.690" Score="1" ViewCount="0" Body="&lt;p&gt;$$r = d - 2 (d \cdot n) n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d \cdot n$ is the dot product, and&#xA;$n$ must be normalized.&lt;/p&gt;&#xA;" OwnerUserId="4241" LastActivityDate="2010-12-06T17:54:33.690" CommentCount="2" />
  <row Id="13264" PostTypeId="2" ParentId="13260" CreationDate="2010-12-06T17:58:26.030" Score="3" ViewCount="0" Body="&lt;p&gt;The explicit cellular decomposition you're looking for is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bruhat_decomposition&quot; rel=&quot;nofollow&quot;&gt;Bruhat decomposition&lt;/a&gt;.  Counting the number of cells in each dimension automatically gives you the homology of classical groups over $\mathbb{C}$ (since then there are only cells in even dimensions), but I think over $\mathbb{R}$ you have to work harder (although I think you do get the Euler characteristic, in any case).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-06T17:58:26.030" />
  <row Id="13265" PostTypeId="2" ParentId="13259" CreationDate="2010-12-06T17:59:53.647" Score="2" ViewCount="0" Body="&lt;p&gt;Assuming you are talking about the sequence of bitstrings repeating in cycles, and the period of that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: You either increase the length of the bitstring by 1 or reduce it by 1.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-06T17:59:53.647" CommentCount="15" />
  <row Id="13266" PostTypeId="2" ParentId="13261" CreationDate="2010-12-06T18:04:09.657" Score="4" ViewCount="0" Body="&lt;p&gt;Let $\hat{n} = {n \over \|n\|}$. Then $\hat{n}$ is the vector of magnitude one in the same direction as $n$. The projection of $d$ in the $n$ direction is given by $\mathrm{proj}_{n}d = (d \cdot \hat{n})\hat{n}$, and the projection of $d$ in the orthogonal direction is therefore given by $d - (d \cdot \hat{n})\hat{n}$. Note that $-r$ has the same projection onto $d$, with its orthogonal projection given by $-1$ times that of $d$. Thus we have &#xA;$$d = (d \cdot \hat{n})\hat{n} + [d - (d \cdot \hat{n})\hat{n}]$$&#xD;&#xA;$$-r = (d \cdot \hat{n})\hat{n} - [d - (d \cdot \hat{n})\hat{n}]$$&#xD;&#xA;The later equation is exactly&#xA;$$r = -(d \cdot \hat{n})\hat{n} + [d - (d \cdot \hat{n})\hat{n}]$$&#xD;&#xA;Hence one can get $r$ from $d$ via&#xA;$$r = d - 2(d \cdot \hat{n})\hat{n}$$&#xD;&#xA;Stated in terms of $n$ itself, this becomes&#xA;$$r = d - {2 d \cdot n\over \|n\|^2}n$$&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-06T23:44:55.710" LastActivityDate="2010-12-06T23:44:55.710" />
  <row Id="13268" PostTypeId="1" CreationDate="2010-12-06T18:24:39.900" Score="3" ViewCount="129" Body="&lt;p&gt;Let $A$ and $B$ be two symmetric matrices of the same size, and let $\alpha_k$ and $\beta_k$ be the $k$th largest eigenvalues of $A$ and $B$ respectively.  What can we say about $|\beta_k - \alpha_k|$ in terms of $||B - A||_2$?  To be clearer, I want the tightest possible upper bound on $|\beta_k - \alpha_k|$ using only the stated assumptions, and computed as a function of $||B - A||_2$.&lt;/p&gt;&#xA;" OwnerUserId="4370" LastEditorUserId="4370" LastEditorDisplayName="" LastEditDate="2010-12-06T18:42:06.197" LastActivityDate="2010-12-07T18:32:12.300" Title="eigenvalue error" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="13269" PostTypeId="2" ParentId="1676" CreationDate="2010-12-06T18:26:48.483" Score="1" ViewCount="0" Body="&lt;p&gt;The picture on Mumford's Red book for the arithmetic surface $Spec\ \mathbb Z[x]$ is very enlightening. The book itself is a fantastic read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://matrix.cmi.ua.ac.be/DATA2/mumforddrawing.jpg&quot; rel=&quot;nofollow&quot;&gt;Lieven le Bruyn&lt;/a&gt; has a copy of that picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book ought to be read. What is equally enlightening is the functorial point of view emphasized by Grothendieck, that schemes are representing their functor of points and thus can be thought of as the functor instead. This is an easy consequence of Yoneda lemma. A delightful book with this point of view is Mumford, &quot;Lectures on curves on an algebraic surface&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-06T18:26:48.483" CommentCount="2" />
  <row Id="13270" PostTypeId="1" AcceptedAnswerId="13281" CreationDate="2010-12-06T18:29:07.980" Score="1" ViewCount="130" Body="&lt;p&gt;Since I started university, almost every exam I've had in mathematics have ended up as failed mostly due to my lack of using proper &quot;syntax&quot; on paper, written by hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I learn the standard way of writing mathematics, beginning with the basics? Does this make sense? Is there a standard &quot;uniform&quot; syntax, or does it depend on locale - country from country?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; With syntax I mean the way you write down your logic on paper - for instance, how you should put up a solution for display; what format to use, when to use deduction/implication arrows, how your answer should be represented etc. When to tell whether you are you allowed to skip mental steps and so on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very simple example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(x + 3)(x + 7) =                     &#xA;= x(x^2 + 10x + 21) =                              &#xA;= x^3 + 10x^2 + 21x         &#xA;Answer: x^3 + 10x^2 + 21x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;vs&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(x + 3)(x + 7) = x^3 + 10x^2 + 21x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so on... Hope this makes it a bit more clear?&lt;/p&gt;&#xA;" OwnerUserId="470" LastEditorUserId="470" LastEditorDisplayName="" LastEditDate="2010-12-06T19:29:34.293" LastActivityDate="2010-12-06T19:34:44.713" Title="Where to learn the syntax of mathematics?" Tags="&lt;learning&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13271" PostTypeId="2" ParentId="13260" CreationDate="2010-12-06T18:58:10.187" Score="4" ViewCount="0" Body="&lt;p&gt;So we know that $GL_n(\mathbb{R})$ and $SL_n(\mathbb{R})$ are homotopy equivalent to $O_n$&#xD;&#xA;and $SO_n$ respectively (Similar things are true for working over $\mathbb{C}$). Now you can work inductively using a Serre SS. Look at the fibration coming from the inclusion of one classical group into the next higher one, if you set it up right the homogenous space will be a sphere. This should get most of the work done for you, and it doesn't require lie theory. I don't think it will generalize though to exceptional lie groups though, you need to understand the homogenous spaces in order for this to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another way using the Serre SS: Use the fact that we know what $H^*(BG)$ is for $G$ unitary or orthoganl, special or not (with integer or mod 2 coefficients depending on whether or not it is $U$ or $O$). We know these rings because they give us chern and stiefel-whitney classes respectively (I believe the effect of looking a the special groups is just to kill $c_1$ or $w_1$). Also, the dimension of the vector space the group is acting on tells you the index of the top characteristic class. Now use our good old friend the loop path space fibration: $\Omega BG \to PBG \to BG$ and remember that $\Omega BG$ is homotopy equivalent to $G$ and $PBG$ is contractible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these approaches give you the ring structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use a Bockstein SS to get the integral cohomology for the stuff with mod two coefficients ($sq^1$ is the differential!).&#xA;Once you get the cohomology you can compute the euler characteristic by taking the alternating sums of the ranks. &lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-06T18:58:10.187" CommentCount="3" />
  <row Id="13272" PostTypeId="1" AcceptedAnswerId="13304" CreationDate="2010-12-06T19:04:18.277" Score="0" ViewCount="59" Body="&lt;p&gt;How would I create a rotation matrix that rotates X by a, Y by b, and Z by c?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to formulas, unless you're using the ardor3d api's functions/methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrix is set up like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;xx, xy, xz,&#xA;yx, yy, yz,&#xA;zx, zy, zz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A Quaternion is fine too.&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2010-12-06T22:22:23.850" Title="How would I create a rotation matrix that rotates X by a, Y by b, and Z by c?" Tags="&lt;geometry&gt;&lt;matrices&gt;&lt;quaternions&gt;" AnswerCount="2" />
  <row Id="13273" PostTypeId="3" CreationDate="2010-12-06T19:05:01.087" Score="0" ViewCount="0" Body="&lt;p&gt;Let me introduce myself first. My name is Willie Wong, and for what it is worth, I am a mathematician. And (at the risk of the nomination statements reading like broken records) I have also been a long-time participant in MathOverflow, have been lurking on Math.SE since public beta, and have been involved with TeX.SE since its inception. So I have some familiarity with the type of community the StackExchange platform is trying to forge. Compared to the other nominees, my name may be much less familiar to you, but I assure you this is not due to lack of activity: I do visit the website every working day, so I pledge to be usually available for moderating duties. My (relatively) low reputation score should be taken as a reflection of the large number of very capable people providing answers on this website, and of my personal belief in suggesting improvements to existing answers, instead of posting duplicate ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If elected as a moderator, I promise to moderate in moderation :) That is, with the exception of extreme cases of abuse (spammers, trolls, or offensive posts or comments), I will in general defer to the concensus of the community on what is or isn't appropriate, and I ask voters to also help set this website's scope and nature &lt;a href=&quot;http://meta.math.stackexchange.com/questions/1222/please-help-construct-moderation-policies-for-our-newcoming-moderators&quot;&gt;in Meta&lt;/a&gt;. I also promise to not involve myself in personal conflicts against other users, and be generally courteous in tone when dealing with whatever situations that may arise (unfortunately, the necessity of using the written form on the internet sometimes lead to unintended interpretations of words; I will try my best in my communications to avoid this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why I want to be a moderator: I was rather late coming to this website, and initially had some reservations about its purpose. I was afraid that it may be overrun with cranks or homework problems. After lurking for a while during public beta, I've been won over by the consistent level and focus of the discussions that take place here, and I want to do my best to help maintain this standard at Math.SE. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-06T19:05:01.087" LastActivityDate="2010-12-06T19:05:01.087" CommunityOwnedDate="2010-12-06T19:05:01.087" />
  <row Id="13274" PostTypeId="2" ParentId="13272" CreationDate="2010-12-06T19:09:29.097" Score="3" ViewCount="0" Body="&lt;p&gt;I think that 3d rotation is more complicated than this; rotating around each of the axes separately can give different cumulative results depending on the order in which you choose to do the rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rotation_matrix#Three_dimensions&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rotation_matrix#Three_dimensions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:09:29.097" CommentCount="2" />
  <row Id="13275" PostTypeId="2" ParentId="13270" CreationDate="2010-12-06T19:12:45.037" Score="2" ViewCount="0" Body="&lt;p&gt;Read your textbook and then use the syntax in your textbook when you take the exam.&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:12:45.037" />
  <row Id="13276" PostTypeId="1" AcceptedAnswerId="13279" CreationDate="2010-12-06T19:14:09.470" Score="2" ViewCount="76" Body="&lt;p&gt;I recently learned about finding density functions of functions of random variables using the cumulative distribution function.  For example, computing the density function for the difference of two uniform random variables on [0,1] (as in Grinstead &amp;amp; Snell's free online book).  I gave myself the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $X$ and $Y$ are independent random variables with exponential density $f(t) = a e^{-at}$. I'm trying to compute the density function for $Z =  X-Y$, but I'm getting nonsense. Since $z&amp;lt;0$ is ok and the density function will go like $e^{-az}$, trying to check that $\int_{-\infty}^\infty f_{X-Y}(z) dz$ diverges!  What is going wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The calculations are as follows.  If $z &amp;gt; 0$, then &#xA;        $$&#xD;&#xA;		\begin{split}&#xD;&#xA;		  P(Z &amp;lt; z) &amp;amp;= \int_0^\infty dy \int_0^{y+z} dx a^2&#xD;&#xA;		  e^{-a(x+y)} \\&#xD;&#xA;		  &amp;amp;= 1 - \frac 1 2 e^{-a z}&#xD;&#xA;		\end{split}&#xD;&#xA;		$$&#xD;&#xA;I found this by thinking about the (x,y) points satisfying $Y &amp;gt; X - z$, giving a region $y &amp;gt; 0, y+z &amp;gt; x &amp;gt; 0$.  If $ z &amp;lt; 0$, &#xA;$$ \begin{split}&#xD;&#xA;\displaystyle P(Z &amp;lt; z) &amp;amp;= \int_z^\infty dy \int_0^{y+z} dx a^2 e^{-a(x+y)} \\&#xD;&#xA;&amp;amp;= -\frac 1 2 e^{-3 a z } + e^{-a z}.\end{split}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3008" LastEditorUserId="3008" LastEditorDisplayName="" LastEditDate="2010-12-06T20:59:28.930" LastActivityDate="2010-12-06T20:59:28.930" Title="Density of a function of random variables." Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13277" PostTypeId="2" ParentId="13270" CreationDate="2010-12-06T19:14:21.047" Score="1" ViewCount="0" Body="&lt;p&gt;It is not clear what you mean with &quot;syntax&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you by &quot;syntax&quot; mean your way of reasoning, i.e. your &quot;path&quot; via logical connectives to the result, my best tip is to carefully read all proofs in your math book. Not only read through them, but understand them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must be able to prove things correctly, i.e. using only valid deductions. They way they do it in your math book is not a &quot;local&quot; thing, but &quot;global&quot; (no pun!). As long as your reasoning is understandable by anyone else reading it, it should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could be more spesific with what you mean by &quot;syntax&quot;, I'm sure somebody could give a clearer answer.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-06T19:14:21.047" />
  <row Id="13278" PostTypeId="2" ParentId="13207" CreationDate="2010-12-06T19:18:06.587" Score="2" ViewCount="0" Body="&lt;p&gt;If the equation is $z^2=xy\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}$, I would be tempted to see the symmetry in $x$ and $y$ and try solutions of the form $z=(xy)^n$.  What happens then?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T19:18:06.587" />
  <row Id="13279" PostTypeId="2" ParentId="13276" CreationDate="2010-12-06T19:20:29.300" Score="1" ViewCount="0" Body="&lt;p&gt;It is a laplace distibution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace_distribution#Relation_to_the_exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Laplace_distribution#Relation_to_the_exponential_distribution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:20:29.300" />
  <row Id="13280" PostTypeId="2" ParentId="13276" CreationDate="2010-12-06T19:24:36.093" Score="0" ViewCount="0" Body="&lt;p&gt;In general, you are actually interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross-correlation&quot; rel=&quot;nofollow&quot;&gt;Cross-correlation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider your problem in the setting of cross-correlation. We'll begin by considering the distribution function of $X-Y$ for $t&amp;gt;0$ (this is enough, since $X-Y$ is a symmetric random variable, hence having a symmetric density function).&#xA;Then, by the law of total probability (conditioning on $Y$), we have&#xA;$$&#xD;&#xA;{\rm P}(X - Y \le t) = \int_0^\infty  {{\rm P}(X - Y \le t|Y = \tau )f(\tau ){\rm d}\tau } = &#xD;&#xA;\int_0^\infty  {{\rm P}(X \le t + \tau )f(\tau ){\rm d}\tau }.  &#xD;&#xA;$$&#xD;&#xA;Of course, you can now calculate the right-hand side integral directly, and then differentiate with respect to $t$, but it is instructive to first differentiate with respect to $t$. Thus, we get&#xA;$$&#xD;&#xA;f_{X - Y} (t) = \frac{{\rm d}}{{{\rm d}t}}\int_0^\infty  {F(t + \tau )f(\tau ){\rm d}\tau }  = \int_0^\infty  {f(t + \tau )f(\tau ){\rm d}\tau } ,&#xD;&#xA;$$&#xD;&#xA;where $F$ denotes the distribution function of $X$ (and $Y$). According to Wikipedia's link given above, this can be denoted as &#xA;$f_{X-Y}(t) = (f \star f)(t)$ (read &quot;$f_{X-Y}(t)$ is the cross-correlation of $f$ with itself&quot;). Finally, from $f(u)= a {\rm e}^{-a u}$, $u &amp;gt; 0$, we find $f_{X-Y}(t) = (a/2){\rm e}^{-at}$, $t &amp;gt; 0$; then, by the aforementioned symmetry, $f_{X-Y}(t) = (a/2){\rm e}^{ - a|t|}$, for all $t \in \mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-06T20:33:52.693" LastActivityDate="2010-12-06T20:33:52.693" />
  <row Id="13281" PostTypeId="2" ParentId="13270" CreationDate="2010-12-06T19:27:27.310" Score="9" ViewCount="0" Body="&lt;p&gt;Learning to write mathematics properly is much like learning to write any language properly, only more so because such a premium is placed on logical organization and clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematics, both the part that is written in words and the part that is written in symbols, is attempting to convey meaning. When you write and read symbols, think about what they &lt;strong&gt;say&lt;/strong&gt; when you read them out loud. Every time you see the symbol $=$, remember that the symbol has a pronounciation when you read it, and it says &quot;is equal to&quot;. So if you write things like&#xA;$$2x = 4 = \frac{4}{2}=2,$$&#xD;&#xA;(which I see far too often on exams) then you are saying &quot;twice $x$ is equal to four, which is equal to four halves, which is equal to $2$&quot;, which of course is false and liable to cost you points, even though you probably don't &lt;em&gt;actually&lt;/em&gt; think that $4$ and $\frac{4}{2}$ are equal. Remember, first and foremost, that every symbol has a meaning and a pronounciation. Unless you recongize that, you won't be able to get very far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first you need to be clear on what you want to say, and then make sure that what you wrote actually conveys that meaning and not other meanings. If you can do that, even if it is with &quot;plain English&quot;, you will have gotten over more than half the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, mathematics is also a technical language with a number of use conventions and jargon. The very best thing you can to become familiar with, and good at using, those conventions, is to read a lot of mathematics, with an eye towards &lt;em&gt;understanding&lt;/em&gt; what is written and how the language helps that understanding (just like doing a lot of reading is one of the best ways to improve one's spelling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A close second is to read books that are meant to help introduce you to proofs and logical arguments, usually with subtitles like &quot;first course in advanced mathematics&quot; or &quot;introduction ot abstract/advanced mathematics&quot;. Find out what the &quot;Intro to proofs&quot; course is at your school, and look at the textbook they use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important thing is not to simply try to mimic the language you see: that will result in the mathematical equivalent of saying &lt;em&gt;&quot;Buenos días. Yo quiero estación de tren ser, por favor&quot;&lt;/em&gt;  when trying to ask for directions to the train station (&quot;Good morning. I want be train station, please.&quot;) You want to keep an eye on the meaning that the words are conveying, and how that particular choice of words (end even the &lt;em&gt;order&lt;/em&gt; of the words) matters. Notice, for instance, that saying &quot;For every $x$ there is a $y$&quot; is &lt;em&gt;not&lt;/em&gt; the same thing as saying &quot;There is a $y$ such that for every $x$...&quot;, even though they may seem very similar when thought of in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: always think first about what you are trying to say and make sure you say &lt;em&gt;that&lt;/em&gt;. Read what you've written, pronouncing every symbol to yourself to make sure you aren't saying that you &quot;want to be train station&quot;. And read your books and professors' notes to see how the language works and become familiar with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; It seems I rather badly misinterpret the true trust of your question (despite the fact that you seem to have &quot;accepted&quot; my answer). In so far as what steps to add or what steps to skip, as has been pointed out it depends on your level. I would not object to a student in my graduate abstract algebra class going from $x(x+1)(x+2)$ directly to $x^3+3x^2+2x$ (or even not writing out the computations before using it!) but I would definitely request a precalculus student to write out how he got to that final answer. If you have to stop and think carefully about what the answer &lt;em&gt;is&lt;/em&gt;, then you should not skip the step and write it down. For a particular class, look at the textbook and which steps it works out explicitly and which steps it skips. Look at what steps your professor works out explicitly and which steps he skips. You'll want to not stray too far from them as far as skipping more steps (though of course you can always skip &lt;em&gt;fewer&lt;/em&gt; steps if you are unsure about a calculation/argument). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T19:34:44.713" LastActivityDate="2010-12-06T19:34:44.713" CommentCount="2" />
  <row Id="13282" PostTypeId="1" AcceptedAnswerId="13311" CreationDate="2010-12-06T19:32:13.700" Score="7" ViewCount="213" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to create a program, that will decompose a matrix using the Cholesky decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decomposition itself isn't a difficult algorithm, but a matrix, to be eligible for Cholesky decomposition, must be symmetric and positive-definite. Checking whether a matrix is symmetric is easy, but the positive part proves to be more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read about the Sylvester's criterion, but that leads to determinants, and based on what I found on the web, those are quite extensive and hard on computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell - is there something I might be missing? Due to the fact the the matrix is square or something like that, is there possibly a simpler way to determine whether it's positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&#xA;Paul&lt;/p&gt;&#xA;" OwnerUserId="3543" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-06T23:56:43.870" LastActivityDate="2010-12-27T22:55:51.737" Title="Determining whether a symmetric matrix is positive-definite (algorithm)" Tags="&lt;linear-algebra&gt;&lt;algorithms&gt;&lt;matrices&gt;&lt;numerical-methods&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="13286" PostTypeId="2" ParentId="13282" CreationDate="2010-12-06T19:42:35.407" Score="3" ViewCount="0" Body="&lt;p&gt;I don't think there is a simpler way than computing a decomposition or determinant unless your matrix has a special form.  For example if it is a sample covariance matrix then it is positive semidefinite by construction.&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T19:42:35.407" CommentCount="1" />
  <row Id="13289" PostTypeId="2" ParentId="13282" CreationDate="2010-12-06T19:57:27.890" Score="0" ViewCount="0" Body="&lt;p&gt;Perhaps someone has an iterative approach or some other approximate approach, but it looks like to be sure, one has to calculate determinants and apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%27s_criterion&quot; rel=&quot;nofollow&quot;&gt;Sylvester's criterion&lt;/a&gt; (as you pointed out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sylvester's states that iff, for all $k&amp;lt;n$, the $\det(A_k) &amp;gt; 0$, where $A_k$ is the $k$'th principle minor, then the matrix is positive definite.  The only deterministic, efficient, algorithm to calculate determinants that I know of is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bareiss_algorithm&quot; rel=&quot;nofollow&quot;&gt;Bareiss algorithm&lt;/a&gt;, for which you can see &lt;a href=&quot;http://www.jstor.org/stable/2004533&quot; rel=&quot;nofollow&quot;&gt;Bareiss's original paper&lt;/a&gt; or Chee Yap's excellent reference &lt;a href=&quot;http://cs.nyu.edu/~yap/book/alge/ftpSite/l10.ps.gz&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bareiss algorithm, in it's intermediate steps, ends up calculating the determinant of the $k$ minors of the matrix for all $k&amp;lt;n$, so you get a simple test for positive definiteness should you end up implementing the Bareiss algorithm yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be interested to know if there are any iterative methods that would give you some confidence of a matrices positive definiteness.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2010-12-06T19:57:27.890" CommentCount="1" />
  <row Id="13290" PostTypeId="2" ParentId="13257" CreationDate="2010-12-06T20:07:48.427" Score="4" ViewCount="0" Body="&lt;p&gt;I don't know how to solve exactly (or even if it's possible) but maybe this will help: Let $A = e^a$ and $B = e^b$. It suffices to find $A$ and $B$ and then take logarithms. Your two equations can be written as&#xA;$$y_1 = {A(B^{x_1} - 1) \over 1 + AB^{x_1}}$$&#xD;&#xA;$$y_2 = {A(B^{x_2} - 1) \over 1 + AB^{x_2}}$$&#xD;&#xA;You can solve each of these equations for $A$ via some algebra:&#xA;$$A = {y_1 \over B^{x_1}(1 - y_1) - 1}$$&#xD;&#xA;$$A = {y_2 \over B^{x_2}(1 - y_2) - 1}$$&#xD;&#xA;So the reciprocals of the right-hand sides are the same (the original equation ensures that $y_1$ and $y_2$ are nonzero):&#xA;$${B^{x_1}(1 - y_1) - 1 \over y_1} = { B^{x_2}(1 - y_2) - 1 \over y_2}$$&#xD;&#xA;In other words,&#xA;$${1 - y_1 \over y_1}B^{x_1} - {1 - y_2 \over y_2}B^{x_2} = {1 \over y_1} - {1 \over y_2}$$&#xD;&#xA;This doesn't have a general solution you can write out (for example if $x_1$ and $x_2$ are large positive integers it's a polynomial equation of large degree). But it shouldn't be too hard to deal with numerically. Once you know what $B$ is (or have a good enough approximation) you can plug back in above to get $A$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-06T20:07:48.427" CommentCount="4" />
  <row Id="13291" PostTypeId="1" CreationDate="2010-12-06T20:10:50.300" Score="2" ViewCount="128" Body="&lt;p&gt;In logic theory, if we have a finite domain A of k elements, then we can construct only finitely many structures (finite structures) each of which has A as domain. I think the number of structures will be $2^{k}$. How to prove it - if my claim is right? &lt;/p&gt;&#xA;" OwnerUserId="4481" LastEditorUserId="4481" LastEditorDisplayName="" LastEditDate="2010-12-06T23:49:42.587" LastActivityDate="2010-12-06T23:49:42.587" Title="In Logic Theory...." Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13292" PostTypeId="2" ParentId="13291" CreationDate="2010-12-06T20:18:00.417" Score="-1" ViewCount="0" Body="&lt;p&gt;2^k is the number of subsets of a set with k elements&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-06T20:18:00.417" />
  <row Id="13293" PostTypeId="1" AcceptedAnswerId="13295" CreationDate="2010-12-06T20:27:15.833" Score="4" ViewCount="94" Body="&lt;p&gt;I have some kind of &quot;homework&quot; question. I have the following theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; (Transfinite Recursion over the class of of ordinals $\mathbf{ON}$: Let $\mathbf{V}$ be the class of all sets. If $\mathbf{F} : \mathbf{V} \to \mathbf{V}$ then there exists a unique $\mathbf{G} : \mathbf{ON} \to \mathbf{V}$ such that for all $\alpha$ $\mathbf{G}(\alpha) = \mathbf{F}(\mathbf{G}|_{\alpha}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to show that if $\hat{x} = \{y : y &amp;lt; x\}$ and if $\mathbf{F} : \mathbf{V} \to \mathbf{V}$ and $(X, &amp;lt;)$ is a well-ordering then there exists a function $f$ with domain $X$ such that $f(x) = \mathbf{F}(f|_{\hat{x}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I let $0 = \text{min}(X)$, and then I note that I want $f(0) = \mathbf{F(\emptyset})$. I continue like this and I note that $f(1) = \mathbf{F}(f(0))$ and so on. Is this correct? If so, can someone give me a hint how I apply the above theorem to say that this construction works?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-07T04:24:46.697" LastActivityDate="2010-12-07T04:24:46.697" Title="Recursion on ordinals." Tags="&lt;homework&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13294" PostTypeId="1" AcceptedAnswerId="13299" CreationDate="2010-12-06T20:41:31.527" Score="0" ViewCount="66" Body="&lt;p&gt;How would I go about it?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-06T22:40:09.427" LastActivityDate="2010-12-06T22:40:09.427" Title="figuring out the x,y, and z rotation of a right triangle?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13295" PostTypeId="2" ParentId="13293" CreationDate="2010-12-06T20:48:42.050" Score="3" ViewCount="0" Body="&lt;p&gt;If you can use the transfinite recursion theorem you stated, why not simply apply it to $F$, and then use an isomorphism between $X$ and some ordinal to transfer the function $G$ from $ON$ to $X$?  It seems like you are trying to reprove the transfinite recursion theorem rather than just applying it. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-06T20:48:42.050" />
  <row Id="13296" PostTypeId="1" CreationDate="2010-12-06T20:53:59.247" Score="-1" ViewCount="63" Body="&lt;p&gt;Let $\lambda$ denote the Lebesgue measure and let $C$ be a collection of intervals and let $\sigma(C)$ denote the sigma-algebra generated by the collection $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the total variation $v$ of the Lebesgue measure $\lambda$ always absolutely continuous with respect $\lambda$?&lt;/p&gt;&#xA;" OwnerUserId="4482" LastActivityDate="2010-12-06T22:13:14.000" Title="Total variation of a measure absolutely continuous?" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13298" PostTypeId="2" ParentId="13153" CreationDate="2010-12-06T21:45:42.803" Score="0" ViewCount="0" Body="&lt;p&gt;Here's a hint: calculate the derivative of $\sec^n(x)$ for integer n &gt; 0, then think how you can use this , and the fact that $\tan^2(x) = 1 + \sec^2(x)$, to calculate the integral.&lt;/p&gt;&#xA;" OwnerUserId="1470" LastActivityDate="2010-12-06T21:45:42.803" />
  <row Id="13299" PostTypeId="2" ParentId="13294" CreationDate="2010-12-06T21:50:09.790" Score="0" ViewCount="0" Body="&lt;p&gt;So you have the co-ordinates $P_i = (x_i, y_i, z_i), \ \ i=1,2,3$ of the vertices and you want to find the angles of the triangle formed by $P_1, P_2, P_3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the angle at $P_1$ be $\theta_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \vec{a} = (x_3 -x_1, y_3 - y_1, z_3 - z_1)$ &lt;br/&gt;&#xA;$\displaystyle \vec{b} =  (x_2 - x_1, y_2 - y_1, z_2  -z_1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle |\vec{a}| |\vec{b} | \cos \theta_1 = \vec{a} \cdot \vec{b}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $|\vec{v}|$ is the length of $\vec{v}$ and $\cdot $ is the dot product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you can find the other angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, see here: &lt;a href=&quot;http://www.mvps.org/directx/articles/math/dot/index.htm&quot; rel=&quot;nofollow&quot;&gt;Understanding the Dot Product&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-06T22:16:45.867" LastActivityDate="2010-12-06T22:16:45.867" CommentCount="1" />
  <row Id="13300" PostTypeId="1" CreationDate="2010-12-06T21:55:54.887" Score="5" ViewCount="230" Body="&lt;p&gt;I have a function $S(x)$ which is bounded above and below as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) + C_1 + \mathcal{O}(g(x)) &amp;lt; S(x) &amp;lt; f(x) + C_2 + \mathcal{O}(g(x))$ as $x \rightarrow \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I conclude that $$S(x) = f(x) + C + \mathcal{O}(g(x))$$ as $x \rightarrow \infty$? Can anyone give a short proof for this fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To clear up the confusion, I am stating the problem below. I am wondering if what I did was right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that as $x \rightarrow \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle \sum_{n \leq x} \frac{\log n}{n} = \frac{(\log x)^2}{2} + C + \mathcal{O}(\frac{\log(x)}{x})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate the summation as an integral $\int \frac{\log(x)}{x} dx$ from above and from below. (The usual way of coming up with bounds for $p$ series). Then we will get the lower bound as $\frac{\log([x])^2}{2} + C_1$ and the upper bound as $\frac{\log([x])^2}{2} + C_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $\log([x]) = \log(x) + \log(\frac{[x]}{x}) = \log(x) + \log(1-\frac{\{x\}}{x}) = \log(x) + \mathcal{O}(\frac{\{x\}}{x}) = \log(x) + \mathcal{O}(\frac{1}{x})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the above I get the the question I asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this when I was asked on a timed exam since I could not think of other ways to get to the answer. Other answers and suggestions welcome as always.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-07T06:04:14.213" LastActivityDate="2010-12-07T15:54:08.410" Title="Big $\mathcal{O}$ Notation question while estimating $\sum \frac{\log n}{n}$" Tags="&lt;asymptotics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="13301" PostTypeId="2" ParentId="13300" CreationDate="2010-12-06T22:04:29.627" Score="2" ViewCount="0" Body="&lt;p&gt;I don't think so because $g(x)$ might be too small.  Take $f(x)=0, C_1=-1.5, C_2=1.5, g(x)=\exp(-x)$.  Then doesn't $S(x)=\sin(x)$ violate it?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T22:04:29.627" CommentCount="4" />
  <row Id="13302" PostTypeId="2" ParentId="13296" CreationDate="2010-12-06T22:13:14.000" Score="1" ViewCount="0" Body="&lt;p&gt;What do you mean? The Lebesgue measure is a positive measure. But anyway we can show (quite directly) that if $\mu$ is a measure and $\nu$ is a positive measure on $(X, \Omega)$ then $\mu \ll \nu$ is equivalent with $|\mu| \ll \nu$.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-06T22:13:14.000" />
  <row Id="13303" PostTypeId="2" ParentId="13300" CreationDate="2010-12-06T22:19:58.703" Score="0" ViewCount="0" Body="&lt;p&gt;If $a&amp;lt;x&amp;lt;b$, then $|x|\le \max \lbrace |a|,|b|\rbrace$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if $C_1=C_2=C$,  what you have there implies &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S(x)-f(x)-C|\le K |g(x)| $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $K&amp;gt;0$, i.e. $S(x)=f(x)+C+O(g(x))$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-06T22:19:58.703" CommentCount="3" />
  <row Id="13304" PostTypeId="2" ParentId="13272" CreationDate="2010-12-06T22:22:23.850" Score="3" ViewCount="0" Body="&lt;p&gt;The rotation matrices around the x, y, and z axes, respectively, are&#xA;$$R_x(\theta) = \begin{pmatrix} 1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; \cos \theta &amp;amp; -\sin \theta \\ 0 &amp;amp; \sin \theta &amp;amp; \cos \theta \end{pmatrix}$$&#xD;&#xA;$$R_y(\phi) = \begin{pmatrix} \cos \phi &amp;amp; 0 &amp;amp; \sin \phi \\ 0 &amp;amp; 1 &amp;amp; 0 \\ - \sin \phi &amp;amp; 0 &amp;amp; \cos \phi \end{pmatrix}$$&#xD;&#xA;$$R_z(\psi) = \begin{pmatrix} \cos \psi &amp;amp; - \sin \psi &amp;amp; 0 \\ \sin \psi &amp;amp; \cos \psi &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 \end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to rotate in the order specified in your comment on mathcast's answer, then you want&#xA;$$R_z(\psi) \cdot R_y(\phi) \cdot R_x(\theta) = $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix} \cos \phi \cos \psi &amp;amp; \cos \psi \sin \theta \sin \phi - \cos \theta \sin \psi &amp;amp; \cos \theta \cos \psi \sin \phi + \sin \theta \sin \psi \\ \cos \phi \sin \psi &amp;amp; \cos \theta \cos \psi + \sin \theta \sin \phi \sin \psi &amp;amp; \cos \theta \sin \phi \sin \psi - \cos \psi \sin \theta \\ - \sin \phi &amp;amp; \cos \phi \sin \theta &amp;amp; \cos \theta \cos \phi \end{pmatrix}$$&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-12-06T22:22:23.850" CommentCount="3" />
  <row Id="13305" PostTypeId="2" ParentId="13300" CreationDate="2010-12-06T22:41:16.147" Score="8" ViewCount="0" Body="&lt;p&gt;A misconception needs to be cleared up here.  Recall that $f(x) = O(g(x))$ means that there exists a constant $C$ such that $|f(x)| \le C |g(x)|$ in a neighborhood of whatever value of $x$ you're interested in.  It is an abuse of notation that we use the equality symbol here at all; $O(g(x))$ is not a function, so it cannot be &quot;equal&quot; to $f(x)$ in the usual sense.  See the discussion at &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Equals_sign&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this definition, people sometimes say that $f(x) = h(x) + O(g(x))$ if $f(x) - h(x) = O(g(x))$.  This is an additional abuse of notation, but it generally does not create problems and can be extremely convenient.  I guess you could write $f(x) &amp;lt; h(x) + O(g(x))$ if you really wanted to, but this is redundant, since the inequality is already built into the definition of big-O notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have no idea what $f(x) &amp;gt; h(x) + O(g(x))$ is supposed to mean.  If you are trying to say that there is a constant $C$ such that $|f(x) - h(x)| &amp;gt; C |g(x)|$, the correct way to write this is using big-Omega, not big-O, as $f(x) = h(x) + \Omega(g(x))$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-06T22:46:42.087" LastActivityDate="2010-12-06T22:46:42.087" CommentCount="4" />
  <row Id="13306" PostTypeId="2" ParentId="4781" CreationDate="2010-12-06T22:55:46.200" Score="2" ViewCount="0" Body="&lt;p&gt;Your solution is almost correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the definition of RL, we require that the algorithm halt with probability $1$ on any input. As you note, your algorithm runs forever on &quot;No&quot; instances. To fix this, we'll take advantage of the fact that we're allowed a small chance of error. In particular, we're allowed to sometimes reject &quot;Yes&quot; instances, as long as we never accept a &quot;No&quot; instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is to keep a counter of how many iterations we've used so far. If a path exists, we should find it within $2N^N$ iterations (with probability at least $3/4$). So if we've gone that long without finding a valid path, we'll reject the graph. Note that we'll never accept a bad graph this way, and we'll rarely reject a good graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't naively count to $N^N$, however, since this would require $N\lg N$ space. Instead, we'll probabilistically &quot;count&quot; that high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After each iteration, we'll flip $N\lg N + 2$ coins. If every coin lands heads, then we stop (and reject the graph). If at least one coin lands tails, then we begin another iteration. This can be done in log-space, since we only need to count how many flips have occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of getting all heads is $1/2^{N\lg N + 2}=1/4N^N$, so with probability at least $3/4$, this won't happen until at least $2N^N$ iterations have elapsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a union bound over both ways the algorithm could fail, we see that everything works with probability at least $1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that this halts with probability $1$, since we'll flip all heads eventually.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-12-06T22:55:46.200" />
  <row Id="13307" PostTypeId="2" ParentId="13291" CreationDate="2010-12-06T23:01:57.043" Score="3" ViewCount="0" Body="&lt;p&gt;Janice : As Carl pointed out, the answer depends on the signature (language) one uses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose the language consists of a single unary relation symbol $\dot R$. Any structure in this language has the form $(A,R)$, where $R\subseteq A$. If we fix $A$ with $|A|=k$, there are precisely $2^k$ structures of this form (one for each subset of $A$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carl's comment was: Suppose your language consists of a single unary function symbol $\dot f$. Any structure of this language has the form $(A,f)$ where $f:A\to A$. The number of functions from a finite set $A$ to a finite set $B$ is $|B|^{|A|}$, so there are $k^k$ such functions $f$, and therefore we have $k^k$ models in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a language consists of: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For each positive integer $n$, some $n$-ary function symbols $\dot f$. Each of them will be interpreted as a function $f:A^n\to A$. There are $k^{k^n}$ possible interpretations, if $A$ is fixed and $|A|=k$.&lt;/li&gt;&#xA;&lt;li&gt;For each positive integer $n$, some $n$-ary relation symbols $\dot R$. Each of them will be interpreted as a set $R\subseteq A^n$. There are $2^{k^n}$ many possible interpretations.&lt;/li&gt;&#xA;&lt;li&gt;Some constant symbols $\dot c$. Each of them will be interpreted as an element of $A$, and there are $k$ possible interpretations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can interpret different symbols independently of one another, and we understand that order matters, so $(A,R,S)$ and $(A,S,R)$ are interpreted as different structures if $R\ne S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that $\rho_n,\lambda_n,\tau$ are cardinals (they may be $0$, they may be infinite), and that our language has precisely $\rho_n$ $n$-ary function symbols, precisely $\lambda_n$ $n$-ary relation symbols (for each positive integer $n$), and precisely $\tau$ constant symbols. Then the number of structures in this language that have domain $A$ with $|A|=k$ finite is precisely $$ \prod_{n\ge 1}(k^{k^n})^{\rho_n}\cdot\prod_{n\ge 1}(2^{k^n})^{\lambda_n}\cdot k^\tau, $$ where (as usual) we understand that empty products equal 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the products are cardinals in the set theoretic sense. If you are only interested in finite languages, this product gives the right answer, when we interpret products as numbers the usual way.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-06T23:01:57.043" CommentCount="3" />
  <row Id="13308" PostTypeId="1" CreationDate="2010-12-06T23:24:39.220" Score="4" ViewCount="66" Body="&lt;p&gt;If I have the Keplerian elements for two orbits, how do I compute the shortest distance between these two orbits in 3D space? Is there any analytical expression to compute that?&lt;/p&gt;&#xA;" OwnerUserId="4486" LastActivityDate="2011-01-05T23:44:39.050" Title="Analytical Expression to find the Shortest Distance between Two Ellipses?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13309" PostTypeId="1" AcceptedAnswerId="13313" CreationDate="2010-12-06T23:29:35.730" Score="0" ViewCount="72" Body="&lt;p&gt;Hello&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all about maths I don't understand, how can I solve this exercise, someone say that is a 2 dimensional problem, but I can not figure out for myself, I already understand what to do, but I don't know why is the formula, can someone explain this exercise for a dummy? &lt;a href=&quot;http://www.codeforces.com/contest/1/problem/A&quot; rel=&quot;nofollow&quot;&gt;http://www.codeforces.com/contest/1/problem/A&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;why the solution of this problem is ((m+a-1)/a)*((n+a-1)/a)?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help really appreciated&lt;/p&gt;&#xA;" OwnerUserId="4487" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-06T23:57:45.640" LastActivityDate="2010-12-06T23:57:45.640" Title="Covering a plaza with square flagstones" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13310" PostTypeId="2" ParentId="13308" CreationDate="2010-12-06T23:33:28.907" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.mathworks.com/matlabcentral/fileexchange/26329-determine-the-distance-between-two-ellipses-in-3d&quot; rel=&quot;nofollow&quot;&gt;http://www.mathworks.com/matlabcentral/fileexchange/26329-determine-the-distance-between-two-ellipses-in-3d&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-06T23:33:28.907" CommentCount="1" />
  <row Id="13311" PostTypeId="2" ParentId="13282" CreationDate="2010-12-06T23:34:22.823" Score="6" ViewCount="0" Body="&lt;p&gt;Mathcast had it; in fact, in practical work, one uses the Cholesky decomposition $\mathbf G\mathbf G^T$ for &lt;strong&gt;efficiently testing&lt;/strong&gt; if a symmetric matrix is positive definite. The only change you need to make to turn your decomposition program into a check for positive definiteness is to insert a check before taking the required square roots that the quantity to be rooted is positive. If it is zero, you have a positive semidefinite matrix; if not, then your symmetric matrix isn't positive (semi)definite. (Programming-wise, it should be easy to throw an exception within a loop! If your language has no way to break a loop, however, you have my pity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one uses the $\mathbf L\mathbf D\mathbf L^T$ decomposition here (an equivalent approach in the sense that $\mathbf G=\mathbf L\sqrt{\mathbf D}$); if any nonpositive entries show up in $\mathbf D$, then your matrix is not positive definite. Note that one could set things up that the loop for computing the decomposition is broken once a negative element of $\mathbf D$ is encountered, before the decomposition is finished!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, I don't understand why people are shying away from using Cholesky here; the statement is &quot;a matrix is positive definite &lt;strong&gt;if and only if&lt;/strong&gt; it possesses a Cholesky decomposition&quot;. It's a biconditional; exploit it! It's exceedingly cheaper than successively checking minors or eigendecomposing, FWIW.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-06T23:34:22.823" CommentCount="1" />
  <row Id="13312" PostTypeId="2" ParentId="13309" CreationDate="2010-12-06T23:38:53.223" Score="0" ViewCount="0" Body="&lt;p&gt;The answer is (the area to cover) / (the area per flagstone) + (penalty for having to use whole flagstones).  Note that your answer is almost equal to (m*n)/(a*a) so it looks about right.&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-06T23:38:53.223" CommentCount="10" />
  <row Id="13313" PostTypeId="2" ParentId="13309" CreationDate="2010-12-06T23:57:01.380" Score="1" ViewCount="0" Body="&lt;p&gt;So break it into one dimensional problems.  How many flagstones of length a does it take to cover a path of m meters?  If you can trust a and m to be integers, m/a is correct if a divides m.  Otherwise you need floor (m/a) +1.  One way to combine these is floor((m+a-1)/a).  Try it with some small numbers to see how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two dimensions, just multiply two one dimensional problems.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-06T23:57:01.380" CommentCount="1" />
  <row Id="13314" PostTypeId="2" ParentId="13300" CreationDate="2010-12-07T00:41:42.893" Score="6" ViewCount="0" Body="&lt;p&gt;Since you wanted a different proof approach, you can try using &lt;a href=&quot;http://books.google.com/books?id=iJx1lP9ZcIkC&amp;amp;pg=PA25&quot; rel=&quot;nofollow&quot;&gt;Abel's Identity&lt;/a&gt;, which has turned out to be quite useful in analytic number theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance see this: &lt;a href=&quot;http://math.stackexchange.com/questions/12566/proving-sum-limits-p-leq-x-frac1-sqrtp-geq-frac12-logx-log/12588#12588&quot;&gt;An estimate for sum of reciprocals of square roots of prime numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To apply this to your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac1{n} = \log x + \gamma + R(x)$, where $\displaystyle R(x) = \mathcal{O}\left(\frac1{x}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Abel's identity we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = (\log x+ \gamma + R(x))\log x - \int_{1}^{x} \frac{\log t+ \gamma + R(t)}{t} \ \text dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = \frac{\log^2 x}{2} + R(x)\log x - \int_{1}^{x} \frac{R(t)}{t} \ \text dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle R(x) = \mathcal{O}\left(\frac1{x}\right)$ we have that $\displaystyle \int_{1}^{\infty} \frac{R(t)}{t} \ \text dt = \eta$ exists. We also have that $\displaystyle R(x) \log x \to 0 \ \text{as} \ x \to \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = \frac{\log^2 x}{2} -\eta + \mathcal{O}\left(\frac{\log x}{x}\right) \ \ \text{as} \ \ x \to \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another useful approach is to try using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-Maclaurin Summation formula&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-07T15:54:08.410" LastActivityDate="2010-12-07T15:54:08.410" CommentCount="4" />
  <row Id="13315" PostTypeId="1" AcceptedAnswerId="13316" CreationDate="2010-12-07T01:10:55.193" Score="1" ViewCount="57" Body="&lt;p&gt;I'm given a number (n), an interval (i), and the number of steps (s).  I want to distribute these numbers over that criteria as best as possible.  The upper bounds would be n (which is critical) and lower bounds is greater than or equal to 0.75 * n (but divisible by the interval).  So, if n=100, i=5, and s=6, you'd have [100, 95, 90, 85, 80, 75].  This one is easy because 100*.75 is &amp;lt;= s*i.  How would you do n=25, i=2.5, and s=6?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a problem I came across while programming and was wondering if there was some mathematical concept that fit this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I had no idea what to tag this under.&lt;/p&gt;&#xA;" OwnerUserId="4489" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-07T01:21:00.950" LastActivityDate="2010-12-07T01:21:00.950" Title="How do I distribute these numbers?" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;" AnswerCount="2" />
  <row Id="13316" PostTypeId="2" ParentId="13315" CreationDate="2010-12-07T01:18:38.987" Score="1" ViewCount="0" Body="&lt;p&gt;If I understand you, the ideal situation is to have your set be $[n, n-i, n-2i, \ldots,n-(s-1)i]$, which has s members.  To keep this within the top $0.75n$, you need $(s-1)i\leq .25n$, which is just satisfied in your first example.  In your second it is not, so you need to decide what to give up.  You could forget about the 0.75 criterion and keep the interval, giving $[25, 22.5, \ldots,12.5]$  Or you could maintain the 0.75 criterion and reduce the step size, giving an interval of 1.25.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-07T01:18:38.987" CommentCount="4" />
  <row Id="13317" PostTypeId="2" ParentId="13315" CreationDate="2010-12-07T01:20:08.017" Score="0" ViewCount="0" Body="&lt;p&gt;in R:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;seq(from=25, by=-2.5, length=6)&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-07T01:20:08.017" />
  <row Id="13318" PostTypeId="1" CreationDate="2010-12-07T01:49:43.813" Score="-1" ViewCount="115" Body="&lt;p&gt;Alright, so I have a triangle that is part of a mesh (3d model) and I need to get the rotation of it's x,y and z axis. Did I say all that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rotation of y axis&#xA;    (Math.toDegrees(Math.atan(min.getYf()/max.getYf())&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to get min and max:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vector3 min = new Vector3((float)Math.min((float) Math.min(v3[s][0].getXf(), v3[s][1].getXf()), v3[s][2].getXf()),&#xA;                    (float)Math.min((float)Math.min(v3[s][0].getYf(), v3[s][1].getYf()), v3[s][2].getYf()),&#xA;                    (float)Math.min((float)Math.min(v3[s][0].getZf(), v3[s][1].getZf()), v3[s][2].getZf()));&#xA;&#xA;            Vector3 max = new Vector3((float) Math.max((float)Math.max(v3[s][0].getXf(), v3[s][1].getXf()), v3[s][2].getXf()),&#xA;                    (float)Math.max((float)Math.max(v3[s][0].getYf(), v3[s][1].getYf()), v3[s][2].getYf()),&#xA;                    (float)Math.max((float)Math.max(v3[s][0].getZf(), v3[s][1].getZf()), v3[s][2].getZf()));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4464" LastEditorUserId="4464" LastEditorDisplayName="" LastEditDate="2010-12-07T02:10:46.330" LastActivityDate="2010-12-07T02:10:46.330" Title="I am so confused!" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="13319" PostTypeId="1" AcceptedAnswerId="13321" CreationDate="2010-12-07T01:55:21.970" Score="2" ViewCount="198" Body="&lt;p&gt;Okay, so almost 3 months into my abstract algebra, we just started rings. I have a few questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;trivial ring&quot; is a ring with only one element. So $R={0}$ is a trivial ring. Understandable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a definition states: Let $R$ be a ring. If there is an element $x \in R$ s.t. for all $a \in R$ you have $a * x = a = x * a$ then R is called a &quot;ring with identity&quot;. The notation is $1_{R}$. I also know it is possible to have a right identity and not a left identity (in the case of a 2 x 2 matrix). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have a definition for an integral domain. An integral domain is a commutative ring $R$ with identity $1_{R} \neq 0_{R}$. Okay so this is where I am confused now. What does this mean? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it saying that the identity element can not be the zero element? And the identity element could be anything right? For integers its 1. So basically it states that if you have a TRIVIAL RING and if the element (since there can only be one) is not the zero element then its an ID? &lt;/p&gt;&#xA;" OwnerUserId="287" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-07T16:58:46.180" LastActivityDate="2010-12-07T17:17:19.043" Title="What is an Integral Domain?" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="13320" PostTypeId="2" ParentId="13318" CreationDate="2010-12-07T01:59:59.993" Score="0" ViewCount="0" Body="&lt;p&gt;You need some kind of reference.  Where did the triangle start and what kind of transformation was applied to it?  Was it only rotated or was it also translated?&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-07T01:59:59.993" CommentCount="3" />
  <row Id="13321" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T02:01:45.670" Score="4" ViewCount="0" Body="&lt;p&gt;An integral domain is a commutative ring with unit $1\neq 0$ such that if $ab=0$ then either $a=0$ or $b=0$.  The idea that $1\neq 0$ means that the multiplicative unit, the element $x$ such that $xa=a$ for all $a\in R$ is not the same element as the additive unit, the element $y$ such that $a+y=a$ for all $a\in R$.  We denote the multiplicative unit by 1 and the additive unit by 0, since they are similar to the same thing in $\mathbb{Z}$.  The second part says that if $a,b\neq 0$, then $ab\neq 0$.  For instance $\mathbb{Z}/6$ has two elements $[2]$ and $[3]$ such that $[2]\cdot [3]=[0]$.  So, it is not a integral domain.&lt;/p&gt;&#xA;" OwnerUserId="2838" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T02:12:42.227" LastActivityDate="2010-12-07T02:12:42.227" CommentCount="20" />
  <row Id="13322" PostTypeId="2" ParentId="13268" CreationDate="2010-12-07T02:09:57.953" Score="0" ViewCount="0" Body="&lt;p&gt;I am too untrusted to add a comment so i'm spamming this as an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J.M: I think you are considering the converse of the question.  If you have two very different-looking matrices it is possible that they have the same spectrum.  But if you have two similar-looking matrices, then is it possible that their spectra can be very different?&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-07T02:09:57.953" CommentCount="8" />
  <row Id="13323" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T02:10:27.727" Score="6" ViewCount="0" Body="&lt;p&gt;There is only one trivial ring up to isomorphism.  It has one element $x$ which satisfies $x + x = x \cdot x = x$.  That means that $x$ is both the additive identity and the multiplicative identity; symbolically, $x = 1_R = 0_R$.  This is an if-and-only-if, too: if a ring $R$ has the property that $1_R = 0_R$, then every element is equal to $0_R$, since $x \cdot 1_R = x = x \cdot 0_R = 0_R$.  So a ring is the trivial ring if and only if $1_R = 0_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the definition of an integral domain, we require that the ring is nontrivial.  There are several good reasons for this, but they are sort of hard to motivate at the level of a first course in abstract algebra.  (But as Joe Johnson points out, this is not the main part of the definition of an integral domain.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-07T02:10:27.727" CommentCount="1" />
  <row Id="13324" PostTypeId="1" CreationDate="2010-12-07T03:08:11.680" Score="0" ViewCount="72" Body="&lt;p&gt;This might be an easy question, but I haven't been able to up come up with a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of the map $$f : \mathbb{R} \to \mathbb{R}^2, a \mapsto (\frac{2a}{a^2+1}, \frac{a^2-1}{a^2+1})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the unit circle take away the north pole. $f$ extends to a function $$g: \mathbb{C} \backslash \{i, -i \} \to \mathbb{C}^2. $$  Can anything be said about the image of $g$?&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-01-06T04:44:39.347" Title="A stereographic projection related question" Tags="&lt;complex-analysis&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13325" PostTypeId="1" AcceptedAnswerId="13350" CreationDate="2010-12-07T03:21:09.970" Score="-2" ViewCount="52" Body="&lt;p&gt;Alright, so I got two vectors that each contain a x y and z. We'll call them a and b.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;atan2(double y, double x)&#xA;Converts rectangular coordinates (x, y) to polar (r, theta).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I create a new point in the middle of the two points.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vector3 middle = new Vector3(//order x,y,z&#xA;        (max.getX() - min.getX())/2,&#xA;        (max.getY() - min.getY())/2,&#xA;        (max.getZ() - min.getZ())/2&#xA;        );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To get the y angle I use the formula &#xA;&lt;code&gt;Math.toDegrees(Math.atan2(middle.getX(), middle.getZ()))&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2010-12-07T08:21:31.780" Title="getting the x,y, and z angle of two points?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13326" PostTypeId="2" ParentId="13324" CreationDate="2010-12-07T03:24:52.883" Score="0" ViewCount="0" Body="&lt;p&gt;Note that although $a$ is complex, is valid :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{2a}{a^2+1}\right)^2+\left(\frac{a^2-1}{a^2+1}\right)^2= \frac{4a^2}{(a^2+1)^2}+\frac{(a^2-1)^2}{(a^2+1)}=$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{4a^2+a^4-2a^2+1}{(a^2+1)^2}=\frac{(a^4+2a^2+1)}{(a^2+1)^2}=\frac{(a^2+1)^2}{(a^2+1)^2}=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus is also an circle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is say, the points of the set $\{g(a)\in \mathbb{C}^2 :a\in \mathbb{C}/ \{\imath,-\imath\}\}$ meet the above.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-07T03:50:37.113" LastActivityDate="2010-12-07T03:50:37.113" CommentCount="1" />
  <row Id="13327" PostTypeId="2" ParentId="13324" CreationDate="2010-12-07T03:28:28.323" Score="0" ViewCount="0" Body="&lt;p&gt;It is the unit circle. Substituting $\tan t$ for $a$ and simplifying, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x=\sin 2t, y=-\cos 2t $$ which is the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-07T03:53:10.150" LastActivityDate="2010-12-07T03:53:10.150" CommentCount="4" />
  <row Id="13328" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T03:29:46.763" Score="3" ViewCount="0" Body="&lt;p&gt;An integral domain is a ring with no zero divisors, i.e. $\rm\ xy = 0\ \Rightarrow\ x=0\ \ or\ \ y=0\:.\:$ Additionally it is a widespread &lt;em&gt;convention&lt;/em&gt; to disallow as a domain the trivial one-element ring (or, equivalently, the ring with $\: 1 = 0\:$). It is the nonexistence of zero-divisors that is the important hypothesis in the definition. The exclusion of the trivial ring is merely a convention that proves convenient in many contexts (see my &lt;a href=&quot;http://math.stackexchange.com/questions/120/is-1-a-prime-number/5735#5735&quot;&gt;post here&lt;/a&gt; for some reasons why the convention proves convenient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may think of a domain as a ring-theoretic analog of a field, since a ring is a domain iff it is a subring of some field. Indeed, a field clearly has no zero-divisors so ditto for all of its subrings. Conversely, any domain may be enlarged to its field of fractions by adjoining an inverse for every nonzero element. Although domains needn't contain inverses for every $\rm\ c\ne 0\:,\:$ they do retain a trace of this property. Namely, nonzero elements $\rm\:c\:$ are cancellable: $\rm\ c\ x = c\ y\ \Rightarrow\ x = y\:.\ $ This enables transfer of many (but not all) properties of fields to domains. Here is one useful example: $\ $ a polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in the ring $\rm\:D\ $ iff $\rm\ D\:$ is a domain. For the simple proof see my &lt;a href=&quot;http://math.stackexchange.com/questions/4597/simple-explanation-and-examples-of-the-miller-rabin-primality-test/4604#4604&quot;&gt;post here&lt;/a&gt;, where I illustrate it constructively in $\rm\ \mathbb Z/m\ $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T17:17:19.043" LastActivityDate="2010-12-07T17:17:19.043" CommentCount="1" />
  <row Id="13330" PostTypeId="1" AcceptedAnswerId="13331" CreationDate="2010-12-07T03:46:45.337" Score="0" ViewCount="519" Body="&lt;p&gt;how does 1+1 = 3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen it around and I mean the obvious response is that its just wrong but people say stuff like there's just some special way to look at it or something. any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1141" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:04:06.223" LastActivityDate="2010-12-07T10:02:44.423" ClosedDate="2010-12-07T12:02:46.113" Title="I'm totally confused about this addition problem" Tags="&lt;addition&gt;" AnswerCount="4" CommentCount="7" />
  <row Id="13331" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T04:04:52.693" Score="1" ViewCount="0" Body="&lt;p&gt;It follows directly from the elementary theorem 2=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a=b$ be integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides by $a$: $a^2 = ab$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtract $b^2$ from both sides: $a^2 -b^2 = ab - b^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factor: $(a-b)(a+b) = b(a-b)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancel $(a-b)$ on both sides: $a+b = b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the fact $a=b$: $2b = b$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide both sides by $b$: $2 = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have $2=1$, we get $3 = 1 + 1$ as a corollary. And an elementary proof of the Riemann Hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. $a-b = 0$, so canceling is the same as division by $0$.&lt;/p&gt;&#xA;" OwnerUserId="4388" LastActivityDate="2010-12-07T04:04:52.693" CommentCount="3" />
  <row Id="13332" PostTypeId="1" CreationDate="2010-12-07T04:08:53.677" Score="2" ViewCount="57" Body="&lt;p&gt;It is a local constant sheaf, but why the local system on topological space $X$ is one-to-one correspondes to $\tilde{X}\times_G V$, where $G$is fundamental group of $X$ and $\tilde{X}$ is the universal space of $X$, and $V$ is a $G$-module,  I am wondering why $\tilde{X}\times_G V$ is a local system (local constant sheaf)?&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:34:04.393" LastActivityDate="2010-12-07T05:45:05.957" Title="What local system really is" Tags="&lt;algebraic-topology&gt;&lt;coherent-sheaves&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13333" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T04:14:31.683" Score="2" ViewCount="0" Body="&lt;p&gt;I have no idea which people are saying $1+1 = 3$ and why they are saying it.  Certainly, without any additional explanation this sounds like a simply false statement: in standard arithmetic, $1+1 = 2$, which does not equal $3$!  (It is not even the case that $1+ 1 = 3$ &lt;em&gt;modulo n&lt;/em&gt; for any $n &amp;gt; 1$, and considering numbers modulo $1$ is, literally, trivial: I believe there was another recent question about that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there &lt;em&gt;are&lt;/em&gt; real-world phenomena which are somewhat reasonably described as $1 +1 = 3$.  For instance, suppose I am living on a desert island and I have fashioned a very crude scale out of coconuts, bamboo, crab exoskeleta, and so forth.  My scale doesn't have decimals on it: it says that the weight of something is some whole number of pounds.  Now it is entirely possible that I can have two rocks, each of which, according to the scale, weighs one pound, but when I put both rocks on the scale, it tells me that together they weigh three pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is an instance of rounding error: e.g. maybe the true weight of each rock is something like $1.35$ pounds -- which gets rounded to $1$ -- so the true weight of the two rocks is something like $2.7$ pounds -- which gets rounded to $3$.  I have explained this in the context of a somewhat unlikely scenario, but rounding errors are something that one has to take very seriously in applied mathematics and the sciences.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-07T04:14:31.683" />
  <row Id="13334" PostTypeId="1" AcceptedAnswerId="13338" CreationDate="2010-12-07T05:04:19.207" Score="3" ViewCount="103" Body="&lt;p&gt;I am reading Joseph Bernstein's notes on D-modules which are available online &lt;a href=&quot;http://www.math.columbia.edu/~khovanov/resources/Bernstein-dmod.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section 1 (page 1 of the pdf) Bernstein writes &quot;By Serre's theorem this condition is local.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering to what theorem Bernstein is referring and wondering where I could find a statement and proof of this theorem.&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2010-12-07T17:35:15.013" Title="What theorem of Serre's is being referenced?" Tags="&lt;reference-request&gt;&lt;d-modules&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13335" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T05:04:26.910" Score="0" ViewCount="0" Body="&lt;p&gt;More generally, if somewhere in our reasoning we make a mistake (by dividing by zero, assuming that $a^2 = b^2 \implies a = b$, assuming that $\lnot P \implies \lnot Q$ follows from $P \implies Q$), we are able to prove anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $P \implies Q$ is always true if $P$ is false to begin with. Wikipedia calls this being &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuous_truth&quot; rel=&quot;nofollow&quot;&gt;vacuously true&lt;/a&gt;&quot;, although I think the term is usually used when the hypothesis is not satisfied but the result is still true, as in the example given of all cell phones being turned off in a room simply because there are no cell phones.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2010-12-07T05:04:26.910" />
  <row Id="13336" PostTypeId="1" AcceptedAnswerId="13337" CreationDate="2010-12-07T05:06:18.217" Score="1" ViewCount="71" Body="&lt;p&gt;Let $R$ be a non-zero commutative ring. Prove that the ideal $(x)$ of $R[x]$ is prime if and only if $R$ is an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on the left-to-right direction right now. I know that $R[x]/(x)$ is an integral domain since $(x)$ is prime. So I want to fix $r\in R$ and use the evaluation map $e_r:R[x]\to R$ given by $f(x)\mapsto f(r)$, a surjective ring homomorphism, and apply the first isomorphism theorem. But I'm having trouble with the kernel of $e_r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, is it even true that $\ker{(e_r)}=(x)$? If so, any hints you could drop would be great. Thanks! &lt;/p&gt;&#xA;" OwnerUserId="2313" LastActivityDate="2010-12-07T06:17:51.950" Title="Understanding the kernel of the evaluation map" Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13337" PostTypeId="2" ParentId="13336" CreationDate="2010-12-07T05:09:39.603" Score="4" ViewCount="0" Body="&lt;p&gt;No, the kernel of $e_r$ consists of all polynomials that are zero at $r$: remember, $\mathrm{ker}(e_r) = \{ f(x)\in R[x]\mid e_r(f(x)) = f(r) = 0\}$. It will contain $(x-r)$ (assuming your ring has an identity), but in general rings it may be all sorts of things (&lt;em&gt;edit&lt;/em&gt;: as Bill Dubuque points out, the Factor Theorem will be on hand in any ring with $1$, but in rings without $1$ weird things may occur). But that does not really matter here, because you don't want to look at an &lt;em&gt;arbitrary&lt;/em&gt; $r$ in $R$. You want to pick a particular $r$ that gives you the kernel you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for you to get $(x)$ as the kernel, you should look at $e_0$, evaluation at $0$. Trivially, $(x)$ is contained in the kernel; showing that every element of the kernel is in $(x)$ should be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T05:42:54.867" LastActivityDate="2010-12-07T05:42:54.867" CommentCount="5" />
  <row Id="13338" PostTypeId="2" ParentId="13334" CreationDate="2010-12-07T05:12:07.993" Score="3" ViewCount="0" Body="&lt;p&gt;I believe the first theorem in article 1.2 &lt;a href=&quot;http://books.google.com/books?id=ZhzXJHUgcRUC&amp;amp;pg=PA28&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-07T17:35:15.013" LastActivityDate="2010-12-07T17:35:15.013" CommentCount="1" />
  <row Id="13340" PostTypeId="2" ParentId="6381" CreationDate="2010-12-07T05:35:16.680" Score="1" ViewCount="0" Body="&lt;p&gt;If you compute homology with twisted coefficients, where the coefficient system involves vector spaces of dimension $d$, then the Euler characteristic of the resulting homology spaces (i.e. the alternating sum of the $H_i$ with twisted coefficients) is equal to $d$ times the Euler characteristic of the space computed via homology with trivial coefficients.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-07T05:35:16.680" />
  <row Id="13341" PostTypeId="1" AcceptedAnswerId="13364" CreationDate="2010-12-07T05:44:43.540" Score="5" ViewCount="131" Body="&lt;p&gt;I have a helix in parametric equations that wraps around the Z axis and a point in space. I want to determine the shortest distance between this helix and the point, how would i go about doing that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using Pythagorean theorem to get the distance and then taking the derivative of the distance function to find the zeros but I can't seem to get an explicit equation for T and I'm stuck at that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I apologize for the tags, not sure how to tag it and I cant create new ones either)&lt;/p&gt;&#xA;" OwnerUserId="3507" LastActivityDate="2010-12-07T13:37:14.433" Title="Shortest distance between a point and a helix" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13342" PostTypeId="2" ParentId="13332" CreationDate="2010-12-07T05:45:05.957" Score="1" ViewCount="0" Body="&lt;p&gt;The group $G$ acts properly discontinuously on $\tilde{X}$, and so if $x$ is any&#xA;point of $\tilde{X}$, it admits a neighbourhood $U$ s.t. that $U g$ is disjoint&#xA;from $U$ if $g \in G$ is non-trivial.   Thus the natural map from&#xA;$U$ to $\tilde{X}/G = X$ is an embedding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the natural map from $U \times V$ to $\tilde{X}\times_G V$ is also an&#xA;embedding, and so $\tilde{X}\times_G V$ is locally constant (i.e. locally a&#xA;product).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detailed remarks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We should equip $V$ with its discrete topology&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The object $\tilde{X}\times_G V$ is not itself actually a sheaf, but is rather the espace etale of a sheaf.  To get the actual sheaf we consider the natural projection $\tilde{X}\times_G V \to \tilde{X}/G = X$, and form the associated sheaf of sections.  Over the open set $U \hookrightarrow X,$ this restricts to the sheaf of sections&#xA;to the projection $U\times V \to U$, whose sections are precisely the constant sheaf on $U$ attached to the vector space $V$.  (Here is where we see that it is important to equip $V$ with the discrete topology.)  Thus our original sheaf&#xA;of sections is locally constant, as claimed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-07T05:45:05.957" CommentCount="2" />
  <row Id="13343" PostTypeId="2" ParentId="13336" CreationDate="2010-12-07T06:05:47.833" Score="2" ViewCount="0" Body="&lt;p&gt;You are on the right track by applying: $\rm\ A/I\ $ is a domain iff $\rm\ I\ $ is prime. For $\rm\ A = R[x],\ \ I = (x)\ $ the problem reduces to showing that $\rm R[x]/(x)\cong R\:.\:$ As you surmise, this is easily achieved by exhibiting an evaluation map with kernel $\rm\ (x)\:,\:$ i.e. an evaluation which maps $\rm\ x\to \ \ldots$?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you already know the existence (and uniqueness) of such evaluation maps for polynomial rings. In fact this &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_property&quot; rel=&quot;nofollow&quot;&gt;universal mapping property&lt;/a&gt; characterizes polynomial rings. It can be employed to give slick conceptual proofs of results about polynomial rings. For a simple example see &lt;a href=&quot;http://math.stackexchange.com/questions/2514/why-cant-the-polynomial-ring-be-a-field/2523#2523&quot;&gt;this proof&lt;/a&gt; that $\rm\:x\:$ is not invertible in $\rm\: R[x]\:.$ The importance of such universal characterizations will become clearer when you study universal algebra and category theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T06:17:51.950" LastActivityDate="2010-12-07T06:17:51.950" />
  <row Id="13344" PostTypeId="1" AcceptedAnswerId="13346" CreationDate="2010-12-07T06:12:05.373" Score="8" ViewCount="255" Body="&lt;p&gt;I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx=\frac{\pi}{2}.$$&#xD;&#xA;What do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this &lt;a href=&quot;http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2/8350#8350&quot;&gt;post&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-07T06:45:17.673" LastActivityDate="2010-12-07T20:21:55.740" Title="Proof for an integral involving sinc function" Tags="&lt;analysis&gt;&lt;integral&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="13345" PostTypeId="1" CreationDate="2010-12-07T06:20:56.720" Score="1" ViewCount="108" Body="&lt;p&gt;Let $\displaystyle X$ be a Polish space. Let $\displaystyle d,d&amp;#39;$ be two complete compatible metrics. Is $\displaystyle 1:(X,d)\to(X,d&amp;#39;)$ uniformly continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really want to know is: can we unambiguously speak of $\displaystyle f:X\to K$ from one Polish space to a compact Polish space as being uniformly continuous? Clearly it won't depend on the compatible metric on $\displaystyle K$, and clearly if the above question is answered in the positive, then the answer will be „yes“.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have done the following. Suppose $\displaystyle 1:(X,d)\to(X,d&amp;#39;)$ were not uniformly cts. Then there would be an $\displaystyle \epsilon\in\mathbb{R}^{+}$ for which $\displaystyle (\forall{\delta\in\mathbb{R}^{+}})(\exists{x\in X})\ \  \mathcal{B}^{d}_{&amp;lt;\delta}(x)\nsubseteq\mathcal{B}^{d&amp;#39;}_{&amp;lt;\epsilon}(x)$.&#xA;For each $\displaystyle \delta\in\mathbb{R}^{+}$, let $\displaystyle S_{\delta}:=\{x\mid \mathcal{B}^{d}_{&amp;lt;\delta}(x)\nsubseteq\mathcal{B}^{d&amp;#39;}_{&amp;lt;\epsilon}(x)\}$. Then for all $\displaystyle \delta&amp;lt;\delta&amp;#39;$ in $\displaystyle \mathbb{R}^{+}$, we would have $\displaystyle S_{\delta}\neq\emptyset$ and $\displaystyle S_{\delta}\subseteq S_{\delta&amp;#39;}$. We also see that $\displaystyle \overline{S_{\delta}}\subseteq\bigcup_{\delta&amp;#39;&amp;gt;\delta}S_{\delta&amp;#39;}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So were $\displaystyle \bigcap_{n\in\omega}\overline{S_{\delta_{n}}}\neq\emptyset$ where $\displaystyle (\delta_{n})_{n\in\omega}\searrow 0$ in $\displaystyle \mathbb{R}$, then it would be so that $\displaystyle \bigcap_{\delta\in\mathbb{R}}S_{\delta}\neq\emptyset$. But then there would be an $\displaystyle x$ and a sequence $\displaystyle (x_{n}\in\mathcal{B}^{d}_{&amp;lt;\delta_{n}}(x)\setminus\mathcal{B}^{d&amp;#39;}_{&amp;lt;\epsilon}(x))_{n\in\omega}$ converging to $\displaystyle x$ in the space $\displaystyle (X,d)$ and not converging to $\displaystyle x$ in $\displaystyle (X,d&amp;#39;)$, which would contradict that $\displaystyle d,d&amp;#39;$ are compatible for $\displaystyle X$. Thus it would be that the countable intersection $\displaystyle \bigcap_{n\in\omega}\overline{S_{\delta_{n}}}=\emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can show, this müsse not be the case, then I would be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However (1) there might be a completely different way to proving this; (2) the question might have a negative answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be great.&lt;/p&gt;&#xA;" OwnerUserId="4493" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T09:06:05.273" LastActivityDate="2010-12-08T15:18:18.507" Title="Is uniform continuity well defined on Polish spaces?" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="13346" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T06:26:47.087" Score="10" ViewCount="0" Body="&lt;p&gt;Let $f(x)=\max\{0,1-|x|\}$. It is easy to calculate the Fourier transform&#xA;$$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-ix\xi}dx=\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2.$$&#xD;&#xA;Taking the inverse Fourier transform, we get &#xA;$$\int_{-\infty}^{\infty}\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2e^{ix\xi}d\xi=2\pi f(x),$$ &#xA;and the result follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second integral can be computed in a similar way. Just take $f(x)=\chi_{[-1,1]}(x)$ (the indicator function of the interval $[-1,1]$).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; It might be interesting to note that there are analogous formulas for the sinc &#xA;sums&#xA;$$\sum_{n=1}^{\infty}\frac{\sin n}{n}=\sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^2=&#xD;&#xA;\frac{\pi}{2}-\frac{1}{2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned about this from the note &lt;em&gt;&quot;Surprising Sinc Sums and Integrals&quot;&lt;/em&gt; by Baillie, Borwein, and Borwein (can be found through a quick web search).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-07T09:03:27.610" LastActivityDate="2010-12-07T09:03:27.610" CommentCount="6" />
  <row Id="13347" PostTypeId="2" ParentId="13345" CreationDate="2010-12-07T06:47:18.793" Score="1" ViewCount="0" Body="&lt;p&gt;Hmm, yes that is. Also $X=[0,\infty)$ with $d(x,y)=|y-x|$ $d&amp;#39;(x,y)=|y^{2}-x^{2}|$ would work too. Thanks for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay then, scrapping that too general a question\ldots can my other concern be addressed, &lt;em&gt;i.e.&lt;/em&gt; can we unambiguously call a map from a Polish space to a compact Polish space uniformly continuous without referring to the metrics?&lt;/p&gt;&#xA;" OwnerUserId="4494" LastActivityDate="2010-12-07T06:47:18.793" CommentCount="1" />
  <row Id="13348" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T06:59:17.263" Score="9" ViewCount="0" Body="&lt;p&gt;Well, it's not hard to reduce this integral to $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$: Just integrate by parts in $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$, integrating $\displaystyle {1 \over x^2}$ and differentiating $\displaystyle \sin^2(x)$. You're left with $\displaystyle \int_0^{\infty} {\sin(2x) \over x}\,dx$ which reduces to the $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$ integral after changing variables from $\displaystyle x$ to $\displaystyle 2x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any elementary proof that $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx = {\pi \over 2}$ is effectively also an elementary proof that $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$ is also $\displaystyle {\pi \over 2}$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-07T07:05:46.230" LastActivityDate="2010-12-07T07:05:46.230" />
  <row Id="13349" PostTypeId="2" ParentId="13055" CreationDate="2010-12-07T07:10:53.883" Score="0" ViewCount="0" Body="&lt;p&gt;My only experience is applying for research postdocs last year, and a lot of schools started looking at applications shortly after Thanksgiving. Some of my friends had offers the first week of December. So I would say apply by mid-November to be safe. This might only apply to research postdocs though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like the blanket advice &quot;apply as early as possible&quot; because for many people, including me, the late summer and early fall of application season is a very productive time, and you might get a lot of research done. If you're hot on the trail of something good it's better to finish it off than apply early, especially if you're a grad student with not so many results. Of course if you're at a natural stopping point in your research (e.g. because you're essentially stuck) then it's better not to procrastinate and get the applications in early. Professors have access to your file on mathjobs as soon as you apply, and maybe someone in the department where you're applying is bored and browsing...&lt;/p&gt;&#xA;" OwnerUserId="4495" LastActivityDate="2010-12-07T07:10:53.883" />
  <row Id="13350" PostTypeId="2" ParentId="13325" CreationDate="2010-12-07T08:00:58.700" Score="0" ViewCount="0" Body="&lt;p&gt;if you have two points, we can state the following let $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ be your two points and $\alpha, \beta, \gamma$, the angles to the X,Y,Z axes respectively; if we want to find the direction (or the angles two the X,Y,Z axes) of the segment formed by the two points $\overline{P_1 P_2}$ these are obtained from these formulas:&#xA;$$\cos{\alpha}=\frac{\Delta x}{d},\cos{\beta}=\frac{\Delta y}{d},\cos{\gamma}=\frac{\Delta z}{d}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where d is the distance between the two points which is given by the following formula:&#xA;$$d=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've found the values of each cosine, you can use their inverse trigonometric function to find each angle, we would have this:&#xA;$$\alpha = \arccos{\frac{\Delta x}{d}}, \beta = \arccos{\frac{\Delta y}{d}}, \gamma = \arccos{\frac{\Delta z}{d}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you are looking for th angles that each point forms with the three axes, let $O(0,0,0)$ be the origin and use it to find the direction of the two segments $\overline{O P_1}$ and $\overline{O P_2}$ as explained above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that this is what your looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards Tristian.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="4312" LastEditorDisplayName="" LastEditDate="2010-12-07T08:21:31.780" LastActivityDate="2010-12-07T08:21:31.780" CommentCount="2" />
  <row Id="13351" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T08:17:45.853" Score="4" ViewCount="0" Body="&lt;p&gt;More generally, there is a result due to Wolstenholme (I can't find a link) that says&#xA;$$ \int_0^\infty \left( \frac {\sin x}{x} \right)^n dx = &#xD;&#xA;\frac{1}{(n-1)!} \frac{\pi}{2^n} \left\lbrace&#xD;&#xA;n^{n-1} - { n \choose 1 } (n-2)^{n-1} + { n \choose 2 } (n-4)^{n-1} - \cdots&#xD;&#xA;\right\rbrace .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-07T08:17:45.853" CommentCount="2" />
  <row Id="13352" PostTypeId="2" ParentId="13091" CreationDate="2010-12-07T08:35:25.010" Score="0" ViewCount="0" Body="&lt;p&gt;To flesh out Robin's answer a little: The thing that you really want to find is the unit normal vector $n=(A,B,C)^t$ to the sought plane. Once you have that, it's easy to find the equation of the plane. This vector has to satisfy three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Length one: $A^2+B^2+C^2=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Orthogonal to the vector $\overrightarrow{P_1 P_2} = (1,0,3)^t$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;60 degrees to the normal vector of the other plane. If we normalize that vector too, this condition becomes $\frac13 (2,-2,1)^t \cdot (A,B,C) = \cos 60^{\circ}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Conditions 2 and 3 give a linear system with two equations and three unknowns $A$, $B$, $C$; solve it and you will get a family of solutions depending on one parameter $t$. Condition 1 then gives a quadratic eqation for determining the two possible values of $t$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-07T08:35:25.010" />
  <row Id="13353" PostTypeId="1" AcceptedAnswerId="13354" CreationDate="2010-12-07T08:42:26.443" Score="1" ViewCount="46" Body="&lt;p&gt;I have some problems understanding a question in a math class test for 10 year olds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Someone pays in 200€ to a bank. The interest rate is 3.75%. Some time later, he looks at the account balance and he sees that there are 235€ now. How much time passed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would calculate this using logarithms, but the interesting part here is that they don't even know square roots, not to mention logarithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any easy way to calculate this without logarithms that I overlooked?&lt;/p&gt;&#xA;" OwnerUserId="4497" LastActivityDate="2010-12-07T09:05:01.600" Title="Interest of interest calculations without logarithm" Tags="&lt;finance&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13354" PostTypeId="2" ParentId="13353" CreationDate="2010-12-07T09:05:01.600" Score="4" ViewCount="0" Body="&lt;p&gt;It depends on whether the interest is simple or compound.  If the interest is simple, you look at the fact the the 35C gain is a gain of 17.5%.  Then, 17.5% divided by 3.75% is 4 and two third years (assuming the interest is 3.75% per year).  If the interest is compound, I see no other way then using logarithms producing an answer of about 4.38 years.  If this is meant to be an easy problem, they are probably assuming simple interest since 4 and two third years is not a very complicated answer.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-07T09:05:01.600" CommentCount="2" />
  <row Id="13355" PostTypeId="2" ParentId="5804" CreationDate="2010-12-07T09:15:17.707" Score="1" ViewCount="0" Body="&lt;p&gt;I think you will find the proof for this in Mallat 1989, 'Multiresolution approximations and wavelet orthonormal bases of L^2'. Theorem 1 (in particular Equations (23), (36)) is what you are after. It is not trivial, longer to prove than I immediately thought. Perhaps there is a very fast proof but I can't think of it now.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastEditorUserId="4322" LastEditorDisplayName="" LastEditDate="2010-12-09T09:58:54.617" LastActivityDate="2010-12-09T09:58:54.617" CommentCount="6" />
  <row Id="13356" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T09:33:04.820" Score="2" ViewCount="0" Body="&lt;p&gt;Apply Parseval-Plancherel to $\chi_{[-1,1]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the Fourier transform as given by &#xA;$$f\mapsto \hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i x\xi}dx$$&#xD;&#xA;then &#xA;$$\int_{-\infty}^\infty |f(x)|^{2}dx=\int_{-\infty}^\infty |\hat{f}(\xi)|^{2}d\xi$$&#xD;&#xA;for $f\in L^2(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $f(x)=\chi_{[-1,1]}(x)$, the characteristic function on $I=[-1,1]$ (that is $f(x)=1$ for $x\in I$ and $f(x)=0$ otherwise), we have&#xA;$$\hat{f}(x)=\int_{-\infty}^\infty f(x)e^{-2\pi i x\xi}dx =\int_{-1}^1 e^{-2\pi i x\xi}dx=\frac{e^{-2\pi i \xi}-e^{2\pi i \xi}}{{-2\pi i \xi}}=\frac{\sin 2\pi\xi}{\pi\xi}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &#xA;$$\int_{-\infty}^\infty \left(\frac{\sin 2\pi\xi}{\pi\xi}\right)^2d\xi=\int_{-1}^1dx = 2$$&#xD;&#xA;by a change of variables, $y=2\pi \xi$,  and using symmetry we arrive at&#xA;$$2=\int_{-\infty}^\infty \left(\frac{2\sin y}{y}\right)^2\frac{dy}{2\pi}=\frac{8}{2\pi}\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy$$&#xD;&#xA;or &#xA;$$\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy=\frac{\pi}{2}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-07T10:57:26.293" LastActivityDate="2010-12-07T10:57:26.293" />
  <row Id="13357" PostTypeId="1" CreationDate="2010-12-07T09:42:09.163" Score="0" ViewCount="96" Body="&lt;p&gt;What does it mean to say that an algebra is complete with respect to a filtration? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T03:10:02.727" LastActivityDate="2010-12-09T03:38:19.087" Title="What do we mean by an algebra is complete with respect to a filtration" Tags="&lt;algebra&gt;" AnswerCount="2" />
  <row Id="13358" PostTypeId="1" AcceptedAnswerId="13361" CreationDate="2010-12-07T09:50:08.880" Score="2" ViewCount="87" Body="&lt;p&gt;The question is how to calculate for the general problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific problem is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a door code of $n$ numbers on a 10 digit panel, how many combinations do you have to try?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The catch is that the mechanism that senses whether the correct code has been entered has no concept of the start and end of the code. So the sequence&#xA;123456 is actually testing 3 unique door codes - my particular case uses a 4 digit code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, given this modification how many combinations are there given a code length of $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to the comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I am looking for the shortest sequence of digits that tests all combinations.&lt;/p&gt;&#xA;" OwnerUserId="4499" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T16:08:38.100" LastActivityDate="2010-12-07T16:10:20.663" Title="Combinations: security code with no defined start and finish" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13359" PostTypeId="2" ParentId="13357" CreationDate="2010-12-07T09:55:49.700" Score="3" ViewCount="0" Body="&lt;p&gt;A filtration induces a topology on the algebra, by providing a basis of open neighbourhoods of 0. The algebra is then complete if the associated topological space is complete. Concretely, if the filtration is called ${\mathfrak{F}}^i$, indexed by natural numbers, where the subalgebras are decreasing as the index increases, say, then you can define a sequence $a_i$ in the algebra to converge if, after possibly replacing it by some translate, it satisfies that for any $i$ there exists an $N\in\mathbb{N}$ such that $a_n\in {\mathfrak{F}}^i$ for all $n\geq N$. You can define Cauchy sequences analogously and the algebra is complete if every Cauchy sequence converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local rings are a prominent example, where the filtration is given by powers of the maximal ideal.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-07T09:55:49.700" />
  <row Id="13360" PostTypeId="2" ParentId="13330" CreationDate="2010-12-07T10:02:44.423" Score="1" ViewCount="0" Body="&lt;p&gt;It's actually much simpler than that. It's not a maths problem though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, brace yourselves, I'm gonna tell you about the little bees and the pretty flowers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously, 1+1=3 is a colloquial expression for saying that when a man and a woman get together, they usually are soon 3 in the family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No math there. More like a riddle, a word game.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-07T10:02:44.423" CommentCount="1" />
  <row Id="13361" PostTypeId="2" ParentId="13358" CreationDate="2010-12-07T10:09:22.997" Score="5" ViewCount="0" Body="&lt;p&gt;Consider a directed graph whose nodes are all possible strings of $n-1$ digits, and with 10 edges labelled 0, ..., 9 going out from each node; the edge $x$ from node $abc\dots de$ leads to $bc\dots dex$, and should be thought of as &quot;testing the code $abc \dots dex$&quot;. The first $n-1$ digits that you enter put you in an initial state, and for each subsequent digit you follow the corresponding edge to another state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, you start in state 123, and move through states 234, 345, 456.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the in- and out-degree of each node is equal (=10), you can find an &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_path&quot; rel=&quot;nofollow&quot;&gt;Eulerian circuit&lt;/a&gt; that passes every edge exactly once; that is, there is a sequence of length $(n-1)+10^n$ which tests all the $10^n$ codes in the shortest way imaginable (each code is tried exactly once).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-07T10:09:22.997" CommentCount="5" />
  <row Id="13362" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T10:12:57.343" Score="3" ViewCount="0" Body="&lt;p&gt;From squaring the identity&#xA;$$\frac{\sin nx}{\sin x}=\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}&#xD;&#xA;=\sum_{k=0}^{n-1}e^{(2k-n+1)ix}$$&#xD;&#xA;and integrating we get&#xA;$$n\pi=\int_{-\pi/2}^{\pi/2}\frac{\sin^2 nx}{\sin^2 x}\,dx.$$&#xD;&#xA;Let&#xA;$$I_n=\int_{-\pi/2}^{\pi/2}\frac{\sin^2 nx}{nx^2}\,dx&#xD;&#xA;=\int_{-n\pi/2}^{n\pi/2}\frac{\sin^2y}{y^2}\,dy.$$&#xD;&#xA;Then&#xA;$$\pi-I_n=\frac{1}{n}&#xD;&#xA;\int_{-\pi/2}^{\pi/2}\sin^2nx(\csc^2x-x^{-2})\,dx.$$&#xD;&#xA;and so&#xA;$$|\pi-I_n|\le\frac{1}{n}\int_{-\pi/2}^{\pi/2}|\csc^2x-x^{-2}|\,dx&#xD;&#xA;=O(1/n)$$&#xD;&#xA;as $x\mapsto\csc^2x-x^{-2}$ extends to a continuous&#xA;function on $[-\pi/2,\pi/2]$.&#xA;Hence $I_n\to\pi$ as $n\to\infty$ and&#xA;$$\pi=\int_\infty^\infty\frac{\sin^2y}{y^2}\,dy.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-07T20:21:55.740" LastActivityDate="2010-12-07T20:21:55.740" CommentCount="0" />
  <row Id="13363" PostTypeId="2" ParentId="13344" CreationDate="2010-12-07T10:28:33.243" Score="3" ViewCount="0" Body="&lt;p&gt;Integrating by parts:&#xA;$$V_n=\int_{n\pi}^{(n+1)\pi} \frac{\sin^2x}{x^2} \mathrm dx = U_{2n} + U_{2n+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA;$$U_n=\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x} \mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&#xA;$$\sum U_n = \sum V_n$$&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2010-12-07T10:28:33.243" />
  <row Id="13364" PostTypeId="2" ParentId="13341" CreationDate="2010-12-07T10:34:22.757" Score="7" ViewCount="0" Body="&lt;p&gt;Let the helix be given by $(\cos t, \sin t, ht)$ (after scaling). If $P$ is your point $(a,b,c)$, and $Q = (\cos t, \sin t, ht)$ is the nearest point on the helix, then $PQ$ is perpendicular to the tangent at $Q$, which is just $(-\sin t, \cos t, h)$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-(\cos t - a)\sin t + (\sin t - b)\cos t + (ht - c)h = 0  $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplifies to $A \sin(t+B) + Ct + D = 0$ for some constants $A,B,C,D,$ as Moron said. But then you have to solve this numerically. There will be more than one solution in general, but (as Jonas Kibelbek pointed out in the comments) you only need to check the solutions with $z$-coordinate in the interval $[c-\pi h, c+\pi h)$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-07T13:37:14.433" LastActivityDate="2010-12-07T13:37:14.433" CommentCount="7" />
  <row Id="13365" PostTypeId="2" ParentId="13319" CreationDate="2010-12-07T10:58:05.480" Score="0" ViewCount="0" Body="&lt;p&gt;Some examples might be in place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;INTEGRAL DOMAINS:&#xA;$\mathbb{Z},\mathbb{Z}/(p) \mathbb{Q}, \mathbb{R}, k[x], \mathbb{Z}[x], \mathbb{Z}[x_1,x_2,\ldots]$&#xD;&#xA;(where the latter denotes a polynomial ring with countably many indeterminates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NON-INTEGRAL DOMAINS: (i.e. every ring in which you can multiply a non-zero number to get zero)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{Z}/(6), k[x]/(x^2)$ ($x \times x = x^2=0$ and $x \neq 0$)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-07T10:58:05.480" />
  <row Id="13366" PostTypeId="2" ParentId="13150" CreationDate="2010-12-07T11:02:02.000" Score="0" ViewCount="0" Body="&lt;p&gt;Is there a way to get the sign of the calculated angle? Right now the calculation gives me an angle of 90° no matter wether I rotate clockwise or counterclockwise.&lt;/p&gt;&#xA;" OwnerUserId="4500" LastActivityDate="2010-12-07T11:02:02.000" />
  <row Id="13367" PostTypeId="1" AcceptedAnswerId="13374" CreationDate="2010-12-07T11:20:27.727" Score="3" ViewCount="57" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;In a school competition,the probability of hitting the target bu Dick is $\frac{1}{2}$,by Betty is $\frac{1}{3}$ and by Joe is $\frac{3}{5}$.If all of them fire independently,calculate the probability that the target will be hit.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This general approach for solving is to &lt;strong&gt;find the complement of the probability that the target would not be hit&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am not getting is that if the there events are mutually exclusive/independent then why not summing the three probabilities doesn't gives the correct answer?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-12-07T13:17:44.490" LastActivityDate="2010-12-07T13:17:44.490" Title="Doubts on Mutually exclusive and Independent events" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13368" PostTypeId="1" AcceptedAnswerId="13369" CreationDate="2010-12-07T11:24:06.050" Score="2" ViewCount="60" Body="&lt;p&gt;Repeating for my exam in commutive algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G be a topological abelian group, i.e. such that the mappings $+:G\times G \to G$ and $-:G\to G$ are continous. Then we have the following Lemma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let H be the intersection of all neighborhoods of $0$ in $G$. Then $H$ is a subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof in the books is the following one-liner: &quot;follows from continuity of the group operations&quot;. (this is from &quot;Introduction to Commutative Algebra&quot; by Atiyah-MacDonald)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must admit that I don't really see how that &quot;follows&quot;. If there is an easy explanation aimed at someone who has not encountered topological groups in any extent, I'd be happy to read it.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-07T16:20:21.050" Title="Intersection of neighborhoods of 0. Subgroup?" Tags="&lt;group-theory&gt;&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13369" PostTypeId="2" ParentId="13368" CreationDate="2010-12-07T11:42:51.490" Score="6" ViewCount="0" Body="&lt;p&gt;If $U$ is a neighbourhood of $0$ then so is $-U=\{-x:x\in U\}$.&#xA;This shows that if $x\in H$ then $-x\in H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that $H$ is closed under addition, use the fact that&#xA;if $U$ is a neighbourhood of $0$ then there is another&#xA;neighbourhood $V$ of $0$ with $V+V\subseteq U$. The existence&#xA;of $V$ follows from the continuity of addition at $(0,0)$.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T16:20:21.050" LastActivityDate="2010-12-07T16:20:21.050" CommentCount="2" />
  <row Id="13370" PostTypeId="1" AcceptedAnswerId="13372" CreationDate="2010-12-07T11:54:27.837" Score="1" ViewCount="98" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read the formula written on the title on a mathematics' book and it doesn't seem correct for me: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the first part of the formula [A ∨ (B∧C)] I have the following possible values:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A; B and C; A and B and C (the Or is not exclusive)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the second part of the formula [(A ∨ B) ∧ (A ∨ C)] I have the following values:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A and A (A); A and C; B and A; B and C; ...; A and B and A and C (A and B and C)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can have for the second part of the formula &lt;em&gt;A and B; A and C&lt;/em&gt; which I can't obtain with the first part of the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm mistaken can somebody please tell me how and give some examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruno&lt;/p&gt;&#xA;" OwnerUserId="4475" LastActivityDate="2010-12-07T12:04:52.897" Title="Is the formula A∨ (B∧C) = (A ∨ B) ∧ (A ∨ C) correct ?" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13371" PostTypeId="2" ParentId="13370" CreationDate="2010-12-07T11:58:41.747" Score="4" ViewCount="0" Body="&lt;p&gt;Draw a Venn diagram and think of $A,B,C$ as sets. (consisting of True,False). Then you'll see that the formula is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could draw a truth table.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Truth_table&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Truth_table&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-07T11:58:41.747" />
  <row Id="13372" PostTypeId="2" ParentId="13370" CreationDate="2010-12-07T12:04:52.897" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose you have $A \vee (B \wedge C)$.  Then you either have $A$ or $B \wedge C$.  If you have $A$, then you have $A \vee B$ and $A \vee C$.  If, on the other hand, you have $B \wedge C$, then you have both $B$ and $C$, in which case you have both $A \vee B$ and $A \vee C$.  So $A \vee (B \wedge C) \Rightarrow (A \vee B) \wedge (A \vee C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going in the other direction, suppose you have $(A \vee B) \wedge (A \vee C)$.  So you have both $A \vee B$ and $A \vee C$.  Either you have $A$, or you don't.  If you have $A$, then you have $A \vee (B \wedge C)$.  If, on the other hand, you don't have $A$, then the fact that you &lt;em&gt;do&lt;/em&gt; have both $A \vee B$ and $A \vee C$ implies that you must have both $B$ and $C$.  Therefore you have $B \wedge C$, and so you have $A \vee (B \wedge C)$, i.e. $(A \vee B) \wedge (A \vee C) \Rightarrow A \vee (B \wedge C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've established that $A \vee (B \wedge C) \Rightarrow (A \vee B) \wedge (A \vee C)$ and $(A \vee B) \wedge (A \vee C) \Rightarrow A \vee (B \wedge C)$.  Therefore $A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)$.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-12-07T12:04:52.897" />
  <row Id="13373" PostTypeId="1" AcceptedAnswerId="13839" CreationDate="2010-12-07T12:21:44.253" Score="17" ViewCount="289" Body="&lt;p&gt;A friend of mine introduced me to the following question:&#xA;Does there exist a smooth function $f: \mathbb{R} \to \mathbb{R}$ ($f \in C^{\infty}$), such that $f$ maps rationals to rationals and irrationals to irrationals and is nonlinear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has been able to prove that such a polynomial (with degree at least 2) doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem has been asked before at least at &lt;a href=&quot;http://www.artofproblemsolving.com&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastEditorUserId="3776" LastEditorDisplayName="" LastEditDate="2010-12-07T13:52:47.810" LastActivityDate="2010-12-10T19:26:51.563" Title="Smooth functions for which $f(x)$ is rational if and only if $x$ is rational" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="6" />
  <row Id="13374" PostTypeId="2" ParentId="13367" CreationDate="2010-12-07T13:01:43.207" Score="6" ViewCount="0" Body="&lt;p&gt;&quot;Mutually exclusive&quot; and &quot;independent&quot; mean different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two events are &lt;strong&gt;mutually exclusive&lt;/strong&gt; if they can't both happen. For example, &quot;my first name is Steve&quot; and &quot;my first name is Fred&quot; are mutually exclusive. When events are mutually exclusive, you are allowed to add their probabilities to get the probability that one of them occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Independent&lt;/strong&gt; events are events where finding out about one doesn't change the probability of the other. Finding out that &quot;it is raining&quot; doesn't tell you anything about &quot;my car is red&quot; so those events are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the question the events &quot;Dick hits the target&quot;, &quot;Betty hits the target&quot; and &quot;Joe hits the target&quot; are independent, but are not mutually exclusive. For instance, Betty and Dick could both hit the target. Since they are not mutually exclusive adding the probabilities will not give the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-12-07T13:01:43.207" CommentCount="1" />
  <row Id="13375" PostTypeId="1" AcceptedAnswerId="13383" CreationDate="2010-12-07T13:57:23.453" Score="0" ViewCount="234" Body="&lt;p&gt;I had many interaction with young girls when I was a private math teacher. Most of them have an assumption that learning mathematics is not necessary for them, because they just want to be a house wife in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some suggestions but I will not say here now (only to avoid interfering your idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the same experience, could you tell me what is your suggestion to them? &lt;/p&gt;&#xA;" OwnerUserId="3901" LastEditorUserId="3901" LastEditorDisplayName="" LastEditDate="2010-12-07T14:07:04.553" LastActivityDate="2010-12-07T15:26:43.423" ClosedDate="2010-12-07T16:56:50.923" Title="How to suggest young girls who think of mathematics as an unnecessary thing if they will be a house wife in the future?" Tags="&lt;soft-question&gt;&lt;teaching&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="13376" PostTypeId="1" AcceptedAnswerId="13390" CreationDate="2010-12-07T14:18:02.027" Score="6" ViewCount="259" Body="&lt;p&gt;(&lt;strong&gt;Disclaimer:&lt;/strong&gt; I'm an engineer)&lt;br&gt;&#xA;Hi everybody, I found this “riddle” posted on the internet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kLBmn.jpg&quot; alt=&quot;Portal Riddle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's meant as a joke, but I do think it deserves an answer :)&lt;br&gt;&#xA;A bit of background: the orange and blue ellipses are a citation from the videogame Portal. They are &quot;portals&quot; connected so that everything that goes into one of them comes out of the other, mantaining his momentum (this last part is the foundation for the game). For further info, you can also &lt;a href=&quot;http://www.youtube.com/watch?v=TluRVBhmf8w&quot; rel=&quot;nofollow&quot;&gt;watch the trailer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So.. is it a circle, a straight line or what else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(also, feel free to retag)&lt;/p&gt;&#xA;" OwnerUserId="894" LastActivityDate="2010-12-11T19:21:08.693" Title="Have I made a straight line, or a circle?" Tags="&lt;geometry&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="13377" PostTypeId="2" ParentId="13375" CreationDate="2010-12-07T14:21:09.420" Score="2" ViewCount="0" Body="&lt;p&gt;Arguably the most important role of a housewife is the raising of children. This is a difficult, often thankless job but no job is more important in the entire world. The more education a mother has, the better prepared she is to help her children experience the world and understand how it works. Mathematics in particular is so very important to quantifying and comparing the relationships that we see all around us, in nature, in populations, and even in the kitchen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I am not saying that girls should feel obligated to be housewives. My response has the assumption, as stated in the questions, that the girls plan on being housewives.&lt;/p&gt;&#xA;" OwnerUserId="4503" LastEditorUserId="4503" LastEditorDisplayName="" LastEditDate="2010-12-07T14:28:54.653" LastActivityDate="2010-12-07T14:28:54.653" CommentCount="2" />
  <row Id="13378" PostTypeId="2" ParentId="13376" CreationDate="2010-12-07T14:25:03.627" Score="5" ViewCount="0" Body="&lt;p&gt;One way to capture the roundness of a circle is to embed it in a plane and measure distance between two points using the ambient plane. A different way to measure distance in a circle is to pretend you are an ant restricted to move within the circle. Then you measure the distance based on how far you walk in the circle. This is a &quot;flat&quot; circle. It is isometric to the circle depicted by the rope in your picture. (An ant walking on that rope couldn't tell any intrinsic difference between the rope and the circle in the plane.)&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-12-07T14:25:03.627" />
  <row Id="13379" PostTypeId="2" ParentId="13375" CreationDate="2010-12-07T14:27:18.540" Score="3" ViewCount="0" Body="&lt;p&gt;I don't believe k-12 mathematics curriculum is optimized from the point of view of helping students understand the nature and usefulness of mathematics when the students are not likely to go on in math and science. I think girls are especially disadvantaged by current curriculum because there are societal factors that already seem to discourage girls from pursing mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very short look at my views here can be found in this paper about public perceptions of mathematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.education.umd.edu/MathEd/conference/vbook/public-perceptions_Malkevitch.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-07T14:27:18.540" />
  <row Id="13380" PostTypeId="1" CreationDate="2010-12-07T14:36:37.667" Score="0" ViewCount="63" Body="&lt;p&gt;There are several version, what is the most common used one? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2010-12-07T14:36:37.667" Title="What's the definition of homology with local coefficient" Tags="&lt;algebraic-topology&gt;" CommentCount="2" />
  <row Id="13381" PostTypeId="2" ParentId="13013" CreationDate="2010-12-07T14:57:15.787" Score="1" ViewCount="0" Body="&lt;p&gt;I don't understand why this question is asked in the language of generating functions.  The easy way to solve the problem is to take any subset $\{a,b,c,d,e\}$ of $\{1,\dots,n-4\}$, and choose the subset $\{a,b+1,c+2,d+3,e+4\}$.  In particular, there are $\binom{n-4}{5}$  ways to do this.  We can recover the generating function using basic generating function knowledge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n\geq 0} \binom{n-4}{5}x^n =\sum_{n\geq 0} \binom{n-9+5}{5}x^n=x^9\sum_{n\geq 0} \binom{n-9+5}{5}x^{n-9}$$&#xD;&#xA;Changing indices and recognizing the form of the generating function, we have &#xA;$$=x^9\sum_{N\geq -9} \binom{N+5}{5}x^N=\frac{x^9}{(1-x)^6}.$$&lt;/p&gt;&#xA;" OwnerUserId="4504" LastActivityDate="2010-12-07T14:57:15.787" />
  <row Id="13382" PostTypeId="2" ParentId="13345" CreationDate="2010-12-07T15:24:16.563" Score="0" ViewCount="0" Body="&lt;p&gt;Yeah i worked it out on the train. It's not the case. Consider $X=[0,\infty)$ and $d:(x,y)\mapsto|y-x|$, $d&amp;#39;:(x,y)\mapsto|\sqrt{y}-\sqrt{x}|$ and consider the map $f:X\to[0,1]$ a periodic continuous function like $sin$. Then $f:(X,d)\to[0,1]$ is uniformly cts, but $f:(X,d&amp;#39;)\to[0,1]$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I'm looking for this: let $X$ be a polish space, then does it embed continuously to a unique &lt;em&gt;polish&lt;/em&gt; compactification, universal in the sense that every continuous map from $X$ to another compact polish space lifts uniquely to a map from the compactification. Doesn't seem likely. I can't see that the Stone-Cech compactification is polish (it's at least separable compact hausdorff, but that actually does not imply second countable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can do, is take $(X,d)$ where $d$ is a compatible complete metric, and embed this via a uniformly continuous map to a compact polish space, universal in the sense that every uniformly cts map from $(X,d)$ to another compact polish space lifts uniquely to a map from the compactification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that compactification is might depend on $d$. Though it would be nice if it didn't, so that we could talk about „the“ compactification of a polish space.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2010-12-07T15:24:16.563" />
  <row Id="13383" PostTypeId="2" ParentId="13375" CreationDate="2010-12-07T15:26:43.423" Score="9" ViewCount="0" Body="&lt;p&gt;The most important thing you need to do for these girls, more important than teaching them any particular mathematical topic, is to inspire them to have a broader vision of what they might attain in their life. Sure, being a housewife is possible, but also they might aim to be a scientist, doctor, philosopher or engineer; they could be an astronaut or a musical performer or a computer programmer. Why should a young girl decide at a young age on such a narrow and impoverishing plan for their life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it shocking that any young girl would assert such a plan for their life (to say nothing of &quot;most of them&quot; as you mentioned). These girls need urgent direction.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-12-07T15:26:43.423" CommunityOwnedDate="2010-12-07T15:26:43.423" CommentCount="5" />
  <row Id="13384" PostTypeId="1" CreationDate="2010-12-07T15:35:59.840" Score="2" ViewCount="38" Body="&lt;p&gt;Let $(X,d),(Y,d)$ be polish spaces and $Y$ compact. Consider the spaces $C_{u}(X,Y)$ of uniformly continuous functions, and $C(X,Y)$ of continuous functions. We can endow these with the uniform-metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can $C(X,Y)$ be generated from $C_{u}(X,Y)$ using only point-wise limits of $\omega$-sequences? &lt;em&gt;I.e.&lt;/em&gt; Define $\mathcal{B}_{0}:=C_{u}(X,Y)$, and $\mathcal{B}_{\alpha}:=$ the set of point-wise limits of a sequence $(f_{n})_{n\in\omega}$ of functions $f_{n}\in\bigcup_{\xi&amp;lt;\alpha}\mathcal{B}_{\xi}$. Then this question asks whether $C(X,Y)\subseteq\mathcal{B}_{\omega_{1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if it's true, then we get that all borel functions from $X$ to $Y$ can be generated from the uniformly continuous functions using only point-wise limits.&lt;/p&gt;&#xA;" OwnerUserId="4508" LastActivityDate="2010-12-07T15:35:59.840" Title="generating continuous functions from uniformly continuous functions" Tags="&lt;real-analysis&gt;&lt;general-topology&gt;" />
  <row Id="13385" PostTypeId="2" ParentId="13245" CreationDate="2010-12-07T15:38:00.553" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I think my previous answer can be adapted to the case we are interested in.&#xA;So let $E_n=\{ f \geq n\}$.&#xA;Let $m$ be a positive integer.&#xA;If $s = \sup_{F \subset [-m,m] \setminus E_n} \lambda(F)$ (the sets $F$ are taken measurable here), then there is a measurable set $F \subset [-m,m] \setminus E_n$ such that $\lambda(F)=s$ (take a sequence and a union).&#xA;We are going to prove that for any real M, for almost all $x \in F$, $f(x)&amp;lt;-M$, that is, there is a measurable $A(M) \subset F$ of zero measure such that $f(x)&amp;lt;-M$ for all $x \in F \setminus A(M)$.&#xA;For this, it is enough to find, for every positive integer $k$, $A_k(M) \subset F$ such that $f(x)&amp;lt;-M$ for all $x \in F \setminus A_k(M)$ and $\lambda(A_k(M)) \leq 1/k$.&#xA;Since $f$ is not continous and $\mathbb{Q}$-linear, there exists $a_k \in \mathbb{R}$ such that $|a_k| \lt 1/(2k)$ and $f(a_k)&amp;lt;-n-M$.&#xA;Then $a_k+F \subset [-m-1/(2k), m + 1/(2k)] \setminus E_n$, so $\lambda((a_k+F) \cap [-m,m] \setminus F) = 0$ and consequently, $\lambda(F \setminus (a_k+F)) \leq 1/k$, so we can set $A_k(M) = F \setminus (a_k+F)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $f(x)&amp;lt;-M$ for all $x \in F \setminus A(M)$, $F = \cup_{M \in \mathbb{N}} A(M)$ has measure $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that $E_n$ is $\lambda$-thick in $[-m,m]$ for all $m$, so $E_n$ is $\lambda$-thick in $\mathbb{R}$, and this is stronger than $E_n$ being $\lambda$-thick in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old answer&lt;/strong&gt;:&#xA;This is not exactly what you want, but here is a similar proof of your lemma.&#xA;Define $E_n = \{ |f| \geq n \}$. Then $E_n$ is $\lambda$-thick (in $\mathbb{R}$).&#xA;We need to show that if $F$ is measurable and does not intersect $E_n$, then $\lambda(F)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume otherwise, and let $m$ be big enough so that $F \cap [-m,m]$ has positive measure.&#xA;Since $f$ is not continuous and $\mathbb{Q}$-linear, $f_{|[0,1]}$ is not bounded, so there is a sequence $(a_k)_{k \geq 0}$ of elements of $[0,1]$ such that $|f(a_{k+1})| &amp;gt; |f(a_k)|+2n$.&#xA;Then $\cup_{k \geq 0} \left( a_k + (F \cap [-m,m]) \right)$ is included in $[-m,m+1]$ and its measure is equal to $\sum_{k \geq 0} \lambda (F \cap [-m,m]) = + \infty$, this is because the $a_k + F$ are disjoint (look at the images by $f$), so we get a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $E_n$ is $\lambda$-thick, and $E_n \cap [0,1]$ is $\lambda$-thick in $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there is a way to deduce your statement from this one.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastEditorUserId="2660" LastEditorDisplayName="" LastEditDate="2010-12-10T15:17:26.020" LastActivityDate="2010-12-10T15:17:26.020" CommentCount="1" />
  <row Id="13386" PostTypeId="1" AcceptedAnswerId="13391" CreationDate="2010-12-07T15:48:55.947" Score="7" ViewCount="216" Body="&lt;p&gt;I read that any continuous function can be represented as a sum of convex and concave function, meaning for all $f(x)$, $f(x) = g(x) + h(x)$ where $g$ is convex and $h$ is concave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be infinitely many decompositions of that sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone knows where I can see a proof or this, or knows of a proof of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-07T16:38:22.260" LastActivityDate="2010-12-07T17:32:27.657" Title="can any continuous function be represented as a sum of convex and concave function?" Tags="&lt;functions&gt;&lt;convex-analysis&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="1" />
  <row Id="13387" PostTypeId="1" AcceptedAnswerId="13444" CreationDate="2010-12-07T16:01:31.837" Score="3" ViewCount="135" Body="&lt;p&gt;What is the best program to factor large arbitrary-form integers on a single computer, or on a few disjointed computers? &quot;Best&quot; is obviously subjective, but what do you recommend?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on a project to factor general-form numbers that I know are composite, with a scale of 100 - 1000 digits. I have a few computers I can use to process in parallel, but they're nowhere close to a cluster - and they certainly don't have the horsepower / memory needed for something like GNFS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked around a little, but what I've found can be classified as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prove primality of general-form numbers &lt;em&gt;(mine = known composites)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Factor special-form numbers &lt;em&gt;(mine = general-form numbers)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Factor general-form numbers on a huge cluster &lt;em&gt;(no cluster)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm missing the last one, &quot;Factor general-form numbers on one or two sneakernet systems&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final caveats: I'd prefer something free, but I'd be willing to spring for Maple / Mathematica / whatnot if it's my best option. Also I'd prefer something that's already built (binary package), but I can compile from source if that's a better option.&lt;/p&gt;&#xA;" OwnerUserId="4507" LastActivityDate="2010-12-10T18:36:05.197" Title="Factoring large integers without a cluster" Tags="&lt;soft-question&gt;&lt;prime-numbers&gt;&lt;math-software&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13388" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:07:23.797" Score="-2" ViewCount="0" Body="&lt;p&gt;what would this look like for sin(x)&lt;/p&gt;&#xA;" OwnerUserId="4370" LastActivityDate="2010-12-07T16:07:23.797" CommentCount="2" />
  <row Id="13390" PostTypeId="2" ParentId="13376" CreationDate="2010-12-07T16:12:27.923" Score="13" ViewCount="0" Body="&lt;p&gt;Though I (slightly) disagree with Nate's characterization of geometers in his comments, I think there is actually some deep notions worth discussing contained in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heart of the matter is, &lt;strong&gt;how do you define a circle&lt;/strong&gt;, and &lt;strong&gt;how do you define a line&lt;/strong&gt;? And in the apparent paradox of the original question, the resolution is that there are (at least) two different ways of distinguishing circles versus lines, which, when taken together, allows for an object that is simultaneously a line in one definition but a circle in the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;strong&gt;topological&lt;/strong&gt; category, the circle and the line are the two examples of connected smooth one-dimensional manifolds. Since we are dealing with topology, everything is allowed to be stretchy/flabby, so the only difference we really care about between the circle and the line is that the latter is simply connected, while the former is not. More precisely, you can (trivially) find a closed loop in the circle which can not be continuously deformed to a point, while any closed loop on the line must &quot;backtrack&quot; sufficiently that it can be continuously shrunk down to a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second description of the circle and the line comes from the &lt;strong&gt;geometrical&lt;/strong&gt; category. (Here I'll just discuss the distinction between lines and non-lines.) A line in geometry is, intuitively, the straightest possible curve, which we take to mean &quot;a curve that locally minimizes the distance between two points&quot;. It is an interesting special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan-Hadamard_theorem&quot; rel=&quot;nofollow&quot;&gt;Cartan-Hadamard theorem&lt;/a&gt; that, in a simply connected manifold of non-positve curvature (in particular the usual flat Euclidean space), rays (geodesics) emanating from the same point in two different directions will diverge forever and never intersect. (In fact, the divergence of geodesic rays is a characterization of non-positive curvature; also compare this to the case of positive curvature on a sphere, where any two great circles intersect at exactly two points.) So since the world in which we live our usual, everyday lives is more-or-less flat, our common intuition is that a line must be non-self-intersecting. Of course, if you consider traveling on the oblate spheroid that is the earth, you may come to a somewhat different conclusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation of &lt;em&gt;Portal&lt;/em&gt; is precisely at the level of breaking the hypothesis of Cartan-Hadamard theorem in topology. As stated above, the Cartan-Hadamard theorem requires the space to be simply connected. By allowing the Portal, which is an identification two distinct subsets of usual Euclidean space, you pick up a non-trivial topology. And therefore Cartan-Hadamard theorem can fail. Hence your rope is allowed to be simultaneously a &lt;strong&gt;line&lt;/strong&gt; in the sense that it is a straightest possible curve and a &lt;strong&gt;circle&lt;/strong&gt; in the sense that it is a curve that returns to its starting point, and intrinsically has non-trivial topology. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-07T16:12:27.923" CommentCount="2" />
  <row Id="13391" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:22:50.100" Score="9" ViewCount="0" Body="&lt;p&gt;If $f$ is $C^2$ (meaning the derivatives $f&amp;#39;$ and $f&amp;#39;&amp;#39;$ are defined and continuous), and if $f$ is defined on a closed interval like $[0,1]$, then there is an easy solution. Since $f&amp;#39;&amp;#39;$ is continuous, it has a minimal value $m$. Choose $c$ positive and greater than $m$. Then $f(x)+c x^2/2$ is convex, by the double derivative test, and $-c x^2$ is concave. Of course, $[f(x)+c x^2/2] + [-cx^2/2] = f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shouldn't be too much harder to do a general continuous function, but I don't see it right now.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-07T16:22:50.100" CommentCount="2" />
  <row Id="13392" PostTypeId="1" AcceptedAnswerId="13400" CreationDate="2010-12-07T16:48:26.870" Score="2" ViewCount="107" Body="&lt;p&gt;Lets say I take a computation that involves only addition and multiplication:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(a+b)*(c+d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which can be done in many other ways, eg.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a*(c+d) + b*(c+d)&#xA;a*c + a*d + b*c + b*d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In terms of additions and multiplications the number of operations required for each of the three examples shown are (2,1) (3,2) (3,4) respectively. Clearly, if the goal is to reduce the total number of operations the first is superior. Is there a way, given an arbitrary expression to find the computation order that requires the least number of operations? &lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-12-07T18:10:40.200" Title="Reduce the number of operations on a simple expression" Tags="&lt;computer-science&gt;&lt;symbolic-computation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13393" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:50:09.327" Score="2" ViewCount="0" Body="&lt;p&gt;Counterexample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x)=x^2sin(\frac{1}{x^2})$      (with f(0)=0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is a continuous function, whose gradient is unbounded as x goes to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g'(x) and h'(x) are monotonic, and hence bounded on [-1,1], so the gradient of g(x)+h(x) is bounded as x goes to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Looking at it again, this doesn't look fantastically rigorous, but I think it holds up.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-12-07T16:50:09.327" />
  <row Id="13394" PostTypeId="2" ParentId="5226" CreationDate="2010-12-07T16:53:38.637" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/48411/are-there-any-abelian-2-groups-with-complete-holomorphs-other-than-c2-2-and-c&quot; rel=&quot;nofollow&quot;&gt;Derek Holt helped me&lt;/a&gt; figure this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose V is not just normal, but characteristic in A.  Then Out(A) has a normal series with factors N, S, H with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;N ≤ Out(G) consisting of those automorphisms that take V to an isomorphic G-module&lt;/li&gt;&#xA;&lt;li&gt;S a quotient of the group Aut&lt;sub&gt;G&lt;/sub&gt;(V) of G-module automorphisms of V by the normal subgroup of automorphisms induced by the center of G&lt;/li&gt;&#xA;&lt;li&gt;H = H&lt;sup&gt;1&lt;/sup&gt;(G,V), the first cohomology group&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For A to be complete, Out(A) = 1, so N=S=H=1.  In particular,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No (non-identity) outer automorphism can take V to an isomorphic G-module&lt;/li&gt;&#xA;&lt;li&gt;V has to be multiplicity-free and split, and the center of G needs to include all the (block) scalar matrices&lt;/li&gt;&#xA;&lt;li&gt;The first cohomology group has to vanish&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming then that V is irreducible and characteristic, then the second condition is just that V is absolutely irreducible and G contains the scalar matrices Z(GL(V)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If V is not characteristic (say if G is very small and unipotent) then this analysis fails, but I think if G starts out close to being complete, V is likely to be characteristic.  Certainly if V is irreducible.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-07T16:53:38.637" CommentCount="1" />
  <row Id="13395" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T16:55:38.293" Score="16" ViewCount="0" Body="&lt;p&gt;You need more conditions than even absolute continuity. A characterisation of continuous, convex functions defined on open intervals is that they must be the indefinite integral of a monotonically non-decreasing function. (See &lt;a href=&quot;http://eom.springer.de/c/c026240.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for example.) In particular, this means that continuous, convex functions are absolutely continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you take any function that is continuous and not absolutely continuous (Cantor's stair case comes to mind), it cannot be decomposed as a sum of a convex and a concave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more important is that a convex function, by &lt;a href=&quot;http://www.math.sc.edu/~howard/Notes/alex.pdf&quot; rel=&quot;nofollow&quot;&gt;Alexandrov's theorem&lt;/a&gt;, must have a second derivative almost everywhere. Therefore your initial function must be even better than just absolutely continuous, it needs to also admit almost everywhere second derivatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even assuming almost everywhere second derivatives is not enough, if you take the function $x \sin(1/x)$  which is analytic away from the origin, you cannot represent it as the difference of two convex functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is still a gap between twice almost everywhere differentiable and David Speyer's everywhere twice-continuously differentiable condition. I am not entirely sure where the correct boundary lies, or if there is a correct boundary using just classical differentiability notions. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-07T16:55:38.293" CommentCount="3" />
  <row Id="13396" PostTypeId="1" AcceptedAnswerId="13465" CreationDate="2010-12-07T17:08:09.030" Score="1" ViewCount="134" Body="&lt;p&gt;So I have this problem to solve...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let X denote the number of paint defects found in a square yard section of a car body painted by a robot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These data are obtained:&#xA;8, 5, 0, 10, 0, 3, 1, 12, 2, 7, 9, 6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Assume that X has a Poisson distribution with parameter λs.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Find an unbiased estimate for λs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Find an unbiased estimate for the average number of flaws per square yard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Find an unbiased estimate for the average number of flaws per square foot.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have no idea where to begin. I mean, how do I even find ANY unbiased estimate? The textbook is worthless imo and I can't find any good readings on the web either... please help.&lt;/p&gt;&#xA;" OwnerUserId="713" LastActivityDate="2010-12-08T01:14:45.323" Title="Find an unbiased estimate for λs (Poisson distribution)" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="13397" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T17:17:53.003" Score="2" ViewCount="0" Body="&lt;p&gt;A $C^1$ function $f$ on a compact interval is convex (or is it concave?)&#xA;if and only if $f&amp;#39;$ is increasing and is concave (or is it convex?)&#xA;if and only if $f&amp;#39;$ is decreasing. If a function on a compact interval&#xA;is a sum of a convex and a concave function, each of which is $C^1$, then&#xA;$f&amp;#39;$ is the sum of an increasing and a decreasing function and so&#xA;has bounded variation. There are continuous functions which are not&#xA;of bounded variation on a compact interval. Integrating one gives a&#xA;$C^1$ function $f$ which is not the sum of a $C^1$ convex and&#xA;a $C^1$ concave function. Perhaps it could be the sum of some&#xA;more general convex and concave functions, but I wouldn't bet on it :-)&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-07T17:17:53.003" CommentCount="3" />
  <row Id="13398" PostTypeId="2" ParentId="13386" CreationDate="2010-12-07T17:32:27.657" Score="0" ViewCount="0" Body="&lt;p&gt;Assume $f$ to be differentiable. We need to write $f&amp;#39;$ as the sum of nondecreasing and a nonincreasing functions $g$ and $h$. This is possible if and only if $f&amp;#39;$ has bounded variation: for every $a&amp;lt;b$, there exists $M$ such that for any integer $n$ and any $a=x_1 \lt \ldots \lt x_{n+1} = b$, $\sum_{i=1}^n |f(x_{i+1})-f(x_i)| \leq M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we define, for $x \geq 0$, $g(x)=\mathrm{sup}_{0=x_1 \lt \ldots \lt x_{n+1}=x} \sum_{i=1}^n \left(f(x_{i+1})-f(x_i) \right)^+$ where $y^+ = y$ if $y \geq 0$ and $0$ otherwise. Define $g$ similarly on $]-\infty,0]$, and $h=f-g$. It is easy to check that $g$ is nondecreasing, $h$ is nonincreasing, and their primitives give the result.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2010-12-07T17:32:27.657" />
  <row Id="13399" PostTypeId="2" ParentId="13268" CreationDate="2010-12-07T18:03:42.073" Score="3" ViewCount="0" Body="&lt;p&gt;I am slightly unsure about the following approach, so you should double check if that is correct. I'll use $\|\cdot\|_2$ to denote the spectral norm as above, and $\|\cdot \|_F$ for the Frobenius norm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that for $n\times n$ matrices,  $\|\cdot \|_F \leq \sqrt{n} \|\cdot\|_2$ is sharp from just the definition of the norms. Now, I claim that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\beta_k - \alpha_k| \leq \|B - A\|_F $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is also sharp. Fix the eigenvalues $\beta_k$ and $\alpha_k$. Consider orthogonal transformations of the symmetric matrix $B$ by $Q$. It suffices to compute the minimum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \inf_{Q\in O(n)} \| QBQ^T - A\|_F $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But using the definition of the Frobenius norm as an inner product, you see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \| QBQ^T - A\|_F^2 = \mathop{Tr}\left[ (QBQ^T - A)(QBQ^T - A) \right] = \|B\|_F^2 + \|A\|_F^2 - 2 A\cdot_F (QBQ^T)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an exercise to see that $A\cdot_F (QBQ^T)$ is maximized when all the $k$th eigenspace of $A$ lines up with that of $QBQ^T$ (some sort of Cauchy-Schwarz plus re-arrangement inequality). When the eigen-spaces line up, you have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \| QBQ^T - A\|_F^2 = \sum_{k = 1}^n |\beta_k - \alpha_k|^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the desired inequality holds, and is attained with the eigenvalues only differ in one position. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the composition of these two inequalities is not automatically sharp, since the bound of $|\beta_k-\alpha_k|$ by $\|B-A\|_F$ is only sharp with $B-A$ has rank 1, while the first bound can be sharpened $\|C\|_F \leq \sqrt{\mathop{rank}(C)} \|C\|_2$. But it is clear that the most general sharp bound should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\beta_k-\alpha_k| \leq s\|B-A\|_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $1 \leq s \leq \sqrt{n}$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-07T18:32:12.300" LastActivityDate="2010-12-07T18:32:12.300" CommentCount="4" />
  <row Id="13400" PostTypeId="2" ParentId="13392" CreationDate="2010-12-07T18:10:40.200" Score="4" ViewCount="0" Body="&lt;p&gt;Here's a closely related question which is easier to answer.  Suppose you only care about the number of multiplications.  (This is reasonable in many situations where addition is cheap but multiplication is expensive, so you want to minimize the number of multiplications you're going to do.)  So suppose someone gives you an expression like a*c + a*d + b*c + b*d and you want to rearrange using as few multiplications as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case what you're doing is computing the &quot;rank&quot; of a tensor.  (Warning: &quot;rank&quot; of a tensor can mean two totally different things, this is the rank that's analogous to rank of a matrix, not the rank that tells you which kind of tensor you're looking at.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose your expression only ever involves products of two things, then this becomes computing the rank of a matrix.  In your example, a*c + a*d + b*c + b*d, you're looking at a 4x4 matrix (since there are 4 variables: a,b,c,d) which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}0&amp;amp;0&amp;amp;1&amp;amp;1\\ 0&amp;amp;0&amp;amp;1&amp;amp;1 \\ 0&amp;amp;0&amp;amp;0&amp;amp;0 \\ 0&amp;amp;0&amp;amp;0&amp;amp;0 \end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matrix has rank 1 as is easily seen using row-reduction, hence it can be written using only one multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether there are good algorithms for computing ranks of higher tensors the way there are for matrices.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-12-07T18:10:40.200" CommentCount="3" />
  <row Id="13401" PostTypeId="2" ParentId="13345" CreationDate="2010-12-07T18:19:14.177" Score="2" ViewCount="0" Body="&lt;p&gt;Dear Raj,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an answer to the question in one of your answers: is the Stone-Cech Compactification of a Polish space necessarily Polish?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is a resounding no: not even close, in some sense.  Let $X$ be a countably &#xA;infinite set equipped with the discrete metric (and thus the discrete topology).  Its Stone-Cech compactification is the space of ultrafilters on $X$, which has cardinality &#xA;$2^{2^{|X|}}$ (&lt;b&gt;added&lt;/b&gt;: e.g., &lt;a href=&quot;http://planetmath.org/encyclopedia/NumberOfUltrafilters.html&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;). Evidently this space is separable: $X$ itself is a countable dense subspace.  Now, if you consult &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Separable_space#Cardinality&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Separable_space#Cardinality&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will see that a first countable separable space can have at most continuum cardinality.  (Disclosure: I wrote this part of the article.)  So the Stone-Cech &#xA;compactification is not first countable, so is not even close to being metrizable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert on Polish spaces, but I think you just do not have the right structure here to define a reasonable notion of uniform continuity.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-08T15:18:18.507" LastActivityDate="2010-12-08T15:18:18.507" />
  <row Id="13402" PostTypeId="1" CreationDate="2010-12-07T18:23:32.777" Score="1" ViewCount="113" Body="&lt;p&gt;Alright, so I got two points in 3d space, so they have a x,y, and z.&#xA;Now if the line's y - which I get like so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vector3 v = new Vector3();&#xA;v = a.subtract(b, v);&#xA;v.normalizeLocal();&#xA;&#xA;float cosine = (float) v.dot(v);&#xA;float angle = (float) Math.toDegrees(Math.acos( cosine ));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I have a third point (the c point), which is inside the line of a and b.&#xA;now I need to know how to get point a and b to 0 and 1, than I need to find out where c is on the line. my goal is to have something like c.y = b.y - (a.y * c.x and c.z's point on the line (which will be between 0 and 1))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do I do this?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2011-01-07T02:44:39.043" Title="Getting the % that a point is on a line" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13403" PostTypeId="1" CreationDate="2010-12-07T18:39:28.703" Score="2" ViewCount="124" Body="&lt;p&gt;Please help me, I can't find the theorem anywhere that states something like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a bounded set $U\subset\mathbb R$ there exists a non-descreasing sequence $(a_n)_{n\in \mathbb N}$ with $a_n \in U$ with $\lim_{n\to\infty}a_n=\sup U$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4514" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T04:46:03.943" LastActivityDate="2010-12-08T04:46:03.943" Title="Sequence converging to the supremum" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="13404" PostTypeId="1" CreationDate="2010-12-07T18:41:28.003" Score="1" ViewCount="99" Body="&lt;p&gt;I have a camera looking at a computer monitor from varying angles. Since the camera is a grid of pixels, I can define the bounds of the monitor in the camera image as:&#xA;&lt;img src=&quot;http://i.imgur.com/MGwUh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that makes sense. What I want to do is come up with an algorithm to translate points within this shape to this:&#xA;&lt;img src=&quot;http://i.imgur.com/niJYY.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have points within the same domain as ABCD, as determined from the camera, but I need to draw these points in the domain of the monitor's resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that makes sense? Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4515" LastActivityDate="2010-12-10T04:15:35.923" Title="Mapping Irregular Quadrilateral to a Rectangle" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13405" PostTypeId="2" ParentId="13403" CreationDate="2010-12-07T18:48:03.087" Score="5" ViewCount="0" Body="&lt;p&gt;You can prove the theorem (under the assumption that $\sup U$ exists) directly from the definition of supremum. For each $n$ there is some point in $U$ within $1/n$ of the supremum. Use the axiom of choice to choose a sequence $(a_n)$ so that for each $n$, $a_n$ is within $1/n$ of the supremum. Then prove that this sequence converges to the supremum. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-07T18:48:03.087" CommentCount="3" />
  <row Id="13406" PostTypeId="2" ParentId="13403" CreationDate="2010-12-07T18:49:10.693" Score="4" ViewCount="0" Body="&lt;p&gt;Because it follows directly from the definition of supremum. You can create the sequence $\{a_n\}$ easily:&#xA;Let $M = \sup U $.&#xA;Assume that $M \notin U$ (otherwise just let $a_i = M \, \forall i \in \mathbb{N}$).&#xA;Fix $a_0 \in U$.&#xA;Choose $a_i \in (\frac{M + a_{i-1}}{2}, M) \subset U$ for $i = 1, 2, \dots$&#xD;&#xA;Such $a_i$ must exist since $M$ is the supremum. Otherwise $\frac{M + a_{i-1}}{2} &amp;lt; M$ would be a lower upper bound for $U$.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-07T18:55:39.300" LastActivityDate="2010-12-07T18:55:39.300" CommentCount="1" />
  <row Id="13407" PostTypeId="2" ParentId="8946" CreationDate="2010-12-07T18:50:51.227" Score="1" ViewCount="0" Body="&lt;p&gt;Try this approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the partial derivative with respect to $\xi$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{\xi}\phi(t,\xi)=i\mathbb{E}\left[X_t e^{i\xi X_t}\right] \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the partial derivative with respect to $t$, I'll take the differential but only varying t, so as to connect with the SDE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_t\phi(t,\xi)=\mathbb{E}\left[e^{i\xi (X_t+dX_t)} - e^{i\xi X_t}\right]=\mathbb{E}\left[(i\xi dX_t-\frac{\xi^2}{2}dX_t^2)e^{i\xi X_t}\right] \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dX_t = (X_t - \mu) dt + \sigma \sqrt{X_t}dW_t $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dX_t^2 = \sigma^2 X_t dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[dX_t e^{i\xi X_t}\right]-\frac{\xi^2}{2}\mathbb{E}\left[dX_t^2 e^{i\xi X_t}\right] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_t\phi(t,\xi)=i\xi \mathbb{E}\left[(X_t-\mu) e^{i\xi X_t}\right]dt+i\xi \mathbb{E}\left[\sqrt{X_t} dW_t e^{i\xi X_t}\right]-\frac{\sigma^2 \xi^2}{2}\mathbb{E}\left[X_t e^{i\xi X_t}\right]dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the middle term contains $dW_t$ as a consequence, taking its expectation gives zero. You can now recognize the other terms as containing $\phi$ or $\partial_{\xi}\phi$, so you've got yourself an ordinary PDE for $\phi$. That should be the aim of the computation.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-07T21:10:49.700" LastActivityDate="2010-12-07T21:10:49.700" CommentCount="12" />
  <row Id="13408" PostTypeId="2" ParentId="13215" CreationDate="2010-12-07T19:19:27.720" Score="0" ViewCount="0" Body="&lt;p&gt;@Jesse: Thanks a lot! I will read carefully this.&lt;/p&gt;&#xA;" OwnerUserId="4482" LastActivityDate="2010-12-07T19:19:27.720" />
  <row Id="13409" PostTypeId="2" ParentId="13404" CreationDate="2010-12-07T19:33:07.270" Score="3" ViewCount="0" Body="&lt;p&gt;In general there is no affine transformation that maps an arbitrary quadrangle onto a rectangle. But there is (exactly one) projective transformation $T$ that maps a given quadrangle $(A, B, C, D)$ in the projective plane onto a given quadrangle $(A&amp;#39;, B&amp;#39;, C&amp;#39; D&amp;#39;)$ in the same or another projective plane. This $T$ is ${\it collinear}$, i.e., it maps lines to lines. To do the calculations you have to introduce homogeneous coordinates $(x,y,z)$ such that $D=(0,0,1)$, $C=(1,0,1)$, $A=(0,1,1)$, $B=(1,1,1)$ and similarly for $A&amp;#39;$, $B&amp;#39;$, $C&amp;#39;$, $D&amp;#39;$. With respect to these coordinates the map $T$ is linear and its matrix is the identity matrix.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-07T19:33:07.270" CommentCount="2" />
  <row Id="13410" PostTypeId="1" CreationDate="2010-12-07T19:48:45.853" Score="0" ViewCount="42" Body="&lt;p&gt;Hi!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to show the following estimation and am not sure, if it holds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $U\subseteq\mathbb R$ and $f:\mathbb R\times \mathbb R\to\mathbb R$. Its derivative with respect to the second argument exists and is denoted by $f_b(a,b)$:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\displaystyle\frac{d}{db}\left(\sup_{a\in U}f(a,b)\right)\ge \inf_{a\in U}f_b(a,b)$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;My proof:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There exists a sequence $a_n$ in $U$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\lim_{n\to\infty}f(a_n,b)=\sup_{a\in U}f(a,b)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall n\in\mathbb N$ the following holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge f(a_n,b+h)-\sup_{a\in U}f(a,b)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the differentiability of $f$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=f(a_n,b)+h f_b(a_n,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is always greater than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\ge f(a_n,b)+h\;\inf_{a\in U} f_b(a,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this holds for an arbitrary $n$ and thus $f(a_n,n)$ comes arbitrarily close to $\sup_{a\in U}f(a,b)$ it holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge h\;\inf_{a\in U} f_b(a,b)+\mathcal O(h^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deviding by $h$ and performing the limit yields the assertion.&lt;/p&gt;&#xA;" OwnerUserId="4514" LastActivityDate="2010-12-07T19:48:45.853" Title="Estimate of the derivative of a function that is supremum of a function of two variables" Tags="&lt;real-analysis&gt;&lt;optimization&gt;" CommentCount="2" />
  <row Id="13411" PostTypeId="1" AcceptedAnswerId="13419" CreationDate="2010-12-07T19:52:16.753" Score="3" ViewCount="108" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a continuous function $f(x)$ and&#xA;  $\forall a\gt 0, |\int_{0}^{a}f(x)dx| \leq M$, show&#xA;  $\int_{0}^{\infty}f(x^2)dx$ exists.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried substituting $t=x^2$ which gave me $\displaystyle \int_{0}^{\infty}f(x^2)dx=\frac{1}{2}\int_{0}^{\infty}\frac{f(t)}{\sqrt t}dt$ but I can't see how to use the fact that $f(x)$ has a bounded indefinite integral in $[0, \infty)$. Simply doing $\frac{1}{2}|\int_{0}^{\infty}\frac{f(t)}{\sqrt t}dt|\leq \frac{M}{2}|\int_{0}^{\infty}\frac{1}{\sqrt t}dt|$ doesn't really help because $\int \frac{1}{\sqrt t}$ doesn't converge in $[0, \infty)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After substituting I tried integration by parts (using $u = \frac{1}{\sqrt t}, v' = f(t) \implies v = F(t)$) but that seems to complicate things further.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-07T20:04:31.737" LastActivityDate="2010-12-07T20:44:38.420" Title="Show $\int_{0}^{\infty}f(x^2)dx$ exists if $f(x)$ is continuous and $\forall a\gt 0, |\int_{0}^{a}f(x)dx| \leq M$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13412" PostTypeId="1" AcceptedAnswerId="13413" CreationDate="2010-12-07T20:03:57.340" Score="4" ViewCount="187" Body="&lt;p&gt;Given the following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(7y-1)(y-7) \le 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me this inequality implies $y \le 7$ and $y \le \frac{1}{7}$ but the correct expression (from my module) happens to be $\frac{1}{7} \le y \le 7$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where exactly I am wrong?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-12-08T12:56:28.383" Title="An elementary doubt on Inequalities" Tags="&lt;elementary&gt;" AnswerCount="4" CommentCount="9" />
  <row Id="13413" PostTypeId="2" ParentId="13412" CreationDate="2010-12-07T20:07:48.903" Score="6" ViewCount="0" Body="&lt;p&gt;Hint: If the product of two numbers is nonpostive, one of them must be nonpositive and one of them must be nonnegative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: $(7y-1)(y-7)\leq 0$. So either, $7y-1 \leq 0$ and $y-7 \geq 0$ or $7y-1 \geq 0$ and $y-7 \leq 0$. Now, can you eliminate one of these two cases?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-07T20:27:49.470" LastActivityDate="2010-12-07T20:27:49.470" CommentCount="7" />
  <row Id="13414" PostTypeId="1" AcceptedAnswerId="13449" CreationDate="2010-12-07T20:19:17.900" Score="13" ViewCount="245" Body="&lt;p&gt;I want to find a nice simple expression for the definite integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I can numerically compute this integral, and it seems to converge to $\pi/2$ for all real values of $a$. Is this integral actually always equal to $\pi/2$? How can I show this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why does Wolfram Alpha give me something that appears to depend on $a$? Is there a good reason it doesn't eliminate $a$?&lt;/p&gt;&#xA;" OwnerUserId="4519" LastActivityDate="2011-01-10T14:18:49.067" Title="Is this definite integral really independent of a parameter? How can it be shown?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="13415" PostTypeId="2" ParentId="6323" CreationDate="2010-12-07T20:26:39.753" Score="0" ViewCount="0" Body="&lt;p&gt;Sometimes the quotient space is badly behaved. For instance when a Lie group acts smoothly on a manifold, the quotient space does not carry, in general, the structure of a smooth manifold. Then the equivariant (co)homology construction can help out. (This is of course not that much of an issue if you're considering singular homology, but for Morse homology there is a HUGE difference...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it is possible to &quot;resolve&quot; the singularities of a group action and compute a certain (co)homology on the resolved space. The homology obtained in this way is equivalent to equivariant (co)homology, see &quot;equivariant cohomology and resolution&quot; &lt;a href=&quot;http://www-math.mit.edu/~rbm/paper.html&quot; rel=&quot;nofollow&quot;&gt;http://www-math.mit.edu/~rbm/paper.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4520" LastEditorUserId="4520" LastEditorDisplayName="" LastEditDate="2010-12-08T15:45:21.453" LastActivityDate="2010-12-08T15:45:21.453" />
  <row Id="13416" PostTypeId="2" ParentId="4894" CreationDate="2010-12-07T20:33:03.647" Score="0" ViewCount="0" Body="&lt;p&gt;There is a simple &quot;philosophical&quot; answer: Symplectic forms allow you to only measure two-dimensional quantities, not one-dimensional ones (you can measure area infinitesimally). That's the basic difference with Riemannian geometry, where you can measure length infinitesimally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore symplectic geometry pops up in connection with the cotangent bundle. For instance if you have a (scalar) first order differential operator on a manifold, then the principal symbol transforms like a covector (when changing coordinates). Note that this does not require a metric.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-07T20:33:03.647" />
  <row Id="13417" PostTypeId="2" ParentId="9655" CreationDate="2010-12-07T20:37:25.857" Score="1" ViewCount="0" Body="&lt;p&gt;Just minimize the energy of the path in the homotopy class of paths connecting p and q.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-07T20:37:25.857" CommentCount="2" />
  <row Id="13418" PostTypeId="2" ParentId="8808" CreationDate="2010-12-07T20:40:29.613" Score="1" ViewCount="0" Body="&lt;p&gt;Well, my answer to 1. is that you can convienently do standard analysis in a chart. Did you ever try working on more complicated (singular) objects? It's a pain in the neck! (2. and 3. have already been answered)&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-07T20:40:29.613" />
  <row Id="13419" PostTypeId="2" ParentId="13411" CreationDate="2010-12-07T20:44:38.420" Score="2" ViewCount="0" Body="&lt;p&gt;If $f$ is only assumed continuous on $(0,\infty)$, then $\int_0^\infty  {f(x^2 )dx}$ might diverge (consider $f$ such that $f(x)=1/\sqrt{x}$ near $0+$). So, assume that $f$ is continuous on $[0,\infty)$. Then, obviously $\int_0^1 {f(x^2 )dx}$ exists (since $f$ is uniformly bounded on $[0,1]$). So, we only have to prove that $\int_1^\infty  {f(x^2 )dx} $ exists. Now, after substituting and performing integration by parts as you did, it suffices to show, using your notation, that $F(t)/\sqrt{t} \to 0$ as $t \to \infty$, and that $\int_1^\infty  {\frac{{F(t)}}{{t^{3/2} }}} dt$ exists. This is clearly true... &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-07T20:44:38.420" CommentCount="1" />
  <row Id="13420" PostTypeId="1" CreationDate="2010-12-07T20:50:09.967" Score="0" ViewCount="67" Body="&lt;p&gt;Let R be a commutative ring and I and J be two ideals of R, how to show that IJ= the intersection of I and J if R=I+J?&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2010-12-07T21:02:34.433" LastActivityDate="2010-12-07T21:12:51.520" Title="A homework problem about ideal" Tags="&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="13421" PostTypeId="1" CreationDate="2010-12-07T20:53:56.297" Score="1" ViewCount="92" Body="&lt;p&gt;By definition, the weak law states that for a specified large $n$, the average  is likely to be near $\mu$. Thus, it leaves open the possibility that $|\bar{X_n}-\mu| \gt \eta$ happens an infinite number of times, although at infrequent intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong law shows that this almost surely will not occur. In particular, it implies that with probability 1, we have that for any $\eta &amp;gt; 0$ the inequality $|\bar{X_n}-\mu| \lt \eta$ holds for all large enough $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is application of these laws. How do I know which distribution satisfies the strong law vs the weak law. For example, consider a distribution $X_n$ be iid with finite variances and zero means. Does the mean $\frac{\sum_{k=1}^{n} X_k}{n}$ converge to $0$ almost surely (strong law of large numbers) or only in probability (weak law of large numbers)?&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-12-07T22:19:22.923" Title="application of strong vs weak law of large numbers" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13422" PostTypeId="2" ParentId="13420" CreationDate="2010-12-07T20:54:54.353" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ One direction is trivial. The other direction requires further hypotheses, e.g. if $\rm\ I + J = 1\ $ then start from the fact that $\rm\ (I+J)\ \ (I\cap J) \ \subseteq\ I\:J\ $ follows by applying the distributive law. It's an ideal form of the integer law $\rm gcd(i,j)\ lcm(i,j) = i\:j\ \ $ so $\rm\ \ gcd(i,j)=1\ \Rightarrow\ lcm(i,j) = i\:j\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-07T21:12:51.520" LastActivityDate="2010-12-07T21:12:51.520" />
  <row Id="13423" PostTypeId="2" ParentId="13412" CreationDate="2010-12-07T21:06:20.157" Score="4" ViewCount="0" Body="&lt;p&gt;Another way is to analyze as a quadratic function $f(y)=7y^2-50y+7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you want solve : $f(y)=7y^2-50y+7 \leq 0$, whose graph is: &lt;img src=&quot;http://i.imgur.com/ophP5.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intersections with the x-axis are : $x=\frac{1}{7}$ and $x=7$, this graph we say that $f(y) \leq 0$ in $[\frac{1}{7},7]$, this  wanted to see.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-08T02:06:32.777" LastActivityDate="2010-12-08T02:06:32.777" />
  <row Id="13424" PostTypeId="1" AcceptedAnswerId="13428" CreationDate="2010-12-07T21:06:25.647" Score="2" ViewCount="123" Body="&lt;p&gt;How would you expand a moment generating function with a negative fractional power as a power series??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method would you be able to use?&lt;/p&gt;&#xA;" OwnerUserId="4522" LastEditorUserId="4522" LastEditorDisplayName="" LastEditDate="2010-12-14T19:37:22.140" LastActivityDate="2010-12-14T19:37:22.140" Title="Expansion of a function with a negative fractional power." Tags="&lt;statistics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13425" PostTypeId="2" ParentId="13421" CreationDate="2010-12-07T21:11:44.767" Score="1" ViewCount="0" Body="&lt;p&gt;If $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with finite mean $\mu$ (in your example, $\mu = 0$),&#xA;then by the strong law of large numbers, $\frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $\mu$ almost surely. In particular, $\frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $\mu$ in probability. So, you actually don't have to assume finite variance. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-07T21:11:44.767" CommentCount="3" />
  <row Id="13426" PostTypeId="2" ParentId="13424" CreationDate="2010-12-07T21:11:47.043" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_series&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Binomial_series&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-07T21:11:47.043" />
  <row Id="13428" PostTypeId="2" ParentId="13424" CreationDate="2010-12-07T21:23:14.487" Score="2" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/GammaDistribution.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, in particular equations (17)-(24). This is the moment-generating function of a gamma distribution, and so having the moments you are actually done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, take $\theta = 2$ and $\alpha = n/2$. Then, the moments $\mu&amp;#39;_r$ are given by &#xA;$$&#xD;&#xA;\mu&amp;#39;_r = \frac{{2^r \Gamma (n/2 + r)}}{{\Gamma (n/2)}} = n(n+2)(n+4) \cdots (n+2r-2),&#xD;&#xA;$$&#xD;&#xA;where for the last equality we have used $\Gamma(p+1) = p \Gamma(p)$.&#xA;Now, the moment-generating function can be expanded as&#xA;$$&#xD;&#xA;M(t) = 1 + \mu&amp;#39; _1 t + \frac{1}{{2!}}\mu&amp;#39; _2 t^2  + \frac{1}{{3!}}\mu&amp;#39; _3 t^3  +  \cdots,&#xD;&#xA;$$&#xD;&#xA;giving us&#xA;$$&#xD;&#xA;\frac{1}{{(1 - 2t)^{n/2} }} = 1 + nt + \frac{{n(n + 2)}}{{2!}}t^2  + \frac{{n(n + 2)(n + 4)}}{{3!}}t^3  +  \cdots &#xD;&#xA;$$&#xD;&#xA;(for $|t|&amp;lt;1/2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, let us show how to find the moments $\mu&amp;#39;_r$. Your moment-generating function corresponds to a gamma distribution with density&#xA;$$&#xD;&#xA;f(x)= \frac{{x^{\alpha  - 1} e^{ - x/\theta } }}{{\theta ^\alpha  \Gamma (\alpha )}}, \;\; x&amp;gt;0,&#xD;&#xA;$$&#xD;&#xA;where $\theta = 2$ and $\alpha = n/2$. So,&#xA;$$&#xD;&#xA;\mu&amp;#39;_r = \frac{1}{{\theta ^\alpha  \Gamma (\alpha )}}\int_0^\infty  {x^{r + \alpha  - 1} e^{ - x/\theta } \,{\rm d}x} = &#xD;&#xA;\frac{{\theta ^r }}{{\Gamma (\alpha )}}\int_0^\infty  {x^{r + \alpha  - 1} e^{ - x} \,{\rm d}x} = \theta ^r \frac{{\Gamma (r + \alpha )}}{{\Gamma (\alpha )}}.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-08T17:16:26.277" LastActivityDate="2010-12-08T17:16:26.277" CommentCount="1" />
  <row Id="13430" PostTypeId="1" AcceptedAnswerId="13432" CreationDate="2010-12-07T22:01:42.367" Score="2" ViewCount="150" Body="&lt;p&gt;Are there solutions in $\mathbb{Q}-\mathbb{Z}$ to $X^2+2y^2=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that there isnt but Im not sure how to show this.  I don't have much experience (yet) with $p$-adics.&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T03:33:20.517" LastActivityDate="2010-12-08T03:33:20.517" Title="Does there exist $X$, $Y \in\mathbb{Q}-\mathbb{Z}$ such that $X^2+2Y^2=1$?" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13431" PostTypeId="1" AcceptedAnswerId="13435" CreationDate="2010-12-07T22:17:46.847" Score="1" ViewCount="47" Body="&lt;p&gt;I came across an interesting problem in Jacod's probability book. But have no idea how to approach it. Should I approach it using induction? Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X_1, X_2, \cdots$ be an infinite sequence of iid sequence of integrable random variables&#xA;and let $N$ be a positive, integer-valued integrable random variable which is independent&#xA;from the sequence. Define $S_n = \sum_{k=1}^{n} X_k$ and assume that $S_0 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Show that $E[S_N] = E[N]E[X_1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Show that the characteristic function of $S_N$ is given by $E[\phi_{X_{1}}(t)^N]$, where $\phi_{X_{1}}$ is the characteristic function of $X_1$.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-12-07T22:32:15.167" Title="distribution of iid sequence of integrable random variables" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;expectation&gt;&lt;distribution-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13432" PostTypeId="2" ParentId="13430" CreationDate="2010-12-07T22:18:16.143" Score="9" ViewCount="0" Body="&lt;p&gt;Yes, there are solutions. For example $({7 \over 11}, {6 \over 11})$ solves it. In general one can find such solutions to $x^2 + ry^2 = 1$ whenever $r$ is a rational number. One can do this by intersecting a line $ax + by = c$ containing $(1,0)$ ($a,b,$ and $c$ rational) with your ellipse $x^2 + ry^2 = 1$; the second intersection point will also have rational coordinates. Since only finitely many of these will have an integer coordinate the rest of such points will have both their entries in $Q - Z$. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-07T22:18:16.143" />
  <row Id="13433" PostTypeId="2" ParentId="13421" CreationDate="2010-12-07T22:19:22.923" Score="2" ViewCount="0" Body="&lt;p&gt;From section 7.4 of Grimmett and Stirzaker's &lt;em&gt;Probability and Random Processes (3rd edition)&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The independent and identically distributed sequence $(X_n)$, with common distribution function $F$, satisfies $${1\over n} \sum_{i=1}^n X_i\to \mu$$ in probability for some constant $\mu$ if and only if the characteristic function $\phi$ of $X_n$ is differentiable at $t=0$ and $\phi^\prime(0)=i \mu$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the weak law holds but the strong law fails for $\mu=0$ and symmetric random variables with $1-F(x)\sim 1/(x\log(x))$ as $x\to\infty$.&lt;/p&gt;&#xA;" OwnerUserId="940" LastActivityDate="2010-12-07T22:19:22.923" />
  <row Id="13434" PostTypeId="2" ParentId="13430" CreationDate="2010-12-07T22:19:44.350" Score="8" ViewCount="0" Body="&lt;p&gt;Yes.  E.g. $X = 1/3$, $Y = 2/3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added: Carrying out the procedure suggested in Zaricuses's answer, i.e. intersecting the line with slope $-p/q$ through $(1,0)$&#xD;&#xA;with the conic $X^2 + 2 Y^2 =1$, we find that the general solution&#xA;is $X = (2p^2 - q^2)/(2 p^2 + q^2), Y = 2pq/(2 p^2 + q^2).$ &#xA;The solution above comes form taking $p = 1, q = 1$.  The solution in Zaricuse's answer comes from taking $p = 3, q = 2$.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-08T03:32:48.393" LastActivityDate="2010-12-08T03:32:48.393" />
  <row Id="13435" PostTypeId="2" ParentId="13431" CreationDate="2010-12-07T22:32:15.167" Score="2" ViewCount="0" Body="&lt;p&gt;For part (a) use $E[S_N]= \sum\nolimits_n {E[S_N |N = n]P(N = n)}$. This leads straightforwardly to the result.&#xA;For part (b) use $E[e^{tS_N } ] = \sum\nolimits_n {E[e^{tS_N } |N = n]P(N = n)} $. Again, this leads straightforwardly to the result.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-07T22:32:15.167" CommentCount="2" />
  <row Id="13436" PostTypeId="1" AcceptedAnswerId="13445" CreationDate="2010-12-07T22:35:50.357" Score="1" ViewCount="148" Body="&lt;p&gt;I need to find the two roots (x-intercepts) from the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x + 7)(x - 3) = (x + 7)(5 - x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about this? &lt;strong&gt;&lt;em&gt;I would greatly appreciate a step-by-step explanation.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4526" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-07T23:24:42.860" LastActivityDate="2010-12-07T23:50:49.997" Title="How would I solve $(x + 7)(x - 3) = (x + 7)(5 - x)$?" Tags="&lt;algebra-precalculus&gt;&lt;roots&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="13437" PostTypeId="2" ParentId="13436" CreationDate="2010-12-07T22:42:16.557" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ A\ B\ =\ A\ C\ \iff\ A\ (B-C)\ =\ 0\ \iff\ A=0\ \ or\ \ B-C = 0 $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-07T22:42:16.557" />
  <row Id="13438" PostTypeId="1" AcceptedAnswerId="13440" CreationDate="2010-12-07T22:49:22.547" Score="1" ViewCount="87" Body="&lt;p&gt;$$\int_0^2 \dfrac{\mathrm dx}{\sqrt{x}(x-1)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to determine whether this integral converges or diverges. Now usually problems like these are easy, but this one is kind of tricky since it is discontinuous at both 0 and 1. Whereas with one situations where the integral is only discontinuous at 1 value, I could just set up 2 integrals and use the &quot;lim of t&quot; method, but here I cant set up two integrals. The only way that two integrals could be set up is if the first one is from 0 to 1 and the second one is from 1 to 2, but the first one is discontinuous at both values, so not sure what the work around is.  &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T02:09:11.140" LastActivityDate="2010-12-08T02:09:11.140" Title="Question on improper integrals" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13439" PostTypeId="2" ParentId="13438" CreationDate="2010-12-07T22:53:16.013" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the change of variables $t=\sqrt x$. It transforms the interval $[0,2]$ into $[0,\sqrt 2]$ and removes one of the problem zeroes. It is easy to see from the resulting expression that the integral diverges.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-07T22:53:16.013" />
  <row Id="13440" PostTypeId="2" ParentId="13438" CreationDate="2010-12-07T22:53:52.863" Score="5" ViewCount="0" Body="&lt;p&gt;You can break up the integral at &lt;em&gt;any&lt;/em&gt; point or points you like. In this case, you could break it up into &lt;em&gt;three&lt;/em&gt; integrals: pick a point $c$ strictly between $0$ and $1$, and consider:&#xA;\begin{align*}&#xA;\int_0^2 \frac{dx}{\sqrt{x}(x-1)} &amp;amp;= \int_0^c\frac{dx}{\sqrt{x}(x-1)}+\int_c^1\frac{dx}{\sqrt{x}(x-1)} + \int_1^2\frac{dx}{\sqrt{x}(x-1)}\\&#xA;&amp;amp;= \lim_{a\to 0^+}\int_a^c \frac{dx}{\sqrt{x}(x-1)} + \lim_{b\to 1^-}\int_c^b\frac{dx}{\sqrt{x}(x-1)} + \lim_{d\to 1^+}\int_d^2\frac{dx}{\sqrt{x}(x-1)}.&#xA;\end{align*}&#xA;The original improper integral exists if and only if each of the three improper integrals exist.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-07T22:53:52.863" CommentCount="3" />
  <row Id="13441" PostTypeId="1" AcceptedAnswerId="13443" CreationDate="2010-12-07T23:01:02.253" Score="2" ViewCount="49" Body="&lt;p&gt;The book I am studying uses the notation $w_k$ to denote the number of roots of unity contained in $K$, an imaginary quadratic number field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposedly, this is the same as the size of the unit group in $O$, the ring of integers in $K$.  I am well familiar with the standard arguments for finding the units in $O$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im not really sure exactly what the phrase &quot;the number of roots of unity in $K$&quot; means precisely.  Does anyone know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I was thinking that it is probibly the case that the only elements of $K$ with finite multiplicative order are the elements of the unit group.  Is this correct?  If so, then is this the correct line of reasoning to show that $w_k=|U(O)|$?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-07T23:17:50.303" Title="The number of roots of unity in an imaginary quadratic number field." Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="13442" PostTypeId="2" ParentId="13441" CreationDate="2010-12-07T23:05:40.100" Score="2" ViewCount="0" Body="&lt;p&gt;You need Dirichlet's unit theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/unittheorem.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/unittheorem.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-07T23:05:40.100" CommentCount="5" />
  <row Id="13443" PostTypeId="2" ParentId="13441" CreationDate="2010-12-07T23:08:58.147" Score="2" ViewCount="0" Body="&lt;p&gt;A root of unity is a complex number $a$ such that $a^n = 1$ for some positive integer $n$. The &quot;number of roots of unity&quot; is the cardinality of the set of roots of unity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $K=\mathbb{Q}(i)$, then the roots of unity in $K$ are $1$, $-1$, $i$, and $-i$, so the number of roots of unity contained in $K$ is $4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes: a complex number is a root of unity if and only if it has finite multiplicative order in $\mathbb{C}^*$. In fact, the group of roots of unity is the torsion subgroup of $\mathbb{C}^*$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove that in the imaginary quadratic case the unit group $\mathcal{O}_K$ coincides with the group of roots of unity in $K$ you could use Dirichlet's Unit Theorem (as Timothy Wagner suggests), but you don't have to. You can simply show that every unit in $\mathcal{O}_K$ must in fact be a root of unity. This is not hard, since you (probably) know exactly what $\mathcal{O}_K$ for $K=\mathbb{Q}(\sqrt{d})$ ($d$ square free, $d\neq 1$), and you can use the norm map. For example, to show it in the case of $K=\mathbb{Q}(i)$, note that an element in $\mathcal{O}_K = \mathbb{Z}[i]$ is a unit if and only if $N(a+bi) = a^2+b^2 = 1$ (since $N((a+bi)(c+di)) = N(a+bi)N(c+di)$). But since $a$ and $b$ are integers, that requires $a=\pm 1$ and $b=0$ or $a=0$ and $b=\pm 1$, giving you roots of unity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact, imaginary quadratic fields are the only number fields in which the group of units in the number field is finite and torsion; this follows from the aforementioned Dirichlet Unit Theorem)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-07T23:17:50.303" LastActivityDate="2010-12-07T23:17:50.303" CommentCount="2" />
  <row Id="13444" PostTypeId="2" ParentId="13387" CreationDate="2010-12-07T23:13:52.243" Score="4" ViewCount="0" Body="&lt;p&gt;You're asking about a very wide range of numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all numbers, you should do ECM first.  I recommend GMP-ECM (&lt;a href=&quot;http://gilchrist.ca/jeff/factoring/index.html&quot; rel=&quot;nofollow&quot;&gt;Windows&lt;/a&gt;, &lt;a href=&quot;http://gforge.inria.fr/projects/ecm&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For numbers of about 100 digits, your best bet (after doing an appropriate amount of trial factorization and ECM) is to use a SIQS. &lt;a href=&quot;http://www.boo.net/~jasonp/qs.html&quot; rel=&quot;nofollow&quot;&gt;msieve&lt;/a&gt; would be a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For numbers of between 100 and 200 digits, you should do a large amount of ECM, at the very least a CPU-day's worth (and possibly 100+ CPU-days).  Once those complete, if no factors were found, you should move to GNFS.  GGNFS (&lt;a href=&quot;http://gilchrist.ca/jeff/factoring/index.html&quot; rel=&quot;nofollow&quot;&gt;Windows&lt;/a&gt;, &lt;a href=&quot;http://ggnfs.svn.sourceforge.net/viewvc/ggnfs/trunk/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;) is good and does not require a cluster, though for the upper end of the range you'll need at least a year if you're running on a single (multicore, one hopes) machine.  Here are some &lt;a href=&quot;http://gilchrist.ca/jeff/factoring/nfs_beginners_guide_perl.html&quot; rel=&quot;nofollow&quot;&gt;guidelines on using NFS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have no hope of factoring hard numbers of 200 to 300 digits without many machines.  Thus, for large general-form numbers, you should focus entirely on ECM in hopes of finding a small factor (up to about 60 digits, with patience, or about 70 digits, with patience and good fortune).  If you find a factor, the cofactor may well be susceptible to GNFS (or it might be prime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one has a reasonable chance to factor hard numbers of 300 to 1000 digits, unless possibly the NSA.  You should use ECM on such numbers in hopes of finding a factor, but you're unlikely to finish the factorization even if you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you have access to a CUDA-enabled GPU (or several), you can speed the ECM with &lt;a href=&quot;http://eecm.cr.yp.to/&quot; rel=&quot;nofollow&quot;&gt;EECM&lt;/a&gt; .  I don't know of any tools that do NFS on GPUs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short&lt;/strong&gt;: ECM a lot, then finish off with SIQS or NFS if the number is small enough (below 180-200 digits depending on your resources and patience).  If you want to throw money at the problem, buy more hardware, possibly GPUs -- don't buy commercial software, that has no bang for your buck.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2010-12-10T18:36:05.197" LastActivityDate="2010-12-10T18:36:05.197" CommentCount="12" />
  <row Id="13445" PostTypeId="2" ParentId="13436" CreationDate="2010-12-07T23:26:07.253" Score="2" ViewCount="0" Body="&lt;p&gt;Generally speaking, in an equation like yours with polynomial-type expressions (even in factored form, as yours are) on both sides, you probably want to collect everything on one side of the equals sign so that you have some polynomial-type expression equal to zero.  You might do this by multiplying out both sides, then collecting all the terms on one side; or you might rewrite your equation $A=B$ as $A-B=0$ and factor the $A-B$ part, since your $A$ and $B$ are already factored.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-07T23:26:07.253" />
  <row Id="13446" PostTypeId="2" ParentId="4894" CreationDate="2010-12-07T23:29:51.383" Score="1" ViewCount="0" Body="&lt;p&gt;I suppose a question to ask here is : what kinds of applications? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the areas where symplectic geometry is used is Hamiltonian dynamics, where the geometric framework allows one to generalize away from cotangent bundles, but this direction is directly related to the origins of the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another direction comes from algebraic geometry, where the fact that complex and Riemannian structure determine a symplectic one  (see Aaron's answer) makes all complex projective (and affine) algebraic varieties symplectic. Viewing them as such allows one to use differential-geometric methods, while the symplectic form provides some control (in the theory of holomorphic curves, symplectic form gives the crucial energy bounds without which things fall apart; this is why there is no Gromow-Witten theory of almost-complex manifolds). Add a bit of mirror symmetry to the mix and you have interaction with many other areas of math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that the wedge product is anti-symmetric, which leads the space of connections on a principal bundle over 2-d Riemannian manifold to be an (infinite dimensional) symplectic manifold (the gauge group is Hamiltonian, and curvature is the moment map, leading to flat connections being the reduction!). This leads, for G=SO(3) and the surface coming from a Heegard splitting, to Atiyah-Floer conjecture and some related stuff leads to Heegard-Floer theory. Thus connections to low-dimensional topology. The &quot;naive&quot; connection here is the anti-symmetry underlying symplectic form and wedge product, but probably there is a deeper reason known to wise people...&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-12-07T23:29:51.383" />
  <row Id="13447" PostTypeId="1" CreationDate="2010-12-07T23:34:38.683" Score="1" ViewCount="53" Body="&lt;p&gt;Let $K$ be an imaginary quadratic field and $U$ denote the unit group in the ring of integers in $K$.  Are there $\alpha \in K-U$ with finite multiplicative order?  That is, is there $n \in N$ such that $\alpha^n=1$?  &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-08T04:33:38.640" Title="Multiplicative order of elements in an imaginary quadratic field." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13448" PostTypeId="2" ParentId="13402" CreationDate="2010-12-07T23:35:43.293" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not entirely sure that this answers what you're asking--if not, please comment so that I can revise it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two points $A$ and $B$, the set of points $P=(1-t)\cdot A+t\cdot B$, where $t$ is a real number, is the line through $A$ and $B$, parameterized with $A=P|_{t=0}$ and $B=P|_{t=1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know for sure that $C$ is on the line through $A$ and $B$, then set $C=(1-t)\cdot A+t\cdot B$ and solve for $t$ in any one of the components (if $C$ is in fact on the line, then the value of $t$ will be the same, regardless of which component you use; if $C$ is not on the line, then the value of $t$ will not be the same for each component).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note&lt;/em&gt;: edited to fix equation $P=$ in the second paragraph, and correspondingly in the third paragraph.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-08T01:37:37.167" LastActivityDate="2010-12-08T01:37:37.167" CommentCount="15" />
  <row Id="13449" PostTypeId="2" ParentId="13414" CreationDate="2010-12-07T23:36:34.650" Score="17" ViewCount="0" Body="&lt;p&gt;Yes it is true! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the substitution $\displaystyle x = \dfrac{a^2}{t}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^6}{t^4\left(\dfrac{a^4}{t^2} + \left(\dfrac{a^4}{t^2} - a^2\right)^2\right)} \ \text{dt} = \int_{0}^{\infty} \dfrac{a^2}{t^2 + (a^2 - t^2)^2} \ \text{dt}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&#xA;$$\displaystyle 2I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx} + \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2}\ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \int_{0}^{\infty} \dfrac{x^2 + a^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle = \int_{0}^{\infty} \dfrac{1 + \dfrac{a^2}{x^2}}{1 + \left(x-\dfrac{a^2}{x}\right)^2} \ \text{dx}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the substitution $\displaystyle t = x - \dfrac{a^2}{x}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle 2I = \int_{-\infty}^{\infty} \dfrac{\text{dt}}{1 + t^2} = \pi$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-08T07:04:14.350" LastActivityDate="2010-12-08T07:04:14.350" CommentCount="0" />
  <row Id="13450" PostTypeId="2" ParentId="13412" CreationDate="2010-12-07T23:41:11.320" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ If $\rm\ a &amp;lt; b\ $ then $\rm\  (x-a)\ (x-b) &amp;lt; 0\ \iff\ a &amp;lt; x &amp;lt; b $&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-07T23:41:11.320" />
  <row Id="13451" PostTypeId="2" ParentId="13436" CreationDate="2010-12-07T23:50:49.997" Score="1" ViewCount="0" Body="&lt;p&gt;Hint: If $x+7\neq 0$, your equation is equivalent to $x-3=5-x$, because you can divide both sides by a quantity different from zero. And what does happen to your equation when $x+7=0$?&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-12-07T23:50:49.997" />
  <row Id="13452" PostTypeId="1" AcceptedAnswerId="13457" CreationDate="2010-12-07T23:53:35.173" Score="3" ViewCount="192" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to solve this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y&amp;#39;&amp;#39;+2y&amp;#39;-3y=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y&amp;#39; = z$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y&amp;#39;&amp;#39; = z\dfrac{dz}{dy}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z\dfrac{dz}{dy}+2z-3y=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$zdz+2zdy-3ydy=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$zdz=(3y-2z)dy$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=3y-2z$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z=y$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=y&amp;#39;=y&amp;#39;&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, I'm pretty sure I did something wrong. could you please correct.&lt;/p&gt;&#xA;" OwnerUserId="4528" LastEditorUserId="4528" LastEditorDisplayName="" LastEditDate="2010-12-08T00:40:01.777" LastActivityDate="2010-12-09T02:34:00.027" Title="how do you solve $y''+2y'-3y=0$?" Tags="&lt;differential-equations&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="13453" PostTypeId="1" AcceptedAnswerId="13456" CreationDate="2010-12-07T23:54:05.357" Score="2" ViewCount="33" Body="&lt;p&gt;The product of two univariate generating functions is simply given by the Cauchy product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x) = \sum_{n=0} a_n x^n $$&#xD;&#xA;$$ B(x) = \sum_{n=0} b_n x^n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x)B(x) = C(x) = \sum_{n=0} x^n c_n $$&#xD;&#xA;with $c_n = \sum_{k=0}^n a_k b_{n-k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the resulting generating function for the bivariate case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x,y) = \sum_{n=0}\sum_{m=0} a_{nm} x^n y^m $$&#xD;&#xA;$$ B(x,y) = \sum_{n=0}\sum_{m=0} b_{nm} x^n y^m $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x,y)B(x,y) = C(x,y) = \sum_{n=0}\sum_{m=0} x^n y^m c_{nm} $$&#xD;&#xA;What is $c_{nm}$, and how does this generalize to multivariate generating functions?&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-12-08T00:00:50.597" Title="Product of bivariate generating functions" Tags="&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="1" />
  <row Id="13455" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T00:00:44.103" Score="2" ViewCount="0" Body="&lt;p&gt;$zdz=(3y−2z)dy$ does not imply $z=(3y−2z)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-08T00:00:44.103" />
  <row Id="13456" PostTypeId="2" ParentId="13453" CreationDate="2010-12-08T00:00:50.597" Score="2" ViewCount="0" Body="&lt;p&gt;$$c_{nm} = \sum_{i+j=n, k+l=m} a_{ik} b_{jl}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows directly from basic properties of addition and multiplication and you should make sure to thoroughly understand this, for example by writing out the first few coefficients by hand.  The generalization to any finite number of variables should be clear.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T00:00:50.597" CommentCount="1" />
  <row Id="13457" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T00:01:12.683" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ 0\  =\  y&amp;#39;&amp;#39; + 2\ y&amp;#39; - 3\ y\ =\  (D^2 + 2\ D -3)\ y\ =\ (D+3)\ (D-1)\ y, \quad D = d/dx $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. factor the differential operator as you would any polynomial.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-08T01:20:49.247" LastActivityDate="2010-12-08T01:20:49.247" CommentCount="3" />
  <row Id="13458" PostTypeId="1" AcceptedAnswerId="13462" CreationDate="2010-12-08T00:23:35.657" Score="3" ViewCount="73" Body="&lt;p&gt;$\displaystyle\sum_{n=0}^{\infty}\sqrt{n}(\sqrt{n^4+1}-n^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok I was really hesitant to post this because it's such an elementary question, and I really should be able to do this. However, I've been looking at this for over half an hour and I've tried many methods, but I'm not able to get the series into a nice form where I can take the limit. Any ideas on how to tackle this problem?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-08T00:38:35.883" Title="Determining convergence of a series" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13459" PostTypeId="1" AcceptedAnswerId="13461" CreationDate="2010-12-08T00:25:39.050" Score="1" ViewCount="47" Body="&lt;p&gt;Let $\rho$ be a representation of $G$ on $V$.  Why are its eigenvalues roots of unity?&lt;/p&gt;&#xA;" OwnerUserId="314" LastActivityDate="2010-12-08T00:27:40.677" Title="Eigenvalues of representations" Tags="&lt;representation-theory&gt;" AnswerCount="1" />
  <row Id="13460" PostTypeId="1" AcceptedAnswerId="13467" CreationDate="2010-12-08T00:27:29.953" Score="4" ViewCount="210" Body="&lt;p&gt;I hope that this question is on-topic, though it is not quite technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious to hear from people how they approach reading a mathematical paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not asking specific questions on purpose, though at first I had a few. But I want to keep it rather open-ended.&lt;/p&gt;&#xA;" OwnerUserId="4342" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T00:36:40.580" LastActivityDate="2010-12-08T06:34:07.320" CommunityOwnedDate="2010-12-08T06:31:13.470" Title="how to read a mathematical paper?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" />
  <row Id="13461" PostTypeId="2" ParentId="13459" CreationDate="2010-12-08T00:27:40.677" Score="4" ViewCount="0" Body="&lt;p&gt;I assume $G$ is finite.  In that case any $g \in G$ has some finite order $n$, hence $\rho(g)^n = 1$.  It follows that the characteristic polynomial of $\rho(g)$ divides $x^n - 1$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T00:27:40.677" />
  <row Id="13462" PostTypeId="2" ParentId="13458" CreationDate="2010-12-08T00:28:14.810" Score="8" ViewCount="0" Body="&lt;p&gt;Just note that $$\sqrt{n^4 + 1} - n^2 = \frac{( \sqrt{n^4 + 1} - n^2 )( \sqrt{n^4 + 1} + n^2 )}{\sqrt{n^4 + 1} + n^2} = \frac{n^4 + 1 - n^4}{\sqrt{n^4 + 1} + n^2} =  \frac{1}{\sqrt{n^4 + 1} + n^2}$$ then you can use the comparison test.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-08T00:38:35.883" LastActivityDate="2010-12-08T00:38:35.883" CommentCount="4" />
  <row Id="13463" PostTypeId="2" ParentId="13460" CreationDate="2010-12-08T00:46:06.157" Score="4" ViewCount="0" Body="&lt;p&gt;How I read a paper really depends on why I'm reading the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of papers I go to because I have a specific goal.  Maybe they have been cited elsewhere as containing a proof of something I want to understand.  Or a different proof of something I already know how to prove.  Or maybe someone refers to the paper as having a particularly lucid explanation of something.  If that's my goal, I go straight to whatever is in the paper that I want, assuming the organization of the paper makes this possible (not always the case, but very common, at least with articles written after say 1950).  In doing my close reading of the particular thing I want, I often take notes.  Depending on the results of my reading the thing I set out to read, I &lt;em&gt;might&lt;/em&gt; go on to read other parts of the paper.  But I generally don't.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm reading a paper without a specific goal, I generally start at the beginning and read the introduction.  Ideally this contains the statements of the main results, but if it doesn't, I skim the paper for those, and also the statements of any lemmas, corollaries, etc.  19 times out of 20 my reading stops there, because I find that, beyond the statements of the theorems adding to my general awareness of the world, I'm not all that interested in the paper.  That 1 time out of 20, I will probably try to come up with the proofs on my own, using the paper's introduction as a really big hint.  Whether I succeed or not, I usually end up giving the paper a pretty thorough reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most of the time, for me, &quot;reading&quot; a paper does not mean reading the entire paper, or even reading a large contiguous chunk of the paper.  I could probably fit all of the papers I have ever read beginning to end in a regular sized binder with tons of room to spare.  I think I am far from alone in this.  In fact I think it is probably counterproductive to insist on reading things beginning to end, and I would advise that people who are new to the mathematical literature (e.g. students) suppress the desire to do this.  Generally speaking, if you are new to the literature in a given subject, you will waste a lot of time if you read this way.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorUserId="3840" LastEditorDisplayName="" LastEditDate="2010-12-08T00:52:18.337" LastActivityDate="2010-12-08T00:52:18.337" CommunityOwnedDate="2010-12-08T06:31:13.470" />
  <row Id="13464" PostTypeId="2" ParentId="13414" CreationDate="2010-12-08T00:48:54.693" Score="9" ViewCount="0" Body="&lt;p&gt;With apologies to Robin Chapman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrand is an even function of $x$, so we can integrate from $-\infty$ to $\infty$ and take half. The integrand tends to $1/z^2$ for large $z$ and as the length of a large arc is $\pi z$ the contribution of the arc tends to zero. So we just need to integrate over the upper half plane. The residues are solutions to $0=(x^2-a^2)^2+x^2= (x^2-i x -a^2)(x^2+i x-a^2)$. If $x_1$ and $x_2$ are solutions to the first quadratic, then at $x_1$ the second polynomial is equal to $2i x_1$, and the residue at $x_1$ is $x_1^2/2ix_1(x_2-x_1)=x_1/2i(x_2-x_1)$. The sum of residues at $x_1$ and $x_2$ is therefore $1/2i$. Now we just note that the two poles in the upper half plane are indeed the solutions $x_1$ and $x_2$ (which are  $i (\pm\sqrt{-a^2+1/4} + 1/2)$)). Hence the contour integral is $\pi$, and the original integral is $\pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-08T05:45:42.947" LastActivityDate="2010-12-08T05:45:42.947" CommentCount="1" />
  <row Id="13465" PostTypeId="2" ParentId="13396" CreationDate="2010-12-08T01:14:45.323" Score="2" ViewCount="0" Body="&lt;p&gt;In a somewhat more general setting, let $A(R)$ denote the area of region $R$. If the number of flaws found on region $R$ follows a Poisson distribution, then the mean is proportional to $A(R)$. That is, if $X(R)$ denotes the number of flaws found on region $R$, then $X(R)$ is Poisson distributed with mean $\lambda A(R)$, for some fixed $\lambda &amp;gt; 0$.&#xA;Now, if $X_1,\ldots,X_n$ are i.i.d. Poisson$(\lambda)$ rv's (corresponding to the number of flaws found on a unit-area region, say square yard), then $\frac{1}{n}\sum\nolimits_{i = 1}^n {X_i }$ is an unbiased estimator for $\lambda$, and in turn, $\frac{1}{n}\sum\nolimits_{i = 1}^n {A(R) X_i }$ is an unbiased estimator for&#xA;$\lambda A(R) = {\rm E}[X(R)]$. So, in order to estimate ${\rm E}[X(R)]$, it suffices to estimate $\lambda$ (and then just multiply by $A(R)$).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-08T01:14:45.323" />
  <row Id="13466" PostTypeId="2" ParentId="13447" CreationDate="2010-12-08T01:23:24.547" Score="3" ViewCount="0" Body="&lt;p&gt;If $\alpha^n=1$, then $\alpha$ satisfies the polynomial $x^n-1$, hence is an algebraic integer. Thus, $\alpha\in \mathcal{O}_K$; once you have that, you can make the easy observation Qiaochu did in the comments that $\alpha^{-1}=\alpha^{n-1}$ is also in $\mathcal{O}_K$ to show $\alpha\in U$ (or you can use the fact that if an algebraic integer satisfies a monic polynomial with integer coefficients and constant term $1$, then it must be a unit). So the answer is &quot;no.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, every root of unity is integral (satisfying $x^n-1$ for some $n$, or the appropriate cyclotomic polynomial if you insist on getting the minimal polynomial), so roots of unity in a number field are always in the ring of integers.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T04:33:38.640" LastActivityDate="2010-12-08T04:33:38.640" CommentCount="5" />
  <row Id="13467" PostTypeId="2" ParentId="13460" CreationDate="2010-12-08T01:32:04.577" Score="6" ViewCount="0" Body="&lt;p&gt;I have a couple of pieces of advice:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Do not be intimidated by the introduction. This is where the whole paper is laid out,so it typically won't be easy to follow. The author will expand sentences into whole pages later on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take notes. I usually get a legal pad out and start copying down the paper, and fill in things, and make notes to myself or work out small examples.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Talk to people about the material. If you have an asdvisor or someone you are going through the paper with it is often helpful to present portions of the paper if not all of it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Skip confusing bits. Feel free to black box certain parts. If you get stuck on something, don't worry, one sentence should not keep you from progressing through the paper. Box it off, pretend you understand the conclusion and make a note of it. Take that piece as a fact and move on to the next part of the paper. The goal is to get something out of a paper not understand it line for line the first time through.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Come back to the paper. You may want to go over the paper multiple times, take several passes at it. At least this is what I have to do when I am reading something. The amount of time it takes me to assimilate different levels/types of arguments gradually decreases, but it does take time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't spend a lot of time reading papers right now, but this is what I try to do if I really want to understand something.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-08T02:55:38.287" LastActivityDate="2010-12-08T02:55:38.287" CommunityOwnedDate="2010-12-08T06:31:13.470" />
  <row Id="13468" PostTypeId="1" AcceptedAnswerId="13472" CreationDate="2010-12-08T01:33:53.527" Score="2" ViewCount="43" Body="&lt;p&gt;Given two random samples a and b of a normal distribution, how can I calculate the probability that their difference |a-b| is bigger (or lesser) than c?&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T02:01:42.270" LastActivityDate="2010-12-08T02:01:42.270" Title="what is the formula for P(|a-b| &gt; c) in a normal distribution?" Tags="&lt;statistics&gt;&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="13469" PostTypeId="1" CreationDate="2010-12-08T01:35:57.480" Score="4" ViewCount="154" Body="&lt;p&gt;I seem to remember reading somewhere an &quot;abstract nonsense&quot; proof of the well-known fact that the fundamental group of a connected topological group is abelian.  By &quot;abstract nonsense&quot; I mean that the proof used little more than the fact that topological groups are the group objects in the category of topological spaces and the fact that $\pi_1$ is a homotopy functor.  Does anybody remember how this works?  References are fine.&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2010-12-08T02:18:11.170" Title="&quot;Abstract nonsense&quot; proof that the fundamental group of a topological group is abelian" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="13470" PostTypeId="1" CreationDate="2010-12-08T01:46:54.010" Score="1" ViewCount="96" Body="&lt;p&gt;$$\sum_{i=0}^{n-1} (i+1)(n-1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that $O(n^2)$ or $O(n^3)$? Can you explain me how you found it? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4531" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T01:59:35.783" LastActivityDate="2010-12-08T03:56:15.823" Title="is this expression O(n^2) or O(n^3)?" Tags="&lt;sequences-and-series&gt;&lt;discrete-mathematics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13471" PostTypeId="2" ParentId="13470" CreationDate="2010-12-08T01:52:00.793" Score="6" ViewCount="0" Body="&lt;p&gt;Hint: Factor out the (n-1) and then use that the average term in the sum has size n/2.&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2010-12-08T01:52:00.793" CommentCount="4" />
  <row Id="13472" PostTypeId="2" ParentId="13468" CreationDate="2010-12-08T01:58:36.190" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose that $X_1 \sim {\rm N}(\mu_1,\sigma_1^2)$ and $X_2 \sim {\rm N}(\mu_2,\sigma_2^2)$ are independent. Then, $X_1 - X_2 \sim {\rm N}(\mu_1 - \mu_2 , \sigma_1^2 + \sigma_2^2)$. Set $Y = X_1 - X_2$, $\mu = \mu_1 - \mu_2$, and $\sigma^2 = \sigma_1^2 + \sigma_2^2$. Then, for any $c &amp;gt; 0$,&#xA;$$&#xD;&#xA;P(|Y| \le c) = P( - c \le Y \le c) = P\bigg(\frac{{ - c - \mu }}{{\sigma }} \le \frac{{Y - \mu }}{{\sigma }} \le \frac{{c - \mu }}{{\sigma }}\bigg).&#xD;&#xA;$$&#xD;&#xA;Noting that $(Y - \mu)/\sigma \sim {\rm N}(0,1)$, we are actually done.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-08T01:58:36.190" />
  <row Id="13473" PostTypeId="1" AcceptedAnswerId="13499" CreationDate="2010-12-08T02:07:52.857" Score="4" ViewCount="89" Body="&lt;p&gt;A version of the Riesz Representation Theorem says that a continuous linear functional on the space of continuous real-valued mappings on a compact metric space, $C(X)$, can be identified with a signed Borel measure on the set $X$. Are there any similar results when we replace $C(X)$ by the space of continuous functions of $X$ (compact metric) into $Y$ when (1) $Y=R^N$ or in general (2) $Y$ is a Banach space? I suspect the answer is yes, but I would like to find the right reference to start looking at. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastActivityDate="2010-12-08T07:30:37.400" Title="Riesz representation and vector-valued functions" Tags="&lt;real-analysis&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13474" PostTypeId="2" ParentId="13469" CreationDate="2010-12-08T02:18:11.170" Score="5" ViewCount="0" Body="&lt;p&gt;There is a mostly nonsensical proof in Robert M. Switzer's &lt;em&gt;Algebraic topology--homotopy and homology&lt;/em&gt;, at the end of the first chapter. (You can see it in googlebooks)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He proves that if $X$ is an $H$-cogroup, then $[X,Y]$ is a group, and that if $Y$ is an $H$-group, then $[X,Y]$ is also a group, so when $X$ and $Y$ are an $H$-cogroups and an $H$-group, respectively, then $[X,Y]$ is a group in two ways. And so on.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-08T02:18:11.170" CommentCount="2" />
  <row Id="13475" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T02:57:51.870" Score="2" ViewCount="0" Body="&lt;p&gt;For this kind of differential equations Laplace transform is handy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;{\cal L}[y&amp;#39;&amp;#39;] + 2 {\cal L}[y&amp;#39;] -3{\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2{\cal L}[y] -sy(0) - y&amp;#39;(0)) +2(s{\cal L}[y] -y(0)) -3{\cal L}[y] = 0  \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put $F(s) = {\cal L}[y]$ and $a= y(0), b= y&amp;#39;(0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(s^2F(s) -as - b) +2(sF(s) -a) -3F(s) = 0 \ \Longleftrightarrow \ (s^2 +2s -3)F(s) -as -2a -b = 0 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We solve this equation for $F(s)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;F(s) = \dfrac{as + 2a +b}{s^2 +2s -3} = \dfrac{as + 2a +b}{(s-1)(s+3)} = \dfrac{A}{s-1} + \dfrac{B}{s+3} \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are constants depending on $a$ and $b$, which I let you the pleasure to compute  :-)  and we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y(t) = A{\cal L}^{-1}\left[ \dfrac{1}{s-1}\right] + B{\cal L}^{-1}\left[ \dfrac{1}{s+3} \right] = A e^t + B e^{-3t} \ .&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-09T02:34:00.027" LastActivityDate="2010-12-09T02:34:00.027" />
  <row Id="13476" PostTypeId="1" AcceptedAnswerId="13488" CreationDate="2010-12-08T03:01:13.473" Score="2" ViewCount="176" Body="&lt;p&gt;Is there a good explanation of a manifold on the web somewhere?  The wikipedia article isn't really working for me.  I was actually hoping for a whiteboard lecture on youtube, but can't find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My math experience is calculus through differential equations, twenty five years ago.  I also have some computer science related math; discrete structures and numerical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My&lt;/em&gt; problem with the Wikipedia article is that I just can't visualize what they're saying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any sources (not just web based) are appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stumbled upon the Wikipedia article while trying to brush up on my math.  I had been talking to my daughter about primes, we wandered there from her 4th grade homework.  We ventured as far as Mersene primes and perfect numbers.  It was at this point that I knew I had to brush up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow I chased a link to Riemann manifolds and this was completely new to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having looked at some of the suggested material from the answers, the Wikipedia article makes a lot more sense.  In fact I'm trying to figure out why I found it unclear, maybe it can be improved.&lt;/p&gt;&#xA;" OwnerUserId="4181" LastEditorUserId="4181" LastEditorDisplayName="" LastEditDate="2010-12-23T14:57:57.913" LastActivityDate="2010-12-23T14:57:57.913" Title="Manifold Explained" Tags="&lt;manifold&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13477" PostTypeId="1" CreationDate="2010-12-08T03:03:22.490" Score="6" ViewCount="158" Body="&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/G9pG2.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a circle, a point $H$ outside the circle, segments $\overline{HE}$ and $\overline{HT}$ tangent to the circle at $E$ and $T$, respectively, and points $I$ and $G$ on the circle such that $I$, $G$, and $H$ are collinear (all as shown above), knowing the measures of $\angle EHG$ and $\angle GHT$ (call them $\alpha$ and $\beta$, respectively, for convenience) determines the measures of each of the four arcs on the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to compute the measures of the minor arcs $EI$, $IT$, $TG$, and $GE$ in terms of $\alpha$ and $\beta$ without using trigonometry?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-09T21:02:24.340" LastActivityDate="2010-12-09T21:42:59.857" Title="How to determine arc measures from angles between secant and tangents (without trigonometry)" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="13478" PostTypeId="2" ParentId="13476" CreationDate="2010-12-08T03:05:08.180" Score="3" ViewCount="0" Body="&lt;p&gt;I liked Lee's book on Manifolds. You can find the first chapter here on his website linked below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.washington.edu/~lee/Books/Manifolds/c1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.washington.edu/~lee/Books/Manifolds/c1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-08T03:05:08.180" CommentCount="1" />
  <row Id="13479" PostTypeId="1" CreationDate="2010-12-08T03:19:14.593" Score="0" ViewCount="55" Body="&lt;p&gt;I am having issues solving the following equation:&#xA;$$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolution of this is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in the section of parabolas and it should be pretty easy to solve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My steps are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$&#xD;&#xA;Multiply by $2x(4x-6)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 -1(4x-6) = 6x$$&#xD;&#xA;gets me here, remove $6x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^2 - 10x + 6 = 0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this with the parabola formula I ended up getting $\{4,6\}$.&lt;/p&gt;&#xA;" OwnerUserId="3034" LastEditorUserId="3034" LastEditorDisplayName="" LastEditDate="2010-12-08T03:44:24.753" LastActivityDate="2010-12-08T03:57:54.377" Title="Solving an equation of terms containing a variable in their denominator " Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13480" PostTypeId="2" ParentId="13479" CreationDate="2010-12-08T03:44:51.600" Score="1" ViewCount="0" Body="&lt;p&gt;@J. M. :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After checking out the formula I found out that I dividing by $2$ instead of $4a$/$8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{10 \pm \sqrt{100-96}}{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the fixed form I get the results $1.5$ and $1$ of which $1$ is matching result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the kind support!&lt;/p&gt;&#xA;" OwnerUserId="3034" LastActivityDate="2010-12-08T03:44:51.600" CommentCount="2" />
  <row Id="13481" PostTypeId="2" ParentId="13357" CreationDate="2010-12-08T03:46:48.963" Score="2" ViewCount="0" Body="&lt;p&gt;Let $G$ be a filtered commutative group and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\dots \subset F^{s+1} \subset F^s \subset \dots&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;its decreasing filtration. Quoting &lt;a href=&quot;http://hopf.math.purdue.edu/Boardman/ccspseq.pdf&quot; rel=&quot;nofollow&quot;&gt;Boardman&lt;/a&gt; &quot;(...) we topologize $G$ by taking the cosets $x + F^s$ of $F^s$ for all $x \in G$ and all $s$ as basic open sets (...) A Cauchy sequence $n \mapsto x_n$ is one in which $x_m − x_n \mapsto 0$ as $m, n \mapsto \infty$.&quot; And $G$ is said to be &lt;em&gt;complete&lt;/em&gt; if every Cauchy sequence has a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition can be written in more algebraic terms as follows: for each $s$, consider the exact sequence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;0 \longrightarrow F^s \longrightarrow G \longrightarrow G/F^s \longrightarrow 0 &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and apply $\varprojlim_s$ to it to obtain the long exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;0 \longrightarrow \varprojlim_s F^s \longrightarrow G \longrightarrow \varprojlim_s G/F^s \longrightarrow \varprojlim_s{}^1F^s \longrightarrow 0 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the filtration is complete if and only if $\varprojlim_s^1F^s = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover: a filtration is called &lt;em&gt;Hausdorff&lt;/em&gt; if so is the topological space $G$. In Hausdorff spaces convergent sequences have unique limits. This is easily seen to be equivalent to the fact that $\varprojlim_s F^s = \bigcap_s F^s = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in $G$, with the topology induced by the filtration, every Cauchy sequence has a unique limit if and only if $G \cong \varprojlim_s G/F^s$. This last group is called the &lt;em&gt;completion&lt;/em&gt; of $G$. (Beware: Weibel in his introduction to Homological Algebra uses the term with a different meaning; for him, &quot;complete&quot; filtration means complete and Hausdorff.)&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-09T03:38:19.087" LastActivityDate="2010-12-09T03:38:19.087" CommentCount="1" />
  <row Id="13482" PostTypeId="1" AcceptedAnswerId="13675" CreationDate="2010-12-08T03:51:30.113" Score="3" ViewCount="86" Body="&lt;p&gt;Let $X = (X_t: t \in [0,T])$ be a stochastic process satisfying a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cox%E2%80%93Ingersoll%E2%80%93Ross_model&quot; rel=&quot;nofollow&quot;&gt;CIR model&lt;/a&gt;&#xA;$$&#xD;&#xA;dX_t = \beta (X_t - \gamma) dt + \sigma\sqrt{X_t} dB_t,&#xD;&#xA;$$&#xD;&#xA;where $B_t$ is a standard Brownian motion, $\beta$ is a negative constant, $\gamma, \sigma$ are positive constants. In order for the SDE to make sense, assume that $X_t &amp;gt; 0$ for all $t \in [0,T]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider following two ways to simulate the model based on discretization of $t$ with Ito-Taylor expansion:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Euler scheme:&#xA;$$                                                                &#xD;&#xA;X_{t + \Delta} \approx X_t + \beta(X_t - \gamma)\Delta + \sigma \sqrt{X_t} Z \Delta, &#xD;&#xA;$$&#xD;&#xA;where $Z$ is $N(0, 1)$ Gaussian variable. &lt;/li&gt;&#xA;&lt;li&gt;the Milstein scheme&#xA;$$&#xD;&#xA;X_{t + \Delta} \approx X_t + \beta(X_t - \gamma)\Delta + \sigma \sqrt{X_t}Z\sqrt{\Delta} + \frac{1}{4}\sigma^2 \Delta (Z^2-1)&#xD;&#xA;$$&#xD;&#xA;where $Z$ is $N(0, 1)$ Gaussian variable. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was wondering why these two schemes have a positive probability of generating negative values of $X_t$ and therefore cannot be used without suitable modifications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References (book, tutorial and/or paper) will be helpful too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards!&lt;/p&gt;&#xA;" OwnerUserId="3348" LastEditorUserId="3348" LastEditorDisplayName="" LastEditDate="2010-12-09T15:03:28.287" LastActivityDate="2010-12-09T16:13:18.470" Title="Negativity in a CIR model discretized by Ito-Taylor expansion" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13483" PostTypeId="2" ParentId="13470" CreationDate="2010-12-08T03:56:15.823" Score="2" ViewCount="0" Body="&lt;p&gt;ok .. so from what you guys say: that will be: n-1*Σ from 0 to n-1 of (n/2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the inner bracket is O(n^2), and the outer is N, so this is the O(n^3) complexity&#xA;right?&lt;/p&gt;&#xA;" OwnerUserId="4531" LastActivityDate="2010-12-08T03:56:15.823" CommentCount="1" />
  <row Id="13484" PostTypeId="2" ParentId="13479" CreationDate="2010-12-08T03:57:54.377" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad \frac{1}{2\ x}\ =\ \frac{2\ x}{4\ x-6} - \frac{3}{4\ x-6}\ =\ \frac{1}2\ \ \Rightarrow\ \ x\ =\ \ldots\quad$ Note: no quadratic formula needed.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-08T03:57:54.377" />
  <row Id="13485" PostTypeId="2" ParentId="13452" CreationDate="2010-12-08T03:58:11.970" Score="1" ViewCount="0" Body="&lt;p&gt;Start by assuming a solution of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(t)=e^{ct}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $y&amp;#39;(t)=ce^{ct}$ and $y&amp;#39;&amp;#39;(t)=c^2e^{ct}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute into the original equation. You will find a quadratic for $c$. Solve it, and note that both solutions work, and as the ODE is linear and homogeneous, the sum of solutions also works. &lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2010-12-08T03:58:11.970" />
  <row Id="13486" PostTypeId="2" ParentId="13477" CreationDate="2010-12-08T04:26:40.067" Score="1" ViewCount="0" Body="&lt;p&gt;I'm within one, so I hope it prompts your thinking.  Basic geometry was a lot of years ago.  The theorems I remember are $\stackrel{\frown}{EGT} = 2\angle EIT$, $\beta=(\stackrel{\frown}{IT} - \stackrel{\frown}{TG})/2$,  and $ET=\pi-\alpha-\beta$, which you can see by drawing radii from E and T and the line from H to the center of the circle.  Then if $\angle EIG=\gamma$ we have $EG=2\gamma, GT=\pi-\alpha-\beta-2\gamma, TI=\pi-\alpha+\beta-2\gamma, IE=2\gamma+2\alpha$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T04:36:44.243" LastActivityDate="2010-12-08T04:36:44.243" CommentCount="3" />
  <row Id="13487" PostTypeId="1" CreationDate="2010-12-08T04:55:01.303" Score="8" ViewCount="76" Body="&lt;p&gt;Consider the set $A$ of natural numbers which are of the form $k^3-p$, for $k$ a positive integer and $p$ a positive prime. Does $A$ have a density (of any of the usual kinds for sets of natural numbers) and if so, what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ contains a lot of numbers, and it seems somewhat difficult to me to prove that a particular number is not in $A$, except that most cubes are not in $A$.&lt;/p&gt;&#xA;" OwnerUserId="4535" LastActivityDate="2010-12-08T05:34:23.930" Title="Density of set of differences of cubes and primes" Tags="&lt;number-theory&gt;" AnswerCount="1" />
  <row Id="13488" PostTypeId="2" ParentId="13476" CreationDate="2010-12-08T05:31:09.570" Score="5" ViewCount="0" Body="&lt;p&gt;This isn't in any way a thorough or rigorous introduction to manifolds, but as far as visualization it's quite nice: Weeks' &lt;a href=&quot;http://books.google.com/books?id=Lurp6nB4LtQC&amp;amp;printsec=frontcover&amp;amp;dq=shape+of+space&amp;amp;source=bl&quot; rel=&quot;nofollow&quot;&gt;The Shape of Space&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T05:31:09.570" />
  <row Id="13489" PostTypeId="2" ParentId="13487" CreationDate="2010-12-08T05:34:23.930" Score="4" ViewCount="0" Body="&lt;p&gt;A naive approach would be to say that the chance $n$ is equal to $k^3-p$ for a given $k$ is about $\frac{1}{\ln(k^3-n)}$.  Then the chance $n$ is not equal to $k^3-p$ for any $k$ is $$\prod_{k=\sqrt[3]{n}}^\infty {1-\frac{1}{\ln(k^3-n)}}$$  as this goes to zero, &quot;all&quot; numbers should be in A.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-08T05:34:23.930" />
  <row Id="13490" PostTypeId="1" AcceptedAnswerId="13494" CreationDate="2010-12-08T06:14:37.940" Score="4" ViewCount="108" Body="&lt;p&gt;Here is an exercise, on analysis which i am stuck. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="649" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T06:33:03.857" LastActivityDate="2010-12-08T06:54:02.943" Title="Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" />
  <row Id="13491" PostTypeId="2" ParentId="5944" CreationDate="2010-12-08T06:31:59.413" Score="1" ViewCount="0" Body="&lt;p&gt;With $x=\frac{1}{2}\tan\theta$ and $dx=\frac{1}{2}\sec^2\theta d\theta$, $2\pi\int x^2\sqrt{1+4x^2}dx=\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta$.  Now, let&#xA;$$\begin{align}&#xD;&#xA;I_1&amp;amp;=\int\tan^2\theta\sec^3\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\int(\sec^3\theta-\sec\theta)\sec^2\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\tan\theta(\sec^3\theta-\sec\theta)-\int3\sec^3\theta\tan^2\theta-\sec\theta\tan^2\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\tan\theta(\sec^3\theta-\sec\theta)-3I_1+\int\sec\theta\tan^2\theta d\theta&#xD;&#xA;\end{align}$$&#xD;&#xA;(using the integration by parts as you suggested) so that $I_1=\frac{1}{4}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)$.  Let&#xA;$$\begin{align}&#xD;&#xA;I_2&amp;amp;=\int\sec\theta\tan^2\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sec\theta\tan\theta-\int\sec^3\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sec\theta\tan\theta-\int\sec^3\theta-\sec\theta d\theta-\int\sec\theta d\theta&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\sec\theta\tan\theta-I_2+\log|\sec\theta-\tan\theta|&#xD;&#xA;\end{align}$$&#xD;&#xA;(using integration by parts with $u=\tan\theta$ and $dv=\sec\theta\tan\theta d\theta$) so that $I_2=\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)$.  Now,&#xA;$$\begin{align}&#xD;&#xA;\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta&#xD;&#xA;&amp;amp;=\frac{\pi}{4}I_1=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{\pi}{16}\left(\tan\theta\sec^3\theta-\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\log|\sec\theta-\tan\theta|\right)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\frac{\pi}{16}\left(2x(1+4x^2)^{3/2}-x\sqrt{1+4x^2}+\frac{1}{2}\log\left|\sqrt{1+4x^2}-2x\right|\right)&#xD;&#xA;\end{align}$$&#xD;&#xA;which is similar to what you have, but not the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, doing this by hand and covering a sheet of paper with scribbles to do the two instances of over-and-back integration by parts, I may well have made a mistake along the way, but my result does appear to check by the method described in &lt;a href=&quot;http://math.stackexchange.com/questions/5944/lateral-surface-area-by-hand/5978#5978&quot;&gt;aschepler's answer&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T06:31:59.413" />
  <row Id="13492" PostTypeId="2" ParentId="13460" CreationDate="2010-12-08T06:34:07.320" Score="2" ViewCount="0" Body="&lt;p&gt;I generally have a &quot;top down approach&quot; to reading papers. More often than not, I am driven to a particular paper looking for a particular result. So I would start by looking at the statement of the result and try to see what background is needed in order to parse the statements which appear there. Once I have understood the philosophy of the result, I think about/look at the proof. Again, if I come across something unfamiliar, I try to acquaint myself with the relevant material on an as needed basis. I find this approach works better for me if I am interested in a paper for isolated results. Reading such papers from introduction to conclusion might not be of immediate (or even future) value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other kinds of papers I read would include survey papers (exposition of a particular new idea) and papers which essentially develop a new idea/concept/structure for the first time. A lot of these papers, I find, can (and eventually should) be read from front matter to bibliography. So I would essentially read these papers as I would read a textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing which I find especially important while reading papers is to look at the cross references. This can be useful for motivation, alternate proofs and related ideas. It also gets me acquainted with who else is working in the area and the exposure to ideas from different mathematicians achieved thus is a nice side effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I feel it is important to ask a lot of questions while I read papers. Why is hypothesis X necessary. Can condition Y be weakened? Can this result be generalized to Z structure? This is eventually useful when I write my own results, to conceive the most general form that I can.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-08T06:34:07.320" CommunityOwnedDate="2010-12-08T06:34:07.320" />
  <row Id="13493" PostTypeId="2" ParentId="6268" CreationDate="2010-12-08T06:43:12.357" Score="1" ViewCount="0" Body="&lt;p&gt;Note, you also need the fact that $f(\mathbb{R})$ is borel. This fortunately is so, as $\mathbb{R}$ is $\sigma$-compact, and the continuous image of a compact set is compact, thus closed (since we are working in hausdorff spaces), thus borel.&lt;/p&gt;&#xA;" OwnerUserId="4537" LastActivityDate="2010-12-08T06:43:12.357" />
  <row Id="13494" PostTypeId="2" ParentId="13490" CreationDate="2010-12-08T06:47:10.750" Score="6" ViewCount="0" Body="&lt;p&gt;First, let's note that for $x\in(0,2\pi)$&#xD;&#xA;$$\sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\int_{0}^{x}\sum\limits_{k=1}^{n}\cos kt\ dt=&#xD;&#xA;-\frac{x}{2}+\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{2\sin{\frac{t}{2}}}\ dt$$&#xD;&#xA;$$=-\frac{x}{2}+\int_{0}^{x}\left(\frac{1}{2\sin{\frac{t}{2}}}-\frac{1}{t}\right)\sin \frac{(2n+1)t}{2}\ dt +\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{t}dt.$$&#xD;&#xA;Now, the first integral at the right-hand side tends to zero by the Riemann-Lebesgue lemma.&#xA;The second one is equal to (via a substitution $s=(2n+1)t/2$) the integral&#xA;$$\int_{0}^{\frac{(2n+1)x}{2}}\frac{\sin s}{s}\ ds\to\int_{0}^{\infty}\frac{\sin s}{s}\ ds=\frac{\pi}{2}.$$&#xD;&#xA;Therefore&#xA;$$\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=f(x),\qquad x\in(0,2\pi).$$&#xD;&#xA;The series converges on $\mathbb R$ to the periodic extension of $f(x)$. &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-08T06:54:02.943" LastActivityDate="2010-12-08T06:54:02.943" CommentCount="3" />
  <row Id="13495" PostTypeId="1" CreationDate="2010-12-08T06:49:05.960" Score="1" ViewCount="95" Body="&lt;p&gt;Here is a problem in Griffiths Introduction to Electrodynamics as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check the divergence theorem for the function $\mathbf{v} = r^2\mathbf{\hat{r}}$, using as your volume, the sphere of radius R, centered at the origin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my calculation for the surface intergral part, please help me finding the error in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{v} = r^2(\sin\theta \cos\phi \ \mathbf{\hat{x}} + \sin\theta \sin\phi \ \mathbf{\hat{y}} + \cos \theta \ \mathbf{\hat{z}})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\oint_s \mathbf{v} \cdot d\mathbf{a} &amp;amp;= 2 \iint r^2 \cos\theta\ dxdy  \\&lt;br&gt;&#xA;&amp;amp;= 2 \iint r^2 \cdot \frac{z}{r} dxdy \\&lt;br&gt;&#xA;&amp;amp;= 2R \iint \sqrt{R^2 - x^2 - y^2} dxdy \\&lt;br&gt;&#xA;&amp;amp;= 2R \int_{r=0}^{R}\sqrt{R^2 - r^2} rdr \int_{\theta=0}^{2\pi} d\theta \\&lt;br&gt;&#xA;&amp;amp;= 4\pi R\int_{\theta=0}^{\frac{\pi}{2}}\sqrt{R^2 - R^2{\sin^2\theta}} R\sin\theta\ R\cos\theta\ d\theta  \\&lt;br&gt;&#xA;&amp;amp;= \frac{4\pi R^4}{3}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the right answer should be $4\pi R^4$ considering the left part of the divergence theorem $\int_V \nabla\cdot\mathbf{v}\ d\tau$. I have check my answer serveral times but could not find error, could you help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-12-08T07:39:59.953" LastActivityDate="2010-12-08T07:39:59.953" Title="What's wrong with my calculation for checking the divergence law?" Tags="&lt;vector-calculus&gt;" CommentCount="4" />
  <row Id="13496" PostTypeId="2" ParentId="8808" CreationDate="2010-12-08T06:53:06.703" Score="2" ViewCount="0" Body="&lt;p&gt;I'm interested in differential topology, so that baises my answer.  $\mathbb{R}$, and Euclidean spaces generally, are interesting because you can do calculus on them.  Manifolds are interesting because, not only as OrbiculaR says, you can do standard analysis in a chart, but you can sensibly extend the concepts of analysis to the entire manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do analysis in the p-adics, so you might wonder why aren't studying p-adic manifolds.  (I don't actually know.)&lt;/p&gt;&#xA;" OwnerUserId="4074" LastActivityDate="2010-12-08T06:53:06.703" CommentCount="1" />
  <row Id="13497" PostTypeId="2" ParentId="10593" CreationDate="2010-12-08T07:07:36.730" Score="1" ViewCount="0" Body="&lt;p&gt;For the 2-sphere, it isn't hard to show it directly.  However, in general, you'll need more powerful tools.  A manifold $M$ is simply-connected if it is path-connected and if $\pi_1(M) = 1$.  The standard tool to compute $\pi_1$ is the van Kampen theorem.&lt;/p&gt;&#xA;" OwnerUserId="4074" LastActivityDate="2010-12-08T07:07:36.730" />
  <row Id="13498" PostTypeId="2" ParentId="13412" CreationDate="2010-12-08T07:17:14.550" Score="2" ViewCount="0" Body="&lt;p&gt;The Zero-Product Property ($ab=0\implies a=0\text{ or }b=0$) is the mechanics behind going from $(7y-1)(y-7)=0$ to $y=\frac{1}{7}$ or $y=7$.  There is no directly analogous property for inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method that I'd suggest for examining inequalities that compare an expression to 0 is the boundary algorithm—find the values that make the expression equal to 0, which are the boundary points of a set of intervals on the number line, then test each interval to see if it satisfies the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific example, the boundary points are $y=\frac{1}{7}$ and $y=7$, so test some value of $y$ below $y=\frac{1}{7}$ (for example, $y=0$), some value of $y$ between $y=\frac{1}{7}$ and $y=7$ (for example, $y=\frac{1}{2}$), and some value of $y$ above $y=7$ (for example, $y=10$).  The original inequality is false below $y=\frac{1}{7}$ and above $y=7$ and true between $y=\frac{1}{7}$ and $y=7$.  Since the original inequality included = (it was ≤), the solution includes the boundary points that solved the corresponding equation, so the solution is $\frac{1}{7}\le y\le 7$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T07:17:14.550" />
  <row Id="13499" PostTypeId="2" ParentId="13473" CreationDate="2010-12-08T07:23:42.367" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, there are similar results in the vector-valued case.  Dunford and Schwartz is a standard reference for this kind of thing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notation: $X$ is a fixed compact Hausdorff space.  For a Banach space $Y$, the space of continuous functions from $X$ to $Y$, endowed with the supremum norm coming from the norm of $Y$, I denote by $C(X,Y)$.  For a Banach space $Z$, I denote its dual by $Z&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one way to think about the dual of $C(X,Y)$ for $Y$ a Banach space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An element $\phi$ of $C(X,Y)&amp;#39;$ gives rise to a family of measures on $X$ parametrized by $Y$ in the following way.  Fixing $\xi \in Y$, one can define a linear functional $L_{\phi,\xi}$ on $C(X)$ by sending the function $f$ on $X$, to the value of $\phi$ on the function $X \to Y$ given by $x \mapsto f(x) \xi$.  In symbols:&#xA;$$&#xD;&#xA;L_{\phi,\xi}(f) = \phi(x \mapsto f(x) \xi).&#xD;&#xA;$$&#xD;&#xA;From the usual Riesz theorem, there is then a measure $m_{\phi, \xi}$ defined on the Borel subsets of $X$ satisfying&#xA;$$&#xD;&#xA;L_{\xi,\phi}(f) = \int_X f \, dm_{\phi, \xi}.&#xD;&#xA;$$&#xD;&#xA;So from $\phi$ we have produced a family of measures on $X$, one for each $\xi$ in $Y$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define a map $m_{\phi}$ from the Borel subsets of $X$ to $Y&amp;#39;$ as follows: for any Borel subset $E$ of $X$, define $m_{\phi}(E)$ to be the linear functional on $Y$ given by&#xA;$$&#xD;&#xA;m_{\phi}(E)(\xi) = \int_E 1 \, dm_{\phi, \xi}.&#xD;&#xA;$$&#xD;&#xA;The map $m_{\phi}$ has various nice properties (it is a $Y&amp;#39;$-valued analogue of a regular signed Borel measure on $X$).  Since the functions of the form $x \mapsto f(x) \xi$, with $f \in C(X)$ and $\xi \in Y$, are dense in $C(X,Y)$, it is easy to show that $\phi$ is uniquely determined by $m_{\phi}$.  (The intuition is to think of $\phi$ as coming from $m_{\phi}$ as follows: for each $f \in C(X,Y)$, the number $\phi(f)$ is obtained &quot;by integrating, over $X$, the values of $f$ with respect to the $Y&amp;#39;$-valued measure $m_{\phi}$, so that $\phi(f) = \int_X f \, dm_{\phi}$.&quot;  You can think of this just as a formal thing, or, think enough about the integration of vector-valued functions with respect to vector-valued set mappings like $m_{\phi}$ to formalize this and remove the quotation marks.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you can reverse this whole chain of reasoning: starting with a map from the Borel sets of $X$ to $Y&amp;#39;$ with nice enough properties, you can show that it must be $m_{\phi}$ for some $\phi$ in $C(X,Y)&amp;#39;$.  There is a natural notion of norm for these things (a &quot;variation&quot; norm) and it turns out to coincide with the norm you'd get from $C(X,Y)&amp;#39;$.  So the dual of $C(X,Y)$, in this picture, is a space of nicely behaved $Y&amp;#39;$-valued mappings on the Borel subsets of $X$, with a certain variation norm.   When $Y$ is the scalars this is turns into the original Riesz theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally you can think of any bounded linear map from $C(X,Y)$ into a Banach space $Z$ in similar terms, but things get more complicated (the &quot;measure-like&quot; things you integrate over $X$ to represent maps $C(X,Y) \to Z$ take values in the linear operators from $Y$ to $Z''$).  You can also weaken various hypotheses here (e.g. you can drop the compactness hypothesis on $X$, or replace $Y$ with a more general topological linear space, provided that you are willing to make additional complicated hypotheses in order to state a decent theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another direction you can go.  If $X$ is compact Hausdorff and $Y$ is a Banach space, the space $C(X,Y)$ is isometrically isomorphic to a certain tensor product, namely the &lt;em&gt;injective Banach space tensor product&lt;/em&gt;, of $C(X)$ with $Y$.  So identifying the dual of $C(X,Y)$ is a special case of identifying the dual of an injective tensor product $A \otimes_i B$ of Banach spaces $A$ and $B$.  The dual of this tensor product has various characterizations.  One is in terms of Borel measures on the Cartesian product of the compact topological spaces $(A&amp;#39;)_1$ and $(B&amp;#39;)_1$ (the unit balls of the duals of $A$ and $B$, given the weak-$*$ topology).  Any book on Banach spaces that discusses the tensor product theory will have theorems about the injective Banach space tensor product and how duality interacts with it.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorUserId="3840" LastEditorDisplayName="" LastEditDate="2010-12-08T07:30:37.400" LastActivityDate="2010-12-08T07:30:37.400" CommentCount="4" />
  <row Id="13500" PostTypeId="1" AcceptedAnswerId="13527" CreationDate="2010-12-08T07:31:34.580" Score="4" ViewCount="173" Body="&lt;p&gt;I was Looking at another questions title, and given the tag of DG, I thought it would read a little more like this one. Or at least that answers to this question would be answers to that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many different techniques one might use to show a manifold is simply conencted, I am only interested in a specific brand of them: those involving Differential Geometry. I borrowed a book from my friends physics professor, I think it was called Comparison theorems in Riemannian Geometry by Cheeger and Ebin, and the content was foreign to me. The goal of the book, if I recall properly, seemed to be proving theorems about $\pi_1(M)$ by examining geometric quantities like curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is amazing to me, and I am curious what the main idea could be. I currently only know of one relationship between geometric quantities and homotopy invariants, that is the fabled Chern-Weil theory. (fabled because I don't understand it... yet)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the precise statements of such types  of results?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which parts of the hypothesis do what? (for example what does compactness help you do here since it doesn't do a whole lot on the homotopy side(maybe that is wrong))&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other than other evidence/theorems, like Gauss-Bonnet, why would someone expect such results?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we get interesting results going the other way? How does $\pi_1$ affect things like curvature?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we say anything about higher homotopy groups?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;thanks for your patience!&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-09T01:09:15.873" Title="How to prove a manifold is simply connected?... using geometry" Tags="&lt;differential-geometry&gt;&lt;algebraic-topology&gt;&lt;homotopy-theory&gt;&lt;curvature&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13501" PostTypeId="2" ParentId="13500" CreationDate="2010-12-08T08:07:43.473" Score="9" ViewCount="0" Body="&lt;p&gt;One example of how geometry affects $\pi_1$: a compact negatively curve manifold will have infinite fundamental group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Negative curvature implies that the universal cover of such a manifold $M$ is diffeomorphic to $\mathbb R^n$.  If $\Gamma = \pi_1(M)$,&#xA;then $M = \mathbb R^n/\Gamma$, and so since $M$ is compact, the group $\Gamma$ must be infinite.  QED&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking more vaguely, negative curvature makes geodesics diverge, and so tends to create a lot of $\pi_1$ (when the manifold is also compact, so that geodesics are forced to wrap around, and so have a chance to close up).  On the other hand, positive curvature focuses geodesics, and so will tend to make $\pi_1$ smaller.&#xA;A result somewhat related to this philosophy (that positive curvature implies simpler topology) is Gromov's theorem that the Betti numbers of  a non-negatively curved manifold are uniformly bounded (the bound depends on the dimension, but nothing else).  (&lt;a href=&quot;http://www.aimath.org/WWN/nnsectcurvature/WilkingSurvey.pdf&quot; rel=&quot;nofollow&quot;&gt;This survey paper by Wilking&lt;/a&gt; gives&#xA;a nice discussion of this result, and many related results about non-negative curvature and its implications for topology.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-08T08:07:43.473" CommentCount="1" />
  <row Id="13502" PostTypeId="2" ParentId="13345" CreationDate="2010-12-08T08:16:16.730" Score="0" ViewCount="0" Body="&lt;p&gt;Hello Pete,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for the (counter-)example. Yes, there's a few basic results (i) if $X$ is compact dann es sei metrisable $\Leftrightarrow$ es sei hausdorff and second countable; (ii) compact metrisable implies polish (obviously!); (iii) any uncountable polish space has cardinality $2^{\aleph_{0}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your claim runs through, provided $Ult(\omega)$ the space of ultrafilters over $\omega$, does indeed have cardinality $&amp;gt;2^{\aleph_{0}}$. However I'll have to run a few calculations to check this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it is certain that it is consistent to assume that compact hausdorff separable does not imply polish ($\Leftrightarrow$ second countable in this context).&lt;/p&gt;&#xA;" OwnerUserId="4537" LastActivityDate="2010-12-08T08:16:16.730" CommentCount="1" />
  <row Id="13503" PostTypeId="1" CreationDate="2010-12-08T08:36:31.227" Score="1" ViewCount="99" Body="&lt;p&gt;I have a circle of radius $r$, and I have a bounding box with 4 sides, $a$, $b$, $c$ and $d$ as illustrated in the below diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1_TelXrp1fyoK1XExygptKEl3ogy_jKPpFCpAhNKYoaA&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few constraints&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r \geq a$$&#xD;&#xA;$$r \geq b$$&#xD;&#xA;$$c &amp;gt;&amp;gt; a$$&#xD;&#xA;$$d &amp;gt;&amp;gt; a$$&#xD;&#xA;$$c &amp;gt;&amp;gt; b$$&#xD;&#xA;$$d &amp;gt;&amp;gt; b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ and $b$ must be perpendicular to $d$ segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another condition is that the the circular sectors must touch one of the long line segment ( the top line segment in this case), base on another long segment ( the bottom line segment in this case) and must not go out from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the algorithm/equation that allows me to generate all the circular sectors that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Touch one of the line segments&lt;/li&gt;&#xA;&lt;li&gt;Won't go out from it&lt;/li&gt;&#xA;&lt;li&gt;Fill the bounding box ( as defined by $a$, $b$, $c$ and $d$) as compactly as possible?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-09T02:25:03.897" LastActivityDate="2011-01-08T02:44:40.303" Title="Algorithm to Pack as Many Circular Sectors into a Rectangular Box " Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13504" PostTypeId="1" AcceptedAnswerId="13512" CreationDate="2010-12-08T08:39:05.407" Score="26" ViewCount="521" Body="&lt;p&gt;This is a very simple question but I believe it's nontrivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if the following is true: If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and $S$ are isomorphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there isn't a proof (or disproof) of the general result, I would be interested to know if there are particular cases when this claim is true. &lt;/p&gt;&#xA;" OwnerUserId="4457" LastActivityDate="2010-12-11T01:52:34.893" Title="Does $R[x] \cong S[x]$ imply $R \cong S$?" Tags="&lt;abstract-algebra&gt;" AnswerCount="4" FavoriteCount="6" />
  <row Id="13506" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T08:49:46.230" Score="24" ViewCount="0" Body="&lt;p&gt;I found &lt;a href=&quot;http://www.springerlink.com/content/n41677w1xj417462/&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Brewer and Rutter that discusses&#xA;related matters. They cite a forthcoming paper by Hochster&#xA;which proves there are non-isomorphic&#xA;commutative integral domains $R$ and $S$ with $R[x]\cong S[x]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hochster's paper is&#xA;M. Hochster, Nonuniqueness of coefficient rings in a polynomial ring&#xA;Proc. Amer. Math. Soc. 34 (1972), 81-82, and is &lt;a href=&quot;http://www.ams.org/journals/proc/1972-034-01/S0002-9939-1972-0294325-3/home.html&quot; rel=&quot;nofollow&quot;&gt;freely available&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-08T09:34:27.707" LastActivityDate="2010-12-08T09:34:27.707" />
  <row Id="13507" PostTypeId="1" AcceptedAnswerId="13517" CreationDate="2010-12-08T09:28:21.233" Score="5" ViewCount="158" Body="&lt;p&gt;I saw an exercise which was meant to find integers all integers $m,n$ satisfying $2m^2 + 5n^2 = 11(mn-11)$. I found them using the factorization $(m-5n)(2m-n)=-11\cdot 11$. However, what kind of methods there are to solve the original problem, find all $(m,n)\in\mathbb{Z}\times\mathbb{Z}$ satisfying $2m^2+5n^3=11(mn-11)$? I haven't solved many cubic Diophantine equations so I was just wondering if there is some birational transformation to convert the equation to a Weierstrass form of an elliptic curve.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-12-08T13:25:25.780" LastActivityDate="2010-12-13T01:58:47.853" Title="Third degree Diophantine equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="13508" PostTypeId="2" ParentId="13503" CreationDate="2010-12-08T12:18:14.137" Score="0" ViewCount="0" Body="&lt;p&gt;There is quite a bit of information on circle packing on &lt;a href=&quot;http://www.packomania.com/&quot; rel=&quot;nofollow&quot;&gt;packomania&lt;/a&gt;, including packings in rectangles. So some of it may be of use to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://www.critticall.com/SQU_cir.html&quot; rel=&quot;nofollow&quot;&gt;another page&lt;/a&gt; on the same theme.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-08T12:18:14.137" CommentCount="2" />
  <row Id="13509" PostTypeId="1" CreationDate="2010-12-08T13:37:02.357" Score="0" ViewCount="24" Body="&lt;p&gt;Let $X$ be a compact orientable surface, and let $x \in X$. Let $P_{x,x}X$ be the space of loops in $X$ based at $x$.  Then we have the pairing of the Chen complex with the group algebra for the fundamental group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \int: H^0(Ch^{\leq n}(P_{x,x}X), \mathbb Q) \otimes \mathbb{Q}[\pi_1(X,x)]/J_x^{n+1} \to  \mathbb{C} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a nice theorem; but here the choice of a base point is involved. Can a more general statement can be made without preference to any particular basepoint? For instance, considering the fundamental groupoid instead of the fundamental group?&lt;/p&gt;&#xA;" OwnerUserId="4543" LastActivityDate="2010-12-08T13:37:02.357" Title="Base-point free version of Chen's theorem" Tags="&lt;algebraic-topology&gt;" />
  <row Id="13510" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T13:46:14.147" Score="0" ViewCount="0" Body="&lt;p&gt;Along the lines of &quot;particular cases where the claim is true,&quot; if $R$ and $S$ are fields, then they must be isomorphic since they are distinguishable as the units (excepting zero of course) of $R[x]$ and $S[x]$, respectively, and an isomorphism conserves units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally if we only know that one of R or S is a field, the claim is still true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $R$ is a field and $R[X] \cong S[Y]$, where $X,Y$ are distinct indeterminates for clarity, and $f:R[X] \rightarrow S[Y]$ is an isomorphism.  Since the nonzero elements of $f(R)$ are units in $S[Y]$, they must be degree zero, i.e. elements of $S$.  Also the inverse image of $S$ is a subring of Euclidean domain $R[X]$, so $S$ is an integral domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that $f(R) \subseteq S$, we need only show $f$ maps $R$ onto $S$.  Suppose $s \in S$.  Then there exists polynomial $p(X) \in R[X]$ s.t. $f(p(X)) = s$.  Let $p(X) = \sum_{i=0}^n \enspace r_i X^i$ where coefficients $r_i \in R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $f(X) = q(Y)$ for some polynomial $q(Y) \in S[Y]$, and degree of $q(Y)$ must be positive for $f(R[X]) = S[Y]$.  Appy $f$ to $p(X)$ and compare degrees:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s = \sum_{i=0}^n \enspace f(r_i) q(Y)^i$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $S$ is an integral domain, the degrees of $q(Y)^i$ are positive for $i \gt 0$.  Thus the only nonzero coefficient of $p(X)$ is $r_0$, which shows $f(r_0) = s$.  Therefore $f$ maps $R$ onto $S$ and $R \cong S$.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-11T01:52:34.893" LastActivityDate="2010-12-11T01:52:34.893" />
  <row Id="13511" PostTypeId="1" AcceptedAnswerId="13561" CreationDate="2010-12-08T13:52:27.997" Score="1" ViewCount="60" Body="&lt;p&gt;Hello All, I have just discussed with student and he told me him task, and I think it is interesting.&#xA;The task.&#xA;There are file with points like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Point0: x=1; y=4;&#xA;Point1: x=199; y=45;&#xA;Point2: x=42; y=333;&#xA;Point3: x=444; y=444;&#xA;...&#xA;PointN: x=nnn; y=mmm;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You should find polygons and draw them. Each polygon present as internal&#xA;I mean something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;---------&#xA;| -----  |&#xA;| |    | |&#xA;| |----| | &#xA;|        |&#xA;|--------|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And question what algorithm can you advice to use in this case?&#xA;I understand this is from graph theory , but want opinion of other.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4544" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-08T13:57:41.727" LastActivityDate="2010-12-08T20:59:42.807" Title="Algorithm to find figures using input points. " Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="13512" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T13:54:20.233" Score="15" ViewCount="0" Body="&lt;p&gt;Here is a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R=\frac{\mathbb{C}[x,y,z]}{(xy(1-z^2))}$, $S=\frac{\mathbb{C}[x,y,z]}{(x^2y(1-z^2))}$. Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T00:18:05.843" LastActivityDate="2010-12-09T00:18:05.843" />
  <row Id="13513" PostTypeId="2" ParentId="13500" CreationDate="2010-12-08T14:00:17.387" Score="3" ViewCount="0" Body="&lt;p&gt;Brendle and Schoen use Ricci flow to prove a smooth version of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere_theorem&quot; rel=&quot;nofollow&quot;&gt;Sphere Theorem&lt;/a&gt;: An n-manifold with pinched positive curvature is &lt;em&gt;diffeomorphic&lt;/em&gt; to a quotient of the $n$-sphere. (In particular, the fundamental group is finite.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very much in the other direction (ie regarding part four of your question) &lt;a href=&quot;http://www.math.uchicago.edu/~shmuel/&quot; rel=&quot;nofollow&quot;&gt;Shmuel Weinberger&lt;/a&gt; and co-authors have results of the form: Suppose that $M$ is a compact Riemannian manifold so that $\pi_1(M)$ has unsolvable word problem.  Then the set of geodesics in $M$ is &quot;wildly complicated&quot;.  Some of this material is discussed in his book &quot;Computers, Rigidity, and Moduli: The Large Scale Geometry of Riemannian Moduli Space&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastEditorUserId="1307" LastEditorDisplayName="" LastEditDate="2010-12-08T15:35:31.370" LastActivityDate="2010-12-08T15:35:31.370" CommentCount="5" />
  <row Id="13514" PostTypeId="1" AcceptedAnswerId="13516" CreationDate="2010-12-08T14:02:13.813" Score="4" ViewCount="109" Body="&lt;p&gt;In &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568810938&quot; rel=&quot;nofollow&quot;&gt;Goldstern and Judah's The Incompleteness Phenomenon&lt;/a&gt; we are asked to prove that any model of the first two Peano Axioms: &#xA;$$\forall x [Sx\neq0]$$  $$\forall x\forall y[Sx=Sy\implies x=y]$$&#xD;&#xA;must be infinite.  S is a unary function intended as successor.  If the model is finite there must be $m$ and $n$ with $S^m0=S^n0$ where the superscript represents iteration.  Then $|n-m|$ applications of the second axiom lead to a contradiction with the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are clearly arguing in the metatheory, as our axioms have no ideas of infinite sets.  Is there a clear definition of what axioms and logic are allowed here?  Can you suggest a book on the subject?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-08T14:44:14.430" LastActivityDate="2010-12-12T07:14:34.147" Title="Logic in the metatheory" Tags="&lt;logic&gt;&lt;reference-request&gt;&lt;model-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13515" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T14:34:34.553" Score="0" ViewCount="0" Body="&lt;p&gt;This is not an answer, since the problem statement is kind of vague, here is what I assume it is.Assuming a polygon immediately inside a polygon to be the child of the polygon (and similarly for parent)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;maximize the number of points utilized to draw polygons such that&lt;br&gt;&lt;br&gt;&#xA;1. each polygon contains only one child polygon (no siblings)&lt;br&gt;&#xA;2. each polygon has only one immediate parent (doesn't lie in the intersection of 2 or more parents)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, outermost and innermost polygons do not need to satify conditions 2 and 1 respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again J.M.'s question is important - do you mean polygons or quadrilateral?&lt;/p&gt;&#xA;" OwnerUserId="4545" LastEditorUserId="4545" LastEditorDisplayName="" LastEditDate="2010-12-08T14:42:57.177" LastActivityDate="2010-12-08T14:42:57.177" CommentCount="2" />
  <row Id="13516" PostTypeId="2" ParentId="13514" CreationDate="2010-12-08T14:38:09.193" Score="4" ViewCount="0" Body="&lt;p&gt;There are two ways to handle this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common way is simply to assume some form of set theory in the metatheory, e.g. ZFC. Then you can use all the normal semantic and model-theoretic methods that you are used to. The advantages of this method are that it matches the way that mathematicians actually think about things, and that it allows us to use all the theorems we're used to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to reinterpret claims made in the metatheory as syntactic claims. For example, the claim &quot;the model must be infinite&quot; can be reinterpreted to mean that the theory itself (!) proves each of the sentences&#xA;$$ \lnot (\exists c_0) (\forall x) (x = c_0) $$&#xD;&#xA;$$ \lnot (\exists c_0, c_1) (\forall x) (x = c_0 \lor x = c_1) $$&#xD;&#xA;$$ \lnot (\exists c_0, c_1, c_2) (\forall x) (x = c_0 \lor x = c_1 \lor c = c_2) $$&#xD;&#xA;and more generally, for each $k &amp;gt; 0$, &#xA;$$ &#xD;&#xA;\lnot (\exists c_0, \ldots, c_k ) (\forall x) (x = c_0 \lor \cdots \lor c = c_k) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that each of these sentences is provable in the theory can be verified using a very weak metatheory. There's a simple and uniform way, for any fixed $k$, to generate a proof from your two axioms for that particular $k$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reinterpretation process is well known to logicians, and it can be applied very generally. The main limitation is whether the property in question can be stated in the language of the theory itself (possibly as an infinite set of sentences, as with the example above). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gödel's completeness theorem says that if a sentence in the language of a theory is true (semantically) in every model of the theory, then that sentence is provable (syntactically) within the theory itself. So we could, if we wanted, ignore the &quot;true&quot; part in those cases and focus just on the &quot;provable&quot; part. The cost of doing that is that it arguably misses the point, since we usually think semantically about models, and that it does not apply to properties of models that cannot be expressed within the language of the theory itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this reinterpretation process is usually very routine, it's rare for logicians to bother to mention it, even though we're well aware it could be done. We just write in the normal model-theoretic (set theoretic) way unless there is a particular reason that we need to focus on the syntactic interpretation. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-08T14:38:09.193" CommentCount="1" />
  <row Id="13517" PostTypeId="2" ParentId="13507" CreationDate="2010-12-08T14:45:08.237" Score="3" ViewCount="0" Body="&lt;p&gt;The equation $2y^2+5x^3=11(xy−11)$ describes an elliptic curve. Given any elliptic curve, you can perform a change of variables to put it into Weierstrass form, and if the field of definition has characteristic different from 2 or 3, you can even put it into the form $y^2 = x^3 + ax + b$. In this particular case, to find the transformation is easy: first scale $x$ and $y$ to get rid of make the coefficients of $x^3$ and of $y^2$ equal to 1. Then write $y&amp;#39; = y +  \alpha x$ to get rid of the $xy$-term. This will introduce an $x^2$ term. So now, translate $x$ to get rid of the $x^2$ term (I haven't actually done the computation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important thing to note is that while the notion of rational points does not depend on the model over $\mathbb{Q}$ that you are working with (as long as you only change variables over $\mathbb{Q}$), the notion of integral points does depend on the integral model. A theorem of Siegel says that on any Weierstrass model of an elliptic curve, there are only finitely many integral points, and there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve#Integral_points&quot; rel=&quot;nofollow&quot;&gt;bounds on their height&lt;/a&gt; in terms of the coefficients of the model. But as far as I know, to actually find the points comes down to brute force methods: e.g. if you manage to compute generators of the Mordell-Weil group, i.e. of the group of the rational points, and the torsion subgroup, then you just check all possible linear combinations up to the given bound. There are better methods using elliptic logarithms, but they are theoretically more involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: some more info on your concrete curve:&#xA;To get a Weierstrass model you can replace $y$ by $y&amp;#39;/20$ and $x$ by $-x&amp;#39;/10$ in the original equation, resulting in $y&amp;#39;^2+11x&amp;#39;y&amp;#39; = x&amp;#39;^3 - 2^35^211^2$, confirming your hint. The elliptic curve has rank 1, as you say, and no torsion. The point $P=(22,-88)$ generates all the rational points on the curve under the group law. Integral solutions of the original equation correspond to points on the Weierstrass model with integral $x$-coordinate divisible by 10 and integral $y$-coordinate divisible by 20 (see our transformation). So the naive way of ruling out integral solutions to the original equation is to check all multiples of the generator $P=(22,-88)$ under the group law up to the given bound and convince yourself that no point satisfies the divisibility criteria. However, the bound of Baker, referred to in the wikipedia article, is huge and the computation might not actually be feasible. A possibly more promising approach is to write down the polynomial that computes $Q\mapsto Q\oplus P$ and see whether this operation always introduces higher and higher denominators that never cancel the numerators.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-13T01:58:47.853" LastActivityDate="2010-12-13T01:58:47.853" CommentCount="3" />
  <row Id="13518" PostTypeId="1" AcceptedAnswerId="13540" CreationDate="2010-12-08T14:45:19.220" Score="4" ViewCount="161" Body="&lt;p&gt;Can someone suggest an idea for a school project in knot-theory for a 13 year old?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4545" LastActivityDate="2010-12-08T18:44:02.897" Title="School project in knot theory" Tags="&lt;middle-school&gt;&lt;knot-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13519" PostTypeId="1" AcceptedAnswerId="14028" CreationDate="2010-12-08T14:46:15.833" Score="7" ViewCount="179" Body="&lt;p&gt;My basic question is: &quot;Do longer tautologies take longer to prove?&quot; But obviously this is underdetermined. If you are allowed an inference rule &quot;Tautological Implication&quot; then any tautology has a 1 line proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's say we're working in natural deduction, is it true that longer tautologies (tautologies with more letters and connectives in them) take more proof steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is still not quite a good enough question, since we can add superfluous steps to a proof and still prove the result. So the question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Natural Deduction, is the shortest proof of a tautology related to its length? Or perhaps: does a shorter tautology always admit a shorter proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how robust is this result across different sets of connectives/inference rules?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-12-12T08:15:30.170" Title="Is there a connection between length of sentence and length of proof?" Tags="&lt;logic&gt;&lt;proof&gt;&lt;proof-theory&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="13520" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T14:55:20.593" Score="0" ViewCount="0" Body="&lt;p&gt;If what I posted below is a correct assumption, then you could think of a solution on these lines.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;find centroid of all points in set S (expensive)&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;obtain farthest points from the centroid. &lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;use these to draw the outermost polygon. &lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;remove from this points from set S&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;repeat step 1 until (no more inner polygons can be drawn)&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4545" LastActivityDate="2010-12-08T14:55:20.593" />
  <row Id="13521" PostTypeId="2" ParentId="13519" CreationDate="2010-12-08T15:00:13.520" Score="3" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;does a shorter tautology always admit a shorter proof?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it seems like there should be a tautology whose statement is longer than fermats last theorem but whose shortest proof is shorter&lt;/p&gt;&#xA;" OwnerUserId="4488" LastActivityDate="2010-12-08T15:00:13.520" />
  <row Id="13522" PostTypeId="2" ParentId="13504" CreationDate="2010-12-08T15:07:15.930" Score="6" ViewCount="0" Body="&lt;p&gt;There's been much work on this problem since the mentioned seminal work in the early seventies. Searching on the buzzword &quot;stably equivalent&quot; should help locate most of it. Below is a helpful introduction from Jon L Johnson: &lt;a href=&quot;http://books.google.com/books?id=m7MKj5tLAJwC&amp;amp;lpg=PA277&amp;amp;ots=Wc0BuJo04-&amp;amp;dq=brewer%20rutter%20hochster&amp;amp;pg=PA277#v=onepage&amp;amp;q=picard&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Cancellation and Prime Spectra&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/AiI7j.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/IQGPZ.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-08T15:07:15.930" />
  <row Id="13523" PostTypeId="2" ParentId="13518" CreationDate="2010-12-08T15:11:16.050" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Borromean_rings&quot; rel=&quot;nofollow&quot;&gt;Borromean rings&lt;/a&gt;? Can the same thing be done with more than three loops?&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2010-12-08T15:11:16.050" />
  <row Id="13524" PostTypeId="2" ParentId="13519" CreationDate="2010-12-08T15:15:17.363" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose that $\phi$ is any sentence of the form $\lnot (\psi \land \lnot \psi)$. Then there is a pattern for a very short proof of $\phi$ that works for any formula $\psi$. I can't format it here, but the idea is to obtain $\bot$ from $\psi \land \lnot \psi$ in a fixed number of steps, and then apply the rule for proof by contradiction to get $\phi$. The original contradiction will take three steps in natural deduction: two are conjunction elimination and the third derives $\bot$.  So the entire proof pattern has four steps in natural deduction regardless of $\psi$.  If you tweak the setup for natural deduction the number of steps may go up to five or six, but it will be constant in any case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether there is any system where longer tautologies always take longer proofs and where length is measured in proof steps. Certainly it would have to have modified versions of the rules relating to $\bot$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to change the question is to change the way you measure the length of a proof. If the size of the proof is the total symbol count then there is clearly a sense in which longer facts require longer proofs.&lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-09T17:39:42.983" LastActivityDate="2010-12-09T17:39:42.983" CommentCount="1" />
  <row Id="13525" PostTypeId="1" AcceptedAnswerId="13758" CreationDate="2010-12-08T15:21:17.627" Score="3" ViewCount="144" Body="&lt;p&gt;I just attended a seminar on the chromatic filtration of the Burnside ring.  I understood it relatively well, but at no point did anyone give an explicit definition of what a chromatic filtration actually is, and I have had some trouble finding one.  Assuming I have the necessary categorical background, i.e. knowledge of filtrations, etc. can anyone give me an explanation of what makes a filtration &quot;chromatic&quot;?&lt;/p&gt;&#xA;" OwnerUserId="3595" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-13T05:43:07.610" LastActivityDate="2010-12-16T21:53:18.190" Title="Chromatic Filtration of Burnside Ring" Tags="&lt;group-theory&gt;&lt;algebraic-topology&gt;&lt;category-theory&gt;&lt;homotopy-theory&gt;&lt;algebraic-k-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="13526" PostTypeId="2" ParentId="13424" CreationDate="2010-12-08T15:26:44.537" Score="0" ViewCount="0" Body="&lt;p&gt;Im trying to work out a similair question, but I am unsure as to how to expand it as a power series?&#xA;I have taken the log of my function and differentiated it, but what is my final answer for the power series expansion? &lt;/p&gt;&#xA;" OwnerUserId="4547" LastActivityDate="2010-12-08T15:26:44.537" CommentCount="3" />
  <row Id="13527" PostTypeId="2" ParentId="13500" CreationDate="2010-12-08T15:32:46.943" Score="5" ViewCount="0" Body="&lt;p&gt;One also has the classical Bonnet-Myers and Synge theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bonnet-Myers theorem states that if $M$ is complete with Ricci curvature bounded below $\delta &amp;gt; 0$, then $M$ is compact with finite fundamental group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the proof is the opposite of the one in Matt E's answer.  In positive (sectional) curvature, geodesics tend to get stuck together.  One uses this to show that geodesics past a certain length (something like $\pi/\delta^2$) cannot continue to minimize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that if $B_r(0)\subseteq T_p M$ is any closed ball with radius $r &amp;gt; \pi/\delta^2$, then $exp(B_r(0)) = M$.  In particular, you've written $M$ as the compact image of a continuous function so it's compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem about the fundamental group is an immediate corollary:  look at the universal cover $\tilde{M}$.  One can &quot;pull back&quot; the metric so that the covering map is a local isometry.  Hence, the same curvature estimates apply to $\tilde{M}$, so it, too, must be compact.  But a compact manifold can only finitely cover something, so the fundamental group of $M$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To push this to Ricci, one uses a nice trick:  if Ric &gt; 0, then the sectional curvature in some directions must be &gt; 0, and one uses these directions only to show geodesics eventually stop minimizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Synge theorems have a different style of proof.  Here's one version of the theorem:  suppose $M$ is compact with positive curvature, and $f:M\rightarrow M$ is an isometry.  Suppose dim(M) is even and $f$ is orientation preserving OR dim(M) is odd and $f$ is orientation reversing.  Then $f$ has a fixed point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is straightforward:  suppose not.  Choose $p\in M$ with  $d(p,f(p))$ as small as possible.  Choose a minimal geodesic from $p$ to $f(p)$.  One then computes variations of the geodesic (this uses the parity of the dimension, if I recall correctly), and shows that some nearby geodesic is smaller, contradicting your choice of $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a corollary, one learns that the fundamental group of an even dimension positively curved compact manifold is either trivial or $mathbb{Z}/2\mathbb{Z}$.  For, suppose $M$ satisfies all the hypothesis.  Look at the deck group action on $\tilde{M}$.  If it's orientation preserving, it has a fixed point, but the only deck transformation with a fixed point is the identity.  If it's orientation reversing, then it's square is orientation preserving, and hence is the identity, so all elements are of order 2 in $\pi_1$.  If there are 2 orientation reversing maps, then their product is orientation preserving and hence the identity, so there is at most one element of order 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, in odd dimensions, the corollary is that $M$ must be orientable, since the deck group must acts on $\tilde{M}$ by orientation preserving maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to just randomly answer one of your other questions, curvature can affect higher homotopy groups.  As Matt E pointed out, in negative (really, nonpositive) curvature, the universal cover is $\mathbb{R}^n$, implying all higher homotopy groups vanish.  In positive curvature, this is a very important and currently unsolved problem.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-12-09T01:09:15.873" LastActivityDate="2010-12-09T01:09:15.873" CommentCount="1" />
  <row Id="13528" PostTypeId="1" CreationDate="2010-12-08T15:38:54.920" Score="0" ViewCount="77" Body="&lt;p&gt;I have been asked to expand (1-2t)^(-n/2). So far I have found the log of this function and then differentiated twice in order to find my mean and variance as n and 2n respectively. &#xA;How do I write my final answer as a power series in t as far as the term in t^2?&lt;/p&gt;&#xA;" OwnerUserId="4547" LastActivityDate="2010-12-08T17:56:55.463" Title="Expansion of an MGF as a power series" Tags="&lt;statistics&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13529" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T15:56:41.540" Score="0" ViewCount="0" Body="&lt;p&gt;There are several algorithms for curve reconstruction from point samples. See for instance &lt;a href=&quot;http://www.cse.ohio-state.edu/~tamaldey/curverecon.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.cse.ohio-state.edu/~tamaldey/curverecon.htm&lt;/a&gt; and &lt;a href=&quot;http://valis.cs.uiuc.edu/~sariel/research/CG/applets/Crust/Crust.html&quot; rel=&quot;nofollow&quot;&gt;http://valis.cs.uiuc.edu/~sariel/research/CG/applets/Crust/Crust.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-08T15:56:41.540" />
  <row Id="13530" PostTypeId="2" ParentId="13528" CreationDate="2010-12-08T16:08:40.663" Score="0" ViewCount="0" Body="&lt;p&gt;If you want to expand your function in a power series on the argument $t$ you can write its &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Taylor Polynomials&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-12-08T16:08:40.663" />
  <row Id="13531" PostTypeId="2" ParentId="13525" CreationDate="2010-12-08T16:14:48.470" Score="2" ViewCount="0" Body="&lt;p&gt;It appears to be non-technical, and roughly equivalent to &quot;periodic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An explanation is given &lt;a href=&quot;http://books.google.com/books?id=RA18_pxdPK4C&amp;amp;pg=PA23&quot; rel=&quot;nofollow&quot;&gt;by Ravenel&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;original paper&quot; appears to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Miller, Haynes R.; Ravenel, Douglas C.; Wilson, W. Stephen. &quot;Periodic phenomena in the Adams-Novikov spectral sequence.&quot;&#xA;  Ann. Math. (2) 106 (1977), no. 3, 469–516.&#xA;  MR&lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=458423&quot; rel=&quot;nofollow&quot;&gt;458423&lt;/a&gt;&#xA;  &lt;a href=&quot;http://www.jstor.org/stable/1971064&quot; rel=&quot;nofollow&quot;&gt;JSTOR&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but the term is not used there.  These sorts of things were for homotopy of spheres, a major topic in algebraic topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the filtration discussed for the Burnside ring had a similar ambience.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-08T16:14:48.470" />
  <row Id="13533" PostTypeId="2" ParentId="13518" CreationDate="2010-12-08T16:59:27.600" Score="3" ViewCount="0" Body="&lt;p&gt;There are some elementary ideas in Colin Adam's book: The Knot Book that could be adapted for investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem to think about would be the number of sides for regions that occur when the same knot (different drawings) is considered embedded in the plane as a 4-valent (degree 4) graph. The number of faces with k sides (knot drawings can have 1-gons, 2-gons, etc.) for such a drawing obeys a relationship that can be derived from Euler's polyhedral formula using the fact that the graph of a projection of a knot is 4-valent.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-08T16:59:27.600" />
  <row Id="13534" PostTypeId="1" AcceptedAnswerId="13537" CreationDate="2010-12-08T17:47:04.250" Score="2" ViewCount="119" Body="&lt;p&gt;We have a group $G$ of order $2p$, $p$ an odd prime, which has a nontrivial center. We wish to show the group is cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple exercise which can be instantly killed using heavy cannons. However, since I was asked about it by students which barely reached quotients, I wonder what simple ways are to prove this; I couldn't find any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T10:17:29.643" LastActivityDate="2010-12-09T10:17:29.643" Title="Simple proof that group of order $2p$ with nontrivial center is cyclic?" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="13535" PostTypeId="2" ParentId="13528" CreationDate="2010-12-08T17:56:55.463" Score="1" ViewCount="0" Body="&lt;p&gt;You have already done most of the work. If the mean $\mu&amp;#39;_1$ and variance $\sigma^2$ are equal to $n$ and $2n$, respectively, then the second moment $\mu&amp;#39;_2$ is given by $\mu&amp;#39;_2 = \sigma^2 + (\mu&amp;#39;_1)^2 = n(n+2)$. Now, the moment-generating function can be expanded as follows:&#xA;$$&#xD;&#xA;M(t) = 1 + \mu&amp;#39; _1 t + \frac{1}{{2!}}\mu&amp;#39; _2 t^2  + \cdots = 1 + nt + \frac{{n(n + 2)}}{2}t^2  \cdots.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-08T17:56:55.463" />
  <row Id="13536" PostTypeId="1" CreationDate="2010-12-08T17:57:49.010" Score="0" ViewCount="77" Body="&lt;p&gt;I have a 3d object which in its simplest form consists of an origin in 3d space and a set of vertices that are all local to this origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then transform this 3d origin into 2d camera coordinates using a perspective transform, but I need some way to also transform the local vertices of this object which has the effect of moving this 3d object from 3d world coordinates into 3d camera coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How can I determine the scale of an object based on it's distance from the viewer/camera?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense, the title is probably the most concise explanation I can give.  Any help would be very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4549" LastActivityDate="2011-01-10T01:44:42.390" Title="Determine scale factor of an object given its distance from a viewer/camera" Tags="&lt;geometry&gt;&lt;transformation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13537" PostTypeId="2" ParentId="13534" CreationDate="2010-12-08T18:19:40.770" Score="4" ViewCount="0" Body="&lt;p&gt;Let $g$ be a nontrivial element of the center.  If $g$ generates the whole group, we are done; otherwise $g$ generates a cyclic subgroup of order $2$ (resp. $p$).  Hence $G/\langle g \rangle$ is a cyclic subgroup of order $p$ (resp. $2$).  It follows that $G$ is generated by $g$ and one other element $h$ such that $h^p = g^k$ for some $k$ (resp. $h^2 = g^k$ for some $k$).  Since $g$ is in the center, $gh = hg$.  But this implies $h$ is also in the center.  Hence $G$ is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $h$ generates the whole group we are done.  Otherwise $h$ has order $2$ (resp. $p$), since it has order dividing $2p$ and cannot have order $p$ (resp. $2$), and $G \simeq C_2 \times C_p$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T18:19:40.770" CommentCount="3" />
  <row Id="13539" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T18:24:04.203" Score="0" ViewCount="0" Body="&lt;p&gt;I will use &lt;a href=&quot;http://en.wikipedia.org/wiki/Convex_hull&quot; rel=&quot;nofollow&quot;&gt;Convex_hull&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to all for answers.&lt;/p&gt;&#xA;" OwnerUserId="4544" LastActivityDate="2010-12-08T18:24:04.203" CommentCount="1" />
  <row Id="13540" PostTypeId="2" ParentId="13518" CreationDate="2010-12-08T18:31:44.743" Score="3" ViewCount="0" Body="&lt;p&gt;You may wish to take a look at the readable and well-illustrated &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1931914222&quot; rel=&quot;nofollow&quot;&gt;Why Knot?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something to consider about knots is how 'basic' they are -- children will often learn to construct simple knots before learning arithmetic. This demonstrates an ability to manipulate (and distinguish between classes of) closed curves in three dimensions, and amounts to an intuitive theory of topology. We can use this as a staging ground to ask more interesting and difficult questions about knots and their classification.&lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="4300" LastEditorDisplayName="" LastEditDate="2010-12-08T18:44:02.897" LastActivityDate="2010-12-08T18:44:02.897" CommentCount="1" />
  <row Id="13541" PostTypeId="2" ParentId="13534" CreationDate="2010-12-08T18:38:03.697" Score="2" ViewCount="0" Body="&lt;p&gt;Qiaochu's answer is simpler, but the following almost classifies all groups of order $2p$, so I'll leave it here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 1: The order of an element divides order of the group. &#xA;No new proof, but easy enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 2: There exists an element of order $p$. &#xA;Can circumvent usual proof as follows: If there is an element of order $2p$ G is cyclic. If no element of order $p$ or $2p$ then $a^2=e$ for all $a$, so $G$ is abelian ($abab=e$, so $ab=b^{-1}a^{-1}=ba$). Then either use the fact that you know all abelian groups, or the fact that then $G$ is a vector space over the field $Z_2$, so has basis and so has order $2^k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call this element $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lemma 3: Any subgroup of index 2 is normal. &#xA;There are two cosets, and one of them is the subgroup itself. So right and left cosets coincide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this to the subgroup generated by $a$, for $b$ not a power of $a$, we have $bab^{-1}=a^m$.  The group contains only $a^k$ and $ba^l$'s, which are all distinct. If $m=1$ they all commute and $ba$ generates the cyclic group. If $m\neq1$ there is no center. (The only step in classification left is to show that different $m \neq 1$ lead to the same group, which is not hard either.)&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-12-08T18:38:03.697" CommentCount="2" />
  <row Id="13542" PostTypeId="1" AcceptedAnswerId="13544" CreationDate="2010-12-08T19:02:09.660" Score="2" ViewCount="64" Body="&lt;p&gt;Suppose that $f(z)$ is complex analytic on $|z| \leq 1$ and satisfies&#xA;$|f(z)| &amp;lt; 1$ for $|z|=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Prove that the equation $f(z)=z$ has exactly one root&#xA;(counting multiplicities) in $|z|&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Prove that if $|z_0| \leq 1$, then the sequence $z_n$ defined recursively&#xA;by $z_n= f(z_{n-1}) , n=1,2,...$, converges to the fixed point of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to prove (a) using Rouche's theorem, but (b) stumps me.&#xA;I know that (b) is true for analytic fuctions such that $f(0)=0$ or&#xA;$|f&amp;#39;(z)|&amp;lt;1$ on the disc, neither of which are necessarily true in general.&#xA;The farthest I was able to get was $|f(z)-z^*|&amp;lt;\frac{1}{1-|z*|}|z-z^*|$,&#xA;where $z^*$ is the fixed point of $f$, but&#xA;$\frac{1}{1-|z^*|}&amp;gt;1$, so I don't think this helps me.&#xA;Can someone please point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="2029" LastActivityDate="2010-12-08T19:12:14.610" Title="Fixed point iteration for analytic functions on the unit disc" Tags="&lt;homework&gt;&lt;complex-analysis&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13543" PostTypeId="1" AcceptedAnswerId="13943" CreationDate="2010-12-08T19:11:47.310" Score="2" ViewCount="235" Body="&lt;p&gt;This question is motivated by my previous &lt;a href=&quot;http://math.stackexchange.com/questions/13344/proof-for-an-integral-involving-sinc-function&quot;&gt;post&lt;/a&gt; about sinc function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire (i.e. analytic everywhere)&#xA;function $f(x)$ on $\mathbb{R}$ such that $$\int_0^\infty f(x) dx=\int_0^\infty f(x)^2 dx$$&#xD;&#xA;or $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f$ is required only to be continuous, then other examples are possible, e.g. the even extension of the following function:&#xA;$$ f(x)=\left\{\begin{array}{ll} -2(5+\sqrt{65})x^2+(7+\sqrt{65})x-1 &amp;amp; 0\le x\le \frac{1}{2}\\&#xD;&#xA;2(5+\sqrt{65})x^2-(13+3\sqrt{65})x+4+\sqrt{65} &amp;amp; \frac{1}{2}\le x\le 1\\&#xD;&#xA;\frac{1}{x^2} &amp;amp; x\ge 1 \end{array}\right. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As commented below, it turns out that there are easy answers to the above question. AD also showed a function below that also satisfies $$ \sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty f(n)^2=0.  $$&#xD;&#xA;In view of these answers, my question is now revised to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire&#xA;function, $f(x)$ on $\mathbb{R}$ such that $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx=\sum_{-\infty}^\infty f(n) =\sum_{-\infty}^\infty f(n)^2 $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-09T14:55:03.370" LastActivityDate="2010-12-14T02:24:55.977" Title="Are there other analytic functions with this property of sinc function?" Tags="&lt;analysis&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="13544" PostTypeId="2" ParentId="13542" CreationDate="2010-12-08T19:12:14.610" Score="3" ViewCount="0" Body="&lt;p&gt;One can reduce it to the case of $f(0)=0$ (i.e. $0$ is the fixed point) by making a suitable linear fractional transformation of the disk. Namely, if $z^*$ is the fixed point, apply the above argument to $L \circ f \circ L^{-1}$ where $L$ sends $z^* \to 0$ and is a LFT.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-08T19:12:14.610" CommentCount="2" />
  <row Id="13547" PostTypeId="1" AcceptedAnswerId="13550" CreationDate="2010-12-08T19:40:06.827" Score="6" ViewCount="117" Body="&lt;p&gt;When one writes $\zeta_n$ which of the n roots of unity is meant here?  Does it matter? &lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-09T00:52:19.070" Title="When one writes $\zeta_n$ which of the n roots of unity is meant here?" Tags="&lt;number-theory&gt;" AnswerCount="3" />
  <row Id="13548" PostTypeId="2" ParentId="13547" CreationDate="2010-12-08T19:41:45.333" Score="1" ViewCount="0" Body="&lt;p&gt;Probably a primitive one.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-08T19:41:45.333" />
  <row Id="13549" PostTypeId="2" ParentId="13547" CreationDate="2010-12-08T19:42:00.590" Score="5" ViewCount="0" Body="&lt;p&gt;Usually this means $e^{ \frac{2\pi i}{n} }$.  Depending on the context, it can just mean any primitive $n^{th}$ root of unity (and sometimes it doesn't matter which one, since they're all taken to each other under the action of the Galois group).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T19:42:00.590" CommentCount="1" />
  <row Id="13550" PostTypeId="2" ParentId="13547" CreationDate="2010-12-08T19:49:59.853" Score="10" ViewCount="0" Body="&lt;p&gt;The context is important here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is rather standard for $\zeta_n$ to at least denote a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_nth_root_of_unity&quot; rel=&quot;nofollow&quot;&gt;primitive&lt;/a&gt; $n$th root of unity in the algebraic closure of the field $k$ which is currently being considered (this is possible iff the characteristic of $k$ does not divide $n$; in particular it is true for all fields of characteristic zero).  When the characteristic of $k$ does not divide $n$ (i.e., when any primitive $n$th roots of unity exist) there are precisely $\varphi(n)$ primitive $n$th roots of unity, where $\varphi$ is Euler's phi function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a context in which $k$ is a subfield of the complex numbers, it is also rather standard for $\zeta_n$ to denote the specific primitive $n$th root of unity $e^{\frac{2 \pi i}{n}}$, i.e., the one of minimal argument in the complex plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it matter?  For algebraic purposes, probably not: the primitive $n$th roots of unity in $\mathbb{C}$ are algebraically conjugate over $\mathbb{Q}$: i.e., different roots of a common irreducible polynomial over $\mathbb{Q}$, the cyclotomic polynomial $\Phi_n(t)$.  Sometimes in number theory one considers systems of $n$th roots of unity for varying $n$, and in this case it is necessary to make a consistent (in a certain sense) choice of $\zeta_n$'s.  Taking $\zeta_n = e^{\frac{2 \pi i}{n}}$ for all positive integers $n$ is such a consistent choice, but there are (many!) others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are always situations when confusing one complex number for another would lead to trouble, so yes, in principle it might matter, especially in analytic or metric arguments.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-09T00:52:19.070" LastActivityDate="2010-12-09T00:52:19.070" CommentCount="3" />
  <row Id="13551" PostTypeId="1" AcceptedAnswerId="13553" CreationDate="2010-12-08T19:57:48.280" Score="2" ViewCount="138" Body="&lt;p&gt;I have been trying to do this for quite a while, but generally speaking the partially relevant information I could find on the internet only dealt with the question: &quot;How does on convert a recurrence relation into... well, a non-recurrence relation?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start of with a simple example: $f(n) = 34n^3+51n^2+27n+5$. How do we find $f_{n}$? &#xA;I'd really like to see this solved in analogy with the following: Consider $g(n)=n^6$ We can then find the recursion formula: $g_n=((g_{n-1})^{1/6}+1)^6$. What about $f_{n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could generalize this question in a number of ways. For instance, is it (also) possible too turn an infinite polynomial, like the Taylor Series expansion of a trigonometric formula, into a recursion formula? Furthermore, what happens when we allow the coefficients of the polynomial to be real and even complex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus side-question: How, if at all, are &quot;generating functions&quot; useful in this context?&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-08T20:45:40.517" LastActivityDate="2010-12-09T17:48:56.933" Title="Is it possible to convert a polynomial into a recurrence relation? If so, how?" Tags="&lt;functions&gt;&lt;polynomials&gt;&lt;generating-functions&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="13552" PostTypeId="2" ParentId="8808" CreationDate="2010-12-08T20:00:31.727" Score="0" ViewCount="0" Body="&lt;p&gt;In topology people study intensely some naturally occurring abut quite abstract and not &quot;locally Euclidean&quot; spaces. Examples are path spaces(space of paths in a given topological space or manifold) and some natural $H$-spaces in the homotopy category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway there a whole lot of new kind of &quot;geometry&quot; in the field of algebraic geometry, which is not at all Euclidean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But your point is good, that the Euclidean space is quite specially useful in physics. With this in mind, some people are actually doing adelic physics these days.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-08T20:00:31.727" />
  <row Id="13553" PostTypeId="2" ParentId="13551" CreationDate="2010-12-08T20:03:33.653" Score="10" ViewCount="0" Body="&lt;p&gt;Any polynomial $f(n)$ of degree $d$, over an arbitrary commutative ring, satisfies the recursion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{d+1} (-1)^{d+1-k} {d+1 \choose k} f(x+k) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any $x$.  This is an application of the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite differences&lt;/a&gt;.  In terms of generating functions, it says that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} f(n) x^n = \frac{P(x)}{(1 - x)^{d+1}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some polynomial $P(x)$.  This should be thoroughly covered in a book like Wilf's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/DownldGF.html&quot; rel=&quot;nofollow&quot;&gt;generatingfunctionology&lt;/a&gt;; otherwise, I discuss it in my &lt;a href=&quot;http://web.mit.edu/~qchu/Public/TopicsInGF.pdf&quot; rel=&quot;nofollow&quot;&gt;notes on generating functions&lt;/a&gt;.  More generally, if $f(n)$ has a generating function of the form $\frac{P(x)}{Q(x)}$ where $P$ is a polynomial, then the coefficients of $Q$ describe a recurrence that $f(n)$ eventually satisfies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand your more general question.  Do you have a specific application in mind?  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-08T20:11:03.527" LastActivityDate="2010-12-08T20:11:03.527" CommentCount="9" />
  <row Id="13554" PostTypeId="1" AcceptedAnswerId="13565" CreationDate="2010-12-08T20:06:22.827" Score="3" ViewCount="79" Body="&lt;p&gt;If $(X, \lt)$ is a well-ordering I can show by transfinite recursion over the ordinals that the function $f(x) = \text{ran} f |_{\hat{x}}$ exists (where $\hat{x} = \{ y : y \lt x\}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have obtained $f$ this way, Let $V$ be the class of all sets and $F:V \to V$ be a class-function, then there is a unique $G:ON \to V$ where $ON$ is the class of all ordinals such that $F(\alpha) = F(G|_\alpha)$. So I can apply this to get a function $f$ such that $f(x) = F(f|_\hat{x})$. Now I let $F = \{(x, \text{ran} x) : x \in V\}$ and then I get the function as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this should be an isomorphism (order preserving bijection) between $X$ and the set of true initial segments of $X$, $I_X$ ordered by inclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I have $x &amp;lt; y$, then I see that $\text{ran} f|_\hat{x} \subset \text{ran} f|_\hat{y}$. So $f(x) \leq f(y)$. Why do I have $f(x) \neq f(y)$?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T21:04:53.223" LastActivityDate="2010-12-08T21:44:57.627" Title="Recursive definition isomorphism." Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="17" />
  <row Id="13555" PostTypeId="1" AcceptedAnswerId="13588" CreationDate="2010-12-08T20:26:56.343" Score="2" ViewCount="44" Body="&lt;p&gt;The n-dimensional cone, with vertex at the origin, central angle, $\alpha$ and central axis in the direction of the unit vector $\xi$ is defined to be all those points, $x\in {R^n}$ whose dot product with $\xi$ is |$x$|$cos(\alpha)$.  How would I find parametric equations for this surface? &lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-09T00:52:48.167" Title="Parametric Equations for a Hypercone" Tags="&lt;differential-geometry&gt;&lt;parametric&gt;" AnswerCount="1" />
  <row Id="13556" PostTypeId="1" CreationDate="2010-12-08T20:34:38.147" Score="6" ViewCount="127" Body="&lt;p&gt;Given a long exact sequence of the form&#xA;$$&#xD;&#xA;\dots\to A&amp;#39;_n \to B&amp;#39;_n \to C&amp;#39;_n&#xD;&#xA;\,\xrightarrow{\omega_n}\,&#xD;&#xA;A&amp;#39;_{n-1} \to B&amp;#39;_{n-1} \to C&amp;#39;_{n-1}\to \dots\qquad (*)&#xD;&#xA;$$&#xD;&#xA;is there a way to recover a short exact sequence of complexes $\mathcal A=\{A_n,\partial_n^A\}$, $\mathcal B=\{B_n,\partial_n^B\}$, $\mathcal C=\{C_n,\partial_n^C\}$ such that the sequence (*) &quot;is&quot; the long exact sequence in homology induced by&#xA;$$&#xD;&#xA;0\to \mathcal A\to  \mathcal B\to  \mathcal C \to 0&#xD;&#xA;$$&#xD;&#xA;and the morphisms $\omega_n$ are in fact the connection morphisms of that homology? I mean $A&amp;#39;_n\cong H_n(\mathcal A)$ for all $n\ge 0$ and similarly for $B&amp;#39;_n$, $C&amp;#39;_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the answer will be &quot;obviously no&quot;, but then is there a case in which it is possible?&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-09T01:49:07.383" Title="Does every les &quot;in homology&quot; come from a ses of complexes?" Tags="&lt;homological-algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="13557" PostTypeId="2" ParentId="13551" CreationDate="2010-12-08T20:34:50.170" Score="1" ViewCount="0" Body="&lt;p&gt;Expand $f(n+1)$ and you'll get $f(n+1)=f(n)+g(n)$ for some $g$. For $f(n)=34n^3+51n^2+27n+5$, you get $f(n+1)=f(n)+102n^2+204n+112$. &lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-08T20:34:50.170" CommentCount="2" />
  <row Id="13558" PostTypeId="2" ParentId="13551" CreationDate="2010-12-08T20:45:48.547" Score="2" ViewCount="0" Body="&lt;p&gt;Max, I'm just detailing Qiaochu's answer, using Pari/GP.&#xA;First define the function&lt;br&gt;&#xA;$ f(x) = 34*x^3+51*x^2+27*x+5$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the first idea to decompose f into a 2-term-recursion is&lt;br&gt;&#xA;$ f(x)-1*f(x-1) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt; = 102*x^2 + 10 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the result is already a reduced polynomial. Now subtract from this $f(x-1)-f(x-2) $ because this will be a polynomial of the same degree, only &quot;shifted&quot; by some coefficients.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;br&gt;&#xA;$ f(x)-2*f(x-1)+f(x-2) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt;  = 204*x - 102 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again the resulting polynomial is of reduced order. Now we finish by subtracting $ f(x-1)-2*f(x-2)+f(x-3) $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get:&lt;br&gt;&#xA;$ f(x)-3*f(x-1)+3*f(x-2)-f(x-3) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt; = 204 $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a constant expression only. We are finished and write, reordering the terms:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = 3*f(x-1)-3*f(x-2)+f(x-3)+204 $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[update] For completeness one can increase the degree. According to Qiaochu's answer/comment we can subtract again the x-1-shifted polynomial $ f(x-1)-3*f(x-2)+3*f(x-3)-f(x-4) $ and of course getting    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x)-4*f(x-1)+6*f(x-2)-4*f(x-3)+f(x-4) $&lt;br&gt;&#xA;Pari/GP: $ &amp;lt;out&amp;gt; = 0 $     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;completing the answer &#xA;$ f(x) = 4*f(x-1)-6*f(x-2)+4*f(x-3)-f(x-4) $    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[end update]&lt;/p&gt;&#xA;" OwnerUserId="1714" LastEditorUserId="1714" LastEditorDisplayName="" LastEditDate="2010-12-08T21:56:48.620" LastActivityDate="2010-12-08T21:56:48.620" />
  <row Id="13559" PostTypeId="2" ParentId="13519" CreationDate="2010-12-08T20:57:58.697" Score="2" ViewCount="0" Body="&lt;p&gt;The size of proofs is almost totally unrelated to the size of the statements they prove (in the worst case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lower Bound: constant size&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the smallest proof for statements of size N is constant. Just make your statements of the form &quot;True or S&quot;. (Unless you count the input statement as part of the size, in which case it is linear.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upper Bound: incomputable size&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the largest proof proving a statement of size less than N grows so fast it is not computable from N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the size of a proof that a Turing Machine does not halt within S steps. Some turing machines run forever but can't be proven to run forever. Suppose you have such a machine M. Proving M doesn't halt before S steps must take more steps as S grows. Otherwise we have a proof for all S and have contradicted ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call that number-of-steps to proof-size function G. I am going to make one assumption, which is that there is a computable function G' such that G(G'(X)) &gt;= X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have some computable function F you claim is an asymptotic upper bound on proof sizes given the statement size. Then I just generate a sequence of statements where the Nth statement is &quot;M does not halt within G'(N*F(N)) steps&quot;. The statements are all provable (worst case: simulate M in the proof; that's why I assumed G' is computable) and grow in size logarithmically. But the proof sizes grow asymptotically faster than F, which is exponentially faster than F with respect to the statement sizes. So F is not actually an asymptotic upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore no computable function is an asymptotic upper bound on proof size given the statement size, which means the upper bound is incomputable.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditorDisplayName="" LastEditDate="2010-12-09T17:37:30.633" LastActivityDate="2010-12-09T17:37:30.633" />
  <row Id="13560" PostTypeId="2" ParentId="13554" CreationDate="2010-12-08T20:58:44.263" Score="2" ViewCount="0" Body="&lt;p&gt;You need to use the fact that a well-ordering can't be isomorphic to any of its initial segments.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-08T20:58:44.263" />
  <row Id="13561" PostTypeId="2" ParentId="13511" CreationDate="2010-12-08T20:59:42.807" Score="2" ViewCount="0" Body="&lt;p&gt;Use the Convex Layers algorithm of Bernard Chazelle, which is optimal: $\mathcal{O}(n \log n)$ time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here: &lt;a href=&quot;http://www.cs.princeton.edu/~chazelle/pubs/ConvexLayers.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.princeton.edu/~chazelle/pubs/ConvexLayers.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot of part of the first page of the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/a6nnh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-08T20:59:42.807" CommentCount="1" />
  <row Id="13562" PostTypeId="1" CreationDate="2010-12-08T21:10:50.160" Score="2" ViewCount="49" Body="&lt;p&gt;let $S(k), k\geq 0$ a discrete random process. Suppose $S(N)$ is with probability one either 100 or 0 and that $S(0)=50$. Suppose further there is at least a sixty percent probability that the price will at some point dip below 40 and then subsequently rise above 60 before time $N$. &#xA;How do you prove that $S(k)$ cannot be a martingale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By advance, thank you very much for your help.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2010-12-08T21:10:50.160" Title="An application of the Optional Sampling Theorem" Tags="&lt;probability&gt;&lt;stochastic-processes&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="13563" PostTypeId="1" AcceptedAnswerId="13564" CreationDate="2010-12-08T21:19:26.333" Score="1" ViewCount="103" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am finding with some formulas for my web application. I need a formula that counts the number of values between two number, e.g. there are four numbers (3,4,5,6) between 2 and 7&lt;/p&gt;&#xA;" OwnerUserId="4553" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:45:51.647" LastActivityDate="2010-12-09T19:45:51.647" Title="Count number of values between two numbers" Tags="&lt;arithmetic&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13564" PostTypeId="2" ParentId="13563" CreationDate="2010-12-08T21:22:03.390" Score="5" ViewCount="0" Body="&lt;p&gt;larger - smaller - 1. If you don't count the larger and the smaller (like your example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the numbers are $n, n+1, \dots, n+k , n+k+1$, the numbers you want are $n+1, n+2, \dots, n+k$, which are $k$ in number = $n+k+1 - n - 1$ = larger - smaller - 1.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-08T21:22:03.390" CommentCount="4" />
  <row Id="13565" PostTypeId="2" ParentId="13554" CreationDate="2010-12-08T21:35:16.583" Score="2" ViewCount="0" Body="&lt;p&gt;Okay, it looks like you are thinking of your $(X,\lt)$ as an ordinal, rather than an arbitrary set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that for all $y\in X$, if $x\lt y$, then  $f(x)\neq f(y)$ and $f(x)\subseteq f(y)$. You have already shown $f(x)\subseteq f(y)$, so we just need to show the inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $y=\emptyset$, the least element of $X$, then there is nothing to do and the claim holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the claim holds for all $z\lt y$. Let $x\lt y$. If $x^+$, the successor of $x$, is also less than $y$, then $f(x)\subseteq f(x^+)\subseteq f(y)$, and $f(x)\neq f(x^+)$ by the induction hypothesis, so $f(x)\neq f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $y=x^+$, then $\hat{y} = \hat{x}\cup\{x\}$. So $f(y) = f(x)\cup\{f(x)\}$. If $f(x)\cup\{f(x)\} = f(x)$, then $f(x)\in f(x)$, which is impossible since ordinals are well-founded relative to $\in$. Therefore, $f(y)=f(x)\cup\{f(x)\}\neq f(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By transfinite induction, the claim holds for all $y\in X$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-08T21:44:57.627" LastActivityDate="2010-12-08T21:44:57.627" />
  <row Id="13566" PostTypeId="2" ParentId="13543" CreationDate="2010-12-08T21:53:11.060" Score="2" ViewCount="0" Body="&lt;p&gt;In the spirit of Zaricuse's solution without the sum requirement, take any sufficiently  well behaved functions f and g.  Then you should be able to find a linear combination af+bg that satisfies both equations.  If you let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$if=\int_{-\infty}^\infty f(x) dx$$&#xD;&#xA;  $$if2=\int_{-\infty}^\infty f(x)^2 dx$$&#xD;&#xA;  $$sf=\sum_{n=1}^\infty f(n)$$&#xD;&#xA;  $$sf2=\sum_{n=1}^\infty f(n)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly for fg and g, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a*if + b*ig=a^2*if2+2ab*ifg+b^2ig2$$ and&lt;br&gt;&#xA;$$a*sf + b*sg=a^2*sf2+2ab*sfg+b^2sg2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be solved for a and b in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; In response to the new request that the values of the two integrals and two sums all match, I just need enough knobs to turn.  Define $g(k,x)=\exp(-kx^2)$ and take $f(x)=g(1,x)+ag(2,x)+bg(3,x)+cg(4,x)$  The nice thing about this $f$ is that $f^2$ is written in terms of $g(k,x)$, though k goes up to 8.  The integral of $g(k,x)$ is just $\sqrt{\frac{\pi}{k}}$ and the sum is calculated by Wolfram Alpha as $\vartheta_3(0,\exp(-k))$. We can make a table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}k&amp;amp;\int g(k,x)&amp;amp;\sum g(k,x)\\1&amp;amp;1.772453851&amp;amp;1.77264\\2&amp;amp;1.253314137&amp;amp;1.27134\\3&amp;amp;1.023326708&amp;amp;1.09959\\4&amp;amp;0.886226925&amp;amp;1.03663\\5&amp;amp;0.79266546&amp;amp;1.01348\\6&amp;amp;0.723601255&amp;amp;1.00496\\7&amp;amp;0.669924586&amp;amp;1.00182\\8&amp;amp;0.626657069&amp;amp;1.00067\end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the integral of f is $\sqrt{\pi}(1+a/\sqrt{2}+b/\sqrt{3}+c/\sqrt{4})$  The integral of f^2 is $\sqrt{\pi}(1/\sqrt{2}+2a/\sqrt{3}+(a^2+2b)/\sqrt{4}+(2c+2ab)/\sqrt{5}+(b^2+2ac)/\sqrt{6}+2bc/\sqrt{7}+c^2/\sqrt{8})$ with similar expressions for the sum in terms of theta3.  We want to find a,b,c so that the integrals and sums all match.  Unless my matrix of coefficients has a very unlikely dependence it will be available.  Excel claims $a=-3.782590725, b=4.503400057, c=-1.83137936$ is very close to a solution, and there should be more.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T03:41:30.973" LastActivityDate="2010-12-10T03:41:30.973" CommentCount="5" />
  <row Id="13567" PostTypeId="2" ParentId="13543" CreationDate="2010-12-08T21:58:19.353" Score="3" ViewCount="0" Body="&lt;p&gt;Similar to the suggestion of Zaricuse, that is take an entire function $f$ such that $\int_{-\infty}^\infty f(x) dx \ne0$ then solve for &#xA;$$\int_{-\infty}^\infty af(x)dx = \int_{-\infty}^\infty (af(x))^2dx$$&#xD;&#xA;Then $g(z)=af(z)$ solves half of the problem. To reach &#xA;$$\sum g(n)=\sum g(n)^2$$&#xD;&#xA;we may for example start with $f(z)=\sin (\pi z) \cdot h(z)$  where $h$ is an other integrable entire function.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-08T21:58:19.353" CommentCount="2" />
  <row Id="13568" PostTypeId="1" AcceptedAnswerId="13570" CreationDate="2010-12-08T21:58:30.607" Score="1" ViewCount="40" Body="&lt;p&gt;If i know the means and standard deviations of 2 sets of data, and i know the slope of the regression line, how can I find the correlation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SD: 12.37&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SD: 7.00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slope of regression line: 0.789&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="1630" LastEditorDisplayName="" LastEditDate="2010-12-08T22:13:25.183" LastActivityDate="2010-12-08T22:20:53.990" Title="Statistics - Correlation between 2 sets of data" Tags="&lt;statistics&gt;&lt;correlation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13569" PostTypeId="1" AcceptedAnswerId="13571" CreationDate="2010-12-08T22:02:48.490" Score="1" ViewCount="90" Body="&lt;p&gt;Find an equation of the parabola with focus at point $(0,5)$ whose directrix is the line $y=0$. (Derive this equation using the definition of the parabola as a set of points that are equidistant from the directrix and the focus)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok this one is killing me. My textbook has this&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An equation of the parabola with focus $(0,p)$ and directrix $y=-p$ is $x^2=4py$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, my problem has $(0,p)$ with p being 5, yet the directrix is not $p=-y=-5$, it's $0$. And so I tried deriving it manually as they said by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{x^2+(y-p)^2}=|y+p|$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that just takes me back to the original equation I mentioned, $x^2=4py$. Not sure what I'm supposed to do. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-09T02:58:54.970" Title="Finding equation of parabola" Tags="&lt;conic-sections&gt;&lt;parabola&gt;" AnswerCount="3" />
  <row Id="13570" PostTypeId="2" ParentId="13568" CreationDate="2010-12-08T22:06:03.793" Score="3" ViewCount="0" Body="&lt;p&gt;The formula is $$r = b_1 \frac{s_x}{s_y},$$ where $r$ is the correlation, $b_1$ is the slope, and $s_x$ and $s_y$ are the standard deviations of the independent $(x)$ and dependent $(y)$ variables, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference is Wikipedia's page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_linear_regression#Fitting_the_regression_line&quot; rel=&quot;nofollow&quot;&gt;simple linear regression&lt;/a&gt;.  See the formula for $\hat{\beta}$. &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-08T22:20:53.990" LastActivityDate="2010-12-08T22:20:53.990" CommentCount="7" />
  <row Id="13571" PostTypeId="2" ParentId="13569" CreationDate="2010-12-08T22:06:33.837" Score="8" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Edited to try to clarify OP's confusion:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between $(x,y)$ and $(0,5)$ is indeed $\sqrt{x^2+(y-5)^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between $(x,y)$ and the line $y=0$ is $|y|$ (if $y&amp;gt;0$, then it is above the $X$-axis, and the distance is just the $y$ coordinate; if $y&amp;lt;0$, then the distance is $-y = |y|$). So the points on the parabola are exactly the points for which the two distances are equal, that is, all $(x,y)$ for which:&#xA;$$\sqrt{x^2+(y-5)^2} = |y|.$$&#xD;&#xA;Now square both sides, and you'll get the equation of the parabola you want.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;How is this related to the formula you quote? It's really all just the same process. If the directrix is either horizontal (a line of the form $y=k$) or vertical (a line of the form $x=\ell$), then it is very easy to compute the distance from a point to the directrix: the point $(x,y)$ is $|y-k|$ away from the line $y=k$, and is $|x-\ell|$ away from the line $x=\ell$. If the focus of the parabola is at $(a,b)$, then you want the distance from $(x,y)$ to $(a,b)$ to equal the distance to the directrix, so you get:&#xA;\begin{align*}&#xA;\sqrt{(x-a)^2 + (y-b)^2} &amp;amp;= |y-k| &amp;amp;\qquad&amp;amp;\mbox{if the directrix is $y=k$,}\\&#xA;\sqrt{(x-a)^2 + (y-b)^2} &amp;amp;= |x-\ell| &amp;amp;&amp;amp;\mbox{if the directrix is $x=\ell$.}&#xA;\end{align*}&#xA;Then squaring both sides yields the equation of the parabola. Doing it correctly will cancel out the $y^2$ term when the directrix is horizontal, and the $x^2$ term when the directrix is vertical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more general parabolas, when the directrix is neither horizontal nor vertical but an arbitrary line $y=mx+b$, you need to work a bit harder, because the distance from $y=mx+b$ to $(x,y)$ is not so simple to compute. Not &lt;em&gt;too&lt;/em&gt; hard, but not as simple. Once you have a formula for the distance, you set it equal to the distance to the focus, square both sides, and get the equation of the parabola.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T02:58:54.970" LastActivityDate="2010-12-09T02:58:54.970" CommentCount="4" />
  <row Id="13572" PostTypeId="3" CreationDate="2010-12-08T22:10:57.063" Score="0" ViewCount="0" Body="&lt;p&gt;I discovered math.stackexchange.com less than 4 weeks (at the time of writing this) ago and I have been a regular follower/contributer since then. I am relatively new to online mathematics and have not been a particularly active member anywhere else, partly because I am relatively new to mathematics itself (I used to be an engineer and switched to math a couple of years ago). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been especially interested in this community and also math overflow (which I also joined only recently) because of the absence of discussion format which is probably the reason for the high signal to noise ratio. If elected, maintaining and improving this efficiency would be my main goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe in a largely &lt;em&gt;laissez faire&lt;/em&gt; operation of such sites and as such would trust the invisible hand of the majority to take care of most issues. I generally prefer to keep my discussions technical (as can be seen from my posts) and as such would avoid imposing my opinions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to see this site prosper and be of permanent value to active and amateur mathematicians at every level. I will try my best to play my part in this.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-08T22:45:23.930" LastActivityDate="2010-12-08T22:45:23.930" CommunityOwnedDate="2010-12-08T22:10:57.063" />
  <row Id="13573" PostTypeId="2" ParentId="13569" CreationDate="2010-12-08T22:19:44.913" Score="1" ViewCount="0" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Parabola&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation for a general parabola with a focus point $F =(u, v)$, and a directrix in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_1x+n_2y+c=0 \,$$&#xD;&#xA;is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\left|n_1x+n_2y+c\right|}{\sqrt{{n_1}^{2}+{n_2}^{2}}}=\sqrt{\left(x-u\right)^2+\left(y-v\right)^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So make the substitutions and verify that the what you get agrees with &lt;a href=&quot;http://math.stackexchange.com/questions/13569/finding-equation-of-parabola/13571#13571&quot;&gt;Arturo Magidin's derivation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-08T22:19:44.913" />
  <row Id="13574" PostTypeId="2" ParentId="13569" CreationDate="2010-12-08T22:38:26.060" Score="3" ViewCount="0" Body="&lt;p&gt;So, you have a method for writing an equation for a parabola where the focus is at $(0,p)$ and the directrix is $y=-p$, but now you want to write an equation for a parabola with focus $(0,5)$ and directrix $y=0$, which doesn't fit the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the conditions you are given and translate them down 2.5 units, the focus would be at $(0,2.5)$ and the directrix would be at $y=-2.5$, so you should be able to write an equation for that parabola.  Now, translate that equation up 2.5 units (replace $y$ with $(y-2.5)$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T22:38:26.060" />
  <row Id="13575" PostTypeId="1" CreationDate="2010-12-08T22:42:57.350" Score="9" ViewCount="233" Body="&lt;p&gt;I have a hazy notion of some stuff in differential geometry and a better, but still not quite rigorous understanding of basics of differential topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have decided to fix this lacuna once for all. Unfortunately I cannot attend a course right now. I must teach myself all the stuff by reading books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Towards this purpose I want to know what are the most important basic theorems in differential geometry and differential topology. For a start, for differential topology, I think I must read Stokes' theorem and de Rham theorem with complete proofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential geometry is a bit more difficult. What is a connection? Which notion should I use? I want to know about parallel transport and holonomy. What are the most important and basic theorems here? Are there concise books which can teach me the stuff faster than the voluminous Spivak books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also finally I want to read into some algebraic geometry and Hodge/Kähler stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestions about important theorems and concepts to learn, and book references, will be most helpful.&lt;/p&gt;&#xA;" OwnerUserId="4555" LastActivityDate="2010-12-09T08:09:39.660" Title="Teaching myself differential topology and differential geometry." Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;&lt;differential-topology&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="3" />
  <row Id="13576" PostTypeId="2" ParentId="13575" CreationDate="2010-12-08T22:53:23.670" Score="2" ViewCount="0" Body="&lt;p&gt;Guillemin and Pollack's &quot;Differential Topology&quot; is about the friendliest introduction to the subject you could hope for.  It's an excellent non-course book.  Good supplementary books would be Milnor's &quot;Topology from a differentiable viewpoint&quot; (much more terse), and Hirsch's &quot;Differential Topology&quot; (much more elaborate, focusing on the key analytical theorems). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential geometry it's much more of a mixed bag as it really depends on where you want to go.  I've always viewed Ehresmann connections as the fundamental notion of connection.  But it suits my tastes.   But I don't know much in the way of great self-learning differential geometry texts, they're all rather quirky special-interest textbooks or undergraduate-level grab-bags of light topics.  I haven't spent any serious amount of time with the Spivak books so I don't feel comfortable giving any advice on them. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-12-08T22:53:23.670" />
  <row Id="13577" PostTypeId="2" ParentId="13477" CreationDate="2010-12-08T23:00:31.150" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;With&lt;/em&gt; trig, I get that the arcs have measure $\theta \pm \alpha$ and $\pi - \theta \pm \beta$, where &#xA;$$\cos\theta = \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} = \frac{\cos\beta-\cos\alpha}{1-\cos\left(\alpha+\beta\right)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the &lt;a href=&quot;http://www.mathwarehouse.com/geometry/circle/tangents-secants-arcs-angles.php&quot; rel=&quot;nofollow&quot;&gt;&quot;far arc / near arc&quot; principle&lt;/a&gt;, $\stackrel{\frown}{EI}-\stackrel{\frown}{EG} = 2\alpha$, so we can write $\stackrel{\frown}{EI} = \theta+\alpha$ and $\stackrel{\frown}{EG}=\theta-\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go analytical ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the circle to have radius $1$, with its center, $O$, at the origin; and let $H$ lie on the positive $x$-axis. In quadrilateral $OEHT$, angles at $E$ and $T$ are right angles, and the angle at $H$ has measure $\alpha+\beta$, so that the angle at $O$ has measure $2 \gamma := \pi-(\alpha+\beta)$. Thus, $\angle HOE = \gamma$ and $|OH| = \sec\gamma$. Moving around the circle by arcs of $\theta \pm \alpha$ from $E$ gets us to $I$ and $G$, and we determine coordinates for these points, writing &quot;$\rm{cis}\cdot$&quot; for &quot;$(\cos\cdot,\sin\cdot)$&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \rm{cis}\left(\gamma + ( \theta + \alpha )\right) = \rm{cis}\left(\gamma+\alpha+\theta\right)$$ &#xA;$$G = \rm{cis}\left(\gamma-(\theta-\alpha)\right) = \rm{cis}\left(\gamma+\alpha-\theta\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $H$, $G$, and $I$ are collinear, we must have equal slopes for segments $HI$ and $HG$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \sin(\gamma+\alpha+\theta) }{ \cos(\gamma+\alpha+\theta)-\sec\gamma } = \frac{ \sin(\gamma+\alpha-\theta) }{ \cos(\gamma+\alpha-\theta)-\sec\gamma }$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(\gamma+\alpha+\theta)\left(\cos(\gamma+\alpha-\theta)-\sec\gamma\right) = \sin(\gamma+\alpha-\theta) \left( \cos(\gamma+\alpha+\theta)-\sec\gamma\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray*}&#xD;&#xA;&amp;amp;&amp;amp;\sin(\gamma+\alpha+\theta)\cos(\gamma+\alpha-\theta)- \sin(\gamma+\alpha-\theta) \cos(\gamma+\alpha+\theta)  \\&#xD;&#xA;&amp;amp;=&amp;amp; \sec{\gamma} \; \left( \sin(\gamma+\alpha+\theta) -\sin(\gamma+\alpha-\theta) \right) \\&#xD;&#xA;\sin 2\theta &amp;amp;=&amp;amp; 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right) \\&#xD;&#xA;2 \sin\theta \cos\theta &amp;amp;=&amp;amp; 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right)&#xD;&#xA;\end{eqnarray*}$$&#xD;&#xA;If $\theta\ne 0$ and $\theta \ne \pi$ [*], then we can cancel $\sin\theta$ and finish-up:&#xA;$$\begin{eqnarray*}&#xD;&#xA;\cos\theta &amp;amp;=&amp;amp; \frac{\cos\left(\gamma+\alpha\right)}{\cos\gamma} = \frac{\cos\frac{\pi+\alpha-\beta}{2}}{\cos\frac{\pi-(\alpha+\beta)}{2}}= \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} \\&#xD;&#xA;\end{eqnarray*}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's certainly a way to arrive at this result without using coordinates and slopes (I got lazy), but if there's a way to get there &quot;without using trigonometry&quot;, I'm not seeing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt;&#xA;I'll just point out that, if $F$ is the foot of the perpendicular from the circle's center to segment $HI$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{|OF|}{|OE|}=\frac{|OH| \sin\angle FHO}{|OH|\sin\angle EHO}=\frac{ \sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}=\cos\left(\pi-\theta\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are points $P$ and $Q$ on the circle, with $P$ on $\stackrel{\frown}{IE}$ and $G$ on $\stackrel{\frown}{QE}$, such that $\stackrel{\frown}{IP} = \stackrel{\frown}{GQ} = \alpha$, whence $\stackrel{\frown}{PE}=\stackrel{\frown}{QE}=\theta$. So, all the components of the final relation appear in the figure somewhere. I haven't looked hard enough to determine if it's possible to &quot;see&quot; the relation geometrically.&#xA;&lt;strong&gt;End edit.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[*] If $\theta = 0$, then $\alpha$ itself must be $0$ (so that $\theta-\alpha$ is a non-negative arc measure), with $I$ and $G$ coinciding with $E$. Likewise, $\theta = \pi$ implies $\beta = 0$, with $I$ and $G$ coinciding with $T$. Analysis of these cases is straightforward, with the same formula resulting as in the general case.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-12-09T13:18:12.850" LastActivityDate="2010-12-09T13:18:12.850" CommentCount="1" />
  <row Id="13578" PostTypeId="1" AcceptedAnswerId="13579" CreationDate="2010-12-08T23:03:39.250" Score="-4" ViewCount="69" Body="&lt;p&gt;What is a primitive nth root of unity?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastActivityDate="2010-12-08T23:11:24.393" Title="What is a primitive nth root of unity?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13579" PostTypeId="2" ParentId="13578" CreationDate="2010-12-08T23:11:24.393" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Primitive_nth_root_of_unity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An nth root of unity, where n = 1,2,3,··· is a positive integer, is a complex number z satisfying the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^n = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An nth root of unity is primitive if it is not a kth root of unity for some smaller k:&#xA;$z^k \neq 1  (k = 1, 2, 3, \ldots, n-1 )$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-08T23:11:24.393" />
  <row Id="13580" PostTypeId="2" ParentId="13556" CreationDate="2010-12-08T23:14:04.753" Score="2" ViewCount="0" Body="&lt;p&gt;This is not a direct answer, but related. This paper by Jan Stovicek addresses the question &quot;which long exact sequences can arise from the snake lemma&quot; so it might be of help here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathjournals.org/mrl/2010-017-001/2010-017-001-011.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mathjournals.org/mrl/2010-017-001/2010-017-001-011.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T01:49:07.383" LastActivityDate="2010-12-09T01:49:07.383" CommentCount="2" />
  <row Id="13581" PostTypeId="2" ParentId="13575" CreationDate="2010-12-08T23:14:11.237" Score="5" ViewCount="0" Body="&lt;p&gt;For differential topology, I would add Poincare duality to something you may want to know. A good textbook is Madsen and Tornehave's &lt;em&gt;From Calculus to Cohomology&lt;/em&gt;. Another nice book is John Lee's &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential geometry, I don't really know any good texts. Besides the standard Spivak, the other canonical choice would be Kobayashi-Nomizu's &lt;em&gt;Foundations of Differential Geometry&lt;/em&gt;, which is by no means easy going. There is a new book by Jeffrey Lee called &lt;em&gt;Manifolds and Differential Geometry&lt;/em&gt; in the AMS Graduate Studies series. I have not looked at it personally in depth, but it has some decent reviews. It covers a large swath of the differential topology, and also the basic theory of connections. (As a side remark, if you like doing computations, Kobayashi's original paper &quot;Theory of connections&quot; is not very hard to read, and may be a good starting place before you jump into some of the more special-topic/advanced texts like Kolar, Slovak, and Michor's &lt;em&gt;Natural operations in differential geometry&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A book I've enjoyed and found useful (though not so much as a textbook) is Morita's &lt;em&gt;Geometry of differential forms&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't help you with algebraic geometry. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-08T23:14:11.237" CommunityOwnedDate="2010-12-08T23:14:11.237" CommentCount="8" />
  <row Id="13582" PostTypeId="1" AcceptedAnswerId="13584" CreationDate="2010-12-08T23:23:02.103" Score="0" ViewCount="177" Body="&lt;p&gt;Question: Let $a, b \in \mathbb{R}$ with $a &amp;lt; b$ and let $f: [a,b] \rightarrow [a,b]$ continuous. Show: $f$ has a fixed point, that is, there is an $x \in [a,b]$ with $f(x)=x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose this has to do with the basic definition of continuity. The definition I am using is that $f$ is continuous at $a$ if $\displaystyle \lim_{x \to a} f(x)$ exists and if $\displaystyle \lim_{x \to a} f(x) = f(a)$. I must not be understanding it, since I am not sure how to begin showing this... Should I be trying to show that $x$ is both greater than or equal to and less than or equal to $\displaystyle \lim_{x \to a} f(x)$ ?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-08T23:28:13.173" LastActivityDate="2010-12-08T23:43:25.073" Title="Show that a continuous function has a fixed point" Tags="&lt;analysis&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="13583" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:27:17.793" Score="6" ViewCount="0" Body="&lt;p&gt;Consider $x-f(x)$ and use Intermediate Value Theorem.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-08T23:27:17.793" />
  <row Id="13584" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:28:30.680" Score="2" ViewCount="0" Body="&lt;p&gt;Let $g(x) = f(x) - x$.  Then $g$ is continuous and $g(a) \geq 0$ while $g(b) \leq 0$.  By the Intermediate Value Theorem, $g$ has at least one zero on $[a, b]$.&lt;/p&gt;&#xA;" OwnerUserId="506" LastActivityDate="2010-12-08T23:28:30.680" CommentCount="3" />
  <row Id="13585" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:31:03.420" Score="2" ViewCount="0" Body="&lt;p&gt;Consider $g(x)=f(x)-x$.  $f(a)\ge a$ so $g(a)=f(a)-a\ge 0$.  $f(b)\le b$ so $g(b)=f(b)-b\le 0$.  By the Intermediate Value Theorem, since $g$ is continuous and $0\in[g(b),g(a)]$ there exists $c\in[a,b]$ such that $g(c)=f(c)-c=0$, so $f(c)=c$ for some $c\in[a,b]$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-08T23:31:03.420" CommentCount="9" />
  <row Id="13586" PostTypeId="2" ParentId="13582" CreationDate="2010-12-08T23:33:12.140" Score="4" ViewCount="0" Body="&lt;p&gt;For a different approach then the ones above, let us take $a = 0$ and $b = 1$. So assume $f:[0,1] \to [0,1]$ has no fixed point. Then $[0,1] = \{x \in [0,1] : f(x) &amp;lt; x\} \cup \{x \in [0,1] : f(x) &amp;gt; x \}$. Now argue that this is not possible.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-08T23:43:25.073" LastActivityDate="2010-12-08T23:43:25.073" CommentCount="3" />
  <row Id="13587" PostTypeId="2" ParentId="12644" CreationDate="2010-12-09T00:35:28.517" Score="7" ViewCount="0" Body="&lt;p&gt;For the sake of completeness: $\mathbb{C}^{\ast}$ is isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Z}$ in the obvious way, and $\mathbb{C}^{\ast}$ modulo the roots of unity is then isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Q}$.  As a $\mathbb{Q}$-vector space this is abstractly isomorphic to $\mathbb{R}^{+}$, but the construction of such an isomorphism is likely to require the axiom of choice.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T00:35:28.517" CommunityOwnedDate="2010-12-09T00:35:28.517" CommentCount="1" />
  <row Id="13588" PostTypeId="2" ParentId="13555" CreationDate="2010-12-09T00:52:48.167" Score="4" ViewCount="0" Body="&lt;p&gt;The easiest answer is if $\xi$ is along one of the axes, say the first.  Then the cone can be parameterized as  $(t\cos(\alpha ),t\sin(\alpha){\bf\hat{x}})$, where ${\bf\hat{x}}$ is a unit vector in $\mathbb{R^{n-1}}$.  In coordinates, this is $(t\cos(\alpha ),t\sin(\alpha )\cos(\phi_1),t\sin(\alpha )\sin(\phi_1)\cos(\phi_2),t\sin(\alpha )\sin(\phi_1)\sin(\phi_2)\cos(\phi_3),\ldots)$ where you have n-1 angular terms and the last has no cosine.  If n&gt;3, $\phi_1$ runs from 0 to $\pi$ while the others run from 0 to $2\pi$.  Then you can apply a rotation matrix to take $(1,0,0,\ldots)$ to $\xi$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-09T00:52:48.167" />
  <row Id="13589" PostTypeId="1" AcceptedAnswerId="13590" CreationDate="2010-12-09T01:16:45.033" Score="1" ViewCount="50" Body="&lt;p&gt;Let Y1...Yn be independent and identically distributed random vairables such that for 0 &amp;lt; p &amp;lt; 1, P(Yi = 1) = p and P(Yi = 0) = q = 1-p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Find the moment-generating functions for the random variable Y1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Find the moment-generating functions for W = Y1 + ... + Yn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. What is the distribution of W?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have started to try A.&#xA;My book stays that m(t) = E(e^(tY)). But i'm sure sure what that is. I think that expected value of Y1 is p. But i'm not sure where to go from here. I'm completely clueless, statistics is not my area of expertise(I'm a computer science guy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-09T02:50:49.793" Title="statistics - moment-generating function" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="13590" PostTypeId="2" ParentId="13589" CreationDate="2010-12-09T01:30:10.380" Score="3" ViewCount="0" Body="&lt;p&gt;If those are the only two values that $Y_i$ takes on then you are correct that $E[Y_i]=p$. The definition of the moment generating function is what you have described as $M_{Y_i}(t)=E[e^{tY_i}]$. So you compute this by multiplying $e^{ty_i}$ by your density function and summing over all of the appropriate values. So in this case $M_{Y_i}(t)=(e^{t(0)})(P(Y_i=0))+(e^{t(1)})(P(Y_i=1))$ which gives you $M_{Y_i}(t)=1-p+pe^t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part B you should use the fact that the moment generating function of a sum of independent random variables is the product of the moment generating functions. That gives you $M_W(t)=(1-p+pe^t)^n$ which i believe is the moment generating function for a Binomial random variable with parameters $p$ and $n$. This makes sense if we do a quick mental check and note that $Y_i$ can be thought of as the success or failure of the $i$th trial, the indicator functions. So the total number of successes would be $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if there are parts where I need to be more specific.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-09T02:50:49.793" LastActivityDate="2010-12-09T02:50:49.793" CommentCount="2" />
  <row Id="13591" PostTypeId="1" AcceptedAnswerId="13594" CreationDate="2010-12-09T01:36:37.943" Score="14" ViewCount="339" Body="&lt;p&gt;It can be shown by simple induction that $\dfrac{\mathrm d^n}{\mathrm dx^n}\left(\dfrac1{x}\right) =  \frac{(-1)^n n!}{x^{n+1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the nth integral of $\displaystyle \frac1{x}$? Finding the first few primitives, I can't discern a pattern. &lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T02:31:08.277" LastActivityDate="2010-12-09T16:38:49.513" Title="What's the nth integral of $\displaystyle \frac1{x}$?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="13592" PostTypeId="2" ParentId="13591" CreationDate="2010-12-09T01:43:09.317" Score="1" ViewCount="0" Body="&lt;p&gt;Remembre that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{1}{x} dx= \ln (x) + C $$  Now integrate again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint :&lt;/strong&gt;&lt;br&gt;&#xA;$\displaystyle\int \ln(x) dx = x \ln(x)-x +C$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-12-09T01:43:09.317" CommentCount="4" />
  <row Id="13593" PostTypeId="2" ParentId="13591" CreationDate="2010-12-09T01:52:55.703" Score="8" ViewCount="0" Body="&lt;p&gt;If we use the &lt;a href=&quot;http://mathworld.wolfram.com/RepeatedIntegral.html&quot; rel=&quot;nofollow&quot;&gt;repeated integral&lt;/a&gt; formula for the reciprocal function, with 1 as the lower limit, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\underbrace{\int_1^x\int_1^{t_{n-1}}\cdots\int_1^{t_1}}_{n} \frac1{t}\;\mathrm dt\cdots\mathrm dt_{n-2}\mathrm dt_{n-1}=\frac1{(n-1)!}\int_1^x\frac{(x-t)^{n-1}}{t}\mathrm dt=(-1)^n \frac{x^{n-1}}{(n-1)!}B_{1-x}(n,1-n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_x(a,b)$ is the &lt;a href=&quot;http://functions.wolfram.com/GammaBetaErf/Beta3/&quot; rel=&quot;nofollow&quot;&gt;incomplete beta function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_n(x)=(-1)^n \frac{x^{n-1}}{(n-1)!}B_{1-x}(n,1-n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the following more &quot;elementary&quot; representation can be derived:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_n(x)=\frac{x^{n-1}}{(n-1)!}(\ln\;x-H_{n-1})-\sum_{j=1}^{n-1}\frac{(-1)^j}{j\cdot j!}\frac{x^{n-j-1}}{(n-j-1)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_n=\displaystyle \sum_{j=1}^n\frac1{j}$ is a harmonic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Moron mentions in his answer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d^n}{\mathrm dx^n}\left(\frac{x^{n-1}}{(n-1)!}\ln\;x+p_{n-1}(x)\right)=\frac1{x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_{n-1}(x)$ is any polynomial of degree $n-1$; $g_n(x)$, however, has the special property (by virtue of how it was constructed) that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.\frac{\mathrm d^k}{\mathrm dx^k}g_n(x)\right|_{x=1}=0\qquad \mathrm{if}\qquad k &amp;lt; n$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T16:38:49.513" LastActivityDate="2010-12-09T16:38:49.513" CommentCount="4" />
  <row Id="13594" PostTypeId="2" ParentId="13591" CreationDate="2010-12-09T02:03:49.673" Score="14" ViewCount="0" Body="&lt;p&gt;As Isaac noted, repeated integration seems to give the following pattern&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{x^{n-1} \log x}{(n-1)!} - C_n x^{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the value of $\displaystyle C_n$ does not really matter, as differentiating $n$ times nukes it. Also, note that, we can add any arbitrary $\displaystyle (n-1)^{th}$ degree polynomial to this, without changing the $\displaystyle n^{th}$ derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove that the $\displaystyle n^{th}$ derivative of $\displaystyle  \frac{x^{n-1} \log x}{(n-1)!}$ is $\displaystyle \frac{1}{x}$, we can use induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(n-1)!} \frac{d (x^{n-1} \log x)}{dx} = \frac{x^{n-2}}{(n-1)!} + \frac{x^{n-2} \log x}{(n-2)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since adding an arbitrary $\displaystyle (n-2)^{th}$ degree polynomial does not change the $\displaystyle (n-1)^{th}$ derivative of $\displaystyle \frac{x^{n-2} \log x}{(n-2)!}$ we are done using induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if $\displaystyle f(x)$ is another function such that $\displaystyle \frac{d^n f}{dx} = \frac{1}{x}$, then we have that $\displaystyle \phi(x) = f(x) - \frac{x^{n-1} \log x}{(n-1)!}$ has it's $\displaystyle n^{th}$ derivative to be zero, and hence it is a polynomial of degree $\displaystyle n-1$ or lower (can be proved using induction, again).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus all the functions you are looking for are of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x^{n-1} \log x}{(n-1)!} + \sum_{j=0}^{n-1} c_j x^j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\displaystyle c_j$ are arbitrary constants.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-09T03:49:22.833" LastActivityDate="2010-12-09T03:49:22.833" CommentCount="4" />
  <row Id="13595" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T02:14:04.540" Score="1" ViewCount="0" Body="&lt;p&gt;I would like to recommend &lt;em&gt;Modern Differential Geometry of curves and surfaces with Mathematica&lt;/em&gt;, by Alfred Gray, Elsa Abbena, and Simon Salamon.  You can look at &lt;a href=&quot;http://books.google.com/books?id=owEj9TMYo7IC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;it on Google books&lt;/a&gt; to decide if it fits your style.  If you are a Mathematica user, I think this is a wonderful avenue for self-study, for you can see and manipulate all the central constructions yourself.  I use Gray's code frequently; I was a fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Here is how he died: &quot;of a heart attack which occurred while working with students in a computer lab at 4 a.m.&quot;!&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T03:23:11.880" LastActivityDate="2010-12-09T03:23:11.880" CommentCount="1" />
  <row Id="13596" PostTypeId="1" AcceptedAnswerId="13598" CreationDate="2010-12-09T02:25:53.507" Score="0" ViewCount="64" Body="&lt;p&gt;Let Y be a binomial random variable with n trials and probability of success given by p. Show that n-Y is a binomial random variable with n trials and probability of success given by 1-p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why this is true, and can explain in logically, but I just don't know how to prove it in mathematical terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-09T06:09:33.073" Title="binomial random variable" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13597" PostTypeId="2" ParentId="13596" CreationDate="2010-12-09T02:47:22.247" Score="1" ViewCount="0" Body="&lt;p&gt;I am not really sure how I qould go about giving a full prove right now, but here is a suggestion. Rewrite $Y=\Sigma_{i=1}^n Y_i$ where the $Y_i$ are from your other question, they are the indicator functions for each trial. &lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-09T02:47:22.247" CommentCount="1" />
  <row Id="13598" PostTypeId="2" ParentId="13596" CreationDate="2010-12-09T03:17:30.077" Score="2" ViewCount="0" Body="&lt;p&gt;A discrete random variable is determined by the probabilities of all possible outcomes. If $Y$ is $n$ trials of a binomial random variable with probability $p$ of success, then the probability that $Y = i$ is given by ${\displaystyle{n \choose i} p^i(1 - p)^{n-i}}$. Thus the probability that $n - Y = i$ is given by the probability that $Y = n - i$, or ${\displaystyle {n \choose  n - i} p^{n-i}(1 - p)^i}$. Since ${\displaystyle{n \choose n - i} = {n \choose i}}$, this is the same as ${\displaystyle{n \choose  i} (1 - p)^ip^{n-i}}$. But if $Z$ denotes $n$ trials of a binomial random variable with probability $1 - p$ of success, this is exactly the probability that $Z = i$. Thus $n - Y$ and $Z$ have the same distribution (i.e. the same probability of each outcome), which is equivalent to saying that $n - Y$ is also $n$ trials of a binomial random variable with probability $1 - p$ of success.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-09T03:17:30.077" />
  <row Id="13599" PostTypeId="1" CreationDate="2010-12-09T03:38:52.673" Score="1" ViewCount="93" Body="&lt;p&gt;I feel like this should have a rather simple solution using logarithms, but I don't quite know where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, every day I am going to multiply a given quantity by $r$, where $r &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if $r$ is 0.5, then my half-life $n$ is 1 day. (This is also stated to remove any ambiguities on what counts as a day/range)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between $r$ and $n$?&lt;/p&gt;&#xA;" OwnerUserId="51" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T04:50:55.783" LastActivityDate="2010-12-09T04:50:55.783" Title="If I wanted something to decay with a half-life of $n$ days, what should I multiply it by every day?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13600" PostTypeId="2" ParentId="13599" CreationDate="2010-12-09T03:45:40.123" Score="5" ViewCount="0" Body="&lt;p&gt;Suppose you start with a quantity of 1 (whatever that means).  You know that after $n$ days, there should be $\frac{1}{2}$ left.  This should give you an equation that you can solve for $r$ (in terms of $n$ or with a specific value of $n$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-09T03:45:40.123" CommentCount="4" />
  <row Id="13602" PostTypeId="2" ParentId="13599" CreationDate="2010-12-09T03:48:36.093" Score="6" ViewCount="0" Body="&lt;p&gt;Suppose $x$ is the quantity, and you want it to decay to $\frac{1}{2}$ after $n$ discrete steps (whether they be minutes, hours, days, years, eons, or nanoseconds), and you want to model that by multiplication by $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with $x$. After one time interval, you want to have $rx$. After two time intervals, you want $r(rx) = r^2x$. After three time intervals, $r(r^2x)=r^3x$. After $n$ times intervals, $r^nx$. Since you want $r^nx = \frac{1}{2}x$, you can now solve for $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;em&gt;discrete&lt;/em&gt; decay, though, not continuous decay. It is exactly analogous to the difference between continuous interest compounding, and interest compounding in discrete time intervals. &quot;Half-life&quot; is a term usually reserved for continuous decay, so you'll want to be careful with that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T03:54:07.990" LastActivityDate="2010-12-09T03:54:07.990" />
  <row Id="13603" PostTypeId="1" AcceptedAnswerId="13604" CreationDate="2010-12-09T03:57:53.103" Score="2" ViewCount="35" Body="&lt;p&gt;Suppose, $k,k^&amp;#39;$ are fields such that $k\subset k&amp;#39;$ and $x_1,x_2,...,x_n,t$ are indeterminates over $k^&amp;#39;$. Suppose we have a homomorphism of rings given by $f:k[x_1,x_2,...,x_n]\to k[t]$, where $$x_1\to f_1(t)$$ $$x_2\to f_2(t)$$ $$.$$ $$.$$ $$x_n\to f_n(t)$$ Let the kernel of this map be $I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be said about the kernel $I^&amp;#39;$ of the map $f^&amp;#39;:k^&amp;#39;[x_1,x_2,...,x_n]\to k^&amp;#39;[t]$, where restriction of $f^&amp;#39;$ to $k[x_1,...,x_n]$ is $f$ (note this completely determines $f^&amp;#39;$ since it determines its action on the indeterminates). The ideal $I$ represents the &quot;ideal of relations&quot; for the polynomials $f_1,...,f_n$. So it seems there may be a compact description for $I^&amp;#39;$ in terms of the generators of $I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular what can we say when $k=\mathbb{Q}$ and $k^&amp;#39;=\mathbb{C}$. This case is important because Macaulay2 can compute kernels of such maps when the field of coefficients is $\mathbb{Q}$. So we can use those results to compute the kernels of such maps when the field is $\mathbb{C}$. (For the same reasons what can be said when $k$ is a finite field?)&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T04:19:28.770" LastActivityDate="2010-12-09T04:19:28.770" Title="Behavior of kernels of homomorphisms of polynomial rings under a change of the field of coefficients" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" />
  <row Id="13604" PostTypeId="2" ParentId="13603" CreationDate="2010-12-09T04:04:46.047" Score="3" ViewCount="0" Body="&lt;p&gt;If $I$ is the original kernel, you have a short exact sequence $$0\to I\to k[x_1,\dots,x_n]\to k[t]\to0$$ Apply the $(\mathord-)\otimes_k{k&amp;#39;}$ functor, which is exact.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-09T04:04:46.047" CommentCount="1" />
  <row Id="13605" PostTypeId="1" CreationDate="2010-12-09T04:18:44.373" Score="1" ViewCount="66" Body="&lt;p&gt;As I understand, martingale is a stochastic process (i.e., a sequence of random variables) such that the conditional expected value of an observation at some time $t$, given all the observations up to some earlier time $s$, is equal to the observation at that earlier time $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sequence $Y_1, Y_2, Y_3 ...$ is said to be a martingale with respect to another sequence $X_1, X_2, X_3 ...$ if for all $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(Y_{n+1}|X_1,...,X_n) = Y_n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I don't understand how it is defined in terms of filtration. Does filtration discretize the time space of a stochastic process so that we can analyze the process as a martingale? A simple explanation or an example on what is filtration and how it relates to martingale theory would be very helpful. I can then read more detailed content.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-12-09T04:18:44.373" Title="martingale and filtration" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;expectation&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="13606" PostTypeId="1" AcceptedAnswerId="13607" CreationDate="2010-12-09T05:28:05.033" Score="2" ViewCount="59" Body="&lt;p&gt;In a set of lecture notes, there is an example of calculating the group $E_\text{tors}$ of an elliptic curve. This is the example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $E$ be the elliptic curve &#xA;  $$y^2=x^3-5x+4.$$&#xD;&#xA;  The curve has discriminant&#xA;  $$\Delta=16(4\cdot 125 - 27\cdot 16)=2^6\cdot 17.$$&#xD;&#xA;  Therefore, if $[a:b:1]$ is a torsion point, we must have&#xA;  $$b=0,1,2,4,\text{or}\ 8.$$&#xD;&#xA;  Now the case $b=0$ clearly occurs with the point $[1:0:1]$. Let $f(x)$ denote the function $x^3-5x+4$. This function has derivative $3x^2-5$, which is positive for $|x|\geq 2$. Also, we have &#xA;  $$f(-3)=-8,\quad f(5)=104.$$&#xD;&#xA;  Combining these two statements we see the only possible integer values for $x$ for which $f(x)$ is between $0$ and $64$ are&#xA;  $$x=-2,-1,0,1,2,3,4.$$&#xD;&#xA;  Computing each of these two statements we see that the only possible torsion points are &#xA;  $$[0:\pm 2: 1],[3:\pm 4:1],[0:1:0],[1:0:1].$$&#xD;&#xA;  Also, one computes that &#xA;  $$[0:2:1]+[1:0:1]=[3:4:1].$$&#xD;&#xA;  However, if you compute &#xA;  $$[0:2:1]+[0:2:1]$$&#xD;&#xA;  then you find that the $x$-coordinate is $25/16$. As this is not an integer, by the Lutz-Nagell thereom $[0:2:1]$ is not a torsion point, and therefore $[3:4:1]$ is not a torsion point either. &#xA;  This implies that $E_\text{tors}=\{[0:1:0],[1:0:1]\}\simeq\mathbb{Z}/(2)$.&#xA;  This also show that the rank of $E$ is $\geq 1$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I understand why $[0:2:1]+[1:0:1]=[3:4:1]$, but it seems to me that there's the implicit assumption that $[0:1:0]$ and $[1:0:1]$ are already in the torsion group. Thus we know that since $[0:2:1]$ is not in the group, neither is $[3:4:1]$, for when composed with the inverse of $[1:0:1]$, it would give an element not in $E_\text{tors}$, which would mean $E_\text{tors}$ is not closed under composition. My question is, is it obvious that $[0:1:0]$ and $[1:0:1]$ are in the torsion group? Otherwise, I don't see why it would follow from $[0:2:1]+[1:0:1]=[3:4:1]$ that $[3:4:1]$ is not in torsion group, if we don't necessarily already know that $[1:0:1]$ is, and hence it's inverse, is in $E_\text{tors}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it generally true that if $[a:b:1]$ is not a torsion point, then neither is $[a,-b:1]$? This seems to be assumed as well. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-12-09T06:42:22.617" Title="Understanding how to calculate $E_\text{tors}$ of an elliptic curve." Tags="&lt;number-theory&gt;&lt;elliptic-curves&gt;" AnswerCount="2" />
  <row Id="13607" PostTypeId="2" ParentId="13606" CreationDate="2010-12-09T05:37:23.513" Score="5" ViewCount="0" Body="&lt;p&gt;I think that pretty much everything you're asking will become clear if you notice what the neutral element is here (hint: it's the point at infinity). &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-12-09T05:37:23.513" CommentCount="14" />
  <row Id="13608" PostTypeId="1" AcceptedAnswerId="13609" CreationDate="2010-12-09T05:48:24.953" Score="4" ViewCount="168" Body="&lt;p&gt;How can I prove that $\gcd(7^{79}+5,7^{78}+3) = 4$ ? This was a question on a past exam, so the naive euclidean algorithm doesn't seem to suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure where to start with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is exam prep, not homework.&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T06:09:07.073" LastActivityDate="2010-12-09T16:21:36.150" Title="Show that $\gcd(7^{79}+5,7^{78}+3) = 4$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13609" PostTypeId="2" ParentId="13608" CreationDate="2010-12-09T05:52:01.057" Score="7" ViewCount="0" Body="&lt;p&gt;For a first step, $7^{79}+5 - 7*(7^{78}+3) = -16$, which gets you a long way.  Then you only need to study the factors of 2 in the numbers.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T06:02:35.100" LastActivityDate="2010-12-09T06:02:35.100" CommentCount="3" />
  <row Id="13611" PostTypeId="2" ParentId="13608" CreationDate="2010-12-09T06:02:25.320" Score="4" ViewCount="0" Body="&lt;p&gt;Since $\gcd(a,b)=\gcd(a-b,b)$, $$\begin{align}&#xD;&#xA;\gcd(7^{79}+5,7^{78}+3)&#xD;&#xA;&amp;amp;=\gcd(7\cdot 7^{78}+5-7^{78}-3,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(6\cdot 7^{78}+2,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(6\cdot 7^{78}+2-7^{78}-3,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(5\cdot 7^{78}-1,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;\vdots&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(7^{78}-13,7^{78}+3)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(7^{78}-13,7^{78}+3-7^{78}+13)&#xD;&#xA;\\&#xD;&#xA;&amp;amp;=\gcd(7^{78}-13,16)&#xD;&#xA;\\&#xD;&#xA;\end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, I'd determine the remainder when $7^{78}$ is divided by 16 and use that to see how $7^{78}-13$ compares to 16.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-09T06:02:25.320" />
  <row Id="13612" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T06:03:10.137" Score="1" ViewCount="0" Body="&lt;p&gt;I'm doing exactly the same thing as you right now.  I'm self-learning differential topology and differential geometry.  To those ends, I really cannot recommend John Lee's &quot;Introduction to Smooth Manifolds&quot; and &quot;Riemannian Manifolds: An Introduction to Curvature&quot; highly enough.  &quot;Smooth Manifolds&quot; covers Stokes Theorem, the de Rham theorem and more, while &quot;Riemnannian Manifolds&quot; covers connections, metrics, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attention to detail that Lee writes with is so fantastic.  When reading his texts that you know you're learning things the standard way with no omissions.  And of course, the same goes for his proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus, the two books are the second and third in a triology (the first being his &quot;Introduction to Topological Manifolds&quot;), so they were really meant to be read in this order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I also agree that Guilleman and Pollack, Hirsch, and Milnor are great supplements, and will probably emphasize some of the topological aspects that Lee doesn't go into.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-09T06:03:10.137" />
  <row Id="13613" PostTypeId="2" ParentId="13596" CreationDate="2010-12-09T06:09:33.073" Score="1" ViewCount="0" Body="&lt;p&gt;$Y$ counts the number of successes in $n$ trials. $n - Y$ counts the number of failures in $n$ trials. Notion of success and failure is arbitrary. $P[Y = i] = P[n - Y = n - i]$, so $n-Y$ is binomially distributed with &quot;success&quot; probability $1 - p$.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastActivityDate="2010-12-09T06:09:33.073" />
  <row Id="13614" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T06:11:35.527" Score="0" ViewCount="0" Body="&lt;p&gt;Like the other posters, I think Lee's books are fantastic.  I'd start with his &lt;em&gt;Introduction to Smooth Manifolds&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential geometry, I'd go on to his &lt;em&gt;Riemannian Manifolds&lt;/em&gt; and then follow up with do Carmo's &lt;em&gt;Riemannian Geometry&lt;/em&gt;.  (That's what I did.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For differential topology, after Lee's &lt;em&gt;Smooth Manifolds&lt;/em&gt;, I'd suggest &lt;em&gt;Differential Forms in Algebraic Topology&lt;/em&gt; by Bott, Tu and anything (and everything) by Milnor.&lt;/p&gt;&#xA;" OwnerUserId="4074" LastActivityDate="2010-12-09T06:11:35.527" />
  <row Id="13615" PostTypeId="2" ParentId="13606" CreationDate="2010-12-09T06:28:39.203" Score="1" ViewCount="0" Body="&lt;p&gt;The point here is that the elliptic curve is not just the affine line whose equation you wrote; actually it is the projective line whose equation is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y^z = x^3 + axz^2 + bz^3$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose solutions are triples of rational numbers $[a,b,c]$ not all of which are zero, with the additional condition that the two triples $[ta,tb,tc]$ is treated as the same as $[a,b,c]$ for rational $t \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line you wrote is an affine segment of the projective line. The projective point $[a,b,c]$ on the projective line with $c\neq 0$ is equivalent to $[a/c,b/c,1]$ on the affine line. But this transformation works only for $c \neq 0$, and to treat $c = 0$, you must include the equivalence class of one additional point, $[0,1,0]$, which is called the point at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That point happens to be the identity for the addition in the group law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you can write an explicit formula for addition and verify everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the inverse of the element $[a,b,1]$ is $[a,-b,1]$. You can check that in any abelian group $G$, if $x \in G$ is torsion, then $-x$ is also torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, to compute the torsion points on an elliptic curve, you can start with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem&quot; rel=&quot;nofollow&quot;&gt;Nagell-Lutz theorem&lt;/a&gt; for a set of possible points and then do computation on each of those points to make an explicit check. The &lt;a href=&quot;http://pari.math.u-bordeaux.fr/&quot; rel=&quot;nofollow&quot;&gt;PARI&lt;/a&gt; routine &lt;em&gt;elltors&lt;/em&gt; can do it automatically for you; for the syntax just download and check some reference for PARI.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here in your particular example, you have to check manually that $[1,0,1]$ is in the torsion group. Write down the formula and check, or use the appropriate PARI routine. I have explained that $[0,1,0]$ is the identity. The rest, you seem to have got right.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-09T06:42:22.617" LastActivityDate="2010-12-09T06:42:22.617" CommentCount="1" />
  <row Id="13616" PostTypeId="1" CreationDate="2010-12-09T06:42:57.503" Score="5" ViewCount="61" Body="&lt;p&gt;Consider large (100,000+ vertices, say) graphs, which we think of as representing some population with edges representing some form of symmetric relation.  They might be the Friend graph of Facebook, mathematicians with the collaboration relation, or a large computer network.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These networks have the property that they are neither highly structured, nor totally random.  No information about other edges on the graph can tell me for certain whether a given pair of vertices is connected.  That said, if a given pair of vertices have many common neighbors, then it is considerably more likely that they are connected by an edge (so it is not entirely random).  I've seen some lectures on graphs like this, and I understand they are a productive area of research (see, for instance, Kleinberg or Lovasz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious about the following phenomenon (my description is vague, but part of my question is asking for a good definition).  These networks tend to have subsets (which I will call '&lt;strong&gt;clumps&lt;/strong&gt;') which are significantly more connected to each other than to the average vertex in the graph.  Consider a college in Facebook or a research group in mathematics, for example.  If the graphs were small enough to draw in a reasonable way, such clumps would be obvious to the naked eye.  For very large graphs, this is impractical; so instead, I ask,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) What is a graph-theoretic way to characterize these clumps?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly, there won't be a yes-no criterion, but I am hoping for some quantity that measures how much a given subset is a clump.  This should also factor in the statistical significance of the clump.  Very small subsets which are highly connected will happen even in totally random graphs, whereas a large subset which is even moderately well-connected is unlikely in a random graph, and would be interesting to find.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Given a graph (and a definition of a clump), how does one find the clumps?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a definition and an algorithm so robust that it can take networks like Facebook or the collaboration graph, and return the clumps that we know are there, like colleges or research discplines?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Oh, and I am not looking for the Szemeredi partition of a graph, which has some similarities to the kinds of partitions I am looking for, but is explicitly a partition of the graph into similar sized chunks.  The clumps in a graph don't have to be the same size, disjoint, or contain every element.&lt;/p&gt;&#xA;" OwnerUserId="1062" LastActivityDate="2010-12-17T03:08:44.013" Title="How to Characterize Clumps in a Large, Semi-Random Graph" Tags="&lt;graph-theory&gt;&lt;pattern-recognition&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="13617" PostTypeId="1" AcceptedAnswerId="13622" CreationDate="2010-12-09T06:46:57.933" Score="12" ViewCount="253" Body="&lt;p&gt;Background: Let $n$ be an integer and let $p$ be a prime. If $p^{e} || n$, we write $v_{p}(n) = e$. A natural number $n$ is a sum of two integer squares if and only if for each prime $p \equiv 3 \pmod 4$, $v_{p}(n)$ is even. Every natural number is a sum of four squares. A natural number $n$ is a sum of three squares if and only if it is not of the form $4^{k}u$ where $u \equiv 7 \pmod 8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know why it is harder to prove the above result for sums of three squares as opposed to sums of two squares or four squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard somewhere that one way to see this involves modular forms... but I don't remember any details. I would also like to know if there is a formula for the number of ways of representing a natural number n as a sum of three squares (or more generally, $m$ squares) that is similar in spirit to the formulas for the number of ways of representing a natural number as the sum of two squares and four squares. &lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T06:52:19.593" LastActivityDate="2010-12-09T16:39:48.927" Title="Why is it harder to prove which integers are sums of three squares rather than sums of two squares or four squares?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="13618" PostTypeId="2" ParentId="13617" CreationDate="2010-12-09T06:56:30.477" Score="12" ViewCount="0" Body="&lt;p&gt;An explanation from a different direction than you may be expecting: one reason why the sum-of-three-squares problem is so much harder is that there's no normed division algebra of dimension three.  A key element of both the two-squares and four-squares proof is expressing the product of two numbers in the relevant form as another number in that form; these product formulas 'come from' the formulas for multiplying complex and quaternionic numbers respectively.  That is, taking norms of $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$ gives rise to the formula $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$, and the equivalent formula for quaternions gives Euler's four-square identity.  John Baez wrote a particularly interesting article on why you don't keep getting division algebras as you keep doubling the dimension; you should be able to find it on his homepage if you're curious.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-12-09T06:56:30.477" CommentCount="6" />
  <row Id="13619" PostTypeId="2" ParentId="13608" CreationDate="2010-12-09T06:58:34.580" Score="7" ViewCount="0" Body="&lt;p&gt;Modulo the gcd: $\rm\ 7^{78}\ \equiv -3\ \Rightarrow\ 0\ \equiv\ 7^{79}+5\ \equiv\ 7(-3)+5\ \equiv -16\:,\ $ i.e. the gcd dvides $16\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\rm\ mod\ 8:\ 7^{78}+3\ \equiv\ (-1)^{78}+3\ \equiv\ 4\ \equiv\ (-1)^{79}+5\ \equiv\ 7^{79}+5\:,\ $  i.e. $\ $ gcd $\rm\equiv 4\ (mod\ 8)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence:$\ $ gcd $\rm = 4 + 8\ k\ $ divides $16\:$ implies $\rm\ k = 0\:,\ $ so $\ $ gcd $ = 4\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the calculations become more intuitive by working in various rings $\rm\ \mathbb Z/m\:.$ Doing such allows us to reuse our well-honed intuition of arithmetic &lt;em&gt;operations&lt;/em&gt; (ring laws), versus the much more cumbersome $\ $ and $\ $ much less intuitive divisibility &lt;em&gt;relation&lt;/em&gt;, i.e. calculating in equational algebras is simpler than calculating in relational algebras, so whenever a problem can be converted from relational to equational it usually yields a simplification.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-09T16:21:36.150" LastActivityDate="2010-12-09T16:21:36.150" CommentCount="1" />
  <row Id="13620" PostTypeId="2" ParentId="13477" CreationDate="2010-12-09T07:50:36.327" Score="1" ViewCount="0" Body="&lt;p&gt;There is no solution without inverse trigonometric functions, because any elementary geometric argument would, for a given value of the angle between the tangent lines, calculate all 4 arc lengths as linear functions of $\alpha$.  This would imply that, as the (secant) line through $H$ is rotated at uniform angular speed, it sweeps out arclength on the circle at a uniform rate.  But this is impossible because the distance from $H$ to the intersection point, $|HG|$, is not constant.  It is not hard to make this argument rigorous using calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the arcs, in the order listed in the question, are denoted $A, B, b, a$, then  elementary geometry does express $(A-a)$, $(B-b)$, and $(A+B+b+a)$ as linear functions of $\alpha, \beta$ and $1$, as in the other solutions.   I think this set of 3 equations in 4 unknowns is what the OP meant by &quot;not enough equations&quot; to solve everything purely geometrically.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-09T21:42:59.857" LastActivityDate="2010-12-09T21:42:59.857" />
  <row Id="13621" PostTypeId="2" ParentId="13575" CreationDate="2010-12-09T08:09:39.660" Score="0" ViewCount="0" Body="&lt;p&gt;I would recommend Jost's book &quot;Riemannian geometry and geometric analysis&quot; as well as Sharpe's &quot;Differential geometry&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first book is pragmatically written and guides the reader to a lot of interesting stuff, like Hodge's theorem, Morse homology and harmonic maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second book is mainly concerned with Cartan connection, but before that it has an excellent chapter on differential topology. Furthermore it treats Ehresmann connections in appendix A.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-09T08:09:39.660" />
  <row Id="13622" PostTypeId="2" ParentId="13617" CreationDate="2010-12-09T08:14:19.563" Score="12" ViewCount="0" Body="&lt;p&gt;The modular forms explanation is basically due to the fact&#xA;that $3$ is odd and so the generating function for representations&#xA;of sums of three squares is a modular form of half-integer weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general if $r_k(n)$ is the number of representations of $n$&#xD;&#xA;as a sum of $k$ squares then&#xA;$$\sum_{n=0}^\infty r_k(n)q^n=\theta(z)^k$$&#xD;&#xA;where $q=\exp(\pi i z)$ and &#xA;$$\theta(z)=1+2\sum_{n=1}^\infty q^{n^2}.$$&#xD;&#xA;Then $f_k(z)=\theta(z)^k$ is a modular form of weight $k/2$ for the group&#xA;$\Gamma_0(4)$. This means that&#xA;$$f_k((az+b)/(cz+d))=(cz+d)^{k/2}f_k(z)$$&#xD;&#xA;whenever the matrix $\begin{pmatrix}a&amp;amp;b\\\\c&amp;amp;d\end{pmatrix}$&#xD;&#xA;lies in $\Gamma_0(4)$, that is $a$, $b$, $c$ and $d$ are integers, $4\mid c$&#xD;&#xA;and $ad-bc=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition is easy to understand when $k$ is even, but for odd&#xA;$k$ one needs to take the correct branch of $(cz+d)^{k/2}$, and&#xA;this is awkward. The space of modular forms of weight $k/2$ is&#xA;finite-dimensional for all $k$, and is one-dimensional for small enough $k$.&#xA;For these small $k$ the space is spanned by an &quot;Eisenstein series&quot;.&#xA;Computing the Eisenstein series isn't too hard for even $k$, but is&#xA;much nastier for odd $k$ where again square roots need to be&#xA;dealt with. See Koblitz's book on modular forms and elliptic functions&#xA;for the calculation for $k\ge5$ odd. The calculation for $k=3$&#xD;&#xA;is even nastier as the Eisenstein series does not converge absolutely.&#xA;In fact the cases where $k$ is divisible by $4$ are even easier, as&#xA;even weight modular forms behave nicer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large $k$, Eisenstein series are no longer enough, one needs&#xA;also &quot;cusp forms&quot;. While fascinating, cusp forms have coefficients&#xA;which aren't given by nice formulae unlike Eisenstein series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there is a formula for $r_3(n)$, due to Gauss in his&#xA;&lt;em&gt;Disquisitiones Arithmeticae&lt;/em&gt;. It involves class numbers of quadratic&#xA;fields (or to Gauss numbers of classes of integral quadratic forms).&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-09T16:39:48.927" LastActivityDate="2010-12-09T16:39:48.927" CommentCount="4" />
  <row Id="13624" PostTypeId="1" CreationDate="2010-12-09T08:45:34.767" Score="2" ViewCount="71" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $X$ be a topological space such that every open cover has a finite refinement. Then is $X$ compact, or is there a counterexample?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $X$ be a topological space such that every open cover has a locally finite subcover. Then is $X$ compact, or is there a counterexample?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4561" LastActivityDate="2010-12-25T07:01:29.440" Title="Weakening paracompactness condition." Tags="&lt;general-topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13625" PostTypeId="1" CreationDate="2010-12-09T08:45:55.577" Score="0" ViewCount="200" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://math.stackexchange.com/questions/8808/why-the-interest-in-locally-euclidean-spaces&quot;&gt;this question&lt;/a&gt;, and I idly wondered if there could be spaces that were globally Euclidean, but locally not, and couldn't think of any circumstances in which that might be the case.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-12-11T01:47:54.680" Title="Are there spaces that are globally Euclidean, but locally not?" Tags="&lt;general-topology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="13627" PostTypeId="1" AcceptedAnswerId="13668" CreationDate="2010-12-09T09:13:31.177" Score="0" ViewCount="130" Body="&lt;p&gt;&lt;strong&gt;The Question:&lt;/strong&gt; Are there any ways that &quot;applied&quot; mathematicians can use Homology theory?  Have you seen any good applications of it to the &quot;real world&quot; either directly or indirectly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do I care?&lt;/strong&gt;  Topology has appealed to me since beginning it in undergrad where my university was more into pure math.  I'm currently in a program where the mathematics program is geared towards more applied mathematics and I am constantly asked, &quot;Yeah, that's cool, but what can you use it for in the real world?&quot;  I'd like to have some kind of a stock answer for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Full Disclosure.&lt;/strong&gt; I am a first year graduate student and have worked through most of Hatcher, though I am not by any means an expert at any topic in the book.  This is also my first post on here, so if I've done something wrong just tell me and I'll try to fix it.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-09T13:04:18.547" LastActivityDate="2010-12-09T14:57:42.770" Title="Applications for Homology." Tags="&lt;topology&gt;&lt;algebraic-topology&gt;&lt;applications&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13628" PostTypeId="2" ParentId="13111" CreationDate="2010-12-09T09:15:49.477" Score="0" ViewCount="0" Body="&lt;p&gt;Ok..I got something here..John.this is betty from class..let me know what you guys think of this.&#xA;Proof:&#xA;Let G be a simple group of order 300. i.e, |G|=300&#xA;assume ----&gt;&amp;lt;----- that G is simple. &#xA;Since |G|=300=2^2*3*5^2, then there exists a sylow 5-subgroup, say P that has order 25. Now s_5 is congruent to 1(mod 5) and s_5 divides 300/25=12. Hence s_5=1 or 6. But if s_5=1, then P&amp;lt;|G, which is impossible. Thus s_5=6 and |G:N_G(P)|=6. Now recall a thm, proved from class. &quot;Let G be a group. If there exists a subgroup of G s.t [G:H]=n, then there exists a N&amp;lt;|G, s.t N C H, and [G:N] | n!.&quot; We apply this thm to |G:N_G(P)|=6 now. If |G:N_G(P)|=6, then there exists a normal subgroup of G that is contained in N_G(P), say this subgroup is P (again), where P&amp;lt;|G and P C N_G(P), then [G:P] | 6!. But this is impossible, since |G|=300 does not divide 6!. Also, if G were simple, then P would be 1 or G, and we already showed above that P cannot be equal to 1, and P is not equal to G. Thus we have reached a contradiction. Therefore G cannot be simple, and there is no simple group of order 300.&lt;/p&gt;&#xA;" OwnerUserId="4562" LastActivityDate="2010-12-09T09:15:49.477" />
  <row Id="13630" PostTypeId="1" AcceptedAnswerId="13645" CreationDate="2010-12-09T09:41:23.547" Score="0" ViewCount="37" Body="&lt;p&gt;Let $X1,X2$ be independent random variables with $P[Xi = 0] = P[Xi = 1] = \frac{1}{2}$ for $i = 1, 2$. What is the random variable set $A1 = \{X1 = X2\}$ mean? I came across this representation in the book but not sure what this means.&lt;/p&gt;&#xA;" OwnerUserId="957" LastActivityDate="2010-12-09T13:14:25.287" Title="what does this set definition mean defined on independent random variables?" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="3" />
  <row Id="13631" PostTypeId="1" CreationDate="2010-12-09T09:41:46.680" Score="1" ViewCount="128" Body="&lt;p&gt;One of false beliefs in &lt;a href=&quot;http://mathoverflow.net/questions/23478/23640#23640&quot; rel=&quot;nofollow&quot;&gt;this question on Math Overflow&lt;/a&gt; is &quot;If f is a smooth function with df=0, then f is constant&quot;. What is a counterexample to this statement? Can it be made correct by adding some restriction, e.g. that f is a function from reals to reals?&lt;/p&gt;&#xA;" OwnerUserId="947" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T09:46:47.723" LastActivityDate="2010-12-09T18:58:07.143" Title="Example of a smooth function with zero derivative that is not constant." Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13633" PostTypeId="2" ParentId="13630" CreationDate="2010-12-09T09:48:49.330" Score="0" ViewCount="0" Body="&lt;p&gt;It's the event that $X_1 = X_2$.  So $X_1 = 0, X_2 = 0$ or $X_1 = 1, X_2 = 1$. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T09:48:49.330" />
  <row Id="13634" PostTypeId="2" ParentId="13534" CreationDate="2010-12-09T09:56:26.953" Score="1" ViewCount="0" Body="&lt;p&gt;Actually, this can be done very simply using the following two facts, both of which follow from things the students should know at this point (though they may not have seen a centralizer before, it is not too hard to understand).&#xA;First, any subgroup of prime index is maximal. This follows directly from Lagrange.&#xA;Second, the center is never maximal. This follows by observing that if the center is not the entire group, let x be an element not in the center. Then the centralizer of x is not the entire group, since then x would be in the center. But clearly the center is contained in the centralizer of x and does not contain x, so the centralizer of x is a subgroup properly between the center and the entire group.&#xA;Now it is easy to see that the group must be abelian, and then it is easy to show that it must in fact be cyclic (it has an element of order 2, which is simpler to show than Cauchy, and then it just needs to be shown that it cannot have 2 elemens of order 2, since then it would have a subgroup of order 4 contradicting Lagrange).&#xA;(As a side note, Cauchy's theorem for abelian groups is fairly simple to show just from knowledge of the order of a product of subgroups, and in fact the general theorem then follows from the class equation)&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2010-12-09T09:56:26.953" />
  <row Id="13635" PostTypeId="2" ParentId="13624" CreationDate="2010-12-09T09:59:52.223" Score="2" ViewCount="0" Body="&lt;p&gt;I quote from Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Paracompact_space&quot; rel=&quot;nofollow&quot;&gt;paracompact spaces&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;Note the similarity between the definitions of compact and paracompact: for paracompact, we replace &quot;subcover&quot; by &quot;open refinement&quot; and &quot;finite&quot; by &quot;locally finite&quot;. Both of these changes are significant: if we take the above definition of paracompact and change &quot;open refinement&quot; back to &quot;subcover&quot;, or &quot;locally finite&quot; back to &quot;finite&quot;, we end up with the compact spaces in both cases.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The page lists Willard's &lt;em&gt;General Topology&lt;/em&gt; in its references, so that seems like a good place to go hunting for a proof - or an exercise.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-09T09:59:52.223" CommentCount="2" />
  <row Id="13636" PostTypeId="1" CreationDate="2010-12-09T10:04:31.250" Score="0" ViewCount="33" Body="&lt;p&gt;The category of simplicial presheaves on a small Grothendieck site $\mathcal{C}$ can be given a model structure by defining weak equivalences and cofibrations sectionwise. It's called the (global) injective model structure and has a mapping space functor $Hom$, given by $Hom(X, Y)_n = hom(X\times \Delta^n, Y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the given topology, one can actually define 'local' weak equivalences and Jardine showed that the left Bousfield localization at the class of local weak equivalence exists. This model category structure is called local injective model structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's often mentioned that this is also the same as localizing the injective model structure at the class $S := \{X\rightarrow L^2X\}$ (where $L^2$ is the sheafification functor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $W$ denotes the class of local weak equivalences, this amounts to say that $S$-local is the same as $W$-local, that is for an injective fibrant simplicial presheav $A$ the following are equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $Hom(L^2X, A) \cong Hom(X, A)$ for all $X$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $Hom(Y, A) \cong Hom(X, Y)$ for all local weak equivalences $X\rightarrow Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\cong$ stands for weak equivalence of simplicial sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) implies 1), of course. But how does 1) imply 2)? Or am I mistaken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In general 1) does not imply 2), and this is not stated anywhere. I mixed things up.&lt;/p&gt;&#xA;" OwnerUserId="4563" LastEditorUserId="4563" LastEditorDisplayName="" LastEditDate="2010-12-09T19:35:44.637" LastActivityDate="2010-12-09T19:35:44.637" Title="Local injective model structure for simplicial presheaves" Tags="&lt;homotopy-theory&gt;" CommentCount="3" />
  <row Id="13637" PostTypeId="2" ParentId="7949" CreationDate="2010-12-09T10:22:23.303" Score="1" ViewCount="0" Body="&lt;p&gt;Considering the category $\Sigma_R(C) $ of the exat seqences like: $0\to A\to B\to C$ this category is complete and the inclusion $\Sigma(C)  \subset  \Sigma _R(C)$ is coreflexive (the coreflector come from the coker of the last arrow), then limits in $\Sigma(C) $ exixst and are gived by the limit in $\Sigma_R(C) $ followed by the coreflection.&#xA;WHen the inclusion $\Sigma(C)  \subset  \Sigma _R(C)$ preserve (directed) limits is argument of Mittag-Leffler teorem's.&lt;/p&gt;&#xA;" OwnerUserId="4565" LastActivityDate="2010-12-09T10:22:23.303" CommentCount="1" />
  <row Id="13638" PostTypeId="1" CreationDate="2010-12-09T10:25:06.450" Score="3" ViewCount="44" Body="&lt;p&gt;The definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss-Manin_connection&quot; rel=&quot;nofollow&quot;&gt;Gauss-Manin&lt;/a&gt; connection involves de Rham cohomology. Surely, Gauss didn't work with de Rham cohomology as we know it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what was the context in which Gauss came up with this idea?&lt;/p&gt;&#xA;" OwnerUserId="4566" LastActivityDate="2010-12-09T10:25:06.450" Title="What is the Gauss part of Gauss-Manin connection?" Tags="&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;" CommentCount="1" />
  <row Id="13639" PostTypeId="1" CreationDate="2010-12-09T11:12:35.300" Score="1" ViewCount="50" Body="&lt;p&gt;Let Let $E \rightarrow X$ be a vector bundle on a manifold $X$. Let $\cal E$ be the sheaf of sections of $E$. Let $\cal F$ be a subsheaf of $\cal E$, and let $F$ be the etale space of $\cal F$. What is an example that the map $F \rightarrow E$ might not be an injection on all fibers?&lt;/p&gt;&#xA;" OwnerUserId="4561" LastEditorUserId="4561" LastEditorDisplayName="" LastEditDate="2010-12-09T11:16:15.170" LastActivityDate="2010-12-09T16:22:38.233" Title="Subbundles and subsheaves." Tags="&lt;differential-topology&gt;&lt;sheaf-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13640" PostTypeId="1" AcceptedAnswerId="13650" CreationDate="2010-12-09T11:20:43.357" Score="0" ViewCount="74" Body="&lt;p&gt;Classify up to similarity all 3 x 3 complex matrices $A$ such that $A^n$ = $I$.&lt;/p&gt;&#xA;" OwnerUserId="4433" LastActivityDate="2010-12-09T13:19:50.820" Title="classification up to similarity of complex n-by-n matrices" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13641" PostTypeId="1" AcceptedAnswerId="13643" CreationDate="2010-12-09T11:48:55.787" Score="8" ViewCount="194" Body="&lt;p&gt;I'm reading the Algebra book by Knapp and he mentions in passing that an orthonormal set in an infinite dimension vector space is &quot;never large enough&quot; to be a vector-space basis (i.e. that every vector can be written as a &lt;strong&gt;finite&lt;/strong&gt; sum of vectors from the basis; such bases do exist for infinite dimension vector spaces by virtue of the axiom of choice, but usually one works with orthonormal sets for which &lt;strong&gt;infinite&lt;/strong&gt; sums yield all the elements of the vector space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is - how can we prove this claim? (That an orthonormal set in an infinite dimension vector space is not a vector space basis).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit (by Jonas Meyer): Knapp says in a footnote on &lt;a href=&quot;http://books.google.com/books?id=NSXCaGSVaX4C&amp;amp;lpg=PR1&amp;amp;pg=PA92&quot; rel=&quot;nofollow&quot;&gt;page 92&lt;/a&gt; of &lt;em&gt;Basic algebra&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the infinite-dimensional theory the term &quot;orthonormal basis&quot; is used for an orthonormal set that spans $V$ when &lt;em&gt;limits&lt;/em&gt; of finite sums are allowed, in addition to finite sums themselves; when $V$ is infinite-dimensional, an orthonormal basis is never large enough to be a vector-space basis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, without explicitly using the word, Knapp is referring only to &lt;em&gt;complete&lt;/em&gt; infinite-dimensional inner product spaces.&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-10T03:44:17.727" LastActivityDate="2010-12-10T03:44:17.727" Title="An orthonormal set cannot be a basis in an infinite dimension vector space?" Tags="&lt;linear-algebra&gt;&lt;orthonormal&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="13642" PostTypeId="1" AcceptedAnswerId="13670" CreationDate="2010-12-09T11:53:01.720" Score="1" ViewCount="33" Body="&lt;p&gt;I'm modeling a process with a Markov chain like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;Pr(i \rightarrow i) = \alpha,&#xD;&#xA;Pr(i \rightarrow i+1) = \beta,&#xD;&#xA;Pr(i \rightarrow i+2) = \gamma&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other probabilities are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to calculate the expected number of steps it takes to reach state $n$ from state $0$? So far, all I've come up is binomial distribution in the case of $\alpha = 0$, but I'm unsure of even that.&lt;/p&gt;&#xA;" OwnerUserId="4567" LastEditorUserId="4567" LastEditorDisplayName="" LastEditDate="2010-12-09T11:58:36.133" LastActivityDate="2010-12-09T15:42:22.697" Title="Persistent stationary Markov chain properties" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13643" PostTypeId="2" ParentId="13641" CreationDate="2010-12-09T12:01:05.393" Score="10" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Edit.&lt;/strong&gt; The answer assumes that one works in a Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathcal A$ denote the set of orthonormal vectors. Then for any orthonormal sequence $(e_n)\subset\mathcal A $,  the element &#xA;$$u=\sum_{n=1}^{\infty}\frac{1}{n}e_n$$&#xD;&#xA;cannot be represented as a finite sum of vectors from  $\mathcal A$.&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-09T12:41:21.537" LastActivityDate="2010-12-09T12:41:21.537" CommentCount="4" />
  <row Id="13644" PostTypeId="2" ParentId="13641" CreationDate="2010-12-09T12:19:55.093" Score="12" ViewCount="0" Body="&lt;p&gt;Take any infinite dimensional inner product space $V$ and any orthonormal sequence $(w_n : n \in \mathbb{N})$. Let $W$ be the subspace generated by this sequence. Then $W$ is certainly an infinite dimensional vector space (because it has an infinite independent subset). Also $W$ has an orthonormal basis, because the inner product on $W$ is inherited from $V$ and thus $(w_n)$ is still an orthonormal sequence in $W$.  This means that the theorem you have suggested, &quot;an orthonormal set in an infinite dimension vector space is not a vector space basis&quot;, is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I believe might be true is that no infinite dimensional &lt;em&gt;complete&lt;/em&gt; inner product space has a orthonormal basis. This is the question that Andrey Rekalo addressed in another answer.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-09T12:19:55.093" CommentCount="1" />
  <row Id="13645" PostTypeId="2" ParentId="13630" CreationDate="2010-12-09T12:19:57.587" Score="3" ViewCount="0" Body="&lt;p&gt;Although this question may look very simple, a full understanding of it is not. For general purposes, I give the following elaborated answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if $X_1$ and $X_2$ are any two random variables on a common probability space $(\Omega,\mathcal{F},{\rm P})$, then $\lbrace X_1 = X_2 \rbrace$ stands as a shorthand for the event $\lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace$. For your specific example, we can define $(\Omega,\mathcal{F},{\rm P})$ as follows:&#xA;$\Omega = \lbrace (0,0),(0,1),(1,0),(1,1) \rbrace$, $\mathcal{F}=2^\Omega$ (the power set of $\Omega$), and, for any $\omega = (i,j) \in \Omega$, ${\rm P}(\lbrace \omega \rbrace) = 1/4$. Note that, by additivity, this determines (the probability measure) ${\rm P}$ on $\mathcal{F}$; for example,&#xA;$$&#xD;&#xA;{\rm P}(\lbrace (0,0),(1,0),(1,1)\rbrace) = {\rm P}(\lbrace (0,0)\rbrace) + {\rm P}(\lbrace (1,0)\rbrace) + {\rm P}(\lbrace (1,1)\rbrace) = 3/4.&#xD;&#xA;$$&#xD;&#xA;Now, we can define the random variables $X_1$ and $X_2$ as follows. For any $\omega = (i,j) \in \Omega$, $X_1 {(\omega)} = i$ and $X_2 {(\omega)} = j$. Thus, &#xA;$$&#xD;&#xA;{\rm P}(X_1 = 0) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = 0 \rbrace) = {\rm P}(\lbrace (0,0),(0,1) \rbrace) = 1/2&#xD;&#xA;$$&#xD;&#xA;and&#xA;$$&#xD;&#xA;{\rm P}(X_1 = 1) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = 1 \rbrace) = {\rm P}(\lbrace (1,0),(1,1) \rbrace) = 1/2,&#xD;&#xA;$$&#xD;&#xA;and analogously for $X_2$. As for the event in question,&#xA;$$&#xD;&#xA;\lbrace X_1 = X_2 \rbrace : = \lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace = \lbrace (0,0),(1,1) \rbrace,&#xD;&#xA;$$&#xD;&#xA;and so&#xA;$$&#xD;&#xA;{\rm P}(X_1 = X_2) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace) = {\rm P}(\lbrace (0,0),(1,1) \rbrace) = 1/2.&#xD;&#xA;$$&#xD;&#xA;Finally, let's see that $X_1$ and $X_2$ are indeed independent (this should be clear from our construction). This amounts to showing that, for any $i,j \in \lbrace 0,1 \rbrace$, &#xA;$$&#xD;&#xA;{\rm P}(X_1 = i, X_2 = j) = {\rm P}(X_1 = i) {\rm P}(X_2 = j), &#xD;&#xA;$$&#xD;&#xA;that is,&#xA;$$&#xD;&#xA;{\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = i, X_2 {(\omega)} = j \rbrace) = {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} =i  \rbrace ){\rm P}(\lbrace \omega \in \Omega : X_2 {(\omega)} =j  \rbrace ).&#xD;&#xA;$$&#xD;&#xA;Indeed, both sides are equal to $1/4$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T12:19:57.587" CommentCount="1" />
  <row Id="13646" PostTypeId="1" AcceptedAnswerId="13647" CreationDate="2010-12-09T12:42:09.437" Score="1" ViewCount="145" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &lt;a href=&quot;http://books.google.com/books?id=g9tH8M3NvDIC&amp;amp;printsec=frontcover&amp;amp;dq=irresistible+integrals&amp;amp;hl=pt-PT&amp;amp;ei=W8sATYbBLoSXOu2craYB&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Irresistible Integrals by George Boros and Victor Moll&lt;/a&gt; on page 204 has the following identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{1}{1+x}=\prod_{k=1}^{\infty}\left(\frac{k+x+1}{k+x} \times \frac{k}{k+1}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one derive this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-09T16:11:11.700" LastActivityDate="2010-12-09T16:11:11.700" Title="How to derive the following identity?" Tags="&lt;real-analysis&gt;&lt;complex-analysis&gt;" AnswerCount="2" />
  <row Id="13647" PostTypeId="2" ParentId="13646" CreationDate="2010-12-09T12:54:57.213" Score="10" ViewCount="0" Body="&lt;p&gt;Multiply out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{k=1}^N\left(\frac{k+x+1}{k+x} \times \frac{k}{k+1}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and cancel like terms. You will be left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{1+x}\left(\frac{N+x+1}{N+1}\right) .$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-09T12:54:57.213" CommentCount="4" />
  <row Id="13648" PostTypeId="1" AcceptedAnswerId="13649" CreationDate="2010-12-09T12:56:06.743" Score="11" ViewCount="403" Body="&lt;p&gt;How do I show that the product $$\biggl(4 - \frac21\biggr) \cdot \biggl(4 - \frac22\biggr) \cdot \biggl(4 - \frac23\biggr) \cdots \biggl(4 - \frac2{n}\biggr)$$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;a href=&quot;http://www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf&quot; rel=&quot;nofollow&quot;&gt;www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T16:54:59.377" LastActivityDate="2011-01-11T19:30:04.360" Title="Proving $\prod_{j=1}^n \left(4-\frac2{j}\right)$ is an integer" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="13649" PostTypeId="2" ParentId="13648" CreationDate="2010-12-09T13:05:15.317" Score="13" ViewCount="0" Body="&lt;p&gt;Show that it's equal to $ { 2n \choose n } . $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Okay, here's how it goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( 4 - \frac{2}{1} \right)&#xD;&#xA;\left( 4 - \frac{2}{2} \right)&#xD;&#xA;\left( 4 - \frac{2}{3} \right) \cdots&#xD;&#xA;\left( 4 - \frac{2}{n} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \frac{(4 \cdot 1 - 2)}{1}&#xD;&#xA;\frac{(4 \cdot 2 - 2)}{2}&#xD;&#xA;\frac{(4 \cdot 3 - 2)}{3} \cdots \frac{(4n - 2)}{n}$$&#xA;$$= \frac{2^n (2 \cdot 1 - 1)(2 \cdot 2 - 1)( 2 \cdot 3 - 1) \cdots (2n-1)}{n!}$$&#xA;$$= \frac{2^n \cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$$&#xA;$$= \frac{2^n \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdots (2n-1)(2n)}{n! \cdot 2 \cdot 4 \cdots (2n)}$$&#xA;$$ = \frac{2^n (2n)!}{ 2^n (n!)^2} = { 2n \choose n}.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-11T19:30:04.360" LastActivityDate="2011-01-11T19:30:04.360" CommentCount="3" />
  <row Id="13650" PostTypeId="2" ParentId="13640" CreationDate="2010-12-09T13:08:03.540" Score="6" ViewCount="0" Body="&lt;p&gt;If you define $p=X^n-1\in\mathbb C[X]$, then $p(A)=0$. This tells you that the minimal polinomial $m_A$ of $A$ divides $p$ and, in particular, that $m_A$ has all its roots simple, because the same is true of $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $A$ is diagonalizable, so, up to similarity, you can suppose that it is diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see which are the diagonal matrices $A$ which satisfy the condition $A^n=I$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; This argument does not depend on your knowing about Jordan canonical forms.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-09T13:19:50.820" LastActivityDate="2010-12-09T13:19:50.820" CommentCount="2" />
  <row Id="13651" PostTypeId="2" ParentId="13630" CreationDate="2010-12-09T13:14:25.287" Score="3" ViewCount="0" Body="&lt;p&gt;To add to Shai Covo's answer, it's worth noting that the convention of writing something like $\{X_1 = X_2\}$ for $\{\omega \in \Omega : X_1(\omega) = X_2(\omega)\}$, essentially &quot;suppressing the $\omega$s&quot;, is very common in probability, and you will see it a lot as you continue your studies.  Essentially, any &quot;statement&quot; written inside braces should be interpreted as the set of all $\omega$ for which the statement (which is generally something in terms of random variables, which are really functions of $\omega$) is true.  Other common examples are things like $\{X_n \to X\}$, which is the set of all $\omega$ for which $X_n(\omega)$ converges to $X(\omega)$.  At first you may find it helpful to &quot;fill in the $\omega$s&quot; when reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for intuition, it is often easy to understand such expressions.  If you are doing an experiment where you flip two coins, with heads and tables labeled 0 and 1, $\{X_1 = X_2\}$ is just the event that the two coins come up the same.&lt;/p&gt;&#xA;" OwnerUserId="822" LastActivityDate="2010-12-09T13:14:25.287" />
  <row Id="13652" PostTypeId="1" AcceptedAnswerId="13653" CreationDate="2010-12-09T13:19:04.220" Score="3" ViewCount="50" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;From the deck of the 52 cards,cards are drawn randomly without replacement.What is the probability of drawing a king of hearts at the third attempt? If it was drawn at the 15th attempt, what is the probability ?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have solved the first by finding P(getting the card at third attempt) = P(not getting at first) $\times$ P(not getting at second) $\times$ P(getting at third attempt) = $\frac{51}{52} \times \frac{50}{51} \times \frac{1}{50} = \frac{1}{52}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method seems fine when the times to be drawn is less but for 15 or say for $x$ ... I have solved for $15$ by just writing just two lines in &lt;a href=&quot;http://en.wikipedia.org/wiki/C_%28programming_language%29&quot; rel=&quot;nofollow&quot;&gt;C&lt;/a&gt; and got  $\frac{1}{52}$, i do agree that 15 is not much big and it may be possible to use this during exam but does there any tricky method exist?  &lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-09T13:47:02.423" Title="Doubts on a probability problem" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="13653" PostTypeId="2" ParentId="13652" CreationDate="2010-12-09T13:29:44.653" Score="6" ViewCount="0" Body="&lt;p&gt;No calculation is required and certainly no C. As you should observe when you start writing out the same product, the denominator of each next term will cancel the numerator of the previous, so the only terms left will be the denominator of the first and the numerator of the last terms. So the answer will always be $\frac{1}{52}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way to see this is to stand the problem on its head: at any attempt, you are equally likely to draw any given card from the deck (why should the kind of hearts be any different from the 9 of clubs?). Since the are 52 cards, you get your probability.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-09T13:29:44.653" CommentCount="3" />
  <row Id="13654" PostTypeId="1" AcceptedAnswerId="13678" CreationDate="2010-12-09T13:31:37.007" Score="2" ViewCount="128" Body="&lt;p&gt;It seems like I am facing some confusion while handling with inequalities,I was doing some tasks where it is asked to find the interval of the variable,after some steps I deduced the the necessary expression but I am facing some problem while trying to deduce the final expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider for example:&#xA;    $$a^2 \ge 8;\qquad 0 &amp;lt; a &amp;lt; 6; \qquad a &amp;lt; \frac{11}{3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the start I write $a^2 \ge 8$ as $ a \lt -2\sqrt{2} \quad\text{or}\quad a &amp;gt; 2\sqrt{2} $ but the final answer is $ 2\sqrt{2} &amp;lt; a &amp;lt; \frac{11}{3}$. How can this be deduced?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The original problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If both the roots of the equation $x^2-ax+2 = 0$ lie in the interval $(0,3)$, then find the value of $a$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the two roots lie in the same interval, those three conditions follow.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T01:22:08.240" LastActivityDate="2010-12-10T02:59:54.770" Title="Problems with Inequalities " Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="13655" PostTypeId="2" ParentId="13625" CreationDate="2010-12-09T13:35:07.473" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, $\mathbb{Z}$ is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Note&lt;/strong&gt;: As pointed out in the comments, $\mathbb{Z}$ is &lt;em&gt;technically&lt;/em&gt; locally Euclidean as it is discrete.  If you are bothered by such things, replace $\mathbb{Z}$ by something like $\mathbb{Q}$, or the union of copies of the Cantor set.  I think that, despite this flaw, $\mathbb{Z}$ is a better example of the point I wish to make so I'm going to stick with it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as Qiaochu and Alex effectively say in the comments, this all depends on what you mean by &quot;globally&quot;.  But before I say what that means, let me point out that &quot;locally X&quot; is not so straightforward.  In topology, we say that a space is &quot;locally &lt;strong&gt;X&lt;/strong&gt;&quot; if whenever we have $p \in U$ with $U$ open (or just a neighbourhood of $p$) then there is some neighbourhood $V$ of $p$ with $p \in V \subseteq U$ such that $V$ has property &lt;strong&gt;X&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works fine with lots of properties such as connected, path connected, compact, and similar.  When we put the word &quot;Euclidean&quot; in it, something is a bit weird.  What does it mean to say that a space is Euclidean?  We don't tend to define that.  Rather we define &quot;locally Euclidean&quot; all in one go.  It's like the definition of the limit of a sequence.  We write $\lim_{n \to \infty} s_n = s$ but we never decompose that funny symbol &quot;$\lim_{n \to \infty}$&quot; and talk about $n$ &lt;em&gt;actually&lt;/em&gt; going to $\infty$ in the same way that we talk about $x$ going to $0$ in the expression $\lim_{x \to 0} \sin(x) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, except that we do.  There &lt;em&gt;is&lt;/em&gt; a way to make sense of $n$ going to $\infty$, but it comes second, not first.  We could happily deal with limits and the like without ever knowing how to really make sense of $n$ going to $\infty$.  But to know what $n$ going to $\infty$ means, we have to know about limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is with &quot;locally Euclidean&quot;.  We &lt;em&gt;can&lt;/em&gt; decompose this, but it comes second.  And the decomposition, into &quot;locally&quot; + &quot;Euclidean&quot;, is almost determined by what &quot;locally Euclidean&quot; means.  I say almost, and this is the point of the ramble, because there are actually two things that it could mean which both give the same answer for &quot;locally Euclidean&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A space is &quot;Euclidean&quot; if it is homeomorphic to $\mathbb{R}^n$ for some $n$.&lt;/li&gt;&#xA;&lt;li&gt;A space is &quot;Euclidean&quot; if it is homeomorphic to an open subset of $\mathbb{R}^n$ for some $n$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For &quot;locally Euclidean&quot;, both of these give the same answer because any open subset of $\mathbb{R}^n$ contains lots of small open subsets that are homeomorphic to $\mathbb{R}^n$.  But if we replace &quot;Euclidean&quot; by some other space, this equivalence will break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so back to the original question.  Why is $\mathbb{Z}$ globally Euclidean but not locally Euclidean?  Again, we need to know what everything means.  But now on a slightly higher level.  What are we doing when we talk about &quot;locally &lt;strong&gt;X&lt;/strong&gt;&quot;?  We're saying that we can get out our microscopes and examine our space near $p$ and find that property &lt;strong&gt;X&lt;/strong&gt; holds.  By doing this, we effectively ignore a huge part of our space.  In fact, we can ignore almost as much as we like (within reason) and focus down only on a tiny, tiny bit near $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens if we reverse this process?  Instead of taking a microscope and examining our space really close up, let's put it in a rocket, shoot it off to the Moon and look at it through a telescope.  Then, not only do we not see any of the small detail, we can't really tell if all that detail is there in the first place.  The resolution of our telescope means that we cannot examine our space very closely at all.  Indeed, we can't even tell if it is all there or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when put on the Moon, $\mathbb{R}$ and $\mathbb{Z}$ are indistinguishable.  So $\mathbb{Z}$ is Euclidean in the opposite sense to &quot;locally Euclidean&quot; but is (clearly) not locally Euclidean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look for this concept in the literature, you won't find it under the heading of &quot;globally &lt;strong&gt;X&lt;/strong&gt;&quot;.  That's because that would be too confusing a term.  Rather, what I'm talking about is &lt;strong&gt;coarse geometry&lt;/strong&gt;.  And it's a lot of fun, and I wish I knew much, much more about it than I do.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastEditorUserId="2907" LastEditorDisplayName="" LastEditDate="2010-12-10T09:24:43.970" LastActivityDate="2010-12-10T09:24:43.970" CommentCount="8" />
  <row Id="13656" PostTypeId="2" ParentId="13654" CreationDate="2010-12-09T13:35:46.867" Score="5" ViewCount="0" Body="&lt;p&gt;Everything you have done so far is correct. Once you get down to linear constraints, you essentially have to intersect all the permitted ranges. E.g. since $0&amp;lt;a&amp;lt;6$, the negative solutions $a&amp;lt;-2\sqrt{2}$ are out. Thus the only possibility is $a&amp;gt;2\sqrt{2}$. Now the remaining inequalities also give you an upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good strategy is, as I said at the beginnig, to sketch all the ranges and then intersect them to find the range in which &lt;em&gt;all&lt;/em&gt; inequalities are satisfied. If you had some &quot;or&quot;-conditions, you would take unions of ranges, rather than intersections.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-09T13:35:46.867" CommentCount="3" />
  <row Id="13657" PostTypeId="1" CreationDate="2010-12-09T13:36:29.940" Score="1" ViewCount="63" Body="&lt;p&gt;To state the context, I am familiar with the Hopf-Rinow theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My request is three fold,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I would like to know of general classes of geodesically incomplete spaces. I basically want to see lots of examples for this. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to know of techniques of proving and testing for geodesic incompleteness or completeness. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to know of methods of proving and testing for existence of maximal extension of curves. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;{Confusingly in quite a bit of literature I have seen the adjective &quot;inextensible&quot; being used when actually they seem to want to mean &quot;maximally extended&quot;. Is there some subtle point here that I am missing?} &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-01-08T14:44:39.960" Title="Intuition about geodesic incompleteness " Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13658" PostTypeId="1" CreationDate="2010-12-09T13:37:25.883" Score="0" ViewCount="25" Body="&lt;p&gt;I would like to gain intuition how &quot;singular&quot; a distribution can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X \in \mathbb R^n$ be open, $\mathcal D(X)$ be the smooth test functions on $X$ with canonical LF topology, and let $\mathcal D&amp;#39;(X)$ be the topological dual of $D(X)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Psi \in D&amp;#39;(X)$. Can the singular support of $ \Psi$, i.e. the complement in $X$ of the largest open subset of $X$, on which $\Psi$ is a locally integrable function, be open?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Dirac delta has singular support in $0$. Similarly, integrals with respect to Hausdorff measures on finite measure submanifolds of $X$  may be distributions, whose singular support is not open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All examples of distributions I know constitute of a function &quot;disturbed&quot; by a &quot;small&quot; singularity.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-12-09T13:37:25.883" Title="Distribution whose singular support is open" Tags="&lt;measure-theory&gt;&lt;functional-analysis&gt;" CommentCount="4" />
  <row Id="13659" PostTypeId="2" ParentId="13652" CreationDate="2010-12-09T13:47:02.423" Score="1" ViewCount="0" Body="&lt;p&gt;A combinatorial approach is useful here: 51!/52! = 1/52.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T13:47:02.423" CommentCount="2" />
  <row Id="13660" PostTypeId="1" CreationDate="2010-12-09T13:49:03.693" Score="1" ViewCount="73" Body="&lt;p&gt;Suppose that $R$ is a ring, $I$ and $J$ are ideals, and $R=I+J$. Can we conclude that $R/J$ is free?&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T16:22:20.427" LastActivityDate="2010-12-09T16:22:20.427" Title="If $R=I+J$, can we conclude that $R/J$ is free?" Tags="&lt;abstract-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13661" PostTypeId="1" CreationDate="2010-12-09T13:49:11.097" Score="1" ViewCount="58" Body="&lt;p&gt;It's well known that under some &quot;weak&quot; hypothesis, such as finitely generation, the support of an A-module is closed in Spec(A). It is true also in the most general case? &lt;/p&gt;&#xA;" OwnerUserId="4571" LastActivityDate="2010-12-09T19:07:33.443" Title="The support of a module is closed?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="13662" PostTypeId="1" AcceptedAnswerId="13810" CreationDate="2010-12-09T13:49:35.550" Score="2" ViewCount="161" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The probability that a man who is 85 years. old will die before attaining the age of 90 is $\frac13$. A,B,C,D are four person who are 85 years old. what is the probability that A will die before attaining the age of 90 and will be the first to die ?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-09T18:08:06.373" LastActivityDate="2010-12-10T21:07:20.553" Title="Probability Problems" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="13663" PostTypeId="2" ParentId="13661" CreationDate="2010-12-09T13:52:22.613" Score="4" ViewCount="0" Body="&lt;p&gt;Pick a non-closed subset $S$ of $\operatorname{Spec}\mathbb Z$, and let $\displaystyle M=\bigoplus_{\mathfrak p\in S}\:\mathbb Z/{\mathfrak p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; Editing this answer required solving a captcha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks go to Akhil for catching a stupid mistake)&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-09T19:07:33.443" LastActivityDate="2010-12-09T19:07:33.443" CommentCount="2" />
  <row Id="13664" PostTypeId="2" ParentId="13660" CreationDate="2010-12-09T13:53:20.540" Score="4" ViewCount="0" Body="&lt;p&gt;$\mathbb{Z} = 2\mathbb{Z} + 3\mathbb{Z}$. I am sure you can now answer your question yourself.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-09T13:53:20.540" CommentCount="2" />
  <row Id="13665" PostTypeId="1" CreationDate="2010-12-09T14:00:03.397" Score="3" ViewCount="50" Body="&lt;p&gt;Hi &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a procedure/algorithm for calculating sums of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n_1,\ldots,n_r &amp;gt;0} \frac1{L_1(n_1,\ldots,n_r)^{m_1} \ldots L_r(n_1,\ldots,n_r)^{m_r}} $$&#xD;&#xA;where &#xA;$$   L_i(n_1,\ldots, n_r) := a_{1,i} n_1 + \ldots + a_{r,i} n_r $$&#xD;&#xA;for some $a_{i,j} \in \mathbb C$. How would one go about evaluating this sum to say 20 or 30 digits?&lt;/p&gt;&#xA;" OwnerUserId="4572" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T14:03:35.843" LastActivityDate="2010-12-09T14:03:35.843" Title="Shintani cone zeta function" Tags="&lt;number-theory&gt;&lt;sequences-and-series&gt;&lt;numerical-methods&gt;&lt;zeta-function&gt;" CommentCount="1" />
  <row Id="13667" PostTypeId="2" ParentId="13657" CreationDate="2010-12-09T14:23:29.867" Score="0" ViewCount="0" Body="&lt;p&gt;The equation for a geodesic is just an ODE with an initial value. The classical Picard-Lindeloef theorem gives the existence of a maximal solution (defined from 0 to t_+). Thus, for instance, if you can show that some &quot;energy&quot; (for instance the L^2 integral of the derivative from 0 to t_+) is finite, then t_+=\infty.&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-09T14:23:29.867" CommentCount="2" />
  <row Id="13668" PostTypeId="2" ParentId="13627" CreationDate="2010-12-09T14:57:42.770" Score="2" ViewCount="0" Body="&lt;p&gt;There are definite real world applications. I would look at the website/work of Gunnar Carlsson (http://comptop.stanford.edu/) and Robert Ghrist (http://www.math.uiuc.edu/~ghrist/). Both are excellent mathematicians. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following could be completely wrong: Carlsson is one of the main proponents of Persistent Homology which is about looking at what homology can tell you about large data sets, clouds, as well as applications of category theory to computer science. Ghrist works on stuff like sensor networks. I don't understand any of the math behind these things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are some preprints by Phillipe Gaucher you might want to check out. Peter Bubenik at cleveland state might also have some fun stuff on his website.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-09T14:57:42.770" />
  <row Id="13669" PostTypeId="2" ParentId="13660" CreationDate="2010-12-09T15:21:12.400" Score="2" ViewCount="0" Body="&lt;p&gt;You can almost never conclude that $R/J$ is free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is a nonzero commutative ring and $J$ is an ideal of $R$, then $R/J$ is a free $R$-module iff $J = 0$.  Indeed, if $J \neq 0$, let $x$ be a nonzero element of $J$.  Then for all $y \in R/J$, $xy = 0$.  On the other hand, consider $\bigoplus_{i \in I} R$ for some nonempty index set $I$: taking $y$ to be an element of the form $(1,0,\ldots,0)$ -- i.e., which has one coordinate equal to $1$ and all the other coordinates equal to $0$ -- we find that $xy = 0$ implies $x = 0$, contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if $J \neq 0$, then $R/J$ has a nonzero annihilator, which is not the case for any nontrivial free module. &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-09T15:30:35.587" LastActivityDate="2010-12-09T15:30:35.587" />
  <row Id="13670" PostTypeId="2" ParentId="13642" CreationDate="2010-12-09T15:42:22.697" Score="4" ViewCount="0" Body="&lt;p&gt;If $\beta,\gamma &amp;gt; 0$ then the expected number is infinity (unless $n = 0$), because either $n &amp;lt; 0$ and then you never reach $n$, or $n &amp;gt; 0$ and then with some positive probability you &quot;hop&quot; over $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that state $n+1$ is also good, and $n &amp;gt; 0$. Let $E_i$ be the expected number of steps to get to states $n$ or $n+1$ starting at state $i$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_i = \alpha E_i + \beta E_{i+1} + \gamma E_{i+2} + 1. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a non-homogeneous recurrence relation that you can solve by first finding some non-homogeneous solution (try $E_i = a i + b$) and then solving the corresponding homogeneous recurrence (without the $+1$) in the usual way (this amounts to solving the quadratic $(1-\alpha)x^2-\beta x-\gamma$; note the singular solution if this is a square). You will get a two-dimensional solution space which can you collapse by using $$E_n = E_{n+1} = 0.$$ Then the answer is $E_0$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-09T15:42:22.697" />
  <row Id="13671" PostTypeId="2" ParentId="13625" CreationDate="2010-12-09T15:57:23.043" Score="7" ViewCount="0" Body="&lt;p&gt;As the comments have suggested, the question is a bit under-determined, but let me supply some definitions in a plausible way and answer it based on those definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P$ be a property of topological spaces.  (Formally speaking this could be any &quot;predicate&quot;, but it would only be asking for trouble if $P$ were not a &lt;em&gt;topological property&lt;/em&gt; of topological spaces -- i.e., such that if $X$ and $Y$ are homeomorphic topological spaces, then $X$ has property $P$ iff $Y$ has property $P$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we say that $X$ is &lt;strong&gt;locally P&lt;/strong&gt; if for every point of $x$ there is a neighborhood base $\{U_x\}$ each of whose elements $U_x$ has property $P$ in the subspace topology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that $X$ is &lt;strong&gt;globally P&lt;/strong&gt; if it has property $P$.  Why would we say this?  Well, it's a matter of emphasis, presumably by way of comparison to something which locally has property $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define a space to be &lt;strong&gt;Euclidean&lt;/strong&gt; if it is homeomorphic to $\mathbb{R}^n$ for some $n \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I claim that any space which is globally Euclidean is also locally Euclidean.  In other words, I claim that for all $n \in \mathbb{N}$, every $x \in \mathbb{R}^n$ admits a neighborhood base, each of whose members is homeomorphic to Euclidean $m$-space for some $m$.  But indeed, for any $\epsilon &amp;gt; 0$, the open $\epsilon$-ball around $x$ is isomorphic to $\mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in general &quot;globally P&quot; &lt;em&gt;need not&lt;/em&gt; imply &quot;locally P&quot;.  For instance a connected space need not be locally connected, and a quasi-compact (non-Hausdorff!) space need not be locally quasi-compact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fairness I should say that some mathematicians don't like the idea that globally $P$ need not imply locally $P$, so they weaken the definition of locally $P$ to &quot;each point of $X$ admits &lt;em&gt;some&lt;/em&gt; neighborhood which has property $P$&quot;.  I would call this &lt;strong&gt;weakly locally P&lt;/strong&gt; and then it is trivial that globally $P$ implies weakly globally $P$: for &#xA;each $x$, take $X$ itself as a neighborhood which has property $P$.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-09T15:57:23.043" />
  <row Id="13672" PostTypeId="2" ParentId="13639" CreationDate="2010-12-09T15:59:04.790" Score="2" ViewCount="0" Body="&lt;p&gt;Take $E$ to be the trivial bundle $X \times \mathbb{R}$. Let $f: X \to \mathbb{R}$ to be a function vanishing only at $x_0$. Then multiplication by the global function $f$ defines a morphism $E \to E$ which is injective as a morphism of sheaves (because it is injective on $E(U)$ for any open set $U$), but the map on fibers at $x_0$ is not injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To avoid uninteresting cases, assume $x_0$ is not an isolated point of $X$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that for an injection of sheaves $E \to F$, the map on &lt;em&gt;stalks&lt;/em&gt; $E_x \to F_x$ is always injective. The map on &lt;em&gt;fibers&lt;/em&gt; $E(x) \to F(x)$ obtained by tensoring the map on stalks with the residue field is generally not, because tensoring is not an exact functor. If the map on stalks is a &lt;em&gt;split&lt;/em&gt; injection, then the map on fibers will be an injection still.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-09T16:22:38.233" LastActivityDate="2010-12-09T16:22:38.233" />
  <row Id="13673" PostTypeId="2" ParentId="13646" CreationDate="2010-12-09T16:03:36.413" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ It &lt;strong&gt;telescopes&lt;/strong&gt;: $\rm\displaystyle\ \ \prod_{k=a}^b\ \frac{f(k+1)}{f(k)}\ \ = \ \ \frac{\rlap{----}f(a+1)}{f(a)}\frac{\rlap{----}f(a+2)}{\rlap{----}f(a+1)}\ \ \cdots\ \ \frac{f(b+1)}{\rlap{--}f(b)}\ =\ \frac{f(b+1)}{f(a)}\:.\ $&lt;br&gt;&#xA;Apply that to both factors in the product then take the limit as $\rm\ b\to\infty\:.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some other examples of additive/multiplicative telescopy see &lt;a href=&quot;http://math.stackexchange.com/questions/6581/proof-of-inequalities-by-induction/6586#6586&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/7441/summation-of-sum-limits-n1-infty-fracxx1-cdots-xn-1yy1-cdo/7453#7453&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/6805/simple-proof-by-induction-9-divides-n3-n13-n23/6814#6814&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T16:03:36.413" />
  <row Id="13674" PostTypeId="1" AcceptedAnswerId="13684" CreationDate="2010-12-09T16:11:12.667" Score="0" ViewCount="35" Body="&lt;p&gt;We have a random variable Y that has a uniform distribution on the interval[1,5]. The cost of delay is given by U = 2Y^2 + 3. Use the method of transformations to derive the density of function of U.&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-09T16:50:31.370" Title="density function by the method of transformations" Tags="&lt;probability&gt;&lt;homework&gt;&lt;statistics&gt;&lt;distribution-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13675" PostTypeId="2" ParentId="13482" CreationDate="2010-12-09T16:13:18.470" Score="1" ViewCount="0" Body="&lt;p&gt;Take the Euler scheme. Let's assume that you start with $X_0=\epsilon$. Let's take $\beta=-1$, $\gamma =1$ and $\sigma=1$ for the sake of definiteness. Then in the next step you'll have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_{\Delta} = \epsilon + (1-\epsilon) \Delta + \sqrt{\epsilon} Z \Delta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the probability that this is smaller than zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}\left[X_{\Delta}&amp;lt;0\right]= \mathbb{P}\left[\epsilon + (1-\epsilon) \Delta + \sqrt{\epsilon} Z \Delta &amp;lt; 0\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging you'll get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}\left[X_{\Delta}&amp;lt;0\right]= \mathbb{P}\left[ Z &amp;lt; -\frac{\sqrt{\epsilon}}{\Delta}(1+(1-\frac{1}{\epsilon})\Delta)\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Chebyshev for an estimate this is smaller than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}\left[X_{\Delta}&amp;lt;0\right] \leq \frac{\Delta^2}{2\epsilon}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the closer you are to zero, the higher the probability of crossing that line. But taking time steps sufficiently small lowers the chance of crossing quadratically. That should give you an idea of how to show it for the other scheme.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-09T16:13:18.470" CommentCount="1" />
  <row Id="13676" PostTypeId="1" AcceptedAnswerId="13677" CreationDate="2010-12-09T16:14:04.820" Score="4" ViewCount="71" Body="&lt;p&gt;This might be an elementary question, but I am just beginning to learn logic theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From wikipedia article on Godel's incompleteness theorems&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory (Kleene 1967, p. 250).&#xA;  The true but unprovable statement referred to by the theorem is often referred to as “the Gödel sentence” for the theory. The true but unprovable statement referred to by the theorem is often referred to as “the Gödel sentence” for the theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question: Is a Godel statement &lt;a href=&quot;http://en.wikipedia.org/wiki/Validity#Validity_of_statements&quot; rel=&quot;nofollow&quot;&gt;logically valid&lt;/a&gt;?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Carl answers below, if the Godel statement is valid, then by completeness theorem, it is provable, which leads to a contradiction. So there exists a model in which the statement is false. Can we construct such a model?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T16:30:04.613" LastActivityDate="2010-12-09T16:30:04.613" Title="Is a Godel sentence logically valid?" Tags="&lt;logic&gt;" AnswerCount="1" />
  <row Id="13677" PostTypeId="2" ParentId="13676" CreationDate="2010-12-09T16:18:40.247" Score="6" ViewCount="0" Body="&lt;p&gt;No, a Gödel sentence is not logically valid. Because the Gödel sentence for a theory $T$ is unprovable from $T$, it follows from the completeness theorem for first-order logic that there is a model of $T$ in which the Gödel sentence is false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the text you quoted says &quot;true&quot; you should read that as &quot;true in the standard model of arithmetic&quot;. Logical validity would correspond to truth in &lt;em&gt;all&lt;/em&gt; models. An example of a logically valid sentence is $(\forall x) (x=x)$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-09T16:18:40.247" CommentCount="3" />
  <row Id="13678" PostTypeId="2" ParentId="13654" CreationDate="2010-12-09T16:20:42.610" Score="7" ViewCount="0" Body="&lt;p&gt;What you are forgetting is that you have more than &lt;em&gt;one&lt;/em&gt; inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequality $a^2\geq 8$, by &lt;em&gt;itself&lt;/em&gt;, only gives you $|a|\geq \sqrt{8}=2\sqrt{2}$, or equivalently as you (almost) write:&#xA;$$a\geq 2\sqrt{2}\qquad\text{or}\qquad a\leq -2\sqrt{2}.$$&#xD;&#xA;&lt;strong&gt;Added:&lt;/strong&gt; Note that you took $a^2\geq 8$, a &lt;em&gt;weak&lt;/em&gt; inequality, and &quot;translated&quot; it into &quot;$a\gt 2\sqrt{2}$ or $a\lt -2\sqrt{2}$.&quot; Technically, this is incorrect because you have now excluded the possitibility of equality. Since you were starting with with $\geq$, you should have continued with $\geq$ and $\leq$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you have &lt;em&gt;other&lt;/em&gt; inequalities that need to be satisfied as well.The second inequality tells you that $0\lt a \lt 6$ must be satisfied. How does this mesh with the inequalities you already have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0\lt a \lt 6$  &lt;em&gt;and&lt;/em&gt; $a\leq -2\sqrt{2}\lt 0$, then that tells you that $a$ must be both greater than &lt;em&gt;and&lt;/em&gt; less than zero. This is a &lt;em&gt;very&lt;/em&gt; big problem for $a$, which decides to solve this problem by ceasing to exist. There is no solution to the inequality $0\lt a\lt 6$ that &lt;em&gt;also&lt;/em&gt; satisfy $a\lt -2\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting these two together, you know that you have $a\geq 2\sqrt{2}$, and $0\lt a \lt 6$. Since satisfying $a\geq 2\sqrt{2}$ automatically guarantees you that $0\lt a$, you can combine them all by saying that the first two inequalities you have are equivalent to&#xA;$$ 2\sqrt{2} \leq a \lt 6.$$&#xD;&#xA;Finally, we need to add in the condition that $a\lt \frac{11}{3}$. Since $\frac{11}{3}\lt 6$, if $a$ is less than $\frac{11}{3}$ it will also automatically be less than $6$. It &lt;em&gt;still&lt;/em&gt; needs to be greater than or equal to $2\sqrt{2}$, though. So combining them all you get that you must have&#xA;$2\sqrt{2} \leq a \lt 6$ in order to satisfy all three (four, if you count &quot;$0\lt a\lt 6$&quot; as two inequalities) inequalities you have.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Okay, the &lt;strong&gt;original&lt;/strong&gt; problem was: &quot;if the two roots of $x^2-ax+2=0$ lie in the interval $(0,3)$, then find the values of $a$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots are $\frac{a\pm\sqrt{a^2-8}}{2}$. So you assumed that you only needed to ask for $a^2-8\geq 0$ in order for the roots to be real. However, the use of the word &quot;&lt;strong&gt;both&lt;/strong&gt;&quot; suggests that in fact they want there to be two &lt;em&gt;distinct&lt;/em&gt; roots, which would require $a^2-8\gt 0$. If you take $a=2\sqrt{2}$, the lower limit I suggested earlier, then you have a single (repeated) root at $\sqrt{2}$ (which satisfies the condition). If they want &lt;strong&gt;two&lt;/strong&gt; roots, then you need the discriminant &lt;em&gt;positive&lt;/em&gt;, not merely nonnegative.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T02:59:54.770" LastActivityDate="2010-12-10T02:59:54.770" CommentCount="12" />
  <row Id="13679" PostTypeId="1" AcceptedAnswerId="13680" CreationDate="2010-12-09T16:27:55.593" Score="6" ViewCount="213" Body="&lt;p&gt;Is there a general way to show whether the limit $$\lim_{n \to \infty}\frac{\mathrm d^n}{\mathrm dx^n} f(x)$$ converges to some expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about repeatedly integrating an expression $n$ times as $n \to \infty$?&lt;/p&gt;&#xA;" OwnerUserId="4576" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T16:27:19.583" LastActivityDate="2010-12-10T16:27:19.583" Title="Does $\lim_{n \to \infty}\frac{\mathrm d^n}{\mathrm dx^n} f(x)$ converge?" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13680" PostTypeId="2" ParentId="13679" CreationDate="2010-12-09T16:38:54.693" Score="11" ViewCount="0" Body="&lt;p&gt;For the notion of limit, you need a notion of convergence or of &quot;distance.&quot; Here, you are talking about &lt;em&gt;functions&lt;/em&gt;, not numbers, so you need to specify how you measure the &quot;distance&quot; between two functions. Turns out, there are many different ways of doing it, and the answer for a particular $f(x)$ &lt;em&gt;may&lt;/em&gt; depend on which way you do it. So you would need to specify that in general. Among the different kinds of &quot;convergence&quot; you can read about there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniform_convergence&quot; rel=&quot;nofollow&quot;&gt;uniform convergence&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pointwise_convergence&quot; rel=&quot;nofollow&quot;&gt;pointwise convergence&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Convergence_in_measure&quot; rel=&quot;nofollow&quot;&gt;convergence in measure&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lp_space&quot; rel=&quot;nofollow&quot;&gt;convergence in $\mathcal{L}_p$-norm&lt;/a&gt;. But this hardly exhausts the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the notion of convergence of functions, then you are dealing with a sequence of functions, and one checks convergence the &quot;usual&quot; way (that is, in essentially the same ways as you do with sequences of real numbers, substituting the absolute value of the difference with whatever way you have decided to describe distance of functions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;repeated integrals&quot;, that's not going to help much. The $n$th iterated integral of $\frac{d^n}{dx^n}f(x)$ is $f(x)+p(x)$, where $p(x)$ is an arbitrary polynomial of degree $n-1$ (which you get through the constants of integration). Taking &quot;the limit&quot; as $n$ goes to infinity just drops you back into the problem of &quot;what does it mean for a sequence of functions to converge&quot;? Plus, an &quot;infinite polynomial&quot; (a power series) is probably not what you want anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for &lt;em&gt;specific&lt;/em&gt; functions one may be able to discern either a pattern, or an answer. For polynomials, it is plain that $f^{(n)}(x)$ will be zero for all sufficiently large $n$, and in that case no matter how you define your distance between functions, the answer will be the zero function. If you have a function with periodic derivatives, such as $\sin(x)$, you know there will be no limit. And so on. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-09T16:38:54.693" CommentCount="2" />
  <row Id="13681" PostTypeId="2" ParentId="13654" CreationDate="2010-12-09T16:39:30.387" Score="0" ViewCount="0" Body="&lt;p&gt;Any polynomial inequality over the reals can be reduced to testing it on the intervals between its roots - where the polynomial has constant sign. This will yield a &lt;em&gt;finite&lt;/em&gt; number of intervals where the inequality holds true. When you have multiple such inequalities you simply intersect the intervals for each polynomial to obtain the &lt;em&gt;finite&lt;/em&gt; number of intervals where all of the inequalities are true. In fact the same idea generalizes to arbitrary first order multivariate statements about the reals - as Tarski has shown by his celebrated method of &lt;em&gt;quantifier elimination&lt;/em&gt; for real-closed fields. It has an algorithmic form that is essentially a higher-dimensional &lt;em&gt;cylindrical&lt;/em&gt; version of the finite interval decomposition for the above one dimensional case - google &quot;cylindrical algebraic decomposition&quot; or CAD algorithm.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T16:39:30.387" />
  <row Id="13682" PostTypeId="1" CreationDate="2010-12-09T16:45:29.860" Score="0" ViewCount="22" Body="&lt;p&gt;Hi my apologies first, I'm a biologist and not so good in statistics. In my study I'm studying the effect of concentration on growth of a certain specimen. I have with me the different concentrations and the growth data for each concentration. What do you suggest is the best statistical test to find out the best concentration (i.e the one having maximum growth). Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4578" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T17:01:08.247" LastActivityDate="2010-12-09T17:01:08.247" Title="Statistics Question -which test to find out best concentration (the one having maximum effect)?" Tags="&lt;statistics&gt;" CommentCount="2" />
  <row Id="13683" PostTypeId="2" ParentId="13648" CreationDate="2010-12-09T16:48:34.687" Score="2" ViewCount="0" Body="&lt;p&gt;Using Pascal identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\binom{2n}{n}=\left(4-\frac{2}{n}\right)\cdot\binom{2n-2}{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one gets for your product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\displaystyle\frac{\binom{2}{1}}{\binom{0}{0}}\cdot\frac{\binom{4}{2}}{\binom{2}{1}}\cdot\cdot\cdot\frac{\binom{2n}{n}}{\binom{2n-2}{n-1}}=\binom{2n}{n}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is an integer as required.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-12-09T18:45:45.460" LastActivityDate="2010-12-09T18:45:45.460" />
  <row Id="13684" PostTypeId="2" ParentId="13674" CreationDate="2010-12-09T16:50:31.370" Score="2" ViewCount="0" Body="&lt;p&gt;First, if $Y$ takes values in $[1,5]$, then $U$ takes values in the range...? So, for $x$ in that range, you should consider the distribution function $F_U {(x)} := {\rm P}(U \le x)$. Substituting $U=2Y^2+3$ leads straightforwardly to $F_U {(x)} = F_Y {(\varphi(x))}$ for some function $\varphi(x)$, where $F_Y$ is the distribution function of $Y$. What is $\varphi(x)$? Note that $\varphi(x)$ takes values in $[1,5]$. Now, the density $f_U$ of $U$ is given by&#xA;$$&#xD;&#xA;f_U (x) = \frac{{\rm d}}{{{\rm d}x}}F_U (x) = \frac{{\rm d}}{{{\rm d}x}}F_Y (\varphi (x)) =...?&#xD;&#xA;$$ &#xA;Final hint: the density $f_Y$ of $Y$ is constant on $[1,5]$; what is that constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestion: verify that the $f_U {(x)}$ you find integrates to $1$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T16:50:31.370" CommentCount="1" />
  <row Id="13685" PostTypeId="1" CreationDate="2010-12-09T16:51:12.697" Score="1" ViewCount="163" Body="&lt;p&gt;Is the function $\frac{e^z-1}{e^z+1}$ analytic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using $\mathop{\lim}\limits_{z\to a} \frac{f(z)-f(a)}{z-a}$, but I didnt really know what to use for $a$.&lt;/p&gt;&#xA;" OwnerUserId="4577" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T18:29:54.260" LastActivityDate="2010-12-09T18:29:54.260" Title="Is $\frac{e^z-1}{e^z+1}$ analytic?" Tags="&lt;calculus&gt;&lt;complex-analysis&gt;" CommentCount="6" />
  <row Id="13686" PostTypeId="2" ParentId="13563" CreationDate="2010-12-09T17:28:46.113" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The problem admits an obvious shift &lt;strong&gt;symmetry&lt;/strong&gt;, e.g. considering the example that you gave, a shift of $-2\:$ to $[3, 6]$ yields $[1,4]$, which clearly has $4$ integers. Innate symmetries should always be investigated since they may play a key role in simplifying the search for a solution. Moreover, it's best to learn these symmetry-based problem solving techniques for simple problems like this, since you may not be able to see the forest for the trees in much more complex problems.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T17:28:46.113" />
  <row Id="13687" PostTypeId="1" CreationDate="2010-12-09T17:32:18.797" Score="2" ViewCount="86" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a non-linear ODE that I need to linearise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go about linearising the trig terms using Taylor series...but not sure about a square root of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{(A+C)^2 + B -2 * \cos (\alpha - \theta)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4581" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:22:18.267" LastActivityDate="2010-12-09T20:32:20.033" Title="How to linearise a square root expression?" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13688" PostTypeId="2" ParentId="13551" CreationDate="2010-12-09T17:48:56.933" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \Delta\ f \ :=\ f(n+1) - f(n)\ $ reduces $\rm\ d = deg\ f\:,\ $ so $\rm\ \Delta^{d+1}\ f\ =\ 0\ $ yields a recurrence for $\rm\: f\:$. Functions admitting recurrences are known as &lt;strong&gt;P-recursive&lt;/strong&gt;. Google &lt;strong&gt;holonomic function&lt;/strong&gt; to learn about their rich theory - which has widespread applications.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-09T17:48:56.933" />
  <row Id="13689" PostTypeId="1" CreationDate="2010-12-09T17:57:32.630" Score="4" ViewCount="106" Body="&lt;p&gt;Suppose we have a countable set of objects $\{x_i|i \in [1..m]\}$ in a metric space $(\mathbb R^n,d_n)$ and a map ($F$) mapping the objects to objects in a metric space $(\mathbb R^1,d_1)$. For each pair of objects we can define an error:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;  e_{ij} = \frac{d_m(x_i,x_j)}{d_1(F(x_i),F(x_j))}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $e_{ii}$ be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that we can define the countable set in a way that any projection has an error ($e_{ij}\ne 1$) at least for one pair of objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for given set of objects we can define the map trying to minimize the errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had met this problem developing a web site's rating system based on several criteria. At the moment I use the following algorithm for calculating F (images of objects):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;select random $\{y_i|i \in [1..m]\}$ set in space $(\mathbb R^1,d_1)$ and think that $y_i=F(x_i)$ where F is defining map&lt;/li&gt;&#xA;&lt;li&gt;for each dot calculate $\delta_i=\sum\limits_{j}k\frac{1}{2}e_{ij}(y_i-y_j)$ where $k$ is a parameter ($0 \le k \le 1$) and them change items in the set $y_i \to y_i + \delta_i$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I repeat the step two over and over several times, but at each step I make the parameter $k$ smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I infer this algorithm from physics intuition: if two objects are too close they push away, if they are too far away they attract and $k$ is a resistance of an environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get deeper knowledge about this problem, so could you provide me some links to articles and literature about it or the official name of the problem.&lt;/p&gt;&#xA;" OwnerUserId="2016" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T02:53:52.597" LastActivityDate="2011-01-09T17:44:40.127" Title="Projection to space with smaller dimension that saves a distance" Tags="&lt;reference-request&gt;&lt;metric-spaces&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13690" PostTypeId="2" ParentId="13687" CreationDate="2010-12-09T18:07:48.713" Score="0" ViewCount="0" Body="&lt;p&gt;If $(A+C)^2+B$ is large compared with 1 you can divide it out to get $$\sqrt{(A+C)^2+B}\sqrt{1-\frac{2cos(\alpha-\theta)}{(A+C)^2+B}}$$ and use $\sqrt{1-x}\approx 1-\frac{x}{2}$ for $x&amp;lt;&amp;lt;1$.  If you are specifically interested in $\theta$ you can expand the cosine term  and still use the approximation to the square root considering $\alpha$ to be fixed.  Does that help?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T20:32:20.033" LastActivityDate="2010-12-09T20:32:20.033" />
  <row Id="13691" PostTypeId="2" ParentId="13687" CreationDate="2010-12-09T18:22:59.923" Score="1" ViewCount="0" Body="&lt;p&gt;Since you have an ODE, you probably want to be linearizing around an equilbrium point $\theta_0$ (which might not be zero).  Note that I'm assuming that $\theta$ is the variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then think about the general form of a Taylor series expansion -- the linearization is given by&#xA;$$&#xD;&#xA;f(\theta) \approx f(\theta_0) + (\theta-\theta_0) f&amp;#39;(\theta_0)&#xD;&#xA;$$&#xD;&#xA;where&#xA;$$&#xD;&#xA;f(\theta) = \sqrt{(A+C)^2 + B - 2\cos(\alpha - \theta)}&#xD;&#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="4229" LastEditorUserId="4229" LastEditorDisplayName="" LastEditDate="2010-12-09T18:57:03.353" LastActivityDate="2010-12-09T18:57:03.353" />
  <row Id="13692" PostTypeId="2" ParentId="13641" CreationDate="2010-12-09T18:24:11.203" Score="4" ViewCount="0" Body="&lt;p&gt;As already mentioned in the other answers and comments, this is true if the space is assumed to be complete (see Andrey Rekalo's answer), but not necessarily otherwise (see Carl Mummert's answer).  In the complete case, more can be said.  If the Hilbert space dimension (cardinality of a maximal orthonormal set) is infinite, then the linear dimension (cardinality of a maximal linearly independent set) is at least $\mathfrak{c}=2^{\aleph_0}=|\mathbb{R}|$.  Of course, this doesn't directly tell you anything if the Hilbert space dimension is $\mathfrak{c}$ or greater, but in the case of separable Hilbert spaces like $\ell^2$ (as defined &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert_space#Second_example%3a_sequence_spaces&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), it tells you that not only does an orthonormal set not span, but &lt;em&gt;no&lt;/em&gt; subset of cardinality less than $\mathfrak{c}$ can span.  One way to see this is to consider the linearly independent set $\{(1,t,t^2,t^3,\ldots):-1\lt t\lt 1\}\subset\ell^2$.  Since $\ell^2$ imbeds into every infinite dimensional Hilbert space as the closed linear span of any countably infinite orthonormal set, this also demonstrates the general fact.  This and other proofs can be found in the solutions to &lt;a href=&quot;http://books.google.com/books?id=S57XLkgbf0oC&amp;amp;lpg=PP1&amp;amp;dq=hilbert%20space%20problem%20book%20halmos&amp;amp;pg=PA6#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Problem 7&lt;/a&gt; of Halmos's &lt;em&gt;Hilbert Space Problem Book&lt;/em&gt;, which I highly recommend. This statement also &lt;a href=&quot;http://www.jstor.org/stable/2318458&quot; rel=&quot;nofollow&quot;&gt;extends&lt;/a&gt; to Banach spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, what I have said so far hasn't answered your question in the case of Hilbert space dimension greater than or equal to $\mathfrak{c}$.  However, you can get an overkill solution to your question by taking a countably infinite subset $A$ of an orthonormal subset of an arbitrary Hilbert space $H$.   If $M_0$ is the linear span of $A$ and $M=\overline{M_0}$ is the closed linear span of $A$, then $H=M\oplus M^\perp$, while the linear span of your orthonormal set is contained in $M_0\oplus M^\perp$, which is properly contained in $H$ because $M$ has higher dimension than $M_0$.  ($M^\perp$ denotes the set of vectors orthogonal to every element of $M$, and $\oplus$ is being used here to denote internal direct sums.)  All that was really needed here is that the linear dimension of $M$ is not countable, and this also follows from &lt;a href=&quot;http://en.wikipedia.org/wiki/Baire_category_theorem&quot; rel=&quot;nofollow&quot;&gt;Baire's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrey Rekalo gives a better, nonoverkill answer for the complete case.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-09T18:24:11.203" />
  <row Id="13693" PostTypeId="1" AcceptedAnswerId="13714" CreationDate="2010-12-09T18:45:46.253" Score="1" ViewCount="51" Body="&lt;p&gt;Hi all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we have two varieties of algebras $A$ and $B$, whose operators all have arities less than some regular cardinal, and such that every $B$-algebra (please correct me if this is not the usual notation) is also an $A$-algebra. It is known that the forgetful functor from the category of $B$-algebras to that of $A$-algebras has a left adjoint $F$, which takes every $A$-algebra $a$ to the relatively free $B$-algebra $Fa$ generated by $a$. My question is, if $a$ and $b$ are $A$-algebras such that $Fa$ is isomorphic (as a $B$-algebra) to $Fb$, are $a$ and $b$ necessarily isomorphic? If not in general, how about the special case where $A$-algebras are Boolean algebras and $B$-algebras are countably complete Boolean algebras? &lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2010-12-09T21:37:58.207" Title="Can nonisomorphic algebras generate isomorphic relatively free algebras?" Tags="&lt;boolean-algebra&gt;&lt;universal-algebra&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="13694" PostTypeId="2" ParentId="13662" CreationDate="2010-12-09T18:53:43.683" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not sure from the comments above if you've solved this or not already, but I had this idea when I first read your post. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can partition the outcome space, based on when $B,C,D$ die. There are four disjoint cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_0:$ none of them dies before 90,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_1:$ exactly one of them dies before 90,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_2:$ exactly two...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_3:$ exactly three...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$P(B_n)=\binom{3}{n}(1/3)^n(2/3)^{3-n}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $F$ be the event that $A$ dies first, and $E$ be the event that it happens before $90$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$P(F\cap E)=\sum_{n=0}^3 P((F\cap E)|B_n)P(B_n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, $P((F\cap E)|B_0)$ is the probability that $A$ dies first and before $90$, given that the other three die after $90$. This occurs with probability $(1/3)$, since all the others are given to die after $90$. Then $P((F\cap E)|B_1)$ is the probability that $A$ dies first and before $90$, when exactly one other person dies before $90$. This probability is $(1/3)(1/2)$. The other cases are similar, and then one can just add them up.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-10T14:34:32.173" LastActivityDate="2010-12-10T14:34:32.173" />
  <row Id="13695" PostTypeId="1" AcceptedAnswerId="13699" CreationDate="2010-12-09T18:56:58.457" Score="0" ViewCount="276" Body="&lt;p&gt;What is the rationale behind saying that $|x|=1$ implies $x=\pm 1$?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:20:10.653" LastActivityDate="2010-12-14T22:46:22.897" Title="If  $|x| = 1 $ why this implies that $ x = \pm 1 $" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="13696" PostTypeId="2" ParentId="13631" CreationDate="2010-12-09T18:58:07.143" Score="4" ViewCount="0" Body="&lt;p&gt;This is described in the comments; $f$ need only be locally constant, so as a counterexample take $f : (0, 1) \cup (2, 3) \to \mathbb{R}$ which is equal to $0$ on the first component and $1$ on the second.  It is certainly true for functions from $\mathbb{R}$ to $\mathbb{R}$, for example by the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems trite, but it's still an important point to keep in mind.  If you ever feel like you're missing something, then of course you should attempt to carefully prove the result that you thought was true and see where it fails (if it does).&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T18:58:07.143" />
  <row Id="13697" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:02:23.400" Score="1" ViewCount="0" Body="&lt;p&gt;$|1| = 1$ and $|-1| = 1$, and there are no other real numbers $x$ such that $|x|=1$.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastActivityDate="2010-12-09T19:02:23.400" />
  <row Id="13699" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:05:17.157" Score="6" ViewCount="0" Body="&lt;p&gt;Absolute value of a number is its distance from zero on the number line. The absolute value of a number n is denoted by, |n|. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here if $|x| = 1$ implies, in the number line the distance from zero on the number is 1.$\text{ Hence, } x = \pm 1$.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-09T19:33:38.920" LastActivityDate="2010-12-09T19:33:38.920" CommentCount="1" />
  <row Id="13700" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:12:05.107" Score="3" ViewCount="0" Body="&lt;p&gt;You should think of the absolute value as being the distance a point is away from 0. So with $|x| = 1$, we want all of  the points that are one unit away from 0. We see that 1 and -1 are the only values for x that satisfies this. Therefore, with $|x| = 1$ we have that $x \in\{-1,1\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastEditorUserId="3791" LastEditorDisplayName="" LastEditDate="2010-12-14T22:46:22.897" LastActivityDate="2010-12-14T22:46:22.897" CommentCount="2" />
  <row Id="13701" PostTypeId="1" AcceptedAnswerId="13704" CreationDate="2010-12-09T19:15:05.733" Score="4" ViewCount="159" Body="&lt;p&gt;Bill Dubuque has answered several questions by indicating that some form of &quot;telescoping&quot; is taking place.  See &lt;a href=&quot;http://math.stackexchange.com/questions/13646/how-to-derive-the-following-identity&quot;&gt;this&lt;/a&gt; post and the links provided by Bill for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never heard of &quot;telescoping&quot; until I read a few answers on here by Bill which refer to this notion.  It seems fairly straightforward, basically you expand some expression using basic arithmetic, there is a minor miracle and lots of terms cancel out in a systematic way, and we are then able to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose &quot;telescoping&quot; in this sense was something I always took for granted, and considered a low level &quot;trick&quot; to keep in my back pocket.  However, considering the importance Bill seems to attach to this notion of telescoping, and considering that I have a great deal of respect for Bill based on the post's by him I have read, I was wondering if I'm not missing something about telescoping.  There is no wiki article on the subject, and a Google search directs me to Bill's answers on SE!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I would like to ask: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What unintuitive results can I achieve with telescoping? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a good reference which only discusses telescoping and applications, or is this concept too narrow for anyone to write a paper/book like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) More trivially, am I missing something about what telescoping actually means?  If not, then why is this called telescoping, because I don't see what this has to do with a telescope?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-09T19:18:44.900" LastActivityDate="2010-12-09T19:37:25.627" Title="Mathematical Telescoping" Tags="&lt;soft-question&gt;&lt;proof-strategy&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13702" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T19:16:47.120" Score="5" ViewCount="0" Body="&lt;p&gt;As in other's answers, the absolute value of $x$ can be defined as the distance between $x$ and 0.  When $x$ is a real number, we can think of $x$ as being on a number line, and there are two locations on a number line that are 1 unit away from 0: 1 unit in the positive direction (called 1) and 1 unit in the negative direction (called &quot;-1&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ can be a complex number, we can think of $x$ as being a point in the plane, and there are a whole lot more locations that are a distance of 1 from 0, such as $\frac{3}{5}+\frac{4}{5}i$ (1 and -1 are also still 1 unit away from 0, so they are still possible values of $x$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-09T19:16:47.120" CommentCount="2" />
  <row Id="13703" PostTypeId="2" ParentId="13701" CreationDate="2010-12-09T19:19:40.567" Score="2" ViewCount="0" Body="&lt;p&gt;The book &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt; by Petkovsek, Wilf and Zeilberger&#xA;is an extended exposition of the Wilf-Zeilberger method&#xA;for summing series often called &quot;creative telescoping&quot;.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-09T19:19:40.567" />
  <row Id="13704" PostTypeId="2" ParentId="13701" CreationDate="2010-12-09T19:19:48.193" Score="2" ViewCount="0" Body="&lt;p&gt;1, 2)  Telescoping is one of the ideas behind modern algorithms to automatically prove hypergeometric identities.  These algorithms allow you, for example, to automatically prove binomial coefficient identities.  The standard reference here is Petkovsek, Wilf, and Zeilberger's &lt;a href=&quot;http://www.math.upenn.edu/~wilf/AeqB.html&quot; rel=&quot;nofollow&quot;&gt;A=B&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  The name comes from the process of collapsing a telescope, which is analogous to the collapsing of a telescoping sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philosophically telescoping is the same as &quot;discrete integration&quot;: telescoping a sum $\sum f(n)$ is the same as finding $g(n)$ such that $f(n) = g(n+1) - g(n)$.  In that sense it is part of the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite differences&lt;/a&gt;, although people probably don't call it &quot;telescoping&quot; in this context.  The context in which I hear the term &quot;telescoping&quot; being used is high school math competitions.  It's one of those basic ideas that everyone has in the back of their head, I suppose.  It's elementary and effective when it applies, but usually there are more sophisticated methods available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Some specific examples.  The ur-example of a telescoping sum is probably&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{n+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and many people have seen this application, but probably far fewer have seen its generalization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \frac{1}{k(k+1)...(k+r)} = \frac{1}{r} \sum_{k=1}^n \left( \frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)} \right) = \frac{1}{r} \left( \frac{1}{r!} - \frac{n!}{(n+r)!} \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other classic example I remember from my competition days is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=1}^n \frac{k}{k^4 - k^2 + 1} = \frac{1}{2} \sum_{k=1}^n \left( \frac{1}{k^2 - k + 1} - \frac{1}{(k+1)^2 - (k+1) + 1} \right) = \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although I have to admit I always found it a little contrived.  Finally, telescoping was put to good use to solve &lt;a href=&quot;http://math.stackexchange.com/questions/1431/prove-that-this-function-is-bounded&quot;&gt;this math.SE question&lt;/a&gt; I posed.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-09T19:37:25.627" LastActivityDate="2010-12-09T19:37:25.627" CommentCount="1" />
  <row Id="13705" PostTypeId="1" CreationDate="2010-12-09T19:25:29.900" Score="2" ViewCount="67" Body="&lt;p&gt;Let $m$ be a positive integer. Define $N_m:=\{x\in \mathbb{Z}: x&amp;gt;m\}$. I was wondering when does $N_m$ have a &quot;basis&quot; of two elements. I shall clarify what I mean by a basis of two elements: We shall say the positive integers $a,b$ generate $N_m$ and denote $N_m=&amp;lt;a,b&amp;gt;$ if every element $x\in N_m$ can be written as $x=\alpha a+\beta b$ where $\alpha,\beta$ are nonnegative integers (not both zero) &lt;STRIKE&gt;and no nonnegative linear combination (with at least one coefficient nonzero) of $a,b$ gives an element not in $N_m$.&lt;/STRIKE&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I want to understand the set $\{(a,b,m):N_m=&amp;lt;a,b&amp;gt;\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be the triple $(2,3,1)$. Every positive integer greater than 1 can be written as $2\alpha +3\beta$ for nonnegative integers $\alpha,\beta$, for if it is even we can write it as $2\alpha$ for some positive integer $\alpha$ and if it is odd and greater than $3$ then we can write it as $3+2\alpha$ for some positive integer $\alpha$ and of course $3=1\cdot 3$. FInally the smallest integer a positive linear combination of $2,3$ can generate is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Not quite sure what to tag this. Feel free to retag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After Jason's answer, it seems the first version of this question is more interesting, where the last condition in paragraph 1 is relaxed.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-09T20:11:19.547" LastActivityDate="2010-12-09T20:24:38.330" Title="Subsets of natural numbers generated by two generators." Tags="&lt;arithmetic&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="13706" PostTypeId="2" ParentId="13705" CreationDate="2010-12-09T19:32:10.467" Score="2" ViewCount="0" Body="&lt;p&gt;If $m$ and $n$ are coprime postive integers, then every integer&#xA;with $&amp;gt;mn-m-n$ is expressible in the form $am+bn$ where $a$&#xD;&#xA;and $b$ are nonnegative integers, but $mn-m-n$ isn't.&#xA;This is due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coin_problem&quot; rel=&quot;nofollow&quot;&gt;Sylvester&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt; The key to the proof is the lemma&#xA;that if $0\le k&amp;lt;(m-1)(n-1)$ then $k$ is representable iff $mn-n-m-k$&#xD;&#xA;isn't. ISTR this involves a cunning use of the Chinese remainder theorem.&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-12-09T19:51:00.597" LastActivityDate="2010-12-09T19:51:00.597" CommentCount="1" />
  <row Id="13708" PostTypeId="2" ParentId="13705" CreationDate="2010-12-09T19:39:09.323" Score="1" ViewCount="0" Body="&lt;p&gt;(Now that you have relaxed the last condition of paragraph 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you mean non negative $\alpha$, $\beta$? (You seem to be allowing them to be zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case, this looks like the &lt;a href=&quot;http://mathworld.wolfram.com/FrobeniusNumber.html&quot; rel=&quot;nofollow&quot;&gt;Frobenius Problem&lt;/a&gt; which says that for $a,b$ such that $(a,b)=1$, every number greater than $(a-1)(b-1) - 1 $ is representable in the for $a\alpha + b \beta$ for $\alpha \ge 0, \beta \ge 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus given an $m$, if a factorization of $m+1$ as $(a-1)(b-1)$ with $(a,b) = 1$ exists, then $N_m  \subset \lt a, b \gt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are also looking for $m$ such that $m \notin \lt a, b \gt$ (the least such $m$) then there are $m$ for which there are no such $a,b$, as $m$ must necessarily be of the form $(a-1)(b-1) - 1$ with $(a,b) = 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact we can characterize all such $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is even, there is no such representation (as both $a-1$ and $b-1$ must be odd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ is odd, then $m+1 = (2-1)(m+2 - 1)$ and thus $a = 2, b = m+2$ will work.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-09T20:24:38.330" LastActivityDate="2010-12-09T20:24:38.330" CommentCount="13" />
  <row Id="13709" PostTypeId="2" ParentId="13705" CreationDate="2010-12-09T19:41:55.497" Score="3" ViewCount="0" Body="&lt;p&gt;This can only happen for the triple you found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For, if $m=1$, then we must be able to generate 2, so we must take 2 in the basis.  Likewise, we must be able to generate 3, so we must allow 3 in the basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m&amp;gt;1$, then we see by the same reasoning that $N_m$ must be generated by $m+1$ and $m+2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then we can't express $m+3$: since $m&amp;gt;1$, we see if $\alpha &amp;gt;1$, then that $\alpha(m+1)\geq 2m + 2 &amp;gt; m+3$ .  So, we must take $\alpha = 0$ or $1$.  Then same reasoning applies to $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take either of $\alpha$ and $\beta$ equal to $0$, then we just generate $m+1$ or $m+2$.  Hence, we must take them both one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $m+1 + m+2 = 2m+3 &amp;gt; m+3$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-12-09T19:41:55.497" CommentCount="1" />
  <row Id="13710" PostTypeId="2" ParentId="13563" CreationDate="2010-12-09T19:43:32.273" Score="0" ViewCount="0" Body="&lt;p&gt;The bigger number minus the smaller number minus 1.&#xA;Or if $b$ is the bigger number, $s$ the smaller. We have:&#xA;$b - s - 1$&lt;/p&gt;&#xA;" OwnerUserId="2533" LastActivityDate="2010-12-09T19:43:32.273" />
  <row Id="13711" PostTypeId="1" AcceptedAnswerId="13985" CreationDate="2010-12-09T19:49:50.043" Score="2" ViewCount="84" Body="&lt;p&gt;I'm looking for a good book on distribution theory (in the Schwartz sense), I have the basic knowledge as given in Grafakos' Classical Fourier Analysis, but I want to know more about it. Is the reference still Laurent Schwartz' Théorie des distributions? That book is hard to obtain, it seems to be only available in France.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-12T00:38:16.777" Title="Distribution theory book." Tags="&lt;reference-request&gt;&lt;distribution-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="13712" PostTypeId="1" AcceptedAnswerId="13720" CreationDate="2010-12-09T19:59:12.280" Score="1" ViewCount="112" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that the formula A if and only if B is a formulation of the equivalence&#xA;that can be decomposed in A if B (the implication B -&gt; A) and in A only if B (the implication A -&gt; B). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't agree with the second part : for me it looks like that A only if B is the translation of ¬B -&gt; ¬A (I know that it's the contraposition of A -&gt; B and consequently it means the same but it's not the same formulation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can anyone tell me the correct interpretation please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bruno&lt;/p&gt;&#xA;" OwnerUserId="4475" LastEditorUserId="4475" LastEditorDisplayName="" LastEditDate="2010-12-09T20:18:10.257" LastActivityDate="2010-12-12T20:18:57.750" Title="How to decompose A if and only if B ?" Tags="&lt;logic&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="13713" PostTypeId="2" ParentId="13712" CreationDate="2010-12-09T20:09:10.877" Score="0" ViewCount="0" Body="&lt;p&gt;I am not sure I am understanding what you are asking, so my answer my not address what you are asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an if an only if (iff) theorem, you have $A \Leftrightarrow B$. This means that A only occurs when B occurs and B only occurs when A occurs. If you are trying to prove an iff theorem, you must prove it in both directions. That is, you must show that $A \Rightarrow B$ and then show that $B \Rightarrow B$.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-09T20:09:10.877" />
  <row Id="13714" PostTypeId="2" ParentId="13693" CreationDate="2010-12-09T20:16:54.610" Score="1" ViewCount="0" Body="&lt;p&gt;Your left adjoint is not quite correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take the case where $A$ is the variety of all groups, and $B$ is the variety of &lt;em&gt;abelian&lt;/em&gt; groups. The forgetful functor from $\mathcal{A}b$ to $\mathcal{G}roup$  indeed has a left adjoint, but the left adjoint $\mathcal{F}$ takes a group $G$ to $G^{\rm ab}$, the abelianization of $G$. This easily yields examples of objects $a$ and $b$ such that $\mathcal{F}(a)\cong \mathcal{F}(b)$, even though $a\not\cong b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;For the specific case of Boolean algebras I think the answer is still negative as stated: take a boolean algebra $a$ that is not [countably] complete and such that its [countable] completion has strictly larger cardinality than $a$ (I assume such things exist, but if this is not the case, then of course I'm falling flat on my case here). Let $b$ be the image of the [countable] completion under the forgetful functor. Since $a$ and $b$ have distinct cardinality, $a\not\cong b$, but $\mathcal{F}(b) = \mathcal{F}(\mathcal{U}(\mathcal{F}(a)))\cong \mathcal{F}(a)$, with the last equality using the unit and counits of the adjunction. &lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to construct $\mathcal{F}(a)$ if your varieties have a nice underlying set functor with an adjoint is to first consider $U(a)$, the underlying set of $a$, then take $\mathfrak{F}_B(U(a))$, the free $B$-algebra on the underlying set of $a$, and then let $\mathcal{F}(a) = \mathfrak{F}_B(U(a))/\Phi$, where $\Phi$ is the $B$-congruence generated by the relations corresponding to the $A$-operations on $a$. For instance, if $B$ is $\mathcal{G}roups$ and $A$ is $\mathcal{M}onoids$, given a monoid $M$, you first take $U(M)=\{[m]\mid m\in M\}$, the underlying set of $M$ (I'm using $[m]$ to denote the element $m$ in the &lt;em&gt;set&lt;/em&gt; $U(M)$). Then you take the free group on $U(M)$, and you mod out by the the normal subgroup generated by the relations $[m][n]=[mn]$ (and since $[mn]=[nm]$, you end up with an abelian group). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I need to think a bit more about the Boolean algebra case; you are correct that my assertion about $\mathcal{F}(\mathcal{U}(\mathcal{F}(a)))$ does not hold in general, and likely not in this case either. My gut feeling is that you can have $\mathcal{F}(a)\cong\mathcal{F}(b)$ even with $a\not\cong b$, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2:&lt;/strong&gt; Heh; reading the MO question you link to in the comments suggests that I am facing exactly what the long discussion between Joel David Hamkins and Todd Trimble was about, and I was making the same kind of assumptions Joel made to reach my conclusion that my construction would necessarily yield an example. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T21:37:58.207" LastActivityDate="2010-12-09T21:37:58.207" CommentCount="7" />
  <row Id="13715" PostTypeId="2" ParentId="13712" CreationDate="2010-12-09T20:23:40.520" Score="0" ViewCount="0" Body="&lt;p&gt;I would expand &quot;A only if B&quot; to &quot;A can only be true if B is also true&quot;, then go to &quot;if A then B&quot; and represent that as $A \implies B$.  As you say, that is English processing, not logic processing.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-09T20:23:40.520" CommentCount="3" />
  <row Id="13716" PostTypeId="1" AcceptedAnswerId="13724" CreationDate="2010-12-09T20:30:37.130" Score="1" ViewCount="63" Body="&lt;p&gt;I came across this problem. If $X_1,X_2, \ldots, X_n$ is an iid sequence with uniform distribution on [-1,1] and we define a new random variable like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_n = \frac{\sum X_k}{\sum(X_k^2 + X_k^3)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I use the strong law of large numbers to show that $Y_n$ is normal and find its mean and variance? The uniform distribution $X_k$ will converge to a normal distribution with mean 0 and variance 3. But I am having trouble finding expectation and variance for $Y_n$.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-09T22:12:58.353" LastActivityDate="2010-12-09T22:12:58.353" Title="problem on strong law of large numbers" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13717" PostTypeId="2" ParentId="13695" CreationDate="2010-12-09T20:44:20.973" Score="1" ViewCount="0" Body="&lt;p&gt;If thinks in this as $f(x)= |x|$, you need to see which are the preimages of $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphicaly you could to get this:&#xA;&lt;img src=&quot;http://i.imgur.com/9xJ4J.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Blue&lt;/em&gt; : $f(x)=|x|$&lt;br&gt;&#xA;&lt;em&gt;Red&lt;/em&gt; : $g(x)=1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $f(x)=1$ only on the intersection points of these two functions, is say on $x=\pm 1$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-09T21:08:11.733" LastActivityDate="2010-12-09T21:08:11.733" />
  <row Id="13718" PostTypeId="1" CreationDate="2010-12-09T20:50:50.990" Score="0" ViewCount="65" Body="&lt;p&gt;This is part of a chapter from Spivak's Calculus on Manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial conditions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a\in A \subset \mathbb{R}^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $g:\mathbb{R}^n \to \mathbb{R}^n$.  Assume $g&amp;#39;(a)=I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $h:A\to\mathbb{R}^n$ by $h(x)=(g^1(x),\dots,g^{n-1}(x),x^n)$.  Then $h&amp;#39;(a)=I$.  Then there is some open $U&amp;#39;$ with $a\in U&amp;#39; \subset A$ where $h$ is 1-1 and $\det h&amp;#39;(x) \ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $k:h(U&amp;#39;)\to\mathbb{R}^n$ by $k(x)=(x^1,\dots,x^{n-1},g^n(h^{-1}(x)))$.  Then $g = k \circ h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spivak shows that $k&amp;#39;(h(a))=I$.  Then there is some open set $V$ where $h(a)\in V\subset h(U&amp;#39;)$ where $k$ is 1-1 and $\det k&amp;#39;(x)\ne 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U=k^{-1}(V)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where my problem is: supposedly, this is sufficient to show that $g=k \circ h$ where $h:U \to \mathbb{R}^n$ and $k:V \to\mathbb{R}^n$ and $h(U) \subset V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $h(U) \subset V$?  Obviously $h(a)\in V$ and $k(U) \subset V$, but I don't see the connection between the two.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-12-09T20:50:50.990" Title="Problem understanding proof of change of variable in integration" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" CommentCount="1" />
  <row Id="13719" PostTypeId="1" CreationDate="2010-12-09T21:03:13.887" Score="1" ViewCount="93" Body="&lt;p&gt;The pullback is a subset of the cartesian product in the category of commutative rings with unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the pullback in the category of commutative $k$-algebras? Is it the same set as in rings?&lt;/p&gt;&#xA;" OwnerUserId="3789" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-09T21:11:58.597" LastActivityDate="2010-12-09T21:17:27.843" Title="What is the pullback in the category of commutative algebras?" Tags="&lt;abstract-algebra&gt;&lt;algebraic-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13720" PostTypeId="2" ParentId="13712" CreationDate="2010-12-09T21:04:36.877" Score="1" ViewCount="0" Body="&lt;p&gt;If I understand your question correctly, you are bothered by the semantics of the statement &quot;$A$ only if $B$&quot;.  You can take this statement as idiomatic, that is, if taken literally it has one meaning, but social convention (in this case the social convention of the mathematics community) leads us to interpret it as always meaning $A \rightarrow B$.  There is certainly room in English to take certain phrases as being somewhat illogical.  My point is that &quot;$A$ only if $B$&quot; doesn't actually have to stand up to logical dissection of the semantics.  Extracting meaning from English sentences is not linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I always look at that the statement &quot;$A$ only if $B$&quot; as meaning A is true only if B is true.  If we look at a truth table for a statement where A is true only if B is true, then we see that this is equivalent to the statement $A\rightarrow B$. &lt;/p&gt;&#xA;" OwnerUserId="1004" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T21:07:55.053" LastActivityDate="2010-12-09T21:07:55.053" CommentCount="1" />
  <row Id="13721" PostTypeId="2" ParentId="13719" CreationDate="2010-12-09T21:17:27.843" Score="2" ViewCount="0" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, fix a category $\mathcal{C}$ that contains finite limits. Given an object $A \in \mathcal{C}$, we can construct the category $\mathcal{C}_A$ of objects &quot;under $A$.&quot; That is, an object of this category $\mathcal{C}_A$ is a morphism $A \to B$ and a  morphism is a commutative triangle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is that $\mathcal{C}_A$ admits finite limits, and the limits are the same as in $\mathcal{C}$. This is basically formal. Given a functor $F:I \to \mathcal{C}_A$ from a finite category $I$, we get a functor $G: I \to \mathcal{C}$ which has a limit by assumption. Moreover, for each $i \in I$, we have a morphism $A \to Gi$. By the universal property, this becomes a morphism $A \to \lim G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that the object $A \to \lim G$ in $\mathcal{C}_A$ is a limit of $F$. This can be checked directly from the definitions.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-09T21:17:27.843" />
  <row Id="13722" PostTypeId="1" CreationDate="2010-12-09T21:33:37.573" Score="0" ViewCount="94" Body="&lt;p&gt;I need to find the function c(k), knowing that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{c(k)}{k!}=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{c(2k)}{(2k)!}=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{c(2k+1)}{(2k+1)!}=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k+1)}{(2k+1)!}=-1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k)}{(2k)!}=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastEditorUserId="2513" LastEditorDisplayName="" LastEditDate="2010-12-09T21:45:45.250" LastActivityDate="2010-12-09T22:25:53.923" Title="How to solve the following system?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;functional-equations&gt;&lt;generating-functions&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="13724" PostTypeId="2" ParentId="13716" CreationDate="2010-12-09T22:10:25.063" Score="4" ViewCount="0" Body="&lt;p&gt;In order to have the desired convergence to the ${\rm N}(0,3)$ distribution, we must multiply $Y_n$ by $\sqrt{n}$ (otherwise, $Y_n \to 0$ almost surely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. uniform$[-1,1]$ rv's.&#xA;Define a sequence $(Y_n)$ by&#xA;$$&#xD;&#xA;Y_n = &#xD;&#xA;\sqrt{n} \frac{{\sum\nolimits_{k = 1}^n {X_k } }}{{\sum\nolimits_{k = 1}^n {(X_k^2  + X_k^3 )} }}.&#xD;&#xA;$$ &#xA;Then $Y_n$ can be written as&#xA;$$&#xD;&#xA;Y_n = \frac{{n^{ - 1/2} \sum\nolimits_{k = 1}^n {X_k } }}{{n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^2 }  + n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^3 } }}.&#xD;&#xA;$$&#xD;&#xA;Now, $X_k^3$ has mean zero, and so by the strong law of large numbers (SLLN), $n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^3 }  \to 0$ almost surely. $X_k^2$, on the other hand, has mean equal to $\int_{ - 1}^1 {x^2 (1/2){\rm d}x}  = 1/3$; hence, by SLLN, $n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^2 }  \to 1/3$ almost surely. Next, $X_k$ has mean zero and (hence) variance equal to ${\rm E}(X_k^2) = 1/3$. So, by the central limit theorem, &#xA;$$&#xD;&#xA;\frac{{\sum\nolimits_{k = 1}^n {X_k } }}{{\sqrt {1/3} \sqrt n }} \to {\rm N}(0,1)&#xD;&#xA;$$&#xD;&#xA;(in distribution). Combining it all, we see that $Y_n$ converges in distribution to $(3/\sqrt{3})Z$, where $Z \sim {\rm N}(0,1)$. Put it another way, $Y_n \to {\rm N}(0,3)$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-09T22:10:25.063" CommentCount="1" />
  <row Id="13725" PostTypeId="2" ParentId="13722" CreationDate="2010-12-09T22:13:51.950" Score="3" ViewCount="0" Body="&lt;p&gt;Following Rotwang, define $b(k)=c(k)/k!$, then note that both the second and fifth can be satisfied as long as $\sum{b(2k)}=0$.  Then the first and third are redundant and define $d(k)=b(2k+1)$.  Now we have $\sum{d(k)}=1, \sum{(-1)^kd(k)}=-1$.  Adding and subtracting, $\sum{d(2k)}=0, \sum{d(2k+1)}=1.$  So the final is that we must have $$\begin{align} \sum\frac{c(2k)}{k!}&amp;amp;=0\\&#xD;&#xA;   \sum\frac{c(4k+1)}{(4k+1)!}&amp;amp;=0\\&#xD;&#xA;   \sum\frac{c(4k+3)}{(4k+3)!}&amp;amp;=1  \end{align}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and any $c(k)$ that satisfies this will work.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-09T22:19:41.027" LastActivityDate="2010-12-09T22:19:41.027" CommentCount="4" />
  <row Id="13726" PostTypeId="2" ParentId="13722" CreationDate="2010-12-09T22:25:53.923" Score="3" ViewCount="0" Body="&lt;p&gt;You are looking for a function $\displaystyle f(z) = \sum_{k \ge 0} \frac{c(k)}{k!} z^k$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(1) = 1$$&#xD;&#xA;$$f(-1) = -1$$&#xD;&#xA;$$f(i) = -i.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinitely many functions have this property.  There is a unique quadratic polynomial $p(z)$ with this property (for example by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow&quot;&gt;Lagrange interpolation&lt;/a&gt;), and for any entire function $q(z)$ the function $p(z) + (x - 1)(x + 1)(x - i) q(z)$ has this property.  In fact these are all entire functions with this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally I think the theory of interpolation by entire functions is fairly well-understood, but I don't know of a good reference.  If the set of $z$ at which you fix the value of $f$ has a limit point, then $f$ is unique by the identity theorem.  If the set of $z$ at which you fix the value of $f$ is countable and does not have a limit point, then $f$ is non-unique by &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Weierstrass factorization&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T22:25:53.923" />
  <row Id="13727" PostTypeId="1" AcceptedAnswerId="13729" CreationDate="2010-12-09T23:09:50.567" Score="2" ViewCount="65" Body="&lt;p&gt;WolframAlpha is confusing me. I'm working on two different integral problems, and with each one, for some reason, Wolfram switches the signs of the answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;http://www.wolframalpha.com/input/?i=integral+dx/(x^.5(x-1))&lt;/li&gt;&#xA;&lt;li&gt;http://www.wolframalpha.com/input/?i=integral+(x-1)/(x^2-4)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the first one, they get $\log[1 - \sqrt{x}] - \log[(1 + \sqrt{x})]$ (see under alternate forms, thats the one I'm using).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I get:&#xA;$\log[\sqrt{x}-1] - \log[(\sqrt{x}+1)]$ And so it differs by a sign, but its really bothering me, since I'm sure I'm doing it right. And the same applies to the second one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They get:&#xA;$3/4 \log(-x-2)+1/4 \log(x-2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get:&#xA;$3/4 \log(x+2)+1/4 \log(x-2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so it differs by a sign again. (If this is a common mistake, then maybe you can point it out, but if this is just me and you can't see where I could have gone wrong, I can list my steps)&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T00:49:34.383" LastActivityDate="2010-12-10T00:49:34.383" Title="Integration and natural log question" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="1" />
  <row Id="13728" PostTypeId="1" CreationDate="2010-12-09T23:11:39.613" Score="2" ViewCount="78" Body="&lt;p&gt;Is it correct to say that the number of circles in an Apollonian gasket is countable becuase we can form a correspondence with a Cantor set, as their methods of construction are similar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about if we apply the Apollonian gasket construction inside of a fractal like the Koch snowflake? (I think that will still be countable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you did the Apollonian gasket construction between f(x) = sin(1/x) and  g(x) = 2 - sin(1/x) between -1 and 1?(I think that will still be countable too, but it's not matching my intuition... which says &quot;no way is that countable!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any closed curve that would result in the number of circles being uncountable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we consider the Apollonian gasket made of spheres in $\mathbb{R}^3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please keep in mind I have only had two courses in Analysis. My apologies if any of this is too naive.) &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditorDisplayName="" LastEditDate="2010-12-09T23:50:27.130" LastActivityDate="2010-12-09T23:56:46.580" Title="Is the number of circles in the Apollonian gasket countable?" Tags="&lt;geometry&gt;&lt;real-analysis&gt;&lt;infinity&gt;&lt;fractals&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13729" PostTypeId="2" ParentId="13727" CreationDate="2010-12-09T23:15:26.477" Score="3" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The (real) indefinite integral is actually different in $(0, 1)$ and in $(1, \infty)$.  WolframAlpha gave you the answer in the first interval and your answer works for the second.  (Remember that in the real numbers, the logarithm of a non-positive number is undefined.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This time WolframAlpha's answer is not defined for any real $x$.  Your answer works in $(2, \infty)$ but there is a different antiderivative in $(-2, 2)$ and yet another antiderivative in $(-\infty, -2)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a fairly subtle property of the logarithm.  Remember that any two antiderivatives of a function differ by a constant.  The problem in this case is that the constant is complex, so if you aren't working in the complex numbers you get genuinely different answers.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-09T23:15:26.477" CommentCount="5" />
  <row Id="13730" PostTypeId="2" ParentId="13728" CreationDate="2010-12-09T23:30:12.897" Score="6" ViewCount="0" Body="&lt;p&gt;The reason a collection of disjoint open sets in $\mathbb{R}^n$ is at most countable is because $\mathbb{R}^n$ is because $\mathbb{R}^n$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Separable_space&quot; rel=&quot;nofollow&quot;&gt;separable&lt;/a&gt;: it has a countable dense subset given by the points with rational coordinates, and any open set must contain such a point that the others don't contain, so there can be at most countably many of them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a measure-theoretic argument relying on the fact that $\mathbb{R}^n$ with the Lebesgue measure is &lt;a href=&quot;http://en.wikipedia.org/wiki/%CE%A3-finite_measure&quot; rel=&quot;nofollow&quot;&gt;$\sigma$-finite&lt;/a&gt;, which might come closer to agreeing with geometric intuition: if you could fit uncountably many open balls into $\mathbb{R}^n$, some closed $n$-cube (WLOG $[0, 1]^n$) would contain uncountably many of them, each of which would have positive measure.  But a standard argument shows that this is impossible:  if $S_k$ is the set of balls in $[0, 1]^n$ of radius at least $\frac{1}{k}$, then since $\bigcup S_k$ is uncountable it follows that some $S_k$ is uncountable, hence countable unions of its elements have unbounded measure; contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument in your first sentence works to show that the number of circles in the standard construction of an Apollonian gasket (start with some circles and add circles) is countable, since at each step only finitely many circles are constructed.  But maybe you are thinking of a more general construction. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-09T23:56:46.580" LastActivityDate="2010-12-09T23:56:46.580" />
  <row Id="13732" PostTypeId="1" CreationDate="2010-12-09T23:47:15.400" Score="1" ViewCount="59" Body="&lt;p&gt;Suppose I have acquired 103,304 customers for a transaction-based business at 12 months into the business and their number has increased at a monthly rate of 40% (from month 1, in which 2,551 customers were acquired). Assume 103,304 represents the total number of new customers (active or dead) acquired over the 12 month period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is one fixed-price product/service worth $10 that each customer makes 4 purchases of across a 6 month period (1 every 1.5 months) and then ceases to be a customer, how can I calculate my total revenue by month 12? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I'm something of an interloper in this forum, but I'll absolutely accept an answer and up-vote all good ones.&lt;/p&gt;&#xA;" OwnerUserId="4588" LastEditorUserId="4588" LastEditorDisplayName="" LastEditDate="2010-12-10T02:00:38.300" LastActivityDate="2010-12-10T02:00:38.300" Title="Tricky Finance-related Question" Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;&lt;finance&gt;" AnswerCount="1" />
  <row Id="13733" PostTypeId="2" ParentId="13732" CreationDate="2010-12-09T23:58:03.770" Score="1" ViewCount="0" Body="&lt;p&gt;You don't give enough data to find an answer.  Each customer is worth 40 in revenue, so if they all signed up by month 6 (and therefore already spend their 40) you have 4M.  If they all signed up in month 11.5 and haven't bought yet you have zero.  You need something about the rate of increase of customers and about when in the six months they buy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would simplify things to assume they spend evenly over time, so every customer is worth 40/6 per month.  Then if n(i) is the number of customers who join in month i, the number of customers active in month i is the sum over the last six months of n(i) and your revenue will be 4M less what you will get from the current customer base in the next months.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-09T23:58:03.770" CommentCount="2" />
  <row Id="13734" PostTypeId="1" AcceptedAnswerId="13738" CreationDate="2010-12-10T00:18:22.717" Score="0" ViewCount="80" Body="&lt;p&gt;If given 2 lines $\alpha$ and $\beta$, &lt;s&gt;that are created by&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;2 points: A and B &lt;/li&gt;&#xA;&lt;li&gt;2 plane intersection&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I want to find shortest distance between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\begin{array}{c} P_1=x_1X+y_1Y+z_1Z+C=0 \\ P_2=x_2X+y_2Y+z_2Z+C=0\end{array}\right.$$&lt;br&gt;&#xA;$$A=\left(x_3;y_3;z_3\right)$$&lt;br&gt;&#xA;$$B=\left(x_4;y_4;z_4\right)$$&lt;br&gt;&#xA;$$\alpha =n_1\times n_2=\left(\left|\begin{array}{cc} y_1 &amp;amp; z_1 \\ y_2 &amp;amp; z_2\end{array}\right|;-\left|\begin{array}{cc} x_1 &amp;amp; z_1 \\ x_2 &amp;amp; z_2\end{array}\right|;\left|\begin{array}{cc} x_1 &amp;amp; y_1 \\ x_2 &amp;amp; y_2\end{array}\right|\right)$$&lt;br&gt;&#xA;$$\beta =$$ &lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I tried:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question of &quot;shortest distance&quot; is&#xA;  only interesting in the skew case.&#xA;  Let's say p0 and p1 are points on the&#xA;  lines L0 and L1, respectively. Also d0&#xA;  and d1 are the direction vectors of L0&#xA;  and L1, respectively. The shortest&#xA;  distance is (p0 - p1) * , in which *&#xA;  is dot product, and is the normalized&#xA;  cross product. The point on L0 that is&#xA;  nearest to L1 is p0 + d0(((p1 - p0) *&#xA;  k) / (d0 * k)), in which k is d1 x d0&#xA;  x d1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read more: &lt;a href=&quot;http://wiki.answers.com/Q/If_the_shortest_distance_between_two_points_is_a_straight_line_what_is_the_shortest_distance_between_two_straight_lines#ixzz17fAWKFst&quot; rel=&quot;nofollow&quot;&gt;http://wiki.answers.com/Q/If_the_shortest_distance_between_two_points_is_a_straight_line_what_is_the_shortest_distance_between_two_straight_lines#ixzz17fAWKFst&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried, but failed. &lt;/p&gt;&#xA;" OwnerUserId="2317" LastEditorUserId="2317" LastEditorDisplayName="" LastEditDate="2010-12-10T00:54:28.867" LastActivityDate="2010-12-10T01:00:02.923" Title="How to find shortest distance between two skew lines in 3D?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13735" PostTypeId="1" CreationDate="2010-12-10T00:20:41.170" Score="6" ViewCount="218" Body="&lt;p&gt;Hi,&#xA;I'm curious to whether you guys have any tips on book concerning classical euclidean geometry. I'd like somewhat of an advanced treatment, around the same level as Coxeter's &quot;Geometry revisited&quot;. I'd however want something more comprehensive, therefore I'm asking you for tips!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopeful for answers ^^&lt;/p&gt;&#xA;" OwnerUserId="2394" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T00:39:14.967" LastActivityDate="2010-12-10T06:41:28.787" Title="Books on classical geometry" Tags="&lt;geometry&gt;&lt;books&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="2" />
  <row Id="13736" PostTypeId="1" AcceptedAnswerId="13741" CreationDate="2010-12-10T00:21:12.140" Score="3" ViewCount="160" Body="&lt;p&gt;So, I am a graduate student who is certain that he does not want to do analysis (I think...).  What are the most exciting fields in mathematics right now?  It seems to me that very generally, they are algebraic geometry and algebraic topology, with the latter being specifically concerned with homotopy theory (something I find very interesting).  Perhaps this is not the correct place to post such a question and I'm welcome to redirection on that front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I would like to do something highly abstract and categorical.  I love the ideas of sheaves, homological algebra, and various categorical concepts.  Does anyone have any good knowledge of where to begin?  Also, do algebraic geometry and homotopy theory have a great deal of common ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, and sorry if this question belongs somewhere else.  &lt;/p&gt;&#xA;" OwnerUserId="3595" LastActivityDate="2010-12-10T01:49:12.157" Title="Geometry or Topology" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="13737" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T00:38:24.203" Score="2" ViewCount="0" Body="&lt;p&gt;Well I only took one course on elementary geometry so I don't know much about this, but we used Edwin Moise's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201508672&quot; rel=&quot;nofollow&quot;&gt;Elementary Geometry From an Advanced Standpoint&lt;/a&gt;. It is very well written and probably has a level that will suit you. We used the first edition so I'm not sure what may have changed in the other editions and I can't seem to find a google preview of the third edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case I highly recommend you take a look at it to see if you like it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;I found the table of contents for the third edition &lt;a href=&quot;http://www.pearsonhighered.com/educator/product/Elementary-Geometry-from-an-Advanced-Standpoint/9780201508673.page&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-10T04:37:07.693" LastActivityDate="2010-12-10T04:37:07.693" />
  <row Id="13738" PostTypeId="2" ParentId="13734" CreationDate="2010-12-10T01:00:02.923" Score="2" ViewCount="0" Body="&lt;p&gt;Per &lt;a href=&quot;http://en.wikipedia.org/wiki/Skew_lines#Distance_between_two_skew_lines&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt;, if your lines are $\vec{X}=\vec{X_1}+t\vec{D_1}$ and $\vec{X}=\vec{X_2}+t\vec{D_2}$, the distance between them is $$\left|\frac{\vec{D_1}\times\vec{D_2}}{|\vec{D_1}\times\vec{D_2}|}\cdot(\vec{X_1}-\vec{X_2})\right|.$$&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-10T01:00:02.923" />
  <row Id="13739" PostTypeId="1" AcceptedAnswerId="15000" CreationDate="2010-12-10T01:26:21.463" Score="1" ViewCount="102" Body="&lt;p&gt;I'm trying to solve exercise 2614.1 from Demidovich's famous book of exercises on Analysis. Having solved the bit about convergence, I'm now stuck in trying to prove that a series of terms $a_n\geq 0$ is divergent if there exists $N$ such that $(1-\sqrt[n]{a_n})\frac{n}{\log{n}} \leq 1$ for all $n&amp;gt;N$. I've been trying to find another divergent series with which to compare it, or to prove that $a_n$ does not tend to zero. Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2010-12-20T19:54:05.347" Title="Test for divergence of series." Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13740" PostTypeId="1" AcceptedAnswerId="13746" CreationDate="2010-12-10T01:29:02.280" Score="0" ViewCount="58" Body="&lt;p&gt;I've been spinning my wheels on this for an embarrassing amount of time. This is not for an assignment, I'm just studying.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three balls are placed at random in&#xA;  three boxes. There is no restriction&#xA;  on the number of balls per box; list&#xA;  the 27 possible outcomes of this&#xA;  experiment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I could only come up with 10 where order of placing the balls into the boxes doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{(0,0,3), (0,1,2), (0,2,1), (0,3,0), (1,0,2), (1,1,1), (1,2,0), (2,0,1), (2,1,0), (3,0,0)}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-12-10T01:50:22.613" Title="How to Count - 3 balls in 3 boxes" Tags="&lt;counting&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13741" PostTypeId="2" ParentId="13736" CreationDate="2010-12-10T01:40:08.230" Score="5" ViewCount="0" Body="&lt;p&gt;Wow, this is great! I would love to expound a bit on some of the things you have mentioned. While I agree with Ryan that what is exciting and hot is entirely subjective, I too find homotopy theory and algebraic geometry enthralling. I really only know things about homotopy theory though, so I can only talk on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are loads of places to go abstract crazy under the general heading of homotopy theory. A lot of the model category theory is being put to use in many many places, like Algebraic Geometry! (This is the work of Dugger-Isaksen that Ryan mentioned above). There is a lot of beautiful abstract framework that people work with that falls under the umbrella of homotopy theory. Personally I am a bit more on the computational side, or rather old fashioned &quot;how do we compute $\pi_n^S(\mathbb{S})$?&quot; So I am interested in different computational aspects of the stable homotopy category of spectra. It really is a big field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just realized I did not address your question about where to begin. There are some amazing deep results that are really cool that you can get to in a finite amount of time. For homotopy theory I would work on trying to get to come classical results of Adams, like vector fields on spheres or Hopf invariant one. Both of these are addressed in Mosher and Tangora (now a dover book). It is a good book, but you should skip certain bits, like you don't need their construction of the steenrod operations. There is also the theory of formal group laws and how those relate to stable homotopy theory, that stuff is awesome. Being at JHU, I would begin by asking Jack Morava or Andrew Salch. They are both super nice guys that know a whole lot, but they are really smart and might be hard to keep up with. So maybe ask them what got them started in being interested in such things, or what they think is something that is really cool to work towards. I also think that Boardman and Wilson would be excellent people to ask, but I have had no interaction with them. They are also deep vats of information. I learned so much going through the first third of Boardmans CCSS paper, it was great!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the interaction, it is large! As Mathew pointed out there is a whole &quot;new&quot; field called motivic homotopy theory that asks &quot;what can homotopy theory tell us about schemes?&quot; There are other interactions though, a lot of number theory and stackiness is interesting via the chromatic picture. A lot of people study the moduli stack of formal group (laws) in order to get at the stable homotopy groups of spheres. That moduli stack necessarily has a lot of AG in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the best thing to do would be to talk to people around about things that you are learning/want to learn. In fact, feel free to drop me a line at first intial last name at wayne.edu. Seriously drop me a line if you want to get into a bit more detail, it is unclear what will be beneficial for me to say without a bit more background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: there are great people at JHU that know loads about homotopy theory, as well as its interactions with algebraic geometry. Morava pioneered the relationship with class field theory, and now it seems like Salch is taking it all the way to Langlands Land. It would be hard to be at a better place to study homotopy theory!&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-10T01:49:12.157" LastActivityDate="2010-12-10T01:49:12.157" CommentCount="2" />
  <row Id="13744" PostTypeId="2" ParentId="13654" CreationDate="2010-12-10T01:45:30.840" Score="2" ViewCount="0" Body="&lt;p&gt;For your original problem, here's the approach I'd take:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what the set of graphs of $y=x^2-ax+2$ looks like as $a$ varies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/U3UQ8.gif&quot; alt=&quot;animation of graphs as a varies&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, the picture in my head isn't that clear, but I know that they all have y-intercept 2 and move in more or less that fashion as $a$ varies (the locus of the vertices of those parabolas is itself a parabola).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I want the interval from when the equation starts to have real roots up to when the greater of its two roots is 3.  Starting with the latter condition, $3=\frac{a+\sqrt{a^2-8}}{2}\implies(6-a)^2=a^2-8\implies 36-12a=-8\implies a=\frac{11}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earlier condition happens when $x^2-ax+2$ is a perfect square, so when $\frac{a}{2}=\pm\sqrt{2}$.  The negative value of $a$ leads to the parabola that touches the x-axis on the negative side, so the relevant value is $a=2\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these two bounding values and the animated image in my head, the desired values of $a$ are $2\sqrt{2}&amp;lt;a&amp;lt;3$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-10T01:45:30.840" CommentCount="2" />
  <row Id="13745" PostTypeId="2" ParentId="13739" CreationDate="2010-12-10T01:48:28.933" Score="1" ViewCount="0" Body="&lt;p&gt;Try to show that $(1 - \frac{{\log n}}{n})^n $ is asymptotically equal to $b_n$, as $n \to \infty$, where &#xA;$\sum\nolimits_{n = 1}^\infty  {b_n }$ is some very well known divergent series...&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-10T01:48:28.933" />
  <row Id="13746" PostTypeId="2" ParentId="13740" CreationDate="2010-12-10T01:50:22.613" Score="1" ViewCount="0" Body="&lt;p&gt;Using &lt;a href=&quot;http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types/691#691&quot;&gt;the stars-and-bars technique&lt;/a&gt;, for your interpretation of the problem, there are in fact ${3+3-1\choose 3}=10$ possible outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Qiaochu Yuan said in his comment, 27 is the result when the three balls are distinguishable (and I agree with him that it is not clear in the problem statement)—for each of the 3 balls, there are 3 possible locations, so $3^3=27$ possible outcomes.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-10T01:50:22.613" />
  <row Id="13748" PostTypeId="2" ParentId="13525" CreationDate="2010-12-10T02:24:18.543" Score="3" ViewCount="0" Body="&lt;p&gt;Expanding on Jack's answer, there is a whole exciting field known as &quot;chromatic homotopy theory&quot;.  I don't know much about it, I'll say a few slogans since I noticed you also asked the question &quot;geometry or topology&quot;.  Hopefully none of them are wrong.  (If so, somebody please correct me!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Chromatic homotopy theory somehow slices the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adams_spectral_sequence&quot; rel=&quot;nofollow&quot;&gt;Adams-Novikov spectral sequence&lt;/a&gt; based on levels of periodicity (hence the name &quot;chromatic&quot; -- it's supposed to conjure up the image of light refracting through a prism or something).&lt;/s&gt; [cf. Sean's answer for a much better informed account.]&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Level 0 is ordinary cohomology with $\mathbb{Q}$ coefficients.  There, the interaction of geometry and homotopy theory is given by integration.&lt;/li&gt;&#xA;&lt;li&gt;Level 1 is K-theory.  There, the interaction of geometry and homotopy theory is given by the Atiyah-Singer index theorem.&lt;/li&gt;&#xA;&lt;li&gt;Level 2 is elliptic cohomology (and something called topological modular forms, which somehow loosely is the &quot;universal elliptic cohomology theory&quot;).  It is unknown how to bridge the gap with geometry, but this is something people are working on.&lt;/li&gt;&#xA;&lt;li&gt;All the other finite levels are unknown.  A few people are trying to figure out what level 3 should be.&lt;/li&gt;&#xA;&lt;li&gt;Level $\infty$ is complex cobordism.&lt;/li&gt;&#xA;&lt;li&gt;All the levels (so far?) are &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_orientation&quot; rel=&quot;nofollow&quot;&gt;complex-oriented&lt;/a&gt;.  The complex cobordism spectrum $MU$ carries the universal formal group law, and complex orientations of a spectrum $E$ are obtained as multiplicative maps of spectra $MU\rightarrow E$.  (In fact, $(MU_*,MU_*MU)$ corepresents the moduli stack of formal &lt;em&gt;groups&lt;/em&gt;!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to actually learn some things about this, try:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hopkins' notes, &lt;a href=&quot;http://www.math.rochester.edu/u/faculty/doug/otherpapers/coctalos.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Complex Oriented Cohomology Theories and the Language of Stacks&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.math.harvard.edu/~lurie/252x.html&quot; rel=&quot;nofollow&quot;&gt;class on chromatic homotopy theory taught by Lurie&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Lurie: &lt;a href=&quot;http://www.math.harvard.edu/~lurie/papers/survey.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;A Survey of Elliptic Cohomology&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-12-10T18:27:27.723" LastActivityDate="2010-12-10T18:27:27.723" />
  <row Id="13749" PostTypeId="2" ParentId="13679" CreationDate="2010-12-10T02:29:56.603" Score="2" ViewCount="0" Body="&lt;p&gt;In terms of pointwise convergence, you cannot ensure the limit to exist at a given point, even if all the derivatives of $f$ exist (everywhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, by a theorem of Borel, &lt;em&gt;any&lt;/em&gt; sequence $a_0,a_1,\dots$ is $f(0),f&amp;#39;(0),f&amp;#39;&amp;#39;(0),\dots$ for some function $f$, see this &lt;a href=&quot;http://math.stackexchange.com/questions/10643/every-real-sequence-is-the-derivative-sequence-of-some-function/&quot;&gt;question&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-10T02:29:56.603" CommentCount="5" />
  <row Id="13750" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T02:34:09.313" Score="2" ViewCount="0" Body="&lt;p&gt;For a careful and very comprehensive treatment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robin Hartshorne: Geometry: Euclid and Beyond; Springer, 2000.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-10T02:34:09.313" CommentCount="1" />
  <row Id="13751" PostTypeId="1" AcceptedAnswerId="13753" CreationDate="2010-12-10T02:38:53.960" Score="0" ViewCount="51" Body="&lt;p&gt;$x=t^2-3$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=9t$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that a point moves in (x,y) plane according to the above parametric equations. Find for which $t$ the velocity is equal to 9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea about this problem, but I'm not sure how to work with it. If we are given the position functions, then the derivative is the velocity. The derivative of both of those with respect to t is $x=2t, y=9$, but I'm not sure what to do after that. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T04:02:49.940" LastActivityDate="2010-12-10T04:02:49.940" Title="Parametric Equation Question" Tags="&lt;calculus&gt;" AnswerCount="2" />
  <row Id="13752" PostTypeId="2" ParentId="13751" CreationDate="2010-12-10T02:44:27.553" Score="4" ViewCount="0" Body="&lt;p&gt;Hint: position vector $r=&amp;lt;x(t),y(t)&amp;gt;$. Velocity vector is $dr/dt$ and speed is the magnitude of this vector.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-10T02:44:27.553" />
  <row Id="13753" PostTypeId="2" ParentId="13751" CreationDate="2010-12-10T02:46:12.317" Score="4" ViewCount="0" Body="&lt;p&gt;$v(t)=(2t,9)$ ( Vector ), then in magnitude get $\|v(t)\|  = \sqrt{(2t)^2+9^2}$ ( speed ), thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\|v(t)\| = 9 \Rightarrow \sqrt{(2t)^2+9^2}= 9$$ Now solve for $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnitude &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/IiDdP.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{y}$ and $A_{x}$ are the components of vector $A$, the magnitude of vector $A$ is the distance from $ A $ to the origin. By pytagorean theorem the length of $ A $ is given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^2=A_{x}^2+A_{y}^2 \Rightarrow A=\sqrt{A_{x}^2+A_{y}^2}$$  ( In magnitude)&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-10T03:47:58.980" LastActivityDate="2010-12-10T03:47:58.980" CommentCount="3" />
  <row Id="13754" PostTypeId="1" AcceptedAnswerId="13756" CreationDate="2010-12-10T03:32:41.433" Score="2" ViewCount="101" Body="&lt;p&gt;The permanent can be interpreted as the number of perfect matchings in bipartite graphs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a similar graph-theoretic interpretation of the determinant? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2459" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T03:34:20.950" LastActivityDate="2010-12-10T07:49:50.377" Title="Graph-theoretic interpretation of determinant?" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13755" PostTypeId="2" ParentId="12054" CreationDate="2010-12-10T03:42:22.207" Score="2" ViewCount="0" Body="&lt;p&gt;I think there is a problem with the above solution.  In the estimate, $$|f^{(k+1)}(x)| \leq \left(\frac{1}{1-r}\right)^{k+1},$$ there is a dropped $k!$.  Indeed, it should read,&#xA;$$|f^{(k+1)}(x)| \leq \left(\frac{k!}{1-r}\right)^{k+1},$$ &#xA;and thus&#xA;$$&#xD;&#xA;|r_k(x)| \leq \left( \frac{r}{1-r} \right)^{k+1} \cdot \frac{k!}{(k+1)!} = \left( \frac{r}{1-r} \right)^{k+1} \cdot \frac{1}{k+1}.&#xD;&#xA;$$&#xD;&#xA;Unfortunately now this expression won't go to $0$ if $r&amp;gt;0$ (now the exponential term will dominate $\frac{1}{k+1}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above solution does work for x in (-1/2,1).  Here's a way to handle the remaining cases.  In fact, let's just take x in (-1,0).  Now Lagrange's form of the remainder gives:&#xA;$$&#xD;&#xA;r_k(x) = \int_0^x \frac{f^{(k+1)}(t)}{k!} (x-t)^{k} dt &#xD;&#xA;= \int_0^x \frac{(-1)^k}{(1+t)^{k+1}} (x-t)^{k} dt &#xD;&#xA;$$&#xD;&#xA;Note that for $x&amp;lt;0$, the above integrand has the same sign for every $t$.  In particular,&#xA;$$&#xD;&#xA;|r_k(x)| &#xD;&#xA;= \int_x^0 \frac{1}{(1+t)^{k+1}} (t-x)^{k} dt. &#xD;&#xA;$$&#xD;&#xA;Consider $\frac{t-x}{1+t}$ as a function in $t$ with $x$ fixed.  It then is an increasing function on [x,0] with maximal value of $-x$ when $t=0$.  Thus,&#xA;$$&#xD;&#xA;|r_k(x)| \leq \int_x^0 (-x)^k \frac{1}{1+t} dt \leq \int_x^0 (-x)^k \frac{1}{1+x} dt = \frac{(-x)^{k+1}}{1+x}.&#xD;&#xA;$$&#xD;&#xA;As desired, this last expression does go to $0$ as $k \to \infty$ as $-1&amp;lt;x&amp;lt;0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in Spivak's Calculus, the stronger bound of &#xA;$$&#xD;&#xA;|r_k(x)| \leq \frac{(-x)^{k+1}}{(1+x)(1+k)}&#xD;&#xA;$$&#xD;&#xA;is left as Exercise 16 in Chapter 20.  I don't see how to get this, but I'm anxious that I'm just making some mistake in the computations above.&lt;/p&gt;&#xA;" OwnerUserId="4593" LastEditorUserId="4593" LastEditorDisplayName="" LastEditDate="2010-12-10T17:00:14.863" LastActivityDate="2010-12-10T17:00:14.863" CommunityOwnedDate="2010-12-10T15:48:51.773" CommentCount="5" />
  <row Id="13756" PostTypeId="2" ParentId="13754" CreationDate="2010-12-10T03:45:59.277" Score="10" ViewCount="0" Body="&lt;p&gt;I'm aware of a few.  There is the &lt;a href=&quot;http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/&quot; rel=&quot;nofollow&quot;&gt;Lindström-Gessel-Viennot lemma&lt;/a&gt;, and there is also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;matrix-tree theorem&lt;/a&gt;.  If $A$ is the adjacency matrix of a finite graph $G$ then $\frac{1}{\det(I - At)}$ describes a kind of &quot;zeta function&quot; of $G$.  I describe some of how this works in &lt;a href=&quot;http://qchu.wordpress.com/2009/11/04/newtons-sums-necklace-congruences-and-zeta-functions-ii/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also be interested in Kuperberg's &lt;a href=&quot;http://www.combinatorics.org/Volume_5/PDF/v5i1r46.pdf&quot; rel=&quot;nofollow&quot;&gt;An exploration of the permanent-determinant method&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-10T07:49:50.377" LastActivityDate="2010-12-10T07:49:50.377" CommentCount="3" />
  <row Id="13757" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T03:54:51.427" Score="1" ViewCount="0" Body="&lt;p&gt;Try Agricola's Elementary Geometry published by the AMS. It has nice expositions of two-dimensional spherical and hyperbolic geometries as well.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-10T03:54:51.427" />
  <row Id="13758" PostTypeId="2" ParentId="13525" CreationDate="2010-12-10T03:56:39.647" Score="5" ViewCount="0" Body="&lt;p&gt;There is no, as far as i know, general chromatic filtration for object of type X. There is a chromatic filtration of the stable homotopy groups of spheres. You could try to make sense of some directed system that is &quot;chromatic&quot; associated to a space/spectrum by looking at the maps between its localizations with respect to Morava K-theory (or Morava E-theory or Johnson-Wilson, or what have you) and how they fit together, but i think this is essentially the chromatic convergence thm, (there are finite type and p-local hypotheses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important ingredient to note is Carlsson's Thm/the Segal conjecture, which states, i believe, that stable cohomotopy is the completed Burnside ring. Maybe you can feed some of the chromatic ideas through all the hard work that went into the Segal Conjecture and get something that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe that is not it at all. A note about Aarons answer, the chromatic filtration does not come from the ANSS, although it does break up the way you say though. Ravenel spends some time in Nilpotence and Periodicity talking about this stuff. I think it has to do with self maps inducing multiplication by $v_n$ in $K(n)_*(X)$, and you filter the self maps of X that way. The $v_n$ are tied to the theory of formal group laws intimately. The way I think of the chromatic filtration, and this could be off, is by what level of (complex oriented) cohomology theory i need to detect that a self map is not trivial, where by level i mean height of the formal group law corresponding to the particular complex oriented theory. The height of the formal group laws govern the filtration and how it behaves. (This is how I think of it, but in light of the below, it is sort of junk, I will leave it because it is less technical.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the paper Jack linked to is where they develop the chromatic spectral sequence which converges to the $E_2$ term of ANSS which is crazy hard to compute (i dont think anyone knows all the 2 line, and if they do then no one knows all of the 3 line, and by all i mean out to $t-s=50$) So there is a filtration on the $E_2$ term of ANSS that gives rise to this spectral sequence, and I think this is the chromatic filtration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, it is a technical term, and it does not just mean periodic.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;I just found the following sentence while reading the review of &quot;$v_n$ periodic elements in ring spectra and applications to bordism theory&quot; by Hovey (the review is by stong):&#xA;&quot;Let $R$ be a ring spectrum. If $v \in \pi_k R$, $v$ is called a $v_n$ element if $K(n)_*(v) \in K(n)_*(R)$ is a unit and $K(i)_*(v) \in K(i)_*(R)$ is nilpotent for all $i \neq n$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be back in a bit to check it against a reading of Ravenel's orange book.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-16T21:53:18.190" LastActivityDate="2010-12-16T21:53:18.190" />
  <row Id="13759" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T04:03:11.923" Score="2" ViewCount="0" Body="&lt;p&gt;A book I like very much is the classic &quot;College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle&quot; by Nathan Altshiller-Court. The (revised in 1952) book has been reissued recently (2007) by Dover in a very affordable paperback. There is also a companion book on spatial geometry, but I do not know how easy it would be to find a copy. In Colombia, training for the Math Olympiads used for a while this book and Coxeter's, so I have first hand experience of learning from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completely different reasons, another book I like a lot is &quot;Higher Geometry&quot; by N.V. Efimov, originally from Mir eds. The book begins with an interesting discussion of axiomatic systems, talks about the attempts to prove the 5th postulate, Hilbert's axiomatization of Euclidean geometry, and then develops it synthetically. The book then does the same with hyperbolic geometry, and this is the only place where I have seen this done (it is well worth reading this part). There is also a treatment of projective geometry which is very nice, and (I think) of differential geometry. (The description in this book of the foundational approach was one of the reasons I decided I wanted to specialize in logic.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-10T04:17:39.540" LastActivityDate="2010-12-10T04:17:39.540" />
  <row Id="13760" PostTypeId="2" ParentId="13404" CreationDate="2010-12-10T04:05:09.033" Score="0" ViewCount="0" Body="&lt;p&gt;I've been wrestling with a very similar problem in order to determine gradients in an irregular quad grid and needing to map points within arbitrary quadrilaterals to a unit square.  In addition, I require the inverse mapping of the x and y axis at the mapped normalized coordinate location back into the quad so I can determine the orientation of the quad grid at that point.  i.e. if &lt;code&gt;[x',y']&lt;/code&gt; are the transformed coordinates, I need to be able to do an inverse transform on &lt;code&gt;[0,y'],[1,y']&lt;/code&gt; and &lt;code&gt;[x',0],[x',1]&lt;/code&gt;. Here is what I've come up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can divide the quad into two tris, and use affine maps on these individually.  This is not difficult.  This will create a noticable effect at the division between the two tris, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a smooth mapping from a quad to a square (or rectangle), you need to use a non-affine transform such as a projective transform.  There are other transforms other than projective that will also work, and also be colinear (preserve straight lines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;code&gt;[x1,y1],[x2,y2],[x3,y3],[x4,y4]&lt;/code&gt; are the four points in the quad, then the 4x4 matrix &lt;strong&gt;B&lt;/strong&gt; in the the following will yield a mapping into the square (on the RHS) that seems to work and may be easier to compute than the proper 3x3 projective matrix.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%     [x1 y1 x1*y1 1]              [0 0 0 1]&#xA;%     [x2 y2 x2*y2 1]   X  B  =    [1 0 0 1]&#xA;%     [x3 y3 x3*y3 1]              [0 1 0 1]&#xA;%     [x4 y4 x4*y4 1]              [1 1 1 1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The question I have is that if one does this, and then wants to use the inverse of &lt;strong&gt;B&lt;/strong&gt; to do the inverse transform, how do you calculate the third elements of the location vectors for the orthogonal coordinates.  (They are no longer x*y.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: If you want to map into any other (arbitrary) quadrilateral (such as a rectangle), then just replace the RHS of what I have above with the new coordinates.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%     [x1 y1 x1*y1 1]              [x1' y1' x1'*y1' 1]&#xA;%     [x2 y2 x2*y2 1]   X  B  =    [x2' y2' x2'*y2' 1]&#xA;%     [x3 y3 x3*y3 1]              [x3' y3' x3'*y3' 1]&#xA;%     [x4 y4 x4*y4 1]              [x4' y4' x4'*y4' 1]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4592" LastEditorUserId="4592" LastEditorDisplayName="" LastEditDate="2010-12-10T04:15:35.923" LastActivityDate="2010-12-10T04:15:35.923" />
  <row Id="13761" PostTypeId="1" AcceptedAnswerId="13765" CreationDate="2010-12-10T04:06:58.627" Score="4" ViewCount="215" Body="&lt;p&gt;Using only four 8's and only addition, subtraction, multiplication, and division.. how can you come up with the number 24? You may use fractions and decimals! But remember, you can ONLY USE four 8's&lt;/p&gt;&#xA;" OwnerUserId="4595" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T04:17:45.497" LastActivityDate="2010-12-10T08:09:02.427" Title="Using 4 '8's, get the number 24" Tags="&lt;puzzle&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="13762" PostTypeId="2" ParentId="13761" CreationDate="2010-12-10T04:32:17.073" Score="3" ViewCount="0" Body="&lt;p&gt;I usually recommend &lt;a href=&quot;http://mrob.com/pub/ries/index.html&quot; rel=&quot;nofollow&quot;&gt;RIES&lt;/a&gt; for this sort of problem.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2010-12-10T04:32:17.073" CommentCount="3" />
  <row Id="13763" PostTypeId="1" AcceptedAnswerId="13767" CreationDate="2010-12-10T05:22:06.750" Score="7" ViewCount="189" Body="&lt;p&gt;My undergraduate number theory class decided to dip into a bit of algebraic geometry to finish up the semester. I'm having trouble understanding this bit of information that the instructor presented in his notes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is in paraphrase (assume we are over an abstract field k)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take a polynomial in k, $f  =Y^2 - X^3 -aX -b$ and homogenize the polynomial to $F = Y^2Z - X^3 -aXZ^2 - bZ^3$. &lt;em&gt;Note that the points at infinity of V(F) consist of triples&lt;/em&gt; $[\alpha : \beta : 0]$ s.t $ -\alpha^3 = 0$, hence the only point at infinity is $[0 : 1 :0]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I'm confused about is in italics. He introduces the terms &quot;points at infinity&quot; without defining it. After some google time, I understand what a point at infinity means in the context of a projective space/projective line but am having trouble understanding how the professor came to his conclusion about the point at infinity in this particular example &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my question. In general, are all points in the locus of vanishing points for a homogeneous polynomials considered points at infinity? If not, is there a general procedure for calculating these point if we are given an arbitrary polynomial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More abstractly, How do I understand that a finite point in the projective space is a &quot;point at infinity&quot; for this polynomial.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-12-11T07:33:00.487" Title="Eliptic Curves and Points at Infinity" Tags="&lt;algebraic-geometry&gt;&lt;elliptic-curves&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="13765" PostTypeId="2" ParentId="13761" CreationDate="2010-12-10T05:32:13.377" Score="6" ViewCount="0" Body="&lt;p&gt;$ 88 - (8 \cdot 8 ) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much easier with five 8's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 8(8+8+8)/8 $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:)&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-12-10T05:32:13.377" CommentCount="2" />
  <row Id="13766" PostTypeId="2" ParentId="13763" CreationDate="2010-12-10T05:36:53.313" Score="4" ViewCount="0" Body="&lt;p&gt;The term &quot;point at infinity&quot; is not actually a well-defined term from the point of view of your projective model. You should think of it this way: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are given a homogeneous polynomial and you are interested in its zero set as a subset of projective space.&#xA;The projective space is made up of several so-called affine pieces, and accordingly your zero set is made up of several affine pieces. Here is how it works: let's take the concrete polynomial $F=Y^2Z - X^3-aXZ^2 - bZ^3$. Since you can take any point $(x:y:z)$ satisfying this polynomial and scale the coordinates by any non-zero scalar, $\alpha$ say, and the result still represents the same point, you may scale the point by $\alpha=z^{-1}$ and write it as $(x/z,y/z,1)$. But now, you have to be careful: this only works for points, at which $z\neq 0$. For those points, you are setting $z=1$, so the resulting affine model is given by your polynomial $f$ (just set $Z=1$ in $F$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand $z$ happened to be 0, then you have attempted to divide by 0, so in this sense, you obtained a &quot;point at infinity&quot;. Now, what is this point at infinity? Set $Z=0$ in $F$ and see what points satisfy the resulting polynomial. For the polynomial to be 0, you also need $-X^3=0$, so $X=0$. Now, since we are in projective space, $Y$ mustn't be 0 (recall that (0:0:0) is not a point in projective space), so you can scale the coordinates so that $Y=1$. That's your &quot;point at infinity&quot; for this particular affine piece: $(X:Y:Z)=(0:1:0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the procedure was not canonical. Instead, you might have chosen to look at the affine piece $Y\neq 0$, say. Then, you would have written all points in this affine model as $(x/y:1:z/y)$ and the dehomogenised polynomial would have been different. Also, the &quot;points at infinity&quot; would have been different ones, namely all those projective points, for which $Y=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense. Otherwise, feel free to ask for clarifications.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-10T05:41:55.147" LastActivityDate="2010-12-10T05:41:55.147" CommentCount="4" />
  <row Id="13767" PostTypeId="2" ParentId="13763" CreationDate="2010-12-10T06:40:16.760" Score="8" ViewCount="0" Body="&lt;p&gt;Here's another way to think about the &quot;line at infinity&quot; and the &quot;points at infinity&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the usual $XY$-plane as sitting inside of $3$-space, but instead of it sitting in its usual place, $\{(x,y,0) : x,y\in\mathbb{R}\}$, shift it up by $1$ so that it sits as the $z=1$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you are sitting at the origin with a very powerful laser pointer. Whenever you want a point on the $XY$-plane, you shine your laser pointer at that point. So, if you want the point $(x,y)$, you are actually pointing your laser pointer at the point $(x,y,1)$; since you are sitting at the origin, the laser beam describes a (half)-line, joining $(0,0,0)$ to $(x,y,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for example, look at the point $(x,0,1)$, and imagine $x$ getting larger. The angle your laser pointer makes with the $z=0$ plane gets smaller and smaller, until &quot;as $x$ goes to infinity&quot;, your laser pointer is just pointing along the line $x$ axis (at the point $(1,0,0)$), and the same thing happens if you let $x$ go to $-\infty$. More generally, if you start pointing to points that are further and further away from the &quot;origin&quot; in your plane (away from $(0,0,1)$), the laser beam's angle with $Z=0$ gets smaller and smaller, until, &quot;at the limit&quot; as $||(x,y)||\to\infty$, you end up with the laser beam pointing along the $z=0$ plane in some direction. We can represent the direction with the slope of the line, so that we are pointing at $(1,m,0)$ for some $m$ (or perhaps to $(-1,-m,0)$, but that's the same direction), or perhaps to the point $(0,1,0)$. So we &quot;add&quot; these &quot;points at infinity&quot; (so called because we get them by letting the point we are shining the laser beam on &quot;go to infinity&quot;), one for each direction away from the &quot;origin&quot;: $(1,m,0)$ for arbitrary $m$ for nonvertical lines, and $(0,1,0)$ corresponding to the direction of $x=0$, $y\to\pm\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: the &quot;usual&quot;, affine points, are the ones in the $z=1$ plane, and they correspond to laser beams coming from the origin; they are each of the form $(x,y,1)$ for some $x,y$ in $\mathbb{R}$. In addition, for each &quot;direction&quot; we want to include that limiting laser beam which does &lt;em&gt;not&lt;/em&gt; intersect the plane $z=1$; those correspond to points $(1,m,0)$, or the point $(0,1,0)$ when you do it with the line $x=0$. So we get one point for every real $m$, $(1,m,0)$, and another for $(0,1,0)$. You are adding one point for every direction of lines through the origin; these points are the &quot;points at infinity&quot;, and together they make the &quot;line at infinity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, put your elliptic curve/polynomial $F=Y^2 - X^3 - aX-b$, and draw the points that correspond to it on the $z=1$ plane; that's the &quot;affine piece&quot; of the curve. But do you also get any of those &quot;points at infinity&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, even though we are thinking of the points as being on the $XY$-plane, they &quot;really&quot; are in the $Z=1$ plane; so our equation actually has a &quot;hidden&quot; $Z$ that we lost sight of when we evaluated at $Z=1$. We use the homogenization $f = Y^2Z - X^3 - aXZ^2 - bZ^3$ to find it. Why that? Well, for any fixed point $(x,y,1)$ in our &quot;$XY$-plane&quot;, the laser pointer points to all points of the form $(\alpha x,\alpha y,\alpha)$. If we were to shift up our copy of the plane from $Z=1$ to $Z=\alpha$, we'll want to scale everything so that it still corresponds to what I'm tracing from the origin; this requires that every monomial have the same total degree, which is why we put in factors of $Z$ to complete them to degree $3$, the smallest we can (making it bigger would give you the point $(0,0,0)$ as a solution, and we &lt;em&gt;do&lt;/em&gt; need to stay away from that because we cannot point the laser pointer in our eye). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we do that, we find the &quot;directions&quot; that also correspond to our curve by setting $Z=0$ and solving, to find those points $(1,m,0)$ and $(0,1,0)$ that may also lie in our curve. But the only one that works is $(0,1,0)$, which is why the elliptic curve $F$ only has one &quot;point at infinity&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T07:33:00.487" LastActivityDate="2010-12-11T07:33:00.487" CommentCount="2" />
  <row Id="13769" PostTypeId="2" ParentId="13735" CreationDate="2010-12-10T06:41:28.787" Score="0" ViewCount="0" Body="&lt;p&gt;Christopher Gibson's &lt;em&gt;Elementary Euclidean Geometry&lt;/em&gt; is a new (published in 2004) and fairly advanced introduction to the geometry of lines and conics in Euclidean space.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2010-12-10T06:41:28.787" />
  <row Id="13770" PostTypeId="2" ParentId="13761" CreationDate="2010-12-10T06:47:48.550" Score="2" ViewCount="0" Body="&lt;p&gt;The 24game solver found &lt;a href=&quot;http://code.google.com/p/solve24game/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; couldn't find any solutions with 4 8s. I don't think the software uses decimals to create 8.8 etc but its hard to see how that path leads to a solution. The other solution is as friedo suggests (88-8*8).&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-12-10T08:09:02.427" LastActivityDate="2010-12-10T08:09:02.427" />
  <row Id="13771" PostTypeId="1" AcceptedAnswerId="13775" CreationDate="2010-12-10T06:51:20.343" Score="2" ViewCount="47" Body="&lt;p&gt;According to wikipedia the &lt;a href=&quot;http://en.wikipedia.org/wiki/Right_quotient&quot; rel=&quot;nofollow&quot;&gt;right quotient of a regular language&lt;/a&gt; with ANY other language is regular. I have not been able to find a proof of this fact. All the sources talk about quotient with another regular language. Can somebody point me to a proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Definition from Wikipedia: The right quotient (or simply quotient) of a formal language $L_1$ with a formal language $L_2$ is the language consisting of strings $w$ such that $wx$ is in $L_1$ for some string $x$ in $L_2$. In symbols, we write:&#xA;$$L_1 / L_2 = \{w \ | \ \exists x ((x \in L_2) \land (wx \in L_1)) \}$$&#xD;&#xA;In other words, each string in L1 / L2 is the prefix of a string wx in L1, with the remainder of the word being a string in L2.&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2010-12-10T07:39:40.990" Title="Quotient of a regular language." Tags="&lt;computer-science&gt;&lt;computability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13772" PostTypeId="1" CreationDate="2010-12-10T07:01:29.713" Score="0" ViewCount="75" Body="&lt;p&gt;I have two faces ($S_1$ and $S_2$), each bounded by a series of Vertex $V$. These two faces may or may not intersect. In addition to that there are two vertical faces $S_3$ and $S_4$, that connect between $S_1$ and $S_2$ vertically, as shown in the below diagram. The question is how to find the polyhedrons that are formed by the bounding of the four faces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is shown below:&#xA;&lt;img src=&quot;http://i.imgur.com/HXLRD.gif&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above example, there should be two polyhedrons returned, one above $S_2$, another below $S_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $S_1$ and $S_2$ are defined by 3D coordinates. Let's say the coordinates for $S_1$ is ($P_1$,$P_2$, $P_3$, $P_4$) and $S_2$ is $P_5$,$P_6$, $P_7$, $P_8$, and the intersected points are $P_9$ and $P_{10}$. I want a program that returns me two polyhedrons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1,P_2,P_9,P_{10},P_5,P_6$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_3,P_4,P_{10},P_9,P_7,P_8$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do this is to find the intersecting line of the two faces, and then construct polyhedron from there. But is there a more elegant method?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-11T15:27:27.613" LastActivityDate="2011-01-10T15:44:42.130" Title="Find the Polyhedron Enclosed by Multiple Faces" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13773" PostTypeId="1" CreationDate="2010-12-10T07:23:49.810" Score="1" ViewCount="70" Body="&lt;p&gt;I am wondering how to find the Zariski Cover of the Grassmannian over $\mathbb{C}$. I was hoping some reference would go through this calculation. I was recently told that it was not too hard, so I wondered if it was written somewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any reference that does this, or at least gives an introduction for how to calculate Zariski Covers of specific schemes, would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2010-12-10T12:44:03.900" Title="How to find a Zariski Cover of the Grassmannian" Tags="&lt;algebraic-geometry&gt;&lt;reference-request&gt;&lt;grassmannian&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13774" PostTypeId="1" AcceptedAnswerId="13776" CreationDate="2010-12-10T07:38:55.033" Score="3" ViewCount="89" Body="&lt;p&gt;such a RSA algorithm.&lt;/p&gt;&#xA;" OwnerUserId="4599" LastActivityDate="2010-12-10T07:50:59.963" Title="What is the importance of prime numbers in cryptographic algorithms ?" Tags="&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13775" PostTypeId="2" ParentId="13771" CreationDate="2010-12-10T07:39:40.990" Score="2" ViewCount="0" Body="&lt;p&gt;You can find a proof in this book: Introduction to Automata Theory, Languages and Computation by Hopcroft and Ullman, 1979 edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short and non-constructive proof appears on page 63.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot (which I got from google books by searching for quotient in that book):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/gswKZ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the last line of the proof is incomplete in the above snapshot, which reads: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus $M&amp;#39;$ accepts $R/L$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here $R$ is a regular language and $L$ is an arbitrary language.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-10T07:39:40.990" />
  <row Id="13776" PostTypeId="2" ParentId="13774" CreationDate="2010-12-10T07:50:59.963" Score="2" ViewCount="0" Body="&lt;p&gt;The short answer is that what makes primes useful is that it is easy to multiply two primes, but difficult to algorithmically factorise a given number into prime factors (i.e. takes a long time, if the number is big). So multiplying primes is an operation that is easy to perform but difficult to reverse. That makes them a perfect candidate for implementing &lt;a href=&quot;http://en.wikipedia.org/wiki/Public-key_cryptography&quot; rel=&quot;nofollow&quot;&gt;public-private key encryption&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long answer would involve explaining RSA, which has been done in lots of places and you will have no difficulty finding them. You need to know Fermat's little theorem to understand how RSA works.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T07:50:59.963" CommentCount="4" />
  <row Id="13777" PostTypeId="1" CreationDate="2010-12-10T08:05:42.583" Score="4" ViewCount="286" Body="&lt;p&gt;I am trying to understand why $1^{x}=1$ for any $x\in\mathbb{R}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it OK to write $1^{x}$? As the base 1 should not equal 1 for $1^{x}$ to be an exponential function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $1^{x}=1$ just because it is defined to be so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible please refer me to a book or article that discusses this topic.&lt;/p&gt;&#xA;" OwnerUserId="4601" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T02:39:48.217" LastActivityDate="2010-12-11T02:39:48.217" Title="What is the value of $1^x$ ?" Tags="&lt;calculus&gt;&lt;functions&gt;&lt;notation&gt;" AnswerCount="3" />
  <row Id="13778" PostTypeId="2" ParentId="13777" CreationDate="2010-12-10T08:23:29.310" Score="5" ViewCount="0" Body="&lt;p&gt;I think you will at least agree that $1^x$=$1$ if $x$ is any natural number (since this is just 1 times itself x times).  We can extend this to all integer values of $x$ by using the facts $a^0$=$1$ for all non-zero $a$ and $x^{-c}$=$1\over{x^c}$.  Then we have $1^x$=1 for all rational numbers $x$ by the fact $1^{b\over{c}}$ is the cth root of $1^b$ which is 1 by previous results.  Finally we exdend this to all real values of $x$ by using the following definition of the exponential function from calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If $t$ is irational, then $a^t$ is defined to be the limit as n approaches infinity of $a^{t_n}$ where {$t_n$} is any sequence of rational numbers converging to $t$ (assuming this limit exists).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $1^t$ we already have that $1^{t_n}=1$ for all rational $t_n$ so any such sequence as described above is the constant sequence, 1,1,1,..... which converges to 1.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2010-12-10T08:39:20.890" LastActivityDate="2010-12-10T08:39:20.890" CommentCount="1" />
  <row Id="13779" PostTypeId="2" ParentId="13777" CreationDate="2010-12-10T08:34:19.207" Score="12" ViewCount="0" Body="&lt;p&gt;You don't have to specifically define $1^x$ to be something. It follows from the definition of exponentiation in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story is roughly as follows: first there was exponentiation with exponents being natural numbers. That is just repeated multiplication of the base with itself. Then people noticed that this exponentiation obeys the rule&#xA;\begin{eqnarray}&#xA;a^{n+m} = a^na^m.&#xA;\end{eqnarray} This suggests that if we want to define $a^0$, then we want it to satisfy $a^m = a^{m+0} = a^ma^0$, so that it is natural to define $a^0 = 1$. Now, we can use the same rule to extend this operation to all integral exponents, i.e. positive and negative. Since we want $a^na^{-n} = a^{n-n} = a^0 = 1$, we are forced to define $a^{-n} = \frac{1}{a^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this way you have defined exponentiation with arbitrary integral exponents. Now, going back to the original exponentiation with natural numbers, you also notice that&#xA;\[(a^n)^m = a^{nm}&#xA;\] and $a^1 = a$. Applying the same considerations as above, you are forced to define $a^{\frac{1}{n}} = \sqrt[n]{a}$. This way, you have now extended exponentiation to all rational exponents. Note that all these rules &lt;em&gt;imply&lt;/em&gt; that $1^x = 1$ for $x\in \mathbb{Q}$. Now, you extend all this to all real exponents using continuity, so you still have $1^x = 1\;\forall x\in \mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-10T13:48:40.400" LastActivityDate="2010-12-10T13:48:40.400" CommentCount="3" />
  <row Id="13781" PostTypeId="1" CreationDate="2010-12-10T09:53:05.893" Score="3" ViewCount="109" Body="&lt;p&gt;If you sum an expression over an uncountable set&#xA;$\sum_{x\in \mathbb{R}}f(x)$, then do we need $f(x)=0$ on all but a countable subset in order for the sum to have a finite value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not can you give an example of a function everywhere nonzero that has a transfinite sum with a finite value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible keywords: Transseries, Écalle–Borel Summation, analyzable function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/0801.4877&quot; rel=&quot;nofollow&quot;&gt;Transseries for beginners&lt;/a&gt;, GA Edgar, 2009&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T17:40:16.763" LastActivityDate="2010-12-10T17:40:16.763" Title="Transfinite series: Uncountable sums" Tags="&lt;analysis&gt;&lt;real&gt;" AnswerCount="1" />
  <row Id="13783" PostTypeId="2" ParentId="13781" CreationDate="2010-12-10T10:03:05.770" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\varepsilon&amp;gt;0$, can you measure the set $\{x:f(x)&amp;gt;\varepsilon\}$?&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-10T10:21:07.097" LastActivityDate="2010-12-10T10:21:07.097" CommentCount="2" />
  <row Id="13784" PostTypeId="2" ParentId="154" CreationDate="2010-12-10T10:07:37.933" Score="1" ViewCount="0" Body="&lt;p&gt;Re: &quot;What's the best way to explain to a non-mathematician that complex numbers are necessary and meaningful, in the same way that real numbers are?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a non-mathematician...so fully qualified to answer this question...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first explain imaginary numbers&lt;/li&gt;&#xA;&lt;li&gt;then explain that complex numbers are simply numbers that use imaginary numbers&lt;/li&gt;&#xA;&lt;li&gt;then tell 'em about how they are applied in weather forecasting&lt;/li&gt;&#xA;&lt;li&gt;bang!....I geddit&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I got all this from an amazing radio show :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;&lt;a href=&quot;http://www.bbc.co.uk/iplayer/episode/b00tt6b2/In_Our_Time_Imaginary_Numbers/&quot; rel=&quot;nofollow&quot;&gt;In Our Time - Imaginary Numbers&lt;/a&gt;&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Melvyn Bragg and his guests discuss imaginary numbers - important mathematical phenomena which provide us with useful tools for understanding the world.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4603" LastActivityDate="2010-12-10T10:07:37.933" />
  <row Id="13785" PostTypeId="1" AcceptedAnswerId="13786" CreationDate="2010-12-10T10:08:25.777" Score="2" ViewCount="152" Body="&lt;p&gt;If I have term -5 and I want to change its sign to opposite can I say that I'm negating its sign? &lt;/p&gt;&#xA;" OwnerUserId="4366" LastEditorUserId="4366" LastEditorDisplayName="" LastEditDate="2010-12-10T15:10:52.987" LastActivityDate="2010-12-10T15:10:52.987" Title="Does saying &quot;to negate a sign of a const term&quot; has a meaning" Tags="&lt;terminology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="13786" PostTypeId="2" ParentId="13785" CreationDate="2010-12-10T10:30:59.497" Score="1" ViewCount="0" Body="&lt;p&gt;You could say that, you could also say &quot;to negate the constant term&quot; -- both seem grammatically correct. In your expression, 'negate' changes the object of negation, while in the &quot;to negate the constant term&quot; 'negate' changes the sign in front of the object of negation, but it makes sense both ways.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-10T10:30:59.497" />
  <row Id="13787" PostTypeId="2" ParentId="13785" CreationDate="2010-12-10T10:52:39.237" Score="3" ViewCount="0" Body="&lt;p&gt;It sounds alright. A more fanciful way of putting it is 'to change its polarity'. Note that 'polarity' is not to be confused with 'parity', whether a number is odd or even.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-10T10:52:39.237" CommentCount="2" />
  <row Id="13788" PostTypeId="2" ParentId="13785" CreationDate="2010-12-10T10:52:42.153" Score="2" ViewCount="0" Body="&lt;p&gt;I think it is much more standard to say 'Multiplying equation (..) by -1 gives' than 'negating equation (...)'.  As for 'negating the constant term', I'm not even sure of what that would mean, and without some serious context seems a bad idea.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastActivityDate="2010-12-10T10:52:42.153" />
  <row Id="13789" PostTypeId="2" ParentId="13773" CreationDate="2010-12-10T11:28:13.837" Score="2" ViewCount="0" Body="&lt;p&gt;Consider Grassmannian of $d$-dimensional subspaces of $n$-dimensional vector space&#xA;$V$. Then you have Pluker inbeding of grassmanian $G=Gr(d,V)$ of d-planes of $V$ in $\mathbb P(\Lambda ^d V)=\mathbb  P^N, N=\binom{n}{d}-1$,  as follows. For every $d-plane$ D chose base $v_1,...,v_d$ of $D$ and consider exterior product $\omega = v_1\wedge...\wedge v_d \in \wedge ^d V$. Pluker inbeding send $D$ to line:  Span $\omega \in \mathbb P(\Lambda ^d V)$  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For find  Zariski covering, fix base $e_1,...,e_n$ of $V$ and receive base $e_I=e_{i_1}\wedge...\wedge e_{i_d}$ of $\wedge ^d V $ ($I$ is increasing sequence $I=(1\leq i_1 \lt ...\lt i_d \leq n$)). Zariski covering consist of $U_I$  formed by these $d-$ planes $D$ whose Pluker image in $\mathbb P(\wedge ^d V)$ has coordinate $z_I\neq 0$. This mean that when you chose base $v_1,...,v_d$ of $D$ and developp $\omega = v_1\wedge...\wedge v_d \in \wedge ^d V$ in base $e_I$ of $\Lambda ^d V$, then coefficient of $e_I$ is non-null. This Zariski cover $U_I$ of grasmanian $Gr(d,V)$ possesses $N+1=\binom{n}{d}$ open sets (Attention: $\mathbb P^N$&#xD;&#xA;has $N+1$ coordinates, not $N$)&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="2872" LastEditorDisplayName="" LastEditDate="2010-12-10T12:44:03.900" LastActivityDate="2010-12-10T12:44:03.900" CommentCount="1" />
  <row Id="13790" PostTypeId="1" CreationDate="2010-12-10T11:49:02.907" Score="2" ViewCount="127" Body="&lt;p&gt;The problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;               Given a Fibonacci number,find its index.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am aware of the standard solution '&lt;strong&gt;generate-hash-find&lt;/strong&gt;'. I am just curious if there is some inverse system of matrix exponentiation or some other mathematical method that gives the solution.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T13:25:32.490" LastActivityDate="2010-12-20T17:57:11.563" Title="Finding index of a Fibonacci number: any mathematical solution possible?" Tags="&lt;algorithms&gt;&lt;fibonacci-number&gt;" AnswerCount="2" CommentCount="16" />
  <row Id="13791" PostTypeId="1" AcceptedAnswerId="13797" CreationDate="2010-12-10T12:14:51.897" Score="3" ViewCount="29" Body="&lt;p&gt;If a Diophantine equation is: exists v, p(x,v) = 0 (where v is a vector of finitely many integers) for some polynomial p, is there a Turing machine which prints out all values of x?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-10T12:32:47.050" Title="Diophantine equation and Turing Machine" Tags="&lt;diophantine-equations&gt;&lt;computability&gt;" AnswerCount="1" />
  <row Id="13792" PostTypeId="2" ParentId="13790" CreationDate="2010-12-10T12:18:00.210" Score="6" ViewCount="0" Body="&lt;p&gt;From wikipedia:&#xA;&quot;Similarly, if we already know that the number F is a Fibonacci number, we can determine its index within the sequence by&#xA;$n = \bigg\lfloor \log_\varphi \left(F\cdot\sqrt{5}\right) + \frac{1}{2} \bigg\rfloor$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T12:18:00.210" />
  <row Id="13793" PostTypeId="1" AcceptedAnswerId="13796" CreationDate="2010-12-10T12:18:17.553" Score="11" ViewCount="223" Body="&lt;p&gt;A while ago I was dreaming up point-set topology exam questions, and this one came to mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $\mathbb Q\setminus \{0\}$ homeomorphic to $\mathbb Q$? (Where both sets have the subspace topology induced from the standard topology on $\mathbb R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I couldn't figure this out at the time, and I'm curious to see whether anyone has a nice argument. I'm not even willing to take a guess as to whether they are or aren't homeomorphic.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-12-12T21:54:47.093" Title="Are the rationals minus a point homeomorphic to the rationals?" Tags="&lt;topology&gt;" AnswerCount="3" />
  <row Id="13794" PostTypeId="1" AcceptedAnswerId="13828" CreationDate="2010-12-10T12:18:56.747" Score="2" ViewCount="87" Body="&lt;p&gt;There is a direct relationship between equivalence relations and partitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to simply use an equivalence relation's definition to get the matching partition? And what about the other way around?&lt;/p&gt;&#xA;" OwnerUserId="3016" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-10T20:02:58.710" LastActivityDate="2010-12-12T19:48:53.387" Title="&quot;Converting&quot; equivalence relations to partitions" Tags="&lt;elementary-set-theory&gt;&lt;relations&gt;&lt;partitions&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13795" PostTypeId="2" ParentId="13777" CreationDate="2010-12-10T12:20:31.673" Score="5" ViewCount="0" Body="&lt;p&gt;Once you move past elementary topics, definitions become much more fundamental in mathematics. So, in a formal sense, you're right that the reason $1^x = 1$ for all $x \in \mathbb{R}$ is that the definition of $1^x$ makes this so.  The emphasis on definitions comes from the use of mathematical proofs; the only way to make a rigorous proof about exponentiation is to start with a rigorous definition.  So, in a sense, all formal mathematical propositions are true because the definitions have been chosen to make them true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we have a clear motivation behind exponentiation, and if the definition of real number exponentiation did not make $1^x = 1$ for all real $x$, then the definition would have been changed.  We don't make up mathematical definitions at random - they are motivated by our informal ideas about the objects we are studying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The role of this motivation can be seen more clearly by considering complex exponentiation.  Unlike natural number exponentiation, complex number exponentiation is not based on repeated multiplication; it's based on logarithms and the function $\exp(z)$. So the definition of complex exponentiation does not imply that $1^i = 1$, and mathematicians are OK with that. There are several possible values for $1^i$, only one of which is $1$. I explained this in &lt;a href=&quot;http://math.stackexchange.com/questions/3668/what-is-the-value-of-1i/3674#3674&quot;&gt;this answer&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that is often confusing at first is that there are really several different exponentiation functions, with different domains, all of which are denoted with the notation $x^y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you asked whether $f(x) = 1^x$, as a function from $\mathbb{R}$ to itself, is an exponential function. Many calculus books seem to include a special clause in their definitions that makes this not be an exponential function. However, things would work just as well if you did call it an exponential function. It's just a question of terminology. The only downside to calling $1^x$ an exponential function is that, when stating some results, you might have to add an exception to get rid of $1^x$. Instead of saying &quot;Every exponential function&quot; you would say &quot;Every exponential function except $1^x$&quot;.  Of course, students in a class need to adopt the conventions of the class so that everyone can understand them. But if you were writing a math book alone on a desert island you could adopt whatever terminology you wanted. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-10T12:20:31.673" />
  <row Id="13796" PostTypeId="2" ParentId="13793" CreationDate="2010-12-10T12:24:26.370" Score="16" ViewCount="0" Body="&lt;p&gt;A well-known theorem of Cantor says that any two countable dense linear orderings without endpoints are isomorphic as linear orders. So in particular there is an order-preserving bijection between $\mathbb{Q}$ and $\mathbb{Q} - \{0\}$. This bijection will be a homeomorphism if you give each space the order topology, which is the standard topology inherited from $\mathbb{R}$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-10T12:24:26.370" CommentCount="3" />
  <row Id="13797" PostTypeId="2" ParentId="13791" CreationDate="2010-12-10T12:32:47.050" Score="4" ViewCount="0" Body="&lt;p&gt;Given a polynomial $p(\bar{x},\bar{v})$ with integer coefficients, there is indeed a Turing machine that enumerates all tuples of integers $\bar{x}$ for which there exists a tuple of integers $\bar{v}$ with $p(\bar{x},\bar{v}) = 0$. In other words, the set of such tuples $\bar{x}$ is recursively enumerable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do the enumeration is to start by making a machine that enumerates &lt;em&gt;all&lt;/em&gt; pairs of integer tuples $(\bar{x}, \bar{v})$ (where each tuple has the appropriate length). Then feed the output of this machine into a machine that checks whether $p(\bar{x},\bar{v}) = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there is no Turing machine that takes a polynomial $p$ as input, and a tuple $\bar{x}$ as a second input, and decides whether there is a tuple $\bar{v}$ of integers with $p(\bar{x},\bar{v}) = 0$. This is the MRDP theorem, which solved &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's 10th problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-10T12:32:47.050" CommentCount="1" />
  <row Id="13798" PostTypeId="2" ParentId="13794" CreationDate="2010-12-10T12:59:14.883" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose you have an equivalence relation $\sim$ on $X$. For each $x\in X$, the set&#xA;$S_x=\lbrace y\in X: y\sim x \rbrace$ is one of the sets in the partition, and every set in the &#xA;partition arises this way. But you should know that $S_x$ and $S_y$ are identical (the same)&#xA;if and only if $x\sim y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you have a partition $\lbrace C_i: i\in I \rbrace$ of $X$. Then&#xA;the relation $\sim$ defined by $$ x\sim y \mbox{ iff } x,y \in C_i \mbox { for some } i $$&#xD;&#xA;is an equivalence relation on $X$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-10T12:59:14.883" />
  <row Id="13799" PostTypeId="2" ParentId="13794" CreationDate="2010-12-10T13:02:26.770" Score="1" ViewCount="0" Body="&lt;p&gt;In addition to my comment above, see also &lt;a href=&quot;http://www.proofwiki.org/wiki/Fundamental_Theorem_on_Equivalence_Relations&quot; rel=&quot;nofollow&quot;&gt;this post at the ProofWiki&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-10T13:02:26.770" />
  <row Id="13800" PostTypeId="1" AcceptedAnswerId="13837" CreationDate="2010-12-10T13:13:28.150" Score="6" ViewCount="50" Body="&lt;p&gt;I have been looking for a proof of the statement:&#xA;&quot;Assume the Generalized Riemann Hypothesis. Let $d$ be a fundamental discriminant and $\chi_d$ the associated primitive quadratic character. Then, $$L(\tfrac{1}{2},\chi_d)\geq 0.&amp;quot;$$&#xD;&#xA;Can anyone point me in the right direction or give a reference? Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="3244" LastActivityDate="2010-12-10T18:55:45.937" Title="Nonnegativity of the quadratic Dirichlet L-function $L(\tfrac{1}{2},\chi)$ under GRH" Tags="&lt;number-theory&gt;&lt;analytic-number-theory&gt;&lt;riemann-hypothesis&gt;" AnswerCount="1" />
  <row Id="13801" PostTypeId="1" AcceptedAnswerId="13808" CreationDate="2010-12-10T13:25:33.600" Score="2" ViewCount="572" Body="&lt;p&gt;What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2010-12-11T06:42:09.430" Title="What's the thing with $\sqrt{-1} = i$" Tags="&lt;complex-numbers&gt;" AnswerCount="6" CommentCount="14" />
  <row Id="13802" PostTypeId="1" AcceptedAnswerId="13813" CreationDate="2010-12-10T13:32:58.397" Score="5" ViewCount="81" Body="&lt;p&gt;When asked to evaluate $\int_{a}^{\infty}f(x)dx$, you split the interval based on the improper points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is another improper point other than $\infty$, at $b$, we will write: $\int_{a}^{\infty}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx+\int_{c}^{\infty}f(x)dx$ and ask whether each of the integrals on the right hand side converge. If they all converge, so does the original one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if at least one of them diverges, the original one doesn't. What is the justification for this conclusion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that if $\int_{c}^{b}f(x)dx$ diverges, we can assume that the original integral converges and move the integrals around like this: $\int_{a}^{\infty}f(x)dx-\int_{a}^{b}f(x)dx-\int_{c}^{\infty}f(x)dx=\int_{b}^{c}f(x)dx$ then we'll get a contradiction. But what if both $\int_{c}^{b}f(x)dx$ and $\int_{a}^{b}f(x)dx$ diverge? What is the argument then?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-10T14:00:15.800" LastActivityDate="2010-12-10T14:43:01.957" Title="Diverging improper integral" Tags="&lt;calculus&gt;&lt;improper-integrals&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13803" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T13:41:08.610" Score="5" ViewCount="0" Body="&lt;p&gt;Alex it is just a notion. We use the value of $i = \sqrt{-1}$ so that we can work more freely. For example consider the Quadratic Equation $$x^{2}+x+1=0$$ The discriminant for this is $D=b^{2}-4ac=-3$. So the roots have the value $$x = \frac{-1 \pm{\sqrt{-3}i}}{2}$$ which looks better when written with an $i$ notation. That's all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how this is taught in the US but to me, i encountered this when i was at high school, learning how to solve for Quadratic Equations when the &lt;em&gt;Discriminant&lt;/em&gt; is less than $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, note that $\mathbb{C}$ doesn't have the same ordering as $\mathbb{R}$. That is for any 2 real numbers $a,b$ we have either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$a&amp;gt;b$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$a&amp;lt; b$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$a=b$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But for complex number's this is not true. Since if you take $i$ and $0$, we must have either $i &amp;gt; 0$ or $i &amp;lt; 0$, but this isn't true.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-12-10T13:58:56.323" LastActivityDate="2010-12-10T13:58:56.323" CommentCount="5" />
  <row Id="13804" PostTypeId="1" CreationDate="2010-12-10T13:54:17.917" Score="3" ViewCount="64" Body="&lt;p&gt;I was thinking about some problem involving quantifiers (the existencial and universal quantifiers) and I noticed how it might resemble probability in a sense. They both assume a variable and its domain, and put restrictions on some predicate of that variable over that domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example if you didn't get me yet, we can put the following statement involving quantified formula in a probability format:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;( \exists x \in X : g(x) ) \equiv ( Pr_{x \in X} [g(x)=1] &amp;gt; 0 ),&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(x) \in \{0,1\}$ is a predicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly defined as well for the universal quantifier as $Pr=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it gets trickier if we nest the quantifiers, for instance I am not sure if the following is correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: &#xA;This equation is wrong: $ \exists x \in X : ( \forall y \in Y : g(x,y) ) \equiv Pr[g(X,Y)=1 | Y=y] &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I propose this one instead: &#xA;$ \exists x \in X : ( \forall y \in Y : g(x,y) ) \equiv Pr_{x\in X} \left[ Pr_{y\in Y}[g(x,y)=1 | X=x] = 1 \right] &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-end update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that line of intuition was correct as I hope, I was wondering if such relationship has been explored before and if not I was hoping to know if it would be possible or not to put a general framework for converting a quantified formula of any alternation depth into a probability equivalent.&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2010-12-10T15:44:10.907" LastActivityDate="2010-12-11T01:05:04.170" Title="Reducing quantification to probability" Tags="&lt;probability&gt;&lt;quantifiers&gt;&lt;quantifier-elimination&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13805" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T14:11:12.733" Score="3" ViewCount="0" Body="&lt;p&gt;This is answered quite succinctly in the MathWorld article on &lt;a href=&quot;http://mathworld.wolfram.com/PrincipalSquareRoot.html&quot; rel=&quot;nofollow&quot;&gt;Principal Square Root&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4604" LastActivityDate="2010-12-10T14:11:12.733" CommentCount="2" />
  <row Id="13806" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T14:12:10.813" Score="1" ViewCount="0" Body="&lt;p&gt;The proper way to define i is to define the laws for complex addition and multiplication the usual way using ordered pairs of real numbers, and then define i to be (0,1). This distinguishes it from -i which is (0,-1). One can then set the root of -1 to be i, but of course this last step is pretty arbitrary. You can see it either as 'wrong' or as 'not wrong' if you want to work with the last step as an arbitrary definition. The important thing is that we have an unambiguous definition of i using the initial steps specified above.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-10T14:12:10.813" CommentCount="2" />
  <row Id="13807" PostTypeId="1" AcceptedAnswerId="13812" CreationDate="2010-12-10T14:13:02.623" Score="0" ViewCount="17" Body="&lt;p&gt;Let's say I have two numbers that I know to have been normalized- 39, 12.  How do I calculate the percentage difference between them?  Is it the same as if they hadn't been normalized?  ie: (12/39) * 100&lt;/p&gt;&#xA;" OwnerUserId="4608" LastActivityDate="2010-12-10T14:33:24.717" Title="Percentage difference in normalized number set" Tags="&lt;normal-distribution&gt;" AnswerCount="1" />
  <row Id="13808" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T14:15:46.613" Score="11" ViewCount="0" Body="&lt;p&gt;I believe this is a common misconception. Here in Sweden we (or at least I) was taught that $i^2 = -1$, &lt;em&gt;not&lt;/em&gt; that $\sqrt{-1} = i$. They are two fundamentally different statements. Of course most of the time one chooses $\sqrt{-1} = i$ as the principal branch of the square root.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2010-12-10T14:15:46.613" CommentCount="12" />
  <row Id="13809" PostTypeId="2" ParentId="13772" CreationDate="2010-12-10T14:28:33.227" Score="0" ViewCount="0" Body="&lt;p&gt;There is a large literature about how to take a drawing in the plane which tries to give a 3-dimensional view of a polyhedron and extract a polyhedron from this drawing, in contrast to applying Steinitz's Theorem which gives necessary and sufficient conditions that a graph be the vertex-edge graph of a convex 3-dimensional polyhedron:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Steinitz%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Steinitz%27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sample of the other kind of work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&quot; rel=&quot;nofollow&quot;&gt;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&quot; rel=&quot;nofollow&quot;&gt;http://linkinghub.elsevier.com/retrieve/pii/S009784930900020X&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://portal.acm.org/citation.cfm?id=1052355&quot; rel=&quot;nofollow&quot;&gt;http://portal.acm.org/citation.cfm?id=1052355&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-10T14:28:33.227" CommentCount="1" />
  <row Id="13810" PostTypeId="2" ParentId="13662" CreationDate="2010-12-10T14:31:04.390" Score="7" ViewCount="0" Body="&lt;p&gt;The probability that A dies first is 1/4 (since it's the same probability for everyone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that they all die after the age of 90 is $(2/3)^4 = 16/81$. The two events are independent, so the probability that A dies first &lt;em&gt;and&lt;/em&gt; they all die after the age of 90, or equivalently that A dies first and that that happens after he is 90, is $\frac{16}{81}\cdot\frac{1}{4}=\frac{4}{81}$. But if A dies first, then either he dies first and is above 90 or he dies first and is younger than 90, and the two are mutually exclusive. Therefore (or equivalently using the rule that $P(X) = P(X\cap Y) + P(X\cap Y^C)$), we deduce that the probability that A dies first and &lt;em&gt;before&lt;/em&gt; the age of 90 is 1/4 - 4/81 = 65/324.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T14:31:04.390" CommentCount="4" />
  <row Id="13811" PostTypeId="1" CreationDate="2010-12-10T14:32:14.627" Score="2" ViewCount="45" Body="&lt;p&gt;I just read through a proof of the Compact Uniformization Theorem, and I follow it up to the very last line. The proof is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compact Uniformization Threorem. If $X$ is a compact regular space, then the neighborhood filter $\mathcal{N}_\Delta$ of the diagonal $\Delta\subset X\times X$ is a uniform structure on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $\mathcal{N}_\Delta$ does not have the property that for any entourage $E$, there exists an entourage $D$ such that $D\circ D\subseteq E$. Then there exists an open neighborhood $E\in\mathcal{N}_\Delta$ such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(D\circ D)\setminus E\neq\emptyset\qquad (D\in\mathcal{N}_\Delta)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the family&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{C}=\{(D\circ D)\setminus E\ | \ D\in\mathcal{N}_\Delta\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;consists of nonempty subsets and is a filter-base on $X\times X\setminus E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter begin a closed subset of the product of two compact spaces is compact by Tychonoff's Theorem. It follows that $\mathcal{C}$ has a cluster point $(p,q)\in X\times X\setminus E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the neighborhood $E(q)$ of $q$ does not contain $p$, and thus the neighborhood $E^{-1}(p)$ of $p$ does not contain $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by a previous Lemma, there exists an open neighborhood $D$ of $\Delta$ such that $D\circ D$ is disjoint with some neighborhood of $(p,q)$. It follows that $(p,q)$ cannot be a cluster point of $\mathcal{C}$, a contradiction. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just don't see why it immediately follows that $(p,q)$ is not a cluster point of $\mathcal{C}$, simply because there is some open neighborhood of the diagonal that is disjoint from one of the neighborhoods of $(p,q)$. Can someone please explain?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The previous lemma is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p$ and $q$ be a pair of points in a regular topological space $X$ such that $\mathcal{N}_p\neq\mathcal{N}_q$. Then there exists an open cover $\mathcal{U}$ of $X$ and a neighborhood $W$ of $(p,q)$ such that &#xA;$$W\cap(\Delta_\mathcal{U}\circ\Delta_\mathcal{U})=\emptyset.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not clear, $$\Delta_\mathcal{U}=\bigcup_{U\in\mathcal{U}}U\times U.$$&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-10T14:57:57.200" LastActivityDate="2010-12-10T14:57:57.200" Title="On the Compact Uniformization Theorem" Tags="&lt;topology&gt;&lt;general-topology&gt;" CommentCount="6" />
  <row Id="13812" PostTypeId="2" ParentId="13807" CreationDate="2010-12-10T14:33:24.717" Score="0" ViewCount="0" Body="&lt;p&gt;Shouldn't the percentage difference be [(12-39)/39]*100 or [(39-12)/12]*100?  Note that you get different answers, not just the negative-when the two numbers are unequal the percentage difference depends upon which you regard as the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether normalization changes the percentage depends upon what you mean by normalization.  Sometimes it is multiplication by a constant to bring the range to something you like.  This will not impact the percentage difference-try multiplying 39 and 12 by a and note that it divides out of the expression.  Sometimes it involves adding or subtracting a constant-that will make a difference.  Compare [(39-12)/39]*100 to [(30-3)/30]*100.  The answer is different.  Sometimes it is a more complicated expression, which will almost always change the percentages.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-10T14:33:24.717" />
  <row Id="13813" PostTypeId="2" ParentId="13802" CreationDate="2010-12-10T14:36:04.100" Score="5" ViewCount="0" Body="&lt;p&gt;This is a good question. One way to think about the issue is that the &quot;convergence&quot; of your integral is not just about whether the value is finite, but also about whether the finite value is well defined. This is somewhat related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Indeterminate_form&quot; rel=&quot;nofollow&quot;&gt;indeterminant forms&lt;/a&gt; where $+\infty + (-\infty)$ is not a well-defined quantity. (You can't say they cancel out, since, heuristically, $\infty - \infty = (1+\infty)-\infty = 1+\infty -\infty = 1 + (\infty - \infty)$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as long as one of the integrals on the right hand side diverges, the entire algebraic expression becomes indeterminate, and hence we say the integral diverges. (Diverges doesn't necessarily mean that the value must run-off to infinity; it can just mean that the value does not converge to a definite number/expression.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A similar issue also crops up when summing infinite series that doesn't converge absolutely. &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann rearrangement theorem&lt;/a&gt; tells you that, depending on &quot;how&quot; you sum the series you can get the final number to be anything you want.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, however, it is advantageous to try to make sense of an integral which can be split into two divergent improper integrals, but also where one can argue that there should be some natural cancellation. For example, one may want to argue that $\int_{-a}^a \frac{1}{x^3} dx$ evaluates to 0 since it is the integral of an odd function. For this kind of situations, the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_principal_value&quot; rel=&quot;nofollow&quot;&gt;Cauchy principal value&lt;/a&gt; is useful. But notice that the definition is taken in the sense of a limit that rely on some cancellation, and so much in the same way of Riemann rearrangement theorem, &quot;how&quot; you take the limit can affect what value you get as the end result. (This is compatible with the notion that the integral diverges; as I said above, divergence should be taken to mean the lack of a well-defined, unique convergence.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;. Let me add another example of an integral that remains finite but does not converge. What is the value of $\int_{0}^\infty \sin(x) dx$? For every fixed $a &amp;gt; 0$, $\int_0^a\sin(x) dx = 1 - \cos(a) $ is a number between 0 and 2. But the limit as $a\to \infty$ doesn't exist! If you pick a certain way to approach $\infty$, say choose a sequence $a_n = 2\pi n$, then you'll come to the conclusion that the &quot;limit&quot; is 0; but if you choose $a_n = (2n + 1)\pi$, then you get the conclusion that the limit is $2$. The idea here is roughly similar: you take a left limit and a right limit approaching the improper point, and depending on how you choose your representative points (by an algebraic relation between the speed at which the left and right limits approach the improper point, say), you can get different answers. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-10T14:43:01.957" LastActivityDate="2010-12-10T14:43:01.957" CommentCount="4" />
  <row Id="13814" PostTypeId="1" AcceptedAnswerId="13816" CreationDate="2010-12-10T14:55:43.147" Score="1" ViewCount="165" Body="&lt;p&gt;I have to test for convergence of the series:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\displaystyle \sum\limits_{n=1}^{\infty} \sin\Bigl(\frac{\pi}{n}\Bigr)$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What i did was &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\sin\Bigl(\frac{\pi}{n}\Bigr)+ \sin\Bigl(\frac{\pi}{n+1}\Bigr) + \cdots  &amp;amp; &amp;lt; \pi \biggl( \frac{1}{n+1} + \frac{1}{n+2} + \cdots \biggr) \\ &amp;amp;= \pi \biggl( \sum\limits_{r=1}^{\infty} \frac{1}{n+r}\biggr) =\int\limits_{0}^{1} \frac{1}{1+x} \ dx \\ &amp;amp;= \pi\log{2}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this proves the &lt;em&gt;convergence&lt;/em&gt; of the series.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am interesting in knowing some more methods which can be used to prove the convergence so that i can apply them. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt; Note that $$\lim_{n \to \infty} \sum\limits_{r=1}^{n} \frac{1}{n} \cdot f\Bigl(\frac{r}{n}\Bigr) = \int\limits_{0}^{1} f(x) \ \textrm{dx}$$&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-12-10T15:56:16.393" LastActivityDate="2010-12-10T15:56:16.393" Title="Is $\sum \sin{\frac{\pi}{n}}$ convergent?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="13815" PostTypeId="1" AcceptedAnswerId="13819" CreationDate="2010-12-10T14:58:44.203" Score="2" ViewCount="157" Body="&lt;p&gt;How can one find if a function $f$ is infinitely differentiable?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-12-10T15:47:05.983" Title="Infinitely differentiable" Tags="&lt;calculus&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="13816" PostTypeId="2" ParentId="13814" CreationDate="2010-12-10T15:02:46.310" Score="11" ViewCount="0" Body="&lt;p&gt;You observed correctly that for small angles $\frac{\pi}{n}$, sin$\left(\frac{\pi}{n}\right)$ is very close to $\frac{\pi}{n}$. As for convergence of $\pi\sum_{n=1}^\infty\frac{1}{n}$...&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T15:02:46.310" CommentCount="2" />
  <row Id="13817" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:05:46.850" Score="2" ViewCount="0" Body="&lt;p&gt;There is no magic recipe to tell whether a function is infinitely differentiable, or even once differentiable, or even continuous indeed. It really depends on the function itself. Of course there are really general and trivial statements like all polynomials are infinitely differentiable and so on.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-10T15:05:46.850" />
  <row Id="13818" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:11:53.123" Score="4" ViewCount="0" Body="&lt;p&gt;Depending on how your function is given, that can be very easy or very hard. For functions given by expressions in closed form, you can try deriving a general formula for the $n$-th derivative. E.g. this way you can prove that all polynomials are smooth (=infinitely differentiable), so are sin, cos and exponential functions and any concatenation, product and sum of smooth functions is smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, you will have to work with the actual derivative as a limit and differentiate until you find either a non-smooth point or some closed form that you recognise, you maybe some periodicity (i.e. so that some $n$-derivative is equal to some $m$-the derivative, like with trigonometric functions) or you have to get really clever.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-10T15:11:53.123" />
  <row Id="13819" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:13:53.443" Score="9" ViewCount="0" Body="&lt;p&gt;By differentiating it an infinite number of times? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But seriously, that's what you do. Just that usually you can infer what the higher order derivatives will be, so you don't have to compute it one by one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt; To see that $\sin(x)$ is infinitely differentiable, you realize the following: $\frac{d}{dx}\sin(x) = \cos(x)$, and $\frac{d}{dx} \cos(x) = -\sin(x)$. So you see that $(\frac{d}{dx})^4\sin(x) = \sin(x)$, and so the derivatives are periodic. Therefore by continuity of $\sin(x)$ and its first three derivatives, $\sin(x)$ must be infinitely differentiable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt; To see that $(1 + x^2)^{-1}$ is infinitely differentiable, you realize that $\frac{d}{dx}(1+x^2)^{-n} = -2n x (1+x^2)^{-n-1}$. So therefore by induction you have the following statement: all derivatives of $(1+x^2)^{-1}$ can be written as a polynomial in $x$ multplied by $(1 +x^2)^{-1}$ to some power. Then you can use the fact that (a) polynomial functions are continuous and (b) quotients of polynomial functions are continuous away from where the denominator vanishes to conclude that all derivatives are continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general philosophy at work is that in order to show all derivatives are bounded and continuous, you can take advantage of some sort of recursive relationship between the various derivatives to inductively give a general form of the derivatives. Then you reduce the problem to showing that all functions of that general form are continuous. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-10T15:13:53.443" CommentCount="2" />
  <row Id="13820" PostTypeId="1" CreationDate="2010-12-10T15:29:06.957" Score="2" ViewCount="31" Body="&lt;p&gt;Define a &lt;strong&gt;quadtope&lt;/strong&gt; P as a polytope which is the convex hull of a set Q of convex quadrilaterals, any pair of which intersect in at most one vertex, and are in maximally general position, or, a k-fold pyramid over such a convex hull. (So, a pair of quadrilaterals from Q with a common vertex subtends a 4-dimensional subpolytope of P, and a pair of disjoint quadrilaterals from Q subtends a five-dimensional subpolytope of P.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;pure quadtope&lt;/strong&gt; is a quadtope each of whose vertices is a vertex of an element of Q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assert that the number of combinatorial types of pure quadtopes for dimensions 1 through 4 are 0, 1, 0, 1, respectively. How many combinatorial types of pure quadtopes of dimension 5 are there?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2010-12-10T18:38:37.547" LastActivityDate="2010-12-10T18:38:37.547" Title="How many combinatorial types of pure quadtopes of dimension 5 are there?" Tags="&lt;geometry&gt;&lt;combinatorics&gt;" CommentCount="1" />
  <row Id="13821" PostTypeId="2" ParentId="13815" CreationDate="2010-12-10T15:47:05.983" Score="1" ViewCount="0" Body="&lt;p&gt;As Willie Wong points out, that's the only way to do it. By the way $\sin{x},\cos{x}$ are examples of infinitely differentiable functions, and most famous example is the function $f(x)=e^{x}$. This function is infinitely differentiable and it's derivative, is $e^{x}$. To see this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;e^{x}  = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots \\ \frac{\textrm{d}} {\textrm{dx}}e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots &#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-12-10T15:47:05.983" />
  <row Id="13822" PostTypeId="1" CreationDate="2010-12-10T16:24:57.843" Score="6" ViewCount="96" Body="&lt;p&gt;I am trying to figure out the fundamental group (actually simply connected or not will suffice) of the following quotient space of $SO(3)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X = SO(3)/E$, where $E$ is the equivalence relation defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E \equiv M \sim {S_{A}}^{i} * M * {S_{B}}^{j}$ where ${S_{A}}^{i} \in $ Crystallographic point group $A$ and ${S_{B}}^{j} \in $ Crystallographic point group $B$, $M \in SO(3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$*$ represents the multiplication operation (matrix multiplication if rotations are represented as $3 \times 3$ special orthogonal matrices)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystallographic point group is defined here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallographic_point_group&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Crystallographic_point_group&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In crystallography, a crystallographic point group is a set of symmetry operations, like rotations or reflections, that leave the crystal invariant (hence a symmetry).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only consider the point groups with rotational symmetries. There exist 11 crystallographic point groups for three-dimensional crystals. Let us start with just considering cyclic point groups. They are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $C_1 = \{I \}$   where I is the identity rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Z_{\omega}$ is a rotation of angle $\omega$ about the $Z-$axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $C_2 = \{I, Z_{\pi} \}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) $C_3 = \{I, Z_{\frac{2\pi}{3}}, Z_{\frac{4\pi}{3}} \}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) $C_4 = \{I, Z_{\frac{\pi}{2}}, Z_{\pi}, Z_{\frac{3\pi}{2}} \}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) $C_6 = \{I, Z_{\frac{\pi}{3}}, Z_{\frac{2\pi}{3}}, Z_{\pi}, Z_{\frac{4\pi}{3}},  Z_{\frac{5\pi}{3}} \}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if Crystallographic point group of A is $C_2$ and B is $C_3$, the equivalence relations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ M \sim M * Z_{\frac{2\pi}{3}} \sim M * Z_{\frac{4\pi}{3}} \sim Z_{\pi}* M \sim Z_{\pi} * M * Z_{\frac{2\pi}{3}}  \sim Z_{\pi} * M * Z_{\frac{4\pi}{3}} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists literature for cases when one of the point groups is $C_1 = \{ I \}$. In this case it is a group action on $SO(3)$ and the space $X = SO(3) / G$ where $G$ is one of the 11 crystallographic point groups. These spaces fall under the so-called spherical 3-manifolds ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_3-manifold&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spherical_3-manifold&lt;/a&gt; ). When $G$ is one of the cyclic groups above, $X$ is a lens space. $L(2n,1) \cong SO(3)/ C_{n}$. I am not able to figure out how to think of these spaces when there are two point groups involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Progress so far: Even when there are two point groups acting on $SO(3)$, which if we refer to as $ G_{1} \backslash SO(3)/ G_{2}$ where $G_{1}$ is crystallographic point group of $A$ and $G_2$ refers to system $B$, there exists a finite subgroup of $\Gamma$ of $SO(4)$ such that  $ G_{1} \backslash SO(3)/ G_{2} \cong S^{3}/ \Gamma$. In the case $G_{1} = C_{1}$, it turns out that $\Gamma$  acts properly discontinuously and hence the fundamental group of that space is $\Gamma$ itself (from literature). But I am not sure how to check if $\Gamma$ acts &quot;properly discontinuously&quot; or not. And I am not sure how to check if the space is simply connected or not if $\Gamma$  does not act properly discontinuously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer Aaron's question: do you know how to obtain this group $\Gamma$ explicitly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the quaternion representation for rotations. Let $M = (q_0, q_1, q_2, q_3)$. I also use the following fact: Each 4D rotation $R$ is in two ways the product of left- and right-isoclinic rotations $R_L$ and $R_R$. $R_L$ and $R_R$ are together determined up to the central inversion, i.e. when both $R_L$ and $R_R$ are multiplied by the central inversion their product is $R$ again. (from wiki: &lt;a href=&quot;http://en.wikipedia.org/wiki/SO(4&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/SO(4&lt;/a&gt;) ). So for any operation ${S_{A}}^{i} * M * {S_{B}}^{j}$, I can find $\Gamma_{ij}$ such that &#xA;${S_{A}}^{i} * M * {S_{B}}^{j} = \Gamma_{ij}*[q_0 \ q_1 \ q_2 \ q_3]&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collection of all such $\Gamma_{ij}$ (and $\Gamma_{ij} * -I_4$, where $I_4$ is the $4 \times 4$ identity matrix) forms a finite subgroup $\Gamma$ of $SO(4)$. I used the definition provided by Aaron below for free action and found that at least for the cases where crystallographic point groups of $A$ and $B$ are the same (e.g. $C_2 \backslash SO(3) / C_2$), the action of finite subgroup $\Gamma_{C_2,C_2}$ is not properly discontinuous. Even though I tried to generalize the problem I am more interested in the cases when the crystallographic point groups of A and B are the same. So, how can I say whether this space $C_2 \backslash SO(3) / C_2$ is simply connected or not, especially now that i know that the action of $\Gamma$ is not properly discontinuous ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas ?&lt;/p&gt;&#xA;" OwnerUserId="3050" LastEditorUserId="3050" LastEditorDisplayName="" LastEditDate="2010-12-10T19:33:25.893" LastActivityDate="2010-12-10T19:33:25.893" Title="Fundamental group of quotient spaces of $SO(3)$" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="13823" PostTypeId="1" AcceptedAnswerId="13824" CreationDate="2010-12-10T17:02:09.667" Score="2" ViewCount="262" Body="&lt;p&gt;I know this is a silly question, but I've tried to find an answer using my TI-89 calculator, Maple and wolframalpha but none of those could tell me whether&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty |\cos(x^2)| \mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;converges or diverges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I'd be very happy if someone could help me out and tell me, whether the given integral converges or not (and why?). Thanks a lot. &lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-10T22:46:29.037" Title="Does $\int_0^\infty |\cos(x^2)| \mathrm dx$ converge?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13824" PostTypeId="2" ParentId="13823" CreationDate="2010-12-10T17:13:04.687" Score="10" ViewCount="0" Body="&lt;p&gt;It diverges.  You should be able to prove that $|\cos(x^2)|&amp;gt;0.1$ for most x.  If you let x=$\sqrt{\pi}u$ it is easier to assess the range of u where the cosine is close to zero.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-10T17:13:04.687" />
  <row Id="13826" PostTypeId="2" ParentId="13823" CreationDate="2010-12-10T17:19:55.223" Score="10" ViewCount="0" Body="&lt;p&gt;Another method: You could write it as the sum of the integrals on the intervals $\left[\sqrt{\frac{\pi}{2}+k\pi},\sqrt{\frac{\pi}{2}+(k+1)\pi}\right]$, and make a substitution $u=x^2$ to bound the integral on such an interval below by $\frac{1}{\sqrt{\frac{\pi}{2}+(k+1)\pi}}$.  (I'm ignoring the interval $\left[0,\sqrt{\frac{\pi}{2}}\right]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the inequality $\frac{\pi}{2}+(k+1)\pi\leq(k+2)\pi$ along with an integral comparison then leads to the estimate &#xA;$$\int_0^\sqrt{\frac{\pi}{2}+N\pi}|\cos(x^2)|dx\geq\frac{2}{\sqrt{\pi}}(\sqrt{N}-\sqrt{2}).$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-10T21:17:42.263" LastActivityDate="2010-12-10T21:17:42.263" CommentCount="1" />
  <row Id="13827" PostTypeId="2" ParentId="13689" CreationDate="2010-12-10T17:21:04.940" Score="1" ViewCount="0" Body="&lt;p&gt;Your problem is almost exactly &lt;a href=&quot;http://en.wikipedia.org/wiki/Multidimensional_scaling&quot; rel=&quot;nofollow&quot;&gt;multidimensional scaling&lt;/a&gt;: given a set of $n$ objects with given dissimilarities $\delta_{ij}$, find $n$ corresponding points $y_i$ in a low-dimensional space such that $\lVert y_i - y_j\rVert \approx \delta_{ij}$. I'm not intimately familiar with this area, but the Wikipedia article mentions a number of algorithms you could look at. You should be aware that there will often be many locally optimal configurations for a given problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, multidimensional scaling is typically used for &lt;em&gt;visualization&lt;/em&gt; applications, where you want to map your data to 2 or 3 dimensions and see what objects are similar to each other. I'm not sure how you plan to use it for computing &lt;em&gt;ratings&lt;/em&gt;, because replacing all $y_i$ with $-y_i$ is exactly as good a solution but will turn low ratings into high ratings and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-10T17:21:04.940" />
  <row Id="13828" PostTypeId="2" ParentId="13794" CreationDate="2010-12-10T17:39:18.997" Score="4" ViewCount="0" Body="&lt;p&gt;Most of the time, an equivalence relation is hiding an &quot;equality&quot; somewhere; that is, $x\sim y$ if and only if $x$ and $y$ have &lt;em&gt;something&lt;/em&gt; that you are trying to isolate which is &quot;equal&quot;. With integers, the equivalence $a\equiv b\pmod{m}$ means that $a$ and $b$ have &lt;strong&gt;equal&lt;/strong&gt; &lt;em&gt;remainder when divided by $m$&lt;/em&gt;. We usually start from a notion of the thing that we want to be &quot;the same&quot;, and define the equivalence relation accordingly, which makes it easier to think about just what are the equivalence classes: they correspond to all objects with the same &quot;&lt;em&gt;thingie&lt;/em&gt;&quot; that we are focusing on in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose you were walking down the street and you found an equivalence relation lying on the ground. Just how easy is it to figure out what this &quot;equality&quot; that is hinding behind it? How easy is it to describe all elements of the equivalence class of a given $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it depends greatly on the equivalence relation you find, and sometimes on the $x$. For some equivalence reations, it is fairly easy to figure it out, and then to partition the objects into equivalence classes. But for others, it may be more mysterious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you only know about positive integers, their order, and their addition and multiplication. You don't know about negative numbers (as most of humanity did not for most of its history). I can define an equivalence relation on pairs of positive integers by&#xA;$$(a,b)\sim (r,s) \Longleftrightarrow a+s = b+r.$$&#xD;&#xA;It is easy to verify that this is an equivalence relation. But what is the partition it induces? What is that &quot;equality&quot; that is hiding behind that equivalence relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's perhaps not so obvious. In fact, it comes from thinking of an ordered pair $(a,b)$ as corresponding to the equation $a+x=b$, so that the pair $(a,b)$ represents what we want to be a solution to this equation, which may or may not exist in the positive integers. Then $(a,b)\sim(r,s)$ means that the solution of $a+x=b$ should be &quot;the same&quot; as the solution of $r+x=s$. It seems obvious, then, that since $x=b-a=s-r$, this gives the condition I give, but the point is that one makes this definition among positive integers, where $b-a$ may be undefined, e.g., if $a\geq b$. It corresponds to a way of trying to think about negative numbers without having to use subtraction. But if you've never even thought about negative numbers, it would be rather difficult to figure it out, thought it would be pretty easy to check, given two pairs, whether they are equivalent or not; and you may even be able to describe all the pairs that are equivalent to a given one, at least sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you really have little hope of having some general method that will work easily and always. The reason is precisely because equivalence relations correspond to partitions. Pick &lt;strong&gt;any&lt;/strong&gt; partition, and define an equivalence relation by &quot;$a\sim b$ if and only if they are in the same piece of the partition.&quot; That will be an equivalence relation, but if you find it lying on the quad you would be hard pressed to figure out where it could have come from, precisely because it came from an arbitrary partition that happened to catch your fancy at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the moral is that most of the time equivalence relations come from a very specific idea we are trying to isolate, or a specific issue that is somewhat troublesome and we want to avoid (for instance, two functions being identical except for some negligible set of points where they differ, which may lead to some 'obvious' conclusions being technically false but true in 'spirit', so we define an equivalence relation that puts two such functions into the same class so that our &quot;true in spirit&quot; becomes &quot;technically true&quot; by now refering to &lt;em&gt;equivalence classes of functions&lt;/em&gt; instead of functions themselves). In such cases, it is very often easy to figure out the equivalence classes, or at least the bulk of each equivalence class (with some outliers being a bit troublesome from time to time). Practice and experience will let you spot them as they show up.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T19:48:53.387" LastActivityDate="2010-12-12T19:48:53.387" CommentCount="1" />
  <row Id="13829" PostTypeId="1" CreationDate="2010-12-10T17:42:27.263" Score="1" ViewCount="78" Body="&lt;p&gt;I was trying to understand the notion of a connection. I have heard in seminars that a connection is more or less a differential equation. I read the definition of Kozsul connection and I am trying to assimilate it. So far I cannot see why a connection is a differential equation. Please help me with some clarification.&lt;/p&gt;&#xA;" OwnerUserId="4611" LastActivityDate="2010-12-10T18:55:24.790" Title="Connections and differential equations" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="13830" PostTypeId="1" CreationDate="2010-12-10T18:05:57.787" Score="1" ViewCount="72" Body="&lt;p&gt;The Mathematica code is &lt;a href=&quot;http://demonstrations.wolfram.com/TransientHeatConductionWithNoSteadyState/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the heat conduction problem with Neumann (constant flux) at both boundaries of a solid slab. A constant radiant heat flux is imposed on one surface (derivative = -1) and the other surface is thermally insulated (derivative = 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the flux of the heat at one boundary, the temperature inside the system should raise faster over time. However, with the current code, if you increase the flux from 1 to 100, the system behavior doesn't change at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the code, &lt;code&gt;Quiet[NDSolve[...]]&lt;/code&gt; is used to suppress the error message: &lt;code&gt;&quot;NDSolve::ibcinc: Warning: Boundary and initial conditions are inconsistent.&quot;&lt;/code&gt; The simulation might be incorrect due to this error message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have the ability to change the system behavior by changing the value of the flux at the boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important question to me and I couldn't figure it out. Any help will be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4612" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-10T18:09:24.463" LastActivityDate="2010-12-11T06:23:20.343" Title="Help with the Neumann boundary conditions" Tags="&lt;differential-equations&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13831" PostTypeId="2" ParentId="13829" CreationDate="2010-12-10T18:06:37.047" Score="2" ViewCount="0" Body="&lt;p&gt;A connection is not a differential equation-- it is a differential operator.  Where the differential aspect comes from is that it is a derivation.  Remember in calculus the product rule says that $\frac{d}{dx} fg = \frac{df}{dx} g + f\frac{dg}{dx}$.  Well for a vector field $\sigma$, a function $f$, and Koszul connection $\nabla$, the property&#xA;$$&#xD;&#xA;\nabla_V (f\sigma) = df(V) \sigma + f\nabla_V \sigma&#xD;&#xA;$$&#xD;&#xA;is satisfied.  This is what makes it differential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes it a &quot;connection&quot; is that with $\nabla$ you can define parallel translation.  If you have a curve $c$, you can say that a vector field $\sigma(t)$ defined on the curve is parallel if $\nabla_{c&amp;#39;(t)} \sigma = 0$.  Think of this as $\frac{d}{dt} \sigma = 0$.  Now take a path $c$ starting at a point $x$.  Then by basic theorems of differential equations, for any vector at $x$ there is a unique extension to a parallel vector field along $c$.  This allows you to connect the tangent space at $x$ with the tangent space at any point on $c$.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-10T18:06:37.047" />
  <row Id="13832" PostTypeId="1" AcceptedAnswerId="13842" CreationDate="2010-12-10T18:20:06.407" Score="2" ViewCount="61" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an elliptic curve $y^2 = x^3 + ax + b$ over a finite field $\mathbf{F}_p$, how can I retrieve the value of $y$ given the value of $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge in this area is quite limited, so I understand this question might seems silly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently working on an application that uses an elliptic curves library and I saw at some point that this value was retrieved via: &#xA;$$(x^3 + ax + b)^{(p+1) &amp;gt;&amp;gt; 2} \mod p$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the $&amp;gt;&amp;gt;$ means shift right though I am not sure what this means mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4613" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T20:07:27.007" LastActivityDate="2010-12-10T20:15:28.917" Title="Extracting the value of $y$ from $x$ in an elliptic curve over a finite field" Tags="&lt;group-theory&gt;&lt;elliptic-curves&gt;&lt;finite-fields&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13833" PostTypeId="1" AcceptedAnswerId="13835" CreationDate="2010-12-10T18:35:51.550" Score="0" ViewCount="59" Body="&lt;p&gt;I am attempting to write a function (programming, not mathematics) which calculates the shortest distance between two points on a toroid.  I got the following formula from &lt;a href=&quot;http://stackoverflow.com/questions/2123947/calculate-distance-between-two-x-y-coordinates&quot;&gt;StackOverflow&lt;/a&gt;, but I don't recognize the &lt;code&gt;min&lt;/code&gt; portion.  Can someone explain that to me, please?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sqrt(min(|x1 - x2|, w - |x1 - x2|)^2 + min(|y1 - y2|, h - |y1-y2|)^2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4614" LastActivityDate="2010-12-10T18:42:05.147" Title="Can someone tell me what &quot;min&quot; means in this context?  " Tags="&lt;geometry&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="13834" PostTypeId="2" ParentId="13822" CreationDate="2010-12-10T18:40:52.980" Score="2" ViewCount="0" Body="&lt;p&gt;This is cool.  Are you doing chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a slightly more standard notation might be $G_1\backslash SO(3)/G_2$ -- this indicates that you're left-multiplying by elements of $G_1$ and right-multiplying by elements of $G_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;properly discontinuous&quot; action of a group $G$ on a space $X$ is one where every point $x\in X$ has a neighborhood $U$ such that $g(U)\cap U=\emptyset$ unless $g=1$; that is, $G$ not only acts &lt;em&gt;freely&lt;/em&gt; (no nonidentity element has fixed points), but the nonidentity elements of $G$ take every point &quot;sufficiently far away from itself&quot;.  This may seem vacuous, but in fact when you're working with Kleinian groups and things this can be an important condition.  However, in your case what is immediately true (since your space $SO(3)$ is Hausdorff and your groups are finite) is that any free action is automatically properly discontinuous.  So the only thing you need to check is that your actions have no fixed points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess the question becomes, do you know how to obtain this group $\Gamma$ explicitly?  If it's cyclic, it probably will be taken to be a subgroup of the circle, which acts freely on $S^3$.  (This is $S^1\subseteq \mathbb{C}$ acting on $S^3\subseteq \mathbb{C}^2$ by multiplication.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Gamma$ doesn't act properly discontinuously, things are harder.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-12-10T18:40:52.980" CommentCount="3" />
  <row Id="13835" PostTypeId="2" ParentId="13833" CreationDate="2010-12-10T18:42:05.147" Score="2" ViewCount="0" Body="&lt;p&gt;As explained in the question you reference, this toroid is created by identifying the sides of a rectangle w units wide in the x direction and h units high in the y direction.  So assume you have two points (x1,y) and (x2,y), at the same height and let x2&gt;x1.  To get from (x2,y) to (x1,y) you can either go left a distance x2-x1, or you can go right (off the edge) and come around a distance of (w-x2)+x1.  The min just picks the smaller of these.  Similarly in y.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-10T18:42:05.147" CommentCount="1" />
  <row Id="13836" PostTypeId="2" ParentId="13829" CreationDate="2010-12-10T18:55:24.790" Score="1" ViewCount="0" Body="&lt;p&gt;I'm going to be bold and go against the wisdom of the rest of the posters here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A connection &lt;em&gt;can&lt;/em&gt; be regarded as a differential equation, through the notion of parallel transport and geodesics. In the parallel transport formulation, a connection can be associated with the following map: give a point $p$ in your manifold $M$, a tangent vector $v \in T_pM$ in the tangent space at $p$, and a curve $\gamma:[0,1] \to M$ such that $\gamma(0) = p$, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (p,v,\gamma) \mapsto \tilde{v}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\tilde{v}$ is a vector field on $M$ defined along $\gamma$. That is, $\tilde{v}(\gamma(s)) \in T_{\gamma(s)}M$ for every $s\in [0,1]$. The specification of $\tilde{v}$ is given by the ordinary differential equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla_{\dot{\gamma}}\tilde{v} = 0, \qquad \tilde{v}(0) = v$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, more explicitly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{ds} \tilde{v}(\gamma(s)) = 0, \qquad \tilde{v}(\gamma(0)) = v$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the geodesic formulation, in local coordinates, a specification of a parallel transport is equivalent to the specification of the geodesics for that connection. More precisely, fixing a coordinate system $\{x_1, \ldots, x_n\}$ for your manifold $M$, the connection can be represented by its Christoffel symbols (relative to the coordinate system) $\Gamma_{ij}^k$, such that the geodesics satisfy &lt;strong&gt;Newton's Equations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d^2 x_i}{dt^2} = -\sum_{k,j} \Gamma^i_{jk} \frac{d x_j}{dt} \frac{d x_k}{dt} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in this way the connection precisely is associated to a particular set of second order differential equations, such that for each prescribed initial data $x_i(0) = x_{i,0}$ and $dx_i/dt (0) = y_{i,0}$ you have a corresponding geodesic. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a more abstract setting, a connection on the tangent space $TM$ of a manifold $M$ can be associated to a vector field $v$ define &lt;em&gt;over&lt;/em&gt; the total space $TM$ (so that $v$ is a section of $TTM$) so that the natural projection $\pi_*v(x) = x$ where $\pi$ is the projection map from $TM$ to $M$. The vector $v$ defines what is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic#Geodesic_spray&quot; rel=&quot;nofollow&quot;&gt;geodesic spray&lt;/a&gt; of the connection, which can be written as a first order ordinary differential equation on $TM$ (or a second order ordinary differential equation on $M$). So yes, it is possible to associate a connection to a differential equation. (The reverse association, however, is more delicate.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-10T18:55:24.790" CommentCount="5" />
  <row Id="13837" PostTypeId="2" ParentId="13800" CreationDate="2010-12-10T18:55:45.937" Score="3" ViewCount="0" Body="&lt;p&gt;For real $s$, $L(s,\chi_d)$ is real, and it is certainly positive for&#xA;large $s$. It is nonzero for $s&amp;gt;1$ by the Euler product and $L(1,\chi_d)\ne0$.&#xA;If $L(1/2,\chi_s)&amp;lt;0$ what would happen for some $1/2&amp;lt;s&amp;lt;1$?&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-10T18:55:45.937" CommentCount="1" />
  <row Id="13838" PostTypeId="1" AcceptedAnswerId="13871" CreationDate="2010-12-10T18:57:10.077" Score="1" ViewCount="85" Body="&lt;p&gt;I'm doing a multiple choice exercise and I'm stuck at a question, where there are two possible answers (may be both or none correct).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the differential equation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$y&amp;#39;&amp;#39;+y&amp;#39;+y=0$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for $y(x)$. Which of the following statements are true?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;The set of complex solutions forms a two-dimensional complex vector space.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The set of real solutions forms a two-dimensional real vector space.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm really not comfortable with vector spaces and the differentiation of real and complex vector space makes me even more unconfident. I solved the differential equation with the ansatz $y(x) = e^{\lambda x}$ and got to $\lambda^2 + \lambda + 1 = 0$, an equation which does not possess any real solutions. However, when I proceeded, I figured that I could rewrite the exponential function (with complex exponent) with the help of the sine and the cosine, so in the end I got a general solution with no complex coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of vector space is this?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-11T04:21:36.920" Title="What kind of vector space does the set of (complex/real) solutions of the following differential equation form?" Tags="&lt;differential-equations&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="13839" PostTypeId="2" ParentId="13373" CreationDate="2010-12-10T19:26:51.563" Score="3" ViewCount="0" Body="&lt;p&gt;Sergei Ivanov has given a positive answer for the existence of such functions on MO:&#xA;&lt;a href=&quot;http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2010-12-10T19:26:51.563" />
  <row Id="13840" PostTypeId="2" ParentId="886" CreationDate="2010-12-10T19:52:34.887" Score="1" ViewCount="0" Body="&lt;p&gt;22 boxes. 2 at the final stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Assume you will definately swap.&#xA;If you've picked 250k, which is 1/22 chance, you lose.&#xA;If you've not picked 250k, which is 21/22 chance, you win.&#xA;Every 22 times you play, you would expect to pick 250k once and expect to have 250k in the last 2 boxes twice (22 x 2 = 44).&#xA;Of the two times that 250k is in the last 2 boxes, 1 time you will swap (and lose), 1 time you will swap (and win). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Assume you will definately not swap.&#xA;If you've picked 250k, which is 1/22 chance, you win.&#xA;If you've not picked 250k, which is 21/22 chance, you lose.&#xA;Every 22 times you play, you would expect to pick 250k once and expect to have 250k in the last 2 boxes twice (22 x 2 = 44).&#xA;Of the two times that 250k is in the last 2 boxes, 1 time you will not swap (and win), 1 time you will not swap (and lose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you swap (or not), half the time you will win, half the time you will lose.&#xA;With the monty hall problem, 2/3 or the time if you swap you will win, 1/3 of the time you will lose. &#xA;The difference is that in EVERY game of the MH problem, a losing box is revealed and the player reaches the final stage, even though there are more than 2 boxes to start with.&#xA;In deal or no deal, 250k would only be at the final stage 2/22 or 1/13 times.&lt;/p&gt;&#xA;" OwnerUserId="4617" LastActivityDate="2010-12-10T19:52:34.887" />
  <row Id="13841" PostTypeId="1" AcceptedAnswerId="13843" CreationDate="2010-12-10T20:12:06.117" Score="7" ViewCount="113" Body="&lt;p&gt;Anyone know of an online tool available for making graphs (as in graph theory - consisting of edges and vertices)? I have about 36 vertices and even more edges that I wish to draw. &#xA;(why do I have so many? It's for pathing in a game)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only tool available to me right now is MS Paint and it would be very messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not actually looking for a 'standard' - just some way to neatly draw a set of vertices and edges, preferably without having to write code. I was planning to screenshot the pic and use it in a question on gamedev but I managed to solve the problem for now. Thanks for all the answers. I still prefer GeoGebra because it accessible online, neat, accurate, aligns elements to a grid, has consistent shapes, and has a high degree of customization available.&lt;/p&gt;&#xA;" OwnerUserId="4618" LastEditorUserId="4618" LastEditorDisplayName="" LastEditDate="2010-12-12T05:20:10.357" LastActivityDate="2010-12-12T05:20:10.357" Title="Online tool for making graphs (vertices and edges)?" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="13842" PostTypeId="2" ParentId="13832" CreationDate="2010-12-10T20:15:28.917" Score="3" ViewCount="0" Body="&lt;p&gt;Call C = x&lt;sup&gt;3&lt;/sup&gt;+ax+b.  You want to solve y&lt;sup&gt;2&lt;/sup&gt; = C, which is a quadratic equation, specifically extracting a square root.  There are lots of ways to extract square roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the way you mention is the first method described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_residue#Prime_or_prime_power_modulus&quot; rel=&quot;nofollow&quot;&gt;this section of the wikipedia article&lt;/a&gt; on numbers that are squares (quadratic residues).  This only works if p+1 is divisible by 4, but is a consequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_little_theorem&quot; rel=&quot;nofollow&quot;&gt;Fermat's little theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also links to a few efficient algorithms, including a &lt;a href=&quot;http://en.wikipedia.org/wiki/Shanks-Tonelli_algorithm&quot; rel=&quot;nofollow&quot;&gt;full algorithm&lt;/a&gt; that uses the one you mentioned if p+1 is divisible by 4.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-10T20:15:28.917" />
  <row Id="13843" PostTypeId="2" ParentId="13841" CreationDate="2010-12-10T20:20:50.077" Score="9" ViewCount="0" Body="&lt;p&gt;For drawing graph,the best is &lt;a href=&quot;http://www.texample.net/tikz/examples/prims-algorithm/&quot; rel=&quot;nofollow&quot;&gt;TIKZ&lt;/a&gt; but it does take some time to study,an alternate is &lt;a href=&quot;http://www.graphviz.org/&quot; rel=&quot;nofollow&quot;&gt;graphviz&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rollbacked:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For drawing 2D geometry, try &lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;GeoGebra&lt;/a&gt; or &lt;a href=&quot;http://www.dynamicgeometry.com/&quot; rel=&quot;nofollow&quot;&gt;Sketchpad&lt;/a&gt; (licensed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For drawing 3D geomtry, try &lt;a href=&quot;http://www.cabri.com/download-cabri-3d.html&quot; rel=&quot;nofollow&quot;&gt;Cabri 3d&lt;/a&gt; (licensed).&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-10T20:49:28.253" LastActivityDate="2010-12-10T20:49:28.253" CommentCount="4" />
  <row Id="13844" PostTypeId="1" AcceptedAnswerId="13851" CreationDate="2010-12-10T20:21:34.880" Score="1" ViewCount="41" Body="&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a prior on a random variable $X\sim \beta(\alpha,\beta)$ but I need to transform the variable to $Y=\frac{X}{1-X}$, for use in an analysis and I would like to know the distribution of $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_prime_distribution#Related_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $X\sim\beta(\alpha,\beta)$ then $\frac{X}{1-X} \sim\beta^\prime(\alpha,\beta)$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, the distribution is $Y\sim\beta^\prime(\alpha,\beta)$. The software that I am using, JAGS, does not support the $\beta^\prime$ distribution. So I would like to find an equivalent of a distribution that is supported by JAGS, such as the $F$ or $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the above relationship between the $\beta$ and $\beta^\prime$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_prime_distribution#Related_distributions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if $X\sim\beta^\prime(\alpha,\beta)$ then $\frac{X\beta}{\alpha}\sim F(2\alpha, 2\beta)$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, neither of these statements are referenced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;1) Can I find $c$, $d$ for $Y\sim\beta^\prime(\alpha,\beta)$ where $Y\sim\beta(c,d)$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2) Are these transformations correct? If so, are there limitations to using them, or a reason to use one versus the other (I presume $\beta$ is a more direct transformation, but why)? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;3) Where can I find such a proof or how would one demonstrate the validity of these relatively simple transformations?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2010-12-13T19:34:07.110" LastActivityDate="2010-12-13T19:34:07.110" Title="Are these transformations of the $\beta^\prime$ distribution from $\beta$ and to $F$ correct?" Tags="&lt;probability&gt;&lt;random-variable&gt;&lt;distribution-theory&gt;" AnswerCount="2" />
  <row Id="13845" PostTypeId="2" ParentId="13841" CreationDate="2010-12-10T20:23:09.470" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://creately.com/&quot; rel=&quot;nofollow&quot;&gt;http://creately.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dabbleboard.com/draw&quot; rel=&quot;nofollow&quot;&gt;http://dabbleboard.com/draw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-12-10T20:23:09.470" CommentCount="2" />
  <row Id="13846" PostTypeId="2" ParentId="13823" CreationDate="2010-12-10T20:29:03.693" Score="4" ViewCount="0" Body="&lt;p&gt;There is a standard technique we learned for dealing with integrals of the form $(*) \int_{a}^{\infty}f(x)\cos (\alpha x)dx$ or $\int_{a}^{\infty}f(x)\sin (\alpha x)dx$ when $f(x)$ is continuous and positive in $[a, \infty)$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\int_{a}^{\infty}f(x)dx$ converges, then (*) converges absolutely by the comparison test because $|f(x)\sin (\alpha x)|\leq f(x)$ (or $|f(x)\cos (\alpha x)|\leq f(x)$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\int_{a}^{\infty}f(x)dx$ diverges while $f(x)$ is decreasing and $\lim_{x\to\infty}f(x)=0$ then the integrals (*) converge conditionally by &lt;a href=&quot;http://math.nyu.edu/student_resources/wwiki/index.php/Category%3aDirichlet_Test#Dirichlet_Test_for_Integrals#Warning_About_Improper_Integrals&quot; rel=&quot;nofollow&quot;&gt;Dirichlets test&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll illustrate the technique on your integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, lets substitute $t=x^2$, then $(**) \int_{0}^{\infty} \cos(x^2) \mathrm dx=\int_{0}^{\infty} \frac{\cos(t)}{2 \sqrt t} \mathrm dt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x)=\frac{1}{2\sqrt x}$ then $f(x)$ is decreasing and $\lim_{x\to\infty}f(x)=0$. Also $cos(x)$ has a bounded anti-derivative. Therefore (**) converges by Dirichlets test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we observe that $|\frac{\cos(x)}{2 \sqrt x}|\geq \frac{\cos^2(x)}{2 \sqrt x}=\frac{1}{4\sqrt x}+\frac{\cos(2x)}{4\sqrt x}$. By the same arguments above $\int_{0}^{\infty}\frac{\cos(2x)}{4\sqrt x}dx$ converges. But then if we assume that $\int_{0}^{\infty}\frac{\cos^2(x)}{2 \sqrt x}$ converges, we get that $\int_{0}^{\infty}\frac{1}{4\sqrt x}$ converges and that's obviously not true. &lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-10T22:46:29.037" LastActivityDate="2010-12-10T22:46:29.037" />
  <row Id="13847" PostTypeId="1" AcceptedAnswerId="13848" CreationDate="2010-12-10T20:31:19.603" Score="4" ViewCount="108" Body="&lt;p&gt;It seems so. $\mathbb{Q}_p(\zeta_p)$ is a $p-1^{th}$ extension of $\mathbb{Q}_p$ which doesn't extend the residue field; and so is $\mathbb{Q}_p(p^{\frac{1}{p-1}})$. However I can't see how to express $\zeta_p$ in $\mathbb{Q}_p(p^{\frac{1}{p-1}})$ or how to express $p^{\frac{1}{p-1}}$ in $\mathbb{Q}_p(\zeta_p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help?&lt;/p&gt;&#xA;" OwnerUserId="4619" LastActivityDate="2010-12-11T02:36:12.933" Title="Is $\mathbb{Q}_p(\zeta_p)$ the same as $\mathbb{Q}_p(p^{\frac{1}{p-1}})$?" Tags="&lt;number-theory&gt;&lt;field-theory&gt;&lt;finite-fields&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="13848" PostTypeId="2" ParentId="13847" CreationDate="2010-12-10T20:45:33.260" Score="9" ViewCount="0" Body="&lt;p&gt;The right result is that $\mathbb{Q}_p(\zeta_p) = \mathbb{Q}_p((-p)^{1/(p-1)})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show that $\zeta_p$ is in $\mathbb{Q}_p((-p)^{1/(p-1)})$. Since both fields are of dimension $p-1$ over $\mathbb{Q}_p$ (compute ramification degrees), this shows the fields are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K=\mathbb{Q}_p((-p)^{1/(p-1)})$ and let $\pi$ be a $(p-1)$st root of $-p$ in $K$. We want to show that the equation $x^p=1$ has $p$ roots in $K$. Make the change of variable $x=1+\pi y$. So $\pi^{p} y + p \pi^{p-1} y^{p-1} + \cdots + p \pi y + 1 =1$. &#xA;Dividing out $\pi^{p} = -p \pi$, we get&#xA;$$y^{p} - \pi^{p-1} y^{p-1} - \frac{1}{p} \binom{p}{2} \pi^{p-2} y^{p-2} - \frac{1}{p} \binom{p}{3} \pi^{p-3} y^{p-3} - \cdots - y=0 \quad (*).$$&#xD;&#xA;Notice that $(1/p) \binom{p}{k}$ is an integer for $1 \leq k \leq p-1$. So, when we reduce $(*)$ modulo $\pi$, we get&#xA;$$y^p - y =0.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation $y^p-y=0$ has $p$ distinct roots in $\mathbb{F}_p$. So, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hensel%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Hensel's lemma&lt;/a&gt;, equation $(*)$ has $p$ roots in $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that $K$ contains a nontrivial $p$-th root of unity, as desired.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a direct argument for the other direction, when $p &amp;gt;2$. Let $1-\zeta = \lambda$ (where $\zeta$ is a primitive $p$th root of unity.) We know that&#xA;$$\prod_{j=1}^{p-1} (1-\zeta^j) = p.$$&#xD;&#xA;The left hand side is&#xA;$$\prod_{j=1}^{p-1} (j \lambda - \binom{j}{2} \lambda^2 + \cdots) = (p-1)! \lambda^{p-1} \cdot \prod_{j=1}^{p-1} \cdot \left( 1 - \frac{j}{2} \lambda + \cdots \right) =$$ &#xA;$$(p-1)! \cdot \lambda^{p-1} \cdot  \left( 1 - \frac{p(p-1)}{4} \lambda + \cdots \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Wilson's theorem&lt;/a&gt;, $(p-1)! \equiv -1 \mod p$. And, since $p$ is odd, $p(p-1)/4 \equiv 0 \mod p$. So we deduce that&#xA;$$p = (-1) \lambda^{p-1} \left( 1+ \mbox{something divisible by} \lambda^2 \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything of the form $ \left( 1+ \mbox{something divisible by} \lambda^2 \right)$ has a $(p-1)$st root in $\mathbb{Q}_p(\zeta)$, as the Taylor series for $(1+u)^{1/(p-1)}$ will converge for $u$ that small. So we see that &#xA;$$(-p)^{1/(p-1)} = \lambda  \left( 1+ \mbox{something divisible by} \lambda^2 \right)^{1/(p-1)}$$&#xD;&#xA;is in $\mathbb{Q}_p(\zeta)$ as desired.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2010-12-11T02:36:12.933" LastActivityDate="2010-12-11T02:36:12.933" />
  <row Id="13850" PostTypeId="2" ParentId="13662" CreationDate="2010-12-10T21:07:20.553" Score="5" ViewCount="0" Body="&lt;p&gt;This is essentially the same approach as Alex Bartel's, expressed perhaps more simply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that &lt;em&gt;at least one&lt;/em&gt; dies by age 90 is $1 - (2/3)^4 = 65/81$, and when that happens the probability that it's A who dies first is (by symmetry) a quarter of that, or $65/324$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-12-10T21:07:20.553" CommentCount="1" />
  <row Id="13851" PostTypeId="2" ParentId="13844" CreationDate="2010-12-10T21:08:56.700" Score="2" ViewCount="0" Body="&lt;p&gt;2 and 3) Both of these transformations are correct; you can prove them with the cdf (cumulative distribution function) technique.  I don't see any limitations on using them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the derivation for the first transformation using the cdf technique.  The derivation for the other will be similar.  Let $X \sim \beta(\alpha, \beta)$.  Let $Y = \frac{X}{1-X}.$  Then&#xA;$$P\left(Y \leq y\right) = P\left(\frac{X}{1-X} \leq y\right) = P\left(X \leq y(1-X)\right) = P\left(X(1+y) \leq y\right) = P\left(X \leq \frac{y}{1+y}\right)$$&#xD;&#xA;$$= \int_0^{\frac{y}{1+y}} \frac{x^{\alpha - 1} (1-x)^{\beta-1}}{B(\alpha,\beta)} dx.$$&#xD;&#xA;Differentiating both sides of this equation then yields the pdf (probability density function) of $Y$.  We have &#xA;$$f_Y(y)=\frac{\left(\frac{y}{1+y}\right)^{\alpha - 1} \left(1-\frac{y}{1+y}\right)^{\beta-1}}{B(\alpha,\beta)} \frac{d}{dy} \left(\frac{y}{1+y}\right)$$&#xD;&#xA;$$= \frac{\left(\frac{y}{1+y}\right)^{\alpha - 1} \left(\frac{1}{1+y}\right)^{\beta-1}}{B(\alpha,\beta)} \left(\frac{1}{1+y}\right)^2$$&#xD;&#xA;$$= \frac{y^{\alpha - 1} \left(1+y\right)^{-\alpha-\beta}}{B(\alpha,\beta)},$$&#xD;&#xA;which is the pdf of a $\beta&amp;#39;(\alpha,\beta)$ random variable.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-10T21:15:13.983" LastActivityDate="2010-12-10T21:15:13.983" CommentCount="2" />
  <row Id="13852" PostTypeId="1" AcceptedAnswerId="13853" CreationDate="2010-12-10T21:19:13.323" Score="2" ViewCount="103" Body="&lt;p&gt;I am Computer science student so pardon me if I am asking this in a wrong place!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My inspiration of learning maths is mainly due to algorithms and its feels to me that without mathematics I am not justifying my computer science degree, I am  trying to work on my rusty math since last two years but until now I am not much confident enough the main problem that I am facing is that I no matter whatever I do I am making silly mistakes, I will be presenting a college level maths entrance exams next year, and am preparing for this for two years; but now there is a feeling that I might lack ingenuity, which is required to be able to crack any competitive exam. So, is there any way to develop mathematical thinking? Does a mathematician usually/(at some part of their life) follow any exercise to concentrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also mention that I am almost alone in my home and I am rather compelled to learn maths all my own. But is this even  possible? I have taken some correspondence classes and I am trying with all my efforts but is it really possible to learn on my own? Please help me!&lt;/p&gt;&#xA;" OwnerUserId="4622" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T21:49:28.440" LastActivityDate="2010-12-10T21:58:25.323" Title="Advice: How can one prepare for a maths entrance exam? How can one develop mathematical thinking?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13853" PostTypeId="2" ParentId="13852" CreationDate="2010-12-10T21:30:54.360" Score="6" ViewCount="0" Body="&lt;p&gt;Whether or not &quot;ingenuity&quot; is &quot;required to be able to crack&quot; (do well?) the exam you will present depends enormously on the kind of exam it is. Ingenuity is not particularly required, say, in the United States' GRE examination (at least, when I took it, even the Mathematics Subject test). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ability to concentrate has little to do with mathematics per se; it is a requirement of &lt;strong&gt;any&lt;/strong&gt; serious field of study or endeavor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to develop &quot;ingenuity&quot; (really, ability to solve certain kinds of problems) is practice. Practice, practice, practice. And then, more practice. The vast majority of problems you will encounter are problems that have something in common with other problems, that can be analyzed with the same kind of ideas, the same kind of methods. So what you &lt;em&gt;really&lt;/em&gt; want is not ways to concentrate, or ways to become &quot;ingenious&quot;, but a big &quot;portfolio&quot; of problems that you know how to solve and that you can refer to when you encounter a new problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is few advice better than that given by Polya decades ago in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/4871878309&quot; rel=&quot;nofollow&quot;&gt;his book &lt;strong&gt;How to Solve It&lt;/strong&gt;&lt;/a&gt;. See some of the reviews there for some of the questions to ask. And always ask: &quot;Is this like something I've seen before? If so, can I try doing things similar to what I did then?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; Yes, it is possible to learn on your own, though it is always much easier to learn with immediate feedback and help from others. From your description of the exam (&quot;only 30 people selected in all of India&quot;) things may be very hard &lt;em&gt;even if&lt;/em&gt; you had immediate feedback and the best tutor money could rent available 24 hours a day at your beck and call. If what you want is reassurance that you are doing things correctly when you are in doubt, there are sites such as this one that can help; try not to get carried away and flood the site with all the questions that have accummulated over the past two years (you'll turn off people), but reasoned questions showing both interest and that you've tried to work them through on your own and are genuinely confused are likely to be well received. None of this, however, has to do with developing &quot;ingenuity&quot;, or the ability to concentrate, or &quot;mathematical thinking&quot;. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-10T21:58:25.323" LastActivityDate="2010-12-10T21:58:25.323" CommentCount="7" />
  <row Id="13854" PostTypeId="1" CreationDate="2010-12-10T21:32:00.027" Score="2" ViewCount="48" Body="&lt;p&gt;Suppose I know the values of a continuous function on $[a,b]$ in some finite number of points $x_0,x_1 \ldots x_n$. I can form the Lagrange interpolating polynomial, $p$. I am curious if there is any interesting estimate of the expression $|f(x)-p(x)|$ for an arbitrary point $x$ that does not assume $f$ is smooth? (Perhaps it would involve moduli of continuity of higher order.) The textbooks on approximation theory I have seen only treat differentiable case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify what I mean by interesting: I can rather trivially bound $|f(x)-p(x)|$ from above by $\sum |\omega(x-x_i)||L_i(x)|$, where $\omega$ is the modulus of continuity and $L_i$ are the normal Lagrange basis polynomials. I do not view this as interesting, among other things because it does not reduce to the normal Lagrange error term when $f$ is smooth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now further that I know that the function is in fact nowhere differentiable and I want to interpolate between the given values of the function. Could one give lower bounds for the error now when using Lagrange interpolation of various orders?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background to the question is that I been thinking about if one could interpolate in a table of the Weierstrass function, such as the one given here: &lt;a href=&quot;http://www.ams.org/journals/mcom/1961-15-074/S0025-5718-1961-0122011-X/home.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/mcom/1961-15-074/S0025-5718-1961-0122011-X/home.html&lt;/a&gt;. (Do not ask me why I have been thinking about that.) &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2010-12-10T21:32:00.027" Title="Error term when Lagrange interpolating continuous non-differentiable functions" Tags="&lt;approximation-theory&gt;" CommentCount="1" />
  <row Id="13855" PostTypeId="2" ParentId="13844" CreationDate="2010-12-10T21:50:36.057" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Answer to 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no $c$, $d$ for $Y~\beta^\prime(\alpha,\beta)$ where $Y~\beta(c,d)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because $Y&amp;gt;0$ but $X\in(0; 1)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the $F$ distribution which has support for $Y$ which is defined for positive real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer to 2 (simulation approach):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;although this is not a mathematical proof, it provides an example of how to sufficiently support the postulate that $X\frac{\beta}{\alpha}\sim F(2*\alpha,2*\beta)$. Feedback appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose arbitrary parameter values&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;alpha &amp;lt;- 2&#xA;beta  &amp;lt;- 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Create a vector of a $\beta^\prime(\alpha, \beta)$ variate &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X &amp;lt;- rbeta(1000000,alpha, beta)&#xA;Y1 &amp;lt;- X/(1-X)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Create a vector of an $F(2*\alpha, 2*\beta)$ variate and divide by $\frac{\beta}{\alpha}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Y2 &amp;lt;- rf(1000000, 2*alpha, 2*beta) / (beta / alpha)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Compare the variables at decile intervals across their range.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;testY1 &amp;lt;- quantile(Y1, seq(0.1,0.9,0.1))&#xA;testY2 &amp;lt;- quantile(Y2, seq(0.1,0.9,0.1))&#xA;&#xA;signif(testY1,2) == signif(testY2,2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer to 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.wm.edu/~leemis/2008amstat.pdf&quot; rel=&quot;nofollow&quot;&gt;Leemis and McQuestion (2008)&lt;/a&gt; is an excellent peer-reviewed reference detailing the inter-relationships among distributions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thanks Mike for providing the proper answer to 2 and 3.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leemis, L.M. and J.T. McQuestion. 2008 Univariate Distribution Relationships. The American Statistician 62(1) 45:53&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="3733" LastEditorDisplayName="" LastEditDate="2010-12-10T23:04:31.480" LastActivityDate="2010-12-10T23:04:31.480" CommentCount="1" />
  <row Id="13856" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T21:50:36.603" Score="7" ViewCount="0" Body="&lt;p&gt;As even the capitalistic &lt;a href=&quot;http://en.wikipedia.org/wiki/Dialectical_materialism%22Wikipedia%22&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; acknowledges, there are three laws of dialectics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The law of the unity and conflict of opposites;&lt;/li&gt;&#xA;&lt;li&gt;The law of the passage of quantitative changes into qualitative changes;&lt;/li&gt;&#xA;&lt;li&gt;The law of the negation of the negation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, from law 1, we see that i and -i are both equal and opposite. From law 2, we learn that $\mathbb C$ is qualitatively different from $\mathbb R$. From law 3, we attain the dynamic equilibrium between i and -i that was so powerfully expounded by the revolutionary hero Évariste Galois.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This at least is how we taught such matters in the USSR.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-10T21:50:36.603" CommentCount="2" />
  <row Id="13857" PostTypeId="2" ParentId="13801" CreationDate="2010-12-10T22:12:45.890" Score="10" ViewCount="0" Body="&lt;p&gt;$\sqrt{-1}$ is an imprecise notation.  There are several ways of making it precise, some of which involve what comes down to an arbitrary choice of square root of $-1$ and some of which don't.  This is because there are several related ways to construct $\mathbb{C}$ from $\mathbb{R}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As the ring $F = \mathbb{R}[x]/(x^2 + 1)$ (edit: following in the discussion in the comments, perhaps it would be better to say &quot;an algebraic closure of $\mathbb{R}$.&quot;).  The Galois group $\text{Gal}(F/\mathbb{R})$ has order $2$, and complex conjugation $x \mapsto -x$ is its only nontrivial element.  The Galois symmetry here forbids us from distinguishing between $x$ and $-x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As the ring $\mathbb{C} = \mathbb{R}[\imath]/(\imath^2 + 1)$, where we fix the choice $\imath$ of generator.  This is what is typically meant by $\mathbb{C}$.  This field has a distinguished element $\imath$ satisfying $\imath^2 = -1$, and this is what we usually mean by $i$.  $\mathbb{C}$ is isomorphic to $F$, but &lt;strong&gt;not canonically so&lt;/strong&gt;, since $\imath$ can be sent to either $x$ or $-x$ and there is no way to choose between these.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As the ring $R$ of linear endomorphisms $\mathbb{R}^2 \to \mathbb{R}^2$ of non-negative determinant preserving an inner product $\langle \cdot, \cdot \rangle$.  There are two elements of $R$ squaring to $-1$, corresponding to a rotation and its inverse, and there is no way to choose between them unless $\mathbb{R}^2$ is also equipped with an &lt;strong&gt;orientation&lt;/strong&gt;, which is an identification of its exterior square with $\mathbb{R}$.  This choice of orientation corresponds to the difference between the first and the second constructions above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But most people will not bother to stop and talk about such subtleties as $F, \mathbb{C}, R$ being non-canonically isomorphic, so they say $\sqrt{-1} = i$ because the truth is (for practical purposes) unnecessarily complicated.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-11T06:42:09.430" LastActivityDate="2010-12-11T06:42:09.430" CommentCount="5" />
  <row Id="13858" PostTypeId="2" ParentId="13838" CreationDate="2010-12-10T22:38:02.397" Score="2" ViewCount="0" Body="&lt;p&gt;Both statements are true.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-10T22:38:02.397" CommentCount="1" />
  <row Id="13859" PostTypeId="2" ParentId="13793" CreationDate="2010-12-10T23:35:11.630" Score="8" ViewCount="0" Body="&lt;p&gt;Since the technique proving the homeomorphism is useful in many other situations, it may be worth adding some details to Carl's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A,B$ are two countable, dense linear orders without end points. We show that they are isomorphic by building an isomorphism $f:A\to B. This is done by what we call a &lt;em&gt;back-and-forth&lt;/em&gt; argument. Say that $A=\{a_n\mid n\in{\mathbb N}\}$ and $B=\{b_n\mid n\in{\mathbb N}\}$. We build $f$ by stages. At the end of stage $2n$ we have ensured that $a_n\in{\rm dom}(f)$, and at the end of stage $2n+1$, we have ensured that $b_n\in{\rm ran}(f)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction is simple. Begin by picking any $b\in B$ and letting $f(a_0)=b$. This completes stage 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we do stage 1: If $b=b_0$ we are done and go to stage 2. If $b&amp;lt;b_0$, we pick an $a\in A$ and set $f(a)=b_0$. Of course, since $f$ is to be an isomorphism, we better ensure that $a_0&amp;lt;a$. But this is trivial to accomplish, since $A$ has no endpoints. Similarly, if $b_0&amp;lt;b$, then we pick $a$ so that $a&amp;lt;a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, at stage $2n$ do the following: If $a_n$ is already in the domain we have built, we are done with this stage. Otherwise, if $a_n$ is larger than all the elements in the domain of $f$ so far, pick an element $c$ of $B$ larger than all the elements in the range of $f$ so far and set $f(a_n)=c$; this is possible since $B$ has no largest element. If $a_n$ is smaller than all elements in the current domain of $f$, pick $c$ in $B$ smaller than all elements in the current range of $f$, and set $f(a_n)=c$. Again, this is possible, since $B$ has no smallest element. Finally, if $a_n$ is between elements of the current domain of $f$, pick $d,e$ in the current domain of $f$ so $d&amp;lt;a_n&amp;lt;e$, $d$ is largest below $a_n$, and $e$ is smallest above $a_n$. Then pick in $B$ some $c$ between $f(d)$ and $f(e)$ and set $f(a_n)=c$. This is possible, since $B$ is dense in itself. This completes this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At stage $2n+1$ we do the same, but now ensuring that $b_n$ is put in the range of $f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construction gives us an isomorphism $f$ at the end: Even stages ensure the domain of $f$ is all of $A$, odd stages that the range is all of $B$. The construction is designed so for $\alpha,\beta$ in the domain of $f$, $\alpha&amp;lt;\beta$ iff $f(\alpha)&amp;lt;f(\beta)$. But this is precisely what it means to be an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of back-and-forth is very flexible. For example, it shows that any two countable random graphs are isomorphic. There are plenty of applications of this technique.  &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-10T23:35:11.630" CommentCount="2" />
  <row Id="13860" PostTypeId="2" ParentId="13830" CreationDate="2010-12-10T23:43:06.470" Score="1" ViewCount="0" Body="&lt;p&gt;This is really a question about Mathematica, but the short answer is that the code that you're looking at is not correctly handling the boundary and initial conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PDE in question is the 1-d heat equation on the unit interval,&#xA;$$\theta_t = \theta_{xx}$$&#xD;&#xA;with Neumann boundary conditions $\theta_x(0,t) = -1$ and $\theta_x(1,t) = 0$ and with initial condition $\theta(x,0) = e^{-1000 x}/1000$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the default method that Mathematica is using to numerically solve this problem does not like the fact that the boundary conditions are inconsistent with the initial condition in the sense that the initial temperature distribution does not satisfy the Neumann condition at $x=1$.  The problem is compounded if one changes the boundary condition at $x=0$, as then the initial temperature satisfies neither boundary condition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This issue is addressed &lt;a href=&quot;http://reference.wolfram.com/mathematica/tutorial/NDSolvePDE.html#51169332&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastEditorUserId="4229" LastEditorDisplayName="" LastEditDate="2010-12-11T06:23:20.343" LastActivityDate="2010-12-11T06:23:20.343" />
  <row Id="13861" PostTypeId="1" AcceptedAnswerId="16977" CreationDate="2010-12-11T00:04:31.593" Score="8" ViewCount="157" Body="&lt;p&gt;Given a collection of points $P \subset \mathbb R^3$, a crude characterization of the &quot;shape&quot; of $P$ is sometimes given by the principal components. We construct a covariance matrix, e.g., if $P$ is discrete, $C = \displaystyle\sum_{p\in P} (p - \mu)(p - \mu)^\intercal$, where $\mu = \displaystyle\frac1{|P|}\sum_{p\in P}\, p$ is the centre of mass. This defines an ellipsoid whose semi-axis are defined by the unit eigenvectors of $C$, scaled by the associated eigenvalues.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is concerned with the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ellipsoid described by the principal components is the best fit&#xA;  ellipsoid for $P$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't know of any author or resource that I can accuse of explicitly making such a claim$^*$.  Anyway, my question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a natural geometric definition of &quot;best fit ellipsoid&quot; for&#xA;  which the above statement becomes true? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, some kind of least squares or other variational characterization of this same ellipsoid is what I am looking for. I would also accept an answer that convinces me that this is the wrong way to be looking at the principal components, but that will be a tough sell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do a coordinate translation, so that $v_p = (p - \mu)$, and let&#xA;$\hat{v}_p = \frac{v_p}{\left\|v_p\right\|}$, and look at $C$ as a&#xA;linear transformation which is the sum of the rank one operators in&#xA;this coordinate system, $C = \displaystyle\sum_{p \in P} \left\|v_p\right\|^2&#xD;&#xA;\hat{v}_p\hat{v}_p^\intercal$, then the ellipsoid in question is the&#xA;image of the unit ball. From this characterization I gain some&#xA;intuition as to why this particular ellipsoid is a good one. I am&#xA;looking for a better understanding, preferably from a geometric perspective.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* Wikipedia comes close to such a claim in the description of the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;moments&lt;/a&gt;: &quot;The 'second moment', ... in higher dimensions measures the shape of a cloud of points as it could be fit by an ellipsoid.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Although I feel that the observation that the ellipsoid reflects the variance of the  Gaussian distribution that has maximal likelihood to produce $P$ (I haven't rolled up my sleeves and checked), this is not the kind of answer I am looking for. Perhaps I should remove all tags that refer to probability or regression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will make the question very specific then. From stuff I've seen elsewhere on the web, I get the feeling that this ellipsoid is different from the one that minimizes the sum of squared distances to the points, but I don't know for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about this then: the &lt;em&gt;radial distance&lt;/em&gt; from a point $p$ to the ellipsoid is the distance as measured along the line that contains $p$ and $\mu$ (the latter being the origin in our new coordinate system). So let this be my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the ellipsoid defined above minimize the sum of the squared radial distances?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditorDisplayName="" LastEditDate="2011-01-10T06:28:15.970" LastActivityDate="2011-01-10T13:09:54.847" Title="Best fit ellipsoid" Tags="&lt;geometry&gt;&lt;probability-theory&gt;&lt;optimization&gt;&lt;regression&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13862" PostTypeId="2" ParentId="13804" CreationDate="2010-12-11T01:05:04.170" Score="2" ViewCount="0" Body="&lt;p&gt;There's a recent book on the subject by Jan Krajicek, &lt;a href=&quot;http://www.karlin.mff.cuni.cz/~krajicek/fdraft.pdf&quot; rel=&quot;nofollow&quot;&gt;draft here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of defining the probabilities directly, the truth value should be the corresponding event itself, i.e. the points in which the statement is true. So for a proposition $P(x)$, the truth value is the points in which $x$ is true. The Boolean connectives correspond to set operations: negation to complementation, and to intersection, or to union. Quantifiers correspond to supremum (exists) and infimum (forall). Having defined the truth values, you can calculate the probability at the end. You then get that a statement is tautological iff it holds with probability $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding has assumed that the probability space is finite; for infinite spaces the picture is more complicated, and you can consult the draft to see how to do it with non-standard models of the naturals (endowed with the Loeb measure, which is an extension of the counting measure).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-11T01:05:04.170" />
  <row Id="13864" PostTypeId="2" ParentId="13841" CreationDate="2010-12-11T01:31:01.330" Score="3" ViewCount="0" Body="&lt;p&gt;You can also use &lt;a href=&quot;http://www.lsv.ens-cachan.fr/~gastin/gastex/gastex.html&quot; rel=&quot;nofollow&quot;&gt;GasTeX&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-11T01:31:01.330" />
  <row Id="13865" PostTypeId="2" ParentId="13536" CreationDate="2010-12-11T01:31:49.953" Score="0" ViewCount="0" Body="&lt;p&gt;This is an interesting optics/physics/math problem.  I believe that your question could be asking for a number of distinct answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Scale in image space is determined according to the magnification of the camera: &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnification#Calculating_the_magnification_of_optical_systems&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnification#Calculating_the_magnification_of_optical_systems&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the distance from the 'lens' to the image using the distance from the object to the image, information about the camera, and lens equations - note that complicated optical systems can be represented by a single lens with a defined focal length.  (Also, note that the focal length may vary as the camera is focused.)  This is most easily accomplished for the pinhole camera, for which the distance from the 'lens' (the pinhole) to the image is defined entirely by the placement of a surface behind the pinhole.  More complicated cameras can be translated into a pinhole camera for the purposes of performing a raytrace.  To calculate the 'scale' of features, project each of the vertices through the pinhole and calculate their intersections with the image surface - when using a simplified representation of a camera, this will be a plane a defined distance beyond the pinhole, but a rotated image can be calculated by placing the plane between the object and the pinhole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, it is possible to image each of the vertices through an optical system individually, taking into account the depth of the scene to account for depth of the image, but typically image sensors are planar and this type of analysis is only useful for depth-of-field type calculations.  I recommend #1 or #2 unless there is a compelling reason to do this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4628" LastActivityDate="2010-12-11T01:31:49.953" />
  <row Id="13866" PostTypeId="2" ParentId="9594" CreationDate="2010-12-11T01:37:40.340" Score="2" ViewCount="0" Body="&lt;p&gt;There are 7,618,438 diferent positions after six movements according to &lt;a href=&quot;http://cube20.org/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; site, but they use face movements. By the way they show that the Rubik's cube can be solved in 20 face movements or less. &lt;/p&gt;&#xA;" OwnerUserId="3274" LastEditorUserId="3274" LastEditorDisplayName="" LastEditDate="2010-12-11T02:01:14.983" LastActivityDate="2010-12-11T02:01:14.983" />
  <row Id="13867" PostTypeId="2" ParentId="13625" CreationDate="2010-12-11T01:47:54.680" Score="0" ViewCount="0" Body="&lt;p&gt;(disclaimer: I'm not writing anything deep in this answer, but it might be useful for people starting to learn manifolds/topology)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it would be safe to say no, globally euclidean always implies locally euclidean, although of course it all depends on how you define these concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (not very useful) definition could be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;X is globally euclidean if it is homeomorphic to $\mathbb{R}^n$ for some n.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;X is euclidean at x $\in X$ if there exists an open neighbourhood of x homeomorphic to an open subset of a globally euclidean space.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;X is locally euclidean if it is euclidean at every point $x \in X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this (tautological) definition then it is obvious that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If X is globally euclidean then X is locally euclidean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The converse is of course false: take the real line $\mathbb{R}$ and the circle $S^1$, the first is globally euclidean, the second is locally euclidean but not globally. If $S^1$ were globally euclidean then (by a general result of algebraic topology &quot;invariance of dimension&quot;) it would be the one-dimensional euclidean space $\mathbb{R}$. But $\mathbb{R}$ is contractible and $S^1$ is not (or $S^1$ is compact and $\mathbb{R}$ is not) (or if you remove a point from $S^1$ it stays connected but if you remove a point from $\mathbb{R}$ it splits in two).&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2010-12-11T01:47:54.680" />
  <row Id="13868" PostTypeId="2" ParentId="460" CreationDate="2010-12-11T01:58:42.147" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a cheap answer which takes the stance that semigroups are inherently interesting :)&#xA;Consider the following definition of a group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Definition}$ A semigroup $S$ is said to be a &lt;em&gt;group&lt;/em&gt; if the following hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is an $e \in S$ such that $ea=a$ for all $a\in S$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;For each $a \in S$ there is an element $a^{-1} \in S$ with $a^{-1}a=e$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At one point in my life, it seemed natural to ask what happens if we replace axiom 2 with the very similar axiom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2$^\prime$. For each $a \in S$ there is an element $a^{-1} \in S$ with $aa^{-1} = e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fun exercise to work out some of the consequences that result from this. Here are a few facts about a semigroup $S$ which satisfies $1$ and $2^\prime$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $e$ is the unique element of $S$ satisfying axiom 1, then $S$ is a group&lt;/li&gt;&#xA;&lt;li&gt;If $S$ has an identity (in the usual sense) then $S$ is a group&lt;/li&gt;&#xA;&lt;li&gt;The principal left ideal $Sa = \{sa \mid s \in S\}$ is a group for all $a \in S$, and in fact all principal left ideals of $S$ are isomorphic as groups.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is not difficult to find examples of such semigroups that are not groups. For example,&#xA;consider the following set of $2\times 2$ matrices (with matrix multiplication as the operation):&#xA;$$\left\{\begin{pmatrix} a &amp;amp; b \\ 0 &amp;amp; 0\end{pmatrix} \mid a,b \in \mathbb{R}, a \neq 0\right\}$$&#xD;&#xA;Or, an example that appears as exercise 30 in section 4 of Fraleigh's abstract algebra text:&#xA;the nonzero real numbers under the operation $\ast$ defined by $a\ast b = |a|b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly it is debatable whether or not semigroups satisfying axioms 1 and 2$^\prime$ are &quot;interesting&quot; or &quot;natural&quot;. But I guess I think they are. And, I am not the only one (or the first one, by a long shot!) to think this. See&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/journals/bull/1944-50-12/S0002-9904-1944-08256-6/S0002-9904-1944-08256-6.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/bull/1944-50-12/S0002-9904-1944-08256-6/S0002-9904-1944-08256-6.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A bit of googling will turn up more results if you are interested.)&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2010-12-11T01:58:42.147" CommunityOwnedDate="2010-12-11T01:58:42.147" />
  <row Id="13869" PostTypeId="2" ParentId="13861" CreationDate="2010-12-11T03:13:59.843" Score="0" ViewCount="0" Body="&lt;p&gt;Just fleshing out Rahul's comment: probably this choice of $\mu$ and $\Sigma$ maximizes the likelihood of the points given a multivariate Gaussian distribution $N(\mu,\Sigma)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-11T03:13:59.843" CommentCount="1" />
  <row Id="13870" PostTypeId="1" AcceptedAnswerId="13872" CreationDate="2010-12-11T04:19:17.997" Score="1" ViewCount="65" Body="&lt;p&gt;A question I recall from Carl Linderholm's &quot;Mathematics Made Difficult&quot;, Chapter 3 Exercise 8.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A farmer acquires and algebraically closed field by extending his field finitely. &#xA;What can be said about the original field?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What can be said? I immediately see two fields with this property, ${\Bbb R}$, and $\overline{\Bbb Q}\cap {\Bbb R}$. Is this it? or is there a better characterization? Do these fields have to contain ${\Bbb Q}$?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-12-11T04:31:48.447" Title="Field extension question" Tags="&lt;group-theory&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" />
  <row Id="13871" PostTypeId="2" ParentId="13838" CreationDate="2010-12-11T04:21:36.920" Score="2" ViewCount="0" Body="&lt;p&gt;As Huy says, both statements are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you are a hammer, all the problems seem nails to you, but since this is a linear differential equation, Laplace transform &lt;a href=&quot;http://math.stackexchange.com/questions/13452/how-do-you-solve-y2y-3y0/13475#13475&quot;&gt;again&lt;/a&gt; gives us the general solution in both, real and complex, cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;{\cal L}[y&amp;#39;&amp;#39;] + {\cal L}[y&amp;#39;] + {\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2 F(s) -sy(0) -y&amp;#39;(0) ) + (sF(s) -y(0)) + F(s) = 0 \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F(s) = {\cal L}[y]$. Now, put $a= y(0)$ and $b= y&amp;#39;(0)$ and solve this equation for $F(s)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;F(s) = \dfrac{as + (a+b)}{s^2+s+1} \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the polynomial $s^2+s+1$ has no real roots, the distinction between the real and complex cases arrives at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the complex case&lt;/strong&gt;, you write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;s^2 + s + 1 = (s-\alpha)(s-\beta)  \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha = \dfrac{-1+i\sqrt{3}}{2}$ and $\beta = \dfrac{-1-i\sqrt{3}}{2}$ are the roots of your characteristic polynomial. Hence the solution of your differential equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y(t) = A{\cal L}^{-1} \left[\dfrac{1}{s-\alpha} \right] + B {\cal L}^{-1} \left[\dfrac{1}{s-\beta} \right] = A e^{\alpha t} + B e^{\beta t} \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are complex numbers, depending on $a , b$, which again I let you the pleasure to compute  :-) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the fact is that the solution of your differential equation is the $\mathbb{C}$-linear span of two linearly independent complex functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;e^{\alpha t} \qquad \text{and} \qquad e^{\beta t} \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a $\mathbb{C}$-vector space of dimension two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the real case&lt;/strong&gt;, you write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;s^2 + s + 1 = \left(s + \frac{1}{2} \right)^2 + \frac{3}{4}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this time, the solution set of your differential equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y(t) = &#xD;&#xA;A {\cal L}^{-1} \left[ \frac{s + \frac{1}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} }  \right] +&#xD;&#xA;B {\cal L}^{-1} \left[ \frac{ \frac{\sqrt{3}}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} }  \right] &#xD;&#xA;= A e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2}t\right) +&#xD;&#xA;B e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2}t\right)  \ ,&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are real numbers, depending on $a,b$, which... Anyway again: the solution set is the $\mathbb{R}$-linear span of two linearly independent real functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2} t \right) \qquad \text{and} \qquad e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2} t \right)  \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a real vector space of dimension two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternatively&lt;/strong&gt;, you could read what &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_differential_equation#Summary_of_General_Solutions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says about linear differential equations.&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-12-11T04:21:36.920" />
  <row Id="13872" PostTypeId="2" ParentId="13870" CreationDate="2010-12-11T04:31:48.447" Score="5" ViewCount="0" Body="&lt;p&gt;Your question was asked and answered &lt;a href=&quot;http://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index&quot; rel=&quot;nofollow&quot;&gt;on MO&lt;/a&gt; before. The examples you give are essentially it, in the sense that the field has to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_closed_field&quot; rel=&quot;nofollow&quot;&gt;real closed&lt;/a&gt; and the algebraic closure is obtained by adding a square root of -1. This is the Artin-Schreier theorem.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-11T04:31:48.447" CommentCount="1" />
  <row Id="13873" PostTypeId="1" AcceptedAnswerId="13875" CreationDate="2010-12-11T04:53:43.507" Score="1" ViewCount="116" Body="&lt;p&gt;I need help with this recursion question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/njBJi.jpg&quot; alt=&quot;help&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4351" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-11T05:26:20.217" LastActivityDate="2010-12-11T15:50:31.047" Title="Homework question dealing with recursion" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;recursive&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="13874" PostTypeId="2" ParentId="12362" CreationDate="2010-12-11T04:54:58.200" Score="1" ViewCount="0" Body="&lt;p&gt;Recall that $\sqrt{4} = \pm 2$. Since $-4 = (4)(-1)$, we have $\sqrt{-4} = \sqrt{4} \sqrt{-1} = \pm 2 i$, where we define $i = \sqrt{-1}$. Note that these are the two roots of the quadratic $x^{2} + 4 = 0$, which has no real roots (positive or negative) by the Descartes Rule of Signs. This last statement rules out the case that the root is $-2$, which is a common mistake and misconception.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-12-11T04:54:58.200" CommentCount="1" />
  <row Id="13875" PostTypeId="2" ParentId="13873" CreationDate="2010-12-11T04:59:58.960" Score="6" ViewCount="0" Body="&lt;p&gt;Don't code it! Instead compute the first 4 or so values by hand and see if you understand what is going on. You don't even have to look at the numbers, you just need to see what you are doing when you evaluate the code. E.g. try to say in words what you do when you follow the code. Pretend that you are explaining the program to a friend: &quot;so this thing first checks whether the second parameter is less than 100...&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't see what you are doing, then do some more iterations (by hand) and look at the numbers, you will easily recognise them. If you don't recognise them, please leave a comment here, telling us what the numbers are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let me repeat: if you want to practice reading code, try to &lt;em&gt;understand&lt;/em&gt; what you are doing, rather than guess a pattern from the numbers. Therefore, &lt;em&gt;do not code it&lt;/em&gt;!&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-11T04:59:58.960" CommentCount="26" />
  <row Id="13876" PostTypeId="1" CreationDate="2010-12-11T05:50:04.670" Score="2" ViewCount="168" Body="&lt;p&gt;If $E$ is Lebesgue measurable in $\mathbb{R}^n$ and $I=[a,b]$ how do I show that $E\times I$ is measurable in $\mathbb{R}^{n+1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jonas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using $\mu^*(E)=\inf \{ \sum \mathrm{Vol}(I_k) \mid E\subseteq \cup I_k\}$ and for every $\epsilon \gt 0$ there exists an open set $G$ containing $E$ such that $\mu^*(G-E)\lt\epsilon$ ($\mu^*$ is the outer measure).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using the first definition since I think it would be easier, but I don't know how to make it fit together.&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-17T00:00:11.400" LastActivityDate="2010-12-17T00:00:11.400" Title="Lebesgue Measure of the Cartesian Product" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="13877" PostTypeId="1" AcceptedAnswerId="13879" CreationDate="2010-12-11T05:58:36.097" Score="2" ViewCount="53" Body="&lt;p&gt;&lt;a href=&quot;http://my.unix-center.net/~martian/math%20album/%E6%95%B0%E5%AD%A6%E4%B8%93%E8%BE%91%202%20-%20%E5%86%B0%E5%B2%9B%E7%81%AB%E5%B1%B1/%E7%9B%B4%E5%B8%83%E7%BD%97%E9%99%80%E7%9A%84%E5%B7%A8%E5%B2%A9%20%28On%20twin%20prime%29.html&quot; rel=&quot;nofollow&quot;&gt;a sequence on twin primes&lt;/a&gt; In the diagram, How do the stripes come from? Can the prime numbers also interfere  like light and wave? When zoom in or zoom out the diagram in Mathematica, the stripes are changing.&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-12-11T06:37:18.643" Title="numbers interference" Tags="&lt;sequences-and-series&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13878" PostTypeId="2" ParentId="13877" CreationDate="2010-12-11T06:32:04.913" Score="3" ViewCount="0" Body="&lt;p&gt;The stripes come mostly because there are more primes among the small numbers than large.  The density goes as about n/log(n).  Also because people are good at seeing patterns, whether they are present or not (that is, distinguishing true patterns from clumps in randomness.)  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-11T06:32:04.913" />
  <row Id="13879" PostTypeId="2" ParentId="13877" CreationDate="2010-12-11T06:37:18.643" Score="3" ViewCount="0" Body="&lt;p&gt;The fact that the stripes change when you change the zoom level supports the idea that this is a graphical problem with your implementation of Mathematica.  Notice that, for instance, every value between 1 and 20 is attained numerous times for n between 1 and 10000, so there should be no &quot;blank stripe&quot; for $0 \le y \le 20$ in your picture.  It is just a graphical anomaly.  This explains the white stripes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stripes that occur as a result of higher concentrations occur where $k$ has a significant number of small prime factors besides 2 and 3, as such a $k$ makes $6pk+1$ and $6pk-1$ more likely to be prime, as they will be relatively prime to a higher density of numbers.  So for instance, you see clumping at $y=35=(5)(7)$, and multiples of 5 generally, to varying degrees.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-12-11T06:37:18.643" />
  <row Id="13880" PostTypeId="1" AcceptedAnswerId="13882" CreationDate="2010-12-11T07:20:21.453" Score="2" ViewCount="161" Body="&lt;p&gt;What does 'general' in 'general topology' really refer to?&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-11T10:54:06.617" Title="What does 'general' in 'general topology' really refer to?" Tags="&lt;general-topology&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13881" PostTypeId="2" ParentId="13876" CreationDate="2010-12-11T07:24:21.627" Score="5" ViewCount="0" Body="&lt;p&gt;An equivalent criterion for measurability of a set $E$ is the existence of a $G_\delta$ set $G$ containing $E$ such that $\mu^*(G\setminus E)=0$.  (If you haven't already seen this, you can prove it.)  You can use this along with the fact that $(G_1\times G_2)\setminus(E_1\times E_2)=((G_1\setminus E_1)\times G_2)\cup(G_1\times(G_2\setminus E_2))$ to show that if $E_1$ is measurable in $\mathbb{R^n}$ and $E_2$ is measurable in $\mathbb{R^m}$, then $E_1\times E_2$ is measurable in $\mathbb{R^{n+m}}$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-11T07:42:02.083" LastActivityDate="2010-12-11T07:42:02.083" />
  <row Id="13882" PostTypeId="2" ParentId="13880" CreationDate="2010-12-11T07:33:40.323" Score="4" ViewCount="0" Body="&lt;p&gt;When I use it I mean to distinguish the subject matter from &quot;specific&quot; applications of topology where there is more structure than just that specified by the axioms for a topological space.  (For example, a metric space, or a manifold or Riemannian manifold, or an algebraic variety, or you name it...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add that in general, investigating terms like these for what they &quot;really refer to&quot; is a little misleading, because it sort of presumes that these terms were consciously designed by somebody (or a group of people) with a specific purpose in mind, instead of arrived at over the years sort of by accident, with no precise meaning--- only a general, vague, descriptive intent--- behind them.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2010-12-11T07:33:40.323" />
  <row Id="13883" PostTypeId="2" ParentId="13880" CreationDate="2010-12-11T07:42:43.327" Score="4" ViewCount="0" Body="&lt;p&gt;To distinguish &lt;a href=&quot;http://en.wikipedia.org/wiki/General_topology&quot; rel=&quot;nofollow&quot;&gt;(general) topology&lt;/a&gt; from more specific branches of topology? -&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_topology&quot; rel=&quot;nofollow&quot;&gt;Differential topology&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_topology&quot; rel=&quot;nofollow&quot;&gt;algebraic topology&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_topology&quot; rel=&quot;nofollow&quot;&gt;geometric topology&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Low-dimensional_topology&quot; rel=&quot;nofollow&quot;&gt;low-dimensional topology&lt;/a&gt;...&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-11T09:39:12.747" LastActivityDate="2010-12-11T09:39:12.747" />
  <row Id="13884" PostTypeId="1" CreationDate="2010-12-11T07:42:51.283" Score="8" ViewCount="494" Body="&lt;p&gt;I've noticed when sweeping the floor that dust gathers particularly in the corners. I assume there is a fluid mechanics reason for this. Does anyone know what it is?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; No, really, this is a mathematical question. Air blows around the room, which constitutes a vector field. Let's say it blows through a square room with doors near bottom-left and top-right. Air blows from bottom-left to top-right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are dust particles in the room, let's say uniformly distributed at first. But after application of this flow they are not uniformly distributed. They pile in the corners. Maybe that's because vortices form more in the corners, maybe some other reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's &lt;em&gt;not&lt;/em&gt; because of where I start sweeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a physics question, but obviously knowledge of Newtonian mechanics won't solve it. It comes down to fluid flow and vector fields, which is math.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-11T08:05:01.997" LastActivityDate="2010-12-14T02:30:34.100" Title="Why does dust gather in corners?" Tags="&lt;physics&gt;&lt;applications&gt;&lt;vector-calculus&gt;" AnswerCount="5" CommentCount="12" FavoriteCount="4" />
  <row Id="13885" PostTypeId="1" CreationDate="2010-12-11T08:22:07.070" Score="4" ViewCount="69" Body="&lt;p&gt;I was thinking about Feynman integrals the other day and in particular about discretizing the paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the lay of the land about what happens when you do path integrals over, say, a lattice or graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry this is not a very specific question.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-12-11T08:22:07.070" Title="Path (Feynman) Integrals over Graphs" Tags="&lt;integral&gt;&lt;graph-theory&gt;&lt;lattices&gt;" CommentCount="3" />
  <row Id="13886" PostTypeId="1" AcceptedAnswerId="13973" CreationDate="2010-12-11T08:24:40.833" Score="4" ViewCount="54" Body="&lt;p&gt;&lt;a href=&quot;http://my.unix-center.net/~martian/math%20album/%E6%95%B0%E5%AD%A6%E4%B8%93%E8%BE%91%202%20-%20%E5%86%B0%E5%B2%9B%E7%81%AB%E5%B1%B1/%E7%9B%B4%E5%B8%83%E7%BD%97%E9%99%80%E7%9A%84%E5%B7%A8%E5%B2%A9%20%28On%20twin%20prime%29.html&quot; rel=&quot;nofollow&quot;&gt;another sequence on twin primes&lt;/a&gt; The maximal prime factor of average of twin prime pair:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 100000;&#xA;averageList = Select[Prime[Range[n]], PrimeQ[# + 2] &amp;amp;] + 1;&#xA;mpfList = FactorInteger[#][[-1, 1]] &amp;amp; /@ averageList;&#xA;ListPlot[%]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this diagram look like a series of trajectories? &#xA;&lt;img src=&quot;http://i.imgur.com/SouNY.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1498" LastActivityDate="2010-12-12T07:39:14.873" Title="trajectories in ListPlot[the maximal prime factor of average of twin prime pair]" Tags="&lt;sequences-and-series&gt;&lt;prime-numbers&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13887" PostTypeId="2" ParentId="2448" CreationDate="2010-12-11T08:50:59.150" Score="8" ViewCount="0" Body="&lt;p&gt;Here is the full question from Halmos's &lt;em&gt;Problems for mathematicians, young and old&lt;/em&gt;:&#xA;&lt;img src=&quot;http://i.imgur.com/6JWk6.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/RhFoy.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Jt39W.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T08:50:59.150" CommentCount="2" />
  <row Id="13888" PostTypeId="1" AcceptedAnswerId="13894" CreationDate="2010-12-11T09:39:01.593" Score="4" ViewCount="255" Body="&lt;p&gt;&lt;strong&gt;The Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \frac{16}{7 \cdot 8 \cdot 9 \cdot 10} + \cdots$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any smarter way to solve this may be within a minute or two?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am adding my solution,If you are a student please don't read-on I find this particular problem to some sort of interesting so just try it once :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started off with trying to find the $T_n$ th term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_n = \frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}$$&#xD;&#xA;$$ = \frac1{24} \times \biggl( \frac1{2n-1} + \frac3{2n+1} - \frac4{2n+2} \biggr)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac1{24} \times \biggl[\biggl( \frac1{2n-1} + \frac1{2n+1}\biggr) + 4 \times \biggl(\frac1{2n+1} - \frac1{2n+2} \biggr) \biggr]$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this this becomes really easy,&#xA;  $$ S_\infty =   \frac1{24} \times \biggl[( 1 - \frac13 + \frac13 -  \frac15 +  \frac15 + \cdots ) + 4 \times (\ln 2 - \frac12)\biggr]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac16 \log_e 2 - \frac1{24}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you can see this approach is a bit tedious and it took me some 30 minutes to reduce the $T_n$ to that form. Any other smart approaches?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-11T11:04:41.360" LastActivityDate="2010-12-17T01:23:33.943" Title="How to sum $\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \cdots$ quickly?" Tags="&lt;sequences-and-series&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="13889" PostTypeId="1" AcceptedAnswerId="13906" CreationDate="2010-12-11T09:46:14.097" Score="8" ViewCount="194" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Are there intuitive ways to introduce cohomology?  Pretend you're talking to a high school student; how could we use pictures and easy (even trivial!) examples to illustrate cohomology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do I care:&lt;/strong&gt; For a number of math kids I know, doing algebraic topology is fine until we get to homology, and then it begins to get a bit hazy: why does all this quotienting out work, why do we make spaces up from other spaces, how do we define attaching maps, etc, etc.  I try to help my peers do basic homological calculations through a sequence of easy examples (much like the ones Hatcher begins with: taking a circle and showing how &quot;filling it in&quot; with a disk will make the &quot;hole&quot; disappear --- ) and then begin talking about what kinds of axioms would be nice to have in a theory like this.  I have attempted to begin studying co-homology through &quot;From Calculus to Cohomology&quot; and Hatcher's text, but I cannot see the &quot;picture&quot; or imagine easy examples of cohomology to start with.  &lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2010-12-11T18:57:55.747" Title="Motivating Cohomology." Tags="&lt;topology&gt;&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13890" PostTypeId="1" CreationDate="2010-12-11T09:57:00.060" Score="4" ViewCount="134" Body="&lt;p&gt;How to prove that in every power of 3, with natural exponent, the tens digit&#xA;is an even number?&#xA;(For example, 3 ^ 5 = 243, 4 is even.)&lt;/p&gt;&#xA;" OwnerUserId="2938" LastEditorUserId="2938" LastEditorDisplayName="" LastEditDate="2010-12-11T10:02:30.727" LastActivityDate="2010-12-12T23:12:10.840" Title="In every power of 3 the tens digit is an even number" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13891" PostTypeId="1" AcceptedAnswerId="13927" CreationDate="2010-12-11T10:02:52.437" Score="5" ViewCount="129" Body="&lt;p&gt;$A$ and $B$ are finite groups.&#xA;$H \leq A \times B$.&#xA;Can we find some $C \leq A$, $D \leq B$ such that $H \cong C \times D$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case the statement is not true: is it true under further assumptions about A and B, such as solvability, nilpotency, etc? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special cases I can prove:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$A$ and $B$ are abelian (following ideas from another discussion: &lt;a href=&quot;http://math.stackexchange.com/questions/10687/g-finite-abelian-a-times-b-embedded-in-g-is-g-c-times-d-such-that-a&quot;&gt;$G$ finite abelian. $A \times B$ embedded in $G$. Is $G=C \times D$ such that $A$ embedded in $C$, $B$ embedded in $D$?&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(|A|,|B|)=1$. In this case we even have $H = C \times D$. By using the Chinese remainder theorem for instance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="3533" LastEditorDisplayName="" LastEditDate="2010-12-11T12:15:31.550" LastActivityDate="2010-12-11T18:42:09.083" Title="Finite groups: $H \leq A \times B$. Is $H \cong C \times D$ for some $C \leq A$, $D \leq B$?" Tags="&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="13892" PostTypeId="2" ParentId="13890" CreationDate="2010-12-11T10:05:41.607" Score="11" ViewCount="0" Body="&lt;p&gt;Use induction on the exponent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last digit is either 1, 3, 7 or 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 1 or 3, multiplying a power of 3 will not change the evenness of the second last digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 7 or 9, multiplying will result in a carry over of 2, which again does not change the evenness of the second last digit.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-11T10:05:41.607" />
  <row Id="13894" PostTypeId="2" ParentId="13888" CreationDate="2010-12-11T10:34:24.020" Score="9" ViewCount="0" Body="&lt;p&gt;Problems like these are usually amenable to partial fraction decompositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my experience with exam problems like these, once we do the partial fractions (which can be done quite quickly as described below), either we can do some sort of a telescoping sum, or write the resulting as a combination of well known series. In some cases, you can quickly and mechanically apply known estimates for well known series and find the sum. Of course, each problem is different and you might have to get creative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us try that approach to your problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step1: Splitting into Partial Fractions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try the following which should give you the partial fraction decomposition pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{n}{2(2n-1)(2n+1)(2n+2)} = \dfrac{A}{2n-1} + \dfrac{B}{2n+1} + \dfrac{C}{2n+2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $\displaystyle 2n-1$ and setting $\displaystyle n=1/2$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle A = \dfrac{1/2}{2 \times 2 \times 3} = \dfrac{1}{24}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the original by $\displaystyle 2n+1$ and set $\displaystyle n=-1/2$. This gives us $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle B = \dfrac{-1/2}{2 \times -2 \times 1} = \dfrac{1}{8}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the original by $\displaystyle 2n+2$ and set $\displaystyle n = -1$. This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle C = \dfrac{-1}{ 2 \times -3 \times -1 } = \dfrac{-1}{6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{n}{2(2n-1)(2n+1)(2n+2)} = \dfrac{1}{24}(\dfrac{1}{2n-1} + \dfrac{3}{2n+1} - \dfrac{4}{2n+2})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this is due to Heaviside: &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_cover-up_method&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Heaviside_cover-up_method&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step2: Summing the series&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to find the sum, you can use the following, which does not require any clever algebraic manipulations and so can be done quite quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_n = \sum_{j=1}^{n} \dfrac{1}{j} = \log n + \gamma + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n} \dfrac{1}{2j - 1} = H_{2n} - \dfrac{1}{2} H_n = \log 2 + \frac{\log n}{2} + \gamma/2 + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your sum to $n$ terms is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{24}(\log 2 + \dfrac{\log n}{2} + \gamma/2 + 3\log 2 + 3 \dfrac{\log (n+1)}{2} - 3 + 3 \gamma/2  - 2\log (n+1)  - 2\gamma +2) + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\log(n+1)  = \log n + \mathcal{O}(\dfrac{1}{n})$ we get that the sum to n terms is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{24}(4 \log 2  -1) + \mathcal{O}(\dfrac{1}{n})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\displaystyle n \to \infty$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the limit is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{1}{24}(4 \log 2 - 1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: As &lt;a href=&quot;http://math.stackexchange.com/users/498/j-m&quot;&gt;J.M&lt;/a&gt; points out in the comments below, a very general method similar to the above can be found in Abramowitz and Stegun's book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The technique you used for summing the&#xA;  series is described in &lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/page_264.htm&quot; rel=&quot;nofollow&quot;&gt;Abramowitz and&#xA;  Stegun&lt;/a&gt;. (Harmonic numbers and&#xA;  digamma functions are trivially&#xA;  related).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note1: As &lt;a href=&quot;http://math.stackexchange.com/users/2370/mike-spivey&quot;&gt;Mike&lt;/a&gt; points, out, there is an updated version of the book available here: &lt;a href=&quot;http://dlmf.nist.gov/&quot; rel=&quot;nofollow&quot;&gt;http://dlmf.nist.gov/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:23:33.943" LastActivityDate="2010-12-17T01:23:33.943" CommentCount="13" />
  <row Id="13895" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T10:37:03.693" Score="0" ViewCount="0" Body="&lt;p&gt;If there is a net airflow in any particular direction at floor-level for any reason (doors/windows/radiators etc.), then the dust will tend to accumulate in the corner that is furthest 'down-stream'.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-11T12:00:33.450" LastActivityDate="2010-12-11T12:00:33.450" CommentCount="1" />
  <row Id="13896" PostTypeId="2" ParentId="13880" CreationDate="2010-12-11T10:54:06.617" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/General_topology&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, general topology or point-set topology is the branch of topology which studies properties of topological spaces and structures defined on them. It is distinct from other branches of topology in that the topological spaces may be very general, and do not have to be at all similar to manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other main branches of topology are algebraic topology, geometric topology, and differential topology. As the name implies, general topology provides the common foundation for these areas.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2010-12-11T10:54:06.617" />
  <row Id="13897" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T12:13:59.460" Score="9" ViewCount="0" Body="&lt;p&gt;Consider the life of a single dust particle that randomly gets kicked up into the air, at which point it floats around for a while before being deposited on the floor again. One (very) simple model of this process might be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the probability of the particle getting kicked up in 1 second increases with the airspeed at it's location, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) once it's in the air, where it falls is random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one would expect the particle to spend more time on the ground where the airspeed is low, and less where the airspeed is high, therefore accumulating in low-speed locations. If this model is legit, the question becomes &quot;why does the air move slower in corners?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Empirically, (at least from walking around my apartment) it seems that the airflow is indeed slower in the corners, so this is a good sanity-check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really do this justice, you'd have to consider the full Navier-stokes equations, but those are hard, so lets consider a simplified model. Suppose the flow is an incompressible, slow, pressure-driven flow with a few sources and sinks (air conditioner, cracks under the door, etc), and no-flow boundary conditions at the walls. In this case, the pressure field should solve the Laplace equation, which we have some intuition about it as some sort of smooth rubber surface trying to be &lt;a href=&quot;http://www.math.hmc.edu/~jacobsen/sirev-flat-as-possible.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;as flat as possible&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The no-flow condition means that the normal derivative of the pressure at a wall is zero. Now consider a corner between one wall along x=0, and another wall along y=0. If we are near enough to the corner, we expect to be within the realm of influence of both walls, so we expect both derivatives dP/dx and dP/dy to be small, since they are zero at the y=0 and x=0 walls respectively. Therefore the pressure gradient is small, so the velocity is small in the corner, as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; a simple particle model predicts dust will accumulate where the air is moving slowly, and a simple pressure-driven flow model predicts flow to be slow near corners. Whether these models are legit or not is worthy of debate, but  they do predict the right behavior in this case.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="3060" LastEditorDisplayName="" LastEditDate="2010-12-11T14:51:39.327" LastActivityDate="2010-12-11T14:51:39.327" />
  <row Id="13898" PostTypeId="1" AcceptedAnswerId="13903" CreationDate="2010-12-11T12:37:56.767" Score="3" ViewCount="234" Body="&lt;p&gt;Here is a problem from Griffith's book Introduction to E&amp;amp;M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\theta(x)$ be the step function&#xA;$$\theta = \begin{cases}&#xD;&#xA;0 &amp;amp; x \leq 0 \\&#xD;&#xA;1 &amp;amp; x &amp;gt; 0 &#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how to prove $\frac{d\theta}{dx} = \delta(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think since the function is discontinous at x = 0, there is no definition of $\frac{d\theta}{dx}$ at the point x = 0 at all. Thus, how could we show the equivalence of a equation if the left part of equation is not defined at the point x = 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2010-12-11T14:49:12.323" LastActivityDate="2011-01-06T16:40:44.250" Title="How to prove $\frac{d\theta}{dx} = \delta(x)$?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13899" PostTypeId="1" AcceptedAnswerId="13932" CreationDate="2010-12-11T12:45:51.953" Score="3" ViewCount="94" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hypercube&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says, A hypercube is an n-dimensional analog of a square/cube. What is a hypercylinder then? An n-dimensional analog of a cylinder?&#xA;Constant Approximation Algorithm for MST in Resource Constrained Wireless&#xA;Sensor Networks gives a definition,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Isolation Property&lt;/strong&gt;. Let c&gt; 0 be a constant. Let E be a set of edges in k−dimensional space, and let e ∈ E be an edge of length l. If it is possible to place a hypercylinder B of radius and height c .l each, such that the axis of B is a subedge of e and B ∩ (E − e)= φ, then e is said to be isolated. If all the edges in E are isolated, then E is said to satisfy the isolation property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does it mean, if I have edge sets E and then take a subset e out E, then if I'm able to place, a some sorts of martian object, B in the so called k space along with an element of e, then B cannot contain any elements from E-e ?! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="1869" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-11T12:49:16.453" LastActivityDate="2010-12-12T02:58:53.483" Title="What is a hypercylinder ?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13900" PostTypeId="2" ParentId="13889" CreationDate="2010-12-11T12:55:26.090" Score="3" ViewCount="0" Body="&lt;p&gt;For simplicial/cellular cohomology, one way to think of it is in terms of dual cell structures: if $a_1,a_2,\dotsc$ are your $k$-simplices or $k$-cells, then they generate the $k$th chain group, and the $k$th cochain group is generated by their duals $\alpha_1,\alpha_2,\dotsc$, where $\alpha_i(a_j)=\delta_{ij}$.  You can then draw a dual cell structure which has an $n-k$-cell for every $k$-cell in your original space, with each cell representing a cochain, and with that cochain sending the chains it intersects to $1$ and the rest to $0$ (and extending linearly).  So if your space is a surface, you put a vertex inside every face, draw an edge between two of those vertices if there's an edge between the corresponding faces, and add a face between a set of edges if the dual edges all intersect in a vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the homology of the new cell structure is the cohomology of the original structure.  With field coefficients on a manifold, at least.  But it does allow you to visualize, for example, the coboundary map: in the case of our surface, it sends $C^1$ to $C^2$, but $C^2$ &quot;looks 0-dimensional&quot;, and so the coboundary map &quot;looks like a boundary map,&quot; which is something your students are hopefully familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you do it with Platonic polyhedra, you get other Platonic polyhedra.  A cube becomes an octahedron, etc.  Of course, they all have the same (co)homology but they have different (co)chain groups so they're nice trivial examples, and more interesting than just an arbitrary sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hatcher goes over this very briefly in the beginning of his chapter on cohomology.  He also gives a thing you can do with $\mathbb{Z}$ coefficients that's similar, though in this case you can run into torsion, so I don't know if it's as good an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing you can do is just take the no-nonsense algebraic tack.  We like cohomology because sometimes we want maps going the wrong direction.  For example, its ring structure is easier to work with than homology's coalgebra structure (if your kids are familiar with homology, at this point you show them the coalgebra structure induced by the diagonal map and how it's a bitch to work with).  And you get so much information for free just by knowing that certain maps are ring homomorphisms rather than graded abelian group homomorphisms.  I can't think of a good example off the top of my head, but I know there is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, I was taught de Rham cohomology before I even knew what homology ways.  I think it's pretty easy to understand.  That's another option.&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2010-12-11T12:55:26.090" CommentCount="3" />
  <row Id="13901" PostTypeId="2" ParentId="13891" CreationDate="2010-12-11T13:51:26.913" Score="1" ViewCount="0" Body="&lt;p&gt;It's not possible if you want the isomorphism to be compatible with the product structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, choose $A=B$ and $H = \lbrace (a,a) : a\in A\rbrace$ the &quot;diagonal subgroup&quot;. Clearly, the subgroup $H$ does &lt;em&gt;not&lt;/em&gt; have the form $H = C\times D \subseteq A\times A$.&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2010-12-11T13:51:26.913" CommentCount="4" />
  <row Id="13902" PostTypeId="1" AcceptedAnswerId="13952" CreationDate="2010-12-11T14:00:30.140" Score="6" ViewCount="78" Body="&lt;p&gt;Hello community,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Fourier transformation of a scalar function with respect to one variable might be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{F}\left[w\right](\omega )\equiv \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}w(t)e^{-\mathrm{i}\omega t}dt$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, this transformation along with its generalization, the Laplace transformation, has a tremendous importance because of its feature to turn linear partial differential equations into algebraic ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that we have a pseudo-Riemannian manifold $\mathcal{M}$ where $\det{g_{\mu\nu}} = -1$ holds like in special and general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering, what would be the generalization of the Fourier transformation of scalar functions or forms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty I have with this question is that a three-dimensional slicing of $\mathcal{M}$ is not unique, so how to take the integral in an invariant form? What will happen to the differntials $dx^{\mu}$, e.g. $dt,dx^i\rightarrow d\omega dx^i$ in some sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight would be well appreciated.&lt;br&gt;&#xA;Thank you in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2010-12-11T20:10:48.430" Title="Does a Fourier transformation on a (pseudo-)Riemannian manifold make sense?" Tags="&lt;differential-geometry&gt;&lt;functional-analysis&gt;&lt;integral-transforms&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="13903" PostTypeId="2" ParentId="13898" CreationDate="2010-12-11T14:10:18.957" Score="8" ViewCount="0" Body="&lt;p&gt;This is a place where physicists and mathematicians would phrase the question differently. A mathematician would say that $d \theta/dx$ is undefined at $0$, and that $\delta$ is not a function. However, as Mariano says, the statement is true &quot;in the sense of distributions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that mean? A distribution is a gadget which takes as input a smooth function $g(x)$, which is zero for $|x|$ sufficiently large, and returns some real number. When we think of an ordinary function as a distribution, that makes we think of $f$ as corresponding to the gadget $F: g \mapsto \int f(x) g(x) dx$. Notice that changing $f$ at a finite number of points leaves $F$ unaltered. From a physical point of view, if $f$ is something like the value of an electric field at a point, we would never know if it had a finite discontinuity at some point, so the gadget $F$ captures everything that is physically measurable about $f$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how can we see differentiation in terms of distributions? By integration by parts, we have $\int f&amp;#39;(x) g(x) dx = - \int f(x) g&amp;#39;(x)$. So, for any distribution $F$, we define the derivative of $F$ to be the gadget $g \mapsto -F(g&amp;#39;)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let $F$ correspond to $\theta$, so $F(g) = \int_{- \infty}^0 g(x) dx$. The Dirac delta distribution is $\delta(g) = g(0)$. I leave it to you to show that $F&amp;#39;(g) = \delta(g)$, with the definitions above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be curious to see how a physicist would answer this question. I suspect that I have acted &lt;a href=&quot;http://thinkexist.com/quotation/mathematicians_are_like_frenchmen-whatever_you/179777.html&quot; rel=&quot;nofollow&quot;&gt;like a Frenchman&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditorDisplayName="" LastEditDate="2011-01-06T16:40:44.250" LastActivityDate="2011-01-06T16:40:44.250" CommentCount="4" />
  <row Id="13904" PostTypeId="1" AcceptedAnswerId="13908" CreationDate="2010-12-11T14:11:41.577" Score="1" ViewCount="40" Body="&lt;p&gt;Is it possible to characterize the distributional derivative as some sort of &quot;best linear approximation&quot; of a distribution (&lt;em&gt;a la&lt;/em&gt; the Fréchet/Gâteaux derivatives), if viewed in the appropriate spaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let $f \in D&amp;#39;$, $\phi \in D$, and let $\psi_x \in D$ be a mollified ramp function so that $\psi_x(x,y,z,...)=x$ on the support of $\phi$. What can we say about the quantity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f(\phi)-f(\phi+s\psi_x)}{s}-\frac{\partial f}{\partial x}(\phi)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\frac{\partial f}{\partial x}$ is the usual distributional derivative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going further, if we relax possibilities for &quot;derivative direction&quot;, so that simply $\psi \in D$, does it even make sense to talk about the quantity &quot;$Df:D'\rightarrow D'$&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Df(\phi)(\psi):=\lim_{s\rightarrow 0} \frac{f(\phi)-f(\phi+s\psi)}{s}$$&#xD;&#xA;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the space of distributions we don't have a norm, but we still have a vector space structure and a notion of convergence, so it seems that something like this is at least plausible.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-11T14:49:35.773" LastActivityDate="2010-12-11T14:59:49.913" Title="Distributional derivative as linear approximation" Tags="&lt;distribution-theory&gt;" AnswerCount="1" />
  <row Id="13905" PostTypeId="1" AcceptedAnswerId="13934" CreationDate="2010-12-11T14:23:03.347" Score="2" ViewCount="106" Body="&lt;p&gt;Does anybody know how to derive &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\det(\mathbf A)\cdot \det(\mathbf D + \mathbf E \cdot \mathbf A^{-1} \cdot \mathbf B) = \det(\mathbf D)\cdot \det(\mathbf A + \mathbf B \cdot \mathbf D^{-1} \cdot \mathbf E)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(most likely using &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem&quot; rel=&quot;nofollow&quot;&gt;Sylvester determinant&lt;/a&gt; theorem), where $\mathbf A$,$\mathbf B$,$\mathbf C$,$\mathbf D$,$\mathbf E$ are non-singular matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: corrected &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE2:&#xA;Thank you all for your answers and suggestions&lt;/p&gt;&#xA;" OwnerUserId="1996" LastEditorUserId="1996" LastEditorDisplayName="" LastEditDate="2010-12-11T18:12:23.710" LastActivityDate="2010-12-19T22:18:11.457" Title="How to derive a certain determinantal identity" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13906" PostTypeId="2" ParentId="13889" CreationDate="2010-12-11T14:31:36.520" Score="5" ViewCount="0" Body="&lt;p&gt;If one operates on an open subset $U$ of Euclidean space $\mathbb{R}^n$, then de Rham cohomology falls out of trying to solve some differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The starting observation is that a locally constant smooth function will be constant on connected components, so the dimension of the vector space of locally constant smooth functions is the number of connected components. This space is the 0th de Rham cohomology group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step, as is the key in a lot of topology, is to start throwing paths all over the place: replace the space $U$ with the paths inside it (not required to be loops in this case). So we view a 1-form $\omega$ as a function $P_{\omega}$ on paths via the integration map which sends a path to the integral of the 1-form $\omega$ along it. And again we look for those functions $P_{\omega}$ which are locally constant, meaning they remain unchanged under small deformations of the path leaving the endpoints fixed. We know that if $\omega = df$ for $f$ a smooth function, then $P_{\omega}$ satisfies this property trivially because it sends a path to the difference of the values of $f$ evaluated at the path's endpoints. We consider two locally constant path functions $P_{\omega_1}, P_{\omega_2}$ the same if their difference is $df$ for some smooth function $f$. This space is the first de Rham cohomology group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. For the $k$th group, we consider locally constant $k$-dimensional integrals modulo trivially constant ones. The precise definition is that $H^k(U)$ is the space of closed $k$-forms modulo exact $k$-forms, where a form is closed if its exterior derivative is zero and it's exact if it's the exterior derivative of a $(k-1)$-form. For instance, the statement that on a space $U$ every closed form is exact now becomes the statement that all the higher cohomology groups are trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finally phrase this in the standard language of homological algebra, you can define de Rham cohomology as the cohomology of the chain complex $(\Omega^k(U), d)$ where $\Omega^k(U)$ is the set of $k$-forms on $U$ and $d$ is the exterior derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that this could be explained to anybody who has studied multivariable calculus. It has the advantage that computing some basic examples of cohomology groups requires one to think only about some calculus, nothing hard to visualize.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-12-11T18:57:55.747" LastActivityDate="2010-12-11T18:57:55.747" />
  <row Id="13907" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T14:41:46.960" Score="3" ViewCount="0" Body="&lt;p&gt;Air flow in a room will change direction from time to time.  (Just the simple act of walking through the room in random directions will shift the air flow.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust particle will be carried in the direction of the air current each time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually the dust particle will hit a wall and can obviously be carried no further in that direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;New air currents, in slightly different directions, will then carry the dust particle in the direction of other walls.  Because the dust particle is next to the wall, it will be difficult to create an air current that will move the dust particle in a direction 180 degrees in the opposite direction; therefore, the air current will have a tendency to move in the direction of one of the two perpendicular walls (assuming a square room for example's sake).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interplay of air currents will continue to push the dust particle back and forth between the two perpendicular walls into it eventually arrives in the corner.  This will happen because the closer the dust particle gets to the corner (as when it is pushed directly against a wall), the surface area available to capture air currents from the general direction of the corner DECREASES, while the sufrace area available to capture air currents from the perpendicular corner INCREASES.  To say another way, the ability to capture an air current that carries the dust particle 180 degrees away from the corner exponentially decreases the closer it gets closer to the corner, while the ability to capture an air current that carries the dust particle 180 degrees towards the corner exponentially increases as it gets closer to the corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EVERYTHING in life is a matter of mathematics.  There is not an organic or inorganic process anywhere in the known universe that can not be illustrated by mathematics (of course, there are many processes we do not yet know HOW to quantify mathematically, but that is a limitation on our part; not within the inherent reach of mathematics to explain it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, that makes for an interesting Saturday morning ... cheers.&lt;/p&gt;&#xA;" OwnerUserId="4639" LastActivityDate="2010-12-11T14:41:46.960" CommentCount="3" />
  <row Id="13908" PostTypeId="2" ParentId="13904" CreationDate="2010-12-11T14:43:17.737" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;(I use below $f$ for the distribution and $\phi$ for the test function...)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi\in D&amp;#39;$ and $f\in D$. Let $\tau_x:D\to D$ be translation, so that $\tau_x(f)(y)=f(y+x)$, and let $\tau^t_x:D&amp;#39;\to D&amp;#39;$ be the transpose of $\tau_x$. Then $\frac{\tau_{h}f-f}h\to f&amp;#39;$ in $D$ when $h\to0$, so $$\langle\frac{\tau^t_h\phi-\phi}h,f\rangle=\langle\phi,\frac{\tau_{h}f-f}h\rangle\to\langle\phi,f&amp;#39;\rangle=\langle \phi&amp;#39;,f\rangle$$ as $h\to 0$. Since the topology in $D&amp;#39;$ is that of pointwise convergence on $D$, it follows that $$\lim_{h\to0}\frac{\tau^t_h\phi-\phi}h=\phi&amp;#39;,$$ which is something along the lines of what you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; Differentialting the actual map $\phi:D\to\mathbb R$ is not going to produce anything interesting, because the map is linear itself.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-11T14:59:49.913" LastActivityDate="2010-12-11T14:59:49.913" CommentCount="1" />
  <row Id="13909" PostTypeId="2" ParentId="13898" CreationDate="2010-12-11T14:54:06.890" Score="3" ViewCount="0" Body="&lt;p&gt;By definition, $\delta(x)$ satisfies $$\int_{-\infty}^\infty f(x)\delta(x) dx=f(0)$$ for any continuous function $f$ on $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of derivative in distribution theory,  $$ \int_{-\infty}^\infty f(x)\theta&amp;#39;(x) dx=-\int_{-\infty}^\infty f&amp;#39;(x) \theta(x) dx$$ for any $C^1$ function $f(x)$ which vanishes outside of a bounded interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$-\int_{-\infty}^\infty f&amp;#39;(x) \theta(x) dx=-\int_0^\infty f&amp;#39;(x) dx= -f(\infty)+f(0)=f(0)=\int_{-\infty}^\infty f(x)\delta(x) dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\theta&amp;#39;(x)=\delta(x)$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-11T14:54:06.890" CommentCount="2" />
  <row Id="13910" PostTypeId="1" AcceptedAnswerId="13913" CreationDate="2010-12-11T15:09:24.527" Score="2" ViewCount="67" Body="&lt;p&gt;These two appear in my module (without any proof):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{r = 1}^{n} C(n-2,r-2) = 2^{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{r = 0}^{n-1} C(n-2,r) = 2^{n-2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first one when when $r=1 \Rightarrow  C(n-2,1-2) =  C(n-2,-1)$ ?!The same thing for $r=n-1$ in the second one,are they valid if yes,please explain how?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-11T17:23:56.513" LastActivityDate="2010-12-11T20:06:58.583" Title="Are these two expressions for sums of binomial coefficients valid?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13911" PostTypeId="2" ParentId="13376" CreationDate="2010-12-11T15:11:24.263" Score="2" ViewCount="0" Body="&lt;p&gt;Is the equator of the Earth a line or a circle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it's a circle of course. But as you know by now, from Willie Wong's answer and from others, it's also a geodesic, which is very much like a straight line: unless you are ready for some serious digging, the equator is the shortest path between any two places on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Portal picture is nothing like the Earth, mind you. But in my kitchen I have a cutting board that's a flat rectangle with a thin rectangular hole cut out near one side, as a handle for carrying. The corners are rounded somewhat. Now an ant walking along the top side, if it passed through the handle, would wind up on the bottom side with its momentum reflected. The same if it walked over an edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an orthographic projection of my cutting board. (By the way I'm also an engineer. I learned drafing the old-school way, with a T-square.) I draw a top view which is a rectangle with a thin vertical rectangular hole near the right side. To the right I place a side view which is just a vertical rectangle with a couple of hidden lines, the thickness of the cutting board. It's not normal &quot;third-angle&quot; to also draw the bottom view, but I will, so it is a rectange with a thin vertical rectangular hole near the left side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ant walking over the right edge would, on the drawing, pass from the top view to the side view to the bottom view with its momentum unchanged. (From a 2-D perspective, the edges are not actually special. The corners are another matter.) If it walked through the handle, it would pass from one rectangle on the top view through an &quot;interspace&quot; - hidden lines on the side view - to the other rectangle on the bottom view, again with its momentum again unchanged. Imagine if we tied a string through around the handle and the right edge of the cutting board. On the drawing, it would be a straight line between the two rectangles passing through the three views. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the area around the handle and right edge of my cutting board is very much like a 2-D version of Portal. 3-D Portal can then be thought of as being played on the hypersurface of a higher dimensional cutting board. (To visualise a higher dimensional cutting board, ponder an ordinary cutting board while enjoying a &quot;herbal cigarette&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="4640" LastEditorUserId="4640" LastEditorDisplayName="" LastEditDate="2010-12-11T19:21:08.693" LastActivityDate="2010-12-11T19:21:08.693" CommentCount="2" />
  <row Id="13912" PostTypeId="1" AcceptedAnswerId="13977" CreationDate="2010-12-11T15:17:16.833" Score="20" ViewCount="470" Body="&lt;p&gt;My algebra background: I've had 2 undergrad semesters of algebra, a reading course in Galois Theory, a graduate course in commutative algebra and one in algebraic geometry, and I've done (most of) MacLane and Birkhoff's Algebra on my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I feel like I still don't have any idea how to do algebra. I do well in my classes and don't have any problem with most exercises in M&amp;amp;B I do. But my process consists pretty much entirely of fiddling around with symbols until I figure out how to apply theorems I know in a completely straightforward way. I'm able to do exercises from a book, but rarely able to prove theorems in the text on my own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a complete contrast to how I feel in topology and analysis (not that I really know any topology or analysis), where I have a halfway decent intuition and really think I know why things are true (and moreover, why anyone should care). I'm able to prove theorems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In topology and analysis, I am able to visualize things pretty directly in a way that I can get insight into how things work. For algebra, I have no picture. I've tried learning about Cayley graphs to visualize groups. I think these are neat, but I have yet to successfully apply any insight from them. I hoped learning about algebraic geometry would help me visualize rings. But the geometry in algebraic geometry is dictated entirely by the algebra. So how can you use geometry to help you with the algebra, when you have to do the algebra first to figure out what the geometry looks like? I don't get it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question I'm trying to get at is: How do I develop some insight or intuition about algebra? I don't really know what form answers might take; maybe a reading suggestion, or just a general way to look at things. Maybe this isn't a good question, but I'm kind of at the end of my rope with this stuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular user on MathOverflow said he fell in love with algebra the first time he saw the axioms for a group. When I first saw the axioms for a group, I spent the next year trying to figure out why the heck anybody cared about groups (and frankly still only know this in a detached and academic way). So it's possible the only answer is: I'm barking up the wrong tree; algebra isn't for me and I should move on to something that comes more naturally. &lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-11T15:19:29.743" LastActivityDate="2010-12-14T07:59:37.877" Title="Intuition in algebra? " Tags="&lt;abstract-algebra&gt;&lt;soft-question&gt;&lt;intuition&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="11" />
  <row Id="13913" PostTypeId="2" ParentId="13910" CreationDate="2010-12-11T15:17:47.063" Score="5" ViewCount="0" Body="&lt;p&gt;Hint: Use the binomial expansion for $(1+x)^{n-2}$ and choose an appropriate value for $x$. Binomial coefficients with a negative value of $r$ are zero. You should see what the LHS ought to be. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-11T15:17:47.063" CommentCount="3" />
  <row Id="13914" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T15:21:30.693" Score="2" ViewCount="0" Body="&lt;p&gt;It seems you have a kind of geometrical intuition. If that's the case, why don't you construct geometrical representations of the algebraic entities you encounter? Have you had any representation theory till now? If not, that might help you figure out why algebra is &quot;important&quot; or why anyone, even outside the field of algebra, cares.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-11T15:21:30.693" />
  <row Id="13915" PostTypeId="1" AcceptedAnswerId="13921" CreationDate="2010-12-11T15:24:59.783" Score="3" ViewCount="96" Body="&lt;p&gt;Let $$a_n=\left(\frac{1+\sqrt{5}}{2}\right)^n.$$&#xD;&#xA;For a real number $r$, denote by $\langle r\rangle$ the fractional part of $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the sequence $$\langle a_n\rangle$$ not equidistributed in $[0,1]$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-11T17:59:33.460" Title="A sequence not equidistributed in [0,1]" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;dynamical-systems&gt;&lt;classical-analysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="13916" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T15:26:24.353" Score="6" ViewCount="0" Body="&lt;p&gt;A lot of intuition about abstract algebra and commutative algebra can be developed by visualizing by means of &lt;a href=&quot;http://en.wikipedia.org/wiki/Commutative_diagram&quot; rel=&quot;nofollow&quot;&gt;commutative diagrams&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-11T15:26:24.353" />
  <row Id="13917" PostTypeId="2" ParentId="309" CreationDate="2010-12-11T15:32:59.133" Score="1" ViewCount="0" Body="&lt;p&gt;In Top (topological speces) take the subspaces of the real line:  $B=[0,1]$ and $A=[0,1]\cup \{3/4\}$ considering the inclusion $A \subset B$ and imbedding map $f: B\to A: x \mapsto x/2$. Then $A$ isnt isomorphic  (i.e. homeomorphic) to $B$&lt;/p&gt;&#xA;" OwnerUserId="4565" LastActivityDate="2010-12-11T15:32:59.133" CommentCount="1" />
  <row Id="13918" PostTypeId="2" ParentId="13873" CreationDate="2010-12-11T15:50:31.047" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Show the sequence of Current values has form $\rm\ 1,\ 1,\ 2,\ \cdots\ a,\ b,\ a+b$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-11T15:50:31.047" />
  <row Id="13919" PostTypeId="2" ParentId="13915" CreationDate="2010-12-11T15:56:03.630" Score="3" ViewCount="0" Body="&lt;p&gt;If you just try it, the fractional part becomes very close to 0 or 1 quickly.  This is because $\phi ^2=\phi +1$.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-11T15:56:03.630" />
  <row Id="13920" PostTypeId="2" ParentId="13884" CreationDate="2010-12-11T15:56:51.087" Score="3" ViewCount="0" Body="&lt;p&gt;First, we must assume a more or less constant airflow in the room, for as everyone knows,in a room that is closed up for a long time with no airflow, dust accumulates evenly throughout the room. The simplest air flow type is random. Like water, or anything tbat flows, air flows through the path of least resistance. For air, the path of least resistance is a straight line. Resistance to flow is proportional to the angular change of direction produced by an obstacle; in this case a corner, which does not neccessarily imply a 90 degree corner. There are two types of corners: inside corners (less than 180 degrees and outside corners (greater than 180 degrees). Outside corners do not gather dust, simply because they do not force the air flow to change its direction. Inside corners, however present a physical barrier to straight-line air flow. The smaller the angle of deflection, the greater the resistance. The air simply avoids the highly flow resistant corner by curving around it. In doing so, it loses some of its energy, slowing its flowrate or speed.  Of course, some of the air particles pass closer to the corner than others. The closer particles have to turn more sharply to avoid the black hole of the corner, thus slowing down more, thus losing more of their ability to hold a dust particle aloft, thus losing more dust particles before returning to the (assumed) random air flow of the room. In a simple rectangular or square room, this action results in dusty corners. This is the simplest and least technical answer I can think of. Such things as  Navier-stokes equations, Laplace equations and pressure derivatives have their place, but are superflous, irrelevant overkill for answering such a simple question. &quot;KISS&quot; always applies. If you want a solution to the dusty corner problem (and seeking the cause of a problem is, after all, the first step in seeking a solution), simply place a contantly running vaccumn inlet in each corner with the (filtered) outlet in the center of the room, directed in all directions, and you will have constantly self-dusting corners. Then, (not so simply) eliminate all dust generators and dust attractors, and replace all objects that have inside corners (or modify them so that they have inside curves instead), and you will have a constantly self-dusting room. However, you should not complain if, upon enterimg this room, you friends declare, &quot;This room really sucks!&quot;. I hope this clears the air (and the corners, of course).&lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2010-12-11T15:56:51.087" CommentCount="3" />
  <row Id="13921" PostTypeId="2" ParentId="13915" CreationDate="2010-12-11T16:07:32.803" Score="9" ViewCount="0" Body="&lt;p&gt;Let $\phi&amp;#39;=(1-\sqrt 5)/2$ denote the Galois conjugate of the golden mean $\phi$. &#xA;Then $\phi^n+\phi&amp;#39;^{n}$ is an integer for every $n\in\mathbb N$, i.e.&#xA;$$\phi^n+\phi&amp;#39;^{n}\equiv 0\ (\mbox{mod}\ 1).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $|\phi&amp;#39;|&amp;lt;1$, so $\phi&amp;#39;^{n}\to 0$. This implies that $\phi^n\to 0\ (\mbox{mod}\ 1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property that the sequence  $\{ \mbox{frac} (x^n)\}$ is not equidistributed is shared by other &lt;a href=&quot;http://mathworld.wolfram.com/PisotNumber.html&quot; rel=&quot;nofollow&quot;&gt;Pisot numbers&lt;/a&gt;. There is quite a lot of research publications devoted to them. &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-11T17:59:33.460" LastActivityDate="2010-12-11T17:59:33.460" />
  <row Id="13922" PostTypeId="1" AcceptedAnswerId="13923" CreationDate="2010-12-11T16:10:58.210" Score="2" ViewCount="103" Body="&lt;p&gt;How to sum up this series :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2C_o + \frac{2^2}{2}C_1 +  \frac{2^3}{3}C_2 + \cdots +  \frac{2^{n+1}}{n+1}C_n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hint that will lead me to the correct solution will be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here $C_i = ^nC_i $&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-11T17:24:23.423" LastActivityDate="2010-12-11T19:07:33.640" Title="How to sum up this series?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="13923" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T16:19:39.607" Score="5" ViewCount="0" Body="&lt;p&gt;Hint: Use binomial expansion for $(1+x)^n$ and integrate once. Then choose an appropriate value for $x$.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-11T16:19:39.607" CommentCount="3" />
  <row Id="13924" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T16:23:32.557" Score="3" ViewCount="0" Body="&lt;p&gt;Let's assume $C_i=\binom ni$. I'll give a solution that is not precalculus level. Consider first the equality&#xA;$$ (1+x)^n=C_0+xC_1+x^2C_2+\dots+x^nC_n. $$&#xD;&#xA;This is the &lt;em&gt;binomial theorem&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate from 0 to t. On the left hand side we get $\frac{(1+t)^{n+1}-1}{n+1}$ and on the right hand side $\sum \frac1{i+1}t^{i+1}C_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now set $t=2$, and a bit of algebra gives you the answer you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty sure there is an elementary approach as well. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T16:23:32.557" CommentCount="3" />
  <row Id="13925" PostTypeId="1" CreationDate="2010-12-11T16:55:39.097" Score="4" ViewCount="210" Body="&lt;p&gt;In other words, is their brain state such that its similar to taking a shower? Or is it always in a state of trying new things which mentally exhausts the brain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when one drives a car he doesn't have to think about all the mechanics of it. Is this true for mathematicians with regards to mathematics?&lt;/p&gt;&#xA;" OwnerUserId="4644" LastActivityDate="2010-12-11T17:51:03.980" ClosedDate="2010-12-11T20:06:17.490" Title="Do mathematicians think hard?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13926" PostTypeId="2" ParentId="2687" CreationDate="2010-12-11T17:25:19.837" Score="0" ViewCount="0" Body="&lt;p&gt;show that a subset of a metric space X is  closed iff its intersection with every compact subset of X is closed&lt;/p&gt;&#xA;" OwnerUserId="4646" LastActivityDate="2010-12-11T17:25:19.837" />
  <row Id="13927" PostTypeId="2" ParentId="13891" CreationDate="2010-12-11T17:28:21.923" Score="7" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;No, a subgroup of a direct product need not be a direct product of subgroups, even up to isomorphism, even if the factors are almost abelian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples are plentiful.  Here is a small nearly-abelian example (nilpotent and solvable and hamiltonian) and a small easy to write down example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take A=B to be quaternion of order 8. A×B has 15 maximal subgroups of order 32, 9 of which are directly indecomposable, and so no such C or D exist.  Hence nilpotent of class at most 2 with all subgroups normal (just barely not-abelian) is not sufficient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take A=B to be non-abelian of order 6. A×B has a maximal subgroup H of order 18 that is directly indecomposable, so again no such C or D exist. H can be generated by (1,2,3), (4,5,6) and (1,2)(4,5).  Its elements of order 2 are all of the form (a,b)(c,d) where a≠b in {1,2,3} and c≠d in {4,5,6}, and so all of them are self-centralizing, and so none of them can be part of a proper direct factor (which is centralized by the entire other non-identity factor).  One of the direct factors has to have even order, and so that one has to be all of H.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-11T18:42:09.083" LastActivityDate="2010-12-11T18:42:09.083" CommentCount="2" />
  <row Id="13928" PostTypeId="2" ParentId="13925" CreationDate="2010-12-11T17:35:11.777" Score="1" ViewCount="0" Body="&lt;p&gt;Well, imagine you are driving on a road with a lot of forks (that you only see after you solve a puzzle), and you have a general idea where you're driving at, but only the vaguest idea of how do you get there. That's your brain state, I think. Sometimes the road itself is rough, too, with an occasional pit.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-11T17:51:03.980" LastActivityDate="2010-12-11T17:51:03.980" CommunityOwnedDate="2010-12-11T17:35:32.150" />
  <row Id="13929" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T17:41:51.173" Score="6" ViewCount="0" Body="&lt;p&gt;For an elementary method which does not use calculus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that $\displaystyle \dfrac{{n \choose k}}{k+1} = \dfrac{1}{n+1} {n+1 \choose k+1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your sum is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{k=0}^{n} \dfrac{1}{n+1} {n+1 \choose k+1} 2^{k+1} = \dfrac{\sum_{k=0}^{n+1} {n+1 \choose k} 2^k -1}{n+1} = \dfrac{3^{n+1} - 1}{n+1}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-11T17:41:51.173" CommentCount="4" />
  <row Id="13930" PostTypeId="2" ParentId="1132" CreationDate="2010-12-11T17:47:37.827" Score="1" ViewCount="0" Body="&lt;p&gt;In the following, $W_t$ is a Wiener process with increments having mean zero and standard deviation 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the Stratonovich formula does the following in terms of the Itô formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)+f(W_t)}{2}dW_t \; ,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)-f(W_t)}{2}dW_t + f(W_t)dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Taylor expansion on the first term and Itô calculus rules, this can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(W_t) \circ dW_t = \frac{f&amp;#39;(W_t)}{2}dt + f(W_t)dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can replace $f$ with $f&amp;#39;$ in the formula to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(W_t) \circ dW_t = \frac{f&amp;#39;&amp;#39;(W_t)}{2}dt + f&amp;#39;(W_t)dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can easily check that the right hand side is nothing but $df(W_t)$ according to Itô calculus rules, therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$df(W_t) = f&amp;#39;(W_t) \circ dW_t \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is just a special case of the chain rule. I guess from here on out you can generalize the argument.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-11T17:47:37.827" />
  <row Id="13931" PostTypeId="2" ParentId="13905" CreationDate="2010-12-11T17:48:34.737" Score="5" ViewCount="0" Body="&lt;p&gt;A neat way to prove this is by observing that these are two ways to write the determinant of the block matrix&#xA;$$ \pmatrix{ A &amp;amp; B \\ -E &amp;amp; D } $$&#xD;&#xA;(see &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant#Block_matrices&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Determinant#Block_matrices&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The proof using Sylvester's determinant theorem is done by simply plugging in the right &quot;pairs&quot; of matrices into the right spots.&#xA;That is, given the theorem as&#xA;$$ \det(I_p + XY ) = \det(I_n + YX) $$&#xD;&#xA;Just let $X = EA^{-1}$ and $Y = BD^{-1}$. From there I think you can figure it out on your own.&lt;/p&gt;&#xA;" OwnerUserId="3638" LastEditorUserId="3638" LastEditorDisplayName="" LastEditDate="2010-12-19T22:18:11.457" LastActivityDate="2010-12-19T22:18:11.457" CommentCount="2" />
  <row Id="13932" PostTypeId="2" ParentId="13899" CreationDate="2010-12-11T17:51:15.633" Score="2" ViewCount="0" Body="&lt;p&gt;First, in n dimensional geometry, it is common to define an object using it's features in 2 or 3 dimensions (i.e., by writing an n dimensional definition, which,when restricted to the case of $n=2$ or $n=3$ is constant with the 2 and 3 dimensional objects we know).  In the case of a cylinder, this object, in 3D is a circular disk or radius $r$, lying on a plane, that has been extended by a distance, $h$, along a vertical, perpendicular axis.  In n dimensions, we proceed as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an n dimensional plane $a_1x_1 +a_2x_2+...+a_nx_n=c$ (where the x's are variables and all the other letters are constants).  Choose a point,$t$ on this plane to be the centre of the base.  The base will be all those points on the plane whose distance from $t$ is less than or equal to $r$ (where distance is measured using the Euclidean metric).  From linear algebra, we can construct a vector, $v$ which is perpendicular to this plane.  I will refer to the &quot;top&quot; side of the plane as the side in the direction of $v$.  The cylinder is all those $x\in{R^n}$ that are &quot;above&quot; the plane, whose projection onto the plane, along $v$ is a point on the base, and the minimum distance from $x$ to a point on the base is less than or equal to $h$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For you second question, the &quot;Isolation Property&quot; seems like a topological definition.  It sounds very similar to the following:&#xA; &quot;For every $e\in E$ there exists an open neighborhood of $e$ not intersecting any other element of $E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an n dimensional way of saying that each edge has a little bit of space around it on all sides (while this definition is clear in 3 dimensions, you need a rigorous description in higher dimensions).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-12T02:58:53.483" LastActivityDate="2010-12-12T02:58:53.483" CommentCount="3" />
  <row Id="13933" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T17:57:45.880" Score="1" ViewCount="0" Body="&lt;p&gt;You could try looking at axiomatic planar geometry for a new perspective on rings and fields. Given enough axioms of abstract planar geometry, one can construct a model of that geometry as $R^2$ for some ring $R$ (it's the same thing as happens when we construct vectors out of the affine plane). It turns out that various geometric theorems are then equivalent to algebraic statements about the ring $R$, like that it be commutative, or finite, or a field, etc. For example, if I remember correctly, then Desargue's theorem holds iff the ring $R$ is a division ring. I don't have references for this at hand, but I can look some up at work next week.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-12-11T17:57:45.880" />
  <row Id="13934" PostTypeId="2" ParentId="13905" CreationDate="2010-12-11T18:02:33.177" Score="7" ViewCount="0" Body="&lt;p&gt;$$\begin{align}&#xD;&#xA;\det(A)\cdot\det(D+E\cdot A^{-1}\cdot B) &amp;amp; = \det(A)\cdot\det(D)\cdot\det(1+D^{-1}\cdot E\cdot A^{-1}\cdot B) \\&#xD;&#xA;&amp;amp; = \det(A)\cdot\det(D)\cdot\det(B)\cdot\det(B^{-1}+D^{-1}\cdot E\cdot A^{-1}) \\&#xD;&#xA;&amp;amp; = \det(A)\cdot\det(D)\cdot\det(1+B\cdot D^{-1}\cdot E\cdot A^{-1}) \\&#xD;&#xA;&amp;amp; = \det(D)\cdot\det(A+B\cdot D^{-1}\cdot E)&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-11T18:02:33.177" />
  <row Id="13935" PostTypeId="1" AcceptedAnswerId="13966" CreationDate="2010-12-11T18:03:24.703" Score="4" ViewCount="93" Body="&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What conclusions can be drawn about the relations between two objects with the same group of endomorphism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we tell from End(A) if A is Abelian or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does End(A) contain information about the sub-objects of A?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any information or references to information about this is highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4643" LastActivityDate="2010-12-11T22:24:20.320" Title="What does the endomorphism group of an object tell us about the object in question?" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" />
  <row Id="13936" PostTypeId="2" ParentId="13935" CreationDate="2010-12-11T18:13:24.643" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a partial answer (although it uses the ring structure on the set of endomorphisms). A torsion abelian group is cyclic if and only if any two elements in $End(A)$ commute with respect to composition (in other words, $End(A)$ is a commutative ring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/m72222448q6j7327/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/m72222448q6j7327/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a whole book on this subject: &lt;a href=&quot;http://books.google.com/books?id=iy4sVSgzrvYC&amp;amp;printsec=frontcover&amp;amp;dq=endomorphism+rings+of+abelian+groups&quot; rel=&quot;nofollow&quot;&gt;Endomorphism rings of abelian groups&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-11T18:31:55.493" LastActivityDate="2010-12-11T18:31:55.493" CommentCount="3" />
  <row Id="13937" PostTypeId="1" AcceptedAnswerId="14319" CreationDate="2010-12-11T18:21:07.583" Score="2" ViewCount="215" Body="&lt;p&gt;Prove that $$\sum_{\substack{x,y \in \mathbb N \\ ax-by \ne 0 }}\frac1{|ax-by|xy}$$ converges, where $a, b \in \mathbb N$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 2-D problem can be proved by an integral test. I'm looking for some other proofs that can be easily generalized to higher dimensional cases like below,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$\sum_{\substack{x_1,x_2, \cdots, x_d \in \mathbb N \\ \bf a \cdot \bf x \ne 0 }}\frac1{|{\bf a} \cdot {\bf x}|x_1 x_2 \cdots x_d}$$ converges, with ${\bf a} \in \mathbb Z^d$, and $\forall \; 1 \le i \le d, a_i \ne 0$.&lt;/p&gt;&#xA;" OwnerUserId="4649" LastEditorUserId="4649" LastEditorDisplayName="" LastEditDate="2010-12-13T14:03:34.303" LastActivityDate="2010-12-15T15:09:53.383" Title="Prove that $\sum_{\substack{x,y \in N \\ ax-by \ne 0 }}\frac{1}{|ax-by|xy}$ (and its generalization to higher dimensions) converges." Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="13938" PostTypeId="2" ParentId="13888" CreationDate="2010-12-11T18:29:56.433" Score="2" ViewCount="0" Body="&lt;p&gt;Again this is trivial employing &lt;strong&gt;telescopy&lt;/strong&gt;. After partial fraction decomposition the &#xA;summand has the form $\rm\ a_1(n-1) - a_1(n) + c\ (a_1(n)+a_0(n))\ $ where $\rm\ a_0(n),\ a_1(n)\ $ are the coefficients of the even, odd part of the  $\rm log\ x\ $ power series, for $\rm\ x = 2\:.\:$ The sum of the first part $\rm\ a_1(n-1) - a_1(n)\ $ telescopes to $\rm\:a(0)\:$ and the sum of second part combines all the odd and even terms into the complete power series for $\rm\: log\ 2\:.\ $  Notice how very simple this approach is. In particular it does not require any knowledge of asymptotics of harmonic series  - as in the approach in Moron's answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally no ingenuity is required to find such telescopy. It can be done  mechanically as follows.  Let $\rm\:S\:$ be the shift $\rm\: S\ a(n)\: \to\: a(n+1)\:,\: $ and let $\rm\ P(S)\ $ be a polynomial in $\rm\:S\:$ whose coefficients $\rm\:c\:$ are constant w.r.t. $\rm\:n\:,\ $ i.e. $\rm\ S\ c = c\:.\: $ Then $\rm\ \sum\ P(S)\ a(n)\ =\ \sum\ (P(S)-P(1))\ a(n)\ +\ P(1)\: \sum\: a(n)\:.\ $ The first sum telescopes since the Factor Theorem $\rm\: \Rightarrow\ S-1 \:|\: P(S)-P(1)\:.\ $ Therefore the sum computation reduces to that of the simpler $\rm\ \sum\: a(n)\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-11T18:29:56.433" CommentCount="0" />
  <row Id="13940" PostTypeId="2" ParentId="13891" CreationDate="2010-12-11T18:40:09.697" Score="3" ViewCount="0" Body="&lt;p&gt;You might be interested in the expository article &quot;&lt;a href=&quot;http://books.google.com/books?id=J-tyGidCI-UC&amp;amp;lpg=PP1&amp;amp;pg=PA1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Subgroups of direct products of groups, ideals and subrings of direct products of rings, and Goursat's lemma&lt;/a&gt;&quot; by Anderson and Camillo.  A couple of excerpts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vE50x.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/GIVMC.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T18:40:09.697" CommentCount="1" />
  <row Id="13941" PostTypeId="1" AcceptedAnswerId="13962" CreationDate="2010-12-11T18:58:39.313" Score="3" ViewCount="35" Body="&lt;p&gt;I am reading Andrew Baker's notes on p-adic number theory. I have a question regarding a result he is using. In Lemma 2.19 he says in the third line the &#xA;limit&#xA;$$\lim_{n\to\infty}N(a_n-b)$$&#xD;&#xA;is positive, where $\{a_n\}$ is Cauchy, $N$ is a non-archimedean norm, and $b$ is not the limit of $a_n$ in the $N$-norm. I am not quite sure why this is true. The notes are obtainable &lt;a href=&quot;http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T21:36:24.827" LastActivityDate="2010-12-11T21:38:37.213" Title="If ${a_n}$ is Cauchy and $N$ is a nonarchimedean norm, why is the limit of $N(a_n-b)$ positive." Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13942" PostTypeId="2" ParentId="13922" CreationDate="2010-12-11T19:07:33.640" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;REMARK&lt;/strong&gt; $\ $ The various approaches are all equivalent. Namely, suppose that we desire to prove without calculus the identity arising from integrating the binomial formula, viz.  $$\rm (1 + x)^{n+1}\ =\ 1 + \sum_{k=1}^{n+1}\: \frac{n+1}{k+1} {n\choose k}\ x^{k+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing coefficients reduces it to the identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \quad\quad\  {n+1 \choose k+1}\ =\ \frac{n+1}{k+1} {n\choose k} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is precisely the identity employed in Moron's &quot;calculus free&quot; approach.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-11T19:07:33.640" CommentCount="6" />
  <row Id="13943" PostTypeId="2" ParentId="13543" CreationDate="2010-12-11T19:16:25.657" Score="0" ViewCount="0" Body="&lt;p&gt;The revised question has been answered in this &lt;a href=&quot;http://mathoverflow.net/questions/48761/continuous-or-analytic-functions-with-this-property-of-sinc-function&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt; at MO. In particular, it was shown that $\frac{\sin ax}{ax}$ satisfies this equality for each $0&amp;lt;a\le \pi$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-14T02:24:55.977" LastActivityDate="2010-12-14T02:24:55.977" />
  <row Id="13944" PostTypeId="2" ParentId="9999" CreationDate="2010-12-11T19:19:22.690" Score="2" ViewCount="0" Body="&lt;p&gt;I can only answer a basic part of it. Namely that Fibonacci numbers have indeed this property. This can be shown by using Lucas numbers. Lucas numbers have the same &#xA;recurrence relation like Fibonacci numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{n+1} = L_n + L_{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But their initial conditions are $L_0 = 2$ and $L_1 = 1$. As result we get a closed form for Lucas numbers similar to Binet's formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_n = \alpha^n + \beta^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_n = {\alpha^n - \beta^n \over \alpha - \beta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\alpha$, $\beta$ are the roots of $x^2 - x - 1 = 0$. Now oberve by some basic polynom arithmetic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\alpha - \beta)^2 F_n^2 + 4(\alpha^n \beta^n) = L_n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $(\alpha - \beta) = \sqrt{5}$ and $\alpha \beta = -1$ we are done. But &#xA;I am stuck with the question whether only Fibonacci numbers have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credits: J. H. E. Cohn. &quot;On Square Fibonacci Numbers.&quot; J. London Math Soc. 39 (1964):537-40.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-26T23:51:24.940" LastActivityDate="2010-12-26T23:51:24.940" CommentCount="4" />
  <row Id="13945" PostTypeId="1" AcceptedAnswerId="13950" CreationDate="2010-12-11T19:19:26.350" Score="2" ViewCount="52" Body="&lt;p&gt;I saw to prove that a^2+b^2=c^2 you can produce the algebraic solutions a = x^2-y^2, b = 2xy, c = x^2+y^2. To check that works just multiply it out but it produces infinitely many solutions when you set x and y to be integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please show me more examples of this, especially if they show how to find the polynomials!&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-11T20:08:10.757" Title="Examples of finding algebraic solutions to diophantine equations." Tags="&lt;algebra&gt;&lt;diophantine-equations&gt;" AnswerCount="2" />
  <row Id="13946" PostTypeId="2" ParentId="13890" CreationDate="2010-12-11T19:38:22.357" Score="5" ViewCount="0" Body="&lt;p&gt;Moron's answer is very elegant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an easy but tedious way of dealing with many similar questions: For any $n$ and $k$, the last $k$ digits of the numbers in the sequence $n^1,n^2,n^3,\dots$ eventually repeat, so we only need to check a few cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question at hand, note that $3^2=9=10-1$, so $3^{20}=(10-1)^{10}$ has the form $100k+1$ for some $k$. This means that the last 2 digits of powers of 3 repeat each 20 numbers. So we just need to check the last two digits of $3^0,3^1,\dots,3^{19}$, and this can be done very quickly even by hand: $$01, 03, 09, 27, 81, 43, 29, 87,\dots,89,67.$$&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T19:38:22.357" />
  <row Id="13947" PostTypeId="1" AcceptedAnswerId="13960" CreationDate="2010-12-11T19:53:28.010" Score="2" ViewCount="65" Body="&lt;p&gt;Let $L \subseteq M$ be A-modules and let $\mathfrak{p}$ be an associated prime of $M/L$ and assume that the annihilator of $L$ is not contained in $\mathfrak{p}$. How would one go about proving that $\mathfrak{p}$ is an associated prime of $M$?&lt;/p&gt;&#xA;" OwnerUserId="3620" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T21:07:08.420" LastActivityDate="2010-12-11T21:19:02.183" Title="Proving that $\mathfrak{p} \in Ass(M/L)$ and $Ann(L) \nsubseteq\mathfrak{p}$ implies $\mathfrak{p} \in Ass(M)$" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" />
  <row Id="13948" PostTypeId="2" ParentId="13945" CreationDate="2010-12-11T20:00:46.320" Score="2" ViewCount="0" Body="&lt;p&gt;This site: &lt;a href=&quot;http://sites.google.com/site/tpiezas/Home&quot; rel=&quot;nofollow&quot;&gt;A Collection of Algebraic Identities&lt;/a&gt;, seems to have a very good list of such solutions. Unfortunately, it does not seem to have any derivations, but it does give the name of the author in most cases (and sometimes, pdf files), so you should be able to track down the derivation if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a general way to find such solutions, you are out of luck, I think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's tenth problem&lt;/a&gt; has been answered in the negative: there is no algorithm which determines if a solution to a general Diophantine equation exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-11T20:00:46.320" />
  <row Id="13949" PostTypeId="1" AcceptedAnswerId="13969" CreationDate="2010-12-11T20:02:42.387" Score="0" ViewCount="147" Body="&lt;p&gt;If one case is true but the other false, does that mean that the statement is true or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists a positive integer $n$ such that $n^2-79n+1601$ is composite&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Case 1 - $n$ is even - false.&lt;/li&gt;&#xA;&lt;li&gt;Case 2 - $n$ is odd  - true. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3691" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T20:39:02.373" LastActivityDate="2010-12-11T22:53:10.060" Title="Proof by cases..." Tags="&lt;proof&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="13950" PostTypeId="2" ParentId="13945" CreationDate="2010-12-11T20:03:08.173" Score="4" ViewCount="0" Body="&lt;p&gt;Read the section &quot;Rational points on conics&quot; in the motivational chapter in the very beautiful book &lt;a href=&quot;http://books.google.com/books?id=mAJei2-JcE4C&amp;amp;pg=PA9&quot; rel=&quot;nofollow&quot;&gt;Rational points on elliptic curves&lt;/a&gt; by Silverman and Tate and/or google &quot;rational parametrization of the circle&quot;. Also, once you have some knowledge of elliptic curves don't miss Franz Lemmermeyer's &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~franz/publ/conics.pdf&quot; rel=&quot;nofollow&quot;&gt;Conics - a poor man's ellitpic curves&lt;/a&gt; which gives a beautiful treatment of Pell conics unified with theory of elliptic curves. Previously fragments of this theory were folklore scattered in pieces throughout the literature. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-11T20:08:10.757" LastActivityDate="2010-12-11T20:08:10.757" />
  <row Id="13951" PostTypeId="2" ParentId="13910" CreationDate="2010-12-11T20:06:58.583" Score="3" ViewCount="0" Body="&lt;p&gt;In addition to the calculation implied by Timothy's answer, let me describe a way to see it directly.  Consider $\sum_{r=0}^{n-1}C(n-2,r)$.  This is the sum of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the number of ways of choosing a subset of $0$ elements out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+ (the number of ways of choosing a subset of $1$ element out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+ (the number of ways of choosing a subset of $2$ elements out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;. . . &#xA;+ (the number of ways of choosing a subset of $n-1$ elements out of a set of $n-2$ elements)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any subset of $n-2$ elements has either 0, 1, 2, . . . , or $n-2$ elements, the total is just the number of ways of choosing &lt;em&gt;any&lt;/em&gt; subset of $n-2$ elements.  And this is just $2^{n-2}$.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-12-11T20:06:58.583" />
  <row Id="13952" PostTypeId="2" ParentId="13902" CreationDate="2010-12-11T20:10:48.430" Score="8" ViewCount="0" Body="&lt;p&gt;There are three main ways of interpreting the Fourier transform. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Decomposition relative to eigenfunctions of the Laplacian&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;On $\mathbb{R}^n$, the plane waves $E_\xi(x) = \exp( i \xi\cdot x)$ can be interpreted as generalized eigenfunctions of the Laplacian. That is, let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \triangle = \sum_{i=1}^n \left(\frac{\partial}{\partial x_i}\right)^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \triangle E_\xi(x) = -|\xi|^2 E_\xi(x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I say &quot;generalized eigenfunctions&quot;. This is because usually it is preferred to do linear algebra/functional analysis on a complete inner product space (a Hilbert space), and in the case of functions on Euclidean space, it is usually preferred to work over the Hilbert space of square integrable functions. The functions $E_\xi(x)$, however, do not belong in the Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But roughly speaking, the Fourier representation of a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x) = \int_{\mathbb{R}^n} \hat{f}(\xi) E_\xi(x) d\xi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be thought of as a decomposition of $f(x)$ in terms of a linear combination of eigenfunctions of the Laplacian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form, as Raskolnikov mentioned in the comment, there is a natural generalization to &lt;strong&gt;compact Riemannian manifolds&lt;/strong&gt;. Let $(M,g)$ be a compact Riemannian manifold, and let $\triangle_g$ denote the associated Laplace-Beltrami operator. One has that $\triangle_g$ is a densely defined, unbounded, self-adjoint operator on $L^2(M)$, the space of square integrable functions on the manifold $M$. As it turns out (using a bit of linear partial differential equations theory and a bit of functional analysis), $\triangle_g$ is invertible if we restrict the co-domain to functions of mean 0 on $M$. The operator $\triangle_g^{-1}$ furthermore turns out to be a compact operator, and thus has discrete eigenvalues with finite dimensional eigen-spaces in $L^2$ by the spectral theorem for compact operators. It is clear by definition that the eigenfunctions of $\triangle_g^{-1}$ will be also eigenfunctions for $\triangle_g$ (just change the eigenvalue $\lambda \to \lambda^{-1}$. Using the spectral theorem we can write an arbitrary square integrable function on $M$ as a unique linear combination of the set of eigenfunctions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one interpretation of the Fourier transform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can further refine this theory in two directions. For functions (and to a small extent tensors), instead of explicitly decomposing using an eigenbasis of $L^2$, you can study &quot;smooth frequency cut-offs&quot; by considering the heat-flow on the manifold. That is, given a function $f$, consider the solution to the heat equation $u(t,x)$ defined on $\mathbb{R}_+ \times M$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_t u = \triangle_g u, \quad u(0,x) = f(x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and study, the cut-off of $f$ by a bump-function $\zeta$ in frequencies in the following sense: take a non-negative function $\zeta\in \C^\infty_0(\mathbb{R}_+)$ and consider the one-parameter family of functions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_\lambda(x) = \int_0^\infty \zeta(\lambda t) u(t,x) dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;roughly speaking $f_\lambda(x)$ here corresponds to, in the Euclidean space case, the restriction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\frac12 \lambda &amp;lt;|\xi|&amp;lt; 2\lambda} \hat{f}(\xi) E_\xi(x) d\xi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the function to the spatial frequencies that are close, in absolute value, to $\lambda$. This theory is developed more in Eli Stein's book &quot;Topics in Harmonic Analysis related to Littlewood-Paley Theory&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another generalization of this decomposition is applicable to not just functions, but also differential forms. In this generalization, instead of looking at the Laplace-Beltrami operator, we consider the Hodge-Laplace operator defined for differential forms. As it turns out, essentially the same argument as for the scalar case give rise to a decomposition of square-integrable differential forms into three categories, the harmonic forms, the exact forms, and the co-exact forms. This goes under the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_theory&quot; rel=&quot;nofollow&quot;&gt;Hodge theory&lt;/a&gt;, and I refer you to Wikipedia, and also S. Agmon's &quot;Lectures on Elliptic Boundary Value Problems&quot;, among many other very good books in this subject. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fourier transform and representation theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another way of treating the Fourier transform on Euclidean space is from the point of view of Euclidean space as an abelian Lie group (in particular a locally compact topological group) which is &quot;self-dual&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very roughly speaking, for an abelian locally compact topological group $G$, you can consider the set of group homomorphisms from $G\to T$, where $T$ is the one-dimensional circle group. In the case $G = \mathbb{R}^n$, we see that each of the functions $E_\xi(x)$ is one such group homomorphism. This set of group homomorphisms can be made into another topological group, known as the dual group of $G$. In the case of Euclidean space, its dual group has the same structure as another copy of the Euclidean space, hence we say it is self-dual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given an invariant measure $\mu$ on $G$, in the sense that group translations in $G$ leaves the measure the same (I am being very rough here and sweeping a lot of stuff under the rug), it turns out that we can define the analogue of the Fourier transform by the same formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_G f(x) E_{-\xi}(x) d\mu(x) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\xi$ is a natural parameter of the dual group of $G$. So in particular, this theory is applicable to the case when we start with a Riemannian manifold $(M,g)$ that is either a Lie group or a symmetric space. This notion is widely applicable in representation theory and analytic number theory, among other things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Pontryagin_duality&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;, Walter Rudin's &quot;Fourier analysis on groups&quot;, S. Helgason's &quot;Differential geometry, Lie groups, and symmetric spaces&quot;, and the article by Atle Selberg called &quot;Harmonic analysis and discontinuous groups in weakly symmetric riemannian spaces, with applications to Dirichlet series&quot; for some standard references. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Symmetries on the manifold and partial Fourier transforms&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A much more pedestrian way of getting the Fourier transform for functions on manifolds is to just integrate using the usual formula. But as you noted in your question, depending on how you &quot;slice&quot; the manifold, you can get different representations, and whether those representations are meaningful can be questionable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, in one special case we can have a meaningful formula. Let $(M,g)$ be a (pseudo)Riemannian manifold (note that this is the only case I am familiar with that has a natural extension to the case of pseudo-Riemannian manifolds), and assume that it admits a continuous symmetry. That is, there exists a nowhere vanishing vector field $T$ such that the Lie derivative $\mathcal{L}_Tg = 0$; in other words $T$ is a Killing vector. Assume further that the integral curves of $T$ are diffeomorphic to $\mathbb{R}$ (so in particular $T$ does not have closed orbits; you can also similarly assume all integral curves are closed, so $T$ does not have open orbits, in this latter case you will deal with Fourier series instead of Fourier transform, I'll leave the obvious changes necessary as an exercise). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the action of $T$ defines an equivalence relation on $M$, and we can consider the quotient manifold $Q = M / T$. (Here's where we used the condition on the topology of the orbits of $T$. If $T$ has fixed points of orbits of different topologies, $Q$ will in general be a manifold with boundary or an orbifold.) The quotient manifold also inherits a Riemannian structure, but that is not too important right now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is the following, we can write a point $x\in M$ as $(t,q)\in \mathbb{R}\times Q$, where $t$ is a natural coordinate on $T$ so that $\partial_t$ corresponds to $\mathcal{L}_T$ in $M$. Using this splitting, you can take the Fourier transform in $t$ the usual way, writing $f(x) = f(t,q)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\tau,q) = \int f(t,q) \exp( - it\tau) dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $T$ is a Killing vector is used here so that if you pick a coordinate system adapted to the coordinate $t$, the actual volume form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{|g|} dt dq $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will be independent of $t$, so when you try to do analysis with this notion of the Fourier transform, the formula above will not look like it is missing a factor coming from the volume form (since anything independent of $t$ can be factored out, and taken outside of the integral, so the worst-case scenario is the entire expression is off be a fixed factor depending on $q$, which can always be inserted back later). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula can also be easily generalized to the case where $(M,g)$ admits several mutually commuting Killing vector fields $T_1, T_2, \ldots, T_k$. In the case of $\mathbb{R}^n$, the standard coordinate vector fields form a complete set of $n$ mutually commuting Killing vector fields that span the tangent space at every point, and this allows us to take the Fourier transform &quot;in all directions&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last formulation is useful especially in studying linear partial differential equations on a space-time that is stationary (for example, a stationary black-hole background). See the papers of Dafermos and Rodnianski for example (&lt;a href=&quot;http://arxiv.org/abs/1010.5132&quot; rel=&quot;nofollow&quot;&gt;paper 1&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/1010.5137&quot; rel=&quot;nofollow&quot;&gt;paper 2&lt;/a&gt; ).&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-11T20:10:48.430" CommentCount="6" />
  <row Id="13953" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:15:12.527" Score="4" ViewCount="0" Body="&lt;p&gt;The statement you give in your example doesn't really have cases.  It is true if some such integer exists, and false otherwise.  I don't know what your cases are supposed to mean, because there are both odd and even examples showing that the statement is true.  However, you only need to give one example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for actually proving this, a hint is that you can look for a value of $n$ such that the 3 terms in your expression have a common factor.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-11T20:20:43.470" LastActivityDate="2010-12-11T20:20:43.470" />
  <row Id="13954" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:15:54.210" Score="4" ViewCount="0" Body="&lt;p&gt;The statement is that there exists &lt;strong&gt;some&lt;/strong&gt; positive integer $n$ such that $n^2 - 79n + 1601$ is composite.  You need to find only one $n$ for this to be a true statement.  To prove the statement is false, you would need to show that $n^2 - 79n + 1601$ is prime for all $n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out the statement is true, but not for all odd $n$.  Take $n = 1$, for example.  $1 - 79 + 1601 = 1523$, which is prime.  I'll leave it to you to find $n$ such that the expression is composite.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-11T20:15:54.210" CommentCount="6" />
  <row Id="13955" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:34:07.060" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The polynomial takes many successive prime values because it is &lt;a href=&quot;http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html&quot; rel=&quot;nofollow&quot;&gt;Euler's famous prime generating polynomial&lt;/a&gt; $\rm\ x^2 + x + 41\ $ shifted by $\rm\ x\to x - c\ $ for $\rm\ c = \ldots$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-11T20:39:27.270" LastActivityDate="2010-12-11T20:39:27.270" />
  <row Id="13956" PostTypeId="1" AcceptedAnswerId="13957" CreationDate="2010-12-11T20:40:39.543" Score="4" ViewCount="74" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the domain of the Gamma function, that is to say all $z \in \mathbb{C}$, for which the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \mathrm dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;converges. I started by splitting up the integral into an integral running from $0$ to $1$ and another one from $1$ to $\infty$. I first tried to figure out for what $z \in \mathbb{C}$ the integral from $0$ to $1$ converges and I came to the conclusion, that $Re(z) &amp;gt; 0$ is the condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other integral, I believe, converges for every $z$, as the exponential function dominates the monomial eventually. So I concluded:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exists \Gamma(z) \iff Re(z) &amp;gt; 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I just learnt that this is wrong. I found out that the integral only diverges for non-positive integers. What did I do wrong or what is a better way to find the domain of the Gamma function?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-11T21:47:03.427" Title="Domain of the Gamma function" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13957" PostTypeId="2" ParentId="13956" CreationDate="2010-12-11T20:48:41.113" Score="4" ViewCount="0" Body="&lt;p&gt;The integral does converge iff $\Re z &amp;gt; 0$. However, it defines a function that can be analytically extended to the whole complex plane except the non-positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are analogous integral representations for $\Gamma(z)$ which hold true for $\Re z &amp;lt; 0$. For instance, it is not difficult to show that for any $k\in\mathbb N$&#xD;&#xA;$$\Gamma(z)=\int_{0}^{\infty} t^{z-1}\left(e^{-t}-1+t-\frac{t^2}{2!}+\dots+&#xD;&#xA;(-1)^{k+1}\frac{t^k}{k!}\right)dt,$$&#xD;&#xA;where $-k &amp;lt; \Re z &amp;lt; - k+1$ (the Cauchy–Saalschütz integral). &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-11T21:47:03.427" LastActivityDate="2010-12-11T21:47:03.427" CommentCount="3" />
  <row Id="13958" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T20:56:27.453" Score="5" ViewCount="0" Body="&lt;p&gt;Trevor, you are confusing what is generally thought of as &quot;Proof by cases&quot;, and a search for an example that proceeds by narrowing the possibilities/examining different cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;Proof by Cases&lt;/strong&gt; is one that proceeds along the following lines: you want to show that a statement is true for all $x$, but you do so by first breaking up all the possible $x$'s into different &quot;camps&quot;, and then showing the property holds in each &quot;camp&quot;. For instance, take&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For all real numbers $x$, $|x|\geq 0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, since the absolute value is defined as&#xA;$$|x|=\left\{\begin{array}{ll}&#xA;x &amp;amp; \mbox{if $x\gt 0$,}\\&#xA;-x &amp;amp; \mbox{if $x\leq 0$.}\\&#xA;\end{array}\right.$$&#xA;then we can proceed in the proof &lt;em&gt;by cases&lt;/em&gt;: every real number $x$ is &lt;em&gt;either&lt;/em&gt; positive, or less than or equal to zero. We examine each possibility separately, and if the conclusion holds in each of them, we'll know that the conclusion always holds (since every real number is in one of the two camps, and everyone in each of the camps has the desired property). If $x\gt 0$. then $|x|=x\gt 0$, and if $|x|\gt 0$, then $|x|\geq 0$ is also true. So the conclusion holds for the gang of positive numbers. If $x\leq 0$, then $|x|=-x$, and since $x\leq 0$, then $-x\geq 0$, so $|x|\geq 0$. Thus, the conclusion &lt;em&gt;also&lt;/em&gt; holds for elements of the &quot;other&quot; gang, the nonpositives.  Since every real number belongs to either the Positives or the Nonpositives, and every positive has the desired property and every nonpositive has the desired property, then every real number has the desired property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a &lt;strong&gt;Proof by cases.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do a &quot;proof by cases&quot; to show that something is &lt;em&gt;false&lt;/em&gt;; but in that case, you are usually trying to negate an &lt;em&gt;existential&lt;/em&gt; statement, such as &quot;There exists an integer $n$ such that $2n=1$.&quot; You could say: &quot;every integer is either even or odd; if $n$ is even, then $2n$ is a multiple of an even number $n$, so it is even, hence not equal to $1$. If $n$ is odd, then $2n$ is a multiple of $2$, which is even, so $2n$ is even, hence not equal to $1$.  Since no even number has the property, no odd number has the property, and every integer is either even or odd, then no integer has the property; the statement is false&quot;  (It's a bit convoluted, and you don't &lt;em&gt;need&lt;/em&gt; to do it by cases, but hopefully you see the idea). In fact, you are really proving the negation, &quot;For all integers $n$, $2n\neq 1$,&quot; so it is the same as the previous example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, what you have here is an &lt;strong&gt;existential&lt;/strong&gt; statement: it says &quot;there is an integer $n$ with a property.&quot; In order to prove that the statement is &lt;em&gt;false&lt;/em&gt;, you could proceed by cases as above; but in order to prove that the statement is &lt;em&gt;true&lt;/em&gt;, what you need to do is produce a single integer $n$ for which the condition holds. Your answer, if the statement is true, should be of the form: &quot;Take $n=xx$; this integer has the property&quot;, followed by the computations to justify the assertion that it has the property (in this case, you would plug it into $n^2 -79n + 1601$, compute the answer, and then factor the answer to show it is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you search for an $n$ for which the property will hold, you don't want to examine every possible integer until you hit one that works; because you may run the risk that it takes a long time to hit the first example (or maybe there isn't  one? Just because you haven't found one after searching for a long time, you don't know if that's because there isn't an example, or because there &lt;em&gt;is&lt;/em&gt; an example but is much bigger than what you've checked so far). So you are certainly smart to consider different possibilities for $n$ to see how they behave; e.g.,  if $n$ is even, then the expression does not have any obvious factors; if $n$ is a multiple of $3$, does it have any obvious factors? And so on. But that's not a &quot;proof by cases&quot;, because in the end your proof will be of the form: &quot;Here is an $n$ with the property&quot;, and not a case-by-case analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did that make sense and help? Feel free to ask questions, I can add to the answer to clarify or give other examples.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-11T20:56:27.453" />
  <row Id="13959" PostTypeId="1" AcceptedAnswerId="13972" CreationDate="2010-12-11T20:59:07.650" Score="17" ViewCount="332" Body="&lt;p&gt;I got this question as a mathematical puzzle from a friend. It looks similar to this one &lt;a href=&quot;http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this case, I have to find the expected length of the smallest segment. The two points where the rope is cut are selected uniformly at random. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried simulating it and I got an average value of 0.1114. I suspect the answer is 1/9 but I don't have any rigorous math to back it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-12-14T00:40:50.273" Title="If a 1 meter rope is cut at two uniformly randomly chosen points (to give three pieces), what is the average length of the smallest piece?" Tags="&lt;probability&gt;&lt;expectation&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="4" />
  <row Id="13960" PostTypeId="2" ParentId="13947" CreationDate="2010-12-11T21:10:30.253" Score="4" ViewCount="0" Body="&lt;p&gt;Recall that $\mathfrak p$ being an associated prime of $M/L$ precisely means that&#xA;$\mathfrak p$ is a prime ideal which is the annihilator of some element of $M/L$.&#xA;If we think what this means in terms of the module $M$, it means that we can&#xA;find an element $m \in M$ such that $\mathfrak p = \{a \in A \,| \, a m \in L\}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you want to replace $m$ by some other element $m&amp;#39; \in M$ with the property&#xA;that $\mathfrak p = \{ a\in A \, | \, a m&amp;#39; = 0\},$ i.e. such that $\mathfrak p$ is the annihilator of $m&amp;#39;$.  How do we find $m&amp;#39;$?  Well, what we have to work with&#xA;is the fact that $\mathfrak p$ does not contain the annihilator of $L$, so&#xA;we may find $a \in A \setminus \mathfrak p$ such that $a L = 0.$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you see how to use $a$ and $m$ together to find the required $m&amp;#39;$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  From the wording of the question I wasn't sure if you wanted a hint or if you needed the complete solution, so to be on the safe side I'm giving a hint.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-11T21:19:02.183" LastActivityDate="2010-12-11T21:19:02.183" CommentCount="3" />
  <row Id="13961" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T21:30:21.653" Score="2" ViewCount="0" Body="&lt;p&gt;If you think of this as uniformly randomly choosing points from $[0,1]\times[0,1]$ and break the square into the triangles sharing the diagonal $y=x$, the expected value can be expressed as twice the integral of the function $f(x,y)=\min(x,y-x,1-y)$ over the region $0\leq x\leq y\leq 1$.  You can use &lt;a href=&quot;http://www.missingbrace.com/2009/03/07/mathematical-formulas-for-max-and-min/&quot; rel=&quot;nofollow&quot;&gt;explicit formulas&lt;/a&gt; for the minimum value of 2 numbers to compute the integral.  There is probably a lot more simplifying that can be done.  If this seems too tedious, it is at least something you could put into Mathematica to find the answer.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T21:30:21.653" />
  <row Id="13962" PostTypeId="2" ParentId="13941" CreationDate="2010-12-11T21:38:37.213" Score="4" ViewCount="0" Body="&lt;p&gt;I think you are missing the fact that $b$ is not the $N$-limit of the $a_n$, which you did not include in your original post (I just added it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $N(a_n-b)$ is greater than or equal to $0$ (it is a norm). Since you are also assuming that $b\neq{\lim\limits_{n\to\infty}}^{(N)}a_n$, that means that $N(a_n-b)\not\to 0$ as $n\to\infty$; that is, the distance from $a_n$ to $b$ is bounded away from $0$; and since the terms are positive and bounded away from zero, the limit (which exists since we are dealing with a Cauchy sequence) must be strictly positive.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-11T21:38:37.213" CommentCount="1" />
  <row Id="13963" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T21:39:17.707" Score="9" ViewCount="0" Body="&lt;p&gt;If we divide the rope into $n$ segments then the $k$th largest segment will have expected length $$\frac{1}{n} \left( \frac{1}{n} + \cdots + \frac{1}{k} \right),$$ but I don't remember why. In particular, the smallest segment has expected length $1/n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that equivalently we could throw $n$ random points on a circle of unit circumference and consider lengths of segments on the circle.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Derivation of the general formula from the formula for the smallest segment&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Len $n$ points be thrown uniformly on the circle. Suppose the smallest segment has length $x$. We can remove an initial prefix of length $x$ from each segment to get a uniformly generated ensemble of $n-1$ segments on a circle of length $1-nx$. In this new arrangement, the smallest segment has expected size $(1-nx)/(n-1)^2$. Therefore the expected size of the second smallest segment is $$Ex + \frac{1-nEx}{(n-1)^2} = \frac{1}{n^2} + \frac{1-1/n}{(n-1)^2} = \frac{1}{n^2} + \frac{1}{n(n-1)}.$$&#xD;&#xA;We can get the general formula in the same way.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Derivation of the formula for the smallest segment&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Use the same method and come up with one more equation that will enable you to find this unknown.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-11T22:27:55.627" LastActivityDate="2010-12-11T22:27:55.627" CommentCount="3" />
  <row Id="13965" PostTypeId="1" AcceptedAnswerId="13984" CreationDate="2010-12-11T22:14:59.453" Score="1" ViewCount="42" Body="&lt;p&gt;A finite abstract simplicial complex is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the name for the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$D=(S,D)$ defined as above except that $D$ is closed under the superset operation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crossposted from &lt;a href=&quot;http://mathoverflow.net/questions/49084/dual-of-an-abstract-simplicial-complex&quot; rel=&quot;nofollow&quot;&gt;MO&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-12-11T23:41:32.813" LastActivityDate="2010-12-12T00:41:02.067" Title="What's the name for a dual of an abstract simplicial complex" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13966" PostTypeId="2" ParentId="13935" CreationDate="2010-12-11T22:16:38.487" Score="3" ViewCount="0" Body="&lt;p&gt;For abelian groups, the ring End(A) is very important.  As far as non-abelian groups A go, End(A) is not even (usually considered) a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Adding&quot; homomorphisms doesn't work in the non-abelian case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define (f+g)(x) = f(x) + g(x), then (f+g)(x+y) = f(x+y) + g(x+y) = f(x) + f(y) + g(x) + g(y), but (f+g)(x) + (f+g)(y) = f(x) + g(x) + f(y) + g(y).  To conclude that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;    f(y) + g(x) = g(x) + f(y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are equal, you use that + is commutative, that A is abelian.  More precisely, if you take f=g to be the identity endomorphism, then f+g is an endomorphism iff A is abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Composing&quot; homomorphisms doesn't work to form a group, since they are not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aut(A), the group of invertible endomorphisms, does form a group.  Aut(A) does not determine if a group is abelian or not: 4×2 and the dihedral group of order 8 have isomorphic automorphism groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a ring, End(A) sits inside the &quot;near-ring&quot; of self-maps.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Nearring#Mappings_from_a_group_to_itself&quot; rel=&quot;nofollow&quot;&gt;the wikipedia article on nearring&lt;/a&gt; for an explanation.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-11T22:24:20.320" LastActivityDate="2010-12-11T22:24:20.320" CommentCount="1" />
  <row Id="13967" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T22:30:01.707" Score="6" ViewCount="0" Body="&lt;p&gt;Here you have two independent random variables $X,Y$, both uniform on $[0,1]$. Let &#xA;$A=\min (X,Y), B=\max (X,Y)$ and $C=\min (A, 1-B, B-A)$. First you want to find the probability&#xA;density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then&#xA;$$ 1- F_C(a) = P(C\ge a)=P(A\ge a, 1-B\ge a, B-A\ge a)$$ $$=P(a\le X\le 1-a, a\le Y\le 1-a, |X-Y|\ge a)$$&#xD;&#xA;This is equal to the total area of the two sets&#xA; $$\lbrace (x,y): 0\le x,y\le 1, y\le x-a, x\le 1-a, y\ge a\rbrace$$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lbrace (x,y): 0\le x,y\le 1, y\ge x+a, x\ge a, y\le 1-a\rbrace$$&#xD;&#xA;which is $(1-3a)^2$. Note that for these sets to be nonempty, one must have $0\le a\le 1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $F_C(a)=1-(1-3a)^2$, from which it follows $f_C(a)=F_C&amp;#39;(a)=6(1-3a)$. Therefore&#xA;the expected value of $C$ is&#xA;$$\int_0^{1/3} 6a(1-3a) da= \frac{1}{9}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-11T22:30:01.707" CommentCount="1" />
  <row Id="13968" PostTypeId="1" CreationDate="2010-12-11T22:35:29.530" Score="4" ViewCount="94" Body="&lt;p&gt;The proof that I know of the theorem goes like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any module $M$ is a quotient of a free module $F$ (over any ring).&lt;/li&gt;&#xA;&lt;li&gt;Any submodule $K$ of a free module $F$ over a PID $R$ is a free module, so in particular the kernel of the above quotient map is free.&lt;/li&gt;&#xA;&lt;li&gt;For any free submodule $K$ of a free module $F$ over a PID $R$ we can find a $y\in K, x\in F$ and $a\in R$ such that $F=\left&amp;lt;x\right&amp;gt;\oplus F&amp;#39;$ and $K=\left&amp;lt;y\right&amp;gt;\oplus K&amp;#39;$ with $K&amp;#39;=K\cap F&amp;#39;$. Furthermore, $\left&amp;lt;a\right&amp;gt;$ is an ideal maximal among images of $K$ under homomorphisms $F\to R$.&lt;/li&gt;&#xA;&lt;li&gt;If $K$ is finitely generated (which would follow from $F$ being finitely generated), we iterate this construction, which gives us sequences $x_i\in F$, $y_i\in K$ and $a_i\in R$ such that $F=F&amp;#39;\oplus_{i} \left&amp;lt;x_i\right&amp;gt;$ and $K=\oplus_{i}\left&amp;lt;y_i\right&amp;gt;$ where $a_ix_i=y_i$ and $a_i$ divides $a_j$ for $i&amp;lt;j$ (which follows from choosing a module homomorphism $f$ from $F$ to $R$ that is $1$ on all the $x_i$ and on the basis elements of the $F&amp;#39;$, then exploiting the fact that $f(y_j)=f(a_jx_j)=a_jf(x_j)=a_j\in f(\oplus_{i\leq j}y_j)\subset\left&amp;lt;a_i\right&amp;gt;$ by $\left&amp;lt;a_i\right&amp;gt;$ being maximal among images of $\oplus_{i\leq j}a_j$ under homomorphisms from $\oplus_{i\leq j}\left&amp;lt;x_j\right&amp;gt;\oplus F&amp;#39;$ to $R$).&lt;/li&gt;&#xA;&lt;li&gt;Taking quotients, we obtain that $M=F&amp;#39;\oplus_i R/(a_i)$ where $a_i$ divides $a_j$ if $i&amp;lt;j$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is why does this break down if we drop the assumption in 4 that $K$ is finitely generated? Does iterating the process transfinitely no longer give us a basis for $K$? If so, why not?&lt;/p&gt;&#xA;" OwnerUserId="400" LastEditorUserId="400" LastEditorDisplayName="" LastEditDate="2010-12-11T22:46:41.330" LastActivityDate="2010-12-12T04:17:00.763" Title="Why does the structure theorem for finitely generated modules over PIDs fail for arbitrary modules over a PID?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13969" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T22:38:59.193" Score="2" ViewCount="0" Body="&lt;p&gt;It is not appropriate to use proof by cases here.  The statement, &quot;There exists a positive integer n such that $n^2−79n+1601$ is composite&quot; means exactly what it says.  It states that there is at least one positive integer that makes this statement true.  It is no different from any other similar statement such as &quot;I have at least one dollar in my pocket&quot;.  All I have to do to prove this is reach into my pocket and show you the dollar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you have to do to show your original proposition is to find one number, that makes it true.  Being too abstract is counterproductive here.  You can find one number such as 1601. Then, you just have to do the computations that show that the number produced is composite. Literally, the same computations you might do in grade school.  In the case of 1601,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1601^2-79\times1601+1601 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives a number that is clearly composite, which is why I picked it.&lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-12-11T22:38:59.193" CommentCount="2" />
  <row Id="13970" PostTypeId="1" AcceptedAnswerId="13982" CreationDate="2010-12-11T22:44:42.893" Score="1" ViewCount="53" Body="&lt;p&gt;Having: $P = N^{CN + 1}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I simplify this equation to $N = \cdots$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using logarithms but I'm stucked...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4652" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T04:00:15.340" LastActivityDate="2010-12-12T04:00:15.340" Title="Is possible to simplify $P = N^{ CN + 1}$ in terms of $N$?" Tags="&lt;analysis&gt;&lt;logarithms&gt;" AnswerCount="1" />
  <row Id="13971" PostTypeId="1" CreationDate="2010-12-11T22:45:43.280" Score="0" ViewCount="31" Body="&lt;p&gt;How do I apply the Doleans-Dade exponential formula for the following levy stochastic differential equation:&#xA;$$dZ_t=Z_t\left(\theta_1(t)dW_t^{(1)} +\theta_2(t)dW_t^{(2)}+\int_\mathbb R \theta(s,x)\mu(ds,st)-v(ds,dt)\right),$$ where $W_t^{(1)}$ and $W_t^{(2)}$ are two independent Brownian motions, $\mu(ds,st)$ is the jump measure and $v(ds,dt)$ is the levy measure&lt;/p&gt;&#xA;" OwnerUserId="626" LastActivityDate="2010-12-11T22:45:43.280" Title="Doleans-Dade exponential formula" Tags="&lt;stochastic-integrals&gt;" />
  <row Id="13972" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T22:49:43.200" Score="9" ViewCount="0" Body="&lt;p&gt;Update: The current version of this answer is more intuitive (IMHO) than the previous one.  See also &lt;a href=&quot;http://math.stackexchange.com/questions/14190/average-length-of-the-longest-segment/14194#14194&quot;&gt;this similar answer&lt;/a&gt; to a similar question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization of this problem to $n$ pieces is discussed extensively in David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bdhzFXg6xFkC&amp;amp;printsec=frontcover&amp;amp;dq=david+and+nagaraja+order+statistics&amp;amp;source=bl&amp;amp;ots=OaK_k-BbTf&amp;amp;sig=vAtEAURNfZTe3NDV9WodQZ-3ip4&amp;amp;hl=en&amp;amp;ei=DfsDTdiXPML88AafsKnqAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBgQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt; (pp. 133-135, and p. 153).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$.  So the $V_i$'s are the lengths of the pieces of rope.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$  This is proved formally in David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bdhzFXg6xFkC&amp;amp;pg=PA133&amp;amp;lpg=PA133&amp;amp;dq=david+and+nagaraja+order+statistics+division+of+random+interval&amp;amp;source=bl&amp;amp;ots=OaK_l7B9-j&amp;amp;sig=por_GsntbxBDII72xTWISQ9Keas&amp;amp;hl=en&amp;amp;ei=urkGTevVIZTmsQOM_7HsBw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt;, p. 155.  Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$.  For example, $P(V_1 &amp;gt; c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;If $V_{(1)}$ denotes the shortest piece of rope, then for $x \leq \frac{1}{n}$, (following Raskolnikov's comment) $$P(V_{(1)} &amp;gt; x) = P(V_1 &amp;gt; x, V_2 &amp;gt; x, \ldots, V_n &amp;gt; x) = (1 - nx)^{n-1}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &#xA;$$E[V_{(1)}] = \int_0^{\infty} P(V_{(1)} &amp;gt; x) dx = \int_0^{1/n} (1-nx)^{n-1} = \frac{1}{n^2}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David and Nagaraja also give the formula Yuval Filmus mentions (as Problem 6.4.2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E[V_{(r)}] = \frac{1}{n} \sum_{j=1}^r \frac{1}{n-j+1}.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-14T00:40:50.273" LastActivityDate="2010-12-14T00:40:50.273" CommentCount="1" />
  <row Id="13973" PostTypeId="2" ParentId="13886" CreationDate="2010-12-11T22:50:28.987" Score="0" ViewCount="0" Body="&lt;p&gt;The nth twin prime pair appears to be around $c n \log^2 n$ for a constant $c$, large $n$.  The largest prime factor of any non-prime $m$ is equal to $n/m&amp;#39;$ where $m&amp;#39;\ge2$.  What you see in the plot are approximations of the curves $y=\frac{c}{k} x \log^2 x$ for $k=6,12,24,...$, since the mean of a twin pair is a multiple of 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Notice the same effect when plotting simply the maximum prime factor of 6n versus n:&#xA;&lt;img src=&quot;http://i.imgur.com/b5Ldm.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2937" LastEditorUserId="2937" LastEditorDisplayName="" LastEditDate="2010-12-12T07:39:14.873" LastActivityDate="2010-12-12T07:39:14.873" CommentCount="5" />
  <row Id="13974" PostTypeId="2" ParentId="13949" CreationDate="2010-12-11T22:53:10.060" Score="1" ViewCount="0" Body="&lt;p&gt;It may actually be worth pointing out that if $p(x)$ is a polynomial with integer coefficients, the only way $p(n)$ is never composite as $n$ varies over the integers is if $p$ is constant (and $p(x)=c$ for some $c$ that is either prime, the negative of a prime, $1$, or $-1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For suppose that $p(x)=ax^n+bx^{n-1}+\dots+c$. Then clearly $c$ divides $p(c)$. Also, $c$ divides $p(2c)$, $p(3c)$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose first $|c|&amp;gt;1$. The only way we do not find a composite number in the list $p(c),p(2c),p(3c),\dots$ is if all these numbers are $c$ or $-c$ and $c$ or $-c$ itself is prime. But then infinitely many of the $p(nc)$ are equal to each other. But polynomials that take the same value too often are constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $c=0$, we are done because $p(0)=0$ is composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finally, suppose that $|c|=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $p(x)=\alpha(x-1)^n+\beta(x-1)^{n-1}+\dots$, $p(x)=A(x-2)^n+B(x-2)^{n-1}+\dots$, etc. If at least one of these expansions has constant term of absolute value larger than 1, and we are done by the previous analysis, or else all of them have constant term $1$, $-1$, or $0$. Note that the constant term in the first case is $p(1)$, in the second is $p(2)$, etc. So again we have that either $p(n)=0$ for some $n$ (and so it is composite), or else $p$ takes the same value ($1$ or $-1$) infinitely often, and we conclude that $p$ is constant.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T22:53:10.060" />
  <row Id="13975" PostTypeId="1" CreationDate="2010-12-11T22:56:55.000" Score="2" ViewCount="87" Body="&lt;p&gt;Prove that $\left \{ 0,1 \right \}^{\mathbb{N}}\sim \left \{ 0,1,2,3 \right \}^{\mathbb{N}}$ and find a direct bijection function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the first part by showing that $\left \{ 0,1 \right \}^{\mathbb{N}} \subseteq \left \{ 0,1,2,3 \right \}^{\mathbb{N}} \subseteq {\mathbb{N}}^{\mathbb{N}}$, which implies that $|\left \{ 0,1 \right \}^{\mathbb{N}}| \leq |\left \{ 0,1,2,3 \right \}^{\mathbb{N}}| \leq |{\mathbb{N}}^{\mathbb{N}}|$ and since $|{\mathbb{N}}^{\mathbb{N}}| = |\left \{ 0,1 \right \}^{\mathbb{N}} | = 2^{\aleph_0} $ and Cantor-Bernstein you get that $\left \{ 0,1 \right \}^{\mathbb{N}}\sim \left \{ 0,1,2,3 \right \}^{\mathbb{N}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm stuck with formulating a bijection function. More generally, what approach do you use when you need a formulate an exact function?&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-11T23:04:17.270" LastActivityDate="2010-12-11T23:08:37.287" Title="$\left \{ 0,1 \right \}^{\mathbb{N}}\sim \left \{ 0,1,2,3 \right \}^{\mathbb{N}}$ bijection function" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="13976" PostTypeId="2" ParentId="13975" CreationDate="2010-12-11T23:02:10.110" Score="6" ViewCount="0" Body="&lt;p&gt;Think in terms of how you convert between binary and base 4 representations of numbers.  Break up a sequence of 0s and 1s into pairs, and to each pair (for which there are the 4 possibilities) assign one of 0, 1, 2, or 3.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-11T23:02:10.110" CommentCount="2" />
  <row Id="13977" PostTypeId="2" ParentId="13912" CreationDate="2010-12-11T23:05:52.207" Score="10" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But the geometry in algebraic geometry is dictated entirely by the algebra. So how can you use geometry to help you with the algebra, when you have to do the algebra first to figure out what the geometry looks like? I don't get it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, geometric intuition suggests to you what should be true, and suggests a strategy for proving it; and algebra is how you carry out the proofs.  For example, it's geometry that should inform your intuitions about things like intersections of varieties, but algebra that you use to actually rigorously define intersections and prove things about them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the &quot;principle of continuity.&quot;  If you intersect a circle and an ellipse, you'll generically get four points, although sometimes you might get two or one.  If you intersect a circle and a parabola the same thing happens.  In a perfect world, you might suspect that the intersection of two conic sections is always &quot;four points,&quot; but this clearly isn't true for the usual definition of &quot;four points.&quot;  But if you broaden your definitions (count the points with multiplicity; count the complex points; count the points at infinity), you will eventually be led to complex projective varieties, where something like the &quot;principle of continuity&quot; holds: if you have two varieties that intersect in $m$ points and you nudge one of them or the other continuously, they will still intersect in $m$ points, if you count the points properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to keep in mind that historically a major incentive for developing commutative algebra was to rigorize parts of algebraic geometry; some algebraic geometers (the Italian school) had begun relying too heavily on geometric intuition and had been ignoring special cases, etc. and commutative algebra was one way to fix their proofs.  But the point is that they were doing algebraic geometry &lt;em&gt;first&lt;/em&gt;!  There is a long and interesting history here which I think it is very instructive to learn; you should try to find Dieudonne's &lt;em&gt;History of Algebraic Geometry&lt;/em&gt;, as well as (for a personal perspective) Parikh's &lt;em&gt;The Unreal Life of Oscar Zariski.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I first saw the axioms for a group, I spent the next year trying to figure out why the heck anybody cared about groups (and frankly still only know this in a detached and academic way).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The group axioms are an abstraction of the notion of symmetry, and symmetry is a natural and beautiful idea, and symmetries are everywhere in mathematics.  Perhaps you aren't acquainted with enough examples; it's hard for me to give more specific advice here without knowing what you find unsatisfying about groups.  But you might be interested in Conway, Burgiel, and Goodman-Strauss's &lt;em&gt;The Symmetries of Things&lt;/em&gt;, as well as in Mumford, Series, and Wright's &lt;em&gt;Indra's Pearls: the Vision of Felix Klein.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-12T02:52:52.363" LastActivityDate="2010-12-12T02:52:52.363" CommentCount="6" />
  <row Id="13978" PostTypeId="2" ParentId="13975" CreationDate="2010-12-11T23:08:37.287" Score="4" ViewCount="0" Body="&lt;p&gt;There are at least two ways to proceed: Either you start as you did, and then you follow the argument of Cantor-Bernstein, which explicitly gives you how to build a bijection from the two given injections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way is to directly argue in the case at hand. For example, identify the sequence $(a_0,a_1,a_2,a_3,...)$ in $\{0,1\}^{\mathbb N}$ with the sequence $(b_0,b_1,b_2,\dots)$ in $\{0,1,2,3\}^{\mathbb N}$ as follows: Replace $a_{2n},a_{2n+1}$ with $b_n$, where $0,0$ is replaced with $0$; $0,1$ is replaced with $1$; $1,0$ with $2$; and $1,1$ with $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Edit&lt;/strong&gt;: I see Jonas wrote the same explicit bijection as I was typing this.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a slightly more challenging exercise, pick any two positive integers $n&amp;lt;m$, and build a &quot;combinatorial&quot; bijection between $\{0,1,\dots,n\}^{\mathbb N}$ and $\{0,1,\dots,m\}^{\mathbb N}$. &lt;em&gt;Combinatorial&lt;/em&gt; meaning here something in the same spirit of the explicit bijection above.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-11T23:08:37.287" CommentCount="1" />
  <row Id="13979" PostTypeId="2" ParentId="13959" CreationDate="2010-12-11T23:38:23.940" Score="8" ViewCount="0" Body="&lt;p&gt;My approach is maybe more naive than the others posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break the unit interval at $x$ and $y$ where $x &amp;lt; y$.  Our lengths are then $x$, $y - x$, and $1 - y$.  It's not hard to show that they all have probability $1/3$ of being the shortest.  In any case, our joint PDF is given by $f(x,y) = 6$ (since $x$ and $y$ remain uniform random variables on $1/6$th of the square $[0,1] \times [0,1]$).  Each triangle in the diagram below corresponds to the domain of the PDF for one of the three cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/obVJs.png&quot; alt=&quot;Each triangle corresponds to which length is shortest&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll take care of the case when $x$ is shortest, that is, $x \leq y - x$ and $x \leq 1 - y$.  This is the leftmost triangle.  Since we're assuming $x$ is least, we are looking for&#xA;$$E[x] = \int_0^{1/3} \int_{2x}^{1 - x} 6x \;dy \;dx = 1/9$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cases when $y - x$ and $1 - y$ are shortest are similar.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-12-11T23:53:34.367" LastActivityDate="2010-12-11T23:53:34.367" CommentCount="1" />
  <row Id="13980" PostTypeId="1" AcceptedAnswerId="13994" CreationDate="2010-12-11T23:39:37.010" Score="1" ViewCount="42" Body="&lt;p&gt;Currently I am implementing a DTW (Dynamic Time Warping) algorithm for my project. As some will know it has the complexity of O(n^2). Considering a sound file of length 1-h with 44100 sampling. How long will it take to compute it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know we need some other knowledge such as cpu speed, length of the word we are trying to match, our library etc. but you can assume them with your own variables.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am aware of the fact that Fast DTW exists and it is way lighter in terms of O complexity but using it is not a case. &lt;/p&gt;&#xA;" OwnerUserId="4653" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-12T00:18:17.663" LastActivityDate="2010-12-12T02:06:32.813" Title="Question regarding Dynamic Programming for Audio Recognition" Tags="&lt;algorithms&gt;&lt;pattern-recognition&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13981" PostTypeId="1" AcceptedAnswerId="13983" CreationDate="2010-12-12T00:12:07.433" Score="4" ViewCount="94" Body="&lt;p&gt;Let $\omega$ denote the countably infinite cardinality and $A$ a cardinality that is strictly larger (i.e., an uncountable one). Is it true that $A^\omega$ has strictly larger cardinality than $A$?&lt;/p&gt;&#xA;" OwnerUserId="4654" LastEditorUserId="4654" LastEditorDisplayName="" LastEditDate="2010-12-12T00:55:31.047" LastActivityDate="2010-12-12T02:47:03.183" Title="$\aleph_0$ to $\aleph_0$" Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="13982" PostTypeId="2" ParentId="13970" CreationDate="2010-12-12T00:13:41.487" Score="3" ViewCount="0" Body="&lt;p&gt;The equation $P=N^N$ doesn't have an elementary solution, but taking logarithms you can find approximate solutions (or even solutions in the form of transseries). In your case, take logarithms to find $(cn+1)\log n = p$. Assuming $p$ and so $n$ are large, $n\log n \approx p/c$. Therefore $n \approx p/c$, and so $$n \approx \frac{p/c}{\log (p/c)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the following is wrong, as J.M. pointed out.&#xA;&lt;i&gt;Again ignoring the $1$ in the exponent, we have $P^{1/c} = N^N$ and so $N = W(P^{1/c})$, where $W$ is the Lambert function (see Wikipedia).&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-12T02:46:12.803" LastActivityDate="2010-12-12T02:46:12.803" CommentCount="1" />
  <row Id="13983" PostTypeId="2" ParentId="13981" CreationDate="2010-12-12T00:29:00.183" Score="8" ViewCount="0" Body="&lt;p&gt;No, it is not necessarily true. For example, let $A$ be the cardinality of the Cantor set, $2^{\aleph_0}$. Then $A^\omega=(2^{\aleph_0})^{\aleph_0}=2^{(\aleph_0\times\aleph_0)}=2^{\aleph_0}=A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is not necessarily false. For example, if $A$ is the supremum of the sequence $\aleph_0,2^{\aleph_0},2^{2^{\aleph_0}},\dots$, then $A^\omega&amp;gt;A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because of a general fact due to König: The &lt;em&gt;cofinality&lt;/em&gt; ${\rm cf}(A)$ of a cardinal $A$ is the smallest $\kappa$ such that $A$ can be written as a union of $\kappa$ sets, each of size smaller than $A$. König showed that $A^{{\rm cf}(A)}&amp;gt;A$ for any $A$ infinite--this is a generalization of Cantor's easier result that $2^\kappa&amp;gt;\kappa$. The $A$ in the previous paragraph explicitly has cofinality $\omega$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-12T00:29:00.183" CommentCount="5" />
  <row Id="13984" PostTypeId="2" ParentId="13965" CreationDate="2010-12-12T00:33:01.930" Score="0" ViewCount="0" Body="&lt;p&gt;I've found &lt;a href=&quot;http://www.albany.edu/~lenart/articles/ucpol.ps&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for simplicial compleces - down closed set system and, respectively, for the objects I asked about - up closed set system.&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-12-12T00:41:02.067" LastActivityDate="2010-12-12T00:41:02.067" />
  <row Id="13985" PostTypeId="2" ParentId="13711" CreationDate="2010-12-12T00:38:16.777" Score="0" ViewCount="0" Body="&lt;p&gt;Since I want to close this question, I will post an answer myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Guide to Distribution Theory and Fourier Transforms - Strichartz is a nice introduction but it contains almost nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Théorie des distributions by Laurent Schwartz is written by the master and father on the subject and therefore I say it is a good book.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-12T00:38:16.777" CommentCount="2" />
  <row Id="13986" PostTypeId="1" AcceptedAnswerId="13989" CreationDate="2010-12-12T00:44:15.030" Score="4" ViewCount="79" Body="&lt;p&gt;Greetings!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a test recently I ended up having to solve this for y:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = \frac{2y}{y + 1} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I kept getting stuck in cirlces...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xD;&#xA;\begin{aligned}&#xD;&#xA;x(y + 1 ) = 2y  \\&#xD;&#xA;xy + x = 2y \\&#xD;&#xA;\frac{xy + x}{2} = y \\&#xD;&#xA;\end{aligned} &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That didn't get me anywhere, so then I started over and tried multiplying both sides by the reciprocal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{aligned}&#xD;&#xA;(\frac{y+1}{2y})(x) = 1 \\ &#xD;&#xA;\frac{xy+x}{2y} = 1 \\&#xD;&#xA;\end{aligned}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And still I can't see a way to isolate y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst part of this is that I remember being specifically taught a trick for this particular conundrum, but I can't remember the trick!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram gives &lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+for+y%3A+x+%3D+2y%2F%28y%2B1%29&quot; rel=&quot;nofollow&quot;&gt;the answer&lt;/a&gt; as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;y = - \frac{x}{x-2} &#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it doesn't show the steps.&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-12-12T02:09:25.317" Title="How to solve this rational equation for y" Tags="&lt;algebra-precalculus&gt;&lt;fractions&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="13987" PostTypeId="1" AcceptedAnswerId="13988" CreationDate="2010-12-12T01:03:08.233" Score="2" ViewCount="52" Body="&lt;p&gt;I have:&#xA;$$ln(ax + b)^4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After calculating (u'/u) I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4/(ax+b)$$ but the solution is $$4a/(ax+b)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: a and b are IR+.&lt;/p&gt;&#xA;" OwnerUserId="2950" LastActivityDate="2010-12-12T01:06:21.750" Title="Derivative log problem" Tags="&lt;calculus&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13988" PostTypeId="2" ParentId="13987" CreationDate="2010-12-12T01:06:21.750" Score="1" ViewCount="0" Body="&lt;p&gt;You need to apply the chain rule, using $\frac{d}{dx}(ax+b)=a$.  (That is your missing factor of $a$.) I can't tell whether you did so, but you can also simplify the first expression to $4\log(ax+b)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-12T01:06:21.750" CommentCount="1" />
  <row Id="13989" PostTypeId="2" ParentId="13986" CreationDate="2010-12-12T01:08:03.880" Score="5" ViewCount="0" Body="&lt;p&gt;Thanks to Timothy and Yuan for pointing out how to factor it. Despite my best efforts, I have never, ever, ever, ever been able to magically see factors appear the way everybody else seems to do so immediately. Sigh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{aligned}&#xD;&#xA;xy + x = 2y \\&#xD;&#xA;x = 2y - xy \\&#xD;&#xA;x = y( 2 - x ) \\&#xD;&#xA;y = \frac{x}{2-x} \\&#xD;&#xA;\end{aligned}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the same as the answer given by Wolfram. All is right with the world again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3384" LastActivityDate="2010-12-12T01:08:03.880" CommentCount="2" />
  <row Id="13990" PostTypeId="2" ParentId="13514" CreationDate="2010-12-12T01:08:26.520" Score="0" ViewCount="0" Body="&lt;p&gt;I guess an alternative standard way to proof it, is to proof sentences that express that there at least n elements. Let denote these entences by $\phi_{\geq n}$ where $\phi_{\geq n} :\Leftrightarrow \exists x_1 ... \exists x_n \rho_n$. The matrix $\rho_n$ can be inductively defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_2 :\Leftrightarrow x_1 \neq x_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_{n+1} :\Leftrightarrow \rho_n \land x_1 \neq x_{n+1} \land ... \land x_n \neq x_{n+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each existence sentence $\phi_{\geq n}$ for $n \geq 2$ can be proved without induction via some trivial witness numerals. Therefore the models are infinite. Induction happens on the meta level, but not on the object level.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-12T01:32:26.430" LastActivityDate="2010-12-12T01:32:26.430" />
  <row Id="13991" PostTypeId="1" AcceptedAnswerId="14349" CreationDate="2010-12-12T01:21:30.017" Score="7" ViewCount="345" Body="&lt;p&gt;A lot of &quot;big&quot; names in analysis (and other fields) seem to be doing some form of combinatorics (without any order, some examples are Tim Gowers, Terence Tao and Jean Bourgain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, looking a bit around makes me conclude that combinatorics is a huge field. There must be one &quot;kind&quot; which is the most fruitful in analysis. What kind is this? What is a good introduction to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I forgot, analysis is also a big field. I mean more in the direction of harmonic analysis and PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-12T01:28:03.063" LastActivityDate="2010-12-14T23:50:58.017" Title="Combinatorics and analysis." Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="13992" PostTypeId="2" ParentId="13991" CreationDate="2010-12-12T01:34:16.597" Score="6" ViewCount="0" Body="&lt;p&gt;Your question seems to ask what combinatorics is fruitful in analysis.  I think the trend (at least with those you have mentioned) is the application of analysis to combinatorics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be most interested in arithmetic combinatorics.  I'd recommend the page from Tao's course &lt;a href=&quot;http://www.math.ucla.edu/~tao/254a.1.03w/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  Alan Frieze has a page with many good links &lt;a href=&quot;http://www.math.cmu.edu/~af1p/Teaching/AdditiveCombinatorics/AC.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also heard good things (and keep intending to read more closely) Richard Stanley's text &lt;a href=&quot;http://www-math.mit.edu/~rstan/ec/&quot; rel=&quot;nofollow&quot;&gt;Enumerative Combinatorics&lt;/a&gt;, which is different &quot;flavor&quot; of combinatorics than the links above.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-12T01:34:16.597" CommentCount="1" />
  <row Id="13993" PostTypeId="1" CreationDate="2010-12-12T01:46:17.637" Score="1" ViewCount="86" Body="&lt;p&gt;Hi! I'm studying physics, so I'm sorry if I'll write some inexact things in this post. I wish you can understand me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have 1D wave equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we say that it's always possible (for physics situations...) to decompose the generic solution $f(x+ct)+g(x-ct)$ using Fourier Transforms in $e^{i\omega(x/c\mp t)}.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the FT is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\pi} \int_R \psi(\omega)e^{i\omega t}d\omega$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have two variables, $x$ and $t$, how can we use the FT with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_R \psi(\omega)e^{i(k(\omega)x\mp\omega t)}d\omega \; ?$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two different types of temporal evolution, $+ct$ and $-ct$... &#xA;I have seen situations with no dispersion law $\omega=kc$, where we consider only a wave $\psi=f(x+ct)$  and use the point $x=0$ to find the spectrum of waves $f(\omega)$, : &#xA;$$&#xD;&#xA;f(\omega)=\frac{1}{2\pi}\int_R f(0,t)e^{-i\omega t}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;f(x,t)=\int_R f(\omega) e^{i\omega(x/c+t)}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but can someone explain me the general method? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2D or 3D I haven't any idea about how I can do all this. Someone can explain me this too?&#xA;Can you link me a .pdf online where this is explained?&#xA;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-12T04:11:47.807" LastActivityDate="2010-12-12T16:10:11.210" Title="Fourier analysis for waves" Tags="&lt;differential-equations&gt;&lt;fourier-analysis&gt;&lt;physics&gt;" AnswerCount="3" />
  <row Id="13994" PostTypeId="2" ParentId="13980" CreationDate="2010-12-12T02:06:32.813" Score="0" ViewCount="0" Body="&lt;p&gt;The asymptotic complexity of the algorithm can not provide any information about the time it will take absolutely, but instead shows how running time scales with input size. What I mean is that if you know that it takes $x$ seconds to process $y$ samples of audio, then it will take $4x$ seconds to process $2y$ samples of audio given that the algorithm runs in $O(n^2)$ with $n$ being number of samples.&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2010-12-12T02:06:32.813" CommentCount="1" />
  <row Id="13995" PostTypeId="2" ParentId="13986" CreationDate="2010-12-12T02:09:25.317" Score="2" ViewCount="0" Body="&lt;p&gt;Another way to arrive at the conclusion, if you don't &quot;see&quot; factors jumping out at you is this: When you see a fraction like $\frac{2y}{y+1}$, I am sure you feel the urge to split into summands. Now, of course, you cannot do it with the sum in the denominator, so that suggests inverting the whole thing (assuming that everything that you want to be non-zero is non-zero):&#xA;$$&#xD;&#xA;\frac{1}{x} = \frac{y+1}{2y} = \frac{y}{2y} + \frac{1}{2y} = \frac{1}{2} + \frac{1}{2y}.&#xD;&#xA;$$&#xD;&#xA;Now subtract 1/2 and invert back again to get the result.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-12T02:09:25.317" />
  <row Id="13996" PostTypeId="1" AcceptedAnswerId="14003" CreationDate="2010-12-12T02:41:21.833" Score="3" ViewCount="207" Body="&lt;p&gt;Hello, which ways to compute this integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} dx&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for different cases of t. t can be any value in $\mathbb{R}$.  I was stuck at how to get the antiderivative of the integrand. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4658" LastEditorUserId="4658" LastEditorDisplayName="" LastEditDate="2010-12-12T19:25:36.333" LastActivityDate="2010-12-12T19:25:36.333" Title="How to compute this integral?" Tags="&lt;integral&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="0" />
  <row Id="13997" PostTypeId="2" ParentId="13912" CreationDate="2010-12-12T02:44:31.120" Score="5" ViewCount="0" Body="&lt;p&gt;Dear Mike,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One suggestion is that you learn some algebraic geometry from a different perspective, and, having done so, return to commutative algebra and see if&#xA;you have more insight by virtue of the intuition in algebraic geometry that you (hopefully) will have developed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wrote in your question that &quot;the geometry in algebraic geometry is dictated entirely by the algebra&quot;, but this need not be true (and certainly is not true historically).  If you find it is this way because of the particular books you've been reading, I suggest that you try some other books.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that you are a graduate student (based on your posts), you might try Mumford's book on complex projective varieties.  (I forget the precise title, but it won't be hard to find.)  It is quite geometric.  You could also try looking at Mumford's Red Book, which tries to unify the geometry and algebra.&#xA;There are other books, too: Griffith's and Harris is one, although it requires&#xA;a reasonably strong background in differential and algebraic topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I've tried to emphasize in some other posts of mine here and on MO, algebraic geometry needn't be done via commutative algebra at all; some of the most important recent work in algebraic geometry has relied almost entirely on analytic methods.  There are also topological approaches, via Morse theory and its generalizations and variations.  (Lefschetz, who was one of the founders of algebraic topology, was motivated by the applications to algebraic geometry.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you needn't focus your study on algebraic geometry at all: I am mentioning it mainly because it came up in your post, and it is an area of mathematics that is very central, and very multifaceted, so that there are lots of ways to approach it, which in turn makes it a good way to build up intuition in lots of different areas of mathematics.  (In your case I am suggesting it as a way to build up intuition in algebra; for myself, algebra was always one of my stronger suits, but through studying algebraic geometry I was able to strengthen&#xA;my geometric and topological intuition.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what other answers appear here, I might write some more later which connects with some of other aspects of your post.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-12T02:44:31.120" />
  <row Id="13999" PostTypeId="2" ParentId="13996" CreationDate="2010-12-12T02:52:18.840" Score="1" ViewCount="0" Body="&lt;p&gt;First, check whether they converge.  You will find that the first one doesn't unless $t=0$, and the second one doesn't unless $t\leq0$.  For the first integral, when $t=0$, you can use the arctangent function.  For the second, when $t\lt0$, you can try to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28I.29&quot; rel=&quot;nofollow&quot;&gt;complex contour integration&lt;/a&gt; methods as seen in many texts on complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was hasty in posting this, and hadn't actually thought through how the contour integration would go.  Based on the other answers, it looks like it wouldn't be so straightforward.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-12T08:45:47.537" LastActivityDate="2010-12-12T08:45:47.537" CommentCount="8" />
  <row Id="14000" PostTypeId="1" AcceptedAnswerId="14024" CreationDate="2010-12-12T02:56:23.223" Score="2" ViewCount="168" Body="&lt;p&gt;Is there any intuition behind surface integrals and their applications? I get that there's some stuff with fluid flow but it doesn't really stick that well, especially when it comes to surface orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any surefire way to figure out the orientation of a surface? And is there any sort of intuition behind that or surface integrals?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastActivityDate="2010-12-17T09:21:24.920" Title="Intuition for Surface Integrals" Tags="&lt;calculus&gt;&lt;soft-question&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="14001" PostTypeId="1" CreationDate="2010-12-12T03:07:11.037" Score="0" ViewCount="38" Body="&lt;p&gt;If E={E1, E2, E3} is an orthonormal frame defined on some open set in R^3&lt;br&gt;&#xA;Show that if the dual frame O={O1, O2, O3} satisfies:&lt;br&gt;&#xA;dOi = 0 for i = 1, 2, 3&lt;br&gt;&#xA;Then O is a constant rotation or reflection of E on R^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that since the frame is orthonormal, that means the connection matrix w is skew symmetric. I really have no idea how to start this problem.&lt;/p&gt;&#xA;" OwnerUserId="4660" LastActivityDate="2010-12-12T03:07:11.037" Title="Working with orthonormal frames and dual frames" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" CommentCount="2" />
  <row Id="14003" PostTypeId="2" ParentId="13996" CreationDate="2010-12-12T03:36:23.157" Score="4" ViewCount="0" Body="&lt;p&gt;When $t \neq 0$, the first integral diverges and the second one converges only if $t &amp;lt; 0$. In the latter case, if you call your integral $f(t)$, then differentiating under the integral sign gives that &#xA;$$f&amp;#39;&amp;#39;(t) + f(t) = \int_0^{\infty} {e^{tx} + x^2e^{tx} \over \pi(1 + x^2)}$$&#xD;&#xA;$$ = {1\over \pi} \int_0^{\infty}e^{tx}$$&#xD;&#xA;The latter integral integrates to ${\displaystyle-{1 \over t}}$, so that you have ${\displaystyle f&amp;#39;&amp;#39;(t) + f(t) = -{1 \over \pi t}}$. This can be solved via variation of parameters, and (according to wolframalpha at least) it does not have an elementary expression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the nonelementary functions in the solution just involve indefinite integrals of ${\displaystyle {\sin(t) \over t}}$ and ${\displaystyle{\cos(t) \over t}}$, so if you want to do something this advanced you might be able to get something satisfactory using initial conditions for this differential equation. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-12T03:41:43.667" LastActivityDate="2010-12-12T03:41:43.667" />
  <row Id="14004" PostTypeId="1" AcceptedAnswerId="14008" CreationDate="2010-12-12T03:44:00.213" Score="13" ViewCount="234" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&quot;&gt;Convergence of $\\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to show the convergence or divergence of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{n=1}^\infty \frac{\sin^n 1}{n} = \frac{\sin 1}{1} + \frac{\sin \sin 1}{2} + \frac{\sin \sin \sin 1}{3} + \ \cdots $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the superscript means iteration (&lt;em&gt;not&lt;/em&gt; multiplication, so it's not simply less than a geometric series -- I couldn't find the standard notation for this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \sin^n 1 \to 0 $ as $ n \to \infty $ (which I eventually proved by assuming a positive limit and having $ \sin^n 1 $ fall below it, after &lt;a href=&quot;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;getting its existence&lt;/a&gt;) helps the series to converge,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but at the same time&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$ \sin^{n+1} 1 = \sin \sin^n 1 \approx \sin^n 1 $ for large $ n $ makes it resemble the divergent harmonic series.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would appreciate it if someone knows a helpful convergence test or a proof (or any kind of advice, for that matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it's useful, here are some things I've tried:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Show $ \sin^n 1 = O(n^{-\epsilon}) $ and use the p-series.  I'm not sure that's even true.&lt;/li&gt;&#xA;&lt;li&gt;Computer tests and looking at partial sums.  Unfortunately, $ \sum 1/n $ diverges very slowly, which is hard to distinguish from convergence.&lt;/li&gt;&#xA;&lt;li&gt;Somehow work in the related series&#xA;$$ \sum_{n=1}^\infty \frac{\cos^n 1}{n} = \frac{\cos 1}{1} + \frac{\cos \cos 1}{2} + \frac{\cos \cos \cos 1}{3} + \ \cdots $$&#xD;&#xA;which I know diverges since the numerators approach a fixed point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="336" LastActivityDate="2010-12-12T04:48:25.280" Title="Convergence of series involving iterated $ \sin $" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;tag-removed&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="14005" PostTypeId="2" ParentId="13968" CreationDate="2010-12-12T03:58:56.323" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the case of $R=\mathbb{Z}$ and $M=\mathbb{Z}_{p^{\infty}}$, the Prufer group (the subgroup of $\mathbb{Q}/\mathbb{Z}$ generated by $\frac{1}{p^n}$ for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $F$ be freely generated on countably many generators $x_n$, with $x_n$ mapping to $\frac{1}{p^n}$. Then $K$ is freely generated by $px_1$, $x_1-px_2$, $x_2-px_3,\ldots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that as you try to perform the procedure in step $4$, at every step you get $a_i = 1$. To see this, simply notice $M$ is divisible, so you can never have a finite direct summand. The subgroup generated by the $y_i$ does not generate $K$, but only a proper submodule of $K$. The process does not capture all of $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the real question is what is &quot;morally&quot; going wrong? I'm not sure I can give a good answer to that right now, other than to point out that it &lt;em&gt;does&lt;/em&gt; go wrong. My feeling is that the process described is trying to arrange things so that we can describe $K$ as a &quot;nice&quot; submodule by a good change of basis; we arrange things so that the first coordinate is chosen nicely, and then &quot;push&quot; the rest of the problem to the remaining coordinates. Because at each step we can retain the things we already fixed, and we can only take so many steps before we run out of room, we eventually have fixed the problem. But in the infinitely generated case, not only do we not &quot;run out of room&quot;, we may be unable to even fix that first coordinate in the first place! I think it comes down to the fact that even though $\mathbb{Z}$ is a PID, you can still have an infinite &lt;em&gt;descending&lt;/em&gt; chain of ideals, like $(p)\supset (p^2)\supset (p^3)\supset\cdots$, which in a sense is what we have with $M$ above. Trying to go &quot;transfinite&quot; doesn't solve the problem because $\cap (p^n) = (0)$ loses all the information about the torsion of $M$. In the finitely generated case this cannot occur, because we have a definite &quot;floor&quot; below which we cannot go; the infinitely generated case does not have that reassurance, which is what is going wrong here and why you keep getting $a_i = 1$, so you are never actually finding a way to describe $K$ effectively this way.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T04:17:00.763" LastActivityDate="2010-12-12T04:17:00.763" />
  <row Id="14006" PostTypeId="1" AcceptedAnswerId="14011" CreationDate="2010-12-12T04:02:26.333" Score="2" ViewCount="58" Body="&lt;p&gt;I don't have enough memory to simply create diagonal matrix D x D, getting an 'out of memory' error.&#xA;Instead of performing M x D x D operations in the first multiplication, I do M x D operations, but still my code takes ages to run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody find a more effective way to multiply these matrices in matlab?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D=20000&#xA;M=25&#xA;&#xA;A = floor(rand(D,M)*10);&#xA;B = floor(rand(1,M)*10);&#xA;&#xA;&#xA;for i=1:D&#xA; for j=1:M&#xA;  result(i,j) = A(i,j) * B(1,j);&#xA; end&#xA;end&#xA;&#xA;&#xA;manual = result * A';&#xA;auto = A*diag(B)*A';&#xA;isequal(manual,auto)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1996" LastEditorUserId="1996" LastEditorDisplayName="" LastEditDate="2010-12-12T18:01:39.040" LastActivityDate="2010-12-12T18:01:39.040" Title="Efficient multiplication of super-size matrices in matlab" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;matlab&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14007" PostTypeId="2" ParentId="13993" CreationDate="2010-12-12T04:11:54.713" Score="1" ViewCount="0" Body="&lt;p&gt;The general idea behind transform methods for solving PDEs is that they turn the differential equation into an ODE.  In the case of the wave equation, the Fourier transform acts on the space variable $x$ only, not $t$.  Since derivatives with respect to $x$ get transformed into multipliation by factors of $2\pi i \xi$ under the action of the transform, the result is a second order ODE in the variable $t$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are multiple conventions about the definition of the Fourier transform, having to do with where the factor of $2\pi$ goes, but for instance, one can define it by:&#xA;$$&#xD;&#xA;F(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i \xi x}\, dx&#xD;&#xA;$$&#xD;&#xA;In particular, notice that the transform is a function of $\xi$ (usually thought of as frequency).  Compare this with the definition you stated, in which the independent variable in the transformed function is $t$.  I think this may be the source of your confusion.&#xA;The method is basically the same in higher dimensions, letting $\xi$ be an $n$-vector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference (although not online) is Strauss's text on Partial Differential Equations.  I also feel compelled to point out that one does not need Fourier transforms to derive the general solution of the wave equation.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-12T04:11:54.713" />
  <row Id="14008" PostTypeId="2" ParentId="14004" CreationDate="2010-12-12T04:12:48.337" Score="10" ViewCount="0" Body="&lt;p&gt;A Google search has turned up an analysis of the asymptotic behavior of the iterates of $\sin$ on &lt;a href=&quot;http://books.google.com/books?id=_tnwmvHmVwMC&amp;amp;lpg=PA157&amp;amp;pg=PA157#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 157&lt;/a&gt; of de Bruijn's &lt;em&gt;Asymptotic methods in analysis&lt;/em&gt;.  Namely, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin^n(1)=\frac{\sqrt{3}}{\sqrt{n}}\left(1+O\left(\frac{\log(n)}{n}\right)\right),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies that your series converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Moron has pointed out in a comment that the problem of showing that $\sqrt{n}\sin^n(1)$ converges to $\sqrt{3}$ already appeared in the &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n&quot;&gt;question&lt;/a&gt; &quot;Convergence of $\sqrt{n}x_n$ where $x_{n+1}=\sin(x_n)$&quot; (asked by Moron in August).  I had missed or forgot about it.  David Speyer gave a great self contained answer, and he also referenced de Bruijn's book.  De Bruijn gives a reference to a 1945 work of Pólya and Szegő for this result.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-12T04:48:25.280" LastActivityDate="2010-12-12T04:48:25.280" CommentCount="2" />
  <row Id="14009" PostTypeId="1" CreationDate="2010-12-12T04:37:19.343" Score="2" ViewCount="90" Body="&lt;p&gt;Let $\mu$ be a finite positive measure on the borel sets in $\mathbb{R}$ and suppose $\mathcal{L}^{1}(\mathbb{R},\mu) \subset \mathcal{L}^{\infty}(\mathbb{R},\mu)$. &#xA;Show that there exists $c&amp;gt;0$ such that if A is a Borel set with $\mu(A) &amp;gt; 0$ then $\mu(A) \geq c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea at all. Can you please help?&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2010-12-16T06:11:51.387" Title="Integration problem, existence of a constant in a inclusion between integrable functions and essentially bounded functions" Tags="&lt;measure-theory&gt;" AnswerCount="3" />
  <row Id="14010" PostTypeId="1" AcceptedAnswerId="14015" CreationDate="2010-12-12T04:41:37.297" Score="1" ViewCount="56" Body="&lt;p&gt;What interesting properties of convex sets are retained by star-convex sets?&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-12-12T05:19:17.253" Title="How close are star-convex sets to convex sets?" Tags="&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;&lt;discrete-geometry&gt;&lt;convex-sets&gt;" AnswerCount="1" />
  <row Id="14011" PostTypeId="2" ParentId="14006" CreationDate="2010-12-12T04:51:26.253" Score="5" ViewCount="0" Body="&lt;p&gt;Your problem is discussed &lt;a href=&quot;http://www.mathworks.com/matlabcentral/newsreader/view_thread/248346&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. They suggest one of the following solutions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A*bsxfun(@times,diag(D),A')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A*sparse(D)*A'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Matlab is an interpreted language and so explicit loops are very slow (even though in the latest version there's supposedly in-place compilation). So everything must be vectorized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, did you search the internet at all? It took Google 0.16s to come up with that web page as the first result (for &lt;i&gt;matlab multiplying by a diagonal matrix&lt;/i&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-12T04:51:26.253" CommentCount="3" />
  <row Id="14012" PostTypeId="1" AcceptedAnswerId="14819" CreationDate="2010-12-12T04:58:08.653" Score="1" ViewCount="85" Body="&lt;p&gt;If you dropped &lt;strong&gt;two rocks in a pond&lt;/strong&gt;, the concentric circles emanating from the two spots would osculate $\infty$ times. The locus of osculating points would form a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that instead of concentric circles, &lt;strong&gt;concentric ellipses emanate&lt;/strong&gt; from the rocks' entry points. Again the concentric ellipses will osculate and again let's consider the locus of points where they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the shape of the locus?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-13T00:59:14.747" LastActivityDate="2010-12-18T23:52:59.840" Title="Locus of osculation of concentric ellipses (elliptic pond ripples)" Tags="&lt;calculus&gt;&lt;differential-geometry&gt;&lt;euclidean-geometry&gt;&lt;analytic-geometry&gt;&lt;vector-calculus&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="14013" PostTypeId="2" ParentId="14009" CreationDate="2010-12-12T04:58:11.897" Score="4" ViewCount="0" Body="&lt;p&gt;The rough idea is that if there are sets of arbitrarily small measure, then you can define unbounded integrable functions by making them get larger and larger but on subsets that are more quickly getting smaller and smaller.  This is why $L^1$ is not contained in $L^\infty$ in the case of $(0,1)$ with Lebesgue measure, for example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to give you a little more of the idea without giving everything away.  You can prove this with contraposition, so your assumption will be that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$.  Use this to construct an infinite sequence of disjoint Borel sets with positive measure converging to zero, say at a rate faster than $2^{-n}$.  Then define a function whose value on the $n^\text{th}$ set is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the work is in constructing the sequence of sets.  I'd rather not spoil your fun by saying more, unless you get stuck along the way and have a specific question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;December 15 update:  To enjoy the new &lt;a href=&quot;http://meta.stackoverflow.com/questions/1191/add-markdown-support-for-hidden-until-you-click-text-aka-spoilers/71396#71396&quot;&gt;spoiler feature&lt;/a&gt;, I decided to post a sketch of one way to construct the sequence in the box below.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Suppose that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$.  Start with any Borel set $A_0$ such that $0\lt\mu(A_0)\lt \infty$.  (Technically the finiteness is redundant here, because $\mu$ is assumed to be finite.)  For each positive integer $n$, the hypothesis guarantees that given $A_{n-1}$ with positive measure, there is a Borel set $A_n$ such that $0\lt\mu(A_n)\lt \frac{1}{3}\mu(A_{n-1})$.  Do this for every positive integer (countable AC).  Then define a sequence of Borel sets $B_0,B_1,\ldots$ by $B_n=A_n\setminus\cup_{k\gt n}A_k$.   The inequality $\mu(A_k)\lt 3^{n-k}\mu(A_n)$ for all $k\gt n$ implies that each $B_n$ has positive measure (at least half of $\mu(A_n)$) and that $\sum_{n\geq 0}n\mu(B_n)$ converges.  If $m\gt n$, then $B_m$ is contained in $A_m$, while $B_n$ is contained in $A_n\setminus A_m$, so $B_m$ and $B_n$ are disjoint.  The function $\sum_{n\geq1}n\chi_{B_n}$ is thus in $L^1\setminus L^\infty$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-16T06:11:51.387" LastActivityDate="2010-12-16T06:11:51.387" />
  <row Id="14014" PostTypeId="2" ParentId="13959" CreationDate="2010-12-12T05:05:43.503" Score="0" ViewCount="0" Body="&lt;p&gt;Hans and Jonas pretty much tell you how to do it. You are basically trying to calculate the volume on the square with side 0:1, and coordinates of each of the random variables, and height equal to the minimum segment. So you have several regions, as shown by Jonas, each of which you could obtain an analytic integral for. There will be symmetry above/below the diagonal, so you can compute just one and double it. The problem is it is quite a bit of algebra, and it is just too easy to make a mistake. A wrote a simple monte-carlo program, pich two random values, r1,r2, compute the minimum, and do this N times. My results for the differing values of N:&#xA; n=1K .111988&#xA; n=10k .111647&#xA; n=100k .111101&#xA; n=1m .111298&#xA; n=10m .111121&#xA; n=100m .111129&#xA; n=1G .111111&#xA;   So it sure seems to be converging on 1/9.&#xA;  Monte Carlo, is a good way to gain confidence in your result, i.e. if you did&#xA;an analytic solution and it agrees with the Monte Carlo result, it is probably correct. If it doesn't agree you have probably made a mistake. Programming the Monte Carlo for this problem only took a few minutes.&lt;/p&gt;&#xA;" OwnerUserId="4664" LastActivityDate="2010-12-12T05:05:43.503" />
  <row Id="14015" PostTypeId="2" ParentId="14010" CreationDate="2010-12-12T05:05:47.537" Score="5" ViewCount="0" Body="&lt;p&gt;&quot;Star-convex set&quot; is a bit of a misnomer. The prevailing term probably should have been &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Star_domain&quot; rel=&quot;nofollow&quot;&gt;star domain&lt;/a&gt;&quot; instead (but star-convex is so common that the ship has sailed on that one) since that is what you use it for: An open star domain is a simply connected domain - a handy fact for proving simple cases of theorems in e.g. complex analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, they have very few of the properties of convex sets. For example: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The intersection of two star-convex sets need not be connected (and thus in particular not star-convex). You can take two L-shaped sets and their intersection will be two disjoint squares.&lt;/li&gt;&#xA;&lt;li&gt;The interior of a star-convex set need not be connected. An example is $\mathbb{C} \setminus \lbrace x+iy \;\vert\; x = 0 \text{ and } y \neq 0 \rbrace$, i.e. the right half-plane plus the left half-plane plus the origin.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-12T05:19:17.253" LastActivityDate="2010-12-12T05:19:17.253" CommentCount="2" />
  <row Id="14016" PostTypeId="2" ParentId="14009" CreationDate="2010-12-12T05:26:23.250" Score="3" ViewCount="0" Body="&lt;p&gt;What Jonas Meyer did is definitely the most natural approach, but there's also a functional analysis approach to such problems I find amusing. In this case, for each $A$ with $\mu(A) &amp;gt; 0$ define&#xA;$$T_A(f) = {1 \over \mu(A)} \int_A f$$&#xD;&#xA; Then $T_A$ is a bounded linear map from $L^1({\mathbb R},\mu)$ to ${\mathbb C}$, and by your condition that $L^1({\mathbb R},\mu) \subset L^{\infty}({\mathbb R},\mu)$, for each $f$ there is a constant $C_f = ||f||_{\infty}$ such that  $||T_A f|| \leq C_f$ for all $A$. Hence by the uniform boundedness principle there is a single constant $C$ such that $||T_A f|| \leq C||f||_1$ for all $A$. Taking $f$ to be the characteristic function of $A$ this leads to $1 \leq C\mu(A)$, which is equivalent to what you are trying to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And those of you who are familiar with the proof of the uniform boundedness principle may see the connection between this proof and Jonas's.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-12T05:26:23.250" CommentCount="1" />
  <row Id="14017" PostTypeId="2" ParentId="13993" CreationDate="2010-12-12T05:34:01.760" Score="2" ViewCount="0" Body="&lt;p&gt;The things that you wanted to know is on pages 184--189 of Stein-Shakarchi 's Fourier Analysis. In particular, you will see how the solution is written as $$f(x+ct)+g(x-ct)$$ on p. 188. (Substitute $\frac{1}{2}(f(x+t)+G(x+t))$ and $\frac{1}{2}(f(x-t)-G(x-t))$ in the book for your $f(x+ct)$ and $g(x-ct)$ respectively), where $G$ denotes the integral (primitive) of $g$. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-12T16:10:11.210" LastActivityDate="2010-12-12T16:10:11.210" CommentCount="1" />
  <row Id="14018" PostTypeId="2" ParentId="3399" CreationDate="2010-12-12T06:33:58.330" Score="2" ViewCount="0" Body="&lt;p&gt;Instead of thinking about the graph of $f: \mathbb{R} \to \mathbb{R}$ as a 2-D object in the plane, think about $f$ mapping one number line onto another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/hFH2J.jpg&quot; alt=&quot;drawing of a convex function from ℝ to ℝ&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll return to that picture, just notice that there is plenty of &quot;room&quot; so to speak and that the arrows mapping point to image will never &quot;fall back&quot; on each other, overlap, or &quot;crowd in&quot; so long as the function is convex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a closed loop in the plane whose interior is non-convex. It's like a deflated balloon. You need to &quot;blow it up&quot; until it's at least modestly ($\leq$) full of air for the interior to be convex. Similarly if you had a non-convex polygon and &quot;blew air&quot; inside of it, you would get a convex shape. So it's like convex shapes have to be sufficiently &quot;inflated&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the arrows in an $\mathbb{R} \to \mathbb{R}$-type picture like the above would be &quot;flopping&quot; inward onto each other, deflated if you will. In the convex mapping the arrows don't overlap at all -- they've got &quot;pressure&quot; or &quot;energy&quot; pushing them outward enough so that they don't overlap. So the image is properly inflated, if you will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So @NateEldridge, the epigraph being a convex set is a red herring. Think about just the right-most point of a graph as it's being generated by a s-l-o-w graphing calculator. The image has to &quot;outrun&quot; the domain it comes from by $\geq$ each $dt$. And there you have your $f(\mathrm{interior\ of\ domain}) \leq \mathrm{image}_1 + \mathrm{image}_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is meant as an elaboration on @whuber's answer.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-12T06:39:39.963" LastActivityDate="2010-12-12T06:39:39.963" CommunityOwnedDate="2010-12-12T06:33:58.330" CommentCount="3" />
  <row Id="14019" PostTypeId="1" CreationDate="2010-12-12T06:42:25.387" Score="0" ViewCount="42" Body="&lt;p&gt;Is the singular locus of the singular locus of a variety somehow &quot;more singular&quot; in the original variety than the rest of the singular locus?  I know there are many complicated tools to measure singularity, but is there any elementary way to see it in this case?&lt;/p&gt;&#xA;" OwnerUserId="4665" LastActivityDate="2010-12-12T06:42:25.387" Title="Singular locus of the singular locus of a variety" Tags="&lt;algebraic-geometry&gt;" CommentCount="2" FavoriteCount="0" />
  <row Id="14020" PostTypeId="1" AcceptedAnswerId="14021" CreationDate="2010-12-12T06:58:44.907" Score="4" ViewCount="45" Body="&lt;p&gt;Is there a good online reference that lists known bounds on Ramsey numbers (and is relatively up to date)? The &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramsey%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; only has numbers for $R_2(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically interested in known bounds and values for hypergraph Ramsey Numbers, i.e. 2 colorings of k-subsets (or 2 colorings of the edges of a complete k-uniform hypergraph).  These are commonly denoted $R_k(m,n)$.  A shallow internet search has yielded only a couple sets of papers and notes on bounds.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-12-12T07:08:49.740" Title="Known bounds and values for Ramsey Numbers." Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14021" PostTypeId="2" ParentId="14020" CreationDate="2010-12-12T07:08:49.740" Score="6" ViewCount="0" Body="&lt;p&gt;The best resource online is the (frequently updated) survey &quot;Small Ramsey Numbers&quot; by Stanisław Radziszowski, in The Electronic Journal of Combinatorics. Go to &lt;a href=&quot;http://www.combinatorics.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.combinatorics.org/&lt;/a&gt; and click on the link for &quot;dynamic surveys&quot;. The most recent update on the Ramsey numbers paper is from August 2009. The paper gives extensive references where one can find complete proofs or details of the computations involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some recent work on asymptotic bounds that is of significance, but the survey has not addressed yet. For that, you may want to look at the papers in the pages of Jacob Fox, David Conlon, and their collaborators.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-12T07:08:49.740" CommentCount="1" />
  <row Id="14022" PostTypeId="2" ParentId="13514" CreationDate="2010-12-12T07:14:34.147" Score="0" ViewCount="0" Body="&lt;p&gt;You are doing it indirectly, the direct way to do this is by defining a one to one function from $N$ to the members of the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $M$ be a model of the first two axioms. (To able to to talk about the model you need a theory that can talk about sets to some extent or a theory that can interpret that amount of set theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing is defining the function inductively in the obvious way and then use induction and the fact that $M$ satisfies the first and second axiom to prove that the function is one to one.&lt;/p&gt;&#xA;" OwnerUserId="468" LastActivityDate="2010-12-12T07:14:34.147" />
  <row Id="14023" PostTypeId="2" ParentId="13898" CreationDate="2010-12-12T07:17:19.820" Score="1" ViewCount="0" Body="&lt;p&gt;I'd like to add an answer that will try to justify the statement with an example. Consider the function sequence&#xA;$$&#xD;&#xA;f_n = \left\{\begin{align} 0 \ \ &amp;amp;\text{ if } |x|&amp;gt;1/n \\&#xD;&#xA;n^2x^2/2 + nx + 1/2 &amp;amp;\text{ if } -1/n\leq x \leq 0\\&#xD;&#xA;-n^2x^2/2 + nx + 1/2 &amp;amp;\text{ if } 0&amp;lt;x\leq 1/n&#xD;&#xA;\end{align}\right..&#xD;&#xA;$$&lt;br&gt;&#xA;These functions are differentiable everywhere, and they approximate the unit step function better and better as $n$ gets larger. If we look at the derivatives of these functions, we'll see that they are a sequence of &lt;em&gt;tent functions&lt;/em&gt; that are getting narrower and taller, as Griffith uses to introduce the delta function.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/I79Ct.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, with an appropriate mode of convergence, when a sequence of differentiable functions converge to the unit step, it can be shown that, their derivatives converge to the delta function. That's why, one may take the derivative of the unit step function to be &lt;em&gt;defined&lt;/em&gt; as the limit of the derivatives, which is the delta function. &lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-12-12T07:50:04.937" LastActivityDate="2010-12-12T07:50:04.937" />
  <row Id="14024" PostTypeId="2" ParentId="14000" CreationDate="2010-12-12T07:36:34.253" Score="5" ViewCount="0" Body="&lt;p&gt;The surface integral of a vector field, $F$ over a surface $S$ (or flux as it is called) is, as stated above, the net flow of fluid through the surface.  The purpose of orientation is to simply indicate which side of the surface is the &quot;positive&quot; side.  For instance, if I calculate flux to be +10, this means there is a net movement of 10 units of fluid towards whichever side of the surface is indicated as the positive side.  The answer would be -10 if the orientation was reversed (since the &quot;positive&quot; side is now losing 10 units of fluid).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asked to find the flux of $F$ over $S$, they have to tell you the orientation of $S$ in the question.  For instance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Find the flux of $F(x,y,z) = &amp;lt;2xz,z-y,y^3&amp;gt;$ DOWNWARD through the section of the plane $x+y+z=6$ in the first octant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question, the region under the plane is the positive side, and a positive flux would indicate a net movement of fluid downward through the plane.  A negative answer indicates a net upward movement of fluid.  So, how do you figure out orientation for a surface?  You don't.  It's usual stated in the question. In some cases they don't state orientation because they will accept either the positive or negative answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some theorem's such as Stokes' and Gauss' Divergence Theorem require that the surface in question has a particular orientation.  Stokes' theorem for instance, allows us to calculate a line integral along a closed curve by first choosing any surface enclosed by this curve.  If the curve is traversed counterclockwise as viewed from above, the orientation of the surface must be &quot;Upward&quot; to use this theorem.  So, whatever surface you choose, it's normal vector must point upward. In this case, you are the one choosing the surface so you so have to figure out the orientation yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should finally point out that I am assuming you are only working with surfaces in $R^3$.  Orientation of a surface in $R^n$ (or hypersurfaces in $R^n$) is a very difficult concept since these are not real world objects that you can look at and decide the orientation of.  Also, some 2d surfaces such as a mobius strip, have no orientation since they only have one side.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="3971" LastEditorDisplayName="" LastEditDate="2010-12-17T07:49:42.317" LastActivityDate="2010-12-17T07:49:42.317" />
  <row Id="14026" PostTypeId="1" AcceptedAnswerId="14029" CreationDate="2010-12-12T07:57:52.330" Score="3" ViewCount="250" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Which computer programs are useful for a pure mathematician to familiarize themselves with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Less Briefly:&lt;/strong&gt; I was once told that, now-a-days, any new mathematician worth his beans knows how to TeX up work; when I began my graduate work one of the fourth year students told me he couldn't TeX, I was horrified!  Similarly, a number of my peers were horrified when I told them I'd never used Matlab or Mathematica before.  Currently, I can &quot;get by&quot; in all three of these programs, but it made me think: what programs are popular with pure mathematicians?  I don't mean to limit this to computing things: programs which help to draw pictures and things can also be included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lest I Start A Flame War:&lt;/strong&gt; This is not meant to be a &quot;what is your favorite computer language&quot; question or a poll on who thinks Mathematica is worse than Sage.  This is meant to be a survey on what programs and resources are currently being used by pure mathematicians and may be good to look at and read up on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also unsure of how to tag this.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-12T08:02:21.490" LastActivityDate="2010-12-12T22:46:48.310" Title="Computer Programs for Pure Mathematicians." Tags="&lt;math-software&gt;&lt;computer-algebra-systems&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="2" />
  <row Id="14027" PostTypeId="1" CreationDate="2010-12-12T08:04:12.560" Score="0" ViewCount="55" Body="&lt;h2&gt;Some background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have the following homogeneous transformation matrix (transpose by 2 on Y axis) in raytracing calculations to transform vectors from object space to world space after a ray-sphere intersection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{OW} = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 2  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corresponding matrix to transform from world space to object space is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{WO}=M_{OW}^{-1} = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; -2  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas regular vectors are transformed from object space to world space like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{WS} = M_{WO} v_{OS}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normals to the surface are transformed different way using a plane perpendicular to the normal instead of the normal itself, see &lt;a href=&quot;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-837-computer-graphics-fall-2003/lecture-notes/5_1_trans_hier.pdf&quot; rel=&quot;nofollow&quot;&gt;MIT's lecture slides on computer graphics&lt;/a&gt; [PDF warning].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{WS} = (M_{OW}^{-1})^T  n_{OS} = M_{WO}^T  n_{OS}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us a transformation matrix for normals from object space to world space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{WO}^T = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Because this particular transformation is translation-only, I expect all vectors represented as points to infinity (where $w = 0$) to be unaffected, including normals (since planes perpendicular to normals are unaffected). But when I transform the following normal, it gets a non-zero $w$ component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{OS} = &#xD;&#xA;\left[ {\begin{array}{*{10}c}&#xD;&#xA;   -0.9 \\&#xD;&#xA;   0.99 \\&#xD;&#xA;   -0.17 \\&#xD;&#xA;   0 \\&#xD;&#xA; \end{array} } \right]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{WS} = &#xD;&#xA;\left[ {\begin{array}{*{40}c}&#xD;&#xA;   1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0  \\&#xD;&#xA;   0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 1  \\&#xD;&#xA;\end{array} } \right]&#xD;&#xA;\left[ {\begin{array}{*{10}c}&#xD;&#xA;   -0.9 \\&#xD;&#xA;   0.99 \\&#xD;&#xA;   -0.17 \\&#xD;&#xA;   0 \\&#xD;&#xA; \end{array} } \right] = &#xD;&#xA;\left[ {\begin{array}{*{10}c}&#xD;&#xA;   -0.9 \\&#xD;&#xA;   0.99 \\&#xD;&#xA;   -0.17 \\&#xD;&#xA;   -1.98 \\&#xD;&#xA; \end{array} } \right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fail to visualize how a normal in a ray-sphere intersection would get a non-zero $w$. Intuitively it should remain the same if I only translate along the Y axis, but obviously this is not the case here.&lt;/p&gt;&#xA;" OwnerUserId="2360" LastActivityDate="2010-12-12T08:04:12.560" Title="How to interpret normals transformation in homogeneous coordinates?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" />
  <row Id="14028" PostTypeId="2" ParentId="13519" CreationDate="2010-12-12T08:06:50.750" Score="2" ViewCount="0" Body="&lt;p&gt;This question is one of the main topics studied in &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_complexity&quot; rel=&quot;nofollow&quot;&gt;proof complexity&lt;/a&gt;. The measure you are talking about is called the number of &lt;em&gt;steps&lt;/em&gt; in the proof or proof-length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of a tautology is related to the number of steps of its shortest proof, but this does not mean that the number of steps is a monotone function of size of tautology. As mentioned by others, the length of a long tautology can be very short. Coming up with arbitrary long tautologies having proofs with constant (i.e. independent from the size of tautology) number of steps is not difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the task of coming up with tautologies with large number of steps is not obvious and is open for many propositional proof systems. Note that it follows from the (constructive) proof of completeness theorem that every tautology of size $s$ has a proof of height $O(s)$ with $O(2^s)$ steps and size $O(s2^s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have some very weak lowerbounds for maximum number of steps needed for a tautology of size $s$ and the question that this upperbound is tight is wide open for sequent calculus and also for natural deduction which are essentially equivalent to Hilbert style proof systems. The bound is tight if we remove cut rule from sequent calculus, i.e. there are tautologies which require exponential number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find out more about the relation between different measure studied in proof complexity and the relation between different propositional proof systems by checking surveys/books on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For first order logic, note that the first order logic is undecidable, i.e. given a formula there is no algorithm to check if it is a valid formula. Any computable upperbound on the proof size would give an algorithm for deciding first order logic, check all possible proofs up to that size. (Note that an upper bound on proof length in natural deduction/sequent calculus will give an upper bound on the size because of normalization/cut elimination). Since there is no such algorithm, there cannot be any computable upperbound.&lt;/p&gt;&#xA;" OwnerUserId="468" LastEditorUserId="468" LastEditorDisplayName="" LastEditDate="2010-12-12T08:15:30.170" LastActivityDate="2010-12-12T08:15:30.170" />
  <row Id="14029" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T08:46:43.760" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Mathematica&lt;/strong&gt; is the de-facto standard for symbolic computation. I would argue that there is no other CAS that matches the sheer amount of stuff that Mathematica can do (this is not to say that there aren't other systems out there that do certain things better - like graphing/plotting). Part of the power of Mathematica's strength is the ability to write and evaluate expressions just like you would in your notebook (hence the &quot;notebook&quot; UI paradigm) , however, its reliance on symbolic computation also is its greatest weekness (in terms of strength as a programming language, portability, and speed). Just because of its ubiquity, if there is one thing you should learn, its this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Matlab&lt;/strong&gt; is the de-facto standard for numerical computations, under the broad category of numerical analysis. It is less designed for symbolic analysis, but its numerical analysis tools are industry standard. With countless additional packages it can seamlessly (well almost seamlessly, and for several thousand dollars) incorporate data capture from everything from webcams to particle physics experiments. Its usefulness as a &lt;em&gt;pure mathematics&lt;/em&gt; tool might not be as as obvious as Mathematica, but if you ever plan on collaborating with applied mathematicians, physicists, or engineers, working knowledge of Matlab is a must.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Python + Scipy + numpy&lt;/strong&gt; . By combining the free programming/scripting language with powerful (C-based) packages for scientific computation, this suite can rival the numerical analysis tools of Matlab. If you extend this combination with Matlplotlib (numpy based package for plotting) you can get publication quality plots and graphs with a familiar syntax (matlab) for $0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GnuPlot&lt;/strong&gt;. With the ability to graph/plot pretty much anything you want and a very powerful text based input, this software allows you to create publication quality plots with ease (after learning the syntax of course). Plus its, free and open source.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2010-12-12T08:46:43.760" CommentCount="6" />
  <row Id="14030" PostTypeId="2" ParentId="13996" CreationDate="2010-12-12T08:48:51.507" Score="1" ViewCount="0" Body="&lt;p&gt;Correspondingly to Zaricuse's second paragraph, according to &lt;a href=&quot;http://www.efunda.com/math/laplace_transform/forward.cfm?FuncName=Basic&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (see the formula before the last one), for any $s &amp;gt; 0$,&#xA;$$&#xD;&#xA;\int_0^\infty  {e^{ - sx} \frac{1}{{\pi (1 + x^2 )}}\,{\rm d}x}  = \frac{1}{\pi } \Big \lbrace \cos (s)\Big[\frac{\pi }{2} - {\rm Si}(s)\Big] - \sin (s){\rm Ci}(s)\Big \rbrace,&#xD;&#xA;$$&#xD;&#xA;where ${\rm Si}(s) = \int_0^s {\frac{{\sin u}}{u}} \,{\rm d}u$ is the Sine Integral, and ${\rm Ci}(s) = \int_s^\infty  {\frac{{\cos u}}{u}\,{\rm d}u}$ the Cosine Integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Probabilistic interpretation of the integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function $f(x) = \frac{1}{{\pi (1 + x^2 )}}$, $x \in \mathbb{R}$, is the density function of a standard Cauchy random variable $X$. The divergent integral&#xA;$$&#xD;&#xA;{\rm E}[e^{tX}] = \int_{ - \infty }^\infty  {e^{tx} f(x)\,{\rm d}x}  = \infty, \;\; t \neq 0, &#xD;&#xA;$$&#xD;&#xA;corresponds exactly to the (elementary) fact that the moment-generating function of the Cauchy distribution does not exist.&#xA;On the other hand, the characteristic function is well known to be&#xA;$$&#xD;&#xA;{\rm E}[e^{{\rm i}{\rm t}X}] = \int_{ - \infty }^\infty  {e^{{\rm i}tx} f(x)\,{\rm d}x}  = e^{ - |t|}, \;\; t \in \mathbb{R}.&#xD;&#xA;$$&#xD;&#xA;Finally, the function $\tilde f(x) = \frac{2}{{\pi (1 + x^2 )}}$, $x &amp;gt; 0$, is the density function of $|X|$. Then, the Laplace transform of $|X|$ is &#xA;$$&#xD;&#xA;{\rm E}[e^{-t|X|}] = \int_0^\infty  {e^{ - tx} \tilde f(x)\,{\rm d}x},\;\; t &amp;gt; 0,&#xD;&#xA;$$&#xD;&#xA;and, according to the first paragraph, can be expressed in terms of the Sine Integral and Cosine Integral functions.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-12T10:11:45.370" LastActivityDate="2010-12-12T10:11:45.370" CommentCount="2" />
  <row Id="14031" PostTypeId="2" ParentId="14000" CreationDate="2010-12-12T09:55:08.667" Score="-1" ViewCount="0" Body="&lt;p&gt;Maybe some trivial examples could show you the importance of orientation. Try to compute the fluid flow through the unit square&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;S = \left\{(x,y,z) \ | \ 0\leq x,y \leq 1 , z=0 \right\}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of some fluid moving upwards at constant speed of $1$m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;F(x,y,z) = (0,0,1) \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you parametrize your surface, for instance with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\varphi (x,y) = (x,y,0) , \quad (x,y) \in D = \left\{ (x,y) \ | \ 0\leq x,y \leq 1  \right\}&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal field to this parametrization is $\varphi_x \times \varphi_y = (0,0,1)$ and we compute the fluid flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_\varphi \langle F, dS\rangle = \int_0^1\int_0^1 \langle (0,0,1) , (0,0,1) \rangle dxdy = 1 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why using that particular parametrization? Why not trying with this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\psi (u,v) = (v,u,0) , \quad (u,v) \in D = \left\{ (u,v) \ | \ 0\leq x,y \leq 1  \right\} \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We compute the normal field again: $\psi_u \times \psi_v = (0,1,0)\times (1,0,0) = (0,0,-1)$. And the fluid flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_\psi \langle F, dS \rangle = \int_0^1\int_0^1 \langle (0,0,1) , (0,0,-1) \rangle dudv = -1 \ .&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astonishing, isn't it? -The fluid flow seems to depend on something secondary: the coordinates you use on the surface $S$. This could be even worse, since some surface integrals are used to compute the area of the surface: if the area of a surface depended on the coordinates you put in it, this would be a nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, this isn't true: you can prove that surface integrals do NOT depend on the coordinates you put on the surface EXCEPT (in the case of fluid flows) on the &lt;em&gt;orientation&lt;/em&gt; of the coordinates -parametrizations. There are two kinds of orientations and before computing a fluid flow you must choose one of them. For instance, in our examples, if we have said that the fluid is going &lt;em&gt;upwards&lt;/em&gt; we have to choose $(0,0,1)$ as the orientation of our surface (the $\varphi$ parametrization). Otherwise, the result we get -with the $\psi$ parametrization- makes no sense: in one second $-1$ litre has gone trhough the surface!?&lt;/p&gt;&#xA;" OwnerUserId="664" LastActivityDate="2010-12-12T09:55:08.667" CommentCount="1" />
  <row Id="14032" PostTypeId="1" CreationDate="2010-12-12T10:13:04.810" Score="2" ViewCount="164" Body="&lt;p&gt;I'm having trouble with this integral&lt;br&gt;&#xA;$$\int\arcsin(\sin x)\,\mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is with the intervals of definition for each function :/ if someone could dumb it down for me.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\arcsin\colon [-1, 1] \to [-\pi/2, \pi/2]$.  &lt;/li&gt;&#xA;&lt;li&gt;$\sin:\mathbb{R}\to [-1, 1]$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about $\arcsin(\sin x)$ ?&lt;/p&gt;&#xA;" OwnerUserId="2227" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T05:52:05.243" LastActivityDate="2010-12-15T05:52:05.243" Title="integral of $\arcsin(\sin(x))$" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14033" PostTypeId="2" ParentId="14032" CreationDate="2010-12-12T10:20:32.083" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;My previous hint was a bit misleading - hence here is a (the comments are still appropriate)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CORRECTED HINT&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On $I= [-\pi/2,\pi/2]$ the sine function grows from $\sin(-\pi/2)=-1$ to $\sin(\pi/2)=1$, hence it is invertible on $I$. The inverse, $\arcsin$, is defined on $[-1,1]$ and has the property $\arcsin(\sin x) = x$ if $-\pi/2\le x\le \pi/2$.&#xA;Since $\sin$ is $2\pi$-peiodic we must have $\arcsin(\sin (x+2n\pi))=\arcsin(\sin(x))$ for all $x$ and integers $n$. &#xA;Hence it is sufficient to understand $\arcsin(\sin x)$ for $\pi/2\le x\le 3\pi/2$, to achieve that use $\sin(x+\pi)=\sin (-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you see the function there will be not problem to perform any integration.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-14T21:16:34.130" LastActivityDate="2010-12-14T21:16:34.130" CommentCount="3" />
  <row Id="14034" PostTypeId="2" ParentId="14032" CreationDate="2010-12-12T10:23:29.983" Score="1" ViewCount="0" Body="&lt;p&gt;Since $\sin(x)$ can take $ \mathbb R$ and outputs $[-1, 1]$ and $\arcsin(x)$ can take $[-1, 1]$ and output $[-\pi/2, \pi/2]$ , then it stands to reason that the constraints for $\arcsin(\sin(x))$ would be $\mathbb R \rightarrow [-\pi/2, \pi/2]$.&lt;/p&gt;&#xA;" OwnerUserId="3384" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-12T10:28:42.903" LastActivityDate="2010-12-12T10:28:42.903" />
  <row Id="14035" PostTypeId="1" CreationDate="2010-12-12T10:44:50.737" Score="4" ViewCount="56" Body="&lt;p&gt;I am able to measure my distance to a set of (about 6 or 7) fixed but unknown points from many positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in position between measurements is also unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that I should be able to work out the relative position of the fixed points, and therefore where I measured from and the path I took.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at the wiki page for trilateration, but it only gives examples working from known points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;" OwnerUserId="4668" LastActivityDate="2011-01-11T21:44:43.497" Title="Trilateration with unknown fixed points" Tags="&lt;euclidean-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14036" PostTypeId="1" CreationDate="2010-12-12T11:01:39.693" Score="0" ViewCount="58" Body="&lt;p&gt;&lt;strong&gt;Short version:&lt;/strong&gt; if two spherical caps of the same sphere intersect, how can I determined the coordinates of the two &quot;singular points&quot; of this intersection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long version:&lt;/strong&gt; On a unit sphere, centered at the original, we consider two spherical cap. Each cap is defined by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the (unit) vector indicating the top of the spherical cap, $\mathbf{n}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;the angle $\theta$, defined as is &lt;a href=&quot;http://wapedia.mobi/en/File%3aSteradian_cone_and_cap.svg&quot; rel=&quot;nofollow&quot;&gt;done there&lt;/a&gt;; any point $\mathbf{r}$ on the sphere belongs to the cap iff $\mathrm{angle}(\mathbf{n},\mathbf{r}) \le \theta$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The question is: given two spherical caps $(\mathbf{n}_1,\theta_1)$ and $(\mathbf{n}_2,\theta_2)$, their intersection can be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;empty&lt;/li&gt;&#xA;&lt;li&gt;a single point&lt;/li&gt;&#xA;&lt;li&gt;a surface delimited by two circular arcs, intersecting in two points&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The question is: in the third case, how can I determine the coordinates of the two points of interest? I'm sure it's not so hard if you find the right way to go at it.&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-12-12T13:12:07.270" Title="Intersection of spherical caps" Tags="&lt;geometry&gt;&lt;intersection&gt;&lt;3d&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14037" PostTypeId="1" CreationDate="2010-12-12T11:20:19.660" Score="3" ViewCount="167" Body="&lt;p&gt;In a And\Or-graph induced by the transition function, each node of G corresponds to a state q belonging to set Q of the state of the Automaton, for q with $\delta(q,a)=q1*q2$, the node is a $*-node$ with two successor q1 and q2. For q={true,false}, the node q is a sink-node. Hence, if * is $\vee$ then q is $\vee-node$, else q is $\wedge-node$. My problem is this : since that the result of transition function of a alternating automata includes more nodes in $\wedge$ and $\vee$ (example: $\delta(q_0,a)=(q1 \vee q2) \wedge q3 \wedge q4 )$ ), how to build the graph And\Or on transition function of an alternating automaton?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Graph And/Or graph is defined as following : A form of graph or tree used in problem solving and problem decomposition. The nodes of the graph represent states or goals and their successors are labeled as either AND or OR branches. The AND successors are subgoals that must all be achieved to satisfy the parent goal, while OR branches indicate alternative subgoals, any one of which could satisfy the parent goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead the alternating automata is a automata with transition function defined  as following : $\delta: S \times \Sigma \longrightarrow B^+(S)$ where S are the states of the automata ,$\Sigma$ the alphabet and $B^+(S)$ is the set of positive Boolean formulas over S.&lt;/p&gt;&#xA;" OwnerUserId="4670" LastEditorUserId="4670" LastEditorDisplayName="" LastEditDate="2010-12-19T15:02:56.680" LastActivityDate="2010-12-19T15:02:56.680" Title="Constructor And\Or-graph on function transition of the alternating automata" Tags="&lt;graph-theory&gt;&lt;computer-science&gt;&lt;propositional-calculus&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="14039" PostTypeId="2" ParentId="12091" CreationDate="2010-12-12T11:43:56.160" Score="0" ViewCount="0" Body="&lt;p&gt;You can try to treat your fuzzy logic algebraically, so as to have Tarskian non-boolean truth definition based on some fuzzy connectives. Some interesting choice is to use a t-norm for conjunction, and then view implication as a residuum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x * y : [0,1] \times [0,1] \rightarrow [0,1]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x \rightarrow y := max\{z| x*z \leq y\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above allows an easy comparison of fuzzy algebras with boolean algebras by the usual methods. But I am not sure whether fuzzy logic provides more or less insight or does use more or less parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original motivation for fuzzy logic was a linguistic observation, namely that in every day language we typically use attributions that express a degree of agreement, and that we even can reason with such degrees. So fuzzy logic is more a convenience for this machinery.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastActivityDate="2010-12-12T11:43:56.160" />
  <row Id="14040" PostTypeId="1" AcceptedAnswerId="14050" CreationDate="2010-12-12T11:44:57.033" Score="1" ViewCount="75" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find out for what $\alpha, \beta$ the following sum converges:&#xA;$$\sum_{n=2}^\infty n^\alpha (\log n)^\beta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I'd do that with the help of the integral criterion, that is to say I considered (after I had substituted $x = e^u$):&#xA;$$\int_2^\infty e^{u(\alpha+1)} \cdot u^\beta \mathrm du$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I realised that for $\alpha + 1 &amp;gt; 0$, the exponential function will always dominate the monomial, so I only considered $\alpha + 1 \leq 0$. First, I let $\alpha + 1=0$, then I would be left with:&#xA;$$\int_2^\infty u^\beta \mathrm du$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved the integral and substituted back and ended up with:&#xA;$$\lim_{b \to \infty} \frac{(\log b)^{\beta+1}}{\beta+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured this would only converge for $\beta+1 \leq 0$ and thought: &quot;Ok, now I have some cases, where the sum converges: $\alpha = -1 \land \beta \leq -1$&quot;. To be sure, I checked the integral with the help of wolframalpha, and it told me the integral would not converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I do wrong? Is there an easier way to find $\alpha$ and $\beta$ so that the sum converges?&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-12T12:29:24.793" Title="When does $\sum_{n=2}^\infty n^\alpha (\log n)^\beta$ converge?" Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="14041" PostTypeId="1" CreationDate="2010-12-12T11:50:34.047" Score="8" ViewCount="132" Body="&lt;p&gt;Given a bijection $f:\mathbb Z \to \mathbb Z$ where $\mathbb Z$ is the set of all integers, does there always exist two bijections $g:\mathbb Z \to \mathbb Z$ and $h:\mathbb Z \to \mathbb Z$ which satisfy $f(n) = g(n) + h(n), ~ \forall n \in \mathbb Z$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently it does while $\mathbb Z$ is replaced with the set of all rational numbers $\mathbb Q$, since $f(q) = 2f(q) - f(q), ~ \forall q \in \mathbb Q$. Nevertheless, $2f$ doesn't maps $\mathbb Z$ onto $\mathbb Z$, thus it's not a bijection on $\mathbb Z$; therefor we need a new construction&#xA;for the $\mathbb Z$ case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly we note that &amp;nbsp; $0 = 0 + 0$ &amp;nbsp; and&#xA;\begin{align*}&#xA;1 &amp;amp;= (-2) + 3, &amp;amp;\qquad -1 &amp;amp;= 2 + (-3),\\&#xA;2 &amp;amp;= 3 + (- 1), &amp;amp;-2 &amp;amp;= (-3) + 1,\\&#xA;3 &amp;amp;= 1 + 2, &amp;amp;-3 &amp;amp;= (-1) + (-2).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to show that any bijections on the symmetric integer interval $[-3,3]$ can be represented as a sum of two bijections; or equivalently, all integers $a \in [-3,3]$ can be represented as a sum of two others, i.e., $a = b + c$ where $b, c \in [-3, 3]$, while $a,b,c$ exhausted the integer interval $[-3,3]$ respectively, and keeps $0 = 0 + 0$. For convenience, let's denote this fact as $[-3,3] = [-3,3] + [-3,3]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now we have $[-m, m] = [-m, m] + [-m, m]$, then we may obtain&#xA;$$[-m + aL, ~ m + aL] = [-m + bL, ~ m + bL] + [-m + cL, ~ m + cL],\qquad L = 2m+1,$$&#xD;&#xA;where $a,b,c \in [-n,n]$ for some positive integers $n$, provided $a = b + c$. Well, if $a,b,c$ form a representation, say, $[-n,n] = [-n,n] + [-n,n]$, then we may get a &quot;larger&quot; representation&#xA;$$[-m - nL, ~ m + nL] = [-m - nL, ~ m + nL] + [-m - nL, ~ m + nL], \qquad L = 2m+1,$$&#xD;&#xA;through the repeatedly use of &quot;small&quot; representations; and obviously, the larger representation &quot;contains&quot; the small representation as a &quot;subrepresentation&quot;, since $0 = 0 + 0$. From now on, we may construct a representation $\mathbb Z = \mathbb Z + \mathbb Z$ by induction, start from $[-3,3] = [-3,3] + [-3,3]$. Thus we've proved that any bijections on $\mathbb Z$ can be represented as a sum of two bijections on $\mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a problem rises:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;for any integers&lt;/strong&gt; $n &amp;gt; 2$, &lt;strong&gt;is there always a representation&lt;/strong&gt; $[-n, n] = [-n, n] + [-n, n]$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that by our means of &quot;representation&quot;, $0 = 0 + 0$ is required; otherwise it's trivial; note also the above approach we've used for resolving the &quot;$\mathbb{Z}$-representation problem&quot; can't reach all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone can answer this question? Please help me.&lt;/p&gt;&#xA;" OwnerUserId="4666" LastEditorUserId="4666" LastEditorDisplayName="" LastEditDate="2010-12-13T07:47:39.280" LastActivityDate="2010-12-13T11:39:31.333" Title="bijection = bijection + bijection on symmetric integer intervals" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;algebra&gt;&lt;discrete-mathematics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="14042" PostTypeId="1" AcceptedAnswerId="14047" CreationDate="2010-12-12T11:52:32.550" Score="3" ViewCount="76" Body="&lt;p&gt;Is $\mathbf{Grp}$ a concrete category? I thought it is, but then the group of symmetries of a square and the quaternion group are both of the order 8, and they are not isomorphic as groups. But sets of the same cardinality are isomorphic as sets, so a standard forgetful functor $(G, \cdot) \mapsto G$ is not faithful! Am I missing something here?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2010-12-12T12:03:02.947" Title="Is $\mathbf{Grp}$ a concrete category?" Tags="&lt;category-theory&gt;" AnswerCount="1" />
  <row Id="14043" PostTypeId="1" AcceptedAnswerId="14046" CreationDate="2010-12-12T11:56:46.117" Score="1" ViewCount="93" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Problem:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Alex flip a fair coin three times. what is the probability that she gets two heads given that the first is a head?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My solution is based on the argument that from the given information the sample space can be limited to four elements {$HHH,HHT,HTH,HTT$} Of these {$HHH,HHT,HTH$} satisfies the required event.Thus the probability is equal to $\frac{3}{4} = 0.75$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the given answer is $0.5$,where exactly is the problem in my approach ?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-12T12:08:30.603" LastActivityDate="2010-12-12T12:14:16.823" Title="Doubt in a probability problem" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14044" PostTypeId="1" AcceptedAnswerId="14057" CreationDate="2010-12-12T11:58:18.590" Score="5" ViewCount="216" Body="&lt;p&gt;I'm going through an &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-781-theory-of-numbers-spring-2003/exams/final.pdf&quot; rel=&quot;nofollow&quot;&gt;old final&lt;/a&gt; from 2003 on MIT's Opencourseware, and problem 6b is giving me a little trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It asks for which primes $p$ is $34$ a square modulo $p$. I approached it like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{34}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{17}{p}\right)=(-1)^{(p^2-1)/8}\left(\frac{p}{17}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure I can break it down into cases where $p\equiv 1,3,5,7\pmod{8}$. So if $p\equiv 1\pmod{8}$, then $(2|p)=1$, and thus I want $(p|17)=1$ as well. I calculated all the squares modulo $17$, and found them to be $1,2,4,8,9,13,15,16$. I suppose I could then go through all cases where $p\equiv 1\pmod{8}$, and $p\equiv 1,2,4,8,\dots\pmod{17}$, and then use the Chinese remainder theorem to find what $p$ is congruent to modulo $8\cdot 17$, but this seems very tedious to do for each case. First of all, is my method correct, and also, is there a better way to solve this question? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2011-01-08T01:37:39.447" LastActivityDate="2011-01-08T01:37:39.447" Title="Trick to showing for which primes $p$ is $34$ a square?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="14045" PostTypeId="2" ParentId="14040" CreationDate="2010-12-12T12:01:40.817" Score="1" ViewCount="0" Body="&lt;p&gt;For $\alpha = \beta = -1$, you have something like $\int \frac{1}{x \log x}$, which is $\log \log x$, which diverges. For $\beta &amp;lt; -1$, though, it's safe.&lt;/p&gt;&#xA;" OwnerUserId="3406" LastActivityDate="2010-12-12T12:01:40.817" CommentCount="1" />
  <row Id="14046" PostTypeId="2" ParentId="14043" CreationDate="2010-12-12T12:02:41.690" Score="2" ViewCount="0" Body="&lt;p&gt;Given that the first flip is a head, the problem is equivalent to asking, what is the probability of getting exactly $1$ head in $2$ flips. There are two possibilities, that the first flip is a head, and the second is tails, and vice versa. This gives the probability the given answer says.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To address your question more directly, you should use conditional probability. Let $A$ be the event that Alex flips exactly $2$ heads, and $B$ be the event that the first flip is heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the sample space you wrote out, note $P(A\cap B)=2/8$, as the two possibilities are $\{HHT,HTH\}$. Also, $P(B)=1/2$, as you listed the four possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(A|B)=\frac{P(A\cap B)}{P(B)}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your main error is that $\{HHH\}$ does not satisfy the condition that the first flip is a head, and that there are exactly two heads, and thus should not be included when calculating the probability $P(A\cap B)$.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastEditorUserId="1583" LastEditorDisplayName="" LastEditDate="2010-12-12T12:10:00.997" LastActivityDate="2010-12-12T12:10:00.997" CommentCount="6" />
  <row Id="14047" PostTypeId="2" ParentId="14042" CreationDate="2010-12-12T12:03:02.947" Score="5" ViewCount="0" Body="&lt;p&gt;Yes, you have the wrong definition of a faithful functor. A functor &lt;em&gt;F&lt;/em&gt; is faithful precisely when &lt;em&gt;F&lt;/em&gt; sends &lt;strong&gt;morphisms&lt;/strong&gt; injectively, but nowhere is there required any kind of injectivity on &lt;strong&gt;objects&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-12-12T12:03:02.947" CommentCount="1" />
  <row Id="14048" PostTypeId="2" ParentId="14043" CreationDate="2010-12-12T12:14:16.823" Score="2" ViewCount="0" Body="&lt;p&gt;The problem here is that if the answer is half, then what the question means is exactly two heads, so three heads are not allowed. In your solution you included the case with three heads.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-12T12:14:16.823" CommentCount="5" />
  <row Id="14049" PostTypeId="1" CreationDate="2010-12-12T12:22:56.877" Score="4" ViewCount="78" Body="&lt;p&gt;Many years ago I wrote this fractal generator:&#xA;&lt;a href=&quot;http://uberto.fractovia.org/&quot; rel=&quot;nofollow&quot;&gt;http://uberto.fractovia.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was shareware but then I put it as open source.&#xA;It's written in Delphi, a language that I don't use anymore. So I'm playing lately with the idea of rewriting on some other lang (scala maybe?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the mathematical part I'm not expert at all, so I don't know if there is something a mathematician would find interesting. Maybe none because of software like Mathematica. I really don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the graphic oriented people I implemented some algos for continuous coloring, including the ones listed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/4035/continuous-coloring-of-a-mandelbrot-fractal&quot;&gt;Continuous coloring of a Mandelbrot fractal&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'd like to add some new ones.&lt;/p&gt;&#xA;" OwnerUserId="4615" LastActivityDate="2011-01-10T17:52:36.040" Title="Which features would be interesting for a mathematician in a fractal program?" Tags="&lt;algorithms&gt;&lt;applications&gt;&lt;fractals&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14050" PostTypeId="2" ParentId="14040" CreationDate="2010-12-12T12:29:24.793" Score="4" ViewCount="0" Body="&lt;p&gt;Since, for any $\varepsilon &amp;gt; 0$ fixed, $(\log n)^\beta / n^\varepsilon \to 0$ as $n \to \infty$, it is obvious that the sum converges for $\alpha &amp;lt; -1$ and diverges for $\alpha &amp;gt; -1$ (regardless of $\beta$). When $\alpha = -1$ and $\beta = - 1$ the sum diverges, as indicated by FX (hence also for all $\beta &gt; -1$). We are left with the case $\alpha =-1$ and $\beta &amp;lt; -1$. Then you should compare with the integral&#xA;$$&#xD;&#xA;\int_2^\infty  {\frac{1}{{x(\log x)^{1 + \delta } }}\,{\rm d}x} ,&#xD;&#xA;$$&#xD;&#xA;where $\delta &amp;gt; 0$ fixed. Make a change of variable $x&amp;#39; = \log x$ to find out that the integral (and hence the sum) converges.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-12T12:29:24.793" />
  <row Id="14051" PostTypeId="1" AcceptedAnswerId="14061" CreationDate="2010-12-12T13:11:01.017" Score="3" ViewCount="135" Body="&lt;p&gt;I want to write&#xA;$P(x,y,z)=yx^{3}+zx^{3}+xy^{3}+zy^{3}+xz^{3}+yz^{3}$&#xD;&#xA;in terms of elementary symmetric polynomials, but I'm getting stuck at the first step. I know I should follow the proof of the fundamental theorem of symmetric polynomials using the Newton identities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I pick out the 'biggest' monomial according to the lexicographical ordering: $yz^{3}$. Now I want to rewrite this as a polynomial in the elementary symmetric polynomials. I don't quite understand how to do this.&lt;/p&gt;&#xA;" OwnerUserId="2298" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-12T22:00:58.757" LastActivityDate="2010-12-12T22:00:58.757" Title="symmetric polynomials and the Newton identities" Tags="&lt;algebra&gt;&lt;polynomials&gt;" AnswerCount="2" />
  <row Id="14052" PostTypeId="2" ParentId="14036" CreationDate="2010-12-12T13:12:07.270" Score="1" ViewCount="0" Body="&lt;p&gt;Put ${\bf n}:={\bf n}_1\times{\bf n}_2/|{\bf n}_1\times{\bf n}_2|$ and write ${\bf x}:=\xi_1{\bf n}_1+ \xi_2 {\bf n_2} + \xi_3 {\bf n}$ with unknown $\xi_i$. Then&#xA;$$\cos\theta_1={\bf x}\bullet {\bf n}_1=\xi_1 +\xi_2({\bf n}_1\bullet{\bf n}_2)$$ and similarly $\cos\theta_2=\ldots\ $. After $\xi_1$ and $\xi_2$ have been computed the value of $\xi_3$ follows from the condition $|{\bf x}|=1$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-12T13:12:07.270" />
  <row Id="14053" PostTypeId="2" ParentId="13993" CreationDate="2010-12-12T13:51:05.770" Score="2" ViewCount="0" Body="&lt;p&gt;There are actually two separate questions in your post.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fourier transform of the travelling waves&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first question is about Fourier decomposition of the travelling wave solutions $\psi(x,t) = f(x + ct) + g(x-ct)$. It suffices to notice the following change of variables: let $u = x + ct$ and $v = x - ct$. Then inverting the change of variables you get $x = \frac12 (u+v)$ and $t = \frac1{2c} (u - v)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can re-write $\psi$ in this new coordinate system as $f(u) + g(v)$. Notice that $f$ is constant along $v$ direction, and $g$ along the $u$ direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since now $f$ is a function of one variable only, you can take the Fourier transform in the $u$ variable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(u) = \int \hat{f}(\omega_u) \exp ( i u \omega_u ) d\omega_u $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with a further change of variables, you have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x,t) = f(x + ct) = \int \hat{f}(\omega) \exp (i\omega(x + ct)) d\omega $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as desired (notice that I normalized the exponential differently. You can rescale coordinates once more to get it to be the form with $x/c + t$ if you so wish). By similarly playing with the $v$ variable you get a Fourier representation of $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to find $\hat{f}(\omega_u)$, we take the Fourier transform in the $u$ variable&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(u) \exp (i u \omega_u) du $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now remember that $f$ is independent of $v$! So writing $f = f(u,v) = f(u)$, for any curve $v = v(u)$ the integral is the same, that is, we can re-write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(u,v(u)) \exp(i u \omega_u) du $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, choose the curve to be $v = -u$, which translates to $x = 0$ by our change of variable formula before. And do a change of variable back to $t,x$, we have that along the curve $v = -u$, $u = ct$. So the integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(0,t) \exp (i ct \omega_u) c dt$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;again, the extra factor of $c$ can be removed by a further change of variables on $\omega_u$. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Two and higher dimensions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In higher dimensions, the situation is analogous, but we do not have the method of characteristics for us to start the first step. So instead, we start with the general high dimensional Fourier transform. (What I am about to write is mathematically non-rigorous, but morally correct.) Consider space-time coordinates $(x_1,\ldots,x_n,t)$ and space-time frequencies $(\xi_1,\ldots,\xi_n, \omega)$. We take the full Fourier transform of a solution to the wave equation in the following way. Suppose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_t^2 \psi = c\triangle \psi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\triangle$ is the Laplacian in $n$ dimensions. Consider the Fourier transform&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{\psi}(\xi_1,\ldots,\xi_n,\omega) = \int_{\mathbb{R}^{n+1}} f(x_1,\ldots,x_n,t) \exp [ -2\pi i (x_1\xi_1 + \cdots + x_n\xi_n + ct\omega) ] c dx_1\cdots dx_n dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is the slightly questionable step. In general this integral will not converge; but it captures the idea of the thing.) Using the properties of the Fourier transform to replace derivatives by multiplication of the frequencies, you get that $\hat{\psi}$ solves the algebraic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - \omega^2 \hat{\psi} = -(\xi_1^2 + \cdots + \xi_n^2) \hat{\psi} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\xi_1^2 + \cdots + \xi_n^2 - \omega^2) \hat{\psi} = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an algebraic equation, this means that if the number inside the first parenthesis is not zero, we must have $\hat{\psi} = 0$. And that $\hat{\psi}$ is only non-zero where the term in the parenthesis vanishes. (In mathematical jargon, this indicates that $\hat{\psi}$ is a distribution/measure, and not strictly speaking a function.) And in particular, we have the &quot;dispersion law&quot; that the only space-time frequencies contributing to $\psi$ are those where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \omega = \pm \sqrt{ \xi_1^2 + \cdots + \xi_n^2 } $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can write $\hat{\psi} = \hat{\psi}_+ + \hat{\psi}_-$, the forward and backward waves, with $\hat{\psi}_\pm$ only functions of $\xi_1,\ldots, \xi_n$. This gives us the &lt;em&gt;plane wave decomposition&lt;/em&gt; of a solution to the wave equation by taking the inverse Fourier transform (to save space, I write $x$ for the vector $(x_1,\ldots,x_n)$ and ditto $\xi$, using $\cdot$ to denote the scalar product)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \psi(x,t) = \int_{\mathbb{R}^n} \hat{\psi}_+(\xi) e^{2\pi i(\xi\cdot x + c|\xi|t)} + \hat{\psi}_-(\xi) e^{2\pi i(\xi\cdot x - c|\xi| t)} d\xi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remark&lt;/strong&gt; this is only one of the very many ways of solving the linear wave equation. (In a recent PDE course where I work, the lecturer allegedly gave 6 different methods...) Maths textbook I know tend not to want to go into the above Fourier transform derivation, because it is not very rigorous (it can be made rigorous with &lt;em&gt;a lot&lt;/em&gt; of additional preparations). The above idea should be treated more as a morally correct guiding principle.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-12T13:51:05.770" CommentCount="1" />
  <row Id="14054" PostTypeId="2" ParentId="14051" CreationDate="2010-12-12T14:13:14.493" Score="6" ViewCount="0" Body="&lt;p&gt;Edit: As per Bill's comment I would like to clarify that this is not related to Gauss' proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(x,y,z)=yx^{3}+zx^{3}+xy^{3}+zy^{3}+xz^{3}+yz^{3}$$&#xD;&#xA;$$=x^3(y+z+x-x)+y^3(x+z+y-y)+z^3(x+y+z-z)$$&#xD;&#xA;$$=x^3(x+y+z)+y^3(x+y+z)+z^3(x+y+z)-x^4-y^4-z^4$$&#xD;&#xA;$$=(x+y+z)(x^3+y^3+z^3)-(x^4+y^4+z^4)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can use identities for &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_identities#Formulation_in_terms_of_symmetric_polynomials&quot; rel=&quot;nofollow&quot;&gt;power sums&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-12T19:25:29.377" LastActivityDate="2010-12-12T19:25:29.377" CommentCount="11" />
  <row Id="14055" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T14:39:14.803" Score="3" ViewCount="0" Body="&lt;p&gt;If you do any commutative algebra or algebraic geometry, &lt;a href=&quot;http://www.math.uiuc.edu/Macaulay2/&quot; rel=&quot;nofollow&quot;&gt;Macaulay2&lt;/a&gt; is amazing. It requires minimal system resources and is free. There is a springer book about computations in Macaulay2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uiuc.edu/Macaulay2/Book/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uiuc.edu/Macaulay2/Book/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a comparison of computer algebra systems on wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_of_computer_algebra_systems&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Comparison_of_computer_algebra_systems&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-12T14:39:14.803" />
  <row Id="14056" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T15:23:49.193" Score="6" ViewCount="0" Body="&lt;p&gt;I found &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt; pretty useful for number theory and algebraic geometry. But it can do lots of other stuff, too.&lt;/p&gt;&#xA;" OwnerUserId="1766" LastActivityDate="2010-12-12T15:23:49.193" />
  <row Id="14057" PostTypeId="2" ParentId="14044" CreationDate="2010-12-12T15:33:12.500" Score="7" ViewCount="0" Body="&lt;p&gt;You have all that you need right there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{34}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{17}{p}\right)=(-1)^{(p^2-1)/8}\left(\frac{p}{17}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $p \equiv \pm 1 \pmod{8}$ then $p$ must be a quadratic residue mod 17, otherwise it must be a quadratic non-residue mod 17. (But delete 14 from your list first!)&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T23:11:34.400" LastActivityDate="2010-12-12T23:11:34.400" CommentCount="1" />
  <row Id="14058" PostTypeId="1" CreationDate="2010-12-12T15:52:50.423" Score="0" ViewCount="123" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/283/is-0-a-natural-number&quot;&gt;Is 0 a natural number?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;There seems to be no consensus, although perhaps one is gathering over the centuries to say yes to the first question and identify $\mathbb{N}$ with $\omega$ : the set of finite Von Neumann ordinal numbers. That leaves $\textbf{N}$ = {$1$, $2$, ...} for the counting numbers (who counts with $0$?). You can dodge the issue by writing $\mathbb{Z}_{\geq0}$, $\mathbb{Z}_+$, etc., but it seems awkward to use derived and complicated notation for something so basic.&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2010-12-12T16:12:23.177" ClosedDate="2010-12-12T17:45:56.347" Title="Do the natural numbers include zero? And what should we call them, with or without zero?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="14059" PostTypeId="2" ParentId="14058" CreationDate="2010-12-12T16:12:23.177" Score="1" ViewCount="0" Body="&lt;p&gt;It seems convenient to include zero - one can then use the term 'positive integers' to exclude zero and 'natural numbers' to include zero without resorting to further notational complexities. It also seems natural to include zero - when constructing numbers axiomatically we start from the empty set so that zero is defined by the empty set and one is defined by the set containing the empty set and so on.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-12T16:12:23.177" CommentCount="4" />
  <row Id="14060" PostTypeId="1" CreationDate="2010-12-12T16:25:23.250" Score="3" ViewCount="97" Body="&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$ \qquad \qquad $ &lt;strong&gt;&lt;em&gt;The greatest term in $\left(1+x\right)^{2n}$ has the greatest cofficient if $\frac{n}{n+1} \lt x \lt \frac{n+1}{n}$&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can we derive something like this for $n$ in general? Please give me some hints.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-12T16:33:13.587" LastActivityDate="2010-12-12T16:52:54.347" Title="How can we prove this property?" Tags="&lt;algebra-precalculus&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14061" PostTypeId="2" ParentId="14051" CreationDate="2010-12-12T16:46:28.017" Score="4" ViewCount="0" Body="&lt;p&gt;By Gauss's algorithm, if  $\rm\ z^a\ y^b\ x^c\ $ is the highest w.r.t. lex order $\rm\ z &amp;gt; y &amp;gt; x\ $ then you subtract $\rm\ s_1^{a-b}\ s_2^{b-c}\ s_3^c\:.\:$ Thus since $\rm\ z^3\ y $ is highest you subtract $\rm s_1^{3-1}\ s_2^{1-0}\ s_3^0\ = (x+y+z)^2\ (xy+yz+zx)$ from $\rm\:P\:$. The result is smaller in lex-order, so iterating this reduction yields a representation of $\rm\:P\:$ in terms of elementary symmetric polynomials $\rm\:s_i\:.\:$ Here the algorithm terminates in two more steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned in a &lt;a href=&quot;http://math.stackexchange.com/questions/3550/using-grobner-bases-for-solving-polynomial-equations/3562#3562&quot;&gt;prior post&lt;/a&gt;, Gauss's algorithm is the earliest known example of using lex-order reduction as in the Grobner basis algorithm. For a nice exposition see Chapter 7 of Cox, Little, O'Shea: Ideals, Varieties and Algorithms. They also give generalizations to  the ring of invariants of a finite matrix group $\rm G \subset  GL(n,k)$. Here's an excerpt which, coincidentally, presents this example. You might find it helpful to first read the example at the end before reading the proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tx23R.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/cGSvO.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-12T17:32:13.807" LastActivityDate="2010-12-12T17:32:13.807" CommentCount="4" />
  <row Id="14062" PostTypeId="2" ParentId="14060" CreationDate="2010-12-12T16:52:54.347" Score="4" ViewCount="0" Body="&lt;p&gt;Hint: I think you can solve this by looking at the adjacent terms of the middle term. If the middle term is the biggest term, it has to be bigger than both adjacent terms. This will give you the bounds.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-12T16:52:54.347" />
  <row Id="14063" PostTypeId="1" CreationDate="2010-12-12T17:28:26.870" Score="2" ViewCount="141" Body="&lt;p&gt;How can I figure out the classical construction (direct sum, product, pullbacks, and in general direct and inverse limits) in the category made by chain complexes and chain maps (of abelian groups or any abelian stuff)? Because of this category is abelian it &lt;em&gt;must&lt;/em&gt; have (co)limits, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular take&#xA;$$&#xD;&#xA;\begin{gather*}&#xD;&#xA;\dots\to A_n\to A_{n-1}\to\dots\\&#xD;&#xA;\dots\to B_n\to B_{n-1}\to\dots&#xD;&#xA;\end{gather*}&#xD;&#xA;$$&#xD;&#xA;I would like to say that $\mathcal A\oplus\mathcal B$ is &quot;what I want it to be&quot;, $A_n\oplus B_n$ with obvious maps. But a standard argument doesn't allow me to conclude it: maybe it is false? A single word with the right reference will be enough to close the topic; I am now reading Hilton &amp;amp; Stammbach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I would like to add what I've tried to do, but seems difficult not to invoke some diagrams, which I'm not able to draw without a suitable package here. However, first of all both complexes inject in the sum by maps which are part of a chian complex, $\iota_n^A,\iota_n^B$. Then, consider another complex $\{C_n,\partial_n^C\}$ and a couple of chain maps $\{a_n\colon A_n\to C_n\}$, $\{b_n\colon B_n\to C_n\}$. For each $n$ there exists a map $\alpha_n\colon A_n\oplus B_n\to C_n$ factoring the $a_n$ and the $b_n$s. Then I would like to show that the maps $\alpha_n$ are part of a chain map between the sum and the complex $\mathcal C$, but trying to prove it I can only conclude that in the diagram (vertical rows are the $\alpha_n$s)&#xA;$$&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;A_n\oplus B_n &amp;amp;\xrightarrow{\partial_n^\oplus}&amp;amp; A_{n-1}\oplus B_{n-1} \\&#xD;&#xA;\downarrow &amp;amp;&amp;amp; \downarrow\\&#xD;&#xA;C_n &amp;amp;\xrightarrow[\partial_n^C]{}&amp;amp; C_{n-1}&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xD;&#xA;which I want to be commutative, aka $\alpha_{n-1}\partial_n^\oplus=\partial_n^C\alpha_n$, I have $\alpha_{n-1}\partial_n^\oplus\iota_n^A=\partial_n^C\alpha_n\iota_n^A$. How can I remove the &lt;em&gt;iotas&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="685" LastEditorDisplayName="" LastEditDate="2010-12-12T19:16:43.190" LastActivityDate="2010-12-13T06:00:35.267" Title="Direct sum of complexes" Tags="&lt;homological-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14064" PostTypeId="1" CreationDate="2010-12-12T18:11:10.360" Score="1" ViewCount="23" Body="&lt;p&gt;I'm computing an LDA (linear discriminant analysis) transform, for an application I'm working on, and I've been following these &lt;a href=&quot;http://www.ed.uiuc.edu/courses/EdPsy494/lectures/DiscriminantAnalysis-online.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; (starting at page 36, especially slide 47 in green).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing this in Python (with numpy and scipy), and this is what I have come up with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;from scipy.linalg import sqrtm&#xA;...&#xA;sw_inv_sqrt = np.linalg.inv(sqrtm(self.sigma_within))&#xA;self.d, self.v = np.linalg.eig(&#xA;    np.dot(&#xA;        np.dot(sw_inv_sqrt, self.sigma_between),&#xA;        sw_inv_sqrt&#xA;        ))&#xA;self.v = np.dot(sw_inv_sqrt, self.v)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I know that this implementation is correct as I have compared it to others.  My concern is whether this is &lt;strong&gt;good&lt;/strong&gt; solution in the &lt;strong&gt;numerical sense&lt;/strong&gt;.  In comparing my solution to others, they match only to about 6 decimal places.  Is there a better way to do this numerically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. I originally posted this to Stack Overflow and it was recommended to me to post it here.&lt;/p&gt;&#xA;" OwnerUserId="4676" LastEditorUserId="4676" LastEditorDisplayName="" LastEditDate="2010-12-12T18:19:30.617" LastActivityDate="2010-12-12T18:19:30.617" Title="good numerical solution for LDA transformation" Tags="&lt;numerical-methods&gt;&lt;numerical-optimization&gt;" />
  <row Id="14065" PostTypeId="1" AcceptedAnswerId="14067" CreationDate="2010-12-12T18:11:23.713" Score="3" ViewCount="143" Body="&lt;p&gt;I have a question about whether the following two sets on the two sides are the same:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\lim_{a \rightarrow \infty} \, \limsup_{n \rightarrow \infty} \, \{ x \in S \, | \, f_n(x)  &amp;gt; a \}&#xD;&#xA;=&#xD;&#xA;\left\{ x \in S \, \Bigm| \, \lim_{n \rightarrow \infty} f_n(x) =\infty \right\}\quad ?&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\{f_n\}$ is a sequence of real-valued functions defined on a set $S$. How can you explain it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="4674" LastEditorUserId="4674" LastEditorDisplayName="" LastEditDate="2010-12-12T23:28:34.800" LastActivityDate="2010-12-13T02:09:21.893" Title="Are these two sets same" Tags="&lt;functions&gt;&lt;elementary-set-theory&gt;&lt;limit&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="14066" PostTypeId="1" CreationDate="2010-12-12T19:06:36.160" Score="13" ViewCount="203" Body="&lt;p&gt;Does there exist a good reference on the algorithms used by calculators, especially on the trig and transcendental funtions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would still like to know how Casio generates its random numbers. I still wonder if they are any good.&lt;/p&gt;&#xA;" OwnerUserId="2809" LastActivityDate="2010-12-13T14:51:16.730" Title="Calculator algorithms" Tags="&lt;algorithms&gt;&lt;calculator&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="14067" PostTypeId="2" ParentId="14065" CreationDate="2010-12-12T19:12:04.777" Score="4" ViewCount="0" Body="&lt;p&gt;Consider a sequence of functions $f_n:\mathbb{R} \to \mathbb{R}$ defined by&#xA;$f_n {(x)} = n$ if $x=0$ and $n$ is even, and $f_n {(x)} = 0$ otherwise. The right-hand side is obviously $\emptyset$, while the left-hand should  be $\lbrace 0 \rbrace$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Answering the additional questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that  $f_n{(x)}$ is an increasing sequence for each $x$. If $x$ belongs to the left set, then it must belong to the set $\lim \sup _{n \to \infty } \{ x \in S|f_n (x) &amp;gt; a\}$ for any $a &amp;gt; 0$ fixed. This means that for any $a&amp;gt;0$, there are infinitely many $n$ such that $f_n (x) &amp;gt; a$. Since $f_n (x)$ in increasing, it must converge to a positive number, or diverge to $\infty$. But $a&amp;gt;0$ is arbitrary, hence $f_n (x)$ must diverge to infinity. That is, $x$ belongs to the right set. Since the right set is contained in the left one, we conclude that both sets are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The fact that the right set is contained in the left one, is proved as follows. Suppose that $x$ belongs to the right set, and let $a&amp;gt;0$ be arbitrary but fixed. Then, for all sufficiently large $n$, $f_n {(x)} &amp;gt; a$. In particular, $x \in \lim \sup _{n \to \infty } \{ x \in S|f_n (x) &amp;gt; a\}$. Since this is true for any $a &amp;gt; 0$, $x$ belongs to the left set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The following point should be stressed. Denote by $E_a$ the set $\lim \sup _{n \to \infty } \{ x \in S|f_n (x) &amp;gt; a\}$.&#xA;If $x \in E_a$, then $x \in E_{a&amp;#39;}$ for any $a&amp;#39; &amp;lt; a$. It follows that $\lim \sup _{a \to \infty} E_a = \lim \inf _{a \to \infty} E_a$; hence, by definition, the limit $\lim  _{a \to \infty} E_a$ exists, and is equal to $\lim \sup _{a \to \infty} E_a = \lim \inf _{a \to \infty} E_a$. So, the left set in the question is indeed properly defined, and $x \in \lim  _{a \to \infty} E_a$ means, in our case, that $x \in E_a$ for every $a$. Finally, note that always the $\lim \inf$ is a subset of the $\lim \sup$ (in analogy with the case of sequences of real numbers, where $\leq$ plays the role of $\subseteq$). &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-13T02:09:21.893" LastActivityDate="2010-12-13T02:09:21.893" CommentCount="25" />
  <row Id="14068" PostTypeId="2" ParentId="14063" CreationDate="2010-12-12T19:26:21.450" Score="2" ViewCount="0" Body="&lt;p&gt;In a preadditive category (it's clear that chain complexes are preadditive), the idea of a  direct sum can be described directly. Given two objects $A, B$, a direct sum $C$ has morphisms $i_1: A \to C, i_2: B \to C$ and projections $p_1: C \to A,  p_2: C \to B$ that satisfy $i_1 p_1 + i_2 p_2 = 1_C$ and $p_1 i_1 = 1_A, p_2 i_2 = 1_B$. Indeed, it is easy to get such maps in a direct sum (inclusion and projection); conversely if we get such maps one can check that $C$ is a coproduct and a direct product by explicitly using these maps. Namely, one checks that such maps guarantee a direct sum in the case of abelian groups, and then applies hom-sets and Yoneda nonsense (if we have $\hom(X, C) = \hom(X, A) \oplus \hom(X, B)$ for all $X$ naturally, then we're done). I'm being a little vague here and only sketching it; if it's not clear, I can clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly the case for the construction of chain complexes that you described above. The maps $p_1, p_2, i_1, i_2$ are the natural projections and  inclusions piecewise. They are maps of complexes, as one can check, and they satisfy the required identities because they do for abelian groups.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-13T06:00:35.267" LastActivityDate="2010-12-13T06:00:35.267" />
  <row Id="14069" PostTypeId="1" CreationDate="2010-12-12T19:46:55.357" Score="4" ViewCount="146" Body="&lt;p&gt;If a Skoda sets off at an average speed of 20 miles per hour at 7 am and a Porsche sets off at an average speed of 40 miles per hour at 4 pm in the same direction, at what time will the Porsche catch the Skoda?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please explain how it works?&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-13T07:36:16.417" LastActivityDate="2010-12-13T09:02:42.463" Title="If a second car leaves later at a higher speed, how long will it take to catch up?" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;&lt;quantitative-aptitude&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="14070" PostTypeId="2" ParentId="14066" CreationDate="2010-12-12T19:47:06.243" Score="5" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/CORDIC&quot; rel=&quot;nofollow&quot;&gt;CORDIC&lt;/a&gt;. See also this question: &lt;a href=&quot;http://stackoverflow.com/questions/2169641/where-to-find-algorithms-for-standard-math-functions/2169666&quot;&gt;http://stackoverflow.com/questions/2169641/where-to-find-algorithms-for-standard-math-functions/2169666&lt;/a&gt; and this review: &lt;a href=&quot;http://mathdl.maa.org/mathDL/19/?pa=reviews&amp;amp;sa=viewBook&amp;amp;bookId=65790&quot; rel=&quot;nofollow&quot;&gt;http://mathdl.maa.org/mathDL/19/?pa=reviews&amp;amp;sa=viewBook&amp;amp;bookId=65790&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditorDisplayName="" LastEditDate="2010-12-12T19:52:18.860" LastActivityDate="2010-12-12T19:52:18.860" CommentCount="1" />
  <row Id="14071" PostTypeId="2" ParentId="14063" CreationDate="2010-12-12T19:54:35.643" Score="4" ViewCount="0" Body="&lt;p&gt;The category of complexes in an abelian category $\mathcal{A}$ is a full subcategory of $\text{Fun}({\mathbb{Z}},\mathcal{A})$, where $\mathbb{Z}$ is partially ordered under reverse inequality.  So if we know how these constructions are performed in the category of functors from $\mathbb{Z}$ to $\mathcal{A}$, we'll have the natural candidates to the category of complexes. The standard result is that (co)limits in $\text{Fun}({\mathcal{D}},\mathcal{C})$, where $\mathcal{D}$ is a small category and $\mathcal{C}$ is a category, are computed pointwise. Take a look at Borceux's 'Handbook of Categorical Algebra, Vol.$1$' section $2.15$. There he explains the precise meaning of being computed pointwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since any abelian category is finitely (co)complete, we can compute any finite (co)limit in $\text{Fun}({\mathbb{Z}},\mathcal{A})$ pointwise. If we consider a (co)complete category, e.g., the category of modules over a ring, we can compute any (co)limit pointwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think Borceux's book is too terse, there is a similar discussion in Rotman's 'An Introduction to Homological Algebra' on page $317$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; In order to remove the iotas you will need to prove that $\alpha_{n-1}\partial_n^\oplus\iota_n^B=\partial_n^C\alpha_n\iota_n^B$. Now use the fact that there is only one morphism $\varphi: A_n \oplus B_n \rightarrow C_{n-1}$ such that $\varphi \iota_n^A = \alpha_{n-1}\partial_n^\oplus\iota_n^A$ and $\varphi \iota_n^B = \alpha_{n-1}\partial_n^\oplus\iota_n^B $.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-12T22:26:45.837" LastActivityDate="2010-12-12T22:26:45.837" CommentCount="4" />
  <row Id="14072" PostTypeId="2" ParentId="14069" CreationDate="2010-12-12T19:55:28.513" Score="5" ViewCount="0" Body="&lt;p&gt;Hint: Distance covered is speed$\times$ time traveled. If you assume the time at which the Skoda leaves to be $t=0$, distance covered by Skoda after $t$ hours is $20t$ and distance covered by Porsche after 4pm is $40(t-9)$. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-12T19:55:28.513" CommentCount="5" />
  <row Id="14073" PostTypeId="2" ParentId="14069" CreationDate="2010-12-12T20:06:38.077" Score="2" ViewCount="0" Body="&lt;p&gt;Well for this problem we could look at it slightly differently, not much though, without resorting to algebraic variables. From 7 am to 4 pm the Skoda would have travelled 180 miles. Thus the Porsche has to cover 180 miles extra to overtake the Skoda. In each hour the Porsche can only cover 20 miles extra, so the Porsche would need another extra 9 hours to do that, thus finishing at 1 am.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-12T20:06:38.077" />
  <row Id="14074" PostTypeId="2" ParentId="14065" CreationDate="2010-12-12T20:07:33.520" Score="1" ViewCount="0" Body="&lt;p&gt;For question number two, construct the following sequence of $f_n(x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;f_n(x)  = \left\{\begin{array}{ll} &#xA;k+\epsilon_n, &amp;amp; \mbox{for $x=1/k$ with $k=1,\ldots,n$,} \\&#xA;\epsilon_n, &amp;amp; \mbox{otherwise.}&lt;br&gt;&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the sequence of $\epsilon_n=1-\frac{1}{2^n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I did not make any mistake, the right-hand side should be empty, while the left-hand side contains $0$. (This example was constructed with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#General_set_convergence&quot; rel=&quot;nofollow&quot;&gt;general set convergence&lt;/a&gt; in mind.)&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-12T21:27:15.213" LastActivityDate="2010-12-12T21:27:15.213" CommentCount="1" />
  <row Id="14075" PostTypeId="1" AcceptedAnswerId="14079" CreationDate="2010-12-12T20:08:55.623" Score="6" ViewCount="124" Body="&lt;p&gt;I have two $2\times 2$ matrices, $A$ and $B$, with the same determinant. I want to know if they are similar or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved this by using a matrix called $S$:&#xA;$$\left(\begin{array}{cc}&#xD;&#xA;a&amp;amp; b\\&#xD;&#xA;c&amp;amp; d&#xD;&#xA;\end{array}\right)$$&#xD;&#xA;and its inverse in terms of $a$, $b$, $c$, and $d$, then showing that there was no solution to $A = SBS^{-1}$. That worked fine, but what will I do if I have $3\times 3$ or $9\times 9$ matrices? I can't possibly make system that complex and solve it.  How can I know if any two matrices represent the &quot;same&quot; linear transformation with different bases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, how can I find $S$ that change of basis matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried making $A$ and $B$ into linear transformations... but without the bases for the linear transformations I had no way of comparing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have read that similar matrices will have the same eigenvalues... and the same &quot;trace&quot; --but my class has not studied these yet. Also, it may be the case that some matrices with the same trace and eigenvalues are not similar so this will not solve my problem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one idea. Maybe if I look at the reduced col. and row echelon forms that will tell me something about the basis for the linear transformation? I'm not really certain how this would work though? Please help.&lt;/p&gt;&#xA;" OwnerUserId="4681" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-13T06:04:12.430" LastActivityDate="2010-12-13T06:04:12.430" Title="How do I tell if matrices are similar?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14076" PostTypeId="2" ParentId="14069" CreationDate="2010-12-12T20:10:20.537" Score="1" ViewCount="0" Body="&lt;p&gt;Let's say Ds = distance traveled by the Skoda and Dp = distance traveled by the Porsche.&#xA;If we set t to be the time after 4pm that the cars are traveling, then we set up the formulas as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dp = 40t&#xA;Ds = 20t + 20(9) *the Skoda has been travelling from 7am to 4pm, or 9 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the cars meet, the distance traveled should be the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;40t = 20t + 180&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtract 20t from both sides:  20t = 180&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide both sides by 20: t = 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after 4pm, the Porsche should catch up with the Skoda 9 hours later, or at 1pm.&lt;/p&gt;&#xA;" OwnerUserId="4680" LastActivityDate="2010-12-12T20:10:20.537" CommentCount="0" />
  <row Id="14077" PostTypeId="2" ParentId="13912" CreationDate="2010-12-12T20:14:27.150" Score="1" ViewCount="0" Body="&lt;p&gt;I have one small insight just for finite groups which works for my own personal intuition.  Since you seem to feel like topology and analysis make sense and are well motivated, then why not think about the elements of a finite group as points on a continuous manifold, in other words, represent the elements of a finite group using $N\times N$ matricies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lie groups are continuous manifolds with a group structure, and they are manifestly topological.  Furthermore, a finite subgroup of $GL(\mathbb{R}^N)$ is, by definition, a &lt;em&gt;representation&lt;/em&gt; of some finite group.  So the elements of a finite group are represented by points on a manifold associated with some subgroup of $GL(\mathbb{R}^N)$, and these elements are connected to each other by continuous paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if anyone is looking for a good introduction to linear groups and representation theory, I highly recommend &lt;a href=&quot;http://books.google.com/books?id=m1VQi8HmEwcC&amp;amp;printsec=frontcover&amp;amp;dq=brian+hall+lie+groups&amp;amp;hl=en&amp;amp;ei=lzIFTYbnN4OclgeS5t2DCA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCsQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;B. Hall&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-12-12T21:40:39.370" LastActivityDate="2010-12-12T21:40:39.370" />
  <row Id="14078" PostTypeId="1" CreationDate="2010-12-12T20:18:36.517" Score="4" ViewCount="121" Body="&lt;p&gt;If your salary is increased by 10% and then after 6 months by a further 50%, what is the overall % increase?&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-13T07:36:58.180" LastActivityDate="2010-12-13T09:17:58.227" Title="Word based problem with percentages" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;&lt;quantitative-aptitude&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14079" PostTypeId="2" ParentId="14075" CreationDate="2010-12-12T20:18:57.080" Score="4" ViewCount="0" Body="&lt;p&gt;There is something called &quot;canonical forms&quot; for a matrix; they are special forms for a matrix that can be obtained intrinsically from the matrix, and which will allow you to easily compare two matrices of the same size to see if they are similar or not. They are indeed based on eigenvalues and eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, without the necessary machinery having been covered, the answer is that it is difficult to know if the two matrices are the same or not. The simplest test you can make is to see whether their &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_polynomial&quot; rel=&quot;nofollow&quot;&gt;characteristic polynomials&lt;/a&gt; are the same. This is &lt;em&gt;necessary&lt;/em&gt;, but &lt;strong&gt;not sufficient&lt;/strong&gt; for similarity (it is related to having the same eigenvalues).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have learned about canonical forms, one can use either the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_canonical_form&quot; rel=&quot;nofollow&quot;&gt;Jordan canonical form&lt;/a&gt; (if the characteristic polynomial splits) or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_normal_form&quot; rel=&quot;nofollow&quot;&gt;rational canonical form&lt;/a&gt; (if the characteristic polynomial does not split) to compare the two matrices. They will be similar if and only if their rational forms are equal (up to some easily spotted differences; exactly analogous to the fact that two diagonal matrices are the same if they have the same diagonal entries, though the entries don't have to appear in the same order in both matrices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduced row echelon form and the reduced column echelon form will not help you, because any two invertible matrices have the same forms (the identity), but need not have the same determinant (so they will not be similar). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T20:28:10.920" LastActivityDate="2010-12-12T20:28:10.920" CommentCount="3" />
  <row Id="14080" PostTypeId="2" ParentId="13712" CreationDate="2010-12-12T20:18:57.750" Score="0" ViewCount="0" Body="&lt;p&gt;In classical logic, $B\rightarrow A$ is equivalent to $\lnot A \rightarrow \lnot B$, so you will simply use the simplest definition of $A\leftrightarrow B$, which is to use $B\rightarrow A$. In intuitionistic logic, the contraposition is not generally derivable. You have $(B\rightarrow A)\rightarrow(\lnot A\rightarrow \lnot B)$, but not $(\lnot A\rightarrow \lnot B)\rightarrow(B\rightarrow A)$. Hence, your definition of equivalence is weaker - and doesnt really make sense, it only makes sense if you assume $A\vee \lnot A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you look at it computationally, $A\leftrightarrow B$ means you have terms $u^{A\rightarrow B}$, $v^{B\rightarrow A}$, such that if you have an object $a^A$ you always get an object $b^B$ and vice versa. With your definition, you would have $u^{A\rightarrow B}$, but $v^{(A\rightarrow\bot)\rightarrow(B\rightarrow\bot)}$ where $\lnot A = A\rightarrow\bot$ and $\bot$ can be defined as $0=1$ (or any other type of which no term is derivable), hence, $v$ maps terms $t^{A\rightarrow\bot}$ that map terms of type $A$ on something not derivable onto terms $u^{B\rightarrow\bot}$ mapping terms of type $B$ on something not derivable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isnt what we usually want from equivalence.&lt;/p&gt;&#xA;" OwnerUserId="1858" LastActivityDate="2010-12-12T20:18:57.750" CommentCount="1" />
  <row Id="14081" PostTypeId="1" CreationDate="2010-12-12T20:25:45.437" Score="1" ViewCount="33" Body="&lt;p&gt;The current price of a stock can be modeled by $P_0 = \frac{D_{1}}{r-g}$ where $D_1$ is the expected dividend, $r$ is the rate of return, and $g$ is the expected growth rate in the perpetuity. If $r&amp;lt;g$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle P_0 = \sum_{t=1}^{N} \frac{D_{0}(1+g)^{t}}{(1+r)^{t}} + \frac{P_N}{(1+r)^{N}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyway to simplify this?  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-12T20:25:45.437" Title="Dividend Discount Model" Tags="&lt;finance&gt;" CommentCount="3" />
  <row Id="14082" PostTypeId="2" ParentId="14078" CreationDate="2010-12-12T20:28:35.987" Score="5" ViewCount="0" Body="&lt;p&gt;When it is first increased you are multiplying it by 110/100. When it is further increased you are then multiplying the new salary by 150/100. Find out how much you are overall multiplying the original by and from there deduce the increase. (Here we have assumed that in the question the further increase is based on the new salary and not the original one.)&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-12T20:28:35.987" CommentCount="4" />
  <row Id="14083" PostTypeId="2" ParentId="14066" CreationDate="2010-12-12T20:57:49.140" Score="3" ViewCount="0" Body="&lt;p&gt;The GNU Multiple Precision Arithmetic Library (GMP) has a very good documentation describing how they implemented all the arithmetic functions for their multiple precision library:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gmplib.org/manual/&quot; rel=&quot;nofollow&quot;&gt;http://gmplib.org/manual/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give it a try!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I know the GMP is used for big number computation, but arithmetic is arithmetic, and to make big number computations you must understand small number computations, so I am sure (or at least &quot;almost&quot; sure) the documentation mention what is enough to for calculator arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="4671" LastEditorUserId="4671" LastEditorDisplayName="" LastEditDate="2010-12-13T14:51:16.730" LastActivityDate="2010-12-13T14:51:16.730" CommentCount="5" />
  <row Id="14084" PostTypeId="1" AcceptedAnswerId="14086" CreationDate="2010-12-12T21:00:57.780" Score="3" ViewCount="104" Body="&lt;p&gt;Hi all,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermat's Little Theorem, in Fermat-Euler form, states that: if $\gcd(a,m)=1$, then $a^{\varphi(m)} \equiv 1 \pmod{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've been asked to prove it via modular arithmetic. In order to do this I understand that I'm to use two lemmas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\varphi(p^n) = p^{n} - p^{n-1}$. This I can prove by working out that there are $p^n$ numbers less than $p^n$ of which $p^{n-1}$ of them are divisible by $p$.&lt;/li&gt;&#xA;&lt;li&gt;Given $gcd(r,s) = 1$, $\varphi(rs) = \varphi(r)\varphi(s)$. This I'm having problems with.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My lecturer suggested the proposition that  $\varphi(n) = n\prod_{p|n}\left(1-\frac{1}{p}\right)$. I can re-arrange this to equal lemma 1, but what I can't understand how that proposition proves 2, which my lecturer claims it does, and why these two points together prove the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I might follow the argument correctly if I fully understood lemma 2. &lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T21:17:47.100" LastActivityDate="2010-12-13T23:54:55.527" Title="Proof of the Euler Generalisation of Fermat's Little Theorem using modular arithmetic" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14085" PostTypeId="2" ParentId="14078" CreationDate="2010-12-12T21:03:10.087" Score="2" ViewCount="0" Body="&lt;p&gt;If x is your original salary,&#xA;10 % increase in salary is obtained by multiplying 1.1 (since 0.1 is 10% of 1)&#xA;Similarly 50 % increase in salary is obtained by multiplying 1.5 ( since 0.5 is 50% of 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so with two consecutive increments of 10% and 50%. your new salary becomes 1.5 * (1.1*(x)) = 1.65*x.&lt;BR&gt;&#xA;Work from here on...&lt;/p&gt;&#xA;" OwnerUserId="3964" LastActivityDate="2010-12-12T21:03:10.087" />
  <row Id="14086" PostTypeId="2" ParentId="14084" CreationDate="2010-12-12T21:10:12.927" Score="4" ViewCount="0" Body="&lt;p&gt;Part 2 can be proven in many different ways; what your lecturer suggests, if you already know the identity, will do it.  Just notice that if $p|rs$, then &lt;em&gt;either&lt;/em&gt; $p|r$ &lt;em&gt;or&lt;/em&gt; $p|s$, &lt;em&gt;and not both&lt;/em&gt;, since $\gcd(r,s)=1$. So we have:&#xA;$$\varphi(rs) = rs\prod_{p|rs}\left(1 - \frac{1}{p}\right) = rs\left(\prod_{p|r}\left(1 - \frac{1}{p}\right)\right)\left(\prod_{p|s}\left(1-\frac{1}{p}\right)\right),$$&#xD;&#xA;and go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One you have this, the proof of the Fermat-Euler Theorem reduces to proving that if $\gcd(a,p)=1$, then $a^{\varphi(p^n)} \equiv 1 \pmod{p^n}$ for every positive integer $n$; you can use 2 and the Chinese Remainder Theorem to prove that this suffices, and then you can use 1 to try to establish the condition modulo a prime power.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T21:30:10.610" LastActivityDate="2010-12-12T21:30:10.610" CommentCount="1" />
  <row Id="14087" PostTypeId="2" ParentId="14035" CreationDate="2010-12-12T21:19:39.513" Score="2" ViewCount="0" Body="&lt;p&gt;For each distance measured, you can write an equation for the distance between that point on your path and that fixed point, both of which are unknown.  In two dimensions (on a flat plane), each unknown point has two unknown coordinates.  If you have $n$ fixed points and you measure from $k$ points on your path, you will have $nk$ distances and equations and $2n+2k$ unknowns.  As long as $nk\ge 2n+2k$, this should yield a system that can be solved (if the distances are exact), but it probably helps to have $nk$ be more than $2n+2k$ because the system when $nk=2n+2k$ might have multiple solutions (because the equations are nonlinear).  (If you are working in three dimensions, then $2n+2k$ changes to $3n+3k$.)&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-12T21:19:39.513" />
  <row Id="14088" PostTypeId="2" ParentId="3399" CreationDate="2010-12-12T21:27:05.663" Score="3" ViewCount="0" Body="&lt;p&gt;The primary concept is convexity, not concavity. It applies to geometric figures, originally lenses, and this usage was adapted to functions.  There is no comparable concept of concavity for, say, 2-dimensional regions, except as the absence of the property of convexity.  There is also no property for figures in general corresponding to the anti-convexity inequality, because most non-convex figures will be locally convex.  It is a matter of historical convention that a function is called &quot;convex&quot; if the region &lt;em&gt;above&lt;/em&gt; the graph of the function is convex, and it would have caused no mathematical problem to use the opposite convention based on the region &lt;em&gt;below&lt;/em&gt;, but concavity is a more limited concept that is defined in terms of convexity (or only defined for functions) and not the other way around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms &quot;concave up&quot; and &quot;concave down&quot; appear mainly in non-specialist US college textbooks on calculus.  They are nonstandard terminology and, I think, bad practice that should be discouraged (with luck and sufficient ruthlessness maybe they can be squelched in a generation...).  As far as I know the etymology went as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Like &quot;convex&quot;, the word &quot;concave&quot; has a prior use in optics. Concave (inward-curved) lenses are the opposite of convex lenses, so there is a pre-existing word for &quot;not convex&quot; or &quot;convex in the opposite direction&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Convex has an absolutely entrenched mathematical use to denote convex figures as well as functions (and sequences) with increasing derivative.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Functions whose negative is convex occur frequently and &quot;concave [function]&quot; came into use as a convenient description of this situation.  The linguistic logic was clear enough to make this immediately understandable. It's not clear whether it was more or less favored compared to statements involving the negative, such as saying that $-f(u)$ is convex, or $f$ is anti-convex, or that is it the negative of a convex function.  I don't have data at hand from web searches or anything like that, but I think concavity is less common as a description of negatively convex sequences.  For functions the ability to draw a graph makes the resemblance to lenses clearer so that both words seem sensible.&#xA;(added: concavity as a counterpart to convexity for functions and sequences also gained momentum as its own term once log-convexity and log-convex became standard usage. Because the relationship between log-convex and log-concave functions is not simply change of sign but a multiplicative inverse, using only the words based on convexity might lead to confusion or circumlocution.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Authors of US college calculus textbooks, writing for an audience not familiar with or necessarily interested in convex figures and optics, and aware of potential for confusion (e.g., the graph of a concave function still bounds a convex-shaped region, or the subsequent use of convex to describe functions of several variables and the regions on which those are defined) cooked up a terminology based on &quot;concavity&quot; as a stand-alone concept, limited to the one-variable context where $f(x)$ is graphed with the $y$-axis direction being upward.  It's not clear how consistent this concave-up and concave-down terminology is between books and whether it agrees with the earlier, non-confusing use of concave to denote negative convexity.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-13T05:43:05.053" LastActivityDate="2010-12-13T05:43:05.053" CommunityOwnedDate="2010-12-12T21:27:05.663" CommentCount="6" />
  <row Id="14089" PostTypeId="2" ParentId="13793" CreationDate="2010-12-12T21:54:47.093" Score="5" ViewCount="0" Body="&lt;p&gt;The ordered approach is fine. A classical theorem by Sierpinski says that all countable metric spaces without isolated points are homeomorphic. Q is such a space.&#xA;It also implies $\mathbf{Q} \setminus \{0\}$ is homeomorphic to $\mathbf{Q}$ and $\mathbf{Q} \times \mathbf{Q}$ e.g., or any finite product for that matter. &#xA;A proof is at &lt;a href=&quot;http://at.yorku.ca/p/a/c/a/25.htm&quot; rel=&quot;nofollow&quot;&gt;the topology atlas, topology explained&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-12T21:54:47.093" CommentCount="2" />
  <row Id="14090" PostTypeId="2" ParentId="1064" CreationDate="2010-12-12T21:57:02.770" Score="0" ViewCount="0" Body="&lt;p&gt;It is well-known that $C$ is homeomorphic to $C \times C$, where $C$ is the Cantor set, as both are zero-dimensional compact metric spaces without isolated points. So $C$ contains uncountably many disjoint homeomorphic copies of $C$ and all but countably many of them can contain rationals...&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-12T21:57:02.770" />
  <row Id="14091" PostTypeId="1" CreationDate="2010-12-12T22:12:46.547" Score="1" ViewCount="55" Body="&lt;p&gt;$Z(t) = \int_0^t  g(s)\,dW(s)$, where $g$ is an adapted stochastic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find $dZ$ ?&lt;/p&gt;&#xA;" OwnerUserId="3964" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-12T22:38:17.607" LastActivityDate="2010-12-12T22:38:17.607" Title="Ito's Lemma application" Tags="&lt;stochastic-integrals&gt;" CommentCount="3" />
  <row Id="14092" PostTypeId="2" ParentId="9523" CreationDate="2010-12-12T22:14:13.440" Score="0" ViewCount="0" Body="&lt;p&gt;According to my professor the approach I've laid out here is fine.  It does appear, though, that the more complicated technique that involves splitting (fg)'/fg into f'/f + g'/g has some uses, so I can see why they'd bring that up.&lt;/p&gt;&#xA;" OwnerUserId="3296" LastActivityDate="2010-12-12T22:14:13.440" />
  <row Id="14093" PostTypeId="2" ParentId="14026" CreationDate="2010-12-12T22:46:48.310" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://code.google.com/p/mpmath/&quot; rel=&quot;nofollow&quot;&gt;mpmath&lt;/a&gt;, which is a part of sage has great special functions support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nickle.org&quot; rel=&quot;nofollow&quot;&gt;nickle&lt;/a&gt; is good for quick things, and has C like syntax.&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-12-12T22:46:48.310" />
  <row Id="14094" PostTypeId="2" ParentId="14078" CreationDate="2010-12-12T22:51:07.327" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; $\text{orig-salary}$ represents the salary you had before it was changed.&#xA;$\text{next-salary}$ represents the salary you had after it was increased by 10%.&#xA;$\text{final-salary}$ represents the salary you had after it was increased by 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever we say &quot;x was increased by y%&quot; we mean this:&#xA;$$x + (x * y\%) = (x * 100\%) + (x * y\%) = x * (100\% + y\%)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, using the information in your problem, we have the following two equations:&#xA;\begin{align*}&#xA;\text{next-salary} &amp;amp;= \text{orig-salary} * (100\% + 10\%)\\&#xA;\text{final-salary} &amp;amp;= \text{next-salary} * (100\% + 50\%)&#xA;\end{align*}&#xA;Replace $\text{next-salary}$ with $\text{orig-salary} * (100\% + 10\%)$&#xD;&#xA;\begin{align*}&#xA;\text{final-salary} &amp;amp;= \text{next-salary} * (100\% + 50\%)\\&#xA;\text{final-salary} &amp;amp;= (\text{orig-salary} * (100\% + 10\%)) * (100\% + 50\%)\\&#xA;\text{final-salary} &amp;amp;= \text{orig-salary} * (110\%) * (150\%)\\&#xA;\text{final-salary} &amp;amp;= \text{orig-salary} * 1.10 * 1.50\\&#xA;\text{final-salary} &amp;amp;= \text{orig-salary} * 1.650&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall increase would be $1.650 - 100\%$&#xD;&#xA;OR&#xA;$165.0\% - 100\% = 65\%$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall increase is $65\%$.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T06:04:17.850" LastActivityDate="2010-12-13T06:04:17.850" />
  <row Id="14095" PostTypeId="2" ParentId="13890" CreationDate="2010-12-12T23:12:10.840" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad 3^4 = 1\ (mod\ 20)\ \Rightarrow\ 3^n\ =\ 20\ k + \{1,3,9,27\}$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-12T23:12:10.840" />
  <row Id="14096" PostTypeId="1" CreationDate="2010-12-12T23:14:23.163" Score="1" ViewCount="57" Body="&lt;p&gt;Hello.&#xA;We know that $H^p(S^2 \vee S^4) = H^p(S^2)\oplus H^p(S^4)$ for $p\neq 0$. I want to show that this space has different ring structure than $CP^2$. So, given a generator in $H^2(S^2 \vee S^4)$ I want to cup it with itself and get 0. My idea is to use the generator from $H^2(S^2)$(which obviously is zero when squared). How should I go from here? Is this even the correct idea? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\vee$ here is one-point intersection).&lt;/p&gt;&#xA;" OwnerUserId="2900" LastEditorUserId="2900" LastEditorDisplayName="" LastEditDate="2010-12-13T00:34:35.993" LastActivityDate="2010-12-13T00:34:35.993" Title="Ring structure of  $H^2(S^2 \vee S^4)$ " Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14097" PostTypeId="1" AcceptedAnswerId="14099" CreationDate="2010-12-12T23:16:23.130" Score="3" ViewCount="82" Body="&lt;p&gt;Define $(a,b) &amp;lt; (a&amp;#39;,b&amp;#39;)$ if $\max(a,b) &amp;lt; \max(a&amp;#39;,b&amp;#39;)$ or $\max(a,b) = \max(a&amp;#39;,b&amp;#39;)$ and $b &amp;lt; b&amp;#39;$ or $\max(a,b) = \max(a&amp;#39;,b&amp;#39;)$ and $b = b&amp;#39;$ and $a &amp;lt; a&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to prove that the order-type of $(\{(b,c) : \max(b,c) = a\}, &amp;lt;)$ is equal to $a + a + 1$, does someone have a hint how to do this? I can't find the bijection. All elements are ordinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T04:26:14.790" LastActivityDate="2010-12-13T04:26:14.790" Title="Ordering on pairs." Tags="&lt;set-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14098" PostTypeId="2" ParentId="14096" CreationDate="2010-12-12T23:19:38.023" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, that is the right idea. Use that the cup product is &quot;natural&quot; with respect to pull backs.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2010-12-12T23:19:38.023" CommentCount="6" />
  <row Id="14099" PostTypeId="2" ParentId="14097" CreationDate="2010-12-12T23:30:22.830" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; The set naturally breaks up into three disjoint parts: $\{(a,a)\}$, the set $\{(b,a)\mid b\lt a\}$; and the set $\{(a,b)\mid b\lt a\}$. Every element of the latter is less than each element of the middle one, and each element of the middle one is stricly smaller than $(a,a)$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-12T23:30:22.830" CommentCount="1" />
  <row Id="14100" PostTypeId="1" AcceptedAnswerId="14110" CreationDate="2010-12-12T23:30:32.390" Score="7" ViewCount="135" Body="&lt;p&gt;Here is what I need to prove: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $d_1,d_2,...,d_n$ be a sequence of natural numbers (&gt;0). Show that $d_i$ is a degree sequence of some tree if and only if $\sum d_i = 2(n-1)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that:&lt;br&gt;&#xA;1. for any graph $\sum_{v \in V}\ deg(v) = 2e$;&lt;br&gt;&#xA;2. for any tree $e=v-1$.&lt;br&gt;&#xA;From 1 and 2 it follows that for any tree $\sum_{v \in V}\ deg(v) = 2(v-1)$.&lt;br&gt;&#xA;If I understand it correctly, this is only a half of the proof ($\rightarrow$), isn't it? &#xA;Any hints on how to prove it the other way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit (induction attempt):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$n=1$, we have $d_1 = 2(1-1) = 0$ and $d_1$ is a degree sequence of a tree. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's assume the theorem holds for all $k&amp;lt;n$. So we know that $d_1 + d_2 + ... + d_{n-1} = 2(n-2) $ and that it's a degree sequence of a tree. In order to add one vertex to this tree, so that it remains a tree we only can add a vertex of degree one (we connect it to any existing vertex with a single edge). By doing this our new sequence is $1,d_1,...,d_j+1,...,d_{n-1}$, where $j$ is the vertex to which we attached the new vertex (it's degree increments). Clearly this sums to $2(n-1)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is this proof correct or am I missing the point?&lt;/p&gt;&#xA;" OwnerUserId="4687" LastEditorUserId="4687" LastEditorDisplayName="" LastEditDate="2010-12-13T16:52:17.140" LastActivityDate="2010-12-13T16:52:17.140" Title="Tree related proof problem" Tags="&lt;graph-theory&gt;&lt;trees&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14101" PostTypeId="1" CreationDate="2010-12-12T23:30:47.807" Score="0" ViewCount="229" Body="&lt;p&gt;Can you tell me why the following is true? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A ring $R$ is local if and only if every principal left $R$-module is indecomposable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Edit by KennyTM: The above is OP's original question. The latest, completely changed question follows:)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the relation between regular ring and this property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ab=1$ then $ba=1$ when $a,b\in R$&lt;/p&gt;&#xA;" OwnerUserId="4688" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-16T09:24:42.660" LastActivityDate="2010-12-16T09:24:42.660" Title="Ring Theory (Local Rings)" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14102" PostTypeId="2" ParentId="14041" CreationDate="2010-12-12T23:32:44.373" Score="0" ViewCount="0" Body="&lt;p&gt;I think you're over-thinking it slightly. With regards to your original question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a bijection $f:\mathbb{Z}\to\mathbb{Z}$ where $\mathbb{Z}$ is the set of all integers, does there always exist two bijections $g:\mathbb{Z} \to \mathbb{Z}$ and $h:\mathbb{Z} \to \mathbb{Z}$ which satisfy $f(n)=g(n)+h(n), \forall n∈\mathbb{Z}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes. Note that the function $g:\mathbb{Z}\to\mathbb{Z}$ defined by $g(n) = n$ is a bijection; moreover, since $f$ and $g$ are both bijections, it's pretty easy to show that the function $h:\mathbb{Z}\to\mathbb{Z}$ defined by $h(n) = f(n) - g(n)$ is also a bijection; just go through the usual steps to show it's both injective and surjective. Finally, $g(n) + h(n) = g(n) + f(n) - g(n) = f(n)$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: It turns out that my argument doesn't hold in general. Apologies!&lt;/p&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4689" LastEditorDisplayName="" LastEditDate="2010-12-12T23:45:17.897" LastActivityDate="2010-12-12T23:45:17.897" CommentCount="2" />
  <row Id="14103" PostTypeId="1" CreationDate="2010-12-12T23:33:53.090" Score="2" ViewCount="69" Body="&lt;p&gt;Let $K$ be a field and $f(x)$ be an irreducible polynomial over $K$. Suppose, $f(x)$ has degree at least $2$. Is it possible that if $a,b$ are two roots of $f(x)$ with $a\neq b$, then $K(a)=K(b)$. Note I need equality, not isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-12-13T04:06:57.683" Title="Can two different roots of an irreducible polynomial generate the same extension?" Tags="&lt;field-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14104" PostTypeId="2" ParentId="14103" CreationDate="2010-12-12T23:40:07.140" Score="5" ViewCount="0" Body="&lt;p&gt;For the question to &lt;em&gt;really&lt;/em&gt; make sense we need to fix an algebraic closure of $K$ and work always inside of it; otherwise, we can even make two constructions of $K(a)$ that would not be &quot;equal&quot; (e.g., constructing $\mathbb{Q}(i)$ as the set of pairs $(a,b)\in\mathbb{Q}^2$ with suitable addition/multiplication; or as $\mathbb{Q}[x]/(x^2+1)$; or as the set of formal symbols $a+bi$ with $a,b\in\mathbb{Q}$, etc.) &lt;strong&gt;Added:&lt;/strong&gt; See Qiaochu's answer for a much better explanation of this issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, yes, it is possible: for a trivial example, take $K$ to be a finite field with $q$ elements. Then both $K(a)$ and $K(b)$ are extensions of degree $\deg(f)$ over $K$; if $\deg(f)=n$, then this is a field with $q^n$ elements, and since there is a unique field of $q^n$ elements sitting inside the algebraic closure of $K$, $K(a)=K(b)$. In fact, in this case you &lt;em&gt;always&lt;/em&gt; have that situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a nonfinite example in a field of characteristic $0$, take either $\mathbb{Q}(\sqrt{d})$: the irreducible polynomial is $x^2-d$, which has roots $\sqrt{d}$ and $-\sqrt{d}$; naturally, $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(-\sqrt{d})$ (always inside the same fixed algebraic closure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example, take $f(x)$ to the cyclotomic polynomial $\Phi_n(x)$ with $K=\mathbb{Q}$; the roots are precisely the primitive $n$-roots of unity, and all the extensions are equal to $\mathbb{Q}(e^{2\pi i/n})$ (again, always sitting inside the same fixed algebraic closure of $\mathbb{Q}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; More generally, any simple extension $K(a)$ which is Galois over $K$ will have the property, since the irreducible polynomial of $a$ necessarily splits in $K(a)$. That means that if $f(x)$ is the irreducible polynomial of $a$, then for any root $r$ of $f(x)$ in the fixed algebraic closure of $K$ containing $K(a)$ we have $r\in K(a)$, so $K(r)\subseteq K(a)$; but applying the usual isomorphism we get that $f(x)$ must also split in $K(r)$, so $a\in K(r)$ hence $K(a)\subseteq K(r)$, giving equality. In particular, this holds for any finite Galois extension of a perfect field. This includes all the examples I gave above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, I claim that if $f(x)$ is an irreducible polynomial over $K$ with degree at least $2$, and for any two roots $a$ and $b$ of $f(x)$ in a fixed algebraic closure of $K$ we have that $K(a) = K(b)$, then $K(a)$ is the splitting field of $f(x)$ (and so if $f(x)$ is separable, $K(a)$ is a finite Galois extension of $K$). Indeed, this is trivial, since for any root $b$ of $f(x)$ we have $b\in K(b)=K(a)$, so $K(a)$ contains all the roots of $f(x)$; since $K(a)$ is the least field containing $a$, $K(a)$ is contained in the splitting field of $f(x)$ over $K$, thus showing $K(a)$ &lt;em&gt;is&lt;/em&gt; the splitting field.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T04:06:57.683" LastActivityDate="2010-12-13T04:06:57.683" CommentCount="2" />
  <row Id="14105" PostTypeId="1" CreationDate="2010-12-12T23:47:03.433" Score="3" ViewCount="51" Body="&lt;p&gt;It is possible to find the shortest route thanks to algorithms like A*, bread first search, depth first search, etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any known algorithm to find how many routes are available if there are more than 1 optimum solution (for example routes with same costs) or any solution at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on some graph algorithms but I don't think I'm on the right direction. Any advices?&lt;/p&gt;&#xA;" OwnerUserId="4653" LastActivityDate="2010-12-12T23:47:03.433" Title="Question about path-finding" Tags="&lt;graph-theory&gt;&lt;optimization&gt;" CommentCount="2" />
  <row Id="14106" PostTypeId="2" ParentId="14100" CreationDate="2010-12-12T23:57:08.307" Score="3" ViewCount="0" Body="&lt;p&gt;Hints for the other way round: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do you know about mathematical induction?&lt;/li&gt;&#xA;&lt;li&gt;Can every $d_i$ be $\gt 1$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-12T23:57:08.307" CommentCount="2" />
  <row Id="14107" PostTypeId="1" AcceptedAnswerId="14114" CreationDate="2010-12-13T00:06:34.620" Score="2" ViewCount="52" Body="&lt;p&gt;Say I have two algorithms that I don't know how they work but I know what they are meant to achieve. I tried algorithm A once and it succeeded. And tried algorithm B 100 times and succeeded 99 times. So A succeeded 100% times and B succeeded 99% times, but my common sense tells me B is still more &quot;reliable&quot; than A. How can I express this mathematically, or, how can I calculate which algorithm should I choose if I have to choose one when all I know about each of them is how many times it has been tested and how many times it succeeded (if it didn't that just means it failed for what I need)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be as simple as possible I'm a newbie at math.&lt;/p&gt;&#xA;" OwnerUserId="1542" LastEditorUserId="1542" LastEditorDisplayName="" LastEditDate="2010-12-13T00:31:36.557" LastActivityDate="2010-12-13T01:26:32.397" Title="How to calculate likelihood to succeed knowing attempts and successful attempts amounts?" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14108" PostTypeId="1" CreationDate="2010-12-13T00:12:43.527" Score="2" ViewCount="77" Body="&lt;p&gt;Let $R$ be a commutative ring with unity and let $B(R)$ be the set of all idempotent elements in $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show for $b\in B(R)$, the $R$-modules $R$ and $Rb \times R(1-b)$ are isomorphic to one another. &lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T00:16:56.580" LastActivityDate="2010-12-13T01:15:13.963" Title="Show the $R$-module $R$ is isomorphic to $Rb \times R(1-b)$ where $b$ is an idempotent of a commutative ring with unity " Tags="&lt;abstract-algebra&gt;&lt;modules&gt;" AnswerCount="0" CommentCount="7" />
  <row Id="14109" PostTypeId="1" AcceptedAnswerId="14122" CreationDate="2010-12-13T00:16:06.113" Score="4" ViewCount="72" Body="&lt;p&gt;Let $R$ be a commutative ring with identity and $I,J,K$ be ideals of $R$. If $I\supseteq J$ or $I\supseteq K$, we have the following modular law &#xA;$$ I\cap (J+K)=I\cap J + I\cap K$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are situations in which the modular law holds in which the hypothesis that $I$ contains at least one of $J,K$ is relaxed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example is when $R$ is a polynomial ring or power series ring and $I,J,K$ are monomial ideals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one containment always holds $I\cap (J+K)\supseteq I\cap J +I\cap K$. In what other situations does the other containment hold?&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-13T01:43:10.753" LastActivityDate="2010-12-13T04:52:15.147" Title="When does the modular law apply to ideals in a commutative ring" Tags="&lt;abstract-algebra&gt;&lt;commutative-algebra&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="14110" PostTypeId="2" ParentId="14100" CreationDate="2010-12-13T00:37:56.883" Score="2" ViewCount="0" Body="&lt;p&gt;From your fact (1) and the hypothesis $\sum d_i = 2(n-1)$, you know $e = n - 1$.  You need now only prove that any connected graph with $n - 1$ edges is a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - Since you are going down an alternate path, I have a new suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Order the $d_i$ so that $d_1 \leq d_2 \leq \ldots \leq d_n$.  Construct a tree with degree sequence $d_2, \ldots, d_n$. You know $\sum_{i \neq 1} d_i = 2(n - 2)$ (why?).  Now try to construct your tree with degree sequence $d_1, \ldots, d_n$.  You have to be careful about how you add in your vertex with degree $d_1$ so that it remains a tree, but this can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - The above is incorrect.  The idea remains to add in a vertex of degree $1$ to a tree with a specific degree sequence related to $d_1, \ldots, d_n$. But $\sum_{i \neq 1} d_i = 2n - 3$, so we cannot claim $d_2, \ldots, d_n$ is a degree sequence of a tree by the inductive hypothesis.  Of course, when we add in the vertex of degree $1$, we alter the degree of another vertex, right?&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-12-13T16:30:02.830" LastActivityDate="2010-12-13T16:30:02.830" CommentCount="8" />
  <row Id="14112" PostTypeId="2" ParentId="14041" CreationDate="2010-12-13T00:57:38.260" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Observation&lt;/em&gt;:  It is sufficient to resolve the question for $f(n)=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof idea&lt;/em&gt;:  If $n=g(n)+h(n)$ for all $n$, then $f(n)=g(f(n))+h(f(n))$.  Conversely, if $f(n)=g(n)+h(n)$ then $n=f(f^{-1}(n))=g(f^{-1}(n))+h(f^{-1}(n))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if &lt;em&gt;some&lt;/em&gt; $f$ can be decomposed into two bijections $h$ and $g$, then &lt;em&gt;every&lt;/em&gt; $f$ can be decomposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So resolving this question amounts to finding a permutation $g$ of $\mathbb{Z}$ for which $n \mapsto g(n)-n$ is also a permutation.  Such permutations are called &lt;em&gt;orthomorphisms&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;UPDATE&lt;/em&gt;: I asked both Ian Wanless and Tony Evans about this, and they simultaneously pointed out that the existence of such an orthomorphism was proved in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MR0040293 (12,670f) &#xA;Bateman, P. T.&#xA;&lt;a href=&quot;http://www.jstor.org/stable/2308116&quot; rel=&quot;nofollow&quot;&gt;A remark on infinite groups&lt;/a&gt;. &#xA;Amer. Math. Monthly 57, (1950). 623–624. &#xA;20.0X &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can conclude that, yes, it's always possible.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-13T02:43:20.050" LastActivityDate="2010-12-13T02:43:20.050" CommentCount="7" />
  <row Id="14113" PostTypeId="1" CreationDate="2010-12-13T00:58:16.170" Score="4" ViewCount="386" Body="&lt;p&gt;I know the series, $4-{4\over3}+{4\over5}-{4\over7}...$ converges to $\pi$ but I have heard many people say that while this is a classic example, there are series that converge much faster.  Does anyone know of any?&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T01:32:31.180" LastActivityDate="2010-12-15T09:07:35.300" Title="Series that converge to $\pi$ quickly" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;&lt;pi&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="4" />
  <row Id="14114" PostTypeId="2" ParentId="14107" CreationDate="2010-12-13T01:11:40.313" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps what you are looking for is &quot;Confidence Interval for Proportion&quot;. There are calculators for that online, see e.g. &lt;a href=&quot;http://www.dimensionresearch.com/resources/calculators/conf_prop.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://www.causascientia.org/math_stat/ProportionCI.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;a href=&quot;http://www.dimensionresearch.com/resources/calculators/sample_size.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is also a calculator that determines the minimum sample size you would need in order for your sample to meet the desired precision requirements for a study .&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-13T01:26:32.397" LastActivityDate="2010-12-13T01:26:32.397" />
  <row Id="14115" PostTypeId="1" AcceptedAnswerId="14422" CreationDate="2010-12-13T01:28:27.437" Score="9" ViewCount="317" Body="&lt;p&gt;The following formula for $\pi$ was discovered by Ramanujan:&#xA;$$\frac1{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\!$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how it works, or what the motivation for it is?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-13T02:53:56.663" LastActivityDate="2010-12-15T16:22:08.093" Title="Motivation for Ramanujan's mysterious $\pi$ formula" Tags="&lt;sequences-and-series&gt;&lt;approximation&gt;&lt;pi&gt;&lt;ramanujan&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="14116" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T01:36:46.327" Score="9" ViewCount="0" Body="&lt;p&gt;The series $$ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^n = \frac{\pi}{2}$$ converges quickly. Here $!!$ is the double factorial defined by $0!! = 1!! = 1$ and $n!! = n (n-2)!!$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is series is not too hard to derive. Start by defining $$f(t) = \sum _{n=0}^{\infty } \frac{(-1)^n}{(2n+1)}t^n.$$ Note that $f(1) = \pi/4$ is the series you referenced. Now we take what is called the Euler Transform of the series which gives us $$ \left(\frac{1}{1-t}\right)f\left(\frac{t}{1-t}\right) = \sum _{n=0}^{\infty } \left(\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)}\right)t^n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)} = \frac{(2n)!!}{(2n+1)!!}$$ for hints on how to prove this identity see &lt;a href=&quot;http://math.stackexchange.com/questions/4261/proving-a-binomial-sum-identity&quot;&gt;Proving a binomial sum identity&lt;/a&gt;. Now put $t = 1/2$ and the identity follows. Showing the error term for the nth partial sum is less than $(1/2)^n$ is not too difficult.&lt;/p&gt;&#xA;" OwnerUserId="899" LastEditorUserId="899" LastEditorDisplayName="" LastEditDate="2010-12-13T01:57:49.257" LastActivityDate="2010-12-13T01:57:49.257" CommentCount="2" />
  <row Id="14117" PostTypeId="2" ParentId="14066" CreationDate="2010-12-13T01:40:15.823" Score="4" ViewCount="0" Body="&lt;p&gt;I would recommend reading Gerald Rising's &lt;a href=&quot;http://books.google.com/books?id=ok_PE_WHCREC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Inside your Calculator&lt;/em&gt;&lt;/a&gt; (which has a &lt;a href=&quot;http://www.acsu.buffalo.edu/~insrisg/InsideYourCalculator.htm&quot; rel=&quot;nofollow&quot;&gt;supplementary website&lt;/a&gt;); there is a nice discussion of the methods used by some calculators that is suitable at the undergraduate level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, to really figure out what methods they are using, it might help to search the technical notes of the manufacturer's websites. For instance, Texas Instruments has notes like &lt;a href=&quot;http://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=74414&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; on their &lt;a href=&quot;http://epsstore.ti.com/OA_HTML/cskmbasicsrch.jsp&quot; rel=&quot;nofollow&quot;&gt;&quot;knowledge base&quot;&lt;/a&gt; that discuss &quot;what's under the hood&quot;, though not in detail of course. (Sometimes, hobbyist sites like &lt;a href=&quot;http://tibasicdev.wikidot.com/68k%3arandseed&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; also discuss calculator algorithms.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-13T01:40:15.823" CommentCount="2" />
  <row Id="14118" PostTypeId="1" AcceptedAnswerId="14121" CreationDate="2010-12-13T01:40:48.730" Score="4" ViewCount="74" Body="&lt;p&gt;This sounds like a simple problem, but I can't get it done. Given the general equation&#xA;$ax^2 + by^2 + cz^2 + dxy + exz + fyz + gx + hy + iz + j = 0,$ &#xA;what are the requirements on the coefficients so that this equation represents an ellipsoid? All coefficients are real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more context. To make the representation unique, we establish another constraint: $a + c = 1$. I'm not sure how important that might be. The question is in homework where we have to fit this equation to a set of points, using QR-decomposition. That isn't a problem, the small followup question &quot;how can you see the coefficients determine an ellipsoid&quot; is the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried working on the equation to get something usable, from basic term regrouping to putting stuff in matrices and decomposing these, which didn't really help. One theoretic way to do it, I think, is to find the center of mass of the surface, translate to the origin, then find the principal axes, transform again, and show the radii are all larger than 0. Sounds like a lot of work for a simple question, and I'm kind of stuck on the &quot;center of mass&quot; part, I don't feel the idea here is to integrate the formula over $\mathbb{R}^3$. I've also been thinking about calculating curvature or something with the Frenet-Serret formulas, but my geometry courses have been too long ago to know what I want here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple test to determine this? Something along the lines of &quot;the eigenvalues of matrix foo are all positive&quot; or &quot;the value of bar is real&quot; or I don't know what...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for any insights you can give me!&lt;/p&gt;&#xA;" OwnerUserId="4692" LastActivityDate="2010-12-13T03:22:39.760" Title="Is a general (non-homogeneous) quadratic equation in $\mathbb{R}^3$ an ellipsoid?" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;&lt;quadratic-forms&gt;" AnswerCount="1" />
  <row Id="14119" PostTypeId="2" ParentId="14084" CreationDate="2010-12-13T01:45:14.123" Score="4" ViewCount="0" Body="&lt;p&gt;Here is another way to prove Euler's generalization. You do not need to know the formula of $\varphi(n)$ for this method which I think makes this method elegant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of all numbers less than $n$ and relatively prime to it. Let $\{a_1,a_2,...,a_{\varphi(n)}\}$ be this set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a number $c &amp;lt; n$ and relatively prime to it i.e. $c \in \{a_1,a_2,\ldots,a_{\varphi(n)}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observe that for any $a_i$, $c a_{i} \equiv a_{j} (\bmod n)$ for some $j$.&#xA;(True since $c$ and $a_i$ are themselves relatively prime to $n$, their product has to be relatively prime to $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if $c a_{i} \equiv c a_{j} (\bmod n)$ then $a_i = a_j$.&#xA;(True as cancellation can be done since $c$ is relatively prime to $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if we now consider the set $\{ca_1,ca_2,...,ca_{\varphi(n)}\}$ this is just a permutation of the set $\{a_1,a_2,...,a_{\varphi(n)}\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thereby, we have $\displaystyle \prod_{k=1}^{\varphi(n)} ca_k \equiv \prod_{k=1}^{\varphi(n)} a_k (\bmod n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $\displaystyle c^{\varphi(n)} \prod_{k=1}^{\varphi(n)} a_k \equiv \prod_{k=1}^{\varphi(n)} a_k (\bmod n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that $\displaystyle \prod_{k=1}^{\varphi(n)} a_k$ is relatively prime to $n$ and hence you can cancel them on both sides to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c^{\varphi(n)} \equiv 1 (\bmod n)$$ whenever $(c,n) = 1$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-13T23:54:55.527" LastActivityDate="2010-12-13T23:54:55.527" CommentCount="0" />
  <row Id="14121" PostTypeId="2" ParentId="14118" CreationDate="2010-12-13T01:59:32.950" Score="6" ViewCount="0" Body="&lt;p&gt;There are two steps involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1: translate to centre of mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of translation to the centre of mass is to remove the &lt;em&gt;linear terms&lt;/em&gt; in your coefficients. You want to turn your expression into the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ AX^2 + BY^2 + CZ^2 + D XY + E YZ + F XZ + G = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $X = (x - x_0)$, $Y = (y - y_0)$ and $Z = (z-z_0)$, where $x_0, y_0, z_0$ are the coordinates for the centre of mass. To do so you can solve forward by completing the square, or solve backwards by plugging in the expressions for $X,Y,Z$ and matching coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2: check for positivity of eigenvalues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to the centre of mass, you can write your equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q(v) = -G $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q$ is the quadratic form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q(v) = AX^2 + BY^2 + \cdots + F XZ, \qquad v = (X,Y,Z) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then recall that the level surfaces of a quadratic form define ellipsoids if and only if it is positive definite. So computing the eigenvalues (among the many ways to check that a quadratic form is positive definite) will get you the answer. (Of course, this will also require $G$ to be a negative number for there to be a solution.)&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-13T03:22:39.760" LastActivityDate="2010-12-13T03:22:39.760" CommentCount="8" />
  <row Id="14122" PostTypeId="2" ParentId="14109" CreationDate="2010-12-13T02:02:25.497" Score="8" ViewCount="0" Body="&lt;p&gt;Such domains are known as Prüfer domains. They are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations. For example, they are those domains satisfy either the Chinese Remainder Theorem for ideals, or Gauss's Lemma for polynomial content ideals, or for ideals: $\rm\ A\cap (B + C) = A\cap B + A\cap C\:,\  $ or $\rm\ (A + B)\ (A \cap B) = A\ B\:,\ $ or $\rm\ A\supset B\ \Rightarrow\ A\:|\:B\ $ for fin. gen. $\rm\:A\:$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/ba5ac69dddfc4a67&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for 30 odd characterizations. Below is an excerpt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt;  $\ \ $ Let $\rm\:D\:$ be a domain. The following are equivalent:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\rm\:D\:$ is a Prufer domain, i.e. every nonzero f.g. (finitely generated) ideal is invertible.&lt;br&gt;&#xA;(2) Every nonzero two-generated ideal of $\rm\:D\:$ is invertible.&lt;br&gt;&#xA;(3) $\rm\:D_P\:$ is a  Prufer   domain for every  prime  ideal $\rm\:P\:$ of $\rm\:D.\:$&lt;br&gt;&#xA;(4) $\rm\:D_P\:$ is a valuation domain for every  prime  ideal $\rm\:P\:$ of $\rm\:D.\:$&lt;br&gt;&#xA;(5) $\rm\:D_P\:$ is a valuation domain for every maximal ideal $\rm\:P\:$ of $\rm\:D.\:$&lt;br&gt;&#xA;(6) Every nonzero f.g. ideal $\rm\:I\:$ of $\rm\:D\:$ is cancellable, i.e. $\rm\:I\:J = I\:K\ \Rightarrow\ J = K\:$&lt;br&gt;&#xA;(7) $\: $ (6) restricted to f.g. $\rm\:J,K.$&lt;br&gt;&#xA;(8) $\rm\:D\:$ is integrally closed and there is an $\rm\:n &amp;gt; 1\:$ such that for all $\rm\: a,b \in D,\ (a,b)^n = (a^n,b^n).$&lt;br&gt;&#xA;(9) $\rm\:D\:$ is integrally closed and there is an $\rm\: n &amp;gt; 1\:$ such that for all $\rm\:a,b \in D,\ a^{n-1} b \ \in\ (a^n, b^n).$&lt;br&gt;&#xA;(10) Each ideal $\rm\:I\:$ of $\rm\:D\:$ is complete, i.e. $\rm\:I = \cap\ I\: V_j\:$ as $\rm\:V_j\:$ run over all the valuation overrings of $\rm\:D.\:$&lt;br&gt;&#xA;(11) Each f.g. ideal of $\rm\:D\:$ is an intersection of valuation ideals.&lt;br&gt;&#xA;(12) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then  $\rm\:I \cap (J + K) = I\cap J + I\cap K.$&lt;br&gt;&#xA;(13) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then  $\rm\:I\ (J \cap K) = I\:J\cap I\:K.$&lt;br&gt;&#xA;(14) If $\rm\:I,J\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:(I + J)\ (I \cap J) = I\:J.\ $  ($\rm LCM\times GCD$ law)&lt;br&gt;&#xA;(15) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ with $\rm\:K\:$ f.g. then $\rm\:(I + J):K = I:K + J:K.$&lt;br&gt;&#xA;(16) For any two elements $\rm\:a,b \in D,\ (a:b) + (b:a) = D.$&lt;br&gt;&#xA;(17) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D\:$ with $\rm\:I,J\:$ f.g. then $\rm\:K:(I \cap J) = K:I + K:J.$&lt;br&gt;&#xA;(18) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of localizations of $\rm\:D.\:$&lt;br&gt;&#xA;(19) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection  of quotient rings of $\rm\:D.\:$&lt;br&gt;&#xA;(20) Each overring of $\rm\:D\:$ is integrally closed.&lt;br&gt;&#xA;(21) Each overring of $\rm\:D\:$ is flat over $\rm\:D.\:$&lt;br&gt;&#xA;(22) $\rm\:D\:$ is integrally closed and prime ideals of overrings of are extensions of  prime ideals of $\rm\:D.$&lt;br&gt;&#xA;(23) $\rm\:D\:$ is integrally closed and for each prime ideal $\rm\:P\:$ of $\rm\:D,\:$ and each overring $\rm\:S\:$ of $\rm\:D,\:$ there is at most one prime ideal of $\rm\:S\:$ lying over $\rm\:P.\:$&lt;br&gt;&#xA;(24) For polynomials $\rm\:f,g \in D[x],\ c(fg) = c(f)\: c(g)\:$ where for a polynomial $\rm\:h \in D[x],\ c(h)\:$ denotes the &quot;content&quot; ideal of $\rm\:D\:$ generated by the coefficients of $\rm\:h.\:$ (Gauss' Lemma)&lt;br&gt;&#xA;(25) Ideals in $\rm\:D\:$ are integrally closed.&lt;br&gt;&#xA;(26) If $\rm\:I,J\:$ are ideals with $\rm\:I\:$ f.g. then $\rm\: I\supset J\ \Rightarrow\ I|J.$ (contains $\:\Rightarrow\:$ divides)&lt;br&gt;&#xA;(27) the Chinese Remainder Theorem $\rm(CRT)$ holds true in $\rm\:D\:,\:$ i.e. a system of congruences $\rm\:x\equiv x_j\ (mod\ I_j)\:$ is solvable iff $\rm\:x_j\equiv x_k\ (mod\ I_j + I_k).$  &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-13T04:52:15.147" LastActivityDate="2010-12-13T04:52:15.147" CommentCount="9" />
  <row Id="14123" PostTypeId="1" CreationDate="2010-12-13T02:14:51.970" Score="1" ViewCount="85" Body="&lt;p&gt;Whilst answering my homework, I have noticed that a question has said that I need to consider the subgroups of $A=\mathbb{Z}_{135}$, including $W=\langle 18,81\rangle$. I was under the impression that for a subset to be a subgroup $a+b$ in the subgroup had to be in the subgroup as well. $18+81 = 99\mod{135}$ which isn't in the subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it my reasoning or the problem sheet that is wrong? Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T02:26:21.880" LastActivityDate="2011-01-12T02:44:44.073" Title="Is $\langle 18,81\rangle$ a subgroup of the abelian group $\mathbb{Z}_{135}$?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14124" PostTypeId="1" AcceptedAnswerId="14129" CreationDate="2010-12-13T02:20:00.453" Score="2" ViewCount="180" Body="&lt;p&gt;I am working through a review problem asking to find the inverse of &#xA;$4\bmod 9 $.&#xA;Through examples I know that I first need to verify that the gcd is equal to 1 and write it as a linear combination of 4 and 9 to find the inverse.    I can do this in just one step:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gcd(4,9)&#xA;9 = 2 * 4 + 1&#xA;1 = 9 - 2 * 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This would suggest that the inverse is 1 if I am understanding this correctly.  However, the solution manual  doesn't show the work but says the LC should actually be&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 = 7 * 4 - 3 * 9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;making the answer to the question 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain to me what is going on here and how to properly find the inverse?&#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  wish I could add tags for congruency, gcd, and inverse.  I can't believe their isn't an inverse tag already :(&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T19:29:44.990" LastActivityDate="2010-12-13T19:29:44.990" Title="finding inverse of $x\bmod y$" Tags="&lt;modular-arithmetic&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="22" FavoriteCount="1" />
  <row Id="14125" PostTypeId="2" ParentId="14123" CreationDate="2010-12-13T02:21:34.477" Score="0" ViewCount="0" Body="&lt;p&gt;Thank you! The question makes far more sense now. I'm glad it's just notation and not all my beliefs about group theory being taken away!&lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T02:26:39.540" LastActivityDate="2010-12-13T02:26:39.540" CommentCount="1" />
  <row Id="14126" PostTypeId="2" ParentId="14124" CreationDate="2010-12-13T02:22:23.010" Score="5" ViewCount="0" Body="&lt;p&gt;$-2$ is congruent to $7$ mod $9$.  Your arithmetic steps are correct, but the conclusion should be that the congruence class of $-2$ is the inverse of $4$ mod $9$, and you probably want to use representatives in $\{1,2,3,4,5,6,7,8\}$ for your inverses, for simplicity, by adding appropriate multiples of $9$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T02:29:17.203" LastActivityDate="2010-12-13T02:29:17.203" CommentCount="8" />
  <row Id="14127" PostTypeId="2" ParentId="14101" CreationDate="2010-12-13T02:33:23.823" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT for $\Leftarrow$&lt;/strong&gt;: Do the contrapositive; if $\mathfrak{m}_1$ and $\mathfrak{m}_2$ are two distinct maximal ideals of $R$, then look at the left $R$-module $R/(\mathfrak{m}_1\cap\mathfrak{m}_2)$; it is principal (generated by $1+(\mathfrak{m}_1\cap\mathfrak{m}_2)$. Is it indecomposable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT for $\Rightarrow$&lt;/strong&gt;: Suppose $M\neq\{0\}$ is principal, and $M=N_1\times N_2$. Let $m=(n_1,n_2)$ generate $M$. Then there are $a,b\in R$ such that $am = (n_1,0)$ and $bm=(0,n_2)$; so $(a+b)m = m$. Then $a+b-1\in\mathrm{Ann}(m)$. Can $a$ and $b$ both lie in the maximal ideal of $R$? What does that mean?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-13T02:33:23.823" />
  <row Id="14128" PostTypeId="1" CreationDate="2010-12-13T02:47:00.883" Score="1" ViewCount="78" Body="&lt;p&gt;I want to create a function $f : [0,255]^3 \rightarrow [0,255]^3$ over the integers such that if you pass in three RGB color values, a triple with a darker or lighter color will be returned, and I want the values to increase in some sort of proportion to how dark or light they are already; lightness and darkness being determined by, say, taking the average of the values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a relatively simple function that will do this? If not, is there a way to emulate it?&lt;/p&gt;&#xA;" OwnerUserId="1562" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-13T03:16:27.213" LastActivityDate="2010-12-13T03:16:27.213" Title="Creating a function to brighten or darken colors" Tags="&lt;functions&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="14129" PostTypeId="2" ParentId="14124" CreationDate="2010-12-13T03:02:25.813" Score="3" ViewCount="0" Body="&lt;p&gt;I know you've already gotten lots of responses, but instead of trying to fit responses into the comments under an answer I thought I'd just reiterate exactly &lt;em&gt;why&lt;/em&gt; you need to do what you're doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inverse to $4 \mod 9$ is an integer $a$ such that $4a \equiv 1 \mod 9$. If we rewrite this, it means precisely that $9| (4a-1)$, or that there is another integer $b$ so that $9b = 4a-1$. What this says is that you need to find integer solutions to $4x-9y=1$. If you find that $x=a$ and $y=b$ works then $a$ is an inverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you do to solve something like this is run the Euclidean algorithm for $9$ and $4$, and then &quot;reverse&quot; the steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$9=4\cdot 2 + 1$, so it is actually over in 1 step in this case. Just subtract over to get $9-4\cdot 2 = 1$, or in the form we wrote it above $4\cdot (-2)-9\cdot (-1)=1$. So a solution is $x=-2$ and $y=-1$ and we said the $x$-value was the inverse, so the inverse is $-2\mod 9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully if you understand the process of why you're doing these things, then if you get confused about which one is which on the final you can always just rederive it from these steps.&lt;/p&gt;&#xA;" OwnerUserId="2000" LastActivityDate="2010-12-13T03:02:25.813" CommentCount="1" />
  <row Id="14130" PostTypeId="2" ParentId="14103" CreationDate="2010-12-13T03:08:56.810" Score="4" ViewCount="0" Body="&lt;p&gt;I think it's worth elaborating on this distinction between equality and isomorphism.  The problem occurs if the extension you're considering isn't &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_extension&quot; rel=&quot;nofollow&quot;&gt;normal&lt;/a&gt; and can therefore embed into an algebraic closure in more than one way.  For example, the abstract field $\mathbb{Q}[x]/(x^3 - 2)$ embeds in three ways into $\mathbb{C}$, corresponding to the three roots of $x^3 - 2$.  So it doesn't make sense to ask whether $\mathbb{Q}[x]/(x^3 - 2)$ is &quot;equal to&quot; $\mathbb{Q}[y]/(y^3 - 2)$ without specifying an embedding into a larger field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly subtle point which I don't think is addressed particularly well in introductions to Galois theory (at least the ones I've seen).  There are three categories one might work in when studying fields (where $K$ is a fixed field):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The category of fields.  Here it is &quot;evil&quot; (really, impossible) to speak of the &quot;equality&quot; of two fields and one can only speak of isomorphism.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The category of (say, algebraic) field extensions $K \to L$.  (The morphisms are morphisms $\phi : L \to L&amp;#39;$ making the obvious triangle commute.)  Here it is still &quot;evil&quot; to speak of the &quot;equality&quot; of two extensions, and one can only speak of isomorphism, but the isomorphism type of an extension $K \to L$ is not determined by the isomorphism type of $L$ (that is, $L$ can be an extension of $K$ in more than one way).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The category of subfields of $\bar{K}$ containing $K$ (for a fixed embedding $K \to \bar{K}$).  This is the category in which it makes sense to take intersections and composita of fields; you can't do either of these constructions in the above category.  Here, at last, one can speak of equality (as subfields of $\bar{K}$), and it is not the same as isomorphism of $K$-extensions, which is in turn not the same as isomorphism of abstract fields.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;People sometimes don't specify which of the above categories they're working in, and until you do this you can't be precise: there are three different notions of equality or isomorphism in the last category.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-13T03:32:14.100" LastActivityDate="2010-12-13T03:32:14.100" />
  <row Id="14132" PostTypeId="2" ParentId="14124" CreationDate="2010-12-13T03:16:16.000" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Congruences preserve inverses, i.e. $\rm\ A\equiv a\ \Rightarrow\ A^{-1}\equiv a^{-1}\:.\ $ This follows from the fact that congruences preserve products, i.e. it's the special case $\rm\: AB \equiv 1\: $ in this &lt;strong&gt;congruence product rule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LEMMA&lt;/strong&gt; $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \ (mod\ m)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\rm\ \ m\: |\: A-a,\:\:\ B-b\ \Rightarrow\ m\ |\ (A-a)\ B + a\ (B-b)\ =\ AB - ab $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This congruence product rule is at the heart of many other similar product rules, for example Leibnitz's product rule for derivatives in calculus, e.g. see my &lt;a href=&quot;http://math.stackexchange.com/questions/3101/derivative-of-a-product-and-derivative-of-quotient-of-functions-theorem-i-dont/3178#3178&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T03:16:16.000" CommentCount="3" />
  <row Id="14133" PostTypeId="1" AcceptedAnswerId="14136" CreationDate="2010-12-13T03:22:39.953" Score="4" ViewCount="207" Body="&lt;p&gt;One of the rules of logarithms that I've been taught is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_b(x)$ is equivalent to $\frac{\log_k(x)}{\log_k(b)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently I've also seen another rule that says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log_b(x)$ is equivalent to $\log_b(k)\log_k(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these equivalent (perhaps via some refactoring) ?&lt;/p&gt;&#xA;" OwnerUserId="4247" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T03:29:50.413" LastActivityDate="2010-12-13T15:37:34.063" Title="Clarify why all logarithms differ by a constant" Tags="&lt;logarithms&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="14134" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T03:27:55.283" Score="5" ViewCount="0" Body="&lt;p&gt;Yes. Your second equation is $\log_b k\cdot \log_k x = \log_b x$. Dividing by $\log_b k$ we get $\log_k x=\frac{\log_b x}{\log_b k}$. This is the same as your first equation with different symbols (an interchange of $k$ and $b$).&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T03:36:49.220" LastActivityDate="2010-12-13T03:36:49.220" CommentCount="1" />
  <row Id="14136" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T03:30:00.700" Score="4" ViewCount="0" Body="&lt;p&gt;Divide the second equation by $\log_b(k)$ and shuffle the variables to see that these are the same thing.  The reason they're true can be expressed in terms of properties of exponents. Start with the observation that $\log_b(x)=y$ means that $b^y=x$.  Then use $b=a^{\log_a(b)}$ and properties of exponents to get $b^y=a^{y\log_a(b)}=x$.  The last equation says that $\log_a(x)=y\log_a(b)$.  Now substitute $\log_b(x)$ back in for $y$ to get the desired identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fits into the general principle that each of the fundamental properties of logarithms can be seen as a translation of a property of their inverses, the exponential functions.  For example, logarithms convert products to sums ($\log_b(xy)=\log_b(x)+\log_b(y)$) because exponential functions convert sums to products ($b^{x+y}=b^xb^y$).  Here, the fact that $b^x=a^{x\log_a(b)}$ says that you can change bases of your exponential function by multiplying the input variable by the constant $\log_a(b)$.  In other words, exponential functions with different bases are related by horizontal stretches or shrinks of the graphs.  When expressed in terms of the inverse functions, input and output switch, and the horizontal stretch (or shrink) changes to a vertical stretch (or shrink).  That is, instead of multiplying the input by $\log_a(b)$, you divide the output by $\log_a(b)$ to get $\log_b(x)=\frac{\log_a(x)}{\log_a(b)}$.  More succinctly, $g(x)=f(cx)$ implies $g^{-1}(x)=\frac{1}{c}f^{-1}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T04:00:11.730" LastActivityDate="2010-12-13T04:00:11.730" CommentCount="1" />
  <row Id="14137" PostTypeId="1" CreationDate="2010-12-13T03:33:56.597" Score="0" ViewCount="116" Body="&lt;p&gt;Can anyone suggest at least verbally how it is that a torus is constructed from two cells?  (Total beginner to topology, been away from math as a whole for awhile...)  Thanks for any insights!&lt;/p&gt;&#xA;" OwnerUserId="4696" LastActivityDate="2010-12-13T23:26:55.440" Title="Please demonstrate decomposition of a torus into two cells" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="14138" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T03:36:25.383" Score="8" ViewCount="0" Body="&lt;p&gt;First: They are not &quot;equivalent&quot;, they are &lt;em&gt;equal&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said: remember the meaning of &quot;$\log_b(a) = r$&quot;. It means that $r$ is the exponent to which you must raise $b$ in order to get $a$; that is, $\log_b(a)=r$ is equivalent to $b^r=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... why is $\log_b(x)=\frac{\log_k(x)}{\log_k(b)}$? Because if you let $r=\log_b(x)$, $s=\log_k(x)$ and $t=\log_k(b)$, then that means that $b^r = x$, $k^s = x$, and $k^t=b$. Therefore, $$k^s = x = b^r = (k^t)^r = k^{tr}.$$ &#xA;So $\log_k(x) = s = rt = \log_b(x)\log_k(b)$, from which you get the equality you want by dividing through by $\log_k(b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second equality, $\log_b(x) = \log_b(k)\log_k(x)$, let $r=\log_b(x)$, $s=\log_b(k)$, and $t=\log_k(x)$. Then $b^r = x$, $b^s = k$, and $k^t=s$. So we have:&#xA;$$b^r = x = k^t = (b^s)^t = b^{st},$$&#xD;&#xA;which means we must have $\log_b(x) = r=st=\log_b(k)\log_k(x)$, as desired.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T15:37:34.063" LastActivityDate="2010-12-13T15:37:34.063" CommentCount="1" />
  <row Id="14139" PostTypeId="2" ParentId="14069" CreationDate="2010-12-13T03:53:32.863" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ The Porsche travels $\rm 20\ t$ more miles than the Skoda during the time $\rm\:t\:$ they're both driving. This must be equal to the $\ 20\cdot 9$ miles of the Skoda's head start. $\ $ Hence $\rm\ t\ =\ \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T03:53:32.863" />
  <row Id="14140" PostTypeId="1" CreationDate="2010-12-13T04:11:12.493" Score="1" ViewCount="61" Body="&lt;p&gt;Hello &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help me in this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 3 independent variables, $(T,H,t)$, as inputs and&#xA;one output $P$  ( I have all data for these inputs and the output, done&#xA;experimentally measured every hour during one year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find a formula of this form:&#xA;$$P = f ( T , H , t )$$&#xD;&#xA;where $t$ is the time in hours and it is always in the $x$-axis (index),&#xA;$T$ is temperature, $H$ is humidity, and $P$ is power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have all the data, and when I draw them in the same graph during one year, meaning that $P$, $T$,  and $H$ vs. hours. I found that the behavior of $P$ is oscillating, making a sinusoidal shape over the entire, year as you can see from the following ($P$ and time only):&lt;img src=&quot;http://i.imgur.com/3FdYF.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I make a zoom view to this figure, for example from&#xA;the $2000$th hour of the year to the $3000$th hour, it is clear that it almost has the same shape but it is oscillating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it keeps oscillating and increasing up to the peak point and then&#xA;it starts decreasing till the end of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is only for only one independent variable which is time in hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how if we include the effect of temperature and humidity and draw &#xA;these vectors along with the power vs. time, to see how power is changing&#xA;with respect to $T$, $H$, and $t$  rather than $t$ only:&#xA;&lt;img src=&quot;http://i.imgur.com/GwBxw.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can I predict the structure of the formula that relates $P$ with $T$, $H$, and $t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any approach that you advise me to follow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for this long question and any help from you is highly appreciated. I read many papers but I could not know how to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="4700" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T04:16:53.430" LastActivityDate="2010-12-13T18:41:02.080" Title="Find formula to changing cyclic behavior" Tags="&lt;statistics&gt;&lt;functions&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14141" PostTypeId="1" AcceptedAnswerId="14143" CreationDate="2010-12-13T04:30:04.207" Score="3" ViewCount="96" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_rotation_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler's rotation theorem&lt;/a&gt; (very informally stated) says that an rotation of a sphere in three dimensional space always has two fixed points. I believe this claim generalizes to (&lt;em&gt;n&lt;/em&gt;-1)-hyperspheres in &lt;em&gt;n&lt;/em&gt;-dimensional space for odd &lt;em&gt;n&lt;/em&gt;, but I am unable to patch the final gap in the proof. My strategy so far is as follows; any ideas on how to proceed would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that an arbitrary rotation of our hypersphere can be characterized by rotations about n-1 different axes. Since a rotation by theta in $\mathbb{R}^2$ is characterized by $\left[\begin{array}{cc}&#xD;&#xA;\cos\theta &amp;amp; -\sin\theta\\&#xD;&#xA;\sin\theta &amp;amp; \cos\theta\end{array}\right]$, it is not hard to see that a rotation about &lt;em&gt;n&lt;/em&gt;-1 different axes can be characterized by the matrix product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\left[\begin{array}{ccccccc}&#xD;&#xA;\cos\theta_{1} &amp;amp; -\sin\theta_{1} &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;\sin\theta_{1} &amp;amp; \cos\theta_{1} &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 1\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \vdots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; 1\end{array}\right]\left[\begin{array}{ccccccc}&#xD;&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; \cos\theta_{2} &amp;amp; -\sin\theta_{2} &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; \sin\theta_{2} &amp;amp; \cos\theta_{2} &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;\vdots &amp;amp; \vdots &amp;amp; \vdots &amp;amp; 1 &amp;amp;  &amp;amp;  &amp;amp; \vdots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xD;&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; 1\end{array}\right]&#xD;&#xA;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\cdots\left[\begin{array}{ccccccc}&#xD;&#xA;\cos\theta_{n-1} &amp;amp; 0 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; -\sin\theta_{n-1}\\&#xD;&#xA;0 &amp;amp; 1 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp;  &amp;amp; 0\\&#xD;&#xA;\\\vdots &amp;amp; \vdots &amp;amp;  &amp;amp; \ddots &amp;amp;  &amp;amp;  &amp;amp; \vdots\\&#xD;&#xA;\\0 &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 1 &amp;amp; 0\\&#xD;&#xA;\sin\theta_{n-1} &amp;amp; 0 &amp;amp;  &amp;amp; \cdots &amp;amp;  &amp;amp; 0 &amp;amp; \cos\theta_{n-1}\end{array}\right]&#xD;&#xA;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then notice that we have two fixed points if the matrix product just described has an eigenvalue of 1. The eigenvalues of a matrix are given by the roots of the characteristic polynomial; the characteristic polynomial here has real coefficents; therefore any complex roots of the characteristic polynomial must come in conjugate pairs; therefore if &lt;em&gt;n&lt;/em&gt; is odd, we must have a real eigenvalue. But since rotations preserve length, any real eigenvalues must have an absolute value of one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've shown that for odd &lt;em&gt;n&lt;/em&gt;, there must be an eigenvalue of 1 or -1, but &lt;strong&gt;to complete the proof, we need to show that there can't be a lone eigenvalue of -1&lt;/strong&gt;, and I haven't been able to figure out how to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I might be able to glean some intuition from the three-dimensional case, where a half-rotation about one axis gives us two eigenvalues of -1. Perhaps eigenvalues of -1 must also come in pairs? That seems plausible---dare I say even obvious?---on intuitive and physical grounds, but how might one prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This excursion was inspired by an exercise in &lt;a href=&quot;http://wps.prenhall.com/esm_bretscher_linear_3/&quot; rel=&quot;nofollow&quot;&gt;Bretscher &lt;em&gt;Linear Algebra&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastEditorUserId="2773" LastEditorDisplayName="" LastEditDate="2010-12-13T04:37:43.870" LastActivityDate="2010-12-14T05:22:12.550" Title="Ruling out a lone eigenvalue of -1 to complete this proof?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14142" PostTypeId="2" ParentId="14009" CreationDate="2010-12-13T04:34:41.023" Score="0" ViewCount="0" Body="&lt;p&gt;Beautiful, thank you both!&lt;/p&gt;&#xA;" OwnerUserId="4701" LastActivityDate="2010-12-13T04:34:41.023" CommentCount="1" />
  <row Id="14143" PostTypeId="2" ParentId="14141" CreationDate="2010-12-13T04:39:47.363" Score="5" ViewCount="0" Body="&lt;p&gt;You need to show that the determinant of your transformation is equal to $1$ (or at least is positive; but then it will necessarily equal $1$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this:&#xA;Recall that this determinant is the product of the eigenvalues.&#xA;Now the (possibly complex) eigenvalues are all of absolute value 1, and there are an odd number of them.  The non-real eigenvalues come in conjugate pairs, which have product 1, so there are also an odd number of real eigenvalues, which are $\pm 1$, and whose product equals $1$.  It follows that at least one of them equals $1$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let you work out the determinant fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added: Several people wrote down the same argument in comments, reflecting (I think) the fact that this is the standard approach to this quesion.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-13T04:46:21.567" LastActivityDate="2010-12-13T04:46:21.567" CommentCount="2" />
  <row Id="14144" PostTypeId="2" ParentId="14141" CreationDate="2010-12-13T05:01:36.477" Score="9" ViewCount="0" Body="&lt;p&gt;Here is an approach which gives a stronger result.  Let $g$ be a rotation of $\mathbb{R}^n$ and suppose that $g$ has a complex eigenvalue $\lambda$.  By the spectral theorem, there is a complex eigenvector $v$ associated to $\lambda$, hence a conjugate eigenvector $\bar{v}$ associated to $\bar{\lambda}$.  It follows that on the real subspace $V$ of $\mathbb{R}^n$ spanned by $v + \bar{v}$ and $\frac{v - \bar{v}}{i}$, $g$ acts as a 2-dimensional rotation, and we can break $\mathbb{R}^n$ up into a direct sum $V \oplus V^{\perp}$, and this respects the action of $g$ (so $g$ acts as a rotation on $V^{\perp} \simeq \mathbb{R}^{n-2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By induction, we obtain a structure theorem for rotations: any rotation is a direct sum of 2-dimensional rotations and 1-dimensional rotations (multiplication by $\pm 1$).  The determinant condition then tells us that the multiplications by $-1$ must occur in pairs, hence must pair up as 2-dimensional rotations by $\pi$.  So we can say more: any rotation has an invariant subspace $V$ of even codimension, and on $V^{\perp}$ it acts as a direct sum of nontrivial 2-dimensional rotations.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-14T05:22:12.550" LastActivityDate="2010-12-14T05:22:12.550" />
  <row Id="14145" PostTypeId="2" ParentId="14133" CreationDate="2010-12-13T05:10:44.343" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\: $ The first equation is $\rm\ log_K\ $ of $\rm\ B^{log_B\ X}\ =\ X$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the $\:$ second $\:$ equation is $\rm\  log_B\ $ of $\rm\ K^{log_K\ X}\ =\ X$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T05:10:44.343" />
  <row Id="14146" PostTypeId="1" AcceptedAnswerId="14192" CreationDate="2010-12-13T06:23:07.577" Score="3" ViewCount="122" Body="&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone could explain me the following. If I understand the &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-1-the-geometry-of-linear-equations/&quot; rel=&quot;nofollow&quot;&gt;lecture&lt;/a&gt; correctly, given a system of, say, following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x+2y-2z=1$ &lt;br&gt;&#xA;$2x+4y+z=3$ &lt;br&gt;&#xA;$4x+8y+4z=10$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that the they do not describe the whole 3D space because the vectors formed by the first and second columns ( $\begin{pmatrix}&#xD;&#xA; 1 \\ &#xD;&#xA; 2 \\ &#xD;&#xA; 4 &#xD;&#xA; \end{pmatrix}&#xD;&#xA;$ and $\begin{pmatrix}&#xD;&#xA; 2 \\ &#xD;&#xA; 4 \\ &#xD;&#xA; 8 &#xD;&#xA; \end{pmatrix}&#xD;&#xA;$ respectively) are not linearly independent (meaning we can't solve the given equations for &lt;em&gt;all possible&lt;/em&gt; RHS). From what I understand, it makes sense geometrically -- if we start at an arbitrary point in 3D, we can cover only a plane as we have only two directions we could move in. However, could someone please explain the meaning of the statement above in algebraic terms? What is it about the coefficients of a given variable that when they can be expressed via multiplication of some $c$ (where $c=2$ for the two aforementioned columns) by coefficients of some &lt;em&gt;other&lt;/em&gt; variable, not all RHS values generate solutions? Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-13T17:50:26.970" LastActivityDate="2010-12-13T17:58:53.777" Title="Linear independence of columns and covering the whole space (algebraic rationale)" Tags="&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="14147" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T06:34:44.887" Score="3" ViewCount="0" Body="&lt;p&gt;If you understand what it means in geometric terms, then you are most of the way there.  I'll try to elaborate a little on the algebra involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation can be rewritten as $xv_1+yv_2+zv_3=b$, where $v_1$, $v_2$, and $v_3$ are the column vectors.  The question is, based on the relationship among the columns, which vectors $b$ can you plug in to make the equation have a solution.  Since there is a linear relationship among the columns, the answer, as you said, is that not all choices of $b$ will work.  You have $v_2=2v_1$, so the equation can also be written as $(x+2y)v_1+zv_3=b$.  Thus, $b$ must lie in the $2$-dimensional subspace spanned by $v_1$ and $v_3$, your first and third columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if there is some nontrivial linear relationship among the columns of the form $a_1v_1+a_2v_2+a_3v_3=0$, where $a_1$, $a_2$, and $a_3$ are scalars and at least one is nonzero, then you can write one of your vectors as a linear combination of the other $2$, and hence any vector in the span (possible &quot;right-hand sides&quot; in your question) will have to be in the span of these $2$.  The same idea generalizes to all (finite) dimensions.  If the set of column vectors is not linearly independent, then the span of the columns has lower dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see explicitly what choices of $b$ will not work in your example, you can write out what the span of $v_1$ and $v_3$ looks like, and choose any vectors not in this span.  For a quick example in this case, you could choose the cross product $v_1\times v_3$, since it is orthogonal to both $v_1$ and $v_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at this is that if the columns are not linearly independent, then neither are the rows, and a linear relationship among the row vectors translates to a linear relationship among the entries of the right hand side.  That is, if $A$ has row vectors $r_1$, $r_2$, and $r_3$, then &#xA;$A\mathbf{x}=&#xD;&#xA;\begin{bmatrix}&#xD;&#xA;r_1\cdot\mathbf{x}\\&#xD;&#xA;r_2\cdot\mathbf{x}\\&#xD;&#xA;r_3\cdot\mathbf{x}&#xD;&#xA;\end{bmatrix}&#xD;&#xA;$, and if there are scalars $a_1$, $a_2$, and $a_3$, not all zero, such that $a_1r_1+a_2r_2+a_3r_3=0$, then it follows from the linearity of the dot product in each variable that $a_1(r_1\cdot\mathbf{x})+a_2(r_2\cdot\mathbf{x})+a_3(r_3\cdot\mathbf{x})=0$.  That is, the right-hand side must lie in the plane described by the equation $a_1x+a_2y+a_3z=0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T06:55:05.547" LastActivityDate="2010-12-13T06:55:05.547" />
  <row Id="14148" PostTypeId="1" AcceptedAnswerId="14150" CreationDate="2010-12-13T06:49:08.700" Score="0" ViewCount="41" Body="&lt;p&gt;I am looking at the first two examples in Paul Garretts&lt;a href=&quot;http://www.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf&quot; rel=&quot;nofollow&quot;&gt; notes&lt;/a&gt; on p-adic number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first example is computing the Newton polygon of $x^5+2x^2+5$ over $\mathbb{Q}_2$. I think this is the lower convex hull of $(0,1),(1,0),(2,0.5),(3,0),(4,0),(5,1)$. So I thought the Newton polygon should just be a horizontal line segment going from $(0,1)$ to $(5,1)$. Of course, I am completely wrong, but I am not sure what is the source of error. Are the points whose convex hull I am supposed to consider correctly identified?&lt;/p&gt;&#xA;" OwnerUserId="3947" LastActivityDate="2010-12-13T07:09:14.213" Title="Computing Newton polygons" Tags="&lt;p-adic-number-theory&gt;" AnswerCount="2" />
  <row Id="14149" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T06:53:36.493" Score="1" ViewCount="0" Body="&lt;p&gt;With $\rm\ c_i = i$'th column, the system is $\rm\ c_1\ x + c_2\ y + c_3\ z\ =\ d\:.\:$ This has a solution iff $\rm\:d\:$ lies in the space spanned by the columns. Since two of the three columns are linearly dependent, the columns can span at most a two-dimensional space. Therefore the  system is not solvable for all $\rm\:d\:,\:$ i.e. the associated map is not onto.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T06:53:36.493" />
  <row Id="14150" PostTypeId="2" ParentId="14148" CreationDate="2010-12-13T07:03:19.400" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Below is the Newton polygon graphed by Filaseta's &lt;a href=&quot;http://www.math.sc.edu/~filaseta/newton/newton.html&quot; rel=&quot;nofollow&quot;&gt;Java applet&lt;/a&gt;. His web pages have much of interest on Newton polygons.&#xA;&lt;img src=&quot;http://i.imgur.com/N4HPL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T07:03:19.400" CommentCount="1" />
  <row Id="14151" PostTypeId="2" ParentId="14148" CreationDate="2010-12-13T07:09:14.213" Score="2" ViewCount="0" Body="&lt;p&gt;You have miscalculated the $2$-adic orders of the coefficients, and probably confused the concept of $p$-adic order with something else. In particular, the $p$-adic order of a $p$-adic number is an integer, but your third point has $y$-coordinate 0.5. As a reminder: the $p$-adic order of an integer $x$ in $\mathbb{Z}_p$ is the highest integer $v$ such that $x\in {\mathfrak m}^v$, where ${\mathfrak{m}}=p\mathbb{Z}_p$ is the maximal ideal of $\mathbb{Z}_p$. E.g. the $2$-adic order of 5 is 0, since 5 is a 2-adic unit, so the first point of the Newton polygon has coords (0,0). The $p$-adic order of 0 is $\infty$ by convention. The $2$-adic order of 2 is 1, since 2 is a uniformiser. And so on.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-13T07:09:14.213" />
  <row Id="14152" PostTypeId="2" ParentId="14078" CreationDate="2010-12-13T07:16:13.323" Score="2" ViewCount="0" Body="&lt;p&gt;I would like to suggest three quick formulas for tackling similar kind of problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there is successive &lt;strong&gt;increase&lt;/strong&gt; of $x\%$ and $y\%$, then the net change will be $$x + y + \frac{x \cdot y}{100}\%$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;If there is successive &lt;strong&gt;decrease&lt;/strong&gt; of  $x\%$ and  $y\%$, then the net change will be $$x + y - \frac{x \cdot y}{100}\%$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;If there is  &lt;strong&gt;$x\%$ increase and then  $y\%$ decrease&lt;/strong&gt;, then the net change is $$ x - y -  \frac{x \cdot y}{100}\%$$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Exercise:  How does the last result change if there is &lt;strong&gt;$x\%$ decrease followed by $y\%$ increase&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results can  be  easily verified, but it's always better to memorize them before  any quantitative aptitude test.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-13T09:17:58.227" LastActivityDate="2010-12-13T09:17:58.227" CommentCount="1" />
  <row Id="14153" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T07:18:20.837" Score="2" ViewCount="0" Body="&lt;p&gt;It seems that whatever algebraic way you try to look at it is, upon some thought, really just a paraphrase of the geometric one. One way of answering your question is that to generate all RHS values, we must have sufficient variation of the variables. However since one of the variables in this case has coefficients a multiple of another, as it varies to try to change the RHS it really is just doing the work of the other variable and so there is just not enough variation to go around to give what you want on the RHS.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-13T07:18:20.837" />
  <row Id="14154" PostTypeId="1" CreationDate="2010-12-13T07:53:29.960" Score="2" ViewCount="105" Body="&lt;p&gt;I have a list of several entities, all of each have a numerical relationship to each other that defines an abstract distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mathematical way to plot all of these on a 2D space, turning the abstract distance into a 2D euclidian distance, and preserving all of their distances?&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-12-13T23:10:38.717" Title="Given a metric function between a set of abstract points, what is the best way to plot them on a 2D space?" Tags="&lt;metric-spaces&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14155" PostTypeId="1" AcceptedAnswerId="14203" CreationDate="2010-12-13T07:54:27.030" Score="3" ViewCount="166" Body="&lt;p&gt;Can it be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an arbitrary quadratic field $Q[\sqrt{d}]$, it's easy to show the equations are simply $ f_x = -\sqrt{d} f_y $, where $ f : Q[\sqrt{d}] \to Q[\sqrt{d}]$. I'm working on the case of $Q[\theta]$, when $\theta$ is a root of $\theta^3 - a\theta - b$, but I'm not sure if it's even possible. Has there been any mathematical research done on this topic? What do you think about it?&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-12-13T22:58:55.377" Title="Generalizing Cauchy-Riemann Equations to Arbitrary Algebraic Fields" Tags="&lt;algebraic-number-theory&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="14156" PostTypeId="2" ParentId="14154" CreationDate="2010-12-13T08:01:10.377" Score="0" ViewCount="0" Body="&lt;p&gt;If you want to keep the distances exactly, then no, not really. If you fix 2 points in the plane, and want a third point with prescribed distances from the first two, you only have two options for that third point (symmetric across the line connecting the first two points). This severely constrains even three points; more will almost surely be impossible to graph exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to alter the distances in a way to preserve an order on them (in which I mean: if the distance between a pair of points is greater than that of a second pair of points, this will remain the case after altercation), or see them in more than three dimensions but from various angles or perspectives, then there might be mathematics to help you out there.&lt;/p&gt;&#xA;" OwnerUserId="3122" LastActivityDate="2010-12-13T08:01:10.377" />
  <row Id="14157" PostTypeId="2" ParentId="14154" CreationDate="2010-12-13T08:19:03.233" Score="3" ViewCount="0" Body="&lt;p&gt;The question may be paraphrased as whether there is an isometry of the metric space into the plane. The answer is in general, no. Consider the counter-example of a metric space with 4 points in which the distance between any 2 distinct points is 1. After placing 2 points in the plane, the third must lie in the intersection of the circles of radius 1 centred at these 2 points. The fourth must lie in the intersection of the circles of radius 1 centred at the other 3 points, and there is no way to place it.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-13T08:19:03.233" />
  <row Id="14158" PostTypeId="1" CreationDate="2010-12-13T08:51:06.133" Score="0" ViewCount="41" Body="&lt;p&gt;As we define convolution for elements of a quantum group, Can we heve a convolution between the elements of our locally compact quantum group and it's dual or even it's predual(when it is a Von Neuman algebra)?&lt;/p&gt;&#xA;" OwnerUserId="4703" LastActivityDate="2010-12-13T08:51:06.133" Title="convolution of quantum groups" Tags="&lt;functional-analysis&gt;" CommentCount="2" />
  <row Id="14159" PostTypeId="2" ParentId="14069" CreationDate="2010-12-13T09:02:42.463" Score="1" ViewCount="0" Body="&lt;p&gt;If you are geometrically inclined, like I am, you can visualize the path of the Skoda as a line on a graph where the X axis is the time of day (in military time) and the Y axis is distance traveled in miles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some points for the Skoda -- &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;(7,0)&lt;/code&gt;, because at 7 am, the Skoda has traveled 0 miles.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;(8,20)&lt;/code&gt;, because at 8 am, the Skoda has traveled 20 miles.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;(12,100)&lt;/code&gt;, because at 12 pm, the Skoda has traveled 100 miles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you plot these points on your graph, you'll see that it makes a straight line with slope 20 and X-intercept of 7.  From that you could figure out that the Y intercept is -140.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the Porsche would have a slope of 40 and an X-intercept of 16 (we're using military time).  From there you could find out that the Y intercept is -640.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Skoda's traveled distance = $20*t - 140$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;Porsche traveled distance = $40*t - 640$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can graph these on a chart -- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img718.imageshack.us/img718/6264/msp136919dbih50965e08ac.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to know when their traveled distance is equal.  So --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Skoda \, distance = Porsche \, distance$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20*t - 140 = 40*t - 640$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20*t = 40*t - 500$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20*t = 500$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = 25$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at 25 o' clock, or 1 am the next day, their traveled distances will be the same.&lt;/p&gt;&#xA;" OwnerUserId="51" LastActivityDate="2010-12-13T09:02:42.463" />
  <row Id="14160" PostTypeId="1" AcceptedAnswerId="14161" CreationDate="2010-12-13T09:11:09.973" Score="0" ViewCount="53" Body="&lt;p&gt;I need a function (a curve -- preferably a simple one) that, given $n$ points of a 2D space ($R^2$) passes (interpolates) through all points in a smooth/continuous way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found out that what I need is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spline_curve&quot; rel=&quot;nofollow&quot;&gt;spline&lt;/a&gt;, however cannot find one that behaves how I need. $n$-degree Bezier curves and B-splines don't pass through the $n$ points, just move from the first to the last one using the others as control points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Bezier spline of $n-1$ Bezier curves (defined between each pair of adjacent points) should do the trick, but to have a smooth result I would need some control points which I don't have and don't know how to compute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2010-12-13T09:20:43.710" Title="Continuous curve interpolating a list of points" Tags="&lt;functions&gt;&lt;plane-curves&gt;&lt;bezier-curve&gt;" AnswerCount="1" />
  <row Id="14161" PostTypeId="2" ParentId="14160" CreationDate="2010-12-13T09:20:43.710" Score="0" ViewCount="0" Body="&lt;p&gt;A parametric cubic spline is what you want; I've talked about them at some length in &lt;a href=&quot;http://math.stackexchange.com/questions/5166/5192#5192&quot;&gt;this answer&lt;/a&gt; (and in probably a few other places on this site, since it seems to be a common question), so I won't repeat myself now. Suffice it to say that yes, Bézier isn't what you need here.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-13T09:20:43.710" CommentCount="2" />
  <row Id="14162" PostTypeId="1" CreationDate="2010-12-13T10:14:25.650" Score="0" ViewCount="80" Body="&lt;p&gt;This image show a histogram (200 bins) of accumulated distances from a radar distance meter (very noisy).&#xA;The peak around 7 meters is an object. At thought this looked kind of like a normal distribution, at least if you ignore values &amp;lt;4m (which for this application is reasonable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am trying to do is to filter out true distances based on the probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/tYgBf.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2367" LastActivityDate="2010-12-13T12:24:03.593" Title="Is this a probability distribution?" Tags="&lt;probability&gt;&lt;signal-processing&gt;&lt;normal-distribution&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="14163" PostTypeId="2" ParentId="14137" CreationDate="2010-12-13T10:27:57.897" Score="2" ViewCount="0" Body="&lt;p&gt;The torus can be made from a single 0-cell, two 1-cells and one 2-cell. The 1-cells are attached along their boundary to the 0-cell, making two circles identified at a point. Then the 2-cell is attached to this 1-skeleton, as illustrated in the comments above. &lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-12-13T10:27:57.897" />
  <row Id="14164" PostTypeId="2" ParentId="14115" CreationDate="2010-12-13T10:30:08.723" Score="2" ViewCount="0" Body="&lt;p&gt;The explanation for the existence of this series is given &lt;a href=&quot;http://mathworld.wolfram.com/PiFormulas.html&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt; Search for the phrase&#xA;&quot;The general form of the series is&quot; to locate it. The series cited in the question appears immediately before the explanation.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-13T10:30:08.723" CommentCount="1" />
  <row Id="14165" PostTypeId="2" ParentId="14041" CreationDate="2010-12-13T10:30:27.317" Score="1" ViewCount="0" Body="&lt;p&gt;Firstly, consider the case when we are not obliged to have the equation 0=0+0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question becomes: given $n \geq 1$ does there exist two permutations $g,h$ of $A:=\{-n,-n+1,\ldots,n\}$ such that $n=g(i)+h(i)$ where addition is addition over the integers (and therefore is not closed on $A$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can interpret an instance of this question as a set of $n$ non-attacking semiqueens on an $(2n+1) \times (2n+1)$ chessboard (with rows and columns indexed by $A$) such that no semiqueen is in cell $(i,j)$ whenever $i+j \not\in A$.  [semiqueens move up-and-down, left-and-right and along one diagonal]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the instance you cite above is illustrated by this figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XEf0N.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as with orthomorphisms of $\mathbb{Z}_n$ we can find linear constructions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/u9G5X.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can take the equations&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0=n+(-n)$, $1=(n-1)+(-n+2)$, up to $n=0+n$, along with&lt;/li&gt;&#xA;&lt;li&gt;$-1=-n+(n-1)$, $-2=(-n+1)+(n-3)$, up to $-n=-1+(-n+1)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, g(n) takes the values $\{n,n-1,\ldots,0\}$ in the first step and $\{-n,-n+1,\ldots,-1\}$ in the second and h(n) takes the values $\{-n,-n+2,\ldots,n\}$ in the first step and $\{n-1,n-3,\ldots,-n+1\}$ in the second step.  Hence both g and h are bijections of $A$, and we can clearly see that $i \mapsto g(i)-h(i)$ is also a bijection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, I guess it is easy in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Warning&lt;/em&gt;: Although the motivation for this question was to prove the existence of an orthomorphism of $\mathbb{Z}$, this is insufficient to prove it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now lets consider the case when the equation 0=0+0 is required.  We can generate some small random examples, for example, in GAP.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RandomPermutationList:=function(n)&#xA;  local A,i,j,temp;&#xA;  A:=List([1..n],k-&amp;gt;k);&#xA;  i:=n;&#xA;  while(i&amp;gt;=2) do&#xA;    j:=Random([1..i]);&#xA;    if(j&amp;lt;i) then&#xA;      temp:=A[i];&#xA;      A[i]:=A[j];&#xA;      A[j]:=temp;&#xA;    fi;&#xA;    i:=i-1;&#xA;  od;&#xA;  return A;&#xA;end;;&#xA;&#xA;TripleJQuestion2:=function(n)&#xA;  local A,B,p,q;&#xA;  A:=Concatenation([-n..-1],[1..n]);&#xA;  B:=List([1..2*n+1],i-&amp;gt;-n-1+i);&#xA;  while(true) do&#xA;    q:=RandomPermutationList(2*n);&#xA;    p:=Concatenation(List([1..n],i-&amp;gt;A[q[i]]),[0],List([1..n],i-&amp;gt;A[q[n+i]]));&#xA;    if(Set(List([1..2*n+1],i-&amp;gt;p[i]-B[i]))=B and p[n+1]=0) then return p; fi;&#xA;  od;&#xA;  return fail;&#xA;end;;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This was successful for $4 \leq n \leq 8$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mwRrJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/1YYqR.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FjR7k.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/z3EZB.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/qxNAa.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, combined with the fact that there are many orthomorphisms of $\mathbb{Z}_n$, it seems very likely that an instance of the desired construction exists for all $n \geq 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, continuing further (i.e. finding a constructions for all $n \geq 3$) could require considerable effort.  If you're really keen on resolving this problem, you could generate lots of random examples, and hope you could generalise enough of them to cover all $n \geq 3$.  But why would you want to do this?  It's not likely to give a surprising result, and it seems to be a fairly contrived problem (yes, the problem is harder if we assume the equation 0=0+0 is required, but why would anyone be interested in this problem?).  And also, unless you somehow build these constructions recursively, you're probably not going to solve the orthomorphisms of $\mathbb{Z}$ existence question.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-13T11:39:31.333" LastActivityDate="2010-12-13T11:39:31.333" CommentCount="1" />
  <row Id="14167" PostTypeId="1" AcceptedAnswerId="14169" CreationDate="2010-12-13T10:55:24.610" Score="2" ViewCount="201" Body="&lt;p&gt;I randomly pick a natural number &lt;em&gt;n&lt;/em&gt;. Assuming that I would have picked each number with the same probability, what was the probability for me to pick &lt;em&gt;n&lt;/em&gt; before I did it?&lt;/p&gt;&#xA;" OwnerUserId="4707" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-12-18T21:50:47.030" LastActivityDate="2010-12-18T21:50:47.030" Title="Probability of picking a random natural number" Tags="&lt;probability&gt;" AnswerCount="3" />
  <row Id="14168" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T11:00:10.657" Score="5" ViewCount="0" Body="&lt;p&gt;Here is a really nice one due to &lt;a href=&quot;http://www.lacim.uqam.ca/~plouffe/inspired2.pdf&quot; rel=&quot;nofollow&quot;&gt;Simon Plouffe.&lt;/a&gt; There are many similar examples in his linked paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi = 72\sum_{n=1}^\infty \frac{1}{n(e^{n\pi} - 1)} - &#xD;&#xA;96\sum_{n=1}^\infty \frac{1}{n(e^{2n\pi} - 1)} +&#xD;&#xA;24\sum_{n=1}^\infty \frac{1}{n(e^{4n\pi} - 1)} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I like about it is that I can see at a glance that the series converge rapidly without having to make some mental estimate of the size of factorials.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-13T11:00:10.657" CommentCount="2" />
  <row Id="14169" PostTypeId="2" ParentId="14167" CreationDate="2010-12-13T11:07:47.227" Score="6" ViewCount="0" Body="&lt;p&gt;There is no uniform distribution on the natural numbers. So the question itself is self-contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that each number is chosen with the same probability. Say probability c.&#xA;Let's denote the probability to choose n by p(n). So you say p(n)=c for any natural n.&#xA;For any distribution, the sum of probabilities always equals 1: p(1)+p(2)+p(3)+... = 1.&#xA;So c+c+c+... = 1.&#xA;But if c &gt; 0 you get c+c+c+... = $\infty$.&#xA;And if c=0 you get c+c+c+... = 0.&#xA;There's no value of c for which this sum equals 1.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="3533" LastEditorDisplayName="" LastEditDate="2010-12-13T11:17:45.897" LastActivityDate="2010-12-13T11:17:45.897" CommentCount="4" />
  <row Id="14171" PostTypeId="2" ParentId="14167" CreationDate="2010-12-13T11:24:12.210" Score="4" ViewCount="0" Body="&lt;p&gt;If you want to choose a natural number from the set $\lbrace 1,2,3,\ldots \rbrace$ of all natural numbers, then user3533 is right. But if you pick a natural number from a finite set, say $A = \lbrace 1,2,\ldots,N \rbrace$, each number with the same probability, then this probability must be the reciprocal of the number of elements in that set. So, if $A = \lbrace 1,2,\ldots,N \rbrace$, then $p=1/N$. This follows immediately from &#xA;$\sum\nolimits_{i = 1}^N {p_i }  = 1$ with $p_1 = p_2 = \cdots = p_N = p$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-13T11:44:13.823" LastActivityDate="2010-12-13T11:44:13.823" />
  <row Id="14172" PostTypeId="1" CreationDate="2010-12-13T11:44:32.107" Score="0" ViewCount="95" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/12328/rsa-fast-factorization-of-n-if-d-and-e-are-know&quot;&gt;RSA: Fast factorization of N if d and e are know&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I factor n that is a product of p and q which are large primes? The setting is an RSA encryption/decryption where you know the multiple of (p-1)(q-1), because the n has been shared through a group of users.&#xA;Also if someone can provide more info about this kind of attack and it's usage would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;best regards,&#xA;Rok&lt;/p&gt;&#xA;" OwnerUserId="4708" LastActivityDate="2010-12-13T11:44:32.107" ClosedDate="2010-12-17T02:49:50.683" Title="RSA: Factorization of n=pq when we know multiple of (p-1)(q-1)" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" CommentCount="4" />
  <row Id="14173" PostTypeId="1" AcceptedAnswerId="14184" CreationDate="2010-12-13T11:56:51.370" Score="2" ViewCount="139" Body="&lt;p&gt;In high school, when we talked about &quot;combinatorics,&quot; we solely meant &quot;mathematics of choice.&quot; For instance:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are 10 people who want to sit around a table. In how many ways is this possible?&lt;/li&gt;&#xA;&lt;li&gt;We have 50 balls and 20 boxes. In how many ways can we distribute balls into the boxes?&lt;/li&gt;&#xA;&lt;li&gt;There are 10 apples and 5 oranges. In how many ways can we select 7 fruits?&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I did have this certain wrong view of combinatorics unless I read about combinatorial objects (lists, sets, graphs, etc.) and combinatorial proofs. I tried reading several sources (Wikipedia, books, papers), but still don't have a clear understanding of when something is called &lt;strong&gt;combinatorial&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could you please elaborate it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2010-12-13T19:11:13.803" Title="What's Combinatorial Proof/Object/etc.?" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="14174" PostTypeId="2" ParentId="14162" CreationDate="2010-12-13T12:24:03.593" Score="0" ViewCount="0" Body="&lt;p&gt;The density function of distance-from-mean of a bivariate normal distribution takes the form $Ax e^{-Bx^2}$, which look pretty much like your noisy graph.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-13T12:24:03.593" CommentCount="3" />
  <row Id="14175" PostTypeId="1" AcceptedAnswerId="14176" CreationDate="2010-12-13T13:11:47.453" Score="1" ViewCount="32" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reading a text on maths applied to naturally occurring networks, eg. social networks. The section I am on is &quot;Walks&quot; networks. The text says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Walks:  which allow both nodes and&#xA;  edges to be used more than once. For&#xA;  example, in Figure 1,&#xA;  Mary-Bob-Sue-Ramona-Bob-Mary-Joe is a&#xA;  walk, but not a trail or a path. A&#xA;  particular dollar bill flows through&#xA;  the person-person network along walks.&#xA;  Mary might give the bill to Bob in one&#xA;  transaction, and Bob might return it&#xA;  to Mary in another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A trail is when a node i may be visited multiple times, but a particular edge connecting it with a node j, can be visited no more than 1 time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An identity from linear algebra is used which I cannot find is used &#xA;Suppose an undirected network has an adjacency matrix A, where $a_{i,j}=1$ if nodes i and j are connected, and $a_{i,j}=0$ otherwise. Then the basic identity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A^{n})_{ij} = \sum^{n}_{k_{1} = 1}\sum^{n}_{k_{2}=1}\ldots\sum^{n}_{k_{n-1}=1}a_{i,k_{1}}a_{k_{1},k_{2}}\ldots a_{k_{n-1},j}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shows that $(A^{n})_{ij}$ counts the total number of walks of length n from node i to node j.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The indice k is not explicitly defined in the text or given more description. What makes it hard to understand is that the index k is not being summed over the possible nodes the matrix since n is the walk length as it is written, unless there is a mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see if I can be bold and say that the equation should have been better written as (making N be the node set or square matrix dimension):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A^{n})_{ij} = \sum^{N}_{k_{1}=1}(\sum^{N}_{k_{2}=1}\ldots(\sum^{N}_{k_{n-1}=1}a_{i,k_{1}}a_{k_{1},k_{2}}\ldots a_{k_{n-1},j}))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I read the equation as a brute force summation over the total set of combinations of length n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this interpretation correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and is $(A^{n})_{ij}$ the result of the entry for ij in doing the matrix exponentiation of the adjacency matrix n times because in simple examples I tried with small graphs I did not get the correct answer (tried simple fully connected graph with n=2,3,4)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="1587" LastActivityDate="2010-12-13T13:29:51.353" Title="calculating walks in an undirected graph using linear algebra identity" Tags="&lt;linear-algebra&gt;&lt;sequences-and-series&gt;&lt;graph-theory&gt;&lt;searching&gt;&lt;network-flow&gt;" AnswerCount="1" />
  <row Id="14176" PostTypeId="2" ParentId="14175" CreationDate="2010-12-13T13:29:51.353" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, you're right (the $k$'s should run from 1 to $N$, the size of the matrix), and yes, $(A^n)_{ij}$ is the $(i,j)$ entry in $A^n=AA\dots A$ ($n$ factors).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-13T13:29:51.353" />
  <row Id="14177" PostTypeId="1" AcceptedAnswerId="14180" CreationDate="2010-12-13T13:31:17.203" Score="0" ViewCount="31" Body="&lt;p&gt;I know the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;my own position&lt;/li&gt;&#xA;&lt;li&gt;my own facing (the angle im turned)&lt;/li&gt;&#xA;&lt;li&gt;my targets position&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What i would like some help with is how i calculate the shortes way to turn and the angle to turn. If i my position was (4,4) and i faced 90 degrees and my target was to face (1,1) where my target is, i know by drawing it on some paper that the shortest way to rotate is counter-clockwise 135 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do i calulate this? Ive tried using tangent, but using sine seems just as easy. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4710" LastActivityDate="2010-12-13T19:08:28.223" Title="Help with calculating the angle to turn towards a target in a coordinatesystem" Tags="&lt;trigonometry&gt;" AnswerCount="1" />
  <row Id="14178" PostTypeId="2" ParentId="14167" CreationDate="2010-12-13T13:40:11.490" Score="0" ViewCount="0" Body="&lt;p&gt;In case you  pick numbers every-time and thus when the number of trials are indefinitely large we use &lt;strong&gt;statistical or empirical&lt;/strong&gt; definition of probability&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If trials are to be repeated a great number of times under essentially the same condition then the limit of the ratio of the number of times the an event happens to the total number of trials number of trials, as the number of trials increases indefinitely is called the probability of the happening of the even. In is assumed that the limit exists and finite uniquely. Symbolically&lt;/strong&gt;&#xA;$$\lim_{n-&amp;gt; \infty} \frac{m}{n} \text{ provided it is finite and unique}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is worth of mentioning that this definition might appear to be different from the classical definition but both of them can be reconciled to the same sense.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-14T07:48:40.170" LastActivityDate="2010-12-14T07:48:40.170" CommentCount="2" />
  <row Id="14179" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T13:55:58.900" Score="6" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula&quot; rel=&quot;nofollow&quot;&gt;BBP formula&lt;/a&gt; is another nice one:&#xA;$$&#xD;&#xA;  \pi = \sum_{k=0}^\infty \left[&#xD;&#xA;  \frac{1}{16^k} \! \left(&#xD;&#xA;  \frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}&#xD;&#xA;  \right) \right]&#xD;&#xA;$$&#xD;&#xA;It can be used to compute the $n$th hexadecimal digit of $\pi$ without computing the preceding $n{-}1$ digits.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-13T13:55:58.900" />
  <row Id="14180" PostTypeId="2" ParentId="14177" CreationDate="2010-12-13T14:03:50.187" Score="2" ViewCount="0" Body="&lt;p&gt;Most computer systems have a function Atan2(x,y) which returns the polar angle of a point in the range $(-\pi,\pi]$ (check the endpoints-I think you get $+\pi$ for (-1,0).  If you subtract your position from the target position and take Atan2 of the difference it will be the angle from you to target.  Now subtract your heading and you have the angle to turn.  If the result is outside  $(-\pi,\pi]$, add or subtract $2\pi$ to get into the range.  The nice thing of Atan2, as opposed to the regular atan, is it sorts out for you the signs and the branch of tangent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use arcsin as well, but that requires measuring the distance to the target, which has an extra square root in it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-13T19:08:28.223" LastActivityDate="2010-12-13T19:08:28.223" CommentCount="2" />
  <row Id="14181" PostTypeId="2" ParentId="14155" CreationDate="2010-12-13T14:05:12.757" Score="2" ViewCount="0" Body="&lt;p&gt;I don't know if this will help, but I thought about something like this when I was an undergrad. I was thinking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_Jugendtraum&quot; rel=&quot;nofollow&quot;&gt;Jugendtraum&lt;/a&gt;: the fact that abelian extensions of imaginary quadratic fields can be described by values of analytic functions on $\mathbb{C}$. My thought was the following: Let $K=\mathbb{Q}(\sqrt{-D})$ be a quadratic imaginary field. Then $\mathbb{C} \cong K \otimes \mathbb{R}$ and we can write the Cauchy-Riemmann equations as $(D \partial_x^2 + \partial_y^2) f=0$, which seems to be built from the norm form $K \to \mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore (I thought), if we want to generalize the Jugendtraum to a real quadratic field $\mathbb{Q}(\sqrt{D})$, we should consider functions on $K \otimes \mathbb{R}$ which obey $(D \partial_x^2 - \partial_y^2) f=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this didn't get anywhere. But I did show that a function $f: \mathbb{R}^2 \to \mathbb{R}$ obeying $(D \partial_x^2 - \partial_y^2) f=0$ is of the form $g(x+\sqrt{D}y) + h(x-\sqrt{D}y)$. So, if that's the road you're gong down, I can tell you where it ends.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-13T14:05:12.757" CommunityOwnedDate="2010-12-13T14:05:12.757" CommentCount="5" />
  <row Id="14182" PostTypeId="2" ParentId="13937" CreationDate="2010-12-13T14:15:22.797" Score="1" ViewCount="0" Body="&lt;p&gt;In the two dimensional case there is a simple proof. consider the set $U = \{ ax-by = 0 : x,y\in\mathbb{N}\}$. Let $A = (a,-b)$ and $v \in \mathbb{N}^2$, then for any $u\in U$, $A\cdot (v + u) = A\cdot v$. Furthermore, we have that, since $A\in\mathbb{Z}^2$, the inner product of $A$ against any vector in $\mathbb{N}^2$ is integral. So we can foliate $\mathbb{N}^2$ by lines $U_k$ such that $A\cdot u_k = k$ for $u_k \in U_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{U_k} \frac{1}{|A\cdot u| u_1 u_2} = \frac{1}{|k|} \sum_{U} \frac{1}{(x_k + u_1)(y_k + u_2)} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where to get to the right hand side we used the fact that on each $U_k$, as long as $a,b &amp;gt; 0$, we have a linear ordering of the elements by distance to the origin, and we can pick $x_k, y_k$ to be the infimum. Now, as long as $a,b \neq 0$, we have that $u_1,u_2$ are both increasing, so the inner sum is summable. Furthermore, a little bit of geometry shows you that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} &amp;lt; \frac{2}{\sqrt{x_ky_k}} \sum_U \frac{1}{u_1u_2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit for clarification: Using the fact that $(u_1,u_2)$ are integer multiples of a unique vector $(\hat{u}_1,\hat{u}_2)$, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} \leq \frac{1}{x_ky_k}\sum_j \frac{1}{1 + j^2\frac{\hat{u}_1\hat{u}_2}{x_ky_k}} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[here we also need to use that $\hat{u}_1\hat{u}_2 \neq 0$] then you use the fact that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_j \frac{1}{1 + \alpha j^2} \leq 1 + (\alpha)^{-1/2} \sum_k \frac{1}{1 + k^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I leave as an exercise [if you run out of ideas, you can always prove this using integral comparison].)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing to note is that $\frac{1}{x_ky_k} = O(|k|^{-1})$ using that the points $(1,k)$ and $(k,1)$ are the infimum points for the slices $U_{-bk}$ and $U_{ak}$ respectively. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{k\neq 0} \sum_{U_k} \frac{1}{|A\cdot u| u_1u_2} \leq \sum_{k\neq 0} \frac{C}{k^{3/2}} \sum_U \frac{1}{u_1 u_2} &amp;lt; \infty$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With some addition condition on what $A$ you allow, I think the claim may also be true in higher dimensions. I'll add a proof to that effect if I can think of it. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-14T16:57:55.243" LastActivityDate="2010-12-14T16:57:55.243" CommentCount="3" />
  <row Id="14183" PostTypeId="1" AcceptedAnswerId="14185" CreationDate="2010-12-13T14:33:18.303" Score="5" ViewCount="260" Body="&lt;p&gt;I'm a lazy type of person. I love experimenting with things before learning the theory. I like to practice, and learn from my mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3-4 months I have a basic University exam on Calculus, and I'd like to learn it in a fun way. I have found these videos to be quite entertaining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.calculus-help.com/tutorials&quot; rel=&quot;nofollow&quot;&gt;http://www.calculus-help.com/tutorials&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However they are not exhaustive enough for my needs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any references you would like to share? I'm not sure if it's a valid type of question for this group, so I'm sorry in advanced if it's off-topic.&lt;/p&gt;&#xA;" OwnerUserId="4713" LastActivityDate="2011-01-11T04:24:58.707" Title="Can I learn Calculus on the web, for free, using simple tutorials?" Tags="&lt;calculus&gt;&lt;learning&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="14184" PostTypeId="2" ParentId="14173" CreationDate="2010-12-13T14:44:41.630" Score="5" ViewCount="0" Body="&lt;p&gt;There are several different branches of combinatorics but in general they deal with discrete structures. Enumerative combinatorics, as the name suggests, deals with counting, so the combinatorics you learn in school mostly falls into this category, asking you for the number of permutations or combinations in a particular situation. Extremal combinatorics, for another example, asks for the largest or smallest structure satisfying certain properties. These terms are deliberately vague to allow for generality. A combinatorial proof is simply a proof using a combinatorial argument. For example, one can prove the binomial theorem using mathematical induction or using a combinatorial argument, in which case what is to be justified is the coefficient of the various terms which is to be obtained by counting in some way.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-13T14:44:41.630" CommentCount="1" />
  <row Id="14185" PostTypeId="2" ParentId="14183" CreationDate="2010-12-13T15:05:05.647" Score="16" ViewCount="0" Body="&lt;p&gt;The MIT courseware is one of the first places to look for such materials. Here is a video series on a first course in Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a video series on multivariable calculus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a video series on differential equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2006/video-lectures/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2006/video-lectures/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I found a few more such video series. I haven't personally looked at these though.&#xA;&lt;a href=&quot;http://press.princeton.edu/video/banner/&quot; rel=&quot;nofollow&quot;&gt;http://press.princeton.edu/video/banner/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.online.math.uh.edu/HoustonACT/videocalculus/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.online.math.uh.edu/HoustonACT/videocalculus/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tutor-homework.com/Math_Help/Calculus.html&quot; rel=&quot;nofollow&quot;&gt;http://www.tutor-homework.com/Math_Help/Calculus.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-13T19:17:26.223" LastActivityDate="2010-12-13T19:17:26.223" CommentCount="3" />
  <row Id="14186" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T15:16:48.213" Score="2" ViewCount="0" Body="&lt;p&gt;The convergence can be arbitrary fast unless you don't specify what kind of series you are looking. Let $k$ be a positive integer, $a_n=\pi/k$ if $n\leq k$ and zero elsewhere. Then $\sum_{n=1}^\infty a_n$ converges to $\pi$ after $k$ summands.&lt;/p&gt;&#xA;" OwnerUserId="2856" LastActivityDate="2010-12-13T15:16:48.213" CommentCount="13" />
  <row Id="14187" PostTypeId="1" AcceptedAnswerId="14188" CreationDate="2010-12-13T15:35:43.477" Score="1" ViewCount="55" Body="&lt;p&gt;Can anybody give me a hint how to approach the following problem, please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually am having a hard time stating the problem. I think an example would help you understand what the problem is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have 5 numbers each can be max 3 digits. I sum them together. What's the probability of the sum having the last 3 digits 500?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of 5 numbers having the last 3 digits 500 are (100, 100, 100, 100, 100) and (0, 0, 0, 0, 1500).&lt;/p&gt;&#xA;" OwnerUserId="4102" LastActivityDate="2010-12-13T15:45:37.813" Title="Probability of the sum of n numbers giving the same last d digits" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="14188" PostTypeId="2" ParentId="14187" CreationDate="2010-12-13T15:45:37.813" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ For $\rm\ x + y + z\equiv 500\ \ (mod\ 1000) $ you have $1000$ choices for $\rm\:x\:$, $1000$ choices for $\rm\:y\:$, and then $\rm\:z\:$ is uniquely determined by the linear equation.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T15:45:37.813" CommentCount="9" />
  <row Id="14189" PostTypeId="1" CreationDate="2010-12-13T16:22:04.127" Score="2" ViewCount="125" Body="&lt;p&gt;I am a scientist writing my first manuscript with a substantial amount of mathematical methodological documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using LaTeX, but this is not my question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find a list of usage rules similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Elements_of_Style&quot; rel=&quot;nofollow&quot;&gt;Strunk and White&lt;/a&gt;. I can follow the practices that I see in my field, but it would be helpful to know the underlying rules and recommendations that will help me explain mathematical concepts in a reproducible and understandable way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3733" LastActivityDate="2010-12-13T19:25:01.033" Title="What are or where can I find style guidelines for writing math?" Tags="&lt;reference-request&gt;&lt;teaching&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="14190" PostTypeId="1" AcceptedAnswerId="14194" CreationDate="2010-12-13T16:41:43.873" Score="2" ViewCount="79" Body="&lt;p&gt;This post is related to a previous SE post &lt;a href=&quot;http://math.stackexchange.com/questions/13959/if-a-1-meter-rope-is-cut-at-two-uniformly-randomly-chosen-points-to-give-three-p/13967#13967&quot;&gt; If a 1 meter rope ….   &lt;/a&gt; concerning average length of a smallest segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.&#xA;My answer is 11/18.  Here is how I do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have two independent random variables $X,Y$, both uniform on $[0,1]$. Let &#xA;$A=\min (X,Y), B=\max (X,Y)$ and $C=\max (A, 1-B, B-A)$. First we want to find the probability&#xA;density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then&#xA;$$ F_C(a) = P(C\le a)=P(A\le a, 1-B\le a, B-A\le a).$$  By rewriting this probability as area in the unit square, I get&#xA;$$F_C(a)=\left\{\begin{array}{ll} (3a-1)^2 &amp;amp; \frac{1}{3}\le a\le \frac{1}{2}\\ 1-3(1-a)^2 &amp;amp; \frac{1}{2}\le a\le 1\end{array}\right.$$ from which it follows that &#xA;$$f_C(a)=\left\{\begin{array}{ll} 6(3a-1) &amp;amp; \frac{1}{3}\le a\le \frac{1}{2}\\ 6(1-a) &amp;amp; \frac{1}{2}\le a\le 1\end{array}\right.$$ Therefore&#xA;the expected value of $C$ is&#xA;$$\int_{1/3} ^{1/2}6a(3a-1) da+\int_{1/2} ^{1}6a(1-a) da= \frac{11}{18}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Is there a &quot;clever&quot; way to figure out this number 11/18?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) What is the answer if the rope is divided into $n&amp;gt;3$ segments?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-14T00:42:04.233" Title="Average length of the longest segment" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="14191" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T16:43:51.480" Score="1" ViewCount="0" Body="&lt;p&gt;This may not be exactly what you're looking for, but &lt;a href=&quot;http://terrytao.wordpress.com/advice-on-writing-papers/&quot; rel=&quot;nofollow&quot;&gt;Tao&lt;/a&gt; has good advice and includes on his page any other source I would consider referencing.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-13T16:43:51.480" />
  <row Id="14192" PostTypeId="2" ParentId="14146" CreationDate="2010-12-13T16:46:16.147" Score="4" ViewCount="0" Body="&lt;p&gt;You seem to want some explanation that does not involve vectors, subspaces spanned by columns, etc. I have to say that, in my personal opinion, vectors, subspaces, spans, etc. are the best way to think about it, but let me give it a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a $3\times 3$ linear system in $x$, $y$, and $c$,&#xA;$$\begin{array}{rcccccl}&#xD;&#xA;a_1x &amp;amp; + &amp;amp; b_1y &amp;amp; + &amp;amp; c_1z &amp;amp; = &amp;amp; r_1\\&#xD;&#xA;a_2x &amp;amp; + &amp;amp; b_2y &amp;amp; + &amp;amp; c_2z &amp;amp; = &amp;amp; r_2\\&#xD;&#xA;a_3x &amp;amp; + &amp;amp; b_3y &amp;amp; + &amp;amp; c_3z &amp;amp; = &amp;amp; r_3&#xD;&#xA;\end{array}$$&#xD;&#xA;and that there is a constant $k$ such that $b_1=ka_1$, $b_2=ka_2$, and $b_3=kc_3$. Why does this mean that the system will not necessarily have solutions for any given $r_1$, $r_2$, and $r_3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $w=x+ky$. Then we can rewrite $a_ix + b_iy = a_ix + ka_iy = a_i(x+ky) = a_iw$. So we can rewrite the system as a system of three equations in &lt;em&gt;two&lt;/em&gt; unknowns:&#xA;$$\begin{array}{rcccl}&#xD;&#xA;a_1w &amp;amp; + &amp;amp; c_1z &amp;amp; = &amp;amp; r_1\\&#xD;&#xA;a_2w &amp;amp; + &amp;amp; c_2z &amp;amp; = &amp;amp; r_2\\&#xD;&#xA;a_3w &amp;amp; + &amp;amp; c_3z &amp;amp; = &amp;amp; r_3&#xD;&#xA;\end{array}$$&#xD;&#xA;This is an overdetermined system (more equations than unknowns), so we know that there are ways to make the right hand side inconsistent and for the system to have no solutions. So it is not the case that the system has solutions for all possible choices of $r_1$, $r_2$, and $r_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a solution to the original system gives a solution to this system by simply letting $w=x+ky$. Does a solution to this system correspond to a solution of the original system? Yes. Suppose that $w=a$, $z=c$ is a solution. Set $x=a$ and $y=0$ to get a solution to the original system (in fact, you can let $x=u$ and $y = (a-u)/k$ if $k\neq 0$ and get a solution, for any real $u$; if $k=0$, then you take $x=a$ and $y$ arbitrary). So solutions to this system are equivalent to solutions of the original system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens if what you have is that there exist nonzero $k$ and $\ell$ such that $a_i = kb_i+\ell c_i$ (another way for the columns to be linearly independent). Then we can rewrite &#xA;$$a_ix + b_iy + c_iz = b_i(kx+y) + c_i(\ell x + z)$$&#xD;&#xA;so letting $w_1 = kx+y$ and $w_2=\ell x+z$ you get an overdetermined system again. And again, solutions of this system correspond to solutions of the original, in the sense that the original has a solution if and only if this one has a solution.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T17:58:53.777" LastActivityDate="2010-12-13T17:58:53.777" CommentCount="1" />
  <row Id="14193" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T16:50:58.560" Score="4" ViewCount="0" Body="&lt;p&gt;An excellent reference is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Krantz, Steven. A primer of mathematical writing. AMS (1997).    &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It covers much more than just papers, and I recommend it even to people with some experience. (There is a more technical book by Krantz in the same vein that treats books more extensively, &quot;Mathematical publishing, a guidebook&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another excellent source, particularly because of Halmos's essay, is the short pamphlet &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Steenrod, Norman; Halmos, Paul; Schiffer, Menahem; Dieudonné, Jean. How to write mathematics. AMS (1973).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, some colleagues have recommended the following as well: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Higham, Nicholas. Handbook of writing for the mathematical sciences. SIAM (1993).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-13T16:50:58.560" />
  <row Id="14194" PostTypeId="2" ParentId="14190" CreationDate="2010-12-13T16:51:17.200" Score="4" ViewCount="0" Body="&lt;p&gt;The answer to (B) is actually given in both Yuval Filmus' and my answers to the question about the &lt;a href=&quot;http://math.stackexchange.com/questions/13959&quot;&gt;average length of the shortest segment&lt;/a&gt;.  It's $$\frac{1}{n} H_n,$$ where $H_n = \sum_{k=1}^n \frac{1}{k},$ i.e., the $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number&quot; rel=&quot;nofollow&quot;&gt;harmonic number&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&#xA;&quot;Clever&quot; is of course subjective, but here's an argument for (A) in the $n$-piece case.  At least there's only one (single-variable) integration in it. :)  &lt;BR&gt; &lt;BR&gt;&#xA;If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$.  So the $V_i$'s are the lengths of the pieces of rope.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$  This is proved formally in David and Nagaraja's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bdhzFXg6xFkC&amp;amp;pg=PA133&amp;amp;lpg=PA133&amp;amp;dq=david+and+nagaraja+order+statistics+division+of+random+interval&amp;amp;source=bl&amp;amp;ots=OaK_l7ycZk&amp;amp;sig=BADI1rALE7hGm3-tvR6du1XKKzk&amp;amp;hl=en&amp;amp;ei=G68GTcqPEIeusAOZmdC4Bw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Order Statistics&lt;/a&gt;&lt;/em&gt;, p. 155.  Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$.  For example, $P(V_1 &amp;gt; c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$.&#xA;&lt;BR&gt;&lt;BR&gt;&#xA;If $V_{(n)}$ denotes the largest piece of rope, then &#xA;$$P(V_{(n)} &amp;gt; x) = P(V_1 &amp;gt; x \text{ or } V_2 &amp;gt; x \text{ or } \cdots \text{ or } V_n &amp;gt; x).$$  This calls for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#In_probability&quot; rel=&quot;nofollow&quot;&gt;principle of inclusion/exclusion&lt;/a&gt;.  Thus we have, using the &quot;key idea&quot; above,&#xA;$$P(V_{(n)} &amp;gt; x) = n(1-x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \cdots + (-1)^{k-1} \binom{n}{k} (1 - kx)^{n-1} + \cdots,$$ &#xA;where the sum continues until $kx &amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &#xA;$$E[V_{(n)}] = \int_0^{\infty} P(V_{(n)} &amp;gt; x) dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \int_0^{1/k} (1 - kx)^{n-1} dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \frac{1}{nk} $$&#xD;&#xA;$$= \frac{1}{n} \sum_{k=1}^n \frac{\binom{n}{k}}{k} (-1)^{k-1} = \frac{H_n}{n},$$&#xD;&#xA;where the last step applies a known &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_number#Calculation&quot; rel=&quot;nofollow&quot;&gt;binomial sum identity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-14T00:42:04.233" LastActivityDate="2010-12-14T00:42:04.233" CommentCount="1" />
  <row Id="14195" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T16:53:47.197" Score="1" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://math.stackexchange.com/questions/7011/book-about-technical-and-academical-writting/7032#7032&quot;&gt;this&lt;/a&gt; previous post.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-12-13T16:53:47.197" />
  <row Id="14196" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T17:19:23.793" Score="7" ViewCount="0" Body="&lt;p&gt;Here's what is in my shelf, in the order in which I picked them up:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mathematics into Type&lt;/em&gt; (Updated Edition). American Mathematical Society, Providence, RI, 1999, ISBN 0-8218-1961-5. It includes what the usual submission/refereeing process is, how to mark manuscripts, how to space symbols, in-line equations, and display equations (and how to break long equations across lines). It does not discuss matters of &lt;em&gt;style&lt;/em&gt; (how a mathematician usually &lt;em&gt;says&lt;/em&gt; certain things), however.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Handbook of Writing for the Mathematical Sciences&lt;/em&gt; by Nicholas J. Higham. Society for Industrial and Applied Mathematics, Philadelphia PA, 1993. ISBN 0-89871-314-5. It includes a section on Mathematical Writing (Chapter 3) and one on English Usage (Chapter 4), as well as tips on organizing a paper and the like. I'll add that Chapter 2 (&quot;Writer's Tools and Recommended Reading&quot;) contains a wealth of references.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;A Primer of Mathematical Writing. Being a disquisition on having your ideas recorded, typeset, published, read, and appreciated&lt;/em&gt; by Steven G. Krantz. American Mathematical Society, Providence RI, 1997, ISBN 0-8218-0635-1. One full chapter devoted specifically to mathematical writing (&quot;How to State a Theorem&quot;, &quot;How to Prove a Theorem&quot;, &quot;How to State a Definition&quot;, etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Paul R. Halmos. &lt;em&gt;How to write mathematics&lt;/em&gt;. Enseign. Math. &lt;strong&gt;16&lt;/strong&gt; (1970), pp. 123-152. A very good read, cited by every other work mentioned above. Also available on the web, as a quick google search will reveal; e.g., &lt;a href=&quot;http://www.math.uga.edu/~azoff/courses/halmos.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;How to write mathematics&lt;/em&gt;, corrected edition, by Norman E. Steenrod, Paul R. Halmos, Menahem M. Schiffer, and Jean A. Dieudonn&amp;eacute; American Mathematical Society, Providence RI 1981, ISBN 0-8218-0055-8. A reprint of four papers on writing mathematics, including Halmos's paper mentioned above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Writing Mathematics Well: A Manual for Authors&lt;/em&gt; by Leonard Gillman. Mathematical Association of America, 1987, ISBN 0-88385-443-0 (page ix reads: &quot;This manuscript was prepared by the author on an Apple Macintosh with the help of Mac$\Sigma$qn, a program for symbols and equations&quot;; how times have changed...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mathematical Writing&lt;/em&gt;, by Donald E. Knuth, Tracy Larrabee, and Paul M. Roberts. MAA Notes no. 14, The Mathematical Association of America, 1989, ISBN 0-88385-063-X; this is a bit of an odd duck, in my opinion. They are the class notes for a course on mathematical writing taught by Knuth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are other, more general guides; I keep a copy of Fowler's &quot;Modern English Usage&quot; and the &quot;Oxford Guide to English Usage&quot; always available, as well as a copy of the Chicago Manual of Style (14th edition) and of Strunk and White at home (recently joined by &lt;em&gt;Eats, Shoots and Leaves: the zero tolerance approach to punctuation&lt;/em&gt; by Lynne Truss; now I can tell my students that the semi-colon was invented by the same guy who invented italics). Mary-Claire van Leunen's &lt;em&gt;A Handbook for Scholars&lt;/em&gt; (revised edition) is a classic as well. But these are not specific to mathematics, and a lot of the advice has to be actively disobeyed to follow standard mathematical phraseology or uses (many a copywriter unaccostumed to mathematics has choked on &quot;a Green's function&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added:&lt;/em&gt; I don't think you will necessarily pick up the traditional jargon/cadence of mathematics from any of the above sources, however. That's something that is picked up by osmosis, through hearing and reading a lot of mathematics. But this even happens across fields, not just across disciplines; one can often spot that someone is new to a particular field within mathematics by how he or she phrases certain statements, which is at odds with the usual practices of the sub-field in question. But the advice you get in any of the above (my recommendations, in no particular order, are Highman, Krantz, and Halmos for writing content, and &lt;em&gt;Mathematics into Type&lt;/em&gt; for final preparation of the manuscript for submission, especially if you are submitting to a mathematics journal) should carry you through.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T19:25:01.033" LastActivityDate="2010-12-13T19:25:01.033" CommentCount="4" />
  <row Id="14197" PostTypeId="2" ParentId="14189" CreationDate="2010-12-13T17:21:37.317" Score="3" ViewCount="0" Body="&lt;p&gt;And if you have time, you should watch J.P. Serre's lecture &lt;a href=&quot;http://wstein.org/edu/basic/serre/&quot; rel=&quot;nofollow&quot;&gt;&quot;How to Write Mathematics Badly&quot;&lt;/a&gt; for what &lt;em&gt;not&lt;/em&gt; to do. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-13T17:21:37.317" CommentCount="1" />
  <row Id="14199" PostTypeId="2" ParentId="14155" CreationDate="2010-12-13T17:51:14.490" Score="3" ViewCount="0" Body="&lt;p&gt;To answer the three literal questions asked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wise man said once, &quot;define PDE over a number field and you will be a rich man&quot;.  This was meant not as research advice but to indicate that everyone would love to have such a thing, many have tried to create it, and obvious approaches don't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(p.s.  if you search this site for &quot;arithmetic Kodaira-Spencer class&quot; you will find an earlier discussion with a pointer from Matt E to a paper by Faltings on the nonexistence of (any linear version of) such an object over number fields.  In other words, the elusiveness of arithmetic differentiation is not just a sociological observation.)&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-13T18:02:21.600" LastActivityDate="2010-12-13T18:02:21.600" CommentCount="1" />
  <row Id="14200" PostTypeId="1" AcceptedAnswerId="16198" CreationDate="2010-12-13T18:19:24.790" Score="1" ViewCount="91" Body="&lt;p&gt;Is there a way to (efficiently) enumerate all simple graphs with a given number of vertices up to isomorphism? That is, I don't care about the degeneracy of a graph, only the set of isomorphicly isomorphically unique ones of a given size.&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2011-01-03T03:13:41.813" Title="Enumerating all simple graphs" Tags="&lt;graph-theory&gt;&lt;enumeration&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14201" PostTypeId="2" ParentId="14140" CreationDate="2010-12-13T18:41:02.080" Score="0" ViewCount="0" Body="&lt;p&gt;It seems you have one independent variable (t) and three dependent variables (P, H, and T) in that you do not have independent control of H and T.  This does not prevent finding a relation that would allow you to predict P based on H and T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at your data from year 2000, P and T overlay very closely.  A naive point of view would be to predict that P is a linear function of T and H doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you need to decide if you think the oscillations are data or noise.  If they are noise, you should filter them out.  The easiest, but far from the best, approach is just to average a number of data points.  There is discussion of filtering in &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; as well as any numeric analysis book.  As suggested by J.M., doing an FFT makes this much easier, and there are routines to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After any filtering, you can write P in some functional form involving T, H, and t and use a multidimensional minimizer to find the best fit.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-13T18:41:02.080" />
  <row Id="14202" PostTypeId="2" ParentId="14173" CreationDate="2010-12-13T19:11:13.803" Score="1" ViewCount="0" Body="&lt;p&gt;A combinatorial argument often consists in giving a bijection between two sets or at least has such an observation as its key step. More precisely, one often starts with one set $A$ whose elements one wishes to count. You then find a bijection from $A$ to some other set $B$ whose elements are easily counted, leading to some expression for the number of elements of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very simple example would be the problem of counting the number of subsets of some finite set of size $n$; call this set of subsets $A$. We first notice there's a bijection from $A$ to the set $B$ of length $n$ binary strings. We can easily count the number of elements of $B$; it's just $2^n$. Hence $A$ also has $2^n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, a combinatorial argument proceeds by noticing that some statement to be proved is equivalent to the assertion that two sets are in bijection and then by constructing an explicit bijection.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-13T19:11:13.803" />
  <row Id="14203" PostTypeId="2" ParentId="14155" CreationDate="2010-12-13T19:40:21.540" Score="5" ViewCount="0" Body="&lt;p&gt;If you want to take derivatives in a rather general context, you can.  For instance, let &#xA;$K$ be any topological field.  Then for any function $f: K \rightarrow K$ and any &#xA;$x \in K$, we say the derivative of $f$ exists at $x$ if the usual limit &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;exists.  So if your topology on $\mathbb{Q}$ is the usual Archimedean one coming from the restricting the Euclidean metric on $\mathbb{R}$, you can speak of continuous and differentiable functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt; these functions lack most of the nice properties of the corresponding functions on $\mathbb{R}$ or $\mathbb{C}$, due to the lack of completeness of $\mathbb{Q}$.  That is, a continuous (and even differentiable) function on a closed &#xA;interval in $\mathbb{Q}$ need not satisfy the intermediate value property, need not be bounded, if it is bounded need not assume a maximum or minimum value, and need not be uniformly continuous, need not satisfy the Mean Value Theorem or Taylor's Theorem, and so forth.  So it is fair to ask why one would want to study differentiable functions on $\mathbb{Q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I should say that it's not completely clear that there is no good answer to this.  For instance, in the case of the $p$-adic field $\mathbb{Q}_p$, it is not so common to speak of or study differentiable functions.  However, there &lt;em&gt;is&lt;/em&gt; a nontrivial theory here, as I learned from Alain Robert's book on $p$-adic analysis.  While it is not as essential as in the real or complex case, it has definitely been studied and written about.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue of defining partial derivatives over number fields is a bit more subtle, and here I think there are problems that the OP has yet to appreciate.  Think about the Cauchy-Riemann equations on $\mathbb{C}$: Step 0 here is identifying $\mathbb{C}$ with $\mathbb{R}^2$ and a function $f: \mathbb{C} \rightarrow \mathbb{C}$ as a function &#xA;$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, i.e., a real function in several variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, let $(K,| \ |)$ be a normed field, i.e., $| \ |: K \rightarrow \mathbb{R}^{\geq 0}$ is such that $\rho(x,y) := |x-y|$ is gives a metric on $K$ with the additional property that $|xy| = |x||y|$ for all $x,y \in K$.  And let $V$ be a finite-dimensional normed $K$-vector space, say of dimension $n$  Then a basic but perhaps underappreciated result is that the &lt;strong&gt;completeness&lt;/strong&gt; of $K$ is essential to make an identification of normed $K$-linear spaces $V \cong K^n$.  So, for example, the Euclidean norm $| \ |$ on $\mathbb{C}$ is &lt;em&gt;equivalent&lt;/em&gt; to any product metric on $\mathbb{R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This property does not hold in general when we extend the Archimedean norm $| \ |$ on $\mathbb{Q}$ to an arbitrary number field $K$.  In fact, things work out okay exactly if $K = \mathbb{Q}$ or $K$ is an imaginary quadratic field.  So let's look at the next &#xA;simplest case, that of a real quadratic field $K = \mathbb{Q}(\sqrt{D})$.  In this case &#xA;there are two norms on $K$ extending the Euclidean norm, say $| \ |_1$ and &#xA;$| \ |_2$ corresponding to the two different embeddings of $K$ into $\mathbb{R}$.  (So, for instance, if $|\sqrt{D}|_1$ is the positive square root of $D$, $|\sqrt{D}|_2$ is &#xA;the negative square root of $D$.)  Neither $(K,| \ |_1)$ nor $(K,| \ |_2)$ is equivalent, as a normed $\mathbb{Q}$-vector space, to $\mathbb{Q}^2$ with the product norm.  Indeed, here is an even stronger statement: consider the (unique) embedding $\iota$ of $\mathbb{Q}$ into $K$.  Then, with respect to the topology induced by either $| \ |_1$ or $|\ |_2$ $\iota(\mathbb{Q})$ is dense, since indeed both are dense in their completions, which are isomorphic to $\mathbb{R}$.  On the other hand, the embedding of $\mathbb{Q}$ into $\mathbb{Q}^2$ via the diagonal, $x \mapsto (x,x)$, has closed image.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the very idea of partial derivatives here makes me nervous.  An upshot of the above discussion is that choosing the basis $\{1,\sqrt{D} \}$ for $\mathbb{Q}(\sqrt{D})$ over $\mathbb{Q}$, the ``directions'' $1$ and $\sqrt{D}$ are not metrically/topologically independent, even though they are independent in the sense of linear algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final remark to make is that, to a number theorist like myself, it is very unnatural to choose a particular Archimedean norm $| \ |$ on a number field $K$.  Rather, there is a finite set of equivalence classes of such norms (&quot;Archimedean places&quot;) which can be determined by looking at the factorization of any polynomial $P(t) \in \mathbb{Q}[t]$ &#xA;such that $K \cong \mathbb{Q}[t]/(P)$ over $\mathbb{R}$: if $P$ has $r$ real roots and &#xA;$s$ complex-conjugate pairs of complex roots, then there are $r + s$ Archimedean places of $K$, and one needs to work with all of them at once in order to do topologically useful things.  In particular, the natural embedding here is really from $K$ into $K \otimes_{\mathbb{Q}} \mathbb{R} \cong \mathbb{R}^r \oplus \mathbb{C}^s$.  Note that this latter object is a field in exactly two cases: when $(r,s) = (1,0)$ (i.e., $K = \mathbb{Q})$ or when $(r,s) = (0,1)$ (i.e., $K$ is an imaginary quadratic field).&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-13T22:58:55.377" LastActivityDate="2010-12-13T22:58:55.377" />
  <row Id="14204" PostTypeId="1" AcceptedAnswerId="14205" CreationDate="2010-12-13T19:45:25.653" Score="2" ViewCount="64" Body="&lt;p&gt;What is the rate of growth of the partial sums of the reciprocals of the odd numbers?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1016" LastEditorDisplayName="" LastEditDate="2010-12-13T19:50:21.273" LastActivityDate="2010-12-13T19:54:17.653" Title="The rate of growth of the partial sums of the reciprocals of the odd numbers" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14205" PostTypeId="2" ParentId="14204" CreationDate="2010-12-13T19:54:17.653" Score="6" ViewCount="0" Body="&lt;p&gt;$\sum_{1}^{n} \frac{1}{2i-1} = \sum_{1}^{2n} \frac{1}{i} - \frac{1}{2}\sum_{1}^{n} \frac{1}{i}$, and this is approximately $\ln(2n) - \frac{1}{2}\ln(n)+\frac{1}{2} \gamma = \frac{1}{2} \ln(n) + \ln(2) + \frac{1}{2} \gamma$ for large $n$.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2010-12-13T19:54:17.653" CommentCount="2" />
  <row Id="14206" PostTypeId="1" CreationDate="2010-12-13T20:14:46.387" Score="2" ViewCount="90" Body="&lt;p&gt;&lt;strong&gt;How does a technical and theoretical study of a project of implementing an algorithm on a specific hardware architecture?&lt;/strong&gt;&#xA;Function example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%Expresion 1:&#xA;y1 = exp (- (const1 + x) ^ 2 / ((conts2 ^ 2)),&#xA;y2 = y1 * const3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where x is the input variable,&#xA;y2 is the output and const1, const3 const2 and are constant.&#xA;I need to determine the error you get in terms of architecture that decides to develop, for example it suppose that is not the same architecture with 11 bits for the exponent and 52 bits for mantissa.&#xA;This is the concept of error I handle:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Relative Error = (Real Data - Architecture Data) / (Real Data) * 100&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I consider as 'Real Data' the ouput of my algorithm I get from Matlab  (Matlab use double precission float point, IEEE 754, 52 bits to mantisa, 11 bits for exponent, one bit fo sign ) with the expression 1 and I consider as 'Architecture Data' the ouput of my algorithm running in an particular architecture (For instance a architecture that uses 12 bits for mantisa, 5 bits for exponent and 1 bit for sign)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;NOTE: The kind of algorithm I am refering to are all those which use matemathical functions that can be descompose in terms of addition, multiplication, subtrations and division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4255" LastEditorUserId="4255" LastEditorDisplayName="" LastEditDate="2010-12-16T17:39:44.323" LastActivityDate="2010-12-23T11:02:30.000" Title="Development of a specific hardware  architecture for a particular algorithm" Tags="&lt;analysis&gt;&lt;algorithms&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14207" PostTypeId="1" CreationDate="2010-12-13T20:35:20.547" Score="1" ViewCount="126" Body="&lt;p&gt;Suppose $g:\mathbb{R} \rightarrow [0,\infty)$ is a strictly increasing function such that $\lim g(x) = \infty$ as $x \rightarrow \infty$.&#xA;Suppose $h:\mathbb{R} \rightarrow [0,\infty)$ is a strictly decreasing function such that $\lim h(x) = 0$ as $x \rightarrow \infty$.&#xA;Consider the product function $f(x)=g(x)h(x)$. Is it possible to construct such $g,h$ such that $f$ does not have a limit as $x \rightarrow \infty$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better, can such an $f$ be constructed if we insist that $g(x)=x$?&lt;/p&gt;&#xA;" OwnerUserId="4719" LastActivityDate="2010-12-14T05:48:53.420" Title="Product of a monotone increasing function and a monotone decreasing function" Tags="&lt;calculus&gt;&lt;functions&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="14208" PostTypeId="1" AcceptedAnswerId="14223" CreationDate="2010-12-13T20:35:23.107" Score="1" ViewCount="75" Body="&lt;p&gt;I will be using relativistic terminology for pseudo-Riemannian manifolds as in the book by Barret O' Neil. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If one can show that the chronal future and the chronal past of some set are disjoint then why is this enough for showing that the set is achronal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean one can have a set such that there is precisely one time-like curve intersecting it at just two distinct points. Then clearly this set is not achronal (because this one curve exists) but chronal future and past of this set will include one each of these points and will be disjoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does something rule out this possibility? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let $p$ be a point on the boundary of some future set. Why is showing that chronal future and chronal past of $p$ are disjoint enough to prove that this future set is edgless? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How generic is the situation that Cauchy Horizon of a set separates that part of the causal future of a set which is not a part of its domain of dependence? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this something on which intuition can be built or is this a very special scenario?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider the interval $(0,1)$ on the x-axis on the $2-$dimensional Minkowski space where time is the y-axis. For this interval is the Cauchy Horizon the union of the set of points on the lines, $y=-x$ and $y=x-1$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If yes, then I guess this is an example where the Cauchy Horizon is not a part of the domain of dependence of the interval which is the set of points above the given interval and contained in between the above two straight lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right? &lt;/p&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-12-13T21:15:40.937" LastActivityDate="2010-12-13T22:55:51.647" Title="Some questions  about geometry on pseudo-Riemannian manifolds " Tags="&lt;differential-geometry&gt;" AnswerCount="1" />
  <row Id="14209" PostTypeId="1" AcceptedAnswerId="14323" CreationDate="2010-12-13T21:03:27.033" Score="5" ViewCount="88" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get familiar with the Newton-iteration over here but I got stuck at the proof of the error estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: [a,b] \rightarrow \mathbb{R}$ be continuously differentiable twice, concave or convex and $f&amp;#39; \neq 0 \;\; \forall x \in [a,b]$. Let $\xi$ be the root of $f$. We define the Newton-iteration for $k \in \mathbb{Z}_{\geq 0}$:&#xA;$$x_{k+1} = x_k - \frac{f(x_k)}{f&amp;#39;(x_k)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we assume $x_1 \in [a,b]$ for $x_0 = a$ and $x_0 = b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already showed that the sequence $(x_k)_{k \in \mathbb{N}}$ converges to $\xi$. Now, I want to show the following error estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\xi - x_{k+1}| \leq \frac{\max_{a \leq x \leq b} |f&amp;#39;&amp;#39;(x)|}{2 \min_{a \leq x \leq b} |f&amp;#39;(x)|} (x_{k+1}-x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite sure I will have to combine the mean value theorem and Taylor's theorem (and Lagrange's remainder), but I have no idea, how to. I don't quite know at what point I should use Taylor's theorem, also, I don't know between which two points I should apply the mean value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be very happy if somebody could give me a little hint so that I can proceed.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-14T16:34:17.730" LastActivityDate="2010-12-14T19:52:28.307" Title="How to prove the error estimate of the Newton-iteration?" Tags="&lt;analysis&gt;&lt;numerical-methods&gt;" AnswerCount="2" />
  <row Id="14210" PostTypeId="1" AcceptedAnswerId="14214" CreationDate="2010-12-13T21:07:54.803" Score="5" ViewCount="167" Body="&lt;p&gt;Every finitely generated free group is a subgroup of $F_2$, the free group on two generators.  This is an elementary fact, as is the fact that $G$, finitely presented, is the quotient of $F(|S|)$ the free group on some set of generators $S$ for $G$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether $F_3$, and hence any finitely presented group, is a quotient of $F_2$.&lt;/p&gt;&#xA;" OwnerUserId="4721" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-13T21:34:52.970" LastActivityDate="2010-12-15T23:27:39.507" Title="The free group $F_3$ being a quotient of $F_2$" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14212" PostTypeId="1" CreationDate="2010-12-13T21:16:26.733" Score="1" ViewCount="48" Body="&lt;p&gt;I am interested in learning (and implementing) an alternative to polynomial interpolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having trouble finding a good description of how these methods work, how they relate, and how they compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate your input on the pros/cons/conditions under which these methods or alternatives would be useful, but some good references to texts, slides, or podcasts would be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note:&lt;/em&gt;&#xA;I asked this &lt;a href=&quot;http://stats.stackexchange.com/q/5015/1381&quot;&gt;question&lt;/a&gt; on stats.SE a few weeks ago and received the recommendation to ask here since I have not received an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T00:49:09.053" LastActivityDate="2010-12-14T00:49:09.053" Title="What are the advantages / disadvantages of using splines, smoothed splines, and gaussian process emulators?" Tags="&lt;statistics&gt;&lt;approximation&gt;" CommentCount="3" />
  <row Id="14213" PostTypeId="1" CreationDate="2010-12-13T21:29:56.237" Score="2" ViewCount="80" Body="&lt;p&gt;I found a contradiction I couldn't resolve by my self.&#xA;It's about a &quot;Uniform White Noise&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let ${x}_{t}$ be a &quot;White Noise&quot; i.i.d. Random Process:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \forall t \in  \mathbb{R}, \ {x}_{t} \sim U[-1, \  1] $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we chose to go by the PSD definition of &quot;White Noise&quot; (Constant all over the Frequencies) we get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ {R}_{xx}( \tau ) = var({x}_{t}) \delta ( \tau ) $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, Clearly:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E[{x}_{t} {x}_{t +  \tau}] \underset{ \tau = 0}{=}E[{x}_{t} {x}_{t}]= \frac{1}{3} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, a Process with bounded variance and values can't be &quot;White Noise&quot;.&lt;br&gt;&#xA;Please mind this is a Continuous Random Process. We don't have such problem in the Discrete case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&#xA;Either there's no such &quot;White Noise&quot; (Why?) or There's a good explanation (Could someone derive it Mathematically) how to get the Delta in The Variance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2010-12-14T00:02:49.307" Title="Uniform White Noise" Tags="&lt;random-variable&gt;" AnswerCount="1" />
  <row Id="14214" PostTypeId="2" ParentId="14210" CreationDate="2010-12-13T21:30:50.160" Score="10" ViewCount="0" Body="&lt;p&gt;If $F_3$ were a quotient of $F_2$, then $\mathbb{Z}^3$ would be, but $\mathbb{Z}^3$ cannot be generated by fewer than $3$ elements. To me it seems easier to see directly that $\mathbb{Z}^3$ needs at least $3$ generators than the corresponding statement for $F_3$, perhaps because it's easy to visualize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rank of a group is the smallest cardinality of a generating set.  Here's a list of some facts about ranks of groups (including that the rank of $F_3$ is $3$) on &lt;a href=&quot;http://en.wikipedia.org/wiki/Rank_of_a_group#Known_facts_and_examples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-13T21:37:59.263" LastActivityDate="2010-12-13T21:37:59.263" CommentCount="1" />
  <row Id="14215" PostTypeId="2" ParentId="14113" CreationDate="2010-12-13T21:51:02.590" Score="3" ViewCount="0" Body="&lt;p&gt;I think you may find interesting to browse the webpage of &lt;a href=&quot;http://www.carma.newcastle.edu.au/~jb616/&quot; rel=&quot;nofollow&quot;&gt;Jon Borwein&lt;/a&gt;, which I would call the standard reference for your question. In particular, take a look at the latest version of his talk on &quot;&lt;a href=&quot;http://www.carma.newcastle.edu.au/~jb616/pi-2010.pdf&quot; rel=&quot;nofollow&quot;&gt;The life of pi&lt;/a&gt;&quot; (and its references!), which includes many of the fast converging algorithms and series used in practice for high precision computations of $\pi$, such as the one from this &lt;a href=&quot;http://www.google.com/hostednews/afp/article/ALeqM5i8R8sLPPRJvax-BjAtH0-lsV3kGw&quot; rel=&quot;nofollow&quot;&gt;Summer&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-13T21:51:02.590" />
  <row Id="14216" PostTypeId="2" ParentId="14209" CreationDate="2010-12-13T22:01:07.430" Score="5" ViewCount="0" Body="&lt;p&gt;Check out the proof &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method#Proof_of_quadratic_convergence_for_Newton.27s_iterative_method&quot; rel=&quot;nofollow&quot;&gt;on wikipedia&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want a hint: call $g(x) = x - \frac{f(x)}{f&amp;#39;(x)}$. Consider $g(x) - g(\xi)$ with the Taylor expansion. You will get something related to $g&amp;#39;&amp;#39;(c)$, which is your desired result, upon expanding $g&amp;#39;&amp;#39;$. &lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-13T22:44:34.117" LastActivityDate="2010-12-13T22:44:34.117" CommentCount="8" />
  <row Id="14217" PostTypeId="1" AcceptedAnswerId="14225" CreationDate="2010-12-13T22:08:06.657" Score="0" ViewCount="63" Body="&lt;p&gt;I'm confused about the concept of &lt;em&gt;pivoting&lt;/em&gt; a Matrix element.  This operation seems to be a fundamental operation which is used as part of more complex operations such as inverting a matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the definition of pivoting, and what it entails, seems to vary depending on the context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of inverting a matrix, for example, pivoting entails changing the pivot element to &lt;code&gt;1&lt;/code&gt;, and then all other elements in the same column to &lt;code&gt;0&lt;/code&gt; (and appropriately adjusting the other elements in the same row/column.)  However, other times the definition of pivoting only involves changing other elements in the same column to &lt;code&gt;0&lt;/code&gt;, while leaving the pivot element as it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a universal definition of &lt;em&gt;pivoting&lt;/em&gt;, or is it simply a casual term which changes depending on context?  Or, to put it another way: must the pivot element always be a &lt;code&gt;1&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4724" LastActivityDate="2010-12-13T23:06:21.920" Title="Pivoting on a matrix element" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14218" PostTypeId="1" CreationDate="2010-12-13T22:17:40.857" Score="5" ViewCount="103" Body="&lt;p&gt;Let's say, we have an array/matrix $n \times m$ and we need to find, how many ways can we fill this array with numbers from $\{ 1, \ldots , m\cdot n \}$, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) every number can be used only 1 time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) every column and every row should be sorted in increasing order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we have $n = m = 3$, there are 42 different ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about this problem, but I can't find out any &lt;em&gt;simple&lt;/em&gt; formula to compute this for every $n, m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve this problem by writnig a program that uses backtracking method/algorithm [1] but it has huge complexity and problem is solved by computer, and I want to know, how to do this &lt;em&gt;by hand&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sombody told me, that in this problem I can use some kind of Catalan numbers [2] [3], but I'm not sure about this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what do You think about this problem and its solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Backtracking (Wikipedia): &lt;a href=&quot;http://en.wikipedia.org/wiki/Backtracking&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Backtracking&lt;/a&gt; &lt;br/&gt;&#xA;[2] Catalan numbers (Wikipedia): &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Catalan_number&lt;/a&gt; &lt;br/&gt;&#xA;[3] Catalan numbers (OEIS): &lt;a href=&quot;http://oeis.org/A000108&quot; rel=&quot;nofollow&quot;&gt;http://oeis.org/A000108&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-14T01:21:18.443" LastActivityDate="2010-12-14T01:21:18.443" Title="Array of numbers, how many solutions/ways?" Tags="&lt;combinatorics&gt;&lt;matrices&gt;&lt;permutations&gt;" AnswerCount="2" />
  <row Id="14219" PostTypeId="2" ParentId="14207" CreationDate="2010-12-13T22:28:18.940" Score="5" ViewCount="0" Body="&lt;p&gt;Take&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(x) = x 2^{\lfloor x/2\rfloor}, h(x) = \frac{1}{x} 2^{-\lfloor (x+1)/2\rfloor} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product oscillates between 1 and 1/2: let $f = g\cdot h$, $f(2n) = 1$, $f(2n+1) = 1/2$ for $n$ an integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;: If you want $g(x) = x$. Take $h(x)$ to be the following (continuous) function: let $h(e^{2n}) = e^{-2n}$, and let $h(e^{2n+1}) = e^{-2n - \frac12}$, and linearly interpolate. Then $f(e^{2n}) = 1$ and $f(e^{2n+1}) =\sqrt{e} \neq 1$ for all natural numbers $n$. So $f$ does not have a limit. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-13T22:35:09.043" LastActivityDate="2010-12-13T22:35:09.043" CommentCount="3" />
  <row Id="14220" PostTypeId="2" ParentId="14218" CreationDate="2010-12-13T22:33:54.450" Score="6" ViewCount="0" Body="&lt;p&gt;These are Young tableaux.  OEIS has the numbers.  For the square case, see &lt;a href=&quot;http://oeis.org/A039622&quot; rel=&quot;nofollow&quot;&gt;A039622&lt;/a&gt;.   For the rectangular case, see &lt;a href=&quot;http://oeis.org/A060854&quot; rel=&quot;nofollow&quot;&gt;A060854&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-13T22:33:54.450" CommentCount="1" />
  <row Id="14221" PostTypeId="2" ParentId="14218" CreationDate="2010-12-13T22:36:35.277" Score="5" ViewCount="0" Body="&lt;p&gt;You are counting the number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Young_tableau&quot; rel=&quot;nofollow&quot;&gt;standard Young tableaux&lt;/a&gt; of shape $(m, m, m, ...)$ (where $m$ occurs $n$ times).  The formula that counts these is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Young_tableau#Dimension_of_a_representation&quot; rel=&quot;nofollow&quot;&gt;hook length formula&lt;/a&gt;.  It correctly reproduces the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{9!}{5 \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2} = 42$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the example you gave.  The general formula seems annoying to write down.  In the special case $n = 2$ you get Catalan numbers; this is a nice exercise and it is worth trying to prove this with and then without the hook length formula.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-13T22:36:35.277" CommentCount="6" />
  <row Id="14222" PostTypeId="2" ParentId="14207" CreationDate="2010-12-13T22:38:42.797" Score="2" ViewCount="0" Body="&lt;p&gt;Define $h(x) = 3e^{-x}$ for $x &amp;lt; 1$, and $h(x) = 1/x$ for $x \geq 1$. Hence, $h$ satisfies the conditions in the question. Define $g(x) = (1/3) e^x$ for $x&amp;lt;1$, and for $x \geq 1$ define it piecewise as follows:&#xA;$g(x) = x + a_n$ for $n \leq x &amp;lt; n+1$, $n=1,2,3,\ldots$, where $a_n$ is a strictly increasing sequence with $a_1 = 0$. Hence, $g$ satisfies the conditions in the question. Now define $f(x)=g(x)h(x)$. Thus, for any $n \in \mathbb{N}$, $f(n)=(n+a_n)/n = 1 + a_n / n$. We can choose $a_n$ so that $a_n / n$ does not have a limit. For example, consider arbitrarily large $n$, and set $a_n = 2n$. Then, let $a_{n+k} = (2n+k)$, for $k = 1,\ldots,n^2$. Then, $a_n / n = 2$, but $a_{n+n^2}/(n+n^2) \approx 1$. Continue the same way to construct a sequence with $\lim \sup _{n \to \infty}(a_n/n) = 2$ but $\lim \inf _{n \to \infty}(a_n/n) = 1$. We conclude that $f(x)$ does not have a limit as $x \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-13T22:38:42.797" />
  <row Id="14223" PostTypeId="2" ParentId="14208" CreationDate="2010-12-13T22:55:51.647" Score="2" ViewCount="0" Body="&lt;p&gt;For the first question, I fear that you may have not understood the definition completely. Let $p,q\in U$ be the two points. Let $\gamma$ be the future time-like curve joining $p = \gamma(0)$ and $q= \gamma(1)$. I claim that for any $s\in (0,1)$, $\gamma(s)$ is in both the future and past of $U$. Indeed, $\gamma(s)$ is in the future of $p\in U$, hence is in the chronal future of $U$. And $\gamma(s)$ is in the past of $q\in U$, hence is in the chronal past of $U$. Hence the chronological future and the chronological past of $U$ must intersect non-emptily, i.e. they are not disjoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, this shows that if a set $U$ is &lt;em&gt;not&lt;/em&gt; achronal, then $I^+(U)$ and $I^-(U)$ must intersect. Hence if the chronological future and past are disjoint, the set $U$ must be achronal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question two, I don't think your statement is right. Can you give me a reference (page number in O'Neill will do)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question three, isn't that just the definition? The Cauchy Horizon is generally regarded as the boundary of the domain of dependence, no? If that's not the case, can you provide the definition you are using for the Cauchy horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question four: not quite. The future Cauchy horizon of the set is the union $ \{ y = x: x \in [0,1/2] \} \cup \{ y + x = 1: x \in [1/2,1]\} $. The past Cauchy horizon is $\{ y = -x: x \in[0,1/2] \} \cup \{ x-y = 1: x\in[1/2,1]\} $. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-13T22:55:51.647" CommentCount="7" />
  <row Id="14224" PostTypeId="2" ParentId="14154" CreationDate="2010-12-13T22:57:53.267" Score="2" ViewCount="0" Body="&lt;p&gt;If your purpose is visualization, then often a good approximate solution is useful.  This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Multidimensional_scaling&quot; rel=&quot;nofollow&quot;&gt;Multidimensional Scaling&lt;/a&gt; (MDS).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-12-13T23:10:38.717" LastActivityDate="2010-12-13T23:10:38.717" CommentCount="1" />
  <row Id="14225" PostTypeId="2" ParentId="14217" CreationDate="2010-12-13T23:06:21.920" Score="2" ViewCount="0" Body="&lt;p&gt;One can better understand the role that pivoting plays in Gaussian elimination by viewing it from more general perspectives. For example one can compare analogous elimination algorithms over rings (vs. fields), e.g. Hermite / Smith normal forms. Additionally one can compare the the more general choice of &quot;critical pair&quot; pivots in non-linear elimination algorithms such as &lt;a href=&quot;http://chiragwiki.wikispaces.com/file/view/publication_190.pdf&quot; rel=&quot;nofollow&quot;&gt;Grobner basis algorithms&lt;/a&gt;, or the more general &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%E2%80%93Bendix_completion_algorithm&quot; rel=&quot;nofollow&quot;&gt;Knuth-Bendix equational completion&lt;/a&gt;, etc. Here an optimal choice of &quot;pivoting&quot; / &lt;a href=&quot;http://dx.doi.org/10.1016/S0747-7171%2888%2980018-X&quot; rel=&quot;nofollow&quot;&gt;critical pair strategy,&lt;/a&gt; can prove crucial to tractable computation (e.g.  to avoid combinatorial explosion). The Knuth-Bendix algorithm provides a fairly universal point of view that encompasses all these elimination algorithms.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-13T23:06:21.920" />
  <row Id="14226" PostTypeId="2" ParentId="14206" CreationDate="2010-12-13T23:09:22.030" Score="2" ViewCount="0" Body="&lt;p&gt;You have several issues to consider.  First, your output format allows very many fewer output values than the double standard.  If your exponent is 2-based, your output will be of the form $\pm m*2^e$ where m has 2048 values available and e has 32 values (maybe they range from -15 to +16).  The restricted range of e means that you will underflow/overflow quite often.  If you are within range, you will still have an error as large as one part in 4096 assuming your calculation is perfect and the only error is representing the output value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second issue is precision of the input and error propagation.  If your input values are represented in the same format, they will be in error by up to 1 part in 4096 for the same reason.  Then when you calculate $f(x)=c_3 \exp {\frac{-(c_1+x)^2}{c_2}}$ (where I absorbed the squaring of $c_2$ into its definition) the error will propagate.  You can assess this by evaluating $\frac{\text{d}\ln f(x)}{\text{d}x}=\frac{\text{d}\ln (c_3 \exp {\frac{-(c_1+x)^2}{c_2}})}{\text{d}x}$ to get the fractional error in $f(x)$ depending on the error in $x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this assumes your calculation is perfect and the only problem is the accuracy of representing the data.  If not, you also need to assess the error that is inherent in the computation.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-13T23:09:22.030" CommentCount="2" />
  <row Id="14227" PostTypeId="2" ParentId="14137" CreationDate="2010-12-13T23:26:55.440" Score="0" ViewCount="0" Body="&lt;p&gt;The point of view offered by Jim is good because it can be generalized: if you take any surface with $n$ holes (say a connected sum of $n$ tori), it has a CW-complex structure made by a single 0-cell, $2n$ 1-cells 'wedged' to the 0-cell and a single 2-cell applied on the 1-cells. It is page 5 of Hatcher's &lt;em&gt;Algebraic Topology&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-13T23:26:55.440" />
  <row Id="14228" PostTypeId="1" CreationDate="2010-12-13T23:48:53.223" Score="1" ViewCount="185" Body="&lt;p&gt;A module $M_{R}$ is called semi-artinian if every nonzero image of $M$ contains a simple submodule. Given $m\in M$ and $a_1,a_2,...$ in J(R). Why $ma_1a_2...a_{n-1}a_n=0$ for some $n\geq 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Edit by KennyTM: The above is OP's original question. The latest, completely changed question follows:)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $R$ is regular what is the relation between:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) J(R)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) a left $R$-module has a projective cover?&lt;/p&gt;&#xA;" OwnerUserId="4688" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-16T09:23:08.807" LastActivityDate="2010-12-16T09:23:08.807" Title="Ring Theory (semi-artinian)" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14229" PostTypeId="1" AcceptedAnswerId="14242" CreationDate="2010-12-13T23:53:53.063" Score="3" ViewCount="173" Body="&lt;p&gt;I was rather confused about this proof I came across in Rudin Chapter 7.  The premise is: If K is a compact metric space, if $f_n \in C(K)$ for $n =  1, 2, 3, ...,$ ($C(K)$ being a set of complex-valued, continuous and bounded functions), and if $\{f_n\}$ converges uniformly on $K$, then $\{f_n\}$ is equicontinuous on $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&#xA;Let $\epsilon &amp;gt; 0$ be given.  Since $\{f_n\}$ converges uniformly, there is an integer $N$ such that $$||f_n - f_N|| &amp;lt; \epsilon (n &amp;gt; N).$$ (See Definition 7.14).  Since continuous functions are uniformly continuous on compact sets, there is a $\delta &amp;gt; 0$ such that $$|f_i(x) - f_i(y)| &amp;lt; \epsilon$$ if $1 \leq i \leq N$ and $d(x,y) &amp;lt; \delta$.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n &amp;lt; N$ and $d(x,y) &amp;lt; \delta$, it follows that $$|f_n(x) - f_n(y)| \leq |f_n(x) - f_n(y)| + |f_n(x) - f_N(y)| + |f_N(x) - f_N(y)| &amp;lt; 3\epsilon.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$QED$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is the line ending with *.  I understand that when a family of functions are uniformly continuous, for every $\epsilon &amp;gt; 0$, there exists an integer $N$ such that $n \geq N$ implies $|f_n(x) - f_n(y)| \leq \epsilon$, but it doesn't say anything about the case when $n \leq N$.  Perhaps it has something to do with the finiteness of the number of functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be something really obvious I'm overlooking.&lt;/p&gt;&#xA;" OwnerUserId="2919" LastEditorUserId="2919" LastEditorDisplayName="" LastEditDate="2010-12-14T01:36:27.093" LastActivityDate="2010-12-14T02:06:09.250" Title="Don't Understand Proof in Rudin Principles of Mathematical Analysis Book Thm 7.24?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14230" PostTypeId="2" ParentId="14213" CreationDate="2010-12-14T00:02:49.307" Score="1" ViewCount="0" Body="&lt;p&gt;Apparently, saying that $(x_t)$ is a continuous white noise process simply refers to the fact that $(x_t)$ is a continuous time process, that is $(x_t)$ is indexed by a continuous parameter set. That is, the sample paths of $(x_t)$ are not assumed to be continuous, and in fact may be expected to be discontinuous at every fixed point (almost surely). Indeed, consider a continuous Gaussian white noise process. Then, $E[x_s x_t]=0$, $s \neq t$, implies that $x_s$ and $x_t$ are independent, and hence the sample paths must be discontinuous at every point (almost surely). The case of &quot;Uniform White Noise&quot; is essentially the same.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-14T00:02:49.307" CommentCount="9" />
  <row Id="14231" PostTypeId="1" CreationDate="2010-12-14T00:07:11.100" Score="1" ViewCount="143" Body="&lt;p&gt;The question asks me to determine if $4\mathbb{Z}$ and $5\mathbb{Z}$ (with standard addition) are isomorphic and if so to give the isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts:&#xA;What I am having difficulty with is showing a mapping that preserves the operation.  i.e., $\phi(a+b) = \phi(a) + \phi(b)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have so far is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(a+b) = \phi(5a/4) + \phi(5b/4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(a+b) = 5/4(a+b) = 5/4a + 5/4b = \phi(a) + \phi(b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my conclusion is $4\mathbb{Z}$ and $5\mathbb{Z}$ are isomorphic and the isomorphism is $\phi(n) = \frac{5}{4}n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone either confirm this or point me in the right direction?  I really appreciate everyone willingness to help one another!  Thank you! &lt;/p&gt;&#xA;" OwnerUserId="4179" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T03:11:26.030" LastActivityDate="2010-12-14T03:14:28.303" Title="$(4\mathbb{Z}, +)$   isomorphic to  $(5\mathbb{Z}, +)$" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14233" PostTypeId="2" ParentId="14183" CreationDate="2010-12-14T00:36:55.203" Score="3" ViewCount="0" Body="&lt;p&gt;I don't have a full calculus 1 course, but I have a lot  on youtube at  Professor Elvis Zap . You can learn to rap the laws of differentiation from the calculus rap. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-12-14T00:36:55.203" />
  <row Id="14234" PostTypeId="2" ParentId="14228" CreationDate="2010-12-14T00:49:26.227" Score="2" ViewCount="0" Body="&lt;p&gt;This is proved in Askar A. Tuganbaev' s &lt;em&gt;Semidistributive modules and rings&lt;/em&gt;, page 153, which you can read on googlebooks.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T00:49:26.227" CommentCount="2" />
  <row Id="14235" PostTypeId="1" AcceptedAnswerId="14247" CreationDate="2010-12-14T00:59:01.793" Score="6" ViewCount="347" Body="&lt;p&gt;I was reading over this in my notes, but it doesnt make any sense to me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, suppose we ﬁx an alphabet for&#xA;  writing our programs (e.g. 8- bit&#xA;  ASCII). Since each individual program&#xA;  is ﬁnite in length, we can put all&#xA;  possible programs into a (very long)&#xA;  ordered list. For any ﬁxed character&#xA;  length k, there are only a ﬁnite set&#xA;  of possible programs. So, we can write&#xA;  down all programs by ﬁrst ﬁrst writing&#xA;  down all the 1-character programs,&#xA;  then all the 2-character programs, and&#xA;  so forth. In other words, there’s a&#xA;  bijection between the integers and the&#xA;  total set of programs. But this means&#xA;  that the number of functions is&#xA;  uncountable, whereas the number of&#xA;  programs is only countably inﬁnite. So&#xA;  there must be mathematical functions&#xA;  that we can’t compute with any&#xA;  (ﬁnite-length) program.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What exactly is this saying? Why does it hold? It seems like it shouldn't. In some cases a program of only 30-40 characters can compute billions of numbers! Maybe with current hardware it may not be possible to computer some numbers, but theoretically, this shouldn't be a problem, right? &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T03:31:16.113" LastActivityDate="2010-12-14T17:04:00.980" Title="Mathematical functions that can't be computed" Tags="&lt;counting&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="14236" PostTypeId="1" CreationDate="2010-12-14T01:10:17.427" Score="0" ViewCount="41" Body="&lt;p&gt;Suppose I have a list of keywords, and number of times that keywords were mentioned this week and last week.  How can I rank them by &quot;percent increase&quot; if some of the keywords were &quot;new&quot; this week (and so were mentioned zero times last week)?  Is there a &quot;proper&quot; way to normalize the numbers so that I can rank them by some metric that is similar to &quot;percent increase&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4725" LastActivityDate="2010-12-14T01:10:17.427" Title="Ranking by percent increase" Tags="&lt;statistics&gt;" CommentCount="1" />
  <row Id="14237" PostTypeId="2" ParentId="14231" CreationDate="2010-12-14T01:11:18.910" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ \ \mathbb Z\ \cong n\ \mathbb Z\ $ via $\rm\ x\to\ n\ x\ $ for $\rm\: n\ne 0\:.\ $ Thus $\rm\ n\ \mathbb Z\ \cong\ \mathbb Z\ \cong\: n&amp;#39;\ \mathbb Z\ $ by transitivity of $\rm\: \cong\:$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-14T02:24:03.843" LastActivityDate="2010-12-14T02:24:03.843" />
  <row Id="14238" PostTypeId="1" CreationDate="2010-12-14T01:19:18.400" Score="0" ViewCount="61" Body="&lt;p&gt;Alright, so this is how I am doing it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;float xrot = 0;&#xA;           float yrot = 0;&#xA;           float zrot = 0;&#xA;&#xA;           Quaternion q = new Quaternion().fromRotationMatrix(player.model.getRotation());&#xA;           if(q.getW() &amp;gt; 1){&#xA;            q.normalizeLocal();&#xA;           }&#xA;&#xA;           float angle = (float) (2 * Math.acos(q.getW()));&#xA;           double s = Math.sqrt(1-q.getW()*q.getW());&#xA;           if (s &amp;lt; 0.001) { // test to avoid divide by zero, s is always positive due to sqrt&#xA;            // if s close to zero then direction of axis not important&#xA;            xrot = q.getXf(); // if it is important that axis is normalised then replace with x=1; y=z=0;&#xA;            yrot = q.getYf();&#xA;            zrot = q.getZf();&#xA;           // z = q.getZ();&#xA;            } else {&#xA;             xrot = (float) (q.getXf() / s); // normalise axis&#xA;             yrot = (float) (q.getYf() / s);&#xA;             zrot = (float) (q.getZf() / s);&#xA;            }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But it doesn't seem to work when I try to put it into use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;player.model.addTranslation(zrot * player.speed, 0, xrot * player.speed);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;AddTranslation takes 3 numbers to move my model by than many spaces (x, y, z), but hen I give it the numbers above it doesn't move the model in the direction it has been rotated (on the XZ plane)    &lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2010-12-14T04:11:28.753" Title="quaternion to angle" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;vector-space&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14239" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T01:19:53.390" Score="13" ViewCount="0" Body="&lt;p&gt;Cantor's &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;diagonal argument&lt;/a&gt; can be used to show that the number of functions from $\mathbb{N}$ to $\mathbb{N}$ is uncountable.  However, of those functions, the number of functions which can be computed by any computer program is countable because there are only countably many computer programs.  So there exist functions which cannot be computed by a computer program - in fact, &quot;most&quot; functions have this property.  An explicit example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Busy_beaver&quot; rel=&quot;nofollow&quot;&gt;busy beaver function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a matter of computing numbers but of computing &lt;em&gt;sequences&lt;/em&gt; of numbers, and is a fundamental limitation of computation.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-14T01:19:53.390" />
  <row Id="14240" PostTypeId="1" AcceptedAnswerId="14243" CreationDate="2010-12-14T01:26:02.057" Score="1" ViewCount="88" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Justin needs to pick 17 toy animals to&#xA;  give to children at a party. The&#xA;  animals come in 5 kinds: dogs,&#xA;  dinosaurs, cows, lizards, and ﬁsh. How&#xA;  many diﬀerent ways can he choose his&#xA;  set of toys&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok the answer for this is 21 choose 4. However, I dont understand how they arrived at this answer. If there are 17 slots, and for each slot, you can choose between 5 animals, then you should have $5^{17}$. Where has my reasoning gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-14T16:12:06.500" Title="Simple combinations question" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14241" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T01:37:32.357" Score="8" ViewCount="0" Body="&lt;p&gt;Each string (that parses) represents a program that computes a function on $\rm\:\mathbb N$. The argument shows that the number of functions representable by such a language is countable because the number of program strings in the language is countable (note that some programs may compute the same function). But there are uncountably many functions on $\rm\:\mathbb N\:$ by  diagonalization (du-Bois-Reymond, Cantor). Thus there are functions not computable by any program in the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find it helpful to consider a more explicit language. For example, consider a language that represents polynomials with natural coefficients. An analogous argument shows that such polynomials are countable, so we can enumerate them, i.e. we can index the polynomials by naturals: $\rm\ f_i(x),\ \:i\in \mathbb N\:.\ $ By diagonalization we may construct a non-polynomial function on $\rm\:\mathbb N\:,\:$ e.g. $\rm\;\ g(n)\ :=\ f_n(n)+1\:.\ $ Notice that if $\rm\ f_i = g\ $ then $\rm\ f_i(i) = g(i) = f_i(i) + 1\:,\ $ a contradiction. Therefore  $\rm\:g\:$ is not equal to any polynomial function $\rm\:f_i\:.$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-14T02:18:41.650" LastActivityDate="2010-12-14T02:18:41.650" />
  <row Id="14242" PostTypeId="2" ParentId="14229" CreationDate="2010-12-14T01:42:24.153" Score="6" ViewCount="0" Body="&lt;p&gt;Now that you have updated with the filled in lines, it seems the confusion is mixing up the conditions for continuity and convergence. Uniform continuity has nothing to do with taking $n≥N$; that would be convergence. And the line ending with an asterisk has nothing to do with the convergence. What you can do is take $N$ different $\delta$s corresponding to the uniform continuity of each of the first $N$ functions, then take their minimum. As you said, it is possible because there are only finitely many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that for convergence, you are talking about functions getting closer together &lt;em&gt;at each point&lt;/em&gt; (uniformly in this case) as $n$ gets larger, whereas for continuity you are talking about &lt;em&gt;each function&lt;/em&gt; taking close values at nearby points.  Equicontinuity says that the you can choose your definition of &quot;nearby&quot; the same for all functions in the family simultaneously.  The step you are asking about is basically saying that a finite family of uniformly continuous functions is equicontinuous.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-14T02:06:09.250" LastActivityDate="2010-12-14T02:06:09.250" />
  <row Id="14243" PostTypeId="2" ParentId="14240" CreationDate="2010-12-14T01:43:37.693" Score="3" ViewCount="0" Body="&lt;p&gt;As Jonas Meyer indicated, the way you counted will include separately different orderings of the same numbers of each animal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you can have 21 slots. Four slots are to be filled with &quot;dividers&quot;. You put dogs to the slots (if any) to the left of divider 1, dinosaurs in slots (if any) between divider 1 and divider 2, and so on until you place fish to the right of the 4th divider. There are ${\displaystyle {21 \choose 4}}$ ways to place the four dividers, and each placement of the four dividers corresponds to exactly one way of choosing the number of each animal brought to the party.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-14T01:43:37.693" />
  <row Id="14244" PostTypeId="2" ParentId="14240" CreationDate="2010-12-14T02:06:12.887" Score="1" ViewCount="0" Body="&lt;p&gt;Following the advice from Jonas, there are $A_{r,n}=\displaystyle \binom{n+r-1}{r}$ ways to obtain indistinguishable distributions where each distribution is described in terms of n-tuples $(r_{1},...,r_{n})$  satisfying $$r_{1}+r_{2}+...+r_{n}=r$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular problem, you want to enumerate the number of solutions for the equation $$r_{1}+r_{2}+...+r_{5}=17$$ where $r_{i}$ represents the number of elements contained in a box/cell labeled as $i$ where $i=1=\text{dogs}$, $i=2=\text{dinosaurs}$,..., $i=5=\text{fishes}$. For example, one such a way is to pick $(17,0,0,0,0)$ (say, 17 dogs and nothing else), other choice would be $(13,1,1,1,1)$ (13 dogs, 1 dinosaur, 1 cow, 1 lizard and 1 ﬁsh), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you have $\displaystyle \binom{5+17-1}{17}$ which is equivalent to $\displaystyle \binom{21}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Feller. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471257087&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Probability Theory and Its Applications&lt;/a&gt;. Chapter II.5&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-14T16:12:06.500" LastActivityDate="2010-12-14T16:12:06.500" CommentCount="2" />
  <row Id="14245" PostTypeId="1" AcceptedAnswerId="14251" CreationDate="2010-12-14T02:24:43.343" Score="3" ViewCount="199" Body="&lt;p&gt;I tried to prove this, but was unsuccessful for a long time.. &#xA;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $(\exp(x))^y=\exp(xy)$ using the identities, &#xA;$$\exp(x)=\sum_{n\geq0} \frac{x^n}{n!}, \quad (1+x)^\alpha=\sum_{n\geq0} {\alpha \choose n} x^n$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T03:08:18.887" LastActivityDate="2010-12-14T05:10:48.647" Title="Formal Proof of Exponential rule" Tags="&lt;sequences-and-series&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="14246" PostTypeId="2" ParentId="13884" CreationDate="2010-12-14T02:30:34.100" Score="0" ViewCount="0" Body="&lt;p&gt;It probably relates to standing wave. At the corner there must be a node as opposed to anti-node. Nodes cannot move but anti-nodes can move freely in a certain direction.&#xA;For normal air flow which in turn producing standing wave, the wave become so &quot;friendly&quot; to trap dust  at the corner. But the story will be different if you make &quot;smart&quot; air flow.&lt;/p&gt;&#xA;" OwnerUserId="3901" LastActivityDate="2010-12-14T02:30:34.100" CommentCount="1" />
  <row Id="14247" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T02:32:16.373" Score="3" ViewCount="0" Body="&lt;p&gt;Here is one analogy which may prove helpful: there are rational numbers, which can be represented as the ratio of two integers: 1/2, 3, 5/6 etc. This can be seen as all the numbers which can be printed out by &quot;programs&quot; which only have access to the &quot;division&quot; operator. Clearly some numbers cannot be printed out by a program with only this operation - $\pi$, for example cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we add in the full power of arithmetic: you can specify a number like &quot;the root of $x^2-2x+3$&quot;. These are the algebraic numbers. We can print out more numbers this way, but not all of them. $\pi$ again cannot be printed out by this type of program, as it is transcendental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a last hurrah, we allow any operation which can print out an arbitrary number of digits in a finite amount of time. We can now find values like $\pi$ and $e$, since there are formulae for these. But can we print &lt;em&gt;all&lt;/em&gt; values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no. As you have indicated, there are &quot;too many&quot; numbers and too few programs. It is not merely that we have not discovered (or created) some operation - even if we double the number of symbols in our language that only increases the amount of numbers we can write by a finite amount, and there is an infinite amount of space to make up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than talking about programs, another way to phrase it is: &quot;for every real number, is there some &lt;em&gt;finitary&lt;/em&gt; description of it?&quot; E.g. $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ - even though $e$ has an infinite number of digits, we can describe it using only a finite number of symbols. The proof you are discussing shows that there must be some numbers which admit no finitary description.&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2010-12-14T02:32:16.373" CommentCount="4" />
  <row Id="14249" PostTypeId="2" ParentId="14231" CreationDate="2010-12-14T03:14:28.303" Score="3" ViewCount="0" Body="&lt;p&gt;Bill's argument is of course the simplest, but to discuss your effort...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two comments: your argument only shows that $\phi$ is a homomorphism; you haven't shown it is an isomorphism (one-to-one and onto). This is a major thing: you cannot conclude it is an isomorphism just form knowing it is a homomorphism. You &lt;em&gt;need&lt;/em&gt; to show it is invertible, or show it is one-to-one and show it is onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second comment is more of a nit-pick: you should not say that $\phi$ is &lt;strong&gt;the&lt;/strong&gt; isomorphism; there may be more than one isomorphism (in fact, there is; you can also take $\psi(4n) = -5n$). You should say (once you establish it, of course), that $\phi$ is &lt;em&gt;an&lt;/em&gt; isomorphism.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-14T03:14:28.303" CommentCount="1" />
  <row Id="14250" PostTypeId="1" CreationDate="2010-12-14T03:22:08.340" Score="5" ViewCount="80" Body="&lt;p&gt;I've been studying the finite-dimensional representations of the lie algebra sl(2,C). I've read that these representations are related to the harmonic oscillator and the associated raising and lowering operators, but I'm not really sure how they are related. I've read you can generate a new energy state from an old one, but I'm not really sure how it works, and the energy states do not differ by 2 like the eigenspaces do for sl(2,C). I'm having trouble finding references about this, especially ones that are accessible. If anyone knows where I can find more info, or has some knowledge of the material, I would greatly appreciate their help.&lt;/p&gt;&#xA;" OwnerUserId="4238" LastActivityDate="2010-12-14T04:30:52.000" Title="sl(2,C) and the harmonic oscillator" Tags="&lt;reference-request&gt;&lt;physics&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="14251" PostTypeId="2" ParentId="14245" CreationDate="2010-12-14T03:44:53.303" Score="10" ViewCount="0" Body="&lt;p&gt;I'll assume you want an equality of formal power series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n \ge 0} \frac{x^n y^n}{n!} = \sum_{n \ge 0} {y \choose n} (\exp(x) - 1)^n.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of this is easier to explain combinatorially than via series manipulation, if you allow me to first assume that $y$ is a non-negative integer.  In that case I claim that the coefficient of $\frac{x^n}{n!}$ on the RHS is the number of functions from $[n]$ to $[y]$ (where $[k] = { 1, 2, ... k }$), which is clearly $y^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, note that $\exp(x) - 1$ is the exponential generating function for non-empty sets, hence $(\exp(x) - 1)^k$ is the exponential generating function for partitions of a set into $k$ non-empty blocks, and ${y \choose k}$ is the number of ways to choose $k$ elements out of $y$ elements.  But a function from $[n]$ to $[y]$ is nothing more than a partition of $[n]$ into $k$ non-empty blocks (the preimages of each element of the range), together with a choice of $k$ elements of $[y]$ (the image of each of the blocks).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this implies the identity for arbitrary $y$, note that the coefficient of $\frac{x^n}{n!}$ on the RHS (regarded as a formal power series in $x$ and $y$) must be a polynomial of degree at most $n$, and this polynomial agrees with $y^n$ for all non-negative integers, so it must be identically equal to $y^n$.  (What I've proven essentially amounts to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_number&quot; rel=&quot;nofollow&quot;&gt;inversion formulas&lt;/a&gt; relating the Stirling numbers of the first and second kinds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a proof without this extra simplification then you need to do some messy inclusion-exclusion or some messy series manipulation and I don't really see the point either way.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-14T03:50:03.880" LastActivityDate="2010-12-14T03:50:03.880" CommentCount="1" />
  <row Id="14252" PostTypeId="1" AcceptedAnswerId="14256" CreationDate="2010-12-14T03:56:50.647" Score="3" ViewCount="69" Body="&lt;p&gt;A two dimensional space (eg. $\mathbb{R}^2$ ) could be a flat or it could be the surface of sphere (or of any shape) in a 3-dimensional space. How do we distinguish between such spaces without invoking higher dimensions.(here I am using the term 'space' in a general term and not strictly in a manner used in mathematics notations). What is the subject of math which deals with such properties ? Also please give the mathematical term closer to the term 'space'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: please feel free to edit this post if it could make it more meaningful and clear.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastActivityDate="2010-12-14T04:17:03.600" Title="property of a space." Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14253" PostTypeId="1" CreationDate="2010-12-14T04:04:54.497" Score="4" ViewCount="144" Body="&lt;p&gt;Let $X$ be a subspace of $\ell_1^4$. Can one always extend a linear operator $l:X\rightarrow \ell_1^4$&#xD;&#xA;to $L:\ell_1^4\rightarrow \ell_1^4$ such that $L$ has the same operator norm as $l$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4730" LastEditorUserId="4730" LastEditorDisplayName="" LastEditDate="2010-12-14T13:14:27.113" LastActivityDate="2010-12-14T16:30:32.727" Title="Can a linear operator from a normed space to itself be extended with operator norm preserved?" Tags="&lt;real-analysis&gt;&lt;linear-algebra&gt;&lt;functional-analysis&gt;&lt;banach-spaces&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="14255" PostTypeId="2" ParentId="14238" CreationDate="2010-12-14T04:11:28.753" Score="0" ViewCount="0" Body="&lt;p&gt;You first need to determine which local (i.e. body-fixed) axis points out of the front of the front of your model.  It is likely the positive X or positive Y axis (depending on your convention).  You then take the first column if it is the X or the second column if it is the Y.  These will give you a unit vector pointing in the direction that you want to travel.  You can then update your position in time with &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P_{k+1} = P_k + U*speed*dt&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where P is your position and U is the unit vector.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-14T04:11:28.753" />
  <row Id="14256" PostTypeId="2" ParentId="14252" CreationDate="2010-12-14T04:17:03.600" Score="2" ViewCount="0" Body="&lt;p&gt;I recommend reading up on differential geometry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what you are asking is how to determine if a space is curved without access to the embedding space.  That depends on curvature of the space.  There are some spaces that are equivalent intrinsically but different extrinsically (i.e. a cylinder or cone to a flat plane).  In these spaces you cannot tell the difference from within the space. &lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-14T04:17:03.600" />
  <row Id="14257" PostTypeId="2" ParentId="14250" CreationDate="2010-12-14T04:30:52.000" Score="4" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The Quantum Mechanical Harmonic Oscillator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; on the quantum mechanical harmonic oscillator, you will see that it is not the Lie algebra $\mathfrak{sl}_2$ that enters the picture, but rather an algebra generated by&#xA;two operators $a$ and $a^{\dagger}$ satisfying $[a,a^{\dagger}] = 1$ (see also Eric's comment above).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a familiar algebra in disguise, the so called Weyl algebra of differential operators with polynomial coefficients $\mathbb C[x,\partial_x]$.  (Note that $[\partial_x,x] = 1$, by the Leibniz rule.)  The product $x\partial x$ is the so-called Euler operator (it acts on $x^n$ as multiplication by $n$); you can see that in the harmonic oscillator interpretation, it acts as the Hamiltonian (up to a shift of $1/2$, to add non-zero groundstate energy, and a rescaling by $\hbar \omega$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algebra has a natural representation, namely on the space of polynomials $\mathbb C[x]$, and we see that this is isomorphic to the representation on the span of the eigenstates that occurs in the harmonic oscillator picture.  (The polynomial $x^n$ corresponds to the eigenstate with energy $\hbar \omega(n + 1/2).$)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Lie algebra $\mathfrak sl_2$ in quantum mechanics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If my memory serves, the first time $\mathfrak sl_2$ entered the picture (in my undergrad QM class, at least) is in the analysis of the hydrogen atom.  The point is that&#xA;the Schrodinger equation in this case has a rotational symmetry, and differentiating this action, one gets an action of the Lie algebra of $SO(3)$ on the space of states.  Complexifying, this gives an action of $\mathfrak sl_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you separate variables in the equation (using spherical coordinates, which are natural in view of the rotational symmetry), the radial part and the spherical part separate, the radial part is easily dealt with, and one is left&#xA;studying a certain differential equation on the space of functions of the&#xA;sphere $S^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, one is looking at $L^2(S^2)$, with its natural action of $\mathfrak sl_2$ (in the guise of the complexified Lie algebra of $SO(3)$).&#xA;The Shrodinger equation involves the Laplacian, which is also the Casimir in the&#xA;enveloping algebra of $\mathfrak sl_2$, and so if you fix the energy (eigenvalue of the Laplacian), one is trying to understand the corresponding eigenspace of the Casimir on $L^2(S^2)$.  This is a special case of the Peter--Weyl theorem,&#xA;which (in this case) also goes under the name of the the theory of spherical harmonics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea, roughly, is:  $S^2 = SO(3)/SO(2)$.  Now by Peter--Weyl,&#xA;$L^2(SO(3))$ is the Hilbert space direct sum of $V\otimes V^*$, where&#xA;$V$ runs over the irreps. of $SO(3)$.&#xA;Thus $L^2(S^2)$ is the Hilbert space direct sum of $V\otimes (V^*)^{SO(2)}$.&#xA;The space of $SO(2)$ invariants in each irrep. is one-dimensional, so in&#xA;fact $L^2(S^2)$ is the Hilbert space direct sum of $V$, as $V$ runs over&#xA;the irreps. of $SO(3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $SO(3) = SU(2)/\{\pm 1\}$, these correspond on the Lie algebra level&#xA;to the &lt;em&gt;odd-dimensional&lt;/em&gt; irreps. of $\mathfrak sl_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $V$ is an eigenspace for the Casimir (if $V$ has dimension $n$ then&#xA;the Casimiar eigenvalue is some quadratic expression in $n$ that you can&#xA;easily figure out, or look up), and distinct $V$ give distinct eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going back to the hydrogen atom, if you fix the energy of the electron,&#xA;then the collection of states has dimension equal to the dimension of the&#xA;corresponding $V$. (The fact that this is typically greater than one-dimensional is referred to in physics as &lt;em&gt;degeneracy&lt;/em&gt;; the energy does not uniquely determine the state.) This has a natural basis corresponding to the $SO(2)$&#xD;&#xA;eigenvectors.  Physically, the derivative of this $SO(2)$ (the traditional&#xA;element $H$ in the Lie algebra $\mathfrak sl_2$, perhaps up some scaling,&#xA;and often denoted $L_z$ in physics literature) is the operator corresponding to angular momentum around the axis in $S^2$ which is fixed by $SO(2)$ (often taken to be the $z$-axis, by convention, in the physics literature), and so the different eigenvectors for $SO(2)$ in the given $V$ are states with the same energy, but with different angular momenta around the $z$-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, one can apply a magnetic field along this axis which will then affect these different $L_z$-eigenstates differently (because its effect on&#xA;the electron depends on the angular momentum of the electron around the field&#xA;lines), and so one can &lt;em&gt;split the degeneracy&lt;/em&gt;.  This is the Zeeman effect.  One sees the different energy lines in the spectrum of hydrogen split into a number of lines ($n$ lines when the energy is proportional to $n^2$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't looked, but this must be discussed in many, many places online, and is also in any standard intro. to quantum text.  (But probably with less appearance of the symbols $L^2(S^2)$ and $\mathfrak sl_2$ then a mathematician would use!)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-14T04:30:52.000" CommentCount="1" />
  <row Id="14258" PostTypeId="2" ParentId="14245" CreationDate="2010-12-14T05:00:42.380" Score="5" ViewCount="0" Body="&lt;p&gt;One can prove the sought power series equality via the uniqueness theorem for first-order ODE's. It's easy to show that both power series satisfy $\rm\ f&amp;#39;(x) = y\ f(x),\ \ f(0) = 1\:.\ $ Put $\rm\ f(x) = \sum_{n \ge 0} \frac{x^n y^n}{n!},\:$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle\ \ g(x)\ =\ \sum_{n \ge 0}\ \ {y \choose n}\ (\exp(x) - 1)^n\:.\ \ $ Clearly $\rm\:f\:$ satisfies the ODE. $\ $ For $\rm\:g\:$ we have  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\displaystyle\ g&amp;#39;(x)\ =\ \sum_{n \ge 1}\ n\: {y \choose n}\ exp(x)\ (\exp(x) - 1)^{n-1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ \exp(x)\ \sum_{n \ge 1}\ {y-1\choose n-1}\ (\exp(x)-1)^{n-1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ \exp(x)\ \sum_{n \ge 0}\ {y-1\choose n}\   (\exp(x)-1)^n $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ \exp(x)\ {\exp(x)}^{y-1}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad =\ y\ g(x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; $\ $ The above is simply $\rm\ (exp(x)^y)&amp;#39; =\ y\ exp(x)\ exp(x)^{y-1}\ $ expressed in power series form. Note that the coefficient identity at the core is the same binomial identity as a &lt;a href=&quot;http://math.stackexchange.com/questions/13922/how-to-sum-up-this-series/13942#13942&quot;&gt;couple days ago&lt;/a&gt;, viz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\displaystyle\quad\quad\quad\quad\quad\  {y \choose n}\ =\ \frac{y}{n}\ {y-1\choose n-1}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-14T05:10:48.647" LastActivityDate="2010-12-14T05:10:48.647" CommentCount="3" />
  <row Id="14259" PostTypeId="1" AcceptedAnswerId="14260" CreationDate="2010-12-14T05:00:53.180" Score="4" ViewCount="115" Body="&lt;p&gt;There are Artin $L$-series and Dirichlet $L$-series, and zeta functions for varieties and for number fields; there are a slew of objects named after Hecke... There are also various kinds of characters in these areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always had a hard time keeping all of them straight. I would be very grateful if someone could put them in some sort of perspective that would make it clear what the role of each of those objects is with respect to the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to use the language of Langlands, that would be acceptable. (perhaps it is best to see it in those terms?)&lt;/p&gt;&#xA;" OwnerUserId="4732" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:33:02.280" LastActivityDate="2010-12-14T17:33:02.280" Title="Getting my number theoretic series straight" Tags="&lt;number-theory&gt;&lt;zeta-function&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="14260" PostTypeId="2" ParentId="14259" CreationDate="2010-12-14T05:33:35.840" Score="9" ViewCount="0" Body="&lt;p&gt;First of all, one shouldn't distinguish too much between the terms $L$-series and $\zeta$-function; it is more or less a matter of history which term you use in a given context.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true distinction is between objects on the &quot;motivic&quot; (Diophantine equation or algebraic number theory) side, and objects on the &quot;automorphic&quot; side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Artin $L$-functions, and $\zeta$-functions and $L$-functions attached to varieties, are on the motivic side.   They don't have obvious analytic continuations or functional equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dirichlet $L$-functions (i.e. the ones built out of characters of the multiplicative group mod $N$ for some $N$), Hecke $L$-functions (built out &#xA;of ideal class group characters), $L$-functions of modular forms, or more generally automorphic forms, are automorphic $L$-functions.  Except for the&#xA;last (general automorphic $L$-functions) these &lt;em&gt;can&lt;/em&gt; be proved to have analytic continuation with functional equations.  The natural limit of these analytic continuation/functional equation arguments (which begin with Riemann, with Tate's thesis being another major point en route) is the fact that &lt;em&gt;standard $L$-functions&lt;/em&gt; for automorphic forms on $GL_n$ have analytic continuation/functional equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Langlands's &lt;em&gt;functoriality conjecture&lt;/em&gt; in particular conjectures that any (&lt;em&gt;a priori&lt;/em&gt; more general) automorphic $L$-function is in fact a standard $L$-function.  Many special cases are known (and Ngo got the Fields medal for proving the fundamental lemma, which is one tool in proving certain cases, namely those arising from &lt;em&gt;endoscopy&lt;/em&gt;), but it is wide open in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;reciprocity conjecture&lt;/em&gt; then states that any of the motivic $L$-functions are also actually standard $L$-functions.  (E.g. for $L$-functions of elliptic curves over $\mathbb Q$, this becomes the statement that any such elliptic curve is associated to a weight 2 modular form, which was proved by Wiles et. al.)  This is also wide open in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The special case of Artin $L$-functions, and some other special cases, are actually covered by both conjectures.  (This is because, if you broaden your perspective sufficiently, as Langlands did, by allowing arbitrary reductive algebraic groups into the picture, then Artin $L$-functions can be thought of as&#xA;a particular kind of automorphic $L$-functions.  This is not so easy to see when you are just entering the subject, but one can think of just the Riemann $\zeta$-function: this is certainly motivic, being the $\zeta$-function of the&#xA;variety Spec $\mathbb Q$, but is also automorphic, being the Dirichlet $L$-function for the trivial character.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in general, the conjectures have a slightly different nature: functoriality can be thought of (if you want) purely in terms of harmonic analysis on adelic groups, whereas by its very nature reciprocity involves arithmetic geometry.  In practice, at least at the moment, the two seem to be fairly intertwined. Functoriality certainly is a tool that can be very helpful in proving reciprocity; also, of those cases of the Artin conjecture which are known, some are proved using the functoriality view-point (e.g. Langlands--Tunnell) and some using the reciprocity view-point (e.g. Buzzard--Dickinson--Shepherd-Barron--Taylor).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-14T05:33:35.840" />
  <row Id="14261" PostTypeId="2" ParentId="14235" CreationDate="2010-12-14T05:42:16.173" Score="0" ViewCount="0" Body="&lt;p&gt;You are looking for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Halting_problem&quot; rel=&quot;nofollow&quot;&gt;Halting Problem&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Higher-order_logic&quot; rel=&quot;nofollow&quot;&gt;Higher-order logic&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't really count the functions via their representations as programs, because you can neither compare all program behaviours nor find out all those programs that won't halt at all. You could define a function which is not only undefined on some or many of uncountably many points, but also can't be proven to be undefined at those points. If you could, then you could count the functions of First-Order Logic, but still not those of Higher Order Logic (Second Order Logic?) and above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aditionally, there are functions, that cannot be expressed as programs at all, like the &quot;busy beaver function&quot;: It describes for every n the largest finite number of steps after which any ever halting program of length n halts. The program of this function itself would need to have an infinite description length to be correct. The busy beaver function could solve the Halting Problem via timeout, iff it would be expressible as a (finite) program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can count, that are the source codes of all programs. In a similar way can all functions be represented as a finite amount of chalk – in infinite different ways. But even with this, they can only be described but not always in all their properties proven, because there exist theorems that are true even though they can't be proven, and because the possibility of a prove does not always imply the existance of a proven example. To compare it with the busy beaver function: There exist functions that would need an infinite amount of chalk to be proven (in other words: they can't be proven) but not to be described; to map those functions to programs would either require to prove them first (→ impossible or program won't halt) or imply to prove them (→ infinite program description). As an example: Think about a program H that receives the description of another program P as input and needs to calculate an existing but unprovable property of that program P (like halting or not) and therefore would not halt itself; the function could simply be defined on that existing property, which makes it to something else than this non-halting program H.&lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="3226" LastEditorDisplayName="" LastEditDate="2010-12-14T17:04:00.980" LastActivityDate="2010-12-14T17:04:00.980" CommentCount="4" />
  <row Id="14262" PostTypeId="2" ParentId="14207" CreationDate="2010-12-14T05:48:53.420" Score="0" ViewCount="0" Body="&lt;p&gt;You are greatly helped by the fact that $f$ and $g$ are not required to be analytic so you can easily analize $f$ and $g$ in some neighbourhood of infinity and don't really care about their behaviour outside that neighbourhood :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you need a function $f(x)$ such that $\lim_{x\to\infty} f(x)g(x) &amp;lt; \infty$, which is to say that $f(x) = \frac{\alpha(x)}{g(x)}$, where $\alpha$ is bounded ($\alpha = h/g^2$). It is not very hard to prove that such $f$ exists for any continuous $g$. Just construct $f$ as $\frac{\alpha(x)}{g(x)}$ in the neighbourhood of infinity where $g(x) \neq 0$ and then 'glue' this $f$ to some other function on the rest of the domain (it is possible to do this in $C^\infty$ way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This covers all monotonic functions in $C^0, C^1, \ldots, C^\infty$. The most interesting stuff is about the rest of them :)&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2010-12-14T05:48:53.420" />
  <row Id="14263" PostTypeId="1" CreationDate="2010-12-14T06:28:51.217" Score="0" ViewCount="132" Body="&lt;p&gt;Are all groups of order $156$ solvable?&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-14T06:39:03.667" LastActivityDate="2010-12-14T14:35:12.017" Title="Groups of order 156" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="14264" PostTypeId="1" AcceptedAnswerId="14460" CreationDate="2010-12-14T06:44:38.780" Score="5" ViewCount="88" Body="&lt;p&gt;I have an epimorphism $f:B_4\longrightarrow S_4$, from the braid group on 4 strands &lt;em&gt;onto&lt;/em&gt; the symmetric group on 4 elements.  Is it possible the kernel is not isomorphic to $P_4$, the pure braid group on 4 strands? &lt;/p&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2010-12-15T23:16:17.980" Title="Braid Group, B_4-&gt;&gt; S_4 onto, do I know kernel is P_4, pure braid group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="14265" PostTypeId="1" CreationDate="2010-12-14T07:39:53.203" Score="1" ViewCount="76" Body="&lt;p&gt;I want to construct a nowhere dense set of positive measure by using the Smith-Volterra-Cantor method but subtracting closed sets intead of open sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we start with the open interval (0,1).  Let $\alpha=1/2$.  Now remove $U_1=[1/2-\alpha/4,1/2+\alpha/4]$.  The measure $\mu(U_1)=\alpha/2$.  For each of the two remaining open sets, subtract a closed from the center of width $\alpha/8$, thus a total measure of $\alpha/4$.  Repeating this process results in removing a total measure of $\alpha/2+\alpha/4+\dots=\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this give me a set of positive measure just as with subtracting closed intervals?  I suspect not, since I have never heard of this, but what's the exact function of removing a closed set instead of an empty set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if I can do this, can I now consider this to be an open set that I can choose an admissible partition of unity for?&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-12-14T14:01:15.163" Title="Constructing a Cantor-like set by subtracting closed intervals." Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14266" PostTypeId="2" ParentId="14265" CreationDate="2010-12-14T07:52:43.717" Score="6" ViewCount="0" Body="&lt;p&gt;&quot;Does this give me a set of positive measure just as with subtracting closed intervals?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, as follows from the basic property of &lt;a href=&quot;http://en.wikipedia.org/wiki/Measure_%28mathematics%29#Measures_of_infinite_intersections_of_measurable_sets&quot; rel=&quot;nofollow&quot;&gt;continuity of measure&lt;/a&gt;: if $E_1 \supset E_2 \supset \ldots \supset E_n \supset \ldots$ is a nested sequence of subsets with $\mu(E_1) &amp;lt; \infty$, then if $E = \bigcap_{n=1}^{\infty} E_n$, we have $\mu(E) = \lim_{n \rightarrow \infty} \mu(E_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;[C]an I now consider this to be an open set...?&quot; No.  Here the variation you've made on the usual construction affects the answer: any intersection of closed sets is necessarily closed, whereas (even) a countably infinite intersection of open sets need not be open.  I think it would be good for you to figure out for yourself exactly why the intersection $E$ is not open here.  Think for instance about the fact that if you do the usual construction by removing open intervals you get a set which is nowhere dense and how your modified set $E$ compares to that closed, nowhere dense set.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-14T08:01:47.763" LastActivityDate="2010-12-14T08:01:47.763" />
  <row Id="14267" PostTypeId="2" ParentId="13912" CreationDate="2010-12-14T07:59:37.877" Score="0" ViewCount="0" Body="&lt;p&gt;This may be a bit fundamental, but i found that Exploring Abtract Alegebra Package for Mathematica had some nice visualizations that helped me develop an intuitive understanding.&lt;/p&gt;&#xA;" OwnerUserId="4735" LastActivityDate="2010-12-14T07:59:37.877" />
  <row Id="14268" PostTypeId="1" CreationDate="2010-12-14T08:47:09.250" Score="4" ViewCount="300" Body="&lt;p&gt;How to check if two multiplications are equal to each other or greater or lesser without actually multiplying them?&lt;br&gt;&#xA;For example, compare (254)(847) and (383)(536)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;While trying to find a rule i got one&lt;br&gt;&#xA;(5)(11) &amp;lt; (6)(10)&lt;br&gt;&#xA;or&lt;br&gt;&#xA;(x)(y) &amp;lt; (x+1)(y-1) when y &gt; x &gt; 0&lt;br&gt;&#xA;and another rule is  that if adding and subtracting 1 equates them the difference is one&lt;br&gt;&#xA;(3)(5) + 1 = (4)(4)&lt;br&gt;&#xA;(x)(y) + 1 = (x+1)(y-1) when y + 2 = x , y &gt; x &gt;= 0&lt;/p&gt;&#xA;" OwnerUserId="1515" LastEditorUserId="1515" LastEditorDisplayName="" LastEditDate="2010-12-15T06:36:07.780" LastActivityDate="2010-12-31T00:52:33.823" Title="How to Compare two multiplications without multiplying?" Tags="&lt;arithmetic&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="2" />
  <row Id="14269" PostTypeId="2" ParentId="13525" CreationDate="2010-12-14T08:53:05.007" Score="2" ViewCount="0" Body="&lt;p&gt;There are joint papers by Benedict Gross (a number theorist) and Mike Hopkins (a topologist) on formal groups and the chromatic tower, that may be more accessible as a quick introduction to the concepts. They try to explain some of this material to non-topologists, and their exposition from almost 20 years ago does not rely on the machinery of stacks, &quot;brave new algebra&quot; or infinity-categories.  The summary paper in the AMS Bulletin has a dictionary relating stable homotopy and quasicoherent sheaves, in particular they explain that the chromatic filtration is analogous to the Cousin complex.  See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ams.org/journals/bull/1994-30-01/S0273-0979-1994-00438-0/&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/journals/bull/1994-30-01/S0273-0979-1994-00438-0/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-14T08:53:05.007" CommentCount="1" />
  <row Id="14270" PostTypeId="1" AcceptedAnswerId="14274" CreationDate="2010-12-14T08:57:10.370" Score="4" ViewCount="122" Body="&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If R is a commutative ring with identity, and a, b are its elements that are divisible by each other, is it true that they must be associates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, a being divisible by b means there exists an r \in R, such that a=rb;&#xA;and a and b being associates means there exists a invertible element u \in R such that a=ub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that this not always true. But I encountered some difficulties in finding a counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="4736" LastActivityDate="2010-12-14T16:43:09.417" Title="If R is a commutative ring with identity, and a, b are its elements that are divisible by each other, is it true that they must be associates?" Tags="&lt;divisibility&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="14272" PostTypeId="1" CreationDate="2010-12-14T09:07:27.543" Score="0" ViewCount="62" Body="&lt;p&gt;Let $f, g$ be two paths in a space such that their concatenation $f * g$ is nullhomotopic. Prove that $f$ is homotopic to $g$ rel $\{0,1\}$.&lt;/p&gt;&#xA;" OwnerUserId="4737" LastActivityDate="2010-12-14T11:20:16.050" Title="Homotopy of two paths whose composition is nullhomotopic" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14273" PostTypeId="1" AcceptedAnswerId="14277" CreationDate="2010-12-14T10:03:12.010" Score="2" ViewCount="57" Body="&lt;p&gt;I'm trying to prove an identity involving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Digamma_function&quot; rel=&quot;nofollow&quot;&gt;digamma function&lt;/a&gt; $\psi(z)$, but I can't seem to figure out a way to do it. Can anyone help me out? The identity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(\frac{n}{2} + iy\right) + \psi\left(\frac{n}{2} - iy\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ and $n$ are integers satisfying $m + n = 2$, and $y$ is any nonzero real number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried looking at a couple of the integral representations of $\psi(z)$ listed on the Wikipedia page, but I haven't been able to figure it out from those. I think there's probably some simple integral identity I'm forgetting that would make the whole thing work out - for instance, if I could show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^\infty \frac{e^{nt/2}\cos(yt)}{\sinh(t/2)}\mathrm{d}t = \int_0^\infty \frac{e^{-nt/2}\cos(yt)}{\sinh(t/2)}\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n \in \mathbb{Z}, y \neq 0$, I think I'd be set. So what I'm hoping to get is a pointer to some relation like that which I could use. Of course I'd be happy with a full proof if you prefer ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I know this sounds kind of &quot;homework-y,&quot; but it's not for a homework assignment; I'm trying to verify a calculation in a physics paper.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastEditorUserId="1190" LastEditorDisplayName="" LastEditDate="2010-12-14T19:09:53.947" LastActivityDate="2010-12-14T19:09:53.947" Title="How can I prove this identity involving the digamma function?" Tags="&lt;special-functions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="14274" PostTypeId="2" ParentId="14270" CreationDate="2010-12-14T10:18:18.883" Score="9" ViewCount="0" Body="&lt;p&gt;See the following paper,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When are Associates Unit Multiples?&#xA;D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles&#xA;Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;a href=&quot;http://projecteuclid.org/DPubS?verb=Display&amp;amp;version=1.0&amp;amp;service=UI&amp;amp;handle=euclid.rmjm/1181069828&amp;amp;page=record&quot; rel=&quot;nofollow&quot;&gt;freely available&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is mostly concerned with finding sufficent conditions on&#xA;commutative ringsthat ensure that $Ra=Rb$ imply $a$ and $b$ associates,&#xA;but they do give some examples of $R$ where this fails. In particular&#xA;this simple example of Kaplansky. Let $R=C[0,3]$, the set of continuous&#xA;function from the interval $[0,3]$ to the reals. Let $f(t)$ and $g(t)$&#xD;&#xA;equal $1-t$ on $[0,1]$, zero on $[1,2]$ but let $f(t)=t-2$&#xD;&#xA;on $[2,3]$ and $g(t)=2-t$ on $[2,3]$. Then $f$ and $g$ are non-associates&#xA;in $R$ but each divides the other.&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-14T10:18:18.883" CommentCount="2" />
  <row Id="14275" PostTypeId="1" AcceptedAnswerId="14330" CreationDate="2010-12-14T10:23:42.220" Score="4" ViewCount="95" Body="&lt;p&gt;Let $a$ and $b$ be elements of the polynomial ring $\mathbb{Z}/n\mathbb{Z}[x]$. If $a$ and $b$ generate the same ideal, must they be associates? If $n$ is prime, then it is easy to see that the answer is &quot;yes&quot;. But what if $n$ is not prime?&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T20:23:16.053" LastActivityDate="2010-12-15T06:16:19.003" Title="Associates in the Ring of Polynomials mod n" Tags="&lt;ring-theory&gt;&lt;elementary-number-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14276" PostTypeId="1" CreationDate="2010-12-14T10:32:07.717" Score="3" ViewCount="119" Body="&lt;p&gt;Is the functor from the category of projective varieties over a field $k$ to the category of pure motives over $k$, faithful? (Perhaps it is not full).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ditto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the functor from the category of &lt;em&gt;affine&lt;/em&gt; varieties over a field $k$ to the category of mixed motives over $k$, faithful? (Perhaps it is not full).&lt;/p&gt;&#xA;" OwnerUserId="4739" LastActivityDate="2010-12-15T00:05:45.690" Title="Does a motive capture everything about an algebraic variety?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="4" />
  <row Id="14277" PostTypeId="2" ParentId="14273" CreationDate="2010-12-14T10:39:01.750" Score="4" ViewCount="0" Body="&lt;p&gt;The first relation can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(\frac{2-m}{2} + iy\right) + \psi\left(\frac{2-m}{2} - iy\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(1-\left(\frac{m}{2} + iy\right)\right) + \psi\left(1-\left(\frac{m}{2} - iy\right)\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the reflection formula for the digamma function (which can be derived from the reflection formula for the gamma function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(1-z)=\psi(z)+\pi\cot\pi z$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your equation simplifies as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cot\left(\frac{m\pi}{2}+i\pi y\right)+\cot\left(\frac{m\pi}{2}-i\pi y\right)=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or since the cotangent is an odd function,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cot\left(\frac{m\pi}{2}+i\pi y\right)=\cot\left(-\frac{m\pi}{2}+i\pi y\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cot\left(\frac{m\pi}{2}+i\pi y\right)=\cot\left(\frac{m\pi}{2}+i\pi y-m\pi\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it is clear that both sides of the equation will agree only if $m\in\mathbb Z$ due to the periodicity of the cotangent.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T12:37:08.963" LastActivityDate="2010-12-14T12:37:08.963" CommentCount="1" />
  <row Id="14278" PostTypeId="2" ParentId="14272" CreationDate="2010-12-14T11:11:19.413" Score="1" ViewCount="0" Body="&lt;p&gt;As Rasmus points out, once you state the result correctly, it's easy -assuming you've already proved some elementary properties of homotopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, let $f,g$ be two loops with $f(0) = g(0) = f(1)= g(1) = x_0$ and such that $g^{-1}*f \sim 1$, where $1$ is the constant path at $x_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f\sim 1* f \sim (g*g^{-1})*f \sim g* (g^{-1}*f) \sim g*1  \sim g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise.&lt;/strong&gt; Which &quot;elementary properties of homotopy&quot; have we used and where?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditorDisplayName="" LastEditDate="2010-12-14T11:20:16.050" LastActivityDate="2010-12-14T11:20:16.050" />
  <row Id="14279" PostTypeId="1" CreationDate="2010-12-14T11:13:54.933" Score="0" ViewCount="50" Body="&lt;p&gt;Let $\psi_{j,k}(t)=\{ \begin{array}{ll} 2^{j/2} &amp;amp;, 2^{-j}k &amp;lt; t &amp;lt; 2^{-j}(k+1/2) \\-2^{j/2} &amp;amp;,  2^{-j}(k+1/2) &amp;lt; t &amp;lt; 2^{-j}(k+1) \\ 0, \textrm{otherwise} \end{array} $. How to prove that it is orthogonal? In other words how to prove that $\langle \psi_{j_1,k_1}(t), \psi_{j_2, k_2}(t) \rangle = 0$, $j_1&amp;lt;j_2$ and $k_1&amp;lt;k_2$. I have already proved that it's norm is one $|| \psi_{j,k} ||_2 = 1$, but now I'm stuck in this orthogonality issue, because it involves many nontrivial computations. &lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="2723" LastEditorDisplayName="" LastEditDate="2010-12-14T11:32:48.233" LastActivityDate="2010-12-14T13:43:31.877" Title="Let $\{ \psi_{j,k}(t)\}$ haar system. How to prove that it is orthogonal?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" />
  <row Id="14280" PostTypeId="2" ParentId="14113" CreationDate="2010-12-14T11:22:39.840" Score="3" ViewCount="0" Body="&lt;p&gt;Just to give people an idea on convergence rates, here is a plot of $-\log_{10}\left|\frac{S_n-\pi}{\pi}\right|$ versus $n$ , where $S_n$ is the nth partial sum of the series in question, for three of the series featured in the answers to this question (note the vertical scale):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PdR9f.png&quot; alt=&quot;partial sum plots&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three series are, from top to bottom, $\arctan(1)$ (the series mentioned by the OP), $2\arcsin\left(\sqrt{\frac12}\right)$ (the series mentioned by yjj in his answer), and the series by Ramanujan I mentioned in the comments (I didn't include the series by the Chudnovsky brothers, since that converges even faster than the Ramanujan series, and that makes for boring plots).&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-14T11:22:39.840" CommunityOwnedDate="2010-12-14T11:22:39.840" />
  <row Id="14281" PostTypeId="1" CreationDate="2010-12-14T11:26:04.487" Score="3" ViewCount="47" Body="&lt;p&gt;Say $M := \begin{pmatrix} A &amp;amp; B\\ C &amp;amp; D \end{pmatrix}$ is a block matrix with $A, D$ being square matrices and this $B$ and $C^T$ having the same shape. Is there any norm characterizing the collection of $B$ and $C$ that is invariant under all block diagonal similarity transformations $M\to S M S^{-1}$ with $S=\begin{pmatrix}E &amp;amp; 0\\ 0 &amp;amp; F\end{pmatrix}$ of the whole block matrix? Were $B$ and $C$ square matrices I'd use e.g. the sum of squared eigenvalues, but is there something similar for non-square matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, the requirements for the block norms I seek are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the norm of the square blocks $A$ and $D$ must be invariant (under these similarity transformations of $M$) respectively&lt;/li&gt;&#xA;&lt;li&gt;either there are individual norms for $B$ and $C$ which are invariant, or the is one combined norm depending on $B$ and $C$ which is invariant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is the latter requirement possible?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Due to the block diagonal structure of $S$, all blocks transform independently, so there should be independent norms for $B$ and $C$. But since $B\to F B E^{-1}$ and $C\to E C F^{-1}$ are no similarity transformations, I couldn't even use eigenvalues if $B$ and $C$ &lt;em&gt;were&lt;/em&gt; square matrices. However, $BC \to F BC F^{-1}$ and $CB \to E CB E^{-1}$ are similarity transformations, the remaining question is what kind of norm to use and maybe also how to decide whether $BC$ or $CB$ or both are &quot;significant&quot;...&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:29:03.517" LastActivityDate="2010-12-15T21:20:12.937" Title="What are norms of sub-matrices invariant under a block diagonal similarity transformation of a block matrix?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;norm&gt;&lt;tag-removed&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14282" PostTypeId="1" AcceptedAnswerId="14289" CreationDate="2010-12-14T11:54:20.500" Score="6" ViewCount="308" Body="&lt;p&gt;Let $G \in \mathbf{Grp}$, $H \leq G$, $G/H := \lbrace gH: g \in G \rbrace$. We can then introduce group operation on $G/H$ as $(xH)*(yH) := (xy)H$, so that $G/H$ becomes a quotient group when $H$ is a normal subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do we only work with quotient groups by normal subgroups? If we introduce the notion of &lt;em&gt;left quotient group&lt;/em&gt; in the above manner, how much good properties of a quotient group do we lose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD: fixed confusing notation: from $(xH)(yH)$ to $(xH)*(yH)$.&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditorDisplayName="" LastEditDate="2010-12-14T13:07:53.323" LastActivityDate="2010-12-16T18:18:38.393" Title="Why do we define quotient groups for normal subgroups only?" Tags="&lt;group-theory&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="3" />
  <row Id="14283" PostTypeId="1" CreationDate="2010-12-14T12:08:11.267" Score="0" ViewCount="38" Body="&lt;p&gt;I'm working on data structure for graph cut algorithm. Problem is to make different cuts on shortest paths. I made data structure for which I'm not sure about properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input is directed graph of shortest paths, which is bounded lattice, partially ordered set with minimum and maximum element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define &lt;em&gt;next node N(n)&lt;/em&gt; of node n as a set of nodes b for which a &amp;lt; b and there is no c with a &amp;lt; c &amp;lt; b. Similar define &lt;em&gt;previous node P(n)&lt;/em&gt;. Extend definitions on sets, N(S) union of N(n) for $n \in S$, similar for P(S).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to make different cuts on list of set of nodes $L, N(L), N(N(L)), ...$ where for each neighboring pair of sets A, N(A)=B holds that there is no partition of $A=A_1 \cup A_2$ and $B=B_1 \cup B_2$ with $B_i = N(A_i)$, $A_i = P(B_i)$ for $i=1,2$.&#xA;With this property create new lattice with mapping:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sub-lattice to one node&lt;/li&gt;&#xA;&lt;li&gt;if upper partition is found create partition cardinality number of edges.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In simple, algorithm for lattice -&gt; lattice mapping is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = {minimum node}&#xA;new_node = [A]&#xA;1:&#xA;while A, N(A) don't have partitions&#xA;  append N(A) to new_node&#xA;  A = N(A)&#xA;for each partition $B_i$&#xD;&#xA;  last_new_node = new_node&#xA;  create new_node = [B_i]&#xA;  create edge last_new_node to new_node&#xA;  go to 1&#xA;At the end fix maximum node in new lattice if needed&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This algorithm can be called repeatedly on new lattices. I'm concern about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there guaranty to reach one-node lattice?&lt;/li&gt;&#xA;&lt;li&gt;Is there any measure of number of iterations to reach one-node lattice? It seams to me that bound is diameter of input graph. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I appreciate link to any similar data structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tnx&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2010-12-14T12:08:11.267" Title="Lattice simplification" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;&lt;lattices&gt;" />
  <row Id="14284" PostTypeId="1" CreationDate="2010-12-14T12:15:13.913" Score="1" ViewCount="39" Body="&lt;p&gt;We are talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-uniform_rational_B-spline&quot; rel=&quot;nofollow&quot;&gt;Non-uniform rational B-spline&lt;/a&gt;. We have some simple 3 dimensional array like &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{1,1,1}&#xA;{1,2,3}&#xA;{1,3,3}&#xA;{2,4,5}&#xA;{2,5,6}&#xA;{4,4,4}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which are points from a plane created by some B-spline&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find control points of spline that created that plane? (I know its a hard task because of weights that need to be calculated but I really hope it is solvable)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lOuL8.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-14T17:45:38.743" LastActivityDate="2010-12-14T17:45:38.743" Title="How to get NURBS control points from an array of points that should be part of its solution from controll points we are searching for?" Tags="&lt;geometry&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" CommentCount="1" />
  <row Id="14285" PostTypeId="2" ParentId="14282" CreationDate="2010-12-14T12:34:45.183" Score="3" ViewCount="0" Body="&lt;p&gt;(Sorry, new here, can't comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To expand on Alex's answer, if we demand that $(xh)(yh) \in (xy)H$ for any coset representative $h$, then this is equivalent to asking that $y^{-1}hyh \in H$ for all $h \in H$. But the conjugate $y^{-1}hy$ need not lie in $H$ if $H$ isn't normal, hence the claim that the operation in your &quot;left quotient group&quot; isn't well defined.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-14T12:34:45.183" CommentCount="8" />
  <row Id="14286" PostTypeId="2" ParentId="14279" CreationDate="2010-12-14T12:48:42.340" Score="1" ViewCount="0" Body="&lt;p&gt;$(j_1,k_1)\ne(j_2,k_2)$ means (a) $j_1\ne j_2$ or (b) $j_1=j_2$ and $k_1\ne k_2$. In each of these cases the statement follows immediately by looking at the graphs of two such $\psi$'s.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-14T12:48:42.340" CommentCount="2" />
  <row Id="14287" PostTypeId="2" ParentId="14279" CreationDate="2010-12-14T13:13:17.203" Score="0" ViewCount="0" Body="&lt;p&gt;If you are working in $L_2(0,1)$ space it is then quite easy. The scalar product then is the integral&#xA;\begin{align*}&#xA;\int_0^1\psi_{j_1,k_1}(t)\psi_{j_2,k_2}(t)dt&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite $\psi_{j,k}(t)$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\psi_{j,k}(t)=2^j1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\right\}-&#xA;2^{-j}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\right\}&#xA;\end{align*}&#xA;then&#xA;\begin{align*}&#xA;\psi_{j,k}(t)\psi_{l,m}(t)&amp;amp;=2^{j+l}1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\cap\left(\frac{m}{2^l},\frac{m+1/2}{2^l}\right)\right\}\\&#xA;&amp;amp;-2^{j+l}1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\cap\left(\frac{m+1/2}{2^l},\frac{m+1}{2^l}\right)\right\}\\&#xA;&amp;amp;-2^{j+l}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\cap\left(\frac{m}{2^l},\frac{m+1/2}{2^l}\right)\right\}\\&#xA;&amp;amp;+2^{j+l}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\cap\left(\frac{m+1/2}{2^l},\frac{m+1}{2^l}\right)\right\}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $j&amp;lt;l$, we have that $2^{l-j-1}\ge 1$ and we need to investigate the following cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$m&amp;lt;2^{l-j}k$. The product is immediately zero, since the intervals do not overlap&lt;/li&gt;&#xA;&lt;li&gt;$2^{l-j}k\le m &amp;lt;2^{l-j}k+2^{l-j-1}$. Then the 3rd and 4th terms in the sum are zero and the product is equal to $2^j\psi_{l,m}$, whose integral is zero.&lt;/li&gt;&#xA;&lt;li&gt;$2^{l-j}k+2^{l-j-1}\le m&amp;lt;2^{l-j}(k+1)$. Then the 1st and 2nd terms in the sum are zero and the product is equal to $-2^{j}\psi_{l,m}$, whose integral is again zero.&lt;/li&gt;&#xA;&lt;li&gt;$m\ge 2^{l-j}(k+1)$. The product is zero, since the intervals do not overlap.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-14T13:43:31.877" LastActivityDate="2010-12-14T13:43:31.877" CommentCount="2" />
  <row Id="14288" PostTypeId="1" AcceptedAnswerId="14291" CreationDate="2010-12-14T13:17:36.510" Score="0" ViewCount="50" Body="&lt;p&gt;I need to understend how  black box optimizer work. I need a real life example of how it hlps. So what I need: What was the task, what were parameters, what was the function to minimize and how it all worked toogether. I need deteiled info on this topic, please - &lt;a href=&quot;http://en.wikipedia.org/wiki/Optimization_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; does not give lot of help...&lt;/p&gt;&#xA;" OwnerUserId="2065" LastActivityDate="2010-12-14T13:53:42.287" Title="Wanted: Examples of how  black box optimizer works - step by step?" Tags="&lt;optimization&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14289" PostTypeId="2" ParentId="14282" CreationDate="2010-12-14T13:33:39.783" Score="8" ViewCount="0" Body="&lt;p&gt;I cannot comment yet, but I thought I would elaborate on what is meant by being well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your definition of the multiplication in $G/H$, you take cosets $xH$ and $yH$ and multiply by setting thir product equal to $(xy)H$. But this means that you have chosen $x$ and $y$, and there will in general be other choices of $x$ and $y$ that give the same cosets. So if you have $xH$ = $x&amp;#39;H$ and $yH$ = $y&amp;#39;H$, you want to be sure that you also have $(xy)H = (x&amp;#39;y&amp;#39;)H$, and it can be shown that this holds for all elements of $G$ iff $H$ is normal in $G$, so this is the only case where this multiplication is well-defined.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T17:49:31.490" LastActivityDate="2010-12-15T17:49:31.490" />
  <row Id="14290" PostTypeId="2" ParentId="14268" CreationDate="2010-12-14T13:43:25.170" Score="6" ViewCount="0" Body="&lt;p&gt;So, you want to compare (254)(847) and (383)(536) without actually computing the products, look at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{847}{383} \; ? \; \frac{536}{254}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying both denominators by 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{847}{766} \; ? \; \frac{536}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+\frac{81}{766} \; ? \; 1+\frac{28}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are now left with comparing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{81}{766} \; ? \; \frac{28}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the same trick as before, consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{81}{28} \; ? \; \frac{766}{508}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side is obviously bigger than $2$ while the right hand side is smaller, therefore, you can conclude that the original left hand side product was the bigger one, since none of the operations performed inverted the order of the inequalities.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-14T13:46:11.407" LastActivityDate="2010-12-14T13:46:11.407" CommentCount="4" />
  <row Id="14291" PostTypeId="2" ParentId="14288" CreationDate="2010-12-14T13:53:42.287" Score="2" ViewCount="0" Body="&lt;p&gt;Chapter 10 of &lt;a href=&quot;http://nr.com/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; has a good description of how the golden section search works in 1D, as well as description of multidimensional cases.  You could see figure 10.1.1 in the obsolete C version (available free online)&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-14T13:53:42.287" />
  <row Id="14293" PostTypeId="1" AcceptedAnswerId="14307" CreationDate="2010-12-14T14:11:16.690" Score="1" ViewCount="46" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm developing a computer program, and need an algorithm to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to select the optimum combination of numbers from a random list that add to up to a certain total (or as close to as possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;List of random numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.0,&#xA;2.1,&#xA;5.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required number = 6.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the answer would be 6.3 (from 5.3 + 1.0) I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program needs to be able to operate on much bigger number sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some maths qualifications (up to a-level), but not expert, so if you could temper your respose to my level if possible, that would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="4745" LastActivityDate="2010-12-14T16:10:17.577" Title="How to select the optimum combination of numbers from a random list that add to up to a certain total (or as close to)." Tags="&lt;computer-science&gt;&lt;random-variable&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14294" PostTypeId="2" ParentId="14293" CreationDate="2010-12-14T14:18:50.857" Score="3" ViewCount="0" Body="&lt;p&gt;This is close to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;subset sum problem&lt;/a&gt;, which is not easy.  But some approaches are in the Wikipedia article.  The section on the approximate algorithm seems quite applicable.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-14T14:18:50.857" CommentCount="3" />
  <row Id="14295" PostTypeId="2" ParentId="14263" CreationDate="2010-12-14T14:35:12.017" Score="2" ViewCount="0" Body="&lt;p&gt;Ask GAP:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gap&amp;gt; AllSmallGroups(156, IsSolvable, false);&#xA;[  ]&#xA;gap&amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This gives the list of all groups of order 156 which are not solvable.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T14:35:12.017" />
  <row Id="14296" PostTypeId="1" AcceptedAnswerId="14321" CreationDate="2010-12-14T15:14:31.947" Score="1" ViewCount="42" Body="&lt;p&gt;I came across this problem in the context of spectroscopy today.  Because of it's simplicity, I'm sure it's a question that's been posed by some mathematician ages ago, I just can't figure out how to formulate it mathematically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a mirror that forms a perfect ring and has an aperture in it with width $\delta w$.  A ray of light enters the aperture at angle $\theta$ relative to the tangent at the aperture.  There are several questions that I have regarding this system, some of which would be of practical use and some of which are mere curiosity:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this a chaotic system?  I know a little about chaotic systems from my undergrad math major, and this looks like one to me, but I don't know where to start working on it if it is.&lt;/li&gt;&#xA;&lt;li&gt;How can this problem be formulated mathematically?  I would assume that there is some sort of iterated function approach, but I can't figure out what the iterated function would be (probably because as a chemist I'm more comfortable with cartesian coordinates than polar coordinates).&lt;/li&gt;&#xA;&lt;li&gt;What is the optimal angle for achieving the most passes through the area within some $\epsilon$ of the center of the ring before exiting through the aperture, assuming (a) $\epsilon$ is small compared to $r$, the radius of the ring, or (b) $\epsilon$ is not small compared to the radius of the ring?  (This is one of the questions of practical importance)&lt;/li&gt;&#xA;&lt;li&gt;Bonus:  Suppose there is a second aperture placed at an angle $\phi$ away from the first (with $\phi$ relative to the center of the circle and to the first aperture).  Can this angle be optimized in conjunction with the variables in $\phi$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To summarize, the variables are $r$, $\theta$, $\delta w$, $\epsilon$, and (if we're lucky) $\phi$.  Assume that $\epsilon$ and $\delta w$ are given (obviously the optimal size for $\delta w$ is infinitesimal, but this isn't practical of course).  Also (for the physicists out there), assume $\delta w$ is wide enough that the beam can be treated classically, i.e. no diffraction.  Also, there needs to be a reasonable way to calculate the exit angle based on the entry angle (if this is a truly chaotic system, I realize this will be practically impossible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help, and let me know if I've left out any important details.  I'm comfortable with any math that a typical undergraduate math major should know by the time they graduate; anything more and I'll need a reference to read up on.&lt;/p&gt;&#xA;" OwnerUserId="1590" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-14T15:23:27.007" LastActivityDate="2010-12-14T19:10:04.507" Title="Ring-shaped mirror chaos points" Tags="&lt;geometry&gt;&lt;chaotic-systems&gt;&lt;dynamical-systems&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14297" PostTypeId="1" AcceptedAnswerId="14316" CreationDate="2010-12-14T15:34:15.323" Score="7" ViewCount="176" Body="&lt;p&gt;You are allowed to roll a die up to six times. Anytime you stop, you get the dollar amount of the face value of your last roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What is the best strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my calculation, for the strategy 6,5,5,4,4, the expected value is $142/27\approx 5.26$, which I consider quite high. So this might be the best strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here,  6,5,5,4,4 means in the first roll you stop only when you get a 6; if you did not get a 6 in the first roll, then in the second roll you stop only when you roll a number 5 or higher (i.e. 5 or 6), etc.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-14T17:39:57.733" Title="Best Strategy for a die game" Tags="&lt;probability&gt;&lt;game-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="14298" PostTypeId="1" CreationDate="2010-12-14T15:38:22.707" Score="2" ViewCount="54" Body="&lt;p&gt;We solve the pell equation using the continued fraction for square root of 2. What equations can we solve using the continued fraction of cube roots (and other numbers too)?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-14T16:27:43.370" Title="Solving equations using continued fractions?" Tags="&lt;diophantine-equations&gt;&lt;continued-fractions&gt;" AnswerCount="1" />
  <row Id="14299" PostTypeId="2" ParentId="14297" CreationDate="2010-12-14T15:45:59.113" Score="1" ViewCount="0" Body="&lt;p&gt;On the next to last roll, the expectation on the last roll is 3.5, so accepting 4 and rejecting 3 is clearly right.  On the third to last roll, you should calculate the expectation if you decline.  You have 1/2 chance of rolling 4 or better and accepting it, with an expectation of 5.  You have 1/2 chance of rolling 3 or worse and rejecting it, with an expectation of 3.5.  So if you decline the third to last, you expect 4.25.  So you should reject a 4 and accept a 5.  If you keep working backwards like this, you may find you should bump at least one of your 5's to a 6.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-14T15:45:59.113" />
  <row Id="14300" PostTypeId="2" ParentId="11703" CreationDate="2010-12-14T15:47:57.007" Score="0" ViewCount="0" Body="&lt;p&gt;Since $$\frac1{1-x}=1+x+x^2+x^3+\cdots,$$ we have&#xA;$$&#xD;&#xA;\frac{1-x^{n+1}}{1-x}=(1+x+x^2+x^3+\cdots)-(x^{n+1}+x^{n+2}+x^{n+3}+x^{n+4}+\cdots)&#xD;&#xA;$$&#xD;&#xA;And on the right hand side every thing cancels except $1+x+x^2+\cdots+x^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This argument is probably circular! :) )&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T15:47:57.007" CommentCount="1" />
  <row Id="14301" PostTypeId="2" ParentId="14296" CreationDate="2010-12-14T15:54:28.700" Score="1" ViewCount="0" Body="&lt;p&gt;I'm not an expert, but circular billiards is a special case of elliptic billiards, and what you want to know can probably be found in &lt;a href=&quot;http://www.math.psu.edu/tabachni/Books/billiardsgeometry.pdf&quot; rel=&quot;nofollow&quot;&gt;Billiards and Geometry (pdf file)&lt;/a&gt; by Serge Tabachnikov.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-14T15:54:28.700" />
  <row Id="14302" PostTypeId="2" ParentId="14297" CreationDate="2010-12-14T15:54:48.510" Score="9" ViewCount="0" Body="&lt;p&gt;Just work backwards. At each stage, you accept a roll that is &gt;= the expected gain from the later stages:&lt;br&gt;&#xA;Expected gain from 6th roll: 7/2&lt;br&gt;&#xA;Therefore strategy for 5th roll is: accept if &gt;= 4&lt;br&gt;&#xA;Expected gain from 5th roll: (6 + 5 + 4)/6 + (7/2)(3/6) = 17/4&lt;br&gt;&#xA;Therefore strategy for 4th roll is: accept if &gt;= 5&lt;br&gt;&#xA;Expected gain from 4th roll: (6 + 5)/6 + (17/4)(4/6) = 14/3&lt;br&gt;&#xA;Therefore strategy for 3rd roll is: accept if &gt;= 5&lt;br&gt;&#xA;Expected gain from 3rd roll: (6 + 5)/6 + (14/3)(4/6) = 89/18&lt;br&gt;&#xA;Therefore strategy for 2nd roll is: accept if &gt;= 5&lt;br&gt;&#xA;Expected gain from 2nd roll: (6 + 5)/6 + (89/18)(4/6) = 277/54&lt;br&gt;&#xA;Therefore strategy for 1st roll is: accept only if 6&lt;br&gt;&#xA;Expected gain from 1st roll: 6/6 + (277/54)(5/6) = 1709/324  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your strategy is 6,5,5,5,4 for an expectation of $5.27469...&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-14T15:54:48.510" CommentCount="4" />
  <row Id="14303" PostTypeId="2" ParentId="11703" CreationDate="2010-12-14T15:58:54.113" Score="1" ViewCount="0" Body="&lt;p&gt;Since $1-x^{n+1}$ has $1$ as a root, the quotient $\frac{1-x^{n+1}}{1-x}$ is a polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbb F_q$ is a finite field with $q$ elements and $V$ is a $\mathbb F_q$-vector space of dimension $n+1$, then $\frac{1-q^{n+1}}{1-q}=|P(V)|$ is the cardinal of the projective space attached to $V$. Now $P(V)$ can be described as a disjoint union $$P(V)=\mathbb A^0\sqcup\mathbb A^1\sqcup \mathbb A^2\sqcup\cdots\sqcup\mathbb A^n$$ where $\mathbb A^k$ is, for each $k$, an affine space of dimension $k$ over $\mathbb F_q$ (which is a complicated way of saying, as far as our purposes go, a vector space over $\mathbb F_q$ of dimension $k$) Since $|\mathbb A^k|=q^k$, we find that&#xA;$$\frac{1-q^{n+1}}{1-q}=1+q+q^2+q^3+\cdots+q^n$$&#xD;&#xA;for all numbers $q$ which are powers of prime numbers. It follows that&#xA;$$\frac{1-x^{n+1}}{1-x}=1+x+x^2+x^3+\cdots+x^n$$&#xD;&#xA;as &lt;em&gt;polynomials&lt;/em&gt;, because the equality holds for infinitely many values of $x$ (and we are working over $\mathbb Z$...)&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T15:58:54.113" CommentCount="3" />
  <row Id="14304" PostTypeId="2" ParentId="14268" CreationDate="2010-12-14T16:01:50.173" Score="1" ViewCount="0" Body="&lt;p&gt;In the worst case, where the comparison is close, you will always end up having to do as much work as multiplication. The performance gains from any non-worst-case improvements would have to be pretty good to make the programming effort worthwhile. Why do you need this?&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-14T16:01:50.173" />
  <row Id="14305" PostTypeId="2" ParentId="14270" CreationDate="2010-12-14T16:04:20.793" Score="3" ViewCount="0" Body="&lt;p&gt;This is true if $R$ is a domain. More generally this is true if $a$ or $b$ is not a zero divisor in $R$. Suppose, $a$ is not a zero divisor and there exist $r,s\in R$ such that $a=rb$ and $b=sa$, then $a=rsa$, so $a(1-rs)=0$. Since $a$ is not a zero divisor, $1-rs=0$, so $rs=1$. So, $r,s$ are units. So $a$ and $b$ are associates.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-14T16:04:20.793" CommentCount="1" />
  <row Id="14306" PostTypeId="1" AcceptedAnswerId="14336" CreationDate="2010-12-14T16:07:00.133" Score="0" ViewCount="76" Body="&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to prove Simpson's rule including the error with the help of the integral remainder. However, I have practically no idea how to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f: [a,b] \rightarrow \mathbb{R}$ be continuously differentiable four times. Let $n \in 2\mathbb{N}$ and $h = (b-a)/n$. Let $y_k = f(a+kh)$ for $0 \leq k \leq n$. Show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\int_a^b f(x) \mathrm dx - \frac{h}{3} [(y_0 + y_n) + 4(y_1+y_3+...+y_{n-1})+2(y_2+y_4+...+y_{n-2})]\right|$$&#xD;&#xA;$$\leq \frac{1}{180} \max_{a \leq x \leq b} |f^{(4)}(x)|(b-a)h^4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I'd do a Taylor expansion for $\int_a^b f(x) \mathrm dx$ and then apply the integral remainder, but then all terms vanish and I get $0=0$ which is not very helpful...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only need a good advice how to start! Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-14T23:01:21.270" Title="Prove Simpson's rule (including error) using the integral remainder" Tags="&lt;analysis&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14307" PostTypeId="2" ParentId="14293" CreationDate="2010-12-14T16:10:17.577" Score="0" ViewCount="0" Body="&lt;p&gt;As Ross Millikan notes, this is a standard &lt;a href=&quot;http://en.wikipedia.org/wiki/Subset_sum_problem&quot; rel=&quot;nofollow&quot;&gt;Subset Sum&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_problem&quot; rel=&quot;nofollow&quot;&gt;Number Partition Problem&lt;/a&gt; which are both &lt;a href=&quot;http://en.wikipedia.org/wiki/NP-complete&quot; rel=&quot;nofollow&quot;&gt;NP-Complete&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your Subset Sum instance is $(a_0, a_1, \dots, a_{n-1})$, where each $a_k \in \mathbb{Z}$, and the numbers in the list are small compared with the list size (i.e. $\log_2 ( \max(a_k) ) &amp;lt;&amp;lt; n$), then you can either use some brute force search, as in the Complete Karmarkar Karp algorithm (a review of which can be found &lt;a href=&quot;http://arxiv.org/pdf/0802.4040v2&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) or some dynamic programming method (&lt;a href=&quot;http://www.cse.ohio-state.edu/~tamaldey/course/794/subset-sum.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always create an instance of Subset Sum/Number Partition out of a list that has it's numbers drawn from the reals by multiplying each element in the list by some large number to make it integer (for example, multiplying each number in your above list by 10 would turn this into a standard Subset Sum/Number Partition Problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being NP-Complete means, in general, this problem is probably hard, in that there is nothing better than an exponential/brute force approach.  Not all brute force algorithms are created equal, though, and depending on what your needs are, you might find a better algorithm suited to the particular type of Subset Sum/Number Partition Problem instances you're creating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some researchers have had some luck &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.57.3331&amp;amp;rep=rep1&amp;amp;type=ps&quot; rel=&quot;nofollow&quot;&gt;with lattice reduction techniques&lt;/a&gt;, so that's another avenue of attack, though lattice reduction techniques (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm&quot; rel=&quot;nofollow&quot;&gt;LLL&lt;/a&gt; and &lt;a href=&quot;http://arminstraub.com/files/pslq.pdfhttp://arminstraub.com/files/pslq.pdf&quot; rel=&quot;nofollow&quot;&gt;PSLQ&lt;/a&gt; for example) are a bit advanced and, from my understanding, don't fare all that well for Subset Sum/Number Partition Problem instances past a certain size.&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2010-12-14T16:10:17.577" CommentCount="1" />
  <row Id="14308" PostTypeId="2" ParentId="14298" CreationDate="2010-12-14T16:27:43.370" Score="4" ViewCount="0" Body="&lt;p&gt;Voronoi developed a generalized continued fraction algorithm for computing  units in cubic fields. You can find a beautiful geometric interpretation of Voronoi's algorithm in Delone and Faddeev's: &lt;em&gt;Theory of Irrationalities of Third Degree&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-14T16:27:43.370" CommentCount="2" />
  <row Id="14309" PostTypeId="2" ParentId="14253" CreationDate="2010-12-14T16:30:32.727" Score="0" ViewCount="0" Body="&lt;p&gt;If $X$ is a closed subspace of $\ell_1^4$, then it is true if the latter has the metric extension property.  If you are working over the reals, then Google :&quot;binary intersection property Nachbin&quot;. I think Grothendieck proved that if a space has the metric extension property, then it is either finite dimensional or non-separable. I hope this helps!.&lt;/p&gt;&#xA;" OwnerUserId="4749" LastActivityDate="2010-12-14T16:30:32.727" CommentCount="1" />
  <row Id="14310" PostTypeId="2" ParentId="14270" CreationDate="2010-12-14T16:43:09.417" Score="2" ViewCount="0" Body="&lt;p&gt;Related questions arise frequently on sci.math, where I &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/d9512c74d9ce4acf&quot; rel=&quot;nofollow&quot;&gt;often&lt;/a&gt; point out the following standard reference on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware that this equivalence, i.e. $\rm\ aR = bR \iff a/b\ $ is a unit in $\rm\:R\:$, &#xA;generally fails when $\rm\:R\:$ has zero-divisors, so that there are at least a few different notions of &quot;associate&quot; that are of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When are Associates Unit Multiples?&lt;br&gt;&#xA;D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles&lt;br&gt;&#xA;Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828. &#xA;&lt;a href=&quot;http://math.la.asu.edu/~rmmc/rmj/vol34-3/andepage1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://math.la.asu.edu/~rmmc/rmj/vol34-3/andepage1.pdf&lt;/a&gt; &#xA;&lt;a href=&quot;http://projecteuclid.org/handle/euclid.rmjm/1181069828&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/handle/euclid.rmjm/1181069828&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-14T16:43:09.417" CommentCount="1" />
  <row Id="14311" PostTypeId="1" CreationDate="2010-12-14T16:59:42.010" Score="5" ViewCount="407" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a CFG over large alphabet describe a language, where each terminal appears even number of times?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If yes, would the Chomsky Normal Form be polynomial in |Σ| ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;EDIT: What about a language where each terminal appears 0 or 2 times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Solving this may give interesting extension to &lt;a href=&quot;http://math.stackexchange.com/questions/14506/restricted-read-twice-bdds-and-context-free-grammars&quot;&gt;&quot;Restricted read twice BDDs and context free grammars&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a context-free grammar (CFG) for the language L of all&#xA;words such that each terminal in a word occurs even number of times over a possibly large alphabet Σ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My long aproach is (the only nonterminal is S):&#xA;&lt;strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;S ⟶ ε | SS&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x ∈ Σ : S ⟶ xSx&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;x,y ∈ Σ : S ⟶ xxSyy | yySxx | xySxy | xySyx | yxSyx | yxSxy&#xA;  &lt;/strike&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;Second try, incremental approach.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;S ⟶ ε | SS&lt;br&gt;&#xA;  Terminal productions (suboptimal): x,y ∈ Σ : S ⟶ xx | yy |xxyy | yyxx | xyxy | xyyx | yxyx | yxxy&lt;br&gt;&#xA;  Incremental: x ∈ Σ : S ⟶ SxSxS &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the way, the homework is over now.&#xA;&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Technically&lt;/em&gt; it wasn't a homework assigned to me. I don't know the answer and thought the question was easy. &#xA;Best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I didn't award the bounty :-)&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-28T10:23:38.807" LastActivityDate="2010-12-31T21:58:47.790" Title="Can a polynomial size CFG over large alphabet describe a language, where each terminal appears even number of times?" Tags="&lt;homework&gt;&lt;computer-science&gt;&lt;computability&gt;&lt;formal-systems&gt;" AnswerCount="5" CommentCount="21" FavoriteCount="2" />
  <row Id="14312" PostTypeId="2" ParentId="14276" CreationDate="2010-12-14T17:06:21.730" Score="3" ViewCount="0" Body="&lt;p&gt;What equivalence relation are you using to define motives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finest that people normally consider is rational equivalence, in which case&#xA;two morphisms whose graphs are rationally equivalent (which is to say, roughly speaking, two morphisms which can be deformed one into the other via a family parameterized by $\mathbb P^1$) will give the same morphism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trivial, but concrete, example: all constant morphisms from any variety to&#xA;$\mathbb P^1$ will be identified in the category of motives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a coarser relation (homological equivalence or numerical equivalence, which are conjecturally the same, and lead to what are called &lt;em&gt;Grothendieck motives&lt;/em&gt;) than even more morphisms of varieties will be identified in the category of motives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, these functors are certainly not full, since one takes a Karoubian closure as part of the construction of the category of motives, and then furthermore inverts the Lefschetz motive.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-14T17:06:21.730" />
  <row Id="14313" PostTypeId="1" CreationDate="2010-12-14T17:16:32.840" Score="0" ViewCount="161" Body="&lt;p&gt;Let $R$ be a ring, $A$ is an ideal of $R$, and assume that $A\subseteq J(R)$.&#xA;Let $r\longmapsto \overline{r}$  denote the coset map $R\rightarrow \overline{R}=R/A$. The following are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $A$ is lifting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If $\overline{R}=\overline{K_1}\oplus...\oplus \overline{K_n}$ where the $\overline{K_i}$ are left ideals of $\overline{R}$, there exists a decomposable $R=L_1\oplus...\oplus L_n$ where the $L_i$ are left ideals of $R$ such that $\overline{L_i}=\overline{K_i}$ for each $i=1,2,...,n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Same as $(2)$ except $n=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any idea or any source for finding a prove?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;em&gt;Edit by Arturo Magidin:&lt;/em&gt; Above is the original question. The new phrasing is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $R$ be a ring, $A$ is an ideal of $R$, and assume that $A\subseteq J(R)$. Let $r\longmapsto \overline{r}$  denote the coset map $R\rightarrow \overline{R}=R/A$&#xD;&#xA;If $A$ is lifting, for $\overline{R}=\overline{K_1}\oplus \overline{K_2}$ where the $\overline{K_i}$ are left ideals of $\overline{R}$, there exists a decomposable $R=L_1\oplus L_2$ where the $L_i$ are left ideals of $R$ such that $\overline{L_i}=\overline{K_i}$ for each $i=1,2$.&lt;/p&gt;&#xA;" OwnerUserId="4688" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T17:24:42.533" LastActivityDate="2010-12-16T17:24:42.533" Title="Ring Theory (Lift)" Tags="&lt;ring-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14314" PostTypeId="2" ParentId="14268" CreationDate="2010-12-14T17:16:59.610" Score="11" ViewCount="0" Body="&lt;p&gt;Generally such comparisons can be done efficiently via continued fractions, e.g.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle a = \frac{847}{383}\ =\ 2 + \cfrac{1}{4 + \cfrac{1}{1 + \cdots}}\ \ \Rightarrow\ \ 2+\frac{1}{4+1} &amp;lt; a &amp;lt; 2+\frac{1}4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle b = \frac{536}{254}\ =\ 2 + \cfrac{1}{9 + \cdots}\ \ \Rightarrow\ \ 2 &amp;lt; b &amp;lt; 2 + \frac{1}9 &amp;lt; 2+\frac{1}5 &amp;lt; a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comparison of the continued fraction coefficients can be done in parallel with the computation of the continued fraction. Namely, compare the integer parts. If they are unequal then that will determine the inequality. Otherwise, recurse on the (inverses of) the fractional parts (and note that inversion reverses the inequality). For the above example this yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm\displaystyle\ \frac{847}{383} &amp;gt; \frac{536}{254}\ \iff\  \frac{81}{383}&amp;gt;\frac{28}{254}\ \iff\ \frac{254}{28}&amp;gt;\frac{383}{81}\ \Leftarrow\ \ 9 &amp;gt; 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words: to test if $\rm\:847/383 &amp;gt; 536/254\: $ we first compare their integer parts (floor). They both have integer part $\:2\:$ so we subtract $\:2\:$ from both and reduce to comparing their fractional parts $\rm\  81/383,\ \ 28/254\:.$ To do this we invert them and recurse. But since  inversion reverses inequalities $\rm\ x &amp;lt; y\ \iff\ 1/y &amp;lt; 1/x\ $ (for $\rm\:xy&gt;0\:$),  the equivalent inequality to test is if $\rm\ 254/28 &amp;gt; 383/81\:.\ $ Comparing their integer parts $\rm\:m,\:n\:$ we see since $\rm\ m &amp;gt; 5 &amp;gt; n\:$ so the inequality is true. This leads to the following simple algorithm that essentially compares any two real numbers via the lex order of their continued fraction coefficients (note: it will not terminate if given two equal reals with infinite continued fraction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm compare\_reals\:(A,\: B)\ := \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\color{blue}{\ // \ computes\ \  sgn(A - B) }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ let\ [\ n_1\leftarrow \lfloor A\rfloor\:;\ \ \ n_2\leftarrow \lfloor B\rfloor\ ] \ \quad\quad\quad\quad\quad\quad\ \color{blue}{\ //\ compare\  integer\ parts}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad if\ \ n_1 \ne n_2\ \ then\ \ return\ \ sgn(n_1 - n_2)\:;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad let\ [\ a \leftarrow A - n_1\:;\ \ \ b \leftarrow B - n_2\ ] \quad\quad\quad\quad \color{blue}{//\ compute\ fractional\ parts\ \ a,\: b }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad if\ \ a\:b=0\ \ then\ \ return\ \ sgn(a-b)\:;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad compare\_reals(b^{-1}\:, a^{-1})\:;\ \quad\quad\quad\quad\quad\quad\color{blue}{\ //\ \text{recurse on inverses of fractional parts}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently one can employ Farey fractions and mediants. Generally such approaches will be quite efficient due to the best approximations properties of the algorithm. For a nice example see my &lt;a href=&quot;http://groups.google.com/group/sci.math/msg/51d8687167920ebd?hl=en&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; using continued fractions to solve the old chestnut of comparing $\ 7^\sqrt 8\ $ to $\ 8^\sqrt 7\ $ and see also &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/2182a45f22ed5d48&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; where some folks struggled to prove this by calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-30T20:21:31.627" LastActivityDate="2010-12-30T20:21:31.627" CommentCount="2" />
  <row Id="14315" PostTypeId="2" ParentId="14282" CreationDate="2010-12-14T17:26:47.587" Score="15" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; So, what is the problem? Let's look at the simplest example of a non-normal subgroup. Take $G=S_3$, and $H=\{e,(1,2)\}$. If we compose permutations right to left, the left cosets of $H$ in $G$ are:&#xA;\begin{align*}&#xA;eH = (1,2)H &amp;amp;= \{ e, (1,2)\};\\&#xA;(1,2,3)H = (1,3)H &amp;amp;= \{(1,2,3), (1,3)\};\\&#xA;(1,3,2)H = (2,3)H &amp;amp;= \{(1,3,2), (2,3)\}.&#xA;\end{align*}&#xA;If we try multiplying cosets term-by-term, we run into problems. Multiplying by $eH$ is not a problem, but take $(1,2,3)H$ multiplied by itself. The products are:&#xA;$$\{ (1,2,3)(1,2,3), (1,2,3)(1,3), (1,3)(1,2,3), (1,3)(1,3)\} = \{(1,3,2), (2,3), (1,2), e\}$$&#xD;&#xA;which is not a coset. If we multiply using representatives, as in the original question,we also run into problems: if we multiply $(1,2,3)H\cdot(1,3,2)H$ as $(1,2,3)(1,3,2)H$, we get $eH$. But $(1,2,3)H = (1,3)H$, and $(1,3,2)H = (2,3)H$, and if we multiply them by looking at these alternative representatives/names, we get $(1,3)H\cdot (2,3)H = (1,3)(2,3)H = (1,3,2)H\neq eH$. That is, the multiplication rule depends on the &lt;em&gt;name&lt;/em&gt; we give the coset, rather than on what the coset &lt;em&gt;is&lt;/em&gt;. This means that the rule is not well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can let $G$ act on the set of left cosets of $H$, by $g(xH) = gxH$; this is not a group, but a group action. We can also, separtely, put group structures on the set of cosets of $H$ in $G$, since it will be a set, and every nonempty set can be given a group structure (assuming the Axiom of Choice; in fact, equivalent to the Axiom of Choice). But that structure will generally have nothing to do with the operation in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want an operation on the set of cosets, and have the operation in the cosets to be induced by the operation in $G$, then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $G$ be a group and $H$ a subgroup. Let $G/H = \{ gH\mid g\in G\}$ be the set of left cosets of $H$ in $G$. The operation&#xA;$$\cdot \colon G/H \times G/H \to G/H\qquad\text{defined by}\qquad (xH)\cdot(yH) = (xy)H$$&#xD;&#xA;is &lt;strong&gt;well defined&lt;/strong&gt; (that is, if $xH = x&amp;#39;H$ and $yH=y&amp;#39;H$ then $xyH = x'y'H$) if and only if $H$ is normal in $G$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Suppose $H$ is normal. Then $gH=Hg$ for every $g\in G$. If $xH=x&amp;#39;H$ and $yH=y&amp;#39;H$, then &#xA;$$xyH = x(yH) = x(Hy) = x(Hy&amp;#39;) = (xH)y&amp;#39; = (x&amp;#39;H)y&amp;#39; = x&amp;#39;(Hy&amp;#39;) = x&amp;#39;(y&amp;#39;H) = x&amp;#39;y&amp;#39;H$$&#xD;&#xA;so the operation is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose the operation is well defined so that whenever $xH=x&amp;#39;H$ and $yH=y&amp;#39;H$, then $xyH = x&amp;#39;y&amp;#39;H$. We want to show that $g^{-1}Hg\subseteq H$ for all $g\in G$. For each $h\in H$, since $hH = eH$, we have $eHgH = egH=gH$ is equal to $hHgH = hgH$. So $gH = hgH$, hence $H=g^{-1}hgH$, so $g^{-1}hg\in H$ for each $h\in H$. Thus, $g^{-1}Hg\subseteq H$. This holds for any $g\in G$, so $g^{-1}Hg\subseteq H$ for all $g\in G$, hence $H\triangleleft G$, as claimed. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I'm going to hijack your question a bit and go on a long excursion to a closely related subject which is not necessary to answer your question, but which I think is interesting. (Sorry!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really connected to the notion of a &lt;em&gt;congruence&lt;/em&gt; on an algebra (in the sense of universal algebra). A subgroup $H$ defines an equivalence relation on $G$ by $x\sim y$ if and only if $xH = yH$ (it also defines the equivalence relation of right congruence, but ignore that for now). Being an equivalence relation, it corresponds to a subset $\Phi$ of $G\times G$, with $(x,y)\in\Phi$ if and only if $x\sim y$, if and only if $xH=yH$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $G\times G$ also has a natural group structure inherited from $G$. Since $\Phi$ is a subset of the group $G\times G$, it makes sense to ask whether $\Phi$ is a subgroup of $G\times G$ or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;$\Phi$ is a subgroup of $G\times G$ if and only if $H$ is normal in $G$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Note that $\Phi$ always contains $(e,e)$ (in fact, the entire diagonal). Assume first that $H$ is normal. If $(x,y)\in \Phi$, then $xH=yH$. Hence, $Hx = Hy$, and taking inverses of each element we get $x^{-1}H = y^{-1}H$, so $(x^{-1},y^{-1})\in\Phi$; thus, $\Phi$ is closed under inverses. If $(x,y),(z,w)\in\Phi$, then $xH=yH$, $zH=wH$, so by what we proved above we have $xzH = ywH$, so $(xz,yw)=(x,y)(z,w)\in\Phi$. Thus, $\Phi$ is closed under products, so $\Phi$ is a subgroup of $G\times G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $\Phi$ is a subgroup. If $xH=yH$ and $zH=wH$, then $(x,y),(z,w)\in \Phi$, hence $(x,y)(z,w)=(xz,yw)\in\Phi$; thus, $xzH = ywH$. This proves that the operation $xH\cdot zH \mapsto xzH$ is well-defined, so by the theorem above $H$ is a normal subgroup of $G$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;Added.&lt;/strong&gt; (Stuff between the two horizontal lines was added in this edit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, how about this? Can we define &lt;em&gt;some&lt;/em&gt; equivalence relation $\sim$ on a group $G$ so that we can multiply equivalence classes through representatives, have it be well-defined, but not have $\sim$ &quot;come from&quot; a normal subgroup?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: suppose $\sim$ is an equivalence relation on the group $G$ (&lt;em&gt;any&lt;/em&gt; equivalence relation; all we require is that it be reflexive, symmetric, and transitive on the underlying set of $G$), and let $[x]$ denote the equivalence class of $x$ for each $x$ in $G$. We want to define a multiplication on the quotient set $G/\sim$ by $[x][y] = [xy]$. When is this well defined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $G$ be a group, and let $\sim$ be an equivalence relation on $G$. Then the multiplication on $G/\sim$ defined by $[x][y]=[xy]$ is well defined if and only if $\sim$ is a subgroup of $G\times G$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If $\sim$ is a subgroup of $G\times G$ and $[x]=[z]$, $[y]=[w]$, then $(x,z),(y,w)\in \sim$, so $(x,z)(y,w) = (xy,zw)\in \sim$, hence $[xy]=[zw]$, so multiplication is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose the multiplication is well defined. Since $\sim$ is an equivalence relation, $(e,e)\in\sim$. Also, if $(x,z), (y,w)\in\sim$, then $[x]=[z]$, $[y]=[w]$, so $[x][y]=[xy]$ equals $[z][w]=[zw]$, so $(xy,zw)=(x,y)(z,w)\in\sim$. Thus, $\sim$ is a submonoid of $G\times G$. It only remains to show that $\sim$ is closed under inverses. Suppose that $(x,y)\in\sim$. Multiplying by $(x^{-1},x^{-1})$ on the left and $(y^{-1},y^{-1})$ on the right (both elements of $\sim$) we have $(y^{-1},x^{-1})\in\sim$. Since $\sim$ is an equivalence relation, it is reflexive, so $(x^{-1},y^{-1})\in\sim$. Thus, $\sim$ is a subgroup of $G\times G$, as claimed. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so we can define the multiplication on the quotient set if and only if the equivalence relation is a subgroup of $G\times G$. Do all such equivalence relations &quot;come&quot; from subgroups? Yes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $\sim$ be an equivalence relation on $G$ that is a subgroup of $G\times G$, and let $H=\{g\in G\mid g\sim e\}$. Then $H$ is a normal subgroup of $G$, and $x\sim y$ if and only if $xH=yH$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; We can identify $G$ with the subgroup $G\times\{e\}$ of $G\times G$; then $H$ corresponds to the intersection of the subgroup $\sim$ and the subgroup $G\times\{e\}$, hence is a subgroup of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\sim$ is in fact left congruence modulo $H$, simply note that $xH=yH$ if and only if $y^{-1}x\in H$, if and only if $(y^{-1}x,e)\in \sim$, if and only if $(y,y)(y^{-1}x,e)\in\sim$, if and only if $(x,y)\in \sim$, if and only if $x\sim y$. That $H$ is normal now follows because multiplication of cosets (equivalently, of equivalence classes modulo $\sim$) is well defined. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;When we generalize this to an arbitrary &lt;em&gt;algebra&lt;/em&gt; (in the sense of universal algebra; examples include magmas, monoids, semigroups, groups, rings, $K$-algebras, lattices, etc), we get the notion of &lt;em&gt;congruence&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition.&lt;/strong&gt; Let $A$ be an &lt;em&gt;algebra&lt;/em&gt; (in the sense of universal algebra). A &lt;em&gt;congruence on $A$&lt;/em&gt; is an equivalence relation $\Phi$ on $A$ such that $\Phi$ is a subalgebra of $A\times A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; &lt;em&gt;Let $A$ and $B$ be algebras of the same type, and let $f\colon A\to B$ be a homomorphism; define $\Phi_f = \{(x,y)\in A\times A\mid f(x)=f(y)\}$. Then $\Phi_f$ is a congruence on $A$. Conversely, if $\Psi$ is a congruence on $A$, then there exists an algebra $\mathcal{B}$ and a homomorphism $\pi\colon A\to \mathcal{B}$ such that $\Psi=\Phi_{\pi}$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Special case.&lt;/em&gt; Let me do the proof with semigroups. First, let $f\colon A\to B$ be a semigroup homomorphism; then $\Phi_f$ is an equivalence relation on $A$. To show it is a subsemigroup of $A\times A$, suppose $(x,y),(z,w)\in \Phi_f$. Then $f(xz) = f(x)f(z) = f(y)f(w) = f(yw)$, so $(xz,yw)\in\Phi_f$. Thus, $\Phi_f$ is a subsemigroup of $A\times A$, hence a congruence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $\Psi$ is a congruence. Let $\mathcal{B}=A/\Psi$, the set of equivalence classes of $\Psi$ in $A$. Let $[x]$ denote the class of $x$. We define a multiplication on $\mathcal{B}$ by $[a][b]=[ab]$. We must prove that this is well defined: suppose $[x]=[y]$ and $[z]=[w]$. Then $(x,y),(z,w)\in\Psi$, so $(xz,yw)\in\Psi$ (since $\Psi$ is a congruence). Therefore, $[xz]=[yw]$, so multiplication is well-defined. It is now easy to verify that multiplication is associative:&#xA;$$[x]([y][z]) = [x][yz] = [x(yz)]=[(xy)z] = [xy][z] = ([x][y])[z].$$&#xD;&#xA;Thus, $\mathcal{B}$ is a semigroup. The map $\pi\colon A\to \mathcal{B}$ given by $\pi(a)=[a]$ is easily verified to be a homomorphism, and $\Phi_{\pi}=\Psi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof for an arbitrary algebra with operations $f_1,\ldots,f_k$ of arities $a_1,\ldots,a_k$ (or even an infinite number of operations, each with finite arities) is essentially the same.&#xA;&lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can then prove the Isomorphism Theorems for &quot;congruences&quot; in exactly the analogous manner as for groups and rings, for any class of algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Groups and Rings, we can &quot;get away&quot; with not studying congruences because if $\Phi$ is a congruence on $G$, then we can let $N = \{ g\in G\mid (g,e)\in\Phi\}$, and it turns out that $\Phi$ is the equivalence relation induced by $N$. Essentially, a congruence tells you that certain things are equal, but in groups (and rings), instead of saying &quot;$x$ is equal to $y$&quot;, we can say &quot;$xy^{-1}$ is the identity&quot;. The normal subgroup (resp. the ideal) is the set of elements equivalent to the identity. In more general classes, such as semigroups, we don't have the luxury of expressing any identity in the form $w=e$, so we must actually keep track of both sides of the equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all part of &lt;em&gt;Universal Algebra&lt;/em&gt;. I cannot recommend &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/&quot; rel=&quot;nofollow&quot;&gt;George Bergman's book&lt;/a&gt; highly enough if you want a great introduction to the subject (with a Category Theory flavor).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T18:18:38.393" LastActivityDate="2010-12-16T18:18:38.393" CommunityOwnedDate="2010-12-16T18:18:38.393" CommentCount="4" />
  <row Id="14316" PostTypeId="2" ParentId="14297" CreationDate="2010-12-14T17:31:08.380" Score="3" ViewCount="0" Body="&lt;p&gt;Let $X_n$ be your winnings in a game of length $n$ (in your case $n = 6$), if you are playing optimally.  Here, &quot;optimally&quot; means that at roll $m$, you will accept if the value is greater than $\mathbb{E} X_{n-m}$, which is your expected winnings if you continued to play with this strategy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X \sim \mathrm{Unif}(1,2,3,4,5,6) $ (you can also insert any distribution you like here).  Then $X_n$ can be defined as $X_1 = X$ and for $n \geq 2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ X_n = \begin{cases}  X_{n-1}, \quad \mathrm{if} \quad X &amp;lt; \mathbb{E}X_{n-1} \\&#xD;&#xA;X \quad \mathrm{if} \quad X \geq \mathbb{E}X_{n-1} \end{cases} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your decisions can be determined by computing $\mathbb{E} X_n$ for each $n$ recursively.  For the dice case,  $\mathbb{E} X_1 = \mathbb{E}X = 7/2$ (meaning on the fifth roll, accept if you get &gt;7/2, or 4,5 or 6), and so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{E} X_2 = \mathbb{E} X_1 \mathrm{P}[X = 1,2,3] + \mathbb{E} [X | X \geq 4] \mathrm{P}[X = 4,5,6]$$&#xD;&#xA;$$ = (7/2)(3/6) + (4 + 5  + 6)/3 (1/2) = 17/4 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on the fourth roll, accept if you get &gt; 17/4, or 5 or 6,  and so on (you need to round the answer up at each step, which makes it hard to give a closed form for $\mathbb{E} X_n$ unfortunately).  &lt;/p&gt;&#xA;" OwnerUserId="2515" LastEditorUserId="2515" LastEditorDisplayName="" LastEditDate="2010-12-14T17:39:57.733" LastActivityDate="2010-12-14T17:39:57.733" />
  <row Id="14317" PostTypeId="2" ParentId="14313" CreationDate="2010-12-14T18:03:02.987" Score="3" ViewCount="0" Body="&lt;p&gt;Suppose you can lift idempotents modulo $A$. If $\bar R=I_1\oplus I_2$ as a left $\bar R$-module, there is an idempotent $e\in\operatorname{End}_{\bar R}(\bar R)=\bar R$, given by projecting onto $I_1$, and we have $I_1=\bar R\cdot e$ and $I_2=\bar R\cdot(1-e)$. Lift $e$ to an idempotent $\tilde e\in R$, and let $J_1=R\cdot \tilde e$ and $J_2=R\cdot(1-\tilde e)$. Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give $(1)\implies(3)$. The rest of the implications are of the same spirit.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-14T18:03:02.987" />
  <row Id="14318" PostTypeId="2" ParentId="14276" CreationDate="2010-12-14T18:09:00.090" Score="1" ViewCount="0" Body="&lt;p&gt;Different varieties with the same zeta function would likely (and maybe provably in the case of varieties over a finite field) provide examples of non-faithfulness on objects.  The zetas are defined without direct reference to motives or &quot;motivic structures&quot; on cohomology.     &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-14T18:14:03.597" LastActivityDate="2010-12-14T18:14:03.597" />
  <row Id="14319" PostTypeId="2" ParentId="13937" CreationDate="2010-12-14T18:19:08.903" Score="1" ViewCount="0" Body="&lt;p&gt;Let $x_{d+1}$ denote $|a \cdot x|$, so that ${\displaystyle {1 \over |a \cdot x| x_1....x_d} = {1 \over x_1....x_{d+1}}}$. &#xA;Then by the arithmetic-geometric inequality, for any $x_1,...,x_{d+1}$ one has&#xA;$${1 \over x_1....x_{d+1}} \leq {1 \over d+1} \sum_{i = 1}^{d+1} \prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;So summing one has&#xA;$$\sum_{x_1,...,x_d}{1 \over x_1....x_{d+1}} \leq {1 \over d+1} \sum_{i = 1}^{d+1} \sum_{x_1,...,x_d}&#xD;&#xA;\prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;We examine a given term ${\displaystyle \sum_{x_1,...,x_d}\prod_{j \neq i} x_j^{-{d+1 \over d}}}$. If $i = d+1$, then one is &#xA;adding ${\displaystyle \prod_{j = 1}^d x_j^{-{d+1 \over d}}}$ over positive integers $x_1,..x_d$ and one gets a finite result since &#xA;${d + 1 \over d} &amp;gt; 1$. If $i \neq d+1$, then we sum ${\displaystyle \sum_{x_1,...,x_d}\prod_{j \neq i} x_j^{-{d+1 \over d}}}$&#xD;&#xA;in the $x_i$ variable first, fixing the other variables. Only the $x_{d+1}$ &#xA;factor depends on $x_i$, and this $x_i$ variable sum is&#xA;$$\sum_{x_i} {1 \over |a_1x_1 + ... + a_nx_n|^{d+1 \over d}} \prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;The sum is a portion of the sum of $|x|^{-{d+1 \over d}}$ over nonzero integers $x$, so the result is bounded by some fixed &#xA;constant $C$. So the overall sum is at most&#xA;$$C \sum_{x_1,...,x_d}  \prod_{j \neq i} x_j^{-{d+1 \over d}}$$&#xD;&#xA;In the sum above the $x_i$ variable is excluded.&#xA;Like above in the $i = d+1$ case, since the exponent ${d+1 \over d}$ is greater than one the sum is finite and we are done.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-15T15:09:53.383" LastActivityDate="2010-12-15T15:09:53.383" CommentCount="6" />
  <row Id="14320" PostTypeId="1" CreationDate="2010-12-14T18:25:58.163" Score="17" ViewCount="251" Body="&lt;p&gt;It is known that, for $n \geq 5$, it is possible to partition the integers $\{1, 2, \ldots, n\}$ into two disjoint subsets such that the product of the elements in one set equals the sum of the elements in the other.  One solution is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N = \{1, 2, \ldots, n\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is even, take $P = \{1, \frac{n-2}{2}, n\}$ and $N-P$ as the two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is odd, take $P = \{1, \frac{n-1}{2}, n-1\}$ and $N-P$ as the two sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this partition unique for infinitely many $n$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One might be able to prove an even stronger result, as I don't know of any other solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: The problem of proving that the partition is possible was posed several years ago as Problem 2826 in the journal &lt;em&gt;Crux Mathematicorum&lt;/em&gt;, with solutions in the &lt;a href=&quot;http://math.ca/crux/v30/n3/page177-192.pdf&quot; rel=&quot;nofollow&quot;&gt;April 2004 issue&lt;/a&gt;.  Every one of the 20 or so solvers (including me, which is why I'm interested in the question) came up with the partition given here.  The person who originally posed the problem also asked if the partition is unique for infinitely many $n$.  I don't think anyone ever submitted an answer to that latter question to &lt;em&gt;Crux&lt;/em&gt; (although I cannot verify that, as I no longer have a subscription).  I thought someone here might be able to give an answer.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-15T22:04:01.737" Title="Partitioning the integers $1$ through $n$ so that the product of the elements in one set is equal to the sum of the elements in the other" Tags="&lt;number-theory&gt;&lt;discrete-mathematics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="3" />
  <row Id="14321" PostTypeId="2" ParentId="14296" CreationDate="2010-12-14T19:10:04.507" Score="3" ViewCount="0" Body="&lt;p&gt;The dynamics are those of a repeated constant rotation, $\theta \to \theta + c$ where $c$ is the angle on the circle (completed to fill in the aperture) cut out by the line on which the light entered.  Exit will occur after $n$ reflections when $\theta_1 + nc$ is in the aperture interval, where $\theta_1$ is the point of first impact between light and mirror.  This type of dynamics is integrable, not chaotic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ring-shaped mirror is not a perfect circle but a more general convex curve this goes under the name &quot;convex billiards&quot;, or elliptical billiards in the case of an ellipse-shaped mirror.  The dynamics can be more interesting (e.g., for polygonal mirrors) in those cases and the book by Tabachnikov cited in the other answer discusses all of these more general problems. &lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-14T19:10:04.507" CommentCount="1" />
  <row Id="14322" PostTypeId="2" ParentId="14276" CreationDate="2010-12-14T19:49:21.290" Score="5" ViewCount="0" Body="&lt;p&gt;First of all, I wouldn't equate &quot;captures everything&quot; with &quot;faithful functor&quot;.  The functor from groups to sets which forgets the multiplication structure is faithful, but I wouldn't say that knowing a group has 512 elements tells you everything about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of nonisomorphic varieties with isomorphic motives: Consider $\mathbb{P}^2$ blown up at $5$ points. Blowing up four points (in general position) rigidifies the $\mathbb{P}^2$. There is then a two dimensional space of moduli for where you choose the fifth point. All of these nonisomorphic surfaces have isomorphic motives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of the functor to motives not being faithful: The automorphism group of $\mathbb{P}^1$ is $PGL_2$. All of these nontrivial automorphisms act trivially on the motive.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-14T19:49:21.290" />
  <row Id="14323" PostTypeId="2" ParentId="14209" CreationDate="2010-12-14T19:52:28.307" Score="1" ViewCount="0" Body="&lt;p&gt;We first apply the Taylor-expansion for $f(x_{k+1})$ around $x_k$:&#xA;$$f(x_{k+1})=f(x_k) + f&amp;#39;(x_k)(x_{k+1}-x_k)+R_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_2$ is the remainder. We'll take Lagrange's remainder and we get to:&#xA;$$f(x_{k+1}) = f&amp;#39;(x_k) \cdot (\frac{f(x_k)}{f&amp;#39;(x_k)} - x_k + x_{k+1}) + \frac{1}{2} f&amp;#39;&amp;#39;(x_0)(x_{k+1} - x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where per definition $\frac{f(x)}{f&amp;#39;(x)}-x_k = -x_{k+1}$ and $x_0$ is some point between $x_{k+1}$ and $x_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first term vanishes and we can write:&#xA;$$|f(x_{k+1})| \leq \frac{1}{2} \max_{a \leq x \leq b} |f&amp;#39;&amp;#39;(x)| (x_{k+1}-x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from the mean value theorem we know that:&#xA;$$\min_{a \leq x \leq b} |f&amp;#39;(x)| \leq \frac{|f(x_{k+1} - f(\xi)|}{|x_{k+1} - \xi|} = \frac{|f(x_{k+1}|}{|x_{k+1} - \xi|}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows:&#xA;$$|\xi - x_{k+1}| \leq \frac{1}{2} \frac{\max_{a \leq x \leq b} |f&amp;#39;&amp;#39;(x)|}{\min_{a \leq x \leq b} |f&amp;#39;(x)|} (x_{k+1}-x_k)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;q.e.d.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-14T19:52:28.307" CommentCount="1" />
  <row Id="14324" PostTypeId="1" CreationDate="2010-12-14T20:09:09.143" Score="0" ViewCount="28" Body="&lt;p&gt;If i have a graph in this form (adjacency matrix): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 0 1 0 0 0 1&lt;br /&gt;&#xA;0 0 0 1 0 1 0&lt;br /&gt;&#xA;1 0 0 1 0 0 0&lt;br /&gt;&#xA;0 1 1 0 0 0 0&lt;br /&gt;&#xA;0 0 0 0 0 1 1&lt;br /&gt;&#xA;0 1 0 0 1 0 0&lt;br /&gt;&#xA;1 0 0 0 1 0 0&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a 2-regular graph with 7 vertices. What are examples of vertex invariants? Usually i would look at the degrees of each vertex, but they are all the same in this case.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2010-12-14T22:10:56.043" Title="Vertex Invariants for a 2-regular Graph" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14325" PostTypeId="2" ParentId="13053" CreationDate="2010-12-14T20:09:55.413" Score="1" ViewCount="0" Body="&lt;p&gt;This seems to work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dzeta[x_] := Derivative[1, 0][Zeta][y, x]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot[{Exp[dzeta[0, x] + 1/2 Log[2 Pi]] , Gamma[x]}, {x, -3, 3}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of a function returns a function. I think that in your version, maybe Mathematica binds x to specific numerical values during the plot and then tries to take the derivative. The version above computes the derivative once and for all and then does the plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oops, Annix is right. This is not the first time I have been undone by mistaking the graph color for two graphs instead of one. Anyway, I'm sorry for the false starts, but the following code properly displays both functions and the graphs coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear[x, y, dzeta]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dzeta[y_, x_] := Derivative[1, 0][HurwitzZeta][y, x]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot[{Exp[dzeta[0, x] + 1/2 Log[2 Pi]], Gamma[x]}, {x, -3, 3}]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing &quot;Zeta&quot; to &quot;HurwitzZeta&quot; in your original expression also works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot[{Exp[&#xA;   Derivative[1, 0][HurwitzZeta][0, x] + 1/2 Log[2 Pi]]}, {x, -3, 3}]&lt;/p&gt;&#xA;" OwnerUserId="1123" LastEditorUserId="1123" LastEditorDisplayName="" LastEditDate="2010-12-15T07:10:54.083" LastActivityDate="2010-12-15T07:10:54.083" CommentCount="4" />
  <row Id="14326" PostTypeId="2" ParentId="14311" CreationDate="2010-12-14T20:14:23.573" Score="4" ViewCount="0" Body="&lt;p&gt;There is a deterministic finite state machine that accepts a word if and only if each terminal appears an even number of times. Assume that there are $k$ symbols in the alphabet. The machine has $2^k$ states, each of which is associated with a subset of the alphabet, and we identify the states with the subsets. The machine is defined so that: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The initial state corresponds to the empty set, and this is the only accepting state.&lt;/li&gt;&#xA;&lt;li&gt;The transition rules are defined as follows. From state $A$, on input $l$, if $l \in A$ then let $B = A - \{l\}$ and move to state $B$. Otherwise let $B = A \cup \{l\}$ and move to state $B$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The idea is that a state $B$ corresponds to the set of terminals that have appeared an odd number of times so far. Being in state $\emptyset$ corresponds to every terminal appearing an even number of times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the language is accepted by a DFSM, the language is regular.  This means that there is, trivially, a CFG that accepts the language. The grammar has one nonterminal for each state of the machine, and the rules of the grammar match up with transitions of the machine. I don't know yet whether there is a context free grammar for this language that is polynomial size compared to the size of the alphabet, though. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-14T21:20:47.177" LastActivityDate="2010-12-14T21:20:47.177" CommentCount="10" />
  <row Id="14327" PostTypeId="1" AcceptedAnswerId="14468" CreationDate="2010-12-14T20:31:12.427" Score="7" ViewCount="126" Body="&lt;p&gt;Consider the game described &lt;a href=&quot;http://math.stackexchange.com/questions/14297/best-strategy-for-a-die-game&quot;&gt;here&lt;/a&gt;, but for a sequence $X_1,\ldots,X_n$ of i.i.d. uniform rv's on $\lbrace 1,\ldots,n \rbrace$ (in the original game $n=6$). Using the original notation, let $a_n$ denote the first element in the best strategy $a_n,\ldots,a_2$. We saw that for $n=6$, $a_n = n$. Can you provide a heuristic explanation as to why $a_n &amp;lt; n$ for all sufficiently large $n$ (this is indicated by numerical results), or even much better, can you determine the behavior of $n - a_n$ as $n \to \infty$? No rigorous proof is required, only heuristic ideas.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-15T05:18:46.987" LastActivityDate="2010-12-16T03:54:30.380" Title="Asymptotic behavior of the first step in a best strategy" Tags="&lt;probability&gt;&lt;sequences-and-series&gt;&lt;asymptotics&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="14328" PostTypeId="1" AcceptedAnswerId="14344" CreationDate="2010-12-14T20:47:46.293" Score="7" ViewCount="93" Body="&lt;p&gt;A Lasker ring is a ring in which every ideal has a primary decomposition. The Lasker-Noether theorem states that every commutative Noetherian ring is a Lasker ring (as an easy consequence of the ascending chain condition). And I've found the statement that there are non-Noetherian Lasker rings, but I can't find an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: As the tag already suggested, I'm particularly interested in a commutative non-Noetherian Lasker ring, but noncommutative examples are also welcome. It never hurts to know more counterexamples.&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditorDisplayName="" LastEditDate="2010-12-14T21:59:17.573" LastActivityDate="2010-12-16T16:21:43.283" Title="A non-Noetherian Lasker ring" Tags="&lt;commutative-algebra&gt;&lt;ring-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14329" PostTypeId="2" ParentId="13991" CreationDate="2010-12-14T21:01:42.240" Score="2" ViewCount="0" Body="&lt;p&gt;Phillipe Flajolet's &lt;a href=&quot;http://algo.inria.fr/flajolet/Publications/books.html&quot; rel=&quot;nofollow&quot;&gt;Analytic Combinatorics&lt;/a&gt; is a good resource for this. (first book on the page, pdf)&lt;/p&gt;&#xA;" OwnerUserId="890" LastActivityDate="2010-12-14T21:01:42.240" CommentCount="1" />
  <row Id="14330" PostTypeId="2" ParentId="14275" CreationDate="2010-12-14T21:31:17.577" Score="3" ViewCount="0" Body="&lt;p&gt;Looking at &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.rmjm/1181069828&quot; rel=&quot;nofollow&quot;&gt;&#xA;the paper&lt;/a&gt; referenced by Robin Chapman and Bill Dubuque in response to &lt;a href=&quot;http://math.stackexchange.com/questions/14270/if-r-is-a-commutative-ring-with-identity-and-a-b-are-its-elements-that-are-divi&quot;&gt;the recent closely related question&lt;/a&gt;, it seems to me that the answer is &quot;yes&quot;, though it may very well fall into mosquito-nuking territory. I hope what follows is not in error!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commutative ring with identity $R$ is called &lt;em&gt;pr&amp;eacute;simplifiable&lt;/em&gt; if and only if $xy=x$ implies $x=0$ or $y\in U(R)$ ($y$ is a unit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is pr&amp;eacute;simplifiable, then $(a)=(b)$ implies $a$ and $b$ are associates: we know that $a=bx$ for some $x$, and $b=ay$ for some $y$, so $a=bx=a(xy)$. Since $R$ is pr&amp;eacute;simplifiable, either $a=0$ or $xy$ is a unit. If $a=0$, then $(a)=(0)=(b)$, so $b=0$, hence $b=a1$. If $xy$ is a unit, then $y$ is a unit, so $b=ay$ is an associate of $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper cited above cites another paper for the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; $R[X]$ is pr&amp;eacute;simplifiable if and only if $R$ is pr&amp;eacute;simplifiable and $(0)$ is a primary ideal of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting back to your question. First, consider the case in which $n$ is a prime power, $n=p^n$. Note that $R=\mathbb{Z}/p^n\mathbb{Z}$ is pr&amp;eacute;simplifiable: if $ab\equiv a \pmod{p^n}$, then $p^n|a(b-1)$. If $p^n|a$, then $\overline{a}=0$ (where $\overline{a}$ is the image of $a$ in $R$). If $p^n$ does not divide $a$, then $b-1$ must be a multiple of $p$. But if $b-1$ is a multiple. of $p$, then $b$ is relatively prime to $p$, hence $\overline{b}$ is a unit in $R$. Thus, if $\overline{a}\overline{b}=\overline{a}$ in $R$, then either $\overline{a}=0$ or $\overline{b}$ is a unit. Also, $(0)$ is a primary ideal of $\mathbb{Z}/p^n\mathbb{Z}$. If $\overline{a}\overline{b}\in(0)$ and $\overline{a}\notin (0)$, then $p|b$, hence some power of $b$ is a multiple of $p^n$, so there exists $j\gt 0$ with $\overline{b}^j\in(0)$. By the theorem quoted above, it follows that $\mathbb{Z}/p^n\mathbb{Z}[X]$ is pr&amp;eacute;simplifiable. In particular, if $a$ and $b$ in $\mathbb{Z}/p^n\mathbb{Z}[x]$ generate the same ideal, then they are associates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the property holds for $n$ a prime power, because then the polynomial ring is pr&amp;eacute;simplifiable, which is a stronger condition. However, when trying to go from prime powers to arbitrary composites, we are not so lucky. For $n$ composite but not a prime power, the ideal $(n)$ of $\mathbb{Z}$ is not primary, so the ideal $(0)$ of $\mathbb{Z}/n\mathbb{Z}$ is not primary, hence $\mathbb{Z}/n\mathbb{Z}[X]$ is not pr&amp;eacute;simplifiable. For instance, $2=2(4)$ in $\mathbb{Z}/6\mathbb{Z}[X]$, but we have neither $2=0$ nor $4$ a unit. But nonetheless I believe the result holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try to get the result for arbitrary $n$, the Chinese Remainder Theorem suggests itself instead. Factor $n$ as a product of prime powers, $n=p_1^{a_1}\cdots p_r^{a_r}$. Suppose that $(a)=(b)$ in $\mathbb{Z}/n\mathbb{Z}[X]$. Moding out by the ideal $(p_i^{a_i})$, we get $(\overline{a})=(\overline{b})$ in $\mathbb{Z}/p_i^{a_i}\mathbb{Z}[X]$, so by the result for prime powers we know that there is a unit $\overline{u}_i$ of $\mathbb{Z}/p_i^{a_i}\mathbb{Z}[X]$ such that $\overline{a}\overline{u}_i=\overline{b}$. Using the Chinese Remainder Theorem we obtain a unit $u\in\mathbb{Z}/n\mathbb{Z}[X]$ such that $u\equiv u_i \pmod{(p_i^{a_i})}$ for each $i$, hence $au=b$ (to see it is a unit, just lift the inverses to get an inverse for $u$; to see that we have equality $au=b$, note that the Chinese Remainder Theorem guarantees it holds modulo the gcd of the ideals, but $(p_1^{a_1})\cdots(p_r^{a_r}) = (0)$ in $\mathbb{Z}/n\mathbb{Z}[X]$, so we actually have equality). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would seem, if I did not make some mistake, that for these rings you do have that $(a)=(b)$ if and only if $b=au$ for some unit $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;I think we can avoid using the sledgehammer theorem quoted above to show that if $p$ is a prime, then $\mathbb{Z}/p^n\mathbb{Z}[X]$ is pr&amp;eacute;simplifiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $a,b\in\mathbb{Z}/p^s\mathbb{Z}[X]$ are such that $ab=a$ and $a\neq 0$. We have $a(b-1)=0$; lift them to $\mathbb{Z}[X]$ and write:&#xA;\begin{align*}&#xA;a &amp;amp;= a_0 + a_1x + \cdots +a_mx^m\\&#xA;b-1 &amp;amp;= b_0 + b_1x + \cdots b_mx^m.&#xA;\end{align*}&#xA;I claim that $p|b_i$ for each $i$. Indeed, since the content of $a(b-1)$ is a multiple of $p^n$, but the content of $a$ is not a multiple of $p^n$, then the content of $b-1$ must be a multiple of $p$ (since the content of the product is the product of the contents). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the content of $(b-1)^{n}$ is divisible by $p^n$, hence projecting down to $\mathbb{Z}/p^n\mathbb{Z}[X]$ we have that $(b-1)^n = 0$. Therefore,&#xA;$$b\left(b^{n-1} - \binom{n}{1}b^{n-2} + \cdots + (-1)^{n-1}\binom{n}{n-1}\right) = (-1)^{n}$$&#xD;&#xA;hence $b$ is a unit in $\mathbb{Z}/p^n\mathbb{Z}[X]$. Thus, if $ab=a$, then either $a=0$ or $b$ is a unit, and we conclude that $\mathbb{Z}/p^n\mathbb{Z}[X]$ is pr&amp;eacute;simplifiable.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T06:16:19.003" LastActivityDate="2010-12-15T06:16:19.003" CommentCount="4" />
  <row Id="14331" PostTypeId="2" ParentId="14324" CreationDate="2010-12-14T21:56:28.920" Score="0" ViewCount="0" Body="&lt;p&gt;2-regular (simple) graphs on n vertices are determined (up to isomorphism) by the partition of n induced by the size of its connected components.  This is because connected components of 2-regular graphs must be cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, given a partition $d_1,d_2,\ldots,d_k$ of n into parts of size at least 3 (you can't have cycles of length 1 or 2, which would induce loops and multiedges) then you can construct a 2-regular graph consisting of a $d_i$-cycle for each $d_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above case, the partition is 7 (since the graph is a 7-cycle).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 0 1 0 0 0 1&#xA;0 0 0 1 0 1 0&#xA;1 0 0 1 0 0 0&#xA;0 1 1 0 0 0 0&#xA;0 0 0 0 0 1 1&#xA;0 1 0 0 1 0 0&#xA;1 0 0 0 1 0 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There exist an automorphism of the graph that maps vertex u to vertex v provided u and v both belong to cycles of the same length.  To illustrate, you can fix everything outside of the components containing u and v, then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if u and v are in the same cycle, just rotate the cycle, or otherwise,&lt;/li&gt;&#xA;&lt;li&gt;swap the cycles containing u and v, then rotate each appropriately.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-14T22:10:56.043" LastActivityDate="2010-12-14T22:10:56.043" />
  <row Id="14333" PostTypeId="1" CreationDate="2010-12-14T22:09:06.060" Score="3" ViewCount="108" Body="&lt;p&gt;I am truly lost as to what this problem is asking.  I did post this on another forum and received what my have been wonderful advice.  However, even after multiple hours and many &quot;Google&quot; searches I am still confused.  Please anyone....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem states:&#xA;If $\sigma$ is a permutation of a set $A$, we say that &lt;em&gt;$\sigma$ moves $a\in A$&lt;/em&gt; iff $\sigma(a)\neq a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the symmetric group $S_{36}$ of all permutations of 36 elements, let $H$ be a subset of $S_{36}$ containing all permutations that move no more than four elements.  Is $H$ a subgroup of $S_{36}$?  Prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for an answer but more an explanation of what the problem is saying.  I do not know what &quot;move no more than four elements&quot; means. I am feeling really stupid.  Any help appreciated.   &lt;/p&gt;&#xA;" OwnerUserId="4755" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-15T04:32:26.823" LastActivityDate="2010-12-15T04:32:26.823" Title="Is the set of permutations in $S_{36}$ that move at most 4 elements a subgroup of $S_{36}$?" Tags="&lt;homework&gt;&lt;group-theory&gt;&lt;permutations&gt;" AnswerCount="2" />
  <row Id="14334" PostTypeId="2" ParentId="14333" CreationDate="2010-12-14T22:14:47.227" Score="2" ViewCount="0" Body="&lt;p&gt;Imagine you have 36 volumes of an encyclopedia in front of you. A permutation $\sigma\in S_{36}$ is a set of instructions of how to swap positions of the volumes of that encyclopedia. For instance, the permutation $\sigma=(1,2,3,4)$  (this is &lt;em&gt;cycle notation&lt;/em&gt;: it says element 1 is sent to element 2, element 2 is sent to element 3, 3 to 4, and 4 to 1; if you are used to two-line or abbreviated two-line notation, then this is the permutation that has&#xA;$$\left(\begin{array}{cccccccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 5 &amp;amp; 6 &amp;amp; \cdots &amp;amp; 36\\&#xD;&#xA;2 &amp;amp; 3 &amp;amp; 4 &amp;amp; 1 &amp;amp; 5 &amp;amp; 6 &amp;amp; \cdots &amp;amp; 36&#xD;&#xA;\end{array}\right).$$&#xD;&#xA;as its two-line description) &#xA;tells you to take volumes $1$ through $4$, and move volume $1$ to the second position, volume $2$ to the third position, volume $3$ to the fourth position, and volume $4$ to the first position. Even though $\sigma$ is a permutation of all 36 volumes, in fact you only need to move four of the volumes to perform the permutation $\sigma$. So we say that $\sigma$ &lt;em&gt;fixes&lt;/em&gt; 32 volumes (volumes 5 through 36 are left untouched) and &lt;em&gt;moves&lt;/em&gt; 4 volumes (volumes 1 through 4). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each permutation $\sigma\in S_n$, which acts on the set $[n]=\{1,2,\ldots,n\}$, you can ask for the set $\mathrm{Fix}(\sigma) = \{ a\in[n]\mid \sigma(a)=a\}$, the set of elements fixed by $\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the elements &quot;moved&quot; by $\sigma$? They are the elements of $[n]-\mathrm{Fix}(\sigma)$, the complement of $\mathrm{Fix}(\sigma)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\sigma\in S_{36}$ will lie in $H$ if and only if $[36]-\mathrm{Fix}(\sigma)$ has at most $4$ elements; that is, if and only if $\sigma$ fixes &lt;em&gt;at least&lt;/em&gt; 32 of the elements of the numbers in $[36]=\{1,2,3,\ldots,36\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether $H$ is a subgroup. It is easy to check that it is nonempty and closed under inverses (though you should do that). The trick is going to be figuring out if it is closed under products: if you have $\sigma$ and $\tau$, and each of them fix at least $32$ elements (though possibly not the &lt;em&gt;same&lt;/em&gt; 32 elements), can $\sigma\tau$ fix &lt;em&gt;fewer&lt;/em&gt; than 32 elements?&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-14T22:20:49.527" LastActivityDate="2010-12-14T22:20:49.527" CommentCount="4" />
  <row Id="14335" PostTypeId="2" ParentId="14333" CreationDate="2010-12-14T22:18:12.420" Score="3" ViewCount="0" Body="&lt;p&gt;Let's look at $S_3$ instead.  There are six permutations:  $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 3 &amp;amp; 2&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 1 &amp;amp; 3&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 3 &amp;amp; 1&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 1 &amp;amp; 2&#xD;&#xA;\end{array}\right)$, and $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 2 &amp;amp; 1&#xD;&#xA;\end{array}\right)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3&#xD;&#xA;\end{array}\right)$ moves no elements.  $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;1 &amp;amp; 3 &amp;amp; 2&#xD;&#xA;\end{array}\right)$, $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 2 &amp;amp; 1&#xD;&#xA;\end{array}\right)$, and $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 1 &amp;amp; 3&#xD;&#xA;\end{array}\right)$ move two elements.  $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;3 &amp;amp; 1 &amp;amp; 2&#xD;&#xA;\end{array}\right)$ and $\left(\begin{array}{ccc}&#xD;&#xA;1 &amp;amp; 2 &amp;amp; 3  \\&#xD;&#xA;2 &amp;amp; 1 &amp;amp; 3&#xD;&#xA;\end{array}\right)$ move three elements.  Just look at how many are left where they started.  With Arturo's answer, you should be on your way.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T00:21:02.670" LastActivityDate="2010-12-15T00:21:02.670" CommentCount="6" />
  <row Id="14336" PostTypeId="2" ParentId="14306" CreationDate="2010-12-14T23:01:21.270" Score="3" ViewCount="0" Body="&lt;p&gt;A nice, very readable reference for this question is a recent paper in the Monthly, &quot;Simpson's rule is exact for quintics&quot;. It is available for download at the author's (&lt;a href=&quot;http://rowdy.mscd.edu/~talmanl/&quot; rel=&quot;nofollow&quot;&gt;Louis A. Talman&lt;/a&gt;) webpage. It appeared American Mathematical Monthly, 113(2006), 144–155. I recommend it; I have used it a couple of times when covering this material in class. Here is the abstract: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this article, we use tools accessible to freshman calculus students to develop exact—though usually uncomputable—expressions for the error that results in replacing a definite integral with its midpoint rule, trapezoidal rule, or Simpson's rule approximation. Among the tools we use is an extended version of the first mean value theorem for integrals. We obtain not only the classical estimates that appear in calculus books, but estimates for functions less smooth than the classical results require. We show, in particular, how to compute the exact error for a Simpson's rule approximation to an integral of a quintic polynomial. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you visit Talman's page, you may also enjoy taking a look at the nice &quot;The Mother of All Calculus Quizzes.&quot;&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-14T23:01:21.270" />
  <row Id="14337" PostTypeId="1" CreationDate="2010-12-14T23:08:38.190" Score="0" ViewCount="195" Body="&lt;p&gt;My teacher says that the answer is 2, but I don't get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please take a moment to consider the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have a container. The container has water. I add water. The result? Water!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the sum of the two waters... is water! And by the same token, 1 + 1 must equal 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my teacher doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S. Don't buy anything from that 'Leibniz' guy, my Calculus books are waaaay better than his!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I can't accept the answer below... my browser isn't quite working right... but thanks anyway. I &lt;em&gt;knew&lt;/em&gt; my teacher goofed up.&lt;/p&gt;&#xA;" OwnerUserId="4757" LastEditorUserId="4757" LastEditorDisplayName="" LastEditDate="2010-12-14T23:18:36.843" LastActivityDate="2010-12-14T23:51:32.343" ClosedDate="2010-12-15T03:09:49.500" Title="What is the sum of 1 + 1?" Tags="&lt;elementary&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="1" />
  <row Id="14338" PostTypeId="2" ParentId="14337" CreationDate="2010-12-14T23:16:09.137" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://quod.lib.umich.edu/cgi/t/text/text-idx?c=umhistmath;idno=AAT3201.0001.001&quot; rel=&quot;nofollow&quot;&gt;http://quod.lib.umich.edu/cgi/t/text/text-idx?c=umhistmath;idno=AAT3201.0001.001&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page 378 or so is where they do the whole 1+1 = 2 thing.&lt;/p&gt;&#xA;" OwnerUserId="1582" LastActivityDate="2010-12-14T23:16:09.137" />
  <row Id="14339" PostTypeId="2" ParentId="14337" CreationDate="2010-12-14T23:17:32.480" Score="0" ViewCount="0" Body="&lt;p&gt;Note that if $1+1 = 1$ then substract of each side $1$, get $1=0$  (contradicts an axiom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore,  water add water is more water ( not only water ) :) :) &lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2010-12-14T23:17:32.480" />
  <row Id="14340" PostTypeId="2" ParentId="14337" CreationDate="2010-12-14T23:19:11.173" Score="4" ViewCount="0" Body="&lt;p&gt;Welcome to the site, Sir Isaac.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I submit to you good Sir that the sum of two waters is the sum of the two waters, not one water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For you see, water is made up of tiny Atoms; it is therefore to be supposed that they add in the same way as many apples add to many apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we only see the world with out mediocre instruments, we some times have to look at it with different eyes. You should go to sit in the orchard to contemplate this.&lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2010-12-14T23:19:11.173" CommentCount="2" />
  <row Id="14341" PostTypeId="2" ParentId="14337" CreationDate="2010-12-14T23:20:33.117" Score="0" ViewCount="0" Body="&lt;p&gt;Well, your error is that you are comparing two different types of objects. You write &quot;The the sum of two waters... is water!&quot; &#xA;The problem here is that you aren't adding water, but &lt;em&gt;amounts&lt;/em&gt; of water. &lt;/p&gt;&#xA;" OwnerUserId="4758" LastActivityDate="2010-12-14T23:20:33.117" />
  <row Id="14342" PostTypeId="2" ParentId="14337" CreationDate="2010-12-14T23:25:08.873" Score="-2" ViewCount="0" Body="&lt;p&gt;I have constructed a proof below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin^\theta/\cos(\f*\theta^2)$&lt;/p&gt;&#xA;" OwnerUserId="4757" LastActivityDate="2010-12-14T23:25:08.873" CommentCount="1" />
  <row Id="14343" PostTypeId="1" CreationDate="2010-12-14T23:32:09.080" Score="0" ViewCount="96" Body="&lt;p&gt;I have a triangle ABC in the x-y coordinate plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corners A and B lie on x-axis. Length of AB is known. Angles A and B are also known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Find (x,y) coordinates of corner C.&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2010-12-15T19:34:56.980" Title="Finding X and Y coordinates from two angles" Tags="&lt;trigonometry&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14344" PostTypeId="2" ParentId="14328" CreationDate="2010-12-14T23:35:34.900" Score="5" ViewCount="0" Body="&lt;p&gt;I don't know of any easier examples, but here is one. Let $k$ be a field and $R$ be the set equivalence classes of elements of the form $\frac{f(x,y)}{g(x,y)}$ where $x,y$ are indeterminates over $k$, $f,g\in k[x,y]$, $x$ does not divide $g$ (in $k[x,y]$) and $\frac{f(0,y)}{g(0,y)}\in k$. Make $R$ into a ring by the usual addition and multiplication of rational functions. Then, it can be shown that $R$ is a commutative ring that is not noetherian and every ideal of $R$ has a primary decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example from Gilmer's paper linked here &lt;a href=&quot;http://www.jstor.org/pss/2034992&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2034992&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also gives a characterization of rings in which every ideal has a unique primary decomposition.&lt;/p&gt;&#xA;" OwnerUserId="3431" LastEditorUserId="3431" LastEditorDisplayName="" LastEditDate="2010-12-16T16:21:43.283" LastActivityDate="2010-12-16T16:21:43.283" />
  <row Id="14345" PostTypeId="1" AcceptedAnswerId="14351" CreationDate="2010-12-14T23:39:04.457" Score="2" ViewCount="59" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Determine sign$(\tau)$ for $$\left( \begin{array}{ccccccc} 1&amp;amp;2&amp;amp;3&amp;amp;4&amp;amp;5&amp;amp;6&amp;amp;7 \\ 2&amp;amp;3&amp;amp;5&amp;amp;7&amp;amp;1&amp;amp;6&amp;amp;4 \end{array} \right )$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Let $A_{n} = \{\tau \in S_{n} | \mbox{sign}(\tau) = 1\}$. Show that $A_{n}$ is a sub group of $S_{n}$ and calculate the number of elements (Tip: Show that for any transposition $\tau$, $S_{n}$ is the disjoint union of $A_{n}$ and $A_{n}\circ\tau$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt so far:&#xA;(a) So I have this formula for sign$(\sigma) = \displaystyle \prod_{i &amp;lt; j} \frac{\sigma(j)-\sigma(i)}{j-i}$. This wouldn't be so difficult, however it's apparently ok to count the number of elements for which $i&amp;lt;j$ but $\sigma(i)&amp;gt;\sigma(j)$ (these are called inversions, I believe). If this number is even, sign$(\sigma)=1$ if odd, $-1$. In this case I named $(1,5),(2,5),(3,5),(3,7),(4,5),(4,6),(4,7),(6,7) \Rightarrow \mbox{sign}(\tau) =1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) To begin with, if I want to show that something is a subgroup, I have to show that: (i)$ a,b \in U \Rightarrow a \circ b \in U$ and that (ii)$a \in U \Rightarrow a^{-1} \in U$ correct? (I'm actually a little confused at why we can just assume that the identity element is still part of a subgroup, for instance if the exercise were a little different and I was checking if $A_{n} = \{\tau \in S_{n} | \mbox{sign}(\tau) = -1\}$ was a sub group, I would say that the identity element for permutations $123...n$ has no inversions $\Rightarrow$ sign$(\tau) = 1 \Rightarrow \tau \notin A_{n}$... I mean, subgroups are groups themselves, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well anyway, for (i) I tried to show it by writing: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\tau, \sigma \in A_{n}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ sign$(\tau) = 1$ and sign$(\sigma)=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a polynomial (I honestly don't totally understand this or know if I need to put this stuff with 'g' in...) $g=g(x_{1},...,x_{n})$, sign$(\tau \circ \sigma)g= (\tau \circ \sigma)(g)=\tau(\sigma(g))=\tau((\mbox{sign}\sigma)g)=(\mbox{sign}\tau)(\mbox{sign}\sigma)g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ sign$(\tau \circ \sigma) = (1)(1) = 1 \Rightarrow (\tau \circ \sigma) \in A_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with (ii) my attempt was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\sigma \in A_{n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\sigma \circ \sigma^{-1} = \epsilon$ (this should represent the identity element)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ sign$(\sigma \circ \sigma^{-1}) = \mbox{sign}\epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mbox{sign}\sigma)(\mbox{sign}\sigma^{-1})=1$ (Because sign$(\epsilon)=1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)(\mbox{sign}(\sigma^{-1}))=1$ (Because $\sigma \in A_{n}\Rightarrow \mbox{sign}(\sigma) = 1$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sign$(\sigma^{-1})=1 \Rightarrow \sigma^{-1} \in A_{n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of (i) and (ii) follows the conclusion that $A_{n}$ is a subgroup of $S_{n}$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So hopefully that would all be on the right track so far... With the calculate the number of elements part, I don't know what to do or why. Following the tip I would try to show that $S_{n}$ is the disjoint union (not the same as showing that the intersection is empty, right?) of $A_{n}$ and $A_{n} \circ \tau$ (where $\tau$ is a transposition). I don't understand this, but directly out of the book is the statement &quot;We call $\tau$ a transposition. If $i&amp;lt;j$, then there are $2(j-i=1)+1$ inversions in $\tau$, and hence the transposition $\tau$ is odd.&quot; Accordingly I think I could use sign$(\tau \circ \sigma)=(\mbox{sign}\tau)(\mbox{sign}\sigma)$ to show that the product of an odd and even permutation is odd and therefore $A_{n} \circ \tau$ is odd and $A_{n}$ has no elements in common with $A_n \circ \tau$. But even then how does one calculate a number of elements?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2010-12-15T00:19:49.117" LastActivityDate="2010-12-15T00:19:49.117" Title="Determine sign of a permutation, calculate number of elements in the subgroup of permutations with sign = 1" Tags="&lt;homework&gt;&lt;group-theory&gt;&lt;permutations&gt;&lt;symmetric-groups&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="14346" PostTypeId="2" ParentId="14327" CreationDate="2010-12-14T23:45:31.347" Score="2" ViewCount="0" Body="&lt;p&gt;If we let $E(n,m)$ be the expectation with m throws of an n sided die, we will accept the $m^{th}$ to last throw if it is $\geq E(n,m-1)$, so $E(n,1)=\frac{n+1}{2}$ and $$E(n,m)=\frac{(n-\lceil E(n,m-1)\rceil +1)}{n}\frac{(n+\lceil E(n,m-1)\rceil )}{2}+\frac{\lceil E(n,m-1)\rceil -1}{n}E(n,m-1)$$  where the first term comes from accepting the throw and the second is from rejecting it.  Letting $n-\lceil E(n,m-1)\rceil=d$ (the whole part of the difference) and $\lceil E(n,m-1)\rceil-E(n,m-1)=e$ (the fractional part) we find $$E(n,m)-E(n,m-1)=\frac{d^2+(1+2e)d+2e}{2n}$$  If we figure that e averages about 1/2, this means we increase E by $\frac{(d+1)^2}{2n}$ per step.  The number of steps to get the cut off up to n is about $$\sum_{d=1}^{n}{\frac{2n}{(d+1)^2}}$$  For n=100, I find by explicit calculation we hit a cutoff of 99 (E&gt;98) at m=75 and a cutoff of 100 (E&gt;99) at m=129.  The sum comes out just about 128.  We can find lower cutoffs by raising the lower limit of the sum.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T04:16:41.570" LastActivityDate="2010-12-15T04:16:41.570" CommentCount="6" />
  <row Id="14347" PostTypeId="1" CreationDate="2010-12-14T23:46:29.380" Score="2" ViewCount="81" Body="&lt;p&gt;Can anyone give a nonlinear regular function from C^2 to C^2 with a constant nonzero Jacobian? It seems to me that the only such functions are linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Jacobian conjecture, a function from C^2 to C^2 with a constant nonzero Jacobian must have an inverse. &lt;/p&gt;&#xA;" OwnerUserId="3163" LastActivityDate="2010-12-15T03:02:51.720" Title="Example of nonlinear regular function with constant nonzero Jacobian" Tags="&lt;analysis&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14348" PostTypeId="2" ParentId="14343" CreationDate="2010-12-14T23:46:53.607" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With law of sines you can find the other sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A= (x_{1},y_{1})$ and $B=(x_{2},y_{2})$, then $|AB|=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $C=(x,y)$ get :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|AC| = \sqrt{(x-x_{1})^2+(y-y_{1})^2}$$  $$|BC| = \sqrt{(x-x_{2})^2+(y-y_{2})^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that you know $|AB|,|AC|,|AB|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can find $x,y$&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-15T18:52:16.270" LastActivityDate="2010-12-15T18:52:16.270" />
  <row Id="14349" PostTypeId="2" ParentId="13991" CreationDate="2010-12-14T23:50:58.017" Score="1" ViewCount="0" Body="&lt;p&gt;With regard to combinatorics &amp;amp; harmonic analysis, you might find &lt;a href=&quot;http://books.google.com/books?id=qjVaD1OQbxEC&amp;amp;lpg=PP1&amp;amp;pg=PA13#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; interesting. &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0008/0008098v1.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; work by Terence Tao (whom you mentioned) sheds light on all three of combinatorics, analysis, and PDEs. Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-14T23:50:58.017" />
  <row Id="14350" PostTypeId="2" ParentId="14337" CreationDate="2010-12-14T23:51:32.343" Score="3" ViewCount="0" Body="&lt;p&gt;Where should I being, Mr. Newton?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My teacher says that the answer is 2, but I don't get it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;She's right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore, the sum of the two waters... is water! And by the same token, 1 + 1 must equal 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;True... my brain + yours = my brain.&lt;/p&gt;&#xA;" OwnerUserId="4762" LastActivityDate="2010-12-14T23:51:32.343" CommentCount="1" />
  <row Id="14351" PostTypeId="2" ParentId="14345" CreationDate="2010-12-15T00:00:19.687" Score="1" ViewCount="0" Body="&lt;p&gt;Your work on (a) got cut off somehow.  For your comment on (b), if you have $a,b \in U \Rightarrow a \circ b \in U$ and $a \in U \Rightarrow a^{-1} \in U$ then you have $a \circ a^{-1} \in U$ so you don't have to check the identity (as long as $U$ is nonempty).  Showing the identity is not in U is another way to prove U is not a subgroup.  You are right that subgroups are groups themselves, but I think you would have trouble showing the product to two odd permutations is odd.  To say $S_n$ is a disjoint union of $A_n$ and $A_{n} \circ \tau$, you need the intersection to be empty and that every element of $S_n$ is in one of them.  In calculating the number of elements, if every element is either odd or even and there are the same number of odd ones and even ones...&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-15T00:00:19.687" CommentCount="7" />
  <row Id="14352" PostTypeId="2" ParentId="14276" CreationDate="2010-12-15T00:05:45.690" Score="0" ViewCount="0" Body="&lt;p&gt;If you work with Grothendieck motives modulo rational equivalence, you even have simpler examples : the motive of any split (projective) quadric of odd dimension $n$ is the same as the motive of $\mathbb{P}^n$.&lt;/p&gt;&#xA;" OwnerUserId="4763" LastActivityDate="2010-12-15T00:05:45.690" />
  <row Id="14353" PostTypeId="1" CreationDate="2010-12-15T00:29:42.653" Score="4" ViewCount="130" Body="&lt;p&gt;Twin, cousin, and sexy primes are of the forms $(p,p+2)$, $(p,p+4)$, $(p,p+6)$ respectively, for $p$ a prime.&#xA;The Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Cousin_prime&quot; rel=&quot;nofollow&quot;&gt;article on cousin primes&lt;/a&gt; says that,&#xA;&quot;It follows from the first Hardy–Littlewood conjecture that cousin primes have the same asymptotic density as twin primes,&quot; but the analogous &lt;a href=&quot;http://en.wikipedia.org/wiki/Sexy_prime&quot; rel=&quot;nofollow&quot;&gt;article on sexy primes&lt;/a&gt; does not make a similar claim. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;b&gt;Q1&lt;/b&gt;. Are the sexy primes expected to have the same density as twin primes?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;b&gt;Q2&lt;/b&gt;. Is it conjectured that there are an infinite number of cousin and sexy prime&#xA;  pairs?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;b&gt;Q3&lt;/b&gt;. Have prime pairs of the form $(p,p+2k)$ been studied for $k&amp;gt;3$?&#xA;  If so, what are the conjectures?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks for information or pointers!&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-15T00:29:42.653" Title="Twin, cousin, sexy, ... primes" Tags="&lt;prime-numbers&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="14354" PostTypeId="1" CreationDate="2010-12-15T00:46:02.637" Score="4" ViewCount="60" Body="&lt;p&gt;I've just been thinking about for what values of n we can place n points in the plane so that any three of those points define an isosceles triangle. A triangle, square and pentagon work for 3,4 and 5, and to get 6 just place a point in the centre of the pentagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is can we do this for 7 points in the plane?&lt;/p&gt;&#xA;" OwnerUserId="4764" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T01:03:15.850" LastActivityDate="2010-12-15T02:42:12.390" Title="Collections of points containing only isosceles triangles" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14355" PostTypeId="2" ParentId="14347" CreationDate="2010-12-15T00:47:09.967" Score="3" ViewCount="0" Body="&lt;p&gt;Some trial and error should get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured $\left[\begin{array}{cc} x &amp;amp; 1\\ x-1 &amp;amp; 1\end{array}\right]$ was a good target Jacobian matrix, and so got $f(x,y)=(\frac 1 2 x^2 + y,\frac 1 2 x^2-x + y)$.  Is that what you're looking for?  Evidently it has an inverse, $g(a,b)=(a-b,a-\frac 1 2(a-b)^2)$.&lt;/p&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2010-12-15T00:47:09.967" CommentCount="1" />
  <row Id="14356" PostTypeId="1" CreationDate="2010-12-15T00:57:04.653" Score="1" ViewCount="88" Body="&lt;p&gt;I am in graduate school for chemical engineering and the method of characteristics appears in many journal articles.  Does anyone have a good reference: either online or a book that provides a good explanation with examples of the method of characteristics.&lt;/p&gt;&#xA;" OwnerUserId="4766" LastActivityDate="2010-12-15T15:17:11.550" Title="Good reference for method of characteristics" Tags="&lt;differential-equations&gt;&lt;differential&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14357" PostTypeId="2" ParentId="14343" CreationDate="2010-12-15T01:10:50.570" Score="1" ViewCount="0" Body="&lt;p&gt;Another possibility is as follows: this solution to your problem involves using the fact that the slope of a line making an angle $\theta$ with the positive x-axis is $\tan\;\theta$; now, what you have are two lines with slopes $\tan\;A$ and $\tan(\pi-B)$ in one case, and slopes $\tan\;B$ and $\tan(\pi-A)$ in the other. Construct the point-slope form for the two lines through $A$ and $B$, and find their intersection through the usual methods.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-15T01:10:50.570" />
  <row Id="14358" PostTypeId="1" AcceptedAnswerId="14860" CreationDate="2010-12-15T01:19:40.527" Score="3" ViewCount="215" Body="&lt;p&gt;Let $G$ be a $n \times n$ matrix with real entries and let $\Lambda = \{x^n \colon \exists i^n \in \mathbb{Z}^n \text{ such that } x^n = G \cdot i^n\}$ define a lattice. I am interested in projecting the lattice points onto a $k$-dimensional subspace $U$ with $k &amp;lt; n$. Let $A$ be a $n \times k$ matrix of vectors that span $U$. Then, the projection of $\Lambda$ onto $U$ has the generator matrix $P_A \cdot G$ where $P_A = A (A^TA)^{-1}A^T$ is the projection operator. Lets also assume that the entries of $G$ have no structure between them - i.e., they are randomly chosen from the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in finding out when the projection of $\Lambda$ onto $U$ (call it $\Lambda_U$) is also a $k$-dimensional lattice. It seems to me that in most cases, $\Lambda_U$ will fill out the $k$-dimensional space $U$ completely, i.e., the shortest distance between neighboring points in $\Lambda_U$ will be arbitrarily small (obs. 1). For certain subspaces $U$ that seem to be &quot;aligned&quot; correctly with $\Lambda$, we get a well-defined $\Lambda_U$. By well-defined, I mean that the spacing between the nearest neighbors of $\Lambda_U$ is bounded away from $0$. As far as I can tell, this happens when $A$ is chosen such that the columns of $P_A \cdot G$ are rational multiples of one another (obs. 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are the observations (obs. 1) and (obs. 2) correct?&lt;/li&gt;&#xA;&lt;li&gt;If so, is this a well-studied concept? I didn't have much success googling with the obvious keywords such as lattice &quot;projection&quot; or &quot;matrix column rational multiples&quot; etc.&lt;/li&gt;&#xA;&lt;li&gt;Assuming the observations are correct, given $G$, is there a way to choose $A$ such that the columns of $P_A \cdot G$ are rational multiples of one another?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-12-19T18:23:31.823" Title="Projection of a lattice onto a subspace" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14359" PostTypeId="1" CreationDate="2010-12-15T01:23:29.147" Score="2" ViewCount="63" Body="&lt;p&gt;The problem is stated-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the following two sets of points have the same cardinality and if so, establish a bijection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A line segment of length four and half of the circumference of radius one (including both endpoints).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is they do have the same cardinality and my bijection is a picture in which I drew a horizontal line and a semicircle under it  (separated by approx. 2cm with the semicircle's open side facing down).  At what would be the center of the circle made by the semi-circle I drew a point P.  I then drew lines vertical from the original horizontal line to point P.  This shows, goes my reasoning, that for every point on the line there is a corresponding point on the semi-circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is is this drawing enough to show a bijection or do I need to do more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your thoughts.&lt;/p&gt;&#xA;" OwnerUserId="4179" LastActivityDate="2010-12-15T01:38:42.587" Title="Cardinality of Two Sets of Points" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14360" PostTypeId="2" ParentId="14328" CreationDate="2010-12-15T01:25:08.383" Score="2" ViewCount="0" Body="&lt;p&gt;Many interesting examples are in: $\ $ Barucci and Montana: &lt;a href=&quot;http://www.springerlink.com/content/7752947723p76557/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;When are $\rm\ D +  \frak M\ $ rings Laskerian?&lt;/a&gt; They construct (non-Noetherian) Laskerian or strongly Laskerian rings and domains either integrally closed or not, of any dimension. They employ a frequently used powerful tool for constructing counterexamples - the  $\rm\ D +  \frak M\ $ construction. For more on this construction see Anderson: &lt;a href=&quot;http://rmmc.asu.edu/TO%20DOUGLAS/JCA/anderson/anderson.pdf&quot; rel=&quot;nofollow&quot;&gt;Some remarks on the  $\rm\ D +  \frak M\ $ construction&lt;/a&gt; and see this very informative survey on more general constructions: $\ $ Zafrullah: &lt;a href=&quot;http://www.lohar.com/researchpdf/axbx.pdf&quot; rel=&quot;nofollow&quot;&gt;Various facets of rings between $\rm\:D[X]\:$ and $\rm\:K[X]\:$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-15T01:25:08.383" />
  <row Id="14361" PostTypeId="2" ParentId="14347" CreationDate="2010-12-15T01:30:06.773" Score="1" ViewCount="0" Body="&lt;p&gt;If $u(x,y)$ is of the form $af(x-y)+bx$, where $a$ and $b$ are constants and $f$ is a one-variable differentiable function, and if $v(x,y)=y-x$, then&#xA;$$&#xD;&#xA;\frac{{\partial u(x,y)}}{{\partial x}} \frac{{\partial v(x,y)}}{{\partial y}} - \frac{{\partial u(x,y)}}{{\partial y}}\frac{{\partial v(x,y)}}{{\partial x}} = [af&amp;#39;(x-y)+b]+[-af&amp;#39;(x-y)] = b.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: More simply, letting $u(x,y)=af(x+y)+bx$ and $v(x,y)=x+y$ gives the same result as above.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-15T03:02:51.720" LastActivityDate="2010-12-15T03:02:51.720" />
  <row Id="14362" PostTypeId="2" ParentId="14359" CreationDate="2010-12-15T01:33:05.273" Score="2" ViewCount="0" Body="&lt;p&gt;I think your lines are not vertical, right?  With your construction, there are points on the semi-circle that do not have a corresponding point on the line.  But you are quite close.  You can 1) modify the construction so that every line from P through the semicircle hits the line (and still have every line from the line segment hit the semi-circle.  What does that tell you about the endpoints of each?) Or 2)find another construction that injects the semi-circle into the line, then argue from the Cantor–Bernstein–Schroeder theorem. &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T01:38:42.587" LastActivityDate="2010-12-15T01:38:42.587" />
  <row Id="14363" PostTypeId="1" AcceptedAnswerId="14385" CreationDate="2010-12-15T01:36:08.573" Score="1" ViewCount="43" Body="&lt;p&gt;Given a multi-undirected graph, how can I find the number of different paths from Node A to B using every node in the graph (means shortest path algorithms are useless)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases there may be no way to achieve this. I am looking for the number of available different paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: You have to use every node and use them once.&lt;/p&gt;&#xA;" OwnerUserId="4653" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T01:54:22.407" LastActivityDate="2010-12-15T06:16:42.303" Title="Find the number of different paths across two nodes in a multi-undirected graph" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14364" PostTypeId="1" CreationDate="2010-12-15T01:40:11.670" Score="1" ViewCount="157" Body="&lt;p&gt;Given $R$ is a ring, and $b$ is some positive integer such that $x + x^{2b + 1} = x^{2b} + x^{10b + 1}$ for all $x \in R$, prove that $R$ is Boolean, i.e. $x =x^2$ for all $x$ in $R$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure where to begin with the problem, and what avenue of approach I should take such that my methodology is inclusive to all $x$ in R. &#xA;Would it be feasible to convert this problem into matrix form, and then prove $A = A^2$? &lt;/p&gt;&#xA;" OwnerUserId="4767" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T03:11:28.173" LastActivityDate="2010-12-15T18:01:59.230" Title="Proving a ring is Boolean " Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="14365" PostTypeId="2" ParentId="14347" CreationDate="2010-12-15T01:53:13.373" Score="0" ViewCount="0" Body="&lt;p&gt;Any linear function has constant Jacobian determinant, as does any map of the form $(z_1,z_2) \rightarrow (z_1, z_2 - f(z_1))$ or $(z_1,z_2) \rightarrow (z_1 - f(z_2), z_2)$. As a result, any finite composition of maps of these forms will have constant Jacobian. These include the examples that Shai Covo and Andrew Marshall listed. I forget if there are known examples outside this category.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-15T01:53:13.373" />
  <row Id="14367" PostTypeId="1" AcceptedAnswerId="14370" CreationDate="2010-12-15T02:39:37.620" Score="2" ViewCount="65" Body="&lt;p&gt;So, we have a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \frac{3\sin(x)}{2 + \cos(x)} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we fix a point $$ x_0 \in \left(0, \frac{2\pi}{3}\right] $$ and define a sequence by setting $$ x_{n+1} = f(x_n) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to show that $$ \lim_{n\to\infty} n^{1/4}x_n = 45^{1/4}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure this has something to do with Cesàro means. In a previous part of the problem we proved that the sequence is strictly decreasing and has a limit of zero, and we also found the 5th order Taylor approximation of the function (centered at 0), which happens to be $$ p_5(f, x) = x - \frac{x^5}{180} $$ but we have not proved very many theorems involving Taylor polynomials (and none that seem to apply), and the section in Rudin (our textbook) is quite brief. Any help or suggestions would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T02:48:26.430" LastActivityDate="2010-12-15T03:33:27.910" Title="recursively defined trigonometric sequence and Cesàro means." Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;sequences-and-series&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14368" PostTypeId="2" ParentId="14354" CreationDate="2010-12-15T02:42:12.390" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a recent paper related to this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cims.nyu.edu/~pach/publications/isosceles.ps&quot; rel=&quot;nofollow&quot;&gt;http://www.cims.nyu.edu/~pach/publications/isosceles.ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-15T02:42:12.390" />
  <row Id="14370" PostTypeId="2" ParentId="14367" CreationDate="2010-12-15T02:47:05.987" Score="2" ViewCount="0" Body="&lt;p&gt;Check David Speyer's excellent answer here: &lt;a href=&quot;http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n/3220#3220&quot;&gt;Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoted from that answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f$ be a function $0 \leq f(u)&#xD;&#xA;&amp;gt; \leq u$ on an interval $[0,c]$ then&#xA;  the sequence $x_n:=f(f(f(\cdots&#xD;&#xA;&amp;gt; f(c)\cdots)$ approaches $0$. If&#xA;  $f(u)=u-a u^{k+1} + O(u^{k+2})$ (with&#xA;  $a&gt;0$) then $x_n \approx \alpha n^{-1/k}$ and you can prove that by&#xA;  the same methods here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-15T03:33:27.910" LastActivityDate="2010-12-15T03:33:27.910" />
  <row Id="14372" PostTypeId="2" ParentId="8933" CreationDate="2010-12-15T03:17:33.547" Score="1" ViewCount="0" Body="&lt;p&gt;By the way, your equation can be written in form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H^{(x)}_3=11$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where H is the generalized harmonic number: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=HarmonicNumber%5b3%2C+x%5d%3D%3D11&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=HarmonicNumber[3%2C+x]%3D%3D11&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to find x you should investigate the inverse function of generalized harmonic number.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-12-15T03:17:33.547" />
  <row Id="14373" PostTypeId="2" ParentId="14343" CreationDate="2010-12-15T04:08:42.227" Score="0" ViewCount="0" Body="&lt;p&gt;A system of linear equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume A is at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can form basis vectors for AC and BC using the angles A and B&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;U_{AC} = [cosA  sinA]^T&#xA;U_{BC} = [cosB  sinB]^T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then we know geometrically&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;w*U_{AC} - x*U_{BC} = [ AB  0 ]^T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is a 2x2 linear system.  Solve for w and x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = w*cosA&#xA;Y = w*sinA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;best regards.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-15T04:08:42.227" />
  <row Id="14374" PostTypeId="2" ParentId="14356" CreationDate="2010-12-15T04:14:47.350" Score="2" ViewCount="0" Body="&lt;p&gt;Off the top of my head, I've found both of these to be well-written and easy to follow.  Cavet:  I don't have them in front of me, so my recommendation is based on general impressions rather than their particular treatment of characteristics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Partial Differential Equations&lt;/em&gt;, by L. C. Evans. Aimed at the intro grad level.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Partial Differential Equations&lt;/em&gt;, by Walter Strauss.  Aimed at an advanced undergrad audience, but good for someone new to the field.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The External Links listed at the end of the wikipedia article on characteristics also look like they might be worth your time.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-15T04:14:47.350" CommentCount="1" />
  <row Id="14375" PostTypeId="1" AcceptedAnswerId="14377" CreationDate="2010-12-15T04:30:50.603" Score="2" ViewCount="80" Body="&lt;p&gt;I'm trying to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The probability $p_{i}$ that a given cell contains exactly i balls is given by the binomial distribution $p_{i}=\displaystyle \frac{\binom{r}{i}(n-1)^{r-i}}{n^{r}}$. Prove that the most probable number is the integer $v$ such that $\displaystyle \frac{(r-n+1)}{n}&amp;lt; v \leq \displaystyle \frac{(r+1)}{n}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've tried to get the inequality using first $\displaystyle \frac{p_{i}}{p_{i-1}}$ (which didn't work quite well) and later $\displaystyle \frac{p_{i}}{p_{i+1}}$. For example, for the first attempt the result I obtained was: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p_{i}}{p_{i-1}}=\frac{r-i+1}{i(n-1)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this kind of problem, I'm expecting to obtain a $1$ in the resulting expression in order to figure out for which values of $i$ there is a maximum in the function, however, I can't get the correct result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-15T22:46:09.993" LastActivityDate="2010-12-15T22:46:09.993" Title="Binomial distribution $\binom{r}{i}(n-1)^{r-i}/n^{r}$" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="14376" PostTypeId="1" CreationDate="2010-12-15T04:39:15.133" Score="7" ViewCount="111" Body="&lt;p&gt;Suppose that every open interval in $\mathbb{R}$ is a connected set. Does this implies the least upper bound axiom? (i.e every non-empty subset of $\mathbb{R}$ which is bounded above has a least upper bound) Is this true? in such case, how would you prove this? &lt;/p&gt;&#xA;" OwnerUserId="4771" LastEditorUserId="4771" LastEditorDisplayName="" LastEditDate="2010-12-15T05:11:03.687" LastActivityDate="2010-12-15T05:59:53.867" Title="Does the fact that every interval in $\mathbb{R}$ is connected implies that $\mathbb{R}$ is order-complete?" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14377" PostTypeId="2" ParentId="14375" CreationDate="2010-12-15T05:07:56.887" Score="2" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://mathforum.org/library/drmath/view/52222.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (also for a better formulation of the problem, using common notation).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-15T05:25:34.197" LastActivityDate="2010-12-15T05:25:34.197" CommentCount="7" />
  <row Id="14378" PostTypeId="2" ParentId="14376" CreationDate="2010-12-15T05:14:11.297" Score="6" ViewCount="0" Body="&lt;p&gt;The answer is yes, in the following sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: For a linear order on a set $X$, the following are equivalent:&lt;br&gt;&#xA;(i) $X$ is connected in the order topology.&lt;br&gt;&#xA;(ii) The ordering is dense and $X$ is Dedekind complete -- every subset which is bounded above admits a least upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Theorem 13 of &lt;a href=&quot;http://www.math.uga.edu/~pete/realinduction.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; for a proof which also explores the connection to induction in totally ordered sets.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-15T05:14:11.297" />
  <row Id="14379" PostTypeId="1" CreationDate="2010-12-15T05:24:11.643" Score="6" ViewCount="91" Body="&lt;p&gt;I am a bit confused on how to go about this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let G be a finite group and $p$ the smallest prime dividing its order.  Assume $P$ is a $p$-Sylow subgroup which is cyclic.  Show $P$ is contained in the center of its normalizer.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T05:30:24.130" LastActivityDate="2010-12-19T20:44:46.580" Title="Containment of small P-sylow subgroups in their Normalizers." Tags="&lt;group-theory&gt;" AnswerCount="3" />
  <row Id="14380" PostTypeId="2" ParentId="14379" CreationDate="2010-12-15T05:39:12.253" Score="4" ViewCount="0" Body="&lt;p&gt;You have to show that if an element $g$ of $G$ normalises $P$, then in fact it commutes with all the elements in $P$, i.e. if the conjugation by $g$ preserves the set, then it actually preserves all the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $g$ normalises $P$, then conjugation by $g$ gives you an automorphism of $P$. What is the automorphism group of a cyclic $p$-group? (Hint: an automorphism is determined by what it does to a generator.) Don't forget the restriction, that $p$ is the smallest prime divisor of $|G|$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-15T05:39:12.253" />
  <row Id="14381" PostTypeId="2" ParentId="14379" CreationDate="2010-12-15T05:40:07.050" Score="4" ViewCount="0" Body="&lt;p&gt;Let $N$ be the normalizer of $P$, and let $N$ act on $P$ by conjugation; this induces a homomorphism $N\to\mathrm{Aut}(P)$; since $P$ is abelian, it is contained in the kernel of that map, so we have that $N/P$ maps into $\mathrm{Aut}(P)$. But what is the order of the automorphism group of a cyclic group of order $p^n$, and what is the order of $N/P$? What does that tell you about the action of $N$ on $P$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-15T05:40:07.050" CommentCount="0" />
  <row Id="14382" PostTypeId="2" ParentId="14376" CreationDate="2010-12-15T05:59:53.867" Score="2" ViewCount="0" Body="&lt;p&gt;Suppose not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take two intervals which witness that fact - namely $(a,b)$ and $(c,d)$ such that $b&amp;lt;c$ however $(b,c)=\emptyset$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose $\varepsilon$ small enough and look at the interval $(b-\varepsilon , c+\varepsilon)$ which is nonempty. By our assumption this interval is also connected, there is someone between $b$ and $c$ otherwise you'd have a non-empty interval which is not connected. A contradiction.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-15T05:59:53.867" />
  <row Id="14383" PostTypeId="1" AcceptedAnswerId="14394" CreationDate="2010-12-15T06:13:49.100" Score="4" ViewCount="198" Body="&lt;p&gt;Here's what I recall of the question from CNML Grade 11, 2010/2011 Contest #3, Question 7:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;There are 2010 points on a circle,&#xA;  evenly spaced.&lt;/strong&gt; Ford Prefect will*&#xA;  randomly choose three points on the&#xA;  circle. He will* connect these points&#xA;  to form a shape. What is the&#xA;  probability that the resulting shape&#xA;  will* form a right angled triangle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9isGJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answered $\frac{1}{4} = 25\%$, but that's probably incorrect. (Right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I got home, I thought it out in my head, and I got this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2010 * (2010/1005)}{2010 \choose 3}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2020050}{1351414120} = \frac{3015}{2017036} = 0.149476756984010201\%$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm probably wrong ...again. Can anyone tell how to get the right answer (if I'm not wrong :) )?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*in the past of the future of the perfect present present time double into ripple fluctuater byer doininger of the past future continuum...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Realized my mistake in copying the question.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-12-15T07:32:32.370" LastActivityDate="2010-12-15T12:23:19.443" Title="How many right angled triangles can a circle have?" Tags="&lt;geometry&gt;&lt;probability&gt;&lt;circle&gt;" AnswerCount="3" CommentCount="19" FavoriteCount="2" />
  <row Id="14384" PostTypeId="1" AcceptedAnswerId="14397" CreationDate="2010-12-15T06:13:51.210" Score="1" ViewCount="64" Body="&lt;p&gt;I was talking with a friend and we were discussing a math problem disguised as a social situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the chance that someone to accept your request to go out with them was 1% and you asked 1 person only, then the probability they would say yes is .01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to ask 100 people, then your chances are still .01 according to my friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, why does it seem intuitive that the chances would increase when asking 100 people rather than 1?  Is it the expected value that's causing this intuition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this situation be modeled using a balls &amp;amp; bins problem?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-12-15T07:55:13.637" Title="Probability increases as sample size increases?" Tags="&lt;probability&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="14385" PostTypeId="2" ParentId="14363" CreationDate="2010-12-15T06:16:42.303" Score="1" ViewCount="0" Body="&lt;p&gt;This problem is #P-complete. Informally, that essentially means that the best you can do is try every possible path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://cstheory.stackexchange.com/questions/2396/counting-the-number-of-hamiltonian-cycles-in-cubic-hamiltonian-graphs&quot;&gt;http://cstheory.stackexchange.com/questions/2396/counting-the-number-of-hamiltonian-cycles-in-cubic-hamiltonian-graphs&lt;/a&gt; for a more complete discussion of this problem.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-12-15T06:16:42.303" />
  <row Id="14386" PostTypeId="1" AcceptedAnswerId="14398" CreationDate="2010-12-15T06:17:50.893" Score="0" ViewCount="113" Body="&lt;p&gt;I have the following overdetermined linear equation system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ax=b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is a matrix of $n \times k$, $x$ is of $k \times 1$,$b$ is of $n \times 1$, where $n&amp;gt;k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know this is an overdetermined linear equation system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is how to check whether the solve for $x$, and check that the vector is consistent in this case? Consistent as in the sense that when we plug in the $x$ vector value into the above linear equation systems, then the above matrix will be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can separate out the $k$ linear equations and find $x$ from $x_1$ to $x_k$, and then substitute in the remaining equations to check for consistency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I afraid that  this method can be numerically unstable; I would like to implement this on a computer, so I would prefer a solution that fully works here. Let us consider one pitfall of my above solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=\begin{bmatrix} 10 &amp;amp; 10 \\ 0 &amp;amp; 0 \\0 &amp;amp; 10 \end{bmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you separate the $1$ and $2$ rows out, and compute the solution, you may not be able to even solve it ( equation $2$ is an equation here with no unknown terms, after you times in the $0$ factor)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there other method?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-16T07:07:08.177" LastActivityDate="2010-12-16T10:50:17.730" Title="Check Whether an Overdetermined Linear Equation System is Consistent: General Approach?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="14387" PostTypeId="1" AcceptedAnswerId="14391" CreationDate="2010-12-15T06:18:14.847" Score="1" ViewCount="75" Body="&lt;p&gt;Hi there, just a quick question dealing with probability.  The annual returns on stocks and treasury bonds over the next 12 months are uncertain. Suppose that these returns can be described by normal distributions with stocks having a mean of 15% and a standard deviation of 20%, and bonds having a mean of 6% and a standard deviation of 9%. Which asset is more likely to have a negative return?  Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="4773" LastActivityDate="2010-12-15T06:54:29.893" Title="Normal distribution probability " Tags="&lt;probability&gt;&lt;homework&gt;&lt;economics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14389" PostTypeId="1" CreationDate="2010-12-15T06:32:46.160" Score="6" ViewCount="216" Body="&lt;p&gt;I was asked to compute the Fourier series for $\sin^2(x)$ on $[0,\pi]$. Now this is what I did and I'd like to know if I'm right.  $\sin^2(x)=\frac12-\frac12\cos(2x)$ . I got the right hand side using trig identities. I'm wondering If I can do this without using the formulas. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4774" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T09:31:49.103" LastActivityDate="2010-12-16T07:09:59.407" Title="Fourier series for $\sin^2(x)$" Tags="&lt;homework&gt;&lt;fourier-series&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="14390" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T06:50:03.237" Score="2" ViewCount="0" Body="&lt;p&gt;Sure. That is the Fourier series for the function.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-15T06:50:03.237" CommentCount="8" />
  <row Id="14391" PostTypeId="2" ParentId="14387" CreationDate="2010-12-15T06:54:29.893" Score="5" ViewCount="0" Body="&lt;p&gt;I don't know if that's what you are looking for, but if $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma &amp;gt; 0$, then &#xA;$$&#xD;&#xA;{\rm P}(X&amp;lt;0) = {\rm P}\bigg(\frac{{X - \mu }}{\sigma } &amp;lt; \frac{{0 - \mu }}{\sigma }\bigg) = {\rm P}\bigg(Z &amp;lt; \frac{{ - \mu }}{\sigma }\bigg),&#xD;&#xA;$$&#xD;&#xA;where $Z$ is a standard normal random variable, i.e. having mean zero and unit variance (or standard deviation). The probability ${\rm P}(Z&amp;lt;z)$ can be found using a Normal Distribution Calculator, and many are available online (check, e.g., &lt;a href=&quot;http://davidmlane.com/hyperstat/z_table.html&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-15T06:54:29.893" CommentCount="3" />
  <row Id="14392" PostTypeId="2" ParentId="14386" CreationDate="2010-12-15T07:35:31.143" Score="0" ViewCount="0" Body="&lt;p&gt;I am not sure why I cannot comment, but anyway, you can always do an RREF and then you should be fine.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-15T07:35:31.143" CommentCount="4" />
  <row Id="14393" PostTypeId="2" ParentId="14383" CreationDate="2010-12-15T07:45:32.823" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: You can choose $3$ points in ${2010}\choose {3}$ ways. To form a right angled triangle, we need to choose a pair of diametrically opposite points (how many ways to do this?) and then any third point will form a right angled triangle with the chosen pair. &lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-15T07:45:32.823" CommentCount="8" />
  <row Id="14394" PostTypeId="2" ParentId="14383" CreationDate="2010-12-15T07:45:44.583" Score="10" ViewCount="0" Body="&lt;p&gt;Let the points be $P_1$, $P_2$, and $P_3$.  Letting $P_1$ be arbitrary, we win if $P_2$ is the point diametrically opposite to $P_1$, probability $1/2009$.  Otherwise, with probability $2008/2009$, we still win if $P_3$ is diametrically opposite either $P_1$ or $P_2$, which it is with probability $2/2008$.  So the probability of a win is $1/2009 + (2008/2009)(2/2008) = 3/2009$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get this as follows:  The number of possible wins is the number of diameters times the number of remaining points, or 1005*2008. The number of possible triples is $2010\choose 3$. Dividing the first by the second gives $3/2009$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even easier:  The probability that a given pair of points lie on a diameter is $1/2009$.  With three points, you have three pairs, hence $3/2009$.  Here it's OK to add probabilities because it's not possible to have overlap; if one of the pairs lies on a diameter, no other pair can lie on a diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I do not agree with the comment thread that says we must consider the possibility that the points are not distinct.  Sometimes there can be an ambiguity, but in common language &quot;choose three points from 2010 points&quot; means &quot;without replacement&quot;.  For example, I am not incorrect in saying that $n\choose r$ is the number of ways of choosing $r$ objects out of a set of $n$.  Although I admit that if I were grading the test and this issue were pointed out to me, I might accept the &quot;with replacement&quot; answer as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other alternative answer I might accept is 1, i.e., that the points are certain to lie on a right triangle.  After all, it is Ford Prefect, so I have no way of knowing whether the Infinite Improbability Drive has been activated.&lt;/p&gt;&#xA;" OwnerUserId="2682" LastEditorUserId="2682" LastEditorDisplayName="" LastEditDate="2010-12-15T12:23:19.443" LastActivityDate="2010-12-15T12:23:19.443" CommentCount="3" />
  <row Id="14395" PostTypeId="2" ParentId="14383" CreationDate="2010-12-15T07:51:18.967" Score="1" ViewCount="0" Body="&lt;p&gt;As you know, the triangle will be right-angled if and only if two of the three points lie on opposite ends of a diameter. There are $2010^3$ ways of choosing three points (your problem didn't specify that the points are distinct). Let's count, how many of the choices give a right-angled triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of ways of doing this counting, here is one of them: I will first count the right-angled triangles whose right angle lies on vertex number 1. All the others will be obtained from these by rotations, so the number will have to by multiplied by 2010. Next, I can choose the vertices of any triangle in any order, so I will have to multiply by 3!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, once I have fixed the vertex 1, there are 2008 ways of choosing the second vertex (I am not allowed to take the same vertex, nor the opposite one, since I want my right angle to be at vertex 1). But that determines the third vertex uniquely, since it has to lie opposite the second, so I get 2008 possibilities. But wait a minute, I have counted each triangle twice, since putting the second vertex at 2 and the third at the opposite end gives the same triangle as swapping the second and the third vertex. So I get 1004 right-angled triangles with the right angle at vertex 1, so by rotating I get 1004*2010 right angled triangles altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any such triangle can be obtained by 6 different choices of vertices, since I can choose the vertices in any order, so the probability is $\frac{6*1004*2010}{2010^3}=\frac{6024}{2010^2}$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-15T07:51:18.967" CommentCount="1" />
  <row Id="14397" PostTypeId="2" ParentId="14384" CreationDate="2010-12-15T07:55:13.637" Score="4" ViewCount="0" Body="&lt;p&gt;The problem stems from ambiguous use of term chance and probability. Probability can be defined as a ratio of successful outcomes and number of trials. So if we increase the number of trials and let the probability remain the same, the number of successful outcomes must increase in order that the ratio remains the same. So if we say that number of successful outcomes are chances, then yes, chances are increasing, yet the probability remains the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at that is to look at the probability of getting a no. With one person the probability is $0.99$. If you ask 100 persons, then the probability that they all say no is $0.99^{100}\approx 0.36$. So the probability that you will get at least one positive answer after asking 100 persons is actualy $1-0.99^{100}\approx0.64$. If you ask 1000 persons, then the probability of getting a no from all of them is $0.99^{1000}\approx 4.31\times 10^{-5}$, so probability of one yes is $1-0.99^{1000}\approx0.99999$. But that is not surprising, since saying that probability of a yes is 0.01 you are effectively saying that 1% of population will say yes. So among 1000 persons there should be 10 who say yes, so getting only one yes from 1000 persons is less likely (the event has smaller probability) than getting 10.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-15T07:55:13.637" CommentCount="1" />
  <row Id="14398" PostTypeId="2" ParentId="14386" CreationDate="2010-12-15T08:09:12.293" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Updated according to comments.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are worried about the numerical stability do &lt;a href=&quot;http://en.wikipedia.org/wiki/QR_decomposition&quot; rel=&quot;nofollow&quot;&gt;QR decomposition&lt;/a&gt; of matrix $A$. Then $A=QR$, where $Q$ is orthogonal and $R$ is triangular. Then you need to check whether $x$ satisfies the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Rx=Q^Tb$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $R$ is triangular and $n&amp;gt;k$ we will have that the last $n-k$ rows of $R$ are zero. Since $Ax=b$ it follows that the last $n-k$ elements of $Q^Tb$ should be zero also. If they are not, then $x$ is not a solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore since we have an overdetermined matrix the solution exists only if $b$ lies in the linear space spanned by columns of $A$. So the real question is, how do we reliably check whether $b$ is in linear space spanned by columns of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n&amp;gt;k$, the $R$ matrix will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;R=\begin{bmatrix}&#xA;R_1\\&#xA;0&#xA;\end{bmatrix}&#xA;\end{align*}&#xA;where $R_1$ is $k\times k$ upper triangular matrix. If the solution exists, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;Q^Tb=\begin{bmatrix}&#xA;b_1\\&#xA;0&#xA;\end{bmatrix}&#xA;\end{align*}&#xA;where $b_1$ is $k\times 1$ vector. The solution for our system is then&#xA;$$x=R_1^{-1}b_1$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-15T11:41:08.443" LastActivityDate="2010-12-15T11:41:08.443" CommentCount="11" />
  <row Id="14399" PostTypeId="1" AcceptedAnswerId="14404" CreationDate="2010-12-15T08:10:26.397" Score="0" ViewCount="66" Body="&lt;p&gt;I am using math as a tool and for entertainment, I am not a pro, I do not understand most of the posts here but I am glad such site exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, a friend asked me a question the other day, and I was able to redefine it in terms of a directed graph where all nodes in and out degree equal to 1. I know that If if each node's indegree equals to outdegree that is called an Eulerian graph, what I am asking is a special case of these graphs. Do they have a name or are their properties known?&lt;/p&gt;&#xA;" OwnerUserId="4776" LastActivityDate="2010-12-15T09:28:19.107" Title="Directed graph with all indegree = outdegree = 1" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14401" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T08:26:52.053" Score="1" ViewCount="0" Body="&lt;p&gt;@AD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the question has been answered... the Fourier expansion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[a_n\cos(nx) + b_n\sin(nx)\right]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a decomposition over the set of orthogonal trigonometric functions $\{1, \sin(kx), \cos(kx)\}, \forall k \in \mathbb{Z}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that means that really the Fourier expansion of a function is unique. And so, simply by identifying $a_0 = \frac{1}{2}$, $a_2 = -\frac{1}{2}$ and all of the other coefficients 0, we find a pairing that has the desired form of a Fourier Series, and which therefore &lt;em&gt;is&lt;/em&gt; the Fourier expansion of $f(x)$.&lt;/p&gt;&#xA;" OwnerUserId="4139" LastEditorUserId="4139" LastEditorDisplayName="" LastEditDate="2010-12-15T09:08:05.217" LastActivityDate="2010-12-15T09:08:05.217" CommentCount="3" />
  <row Id="14402" PostTypeId="2" ParentId="14113" CreationDate="2010-12-15T08:36:00.337" Score="1" ViewCount="0" Body="&lt;p&gt;You should take a look at the paper: &lt;a href=&quot;http://projecteuclid.org/euclid.em/1087568020&quot; rel=&quot;nofollow&quot;&gt;Some New Formulas for π&lt;/a&gt; by Gert Almkvist, Christian Krattenthaler, and Joakim Petersson, &lt;em&gt;Experiment. Math.&lt;/em&gt; Volume 12, Number 4 (2003), 441-456.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T09:07:35.300" LastActivityDate="2010-12-15T09:07:35.300" />
  <row Id="14403" PostTypeId="1" AcceptedAnswerId="14641" CreationDate="2010-12-15T09:00:15.290" Score="0" ViewCount="93" Body="&lt;p&gt;It is referred to in &lt;a href=&quot;http://ifile.it/niok3wy/jeffreys-an_invariant_form-1946.pdf&quot; rel=&quot;nofollow&quot;&gt;An invariant form for the prior probability in estimation problems (Jeffreys)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="774" LastActivityDate="2010-12-17T07:17:42.670" Title="What is the $d\sigma/\sigma$ rule?" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14404" PostTypeId="2" ParentId="14399" CreationDate="2010-12-15T09:28:19.107" Score="3" ViewCount="0" Body="&lt;p&gt;These types of graphs are equivalent to &lt;a href=&quot;http://en.wikipedia.org/wiki/Permutation&quot; rel=&quot;nofollow&quot;&gt;permutations&lt;/a&gt; (or permutations without fixed points, if you're not allowed a vertex to have an edge to itself).  If there is an edge from a to b, then a is mapped to b.  These are extremely well-studied in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[As Prometheus remarks, permutations are equivalent to a collection of cycles]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-15T09:28:19.107" CommentCount="2" />
  <row Id="14406" PostTypeId="1" AcceptedAnswerId="14414" CreationDate="2010-12-15T11:27:25.777" Score="11" ViewCount="275" Body="&lt;p&gt;I was watching this &lt;a href=&quot;http://www.ted.com/talks/nicholas_christakis_how_social_networks_predict_epidemics.html&quot; rel=&quot;nofollow&quot;&gt;TED talk&lt;/a&gt;, which suggested that on average your friends tend to individually have more friends than you do. To define this more formally, we are comparing the average number of friends with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;average over each person p of:&#xA;    friend popularity, defined as:&#xA;    average over each friend f of p:&#xA;        number of friends f has&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Intuitively, this seems to make sense. After all, if someone has a high number of friends, they will tend to increase friend popularity and affect a high number of people, while those people who decrease friend popularity only affect a low number of people. Does this result hold for all graphs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a person &lt;code&gt;p&lt;/code&gt;, let &lt;code&gt;t&lt;/code&gt; stand for:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sum over each friend f of p:&#xA;    number of friends f has&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is pretty clear that &lt;code&gt;sum(t)=sum(f^2)&lt;/code&gt; as a person with &lt;code&gt;f&lt;/code&gt; friends has value of &lt;code&gt;f&lt;/code&gt; towards their &lt;code&gt;f&lt;/code&gt; friends value of &lt;code&gt;t&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are then trying to determine whether: &lt;code&gt;sum(t/f)&amp;gt;sum(f)&lt;/code&gt; holds for all graphs.&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="123" LastEditorDisplayName="" LastEditDate="2010-12-15T11:47:45.143" LastActivityDate="2010-12-15T14:12:12.140" Title="Do your friends on average have more friends than you do?" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="14407" PostTypeId="2" ParentId="13053" CreationDate="2010-12-15T11:34:42.273" Score="3" ViewCount="0" Body="&lt;p&gt;To expand on the comments I made (and Anixx has confirmed that it works):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem, as noted in the &lt;em&gt;Mathematica&lt;/em&gt; documentation for &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/Zeta.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;Zeta[]&lt;/code&gt;&lt;/a&gt; and &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/HurwitzZeta.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;HurwitzZeta[]&lt;/code&gt;&lt;/a&gt; is that those two functions are defined slightly differently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Zeta}(s,a):=\sum_{j=0}^\infty\frac1{((j+a)^2)^{\frac{s}{2}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta(s,a):=\sum_{j=0}^\infty\frac1{(j+a)^s}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the two expressions are equivalent for $\Re a &amp;gt; 0$, they behave rather differently in the left half-plane:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N[Zeta[1/3, -1/5], 20]&#xA;0.99113858338730120895&#xA;&#xA;N[HurwitzZeta[1/3, -1/5], 20]&#xA;0.1361506100489527143 - 1.4808826096823642385 I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and this manifests itself when you use &lt;code&gt;Zeta[]&lt;/code&gt; instead of &lt;code&gt;HurwitzZeta[]&lt;/code&gt; for your gamma function expression in the left half-plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not knowing how they implemented &lt;code&gt;HurwitzZeta[]&lt;/code&gt; internally, however, (it seems to have only been implemented starting with version 7 of &lt;em&gt;Mathematica&lt;/em&gt;), I cannot say what, exactly, are the differences in the behavior of those two functions in the left half of the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, a similar issue arises with the Lerch transcendent $\Phi(z,s,a)$, the trivariate generalization of the zeta function and the related polylogarithms; &lt;em&gt;Mathematica&lt;/em&gt; from version 7 onwards provides for two different functions, &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/LerchPhi.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;LerchPhi[]&lt;/code&gt;&lt;/a&gt; and &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/HurwitzLerchPhi.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;HurwitzLerchPhi[]&lt;/code&gt;&lt;/a&gt; which also agree for $\Re a &amp;gt; 0$ but behave differently otherwise.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-15T11:34:42.273" CommentCount="5" />
  <row Id="14408" PostTypeId="1" AcceptedAnswerId="14419" CreationDate="2010-12-15T12:02:59.000" Score="1" ViewCount="49" Body="&lt;p&gt;It's possible, that count of unique algebras (up to isomorphism) with one unary operation on set with $n$ elements is $2^n-1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1,2,3,4$ is this hypothesis true (I still have not verified it on the computer). Is there any counterexample, or idea for proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4195" LastEditorUserId="4195" LastEditorDisplayName="" LastEditDate="2010-12-15T12:10:53.023" LastActivityDate="2010-12-15T15:32:07.513" Title="Count of unique algebras with one unary operation (on finite set)" Tags="&lt;algebra&gt;&lt;universal-algebra&gt;" AnswerCount="1" />
  <row Id="14409" PostTypeId="1" CreationDate="2010-12-15T12:41:19.163" Score="0" ViewCount="117" Body="&lt;p&gt;I was trying to expand $\displaystyle \left(1+x\right)^{y}$ as a power series in terms of $y$ using the Generalized Binomial Theorem, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}\binom{y}{k}x^{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $x,y \in \mathbb{R}$ and $\left | x \right | &amp;lt; 1$. With this in mind I would get something like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}a_{k}\cdot y^{k}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the general form of the $a_{k}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastActivityDate="2010-12-15T12:41:19.163" Title="Generalized Binomial Expansion of $\left(1+x \right )^{y}$" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;power-series&gt;" CommentCount="10" />
  <row Id="14410" PostTypeId="2" ParentId="14406" CreationDate="2010-12-15T13:13:49.847" Score="4" ViewCount="0" Body="&lt;p&gt;Trivial comment, but I'm new so I have to post it as an answer: The strict inequality definitely doesn't hold in general, since the sums are equal for regular graphs.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-15T13:13:49.847" />
  <row Id="14411" PostTypeId="1" AcceptedAnswerId="14413" CreationDate="2010-12-15T13:37:48.860" Score="1" ViewCount="92" Body="&lt;p&gt;Hi,&#xA;It is well known that any conformal mapping of the unit disk onto itself extends to the unit cirle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is there an homeomorphism of the unit disk onto itself which does not extend to a continuous function on the closed unit disk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, can you give an explicit one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-12-15T18:23:27.797" Title="Homeomorphism of the unit disk onto itself which does not extend to the boundary" Tags="&lt;analysis&gt;" AnswerCount="2" />
  <row Id="14412" PostTypeId="1" CreationDate="2010-12-15T13:54:55.610" Score="4" ViewCount="101" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Omega$ be a domain of the complex plane. The Hardy space $H^p(\Omega)$ is defined, for $1 \leq p&amp;lt;\infty$, as the class of functions $f$ that are holomorphic on $\Omega$ such that $|f|^p$ has a harmonic majorant on $\Omega$, i.e. there is a function $u$ harmonic on $\Omega$ such that&#xA;$$|f(z)|^p \leq u(z) $$&#xD;&#xA;for all $z \in \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=\infty$, $H^\infty(\Omega)$ is the class of bounded holomorphic functions on $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in cases when $H^p(\Omega)$ consist only of constant functions. For example, this is the case when $\Omega$ is the whole plane, because positive harmonic functions on $\mathbb{C}$ are constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came upon the following question :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be a compact subset of the real line, and suppose that $E$ has zero length. Let $\Omega$ be the complement of $E$. Does $H^p(\Omega)$ consist only of the constant functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $p=\infty$, the answer is yes : one can use Cauchy's formula to extend any bounded holomorphic function on $\Omega$ to a bounded holomorphic function on $\mathbb{C}$, and that function is now constant by Liouville's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $1 \leq p&amp;lt;\infty$, I am pretty sure the answer is also yes. However, I can't seem to find a way to extend $f$ or the harmonic majorant of $|f|^p$ to the whole plane. Is there any way to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-12-26T15:43:06.340" Title="Removable sets for harmonic functions and hardy spaces of general domains" Tags="&lt;complex-analysis&gt;&lt;harmonic-analysis&gt;" FavoriteCount="1" />
  <row Id="14413" PostTypeId="2" ParentId="14411" CreationDate="2010-12-15T14:06:29.490" Score="10" ViewCount="0" Body="&lt;p&gt;Yes, suppose we have in polar coordinates $f(r,\theta) = (r,\theta + r/(1-r))$.  In effect the disk is rotated by increasing amounts as $r$ tends to 1.  It is still a homeomorphism on the interior of the unit disk, but it's impossible to extend continuously to the boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; This map cannot be extended continuously to any point of the boundary.  If it could be, then for any &quot;target&quot; open neighborhood of the image of that point, we should be able to find an open neighborhood of the point whose image is contained in the &quot;target&quot;.  But the open neighborhood of any boundary point will be mapped to points arbitrarily close to every point on the boundary (and so cannot be contained in any sufficiently small &quot;target&quot;).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-15T18:23:27.797" LastActivityDate="2010-12-15T18:23:27.797" CommentCount="2" />
  <row Id="14414" PostTypeId="2" ParentId="14406" CreationDate="2010-12-15T14:12:12.140" Score="13" ViewCount="0" Body="&lt;p&gt;The answer is yes, this holds for any graph (with weak inequality, as Jon points out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's set up some notation. The graph of friendships is $G$. The set of vertices of $G$ (the people) is $V$; the set of edges (the friendships) is $E$. For a person $v$, the number of friends that person has is $\deg v$. The total number of people is $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to show that&#xA;$$\frac{1}{n} \sum_{v \in V} \deg v \leq \frac{1}{n} \sum_{v \in V} \frac{1}{\deg v} \sum_{(u,v) \in E} \deg(u).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancel the $1/n$'s from both sides. After a little rewriting, we want to show that&#xA;$$\sum_{v \in V} \sum_{(u,v) \in E} 1 \leq \sum_{v \in V} \sum_{(u,v) \in E} \frac{\deg u}{\deg v}. \quad (*)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider what a given edge $(u,v)$ contributes to each side of $(*)$. On the left, it contributes $1+1=2$. On the right, it contributes $(\deg u)/(\deg v) + (\deg v)/(\deg u)$. For any two positive numbers $x$ and $y$, we have $2 \leq x/y+y/x$. So every edge contributes more to the right hand side of $(*)$ than to the left, and we have the claimed result.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-15T14:12:12.140" CommentCount="2" />
  <row Id="14415" PostTypeId="2" ParentId="14411" CreationDate="2010-12-15T14:20:34.210" Score="0" ViewCount="0" Body="&lt;p&gt;This following gives an example, &lt;em&gt;probably&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H=\{(x,y)\in\mathbb R^2:y&amp;gt;0\}$ be the upper half plane in $\mathbb R^2$ and let $\bar H$ be its closure. Let me give you an homeo $f:H\to H$ which does not extend to an homeo $\bar f:\bar H\to\bar H$, and which fixes the point of infinity on the $x$-axis. Then you can conjugate with an homeo $D\to H$ from the open unit disk which extends to the boundary (apart from one point...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\phi:\mathbb R\times(0,\infty)\to\mathbb R$ be given by $\phi(x,t)=\frac1{\sqrt{4\pi t}}\exp\left(-\frac{x^2}{4t}\right)$, and let $h:\mathbb R\to\mathbb R$ be a positive smooth function with support on $[-2,2]$ and such that it is constantly $1$ on $[-1,1]$. Define now $g:H\to H$ so that when $x\in\mathbb R$ and $y&amp;gt;0$ we have $$&#xD;&#xA;g(x,y)=\int_{-\infty}^\infty\phi(x-\xi,y)(1-h(\xi))\,\mathrm d\xi.&#xD;&#xA;$$ This is a solution of the one-dimensional head equation. Properties of that equation (which one can easily show in this case!) imply that $f$ extends continuously to a map $\bar g:\bar H\to\mathbb H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the map $f:(x,t)\in H\mapsto (xg(x,t),t)\in H$. This is an homeomorphism, and it extends to the map $\bar f:(x,t)\in\bar H\mapsto (x\bar g(x,t),t)\in\bar H$, which is not an homeo.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-15T14:27:04.660" LastActivityDate="2010-12-15T14:27:04.660" />
  <row Id="14416" PostTypeId="2" ParentId="14327" CreationDate="2010-12-15T14:33:17.497" Score="2" ViewCount="0" Body="&lt;p&gt;Thinking more, as n gets large we can pass to the continuous case and get rid of the ceilings and e.  So imagine we are pulling numbers from [0,1] n times and each time you can accept it and take the value or decline it and take your future chances.  Let E(n) be the expectation value of n pulls.  Again, when you have n pulls left you should accept anything greater than E(n-1).  So the recurrence is $E(n)=(1-E(n-1))\frac{1+E(n-1)}{2}+E(n-1)^2$ or $E(n)=\frac{1+E(n-1)^2}{2}$  I think we were told &lt;a href=&quot;http://math.stackexchange.com/questions/12485/whats-known-about-recurrences-involving-a-n2&quot;&gt;here&lt;/a&gt;  that sequences involving E(n-1)^2 are very hard.  In the current case the E(n-1)^2 is divided by 2 but we can take that out by rescaling to D=2E and we have $D(n)=D(n-1)^2+\frac{1}{2}$ with D(1)=1&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-15T14:33:17.497" CommentCount="2" />
  <row Id="14417" PostTypeId="1" AcceptedAnswerId="14539" CreationDate="2010-12-15T15:16:28.970" Score="5" ViewCount="352" Body="&lt;p&gt;Can this sum be simplified: $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or at least is there a simple fairly tight upperbound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;So I think this sum is more easily bounded than I previously thought:&#xA;Clearly, $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \leq 2^{\sqrt{n}} \sum_{k=0}^{n} \binom{n}{k} = 2^{n+\sqrt{n}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, along the same lines as Shai Covo's answer, $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \geq \binom{n}{n/2} 2^{\sqrt{n/2}}$. The central binomial coefficient: $\binom{n}{n/2}$ is at least $\frac{2^n}{\sqrt{2n}}$ hence $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \geq 2^{\sqrt{n/2}} \cdot \frac{2^n}{\sqrt{2n}} = \frac{2^{n + \sqrt{n/2}}}{2n}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ is $O(2^{n + \sqrt{n}})$ and $\Omega(\frac{2^{n+\sqrt{n/2}}}{n})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about expressions of the form:&#xA;$\sum_{k=0}^{n} \binom{n}{k} a^{\sqrt{k}} b^{\sqrt{n-k}}$&lt;/p&gt;&#xA;" OwnerUserId="1487" LastEditorUserId="1487" LastEditorDisplayName="" LastEditDate="2010-12-15T19:28:07.063" LastActivityDate="2010-12-17T06:45:13.783" Title="Simplify $\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="14418" PostTypeId="2" ParentId="14356" CreationDate="2010-12-15T15:17:11.550" Score="1" ViewCount="0" Body="&lt;p&gt;Since you have more of an engineering background, I would suggest the textbook of Robert McOwen, &lt;em&gt;Partial Differential Equations: Methods and Applications&lt;/em&gt;. Chapter 1 is about solving first order partial differential equations, where the method of characteristics is most applicable. Chapters 2 and 3 also contains some application of the notion of characteristics to higher order equations and to the wave equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare to the treatments in other standard, more theory-oriented textbooks, I find McOwen's treatment of the method of characteristics to be more detailed and complete, and easier to teach from. As a remark: section 1.3, the application to fully non-linear first order equations, is very difficult to wrap your head around. So don't worry if you feel like you suddenly hit a wall when you get to that section. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-15T15:17:11.550" />
  <row Id="14419" PostTypeId="2" ParentId="14408" CreationDate="2010-12-15T15:32:07.513" Score="3" ViewCount="0" Body="&lt;p&gt;In brief: no. There are 19 algebras on 4 elements, and the pattern breaks down after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of algebras with one unary operation, we might as well speak of nonisomorphic functions on a set of $n$ elements. The answer is then given by Theorem 6 of &quot;The Number of Structures of Finite Relations&quot; by Davis, R.L. A pdf is available &lt;a href=&quot;http://www.ams.org/journals/proc/1953-004-03/S0002-9939-1953-0055294-2/S0002-9939-1953-0055294-2.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-15T15:32:07.513" />
  <row Id="14420" PostTypeId="2" ParentId="639" CreationDate="2010-12-15T16:00:57.083" Score="1" ViewCount="0" Body="&lt;p&gt;Pour all the wine down the drain and nip down to the nearest liquor store. Result, 1000 cups of untainted wine and 0 servant related deaths.&lt;/p&gt;&#xA;" OwnerUserId="4780" LastActivityDate="2010-12-15T16:00:57.083" CommentCount="1" />
  <row Id="14421" PostTypeId="1" AcceptedAnswerId="14425" CreationDate="2010-12-15T16:16:13.607" Score="2" ViewCount="119" Body="&lt;p&gt;Hello.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are some encrypting algorithm that once the text (the message) is encrypted there is no way to decrypt it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think one way of doing this: using as encryption key the current date at the encryption moment. But there is one problem wiht that: Each time you encrypt the same message the result will be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there must be some &quot;math-algorithm&quot; to encrypt without out the posibility to decrypt and that always the same message turns out with the same encrypted message, but I realy can't imagine how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I'm posting this question in this forum and not in &lt;a href=&quot;http://stackoverflow.com&quot;&gt;stackoverflow.com&lt;/a&gt; intentionally, it is not referred to &quot;how to programm the algorithm&quot;, it is refered to &quot;how is the math algorithm&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3314" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T16:18:35.803" LastActivityDate="2010-12-16T15:16:06.130" Title="Encrypting algorithm" Tags="&lt;algorithms&gt;&lt;cryptography&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="14422" PostTypeId="2" ParentId="14115" CreationDate="2010-12-15T16:22:08.093" Score="1" ViewCount="0" Body="&lt;p&gt;Here's an easy introduction to the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Pi Formulas and the Monster Group&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sites.google.com/site/tpiezas/0013&quot; rel=&quot;nofollow&quot;&gt;http://sites.google.com/site/tpiezas/0013&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4781" LastActivityDate="2010-12-15T16:22:08.093" />
  <row Id="14423" PostTypeId="1" CreationDate="2010-12-15T16:38:40.067" Score="4" ViewCount="142" Body="&lt;p&gt;I am studing for exams and am stuck on this problem.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt; &#xA;Suppose $f$ is an entire function s.t. $f(z) =f(z+1)$ and $|f(z)| &amp;lt; e^{|z|}$.&#xA;Show $f$ is constant.&#xA;&lt;/blockquote&gt; &#xA;&#xA;&lt;p&gt;I've deduced so far that:&#xA;a) $f$ is bounded on every horizontal strip &#xA;b) for every bounded horizontal strip of length greater than 1 a maximum modulus must occur on a horizontal boundary.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-15T18:01:33.990" Title="Entire &quot;periodic&quot; function" Tags="&lt;complex-analysis&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="14424" PostTypeId="2" ParentId="14421" CreationDate="2010-12-15T16:43:49.993" Score="5" ViewCount="0" Body="&lt;p&gt;As J.M. commented above, the only totally secure cryptographic algorithm is the one-time pad. Because this relies on both the sender and receiver of a message to be able to have access to the &lt;em&gt;same&lt;/em&gt; arbitrarily long random sequence, it is often not feasible. Cryptography originally relied on shared keys, from which simple ciphers, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher&quot; rel=&quot;nofollow&quot;&gt;Vigenere&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesar_cipher&quot; rel=&quot;nofollow&quot;&gt;Caesar&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Substitution_cipher&quot; rel=&quot;nofollow&quot;&gt;substitution&lt;/a&gt;, come from. Unfortunately, all of these algorithms are so simple that using frequency analysis can often be used to break them very quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern cryptography is generally split into two core areas: symmetric and asymmetric algorithms. Symmetric seems to be what you are describing, so I will describe that first. In symmetric algorithms, both users must share a key, which they often pass to one another via asymmetric cryptography or using &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange&quot; rel=&quot;nofollow&quot;&gt;Diffie-Hellman key exchange&lt;/a&gt;. Once the key is known between the two parties, eavesdroppers are generally going to have a very difficult time breaking the encryption. As with all things, certain algorithms are stronger than others. Older encryption schemes like &lt;a href=&quot;http://en.wikipedia.org/wiki/Data_Encryption_Standard&quot; rel=&quot;nofollow&quot;&gt;DES&lt;/a&gt; are not as secure as some of the newer ones like &lt;a href=&quot;http://en.wikipedia.org/wiki/Advanced_Encryption_Standard&quot; rel=&quot;nofollow&quot;&gt;AES&lt;/a&gt;, and it's not infeasible to try to break the weaker ones with sufficient computing power. Asymmetric cryptography, or public key cryptography, eliminates the need to share keys between users by admitting a public key and private key for each user. To send something to someone else, you encrypt using their public key (which they publish) and only they are able to decrypt using their private key. &lt;a href=&quot;http://en.wikipedia.org/wiki/RSA&quot; rel=&quot;nofollow&quot;&gt;RSA&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_curve_cryptography&quot; rel=&quot;nofollow&quot;&gt;ECC&lt;/a&gt; are two of the more common public key protocols.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastActivityDate="2010-12-15T16:43:49.993" CommentCount="3" />
  <row Id="14425" PostTypeId="2" ParentId="14421" CreationDate="2010-12-15T16:50:01.953" Score="6" ViewCount="0" Body="&lt;p&gt;Such algorithms do exist. Which one you mean depends on what you mean by &quot;no way to decrypt it&quot;. You refer to either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A function which takes an input and generates a unique output, with there being ideally no known way to derive the input from the output. These are called hashing algorithms; examples of which include &lt;a href=&quot;http://en.wikipedia.org/wiki/MD5&quot; rel=&quot;nofollow&quot;&gt;MD5&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/SHA256&quot; rel=&quot;nofollow&quot;&gt;SHA256&lt;/a&gt;. Those links provide more detail than I possibly can.&lt;/li&gt;&#xA;&lt;li&gt;A cryptosystem that is impossible for any other person but the keyholder to decrypt. This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/One-time_pad&quot; rel=&quot;nofollow&quot;&gt;one time pad&lt;/a&gt; &lt;a href=&quot;http://math.stackexchange.com/users/498/j-m&quot;&gt;J.M.&lt;/a&gt; mentions. This is the only currently &lt;a href=&quot;http://netlab.cs.ucla.edu/wiki/files/shannon1949.pdf&quot; rel=&quot;nofollow&quot;&gt;proven-secure algorithm&lt;/a&gt; and even that has been &lt;a href=&quot;http://arxiv.org/pdf/0709.4420&quot; rel=&quot;nofollow&quot;&gt;thrown into doubt&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edited in because this is probably too long for comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diego, no problem. They're not, as they're two different things and you're comparing apples and oranges, so to speak. A cryptosystem is designed to be hard to undo without the key, but easy if you have the key. By contrast, a hashing function should only require 1 input, be one way and give a unique output for each input. The idea is to be able to generate a unique representation of the output without being able to deduce what the output is. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well aside from verifying the unique nature of your input, a hashing function is used as part of a digital signature. In a public key cryptosystem, you may well be aware that Alice can send Bob an encrypted message using Bob's public key. However, anyone can send Bob a message using Bob's public key. If Eve has broken the cryptosystem, she could intercept Alice's message, decrypt it, change it, re-encrypt it and send it on to Bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we prevent this? Well, let $H: x\rightarrow y$ be a map from $x$ some input to $y$ some output. It is trivial to compute $H$ but impossible (ideally, the truth is just very very unlikely) to compute $H^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let $C:x,k$ be some assymetric cryptosystem Alice and Bob know, with Alice's keys $a_p$ and $a_s$ (public and secret) and likewise Bob's keys $b_p$ and $b_s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that as this cryptosystem is assymetric, that $C:x,b_p \rightarrow z$ where $z$ is some ciphertext (encrypted secret) and that $C:z,b_s \rightarrow x$ i.e. if you use Bob's public key to encrypt, Bob's private key using exactly the same algorithm will decrypt the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This we know already. However, as discussed, a man in the middle attack is possible if Eve knows Bob's secret. However, Alice can do two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Compute $C(H(x), a_s)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;Compute $C(x+H(x), b_p)$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And send the second one to Bob. What's the point in this? Well, Bob can reverse both &quot;encryptions&quot;, because he holds the opposite keys (his own private and Alice's public. He is left with two things: the ciphertext and a unique hash of the ciphertext. He can also compute H(x) and compare results. If they differ, he knows the message has been tampered with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Well, consider Eve sat gleefully with Bob's keys. She intercept's Alices message and sees it contains the two parts (she can decrypt the outer layer, Bob's message). However, now she's left with an impossible situation. She can compute the hash of her newly inserted message, but she does not know Alice's private key (if she does, the cryptosystem is truly broken). Therefore she cannot encrypt the new hash. So, if she tries to send the new message onwards, Bob will immediately be aware the system has been tampered with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called a digital signature or digitally signing and is used to identify the message author as who they truly say they are. After all, only they can encrypt with their private key (allegedly). The importance of hashing in this scenario is that it generates a unique output for each given output. However, not being reversible is also desired so that it is impossible to determine what a secret is given the output. This also makes it a highly used password-storage mechanism (because in theory, you're not storing the passwords, just a unique representation of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always in cryptography, someone is always trying to break hash functions. &lt;a href=&quot;http://en.wikipedia.org/wiki/Rainbow_table&quot; rel=&quot;nofollow&quot;&gt;Rainbow tables&lt;/a&gt; are the technique used to pre-compute parts of hashes to make hash reversing easier. Note however that the longer the input, the larger the tables need to be. Hence why passwords should always be greater than 6 characters...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you may be interested to know in Hash weaknesses. If two hashes with unique input produce the same output, this is called a collision and is a serious weakness, especially if someone can work out any pattern to said collisions. MD5 is thought to suffer from some, as is SHA1. SHA2 may, also. In any case, NIST has launched SHA-3, a competition to find future hash functions offering security now and in many years to come. &lt;a href=&quot;http://csrc.nist.gov/groups/ST/hash/sha-3/Round3/submissions_rnd3.html&quot; rel=&quot;nofollow&quot;&gt;The finalists&lt;/a&gt; are available to view here.&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="892" LastEditorDisplayName="" LastEditDate="2010-12-16T15:16:06.130" LastActivityDate="2010-12-16T15:16:06.130" CommentCount="2" />
  <row Id="14428" PostTypeId="1" CreationDate="2010-12-15T17:01:22.650" Score="-6" ViewCount="150" Body="&lt;p&gt;Can you find the solution of the following code?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BDHBYJW / QIEKLJEsHUKHQFEuUILBD: .ZFUUIJ / WREJQKX / Bc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indexes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- There are more than one level of encryption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- The final solution is in French.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- The following characters / @. ,;: are not encrypted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- Letters must remain either in uppercase or lowercase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nobody knows, I will bounty my question and give away all my points!&lt;/p&gt;&#xA;" OwnerUserId="4784" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-15T17:31:15.387" LastActivityDate="2010-12-15T17:31:15.387" ClosedDate="2010-12-15T22:13:30.443" Title="Need the solution for the following code" Tags="&lt;homework&gt;&lt;cryptography&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="14429" PostTypeId="1" AcceptedAnswerId="14434" CreationDate="2010-12-15T17:15:43.347" Score="3" ViewCount="109" Body="&lt;p&gt;My motivation for asking this question is that a classmate of mine asked me some kind of question that made me think of this one. I can't recall his exact question because he is kind of messy (both when talking about math and when thinking about math). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm kind of stuck though. I feel like the set $A^{\mathbb{N}} = \{f: \mathbb{N} \rightarrow A, f \text{ is a function} \}$ should have the same cardinality as the power set of A, if A is infinite. On the other hand, in this &lt;a href=&quot;http://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions&quot;&gt;post&lt;/a&gt;, it is stated that the sequences with real coefficients have the same cardinality as the reals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that $A^{\mathbb{N}} \subseteq P(A)$, but (obviously) I got stuck on the other inclusion. Is there any general result that says anything else? References would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT To clarify the intetion of this question: I want to know if there are any general results on the cardinality of $A^{\mathbb{N}}$ other that it is strictly less than that of the power set of A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I was aware that the other inclusion isn't true in general (as the post on here I linked to gave a counterexample), but thanks for pointing out why too. :)&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-15T18:19:07.377" LastActivityDate="2010-12-15T18:31:56.200" Title="What's the cardinality of all sequences with coefficients in an infinite set?" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="14430" PostTypeId="2" ParentId="14423" CreationDate="2010-12-15T17:23:16.173" Score="0" ViewCount="0" Body="&lt;p&gt;The following lemma may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If $f$ is an entire function and and $Re(f)$ is bounded, then $f$ is constant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $f(z)=g(z)+ih(z)$ where $g$ and $h$ are real valued functions.  Now use the information in the question to prove that $g$ is bounded which completes the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the lemma consider $e^{f(z)}$ and use Liouville's theorem.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastActivityDate="2010-12-15T17:23:16.173" CommentCount="1" />
  <row Id="14431" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T17:28:05.530" Score="3" ViewCount="0" Body="&lt;p&gt;Take $A = \mathbb{R}$ then you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|A| = 2^{\aleph_0}$ and so $|A^\mathbb{N}| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0} = 2^{\aleph_0} = |A| &amp;lt; |\mathcal{P}(A)|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the comment, as well as a small &quot;obvious&quot; remark that should probably be mentioned - if $\kappa&amp;gt;1$ and $\lambda\ge\aleph_0$ are two cardinal numbers, to &quot;generate&quot; a relatively small (although this is a bad term, as it is consistent that this size is pretty much unbounded) cardinal which is invariant under powers of $\lambda$, one can always take $\kappa^\lambda$. The above computation is as good for this case to show that the result has this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also obvious that if $\kappa$ is already invariant under powers of $\lambda$ then $\kappa^\lambda = \kappa$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-15T18:31:56.200" LastActivityDate="2010-12-15T18:31:56.200" CommentCount="2" />
  <row Id="14432" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T17:39:36.960" Score="2" ViewCount="0" Body="&lt;p&gt;If $A$ is countably infinite, then the cardinality of $A^{\Bbb N}$ equals the cardinality of the reals: $|A^{\Bbb N}|=|A|^{|\Bbb N|}=\aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\aleph_0}=2^{\aleph_0}=|\Bbb R|$ and $\aleph_0^{\aleph_0}\geq 2^{\aleph_0}$ together gives $|A^{\Bbb N}|=|\Bbb R|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Cardinal_number&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page on cardinal numbers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown by Asaf in his answer, $|A^{\Bbb N}|=2^{\aleph_0}$ if $|A|=2^{\aleph_0}$. It follows easily that if $\aleph_0\leq|A|\leq2^{\aleph_0}$, then still $|A^{\Bbb N}|=2^{\aleph_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $A&amp;gt;2^{\aleph_0}$, things start to become complicated, and depends on whether you assume the Generalised Continuum Hypothesis or not. See for example &lt;a href=&quot;http://www.ucl.ac.uk/~ucahcjm/ast/ast_notes_4.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes by Charles Morgan&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1021" LastEditorUserId="1021" LastEditorDisplayName="" LastEditDate="2010-12-15T17:49:55.753" LastActivityDate="2010-12-15T17:49:55.753" />
  <row Id="14433" PostTypeId="2" ParentId="14417" CreationDate="2010-12-15T17:47:00.183" Score="8" ViewCount="0" Body="&lt;p&gt;I will only give some simple and useful estimate (confirmed by numerical results, based on some mathematical idea). It is essentially a lower bound, but can probably be easily modified to be an upper bound. Moreover, it might give you further ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote your expression by $\varphi_n$, and consider a random variable $X$ with binomial$(n,1/2)$ distribution. Then&#xA;$$&#xD;&#xA;{\rm E}\big[2^{\sqrt X } \big] = \sum\limits_{k = 0}^n {2^{\sqrt k } {n \choose k}\frac{1}{{2^n }}}  = \frac{1}{{2^n }}\varphi _n. &#xD;&#xA;$$&#xD;&#xA;The left-hand side is actually the moment-generating function $M(t)$ of $\sqrt{X}$ at $t=\ln2$, but we will not use this fact.&#xA;I have plotted the function $2^{\sqrt{x}}$, $x &amp;gt; 0$, and it seems that this function is typically convex (it is not convex near $0$). So, heuristically, Jensen's inequality suggests the following result:&#xA;$$&#xD;&#xA;{\rm E}\big[2^{\sqrt X } \big] &amp;gt; 2^{\sqrt {{\rm E}[X]} }, &#xD;&#xA;$$&#xD;&#xA;for all sufficiently large $n$. (If $n$ is large, then $X$ typically takes values in a set on which $2^{\sqrt{x}}$ is convex.) Finally, ${\rm E}[X]=n/2$ leads to $2^{ - n} \varphi _n  &amp;gt; 2^{\sqrt {n/2} }$, that is&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^n {{n \choose k} 2^{\sqrt k } }  &amp;gt; 2^{n + \sqrt {n/2} }, &#xD;&#xA;$$&#xD;&#xA;for all sufficiently large $n$. Numerical results indicate that this inequality holds for all $n \geq 6$. Moreover, the results suggest that the (lower-)bound is quite tight (I checked for $n \leq 150$). So, a tight upper-bound is likely to be something not far from $2^{n + \sqrt {n/2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: We can also employ the strong law of large numbers (SLLN) to show that $\varphi_n$ may be expected to behave something like $2^{n + \sqrt {n/2}}$ for large $n$. Indeed, writing $X$ as $Y_1 + \cdots + Y_n$, where $Y_i$ are i.i.d. rv's with ${\rm P}(Y_i = 0) = {\rm P}(Y_i = 1) = 1/2$, we have&#xA;$$&#xD;&#xA;\varphi _n  = 2^n {\rm E}\bigg[2^{\sqrt n \sqrt {\frac{{Y_1  +  \cdots  + Y_n }}{n}} } \bigg].&#xD;&#xA;$$&#xD;&#xA;By SLLN, $\frac{{\sum\nolimits_{i = 1}^n {Y_i } }}{n}$ converges to $1/2 \,(= {\rm E}[Y_1])$ almost surely, and the conclusion follows.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-15T19:52:21.513" LastActivityDate="2010-12-15T19:52:21.513" />
  <row Id="14434" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T17:48:30.820" Score="6" ViewCount="0" Body="&lt;p&gt;From Jech's &lt;strong&gt;Set Theory&lt;/strong&gt;, we have the following theorems on cardinal exponentiation (a Corollary on page 49):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $\alpha,\beta$, the value of $\aleph_{\alpha}^{\aleph_{\beta}}$  is always either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2^{\aleph_{\beta}}$; or&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_{\alpha}$; or&lt;/li&gt;&#xA;&lt;li&gt;$\aleph_{\gamma}^{\mathrm{cf}\;\aleph_{\gamma}}$ for some $\gamma\leq\alpha$ where $\aleph_{\gamma}$ is such that $\mathrm{cf}\;\aleph_{\gamma}\leq\aleph_{\beta}\lt\aleph_{\gamma}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, $\mathrm{cf}\;\aleph_{\gamma}$ is the cofinality of $\aleph_{\gamma}$: the cofinality of a cardinal $\kappa$ (or of any limit ordinal) is the least limit ordinal $\delta$ such that there is an increasing $\delta$-sequence $\langle \alpha_{\zeta}\mid \zeta\lt\delta\rangle$ with $\lim\limits_{\zeta\to\delta} = \kappa$. The cofinality is always a cardinal, so it makes sense to understand the operations above as cardinal operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; If the Generalized Continuum Hypothesis holds, then&#xA;$$\aleph_{\alpha}^{\aleph_{\beta}} = \left\{\begin{array}{lcl}&#xA;\aleph_{\alpha} &amp;amp;\quad &amp;amp; \mbox{if $\aleph_{\beta}\lt\mathrm{cf}\;\aleph_{\alpha}$;}\\&#xA;\aleph_{\alpha+1} &amp;amp;&amp;amp;\mbox{if $\mathrm{cf}\;\aleph_{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$;}\\&#xA;\aleph_{\beta+1} &amp;amp;&amp;amp;\mbox{if $\aleph_{\alpha}\leq\aleph_{\beta}$.}&#xA;\end{array}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, under GCH, for all cardinals $\kappa$ with cofinality greater than $\aleph_0$ have $\kappa^{\aleph_0} = \kappa$, and for cardinals $\kappa$ with cofinality $\aleph_0$ (e.g., $\aleph_0$, $\aleph_{\omega}$), we have $\kappa^{\aleph_0} = 2^{\kappa}$. (In particular, it is &lt;em&gt;not&lt;/em&gt; the case the cardinality of $A^{\mathbb{N}}$ is necessarily less than the cardinality of $\mathcal{P}(A)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, GCH is usually considered &quot;boring&quot; by set theorists, from what I understand. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T18:28:33.157" LastActivityDate="2010-12-15T18:28:33.157" CommentCount="2" />
  <row Id="14435" PostTypeId="2" ParentId="14423" CreationDate="2010-12-15T18:01:33.990" Score="4" ViewCount="0" Body="&lt;p&gt;I'm a little wary of Liouville Theorem approaches... if you choose $f(z) = {1 \over 2}\sin(2\pi z)$ then it satisfies the conditions of the problem except $|f(z)| &amp;lt; e^{2\pi |z|}$ instead of $|f(z)| &amp;lt; e^{|z|}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A suggestion: try showing $f(z) = g(e^{2\pi iz})$ where $g(z)$ is analytic except at $z = 0$. Then translate the condition $|f(z)| &amp;lt; e^{|z|}$ into growth conditions of $|g(z)|$ as $z \rightarrow \infty$ and $z \rightarrow 0$ and show that if they occur $g(z)$ must be constant.  &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-15T18:01:33.990" />
  <row Id="14436" PostTypeId="2" ParentId="14364" CreationDate="2010-12-15T18:01:59.230" Score="1" ViewCount="0" Body="&lt;p&gt;This kind of problem usually succumbs to some sort of &quot;clever choice&quot; of $x$'s to plug in and manipulate. I don't have a full answer for you, but here's Bill's suggestion for showing that such a ring has characteristic $2$ (a necessary condition for being Boolean). Let $a\in R$. Evaluating the identity at $a$ and at $-a$, we have:&#xA;\begin{align*}&#xA;a + a^{2b+1} &amp;amp;= a^{2b} + a^{10b+1}\\&#xA;-a -a^{2b+1} &amp;amp;= a^{2b} - a^{10b+1}&#xA;\end{align*}&#xA;Adding both equalities we obtain $2a^{2b}=0$. Now, using the fact that $a=a^{2b}+a^{10b-1}-a^{2b+1}$, we have:&#xA;$$2a = 2a^{2b} + 2a^{10b-1}-2a^{2b+1} = 0 + 2a^{2b}a^{8b-1} - 2a^{2b}a = 0,$$&#xD;&#xA;so for all $a\in R$ we have $a+a=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any clever manipulations to suggest for proving $a=a^2$ for all $a$, alas. I tried a few last night, and didn't get anywhere.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-15T18:01:59.230" CommunityOwnedDate="2010-12-15T18:18:18.583" CommentCount="1" />
  <row Id="14437" PostTypeId="2" ParentId="14429" CreationDate="2010-12-15T18:08:26.803" Score="3" ViewCount="0" Body="&lt;p&gt;Arturo Magidin's answer has the general theorem. Here are two more facts that can be useful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\aleph_0 \leq \lambda$ and $2 \leq \kappa \leq \lambda$ then $\kappa^\lambda = 2^\lambda = |P(\lambda)|$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\aleph_0 \leq \lambda \leq \kappa$ then $\kappa^\lambda = |\{ X \subseteq \kappa : |X| = \lambda \}|$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-15T18:08:26.803" />
  <row Id="14438" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T19:03:25.327" Score="2" ViewCount="0" Body="&lt;p&gt;One small point...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the question is stated, there may be a slight ambiguity.  One way (and almost certainly the intended way) to read the question is:  given the (periodic) function $\sin^2(x)$, find its Fourier series on the interval $[0, \pi]$.  In this case, $(1 - \cos(2x))/2$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we could also read it as follows:  given the function $\sin^2(x)$ defined on the interval $[0, \pi]$, find its Fourier series.  In this case, we must first decide how to extend the function to be periodic.  Of course the natural choice is to extend it to equal $\sin^2(x)$ for all $x$.  Again, the cosine series is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... we do have the freedom to extend $\sin^2(x)$ to an &lt;em&gt;odd&lt;/em&gt; function on $[-\pi, \pi]$ instead, in which case the Fourier series will contain only sine functions (with the coefficients computed in the usual way). The point being that there is in fact another series, featuring only sines, that converges to $\sin^2(x)$ on the interval $[0, \pi]$.  Of course, the convergence isn't as fast ;-).&lt;/p&gt;&#xA;" OwnerUserId="4229" LastEditorUserId="4229" LastEditorDisplayName="" LastEditDate="2010-12-16T07:09:59.407" LastActivityDate="2010-12-16T07:09:59.407" CommentCount="1" />
  <row Id="14439" PostTypeId="2" ParentId="14343" CreationDate="2010-12-15T19:34:56.980" Score="0" ViewCount="0" Body="&lt;p&gt;$c = || AB ||;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_c = 360 - \theta_a - \theta_b;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b = {\sin(\theta_b) \over \sin(\theta_c)} c;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C.x = b \cos(\theta_a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C.y = b \sin(\theta_a)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2010-12-15T19:34:56.980" CommentCount="1" />
  <row Id="14440" PostTypeId="2" ParentId="14320" CreationDate="2010-12-15T19:42:39.503" Score="5" ViewCount="0" Body="&lt;p&gt;This does not answer the question but it is progress towards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=4m$ we can obtain a partition with the required property by taking $P=\lbrace 8,m-1,m+1 \rbrace $ for $m&amp;gt;1 \textrm{ and } m \ne 7 \textrm{ or } 9.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This partition is distinct from the one given in the question for all $m&amp;gt;2$ and hence the partition in the question is not unique for $n$ a multiple of $4$ and greater than $8.$&#xD;&#xA;(Other solutions exist for $n=28$ and $n=36.$)&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-15T22:04:01.737" LastActivityDate="2010-12-15T22:04:01.737" CommentCount="3" />
  <row Id="14441" PostTypeId="1" CreationDate="2010-12-15T19:42:50.960" Score="0" ViewCount="42" Body="&lt;p&gt;I've been assigned to do a computer model of a simple pendulum. The model uses Runge-Kutta to do the integration to find the position and velocity of the pendulum. I am currently having problems with using the correct intial conditions that would allow me to achieve chaos, was wondering if any one has any pointers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pendulum equation is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = F cos(omega*t) - c*v - (x-x^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where: &#xA;a is acceleration&#xA;F is the driving force&#xA;c*v is the damping force&#xA;(x-X^3) is the duffing oscillator eq&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4789" LastActivityDate="2010-12-15T19:42:50.960" Title="What are the intial conditions for Chaotic Pendulum behavior?" Tags="&lt;calculus&gt;&lt;chaotic-systems&gt;" CommentCount="5" />
  <row Id="14442" PostTypeId="1" AcceptedAnswerId="14603" CreationDate="2010-12-15T19:49:28.513" Score="7" ViewCount="196" Body="&lt;p&gt;It's pretty easy to construct an (R-S) bi-module M and a left S-module N such that neither M_S nor N is a projective S-module, but the tensor product is a non-zero projective R-module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, taking R=S=Z defies my zoo of examples.  Here projective=free is closed under direct sums and summands, and so it seems like M and N can be chosen indecomposable.  The indecomposable abelian groups I know don't seem to fit.  I don't know how to &quot;uninvert&quot; using a tensor product.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the tensor product of two non-free abelian groups be non-zero free?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inspired by &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&amp;amp;t=382311&quot; rel=&quot;nofollow&quot;&gt;an AoPS question&lt;/a&gt; and finals week.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-17T02:02:36.880" Title="Can the tensor product of two non-free abelian groups be non-zero free?" Tags="&lt;abstract-algebra&gt;&lt;modules&gt;&lt;abelian-groups&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="14443" PostTypeId="2" ParentId="14389" CreationDate="2010-12-15T19:57:57.493" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Regarding the uniqueness theorem for Fourier series.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I felt the obligation to say something due to all comments and the directed answer. What I meant is that it is in general one can not just say that a trigonometric expansion is a Fourier expansion, you must relay on something (for example that you have a &lt;em&gt;trigonometric polynomial&lt;/em&gt; of basis elements, hence it is the Fourier series of the function). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we look at $I=(0,2\pi)$, then $\Sigma=\{1,\sin(nx),\cos(nx)\}$ makes up an &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orthonormal_basis&quot; rel=&quot;nofollow&quot;&gt;orthonormal basis&lt;/a&gt;&lt;/em&gt; for $L^2(I)$ &#xA;(or piecewise continuous complex valued functions on $I$ below denoted by $C_p(I)$) in the sense&#xA;$$(1)\qquad\langle e,f\rangle=0$$&#xD;&#xA;for all $e_1,e_2\in\Sigma$ where $e_1\ne e_2$, and if $f\in L^2(I)$ (or $f\in C_p(I)$ satisfy) &#xA;$$(2)\qquad \langle e,f\rangle=0$$&#xD;&#xA;for all $e\in\Sigma$ then $f=0$. &#xA;Here &#xA;$$ \langle e,f\rangle =\frac{1}{\pi}\int_0^{2\pi}e(x)\bar{f(x)}dx.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The property (2) is called the uniquness of Fourier expansion for $L^2(I)$ (or $C_p(I)$) and is a cosecuence of the more general statement &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Theorem.&lt;/em&gt; If $f\in L^1(I)$ then $\langle f,e\rangle=0$ for all $e\in\Sigma$ if and only if $f=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem disurves some comment even though it seams easy. Let us write &#xA;$$ a_n=\frac{1}{2\pi}\int_0^{\pi}f(x)\cos(nx)dx\qquad\text{and}\qquad b_n=\frac{1}{\pi}\int_0^{2\pi}f(x)\sin(nx)dx.$$&#xD;&#xA;The difficult part of the theorem is that in general we do not know if the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series&quot; rel=&quot;nofollow&quot;&gt;Fourier series&lt;/a&gt; $\frac{a_0}{2}+\sum_k=1^\infty a_n\cos(nx)+b_n\sin(nx)$ &#xA;converges to $f$, and also if it is not obvious that it is it possible to interchange integration and summation like this&#xA;$$\int_0^{2\pi}f(x)\sin(kx)dx=\int_0^{2\pi}(\frac{a_0}{2}+\sum_{k=1}^\infty a_n\cos(nx)+b_n\sin(nx))\sin(kx)dx =$$&#xD;&#xA;$$\qquad\sum_{k=1}^\infty \int_0^{2\pi}(a_n\cos(nx)+b_n\sin(nx))\sin(kx).$$&#xD;&#xA;Also, there are convergent trigonometric series that are &lt;em&gt;not&lt;/em&gt; Fourier series an example is (see Katznelson, Yitzhak (1976) &lt;em&gt;An introduction to harmonic analysis&lt;/em&gt;.) $$\sum_{k=2}^\infty \frac{\sin(nx)}{\log n}$$&#xD;&#xA;A way to prove the theorem is to use that $K_n*f\to f$ in $L^1$-norm where $K_n$ is a trigonometric summation kernel such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fej%C3%A9r_kernel&quot; rel=&quot;nofollow&quot;&gt;Fejér kernel&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="1154" LastEditorDisplayName="" LastEditDate="2010-12-16T05:13:55.617" LastActivityDate="2010-12-16T05:13:55.617" />
  <row Id="14444" PostTypeId="2" ParentId="14421" CreationDate="2010-12-15T20:25:05.040" Score="1" ViewCount="0" Body="&lt;p&gt;I can give an answer, which may seem like a joke. But you can map all text (the set of all codes) to the same result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for example I map the codes &#xA;ABC to 1&#xA;BDCD to 1 &#xA;ABDGR to 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is no opportunity for decoding here. :)&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-15T20:25:05.040" CommentCount="3" />
  <row Id="14445" PostTypeId="1" CreationDate="2010-12-15T20:30:43.287" Score="5" ViewCount="69" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this question may be shameful, but nevertheless I can't help myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U \subset \mathbb R^n$ be arbitrary, in particular not the whole of the space itself. I wonder about the dual of the space $W^{1,p}(U)$, for $p &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $U = \mathbb R^n$, we have $(W^{1,p})&amp;#39; = (W^{1,p&amp;#39;})$ with $p&amp;#39; = \frac{p}{p-1}$. How about different $U$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in case $U = B_1(0)$ being the closed $1$-Ball, it seems the dual is not a function space. Just recall that the trace is well-defined, linear and continuous on $W^{1,p}(U)$ and, with $S_1$ the boundary of $B_1(0)$ and $w \in L^p(S_1)$, we are given are continuous linear functional by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ W^{1,p}(B_1(0)) \longrightarrow \mathbb C \, , f \mapsto \int_{S_1} w \cdot tr f dx $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I wouldn't be surprised if the above example were somehow prototypical, but I have no clue how to proceed from this point. I regard this relevant, as these spaces are ubiquitous in analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-12-15T20:40:35.800" LastActivityDate="2010-12-16T18:25:22.333" Title="Dual of Sobolev space $W^{1,p}(U)$ for $U$ an arbitrary subset of $\mathbb R^n$" Tags="&lt;functional-analysis&gt;&lt;distribution-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14446" PostTypeId="1" CreationDate="2010-12-15T20:38:16.443" Score="4" ViewCount="98" Body="&lt;p&gt;Prove that if $\kappa$ is an inaccessible cardinal, then $V_{\kappa}$ satisfies all the axioms of ZFC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this done for the axiom of choice and for regularity?&lt;/p&gt;&#xA;" OwnerUserId="4792" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-15T23:05:51.700" LastActivityDate="2010-12-16T13:37:04.983" Title="Proving $V_{\kappa}$ is a model of ZFC for inaccessible $\kappa$" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14447" PostTypeId="2" ParentId="14445" CreationDate="2010-12-15T20:48:30.817" Score="2" ViewCount="0" Body="&lt;p&gt;I think you can find the answer yourself - I will just give you an hint: You are considering the wrong pairing between W^{1,p}(U) and W^{1,p'}(U). (The pairing you consider doesn't give you the desired isomorphism even in the R^n case.)&lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2010-12-15T20:48:30.817" CommentCount="1" />
  <row Id="14448" PostTypeId="1" CreationDate="2010-12-15T20:59:24.857" Score="0" ViewCount="131" Body="&lt;p&gt;Is there an explicit expression for &#xA;$$\int_a^b \sqrt{(k-x)^2 + (f(x))^2} dx$$ &#xA;in terms of $\int_a^b f(x)dx$ for constant k ?&lt;/p&gt;&#xA;" OwnerUserId="4793" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T21:02:04.637" LastActivityDate="2010-12-15T21:12:29.207" Title="Integration question" Tags="&lt;calculus&gt;" AnswerCount="0" CommentCount="7" />
  <row Id="14449" PostTypeId="2" ParentId="14446" CreationDate="2010-12-15T21:03:35.287" Score="2" ViewCount="0" Body="&lt;p&gt;The idea is that you need to remember that whenever ZFC requires you something to be a set, you want it to be an element of $V_\kappa$, as this is the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to expand Arturo's comment, the axiom of choice says that when you have a set of non-empty sets the product is non-empty. Since the sets of $V_\kappa$ are elements of it in the world, you can take a product of them, from the inaccessibility of $\kappa$ you'd have that the rank of the product (in the big universe) is still less than $\kappa$ and thus the product remains within $V_\kappa$ as you'd like it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for regularity, well... basically the relation $\in$ is the same as in the world, so it's well-founded and everything holds.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-15T21:03:35.287" />
  <row Id="14450" PostTypeId="1" AcceptedAnswerId="14453" CreationDate="2010-12-15T21:10:27.277" Score="1" ViewCount="120" Body="&lt;p&gt;I'm trying to understand this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working in $\mathbb{Z}_p[x]$, $(x-c)^p \equiv x-c \pmod{x^p-x}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Fermat Little theorem we know that $(x-c)^p \equiv (x-c) \pmod{p}$ but I don't see a connection. Actually, I'm not sure if I'm mixing things up because $\mathbb{Z}_p[x]$ is a polynomial ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I tried to show it from: $x^p \equiv x \pmod{x^p-x}$ holds because $x^p-p \mid x^p-x$. But I don't know how substitute $x-c$ instead of $x$.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-15T21:23:46.763" LastActivityDate="2010-12-16T03:40:22.387" Title="Why is $(x-c)^p \equiv x-c \pmod{x^p-x}$?" Tags="&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14451" PostTypeId="1" CreationDate="2010-12-15T21:11:02.407" Score="4" ViewCount="59" Body="&lt;p&gt;Given four coordinates that define the corners of an irregular quadrilateral and a point defined by its coordinates, what is the simplest way to determine if the point is within or outside of the quadrilateral?&lt;/p&gt;&#xA;" OwnerUserId="4794" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-16T00:38:28.717" LastActivityDate="2010-12-20T15:24:48.403" Title="Determine if  the coordinates of a point are within an irregular quadrilateral whose corners are defined by coordinates" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14453" PostTypeId="2" ParentId="14450" CreationDate="2010-12-15T21:19:34.880" Score="8" ViewCount="0" Body="&lt;p&gt;Working in $\mathbb{Z}_p[x]$, you have the &quot;Freshman's Dream&quot; binomial theorem:&#xA;$$(a+b)^p = a^p + b^p.$$&#xD;&#xA;(In fact, this holds in any commutative ring of characteristic $p$, simply by the Binomial Theorem, as mentioned by Qiaochu: all the mixed coefficients are multiples of $p$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $(x-c)^p = x^p - c^p$. By Fermat's Little Theorem, you also know that $c^p \equiv c \pmod{p}$, so $(x-c)^p = x^p - c$ in $\mathbb{Z}_p[x]$. That's pretty much all you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;Addendum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add that you were very perceptive to notice that even though the &lt;em&gt;function&lt;/em&gt; $x\mapsto (x-c)^p$ is the same as the function $x\mapsto x-c$ in $\mathbb{Z}_p$, this does not mean that the polynomials $(x-c)^p$ and $x-c$ are equal. We are used to (in the integers, rationals, reals, etc) the fact that two polynomials give the same function if and only if they are identical, but this does not occur over finite rings; for example, Fermat's Little Theorem tells you that the polynomial $x^p$ and the polynomial $x$ both define the same function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what you &lt;em&gt;can&lt;/em&gt; prove is that if $p(x)$ and $q(x)$ are two polynomials with coefficients in $\mathbb{Z}_p$, and they define the same function (that is, $p(a) = q(a)$ for all $a\in\mathbb{Z}_p$), then they differ by a multiple of the polynomial $x^p-x$; i.e., $p(x)\equiv q(x)\pmod{x^p-x}$ in $\mathbb{Z}_p[x]$. What you have here is a special case. See if you can figure out how to prove the more general statement.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T02:36:00.890" LastActivityDate="2010-12-16T02:36:00.890" CommentCount="1" />
  <row Id="14454" PostTypeId="2" ParentId="14281" CreationDate="2010-12-15T21:20:12.937" Score="2" ViewCount="0" Body="&lt;p&gt;Let $X$ be an $m \times n$ matrix. A norm $\|X\|$ is called &lt;em&gt;unitarily invariant&lt;/em&gt; if $\|UXV\| = \|X\|$ for appropriately shaped unitary matrices $U$ and $V$ (i.e., $U^*U=I_m$ and $V^*V=I_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the singular value decomposition of $X$ tells us that if $\|\cdot\|$ is a unitarily invariant norm, it will be a function &lt;em&gt;only&lt;/em&gt; of the singular values of $X$ (a so-called spectral function). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple observation is that if $X$ is square and diagonalizable, then we can select $S=V$, the matrix of eigenvectors of $X$, and consequently the norms that you are looking for again must be unitarily (here orthogonally) invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please have a look in the book &lt;em&gt;Matrix Analysis&lt;/em&gt; by Horn and Johnson, who devote an entire chapter (or more) to matrix norms.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-15T21:20:12.937" CommentCount="1" />
  <row Id="14455" PostTypeId="2" ParentId="14451" CreationDate="2010-12-15T21:30:43.137" Score="5" ViewCount="0" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_in_polygon&quot; rel=&quot;nofollow&quot;&gt;this wikipedia entry&lt;/a&gt;, or an introductory book on computational geometry.&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2010-12-15T21:30:43.137" CommentCount="1" />
  <row Id="14456" PostTypeId="2" ParentId="14442" CreationDate="2010-12-15T21:38:39.833" Score="5" ViewCount="0" Body="&lt;p&gt;Any example must arise from infinitely generated groups.  Suppose $A$ and $B$ are finitely generated, non-zero, non-free, abelian groups.  Being non-free means there is torsion (since we're in the finitely generated case).  If every element of $B$ has finite order, then every element of the tensor product has finite order.  So, assume $b \in B$ has infinite order, and let $a \neq 0$ be an element of order $n$.  Then $a \otimes b$ is non-zero and has order dividing $n$.  Hence, $A \otimes B$ has torsion and, consequently, it is non-free.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-15T21:38:39.833" CommentCount="1" />
  <row Id="14457" PostTypeId="1" AcceptedAnswerId="14459" CreationDate="2010-12-15T22:56:09.143" Score="2" ViewCount="34" Body="&lt;p&gt;While I tagged this question homework, it's not really homework. I'm really studying for a test and don't understand how to do something. I have the answer to the question, so the important part is for my to understand how to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X has distribution &lt;code&gt;Gamma(2,4)&lt;/code&gt;&#xA;&lt;code&gt;X = Y-3&lt;/code&gt; Find: &lt;code&gt;P(Y&amp;gt;11|Y&amp;gt;7)&lt;/code&gt; Which I know is equals to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(1-P(X&amp;lt;8))/(1-P(X&amp;lt;4)) = (1-Fx(8))/(1-Fx(4))&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that &lt;code&gt;Fx(a) = 1-e^(-a/4)-(a/4)*e^(-a/4)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can clearly solve the problem now it's just algebra. &lt;strong&gt;But how do we know what Fx(a) equals???&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2010-12-15T23:16:04.047" Title="Finding conditional probability with Gamma distribution" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" />
  <row Id="14458" PostTypeId="1" AcceptedAnswerId="14607" CreationDate="2010-12-15T23:06:28.583" Score="3" ViewCount="151" Body="&lt;p&gt;Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be strictly monotonically increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Is $f$ not continuous at $p \in \mathbb{R}$, there exists a non-empty, open interval $(a_p, b_p) \subset \mathbb{R}$ such that $f(x)\leq a_p$ for all $x &amp;lt; p$ and $f(x) \geq b_p$ for all $x &amp;gt; p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The set of discontinuity points $$ \{ p \in \mathbb{R} | f \; \mbox{is not continuous at} \; p \}$$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (i) I could be way off, but I am picturing a graph with $p$ on the $x$-axis for which the value $f(p)$ on the $y$-axis is undefined. Am I correct to interpret the open interval $(a_p, b_p)$ as an interval on the $y$-axis which should be contained within the distance between two $f(x)$'s (one for $x&amp;lt;p$ and the other for $x&gt;p$)? If that is so far correct, there could be 2 types of discontinuous points $p$, a jump or removable type. For the jump it would be easier to show that somehow the interval $(a_p, b_p)$ is smaller than the vertical jump... For a removable discontinuity $c$, I would think that the interval could contain just the $y$-axis value $\displaystyle \lim_{x\to c}f(x)$ but I don't really know how to express that the upper and lower bounds would be just above and below that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With (ii) I am currently trying to understand a proof, what exactly does the notation $f(p-)$ or $f(p+)$ mean in this context? Is it simply the value when approached from the left or the right(respectively)?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-04T06:17:10.633" LastActivityDate="2011-01-04T06:17:10.633" Title="Showing properties of discontinuous points of a strictly increasing function" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14459" PostTypeId="2" ParentId="14457" CreationDate="2010-12-15T23:16:04.047" Score="3" ViewCount="0" Body="&lt;p&gt;$F_x(a)$ is the cumulative distribution function of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_distribution#Probability_density_function&quot; rel=&quot;nofollow&quot;&gt;gamma$(2,4)$ distribution&lt;/a&gt;.  This is given by $$F_x(a) = \int_0^a \frac{t e^{-t/4}}{(4^2)\Gamma(2)} dt.$$ &#xA;Integrating (by parts) yields the given expression for $F_x(a)$, since $\Gamma(2) = (2-1)! = 1$.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-15T23:16:04.047" CommentCount="5" />
  <row Id="14460" PostTypeId="2" ParentId="14264" CreationDate="2010-12-15T23:16:17.980" Score="6" ViewCount="0" Body="&lt;p&gt;By Ryan Budney's suggestion, I went ahead and proved the general case.  When $B_n$ onto $S_n$ the kernel is isomorphic to $P_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof sketch is this: relations for the Artin generators in $B_n$ must be satisfied in the image.  Relations $b_ib_{i+1}b_i=b_{i+1}b_ib_{i+1}$ can be rewritten as in terms of conjugation so that every $b_i$ has image of a fixed cycle structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relations which impose commutativity of non-adjacent generators imply that non-adjacent generators get sent to permutations with cycles either coincidental or disjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that for $n&amp;gt;4$ the images of non-adjacent generators must actually be disjoint: mildly technical, but not difficult. (The $n=4$ case being easily solved by hand or GAP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then counting every other odd generator $b_1,b_3,\ldots$, of which there are $\lceil \frac n 2\rceil$ we have that they must be transpositions, since $3\lceil \frac n 2\rceil&amp;gt;n$.  Essentially that's it: up to isomorphism of $S_n$ the images of the generators for $B_n$ are the usual transpositions they induce, so kernel is $P_n$.&lt;/p&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2010-12-15T23:16:17.980" CommentCount="1" />
  <row Id="14461" PostTypeId="1" AcceptedAnswerId="14464" CreationDate="2010-12-15T23:18:58.030" Score="4" ViewCount="48" Body="&lt;p&gt;Is there a closed form expression for the number of terms in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_polynomial#Monomial_symmetric_polynomials&quot; rel=&quot;nofollow&quot;&gt;monomial symmetric polynomial&lt;/a&gt; in a given number of variables for a particular partition of exponents, in terms of which/how many exponents are distinct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel the answer should be straightforward, and it's probably just a contrived statement of a more elementary number theoretic question, but I'm just drawing a blank at the moment. I at first thought that the answer might be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{N!}{(N-k+1)!}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $N$ variables and $k$ distinct exponents, which works for all $N=2$ and $N=3$ cases, but predicts 4 terms for a partition $\alpha=(1,1,0,0)$, while there are in fact 6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_\alpha(a,b,c,d) = ab+ac+ad+bc+bd+cd$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being an idiot?&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="67" LastEditorDisplayName="" LastEditDate="2010-12-15T23:29:28.790" LastActivityDate="2010-12-15T23:45:01.463" Title="Number of terms in a monomial symmetric polynomial" Tags="&lt;combinatorics&gt;&lt;polynomials&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14462" PostTypeId="2" ParentId="14210" CreationDate="2010-12-15T23:27:39.507" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a slightly different way, perhaps a little more sophisticated, to see this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free groups are &lt;em&gt;Hopfian&lt;/em&gt;, meaning that every surjective endomorphism is an isomorphism.  There are a variety of ways to prove this.  It's proved in Lyndon &amp;amp; Schupp, using Nielsen transformations. Alternatively, you can appeal to an (easy) result of Malcev, which states that every finitely generated, residually finite group is Hopfian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is an obvious epimorphism $F_3\to F_2$ with non-trivial kernel, given by killing a generator.  If $F_3$ were a quotient of $F_2$, the composition of these two maps would give an epimorphism $F_2\to F_2$ with non-trivial kernel.&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2010-12-15T23:27:39.507" />
  <row Id="14463" PostTypeId="2" ParentId="14461" CreationDate="2010-12-15T23:42:01.457" Score="4" ViewCount="0" Body="&lt;p&gt;Suppose the number of indeterminates is $n$. Suppose the monomial symmetric polynomial is of type $(a_1,a_2,...,a_n)$ where the first $b_1$ exponents are identical, the next $b_2$ are identical and so on until $b_r$, where $1\leq b_j \leq n$ and $\sum_{1}^{r} b_j = n$. Then, number of terms in a monomial symmetric polynomial is the same as number of permutations of the set of exponents. The number of permutations when they are all distinct is $n!$. If not, the every permutations of any collection of identical exponents amongst themselves should be treated the same. So factoring out such permutations, we get the total number to be $\frac{n!}{b_1!\cdot b_2!\cdot...\cdot b_r!}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the examples on the wikipedia page.&#xA;We have, $n=3$. In the first case, we have 2 identical exponents, so number of terms is $\frac{3!}{2!}=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, all exponents are distinct, so the number of terms is $3!=6$&lt;/p&gt;&#xA;" OwnerUserId="3431" LastActivityDate="2010-12-15T23:42:01.457" CommentCount="3" />
  <row Id="14464" PostTypeId="2" ParentId="14461" CreationDate="2010-12-15T23:45:01.463" Score="5" ViewCount="0" Body="&lt;p&gt;If the smallest exponent appears $k_1$ times, the second smallest appears $k_2$ times, etc. then the answer is the multinomial coefficient: ${n \choose k_1,k_2,...,k_n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multinomial coefficient $n \choose k_1,k_2,...,k_n$ counts how many ways you have to partition n objects into disjoint subsets of sizes $k_1,...,k_n$ which is just what you need.&lt;/p&gt;&#xA;" OwnerUserId="3533" LastActivityDate="2010-12-15T23:45:01.463" />
  <row Id="14465" PostTypeId="1" CreationDate="2010-12-16T00:11:59.877" Score="-1" ViewCount="111" Body="&lt;p&gt;This is a question on my homework could you help me figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write a non-recursive procedure that takes an array and its length and prints its elements in the reverse order. For example, if the array has elements 77, 12, 90, 34, 45, the procedure should print the values 45, 34, 90, 12, 77. Here is the header for the procedure:&#xA;Procedure printReverse(array[], length)&lt;/p&gt;&#xA;" OwnerUserId="4351" LastActivityDate="2010-12-16T00:54:24.990" ClosedDate="2010-12-16T03:57:25.890" Title="help writing a non-recursive procedure" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;recursive&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14466" PostTypeId="2" ParentId="14465" CreationDate="2010-12-16T00:23:30.340" Score="0" ViewCount="0" Body="&lt;p&gt;Can you just print array[i] for i counting down from length to 1?  Python can do it in one line.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-16T00:23:30.340" CommentCount="2" />
  <row Id="14467" PostTypeId="1" AcceptedAnswerId="14469" CreationDate="2010-12-16T00:26:52.773" Score="1" ViewCount="81" Body="&lt;p&gt;In Hatcher's &quot;Algebraic Topology&quot; in the Poincare Duality section he introduces the subject by doing orientable surfaces.  He shows that there is a dual cell structure to each cell structure and it's easy to see that the first structure gives the cellular chain complex, while the other gives the cellular cochain complex.  He goes on to say that this generalizes for manifolds of higher dimension, but that &quot;requires a certain amount of manifold theory&quot;.  Is there a good book or paper where I can read about this formulation of Poincare Duality?&lt;/p&gt;&#xA;" OwnerUserId="2838" LastActivityDate="2010-12-16T01:16:07.813" Title="Poincare Duality Reference" Tags="&lt;reference-request&gt;&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="14468" PostTypeId="2" ParentId="14327" CreationDate="2010-12-16T00:44:28.743" Score="3" ViewCount="0" Body="&lt;p&gt;Update: See the last section for a possible proof (**) that $a_n &amp;lt; n$ for all $n \geq 18$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some bounds on $E(n,m)$ that turn out to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $E_{n,m}$ for $E(n,m)$ in Ross Millikan's notation, let $R_{n,m} = \lceil E_{n,m} \rceil - E_{n,m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying the expression for $E_{n,m}$ in Ross's first answer yields (thanks to some nice cancellation)&#xA;$$E_{n,m} = \frac{n+1}{2} + \frac{E_{n,m-1}(E_{n,m-1}-1)}{2n} - \frac{R_{n,m-1}(R_{n,m-1}-1)}{2n}.$$&#xD;&#xA;Since $0 \leq R_{n,m-1} &amp;lt; 1$, we have $$0 \leq - \frac{R_{n,m-1}(R_{n,m-1}-1)}{2n} \leq \frac{1}{8n}.$$  This can be seen easily by the fact that the expression being bounded is quadratic in $R_{n,m-1}$ with vertex at $R_{n,m-1} = \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $F_{n,m} \leq E_{n,m} \leq G_{n,m}$, where $F_{n,1} = G_{n,1} = \frac{n+1}{2}$, and&#xA;$$F_{n,m} =  \frac{n+1}{2} + \frac{F_{n,m-1}(F_{n,m-1}-1)}{2n},$$&#xD;&#xA;$$G_{n,m} =  \frac{n+1}{2} + \frac{G_{n,m-1}(G_{n,m-1}-1)}{2n} + \frac{1}{8n}.$$&#xD;&#xA;Thus we have recurrences that give upper and lower bounds on $E_{n,m}$ without having to deal with the problem of taking ceilings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical experiments indicate that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$F_{n,n}$ and $G_{n,n}$ are very close to each other,&lt;/li&gt;&#xA;&lt;li&gt;$G_{n,n} - F_{n,n}$ is decreasing,&lt;/li&gt;&#xA;&lt;li&gt;$n - G_{n,n}$ is increasing, &lt;/li&gt;&#xA;&lt;li&gt;$n - G_{n,n} &amp;gt; 1$ for $n \geq 15$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A proof of 3 or 4 would imply $a_n &amp;lt; n$ for $n \geq 15$.  A close analysis of $G_{n,n} - F_{n,n}$, together with an asymptotic estimate of $F_{n,n}$ or $G_{n,n}$, would help with the requested behavior of $n - a_n$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is easy to see that $F_{n,m} = n$ is an equilibrium solution for the $F_{n,m}$ recurrence.  That should be helpful as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;HR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that $G_{n,m} = \frac{1}{2} + n a_m$, where $a_1 = \frac{1}{2}$ and $a_m$ satisfies the recurrence $$a_m = \frac{a^2_{m-1}+1}{2}.$$  This is easy to verify once one has the conjectured expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also turns out that &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1710384#p1710384&quot; rel=&quot;nofollow&quot;&gt;the $a_m$ recurrence has been studied&lt;/a&gt; (**), with the following bounds:&#xA;$$ 1 - \frac{2}{m} + \frac{2}{m^2} \ln \frac{m}{3} + \frac{417}{128m^2} \leq a_m \leq 1 - \frac{2}{m} + \frac{5 \ln m + 3}{2m^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper bound implies &#xA;$$G_{n,n} \leq \frac{1}{2} + n \left(1 - \frac{2}{n} + \frac{5 \ln n + 3}{2n^2}\right) = n - \frac{3}{2} + \frac{5 \ln n + 3}{2n} (*) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $E_{n,n} \leq G_{n,n}$, $G_{n,n} &amp;lt; n-1$ implies $a_n &amp;lt; n$.  The expression on the right in $(*)$ is less than $n-1$ when $$\frac{5 \ln n + 3}{2n} &amp;lt; \frac{1}{2},$$&#xD;&#xA;which is true for all $n \geq 18$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) The bounds required for my argument are given in a post in the &quot;Real Analysis Unsolved and Proposed Problems&quot; forum at the Art of Problem Solving.  I cannot tell whether the bounds are conjectured and the poster is asking for a proof, or whether the poster has a proof and is merely posing the problem for others to solve.  So I cannot claim that this is a complete proof.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-16T03:54:30.380" LastActivityDate="2010-12-16T03:54:30.380" CommentCount="3" />
  <row Id="14469" PostTypeId="2" ParentId="14467" CreationDate="2010-12-16T00:47:42.670" Score="5" ViewCount="0" Body="&lt;p&gt;As far as I know, the only textbook reference for this approach, which is Poincare's original approach, is Seifert and Threlfall's text &quot;A textbook of topology&quot;.  It's available in English translation but the original was in German.  Moreover, Seifert and Threlfall's proof isn't as efficient as it could be, since they're working &lt;em&gt;entirely&lt;/em&gt; with simplicial homology.  It's much more efficient to work with both simplicial homology and CW-cohomology (together with the knowledge that simplicial and CW homology / cohomology are canonically isomorphic via the relation to singular homology / cohomology).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This version of the proof only works in the context where your manifold is triangulable, and here it goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of the dual cell decomposition in general goes like this. Let $\Delta_n$ be an $n$-simplex, and let $F$ be a facet of $\Delta_n$, meaning the convex hull of some collection of $\Delta_n$'s vertices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;dual polyhedral bit&lt;/em&gt; corresponding to $F$ is the convex hull of the barycentres of &lt;em&gt;all&lt;/em&gt; facets $F&amp;#39;$ of $\Delta_n$ which contain $F$ (including $\Delta_n$ itself).  So in a tetrahedron $\Delta_3$, if $F$ was an edge, the dual polyhedral bit would be a quadrilateral that intersects $F$ in a single point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a triangulated manifold $M$, if $F$ is a simplex of the triangulation, the &lt;em&gt;dual cell&lt;/em&gt; corresponding to $F$ is the union of all the dual polyhedral bits to $F$ in all the top-dimensional simplices containing $F$.  If $M$ is $m$-dimensional and $F$ is $k$-dimensional, a little geometry later and you'll see the dual cell is an $(m-k)$-dimensional cell in a genuine CW-decomposition of $M$.  Again, in the 3-manifold case, if $F$ were an edge, the dual cell would be a $2$-cell with a single vertex at its centre, decomposed into squares. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the basic idea.  From there the proof of Poincare duality is very much &quot;follow your nose&quot;.  It's a fun chase and I encourage you to try to work it out on your own, rather than looking it up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, spend as much time as you can thinking about evaluating a homology class $X$ on the dual of a homology class $Y$ (provided $X$ and $Y$ have complementary dimensions).  You'll have to be careful about thinking of the simplicial vs. CW-homology when thinking this through, of course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2G38W.png&quot; alt=&quot;dual polyhedral bits in a tetrahedron&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-12-16T01:16:07.813" LastActivityDate="2010-12-16T01:16:07.813" />
  <row Id="14470" PostTypeId="2" ParentId="14465" CreationDate="2010-12-16T00:54:24.990" Score="1" ViewCount="0" Body="&lt;p&gt;Ross is correct.  Remember that a recursive function is one that calls itself.  Writing a for loop as follows involves no such recursion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for(int i = length - 1; i &gt;= 0; i--) cout &amp;lt;&amp;lt; array[i] &amp;lt;&amp;lt; &quot; &quot;;&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2010-12-16T00:54:24.990" CommentCount="1" />
  <row Id="14473" PostTypeId="1" CreationDate="2010-12-16T03:17:14.927" Score="0" ViewCount="53" Body="&lt;p&gt;Let $R$ be a ring and assume that for every left $R$-module ${}_RM$, if $M$ has a projective cover, then $M$ is projective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me prove that in that case, $J(R)=0$?&lt;/p&gt;&#xA;" OwnerUserId="4797" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T03:56:02.163" LastActivityDate="2010-12-16T03:56:02.163" Title="projective-projective cover" Tags="&lt;ring-theory&gt;" CommentCount="2" />
  <row Id="14474" PostTypeId="2" ParentId="14450" CreationDate="2010-12-16T03:40:22.387" Score="2" ViewCount="0" Body="&lt;p&gt;Alternatively, by Fermat's little Theorem every $\rm\ a\in \mathbb F_p\ $ is a root of $\rm\  f(x)^p - f(x)\:,\ $ therefore $\rm\ x\ (x-1)\ (x-2)\cdots (x-p+1)\ |\ f(x)^p-f(x)\:.\ $ Making the specialization $\rm\ f(x) = x\ $ this yields $\rm\ x \ (x-1)\ (x-2)\cdots (x-p+1)\ |\ x^p-x\:.\ $ This divisibility must be an equality since both are monic polynomials of equal degree. Therefore we deduce that $\rm\: x^p - x\ |\ f(x)^p - f(x),\ \forall\: f\in \mathbb F_p\:[x]\:.\ $ The sought result is the special case $\rm\ f(x) = x-c\:. $ &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-16T03:40:22.387" />
  <row Id="14475" PostTypeId="1" AcceptedAnswerId="14487" CreationDate="2010-12-16T03:50:59.870" Score="1" ViewCount="183" Body="&lt;p&gt;I have already taken up to Multivariable Calculus, Linear Algebra and Diff Eq.&#xA;I want to learn Calculus on Manifolds by myself, could you recommend a good introductory book on this subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I look at other things first, like topology, to get a better background?&lt;/p&gt;&#xA;" OwnerUserId="4800" LastActivityDate="2010-12-16T18:29:46.027" Title="Good introductory book on Calculus on Manifolds" Tags="&lt;manifold&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="14476" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T03:58:29.927" Score="4" ViewCount="0" Body="&lt;p&gt;May be this :  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805390219&quot; rel=&quot;nofollow&quot;&gt;Spivak,Calculus-Manifolds-Approach&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an reprint from year 1995, although maybe an more new could also you serve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TQkRk.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-16T18:29:46.027" LastActivityDate="2010-12-16T18:29:46.027" CommentCount="1" />
  <row Id="14477" PostTypeId="1" AcceptedAnswerId="14479" CreationDate="2010-12-16T04:13:31.533" Score="6" ViewCount="161" Body="&lt;p&gt;Any hints about how to prove &#xA;$$!n = n!- \sum_{i=1}^{n} {{n} \choose {i}} \quad!(n-i)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;http://en.wikipedia.org/wiki/Subfactorial&quot; rel=&quot;nofollow&quot;&gt;&#xA;Wikipedia's article on derangements&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $!n$ is the number of derangements of a set with $n$ elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for proofs, just nudges in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T04:29:26.553" LastActivityDate="2010-12-16T05:35:46.160" Title="Proof of subfactorial formula $!n = n!- \sum_{i=1}^{n} {{n} \choose {i}} \quad!(n-i)$" Tags="&lt;combinatorics&gt;&lt;proof&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14478" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T04:22:41.170" Score="0" ViewCount="0" Body="&lt;p&gt;If you are looking for a book with tons of proofs, I would strongly recommend J.H. Hubbard and B.B. Hubbard. Vector Calculus, Linear Algebra, and Differential Forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Super good as an introduction book if you can find someone to go to when you are stuck. &#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130414085&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Vector-Calculus-Linear-Algebra-Differential/dp/0130414085&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-16T04:37:12.523" LastActivityDate="2010-12-16T04:37:12.523" />
  <row Id="14479" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:29:34.890" Score="7" ViewCount="0" Body="&lt;p&gt;Hint: ${{n} \choose {i}} \cdot!(n-i)$ counts the number of permutations that fix exactly $i$ elements.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-16T04:29:34.890" CommentCount="2" />
  <row Id="14480" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:29:35.467" Score="4" ViewCount="0" Body="&lt;p&gt;I actually prove a generalization of this in my paper &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/DT.pdf&quot; rel=&quot;nofollow&quot;&gt;Deranged Exams&lt;/a&gt;&quot; (&lt;em&gt;College Mathematics Journal&lt;/em&gt;, 41 (3): 197-202, 2010).  See Theorem 7.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization is the following.  Let $S_{n,k}$ be the number of permutations on $n$ elements in which none of the first $k$ elements remains in its original position.  Thus $S_{n,0} = n!$, and the number of derangements on $n$ elements, $D_n$, is $S_{n,n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{n+k,k} = \sum_{j=0}^n \binom{n}{j} D_{k+j}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP's question is the case $k = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;I'll extract the essence of the proof and post it in the next few minutes.&lt;/strike&gt;&#xA;Since you want hints rather than a full proof, I'll just leave this as a reference in case you (or anyone else reading this) is interested.  Jonas Meyer's answer gives a good hint.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-16T04:46:46.327" LastActivityDate="2010-12-16T04:46:46.327" CommentCount="2" />
  <row Id="14481" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:30:45.367" Score="1" ViewCount="0" Body="&lt;p&gt;The number of derangements is the number of permutations less the number that leave some numbers fixed.  So for example the term i=5 in the sum is all the ways to pick 5 out of n to leave fixed and derange all the rest.  Then sum over all the numbers of ones that can be fixed.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-16T04:30:45.367" />
  <row Id="14482" PostTypeId="2" ParentId="14477" CreationDate="2010-12-16T04:32:43.367" Score="3" ViewCount="0" Body="&lt;p&gt;Here's a proof, obscured using spoiler space.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If $d_n$ is the number of derangements on $n$ elements, then the number of permutations on $n$ elements with &lt;em&gt;exactly&lt;/em&gt; $i$ fixed points is ${n \choose i} d_{n-i}$ (choose i points to fix, then any permutation that fixes exactly those $i$ points (and nothing else) determines a derangement on the non-fixed points, and there are $d_{n-i}$ such derangements).  Hence, $n!=\sum_{i=0}^n {n \choose i} d_{n-i}$, which can be rearranged to give the above formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PS. I'm not a fan of the $!n$ notation, I'm pretty sure it's not standard in combinatorics.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-12-16T05:35:46.160" LastActivityDate="2010-12-16T05:35:46.160" CommentCount="2" />
  <row Id="14483" PostTypeId="1" CreationDate="2010-12-16T04:44:25.077" Score="10" ViewCount="185" Body="&lt;p&gt;I have a question about Hatcher's proof that the fundamental group of a circle is Z. Specifically, halfway through, ( &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf&lt;/a&gt; , page 30), he proves an important a lemma stated rather generally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a map $Y\times I\rightarrow S^1$ (Y is any space, and I is an interval), there is a unique lifting to  $Y\times I\rightarrow \mathbb{R}$ once we've specified an initial condition  $Y\times \{0\}\rightarrow \mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the proof of this, he breaks $Y\times I$ into smaller pieces $N\times \{y_0\}$, and keeps adjusting the size of the $N$. In the corresponding proofs in Munkres or Fulton, $Y$ is fixed as just another copy of $I$, and there is no need to modify the pieces we're looking at as the proof progresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: what is different in Hatcher's more general proof? Specifically, why does he keep modifying the $N$? My feeling is that this is because $Y$ could be disconnected or bad in some other sense. My question seems a little vague, but (a) if you look at Hatcher's proof, you'll see what I mean (near the bottom of p. 30), and (b) I'm hoping that maybe this situation, and the difficulties with it, are well known enough for someone to answer anyway.&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-12-19T21:15:11.053" LastActivityDate="2010-12-19T21:15:11.053" Title="Question about the first proof in Hatcher's Algebraic Topology" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="14484" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T04:50:32.173" Score="4" ViewCount="0" Body="&lt;p&gt;Munkres (the author of the very clear text on Topology) wrote a book called &lt;em&gt;Analysis on Manifolds&lt;/em&gt;. This is basically like Spivak, but twice as long and with more pictures. Everyone has fond memories of Spivak &lt;em&gt;after&lt;/em&gt; they've used it, because of how small and &quot;cool&quot; it is. But at the end of the day (especially for self-learning) I think Munkres is probably more straightforward and less frustrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201315963&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Analysis-Manifolds-Advanced-Books-Classics/dp/0201315963&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-12-16T05:01:35.647" LastActivityDate="2010-12-16T05:01:35.647" />
  <row Id="14485" PostTypeId="1" CreationDate="2010-12-16T05:18:40.600" Score="1" ViewCount="110" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2&quot;&gt;Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I just got the &quot;New and Revised&quot; edition of &quot;Mathematics: The New Golden Age&quot;, by Keith Devlin. On p. 64 it says the sum is $\pi^2/6$, but that's way off. $\pi^2/6 \approx 1.64493406685$ whereas the sum in question is $\approx 1.29128599706$. I'm expecting the sum to be something interesting, but I've forgotten how to do these things. &lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T06:12:26.580" LastActivityDate="2010-12-16T06:12:26.580" ClosedDate="2010-12-16T05:32:57.037" Title="Sum of 1/n*n for n = 1 ,2 ,3, ... ?" Tags="&lt;sequences-and-series&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14486" PostTypeId="2" ParentId="14485" CreationDate="2010-12-16T05:23:10.987" Score="2" ViewCount="0" Body="&lt;p&gt;The answer is indeed pretty interesting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proven using complex analysis or calculus, or probably in many hundreds of other ways. One example of how to prove this is given here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.uu.se/~bjorklund/euler.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uu.se/~bjorklund/euler.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-16T05:23:10.987" />
  <row Id="14487" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T05:37:27.780" Score="1" ViewCount="0" Body="&lt;p&gt;A good place to start learning about manifolds is to read a book by Spivak called &quot;Calculus on Manifolds.&quot; This book is very small, and the first three chapters are a short review on multivariate calculus, but I should say that the focus is much more on mathematical rigor than a normal calculus course so it would be worth your time to look through that I imagine. The last chapter introduces manifolds, how to integrate on them, and eventually culminates in the modern version of Stoke's Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want however to get a much more in depth view on manifolds, you will have to learn some topology. A good free online book to learn from, that I myself originally used, is called &quot;Topology Without Tears.&quot; It can be downloaded for free at: &lt;a href=&quot;http://uob-community.ballarat.edu.au/~smorris/topology.htm&quot; rel=&quot;nofollow&quot;&gt;http://uob-community.ballarat.edu.au/~smorris/topology.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It covers about everything a first course in topology covers, while being fairly straightforward and easy to understand. It is especially good for people who have not dealt with mathematical proof very much yet, because the author is very clear and detailed in his explanations and often explains about common proof techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer a different book, or a more commonly used book than &quot;Topology Without Tears,&quot; Munkre's Topology book is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can recommend two books on introductory manifolds which will be a lot more in depth than Spivak, but are still at the introductory level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John Lee's &quot;Introduction to Smooth Manifolds&quot; and&#xA;Boothby's &quot;Introduction to Differentiable manifolds&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These books are about Smooth Manifolds, which are a type of manifold in which calculus can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, if you learn topology and think you want to learn about manifolds from a topological point of view&lt;/p&gt;&#xA;&#xA;&lt;p&gt;William Fulton's &quot;Algebraic Topology&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a good place to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps! I've read a lot of books, so if you are looking for anything in particular that is slightly different from what I've said please feel free to email me.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-16T05:37:27.780" />
  <row Id="14488" PostTypeId="1" AcceptedAnswerId="14490" CreationDate="2010-12-16T05:44:15.843" Score="1" ViewCount="60" Body="&lt;p&gt;Hi, we are a group of three, and we've got this question in an assignment (the question is originally in French, so bear with me) :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $X_1 \dots X_4$ be a random sample drawn from a population of average $\mu$ and variance $\sigma^2$. We define the following estimators:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\hat{\mu}_1=\frac{X_1+2X_2-2X_3+5X_4}{6} \qquad \text{and} \qquad \hat{\mu}_2=\frac{X_1-2X_2+X_3+5X_4}{5}$$&#xD;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Show that $\hat{\mu}_1$ and $\hat{\mu}_2$ are both unbiased estimators&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Which one is the best? Justify.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the only number (out of 9) that we still have to complete, and we have no clue on how to start this up. Please, none of us are actually studying in a math degree (we're all programmers), so please give examples in your answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4803" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-16T06:32:13.317" LastActivityDate="2010-12-16T06:32:13.317" Title="Unbiased estimators ($\hat{\mu}_1$ and $\hat{\mu}_2$); proof and which is best" Tags="&lt;homework&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14489" PostTypeId="1" AcceptedAnswerId="14492" CreationDate="2010-12-16T05:49:31.273" Score="4" ViewCount="62" Body="&lt;p&gt;This is from one of the exercises in &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201558025&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;&quot;, and is something I'm doing privately, not homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a variant on the classic tower of hanoi, where all moves must be made clockwise only. The recursive solution is to consider Qn to be the number of moves to move a tower size n one step across(e.g. from start-&gt;end post), and Rn to be the number of moves to move it two(e.g start-&gt;end-&gt;spare)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recursive solution is(sorry about the formatting, just getting to grips with tex)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_n = 0 \quad \text{if n = 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_n = 2R_{n-1} + 1\quad \text{if n &amp;gt; 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = 0 \quad \text{if n = 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = Q_n + Q_{n-1} + 1 \quad \text{if n &amp;gt; 0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prove these? I presume an inductive proof is required but everything I've tried to do just gets garbled into a mess of further decreasing n values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explaining why these work is simple enough: $Q_n$ can be explained by the fact that you need to move a stack of size n-1 two steps forward followed by the n piece, followed by moving the n-1 stack two steps again onto the n piece=&gt; $Q_n = R_{n-1} + 1 + R_{n-1} = 2R_{n-1}+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wordy proof for $R_n$ is similar: It consists of moving $R_{n-1}$ then 1 followed by $Q_n$, 1 and $R_{n-1}$ which will end us up with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_n = 2R_{n-1}+1+Q_{n-1}+1 = Q_n+Q_{n-1}+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these are all just words that don't form any real proof... I've gone over the leading chapter and looked up some resources on proof by induction, but doing it with a recursion with two components has put me for a spin(Having forgotten a lot of the math studied in the past certainly isn't helping either). If someone could run me through a rigorous proof for n &gt; 0 it would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, any recommended learning resources on proof by induction would be appreciated as I'm still barely &quot;getting it&quot;&lt;/p&gt;&#xA;" OwnerUserId="4801" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-16T06:00:23.877" LastActivityDate="2010-12-17T21:53:23.007" Title="Proof of clockwise tower of hanoi variant recursive solution" Tags="&lt;discrete-mathematics&gt;" AnswerCount="1" />
  <row Id="14490" PostTypeId="2" ParentId="14488" CreationDate="2010-12-16T05:52:04.063" Score="4" ViewCount="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Show that $E[\hat{\mu_{1}}] = \mu$ and $E[\hat{\mu_2}] = \mu$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find $\min \left[\text{Var}(\hat{\mu_{1}}), \text{Var}(\hat{\mu_{2}}) \right]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-16T05:52:04.063" CommentCount="5" />
  <row Id="14492" PostTypeId="2" ParentId="14489" CreationDate="2010-12-16T06:09:36.083" Score="5" ViewCount="0" Body="&lt;p&gt;The &quot;word&quot; description is in fact the guide to a proof. The only problem is showing that the process you describe is in fact optimal. This is done by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we want to prove the formulas you give are the optimal solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $n=0$ is immediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: to prove the $k+1$ case assuming the $k$th case (so our induction hypothesis is that the formula you give is optimal for $R_k$ and for $Q_k$), you take a stack of $k+1$ disks. To move them all forward (to perform $Q_{k+1}$) you must get the top $k$ out of the way; that is done, by the induction assumption, with $R_k$ as optimal. Then you move the $k+1$ disk to the target, taking one move; then you move the disks from where they are to the target. Again, by the induction assumption, the best possible way of doing this is $R_k$. So you get that $Q_{k+1}=R_k+1+R_k=2R_k+1$, on the assumption that $Q_k$ and $R_k$ are the optimal moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument works for $R_{k+1}$; the only thing that your &quot;just words&quot; is missing is to be clear about what the induction hypothesis is, and to apply it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Why is the given strategy actually the best? In order to move all the disks from the original rod to the target rod, we must move the largest disk at least once; note that the position of the largest disk does not affect the validity or lack thereof of any move, so the position of the largest disk is immaterial for any other moves (as opposed to any other disk). So the optimal solution will only require us to move the largest disk exactly once, from the original rod to the target rod. In order to do this, we must first remove all disks from above the largest disk, and have no disks on the target rod so we can perform the move; that is $R_k$, because the only configuration that allows us to move the largest disk from the original rod to the target rod is to have all other disks in the non-target rod. Then we move the largest disk, which is the $+1$. And now, we just want to move the remaining disks to their final position in the quickest possible way, which again is $R_k$. A similar argument focusing on placing the largest disk in its final position, works for the computation of $R_{k+1}$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T21:53:23.007" LastActivityDate="2010-12-17T21:53:23.007" CommentCount="2" />
  <row Id="14493" PostTypeId="2" ParentId="14268" CreationDate="2010-12-16T06:26:08.117" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a simple example that does the job without any multiplication:&#xA;$5 \times 12$ vs $6 \times 11$ , rewrite both sides as $5\times(11+1)$ and $(5+1)\times11$, &#xA;then we get $5\times11+5\times1$ vs $5\times11 + 1 \times 11$ subtracting $5\times11$ from both sides we get : $1\times5 vs 1\times11$ which does not require any multiplication at all. The same technique can be used on (254)(847) and (383)(536).&#xA;$ 254 \times 847 vs 383 \times 536 $&lt;br&gt;&#xA;$ 254 \times ( 536 + 310 ) vs (254+129)\times536$&lt;br&gt;&#xA;$ 254 \times 310 vs 129\times536$ continuing in similar fashion we end up with&lt;br&gt;&#xA;$125\times 80 vs 4 \times 101$ although obvious but just for fun :&lt;br&gt;&#xA;$ (101+24)\times(76+4) vs 4 \times 101$ where in the next line we get :&lt;br&gt;&#xA;$ 4 \times 101 + 101\times 76 + 24\times76 + 24\times 4 vs 4\times 101 $  &lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2010-12-16T07:37:02.000" LastActivityDate="2010-12-16T07:37:02.000" />
  <row Id="14494" PostTypeId="1" AcceptedAnswerId="14535" CreationDate="2010-12-16T08:02:15.070" Score="4" ViewCount="94" Body="&lt;p&gt;I am trying to work through Boothby's An Introduction to Differentiable Manifolds on my own and, embarassingly, have got stuck at the very first chapter. At the end of section 4, chapter 1 (called: Further examples of manifolds: cutting and pasting), there's this question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that adding a handle to a 2-manifolds in the fashion described above for $S^2$ and $T^2$ actually does give a 2 manifold&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, Boothby gives no formal definitions for cutting and pasting so I have no idea about how to show that a sphere with a handle will be locally euclidean. For example, what if the handle is attached to the surface such that at the joint (a circle) the two surfaces meet sharply, so that for points on the joint we cannot find a normal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where can I find the operations of cutting formalised (I realise that pasting is connected to quotient topology, so I am studying that now). And, how does one answer this question rigorously?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&#xA;Upon further thought, 'cutting' could be this: since a manifold $M$ is metrizable, we can remove an open disk (open so that I get a boundary in the resulting manifold) around a point, i.e., we consider $M - B(p,\epsilon)$. So, given two dissimilar points, we remove two non-intersecting open disks (we can do this because a manifold is hausdorff). Now, how do I 'paste' the ends of the cylinder 'smoothly'? &lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-16T08:54:38.937" LastActivityDate="2010-12-16T17:25:55.777" Title="Adding handles to a sphere" Tags="&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="14495" PostTypeId="1" AcceptedAnswerId="16689" CreationDate="2010-12-16T08:11:18.893" Score="0" ViewCount="77" Body="&lt;p&gt;Let's say I have a list of points( in 2D coordinates!) with the vertex $V$, how to find all the triangular faces $F$ ( which has three edges $E$) that covers &lt;strong&gt;all&lt;/strong&gt; of the points? In other words, I'm looking for a mesh that connects all the points together with well behaved triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions are that the faces must not overlap or cut through one another. And it would help if all the $F$ has not-too-narrow angles in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the meshing algorithm above &lt;em&gt;must not introduced additional points&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Delaunay_triangulation&quot; rel=&quot;nofollow&quot;&gt;delaunay algorithm&lt;/a&gt; is meant for something like this. But I am not too clear about this. Am I right? Or is it that there are other algorithms that work for this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: A point cloud is just a list of points, scattered in 2D. The condition is all the points must have distinct ${x,y}$ coordinates.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-17T01:16:45.627" LastActivityDate="2011-01-07T16:00:35.493" Title="Given a Set of Points, Find All the Triangular Faces that Connect Them Together" Tags="&lt;geometry&gt;&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="14497" PostTypeId="1" CreationDate="2010-12-16T08:53:01.637" Score="1" ViewCount="71" Body="&lt;p&gt;How can I construct a CW complex A with H_0(A) = Z, H_2(A) = Z/4Z, H_4(A) =&#xA;Z ⊕ Z and all other homology groups trivial? &#xA;Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4808" LastActivityDate="2010-12-16T13:13:36.577" Title="construct CW complex " Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14499" PostTypeId="1" AcceptedAnswerId="14545" CreationDate="2010-12-16T10:10:01.340" Score="5" ViewCount="177" Body="&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: n is said to be a highly composite number if and only if $d(n)&amp;gt;d(m)$ for all $m&amp;lt;n$, where $d(n)$ denotes number of divisors of n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;:&lt;br&gt;&#xA;1) Are there any theorems about highest prime that divides n, assuming that n is highly composite?. It seems that the highest prime is of the order of ln(n).&lt;br&gt;&#xA;2) Are there any theorems about the highest power of 2 which divides n, assuming that n is highly composite?&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-16T14:44:09.167" LastActivityDate="2010-12-17T06:07:06.710" Title="Highly composite number" Tags="&lt;number-theory&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="14500" PostTypeId="2" ParentId="14386" CreationDate="2010-12-16T10:50:17.730" Score="0" ViewCount="0" Body="&lt;p&gt;Consider the related least squares question: find x minimizing $||Ax-b||_2$. In the unlikely scenario that the overdetermined system has a solution, then we in fact have $||Ax-b||_2=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The x solving the least squares problem can be found via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moore-Penrose_pseudoinverse&quot; rel=&quot;nofollow&quot;&gt;Moore-Penrose pseudoinverse&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=(A^tA)^{-1}A^tb$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one algorithm would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Solve $(A^tA)x=A^tb$ via your favorite linear solver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Check if $Ax-b=0$.&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2010-12-16T10:50:17.730" CommentCount="3" />
  <row Id="14501" PostTypeId="1" CreationDate="2010-12-16T11:44:08.030" Score="1" ViewCount="176" Body="&lt;p&gt;You are given three non-negative integers $A$, $B$ and $C$, find a number $X$ (say) satisfy&#xA;$X^A \equiv B\pmod{2C + 1}$ and $0 \le X \le 2C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am inquisitive about how to approach this one?&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T18:07:37.533" LastActivityDate="2010-12-16T18:07:37.533" Title="A problem on Number theory" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14502" PostTypeId="1" AcceptedAnswerId="14504" CreationDate="2010-12-16T12:48:37.807" Score="0" ViewCount="33" Body="&lt;p&gt;Common form of system of linear equations is A*X = B, X is unknown. But how to find A, if X and B are known?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is MxN matrix, X is column vector(N), B is column vector(M)   &lt;/p&gt;&#xA;" OwnerUserId="3532" LastEditorUserId="3532" LastEditorDisplayName="" LastEditDate="2010-12-16T13:04:20.533" LastActivityDate="2010-12-16T13:08:40.593" Title="System of linear equations" Tags="&lt;linear-algebra&gt;&lt;elementary&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="14503" PostTypeId="2" ParentId="14497" CreationDate="2010-12-16T12:55:13.287" Score="5" ViewCount="0" Body="&lt;p&gt;Let $f:S^2\to S^2$ be a map of degree $4$. Attach a $3$-cell $B^3$ to $Y=S^2$ using $f$ to glue the boundary $\partial B^3=S^2$ to $Y$. Then $Z=Y\cup_f B^3$ has homology given by $$H_q(Z)=\begin{cases}&#xD;&#xA;\mathbb Z, &amp;amp; \mbox{if }q=0; \\&#xD;&#xA;\mathbb Z/\mathbb 4Z, &amp;amp; \mbox{if }q=2; \\&#xD;&#xA;0, &amp;amp; \mbox{otherwise.}&#xD;&#xA;\end{cases}&#xD;&#xA;$$&#xD;&#xA;Now consider the connected sum $Z\vee S^4\vee S^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; The interesting part is of course the construction of $Z$. If I recall correctly, you can find in Hatcher's book information regarding &lt;em&gt;Moore spaces&lt;/em&gt;, of which $Z$ is a simple example.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-16T13:13:36.577" LastActivityDate="2010-12-16T13:13:36.577" />
  <row Id="14504" PostTypeId="2" ParentId="14502" CreationDate="2010-12-16T12:57:57.390" Score="2" ViewCount="0" Body="&lt;p&gt;If all matrices are square, then $A=BX^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-16T12:57:57.390" />
  <row Id="14505" PostTypeId="2" ParentId="14502" CreationDate="2010-12-16T13:08:40.593" Score="1" ViewCount="0" Body="&lt;p&gt;Do it row by row. Row $k$ in $A$ multiplied by the column vector $X$ equals the $k$th entry in the vector $B$. This is a single equation for the $N$ entries in that row of $A$ (so unless $X$ is zero, you get an $(N-1)$-parameter set of solutions for each row).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-16T13:08:40.593" />
  <row Id="14506" PostTypeId="1" CreationDate="2010-12-16T13:18:59.937" Score="3" ViewCount="132" Body="&lt;p&gt;Several papers give poly-time algorithms for constrained paths on labelled graphs, e.g. [1]&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quote:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Given an alphabet Σ, a (directed) graph G whose edges are weighted and Σ-labeled,&#xA;and a formal language L ⊆ Σ∗ , the formal-language-constrained shortest/simple path problem con-&#xA;sists of finding a shortest (simple) path p in G complying with the additional constraint that l(p) ∈ L.&#xA;Here l(p) denotes the unique word obtained by concatenating the Σ-labels of the edges along the&#xA;path p.&#xA;(1) We show that the formal-language-constrained shortest path problem is solvable efficiently&#xA;in polynomial time when L is restricted to be a context-free language (CFL).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can this be used for restricted read twice BDDs (Binary Decision Diagram)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider BDD where every variables occurs twice and the order of all paths root-&gt;T is restricted to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 .. n n .. 1 (this is palindrome)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat the BDD as edge labelled digraph where the true edge of variable is labelled as +variable and the false edge as -variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some paths in the digraph are &quot;BDD inconsistent&quot; - if a variable takes both values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like CFG constraint can help to find &quot;consistent&quot; paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to construct CFG that accepts all palindromes and only palindromes (over the alphabet U {+variable,-variable}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding a path from the root to &quot;true&quot; constrained by the above CFG will give a consistent satisfying assignment - the first half of the path is correct because a variable occurs only once and the second part is the reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct (computer experiments suggest it is)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different grammar may give different restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this published somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my humble knowledge this is one of the very few examples of read twice BDD supporting polytime SAT query.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The problem I am trying to solve is to find types of read twice BDDs which allow polytime SAT query with the help of CFGs. The above construction gives example for the case when the order of the variables is restricted to a palindrome, hence the (possibly incorrect) description &quot;restricted&quot; (because of the restriction of the order). I am not familiar with CFGs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]&#xA;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.105.446&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;FORMAL-LANGUAGE-CONSTRAINED PATH PROBLEMS∗&#xA;CHRIS BARRETT† , RIKO JACOB‡ , AND MADHAV MARATHE†&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Simple read twice BDD on x,y. The concatenation of edge labels give the palindrome word: &quot;+x -y -y +x&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://ll0x.wdfiles.com/local--files/admin%3Amanage/aa.png&quot; alt=&quot;BDD&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph is at: &lt;a href=&quot;http://ll0x.wdfiles.com/local--files/admin%3Amanage/aa.png&quot; rel=&quot;nofollow&quot;&gt;http://ll0x.wdfiles.com/local--files/admin%3Amanage/aa.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the graph may need copy/paste of the link)&lt;/p&gt;&#xA;" OwnerUserId="4813" LastEditorUserId="4813" LastEditorDisplayName="" LastEditDate="2011-01-03T11:33:21.777" LastActivityDate="2011-01-03T11:33:21.777" Title="Restricted read twice BDDs and context free grammars" Tags="&lt;graph-theory&gt;&lt;computer-science&gt;&lt;computability&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="14507" PostTypeId="1" AcceptedAnswerId="14508" CreationDate="2010-12-16T13:27:13.893" Score="3" ViewCount="177" Body="&lt;p&gt;I am helping my teenager son. So far I have been able to answer all his questions, and to solve all the problems he has had problems with. Except this one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int{{3^x}{e^x}dx}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, help me keep my success track and my son's confidence.&lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2010-12-17T00:52:02.463" Title="solving integral $\int{{3^x}{e^x}dx}$" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;integral&gt;" AnswerCount="2" />
  <row Id="14508" PostTypeId="2" ParentId="14507" CreationDate="2010-12-16T13:28:35.190" Score="9" ViewCount="0" Body="&lt;p&gt;Hint: try rewriting $3^x$ as $e^{x \ln 3}$.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2010-12-16T13:28:35.190" CommentCount="8" />
  <row Id="14509" PostTypeId="2" ParentId="14501" CreationDate="2010-12-16T13:36:00.197" Score="1" ViewCount="0" Body="&lt;p&gt;Assuming X coprime to 2C+1. Then Euler totient theorem gives $X^{\varphi(2C+1)} \equiv 1 \mod 2C+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming further that A coprime to $\varphi(2C+1)$. Then (By Bezout's identity) there's a D such that $AD \equiv 1 \mod \varphi(2C+1)$. In which case $(X^A)^D \equiv X^{(AD)} \equiv X \equiv B^D \mod 2C+1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Compute $\varphi(2C+1)$ (Which should be very time consuming if it's a product of large primes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Obtain D (through extended euclidean algorithm!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Compute $B^D \mod 2C+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, if a fast solution to your problem exists, then RSA cryptography would be insecure.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-12-16T14:27:22.737" LastActivityDate="2010-12-16T14:27:22.737" CommentCount="0" />
  <row Id="14510" PostTypeId="2" ParentId="14446" CreationDate="2010-12-16T13:37:04.983" Score="6" ViewCount="0" Body="&lt;p&gt;Let's solve the general problem surrounding the question,&#xA;with a few observations, each of them easy to see.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ for any ordinal $\alpha$ satisfies&#xA;Extensionality and Foundation, since all transitive sets satisfy&#xA;Extensionality and Foundation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ satisfies Separation, for the simple reason that&#xA;$A\subset B\in V_\alpha\implies A\in V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ satisfies Union, since $A\in&#xD;&#xA; V_\alpha\implies \bigcup A\in V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\lambda$ for a limit ordinal $\lambda$&#xD;&#xA;satisfies Pairing and Powerset, since the required set is&#xA;added at the next stage below $\lambda$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ satisfies the Axiom of Choice (assuming&#xA;this holds in $V$) in the choice-set version, since if&#xA;${\cal A}\in V_\alpha$ is a family of disjoint sets , then all choice&#xA;sets $B\subset \bigcup {\cal A}$ selecting one element from each $A\in{\cal A}$&#xD;&#xA;have the same or lower rank than ${\cal A}$, and so $B\in&#xD;&#xA; V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every $V_\alpha$ with $\omega\lt\alpha$ satisfies&#xA;Infinity, since $\omega\in V_\alpha$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only remaining axiom is Replacement, and this is the&#xA;only one that makes use of inaccessibility. But if&#xA;$\kappa$ is inaccessible, then $V_\kappa$ satisfies&#xA;Replacement, since if $A\in V_\kappa$ and $F:A\to&#xD;&#xA; V_\kappa$ is definable over $V_\kappa$, then $F&amp;#39;&amp;#39;A$ has&#xA;bounded rank below $\kappa$, since $|A|\lt\kappa$ and&#xA;$\kappa$ is regular. Thus, $F&amp;#39;&amp;#39;A\in V_\kappa$ as desired.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, one can also consider the question about&#xA;$H_\delta$, the sets of hereditary size less than $\delta$,&#xA;and things are a bit nicer here in several ways.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For any regular uncountable cardinal $\delta$, the set $H_\delta$ of&#xA;sets having hereditary size less than $\delta$, satisfies&#xA;$ZFC^-$, that is, all of ZFC except the Powerset axiom.&#xA;One gets the easy axioms easily; Separation is easy since the subset also has small hereditary size;&#xA;and Replacment follows from the fact that the union of fewer&#xA;than $\delta$ many sets in $H_\delta$ still has size less&#xA;than $\delta$ by the regularity of $\delta$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In particular, this shows that ZFC proves that there are&#xA;numerous transitive models of $ZFC^-$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $\kappa$ is inaccessible, then $V_\kappa=H_\kappa$,&#xA;and this satisfies full ZFC, since we get $ZFC^-$ in&#xA;$H_\kappa$, and we get power set since $\kappa$ is a&#xA;strong limit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="413" LastActivityDate="2010-12-16T13:37:04.983" />
  <row Id="14511" PostTypeId="1" AcceptedAnswerId="14529" CreationDate="2010-12-16T13:37:16.167" Score="1" ViewCount="62" Body="&lt;p&gt;Let A be a 4 × 4 matrix with eigenvalues -5, -2, 1, 4. Which of the following is an&#xA;eigenvalue of&lt;br&gt;&#xA;A  I&lt;br&gt;&#xA;I  A  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I is the 4 × 4 identity matrix?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) -5 (B) -7 (C) 2 (D) 1  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to find out relation between eigenvalues of matrix A and of the given block matrix.&lt;/p&gt;&#xA;" OwnerUserId="4814" LastActivityDate="2010-12-16T15:34:25.197" Title="Eigenvalue of Block matrix" Tags="&lt;matrices&gt;&lt;eigenvectors&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="14512" PostTypeId="1" AcceptedAnswerId="14513" CreationDate="2010-12-16T13:39:42.907" Score="0" ViewCount="36" Body="&lt;p&gt;Let x,y,z be three random variables. How can you show that cov(x+y,z) = cov(x,z) + cov(y,z) by using the definition of covariance.&lt;/p&gt;&#xA;" OwnerUserId="4815" LastActivityDate="2010-12-16T13:54:02.280" Title="simple question about covariance" Tags="&lt;probability&gt;" AnswerCount="1" />
  <row Id="14513" PostTypeId="2" ParentId="14512" CreationDate="2010-12-16T13:46:44.303" Score="2" ViewCount="0" Body="&lt;p&gt;Use the fact that $cov(X,Y)=E(X-EX)(Y-EY)$ and rearrange the terms.&#xA;I've included the full solution below, just move the mouse on the grey area.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  \begin{align*}&#xA;   cov(x+y,z)&amp;amp;=E[(x+y-E(x+y))(z-Ez)]\\&#xA;   &amp;amp;=E[((x-Ex)+(y-Ey))(z-Ez)]\\&#xA;   &amp;amp;=E(x-Ex)(z-Ez)+E(y-Ey)(z-EZ)=cov(x,z)+cov(y,z)&#xA;   \end{align*}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-16T13:54:02.280" LastActivityDate="2010-12-16T13:54:02.280" />
  <row Id="14514" PostTypeId="2" ParentId="14475" CreationDate="2010-12-16T14:06:41.207" Score="0" ViewCount="0" Body="&lt;p&gt;Differential forms and connections by Darling is not too bad. I can't remember how basic it is though. He does spend a lot of time going through what differential forms are. My approach to learning is to grab a bunch of books and read each one in turn until i get confused or stuck and then to go to the other to see if it makes more sense. There is also topology from the differentiable viewpoint by milnor. but this is more on smooth manifolds than calculus.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-16T14:06:41.207" />
  <row Id="14515" PostTypeId="1" AcceptedAnswerId="14518" CreationDate="2010-12-16T14:07:41.617" Score="3" ViewCount="158" Body="&lt;p&gt;My problem is:&#xA;What is the expression in $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much~&lt;/p&gt;&#xA;" OwnerUserId="4736" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-16T14:11:44.813" LastActivityDate="2010-12-16T15:52:52.927" Title="What is the expression of $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?" Tags="&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14516" PostTypeId="2" ParentId="14511" CreationDate="2010-12-16T14:24:43.320" Score="0" ViewCount="0" Body="&lt;p&gt;Show that if $\alpha$ is an eigenvalue of $A$ and $\beta$ is&#xA;an eigenvalue of&#xA;$$B=\begin{pmatrix}&#xD;&#xA;\alpha&amp;amp;1\\&#xD;&#xA;1&amp;amp;\alpha&#xD;&#xA;\end{pmatrix}$$&#xD;&#xA;then $\beta$ is also an eigenvalue of&#xA;$$C=\begin{pmatrix}&#xD;&#xA;A&amp;amp;I\\&#xD;&#xA;I&amp;amp;A&#xD;&#xA;\end{pmatrix}.$$&#xD;&#xA;(you should be able to write down an eigenvector for $C$in terms&#xA;of ones for $A$ and for $B$).&lt;/p&gt;&#xA;" OwnerUserId="226" LastActivityDate="2010-12-16T14:24:43.320" CommentCount="1" />
  <row Id="14518" PostTypeId="2" ParentId="14515" CreationDate="2010-12-16T14:31:08.927" Score="6" ViewCount="0" Body="&lt;p&gt;I don't think there is a &quot;closed&quot; form. You can give a good approximation using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;nofollow&quot;&gt;Euler-McLaurin Summation&lt;/a&gt; formula though:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n} \dfrac{1}{j^2}  = \dfrac{\pi^2}{6} - \dfrac{1}{n} - \dfrac{1}{2n^2} + \mathcal{O}(\dfrac{1}{n^3})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you need more accuracy you can include more terms from the summation formula to give the coefficients of the lower order terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The Euler McLaurin Summation formula only tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{n} \dfrac{1}{j^2}  = C - \dfrac{1}{n} - \dfrac{1}{2n^2} + \mathcal{O}(\dfrac{1}{n^3})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $\displaystyle C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know by other means that $\displaystyle C = \dfrac{\pi^2}{6}$, for instance, see this for a multitude of ways: &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2&quot;&gt;Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-16T14:48:17.657" LastActivityDate="2010-12-16T14:48:17.657" CommentCount="1" />
  <row Id="14520" PostTypeId="2" ParentId="14483" CreationDate="2010-12-16T14:39:14.627" Score="2" ViewCount="0" Body="&lt;p&gt;This steps seems to be unnecessary.  In what follows, I'll assume you have p. 30 in front of you (since it's available freely online).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the part where he has constructed neighborhood $N$ of $y_0 \in Y$ and a partition $0 = t_0 &amp;lt; \dots t_n = 1$ such that $F(N \times [t_i,t_{i+1}]) \subset U_i$ for each $i = 0, \dots, n-1$.  The lift is constructed by defining&#xA;$\widetilde{F}:N \times I \to \mathbb{R}$ to be $h_{i} \circ F$, where $h_{i}$ is the homeomorphism from the evenly covered neighborhood $U_i$ to the sheet $\widetilde{U_i}$ which contains $h_{i-1} \circ F(y_0,t_i)$.  (The homeomorphism $h_0$ is determined by the given lift $\widetilde{F}$ at time $t=0$.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only question is whether this is well-defined.  One might worry that $N \times \{t_i\}$ is mapped by the $(i-1)$-st lift into the sheet $\widetilde{U}_{i-1}$ but that the neighborhood $N$ needs to be modified so that the $i$-th lift glues to this.  In other words, could it be that we need to replace $N$ with a smaller neighborhood $N&amp;#39;$ so that $h_{(i-1)} \circ F(N&amp;#39; \times \{t_i\}) \subset \widetilde{U}_{i-1} \cap \widetilde{U}_{i}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the case because, by design, $F(N \times \{t_i\}) \subset U_{i-1} \cap U_i$. Therefore, the map above is well-defined.  The lift is continuous by the glueing lemma for open sets.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-16T14:39:14.627" />
  <row Id="14521" PostTypeId="2" ParentId="14515" CreationDate="2010-12-16T14:39:40.013" Score="4" ViewCount="0" Body="&lt;p&gt;I not sure of the thrust of your question but maybe the generalised harmonic numbers are what you want&#xA;$$ H_{n,r} = \sum_{k=1}^n \frac{1}{k^r} , $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in particular $H_{n,2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information &lt;a href=&quot;http://mathworld.wolfram.com/HarmonicNumber.html&quot; rel=&quot;nofollow&quot;&gt;here,&lt;/a&gt; including a very nice identity for $H_{n,2}$ by B. Cloitre.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-16T14:39:40.013" CommentCount="12" />
  <row Id="14522" PostTypeId="1" AcceptedAnswerId="14532" CreationDate="2010-12-16T14:49:01.743" Score="10" ViewCount="103" Body="&lt;p&gt;In our analysis course, the following question came up and could, up to now, not be solved:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $a: \mathbb{N} \to \mathbb{C}$ be a sequence of complex numbers. What are necessary and sufficient conditions for the existence of a function $\mu: 2^\mathbb{N} \to \mathbb{C}$ satisfying the properties&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$\mu(\{i\}) = a_i$&#xD;&lt;/li&gt;&#xA;  &lt;li&gt;$\mu$ is finitely additive, i.e. $A\cap B = \emptyset\implies \mu(A\cup B) = \mu(A)+\mu(B)$?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Partial Results&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we are given such a function defined on a subset of $2^\mathbb{N}$, we can of course extend it to all sets of the form $A\cup B, A\cap B = \emptyset$ and $A\setminus B, B\subset A$. This invites the use of Zorn's Lemma; but it seems impossible to prove that a maximal set closed under these operations must be $2^\mathbb{N}$. However, this approach strongly suggests that $\mu$ exists for all $(a_i)$, as the problem only depends on $2^\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if $a_i$ converges absolutely, one can set $\mu(I) = \sum_{i\in I} a_i$ which fulfills the required properties, but this approach does not generalise at all.&lt;/p&gt;&#xA;" OwnerUserId="4817" LastActivityDate="2010-12-16T16:30:05.913" Title="Finitely additive measures on $2^\mathbb{N}$" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="14523" PostTypeId="1" CreationDate="2010-12-16T14:57:20.950" Score="0" ViewCount="113" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is ; How can I solve the following integral question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\tan\;x\;\cos\;2x\;\mathrm dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T01:46:08.627" LastActivityDate="2010-12-17T01:46:08.627" Title="How can I find $\int\tan\;x\;\cos\;2x\;\mathrm dx$?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14524" PostTypeId="2" ParentId="14523" CreationDate="2010-12-16T15:03:58.120" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $\cos 2x = \cos^2x-\sin^2x=2\cos^2x-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $2\sin x\cos x = \sin 2x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\frac{d}{dx}\log(f(x))=?$&lt;/p&gt;&#xA;" OwnerUserId="1154" LastActivityDate="2010-12-16T15:03:58.120" CommentCount="3" />
  <row Id="14525" PostTypeId="2" ParentId="14523" CreationDate="2010-12-16T15:08:27.337" Score="1" ViewCount="0" Body="&lt;p&gt;Suppose I gave you an integral of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int \cot x \ \ f(\sin x) \ \text{dx}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of a substitution to get rid of the $\cot x$ term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a concrete example, can you try evaluating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int \cot x  \ \ (1 + \sin^5 x) \ \ \text{dx}$ ?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-16T15:20:49.363" LastActivityDate="2010-12-16T15:20:49.363" />
  <row Id="14526" PostTypeId="2" ParentId="14523" CreationDate="2010-12-16T15:11:13.987" Score="-1" ViewCount="0" Body="&lt;p&gt;Thank you for all the answers.&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="3669" LastEditorDisplayName="" LastEditDate="2010-12-16T15:32:26.160" LastActivityDate="2010-12-16T15:32:26.160" CommentCount="0" />
  <row Id="14527" PostTypeId="1" CreationDate="2010-12-16T15:13:48.093" Score="0" ViewCount="21" Body="&lt;p&gt;If $N$ is even, how to prove that $(D(z))\hat{}(n)=1/2(\hat{z}(n)+\hat{z}(n+N/2))$, for all $n$. Hint: Show that the discrete Fourier transformation(DFT) on $N/2$ points of $D(z)$ agrees with the DFT on $N$ points of $(z+z\ast)/2$,  for $z\ast(n)=(-1)^nz(n)$, DFT: $\hat{z}(m)=\sum_{n=0}^{N-1} z(n)e^{-2\pi imn/N}$, $D(z)(n) = z(2n)$(i.e. if $z=(z(0),z(1),\dots,z(N-1))$ then $D(z)=(z(0),z(2),\dots,z(N-2))$) also remember $(z+z\ast)(n)=z(n)(1+(-1)^n))= { \begin{array}{ll} 2 z(n) &amp;amp;, n \textrm{is even} \0 &amp;amp;, \textrm{is odd}\end{array} &lt;/p&gt;&#xA;" OwnerUserId="2723" LastActivityDate="2010-12-16T15:13:48.093" Title="If $N$ is even, how to prove that $(D(z))\hat{}(n)=1/2(\hat{z}(n)+\hat{z}(n+N/2))$" Tags="&lt;linear-algebra&gt;" />
  <row Id="14528" PostTypeId="2" ParentId="14523" CreationDate="2010-12-16T15:32:40.513" Score="1" ViewCount="0" Body="&lt;p&gt;I'm going to tell you that by parts done directly isn't the way to approach this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \tan(x)\cos(2x)dx = -\ln(\cos(x))\cos(2x) - 2\int \ln(\cos(x))\sin(2x)dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this expression is not likely to become any more manageable by solving the next integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, your problem comes down to simplifying the expression $\tan(x)cos(2x)$. Big hint. The other answers have shown you how to do this. Once you simplify it, you will have a much easier job of integrating said expression and you most certainly won't need integration by parts.&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="892" LastEditorDisplayName="" LastEditDate="2010-12-16T15:50:42.777" LastActivityDate="2010-12-16T15:50:42.777" />
  <row Id="14529" PostTypeId="2" ParentId="14511" CreationDate="2010-12-16T15:34:25.197" Score="1" ViewCount="0" Body="&lt;p&gt;The block matrix can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=A \otimes I_2 + I_4 \otimes J_2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J_2=\left (\begin{array}{cc} 0 &amp;amp;\ 1\\ 1 &amp;amp;\ 0 \end{array} \right )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denote the eigenvectors of $A$ by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Av_{\lambda} = \lambda v_{\lambda}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $J_2$ by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_2w_{\pm} = \pm w_{\pm}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The eigenvalues of $J_2$ are $1$ and $-1$). Clearly, the eight combinations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\lambda} \otimes w_{\pm}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are all eigenvectors of $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the eigenvalues of C are $\lambda \pm 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the right answer is C corresponding to $\lambda = 1$&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-12-16T15:34:25.197" CommentCount="1" />
  <row Id="14530" PostTypeId="1" CreationDate="2010-12-16T15:39:16.890" Score="1" ViewCount="53" Body="&lt;p&gt;Looking on Internet, I mostly found this definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given quantities on a cyclic domain D, first rescale the domain to [0;2$\pi$[, then, find $z_n$ the point on the unit circle corresponding to the $n$th value, and compute the average by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z_m = \sum_{n=1}^N z_n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average angle is then $\theta_m = \arg z_m$ and to obtain the average value you scale back to your original domain D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must say, I have problems with this definition. For simplicity, I will use oriented angles in degree for my examples (i.e. D = [0;360[). With this formula, having angles -90, 90 and 40 will give 40 as mean angle, when I would expect 13.33 as an answer (i.e. (90-90+40)/3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my own problems, I usually use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_m = \mathop{\rm arg\,min}_{v\in D} \sum_{n=1}^{N} d(v_n,v)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $d(x,y)$ is the distance in the cyclic domain, and $\{v_1, v_2, \ldots, v_n\}$ is the set of cyclic data I want to average of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has the advantage to work the same way whatever the domain (replace D by a non-cyclic domain and $d$ with the usual euclidean distance, and you find the usual definition of an average). However, it is expensive to compute and I don't know any exact method to do it in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: what is the appropriate way to deal with average of cyclic data? And do you have good pointers that explain the problem and its solutions?&lt;/p&gt;&#xA;" OwnerUserId="4709" LastEditorUserId="4709" LastEditorDisplayName="" LastEditDate="2010-12-16T17:53:09.050" LastActivityDate="2010-12-16T19:06:07.320" Title="How to average cyclic quantities?" Tags="&lt;algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14531" PostTypeId="2" ParentId="14515" CreationDate="2010-12-16T15:52:52.927" Score="0" ViewCount="0" Body="&lt;p&gt;I am not sure if this will work or not, but maybe you could try writing the expression in terms of falling factorials[1]. Then maybe use summation by parts[2]. I am not sure how nicely this will work, but you could try it. Let me know what you find out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://mathworld.wolfram.com/FallingFactorial.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/FallingFactorial.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://mathworld.wolfram.com/SummationbyParts.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/SummationbyParts.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-16T15:52:52.927" CommentCount="2" />
  <row Id="14532" PostTypeId="2" ParentId="14522" CreationDate="2010-12-16T15:58:20.213" Score="6" ViewCount="0" Body="&lt;p&gt;Given any such sequence $a_i$, you can use it to define a finitely additive measure on the collection of &lt;em&gt;finite&lt;/em&gt; subsets of $\mathbb N$.  Now consider the vector&#xA;space $V:= \mathbb C^{\oplus \mathbb N}$, i.e. the direct sum of countably many copies of $\mathbb C$, with the copies of $\mathbb C$ being indexed by elements $i \in \mathbb N$.  Alternatively, this is the space of sequences $(z_i)_{i \in \mathbb N}$ with $z_i =0$ for all but finitely many $i$.&#xA;Or, if you want to think a little more analytically, you can think&#xA;of this as the space of $\mathbb C$-valued functions on $\mathbb N$ with &lt;em&gt;finite support&lt;/em&gt;, i.e. which vanish outside a finite set.   (The function attached to a sequence is just $i \mapsto z_i$,  of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our choice of finitely addivite measure defines a functional on $V$, given by&#xA;$(z_i) \mapsto \sum_{i \in \mathbb N} a_i z_i.$ (More analytically, this is &lt;em&gt;integration&lt;/em&gt; of the finitely supported function corresponding to $(z_i)$ against our finitely addivite measure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $W$ be the vector space of &lt;em&gt;all&lt;/em&gt; $\mathbb C$-valued functions on $V$.&#xA;Certainly $V \subset W$, and we may always extend linear functionals from a subspace to the whole space; thus we may extend our given functional to a&#xA;functional $I: W \to \mathbb C$.  (The label $I$ is chosen to suggest &lt;em&gt;integration&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $S$ is any subset of $\mathbb N$, let $\chi_S$ be the characteristic function of $S$.  Define $\mu(S) = I(\chi_S)$.  Then $\mu$ is a finitely additive measure on $\mathbb N$ satisfying the required properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is essentially the Zorn's lemma argument suggested in the original posting, but reformulated in terms of extending functionals on vector spaces,&#xA;which makes it more transparent.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-16T15:58:20.213" CommentCount="3" />
  <row Id="14533" PostTypeId="2" ParentId="14507" CreationDate="2010-12-16T16:06:34.150" Score="6" ViewCount="0" Body="&lt;p&gt;Why not simply $\int 3^x e^x \mathrm dx = \int (3e)^x \mathrm dx = \frac{(3e)^x}{\ln(3e)} + Constant$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that $\int (3e)^x \mathrm dx = \frac{(3e)^x}{\ln(3e)}+C$,  You can differentiate $(3e)^x$ i.e. Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = (3e)^x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln y = x \ln(3e)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm dy/\mathrm dx = y \ln(3e) = (3e)^x \ln(3e)$$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-17T00:52:02.463" LastActivityDate="2010-12-17T00:52:02.463" CommentCount="2" />
  <row Id="14534" PostTypeId="1" CreationDate="2010-12-16T16:06:37.453" Score="2" ViewCount="49" Body="&lt;p&gt;Is it true that $O(M^3 + NM^2) \, = \, O(M^3 + N)$, where $M$ and $N$ are variables of the function?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastActivityDate="2010-12-16T18:00:31.240" Title="Big-oh for function of two variables" Tags="&lt;computer-science&gt;&lt;big-o&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14535" PostTypeId="2" ParentId="14494" CreationDate="2010-12-16T16:17:12.193" Score="4" ViewCount="0" Body="&lt;p&gt;This is a well-known issue, which one resolves as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it is better not to work with some given metric on the manifold, because you don't know precisely how the metric structure (e.g. the balls in the given metric) interact with the locally Euclidean structure.  So it is better to &lt;em&gt;first&lt;/em&gt; choose a n.h. of a point $p$ which is diffeomorphic to an open disk in $\mathbb R^2$, and then to work with that n.h., which one can now think of very&#xA;concretely as an open disk in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concreteness, suppose it is the open disk of radius $r$ around $0$.  Now &#xA;delete the open disk of radius $r- \epsilon$, for some positive $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are in a situation very similar to you one you described (i.e. $M\setminus B(p,\epsilon)$) except that we know exactly what it is that we deleted (i.e. we are certain that it is a small disk), &lt;em&gt;and&lt;/em&gt; we know that there is an open annulus of inner radius $r-\epsilon$ and outer radius $r$ left over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do the same thing at the other point (using Hausdorffness, as you point out).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when we glue in the cylinder, think of the cylinder as a circle cross an open interval, say $I$.  Now we can choose  small open subintervals $I_1$ and $I_2$ at either end of the interval $I$ (so that altogether $I = I_1 \cup \text{ a closed interval } \cup I_2$, with the unions being disjoint).&#xA;Notice that a circle cross $I_1$ is diffeomorphic to our open annulus from above,&#xA;and the same for a circle cross $I_2$.  (This is just using the fact that the $I_i$ are open intervals, and I leave it as an exercise for you to find the&#xA;diffeomorphism.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can glue our cylinder to $M$ minus the two disks by identifying&#xA;the circle cross $I_1$ with the annulus around one of our deleted points,&#xA;and the circle cross $I_2$ with the other annulus, using the diffeos of the preceding paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: we have carefully glued in our cylinder to that there are no sharp corners, by making sure that each end of the cylinder is glued smoothly to an annulus around each of the points that is being deleted.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-16T16:17:12.193" />
  <row Id="14537" PostTypeId="1" CreationDate="2010-12-16T16:31:02.247" Score="-1" ViewCount="234" Body="&lt;p&gt;E.g. if one only wants to get an M.S?&lt;/p&gt;&#xA;" OwnerUserId="4821" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-17T06:07:41.013" LastActivityDate="2010-12-17T06:07:41.013" ClosedDate="2010-12-16T20:46:46.227" Title="Is Montclair State University good for math for grad school?" Tags="&lt;tag-removed&gt;" CommentCount="2" />
  <row Id="14538" PostTypeId="2" ParentId="14530" CreationDate="2010-12-16T16:33:15.537" Score="0" ViewCount="0" Body="&lt;p&gt;The angle is supposed to be the independent variable, not the dependent variable.  If your function is cyclic (using degrees), z(-90)=z(270) so it doesn't matter which you use.  Then the average value of the function is $$z_m=\sum_{n=1}^Nz(\theta_n)$$  You are right that if you average the angles, it matters which lap of the circle you use (-90 vs 270) because the difference of 360 gets divided by N.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-16T16:33:15.537" CommentCount="3" />
  <row Id="14539" PostTypeId="2" ParentId="14417" CreationDate="2010-12-16T16:47:49.827" Score="4" ViewCount="0" Body="&lt;p&gt;For the more general question at the end, let's first consider the weighted inner product and weighted norm on $\mathbb{R}^n$, which are defined by &#xA;$$&#xD;&#xA;\langle u,v\rangle  = \sum\limits_{i = 1}^n {c_i u_i v_i }, \;\; ||u|| = \sqrt {\langle u,u\rangle }  = \sqrt {\sum\limits_{i = 1}^n {c_i u_i^2 } },  &#xD;&#xA;$$&#xD;&#xA;where $c_1,\ldots,c_n$ are positive constants (called weights). By Cauchy-Schwartz inequality, $|\langle u,v\rangle | \le ||u||||v||$. Hence, letting ${n \choose k}$ play the role of weights (and using the symmetry of binomial coefficients), we have&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^n {{n \choose k}a^{\sqrt k } b^{\sqrt {n - k} } }  \le \sqrt {\sum\limits_{k = 0}^n {{n \choose k}(a^2 )^{\sqrt k } } } \sqrt {\sum\limits_{k = 0}^n {{n \choose k}(b^2 )^{\sqrt k } } } .&#xD;&#xA;$$&#xD;&#xA;So, if we assume that $a,b \geq 1$, the problem essentially reduces to the original one (as far as an upper-bound is concerned). In particular, when $a=b$, we get&#xA;$$&#xD;&#xA;\sum\limits_{k = 0}^n {{n \choose k} a^{\sqrt k } a^{\sqrt {n - k} } }  \le \sum\limits_{k = 0}^n {{n \choose k} a^{2\sqrt k } }.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-16T16:47:49.827" />
  <row Id="14540" PostTypeId="1" CreationDate="2010-12-16T16:51:19.060" Score="2" ViewCount="95" Body="&lt;p&gt;Find Fourier cosine transform of $e^{-a^2 x^2}$ and hense evaluate Fourier sine transform of $x\cdot e^{-a^2x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can solve this question only if there is $x$ instead of $x^2$ in the exponential function $e^{-a^2x^2}$. Because in this situation i can use integral formula :- &#xA;$$\int e^{-ax} \cos sx\,dx = \frac{e^{-ax}}{a^2+x^2} \left( s\sin sx-a\cos sx \right)$$&#xD;&#xA;but what should i do if there is $x^2$ in exponential function $e$??&lt;/p&gt;&#xA;" OwnerUserId="4822" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-16T18:52:44.780" LastActivityDate="2010-12-16T20:37:59.313" Title="Fourier cosine transform" Tags="&lt;fourier-analysis&gt;&lt;fourier-series&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14541" PostTypeId="1" CreationDate="2010-12-16T17:05:07.160" Score="1" ViewCount="53" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a quite general question , but I've always heard about creating mathematical models  in economics, social sciences, engineering,... And I would want to know what are the starting points and roadmap to understand what a mathematical model is and how to create them. Thanks&lt;/p&gt;&#xA;" OwnerUserId="4824" LastActivityDate="2010-12-16T18:09:21.253" Title="How to start creating Mathematical models?" Tags="&lt;mathematical-modeling&gt;" AnswerCount="2" />
  <row Id="14542" PostTypeId="2" ParentId="14495" CreationDate="2010-12-16T17:10:35.577" Score="3" ViewCount="0" Body="&lt;p&gt;Typically, in modeling problems of the kind you are suggesting there are a variety of ways that the triangles could be selected, and one's goal would govern how the triangles might be chosen. Sometimes these are referred to as a triangular mesh:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle_mesh&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Triangle_mesh&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the analogous problem of selecting triangles based on a point set in the plane you can find some ideas here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Point_set_triangulation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Point_set_triangulation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-16T17:10:35.577" CommentCount="1" />
  <row Id="14543" PostTypeId="2" ParentId="14541" CreationDate="2010-12-16T17:13:49.380" Score="1" ViewCount="0" Body="&lt;p&gt;COMAP, the Consortium for Mathematics and Its Applications has developed lots of materials at different levels that address your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.comap.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.comap.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-16T17:13:49.380" CommentCount="2" />
  <row Id="14544" PostTypeId="2" ParentId="14494" CreationDate="2010-12-16T17:25:55.777" Score="0" ViewCount="0" Body="&lt;p&gt;I cannot add a comment yet, so I am posting this as a separate answer.&#xA;Firstly, thank you Matt E for the response. &#xA;Now, supposing I want to find a diffeomorphism from $S^1\times I_1$ to the annulus, shouldn't the annulus itself be open, i.e., I have to remove a closure of $B(p,r-\epsilon)$. In which case, I can think of the annulus as $S^1 \times (r-\epsilon,r)$ and then the diffeomorphism is just the lift of the usual diffeomorphism between open intervals.&lt;/p&gt;&#xA;" OwnerUserId="4828" LastActivityDate="2010-12-16T17:25:55.777" CommentCount="1" />
  <row Id="14545" PostTypeId="2" ParentId="14499" CreationDate="2010-12-16T17:41:16.957" Score="7" ViewCount="0" Body="&lt;p&gt;This &lt;a href=&quot;http://www.ams.org/journals/tran/1944-056-00/S0002-9947-1944-0011087-2/S0002-9947-1944-0011087-2.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Alaoglu and Erdős proves that if $p$ is the largest prime divisor of a highly composite number (HCN) $n$ then the exponent $k_q$ on another prime $q$ dividing&#xA;$n$ satisfies &#xA;$$\log\left(1 + \frac{1}{k_q}\right) &amp;gt; \frac{\log q \log 2}{\log p} + O\left(\frac{(\log \log p)^3}{(\log p)^3}\right),$$&#xD;&#xA;$$\log\left(1 + \frac{1}{k_q+1}\right) &amp;lt; \frac{\log q \log 2}{\log p} + O\left(\frac{(\log \log p)^3}{(\log p)^3}\right).$$&#xD;&#xA;(See Theorem 12.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sufficiently large $q$ they claim that $k_q$ is either $\Lambda$, $\Lambda + 1$, or $\Lambda - 1$, where $$\Lambda = \left\lfloor \left(2^{\log q/\log p} - 1\right)^{-1} \right\rfloor.$$  (See Theorem 13.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $q = 2$ this latter formula appears to be close, although it is not exactly correct, as there are more than three possible values for the power of 2 that divides a highly composite number with largest prime factor $p$.  See, for example, &lt;a href=&quot;http://wwwhomes.uni-bielefeld.de/achim/highly.txt&quot; rel=&quot;nofollow&quot;&gt;this list of the first 1200 HCN's&lt;/a&gt;, which includes their prime factorizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, together with Derek Jennings' answer on the order of $p$, should give you the estimate of the highest power of 2 that divides $n$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-17T06:07:06.710" LastActivityDate="2010-12-17T06:07:06.710" CommentCount="1" />
  <row Id="14546" PostTypeId="2" ParentId="14540" CreationDate="2010-12-16T17:42:28.810" Score="2" ViewCount="0" Body="&lt;p&gt;HINT 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_{ - \infty }^\infty  {\cos (sx)e^{ - a^2 x^2 } \,{\rm d}x}  = \int_{ - \infty }^\infty  {[\cos (sx) + {\rm i}\sin (sx)]e^{ - a^2 x^2 } \,{\rm d}x}.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Characteristic function of the (centered) normal distribution: see &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_distribution#Characteristic_function_and_moment_generating_function&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: The first hints didn't help. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_{ - \infty }^\infty  {\cos (sx)e^{ - a^2 x^2 } \,{\rm d}x}  = \int_{ - \infty }^\infty  {[\cos (sx) + {\rm i}\sin (sx)]e^{ - a^2 x^2 } \,{\rm d}x} = \int_{ - \infty }^\infty  {e^{{\rm i}sx} e^{ - a^2 x^2 } {\rm d}x} = ? &#xD;&#xA;$$&#xD;&#xA;In order to calculate the right-most integral, bring it to the form(s) &#xA;$$&#xD;&#xA;b \int_{ - \infty }^\infty  {e^{ - a^2(x - {\rm i}c )^2  } \, {\rm d}x} = &#xD;&#xA;b \int_{ - \infty }^\infty  {e^{ - a^2 x^2 } \,{\rm d}x} = b&amp;#39; \int_{ - \infty }^\infty  {e^{ - x^2 } \,{\rm d}x} =...&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-16T20:37:59.313" LastActivityDate="2010-12-16T20:37:59.313" CommentCount="5" />
  <row Id="14547" PostTypeId="2" ParentId="14540" CreationDate="2010-12-16T17:43:15.990" Score="2" ViewCount="0" Body="&lt;p&gt;Find the Fourier transform of $e^{-a^2x^2}$. The real part of that is relevant to your cosine transform. &lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-16T17:43:15.990" CommentCount="2" />
  <row Id="14548" PostTypeId="2" ParentId="14501" CreationDate="2010-12-16T17:49:46.497" Score="0" ViewCount="0" Body="&lt;p&gt;Generally it's difficult to compute such $\rm\:A$'th roots in $\rm \mathbb Z/m\ $ except for simple special cases, e.g. when $\rm\ gcd(A,\phi(m)) = 1\:.\ $ Otherwise one generally has to resort to factoring $\rm\:m\:$ to get $\rm\: \phi(m)\: $ and hence the factorization of $\rm\ Z/m^*\ $ into a product of cyclic groups. Then one may apply generalizations of Shanks's  square-root algorithm. See for example Section 3.2: Extensions of Shanks's algorithm to r'th roots in cyclic groups in Lindhurst: &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.144.2770&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;Computing roots in finite fields and groups, with a jaunt through sums of digits&lt;/a&gt;. One can also generalize the Euler criterion for square-roots - see my &lt;a href=&quot;http://math.stackexchange.com/questions/5424/how-can-i-find-out-whether-a-number-is-a-quadratic-residue-in-a-large-modulo/5433#5433&quot;&gt;post here&lt;/a&gt; and Chapter 4 of Ireland and Rosen, A Classical Introduction to Modern Number Theory.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-16T17:49:46.497" CommentCount="1" />
  <row Id="14549" PostTypeId="1" AcceptedAnswerId="14636" CreationDate="2010-12-16T17:56:43.983" Score="1" ViewCount="50" Body="&lt;p&gt;Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chernoff_bound&quot; rel=&quot;nofollow&quot;&gt;Chernoff bound&lt;/a&gt;: It assumes $X_1, \ldots, X_n$ are independent Bernoulli random variables, each having probability $p &amp;gt; 1/2$, and seeks the probability of simultaneous occurrence of more than $n/2$ of the events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once heard that, in the proof of the Chernoff bound, parameters $p$ and $n$ must be constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? for instance, can we let $p = 1/2 + n^{-3}$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Edit (As per moron's request)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'll elaborate on how I wan to use it. The &lt;a href=&quot;http://en.wikipedia.org/wiki/BPP&quot; rel=&quot;nofollow&quot;&gt;complexity class BPP&lt;/a&gt; is defined as below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A language L is in BPP if and only if there exists a probabilistic Turing machine M, such that&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;M runs for polynomial time on all inputs;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Completeness:&lt;/strong&gt; For all $x \in L$, M outputs 0 with probability less than or equal to 1/3;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Soundness:&lt;/strong&gt; For all $x \not \in L$, M outputs 1 with probability less than or equal to 1/3.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In practice, an error probability of 1/3 might not be acceptable, ... but if the algorithm is run many times, the chance that the majority of the runs are wrong drops off exponentially as a consequence of the &lt;strong&gt;Chernoff bound&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to be true as long as the error probability is a constant. However:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It does not even have to be constant: the same class of problems is defined by allowing error as high as $1/2 − n^{−c}$ on the one hand, or requiring error as small as $2^{−n^c}$ on the other hand, where $c$ is any positive constant, and $n$ is the length of input.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is Chernoff bound applied in such cases?&lt;/p&gt;&#xA;" OwnerUserId="2420" LastEditorUserId="2420" LastEditorDisplayName="" LastEditDate="2010-12-16T21:19:49.450" LastActivityDate="2010-12-17T04:52:40.090" Title="Non-constant parameters in Chernoff bound" Tags="&lt;probability&gt;&lt;inequality&gt;&lt;random-variable&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14550" PostTypeId="1" AcceptedAnswerId="14587" CreationDate="2010-12-16T17:59:22.947" Score="1" ViewCount="55" Body="&lt;p&gt;Let $G$ be a group of permutations of  $n$ objects. Let subsets $A$,$B$ of the objects be equivalent if $A=g(B)$ for some $g \in G$. &#xA;Is there a way to show that a bijection exists between the equivalence classes and the $2$-colorings of the objects that are invariant under those permutations in $G$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Far, I have verified that it works(their cardinalities are the same) for a few cases I tried, but I do not see how it can be generalized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will post an example, the definition should be clear from the example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $G=\{e,(2 3)\}$, $n=3$&#xD;&#xA;then, we have 8 subsets and 6 equivalent subsets. namely $\{\}, \{1\},\{\{2\},\{3\}\}, \{\{1,2\},\{1,3\}\},\{2,3\},\{1,2,3\}$ &#xA;and the invarient transformations are. &#xA;$\{1,1,1\},\{1,1,0\} \sim \{1,0,1\},\{0,1,1\},\{1,0,0\},\{0,0,1\} \sim \{0,1,0\},\{0,0,0\}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example applying (2 3) to {0,1,0} changes it to {0,0,1} and applying e, leaves it as it is. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-16T19:38:54.607" LastActivityDate="2010-12-16T21:31:07.990" Title="Bijection of 2-colorings and permutation equivalence classes" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="14551" PostTypeId="2" ParentId="14534" CreationDate="2010-12-16T18:00:31.240" Score="4" ViewCount="0" Body="&lt;p&gt;[Throughout this answer, I'm going to use set notation in order to be pedantic.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that $O(M^3+N)\subset O(M^3+NM^2)$, since for $M$ sufficiently large, $c(M^3+N)&amp;lt;c(M^3+NM^2)$. It's worth noting that for $M$ sufficiently close to $0$, $M^3+N&amp;gt;M^3+NM^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reverse inclusion, consider $f(M,N)=M^3+NM^2$. Clearly $f\in O(M^3+NM^2)$. We now need to show that for all $c$ and arbitrarily large $M,N$ that $f(M,N)&amp;gt;c(M^3+N)$. We can do this by setting $M=\log N$. We then have $f(\log N,N)=\log^3 N + N\log^2 N$, which is clearly larger than $c(\log^3 N + N)$ for all sufficiently large $N$. So $f(M,N)\not\in O(M^3+N)$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2010-12-16T18:00:31.240" />
  <row Id="14552" PostTypeId="2" ParentId="14499" CreationDate="2010-12-16T18:06:27.230" Score="4" ViewCount="0" Body="&lt;p&gt;In this paper &lt;a href=&quot;http://www.renyi.hu/~p_erdos/1944-04.pdf&quot; rel=&quot;nofollow&quot;&gt;On Highly Composite Numbers&lt;/a&gt; P. Erdös uses Bertrand's postulate and the Prime Number Theorem to show that if $p$ is the highest prime that divides $n,$ a highly composite number, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_1 \log n &amp;lt; p &amp;lt; c_2 \log n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constants $c_1$ and $c_2 .$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-16T18:06:27.230" CommentCount="1" />
  <row Id="14553" PostTypeId="2" ParentId="14541" CreationDate="2010-12-16T18:09:21.253" Score="1" ViewCount="0" Body="&lt;p&gt;The last time I taught mathematical modeling we used &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=XPxrItVsXpQC&amp;amp;printsec=frontcover&amp;amp;dq=a+first+course+in+mathematical+modeling&amp;amp;source=bl&amp;amp;ots=4EcHhIII_M&amp;amp;sig=K1KZ1lq9OmZWj-aMKMdJiv5wg64&amp;amp;hl=en&amp;amp;ei=pVQKTeTAI4nUtQPioODDCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=6&amp;amp;ved=0CEMQ6AEwBQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Introduction to Mathematical Modeling&lt;/a&gt;,&lt;/em&gt; by Giordano, Fox, Horton, and Weir.  The first part of Chapter 2 contains a nice introduction to the modeling process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.google.com/search?tbs=bks%3A1&amp;amp;tbo=1&amp;amp;q=mathematical+modeling&amp;amp;btnG=Search+Books&quot; rel=&quot;nofollow&quot;&gt;search on Google books for mathematical modeling texts&lt;/a&gt; should yield some more good results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;a href=&quot;http://math.stackexchange.com/questions/11231/probability-in-single-lane-traffic-flow-what-are-the-odds-of-choke-points-bein/11316#11316&quot;&gt;answer to a recent math.SE question&lt;/a&gt; also contains some discussion of the modeling process.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-16T18:09:21.253" />
  <row Id="14554" PostTypeId="1" AcceptedAnswerId="14555" CreationDate="2010-12-16T18:17:18.190" Score="4" ViewCount="181" Body="&lt;p&gt;I am a calculus TA a bit stumped at a question a student has posed me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the method for finding the sum of the series $$\sum_{n=0}^\infty\frac{1+n}{3^n}$$ If you split the numerator up, one becomes a geometric series summing to 3/2, and the other sums to 3/4 by Wolfram Alpha, but I can't explain why.&lt;/p&gt;&#xA;" OwnerUserId="541" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-16T19:32:24.853" LastActivityDate="2010-12-16T19:32:24.853" Title="Sum of a Series (Calculus)" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="14555" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T18:21:24.257" Score="8" ViewCount="0" Body="&lt;p&gt;Using the usual geometric series, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{1-x} = \sum_{n=1}^{\infty} x^n $$ thus differentiating you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$$ thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$ Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your sum is just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{1/3}{(1-1/3)^2} = \frac{3}{4}$$&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-16T18:30:26.570" LastActivityDate="2010-12-16T18:30:26.570" CommentCount="4" />
  <row Id="14556" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T18:24:37.510" Score="3" ViewCount="0" Body="&lt;p&gt;Consider for $|x| &amp;lt; 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^\infty x^n = \frac{x}{1-x} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate and set $x=1/3.$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-16T18:24:37.510" />
  <row Id="14557" PostTypeId="2" ParentId="14445" CreationDate="2010-12-16T18:25:22.333" Score="1" ViewCount="0" Body="&lt;p&gt;How about embedding $W^{1,p}$ in $L^p\times (L^p)^n$ and using Hahn-Banach and Riesz' representation theorem to get a nice characterization of elements in the dual &lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2010-12-16T18:25:22.333" CommentCount="3" />
  <row Id="14559" PostTypeId="2" ParentId="14530" CreationDate="2010-12-16T19:06:07.320" Score="0" ViewCount="0" Body="&lt;p&gt;For angles, one can adapt the iterative way of computing means to angles, that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given angles v[1] .. v[n]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m[1] = v[1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m[i] = remainder( m[i-1] + remainder( v[i]-m[i-1], C)/i, C)   (i=2..n)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where remainder(x,y) gives the signed remainder on dividing x by y and C is the measure of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this gives your vm, but haven't been able to prove it.&lt;/p&gt;&#xA;" OwnerUserId="4830" LastActivityDate="2010-12-16T19:06:07.320" CommentCount="1" />
  <row Id="14560" PostTypeId="1" AcceptedAnswerId="14565" CreationDate="2010-12-16T19:11:32.010" Score="7" ViewCount="222" Body="&lt;p&gt;I understand the basic idea of Conditionally Convergent (some infinitely long series can be made to converge to any value by reordering the series). I just do not understand how this could possibly be true. I think it defies common sense and seems like a clear violation of the Commutative Law.&lt;/p&gt;&#xA;" OwnerUserId="4831" LastActivityDate="2010-12-17T03:01:11.483" Title="Please explain how Conditionally Convergent can be valid?" Tags="&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="14561" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T19:12:37.740" Score="4" ViewCount="0" Body="&lt;p&gt;Michael, there is a general trick here that can be very useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of $xD$ as an operator on differentiable functions $p(x)$ that &quot;differentiates and then multiplies by $x$&quot;. Then polynomials in $xD$ are also operators on differentiable functions, if one interprets powers of $xD$ by composition. For example if $f(x)=2x^{2}+3$ then $f(xD)$ applied to a differentiable function p(x) gives&#xA;$2xD(xD(p(x)))+3p(x)$. (Think of 3 as the operator &quot;multiplication by 3&quot;. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should verify the following very useful fact: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operator $p(xD)$ applied term by term to the power series $\sum_{n}a_{n}x^{n}$ gives the power series $\sum_{n}p(n)a_{n}x^{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, your series has the form $\sum f(n)x^{n}$ evaluated at $x=1/3$, where $f(n)=n+1$. So to compute the sum, apply $1+xD$ to the function $1/(1-x)=1+x+x^{2}+\ldots$ and then evaluate at $x=1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I learned this from Polya and Szego's &quot;Problems and Theorems in Analysis&quot;, which is absolutely stuffed with mathematical wisdom.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditorDisplayName="" LastEditDate="2010-12-16T19:25:55.877" LastActivityDate="2010-12-16T19:25:55.877" CommentCount="1" />
  <row Id="14562" PostTypeId="2" ParentId="14554" CreationDate="2010-12-16T19:25:06.383" Score="5" ViewCount="0" Body="&lt;p&gt;I believe the other answers about differentiating and such are much cleaner, but I thought I would add that there is an even more elementary way of doing the sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \sum_{k=1}^{\infty} k b^{-k} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by using the standard trick of multiplying by $b^{-1}$ then subtracting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ b^{-1} S = \sum_{k=2}^{\infty} (k-1) b^{-k} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S - b^{-1} S = S (1 - b^{-1}) = \sum_{k=1}^{\infty} b^{-k} $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4143" LastActivityDate="2010-12-16T19:25:06.383" />
  <row Id="14563" PostTypeId="2" ParentId="14560" CreationDate="2010-12-16T19:26:26.810" Score="15" ViewCount="0" Body="&lt;p&gt;When it comes to infinity, most common sense goes out the window. Cantor had a hard time convincing some of his peers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me what was surprising is that absolutely convergent series can be reordered in any fashion and still give the same result!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When begining to learn, one naturally considers the series $\displaystyle 1 - 1 + 1 - 1 \dots$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle  1 - (1-1) - (1-1) \dots = 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle (1-1) + (1-1) + \dots = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it should not be a (big) surprise if we could reorder some series differently to give multiple possible values. From there it should not be a big surprise if there were some series which could give &lt;em&gt;any&lt;/em&gt; possible value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_series_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann Rearrangement Theorem&lt;/a&gt; might help reduce the surprise :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-16T19:26:26.810" />
  <row Id="14564" PostTypeId="2" ParentId="14560" CreationDate="2010-12-16T19:31:57.077" Score="9" ViewCount="0" Body="&lt;p&gt;The commutative law holds for finitely many terms.  There's no reason to expect it to hold for infinitely many terms unless the series converges absolutely.  That's what absolute convergence is for.  In mathematics, you can't assume that the infinite analogues of finite statements are always true; you have to prove it, because often they aren't.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-16T19:31:57.077" CommentCount="1" />
  <row Id="14565" PostTypeId="2" ParentId="14560" CreationDate="2010-12-16T19:38:49.387" Score="17" ViewCount="0" Body="&lt;p&gt;What Riemann's Rearrangement Theorem on convergent series tells you is &lt;em&gt;precisely&lt;/em&gt; that these &quot;infinite sums&quot; don't behave in the same way as the &quot;usual&quot; sums do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to understand that just because we can add two numbers together, and from that (through the use of the associative law) we can add any &lt;em&gt;finite&lt;/em&gt; quantity of numbers together, we do &lt;em&gt;not&lt;/em&gt; have a way to &quot;add an infinite quantity of numbers&quot; together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So series are &lt;em&gt;not&lt;/em&gt; really sums in the usual sense, they are &lt;em&gt;limits&lt;/em&gt;, and they are limits of &lt;em&gt;sequences&lt;/em&gt;. So commutativity and associativity of sums are not really at play, because series are not really sums in the usual sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, $\sum_{i=1}^{\infty}a_n = L$ does not mean that you add infinitely many numbers and you get $L$, what it means is that the sequence&#xA;$s_1,s_2,s_3,\ldots,s_n,\ldots$ converges to $L$, where $s_k = a_1+a_2+\cdots+a_k$ is the $k$th partial sum. So, for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n\geq N$, $|s_n-L|\lt \epsilon$. We are not adding up the infinitely  many terms, we are saying that we can get arbitrarily close to $L$ if we add enough (but still only finitely many) of the terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can certainly play the usual games of commutativity and associativity so long as we only modify finitely many of the terms of the sequence $s_1,s_2,s_3,\ldots$, without modifying the limit: remember that the limit of the sequence only depends on what happens &quot;eventually&quot;, not any changes you make to the beginning of the sequence. But that means that you need to keep the sequence intact from some point on. So you can only move around your original terms $a_n$ up to some point (up to some $n_0$) without having to worry about potentially changing the limit. But if you allow yourself to move the numbers around so much that you change the values of arbitrarily large partial sums, then you no longer have any reason to expect the new sequence to have the same limit as the old one (it &lt;em&gt;could&lt;/em&gt;, but it doesn't &lt;em&gt;have to&lt;/em&gt;, because you are now looking at an entirely different sequence). Which means that the usual ideas of &quot;associativity&quot; and &quot;commutativity&quot; of &lt;strong&gt;finite&lt;/strong&gt; sums (you can add them up in any order and still get the same answer) no longer necessarily work for &quot;&lt;strong&gt;infinite&lt;/strong&gt; sums&quot; (because they are &lt;em&gt;not really&lt;/em&gt; sums, they are limits). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is how Riemann's Rearrangement Theorem works: it modifies &lt;em&gt;all&lt;/em&gt; the terms of the sequence so that you get a new limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Moron has given you the standard example of why you cannot play associative and commutative games with infinite sums the way you do with finite sums, so I don't have to)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:35:20.050" LastActivityDate="2010-12-16T20:35:20.050" />
  <row Id="14566" PostTypeId="1" CreationDate="2010-12-16T19:46:05.943" Score="0" ViewCount="62" Body="&lt;p&gt;I am running an image classification with 5 classes. There is a large set of training data from which I randomly select a training subset, yielding slightly different classifications. I run the classification three times, and obtain the following user's accuracies for each class (the user's accuracy is the probability that a certain pixel of a given class is correctly classified):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 | .794 | .802 | .796 &lt;br&gt;&#xA;2 | .511 | .506 | .503 &lt;br&gt;&#xA;3 | .938 | .941 | .940 &lt;br&gt;&#xA;4 | .909 | .907 | .912 &lt;br&gt; &#xA;5 | .768 | .765 | .765 &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each pixel, there are three outcomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) All runs have the same class (either none are wrong or all are wrongly classified)&lt;br&gt;&#xA;2) 2 runs have the same class (either 1, 2 or all were wrongly classified)&lt;br&gt;&#xA;3) All runs are different (either 2 or all were wrongly classified)&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to calculate the probability for each combination of results, of each of these outcomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1, 1, 1:  P none wrong, P all wrong &lt;br&gt;&#xA;1, 1, 2:  P 1 one wrong, P 2 wrong, P all wrong&lt;br&gt;&#xA;1, 2, 3:  P 2 wrong, P all wrong&lt;br&gt;&#xA;etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to approach the problem? Could any one lead me in the right direction? Is it simpler to assume that the accuracies are the same for each run?&lt;/p&gt;&#xA;" OwnerUserId="4833" LastEditorUserId="4833" LastEditorDisplayName="" LastEditDate="2010-12-17T18:41:10.827" LastActivityDate="2010-12-17T18:41:10.827" Title="Probability of multiple classification results being incorrect" Tags="&lt;probability&gt;" CommentCount="2" />
  <row Id="14567" PostTypeId="1" AcceptedAnswerId="14571" CreationDate="2010-12-16T19:51:19.853" Score="3" ViewCount="49" Body="&lt;p&gt;I'm working my way through &lt;em&gt;Linear Algebra Done Right&lt;/em&gt;. To help with one proof, I want to prove the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\mathbf{V}$, a vector space and $T$, a linear operator on it, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathbf{W}_1$ and $\mathbf{W}_2$ are subspaces of $\mathbf{V}$ such that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathbf{V}$ is a direct sum of $\mathbf{W}_1$ and $\mathbf{W}_2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathbf{W}_1$ and $\mathbf{W}_2$ are invariant under $T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The restriction of $T$ to $\mathbf{W}_1$ has at most $k$ eigenvalues.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The restriction of $T$ to $\mathbf{W}_2$ has at most $p$ eigenvalues.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then $T$ has at most $k+p$ eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a sketch of a proof using determinants, but it was based on old knowledge about the properties of determinants with regards to eigenvalues, so it may not be correct. The book doesn't emphasize using them though, and maybe there's a proof of this without using determinants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried a proof by contradiction, trying to find something weird by assuming that T can have more than $k+p$ eigenvalues, but I haven't been able to find anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4832" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:03:48.433" LastActivityDate="2010-12-16T20:06:43.757" Title="Help with proof about maximum number of eigenvalues." Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14568" PostTypeId="2" ParentId="14550" CreationDate="2010-12-16T19:59:44.290" Score="0" ViewCount="0" Body="&lt;p&gt;So, what you are actually doing is this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a set, and let $G$ be a subgroup of $S_X$, the group of permutations on $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define two equivalence relations, one, let's call it $\sim$, on $\mathcal{P}(X)$, the power set of $X$, and one, let's call it $\equiv$, on the set of functions $2^X = \{f\colon X\to \{0,1\}\}$, which we think of as the $2$-colorings of $X$. We define:&#xA;\begin{align*}&#xA;A\sim B &amp;amp;\Longleftrightarrow \text{there exists $\sigma\in G$ such that $\sigma(A)=B$}\\&#xA;f\equiv g &amp;amp;\Longleftrightarrow \text{there exists $\sigma\in G$ such that $f\circ\sigma = g$.}&#xA;\end{align*}&#xA;The question is whether there is a bijection between $\mathcal{P}(X)/\sim$ and $2^X/\equiv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that there is a natural identification of $2^X$ with $\mathcal{P}(X)$, by mapping $f\colon X\to\{0,1\}$ to the subset $f^{-1}(1) = \{x\in X\mid f(x)=1\}$, and mapping $A\subseteq X$ to $\chi_A$, the characteristic function of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\mathcal{F}\colon \mathcal{P}(X)/\sim \to 2^X/\equiv$ as follows: given a class $[A] = \{ B\in\mathcal{P}(X)\mid A\sim B\}$, we let&#xA;$$\mathcal{F}([A]) = \{ \chi_B\mid B\in [A]\}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I claim this is well defined: that is, I claim that $\mathcal{F}([A])$ is an equivalence class modulo $\equiv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A\sim B$. Then there exists $\sigma\in G$ such that $\sigma(A)=B$. I claim that $\chi_B\circ\sigma = \chi_A$. Indeed, let $x\in X$. If $x\in A$, then $\sigma(x)\in B$, so $\chi_B\circ\sigma(x) = 1 = \chi_A(x)$. And if $x\notin A$, then $\sigma(x)\notin B$ (since $A=\sigma^{-1}(B)$), so $\chi_B\sigma(x) = 0 = \chi_A(x)$. Thus, $\chi_B\equiv \chi_A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, suppose that $f\in 2^X$ is equivalent to $\chi_A$. Then there exists $\tau\in G$ such that $f\circ\tau = \chi_A$. Let $B=\tau(A)$. I claim that $\chi_B=f$. Indeed, let $x\in X$. Then $f(\tau(x))=1$ if and only if $x\in A$, if and only if $\tau(x)\in B$. So $f(y)=1$ if and only if $y\in B$. Note that $B\sim A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, every element of $[A]$ gets mapped to an element of&#xA;$$\langle \chi_A \rangle = \{ f\in 2^X\mid f\equiv \chi_A\}$$&#xD;&#xA;and every element of $\langle\chi_A\rangle$ comes from an element of $[A]$. Thus, $\mathcal{F}$ is well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathcal{F}$ is one-to-one: if $\mathcal{F}([A]) = \mathcal{F}([B])$, then &#xA;$$\chi_B\in \langle \chi_B\rangle = \mathcal{F}([B]) = \mathcal{F}([A])=\langle \chi_A\rangle$$&#xD;&#xA;so there exists $\sigma\in G$ such that $\chi_B\sigma = \chi_A$. that means that $x\in A$ if and only if $\sigma(x)\in B$, so $B=\sigma(A)$, hence $B\sim A$, so $[A]=[B]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\mathcal{F}$ is onto: let $\langle f\rangle$ be an equivalence class of $2^X/\equiv$. Let $A = f^{-1}(1)$. I claim that $\mathcal{F}([A])=\langle \chi_A\rangle = \langle f\rangle$. Indeed, $\chi_A = f$ by construction, so $f\in\langle\chi_A\rangle$, hence $\langle f\rangle = \langle \chi_A\rangle = \mathcal{F}([A])$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\mathcal{F}$ is a bijection between $\mathcal{P}(X)/\sim$ and $2^X/\equiv$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;However, I want to add that I think it is fair to say that your nomenclature is misleading/nonstandard. I would take that a $2$-coloring of $X$ is &lt;em&gt;invariant under $G$&lt;/em&gt; to mean that if $f\colon X\to\{0,1\}$ is the $2$-coloring, then $f\circ \sigma = f$ for all $\sigma\in G$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T21:05:22.547" LastActivityDate="2010-12-16T21:05:22.547" CommentCount="2" />
  <row Id="14569" PostTypeId="1" AcceptedAnswerId="14590" CreationDate="2010-12-16T20:00:26.193" Score="7" ViewCount="227" Body="&lt;p&gt;The Collatz Conjecture is a famous conjecture in mathematics that has lasted for over 70 years. It goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $f(n)$ to be as a function on the natural numbers by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(n) = n/2$ if $n$ is even and&#xA;$f(n) = 3n+1$ if $n$ is odd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjecture is that for all $n \in \mathbb{N}$, $n$ eventually converges under iteration by $f$ to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the &quot;5n+1&quot; problem has been solved. This problem is the same as the Collatz problem except that in the above one replaces $3n+1$ with $5n+1$.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-16T21:44:46.947" Title="The $5n+1$ Problem" Tags="&lt;analysis&gt;&lt;combinatorics&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14570" PostTypeId="2" ParentId="14567" CreationDate="2010-12-16T20:01:30.143" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;: $ker(T) = ker(T|W_1)\ \oplus\ ker(T|W_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $ker$ stands for kernel, $T|W_i$ stands for $T$ restricted to $W_i$, etc.. Actually you have to do this with the transformation $T -\lambda I$ instead of $T$.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-16T20:06:43.757" LastActivityDate="2010-12-16T20:06:43.757" CommentCount="4" />
  <row Id="14571" PostTypeId="2" ParentId="14567" CreationDate="2010-12-16T20:05:37.293" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Suppose $\lambda$ is an eigenvalue. Then there exists an eigenvector $\mathbf{v}$ corresponding to $\lambda$. Write $\mathbf{v}=\mathbf{w}_1+\mathbf{w}_2$; since $\mathbf{v}$ is nonzero, at least one of $\mathbf{w}_1$ and $\mathbf{w}_2$ is nonzero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, evaluate $T(\mathbf{v})$; since $\mathbf{V}$ is a &lt;em&gt;direct sum&lt;/em&gt;, and each $\mathbf{W}_i$ is invariant, what can you say about $T(\mathbf{w}_1)$ and $T(\mathbf{w}_2)$?&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-16T20:05:37.293" CommentCount="4" />
  <row Id="14573" PostTypeId="1" AcceptedAnswerId="14578" CreationDate="2010-12-16T20:16:25.723" Score="2" ViewCount="54" Body="&lt;p&gt;Let's say that I have a tax of 17% over my services and I want to receive $20,000.00 regardless (after paying) the tax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I want something like: &#xA;$$X - 17\% = \$20,000.00$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $X$ is the final value that I need to charge to receive $\$20,000$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate this?&lt;/p&gt;&#xA;" OwnerUserId="1425" LastEditorUserId="1425" LastEditorDisplayName="" LastEditDate="2010-12-16T22:53:50.477" LastActivityDate="2010-12-16T22:53:50.477" Title="Calculating percentages for taxes." Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14574" PostTypeId="2" ParentId="14569" CreationDate="2010-12-16T20:21:10.077" Score="4" ViewCount="0" Body="&lt;p&gt;you have the wrong problem, it is solved by 13. The generalization next is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if n = 3k then k&#xA;if n = 2k then k&#xA;else 5n+1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4290" LastActivityDate="2010-12-16T20:21:10.077" />
  <row Id="14575" PostTypeId="2" ParentId="14573" CreationDate="2010-12-16T20:25:38.580" Score="4" ViewCount="0" Body="&lt;p&gt;Removing 17% leaves (100-17)%=83%, so you want 83%*x=20.000.  Divide both sides by 83% to get your answer.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-16T20:25:38.580" CommentCount="1" />
  <row Id="14578" PostTypeId="2" ParentId="14573" CreationDate="2010-12-16T20:31:28.087" Score="6" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Rule_of_three_%28mathematics%29#Rule_of_Three&quot; rel=&quot;nofollow&quot;&gt;Rule of Three&lt;/a&gt; is your friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want the 83% that is left to be $\$20,000$. If $X$ is 100% of what you need to charge, then you have:&#xA;$$\begin{array}{rcl}&#xD;&#xA;X &amp;amp;\longrightarrow &amp;amp;100\%\\\&#xD;&#xA;20,000 &amp;amp;\longrightarrow&amp;amp;83\%&#xD;&#xA;\end{array}$$&#xD;&#xA;By the Rule of Three, to get the value of $X$ you multiply across the known diagonal and divide by the remaining quantity; so $X = \frac{(20,000)(100)}{83}\approx \$24,096.39$.  &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:57:47.867" LastActivityDate="2010-12-16T20:57:47.867" CommentCount="1" />
  <row Id="14579" PostTypeId="1" CreationDate="2010-12-16T20:46:49.297" Score="1" ViewCount="67" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/14124/finding-inverse-of-x-bmod-y&quot;&gt;finding inverse of $x\bmod y$&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hello all  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Me and some friends are studying for a discrete exam and we are having some troubles finding the inverse modulus of things. The question we are suck on is &quot;what is the inverse of 8 mod 29&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-16T20:52:22.123" LastActivityDate="2010-12-16T23:37:46.680" ClosedDate="2010-12-17T02:49:35.270" Title="Troubles finding inverse modulus " Tags="&lt;discrete-mathematics&gt;&lt;elementary-number-theory&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="14580" PostTypeId="2" ParentId="14579" CreationDate="2010-12-16T20:51:52.103" Score="4" ViewCount="0" Body="&lt;p&gt;The inverse of $8$ modulo $29$ is an integer $k$ such that $8k \equiv 1\pmod{29}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways of finding such an integer. The standard way is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;Extended Euclidean Algorithm&lt;/a&gt;, but sometimes you can do it without using this algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, since $8\times 3 = 24 = -5\bmod 29$, and $-5\times -6 =30=1\bmod 29$, then $8\times(3\times-6) = 1\bmod 29$. Now just adjust by adding multiples of $29$ to get an answer between $0$ and $28$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-16T20:51:52.103" CommentCount="1" />
  <row Id="14581" PostTypeId="1" AcceptedAnswerId="14609" CreationDate="2010-12-16T20:52:49.327" Score="2" ViewCount="156" Body="&lt;p&gt;Draw two balls centered at the origin in R^3, where $r_1 &amp;lt; r_2$. What is the topological name for the boundary created? Does it even make sense to talk about a figure with an inside boundary and an outside one?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditorDisplayName="" LastEditDate="2010-12-16T21:48:23.277" LastActivityDate="2010-12-16T23:32:47.307" Title="What is topological name of a sphere with a hole in it?" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14582" PostTypeId="1" AcceptedAnswerId="14596" CreationDate="2010-12-16T21:11:09.873" Score="3" ViewCount="216" Body="&lt;p&gt;I have a problem with solution of this limit:&#xA;$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}$$&#xD;&#xA;Of course, it's a very easy to solve, using (twice) L'Hôpital's rule, but I need to find out, how to do this without this rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stuck in this point:&#xA;$$\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$$&#xD;&#xA;Everything I need to know is how to eliminate $\frac{\sin^2{x}}{x^2}$, because - as my tutor said - I can't simply substitute $1$ for this expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: It's not a homework. My tutor showed this problem as a puzzle and said, that it would be a good exercise to solve this without L'Hôpital's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is a way I got to the point $\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\lim_{x-&gt;0}{\frac{\tan{x}-x}{x^2}} &amp;amp;= \lim_{x-&gt;0}{\frac{(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}-x)\cdot (\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}{x^2(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}}\\&#xA; &amp;amp;= \lim_{x-&gt;0}{\frac{\frac{\sin^2{x}}{x^2\cos^2{x}}-1}{x(\frac{\sin{x}}{x}\frac{1}{\cos{x}}+1)}}= \lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T04:34:27.280" LastActivityDate="2010-12-17T04:39:41.483" Title="Limit, solution in unusual way" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;limit&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="14583" PostTypeId="2" ParentId="14569" CreationDate="2010-12-16T21:14:15.133" Score="7" ViewCount="0" Body="&lt;p&gt;In Part I of Lagarias' &lt;a href=&quot;http://arxiv.org/abs/math/0309224&quot; rel=&quot;nofollow&quot;&gt;extensive, annotated bibliography&lt;/a&gt; of the 3x+1 problem, he notes a 1999 paper by Metzger (reference 112) regarding the 5x+1 problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the 5x + 1 problem he shows that on the positive integers there is no cycle of size 1, a unique cycle of size 2, having smallest element n = 1, and exactly two cycles of size 3, having smallest elements n = 13 and n = 17, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is unclear from the notes whether the paper shows that these are the &lt;em&gt;only&lt;/em&gt; cycles of the 5x+1 problem or whether there may exist longer cycles.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2010-12-16T21:14:15.133" CommentCount="2" />
  <row Id="14585" PostTypeId="1" AcceptedAnswerId="14599" CreationDate="2010-12-16T21:15:53.010" Score="5" ViewCount="85" Body="&lt;p&gt;Ok, so I just finished a Discrete Math class where we learned about all types of proofs and counting and graphs, etc. So I know the basics, but never thought I'd need to use it in the real world (not anytime soon at least). However, a friend of mine showed me a puzzle game, and he said that it's &lt;em&gt;always&lt;/em&gt; possible to win this game, even though it seems unlikely to me. I will explain the rules, which are pretty simple, and I want to know how I could prove this, or if the answer is evident using combinatorics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how the game works:&#xA;The game is a slight variation of a game called 'Poker Squares'. Anyway, you lay down (all at once) 25 &lt;strong&gt;random&lt;/strong&gt; cards arranged in a 5x5 grid (no jokers). The goal of the game is to arrange all the cards so that each row is a poker hand of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A straight: 5 cards arranged in order of their number (i.e. 9,10,J,Q,K, irregardless of their suite.)&lt;/li&gt;&#xA;&lt;li&gt;A flush: any 5 cards in any order of the same suite (i.e. five hearts)&lt;/li&gt;&#xA;&lt;li&gt;A full house: A triple and a double (i.e. three 10's and two J's.)&lt;/li&gt;&#xA;&lt;li&gt;A straight flush: 5 cards of the same suite arranged in increasing or decreasing order (i.e. 4,5,6,7,8 all diamonds)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And thats it. The claim is that given any random 25 cards, these cards can ALWAYS be arranged so that each row is a winning poker hand (of the above mentioned hands only). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditorDisplayName="" LastEditDate="2010-12-16T21:47:32.450" LastActivityDate="2010-12-16T22:24:11.657" Title="Proving a certain type of poker game can always be won" Tags="&lt;combinatorics&gt;&lt;card-games&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14586" PostTypeId="2" ParentId="14549" CreationDate="2010-12-16T21:23:08.530" Score="3" ViewCount="0" Body="&lt;p&gt;The Chernoff bound states that a certain function $f(p, n)$ of $p$ and $n$ is greater than a certain other function $g(p, n)$ of $p$ and $n$, subject to the condition that $p &amp;gt; \frac{1}{2}$.  I see no reason why you can't pick $p$ and $n$ to be whatever you want; in particular, I see no reason why you can't let $p$ depend on $n$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-16T21:23:08.530" />
  <row Id="14587" PostTypeId="2" ParentId="14550" CreationDate="2010-12-16T21:31:07.990" Score="0" ViewCount="0" Body="&lt;p&gt;I think this will work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the bijection where coloring $\{...,x_i=1,...\}$ means that object $i$ is included in our set, then the rest follows. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-16T21:31:07.990" CommentCount="3" />
  <row Id="14588" PostTypeId="1" AcceptedAnswerId="14732" CreationDate="2010-12-16T21:33:15.500" Score="2" ViewCount="90" Body="&lt;p&gt;If we have a two lattices (partially ordered) - one for subgroups, one for factor groups, and we know order of the group we want to have these subgroup and factor group lattices, is such a group unique up to isomorphism (if exists)? Or is there a counterexample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true, are sufficient conditions on the order and subgroup lattices to guarantee uniqueness? Another way - what if we now lattice for subgroup and group of automorphism of group; is that group uniquely determined by that information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for help. (sorry for english)&lt;/p&gt;&#xA;" OwnerUserId="4195" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T02:58:22.733" LastActivityDate="2010-12-18T06:23:09.310" Title="Does the order, lattice of subgroups, and lattice of factor groups, uniquely determine a group up to isomorphism?" Tags="&lt;group-theory&gt;&lt;algebra&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14590" PostTypeId="2" ParentId="14569" CreationDate="2010-12-16T21:44:46.947" Score="14" ViewCount="0" Body="&lt;p&gt;You shouldn't expect this to be true. Here is a nonrigorous argument. Let $n_k$ be the sequence of odd numbers you obtain. So (heuristically), with probability $1/2$, we have $n_{k+1} = (5n_k+1)/2$, with probability $1/4$, we have $n_{k+1} = (5 n_k+1)/4$, with probability $1/8$, we have $n_{k+1} = (5 n_k+1)/8$ and so forth. Setting $x_k = \log n_k$, we approximately have $x_{k+1} \approx x_k + \log 5 - \log 2$ with probability $1/2$, $x_{k+1} \approx x_k + \log 5 - 2 \log 2$ with probability $1/4$, $x_{k+1} \approx x_k + \log 5 - 3 \log 2$ with probability $1/8$ and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the expected change from $x_{k}$ to $x_{k+1}$ is &#xA;$$\sum_{j=1}^{\infty} \frac{ \log 5 - j \log 2}{2^j} = \log 5 - 2 \log 2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is positive! So, heurisitically, I expect this sequence to run off to $\infty$. This is different from the $3n+1$ problem, where $\log 3 - 2 \log 2 &amp;lt;0$, and so you heurisitically expect the sequence to decrease over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a numerical example. I started with $n=25$ and generated $25$ odd numbers. Here is a plot of $(k, \log n_k)$, versus the linear growth predicted by my heuristic. Notice that we are up to 4 digit numbers and show no signs of dropping down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8nRFy.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-16T21:44:46.947" />
  <row Id="14591" PostTypeId="1" AcceptedAnswerId="14623" CreationDate="2010-12-16T21:50:55.660" Score="5" ViewCount="83" Body="&lt;p&gt;Is it possible to construct a non-measurable set in $[0,1]$ of a given outer measure $x \in [0,1]$? This will probably require the axiom of choice. Does anyone have a suggestion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I forgot to mention that the set has to be in $[0,1]$ which seemed &quot;clear&quot; in my head, but which obviously isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; Now it is clear that I just need a non-measurable set with outer measure 1, from that I can make any other one. I have once seen somewhere that they do this with transfinite recursion (if I recall correctly), but I don't remember where and how because I didn't know that transfinite recursion was back then.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-16T23:32:07.563" LastActivityDate="2010-12-17T03:37:34.670" Title="Vitali-type set with given outer measure." Tags="&lt;measure-theory&gt;" AnswerCount="2" />
  <row Id="14592" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T21:51:52.417" Score="5" ViewCount="0" Body="&lt;p&gt;The Maclaurin series for $\tan x$ begins with $x$, and there's no $x^2$ term since it's an odd function. Thus $\tan x = x + O(x^3)$, and therefore $(\tan x - x)/x^2 = O(x) \to 0$ as $x\to 0$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-16T21:51:52.417" CommentCount="13" />
  <row Id="14593" PostTypeId="1" CreationDate="2010-12-16T21:55:16.813" Score="0" ViewCount="38" Body="&lt;p&gt;Anyone know how to convert from lat/lng to EPSG:27700?&lt;/p&gt;&#xA;" OwnerUserId="4840" LastActivityDate="2010-12-17T08:28:00.570" Title="Formula to convert from lat/lng to EPSG:27700?" Tags="&lt;geometry&gt;" AnswerCount="2" />
  <row Id="14594" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T21:59:09.557" Score="1" ViewCount="0" Body="&lt;p&gt;I am not sure what you can use; if you know of Maclaurin series, the argument is very easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that what your tutor meant is that you cannot rewrite $\displaystyle \frac{\tan x-x}{x^2}=\frac{\frac{\sin x}x-\cos x}{x\cos x}$ as $\displaystyle \frac{1-\cos x}{x\cos x}$. The argument below uses that $\sin x/x\to 1$, but not by substituting it this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\displaystyle \frac{1-\cos x}{x^2}=\frac{1-\cos^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}x\right)^2\frac1{1+\cos x}\to\frac12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to find the limit of $\displaystyle \frac{\frac{\sin x}x-\cos x}x$. Using the limit above, all you need is to find the limit of $\displaystyle\frac{\sin x -x}{x^2}$, because $$\frac{\frac{\sin x}x-\cos x}x=\frac{\frac{\sin x}x+\left(\frac{1-\cos x}{x^2}\right)x^2-1}x.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit of $\displaystyle\frac{\sin x -x}{x^2}$ is 0. This is fairly easy to evaluate using the Maclaurin expansion of $\sin x$. All you really need is that $\sin x=x+O(x^3)$, but I am not sure you are familiar with this notation. Also, you could define $f(x)=\sin x/x$ if $x\ne 0$ and $f(0)=1$, and verify that this function is differentiable at 0. But I am not certain you have the tools to do that. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-16T21:59:09.557" />
  <row Id="14595" PostTypeId="2" ParentId="14417" CreationDate="2010-12-16T22:11:06.073" Score="1" ViewCount="0" Body="&lt;p&gt;The mass in the sum is approximately proportional to a (shifted) Gaussian centered at $k = n/2 + C\sqrt{n}$ and with standard deviation $A \sqrt{n}$ for explicitly calculable constants $C$ and $A$.  This implies asymptotics of the form $(M + o(1))2^{n + \sqrt{n/2}}$ where $M$ is another computable constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for the $a,b$ version of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit: my calculations give $M = 2^{(\ln 2)/8} = 1.0618966...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit-2: similar calculations give $M=(a/b)^{\ln (a/b)}/8$ and asymptotics $(M+o(1))2^n (ab)^\sqrt{n/2}$ for the $a,b$ sum.]  &lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-17T06:45:13.783" LastActivityDate="2010-12-17T06:45:13.783" CommentCount="6" />
  <row Id="14596" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T22:16:48.873" Score="13" ViewCount="0" Body="&lt;p&gt;Here is an attempt at a geometric proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8AjCc.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Figure thanks to J.M)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\triangle ABC$ such that $\angle{BCA} = x$. Let $BC=1$ and so $AB = \tan x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $BE$ be the arc of radius 1 and angle $x$ drawn with $C$ as the center (note that $E$ is on AC, between $A$ and $D$ and is kind of hidden in the brown region). Note that $CE = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the area of the gray region is $\dfrac{\tan x}{2} - \dfrac{x}{2}$ (area of $\triangle ABC$ - area of the sector $CBE$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D$ be the perpendicular on the hypotenuse $AC$ from $B$. It can be seen that $CD = \cos x$ and thus distance from $D$ to $C$ is less than distance from $E$ to $C$ (which is $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the area of the gray region is less than the area of $\triangle BAD$ (gray + brown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $AD = \dfrac{\sin^2 x}{\cos x}$ and thus we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; \dfrac{\tan x - x}{2} \lt \dfrac{\sin^3 x}{2\cos x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 &amp;lt; \dfrac{\tan x - x}{2x^2} \lt \dfrac{\sin^3 x}{2x^2 \cos x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that $\lim_{x \to 0+} \dfrac{\sin x}{x} = 1$, and that $\dfrac{\tan x - x}{x^2}$ is an odd function, that the limit is $0$, follows.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Previous answer, which was a feeble attempt at being pedantic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a way to find the limit without using more advanced concepts like McLaurin series etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\tan(2x) - 2x}{(2x)^2} = \dfrac{ \frac{2\tan x}{1-\tan^2 x} - 2x}{4x^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \dfrac{(2\tan x - 2x) + 2x \tan^2 x}{4x^2(1 - \tan^2 x)} = (\dfrac{\tan x - x}{2x^2} +  \dfrac{x\tan^2 x}{2x^2}) \dfrac{1}{1-\tan^2 x}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, taking limits as $\displaystyle x \to 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = (\dfrac{L}{2} + 0)\dfrac{1}{1-0}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one problem with the above, though. Can you tell what that is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or rather more simply, replace $\displaystyle x$ with $\displaystyle -x$)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T04:39:41.483" LastActivityDate="2010-12-17T04:39:41.483" CommentCount="8" />
  <row Id="14597" PostTypeId="2" ParentId="14593" CreationDate="2010-12-16T22:19:26.777" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://osgeo-org.1803224.n2.nabble.com/Mouse-position-in-lat-long-td5567114.html&quot; rel=&quot;nofollow&quot;&gt;http://osgeo-org.1803224.n2.nabble.com/Mouse-position-in-lat-long-td5567114.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be useful&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-16T22:19:26.777" />
  <row Id="14598" PostTypeId="2" ParentId="14591" CreationDate="2010-12-16T22:23:42.357" Score="4" ViewCount="0" Body="&lt;p&gt;This was my answer to the question as originally phrased: Once you have one nonmeasurable set with finite outer measure, you can get all positive values by rescaling.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the additional requirement that the set is contained in $[0,1]$, this argument will work as long as you have a nonmeasurable subset of outer measure $1$.  Sierpiński and Lusin showed that $[0,1]$ can be decomposed into uncountably many nonmeasurable sets of outer measure $1$ in an &lt;a href=&quot;http://www.emis.de/cgi-bin/JFM-item?46.0294.01&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; published in 1917, but I do not know the details.  Robert Israel sketches a construction of a Vitali subset of [0,1]  with outer measure 1  in a &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/a1f91aa3b8ae80d8?pli=1&quot; rel=&quot;nofollow&quot;&gt;1997 sci.math post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Oh, and strictly speaking the answer is no, you can't have $x=0$; I suppose you meant $x\in(0,1]$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the axiom of choice, I'm no expert but have been told that it is consistent with ZF that all subsets of the real line are Lebesgue measurable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter LeFanu Lumsdaine points out in a comment that the result is due to Robert Solovay.  The &lt;a href=&quot;http://www.jstor.org/stable/1970696&quot; rel=&quot;nofollow&quot;&gt;1970 article&lt;/a&gt; is on JSTOR, and begins as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/b5H9D.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-16T23:53:23.420" LastActivityDate="2010-12-16T23:53:23.420" CommentCount="14" />
  <row Id="14599" PostTypeId="2" ParentId="14585" CreationDate="2010-12-16T22:24:11.657" Score="10" ViewCount="0" Body="&lt;p&gt;Suppose you have one two of clubs, and the rest of the cards are all four and up, and all diamonds, hearts and spades (there are 33 such cards, which is more than enough). You can't use the two of clubs in any winning hand. So the claim is not true. In fact, this counterexample works even if you allow four of a kind.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastActivityDate="2010-12-16T22:24:11.657" CommentCount="2" />
  <row Id="14600" PostTypeId="1" AcceptedAnswerId="14619" CreationDate="2010-12-16T22:27:28.097" Score="5" ViewCount="140" Body="&lt;p&gt;In multivariable calculus, I was taught to compute the chain rule by drawing a &quot;tree diagram&quot; (a directed acyclic graph) representing the dependence of one variable on the others. I now want to understand the theory behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&#xA;Let $y$ and $x$ both be functions of $t$.&#xA;Let $z$ be a function of both $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of z with respect to t is:&#xA;$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute this derivative, I was taught to draw a graph with the following edges:&#xA;$x \to z$, $y \to z$, $t \to x$, and $t \to y$.&#xA;Source: &lt;a href=&quot;http://www.math.hmc.edu/calculus/tutorials/multichainrule/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.hmc.edu/calculus/tutorials/multichainrule/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/lwPwl.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These tree diagrams can be constructed for arbitrarily complex functions with many variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, to find a derivative of a dependent variable with respect to an independent variable, you need to take the sum of all of the different paths to reach the dependent variable from the independent variable. Traveling down a path, you multiply the functions (e.g. $\frac{\partial z}{\partial x} \cdot \frac{dx}{dt}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this work? What is the connection to the sum-product algorithm for semi-rings? If there's a connection, what is the semi-ring? What group operation does function composition play?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, I'm looking for an algebra (group, semi-ring, etc.) based explanation. Does one exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping for something along the lines of a semiring $(R, \cdot, +)$, where $R$ is the set of functions (invariant under some transformation (possibly scaling?)), $\cdot$ is function composition, $+$ does something like combine the arguments of functions: $f(x) + f(y) = f(x, y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $0$ element would be the constant function $f_0(x) = c$, and the $1$ element would be the identity function: $f_1(x) = x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, $f_0(x)$ is typically unique, so we need to treat $R$ such that all constant functions are in the same equivalence class (hence the invariance mentioned earlier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we can see differentiation as an operation that maps between $(R, \cdot, +)$ and $(S, \cdot&amp;#39;, +&amp;#39;)$, where $\cdot&amp;#39;$ and $+&amp;#39;$ are the standard times and plus over functions that we learn in grammar school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a pipe-dream, or is there an explanation along these lines?&lt;/p&gt;&#xA;" OwnerUserId="4841" LastEditorUserId="4841" LastEditorDisplayName="" LastEditDate="2010-12-17T04:32:19.187" LastActivityDate="2010-12-17T04:32:19.187" Title="chain rule using tree diagram, why does it work?" Tags="&lt;calculus&gt;&lt;graph-theory&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="14601" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T22:31:14.377" Score="0" ViewCount="0" Body="&lt;p&gt;My inclination is to rewrite the original limit as:$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}=\lim_{x\to 0}{\frac{\frac{\tan{x}}{x}-1}{x-0}},$$ with the thinking that if $f(x)=\frac{\tan x}{x}$, $\lim_{x\to 0}f(x)=1$ and $\lim_{x\to 0}f&amp;#39;(x)=0$ and $$f&amp;#39;(a)=\lim_{x\to a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\to a}{\frac{\frac{\tan x}{x}-\frac{\tan a}{a}}{x-a}}.$$&#xD;&#xA;Unfortunately, to proceed from there, I'm getting stuck in some ugliness with double limits.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-16T22:31:14.377" />
  <row Id="14603" PostTypeId="2" ParentId="14442" CreationDate="2010-12-16T22:48:24.677" Score="6" ViewCount="0" Body="&lt;p&gt;The tensor product $M\otimes N$ of abelian groups cannot be non-zero and free unless both $M$ and $N$ are free. This follows from the following facts.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Any subgroup of a free abelian group is free. This holds even in the infinitely generated case. Wikipedia has a proof in its article on &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Free_abelian_group&amp;amp;oldid=400178458#Subgroup_closure&quot; rel=&quot;nofollow&quot;&gt;free abelian groups&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Under certain conditions on $M$ and $N$, for non-zero $n\in N$ the homomorphism&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp;M\to M\otimes N,\\&#xD;&#xA;&amp;amp;m\mapsto m\otimes n,&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;is injective. It is easy to see that this holds when $N$ is free and $n$ is a basis element. It also holds when $M$ and $N$ are both torsion free. You could prove this by &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Extension_of_scalars&amp;amp;oldid=303425028&quot; rel=&quot;nofollow&quot;&gt;extension of scalars&lt;/a&gt; to reduce it to the case of vector spaces over the rationals (and vector spaces are always free). This &lt;em&gt;does not&lt;/em&gt; hold if you merely assume that $N$ is torsion free, as pointed out by Jack Schmidt in a comment (and I must apologize for making the mistake of assuming this in my original answer).&lt;/li&gt;&#xA;&lt;li&gt;Tensor products are &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Exact_functor&amp;amp;oldid=377038315&quot; rel=&quot;nofollow&quot;&gt;right exact&lt;/a&gt;. That is, if $0\to A\to B\to C\to 0$ is an exact sequence of abelian groups, then&#xA;$$&#xD;&#xA;M\otimes A\to M\otimes B\to M\otimes C\to 0&#xD;&#xA;$$&#xD;&#xA;is exact. In particular, if $M\otimes A$ maps to zero, this shows that $M\otimes B$ and $M\otimes C$ are isomorphic, which I will use to quotient out the torsion subgroups before applying 2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that $M\otimes N$ is non-zero and free.&#xA;Let $T$ be the torsion subgroup of $N$, so that $N/T$ is torsion free and $0\to T\to N\to N/T\to 0$ is an exact sequence. By 3,&#xA;$$&#xD;&#xA;M\otimes T\to M\otimes N\to M\otimes(N/T)\to 0&#xD;&#xA;$$&#xD;&#xA;is exact. However, as $T$ is torsion, $M\otimes T$ will also be torsion, so its image in the torsion free group $M\otimes N$ is zero. This gives an isomorphism $M\otimes N\cong M\otimes(N/T)$. Applying the same argument with the torsion subgroup $S$ of $M$ shows that $(M/S)\otimes(N/T)$ is isomorphic to the free group $M\otimes N$. Then, as $M/S$, $N/T$ are torsion free, picking any non-zero $n\in N/T$ and using 2 gives an injection&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp;M/S\to (M/S)\otimes(N/T),\\&#xD;&#xA;&amp;amp; m\mapsto m\otimes n.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;So $M/S$ is isomorphic to a subgroup of the free group $(M/S)\otimes(N/T)$ which, as stated in 1, means that $M/S$ is free. Similarly, $N/T$ is free. Applying 2 again, the homomorphism&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;&amp;amp;M\to M\otimes(N/T),\\&#xD;&#xA;&amp;amp; m\mapsto m\otimes n&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xD;&#xA;is injective for any basis element $n\in N/T$. So $M$ is isomorphic to a subgroup of the free abelian group $M\otimes(N/T)\cong M\otimes N$ and, using 1 once more, must be free. Similarly, $N$ is free.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="1321" LastEditorDisplayName="" LastEditDate="2010-12-17T02:02:36.880" LastActivityDate="2010-12-17T02:02:36.880" CommentCount="7" />
  <row Id="14604" PostTypeId="1" AcceptedAnswerId="14616" CreationDate="2010-12-16T23:04:53.837" Score="2" ViewCount="84" Body="&lt;p&gt;Is the projective closure of an infinite affine variety (over an algebraically closed field, I only care about the classical case right now) always strictly larger than the affine variety? I know it is an open dense subset of its projective closure, but I don't think it can actually be its own projective closure unless it is finite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess my intuition has me worried about cases like the plane curve $X^2 + Y^2 - 1$, since the real part is compact, but such a curve must still &quot;escape to infinity&quot; over an algebraically closed field, right?&lt;/p&gt;&#xA;" OwnerUserId="4842" LastActivityDate="2010-12-18T05:22:05.953" Title="Projective closure" Tags="&lt;algebraic-geometry&gt;" AnswerCount="2" />
  <row Id="14605" PostTypeId="2" ParentId="14604" CreationDate="2010-12-16T23:10:43.033" Score="1" ViewCount="0" Body="&lt;p&gt;For your own example: the projectivization of your curve is given by $X^2 + Y^2 - Z^2= 0$. There is the point corresponding to the projective equivalence class of $(1,i,0)$, which does not belong to the affine curve.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For more general varieties given by a single equation, you can always similarly find a projective point that is part of the affine curve. For more general varieties, I should imagine that it is still the case; but for a proper argument a better person in algebraic geometry is needed.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-16T23:22:20.393" LastActivityDate="2010-12-16T23:22:20.393" CommentCount="1" />
  <row Id="14606" PostTypeId="2" ParentId="14582" CreationDate="2010-12-16T23:11:37.463" Score="2" ViewCount="0" Body="&lt;p&gt;Note that ${\displaystyle {\tan(x) - x \over x^2}}$ is an odd function, so it suffices to show the limit from either side is zero. So we focus on the right limit, and changing $x$ to $\sqrt{x}$ it suffices to show that&#xA;$$\lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0$$&#xD;&#xA;We apply the mean-value theorem to $f(y) = \tan(\sqrt{y}) - \sqrt{y}$ on $[0,x]$, which we can do since the mean-value theorem only requires that $f(x)$ is differentiable on the interior of the interval. We obtain that there is a $y \in (0,x)$ such that&#xA;$$f&amp;#39;(y) = {\tan(\sqrt{x}) - \sqrt{x} \over x}$$&#xD;&#xA;But using the chain rule we have&#xA;$$f&amp;#39;(y) = {\sec^2(\sqrt{y}) - 1 \over 2 \sqrt{y}}$$&#xD;&#xA;$$= {\tan^2(\sqrt{y}) \over \sqrt{y}}$$&#xD;&#xA;Note that we have&#xA;$$\lim_{y \rightarrow 0^+} {\tan^2(\sqrt{y}) \over \sqrt{y}} = \lim_{y \rightarrow 0^+}\tan(\sqrt{y})\,\,\,\times \,\,\lim_{y \rightarrow 0^+}{\tan(\sqrt{y}) \over \sqrt{y}}$$&#xD;&#xA;$$ = 0*1 = 0$$&#xD;&#xA;Thus we conclude that ${\displaystyle \lim_{x \rightarrow 0^+} {\tan(\sqrt{x}) - \sqrt{x} \over x} = 0}$ as needed.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-16T23:11:37.463" />
  <row Id="14607" PostTypeId="2" ParentId="14458" CreationDate="2010-12-16T23:27:20.947" Score="0" ViewCount="0" Body="&lt;p&gt;Revised answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)$x &amp;lt; p \Rightarrow f(x) &amp;lt; f(p)$ (because $f$ is strictly monotonically increasing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this set of values is bounded from above there exists a supremum, namely $f(p-)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, $x &amp;gt; p \Rightarrow f(x) &amp;gt; f(p)$ (because $f$ is strictly monotonically increasing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this set of values is bounded from below there exists an infinimum, namely $f(p+)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_p = f(p-)$ and $b_p = f(p+)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow f(x) &amp;lt; a_p \forall x &amp;lt; p$ and $f(x) &amp;gt; b_p \forall x &amp;gt;p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, $p$ discontinuous $\Rightarrow f(p-) \neq f(p+)$, $f$ strictly monotonically increasing $\Rightarrow f(p+) - f(p-) &amp;gt; 0 \Rightarrow (a_p, b_p) \neq \emptyset$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)For every point of discontinuity $p_i$, there exists (after (i)) an interval $(a_{p_i}, b_{p_i})$. These intervals are by construction disjoint. Because $\mathbb{R}$ is dense, in every interval there exists an element from $\mathbb{Q} \Rightarrow$ the set of discontinuity points has cardinality at most equal to that of $\mathbb{Q} \Rightarrow$ the set of discontinuity points is countable.&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="3711" LastEditorDisplayName="" LastEditDate="2011-01-04T06:12:50.687" LastActivityDate="2011-01-04T06:12:50.687" CommentCount="12" />
  <row Id="14608" PostTypeId="2" ParentId="14579" CreationDate="2010-12-16T23:31:22.633" Score="1" ViewCount="0" Body="&lt;p&gt;Using &lt;a href=&quot;http://math.stackexchange.com/questions/2991/not-understanding-simple-modulus-congruency/3230#3230&quot;&gt;Gauss's algorithm&lt;/a&gt; we have $\rm\displaystyle\ \frac{1}8\ \to\ \frac{4}{32}\ \equiv\ \frac{4}{3}\ \to\ \frac{40}{30}\ \equiv\ \frac{11}{1}\ \ (mod\ 29)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. scale $\rm A/B\ \to AN/BN\ $ by the least $\rm\:N\:$ so that $\rm\ BN &amp;gt; 29\:.\ $ Then reduce the numerator and denominator $\rm\ mod\ 29,\:$ and iterate. You will eventually obtain a denominator of $1$ since each step reduces the denominator. Isn't that sweet? That's they key idea that led Gauss to the first proof of the Fundamental Theorem of Arithmetic, i.e. unique factorization of integers.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-16T23:37:46.680" LastActivityDate="2010-12-16T23:37:46.680" />
  <row Id="14609" PostTypeId="2" ParentId="14581" CreationDate="2010-12-16T23:32:47.307" Score="2" ViewCount="0" Body="&lt;p&gt;Please let me know if the following answer is at the correct level. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a &quot;nice&quot; subset $X \subset R^3$ then it makes sense to talk about the outer boundary of $X$ and the inner boundaries.  Here &quot;nice&quot; means a connected, compact submanifold of dimension three.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's break this down into easy pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $F \subset R^3$ is a compact, connected surface without boundary then $F$ cuts $R^3$ into exactly two pieces $A$ and $B$ so that the closure of $A$ is compact and the closure of $B$ is not compact. Let's refer to $A$ as the piece &quot;inside&quot; of $F$ and to $B$ as the piece &quot;outside&quot; of $F$.  (For help with the exercise please see &lt;a href=&quot;http://www.math.cornell.edu/~hatcher/3M/3Mfds.pdf&quot; rel=&quot;nofollow&quot;&gt;Hatcher's notes&lt;/a&gt; on three-manifolds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Useful Lemma: If $F$ and $F&amp;#39;$ are surfaces (as above), and $F$ meets the outside of $F&amp;#39;$, then $F&amp;#39;$ meets the inside of $F$.  Proof: Exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $X$ is as above.  Let $F_0, F_1, \ldots, F_k$ be the connected components of $\partial X$, the boundary of $X$.  Let $A_i$ and $B_i$ be the inner and outer pieces of $R^3 - F_i$.  Say that $F_i$ is &quot;inside&quot; $X$ if $X$ meets $B_i$ (and so is disjoint from $A_i$).  Otherwise say that $F_i$ is &quot;outside&quot; of $X$.  (Exercise: check everything is well-defined using connectedness of $X$!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: exactly one of the $F_i$ is outside $X$ and the rest are inside.  Proof: Exercise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, this is a long row to hoe.  But this gives precise meanings to inside and outside and then verifies your intuition about those words.  It is also comforting that it all comes down to understanding just a few facts -- the Schoneflies theorem (Alexander's theorem in dimension three) and properties of connected sets. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-12-16T23:32:47.307" CommentCount="1" />
  <row Id="14610" PostTypeId="2" ParentId="14600" CreationDate="2010-12-17T01:26:50.893" Score="1" ViewCount="0" Body="&lt;p&gt;Think about the differentiation as the derivatives along different axes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you have is in essense &#xA;$\frac{dz}{dt} = \frac{\partial z}{\partial  x}|_y \frac{dx}{dt}+ \frac{\partial  z}{\partial  y}|_x \frac{dy}{dt}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum exists when you are not travelling in either of those axes, then you are travelling along a path that is 'shared' by the two axes, and their sum tells you the gradient of that path. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-17T01:26:50.893" CommentCount="1" />
  <row Id="14611" PostTypeId="2" ParentId="13137" CreationDate="2010-12-17T01:31:58.707" Score="4" ViewCount="0" Body="&lt;p&gt;For the sake of completeness, the answer is no; a counterexample is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set&quot; rel=&quot;nofollow&quot;&gt;Smith-Volterra-Cantor set&lt;/a&gt;, or fat Cantor set.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-17T01:31:58.707" CommunityOwnedDate="2010-12-17T01:31:58.707" />
  <row Id="14612" PostTypeId="1" CreationDate="2010-12-17T01:37:52.070" Score="2" ViewCount="52" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in standard text books on (smooth) manifolds, for example the known series by John M. Lee or Jeffrey Lee, you either deal with continuous manifolds, or with smooth manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, neither in these books nor in lectures I have encountered real examples when a manifold may be $C^k$, but not $C^{k+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I would suppose the $|\cdot|_\infty$-Ball with radius $1$ to be a merely continuous, non-smooth manifold, because smoothness fails at the edges of the cube. In contrast to this, polar coordinates show the $|\cdot|_{2}$ with radius $1$ is in fact a smooth manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be thankful for some examples with clues to basic techniques, how the different degrees of smoothnesses manifest 'in real life'.&lt;/p&gt;&#xA;" OwnerUserId="3557" LastActivityDate="2010-12-17T01:56:58.863" Title="Intuition for not-so-smooth manifolds" Tags="&lt;intuition&gt;&lt;differential-topology&gt;&lt;manifold&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="14613" PostTypeId="1" CreationDate="2010-12-17T01:42:34.683" Score="0" ViewCount="42" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the best way to calculate some sort of correlation or similarity factor between two lists of time intervals :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;List1 :&lt;br&gt;&#xA;1. 2010-06-06 to 2010-12-12&lt;br&gt;&#xA;2. 2010-05-04 to 2010-11-02&lt;br&gt;&#xA;3. 2010-02-04 to 2010-10-08&lt;br&gt;&#xA;4. 2010-04-01 to 2010-08-02&lt;br&gt;&#xA;5. 2010-01-03 to 2010-02-02&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and List2 :&lt;br&gt;&#xA;1. 2010-06-08 to 2010-12-14&lt;br&gt;&#xA;2. 2010-04-04 to 2010-10-10&lt;br&gt;&#xA;3. 2010-02-02 to 2010-12-16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question was previously asked at &lt;a href=&quot;http://stackoverflow.com/questions/4466255/how-to-calculate-correlation-between-time-periods&quot;&gt;StackExchange&lt;/a&gt; and the answer provided was a link to &lt;a href=&quot;http://downloads.hindawi.com/journals/asp/2007/089150.pdf&quot; rel=&quot;nofollow&quot;&gt;this article about Feature-List Cross-Correlation&lt;/a&gt; in the field of medical imaging. While it is plenty of useful information, it is a bit daunting for me so it was suggested to try and find a more simple intro to &lt;strong&gt;Vector Cross Correlation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone could provide some help on this please?&lt;/p&gt;&#xA;" OwnerUserId="4845" LastEditorUserId="4845" LastEditorDisplayName="" LastEditDate="2010-12-17T02:13:00.553" LastActivityDate="2010-12-17T17:35:00.883" Title="How to calculate correlation between time periods" Tags="&lt;sequences-and-series&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14614" PostTypeId="2" ParentId="14612" CreationDate="2010-12-17T01:55:32.360" Score="5" ViewCount="0" Body="&lt;p&gt;One can show that any $C^k$-manifold, for $k \geq 1$, has a unique enrichment to a $C^{\infty}$-manifold.  (I.e. given $M$ with its $C^k$-atlas, we can find a $C^{\infty}$-atlas on $M$, compatible with the given $C^k$-atlas, and this $C^{\infty}$-atlas is unique up to equivalence; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Differentiable_manifold&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; for more details.)&#xA;So there is not much point in considering $C^k$-manifolds other than for $k = 0$ or $\infty$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to your unit ball examples, note that the $| \cdot |_{\infty}$-unit ball, although it has&#xA;corners, is homeomorphic to the $| \cdot |_2$-unit ball; one says that it can be &lt;em&gt;smoothed&lt;/em&gt;.   There &lt;em&gt;are&lt;/em&gt; topological manifolds that cannot be smoothed (in dimension 4 and higher), in the sense that they are not homeomorphic to a smooth manifold.  There are also smooth manifolds that are homeomorphic, but not diffeomorphic.  (E.g. when $n \geq 7,$ one can find smooth manifolds that are homeomorphic to $S^n$, but not diffeomorphic to it; these are so-called &lt;em&gt;exotic spheres&lt;/em&gt;.)  Again, the wikipedia entry has more details.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-17T01:55:32.360" CommentCount="1" />
  <row Id="14615" PostTypeId="2" ParentId="14612" CreationDate="2010-12-17T01:56:58.863" Score="4" ViewCount="0" Body="&lt;p&gt;Take a look at Morris Hirsch's text &quot;Differential Topology&quot;, in particular Theorem 2.9 on page 51.  The upshot is that if $M$ is a $C^k$-manifold for $k \geq 1$, then it admits the structure of a $C^j$-manifold for any $j \geq k$, and that structure is unique up to diffeomorphism.   So in that sense the different degrees of smoothness is largely just an artefact of how your manifold is constructed -- you can always &quot;do better&quot; provided the manifold is at least $C^1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I finished this as Emerton's answer appeared, anyhow, take this as a reference for the result. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-12-17T01:56:58.863" />
  <row Id="14616" PostTypeId="2" ParentId="14604" CreationDate="2010-12-17T02:00:04.150" Score="6" ViewCount="0" Body="&lt;p&gt;One can show that if the dimension of the affine variety is positive (equivalently, if it is infinite in the sense of our question) then the projective closure is strictly larger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably lots of ways to prove this, but one is by a version of Noether normalization.   If you would like a detailed proof, let me know by a comment and I can give one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another way to phrase this result is to say that a variety that is simultaneously affine and projective is necessarily finite.  Scheme-theory mavens will recognize this as a special case of the more general statement that a morphism which is simultaneously affine and proper is finite.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Here is a sketch of a proof, as promised:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me begin with Noether normalization in a geometric form.  Fix an afffine variety $V$ contained in $\mathbb A^n$, with projective closue $\overline{V}$.  (Here and below I am always working over an algebraically closed field $k$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $V$ is not all of $\mathbb A^n$, we see that $\overline{V}$ does not contain the hyperplane at infinity, and so we may choose a point $P$ lying in the hyperplane at infinity, but not lying in $\overline{V}$.  We may also choose a different hyperplane $H$ (i.e. not the hyperplane at infinity) which doesn't contain $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $P$ and $H$ in hand, we may define the projection map  $\pi: \mathbb P^n \setminus P \to H$, which maps any $Q \neq P$ to the intersection of the line $\ell$ joining $P$ and $Q$ with the hyperplane $H$.  Restricting $\pi$ to $\overline{V}$, we obtain a map $V \to H$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $P$ is not contained in $\overline{V}$, none of the lines $\ell$ appearing in the projection map are contained in $V$, and so each of them meets $\overline{V}$ in only finitely many points.  In particular, if $\overline{V}$ is infinite, so is its image under $\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by elimination theory, i.e. the fact that projective varieties are proper, we know that $\pi(\overline{V})$ is closed in $H$, i.e. is a projective variety in $H$,&#xA;which is a projective space of dimension $n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, our choice of $P$ ensures that for any $Q \in \overline{V}$, the image $\pi(Q)$ lies at infinity if and only if $Q$ itself does.  So $\pi(V)$ is an affine variety (in the affine space $H \cap \mathbb A^n$ of dimension $n-1$), and $\pi(\overline{V})$ is its projective closure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have just done is prove Noether normalization, in a geometric form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result we want now follows immeidately, by induction on $n$.  (Basically, the case when&#xA;$V = \mathbb A^n$ is clear, and if $V$ is not all of $\mathbb A^n$, the preceding argument allows us to reduce the dimension of the ambient affine space&#xA;by one.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-18T05:22:05.953" LastActivityDate="2010-12-18T05:22:05.953" CommentCount="7" />
  <row Id="14617" PostTypeId="1" AcceptedAnswerId="14620" CreationDate="2010-12-17T02:19:26.773" Score="5" ViewCount="123" Body="&lt;p&gt;There's a question in Spivak's &lt;em&gt;Calculus&lt;/em&gt; (&lt;strike&gt;I don't happen to have the question number in front of me&lt;/strike&gt; in the 2nd Edition, it's Chapter 21, Problem 7) that develops the concept of continued fraction, specifically the continued fraction representation of $\sqrt{2}$.  Having only recently become aware of continued fractions, I'm trying to work my way through this problem, but I'm stuck at the very last stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've got so far:  Let $\{a_n\}$ be the sequence defined recursively as $$a_1 = 1, a_{n + 1} = 1 + \frac{1}{1 + a_n}$$  Consider the two subsequences $\{a_{2n}\}$ and $\{a_{2n - 1}\}$.  I've already shown that $\{a_{2n}\}$ is monotonic, strictly decreasing, and bounded below by $\sqrt{2}$, and similarly I've shown that $\{a_{2n - 1}\}$ is monotonic, strictly increasing, and bounded above by $\sqrt{2}$.  Obviously, both of these subsequences converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although of course in general, if two subsequences of a sequence happen to converge to the same value, that doesn't guarantee that the sequence itself converges at all (much less to that same value), in the case where the subsequences are $\{a_{2n}\}$ and $\{a_{2n - 1}\}$, it's easy to show that if they both converge to the same value, then so will $\{a_n\}$ (since every term of $\{a_n\}$ is a term of one of the two subsequences).  So no problem there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it remains only to show that not only do the subsequences converge, but they converge to $\sqrt{2}$ in particular.  Take, for starters, $\{a_{2n - 1}\}$ (if I can get $\{a_{2n - 1}\}$ to converge to $\sqrt{2}$, I'm sure getting $\{a_{2n}\}$ to converge to $\sqrt{2}$ won't be very different).  Because it's strictly increasing and bounded above by $\sqrt{2}$, it converges to some number $x \leq \sqrt{2}$.  Suppose that $x &amp;lt; \sqrt{2}$.  We want to show this doesn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is where I'm getting stuck.  I feel like I want to take $\epsilon = \sqrt{2} - x$ and show that there exists some $N$ such that $a_{2N - 1} &amp;gt; x$, which would finish the problem due to the monotonicity of $\{a_{2n - 1}\}$.  But this isn't working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints?  Thanks a ton.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditorDisplayName="" LastEditDate="2010-12-17T15:12:53.237" LastActivityDate="2010-12-17T15:12:53.237" Title="Proving the continued fraction representation of $\sqrt{2}$" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;continued-fractions&gt;" AnswerCount="3" />
  <row Id="14618" PostTypeId="2" ParentId="13137" CreationDate="2010-12-17T02:22:19.370" Score="3" ViewCount="0" Body="&lt;p&gt;What is actually true is this: for every set $C$ of positive measure and every $\epsilon &amp;lt; 1$ there is some open interval $(a,b)$ such that $\mu(C \cap (a,b)) \geq \epsilon |b-a|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always viewed this as an instance of one of &lt;a href=&quot;http://en.wikipedia.org/wiki/Littlewood%27s_three_principles_of_real_analysis&quot; rel=&quot;nofollow&quot;&gt;Littlewood's three principles&lt;/a&gt; for analysis: a measurable set is almost an open set. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-17T02:27:25.063" LastActivityDate="2010-12-17T02:27:25.063" />
  <row Id="14619" PostTypeId="2" ParentId="14600" CreationDate="2010-12-17T02:23:01.320" Score="6" ViewCount="0" Body="&lt;p&gt;The point of derivatives in one variable is to provide linear approximations $f(x) = f(p) + f&amp;#39;(p) (x - p) + o(|x - p|)$ to nice functions.  Multivariate derivatives work the same way, except &quot;linear approximation&quot; here means approximation by a general linear transformation (a matrix) instead of a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is made precise by the following definition: we say that a function $f : \mathbb{R}^n \to \mathbb{R}^m$ has &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_derivative&quot; rel=&quot;nofollow&quot;&gt;total derivative&lt;/a&gt; a linear transformation $df_p : \mathbb{R}^n \to \mathbb{R}^m$ at a point $p$ if there exists $\epsilon &amp;gt; 0$ and a function $E_p(h)$ defined for $|h| &amp;lt; \epsilon$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(p + h) = f(p) + df_p(h) + |h| E_p(h)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lim_{h \to 0} E_p(h) = 0$.  The matrix $df_p$ is sometimes called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;Jacobian&lt;/a&gt;.  In &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation&quot; rel=&quot;nofollow&quot;&gt;little-o notation&lt;/a&gt;, we write this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(p + h) = f(p) + df_p(h) + o(|h|).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might seem unnecessarily complicated, but it is the key to understanding the multivariate chain rule.  Suppose that in addition to $f$ we have another function $g : \mathbb{R}^m \to \mathbb{R}^k$ with a total derivative $dg_q$ at some point $q$, and suppose that $f(p) = q$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$gf(p + h) = g \left( f(p) + df_p(h) + o(|h|) \right) = gf(p) + dg_q df_p(h) + o(|h|)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, in other words, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The total derivative $d(gf)_p$ of $gf$ at $p$ is the (matrix) product of the total derivatives $dg_q$ and $df_p$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the most general statement of the multivariate chain rule.  The relationship to tree diagrams is that one can model matrix multiplication using composition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Incidence_matrix&quot; rel=&quot;nofollow&quot;&gt;incidence matrices&lt;/a&gt;, which come from graphs depicting incidence relationships between sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular example, you have a function $t \mapsto (x, y) : \mathbb{R}^1 \to \mathbb{R}^2$ and another function $(x, y) \mapsto z : \mathbb{R}^2 \to \mathbb{R}^1$.  The total derivative of the first function is $\left[ \begin{array}{c} \frac{dx}{dt} \\\ \frac{dy}{dt} \end{array} \right]$ and the total derivative of the second function is $\left[ \frac{dz}{dx}, \frac{dz}{dy} \right]$, so the total derivative of their composition is the product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dz}{dt} = \left[ \frac{dz}{dx}, \frac{dz}{dy} \right] \left[ \begin{array}{c} \frac{dx}{dt} \\\ \frac{dy}{dt} \end{array} \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is precisely the formula you give.  The connection to diagrams is that one can represent a composition of linear transformations $\mathbb{R}^1 \to \mathbb{R}^2$ and $\mathbb{R}^2 \to \mathbb{R}^1$ using a pair of incidence matrices, one to represent incidences between a $1$-element set and a $2$-element set, and the other to represent incidences between that $2$-element set and another $1$-element set.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-17T02:29:46.433" LastActivityDate="2010-12-17T02:29:46.433" CommentCount="3" />
  <row Id="14620" PostTypeId="2" ParentId="14617" CreationDate="2010-12-17T02:31:57.760" Score="6" ViewCount="0" Body="&lt;p&gt;You have shown that the sequence of $a_{2n}$ converges. Call $e$ (for &lt;em&gt;even&lt;/em&gt;) its limit, and call $o$ the limit of $a_{2n+1}$. Then the recursive relation giving $a_{n+1}$ in terms of $a_n$ implies that $$e=1+1/(1+o)$$ and $$o=1+1/(1+e),$$ from which it follows that $e=o=\sqrt2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[For example, take the limit as $n\to\infty$ of $a_{2n}=1+1/(1+a_{2n-1})$ to get the first relation, and similarly you get the second exchanging odd and even. From the first equation, you get $2+o=e+eo$ and from the second $2+e=o+oe$. Subtracting, you get $o-e=e-o$, so $o=e$. From the first equation, you now get $2=o^2$.]&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-17T02:48:12.880" LastActivityDate="2010-12-17T02:48:12.880" CommentCount="0" />
  <row Id="14621" PostTypeId="2" ParentId="14617" CreationDate="2010-12-17T02:34:59.863" Score="3" ViewCount="0" Body="&lt;p&gt;As I explained in a &lt;a href=&quot;http://math.stackexchange.com/questions/7266/why-does-this-process-when-iterated-tend-towards-a-certain-number-the-golden/7275#7275&quot;&gt;previous similar question&lt;/a&gt;, one can use the Banach fixed point theorem to solve these kinds of problems.  For the special case of iterating a fractional linear transformation you can also (as I think is also explained in answers to that question) write down an explicit closed form for the $n^{th}$ term by diagonalizing a $2 \times 2$ matrix.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-17T02:34:59.863" CommentCount="1" />
  <row Id="14622" PostTypeId="1" CreationDate="2010-12-17T02:37:39.030" Score="4" ViewCount="70" Body="&lt;p&gt;For a two-sphere embedded in $\mathbb{R}^4$,how can you check whether or not there is an ambient isotopy to the &quot;standard&quot; 2-sphere (the set of points $(x,y,z,0)$ in $\mathbb{R}^4$ distance 1 from the origin)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knot theory was discussed in an intro topology course I took and I'm wondering about further generalizations of the concept.  It seems to me that such &quot;knotted&quot; 2-spheres could be created by rotating a knotted arc about a plane in $\mathbb{R}^4$; and knotted tori by rotating a standard knot about a nonintersecting plane.  However I cannot think of a way to show which of these constructions, if any, are indeed nonisotopic to their standard counterparts.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T02:43:39.507" LastActivityDate="2010-12-17T12:36:44.767" Title="The construction of knotted surfaces in $\mathbb{R}^4$" Tags="&lt;topology&gt;&lt;knot-theory&gt;" AnswerCount="2" />
  <row Id="14623" PostTypeId="2" ParentId="14591" CreationDate="2010-12-17T02:42:18.473" Score="5" ViewCount="0" Body="&lt;p&gt;One can proceed directly with the Vitali construction, without need for any scaling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, just carry out the Vitali construction, but ensure that the resulting Vitali set is contained in the interval $[0,a]$. That is, declare that two reals are equivalent if their difference is rational, and observe that every real is equivalent to a real in the interval $[0,a]$. Let $V\subset [0,a]$ select exactly one element from each equivalence class. Observe that the rational translations $V+q$, working modulo $1$ so as to regard $V+q\subset [0,1]$, are disjoint and union up to the whole interval $[0,1]$. It follows easily that $V$ is not measurable and has inner measure $0$, since otherwise the translates modulo $1$ would have infinite measure inside $[0,1]$, which is impossible. Thus, the complement $[0,a]-V$ is non-measurable and has outer measure $a$, as desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to note that this argument shows that the outer measure of the classical Vitali set is not determined by the usual features of that set. For example, if all you know about a Vitali set $V$ is that it is contained in $[0,1]$ and contains exactly one element of each equivalence class, then it follows that $V$ is non-measurable and has inner measure $0$, but for all you know, $V$ is actually contained in a very tiny interval $[0,\epsilon]$, and could have tiny outer measure. &lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-12-17T03:37:34.670" LastActivityDate="2010-12-17T03:37:34.670" CommentCount="0" />
  <row Id="14624" PostTypeId="1" CreationDate="2010-12-17T02:56:34.453" Score="2" ViewCount="159" Body="&lt;p&gt;The adelics seem counter-intuitive. I wonder how they came up originally, and what was the immediate reward for introducing them. What was the politics of introducing the adelics into mathematical mainstream?&lt;/p&gt;&#xA;" OwnerUserId="4732" LastActivityDate="2010-12-17T08:58:11.977" Title="Politics of the Adelics" Tags="&lt;number-theory&gt;&lt;math-history&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14625" PostTypeId="1" CreationDate="2010-12-17T02:59:16.927" Score="15" ViewCount="366" Body="&lt;p&gt;Consider the curve $\frac{1}{x}$  where $x \geq 1$. Rotate this curve around the x-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Dimension - Clearly this structure is infinitely long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two Dimensions - Surface Area = $2\pi\int_∞^1\frac{1}{x}dx = 2\pi(\ln ∞ - \ln 1) = ∞$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three Dimensions -  Volume = $\pi\int_∞^1{x}^{-2}dx = \pi(-\frac{1}{∞} + \frac{1}{1}) = \pi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this structure has infinite length and infinite surface area. However it has finite volume, which just does not make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more interesting, the &quot;walls&quot; of this structure are infinitely thin. Since the volume is finite, we could fill this structure with a finite amount of paint. To fill the structure the paint would need to cover the complete surface area of the inside of this structure. Since the &quot;walls&quot; are infinitely thin, why would a finite amount of paint not be able to cover the outside of the &quot;walls&quot; too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me make sense of this whole thing.&lt;/p&gt;&#xA;" OwnerUserId="4831" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-17T16:41:06.983" LastActivityDate="2010-12-17T16:41:06.983" Title="How can a structure have infinite length and infinite surface area, but have finite volume?" Tags="&lt;calculus&gt;&lt;infinity&gt;&lt;paradox&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="14626" PostTypeId="2" ParentId="14560" CreationDate="2010-12-17T03:01:11.483" Score="3" ViewCount="0" Body="&lt;p&gt;It deserves to be better known that there are simple cases where one can give closed forms for some rearrangements of alternating series. Here are a couple of interesting examples based on results of Schlömilch in 1873. Many further results can be found in classical textbooks on infinite series, e.g. those by Bromwich and Knopp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\rm\ H^m_n\ $ be the rearrangement of the alternating harmonic series $\rm\ 1 - 1/2 + 1/3 - 1/4 +\: \cdots\ $ obtained by taking consecutive groups of $\rm\:m\:$ positive terms and $\rm\:n\:$ negative terms. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm H^m_n\ =\ log\ 2 + \frac{1}2\ \lim_{k\to\infty}\ \int^{\:mk}_{nk}\frac{1}x\ dx\ =\ \log 2 + \frac{1}2 \log\frac{m}n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly rearranging Lebniz's sum $\rm\ L\ =\ \pi/4\ =\ 1 - 1/3 + 1/5 - 1/7 +\: \cdots\ $ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm L^m_n\ =\ \frac{\pi}4 + \frac{1}2\ \lim_{k\to\infty}\ \int^{\:mk}_{nk}\frac{1}{2x-1}\ dx\ =\ \frac{\pi}4 + \frac{1}4 \log\frac{m}n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus as $\rm\:m\:$ varies we obtain infinitely many rearrangements with distinct sums. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof of the general theorem underlying these results is quite simple - using nothing deeper than the integral test. See Beigel: &lt;a href=&quot;http://www.jstor.org/stable/2689984&quot; rel=&quot;nofollow&quot;&gt;Rearranging Terms in Alternating Series&lt;/a&gt;, Math. Mag. 1981.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T03:01:11.483" />
  <row Id="14627" PostTypeId="2" ParentId="13616" CreationDate="2010-12-17T03:08:44.013" Score="2" ViewCount="0" Body="&lt;p&gt;Here is one example I stumbled upon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random walks have been used to identify clustering in networks in Rosvall and Bergstrom, &lt;a href=&quot;http://octavia.zoology.washington.edu/publications/RosvallAndBergstrom08.pdf&quot; rel=&quot;nofollow&quot;&gt;Maps of random walks on complex networks reveal community structure&lt;/a&gt;.  In particular, this technique was used by &lt;a href=&quot;http://www.eigenfactor.org/map/methods.htm&quot; rel=&quot;nofollow&quot;&gt;Eigenfactor&lt;/a&gt; (who publish journal rankings) to deduce clusterings in research communities.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T03:08:44.013" />
  <row Id="14628" PostTypeId="2" ParentId="14622" CreationDate="2010-12-17T03:09:10.223" Score="4" ViewCount="0" Body="&lt;p&gt;The Alexander polynomial is one of the more interesting isotopy-invariants for co-dimension two embedded things in spheres. So it applies to both of the things you're interested in -- knotted $2$-spheres and tori in $S^4$.  Take a look at Akio Kawauchi's &lt;em&gt;Survey of Knot Theory&lt;/em&gt;, or Jonathan Hillman's &lt;em&gt;Algebraic Invariants of Knots and Links&lt;/em&gt; for some details.  The Alexander polynomials have more subtle, associated invariants like Tristram-Levine invariants and such, these are induced from Poincare duality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;rotating about a plane&quot; construction you mention is called Artin spinning.  It produces non-trivial knots for fairly simple reasons.  Given a knot $K$ in $\mathbb S^3$, let $G$ be the fundamental group of its complement.  Then if you do Artin spinning on $K$ to produce a knotted $S^2$ in $S^4$, call it $K&amp;#39;$, then the fundamental group of the complement of $K&amp;#39;$ is also isomorphic to $G$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-12-17T07:51:26.263" LastActivityDate="2010-12-17T07:51:26.263" />
  <row Id="14629" PostTypeId="1" AcceptedAnswerId="14893" CreationDate="2010-12-17T03:11:51.767" Score="5" ViewCount="118" Body="&lt;p&gt;I'm having some trouble proving the following statement: let $g:S^{2n-1} \to S^{2n-1}, f: S^{2n-1} \to S^n$.  Then $H(f\circ g) = \deg g H(f)$ where $H(f)$ is the Hopf invariant.  The definition I am using for Hopf invariant is as follows:  let $C_f = D^{2n} \sqcup_f S^{n}$ where $D^{2n}$ is the $2n$ disk which we attach to $S^{n}$ via $f$.  Let $\alpha, \beta$ be generators for $H^n(C_f)$ and $H^{2n}(C_f)$ respectively.  Then $H(f)$ is defined by $\alpha \cup \alpha = H(f) \beta$.  I can prove the statement using the integral formula for $H(f)$ but would like to prove it cohomologically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should consider the map $G: C_{f\circ g} \to C_f$ that takes $x\in \partial D^{2n}$ to $g(x)$ and acts as the identity on everything else.  But then I don't know how to compute what $G^* \alpha$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-12-19T14:45:06.080" LastActivityDate="2010-12-20T06:17:41.787" Title="Hopf invariant formula" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14630" PostTypeId="1" CreationDate="2010-12-17T03:18:07.800" Score="2" ViewCount="52" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a random variable X that is a mixture of a binomial and two normals (see what the &lt;a href=&quot;http://stackoverflow.com/questions/4454513/combining-two-normal-random-variables/4454941#4454941&quot;&gt;probability density function would look like&lt;/a&gt; (first chart))&lt;br&gt;&#xA;and I have another random variable Y of similar shape but with different values for each normally distributed side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X and Y are also correlated, here's an example of data that could be plausible :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    X     Y&#xA;1.  0    -20&#xA;2. -5     2&#xA;3. -30    6&#xA;4.  7    -2&#xA;5.  7     2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you can see, that was simply to represent that my random variables are either a small positive (often) or a large negative (rare) and have a certain covariance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is : I would like to be able to sample correlated and random values from these two distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could use Cholesky decomposition for generating correlated &lt;em&gt;normally distributed&lt;/em&gt; random variables, but the random variables we are talking here are not normal but rather a mixture of a binomial and two normals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="4845" LastActivityDate="2010-12-17T07:21:20.997" Title="Generating random values from non-normal and correlated distributions" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;probability-theory&gt;&lt;distribution-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14631" PostTypeId="1" CreationDate="2010-12-17T03:43:38.053" Score="3" ViewCount="41" Body="&lt;p&gt;Let $S$ be a ruled surface over a curve of genus $g$. Is it possible to compute the second Chern class of $S$ in terms of $g$? &lt;/p&gt;&#xA;" OwnerUserId="4654" LastActivityDate="2010-12-17T03:43:38.053" Title="Second Chern class of a ruled surface" Tags="&lt;algebraic-geometry&gt;" FavoriteCount="1" />
  <row Id="14632" PostTypeId="2" ParentId="14625" CreationDate="2010-12-17T03:45:41.110" Score="31" ViewCount="0" Body="&lt;p&gt;It seems you already have made sense of this whole thing from a mathematical point of view (it has infinite surface area and finite volume -- there's no contradiction here).  The paradox is that infinity, at times, does not match well with our day-to-day experiences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a related example (and hopefully a little bit easier to picture):  Take a unit square.  Cut it in half vertically down the middle -- forming two pieces.  The area clearly remains the same, but the total perimeter (counting both pieces) has increased by 2.  Keep cutting the right-most piece vertically down the middle forever.  The area remains unchanged throughout, whereas the perimeter is $4+\lim_{n \rightarrow \infty} \sum_{i=1}^n 2=\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/s22uq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you wanted to make it look more like a horn, you could move the pieces around, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2CgVv.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which still has area 1 and perimeter infinity.  If you wanted to make it have volume 1 and surface area infinity, replace &quot;square&quot; with &quot;cube&quot;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T03:45:41.110" CommentCount="3" />
  <row Id="14633" PostTypeId="2" ParentId="14625" CreationDate="2010-12-17T03:50:24.493" Score="5" ViewCount="0" Body="&lt;p&gt;The answer to the paint question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define terms, let's say you are rotating about the $x$ axis and the volume is between the surface and the $y-z$ plane.  If we take a small (but finite) patch of the $y-z$ plane of $\Delta A$ area then the area above it is roughly $\Delta A$ as $\sqrt{y^2 + z^2} \rightarrow \infty$ and the volume is roughly $\frac{\Delta A}{\sqrt{y^2 + z^2}}$.  So as we move away from the $x$ axis, the trapped volume goes to zero but the surface area does not.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now back to the paint analogy.  It takes infinite paint because a fixed volume of paint covers a fixed surface area (i.e. one gallon / 400 sq.ft.).  This is true even as $\sqrt{y^2 + z^2} \rightarrow \infty$, so we find ourselves having to paint a surface area equivalent to the $y-z$ plane.  Now if we tried to fill the volume with the paint, we are no longer required to cover at a given rate of 1 gal. / 400 sq.ft.  In other words our film thickness is $\frac{1}{\sqrt{y^2 + z^2}}$ which tends to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, your assumption regarding the walls being infinitely thin is true, but you relate volume and surface area using the metaphor of paint.  Paint always has a finite spread rate no matter how thin the film.  This film does not get thinner as we move away from the $x$ axis.  The volume between the curve and the $y-z$ plane however does.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-17T03:50:24.493" />
  <row Id="14634" PostTypeId="2" ParentId="14625" CreationDate="2010-12-17T04:05:57.900" Score="10" ViewCount="0" Body="&lt;p&gt;The mathematics has already been explained; when the mathematics contradicts your intuition the answer is to ignore your old intuition and find new intuition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the physics, Gabriel's horn is in no meaningful sense a physically realistic object.  One way to summarize the lesson here is that the assumption that very thin objects have zero thickness is only reasonable if this causes a reasonable amount of error in one's computations, but in the construction of Gabriel's horn the thickness of the membrane comes to dominate the computation and it is no sense reasonable to ignore this error.  Any time a mathematical model fails to accurately reflect reality one should question the assumptions that go into building that model, and to the extent that this is a mathematical model of anything, the assumption that thickness is negligible is the one that should be thrown out.  If you try to repeat the construction of Gabriel's horn with a fixed finite thickness you will, of course, find that the resulting object has infinite volume.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-17T04:05:57.900" />
  <row Id="14635" PostTypeId="1" CreationDate="2010-12-17T04:40:10.847" Score="0" ViewCount="46" Body="&lt;p&gt;Just wondering:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to solve ultrahyperbolic PDEs? Is there any analytical solution for linear ultrahyperbolic PDEs?&lt;/li&gt;&#xA;&lt;li&gt;If there are only numerical solutions, are the solutions' behavior similar to those of nonlinear eqns? I mean, like an anharmonic oscillator, chaotic but deterministic?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Information of reference books is also welcomed - but better not too specialized in math. I am an engineering student with very very limited math skills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="4846" LastActivityDate="2010-12-17T12:47:17.690" Title="ultrahyperbolic PDE" Tags="&lt;differential-equations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14636" PostTypeId="2" ParentId="14549" CreationDate="2010-12-17T04:52:40.090" Score="2" ViewCount="0" Body="&lt;p&gt;the wikipedia link you give has an inequality that allows the success probabilities for the different bernoulli trials to be different - and the bound does not depend explicitly on the number of trials n. rather, n is replaced by $\mu = \sum_1^n p_i$, where $p_i$ is the success probability for the i$^{th}$ trial. look at the section &quot;Theorem for multiplicative form of Chernoff bound (relative error)&quot;. in this form, it is clear, as qiaochu states, that the $\{p_i\}$ can depend on n - and can all be the same - or not.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-12-17T04:52:40.090" />
  <row Id="14637" PostTypeId="1" AcceptedAnswerId="14640" CreationDate="2010-12-17T05:31:20.400" Score="0" ViewCount="83" Body="&lt;p&gt;I've always had the following silly(?) doubt about convergent sequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a finite sequence $A_{1:n}=(a_1,a_2\ldots ,a_n)$, define its &lt;i&gt; reverse &lt;/i&gt; as $rev(A) = (a_n,a_{n-1},..a_1)$.  Further, if $A$ is infinite, define its reverse as the limit of the reverses of its initial subsequences, i.e. as the limit of the sequence of sequences $ \{ (a_1), (a_2,a_1), (a_3,a_2,a_1), ..\}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q:&lt;i&gt; Is this notion well defined? Why or why not? &lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect not, because the reverse of $(1,\frac{1}{2},\frac{1}{3},...)$ would be $(0,0,... , \frac{1}{3},\frac{1}{2},1)$? This seems absurd!&lt;/p&gt;&#xA;" OwnerUserId="1585" LastActivityDate="2010-12-17T07:12:14.193" Title="Reversing a convergent sequence" Tags="&lt;real-analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14638" PostTypeId="2" ParentId="14624" CreationDate="2010-12-17T06:01:40.247" Score="8" ViewCount="0" Body="&lt;p&gt;The ideles came first, and were introduced by Chevalley as part of his program of giving a (then) new proof of class field theory which avoided the analysis of $L$-function.  The adeles came a little later; they were originally called &lt;em&gt;valuation vectors&lt;/em&gt;, and were introduced by Artin and Whaples.  It took a little time for the&#xA;details to be sorted out (e.g. Chevalley orginally used a slightly different topology on the ideles to the one used currently; see &lt;a href=&quot;http://mathoverflow.net/questions/41253/who-fixed-the-topology-on-ideles&quot; rel=&quot;nofollow&quot;&gt;this MO question&lt;/a&gt; for more detailds); when Tate wrote his thesis in 1950, he used the current definitions (although I think he still used the term valuation vectors,&#xA;rather than adeles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the name ideles was invented by Chevalley, as a pun on the word ideals, since from the ideles one could recover all the generalized ideal class groups of traditional algebraic number theory.  The term adeles (when it was finally settled on, maybe by Weil?) was then a pun on ideles, together with the fact that the adeles have an additive structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogue of the adeles in which the number field is replaced by the function field of a smooth projective irreducible algebraic curve (so that the places of the number field are replaced by the points of the curve) appeared in Weil's approach to proving the Riemann--Roch theorem.  (In this context the adeles were also called &lt;em&gt;repartitions&lt;/em&gt;, although I doubt anyone uses that terminology anymore.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one looks in Tate's thesis, there is a result there called the Riemann--Roch theorem, which really does reduce to the usual result with that name in the function field case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consideration of adelic points of algebraic groups was first undertaken by Weil, I think, in his reformuation of the work of Tamagawa in the study of what are now called Tamagawa numbers.   The use of this point of view in the study of automorphic forms was brought to the fore in the work of Langlands in the late 1960s and early 1970s, although I don't know whether he was the first one who realized that it was the natural way to encode the notion of Hecke eigenform in representation theoretic terms.  Chevalley's original concept of idele is subsumed in this general formalism; they are precisely the adelic points of the group $GL_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overarching reason for introducing the adelic viewpoint, beginning with Chevalley's original work and continuing through to the modern treatment of automorphic forms, has always been the same: they provide the most convenient and flexible framework for studying the interaction of local and global phenomena.  All the traditional concepts (e.g. finiteness of the class number, Dirichlet's unit theorem, finite volume of $\Gamma\backslash G$ for semi-simple groups $G$ and congruence subgroups $\Gamma$) are easily expressed in adelic terms in a natural and uniform way, and for studying certain concepts (local reciprocity laws and their interaction with global reciprocity) it is hard to imagine how one could sensibly formulate things in anything other than the adelic manner. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-17T08:47:13.073" LastActivityDate="2010-12-17T08:47:13.073" CommentCount="2" />
  <row Id="14640" PostTypeId="2" ParentId="14637" CreationDate="2010-12-17T07:12:14.193" Score="4" ViewCount="0" Body="&lt;p&gt;This &quot;reverse sequence&quot; notion is either going to be meaningless, or be identical to sequences (but just written backwards).  For it to be well-defined, you need somehow to index the entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is defined, the sequence of sequences $(a_1)$, $(a_2,a_1)$, etc., would converge pointwise to the sequence $(\ell,\ell,\ldots)$, where $\ell = \lim_{n \rightarrow \infty} a_n$ (if it exists).  So this is not particularly exciting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the $(\ell,\ell,\ldots,a_N,a_{N-1},\ldots,a_1)$ idea, is that you need to take $\infty-\infty=N$ (which will give rise to nonsense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you defined the sequence $\mathrm{rev}(A)$ such that the $(-i)$-th entry of $\mathrm{rev}(A)$ is the $i$-th entry of $A$, then it will also be well-defined, but identical to $A$ itself (just written backwards).  So this is also not particularly exciting.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T07:12:14.193" />
  <row Id="14641" PostTypeId="2" ParentId="14403" CreationDate="2010-12-17T07:17:42.670" Score="1" ViewCount="0" Body="&lt;p&gt;don't know exactly what you want to know about $d\sigma/\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mathematically, it is like lebesgue measure $dx$ on $\mathbb{R}$ - in the following sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dx$ does not change under a simple additive change [or shift] of a variable: $y= x+a$, where $a$ is a constant. that is, $dy=dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;similarly, $d\sigma/\sigma$ does not change under a multiplicative change [or rescaling] of a variable: $y=c\sigma$, where $c&amp;gt;0$. that is, $dy/y=d\sigma/\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;both $dx$ on $\mathbb{R}$ and $d\sigma/\sigma$ on $\mathbb{R}^+$ are examples of haar measures [&lt;em&gt;i.e.&lt;/em&gt;, invariant measures] on $\mathbb{R}$ and $\mathbb{R}^+$, respecctively, considered as groups acting on themselves, as indicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as for jeffries: many [non-orthodox] bayesians consider $dx$ to be like a uniform [prior] distribution on $\mathbb{R}$. [it is the limit - in some sense - of the uniform distribution on [-A, A], as A $\to\infty$.]  it is considered to be an 'objective prior',&#xA;in that everyone [supposedly] agrees that it is THE uniform distribution on $\mathbb{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that $dx$ is improper - it is not actually a probability distribution on $\mathbb{R}$ - doesn't seem to faze these folks; as long as one comes up with a proper posterior distribution when using it [as is often the case], it can be taken as a 'prior'. an example is inference about the unknown mean $\mu$ of a normal distribution with $\sigma = 1$. if $X$ is one observation from this distribution, the conditional pdf of $X|\mu$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{X|\mu}(x|\mu) = \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2},\quad -\infty &amp;lt; x &amp;lt; \infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying by the 'uniform' prior 'pdf' of $\mu$, the [improper] joint pdf of $(X,\mu)$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{X,\mu}(x,\mu) = f_{X|\mu}(x|\mu){\mathbb{1}}_{(-\infty.\infty)}(\mu) = \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2},\quad -\infty &amp;lt; x, \mu &amp;lt;\infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note that the [improper] prior 'pdf' for $\mu$ is $f_\mu(\mu)={\mathbb{1}}_{(-\infty,\infty)}(\mu)$, in that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{&amp;#39;prob&amp;#39;}(a&amp;lt; \mu&amp;lt; b) = \int_a^b f_\mu(\mu) d\mu = \int_a^b d\mu = b-a.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, integrating out the $\mu$ in the joint pdf, one gets that the marginal 'pdf' of $X$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_X(x) = \int_{-\infty}^{\infty}  f_{X|\mu}(x|\mu)d\mu = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2}d\mu = {\mathbb{1}}_{(-\infty.\infty)}(x),$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that $X$ marginally has an improper distribution as well. finally, the posterior pdf for $\mu$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{\mu|X}(\mu|x)=  f_{X,\mu}(x,\mu)/f_{X}(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}(x-\mu)^2},\quad -\infty &amp;lt; \mu &amp;lt; \infty,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that $\mu|X$ is normal with mean $X$ and SD = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with this example in mind, perhaps you can persuade yourself that $d\sigma/\sigma$ is some sort of analog of a uniform distribution on $\mathbb{{R}^+}$. [if one considers the parameter $\nu=\log\sigma $, this turns into the [lebesgue] uniform distribution for $\nu$, since [as pointed out by @hardmath], $d\nu=d(\log\sigma)=d\sigma/\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whether or not you buy into this depends, i guess, on your propensity for being a 'true believer'.&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-12-17T07:17:42.670" CommentCount="1" />
  <row Id="14642" PostTypeId="2" ParentId="14630" CreationDate="2010-12-17T07:21:20.997" Score="0" ViewCount="0" Body="&lt;p&gt;If you want to sample $(X,Y)$ you need to find out what is their distribution function. If you know the the conditional densities $p(X|Y)$ and $p(Y|X)$ you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_sampling&quot; rel=&quot;nofollow&quot;&gt;Gibbs sampling&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-17T07:21:20.997" CommentCount="3" />
  <row Id="14643" PostTypeId="2" ParentId="3477" CreationDate="2010-12-17T07:46:14.033" Score="0" ViewCount="0" Body="&lt;p&gt;I think it is possible to introduce the new set of data periodically by assigning its initial value with the interpolated value of its neighbor while the possible set of neighbor for each new data point could be the points in the triangle that data point resides in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what is more important is the order of assigning the initial values for this set of data point since the data point that is closer to CSO will be updated first. To determine the closeness of data point to CSO, we can use Euclidean distance, weighted link or both. &lt;/p&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2010-12-17T07:46:14.033" />
  <row Id="14644" PostTypeId="2" ParentId="3477" CreationDate="2010-12-17T08:05:08.713" Score="0" ViewCount="0" Body="&lt;p&gt;Note:something to think about.&#xA;1. Data point that lie outside the contour of current data set.&#xA;2. The number of update recurring. The data point tends to converge to CSO as the number of update increases. This can cause saturation problem. To solve this problem, subsidizing current data points or levitating new data points is necessary. &lt;/p&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2010-12-17T08:05:08.713" />
  <row Id="14645" PostTypeId="1" CreationDate="2010-12-17T08:12:22.367" Score="3" ViewCount="49" Body="&lt;p&gt;Is there an algorithm to find out all combinations of series, parallel, and series-parallel arrangements given $n$ identical resistors, $R$? All of them must be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might even extend to differently-valued resistors, but I'll just focus on identical resistors.&lt;/p&gt;&#xA;" OwnerUserId="1323" LastActivityDate="2010-12-17T13:38:26.023" Title="Given $n$ identical resistors $R$, find combinations of series, parallel, and series-parallel arrangements" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14646" PostTypeId="1" CreationDate="2010-12-17T08:17:13.533" Score="2" ViewCount="66" Body="&lt;p&gt;I am studying QR decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you explain the geometric intuition for what the Householder transformation does in that context, and why it's sometimes referred to as the Householder &lt;em&gt;reflection&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2010-12-17T14:39:25.303" Title="Geometric intuition for the Householder transformation" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;reflection&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14648" PostTypeId="2" ParentId="14593" CreationDate="2010-12-17T08:28:00.570" Score="0" ViewCount="0" Body="&lt;p&gt;hi, you can refer this link &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ehow.com/how_5169102_convert-latitude-longitude-gps-coordinates.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ehow.com/how_5169102_convert-latitude-longitude-gps-coordinates.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4851" LastActivityDate="2010-12-17T08:28:00.570" />
  <row Id="14649" PostTypeId="1" CreationDate="2010-12-17T08:30:54.483" Score="2" ViewCount="73" Body="&lt;p&gt;In my lecture today, the lecturer mentioned in passing that there are unusual results when you look at combination of elements of a ring, and whether they form a subring or not.&lt;br&gt;&#xA;More specifically he said that if a and b are subrings of a ring, this implies that a-2b, 2a-b and 3a-b are subrings, while a-3b is not.  He didn't go into any detail why this is the case (perhaps we haven't learnt enough to explain it yet) but I was wondering whether anyone could shed any light on why this is the case?&lt;br&gt;&#xA;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="4697" LastActivityDate="2010-12-17T08:30:54.483" Title="When is a combinations of subrings a subring?" Tags="&lt;ring-theory&gt;" CommentCount="3" />
  <row Id="14650" PostTypeId="2" ParentId="14624" CreationDate="2010-12-17T08:58:11.977" Score="1" ViewCount="0" Body="&lt;p&gt;Recall the process of completing $\mathbb Q$ to obtain $\mathbb R$. Here the metric is coming from the absolute value norm. It turns out that in addition to the usual absolute value, one can put the $p$-adic valuations on $\mathbb Q$. One can do a similar completion and complete $\mathbb Q$ to $\mathbb Q_p$. You can also prove that these are the only valuations on $\mathbb Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, on number fields and function fields(together called global fields), one can similarly classify all the valuations,aka &lt;em&gt;places&lt;/em&gt;, and some of them are Archimedean and others non-Archmidean. Now according to Weil in his book &lt;em&gt;Basic Number Theory&lt;/em&gt;, the theory of adèles come up with the point of view that from the arithmetician's perspective, no valuation is preferred over the other; and the global field must be seen at once as completing to infinitely many places all at once, with $\mathbb R$ being no better or worse than any other. He uses the theory of adèles to achieve this end. The nomenclature was perhaps influenced by Chevalley's &lt;em&gt;ideles&lt;/em&gt;, which can be realized as a subset of adèles. The name is supposed to be a pun on a French girls's name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consult the preface of Weil's &lt;em&gt;Basic Number Theory&lt;/em&gt;, you can read exactly what he says. My paraphrase here might be a bit jaded by memory.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-17T08:58:11.977" />
  <row Id="14651" PostTypeId="2" ParentId="14000" CreationDate="2010-12-17T09:12:41.620" Score="1" ViewCount="0" Body="&lt;p&gt;A surface integral measures the &quot;average outflow&quot; of the vector field from the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One illuminating application is &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law&quot; rel=&quot;nofollow&quot;&gt;Gauss's Law&lt;/a&gt; for electric fields, i.e.,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The electric flux through any closed surface is proportional to the enclosed electric charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More precisely, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A}  = \frac{Q}{\varepsilon_0},$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the left-hand side of the equation is a surface integral denoting the electric flux through a closed surface $S$, and the right-hand side of the equation is the total charge $Q$ enclosed by $S$, divided by the electric constant $\varepsilon_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gauss's Law itself has very interesting physical consequences. See the wikipedia page for examples. For instance, the renowned &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;nofollow&quot;&gt;Coulomb's law&lt;/a&gt; is equivalent to it. For another: all the charges on a conductor migrate to its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law_for_magnetism&quot; rel=&quot;nofollow&quot;&gt;Gauss's Law for magnetism&lt;/a&gt;. It states that the net flux of magnetic fields through any surface is always zero. The difference with electric fields is that there are no magnetic &quot;monopoles&quot;; you always find dipoles. The equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \oint_S \mathbf{B} \cdot \mathrm{d}\mathbf{A} = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S$ is any closed surface and $B$ is the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both these Gauss's laws are part of the important &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations&lt;/a&gt;. Many applications are through integration.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-17T09:21:24.920" LastActivityDate="2010-12-17T09:21:24.920" />
  <row Id="14652" PostTypeId="2" ParentId="14646" CreationDate="2010-12-17T09:57:17.230" Score="1" ViewCount="0" Body="&lt;p&gt;Assume that $V$ is an $n$-dimensional Euclidean space. Then, given an orthonormal basis $(e_i)_{i=1}^n$ and an arbitrary n-tuple of vectors $(u_i)_{i=1}^n$, there exists a sequence of $n$ isometries $H_1,...,H_n$ such that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$H_i$ is either &lt;em&gt;a hyperplane reflection&lt;/em&gt; or the identity;&lt;/li&gt;&#xA;&lt;li&gt;every vector of the form &#xA;$$r_j=H_nH_{n-1}\dots H_1(u_j),\qquad j=1,\dots, n,$$&#xD;&#xA;is a linear combination of the vectors $e_1,\dots, e_j$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, the matrix $R$ whose columns are the components of the vectors $r_j$ written over the basis $(e_i)_{i=1}^n$ is upper triangular. The isometries $H_1,\dots, H_n$ which are not identities (i.e. the hyperplane reflections) correspond to Householder matrices. The composition &#xA;$$R=H_n\dots H_1A$$ directly yields the $QR$-decomposition of a given matrix $A$ (whose columns are the components of $u_j$ over the given basis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric interpretation of the $QR$-decomposition can be probably found in several places. I like the presentation in &lt;a href=&quot;http://books.google.com.ua/books?id=B4JtblR1lkMC&amp;amp;printsec=frontcover&amp;amp;dq=books%09+++Geometric+Methods+and+Applications%3a+For+Computer+Science+and+Engineering&amp;amp;source=bl&amp;amp;ots=clsb3DcEQF&amp;amp;sig=-LHtZ9JiiwlgZKWtcgILchZuhHY&amp;amp;hl=en&amp;amp;ei=1zcLTaaLG4Wq8QO6k7H-DQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CCAQ6AEwAQ#v=onepage&amp;amp;q=books%09%20%20%20Geometric%20Methods%20and%20Applications%3A%20For%20Computer%20Science%20and%20Engineering&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Geometric Methods and Applications&lt;/em&gt;&lt;/a&gt; by Jean Gallier (have a look at Section 7.3).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-17T10:17:37.937" LastActivityDate="2010-12-17T10:17:37.937" />
  <row Id="14653" PostTypeId="1" CreationDate="2010-12-17T10:10:50.603" Score="1" ViewCount="92" Body="&lt;p&gt;I've been looking for less trivial examples of computing Ext than finitely generated abelian groups, which tends to be the standard example (and often the only example). Here's an interesting exercise I found in some notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M = \mathbb{C}[x,y] / (x,y), N = \mathbb{C}[x,y] / (x-1)$. My question is how to compute $\text{Ext}_v(M,N)$ in the category of $\mathbb{C}[x,y]$-modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, first of all $\text{Ext}_0(M,N) = \text{Hom}(M,N)$. However, I'm not sure how to identify what this $\text{Hom}$ is! More generally, we have the short exact sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0\rightarrow K \rightarrow \mathbb{C}[x,y] \rightarrow M \rightarrow 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the second map is the inclusion, the third map is the quotient projection, and $K$ is the kernel of the projection. This sequence gives the exact sequence (a piece of the long exact sequence)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0\rightarrow \text{Ext}_1(M,N) \rightarrow \text{Hom}(K,N) \rightarrow \text{Hom}(\mathbb{C}[x,y],N) \rightarrow \text{Hom}(M,N) \rightarrow 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that $\text{Ext}_1(M,N)$ is the kernel of the map $\text{Hom}(K,N) \rightarrow \text{Hom}(\mathbb{C}[x,y], N)$. But again I'm having trouble determining this kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I think the projective resolution $0 \rightarrow \mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y]/(x-1)$ shows that the higher Ext's are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2010-12-17T12:58:53.483" Title="An example of computing Ext" Tags="&lt;homological-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14654" PostTypeId="1" AcceptedAnswerId="14677" CreationDate="2010-12-17T10:21:50.490" Score="0" ViewCount="161" Body="&lt;p&gt;Concerning &lt;a href=&quot;http://math.stackexchange.com/questions/13646&quot;&gt;this&lt;/a&gt; question of mine which involves telescoping in the solution, I was wondering if it is possible to express&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln(x)=\lim_{n\to\infty}n\left(x^{\frac1{n}}-1\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as infinite telescoping products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-17T14:19:32.337" LastActivityDate="2010-12-17T16:29:56.530" Title="Can $\ln(x)=\lim_{n\to\infty} n\left(x^{\frac1{n}}-1 \right)$ be expressed as an infinite telescoping product?" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14655" PostTypeId="1" CreationDate="2010-12-17T10:33:49.400" Score="2" ViewCount="81" Body="&lt;p&gt;Can a polynomial size CFG over large alphabet describe any of these languages:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each terminal appears 0 or 2 times&lt;/li&gt;&#xA;&lt;li&gt;Word repetition {www* | w \in \Sigma *} (word repetition of an arbitrary word w)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&quot;Polynomial size&quot; Context Free Grammar is defined as Chomsky Normal Form polynomial in |Σ|.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References to poly-size CFGs over large alphabets will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is not a homwork. Positive answer can give extension to &lt;a href=&quot;http://math.stackexchange.com/questions/14506/restricted-read-twice-bdds-and-context-free-grammars&quot;&gt;Restricted read twice BDDs and context free grammars&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-17T15:15:07.847" LastActivityDate="2010-12-21T07:06:05.003" Title="Can a polynomial size CFG over large alphabet describe any of these languages:" Tags="&lt;computer-science&gt;&lt;computability&gt;&lt;formal-systems&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14656" PostTypeId="2" ParentId="14654" CreationDate="2010-12-17T11:24:05.797" Score="2" ViewCount="0" Body="&lt;p&gt;Well, we could produce a very contrived example for $e^x$ by defining&#xA;$S_m(x) = \sum_{k=0}^m x^k/k! $ and let&#xA;$$ P_N(x) = \prod_{k=0}^N \frac{S_{k+1}(x)}{S_k(x)} = S_{N+1}(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $\lim_{N \rightarrow \infty} P_N(x) = e^x .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concretely, $e^x$ cannot be represented as an infinite product, such as we can do for $\sin x$ for example, because any zeros of the factors would also be zeros of $e^x$ but $e^x$ is positive for all $x \in \mathbb{R}.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's a slightly less contrived product formula for $e^x .$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_n(x) = \prod_{k=1}^n \left( 1+ \frac{x}{n} \right)^{n/k(k+1)} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $e^x = \lim_{n \rightarrow \infty} P_n(x),$ although there's no telescoping going on here.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-17T14:21:04.350" LastActivityDate="2010-12-17T14:21:04.350" />
  <row Id="14657" PostTypeId="2" ParentId="14645" CreationDate="2010-12-17T11:27:01.927" Score="2" ViewCount="0" Body="&lt;p&gt;Following the Wikipedia link given by Douglas, and the references from there to OEIS, leads to this paper which seems to answer your question:&#xA;&lt;a href=&quot;http://dx.doi.org/10.1119/1.19396&quot; rel=&quot;nofollow&quot;&gt;Antoni Amengual, The intriguing properties of the equivalent resistances of n equal resistors combined in series and in parallel, American Journal of Physics, 68(2), 175-179 (February 2000)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author points out that asking for the possible values of the resistance is not the same as asking for the total number of networks. The answer to the former question is claimed to be approximately $2.55^n$ (for large $n$, I presume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT: A look in the paper reveals that the formula $2.55^n$ was simply obtained by straight-line fitting of the logarithm of the exact values for $6 \le n \le 16$, so (being a mathematician) I'm not quite convinced about its validity for large $n$...]&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-12-17T13:38:26.023" LastActivityDate="2010-12-17T13:38:26.023" CommentCount="2" />
  <row Id="14658" PostTypeId="1" AcceptedAnswerId="14661" CreationDate="2010-12-17T11:46:28.397" Score="3" ViewCount="126" Body="&lt;p&gt;The infinite-length sequence $x_1[n]$ defined by &#xA;\begin{multline}&#xA;x_1[n]=&#xA;\begin{cases}&#xA;\dfrac{1}{n}&amp;amp; \text{if $n \geq $1},\&#xA;0&amp;amp; \text{if $n \leq $0}.&#xA;\end{cases}&#xA;\end{multline}&#xA;has an energy equal to &lt;br/&gt;&#xA;$\mathcal{E _x {_1}} = \sum^\infty_{n=1}(\dfrac{1}{n})^{2}$&#xD;&#xA;&lt;br/&gt;which converges to $\pi^2/6$ indicating that $x_1[n]$ has finite energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get where we find $\pi^2/6$. It would be great if anyone can help me out.&lt;/p&gt;&#xA;" OwnerUserId="4653" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-18T02:59:26.087" LastActivityDate="2010-12-18T02:59:26.087" Title="Easy question about finite energy due to convergence" Tags="&lt;sequences-and-series&gt;&lt;signal-processing&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="14659" PostTypeId="2" ParentId="14617" CreationDate="2010-12-17T11:56:58.907" Score="1" ViewCount="0" Body="&lt;p&gt;I don't think you need to break it up into two cases. In stead we can do this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Establish that the sequence $\langle a_i \rangle$ is cauchy and therefore convergent to some $L$. For this establish the sequence $\langle f^i(0)\rangle$ converges to zero, where $f^i$ is the 'i'th iterate of $\frac{1}{2+x}$.&lt;/li&gt;&#xA;&lt;li&gt;Notice that the relation: $a_{n+1}=1+\frac{1}{1+a_n}$ is the same as $a_{n+1}a_n + a_{n+1} - a_n -2 = 0$.&lt;/li&gt;&#xA;&lt;li&gt;Take limits at both ends to get: $L^2+L-L-2=0$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-17T11:56:58.907" />
  <row Id="14660" PostTypeId="1" AcceptedAnswerId="14664" CreationDate="2010-12-17T12:01:09.650" Score="6" ViewCount="74" Body="&lt;p&gt;One partition of 8 is 5 + 3, but if we then partition each of the 5 and 3 we could get (3+2) + (2+1), and then partition again to get ((2+1)+(1+1)) + ((1+1)+1) and finally (((1+1)+1)+(1+1)) + ((1+1)+1). 5+3 could also be expanded as (4+1)+(2+1), then ((2+2)+1)+((1+1)+1), then (((1+1)+(1+1))+1)+((1+1)+1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is about viewing &quot;+&quot; as a binary operation , so 1+1+1+1 would have to written as either (1+1)+(1+1) or ((1+1)+1)+1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every partition can be written as a such nested partition of 1s. It is still order independent, but associative dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given number $n$, how many associative dependent binary-operation nested partitions of 1s are there of $n$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is known about this function ?&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2010-12-17T12:57:47.660" Title="Counting nested integer partitions" Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="14661" PostTypeId="2" ParentId="14658" CreationDate="2010-12-17T12:02:42.330" Score="5" ViewCount="0" Body="&lt;p&gt;The sum of the series $\displaystyle\sum^\infty_{n=1}(\dfrac{1}{n})^{2}=\dfrac{\pi^2}{6}$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Basel_problem&quot; rel=&quot;nofollow&quot;&gt;classical result due to Euler&lt;/a&gt;. Several proofs are given in the answers to &lt;a href=&quot;http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PS. &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; Robin Chapman collects 14 proofs.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PPS. The improper double integral $$\int_{0}^{1}\int_{0}^{1}\left(\dfrac{1}{1-xy}\right) \mathrm{d}x\mathrm{d}y=\int_{0}^{1}\int_{0}^{1}\left(\sum_{n=1}^{\infty }\left( xy\right)^{n-1}\right) \mathrm{d}x\mathrm{d}y=\sum^\infty_{n=1}\dfrac{1}{n^2} =\dfrac{\pi^2}{6}=\zeta(2)$$ is finite, as pointed out in &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642008550&quot; rel=&quot;nofollow&quot;&gt;Proofs from THE BOOK&lt;/a&gt;&lt;/em&gt; by Martin Aigner and Günter Ziegler. The original article by Tom Apostol is &lt;a href=&quot;http://www.springerlink.com/content/045l250v885v033u/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-18T01:46:30.120" LastActivityDate="2010-12-18T01:46:30.120" CommentCount="8" />
  <row Id="14662" PostTypeId="2" ParentId="14622" CreationDate="2010-12-17T12:36:44.767" Score="2" ViewCount="0" Body="&lt;p&gt;Kawauchi's and Hillman's books are indeed good references. Many known knotted surfaces have non-trivial and easily computable fundamental groups. You should also look at R.H. Fox's &quot;Quick Trip through Knot Theory,&quot; in Fort's Georgia Topology Conference Proceedings, republished by Dover. In addition, there are three books dedicated to studying knotted surfaces from a diagrammatic point of view. Kamada's book on Surface Braids published by the AMS, my book with Saito, &quot;Knotted Surfaces and Their Diagrams,&quot; and my book with Kamada and Saito, &quot;Surfaces in 4-space.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quandle cocycle invariants discovered by Jelsovsky, Langford, Kamada, and me are known to be powerful invariants of knotted surfaces. &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-12-17T12:36:44.767" />
  <row Id="14663" PostTypeId="2" ParentId="14635" CreationDate="2010-12-17T12:47:17.690" Score="2" ViewCount="0" Body="&lt;p&gt;The answer depends on what you mean by &quot;solving&quot; the PDE. The initial value problem for the ultrahyperbolic PDEs are ill-posed. In particular, there is a theorem (the version I know is due to Hormander, you can find it in his &lt;em&gt;Analysis of Linear Partial Differential Operators&lt;/em&gt;; but presumably some versions go back earlier) which states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem&lt;/strong&gt; Let $L$ be a linear partial differential operator with smooth coefficients of order $m$ on $\mathbb{R}^{1+n}$. Consider the Cauchy problem for $Lu = F$ on the upper-half-space $\{ x_0 \geq 0 \}$ with initial data $u_0, u_1, \ldots, u_{m-1}$ (such that $(\partial_0)^k u|_{x_0 = 0} = u_k$ ). Then the following are equivalent. (a) The Cauchy problem has a unique smooth solution $u$ for every prescribed smooth data $u_0,\ldots,u_{m-1}$ and source $F$ and (b) $L$ is a hyperbolic operator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in particular, in general there cannot be well-defined solutions to the initial value problem for ultrahyperbolic PDEs. (They either don't exist or aren't unique. And in the case you do have a solution, the solution is unstable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you cannot even construct approximate solutions to the initial value problem reliably using numerical methods, since ultrahyperbolic equations do not have finite-speed of propagation, so you cannot &quot;localise&quot; the problem, and small changes around a point $x$ may almost instantaneously affect the solution at a far-away point $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In certain special cases you can produce some semblance of a solution. In the constant coefficient case in second order, where the equation can be written as $(\triangle_X - \triangle_Y)u = 0$, you have what is known as Asgeirsson's Mean Value Theorem which is sort of a generalisation of the mean value theorem for harmonic functions, and also a generalisation for the Green's function formula for the linear, constant coefficient wave equation. In this particular case you can also consider solutions using Fourier analytic methods. From there one sees that if one were to assume certain restrictions on the allowed wave-numbers (which leads to a non-local constraint on the initial data), one can &lt;a href=&quot;http://dx.doi.org/10.1098/rspa.2009.0097&quot; rel=&quot;nofollow&quot;&gt;recover well-posedness of the initial value problem&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to references, perhaps an easy way to do a reverse search on the classical paper of Fritz John &lt;a href=&quot;http://projecteuclid.org/euclid.dmj/1077490637&quot; rel=&quot;nofollow&quot;&gt;on the subject&lt;/a&gt;. John's various textbooks also contain some information about it; in particular you may want to consult his &lt;em&gt;Partial Differential Equations&lt;/em&gt; book, and his book on &lt;em&gt;Plane Waves and Spherical Means Applied to Partial Differential Equations&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-17T12:47:17.690" />
  <row Id="14664" PostTypeId="2" ParentId="14660" CreationDate="2010-12-17T12:57:47.660" Score="8" ViewCount="0" Body="&lt;p&gt;Every nested partition defines a full binary tree (all nodes have degree 0 or 2) with $n$ leaves (the 1s) and $n-1$ internal nodes (the binary operators). The number of different full binary trees with $n+1$ leaves is related to the $n$-th &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalan_number&quot; rel=&quot;nofollow&quot; title=&quot;Catalan Number&quot;&gt;Catalan Number&lt;/a&gt;. This is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_n = \frac{1}{n+1} {2n \choose n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Catalan Numbers occur in many counting problems (see the wikipedia article for more details).&lt;/p&gt;&#xA;" OwnerUserId="4857" LastActivityDate="2010-12-17T12:57:47.660" />
  <row Id="14665" PostTypeId="2" ParentId="14653" CreationDate="2010-12-17T12:58:53.483" Score="1" ViewCount="0" Body="&lt;p&gt;First, as Aaron mentioned, a homomorphism from $M$ to $N$ is uniquely defined by its image on the generator 1 of $M$, and it must commute with the action $x\cdot 1=y\cdot 1=0$. In particular, if $f$ is such a homomorphism and $\bar{f(1)}$ is a coset representative of $f(1)$, then you must have $x\cdot \bar{f(1)} \in (x-1)$. Since $\mathbb{C}[x,y]$ is a UFD and $x$ and $x-1$ are both irreducible, this implies that $\bar{f(1)}\in (x-1)$, i.e. that $f(1) = 0\in N$. So, that hom-space is 0, and that's good news for the next computation (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your computation of $\text{Ext}^1$, you have forgotten that $N\mapsto \text{Hom}(,N)$ is a&#xA;contraviariant functor and you have to turn your long exact sequence around accordingly (indeed,&#xA;there is no obvious way of defining the map $\text{Hom}(K,N)\rightarrow \text{Hom}(\mathbb{C}[x,y],N)$,&#xA;while it's perfectly clear how to define the map the other way round - namely by restriction; similarly $\text{Hom}(M,N)\rightarrow \text{Hom}(\mathbb{C}[x,y],N)$ should be defined by composing homs with projection). See if that gets you anywhere&#xA;and feel free to report back.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-17T12:58:53.483" />
  <row Id="14666" PostTypeId="1" CreationDate="2010-12-17T13:05:03.393" Score="5" ViewCount="94" Body="&lt;p&gt;I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, 3,1,5,2,4 is an acceptable permutation where 3,1,2,4,5 is not because 5 is in position 5.  I know that the number of total permutations is n!. Is there a formula for how many are acceptable given the case that no position holds its own number?&lt;/p&gt;&#xA;" OwnerUserId="4858" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-17T14:31:46.657" LastActivityDate="2010-12-17T14:31:46.657" Title="Number of permutations where n ≠ position n" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14667" PostTypeId="1" AcceptedAnswerId="14680" CreationDate="2010-12-17T13:09:08.610" Score="2" ViewCount="96" Body="&lt;p&gt;Suggest an algorithm for factorizing an integer such that each of its factor is &lt;a href=&quot;http://en.wikipedia.org/wiki/Square-free_integer&quot; rel=&quot;nofollow&quot;&gt;square-free&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T17:08:01.613" LastActivityDate="2010-12-17T17:08:01.613" Title="Square-free Integers Factorization" Tags="&lt;number-theory&gt;&lt;algorithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14668" PostTypeId="2" ParentId="14666" CreationDate="2010-12-17T13:16:54.020" Score="6" ViewCount="0" Body="&lt;p&gt;Check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Derangement&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2010-12-17T13:16:54.020" />
  <row Id="14669" PostTypeId="2" ParentId="14666" CreationDate="2010-12-17T13:19:41.723" Score="8" ViewCount="0" Body="&lt;p&gt;What you are looking for is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement&quot; rel=&quot;nofollow&quot;&gt;derangement&lt;/a&gt;.However,for counting the number of derangement for say $n$-elements you could possible use a trick,compute$\frac{n!}{e}$ and then round off to an integer and this will give you the desired result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually another application of $e$,which was discovered by Jacob Bernoulli in the problem of derangement,also known as the &lt;a href=&quot;http://www.jstor.org/pss/2315337&quot; rel=&quot;nofollow&quot;&gt;hat check problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-17T13:41:13.947" LastActivityDate="2010-12-17T13:41:13.947" />
  <row Id="14670" PostTypeId="1" CreationDate="2010-12-17T13:23:46.963" Score="0" ViewCount="39" Body="&lt;p&gt;Hello! I would like to know a way to check if my version of Matlab support object oriented programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By, the way this question is related with other I asked before in a stackexchange site.&#xA;&lt;a href=&quot;http://stackoverflow.com/questions/4462441/matlab-creating-customized-numeric-data-types&quot;&gt;Matlab: Creating customized numeric data types&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2010-12-17T13:23:46.963" ClosedDate="2010-12-17T15:01:40.127" Title="How can I check if my version of matlab support object oriented programming?" Tags="&lt;matlab&gt;" CommentCount="6" />
  <row Id="14671" PostTypeId="1" AcceptedAnswerId="14684" CreationDate="2010-12-17T13:34:18.817" Score="1" ViewCount="97" Body="&lt;p&gt;Recall that complex topological $K$-theory is representable on reasonable spaces by the space $BU \times \mathbb{Z}$ (where $BU$ is a colimit of various infinite Grassmannians), and that the total Chern class provides  a natural map $\mathrm{Vect}(B) \to H^*(B)$ for every such space $B$. By the multiplicativity property, this map factors through the K-group and leads to a natural transformation $K(B) \to H^*(B)$. $H^*(B)$ is also representable by a product of Eilenberg-MacLane spaces $K(\mathbb{Z}, n)$ over all $n$. There is thus a map, unique up to homotopy &#xA;$$BU \times\mathbb{Z} \to \prod_n K(\mathbb{Z},n).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this map?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Mariano observes, one can simply define the individual Chern classes on the $K$-group as well, albeit not immediately through the universal property, so the question reduces to the determination of the (homotopy class of the) map $BU \times \mathbb{Z} \to K(\mathbb{Z},n)$ induced by each Chern class.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-17T15:29:45.497" LastActivityDate="2010-12-17T19:58:39.387" Title="What's the map $BU \times \mathbb{Z} \to \prod K(\mathbb{Z},n)$ representing the total Chern class?" Tags="&lt;algebraic-topology&gt;&lt;k-theory&gt;&lt;characteristic-classes&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="14672" PostTypeId="1" CreationDate="2010-12-17T13:37:50.330" Score="4" ViewCount="114" Body="&lt;p&gt;Hi,&lt;br /&gt;&#xA;Given two positive rational number $\frac{a_1}{b_1}$ and $\frac{a_3}{b_3}$ (written in lowest terms) such that $$\frac{a_1}{b_1} &amp;lt; \frac{a_3}{b_3},$$&#xD;&#xA;I want to find a rational number&#xA;$\frac{a_2}{b_2}$ such that $$\frac{a_1}{b_1} &amp;lt; \frac{a_2}{b_2} &amp;lt; \frac{a_3}{b_3}$$&#xD;&#xA;and $a_2$ and $b_2$ are the smallest possible numbers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come with the solution $a_2 = a_1 + a_3$ and $b_2 = b_1 + b_3$ then dividing $a_2$ and $b_2$ by their greatest common divisor but there must be better solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks by advance for your suggestions.&lt;/p&gt;&#xA;" OwnerUserId="4861" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T19:56:39.953" LastActivityDate="2010-12-18T07:36:31.947" Title="Computing a rational number between two others, minimizing numerator and denominator" Tags="&lt;number-theory&gt;&lt;fractions&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14673" PostTypeId="2" ParentId="14672" CreationDate="2010-12-17T13:57:24.657" Score="3" ViewCount="0" Body="&lt;p&gt;See Wikipedia's article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mediant_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;mediant&lt;/a&gt;, which explains when your solution is the best.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-17T16:03:10.700" LastActivityDate="2010-12-17T16:03:10.700" CommentCount="2" />
  <row Id="14674" PostTypeId="2" ParentId="14646" CreationDate="2010-12-17T14:39:25.303" Score="1" ViewCount="0" Body="&lt;p&gt;We start with a square matrix $M$ of dimension $n$. We can think of its $n$ columns as vectors in $\mathbb{R}^n$. We consider the hyperplane generated by the first column (for example the orthogonal complement of that vector). Next, we reflect each of the columns about this hyperplane. In symbols: $H_1M= [ H_1(v_1) \ldots H_1(v_n)]$, where on the RHS we use functional notation for $H_1$. Now, because $v_1$ is normal to the hyperplane, $H_1(v_1)$ looks simple. The rest of the vectors transform like: &lt;img src=&quot;http://i.imgur.com/Arm68.png&quot; alt=&quot;alt text&quot;&gt;     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is we subtract twice their projections onto $v_1$ (this gives me the formula for householder reflections).&#xA;Then we consider the $n-1$ dimensional submatrix of $H_1M:=M_2$, and repeat. The submatrix takes me into the hyperplane, since the first reflection leaves that plane invariant. What we are doing is changing the basis (since Reflections have $det \neq 0$) of the underlying space progressively so that the vectors have a nice representation (Thats what QR decomposition is, The Q contains the orthonormal vectors, while the R tracks all the changes we have made).  &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-17T14:39:25.303" />
  <row Id="14675" PostTypeId="1" AcceptedAnswerId="14676" CreationDate="2010-12-17T14:44:52.280" Score="1" ViewCount="32" Body="&lt;p&gt;I'm trying to compute the variational distance between two negative exponential distributions. If $a$ denotes the value of $x$ where both curves are equal, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{+\infty}\left|\lambda_1e^{-\lambda_1x}-\lambda_2e^{-\lambda_2x}\right|dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=-\int_{0}^a(\lambda_1e^{-\lambda_1x}-\lambda_2e^{-\lambda_2x})dx+\int^{+\infty}_a(\lambda_1e^{-\lambda_1x}-\lambda_2e^{-\lambda_2x})dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\left[e^{-\lambda_2x} - e^{-\lambda_1x}\right]_{0}^a +\left[e^{-\lambda_1x} - e^{-\lambda_2x}\right]^{+\infty}_a$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=e^{-\lambda_2a} - e^{-\lambda_1a} -1 +1  +e^{-\infty} - e^{-\infty}-e^{-\lambda_1a}+e^{-\lambda_2a}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=2\left(e^{-\lambda_2a}-e^{-\lambda_1a}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing $a$ is not difficult:&#xA;\begin{eqnarray*}&#xA; \lambda_1e^{-\lambda_1x}=\lambda_2e^{-\lambda_2x} \Leftrightarrow\ln(\lambda_1)-\lambda_1x&amp;amp;=&amp;amp;\ln(\lambda_2)-\lambda_2x\&#xA; \Leftrightarrow x&amp;amp;=&amp;amp;\frac{\ln(\lambda_2)-\ln(\lambda_1)}{\lambda_2-\lambda_1}.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what puzzles me is that I am integrating a positive function, yet depending on how parameters are related to each other, the value of the above result may be negative ... what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry for the messy layout, it seems that my &amp;amp;'s are ignored ...)&lt;/p&gt;&#xA;" OwnerUserId="4863" LastActivityDate="2010-12-17T15:17:59.317" Title="Variational distance between negative exponential distributions" Tags="&lt;integral&gt;" AnswerCount="1" />
  <row Id="14676" PostTypeId="2" ParentId="14675" CreationDate="2010-12-17T15:17:59.317" Score="3" ViewCount="0" Body="&lt;p&gt;In the change from the first line of your equation to the second line, you implicitly used the fact that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_1 e^{-\lambda_1x} &amp;gt; \lambda_2 e^{-\lambda_2x} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $x &amp;gt; a$. (Which is equivalent to assuming $0 &amp;lt; \lambda_1 &amp;lt; \lambda_2$.) You'll see that the parameter ranges that gives you a negative answer from your computation precisely corresponds to the case where the above assumption is false&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-17T15:17:59.317" CommentCount="1" />
  <row Id="14677" PostTypeId="2" ParentId="14654" CreationDate="2010-12-17T15:34:27.120" Score="4" ViewCount="0" Body="&lt;p&gt;Let $A_n(x) = n(x^{1 \over n} - 1)$. Then if $B_n$ denotes $A_{2^{n}}$, then $B_n(x) = 2^n(x^{1 \over 2^n} - 1)$, and $\lim_{n \rightarrow \infty} B_n(x)$ equals $\ln(x)$ because $B_n(x)$ is a subsequence of $A_n(x)$. Note that &#xA;$${B_n(x) \over B_{n-1}(x)} = {2^n(x^{1 \over 2^n} - 1)   \over 2^{n-1}(x^{1 \over 2^{n-1}} - 1)}$$&#xD;&#xA;$$= {2 (x^{1 \over 2^n} - 1) \over (x^{1 \over 2^n} - 1)(x^{1 \over 2^n} + 1)}$$&#xD;&#xA;$$= {2 \over x^{1 \over 2^n} + 1}$$&#xD;&#xA;So $B_n$ can be written as a finite telescoping product&#xA;$$B_n = (x- 1)\prod_{i = 1}^{n} {2 \over x^{1 \over 2^i} + 1}$$&#xD;&#xA;So taking the limits as $n$ goes to $\infty$ one has&#xA;$$\ln(x) = (x-1)\prod_{i = 1}^{\infty} {2 \over x^{1 \over 2^i} + 1}$$&#xD;&#xA;So at least a subsequence of the partial products gives a telescoping product.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-17T15:34:27.120" />
  <row Id="14678" PostTypeId="2" ParentId="14672" CreationDate="2010-12-17T15:46:11.297" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\displaystyle\quad \frac{A}B &amp;lt; \frac{X}Y &amp;lt; \frac{C}D\ \iff\ X = \frac{MA+NC}{BC-AD},\ Y = \frac{MB+ND}{BC-AD}\quad$ for some integers $\rm\:M,N&amp;gt;0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a nice geometric interpretation in terms of plane lattices - which provides a vivid geometric interpretation of good approximations by Farey mediant searching - see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zpsy7ixpf.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T15:46:11.297" />
  <row Id="14679" PostTypeId="1" AcceptedAnswerId="14681" CreationDate="2010-12-17T15:57:18.877" Score="2" ViewCount="128" Body="&lt;p&gt;Let $f(x)$ be two times differentiable in $[a, b]$. The Mean Value Theorem tells us that there's such a point $a\lt c\lt b$, such that ${f}&amp;#39;(c)=\frac{f(b)-f(a)}{b-a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\forall x \in [a, b], {f}&amp;#39;&amp;#39;(x)\gt 0$ and $f&amp;#39;&amp;#39;(x)$ is strictly increasing then $c\gt \frac{a+b}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time figuring this one out. More so, I can't seem to understand why this is true, intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried the following:&#xA;Write $f(x)$ Taylor's expansion of degree 1, around $a$. Then, for $x=b$ we get: $f(b)-f(a)=f&amp;#39;(a)(b-a)+\frac{1}{2}f&amp;#39;&amp;#39;(\xi)(b-a)^2 \implies f&amp;#39;(c)=f&amp;#39;(a)+\frac{1}{2}f&amp;#39;&amp;#39;(\xi)(b-a)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I still haven't used the fact that $f&amp;#39;&amp;#39;(x)$ is positive and strictly increasing in $[a, b]$ (which btw also implies that $f&amp;#39;(x)$ is strictly increasing). I tried playing with the expression above but I just don't see which direction to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't give a full solution. Hints and intuition are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-17T16:06:56.747" LastActivityDate="2010-12-18T08:33:46.127" Title="Mean value theorem and a strictly increasing, positive second derivative" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;taylor-expansion&gt;" AnswerCount="3" />
  <row Id="14680" PostTypeId="2" ParentId="14667" CreationDate="2010-12-17T16:03:10.540" Score="5" ViewCount="0" Body="&lt;p&gt;It's trivial given the complete factorization into primes. Currently we do not know any other way to compute squarefree parts and it is widely suspected that it cannot be computed in any simpler way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is important because one of the main tasks of computational algebraic number theory reduces to it (in deterministic polynomial time). Namely the problem of computing the ring of integers of an algebraic number field depends upon the square-free decomposition of the polynomial discriminant when computing an integral basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrast this difficulty with the trivial squarefree decomposition of polynomials by way of gcd with its derivative. The availability of derivatives for polynomials opens up a powerful toolbox that is not available for integers. For example once derivatives are available so are Wronskians - which provide powerful measures of dependence in transcendence theory and diophantine approximation. A simple yet stunning example is the elementary proof of the polynomial case of Mason's ABC theorem, which yields as a very special case a high-school-level proof of FLT for polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For references see my &lt;a href=&quot;http://math.stackexchange.com/questions/2239/importance-of-determining-whether-a-number-is-squarefree-using-geometry/2249#2249&quot;&gt;post here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T16:03:10.540" />
  <row Id="14681" PostTypeId="2" ParentId="14679" CreationDate="2010-12-17T16:07:06.463" Score="1" ViewCount="0" Body="&lt;p&gt;Try sketching a graph, since you are looking for intuition only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sketch of the f' will be easier as then you can use the Intermediate Value Theorem too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we want to see that the MVT holds for the case of $f(x)$ in [a,b] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to prove that the slope $\frac{f(b)-f(a)}{b-a}$ is inside $[f&amp;#39;(a),f&amp;#39;(b)]$&#xD;&#xA;This is not too difficult to prove.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-17T19:00:45.303" LastActivityDate="2010-12-17T19:00:45.303" CommentCount="7" />
  <row Id="14682" PostTypeId="2" ParentId="14654" CreationDate="2010-12-17T16:29:56.530" Score="2" ViewCount="0" Body="&lt;p&gt;This is a comment on generalities(therefore just community wiki).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For infinite products of expressions of the form $(1 + \mathrm{something})$, it is often more productive to take the logarithm and convert it into an infinite sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a little bit of an issue of convergence here. If one of the expressions is $0$, then there would be trouble, etc.. A precise handling of such stuff can be found in the book of Ahlfors on Complex Analysis.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-17T16:29:56.530" CommunityOwnedDate="2010-12-17T16:29:56.530" />
  <row Id="14684" PostTypeId="2" ParentId="14671" CreationDate="2010-12-17T16:44:42.327" Score="3" ViewCount="0" Body="&lt;p&gt;One way, my favorite, of defining the chern classes goes as follows: first compute $H^*(BU; \mathbb{Z})=\mathbb{Z}[c_1, c_2, ...]$ by using your favorite method. I know two methods, one is using some cellular description of the grassmanians (see Milnor and Stasheff), the other is to first compute the cohomology of the lie groups $U(n)$ and then use a path-loop space fibration and the Serre SS to get at the cohomology of $H^*(BU(n);\mathbb{Z})$ (see &lt;a href=&quot;http://math.stackexchange.com/questions/13260/homology-and-euler-characteristics-of-the-classical-lie-groups/13271#13271&quot;&gt;Homology and Euler characteristics of the classical Lie groups&lt;/a&gt;). Next you notice that $BU$ and $BU(n)$ have the same cell structure through a range and so you essentially take the limit (colimit, and there are some subtleties here, the place I recall seeing these is in Jacob Lurie's survey on elliptic cohomology, at the beginning). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next suppose you have a complex vector bundle $\xi : E \to B$ then it is classified by a map $f: B \to BU(n)$ where n is the dimension of $\xi$. You can get a map on all of $BU$ by just adding on trivial bundles to get a map out of $BU(k)$ for all $k$ larger than $n$. This gives a map out of BU (although technically you don't really need this, $BU(n)$ works fine for defining the chern classes of an $n$ dimensional complex vector bundle). Now define $c_n(\xi):=f^*(c_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this perspective we started with the universal case, so maybe it is a bit of a cheat. For me this even clarifies how I should think about characteristic classes in general. Suppose you want to look at $G$-bundles and see what you can tell about them from $E$-theory ($E$ some ring spectrum and $G$ some compact lie group or whatever you need for $BG$ to be nice, I am not sure if there are other restrictions for this to work). Now compute $E^*BG$ and use the fact that $G$-bundles over $X$ are classified by homotopy classes $X \to BG$. You should check out some of the threads on MO about characteristic classes, I think I can learn something from each of Rezk's answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please let me know if I can make some of the above clearer or if there are any mistakes.&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="627" LastEditorDisplayName="" LastEditDate="2010-12-17T19:58:39.387" LastActivityDate="2010-12-17T19:58:39.387" CommentCount="11" />
  <row Id="14685" PostTypeId="1" CreationDate="2010-12-17T17:15:40.463" Score="1" ViewCount="57" Body="&lt;p&gt;This is a quick (not to scale) drawing of what I mean: the floor gradually slopes from the wall to the drain at about 7/16( 0.4375) inch per foot from the short sides of the wall (the 7 ft from wall to the drain) and exactly 3/16(.1875&quot;) of a inch per foot on the long sides ( the 16ft from the wall to the drain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the base is a rectangle 14'x32' but the shape is most related to a cone ,.... but all cone formulas assume the base is circular. I'm at a loss of how to pull out only the rectangular portion out from the middle of a cone shape. I hope this made it more visual &lt;img src=&quot;http://i119.photobucket.com/albums/o141/Elder_Paul/area.jpg&quot; alt=&quot;&quot;&gt; it has 3&quot; of Drop in the Center, length is 32' and width is 14'. just looking for a formula to calculate it. Any help in the right direction would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4870" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T18:15:12.607" LastActivityDate="2010-12-17T19:14:59.990" Title="How do I find the area or the volume (or a formula) for a region that is rectangular but has a cone slope to a drain ?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="14686" PostTypeId="1" CreationDate="2010-12-17T17:18:23.787" Score="1" ViewCount="186" Body="&lt;p&gt;Why is integral of $x$ from $-1$ to $1$ $x^2/2$ rather than $x^2$?&lt;/p&gt;&#xA;" OwnerUserId="4871" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T20:09:46.770" LastActivityDate="2010-12-18T03:27:09.293" Title="Why is the integral of $x$ from $-1$ to $1$  $x^2/2$ rather than $x^2$?" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14687" PostTypeId="1" AcceptedAnswerId="14690" CreationDate="2010-12-17T17:19:34.970" Score="1" ViewCount="63" Body="&lt;p&gt;Suppose $f$ is analytic on the domain $D = \{z \big| |z|&amp;gt;1\}$ in $\mathbb{C}$.   I know that if the line integral of $f$ along any closed path is 0 then $f$ has an anti-derivative.  Is this also a necessary condition.&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-12-17T17:26:56.103" LastActivityDate="2010-12-17T17:29:04.120" Title="Necessary condition for existence of anti-derivative" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14688" PostTypeId="2" ParentId="14686" CreationDate="2010-12-17T17:23:49.733" Score="2" ViewCount="0" Body="&lt;p&gt;Integration is the opposite of differentiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you differentiate  x^2/2 you get x (you bring down the power of 2, which cancels with the 1/2 out front) which means that when you integrate x you must get x^2/2.&lt;/p&gt;&#xA;" OwnerUserId="4873" LastActivityDate="2010-12-17T17:23:49.733" />
  <row Id="14689" PostTypeId="1" CreationDate="2010-12-17T17:27:12.103" Score="7" ViewCount="229" Body="&lt;p&gt;Two players in a dice game to see who can roll a total of 60 first, taking turns each rolling 2 dice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one player, one die is 4 sided and one is 6 sided. Therefore the average roll for this player will be a 6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other player has two 6 sided dice. So this players average roll will be a 7. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they take turns, each rolling both dice, what % of the time will the first player (a 4-6 dice combo) get to 60 first and what % of the time will the second player (6-6 dice combo) get to 60 first. If they get to 60 on the same roll, the player who goes the furthest wins.&lt;/p&gt;&#xA;" OwnerUserId="4872" LastActivityDate="2010-12-20T00:34:01.337" Title="Thought experiment for dice game.." Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="14690" PostTypeId="2" ParentId="14687" CreationDate="2010-12-17T17:29:04.120" Score="3" ViewCount="0" Body="&lt;p&gt;Yes.  If $F&amp;#39;=f$ and $\gamma:[a,b]\to D$ is a curve, then $\int_\gamma f = F(\gamma(b))-F(\gamma(a))$.  For example, see Theorem 1.18 on &lt;a href=&quot;http://books.google.com/books?id=9LtfZr1snG0C&amp;amp;lpg=PP1&amp;amp;ots=OePFw5q3Ex&amp;amp;dq=conway%20complex&amp;amp;pg=PA65#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 65&lt;/a&gt; of John B. Conway's &lt;em&gt;Functions of one complex variable&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-17T17:29:04.120" />
  <row Id="14691" PostTypeId="2" ParentId="14679" CreationDate="2010-12-17T17:31:44.477" Score="0" ViewCount="0" Body="&lt;p&gt;I prefer words, really. What the mean value theorem tells us is that at some point on the interval $[a,b]$ the function $f(x)$ (continuous and differentiable on this interval) has a gradient equal to the slope of the line connecting $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should this be? Well, the trivial case we can handle, namely that if $f\prime(x) = k$ some constant and passes through $a,b$ then k can only be that slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's take your case. $f\prime\prime(x) &amp;gt; 0$ and strictly increasing, so the rate of change of gradient is positive and going up. So this means the gradient is increasing in increasingly big chunks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sketch what that means, then take the line from $a$ to $b$ and find out where it hits your sketched curve. Why can it not appear before $\frac{a+b}{2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ask a different question - the rate of change of gradient is strictly increasing. Given then that the gradient between $a,b$ gets bigger in bigger chunks, where do you expect the gradient to change most between? If I said compare this to something like a probability distribution, would it make sense why it's called the mean value theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help at all?&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-12-17T17:31:44.477" />
  <row Id="14692" PostTypeId="2" ParentId="14613" CreationDate="2010-12-17T17:35:00.883" Score="0" ViewCount="0" Body="&lt;p&gt;Here's something off the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Pick a statistic for your data which might measure how correlated they are (traditional notions of correlation don't apply here, because you don't have any observations -- you only have trade times). For example, you might calculate the average difference between each point in series A and its closest point in series B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Generate some new data from a known distribution. In your example, you might generate series A' as five random draws from a uniform distribution and series B' as three random draws from a uniform distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Compute your chosen statistic on your artificial data. The idea is to work out how the statistic behaves on 'random' data with no correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Repeat steps 2 and 3 until you have a good idea what the distribution of your statistic is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Compare your result to the generated distribution. If it's lying way out in the tails, you can be relatively certain that there is significant correlation between your two strategies. If it lies near the center of the distribution, then you can't reject the hypothesis that there is no correlation.&lt;/p&gt;&#xA;" OwnerUserId="4873" LastActivityDate="2010-12-17T17:35:00.883" />
  <row Id="14693" PostTypeId="1" CreationDate="2010-12-17T17:44:24.470" Score="3" ViewCount="151" Body="&lt;p&gt;I have this equation $3-x-x^2$ which I want to convert into the form $a^2 - y^2$, so that then I can suppose $y$ = $a\sin\theta$, differentiate and then integrate it. But I can't convert that equation into this $a^2 - y^2$ form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. this is a homework question. My question here is not to solve this for me, but tell me how to do such (if there exists a formula) so that I may learn and prosper :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; $a^2 - y^2$, in this equation $y$ may be anything which involves $x$ e.g. $(x-2)^2$&lt;/p&gt;&#xA;" OwnerUserId="4875" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T19:40:03.017" LastActivityDate="2010-12-17T20:19:39.597" Title="Rewrite a general quadratic as $a^2-x^2$" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14694" PostTypeId="2" ParentId="14693" CreationDate="2010-12-17T17:55:17.950" Score="11" ViewCount="0" Body="&lt;p&gt;What you are trying to do is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Completing_the_square&quot; rel=&quot;nofollow&quot;&gt;completing the square&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-17T17:55:17.950" CommunityOwnedDate="2010-12-17T17:55:17.950" />
  <row Id="14696" PostTypeId="2" ParentId="14686" CreationDate="2010-12-17T17:59:02.580" Score="7" ViewCount="0" Body="&lt;p&gt;Consider $f(x)=x$ for x equals 0 to a, it is a right-angled triangle (the blue region in figure below). &#xA;&lt;img src=&quot;http://i.imgur.com/M8vb6.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;The area of the square (having sides equal a) is $a^2$. Since there are two identical triangles each filling half the area, the area under the curve of f(x)=x for x=0 to a is $\frac{a^2}{2}$, which is precisely what you get from computing the definite integral, since a definite integral calculates area under the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, consider f(x)=x from a to b, with a&gt;=0. The definite integral from a to b calculated the area of the blue region illustrated below. It can be reduced to the previous case as the area of the bigger triangle from 0 to b minus the area of the smaller triangle from 0 to a. Hence $\frac{b^2}{2} -\frac{a^2}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KnyNM.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I should also mention, when the area is below the x-axis, such as f(x) for x=a to 0 (a&amp;lt;0), a change of sign is needed, i.e $\frac{-a^2}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-12-17T18:40:33.527" LastActivityDate="2010-12-17T18:40:33.527" />
  <row Id="14697" PostTypeId="1" AcceptedAnswerId="14700" CreationDate="2010-12-17T18:02:21.377" Score="4" ViewCount="155" Body="&lt;p&gt;How to prove $\{t\} \notin t$ using the axiom of foundation (aka axiom of regularity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \emptyset \vee \exists x \in A \forall y \in x : y \not\in A$&lt;/p&gt;&#xA;" OwnerUserId="4876" LastActivityDate="2010-12-17T18:12:45.700" Title="How to prove $\{t\} \notin t$" Tags="&lt;logic&gt;&lt;set-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="14698" PostTypeId="2" ParentId="14693" CreationDate="2010-12-17T18:06:02.667" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad A^2 - (X-B)^2\ =\ 3 - X - X^2\ \Rightarrow\ B = \ldots\ \Rightarrow\ A = \ldots\ $ (expand and compare coeff's)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally if you want to check that some specific polynomial is an instance of some general polynomial form  you may treat the coefficients of the general form as variables or indeterminates (a.k.a. &quot;undetermined coefficients&quot;). With luck, the system of equations obtained by comparing coefficients might be easily solvable (e.g. it might be triangular). This is sometimes referred to as the &lt;em&gt;method of undetermined coefficients&lt;/em&gt; and it works more generally for objects that are uniquely determined by their &quot;coefficients&quot;, e.g. vectors, matrices, (formal) power series, radix notation, partial fraction expansions, etc. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-17T20:19:39.597" LastActivityDate="2010-12-17T20:19:39.597" CommentCount="2" />
  <row Id="14699" PostTypeId="2" ParentId="14672" CreationDate="2010-12-17T18:08:37.467" Score="2" ViewCount="0" Body="&lt;p&gt;Your solution is not the best. (Take, for instance, 1/2 and 9/10.) One way to solve your problem is to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continued_fraction&quot; rel=&quot;nofollow&quot;&gt;continued fractions&lt;/a&gt; of the two numbers until they differ. Say the continued fractions start $[x_0;x_1,x_2,x_3]$ and $[x_0;x_1,x_2,y_3]$, with $x_3 \ne y_3$. Then you can just try all values $z_3$ between $x_3$ and $y_3$ inclusive to get your answer. There might be a cleverer way to find the best $z_3$, but this way should work: it relies on the fact that continued-fraction convergents are alternately greater and less than the value that they are approximating, so one of $[x_0;x_1,x_2,x_3]$ and $[x_0;x_1,x_2,y_3]$ will lie between $a1/b1$ and $a3/b3$.&lt;br&gt;&#xA;There is a slight complication: suppose $x_3$ and $y_3$ differ by one, and one of the convergents is exact. Then you might not be able to find $z_3$ such that $[x_0;x_1,x_2,z_3]$ is &lt;em&gt;strictly&lt;/em&gt; between $a1/b1$ and $a3/b3$. But in this case you can use your formula on the two convergents (instead of on $a1/b1$ and $a3/b3$) to get the answer, because the condition that $x_3$ and $y_3$ differ by one is sufficient to guarantee that your mediant is the best approximation.&lt;br&gt;&#xA;For example, 1/2=[0;2] and 9/10=[0;1,9]. So they differ at $x_1$, which means we should try [0;1] = 1 and [0;2] = 1/2. Neither is satisfactory, so we take their mediant, which is 2/3.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-18T07:36:31.947" LastActivityDate="2010-12-18T07:36:31.947" />
  <row Id="14700" PostTypeId="2" ParentId="14697" CreationDate="2010-12-17T18:12:45.700" Score="12" ViewCount="0" Body="&lt;p&gt;If you are okay with Choice, the Axiom of Foundation implies that there can be no infinite descending chain of sets ordered by $\in$. You would have $t\in\{t\}\in t$, leading to such an infinite chain, and thus contradicting AF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can do it directly. Let $t$ be a set, and suppose that $\{t\}\in t$. Let&#xA;$$X = \{t,\{t\}\}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by Foundation there exists $x\in X$ such that $x\cap X=\emptyset$. So either $t\cap X=\emptyset$ or $\{t\}\cap X=\emptyset$. Now, $t\in \{t\}\cap X$, so the latter cannot hold. And since $\{t\}\in t$ and $\{t\}\in X$, then $\{t\}\in t\cap X\neq\emptyset$. So neither holds, hence $X$ contradicts Foundation.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-17T18:12:45.700" CommentCount="1" />
  <row Id="14701" PostTypeId="2" ParentId="14689" CreationDate="2010-12-17T18:19:49.413" Score="6" ViewCount="0" Body="&lt;p&gt;A rough answer, from Monte Carlo simulation, is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Player 1 wins  83.85 %, Player 2 wins  14.28 %, and  1.87 % are draws.&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The following R code generates this output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N &amp;lt;- 60&#xA;&#xA;p1Wins  &amp;lt;- 0&#xA;p2Wins  &amp;lt;- 0&#xA;draws   &amp;lt;- 0&#xA;&#xA;n.sim   &amp;lt;- 10000&#xA;&#xA;p1Roll &amp;lt;- function() {&#xA;    d1 &amp;lt;- sample(1:6,1)&#xA;    d2 &amp;lt;- sample(1:6,1)&#xA;    return(d1+d2)&#xA;}&#xA;&#xA;p2Roll &amp;lt;- function() {&#xA;    d1 &amp;lt;- sample(1:6,1)&#xA;    d2 &amp;lt;- sample(1:4,1)&#xA;    return(d1+d2)&#xA;}&#xA;&#xA;&#xA;for (i in 1:n.sim) {&#xA;    p1Total &amp;lt;- 0&#xA;    p2Total &amp;lt;- 0&#xA;    while (p1Total &amp;lt; 60 &amp;amp; p2Total &amp;lt; 60) {&#xA;        p1Total &amp;lt;- p1Total + p1Roll()&#xA;        p2Total &amp;lt;- p2Total + p2Roll()&#xA;    }&#xA;    if (p1Total &amp;gt; p2Total) {&#xA;        p1Wins &amp;lt;- p1Wins + 1&#xA;    } else&#xA;    if (p2Total &amp;gt; p1Total) {&#xA;        p2Wins &amp;lt;- p2Wins + 1&#xA;    } else {&#xA;        draws &amp;lt;- draws + 1&#xA;    }&#xA;}&#xA;&#xA;print(paste(&quot;Player 1 wins &quot;,p1Wins/n.sim*100,&quot;%, Player 2 wins &quot;,p2Wins/n.sim*100,&quot;%, and &quot;,draws/n.sim*100,&quot;% are draws.&quot;))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4873" LastActivityDate="2010-12-17T18:19:49.413" CommentCount="3" />
  <row Id="14702" PostTypeId="2" ParentId="14685" CreationDate="2010-12-17T18:21:06.087" Score="0" ViewCount="0" Body="&lt;p&gt;I am not completely sure about the actual geometry of the problem, so I'll make a few assumptions. Correct me if the assumptions are wrong.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have a rectangle of 168&quot; by 384&quot;. &lt;/li&gt;&#xA;&lt;li&gt;The center of the rectangle is depressed by 3&quot;&lt;/li&gt;&#xA;&lt;li&gt;(most importantly) The shape is such that if you draw a line segment from the &quot;drain&quot; to the &quot;wall&quot;, it will be straight (not curved in the vertical direction). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The third assumption may be different from what you want, since this assumption will require there be something like a &quot;fold/corner&quot; along the diagonals. To get more precise answers, you need to specify exactly what the shape of the floor is (how it curves towards the drain). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in this case you don't need calculus at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each &quot;trianglular piece&quot; in your drawing can be computed directly. For the triangles with base on the short sides, the base is 168&quot;. The &quot;height&quot; can be found using Pythagorean theorem applied to the vertical dip:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h^2 = 192^2 + 3^2 \implies h \sim 192.0234 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the area of the triangle piece is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = h \times 168 \sim 16130.0 \textrm{sq in}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the area of the triangle piece based on the long side of the wall is roughly 16130.6 sq in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total area is roughly 64521 sq in, which is 9 square inches more than if you don't factor in the vertical drop to the drain hole. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-17T18:21:06.087" CommentCount="2" />
  <row Id="14703" PostTypeId="2" ParentId="14685" CreationDate="2010-12-17T18:30:58.267" Score="1" ViewCount="0" Body="&lt;p&gt;It looks like this is a truncated rectangular pyramid or &lt;em&gt;pyramidal frustum&lt;/em&gt;. This is probably what frustrated your searches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume for a square base is given in &lt;a href=&quot;http://mathworld.wolfram.com/TruncatedSquarePyramid.html&quot; rel=&quot;nofollow&quot;&gt;mathworld&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the rectangular case, you can compute the area of the full pyramid and then of the top that you cut off; the formulas are also available at &lt;a href=&quot;http://mathworld.wolfram.com/RegularPyramid.html&quot; rel=&quot;nofollow&quot;&gt;mathworld&lt;/a&gt;. A quick Google search also provided &lt;a href=&quot;http://wiki.answers.com/Q/Volume_of_a_truncated_rectangular_based_pyramid&quot; rel=&quot;nofollow&quot;&gt;this page.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be done with Calculus fairly straightforward (a standard Calculus II exercise/problem); but you don't need it. The egyptians already knew the formula. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-17T18:30:58.267" CommentCount="1" />
  <row Id="14704" PostTypeId="2" ParentId="9524" CreationDate="2010-12-17T18:33:04.570" Score="2" ViewCount="0" Body="&lt;p&gt;Good question, and one that is timely for me as I'm teaching calc II next semester for the first time in several years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An application of Taylor series that sometimes comes up in an intro to ODEs course is that of series solutions to differential equations.  I usually don't spend more than a day or two on the topic, but I think the students appreciate seeing how a series representation of a solution can be found in cases that can't be solved by any of the other methods that they've learned.  I especially like the example of Airy's equation&#xA;$$&#xD;&#xA;y&amp;#39;&amp;#39; - t \, y = 0&#xD;&#xA;$$&#xD;&#xA;because it's simple enough that students feel like there ought to be an explicit solution plus the series coefficients are fairly easy to express.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this kind of application would be helpful or meaningful for calc II students, but maybe if they've seen some simple differential equations (separation of variables, for example) already?&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-17T18:33:04.570" CommentCount="1" />
  <row Id="14705" PostTypeId="2" ParentId="14686" CreationDate="2010-12-17T18:36:41.083" Score="11" ViewCount="0" Body="&lt;p&gt;First, the integral of $x$ from $-1$ to $1$ is &lt;strong&gt;a number&lt;/strong&gt;, not a function. Definite integrals give you the net signed area between the $X$-axis and the function. So the integral of $x$ from $-1$ to $1$ is &lt;em&gt;neither&lt;/em&gt; $x^2$  &lt;em&gt;nor&lt;/em&gt; $x^2/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the &lt;em&gt;indefinite&lt;/em&gt; integral of $x$ is a &lt;em&gt;family&lt;/em&gt; of functions, namely, the family of all functions $F(x)$ such that $F&amp;#39;(x)=x$; that is, the family of antiderivatives of $x$. Since the indefinite integral is a &lt;em&gt;family&lt;/em&gt; of functions and not a single function, the indefinite integral of $x$ is &lt;em&gt;neither&lt;/em&gt; $x^2/2$  &lt;em&gt;nor&lt;/em&gt; $x^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by the Fundamental Theorem of Calculus, if you are trying to compute the definite integral of a function $f(x)$ that is continuous on $[a,b]$, and you are lucky enough to have &lt;em&gt;any&lt;/em&gt; antiderivative $F(x)$ of $f(x)$, then you can compute the definite integral as the Total Change of $F(x)$ on $[a,b]$. That is,&#xA;$$\int_a^b f(x)\,dx = F(b)-F(a).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to compute &#xA;$$\int_{-1}^1 x\,dx$$&#xD;&#xA;you have several choices:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the net signed area geometrically. This is very easy in this case, because all you have is two triangles that cancel each other, so the net signed area is $0$. That is, $\int_{-1}^1 x\,dx = 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the definition of the integral as a limit of Riemann sums. Again, not hard to do: the function $y=x$ is increasing, the left hand and right hand sums are easy to compute, and the limit turns out to be $0$. For example, the left hand Riemann sum with $n$ intervals, each of equal length $\frac{2}{n}$, is:&#xA;\begin{align*}&#xA;\mathrm{LHS}(n) &amp;amp;= \frac{-n}{n}\left(\frac{2}{n}\right) + \frac{-n+2}{n}\left(\frac{2}{n}\right) + \cdots + \frac{n-2}{n}\left(\frac{2}{n}\right)\\&#xA;&amp;amp;= \frac{2}{n^2}\Bigl(-n + (-n+2) + (-n+4)+\cdots+\bigl(-n+(2n-2)\bigr)\Bigr)\\&#xA;&amp;amp;= \frac{2}{n^2}\Bigl( -n(n) + 2\bigl(0 + 1 + 2 + \cdots + (n-1)\bigr)\Bigr)\\&#xA;&amp;amp;= \frac{2}{n^2}(-n^2 + n^2 - n) = \frac{-2n}{n^2} = -\frac{2}{n}.&#xA;\end{align*}&#xA;Thus, &#xA;$$\lim\limits_{n\to\infty}\mathrm{LHS}(n) = \lim_{n\to\infty}\left(-\frac{2}{n}\right) = 0.$$&#xD;&#xA;The right hand sums also converge to $0$, so the integral is equal to $0$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the Fundamental Theorem of Calculus and find some function $F(x)$ whose derivative is $x$ on $[-1,1]$. Now, $F(x)=x^2$  &lt;strong&gt;does not work&lt;/strong&gt; because $F&amp;#39;(x) = 2x\neq x$. But $\mathcal{F}(x) = \frac{1}{2}x^2$  &lt;em&gt;does&lt;/em&gt; work because $\mathcal{F}&amp;#39;(x) = \frac{1}{2}(2x) = x$. That means (by the Constant Function Theorem) that&#xA;$$\int x\,dx = \frac{1}{2}x^2 + C,$$&#xD;&#xA;(the family of functions that are the antiderivatives of $x$ are the vertical translates of $\frac{1}{2}x^2$) so you can use any one of them (say, the one with $C=0$) to compute $\int_{-1}^1x\,dx$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T03:27:09.293" LastActivityDate="2010-12-18T03:27:09.293" />
  <row Id="14706" PostTypeId="1" CreationDate="2010-12-17T19:14:30.613" Score="1" ViewCount="40" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand a bit about chern classes/chern characters (in the algebraic setting, for varieties over C say) and was hoping to find some notes describing some well-known properties of ch (e.g. the relation between the support of a coherent sheaf and the component of ch of the same dimension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really hoping of something in the middle between Hartshorne's appendix and say Fulton's intersection theory (or Fulton-Lang RR algebra), which concentrates more on coherent sheaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know, I'm asking too much)&lt;/p&gt;&#xA;" OwnerUserId="1007" LastActivityDate="2010-12-17T19:14:30.613" CommunityOwnedDate="2010-12-17T19:21:42.383" Title="notes on properties of chern character of coherent sheaves" Tags="&lt;soft-question&gt;&lt;algebraic-geometry&gt;&lt;reference-request&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="14707" PostTypeId="2" ParentId="14685" CreationDate="2010-12-17T19:14:59.990" Score="1" ViewCount="0" Body="&lt;p&gt;The volume of a pyramid is one-third the height times the area of the base. In your case you have sliced off a tiny (approximate) pyramid from the top, to create the drain hole, so subtract that volume from the total.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-17T19:14:59.990" />
  <row Id="14708" PostTypeId="2" ParentId="14679" CreationDate="2010-12-17T19:28:05.547" Score="0" ViewCount="0" Body="&lt;p&gt;After changing $f$ by a suitable linear function you may assume $f(a)=f(b)=f&amp;#39;(c)=0$. Expanding $f$ at $c$ by Taylor's theorem you get $f(x)=f(c)+{1\over2}(x-c)^2 f&amp;#39;&amp;#39;(\xi)$ where $\xi$ lies between $c$ and $x$. Now put $x:=a$ and $x:=b$ and draw conclusions.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-12-18T08:33:46.127" LastActivityDate="2010-12-18T08:33:46.127" />
  <row Id="14709" PostTypeId="1" AcceptedAnswerId="14710" CreationDate="2010-12-17T19:31:27.710" Score="3" ViewCount="109" Body="&lt;p&gt;Same as title. &#xA;What is the difference between Godel's Completeness and Incompleteness Theorems?&lt;/p&gt;&#xA;" OwnerUserId="4878" LastActivityDate="2010-12-17T19:47:33.163" Title="What is the difference between Godel's Completeness and Incompleteness Theorems?" Tags="&lt;logic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14710" PostTypeId="2" ParentId="14709" CreationDate="2010-12-17T19:47:33.163" Score="7" ViewCount="0" Body="&lt;p&gt;First, note that, in spite of their names, one is not a negation of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The completeness theorem applies to any first order theory: If $T$ is such a theory, and $\phi$ is a sentence (in the same language) and any model of $T$ is a model of $\phi$, then there is a (first-order) &lt;em&gt;proof&lt;/em&gt; of $\phi$ using the statements of $T$ as axioms. One sometimes says this as &quot;anything true is provable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The incompleteness theorem is more technical. It says that if $T$ is a first-order theory that is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Recursively enumerable (i.e., there is a computer program that can list the axioms of $T$),&lt;/li&gt;&#xA;&lt;li&gt;Consistent, and &lt;/li&gt;&#xA;&lt;li&gt;Capable of interpreting some amount of Peano arithmetic (typically, one requires the fragment known as Robinson's Q), &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;then $T$ is not complete, i.e., there is at least one sentence $\phi$ in the same language as $T$ such that there is a model of $T$ and $\phi$, and there is also a model of $T$ and $\lnot\phi$. Equivalently (by the completeness theorem), $T$ cannot prove $\phi$ and also $T$ cannot prove $\lnot\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One usually says this as follows: If a theory is reasonable and at least modestly strong, then it is not complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second incompleteness theorem is more striking. If in fact we require that $T$ interprets Peano Arithmetic, then in fact $T$ cannot prove its own consistency. So: There is no way of proving the consistency of a reasonably strong mathematical theory, unless we are willing to assume an even stronger setting to carry out the proof. Or: If a reasonably strong theory can prove its own consistency, then it is in fact inconsistent. (Note that any inconsistent theory proves anything, in particular, if its language allows us to formulate this statement, then it can prove that it is consistent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement that $T$ is recursively enumerable is reasonable, I think. Formally, a theory is just a set of sentences, but we are mostly interested in theories that we can write down or, at least, for which we can recognize whether something is an axiom or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretability requirement is usually presented in a more restrictive form, for example, asking that $T$ is a theory about numbers, and it contains Peano Arithmetic. But the version I mentioned applies in more situations; for example, to set theory, which is not strictly speaking a theory about numbers, but can easily interpret number theory. The requirement of interpreting Peano Arithmetic is two-fold. First, we look at theories that allows us (by coding) to carry out at least some amount of common mathematical practice, and number theory is singled out as the usual way of doing that. More significantly, we want some amount of &quot;coding&quot; within the theory to be possible, so we can talk about sentences, and proofs. Number theory allows us to do this easily, and this is way we can talk about &quot;the theory is consistent&quot;, a statement about proofs, although our theory may really be about numbers and not about first order formulas.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-17T19:47:33.163" CommentCount="1" />
  <row Id="14711" PostTypeId="1" CreationDate="2010-12-17T19:57:28.873" Score="-4" ViewCount="160" Body="&lt;p&gt;In &lt;a href=&quot;http://www.cs.dal.ca/~eem/fea-knuth.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; Knuth talks about the bottom 10 problems rather than top 10. The problems which are not solved but they are most ready to be solved of all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are they?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T20:02:23.957" LastActivityDate="2010-12-17T23:10:27.420" ClosedDate="2010-12-18T03:06:13.463" Title="What are the bottom 10 theorems of number theory?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="14712" PostTypeId="2" ParentId="14693" CreationDate="2010-12-17T20:13:32.510" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt;.  step one is to change the 3 into 3.25  minus   1/4&lt;/p&gt;&#xA;" OwnerUserId="4879" LastActivityDate="2010-12-17T20:13:32.510" />
  <row Id="14713" PostTypeId="2" ParentId="14658" CreationDate="2010-12-17T20:16:16.860" Score="3" ViewCount="0" Body="&lt;p&gt;Since you are into signal processing, you might like the proof using Parseval's theorem. I have paraphrased this into the signal processing language from the original proof #4 of &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;Robin Chapman's collection of proofs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $e_n = e^{2\pi inx}$ where $n \in \mathbb Z$. Let $f(x) = x$ in the interval $[0,1]$ and we compute its Fourier series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f (x) = \sum_n a_n e^{2\pi inx}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in your terms, Parseval's theorem would mean that the energy computation in the time domain is identical to the energy computation in the frequence domain. To compute energy in the time domain, we integrate the square of the abs. value of the function, and to compute energy in the Frequency domain, we sum the squares of the abs. values of the Fourier coefficients. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^1 x^2 dx\ =\ \sum |a_n|^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As R. Chapman remarks, the left side is $1/3$, and we have $a_0 = 1/2$ and $a_n = 1/2\pi in$ for $n \neq 0$. So the above simplifies to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{3} = \frac{1}{4} + \underset{n\in \mathbb Z , n \neq 0}{\sum} \frac{1}{4\pi n^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the result follows.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Incidentally, note here that the explicit calculation was not necessary to prove finite energy. The much simpler way is to note that your signal can be bounded by the Fourier series of some other signal, and then observe that energy of that signal as seen in the time domain is finite.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="1119" LastEditorDisplayName="" LastEditDate="2010-12-17T20:24:22.643" LastActivityDate="2010-12-17T20:24:22.643" CommentCount="3" />
  <row Id="14714" PostTypeId="1" AcceptedAnswerId="14720" CreationDate="2010-12-17T20:47:52.980" Score="2" ViewCount="73" Body="&lt;p&gt;I've recently studied characteristic functions in my probability course and I can't get &lt;i&gt;why&lt;/i&gt; we define it to be the Fourier transform of the distribution (if the random variabile is continuous). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean that if $X$ is a random variable, $\varphi_X (t) = \mathbb{E}(e^{i t X}) = \int_{-\infty}^{+\infty} e^{i t x}f_X(x) dx$ where $f_X(x)$ is the distrubution function of $X$, and I can't see any motivation for doing this. I asked my professor but he wasn't clear at all; he said something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Since we proved the theorem that if $\varphi_X (t) = \varphi_Y (t)$ then $X \sim Y$ (or $P_X \equiv P_Y)$, it is natural to define it this way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, to prove that we need the definition! So I couldn't really make up my mind about it, if you could provide some help in this sense (motivation for defining the characteristic function of a random variable as the Fourier transform of its distribution) it would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditorDisplayName="" LastEditDate="2010-12-17T20:53:33.617" LastActivityDate="2010-12-18T00:18:27.023" Title="Characteristic Functions and motivations" Tags="&lt;probability-theory&gt;" AnswerCount="2" />
  <row Id="14715" PostTypeId="1" AcceptedAnswerId="14716" CreationDate="2010-12-17T20:56:19.790" Score="3" ViewCount="140" Body="&lt;p&gt;First,&#xA;to understand my question, checkout this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/14573/calculating-percentages-for-taxes&quot;&gt;Calculating percentages for taxes.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, &#xA;consider that I'm a layman in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after trying to understand the logic used to get the final result. I was wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why multiply $20,000 by 100 and then divide by 83? I know this is the rule of three, but, I can't understand the &quot;internals&quot;  of this approach. It isn't intuitive as think in this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say 100% of one value, is the same of divide this value by 100. In other words: I have 100 separeted parts of this integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's intuitive think about the taxes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X - 17\% = \$20.000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\$20.000 = 83\%$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, the most easy and compreensive way to solve this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\$20.000 / 83 = 240.96$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the same as think, if 100% is 100 parts of one integer, 83% of that integer is the same of divide this integer by 83.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally to get the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\$20.000 + 17 * 240.96$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My final question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I think intuitively like this using the Rule of Three?&#xA;In other words, why multiply 20.000 by 100 and then divide by 83 is a shorcut to get the result? &lt;/p&gt;&#xA;" OwnerUserId="1425" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-17T21:36:21.000" LastActivityDate="2010-12-20T01:10:44.737" Title="The logic behind the rule of three on this calculation." Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="14716" PostTypeId="2" ParentId="14715" CreationDate="2010-12-17T21:07:33.343" Score="6" ViewCount="0" Body="&lt;p&gt;The Rule of Three is just doing the same thing you are doing: if $\$20,000$ is $83$%, then $\frac{20000}{83}$ is $1$% of the original. Since $X$ is the original, then $\frac{X}{100}$ is &lt;strong&gt;also&lt;/strong&gt; $1$% of the original. That means that the two quantities are equal, so we have:&#xA;$$\frac{20,000}{83} = \frac{X}{100}.$$&#xD;&#xA;From here it is easy to solve for $X$:  &lt;em&gt;Edited:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both sides are equal, if you do the same thing to both sides you will get equal quantities. Or, intuitively, since each of these quantities is 1% of what we want, multiplying them by 100 will give us 100% (which is what we want). So we have&#xA;$$100\left(\frac{20,000}{83}\right) = 100\left(\frac{X}{100}\right).$$&#xD;&#xA;Now, on the right hand side, the $100$ cancels with the $100$ on the denominator (which again, makes sense: remember that $X$ was the total, $\frac{X}{100}$ was 1% of $X$, so a hundred times 1% of $X$ gives us back $X$), so we are left with&#xA;$$\frac{(100)(20,000)}{83} = X.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the Rule of Three works because you have proportionality: if $a$ is to $b$ like $x$ is to $c$, (if $\$20,000$ is to $83$% like $X$ is to $100$%)&#xA;$$\begin{array}{ccc}&#xD;&#xA;a &amp;amp; \longrightarrow &amp;amp;b\\\&#xD;&#xA;x &amp;amp; \longrightarrow &amp;amp;c&#xD;&#xA;\end{array}$$&#xD;&#xA;then that means that&#xA;$$\frac{a}{b} = \frac{x}{c}.$$&#xD;&#xA;which is solved by multiplying through by $c$, which cancels the $c$ in the denominator on the right hand side; it comes out to exactly the same as the Rule of Three says: multiply across the diagonal of known quantities (in this case, $a$ and $c$) and divide by the remaining one (in this case, $b$). The point of the Rule of Three is to do it by rote and not having to think about setting everything up carefully to get the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also works if your unknown were the percentage: if $\$20,000$ is $87$%, then what percentage of the original is $\$15,575$? (I just made that number up) Well, using the logic from above, since $\frac{20,000}{87}$ is $1$%, and $\frac{15,575}{p}$ is $1$% (where $p$ is the percentage we are looking for) then we have&#xA;$$\frac{20,000}{87} = \frac{15,575}{p}$$&#xD;&#xA;which solving for $p$ by first multiplying across by $p$, then multiplying across by $87$, and finally  dividing through by $20,000$ (aka &quot;cross-multiplying&quot;):&#xA;\begin{align}&#xA;\frac{20,000}{87} &amp;amp;= \frac{15,575}{p}\\&#xA;\frac{20,000p}{87} &amp;amp;= \frac{15,575p}{p}\\&#xA;\frac{20,000p}{87} &amp;amp;= 15,575\\&#xA;\frac{20,000p(87)}{87} &amp;amp;= (15,575)(87)\\&#xA;20,000p &amp;amp;= (15,575)(87)\\&#xA;\frac{20,000p}{20,000} &amp;amp;= \frac{(15,575)(87)}{20,000}\\&#xA;p &amp;amp;= \frac{(15,575)(87)}{20,000} \approx 67.75&#xA;\end{align}&#xA;so the percentage here is approximately $67.75$%. Using the Rule of Three, you have that $20,000$ is to $87$ like $15,575$ is to $p$, so&#xA;$$\begin{array}{ccc}&#xD;&#xA;20,000 &amp;amp;\longrightarrow &amp;amp; 87\\\&#xD;&#xA;15,575 &amp;amp; \longrightarrow &amp;amp; p&#xD;&#xA;\end{array}$$&#xD;&#xA;so to get $p$, you multiply across the known diagonal ($15,575$ times $87$) and divide by the remaining quantity ($20,000$), same operation as before.&#xA;&lt;em&gt;Added:&lt;/em&gt; Again, intuitively, what we are doing is noting that if $20,000$ is 87%, then $\frac{20,000}{87}$ is 1% of the total; we want to know the $p$ such that $15,575$ is $p$%; whatever it is, $\frac{15,575}{p}$ will also be $1$%, so the two quantities are equal. Multiplying both sides by $p$ makes both sides equal to $p$%. of the total. Multiplying both sides by 87 makes both sides $87p$&amp;amp;%. Since $20,000$ is $87$%, dividing both sides by $20,000$ will give $p$, the percentage we are looking for (since we have $87p$% divided by $87$%). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But it's important that the Rule of Three only works when you &lt;em&gt;do&lt;/em&gt; have proportionality. It works for percentages, it works for linear aggregates, but it does not work for things like exponential growth or decay, because then you don't have proportionality between growth and time elapsed.)&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T01:10:44.737" LastActivityDate="2010-12-20T01:10:44.737" CommentCount="6" />
  <row Id="14717" PostTypeId="2" ParentId="14715" CreationDate="2010-12-17T21:10:33.163" Score="3" ViewCount="0" Body="&lt;p&gt;I think that the form to see this more easy,is to see it as proportionally; is say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{X}{100}=\frac{20000}{83}$$ (Are equals as they have the same proportion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus only, you must solve the equation for $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $$X= \frac{(20000)(100)}{83}$$ Passing to multiply to 100, remember this is solved in cross multiplying .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arturo had been clear in the other post.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-17T21:18:58.150" LastActivityDate="2010-12-17T21:18:58.150" />
  <row Id="14718" PostTypeId="1" CreationDate="2010-12-17T21:12:37.750" Score="3" ViewCount="67" Body="&lt;p&gt;I am looking for the the method of constructing the midpoint of two points in spherical geometry. The only tools allowed for the construction are a pair of spherical compasses and a spherical ruler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Eclidean geometry constructing the midpoint is relatively easy. We are looking for the midpoint of points A and B. We construct two circles on A and B with the radiuses of AB. Then we construct two straight lines. One is through the two intersections of the two circles and one is through A and B. The intersection of these two lines will give the midpoint of A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the Euclidean method of construction does not work in Spherical geometry. The circles do not intersect when the distance of our two points exceeds 120°. There is also no solution when their distance is exactly 90°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you construct the midpoint of two points in spherical geomerty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="4881" LastActivityDate="2010-12-17T22:40:25.707" Title="How to construct the midpoint in spherical geometry?" Tags="&lt;geometry&gt;&lt;spherical-geometry&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="14719" PostTypeId="2" ParentId="5510" CreationDate="2010-12-17T21:19:40.990" Score="1" ViewCount="0" Body="&lt;p&gt;I have used the KKT conditions numerous times, and am working on a project now that involves using the KKT conditions along with convexity in non-negative matrix factorization, but I'm having trouble finding a good, two-dimensional, image of why the KKT conditions work.  Although I understand what is going on, I have never been one who is able to draw what makes sense.  Does anyone know where I can find a good image to show why these conditions work with a convex function?&lt;/p&gt;&#xA;" OwnerUserId="4882" LastActivityDate="2010-12-17T21:19:40.990" CommentCount="2" />
  <row Id="14720" PostTypeId="2" ParentId="14714" CreationDate="2010-12-17T21:20:48.227" Score="4" ViewCount="0" Body="&lt;p&gt;The reason we care about the Fourier transform of a distribution is that it has some useful properties.  The term &quot;characteristic function&quot; is, of course, just a label.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a comparison with some of the other kinds of generating functions for random variables would be helpful.  These include:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Probability-generating_function&quot; rel=&quot;nofollow&quot;&gt;Probability generating function&lt;/a&gt;, $E[t^X]$ (also known as the factorial moment generating function),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Moment-generating_function&quot; rel=&quot;nofollow&quot;&gt;Moment generating function&lt;/a&gt;, $E[e^{tX}]$ (which is the Laplace transform if $X$ is nonnegative),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;Characteristic function&lt;/a&gt;, $E[e^{itX}]$ (which, as you have stated, is the Fourier transform).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To quote from Casella and Berger's &lt;em&gt;Statistical Inference&lt;/em&gt; (first edition, p. 84), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Perhaps the most useful of all these types of functions is the characteristic function...  The characteristic function does much more than the mgf [moment generating function] does.  When the moments of $F_X$ exist, $\phi_X$ can be used to generate them, much like an mgf.  The characteristic function always exists [unlike the mgf] and it completely determines the distribution.  That is, every cdf has a unique characteristic function.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then give an example of a theorem that, for mgf's, has some qualifications to it, but that doesn't require any for characteristic functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate on the examples in the quote, the characteristic function has the two following useful properties (although it has more):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It can be used to find the moments $E[X^n]$,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It can be used to show that a transformation of a random variable from a particular distribution has some other known distribution.  This method is often much easier than using the cdf technique for transformations.  (See, for example, &lt;a href=&quot;http://math.stackexchange.com/questions/10318/sum-of-two-independent-geometric-random-variables/10371#10371&quot;&gt;this answer&lt;/a&gt; to a recent math.SE question.  The answer uses mgf's rather than characteristic functions, but the idea is similar.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See also the Wikipedia page for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;characteristic function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-17T21:20:48.227" />
  <row Id="14721" PostTypeId="1" AcceptedAnswerId="14727" CreationDate="2010-12-17T22:21:55.943" Score="6" ViewCount="123" Body="&lt;p&gt;I haven't found any useful method to solve the following problem: Prove that if $x,y,z\in\mathbb{Z}$ and $x^3+y^3=3z^3$ then $xyz=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&amp;amp;t=382377&quot; rel=&quot;nofollow&quot;&gt;http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&amp;amp;t=382377&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2856" LastEditorUserId="2856" LastEditorDisplayName="" LastEditDate="2010-12-17T22:27:55.223" LastActivityDate="2010-12-17T23:34:26.267" Title="Three variable, third degree Diophantine equation" Tags="&lt;number-theory&gt;&lt;diophantine-equations&gt;" AnswerCount="1" />
  <row Id="14722" PostTypeId="2" ParentId="14718" CreationDate="2010-12-17T22:40:25.707" Score="2" ViewCount="0" Body="&lt;p&gt;Draw a great circle $L$ between $A$ and $B$. Set your compass to a small distance, say $r$.  Mark off distance $r, 2r, 3r, \ldots$ along $L$, moving from $A$ to $B$.  Now do the same from $B$ to $A$.  Count to find the middle interval and subdivide that. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2010-12-17T22:40:25.707" CommentCount="3" />
  <row Id="14723" PostTypeId="1" AcceptedAnswerId="14726" CreationDate="2010-12-17T22:44:06.563" Score="4" ViewCount="138" Body="&lt;p&gt;There's a puzzle where you have 3 houses and 3 utilities. You must draw lines so that each house is connected to all three utilities, but the lines cannot overlap. However, I'm fairly sure that the puzzle is impossible. How is this proved?&lt;/p&gt;&#xA;" OwnerUserId="4885" LastEditorUserId="4885" LastEditorDisplayName="" LastEditDate="2010-12-17T23:12:03.403" LastActivityDate="2010-12-18T02:28:38.790" Title="3 Utilities | 3 Houses puzzle?" Tags="&lt;graph-theory&gt;&lt;puzzle&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="14724" PostTypeId="2" ParentId="14711" CreationDate="2010-12-17T23:10:27.420" Score="2" ViewCount="0" Body="&lt;p&gt;Well... this is somewhat subjective but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Knuth is hinting that most mathematicians (and I'm guessing most computer scientists) work on problems that are not particularly glamorous and not immediately important, nor do they have particularly elegant proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of the phrase &quot;bottom ten&quot; appears to be a metaphor, referring to problems that could be solved (using well-established techniques), but haven't yet been solved.  There's no actual list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;stones that make up the wall&quot; refers to that important mathematical discoveries are typically not made by one person working hard on a problem, but by a host of contributors over many years, each making incremental progress on the problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have seen a little further it is by standing on the shoulders of Giants. -- Isaac Newton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PS.  I can understand that Knuth's response might be surprising to people without research experience.  Why would he want to work on the least interesting problems?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T23:10:27.420" />
  <row Id="14725" PostTypeId="2" ParentId="9524" CreationDate="2010-12-17T23:22:30.023" Score="1" ViewCount="0" Body="&lt;p&gt;Some CS / computer engineering things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Division_%28digital%29#Newton.E2.80.93Raphson_division&quot; rel=&quot;nofollow&quot;&gt;Newton-Raphson&lt;/a&gt; division is a clever way that you can do division with only using multiplication / addition. Telling people that that's how calculators do division might be interesting, and it is obviously relevant to ECE majors. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;the analysis of its error&lt;/a&gt; for how it relates to Taylor series.)&lt;/li&gt;&#xA;&lt;li&gt;You can use infinite sums to calculate pi, e, etc.&lt;/li&gt;&#xA;&lt;li&gt;More relevant to calc is the fact that numerical integration methods use e.g. Simpson's method, the error analysis of which is related to Taylor series.&lt;/li&gt;&#xA;&lt;li&gt;(Also, here Calc II is the same course people get introduced to linear algebra, and I wish someone had told me that PageRank was an eigenvector problem. So some unasked-for advice if it's the same where you are is to introduce eigenvectors this way)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-19T02:34:08.720" LastActivityDate="2010-12-19T02:34:08.720" CommentCount="4" />
  <row Id="14726" PostTypeId="2" ParentId="14723" CreationDate="2010-12-17T23:25:31.300" Score="4" ViewCount="0" Body="&lt;p&gt;It's not hard to see this is impossible. Connect two houses to the three utilities, and you will essentially have a square with one diagonal drawn. The two corners joined are two of the houses, the other two corners and the midpoint of the diagonal are the utilities. (The actual shape may look distorted from this, but it is &lt;em&gt;essentially&lt;/em&gt; this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/kF0IT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the third house? If the house is &quot;outside&quot; the square, you cannot connect it to the utility in the middle. If the house is inside the square, then it is on one of the two sides of the diagonal, and therefore you cannot connect it to the vertex (utility) that is &quot;across&quot; the diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Postscript.&lt;/em&gt; As Moron points out, I am implicitly using &lt;a href=&quot;http://en.wikipedia.org/wiki/Jordan_curve_theorem&quot; rel=&quot;nofollow&quot;&gt;the Jordan curve theorem&lt;/a&gt; which says that a simple closed curve divides the plane into two disjoint regions, so that any path joining a point from the &quot;outside&quot; to a point &quot;inside&quot; has to cross the boundary. I use it when I argue that the house &quot;outside&quot; cannot be connected to the utility &quot;inside&quot; (without crossing the lines), or that you cannot go from one side of the diagonal to the other without crossing the diagonal.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T02:28:38.790" LastActivityDate="2010-12-18T02:28:38.790" CommentCount="7" />
  <row Id="14727" PostTypeId="2" ParentId="14721" CreationDate="2010-12-17T23:25:52.437" Score="7" ViewCount="0" Body="&lt;p&gt;The technique used to prove that $x^3 + y^3 + z^3 = 0$ has no non-trivial solutions in $\mathbb{Z}(\sqrt{-3})$ is also applicable to showing that $x^3 + y^3 = 3z^3$ has no non-trivial solutions (in $\mathbb{Z}(\sqrt{-3})$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this appears (with proof) in section 13.5 as Theorem 232 in the excellent book, &quot;An Introduction to the Theory of Numbers&quot;, by Hardy &amp;amp; Wright, 5th Edition (I have the Indian Edition, so might be a bit different from yours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot I managed to scrape (though the notation is quite old, and you would need parts of the rest of the book to make sense of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/Xxz8L.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T23:34:26.267" LastActivityDate="2010-12-17T23:34:26.267" />
  <row Id="14728" PostTypeId="2" ParentId="14723" CreationDate="2010-12-17T23:41:42.137" Score="2" ViewCount="0" Body="&lt;p&gt;So, the question seems to be asking for an embedding of $K_{3,3}$ in the plane.  This is well-known to be impossible.  But, we live on a sphere, rather than a plane.  However, it's also impossible in this case, which we can visualise using &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereographic_projection&quot; rel=&quot;nofollow&quot;&gt;stereographic projection&lt;/a&gt;.  You can get pretty close, $K_{3,3} \setminus e$ is planar (i.e. delete an edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... thinking outside the box, there are situations where it is possible:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We live in three dimensions, where an embedding of $K_{3,3}$ is easily possible.&lt;/li&gt;&#xA;&lt;li&gt;There doesn't seem to be anything to prevent one of the utilities also being a house.  So in the following graph, red vertices are utilities, blue vertices are houses and the green vertex is both a house and a utility.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vQQul.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-17T23:41:42.137" />
  <row Id="14729" PostTypeId="2" ParentId="14723" CreationDate="2010-12-17T23:42:40.623" Score="1" ViewCount="0" Body="&lt;p&gt;Being able to draw in the plane without edges crossing is the property of planar graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That you cannot solve the puzzle follows readily from the following theorem on planar graphs&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $G(V,E)$ is a planar graph and has&#xA;  no cycles of length $3$, then $|E| \le 2|V| - 4$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In our case $|V| = 6$ (number of nodes) and we need $|E| = 9$ (number of edges). The graph we have cannot have any triangles, as it is bipartite, and any bipartite graph can only have even length cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above theorem can be proven using &lt;a href=&quot;http://www.ics.uci.edu/~eppstein/junkyard/euler/&quot; rel=&quot;nofollow&quot;&gt;Euler's Formula&lt;/a&gt; very easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph you have is known as $K_{3,3}$ and appears in the statement of &lt;a href=&quot;http://mathworld.wolfram.com/KuratowskiReductionTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Kuratowski's Theorem&lt;/a&gt; on planar graphs. This theorem characterizes all planar graphs in terms on forbidding $K_5$ (complete graph on 5 vertices) and $K_{3,3}$ as minors.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-17T23:49:53.640" LastActivityDate="2010-12-17T23:49:53.640" />
  <row Id="14730" PostTypeId="2" ParentId="14723" CreationDate="2010-12-18T00:00:46.380" Score="2" ViewCount="0" Body="&lt;p&gt;Here's a short proof that $K_{3,3}$ is nonplanar.  (It's borrowed from Bondy and Murty, &lt;em&gt;Graph Theory with Applications&lt;/em&gt;, pp. 144-145.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some notation: Let $F$ and $\phi$ denote the set and number, respectively, of faces of a planar embedding of a graph.  Let $V$ and $\nu$, denote the set and number, respectively, of vertices of a graph.  Let $\epsilon$ denote the number of edges in the graph.  Let $d(f)$ denote the degree of the face $f$; i.e., the number of edges incident on $f$.  Let $G^*$ denote the dual graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim: $\sum_{f \in F} d(f) = 2\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: $\sum_{f \in F} d(f) = \sum_{f^* \in V(G^*)} d(f^*) = 2\epsilon^* = 2\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The first step follows from the fact that faces in $G$ correspond to vertices in $G^*$, the second from the well-known fact that the sum of the vertex degrees of a graph is twice the number of edges, and the last from the fact that edges in $G$ correspond to edges in $G^*$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $K_{3,3}$ is planar, and let $G$ be a planar embedding of $K_{3,3}$.  Since $K_{3,3}$ has no cut edges and no cycles of length less than four, every face of $G$ must have degree at least four.  Thus &#xA;$$4\phi \leq \sum_{f \in F} d(f) = 2\epsilon  = 18.$$&#xD;&#xA;Thus $\phi \leq 4$.  But, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_characteristic&quot; rel=&quot;nofollow&quot;&gt;Euler's theorem&lt;/a&gt;, this means $2 = \nu - \epsilon + \phi \leq 6 - 9 + 4 = 1$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-18T00:06:24.820" LastActivityDate="2010-12-18T00:06:24.820" />
  <row Id="14731" PostTypeId="2" ParentId="14714" CreationDate="2010-12-18T00:18:27.023" Score="2" ViewCount="0" Body="&lt;p&gt;Practically speaking, the short answer is that it's convenient.  The characteristic function has better analytic properties than the moment generating function, lets you study all of the moments of a random variable at once, and has the extremely convenient property that $\phi_{X+Y}(t) = \phi_X(t) \phi_Y(t)$ if $X, Y$ are independent.  This makes the characteristic function an amazing tool for understanding sums of independent random variables, and indeed a standard proof of the central limit theorem proceeds via a computation of characteristic functions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many constructions in mathematics translate problems in one domain (understanding distribution functions) to problems in another domain (understanding characteristic functions), and these constructions are useful because different tools apply in the second domain.  That is exactly what happens in the Fourier-theoretic proof of the central limit theorem.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T00:18:27.023" />
  <row Id="14732" PostTypeId="2" ParentId="14588" CreationDate="2010-12-18T00:24:45.910" Score="4" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, the lattice of subgroups, the lattice of normal subgroups, the order of the group, and the automorphism group do not (even taken together) determine the isomorphism type of a finite group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Take G = SmallGroup(243,19) and H = SmallGroup(243,20).  There is a bijection f:L(G)→L(H) between their lattices of subgroups such that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;|X| = |f(X)|&lt;/li&gt;&#xA;&lt;li&gt;X ≅ f(X) unless X = G&lt;/li&gt;&#xA;&lt;li&gt;X ≤ Y iff f(X) ≤ f(Y)&lt;/li&gt;&#xA;&lt;li&gt;X ⊴ G iff f(X) ⊴ f(G) = H&lt;/li&gt;&#xA;&lt;li&gt;G/X ≅ H/f(X) whenever X≠1 is normal&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additionally Aut(G) ≅ Aut(H).  The fourth bullet shows in particular, that f induces an isomorphism between the lattice of quotient groups of G and the lattice of quotient groups of H.  The second and fifth bullets show the isomorphism respects everything about the subgroups' properties as abstract groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The groups have presentations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA; G &amp;amp;= \bigl\langle a,b,c \mid a^{27} = b^{3} = c^{3} = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where }z = a^9\\&#xA;H &amp;amp;= \bigl\langle a,b,c \mid a^{27} = b^3 = c^{3} = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where }z = a^{-9}&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function is induced by a bijection of the underlying sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;f(a&lt;sup&gt;i&lt;/sup&gt; b&lt;sup&gt;j&lt;/sup&gt; c&lt;sup&gt;k&lt;/sup&gt;) = a&lt;sup&gt;i&lt;/sup&gt; b&lt;sup&gt;j&lt;/sup&gt; c&lt;sup&gt;k&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are no such groups of order dividing 64 (even just having an isomorphism of subgroup lattices respecting normal subgroups).&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T06:23:09.310" LastActivityDate="2010-12-18T06:23:09.310" CommentCount="5" />
  <row Id="14733" PostTypeId="2" ParentId="14689" CreationDate="2010-12-18T00:25:23.700" Score="1" ViewCount="0" Body="&lt;p&gt;Here's how I &lt;strong&gt;approximated&lt;/strong&gt; an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know player-2 usually wins. The expected value of player-2 after 9 games is 63, just over 60. So the expected number of turns the game will last is about 9.&#xA;Let's consider another version of this game which is played for 9 turns exactly and the winner is the one who got the higher score.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the score of each player distributed after 9 games? We need to get the distribution of a sum of 18 random variables (because 9 turns correspond to 18 dice throws). The distribution of a sum of a random variable is exactly the &lt;strong&gt;convolution&lt;/strong&gt; of the distributions. This can be efficiently computed by the &lt;strong&gt;convolution theorem&lt;/strong&gt; which says: compute the &lt;strong&gt;Discrete Fourier Transform&lt;/strong&gt; of each of the distributions, multiply the resulting distributions element by element, then compute the Inverse Discrete Fourier Transform of the resulting distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I computed that in python. Now that we got the distribution of the scores for each player after 9 turns. To compute the probability player-2 wins, I computed the cumulative distribution of the score for each player (i.e: the probability of getting score &gt;= i for each i). Now I just had to sum of each score i, the probability of player-1 getting score &amp;lt; i, multiplied by the probability of player-2 getting score &gt;= i.&#xA;The probability for a draw is computed similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating the calculation I get about: 81%,16%,3% (but it's just an approximation anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the accurate numbers you should do this calculation for any number of turns, and then compute the average of these numbers, weighted by the probability of the game to end in any specific number of turns.&#xA;But I expect the calculation for 9 turns to give a very good approximation, because the distribution of the number of turns is centered around 9 and I expect to get similar results of 8 or 10 turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some notes in case you're not familiar with convolutions and Fourier Transform&lt;/strong&gt;:&#xA;Look at a random variable X over some finite probability space of n elements. Think of X as a vector of n real numbers (representing its distribution). Then the operator T, defined as T(Y)=X+Y for any Y (this is &lt;strong&gt;not&lt;/strong&gt; element-wise addition. this is addition of random variables, also called &lt;strong&gt;convolution&lt;/strong&gt;). Then T is a linear operator in $\mathbb{R}^n$. It happens to be diagonizable. The Discrete Fourier Transform is just a change of basis, after which T is a diagonal operator. It may be a little surprising, but this basis does not depend of X (although T does). The i,i element in the diagonal of the matrix representing T in this new basis happens to be Pr(X=i). Computing the compositions of diagonal operators in very easy - just multiply the diagonals element-wise. This is the justification for the convolution theorem. After you're done adding random variables in this new basis, just return to the original basis (i.e: do the inverse Fourier transform) and get the desired probability vector of the addition of all the random variables you added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how the probabilities are calculated in python (I'm supplying all the routines you need):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def dft(x):&#xA;     N = len(x)&#xA;     ret = [0]*N&#xA;     for k in range(N):&#xA;         for n in range(N):&#xA;             ret[k] += x[n]*math.e**((-2*math.pi*(1j)/N)*k*n)&#xA;     return ret&#xA;def idft(x):&#xA;     N = len(x)&#xA;     ret = [0]*N&#xA;     for k in range(N):&#xA;         for n in range(N):&#xA;             ret[k] += x[n]*math.e**((2*math.pi*(1j)/N)*k*n)&#xA;         ret[k] /= N&#xA;     return ret&#xA;def mul_vec(x1,x2):&#xA;     return [a1*a2 for (a1,a2) in zip(x1,x2)]&#xA;def pow_vec(x,n):&#xA;     ret = [1]*len(x)&#xA;     for i in range(n):&#xA;         ret = mul_vec(ret, x)&#xA;     return ret&#xA;def pad(x,n):&#xA;     return x+[0]*(n-len(x))&#xA;p1_probs=idft(pow_vec(dft(pad([0]+[1./6]*6,200)),18))&#xA;p2_probs=idft( mul_vec(pow_vec(dft(pad([0]+[1./6]*6,200)),9) ,  pow_vec(dft(pad([0]+[1./4]*4,200)),9)    )  )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3533" LastEditorUserId="3533" LastEditorDisplayName="" LastEditDate="2010-12-18T12:53:04.927" LastActivityDate="2010-12-18T12:53:04.927" />
  <row Id="14734" PostTypeId="1" CreationDate="2010-12-18T00:35:17.777" Score="0" ViewCount="27" Body="&lt;p&gt;Alright, so I'm trying to check if Quaternion a's y axis is &gt; quaternion b's y axis. How would I do this?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2010-12-18T00:35:17.777" Title="Check if Quaternion a's y axis angle is greater than Quaternion b?" Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;quaternions&gt;" CommentCount="1" />
  <row Id="14735" PostTypeId="1" AcceptedAnswerId="14739" CreationDate="2010-12-18T00:38:05.610" Score="2" ViewCount="166" Body="&lt;p&gt;Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be uniformly continous with $g(0)=0,c\geq 0, c \in \mathbb{R}$. Show: $\exists a\geq 0 \in \mathbb{R}: \forall x \in \mathbb{R}: |g(x)| \leq a \cdot |x|+c$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could also say $g(x) \in \mathcal{O}(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: I could not make up any counterexample so I guess it could be true, all uniformly continous functions I know grow to slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an $\epsilon &amp;gt; 0$, we have because of the continuity of $g$, that: $\exists \delta(\epsilon): |x-y|&amp;lt;\delta=&amp;gt; |g(x)-g(y)|&amp;lt;\epsilon$. Now choose $n=\text{max}\{n \in \mathbb{N}: (n-1)\delta/2\leq|x|\}$. Obviously such a $n$ exists, and $n &amp;gt; 0$. We also easily see that an upper bound for $n$ is $n \leq \frac{2}{\delta}|x|+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use this to seperate $|x|$ into $n-1$ distinct parts of size $s&amp;lt;\delta/2$, and the last part thats smaller than $\delta$ :&#xA;$|x|=|x_1-x_0|+|x_2-x_1|+|x_3-x_2|+...+|x_n-x_{n-1}| &amp;lt; (n-1)\delta/2 + \delta = (n+1)\delta/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow |g(x)|=|g(x_1)-g(x_0)+g(x_2)-g(x_1)+g(x_3)-g(x_2)+...+g(x_n)-g(x_{n-1})| $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\leq |g(x_1)-g(x_0)|+|g(x_2)-g(x_1)|+|g(x_3)-g(x_2)|+...+|g(x_n)-g(x_{n-1})|$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&amp;lt;n \cdot \epsilon \leq (\frac{2}{\delta}|x|+1) \cdot \epsilon = \frac{2\epsilon}{\delta} \cdot |x|+\epsilon$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can see that the constant $c$ we were given can be set as the $\epsilon := c$, and that was also the reason why general speaking $c&amp;gt;0$. Then we can choose $a := \frac{2\epsilon}{\delta}$, as our $\delta$ only depends on the $\epsilon$ and we have that $|g(x)| \leq a \cdot |x| + c$ for $c &amp;gt; 0$. $\square$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="3123" LastEditorUserId="3123" LastEditorDisplayName="" LastEditDate="2010-12-18T22:25:35.347" LastActivityDate="2010-12-18T22:25:35.347" Title="Is a uniformly continuous function vanishing at $0$ bounded by $a|x|+c$?" Tags="&lt;real-analysis&gt;&lt;homework&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="14736" PostTypeId="1" AcceptedAnswerId="14737" CreationDate="2010-12-18T00:39:09.733" Score="2" ViewCount="64" Body="&lt;p&gt;The solution for the number of distributions leaving none of the $n$ cells empty (with unlike cells and $r$ unlike objects) is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A(r,n)=\sum_{\nu=0}^{n-1}(-1)^{\nu}\binom{n}{\nu}(n-\nu)^{r}$$ (although I have seen the same expression summing from $\nu=0$ to $n$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;By the way, if anyone knows which one is correct, please let me know. However, that's not the question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that this expression provides a solution to a famous problem. I'd like to know where can I find more information about this solution and which famous problem solves. I already tried &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691023654&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Combinatorial Analysis&lt;/a&gt; by John Riordan and &lt;a href=&quot;http://www.jstor.org/pss/2688836&quot; rel=&quot;nofollow&quot;&gt;Certain distributions of unlike objects into cells&lt;/a&gt; by Morton Abramson but their treatment of this particular expression is very brief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-18T00:44:35.440" LastActivityDate="2010-12-18T00:58:19.017" Title="Number of distributions leaving none of $n$ cells empty" Tags="&lt;combinatorics&gt;&lt;partitions&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14737" PostTypeId="2" ParentId="14736" CreationDate="2010-12-18T00:58:19.017" Score="3" ViewCount="0" Body="&lt;p&gt;These are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt;.  I am not sure what you mean by a famous problem that this sequence solves; you seem to already have given the standard combinatorial interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stirling numbers of the second kind occur in a certain formula for the sum $\sum_{k=1}^n k^r$; perhaps that's the problem you're thinking of?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T00:58:19.017" CommentCount="10" />
  <row Id="14739" PostTypeId="2" ParentId="14735" CreationDate="2010-12-18T01:18:16.817" Score="4" ViewCount="0" Body="&lt;p&gt;It is false if $c=0$.  To see this, try to think of a continuous function that grows very rapidly near $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true if $c\gt 0$.  One way to show it is by taking a number of very small steps from $0$ to $x$, small enough to guarantee (using uniform continuity) that the function changes no more than a certain fixed amount at each step.  Trying to write out the details should lead you to what this fixed amount is, and to what value of $a$ will work.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-18T01:18:16.817" CommentCount="12" />
  <row Id="14740" PostTypeId="1" CreationDate="2010-12-18T02:36:11.697" Score="0" ViewCount="156" Body="&lt;p&gt;Linear Algebra - True or False if $Ay = B$ and $Az=B$ then y = z? Why or why not? you may give a counter example to prove it false.&lt;/p&gt;&#xA;" OwnerUserId="4887" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T02:38:43.253" LastActivityDate="2010-12-18T09:34:23.940" Title="Linear Algebra - True or False if Ay = B and Az=B then y = z? Why or why not?" Tags="&lt;linear-algebra&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="14741" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T02:40:21.847" Score="1" ViewCount="0" Body="&lt;p&gt;It is not true if the null space of $A$ is nontrivial: If $v$ is any nonzero vector such that $Av=0$, then $Av=A0$ and, more generally, $Ay$ and $A(y+v)$ are equal, no matter what $y$ is. On the other hand, it should be easy to check now that the question is true if the null space of $A$ is trivial (i.e., if $Av=0$ only for $v=0$).&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-18T02:40:21.847" />
  <row Id="14742" PostTypeId="1" AcceptedAnswerId="14744" CreationDate="2010-12-18T02:41:36.230" Score="1" ViewCount="195" Body="&lt;p&gt;I am looking at the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find all values of $m$ for which the equation $(3m-2)x^2 + 2mx + 3m = 0$ shas only one root between $-1$ and $0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know what only one root between -1 and 0 actually means. I count these possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$-1&amp;lt;x_1=x_2&amp;lt;0$ &lt;/li&gt;&#xA;&lt;li&gt;$x_1&amp;lt;-1&amp;lt;x_2&amp;lt;0$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$-1&amp;lt;x_1&amp;lt;0&amp;lt;x_2$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let $a=3m-2$, $\Delta$ be the discriminant, and $S/2$ the $x$-coordinate of the vertex ($S/2 = -m/(3m-2)$) and also the arithmetic mean value of the roots $(x_1+x_2)/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions to solve 1 alone would be:&#xA;$af(-1)&amp;lt;0$,&#xA;$af(0)&amp;lt;0$ and&#xA;$\Delta \geq 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions to solve 2 alone would be:&#xA;$af(-1)&amp;lt;0$,&#xA;$af(0)&amp;gt;0$,&#xA;$\Delta&amp;gt;0$ and&#xA;$S/2&amp;lt;0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions to solve 3 alone would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$af(-1)&amp;gt;0$,&#xA;$af(0)&amp;lt;0$,&#xA;$\Delta&amp;gt;0$, and&#xA;$S/2&amp;gt;-1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it could also mean that it is all conditions intersected $ 1 \cap 2 \cap 3$&#xD;&#xA;or just $2 \cap 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? the correct answer for this problem according to the book is $0&amp;lt;m&amp;lt;1/2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: Although this question was already answered correctly I would like to see how to use the theorems that I brought in, and the book as well, to solve this. Solving questions any way or another is always welcomed and helpful, what should certainly be done. Yet the purpose in the book is not a general math challenge, it's intended to teach the theorems and aplly them. So if I can, let me try it also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will put the theorems here because no one seems to know what theorems I am refering to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(x)=ax^2+bx+c$ presents two real roots $x_1 \leq x_2$ and $ \alpha $ is the real number that will be compared to $x_1$ and $x_2$, we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$af(\alpha)&amp;lt;0 \implies x_1 &amp;lt; \alpha &amp;lt;x_2$.&lt;/li&gt;&#xA;&lt;li&gt;$af(\alpha)=0 \implies \alpha$ is one of the roots.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $af(\alpha)&amp;gt;0$ and $\Delta \geq 0$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)  $\alpha &amp;lt;x_1 \leq x_2$ if $\alpha &amp;lt; S/2$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) $x_1 \leq x_2 &amp;lt; \alpha$ if $\alpha &amp;gt; S/2$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is two situations in this problem: $-1&amp;lt;x_1&amp;lt;0&amp;lt;x_2$ and $x_1&amp;lt;-1&amp;lt;x_2&amp;lt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I.&lt;/strong&gt;&#xA;$$-1&amp;lt;x_1&amp;lt;0&amp;lt;x_2 \implies af(-1)&amp;gt;0 \wedge af(0)&amp;lt;0 \wedge \Delta &amp;gt; 0 \wedge S/2 &amp;gt; -1.$$&#xD;&#xA;\begin{align*}&#xA;af(-1)&gt;0 &amp;amp;\implies (3m-2)[(3m-2)(-1)^2+2m(-1)+3m]&gt;0\\&#xA; &amp;amp;\implies (3m-2)(4m-2)&gt;0 \implies 12m^2-14m+4&gt;0\\&#xA; &amp;amp;\implies m&amp;lt;1/2\text{ or }m&gt;2/3.\\&#xA;af(0)&amp;lt;0 &amp;amp;\implies (3m-2)[(3m-2)0^2+2m0+3m]&amp;lt;0 \\&#xA;&amp;amp;\implies 9m^2-6m\lt 0 \implies 0\lt m\lt 2/3.\\&#xA;\Delta \gt 0 &amp;amp;\implies (2m)^2-4(3m-2)3m\gt 0 \implies 4m^2-12m(3m-2)\gt 0\\&#xA; &amp;amp;\implies 4m^2-36m^2+24\gt 0 \implies -32m^2+24\gt 0 \implies 0 \lt m \lt 3/4.\\&#xA;S/2\gt -1 &amp;amp;\implies -b/2a\gt -1 \implies -m/(3m-2) \gt -1\\&#xA; &amp;amp;\implies (-m+3m-2)/(3m-2) \gt 0 \implies (2m-2)/(3m-2) \gt 0\\&#xA; &amp;amp;\implies m \lt 2/3 \text{ or } m \gt 1.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(af(-1) \gt 0) \wedge (af(0) \lt 0) \wedge (\Delta \gt 0) \wedge (S/2 \gt -1) \implies 0 \lt m \lt 1/2.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;II.&lt;/strong&gt;&#xA;$$x_1&amp;lt;-1&amp;lt;x_2&amp;lt;0 \implies af(-1)&amp;lt;0 \wedge af(0)&amp;gt;0 \wedge \Delta &amp;gt; 0 \wedge S/2&amp;lt;0.$$&#xD;&#xA;\begin{align*}&#xA;af(-1)&amp;lt;0 &amp;amp;\implies 1/2 \lt m \lt 2/3.\\&#xA;af(0)&gt;0 &amp;amp;\implies m \lt 0\text{ or }m \gt 2/3.\\&#xA;\Delta \gt 0 &amp;amp;\implies 0 \lt m \lt 3/4.\\&#xA;S/2 \lt 0 &amp;amp;\implies -b/2a&amp;lt;0 \implies -m/(3m-2)&amp;lt;0 \\&#xA;&amp;amp;\implies m \lt 0 \text{ or }m \gt 2/3.&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(af(-1)&amp;lt;0 \wedge af(0)&amp;gt;0 \wedge \Delta \gt 0 \wedge S/2 \lt 0) = \emptyset$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As both set of solutions are possible answers, non-exclusory, I think the answer is to unite the two sets: $I \cup II = (0 \lt m \lt 1/2) \cup \emptyset = 0 \lt m \lt 1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final answer $0 \lt m \lt 1/2$.&lt;/p&gt;&#xA;" OwnerUserId="3509" LastEditorUserId="3509" LastEditorDisplayName="" LastEditDate="2010-12-20T15:02:28.303" LastActivityDate="2010-12-20T15:02:28.303" Title="Find the values of  $m$ in the 2nd degree equation $(3m-2)x^2+2mx+3m=0$ so that  it has only one root between -1 and 0." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="14743" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T02:50:05.127" Score="1" ViewCount="0" Body="&lt;p&gt;If $A=B=0$, then we can find many $y\ne z$ such that the equations are satisfied!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-12-18T02:50:05.127" />
  <row Id="14744" PostTypeId="2" ParentId="14742" CreationDate="2010-12-18T03:13:11.230" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\ \ 0\ &amp;gt;\ f(-1)\ f(0)\ =\ 3\:m\:(4m-2)\ \iff\ a &amp;lt; m &amp;lt; b\ $ for $\rm\ a = \ldots,\ b = \ldots$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T03:13:11.230" CommentCount="5" />
  <row Id="14745" PostTypeId="2" ParentId="14742" CreationDate="2010-12-18T03:13:33.220" Score="1" ViewCount="0" Body="&lt;p&gt;Some suggestions... If $ax^2 + bx + c$ is any quadratic with $a \neq 0$, it's a parabola (possibly upside down). The center of the parabola is at $x = -{b \over 2a}$, corresponding to $x = {m \over 2 - 3m}$ in your case. So if ${m \over  2 - 3m}$ lies outside the range $(-1,0)$ then the function is monotone on $[-1,0]$. As a result it has exactly one root in $(-1,0)$ if and only if one of $f(0)$ and $f(-1)$ is positive and one is negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $-1 &amp;lt; {m \over  2 - 3m} &amp;lt; 0$ on the other hand then your function is monotone from $-1$ to ${m \over  2 - 3m}$ and also monotone from ${m \over  2 - 3m}$ to $0$. So you just have to look at the signs of $f(-1)$, $f({m \over  2 - 3m})$, and $f(0)$ and consider the various possibilities.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-18T03:28:01.710" LastActivityDate="2010-12-18T03:28:01.710" CommentCount="3" />
  <row Id="14746" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T03:33:12.267" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad A\ y\ =\ A\ z\ \Rightarrow\ y = z\ \ $ is equivalent to $\rm\ A\ x = 0\ \Rightarrow\ x = 0,\ \ for\ \ \ x = y-z\:$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T03:33:12.267" />
  <row Id="14747" PostTypeId="1" CreationDate="2010-12-18T03:42:25.863" Score="1" ViewCount="96" Body="&lt;p&gt;Is the intersection of a decreasing sequence of countably many stationary subsets of $\omega_1$ always stationary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems elementary, but I fail to find the answer in textbooks.&lt;/p&gt;&#xA;" OwnerUserId="4892" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T04:30:22.303" LastActivityDate="2010-12-18T06:18:33.140" Title="Is the intersection of a decreasing sequence of countably many stationary subsets of $\omega_1$ always stationary?" Tags="&lt;set-theory&gt;" AnswerCount="1" />
  <row Id="14748" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T03:43:03.883" Score="3" ViewCount="0" Body="&lt;p&gt;If we have $Ay = Az$, all we could conclude is $A(y-z) = 0 \Rightarrow y-z \in $Nullspace$(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit in terms of matrices, if $A$ is a square or a skinny matrix, and if $A$ is full rank, then we can conclude that $y=z$. If not, $y$ need not be equal to $z$. All we can say is $y-z \in $Nullspace$(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplest example is $2 + 3 = 1 + 4$ i.e. $[1,1][2,3]^T = [1,1][1,4]^T$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-18T03:48:49.857" LastActivityDate="2010-12-18T03:48:49.857" />
  <row Id="14749" PostTypeId="1" CreationDate="2010-12-18T03:48:20.150" Score="2" ViewCount="43" Body="&lt;p&gt;I'm preparing a talk on lattice point enumeration in polytopes (Ehrhart-Macdonald Theory), and I'd like to have an introduction with a few motivational problems/results which arise from the application of lattice point methods. (The audience consists of junior and senior faculty across the mathematical disciplines.) A few that immediately come to mind are the following: Pick's theorem, Ehrhart reciprocity, Minkowski's bound in algebraic number theory and Hilbert series in Stanley-Reisner rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What kinds of interesting problems have been tackled (or generated some widespread interest) by counting points in lattices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2010-12-18T04:39:52.497" Title="Mathematical Results from Counting Points in Lattices" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;&lt;soft-question&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14750" PostTypeId="2" ParentId="14747" CreationDate="2010-12-18T03:57:42.387" Score="2" ViewCount="0" Body="&lt;p&gt;The answer is no. Here is an example: Take $T_n$ for $n&amp;lt;\omega$ disjoint stationary sets. Let $S_m$, for $m&amp;lt;\omega$, be the union of $T_n$ with $n\ge m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That there are $\omega$ (in fact, $\omega_1$) disjoint stationary subsets of $\omega_1$ is an immediate consequence of the existence of Ulam matrices. (See for example this &lt;a href=&quot;http://caicedoteaching.wordpress.com/2009/02/13/580-cardinal-arithmetic-5/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; of mine. You can also find this in Kunen's book or in Jech's book.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the thing is that the construction suggests itself: Suppose $S_0\supseteq S_1\supseteq\dots$ are stationary sets. If each $S_i\setminus S_{i+1}$ is non-stationary, then $\bigcap_n S_n$ is stationary; in fact, $S_0\setminus \bigcap_n S_n$ is a non-stationary set. So we would need that infinitely often we have $S_i\setminus S_{i+1}$ stationary. By passing to a subsequence if necessary, we may as well assume that it happens for all $i$. But now the construction above is immediate.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-18T06:18:33.140" LastActivityDate="2010-12-18T06:18:33.140" CommentCount="1" />
  <row Id="14752" PostTypeId="2" ParentId="14749" CreationDate="2010-12-18T04:39:52.497" Score="1" ViewCount="0" Body="&lt;p&gt;You may well know this paper already but I think it is very nice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.87.2267&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.87.2267&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-18T04:39:52.497" />
  <row Id="14753" PostTypeId="2" ParentId="14715" CreationDate="2010-12-18T05:16:50.170" Score="7" ViewCount="0" Body="&lt;p&gt;The &quot;rule of three&quot; is an ancient ad-hoc mindless rote rule of inference that is best ignored. Instead, you should strive to learn the general principles behind it - namely, the laws of fraction arithmetic. Let's consider the example at hand. You  seek the number of dollars $\rm\:X\:$ such that when decremented by $\:17\%\:$ yields $\rm\:N\:$ dollars. First, recall that $\rm\  17\%\:$ of $\rm\:X\:$ means $\rm\displaystyle \frac{17}{100}\ X\:.\ $ Thus your equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm N\ =\ X - \frac{17}{100}\ X\ =\ \bigg(1 - \frac{17}{100}\bigg)\ X\ =\ \bigg(\frac{100}{100}-\frac{17}{100}\bigg)\ X\ =\ \frac{83}{100}\ X $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\rm\displaystyle\ \frac{83}{100}\ X\ =\ N\ \ \Rightarrow\ \ X\ =\ \frac{100}{83}\ N\ $ follows by multiplying both sides by $\rm\displaystyle\ \frac{100}{83}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we applied no ad-hoc rules above - just the basic laws of the arithmetic of fractions. &lt;em&gt;These&lt;/em&gt; are the laws that are worthy of mastering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's interesting to look at the decline of the use of the &quot;rule of three&quot; over the last two centuries as the knowledge of general (abstract) algebra evolved. This is very easy using the recently-released &lt;a href=&quot;http://booksearch.blogspot.com/2010/12/find-out-whats-in-word-or-five-with.html&quot; rel=&quot;nofollow&quot;&gt;Google Books Ngram viewer&lt;/a&gt; - which searches for phrases over 5 million books back to 1500. Browsing one of the &lt;a href=&quot;http://books.google.com/books?id=1P0nX_ld-nMC&amp;amp;dq=%22rule%20of%20three%22&amp;amp;pg=PA60#v=onepage&amp;amp;q=%22rule%20of%20three%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;earliest textbooks&lt;/a&gt; in the Google corpus containing the rule of three I noticed that it is immediately followed by a section titled &quot;method of making taxes&quot;. So it seems this was a big application in the old days. Also notice how &quot;fraction arithmetic&quot; really ramped up circa 1960 (perhaps due to &quot;new math&quot; programs?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/FMJK8.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/bHyZw.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-18T07:22:28.113" LastActivityDate="2010-12-18T07:22:28.113" CommentCount="1" />
  <row Id="14754" PostTypeId="1" CreationDate="2010-12-18T05:59:09.357" Score="1" ViewCount="90" Body="&lt;p&gt;Is it true or false that a group of order 12 always has a normal 2-sylow subgroup? &#xA;I have a hunch it is false.. &lt;/p&gt;&#xA;" OwnerUserId="4896" LastActivityDate="2010-12-18T06:46:48.983" Title="Group of order 12" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="14755" PostTypeId="1" AcceptedAnswerId="14758" CreationDate="2010-12-18T06:30:41.093" Score="0" ViewCount="30" Body="&lt;p&gt;Consider an sequence of unique values. Each value is composed of a surname and a first name in that order of significance. Based on that, I can define a complete ordering of values in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can also usefully define less fine-grained equivalence classes of items that have a &quot;consistent&quot; ordering, but where some values compare as equivalent (and have no particular ordering relative to each other) even though in the &quot;full&quot; ordering they compare as greater and lesser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, sometimes, I may want to search an index for a particular surname, not caring about the first name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the most general case, there are finer-grain and coarser-grain but otherwise consistent ways to divide a set of values into equivalence classes and order those classes. There are generally many (mutually incompatible) finer grained orderings for any one coarser-grained ordering. It is possible to have a partitioning that is neither finer-grain nor coarser-grain than another different partitioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is that when I use the terminology of being &quot;consistent&quot;, people get confused. A lot of people won't accept that an ordering in which $a=b$ is consistent with (but coarser grained) than one in which $a&amp;lt;b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that there are likely to be official terms to describe these relationships in abstract algebra, but my abstract algebra is pretty limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? If so, which is the relevant field of abstract algebra, and what are the relevant terms?&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="510" LastEditorDisplayName="" LastEditDate="2010-12-18T07:07:00.830" LastActivityDate="2010-12-18T07:07:00.830" Title="Are there standard terms for more fine/coarse grained (but otherwise consistent) ways of ordering values?" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14756" PostTypeId="2" ParentId="14755" CreationDate="2010-12-18T06:41:11.020" Score="3" ViewCount="0" Body="&lt;p&gt;I would call the relevant maps here &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_theory#Functions_between_orders&quot; rel=&quot;nofollow&quot;&gt;order-preserving surjections&lt;/a&gt;.  Alternately, if instead of using the language of partial orders you want to use the language of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pre-order&quot; rel=&quot;nofollow&quot;&gt;preorders&lt;/a&gt;, you could say that a preorder $\le$ &lt;em&gt;refines&lt;/em&gt; a preorder $\le&amp;#39;$ on the same set if whenever $x \le y$ it is also true that $x \le&amp;#39; y$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T06:41:11.020" />
  <row Id="14757" PostTypeId="2" ParentId="14754" CreationDate="2010-12-18T06:46:48.983" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint&lt;/strong&gt; : Use the second of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sylow_theorems&quot; rel=&quot;nofollow&quot;&gt;Sylow theorems&lt;/a&gt;. All the $p$-Sylow subgroups are conjugate to each other and a group is normal iff it is equal to each of its conjugates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some semidirect product would give you an explicit example.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-18T06:46:48.983" />
  <row Id="14758" PostTypeId="2" ParentId="14755" CreationDate="2010-12-18T06:53:54.980" Score="3" ViewCount="0" Body="&lt;p&gt;The concepts you want are those associated with the notion of partial orders, part of &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_theory&quot; rel=&quot;nofollow&quot;&gt;Order Theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When everything is ordered, you have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_order&quot; rel=&quot;nofollow&quot;&gt;totally ordered set&lt;/a&gt; (also called &lt;em&gt;linearly ordered set&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you only worry about, say, last names, if you do not specify any order relations between the people with same last name (neither is greater than or smaller than any other), then you have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_order&quot; rel=&quot;nofollow&quot;&gt;partially ordered set&lt;/a&gt;. The fact that your total order is &quot;consistent&quot; with the partial order is expressed saying that the total order is a &quot;linear extension&quot; of the partial order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are considering people with the same last name but different first name &quot;equal&quot;, then you are dealing with either a &lt;a href=&quot;http://www.wikipedia.org/wiki/Preorder&quot; rel=&quot;nofollow&quot;&gt;preorder&lt;/a&gt;, or a (partial or total) order on the set of &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_class&quot; rel=&quot;nofollow&quot;&gt;equivalence classes&lt;/a&gt; induced by the preorder. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-18T06:53:54.980" />
  <row Id="14759" PostTypeId="1" AcceptedAnswerId="14760" CreationDate="2010-12-18T06:58:07.130" Score="2" ViewCount="112" Body="&lt;p&gt;This may seem like an overly trivial question, but I've just recently become confused about Langrange's 'prime' notation for derivatives (for example $f'(x)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know for sure that $f&amp;#39;(x) = \frac{\delta f(x)}{\delta x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose we replace x with an expression, like 2x+1. Do we write $f&amp;#39;(x^2+1) = \frac{\delta f(x^2+1)}{\delta x}$ or $f&amp;#39;(x^2+1) = \frac{\delta f(x^2+1)}{\delta (x^2+1)}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does putting the prime around the function instead of between its letter and parentheses make a difference? For example what does $(f(x^2+1))&amp;#39;$ mean?&lt;/p&gt;&#xA;" OwnerUserId="181" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T07:12:54.427" LastActivityDate="2010-12-18T09:14:38.813" Title="Prime notation for derivatives" Tags="&lt;calculus&gt;&lt;notation&gt;&lt;elementary&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14760" PostTypeId="2" ParentId="14759" CreationDate="2010-12-18T07:01:34.240" Score="7" ViewCount="0" Body="&lt;p&gt;$f&amp;#39;$ is a function, so $f&amp;#39;(2x + 1)$ denotes $f&amp;#39;$ applied to $2x + 1$, or $\frac{df}{dx}(2x + 1)$.  For example if $f = x^2$ then $f&amp;#39; = 2x$ and $f&amp;#39;(2x + 1) = 4x + 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f(x^2 + 1))&amp;#39;$ is the derivative of the function $f(x^2 + 1)$, which is $2x f&amp;#39;(x^2 + 1)$ by the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question highlights a weakness of the $&amp;#39;$ notation, which is that it always comes with an implied variable with respect to which you're differentiating.  If this variable is clear from context there's no problem, but sometimes it isn't.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-18T07:01:34.240" CommentCount="2" />
  <row Id="14761" PostTypeId="2" ParentId="2293" CreationDate="2010-12-18T07:52:59.350" Score="2" ViewCount="0" Body="&lt;p&gt;Kakutani's fixed point theorem allowed John Nash to prove his crucial result that started the whole field of game theory in like 9 pages or something -- a very sexy maths thesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brouwer's theorem appears in economics as well in &quot;satisficing&quot; solutions (behavior based on choosing &quot;better&quot; options rather than &quot;the best&quot;), for example &lt;a href=&quot;http://www.springerlink.com/index/39Y2VUHUDC749Q8D.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/index/39Y2VUHUDC749Q8D.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, Brouwer's fixed-point theorem was used to Arrow &amp;amp; Debreu to establish general equilibrium theory. That theory is widely despised but mathematically beautiful, however nowadays it's considered more believable if your paper says &quot;I rely on no fixed-point theorems&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1457" LastActivityDate="2010-12-18T07:52:59.350" CommunityOwnedDate="2010-12-18T07:52:59.350" />
  <row Id="14762" PostTypeId="2" ParentId="14759" CreationDate="2010-12-18T09:14:38.813" Score="4" ViewCount="0" Body="&lt;p&gt;Often people write something like $(x^2)&amp;#39;=2x$ or $(e^x)&amp;#39;=e^x$, but as you noticed yourself, this is ambiguous, and I never use this notation. You can write $\frac{d}{dx}x^2=2x$ instead, i.e., you don't have to put the $x^2$ into the numerator. This way it looks much clearer, I think (though not shorter, unfortunately). So for me it would be&#xA;$$&#xD;&#xA;  \tfrac{d}{dx}f(x^2+1) = 2xf&amp;#39;(x^2+1).&#xD;&#xA;$$&#xD;&#xA;(However, $\tfrac{df}{dx}(x^2+1) = f&amp;#39;(x^2+1)$.)&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-18T09:14:38.813" />
  <row Id="14764" PostTypeId="2" ParentId="14740" CreationDate="2010-12-18T09:34:23.940" Score="0" ViewCount="0" Body="&lt;p&gt;You'd have y=z only if A is an injective linear map. In fact, that's the definition of being injective: different values in the domain map through A to different values in the codomain or if two values in the domain have identical images through A then they are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, this does not hold. As an example, consider the operator T on $\mathbb{R}^3$ &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(x,y,z) = (0,y,z)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;T maps, for example, every triple of the form (a,2,2) to (0,2,2). Therefore, you not only have two, but the whole infinite set {a * (1,2,2) | a $\in$ $\mathbb{R}$} mapping to (0,2,2) through T.&lt;/p&gt;&#xA;" OwnerUserId="4832" LastActivityDate="2010-12-18T09:34:23.940" />
  <row Id="14765" PostTypeId="2" ParentId="3977" CreationDate="2010-12-18T09:53:15.143" Score="0" ViewCount="0" Body="&lt;p&gt;Try kd3-surf if you haven't yet found it at &lt;a href=&quot;http://k3dsurf.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;http://k3dsurf.sourceforge.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free software and small file size, easy to manipulate.&lt;/p&gt;&#xA;" OwnerUserId="4899" LastActivityDate="2010-12-18T09:53:15.143" CommunityOwnedDate="2010-12-18T09:53:15.143" />
  <row Id="14766" PostTypeId="1" CreationDate="2010-12-18T10:24:50.720" Score="2" ViewCount="42" Body="&lt;p&gt;I am mostly asking about terminology here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt; A generalized linear eigenvalue problem is of the form $Ax=\lambda B x$, where $A$ and $B$ are square matrices, $\lambda$ is the eigenvalue and $x$ is the eigenvector. We can write this in the form that makes it evident that it is linear in $\lambda$: $(A-\lambda B) x = 0$. A polynomial eigenvalue problem is of the form&#xA;$$ \left[\sum_{k=0}^N \lambda^k A_k\right] x = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Now, I am wondering what you would call an eigenvalue problem which is &quot;bilinear&quot; in its eigenparameters:&#xA;$$ \left[ A\lambda\mu + B\lambda + C\mu + D \right] x = 0 $$&#xD;&#xA;where $\lambda$ and $\mu$ are the eigenvalues, $x$ is the eigenvector and $A,B,C,D$ are square matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this kind of problem been studied? I know this is generally a nonlinear eigenvalue problem of the form $M(\lambda,\mu)x = 0$, but it seems like there is important additional structure here.&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2010-12-18T10:24:50.720" Title="Multiparameter eigenvalue problem" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" CommentCount="2" />
  <row Id="14767" PostTypeId="1" AcceptedAnswerId="14799" CreationDate="2010-12-18T10:55:01.493" Score="4" ViewCount="151" Body="&lt;p&gt;&lt;strong&gt;Let $F(n)$ be $n$-th Fibonacci number.$$F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, F(4) = 3 \text{ and so on. }$$Given a positive integer $n \gt 2$,Find the smallest prime number $P$ such that $P$ divides $F(n)$ but it does not divide any $F(k)$ smaller than $F(n)$ ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,my question: Is it possible to find the answer  without actually computing the value of $F(n)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an interesting algorithm for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-18T21:17:26.417" LastActivityDate="2010-12-19T16:09:18.567" Title="Finding the  smallest prime number that divides $Fib(n)$ but not any other $Fib(k)$ smaller than $Fib(n)$" Tags="&lt;number-theory&gt;&lt;algorithms&gt;&lt;fibonacci-number&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14768" PostTypeId="1" CreationDate="2010-12-18T11:51:55.097" Score="1" ViewCount="23" Body="&lt;p&gt;Suppose we have a Hadamard matrix $H$ with rows $H_1,H_2,\ldots$ and look at the following system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_j \cdot \mathbf{x} \le 1 \ \forall j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i x_i=1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for polytopes defined in this way?&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-12-18T11:51:55.097" Title="Polytopes defined by Hadamard matrices?" Tags="&lt;geometry&gt;" />
  <row Id="14771" PostTypeId="1" AcceptedAnswerId="14788" CreationDate="2010-12-18T12:59:01.753" Score="2" ViewCount="81" Body="&lt;p&gt;The invariant subspace problem states something along the following lines: Let $H$ be a Banach space of dimension greater than one. Does every bounded linear operator $T:H \to H$ have a non-trivial closed $T$-invariant subspace?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is: Why is this a big problem? What does the solution give us? It has been solved in most of the special cases which occur in practice.&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-18T13:26:47.257" LastActivityDate="2010-12-18T17:46:17.463" Title="Invariant Subspace Problem." Tags="&lt;reference-request&gt;&lt;functional-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14772" PostTypeId="2" ParentId="14771" CreationDate="2010-12-18T13:17:46.513" Score="1" ViewCount="0" Body="&lt;p&gt;The invariant subspace problem asks whether &lt;strong&gt;every&lt;/strong&gt;  bounded linear operator has a non-trivial invariant subspace. The existence of some operator with invariant subspaces is trivial. (Consider finite-rank operators.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is known that the theorem is not true for all Banach spaces so the current formulation asks whether it is true for Hilbert spaces. I do not have any good references however. &lt;/p&gt;&#xA;" OwnerUserId="4623" LastActivityDate="2010-12-18T13:17:46.513" CommentCount="1" />
  <row Id="14774" PostTypeId="2" ParentId="12526" CreationDate="2010-12-18T13:35:44.517" Score="1" ViewCount="0" Body="&lt;p&gt;It's wrong: Take $x=0$ and $y=e_1=(1,0,0,\dots)$. Then $f(x)=e_1$ and $f(y)=e_2=(0,1,0,\dots)$, so $r(x) = 1/2$ and $r(y) = \sqrt2/2$. Now it is not hard to see that $p = (-\sqrt7/4,3/4,0,\dots)$ is the point in $Ty = B(f(y),r(y))\cap B$ farthest away from $f(x)$. From this one deduces that&#xA;$$&#xD;&#xA;  D(Tx,Ty) = \sqrt{(1+\sqrt7/4)^2+(3/4)^2} - 1/2 = \sqrt{2+\sqrt7/2} - 1/2  &amp;gt; 1 = \|x-y\|.&#xD;&#xA;$$&#xD;&#xA;(As $Ty$ is the larger chunk, geometric considerations show that $Ty \subseteq N_r(Tx)$ implies $Tx \subseteq N_r(Ty)$, for all $r&amp;gt;0$.)&lt;/p&gt;&#xA;" OwnerUserId="2575" LastEditorUserId="2575" LastEditorDisplayName="" LastEditDate="2010-12-18T17:38:16.397" LastActivityDate="2010-12-18T17:38:16.397" CommentCount="4" />
  <row Id="14775" PostTypeId="1" AcceptedAnswerId="14784" CreationDate="2010-12-18T13:50:48.087" Score="5" ViewCount="269" Body="&lt;p&gt;Fiction &quot;Division by Zero&quot; By Ted Chiang&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the fiction story &lt;a href=&quot;http://www.fantasticmetropolis.com/i/division/full/&quot; rel=&quot;nofollow&quot;&gt;&quot;Division by Zero&quot; By Ted Chiang&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is the character finds a proof that arithmetic is inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formal proof the fiction can't come true? (I don't suggest the fiction can come true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I see someone &lt;a href=&quot;http://mathoverflow.net/questions/40920&quot; rel=&quot;nofollow&quot;&gt;tried&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-19T08:07:29.193" LastActivityDate="2010-12-20T02:43:19.067" Title="Fiction &quot;Division by Zero&quot; By Ted Chiang" Tags="&lt;meta-math&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="14776" PostTypeId="2" ParentId="14771" CreationDate="2010-12-18T13:54:33.493" Score="4" ViewCount="0" Body="&lt;p&gt;A similar question was asked on MathOverflow a week ago. Take a look at the enlightening answer by Bill Johnson and the discussion in the comments &lt;a href=&quot;http://mathoverflow.net/questions/48908/is-the-invariant-subspace-problem-interesting&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-18T13:54:33.493" CommentCount="1" />
  <row Id="14777" PostTypeId="1" AcceptedAnswerId="14779" CreationDate="2010-12-18T14:08:29.123" Score="1" ViewCount="163" Body="&lt;p&gt;Apparently, the solution to the &lt;a href=&quot;http://math.stackexchange.com/questions/234/card-doubling-paradox&quot;&gt;Card Doubling Paradox&lt;/a&gt; is that a uniform probability distribution over the positive real numbers doesn't exist. Can anyone explain why this is the case and what probability distributions can exist over the positive real numbers (it seems that this would be quite limited, given that such a simple distribution is impossible)?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-18T14:19:27.770" LastActivityDate="2010-12-18T21:04:13.963" Title="Why isn't there a uniform probability distribution over the positive real numbers?" Tags="&lt;probability-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14779" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T15:40:34.400" Score="6" ViewCount="0" Body="&lt;p&gt;A uniform probability over the positive real numbers would not satisfy all three axioms of probability. Specifically, there is a conflict between the second axiom ($P(\Omega) = 1$) and the third axiom (countable additivity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the wikipedia entry &lt;a href=&quot;http://en.wikipedia.org/wiki/Axioms_of_probability&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Axioms_of_probability&lt;/a&gt;. I think it should be clear how a uniform probability over the positive reals would cause problems.&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2010-12-18T15:40:34.400" />
  <row Id="14780" PostTypeId="1" AcceptedAnswerId="14781" CreationDate="2010-12-18T16:19:37.773" Score="2" ViewCount="119" Body="&lt;p&gt;I recently took my Honors Pre-Calc. exam. Wanting to get as many points as I possibly could, I circled and skipped several problems. However, I forgot to do this one, and wanted to know if I'm on the right track, because the answer I get for x doesn't work correctly. I've come up with a separate solution that comes close, but is off by .4 as well. &#xA;\begin{align*}&#xA;e^{2x} - 7e^x + 12 &amp;amp;= 0\\&lt;br&gt;&#xA;e^{2x} - 7e^x &amp;amp;= -12 \\ &#xA;e^x(1^2 - 7) &amp;amp;= -12\\&lt;br&gt;&#xA;e^x &amp;amp;= \frac{-12}{1^2 - 7}\\&lt;br&gt;&#xA;e^x &amp;amp;= 2\\&lt;br&gt;&#xA;x &amp;amp;= \ln 2&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this back into the equation doesn't work. I think I might have a problem with how I'm factoring, but I can't seem to fix it. The only other way I've tried comes closer, but still gives me a wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your help and suggestions are appreciated :).&lt;/p&gt;&#xA;" OwnerUserId="4888" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-18T17:20:32.097" LastActivityDate="2010-12-18T17:21:49.083" Title="Solving an equation with $e^x$" Tags="&lt;algebra-precalculus&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14781" PostTypeId="2" ParentId="14780" CreationDate="2010-12-18T16:24:00.077" Score="12" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\rm\quad  e^{\:2\:x} = (e^x)^2\:.\ $ Thus for $\rm\ y = e^x\ $ the equation is $\rm\ y^2 - 7\ y + 12 = 0\:$&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T16:24:00.077" CommentCount="1" />
  <row Id="14782" PostTypeId="1" CreationDate="2010-12-18T17:03:47.753" Score="1" ViewCount="34" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a finite group which is contained in infinitely many core-free finite groups as the centralizer of a central involution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Brauer–Fowler results show that if a finite group has no non-identity normal subgroups of odd order (so is core-free), then its order is bounded by a function of the size of a few centralizers (of involutions, of strongly real elements, etc.).  In particular, the size of a core-free group is bounded by a function of the size of its largest and smallest involution centralizers, and a simple group by a function of the size of any of its involution centralizers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes one can apparently do better: there are only finitely many core-free groups with a central involution centralizer of size 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an explicit example where one cannot do better?  In other words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the some finite group X such that there are infinitely many finite groups G with no non-identity normal subgroups, and with a Sylow 2-subgroup P and involution t in Z(P) such that C&lt;sub&gt;G&lt;/sub&gt;(t) = X?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-18T17:03:47.753" Title="Involution centralizer does not determine the group" Tags="&lt;group-theory&gt;" />
  <row Id="14783" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T17:03:59.580" Score="3" ViewCount="0" Body="&lt;p&gt;For every probability density $f$, $\lim_{x\to \pm \infty} f(x) = 0$ must hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that for a uniform distribution, the density has to be constant on the considered interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining both requirements, only $g(x) = 0$ remains as a choice for a uniform density on $(-\infty, \infty)$. But $\int_{-\infty}^\infty g(x)dx = 0 \neq 1$, that is $g$ is no proper probability density.&lt;/p&gt;&#xA;" OwnerUserId="3330" LastActivityDate="2010-12-18T17:03:59.580" />
  <row Id="14784" PostTypeId="2" ParentId="14775" CreationDate="2010-12-18T17:16:51.837" Score="4" ViewCount="0" Body="&lt;blockquote&gt;Is there a formal proof the fiction can't come true?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, by &lt;a href=&quot;http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem&quot; rel=&quot;nofollow&quot;&gt;Gödel's second incompleteness theorem&lt;/a&gt;, formal systems can prove their own consistency if and only if they are inconsistent. So given that arithmetic is consistent, we'll never be able to &lt;i&gt;prove&lt;/i&gt; that it is. (&lt;b&gt;EDIT&lt;/b&gt;: Actually not quite true; see Alon's clarification below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, if you liked &quot;Division by Zero,&quot; you might also like Greg Egan's pair of stories in which arithmetic isn't consistent: &quot;Luminous&quot; and &quot;&lt;a href=&quot;http://www.asimovs.com/_issue_0805/DarkINtegers.shtml&quot; rel=&quot;nofollow&quot;&gt;Dark Integers&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastEditorUserId="2773" LastEditorDisplayName="" LastEditDate="2010-12-20T02:43:19.067" LastActivityDate="2010-12-20T02:43:19.067" CommentCount="2" />
  <row Id="14785" PostTypeId="2" ParentId="14780" CreationDate="2010-12-18T17:21:49.083" Score="4" ViewCount="0" Body="&lt;p&gt;Your error is in the third line. $e^{2x}$ is not equal to $e^x$ times $1^2$ (that would make $e^{2x}$ equal to $e^x\times 1 = e^x$). That line would read&#xA;$$e^x(e^x - 7) = -12,$$&#xD;&#xA;and that won't really help you. Bill's solution is the correct approach.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-18T17:21:49.083" />
  <row Id="14786" PostTypeId="1" CreationDate="2010-12-18T17:31:06.293" Score="7" ViewCount="276" Body="&lt;p&gt;I was looking at movie times today and was struck by the oddly-spaced showing times. For example, at the local Loew's Theater &quot;Tron: Legacy 3D&quot; (127 min.) is playing on two screens at the following interlaced times: 1:00 pm, 1:45 pm, 4:00 pm, 4:45 pm, 7:00 pm, 7:45 pm, 10:00 pm and 10:45 pm. Why not space the times equally? Is there an algorithm at work here? Other than optimizing food sales by cleverly keeping a pool of waiters, the strange times might have to do with overbooking and accommodating johnnys-come-lately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following idealized scenario. Suppose only $1$ movie is a playing at a theater with $n$ screens and free popcorn and refreshments is given upon sitting in the theater, so no other factors are relevant for spacing movie times but ticket sales. Suppose each showing can accommodate at most $N$ people. Suppose $N \pm M$ arrive at the kiosk reasonably before any particular showing time, where $0 &amp;lt; M &amp;lt; N$, and $0 &amp;lt; L &amp;lt; M$ people show up just a little too late for any particular show -- the same number of latecomers come by each time. If any person has to wait for more than some fraction $0 &amp;lt; R &amp;lt; 1$ of the time $t$ of the movie in question to watch the next movie in the cue, then he/she returns the ticket and goes home. Suppose the $\pm$ sign above is governed by tossing a fair coin, $+$ for heads, $-$ for tails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Given the above data, what is the optimal spacing of $X$ movie times, each movie of the same length $t$, on $n$ different screens that maximizes the total number of ticket purchases &lt;em&gt;and&lt;/em&gt; (happy) moviegoers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this question is too easy, then generalize the above scenario to multiple movies showing at the same theater. If this question is too hard, then simplify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, feel free to edit and improve.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Added Thoughts&lt;/em&gt;) The constraints above are in place to try to model the scenario as closely as possible while keeping the mathematics simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to account for a little randomness, and the simplest truly non-trivial random event is the tossing of a fair coin. If $N - M$ or $N + M$ people come every time, then the problem is trivial or cumulatively impossible, respectively. What makes this problem tractable is that there are &lt;em&gt;some&lt;/em&gt; occasions when some people are left out of a showing. These people are either at the end of a long cue or literally late; either way they must wait but few, if any, will wait longer than the length of the movie. &lt;em&gt;I believe the answer of spacing depends heavily on the amount of wait time.&lt;/em&gt; That is, if $R = 0$, $L + M &amp;gt; 0$ people go home every time (not optimal). If $R = 1$, then any reasonable spacing should suffice to accommodate the extremely patient moviegoers. I think this possibility oversimplifies the problem, unless I'm missing something crucial or obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose also that the condition $L &amp;lt; M$ could be relaxed to $L &amp;lt; N$, but my reasoning is that latercomers seem to be rarer than overbookers. Are these constraints reasonable?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-12-19T21:24:26.570" LastActivityDate="2010-12-19T21:24:26.570" Title="Going to the Movies!" Tags="&lt;probability&gt;&lt;recreational-mathematics&gt;" CommentCount="10" FavoriteCount="2" />
  <row Id="14787" PostTypeId="1" AcceptedAnswerId="14790" CreationDate="2010-12-18T17:34:06.200" Score="4" ViewCount="135" Body="&lt;p&gt;I'm investigating an idea in cryptography that requires irreducible polynomials with coefficients of either 0 or 1 (e.g. over GF[2]). Essentially I am mapping bytes to polynomials. For this reason, the degree of the polynomial will always be an integer multiple of 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to build a table of such irreducible polynomials for various degrees (e.g. degrees from 8, 16, 24, ..., 1024). Because there are multiple irreducible polynomials for a given degree, I'd like the one with the fewest number of terms since I will hard code the non-zero terms. For example, for degree 16, both of these polynomials are irreducible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{16} + x^{14} + x^{12} + x^7 + x^6 + x^4 + x^2 + x + 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{16} + x^5 + x^3 + x + 1$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the latter one is preferred because it requires less space in code and is more likely to be right (e.g. that I wouldn't have made a copy/paste error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I've noticed that to at least degree 1024 where the degree is a multiple of 8, there are irreducible polynomials of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^n + x^i + x^j + x^k + 1$ where $n = 8*m$ and $0 &amp;lt; i,j,k &amp;lt; 25$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an good algorithmic way of finding these polynomials (or ones that have even fewer terms)? Again, the purpose is to keep the non-zero terms in a look-up table in code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Mathematica code generates all pentanomials for degrees that are multiples of 8 up to degree 1024:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;IrreducibleInGF2[x_] := IrreduciblePolynomialQ[x, Modulus -&amp;gt; 2]&#xA;&#xA;ParallelTable[&#xA; Select[&#xA;   Sort[&#xA;    Flatten[&#xA;        Table[&#xA;           x^n + x^a + x^b + x^c + 1, &#xA;           {a, 1, Min[25, n - 1]}, {b, 1, a - 1}, {c, 1, b - 1}&#xA;             ]&#xA;           ]&#xA;       ], &#xA;       IrreducibleInGF2, 1], &#xA;       {n, 8, 1024, 8}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(I sorted the list of polynomials to make sure I always got the one with the overall smallest degrees first). However, it takes quite a bit of time to run. For example, it took over 26 minutes for the case of $x^{984} + x^{24} + x^9 + x^3 + 1$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HP paper &quot;&lt;a href=&quot;http://www.hpl.hp.com/techreports/98/HPL-98-135.pdf&quot; rel=&quot;nofollow&quot;&gt;Table of Low-Weight Binary Irreducible Polynomials&lt;/a&gt;&quot; has been incredibly helpful. It lists up to $x^{10000}$ and reiterates a proof by Swan that there are no trinomials when the degree is a multiple of 8 (which matches my findings). I've spot checked that their results match mine up to $x^{1024}$ so I'll just need to double check their results up to 10000 which should be much easier than finding them myself.&lt;/p&gt;&#xA;" OwnerUserId="872" LastEditorUserId="872" LastEditorDisplayName="" LastEditDate="2010-12-20T04:01:42.080" LastActivityDate="2010-12-20T17:05:20.637" Title="Finding irreducible polynomials over GF(2) with the fewest terms" Tags="&lt;algorithms&gt;&lt;polynomials&gt;&lt;finite-fields&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14788" PostTypeId="2" ParentId="14771" CreationDate="2010-12-18T17:46:17.463" Score="3" ViewCount="0" Body="&lt;p&gt;One of the primary reasons for studying the invariant subspace problem is the hope that it will lead to useful structural theories. For example, in their interesting expository paper $\ $ &lt;a href=&quot;http://www.springerlink.com/content/9393t358126466x4/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;The invariant subspace problem,&lt;/a&gt; $\ $ Radjavi and Rosenthal wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is hoped  that knowledge of the  invariant subspaces of operators will shed light on their structure.  In  the case of operators on finite-dimensional spaces,  for example, the Jordan canonical form theorem shows that operators &#xA;  are direct sums of particularly well-behaved operators on certain invariant subspaces.  Also, normal operators on infinite-dimensional spaces can be represented as integrals with respect to some invariant subspaces, (their spectral subspaces -  this is the  spectral theorem). To prove that every operator has a non-trivial invariant subspace might be the beginning of a general structure theory. $\ \:$ On the other hand, $\ $ a concrete counterexample would also be very interesting. To say that $A$ has no non-trivial invariant subspace  is equivalent to saying that linear combinations of $\ {A^n\:f\:: n \in \mathbb N}\ $ are dense in the space for each nonzero vector $f$. Representing such an $A$  on various spaces would give a number of approximation theorems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This paper, combined with Yadav's $\ $ &lt;a href=&quot;http://www.math.leidenuniv.nl/~naw/serie5/deel06/jun2005/pdf/yadav.pdf&quot; rel=&quot;nofollow&quot;&gt;The Invariant Subspace Problem&lt;/a&gt; $\ $provide nice expositions on the problem and its rich history. Neither of these introductions are mentioned on said MO page. IMO they provide a much nicer overview of the problem than the little that is said there.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T17:46:17.463" />
  <row Id="14789" PostTypeId="1" CreationDate="2010-12-18T17:58:17.487" Score="3" ViewCount="91" Body="&lt;p&gt;In numerical analysis, I was asked whether Bairstow's algorithm convergence rate is quadratic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial feeling was that it does, since it is essentially Newton's method for a system of non-linear equations, and Newton's method converges quadratically in one dimension (When f is from R to R).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick search on the internet seemed to indicate that indeed, the rate is quadratic as long as the roots are of multiplicity 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have real trouble proving this. I was trying to imitate the proof of Newton's method in one dimesnsion found in Kincaid and Cheney's book &quot;Numerical Analysis&quot;, but ran into some trouble since there isn't a simple &quot;second derivative&quot; for a vector valued function (unless I want to use tensors, and that's way beyond our course material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm stuck. How do I show that the rate of convergence is quadratic (or that it's not if the roots aren't simple)?&lt;/p&gt;&#xA;" OwnerUserId="4209" LastActivityDate="2011-01-07T02:52:52.200" Title="Bairstow's method: Rate of convergence" Tags="&lt;homework&gt;&lt;numerical-methods&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14790" PostTypeId="2" ParentId="14787" CreationDate="2010-12-18T18:00:50.213" Score="4" ViewCount="0" Body="&lt;p&gt;According to a paper &quot;Optimal Irreducible Polynomials for GF(2m) Arithmetic&quot; by M. Scott,&#xA;&quot;it is in practise always possible to chooose as an irreducible polynomial either a trinomial... or a pentanomial.&quot; &lt;a href=&quot;http://www.hyperelliptic.org/SPEED/slides/Scott_optimal_polynomials.pdf&quot; rel=&quot;nofollow&quot;&gt; [talk slides]&lt;/a&gt; &lt;a href=&quot;http://eprint.iacr.org/2007/192&quot; rel=&quot;nofollow&quot;&gt; [PDF link]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In random number generators irreducible trinomials of various degrees with three nonzero binary coefficients are associated with the names &lt;a href=&quot;http://www.ciphersbyritter.com/RES/RNGENS.HTM&quot; rel=&quot;nofollow&quot;&gt; Tausworthe-Lewis-Payne&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; It has been known &lt;a href=&quot;http://arxiv.org/abs/1001.0409&quot; rel=&quot;nofollow&quot;&gt; since Gauss&lt;/a&gt; that there are lots of irreducible polynomials over a finite field, basically &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem#Analogue_for_irreducible_polynomials_over_a_finite_field&quot; rel=&quot;nofollow&quot;&gt; the analog of the Prime Number Theorem&lt;/a&gt; for such polynomials.  Among the $2^m$ (monic) polynomials over Z/2Z of degree $m$, approximately $1/m$ of them are irreducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can eliminate the possibility of first degree factors by inspection, for divisibility by $x$ or $x+1$ would imply respectively a zero constant term or an even number of nonzero terms in the polynomial.  So the first case to test for irreducibility is the trinomials of degree $m$.  With leading and constant coefficients accounting for two of the three nonzero terms, there are but $m-1$ possibilities to test, and by symmetry of $x$ &amp;rarr; $1/x$ substitution, we can restrict the middle terms to degree &amp;le; $m/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If none of those pan out, we have the richer supply of pentanomials to test.  Indeed you seem to have hit upon a seam of cases where &lt;strong&gt;trinomials will never work out&lt;/strong&gt;, namely degree $m$ &lt;a href=&quot;http://u.math.biu.ac.il/~vishne/publications/trinom.ps&quot; rel=&quot;nofollow&quot;&gt; a multiple of 8 [PS]&lt;/a&gt; (Swan, 1962).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work then comes down to testing all the $\binom{m-1}{3}$ binary pentanomials $p(x)$ until we find one that's irreducible.  Your application might make other conditions, perhaps similar to those considered in Scott's paper above, attractive.  Given the modest degrees you are working with, trial division (taking $p(x) \; mod \; q(x)$ for all $q(x)$ of degree &amp;le; $m/2$) should be fast enough. [Remember, we shouldn't have to test more than O(m) possibilities before we find success.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;a href=&quot;http://wwwmaths.anu.edu.au/~brent/pd/ANZMC08t4.pdf&quot; rel=&quot;nofollow&quot;&gt; a fancier way [PDF]&lt;/a&gt; to test polynomials for irreducibility over GF(2).  A &lt;strong&gt;necessary&lt;/strong&gt; condition for binary polynomial $p(x)$ to be irreducible over $GF(2)$ is that:&lt;br&gt;&#xA;$$x^{2^m} = x \mod p(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact Gauss showed for prime q that $x^{q^m} - x$ is precisely the product&#xA;of all monic irreducible polynomials over $GF(q)$ whose degrees divide&#xA;$m$.  [From this he deduced the count of monic irreducible polynomials of&#xA;degree &lt;strong&gt;exactly&lt;/strong&gt; $m$ is asymptotically $q^m/m$ as $m \rightarrow \infty$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $q = 2$ it follows that if $p(x)$ is irreducible of degree $m$,&#xA;it divides $x^{2^m} - x$ over $GF(2)$, i.e. the congruence above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rubin (1980) proposed a necessary and sufficient test for irreducibility,&#xA;combining the above with some additional steps to rule out the possibility&#xA;that $p(x)$ might be the product of some irreducible factors whose degrees&#xA;properly divide $m$.  [While the degrees of the irreducible factors would&#xA;naturally &lt;em&gt;sum&lt;/em&gt; to $m$, having all the irreducible factors' degrees &lt;em&gt;divide&lt;/em&gt;&#xA;$m$ would be somewhat special, unless of course there is only one factor.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional &quot;sufficiency&quot; steps are to check for each &lt;em&gt;prime&lt;/em&gt; factor&#xA;$d_i$ of $m$ that:&#xA;$$GCD(x^{2^{m/d}} - x, p(x)) = 1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if $p(x)$ were to have an irreducible factor of degree $k$ properly&#xA;dividing $m$, it would crop up when taking the gcd of $x^{2^{m/d}} - x$ and&#xA;$p(x)$ if $k$ divides $m/d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then a lot of ingenuity has been applied to efficiently doing these&#xA;steps.  Of course the necessary condition lends itself to &lt;strong&gt;repeated squaring&lt;/strong&gt;,&#xA;computing $x^{2^{k+1}} = (x^{2^k})^2$ mod $p(x)$, for $k$ up to $m-1$.  We&#xA;can take advantage here of the fact that the multiplication we're doing&#xA;is a squaring and advantage of the sparsity of our $p(x)$ as a pentanomial&#xA;when doing reduction mod $p(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the report by Brent of work with Zimmerman (linked above) points out,&#xA;this repeated squaring gives with each (fairly inexpensive step) linear&#xA;progress toward the &quot;exponent of the exponent&quot; $m$.  There is also a way&#xA;to progress farther with greater computational effort by &lt;strong&gt;modular&#xA;composition&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, suppose we've already arrived at:&#xA;$$f(x) = x^{2^k} \mod p(x)$$&#xD;&#xA;and&#xA;$$g(x) = x^{2^j} \mod p(x)$$&#xD;&#xA;Then:&#xA;$$f(g(x)) = x^{2^{k+j}} \mod p(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus composition of two polynomials $f(x)$ and $g(x)$ mod $p(x)$ can&#xA;replace a number of repeated squaring steps.  But composition mod $p(x)$&#xD;&#xA;is more expensive than squaring or even than multiplication generally&#xA;mod $p(x)$.  So as Brent points out, practical advantage for using&#xA;modular composition lies at the final stage(s) of evaluating the&#xA;necessary condition.  E.g. at the end one modular composition might&#xA;replace $m/2$ repeated squarings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the &quot;sufficiency&quot; conditions go, Gao and Panario outlined&#xA;an improvement over a naive implementation of Rubin's tests in this&#xA;&lt;a href=&quot;http://www.math.clemson.edu/faculty/Gao/papers/GP97a.pdf&quot; rel=&quot;nofollow&quot;&gt;&#xA;1997 paper [PDF]&lt;/a&gt;, basically sequencing the gcd computations in&#xA;an expedient order.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-20T17:05:20.637" LastActivityDate="2010-12-20T17:05:20.637" CommentCount="3" />
  <row Id="14791" PostTypeId="2" ParentId="14787" CreationDate="2010-12-18T18:10:42.423" Score="0" ViewCount="0" Body="&lt;p&gt;The abstract of the following paper indicates that it may be of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An Improved Method for the Table Lookup Multiplication Algorithm over $\rm\: GF(2^m\:)$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Author: Hadama Yoichi (Kobe Univ., Graduate School of Sci. and Technol., JPN)&lt;br&gt;&#xA;Nakamura Daisuke (Kobe Univ., Graduate School of Sci. and Technol., JPN)&lt;br&gt;&#xA;Hirotomo Masanori (Hyogojohokyoikukiko)&lt;br&gt;&#xA;Morii Masakatsu (Kobe Univ., Fac. Engineering, JPN)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journal Title: IEIC Technical Report (Institute of Electronics, Information and Communication Engineers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journal Code: S0532B; ISSN:0913-5685 VOL.106;NO.175(ISEC2006 9-39);PAGE.95-102(2006)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abstract:&lt;/em&gt; As effective arithmetic methods in the finite field $\rm\: GF(2^m\:)$, M.A. Hasan has presented a look-up table-based algorithm for $\rm\: GF(2^m\:)$ multiplication. In Hasan's method, the number of memory access and the table size are depend on the form of the irreducible polynomial which is used in the $\rm\: GF(2^m\:)$ multiplication. In this paper, we propose a lookup-up table-based algorithm which can be performed as an efficient $\rm\: GF(2^m\:)$ multiplication using the irreducible polynomial which has $m$ degree and $m$ terms. In this method, the algorithm is performed as the multiplication modulo quadnomial which is obtained by the irreducible polynomial which has m degree and m terms. Using this method, the number of memory access and the table size in the algorithm is smaller than that in Hasan's method, and $\rm\: GF(2^m\:)$ multiplication is faster than Hasan's method.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T18:10:42.423" CommentCount="4" />
  <row Id="14792" PostTypeId="1" AcceptedAnswerId="14941" CreationDate="2010-12-18T18:14:19.313" Score="2" ViewCount="54" Body="&lt;p&gt;This is part of Exercise III.1.16(c) of Bourbaki's set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E$ be a lattice. The condition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x\vee y)\wedge(z\vee(x\wedge y))=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all $x,y,z\in E$ implies that $E$ is distributive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x\vee y)\wedge(y\vee z)\wedge(z\vee x)=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implies distributivity, so it suffices to prove&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x\vee y)\wedge(z\vee(x\wedge y))=(x\vee y)\wedge(y\vee z)\wedge(z\vee x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find a proof of this last statement. Either it is true in any lattice, then the proof should be so simple I must be blind not to see it, or I have to use the condition above again, but I can't imagine in what way. &lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2010-12-20T12:37:45.443" Title="Condition for a lattice to be distributive" Tags="&lt;lattices&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14793" PostTypeId="1" CreationDate="2010-12-18T18:34:42.893" Score="1" ViewCount="85" Body="&lt;p&gt;Let me try to split the question in a few parts,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I would like to understand the claim that all non-degenerate bilinear symmetric forms are equivalent over the complex while for the reals they can be distinguished by the signature. Hence there is just one Clifford algebra over $\mathbb{C}^n$ but over $\mathbb{R}^n$ there are as many Clifford algebras as integral solutions to $p+q = n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Is the above thing that I am puzzled about also related to another fact that I want to understand that any complex square matrix can be similarity transformed to a symmetric complex matrix?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I have a bilinear form on a real vector space with a non-trivial signature and then I complexify the vector space then why does the bilinear form always lift to something with a positive definite signature? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How does one see that the complexified Clifford algebra over a real vector space with a non-trivial signature is isomorphic to the Clifford algebra over the complexified vector space with a positive definite norm? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above something special about the Clifford algebra or is there something more general?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I want to understand how on this complexified algebra there exists a natural automorphism which leaves a subalgebra invariant and hence gives a definition for ``real subalgebra&quot; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In this context I would want to know about the notion of &quot;even subalgebras&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to understand the constraints between dimensions and signatures about taking real subalgebras? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2010-12-18T19:20:38.907" LastActivityDate="2010-12-18T20:40:14.803" Title="Complexifying representations" Tags="&lt;abstract-algebra&gt;&lt;physics&gt;&lt;representation-theory&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14794" PostTypeId="1" AcceptedAnswerId="14796" CreationDate="2010-12-18T18:42:25.937" Score="11" ViewCount="261" Body="&lt;p&gt;This question came up when trying to trying to find a $3\times3$ real matrix $A$ such that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$Ax$ is nonzero for nonzero $x$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$Ax$ is orthogonal to $x$ for any $x$ in $\mathbb{R}^3$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We know such a matrix cannot exist because $A$ must have an eigenvalue (thus there is some $x$ such that either $Ax = 0$ or $Ax$ is parallel to $x$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a nice, &lt;strong&gt;purely geometric&lt;/strong&gt; way to justify that every operator on $\mathbb{R}^3 $ has a (real) eigenvalue?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To clarify: I'm looking an intuitive way to &lt;em&gt;visualize&lt;/em&gt; why an eigenvalue must exist in this case.  In particular, no polynomials and no determinants are allowed!&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-19T04:12:52.063" LastActivityDate="2010-12-20T18:57:36.887" Title="Geometric argument that operators on $\mathbb{R}^3$ have an eigenvalue?" Tags="&lt;linear-algebra&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="14795" PostTypeId="2" ParentId="14789" CreationDate="2010-12-18T18:43:39.587" Score="0" ViewCount="0" Body="&lt;p&gt;You can find a nice concise exposition of Newton's method for algebraic systems of equations in the paper below. There it is applied to the algebraic equations that arise from (multivariate) polynomial factorization - but the ideas are quite general (Newton-Hensel methods are fundamental for effective algebraic computation). For a more general and more detailed presentation see the chapter on &quot;Hensel algorithms&quot; in Zippel's textbook &lt;em&gt;Effective Polynomial computation,&lt;/em&gt; and see also my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zohi1mgg1.fsf_-_@martigny.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for references to expositions of Mora, Ribenboim, von zur Gathen et. al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zippel, R. $\ $ &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=806372&quot; rel=&quot;nofollow&quot;&gt;Newton's Iteration and the Sparse Hensel Algorithm&lt;/a&gt;,&lt;br&gt;&#xA;Proc. 1981 ACM Sympos. Symbolic and Algebraic Computation, Utah, 1981, pp. 68-72&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-18T18:43:39.587" />
  <row Id="14796" PostTypeId="2" ParentId="14794" CreationDate="2010-12-18T19:10:31.177" Score="19" ViewCount="0" Body="&lt;p&gt;If $A\colon\mathbb{R}^3\to\mathbb{R}^3$ was a linear map with no eigenvectors, then $x\mapsto Ax/\Vert Ax\Vert$ would give a map on the unit sphere with no fixed points and without taking any point to its antipode, contradicting the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Hairy_ball_theorem&amp;amp;oldid=402767345#Corollary&quot; rel=&quot;nofollow&quot;&gt;hairy ball theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-12-18T19:10:31.177" />
  <row Id="14797" PostTypeId="2" ParentId="14794" CreationDate="2010-12-18T19:55:12.590" Score="3" ViewCount="0" Body="&lt;p&gt;Just using linear algebra but no determinants, the closer to what you ask that I know uses that polynomials of degree 3 with real coefficients have roots: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's argue by contradiction, and assume that $A$ has no eigenvalues. First, I claim that any $v$ is contained in a proper $A$-invariant subspace of ${\mathbb R}^3$. This is clear if $v=0$. Otherwise, consider $v,Av,A^2v,A^3v$. They are linearly dependent, so there are real coefficients $a,b,c,d$ not all zero with $(aA^3+bA^2+cA+dI)v=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we examine the polynomial $p(x)=ax^3+bx^2+cx+d$. If $a\ne0$, $p$ has a real root $r$, and we can write $p(x)=(x-r)(ax^2+ex+f)$ for some real numbers $e,f$. This gives us $(A-rI)(aA^2+eA+fI)v=0$, but $A-rI$ is invertible (since $r$ is not an eigenvalue of $A$), so in fact $(aA^2+eA+fI)v=0$. If $a=0$, then directly we have $(bA^2+cA+dI)v=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that $v$ is contained in a proper $A$-invariant subspace (consider the span $S$ of $v,AV$, and note that the argument above shows that either $A^2v$ is in this span, from which it follows that $S$ is $A$-invariant, or else, the coefficient of $A^2$ above is 0, and in fact $v$ is an eigenvalue, a contradiction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may assume that any non-zero $v$ is in an $A$-invariant &lt;em&gt;plane&lt;/em&gt; $P_v$. Now, if $w$ is a vector not in $P_v$, then any vector in $P_v\cap P_w$ is mapped by $A$ to another vector in the same line, so $P_v\cap P_w$ is an $A$-invariant line, i.e., $A$ has a real eigenvalue after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that I presented the argument as a contradiction for brevity, but it can be rearranged as a direct proof.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Milovich found a nice way of extending this argument so we have a short nice &quot;determinant free&quot; proof that $n\times n$ matrices with real entries and $n$ odd admit a real eigenvalue, see this &lt;a href=&quot;http://caicedoteaching.wordpress.com/2010/02/28/403503-eigenvectors-for-operators-on-real-odd-dimensional-spaces/&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; of mine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was interested in this because this is the base case of a nice inductive argument (that actually can be traced back to one of Gauss' first proofs of the fundamental theorem of algebra) that allows us to show that &lt;em&gt;any&lt;/em&gt; square matrix with real coefficients admits a (perhaps complex) eigenvalue, from which we can deduce the fundamental theorem of algebra. I refer to this in the post above, but it comes from &quot;The fundamental theorem of algebra and linear algebra&quot; by Harm Derksen, American Mathematical Monthly, 110 (7) (2003), 620-623. (The issue is that in that paper, the odd-dimensional case is done by appealing to determinants.) &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-18T19:55:12.590" CommentCount="2" />
  <row Id="14798" PostTypeId="1" AcceptedAnswerId="14805" CreationDate="2010-12-18T20:14:29.047" Score="2" ViewCount="113" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not so well up on my math, but I was wondering if the set of &quot;[EDIT: &lt;em&gt;finitely&lt;/em&gt;] decimal expressible&quot; real numbers $R$ between $[0,1]$ is countable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia gives the definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A set $S$ is called countable if there exists an injective function $f$ from $S$ to the natural numbers $\mathbb{N}$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would seem to my mathematically na&amp;iuml;ve brain that such an injective function exists&amp;mdash;for any element $r \in R$ you can simply &quot;reverse the number&quot; and remove the decimal point to end up with, e.g.,:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(0.0001) = 10000$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to three questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this intuition correct?&lt;/li&gt;&#xA;&lt;li&gt;Would this constitute a proof? &lt;/li&gt;&#xA;&lt;li&gt;Can the result be extended to general real numbers between $[0,1]$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any help with terminology would also be great... E.g., I don't know a mathematical function for &quot;reversing&quot; a number...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S., I've just noted a specific question &lt;a href=&quot;http://math.stackexchange.com/questions/12088/real-numbers-and-countability&quot;&gt;here&lt;/a&gt; which is related and gives the general principle, but still doesn't answer my specific examples.)&lt;/p&gt;&#xA;" OwnerUserId="2331" LastEditorUserId="2331" LastEditorDisplayName="" LastEditDate="2010-12-18T21:19:36.463" LastActivityDate="2011-01-10T05:16:00.333" Title="(Finitely) decimal expressible real numbers between [0,1] countable?" Tags="&lt;elementary-set-theory&gt;&lt;discrete-mathematics&gt;" AnswerCount="6" CommentCount="7" />
  <row Id="14799" PostTypeId="2" ParentId="14767" CreationDate="2010-12-18T20:14:56.643" Score="5" ViewCount="0" Body="&lt;p&gt;According to this article &lt;a href=&quot;http://www.fq.math.ca/Scanned/8-1/daykin-a.pdf&quot; rel=&quot;nofollow&quot;&gt;Factorisation of Fibonacci Numbers,&lt;/a&gt; if $F_n$ is the entry point in the Fibonacci sequence of the prime $p$ (i.e. $F_n$ is the smallest Fibonacci number divisible by the prime $p$) then for $p&amp;gt;5$ we have: for odd $n$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=4kn+1 \quad \textrm{ or } \quad p=(4k+2)n-1,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n \equiv 2 \pmod{4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=kn+1,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for $n \equiv 0 \pmod{4}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=2kn+1 \quad \textrm{ or } \quad p=(2k+1)n-1 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should significantly reduce the search of possible candidates.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-18T20:14:56.643" />
  <row Id="14800" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:24:33.560" Score="1" ViewCount="0" Body="&lt;p&gt;No they are not countable. Any real number can be expressed in decimal notation, and this representation may be infinite (non-terminating). And that's where your problem lies: how can you reverse an infinite decimal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my interpretation of your question there is no difference between what you are calling &quot;decimal expressible&quot; and real numbers, since any real number can be expressed in decimal notation. So the answer to part 3 is no, since the argument doesn't work as you can't reverse an infinite representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In the light of your recent comment, if you are taking your expression &quot;decimal expressible&quot; to mean a terminating decimal, then yes these are countable. But unfortunately your argument cannot be extend to the reals in $[0,1]$ because of the problem of reversing a non-terminating decimal.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-18T20:29:43.260" LastActivityDate="2010-12-18T20:29:43.260" CommentCount="3" />
  <row Id="14801" PostTypeId="1" AcceptedAnswerId="14818" CreationDate="2010-12-18T20:29:40.780" Score="3" ViewCount="64" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a commutative integral domain R in which:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;every nonzero prime ideal Q is maximal, and&lt;/li&gt;&#xA;  &lt;li&gt;there are maximal ideals Q with R/Q of sizes 3, 11, and 27?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't happen with number rings of Galois extensions of &lt;b&gt;Q&lt;/b&gt;, as far as I know, since what happens to 3 seriously happens to 3.  You cannot have residue fields of sizes both 3 and 27.  Now &lt;b&gt;Z&lt;/b&gt;/3&lt;b&gt;Z&lt;/b&gt;[X] has residue fields of sizes 3 and 27, but not 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I'll separate the next into its own question.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-20T14:42:58.057" LastActivityDate="2010-12-20T14:42:58.057" Title="A variety of residue fields of size 3 ≡ q mod 8" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" AnswerCount="1" />
  <row Id="14802" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:32:38.050" Score="1" ViewCount="0" Body="&lt;p&gt;When I was taking my first course in set theory I asked my prof. (now my advisor) the same thing really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the function is not well-defined, since there are numbers with several different decimal expansions (for example $\frac{1}{2} = 0.49999\ldots = 0.5$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that we only take the shortest finite expansion, it is possible but only for a very small portion of the numbers, for example $\frac{1}{3}$ has no finite decimal expansion, despite being a very simple rational number. So in fact, you don't really use even all the rational numbers in the interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume that a number $x$ has a finite representation at all, that means that for some $n\in\mathbb{N}$ we have that $x$ has $n$ digits after the decimal point, namely $x\cdot 10^n\in\mathbb{N}$, so you only define your function on a small portion of the rational numbers in the interval $[0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed by Derek Jennings, some (in fact almost all) numbers doesn't have a finite decimal representation (and actually most numbers has no finite representation in any base you can choose), so even if you do define this function it will only cover a countable subset of the interval, whereas the interval itself is uncountable and much much bigger than $|\mathbb{N}|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, yes it is countable and your function (after correcting the minor problem of the uniqueness in the expression) is a proof for that.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-18T20:32:38.050" CommentCount="4" />
  <row Id="14803" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:33:04.130" Score="1" ViewCount="0" Body="&lt;p&gt;No, they are not countable. The proof uses an argument by contradiction. If you at all manage to list the real numbers into a countable list, you can always construct one which is not in the list. You can find a very neat exposition of the argument at the &lt;a href=&quot;http://planetmath.org/encyclopedia/CantorsDiagonalArgument.html&quot; rel=&quot;nofollow&quot;&gt;planetmath page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-18T20:33:04.130" CommentCount="1" />
  <row Id="14804" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T20:36:01.383" Score="1" ViewCount="0" Body="&lt;p&gt;&quot;Uniform probability distribution&quot; always refers to a pre-existent natural measure on a space with a transitive group of &quot;translations&quot;, so that any two points resp. neighbourhoods of two points are comparable which each other. The simplest space of this sort would be the set $\mathbb Z$ of integers, and already there the problem you address becomes manifest. Of course it is easy to prove that there doesn't exist a uniform probability distribution on $\mathbb Z$; but you want an intuitive reasoning. While it is easy to come up with a random mechanism that selects any number between $-10^6$ and $10^6$ with equal probability, it is just unimaginable to fathom a mechanism that selects 5 with the same probability as any power of the prime with number $67212345$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-18T20:36:01.383" />
  <row Id="14805" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:38:55.660" Score="3" ViewCount="0" Body="&lt;p&gt;Your intuition is correct: the set of &quot;decimal expressible&quot; real numbers in [0, 1] such that the decimal expression has only finitely many nonzero digits is &lt;strong&gt;countable&lt;/strong&gt;, and the &quot;injective function&quot; you gave works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you allow decimal expressions with an infinite number of decimals (e.g. 0.333...), then this no longer works.  First of all, which natural number are you going to associate with 0.333...?  In fact, if you allow decimal expressions with an infinite number of decimals, then the set you described becomes uncountable; for the justification of that, I'll point you to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Cantor's Diagonal Argument&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your three questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, all you needed to do is associate every element of your set with an element of the natural numbers (and vice versa), which you've done.  If I give you a decimal expression with finitely many terms, you can give me the unique natural number that you've associated with that decimal expression, and if I give you a natural number, you can give me the unique (finite) decimal expression it corresponds to.  This is exactly what it means for a set to be countable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No; again, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Cantor's Diagonal Argument&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2010-12-18T20:51:31.583" LastActivityDate="2010-12-18T20:51:31.583" CommentCount="2" />
  <row Id="14806" PostTypeId="2" ParentId="14793" CreationDate="2010-12-18T20:40:14.803" Score="0" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The signature of a real non-degenerate bilinear form on an $n$-dimensional vector space can be defined as $(p,n-p)$ where $p$ is the largest dimension of a subspace on which the form restricts to be positive definite.  But over the complex numbers you can never have a positive definite symmetric form:  if $\langle v ,v \rangle &amp;gt; 0$ then $\langle iv ,iv\rangle &amp;lt; 0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To address your third question, let $Cl(p,q)$ be the Clifford algebra associated to a real inner product of signature $(p,q)$.  To show that $Cl(p,q) \otimes C \simeq Cl(C^{p+q})$, we just need to show that $Cl(p,q) \otimes C$ has elements $\{E_1, \ldots E_{p+q}\}$ that anti-commute and satisfy $E_i^2 = -1$.  But $\{e_1 \otimes 1, \ldots, e_p \otimes 1, e_{p+1} \otimes i, \ldots, e_{p+q}\otimes i\}$ is such a set, where $\{e_i\}$ is an orthonormal basis for $R^{p,q}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think the automorphism is just complex conjugation: $u \otimes z \mapsto u \otimes \bar z$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not sure what is meant by even algebra in this context.  Clifford algebras are $Z_2$ graded so there is an even subalgebra.  This is the subalgebra generated by even products of vectors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-18T20:40:14.803" />
  <row Id="14807" PostTypeId="2" ParentId="2487" CreationDate="2010-12-18T20:45:40.807" Score="0" ViewCount="0" Body="&lt;p&gt;The answer by Akhil Mathew is incomplete. It remains to check that $\sum_{n&amp;gt;k} x_n/n!\neq 0$ for infinitely many $k$. This is not obviously true here, it might even be false for some sequence $(x_n)$. Akhil Mathew's argument could be used for the series $\sum_{n\ge 0} n/(n+1)!$ which is similar to $\sum_{n\ge 0} 1/n!$. But its sum is equal to 1.&lt;/p&gt;&#xA;" OwnerUserId="4906" LastActivityDate="2010-12-18T20:45:40.807" CommentCount="1" />
  <row Id="14808" PostTypeId="2" ParentId="14798" CreationDate="2010-12-18T20:46:41.243" Score="2" ViewCount="0" Body="&lt;p&gt;Yes, that works for terminating decimals (which does not include all rationals). More generally, it's simpler to note that the rationals are countable by mapping the reduced rational $\rm\ m/n\ \to\ 2^m\ 3^n\:.$ The same idea easily extends from pairs of naturals to all eventually zero sequences of naturals. Such arguments cannot be extended to all reals since the well-know diagonalization argument due to du-Bois-Reymond (and, later, Cantor) easily proves that that they are uncountable.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-18T20:51:46.167" LastActivityDate="2010-12-18T20:51:46.167" CommentCount="1" />
  <row Id="14809" PostTypeId="2" ParentId="14777" CreationDate="2010-12-18T21:04:13.963" Score="4" ViewCount="0" Body="&lt;p&gt;If a uniform probability distribution did exist, then for any integer $n$ the probability a real number $x$ satisfies $n \leq x &amp;lt; n+1$ would have to be the same for all $n$. Call this probability $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the rules that a probability distribution has to satisfy is that if $\{E_n\}$ are a countably infinite collection of disjoint events then $P(\cup_n E_n) = \sum_n P(E_n)$. Let $E_n$ be the event &quot;the real number $x$ satisfies $n \leq x &amp;lt; n+1$&quot;. Since every real number is between some $n$ and $n + 1$, $P(\cup_n E_n) = 1$. On the other hand, $\sum_n P(E_n) = p + p + p + ....$. If $p &amp;gt; 0$, this gives infinity. If $p = 0$ it gives zero. In either case, you'll never add up to $1$. Hence you cannot have a uniform probability distribution over the reals. &lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-18T21:04:13.963" />
  <row Id="14811" PostTypeId="1" AcceptedAnswerId="14812" CreationDate="2010-12-18T21:37:56.483" Score="11" ViewCount="243" Body="&lt;p&gt;Suppose that $g$ is a bijection on the real line, and $g^{-1} \circ f \circ g$ is a $C^\infty$ function whenever $f$ is $C^\infty$.  It seems howlingly obvious that this can only happen if $g$ is itself $C^\infty$.  But I can't figure out how to prove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me, Internets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Context: I want to show that in a manifold of dimension at least 2, the facts about which unparameterized curves are smooth suffice to determine the differential structure.  I think I can sort of see how to prove this provided I can rely on the claim above, but that's where I'm stuck.  However, if anyone here knows of a proof of the geometric claim, pointers would be most welcome.)&lt;/p&gt;&#xA;" OwnerUserId="4909" LastEditorUserId="4909" LastEditorDisplayName="" LastEditDate="2010-12-18T22:09:15.430" LastActivityDate="2010-12-20T08:35:34.280" Title="If $g^{-1} \circ f \circ g$ is $C^\infty$ whenever $f$ is $C^\infty$, must $g$ be $C^\infty$?" Tags="&lt;real-analysis&gt;&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="5" />
  <row Id="14812" PostTypeId="2" ParentId="14811" CreationDate="2010-12-18T21:53:40.380" Score="13" ViewCount="0" Body="&lt;p&gt;This is a result due to Floris Takens:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $\Phi \colon M_1 \to M_2$ be a bijection between two smooth $n$-manifolds such that $\lambda \colon M_2 \to M_2$ is a diffeomorphism iff $\Phi^{-1} \circ \lambda \circ \Phi$ is a diffeomorphism.  Then $\Phi$ is a diffeomorphism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be found in &lt;em&gt;Characterization of a differentiable structure by its group of diffeomorphisms&lt;/em&gt; (Math Reviews number: MR552032).  I remember that I found an online version of it (but don't remember where off the top of my head, probably via MathSciNet).  I don't have it in front of me, but I do recall that the argument was fairly involved.  (I found it for a fairly similar reason: I wanted to show that the only Fr&amp;ouml;licher space with endomorphism monoid $C^\infty(\mathbb{R},\mathbb{R})$ was $\mathbb{R}$ itself, this is Proposition 8.4 of &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Research/Preprints/smthcat.html&quot; rel=&quot;nofollow&quot;&gt;Comparative Smootheology&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: After reading the comments, I realised that the quoted theorem is not quite what is wanted.  Actually, what you need is Proposition 8.4 of &lt;a href=&quot;http://www.math.ntnu.no/~stacey/Research/Preprints/smthcat.html&quot; rel=&quot;nofollow&quot;&gt;Comparative Smootheology&lt;/a&gt;, which uses Takens' result at the key stage but nonetheless is about a page long in the proving.  The result in the paper is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only Fr&amp;ouml;licher structures on $\mathbb{R}$ whose endomorphism monoid contains $C^\infty(\mathbb{R},\mathbb{R})$ are the standard, the discrete, and the indiscrete structures.  In particular, the only Fr&amp;ouml;licher structure on $\mathbb{R}$ whose endomorphism monoid is precisely $C^\infty(\mathbb{R},\mathbb{R})$ is the standard structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can almost replace &quot;Fr&amp;ouml;licher structure&quot; by &quot;smooth structure&quot; in the above, except that there's no such thing as the &quot;indiscrete smooth structure&quot;.  The &lt;em&gt;discrete&lt;/em&gt; smooth/Fr&amp;ouml;licher structure on $\mathbb{R}$ views $\mathbb{R}$ as a load of disjoint points: a zero-dimensional manifold.  This is characterised by the fact that the only smooth functions $\mathbb{R} \to \mathbb{R}_{disc}$ are the constant functions (here the unadorned $\mathbb{R}$ is the usual manifold $\mathbb{R}$).  The indiscrete structure is the opposite: the only smooth functions $\mathbb{R}_{indisc} \to \mathbb{R}$ are the constant functions.  So we could rephrase the above result as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only smooth structures on $\mathbb{R}$ whose endomorphism monoid contains $C^\infty(\mathbb{R},\mathbb{R})$ are the standard and the discrete structures.  In particular, the only smooth structure on $\mathbb{R}$ whose endomorphism monoid is precisely $C^\infty(\mathbb{R},\mathbb{R})$ is the standard structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us connect that to your question.  You ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $g \colon \mathbb{R} \to \mathbb{R}$ is a bijection such that $g^{-1} \circ f \circ g$ is $C^\infty$ whenever $f$ is $C^\infty$, must $g$ be $C^\infty$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, we can use $g$ to put a smooth structure on $\mathbb{R}$ by post-composing $g$ with charts (or pre-composing with $g^{-1}$ if your charts go &lt;em&gt;from&lt;/em&gt; your manifold).  Let us write this as $\mathbb{R}_g$.  For clarity, let us write $\mathbb{R}_s$ for the standard smooth structure on $\mathbb{R}$.  Then $g \colon \mathbb{R}_s \to \mathbb{R}_g$ is a diffeomorphism, by construction.  Given a smooth function $f \colon \mathbb{R}_s \to \mathbb{R}_s$ we get a smooth function $f_g \colon \mathbb{R}_g \to \mathbb{R}_g$ by $g \circ f \circ g^{-1}$.  Conversely, given a smooth function $h \colon \mathbb{R}_g \to \mathbb{R}_g$, $g^{-1} \circ h \circ g$ is smooth on $\mathbb{R}_s$ and thus is $C^\infty$.  Hence the endomorphism monoid of $\mathbb{R}_g$, $C^\infty(\mathbb{R}_g,\mathbb{R}_g)$, is $g C^\infty(\mathbb{R}_s,\mathbb{R}_s) g^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition imposed is that $g^{-1} \circ f \circ g \in C^{\infty}(\mathbb{R},\mathbb{R})$ for all $f \in C^\infty(\mathbb{R},\mathbb{R})$.  The $\mathbb{R}$s here are $\mathbb{R}_s$ in our notation.  Translated into a single statement (rather than one for each $f \in C^\infty(\mathbb{R},\mathbb{R})$, this is: $g^{-1} C^\infty(\mathbb{R}_s,\mathbb{R}_s) g \subseteq C^\infty(\mathbb{R}_s,\mathbb{R}_s)$.  Now here's the sneaky bit; the inclusion here is going the wrong way.  So we simply apply $g \circ  - \circ g^{-1}$ to both sides to get $C^\infty(\mathbb{R}_s,\mathbb{R}_s) \subseteq g C^\infty(\mathbb{R}_s,\mathbb{R}_s) g^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply my result to deduce that $\mathbb{R}_g$ must be either the discrete smooth structure (which it isn't) or the standard one: $\mathbb{R}_g = \mathbb{R}_s$.  But then $g$ is a diffeomorphism $\mathbb{R}_s \to \mathbb{R}_g = \mathbb{R}_s$ and hence is $C^\infty$.&lt;/p&gt;&#xA;" OwnerUserId="2907" LastEditorUserId="2907" LastEditorDisplayName="" LastEditDate="2010-12-20T08:35:34.280" LastActivityDate="2010-12-20T08:35:34.280" CommentCount="8" />
  <row Id="14814" PostTypeId="2" ParentId="1037" CreationDate="2010-12-18T22:36:49.877" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.friesian.com/calculus.htm&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; page, (which quotes from &quot;Yearning for the Impossible, The Surprising Truths of Mathematics&quot;) has:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;...It has to be explained that dy/dx is not the ratio of infinitesimal differences dy and dx -- since infinitesimals do not exist -- but is rather a symbol for the limit of the ratio y/x as x tends to zero...&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2010-12-18T22:36:49.877" />
  <row Id="14815" PostTypeId="1" AcceptedAnswerId="14817" CreationDate="2010-12-18T22:58:09.783" Score="5" ViewCount="303" Body="&lt;p&gt;The infinite series...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi/4 = 1 - 1/3 + 1/5 - 1/7 ...$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...is very intriguing to me and seems like a crazy coincidence (its relationship to $\pi$).  Is it actually crazy or does it have an easy to explain, logical reasoning behind it that would make it seem not so magical?&lt;/p&gt;&#xA;" OwnerUserId="4914" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-18T23:06:22.330" LastActivityDate="2010-12-19T14:26:45.343" Title="Why does this converge to $\pi/4$?" Tags="&lt;sequences-and-series&gt;&lt;pi&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="2" />
  <row Id="14816" PostTypeId="2" ParentId="14815" CreationDate="2010-12-18T23:09:26.727" Score="3" ViewCount="0" Body="&lt;p&gt;It comes from evaluating &lt;a href=&quot;http://en.wikipedia.org/wiki/Leibniz_formula_for_pi&quot; rel=&quot;nofollow&quot;&gt;the arctan of 1&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-18T23:09:26.727" CommentCount="1" />
  <row Id="14817" PostTypeId="2" ParentId="14815" CreationDate="2010-12-18T23:14:02.217" Score="16" ViewCount="0" Body="&lt;p&gt;The standard proof is that this follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow&quot;&gt;Taylor series&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} \mp ...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the arctangent.  This Taylor series is closely related to the Taylor series of the logarithm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} \mp ...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is because the tangent function can be written in terms of complex exponentials, so the arctangent function can be written in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_logarithm&quot; rel=&quot;nofollow&quot;&gt;complex logarithms&lt;/a&gt;.  So the appearance of $\pi$ in this formula is morally due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_formula&quot; rel=&quot;nofollow&quot;&gt;Euler's formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is also the following beautiful proof, which I learned from Gabor Toth's &lt;a href=&quot;http://books.google.com/books?id=lJxOgVcHBlkC&amp;amp;printsec=frontcover&amp;amp;dq=toth+algebra+and+geometry&quot; rel=&quot;nofollow&quot;&gt;Glimpses of Algebra and Geometry&lt;/a&gt;.  Consider the number $N(r)$ of integer lattice points inside the circle of radius $r$ centered at the origin, or in other words the number of pairs of integers $x, y$ satisfying $x^2 + y^2 \le r^2$.  It is not hard to see that $N(r) \sim \pi r^2$ for large $r$; in fact, it is not hard to see that $N(r) = \pi r^2 + O(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r_2(n)$ denote the number of pairs of integers $(x, y)$ such that $x^2 + y^2 = n$.  Then $N(r) = 1 + r_2(1) + ... + r_2(r^2)$ (if $r$ is an integer).  On the other hand, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares&quot; rel=&quot;nofollow&quot;&gt;classic result of number theory&lt;/a&gt; implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_2(n) = 4(d_1(n) - d_3(n))$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d_k(n)$ is the number of divisors of $n$ congruent to $k \bmod 4$.  It follows that we can evaluate $N(r)$ by counting how many numbers between $1$ and $r^2$ are divisible by each number congruent to $1, 3 \bmod 4$ with the appropriate sign.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{N(r) - 1}{4} = r^2 - \left\lfloor \frac{r^2}{3} \right\rfloor + \left\lfloor \frac{r^2}{5} \right\rfloor \mp ...$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the result follows by taking the limit as $r \to \infty$.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-19T07:31:45.060" LastActivityDate="2010-12-19T07:31:45.060" CommentCount="1" />
  <row Id="14818" PostTypeId="2" ParentId="14801" CreationDate="2010-12-18T23:35:04.387" Score="4" ViewCount="0" Body="&lt;p&gt;The answer to your first question is yes.  Indeed, for any $d \geq 4$, you can construct infinitely many degree $d$ algebraic number fields $K/\mathbb{Q}$ such that the ring of &#xA;integers $\mathbb{Z}_K$ has this property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a basic philosophy here that you can build a global object that has prescribed local behavior at any finite set of primes.  In this particular case, Theorem 23 of &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter3.pdf&quot; rel=&quot;nofollow&quot;&gt;these notes&lt;/a&gt; does what you want: you want to take as a $\mathbb{Q}_3$ algebra &#xA;$L \oplus \mathbb{Q}_3 \oplus A&amp;#39;$ where $L/\mathbb{Q}_3$ is an unramified cubic extension (and $A&amp;#39;$ is any separable $\mathbb{Q}_p$-algebra of dimension $d-4$), and similarly you &#xA;want to choose your $\mathbb{Q}_{11}$-algebra  to have a $\mathbb{Q}_{11}$ factor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that these conditions cannot be realized by a global Galois extension, since then the local invariants of ramification index and inertial degree would have to agree for each of the factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still parsing the rest of your question...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-18T23:35:04.387" CommentCount="3" />
  <row Id="14819" PostTypeId="2" ParentId="14012" CreationDate="2010-12-18T23:52:59.840" Score="1" ViewCount="0" Body="&lt;p&gt;Just a remark on how to find the general answer to problems of this type. The &quot;ripples&quot; you are describing are level sets of some function $f_1, f_2$. Corresponding to them are their gradient vector fields $\nabla f_1, \nabla f_2$. The set of &quot;osculating points&quot; that you are looking for in your question is the set of points where $\nabla f_1$ and $\nabla f_2$ are parallel. So we can express it using the two-dimensional cross product $v \times w = v_x w_y - v_y w_x$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C = \{ p\in\mathbb{R}^2 | \nabla f_1 \times \nabla f_2 = 0 \} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For families of confocal ellipses, the level function $f$ is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(x,y) = \sqrt{(x-x_1)^2 + (y-y_1)^2} + \sqrt{(x-x_2)^2 + (y-y_2)^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the family of ellipses are the loci of points where sum of distances to the two foci are constant. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla f =  \frac{(x-x_1, y-y_1)}{\sqrt{(x-x_1)^2 + (y-y_1)^2}} + \frac{(x-x_2, y-y_2)}{\sqrt{(x-x_2)^2 + (y-y_2)^2}} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see that for generic focal points, the expression defining the algebraic curve $C$ is often very complicated. You can try plotting some examples in Mathematica to see what they will look like.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In your final comment, however, it looks like you are not considering confocal ellipses. Instead, it appears you are considering functions $f$ of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(z) = Q(z-z_0) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $z = (x,y)$ is a point in the plane, and $Q$ is a positive definite quadratic form. $z_0$ is the centre of the ellipse. (In other words, you are considering a family of ellipses that all share a centre and are dilations of each other.) In this case the answer is easier. Remember that a quadratic form can be represented by a matrix: $Q(z-z_0) = (z-z_0)^T A (z-z_0)$. Let $\epsilon$ denote the totally antisymmetric matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \epsilon = \begin{pmatrix} 0 &amp;amp; 1 \\ -1 &amp;amp; 0 \end{pmatrix} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and take $f_1(z) = (z-z_1)^T A_1 (z-z_1)$, $f_2(z) = (z-z_2)^T A_2(z-z_2)$, we have easily that $\nabla f_1 = 2 A_1(z-z1)$ and $\nabla f_2 = 2 A_2(z-z_2)$. So the set $C$ is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla f_1 \times \nabla f_2 = 0 \iff (z-z_2)^T A_2^T \epsilon A_1 (z-z_1) = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies that $C$ is described by a quadratic expression in $z$. That is, $C$ is in general given by a conic section. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-18T23:52:59.840" CommentCount="8" />
  <row Id="14820" PostTypeId="2" ParentId="14689" CreationDate="2010-12-18T23:58:40.130" Score="6" ViewCount="0" Body="&lt;p&gt;Here is the exact answer, which I used Mathematica to compute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(player\ 1\ wins) = \frac{118598889714523902216022358617928917633636253614645787377890526452587094871641057161197}{778560366535929033488842048429259732340012411410736514701931793474234814991339289051136}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(player\ 2\ wins) = \frac{858777828556435297558093986193070674512712644399583464104126713012616584348576395258605}{1038080488714572044651789397905679643120016548547648686269242391298979753321785718734848}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(players\ tie) = \frac{63512421616314632416996800666111235287366697985612516983784929048741127433063741783941}{3114241466143716133955368193717038929360049645642946058807727173896939259965357156204544}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically, that's &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.15233101351178373415...&#xA;.82727479987591082853...&#xA;.02039418661230543732...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are the Mathematica rules I used to compute the probability of player 1 winnng. The other two cases are similar.&#xA;&lt;br /&gt;&#xA;(sorry for the image) &lt;br /&gt;&lt;br /&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/C2eot.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these rules in place I then computed &lt;code&gt;p[60,60]&lt;/code&gt;. I thought it was simpler to start the players at 60 and count down to 0, rather than the other way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first rule just says if player 1 has reached 0 and he player 2 hasn't, then the probability of player 1 winning is 1. The second rule is similar for player 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third rule handles the case where both players have crossed the finish line. Note that despite the game description, there aren't really turns involved—there are rounds. It doesn't matter which player goes first, so me might as well consider all four dice being rolled simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth and final rule is where all the action is. The recursive sum is over all possible outcomes of the four dice (two for each player). The constant $864$ is merely $4\cdot6\cdot 6\cdot 6$. The &lt;code&gt;p[x,y] =&lt;/code&gt; thing in the center is a kind of memoization (caching). Without it, the simplistic recursion would take forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Here is a plot of $P(n)$, the probability that the weaker player wins, when playing to a total of $n$ (instead of $60$), for $n = 1\dots80$. To my surprise, it's not quite monotonic! But a little thought will explain why.&#xA;&lt;img src=&quot;http://i.imgur.com/9Rk55.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="130" LastEditorUserId="130" LastEditorDisplayName="" LastEditDate="2010-12-20T00:34:01.337" LastActivityDate="2010-12-20T00:34:01.337" CommentCount="3" />
  <row Id="14821" PostTypeId="1" AcceptedAnswerId="14822" CreationDate="2010-12-19T00:41:01.390" Score="4" ViewCount="169" Body="&lt;p&gt;Hello, I was given the question:what is 9+99+999+9999+...+999..99(30 digits) After noticing a trend, I came with the conclusion that the answer would be 28 1's 080. Can anyone confirm my answer and give a reason as to why?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T01:33:26.313" LastActivityDate="2010-12-19T17:17:22.760" Title="help confirm my answer for the following question" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="14822" PostTypeId="2" ParentId="14821" CreationDate="2010-12-19T00:45:08.607" Score="16" ViewCount="0" Body="&lt;p&gt;Note that&#xA;$$\underbrace{99\cdots 9}_{k\text{ digits}} = 10^k - 1.$$&#xD;&#xA;So your sum is the same as&#xA;$$(10-1) + (10^2-1) + (10^3-1) + \cdots + (10^{30}-1),$$&#xD;&#xA;which is equal to &#xA;$$(10 + 10^2 + 10^3 + \cdots + 10^{30}) - 30.$$&#xD;&#xA;The first sum is easy to do, the difference is easy to do, and it gives your answer.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-19T00:45:08.607" CommentCount="6" />
  <row Id="14823" PostTypeId="1" CreationDate="2010-12-19T01:22:07.820" Score="0" ViewCount="72" Body="&lt;p&gt;Imagine the varying frequency generator, which feeds this frequency to the oscilloscope.&#xA;Oscilloscope is turned to show signal spectra with FFT.&lt;br&gt;&#xA;The image on the screen of the oscilloscope is as expected: single peak, traveling from 1 kHz to 50 kHz. Of course, my generator is not ideal, it gives a number of harmonics with less magnitude. But the strange thing happens: if my peak goes from 1 kHz to 50 kHz, the other peak appears at 50-60 kHz, and goes to lower frequencies towards mine peak. At time of appear, it's hardly noticeable, but in the end it's almost half of main frequency. Whenever varying frequency comes to it's maximum (50 kHz), ghost frequency almost approaches it. At this moment my generator drops back frequency to 1 kHz, ghost frequency disappears and process starts again.&lt;br&gt;&#xA;What causes such a behaviour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I think it's more mathematical question, rather then physical or electronic.&lt;br&gt;&#xA;PPS: I have some decisions, but don't want to spoil the question.&lt;/p&gt;&#xA;" OwnerUserId="2715" LastEditorUserId="2715" LastEditorDisplayName="" LastEditDate="2010-12-19T04:19:13.937" LastActivityDate="2010-12-23T05:39:07.410" Title="FFT gives ghost frequency." Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14824" PostTypeId="1" CreationDate="2010-12-19T01:47:14.767" Score="0" ViewCount="47" Body="&lt;p&gt;I have an old Spanish CG book that calls Perlin Noise a &quot;fractal structure&quot;. After reading &lt;a href=&quot;http://webstaff.itn.liu.se/~stegu/TNM022-2005/perlinnoiselinks/perlin-noise-math-faq.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; I couldn't deny it or confirm it. Is it a fractal structure? What would it Hausdorff dimension be?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastActivityDate="2010-12-19T01:47:14.767" Title="Is Perlin Noise a &quot;fractal&quot;? " Tags="&lt;fractals&gt;&lt;noise&gt;" CommentCount="2" />
  <row Id="14825" PostTypeId="1" AcceptedAnswerId="14829" CreationDate="2010-12-19T02:05:00.693" Score="7" ViewCount="66" Body="&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;G = { x,y : x&lt;sup&gt;2&lt;sup&gt;n+1&lt;/sup&gt;&lt;/sup&gt; = 1, y&lt;sup&gt;4&lt;/sup&gt; = 1, xy = yx&lt;sup&gt;−1&lt;/sup&gt; }&lt;/li&gt;&#xA;  &lt;li&gt;H = { x,y : x&lt;sup&gt;2&lt;sup&gt;n+1&lt;/sup&gt;&lt;/sup&gt; = 1, y&lt;sup&gt;4&lt;/sup&gt; = 1, xy = yx&lt;sup&gt;2&lt;sup&gt;n&lt;/sup&gt;−1&lt;/sup&gt; }&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Can these groups of order 2&lt;sup&gt;n+3&lt;/sup&gt; and nilpotency class n+1 be distinguished by any reasonable (computable) set of invariants?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While for any set of reasonable invariants, there is surely a pair of groups that share the invariants, it is also surely true that for any pair of groups there is some invariant that distinguishes them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally have not found any significant differences in the conjugacy classes, centralizers, proper subgroups, normalizers, normal subgroups, etc., but perhaps I overlooked something.  I've been looking at simple groups lately, and have forgotten how muddy p-groups can be.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-19T05:29:38.717" Title="Can invariants distinguish these two 2-groups?" Tags="&lt;group-theory&gt;" AnswerCount="2" />
  <row Id="14826" PostTypeId="1" AcceptedAnswerId="14838" CreationDate="2010-12-19T02:16:48.600" Score="7" ViewCount="157" Body="&lt;p&gt;Let $X$ be a CW complex. The (isomorphism classes of) complex line bundles on $X$ are classified by the homotopy classes of maps $X \to \mathbb{CP}^\infty$, that is by the elements of $H^2(X, \mathbb{Z})$. &#xA;It is also true that the tensor product of line bundles corresponds to adding cohomology classes. It follows that if $n \in \mathbb{N}$, then the line bundles of $\mathbb{CP}^n$ are generated by the tensor powers of the tautological line bundle, or, equivalently, by the tensor powers of the sheaf typically denoted  $\mathcal{O}(1)$ in algebraic geometry (because the dual of $\mathcal{O}(1)$ is the tautological bundle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also true that the isomorphism classes of algebraic line bundles on $\mathbb{CP}^n$ forms  a group isomorphic to $\mathbb{Z}$, given by the powers of $\mathcal{O}(1)$, as one can see by studying the Weil class group. It follows that the isomorphism classes of line bundles are the same in both the algebraic and the topological category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above observation is also true for affine $n$-space, because the topological line bundles are trivial ($\mathbb{C}^n$ being contractible), and the algebraic ones are as well (the polynomial ring being a UFD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent is this true in general?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-19T18:37:05.063" Title="Algebraic versus topological line bundles" Tags="&lt;algebraic-geometry&gt;&lt;algebraic-topology&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" />
  <row Id="14827" PostTypeId="2" ParentId="14794" CreationDate="2010-12-19T03:02:30.210" Score="-1" ViewCount="0" Body="&lt;p&gt;Your first condition &quot;(1) $Ax$ is nonzero for nonzero $x$&quot; implies that the matrix is full rank.  We hence conclude that $A$ has non-vanishing eigenvalues.  We then have two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) There exist three non-vanishing real eigenvalues (possibly degenerate).&#xA;(b) There exist a complex conjugate pair of eigenvalues and one real non-vanishing eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case you have you have at least one real eigenvalue with an associated eigenvector.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also reference &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_rotation_theorem&quot; rel=&quot;nofollow&quot;&gt;Euler's Rotation Theorem&lt;/a&gt; which states that in ${\mathbb R}^3$ there will always be one fixed axis (i.e. eigen value of 1) under any rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, your objective is possible in ${\mathbb R}^4$.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-19T03:02:30.210" CommentCount="2" />
  <row Id="14828" PostTypeId="1" AcceptedAnswerId="14842" CreationDate="2010-12-19T03:31:32.513" Score="8" ViewCount="184" Body="&lt;p&gt;I am reading the book by Goldrei on Classic Set Theory. My question is more of a clarification. It is on if we are overloading symbols in some cases. For instance, when we define $2$ as a natural number, we define $$2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \}$$ When we define $2$ as an integer, $2_{\mathbb{Z}}$ is an equivalence class of ordered pair $$2_{\mathbb{Z}} = \{(n+_{\mathbb{N}}2_{\mathbb{N}},n):n \in \mathbb{N}\}$$ Similarly, when we define $2$ as a rational number, $2_{\mathbb{Q}}$ is an equivalence class of ordered pair $$2_{\mathbb{Q}} = \{(a \times_{\mathbb{Z}} 2_{\mathbb{Z}},a):a \in \mathbb{Z}\backslash\{0\}\}$$ and as a real number we define it as the left Dedekind cut of rationals less than $2_{\mathbb{Q}}$, i.e. $$2_{\mathbb{R}} = \{q \in \mathbb{Q}: q &amp;lt;_{\mathbb{Q}} 2_{\mathbb{Q}}\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clarification is each of the above are different objects right? So when we say $2$, it depends on the context? Also, if the above is true, is it correct or incorrect to say that &quot;The set of natural numbers is a subset of reals&quot;? Should we take the statement with a pinch of salt and understand accordingly?&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-19T04:14:00.830" LastActivityDate="2010-12-24T23:17:39.900" Title="Set Theoretic Definition of Numbers" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="14829" PostTypeId="2" ParentId="14825" CreationDate="2010-12-19T03:47:00.457" Score="5" ViewCount="0" Body="&lt;p&gt;I have played around with these two groups for $n=3$ and $n=4$, and for those, one distinction is that the first group has no 2-dimensional irreducible representations with non-real characters (i.e. the Frobenius-Schur indicators of all the 2-dim irreducibles are $\pm1$), while the second has plenty 2-dim irreducible non-real characters. That should be easily provable for all $n$, although I haven't tried yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether this can be turned into a purely group theoretic criterion. I briefly tried something in the spirit of &lt;a href=&quot;http://mathoverflow.net/questions/41784/roots-of-permutations/41788#41788&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and the answer to &lt;a href=&quot;http://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, but to no avail: both groups have the same number of involutions and for both of them, the square roots counting function assumes its maximum at a non-trivial central element.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-19T05:25:41.597" LastActivityDate="2010-12-19T05:25:41.597" CommentCount="1" />
  <row Id="14830" PostTypeId="2" ParentId="14828" CreationDate="2010-12-19T04:54:22.657" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, in the sense that in ZF, each of the above definitions defines a different collection of sets.  Also yes, in the sense that one should always be aware of context when discussing a mathematical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical fact that allows us to overload the symbol $2$ here is the existence of a sequence of natural injections $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$, and elements in any term in this sequence can be pushed forward under the next injection or, if they lie in the range of the previous injection, pulled back.  Most people do this automatically and so don't usually point it out explicitly.  More generally, $1$ refers to the multiplicative identity in any ring, $0$ to the additive identity in any ring, and $2 = 1 + 1, 3 = 1 + 1 + 1$, and so forth.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-19T04:54:22.657" CommentCount="1" />
  <row Id="14831" PostTypeId="1" CreationDate="2010-12-19T04:56:18.920" Score="3" ViewCount="125" Body="&lt;p&gt;If &lt;em&gt;G&lt;/em&gt; is a free group generated by &lt;em&gt;n&lt;/em&gt; elements, is it possible to find an isomorphism of &lt;em&gt;G&lt;/em&gt; with a free group generated by &lt;em&gt;n-1&lt;/em&gt; (or any fewer number) of elements?&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T06:02:53.410" LastActivityDate="2010-12-20T08:03:16.980" Title="Generators of a free group." Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14832" PostTypeId="2" ParentId="9416" CreationDate="2010-12-19T04:59:38.107" Score="2" ViewCount="0" Body="&lt;p&gt;Remember that the FFT is circular.  Inputs which contain an integer number of cycles will come out clean as a single point, in the corresponding bin.  Those which do not, act as if they are multiplied by a rectangular pulse in the time domain, which creates convolution by a sinc function in the frequency domain.  Since sinc has unlimited support, your supposition that all bins except the closest two would be zero is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding a closed-form analytic solution may be impossible, in which case your best bet would be to start with the center frequency for the two strongest bins and use binary search to find the frequency in-between that most closely corresponds to your actual spectrum.&lt;/p&gt;&#xA;" OwnerUserId="4923" LastActivityDate="2010-12-19T04:59:38.107" />
  <row Id="14833" PostTypeId="2" ParentId="14831" CreationDate="2010-12-19T05:08:03.523" Score="8" ViewCount="0" Body="&lt;p&gt;No, if $F_n$ and $F_m$ are isomorphic, then their abelianizations are also isomorphic, which only happens when n=m (for example, reduce mod p and count points). &lt;/p&gt;&#xA;" OwnerUserId="8" LastEditorUserId="8" LastEditorDisplayName="" LastEditDate="2010-12-19T05:15:46.687" LastActivityDate="2010-12-19T05:15:46.687" />
  <row Id="14834" PostTypeId="1" AcceptedAnswerId="14837" CreationDate="2010-12-19T05:13:02.287" Score="1" ViewCount="98" Body="&lt;p&gt;For example, saying that $T$ is a k-tensor one might see $T\in 3^k(V)$, of course it's not actually a  3.  It looks somewhat like Fraktur font Z: $\frak{Z}$.  I couldn't detexify it, and it doesn't appear to be a standard notation for the set of tensors.  Other resources seem to just use $T$.&lt;/p&gt;&#xA;" OwnerUserId="326" LastActivityDate="2010-12-19T06:09:28.027" Title="In tensor notation in Spivak's Calculus on Manifolds, what is that character that looks like a 3?" Tags="&lt;soft-question&gt;&lt;tensors&gt;" AnswerCount="2" />
  <row Id="14835" PostTypeId="2" ParentId="14831" CreationDate="2010-12-19T05:17:44.923" Score="2" ViewCount="0" Body="&lt;p&gt;To add to Noah Snyder's answer: abelian groups are $\mathbb{Z}$-modules, and the ring $\mathbb{Z}$ has the &lt;a href=&quot;http://en.wikipedia.org/wiki/Invariant_basis_number&quot; rel=&quot;nofollow&quot;&gt;Invariant Basis Number&lt;/a&gt; property. That means that for free $\mathbb{Z}$-modules, the rank is well-defined. More generally, all commutative rings have this property, as well as all left-Noetherian rings ($\mathbb{Z}$ is both). The wikipedia article gives an example of a ring that doesn't have this property, so modules over that ring can exhibit the strange behaviour that you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-19T05:23:51.177" LastActivityDate="2010-12-19T05:23:51.177" CommentCount="2" />
  <row Id="14836" PostTypeId="2" ParentId="14825" CreationDate="2010-12-19T05:29:38.717" Score="0" ViewCount="0" Body="&lt;p&gt;A completely different method of recognition is just a random constructive recognition:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Choose a random element z&lt;/li&gt;&#xA;&lt;li&gt;If z has order |G|/4, call it x [ probability 25% ]&lt;/li&gt;&#xA;&lt;li&gt;If z has order 4 and if it has more than 2 conjugates, call it y [ probability 50% ]&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, go to 1 [ probability 25% ]&lt;/li&gt;&#xA;&lt;li&gt;If we are missing x or y, go to 1.&lt;/li&gt;&#xA;&lt;li&gt;If x^y = x^-1, return &quot;G&quot;, else x^y = x^(|G|/8-1) and so return &quot;H&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is easy to do since the automorphism groups of G and H act quite transitively on the possible generators.  This works well even for large n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is using the normally &quot;cheating&quot; purely-group-theoretic invariant of &quot;how many isomorphisms to G?&quot;  In this case, there are just so many, it is easy to find one by random chance.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-19T05:29:38.717" />
  <row Id="14837" PostTypeId="2" ParentId="14834" CreationDate="2010-12-19T05:34:37.737" Score="1" ViewCount="0" Body="&lt;p&gt;As far as I can tell, it's fraktur I: $\mathfrak{I}$.  I don't think this is standard.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-19T05:34:37.737" />
  <row Id="14838" PostTypeId="2" ParentId="14826" CreationDate="2010-12-19T05:54:31.677" Score="4" ViewCount="0" Body="&lt;p&gt;Let $X$ be a complex analytic space.  There is an exact sequence, the &lt;em&gt;exponential exact sequence&lt;/em&gt;, which is of fundamental importance for analyzing this (and related) questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 \to 2 \pi i \mathbb Z \to \mathcal O_X \buildrel \exp \over \longrightarrow \mathcal O^{\times}_X \to 1 .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume now that $X$ is proper and connected. When we pass to cohomology, the sequence of $H^0$s is then short exact, but we obtain the following crucial long exact sequence:&#xA;$$H^1(X, 2 \pi i \mathbb Z) \to H^1(X,\mathcal O_X) \to Pic(X) \to&#xD;&#xA;H^2(X,2 \pi i \mathbb Z) \to H^2(X,\mathcal O_X).$$&#xD;&#xA;Here I am writing (as is usual) $Pic(X)$ to denote $H^1(X,\mathcal O_X^{\times})$, the group of isomorphism classes of analytic line bundles on $X$.  If $X$ is algebraic, then by GAGA this is the same as the group of algebraic line bundles on $X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary map $Pic(X) \to H^2(X,2 \pi i \mathbb Z)$ is the Chern class map&#xA;(with a $2\pi i$ twist, or &lt;em&gt;Tate twist&lt;/em&gt;; this is natural in the algebraic context, and to get the topological Chern class you just divide through by&#xA;$2 \pi i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that we see that the kernel of the Chern class map can be identified with&#xA;$H^1(X,\mathcal O_X)/H^1(X,2 \pi i\mathbb Z)$, and vanishes when $H^1(X,\mathcal O_X) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image of $Pic(X)$ under the Chern class map is called the Neron--Severi group; its kernel is denoted $Pic^0(X)$ or $Pic^{\tau}(X)$.   When $X$ is algebraic, $Pic(X)$ is naturally an algebraic group,  $Pic^0(X)$ is the connected component of the identity, and $H^1(X,\mathcal O_X)$ is the tangent space to the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is a smooth projective curve, then $Pic^0(X)$ is usually called the &lt;em&gt;Jacobian&lt;/em&gt; of $X$.  You can look at the section of Hartshorne in Chapter IV to get some sense of it, although you may not realize from reading that how fundamental the role of the Jacobian is in the theory of algebraic curves.  If you google Torrelli theorem, Abel--Jacobi theorem, and theta divisor (just to give some sample search terms) you will get some sense of it.  Griffiths and Harris also has a detailed discussion, which gives a better sense of its significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is algebraic but not proper, then you can compacitify it by adding a divisor at infinity.  Let me write $\overline{X}$ for the compacification, and let me assume that $\overline{X}$ is in fact smooth, so then the divisor $D := \overline{X}\setminus X$ is a Cartier divisor, and gives rise to an associated line bundle $\mathcal O(D) \in  Pic(\overline{X}).$  (This is denoted&#xA;$\mathcal L(D)$ in Hartshorne, I think, and in some other texts, especially older ones, but $\mathcal O(D)$ is more common notation these days, and is better notation too.)  If $D$ is reducible (as can happen; in general it can be taken to be a normal crossings divisor, but no better --- e.g. to compactify a curve to a smooth projective curve, we have to add in a finite number of points, but one point will not be enough, typically), write it as &#xA;$D_1 \cup \cdots \cup D_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we also have associated line bundles $\mathcal O(D_i)$ for each $i$, whose product is $\mathcal O(D)$, and note that each of these is trivial when restricted to $X$ (because $X = \overline{X} \setminus D$).  One now sees&#xA;that $Pic(X) = Pic(\overline{X})/\langle \mathcal O(D_1),\ldots,\mathcal O(D_n)&#xD;&#xA;\rangle,$ and so if $Pic^0(\overline{X})$ is non-trivial, then $Pic(X)$ will also have a non-discrete part (because we can't kill a connected algebraic group by quotienting out a finitely generated subgroup).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is, at least for smooth $X$, is: compactify $X$ to $\overline{X}$, and then compute $H^1(\overline{X}, \mathcal O)$; if this is non-trivial, then the Chern class map has a (huge!) kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Added later: As an example, if $X$ is a hypersurface in $\mathbb P^n$ for $n &amp;gt; 2$ (so $X$ has dimension $&gt; 1$), then $H^1(X,\mathcal O_X) = 0$ (exercise!), and so hypersurfaces give interesting examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a surfaces, the dimension $H^1(X,\mathcal O_X)$ was classically (i.e. by the Italians) known as the irregularity of the surface $X$.  (The reason being that they knew formulas, like Riemann--Roch, for surfaces in space, which when they tried to extend to more general surfaces became false unless the extra quantity $\dim H^1(X,\mathcal O_X)$ was introduced --- although of course they didn't describe it this way.)   See &lt;a href=&quot;http://quomodocumque.wordpress.com/2009/08/05/professor-severi-is-a-jerk/&quot; rel=&quot;nofollow&quot;&gt;my comment here&lt;/a&gt;, as well as the notes of Kleiman linked to by Jason Starr, which will tell you a lot about Picard varieties and much more.]&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-19T06:48:43.193" LastActivityDate="2010-12-19T06:48:43.193" CommentCount="10" />
  <row Id="14839" PostTypeId="2" ParentId="14834" CreationDate="2010-12-19T06:09:28.027" Score="2" ViewCount="0" Body="&lt;p&gt;I always thought it was a script T, as is used in John Lee's &quot;Introduction to Smooth Manifolds&quot; (though admittedly for tensor &lt;em&gt;fields&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-19T06:09:28.027" CommentCount="2" />
  <row Id="14840" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T06:16:59.887" Score="4" ViewCount="0" Body="&lt;p&gt;The function $\sec(x)=\frac{1}{\cos(x)}$ also has a series expansion that will give you this result, but it is not its Taylor series.  Namely, its &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_fractions_in_complex_analysis&quot; rel=&quot;nofollow&quot;&gt;partial fraction expansion&lt;/a&gt; is $$\sec(x)=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi }{x^2-\left (n-\frac{1}{2}\right )^2\pi^2},$$ &#xA;as derived in Example 1 at &lt;a href=&quot;http://books.google.com/books?id=DYCOCBCBwoIC&amp;amp;lpg=PA64&amp;amp;dq=partial%20fraction%20series%20csc%20cot%20sec%20complex&amp;amp;pg=PA59#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; to Markushevich and Silverman, for example.  The series converges to $\sec(x)$ everywhere the latter is defined (and uniformly on compact subsets of the domain).  Plugging in $x=0$ and multiplying both sides by $\frac{\pi}{4}$ yields $$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-19T06:16:59.887" />
  <row Id="14841" PostTypeId="1" CreationDate="2010-12-19T06:33:42.950" Score="5" ViewCount="432" Body="&lt;p&gt;What is the geometric, physical or other meaning of the third derivative of a function at a point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Originally &lt;a href=&quot;http://mathoverflow.net/questions/49820/whats-meaning-of-geometrical-meaning-of-third-derivative-of-a-function-at-a-poin&quot; rel=&quot;nofollow&quot;&gt;asked on MO&lt;/a&gt; by AJAY)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have interesting things to say about the meaning of the first, and second derivatives, please do.&lt;/p&gt;&#xA;" OwnerUserId="1594" LastEditorUserId="1594" LastEditorDisplayName="" LastEditDate="2010-12-19T07:20:19.633" LastActivityDate="2010-12-19T09:04:00.387" Title="What is the meaning of the third derivative of a function at a point" Tags="&lt;calculus&gt;&lt;geometry&gt;&lt;physics&gt;&lt;education&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="14842" PostTypeId="2" ParentId="14828" CreationDate="2010-12-19T06:34:11.683" Score="7" ViewCount="0" Body="&lt;p&gt;Yes. And no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with $\mathbb{N}$, and define $+$ and $\times$ (and $\lt$ and so on) appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you define an equivalence relation on $\mathbb{N}\times\mathbb{N}$ given by &#xA;$$(a,b)\sim(c,d) \Longleftrightarrow a+d=b+c,$$&#xD;&#xA;and call the quotient set $(\mathbb{N}\times\mathbb{N})/\sim$ by the name $\mathbb{Z}$. (Behind the scenes, we are thinking of $(a,b)$ as meaning &quot;the solution to $a=x+b$&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then define an addition $+_{\mathbb{Z}}$ and a product $\times_{\mathbb{Z}}$ on $\mathbb{Z}$, as well as an order $\leq_{\mathbb{Z}}$ by&#xA;$$\begin{array}{rcl}&#xD;&#xA;[(a,b)]+_{\mathbb{Z}}[(c,d)] &amp;amp;=&amp;amp; [(a+c,b+d)]\\\&#xD;&#xA;[(c,d)]\times_{\mathbb{Z}}[(c,d)] &amp;amp;=&amp;amp; [(ac+bd,ad+bc)]\\\&#xD;&#xA;[(a,b)] \leq [(c,d)] &amp;amp;\Leftrightarrow&amp;amp; a+d\leq b+c,&#xD;&#xA;\end{array}$$&#xD;&#xA;and show that this is well defined. (I am using $[(a,b)]$ to denote the equivalence class of the pair $(a,b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, $\mathbb{N}$ and $\mathbb{Z}$ are entirely different animals; set-theoretically, you can even show that they are disjoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can define a map $f\colon \mathbb{N}\to\mathbb{Z}$ by $f(n) = [(n,0)]$. This map is one-to-one, and for all $n,m\in\mathbb{N}$,&#xA;$$\begin{array}{rcl}&#xD;&#xA;f(n+m) &amp;amp;=&amp;amp; f(n)+_{\mathbb{Z}}f(m),\\\&#xD;&#xA;f(n\times m) &amp;amp;=&amp;amp; f(n)\times_{\mathbb{Z}}f(m)\\\&#xD;&#xA;n\leq m &amp;amp;\Leftrightarrow&amp;amp; f(n)\leq_{\mathbb{Z}} f(m)&#xD;&#xA;\end{array}$$&#xD;&#xA;That means that even though $\mathbb{N}$ and $\mathbb{Z}$ are disjoint, there is a &quot;perfect copy&quot; of $\mathbb{N}$ (in so far as its operations $+$ and $\times$ are concerned, and as far as the order $\leq$ is concerned) sitting inside of $\mathbb{Z}$. (&lt;em&gt;Added:&lt;/em&gt; In fact, this $f$ not only gives us perfect copy, it is the &lt;em&gt;only&lt;/em&gt; map from $\mathbb{N}$ to $\mathbb{Z}$ that is one-to-one and respects all the operations; we say it is a &quot;canonical embedding&quot;). Since we have this perfect copy, and a very specific map identifying this copy with the original, we can think of $\mathbb{N}$ as being a subset of $\mathbb{Z}$ by &quot;identifying it with its copy&quot;. So we do that. We can then introduce notation by showing that for every $[(a,b)]\in\mathbb{Z}$, either $a=b$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(a,b)]=f(n)=[(n,0)]$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(b,a)]=f(n)=[(n,0)]$; and then using $0$ to denote the class with $a=b$, $n$ to denote the class with $[(a,b)]=[(n,0)]$, and $-n$ to denote the class $[(c,d)]$ with $[(d,c)]=[(n,0)]$. This notation makes the identification clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, once we have $\mathbb{Z}$, we define $\mathbb{Q}$ as the quotient of $\mathbb{Z}\times(\mathbb{Z}-\{0\}$ modulo $\cong$, where &#xA;$$(a,b)\cong (c,d) \Longleftrightarrow ad=bc$$&#xD;&#xA;(behind the scenes, we think of $(a,b)$ as meaning &quot;the solution to $a=xb$&quot;). &#xA;We can then proceed as before, defining&#xA;$$\begin{array}{rcl}&#xD;&#xA;[(a,b)]+_{\mathbb{Q}}[(c,d)] &amp;amp;=&amp;amp; [(ad+bc,bd)]\\\&#xD;&#xA;[(a,b)]\times_{\mathbb{Q}}[(c,d)] &amp;amp;=&amp;amp; [(ac,bd)]&#xD;&#xA;\end{array}$$&#xD;&#xA;and showing this is well defined; defining an order, etc. Again, $\mathbb{Q}$ and $\mathbb{Z}$ (and the original $\mathbb{N}$) are completely different &lt;em&gt;sets&lt;/em&gt;. But we have a function $g\colon\mathbb{Z}\to\mathbb{Q}$ defined by $g(a) = [(a,1)]$. This is one-to-one, $g(a+_{\mathbb{Z}}b) = g(a)+_{\mathbb{Q}}g(b)$, and $g(a\times_{\mathbb{Z}}b) = g(a)\times_{\mathbb{Q}}g(b)$. (&lt;em&gt;Added:&lt;/em&gt; And again, this is the &lt;em&gt;only&lt;/em&gt; map from $\mathbb{Z}$ to $\mathbb{Q}$ that satisfies these conditions.) So again, we have a &quot;perfect copy&quot; of $\mathbb{Z}$ sitting inside of $\mathbb{Q}$ (and so also a perfect copy of the perfect copy of $\mathbb{N}$ that is sitting inside of $\mathbb{Z}$). So once again we &quot;identify&quot; $\mathbb{Z}$ with its image inside $\mathbb{Q}$ (and so we identify $\mathbb{N}$ with its image inside the image of $\mathbb{Z}$). Because, via $f$ and $g$, we have perfect copies of them anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do the same thing with $\mathbb{Q}$ as being &quot;inside of $\mathbb{R}$&quot;, by identifying elements of $\mathbb{Q}$ with specific Dedekind cuts or with specific equivalence classes of Cauchy sequences, showing the identification is one-to-one and respects all the operations (and is essentially unique), and so obtaining a &quot;perfect copy&quot; of $\mathbb{Q}$ inside of $\mathbb{R}$ (and by extension, perfect copies of $\mathbb{N}$ and of $\mathbb{Z}$ also sitting inside of $\mathbb{R}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can keep going, of course: define $\mathbb{C}$ as the set of all pairs $\mathbb{R}\times\mathbb{R}$; then identify $\mathbb{R}$ with the pairs $\mathbb{R}\times\{0\}$, and you have a copy of $\mathbb{N}$ sitting inside a copy of $\mathbb{Z}$ sitting inside a copy of $\mathbb{Q}$ sitting inside a copy of $\mathbb{R}$ sitting inside $\mathbb{C}$. (And then you can stick $\mathbb{C}$ inside the quaternions, the quaternions inside the octonions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though they are actually very different sets, we have copies of each sitting inside the &quot;next one&quot;, copies that respect all the structures we are interested in, so we can still think of them as being &quot;subsets&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You used to do that all the time without the formalism: we think of &quot;fractions&quot; as being made up of an integer, a solidus, and a nonzero integer, so that &quot;$3$&quot; is not a fraction; but when needed, we are perfectly happy writing &quot;$3 = \frac{3}{1}$&quot; and working with either version of $3$ (the integer, or the fraction) depending on context.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T19:02:56.857" LastActivityDate="2010-12-19T19:02:56.857" CommentCount="1" />
  <row Id="14843" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T07:01:47.557" Score="1" ViewCount="0" Body="&lt;p&gt;By asking the question, you can interpret the 1st and 2nd derivatives in some meaningful ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the 3rd as the combination of them. In other words, consider the third derivative as the 'acceration' if the velocity was infact the displacement. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-19T07:01:47.557" CommentCount="2" />
  <row Id="14844" PostTypeId="1" AcceptedAnswerId="14874" CreationDate="2010-12-19T07:18:04.157" Score="4" ViewCount="108" Body="&lt;p&gt;cross-post from &lt;a href=&quot;http://stackoverflow.com/questions/4481973/permutations-distinct-under-given-symmetry-mathematica-8-group-theory&quot;&gt;stack overflow&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a list of numbers like {2,1,1,0} and I'd like to list all permutations of that list that are not equivalent under given symmetry group. So for instance if it was a dihedral group of order 4, result would be {{2, 1, 1, 0}, {2, 1, 0, 1}}. Is there a practical way to do this when the list is too large to generate all permutations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica 8 seems to have a few group theory functions, but I don't have any group theory background, so any pointers are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2010-12-20T23:15:49.550" Title="Listing permutations distinct under given symmetry in Mathematica" Tags="&lt;group-theory&gt;&lt;mathematica&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14845" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T07:25:48.320" Score="7" ViewCount="0" Body="&lt;p&gt;It is a common theme in applied math that you can easily interpret first and second derivative or moment (in case of probability theory), but after that trouble begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the third derivative is used in calculating torsion of a curve, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Torsion_of_a_curve&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Torsion_of_a_curve&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's review an example (rather poor one, I admit). Let's work in $\mathbb{R}^3$ with a Cartesian coordinate system $x$, $y$, $z$ and the associated basis $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$. Then let $\gamma$ be a unit circle given by $x = \cos \varphi$, $y = \sin \varphi$, $z = z_0$, parametrized naturally (by $\varphi$). Then $\dot{\gamma} \times \ddot{\gamma} = \mathbf{e}_3$, and then $\tau = {\dddot{\gamma}}^3 = 0$. I guess this particular curve is not very instructive, but I can't think of a better one off the top of my head :)&lt;/p&gt;&#xA;" OwnerUserId="3793" LastActivityDate="2010-12-19T07:25:48.320" CommentCount="2" />
  <row Id="14846" PostTypeId="2" ParentId="6009" CreationDate="2010-12-19T07:45:59.363" Score="0" ViewCount="0" Body="&lt;p&gt;the second problem is the one i too hv got my idea wud be arranging the men first in order and thn finding derangement of the 8 wommen put the answer dosnt match&lt;/p&gt;&#xA;" OwnerUserId="4924" LastActivityDate="2010-12-19T07:45:59.363" />
  <row Id="14847" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T07:52:34.817" Score="3" ViewCount="0" Body="&lt;p&gt;For the position function $p=f(t)$, you probably know that the first derivative $f&amp;#39;(t)$ gives the instantaneous velocity, and that the second derivative $f&amp;#39;&amp;#39;(t)$ gives the instantaneous acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of change of the acceleration is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jerk_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;&quot;jerk&quot;&lt;/a&gt; (also known as the &quot;surge&quot;, the &quot;jolt&quot;, or the &quot;lurch&quot;). So the third derivative $f&amp;#39;&amp;#39;&amp;#39;(t)$ would give the instantaneous jerk.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-19T07:52:34.817" CommentCount="3" />
  <row Id="14848" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T08:01:57.027" Score="0" ViewCount="0" Body="&lt;p&gt;I think this is relevant, and an explanation of QY's proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+i\,\tan\,\theta=\sec\,\theta(\cos\,\theta+i\,\sin\,\theta)$$&lt;br&gt;&#xA;$$1+i\,\tan\,\theta=\sec\,\theta\cdot e^{i\theta}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking logs, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln(1+i\,\tan\,\theta) = \ln(\sec\,\theta\cdot e^{i\theta})$$&lt;br&gt;&#xA;$$ \ln(1+i\,\tan\,\theta) = i\theta + \ln(\sec\,\theta)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest follows easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other interesting identities which stem from this, and are easily proved.&#xA;$$\arctan(1/2)+\arctan(1/3) = \pi/4$$&#xD;&#xA;$$4\arctan(1/5)-\arctan(1/239) = \pi/4$$&#xD;&#xA;$$4\arctan(1/5)-\arctan(1/70)+\arctan(1/99) = \pi/4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These final equations make it progressively easier to evaluate $\pi$, and that is what they are used for.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-19T09:04:51.323" LastActivityDate="2010-12-19T09:04:51.323" CommentCount="1" />
  <row Id="14849" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T08:37:54.357" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose y is a smooth function of x on the real line. The first derivative represents the gradient of the curve. An everywhere positive gradient implies that y is strictly increasing; an everywhere negative gradient implies that y is strictly decreasing. The second derivative represents the rate of change of the gradient. If the gradient is zero and the second derivative is negative, then we have a local maximum; if the gradient is zero and the second derivative is positive, then we have a local minimum, for example at the origin of the curve x squared. The third derivative represents the rate of change of gradient change. If the second derivative is zero and the third derivative is nonzero, then we have a point of inflection, for example at the origin of the curve x cubed.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-19T08:37:54.357" />
  <row Id="14850" PostTypeId="2" ParentId="14841" CreationDate="2010-12-19T09:04:00.387" Score="7" ViewCount="0" Body="&lt;p&gt;The rate of change of acceleration is studied in various situations in physics, mechanics and engineering design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Jerk_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physics, &lt;strong&gt;jerk&lt;/strong&gt;, also known as &lt;strong&gt;jolt&lt;/strong&gt; (especially in British English), &lt;strong&gt;surge&lt;/strong&gt; and &lt;strong&gt;lurch&lt;/strong&gt;, is the rate of change of acceleration; that is, the derivative of acceleration with respect to time, the second derivative of velocity, or the third derivative of position. Jerk is defined by any of the following equivalent expressions:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ \vec j=\frac {\mathrm{d} \vec a} {\mathrm{d}t}=\frac {\mathrm{d}^2 \vec v} {\mathrm{d}t^2}=\frac {\mathrm{d}^3 \vec s} {\mathrm{d}t^3}$$&#xA;  where&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\vec a$ is acceleration, $\vec v$ is velocity, $\vec s$ is position and $\mathit{t}$ is time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Jerk is a vector, and there is no generally used term to describe its scalar magnitude (e.g. &quot;speed&quot; as the scalar magnitude for velocity). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Think of roller coaster designs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, in mechanics, the fourth derivative is also studied. It is called Jounce or Snap. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Jounce&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physics, &lt;strong&gt;jounce&lt;/strong&gt; or &lt;strong&gt;snap&lt;/strong&gt; is the fourth derivative of the position vector with respect to time, with the first, second, and third derivatives being velocity, acceleration, and jerk, respectively; in other words, the jounce is the rate of change of the jerk with respect to time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\vec s =\frac {d \vec j} {dt}=\frac {d^2 \vec a} {dt^2}=\frac {d^3 \vec v} {dt^3}=\frac {d^4 \vec r} {dt^4}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes even the higher derivatives are nontrivial and come into play. Think of a sudden impact, an earthquake, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Shock_%28mechanics%29&quot; rel=&quot;nofollow&quot;&gt;shock&lt;/a&gt;, or a lightning effect in electrical systems. Constructs like &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;Dirac Delta function&lt;/a&gt; are very convenient for dealing with such situations. For instance, for a lightning, there is approximately a very high surge of current for a very brief instant and for any smooth approximation all the higher derivatives are nonzero. So you take the limit and manipulate it as if everything were concentrated at a single point. &lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2010-12-19T09:04:00.387" />
  <row Id="14851" PostTypeId="1" AcceptedAnswerId="14884" CreationDate="2010-12-19T09:42:27.567" Score="3" ViewCount="87" Body="&lt;p&gt;The recent post &lt;a href=&quot;http://math.stackexchange.com/questions/14540/fourier-cosine-transform/14546#14546&quot;&gt;here&lt;/a&gt; has led to the following question (consider $\cos(sx) = \sum\nolimits_{k = 1}^\infty  {\frac{{( - 1)^k (sx)^{2k} }}{{(2k)!}}}$). I find it instructive, whether it will turn out to be easy/well-known or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose that $(f_k)_{k \geq 1}$ is a sequence of continuous functions on a non-compact interval $I$. Suppose that $&#xD;&#xA;\sum\nolimits_{k = 1}^\infty  {f_k (x)}$ converges uniformly on every compact subinterval of $I$. Further, suppose that both $\sum\nolimits_{k = 1}^\infty  {\int_I {f_k (x)\,{\rm d}x} }$ and $\int_I {[\sum\nolimits_{k = 1}^\infty  {f_k (x)]} \,{\rm d}x}$ converge (i.e., are finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these assumptions, can you find an example where &#xA;$$&#xD;&#xA;\int_I {\bigg[\sum\limits_{k = 1}^\infty  {f_k (x)\bigg]} {\rm d}x} \neq \sum\limits_{k = 1}^\infty  {\int_I {f_k (x)\,{\rm d}x} }?&#xD;&#xA;$$&#xD;&#xA;Consider also the case where we are only given that the right-hand side converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If relevant: What about the case where $\sum\nolimits_{k = 1}^\infty  {f_k (x)}$ and its partial sums are analytic functions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any relevant discussion is welcome here.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-19T19:10:51.173" Title="Term-by-term integration under a certain condition" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14852" PostTypeId="1" AcceptedAnswerId="14856" CreationDate="2010-12-19T10:21:41.080" Score="3" ViewCount="77" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;How many ways of giving out 6 pieces of candy to 3 children if each child must receive at-least one piece ?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, $ 5 \choose 2$ is the answer,consistent to $ n-1 \choose r $ as this is similar to  finding the  number of non-zero positive integer solution of the equation $X_1 + X_2 + X_3 = 6$ ,but the correct answer (as given in my module) is $540$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where exactly am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2010-12-19T11:07:51.667" Title="About a combinatoric problem" Tags="&lt;combinatorics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14853" PostTypeId="1" AcceptedAnswerId="14859" CreationDate="2010-12-19T10:34:05.240" Score="0" ViewCount="114" Body="&lt;p&gt;How to show that:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A) $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i and e are constants&lt;br&gt;&#xA;A is a vector field&lt;br&gt;&#xA;$\nabla$ = vector differential operator&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditorDisplayName="" LastEditDate="2010-12-19T12:02:54.270" LastActivityDate="2010-12-19T12:02:54.270" Title="Cross product of operators" Tags="&lt;differential-geometry&gt;&lt;vector-calculus&gt;&lt;operator-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="14854" PostTypeId="1" CreationDate="2010-12-19T10:39:10.817" Score="1" ViewCount="133" Body="&lt;p&gt;I have a question about abelian group (or rather X-group):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose A is abelian group (or rather X-group) and it's artinian and noetherian. f is an endomorphism (or rather X-endomorphism) on A, i.e. $f\in \mathrm{End}(A)$. Also know that f(f(A))=f(A).&#xA;Prove $A=ker(f)\dot{+}f(A)$ i.e. A is direct sum of those two subgroup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is&#xA;firstly show $A=ker(f)+f(A)$ which I have proved. So next is to prove that $ker(f)\bigcap f(A)$ is trivial, i.e. =0. This part I got astray.&#xA;I thought I should use the noetherian condition here. Since, A is noetherian, then there exists an subgp C of A and $C\subset f(A)$, s.t. C is maximal in the sense of $ker(f)+C=0$, then find an subgp T, s.t. $C&amp;lt;T&amp;lt;f(A)$ satisfy $T\bigcap f(A)=0$ then contradict with maximality of C to reach a contradiction. But I get astray here to find this T. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I use the wrong method. Can anyone help me?&lt;/p&gt;&#xA;" OwnerUserId="4900" LastEditorUserId="4900" LastEditorDisplayName="" LastEditDate="2010-12-19T12:36:15.513" LastActivityDate="2010-12-19T12:36:15.513" Title="a problem about abelian group" Tags="&lt;homework&gt;&lt;abstract-algebra&gt;&lt;group-theory&gt;" CommentCount="12" />
  <row Id="14855" PostTypeId="1" AcceptedAnswerId="14903" CreationDate="2010-12-19T10:40:20.527" Score="3" ViewCount="111" Body="&lt;p&gt;Please give me a link to a reference for an example of a continuous function whose Fourier series diverges at a dense set of points. (given by Du Bois-Reymond). I couldn't find this in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T02:05:15.237" LastActivityDate="2010-12-20T02:05:15.237" Title="an example of a continuous function whose Fourier series diverges at a dense set of points." Tags="&lt;reference-request&gt;&lt;fourier-series&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14856" PostTypeId="2" ParentId="14852" CreationDate="2010-12-19T10:45:50.787" Score="4" ViewCount="0" Body="&lt;p&gt;You have assumed that $6$ candies are identical but that is not the case here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem can be modeled be letting A represent the set of candies and B represent the set of children.Then a function $f \colon A \to B $ can be interpreted as giving candy $a_i$ to $c_J$. Since each child must receive at-least one candy, we are considering only onto functions.So your problem boils down to counting number of onto functions in $f \colon A \to B $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate  this value we could use a formula which is actually derived by using the principle of inclusion-exclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A,B are non-empty sets of cardinality $m$,$n$ with $m \ge n$.Then there are &#xA;$$\sum_{i=0}^{(n-1)} (-1)^i {n \choose i} (n-i)^m \text{ onto functions in } f \colon A \to B $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we have $m = 6$ and $n = 3$ then we get : $3^6 - {3 \choose 1} (3 - 1)^6 + {3 \choose 2} (3 - 2)^6  = 540$ &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-19T11:07:51.667" LastActivityDate="2010-12-19T11:07:51.667" CommentCount="1" />
  <row Id="14857" PostTypeId="2" ParentId="14852" CreationDate="2010-12-19T11:02:53.410" Score="2" ViewCount="0" Body="&lt;p&gt;If all you care about is how many pieces of candy each chuld gets, then you are right. We would say in this case that the pieces of candy are &quot;indistinguishable&quot;. But if you care WHICH piece of candy each child gets, e.g if the individual candies are numbered from 1 to 6, then there are many solutions corresponding to each solution of the equation $X_1+X_2+X_3=6$, In this case the problem is to count the number of surjective (or &quot;onto&quot;) functions from a six-element set to a three-element set.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general there is no &quot;nice&quot; formula (comparable to the one for the binomial coefficients) for the number of surjective functions from an $n$ element set to an $m$ element set, but this number is $m!S(n,m)$, where the expression $S(n,m)$ refers to a Stirling number of the second kind, and is defined recursively for $n,m&amp;gt;1 $ by  $S(n,m)=mS(n-1,m)+S(n-1,m-1)$. The basis cases are a bit messy: $S(n,1)=1$, $S(n,n)=1$, and  $S(n,m)=0$ if $n&amp;lt;m$ or $n=m=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually $S(n,m)$ counts the number of partitions of $n$ distinguishable elements into $m$ non-empty sets. All onto functions from an $n$ element set to an $m$ element set are obtained choosing such a partition $p$ arbitrarily, and then assigning each of the $m$ elements of $p$  to one of the $m$ elements of the $m$-element set. There are $m!$ ways to do this, hence the formula $m!S(n,m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice discussion of all this can be found in Section 1.4, &quot;The Twelvefold Way&quot;, in Stanley's &quot;Enumerative Combinatorics&quot; volume 1.&lt;/p&gt;&#xA;" OwnerUserId="746" LastActivityDate="2010-12-19T11:02:53.410" />
  <row Id="14858" PostTypeId="2" ParentId="14855" CreationDate="2010-12-19T11:05:04.760" Score="4" ViewCount="0" Body="&lt;p&gt;Actually, such an almost-everywhere divergent Fourier series was constructed by Kolmogorov.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an explicit example, you can consider a &lt;a href=&quot;http://books.google.com/books?id=W9AxAjSiIaUC&amp;amp;pg=PA302&quot; rel=&quot;nofollow&quot;&gt;Riesz product&lt;/a&gt; of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \prod_{k=1}^\infty \left( 1+ i \frac{\cos 10^k x}{k}\right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is divergent. For more examples, see &lt;a href=&quot;http://projecteuclid.org/euclid.pja/1195523374&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/journals/proc/1958-009-02/S0002-9939-1958-0096068-0/S0002-9939-1958-0096068-0.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: (response to comment). Yes, you are right, du Bois Reymond did indeed construct the examples of Fourier series diverging at a dense set of points. However the result of Kolmogorov is stronger in that it gives almost everywhere divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The papers of du Bois Reymond are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ueber die Fourierschen Reihen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;available for free download &lt;a href=&quot;http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002515881&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; also another one &lt;a href=&quot;http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN00224361X&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1119" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-19T12:08:26.827" LastActivityDate="2010-12-19T12:08:26.827" CommentCount="10" />
  <row Id="14859" PostTypeId="2" ParentId="14853" CreationDate="2010-12-19T11:11:30.107" Score="4" ViewCount="0" Body="&lt;p&gt;As I suggested in my hint, apply the operator on a test function $f$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-i\nabla-eA)\times(-i\nabla-eA)f = (-i\nabla-eA)\times(-i\nabla f-e A f)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working this out we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-i\nabla-eA)\times(-i\nabla f-e A f)= -\nabla \times \nabla f + ie A\times \nabla f +i e \nabla \times (A f) + e^2 A \times A f$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and last term are zero. The third term can be expanded using &lt;a href=&quot;http://en.wikipedia.org/wiki/Curl_%28mathematics%29#Identities&quot; rel=&quot;nofollow&quot;&gt;the last identity&lt;/a&gt; (or simply working it out in terms of partial derivatives). This will give the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A)$$ &lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-19T11:11:30.107" CommentCount="6" />
  <row Id="14860" PostTypeId="2" ParentId="14358" CreationDate="2010-12-19T11:40:26.837" Score="1" ViewCount="0" Body="&lt;p&gt;I disagree with observation 2. It gives a sufficient condition that is not necessary unless you only consider projections onto a 1-dimensional subspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A weaker sufficient condition, where I am not sure whether its also necessary&#xA;is the following: if there is a decomposition $P_A G = BC$, where $B$ is&#xA;any regular matrix and $C$ is a matrix with only integer entries then $\Lambda_U$&#xD;&#xA;also must be a lattice. This must be because $C$ represents a map $\mathbb{Z}^n \rightarrow \mathbb{Z}^n$, and $B$ represents a vector space isomorphism. (note&#xA;that it does not matter if $C$ can be integer or rational, for any common denominator can be moved into $B$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems clear to me for geometric reasons that for any regular $G$ a suitable&#xA;1-dimensional $U$ can be found. Consider the basis vectors of the lattice, i.e.&#xA;the columns of $G$. There must be a hyperplane through them and a line through&#xA;the origin perpendicular to that hyperplane. An orthogonal projection onto that&#xA;line will move points parallel to the hyperplane. In particular, the basis&#xA;vectors will all be projected to the intersection of the line with the hyperplane. It follows that integer combinations of the basis&#xA;vectors will be mapped to integer multiples of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether there is a simple way to find a base for this&#xA;$U$, but it should always possible to take differences of basis vectors as a&#xA;basis for the hyperplane and use Gramm-Schmidt orthonogonalization to find a&#xA;normal vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested, it can be proved easily, at least in $\mathbb{R}^2$, that there is in fact an (countable) infinity of suitable $U$s.&#xA;This can be done by setting up a sine/cosine parametrization of the unit&#xA;vectors and showing that the ratio of the projection of the basis vectors of the&#xA;lattice is a continuous non-constant function of the angle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt; Actually, it is algebraically not as complicated as I imagined.&#xA;Supposing that $G$ is invertible, let $H = G^{-1}$ and take for $D$ an&#xA;$n \times 1$ matrix of all ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we let $A = H^TD$ then $A^TA$ becomes scalar and we can simplify&#xA;$$&#xD;&#xA;P_A = \frac{H^TDD^TH}{D^THH^TD}&#xD;&#xA;$$&#xD;&#xA;This means that&#xA;$$&#xD;&#xA;P_AG = \frac{H^TD}{D^THH^TD} D^T&#xD;&#xA;$$&#xD;&#xA;which means that the columns are all equal.&#xA;Alternatively, you could of course replace the ones in D with any rationals.&lt;/p&gt;&#xA;" OwnerUserId="3457" LastEditorUserId="3457" LastEditorDisplayName="" LastEditDate="2010-12-19T18:23:31.823" LastActivityDate="2010-12-19T18:23:31.823" CommentCount="1" />
  <row Id="14861" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T12:01:20.497" Score="1" ViewCount="0" Body="&lt;p&gt;For a historical perspective (i.e. if you want to see how geniuses struggled through things that would be natural today, mostly because of things they later discovered), the nice article &lt;a href=&quot;http://mathdl.maa.org/mathDL/46/?pa=content&amp;amp;sa=viewDocument&amp;amp;nodeId=1408&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Discovery of the Series Formula for π by Leibniz, Gregory and Nilakantha&lt;/em&gt;&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; by Ranjan Roy (1990) describes how the formula was discovered independently thrice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The series (2) was obtained independently by Gottfried Wilhelm Leibniz (1646-1716), James Gregory (1638-1675) and an Indian mathematician of the fourteenth century or probably the fifteenth cen­tury whose identity is not definitely known. Usually ascribed to Nilakantha, the Indian proof of (2) ap­pears to date from the mid-fifteenth century and was a consequence of an effort to rectify the circle. […] Leibniz's work, in fact, was primarily concerned with quadrature; the π/4 series resulted (in 1673) when he applied his method to the circle. Gregory, by comparison, was interested in finding an infinite series representation of any given func­tion and discovered the relationship between this and the successive derivatives of the given function. Gre­gory's discovery, made in 1671, is none other than the Taylor series; note that Taylor was not born un­til 1685. […]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, although the proofs of (2) by Leibniz, Gregory and Nilakantha are very different in approach and motivation, they all bear a relation to the modem proof given above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps I'll come back and edit this post for a summary of their methods if I actually read the article. :-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;: Ranjan Roy, &lt;em&gt;Mathematics Magazine&lt;/em&gt;, Vol. 63 (1990), pp. 291-306. I found this while flipping through the book &lt;a href=&quot;http://books.google.com/books?id=BKRE5AjRM3AC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Sherlock Holmes in Babylon: and other tales of mathematical history&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2010-12-19T12:01:20.497" />
  <row Id="14862" PostTypeId="1" AcceptedAnswerId="14871" CreationDate="2010-12-19T12:46:29.773" Score="5" ViewCount="115" Body="&lt;p&gt;I had this problem on my last exam in calculus, I still don't know, how to solve this, so I would appreciate your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $a_{ij} \in \mathbb{R}$ for $i,j \in \mathbb{N}$. Suppose, that for every $j$ series $S_j = \sum_{i=1}^{\infty}{a_{ij}}$ is absolutely convergent and, for every $i$ exists finite limit (limit $\not= \pm\infty$) $c_i=\lim_{j\to\infty}{a_{ij}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Prove, that if exists such absolutely convergent series $\sum_{i=1}^{\infty}{b_i}$ that $|a_{ij}| \leq |b_i|$ for every $i,j$ then $S_j \to S = \sum_{i=1}^{\infty}{c_i}$, $j \to \infty$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Is argument in (a) true without assumption that series $\sum{}b_i$  with properties given in (a) exists?&lt;/p&gt;&#xA;" OwnerUserId="3856" LastActivityDate="2010-12-19T14:47:22.977" Title="Series absolute convergence, proof" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14864" PostTypeId="2" ParentId="14358" CreationDate="2010-12-19T13:23:01.057" Score="1" ViewCount="0" Body="&lt;p&gt;The technique you suggest is sometimes used to construct &lt;a href=&quot;http://en.wikipedia.org/wiki/Aperiodic_tiling#Cut-and-project_method&quot; rel=&quot;nofollow&quot;&gt;quasi-crystaline structures&lt;/a&gt; or in mathematical parlance, aperiodic tilings. If you take a high dimensional lattice and you project on an appropriate subspace U, the result can be the lattice structure of a quasi-crystal. &lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-19T13:23:01.057" CommentCount="1" />
  <row Id="14865" PostTypeId="1" AcceptedAnswerId="14866" CreationDate="2010-12-19T13:25:38.320" Score="0" ViewCount="98" Body="&lt;p&gt;I've looked but cannot find the mathematical way to represent the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r = Max(x1, x2, x3)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I want to mathematically show that r = max value of the set (x1, x2, x3). To make sure I'm being clear, I want to write the math formula of the code equivalent.&lt;/p&gt;&#xA;" OwnerUserId="4930" LastActivityDate="2010-12-19T13:53:35.760" Title="Mathematical representation of the largest element in a set" Tags="&lt;set-theory&gt;&lt;notation&gt;&lt;elementary&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14866" PostTypeId="2" ParentId="14865" CreationDate="2010-12-19T13:29:01.220" Score="2" ViewCount="0" Body="&lt;p&gt;As I responded on stackoverflow...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Infinity_norm&quot; rel=&quot;nofollow&quot;&gt;infinity/uniform norm&lt;/a&gt; of the corresponding tuple, which is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{n \rightarrow \infty} (|x_1|^n + |x_2|^n + |x_3|^n + ...)^{\frac{1}{n}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can just have a maximum/infimum function... what exactly is your problem with using that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can define $f\mbox{ = max}$ recursively as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\{a_1,a_2,a_3,\cdots\}) = \begin{cases} a_1 &amp;amp; \mbox{if the sequence is singleton} \\ f(\{a_2,a_3,a_4,\cdots\}) &amp;amp; \mbox{if } a_1 \leq a_2 \\ f(\{a_1,a_3,a_4,\cdots\}) &amp;amp; \mbox{otherwise} \end{cases}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be made a lot nicer if you allow $f$ to be a two variable function with an accumulator.&lt;/p&gt;&#xA;" OwnerUserId="1507" LastEditorUserId="1507" LastEditorDisplayName="" LastEditDate="2010-12-19T13:52:37.987" LastActivityDate="2010-12-19T13:52:37.987" CommentCount="3" />
  <row Id="14868" PostTypeId="2" ParentId="14865" CreationDate="2010-12-19T13:53:35.760" Score="1" ViewCount="0" Body="&lt;p&gt;I wouldn't write $\max(a,b)$ or $\max(a,b,c)$. For me, max assigns to a &lt;em&gt;set&lt;/em&gt; its maximal element (if it exists), so I use $\max\{a,b\}$ and $\max\{a,b,c\}$. In general, if $A$ is a set of real numbers, I write $\max A := \sup A$ (the supremum of $A$) if $\sup A$ is an element of $A$.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-19T13:53:35.760" CommentCount="1" />
  <row Id="14869" PostTypeId="2" ParentId="14826" CreationDate="2010-12-19T14:10:11.230" Score="2" ViewCount="0" Body="&lt;p&gt;Deat Akhil, I have nothing to add to Matt E's masterful survey on the algebraic/analytic comparison. However since you ask about &lt;em&gt;topological&lt;/em&gt; line bundles, you have to modify his answer in the following way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must replace everywhere $\mathcal O$ by $\mathcal C$, the sheaf of continuous functions. Now things are very easy: since $\mathcal C$ is soft (fine if you prefer), it is acyclic and so the map &#xA;$Pic^{top}(X) \to H^2(X,&#xD;&#xA; \mathbb Z)$ is an isomorphism: a topological line bundle is classified by its Chern class, which lives in the second cohomology group of the space. To put it dramatically: the continuous jacobian is trivial ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example topological line bundles on a compact Riemann surface are classified by  $\mathbb Z$ in stark contrast to the huge Picard variety classifying its algebraic= analytic line bundles. So to answer your question &quot;To what extent is this true in general?&quot; [natural bijection between  algebraic and topological line bundles] I would answer, just for the pleasure of using the anglicism : &quot;once in a blue moon&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3217" LastEditorUserId="3217" LastEditorDisplayName="" LastEditDate="2010-12-19T18:37:05.063" LastActivityDate="2010-12-19T18:37:05.063" CommentCount="4" />
  <row Id="14870" PostTypeId="2" ParentId="14815" CreationDate="2010-12-19T14:26:45.343" Score="1" ViewCount="0" Body="&lt;p&gt;This is a follow-up to the comment about Abel's theorem below Willie Wong's post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need Abel's theorem if you proceed as follows (taken from &quot;What is Mathematics?&quot; by Courant and Robbins):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{k=0}^{n} (-1)^n x^2 = \frac{1+(-1)^nx^{2n+2}}{1+x^2} = \frac{1}{1+x^2} + \frac{(-1)^nx^{2n+2}}{1+x^2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you integrate each side from $x=0$ to $x=1$ and then take a limit, bounding the rightmost term in the right integral by discarding the $x^2$ term in the denominator.  This is nice as it allows a rigorous presentation of this to (a suitably motivated) audience of first year calculus students.&lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-19T14:26:45.343" />
  <row Id="14871" PostTypeId="2" ParentId="14862" CreationDate="2010-12-19T14:47:22.977" Score="5" ViewCount="0" Body="&lt;p&gt;First let us look at part (b) to extract some ideas about proving part (a). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the double sequence $a_{ij} = \delta_{ij}$, that is, $a_{ij} = 1$ if $i = j$ and $a_{ij} = 0$ otherwise. Then $S_j = 1$ for any $j$. On the other hand, $c_i = 0$ for any $i$. So $S = \sum c_i = 0 \neq \lim S_j$. So without the dominating sequence $b_i$ the statement is false.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Like Pete says, this is another reflection of something in the continuous case, namely the principle of weak convergence and also Fatou's lemma.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we prove part (a). Fix $\epsilon &amp;gt; 0$. Let $N$ be chosen large enough such that $\sum_{i = N}^\infty |b_i| &amp;lt; \epsilon/4 $, this $N$ exists since $(b_i)$ is absolutely convergent. Consider the partial sums $S_{j,N} = \sum_{i = 1}^{N-1} a_{ij}$. We can then pick $M$ sufficiently large such that for all $j &amp;gt; M$ and  $i &amp;lt; N$, $|a_{ij} - c_i| &amp;lt; \epsilon / (4N)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in particular, $|S_{j,N} - S_N| &amp;lt; \epsilon / 4$ for $j &amp;gt; M$ where $S_N = \sum_{i = 1}^{N-1} c_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by assumption of convergence, and the domination of $(b_i)$, you have $|c_i| &amp;lt; |b_i|$. So in particular &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S - S_N| = | \sum_{N}^{\infty} c_i | \leq \sum_{N}^\infty |b_i| \leq \epsilon/4 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S_N - S_{j,N} | &amp;lt; \epsilon / 4, \quad j &amp;gt; M $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $M$ implicitly depends on $N$, and hence on $\epsilon$)  and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S_j - S_{j,N} | &amp;lt; \epsilon / 4 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have that, applying the triangle inequality, that &lt;em&gt;for every fixed $\epsilon &amp;gt; 0$, we can pick $M$ sufficiently large such that&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |S_j - S| &amp;lt; \epsilon \qquad \forall j &amp;gt; M$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-19T14:47:22.977" CommentCount="4" />
  <row Id="14872" PostTypeId="2" ParentId="9999" CreationDate="2010-12-19T15:10:23.837" Score="1" ViewCount="0" Body="&lt;p&gt;Using integers only, I would use &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search&quot; rel=&quot;nofollow&quot;&gt;Binary Search&lt;/a&gt;. Certainly you can compute $F_n$ only with integers, the simplest way is matrix exponentiation. Using Binary Search you can find numbers ``near'' your number $x$ and you will find either $x = F_n$ or $F_{n+1} &amp;gt; x &amp;gt; F_n$. I suppose this method is generic for anything monotone you can compute fast. To initiliaze the binary search, just keep doubling $ F_{2n} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Binary search allows you to search for a number x in a sorted &quot;array&quot; F[] (in the programming sense). Use this method to search for your number. When you need F[n] just compute $F_n$. This will work because the sequence is strictly increasing except for the initial 1,1.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-19T15:26:56.887" LastActivityDate="2010-12-19T15:26:56.887" />
  <row Id="14873" PostTypeId="2" ParentId="154" CreationDate="2010-12-19T15:35:30.110" Score="1" ViewCount="0" Body="&lt;p&gt;Ordered pairs exist.&#xA;If you define an operation on something that exists, the thing 'with the operation' exists.&#xA;Certain operations can be defined on ordered pairs&#xA;Complex numbers are ordered pairs with those operations defined on them.&#xA;∴ Complex numbers exist.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2010-12-19T15:35:30.110" />
  <row Id="14874" PostTypeId="2" ParentId="14844" CreationDate="2010-12-19T15:55:19.737" Score="3" ViewCount="0" Body="&lt;p&gt;To do this in (the free, open-source, group-theory software) &lt;a href=&quot;http://www.gap-system.org/&quot; rel=&quot;nofollow&quot;&gt;GAP&lt;/a&gt;, you could use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;g := Group( (1,2)(3,4), (1,3)(2,4), (1,2,3,4) );;&#xA;d := PermutationsList( [2,1,1,0] );;&#xA;o := OrbitsDomain( g, d, Permuted );;&#xA;r := List( o, orbit -&amp;gt; Reversed( AsSet( orbit ) )[1] );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first command defines a dihedral group (acting on the vertices of a square).  The second command defines the domain on which the group will act: the set of all permutations of the list [2,1,1,0].  This list is constructed in memory, so this limits the size of the list.  You can calculate the size of the list with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NrPermutationsList([2,1,1,0]);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and look for a different method if the list is too long (up to ten thousand will be very quick, up to a million will be fine, after ten or twenty million you'll need to try something else).  The third command sorts the domain into &quot;orbits&quot;, the sets of equivalent permutations, with the group acting on the domain via the function &quot;Permuted&quot; which permutes the indices of a list.  The fourth command asks for orbit representatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If g is small compared to d, then you are stuck.  This is simply how hard the problem is.  If g is reasonably large compared to d you can avoid computing d explicitly, using a transversal of an overgroup of g in the symmetric group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, try this version first.  If you get to an example that doesn't work, post a specific group and list and I'll describe how the transversal works (it is often going to be slower, just more efficient in memory usage).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To work more specifically with your problem here are some GAP commands.  For many partitions, GAP takes too long (so memory efficiency is not the problem).  To get a random partition you could use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x:=Random(Partitions(16));;&#xA;Append(x,ListWithIdenticalEntries(16-Size(x),0));;&#xA;x; NrPermutationsList(x);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'll take:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x := [ 2, 2, 2, 2, 2, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0 ];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now construct the symmetry group of the hypercube, the set of permutations of the partition, the orbits, and the orbit representatives as before:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;g := WreathProductProductAction(SymmetricGroup(2),SymmetricGroup(4));;&#xA;d := PermutationsList( x );;&#xA;o := OrbitsDomain( g, d, Permuted );;&#xA;r := List( o, orbit -&amp;gt; Reversed( AsSet( orbit ) )[1] );;&#xA;Size( r );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After about a minute, it should print 1292.  For 16 = 1+1+2+3+4+5 it has run for a day or so, and still isn't finished.  No trouble with memory usage, it is just sifting through 3 million orderings.  So, it is not practical for all the partitions, but for many it will work just fine.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-20T23:15:49.550" LastActivityDate="2010-12-20T23:15:49.550" CommentCount="5" />
  <row Id="14876" PostTypeId="1" CreationDate="2010-12-19T16:58:56.543" Score="4" ViewCount="219" Body="&lt;p&gt;Hi, I am a Software Engineer. With interest on computers I have chosen this path. But I don't know anything about great maths. I want to learn from basic to some level(which would help me in my field..like identifying solution to a problem, designing algorithms, designing shapes and some animation on them). So, please suggest me where do I need to start. Please suggest me/list out the concepts that I need to go through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much&lt;/p&gt;&#xA;" OwnerUserId="4937" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-19T19:15:38.627" LastActivityDate="2010-12-20T14:03:20.613" Title="Suggest me the path to learn Maths" Tags="&lt;soft-question&gt;&lt;self-learning&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="14877" PostTypeId="2" ParentId="14821" CreationDate="2010-12-19T17:17:22.760" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Exploit the linearity of $\Sigma\:$: $\rm\ \Sigma\ (f(k)+c)\ = \ \Sigma\ f(k) + \Sigma\ c\ $ to reduce to a geometric sum.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-19T17:17:22.760" />
  <row Id="14878" PostTypeId="1" AcceptedAnswerId="14916" CreationDate="2010-12-19T17:24:01.417" Score="11" ViewCount="296" Body="&lt;p&gt;My algebra professor gave me this puzzle a while back.  I'm pretty sure I've found the right solution; nonetheless, I wanted to share it and see if you come up with anything really nice or unexpected.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prove that if you take a unit square and cut it into a finite number of smaller squares (in any way you can think of), the side lengths of the smaller squares are all rational.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. The first tag was my professor's hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] Just to be clear, every piece must actually be a square (e.g. no gluing two triangles into a square).&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-20T03:21:53.763" LastActivityDate="2011-01-03T04:53:28.177" Title="Cutting a unit square into smaller squares" Tags="&lt;linear-algebra&gt;&lt;combinatorics&gt;&lt;puzzle&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="3" />
  <row Id="14879" PostTypeId="1" AcceptedAnswerId="14901" CreationDate="2010-12-19T18:13:56.870" Score="-13" ViewCount="901" Body="&lt;p&gt;I know 0.9999999999... =1   , cause of its infinite nines&lt;br&gt;&#xA;But what about 0.9999999..(infinite nines)...999998&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it also 1?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Notice the 8 in the end)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a difference, as we see clearly, but they are also the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As T.. answer says, it seems I have invented new math field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We naturally call these super-numbers for kalle-numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to know what are the generall consequences for this extension of math, or I want a proof that they are useless (Which I highly doubt).&lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-21T17:47:46.460" LastActivityDate="2010-12-21T17:47:46.460" ClosedDate="2010-12-20T19:45:13.227" Title="What are the implications of the discovery of the kalle-numbers ?" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;" AnswerCount="3" CommentCount="34" FavoriteCount="1" />
  <row Id="14880" PostTypeId="2" ParentId="14879" CreationDate="2010-12-19T18:20:46.523" Score="23" ViewCount="0" Body="&lt;p&gt;What does it mean to have &quot;infinitely many 9's and then an 8&quot;?  Well, you're putting the 8 after the last 9, but there is no last 9!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible way to get around this is to consider the sequence 0.98, 0.998, 0.9998, etc., where the n-th term of the sequence has n 9's and then an 8.  Then your question becomes, &quot;is the limit of this sequence, as n goes to infinity, equal to 1?&quot; to which the answer is, &quot;yes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2010-12-19T18:20:46.523" CommentCount="10" />
  <row Id="14881" PostTypeId="1" CreationDate="2010-12-19T18:21:55.633" Score="-2" ViewCount="71" Body="&lt;p&gt;Hi&#xA;there are how many modes between 4 number ? &#xA;thanks&lt;/p&gt;&#xA;" OwnerUserId="4940" LastActivityDate="2010-12-19T18:21:55.633" ClosedDate="2010-12-19T19:19:06.737" Title="there are how many modes between 4 number ? " Tags="&lt;number&gt;" CommentCount="2" />
  <row Id="14882" PostTypeId="2" ParentId="14876" CreationDate="2010-12-19T18:30:24.320" Score="3" ViewCount="0" Body="&lt;p&gt;Perhaps what you mean is that you want a book like Polya's &quot;How to solve it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691023565&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/How-Solve-Aspect-Mathematical-Method/dp/0691023565&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-19T18:30:24.320" />
  <row Id="14883" PostTypeId="1" CreationDate="2010-12-19T18:56:18.737" Score="-3" ViewCount="137" Body="&lt;p&gt;Does this equation define a function with independent variable x?&lt;br&gt;&#xA;xlyl=x+9&lt;/p&gt;&#xA;" OwnerUserId="4942" LastActivityDate="2010-12-19T19:29:57.253" Title="Defining a function" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14884" PostTypeId="2" ParentId="14851" CreationDate="2010-12-19T19:04:22.523" Score="4" ViewCount="0" Body="&lt;p&gt;Here I use notation $\chi_N$ to be the characteristic function of $[N,N+1]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f_1 = \chi_1$. Let $f_k = \chi_k - \chi_{k-1}$ for $k\geq 2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On any compact set of the reals, $\sum f_i$ converges uniformly to $f = 0$. So $\int \sum f_i = 0$. Yet $\int f_1 = 1$ and $\int f_k = 0$ for $k \geq 2$. So $\sum \int f_i = 1\neq \int\sum f_i$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Analyticity doesn't matter. Let $g(x)$ be any arbitrary, positive, integrable function on the real line which decays faster then, say, $|x|^{-1}$. Then if we take $f_1(x) = g(x-1) |_{\mathbb{R}_+}$ and $f_k(x) = g(x-k) - g(x-k+1) |_{\mathbb{R}_+}$ for $k\geq 2$. You get the same result that on any compact subset of the reals $\sum f_i$ converges uniformly to 0. Yet $\sum\int f_i \geq \int_0^\infty g(x) dx$. You can no insert in an arbitrary analytic $g(x)$ (say the Gaussian for example) and get the same conclusion. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-19T19:10:51.173" LastActivityDate="2010-12-19T19:10:51.173" CommentCount="3" />
  <row Id="14885" PostTypeId="2" ParentId="13054" CreationDate="2010-12-19T19:08:35.580" Score="0" ViewCount="0" Body="&lt;p&gt;Perhaps you could do something with the taylor series? Would be pretty nasty but youd only have to worry about the first few terms.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastActivityDate="2010-12-19T19:08:35.580" CommentCount="7" />
  <row Id="14886" PostTypeId="2" ParentId="14883" CreationDate="2010-12-19T19:19:48.600" Score="3" ViewCount="0" Body="&lt;p&gt;assuming that you mean $x|y|=x+9$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $|y|=1+9/x$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $-9 &amp;lt; x &amp;lt; 0 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $y$ is not defined. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-19T19:29:57.253" LastActivityDate="2010-12-19T19:29:57.253" CommentCount="10" />
  <row Id="14887" PostTypeId="1" AcceptedAnswerId="14889" CreationDate="2010-12-19T19:42:51.367" Score="6" ViewCount="149" Body="&lt;p&gt;I seem to have a misunderstanding of how to work with a Taylor series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I want to write $f(x)=x e^x$'s Taylor expansion of $n$ degree around $0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Find the $n$th derivative of $f(x)$, this is quite easy: $f^{(n)} (x)=n e^x+x e^x$.&lt;br&gt;&#xA;And from the formula I get: &#xA;$$f(x)=\sum_{k=1}^{n} \frac{x^k}{(k-1)!} + R_{n}(x).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Since $e^x$ Taylors expansion is already known, it's also possible to do this:&#xA;$$f(x)=xe^x=x\left(\sum_{k=0}^{n-1} \frac{x^k}{k!} + R_{n-1}(x)\right)=\sum_{k=0}^{n-1} \frac{x^{k+1}}{k!} + xR_{n-1}(x)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do I interpret $xR_{n-1}(x)$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm missing something fundamental about the meaning of $R_{n}(x)$.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-19T20:00:18.440" LastActivityDate="2010-12-20T04:59:07.480" Title="Basic Taylor expansion question" Tags="&lt;calculus&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="14888" PostTypeId="2" ParentId="11076" CreationDate="2010-12-19T19:49:05.200" Score="1" ViewCount="0" Body="&lt;p&gt;1) Sean Carroll claims that the essentials of the vector concept emerged with &lt;strong&gt;Galileo Galilei&lt;/strong&gt;. Galileo observed that if you kick a ball forwards off a building, or drop a ball off a building, that the balls land at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore horizontal motion (&lt;strong&gt;kick&lt;/strong&gt;) is &lt;strong&gt;separable&lt;/strong&gt; from vertical motion (&lt;strong&gt;fall&lt;/strong&gt;). From there you get the separation of the force into two parts: the 2 components of the vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Regarding the dot product, it is the most natural way to numerically capture the concept of &quot;angle between two things&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The pattern continues like this: $\sqrt{a^2 + b^2 + c^2 + d^2 + e^2 + \ldots}$ for a multi-dimensional right triangle with sides $a, b, c, d, e, \ldots$. Surprising, isn't it?&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2010-12-20T01:15:40.627" LastActivityDate="2010-12-20T01:15:40.627" />
  <row Id="14889" PostTypeId="2" ParentId="14887" CreationDate="2010-12-19T19:54:45.217" Score="14" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Edited and rewritten, hopefully to make it clearer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is something that the notation does not help and which you need to keep straight (this may be the source of your confusions). $R_n(x)$ is the &quot;remainder&quot; at $x$: the error between the actual value of the function and the value you get from evaluating the Taylor polynomial of degree $n$ instead. That is, the $n$th remainder is equal to&#xA;$$f(x) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$$&#xD;&#xA;(I'm assuming the polynomial is being expanded around $a=0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the remainder really needs three pieces of information to make it precise: the value of $n$, the value of $x$, and the function $f$ in question. It would be best to write it as $R(n,x,f)$, but since the function $f$ is usually clear from context, we ignore it. But I will use $R(n,x,f)$ in what follows. So notice that in your question, in (1) what you have is $R(n,x,xe^x)$, while in (2) what you have is $xR(n-1,x,e^x)$.  And so the question is what is the relation between the two expressions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before addressing this in more detail, let me take a slight detour to Taylor series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have the Taylor &lt;em&gt;series&lt;/em&gt; for $f$,&#xA;$$\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k.$$&#xD;&#xA;We know this power series converges inside the radius of convergence (which can be found using, say, the Ratio Test); for $f(x)=e^x$, the radius of convergence is infinite, so the series converges for &lt;em&gt;all&lt;/em&gt; values of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is still the question of whether the value of the series at a given $x$ &lt;em&gt;equals&lt;/em&gt; the value of the function at that point. This is where the remainders come in. The Taylor series for $f$ converges to $f(x)$ at the point $x_0$ if and only if $R(n,x_0,f)\to 0$ as $n\to\infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $f(x)=e^x$, one can show that the Taylor series not only converges everywhere, but also that it converges &lt;strong&gt;to&lt;/strong&gt; $e^x$. For example, this can be done using the Cauchy estimate for the remainder: given $r\gt 0$, and $x$ in $(-r,r)$, if $M_n$ is a number such that $|f^{(n+1)}(x)|\leq M_n$ for all $x$ in $(-r,r)$, then&#xA;$$|R(n,x,f)|\leq \frac{M_nr^{n+1}}{(n+1)!}.$$&#xD;&#xA;For $f(x) = e^x$, you can take $M=e^r$ (or even $3^r$), so you get an exponential divided by a factorial, and that goes to $0$ as $n\to\infty$. This holds for any $x$ (by changing the $r$), so that $R(n,x,e^x)\to 0$ as $n\to\infty$ for all $x$. So the Taylor series for $e^x$ converges to $e^x$ at every $x$. That is, for each $x$,&#xA;$$e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}+\cdots$$&#xD;&#xA;in the sense that the value of the limit on the right hand side &lt;em&gt;is&lt;/em&gt; exactly equal to $e^x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Taylor series&quot; for $x$ is also very easy: it's just $x$ itself (not even a real series). It also has infinite radius of convergence, and $R(n,x,x) = 0$ for any $n\gt 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a theorem that if the Taylor series for $f$ converges to $f$ in $(-r,r)$, and the Taylor series for $g$ converges to $g$ in $(-R,R)$, then the product of the series will converge to $fg$ in $(-\min(r,R),\min(r,R))$ (that is, on the interval where they both converge). For $e^x$ and $x$, this tells you that indeed you have that &#xA;$$xe^x = x\left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right) = \sum_{k=0}^{\infty}\frac{x^{k+1}}{k!} = \sum_{k=1}^{\infty}\frac{x^k}{(k-1)!}.$$&#xD;&#xA;Because the series converges, and converges to $xe^x$, that means that if you write&#xA;$$xe^x = \sum_{k=1}^n \frac{x^k}{(k-1)!} + R(n,x,xe^x)$$&#xD;&#xA;then you have that &#xA;$$R(n,x,xe^x) = \sum_{k=n+1}^{\infty} \frac{x^k}{(k-1)!}$$&#xD;&#xA;(in the sense that the value of $R(n,x,xe^x)$ is the limit of the partial sums of that series) and moreover that&#xA;$$\sum_{k=n+1}^{\infty}\frac{x^k}{(k-1)!} \to 0\text{ as }n\to\infty.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are doing in (2) is to essentially deal with the Taylor polynomials instead of the series as I did above. We have that the Taylor series for $e^x$ converges to $e^x$ at every $x$, so that if we write&#xA;$$e^x = \sum_{k=0}^{n}\frac{x^k}{k!} + R(n,x,e^x)$$&#xD;&#xA;then&#xA;$$R(n,x,e^x) = \sum_{k=n+1}^{\infty}\frac{x^k}{k!}$$&#xD;&#xA;and $R(n,x,e^x)\to 0$ as $n\to \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying the expression for $e^x$ by $x$, we get&#xA;$$xe^x = x\left(\sum_{k=0}^n\frac{x^k}{k!} + R(n,x,e^x)\right) = \sum_{k=0}^n\frac{x^{k+1}}{k!}+xR(n,x,e^x).$$&#xD;&#xA;Except that note that this time $xR(n,x,e^x)$ is giving the $(n+1)$st remainder, since the polynomial we have is of degree $n+1$. So, equating this with the expression we had before, this suggests that we should have&#xA;$$R(n+1,x,xe^x) = xR(n,x,e^x).$$&#xD;&#xA;And indeed, this is what we find when we express them as series and limits:&#xA;\begin{align}&#xA;xR(n,x,e^x) &amp;amp;= x\left(\sum_{k=n+1}^{\infty}\frac{x^k}{k!}\right)\\&#xA;&amp;amp;=x\left(\lim_{m\to\infty}\sum_{k=n+1}^m\frac{x^k}{k!}\right)\\&#xA;&amp;amp;= \lim_{m\to\infty}\left(x\sum_{k=n+1}^m \frac{x^k}{k!}\right)\\&#xA;&amp;amp;= \lim_{m\to\infty}\sum_{k=n+1}^m \frac{x^{k+1}}{k!}\\&#xA;&amp;amp;=\lim_{m\to\infty}\sum_{k=n+2}^{m-1}\frac{x^k}{(k-1)!}\\&#xA;&amp;amp;= \lim_{m\to\infty}\sum_{k=n+2}^{m}\frac{x^k}{(k-1)!}\\&#xA;&amp;amp;= R(n+1,x,xe^x).&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this tells you is that the $n+1$st remainder at $x$ for the function $xe^x$ equals $x$ times the $n$th remainder at $x$ for the function $e^x$.  So your equations both make sense and they are telling you correct things, once you put back the necessary information into the remainder.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T04:59:07.480" LastActivityDate="2010-12-20T04:59:07.480" CommentCount="8" />
  <row Id="14891" PostTypeId="2" ParentId="14379" CreationDate="2010-12-19T20:44:46.580" Score="1" ViewCount="0" Body="&lt;p&gt;Since P is abelian, when acting upon P by conjugation from P, say you have p,q in P, then pqp^{-1} = q.  Thus P is indeed contained in the kernel of N -&gt; Aut(P), and hence induces a map N/P -&gt; Aut(P).  Examine the order of Aut(P) and this allows you to conclude that the image of N/P in Aut(P) must be trivial.  From this you can conclude the desired result.&lt;/p&gt;&#xA;" OwnerUserId="4943" LastActivityDate="2010-12-19T20:44:46.580" />
  <row Id="14892" PostTypeId="2" ParentId="14876" CreationDate="2010-12-19T20:52:36.003" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;b&gt;program ``sage''&lt;/b&gt; helped me a lot. You can experiment with math objects in a programming environment. It is free software - GPL2. You can try it online in a browser. The site is &lt;a href=&quot;http://sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;http://sagemath.org/&lt;/a&gt;. More information: &lt;a href=&quot;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This will not teach you math, but may help you.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-20T08:37:15.153" LastActivityDate="2010-12-20T08:37:15.153" />
  <row Id="14893" PostTypeId="2" ParentId="14629" CreationDate="2010-12-19T20:54:49.803" Score="2" ViewCount="0" Body="&lt;p&gt;So, if for $x\in \partial D^{2n}$ you take $G(x)=g(x)$, then probably you want to do &quot;the same&quot; map over the whole cone.  It follows from the definition of the union topology (or whatever it's called) that this extends to a continuous map.  Then, look at the induced map on the pairs $(K,K-D^{2n})$ (for either complex $K$), and here the map is just (up to confusing maps on pairs with maps on their quotients) the suspension of $g$.  Note that for both complexes, the inclusion $(K,\emptyset)\rightarrow (K,K-D^{2n})$ is a cohomology isomorphism in degree $2n$, and use the fact that suspension induces an isomorphism $\pi_n(S^n)\rightarrow \pi_{n+1}(S^{n+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram might help:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    (Cfg,o)    ---&amp;gt;     (Cf,o)&#xA;       |                  |&#xA;       |                  |&#xA;       V                  V&#xA;(Cfg,Cfg-D^2n) ---&amp;gt; (Cf, Cf-D^2n)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here, the o's stand for $\emptyset$.  The vertical maps are inclusions of pairs, and the horizontal lines are really all just $G$ (as I defined it).  The top arrow induces $\alpha_{Cf} \mapsto \alpha_{Cfg}$, and the vertical maps are cohomology isomorphisms in degree $2n$.  On the bottom row, since $H^*(X,A)=\tilde{H}^*(X/A)$, this is basically a map $S^{2n}\rightarrow S^{2n}$, which by construction is exactly the suspension $Sg$ of $g$.  Then, $\deg(Sg)=\deg(g)$; that is, $S:\pi_{2n-1}(S^{2n-1})\rightarrow \pi_{2n}(S^{2n})$ is an isomorphism.  So this induces $\beta_{Cf}\mapsto \deg(g)\cdot \beta_{Cfg}$.  By naturality,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H(fg)\cdot \beta_{Cfg} = \alpha_{Cfg}^2=G^*(\alpha_{Cf}^2)=G^*(H(f)\cdot \beta_{Cf})=H(f)\cdot (\deg(g)\cdot \beta_{Cfg}).$$&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="2159" LastEditorDisplayName="" LastEditDate="2010-12-20T06:17:41.787" LastActivityDate="2010-12-20T06:17:41.787" CommentCount="4" />
  <row Id="14894" PostTypeId="1" CreationDate="2010-12-19T21:14:15.917" Score="4" ViewCount="82" Body="&lt;p&gt;Does anyone have a good explanation for why Knot invariants tend to be well organized  as polynomials?  What exactly is going on and why don't we often see polynomial invariants for classifying other geometric objects?  For instance, why don't we just say that the Alexander invariants are a finite set of numbers?  Presumably, the organizing of them into polynomials shows off some inherent geometric properties of knot description and combination which is easily encapsulated in the multiplication and addition of polynomials.  I would be happy (and not surprised) if this also helps me understand why Lie algebra representations also often arise, especially in more recently discovered invariants.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-12-19T21:57:14.447" Title="Why are knot invariants best organized as polynomials?" Tags="&lt;algebraic-topology&gt;&lt;lie-algebra&gt;&lt;knot-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14895" PostTypeId="1" CreationDate="2010-12-19T21:21:17.013" Score="0" ViewCount="59" Body="&lt;p&gt;Here is my set up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am running a simulation of a gambler betting over time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the givens I am using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability of win = 35%   which is in B1 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odds = 2  which is in B3 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kelly # (amount of current bankroll to bet each time) = 2.5%  which is in D1 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bankroll = $10,000  which is in A6 table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to put in formulas I have for other cells in my simulation of bets, but will provide the data that those tables hold as well&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SOME OTHER COLUMNS&#xA;B6:&#xA;=$D$1*A6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C6:&#xA;=randbetween(1,000) column&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D6:&#xA;=if(C6&amp;lt;36,&quot;Win&quot;,&quot;Loss&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E6:&#xA;=if(D6=&quot;Win&quot;,B6*$B$3,(-B6))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A7:&#xA;=A6+E6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I drag these down to run 2000 simulation of random bets &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near the bottom I have a table that has the capital gain&#xA;the final bankroll - the beginning $10,000 bankroll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(is there a way to graph these wins on a line graph just the start of $10,000 to my final bankroll?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also what i need most&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to make a what if analysis... comparing different outcomes if the Kelly # (the percentage of my current bankroll I am betting) was different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.01     0.015   0.02       0.025      0.03     0.035    0.04&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and under each column of the different percentages I want the final capital gain in them... and I want to have like 200 examples of this new capital gain...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I will average each column above for the 200 capital gains of the different percentages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion is this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off what is the value on the left of the percentages that I need to put in that block that will be concrete and unchanging? &#xA;Is it $10,000 my beginning bankroll?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what are the values under that number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I need to highlight the grid beginning with that concrete block (set as $10,000?) and then highlight down the 200 simulations I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I go to data &gt; what if analysis &gt; data table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what do i put in the table for row input cell?&#xA;what about for the column input cell? &amp;lt;--- I believe I am supposed to leave this as a blank box - so I will just chose $G$1  an unused cell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HELP!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also here is a screen cap of my excel sheet (I am unable to link to my second screen cap... here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i6.photobucket.com/albums/y249/spittballz/Capture2.jpg&quot; alt=&quot;Excel screenshot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T00:56:50.070" LastActivityDate="2010-12-20T00:56:50.070" Title="What if analysis of an Excel spreadsheet." Tags="&lt;statistics&gt;&lt;probability-theory&gt;&lt;simulation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14896" PostTypeId="1" CreationDate="2010-12-19T21:24:41.517" Score="2" ViewCount="82" Body="&lt;p&gt;In Wikipedia and MathPlanet an equivalent definition of a symplectic matrix is given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \begin{array}{ccc}&#xD;&#xA;A &amp;amp; B \\&#xD;&#xA;C &amp;amp; D \end{array} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is symplectic if and only if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^TD-C^TB=I, A^TC=C^TA, D^TB=B^TD$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it seems wrong, since, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \begin{array}{ccc}&#xD;&#xA;0 &amp;amp; 0 &amp;amp; -1 &amp;amp; 1 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; -1 \\&#xD;&#xA;1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \end{array} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is symplectic but doesn't satisfy the conditions. Or have I mixed everything up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: this is crazy talk. That matrix isn't symplectic! (not for the form defined in Wikipedia or MathPlanet.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditorDisplayName="" LastEditDate="2010-12-19T21:39:25.147" LastActivityDate="2010-12-19T21:39:25.147" Title="Definition of Symplectic Matrix" Tags="&lt;matrices&gt;&lt;definition&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14897" PostTypeId="2" ParentId="14896" CreationDate="2010-12-19T21:28:52.420" Score="6" ViewCount="0" Body="&lt;p&gt;This is another question which highlights the problems with not thinking about things in a coordinate-free manner.  Symplectic transformations are defined relative to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Symplectic_form&quot; rel=&quot;nofollow&quot;&gt;symplectic form&lt;/a&gt;, and symplectic matrices in turn are defined relative to some &quot;canonical&quot; symplectic form with respect to the standard basis.  The problem is that there are at least two reasonable choices for such a &quot;canonical&quot; form (both of which are described at the Wikipedia article), and the resulting symplectic matrices you get from each form are different.  So you are probably just using a different one.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-19T21:28:52.420" CommentCount="5" />
  <row Id="14898" PostTypeId="2" ParentId="14894" CreationDate="2010-12-19T21:57:14.447" Score="6" ViewCount="0" Body="&lt;p&gt;I disagree with the hypothesis of the question -- knot invariants do not tend to be organized as polynomials.  The Alexander module is not a polynomial.  The fundamental group of the complement is not a polynomial.  The complement of a knot is not a polynomial.  These are some of the most fundamental invariants of a knot, and they're not polynomials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, polynomial invariants do occur in knot theory and there are a few good reasons.  One is that knots up to isotopy have the structure of a commutative monoid under the connect-sum operation. Invariably, when you start talking about functions on an object $O$, you like to talk about the vector space/module structure of such a function space, so you like to talk about formal linear combinations of objects from $O$.  Formal addition and connect-sum turns knots into a commutative ring.  Polynomial rings are free commutative rings, so they're fundamental in discussions about the structure of commutative rings.  In the special case of knots in $\mathbb R^3$, it's a theorem of Schubert's, that knots under these two operations &lt;em&gt;are&lt;/em&gt; a free commutative ring (the free generators being the &quot;prime&quot; knots).   So in some sense knots can be thought of as a polynomial ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the upshot of this is the study of &quot;knot invariants&quot; together with the connect-sum operation on knots prejudices the discussion towards one where you are concerned with commutative rings, and so commutative-ring valued invariants.  So that's why polynomial invariants occur so often. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to study different internal structures on knots, you perhaps prejudice the types of invariants you use to study knots.  Tangle categories and categorical invariants would be another example.  The knot complement and geometrization would be another. &lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2010-12-19T21:57:14.447" />
  <row Id="14900" PostTypeId="2" ParentId="14879" CreationDate="2010-12-19T22:14:10.477" Score="5" ViewCount="0" Body="&lt;p&gt;I'll try and give an answer that needs a bit of calculus. Whenever a mathematician writes something down, he has to define what he means by what he writes. For example, when I write $1.3$, then by definition this is the number $1+\frac{3}{10}$. When I write $5.402$, then this is $5 + \frac{1}{10} + \frac{0}{100} + \frac{2}{1000}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is this with infinitely many digits? Let's assume for simpliciticy that we have $0$ before the decimal point. If I write&#xA;$$&#xD;&#xA;  0.a_1a_2a_3a_4\dots\,,&#xD;&#xA;$$&#xD;&#xA;with $a_n\in\{0,1,2,3,4,5,6,7,8,9\}$ for all $n\in\mathbb N$, then this mean the &lt;em&gt;infinite series&lt;/em&gt;&#xA;$$&#xD;&#xA;  \sum_{n=1}^\infty a_n(\tfrac{1}{10})^n = \lim_{N\to\infty} \sum_{n=1}^N a_n(\tfrac{1}{10})^n\,.&#xD;&#xA;$$&#xD;&#xA;If, for example, $a_n=9$ for all $n\in\mathbb N$, then by the formula for the geometric series you get&#xA;$$&#xD;&#xA;  0.9999\dots = \sum_{n=0}^\infty 9\cdot(\tfrac{1}{10})^n = 9\cdot\frac{1}{1-1/10} = 1.&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, now you want an additional $8$ after all the nines. Well, you'll have to say what you &lt;em&gt;mean&lt;/em&gt; when you write that. I don't know of any definition of such an object, so my answer is &lt;em&gt;it's not defined&lt;/em&gt;. (By the way, I also think that it will be very difficult to give it a sensible definition, but strictly speaking this is only my belief.)&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-19T22:14:10.477" />
  <row Id="14901" PostTypeId="2" ParentId="14879" CreationDate="2010-12-19T22:37:49.423" Score="6" ViewCount="0" Body="&lt;p&gt;You could certainly have a number system that includes decimal expansions with a string of digits that is &quot;transfinitely long&quot;, indexed by an ordinal higher than $\omega$, such as &#xA;$$0.\overline{999}3216\overline{12345}623$$ (where the bar denotes infinite repetition of a block) which has an ordinal length of $\omega + \omega + 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this system there are objects between $0.9999....$ and $1.000...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that other than ordering these &quot;numbers&quot; (i.e., creating them for the sake of having things between 0.999... and 1) it is unclear whether there is anything useful that can be done with them, such as measuring some other type of object, or extending the arithmetic operations on ordinary decimals to the realm of trans-decimals.  It's not that these superobjects can't exist, it's that there is nothing to be done with them except point to their existence.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-19T22:37:49.423" CommentCount="10" />
  <row Id="14902" PostTypeId="2" ParentId="9524" CreationDate="2010-12-19T23:32:32.717" Score="6" ViewCount="0" Body="&lt;p&gt;Thanks for all the responses so far.  I thought I would summarize them for anyone else who might be interested in this question.  Rather than continuing to update the summary in the original question, though, it seems better from an organizational standpoint for the summary to appear in a community wiki answer than I then accept so that it appears at the top of the list of answers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Others seem to agree that focusing on Taylor series is the right approach.  See the answers and comments for justifications.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Examples of the use of infinite series &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. General: Zeno's paradoxes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Physics: Using the first order Taylor approximation $\sin \theta \approx \theta$ in solving the pendulum differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Chemistry: Extending the ideal gas law to apply to high pressure and low temperature situations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d. Economics: Calculating fiscal multipliers involves geometric series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e. Computer science, 1: Several uses for generating functions (see examples by robinhoode and Raphael)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f. Computer science, 2: Taylor series are involved in the error analysis of some numerical methods, such as Newton-Raphson and Simpson's rule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;g. Mathematics, 1: Taylor series show that calculations involving functions like $e^x$ and $\sin x$ can all be computed using just addition, subtraction, multiplication, and division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h. Mathematics, 2: Power series, and Euler products in number theory in particular, as most people find number theory intrinsically interesting whether they have the background or not&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i. Mathematics, 3: Taylor series can be used to solve differential equations.  (Often students will have seen a brief introduction to differential equations earlier in the course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;j. Mathematics, 4: There are infinite series expressions for interesting constants such as $\pi$ and $e$.  Also, any nonterminating decimal representation of a real number is an infinite series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k. Mathematics, 5: Using Taylor polynomials to approximate integrands in definite integrals.  (This fits well in a course like Calculus II that spends a lot of time on the integral.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Links to other resources &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Graphs of Taylor polynomials converging to a function (Brandon Carter's answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Graphs of Taylor polynomials in the complex plane (Hans Lundmark's comment on Brandon's answer)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Emphasize similarity of series and improper integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Emphasize that convergence must be dealt with carefully.  Use geometric series as a way to introduce convergence issues.  Mention that even Euler was not always good about handling convergence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-06T23:22:44.117" LastActivityDate="2011-01-06T23:22:44.117" CommunityOwnedDate="2010-12-19T23:32:32.717" CommentCount="3" />
  <row Id="14903" PostTypeId="2" ParentId="14855" CreationDate="2010-12-19T23:51:28.237" Score="3" ViewCount="0" Body="&lt;p&gt;As I mentioned in comments below, Kolmogorov's example is for a discontinuous function in $L^1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a continuous function whose Fourier series diverges at all rational multiples of $2\pi$ (and hence on a dense set) see &lt;a href=&quot;http://www.amazon.com/s/ref=nb_sb_ss_i_2_28?url=search-alias%3Daps&amp;amp;field-keywords=katznelson+harmonic+analysis&amp;amp;sprefix=katznelson+harmonic+analysis&quot; rel=&quot;nofollow&quot;&gt;Katznelson's book: An Introduction to Harmonic Analysis&lt;/a&gt; Chapter 2, Remark after proof of Theorem 2.1. Note that the Fourier series of such a continuous function still converges almost everywhere by Carleson's theorem.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-19T23:56:34.673" LastActivityDate="2010-12-19T23:56:34.673" />
  <row Id="14904" PostTypeId="1" AcceptedAnswerId="14913" CreationDate="2010-12-20T00:35:42.783" Score="3" ViewCount="67" Body="&lt;p&gt;The vector force field F=(yi,-xj) has a curl of -2. The acceleration of a particle in space is given by:&#xA;ax=y/m&#xA;ay=-x/m&#xA;This vector field has a divergence of 0. Will particles in this vector FORCE field move in circles or will they spiral outwards in a crazy fashion? My conundrum is this, the acceleration of the particles is never towards the center, therefor there is no centripetal acceleration, so how can these particles move in discreet circles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this reasoning make sense or am I doing a nested substitution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r=Sqrt[x^2+y^2]&#xA;r=Sqrt[((1/2)ax*t^2)^2 + ((1/2)ay*t^2)^2]&#xA;r=(1/2)Sqrt[t^4 (ax^2 + ay^2)]&#xA;r=(1/2)Sqrt[t^4 a^2]&#xA;r=(1/2)at^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4951" LastEditorUserId="4951" LastEditorDisplayName="" LastEditDate="2010-12-20T01:57:44.680" LastActivityDate="2010-12-20T02:01:57.297" Title="Vector Force Fields and Their Physical Interpretations" Tags="&lt;calculus&gt;&lt;physics&gt;&lt;vector-space&gt;&lt;simulation&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14905" PostTypeId="2" ParentId="14895" CreationDate="2010-12-20T00:44:59.543" Score="1" ViewCount="0" Body="&lt;p&gt;I am a big fan of programming in Excel, as I find it easier to see what is going on during the debug phase.  But you are stretching what is easy.  Each run of your simulation will use 2000 rows by 6 columns.  You could cut that down to 2 if you just have the current bankroll in one and the random number in the next.  If you want to run 200 cases of each of seven different Kelly numbers, you will fill 6*200*7=8400 columns.  I think it will do that, but it will be unwieldy.  This would be enough to drive me to Python, though I might keep the spreadsheet as a check.  You might be able to set up one tab per Kelly number, and just copy from one tab to the next.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It basically sounds like you are doing the right thing.  I have an older Excel that doesn't have randbetween.  I would use rand() which gives 0-1 and compare with .35.  Column A will then have the bankroll after each play and you can look at the bottom for the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get an excellent prediction of what you should see.  For a bet of 1, your expected return is .35*3=1.05 (it looks like you are paying 2:1 plus returning the bet), so you want to have as much money on the table as possible.  For a run of 2000 plays, you should win about 700 and lose 1300.  The standard deviation in wins is sqrt(2000*.35*.65)=21.33.  And if we write the Kelly number, K,  as a decimal and the number of wins as W, the bankroll at the end should be $10000(1+2K)^W(1-K)^{(2000-W)}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-20T00:44:59.543" CommentCount="2" />
  <row Id="14906" PostTypeId="1" AcceptedAnswerId="14907" CreationDate="2010-12-20T01:05:31.857" Score="0" ViewCount="62" Body="&lt;p&gt;Hi i am trying to understand a proof, I dont understand one of the steps.&#xA;could someone shed some light why the following is true.&#xA;if $f(x)$ is a non-increasing and continuous function and $a&amp;gt;b$ then $\{x\colon f(x) &amp;gt; f(b) \} \subset \{x\colon f(x) &amp;gt; a \} $ &#xA;i dont see why this is true. any insights ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="4952" LastActivityDate="2010-12-20T01:09:41.103" Title="non increasing function" Tags="&lt;real-analysis&gt;&lt;real&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="14907" PostTypeId="2" ParentId="14906" CreationDate="2010-12-20T01:09:41.103" Score="6" ViewCount="0" Body="&lt;p&gt;To say that $f$ is nonincreasing means that for all $x,y$ in the domain, if $x \leq y$ then $f(x) \geq f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given that $b &amp;lt; a$, so $f(b) \geq f(a)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally if $A \leq B$ are two real numbers, then for any function $f$, the set of $x$ in the domain such that $f(x) &amp;gt; B$ is a subset of the set of $x$ in the domain such that &#xA;$f(x) &amp;gt; A$, just because any number which is greater than $B$ is also greater than $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that continuity of $f$ was not used here.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-20T01:09:41.103" />
  <row Id="14908" PostTypeId="1" AcceptedAnswerId="14910" CreationDate="2010-12-20T01:14:20.400" Score="2" ViewCount="185" Body="&lt;p&gt;Hello, I was given the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate &#xA;  $$1-\frac12+\frac14-\frac18+\frac1{16}- \frac1{32}+\frac1{64}-\cdots -\frac1{512}+\cdots;$$&#xD;&#xA;  and express the answer as a simple fraction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My approach was to use the following formula: $\frac1{1-r}$ where $r$ is the common ratio.  In the end I got $2/3$.  Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Edit...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would i be able to explain this by words?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="4918" LastEditorDisplayName="" LastEditDate="2010-12-20T02:04:47.330" LastActivityDate="2010-12-21T00:21:35.053" Title="Does the following series  converge to 2/3?" Tags="&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="14909" PostTypeId="1" CreationDate="2010-12-20T01:15:05.857" Score="1" ViewCount="48" Body="&lt;p&gt;I need to find out $\log_g {-1}$ in $\mathbb{Z}_n$ where $n$ is an odd prime and $g$ is a primitive root mod $n$. How do I do that? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2868" LastActivityDate="2010-12-20T14:39:16.150" Title="Discrete logarithm to a primitive root base" Tags="&lt;number-theory&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14910" PostTypeId="2" ParentId="14908" CreationDate="2010-12-20T01:17:03.483" Score="4" ViewCount="0" Body="&lt;p&gt;Yes, and your reason is correct.  If you want a simple way to check your work on such things, you could use a computer algebra system like Mathematica, or simply Wolfram Alpha:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=sum+%28-1%2F2%29%5En&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=sum+%28-1%2F2%29%5En&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-20T01:22:10.440" LastActivityDate="2010-12-20T01:22:10.440" CommentCount="5" />
  <row Id="14911" PostTypeId="1" CreationDate="2010-12-20T01:32:50.853" Score="3" ViewCount="25" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&#xA;We say that a convex lattice polytope $P\subset \mathbb{R}^d$ is a &lt;em&gt;smooth Fano polytope&lt;/em&gt; if:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The origin is contained in the interior of $P$  &lt;/li&gt;&#xA;&lt;li&gt;The vertices of every facet of $P$ are a $\mathbb{Z}$-basis of $\mathbb{Z}^d$&#xD;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, suppose we have a set $V=\{v_1,\ldots,v_n\}\subset \mathbb{Z}^d$. What is an (preferentially efficient) algorithm for deciding if the convex hull $\text{conv}(V)$ is a smooth Fano polytope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;:&lt;br&gt;&#xA;I've read a paper that gives an algorithm to classify all smooth Fano polytopes given the dimension $d$ as input (&lt;a href=&quot;http://arxiv.org/abs/0704.0049v1&quot; rel=&quot;nofollow&quot;&gt;An algorithm for the classification of smooth Fano polytopes&lt;/a&gt; by Mikkel Øbro), and while trying to implement said algorithm I discovered I don't know how to solve this question.&lt;br&gt;&#xA;For those who are curious, I want to use the algorithm to help me gain some intuition about toric varieties, and use  it to compute known invariants and test conjectures.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-12-20T01:32:50.853" Title="How to check if a polytope is a smooth Fano polytope?" Tags="&lt;geometry&gt;&lt;combinatorics&gt;&lt;algorithms&gt;&lt;computational-geometry&gt;" />
  <row Id="14912" PostTypeId="2" ParentId="14904" CreationDate="2010-12-20T01:57:17.767" Score="2" ViewCount="0" Body="&lt;p&gt;From some basic facts of systems of linear differential equations, it follows that the particles will spiral like your simulation showed.  Indeed, solutions of the differential equation have terms involving things like $e^{t/\sqrt{2}} \sin(t/\sqrt{2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system of differential equations you have can be made a first order system by introducing new variables $w$ and $z$ with $x&amp;#39; = w$ and $y&amp;#39; = z$.  Then $w&amp;#39; = y$ and $z&amp;#39; = -x$ (here I'm taking $m=1$).  We can then write the system in matrix form as&#xA;$$&#xD;&#xA;\left(\begin{array}{c} w \\ x \\ y \\ z \end{array}\right)&amp;#39; = \left(\begin{array}{cccc} 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\ 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 \\ 0 &amp;amp; -1 &amp;amp; 0 &amp;amp; 0 \end{array}\right)\left(\begin{array}{c} w \\ x \\ y \\ z \end{array}\right).&#xD;&#xA;$$&lt;br&gt;&#xA;The type of solutions are then determined by the eigenvalues of the $4 \times 4$ matrix.  If an eigenvalue if $a + ib$, then a solution will be given by $v e^{a+ib} = v e^a + v(\cos b + i \sin b)$ where $v$ is the corresponding eigenvector.  In your system, the eigenvalues are $\frac{1}{\sqrt 2}(\pm 1 \pm i)$, so we get spiral solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its pretty cool that you made a flash simulation and I wish I could give you an answer without invoking ODE theory, but I can't really think of one.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-20T01:57:17.767" CommentCount="1" />
  <row Id="14913" PostTypeId="2" ParentId="14904" CreationDate="2010-12-20T02:01:57.297" Score="1" ViewCount="0" Body="&lt;p&gt;Err, right after I said I coulnd't think of a more elementary way I did.  Suppose a solution is a circle, then we'd have $x^2 + y^2 = const$.  Differentiating both sides gives $2x x&amp;#39; + 2y y&amp;#39; = 0$.  Dividing by 2 and differentiating again gives $x&amp;#39;^2 + xy + y&amp;#39;^2 - xy = 0$.  So $x&amp;#39;^2 + y&amp;#39;^2 = 0$.  This means $x&amp;#39; = y&amp;#39; = 0$.  So the solutions are constant, but only $(x,y) = (0,0)$ is a constant solution to the differential equation.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2010-12-20T02:01:57.297" />
  <row Id="14914" PostTypeId="2" ParentId="14908" CreationDate="2010-12-20T02:06:25.407" Score="0" ViewCount="0" Body="&lt;p&gt;You could also have split it up into two sums. &#xA;$$(1+\frac{1}{4}+\frac{1}{16}+...)-(\frac{1}{2}+\frac{1}{8}+...)$$&#xD;&#xA;$$(1+\frac{1}{4}+\frac{1}{16}+...)-\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$$&#xD;&#xA;$$\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading Moron's comment, I decided to post the alternative way to get to my final equation. &#xA;$$(1-\frac{1}{2})+(\frac{1}{4}-\frac{1}{8})+...$$&#xD;&#xA;$$\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-20T02:44:06.683" LastActivityDate="2010-12-20T02:44:06.683" CommentCount="2" />
  <row Id="14915" PostTypeId="1" AcceptedAnswerId="14923" CreationDate="2010-12-20T02:12:52.180" Score="7" ViewCount="129" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading up on class field theory, and I have a question. On wiki (http://en.wikipedia.org/wiki/Artin_reciprocity), one formulation is that there's &lt;em&gt;some&lt;/em&gt; modulus for which $I^c_K/i(K_{c,1})Nm_{L/K}(I^c_L)$ is isomorphic to $Gal(L/K)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another formulation on the same page is:&#xA;$C_K/N_{L/K}(C_L)$ is isomorphic to $Gal(L/K)$.&#xA;(where $C_{number\,field}$ is the idele class group of that number field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one relate these two formulations? Is it true that for some modulus, $c$, $I^c_K$ is $C_K$? I don't really see how this fits into one picture.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="4953" LastEditorDisplayName="" LastEditDate="2010-12-20T02:18:57.397" LastActivityDate="2010-12-20T07:57:07.620" Title="Different formulations of Class Field Theory" Tags="&lt;number-theory&gt;" AnswerCount="2" />
  <row Id="14916" PostTypeId="2" ParentId="14878" CreationDate="2010-12-20T02:36:01.483" Score="7" ViewCount="0" Body="&lt;p&gt;This theorem was proved by Max Dehn in early 20th century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is some square whose side $s$ is not rational. Pick some $\mathbb{Q}$-homomorphism $h$ such that $h(1) = 0$ and $h(s) = 1$. Define the &lt;i&gt;area&lt;/i&gt; $\sigma(R)$ of a rectangle $R$ with sides $a,b$ by $h(a)h(b)$. We show below that if a rectangle $R$ is partitioned into rectangles $R_i$ then $$\sigma(R) = \sum \sigma(R_i).$$ Take $R$ to be the unit square, and $R_i$ the squares into which it is partitioned. If $R_i$ has side $s_i$ then we get $$0 = h(1)^2 = \sigma(R) = \sum \sigma(R_i) = \sum h(s_i)^2 \geq h(s)^2 = 1.$$ This contradiction shows that all sides must be rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to show that $\sigma$ is additive. Extend all sides in the partition of $R$ to get a grid $G_j$. From additivity of $h$ we easily get $$\sigma(R) = \sum\sigma(G_j) = \sum\sigma(R_i).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using very similar methods, you can show the following slightly stronger theorem: Suppose that a rectangle is partitioned into squares. Then all edge lengths are rationally related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if the outside rectangle had sides $a,b$ not rationally dependent then we could pick a homomorphism $h$ such that $h(a)=1,h(b)=-1$, and then get that $-1$ is a sum of squares. Wlog, $a,b$ are both rational, and the previous proof completes the argument.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-20T02:36:01.483" CommentCount="3" />
  <row Id="14917" PostTypeId="1" CreationDate="2010-12-20T02:38:07.373" Score="3" ViewCount="30" Body="&lt;p&gt;One of the perks of my research topic (Latin squares) is that it's somewhat possible to explain what I do to those with a fairly minimal mathematics background.  I'll pose a typical question that comes up in one of my main research topics as a problem here  -- I'm hoping that this will whet the appetite of someone considering doing postgrad mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Latin_square&quot; rel=&quot;nofollow&quot;&gt;Latin square&lt;/a&gt; is an $n \times n$ matrix with symbols from $[n]:=\{1,2,\ldots,n\}$ such that each row and each column contains every symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $L=(l_{ij})$ be a Latin square.  For any row $i$ and column $j$, We say $(i,j,l_{ij})$ is a &lt;em&gt;entry&lt;/em&gt; of $L$.  The set of entries $O=\{(i,j,l_{ij}):i \in [n], j \in [n]\}$ is called the &lt;em&gt;orthogonal array&lt;/em&gt; of $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;em&gt;automorphism&lt;/em&gt; of $L$ is a permutation $\alpha$ of $\{1,2,\ldots,n\}$ such that $O$ is preserved under $(i,j,l_{ij}) \mapsto (\alpha(i),\alpha(j),\alpha(l_{ij}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition might seem a little complicated, but we can view this property reasonably easily.  Here's an example for order 15:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\begin{array}{cccccccccccc|ccc}&#xD;&#xA;13 &amp;amp; 8 &amp;amp; 14 &amp;amp; 9 &amp;amp; 15 &amp;amp; \mathbf{10} &amp;amp; 4 &amp;amp; 11 &amp;amp; 5 &amp;amp; 12 &amp;amp; 6 &amp;amp; 1 &amp;amp; 7 &amp;amp; 3 &amp;amp; 2 \\&#xD;&#xA;2 &amp;amp; 14 &amp;amp; 9 &amp;amp; 15 &amp;amp; 10 &amp;amp; 13 &amp;amp; \mathbf{11} &amp;amp; 5 &amp;amp; 12 &amp;amp; 6 &amp;amp; 1 &amp;amp; 7 &amp;amp; 3 &amp;amp; 8 &amp;amp; 4 \\&#xD;&#xA;8 &amp;amp; 3 &amp;amp; 15 &amp;amp; 10 &amp;amp; 13 &amp;amp; 11 &amp;amp; 14 &amp;amp; \mathbf{12} &amp;amp; 6 &amp;amp; 1 &amp;amp; 7 &amp;amp; 2 &amp;amp; 5 &amp;amp; 4 &amp;amp; 9 \\&#xD;&#xA;3 &amp;amp; 9 &amp;amp; 4 &amp;amp; 13 &amp;amp; 11 &amp;amp; 14 &amp;amp; 12 &amp;amp; 15 &amp;amp; \mathbf{1} &amp;amp; 7 &amp;amp; 2 &amp;amp; 8 &amp;amp; 10 &amp;amp; 6 &amp;amp; 5 \\&#xD;&#xA;9 &amp;amp; 4 &amp;amp; 10 &amp;amp; 5 &amp;amp; 14 &amp;amp; 12 &amp;amp; 15 &amp;amp; 1 &amp;amp; 13 &amp;amp; \mathbf{2} &amp;amp; 8 &amp;amp; 3 &amp;amp; 6 &amp;amp; 11 &amp;amp; 7 \\&#xD;&#xA;4 &amp;amp; 10 &amp;amp; 5 &amp;amp; 11 &amp;amp; 6 &amp;amp; 15 &amp;amp; 1 &amp;amp; 13 &amp;amp; 2 &amp;amp; 14 &amp;amp; \mathbf{3} &amp;amp; 9 &amp;amp; 8 &amp;amp; 7 &amp;amp; 12 \\&#xD;&#xA;10 &amp;amp; 5 &amp;amp; 11 &amp;amp; 6 &amp;amp; 12 &amp;amp; 7 &amp;amp; 13 &amp;amp; 2 &amp;amp; 14 &amp;amp; 3 &amp;amp; 15 &amp;amp; \mathbf{4} &amp;amp; 1 &amp;amp; 9 &amp;amp; 8 \\&#xD;&#xA;\mathbf{5} &amp;amp; 11 &amp;amp; 6 &amp;amp; 12 &amp;amp; 7 &amp;amp; 1 &amp;amp; 8 &amp;amp; 14 &amp;amp; 3 &amp;amp; 15 &amp;amp; 4 &amp;amp; 13 &amp;amp; 9 &amp;amp; 2 &amp;amp; 10 \\&#xD;&#xA;14 &amp;amp; \mathbf{6} &amp;amp; 12 &amp;amp; 7 &amp;amp; 1 &amp;amp; 8 &amp;amp; 2 &amp;amp; 9 &amp;amp; 15 &amp;amp; 4 &amp;amp; 13 &amp;amp; 5 &amp;amp; 11 &amp;amp; 10 &amp;amp; 3 \\&#xD;&#xA;6 &amp;amp; 15 &amp;amp; \mathbf{7} &amp;amp; 1 &amp;amp; 8 &amp;amp; 2 &amp;amp; 9 &amp;amp; 3 &amp;amp; 10 &amp;amp; 13 &amp;amp; 5 &amp;amp; 14 &amp;amp; 4 &amp;amp; 12 &amp;amp; 11 \\&#xD;&#xA;15 &amp;amp; 7 &amp;amp; 13 &amp;amp; \mathbf{8} &amp;amp; 2 &amp;amp; 9 &amp;amp; 3 &amp;amp; 10 &amp;amp; 4 &amp;amp; 11 &amp;amp; 14 &amp;amp; 6 &amp;amp; 12 &amp;amp; 5 &amp;amp; 1 \\&#xD;&#xA;7 &amp;amp; 13 &amp;amp; 8 &amp;amp; 14 &amp;amp; \mathbf{9} &amp;amp; 3 &amp;amp; 10 &amp;amp; 4 &amp;amp; 11 &amp;amp; 5 &amp;amp; 12 &amp;amp; 15 &amp;amp; 2 &amp;amp; 1 &amp;amp; 6 \\&#xD;&#xA;\hline&#xD;&#xA;12 &amp;amp; 2 &amp;amp; 1 &amp;amp; 3 &amp;amp; 5 &amp;amp; 4 &amp;amp; 6 &amp;amp; 8 &amp;amp; 7 &amp;amp; 9 &amp;amp; 11 &amp;amp; 10 &amp;amp; 15 &amp;amp; 14 &amp;amp; 13 \\&#xD;&#xA;11 &amp;amp; 1 &amp;amp; 3 &amp;amp; 2 &amp;amp; 4 &amp;amp; 6 &amp;amp; 5 &amp;amp; 7 &amp;amp; 9 &amp;amp; 8 &amp;amp; 10 &amp;amp; 12 &amp;amp; 14 &amp;amp; 13 &amp;amp; 15 \\&#xD;&#xA;1 &amp;amp; 12 &amp;amp; 2 &amp;amp; 4 &amp;amp; 3 &amp;amp; 5 &amp;amp; 7 &amp;amp; 6 &amp;amp; 8 &amp;amp; 10 &amp;amp; 9 &amp;amp; 11 &amp;amp; 13 &amp;amp; 15 &amp;amp; 14 \\&#xD;&#xA;\end{array}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which admits the automorphism $\alpha=(1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15)$.  The horizontal and vertical lines indicate where the cycles &quot;break&quot; in $\alpha$.  I highlight an orbit of an entry [i.e. starting with the entry (1,6,10), we apply $\alpha$ repeatedly] -- it traces out a broken diagonal of the upper-left block, and the symbol &quot;increases by one, wrapping around after 12&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here's a sample question from my research:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Let $\alpha=(1,2,3,4,5,6)(7,8,9,10,11,12)(13,14)(15,16)$.  Is $\alpha$ an automorphism of some Latin square of order $16$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a question that has come up in my research, and the more general question &quot;Given $\alpha$, is $\alpha$ an automorphism of some Latin square of order $n$?&quot; is one of my main research topics.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-20T02:38:07.373" Title="Does there exist a Latin square of order 16 that admits a specific automorphism?" Tags="&lt;combinatorics&gt;&lt;latin-square&gt;" />
  <row Id="14918" PostTypeId="2" ParentId="14915" CreationDate="2010-12-20T02:39:19.743" Score="3" ViewCount="0" Body="&lt;p&gt;The generalized ideal and (normal) idele class groups (by the generalized ideal class group I really mean ideals prime to $\mathfrak{c}$ for $\mathfrak{c}$ an &quot;admissible cycle,&quot; which more or less means divisible by enough primes such that anything congruent to 1 mod it is a local norm) are isomorphic. The short answer for why this is is that given an element in the idele class group, you can get an element of the (generalized) ideal class group since an idele always gives you a fractional ideal. By choosing a suitable representative, you can assume that the ideal you get is actually prime to $\mathfrak{c}$. It is then necessary (and nontrivial) to prove that the corresponding norm groups are mapped to each other, but this is done in Lang's &quot;Algebraic number theory.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are (at least) two approaches to class field theory:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Define the Artin map globally, on the generalized &lt;em&gt;ideal&lt;/em&gt; class group. This has the benefit that you only have to define it on ideals, so just on primes which are unramified (in which case you just use the Frobenius). In this case, it is a lot of work to show that it actually factors not just through the norm group, but also through a group of principal ideals.&lt;/li&gt;&#xA;&lt;li&gt;Define the Artin map on ideles by piecing together the local Artin maps.  This then requires some construction of the local Artin maps (which can be done using more-or-less pure group cohomology theory and a little bit of analysis of local fields). One has to check again that it is trivial on principal ideles, which is very nontrivial -- since, again the Artin map was defined by piecing together local things.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, in a sense, 1 is the one which gives you the ideal form of Artin reciprocity, and 2 gives you the idele one.&lt;/p&gt;&#xA;" OwnerUserId="536" LastEditorUserId="536" LastEditorDisplayName="" LastEditDate="2010-12-20T03:00:15.693" LastActivityDate="2010-12-20T03:00:15.693" CommentCount="5" />
  <row Id="14919" PostTypeId="1" AcceptedAnswerId="14944" CreationDate="2010-12-20T02:44:52.707" Score="4" ViewCount="79" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many stripes can you paint on a 2-group of fixed size?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A group of order 2&lt;sup&gt;a&lt;/sup&gt; p&lt;sup&gt;b&lt;/sup&gt; is solvable, by Burnside's theorem, so its chief factors are either abelian 2-groups or abelian p-groups.  Such a group is a &lt;strong&gt;zebra group&lt;/strong&gt; if its chief factors in any chief series alternate between the 2 and the p.  In other words, if you take a chain of normal subgroups 1 = N&lt;sub&gt;1&lt;/sub&gt; ⊲ N&lt;sub&gt;2&lt;/sub&gt; ⊲ &amp;hellip; ⊲ N&lt;sub&gt;n&lt;/sub&gt; = G of maximal length, then the quotient groups N&lt;sub&gt;i+1&lt;/sub&gt;/N&lt;sub&gt;i&lt;/sub&gt; alternate between being abelian 2-groups (the stripes) and abelian p-groups (the background).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we fix a, say a=8, then how many stripes can a zebra group have?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously no more than 8 2-stripes, but with a little work one can see that it can have no more than 4 2-stripes.  Unfortunately, I'm having trouble getting even 3 stripes.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-20T16:02:52.870" Title="Zebra groups and counting stripes" Tags="&lt;group-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14920" PostTypeId="2" ParentId="14878" CreationDate="2010-12-20T03:14:52.960" Score="4" ViewCount="0" Body="&lt;p&gt;I believe there is also a proof by associating the tiling with an electric network!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a section in the book &quot;Modern Graph Theory&quot; by B&amp;eacute;la Bollob&amp;aacute;s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is titled &quot;Squaring the Square&quot; and is numbered section II.2. See this: &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA46&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA46&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snapshot of the page which talks about Dehn's theorem and it's proof (also found here: &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA48&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA48&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vz7kV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, even though the section starts off talking about the tiling squares being of distinct lengths, the proof of Dehn's theorem (and the problem you seek) is essentially there.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-20T03:24:09.157" LastActivityDate="2010-12-20T03:24:09.157" CommentCount="0" />
  <row Id="14921" PostTypeId="1" AcceptedAnswerId="14922" CreationDate="2010-12-20T03:20:36.047" Score="16" ViewCount="295" Body="&lt;p&gt;It is well-known that there is only one &quot;kind&quot; of line, and that there are three &quot;kinds&quot; of quadratic curves (the nature of which depends on the sign of a so-called &quot;discriminant&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is noteworthy that many of the named cubic curves look rather similar: the &lt;a href=&quot;http://mathworld.wolfram.com/FoliumofDescartes.html&quot; rel=&quot;nofollow&quot;&gt;folium of Descartes&lt;/a&gt;, the &lt;a href=&quot;http://mathworld.wolfram.com/MaclaurinTrisectrix.html&quot; rel=&quot;nofollow&quot;&gt;trisectrix of Maclaurin&lt;/a&gt;, the &lt;a href=&quot;http://mathworld.wolfram.com/RightStrophoid.html&quot; rel=&quot;nofollow&quot;&gt;(right) strophoid&lt;/a&gt;, and the &lt;a href=&quot;http://mathworld.wolfram.com/TschirnhausenCubic.html&quot; rel=&quot;nofollow&quot;&gt;Tschirnhausen cubic&lt;/a&gt; look very similar in form; the &lt;a href=&quot;http://mathworld.wolfram.com/SemicubicalParabola.html&quot; rel=&quot;nofollow&quot;&gt;semicubical parabola&lt;/a&gt; and the &lt;a href=&quot;http://mathworld.wolfram.com/CissoidofDiocles.html&quot; rel=&quot;nofollow&quot;&gt;cissoid of Diocles&lt;/a&gt; resemble each other as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have deliberately placed the word &quot;kind&quot; in quotes since there does not seem to me an intuitive way of defining the term, so an answer to my question might have to define &quot;kind&quot; rigorously in the context of cubic curves. (An algebraic invariant, for instance... it is a pity that there does not seem to be an analogue of &quot;eccentricity&quot; for cubics!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://mathworld.wolfram.com/CubicCurve.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, it is noted that Newton classified cubics into 72 &quot;kinds&quot;, and Plücker after him described 219 &quot;kinds&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does one algebraically distinguish one cubic curve from another, and with a rigorous definition of &quot;kind&quot;, how many cubics are there?&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2011-01-01T00:52:56.590" Title="How many cubic curves are there?" Tags="&lt;algebraic-geometry&gt;&lt;plane-curves&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="14922" PostTypeId="2" ParentId="14921" CreationDate="2010-12-20T03:38:29.743" Score="14" ViewCount="0" Body="&lt;p&gt;The most standard notion of &quot;kind&quot; is that of isomorphism in algebraic geometry, that is, there are polynomial maps from one curve to the other which, when composed, are the identity on the curves (they don't have to be the identity in other places).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For real, affine plane curves, this notion reproduces the fact that there is a unique line and exactly three smooth conics (over the complex numbers, ellipses and hyperbolas turn out to be the same, and in the projective plane, all are the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, regardless of ground field (well, regardless of working over the real or complex numbers), there are infinitely many different cubics up to isomorphism, determined by the &lt;a href=&quot;http://en.wikipedia.org/wiki/J-invariant#Algebraic_definition&quot; rel=&quot;nofollow&quot;&gt;j-invariant&lt;/a&gt;, one for every element of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conclusion you can draw from this is that there really isn't a good algebraic way to say that there are only finitely many types of plane cubic, unless you want to lump all smooth cubics together (which you clearly don't, as you consider parabolas and hyperbolas and ellipses to be distinct).  I don't know what method Newton and Plücker used to classify them, but isomorphism is the standard method in modern algebraic geometry.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-12-20T03:38:29.743" CommentCount="2" />
  <row Id="14923" PostTypeId="2" ParentId="14915" CreationDate="2010-12-20T04:26:18.570" Score="9" ViewCount="0" Body="&lt;p&gt;Dear Nicole, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the initial section or two of the chapter by Tate in Cassels and Frolich, he gives a nice explanation of how to pass from the classical formulation in terms of generalized ideal class groups w.r.t a modulus, and the more modern formulation in terms of idele class groups.  As Tate explains,&#xA;the two formulations are indeed equivalent, but it is not quite as simple as saying that $I^c_K = C_K$.   Here is a sketch of the equivalence (of course it is the same is in Akhil Mathew's answer, just slightly more detailed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $I_K^c$ has already been taken to denote the ideals prime to $c$ (at least, this is how I interpret your notation), let me use $J_K$ to denote the ideles for&#xA;$K$.  Then we can consider the subgroup $J_K^c$ of the ideles whose entries are all $1$ at any finite place dividing $c$, and at any infinite place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a natural surjection $J_K^c \to I_K^c$ given by sending any element&#xA;$(a_{\wp})$ of the former to the ideal $\prod_{\wp} \wp^{v_{\wp}(a_{\wp})}$&#xD;&#xA;of the  latter (where the product is over finite places, i.e. prime ideals,&#xA;$\wp$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one can show that $K^{\times} J_K^c$ is dense in $J_K$, so&#xA;the image of $J_K^c$ is dense in $C_K$.  Since $N_{L/K}(C_L)$ is open in $C_K$,&#xA;we see that $J_K^c$ surjects onto $C_K/N_{L/K}(C_L)$. &#xA;Now one checks that this map factors through the surjection $J_K^c \to I_K^c$&#xD;&#xA;described in the preceding paragraph, and in fact induces an isomorphism&#xA;$I_K^c/i(K_{c,1}) N_{L/K}(I_L^c) \buildrel \sim \over \longrightarrow&#xD;&#xA;C_K/N_{L/K}(C_L)$, as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, suppose you want to compute the Artin map on an element of $J_K$:&#xA;the algorithm is you first multiply by a principal idele so that the resulting element is in $J_K^c$ times $N_{L/K}(C_L)$.  (You may not know exactly what this&#xA;group is, but its not hard to at least identify an open subgroup of it: for example, at any complex infinite place $v$ the norm map is surjective, at any real place $v$ the image of the norm map at least contains the positive reals, and at any finite place $\wp$ the image of the norm map will contain elements which are congruent to $1$ modulo the power of $\wp$ dividing the relevant modulus $c$.)&#xA;Now the Artin map on $J_K^c$ factors through the surjection $J_K^c \to I_K^c$,&#xA;and is computed on the target using Frobenius elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, this was the argument via which local class field theory was originally proved; one took a local extension, embedded it into a global context (so that the original local situation was realized as $L_{\wp}/K_{\wp}$ for some abelian&#xA;extension of number fields $L/K$), and then defined the Artin map via the above computation (which means concretely that one passes from the possibly ramified&#xA;situation at $\wp$ to a consideration just at the unramified primes, where everything is easily understood just in terms of ideals and Frobenius elements).&#xA;Of course, one then had to check that the resulting local Artin map was well-defined independent of the choice of &quot;global context&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, one can define the local Artin maps at &lt;em&gt;all&lt;/em&gt; places (unramified or ramified) first.  However, in generalizations to the non-abelian situation (i.e. local and global Langlands) one generally uses the old-fashioned technique of proving certain global results first, and then establishing the precise local results by passing to a well-chosen global context and reducing to a calculation at unramified primes. (This is a bit of an oversimplification, but I think it is correct in spirit.)   So (if one has an eventual aim of understanding modern algebraic number theory and the Langlands program) it is well worth understanding the passage between the idelic and ideal-theoretic view-points on class field theory, and practicing how to use the algorithm described above.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-20T07:57:07.620" LastActivityDate="2010-12-20T07:57:07.620" />
  <row Id="14924" PostTypeId="2" ParentId="14878" CreationDate="2010-12-20T04:48:09.460" Score="5" ViewCount="0" Body="&lt;p&gt;Since your professor's hint was to use linear algebra, the intended proof is probably that of Hadwiger - employing a Hamel basis of $\:\mathbb R\:$ over $\:\mathbb Q\:$ to construct additive &quot;area&quot; functions in order to deduce a contradiction (as in Yuval's answer). Related results were proved by Dehn circa 1900 by different complicated methods. Although Hamel's work on Hamel bases occurred shortly later in 1905, it was not until much later in the 1950's that Hadwiger and his students noticed that Hamel bases could be employed to greatly simplify Dehn's proofs. For a nice exposition see Freiling; Rinne: &lt;a href=&quot;http://www.mathjournals.org/mrl/1994-001-005/1994-001-005-003.pdf&quot; rel=&quot;nofollow&quot;&gt;Tiling a square with similar rectangles&lt;/a&gt;. It's worth remarking that use of the axiom of choice in the construction of a Hamel basis can easily be eliminated for applications of this type, e.g. see the discussion in section 2 of Feshchenko et. al.: &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0809/0809.1883v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Dissecting a brick into bars.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-20T14:37:56.760" LastActivityDate="2010-12-20T14:37:56.760" CommentCount="1" />
  <row Id="14925" PostTypeId="1" AcceptedAnswerId="14935" CreationDate="2010-12-20T06:11:00.490" Score="9" ViewCount="216" Body="&lt;p&gt;Is it true that $\frac1{n!} \int_0^\infty x^{n-k} (x-1)^k e^{-x}\,dx \approx e^{-k/n}$ when $k$ and $n$ are large integers with $k \le n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quantity is the probability that a random permutation of $n$ elements does not fix any of the first $k$ elements.&lt;/p&gt;&#xA;" OwnerUserId="857" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-20T21:30:24.557" LastActivityDate="2010-12-20T21:30:24.557" Title="Probability that a random permutation has no fixed point among the first $k$ elements." Tags="&lt;calculus&gt;&lt;probability&gt;&lt;analysis&gt;&lt;permutations&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="14926" PostTypeId="1" CreationDate="2010-12-20T06:48:39.667" Score="7" ViewCount="94" Body="&lt;p&gt;What are the formal series $f$ with non-negative integer coefficients and constant term equal to $1$ whose multiplicative inverse $1/f$ has all coefficients, apart from a finite subset, all non-positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, assume the series converges in some disk if you want...&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-20T16:02:48.517" LastActivityDate="2010-12-20T17:28:29.210" Title="Multiplicative inverses of formal series with non-negative coefficients" Tags="&lt;sequences-and-series&gt;&lt;power-series&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14927" PostTypeId="2" ParentId="14831" CreationDate="2010-12-20T07:18:57.113" Score="4" ViewCount="0" Body="&lt;p&gt;Here is another approach. Let $G$ be a free group on $m$ generators, and let $H$ be a free group on $n$ generators. There are exactly $2^m$ homomorphisms from $G$ to a group of order two, since each generator can be mapped in two ways. Likewise, there are $2^n$ homomorphisms from $H$ to a group of order two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $G$ and $H$ are isomorphic, then they have the same number of homomorphisms to a group of order two. Therefore $2^m = 2^n$, which implies $m=n$.&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2010-12-20T07:18:57.113" />
  <row Id="14928" PostTypeId="2" ParentId="14831" CreationDate="2010-12-20T07:55:01.663" Score="2" ViewCount="0" Body="&lt;p&gt;Dave R's answer highlights a more general principle: by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Yoneda_lemma&quot; rel=&quot;nofollow&quot;&gt;Yoneda lemma&lt;/a&gt;, an object $X$ in a category is determined up to isomorphism by the behavior of the functor $F_X = \text{Hom}(X, -)$, so to show that two objects $X, Y$ are &lt;em&gt;not&lt;/em&gt; isomorphic it suffices to show that the corresponding functors $F_X, F_Y$ are not isomorphic.  (In particular it suffices to show the existence of an object $Z$ such that $\text{Hom}(X, Z)$ has a different size from $\text{Hom}(Y, Z)$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free groups $F_n$ represent some special functors $\text{Grp} \to \text{Set}$: namely, $\text{Hom}(F_n, G)$ is precisely the set $G^n$ of $n$-tuples of elements of $G$.  This is a manifestation of the adjunction between the free group functor $\text{Set} \to \text{Grp}$ and the underlying set functor $\text{Grp} \to \text{Set}$, and by setting $G$ to any nontrivial finite group (Dave R's answer uses $\mathbb{Z}/2\mathbb{Z}$) it is not hard to see that these functors are all nonisomorphic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, I believe one can say the following.  Let $F : D \to C$ and $G : C \to D$ be an adjunction, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Hom}_C(FX, Y) \simeq \text{Hom}_D(X, GY)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and suppose that the functor $G$ is essentially surjective.  (In this case $G$ is the forgetful functor $\text{Grp} \to \text{Set}$ and $F$ is the free group functor $\text{Set} \to \text{Grp}$; then it is a classical result that $G$ is essentially surjective.)  Then by another application of the Yoneda lemma, I believe the functors $\text{Hom}_C(FX_1, -)$ and $\text{Hom}_C(FX_2, -)$ are isomorphic if and only if the objects $X_1, X_2$ are isomorphic in $D$.  Can anyone confirm this?&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-20T08:03:16.980" LastActivityDate="2010-12-20T08:03:16.980" />
  <row Id="14929" PostTypeId="1" CreationDate="2010-12-20T07:57:17.450" Score="0" ViewCount="25" Body="&lt;p&gt;If $A_n$ is minuend,$B_n$ is subtrahend and $C_{n-1}$ are the three inputs and $D_n$ (difference), $C_n$ (borrow) are the two outputs of a full substractor,we know the $C_n$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_n = \overline{A_n}B_n + \left(\overline{A_n \bigoplus B_n}\right)C_{n-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we reduce it  to $C_n = \overline{A_n} B_n + \overline{A_n}C_{n-1} + B_nC_{n-1}$ ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have reduced upto $C_n = \overline{A_n}B_n + \left(A_n B_n + \overline{A_n} \overline{B_n}\right)C_{n-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried using Karnaugh map for further reduction but not quite there.Any pointer in the correct direction would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Just now I solved it! Earlier it was a mistake in my Karnaugh Map,I also got the same answer using Boolean algebra substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-20T08:29:08.260" LastActivityDate="2010-12-20T08:29:08.260" Title="Boolean Algebra :: Full substactor" Tags="&lt;elementary&gt;&lt;boolean-algebra&gt;" CommentCount="1" />
  <row Id="14930" PostTypeId="1" AcceptedAnswerId="14959" CreationDate="2010-12-20T08:48:43.307" Score="1" ViewCount="147" Body="&lt;p&gt;And how to express $\{ x \} = x - \lfloor x \rfloor$ as function of $sin(x)$ and $sign(x)$?&lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="367" LastEditorDisplayName="" LastEditDate="2010-12-20T15:06:17.143" LastActivityDate="2010-12-20T18:58:39.300" Title="How to express the whole part $\lfloor x \rfloor$ as analytical function or Taylor/Fourier series?" Tags="&lt;fourier-series&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14931" PostTypeId="2" ParentId="14876" CreationDate="2010-12-20T09:10:45.233" Score="2" ViewCount="0" Body="&lt;p&gt;The most useful mathematical areas for a programmer are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;trigonometry and linear algebra, especially to draw and recognise shapes, ray tracing, animations and so on,&lt;/li&gt;&#xA;&lt;li&gt;combinatorics, particularly to cleverly enumerate sets, estimate running times of simple loops and recursions etc.,&lt;/li&gt;&#xA;&lt;li&gt;some elementary group theory and particularly permutation groups, for algorithms that involve tracing out graphs (particularly trees) and also for clever enumeration, as above.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are lots of introductory books on each of the above subjects, although maybe other people can comment on which ones are best suited for non-mathematicians. Note that the three areas are fairly independent, at least at the beginning, and you can learn them in any order or in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I forgot the possibly best way to get started: once you know the definition and some basic facts about groups, you should read the book &lt;a href=&quot;http://books.google.co.uk/books?id=kC5kdUirHHoC&amp;amp;lpg=PP1&amp;amp;ots=2m5l2_shyb&amp;amp;dq=indra%27s%20pearls&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Indra's Pearls&lt;/a&gt;. It is a very beautiful book that explains how to draw certain types of fractals with the computer. It introduces all the relevant mathematics pretty much from zero and it provides the actual programs in pseudo-code, ready for you to implement in your favourite language. The mathematics and the algorithms it introduces are very relevant to other situations, but since you are particularly interested in graphics, you will hardly find a better way to get started! There is even a &lt;a href=&quot;http://www.josleys.com/galleries.php?catid=7&quot; rel=&quot;nofollow&quot;&gt;computer graphics artist&lt;/a&gt; who bases some of his work on that book.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-20T14:03:20.613" LastActivityDate="2010-12-20T14:03:20.613" CommentCount="1" />
  <row Id="14932" PostTypeId="1" CreationDate="2010-12-20T09:29:04.987" Score="5" ViewCount="346" Body="&lt;p&gt;Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, will large number of solutions give moderate rank EC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one drops $-1$ i.e. $xy(x-y)=n$ the answer is &quot;yes&quot; via multiples of rational point(s) and then multiplying by a cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Suppose it is an open question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: What I find intriguing is that &lt;strong&gt;all integral points&lt;/strong&gt; in this model come from factorization/divisors only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;&#xA;EDIT: Current record is n=&lt;strong&gt;179071200&lt;/strong&gt; with 22 solutions with positive x,y. Due to Matthew Conroy.&#xA;Current record is n=&lt;strong&gt;391287046550400&lt;/strong&gt; with 26 solutions with positive x,y. Due to  Aaron Meyerowitz&#xA;Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.&#xA;&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by Tapio Rajala.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2011-01-01T19:36:20.143" LastActivityDate="2011-01-01T19:36:20.143" CommunityOwnedDate="2010-12-28T06:42:53.157" Title="Can the number of solutions $xy(x-y-1)=n$   for   $x,y,n \in Z$ be unbounded as n varies?" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;" CommentCount="18" />
  <row Id="14933" PostTypeId="1" CreationDate="2010-12-20T09:53:38.237" Score="2" ViewCount="65" Body="&lt;p&gt;I need to solve following system of nonlinear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some characteristics of this system:&lt;br&gt;&#xA;It consists of n equation and n variables.&lt;br&gt;&#xA;Every equation is in similar form -&gt; sum of products = constant.&lt;br&gt;&#xA;The lenght of every product is the same (it will be denoted as k).&lt;br&gt;&#xA;The number of elements in sum might be different in each equation.&lt;br&gt;&#xA;In every product in equation $i$ one of elements is $x_{i}$ - this is very important.&lt;br&gt;&#xA;This system is &quot;symmetrical&quot;, it means that if $x_{i} \cdot x_{j} \cdot ...$ is one of elements of equation i, then it is also in equaton j.&lt;br&gt;&#xA;$b_{i} &amp;gt; 0$ - where $b_{i}$ is intercept in equation i.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll write an example of such equation for k=3 and n=6:&lt;br&gt;&#xA;$x_{1} \cdot x_{3} \cdot x_{6} + x_{1} \cdot x_{2} \cdot x_{4} = b_{1}$&lt;br&gt;&#xA;$x_{2} \cdot x_{1} \cdot x_{4} + x_{2} \cdot x_{5} \cdot x_{6} = b_{2}$&lt;br&gt;&#xA;$x_{3} \cdot x_{1} \cdot x_{6} + x_{3} \cdot x_{4} \cdot x_{6} = b_{3}$&lt;br&gt;&#xA;$x_{4} \cdot x_{1} \cdot x_{2} + x_{4} \cdot x_{3} \cdot x_{6} = b_{4}$&lt;br&gt;&#xA;$x_{5} \cdot x_{2} \cdot x_{6} = b_{5}$&lt;br&gt;&#xA;$x_{6} \cdot x_{1} \cdot x_{3} + x_{6} \cdot x_{2} \cdot x_{5} + x_{6} \cdot x_{3} \cdot x_{4} = b_{6}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very easy to transform this equation to following form:&lt;br&gt;&#xA;$x_{i} = b_{i} / something$ , $x_{i}$ is only on left-hand side of ith equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have all equation in such form then the fixed point is solution of it.&lt;br&gt;&#xA;I've experimentally checked that algorithm analogic to Gauss-Seidel is covergent (i've checked ~100 random examples, and in every case it was convergent).&lt;br&gt;&#xA;By analogic to Gauss-Seidel algorithm I mean:&lt;br&gt;&#xA;1) Choose any initial solution $[x_{1}^{0} , ... , x_{n}^{0}]$&lt;br&gt;&#xA;2.1) Calculate value of $x_{1}^{i+1}$ using $[x_{2}^{i} , ... , x_{n}^{i}]$&lt;br&gt;&#xA;2.2) Calculate value of $x_{2}^{i+1}$ using $[x_{1}^{i+1} , ... , x_{n}^{i}]$&lt;br&gt;&#xA;...&lt;br&gt;&#xA;2.n) Calculate value of $x_{n}^{i+1}$ using $[x_{1}^{i+1} , ... , x_{n-1}^{i+1}]$&lt;br&gt;&#xA;3) If solution is good enough stop, otherwise go to 2.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried Banach fixed point theorem, but is hard to say anything about spectral radius.&#xA;Does anyone have a clue how to prove convergence of this algorithm?&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T10:25:45.673" LastActivityDate="2010-12-20T10:25:45.673" Title="System of nonlinear equations" Tags="&lt;numerical-methods&gt;" CommentCount="6" FavoriteCount="1" />
  <row Id="14934" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T10:27:22.940" Score="5" ViewCount="0" Body="&lt;p&gt;This is not an answer but an observation that the quantity is equal to $\sum_{i=0}^{k} \frac{{(-1)}^i}{n!}{k \choose i}(n-i)!$.&lt;/p&gt;&#xA;" OwnerUserId="4959" LastActivityDate="2010-12-20T10:27:22.940" />
  <row Id="14935" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T10:27:52.900" Score="6" ViewCount="0" Body="&lt;p&gt;Here's a heuristic idea, based on the strong law of large numbers (SLLN). A rigorous proof can be made using the central limit theorem (CLT), as in my new answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First write &#xA;$$&#xD;&#xA;I = \frac{1}{{n!}}\int_0^\infty  {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}  = \int_0^\infty  {\bigg(1 - \frac{1}{x}} \bigg)^k \frac{{x^n e^{ - x} }}{{\Gamma (n + 1)}}\,{\rm d}x.&#xD;&#xA;$$&#xD;&#xA;Now, if $X_1,\ldots,X_{n+1}$ are i.i.d. exponential(1) random variables, then their sum $X_1 + \cdots + X_{n+1}$ has gamma density $x^n e^{-x} / \Gamma(n+1)$, $x &amp;gt; 0$. Hence,&#xA;$$&#xD;&#xA;I = {\rm E} \bigg[ 1 - \frac{1}{{X_1  +  \cdots  + X_{n + 1} }}\bigg]^k = {\rm E}\bigg[1 - \frac{{n + 1}}{{X_1  +  \cdots  + X_{n + 1} }}\frac{1}{{n + 1}}\bigg]^k .&#xD;&#xA;$$&#xD;&#xA;By SLLN, $(n + 1)^{ - 1} \sum\nolimits_{i = 1}^{n + 1} {X_i }$ converges almost surely to  ${\rm E}[X_1 ] = 1$ as $n \to \infty$. So this suggests the approximation&#xA;$$&#xD;&#xA;I \approx \bigg(1 - \frac{1}{{n + 1}}\bigg)^k  = \bigg(1 - \frac{1}{{n + 1}}\bigg)^{n(k/n)}  \approx e^{ - k/n}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-20T15:44:22.207" LastActivityDate="2010-12-20T15:44:22.207" />
  <row Id="14936" PostTypeId="1" AcceptedAnswerId="14937" CreationDate="2010-12-20T10:33:53.070" Score="0" ViewCount="51" Body="&lt;p&gt;This may have more to do with computing than Mathematics in its application, however this has been giving me a headache for some time and I see no other recourse than to ask...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a natural integer T &gt; 9 (and probably between 1,000 and 1,000,000,000), and a choice N in {0,1,2,3,4}, provide two algorithms (encoding and decoding):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One which take T and N as inputs and outputs R, a number which has no more digits than T. [encoding]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other which takes an (encoded) number R and outputs the original (unencoded) T and N. [decoding]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The complexity of the algorithms should at best allow to be implemented by modern computing. Apart from that, &lt;em&gt;everything and anything&lt;/em&gt; is allowed :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, do advise me if this is impossible.&lt;/p&gt;&#xA;" OwnerUserId="4958" LastActivityDate="2010-12-20T11:11:19.580" Title="&quot;Encode&quot; a single digit in a multi-digit number in the smallest way possible" Tags="&lt;algorithms&gt;&lt;natural-number&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14937" PostTypeId="2" ParentId="14936" CreationDate="2010-12-20T10:47:23.080" Score="5" ViewCount="0" Body="&lt;p&gt;This is impossible for the same reason a compression algorithm that reduces the size of every file is impossible. Suppose $T$ has $k$ decimal digits. There are $9\times10^{k-1}$ possible values of $T$, and $5$ possible values of $N$, making $45\times10^{k-1}$ possible $(T,N)$ pairs. But since you're only allowed up to $k$ digits in $R$, there are only $10\times10^{k-1}$ possible values of $R$ to encode them in. So no matter what you do, you won't be able to decode $R$ back into $T$ and $N$.&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-20T11:11:19.580" LastActivityDate="2010-12-20T11:11:19.580" CommentCount="2" />
  <row Id="14938" PostTypeId="2" ParentId="11076" CreationDate="2010-12-20T10:52:40.393" Score="1" ViewCount="0" Body="&lt;p&gt;If vectors are thought of as triplets of numbers (a,b,c) then why should these triplets be added by the parallelogram law? It all depends what these triplets are being used to represent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with addition of positive numbers on the number line, say 5+3, this can be seen as laying down two line segments of length 5 and 3 along a line starting at the origin. This can be generalized to line segments in a plane by allowing them to point in different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with two adjoined line segments then you can expand the drawing to a parallelogram and it is due to Euclidean geometry that opposite sides of a parallelogram have the same length and point in the same direction. So the parallelogram gives two different paths to move from the origin to the resulting point which is saying that the paralleogram law is equivalent to the commutative law A+B=B+A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in one dimension, the line segments can be given a direction, so that 5+(-3) means putting down a line of length 5 in the +direction, and then turning around and drawing a line of length 3 in the opposite direction to arrive at the number 2. Being commutative you could also start with a line of length 3 in the negative direction from the origin along the real number line and then turn around and draw a line of length 5 in the positive direction again arriving at 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the parallelogram law of vector addition is a straightforward extension of addition of ordinary numbers as directed line segments and is compatible with addition of negative numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long before cartesian coordinates were developed, people would have used geometry including the pythagorean theorem to calculate the addition of distances and velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then special relativity was discovered and it was found that the parallelogram law no longer holds for velocity addition, so to answer your question: other laws are indeed sometimes used. The question is: how do we know which laws apply to which aspects of nature? We don't know. We make observations and then make assumptions about how things behave based on those observations. We try to use mathematical structures that accurately describe the observed behaviour, and assume that if our models are accurate enough then any calculations derived from those models will also describe nature accurately.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2010-12-20T10:52:40.393" />
  <row Id="14939" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T11:05:36.323" Score="3" ViewCount="0" Body="&lt;p&gt;Continuing Aravind's observation, write it like&#xA;$$&#xD;&#xA;1 - \frac{k}{n} + \frac{1}{2!} \left( \frac{k}{n} \right)^2 \frac{1-1/k}{1-1/n}&#xD;&#xA;- \frac{1}{3!} \left( \frac{k}{n} \right)^3 \frac{(1-1/k)(1-2/k)}{(1-1/n)(1-2/n)}&#xD;&#xA;$$&#xD;&#xA;$$+ \dots&#xD;&#xA;+ (-1)^k \frac{1}{k!} \left( \frac{k}{n} \right)^k \frac{(1-1/k)(1-2/k)\dots (1-(k-1)/k)}{(1-1/n)(1-2/n) \dots (1-(k-1)/n)}.&#xD;&#xA;$$&#xD;&#xA;If $k$ and $n$ are large, this seems to be close to the $k$th degree Maclaurin polynomial&#xA;for $e^z$ evaluated at $z=-k/n$ (although I don't have an estimate of the error).&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-20T11:05:36.323" />
  <row Id="14940" PostTypeId="1" AcceptedAnswerId="14943" CreationDate="2010-12-20T12:22:19.020" Score="3" ViewCount="88" Body="&lt;p&gt;I am unable to understand the motivation behind the set theoretic definition of a natural number. The definition given in the book by Goldrei is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First he defines an inductive set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set $y$ is &lt;em&gt;inductive&lt;/em&gt; if $\emptyset \in y$ and $x^{+} \in y$ whenever $x \in y$, where $x^{+} = x \cup \{x\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then the set of natural number is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of natural numbers $\mathbb{N}$ is the intersection of all inductive subsets of any inductive set $y$, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbb{N} = \cap \{z:z \text{ is an inductive subset of }y\} = \{x: x \in z, \forall \text{ inductive }z \subseteq y\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am actually lost with this definition. I feel any inductive set is isomorphic to $\mathbb{N}$. What is the reason/motivation for defining $\mathbb{N}$ as the intersection of all the &lt;em&gt;inductive&lt;/em&gt; subsets of an &lt;em&gt;inductive&lt;/em&gt; set $y$? What are we missing if we were to define $\mathbb{N}$ as just an inductive set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my statements/questions don't make sense, it is because I am confused. I would appreciate if someone could throw light on this.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-20T12:42:09.263" Title="Set theoretic definition of a Natural Number" Tags="&lt;set-theory&gt;&lt;natural-number&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14941" PostTypeId="2" ParentId="14792" CreationDate="2010-12-20T12:31:36.803" Score="1" ViewCount="0" Body="&lt;p&gt;$(x\vee y)\wedge(z\vee(x\wedge y))=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\vee((x\vee y)\wedge(z\vee(x\wedge y)))=x\vee((x\wedge y)\vee(y\wedge z)\vee(z\wedge x)).$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right side equals $x\vee(y\wedge z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the left side, we use the condition above again, substituting $x\vee y$ for $y$ and $z\vee(x\wedge y)$ for $z$, and get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x\vee((x\vee y)\wedge(z\vee(x\wedge y)))=(x\vee y)\wedge((z\vee(x\wedge y))\vee x)=(x\wedge y)\vee(x\wedge z).$&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditorDisplayName="" LastEditDate="2010-12-20T12:37:45.443" LastActivityDate="2010-12-20T12:37:45.443" />
  <row Id="14942" PostTypeId="1" CreationDate="2010-12-20T12:32:17.803" Score="0" ViewCount="62" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to make one body fly around another, using Coulomb law $F=\frac{q_1 q_2}{r^2}$ and second Newton law $ma=F$. Now I'm doing it this intuitive way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Move body1 according to current speed and $dt$;&lt;/li&gt;&#xA;&lt;li&gt;Calc new Coulomb force then find acceleration $a=\frac{F}{m}$;&lt;/li&gt;&#xA;&lt;li&gt;Add acceleration to current speed;&lt;/li&gt;&#xA;&lt;li&gt;Repeat.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This works fine. I know this is very simple Euler method and it's very inaccurate. I'd like to use Runge-Kutta method, but I can't figure out how I should implement it. &lt;a href=&quot;http://en.wikipedia.org/wiki/Runge%96Kutta_methods&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; it's described, but&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is my $f(x, y)$?&lt;/li&gt;&#xA;&lt;li&gt;What is my $y&amp;#39;$ and $y&amp;#39;&amp;#39;$?&lt;/li&gt;&#xA;&lt;li&gt;Where $dt$ goes?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4961" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-20T13:03:09.233" LastActivityDate="2010-12-20T13:39:30.980" Title="Runge-Kutta method for Newton law" Tags="&lt;differential-equations&gt;&lt;numerical-methods&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14943" PostTypeId="2" ParentId="14940" CreationDate="2010-12-20T12:42:09.263" Score="8" ViewCount="0" Body="&lt;p&gt;Usually in set theory we write $\omega$ instead of $\mathbb{N}$. A concrete example of an inductive set other than $\omega$ is the ordinal $\omega + \omega$, which is the closure of $\omega \cup \{\omega\}$ under ${}^+$. This set has the structure of two copies of $\omega$, one after the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any set $x$, if you take the closure of $x \cup \{\emptyset\}$ under ${}^+$ then you get an inductive set containing every member of $x$. So it is not true that any inductive set is even isomorphic to $\omega$. For example there are uncountable inductive sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you really want is a $\subseteq$-*&lt;em&gt;minimal&lt;/em&gt;* inductive set. There is only one of these, which is $\omega$ (which, as before, you are identifying with $\mathbb{N}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most likely reason that Goldrei says &quot;the intersection of all inductive subsets of a fixed inductive set&quot; rather than &quot;the intersection of all inductive sets&quot; is that the latter of those is an intersection of a proper class worth of sets, which he may not have defined formally. But the result is the same: if you take the intersection of all inductive sets, or the intersection of all inductive subsets of an inductive set, either way you get the unique $\subseteq$-minimal inductive set. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-20T12:42:09.263" CommentCount="1" />
  <row Id="14944" PostTypeId="2" ParentId="14919" CreationDate="2010-12-20T12:45:14.443" Score="2" ViewCount="0" Body="&lt;p&gt;I doubt whether 3 stripes is possible with $a=8$, although I have not tried to prove it. The problem is that, if you construct the group from the top downwards, then each new layer needs to be a faithful module for the group you have already, which means that the dimensions grow rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious way to start is a 2 at the top, followed by a 3 and then $2^2$, so we now have $S_4$. The smallest faithful module for $S_4$ over any field has dimension 3, so we can put a $3^3$ under the $S_4$ to get a group $G$ of order $2^33^4$. Now the smallest faithful irreducible module for $G$ over the field of order 2 has dimension 6 (I did that calculation in Magma, but I expect you can do it in GAP), so we get a group of order $2^93^4$ with three stripes, which is a subgroup of ${\rm AGL}(6,2)$, and I would be surprised if you can do better.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-20T12:45:14.443" CommentCount="3" />
  <row Id="14945" PostTypeId="1" CreationDate="2010-12-20T12:47:58.700" Score="2" ViewCount="49" Body="&lt;p&gt;Thanks for your attention,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an undergraduate. I'm reading the book of Dieter Happel, &lt;em&gt;Triangulated categories in the representation theory of finite dimensional algebras&lt;/em&gt;, &#xA;I cannot prove this sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The maps factoring through an injective object are precisely the null-homotopic maps.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the object is in the complex category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give a completely proof , or some references which take it!&#xA;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="4962" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T18:48:38.533" LastActivityDate="2010-12-20T18:48:38.533" Title="How to prove &quot;The maps factoring through an injective object are precisely the null-homotopic maps&quot;" Tags="&lt;category-theory&gt;&lt;representation-theory&gt;" CommentCount="2" />
  <row Id="14946" PostTypeId="1" AcceptedAnswerId="14950" CreationDate="2010-12-20T12:57:13.097" Score="1" ViewCount="46" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0 \leq \alpha &amp;lt; \beta \leq 1$. I'm looking for an example of a Lebesgue measurable subset $E$ of $\mathbb{R}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\liminf_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \alpha$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\limsup_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \beta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the Lebesgue measure on $\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give an example?&#xA;Thank you,&#xA;Malik&lt;/p&gt;&#xA;" OwnerUserId="1197" LastActivityDate="2010-12-20T13:20:44.350" Title="Lebesgue measurable subset of $\mathbb{R}$ with given metric density at zero" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="14947" PostTypeId="1" CreationDate="2010-12-20T13:02:07.923" Score="1" ViewCount="104" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an ellipsoid of semi-axes a, b, c (possibly prolate, b=c). I am interested in the &quot;shadow&quot; of this solid onto a distant plane, in a given direction d=(k,l,m) orthogonal to that plane. By shadow I mean the projected area onto the plane: each point on the surface of the ellipsoid is translated in the same direction until it intersects with a plane normal to it; the shadow is defined by the envelope of the intersection points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question: Is the projected curve an ellipse? (and what is its equation in terms of a,b,c and the direction vector d)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question: What is the mean area of this shadow when averaged over all orientations of the ellipsoid (or, equivalently, the plane of projection)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing this problem has been solved in the past; any references would be very welcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;baptiste&lt;/p&gt;&#xA;" OwnerUserId="4964" LastActivityDate="2010-12-21T09:05:15.497" Title="Average projected area of an ellipsoid" Tags="&lt;analytic-geometry&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14948" PostTypeId="1" AcceptedAnswerId="14953" CreationDate="2010-12-20T13:09:23.297" Score="2" ViewCount="62" Body="&lt;p&gt;I was asked to prove:&#xA; no group can have a minimal normal subgroup isomorphic to a $\mathrm{Syl}_2(A_7)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should find some property that $\mathrm{Syl}_2(A_7)$ has but not a minimal normal subgroup. So then I thought that it a $\mathrm{Syl}_2(A_7)$ always has characteristic subgroup, i.e. exist $G\ char\ \mathrm{Syl}_2(A_7)$, then the image of G under isomorphism will contradict minimality of &quot;group can have a minimal normal subgroup isomorphic to a $\mathrm{Syl}_2(A_7)$&quot;. But I get stuck then about whether it is true that $\mathrm{Syl}_2(A_7)$ always has characteristic subgroup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is not the correct way to prove it. Can anyone help me?&lt;/p&gt;&#xA;" OwnerUserId="4900" LastActivityDate="2010-12-20T13:28:38.417" Title="Sylow Subgroup problem" Tags="&lt;abstract-algebra&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14949" PostTypeId="2" ParentId="14942" CreationDate="2010-12-20T13:15:13.523" Score="4" ViewCount="0" Body="&lt;p&gt;Write $u(t)=(x(t),v(t)$, where $x$ is the position and $v$ is the velocity. Then Newton's law implies $u&amp;#39; = (v,F/m)$. That's your differential equation, and you can apply RK to it.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-20T13:15:13.523" />
  <row Id="14950" PostTypeId="2" ParentId="14946" CreationDate="2010-12-20T13:16:03.880" Score="2" ViewCount="0" Body="&lt;p&gt;Try the duplication across zero of $$\bigcup_{n \geq 1} \left[\frac{1}{(2n)!}-\alpha\left(\frac{1}{(2n)!} - \frac{1}{(2n+1)!}\right),\frac{1}{(2n)!}\right] \cup \left[\frac{1}{(2n+1)!}-\beta\left(\frac{1}{(2n+1)!} - \frac{1}{(2n+2)!}\right),\frac{1}{(2n+1)!}\right].$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-20T13:16:03.880" CommentCount="1" />
  <row Id="14952" PostTypeId="1" AcceptedAnswerId="14955" CreationDate="2010-12-20T13:25:56.140" Score="2" ViewCount="218" Body="&lt;p&gt;What is Jacobian matrix?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are its applications? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is its physical and geometrical meaning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain with examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! :)&lt;/p&gt;&#xA;" OwnerUserId="705" LastActivityDate="2010-12-21T19:31:37.013" Title="What is Jacobian Matrix?" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="14953" PostTypeId="2" ParentId="14948" CreationDate="2010-12-20T13:28:38.417" Score="2" ViewCount="0" Body="&lt;p&gt;Minimal normal subgroups are direct products of isomorphic simple groups.  If they have order a power of a prime, like 8, then they are abelian of exponent p.  The dihedral group of order 8 is not abelian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea works.  Just take the derived subgroup of the dihedral group; it has order 2 and is normal in any group containing the dihedral subgroup as a normal subgroup.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-20T13:28:38.417" />
  <row Id="14954" PostTypeId="2" ParentId="14942" CreationDate="2010-12-20T13:39:30.980" Score="4" ViewCount="0" Body="&lt;p&gt;I wanted to note that the method you are using is &lt;strong&gt;not&lt;/strong&gt; foward Euler. What you are doing is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t+\delta t) = v(t) + \delta t \cdot a(t) $$&#xD;&#xA;$$x(t+\delta t) = x(t) + \delta t \cdot v(t+\delta t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is called symplectic Euler method, and has much better convergence properties than the standard Forward Euler method. Forward Euler would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t+\delta t) = v(t) + \delta t \cdot a(t) $$&#xD;&#xA;$$x(t+\delta t) = x(t) + \delta t \cdot v(t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your answer, the trick is indeed to consider the velocity as a final variable of your system and write what @lhf said:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(t) = (x(t), v(t))$$&#xD;&#xA;$$u&amp;#39; = (v, F/m)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to solve a single system with six equations per points now. But each equation is only a first order ODE.&lt;/p&gt;&#xA;" OwnerUserId="4709" LastActivityDate="2010-12-20T13:39:30.980" />
  <row Id="14955" PostTypeId="2" ParentId="14952" CreationDate="2010-12-20T14:32:26.163" Score="8" ViewCount="0" Body="&lt;p&gt;Here is an example. Suppose you have two implicit differentiable functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x,y,z,u,v)=0,\qquad G(x,y,z,u,v)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the functions, also differentiable, $u=f(x,y,z)$ and $v=g(x,y,z)$ such&#xA;that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(x,y,z,f(x,y,z),g(x,y,z))=0,\qquad G(x,y,z,f(x,y,z),g(x,y,z))=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you differentiate $F$ and $G$, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;\frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\frac{\partial u}{&#xA;\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x} &amp;amp;=&amp;amp;0\qquad&#xA;\&#xA;\frac{\partial G}{\partial x}+\frac{\partial G}{\partial u}\frac{\partial u}{&#xA;\partial x}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial x} &amp;amp;=&amp;amp;0.&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this system you obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u}{\partial x}=-\frac{\det \begin{pmatrix}&#xD;&#xA;\frac{\partial F}{\partial x} &amp;amp;  &amp;amp; \frac{\partial F}{\partial v} \\ &#xD;&#xA;&amp;amp;  &amp;amp;  \\ \frac{\partial G}{\partial x} &amp;amp;  &amp;amp; \frac{\partial G}{\partial v}%&#xD;&#xA;\end{pmatrix}}{\det \begin{pmatrix}&#xD;&#xA;\frac{\partial F}{\partial u} &amp;amp;  &amp;amp; \frac{\partial F}{\partial v} \\ &#xD;&#xA;&amp;amp;  &amp;amp;  \\ &#xD;&#xA;\frac{\partial G}{\partial u} &amp;amp;  &amp;amp; \frac{\partial G}{\partial v}%&#xD;&#xA;\end{pmatrix}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similar for $\dfrac{\partial u}{\partial y}$, $\dfrac{\partial u}{\partial&#xD;&#xA;z}$, $\dfrac{\partial v}{\partial x}$, $\dfrac{\partial v}{\partial y}$, $%&#xD;&#xA;\dfrac{\partial v}{\partial z}$. The compact notation for the denominator is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial (F,G)}{\partial (u,v)}=\det &#xD;&#xA;\begin{pmatrix}&#xD;&#xA;\frac{\partial F}{\partial u} &amp;amp;  &amp;amp; \frac{\partial F}{\partial v} \\ &#xD;&#xA;&amp;amp;  &amp;amp;  \\ &#xD;&#xA;\frac{\partial G}{\partial u} &amp;amp;  &amp;amp; \frac{\partial G}{\partial v}&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similar for the numerator. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial u}{\partial x}=-\dfrac{\dfrac{\partial (F,G)}{\partial (x,v)}}{%&#xD;&#xA;\dfrac{\partial (F,G)}{\partial (u,v)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\dfrac{\partial (F,G)}{\partial (x,y)},\dfrac{\partial (F,G)}{\partial&#xD;&#xA;(u,v)}$ are &lt;em&gt;Jacobians&lt;/em&gt; (after the 19th century German mathematician Carl Jacobi).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The absolute value of the Jacobian of a coordinate system transformation is also used to convert a multiple integral from one  system into another. In $\mathbb{R}^2$ it measures how much the unit area is distorted by the given transformation, and in $\mathbb{R}^3$ this factor measures the unit volume distortion, etc. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another example: the following coordinate transformation (due to Beukers, Calabi and Kolk)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{\sin u}{\cos v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{\sin v}{\cos u}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;transforms (see &lt;a href=&quot;http://math.stackexchange.com/questions/7807/a-limit-and-a-coordinate-trigonometric-transformation-of-the-interior-points-of-a&quot;&gt;this&lt;/a&gt; question of mine) the square domain $0\lt x\lt 1$ and $0\lt y\lt 1$ into the triangle domain $u,v&amp;gt;0,u+v&amp;lt;\pi /2$ (in &lt;em&gt;&lt;a href=&quot;http://books.google.pt/books?id=CuTpgZaLahkC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+from+the+book&amp;amp;source=bl&amp;amp;ots=swepaSdSxh&amp;amp;sig=j1gB1zxabJE8qxEgbFh1ig8RA3Q&amp;amp;hl=pt-PT&amp;amp;ei=Pnj2TKKcCZOKhQfb67yABg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDIQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs from the BOOK&lt;/a&gt;&lt;/em&gt; by M. Aigner and G. Ziegler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this transformation you get (see Proof 2 in &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; collection of proofs by Robin Chapman) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial (x,y)}{\partial (u,v)}=1-x^2y^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Jacobian sign and orientation of closed curves. Assume you have two small closed curves, one around $(x_0,y_0)$ and another around $u_0,v_0$, this one being the image of the first under the mapping $u=f(x,y),v=g(x,y)$. If the sign of $\dfrac{\partial (x,y)}{\partial (u,v)}$ is positive, then both curves will be travelled in the same sense. If the sign is negative, they will have opposite senses. (See &lt;a href=&quot;http://kr.cs.ait.ac.th/~radok/math/mat9/05a.htm&quot; rel=&quot;nofollow&quot;&gt;Oriented Regions and their Orientation&lt;/a&gt;.) &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-21T17:59:10.257" LastActivityDate="2010-12-21T17:59:10.257" />
  <row Id="14956" PostTypeId="2" ParentId="14909" CreationDate="2010-12-20T14:39:16.150" Score="1" ViewCount="0" Body="&lt;p&gt;Seems you are looking for $x$ such that $g^x = -1$.&lt;br&gt;&#xA;$x = \frac{n-1}{2}$ seems to work because $\phi(n)=n-1$ and g is primitive root and $g^{\phi(n)}=1$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="4750" LastActivityDate="2010-12-20T14:39:16.150" />
  <row Id="14957" PostTypeId="1" CreationDate="2010-12-20T14:46:26.273" Score="3" ViewCount="47" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a commutative integral domain R in which:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;every nonzero prime ideal Q is maximal, and&lt;/li&gt;&#xA;  &lt;li&gt;for every prime power q congruent to 3 mod 8, there is a maximal ideal Q of R with R/Q a field of size q?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am looking for a ring where &quot;q = 3 mod 8&quot; describes finite fields, rather than just finite local rings like in the ring &lt;strong&gt;Z&lt;/strong&gt;.  The only examples I can think of that satisfy the first condition have a finite number of residue fields of each characteristic.  The only examples I can think of that satisfy the second condition have a ton of non-maximal non-proper prime ideals with all sorts of bizarre fields associated to them.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-20T14:46:26.273" Title="Is there an integral domain with a lot of residue fields of the same characteristic?" Tags="&lt;abstract-algebra&gt;&lt;algebraic-number-theory&gt;" CommentCount="4" />
  <row Id="14958" PostTypeId="1" AcceptedAnswerId="14964" CreationDate="2010-12-20T14:47:09.260" Score="1" ViewCount="98" Body="&lt;p&gt;I am looking for a symbol to represent the operation of taking unique values from a vector. So, say the symbol was $\theta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = [0, 0, 0, 1, 1, 1, 3, 1, 2, 0]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta(v) = [0, 1, 3, 2]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this something that just isn't defined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks all for your help here. Your answers and comments have led me to realise that doing this operation on a vector is not really what I want to do; using a set makes much more sense logically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Background: I have many coordinates $\boldsymbol{p}_i = [x_i, y_i, z_i]$ in 3D space, which I then perform a rounding function on so they are equally sampled. However, there will be many duplicates points at each new coordinate, so I remove the duplicated $\boldsymbol{p}$ points to get the final set of points. So it makes sense to define a set $\boldsymbol{P} = \{\boldsymbol{p}_1, ... \boldsymbol{p}_i, \cdot \boldsymbol{p}_n\}$).&lt;/p&gt;&#xA;" OwnerUserId="4965" LastEditorUserId="4965" LastEditorDisplayName="" LastEditDate="2010-12-20T17:36:16.423" LastActivityDate="2010-12-20T17:36:16.423" Title="What is the mathematical symbol for the unique values in a vector?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="14959" PostTypeId="2" ParentId="14930" CreationDate="2010-12-20T14:53:52.777" Score="8" ViewCount="0" Body="&lt;p&gt;To express ⌊x⌋ in terms of Fourier series for $x \in \mathbb{R}/\mathbb{Z}$, let f(x) = x - ⌊x⌋, then the Fourier series for f(x) is quite straightforward, and so we start by showing that f(x) is indeed periodic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any integer $n \in \mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x) = x - ⌊x⌋ = x + n - n - ⌊x⌋ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;= x + n - ⌊x + n⌋ = f(x+n) = f(x + n*1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence f(x) is periodic with fundamental period $T_0 = 1$, so $\omega_0 = \frac{2\pi}{T_0} = 2\pi$, and f(x)=x at $0 \leq x &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/dWIIN.png&quot; alt=&quot;A plot of f(x)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periodic function f(x) can be expressed by the Trigonometric Fourier series,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x) = a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(m\omega_0 x) + b_{m}\sin(m\omega_0 x))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(2\pi mx) + b_{m}\sin(2\pi m x))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficients are calculated as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_0 = \frac{1}{T_0}\int_{&amp;lt;T_0&amp;gt;}f(x)dx = \int_{0}^{1}x dx = \frac{1}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_m = \frac{2}{T_0}\int_{&amp;lt;T_0&amp;gt;}f(x)\cos(m 2\pi  x)dx = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b_m = \frac{2}{T_0}\int_{&amp;lt;T_0&amp;gt;}f(x)\sin(m 2\pi  x)dx = -\frac{1}{\pi m}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint. Use integration by parts...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot of $f(x) = a_0 + \sum_{m=1}^{10} b_m\sin(2\pi m x)$, (summation from m=1 to 10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/XshIE.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we obtained the Fourier series representation of f(x) = x - ⌊x⌋ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then ⌊x⌋ $ = x - f(x) = x - 0.5 + \frac{1}{\pi}\sum_{m=1}^{\infty} \frac{\sin(2\pi m x)}{m}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plot of  ⌊x⌋, where summation from m=1 to m=10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/e4JYv.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this is only valid for $x \in \mathbb{R}/\mathbb{Z}$. Because although ⌊x⌋ is continuous on $\mathbb{R}/\mathbb{Z}$. It has a jump of one unit at each integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;⌊x⌋ consists of an infinite number of steps, and also can be represented as a summation of shifted unit step functions u(x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;⌊x⌋ $ = \sum_{m=1}^{\infty}(u(x-m)-u(-x-m+1))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that u(x) = Sign(x)/2 + 0.5&lt;/p&gt;&#xA;" OwnerUserId="1746" LastEditorUserId="1746" LastEditorDisplayName="" LastEditDate="2010-12-20T18:58:39.300" LastActivityDate="2010-12-20T18:58:39.300" CommentCount="4" />
  <row Id="14960" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T15:09:47.973" Score="6" ViewCount="0" Body="&lt;p&gt;A rigorous proof can be made using the central limit theorem (CLT), as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_0^1  {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x} \to 0  &#xD;&#xA;$$&#xD;&#xA;as $n \to \infty$, it suffices to consider the integral from $1$ to $\infty$. Let $c&amp;gt;0$ be arbitrary but fixed, and let $X_i$ be as in my previous answer. Define $f_n {(x)} = x^n e^{-x} / \Gamma(n+1)$, $x&amp;gt;0$, and put $n&amp;#39;=n+1$. Then, since $f_n$ is the density of $\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }$,&#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_1^{n&amp;#39; - c\sqrt {n&amp;#39;} } {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}  \leq \int_0^{n&amp;#39; - c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x} = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }  - n&amp;#39;}}{{\sqrt {n&amp;#39;} }} \le  - c \bigg).&#xD;&#xA;$$&#xD;&#xA;By the CLT, since the $X_i$ have zero mean and unit variance, the probability on the right-hand side tends to $\Phi(-c)$ as $n \to \infty$, where $\Phi$ denotes the distribution function of the standard normal distribution. &#xA;Similarly,&#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_{n&amp;#39; + c\sqrt {n&amp;#39;} }^\infty  {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}  \leq \int_{{n&amp;#39;} + c\sqrt {n&amp;#39;} }^\infty  {f_n (x)\,{\rm d}x} = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }  - n&amp;#39;}}{{\sqrt {n&amp;#39;} }} &amp;gt; c \bigg),&#xD;&#xA;$$&#xD;&#xA;and, by the CLT, the probability on the right-hand side tends to $1 - \Phi(c)$ as $n \to \infty$.&#xA;Next, note that&#xA;$$&#xD;&#xA;\bigg(1 - \frac{1}{{n&amp;#39; - c\sqrt {n&amp;#39;} }}\bigg)^k \int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x} \leq&#xD;&#xA;\frac{1}{{n!}}\int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x}&#xD;&#xA;$$&#xD;&#xA;and &#xA;$$&#xD;&#xA;\frac{1}{{n!}}\int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {x^{n - k} (x - 1)^k e^{ - x} \,{\rm d}x} \leq \bigg(1 - \frac{1}{{n&amp;#39; + c\sqrt {n&amp;#39;} }}\bigg)^k \int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x}.&#xD;&#xA;$$&#xD;&#xA;By the CLT,&#xA;$$&#xD;&#xA; \int_{n&amp;#39; - c\sqrt {n&amp;#39;} }^{n&amp;#39; + c\sqrt {n&amp;#39;} } {f_n (x)\,{\rm d}x} = {\rm P}\bigg(-c \leq \frac{{\sum\nolimits_{i = 1}^{n&amp;#39;} {X_i }  - n&amp;#39;}}{{\sqrt {n&amp;#39;} }} \leq c \bigg) \to 2 \Phi(c) -1&#xD;&#xA;$$&#xD;&#xA;as $n \to \infty$. Combining it all, and noting that &#xA;$$&#xD;&#xA;\bigg(1 - \frac{1}{{n&amp;#39; \pm c\sqrt {n&amp;#39;} }}\bigg)^k  = \bigg(1 - \frac{1}{{n&amp;#39; \pm c\sqrt {n&amp;#39;} }}\bigg)^{n(k/n)}  \approx e^{ - k/n}, &#xD;&#xA;$$&#xD;&#xA;the result follows by choosing $c$ sufficiently large. [Note that first we choose $c$ sufficiently large, then we let $n \to \infty$.]&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-20T15:09:47.973" />
  <row Id="14962" PostTypeId="2" ParentId="14451" CreationDate="2010-12-20T15:24:48.403" Score="4" ViewCount="0" Body="&lt;p&gt;Although the links provided in some sense answer the question, the specific question can be answered without the full force of a point-in-polygon computation.  I would recommend this.&#xA;Compute whether each angle of your quad $(a,b,c,d)$ is convex or reflex.  If one is reflex&#xA;(say $a$), connect it to the opposite vertex $c$.  If all are convex, choose any diagaonal; e.g., $(a,c)$.  Now you have partitioned your quad into two triangles.  Check if your point is in either triangle, by checking if it is left-of-or-on each of its three edges. &lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-20T15:24:48.403" />
  <row Id="14963" PostTypeId="1" AcceptedAnswerId="14966" CreationDate="2010-12-20T15:43:42.733" Score="2" ViewCount="43" Body="&lt;p&gt;Does there exist a bi-Invariant metric on $SL_n(\mathbb{R})$. I tried to google a bit but I didn't find anything helpful. &lt;/p&gt;&#xA;" OwnerUserId="707" LastActivityDate="2010-12-20T16:20:52.107" Title="Bi invariant metrics on $SL_n(\mathbb{R})$" Tags="&lt;lie-groups&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14964" PostTypeId="2" ParentId="14958" CreationDate="2010-12-20T15:49:05.710" Score="3" ViewCount="0" Body="&lt;p&gt;A common way to express that you are throwing away duplicates is to translate a sequence into a set. A set is commonly understood as an unordered collection of &lt;em&gt;distinct&lt;/em&gt; objects. It is usually denoted using curly braces. So you could write $\theta(v) = \{v_i\}_{i\in\{1,\ldots,n\}}$, where $v=[v_1,\ldots,v_n]$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-20T15:49:05.710" CommentCount="3" />
  <row Id="14965" PostTypeId="2" ParentId="14919" CreationDate="2010-12-20T15:52:15.517" Score="1" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; is the maximum number of stripes possible on a zebra group with a=8.  S stripes require &lt;b&gt; a ≥ C⋅9&lt;sup&gt;S&lt;/sup&gt; &lt;/b&gt; for large S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The number of stripes S of a zebra group gives bounds on the derived length D of a zebra group: 2S - 1 ≤ D ≤ 2S + 1.  The action of G/H on a chief factor H/K must be irreducible and faithful, and so G/H embeds in GL(H/K) = GL(n,q) for q in {2,p} the prime dividing the order q^n of H/K.  However, for large n, the maximal derived length of a solvable subgroup of GL(n,q) is about log(n-2)/log(9), or in other words, n ≥ 9^D.  In particular, as S increases, the &quot;a&quot; from 2&lt;sup&gt;a&lt;/sup&gt; p&lt;sup&gt;b&lt;/sup&gt; increases exponentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small n, the maximal derived length of soluble subgroups of GL(n,2) are: 2, 3, 4, 4.  The maximal derived length of irreducible soluble subgroups of GL(n,2) appear to be 2,2,3,2,6,2,5,4,4.  Hence the minimum dimensions for derived length D=2 is n=2, D=3 is n=4, D=4 is n=6, D=5 is n=6, D=6 is n=6, and D≥7 is n≥11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top stripe has order at least 2^1.  By the time H/K is the second stripe, G/H has derived length 2, and so H/K has order at least 2^2.  By the time H/K is the third stripe, G/H has derived length 4, and so H/K has order at least 2^6.&#xA;  Since 1+2+6 ≥ 8, this shows there is no three-striped zebra group with a Sylow 2-subgroup of order 2^8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three stripes is attainable at a=9.  Four stripes is not attainable for a≤14 and is attainable for a=27.  Roughly speaking, S stripes require a ≥ C⋅9&lt;sup&gt;S&lt;/sup&gt; (though my proof needs S really large).&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="583" LastEditorDisplayName="" LastEditDate="2010-12-20T16:02:52.870" LastActivityDate="2010-12-20T16:02:52.870" />
  <row Id="14966" PostTypeId="2" ParentId="14963" CreationDate="2010-12-20T16:11:52.173" Score="5" ViewCount="0" Body="&lt;p&gt;A connected Lie group has a biinvariant metric iff it is isomorphic to the dirtect product of a compact one and a vector space---see Milnor's &lt;em&gt;Curvatures of left invariant metrics&lt;/em&gt;. Such a factorization would carry over to the Lie algebras, etc.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-20T16:11:52.173" />
  <row Id="14967" PostTypeId="2" ParentId="14963" CreationDate="2010-12-20T16:20:52.107" Score="5" ViewCount="0" Body="&lt;p&gt;It's enough to show the answer is &quot;no&quot; for $SL_2(\mathbb{R})$, because $SL_2(\mathbb{R})$ is naturally embedded in $SL_n(\mathbb{R})$, and if $SL_n(\mathbb{R})$ has a biinvariant metric, then the induced metric on $SL_2(\mathbb{R})$ would be biinvariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's focus on $SL_2(\mathbb{R})$.  Finding a biinvariant metric on $SL_2(\mathbb{R})$ is equivalent to finding an $Ad(SL_2(\mathbb{R})$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, so I'll show there is no such $Ad$ invariant metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathfrak{sl_2(\mathbb{R})}$ is 3 dimensional with basis $E_{12}, E_{21}, E_{11} - E_{22}$ where $E_{ij}$ denotes the matrix with a 1 in the $ij$ slot and a 0 elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $A = diag(2,1/2)$ ,the diagonal matrix with diagonal entries 2 and 1/2.  This is clearly in $SL_2(\mathbb{R})$.  Note then that $AE_{12}A^{-1} = 4 E_{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if an inner product were $Ad$ invariant, it would have to preserve the length of $E_{12}$, giving a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there is no $Ad$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, and hence there is no biinvariant metric on $SL_n(\mathbb{R})$.&lt;/p&gt;&#xA;" OwnerUserId="331" LastActivityDate="2010-12-20T16:20:52.107" />
  <row Id="14968" PostTypeId="2" ParentId="14947" CreationDate="2010-12-20T16:23:54.147" Score="1" ViewCount="0" Body="&lt;p&gt;When a surface element $d\omega$ at a point $P$ in three-space is projected orthogonally into all possible directions then the average area of the image is $\rho\thinspace d\omega$ where $\rho$ is a universal constant. To determine the value of $\rho$ we argue as follows: The unit sphere $S^2$ has a surface area of $4\pi$, and the area of its projection is $2\pi$ in all directions (note that the &quot;shadow&quot; is doubly covered); whence also the average area of the projection is $2\pi$. So $\rho$ must be ${1\over2}$. What we have found out is valid for any surface element $d\omega$ of your ellipsoid $E$. It follows that the average area of the shadow of $E$ is equal to $\omega(E)/4$, where $\omega(E)$ is the total surface area of $E$ and we have taken into account that the shadow is covered twice. The value so obtained can be expressed as an elliptic integral in terms of $a$, $b$, $c$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastEditorUserId="1303" LastEditorDisplayName="" LastEditDate="2010-12-21T09:05:15.497" LastActivityDate="2010-12-21T09:05:15.497" CommentCount="7" />
  <row Id="14969" PostTypeId="2" ParentId="14958" CreationDate="2010-12-20T16:41:04.230" Score="0" ViewCount="0" Body="&lt;p&gt;There is no standard notation for the &lt;em&gt;set&lt;/em&gt; of coefficients of a vector. Indeed, there's not even any standard notation for the set of coefficients of a polynomial (which would be useful when taking the &lt;em&gt;content&lt;/em&gt;, i.e. the gcd / ideal generated by the coefficients).&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-20T16:41:04.230" />
  <row Id="14970" PostTypeId="1" AcceptedAnswerId="14974" CreationDate="2010-12-20T17:14:57.207" Score="6" ViewCount="293" Body="&lt;p&gt;Suppose $A$ is the set of all rational numbers $p$ such that $p^2 &amp;lt;2$ and $B$ is the set of all rational numbers $p$ such that $p^2 &amp;gt; 2$. We want to show that $A$ contains no largest element and $B$ contains no smallest element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt;, he associates $q = p- \frac{p^2-2}{p+2} = \frac{2p+2}{p+2}$. Did he just come up with this using trial and error?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-21T01:16:09.827" Title="Rational Numbers" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14971" PostTypeId="1" AcceptedAnswerId="14993" CreationDate="2010-12-20T17:22:51.870" Score="5" ViewCount="71" Body="&lt;p&gt;I know this seems like a juvenile question, but for some reason I can't recall how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I want to find 87 reduced by 99.75%. What is the equation to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3621" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-20T19:15:03.070" LastActivityDate="2010-12-20T19:15:03.070" Title="What is the equation for finding a number that is X% less than another number?" Tags="&lt;algebra-precalculus&gt;&lt;elementary&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="14972" PostTypeId="2" ParentId="14926" CreationDate="2010-12-20T17:28:29.210" Score="0" ViewCount="0" Body="&lt;p&gt;Here is a non-rigorous attempt at an answer, but perhaps someone can make this into a formal argument.  I will assume your series is centered at $0$.  I will also denote your function by $g$ for reasons you will see as you read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $g(x) = \sum_{n=0}^{\infty} a_n x^n$, where $a_n \geq0$ for every $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We look for a series for $h(x) = \frac{1}{g(x)} = \sum_{n=0}^{\infty} b_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking formal derivatives $b_n = D^n(h(0))/n!$ (as in the Taylor expansion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then do a Google search... Behold the Faá di Bruno's Formula for derivatives of composite functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/FaadiBrunosFormula.html&quot; rel=&quot;nofollow&quot;&gt;http://mathworld.wolfram.com/FaadiBrunosFormula.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f(x) = 1/x$ so that $h(x) = f(g(x))$.  Comparing with the terms in this formula, we see that the signs of the $b_n$ must alternate (or be zero).  This is because the derivatives of $1/x$ alternate in sign and the derivatives of $g(x)$ are non-negative by your hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't trust any of this, however, since I've completely neglected matters of convergence; but when dealing with formal power series, maybe this point is moot?  &lt;/p&gt;&#xA;" OwnerUserId="4009" LastActivityDate="2010-12-20T17:28:29.210" />
  <row Id="14973" PostTypeId="2" ParentId="14971" CreationDate="2010-12-20T17:33:19.017" Score="6" ViewCount="0" Body="&lt;p&gt;If you want $X$% of 87, multiply the latter by $\frac{X}{100}$. (So, for example, to get 25 percent, you multiply by $0.25$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by &quot;reduced by 99.75%&quot; you mean 99.75% of the total, then, multiply by $.9975$. If &quot;reduced by 99.75%&quot; means taking 99.75% &lt;em&gt;off&lt;/em&gt; (so you are left with only 0.25%), then multiply by $0.0025$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T18:18:29.320" LastActivityDate="2010-12-20T18:18:29.320" CommentCount="2" />
  <row Id="14974" PostTypeId="2" ParentId="14970" CreationDate="2010-12-20T17:42:15.953" Score="17" ViewCount="0" Body="&lt;p&gt;The point here is that iterations of the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_transformation&quot; rel=&quot;nofollow&quot;&gt;Mobius transformation&lt;/a&gt; $p \mapsto \frac{2p+2}{p+2}$ converge to $\sqrt{2}$, so every time you apply the transformation you get closer to $\sqrt{2}$.  This can be thought of as describing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction&quot; rel=&quot;nofollow&quot;&gt;generalized continued fraction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 - \frac{2}{4 - \frac{2}{4 - \frac{2}{...}}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\sqrt{2}$.  The dynamics of Mobius transformations in general are fairly well-understood; for the transformation $p \mapsto \frac{ap+b}{cp+d}$ the possible fixed points are the roots of the quadratic polynomial $cp^2 + dp = ap + b$, and using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_fixed_point_theorem&quot; rel=&quot;nofollow&quot;&gt;Banach fixed point theorem&lt;/a&gt; (or a more specific closed form of iterations of Mobius transformations using linear algebra) one can determine which of the fixed points are attractive or repellent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if one wanted to design a Mobius transformation converging to $\sqrt{n}$ for some non-square $n$, this would require that $d = a, c = 1, b = n$, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p \mapsto \frac{ap + n}{p + a}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some $a$, and it should not be hard to find a value of $a$ such that the fixed point $\sqrt{n}$ is attractive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this technique is a nice trick, because the polynomial $cp^2 + dp = ap + b$ is quadratic, it does not generalize to prove the corresponding result for roots of cubic or higher degree polynomials.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-20T17:42:15.953" CommentCount="5" />
  <row Id="14975" PostTypeId="1" CreationDate="2010-12-20T17:51:01.357" Score="3" ViewCount="47" Body="&lt;p&gt;Hi Folks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to track down the source of some wonky data. The data are response times (RTs) collected from humans using a computer keyboard. Here's a histogram of the RTs, binned to 1ms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/54QBH.png&quot; alt=&quot;1hist&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obvious is a skew in the data, which is expected/known for such data. However, also noticeable is a slight &quot;ghosting&quot; of the image. Zooming in we see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/mLVww.png&quot; alt=&quot;2hist&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appears to be some phenomenon such that the probability of observing an RT at a given time point fluctuates across time in a cyclical manner. This is also evident in the empirical cumulative density function (suggesting that the phenomenon is about data, not the binning algorithms of the histograms):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/n9RLH.png&quot; alt=&quot;3ecdf&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the nature of this cyclicality, I computed the 2nd derivative of the ecdf (actually, I've never taken calculus, so I simply computed diff(diff(y)) in R), yielding the following waveform (again, zoomed in to show details):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/q9P1h.png&quot; alt=&quot;4ecdf_deriv&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually encountered a similar phenomenon in the past and at that time my googling seemed to suggest that a periodogram would help me determine the Hz values at which something was happening to cause this phenomenon. I found some R code for a periodogram:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;periodogram &amp;lt;- function(x) {&#xA;    n &amp;lt;- length(x)&#xA;    Mod(fft(x))[1:(n%/%2 + 1)]^2 / (2*pi*n)&#xA;}&#xA;&#xA;plot(&#xA;    x = 1:((length(rt)/2)+1)&#xA;    , y = periodogram(ddy)&#xA;    , type = 'l'&#xA;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/CAwJJ.png&quot; alt=&quot;5periodogram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure what the x-axis is supposed to be. Hz seems reasonable, but doesn't seem to fit with my impression from the histograms that the phenomenon has a cycle of about 100Hz. Any help providing clarification on the x-axis of this plot would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For those with interest in this over-and-above the math, if I do find that the phenomenon has a 100Hz or thereabouts frequency, then I'm pretty sure it can be attributable to the polling rate of the USB keyboard, plus a little random error here and there in other aspects of timing)&lt;/p&gt;&#xA;" OwnerUserId="1361" LastActivityDate="2010-12-20T19:21:33.753" Title="Can you provide help with interpreting this periodogram?" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" />
  <row Id="14976" PostTypeId="2" ParentId="13790" CreationDate="2010-12-20T17:57:11.563" Score="1" ViewCount="0" Body="&lt;p&gt;Using this &lt;a href=&quot;http://math.stackexchange.com/questions/9999/checking-if-a-number-is-a-fibonacci-or-not/14872#14872&quot;&gt;other my answer&lt;/a&gt;&lt;br&gt;&#xA;Using integers only, I would use &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_search&quot; rel=&quot;nofollow&quot;&gt;Binary Search&lt;/a&gt;. Certainly you can compute $F_n$ only with integers, the simplest way is matrix exponentiation. Using Binary Search you can find numbers ``near'' your number $x$ and you will find $x = F_n$ (and $n$). I suppose this method is generic for anything monotone you can compute fast. To initiliaze the binary search, just keep doubling $ F_{2n} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary search allows you to search for a number x in a sorted &quot;array&quot; F[] (in the programming sense). Use this method to search for your number. When you need F[n] just compute $F_n$. This will work because the sequence is strictly increasing except for the initial 1,1.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastActivityDate="2010-12-20T17:57:11.563" />
  <row Id="14977" PostTypeId="1" AcceptedAnswerId="14978" CreationDate="2010-12-20T18:01:57.073" Score="3" ViewCount="80" Body="&lt;p&gt;Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = -\sup(-A)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $-A$ is bounded above. Hence $\sup(-A)$ exists. So to proceed, one would show that $\inf A \leq -\sup(-A)$ and $\inf A \geq -\sup(-A)$? Or $-\inf A &amp;gt; \sup(-A)$ and $-\inf A &amp;lt; \sup(-A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the sense that one can use the following fact as well: (i) Let $\alpha = \sup S$. If $\beta &amp;lt; \alpha$ then $\beta$ is not an upper bound for $S$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-20T18:23:53.573" Title="Supremum and Infimum" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14978" PostTypeId="2" ParentId="14977" CreationDate="2010-12-20T18:23:12.263" Score="5" ViewCount="0" Body="&lt;p&gt;This is just a matter of using the definitions of supremum and infimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just let $x = \inf{A}$ and $y = \sup{(-A)}$. So since $x = \inf{A}$ then for every $a \in A$ you know that $x \leq a$. But then this implies that $-x \geq -a$ for all $a \in A$, so $-x = - \inf{A}$ is an upper bound for the set $-A$. But since $y = \sup{(-A)}$ is the least upper bound for $-A$, then this means that $y \leq -x$ or that $\sup{(-A)} \leq - \inf{A}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prove the other inequality.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-20T18:23:12.263" />
  <row Id="14979" PostTypeId="2" ParentId="14977" CreationDate="2010-12-20T18:23:53.573" Score="1" ViewCount="0" Body="&lt;p&gt;You can do it by inequalities; or you can use the definitions of supremum and infimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that a real number $S$ is the supremum of $A$ if and only if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$a\leq S$ for all $a\in A$ ($S$ is an upper bound for $A$); and&lt;/li&gt;&#xA;&lt;li&gt;For all $\epsilon\gt 0$ there exists $a_{\epsilon}\in A$ such that $S-\epsilon \lt a_{\epsilon}$ (no number strictly smaller than $S$ is an upper bound for $A$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(2 above can be written slightly differently so that it applies to arbitrary partially ordered sets: you require that for all $s$, if $s\lt S$, then there exists $a\in A$ such that $s\lt a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dually, a real number $T$ is the infimum of $A$ if and only if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$a\geq T$ for all $a\in A$ ($T$ is a lower bound for $A$); and&lt;/li&gt;&#xA;&lt;li&gt;For all $\epsilon \gt 0$ there exists $a_{\epsilon}\in A$ such that $T+\epsilon \gt a_{\epsilon}$ (no number strictly larger than $T$ is a lower bound for $A$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, you can use the fact that $T=\inf(A)$ is the infimum to show that $-T$ is the supremum of $-A$ by verifying the properties of the supremum. It should be straightforward.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-20T18:23:53.573" />
  <row Id="14980" PostTypeId="1" AcceptedAnswerId="14983" CreationDate="2010-12-20T18:30:54.757" Score="1" ViewCount="58" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does someone know if the following is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\mathbb{X}$ be some arbitrary Banach space.&#xA;$\{x_k \}_{k=1}^{\infty} \in \mathbb{X}$ is a sequence chosen from $\mathbb{X}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the series &#xA;$$\sum_{k=1}^\infty \|x_k\|_X$$&#xD;&#xA;converges, would the &quot;more generic&quot; series&lt;br&gt;&#xA;$$\sum_{k=1}^\infty x_k$$ &#xA;also converge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, could you please give the proof (or just mark the proof steps) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4971" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T18:36:40.593" LastActivityDate="2010-12-20T18:59:58.737" Title="Converging series in Banach space" Tags="&lt;convergence&gt;&lt;banach-spaces&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14982" PostTypeId="2" ParentId="14952" CreationDate="2010-12-20T18:34:25.600" Score="11" ViewCount="0" Body="&lt;p&gt;The Jacobian $df_p$ of a differentiable function $f : \mathbb{R}^n \to \mathbb{R}^m$ at a point $p$ is its best linear approximation at $p$, in the sense that $f(p + h) = f(p) + df_p(h) + o(|h|)$ for small $h$.  This is the &quot;correct&quot; generalization of the derivative of a function $f : \mathbb{R} \to \mathbb{R}$, and everything we can do with derivatives we can also do with Jacobians.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, when $n = m$, the determinant of the Jacobian at a point $p$ is the factor by which $f$ locally dilates volumes around $p$ (since $f$ acts locally like the linear transformation $df_p$, which dilates volumes by $\det df_p$).  This is the reason that the Jacobian appears in the change of variables formula for multivariate integrals, which is perhaps the basic reason to care about the Jacobian.  For example this is how one changes an integral in rectangular coordinates to cylindrical or spherical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jacobian specializes to the most important constructions in multivariable calculus.  It immediately specializes to the gradient, for example.  When $n = m$ its trace is the divergence.  And a more &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_dual&quot; rel=&quot;nofollow&quot;&gt;complicated construction&lt;/a&gt; gives the curl.  The rank of the Jacobian is also an important local invariant of $f$; it roughly measures how &quot;degenerate&quot; or &quot;singular&quot; $f$ is at $p$.  This is the reason the Jacobian appears in the statement of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Implicit_function_theorem&quot; rel=&quot;nofollow&quot;&gt;implicit function theorem&lt;/a&gt;, which is a fundamental result with applications everywhere.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-20T18:34:25.600" CommentCount="1" />
  <row Id="14983" PostTypeId="2" ParentId="14980" CreationDate="2010-12-20T18:35:18.713" Score="3" ViewCount="0" Body="&lt;p&gt;Yes.  You can use the hypothesis that $\sum\|x_k\|$ converges to show that the sequence of partial sums of the series $\sum x_k$ is a Cauchy sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if this property holds in a normed space, then the space is complete; that direction isn't as straightforward.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-20T18:35:18.713" />
  <row Id="14984" PostTypeId="1" AcceptedAnswerId="14987" CreationDate="2010-12-20T18:47:22.900" Score="5" ViewCount="193" Body="&lt;p&gt;I understand the principle of finite induction, but my book then mentions that there is a variant of the first where requirement b is changed to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If k is a positive integer such that 1,2...,k belong to S, then k + 1 must also be in S.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sample problem is proving that the inequality about the Lucas numbers &#xA;l&lt;sub&gt;n&lt;/sub&gt; &amp;lt; (7/4)&lt;sup&gt;n&lt;/sup&gt;&#xA;I understand the algebra of the proof, but I do no understand how it uses the second principle or how it is different from the first principle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof for the lucas numbers&#xA;$$&#xD;&#xA;a_1 = 1 &amp;lt; \left(\frac{7}{4}\right)^1 = \frac{7}{4}\text{ and }a_2 = 3 \lt \left(\frac{7}{4}\right)^2 = \frac{49}{16}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose an integer $k\geq 3$ and assume that the inequality is valid for $n = 1,2,\ldots,k$.  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{k-1} \lt \left(\frac{7}{4}\right)^{k-1}\text{ and }a_{k-2} \lt \left(\frac{7}{4}\right)^{k-2},$$&#xD;&#xA;so &#xA;\begin{align*}&#xA;a_k &amp;amp;= a_{k-1} + a_{k-2}\\&#xA;&amp;amp;\lt \left(\frac{7}{4}\right)^{k-1} + \left(\frac{7}{4}\right)^{k-2}\\&#xA;&amp;amp;= \left(\frac{7}{4}\right)^{k-2}\left(\frac{7}{4} + 1\right)\\&#xA;&amp;amp;= \left(\frac{7}{4}\right)^{k-2}\left(\frac{11}{4}\right)\\&#xA;&amp;amp;\lt \left(\frac{7}{4}\right)^{k-2}\left(\frac{7}{4}\right)^2\\&#xA;&amp;amp;= \left(\frac{7}{4}\right)^k&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The responses are all helpful, but if someone could point out what the difference is when you go about actually using strong induction instead of weak induction. (You could use the lucas number proof or something else)&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T20:45:28.123" LastActivityDate="2010-12-20T21:04:52.687" Title="What is the second principle of finite induction?" Tags="&lt;proof&gt;&lt;induction&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" />
  <row Id="14985" PostTypeId="1" AcceptedAnswerId="14992" CreationDate="2010-12-20T18:48:58.887" Score="8" ViewCount="157" Body="&lt;p&gt;This problem came from a Putnam problem solving seminar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If each person in a group of n people is a friend of at least half the people in the group, then show that it is possible to seat the n people in a circle so that everyone sits next to friends only.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My idea was to use induction on $n$; if $n$ is odd we can remove a person, note that the remaining $n-1$ people all have at least $n/2$ friends left, use our inductive hypothesis to seat them, and then use the pigeonhole principle to seat the last person.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this doesn't work when $n$ is even because after removing a person, some of the remaining people might have less than $n/2$ friends left.  In fact, my friend who actually participated in the seminar said they had the same issue, but didn't address it because they ran out of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this sort of induction a reasonable approach?  If so, how would we deal with the case when $n$ is even?  If not, what's a better way to think about the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm not sure of the best tags for this question, so please feel free to re-tag if necessary.&lt;/p&gt;&#xA;" OwnerUserId="4904" LastActivityDate="2010-12-20T19:57:11.473" Title="Seating friends around a dinner table" Tags="&lt;combinatorics&gt;&lt;graph-theory&gt;&lt;contest-math&gt;" AnswerCount="2" />
  <row Id="14987" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T18:58:13.483" Score="12" ViewCount="0" Body="&lt;p&gt;There are two basic differences:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In ordinary induction, we need a &lt;strong&gt;base case&lt;/strong&gt; (proving it for $k=1$; that is, proving that $1\in S$); in the second principle of induction (also called &quot;strong induction&quot;) you do not need a base case (but see the &lt;em&gt;caveat&lt;/em&gt; below).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In ordinary induction, the &quot;Induction Hypothesis&quot; is that the $k\in S$; in the second principle, the &quot;Induction Hypothesis&quot; is that &lt;em&gt;all&lt;/em&gt; the natural numbers strictly less than $k+1$ are in $S$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, if you want to prove that every positive integer can be factored into primes, if you use ordinary induction you would have to first show that $1$ can be factored into primes, and then you would show that if $k$ can be factored into primes, then $k+1$ can be factored into primes. This is actually pretty hard to do, because the factorization of $k+1$ has little or nothing to do with the factorization of $k$. This is a very difficult proposition to prove directly using ordinary induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to prove the same proposition using the second principle of induction, you would instead assume that &lt;strong&gt;all&lt;/strong&gt; numbers strictly less than $k$ can be factored into primes, and try to show from this assumption that $k$ can be factored into primes (which is much simpler: if $k=1$ or $k$ is prime, there is nothing to do; otherwise, you can write $k=ab$, each $a$ or $b$ strictly smaller than $k$ and greater than $1$, so you can apply the induction hypothesis to both to get a factorization).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, in the second principle, you get to assume &lt;em&gt;more&lt;/em&gt; in the inductive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Caveat:&lt;/em&gt; Formally, the second principle does not require a &lt;em&gt;base&lt;/em&gt;; this is because if you can actually prove that &quot;if all positive integers strictly smaller than $k$ are in $S$ then $k$ is in $S$&quot;, then for $k=1$ the premise is vacuously true (&lt;em&gt;all&lt;/em&gt; positive integer smaller than $1$ &lt;em&gt;are&lt;/em&gt; in $S$, because there is no such number at all), so the implication would yield that $k=1$ must also be in $S$. However, in most actual applications, the inductive argument does not work for $k=1$ (or for other special cases), so that these have to be proven separately. They amount to &quot;special cases&quot; instead of &quot;bases&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, both principles are &lt;em&gt;equivalent&lt;/em&gt; for the positive integers. That is, if you can do a proof using ordinary induction, then you can translate that into a proof using the second induction principle in a fairly mechanical manner. And if you can do a proof using the second induction principle, you can transform that into a proof using the first induction principle (though in an indirect manner), again in a fairly mechanical way (provided you take for granted some properties of the positive integers; for example, that every positive integer is either equal to $1$, or else is $k+1$ for some positive integer $k$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The reason the second principle is called &quot;strong induction&quot; is that you can use it in other contexts to prove more than ordinary induction would. For example, imagine a situation in which you have a set $T$ that consists of &lt;em&gt;two&lt;/em&gt; copies of the integers: a blue copy and a green copy, with the blue copy going &quot;first&quot;. If you prove that $S$ is a subset of $T$ such that blue $1$ is in $S$, and whenever $k$ ($k$ either a blue positive integer or a green positive integer) is in $S$ then $k+1$ is in $S$ (ordinary induction), then you can conclude that all blue integers are in $S$, but the argument by itself does not tell you whether there are any green integers in $S$ (you would have to separately prove that green-$1$ is in $S$). But if you use strong induction, and prove that whenever all integers, blue or green, that are strictly smaller than $k$ are in $S$ then $k$ is in $S$, then you &lt;em&gt;can&lt;/em&gt; conclude that $S$ will contain all blue integers &lt;em&gt;and&lt;/em&gt; all green integers too. So you can actually get a &lt;em&gt;stronger&lt;/em&gt; conclusion from using strong induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Caveat the Second&lt;/em&gt;: Special cases show up far more often in proofs using strong induction than in proofs using ordinary induction, but they sometimes occur in the latter as well. As a famous example of a &quot;proof&quot; using regular induction that would require a special case (and that in fact is false because that special case fails) is the fake proof of the proposition that &quot;If, in a group of people, at least one of them has blue eyes, then everyone in the group has blue eyes.&quot; The proof is as follows: Let $k$ denote the number of people in the group. We prove it by induction on $k$. If $k=1$ then the result obviously holds. Assume the result holds for any group with $k$ people, and take a group with $k+1$ people, $(p_1,\ldots,p_{k},p_{k+1})$, in which at least one has blue eyes, say $p_1$. Looking at the first $k$, $(p_1,\ldots,p_k)$, we can use the induction hypothesis to conclude that $p_1,p_2,\ldots,p_k$ each have blue eyes. Then looking at the last $k$, $(p_2,\ldots,p_k,p_{k+1})$, since at least one has blue eyes (say $p_k$), then they all have blue eyes by the induction hypothesis. In particular, $p_{k+1}$ has blue eyes. Therefore, they al have blue eyes, QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is incomplete is that the inductive step only works if $k\geq 3$, so that a proof would require the &lt;em&gt;special case&lt;/em&gt; of showing that $1\in S$ implies $2\in S$ (the general argument we have does not cover that particular implication). That's where it all breaks down, of course. But the point is that sometimes ordinary induction also requires &quot;special cases&quot;, though this tends to happen less often than with strong induction.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;The step where you use the inductive hypothesis in the proof above is in the second line of the long display, where you are using that the inequality you want holds for both $a_{k-1}$  &lt;strong&gt;and&lt;/strong&gt; for $a_{k-2}$. This requires you to assume that it holds for two numbers strictly smaller than $k$ (you don't need the fact that it holds for &lt;em&gt;all&lt;/em&gt; numbers less than $k$ here, but you do end up needing it for more than just the immediately previous one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regular induction, your only assumption is that the result holds for the immediately previous number (for $k-1$, when you are trying to prove it for $k$; or for $k$, when you are trying to prove it for $k+1$). You make no assumptions about what happens for $\ell$ with $\ell\lt k-1$. In strong induction, you assume that it holds for &lt;em&gt;all&lt;/em&gt; $\ell\lt k$, not just for $k-1$. Here, because you need the assumption for &lt;em&gt;two&lt;/em&gt; smaller values of $k$, this argument would not work if you were only assuming that $a_{k-1}\lt\left(\frac{7}{4}\right)^{k-1}$ but were making no assumptions about $a_{k-2}$. There &lt;em&gt;are&lt;/em&gt; ways to prove the result using ordinary induction, but this particular argument would not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that since your proof requires you to have at least two smaller values than $k$, the argument will not work for either $k=1$ or $k=2$ (the first has no smaller values, the second has only one); that is why you end up having to prove the &lt;em&gt;two&lt;/em&gt; special cases of $a_1$ and $a_2$. Again, this is not a &quot;basis for the induction&quot;, but rather special cases of the inductive argument.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-20T21:04:52.687" LastActivityDate="2010-12-20T21:04:52.687" />
  <row Id="14989" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T18:59:27.570" Score="4" ViewCount="0" Body="&lt;p&gt;The second principle of induction to which you refer is sometimes called the strong principle of induction, as opposed to the usual one which is also sometimes called the weak principle of induction. This distinction is only apparent at first since they are both equivalent, so there's no harm in using the strong version of it and thinking you're using something much more powerful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in this &lt;a href=&quot;http://mathcircle.berkeley.edu/BMC4/Handouts/induct/node6.html&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt;, the difference is basically that in the weak principle of induction, your inductive hypothesis is just that the property you want to prove is true for some integer $n$, and then you show that it also holds for $n+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the strong version you assume that the property holds for all integers up to $n$, that is, you assume it holds for $0, 1, 2, \dots , n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a theorem that is sometimes proved using the strong version is the so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;Fundamental Theorem of Arithmetic&lt;/a&gt;. The theorem basically consists of two parts, one is showing the existence of a factorization into primes and the other is showing that &quot;the&quot; factorization is unique. So for instance if you're showing that a factorization exists by induction, then in the inductive step you'll have to prove that if $n$ can be factored as a product of primes, the so can $n+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem now is that you have basically two options, that $n+1$ is a prime in which case you're done, or that $n+1$ is composite, thus it can be factored as $n+1 = ab$ where $1 &amp;lt; a \leq n$ and $1 &amp;lt; b \leq n$. Now you can see where the strong principle comes in handy because if you assume not only that $n$ can be factored as a product of primes, but also that all the integers greater than $1$ can (up to $n$ of course), then you immediately know that both $a$ and $b$ are products of primes, say $a = p_1 \cdots p_k$ and $b = q_1 \cdots q_s$ with $p_i$ and $q_j$ primes, then $n+1 = ab = p_1 \cdots p_k \cdot q_1 \cdots q_s$ is also a product of primes, so the induction is complete.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-20T19:28:53.040" LastActivityDate="2010-12-20T19:28:53.040" CommentCount="2" />
  <row Id="14990" PostTypeId="2" ParentId="14980" CreationDate="2010-12-20T18:59:58.737" Score="2" ViewCount="0" Body="&lt;p&gt;It is indeed true. It is in fact almost the same as the statement of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_M-test&quot; rel=&quot;nofollow&quot;&gt;Weierstrass M-test&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-20T18:59:58.737" />
  <row Id="14991" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T19:01:41.140" Score="6" ViewCount="0" Body="&lt;p&gt;That's not a very good example.  This principle usually goes by the name of &lt;strong&gt;strong induction&lt;/strong&gt;, and the point is that while it's equivalent to ordinary induction, it's often more convenient to use in proofs.  For proving a sequence $P(k)$ of identities, it's often straightforward to write $P(k+1)$ in terms of $P(k)$, but for a more complicated sequence of statements one might need the stronger assumption that $P(1), ... P(k)$ all hold in order to readily deduce $P(k+1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if one is proving a sequence of statements $P(k)$ about finite &lt;a href=&quot;http://en.wikipedia.org/wiki/Graph_theory&quot; rel=&quot;nofollow&quot;&gt;graphs&lt;/a&gt; with $k$ vertices, one general proof strategy is to break a graph up into two smaller graphs, and if the statement for each of the smaller graphs implies the statement for the larger graph, we are done by strong induction, but not necessarily by weak induction, since the smaller graphs do not necessarily have sizes $1$ and $k-1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, here's a nice exercise.  Consider the sequence $a_n$ with $a_1 = 1$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{2n} = a_n$$&#xD;&#xA;$$a_{2n+1} = a_n + 1.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try proving by strong induction that $a_n$ is the sum of the digits of $n$ in base $2$.  Can you see why this proof is difficult to phrase using weak induction?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-20T19:01:41.140" CommentCount="2" />
  <row Id="14992" PostTypeId="2" ParentId="14985" CreationDate="2010-12-20T19:06:14.883" Score="11" ViewCount="0" Body="&lt;p&gt;This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_path#Bondy.E2.80.93Chv.C3.A1tal_theorem&quot; rel=&quot;nofollow&quot;&gt;Dirac's theorem&lt;/a&gt;.  I don't know if an inductive proof is possible; the standard proof is extremal.  You consider the longest sequence of friends and show by contradiction that it must be a Hamiltonian circuit.  I can edit with more details if you want the full argument; it's quite nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the argument is as follows.  It can be found, for example, in Bollobas' &lt;em&gt;Modern Graph Theory&lt;/em&gt; (p. 106).  First let's establish the graph-theoretic language.  Let $G$ be a graph on $n$ vertices, $n \ge 3$.  For each $v \in G$ let $d_v$ denote the number of neighbors of $v$.  Then the claim is that if $d_v \ge \frac{n}{2}$ for each $v \in G$, then $G$ is Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We claim that $G$ is connected.  Indeed, if $v, w \in G$ then $v$ and $w$ each have $\frac{n}{2}$ neighbors out of a possible $n-1$, so they must share some neighbor.  (This step works under the weaker assumption that $d_v + d_w \ge n$ for each $v \neq w$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that $G$ is not Hamiltonian.  Let $v_1, ... v_k$ be a path all of whose vertices are distinct with $k$ maximal.  Then $v_1$ and $v_k$ have no neighbors outside the path (by maximality), so all of their neighbors must be one of $v_1, ... v_k$.  In addition, $v_1$ cannot be connected to $v_k$; otherwise we would have a cycle of length $k$, and by connectivity this cycle would be connected to some $w$ not in the cycle, giving a path of length $k+1$, which contradicts maximality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, $v_1$ cannot be connected to a vertex $v_{i+1}$ with the property that $v_i$ is connected to $v_k$, since otherwise $v_{i+1} v_1 v_2 ... v_i v_k v_{k-1} ... v_{i+2}$ is a cycle of length $k$.  It follows that the sets $\{ v_i : v_1 \sim v_i \}$ and $\{ v_{i+1} : v_i \sim v_k \}$ are disjoint subsets of $\{ v_2, ... v_k \}$, but by assumption their union must have size at least $n$ (this step also works under the weaker assumption that $d_v + d_w \ge n$ for each $v \neq w$); contradiction.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-20T19:57:11.473" LastActivityDate="2010-12-20T19:57:11.473" CommentCount="3" />
  <row Id="14993" PostTypeId="2" ParentId="14971" CreationDate="2010-12-20T19:10:11.153" Score="5" ViewCount="0" Body="&lt;p&gt;A simple formula for this is: $$ \text{ New Number } = \text{ Old Number } \times \left(\frac{100 - x}{100}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general if a number is increased or decreased by $x\%$ then the new number is given by: $$ \text{ Old Number } \times \left(\frac{100 \pm x}{100}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-20T19:10:11.153" CommentCount="1" />
  <row Id="14994" PostTypeId="2" ParentId="14975" CreationDate="2010-12-20T19:21:33.753" Score="0" ViewCount="0" Body="&lt;p&gt;What is your sampling interval?  Looking at your second graph, it appears there is something in your acquisition that makes it much more probable to measure times that are at about 7.5 or 8 millisecond intervals than times in between.  One approach is to figure out the spacing and bin at this interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periodogram shows frequencies where there is lots of energy in the data.  It looks like the peak at 1000 would correspond to 1/(8 msec) or about 120 Hz.  Does that correspond to your sampling?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-20T19:21:33.753" CommentCount="1" />
  <row Id="14995" PostTypeId="2" ParentId="14925" CreationDate="2010-12-20T19:40:48.687" Score="6" ViewCount="0" Body="&lt;p&gt;Let $S_{n,k}$ denote the number of permutations in which the first $k$ elements are not fixed.  I published an expository paper on these numbers earlier this year.  See &quot;&lt;a href=&quot;http://math.pugetsound.edu/~mspivey/DT.pdf&quot; rel=&quot;nofollow&quot;&gt;Deranged Exams&lt;/a&gt;,&quot; (College Mathematics Journal, 41 (3): 197-202, 2010).  Aravind's formula is in the paper, as are several others involving $S_{n,k}$ and related numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem 7 (which I also mention in &lt;a href=&quot;http://math.stackexchange.com/questions/14477/&quot;&gt;this recent math.SE question&lt;/a&gt;) is relevant to this question.  It's&#xA;$$S_{n+k,k} = \sum_{j=0}^n \binom{n}{j} D_{k+j},$$&#xD;&#xA;where $D_n$ is the number of derangements on $n$ elements.  See the paper for a simple combinatorial proof of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $D_n$ grows as $n!$ via $D_n = \frac{n!}{e} + O(1)$ (see Wikipedia's page on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Derangement#Counting_derangements&quot; rel=&quot;nofollow&quot;&gt;derangement numbers&lt;/a&gt;), &#xA;the dominant terms in the probability&#xA;$\frac{S_{n+k,k}}{(n+k)!}$ are the $j = n$ and $j = n-1$ terms from the Theorem 7 expression.  Thus we have &#xA;$$\frac{S_{n+k,k}}{(n+k)!} \approx \frac{D_{n+k} + n D_{n+k-1}}{(n+k)!} \approx \frac{1}{e}\left(1  + \frac{n}{n+k}\right) \approx e^{-1} e^{\frac{n}{n+k}} = e^\frac{-k}{n+k},$$&#xD;&#xA;where the second-to-last step uses the first two terms in the Maclaurin series expansion for $e^x$.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-20T19:40:48.687" CommentCount="2" />
  <row Id="14996" PostTypeId="1" AcceptedAnswerId="15032" CreationDate="2010-12-20T19:41:41.883" Score="3" ViewCount="88" Body="&lt;p&gt;Let $C: x=x(t), y=y(t), a\le t\le b$ be a $C^1$ closed curve (not necessarily simple).The isoperimetric inequality says that&#xA; $$ A\le \frac{\ell^2}{4\pi},$$ where $$A=\left|\int_C y(t)x&amp;#39;(t) dt\right|$$ is the area enclosed by $C$, and &#xA;$\ell=\int_a^b \sqrt{(x&amp;#39;(t))^2+(y&amp;#39;(t))^2} dt$ is the arc length of $C$. &#xA;My question is how to use this theorem to prove  Wirtinger theorem: If $f(t)$ is a $T$-periodic $C^1$ real-valued function such that&#xA;$\int_0^T f(t) dt=0,$ then $$\int_0^T |f(t)|^2 dt\le \frac{T^2}{4\pi^2}\int_0^T |f&amp;#39;(t)|^2 dt.$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-20T22:05:33.280" LastActivityDate="2010-12-21T15:33:35.920" Title="Isoperimetric inequality implies Wirtinger's inequality" Tags="&lt;analysis&gt;&lt;fourier-analysis&gt;&lt;inequality&gt;&lt;fourier-series&gt;" AnswerCount="1" />
  <row Id="14997" PostTypeId="1" CreationDate="2010-12-20T19:42:04.367" Score="1" ViewCount="70" Body="&lt;p&gt;Fix $b&amp;gt;1$. Show the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) If $m,n,p,q$ are integers, $n&amp;gt;0, q&amp;gt;0$, and $r = m/n = p/q$, prove that $(b^{m})^{1/n} = (b^p)^{1/q}$. So $(b^r) = (b^m)^{1/n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Prove that $b^{r+s} = b^{r}b^{s}$ for $r,s \in \mathbf{Q}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, there $t$ is rational and $t \leq x$. Prove that $b^r = \sup B(r)$ where $r$ is rational. So $b^x = \sup B(x)$ for every real $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) Show that $b^{x+y} = b^{x}b^{y}$ for all real $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (a), we know that $mq = np$. Thus $b^{mq} = b^{np}$. Then $(b^{m})^{q} = (b^{n})^{p}$. Or $(b^{m})^{q} = (b^{p})^{n}$. So the result follows if we note that $m = np/q$ and $n = mq/p$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) just start with $b^{\frac{mq+pn}{qn}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (c) and (d), we just need to show that $B(x)$ is non-empty. Then use a proof by contradiction? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-20T20:11:34.663" LastActivityDate="2010-12-20T20:23:44.030" Title="Exponents and Suprema" Tags="&lt;real-analysis&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14998" PostTypeId="1" AcceptedAnswerId="14999" CreationDate="2010-12-20T19:47:50.087" Score="1" ViewCount="81" Body="&lt;p&gt;Hello!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was particularly interested in the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I read this proof, everything seemed fine and logical except one detail (the proof is located &lt;a href=&quot;http://planetmath.org/?method=l2h&amp;amp;from=objects&amp;amp;name=ProofOfNecessaryAndSufficientConditionsForANormedVectorSpaceToBeABanachSpace&amp;amp;op=getobj&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right after we prove, that the series $\sum_{k=1}^\infty x_{N_m} - x_{N_m+1}$ converges, there is a statement which tells us that the &lt;em&gt;limit&lt;/em&gt; of that series (let's name it $s$) definitely belongs to the initial space $\mathbb{X}$: $s \in \mathbb{X}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why is that? That could probably be very obvious, but unfortunately I can't get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should a $lim \hspace{2 mm} \sum_{k=1}^\infty (x_{i} - x_{j})$, where $x_i \in \mathbb{X}$ belong to $\mathbb{X}$ itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What am I missing? :)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4971" LastActivityDate="2010-12-20T20:06:22.113" Title="Series sum question" Tags="&lt;functional-analysis&gt;" AnswerCount="1" />
  <row Id="14999" PostTypeId="2" ParentId="14998" CreationDate="2010-12-20T19:52:08.117" Score="5" ViewCount="0" Body="&lt;p&gt;Because that is what it means for a series to converge in $X$.  It means that the sequence of partial sums converges to an element of $X$.  In general, convergence of a sequence means that there is some element of your space to which the sequence converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate a bit, to say that a sequence $(y_n)$ in $X$ converges without adding any qualification is another way to say that the sequence converges &lt;em&gt;in&lt;/em&gt; $X$, which means that it has a limit in $X$.  That is, there is an element $L$ of $X$ such that for all $\varepsilon\gt0$ there exists an $N$ such that $n\gt N$ implies $\|y_n-L\|\lt\varepsilon$.  That is precisely the notion used in the hypothesis at your link, and it is the convergence referred to when we say that Cauchy sequences converge in complete spaces.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-20T20:06:22.113" LastActivityDate="2010-12-20T20:06:22.113" CommentCount="4" />
  <row Id="15000" PostTypeId="2" ParentId="13739" CreationDate="2010-12-20T19:54:05.347" Score="2" ViewCount="0" Body="&lt;p&gt;I'll prove both parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that for some $N$ and some $p$,&#xA;$(1-\sqrt[n]{a_n})\frac{n}{\log{n}}\geq p&amp;gt;1$ for $n&amp;gt;N$. From here it&#xA;follows that it's enough to show the convergence of $\sum&#xD;&#xA;(1-p\frac{\log{n}}{n})^n$. Since&#xA;$(1-p\frac{\log{n}}{n})^n=\exp(n\log(1-p\frac{\log{n}}{n}))$, I'll use&#xA;an order argument using the series for $\log(1-x)$ and $\exp{x}$.&#xA;These are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-...-\frac{x^n}{n} + o(x^n), x\to 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp{x}=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...+\frac{x^n}{n!} + o(x^n), x\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\log(1-p\frac{\log{n}}{n})=-p\frac{\log{n}}{n}-p^2\frac{(\log{n})^2}{n^2}+o(\frac{(\log{n})^2}{n^2})$,&#xA;for $\frac{\log{n}}{n}\to 0$, so&#xA;$n\log(1-p\frac{\log{n}}{n})=-p\log{n}-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})$.&#xA;Therefore&#xA;$\exp(n\log(1-p\frac{\log{n}}{n}))=\exp(-p\log{n}-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))=\exp(-p\log{n})\exp(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))=\frac{1}{n^p}\exp(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))=\frac{1}{n^p}\left(&#xD;&#xA;1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})+o\left(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})&#xD;&#xA;\right) \right)=\frac{1}{n^p}\left(&#xD;&#xA;1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}) \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $(1-p\frac{\log{n}}{n})^n=\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}+o(\frac{(\log{n})^2}{n^{p+1}})$.&#xA;Dividing the LHS of the last equation by&#xA;$\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}$ and evaluating the&#xA;limit, we find that&#xA;$\lim_{n\to\infty}\frac{(1-p\frac{\log{n}}{n})^n}{\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}}=1$.&#xA;So the series of the terms in the denominator converges iff if the&#xA;series of the terms in the numerator does. But since $p&amp;gt;1$, we know&#xA;that $\sum\frac{1}{n^p}$ converges, and&#xA;$\sum\frac{(\log{n})^2}{n^{p+1}}$ converges too by comparing with the&#xA;former. Hence $\sum (1-p\frac{\log{n}}{n})^n$ converges and so does&#xA;$\sum a_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to the divergence part. It's enough to prove that $\sum&#xD;&#xA;(1-\frac{\log{n}}{n})^n$ diverges. All steps above are still valid for&#xA;$p=1$, so $\lim_{n\to\infty}\frac{(1-\log{n}/n))^n}{1/n-(\log{n})^2/n^2}=1$.&#xA;But $\sum(\log{n})^2/n^2$ can be shown to converge, by comparison&#xA;with the series $\sum\frac{1}{n^\frac{3}{2}}$. This implies that the series in the&#xA;denominator diverges and so does that in the numerator, which is what&#xA;we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the key here involves manipulation of $o-$little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this test is known in&#xA;Russian circles as &quot;Zhame test&quot; &lt;/p&gt;&#xA;" OwnerUserId="4786" LastActivityDate="2010-12-20T19:54:05.347" CommentCount="2" />
  <row Id="15001" PostTypeId="2" ParentId="14985" CreationDate="2010-12-20T19:54:39.143" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.math.uri.edu/~eaton/GT4F03.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uri.edu/~eaton/GT4F03.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has a simple proof: Theorem 6&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-20T19:54:39.143" />
  <row Id="15002" PostTypeId="2" ParentId="14984" CreationDate="2010-12-20T19:59:05.193" Score="3" ViewCount="0" Body="&lt;p&gt;There is a common misconception that strong induction is somehow &quot;stronger&quot; than normal induction, but it isn't really. Strong induction is just the special case of normal induction in which the induction hypothesis $P(n)$ takes the form &quot;for all $m &amp;lt; n$, $Q(m)$&quot;. Thus strong induction is actually a &lt;em&gt;limited&lt;/em&gt; form of the principle of induction, because not all induction hypotheses are of that special form. On the other hand, any proof by strong induction can be trivially rephrased as a proof by &quot;weak&quot; induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason for the terminological difficulty is that the only place that people talk about &quot;strong induction&quot; is in introductory courses. There, &quot;use strong induction&quot; can be a hint about what sort of induction hypothesis to choose. In these classes, the principle of induction is stated somewhat informally, so it may not be obvious what counts as a legal induction hypothesis in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contexts where the principle of induction is stated formally, e.g. Peano arithmetic, nobody talks about strong induction at all. The axiom scheme for induction in Peano arithmetic simply includes all axioms of the form&#xA;$$&#xD;&#xA;(P(0) \land (\forall n)(P(n) \to P(n+1)) \to (\forall n)(P(n))&#xD;&#xA;$$&#xD;&#xA;where $P$ is any formula of Peano arithmetic. This includes formulas of the form $(m &amp;lt; n)\to Q(m)$ as well as all other formulas. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-20T19:59:05.193" />
  <row Id="15003" PostTypeId="2" ParentId="14997" CreationDate="2010-12-20T20:23:44.030" Score="0" ViewCount="0" Body="&lt;p&gt;I've been working on this exercise today as well. I managed to prove (I think) (a), but it's a bit complicated. Let's see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 1:&lt;/strong&gt; $(a^x)^{1/x} = a$,  for all $a&amp;gt;0,x\in\mathbb{Q}$. This follows from the uniqueness of positive roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 2:&lt;/strong&gt; $(a^x)^{1/y} = (a^{1/y})^x$, for all $a&amp;gt;0\in\mathbb{Q}$, $x,y\in\mathbb{Z}$. This follows from Theorem 1.2.1 from Rudin's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma 3:&lt;/strong&gt; $(a^x)^y = a^{xy}$, for all $a\in\mathbb{Q}$, $x,y\in\mathbb{Z}$, since $\prod_1^y\prod_1^x a = \prod_1^{xy} a$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove (a), we start from $b = (b^{1/q})^q$ because of (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we raise both sides to the $m$th power to get $b^m = ((b^{1/q})^q)^m= (b^{1/q})^{qm}$, because of (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $m/n = p/q$, then $qm = pn$. Then $b^m = (b^{1/q})^{pn}=((b^{1/q})^p)^n$, once again because of (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $b^m = ((b^p)^{1/q})^n$ because of (2). Taking the $n$th root to both sides we get $(b^m)^{1/n} = (b^p)^{1/q}$, which is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove (b), consider $b^{mq+np}=b^{mq}b^{np}$ (which holds because $m,n,p,q\in\mathbb{Z}$). This is equivalent to $(b^{m/n})^{nq}(b^{p/q})^{qn}$ because of the previously mentioned lemmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the $nq$th root to both sides, we end up with $b^{\frac{mq+np}{nq}} = b^{m/n}b^{p/q}$, which is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) is proved easily since $b^r \geq x\in B(r)$ and $b^r\in B(r)$, therefore $b^r=\sup(B(r))$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't proved (d) yet but will edit this if I do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'm not sure about this so feedback is really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-12-20T20:23:44.030" />
  <row Id="15004" PostTypeId="1" CreationDate="2010-12-20T20:43:42.963" Score="1" ViewCount="39" Body="&lt;p&gt;I completely understand how to find the likelihood functions of simple pdfs. However, how would you attempt to find the likelihood function of a pdf with a negative exponential function with two different means, where the overall mean is just the two means added together??? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method to do this would be appreciated. thanks.&lt;/p&gt;&#xA;" OwnerUserId="4522" LastActivityDate="2010-12-20T21:41:56.650" Title="Likelihood function estimating two different means." Tags="&lt;statistics&gt;&lt;estimation-theory&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15005" PostTypeId="2" ParentId="15004" CreationDate="2010-12-20T21:02:32.067" Score="1" ViewCount="0" Body="&lt;p&gt;I suspect that you are describing &lt;a href=&quot;http://en.wikipedia.org/wiki/Mixture_model&quot; rel=&quot;nofollow&quot;&gt;mixture model&lt;/a&gt;. There are a lot of ways of estimating these types of models.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2010-12-20T21:41:56.650" LastActivityDate="2010-12-20T21:41:56.650" />
  <row Id="15006" PostTypeId="2" ParentId="14970" CreationDate="2010-12-20T21:03:25.887" Score="10" ViewCount="0" Body="&lt;p&gt;Rudin's result is very simply explained: $\ $ given lower and upper approximations to a square-root, we may obtain a better lower approximation $\rm\ p&amp;#39;\ $ by $\:$ &quot;composing&quot; $\:$ them,$\ $ namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM&lt;/strong&gt; $\rm\displaystyle\quad\ \  q\  &amp;gt;\ \sqrt d\  &amp;gt; \ p\ \ \:\Rightarrow\:\ \ \sqrt d\ &amp;gt; \ p&amp;#39;\ &amp;gt;\ p\quad\ \ for\quad\ p&amp;#39; \:=\ \frac{p\:q+d}{p+q} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt; $\rm\quad\displaystyle  0\ \: &amp;gt;\ (q-\sqrt d)\ \ (p-\sqrt d)\ =\  p\:q+d - (p+q)\:\sqrt d\ \ \Rightarrow\ \  \sqrt d\ &amp;gt;\ p&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally $\rm\quad\quad\displaystyle p&amp;#39;-p\ =\ \frac{p\:q+d}{p+q} - p\ =\ \frac{\ d - p^2}{p+q}\: &amp;gt;\ 0\ \ \Rightarrow\ \ p&amp;#39;\ &amp;gt;\ p\quad\quad\ \ $ &lt;strong&gt;QED&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally specializing $\rm\: q = 2 = d\: $ yields Rudin's approximation $\rm\displaystyle\  p&amp;#39;\ =\ \frac{2\:p+2}{\ \:p+2}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-21T01:16:09.827" LastActivityDate="2010-12-21T01:16:09.827" />
  <row Id="15007" PostTypeId="1" CreationDate="2010-12-20T21:14:29.660" Score="3" ViewCount="178" Body="&lt;p&gt;I'm having a little trouble grasping the definition of a topological space in terms of the open-sets. Suppose that we're defining a topology on the real line, in theory could we arbitrarily select the elements of the open-sets, so long as the set of all open sets satisfies the properties of a topological space?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastActivityDate="2010-12-21T09:18:59.600" Title="Topological space definition in terms of open-sets" Tags="&lt;topology&gt;&lt;general-topology&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="15008" PostTypeId="1" AcceptedAnswerId="15026" CreationDate="2010-12-20T22:14:19.137" Score="4" ViewCount="198" Body="&lt;p&gt;Is it possible to find closed form of following sum: $\sum\limits_{k=1}^{n-1}\frac{x^k}{n-k}$ ?&lt;br&gt;&#xA;If not then maybe there is at least some good aproximation?&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-21T00:45:56.973" LastActivityDate="2010-12-21T12:00:30.550" Title="Sum of series $\frac{x^k}{n-k}$" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="15009" PostTypeId="2" ParentId="15007" CreationDate="2010-12-20T22:21:43.153" Score="5" ViewCount="0" Body="&lt;p&gt;The main drive in mathematics is usually to look at some known object and to generalize the notion. Once you defined &quot;open intervals&quot; and saw that the union of them remains open (in the sense that each point has some interval around it) and that a finite intersection is also open, and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have, say, a list of properties that some space you know has, and you want things to look more or less like your known space (in this case $\mathbb{R}$). So you start looking for a &quot;right&quot; generalization. In topology it was found out that it is enough to request that the open sets have these three properties, namely - any union, and a finite intersection of open sets is still open, the empty set and the entire space are open as well. Once you have these properties you can talk about convergence and continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did we define convergence of sequences on the real line? $a_n \to a$ if and only if, for an interval around $a$ which is arbitrarily small we have only finitely many elements outside the interval. So we do the same with open sets. Eventually it was apparent that sequences are not always enough, and you need something stronger (namely &lt;a href=&quot;http://en.wikipedia.org/wiki/Net_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;nets&lt;/a&gt;). The main idea here, that you took something that well-defined to begin with and sought a way to generalize it so it would be both consistent and useful. And when groups, rings and things like function spaces (that is all the functions from some set $X$ to $\mathbb{R}$ or $\mathbb{C}$) are useful to you, in one way or another - you might want to define some general notion that would capture the essence in all the spaces that you encounter, granted that they have certain properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About continuity, it turns out that one can simply require that the preimage of an open set is an open set, and that is enough for a function to be continuous. Combine this with convergence and you have that by defining the open sets you already know a lot about the space at hand (with the given topology), because you know to say when something converges and when functions are continuous, and that is quite a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question, you can define several different topologies on the real line. The standard one is generated by open intervals, while others might have different properties and would be generated by other kind of sets (e.g. sets of the form $[a,b)$) or maybe stated explicitly (e.g. &quot;Every subset is open&quot;), but different topologies might have different characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two topologies, and you have a basis for each one, if you can show that each open set in one topology contains an open set from the other - you can say that the topologies are essentially the same, and would have the same properties. A good example is when you take $\mathbb{R}^2$ and you take the topology generated by open balls (that is given a point and a radius under the familiar Euclidean distance) and by squares, for example. These two families would generate the same topology. That is, both the topologies will have the same open sets and the same properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main issue, when you approach a new space and try to define topology on it, is just one: You want the topology not to be &quot;too loose&quot; (i.e. not enough open sets, so nothing really converges, and nothing is really continuous) and not &quot;too tight&quot; (i.e. too many open sets, so not enough functions are continuous and not enough sequences/nets would converge.)&lt;br&gt;&#xA;This is not always an easy task, although many times there is a natural way to define things, so not all is lost.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-20T22:21:43.153" />
  <row Id="15010" PostTypeId="2" ParentId="14311" CreationDate="2010-12-20T22:31:03.100" Score="1" ViewCount="0" Body="&lt;p&gt;Edit: This solution is wrong, as explained in the comments below. The proof breaks in the italicized phrase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Sigma = \{ \sigma_i \}$. Take the following grammar of size $O(|\Sigma|)$: $$S \longrightarrow \sigma_i S \sigma_i S | \varepsilon$$&#xA;Clearly in every word produced, every terminal appears an even number of times (proof by induction). In order to generate a word $w$ matching this description, suppose that $w = \sigma_i w&amp;#39;$. Then $\sigma_i$ must appear again in $w&amp;#39;$, say $w = \sigma_i z_1 \sigma_i z_2$. &lt;i&gt;Note that $z_1,z_2$ also match the description.&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in CNF: $$\begin{align*} S &amp;amp;\longrightarrow \varepsilon \\ S &amp;amp;\longrightarrow S_i S_i \\ S_i &amp;amp;\longrightarrow \sigma_i | A_i T \\ A_i &amp;amp;\longrightarrow \sigma_i \\ T &amp;amp;\longrightarrow S_i S_i \end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-21T18:36:41.447" LastActivityDate="2010-12-21T18:36:41.447" CommentCount="8" />
  <row Id="15011" PostTypeId="1" CreationDate="2010-12-20T23:11:02.387" Score="7" ViewCount="97" Body="&lt;p&gt;I was sorting out the coins in my loose-change jar the other day, and the following thought crossed my head:  Is it possible to deduce the number of each type of coin in this pile by simply weighing them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coins I were counting were Australian 5c, 10c, 20c and 50c.  According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coins_of_the_Australian_dollar&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, they weights are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5c     2.83 grams&#xA;10c    5.65 grams&#xA;20c   11.30 grams&#xA;50c   15.55 grams&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But this might be a particularly bad choice of weights for the coins.  Since 565 divides 1130, we can't tell the difference (through weighing) between two 10c coins and one 20c coin.  It seems that two 5c coins and one 10c coin would be hard to distinguish also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What would be a better way to designate the weights of these coins so that we could (in most cases) uniquely determine the number of each type of coin in a single weighing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would be subject to some practical constraints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each coin is a reasonably light, but not too light (e.g. between 2 and 20 grams).&lt;/li&gt;&#xA;&lt;li&gt;If A and B are two multisets of coins, and A and B have equal weights, then |A| and |B| should be very large (more than is likely to be in a typical jar).&lt;/li&gt;&#xA;&lt;li&gt;The scales do not measure with infinite precision.  Coins are not minted with infinitely accurate weights.&lt;/li&gt;&#xA;&lt;li&gt;The weights of the coins must differ by a reasonable amount (e.g. by at least 3 grams).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-21T00:35:52.787" Title="How can I assign weights to coins so that they can be counted in a single weighing?" Tags="&lt;number-theory&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15012" PostTypeId="2" ParentId="15008" CreationDate="2010-12-20T23:20:40.730" Score="3" ViewCount="0" Body="&lt;p&gt;Tomek: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a suggestion, perhaps it is not giving you as explicit a formula as you'd like. Let $f_n(x)$ the sum that you are interested in. Then $$g_n(x):=x^{-n}f_n(x)=\sum_{k=1}^{n-1}\frac{x^{k-n}}{n-k}=\sum_{k=1}^{n-1}\frac{x^{-k}}{k}.$$&#xD;&#xA;Let $y=1/x$, and $h_n(y):=g_n(1/y)$. Differentiate to get $$\sum_{k=1}^{n-1}y^{k-1}=\frac{1-y^{n-1}}{1-y}.$$ This gives you a formula for $h_n(y)$ after integrating (if you try to find it explicitly rather than leaving it in terms of the integral, this formula is a bit messy because it depends on $n$). From this, you easily get the formula for $f_n(x)$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-20T23:20:40.730" CommentCount="5" />
  <row Id="15013" PostTypeId="1" AcceptedAnswerId="15014" CreationDate="2010-12-20T23:25:42.787" Score="0" ViewCount="125" Body="&lt;p&gt;I am supposed to prove that $(1/1) + (1/3) + (1/6) + \dots + (1/t_n) &amp;lt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hint is that &#xA;$$&#xD;&#xA;\frac2{n(n+1)} = 2\left(\frac1{n} - \frac1{n+1}\right)&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was thinking that if you take $ 2\left(\frac{1}{t_n} + \frac{1}{t_{n+1}}\right) = \frac1{t_{\frac{n}2}}$ where $n$ is even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my logic is you can subtract 1 from both sides multiply by 2 and you are back where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg&#xA;$$ 1-1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + ... &amp;lt; 2 -1$$&#xD;&#xA;$$ 2\left(\frac{1}{2} + \frac{1}{6} + ... \right) &amp;lt; 1 \times 2 $$&#xD;&#xA;$$ 1 - 1 + \frac{1}{3} + ... &amp;lt; 2 -1$$&#xD;&#xA;etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct logic or am I missing something?  Obviously by the hint this was not the proof they had in mind.  So how would I prove this using the hint?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditorDisplayName="" LastEditDate="2010-12-21T05:12:03.763" LastActivityDate="2010-12-21T15:10:47.817" Title="The sum of the reciprocal of the triangular numbers up to $\frac1{t_n}$ is &lt; 2" Tags="&lt;algebra-precalculus&gt;" AnswerCount="3" />
  <row Id="15014" PostTypeId="2" ParentId="15013" CreationDate="2010-12-20T23:38:54.973" Score="6" ViewCount="0" Body="&lt;p&gt;You're supposed to add the reciprocals of the triangular numbers, which are given by $$t_n = \frac{n(n+1)}{2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the hint is just telling you to write in a more convenient way the reciprocal of $t_n$, so that when you add them something nice will happen. For example, try writing the first few sums using the suggested identity and try to discern a pattern from that.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-20T23:38:54.973" CommentCount="3" />
  <row Id="15015" PostTypeId="2" ParentId="15013" CreationDate="2010-12-20T23:50:50.787" Score="3" ViewCount="0" Body="&lt;p&gt;Your statement that $\frac{1}{t_n} + \frac{1}{t_{n+1}} = \left(\frac{1}{t_{n-1}}\right)\left(\frac{1}{2}\right)$ where n&gt;1 is not correct. $\frac{1}{6}+\frac{1}{10}=\frac{4}{15}$ which is not equal to $\frac{1}{3}\cdot \frac {1}{2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not clear what you mean &quot;you can always get back to the first term 1 as long as you multiply the inequality by 2.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say &quot;$(1/t_4)+(1/t_5)\dots(1/t_n) &amp;lt; 2$. Because $(1/t_1) = 1$ and $(1/t_2) + (1/t_3) = 1$&quot;, in fact $(1/t_2) + (1/t_3) = 1/2$.  If you add the three inequalities together, the right side is $3 \frac{1}{2}$ with this correction, but you were asked to prove it with $2$ on the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Adrian Barquero said, you need to follow the direction of the hint.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-20T23:50:50.787" CommentCount="4" />
  <row Id="15016" PostTypeId="1" AcceptedAnswerId="15020" CreationDate="2010-12-20T23:51:07.680" Score="2" ViewCount="60" Body="&lt;p&gt;Another problem on combinatorics. This time I'm asking for a hint and if it is possible a general strategy when dealing with this kind of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show without using the preceding results * that the probability $p_{m}(r,n)=n^{-1}E_{m}(r,n)$ of finding exactly $m$ cells empty satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{m}(r+1,n)=p_{m}(r,n)\frac{n-m}{n}+p_{m+1}(r,n)\frac{m+1}{n} \qquad \qquad (1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* The results which I can't use must be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{m}(r,n)=\binom{n}{m}A(r,n-m)=\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}(n-m-\nu)^{r}$$ and by association $$A(r,n+1)=\sum_{k=1}^{r}\binom{r}{k}A(r-k,n)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, $A(r,n)$ is equal to $n!S(r,n)$ where $S(r,n)$ are the Stirling numbers of the second kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to proceed since the 'preceding result' I can't use is basically the definition for $E_{m}(r,n)$. Even if I use such definition in the left hand side of equation $(1)$, I end up with the first term $p_{m}(r,n)\frac{n-m}{m}$ and a nasty second term $\binom{n}{m}\sum_{s=0}^{n-m}(-1)^{s}\binom{n-m}{s}(1-\frac{m+s}{n})^{r}(-\frac{s}{n})$which doesn't resemble the equation $(1)$.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-20T23:56:58.023" LastActivityDate="2010-12-21T00:23:23.743" Title="Probability of finding exactly $m$ cells empty" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15017" PostTypeId="1" CreationDate="2010-12-20T23:56:25.030" Score="5" ViewCount="100" Body="&lt;p&gt;$X$ is a random variable, which is not constant. $E[X]=0$. $E[X^4] \leq 2(E[X^2])^2$.&#xA;Let $Y$ be given by: $P(Y=E[X|X \geq 0]) = P(X \geq 0)$ and $P(Y=E[X|X \lt 0]) = P(X \lt 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we necessarily have $E[Y^4] \leq 2(E[Y^2])^2$?&lt;/p&gt;&#xA;" OwnerUserId="4979" LastActivityDate="2010-12-23T15:17:36.993" Title="Effects of condensing a random variable to only 2 possible values" Tags="&lt;probability&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15018" PostTypeId="2" ParentId="14908" CreationDate="2010-12-21T00:21:35.053" Score="1" ViewCount="0" Body="&lt;p&gt;Couldn't resist that the 'solution' that instantly came to mind was to observe that if x is the supplied series, then 2x = 2 - x, from which x = 2/3 follows easily. Also, x = 1 - 2 + 4 - 8 + 16 ... implies 2x + x = 1 implies x = 1/3. BZZZT!!!&lt;/p&gt;&#xA;" OwnerUserId="4980" LastActivityDate="2010-12-21T00:21:35.053" />
  <row Id="15020" PostTypeId="2" ParentId="15016" CreationDate="2010-12-21T00:23:23.743" Score="4" ViewCount="0" Body="&lt;p&gt;So (looking at your last problem) $p_m(r,n)$ is the probability of finding m cells out of n empty when you scatter r objects randomly among the bins.  Now you are trying to find an expression for $p_m(r+1,n)$.  Think about how you can get there with $r+1$ objects.  You can either have $m$ cells empty with $r$ objects and put the new one in an occupied bin, or you can have $m+1$ cells empty with $r$ objects and put the new one in an unoccupied bin.  This should lead you to the equation you want.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T00:23:23.743" CommentCount="3" />
  <row Id="15021" PostTypeId="2" ParentId="15011" CreationDate="2010-12-21T00:35:52.787" Score="1" ViewCount="0" Body="&lt;p&gt;A small observation.  For two coins let's suppose we arbitrarily fix the precision at somewhat better than one gram, so let's make the weights $w_1, w_2$ positive integers.  Then the way to minimize redundancy is to require that the weights $w_1, w_2$ are relatively prime.  This is to maximize the size of the smallest nontrivial solution to $w_1 x + w_2 y = 0$.  Within the constraints you describe this means we should take $w_1 = 17, w_2 = 20$; then we can decide the number of coins unambiguously up to the point where we have either $20$ of the first coin or $17$ of the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we up the precision to somewhat better than a tenth of a gram then we should instead take something like $w_1 = 16.9, w_2 = 20$; then I think we can decide the number of coins unambiguously up to the point where we have either $200$ of the first coin or $169$ of the second, which is surely good enough for all practical purposes.  So it seems to me that the answer to the problem depends strongly on what precision we can measure weights to.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-21T00:35:52.787" CommentCount="2" />
  <row Id="15022" PostTypeId="1" AcceptedAnswerId="15027" CreationDate="2010-12-21T00:44:24.173" Score="0" ViewCount="81" Body="&lt;p&gt;How do I find generators of &lt;a href=&quot;http://en.wikipedia.org/wiki/Octahedral_symmetry&quot; rel=&quot;nofollow&quot;&gt;Octahedral&lt;/a&gt; group when viewing it as a subgroup of S8?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update&lt;/b&gt;: Alex Bartel's suggestion gives the following permutations and Cayley graph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cube.png&quot;&gt;&#xA;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley.png&quot;&gt;&#xA;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley3d.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update 2&lt;/b&gt;: second suggestion by Alex gives following Cayley graph, which &lt;a href=&quot;http://11011110.livejournal.com/116782.html&quot; rel=&quot;nofollow&quot;&gt;seems to be isomorphic&lt;/a&gt; to the Cayley graph of symmetric group given elsewhere, check!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley2.png&quot;&gt;&#xA;&lt;img src=&quot;http://yaroslavvb.com/upload/save/math.se-cayley23d.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-12-21T05:13:00.297" LastActivityDate="2010-12-21T05:13:37.410" Title="How to get generators for a finite point group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15023" PostTypeId="1" AcceptedAnswerId="15036" CreationDate="2010-12-21T00:45:45.700" Score="3" ViewCount="84" Body="&lt;p&gt;There are many examples in which making more &quot;precise&quot; predictions gives worse performance (e.g. Runge's phenomenon). My professor implied that there was a sound basis for choosing &quot;simple&quot; functions over complex ones in the general case, and that it had to do with information theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what he was referring to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example: consider least square's. Obviously we could find a polynomial of very high degree which has zero error, but we prefer a linear equation with higher error. Why should this be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am familiar with some basic notions like entropy, but not much more than that, so simpler explanations would be much preferred. Although I understand that if it's complex, it's complex.)&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-21T06:18:22.123" LastActivityDate="2010-12-21T06:18:22.123" Title="Theoretical basis for overfitting" Tags="&lt;information-theory&gt;&lt;machine-learning&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15024" PostTypeId="1" CreationDate="2010-12-21T01:41:04.207" Score="1" ViewCount="55" Body="&lt;p&gt;I have a series of tasks where when one task finishes the next task runs, until all of the tasks are done. I need to find the probability that everything will be finished at different points in time. How should I approach this? Is there a way to find this in polynomial time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pdfs for how long individual tasks will run have been found experimentally and are not guaranteed to follow any particular type of distribution.&lt;/p&gt;&#xA;" OwnerUserId="4981" LastActivityDate="2010-12-21T03:01:03.877" Title="convolution of multiple probability density funtions" Tags="&lt;probability&gt;&lt;probability-theory&gt;" AnswerCount="1" />
  <row Id="15026" PostTypeId="2" ParentId="15008" CreationDate="2010-12-21T01:54:56.020" Score="4" ViewCount="0" Body="&lt;p&gt;Here's a very simple yet quite good approximation to $\sum\nolimits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}}$, suitable for $|x|&amp;gt;1$ and $n$ sufficiently large (which is the interesting case, in my opinion).&#xA;First write &#xA;$$&#xD;&#xA;S(n,x) := \sum\limits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}} = x^n \sum\limits_{k = 1}^{n - 1} {\frac{{u^k }}{k}},&#xD;&#xA;$$&#xA;where $u=1/x$. If $|x|&amp;gt;1$, then $|u|&amp;lt;1$; hence $\sum\nolimits_{k = 1}^\infty  {\frac{{u^k }}{k}}  =  - \log (1 - u)$. &#xA;So, noting that $ - \log (1 - \frac{1}{x}) = \log (\frac{x}{{x - 1}})$, this gives rise to the approximation&#xA;$$&#xD;&#xA;S(n,x) \approx x^n \log \bigg(\frac{x}{{x - 1}}\bigg),&#xD;&#xA;$$&#xA;provided that $n$ is sufficiently large. It turns out that this simple approximation is quite good. (You can check by yourself.) See the table for an example.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Rounded values of $S(n,x)$ and its approximation, for $n=10$ and several values of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}x&amp;amp;S(n,x)&amp;amp;x^n \log\left(\frac{x}{x-1}\right)\\-5&amp;amp;-1780484.04&amp;amp;-1780483.95\\-3&amp;amp;-16987.42&amp;amp;-16987.34\\2&amp;amp;709.598&amp;amp;709.783\\&#xD;&#xA;4&amp;amp;301656.39&amp;amp;301656.52\\6&amp;amp;1.102428722 \times 10^7&amp;amp;1.102428734 \times 10^7\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-21T12:00:30.550" LastActivityDate="2010-12-21T12:00:30.550" CommentCount="4" />
  <row Id="15027" PostTypeId="2" ParentId="15022" CreationDate="2010-12-21T02:22:13.360" Score="5" ViewCount="0" Body="&lt;p&gt;Observation number one: the group we are talking about is $S_4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-21T05:13:37.410" LastActivityDate="2010-12-21T05:13:37.410" CommentCount="8" />
  <row Id="15028" PostTypeId="2" ParentId="15017" CreationDate="2010-12-21T02:52:00.410" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a start.  Let $f(x)$ be the probability density function of $X$.  Let $p$ be the probability that $X&amp;lt;0,  p=\int_{-\infty }^0f(x)dx$  $Y$ has only two values, $y_1=\frac{1}{p}\int_{-\infty }^0xf(x)dx$ with probability $p$ and $y_2=\frac{1}{1-p}\int_0^{\infty }xf(x)dx$ with probability $1-p$.  The condition that $E[X]=0$ tells us that the integrals in $y_1$ and $y_2$ are the negative of each other, call them $-a$ and $a$.  So $Y$ is $\frac{-a}{p}$ with probability $p$ and $\frac{a}{1-p}$ with probability $1-p$.  $E[Y^2]=\frac{a^2}{p}+\frac{a^2}{1-p}=\frac{a^2}{p(1-p)}$.  $E[Y^4]=\frac{a^4}{p^3}+\frac{a^4}{(1-p)^3}=\frac{a^4(p^3+(1-p)^3)}{p^3(1-p)^3}=\frac{a^4(1-3p+3p^2)}{p^3(1-p)^3}$&lt;br&gt;&#xA;Wolfram Alpha says $E[Y^4]\leq 2(E[Y^2])^2$ if $\frac{5-\sqrt{5}}{10}\leq p \leq \frac{5+\sqrt{5}}{10}$ or about $0.276393 \leq p \leq 0.723607$ .  Presumably the condition $E[X^4] \leq 2(E[X^2])^2$ can show us that, but I need to think more on it.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T02:52:00.410" />
  <row Id="15029" PostTypeId="2" ParentId="11468" CreationDate="2010-12-21T03:00:26.217" Score="0" ViewCount="0" Body="&lt;p&gt;I'm too new here to comment on @Sivaram's answer, but I believe it to be correct and well explained. For further reference see Stanley's &quot;Twelvefold Way&quot; in Combinatorics at either [1] or the Wikipedia page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://mathsci.kaist.ac.kr/~drake/pdf/twelvefold-way.pdf&quot; rel=&quot;nofollow&quot;&gt;http://mathsci.kaist.ac.kr/~drake/pdf/twelvefold-way.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2010-12-21T03:00:26.217" />
  <row Id="15030" PostTypeId="2" ParentId="15024" CreationDate="2010-12-21T03:01:03.877" Score="0" ViewCount="0" Body="&lt;p&gt;I don't know if that is what you are looking for, but if $X_1,\ldots,X_n$ are i.i.d. random variables with mean $\mu$ and (finite) variance $\sigma^2$, then&#xA;$$&#xD;&#xA;{\rm P}\bigg(\sum\limits_{i = 1}^n {X_i }  \le t \bigg) = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^n {X_i }  - n\mu }}{{\sigma \sqrt n }} \le \frac{{t - n\mu }}{{\sigma \sqrt n }}\bigg),&#xD;&#xA;$$&#xD;&#xA;and the random variable $\frac{{\sum\nolimits_{i = 1}^n {X_i }  - n\mu }}{{\sigma \sqrt n }}$ converges to the standard normal distribution as $n \to \infty$. Now, you can estimate the unknown parameters $\mu$ and $\sigma^2$ (assuming the variance is finite), so if $n$ is sufficiently large, you are actually done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details see &lt;a href=&quot;http://en.wikipedia.org/wiki/Central_limit_theorem&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (note the subsection &lt;strong&gt;Density functions&lt;/strong&gt;). &lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-21T03:01:03.877" CommentCount="4" />
  <row Id="15031" PostTypeId="1" CreationDate="2010-12-21T03:06:30.267" Score="2" ViewCount="135" Body="&lt;p&gt;Fix $b&amp;gt;1, y&amp;gt;0$ and prove that there is a unique real $x$ such that $b^x = y$. So for uniqueness, $x_1 &amp;lt; x_2 \Rightarrow b^{x_1} &amp;lt; b^{x_2}$. Then consider the set $S = \{b^t: t \leq x \}$ and work from there? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Chapter 1, Problem 7, &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; by Rudin&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-21T03:56:43.860" LastActivityDate="2010-12-21T16:54:12.400" Title="Logarithm and Uniqueness" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15032" PostTypeId="2" ParentId="14996" CreationDate="2010-12-21T03:24:22.390" Score="1" ViewCount="0" Body="&lt;p&gt;First thing to note that if we re-parametrize $t = ks$, we have, by the chain rule,  $\frac{d}{ds}f = k (\frac{d}{dt}f)\circ t$. Choose $k = T/(2\pi)$, then the change of variable shows that it is sufficient to prove the claim for the period being $2\pi$. That is, if we can show for any $2\pi$ periodic function with mean 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} f^2 ds \leq \int_0^{2\pi} |f&amp;#39;|^2 ds $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we'll be done by a re-scaling argument.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ has mean 0, you can write $f = F&amp;#39;$ for $F$ another $2\pi$ periodic function. So the isoperimetric inequality implies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left|\int_0^{2\pi} f F&amp;#39; ds \right| \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{ (F&amp;#39;)^2 + (f&amp;#39;)^2 } ds \right)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} f^2 ds \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{f^2 + (f&amp;#39;)^2} ds \right)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Holder's inequality on the finite interval $[0,2\pi]$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} \sqrt{f^2 + (f&amp;#39;)^2} ds \leq \sqrt{2\pi} \left( \int f^2 + (f&amp;#39;)^2 ds \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{2\pi} f^2 ds \leq \frac{1}{2}\int_0^{2\pi} f^2 + (f&amp;#39;)^2 ds $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, subtracting $\frac12 \int f^2$ from both sides yields the desired inequality. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, a short remark on why it is necessary to first use the scaling argument. The principle is the following: Wirtinger's inequality, as discussed in the first paragraph above, is scale invariant: changing the scale of parameter $t$ changes the terms $f^2$ and $T^2 (f&amp;#39;)^2$ equally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isoperimetric inequality, however, is not scale invariant in the same way: using the ansatz where $x$ corresponds to $F$ and $y$ corresponds to $F&amp;#39; = f$, you see that a change of parametrization $t \to ks$ will leave $x$ the same while changing $y$ by a multiple factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, this means that the depending on scale, the inequality may not be sharp. In other words, the changing of scale corresponds, morally speaking, to changing the $x$ and $y$ directions in different proportions. So if we start out with a circle, which is a maximizer of the isoperimetric inequality, after this funny reparametrisation which scales $x$ and $y$ differently, we end up with an ellipse, which no longer is a maximizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of scale in the first paragraph allows us to &quot;use a version of the isoperimetric inequality as close to the circle version as possible&quot;. In other words, by isolating the scale you can most efficiently use the isoperimetric inequality, which then allows for a simple proof of Wirtinger's inequality.   &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-21T15:33:35.920" LastActivityDate="2010-12-21T15:33:35.920" CommentCount="5" />
  <row Id="15033" PostTypeId="1" AcceptedAnswerId="15035" CreationDate="2010-12-21T03:29:58.430" Score="1" ViewCount="60" Body="&lt;p&gt;I have one incident edges and multiple outgoing Edges, for which I want to pick an outgoing edge such that the angles between the outgoing edge and the incoming edge is the smallest of all. We know the coordinates for the vertex $V$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle must start from the incoming edge ($e_1$) and ends at another edge ($e_2$, $e_3$, $e_4$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the angle must be form in such a way that the face constructed from $e_1$ to $e_2$ must be in counter-close-wise direction. In other words, the face that you construct when you link all the vertexes $V$ in $e_1$ and $e_i$ together, must be a counter clockwise cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case below, edge $e_2$ must be picked because the $\theta$ is the smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1ducKqeABywKY4A1WS17exNytbvNnDxSAovK0cZnUs3Y&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an algorithm that does this elegantly? The few algorithms I devise are just plain ugly and requires a lot of hacks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: One method that I think of ( and &lt;a href=&quot;http://math.stackexchange.com/questions/15033/find-the-outgoing-edge-with-the-smallest-angles-given-one-incident-edges-and-mul/15035#15035&quot;&gt;proposed below&lt;/a&gt;) is to use the cross product for each edge against the incoming edge $e_1$, and get the minimum $\theta_{i}$. i.e.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{1} \times e_{i} = |e_{1}||e_{i}| \sin\, \theta_{i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem of this is it doesn't really work. Consider the following test cases:&#xA;$$e_1=(1,0)$$&#xA;$$e_2=(-1/\sqrt(2),1/\sqrt(2) )$$&#xA;$$e_3=(1/\sqrt(2),-1/\sqrt(2))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_2 = -\angle45$$&#xA;$$\theta_3=-\angle45$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are equal-- even though we know that $e_2$ should be selected. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-21T07:33:19.250" LastActivityDate="2011-01-02T09:20:04.803" Title="Find the Outgoing Edge with the Smallest Angles, Given one Incident Edges and Multiple Outgoing Edges " Tags="&lt;geometry&gt;&lt;trigonometry&gt;&lt;euclidean-geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15034" PostTypeId="2" ParentId="15031" CreationDate="2010-12-21T04:51:04.693" Score="1" ViewCount="0" Body="&lt;p&gt;Yes, if you prove that $x\mapsto b^x$ is strictly increasing, that will show that there is at most one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your proposal with $S$ does not really work because what you are looking for is $x$; since you do not know ahead of time what $x$ is, you cannot define $S$ in the first place!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, consider the set $S = \{ r\in\mathbb{Q} \mid b^r\leq y\}$ (or $\{r\in\mathbb{R}\mid b^r\leq y\}$). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-21T05:31:24.877" LastActivityDate="2010-12-21T05:31:24.877" />
  <row Id="15035" PostTypeId="2" ParentId="15033" CreationDate="2010-12-21T05:08:32.937" Score="2" ViewCount="0" Body="&lt;p&gt;Assuming that finding the coordinates(of your vectors) is not too difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the definition of dot and or cross products, and then solve for the angles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can for example use $| a \times b | = |a||b| \sin\, \theta$&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="4728" LastEditorDisplayName="" LastEditDate="2010-12-21T07:29:48.597" LastActivityDate="2010-12-21T07:29:48.597" CommentCount="6" />
  <row Id="15036" PostTypeId="2" ParentId="15023" CreationDate="2010-12-21T06:17:28.007" Score="0" ViewCount="0" Body="&lt;p&gt;Look up VC theory (named after Vapnik and Chervonenkis), and especially Structural Risk Minimization (SRM).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T06:17:28.007" />
  <row Id="15037" PostTypeId="2" ParentId="14655" CreationDate="2010-12-21T07:06:05.003" Score="0" ViewCount="0" Body="&lt;p&gt;The language described in (2) is not context free. Take the word $a^nba^nba^nb$ for $n$ large enough, and use the pumping lemma to get a word not in the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (1), consider a grammar in CNF (Chomsky Normal Form) generating that language. Let $n = |\Sigma|$. Consider all $S=(2n)!/2^n$ saturated words containing each character twice. Looking at the tree generating each of these words $w$, there must be a symbol $A_w$ which generates a subword $s_w$ of length between $1/3$ of $2/3$ of the entire word (of length $2n$). Suppose $A_u = A_v$. Then $|s_u| = |s_v|$ and $s_u = s_v$ up to permutation, since otherwise we can replace one occurrence by the other to get a word not in the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times can a given $A_u$ appear? Equivalently, in how many saturated words does $s_u$ appear, up to permutation, somewhere in the word? Any such word can be obtained from $u$ by permuting $s_u$, permuting the rest of the word, any rotating the result, to a total of $2n|s_u|!(2n-|s_u|)! \leq 2n(2n/3)!(4n/3)!$. Therefore the number of different symbols in the language is at least&#xA;$$ \frac{(2n)!/2^n}{2n(2n/3)!(4n/3)!} = O\left(\frac{c^n}{n^{1.5}}\right),$$&#xD;&#xA;where $c = 9/4\sqrt[3]{2} \approx 1.78583$ (asymptotics courtesy of Wolfram Alpha).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, one can transform a grammar $G$ into a CNF grammar of size $O(|G|^2)$. Therefore this language requires exponentially many symbols for any grammar.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T07:06:05.003" CommentCount="2" />
  <row Id="15038" PostTypeId="2" ParentId="15031" CreationDate="2010-12-21T07:13:10.753" Score="2" ViewCount="0" Body="&lt;p&gt;Actually, the proof is quite easy following his outline:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) &lt;strong&gt;Lemma:&lt;/strong&gt; For any positive integer $n$, $b^n -1 \geq n(b-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Suppose this holds for $n$. Then $b^n-1\geq n(b-1)\geq n/b(b-1)$, since $b&amp;gt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $b^{n+1} \geq n(b-1)+b$, and so $b^{n+1}-1 \geq n(b-1)+(b-1)$, therefore $b^{n+1}-1\geq (n+1)(b-1)$. Since $b^n -1 \geq n(b-1)$ holds for $n=1$, by induction it holds for any $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Hence, $b-1 \geq n(b^{1/n}-1)$ if we apply the lemma setting $b&amp;#39; =b^n$ ($b'&gt;1$ since $b&gt;1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $t&amp;gt;1$ and $n&amp;gt;\frac{b-1}{t-1}$, then $b^{1/n} &amp;lt; t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: Since $n&amp;gt;\frac{b-1}{t-1}$, then $n(t-1)&amp;gt;(b-1)\geq n(b^{1/n}-1)$. It follows that $b^{1/n} &amp;lt;t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) If $b^w&amp;lt;y$, then $b^{w+(1/n)}&amp;lt;y$ for sufficiently large $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set $t=y/b^w$, we see that $t&amp;gt;1$ and we can choose a sufficiently large $n$ such that $n &amp;gt;\frac{b-1}{t-1}$. Because of (c) we know that $b^{1/n}&amp;lt;t$. Multiply both sides by $b^w$ to get $b^{w+1/n}&amp;lt;y$, which is what we wanted to prove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) If $b^w&amp;gt;y$, then $b^{w-(1/n)}&amp;lt;y$ for sufficiently large $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set $t=b^w/y$, we see that $t&amp;gt;1$ and we can choose a sufficiently large $n$ such that $n &amp;gt;\frac{b-1}{t-1}$. Because of (c) we know that $b^{1/n}&amp;lt;t$. Then $1&amp;lt;\frac{t}{b^{1/n}}$, so $1&amp;lt;\frac{b^w}{yb^{1/n}}$ and therefore $y&amp;lt;b^{w-(1/n)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f) Let $A = \{w : b^w&amp;lt;y\}$, then if $x=\sup A$, $b^x = y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $b&amp;gt;1, y&amp;gt;0$, choosing a sufficiently small $w$ will get us arbitrarily close to $0$, so $A$ is non-empty. Since $b^w&amp;gt;y$ for a sufficiently large $w$, $A$ is bounded above, and because of $A$ being a subset of $\mathbb{R}$ then $\exists\sup A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix $x = \sup A$. Suppose $b^x &amp;lt; y$, then because of (d), $b^{x+(1/n)}&amp;lt;y$ for some $n$. But then $x+1/n &amp;gt; x$ and $x+1/n \in A$, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $b^x &amp;gt; y$, then because of (e), $b^{x-(1/n)}&amp;gt;n$ for some $n$. But then $x-(1/n)$ is an upper bound of $A$ smaller than $x$, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if $x=\sup A$, then $b^x=y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(g) The uniqueness of $x$ follows from the fact that there can't be two different least upper bounds for the same set $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feedback is appreciated, thanks.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-12-21T16:54:12.400" LastActivityDate="2010-12-21T16:54:12.400" />
  <row Id="15039" PostTypeId="1" CreationDate="2010-12-21T07:41:15.703" Score="-2" ViewCount="587" Body="&lt;p&gt;Dear Professors, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for any &lt;em&gt;&lt;strong&gt;reference articles on the uses of the preposition &quot;of&quot;&lt;/strong&gt;&lt;/em&gt; that are applied most often in mathematics. I once read in the teacher's edition of Pre-algebra: An Accelerated Course by Dolciani that students should be taught, as early as possible, the many uses of the preposition &quot;of&quot; in mathematics, but I have been unable to find anything on this particular question/topic. &lt;em&gt;&lt;strong&gt;Could you please help me find something? (books, articles, etc.)&lt;/strong&gt;&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please note:&#xA; I am &lt;strong&gt;&lt;em&gt;NOT&lt;/em&gt;&lt;/strong&gt; saying that the preposition &quot;of&quot; has a &quot;unique-mathematical-meaning&quot; when used by mathematicians. What I am saying is that of all the general usages that can be found at dictionary.com for &quot;of&quot;, mathematicians &lt;em&gt;CHOOSE&lt;/em&gt; to use a limited subset of them  and then asking &lt;em&gt;WHY&lt;/em&gt; did they choose this specific subset of meanings? This is, in my opinion, a question worth trying to answer because it will help us to more clearly explain things in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The teacher's edition of Pre-algebra: An Accelerated Course by Dolciani that I was referencing is the 2006 edition. I am unsure of the exact page number since I had to borrow it through interlibrary loan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of the uses of this preposition as applied most often in Algebra under the topic heading of &quot;Variables and Equations&quot; include: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) values &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a variable&lt;br&gt;&#xA;2) value &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a numerical expression&lt;br&gt;&#xA;3) sides &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; an equation&lt;br&gt;&#xA;4) domain &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a variable&lt;br&gt;&#xA;5) solution &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; a sentence&lt;br&gt;&#xA;and my personal favourite,&lt;br&gt;&#xA;6) solution set &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; an open sentence: The set &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; all solutions &lt;em&gt;&lt;strong&gt;of&lt;/strong&gt;&lt;/em&gt; the sentence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note these examples have been taken from Chapter 1 sections 1-1 and 1-3 in &quot;Algebra Structure and Method Book 1&quot; by Brown, Dolciani et al. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand, you find it useless that I am trying to explain to others the beauty of mathematics through its choice of vocabulary and why that choice was made as demonstrated by comments about counting occurrences of &quot;of&quot; in mathematical writing, being unduly obsessed, and categorizing what I am trying to do as the subject of a &quot;crank.&quot;  May I respond by a description of mathematics as the art of creating images and establishing relationships, is it not so that the &quot;material&quot;  of this activity are thoughts? And since words convey those thoughts we must &quot;measure&quot; our choice of words carefully for the sake of lucidity. The sheer number of occurrences of &quot;of&quot; in academic writing and in mathematical writing in particular justifies, in my opinion, at least an honest attempt at &quot;looking for any &lt;em&gt;&lt;strong&gt;reference articles on the uses of the preposition &quot;of&quot;&lt;/strong&gt;&lt;/em&gt;...I have been unable to find anything on this particular question/topic. &lt;em&gt;&lt;strong&gt;Could you please help me find something? (books, articles, etc.)&lt;/strong&gt;&lt;/em&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: The primary tag for this thread is a &lt;strong&gt;REFERENCE-REQUEST&lt;/strong&gt;. Reference-requests are for questions that the poster wants to seek for references &lt;strong&gt;(books, articles, etc.)&lt;/strong&gt; in a subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yours respectfully.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="5030" LastEditorDisplayName="" LastEditDate="2010-12-25T09:38:53.360" LastActivityDate="2010-12-25T09:38:53.360" CommunityOwnedDate="2010-12-23T12:10:27.240" ClosedDate="2010-12-24T00:14:28.970" Title="Uses of &quot;of&quot; applied most often in math and why this specific choice was made?" Tags="&lt;reference-request&gt;" AnswerCount="3" CommentCount="12" />
  <row Id="15040" PostTypeId="2" ParentId="15007" CreationDate="2010-12-21T07:45:46.190" Score="0" ViewCount="0" Body="&lt;p&gt;As others have commented and answered, the answer is, roughly, &quot;yes&quot;. I might add only that to see examples of 'other' topologies, ones in which open sets are not what you may be used to, check out (a) the problems in a point-set topology book or (b) &lt;a href=&quot;http://en.wikipedia.org/wiki/Counterexamples_in_Topology&quot; rel=&quot;nofollow&quot;&gt;the book &lt;em&gt;Counterexamples in Topology&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2010-12-21T07:45:46.190" />
  <row Id="15041" PostTypeId="1" CreationDate="2010-12-21T07:59:02.607" Score="-3" ViewCount="200" Body="&lt;p&gt;i need to know which number divides 566?&lt;/p&gt;&#xA;" OwnerUserId="4993" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-21T11:53:53.227" LastActivityDate="2010-12-21T12:34:10.350" ClosedDate="2010-12-21T14:39:18.760" Title="which number divides 566?" Tags="&lt;elementary&gt;&lt;arithmetic&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="15042" PostTypeId="1" CreationDate="2010-12-21T08:17:26.967" Score="3" ViewCount="98" Body="&lt;p&gt;Hi all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a square with some numbers from 1..N in some cells. It's needed to determine if it can be completed to a magic square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;2 _ 6       2 7 6&#xA;_ 5 1  &amp;gt;&amp;gt;&amp;gt;  9 5 1&#xA;4 3 _       4 3 8&#xA;&#xA;7 _ _ &#xA;9 _ _  &amp;gt;&amp;gt;&amp;gt;  NO SOLUTION &#xA;8 _ _&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this problem NP-complete? If yes, how can I prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.:&lt;/strong&gt;&lt;br&gt;&#xA;I know that I should reduce one of the known NP-complete problems to this one to prove its NP-completeness. The main question is which NP-complete probleme to choose and how to buld polynomial reduction algorithm. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4003" LastEditorUserId="4003" LastEditorDisplayName="" LastEditDate="2010-12-21T09:28:02.700" LastActivityDate="2010-12-21T09:28:02.700" Title="Is half-filled magic square problem NP-complete?" Tags="&lt;homework&gt;&lt;np-complete&gt;" CommentCount="11" />
  <row Id="15044" PostTypeId="2" ParentId="15007" CreationDate="2010-12-21T08:52:11.473" Score="5" ViewCount="0" Body="&lt;p&gt;The answer to your specific question is &lt;em&gt;yes&lt;/em&gt;: if you are given a set $X$, then (even in practice!) you may choose any collection $\{U_i\}_{i \in I}$ of subsets of $X$ which (i) contains the empty set and $X$ itself, (ii) is closed under the operation of finite intersections and (iii) is closed under the operation of arbitrary unions, and then this family will be the open sets for a unique topology on $X$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, all of the above is nothing else than the definition of a topology on a set $X$.  But let's be honest: this is quite a profound and general definition, especially at first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can absolutely define &quot;strange&quot; topologies on the real line.  For instance, a famous one is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lower_limit_topology&quot; rel=&quot;nofollow&quot;&gt;lower limit topology&lt;/a&gt; in which the open sets are decreed to be unions of half-open intervals of the form $[a,b)$.  It is easy to see that this gives a strictly larger family of open sets than the usual &quot;Euclidean&quot; topology on $\mathbb{R}$, i.e., every open interval $(c,d)$ is an infinite union of half-open intervals, but no $[a,b)$ is an infinite union of open intervals!  This lower limit topology has all sorts of bizarre properties, many of which are detailed in the wikipedia article linked to above.  In fact it is primarily of interest as a &quot;counterexample&quot;, i.e., an example of a topological space that does not have many of the properties of more familiar spaces -- like metrizable spaces -- do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief discussion of the family of all topologies on a fixed set $X$ can be found &lt;a href=&quot;http://math.uga.edu/~pete/TopSection4.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (although I do not necessarily recommend these notes for a true beginner; they are somewhat more sophisticated than is necessary for a first pass).  In particular, right at the beginning I have left as a (not that easy) exercise that on any infinite set $X$, there are $2^{2^{\# X}}$ different topologies on $X$.  When $X = \mathbb{R}$, this is a really, really large infinite number, larger than one usually encounters outside of set-theoretic considerations.  One is certainly not going to meet most of these topologies personally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though that if you change the topology on &quot;the real numbers&quot;, it loses much of the structure that made it the real numbers in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a first pass through topological spaces, you should definitely keep your eye on the &lt;strong&gt;metric topologies&lt;/strong&gt;: i.e., the topology associated to a metric space $(X,d)$ by decreeing a set to be open if it contains an $\epsilon$-ball around each of its points.  For these types of topologies one can use the spatial and analytic intuition developed in one's previous studies in a useful way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may well ask: why do we speak of topological spaces rather than just metric spaces?  There are (at least) two good reasons for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Even if the spaces that we want to work with are &lt;em&gt;metrizable spaces&lt;/em&gt; -- i.e., the topology is derived from a metric -- often it turns out to be the case that the properties of the space that one wants to study depend only on the topology induced by the metric.  For instance, continuous functions and convergent sequences -- two basic concepts in elementary analysis -- are purely topological in this way.  (Arguably the only important concept of basic analysis which is not purely topological is that of Cauchy sequence and the related notion of completeness.  Even these depend only a structure intermediate between the metric and the induced topology, namely the &lt;strong&gt;uniform structure&lt;/strong&gt; of the space.)  If one is really working with topological notions like convergence, after a certain point it becomes simpler not to carry around the metric structure.  The standard example I like to give of this is that of product spaces: if $(X_1,d_1),\ldots,(X_n,d_n)$ are finitely many metric spaces, one would like to discuss concepts like convergence on the product set $X = X_1 \times \ldots \times X_n$.  Experience shows that we want e.g. a sequence in the product to converge iff each of its components converges.  There is one natural topology on the product which gives this convergence, the &lt;strong&gt;product topology&lt;/strong&gt;.  The product topology on a finite product of metric spaces &lt;em&gt;is&lt;/em&gt; a metric topology, but not in a canonical way: even to put a metric on $\mathbb{R}^N$ there are plenty of choices, e.g. the $L^p$ norm $||(x_1,\ldots,x_n)||_p := (|x_1|^p + \ldots + |x_n|^p)^{\frac{1}{p}}$ for every $1 \leq p &amp;lt; \infty$ and also the infinity norm $||(x_1,\ldots,x_n)||_{\infty} = \max_i |x_i|$.  (Kaplansky in his book &lt;em&gt;Set Theory and Metric Spaces&lt;/em&gt; describes this situation as an &quot;embarrassment of riches&quot;.)  When one wants the product of a countably infinite family of metric spaces, there is again an obvious topology which turns out to be induced by a metric, but now it becomes slightly tricky to write down a single metric.  But the product topology exists -- and is very natural and well-behaved -- for &lt;em&gt;any&lt;/em&gt; family of topological spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Remarkably -- even spookily -- in many parts of mathematics topological spaces show up which are far from being metrizable.  For instance, non-Hausdorff topologies are useful in functional analysis, algebraic geometry and order theory.  A few examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) As above, any family $\{X_i\}_{i \in I}$ of (say nonempty) topological spaces admits a product space $X = \prod_i X_i$.  Even if all of the spaces $X_i$ are metrizable, then as long as uncountably many of them consist of more than a single point, the product topology will not even be first countable, let alone metrizable.  But still any product of Hausdorff spaces is Hausdorff and any product of compact spaces is compact: these are immensely useful facts.  Taking &quot;big&quot; products of spaces -- and in particular of the interval $[0,1]$ -- comes up much more often in topology and related branches of mathematics than one might at first suspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) A space is &lt;strong&gt;Boolean&lt;/strong&gt; if it is compact, Hausdorff and totally disconnected.  From the perspective of classical geometric topology, such spaces are of rather limited interest.  Indeed if such a space has no isolated points and is metrizable, then it must be homeomorphic to the classical Cantor set.  But M. Stone proved that to every Boolean space one can associate a Boolean ring and conversely: there is a categorical (anti-)equivalence between Boolean spaces and Boolean rings.  (More generally, there is an equivalence between locally compact Hausdorff totally disconnected spaces and Boolean rings-without-multiplicative-identity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) A space is &lt;strong&gt;Alexandroff&lt;/strong&gt; if the intersection of any collection of open sets is again open.  The obvious examples are discrete spaces and finite spaces.  Indeed it is not hard to show that a Hausdorff space is Alexandroff iff it is discrete, so in studying interesting Alexandroff spaces we are leaving Hausdorff spaces far behind. Every $T_0$-Alexandroff space determines a partial ordering on the underlying set, and conversely given a partially ordered set one can define an Alexandroff topology.  In this way one gets an equivalence between $T_0$-Alexandroff spaces and partially ordered sets.  (More generally, there is an equivalence between Alexandroff spaces and quasi-ordered sets, i.e., with the axiom of anti-symmetry removed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) There is up to homeomorphism a unique 2-point space in which exactly one of the points is closed.  In general topology this is sometimes called the &lt;strong&gt;Sierpinski space&lt;/strong&gt;.  It is arguably the single most important topological space in arithmetic geometry, since it is the spectrum (in the sense of Zariski) of any discrete valuation ring.  The germ of the idea of &lt;em&gt;specialization&lt;/em&gt; in algebraic geometry is already contained in this two-point non-Hausdorff space!&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-21T09:18:59.600" LastActivityDate="2010-12-21T09:18:59.600" />
  <row Id="15045" PostTypeId="1" AcceptedAnswerId="15056" CreationDate="2010-12-21T09:36:36.690" Score="1" ViewCount="48" Body="&lt;p&gt;I am looking for a generalized normal distribution with mean $\mu$ and variance $\sigma^2$ but with an additional parameter for the kurtosis to add fat-tails. I can't find it - can anybody help?&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-22T04:42:54.460" LastActivityDate="2010-12-22T04:42:54.460" Title="generalized normal distribution with additional kurtosis parameter" Tags="&lt;probability-distributions&gt;&lt;normal-distribution&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15047" PostTypeId="2" ParentId="11650" CreationDate="2010-12-21T10:16:22.013" Score="0" ViewCount="0" Body="&lt;p&gt;About the connection with  Schrodinger equations of quantum mechanics and certain models in QFT: you can have look to the papers: arXiv:hep-th/9812211,  arXiv:hep-th/9906219&lt;/p&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2010-12-21T10:16:22.013" />
  <row Id="15048" PostTypeId="1" CreationDate="2010-12-21T10:44:56.330" Score="4" ViewCount="98" Body="&lt;p&gt;Suppose I have an inscribed simplex which has n+1 vertices, and the diameter of the hypersphere is d. I have a point x inside this simplex, is it true that the distance between x and x's nearest vertex is not larger than d/2 ?&lt;/p&gt;&#xA;" OwnerUserId="4998" LastEditorUserId="4998" LastEditorDisplayName="" LastEditDate="2010-12-21T11:53:35.173" LastActivityDate="2010-12-27T08:55:33.877" Title="inscribed simplex. " Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15050" PostTypeId="2" ParentId="15041" CreationDate="2010-12-21T11:34:39.163" Score="4" ViewCount="0" Body="&lt;p&gt;To find out which number divides a certain number,you need to compute the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_factor&quot; rel=&quot;nofollow&quot;&gt;prime factorization&lt;/a&gt; of that number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to the internet you could try &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;nofollow&quot;&gt;wolfram alpha&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://www.wolframalpha.com/input/?i=factor+566&quot; rel=&quot;nofollow&quot;&gt;$566$&lt;/a&gt;,only $2$ and $283$ divides $566$.Also any number is divisible by $1$ and itself.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-21T11:34:39.163" CommentCount="3" />
  <row Id="15051" PostTypeId="1" AcceptedAnswerId="15054" CreationDate="2010-12-21T11:43:57.410" Score="4" ViewCount="55" Body="&lt;p&gt;Please help me in understanding the difference between this two question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many ways can $4$ men and $4$ women stand in a line if the women and men alternate ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How many ways can $4$ men and $4$ women stand in a line if no two women are together ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We employ a bit of different strategies for solving them and the answer is also different,but I am not understanding why is this differences.Kindly explain.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am in bit of a doubt in another similar problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find the number of ways of arranging $21$ white balls and $3$ black balls in a row so that no two black balls are together.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer should be $21! \times ^{24}P_3$ but this is incorrect.Could somebody explain?&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="3744" LastEditorDisplayName="" LastEditDate="2010-12-21T11:59:10.803" LastActivityDate="2010-12-21T12:07:31.663" Title="Understanding differences " Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="15052" PostTypeId="2" ParentId="15041" CreationDate="2010-12-21T11:53:45.737" Score="2" ViewCount="0" Body="&lt;p&gt;If you are interested what divides other numbers you can use the free software &quot;sage&quot; or &quot;gp/pari&quot;. The &lt;em&gt;function&lt;/em&gt; you are looking for is &lt;strong&gt;divisors&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;? divisors(566)&lt;br&gt;&#xA;  %1 = [1, 2, 283, 566]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Links:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/PARI/GP&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/PARI/GP&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: If the computer fail to answer for other numbers, I would pick &lt;strong&gt;1&lt;/strong&gt; or the number itself: &lt;strong&gt;566&lt;/strong&gt; in your case.&lt;/p&gt;&#xA;" OwnerUserId="4750" LastEditorUserId="4750" LastEditorDisplayName="" LastEditDate="2010-12-21T12:34:10.350" LastActivityDate="2010-12-21T12:34:10.350" />
  <row Id="15053" PostTypeId="1" AcceptedAnswerId="15062" CreationDate="2010-12-21T11:53:48.483" Score="2" ViewCount="120" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are questions given for an assignment and I REALLY need to get a good grade so please could you guys check over them and give some pointers where I've gone wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ed: I hope it's ok to overwrite the original post: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a_n &amp;amp;= \frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}\\&#xA;a_{n+1} &amp;amp;=&#xA;\frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)}.\\&#xA;\left|\frac{a_n}{a_{n+1}}\right| &amp;amp;= \left|\frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)} \times&#xA;\frac{4^{n}(n^{2}+2)}{(2x-8)^{n}}\right|.\\&#xA;\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|&amp;amp;= |2x-8| \lim_{n \to&#xA;\infty}\frac{n^{2}+2}{4(n^{2}+2n+3)}\\&#xA;&amp;amp;= |2x-8| \times\frac{1}{4}.&#xA;\end{align}&#xA;Therefore $L = \frac{1}{4}$ and series converges when&#xA;$|2x-8| \times\frac{1}{4} &amp;lt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radius of convergence is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Endpoints; $|2x-8| = 4 \rightarrow x = 6$ and $x=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing endpoints. &#xA;When $x = 6$:&#xA;$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} = \sum_0^\infty&#xD;&#xA;\frac{1}{n^{2}+2}.$$&#xA;convergent by comparison test with $\sum\frac{1}{n^{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When x = 2:&#xA;$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} =&#xD;&#xA;\sum_0^\infty\frac{(-1)^n}{n^{2}+2},$$&#xA;convergent by alternating sequence test and is absolutely convergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the series $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$&#xA;is convergent for $x\in [2,6]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a_n &amp;amp;= \frac{(-1)^n}{n4^{2n}},\\&#xA;a_{n+1} &amp;amp;= \frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}}.\\&#xA;\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= \left|\frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}} \times&#xA;\frac{n4^{2n}}{(-1)^n}\right|\\&#xA;\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= |x+3|\lim_{n \to&#xA;\infty}\frac{n}{4^2(n+1)}\\&#xA; &amp;amp;= |x+3|\times\frac{1}{16}.&#xA;\end{align} &#xA;Series converges when $|x+3|\times\frac{1}{16} \lt 1$.&#xA;Radius of convergence \is $16$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Endpoints: $|x+3| = 16 \rightarrow x = 13$ or $-19$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing endpoints:&#xA;When $x = 13$, &#xA;$$\sum_1^\infty\frac{(-1)^n(16)^n}{n4^{2n}} =&#xD;&#xA;\sum_1^\infty\frac{(-1)^n)}{n}.$$&#xA;This is an alternating harmonic series and therefore convergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When x = -19, &#xA;$$\sum_1^\infty\frac{(-1)^n(-16)^n}{n4^{2n}}{} =&#xD;&#xA;\sum_1^\infty\frac{(-1)^{2n}}{n}.$$&#xA;Harmonic series and therefore divergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the series $$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$&#xA;is convergent for $x \in(-19, 13]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) c)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$$&#xA;\begin{align}&#xA;a_n &amp;amp;= \frac{(1+3^{n-1})x^n}{(n+1)!}\\&#xA;a_n+1 &amp;amp;=&#xA;\frac{(1+3^{(n+1)-1})x^{n+1}}{((n+1)+1)!}.\\&#xA;\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= \left|\frac{(1+3^n)x^{n+1}}{(n+2)!} \times&#xA;\frac{(n+1)!}{(1+3^{n-1})x^n}\right|\\&#xA;\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &amp;amp;= &#xA;|x|\lim_{n \to \infty}&#xA;\frac{1+3^n}{1+3^{n-1}} \times \lim_{n \to \infty} \frac{1}{n+2}\\&#xA;&amp;amp;= |x|\lim_{n \to \infty} \frac{3^{-n}+ 1}{3^{-n}+3^{-1}} \times 0\\&#xA;&amp;amp;=&#xA;|x|\lim_{n \to \infty} \frac{0 + 1}{0 +3^{-1}} \times 0\\&#xA;&amp;amp;= |x|\times3\times0 = 0.&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $L = 0$ and the series converges absolutely for all $x$. &#xA;The radius of convergence is $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$ converges absolutely for $x\in(-\infty,\infty)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gregg&lt;/p&gt;&#xA;" OwnerUserId="4362" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-22T21:45:22.900" LastActivityDate="2010-12-22T21:45:22.900" Title="radius of convergence of a power series" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15054" PostTypeId="2" ParentId="15051" CreationDate="2010-12-21T11:56:15.890" Score="8" ViewCount="0" Body="&lt;p&gt;In the first point, the problems are different because the second problem would allow e.g. WMW*MM*WMW while the first would not. (The first is essentially saying &quot;no two women are together &lt;em&gt;and no two men are together&lt;/em&gt;&quot;, while the second only says &quot;no two women are together&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second point, I think you're making two mistakes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;By multiplying by $21!$, and by using $P$ instead of $C$, you're assuming you can tell the difference beween the white balls, and can tell the difference between the black balls. (i.e. B1 B2 W1 W2... is different from B2 B1 W1 W2...) Usually when dealing with balls, they're indistinguishable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer $^{24}C_3$ (i.e. choosing any 3 balls out of the 24 to be black) would be incorrect because it doesn't exclude answers such as BWBBW... where two black balls appear next to each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5001" LastEditorUserId="5001" LastEditorDisplayName="" LastEditDate="2010-12-21T12:07:31.663" LastActivityDate="2010-12-21T12:07:31.663" CommentCount="4" />
  <row Id="15055" PostTypeId="1" AcceptedAnswerId="15269" CreationDate="2010-12-21T12:17:16.163" Score="17" ViewCount="799" Body="&lt;p&gt;In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just dipped into a book, &lt;em&gt;The Drunkard's Walk - How Randomness Rules Our Lives&lt;/em&gt;, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2010-12-30T01:54:48.190" Title="In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?" Tags="&lt;probability&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="3" />
  <row Id="15056" PostTypeId="2" ParentId="15045" CreationDate="2010-12-21T12:17:59.453" Score="5" ViewCount="0" Body="&lt;p&gt;I think &lt;a href=&quot;http://en.wikipedia.org/wiki/Pearson_distribution&quot; rel=&quot;nofollow&quot;&gt;Pearson type VII distribution&lt;/a&gt; might be what your are looking for. You will need to reparametrize it, but wikipedia page has the necessary formulas.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-21T12:17:59.453" CommentCount="2" />
  <row Id="15057" PostTypeId="2" ParentId="15048" CreationDate="2010-12-21T12:24:33.703" Score="6" ViewCount="0" Body="&lt;p&gt;Center the hypersphere at the origin, and let $h$ be the (non-negative!) distance between the point $P$ and the center. We can position our hypersphere so that $P$ lies on the $x$ axis with $x$-coordinate $-h$. I'll write $r$ for $d/2$, the radius of the hypersphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case $n=3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the distance from a vertex, $V(x,y,z)$, to $P$ is strictly greater than $r$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{dist}(V,P)^2=(x+h)^2+y^2+z^2&amp;gt;r^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $V$ is on the (hyper)sphere, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 + z^2 = r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;2hx+h^2 &amp;amp;&amp;gt; 0 \\&#xD;&#xA;\Rightarrow h(x+\frac{h}{2}) &amp;amp;&amp;gt; 0&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see, then, that our non-negative $h$ must be strictly positive, and that $x &amp;gt; -\frac{h}{2} &amp;gt; -h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if ALL vertices are further than $r$ away from $P$, then the above shows that $P$ is separated from those vertices --and the simplex they determine-- by the (hyper)plane $x=-\frac{h}{2}$. The point cannot lie inside the simplex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, for a point within the simplex, the distance to &lt;em&gt;at least one&lt;/em&gt; vertex (in particular, the closest) must be no greater than $r$. In the case where $P$ lies at the center of the hypersphere, the distance to any vertex is exactly $r$, so this upper bound on the minimum distance cannot be improved.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2010-12-22T00:38:37.460" LastActivityDate="2010-12-22T00:38:37.460" CommentCount="1" />
  <row Id="15058" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T12:42:27.283" Score="18" ViewCount="0" Body="&lt;p&gt;In a family with 2 children there are four possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the first child is a boy and the second child is a boy (bb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the first child is a boy and the second child is a girl (bg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) the first child is a girl and the second child is a boy (gb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) the first child is a girl and the second child is a girl (gg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg. Out of those three possibilities the only one with two girls is gg. Hence the probability is $\frac{1}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="1487" LastActivityDate="2010-12-21T12:42:27.283" CommentCount="5" />
  <row Id="15059" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T12:47:59.120" Score="7" ViewCount="0" Body="&lt;p&gt;Here's some idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you sample $4n$ families with two children, where $n$ is a very large integer. Then, with a very high probability, about $n$ of them have two boys, about $n$ have two girls, and about $2n$ have one girl and one boy.&#xA;Since you ignore the families with two boys, the desired probability is given by $n/(3n) = 1/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The above solution corresponds to the following situation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you check if one of the kids is a girl. If no, you ignore this family. If yes, you check if both kids are girls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Among all the OP's examples, only one is relevant here. Namely, &quot;I visit this family. I know they have 2 kids. One of them, a girl, comes into the room. The probability that the 2nd kid is also a girl is 1/2, no?&quot;. Well, in this situation the probability is indeed $1/2$, assuming the following interpretation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you wait until one of the kids comes into the room. If it is a boy, you ignore this family; otherwise you check if both kids are girls. It is readily understood that this leads to a probability of $1/2$. &lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-21T13:52:48.960" LastActivityDate="2010-12-21T13:52:48.960" CommentCount="16" />
  <row Id="15060" PostTypeId="1" CreationDate="2010-12-21T12:59:42.207" Score="3" ViewCount="83" Body="&lt;p&gt;Hello,&#xA;we all know that 31,331,3331,33331,333331,3333331,33333331 all are primes. Here we prepend the digit 3 to 31, to get a list of 7 primes.This gives me the following thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D = \{\text{all possible nonnull finite digit strings}\}$,  $D&amp;#39; = \{\text{all things in D that do not start with &amp;quot;0&amp;quot;}\}$.  Define a function $m: D&amp;#39; \times D -&amp;gt; N \cup {\infty}$  by: $m(A,B)= |${all prime members of the list AB, AAB, AAAB, &#xA;...up until but not including the first composite member}| (the size of the set).Then:&#xA;Does m ever take the value $\infty$ ?  If not, is it an unbounded function?&lt;/p&gt;&#xA;" OwnerUserId="2394" LastActivityDate="2010-12-21T12:59:42.207" Title="Prepending strings and primes?" Tags="&lt;elementary-number-theory&gt;&lt;number&gt;" CommentCount="7" />
  <row Id="15061" PostTypeId="1" AcceptedAnswerId="15121" CreationDate="2010-12-21T13:09:11.413" Score="8" ViewCount="167" Body="&lt;p&gt;I could use some help with proving the following irreducibility criterion. (It came up in class and got me interested.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;em&gt;p&lt;/em&gt; be a prime. For an integer $n = p^k n_0$, where &lt;em&gt;p&lt;/em&gt; doesn't divide $n_0$, set: $e_p(n) = k$. Let $f(x) = a_n x^n + \cdots + a_1 x + a_0$ be a polynomial with integer coefficients. If:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$e_p(a_n) = 0$, &lt;/li&gt;&#xA;&lt;li&gt;$e_p(a_i) \geq n - i$, where $i = 1, 2, \ldots, n-1$,&lt;/li&gt;&#xA;&lt;li&gt;$e_p(a_0) = n - 1$,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;then &lt;em&gt;f&lt;/em&gt; is irreducible over the rationals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing mod &lt;em&gt;p&lt;/em&gt; and mimicking the proof of Eisenstein's criterion doesn't cut it (I think). I also tried playing with reduction mod $p^k$, but got stuck since $Z_{p^k}[X]$ is not a UFD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does this criterion has a name?&lt;/p&gt;&#xA;" OwnerUserId="1875" LastActivityDate="2010-12-21T22:56:40.207" Title="An Eisenstein-like irreducibility criterion" Tags="&lt;polynomials&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15062" PostTypeId="2" ParentId="15053" CreationDate="2010-12-21T13:09:19.167" Score="3" ViewCount="0" Body="&lt;p&gt;It seems mostly OK (at a quick glance), but there are few minor issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't write $|a_n/a_{n+1}|=(\dots)=\lim (\dots)$, because the elements of the sequence are not equal to the limit of the sequence. Either you write &quot;lim&quot; in &lt;em&gt;each&lt;/em&gt; step,&#xA;like $\lim (\dots) =\lim (\dots) = A$, or else use the arrow notation $(\dots) = (\dots) \to A$. (But don't mix the two notations as many people do; that is, don't write &quot;$\lim(\dots) \to A$&quot;!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $x\in [2,6]$, not $x=(2,6)$. Also: $x \in (-19,13]$ (the smallest number first).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last problem, you've made a mistake: $(n+1)!/(n+2)! = 1/(n+2)$, not $(n+1)/(n+2)$.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-21T13:09:19.167" CommentCount="1" />
  <row Id="15063" PostTypeId="2" ParentId="15053" CreationDate="2010-12-21T13:11:50.100" Score="2" ViewCount="0" Body="&lt;p&gt;In the first problem, you have the wrong radius of convergence. You forgot that you have a factor of $2$ in the term $|2x-8|$. Second problem is done correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the last problem, I would suggest doing the arithmetic correctly. You have a mistake in the computation of the limits (as well as, though it is moot, a mistake in the computation of the series when $x=\frac{1}{3}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to mention the issues Hans Lundmark raised, about confusing the terms of the sequence with the limits and the value of the variable with the interval in which it may lie for convergence.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-21T13:11:50.100" CommentCount="3" />
  <row Id="15064" PostTypeId="1" CreationDate="2010-12-21T14:00:47.500" Score="2" ViewCount="120" Body="&lt;p&gt;Hello! How to prove that the group of order 28 which have normal subgroup of order 4 is Abelian group?&lt;/p&gt;&#xA;" OwnerUserId="5004" LastEditorUserId="5004" LastEditorDisplayName="" LastEditDate="2010-12-21T15:05:25.610" LastActivityDate="2010-12-21T15:37:55.723" Title="Proof - Abelian group?" Tags="&lt;homework&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15065" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T14:03:29.243" Score="3" ViewCount="0" Body="&lt;p&gt;It really depends on what knowledge you have to restrict the probability space.  (The original question is a little ambiguous, but I agree that 1/3 is the best answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The marble example and the example of the firstborn daughter are the same situation-- the second trial is independent, and the odds are 1/2.  In these cases you know the first trial is independent, and you can ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question intends you to imagine a randomly chosen 2 child family from among all families with at least 1 daughter.  For example, suppose there is a social science study on 2 child families with at least 1 daughter-- in this situation, about 1/3 of the families will be daughter-daughter, 1/3 will be daughter-son, and 1/3 will be son-daughter.  You have to consider the full probability space of two trials (d-d,d-s,s-d,s-s) and eliminate the s-s possibility.  This leaves a 1/3 chance of 2 daughters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other variations on this question, you have to decide the appropriate probability space before answering.  (One common, confusing variation is to suppose the family has two children, one of which is a son born on a Tuesday-- what are the odds of two sons?)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example where the first child in the room is a daughter, if we suppose that the order the children enter has nothing to do with sex, then the probability the other child is a daughter is 1/2.  However, maybe the family encourages their daughters to play inside and their sons outside-- then you could assume the other child is a son, since he hasn't come in yet.  Maybe the culture has a tradition that the oldest daughter in the household (if there is a daughter) offers guests drinks-- then you know the family has at least one daughter, and the odds that they have another daughter is 1/3.  Once you've chosen your assumptions, you can define the probability space and compute the answer.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-12-21T14:03:29.243" CommentCount="6" />
  <row Id="15066" PostTypeId="2" ParentId="15013" CreationDate="2010-12-21T14:07:26.210" Score="1" ViewCount="0" Body="&lt;p&gt;Using your approach, we can write the proof as follows:&#xA;Choose $k$ such that $2^k+1 \geq n$.  Then $$\sum_{i=1}^n \frac{2}{i(i+1)} \leq \sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}$$  Using the summation formula lets us talk about how many terms are involved and gives a name for the dummy variable.  $$\sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}=1+\frac{1}{2}\sum_{i=1}^{2^{k-1}+1} \frac{2}{i(i+1)}$$ Here is where we used your relation cutting the number of terms in half.  I chose the upper limit of the sum so we can continue the process.  Define $$f(2^k+1) = \sum_{i=1}^{2^k+1} \frac{2}{i(i+1)}$$  We have shown $f(2^k+1)=1+\frac{1}{2}f(2^{k-1}+1)$.  We can continue the recursion to find $$f(2^k+1)=\sum_{i=0}^{k-2} 2^{-i} + \frac{1}{2^{k-1}}f(3)$$  As $f(3)=\frac{3}{2}$, we have $$f(2^k+1)=\sum_{i=0}^{k} 2^{-i} =2-\frac{1}{2^k}\lt 2$$ &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-21T15:10:47.817" LastActivityDate="2010-12-21T15:10:47.817" CommentCount="2" />
  <row Id="15067" PostTypeId="2" ParentId="15064" CreationDate="2010-12-21T14:23:05.150" Score="5" ViewCount="0" Body="&lt;p&gt;Let $G$ be the group of order $28$, $N$ the normal subgroup of order $4$, and let $K = G/N$ be the quotient group, which has order $7$ and is therefore cyclic.  By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schur%E2%80%93Zassenhaus_theorem&quot; rel=&quot;nofollow&quot;&gt;Schur-Zassenhaus Theorem&lt;/a&gt;, $G$ is a semidirect product of $N$ and $K$.  It will be abelian iff the semidirect product is actually direct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to give a nontrivial semidirect product here, it is necessary and sufficient to find a nontrivial homomorphism from &#xA;$K$ into the automorphism group $\operatorname{Aut}(N)$ of $N$, or equivalently an element of order $7$ in $\operatorname{Aut}(N)$.  But $N$ is either cyclic of order $4$ -- in which case $\operatorname{Aut}(N)$ has order $2$ -- or is the Klein group $C_2 \times C_2$, in which case $\operatorname{Aut}(N)$ has order $6$.  Either way there is no element of order $7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, consider what Sylow theory has to say about the Sylow $7$-subgroups. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: of course the second method, which is only hinted at, is probably what is intended.  But the two methods interact in an interesting way with regard to the following more general problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $G$ is a finite group of order $np$ with $p$ &#xA;  a prime number which does not divide $n$, and that there is an abelian normal subgroup $N$ of order $n$.  Under what conditions on $N$ and $p$ is $G$ necessarily abelian?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you use Sylow theory, you get that it is enough for $p$ to be sufficiently large compared to $n$.  (One can be explicit; I don't want to be yet to give the OP some time to think about his particular problem.)  Now comparing to the approach of showing that any semidirect product of $N$ with the cyclic group of order $p$ is necessarily direct one gets information about the largest prime $p$ which divides the order of &#xA;the automorphism group $\operatorname{Aut}(N)$.  It could be fun to explore this further.  (More fun than proving something about groups of order $28$, anyway.  This type of overly specific question tends to bore me.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-21T15:37:55.723" LastActivityDate="2010-12-21T15:37:55.723" CommentCount="1" />
  <row Id="15068" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T14:36:32.757" Score="1" ViewCount="0" Body="&lt;p&gt;One possible solution for this problem might be though using conditional probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B_i$ and $G_i$ represent $i$th boy and $i$th girl respectively. Here the sample space could be expressed as $S =$ {$(B_1,B_2),(B_1,G_2),(G_1,B_2),(G_1,G_2)$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $A$ = An event of having both girl chil.&#xA;$B$ =   An event of having at-least one girl child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = ${$(G_1,G_2)$} and $B =${$(B_1,G_2),(G_1,B_2),(G_1,G_2)$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you are asking is $$P(A / B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-21T14:36:32.757" />
  <row Id="15069" PostTypeId="2" ParentId="8140" CreationDate="2010-12-21T14:51:12.283" Score="3" ViewCount="0" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/12488/for-a-planar-graph-find-the-algorithm-that-constructs-a-cycle-basis-with-each-e&quot;&gt; your previous question&lt;/a&gt; better use was made of graph theory terminology:  you refer to the graph as planar and to a cycle basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you understand those terms, perhaps I can point out how Mariano's answer builds on the earlier problem.  In fact here you asked, &quot;Why is it that the algorithm to find the face in the case without coordinate is different from the one with coordinates?&quot;  Although this is probably meant rhetorically to explain your assumption &quot;All the coordinates for the [vertices] are known,&quot; it's important to point out that &quot;faces&quot; you've asked to find will depend on where the edges are as well as where the vertices are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's illustrate with the complete graph on four vertices $K_4$.  Define planar coordinates for the vertices to be corners of the unit square {(0,0),(1,0),(1,1),(0,1)}.  There will be four edges, but we cannot take all of them to be straight line segments in the plane, since doing so produces an intersection (something a planar embedding of graph does not allow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact depending on how you connect the vertices, you will get different faces.  Draw two squares (with four edges each), and connect one pair of &quot;diagonal&quot; vertices inside one square and the other pair inside the other square.  Now complete both graphs with an outside edge that connects the remaining diagonal pairs, looping over the square in one case and under the square in the other.  Two (or more) cases with different arrangments of &quot;faces&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that knowing a planar embedding well enough to make the &quot;faces&quot; cycle basis you're looking for unique requires you to &quot;know the coordinates&quot; of the edges as well as the vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mariano's answer focuses on the edges.  This has the nice effect of allowing us to count each edge as belonging to exactly two cycles or faces, once for each direction that the edge is travelled.  [No such simple statement will hold for the number of faces that a vertex appears in.]  However to make that &quot;two cycles per edge&quot; work, we must include the &quot;unbounded face&quot; or the cycle that runs all around the outside of the planar graph.  In the simplest case, a triangle say, we have a cycle that goes clockwise around it (following the direction you've used in your diagrams at top).  To get each edge appearing twice, once in each direction, we need to get a counterclockwise cycle.  You can think of one being the &quot;finite&quot; face bounded by the triangle, and the other being the face corresponding to the unbounded region outside the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you ask Mariano to tell you how to &quot;sort the edge[s] cyclically,&quot; I guess there are two aspects of this.  One is to understand what he means by that, and the other is to connect this task with how you &quot;know the coordinates&quot; of the graph including its edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Mariano means is just to know, when we arrive at a vertex along one &quot;inbound edge&quot;, what will be the right &quot;outbound edge&quot; to take next?  Let's suppose you want to construct all the &quot;finite&quot; faces going in the &lt;strong&gt;clockwise&lt;/strong&gt; direction, as in your diagrams.  Then you would need to sort the &lt;strong&gt;edges&lt;/strong&gt; around each vertex into a cyclically linked &quot;list&quot; that goes around the vertex in the &lt;strong&gt;counterclockwise&lt;/strong&gt; direction.  This &quot;list&quot; wraps around; it has no specific beginning or ending edge.  It also does not distinguish the directions of travel along the edge.  It just tells us, for all the edges at a given vertex, which follows which going counterclockwise around the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can elaborate Mariano's solution to your question.  For each edge and &lt;em&gt;each direction of travel&lt;/em&gt; along that edge, build a cycle by this algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Algorithm FollowYourNose&lt;/strong&gt;&lt;br&gt;&#xA;(1) Follow the given edge in the given direction to its end vertex.&lt;br&gt;&#xA;(2) Consult the cyclically linked list of edges at that vertex to determine the next edge.&lt;br&gt;&#xA;(3) Repeat steps (1) and (2) until you return to the original edge.&lt;br&gt;&#xA;(4) Stop.  You have built a cycle that uses the given edge and given direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you've done this for each edge and each direction, you will have a list of cycles that contains duplicates.  If a cycle contains $k$ edges, you will have built that same cycle $k$ times.  Discard the duplicates, leaving just one copy of each cycle.  [If it's important for performance reasons, we could keep track of which edges have been travelled in which directions and preemptively avoid creating the duplicates.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will then have one copy left of the cycle that corresponds to the unbounded face, the outside of the planar graph.  You want to &lt;strong&gt;remove&lt;/strong&gt; that also.  Where all the cycles that correspond to the bounded faces are going in the clockwise direction, the cycle that corresponds to the unbounded face goes in the counterclockwise direction.  In fact it is algebraically &lt;em&gt;minus&lt;/em&gt; the sum of all the clockwise/bounded face cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; A cyclically linked list (or some would write, cyclic linked list) is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Linked_list&quot; rel=&quot;nofollow&quot;&gt; a linked list&lt;/a&gt;.  It means that the linked list comes around in cycle or loop.  No confusion with the &quot;cycles&quot; of the graph is intended here.  We are simply asking for some concrete way to systematically determine how to continue a path in the graph with each vertex we come to.  An informal description would be to say &quot;bear right at each intersection&quot;.  Doing so will of course take us around the block in terms of ordinary street traffic, and it has a similar effect in navigating a planar graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Mariano proposed was to answer the question of &quot;which way to go&quot; by sorting the edges that are incident to each vertex in an order corresponding to the angles at which they reach the vertex.  Think of standing in a intersection (vertex) of two or more streets (edges).  List them (the edges or streets) in the order they appear as you turn counterclockwise through one revolution, and use that list/ordering in the algorithm outlined by Mariano, so that when you enter an intersection (vertex) on one street (edge), you leave along the next one on the list (bear right!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should also be a bit careful about the kind of planar graph that the algorithm applies to.  If you are willing to accept certain degeneracies, then planar graph is good enough.  But consider the case of a single edge connecting two vertices.  What is the face?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to avoid such degeneracies, then assume a simple planar graph that is 2-connected.  What 2-connected means is that the graph is connected and remains connected even after any one edge is removed.  &quot;Simple&quot; means that there are no &quot;self-edges&quot; (edge that connects a vertex to itself) nor multiple edges connecting a single pair of vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding Update 3:  It is not suggested that you sort edges &quot;according to the angles made between the incoming edge and the outgoing edge.&quot;  You are asked to make for each vertex a list of the edges ordered going counterclockwise around that vertex (a circular list if you will), without any regard to incoming or outgoing.  In your butterfly graph there is only one vertex with more than two edges, so only the list for that vertex requires care in sorting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Worked Examples Involving Butterfly Graph&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider first a planar embedding of the butterfly graph similar to what was once part of the Question:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;   v2    v3    +e1: (x,v1)  -e1: (v1,x)  &#xA;    |\   /|    +e2: (x,v2)  -e2: (v2,x)  &#xA;    | \ / |    +e3: (x,v3)  -e3: (v3,x)  &#xA;    |  X  |    +e4: (x,v4)  -e4: (v4,x)  &#xA;    | / \ |    +e5: (v1,v2) -e5: (v2,v1)  &#xA;    |/   \|    +e6: (v3,v4) -e6: (v4,v3)  &#xA;   v1    v4  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've used plus and minus signs to (arbitrarily) assign directions for each of the six edges.  Only the center vertex &lt;strong&gt;x&lt;/strong&gt; has more than two incident edges, so I'll be explicit about ordering the edges (without regard to incoming or outgoing direction) in counterclockwise order around &lt;strong&gt;x&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;...e4,e3,e2,e1,e4,...&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be clear that this is a &lt;em&gt;circular&lt;/em&gt; list, that it repeats, looping through the same order of edges over and over again, which in data structures we would represent as a cyclically linked list.  Nothing more than the standard apparatus for linked lists is required to do this.  One simply makes the final node of the linked list point back to the first node of the linked list, and voila, a cyclically linked list.  But let's not confuse the details of the implementation with the details of the algorithm.  All that the algorithm requires is how to find the next edge at a vertex which is counterclockwise from a given edge at that vertex.  A cyclically linked list does that function, in that we would search the list linearly for the given edge at the given vertex, then take the next edge on the list.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are six edges, there are twelve directed edges, and thus we will apply the algorithm &lt;strong&gt;FollowYourNose&lt;/strong&gt; twelve times (once for each directed edge).  Each time we will get a cycle (a loop on edges).  Cycles may visit a vertex more than once but will never use an edge more than once (assuming the planar graph is simple and 2-connected).  However the twelve results will contain duplicates.  Once the duplicates are removed, there are just three unique cycles left:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e1,+e5,-e2]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e3,+e6,-e4]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[-e1,+e4,-e6,-e3,+e2,-e5]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears from the diagrams provided in the Question, that &quot;faces&quot; are meant to be the bounded components of the plane after the edges of the embedded graph are removed (thereby disconnecting the plane into one unbounded and finitely many bounded components).  So we should remove from the list of three cycles above that one which corresponds to the boundary of the unbounded component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distinction can be made algorithmically on the clockwise or counterclockwise orientation of the cycles.  The cycles that give &quot;faces&quot; are going clockwise around their bounded components.  The cycles which is a boundary of the unbounded component goes counterclockwise.  In this case it is the third cycle that should be discarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate how the distinction is made on the basis of clockwise orientation, and not on some other criterion such as which cycle has greatest length, let's look at another embedding of the butterfly graph:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;           _ x _              +e1: (x,v1)  -e1: (v1,x)  &#xA;         _/ / \ \_            +e2: (x,v2)  -e2: (v2,x)  &#xA;       _/  /   \  \_          +e3: (x,v3)  -e3: (v3,x)  &#xA;     _/   v2___v1   \_        +e4: (x,v4)  -e4: (v4,x)  &#xA;    /                 \       +e5: (v1,v2) -e5: (v2,v1)  &#xA;   v4_________________v3      +e6: (v3,v4) -e6: (v4,v3)  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have taken care to make the labelling of the edges with respect to the vertices the same here, as well as the orientations of the two 3-cycles in the plane the same.  The algorithm will now produce a different result based on the change in how the edges incident to vertex &lt;strong&gt;x&lt;/strong&gt; are ordered going counterclockwise around that vertex:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;...e4,e2,e1,e3,e4,...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again we will apply algorithm &lt;strong&gt;FollowYourNose&lt;/strong&gt; to each of the twelve directed edges and obtain twelve cycles, including duplicates.  After eliminating duplicates the following three cycles remain:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e1,+e5,-e2]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e3,+e6,-e4,+e2,-e5,-e1]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[+e4,-e6,-e3]&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is left to the reader to verify which of these has counterclockwise orientation and that it is not the longest of the three cycles.  The one with the counterclockwise orientation is the &quot;outermost&quot; and thus the boundary of the unbounded component of the plane\graph.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2010-12-24T18:49:30.593" LastActivityDate="2010-12-24T18:49:30.593" CommentCount="11" />
  <row Id="15070" PostTypeId="1" AcceptedAnswerId="15071" CreationDate="2010-12-21T15:18:51.957" Score="2" ViewCount="81" Body="&lt;p&gt;Given an alphabet of $k$ symbols, $\lbrace s_1,s_2, \ldots,s_k \rbrace$, how many words of length $n$, $w(n)$, can be generated knowing that, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Only two of the $k$ available symbols participate;&lt;/li&gt;&#xA;&lt;li&gt;The symbol $s_i$ must appear exactly $n_i$ times and the symbol $s_j$ must appear exactly $n_j$ times and $n_i+n_j=n$ (the length of the word);&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To ilustrate the problem suppose that the alphabet is $\lbrace A, B, C\rbrace$ and that I want words of length $4$ with two $A$ and two $B$. With the stated restrictions all of the possible words are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lbrace AABB, ABAB,ABBA, BBAA, BAAB, BABA \rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I need to know what this $w(n)=w(n_i,n_j)$ is. Some references on aproaches on how to solve this kind of problems and related algorithms to generate such words would be apreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2010-12-22T08:42:47.563" LastActivityDate="2010-12-22T08:42:47.563" Title="How many words of length $n$ can be generated from an alphabet of $k$ symbols with restrictions on the number of symbols. " Tags="&lt;combinatorics&gt;" AnswerCount="1" />
  <row Id="15071" PostTypeId="2" ParentId="15070" CreationDate="2010-12-21T15:24:47.177" Score="3" ViewCount="0" Body="&lt;p&gt;Hint:  For this problem, you first have to select two of the symbols from your list.  How many ways can that be done?  Should the order matter? Second, you have to select which $n_1$ positions in the word can take the first of the two symbols.  How many ways can that be done?  As the two selections are independent, you then multiply the two numbers.  You also have to worry about factors of 2-is there some way that interchanging the selected letters in the first choice and interchanging the selected positions in the second can get you to the same word in different ways?  This is particularly a problem when $n_1$=$n_2$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-21T22:55:51.527" LastActivityDate="2010-12-21T22:55:51.527" CommentCount="0" />
  <row Id="15072" PostTypeId="2" ParentId="942" CreationDate="2010-12-21T15:38:30.933" Score="0" ViewCount="0" Body="&lt;p&gt;Sorry for posting again on this old thread but I'd like to know what's wrong, if something, with the following, because I think it should works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have an open affine covering $X = \bigcup_{i} U_{i}$ with $U_{i} = \mbox{Spec}(A_{i})$. Let $\xi_{1}$ be a closed point in $U_{1}$ (just take a maximal ideal). If $\xi_{1}$ is still closed in $X$ we are done. Otherwise, take $\xi_{2} \neq \xi_{1}$ with $\xi_{2} \in \overline{\left\{\xi_{1}\right\}}$. Hence $\xi_{2} \in U_{i}$ for some $i \neq 1$, say $i = 2$. Then repeat the reasoning above again. The process must stop since there are only finitely many $U_{i}$ and so $X$ has a closed point.&lt;/p&gt;&#xA;" OwnerUserId="5006" LastActivityDate="2010-12-21T15:38:30.933" CommentCount="1" />
  <row Id="15073" PostTypeId="1" AcceptedAnswerId="15099" CreationDate="2010-12-21T16:52:48.843" Score="4" ViewCount="52" Body="&lt;p&gt;Consider the case where we have a bag of 'N' unique marbles, and we randomly select all 'S' possible sets of size 'm' from the bag (with replacement) under the following constraints:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) - Any set of 'm' marbles must have all unique elements, i.e. there must be no duplicate copies of a marble in any given set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) - The maximum intersection between any two sets, i.e. the number of unique marbles they have in common, can be of &lt;em&gt;at most&lt;/em&gt; size 'k'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided (1) &amp;amp; (2), what is 'S'?  Equivalently, what is the number of possible sets of size 'm' obeying the aforementioned constraints?&lt;/p&gt;&#xA;" OwnerUserId="5009" LastActivityDate="2010-12-21T20:01:10.813" Title="Counting all possible size 'm' sets of 'N' marbles under intersection and membership constraints" Tags="&lt;probability&gt;&lt;counting&gt;" AnswerCount="1" />
  <row Id="15074" PostTypeId="1" CreationDate="2010-12-21T17:16:04.853" Score="0" ViewCount="74" Body="&lt;p&gt;Let $f:(a+bi) \to (a^2+b)+ai$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an expression for f such that does not use functions Re and Im?&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2010-12-21T17:41:17.410" Title="Expression for function f:(a+bi) -&gt; (a^2+b)+ai" Tags="&lt;complex-analysis&gt;&lt;complex-numbers&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15075" PostTypeId="1" CreationDate="2010-12-21T17:29:04.793" Score="4" ViewCount="405" Body="&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/49647/finite-or-infinite-closed&quot; rel=&quot;nofollow&quot;&gt;Crossposted on Mathoverflow.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a natural number $a$, are there infinitely many natural numbers &lt;i&gt;not&lt;/i&gt; of the form $anm \pm m \pm n$, $n, m$ positive natural?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I give a proof that for $a=6$ the question is equivalent to the twin prime conjecture so it is known that we don't have any proof. But what about other values of $a$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are infinitely many twin primes if and only if there are infinitely many natural numbers that are not of the form $6nm \pm n \pm m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proof:&lt;/b&gt; Every number that is not a multiple of $2$ or $3$ is of the form $6N\pm 1$. So the only pairs that are not divisible by $2$ or $3$ are $(6N-1,6N+1)$ for any $N$. Now are there infinitely many such prime pairs (twin primes)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number $6N-1$ is prime it should not be written as a product of some numbers $6n+1,6m-1$ for any $n,m &amp;gt; 0$. So $(6n+1)(6m-1)=6(6nm-n+m)-1$, which means that $N$ should not be of the form $6nm-n+m$ for any $n,m&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if $6N+1$ is a prime it should not be a product of some numbers $(6n-1)(6m-1) =6(6nm-n-m)+1$, or $(6n+1)(6m+1) =6(6nm+n+m)+1$. Which means that we have a prime couple of the form $(6N-1,6N+1)$ if and only if $N$ is not of the form $6nm \pm n \pm m$ for any $n,m$.&lt;/p&gt;&#xA;" OwnerUserId="5010" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-22T16:27:58.483" LastActivityDate="2010-12-22T19:06:08.187" Title="Do we have a proof of the infiniteness?" Tags="&lt;number-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15076" PostTypeId="2" ParentId="15074" CreationDate="2010-12-21T17:41:17.410" Score="7" ViewCount="0" Body="&lt;p&gt;No expression as an analytic function because it does not satisfy the Cauchy-Riemann equations.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-21T17:41:17.410" CommentCount="0" />
  <row Id="15077" PostTypeId="1" CreationDate="2010-12-21T17:49:35.180" Score="1" ViewCount="131" Body="&lt;p&gt;Suppose $z = a+bi$ and $w = u+iv$. Let $\displaystyle a = \left(\frac{|w|+u}{2} \right)^{1/2}$ and $\displaystyle b = \left(\frac{|w|-u}{2} \right)^{1/2}$. Show that $z^2 = w$ if $v \geq 0$ and $(\bar{z})^{2} = w$ if $v \leq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $|w| = \sqrt{u^2+v^2}$. So this is just a matter of computing $(a+bi)^2$ and $(a-bi)^2$ and substituting in the values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Chapter 1, Problem 10 from &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; by Rudin&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-21T18:01:19.990" LastActivityDate="2010-12-21T20:46:52.317" Title="Complex Numbers and Square Roots" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15078" PostTypeId="2" ParentId="15039" CreationDate="2010-12-21T18:00:22.707" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.scribd.com/doc/3573726/ENGLISH-for-Mathematics&quot; rel=&quot;nofollow&quot;&gt;http://www.scribd.com/doc/3573726/ENGLISH-for-Mathematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the risk of getting voted down,I am posting a reference used to teach English to Mathematics majors. It may be the resource you are looking for, though I highly doubt that &quot;of&quot; is such a big issue that people have sleepless nights over. However, you certainly seem to, so that is the reason for this reply. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-21T18:00:22.707" CommentCount="2" />
  <row Id="15079" PostTypeId="1" CreationDate="2010-12-21T18:20:09.023" Score="10" ViewCount="253" Body="&lt;p&gt;I think it's an interesting coincidence that the &lt;a href=&quot;http://gato-docs.its.txstate.edu/mathworks/Locker_Puzzle.pdf&quot; rel=&quot;nofollow&quot;&gt;locker puzzle&lt;/a&gt; and this &lt;a href=&quot;http://www.msri.org/people/staff/levy/files/Emissary/0308/puzzles.pdf&quot; rel=&quot;nofollow&quot;&gt;puzzle about duplicate array entries (see problem 6b)&lt;/a&gt; have such similar solutions.  &lt;em&gt;Spoiler alert!&lt;/em&gt; Don't read on if you want to solve these puzzles yourself first (they're two of the best puzzles I've ever seen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both solutions we consider a collection of labeled boxes, each with a number inside, and then &quot;traverse through boxes&quot; by starting at a given box and using the number inside to decide where to go next.  For example, we might start at box $1$, find the number $5$ inside, proceed to box $5$, find the number $2$ inside, proceed to box $2$, and so forth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in both solutions we &quot;traverse through boxes&quot; for the same reason: we're interested in finding cycles.  More specifically, for the locker puzzle, we're interested in the question, &lt;em&gt;&quot;If we start at box $n$, how many steps does it take to get back to box $n$?&quot;&lt;/em&gt;, and for the duplicate array puzzle, we're interested in the question, &lt;em&gt;&quot;Does there exist $n$ such that (1) if we start at box $1$, we will eventually get to box $n$, and (2) if we start at box $n$, we will eventually get back to box $n$?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the two puzzles seem quite unrelated at first glance, I'll pose the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a deep reason why &quot;traversing through boxes&quot; (described in the 2nd paragraph above) shows up in the solution for both of these puzzles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there other interesting problems for which &quot;traversing through boxes&quot;, whether to find cycles or for any other reason, shows up in the solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Edit] I mistakenly said that for the locker puzzle, we're interested in the question, &lt;em&gt;&quot;If we start at box $n$, will we eventually get back to box $n$?&quot;&lt;/em&gt;  Instead, the question should be, &lt;em&gt;&quot;If we start at box $n$, how many steps does it take to get back to box $n$?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] Thanks for all the great answers so far!  However, I'm still not completely convinced there's nothing more going on here...&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2010-12-23T17:11:41.603" LastActivityDate="2010-12-23T17:11:41.603" Title="Two seemingly unrelated puzzles have very similar solutions; what's the connection?" Tags="&lt;group-theory&gt;&lt;graph-theory&gt;&lt;puzzle&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15080" PostTypeId="1" AcceptedAnswerId="15084" CreationDate="2010-12-21T18:24:39.137" Score="1" ViewCount="59" Body="&lt;p&gt;What's the reasoning behind assigning a proposition to a set, like $A=p(x)$ and $B=q(x)$? Is it because any set of elements has some property that defines all elements in it?&#xA;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;A \cup (B \cap C) &amp;amp;= \{x|p(x) \vee (q(x) \wedge r(x))\}= \\\&#xD;&#xA;&amp;amp;= \{x| (p(x)\vee q(x)) \wedge (p(x) \vee r(x)) \} \\\&#xD;&#xA;&amp;amp;= \{x| p(x) \vee q(x)\} \cap  \{x|p(x) \vee r(x) \} \\\&#xD;&#xA;&amp;amp;= (A \cup  B) \cap (A \cup C)&#xD;&#xA;&#xD;&#xA;\end{align*}&#xD;&#xA;$$&#xA;How exactly it works to prove things like this?&lt;/p&gt;&#xA;" OwnerUserId="3509" LastActivityDate="2010-12-21T18:58:42.857" Title="What's the reasoning behind assigning a proposition to a set, like $A=p(x)$ and $B=q(x)$?" Tags="&lt;homework&gt;&lt;elementary-set-theory&gt;" AnswerCount="2" />
  <row Id="15081" PostTypeId="1" AcceptedAnswerId="15083" CreationDate="2010-12-21T18:27:53.713" Score="0" ViewCount="76" Body="&lt;p&gt;Suppose $z$ is a complex number. Prove that there exists an $r \geq 0$ and a complex number $w$ (with $|w| = 1$) such that $z = rw$. Does $z$ uniquely determine $r$ and $w$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $z = a+bi$. Then $|z| = \sqrt{a^2+b^2}$. So take $r = |z|$. It seems like one can take $w$ to be any complex number such that $|w| = 1$. So I think $z$ uniquely determines $r$ but not $w$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-22T07:21:04.980" Title="Existence of factors of a complex number" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="15082" PostTypeId="2" ParentId="15081" CreationDate="2010-12-21T18:33:29.047" Score="1" ViewCount="0" Body="&lt;p&gt;No, if you take $w=1+0i$ for example, $wr$ will always be real.  So $w$ must have the same phase as $z$.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T18:33:29.047" CommentCount="2" />
  <row Id="15083" PostTypeId="2" ParentId="15081" CreationDate="2010-12-21T18:34:25.003" Score="2" ViewCount="0" Body="&lt;p&gt;Let $k\in\mathbb{R}, z\in\mathbb{C}, z=a+bi$. Then $kz = ka+kbi$. Therefore $w$ is uniquely determined by $z$, and $w = \frac{a}{r} + \frac{b}{r}i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also think of this using vectors in $\mathbb{R}^2$; any vector can be represented as the unit vector with its same angle multiplied by its length.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-12-22T07:21:04.980" LastActivityDate="2010-12-22T07:21:04.980" CommentCount="2" />
  <row Id="15084" PostTypeId="2" ParentId="15080" CreationDate="2010-12-21T18:34:58.327" Score="2" ViewCount="0" Body="&lt;p&gt;Firstly I want to make clear that the only way to ensure that every set has a property defining it is some trivial property like $A = \{x|x\in A\}$, because there are sets which you cannot define by a simple formula just like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to prove equality of sets, one usually shows that both the left hand side is included in the right hand side, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you take some arbitrary $x\in A\cup (B\cap C)$ and then that means either $x\in A$ or $x\in B\cap C$ that is either $p(x)$ or both $q(x)$ and $r(x)$, that means that $p(x) \vee q(x) \wedge r(x)$ is true, from here you need to either prove it directly or by DeMorgan's laws show that this is equivalent that $(p(x) \vee q(x)) \wedge (p(x) \vee r(x))$ and from here follows that $x\in A\cup B$ and $x\in A\cup C$, that is to say $x\in (A\cup B)\cap (A\cup C)$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-21T18:34:58.327" />
  <row Id="15085" PostTypeId="2" ParentId="15055" CreationDate="2010-12-21T18:36:39.293" Score="6" ViewCount="0" Body="&lt;p&gt;Just like the better-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt;, this is a question about knowledge, and knowledge-of-knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Monty Hall problem, &lt;a href=&quot;http://math.stackexchange.com/questions/886/deal-or-no-deal-does-one-switch-to-avoid-a-goat-should-deal-or-no-deal-be-10/888#888&quot;&gt;the probability depends on whether or not Monty &lt;em&gt;knows&lt;/em&gt; where the car is&lt;/a&gt;.  If Monty knows where the car is, he will &lt;em&gt;always&lt;/em&gt; open a door with a goat, and switching will give you better odds.  If Monty doesn't know where the car is, then he will only open a goat-door 2/3rds of the time, and switching will not give you any better odds.  If you, as the contestant, don't know whether or not &lt;em&gt;Monty&lt;/em&gt; knows, you cannot assign a probability - you must assume one way or the other first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, let's change this problem slightly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your friend tells you that Mrs. Smith has two children, and at least one is a girl.  What is the probability that Mrs. Smith has two girls?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The important question here is:  &lt;strong&gt;Does your friend know whether or not Mrs. Smith has two girls?&lt;/strong&gt;  There are two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Yes, your friend knows the gender of both of Mrs. Smith's children.&lt;/strong&gt;  In this case, there are three equally likely possibilities: {BG, GB, GG} &lt;em&gt;(where BG means she has an older boy and a younger girl)&lt;/em&gt;.  Thus, the probability of two girls is 1/3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;No, your friend does not know the gender of both of Mrs. Smith's children.&lt;/strong&gt;  In this case, let's say he met only one of the children, and she was a girl.  Whether she was the older or younger sibling, the chances of the other sibling also being a girl are 1/2, so the overall probability is 1/2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, we have another question about knowledge-of-knowledge.  Like the Monty Hall problem, without knowledge of the asker's knowledge, there is no way to assign a probability.  Most other posts have assumed the person asking has knowledge of the sex of both of the children; however, NotSuper's counter-examples (in the comments above) assume the asker &lt;em&gt;does not&lt;/em&gt; have knowledge of both children, which is where the apparant contradiction lies.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-12-21T23:19:23.767" LastActivityDate="2010-12-21T23:19:23.767" CommentCount="5" />
  <row Id="15086" PostTypeId="1" AcceptedAnswerId="15089" CreationDate="2010-12-21T18:41:58.840" Score="3" ViewCount="126" Body="&lt;p&gt;Are there any advantages or disadvantages in defining a real number in the following ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 1&lt;/strong&gt;&#xA;A real number is an object of the form $\lim\limits_{n \to \infty} a_n$ where $(a_n)_{n}^{\infty}$ is a cauchy sequence of rational numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition 2&lt;/strong&gt; A real number is a cut in $\mathbb{Q}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-21T18:57:17.607" LastActivityDate="2010-12-21T20:25:09.897" Title="Dedekind Cuts versus Cauchy Sequences" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15087" PostTypeId="2" ParentId="15077" CreationDate="2010-12-21T18:45:31.647" Score="1" ViewCount="0" Body="&lt;p&gt;HINT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is:  if $v \geq 0 \Longrightarrow  (v^{2})^\frac{1}{2}=v$ and if $v \leq 0 \Longrightarrow  (v^{2})^\frac{1}{2}=-v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part Note that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^{2}=a^{2}-b^{2}+2ab \imath = \frac{|w|+u}{2}-\frac{|w|-u}{2}+2\imath \left(\frac{|w|^{2}-u^{2}}{4} \right)^\frac{1}{2} = $$ &#xA;$$ u + 2 \imath \left(\frac{\left(\sqrt{u^{2}+v^{2}}\right)^{2}-u^{2}}{4}\right)^\frac{1}{2}= u+2 \imath \left(\frac{v^{2}}{4}\right)^\frac{1}{2} = u + \imath (v^{2})^\frac{1}{2} = u+v\imath = w$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is same, using $-v$ (in the key) and expanding $\bar{z}$.&lt;/p&gt;&#xA;" OwnerUserId="3421" LastEditorUserId="3421" LastEditorDisplayName="" LastEditDate="2010-12-21T20:36:53.433" LastActivityDate="2010-12-21T20:36:53.433" CommentCount="6" />
  <row Id="15088" PostTypeId="1" AcceptedAnswerId="15093" CreationDate="2010-12-21T18:49:19.413" Score="5" ViewCount="104" Body="&lt;p&gt;I know that if $f$ is a Lebesgue measurable function on $[a,b]$ then there exists a continuous function $g$ such that $|f(x)-g(x)|&amp;lt; \epsilon$ for all $x\in [a,b]\setminus P$ where the measure of $P$ is less than $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to imply that every Lebesgue measurable function on $\mathbb{R}$ is the pointwise limit of continuous functions.  Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-21T19:01:12.357" LastActivityDate="2010-12-21T19:02:42.330" Title="Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions?" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="15089" PostTypeId="2" ParentId="15086" CreationDate="2010-12-21T18:52:12.330" Score="4" ViewCount="0" Body="&lt;p&gt;They are the same (up to order-preserving isomorphism). I defer to Theorem 15 of a &lt;a href=&quot;http://math.stackexchange.com/questions/11923/completion-of-rational-numbers-via-cauchy-sequences/12009#12009&quot;&gt;previous answer of mine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, your first definition is wrong. It's supposed to be &lt;em&gt;equivalence classes&lt;/em&gt; of rational Cauchy sequences - otherwise you have a ring but not a field.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-21T18:52:12.330" CommentCount="3" />
  <row Id="15091" PostTypeId="2" ParentId="15088" CreationDate="2010-12-21T18:55:42.100" Score="6" ViewCount="0" Body="&lt;p&gt;No. Pointwise limits of continuous functions are &lt;a href=&quot;http://en.wikipedia.org/wiki/Baire_function&quot; rel=&quot;nofollow&quot;&gt;Baire functions of class 1&lt;/a&gt; (or 0 it they are simply continuous). The characteristic function of the rationals is measurable but not of Baire class 1. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-12-21T18:55:42.100" />
  <row Id="15092" PostTypeId="2" ParentId="15080" CreationDate="2010-12-21T18:58:42.857" Score="1" ViewCount="0" Body="&lt;p&gt;As Asaf said, not all sets have a good definition.  But many of the ones we care about do.  In your example, you did not make use of the definition of A as $\{x|p(x)\}$, so could bypass that step and not even define it.  But it is often useful, as when we want $A\cap B$ we can look at p(x) and q(x) and ask what would x have to be to satisfy both.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-21T18:58:42.857" />
  <row Id="15093" PostTypeId="2" ParentId="15088" CreationDate="2010-12-21T19:02:42.330" Score="6" ViewCount="0" Body="&lt;p&gt;I thought of a worse example.  A pointwise limit of a sequence of continuous functions is Borel measurable, and there are Lebesgue measurable functions that are not Borel measurable.  The characteristic function of any non-Borel set of measure $0$ will do, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with &quot;this seems to imply&quot; is that &quot;almost everywhere&quot; and &quot;everywhere&quot; are different.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-21T19:02:42.330" CommentCount="5" />
  <row Id="15094" PostTypeId="2" ParentId="14952" CreationDate="2010-12-21T19:31:37.013" Score="1" ViewCount="0" Body="&lt;p&gt;I found the most beautiful usage of jacobian matrices in studying differential geometry, when one abandons the idea that analysis can be done &quot;only on balls of $\mathbb{R}^n$&quot;. The definition of tangent space in a point $p$ of a manifold $M$ can be given via the kernel of the jacobian of a suitable submersion, or via the image of the differential of a suitable immersion from an open set $U\subseteq\mathbb{R}^{\dim M}$. Quite a simple example, but when I was an undergrad four years ago it gave me the &quot;right&quot; idea of what a linear transformation &lt;em&gt;does&lt;/em&gt; in a differential (analytical) framework.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-21T19:31:37.013" CommentCount="1" />
  <row Id="15095" PostTypeId="2" ParentId="14311" CreationDate="2010-12-21T19:35:21.327" Score="2" ViewCount="0" Body="&lt;p&gt;Edit: Here is a more verbose proof. The original proof is retained below.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Introduction&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let $\Sigma$ be some alphabet with $n$ symbols. Let $L$ be the language of words in which each symbol appears an even number of times. Suppose $G$ is a grammar for $L$. Our goal is to given an exponential lower bound on the size of $G$ in terms of $n$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Transition to Chomsky Normal Form&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia (and not difficult to prove), there is a grammar $G&amp;#39;$ for $L$ in Chomsky Normal Form such that $|G&amp;#39;| = O(|G|^2)$. We will show an exponential lower bound on $|G&amp;#39;|$, and this will imply an exponential lower bound on $|G|$ (if $|G&amp;#39;| = \Omega(c^n)$ for $c &amp;gt; 1$ then $|G| = \Omega(\sqrt{c}^n)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes, a grammar is in Chomsky Normal Form (CNF) if the productions are of the following two types: $A \rightarrow BC$ and $A \rightarrow a$, where $A,B,C$ are non-terminals and $a$ is terminal. Some people also place the restriction that $B,C$ cannot equal the starting symbol $S$, but we do not need this condition.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Word derivation in CNF grammars&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let $w \in L$. Choose some derivation of $w$ in $G&amp;#39;$ (there might be several sharing the same &lt;i&gt;derivation tree&lt;/i&gt;). The derivation can be thought of as a binary tree (&lt;i&gt;derivation tree&lt;/i&gt;) where each non-leaf node is labeled by a terminal, and each leaf is labeled by a non-terminal; a non-leaf node either has two non-leaf children or one leaf child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider the derivation $$S \rightarrow AB \rightarrow ACD \rightarrow^* xyz.$$ The corresponding derivation tree can be described as follows (sorry for the missing graphics):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The root is labeled $S$.&lt;/li&gt;&#xA;&lt;li&gt;The root has two children labeled $A,B$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $B$ has two children labeled $C,D$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $A$ has one child labeled $x$, which is a leaf.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $C$ has one child labeled $y$, which is a leaf.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $D$ has one child labeled $z$, which is a leaf.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that in general, several nodes can share a label (indeed, this is how one proves the pumping lemma).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each node in the tree generates a subword of the original word, which is formed from its descendant leaf nodes in order. Continuing our example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The root generates the entire word $xyz$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $A$ generates the subword $x$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $B$ generates the subword $yz$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $C$ generates the subword $y$.&lt;/li&gt;&#xA;&lt;li&gt;The node labeled $D$ generates the subword $z$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;A subword lemma&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Lemma.&lt;/b&gt; Suppose $w \in L$ is of length $m$, and choose some positive $\ell \leq m$. Then there is a node in the derivation tree of $w$ that generates a subword of length $\alpha$ satisfying $\ell \leq \alpha &amp;lt; 2\ell$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Proof.&lt;/b&gt; Run the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Set $v$ to the root of the derivation tree.&lt;/li&gt;&#xA;&lt;li&gt;While the subword generated by $v$ has size at least $2\ell$, replace $v$ by its child whose generated subword is longest.&lt;/li&gt;&#xA;&lt;li&gt;Output the subword generated by $v$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that the node $v$ always has two children, since otherwise it generated a subword of length $1$. If the loop in step (2) never runs then certainly $v$ generates a subword of the requisite length, by our assumption that $\ell \leq m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, consider the last iteration of the loop. Denote by $w(u)$ the subword generated by a node $u$. In the last iteration of the loop, there is a node $v$ with $|w(v)| \geq 2\ell$, with children $v_1,v_2$ such that $|w(v_1)| \geq |w(v_2)|$; we move to $v_1$. Note that $$2\ell \leq |w(v)| = |w(v_1)| + |w(v_2)| \leq 2|w(v_1)|$$ so that $|w(v_1)| \geq \ell$. Since the loop stopped, $|w(v_1)| &amp;lt; 2\ell$ and $w(v_1)$ is of the requisite size.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The proof&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We consider the set $P$ of words in $L$ of the form $\pi(\Sigma) \pi(\Sigma)$, where $\pi$ runs over all permutations of $n$ symbols. For example, if $\Sigma = \{0,1\}$ then $$P = \{0101,1010\}.$$ Note that indeed $P \subseteq L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all words in $P$ are of length $2n$ (recall $n$ is the size of the alphabet $\Sigma$), using the lemma with $\ell = n/3$ (assume for simplicity that $n$ is divisible by $3$) we get that some node in the derivation tree of any $w \in P$ generates a subword $x(w)$ whose size satisfies $$n/3 \leq |x(w)| &amp;lt; 2n/3.$$ Denote by $s(w)$ the non-terminal generating $x(w)$ (it is the label of the node).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $|x(w)| \leq n$, all symbols in $x(w)$ are distinct: indeed, any subword of $w$ of length $n$ is a cyclic rotation of the original permutation (think of all subwords of length $3$ in $abcabc$: $abc,bca,cab,abc$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our strategy now will be to show that any single non-terminal can generate only a relatively small number of the $x(w)$; since $P$ is big, it will follow that $G&amp;#39;$ must have lots of non-terminals. More explicitly, let $$S = \{ s(w) : w \in P\}.$$ We can find a list of representative, distinct words $w_1,\ldots,w_{|S|}$ such that $$S = \{ s(w_i) : 1 \leq i \leq |S| \}.$$ We will show that for each $w_i$, the set of words $w \in P$ with $s(w) = s(w_i)$ consists of at most $B$ words ($B$ will be determined below). It will follow that the number of distinct non-terminals is at least $$\frac{|P|}{B} = \frac{n!}{B}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $w \in P$, and suppose that $s(w) = s(w_i)$ for some fixed $i$. What this means is that we can replace $x(w)$ in $w$ with $x(w_i)$ and still get a word in $L$. The word $w$ has a decomposition $w = lx(w)r$. Every symbol appearing in $x(w)$ appears exactly once in $lr$, and every other symbol appears exactly twice in $lr$. Note that this defines the word $x(w)$ up to permutation, i.e. since we assumed that $lx(w_i)r \in L$, then $x(w_i)$ must be a permutation of $x(w)$ (more accurately, the set of (unique) symbols occurring in $x(w),x(w_i)$ must be the same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will now bound the number of words in $P$ that have a subword of length $x(w_i)$ which is a permutation of $x(w_i)$. Any such word $w$ can be broken as $lyzr$ where $yz$ is a permutation of $x(w_i)$ and $|ly|=|zr|=n$; in fact, since $w \in P$ then $ly = zr$. There are $|x(w_i)|!$ possible choices for $yz$. Together, these define $|x(w_i)|$ points of the permutation $\pi(\Sigma) = ly$. For the rest of the points we have $(n-|x(w_i)|)!$ possible choices. Finally, $yz$ can start in at most $n$ &quot;cyclic&quot; locations inside $\pi$, and so $$B \leq n|x(w_i)|!(n-|x(w_i)|)!.$$ When is this expression maximized, as a function of $\alpha = |x(w_i)|$? Taking the logarithm of the bound, $$\log B \leq \log 2n + \log \alpha! + \log (n-\alpha)!.$$ The factorial function (in fact, the Gamma function) is log-convex, and so subject to the condition $n/3 \leq \alpha \leq 2n/3$, this expression is maximized for $\alpha=n/3$ or $\alpha =2n/3$ (the value is the same). That is $$B \leq n(n/3)!(2n/3)!.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging the value of $B$ that we have laboriously estimated, we conclude that the number of different non-terminals in $G&amp;#39;$ must be at least $$\frac{|P|}{B} = \frac{n!}{n(n/3)!(2n/3)!} = \frac{1}{n} \binom{n}{n/3}.$$ Now we will use Stirling's approximation $$m! \sim \sqrt{2\pi m} (m/e)^m.$$ Plugging the approximation, we get&#xA;$$\frac{n!}{2n(n/3)!(2n/3)!} \sim \frac{\sqrt{2\pi n}(n/e)^n}{2n\sqrt{2\pi(n/3)}(n/3e)^{n/3} \sqrt{2\pi(2n/3)}(2n/3e)^{2n/3}}.$$ All powers of $n$ and $e$ cancel, and we're left with the following $$\frac{n!}{n(n/3)!(2n/3)!} \sim \frac{1}{n\sqrt{4\pi n/9}} 3^{n/3} (3/2)^{2n/3} = \Omega\left(n^{-3/2} c^n \right),$$ where the constant $c$ is $$c = 3^{1/3} (3/2)^{2/3} = \frac{3}{2^{2/3}} &amp;gt; 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the original grammar $G$ must be of size at least $\Omega(n^{-3/4} \sqrt{c}^n)$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Exactly 0 or 2 occurrences&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Another questioned concerned the language $L&amp;#39;$ of all words which contain exactly $0$ or $2$ occurrences of each symbol in $\Sigma$. The proof given in the preceding section applies equally well to $L&amp;#39;$ since the set of words $P$ we used belongs to $L&amp;#39;$; the rest of the proof uses the fact that certain words are not in $L$, and since $L \supset L&amp;#39;$, they are a fortiori not in $L&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Generalization&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We have seen that the lower bound holds for both $L$ and $L&amp;#39;$. When does the proof hold for a language $\Lambda$? We used only two properties of $\Lambda$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\Lambda$ contains $P$.&lt;/li&gt;&#xA;&lt;li&gt;If $w_1,w_2 \in P$, $x_1,x_2$ are the subwords of $w_1,w_2$ chosen using the subword lemma, and $w_1(x_1:=x_2),w_2(x_2:=x_1) \in \Lambda$ (i.e. the word resulting by replacing $x_1$ with $x_2$ in $w_1$, and the word resulting by replacing $x_2$ with $x_1$ in $w_2$) then $x_2$ is a permutation of $x_1$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When will the second condition fail? The words $x_1,x_2$ both have no repeated symbols. If they are not permutations of each other then $w_1(x_1:=x_2)$ will have a symbol repeated $1$ or $3$ times. So words containing such symbols must lie out of $\Lambda$. Considering $w_2(x_2:=x_1)$, we see that it is enough to exclude one of the possibilities. We get the following theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; Suppose $\Lambda$ is a language containing $P$ in which no word contains any symbol exactly once, or in which no word contains any symbol exactly $3$ times. Then any grammar for $\Lambda$ must have size $\Omega(n^{-3/4} C^n)$, where $C = \sqrt{3}/\sqrt[3]{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the theorem applies for the language in which the number of occurrences of each symbol lies in $S$ as long as $2 \in S$ and $1,3 \notin S$. For example, $L$ corresponds to $S = \{0,2,4,6,\ldots\}$ and $L&amp;#39;$ corresponds to $S = \{0,2\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We leave the reader the following easy generalizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem.&lt;/b&gt; Suppose $\Lambda$ is a language containing $\{ \pi(\Sigma)^\ell \}$ for some $\ell &amp;gt; 0$, in which no word contains any symbol exactly $\ell + \epsilon$ times, where $\epsilon$ is either $1$ or $-1$. Then any grammar for $\Lambda$ must have size $\Omega(n^{-3/4} C^n)$, where $C = \sqrt{3}/\sqrt[3]{2}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Original, condensed proof&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let $L$ be the language of words in which each symbol appears an even number of times. We give an exponential lower bound (in $n = |\Sigma|$) on the size of a Chomsky Normal Form (CNF) grammar for $L$. This implies an exponential lower bound on the size of any context-free grammar for $L$, since (according to Wikipedia) the blowup to CNF is at most quadratic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any derivation of a word $w$ of length $m$ and any $\ell &amp;lt; m/2$ we can always find a non-terminal symbol in the derivation which derives a subword $x$ of $w$ of length $\ell \leq |x| &amp;lt; 2\ell$: start from the root of the derivation, and always pick the larger branch. The first time we hit a branch smaller than $2\ell$, it must be at least $\ell$ in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the set of all $S = n!$ words in $L$ of the form $\pi\pi$ (all of size $2n$), where $\pi$ is a permutation of $\Sigma$. For each such word $w$ we can find a symbol $s_w$ generating a subword $x_w$ of size $n/3 \leq |x_w| &amp;lt; 2n/3$. Note that all symbols in $x_w$ are distinct. Therefore if $s_u = s_v$ then $x_u,x_v$ must be permutations of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $w$, for how many words $u$ can it be that $s_w = s_u$? To get from $w$ to $u$ we need to permute $x_w$, to permute the rest of the permutation, and to choose the starting location of $x_u$ within the permutation generating $w_u$. All in all, the number of possibilities is at most $$M = n|x_w|!(n-|x_w|)! \leq n(n/3)!(2n/3)!.$$ Therefore the number of distinct symbols in the grammar is at least $$\frac{S}{M} = \frac{n!}{n(n/3)!(2n/3)!} = O\left(\frac{c^n}{n^{1.5}}\right),$$&#xA;where $c = 3/2^{2/3} \approx 1.88988$ (thanks to Wolfram Alpha).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-31T21:58:47.790" LastActivityDate="2010-12-31T21:58:47.790" CommentCount="4" />
  <row Id="15096" PostTypeId="2" ParentId="15039" CreationDate="2010-12-21T19:42:06.960" Score="2" ViewCount="0" Body="&lt;p&gt;When students are learning to solve word problems, they are usually taught to look for keywords that identify the operation. The word &quot;of&quot; sometimes signals a multiplication, especially multiplication by a fraction. Examples: half of the books are missing; 70% of Earth's surface is covered in water. But it is used in many other ways: a function of x; a set of natural numbers; 4 out of 5 dentists agree.&lt;/p&gt;&#xA;" OwnerUserId="857" LastActivityDate="2010-12-21T19:42:06.960" CommentCount="1" />
  <row Id="15097" PostTypeId="1" AcceptedAnswerId="15103" CreationDate="2010-12-21T19:43:35.007" Score="6" ViewCount="76" Body="&lt;p&gt;When solving questions like these:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $f(x)$ be a real function. Find&#xA;  $f(0.1)$ using its Taylor expansion&#xA;  such that the error is less than&#xA;  $10^{-3}$. Find the lowest degree of&#xA;  Taylor polynomial needed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(if someone can rephrase the question so its more clear that would be great. I didn't quite nail the translation to English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were explained that usually the process of solving this is finding some degree that works. Then, depending on how 'far' you are from the error term, you start trying lower degrees. When some degree doesn't work anymore, you say the one before it is the minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering though, why the implicit assumption that the remainder function is decreasing? i.e. if degree $k$ doesn't suffice, $1, 2, ..., k-2, k-1$ won't work either. Our professor said that we can rely on this because most functions we're dealing with comply with this property. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-21T23:43:42.960" LastActivityDate="2010-12-22T00:17:53.063" Title="Why is the remainder function $R_{n}(x)$ decreasing?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15098" PostTypeId="1" AcceptedAnswerId="15106" CreationDate="2010-12-21T19:50:19.323" Score="5" ViewCount="50" Body="&lt;p&gt;I have an HP 50g graphing calculator and I am using it to calculate the standard deviation of some data. In the statistics calculation there is a type which can have two values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample&#xA;Population&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't change it, but I kept getting the wrong results for the standard deviation. When I changed it to &quot;Population&quot; type, I started getting correct results!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? As far as I know, there is only one type of standard deviation which is to calculate the root-mean-square of the values!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I miss something?&lt;/p&gt;&#xA;" OwnerUserId="4671" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-21T21:05:23.643" LastActivityDate="2010-12-21T21:06:06.467" Title="Sample Standard Deviation vs. Population Standard Deviation" Tags="&lt;statistics&gt;&lt;standard-deviation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15099" PostTypeId="2" ParentId="15073" CreationDate="2010-12-21T20:01:10.813" Score="2" ViewCount="0" Body="&lt;p&gt;You're looking for a family of subsets of $[n]$ of size $m$, the intersection of any two of which is at most $k$. If $k \geq m$ then we can take all subsets of size $m$. If $k &amp;lt; m$ then a well-known theorem of Ray-Chaudhuri and Wilson shows an upper bound of $\binom{n}{k}$. A trivial construction gives a lower bound of $\binom{n-(m-k)}{k}$. So for $k \leq m$ constant, the answer is $\Theta(n^k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the theorem mentioned above, see for example Babai and Frankl's book &quot;Linear Algebra Methods in Combinatorics with Applications to Geometry and Computer Science&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T20:01:10.813" CommentCount="1" />
  <row Id="15100" PostTypeId="1" AcceptedAnswerId="15116" CreationDate="2010-12-21T20:05:30.453" Score="3" ViewCount="119" Body="&lt;p&gt;IF $A$,$B$,$C$ are finite sets then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of elements in exactly one of sets $A$,$B$,$C$:$$n(A)+n(B)+n(C)-2 \times n(A \cap B)-2 \times n(A \cap C)-2 \times n(C \cap B) + 3 \times n(A \cap B \cap C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of elements in exactly two of the sets $A$,$B$,$C$: $$n(A \cap B)+(A \cap C)+n(C \cap B) - 3 \times n(A \cap B \cap C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody explain how can we prove them? I think it's something with inclusion and exclusion but am not getting how to get those results.Also I think we can have a generalized results for these kinds ?!&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T05:57:58.843" LastActivityDate="2010-12-22T05:57:58.843" Title="Finding the cardinality of exactly one or two elements of two or three finite sets" Tags="&lt;elementary-set-theory&gt;&lt;counting&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15101" PostTypeId="1" CreationDate="2010-12-21T20:09:25.470" Score="0" ViewCount="48" Body="&lt;p&gt;I have two images. One is the reference image and the other is the rotated version of the reference image. I would like to find what is the angle by which the second image is rotated with respect to the first one. &#xA;  i want this to work automatic.&lt;/p&gt;&#xA;" OwnerUserId="5018" LastEditorUserId="5018" LastEditorDisplayName="" LastEditDate="2010-12-22T18:00:03.537" LastActivityDate="2010-12-22T18:00:03.537" Title="find the rotation of one image with respect to other in MATLAB" Tags="&lt;matlab&gt;&lt;image-processing&gt;" AnswerCount="2" />
  <row Id="15102" PostTypeId="2" ParentId="15086" CreationDate="2010-12-21T20:25:09.897" Score="3" ViewCount="0" Body="&lt;p&gt;In the real numbers case, the construction is isomorphic, as kahen said in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at this fact closely enough, you realize that this is because the backbone of the real line is actually the integers, which has no accumulation points within itself, that is if a sequence converges it has to be constant from some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact, while seems unrelated, helps us avoid the following case:&#xA;Consider some dense linear order (like the rationals) only bigger. For example, put two copies of the rationals one on top of the other, of course this can be embedded back into the rationals themselves, but consider the Dedekind closure of this order. It would have a point which is the exact point in which you switch from the first copy of the rationals to the second one. (This might not be such a good example, if you have a better example - this post is CW for this very reason)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let us try and prove something (together, I got stuck halfway through and although the intuition is clear to me, I'm uncertain how to put it into words - and this might as well be complete rubbish too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem:&lt;/strong&gt; Suppse $F$ is an ordered field (so it has characteristics 0, and we can assume $\mathbb{Q}\subseteq F$ without the loss of generality) such that its Cauchy completion (denoted by $C(F)$) and Dedekind completion (denoted by $D(F)$), both form a field - and it is the same field (namely there is a ring isomorphism between them).&lt;br&gt;&#xA;Then $F\subseteq\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Suppose not. Since $\mathbb{Q}\subseteq F$ then $\mathbb{R}\subseteq C(F)$ (and from our assumption, $\mathbb{R}\subseteq D(F)$ as well - they are isomorphic). Now let $a\in C(F)\setminus\mathbb{R}$ (without loss of generality take $a&gt;0$), that is it is not a Cauchy limit of rational sequences. Meaning every $a_n\to a$ has only finitely many rational elements, so again without the loss of generality $a_n$ is not rational for all $n$. Since $D(F)$ is an ordered field so is $C(F)$, if there was $r\in\mathbb{R}$ such that $a&amp;lt;r$ then we have that $a\in (0,r)$ and so we can find a rational Cauchy sequence that converges to $a$. That to say that $a&amp;gt;r$ for any real number $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once more, we use the isomorphism, $a$ defines a cut which means the set $A = \{x\in F| x&amp;gt;q,\ \forall q\in\mathbb{Q}\}$ is non-empty, take $z$ to be the cut defined as the number between $\mathbb{Q}$ and $A$ (usually denoted by $z=\langle \mathbb{Q}\big| A\rangle$). Take some $\epsilon&amp;gt;0$ then $z-\epsilon$ is already smaller than some rational number, and therefore can be expressed as a limit of rational numbers, ergo $z-\epsilon\in\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $z_n \to z$ some Cauchy sequence, since $z$ cannot be expressed as a real number, we can assume $z_n$ is not rational for all $z$, and therefore even if we take real numbers we cannot express $z$ as a Cauchy limit of such real numbers. However, for all $\epsilon&amp;gt;0$ we can express $z-\epsilon$ as a Cauchy limit of real numbers, so we can choose some sequence that would converge to $z$ Cauchy wise, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am quite certain that I made a big mess in this proof, and would like it very much if someone will clean it up a bit. It should probably be easier to prove it by some model theoretic method, or maybe by using equivalent topologies of order and metric. I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I completely resent the whole idea that we based metric spaces on the real line and then we say there are no metric-complete ordered fields except $\mathbb{R}$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-21T20:25:09.897" CommunityOwnedDate="2010-12-21T20:25:09.897" />
  <row Id="15103" PostTypeId="2" ParentId="15097" CreationDate="2010-12-21T20:38:56.323" Score="5" ViewCount="0" Body="&lt;p&gt;I presume you have a theorem like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose $f \in C^k(I)$ for some open interval $I$. Let $a,b \in I$ and $M$ be numbers such that $|f^{(n+1)}(t)| \leq M$ for all $t$ between $a$ and $b$. Then we have that&#xA;  $$ |R_n f(b)| \leq \frac{M}{(n+1)!} |b-a|^{n+1}.$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously you'll want $|b-a| &amp;lt; 1$ which you almost always can arrange for. Now if $f$ is analytic (and thus equal to its own Taylor series), we can use the third of these &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytic_function#Alternate_characterizations&quot; rel=&quot;nofollow&quot;&gt;alternate characterizations&lt;/a&gt; to see that the approximations get better. This should give you a good idea of how the approximations might get worse.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-21T20:38:56.323" />
  <row Id="15104" PostTypeId="2" ParentId="15077" CreationDate="2010-12-21T20:46:52.317" Score="0" ViewCount="0" Body="&lt;p&gt;This is related to what Gunnar Magnusson said above. First suppose $|w| = 1$. Then $u = \cos(\theta)$ and $v = \sin(\theta)$ for some $\theta \in (-\pi,\pi]$, so that $w = e^{i\theta}$. Then $a = ({1 + \cos(\theta) \over 2})^{1 \over 2}$ and $b = ({1 - \cos(\theta) \over 2})^{1 \over 2}$. By the half-angle formulas, $a = \cos{\theta \over 2}$. Also, $ b= \sin{\theta \over 2}$ when $v \geq 0$, while $b =&#xD;&#xA;-\sin{\theta \over 2}$ when $v &amp;lt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $v \geq 0$, $z = a + ib = \cos{\theta \over 2}+ i\sin{\theta \over 2} = e^{i{\theta \over 2}}$, while if $v &amp;lt; 0$, $\bar{z} = a - ib = \cos{\theta \over 2}+ i\sin{\theta \over 2} = e^{i{\theta \over 2}}$. Thus in the first case, $z^2 = w$, while in the second case $(\bar{z})^2 = w$. This is what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $|w| \neq 1$, the above proves it for ${w \over |w|}$ in place of $w$. Multiplying everything through by $|w|$ then gives your result.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-21T20:46:52.317" />
  <row Id="15105" PostTypeId="2" ParentId="15061" CreationDate="2010-12-21T20:58:33.197" Score="5" ViewCount="0" Body="&lt;p&gt;One way to prove the irreducibility seems to be to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton_polygon&quot; rel=&quot;nofollow&quot;&gt;Newton Polygon&lt;/a&gt;.&#xA;The condition on the coefficients of $f$ means that the Newton Polygon has a side of slope $\dfrac{1}{n}-1$ and hence that $f$ has a root $\alpha$ in some algebraic closure $F$ of $\mathbf{Q}_p$ having valuation $1 - \dfrac{1}{n}$ (there is a unique way to prolong $e_p$ to a valuation of $F$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then the extension $\mathbf{Q}_p \subset \mathbf{Q}_p(\alpha)$ is totally ramified&#xA;of degree $n$ and $f$ must be irreducible over $\mathbf{Q}_p$ hence &lt;em&gt;a fortiori&lt;/em&gt; irreducible over $\mathbf{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2010-12-21T20:58:33.197" CommentCount="3" />
  <row Id="15106" PostTypeId="2" ParentId="15098" CreationDate="2010-12-21T21:00:41.323" Score="10" ViewCount="0" Body="&lt;p&gt;There are, in fact, two different formulas for standard deviation here: The population standard deviation $\sigma$ and the sample standard deviation $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x_1, x_2, \ldots, x_N$ denote all $N$ values from a population, then the (population) standard deviation is &#xA;$$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2},$$&#xA;where $\mu$ is the mean of the population.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x_1, x_2, \ldots, x_N$ denote $N$ values from a sample, however, then the (sample) standard deviation is &#xA;$$s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x})^2},$$&#xA;where $\bar{x}$ is the mean of the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the change in formula with the sample is this: When you're calculating $s$ you are normally using $s^2$ (the sample variance) to estimate $\sigma^2$ (the population variance).  The problem, though, is that if you don't know $\sigma$ you generally don't know the population mean $\mu$, either, and so you have to use $\bar{x}$ in the place in the formula where you normally would use $\mu$.  Doing so introduces a slight bias into the calculation: Since $\bar{x}$ is calculated from the sample, the values of $x_i$ are on average closer to $\bar{x}$ than they would be to $\mu$, and so the sum of squares $\sum_{i=1}^N (x_i - \bar{x})^2$ turns out to be smaller on average than $\sum_{i=1}^N (x_i - \mu)^2$.  It just so happens that that bias can be corrected by dividing by $N-1$ instead of $N$.  (Proving this is a standard exercise in an advanced undergraduate or beginning graduate course in statistical theory.)  The technical term here is that $s^2$ (because of the division by $N-1$) is an &lt;em&gt;unbiased estimator&lt;/em&gt; of $\sigma^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it is that with a sample you have $N$ independent pieces of information.  However, since $\bar{x}$ is the average of those $N$ pieces, if you know $x_1 - \bar{x}, x_2 - \bar{x}, \ldots, x_{N-1} - \bar{x}$, you can figure out what $x_N - \bar{x}$ is.  So when you're squaring and adding up the residuals $x_i - \bar{x}$, there are really only $N-1$ independent pieces of information there.  So in that sense perhaps dividing by $N-1$ rather than $N$ makes sense.  The technical term here is that there are $N-1$ &lt;em&gt;degrees of freedom&lt;/em&gt; in the residuals $x_i - \bar{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, see Wikipedia's article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_deviation#With_sample_standard_deviation&quot; rel=&quot;nofollow&quot;&gt;sample standard deviation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-21T21:06:06.467" LastActivityDate="2010-12-21T21:06:06.467" />
  <row Id="15107" PostTypeId="1" AcceptedAnswerId="15171" CreationDate="2010-12-21T21:03:40.790" Score="1" ViewCount="55" Body="&lt;p&gt;&lt;strong&gt;What is the integral of a $k$-dimensional (multivariate) Gaussian over a convex $k$-polytope?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am specifically interested in $k\in\{2,3\}$, but insight on the general problem would also be appreciated.  I am guessing that there is no closed form in the general case, but what if we restrict the number of vertices of the polytope to at most $k+1$?  If no closed form exists even under this constraint, are there efficient ways to approximate the integral?&lt;/p&gt;&#xA;" OwnerUserId="5022" LastActivityDate="2010-12-22T08:19:40.100" Title="Integrating a $k$-dimensional (multivariate) Gaussian over a convex $k$-polytope" Tags="&lt;probability-theory&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15108" PostTypeId="1" CreationDate="2010-12-21T21:08:13.777" Score="3" ViewCount="95" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many symmetric relations are there for an $n$-element set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5023" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-21T21:47:08.890" LastActivityDate="2010-12-21T22:10:57.113" Title="How Many Symmetric Relations on a Finite Set?" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15109" PostTypeId="1" AcceptedAnswerId="15117" CreationDate="2010-12-21T21:15:45.270" Score="3" ViewCount="62" Body="&lt;p&gt;If $K_1$ and $K_2$ are subfields of a pre-chosen $\overline{\mathbb{Q}_p}$, and if they're both unramified at $p$, and $[K_1:\mathbb{Q}_p]=[K_2:\mathbb{Q}_p]$, does that imply that $K_1=K_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition says that this is true because all that's happening is that we're extending the residue field, and there's only one way to do that if we fix the degree of the extension. But that's not a proof...&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2010-12-21T22:05:13.403" Title="Is an unramified cover of the p-adics determined by its degree?" Tags="&lt;number-theory&gt;&lt;field-theory&gt;&lt;galois-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15110" PostTypeId="1" AcceptedAnswerId="15111" CreationDate="2010-12-21T21:40:15.733" Score="2" ViewCount="186" Body="&lt;p&gt;Hello, during a test i was given the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is 10+15+20+25+...+1490+1495+1500=?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After process of elimination and guessing, i came up with an answer.  Can anyone tell me a simple way to calculate this problem without having to try to actually add the numbers?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T00:45:46.150" LastActivityDate="2010-12-22T00:45:46.150" Title="How to find $10 + 15 + 20 + 25 + \dots + 1500$." Tags="&lt;homework&gt;&lt;algebra-precalculus&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="15111" PostTypeId="2" ParentId="15110" CreationDate="2010-12-21T21:45:33.230" Score="14" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Hint 1&lt;/strong&gt; -  Is there anything you can factor out of all the numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint 2&lt;/strong&gt; -  Do you know how to sum $1 + 2 + \cdots + n$ in a simple way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; - Your process in the comments is not wrong, but I don't feel like it's very intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do think it's immediately clear that all the numbers in your sum above are divisible by 5.  That is, we can rewrite&#xA;$$10 + 15 + 20 + \cdots + 1500 = 5(2 + 3 + 4 + \cdots + 300)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe not as obvious, but incredibly useful to know, is that&#xA;$$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally notice $2 + 3 + \cdots + 300$ is &lt;em&gt;almost&lt;/em&gt; $1 + 2 + \cdots + 300$, and you can use the above formula to finish the problem.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastEditorUserId="3537" LastEditorDisplayName="" LastEditDate="2010-12-22T00:11:56.420" LastActivityDate="2010-12-22T00:11:56.420" CommentCount="7" />
  <row Id="15113" PostTypeId="2" ParentId="15110" CreationDate="2010-12-21T21:47:23.917" Score="4" ViewCount="0" Body="&lt;p&gt;HINT: &#xA;$(10 + 1500) + (15 + 1495) + (20 + 1490) +  \cdots + (750 + 760) + 755$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-21T21:54:18.843" LastActivityDate="2010-12-21T21:54:18.843" />
  <row Id="15115" PostTypeId="2" ParentId="15110" CreationDate="2010-12-21T21:47:45.443" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;b&gt;Hint 3&lt;/b&gt; (Gauss): $10 + 1500 = 15 + 1495 = 20 + 1490 = \cdots = 1500 + 10$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-21T21:47:45.443" />
  <row Id="15116" PostTypeId="2" ParentId="15100" CreationDate="2010-12-21T21:55:25.377" Score="2" ViewCount="0" Body="&lt;p&gt;You're right about inclusion and exclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the number of elements in exactly one of the sets, you first count all $n(A) + n(B) + n(C)$ elements.  You disregard the $n(A \cap B)$ elements twice, as you need to disregard them from your count for both $A$ and $B$.  The logic is similar for $A \cap C$ and $B \cap C$.  Finally, you've disregarded the $n(A \cap B \cap C)$ elements in all three sets twice, so you have to correct the count by adding this quantity back in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet you can figure out the second one once you've carefully understood the first.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-21T21:55:25.377" CommentCount="1" />
  <row Id="15117" PostTypeId="2" ParentId="15109" CreationDate="2010-12-21T22:05:13.403" Score="3" ViewCount="0" Body="&lt;p&gt;This is slightly outside my domain.. but Proposition 5.4.11 of Gouvea's (excellent) book on p-adics says &quot;For each $f$ there is exactly one unramified extension of degree $f$. It can be obtained by adjoining to ${\mathbb Q}_p$ a primitive $p^f - 1$-st root of unity&quot;. So in your case $K_1$ and $K_2$ would both have to be extensions of ${\mathbb Q}_p$ by a primitive $p^f - 1$-st root of unity. Since any two such extensions contained within the same $\bar{{\mathbb Q}_p}$ must be the same, $K_1$ must be $K_2$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-21T22:05:13.403" CommentCount="5" />
  <row Id="15118" PostTypeId="2" ParentId="15108" CreationDate="2010-12-21T22:10:57.113" Score="1" ViewCount="0" Body="&lt;p&gt;This problem is very similar to &lt;a href=&quot;http://math.stackexchange.com/questions/12139/number-of-relations-that-are-both-symmetric-and-reflexive/&quot;&gt;Number of relations that are both symmetric and reflexive&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetric relation $R$ on a set $A$ is a subset $A\times A$. We can write $R$ as $B\cup C$, where $B$ is a subset of $\{(a,a)\mid a\in A\}$ and $C$ is a subset of $\{(b,c)\in A\times A\mid b\ne c\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note there are as many choices for $B$ as subsets of $A$, namely $2^n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To count the number of possible sets $C$, we use that $C$ is symmetric, meaning, if $(b,c)\in C$ then also $(c,b)\in C$. Now, let ${}[A]^2$ be the collection of subsets of $A$ of size 2. Note ${}[A]^2$ consists of subsets rather than ordered pairs. Given any $D$ subset of ${}[A]^2$, we can associate to it a set $C$ by setting $C=\{(b,c)\mid \{b,c\}\in D\}$. Note that $C$ is symmetric, since $\{b,c\}=\{c,b\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, any $C$ symmetric corresponds to a unique $D\subseteq[A]^2$, namely $\{\{b,c\}\mid (b,c)\in C\}$. This is why in $C$ we only allow pairs $(b,c)$ with $b\ne c$, so the resulting $D$ is a subset of ${}[A]^2$ and we have a correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have shown that there are as many $C$ as there are subsets $D$ of ${}[A]^2$. But $\displaystyle |[A]^2|={n\choose 2}=\frac{n(n-1)}2$, so the number of subsets is $2^{n\choose 2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since we can pair any $B$ with any $C$, we have that the number of binary symmetric relations on a set $A$ of size $n$ is precisely $$ 2^{n+{n\choose 2}}=2^{{n+1}\choose 2}.$$ &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-21T22:10:57.113" CommentCount="9" />
  <row Id="15119" PostTypeId="1" CreationDate="2010-12-21T22:15:13.457" Score="1" ViewCount="86" Body="&lt;p&gt;Taken from Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt;, 3rd edition (Chapter 1, Exercise 16):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $k\geq 3, x,y\in\mathbb{R}^k, |x-y|=d&amp;gt;0$, and $r&amp;gt;0$. Prove:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) If $2r &amp;gt; d$, there are infinitely many $z\in\mathbb{R}^k$ such that $|z-x|=|z-y|=r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If $2r=d$ there is exactly one such $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) If $2r &amp;lt; d$, there is no such $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrically (at least in $\mathbb{R}^3$), if $2r&amp;gt;d$ then there exists a circumference of points $z$ such that $|z-x|=|z-y|=r$. I guess that for $\mathbb{R}^k$ the set of such points forms a $(k-2)$-sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More formally, let $m$ be the midpoint between $x$ and $y$: $m=\frac{x+y}{2}$. Then we define the plane $\Pi$ as the set of $w\in\mathbb{R}^k$ that satisfy $(\lambda(x-y))w = (\lambda&amp;#39;(x-y))m$ for some $\lambda, \lambda&amp;#39;\in\mathbb{R}$. Let $S = \{p\in\Pi : |p-x|=r\}$. Because of its geometric interpretation it's obvious that $|p-x|=|p-y|$ for every $p\in S$, but I can't figure out how to prove it formally. $S$ can also be defined as the intersection of $\Pi$ and a $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) is obvious since $m$ is the only possible point that fulfills what's asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) follows from the fact that, if we set $S$ as the $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$, its radius is non-existent, therefore no such sphere exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach correct? How can I prove (a)? Thanks in advance.&#xA;(c) &lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditorDisplayName="" LastEditDate="2010-12-21T23:12:14.557" LastActivityDate="2010-12-21T23:12:14.557" Title="Proving a statement related to norms in euclidean spaces." Tags="&lt;real-analysis&gt;" CommentCount="5" />
  <row Id="15120" PostTypeId="2" ParentId="15079" CreationDate="2010-12-21T22:42:55.617" Score="3" ViewCount="0" Body="&lt;p&gt;I'm not sure of a deep connection; from what I gather, it is merely the case that the locker puzzle and problem 6 have a common subproblem, which is: prove that &quot;traversing through boxes&quot; produces a cycle, i.e., leads us to visit a node we have already visited. In the locker puzzle, there is an extra restriction that this cycle include the starting node. This restriction holds because the locker puzzle involves a permutation, while problem 6 does not necessarily involve a permutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I'm missing something, problems 6a and 6b can be solved simultaneously using the &quot;traverse through boxes&quot; technique starting at array index 1, despite the comment that &quot;Part (b) seems to be a lot harder than part (a), and it apparently requires entirely different techniques&quot;. Since the hedge words &quot;seems&quot; and &quot;apparently&quot; were used, we can't say that the comment is false, but nonetheless, I believe it would be false without those hedge words. The key idea is that, as explained in the PDF on the locker puzzle, &quot;player $ B_i $ never opens a locker (other than locker i) without first finding its number, so each time he opens a new locker he must find either his own number or the number of another unopened locker&quot;. The same basic reasoning applies to problem 6, with just slight modifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 6 can be made to look a little more like the locker puzzle by taking a[1] out of the picture. We can consider b = {a[2], ... , a[n]} as the &quot;real&quot; array with m = n - 1 elements, and the value a[1] as the index of the start node. We can further relabel {2, ... , n} as {1, ... , m} if desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All we care about is finding a cycle, that is, coming across a value that is an index we have already visited. This is equivalent to finding a duplicate in array a, since the indices we visited are precisely array values in a. We can fulfill the time and memory requirements using a boolean array of size m. The boolean array allows us to see if we have visited an index yet, and &lt;strike&gt;of course has constant memory and&lt;/strike&gt; is fast. (&lt;strong&gt;Correction:&lt;/strong&gt; the boolean array would have linear memory, not constant. Evidently there is a better way to solve this.) We are guaranteed not to visit any array index twice, so the algorithm has linear time. If we didn't think of a boolean array, we might be tempted to use some sort of a list structure with fast insert and find, but a boolean array is faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To tie the problems together further, we can interpret the data as a directed graph where each vertex has outdegree 1. In the locker puzzle, the indegree for each vertex is also 1, but problem 6 has no such restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outdegree being 1 guarantees that we can traverse from one vertex to another without any ambiguity. Beyond that, we need to guarantee a cycle will occur regardless of where we start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A permutation is partitioned into cycles, so this allows us to conclude for the locker puzzle that we will find our cycle precisely when we get back to the starting node. For problem 6, we are guaranteed to find some cycle, because otherwise we would have to go on endlessly, but the array is finite, so that is absurd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe my answer should be made more rigorous, but I hope it is at least satisfactory and accurate. If not, let me know, and I'll try to fix it. I admit that after looking at this problem for an extended period of time, my mind is getting a bit foggy..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Oh I think I figured out how to solve 6b with constant memory. You can follow n - 1 links, thus covering n (not necessarily distinct) array indices of a, and by this time you are guaranteed to be in the middle of a cycle. Note your current array index. Then go around until you see that index again, which will give you the period of the cycle. Then using this you can find the &quot;entry node&quot; of the cycle, which will be your duplicate. (Use modular arithmetic.) I haven't worked it out in detail but I believe that's an accurate sketch.&lt;/p&gt;&#xA;" OwnerUserId="4604" LastEditorUserId="4604" LastEditorDisplayName="" LastEditDate="2010-12-22T00:26:12.907" LastActivityDate="2010-12-22T00:26:12.907" CommentCount="8" />
  <row Id="15121" PostTypeId="2" ParentId="15061" CreationDate="2010-12-21T22:56:40.207" Score="3" ViewCount="0" Body="&lt;p&gt;Apply Eisenstein's criterion to ${1 \over p^{n-1}}x^nf({p \over x})$.&lt;/p&gt;&#xA;" OwnerUserId="3035" LastActivityDate="2010-12-21T22:56:40.207" CommentCount="3" />
  <row Id="15123" PostTypeId="2" ParentId="15110" CreationDate="2010-12-22T00:28:52.340" Score="1" ViewCount="0" Body="&lt;p&gt;This is probably the method C. F. Gauss would have used ;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Factor out 5:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(2+3+4+5+...+297+298+299+300)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notice: $302/2 = 151$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5((2+300)+(3+299)+(4+298)+(5+297)+...(149+153)+(150+152)+(151))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(302+302+302+302+...+302+302+151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There are $(150-2+1)$ 302s, so:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5((150-2+1)(302) + 151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5((149)(302) + 151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(44998 + 151)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(45149)$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer is $225745$.&lt;/p&gt;&#xA;" OwnerUserId="2736" LastActivityDate="2010-12-22T00:28:52.340" CommentCount="2" />
  <row Id="15124" PostTypeId="1" CreationDate="2010-12-22T00:52:10.653" Score="4" ViewCount="83" Body="&lt;p&gt;Let $X \geq 1$ be an integer r.v. with $E[X]=\mu$. Let $X_i$ be a sequence of iid rvs with the distribution of $X$. On the integer line, we start at $0$, and want to know the expected position after we first cross $K$, which is some fixed integer. Each next position is determined by adding $X_i$ to the previous position. So the question is, if we stop this process after the first time $\tau$ for which $Y_{\tau}=\sum_{i=1}^{\tau}X_i &amp;gt; K$, that is, after the first time it crosses $K$, then what is $E[Y_{\tau}-K]$?.   Can we get a bound of $O(\mu)$?&lt;/p&gt;&#xA;" OwnerUserId="5031" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-22T02:18:57.000" LastActivityDate="2010-12-22T15:45:38.377" Title="A type of stochastic jump process" Tags="&lt;stochastic-processes&gt;" AnswerCount="2" />
  <row Id="15125" PostTypeId="1" CreationDate="2010-12-22T01:24:46.087" Score="-1" ViewCount="103" Body="&lt;p&gt;If $f(x) = \sqrt{ 4\sin x + 2 }$, then $f&amp;#39;(0) =$?&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T01:34:32.333" LastActivityDate="2010-12-22T03:20:34.763" Title="Derivative of $\sqrt{4 \sin x + 2}$ at $0$" Tags="&lt;calculus&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="15126" PostTypeId="2" ParentId="15125" CreationDate="2010-12-22T01:28:55.307" Score="5" ViewCount="0" Body="&lt;p&gt;Do you know the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule&quot; rel=&quot;nofollow&quot;&gt;chain rule?&lt;/a&gt;  What have you tried?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T01:28:55.307" />
  <row Id="15127" PostTypeId="2" ParentId="15125" CreationDate="2010-12-22T01:42:41.110" Score="2" ViewCount="0" Body="&lt;p&gt;By definition, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{\sqrt{4\sin(h)+2}-\sqrt2}{h}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can &quot;rationalize&quot; the numerator, use the fact that a limit of a product is the product of the limits, use continuity of the square root and sine functions, and use the fact that $\lim_{h\to0}\frac{\sin(h)}{h}=1$ to finish.  Of course, you would get the same answer by deriving a general formula for $f&amp;#39;(x)$ using the chain rule and plugging in $x=0$, as Ross hints.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T01:42:41.110" CommentCount="4" />
  <row Id="15128" PostTypeId="2" ParentId="15079" CreationDate="2010-12-22T01:44:59.563" Score="1" ViewCount="0" Body="&lt;p&gt;Since you requested for more puzzles, here is one which is an algorithmic puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given an array of $\displaystyle 2n$ elements: $$a_1, a_2, \dots, a_n , b_1, b_2, \dots, b_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to rearrange the elements of the array to look like: $$b_1, a_1, b_2, a_2, \dots, b_n , a_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is trivial by using an extra array, but the hard part of the puzzle is that you need to do it in $\displaystyle \mathcal{O}(1)$ (or to be pedantic, $\displaystyle \mathcal{O}(\log n))$ space and $\displaystyle \mathcal{O}(n)$ time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume each element of the array takes up constant space)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T02:10:36.407" LastActivityDate="2010-12-22T02:10:36.407" CommentCount="2" />
  <row Id="15129" PostTypeId="1" AcceptedAnswerId="15130" CreationDate="2010-12-22T01:45:21.120" Score="1" ViewCount="58" Body="&lt;p&gt;how can be prove that $\max(f(n),g(n)) = \theta(f(n)+g(n))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;though the big O case is simple since $\max(f(n),g(n)) \leq f(n)+g(n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit : where $f(n)$ and $g(n)$ are asymptotically nonnegative functions.&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T02:09:42.757" LastActivityDate="2010-12-22T02:09:42.757" Title="how can be prove that $\max(f(n),g(n)) = \theta(f(n)+g(n))$" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15130" PostTypeId="2" ParentId="15129" CreationDate="2010-12-22T01:58:58.040" Score="2" ViewCount="0" Body="&lt;p&gt;Hint...: $2 + 100 \le 100 + 100$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T01:58:58.040" CommentCount="6" />
  <row Id="15131" PostTypeId="2" ParentId="15124" CreationDate="2010-12-22T02:18:47.747" Score="0" ViewCount="0" Body="&lt;p&gt;Edit: This answer shows what happens when $X$ could be zero (contrary to the question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K = 1$, and $X = M$ ($M &gt; 1$) with probability $p$ and $0$ otherwise. Thus $E[Y_\tau - K] = M-1$ whereas $E[X] = pM$. Choosing $M = n+1$ and $p = 1/n(n+1)$ we get $E[X] = 1/n$ and $E[Y_\tau - K] = n$, so a bound of $O(\mu)$ isn't possible.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-22T15:45:38.377" LastActivityDate="2010-12-22T15:45:38.377" CommentCount="1" />
  <row Id="15132" PostTypeId="2" ParentId="15101" CreationDate="2010-12-22T02:21:54.313" Score="1" ViewCount="0" Body="&lt;p&gt;Enumerate over all possible angles (under some reasonable discretization), rotate one of the images, and calculate the $L_2$ distance (sum of squared differences); the angle at which the distance is minimal is the one you seek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even better idea is to calculate the two-dimensional Fourier transform, and now do the same with shifts instead of rotations (rotation in the time domain corresponds to shift in the frequency domain). Maybe you can reduce the amount of shifts tested by looking at some prominent peeks of the spectrum and trying to match them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to implement these suggestions in Matlab, but I'm sure you can find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both solutions assume that the image is &quot;continuous&quot; with reasonable &quot;modulus of continuity&quot;. If the image is random that the methods wouldn't work.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T02:21:54.313" CommentCount="2" />
  <row Id="15133" PostTypeId="1" AcceptedAnswerId="15140" CreationDate="2010-12-22T02:21:59.323" Score="1" ViewCount="81" Body="&lt;p&gt;Let $G$ be a set of elements and $*$ a binary operation defined on the elements of $G$. Then $G$ is a group with respect to $*$ if the following holds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a*(b*c) = (a*b)*c$ for all $a,b,c \in G$.&lt;/li&gt;&#xA;&lt;li&gt;There exists a left identity in $G$.&lt;/li&gt;&#xA;&lt;li&gt;For each $a \in G$, there exists a left inverse.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We could replace &quot;left&quot; with &quot;right.&quot; Is there any relationship between this formulation and &lt;a href=&quot;http://en.wikipedia.org/wiki/Burnside%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Burnside's Lemma&lt;/a&gt;? In other words, does the &quot;left&quot; or &quot;right&quot; criteria for groups come in handy when dealing with Burnside's Lemma (e.g could the number of orbits be the &quot;left&quot; part and the number of fixed points be the &quot;right&quot; part)?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-22T03:28:47.417" Title="Criteria for Groups and Burnside's Lemma" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="15134" PostTypeId="1" AcceptedAnswerId="15175" CreationDate="2010-12-22T02:31:27.850" Score="2" ViewCount="151" Body="&lt;p&gt;Are there any good algorithms which can be used to construct a prime greater than $n$, for arbitrary $n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some brute force approaches: for example, factoring $n!+1$. However, I'm looking for something significantly more efficient; preferably, polynomial in $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if anyone knows of a proof showing a bound on the speed of such an algorithm (say, $n^4$), I would like to hear about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It appears that polynomial in $ln(n)$ makes more sense. &lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="796" LastEditorDisplayName="" LastEditDate="2010-12-22T17:04:14.773" LastActivityDate="2010-12-22T17:04:14.773" Title="Algorithm for constructing primes" Tags="&lt;prime-numbers&gt;&lt;elementary-number-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="15135" PostTypeId="1" AcceptedAnswerId="15137" CreationDate="2010-12-22T02:39:05.087" Score="3" ViewCount="72" Body="&lt;p&gt;I keep trying to solve it but I seem to be making an error somewhere: &#xA; $$4-\left|\frac{5y}{3}+4\right| &amp;gt; \frac25$$&lt;/p&gt;&#xA;" OwnerUserId="5032" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T03:01:58.733" LastActivityDate="2010-12-22T03:54:47.373" Title="what is the process to solving this absolute value inequality: $4-\left|\frac{5y}{3}+4\right| &gt; \frac25$?" Tags="&lt;algebra-precalculus&gt;&lt;inequality&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15136" PostTypeId="2" ParentId="15134" CreationDate="2010-12-22T02:40:46.210" Score="5" ViewCount="0" Body="&lt;p&gt;Well theoretically speaking, there is a &lt;em&gt;deterministic&lt;/em&gt; $\mathcal{O}(n \log ^{12} n)$ time algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Bertrand's postulate, there is a prime $p$ guaranteed to satisfy $n \le p \lt2n$. (In practice, you should hit one much earlier than $2n$, owing to the prime number theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the AKS primality testing program to test each of the numbers in the range $[n, 2n)$ and you will find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try some sieving etc as suggested in the answers to this other question here: &lt;a href=&quot;http://math.stackexchange.com/questions/8865/the-least-prime-greater-than-2000&quot;&gt;The least prime greater than 2000&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T02:40:46.210" CommunityOwnedDate="2010-12-22T02:44:43.677" CommentCount="2" />
  <row Id="15137" PostTypeId="2" ParentId="15135" CreationDate="2010-12-22T03:14:39.747" Score="2" ViewCount="0" Body="&lt;p&gt;$4-\left|\frac{5y}{3}+4\right| &amp;gt; \frac25$ can be rewritten as $-\left|\frac{5y}{3}+4\right| &amp;gt; \frac{-18}{5}$ or $\left|\frac{5y}{3}+4\right| &amp;lt; \frac{18}{5}$. From here, represent the expression as $\frac{-18}{5} &amp;lt; \frac{5y}{3}+4\ &amp;lt; \frac{18}{5}$ and proceed to solve it using arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-22T03:54:47.373" LastActivityDate="2010-12-22T03:54:47.373" CommentCount="2" />
  <row Id="15138" PostTypeId="2" ParentId="15125" CreationDate="2010-12-22T03:20:34.763" Score="0" ViewCount="0" Body="&lt;p&gt;$$f&amp;#39;(x)=\frac{4\cos x}{2\sqrt{4\sin x+2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(0)=\frac{4\cdot1}{2\sqrt{4\cdot0+2}}=\frac{4}{2\sqrt{2}}=\sqrt{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the difficulty? It does not even involve a $\frac{0}{0}$ like $\sin{x}/x$ does.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastActivityDate="2010-12-22T03:20:34.763" />
  <row Id="15139" PostTypeId="1" AcceptedAnswerId="15168" CreationDate="2010-12-22T03:27:15.800" Score="6" ViewCount="113" Body="&lt;p&gt;What values of $Z_1$ and $Z_2$ make the five regions of the unit circle, shown below, equal in area? $\overline{Z_1}$ and $\overline{Z_2}$ are conjugates of $Z_1$ and $Z_2$; in other words they lie directly across the real axis from their counterparts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8cmP9.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2010-12-22T07:56:20.363" Title="Two points on circle resulting in 5 equal regions " Tags="&lt;geometry&gt;&lt;euclidean-geometry&gt;&lt;complex-numbers&gt;&lt;circle&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15140" PostTypeId="2" ParentId="15133" CreationDate="2010-12-22T03:28:47.417" Score="0" ViewCount="0" Body="&lt;p&gt;First, I assume you know that the two definitions yield objects which are not just isomorphic, but identical: from the 'left' definition it follows that the 'left' identity is also a 'right' identity and ergo it is 'the' identity, and the 'left' inverse is also a 'right' inverse and ergo 'the' inverse; and vice versa follows for the 'right' definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would say that the formulation of the axioms has little to do with Burnside's lemma per se; and instead is intended to provide a minimum set of requirements: only one side, left or right, is required to deduce that the resulting entities (identity, inverse) are actually /both-sided/. This is in contrast to other types of structures; e.g., modules, where a left module is not necessarily a right module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, historically, the notation for the application of a group operation has varied between right to left and left to right over time. For example in W. R. Scott's &quot;Group Theory&quot; (1964), assuming x in S and g,h in G where G acts on S, the author writes &quot;xhg&quot; for what would now more typically would be written as &quot;ghx&quot;. This has other terminological effects such as the precise meaning of &quot;left coset&quot; or &quot;right coset&quot;, but of course the essential theory is the same.&lt;/p&gt;&#xA;" OwnerUserId="4980" LastActivityDate="2010-12-22T03:28:47.417" />
  <row Id="15141" PostTypeId="1" CreationDate="2010-12-22T03:55:20.507" Score="4" ViewCount="56" Body="&lt;p&gt;I'm asking in the spirit of these &lt;a href=&quot;http://math.stackexchange.com/questions/13237&quot;&gt;two&lt;/a&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/13253&quot;&gt;questions&lt;/a&gt;: can bitwise operations (AND, OR, XOR) be expressed in terms of other (more familiar?) functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been playing around with the bitwise operations in &lt;em&gt;Mathematica&lt;/em&gt; a while back, and was at first struck by this identity for bitwise NOT: &lt;code&gt;~ n == -1 - n&lt;/code&gt; (I'll use C-ish notation/syntax for this question, but I'm using &lt;a href=&quot;http://reference.wolfram.com/mathematica/tutorial/IntegerAndNumberTheoreticalFunctions.html#28221&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Mathematica&lt;/em&gt;'s definitions&lt;/a&gt;, which assume a two's complement representation for negative integers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try as I might, I have not managed to figure out alternative ways of expressing the other bitwise operations. I have also tried Using The Fabulous Search Engine to see if other people have looked into this, but I probably am not using the right keywords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed that the implementation of the bitwise operators in &lt;em&gt;Mathematica&lt;/em&gt; satisfies &lt;a href=&quot;http://mathworld.wolfram.com/deMorgansLaws.html&quot; rel=&quot;nofollow&quot;&gt;de Morgan's theorem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i | j == -1 - ((-1 - i) &amp;amp; (-1 - j))&#xA;&#xA;i &amp;amp; j == -1 - ((-1 - i) | (-1 - j))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the following identity for bitwise XOR is satisfied as well:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i ^ j == (i | j) &amp;amp; (-1 - (i &amp;amp; j))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I suppose then that any alternate expression for the bitwise operators would have to satisfy these identities as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am not intending to replace the bitwise expressions in actual programming, of course; they are already optimized, and there is really no practical reason to replace them with &quot;more mathematical&quot; expressions. I am asking merely for curiosity's sake.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References and sundry information will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-22T04:55:32.470" Title="Expressing bitwise operations in terms of other functions" Tags="&lt;functions&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15142" PostTypeId="2" ParentId="15139" CreationDate="2010-12-22T03:57:15.527" Score="2" ViewCount="0" Body="&lt;p&gt;We have segment $Z_2\overline{Z_2}$ is $\frac15$ of the area.  The area of a segment is $\frac{\theta-\sin(\theta)}{2}$.  Setting this to $\frac{\pi}{5}$ gives the angle at the center to be $\theta$=2.113139.  Segment $Z_1\overline{Z_1}\overline{Z_2}$ then has area $\frac{2\pi}{5}$, so the angle is 2.824797.  So $Z_2=\exp{(\pi-1.056569)i}$ and $Z_1=\exp{(.74890i)}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T06:38:33.223" LastActivityDate="2010-12-22T06:38:33.223" CommentCount="3" />
  <row Id="15143" PostTypeId="1" CreationDate="2010-12-22T04:00:48.330" Score="3" ViewCount="40" Body="&lt;p&gt;Differential equations of the form $M\,dx+N\,dy=0$ such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ are said to be &lt;i&gt;exact&lt;/i&gt;, because the left hand side of our equation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exact_differential&quot; rel=&quot;nofollow&quot;&gt;exact differential&lt;/a&gt; of some function &lt;i&gt;f&lt;/i&gt;, and the DE has a solution of the form $f(x,y)=c$, where &lt;i&gt;c&lt;/i&gt; is a constant. Standard textbooks give us a procedure for solving such equations that essentially amounts to using the identity  $f=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$. Specifically, we integrate &lt;i&gt;M&lt;/i&gt; (respectively &lt;i&gt;N, &amp;amp;c.&lt;/i&gt;) with respect to &lt;i&gt;x&lt;/i&gt; to get $f=\int\frac{\partial f}{\partial x}\: dx+C(y)$, and then differentiate this with respect to &lt;i&gt;y&lt;/i&gt; and do algebra to find &lt;i&gt;C'(y)&lt;/i&gt;, which we can then integrate to find &lt;i&gt;C(y)&lt;/i&gt;, which we can substitute back into our expression for &lt;i&gt;f&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It occurred to me that this technique should generalize---that is, when instead of $M\,dx+N\,dy=0$  such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, we have $\sum_{i}M_{i}\,dx_{i}=0$ where the $M_{i}$ satisfy the criteria for being an exact differential, then we can iteratively apply the same procedure to solve the equation. And so I came up with the following---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Theorem (?).&lt;/b&gt; A differential equation of the form $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$ for some function &lt;i&gt;f&lt;/i&gt; has an implicit solution of the form $f=\sum_{i=1}^{n}a_{i}=c$ where $a_{0}=0$, $a_{i}=\int\left(\frac{\partial f}{\partial x_{i}}-\frac{\partial}{\partial x_{i}}\left(\sum_{j=1}^{i-1}a_{j}\right)\right)\: dx_{i}$ for $i\in\mathbb{N}_{+}$, and &lt;i&gt;c&lt;/i&gt; is an arbitrary constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;i&gt;Proof (?).&lt;/i&gt;&lt;/b&gt; By induction on the number of variables &lt;i&gt;n&lt;/i&gt;. The theorem is true for &lt;i&gt;n&lt;/i&gt;=1 because $f=\int\,(\frac{\partial f}{\partial x}-0)\,dx$ by the fundamental theorem of calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the induction, we need to show that if $f(x_{1},...,x_{n})=\sum_{i=1}^{n}a_{i}=c$&#xA;is a solution to $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$, then $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}=c$ is a solution to $\sum_{i=1}^{n+1}\frac{\partial f}{\partial x_{i}}=0$. Think of $f(x_{1},...,x_{n})$ as being the &quot;special case&quot; of $f(x_{1},...,x_{n+1})$ where $x_{n+1}$ is being &quot;treated like a constant.&quot; (My thinking here is not nearly as precise as it should be and I'm overloading the function name &lt;i&gt;f&lt;/i&gt;, but hopefully you understand the intuition to which I am appealing.) So we can say that $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+C(x_{n+1})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we proceed analogously as in the two-variable case. Differentiating by $x_{n+1}$ and applying algebra yields $C&amp;#39;(x_{n+1})=\frac{\partial f}{\partial x_{n+1}}-\frac{\partial}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)$, and then integrating with respect to $x_{n+1}$ yields $C(x_{n+1})=\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substituting into our earlier expression for $f(x_{1},...,x_{n+1})$ we get $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$, which by virtue of the definition of $a_{i}$ is equivalent to $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}$. But by the principle of induction, this is &lt;i&gt;quod erat demonstrandum&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Example&lt;/b&gt;. In the &lt;i&gt;n&lt;/i&gt;=3 case (and naming our variables x, y, and z), we get $f(x,y,z)=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$&#xA;$+\int(\frac{\partial f}{\partial z}-\frac{\partial}{\partial z}(\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy))\, dz$, which is seen to be an identity by performing the integrations. What's going on is that $a_{1}=\int\frac{\partial f}{\partial x}\: dx=f$, and all following terms are zero by design, &lt;i&gt;e.g.&lt;/i&gt;, $a_{2}=f-f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is (and I offer my sincerest apologies if this is not an appropriate question, or if my poor exposition has rendered my intentions virtually unreadable) &lt;b&gt;does this seem basically correct&lt;/b&gt;, or am I doing something wrong? And &lt;b&gt;how do I fix my sloppy reasoning in the inductive step&lt;/b&gt; (&quot;special case&quot; ... &quot;treated like a constant&quot;)? I would be much obliged for any input.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2010-12-22T04:00:48.330" Title="Does the method for solving exact DEs generalize like this?" Tags="&lt;differential-equations&gt;" />
  <row Id="15144" PostTypeId="2" ParentId="15141" CreationDate="2010-12-22T04:55:32.470" Score="1" ViewCount="0" Body="&lt;p&gt;Exclusive OR is addition in $\mathbb{Z}_2^n$. The AND operation corresponds to set intersection, OR to set union, NOT to complementation (and XOR to symmetric difference).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T04:55:32.470" CommentCount="1" />
  <row Id="15145" PostTypeId="1" AcceptedAnswerId="15146" CreationDate="2010-12-22T05:14:48.110" Score="5" ViewCount="241" Body="&lt;p&gt;What is a &quot;simpler&quot; formula for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{3}^{n} \frac{(k-1)(k-2)(k-3)}{6}$$&lt;/p&gt;&#xA;" OwnerUserId="5035" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T06:03:05.513" LastActivityDate="2010-12-22T16:58:59.850" Title="A general formula for  $\sum (k-1)(k-2)(k-3)$?" Tags="&lt;algebra-precalculus&gt;&lt;arithmetic&gt;" AnswerCount="6" CommentCount="10" FavoriteCount="1" />
  <row Id="15146" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T05:23:04.393" Score="12" ViewCount="0" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k(k-1)(k-2)(k-3) - (k-1)(k-2)(k-3)(k-4) = 4(k-1)(k-2)(k-3)$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T05:23:04.393" CommentCount="5" />
  <row Id="15147" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T05:23:51.053" Score="4" ViewCount="0" Body="&lt;p&gt;There is no need for a general formula for this special case, all one needs is to use the formulas for $\sum {k} , \sum{k^2} \text{ and } \sum{k^3}$. Just expand the expression and use the simpler known sums.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-22T06:14:24.790" LastActivityDate="2010-12-22T06:14:24.790" CommentCount="3" />
  <row Id="15148" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T05:28:19.383" Score="6" ViewCount="0" Body="&lt;p&gt;I really like Moron's hint, but another way to simplify the sum is to reindex with $j=k-2$, and use $(j+1)j(j-1)=j^3-j$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T05:28:19.383" CommentCount="7" />
  <row Id="15149" PostTypeId="2" ParentId="15079" CreationDate="2010-12-22T05:29:40.707" Score="2" ViewCount="0" Body="&lt;p&gt;You might also enjoy this one, which is, I think, easier than the two you have linked to: you are starting with a pack of $n$ cards, numbered from 1 to $n$, but in a random order. You now play the following game: you look at the top card, numbered $i$, say, then you put the card into the $i$-th position from top. Repeat the whole procedure. Clearly, if at some point the top card it numbered 1, nothing interesting happens any more and the position of the cards never changes after that point. Prove that, independently of the starting configuration, after finitely many steps, the 1 will turn up on the top.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-22T05:29:40.707" CommentCount="2" />
  <row Id="15150" PostTypeId="1" CreationDate="2010-12-22T05:49:28.787" Score="3" ViewCount="64" Body="&lt;p&gt;I want to show that $G$ is open in $X$ iff $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This direction $\rightarrow$ is easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck in showing that $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$ implies $G$ is open in $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess the game is to find suitable A and apply the hypothesis, I tried taking A as the interior of G and also as the interior of the complement of G but didn't work (at least I didn't see it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2010-12-22T05:54:39.813" Title="Prove G is an open set" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15151" PostTypeId="2" ParentId="15150" CreationDate="2010-12-22T05:54:33.683" Score="3" ViewCount="0" Body="&lt;p&gt;Here are some hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G$ is open if and only if the complement $G^C$ is closed.&lt;/li&gt;&#xA;&lt;li&gt;A set is closed if and only if it contains all its limit points.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, see whether the assumption that $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for all $A$ and that $G^C$ is not closed (i.e. that there exists a limit point of $G^C$ that is not in $G^C$) leads you to a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-22T05:54:33.683" CommentCount="1" />
  <row Id="15152" PostTypeId="2" ParentId="15150" CreationDate="2010-12-22T05:54:39.813" Score="3" ViewCount="0" Body="&lt;p&gt;Trying $A=X\setminus G$ should do the trick.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T05:54:39.813" />
  <row Id="15153" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T06:11:03.543" Score="8" ViewCount="0" Body="&lt;p&gt;Show that ${k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_4 = k$ in non-negative integers.  Then $\sum_{k=0}^{n-4} {k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_5 = n-4$ in non-negative integers, which is...?&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-22T06:11:03.543" CommentCount="1" />
  <row Id="15154" PostTypeId="1" CreationDate="2010-12-22T06:11:50.697" Score="0" ViewCount="54" Body="&lt;p&gt;Ronald's federal income tax for the year 2000 was $34597. Find his exact taxable income m for 2000 by solving the equation 2371.7 + 0.31(m - 10570)=34597. (Please round your answer to two decimal places.) i got M=114522.58  but my online test is saying its incorrect ! HELP !&lt;/p&gt;&#xA;" OwnerUserId="5037" LastActivityDate="2010-12-22T15:44:50.910" Title="algebra 2 solving linear equations" Tags="&lt;algebra&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15155" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T06:12:53.650" Score="3" ViewCount="0" Body="&lt;p&gt;Consider the following sum $$\sum_{m=0}^n \binom{m}{k}.$$ It counts the number of possibilities to select $k$ elements from at most $n$ elements. Some choices are counted multiple times, for example $\{0,\ldots,k-1\}$ is counted once for each $m \in [k-1,n]$. So it's natural to distinguish among those by tagging them with $m$ somehow. The best way to do that is to &lt;i&gt;add&lt;/i&gt; the element $m+1$. The result is a choice of $k+1$ elements from $n+1$, and so $$\sum_{m=0}^n \binom{m}{k} = \binom{n+1}{k+1}.$$ From this formula one can extract (using linear algebra) the usual formulas for $\sum_{m=0}^n m^k$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T06:12:53.650" />
  <row Id="15156" PostTypeId="1" CreationDate="2010-12-22T06:26:09.490" Score="1" ViewCount="39" Body="&lt;p&gt;A question from Dugundji's book which I don't even understand the statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Describe the relative topology of $\{z: |z|=1\}$ as a subspace of ${\mathbb{R}}^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do they mean by &quot;describe&quot;? I don't understand what they are asking for. Can you please clarify?&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2010-12-22T06:33:43.523" Title="Describe the relative topology of the unit circle as a subspace of the plane." Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15157" PostTypeId="2" ParentId="15156" CreationDate="2010-12-22T06:33:43.523" Score="2" ViewCount="0" Body="&lt;p&gt;I agree that the exercise is not very precisely stated, and I don't myself know exactly what Dugundji was intending here (and not, I think, due to any lack of understanding of basic topology).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I can do is to suggest a reasonable problem which &lt;em&gt;could&lt;/em&gt; be what Dugundji had in mind: show that &quot;open intervals&quot; (&lt;b&gt;edit&lt;/b&gt;: Jonas Meyer's terminology &lt;em&gt;open arcs&lt;/em&gt; is better) on the unit circle form a base for the subspace topology.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-22T06:33:43.523" CommentCount="2" />
  <row Id="15158" PostTypeId="1" AcceptedAnswerId="15163" CreationDate="2010-12-22T06:37:48.200" Score="0" ViewCount="33" Body="&lt;p&gt;(1/8)(3m - 5) = (6/5)(6m - 2) + 5...Type your answer as a fraction. Do not use decimals!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;**i got -129/273 is this correct ?&lt;/p&gt;&#xA;" OwnerUserId="5037" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T07:45:40.420" LastActivityDate="2010-12-22T07:45:40.420" Title="algebra 2 fraction and linear equations " Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="15159" PostTypeId="2" ParentId="8140" CreationDate="2010-12-22T06:41:52.297" Score="0" ViewCount="0" Body="&lt;p&gt;This was my project not so long ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved it in more than one way. But I will present you the easiest of the methods I used. Use quick fill inside of a region. Then find the lines that border the region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It fails in certain situations, however, in most cases, those situations are not that serious anyway. If this method fails for you, post back and I will share the other methods that I used. &lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2010-12-22T06:41:52.297" CommentCount="3" />
  <row Id="15160" PostTypeId="1" CreationDate="2010-12-22T06:49:30.653" Score="1" ViewCount="46" Body="&lt;p&gt;Two questions from Dugundji's book (not hw, just practice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Let $Y_{1}, Y_{2}$ be subspaces of $X$ and $A \subset Y_{1} \cap Y_{2}$. Assume that $A$ is open in $Y_{1}$ and open in $Y_{2}$. Prove A is open in $Y_{1} \cup Y_{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please give a hint for this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) a. Let $D$ be dense in $X$. Give an example to show that $D \cap A$ need not to be dense in $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't we take $X = \mathbb{R}$, $D=\mathbb{Q}$ then $D$ in dense in $X$. Now take $A =$ irrationals, since the empty set is closed then it cannot be dense in $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. If $A$ is dense in $B \subset X$ then $A$ is dense in $\overline{B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V \subset \overline{B}$ be an open set, then by definition of subspace topology we have $V = C \cap \overline{B}$ where $C$ is an open subset of $X$. Now consider $C \cap B$ ,this is an open set in $B$ so $A$ intersects this set and hence $A$ is dense in $\overline{B}$. What bothers me, is how do we know that $C \cap B$ is non-empty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5036" LastEditorUserId="5036" LastEditorDisplayName="" LastEditDate="2010-12-22T07:34:27.417" LastActivityDate="2010-12-22T07:34:27.417" Title="Closed, open subspaces, dense subsets" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="15161" PostTypeId="2" ParentId="15160" CreationDate="2010-12-22T07:04:39.743" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; For the first one Recall the definition of $A$ being open in the relative topology: if there exists some $A&amp;#39;$ open in $X$ such that $A&amp;#39;\cap Y_1 = A$.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-22T07:04:39.743" CommentCount="5" />
  <row Id="15162" PostTypeId="1" AcceptedAnswerId="15167" CreationDate="2010-12-22T07:17:27.120" Score="3" ViewCount="94" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off: I barely have any set theoretic knowledge, but I read a bit about cardinal arithmetic today and the following idea came to me, and since I found it kind of funny, I wanted to know a bit more about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A$ is the set of all real positive sequences that either converge to $0$ or diverge to $\infty$, we put an equivalence relation &quot;$\sim$&quot; on $A$ defined as $a \sim b$ iff $\lim \frac a b \in \mathbb R ^+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $B$ is the set of all infinite cardinals, can we associate to every equivalence class $[a]$ in $A/\sim$ a cardinal $p([a])$ or to every cardinal $\lambda$ an equivalence class $[q(\lambda)]$ in such a way that the map $p: A/\sim \to B$, or $q: B \to A/\sim$ is a &quot;homomorphism&quot;? That is, so that we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p([a] + [b]) = p([a]) + p([b]) $$ or $$ [q(\lambda + \mu)] = [q(\lambda)] + [q(\mu)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, could this map even be surjective, injective or an &quot;isomorphism&quot;? (I don0t know how many cardinals there are of course...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It at least superficially seems to make some sense, since for cardinals $\lambda, \mu$ we have $\lambda + \mu = \max\{\lambda, \mu\}$ and the same is true for the classes of sequences, if we order them by $a &amp;lt; b \Leftrightarrow \lim \frac a b = 0$&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-22T07:54:37.783" LastActivityDate="2010-12-25T21:29:36.747" Title="&quot;Homomorphism&quot; from set of sequences to cardinals?" Tags="&lt;set-theory&gt;&lt;cardinals&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15163" PostTypeId="2" ParentId="15158" CreationDate="2010-12-22T07:19:53.417" Score="0" ViewCount="0" Body="&lt;p&gt;Assuming you mean: $\frac{1}{8}(3m - 5) = \frac{6}{5}(6m - 2) + 5$&lt;br&gt;&#xA;You would solve as:&lt;br&gt;&#xA;$$40 * \frac{1}{8}(3m - 5) = 40 * (\frac{6}{5}(6m - 2) + 5)$$&#xA;$$5 * (3m - 5) = (48 * (6m - 2)) + 200$$&#xA;$$15m - 25 = 288m - 96 + 200$$&#xA;$$15m - 25 = 288m + 104$$&#xA;$$-129 = 273m$$&#xA;$$m = -\frac{129}{273} = -\frac{43}{91}$$    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming you mean $\frac{1}{8(3m - 5)} = \frac{6}{5(6m - 2)} + 5$&lt;br&gt;&#xA;You would solve as:&lt;br&gt;&#xA;$$\frac{1}{24m - 40} = \frac{6}{30m - 10} + 5$$&#xA;$$(30m - 10)(24m - 40) * \frac{1}{24m - 40} = (30m - 10)(24m - 40) * (\frac{6}{30m - 10} + 5)$$&#xA;$$30m - 10 = 24m - 40 + (5*(30m - 10)(24m - 40))$$&#xA;$$30m - 10 = 24m - 40 + (5 * (720m^2 - 1140m + 400))$$&#xA;$$30m - 10 = 24m - 40 + 3600m^2 - 7200m + 2000$$&#xA;$$0 = 3600m^2 - 7206m + 1970$$&#xA;$$m = \frac{1201 \pm \sqrt{654401}}{1200}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I would double check the work on both of these though, I'm tired =/) &lt;/p&gt;&#xA;" OwnerUserId="5038" LastEditorUserId="5038" LastEditorDisplayName="" LastEditDate="2010-12-22T07:33:32.613" LastActivityDate="2010-12-22T07:33:32.613" CommentCount="5" />
  <row Id="15164" PostTypeId="1" AcceptedAnswerId="15816" CreationDate="2010-12-22T07:38:13.647" Score="1" ViewCount="52" Body="&lt;p&gt;This &lt;a href=&quot;http://math.stackexchange.com/questions/15033/find-the-outgoing-edge-with-the-smallest-angles-given-one-incident-edges-and-mul&quot;&gt;is a similar question&lt;/a&gt; to this one, but slightly different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is given two edges ($e_1$ and $e_2$, with the vertex coordinates known), how to find the angles from $e_1$ to $e_2$, with the angles measured in anti clock wise direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1qQx7aHPuQv3wX9UULf-6a082MTUlODFhv2iIBogAWyA&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way I can think of is to compute the cross and dot product of the two edge's unit vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin\,\theta=\frac{|e_1\times e_2|}{|e_1||e_2|}$$&#xA;$$\cos\,\theta=\frac{e_1\cdot e_2}{||e_1|| ||e_2||}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And try to find the $\theta$, taken into account of whether $sin\theta$ and $cos\theta$ is $&amp;gt;0$ or $&amp;lt;0$. But this is very, very tedious and error prone. Not to mention I'm not too sure whether the angle I get is &lt;em&gt;always&lt;/em&gt; measured in counter clock wise direction or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a single, clean formula that allows me to do what I want to do?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2010-12-29T13:09:11.507" Title="Find the Angle ( as Measured in Counter Clock Wise Direction)  Between Two Edges" Tags="&lt;geometry&gt;&lt;trigonometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15165" PostTypeId="2" ParentId="15101" CreationDate="2010-12-22T07:42:53.620" Score="0" ViewCount="0" Body="&lt;p&gt;i want this to work automatic.&lt;/p&gt;&#xA;" OwnerUserId="5039" LastActivityDate="2010-12-22T07:42:53.620" CommentCount="1" />
  <row Id="15166" PostTypeId="1" AcceptedAnswerId="15187" CreationDate="2010-12-22T07:45:40.550" Score="4" ViewCount="192" Body="&lt;p&gt;Could you a give a intuitive interpretation of curl, geometrical interpretation, real-world example or physical interpretation would be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Consider a specific vector field &#xA;$$\mathbf{v} = -y\mathbf{\hat{x}} + x\mathbf{\hat{y}}$$&#xA;$$\nabla\times\mathbf{v} = 2\mathbf{\hat{z}}$$&#xA;Could you give a intuitive interpretation about why all points in this field has this curl w? What if &#xA;$\mathbf{v} = y\mathbf{\hat{x}} - x\mathbf{\hat{y}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-12-22T10:29:04.137" LastActivityDate="2010-12-23T01:02:11.163" Title="Could you a give a intutive interpretation of curl?" Tags="&lt;intuition&gt;&lt;vector-calculus&gt;" AnswerCount="3" />
  <row Id="15167" PostTypeId="2" ParentId="15162" CreationDate="2010-12-22T07:52:51.467" Score="5" ViewCount="0" Body="&lt;p&gt;I think that there are a handful of points that might need clarification here.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is no &quot;set of all infinite cardinals&quot;, the family of all infinite cardinals is a proper class, and if it were a set it would imply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burali-Forti_paradox&quot; rel=&quot;nofollow&quot;&gt;Burali-Forti paradox&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Secondly, consider the meaning of this equivalence relation: $a\sim b$ if and only if they behave the same asymptotically, up to some finite constant. You want, now to take this and say &quot;cardinals and sequences behave similarly&quot; in the manner that the &quot;stronger&quot; wins. This intuition is not too bad, although I'd have taken multiplication and not addition. However, when you want some structure to be preserved, it might take a bit more than just &quot;behaving the same&quot;, because certain interpretation of the same idea could have very different properties, as in my next point&lt;/li&gt;&#xA;&lt;li&gt;Cardinals are well ordered. There is a minimal cardinal, and there is a next one after that, and a next one after that. In your equivalence classes take $n^{-x}$ and $n^{-y}$, for choice of different $x,y\in\mathbb{R}^+$ those will give you different equivalence classes, it means you have continuum many equivalence classes and in terms of ordering - you're a lot like the order of $\mathbb{R}^{\ge 0}$ which is dense, which makes the idea of taking a cardinal for each equivalence class (even if you do only limit yourself to the first continuum-many cardinals) a tad problematic. &lt;/li&gt;&#xA;&lt;li&gt;And another point would be infinite summations. Cardinal arithmetic is slightly different when it comes to infinite sums, for example take the first $\aleph$ number which has an uncountable index (denoted by $\aleph_{\omega_1}$) then every addition of the form $\sum_{i\in\mathbb{N}} \lambda_i$ where all the $\lambda_i&amp;lt;\aleph_{\omega_1}$ is still a lot smaller than $\aleph_{\omega_1}$. So unlike the real numbers, adding uncountably many cardinals would have meaning if the &lt;a href=&quot;http://en.wikipedia.org/wiki/Continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Continuum Hypothesis&lt;/a&gt; fails, &lt;/li&gt;&#xA;&lt;li&gt;If however CH holds, and you only speak of the first $\omega_1$ cardinals then summation has a different trick up its sleeve: $\sum_{i\in\mathbb{N}}\lambda_i$ is just the biggest $\lambda_i$ or the first one which is bigger than all of them. I'm not quite certain how that would be handled by the equivalent classes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That been said, &quot;homomorphism&quot; is not dealing with infinite summations so the two points about that might be redundant, but these are things to consider when you go between two structures, especially infinite ones that have a very strong meaning when it comes to sequences and convergence (because what is an infinite sum? It is the limit of its finite partial sums).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is probably a &quot;nice&quot; way disproving the existence of such function, but I will leave this for Future-Asaf as well the other members of the site.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-22T07:52:51.467" CommentCount="2" />
  <row Id="15168" PostTypeId="2" ParentId="15139" CreationDate="2010-12-22T07:56:20.363" Score="7" ViewCount="0" Body="&lt;p&gt;The problem is overdetermined/impossible.  The values of $Z_1$ and $Z_2$ are forced, &lt;a href=&quot;http://math.stackexchange.com/questions/15139/two-points-on-circle-resulting-in-5-equal-regions/15142#15142&quot;&gt;as in Ross's answer&lt;/a&gt;, but the segments with area $\frac{2\pi}{5}$ are not divided into regions of equal area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0&amp;lt;\alpha&amp;lt;\pi$ be the solution to $\frac{\alpha-\sin\alpha}{2}=\frac{\pi}{5}$ (so $\alpha\approx 2.11314$).  $\arg(Z_2)=\pi-\frac{\alpha}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $0&amp;lt;\beta&amp;lt;\pi$ be the solution to $\frac{\beta-\sin\beta}{2}=\frac{2\pi}{5}$ (so $\beta\approx 2.8248$).  $\arg(\overline{Z_1})=\arg(Z_2)-\beta=\pi-\frac{\alpha}{2}-\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any point on the line segment joining $Z_2$ and $\overline{Z_1}$ is a linear combination of these two points, $t\cdot Z_2+(1-t)\overline{Z_1}$ with $0&amp;lt;t&amp;lt;1$.  The intersection of that line segment and the line segment joining $\overline{Z_2}$ and $Z_1$ has imaginary part $0$.  Solving for $t$ in terms of $\alpha$ and $\beta$ yields $t=\frac{1}{2}\csc\frac{\beta}{2}\sec\frac{\alpha+\beta}{2}\sin(\frac{\alpha}{2}+\beta)$ ($t\approx 0.436382$).  The location of the point of intersection of the two lines is at $-\cos\frac{\beta}{2}\sec\frac{\alpha+\beta}{2}+0i$ ($\approx 0.20166$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the area of the region determined by this point of intersection, $Z_1$, $\overline{Z_1}$, and the circle is the area of the segment of the circle plus the area of the triangle determined by those three points.  The area of the segment of the circle is $\frac{2\arg(Z_1)-\sin(2\arg(Z_1))}{2}=\frac{1}{2}(\alpha+2\beta-2\pi-\sin(\alpha+2\beta))$ $\approx 0.241854$; the area of the triangle is $|\sin^2(\frac{\alpha}{2}+\beta)\tan\frac{\alpha+\beta}{2}|\approx 0.361977$ (found using shoelace method with the three vertices).  The total area of the region is thus $\approx 0.603831&amp;lt;0.628319\approx\frac{\pi}{5}$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-22T07:56:20.363" />
  <row Id="15169" PostTypeId="2" ParentId="15166" CreationDate="2010-12-22T08:01:29.577" Score="5" ViewCount="0" Body="&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the vector field represents the flow velocity of a moving fluid, then the curl is the circulation density of the fluid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The circulation is defined as the integral of the velocity around a closed curve. An intuitive way of thinking about this is that if the fluid flows in the same direction everywhere, then integrating the velocity around any closed curve will give you 0. Indeed, you will integrate in the direction of the flow on some interval of the curve, but you will also have to &quot;come back&quot; and integrate against the flow, so the contributions will cancel. If on the other hand the fluid rotates in a vortex, then as you integrate the velocity around a circle centred on the axis of rotation, you pick up a positive contribution to the integral at each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose the vector field describes the velocity field of a fluid flow (maybe a large tank of water or gas) and a small ball is located within the fluid or gas (the centre of the ball being fixed at a certain point). If the ball has a rough surface it will be made to rotate by the fluid flowing past it. The rotation axis (oriented according to the right hand rule) points in the direction of the curl of the field at the centre of the ball, and the angular speed of the rotation is half the value of the curl at this point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-22T08:01:29.577" CommentCount="1" />
  <row Id="15170" PostTypeId="1" CreationDate="2010-12-22T08:12:57.253" Score="0" ViewCount="106" Body="&lt;p&gt;$-2(y-2)= 7(5-y)-9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does y= 1/8 in the equation above&lt;/p&gt;&#xA;" OwnerUserId="5037" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-22T08:18:39.557" LastActivityDate="2010-12-22T15:17:54.037" ClosedDate="2010-12-22T15:27:33.657" Title="Is my answer of $-2(y-2)= 7(5-y)-9$ correct?" Tags="&lt;algebra-precalculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15171" PostTypeId="2" ParentId="15107" CreationDate="2010-12-22T08:19:40.100" Score="1" ViewCount="0" Body="&lt;p&gt;So you want to compute the following integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\frac{1}{(2\pi|\Sigma|)^{n/2}}\int_{V_k}\exp\left(-\frac{1}{2}(x-a)^T\Sigma^{-1}(x-a)\right)dx&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V_k\subset \mathbb{R}^k$ is a $k$-polytope, $a\in \mathbb{R}^k$ is the mean and $\Sigma$ is the $k\times k$ covariance matrix of $k$-variate normal vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach would be doing a change of variables $y=\Sigma^{1/2}(x-a)$ and then integrate by one variable at a time, since the integrand function will be a product of one-variable functions. If $\Sigma^{1/2}V_k$ is a $k$-cube then the integration is straightforward. For the general case it will not be easy. Take $k=2$, normal with zero mean and unit covariance matrix, and $V_2$ a triangle with vertices in $(0,0)$, $(1,0)$, $(0,1)$. Then your integral will be&#xA;\begin{align}&#xA;\frac{1}{2\pi}\int_{V_2}e^{-\frac{x_1^2}{2}}e^{-\frac{x_2^2}{2}}dx_1dx_2&amp;amp;=\frac{1}{2\pi}\int_{0}^{1}e^{-\frac{x_1^2}{2}}dx_1\int_{0}^{1-x_1}e^{-\frac{x_2^2}{2}}dx_2\\&#xA;&amp;amp;=\frac{1}{\sqrt{2\pi}}\int_0^1e^{-\frac{x_1^2}{2}}(\Phi(1-x_1)-\Phi(0))dx_1&#xA;\end{align}&#xA;which I do not think has closed formula. (I might be wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for approximation, I do not see why standard numerical integral solving methods would not work.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-22T08:19:40.100" CommentCount="1" />
  <row Id="15172" PostTypeId="1" CreationDate="2010-12-22T08:47:08.150" Score="0" ViewCount="36" Body="&lt;p&gt;Describe the open sets if all straight lines in the plane parallel to the x-axis are&#xA;used for subbasis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, wouldn't all open sets just be horizontal lines? Or whatelse should I specify?&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2010-12-22T11:36:26.440" Title="Straight lines in the plane parallel to the x-axis as a subbasis for a topology" Tags="&lt;topology&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="15173" PostTypeId="2" ParentId="9851" CreationDate="2010-12-22T09:17:55.817" Score="0" ViewCount="0" Body="&lt;p&gt;I'm not sure if this applies, but for minimum-phase functions, the phase and magnitude of the Fourier Transform are related.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Minimum_phase#Relationship_of_magnitude_response_to_phase_response&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for a brief overview.  I've never actually used this relationship in practice, so can't really give you much more information.&lt;/p&gt;&#xA;" OwnerUserId="5042" LastActivityDate="2010-12-22T09:17:55.817" />
  <row Id="15174" PostTypeId="1" AcceptedAnswerId="15457" CreationDate="2010-12-22T09:21:50.943" Score="2" ViewCount="77" Body="&lt;p&gt;How do I find generators of group of rotations of d-dimensional cube into itself when viewed as a subgroup of symmetric group of order $2^d$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to my previous &lt;a href=&quot;http://math.stackexchange.com/questions/15022/how-to-get-generators-for-a-finite-point-group&quot;&gt;question&lt;/a&gt; but not specializing to $d=3$&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditorDisplayName="" LastEditDate="2010-12-22T09:35:57.387" LastActivityDate="2010-12-24T18:29:15.017" Title="How to get group of hypercube rotations as a subgroup of symmetric group?" Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="15175" PostTypeId="2" ParentId="15134" CreationDate="2010-12-22T09:23:35.793" Score="1" ViewCount="0" Body="&lt;p&gt;Terence Tao has set up a &quot;polymath project&quot; which addresses exactly your question. Here is the relevant page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://michaelnielsen.org/polymath1/index.php?title=Finding_primes&quot; rel=&quot;nofollow&quot;&gt;http://michaelnielsen.org/polymath1/index.php?title=Finding_primes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It contains a list of conjectures, partial results and references. To sum it all up: At the moment the answer to your question is No.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-22T09:23:35.793" />
  <row Id="15177" PostTypeId="1" CreationDate="2010-12-22T10:01:48.507" Score="1" ViewCount="93" Body="&lt;p&gt;I have two numbers $x$ and $y$ which are $\log_2$ of two other numbers $P$ and $V$.  I'm trying to calculate $\log_2 (P-V)$ without transforming $x$ and $y$ back to the linear world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P = AV$, then $P-V = V(A-1)$, and $x-y = \log_2 A$.  So, if I could calculate $\log_2 (A-1)$ I could just add it to $y$ and be done.  Is there a relationship I can use to do this?&lt;/p&gt;&#xA;" OwnerUserId="5042" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T10:36:03.610" LastActivityDate="2010-12-22T10:36:03.610" Title="Given $\log_2 A$, is there an easy way to calculate $\log_2 (A-1)$?" Tags="&lt;logarithms&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15178" PostTypeId="1" AcceptedAnswerId="15182" CreationDate="2010-12-22T10:09:53.333" Score="0" ViewCount="86" Body="&lt;p&gt;Hola, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently creating a program to plot simple graphs in 2D. In order to draw those properly I have to know if a function graph is continuous between 2 points on the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the graph for f(x)=tan(x) and two points on the graph P1(1,tan(1)) and P2(2,tan(2)). Is there an algorithmic way to tell if the graph is continuous for 1 &amp;lt;= x &amp;lt;= 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="5043" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T10:40:26.697" LastActivityDate="2010-12-22T11:04:33.983" Title="How to tell if a function is continuous in an interval" Tags="&lt;analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15179" PostTypeId="2" ParentId="15177" CreationDate="2010-12-22T10:23:34.247" Score="3" ViewCount="0" Body="&lt;p&gt;One way is to write (am using natural logs for now): $\log(A-1) = \log A + \log (1-1/A)$, and to then use the power series for $\log(1-x)$ to approximate the latter term. But this is not really a good solution.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-22T10:23:34.247" CommentCount="1" />
  <row Id="15180" PostTypeId="2" ParentId="15124" CreationDate="2010-12-22T10:24:29.883" Score="4" ViewCount="0" Body="&lt;p&gt;Let $\tau  = \min \{ n \geq 1:X_1  +  \cdots  + X_n  &amp;gt; K \}$. Then $\tau$ is an integer-valued random variable, bounded from above by $K+1$ (since $X_i \geq 1$). Note that $\tau = n$ if and only if $\sum\nolimits_{i = 1}^{n - 1} {X_i }  \le K$ and $\sum\nolimits_{i = 1}^{n} {X_i } &amp;gt; K$. Thus, the event $\lbrace \tau = n \rbrace$ depends only on the values $X_1,\ldots,X_n$. So, by definition, $\tau$ is a stopping time with respect to the sequence $X_1,X_2,\ldots$.  Now, $X_1,X_2,\ldots$ are i.i.d. with finite expectation $\mu$, and $\tau$ is a stopping time for them. Moreover, ${\rm E}(\tau) &amp;lt; \infty$ since $\tau \leq K+1$. Hence, by Wald's identity,&#xA;$$&#xD;&#xA;{\rm E}\bigg(\sum\limits_{i = 1}^\tau  {X_i } \bigg) = {\rm E}(\tau )\mu \leq (K+1)\mu.&#xD;&#xA;$$&#xA;So if we put $Y_\tau = \sum\nolimits_{i = 1}^\tau  {X_i }$, we get&#xA;$$&#xD;&#xA;{\rm E}(Y_\tau - K) = {\rm E}(Y_\tau) - K \leq (K+1)\mu - K.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\tau \geq 1$, we have&#xA;$$&#xD;&#xA;\mu - K \leq {\rm E}(Y_\tau - K) \leq (K+1)\mu - K.&#xD;&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we have seen above, the problem reduces to calculating ${\rm E}(\tau)$. Put $S_n = \sum\nolimits_{i = 1}^n {X_i }$ ($S_0 = 0$).&#xA;Note that &#xA;$$&#xD;&#xA;{\rm P}(\tau  = n) = {\rm P}(S_{n - 1}  \le K,S_n  &amp;gt; K) = {\rm P}(S_{n - 1}  \le K) - {\rm P}(S_n  \le K).&#xD;&#xA;$$&#xA;Hence,&#xA;$$&#xD;&#xA;{\rm E}(\tau) = \sum\limits_{n = 1}^{K + 1} {n{\rm P}(\tau  = n)}  = \sum\limits_{n = 1}^{K + 1} n[{\rm P}(S_{n - 1}  \le K) - {\rm P}(S_n  \le K)] = \sum\limits_{n = 0}^K {{\rm P}(S_n  \le K)}. &#xD;&#xA;$$&#xA;So, we can write &#xA;$$&#xD;&#xA;{\rm E}(\tau) = 1 + \sum\limits_{n = 1}^\infty  {{\rm P}(S_n  \le K)} = 1 + \sum\limits_{n = 1}^\infty  {F^{(n)}(K)},&#xD;&#xA;$$&#xA;where $F^{(n)}$ is the distribution function of $S_n$. For $t&amp;gt;0$ real, define $m(t) = \sum\nolimits_{n = 1}^\infty  {F^{(n)}(t)}$.&#xA;From the theory of renewal processes, we know that $m(t) = {\rm E}(N_t)$, where $\lbrace N_t:t \geq 0 \rbrace$ is a renewal process with inter-arrival times distributed according to the distribution of $X$. $m(t)$ is called the renewal function. It may be worth noting that by the Elementary Renewal Theorem, &#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{t \to \infty } \frac{{m(t)}}{t} = \frac{1}{\mu }.&#xD;&#xA;$$&#xA;Returning to our original setting, we have&#xA;$$&#xD;&#xA;{\rm E}(Y_\tau  ) = {\rm E}(\tau )\mu  = (1 + m(K))\mu.&#xD;&#xA;$$&#xA;So, the problem reduces to calculating $m(K)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;http://www.postech.ac.kr/class/ie272/ie666_temp/Renewal.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is some useful link concerning renewal theory, which is very relevant to this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.postech.ac.kr/class/ie272/ie666_temp/Renewal.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.postech.ac.kr/class/ie272/ie666_temp/Renewal.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-22T14:16:22.937" LastActivityDate="2010-12-22T14:16:22.937" CommentCount="5" />
  <row Id="15181" PostTypeId="1" AcceptedAnswerId="15185" CreationDate="2010-12-22T11:03:30.817" Score="2" ViewCount="56" Body="&lt;p&gt;For example, I have set of points in 3D. Points lie on straight line. Transformed set of points lies on straight line too. How to check if transformation is affine? &lt;/p&gt;&#xA;" OwnerUserId="3532" LastActivityDate="2010-12-22T12:17:17.097" Title="How to check if transformation is affine?" Tags="&lt;linear-algebra&gt;&lt;elementary&gt;" AnswerCount="2" />
  <row Id="15182" PostTypeId="2" ParentId="15178" CreationDate="2010-12-22T11:04:33.983" Score="4" ViewCount="0" Body="&lt;p&gt;You can use interval arithmetic to compute reliable plots. See for instance this paper: Jeff Tupper, &lt;a href=&quot;http://www.dgp.toronto.edu/~mooncake/papers/SIGGRAPH2001_Tupper.pdf&quot; rel=&quot;nofollow&quot;&gt;Reliable Two-Dimensional Graphing Methods for Mathematical Formulae with Two Free Variables&lt;/a&gt;, SIGGRAPH 2001. The excellent &lt;a href=&quot;http://www.peda.com/grafeq/&quot; rel=&quot;nofollow&quot;&gt;GrafEq&lt;/a&gt; software uses this technique.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-22T11:04:33.983" CommentCount="1" />
  <row Id="15184" PostTypeId="2" ParentId="15181" CreationDate="2010-12-22T11:44:40.500" Score="0" ViewCount="0" Body="&lt;p&gt;Dear qutron,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;affine transformation&lt;/a&gt; between two vector spaces $$F:X\rightarrow Y$$ (one might define it more general) is defined as&#xA;$$y = F(x) = Ax + y_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is a constant map (might be represented as matrix) and $y_0\in Y$ is a constant element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to check if a transformation is affine you might try to proof that such $A$ and $y_0$ exist.&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2010-12-22T11:44:40.500" CommentCount="1" />
  <row Id="15185" PostTypeId="2" ParentId="15181" CreationDate="2010-12-22T11:56:16.743" Score="2" ViewCount="0" Body="&lt;p&gt;According to the fundamental theorem of affine geometry, it suffices to check that the transformation is a bijective &lt;em&gt;collineation&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; Let $\mathbb A^{n}$ be an affine space over $\mathbb R$ with $n &amp;gt; 2$ and fix&#xA;  $a \in A$. Let $\phi :\mathbb A^{n}\to \mathbb A^{n}$ be a bijection which takes each three collinear points into collinear points. Then there exists a point $b\in \mathbb A^{n}$ and an&#xA;  invertible linear map $F$ such that $\phi(x) = F(x-a) + b$ for all $x \in\mathbb  A^n$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof can be found in Berger's &lt;a href=&quot;http://books.google.com/books?id=iER421ZjkqcC&amp;amp;pg=PA320&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Geometry 1&lt;/em&gt;&lt;/a&gt;  (Springer, 1987, pp. 52-56).&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T12:17:17.097" LastActivityDate="2010-12-22T12:17:17.097" CommentCount="1" />
  <row Id="15186" PostTypeId="2" ParentId="15134" CreationDate="2010-12-22T13:07:39.887" Score="0" ViewCount="0" Body="&lt;p&gt;Polynomial in $n$ is not efficient because the input length (measured in bits) is $\log_2 n$ and so $n$ is exponentially large in relation to your input length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is easy to find a prime number greater than $n$ in time polynomial in $n$ (e.g. with the answer from Moron). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in a deterministic algorithm than the polymath project gives you the current known results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a randomized algorithm with time polynomial in $\log_2 n$ that gives you a prime greater than $n$. You pick a random number from the set $n, \ldots, 2n$ and check this number with AKS or &lt;a href=&quot;http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test&quot; rel=&quot;nofollow&quot;&gt;Miller Rabin primality test&lt;/a&gt;. With the &lt;a href=&quot;http://en.wikipedia.org/wiki/Prime_number_theorem&quot; rel=&quot;nofollow&quot;&gt;prime number theorem&lt;/a&gt; you can proof that there are enough primes in the set (for large $n$) and if you repeat this algorithm $O(\log_2 n)$ times you find a prime with high probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I cannot find a reference for a detailed proof for the randomized algorithm. But if you are interested I can look for it when I have more time.&lt;/p&gt;&#xA;" OwnerUserId="4857" LastActivityDate="2010-12-22T13:07:39.887" />
  <row Id="15187" PostTypeId="2" ParentId="15166" CreationDate="2010-12-22T13:36:53.063" Score="8" ViewCount="0" Body="&lt;p&gt;Here's a &lt;a href=&quot;http://www.sciencelive.org/component/option,com_mediadb/task,view/idstr,Open-podcast-feeds_mst209_grad_div_curl_rss2_xml/Itemid,97&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; explaining curl.&lt;/p&gt;&#xA;" OwnerUserId="247" LastActivityDate="2010-12-22T13:36:53.063" CommentCount="1" />
  <row Id="15188" PostTypeId="1" CreationDate="2010-12-22T13:50:14.890" Score="1" ViewCount="117" Body="&lt;p&gt;Let $A,B \text{ and } C$ are three sets then if $ A \subset B, B \subset C, C \subset A \Rightarrow B = C $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could we prove this ?&lt;/p&gt;&#xA;" OwnerUserId="5045" LastEditorUserId="5045" LastEditorDisplayName="" LastEditDate="2010-12-22T13:57:10.877" LastActivityDate="2010-12-22T14:01:14.503" Title="How to prove :: $ A \subset B, B \subset C, C \subset A \Rightarrow B = C $" Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="15189" PostTypeId="2" ParentId="15188" CreationDate="2010-12-22T14:01:14.503" Score="3" ViewCount="0" Body="&lt;p&gt;This is effectively asking to prove $B \subseteq C \land C \subseteq B \implies B = C$. The usual way to prove this is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_extensionality&quot; rel=&quot;nofollow&quot;&gt;Axiom of Extensionality&lt;/a&gt; - i.e. take an element $b \in B$ and show that it is in $C$. Then show that $c \in C \implies c \in B$. Extensionality now tells you that the two sets are identical. &lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-22T14:01:14.503" CommentCount="5" />
  <row Id="15190" PostTypeId="1" CreationDate="2010-12-22T14:21:56.950" Score="1" ViewCount="86" Body="&lt;p&gt;Hi,&#xA;i have $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ that lies on a line, $L$,&#xA;and another point $P_0(x_0, y_0, z_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i want to fine the distance between $P_0$ and $L$.&#xA;some helps?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-22T15:38:20.760" LastActivityDate="2010-12-22T15:38:20.760" Title="Distance between a point and a line in space" Tags="&lt;euclidean-geometry&gt;" CommentCount="9" />
  <row Id="15191" PostTypeId="2" ParentId="8263" CreationDate="2010-12-22T14:44:13.733" Score="1" ViewCount="0" Body="&lt;p&gt;For the second question, for single digits $a+b\geq ab$ when a=0 or 1, b=0 or 1, or a=b=2.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T14:44:13.733" />
  <row Id="15192" PostTypeId="1" CreationDate="2010-12-22T14:59:09.063" Score="2" ViewCount="62" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let $X$ be some Banach space. Let $L(X)$ be the set of continuous operators on $X$ into $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(\tau_i)_i$ be a set of topologies on $L(X)$ s.t. $L(X)$ is topological vector space (i.e. addition and scalar multiplication continuous, not necessarily Hausdorff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\tau$ be the initial topology on $L(X)$ induced by the linear identities $f_i : L(X) \longrightarrow (L(X),\tau_i)$. Then $(L(X),\tau)$ is a topological vector space, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of illustration: if we restrict $(\tau_i)_i$ to the topologies that are coarser than the operator norm topology on $X$, $\tau_{op}$, then obviously $\tau = \tau_{op}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Does $\tau = \tau_{op}$ hold even in general? Alternatively, is there at least a &quot;nice&quot; set of restrictions on $(\tau_i)_i$, or a characterization of &quot;reasonable&quot; topologies, for which we have $\tau = \tau_{op}$?&lt;/p&gt;&#xA;" OwnerUserId="3557" LastEditorUserId="3557" LastEditorDisplayName="" LastEditDate="2010-12-22T19:13:52.007" LastActivityDate="2010-12-22T21:16:54.757" Title="Finest topology on a space of banach space operators" Tags="&lt;functional-analysis&gt;&lt;topological-vector-spaces&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15193" PostTypeId="1" CreationDate="2010-12-22T15:07:45.773" Score="0" ViewCount="39" Body="&lt;p&gt;Apologies if the title is unclear, I couldn't think of a good way to express what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the usual 2D representation of the complete graph $K_n$ as a regular $n$-gon with each pair of vertices joined, let $U_n$ be the set of the disjoint regions of space delimited by the edges of the graph (so $|U_3|=1, |U_4|=4, |U_5|=11...$). Further, label the nodes of $K_n$ as $p_1, ..., p_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the size of the largest subset $S_n \subset U_n$ such that $\forall s \in S_n \exists i, j, k \in K_n$ with $s \subset \triangle ijk$ and $t \not\subset \triangle ijk \forall t \in S_n, t \ne s$? That is, How many points can I place inside drawing of $K_n$ such that each triangle formed by three points on the exterior of $K_n$ contains at most one point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it helps clarify what I'm trying to do, I'm working on a puzzle where letters are encoded by giving three points from a discrete set defining a triangle in which the letter is written. For instance, for $K_4$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1-------2&#xA;|\  A  /|&#xA;| \   / |&#xA;|  \ /  | &#xA;|   X   |&#xA;|  / \  |&#xA;| /   \ |&#xA;|/  B  \|&#xA;3-------4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&quot;A&quot; can be encoded as 123 or 124; B as 234 or 134. No further letters can be unambiguously encoded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing around with pencil and paper, it seems as though $K_n$ can define $n-2$ letters unambiguously, but I haven't investigated far, and I can't immediately come up with a proof.&lt;/p&gt;&#xA;" OwnerUserId="2432" LastActivityDate="2011-01-02T01:57:48.957" Title="Largest uniquely-identifiable subset of regions of space within complete graph $K_n$" Tags="&lt;geometry&gt;&lt;graph-theory&gt;" AnswerCount="1" />
  <row Id="15194" PostTypeId="1" CreationDate="2010-12-22T15:11:47.863" Score="11" ViewCount="133" Body="&lt;p&gt;A while back, I wanted to see if the notion of the &lt;a href=&quot;http://mathworld.wolfram.com/Arithmetic-GeometricMean.html&quot; rel=&quot;nofollow&quot;&gt;arithmetic-geometric mean&lt;/a&gt; could be extended to a pair of symmetric positive definite matrices. (I considered positive definite matrices only since the notion of the matrix square root is a bit intricate for other kinds of matrices.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expected that some complications would arise since, unlike scalar multiplication, matrix multiplication is noncommutative. Another complication would be that the product of two symmetric matrices need not be symmetric (though the positive definiteness is retained, so one can still speak of a matrix square root).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy with the scalar AGM, I considered the iteration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf A_0=\mathbf A \; ; \; \mathbf B_0=\mathbf B$$ $$\mathbf A_{i+1}=\frac12(\mathbf A_i+\mathbf B_i) \; ; \; \mathbf B_{i+1}=\sqrt{\mathbf A_i \mathbf B_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cranked up a short &lt;em&gt;Mathematica&lt;/em&gt; routine:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;matAGM[u_, v_] := First[FixedPoint[({Apply[Plus, #]/2, MatrixPower[Apply[Dot, #], 1/2]}) &amp;amp;, {u, v}]] /; MatrixQ[u, InexactNumberQ] &amp;amp;&amp;amp; MatrixQ[v, InexactNumberQ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and decided to try it out on randomly generated SPD matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A numerical note: &lt;em&gt;Mathematica&lt;/em&gt; uses the numerically stable Schur decomposition in computing matrix functions like the matrix square root.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that for all of the randomly generated pairs of SPD matrices I tried, the process was convergent (though the rate of convergence is apparently not as fast as the scalar AGM). As expected, the order matters: &lt;code&gt;matAGM[A, B]&lt;/code&gt; and &lt;code&gt;matAGM[B, A]&lt;/code&gt; are usually not equal (and both results are &lt;em&gt;unnsymmetric&lt;/em&gt;) unless &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt; commute (for the special case of diagonal &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt;, the result is the diagonal matrix whose entries are the arithmetic-geometric means of the corresponding entries of the pair.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now have three questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How do I prove or disprove that this process converges for any pair of SPD matrices? If it is convergent, what is the rate of convergence?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there any relationship between &lt;code&gt;matAGM[A, B]&lt;/code&gt; and &lt;code&gt;matAGM[B, A]&lt;/code&gt; if the two matrices &lt;code&gt;A&lt;/code&gt; and &lt;code&gt;B&lt;/code&gt; do not commute?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there any relationship between this matrix arithmetic-geometric mean and the usual scalar arithmetic-geometric mean? Would, say, arithmetic-geometric means of the eigenvalues of the two matrices have anything to do with this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T00:19:48.887" LastActivityDate="2010-12-23T16:10:41.600" Title="The arithmetic-geometric mean for symmetric positive definite matrices" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15195" PostTypeId="2" ParentId="15170" CreationDate="2010-12-22T15:17:54.037" Score="2" ViewCount="0" Body="&lt;p&gt;No. You can check this by performing the following calculation (Plug in the answer and check for equality):&#xA;$$-2(y - 2) = 7(5 - y) - 9, y = \frac{1}{8}$$&#xA;$$-2(\frac{1}{8} - 2) = 7(5 - \frac{1}{8}) - 9$$&#xA;$$4 - \frac{2}{8} = 35 - \frac{7}{8} - 9$$&#xA;$$\frac{7}{8} - \frac{2}{8} = 35 - 9 - 4$$&#xA;$$\frac{5}{8} = 22$$&#xA;$$5 = 176$$&#xA;$$5 \ne 176 \therefore y \ne \frac{1}{8}$$    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To solve for the correct answer: &#xA;$$−2(y−2)=7(5−y)−9$$&#xA;$$4 - 2y = 35 - 7y - 9$$&#xA;$$5y = 35 - 9 - 4$$&#xA;$$5y = 22$$&#xA;$$y = \frac{22}{5}$$&lt;/p&gt;&#xA;" OwnerUserId="5038" LastActivityDate="2010-12-22T15:17:54.037" />
  <row Id="15196" PostTypeId="1" CreationDate="2010-12-22T15:42:19.283" Score="0" ViewCount="43" Body="&lt;p&gt;In the definition of the complexity class NP, we allow the certificate to be polynomial in the input size. But do we ever need a certificate that is longer than the input? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all the NP problems I could think of, we don't. The &quot;worst&quot; problem was factorization&lt;s&gt;, where the obvious incoding is $n+O(\lg(n))$ (where n is length of the number we wish factor; I just realized that the input also contain a maximal number, as the decision problem is: Do there exists a factor less than some number), because we want to tell where the first factor end and the second begins. However we don't need to tell this, the verifier could just try all possibilities. Furthermore, we could decide not to write the last bit of the factors, and this way we get a certificate that is shorter than the number we wish to factor. &lt;/s&gt; Do'h division is fast, so you could just give one of the factors as certificate.&lt;/p&gt;&#xA;" OwnerUserId="3078" LastEditorUserId="3078" LastEditorDisplayName="" LastEditDate="2010-12-23T15:11:36.573" LastActivityDate="2010-12-23T15:11:36.573" Title="NP-problem that needs superlinear certificates" Tags="&lt;computer-science&gt;" AnswerCount="1" />
  <row Id="15197" PostTypeId="2" ParentId="15154" CreationDate="2010-12-22T15:44:50.910" Score="0" ViewCount="0" Body="&lt;p&gt;Sorry, I'd comment but I'm new here. I'm also getting (once we round to two decimal places) $114522.58$. Perhaps the machine is expecting rounding up (to $114522.59$) instead of rounding to the nearest hundredth?&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2010-12-22T15:44:50.910" />
  <row Id="15198" PostTypeId="2" ParentId="15075" CreationDate="2010-12-22T15:53:50.477" Score="6" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Unfortunately, the answer below as originally written was wrong.  It answered a related, but easier, question.  I have added some bracketed remarks to point out where the blunder happens, and appended some remarks about the expected answer to the actual question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It might help to explicitly reformulate the question.  (This is implicit in the proof that the question is equivalent to the twin prime conjecture for $a = 6$, but it seems worthwhile to make it explicit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k = a m n \pm m \pm n,$ &#xA;then &#xA;$a k \pm 1 = (a m \pm 1) (a n \pm 1).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $a = 2$ then we can write $1 = (2\cdot 1 - 1)$, and so for any $k$,&#xA;we have $2 k - 1 = (2 \cdot 1 - 1)(2 k  -1)$,&#xA;and (as the OP noted) every integer $k$ can be written in the desired form.&#xA;Thus we suppose from now on that $a &amp;gt; 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is then equivalent to asking if we can find infinitely many&#xA;numbers congruent to $\pm 1 \bmod a$ which have no proper factors congruent&#xA;to $\pm 1 \bmod a$.  [&lt;strong&gt;Added&lt;/strong&gt;: This is wrong.  The problem is rather to&#xA;find infintely many multiples $a k$ of $a$ so that &lt;em&gt;both&lt;/em&gt; $a k +1 $ and&#xA;$a k - 1$ have no proper factors congruent to $\pm 1 \bmod a$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the answer to the question depends a lot  on whether or not&#xA;&lt;em&gt;every&lt;/em&gt; residue class mod $a$ which is coprime to $a$ is congruent to $\pm 1 \bmod a$, i.e. on whether or not $\varphi(a) = 2$.  [&lt;strong&gt;Added&lt;/strong&gt;: It is not so clear to me now what difference the condition on $\varphi(a)$ makes, if any.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\varphi(a) = 2$, then we are simply asking whether we can find infinitely many numbers coprime to $a$, which differ by $2$, and which have no proper factors, i.e. we asking for infinitely many twin primes.  This is a well-known open problem!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand,&#xA;suppose that $\varphi(a) &amp;gt; 2$ (which in particular is the case if $a &gt; 6$).  We may then choose &#xA;$b$ and $c$ so that $b, c \not\equiv \pm 1 \bmod a$, and $b c \equiv 1 \bmod a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Dirichlet's theorem, we may choose infinitely primes $p$ and $q$ such that&#xA;$p \equiv b \bmod a$ and $q \equiv c \bmod a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any such $p$ and $q$, the product $pq \equiv 1 \bmod a$, but by construction has no proper factors congruent to $\pm 1 \bmod a$.&#xA;So if we set $k = (p q - 1)/a$, then $k$ cannot be written in the given form.&#xA;This gives infinitely many $k$, as desired.&#xA;[&lt;strong&gt;Added&lt;/strong&gt;: Rather, this gives infinitely many $k$ that are not of the form&#xA;$a m n + m + n$ or $a m n - m - n$, but doesn't address whether these can be written in the form $a m n - m + n$, since we don't have any control over $a k - 1 = p q - 2.] &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;:&#xA;It is not clear to me how to gain control over the factors of $p q - 2$, other than to assume it is prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one way to solve the problem would be to find infinitely many primes $r \equiv -1 \bmod a,$ such that $r + 2$ is either prime, or is of the form&#xA;$r + 2 = p q,$ where $p,q \not\equiv \pm 1 \bmod a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is then related to Chen's theorem, indeed is a strengthening of it.  Thus&#xA;we are led to &lt;a href=&quot;http://mathoverflow.net/questions/49751/chens-theorem-with-congruence-conditions&quot; rel=&quot;nofollow&quot;&gt;MO Scribe's question&lt;/a&gt; on MO which was inspired by the initial posting of the present question on MO.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no doubt that this result (on the existence of such primes $r$) is true. (Indeed, standard conjectures predict that there should be infinitely many twin primes $r, r+2$ with $r\equiv -1 \bmod a$.)  But it is not clear (to me) whether or not it is in reach of current methods; that is the thrust of MO Scribe's question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the conclusion:  Well, there is no doubt that one should be able to find infintely many such $k$, since it follows from standard conjectures on twin primes satisfying congruence conditions.  On the other hand, proving this may be tricky, since it seems to require results that are at the edge of what is currently possible via seiving techniques.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-22T19:06:08.187" LastActivityDate="2010-12-22T19:06:08.187" CommentCount="15" />
  <row Id="15199" PostTypeId="1" CreationDate="2010-12-22T15:56:02.397" Score="1" ViewCount="70" Body="&lt;p&gt;Three fellows accused of stealing CDs make the following statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Ed: “Fred did it, and Ted is innocent.”&lt;br&gt;&#xA;(2) Fred: “If Ed is guilty, then so is Ted.”&lt;br&gt;&#xA;(3) Ted: “I’m innocent, but at least one of the others is guilty.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the innocent told the truth and the guilty lied, who is guilty?&#xA;(Remember that false statements imply anything).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Ed and Ted are innocent and Fred is guilty.&#xA;Is it in contradiction with statement 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you say?&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-22T16:44:14.213" LastActivityDate="2010-12-22T16:44:14.213" Title="Implication of three statements" Tags="&lt;logic&gt;&lt;puzzle&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="15200" PostTypeId="2" ParentId="15196" CreationDate="2010-12-22T15:57:20.663" Score="1" ViewCount="0" Body="&lt;p&gt;In the definition of NP, no attempt is made to shorten the size of the certificate. Since the running time is polynomial and the machine might &quot;guess&quot; at each step, the certificate is polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can artificially create an NP problem which seems to require longer certificates: given an instance $x$, answer some predicate on $G(x)$, where $G$ is some PRNG that takes an input of size $n$ and outputs a string of size $n^2$. Now the certificate would be quadratic (though probably no-one knows whether it has to be that large).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, &quot;factorization&quot; is not an NP predicate. To make it a predicate, you need to ask some &quot;hard-core&quot; question on one of the factors.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-22T15:57:20.663" CommentCount="3" />
  <row Id="15201" PostTypeId="1" CreationDate="2010-12-22T16:03:44.500" Score="3" ViewCount="209" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your recommendation for an in-depth introductory combinatoric book? A book that doesn't just tell you about the multiplication principle, but rather shows the whole logic behind the questions with full proofs. The book should be for a first-year-student in college. Do you know a good book on the subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5023" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-22T16:07:53.297" LastActivityDate="2010-12-23T18:51:20.357" Title="Good Book On Combinatorics" Tags="&lt;combinatorics&gt;&lt;reference-request&gt;" AnswerCount="8" CommentCount="6" />
  <row Id="15202" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T16:20:35.127" Score="4" ViewCount="0" Body="&lt;p&gt;If the student's leaning towards computer science at all, I'd recommend Knuth et. al.'s &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_Mathematics&quot; rel=&quot;nofollow&quot;&gt;Concrete Mathematics&lt;/a&gt;. It's full of solid math and has the aim of building mathematical tools for CS. Other than that, a newer addition that looks promising is Russell Merris's &quot;Combinatorics, 2nd ed.&quot; It gives a pretty broad introduction while also giving in-depth work and examples.&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2010-12-22T16:20:35.127" CommentCount="1" />
  <row Id="15203" PostTypeId="2" ParentId="15199" CreationDate="2010-12-22T16:21:28.173" Score="0" ViewCount="0" Body="&lt;p&gt;You are right.  According to the problem, the guilty lie, so Fred's statement should be false.  If the problem is well constructed (and I think this one is) there should be only one consistent assignment of truth values to the atomic sentences.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a large body of problems of this nature.  One of my favorite authors if Raymond Smullyan.  A couple of his books lead you through the logic of Goedel's incompleteness proof.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T16:21:28.173" CommentCount="2" />
  <row Id="15204" PostTypeId="1" AcceptedAnswerId="15211" CreationDate="2010-12-22T16:24:33.420" Score="1" ViewCount="53" Body="&lt;p&gt;Please forgive me if this is not the right Stack Exchange (I also posted it at &lt;a href=&quot;http://stats.stackexchange.com/questions/5698/can-the-mutual-information-of-a-cell-be-negative&quot;&gt;Cross Validated&lt;/a&gt; and &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3939/can-the-mutual-information-of-a-cell-be-negative&quot;&gt;Theoretical Computer Science&lt;/a&gt;).  Please also forgive me for inventing terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For discrete random variables X and Y, the mutual information of X and Y can be defined as follows:&#xA;$I(X;Y) = \sum_{y \in Y} \sum_{x \in X} &#xD;&#xA;                 p(x,y) \log{ \left( \frac{p(x,y)}{p_1(x)\,p_2(y)}&#xD;&#xA;                              \right) }, \,\!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will define the mutual information of a &quot;cell&quot; $x_0$ to be:&#xA;$CI(x_0,Y) = \sum_{y \in Y} p(x_0,y) \log{ \left( \frac{p(x_0,y)}{p_1(x_0)\,p_2(y)}&#xD;&#xA;                              \right) }, \,\!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this quantity goes by another name.  Essentially I'm restricting focus to a single state of variable X (and then the full MI can be calculated by summing all the cell MIs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: is it guaranteed that $CI(x_0,Y) \ge 0$?  We know $I(X;Y)\ge0$ and we know that the pointwise mutual information can be negative.  I feel like CI should be nonnegative and that I might be missing some obvious proof.&lt;/p&gt;&#xA;" OwnerUserId="5048" LastEditorUserId="5048" LastEditorDisplayName="" LastEditDate="2010-12-22T16:49:28.757" LastActivityDate="2010-12-22T18:16:43.827" Title="Can the mutual information of a &quot;cell&quot; be negative?" Tags="&lt;probability-theory&gt;&lt;information-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15205" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T16:33:04.080" Score="4" ViewCount="0" Body="&lt;p&gt;I had my first intro graph theory and combinatorics class last semester. The book we were using was pretty terrible so I looked around and found a copy of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387797106&quot; rel=&quot;nofollow&quot;&gt;Combinatorics and Graph Theory&lt;/a&gt; by Harris et. al. and I really enjoyed it.  The book contains a lot of topics and the explanations are very to the point.  I especially liked the sections on Ramsey numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are all to the point and illustrate some important concept which is also nice.  For example in the section on the happy ending problem the exercises reconstruct several historical proofs and introduce you to other problems like the empty polygon problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a very positive review I read recently: &lt;a href=&quot;http://www.maa.org/reviews/combinatoricsgraphs.html&quot; rel=&quot;nofollow&quot;&gt;http://www.maa.org/reviews/combinatoricsgraphs.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I'm not sure if this book is appropriate for your situation specifically, but I highly recommend it none the less.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2010-12-22T16:44:33.697" LastActivityDate="2010-12-22T16:44:33.697" CommentCount="2" />
  <row Id="15206" PostTypeId="2" ParentId="15199" CreationDate="2010-12-22T16:38:59.770" Score="1" ViewCount="0" Body="&lt;p&gt;Let us write the following propositions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_g$ means Fred is guilty, and $F_i$ means Fred is innocent, $T_g$ and $T_i$ for Ted and $E_g$ and $E_i$ for Ed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ed says: $F_g \wedge T_i$&lt;/li&gt;&#xA;&lt;li&gt;Fred says: $E_g \rightarrow T_g$&lt;/li&gt;&#xA;&lt;li&gt;Ted says: $T_i \wedge (F_g \vee E_g)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We know that the guilty is lying and the innocent tells the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume Fred did it, he is guilty and lying. Meaning $E_g \rightarrow T_g$ is false, meaning $E_g$ is true and $T_g$ is false, that is Ed is guilty and Ted is innocent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In turn it means that Ed is also lying, that is either Fred is innocent or Ted is guilty. If Fred is innocent then he is telling the truth - contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we know that Ted is guilty as well, another contradiction because it follows from Fred's lie that Ted is innocent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, then we have that Fred is definitely not guilty. So either Ed is innocent or Ted is guilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that Ed is innocent, this is a right out contradiction, as this implies Fred is guilty. So we have that Ed has to be guilty, which means Ted is also guilty - therefore he is lying. Which means he either guilty or both the other guys are innocent. Since he is guilty we have a good solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fred is innocent, Ted and Ed are guilty.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-22T16:38:59.770" CommentCount="2" />
  <row Id="15207" PostTypeId="1" AcceptedAnswerId="15210" CreationDate="2010-12-22T16:44:08.003" Score="8" ViewCount="213" Body="&lt;p&gt;I would like to prove $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \le \frac{1}{\sqrt{n}}$$ for all natural $n \ge 1$. The inequality does seem to be true numerically, but the proof eludes me.&lt;/p&gt;&#xA;" OwnerUserId="5051" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T14:04:40.393" LastActivityDate="2010-12-23T16:16:14.837" Title="A gamma function inequality" Tags="&lt;inequality&gt;&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" />
  <row Id="15208" PostTypeId="2" ParentId="15145" CreationDate="2010-12-22T16:51:39.033" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The sum &lt;strong&gt;telescopes&lt;/strong&gt; since the falling factorial summand is a perfect difference: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad (k+1)^{[n]} - k^{[n]}\ =\ (k+1)\ k\ \cdots\ (k-n+2)\ -\ k\ (k-1)\ \cdots\ (k-n+1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\phantom{(k+1)^{[n]} - k^{[n]}\ }  =\ (k+1 - (k-n+1))\ \ \ k\ (k-1)\ \cdots\ (k-n+2)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\phantom{(k+1)^{[n]} - k^{[n]}\ }\ =\ n\ k^{[n-1]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other examples of additive/multiplicative telescopy see &lt;a href=&quot;http://math.stackexchange.com/q/13701/242&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/13646/how-to-derive-the-following-identity/13673#13673&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/6581/proof-of-inequalities-by-induction/6586#6586&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/7441/summation-of-sum-limits-n1-infty-fracxx1-cdots-xn-1yy1-cdo/7453#7453&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://math.stackexchange.com/questions/6805/simple-proof-by-induction-9-divides-n3-n13-n23/6814#6814&quot;&gt;here.&lt;/a&gt; For much more on the falling factorials see Steven Roman's textbook &lt;em&gt;The Umbral Calculus&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-22T16:58:59.850" LastActivityDate="2010-12-22T16:58:59.850" CommentCount="1" />
  <row Id="15209" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T17:07:49.700" Score="2" ViewCount="0" Body="&lt;p&gt;A good suggestion is Combinatorial Problems and Exercises by Laszlo Lovasz.&lt;/p&gt;&#xA;" OwnerUserId="5010" LastActivityDate="2010-12-22T17:07:49.700" CommentCount="2" />
  <row Id="15210" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T17:08:02.647" Score="7" ViewCount="0" Body="&lt;p&gt;This follows directly from &lt;a href=&quot;http://dlmf.nist.gov/5.6.E4&quot; rel=&quot;nofollow&quot;&gt;Gautschi's inequality&lt;/a&gt;, valid for $0 &amp;lt; s &amp;lt; 1$ and $x &amp;gt; 0$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^{1-s} &amp;lt; \frac{\Gamma(x+1)}{\Gamma(x+s)} &amp;lt; (x+1)^{1-s}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, the link is to NIST's new (as of this year) &lt;a href=&quot;http://dlmf.nist.gov/&quot; rel=&quot;nofollow&quot;&gt;Digital Library of Mathematical Functions&lt;/a&gt;.  The DLMF is an update of the classic Abramowitz and Stegun text, &lt;em&gt;&lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/&quot; rel=&quot;nofollow&quot;&gt;Handbook of Mathematical Functions&lt;/a&gt;&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="2370" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T08:33:47.420" LastActivityDate="2010-12-23T08:33:47.420" CommentCount="2" />
  <row Id="15211" PostTypeId="2" ParentId="15204" CreationDate="2010-12-22T17:17:46.973" Score="1" ViewCount="0" Body="&lt;p&gt;Answered on &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3939/can-the-mutual-information-of-a-cell-be-negative/3943#3943&quot;&gt;Theoretical Computer Science&lt;/a&gt;.  CI cannot be negative.&lt;/p&gt;&#xA;" OwnerUserId="5048" LastEditorUserId="5048" LastEditorDisplayName="" LastEditDate="2010-12-22T18:16:43.827" LastActivityDate="2010-12-22T18:16:43.827" CommentCount="2" />
  <row Id="15212" PostTypeId="1" AcceptedAnswerId="15215" CreationDate="2010-12-22T17:25:47.267" Score="8" ViewCount="254" Body="&lt;p&gt;First some context, I'm not a mathematician, not even close (as you will soon see) I do grasp some things about it but in a need to know basis, so plain english answers are appreciated (too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't figure out this one even though some have tried very hard to explain/convince me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say I'm on a roulette (let's leave out the zero, or make it a coin for that matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roulette hits 5 (or any amount of) times red &lt;strong&gt;on a row&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why should't I bet to black (or better said why it is the same the one I pick).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I DO UNDERSTAND THE BASIC PRINCIPLE, that there is a material reality which makes the ball randomly fall on any of the two colors (which gives the options a 50-50 chance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I also understand that if you &quot;say&quot;: I bet red will come out six times on a row, you do have a very low probability of that happening so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it that if you have seen the ball fall 5 times on a row on red, and you bet on red, you are NOT betting on 6 times on a row on red (that does have lower probability).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I hope I was clear enough with my question, if not, please ask for any clarifications needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the VERY plain English, feel free to modify or suggest a change to anything that may be misleading.&lt;/p&gt;&#xA;" OwnerUserId="4256" LastActivityDate="2010-12-23T03:15:15.713" Title="When the roulette has hit 5 reds why shouldn't I bet to black?" Tags="&lt;probability&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="15213" PostTypeId="2" ParentId="15194" CreationDate="2010-12-22T17:26:24.283" Score="2" ViewCount="0" Body="&lt;p&gt;Nice question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What follows below is not really an answer, but too long to fit as a comment. It would be helpful if you could write your Mathematica algorithm is traditional mathematical notation so it is easier to see what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I am sure you are aware of the standard definitions of matrix means. I recall them below for the benefit of others who consider your question.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The arithmetic mean is: $(A+B)/2$ as expected. It satisfies several of the standard desirable properties of a mean of spd matrices (nonnegative, if $A \le B$, then $A \le AM(A,B) \le B$, etc)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The standard geometric mean is: $$GM(A,B) = A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}$$&#xA;This particular choice of GM satisfies: $GM(A,B)=GM(B,A)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now if you iterate the AGM, using the above two definitions of the means, then I think convergence, etc. can be proven---though you might have to resort to some kind of Fixed-point theorem. If I get time, I will think more about this.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-22T17:26:24.283" CommentCount="3" />
  <row Id="15214" PostTypeId="1" CreationDate="2010-12-22T17:30:46.397" Score="0" ViewCount="64" Body="&lt;p&gt;Greetings,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm performing research that will help determine the size of observed space and the time elapsed since the big bang. Hopefully you can help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have data conforming to a piecewise linear function on which I want to perform two linear regressions. There is a point at which the slope and intercept change, and I need to (write a program to) find this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="5053" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T06:01:21.140" LastActivityDate="2010-12-23T06:01:21.140" Title="Finding the change point in data from a piecewise linear function" Tags="&lt;statistics&gt;&lt;regression&gt;&lt;linear&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15215" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T17:33:58.793" Score="13" ViewCount="0" Body="&lt;p&gt;Lots of people have trouble with this.  If you bet before any spins on six in a row, the chance of winning is 1/2^6=1/64.  But having seen the five reds, as you say, the chances are 1/2 each on red and black for the next spin.  It is also true for &lt;strong&gt;any&lt;/strong&gt; specific series of six spins, at the start the chance of that series is 1/64.  If you were to bet on RBBRRB, that is also 1/64.  Now that you have seen five reds, there are only two series of six that are possible:  RRRRRR and RRRRRB.  They each started out 1/64 and are now each 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T17:33:58.793" CommentCount="1" />
  <row Id="15216" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T17:36:10.977" Score="3" ViewCount="0" Body="&lt;p&gt;The events &quot;given that the roulette hits $5$ reds, the next hit is red&quot; and &quot;the roulette hits $6$ reds&quot; are different.  The first has probability $\frac{1}{2}$ because different spins of the roulette are independent and the second has probability $\frac{1}{64}$ by the multiplication principle.  (This is the same as the probability of &quot;the roulette hits $5$ reds and a black,&quot; which is exactly the point of the first computation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For questions like these the general principle for getting to grips with how your intuition is failing is &quot;when in doubt, list out all the possibilities.&quot;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-22T17:36:10.977" CommentCount="1" />
  <row Id="15217" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T17:36:57.823" Score="3" ViewCount="0" Body="&lt;p&gt;There's an analogy that I like. Suppose I'm afraid to take planes as it's possible that some crazy terrorist brings a bomb onto the plane. Now it's really improbable that there are &lt;em&gt;two&lt;/em&gt; bombs on the plane that I'm going to take. So I'll bring a bomb myself (which I of course won't detonate); then it's virtually impossible that anything happens. Well ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know, but maybe this helps.&lt;/p&gt;&#xA;" OwnerUserId="2575" LastActivityDate="2010-12-22T17:36:57.823" CommentCount="1" />
  <row Id="15218" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T17:40:04.210" Score="4" ViewCount="0" Body="&lt;p&gt;My personal favorites are the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Introduction to Combinatorial analysis --- Riordan&lt;/li&gt;&#xA;&lt;li&gt;Concrete Mathematics --- Graham, Knuth, Patashnik&lt;/li&gt;&#xA;&lt;li&gt;Enumerative Combinatorics vol. 1 --- Richard Stanley (is not always that introductory, but for those who like counting, it is a must have)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you want really easy, but still interesting books, you might like Brualdi's book (though apparently, that book has many mistakes). Also interesting might be some chapters from Feller's book on Probability (volume 1).&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-22T17:40:04.210" CommentCount="1" />
  <row Id="15219" PostTypeId="2" ParentId="15214" CreationDate="2010-12-22T17:48:02.203" Score="1" ViewCount="0" Body="&lt;p&gt;A simple answer would be to consider the break point a variable, call it x.  Collect the points below x and perform a linear fit, returning the error (say sum of squared errors over the points).  Collect the points above x and do the same.  Then you can define f(x) as the total error over the two fits.  Consider this a function in one variable and minimize it.  &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T17:48:02.203" />
  <row Id="15221" PostTypeId="2" ParentId="15212" CreationDate="2010-12-22T18:06:36.810" Score="2" ViewCount="0" Body="&lt;p&gt;Why is it so unlikely to get 6 reds in a row?  Well, you have to cross your fingers and say &quot;Hope the first one's red!&quot;  And if you get lucky and that one comes up red, then you have to cross your fingers and say &quot;Hope the second one's red!&quot;  And if you get really lucky and that one also comes up red, then you have to cross your fingers and say &quot;Hope the third one's red!&quot;  And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if five reds already came up, all you have to do is cross your fingers and say &quot;Hope this one's red!&quot; a single time.  In one case, you're hoping for &quot;red red red red red red&quot;; in the other, you don't care about the first five (they're already done with); you're just hoping to get &quot;red&quot; once.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way: would you rather go for six reds &lt;em&gt;from scratch&lt;/em&gt;, or would you rather go for six reds &lt;em&gt;after five have already come up red&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4904" LastEditorUserId="4904" LastEditorDisplayName="" LastEditDate="2010-12-22T18:13:00.177" LastActivityDate="2010-12-22T18:13:00.177" />
  <row Id="15223" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T18:10:07.150" Score="3" ViewCount="0" Body="&lt;p&gt;Since&#xA;$$&#xD;&#xA;\Gamma \bigg(n + \frac{1}{2}\bigg) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi  &#xD;&#xA;$$&#xA;(see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;),&#xA;we have&#xA;$$&#xD;&#xA;\frac{{\Gamma (n + \frac{1}{2})}}{{\Gamma (n+1)}} = \frac{{{2n \choose n}}}{{4^n }}\sqrt \pi  .&#xD;&#xA;$$&#xA;This might be very useful, but I have to check.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-22T18:44:14.623" LastActivityDate="2010-12-22T18:44:14.623" CommentCount="3" />
  <row Id="15224" PostTypeId="2" ParentId="15194" CreationDate="2010-12-22T18:26:24.563" Score="0" ViewCount="0" Body="&lt;p&gt;I'm sorry that I'm putting this in an answer rather than a comment, but I'm too new to comment on your question. My thoughts for proving this process converges (question 1.) were along the lines of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$A\,\&amp;amp;\,B$ both SPD $\implies AB$ and $A+B$ are also SPD&lt;/li&gt;&#xA;&lt;li&gt;If $A$ and $B$ are symmetric $n\times n$ matrices, $A$'s eigenvalues $\{\lambda_n^A\} \subset [0, a]$, and $B$'s $\{\lambda_n^B\} \subset [0, b]$, then the eigenvalues $\{\lambda_N^{AB}\}$ of $AB$ are $\subset [0, ab]$. See, for instance, &lt;a href=&quot;http://mathforum.org/kb/thread.jspa?threadID=1347224&amp;amp;messageID=4543523&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; discussion. Moreover, $\{\lambda_n^{A+B}\} \subset [0, a + b]$; see Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2010/01/12/254a-notes-3a-eigenvalues-and-sums-of-hermitian-matrices/&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; on this.&lt;/li&gt;&#xA;&lt;li&gt;I think convergence might follow by examining the AGM of the eigenvalues; that is, look at $AGM(\lambda_k^A, \lambda_k^B)$ for $k = 1,\ldots,n$. The difference between the arithmetic and geometric means gets arbitrarily small, and the eigenvalues are all positive; this iteration leads to a sequence of SPD matrices whose largest eigenvalue $\lambda_1 \to 0$, so the AGM in question does converge (perhaps).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Not sure, but thought I'd give my $\$0.02$. As @Slowsolver said, nice question! I also second the request of writing your algorithm in more traditional notation, since I am not as well-versed in Mathematica as others may be. Finally, I might add the tag &quot;numerical-linear-algebra&quot; to your question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: turns out I was wrong in part 1, as others have pointed out to me. My apologies! In light of their corrections, it may make sense to use the geometric mean given by SlowSolver.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4689" LastEditorDisplayName="" LastEditDate="2010-12-23T15:20:10.693" LastActivityDate="2010-12-23T15:20:10.693" CommentCount="3" />
  <row Id="15225" PostTypeId="1" AcceptedAnswerId="15227" CreationDate="2010-12-22T18:34:07.953" Score="5" ViewCount="113" Body="&lt;p&gt;I was trying to write a poker program yesterday, and trying to figure out a way to tell the computer how to detect a straight. I realized that if you add 5 consecutive numbers and divide by 5, you'll get no remainder. I tried for 6 consecutive numbers, but it didn't work. It turns out it only works for odd numbers. Really silly question, but does this 'method' have a name? Has it been proven that for all odd $n$, the sum of any $n$ consecutive numbers divided by $n$ yields a remainder of zero?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2010-12-22T20:20:50.037" Title="Sum of $n$ consecutive numbers divided by $n$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15226" PostTypeId="2" ParentId="15225" CreationDate="2010-12-22T18:41:06.103" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, If $n=2k+1$, write the numbers as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a - k, a-(k-1), \dots,a-1, a, a + 1, \dots, a+(k-1), a+k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding gives you $na$ which is divisible by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try something similar for even $n =2m$ and you see that the sum is $m \mod n$, which is not divisible by $n$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T18:41:06.103" />
  <row Id="15227" PostTypeId="2" ParentId="15225" CreationDate="2010-12-22T18:45:07.433" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.jimloy.com/number/consec0.htm&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; have a proof for the same.It also includes the the prove that when $n$ is even then this is not true.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2010-12-22T18:45:07.433" />
  <row Id="15229" PostTypeId="2" ParentId="15214" CreationDate="2010-12-22T18:59:19.030" Score="1" ViewCount="0" Body="&lt;p&gt;I am assuming that you want a &lt;strong&gt;continuous&lt;/strong&gt; piecewise linear fit of the form&#xA;$$&#xD;&#xA;a(x-\xi)+c\hbox{ if }x\le\xi,\quad b(x-\xi)+c\hbox{ if }x&amp;gt;\xi.&#xD;&#xA;$$&#xA;There are 4 variables to determine: the slopes $a$, $b$, the point $\xi$ and $c$, the ordinate at $\xi$. Let $\{(x_i,y_i)\}$, $1\le i\le N$, be your data. For simplicity I will assume that $x_i&amp;lt;x_{i+1}$. For each $i=1,\dots,N-1$ minimize the expression&#xA;$$&#xD;&#xA;O_i=\sum_{j=1}^i(y_j-a(x_j-\xi)-c)^2+\sum_{j=i+1}^N(y_j-b(x_j-\xi)-c)^2.&#xD;&#xA;$$&#xA;This can be done by solving the system of 4 equations with 4 unknowns&#xA;$$&#xD;&#xA;\frac{\partial O_i}{\partial a}=\frac{\partial O_i}{\partial b}=\frac{\partial O_i}{\partial c}=\frac{\partial O_i}{\partial \xi}=0.&#xD;&#xA;$$&#xA;(This will be nonlinear, and will have more than one solution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, choose the index $i$ that gives the minimum $O_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this in a mock example, and it came out quite right.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastActivityDate="2010-12-22T18:59:19.030" />
  <row Id="15230" PostTypeId="2" ParentId="15192" CreationDate="2010-12-22T19:02:28.960" Score="2" ViewCount="0" Body="&lt;p&gt;Assuming that your $f_i$ is just the identity map with $\tau_i$ on the codomain, $\tau$ is just the coarsest topology finer than all $\tau_i$.  This explains the assertion that restricting to those $\tau_i$ coarser than $\tau_{op}$ yields $\tau=\tau_{op}$.  But then to ask whether $\tau=\tau_{op}$ in general is to ask whether every TVS topology on $L(X)$ is coarser than $\tau_{op}$.  To contradict it you would just need an example of a TVS topology that is not comparable to $\tau_{op}$ or one that is strictly finer than $\tau_{op}$.  If $X$ is infinite dimensional, then there are Banach space norms on $L(X)$ that are not equivalent to the operator norm, which by the open mapping theorem means the topologies are not comparable. So the answer is no.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-22T21:16:54.757" LastActivityDate="2010-12-22T21:16:54.757" />
  <row Id="15231" PostTypeId="1" AcceptedAnswerId="15265" CreationDate="2010-12-22T19:05:31.237" Score="0" ViewCount="114" Body="&lt;p&gt;I have asked this question on &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3592/can-we-partition-np-complete-problem-into-finite-number-of-polynomially-solvable&quot;&gt;cstheory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\Pi$ be NP-complete problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we partition the set of instances of $\Pi$ into finite number of subsets (subproblems) each of which is polynomially solvable (and not necessarily polynomially recognizable)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On cstheory I have obtained two answers on my question: &quot;trivially yes&quot; and no, unless $P=NP$. Trivially yes, since we could just take the sets of instances with equal value of objective function for each of subset. And no, since if we had such subproblems then we could solve the general problem, but that's not possible unless $P=NP$. So the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how to formulate two questions such that for each of them corresponding answers would hold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="1856" LastEditorDisplayName="" LastEditDate="2010-12-22T19:15:12.973" LastActivityDate="2010-12-23T00:51:45.250" Title="Can we partition NP-complete problem into finite number of polynomially solvable problems?" Tags="&lt;computer-science&gt;" AnswerCount="2" CommentCount="17" />
  <row Id="15232" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T19:09:39.843" Score="1" ViewCount="0" Body="&lt;p&gt;Just to add two more to the list already here, try Principles and Techniques in Combinatorics by Chen Chuan Chong and Koh Khee Meng or Combinatorics by Peter Cameron.&lt;/p&gt;&#xA;" OwnerUserId="4594" LastActivityDate="2010-12-22T19:09:39.843" CommentCount="1" />
  <row Id="15233" PostTypeId="2" ParentId="15201" CreationDate="2010-12-22T19:22:44.180" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;A Path to Combinatorics for Undergraduates&lt;/em&gt;, by Titu Adreescu and Zuming Feng introduces the subject by presenting a large number of problems (many from Olympiads and other competitions), and covers a broad range of methods and results.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastActivityDate="2010-12-22T19:22:44.180" CommentCount="1" />
  <row Id="15234" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T19:41:26.827" Score="5" ViewCount="0" Body="&lt;p&gt;To elaborate on Shai Covo's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2n\left[{2n\choose n} 4^{-n}\right]^2&#xD;&#xA;= {1\over2}\,{3\over2}\,{3\over4}\,{5\over4}\cdots {{2n-1}\over{2n-2}}\,{{2n-1}\over{2n}}&#xD;&#xA;={1\over 2}\prod_{j=2}^n\left(1+{1\over4j(j-1)}\right).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Wallis's formula, the middle expression converges to $2\over\pi$.&#xA;The right hand expression shows that it is strictly increasing. Therefore, multiplying by $\pi/2$ and taking square roots shows that&lt;br&gt;&#xA;$$\sqrt{n\pi}{2n\choose n} 4^{-n}\uparrow 1.$$&lt;/p&gt;&#xA;" OwnerUserId="940" LastEditorUserId="940" LastEditorDisplayName="" LastEditDate="2010-12-23T16:16:14.837" LastActivityDate="2010-12-23T16:16:14.837" />
  <row Id="15235" PostTypeId="1" CreationDate="2010-12-22T19:50:01.713" Score="5" ViewCount="39" Body="&lt;p&gt;Suppose we have a square real $n\times n$ matrix $X=[x_1,...,x_n]$, where $x_i$ is $i$-th column of the matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define  $X_k=[x_1,..,x_k]$, i.e. matrix $X_k$ columns are the first $k$ columns of the matrix $X$. Define $\lambda_k$ as the maximal eigenvalue of $(X_k^TX_k)^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to prove that $\lambda_1\le \lambda_2\le ... \le \lambda_n$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $X$ is orthogonal, then the answer is yes. Maybe this holds only for certain matrices? Any pointers would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-22T19:50:01.713" Title="Do the maximal eigenvalues of $(X^TX)^{-1}$ matrix increase when number of columns of $X$ increases?" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" CommentCount="3" />
  <row Id="15236" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T19:58:16.993" Score="9" ViewCount="0" Body="&lt;p&gt;Here's a direct proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gamma function satisfies $\Gamma(n+1) = n \Gamma(n)$,&#xA;so the function $g(x) = \ln \Gamma(x)$ satisfies&#xA;$g(n+1) = \ln n + g(n)$.&#xA;In other words, the line segment between the points $(n,g(n))$&#xA;and $(n+1,g(n+1))$ has slope $\ln n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, $g$ is known to be a convex function for $x&amp;gt;0$&#xA;(cf. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem&quot; rel=&quot;nofollow&quot;&gt;Bohr–Mollerup theorem&lt;/a&gt;),&#xA;so the point $(n+1/2, g(n+1/2))$ lies &lt;em&gt;below&lt;/em&gt; that line segment:&#xA;$g(n+1/2) &amp;lt; g(n+1) - \frac12 \ln n$.&#xA;Exponentiating both sides gives $\Gamma(n+1/2) &amp;lt; \Gamma(n+1) / \sqrt{n}$,&#xA;as desired.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2010-12-22T19:58:16.993" CommentCount="2" />
  <row Id="15237" PostTypeId="2" ParentId="15225" CreationDate="2010-12-22T20:11:59.860" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ The sum is $\rm\ 0 + 1 +\:\cdots\:+\: n-1\ \ (mod\ n)\:.\: $ Pairing each summand $\ne 0\:$ with its negation shows that the sum is $\rm\: \equiv 0\ \ (mod\ n)\:.\ \ $ Note: $\rm\ - x\ \equiv\ x\ \Rightarrow\ 2\ x \equiv 0\ \Rightarrow\ x\equiv 0\ \:$ since $\rm\:n\:$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\rm\:n\:$ is even then the above argument demonstrates simply that the sum reduces to the sum over the &lt;em&gt;fixed points&lt;/em&gt; of the negation map, viz. $\rm\ 0,\ n/2\:.\ $ Therefore the sum is  $\rm\:\equiv n/2\ \ (mod\ n)\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of &lt;strong&gt;Wilson's theorem&lt;/strong&gt; for groups, e.g. see &lt;a href=&quot;http://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-n/2267#2267&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1/9345#9345&quot;&gt;here&lt;/a&gt; -&#xA;which highlight the key role played by the reflection / involution symmetry (here negation $\rm\ x\to -x\:$).&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-22T20:20:50.037" LastActivityDate="2010-12-22T20:20:50.037" />
  <row Id="15238" PostTypeId="1" CreationDate="2010-12-22T20:13:27.520" Score="-7" ViewCount="457" Body="&lt;p&gt;The Green-Tao Theorem states that for any natural number k, there exist k-term arithmetic progressions of primes. Suppose this were not true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there must exist a minimum k such that there are no k-term arithmetic progressions of primes. So if a-(k-2)b,a-(k-3)b,...,a-2b,a-b,a is a (k-1)-term arithmetic progression of primes and a&gt;(k-1)b, then c=a-(k-1)b cannot be prime. Notice that we have determined that c is not prime without knowing anything about the factors of c. This is impossible since the factors of c determine whether c is prime or not. Hence, the Green-Tao theorem must be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the flaw in this proof?&lt;/p&gt;&#xA;" OwnerUserId="3163" LastEditorUserId="3163" LastEditorDisplayName="" LastEditDate="2010-12-23T01:04:04.153" LastActivityDate="2010-12-23T03:28:15.893" ClosedDate="2010-12-23T04:16:36.123" Title="Puzzle - A strange proof of the Green-Tao Theorem" Tags="&lt;number-theory&gt;" AnswerCount="2" CommentCount="27" />
  <row Id="15239" PostTypeId="2" ParentId="15207" CreationDate="2010-12-22T20:17:45.963" Score="7" ViewCount="0" Body="&lt;p&gt;For a completely elementary proof let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n= \int_0^{ \pi/2} \sin^n x \textrm{ d}x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then integration by parts gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_n = \frac{n-1}{n} I_{n-2} . \quad (1) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2n}= \frac{2n-1}{2n} \frac{2n-3}{2n-2} \cdots \frac{1}{2} \frac{\pi}{2}  \quad (2) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2n+1}= \frac{2n}{2n+1} \frac{2n-2}{2n-1} \cdots \frac{2}{3} .  \quad (3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ \sin^{2n+1} x &amp;lt; \sin^{2n} x $ for $ x \in (0,\pi/2) $ we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{2n+1}&amp;lt; I_{2n} . \quad (4) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from $(1)$ we have $(2n+1)I_{2n+1} = 2nI_{2n-1} &amp;gt; 2nI_{2n}$ since $I_{2n-1} &amp;gt;I_{2n}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore using this and $(4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2n+1}{2n} &amp;gt; \frac{ I_{2n} }{ I_{2n+1} } &amp;gt; 1 .  \quad (5) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from $(2)$ and $(3)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \frac{ I_{2n} }{ I_{2n+1} }  = \frac{2n+1}{2} \frac{ (2n!)^2 }{ 4^{2n} (n!)^4 } \pi .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this in $(5)$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{ \sqrt{n} } &amp;gt; \frac{1}{4^n} { 2n \choose n } \sqrt{ \pi } &amp;gt; \frac{1}{ \sqrt{ n + 1/2} }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{ \sqrt{n} } &amp;gt; \frac{ \Gamma(n + 1/2) }{ \Gamma(n+1) } &amp;gt; \frac{1}{ \sqrt{ n + 1/2} }.$$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-22T21:34:55.240" LastActivityDate="2010-12-22T21:34:55.240" CommentCount="3" />
  <row Id="15240" PostTypeId="1" AcceptedAnswerId="15296" CreationDate="2010-12-22T20:26:12.727" Score="9" ViewCount="220" Body="&lt;p&gt;Suppose you have a double sequence $\displaystyle a_{nm}$.  What are sufficient conditions for you to be able to say that $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}$?  Bonus points for necessary and sufficient conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of a sequence where this is not the case, consider $\displaystyle a_{nm}=\left(\frac{1}{n}\right)^{\frac{1}{m}}$.  $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}}=\lim_{n\to \infty}{\left(\frac{1}{n}\right)^0}=\lim_{n\to \infty}{1}=1$, but $\displaystyle \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}=\lim_{m\to \infty}{0^{\frac{1}{m}}}=\lim_{m\to \infty}{0}=0$.&lt;/p&gt;&#xA;" OwnerUserId="781" LastEditorUserId="781" LastEditorDisplayName="" LastEditDate="2010-12-23T07:08:01.947" LastActivityDate="2010-12-27T08:30:21.430" Title="When can you switch the order of limits?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="4" />
  <row Id="15241" PostTypeId="2" ParentId="9659" CreationDate="2010-12-22T20:41:47.783" Score="0" ViewCount="0" Body="&lt;p&gt;The basic idea here is that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuous_truth&quot; rel=&quot;nofollow&quot; title=&quot;Vacuous Truth&quot;&gt;Vacuous Truth&lt;/a&gt;.  The universal quantifier applied to the empty set is, by definition, true.  In symbols, $\forall{x}{\in}\emptyset \, P(x)$ is true, regardless of the statement $P(x)$ (see also &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_quantifier#The_empty_set&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt;).  This is mainly defined this way for convenience, because otherwise you would always have to consider the empty set as a special case every time you used the universal quantifier.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This apples to subsets because a set $A\subseteq B$ by definition means $\forall x\in A \, x\in B$.  If $A=\emptyset$, then here $P(x)=x\in B$, and we see from the above that this must be true, i.e., $\emptyset \subseteq B$ for any set $B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_subset#proper_subset&quot; rel=&quot;nofollow&quot;&gt;proper subset&lt;/a&gt;&quot; almost always means $A\subset B;\,A\neq B$, so of course $\emptyset$ is a proper subset of any set, except for itself.  &lt;/p&gt;&#xA;" OwnerUserId="781" LastActivityDate="2010-12-22T20:41:47.783" CommentCount="2" />
  <row Id="15242" PostTypeId="1" AcceptedAnswerId="15244" CreationDate="2010-12-22T20:51:54.867" Score="3" ViewCount="30" Body="&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Midpoint_method&quot; rel=&quot;nofollow&quot;&gt;says&lt;/a&gt; the midpoint formula for numerical integration has error of order $h^3 f&amp;#39;&amp;#39;(\xi)$. I am trying to replicate this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that I want to use Lagrange's formulation of the remainder for Taylor series. Let $x_0=\frac{a+b}{2}$ (i.e. the midpoint). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The midpoint method says $\int_a^b f(x)dx \approx (b-a)f(\frac{a+b}{2})$, so to get the error I find $(b-a) f(\frac{a+b}{2}) - \int_a^bf(x)dx$. If I expand this using Taylor's theorem I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \begin{aligned}&#xD;&#xA;error &amp;amp; =(b-a) f(x_0) - \int_a^bf(x_0)+\frac{f&amp;#39;(\xi)(x-x_0)}{2}dx \\&#xD;&#xA;&amp;amp; =\frac{f&amp;#39;(\xi)}{2}\int_a^b(x-x_0)dx \\&#xD;&#xA;&amp;amp; =\frac{f&amp;#39;(\xi)}{2}\int_a^b(x-\frac{a+b}{2})dx \\&#xD;&#xA;&amp;amp; = 0&#xD;&#xA;\end{aligned}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So apparently I have just proven that it has zero error? Any hints as to what I did wrong? (I realize that since wikipedia gives it in terms of $f&amp;#39;&amp;#39;$ I probably want to take the expansion one level further to match them, but I don't understand why this doesn't work.)&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="2549" LastEditorDisplayName="" LastEditDate="2010-12-22T21:24:28.677" LastActivityDate="2010-12-22T21:24:28.677" Title="The error of the midpoint rule for quadrature" Tags="&lt;calculus&gt;&lt;numerical-methods&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15244" PostTypeId="2" ParentId="15242" CreationDate="2010-12-22T21:12:36.783" Score="3" ViewCount="0" Body="&lt;p&gt;You are right that you need one more term in your Taylor series.  If you write $f(x)=f(x_0)+(x-x_0)f&amp;#39;(x_0)+(x-x_0)^2f&amp;#39;&amp;#39;(\xi)/2$ and plug that into the midpoint method, the term in $f&amp;#39;&amp;#39;$ will survive.  What you have done essentially shows that if $f$ is linear, the method is exact.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T21:12:36.783" CommentCount="2" />
  <row Id="15245" PostTypeId="1" AcceptedAnswerId="15247" CreationDate="2010-12-22T21:19:30.107" Score="0" ViewCount="88" Body="&lt;p&gt;Suppose $w(z)$ is an analytic function in a domain $G$ which is symmetric with respect to the real axis. Show that $f(z) = \overline{w(\bar{z})}$ is then an analytic function of $z$ in $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is interesting because we are looking at the conjugate of a function of a conjugate. Let $w(z) = u(x,y)+iv(x,y)$. Then $f(z) = \overline{u(x,y)-iv(x,y)}$. So then just see that the Cauchy-Riemann equations are satisfied? But this is a necessary condition. So we need a sufficient condition to show analyticity right?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-22T21:38:48.233" LastActivityDate="2010-12-22T21:38:48.233" Title="Analytic Functions" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15246" PostTypeId="2" ParentId="12167" CreationDate="2010-12-22T21:21:46.343" Score="0" ViewCount="0" Body="&lt;p&gt;Just because the integers are a proper subset of the rationals doesn't mean that the rationals have a higher cardinality than the integers.  Actually, there is a theorem that says that a set is infinite if and only if it has the same cardinality to a proper subset of itself (so your logic would only apply to a finite set).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set is countable, or has the same cardinality as the integers, if you can count the elements.  In other words, you can label each element by a unique positive integer.  We can see from the diagonals argument (see &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aDiagonal_argument.svg&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt; on Wikipedia for a good illustration) that this holds for that rational numbers.  Once you get the hang of it, you can see that a lot of sets that seem to be a lot bigger than the set of integers are in fact the same &quot;size&quot; (have the same cardinality) as the integers.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel&quot; rel=&quot;nofollow&quot;&gt;Hilbert's Paradax of the Grand Hotel&lt;/a&gt; for a good example of this.&lt;/p&gt;&#xA;" OwnerUserId="781" LastActivityDate="2010-12-22T21:21:46.343" />
  <row Id="15247" PostTypeId="2" ParentId="15245" CreationDate="2010-12-22T21:25:24.983" Score="3" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://mathworld.wolfram.com/SchwarzReflectionPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;Reflection Principle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-22T21:25:24.983" />
  <row Id="15248" PostTypeId="1" CreationDate="2010-12-22T21:33:26.163" Score="5" ViewCount="119" Body="&lt;p&gt;Suppose $\Omega$ is a bounded domain in the plane whose boundary consist of $m+1$ disjoint analytic simple closed curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f$ be holomorphic and nonconstant on a neighborhood of the closure of $\Omega$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|f(z)|=1$$ for all $z$ in the boundary of $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m=0$, then the maximum principle applied to $f$ and $1/f$ implies that $f$ has at least one zero in $\Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the general case? I believe that $f$ must have at least $m+1$ zeros in $\Omega$, but I'm not able to prove it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3550" LastActivityDate="2010-12-23T02:42:13.123" Title="Zeros of a holomorphic function" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" />
  <row Id="15249" PostTypeId="1" AcceptedAnswerId="15653" CreationDate="2010-12-22T21:35:11.933" Score="2" ViewCount="208" Body="&lt;p&gt;I read somewhere that a set is infinite if and only if it has a proper infinite subset.  I also remember seeing someones name attached to this theorem on Wikipedia once, but I can't even find that now.  I haven't been able to find a proof of this theorem, nor been able to generate one myself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can prove that if a set has a proper infinite subset then it is itself infinite by proving the contrapositive that if a set is finite then it does not have any proper infinite subsets (this is a simple contradiction proof).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't figure out how to, given an arbitrary infinite set, construct a proper infinite subset.  Does this require the Axiom of Choice?  I can't really figure out how to do it with that either.  A proof or reference to a proof would be much appreciated.  &lt;/p&gt;&#xA;" OwnerUserId="781" LastActivityDate="2010-12-27T18:40:52.170" Title="Proof that a set is infinite if and only if it has an infinite proper subset" Tags="&lt;set-theory&gt;" AnswerCount="3" CommentCount="14" />
  <row Id="15250" PostTypeId="2" ParentId="15249" CreationDate="2010-12-22T21:41:09.400" Score="5" ViewCount="0" Body="&lt;p&gt;It doesn't require the axiom of choice.  Remove one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a &lt;em&gt;countably&lt;/em&gt; infinite subset of every infinite set, I think you need to use (or at least the usual proof uses) the axiom of countable choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, maybe you are thinking of the condition of being Dedekind infinite, as mentioned in Adrián Barquero's comment.  There you want not just an &lt;em&gt;infinite&lt;/em&gt; proper subset, but a proper subset that is in bijection with the whole set.  This can be proved using the existence of a countably infinite subset, so again uses countable choice.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-22T21:41:09.400" CommentCount="6" />
  <row Id="15251" PostTypeId="1" CreationDate="2010-12-22T21:52:35.710" Score="17" ViewCount="364" Body="&lt;p&gt;Doing a little reading over the break (The Probabilistic Method by Alon and Spencer); can't come up with the solution for this seemingly simple (and perhaps even a little surprising?) result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A-S 1.6.3) Prove that for every two independent, identically distributed real random variables X and Y,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pr[|X-Y| \leq 2] \leq 3 Pr[|X-Y| \leq 1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="5056" LastActivityDate="2010-12-28T08:04:38.377" Title="Random Variable Inequality" Tags="&lt;probability&gt;&lt;probability-theory&gt;&lt;measure-theory&gt;&lt;inequality&gt;&lt;random-variable&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="6" />
  <row Id="15252" PostTypeId="1" CreationDate="2010-12-22T21:57:05.210" Score="0" ViewCount="119" Body="&lt;p&gt;Are there any analogues of Galois Theory for complex numbers (with non-zero imaginary part)? This is motivated by &lt;a href=&quot;http://mathworld.wolfram.com/SchwarzReflectionPrinciple.html&quot; rel=&quot;nofollow&quot;&gt;Schwartz Principle&lt;/a&gt;. It says that if $f$ is analytic, $f$ is defined in the upper-half disk, and $f$ extends to a continuous function on the real axis, then $f$ can be extended to an analytic function on the whole disk by the formula $f(\bar{z}) = \overline{f(z)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reminds me of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomorphism_extension_theorem&quot; rel=&quot;nofollow&quot;&gt;Isomorphism Extension Theorem&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-23T02:13:04.323" Title="Analogues of Galois Theory for Complex Numbers" Tags="&lt;complex-analysis&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15253" PostTypeId="2" ParentId="14311" CreationDate="2010-12-22T22:09:33.910" Score="1" ViewCount="0" Body="&lt;p&gt;I have no idea how to comment. But I do know how to answer. Sorry about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered the pumping lemma. And I considered how many words are necessary to pump the language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the language over {0,1}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need the following words: 00, 11, 0101, and 1010. To pump we can combine any words by concatenation or insertion. For this example a grammar would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S =&gt;  | S0S0S | S1S1S | S0S1S0S1S | S1S0S1S0S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick count shows the number of words necessary for the grammar with n terminals is exponential in n. I guess that shows there is no polynomial sized grammar.&lt;/p&gt;&#xA;" OwnerUserId="5055" LastActivityDate="2010-12-22T22:09:33.910" CommentCount="3" />
  <row Id="15254" PostTypeId="1" CreationDate="2010-12-22T22:34:56.003" Score="3" ViewCount="96" Body="&lt;p&gt;I am under the impression that the standard convention for the homology (singular) of the empty set is 0 in all non negative degrees and $\mathbb{Z}$ in degree $-1$. I have no problem with this convention, I am just curious what role it plays. Most conventions helps something sensible remain true in a particular case, what is this convention doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doe this convention depend on which version of cohomology we use? Is there a different convention for Cech or what have cohomology?&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-22T22:46:49.257" Title="Homology of the Empty set" Tags="&lt;algebraic-topology&gt;&lt;cohomology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15255" PostTypeId="2" ParentId="15254" CreationDate="2010-12-22T22:46:49.257" Score="1" ViewCount="0" Body="&lt;p&gt;Here's an idea. The &lt;em&gt;reduced homology&lt;/em&gt;  of the empty set is what you describe: one $\mathbb Z$ in degree $-1$ and $0$ otherwise. This is because the chain complex computing the reduced homology groups has an extra $\mathbb Z$ in degree $-1$, whose role is to kill one $\mathbb Z$ in degree $0$. However, if we are looking at the empty set, all chain groups are $0$ so there is only the extra $\mathbb Z$ in degree $-1$. I'm just guessing here.&lt;/p&gt;&#xA;" OwnerUserId="2941" LastActivityDate="2010-12-22T22:46:49.257" />
  <row Id="15256" PostTypeId="2" ParentId="15055" CreationDate="2010-12-22T22:47:35.800" Score="1" ViewCount="0" Body="&lt;p&gt;This is really a continuation of BlueRaja's answer, where I'm changing the question slightly to give another illustration of the underlying point.  Consider the following two procedures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procedure A: I flip two coins, a nickel and a penny.  After the flip, I&#xA;pick one at random, and report its value to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procedure B: I flip two coins, a nickel and a penny.  If exactly one lands&#xA;heads, I report that one's value to you, i.e. I say either &quot;the nickel&#xA;landed heads&quot; or &quot;the penny landed heads&quot; as appropriate.  If both land&#xA;heads, I report on one of them chosen at random.  If neither lands&#xA;heads, I say &quot;both landed tails.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I follow Procedure A and say &quot;the nickel landed&#xA;heads&quot;, then the probability that the penny landed heads is 50/50.  But&#xA;if I say &lt;em&gt;the very same thing&lt;/em&gt; after following Procedure B, the&#xA;probability that the penny landed heads is 1 in 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take the following problem:  I flip two coins and say to you &quot;the&#xA;nickel landed heads&quot;.  What is the probability that the penny landed&#xA;heads? Clearly there's no way to answer without knowing or assuming something about the procedure I'm following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All this having been said, I agree that the best answer to the original question is 1/3.)&lt;/p&gt;&#xA;" OwnerUserId="2682" LastActivityDate="2010-12-22T22:47:35.800" />
  <row Id="15257" PostTypeId="2" ParentId="15240" CreationDate="2010-12-22T23:00:24.460" Score="7" ViewCount="0" Body="&lt;p&gt;Here's some results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say that a sequence $a_{nm}$ of real numbers indexed by pairs of positive integers $(n,m)$ &lt;em&gt;converges to $L$&lt;/em&gt; if and only if for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L|\lt \epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly strong condition. We have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;For each (sufficiently large) $n_0$, $\lim\limits_{m\to\infty}a_{n_0m}$ exists;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}a_{nm} = L$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If 2 holds, then we must have 1 (otherwise the expression in 2 does not even make sense). Now assume that 1 holds, and let $\lim\limits_{m\to\infty}a_{nm} = L_{n}$. We want to prove that $\lim\limits_{n\to\infty}L_n=L$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\epsilon\gt 0$. Then there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L| \lt \epsilon$. Let $M_N\gt N$ be such that for all $m\geq M_N$, $|a_{M_Nm}-L_{M_N}|\lt \epsilon$. Since $M_n\gt N$, we have&#xA;$$|L_{M_N} - L| \leq |L_{M_N}-a_{M_Nm}| + |a_{M_Nm}-L| \leq 2\epsilon,$$&#xA;so this proves that $L_n\to L$ as $n\to\infty$. In particular, we have for the iterated limit &#xA;$$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{nm} = \lim_{n\to\infty}L_n = L.\ \Box$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetric argument shows that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;For each (sufficiently large) $m_0$, $\lim\limits_{n\to\infty}a_{nm_0}$ exists;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}a_{nm} = L$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, $\lim\limits_{n,m\to\infty}a_{nm}=L$ does not imply the existence of the iterated limits; in fact, existence of the double limit and the existence of &lt;em&gt;one&lt;/em&gt; iterated limit does not suffice to imply that the other iterated limits exists. Take $a_{nm}=\frac{(-1)^n}{m}$. Then $\lim\limits_{n,m\to\infty}a_{nm}=0$ (given $\epsilon\gt 0$, pick $N$ such that $\frac{1}{N}\lt\epsilon$), and the limit as $m\to\infty$ of $\frac{(-1)^n}{m}$ exists for each fixed $n$, but $\lim\limits_{n\to\infty}\frac{(-1)^n}{m}$ does not exist for any $m$. (By taking $a_{nm} = \frac{(-1)^n}{m} + \frac{(-1)^m}{n}$ you get one in which neither iterated limit exists.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a sufficient conditions for the iterated limits to exist is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; &lt;em&gt;Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}=L$. Then the iterated limits&#xA;$$\lim_{n\to\infty}\lim_{m\to\infty}a_{nm}\text{ and }\lim_{m\to\infty}\lim_{n\to\infty}a_{nm}$$&#xA;both exist and are equal to $L$ if and only if $\lim\limits_{n\to\infty}a_{nm}$ exists for almost all $m$ and $\lim\limits_{m\to\infty}a_{nm}$ exists for almost all $n$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here, &quot;almost all&quot; means &quot;all except perhaps for a finite number&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, the condition above is pretty strong. You can have both iterated limits exist &lt;em&gt;and&lt;/em&gt; be equal and yet for the double limit not to exist. To adapt the standard two-variable example, take $a_{nm}=\frac{nm}{n^2+m^2}$. The iterated limits both exist and are equal to $0$, but the double limit does not exist (for any $N\gt 0$ there exist $n,m\geq N$ such that $a_{nm}=\frac{1}{2}$ and there exist $n,m\geq N$ such that $a_{nm}=\frac{2}{5}$; just take $n=m=N$ for the first, and $n=2m=2N$ for the second). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get that the double limit exists and is equal to (one of) the iterated limits if you have some uniformness conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any necessary and sufficient conditions, and I suspect there won't generally be without some other overarching conditions. This is essentially the same problem as the problem of iterated limits in functions of two variables (just as the problem of finding the limit of a real function of real variable is closely connected to the problem of finding limits of sequences of real numbers). They are connected to the double limits, and you often have conditions based on uniform convergence that guarantee good things happen, but to state some general, simple condition for the iterated limits to exist and be equal seems difficult in general.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-22T23:37:36.317" LastActivityDate="2010-12-22T23:37:36.317" CommentCount="3" />
  <row Id="15258" PostTypeId="2" ParentId="15231" CreationDate="2010-12-22T23:02:18.390" Score="4" ViewCount="0" Body="&lt;p&gt;If you allow exponentially many machines and each can know its serial number, try this.  For travelling salesman, you have n! machines.  Each one attempts to verify the cost of a trip for a given permutation, generated in a reasonable way from its serial number.  This can be done in polynomial time.  Each machine raises a bit if it succeeds.  Can't you then combine these in log(n!) time, which is polynomial.  You are still doing exponentially much computation, but spreading it over many machines.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-22T23:02:18.390" CommentCount="3" />
  <row Id="15259" PostTypeId="2" ParentId="15238" CreationDate="2010-12-22T23:30:50.077" Score="5" ViewCount="0" Body="&lt;p&gt;The problem is in the claim &quot;Notice that we have determined that c is not prime without proving that there are other factors of c besides c and 1. This is impossible.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absence of knowledge is not the knowledge of absence.  If you actually show that c has no other factors than 1 and c, you would have a contradiction to c being composite.  But I hope you will agree that you haven't shown that; you just note that you don't know anything specific about other factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, &quot;determining that c is not prime&quot; = &quot;determining that c has other factors than 1 and c.&quot;  The definition of prime numbers is simple, but the study of prime numbers is a deep and active part of mathematics.  It involves many long and (initially) mysterious arguments.  One nice thing about theorems is that you don't have to think about all of the details of the proof to use the result-- if you cite any theorem about prime numbers, you know something about the factors or non-factors of the numbers involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example argument, you assume that no k-term arithmetic progression of primes exists.  Then you automatically know that the terms preceding and following a k-1 term progression of primes are composite, because of your assumption.  In other words, your attempted proof doesn't actually do anything-- you assume something and never get anything more than the assumption you started with.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2010-12-22T23:30:50.077" />
  <row Id="15260" PostTypeId="1" CreationDate="2010-12-22T23:52:40.007" Score="6" ViewCount="190" Body="&lt;p&gt;Hi all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an university assignment I have to visualize some curves in 3 dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now I've implemented Bézier, helix and conical spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you give me some advice about some famous curves in geometry history?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T06:47:42.077" LastActivityDate="2010-12-23T12:48:38.587" Title="famous space curves in geometry history?" Tags="&lt;homework&gt;&lt;differential-geometry&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="15261" PostTypeId="1" AcceptedAnswerId="15462" CreationDate="2010-12-22T23:58:39.480" Score="5" ViewCount="128" Body="&lt;p&gt;For improper single integrals with positive integrands of the first, second or mixed type there are the comparison and the limit tests to determine their convergence or divergence. There is also the absolute convergence theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For multiple integrals I know that the comparison test can be used as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: But can the limit or other tests be generalized to multiple integrals? Could you provide some references?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: I have thought of Riemann integration only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example&lt;/em&gt;: When we evaluate (Proof 1 of &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www.springerlink.com/content/045l250v885v033u/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;Apostol's article&lt;/a&gt;)  this improper double integral $$\int_{0}^{1}\int_{0}^{1}\left(\dfrac{1}{1-xy}\right) \mathrm{d}x\mathrm{d}y,$$ we conclude that it is finite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; 2: Which test should we apply to know in advance that it converges? &lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-23T23:17:13.627" LastActivityDate="2010-12-24T19:30:24.557" Title="Convergence tests for improper multiple integrals" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;&lt;reference-request&gt;&lt;multivariable-calculus&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15262" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T00:01:44.900" Score="8" ViewCount="0" Body="&lt;p&gt;How about &lt;a href=&quot;http://en.wikipedia.org/wiki/Viviani%27s_curve&quot; rel=&quot;nofollow&quot;&gt;Viviani's curve&lt;/a&gt;?  It is the intersection of a sphere with a cylinder. Viviani was a student (or disciple) of Gallileo.&#xA;Wikipedia has a beautiful image:&#xA;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/0/05/Viviani_curve.png/200px-Viviani_curve.png&quot; width=&quot;300&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-23T00:01:44.900" />
  <row Id="15263" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T00:09:38.313" Score="9" ViewCount="0" Body="&lt;p&gt;Space-filling curves, in particular the Hilbert curve (in 3D):&#xA;&lt;img src=&quot;http://i.imgur.com/OoCvm.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-23T00:09:38.313" />
  <row Id="15264" PostTypeId="2" ParentId="15238" CreationDate="2010-12-23T00:12:29.653" Score="2" ViewCount="0" Body="&lt;p&gt;Use [false-proof] or whatever the tag is to ask about proofs whose flaws are the subject of discussion.  This would help prevent the impression of claiming an easy 3-sentence proof of a result that took 40 pages and decades of prior work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is identical, but leads to a false conclusion, if you replace &quot;prime&quot; by &quot;not divisible by $k$&quot;.   Given a $(k-1)$-term arithmetic progression [with difference having no factor in common with $k$] of not-divisible-by-k integers, its extension to a $k$ term progression has to have the last term divisible by $k$, even though we were given no divisibility information in advance about that last term.   The difference between this and the Green-Tao theorem is that &lt;em&gt;sometimes&lt;/em&gt; a string of $(k-1)$ primes can be extended to a string of $k$, and sometimes not. There is a relationship between the first $(k-1)$ items and the $k$th one, but it is more statistical in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Green-Tao proof, loosely speaking, does use the logic you outlined, but in a statistical, on-average sense.  A deterministic guarantee that any $k$ term progression can't have all terms primes (so e.g., if the first (k-1) are prime the last cannot be) would create too much correlation, too much &quot;shared information&quot; about the prime factorization of the different terms.  There are theorems in analytic number theory saying that there is at least a certain amount of pseudorandomness in the primes and in the prime factorizations of integers, and knowing in advance that every string of $(k-1)$ primes in progression cannot be extended to $k$ would enforce too high a degree of structure.  The reason 40 pages of cutting-edge analytic number theory and Fourier analysis are needed, is to convert the simple heuristics about &quot;information&quot; and &quot;correlation&quot; into provable and well-defined mathematical assertions, which is complicated because the primes themselves are deterministically defined, not a sequence of random numbers to which the probabilistic concepts most naturally apply.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="467" LastEditorDisplayName="" LastEditDate="2010-12-23T03:28:15.893" LastActivityDate="2010-12-23T03:28:15.893" CommentCount="17" />
  <row Id="15265" PostTypeId="2" ParentId="15231" CreationDate="2010-12-23T00:45:44.587" Score="5" ViewCount="0" Body="&lt;p&gt;There are two different possible questions here. When you ask for the solution of an NP-complete problem, you can either (a) require the computer to give you a witness in the &quot;yes&quot; cases or (b) just require the computer to give you the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are just asking for the answer, as in (b). Then you can partition the set of instances of $\Pi$ into &quot;yes&quot; instances and &quot;no&quot; instances. For each set of instances, there is a one-line program which gives you the answer for that set of instances. So in this case, the answer to your question is &quot;yes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose, as in (a), you require a witness in the &quot;yes&quot; cases as well as the answer. Let the algorithms for the different subclasses be $A_1$, $A_2$, $A_3$, $\ldots$, $A_k$. Now we give a master algorithm $A$ that, given an instance $x$, runs all $k$ algorithms on it. $A$ checks the witnesses that the algorithms $A_i$ output to see whether one of them verifies that $x$ is a &quot;yes&quot; instances. If one of them does, then $A$ outputs &quot;yes&quot;. Otherwise, $A$ outputs &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a &quot;yes&quot; instance, one of the algorithms $A_i$ must output a valid witness, and so $A$ will accept.  If $x$ is a &quot;no&quot; instance, there are no valid witnesses, and so $A$ will reject. Thus, for case (a), if the answer to your question is &quot;yes&quot;, $A$ solves an NP-complete problem in polynomial time, and we have P$=$NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this argument is original; I'm just consolidating and summarizing the answers that this question got on &lt;a href=&quot;http://cstheory.stackexchange.com/questions/3592/can-we-partition-np-complete-problem-into-finite-number-of-polynomially-solvable&quot;&gt;cstheory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditorDisplayName="" LastEditDate="2010-12-23T00:51:45.250" LastActivityDate="2010-12-23T00:51:45.250" CommentCount="7" />
  <row Id="15266" PostTypeId="2" ParentId="15039" CreationDate="2010-12-23T00:49:55.500" Score="0" ViewCount="0" Body="&lt;p&gt;Falling in to the trap of thinking that mathematical usage should correspond directly with usages in English is very dangerous.  I would advise focusing on understanding the mathematics in whatever way you can and accepting that it might occasionally offend your sense of proper English grammar or semantics.  English is just another language, albeit a very successful global language.  There is no reason it should be perfectly conducive to conveying mathematical meaning.  Note that most English speakers are not mathematicians and yet they are the population that has the largest effect on the evolution of the English language.  Mathematical language is functional, meant to help us get greater insight into mathematical objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examining 'of' as defined by dictionary.com, I was able to clean the following relevant possible usages: used to indicate distance or direction from, separation, deprivation; used to indicate derivation, origin, or source; used to indicate cause, motive, occasion, or reason; used to indicate material, component parts, substance, or contents; used to indicate apposition or identity; used to indicate specific identity or a particular item within a category; used to indicate possession, connection, or association; used to indicate inclusion in a number, class, or whole; used to indicate the objective relation, the object of the action noted by the preceding noun or the application of a verb or adjective; used to indicate reference or respect; used to indicate qualities or attributes; used to indicate a specified time; on the part of; in respect to; and set aside for or devoted to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having looked at the list, I was not convinced that the preposition 'of' had unique meaning in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usage 'f of x' seems like the only usage that has even the slightest hint of uniqueness to mathematics. However, we hear sentences like 'this is the average of', 'this is the maximum of' or &quot;this is the total of&quot;.  This seems to be the sense in which 'f of x' is meant. &lt;/p&gt;&#xA;" OwnerUserId="1004" LastActivityDate="2010-12-23T00:49:55.500" CommentCount="2" />
  <row Id="15267" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T00:54:54.223" Score="7" ViewCount="0" Body="&lt;p&gt;I'd propose the curve of Archytas, the space curve formed by the intersection of a horn torus and a cylinder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/sxAR5.png&quot; alt=&quot;curve of Archytas&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parametric equations for the curve of Archytas are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\cos^2t\qquad\frac12\sin\;2t\qquad\pm\sqrt{2\cos\;t}\sin\frac{t}{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mlahanas.de/Greeks/Bios/Archytas.html&quot; rel=&quot;nofollow&quot;&gt;Archytas of Tarentum&lt;/a&gt; used this curve to help him solve the &lt;a href=&quot;http://mathworld.wolfram.com/CubeDuplication.html&quot; rel=&quot;nofollow&quot;&gt;Delian problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-23T00:54:54.223" />
  <row Id="15268" PostTypeId="2" ParentId="15166" CreationDate="2010-12-23T01:02:11.163" Score="2" ViewCount="0" Body="&lt;p&gt;The whole &quot;paddlewheel&quot; business was explained to me the first time I took multivariable calculus (I was 17), and I confess I found it entirely mysterious.  When I taught multivariable calculus myself for the first time (I was 27), I was delighted to find that with a little thought one could make sense of these geometric and physical explanations in a fairly satisfying way.  (Note: I knew less about physics at age 27 than age 17, so that's not the difference.  In between I had gotten a PhD in mathematics, which probably helped, although perhaps mostly in a psychological way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote these observations up in notes for my class.  See Section 2 &lt;a href=&quot;http://math.uga.edu/~pete/handoutfive.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and Section 3 &lt;a href=&quot;http://math.uga.edu/~pete/handouteight.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-23T01:02:11.163" CommentCount="1" />
  <row Id="15269" PostTypeId="2" ParentId="15055" CreationDate="2010-12-23T01:05:05.577" Score="4" ViewCount="0" Body="&lt;p&gt;I think this question confuses a lot of people because there's a lack of intuitive context -- I'll try to supply that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is a birthday party to which all of the girls (and none of the boys) in a small town are invited. If you run into a mother who has dropped off a kid at this birthday party and who has two children, the chance that she has two girls is $1/3$. Why? $3/4$ of the mothers with two children will have a daughter at the birthday party, the ones with two girls ($1/4$ of the total mothers with two children) and the ones with one girl and one boy ($1/2$ of the total mothers with two children). Out of these $3/4$ of the mothers, $1/3$ have two girls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the birthday party is only for fifth-grade girls, you get a different answer. Assuming there are no siblings who are both in the fifth grade, the answer in this case is $1/2$. The child in fifth grade is a girl, but the other child has probability $1/2$ of being a girl. Situations of this kind arise in real life much more commonly than situations of the other kind, so the answer of $1/3$ is quite nonintuitive.&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="2501" LastEditorDisplayName="" LastEditDate="2010-12-30T01:54:48.190" LastActivityDate="2010-12-30T01:54:48.190" CommentCount="2" />
  <row Id="15270" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T01:18:37.547" Score="2" ViewCount="0" Body="&lt;p&gt;Let $t \in [0,2\pi]$. And let $r(t)=\frac{1}{2-\sin (2t)}[\cos(3t),\sin(3t),\cos(2t)]$. Then $r(t)$ parametrizes a 3-dimensional curve with no three-tangent plane (I must admit I'm not sure of the English terminology here) - that is, every tangent plane of the curve meets the curve in no more than two points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have no clue how to prove such things, however)&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-23T01:18:37.547" CommentCount="2" />
  <row Id="15271" PostTypeId="2" ParentId="15194" CreationDate="2010-12-23T01:49:13.207" Score="3" ViewCount="0" Body="&lt;p&gt;I'll use the characterisation of the geometric mean as given by Slowsolver. As I mentioned in a comment to his answer, the geometric mean $GM(A,B)$ is the geodesic midpoint corresponding to the Riemannian metric on the space of symmetric positive definite matrices with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g(X,Y) = \mathop{tr}(A^{-1}XA^{-1}Y)~,\quad X,Y\in T_A(SPD) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the tangent space $T_A(SPD)$ is the space of symmetric matrices. This interpretation has the following advantage: both the AM and GM are &quot;translation invariant&quot;. Let $P$ be a non-singular matrix, then $AM(PAP^T,PBP^T) = P(AM(A,B))P^T$ trivially, and similarly for $GM$ using that $$g_A(X,Y) = g_{PAP^T}(PXP^T, PYP^T)$$ so that a geodesic $\gamma$ joining $A,B$ becomes a geodesic $P\gamma P^T$ joining $PAP^T$ and $PBP^T$. By first conjugating with $P = A^{-1/2}$, we can WLOG assume $A = I$. By then conjugating with a suitable orthogonal matrix $O$ to diagonalize $B$, we can WLOG assume $A = I$ and $B$ is diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result on convergence now follows from the convergence of the AGM for the single, real variable case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The rate of convergence is the same as the scalar case, up to a constant factor depending on the initial values of $A,B$ (which determines $P$). &lt;/li&gt;&#xA;&lt;li&gt;In this definition $GM(A,B) = GM(B,A)$, so the question is moot. &lt;/li&gt;&#xA;&lt;li&gt;This is in some sense the natural generalisation of the scalar case. Notice that the scalar geometric mean corresponds to the geodesic midpoint for the multiplicative Lie group $(\mathbb{R}_+,\times)$ with the invariant metric. A first generalisation is to the abelian Lie group $\mathbb{R}^k_+$ with componentwise multiplication. (This corresponds to the diagonal case.) What we showed above is that the entire AGM sequence for arbitrary starting SPD matrices $A$ and $B$ is contained, up to conjugation by $P\cdot P^T$, in one such abelian case. Note that this is slightly less related to eigenvalues per se, since the transformation we use above does not preserve eigenvalues. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: It just occurred to me to do a literature search. In &lt;a href=&quot;http://www.jstor.org/stable/2695553&quot; rel=&quot;nofollow&quot;&gt;Lawson, J. D. &amp;amp; Lim, Y. &quot;The geometric mean, matrices, metrics, and more&quot;. Amer. Math. Monthly, 2001, 108, 797-812&lt;/a&gt;, many properties of this geometric mean on SPD matrices are derived in Sections 2 and 3. In particular, the crucial property (as slowsolver remarks below) of being able to set $A = I$ and $B$ diagonal, is the content of Lemma 3.1 of that paper. That is followed by the harmonic-geometric-arithmetic-mean inequality for SPD matrices. By starting with that and playing with monotonicity properties, you presumably also get another proof of the convergence result. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-23T16:10:41.600" LastActivityDate="2010-12-23T16:10:41.600" CommentCount="8" />
  <row Id="15272" PostTypeId="1" CreationDate="2010-12-23T01:51:07.700" Score="2" ViewCount="70" Body="&lt;p&gt;How to prove that the Sorgenfrey line is hereditarily separable?&lt;/p&gt;&#xA;" OwnerUserId="5059" LastActivityDate="2010-12-24T15:56:42.387" Title="Sorgenfrey line is hereditarily separable" Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="15273" PostTypeId="2" ParentId="15252" CreationDate="2010-12-23T02:13:04.323" Score="0" ViewCount="0" Body="&lt;p&gt;Galois theory is useful when you have some algebraic object, and a list of tools you are allowed to use within that object. The purpose of Galois theory is to explain how far one can go only using those tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it is impossible to create, using only the tools of +, -, *, / and nth roots, a formula for the zeroes of a general 5th degree polynomial. This is the most basic application of Galois theory. This is the Abel-Ruffini theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another application is Differential Galois theory, which tells you which integrals can be expressed in terms of elementary functions. But once again, it comes down to having only a few basic tools with which to construct an anti-derivative for a function using only elementary functions, and that is why Galois theory is useful in this application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, its hard to understand what your question is asking about, because you haven't really phrased your question as a Galois Theory question.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-23T02:13:04.323" />
  <row Id="15274" PostTypeId="1" AcceptedAnswerId="15288" CreationDate="2010-12-23T02:18:34.833" Score="2" ViewCount="60" Body="&lt;p&gt;I'm going through Enderton's Elements of Set Theory, as I heard it is a gentle introduction to set theory. I'm a little confused on how the subset axioms are used. In the text, the axiom is given as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Subset Axioms&lt;/strong&gt; For each formula $\underline{\hspace{1cm}}$ not containing $B$, the following is an axiom:&#xA;  $$&#xD;&#xA;\forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge \underline{\hspace{1cm}}).&#xD;&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later, the text asserts the existence of the set $\{A\cup X\ | \ X\in\mathscr{B}\}$, call it $\mathscr{D}$, saying that since $A\cup X\subseteq A\cup\bigcup\mathscr{B}$, $\mathscr{D}$ is a subset of $\mathscr{P}(A\cup\bigcup\mathscr{B})$. A subset axiom produces &#xA;$$&#xD;&#xA;\{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what exactly is that subset axiom? Would it be something of the form&#xA;$$&#xD;&#xA;\forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [(x=t_1\cup t_2)\vee(x=t_1\cup t_3)\vee\cdots\vee(x=t_1\cup t_k)])&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we take $A$ to be a particular instance of $t_1$, $\mathscr{P}(A\cup\bigcup\mathscr{B})$ to be a particular instance of $c$, and $t_2,\dots,t_k$ to be particular instances of members of $\mathscr{B}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps something like&#xA;$$&#xD;&#xA;\forall t_1\forall t_2\forall t_3\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [\exists t_3\in t_2(x=t_1\cup t_3)])&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where in this case $\mathscr{B}$ is a particular instance of $t_2$?, and then $\mathscr{D}$ would be the set $B$ which has been proven to exist. Thanks, right now I'm just not used to what should go in that blank spot for the formula.&lt;/p&gt;&#xA;" OwnerUserId="1583" LastActivityDate="2010-12-23T03:58:44.367" Title="Using the Subset Axioms to prove the existence of a set" Tags="&lt;elementary-set-theory&gt;&lt;axioms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15275" PostTypeId="1" AcceptedAnswerId="15277" CreationDate="2010-12-23T02:32:01.130" Score="4" ViewCount="78" Body="&lt;p&gt;Almost everyone is familiar with the famous Taylor Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, if it converges at more than one point, will converge in some interval about $a$. Has anyone considered the &quot;Reverse&quot; Taylor Series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g(x) = \sum_{n=0}^\infty \frac{f^{(n)}(x-a)}{n!} a^n $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I call it reverse, because its what you get for symbolically taking $a \rightarrow (x-a)$. You might be saying that there is absolutely no reason to believe that this series should converge to $f(x)$, but there are two big examples where it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $e^x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{n=0}^\infty \frac{e^{(x-a)}}{n!} a^n = \frac{e^x}{e^a} \sum_{n=0}^\infty \frac{a^n}{n!} = e^x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x^k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(x) = \sum_{n=0}^k \frac{k(k-1)...(k-n+1)(x-a)^{k-n} a^n}{n!} = \sum_{n=0}^k {k \choose n} (x-a)^{k-n} a^n = ((x-a) + a)^k $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to find a counter-example for when the Reverse Taylor Series does not give back the original function for an analytic function, and I also have yet to think of a way to prove that the Reverse Taylor Series should converge for a given function. Does anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-23T02:57:52.947" Title="Reverse Taylor Series" Tags="&lt;calculus&gt;&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15276" PostTypeId="2" ParentId="15248" CreationDate="2010-12-23T02:42:13.123" Score="-3" ViewCount="0" Body="&lt;p&gt;Since each of the $m+1$ simple closed curves are disjoint, it is clear that $\Omega$ has $m+1$ connected components. Let $U$ be any one of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $f(z)$ restricted to $U$ is a holomorphic function. Furthermore, it is non-constant and holomorphic on some neighborhood of the closure of $U$, because this closure is contained in $\Omega$. Furthermore, the boundary of $U$ is contained in the boundary of $\Omega$, so $|f(z)| = 1$ on the boundary of $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like you outlined in your explanation of the problem, if we assume that $f$ is non-zero in $U$ then both $f(z)$ and $1/f(z)$ are holomorphic in $U$, and thus must achieve their maximums on the boundary of $U$. But its given that $|f(z)| = 1$ on the boundary of $U$, and so we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|f(z)| \leq 1$ and $|1/f(z)| \leq 1$ for $z \in U$. Thus we must have that $|f(z)| \equiv 1$ on $U$, which is a contradiction as we assumed that $f(z)$ is non-constant. Thus $f$ has at least one zero in $U$. We assumed nothing about which connected component we used, so this is true in each component of $\Omega$, and therefore $f(z)$ has at least $m+1$ zeroes.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2010-12-23T02:42:13.123" CommentCount="2" />
  <row Id="15277" PostTypeId="2" ParentId="15275" CreationDate="2010-12-23T02:48:50.303" Score="6" ViewCount="0" Body="&lt;p&gt;Your two series are the same thing pointwise when they converge. $g(x)$ is the Taylor series for $f$ expanded around the point $x-a$. So it will converge as long as the Taylor series for $f$ at $x-a$ has a radius of convergence at least $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in $f(x)$, moving $x$ around may make the Taylor series fail to converge due to your $x$ moving out of the radius of convergence. Whereas in $g(x)$, moving $x$ around will make the series fail to converge due to &lt;em&gt;moving the point&lt;/em&gt;  $x - a$  &lt;em&gt;somewhere with a radius of convergence smaller than&lt;/em&gt;  $a$. That is, once you fix $a$, in the Taylor series for $f$ around $a$, the convergence radius is fixed, independent of $x$. But in the series $g(x)$, the convergence of the series depends on how the radius of convergence at $x-a$ (which depends on $x$) compares with $a$. So that the function $g(x)$ converges at more than one point is insufficient to guarantee that $g(x)$ converges in a small interval about $a$. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-23T02:48:50.303" />
  <row Id="15278" PostTypeId="1" AcceptedAnswerId="15402" CreationDate="2010-12-23T02:49:24.030" Score="9" ViewCount="167" Body="&lt;p&gt;&lt;i&gt;I've revised my question a bit in response to the (very helpful) advice so far--&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an engineering background but am interested in learning abstract harmonic analysis. My interest is rather unstructured; the Fourier transform sparked my interest in mathematics, and I've been following since I first learned of it as a sort of guide through mathematics. I've decided it's time I delved into the abstract, and from my readings so far it's become clear to me that this will require a sojourn into a swathe of unfamiliar mathematics. My background in classical analysis, linear algebra, and probability is rather good (probably the rough equivalent of a 1st year american grad student), and so I'm familiar as well with point-set topology. I'm also familiar with measure theory (via probability, and some real analysis). I have essentially no abstract algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested in three bits of advice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What algebra is essential for the canon of harmonic analysis? Perhaps this is a vacuous question (i.e., that an answer is as simple &quot;just google&quot; or &quot;open a book&quot;)? It strikes me that techniques or background might work their way in unnamed. I'm hoping to prevent the sinking feeling that I don't know something I should. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Specific examples (theorems, objects, counterexamples) which illustrate important applications of harmonic analysis, especially in relation to algebra. I'm not looking for explanations, just statements of what's thought to be important. I'm hoping that this information will help me sift through the different branches of harmonic analysis, especially where I am most unfamiliar with the language. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Canonical/important texts which I might find useful. I'm especially interested in those good for self study (well-edited with many examples and end-of-chapter exercises, and ideally a conversational tone with lengthy remarks and some historical content). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much!&lt;/p&gt;&#xA;" OwnerUserId="5060" LastEditorUserId="5060" LastEditorDisplayName="" LastEditDate="2010-12-24T08:47:25.130" LastActivityDate="2010-12-24T14:02:12.033" Title="learning algebra and harmonic analysis" Tags="&lt;soft-question&gt;&lt;fourier-analysis&gt;&lt;harmonic-analysis&gt;" AnswerCount="3" />
  <row Id="15279" PostTypeId="2" ParentId="15275" CreationDate="2010-12-23T02:49:27.993" Score="3" ViewCount="0" Body="&lt;p&gt;The general idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put $y=x-a$.&#xA;Then, &#xA;$$&#xD;&#xA;g(x) = \sum\limits_{n = 0}^\infty  {\frac{{f^{(n)} (y)}}{{n!}}(x - y)^n }  = f(x).&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-23T02:57:52.947" LastActivityDate="2010-12-23T02:57:52.947" />
  <row Id="15280" PostTypeId="1" CreationDate="2010-12-23T02:53:10.370" Score="2" ViewCount="112" Body="&lt;p&gt;I became interested in how numbers actually work recently. I want a source that covers number theory. Explaining how and why numbers react the way they do. A friend recommended Mathematical Proofs: A Transition to Advanced Mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to learn about the concept of numbers, how they interact etc. I am a high school student, but most college level material will be appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Also, in my search I have found Mathematics in 10 lessons: The Grand Tour, and Foundations and Fundamentals of Mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could confirm that those two resources are adequate, I would appreciate it, and other resources are still welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I do not like the Khan Academy. His presentations do not... serve my needs.&lt;/p&gt;&#xA;" OwnerUserId="5061" LastEditorUserId="5061" LastEditorDisplayName="" LastEditDate="2010-12-23T17:34:53.543" LastActivityDate="2010-12-23T17:57:02.690" Title="Fundamental Math Theory Resources? " Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;learning&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="15281" PostTypeId="2" ParentId="15280" CreationDate="2010-12-23T03:03:38.547" Score="2" ViewCount="0" Body="&lt;p&gt;For an overview of the many parts of mathematics and the goals and accomplishments of various subareas within mathematics I recommend Timothy Gowers (ed.) The Princeton Companion to Mathematics. The book has pointers to more technical issues if you want to go into more depth after having a broad view. &lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-23T03:03:38.547" />
  <row Id="15282" PostTypeId="2" ParentId="15280" CreationDate="2010-12-23T03:11:16.383" Score="2" ViewCount="0" Body="&lt;p&gt;It seems like you're interested in seeing how mathematics is used beyond calculation in a general sort of way. Here are some recommendations - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Euler's Fabulous Formula&lt;br/&gt;&#xA;Paul Nahin &#xA;(and check out his other work as well)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number: The Language of Science &lt;br/&gt;&#xA;Tobias Dantzig&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pleasures of Counting &lt;br/&gt;&#xA;T.W. Korner&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gödel, Escher, Bach&lt;br/&gt;&#xA;Douglas R. Hofstadter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These texts try to tease out interesting ideas in an accessible way. They're not exactly going to pave the road to learning proof (for that, I think a teacher is best), but they might help you see where you're going. &lt;/p&gt;&#xA;" OwnerUserId="5060" LastActivityDate="2010-12-23T03:11:16.383" />
  <row Id="15283" PostTypeId="2" ParentId="15212" CreationDate="2010-12-23T03:15:15.713" Score="3" ViewCount="0" Body="&lt;p&gt;The best explanation I have heard for this is: The roulette wheel has no memory. When you spin the wheel it does not know that there have been 5 reds in a row and it is due for a black. The odds are 1/2 each time.&lt;/p&gt;&#xA;" OwnerUserId="5062" LastActivityDate="2010-12-23T03:15:15.713" />
  <row Id="15284" PostTypeId="2" ParentId="15272" CreationDate="2010-12-23T03:22:44.007" Score="2" ViewCount="0" Body="&lt;p&gt;Hint: Let $S$ be a subset of the Sorgenfrey line. From each interval of the form $[q,r)$ in the original line, where $q,r \in \mathbb{Q}$, pick one point from $S$ if possible. Then characterize the points in $S$ that are not limits of the points you just chose. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-23T03:22:44.007" />
  <row Id="15285" PostTypeId="1" AcceptedAnswerId="15287" CreationDate="2010-12-23T03:24:42.167" Score="4" ViewCount="158" Body="&lt;p&gt;It's famously unknown whether the natural log of 2 is rational or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about the natural log of other numbers. Is it known/unknown whether these are rational?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously ln(1) is 0, and ln(2^n) is n*ln(2) (and is thus rational iff ln(2) is rational), but how about other cases?&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:29:06.960" LastActivityDate="2010-12-23T05:29:06.960" Title="Is the natural log of n rational?" Tags="&lt;real-analysis&gt;&lt;logarithms&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="15287" PostTypeId="2" ParentId="15285" CreationDate="2010-12-23T03:29:21.780" Score="6" ViewCount="0" Body="&lt;p&gt;To shoot a bird with a canon...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Lindemann-Weierstrass theorem&lt;/a&gt;, $e^a$ is transcendental for all $a$ algebraic and non-zero. In particular if $a$ is rational, $e^a$ cannot be rational. Hence $\ln(n)$ is always irrational. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-23T03:29:21.780" CommentCount="4" />
  <row Id="15288" PostTypeId="2" ParentId="15274" CreationDate="2010-12-23T03:32:47.297" Score="4" ViewCount="0" Body="&lt;p&gt;If you want to form the set &#xA;$$&#xD;&#xA;\{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}.&#xD;&#xA;$$&#xA;then a natural instance of the axiom of separation (the &quot;subset axiom&quot;) is&#xA;$$&#xD;&#xA;(\exists B)(\forall t) ( t \in B \Leftrightarrow t\in\mathscr{P}(A\cup\bigcup\mathscr{B}) \land (\exists X)[X \in \mathscr{B} \land (\forall s) (s \in t \Leftrightarrow s \in A \lor s \in X)]).&#xD;&#xA;$$&#xA;This formula is obtained by simply unpacking the original abbreviated formula, for example by expanding the &quot;=&quot; using its set-theoretic definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a general principle that if you have assumed a formula of the form $(\forall z)\phi(z)$ then, for any set $Y$, the formula $\phi(Y)$ will be true as well. This is called (universal) instantiation. The way you get to the separation axiom above is to start with &#xA;$$&#xD;&#xA;(\forall C_1)(\forall C_2)(\forall C)(\exists B)(\forall t) ( t \in B \Leftrightarrow t\in C \land (\exists X)[X \in C_1 \land (\forall s) (s \in t \Leftrightarrow s \in C_2 \lor s \in X)]).&#xD;&#xA;$$&#xA;and then instantiate it with $C_1 = \mathscr{B}$, $C_2 = A$, $C = \mathscr{P}(A\cup\bigcup \mathscr{B}) $. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-23T03:58:44.367" LastActivityDate="2010-12-23T03:58:44.367" CommentCount="1" />
  <row Id="15289" PostTypeId="1" CreationDate="2010-12-23T03:50:32.903" Score="3" ViewCount="68" Body="&lt;p&gt;Let $T: E \rightarrow E$ be an endomorphism of a finite-dimensional vector space, and let $S$ be a circle in the complex plane that does not intersect any eigenvalues of $T$. Now let $Q = \frac{1}{2\pi i} \int_S (z-T)^{-1} \, dz$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $Q$ a projection operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation behind this question is that the above situation occurs in a proof of Bott's periodicity theorem, but it's not clear to me that $Q$ is a projection...&lt;/p&gt;&#xA;" OwnerUserId="423" LastEditorUserId="423" LastEditorDisplayName="" LastEditDate="2010-12-23T04:27:50.780" LastActivityDate="2010-12-23T04:27:50.780" Title="why is the following thing a projection operator?" Tags="&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15290" PostTypeId="2" ParentId="15289" CreationDate="2010-12-23T03:53:53.727" Score="3" ViewCount="0" Body="&lt;p&gt;Show that the integral depends continuously on $T$, and show that $Q^2=Q$ when $T$ is diagonalizable, by finding how $Q$ changes change you change $T$ by a similar matrix, and then reducing to the one dimensional case. Then use the fact that diagonalizable matrices are dense in the space of all matrices, and that $Q^2$ and $Q$ are continuous functions of $Q$.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-23T03:53:53.727" CommentCount="5" />
  <row Id="15291" PostTypeId="2" ParentId="15289" CreationDate="2010-12-23T04:05:14.137" Score="2" ViewCount="0" Body="&lt;p&gt;If $A$ is any Banach algebra (such as the algebra of endomorphisms of a finite dimensional  complex vector space), then for each subset $\Omega$ of the complex plane and each element $T$ of $A$ whose spectrum is contained in $\Omega$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_functional_calculus&quot; rel=&quot;nofollow&quot;&gt;holomorphic functional calculus&lt;/a&gt; yields a homomorphism $f\mapsto f(T)$ from the algebra of functions holomorphic in an open set containing $\Omega$ (identified if they agree on some neighborhood of $\Omega$) into $A$.  Since the function $f:(\mathbb{C}\setminus S)\to\mathbb{C}$ defined by $f(w)=\frac{1}{2\pi i}\int_S(z-w)^{-1}dz$ takes on only the values $0$ and $1$ (it gives the winding number of $S$ about $w$), $f$ is idempotent (i.e., $f(w)^2=f(w)$ for all $w\in\mathbb{C}\setminus S$), and thus $f(T)$ is an idempotent element of $A$ for each $T$ whose spectrum is disjoint from $S$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-23T04:18:29.757" LastActivityDate="2010-12-23T04:18:29.757" CommentCount="5" />
  <row Id="15292" PostTypeId="2" ParentId="13054" CreationDate="2010-12-23T04:17:20.763" Score="1" ViewCount="0" Body="&lt;p&gt;Would &lt;a href=&quot;http://math.stackexchange.com/questions/15285/is-the-natural-log-of-n-rational/15287#15287&quot;&gt;this&lt;/a&gt; help? Willie Wong mentions that by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem&quot; rel=&quot;nofollow&quot;&gt;Lindemann-Weierstrass theorem&lt;/a&gt;, $e^x$ (where $x$ is algebraic) is transcendental (and hence not an integer). We are then left with $e^{e^y}$ where $y$ is transcendental. Can we infer anything from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt; I think we have no need in $e^{e^{e^{79}}}$ anymore, as we can refer to a number like $e^{9999999999999999999999999999999999999999999999999!}$, which is even greater than $e^{e^{e^{79}}}$ (in powers of 10 representation, $e^{9999999999999999999999999999999999999999999999999!} = 10^{10^{10^{50.68632970215017}}}$ while $e^{e^{e^{79}}} = 10^{10^{10^{33.94704838165742}}}$), and $9999999999999999999999999999999999999999999999999!$ is clearly algebraic, so $e^{9999999999999999999999999999999999999999999999999!}$ would be transcendental (hence non-integer), and will serve the purpose of my &lt;a href=&quot;http://math.stackexchange.com/questions/13050/eee79-and-ultrafinitism&quot;&gt;original question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2:&lt;/strong&gt; Hold on, using the aforementioned representation of $e^{e^{e^{79}}}$ as $10^{10^{10^{33.94704838165742}}}$ we can show that $e^{e^{e^{79}}}$ is not an integer because $10^{33.94704838165742}$ &lt;a href=&quot;http://goo.gl/vdIYZ&quot; rel=&quot;nofollow&quot;&gt;isn't one&lt;/a&gt;. So we are left with $10^{10^{x}}$ where $x$ is not an integer, &lt;s&gt; and 10 to any non-integer power $&amp;gt;1$ is not an integer (or could it be one? Can a PMI-based proof be devised to show it can't be one?), so we could claim that the whole expression is not an integer, as it is $10$ continuously being raised to the power of $x$ where $x$ is a non-integer $&amp;gt;1$.&lt;/s&gt; Thoughts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3:&lt;/strong&gt; Counter-example: $10^{9\frac{log(2)}{log(2)+log(5)}} = 512$ and the exponent is clearly $&amp;gt;1$, so I guess stuff from &lt;strong&gt;EDIT 2&lt;/strong&gt; does not apply. Back to the point of $e^{9999999999999999999999999999999999999999999999999!}$ then. :)&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-23T05:54:38.943" LastActivityDate="2010-12-23T05:54:38.943" />
  <row Id="15293" PostTypeId="2" ParentId="15240" CreationDate="2010-12-23T04:28:59.047" Score="1" ViewCount="0" Body="&lt;p&gt;The &lt;a href=&quot;http://www.mathresources.com/products/mathresource/maa/mooreosgood_theorem.html&quot; rel=&quot;nofollow&quot;&gt;Moore-Osgood Theorem&lt;/a&gt; might also be relevant here.&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-23T04:28:59.047" />
  <row Id="15294" PostTypeId="1" CreationDate="2010-12-23T04:47:57.227" Score="5" ViewCount="210" Body="&lt;p&gt;I understand how derivatives work based on the definition, and the fact that my professor explained it step by step until the point where I can derive it myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when it comes to the area under a curve for some reason when you break it up into an infinite amount of rectangles, magically it turns into the anti-derivative. Can someone explain why that is the definition of the integral and how Newton figured this out?&lt;/p&gt;&#xA;" OwnerUserId="5063" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:30:45.383" LastActivityDate="2010-12-23T06:40:29.690" Title="Why is the area under a curve the integral?" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="15295" PostTypeId="1" AcceptedAnswerId="15305" CreationDate="2010-12-23T04:50:35.530" Score="2" ViewCount="42" Body="&lt;p&gt;Suppose $\kappa$ is an uncountable cardinal. Then $S\subseteq\kappa$ is club (CLosed UnBounded) if $S$ is unbounded in $\kappa$ and is a closed subset of $\kappa$ under the order topology. My question is the following: suppose $S\subseteq\kappa$ is club, and $S=S_1\cup S_2$. Is there necessarily some club set $C$ such that either $C\subseteq S_1$ or $C\subseteq S_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that, since the intersection of two clubs is again club, this question amounts to asking whether, for $\kappa$ an infinite cardinal, the set $\{S\subseteq\kappa: S$ is a superset of some club set $\}$ is an ultrafilter.) &lt;/p&gt;&#xA;" OwnerUserId="5064" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-23T06:54:43.563" LastActivityDate="2010-12-23T06:54:43.563" Title="Splitting a club set" Tags="&lt;set-theory&gt;" AnswerCount="2" />
  <row Id="15296" PostTypeId="2" ParentId="15240" CreationDate="2010-12-23T04:57:30.503" Score="3" ViewCount="0" Body="&lt;p&gt;If you want to avoid hypotheses that involve uniform convergence, you can always cheat and use the counting measure on {0, 1, 2,...} and then use either the Monotone or Dominated Convergence Theorem from integration theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monotone_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Monotone Convergence Theorem&lt;/a&gt;, we get the following (perhaps silly) sufficient criterion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition:&lt;/strong&gt; If $a_{mn}$ is monotonically increasing in $n$, and is such that $c_{mn} = a_{m,n+1} - a_{mn}$ is also monotonically increasing in $n$, then $\lim\limits_{m \to \infty} \lim\limits_{n \to \infty} a_{mn} = \lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{mn}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof:&lt;/em&gt; Our two hypotheses really just amount to saying that each $c_{mn} \geq 0$ and that the $c_{mn}$ are monotonically increasing in $n$.  So we can use the Monotone Convergence Theorem with respect to the counting measure:&#xA;$$\lim_{m\to \infty} \int c_{mn} = \int \lim_{m\to \infty} c_{mn}$$&#xA;But really, these integrals are sums, so:&#xA;$$\lim_{m\to \infty} \sum_{n=0}^\infty c_{mn} = \sum_{n=0}^\infty \lim_{m\to \infty} c_{mn}$$&#xA;Since infinite sums are just limits of partial sums, we have:&#xA;$$\lim_{m\to \infty} \lim_{N \to \infty} \sum_{n=0}^N c_{mn} = \lim_{N\to\infty}\lim_{m\to \infty} \sum_{n=0}^N c_{mn}$$&#xA;By our construction of $c_{mn}$, the left side is $\lim\limits_{m\to\infty} \lim\limits_{N\to\infty} a_{mN}$, and the right side is $\lim\limits_{N\to\infty} \lim\limits_{m\to\infty}a_{mN}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Dominated Convergence Theorem&lt;/a&gt;, we could probably get something slightly more useful.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-27T08:30:21.430" LastActivityDate="2010-12-27T08:30:21.430" CommentCount="4" />
  <row Id="15297" PostTypeId="1" CreationDate="2010-12-23T05:09:02.693" Score="0" ViewCount="98" Body="&lt;p&gt;Hi Guys, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of an assignment that I need to get a good mark for - I'd appreciate it if you guys could look over it and give some pointers where I've gone wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(apologies for the italics)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)\; converges \;when&#xD;&#xA;\sum_1^\infty ln\left(1+\frac{2}{n}\right)\;converges$$&#xA; $$\sum_1^\infty ln\left(1+\frac{2}{n}\right)\;=\;\sum_1^\infty ln(n+2)-ln(n)$$&#xA; $$ let\, f(x)=ln(x+2)-ln(x) \rightarrow f&amp;#39;(x)=\frac{1}{x+2} - \frac{1}{x}$$&#xA; $$ = \frac{x-x-2}{x(x+2)} = \frac{-2}{x(x+2)}&amp;lt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)\; is\; a\; decreasing \;function.$$&#xA; $$f(x) \;is \;a \;positive \;function \;for \;x&amp;gt;=1$$ &#xA; $$f(x)\;is \;a \;continuous \;function \;for \;x&amp;gt;=1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;using integration test. \newline&#xA; $$\int_1^\infty ln(x+2) - ln(x) = \lim_{t \to \infty}\int_1^t ln(x+2)dx -&#xD;&#xA; \lim_{t \to \infty}\int_1^t lnx dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int ln(x)dx = xlnx - x + c \;\;it\;follows\;that\; \int ln(x+2) =&#xD;&#xA;(x+2)ln(x+2) - (x+2) + c$$&#xA;Therefore&#xA;$$\int ln(x+2) - ln(x)dx  = (x+2)ln(x+2)-x - 2 - xln(x) + x + c$$&#xA;$$ = xln(\frac{x+2}{x})+ 2ln(x+2)-2 + c $$&#xA;Therefore, &#xA;$$\int_1^\infty ln(x+2) - ln(x)dx = \lim_{t \to&#xD;&#xA;\infty}\left[xln(\frac{x+2}{x}) + 2ln(x+2) - 2\right]_1^t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \lim_{t \to&#xD;&#xA;\infty}\left[t ln(\frac{t+2}{t}) + 2ln(t+2) - 2\right] - \lim_{t \to&#xD;&#xA;\infty}\left[ln(\frac{3}{1}) + 2ln(3) - 2\right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ =\lim_{t \to&#xD;&#xA;\infty}\left[t ln(\frac{t+2}{t}) + 2ln(t+2) - 3ln(3)\right]$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ As\; t\rightarrow\infty, \; \lim_{t \to \infty}t ln\left(\frac{t+2}{t}\right)&#xD;&#xA;+ 2ln(t+2) = \infty. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Therefore\; the \; series\; \sum_1^\infty ln\left(1+\frac{2}{n}\right)\; is\;&#xD;&#xA;divergent$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Similarly\; the\; infinite\; product\;&#xD;&#xA;\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)\; is\; also\; divergent$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gregg&lt;/p&gt;&#xA;" OwnerUserId="4362" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-23T08:36:26.417" LastActivityDate="2010-12-23T09:01:22.193" Title="integral test to show that infinite product $ \prod_{n=1}^\infty\left(1+\frac{2}{n}\right)$ diverges" Tags="&lt;calculus&gt;&lt;homework&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="15298" PostTypeId="2" ParentId="7296" CreationDate="2010-12-23T05:14:58.913" Score="0" ViewCount="0" Body="&lt;p&gt;i would like to know the solution of the abel equation from either the integral form or the differintegral one&lt;/p&gt;&#xA;" OwnerUserId="5065" LastActivityDate="2010-12-23T05:14:58.913" />
  <row Id="15299" PostTypeId="2" ParentId="15285" CreationDate="2010-12-23T05:19:53.260" Score="4" ViewCount="0" Body="&lt;p&gt;We can also use a non-simple continued fraction expansion of $\displaystyle e^{2x/y}$ to prove the irrationality of $\displaystyle e^{2x/y}$ when $\displaystyle x,y$ are positive integers. Thus if $\displaystyle \log n = x/y$, then $\displaystyle e^{2x/y} = n^2 $ is rational, contradicting irrationality of $\displaystyle  e^{2x/y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the first proof of irrationality of $\pi$ by Lambert used a continued fraction expansion (of $\tan x$, I believe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion we use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/RA92t.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the theorem we use to prove irrationality is quoted in the wiki page for Generalized Continued Fractions here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_continued_fraction#Conditions_for_irrationality&quot; rel=&quot;nofollow&quot;&gt;Conditions of Irrationality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this theorem, it is enough that for all sufficiently large positive integers $\displaystyle m$ we have that $\displaystyle (2m+1)y \gt x^2$, which is true for fixed $\displaystyle x,y$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T05:19:53.260" />
  <row Id="15300" PostTypeId="1" CreationDate="2010-12-23T05:20:56.893" Score="0" ViewCount="51" Body="&lt;p&gt;one of the early applications of fractional calculus is the tautochronus problem set up by Abel in the integral form or its fractional derivative one. i wish to know its solution&lt;/p&gt;&#xA;" OwnerUserId="5065" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-23T05:26:58.100" LastActivityDate="2010-12-23T05:26:58.100" Title="a problem in fractional calculus" Tags="&lt;calculus&gt;&lt;analysis&gt;" CommentCount="1" />
  <row Id="15301" PostTypeId="2" ParentId="15294" CreationDate="2010-12-23T05:29:16.647" Score="15" ViewCount="0" Body="&lt;p&gt;First: the integral is &lt;em&gt;defined&lt;/em&gt; to be the (net signed) area under the curve. The definition in terms of Riemann sums is precisely designed to accomplish this. The integral is a limit, a number. There is, a priori, no connection whatsoever with derivatives. (That is one of the things that makes the Fundamental Theorems of Calculus such a potentially surprising things). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the limit of the Riemann sums actually give the area under the graph? The idea of approximating a shape whose area we don't know both from &quot;above&quot; and from &quot;below&quot; with areas we &lt;em&gt;do&lt;/em&gt; know goes all the way back to the Greeks. Archimedes gave bounds for the value of $\pi$ by figuring out areas of inscribed and circumscribed polygons in a circle, knowing that the area of the circle would be somewhere between the two; the more sides to the polygons, the closer the inner and outer polygons are to the circle, the closer the areas are to the area of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way Riemann tried to formalize this was with the &quot;upper&quot; and &quot;lower&quot; Riemann sums: assuming the function is relatively &quot;nice&quot;, so that on each subinterval it has a maximum and a minimum, the &quot;lower Riemann sum&quot; is done by taking the largest &quot;rectangle&quot; that will lie completely under the graph by looking at the minimum value of the function on the interval and using that as height; and the &quot;upper Riemann sum&quot; is done by taking the smallest rectangle for which the graph will lie completely &lt;em&gt;under&lt;/em&gt; it (by taking the maximum value of the function as the height). Certainly, the exact area under the graph on that interval will be somewhere between the two. If we let $\underline{S}(f,P)$ be the lower sum corresponding to some fixed partition $P$ of the interval, and $\overline{S}(f,P)$ be the upper sum, we will have that &#xA;$$\underline{S}(f,P) \leq \int_a^b f(x)\,dx \leq \overline{S}(f,P).$$&#xA;(Remember that $\int_a^bf(x)\,dx$ is just the symbol we use to denote the exact (net signed) area under the graph of $f(x)$ between $a$ and $b$, whatever that quantity may be.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, intuitively, the &lt;em&gt;more&lt;/em&gt; intervals we take, the closer these two approximations (one from below and one from above) will be. This does not always work out if &lt;em&gt;all&lt;/em&gt; we do is take &quot;more&quot; intervals. But one thing we &lt;em&gt;can&lt;/em&gt; show is that if $P&amp;#39;$ is a &lt;em&gt;refinement&lt;/em&gt; of $P$ (it includes all the dividing points that $P$ had, &lt;em&gt;and&lt;/em&gt; possibly more points) then&#xA;$$\underline{S}(f,P)\leq \underline{S}(f,P&amp;#39;)\text{ and } \overline{S}(f,P&amp;#39;)\leq \overline{S}(f,P)$$&#xA;so at least the approximations are heading in the right direction. To see why this happens, suppose you split one of the subintervals $[t_i,t_{i+1}]$ in two, $[t_i,t&amp;#39;]$ and $[t&amp;#39;,t_{i+1}]$. The minimum of $f$ on $[t_i,t&amp;#39;]$ and on $[t&amp;#39;,t_{i+1}]$ are each greater than or equal to the minimum over the whole of $[t_i,t_{i+1}]$, but it may be that the minimum in one of the two bits is actually strictly larger than the minimum over $[t_i,t_{i+1}]$. The areas we get after the split can be no smaller, but they can be larger than the ones we had before the split. Similarly for the upper sums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's consider one particular sequence of partitions: divide the interval into 2 equal parts; then into 4; then into 8; then into 16; then into 32; and so on; then into $2^n$, etc. If $P_n$ is the partition that divides $[a,b]$ into $2^n$ equal parts, then $P_{n+1}$ is a refinement of $P_n$, and so we have:&#xA;$$\underline{S}(f,P_1) \leq\cdots \leq \underline{S}(f,P_n)\leq\cdots \leq\int_a^b f(x)\,dx \leq\cdots \leq\overline{S}(f,P_n)\leq\cdots \leq \overline{S}(f,P_2)\leq\overline{S}(f,P_1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the sequence of numbers $\underline{S}(f,P_1) \leq \underline{S}(f,P_2)\leq\cdots \leq \underline{S}(f,P_n)\leq\cdots$ is increasing and bounded above (by the area). So the numbers have a supremum; call it $\underline{S}$. This number is no more than $\int_a^b f(x)\,dx$. And the numbers $\overline{S}(f,P_1) \geq \overline{S}(f,P_2)\geq\cdots \geq \overline{S}(f,P_n)\geq\cdots$ are decreasing and bounded below, so they have a minimum; call this $\overline{S}$; again, it is no less than $\int_a^bf(x)\,dx$. So we have:&#xA;$$\lim_{n\to\infty}\underline{S}(f,P_n) = \underline{S} \leq \int_a^b f(x)\,dx \leq \overline{S} = \lim_{n\to\infty}\overline{S}(f,P_n).$$&#xA;What if we are lucky? What if actually we have $\underline{S}=\overline{S}$? Then it &lt;em&gt;must&lt;/em&gt; be the case that this common value is the value of $\int_a^b f(x)\,dx$. It just doesn't have a choice! It's definitely trapped between the two, and if there is no space between them, then it's equal to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Riemann proved was several things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $f$ is &quot;nice enough&quot;, then you will &lt;em&gt;necessarily&lt;/em&gt; get that $\underline{S}=\overline{S}$. In particular, continuous functions happen to be &quot;nice enough&quot;, so it will definitely work for them (in fact, continuous functions turn out to be &quot;very nice&quot;, not just &quot;nice enough&quot;). &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $f$ is &quot;nice enough&quot;, then you don't &lt;em&gt;have&lt;/em&gt; to use the partitions we used above. You can use &lt;em&gt;any&lt;/em&gt; sequence of partitions, so lonthe &quot;mesh size&quot; (the size of the largest subinterval in the partition) gets smaller and smaller, and has limit of $0$ as $n\to\infty$; if it works for the partitions &quot;divide-into-$2^n$-equal-intervals&quot;, then it works for &lt;em&gt;any&lt;/em&gt; sequence of partitions whose mesh size goes to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, we can take $P_n$ to be the partition that divides $[a,b]$ into $n$ equal parts, even though $P_{n+1}$ is not a refinement of $P_n$ in this case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In fact, you don't have to do $\underline{S}(f,P)$ and $\overline{S}(f,P)$. For the partition $P$, just pick any rectangle that has as its height &lt;em&gt;any&lt;/em&gt; value of the function in the subinterval (that is, pick an arbitrary $x_i^*$ in the subinterval $[t_i,t_{i+1}]$, and use $f(x_i^*)$ as the height). Call the resulting sum $S(f,P,x_1^*,\ldots,x_n^*)$. Then you have&#xA;$$\underline{S}(f,P) \leq S(f,P,x_1^*,\ldots,x_n^*)\leq \overline{S}(f,P)$$&#xA;because $\underline{S}(f,P)$ is computed using the &lt;em&gt;smallest&lt;/em&gt; possible values of $f$ throughout, and $\overline{S}(f,P)$ is computedusing the &lt;em&gt;largest&lt;/em&gt; possible values of $f$ throughout. But since we already know, from 1 and 2 above, that $\underline{S}(f,P)$ and $\overline{S}(f,P)$ have the same limit, then the sums $S(f,P,x_1^*,\ldots,x_n^*)$  &lt;em&gt;also&lt;/em&gt; get squeezed and must have &lt;em&gt;that&lt;/em&gt; same limit, which equals the integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we can always take the left endpoint (and get a &quot;Left Hand Sum&quot;) or we can always take the right endpoint (and get a &quot;Right Hand Sum&quot;), and you will nevertheless get the same limit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in summary, you can pick &lt;em&gt;any&lt;/em&gt; sequence of partitions, whichever happens to be convenient, so long as the mesh size goes to $0$, and you can pick &lt;em&gt;any&lt;/em&gt; points on the subintervals (say, ones which make the calculations simpler) at each stage, and so long as the function is &quot;nice enough&quot; (for example, if it is continuous), everything will work out and the limit will be the number which &lt;em&gt;must&lt;/em&gt; be the value of the area (because it was trapped between the lower and upper sums, and they both got squeezed together trapping the limit and the integral both between them). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, (1) and (2) above are the hardest part of what Riemann did. Don't be surprised if it sounds a bit magical at this point. But I hope that you agree that &lt;em&gt;if&lt;/em&gt; the lower and upper sums for the special partitions have the same limits then that limit &lt;em&gt;must be&lt;/em&gt; the area that lies under the graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to that work of Riemann, then (at least for continuous functions) we can &lt;em&gt;define&lt;/em&gt; $\int_a^b f(x)\,dx$ to be the limit of, say, the left hand sums of the partitions we get by dividing $[a,b]$ into $n$ equal parts, because these partitions have mesh size going to $0$, we can pick any points we like (say, the left end points), and we know the limit is going to be that common value of $\underline{S}$ and $\overline{S}$, which &lt;em&gt;has&lt;/em&gt; to be the area. So that, under this definition, $\int_a^b f(x)\,dx$ really &lt;em&gt;is&lt;/em&gt; the net signed area under the graph of $f(x)$. It just doesn't have a choice but to be that, when $f$ is &quot;nice enough&quot;. &lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;Second, the area does &lt;em&gt;not&lt;/em&gt; turn into &quot;the&quot; antiderivative. What happens is that it turns out (perhaps somewhat magically) that the area can be &lt;em&gt;computed&lt;/em&gt; using &lt;em&gt;an&lt;/em&gt; antiderivative. I'll go into some more details below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how Newton figured this out, &lt;em&gt;his&lt;/em&gt; teacher, Isaac Barrow, was the one who discovered there was a connection between derivatives and tangents; some of the basic ideas were his. They came from studying some simple functions and some simple formulas for tangents he had discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the tangents to the parabola $y=x^2$ were interesting (there was generally geometric interest in tangents and in &quot;squaring&quot; regions, also known as finding the &quot;quadrature&quot; of a region, that is, finding a way to construct a square or rectangle that had the same area as the region you were considering), and let to associate the parabola $y=x^2$ to lines of the form $y=2x$. It does not take too much experimentation to realize that if you look at the area under $y=2x$ from 0 to a, you end up with $a^2$, establishing a connection. Barrow did this with arguments with infinitesimals (which were a bit fuzzy and not set on entirely correct  and solid logical foundation until well into the 20th century), which were generally laborious, and only for &lt;em&gt;some&lt;/em&gt; curves. When Newton extended Barrow's methods to more general curves and tangents, he also extended the discovery of the connection with areas, and was able to prove what is essentially the Fundamental Theorem of Calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is one way to approach the connection. We want to figure out the value of, say, &#xA;$$\int_0^a f(x)\,dx$$&#xA;for some $a$. This can be done using limits and Riemann sums (Newton and Leibnitz had similar methods, though not set up quite as precisely as Riemann sums are). But here is an absolutely crazy suggestion: suppose you can find a &quot;master function&quot; $\mathcal{M}$, which, when given &lt;em&gt;any&lt;/em&gt; point $b$ between $0$ and $a$, will give you the value of $\int_0^b f(x)\,dx$. If you have such a master function, then you can use it to find the value of the integral you want just by taking $\mathcal{M}(a)$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is the approach Barrow had taken: his great insight was that instead of trying to find the quadratire a particular area, he was trying to solve the problem of squaring several different (but related) areas at the same time. So he was looking for, for instance, a &quot;master function&quot; for the region was like a triangle except that the top was a parabola instead of a line (like the area from $0$ to $a$ under $y=x^2$), and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On its face, this is a completely ludicrous suggestion. It's like telling someone who is trying to know how to get from building A to building B that if he only memorizes the map for the entire city &lt;em&gt;first&lt;/em&gt;, then he can use that knowledge to figure out how to get form A to B. If we are having trouble finding the integral $\int_0^a f(x)\,dx$, then the &quot;master function&quot; seems to require us to find not just &lt;em&gt;that&lt;/em&gt; area, but also &lt;em&gt;all&lt;/em&gt; areas in between! It's like telling someone who is having trouble walking that he should just run very slowly when he wants to walk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, again, the interesting thing is that even though we may not be able to say what the &quot;master function&quot; &lt;em&gt;is&lt;/em&gt;, we &lt;em&gt;can&lt;/em&gt; say how it changes as b changes (remember, $\mathcal{M}(b) = \int_0^b f(x)\,dx$ is a number that depends on $b$, so $\mathcal{M}$ is a function of $b$). Because figuring out how functions change is easier than computing their values (just thing about derivatives, and how we can easily figure out the rate of change of $\sin(x)$, but we have a hard time actually &lt;em&gt;computing&lt;/em&gt; specific values of $\sin(x)$ that are not among some very simple ones). (This is also something Barrow already knew, as did Newton).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;nice functions&quot; (if $f$ is &lt;em&gt;continuous&lt;/em&gt; on an interval that contains $0$ and $a$), we can do it using limits and some theorems about &quot;nice&quot; functions:&#xA;Using limits, we have:&#xA;\begin{align}&#xA;\lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}}{h} &amp;amp;= \frac{1}{h}\left(\int_0^{b+h}f(x)\,dx - \int_0^bf(x)\,dx\right)\\&#xA;&amp;amp;= \frac{1}{h}\int_b^{b+h}f(x)\,dx.&#xA;\end{align}&#xA;Since we are assuming that $f$ is continuous on $[0,a]$, it is continuous on the interval with endpoints $b$ and $b+h$ (I say it this way because $h$ could be negative). So it has a maximum and a minimum (continuous function on a finite closed interval). Same the maximum is $M(h)$ and the minimum is $m(h)$. Then $m(h) \leq f(x) \leq M(h)$ for all $x$ in the interval, so we know, since the integral &lt;em&gt;is&lt;/em&gt; the area, that&#xA;$$hm(h) \leq \int_b^{b+h}f(x)\,dx \leq hM(h).$$&#xA;That means that&#xA;$$m(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq M(h)\text{ if $h\gt 0$}$$&#xA;and&#xA;$$ M(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq m(h)\text{ if $h\lt 0$.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $h\to 0$, the interval gets smaller, the difference between the minimum and maximum value gets smaller. One can prove that both $M$ and $m$ are continuous functions, and that $m(h)\to f(b)$ as $h\to 0$, and likewise that $M(h)\to f(b)$ as $h\to 0$. So we can use the Squeeze Theorem to conclude that since the limit of $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ is squeezed between two functions that both have the same limit as $h\to 0$, then $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ also has a limit as $h\to 0$ and is in fact that same quantity, namely $f(b)$. That is&#xA;$$\frac{d}{db}\mathcal{M}(b) = \lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}(b)}{h} = \lim_{h\to 0}\frac{1}{h}\int_{b}^{b+h}f(x)\,dx = f(b).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: when $f$ is continuous, the &quot;Master function&quot; for areas turns out to have a rate of change equal to $f$. This is not that crazy, if you think about it: how is the area under $y=f(x)$ from $x=0$ to $x=b$ changing? Well, it's changing by whatever $f$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, whatever the &quot;Master function&quot; turns out to be, it will be an antiderivative of $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know, because we are very good with derivatives, that if $\mathcal{F}(x)$ and $\mathcal{G}(x)$ are two functions, and $\mathcal{F}&amp;#39;(x) = \mathcal{G}&amp;#39;(x)$ for all $x$, then $\mathcal{F}$ and $\mathcal{G}$ differ by a constant: there exists a constant $k$ such that $\mathcal{F}(x) = \mathcal{G}(x)+k$ for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we know that the &quot;Master function&quot; is &lt;em&gt;an&lt;/em&gt; antiderivative. If, by some sheer stroke of luck, we happen to find &lt;em&gt;any&lt;/em&gt; antiderivative $F(x)$ for $f(x)$, then we know that the only possible difference between $\mathcal{M}(b)$ and $F(b)$ is a constant. What constant? Well, luckily we know &lt;em&gt;one&lt;/em&gt; value of $\mathcal{M}(b)$: we know that $\mathcal{M}(0) = \int_0^0f(x)\,dx$ &lt;em&gt;should&lt;/em&gt; be $0$.  So, $M(0) = 0 = F(0)-F(0)$, which means the constant has to be $-F(0)$. That is, we must have $M(b) = F(b)-F(0)$ for all $b$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we find &lt;em&gt;any&lt;/em&gt; antiderivative $F$ of $f$, then $\mathcal{M}(b) = F(b)-F(0)$ is in fact the &quot;Master function&quot; we were looking for, the one that gives &lt;em&gt;all&lt;/em&gt; the integrals between $0$ and $a$, including $0$ and including $a$. So that we have that two very different processes (computing areas using limits of Riemann sums, and derivatives) are connected: if $f(x)$ is continuous, and $F(x)$ is &lt;em&gt;any&lt;/em&gt; antiderivative for $f(x)$ on $[0,a]$, then&#xA;$$\int_0^a f(x)\,dx = \mathcal{M}(a) = F(a)-F(0).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the integral did not &quot;magically turn&quot; into an antiderivative. It's that the &quot;Master function&quot; which can be used to keep track of &lt;em&gt;all&lt;/em&gt; integrals of $f(x)$ has rate of change equal to $f$, which gives us a &quot;back door&quot; to computing integrals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton was able to prove this because he had the guide of Barrow's insight that this was happening for the functions he worked with. Barrow's insight was achieved because he had the brilliant idea of trying to come up with a &quot;Master function&quot; instead of trying to rectify lots of different areas one at a time, and he noticed the connection because he had already worked with tangents/derivatives for those functions. Leibnitz likewise had access to Barrow's ideas, so the connection between the two was also known to him.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-23T06:40:29.690" LastActivityDate="2010-12-23T06:40:29.690" CommentCount="4" />
  <row Id="15302" PostTypeId="2" ParentId="15294" CreationDate="2010-12-23T05:31:52.793" Score="11" ViewCount="0" Body="&lt;p&gt;One way you can perhaps &quot;justify&quot;/give an intuitive reason is to consider the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZhB8N.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x)$ is the area under the curve from $0$ to $x$, the brown region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(x+dx)$ is the area under the curve from $0$ to $x + dx$, the brown + gray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for for really small $dx$, we can consider the gray region to be a rectangle of side width $dx$ and height $f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\dfrac{A(x+dx) - A(x)}{dx} = f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus as $dx \to 0$, we see that $A&amp;#39;(x) = f(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is kind of intuitive to define area by approximating by very thin rectangles. The above gives an intuition as to why the derivative of the area gives the curve.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T05:31:52.793" />
  <row Id="15303" PostTypeId="2" ParentId="14823" CreationDate="2010-12-23T05:39:07.410" Score="3" ViewCount="0" Body="&lt;p&gt;@PF4Public, As stated in the comment and also mentioned by @mtrw this is likely aliasing phenomenon.  Your intuition that it is a mathematical issue is correct.  There are many websites that describe the mechanics of aliasing such as this &lt;a href=&quot;http://en.wikipedia.org/wiki/Aliasing&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.  Basically, the phenomenon occurs when an analog signal is sampled at a rate that is too slow as compared to the frequency content.  You will get this phenomenon whenever the sampling frequency is less than 2x the highest frequency component (known as the Nyquist rate).  When aliasing occurs, higher frequencies &quot;appear&quot; to be lower.  Furthermore, when a signal frequency exceeds 1/2 the Nyquist rate its apparent frequency starts to go down, exactly the phenomenon that you are seeing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this problem you need to make sure that the sampling frequency of you oscilloscope is at least 2x your largest harmonic.  I am not sure how the FFT function works on you oscilloscope, but sometimes the sampling rate is tied to the sec/div setting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that this helps.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2010-12-23T05:39:07.410" />
  <row Id="15304" PostTypeId="2" ParentId="15214" CreationDate="2010-12-23T05:43:52.773" Score="0" ViewCount="0" Body="&lt;p&gt;(This was supposed to be a comment, but it got too long.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of piecewise linear regression has been looked into many times before; I do not currently have access to these papers (and thus cannot say more about them), but you might want to look into these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/2759968&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; (published in a &lt;em&gt;physiology&lt;/em&gt; journal, of all places) discusses how to fit a piecewise linear function to certain data sets encountered in neurological research. A FORTRAN routine is included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2285594&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; also includes a FORTRAN routine for fitting piecewise linear functions to data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2287599&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; relies on maximum likelihood to find the best fit piecewise linear function.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-23T05:43:52.773" />
  <row Id="15305" PostTypeId="2" ParentId="15295" CreationDate="2010-12-23T06:10:03.140" Score="3" ViewCount="0" Body="&lt;p&gt;Hi Noah,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no, this is never an ultrafilter. This uses choice. We assume that the cofinality of $\kappa$ is bigger than $\omega$, else, $C$ has a subset of type $\omega$ cofinal (which is automatically club), and this can be split into two disjoint cofinal (therefore club) sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cofinality of $\kappa$ is larger than $\omega_1$, simply note that for any regular $\alpha&amp;lt;\kappa$, $\{\beta&amp;lt;\kappa:\beta$ has cofinality $\alpha\}$ is &lt;em&gt;stationary&lt;/em&gt;, i.e., it meets every club (consider the $\alpha$-th member of the increasing enumeration of the club. These sets are disjoint for different $\alpha$, and if $\kappa&amp;gt;\omega_1$, there are at least two regular $\alpha$ below $\kappa$, namely $\omega$ and $\omega_1$. This shows that we can split $C$ into disjoint stationary sets (and therefore, neither is club, since a stat. set must meet every club).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I've used so far is that $\omega_1$ is regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that clubs on ordinals of cofinality $\omega_1$ can be split into disjoint stationary sets  requires more choice. In fact, it is consistent that the club filter on $\omega_1$ is an ultrafilter. This is the case, for example, in models where the axiom of determinacy holds. (But it is strictly weaker than determinacy. A measurable suffices in consistency strength.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical argument from choice uses the existence of Ulam matrices. See definition 12 and the following paragraph in this &lt;a href=&quot;http://caicedoteaching.wordpress.com/2009/02/13/580-cardinal-arithmetic-5/&quot; rel=&quot;nofollow&quot;&gt;blog entry&lt;/a&gt;. An $\omega\times\omega_1$ Ulam matrix gives us that every stat. subset of $\omega_1$, in particular every club, can be split into $\omega_1$ disjoint stat. sets. So the club filter is not an ultrafilter, and we are done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ulam's argument also shows that any stat. subset of $\kappa^+$ can be split into $\kappa^+$ stat. sets, and Solovay proved the same for any reglar $\lambda$, not just a successor. This, of course, gives the result that no club filter is an ultrafilter, but I wanted to point out that only in cofinality $\omega_1$ we need that much choice to carry out the argument.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-23T06:36:01.103" LastActivityDate="2010-12-23T06:36:01.103" CommentCount="1" />
  <row Id="15306" PostTypeId="2" ParentId="15295" CreationDate="2010-12-23T06:54:20.553" Score="1" ViewCount="0" Body="&lt;p&gt;If you are familiar with the idea of clubs and stationary, then there is a good measure theoretic intuition around it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clubs are of measure $1$. They are almost everywhere.&lt;br&gt;&#xA;Non-stationary sets are of measure $0$ and are almost nowhere.&lt;br&gt;&#xA;Stationary sets are of positive measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question, look at the unit interval, and take some subset of measure $1$, you can split it into two positive measured sets, none of which is of measure $1$, namely intersect it with $[0,\frac{1}{2})$ and with $[\frac{1}{2},1]$, and you have two disjoint subsets of positive measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is important to say that this is just the intuition at heart, and since Andres gave a good answer with a formal backbone I decided to add this one as well.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-23T06:54:20.553" />
  <row Id="15307" PostTypeId="1" AcceptedAnswerId="15308" CreationDate="2010-12-23T07:16:04.083" Score="3" ViewCount="87" Body="&lt;p&gt;OK, I know that Matiyasevich's solution to Hilbert's 10th problem shows that there is no algorithm to decide whether or not a polynomial $p(x_1,\ldots,p_n)$ with integer coefficients has a solution over the integers.  I have no doubts about the truth of that.  However, I have been unable to figure out why the following algorithm doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Program one computer to plug every $(x_1,\ldots,x_n) \in \mathbb{Z}^n$ into $p(x_1,\ldots,p_n)$.  If there is an integer solution, then this will halt eventually.  Program a second computer to enumerate every possible proof.  There are only countably many, so if there is a proof that $p(x_1,\ldots,x_n)$ has no integer solutions, then this computer will eventually find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given Matiyasevich's theorem, the only conclusion I can draw is that there are polynomials that don't have integral zeros, but for which there is no proof that they don't (so it's independent of ZFC).  This seems very odd to me.  What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="5067" LastActivityDate="2010-12-23T07:24:36.993" Title="Algorithms to prove that polynomials don't have integer solutions" Tags="&lt;algorithms&gt;&lt;algebraic-geometry&gt;&lt;computability&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15308" PostTypeId="2" ParentId="15307" CreationDate="2010-12-23T07:24:36.993" Score="6" ViewCount="0" Body="&lt;p&gt;(Usual disclaimers about ZFC being consistent, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are not missing anything, this is precisely the case. For example, there is a polynomial $p(x_1,\dots,x_n)\in{\mathbb Z}[x_1,\dots,x_n]$ for some $n$ so that any integer solution codes a proof that ZFC is inconsistent. There are no integer solutions, and there are no proofs in ZFC of this fact. The point is that the 10th problem shows that the r.e. sets are precisely the Diophantine sets.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-23T07:24:36.993" CommentCount="2" />
  <row Id="15309" PostTypeId="1" AcceptedAnswerId="15311" CreationDate="2010-12-23T07:35:04.897" Score="4" ViewCount="57" Body="&lt;p&gt;I got an assignment to prove certain things about right triangles in Lobachevsky geometry, but so far I don't know where to start. What model is the best for studying these objects? What is the general approach to this kind of task?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T08:49:34.237" LastActivityDate="2010-12-23T09:15:38.290" Title="How to analyze triangles in Lobachevsky geometry?" Tags="&lt;geometry&gt;&lt;soft-question&gt;&lt;hyperbolic-geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15310" PostTypeId="2" ParentId="6090" CreationDate="2010-12-23T07:38:57.383" Score="1" ViewCount="0" Body="&lt;p&gt;De La Vallee-Pousin's theorem has been improved by Korobov and Vinogradov in the 50's and I believe their result is the strongest known asymptotic zero free region, cf. &lt;em&gt;The Riemann zeta-function: Theory and applications&lt;/em&gt; by Alexandar Ivic. One can find more recent papers but my impression is that they don't touch the main exponents and so relatively tractable problems seem to be improving the constants, or giving explicit bounds on the constants etc.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastEditorUserId="2473" LastEditorDisplayName="" LastEditDate="2010-12-24T23:30:36.527" LastActivityDate="2010-12-24T23:30:36.527" CommentCount="2" />
  <row Id="15311" PostTypeId="2" ParentId="15309" CreationDate="2010-12-23T07:48:40.943" Score="2" ViewCount="0" Body="&lt;p&gt;Lobachevsky's geometry also goes by the more common name &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow&quot;&gt;hyperbolic geometry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that we are talking about the 2-dimensional case. There are two particularly convenient models: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model&quot; rel=&quot;nofollow&quot;&gt;Poincaré upper half plane&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_disc_model&quot; rel=&quot;nofollow&quot;&gt;Poincaré disc&lt;/a&gt;. The linked pages have plenty of identities between angles and sides in hyperbolic triangles. Many of the usual trigonometric identities hold (I imagine that you will have to prove that if you want to use it), but it is no longer true that angles of a triangle sum to $\pi$. Indeed, the sum is always strictly smaller (see formula for the area of a triangle).&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-23T08:03:12.220" LastActivityDate="2010-12-23T08:03:12.220" />
  <row Id="15312" PostTypeId="1" AcceptedAnswerId="15336" CreationDate="2010-12-23T07:54:30.557" Score="0" ViewCount="90" Body="&lt;p&gt;Everything is in $\mathbb{Z}$. Let $v_1 &amp;lt; v_2 &amp;lt; ... &amp;lt; v_n = k$, and $v_1 = 1$ for $k &amp;gt;&amp;gt; n$. Let $ P = \Pi_{i &amp;lt; j} (v_j - v_i)$. How can I show that $P \le k^{n^2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $n + (n-1) + ... + 1 = \frac{n(n+1)}{2}$ terms in the product. Starting from $v_n - v_{n-1} = 1$, etc. Clearly, $P = 1(1*2)(1*2*3) ...(k-1)! = \Pi_{i=1}^{i=k-1} i!$. But I'm not sure about this superfactorial(?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I noticed that the product $P$ is very similar to the determinant of a Vandermonde matrix.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastActivityDate="2010-12-23T15:09:23.030" Title="Show that $\Pi_{i &lt; j} (v_i - v_j) \le k^{n^2}$ for $1 \le v_1 &lt; v_2 &lt; ... &lt; v_n = k$" Tags="&lt;sequences-and-series&gt;&lt;factorial&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15313" PostTypeId="2" ParentId="15297" CreationDate="2010-12-23T08:23:11.687" Score="1" ViewCount="0" Body="&lt;p&gt;For $a_n \ge 0 $ the infinite product $\prod_{n=1}^\infty (1+ a_n)$ converges precisely when the infinite sum $\sum_{n=1}^\infty a_n $ converges, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+ \sum_{n=1}^N a_n \le \prod_{n=1}^N (1+ a_n) \le \exp \left( \sum_{n=1}^N a_n \right) . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you only need consider $ \sum_{n=1}^\infty 2/n $ and you can use your integral test for that, or just quote the standard result.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-23T08:23:11.687" />
  <row Id="15314" PostTypeId="1" CreationDate="2010-12-23T08:29:24.130" Score="0" ViewCount="47" Body="&lt;p&gt;Hello! I have n points that determine a polygon. I need to color the interior of this polygon with a color . How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="2777" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T08:46:59.793" LastActivityDate="2010-12-23T08:51:04.703" Title="color polygon interior in different color" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15315" PostTypeId="1" AcceptedAnswerId="15322" CreationDate="2010-12-23T08:37:25.017" Score="4" ViewCount="108" Body="&lt;p&gt;The eigenvalues of a rotation matrix are complex numbers. I understand that they cannot be real numbers because when you rotate something no direction stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt;&#xA;What is the intuition that the eigenvalues are complex? Why do they exist at all for rotation matrices? I mean it is not so that every time a calculation is not possible the result is complex (dividing by 0 is not possible at all - the result is not real, but it is not complex either!). The complex numbers seem to cover some middle ground here, but I don't understand how and why they come into play - there don't seem to be any square roots taken from negative numbers...)&lt;/p&gt;&#xA;" OwnerUserId="346" LastEditorUserId="346" LastEditorDisplayName="" LastEditDate="2010-12-23T08:48:18.520" LastActivityDate="2010-12-23T09:50:30.730" Title="intuition for complex eigenvalues" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;intuition&gt;&lt;complex-numbers&gt;&lt;eigenvalues&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="15316" PostTypeId="2" ParentId="15314" CreationDate="2010-12-23T08:41:14.013" Score="2" ViewCount="0" Body="&lt;p&gt;Here is one naive approach: triangulate the polygon (that's pretty easy to do and I will leave that to you to figure out) and then colour the triangles. That reduces the problem to determining the interior of a triangle that is given by three points. To do that, you could express the triangle as an intersection of three half planes. Then, each half plane is determined by a linear inequality, so given any point, you can determine whether it's in the interior by checking all three inequalities. The rest is accomplished by obvious loops, with judiciously chosen start and end values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can optimise some of that by putting more thought into how to set up the loops. E.g. if the outer loop goes through rows and the inner through the columns, then the starting point of the inner loop can be set to the left most point of the interior.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastEditorUserId="3212" LastEditorDisplayName="" LastEditDate="2010-12-23T08:51:04.703" LastActivityDate="2010-12-23T08:51:04.703" />
  <row Id="15318" PostTypeId="2" ParentId="15297" CreationDate="2010-12-23T09:01:22.193" Score="1" ViewCount="0" Body="&lt;p&gt;If You dont have to use integration test then You can do something like this, it is much easier.&#xA;$\prod\limits_{n=1}^{k}(1+\frac{2}{n})=\frac{(k+1)(k+2)}{2}$&lt;br&gt;&#xA;It is very easy to show since it is equal to: $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdot\frac{4}{6}\cdot\frac{5}{7}\cdot\cdot\cdot\cdot$&lt;br&gt;&#xA;You can reduce almost all this fractions. Denominator of nth fraction can be reduced with numerator of n+2th fraction.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2010-12-23T09:01:22.193" />
  <row Id="15319" PostTypeId="2" ParentId="15309" CreationDate="2010-12-23T09:15:38.290" Score="2" ViewCount="0" Body="&lt;p&gt;There are a few models, and the great thing is that (once you've shown that they're equivalent) different things are easy to show in different models.  If you're clever about it, you can juggle back and forth and do as little work as possible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Alex mentions, there are the Poincare disc and UHP models.  Both of these &quot;respect angles&quot; (e.g. two hyperbolic geodesics are perpendicular iff they look perpendicular in either of these models).  There's another disc model which does &lt;em&gt;not&lt;/em&gt; respect angles: hyperbolic geodesics are now just chords of the disc.  In this model, for instance, it's just a tad easier to prove that the postulate &quot;two lines intersect in at most one point&quot; holds than it is in either of the others (not that it's really so hard in any of them).  I'm sure there are less trivial things that are easiest to show in this model, but none are springing to mind.  But then, for instance, if you want to find a regular octagon whose angles are each 45 degrees (which comes up in topology as the fundamental domain of a 2-holed torus), you're going to be by far best off with the Poincare disc model.  As I said, each model has its own perks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, if you're dealing with right triangles it might be easiest to use the Poincare disc model, because then you can take two of the three edges of your triangle to lie in the x- and y-axes.  Or similarly in the UHP model you could let one edge lie in the y-axis and one edge lie in the unit circle.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-12-23T09:15:38.290" />
  <row Id="15321" PostTypeId="1" CreationDate="2010-12-23T09:47:22.487" Score="0" ViewCount="40" Body="&lt;p&gt;Cartan for beginner book page 6, 7, 8 not have enough examples for understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how to rewrite ODE or PDE as exterior differential equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If further, please give example to solve these exterior differential form in maple.&lt;/p&gt;&#xA;" OwnerUserId="5070" LastActivityDate="2010-12-23T09:47:22.487" Title="Please give exmaples of rewriting ODE as ideal of differential forms and PDE case as well?" Tags="&lt;differential-geometry&gt;&lt;differential-forms&gt;" CommentCount="1" />
  <row Id="15322" PostTypeId="2" ParentId="15315" CreationDate="2010-12-23T09:50:30.730" Score="2" ViewCount="0" Body="&lt;p&gt;Heuristically, I suppose we could see this being because the standard action of the complex numbers on $V = \mathbb R^{2n}$ is by rotation. That is, $(e_1, \ldots, e_{2n})$ is a basis for $V$, then we define multiplication by $i$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ i e_{2k-1} = e_{2k}, \quad i e_{2k} = - e_{2k-1},$ for $k = 1, \ldots, n$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so multiplying a vector $v$ by a complex number $\lambda$ will correspond to a scaling by a real and a rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we have a rotation $A$ on the space $V$ and we want to find a line $l$ &quot;invariant&quot; under $A$, then we can try to look for a complex number $\lambda$ such that rotation of $l$ under $A$ is equivalent to the action of $\lambda$ on $l$. Thus, we can try and look for complex eigenvalues $\lambda$ of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This line of &quot;reasoning&quot; completely breaks down in the odd-dimensional case, because we can't define a complex structure on odd-dimensional spaces, but it might give a hint into why we'd look for complex eigenvalues at all. Then it becomes a matter of algebra to figure out that it actually works, and that it does so in a vector space of any finite dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for the existence, as Alex already pointed out, we look for eigenvalues by finding roots of polynomials. All polynomials admit a root over the complex numbers, which translates into the existence of a complex eigenvalue.&lt;/p&gt;&#xA;" OwnerUserId="3225" LastActivityDate="2010-12-23T09:50:30.730" />
  <row Id="15323" PostTypeId="1" CreationDate="2010-12-23T10:04:39.483" Score="0" ViewCount="59" Body="&lt;p&gt;Given the Right ascension in h m s, Declination in deg ' &quot; and the Trigonometric parallax How can I get the cartesian (x,y,z) coordinates of a star? I'm guessing I need 3 separate formulas to get each x, y and z values.&lt;/p&gt;&#xA;" OwnerUserId="5071" LastActivityDate="2010-12-23T10:04:39.483" Title="How do I calculate the cartesian coordinates of stars" Tags="&lt;trigonometry&gt;" CommentCount="3" />
  <row Id="15324" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T10:35:02.307" Score="0" ViewCount="0" Body="&lt;p&gt;wow! thanks all, everyone got very interesting answer&#xA;i've found also &lt;a href=&quot;http://mathworld.wolfram.com/Slinky.html&quot; rel=&quot;nofollow&quot;&gt;slinky&lt;/a&gt; that is quite funny cause it is also a toy&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2010-12-23T10:35:02.307" />
  <row Id="15326" PostTypeId="1" AcceptedAnswerId="15329" CreationDate="2010-12-23T10:51:38.213" Score="2" ViewCount="353" Body="&lt;p&gt;there is a prove that there is no $2^n$ that is divisible by 5?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-23T20:09:33.843" LastActivityDate="2010-12-23T20:09:33.843" Title="prove: $2^n$ is not divisible by 5 for any $n$" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="15327" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T10:54:20.240" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Use the fact that the powers of $2$ end with $2,4,8,6$ and for divisibility for 5 last digit has to end with either 5 or $0$.&lt;/p&gt;&#xA;" OwnerUserId="649" LastEditorUserId="649" LastEditorDisplayName="" LastEditDate="2010-12-23T11:01:08.457" LastActivityDate="2010-12-23T11:01:08.457" />
  <row Id="15328" PostTypeId="2" ParentId="14206" CreationDate="2010-12-23T11:02:30.000" Score="1" ViewCount="0" Body="&lt;p&gt;I specially recomend you reading &lt;a href=&quot;http://docs.sun.com/source/806-3568/ncg_goldberg.html&quot; rel=&quot;nofollow&quot;&gt;What Every Computer Scientist Should Know About Floating-Point Arithmetic&lt;/a&gt; by David Goldberg. It should illuminate a lot of issues regarding floating-point computations Errors. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This paper is a tutorial on those aspects of floating-point arithmetic (floating-point hereafter) that have a direct connection to systems building. It consists of three loosely connected parts. The first section, Rounding Error, discusses the implications of using different rounding strategies for the basic operations of addition, subtraction, multiplication and division. It also contains background information on the two methods of measuring rounding error, ulps and relative error. The second part discusses the IEEE floating-point standard, which is becoming rapidly accepted by commercial hardware manufacturers. Included in the IEEE standard is the rounding method for basic operations. The discussion of the standard draws on the material in the section Rounding Error. The third part discusses the connections between floating-point and the design of various aspects of computer systems. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I specially recomend you to design your algorithm according The IEEE 754 rounding methods.&#xA;I also recomend you to read the third part because it discussed the implications of floating point arithmetic on the design of computer systems.&lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2010-12-23T11:02:30.000" />
  <row Id="15329" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T11:10:05.613" Score="12" ViewCount="0" Body="&lt;p&gt;This is a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic&quot; rel=&quot;nofollow&quot;&gt;fundamental theorem of arithmetic&lt;/a&gt;: every integer can be &lt;em&gt;uniquely&lt;/em&gt; written as a product of primes. Since both 2 and 5 are primes, an identity like $2^n = 5k$ for $k\in \mathbb{N}$ would contradict the uniqueness of prime factorisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise: more generally, find a necessary and sufficient criterion on integers $m$ and primes $p$ for $m^n$ to be divisible by $p$ for some $n$. Further prove that if divisibility holds for some $n$, then it holds for all natural $n&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-23T11:10:05.613" CommentCount="2" />
  <row Id="15330" PostTypeId="1" CreationDate="2010-12-23T11:25:36.790" Score="0" ViewCount="45" Body="&lt;p&gt;Hello!I´m trying to develop a architecture hardware to make a implementation of an algorithm that can be descompose in terms of sums, multiplications, substrations and exponential functions. I,m trying to modelling, exp(-x) through Taylor series. The domain of my function is bounded between 0 and 1500, but I want to use a particular Taylor approximation whose domain is bounded between 0 and 0.5. &lt;strong&gt;Is there any way to get an approximation using the my tailor series whose domain is bounded between 0 and 0.5 to modelling the function whose domain is bounded between 0 and 1500?&lt;/strong&gt; &#xA;The function I want to modelling for bounded domain is exp(-x).&#xA;Thank you for your help. &lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2010-12-23T11:25:36.790" Title="Development of a specific hardware architecture for a particular algorithm. Modelling fuctions by Taylor Series." Tags="&lt;algorithms&gt;&lt;functions&gt;&lt;approximation&gt;&lt;mathematical-modeling&gt;" />
  <row Id="15331" PostTypeId="2" ParentId="15260" CreationDate="2010-12-23T12:48:38.587" Score="2" ViewCount="0" Body="&lt;p&gt;If I may be allowed to give another example, there is the spherical epitrochoid, the locus of a circle rolling on another circle. It was studied by Johann Bernoulli and Franz Reuleaux in the context of the kinematics of gear systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $p$ be the radius of the fixed circle (&quot;pitch circle&quot;), $r$ the radius of the rolling circle, $hr$ the distance of the tracing point from the center of the rolling circle, and $\omega$ the inclination of the rolling circle from the x-y coordinate plane, the parametric equations for the spherical epitrochoid are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=(p+r\cos\;\omega)\cos\;t-h r\left(\cos\;\omega\cos\left(\frac{p}{r}t\right)\cos\;t-\sin\left(\frac{p}{r}t\right)\sin\;t\right)\\y&amp;amp;=(p+r \cos\;\omega)\sin\;t-h r\left(\cos\;\omega\cos\left(\frac{p}{r}t\right)\sin\;t+\sin\left(\frac{p}{r}t\right)\cos\;t\right)\\z&amp;amp;=r\sin\;\omega\left(1-h\cos\left(\frac{p}{r}t\right)\right)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the case $p=1$, $r=\frac15$, $h=1$, $\omega=\frac{2\pi}{3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/4m4Tf.gif&quot; alt=&quot;three-dimensional epitrochoid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the case $p=1$, $r=\frac15$, $h=1$, $\omega=\frac{\pi}{6}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/BkZdu.gif&quot; alt=&quot;three-dimensional epitrochoid&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-23T12:48:38.587" />
  <row Id="15332" PostTypeId="1" AcceptedAnswerId="15370" CreationDate="2010-12-23T13:27:16.247" Score="7" ViewCount="105" Body="&lt;p&gt;It is well known how to use algebraic geometry (differentials, divisors, and Riemann-Roch) in order to prove the finite dimensionality of the vector space of modular forms of some fixed weight and level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to prove the finite dimensionality of modular forms using representation theory? If so, any references?&lt;/p&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2010-12-23T21:29:30.217" Title="Proving finite dimensionality of modular forms using representation theory?" Tags="&lt;reference-request&gt;&lt;representation-theory&gt;&lt;modular-forms&gt;" AnswerCount="1" />
  <row Id="15333" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T13:43:27.523" Score="5" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ By &lt;a href=&quot;http://math.stackexchange.com/questions/14579/troubles-finding-inverse-modulus/14608#14608&quot;&gt;Gauss's algorithm&lt;/a&gt; for computing inverses modulo a prime $\rm\: 2\:n+1\:$ (here $5$) we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\ \ \displaystyle \frac{1}2\ \equiv\ \frac{n+1}{2\:(n+1)}\ \equiv\ n+1\ \ (mod\:\ 2\:n+1)\:.\quad $ Hence $\rm\ 2^n \equiv 0\ $ times $\ (1/2)^n\ $ yields $\rm\ 1\equiv 0\ \Rightarrow\Leftarrow$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally any integer $\rm\:a\:$ coprime to $\rm\:d\:$ can be constructively inverted $\rm\:(mod\ d)\:$ by employing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm&quot; rel=&quot;nofollow&quot;&gt;extended Euclidean algorithm&lt;/a&gt; to find a Bezout relation $\rm\ a\ b + c\ d = 1\ \Rightarrow\ a\ b \equiv 1\ \:(mod\ d)\:.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very closely related to &quot;rationalizing denominators&quot;, e.g. see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zwtid4b0u.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;sci.math post&lt;/a&gt; for further remarks on rationalizing denominators &#xA;and computing inverses via (minimal) polynomials, the extended Euclidean &#xA;algorithm, Grobner bases, resultants, norms, etc. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-23T18:42:58.197" LastActivityDate="2010-12-23T18:42:58.197" CommentCount="7" />
  <row Id="15334" PostTypeId="1" AcceptedAnswerId="15379" CreationDate="2010-12-23T14:29:07.750" Score="2" ViewCount="122" Body="&lt;p&gt;If we start with $\int\frac{v}{g-2v} dv$ how would we go about integrating it? (g is a constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers to the past exam paper I have tell me to rearrange it to $\frac{1}{2}\int ( -1 + \frac{g}{g-2v} ) dv$ which integrates to $\frac{1}{2}v-\frac{1}{4}g\ln(g-2v) + c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that the rearrangement works 'in reverse', but I'm not sure how I would go about rearranging it the right way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Part of question 3(iii) in January 2006 MEI Differential Equations paper.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1699" LastEditorUserId="1699" LastEditorDisplayName="" LastEditDate="2010-12-23T16:25:17.570" LastActivityDate="2010-12-23T21:11:07.567" Title="Rearranging an Integral" Tags="&lt;algebra&gt;&lt;integral&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15335" PostTypeId="1" AcceptedAnswerId="15348" CreationDate="2010-12-23T15:06:41.863" Score="3" ViewCount="203" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is ; How can I solve the following integral question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{(1 + \cos x)^3\mathrm dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-23T15:16:43.170" LastActivityDate="2010-12-24T03:46:38.150" Title="How can I find  $ \int{(1 + \cos x)^3\mathrm dx} $ ?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="15336" PostTypeId="2" ParentId="15312" CreationDate="2010-12-23T15:09:23.030" Score="2" ViewCount="0" Body="&lt;p&gt;First note that there are a total of $1 + 2 + \cdots + (n-1)$ terms i.e. a total of $\frac{n(n-1)}{2}$ terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\forall n \in \mathbb{N}$, $\frac{n(n-1)}{2} &amp;lt; n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $|v_i - v_j| &amp;lt; k$, $\forall i,j$ since $v_{i} \in [1,k]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we get $\displaystyle \prod_{i&amp;lt;j} |v_i - v_j| &amp;lt; k^{\frac{n(n-1)}{2}} &amp;lt; k^{n^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last inequality follows from the fact that $\frac{n(n-1)}{2} &amp;lt; n^2$ and $k&amp;gt;1$&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-23T15:09:23.030" CommentCount="3" />
  <row Id="15337" PostTypeId="1" AcceptedAnswerId="15351" CreationDate="2010-12-23T15:15:53.787" Score="2" ViewCount="127" Body="&lt;p&gt;Is  $$\left |\int_1^N e^{ix^\alpha} \ \text{dx} \right| = \mathcal{O}(N^\alpha)?$$ &#xA;where $\displaystyle 0\lt \alpha \lt 1$ and $\displaystyle i=\sqrt{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf Edit.}$ As Willie Wong shows below, the correct equation is &#xA;$$\left |\int_1^N e^{ix^\alpha} \ \text{dx} \right| = \mathcal{O}(N^{1-\alpha}).$$&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-23T20:32:09.167" LastActivityDate="2010-12-24T06:12:31.660" Title="Is this integral of order $N^{\alpha}$?" Tags="&lt;analysis&gt;&lt;integral&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="15338" PostTypeId="2" ParentId="15017" CreationDate="2010-12-23T15:17:36.993" Score="2" ViewCount="0" Body="&lt;p&gt;No. Here is a counterexample. Define $X$ as follows. ${\rm P}(X=a) = p_1$, ${\rm P}(X=0) = p_2$, and ${\rm P}(X=\frac{{ - ap_1 }}{{1 - p_1  - p_2 }}) = 1-p_1-p_2$,&#xA;where $a$ is a positive constant. Then, ${\rm E}(X)=0$. Denote ${\rm E}(X^4)-2 [{\rm E}(X^2)]^2$ by $\xi$. Then,&#xA;$$&#xD;&#xA;\xi = a^4 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2 ) - 2\Big[a^2 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2 )\Big]^2.&#xD;&#xA;$$&#xA;To find ${\rm E}(Y^4)-2 [{\rm E}(Y^2)]^2$, which we denote by $\eta$, we first find &#xA;$$&#xD;&#xA;{\rm E}[X|X \ge 0] = a{\rm P}(X = a|X \ge 0) = a\frac{{{\rm P}(X = a)}}{{{\rm P}(X \ge 0)}} = \frac{{ap_1 }}{{p_1  + p_2 }}&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;{\rm E}[X|X &amp;lt; 0] = \frac{{ - ap_1 }}{{1 - p_1  - p_2 }}.&#xD;&#xA;$$&#xA;Hence, by definition, ${\rm P}(Y = \frac{{ap_1 }}{{p_1  + p_2 }}) = p_1  + p_2 $ and ${\rm P}(Y = \frac{{ - ap_1 }}{{1 - p_1  - p_2 }}) = 1 - p{}_1 - p_2$. &#xA;Thus, &#xA;$$&#xD;&#xA;{\rm E}(Y^i) = \Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^i (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^i (1 - p_1  - p_2),&#xD;&#xA;$$&#xA;from which we get an explicit expression for $\eta$. Now, to furnish a counterexample, it suffices to find a triple $(a,p_1,p_2)$ for which $\xi \leq 0$ and $\eta &amp;gt; 0$. This is most easily done using a computer. Here is a concrete example.&#xA;Letting $a=4$, $p_1=0.36$, and $p_2=0.4$, we have $-ap_1/(1-p_1-p_2) = -6$. Here,&#xA;$$&#xD;&#xA;a^4 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2 ) = 403.2&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;2 \Big[a^2 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2 )\Big]^2 = 414.72,&#xD;&#xA;$$&#xA;so ${\rm E}(X^4) \leq 2 [{\rm E}(X^2)]^2$; on the other hand,&#xA;$$&#xD;&#xA;\Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^4 (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2) \approx 320.835&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;2\Big[\Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^2 (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2)\Big]^2 \approx 258.482,&#xD;&#xA;$$&#xA;so ${\rm E}(Y^4) &amp;gt; 2 [{\rm E}(Y^2)]^2$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastActivityDate="2010-12-23T15:17:36.993" />
  <row Id="15339" PostTypeId="2" ParentId="15335" CreationDate="2010-12-23T15:19:55.500" Score="3" ViewCount="0" Body="&lt;p&gt;Expand $(1+\cos(x))^3$. And rewrite $\cos^n(x)$ in terms of $\cos(nx)$ and other similar terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+\cos(x))^3 = 1 + 3 \cos(x) + 3 \cos^2(x) + \cos^3(x) = \frac{\cos(3x)+6 \cos(2x)+15 \cos(x)+10}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, integrate term by term to get the desired answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \cos(nx) = \frac{\sin(nx)}{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, $\int (1+\cos(x))^3 = \frac{\frac{\sin(3x)}{3}+6\frac{\sin(2x)}{2}+15 \sin(x) + 10x}{4} + c = \frac{\sin(3x)}{12} + \frac{3\sin(2x)}{4} + \frac{15 \sin(x)}{4} + \frac{5x}{2} + c$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-23T15:19:55.500" />
  <row Id="15340" PostTypeId="2" ParentId="15337" CreationDate="2010-12-23T15:27:34.460" Score="1" ViewCount="0" Body="&lt;p&gt;$\alpha = 0$ gives the integral as $(N-1)e^{i} = \mathcal{O}(N)$ and $\alpha = 1$ gives the integral to be $\frac{e^{iN}-e^{i}}{i} = \mathcal{O}(1)$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-23T15:50:35.117" LastActivityDate="2010-12-23T15:50:35.117" CommentCount="3" />
  <row Id="15341" PostTypeId="2" ParentId="5377" CreationDate="2010-12-23T15:35:42.033" Score="-1" ViewCount="0" Body="&lt;p&gt;Why not replace &quot;ZFC is inconsistent&quot; by false? Namely you can prove anything this way.&#xA;&lt;hr&gt;&#xA;But this argument is not a proof because it has an extra assumption that A halts iff there is a proof that A does not halt. You mean A is constructible.&#xA;&lt;hr&gt;&#xA;What does consistent mean? Someone defines consistent as ~provable(0 = 1), but this only implies the axioms are not against our intuition. But we could have 0 = 1 as an axiom.&lt;/p&gt;&#xA;" OwnerUserId="5079" LastEditorUserId="5079" LastEditorDisplayName="" LastEditDate="2010-12-24T03:53:16.367" LastActivityDate="2010-12-24T03:53:16.367" CommentCount="2" />
  <row Id="15342" PostTypeId="1" CreationDate="2010-12-23T15:42:33.363" Score="0" ViewCount="110" Body="&lt;p&gt;Let $A \subset X$ be closed and $U \subset A$ open in $A$. Let $V$ be any open&#xA;set in $X$ with $U \subset V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove $U \cup (V \setminus A)$ is open in $X$.&lt;/p&gt;&#xA;" OwnerUserId="5080" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-23T15:44:36.077" LastActivityDate="2010-12-23T16:01:44.070" Title="Prove this set is open" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15343" PostTypeId="2" ParentId="15326" CreationDate="2010-12-23T15:44:00.433" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n&quot; rel=&quot;nofollow&quot;&gt;For every modulus, all relatively prime integers are a multiplicative group.&lt;/a&gt;&lt;br&gt;&#xA;(Without this, number theory would be quite pointless)&lt;/p&gt;&#xA;" OwnerUserId="5078" LastActivityDate="2010-12-23T15:44:00.433" />
  <row Id="15344" PostTypeId="2" ParentId="15055" CreationDate="2010-12-23T15:47:04.217" Score="4" ViewCount="0" Body="&lt;p&gt;I think that the reason that these puzzles are so often confusing is that they rely on the limitations of the English language rather than on any mathematical difficulties.  Of course this is not unique to English, and I think you should be able to find similar similar puzzles in pretty much any natural language.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example that brings out the difficulty more clearly.  First, consider the following similar puzzle: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A family has two children, Robin and Lindsay. Lindsay is a girl. What is the probability that both children are girls? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In elementary probability class, they teach you to answer this by making a table of the four options&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  Robin    Lindsay&#xA;  B        B&#xA;* B        G&#xA;  G        B&#xA;* G        G&#xA;&lt;/pre&gt; &#xA;&#xA;&lt;p&gt;The starred rows are the ones where Lindsay is a girl, and we compute from them that the chance that both children are girls is 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider this puzzle&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A family has two children, Robin and Lindsay. At least one of them is a girl. What is the probability that both children are girls? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The elementary probability method gives the following table, and a probability of 1/3.  The difference is that we gain one more row, compared to the previous puzzle. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;  Robin    Lindsay&#xA;  B        B&#xA;* B        G&#xA;* G        B&#xA;* G        G&#xA;&lt;/pre&gt; &#xA;&#xA;&lt;p&gt;After looking at these, you can see that the difficulty of the original puzzle comes because, in English, &quot;one of the children&quot; can mean several different things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;one&quot; can mean &quot;a particular one&quot;. If you read the original puzzle like this, it becomes analogous to the first puzzle I wrote, and the answer will be 1/2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;one&quot; can mean &quot;at least one&quot;. If you read the original puzzle like this, it become analogous to the second puzzle I wrote, and the answer is 1/3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;one&quot; can mean &quot;exactly one&quot;. If you read the original puzzle like this, the answer is 0.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is a common convention in mathematics that &quot;one&quot; usually means &quot;at least one&quot;. For example, this is the sense intended in the following sentence, which is a typical example of mathematical English: &quot;if a natural number $n$ is a multiple of a prime number $p$, and $n = ab$, then one of $a$ and $b$ is divisible by $p$.&quot;  We would not read this as saying that exactly one of $a$ and $b$ is divisible by $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe this convention is very common in non-mathematical English. If I say, &quot;one of my children is a girl&quot;, in normal English this means that the other is a boy. Similar discrepancies between mathematical and non-mathematical English come up with our use of the word &quot;or&quot; and our use of the phrase &quot;if/then&quot;.  When we teach mathematics, we have to spend time explaining this mathematical argot to students, so they can use the same English conventions that we do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability puzzles like the one you're asking about rely on these differences of English meaning, rather than on any logical or mathematical problem.  In that sense, they aren't really puzzles, they're just tricks. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-23T15:47:04.217" CommentCount="4" />
  <row Id="15346" PostTypeId="2" ParentId="15334" CreationDate="2010-12-23T15:53:02.533" Score="4" ViewCount="0" Body="&lt;p&gt;Here's how you can find your numerator. Set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v \equiv B + A(g-2v).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This immediately gives $A = -1/2 $ and hence $ B = g/2 . $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try $\int \frac{5g - 3v}{g+v} \textrm{ d}v$ to check that you have the idea.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-23T16:01:09.263" LastActivityDate="2010-12-23T16:01:09.263" CommentCount="3" />
  <row Id="15347" PostTypeId="2" ParentId="15342" CreationDate="2010-12-23T16:01:44.070" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; Since $U$ is open in $A$, there is an open set $\mathscr{U}$ of $X$ such that $U = A\cap\mathscr{U}$. Consider the open sets $\mathscr{U}\cap V$ and $V\setminus A$.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-23T16:01:44.070" />
  <row Id="15348" PostTypeId="2" ParentId="15335" CreationDate="2010-12-23T16:06:08.163" Score="7" ViewCount="0" Body="&lt;p&gt;If you don't know how to rewrite $\cos^n(x)$ in terms of $\cos(nx)$, here is an alternate solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand $(1+\cos(x))^3 = 1 + 3\cos(x) + 3\cos^2(x) + \cos^3(x)$.&#xA;Now, you know how to integrate the first two terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third term can be integrated by using the half-angle formula&#xA;$$\cos^2(x) = \frac{1+\cos(2x)}{2},$$&#xA;so that&#xA;$$\int 3\cos^2(x)\,dx = 3\left(\frac{x}{2} + \frac{\sin(2x)}{4}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term can be integrated using the pythagorean identity $\cos^2(x) = 1 - \sin^2(x),$ so&#xA;$$\cos^3(x) = \cos(x) - \cos(x)\sin^2(x).$$&#xA;And now $\int \cos(x)\,dx = \sin x$ and $\int \cos(x)\sin^2(x)\,dx = \frac{\sin^3(x)}{3}$, so&#xA;$$\int \cos^3(x)\,dx = \sin x - \frac{\sin^3(x)}{3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just put the pieces together.&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-23T16:16:46.413" LastActivityDate="2010-12-23T16:16:46.413" CommentCount="1" />
  <row Id="15349" PostTypeId="1" AcceptedAnswerId="15467" CreationDate="2010-12-23T16:10:14.270" Score="0" ViewCount="53" Body="&lt;p&gt;Given a directed graph $G\langle V,E\rangle$, and weight $w[u][v]$ of each $\langle u,v \rangle$ representing an edge of $E$ the question is, try to divide the whole graph into several &quot;rings&quot;, making each vertex $u$ belong to a certain &quot;ring&quot; and a &quot;ring&quot; contains at least two vertices, it's guaranteed that the given graph must be able to be divided, satisfying the restrictions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, please make the total weight of all the &quot;ring&quot;s minimum .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a task assigned by my tutor, I don't know if it's an old problem because I'm new on algorithm, my tutor told me that it could be solved within polynomial time complexity, I wanna know how it can be done, I just have no idea about that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm terribly sorry for my forgetting to post the correct question at the first time... &lt;/p&gt;&#xA;" OwnerUserId="5066" LastEditorUserId="5066" LastEditorDisplayName="" LastEditDate="2010-12-24T14:15:47.630" LastActivityDate="2010-12-24T21:36:32.173" Title="algorithm problem: Prove minimum ring division on directed graph can be solved in polynomial time" Tags="&lt;graph-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15350" PostTypeId="1" AcceptedAnswerId="15424" CreationDate="2010-12-23T16:41:02.683" Score="7" ViewCount="179" Body="&lt;p&gt;(I'll keep this one short)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a Dirichlet series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(s)=\sum_{k=1}^\infty\frac{c_k}{k^s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_k\in\mathbb R$ and $c_k \neq 0$ (i.e., the coefficients are a sequence of arbitrary nonzero real numbers), and assuming that $g(s)$ can be analytically continued, does it follow that $g(s)$ possesses a critical strip containing its nontrivial zeroes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this does not generally hold, what restrictions should there be on the $c_k$ for $g(s)$ to possess a critical strip?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My attempts at searching bring too much stuff on Riemann $\zeta$, with only a quick mention of general Dirichlet series; pointers to the literature would be appreciated.)&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-24T09:13:10.720" Title="On Dirichlet series and critical strips" Tags="&lt;analytic-number-theory&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="15351" PostTypeId="2" ParentId="15337" CreationDate="2010-12-23T16:45:10.073" Score="5" ViewCount="0" Body="&lt;p&gt;I think what you should get is, for each fixed $\alpha$, asymptotically $O(N^{1-\alpha})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why: do the change of variables $y = x^\alpha$. Your integral becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha \int_1^{N^\alpha} y^{\frac{1-\alpha}\alpha} e^{iy} dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now integrate by parts to lower the exponent on $y$. The first step would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{\alpha}{i}\left( N^{1-\alpha} e^{i N^\alpha} - e^i \right) - \frac{1-\alpha}{i} \int_1^{N^\alpha} y^{\frac{1-2\alpha}\alpha} e^{iy} dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating the process $M = \lceil \frac{1}{\alpha} \rceil$ number of times, you get a string of terms that estimates to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = O(N^{1-\alpha} + N^{1-2\alpha} + \ldots + N^{1-(M-1)\alpha}) + O(\int_1^{N^\alpha} y^{\frac{1-M\alpha}{\alpha}} e^{iy} dy) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $M\alpha \geq 1$, the final integral is dominated by a constant. Noting that for large $N$, the first term inside the first $O(\cdot)$ dominates, you get the result. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-23T16:45:10.073" />
  <row Id="15352" PostTypeId="2" ParentId="15350" CreationDate="2010-12-23T16:52:35.590" Score="1" ViewCount="0" Body="&lt;p&gt;Possible keywords include &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_L-function&quot; rel=&quot;nofollow&quot;&gt;Dirichlet L-functions&lt;/a&gt; or the generalized Riemann hypothesis. For the case where $c_k$ are the values of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_character&quot; rel=&quot;nofollow&quot;&gt;Dirichlet character&lt;/a&gt;, in certain situations the existence of a critical strip is known. See the Wikipedia article for references. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-23T16:52:35.590" CommentCount="13" />
  <row Id="15353" PostTypeId="2" ParentId="15350" CreationDate="2010-12-23T17:14:14.833" Score="4" ViewCount="0" Body="&lt;p&gt;$g$ will always have a half-plane free from zeroes (cf. e.g. Titchmarsh, &lt;em&gt;Theory of functions&lt;/em&gt;, Section 9.6). This means a functional equation would be sufficient to guarantee a critical strip (essentially calling the other zeroes trivial by definition).&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-23T17:14:14.833" />
  <row Id="15354" PostTypeId="1" AcceptedAnswerId="15374" CreationDate="2010-12-23T17:45:57.250" Score="6" ViewCount="146" Body="&lt;p&gt;How may one go about proving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\Gamma&amp;#39;(s)}{\Gamma(s)}=O(\log|s|)$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(away from the poles) directly? By a direct proof, I mean not to go through the usual Stirling formula with the exact error term. The use of a rough form of Stirling's formula is welcome.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T00:57:28.590" LastActivityDate="2010-12-24T00:57:28.590" Title="Stirling-type formula for the logarithmic derivative of the Gamma function" Tags="&lt;special-functions&gt;&lt;gamma-function&gt;" AnswerCount="1" />
  <row Id="15355" PostTypeId="2" ParentId="15337" CreationDate="2010-12-23T17:53:03.800" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Corput_lemma_%28harmonic_analysis%29&quot; rel=&quot;nofollow&quot;&gt;Van der Corput's lemma&lt;/a&gt; gives&#xA;$$&#xD;&#xA;\left|\int_1^N e^{ix^\alpha}dx\right|\le \frac{C}{\alpha}\,N^{1-\alpha},&#xD;&#xA;$$&#xA;where the constant $C$ is independent of $\alpha$ (and of $N$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Edited to answer TCL's comment&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Van der Corput's lemma&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\phi\colon(a,b)\to\mathbb{R}$ is smooth and $|\phi^{(k)}(x)|\ge1$ for $a&amp;lt;x&amp;lt;b$, then&#xA;$$&#xD;&#xA;\left|\int_1^N e^{i\lambda\phi(x)}dx\right|\le C_k\lambda^{-1/k}.&#xD;&#xA;$$&#xA;This valid if $k\ge2$ or if $k\ge1$ and $\phi&amp;#39;$ is monotone.&#xA;It is clearly equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\phi\colon(a,b)\to\mathbb{R}$ is smooth and $|\phi^{(k)}(x)|\ge\lambda$ for $a&amp;lt;x&amp;lt;b$, then&#xA;$$&#xD;&#xA;\left|\int_a^b e^{i\phi(x)}dx\right|\le C_k\lambda^{-1/k}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To aply it to $\int_1^N e^{ix^\alpha}dx$, choose $a=1$, $b=N$, $\phi(x)=x^\alpha$, and $k=1$. Then $\phi&amp;#39;(x)=\alpha\,x^{\alpha-1}$ is decrasing, so that $|\phi&amp;#39;(x)|\ge\alpha\,N^{\alpha-1}(=\lambda)$.&lt;/p&gt;&#xA;" OwnerUserId="4791" LastEditorUserId="3035" LastEditorDisplayName="" LastEditDate="2010-12-24T06:12:31.660" LastActivityDate="2010-12-24T06:12:31.660" CommentCount="3" />
  <row Id="15356" PostTypeId="2" ParentId="15201" CreationDate="2010-12-23T17:54:27.570" Score="0" ViewCount="0" Body="&lt;p&gt;There are some excellent combinatorics books which also look at the applicability of combinatorics both within mathematics and outside of mathematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Applied Combinatorics by Fred Roberts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Applied Combinatorics by Alan Tucker&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-23T17:54:27.570" />
  <row Id="15357" PostTypeId="2" ParentId="15280" CreationDate="2010-12-23T17:57:02.690" Score="0" ViewCount="0" Body="&lt;p&gt;You should try reading &quot;Mathematics - form and function&quot; by Saunders MacLane. The author tries to convey what Mathematics is &quot;all about&quot; and at the same time giving a overview over several different areas. (including manifolds, number theory, groups/fiels, algebra, geometry, etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematics,_Form_and_Function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathematics,_Form_and_Function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387962174&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Mathematics-Function-Saunders-Mac-Lane/dp/0387962174&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presentation may at times be too dense for someone who has not encountered the material before, but if you're prepared to read slow and careful at times, the book is excellent.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastActivityDate="2010-12-23T17:57:02.690" />
  <row Id="15358" PostTypeId="2" ParentId="15278" CreationDate="2010-12-23T18:01:43.627" Score="4" ViewCount="0" Body="&lt;p&gt;Harmonic analysis is a very broad field. A nice introduction to harmonic analysis at the level of a beginning graduate student would be Grafakos' Classical and Modern Fourier analysis (two books).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elias Stein also has many nice books, but they are quite hard as he leaves out a lot of details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The books require at least knowledge in measure and integration theory (and so real analysis), some basic functional analysis and for Grafakos you might want to know what a group is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, maybe you should also state what kind of harmonic analysis you want to do. The contents of the first book of Grafakos (Classical Fourier Analysis) is at least something you should know if you want to do harmonic analysis (as I have been told).&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-23T18:01:43.627" CommentCount="1" />
  <row Id="15359" PostTypeId="1" AcceptedAnswerId="15361" CreationDate="2010-12-23T18:26:48.247" Score="0" ViewCount="27" Body="&lt;p&gt;I'm working with percentage splits and I'm trying to add extra players into the split but I'm coming unstuck...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have agreements with a number of parties that set out what type of split we get from a pot of cash.  With some it's 50/50, with others it may be 60/40 in favour of either party or perhaps even 70/30.  In a scenario where there are only 2 players, myself and someone else, this is simple.  What happens however when I add a third party?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets say I am already working with someone that gets an equal cut (50/50) with me, and we have to bring someone else on board whom I have a 70/30 agreement.  So now that adjusts the equation, two parties must make equal amounts and the third party must make some factor less to balance the pot to 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can figure out that this is about ratios - 30%/70% = 0.428571.  By trial and error, I can figure out that each of the parties getting an equal cut now get 41.2% of the pot and the remaining party (30%) gets 17.6% of the pot.  I'd like to figure out a formula that gives me the values but I can only seem to get so far...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the splits are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Me     Other    Ratio&#xA;--     -----    -----&#xA;70%    30%      0.428571&#xA;50%    50%      1&#xA;&#xA;X  = First player   (Me/Player 1)&#xA;Y  = Second player  (Player 2)&#xA;Z  = Third player   (Player 3)&#xA;R1 = Ratio split between X and Y&#xA;R2 = Ratio split between X and Z&#xA;&#xA;Y + Z = X / 0.428571&#xA;&#xA;Given that Y = Z in this scenario, we could say that:&#xA;&#xA;  2Y = X / 0.428571 or &#xA;  Y (or Z) = X / 0.428571 / 2&#xA;&#xA;Or&#xA;&#xA;  Y (or Z) = X / (2 x 0.428571)&#xA;           = X / 0.857143&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems like X / R1 = Y, which divided by R2 = Z, the sum of which adds up to 1.  But I come unstuck when I calculate different splits using this kind of algorithm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Me    Other    Ratio&#xA;--    -----    -----&#xA;60%   40%      0.666667&#xA;30%   70%      2.333333&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When I calculate using the second ratio, I should be multiplying(?)  In the first scenario, R1 is 1, so it makes no difference whether I divide or multiply, so I guess I should be multiplying in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I've take completely the wrong path.  Can anyone show me either where I'm going wrong or show me the next leap I need to make to get to a result?  Given the original splits, is there a formula to quickly tell me what the adjusted split would be when the extra party is added to the equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated, I've been out of school so long, this kind of theoretical math is way outside my normal range of expertise.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2010-12-23T18:35:21.753" Title="How can I factor in an extra player on a percentage split?" Tags="&lt;linear-algebra&gt;&lt;algebra&gt;" AnswerCount="1" />
  <row Id="15360" PostTypeId="1" AcceptedAnswerId="15405" CreationDate="2010-12-23T18:30:52.157" Score="7" ViewCount="156" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Integral_test_for_convergence&quot; rel=&quot;nofollow&quot;&gt;integral test of convergence&lt;/a&gt; states that, if $f:[1,+\infty)\to[0,+\infty)$ is a monotonically decreasing nonnegative function, then the series $\sum_1^\infty f(n)$ converges iff $\int_1^\infty f(n) dn$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the high-dimensional generalization also true? That is, given $f:[1,+\infty)^N \to[0,+\infty)$, and $f(\dotsc,n_i,\dotsc) \ge f(\dotsc,n_i+\epsilon,\dotsc)$ for all $1\le i\le N$ and $n_i\in[1,+\infty)$ and $\epsilon&amp;gt;0$, then the sum&#xA;$$ \sum_{n_1=1}^\infty  \cdots \sum_{n_N=1}^\infty f(n_1,\dotsc,n_N) $$&#xA;converges iff the multiple integral&#xA;$$ \int_1^\infty \cdots \int_1^\infty f(n_1,\dotsc,n_N) dn_1 \dotsm dn_N $$&#xA;is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(This is just for checking if &lt;a href=&quot;http://physics.stackexchange.com/questions/2196/anti-gravity-in-an-infinite-lattice-of-point-masses/2202#2202&quot;&gt;my answer over physics.SE&lt;/a&gt; is reasonable.)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2010-12-23T19:12:45.193" LastActivityDate="2010-12-26T23:51:13.707" Title="Does the integral test work on higher dimensions?" Tags="&lt;real-analysis&gt;&lt;sequences-and-series&gt;&lt;convergence&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15361" PostTypeId="2" ParentId="15359" CreationDate="2010-12-23T18:35:21.753" Score="2" ViewCount="0" Body="&lt;p&gt;If you have a two way deal with all the other players, you can just normalize them all.  Say you have a 70/30 split with A and a 40/60 split with B.  Then if you get 1, A gets 7/3 and B gets 4/6.  That would call for a total of 7/3+1+2/3=4, so divide everybody by 4.  A gets 7/12, you get 1/4=3/12, B gets 1/6=2/12.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-23T18:35:21.753" CommentCount="1" />
  <row Id="15362" PostTypeId="2" ParentId="15201" CreationDate="2010-12-23T18:51:20.357" Score="0" ViewCount="0" Body="&lt;p&gt;I really like &quot;A Course in Enumeration&quot; by Martin Aigner.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2010-12-23T18:51:20.357" />
  <row Id="15363" PostTypeId="2" ParentId="15278" CreationDate="2010-12-23T18:57:02.347" Score="1" ViewCount="0" Body="&lt;p&gt;Worth having a look at is &quot;Harmonic analysis on finite groups&quot; for applications to probability theory (one of your stated strengths). That way you'll be going from what you already know to the new stuff. &lt;a href=&quot;http://books.google.com/books?id=3trud4weQ3AC&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the Google Books link.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T10:53:49.933" LastActivityDate="2010-12-24T10:53:49.933" />
  <row Id="15364" PostTypeId="1" CreationDate="2010-12-23T18:57:16.000" Score="0" ViewCount="32" Body="&lt;p&gt;Frequently a company wants to match its assets and liabilities. However, perfect matching is not practical due to fluctuations in interest rates. So they hedge their risk using &lt;a href=&quot;http://en.wikipedia.org/wiki/Immunization_%28finance%29&quot; rel=&quot;nofollow&quot;&gt;immunization&lt;/a&gt;. This can be achieved using present value matching, duration matching, and greater convexity for assets. In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$PV_{A}(i_0) = PV_{L}(i_0)$&lt;/li&gt;&#xA;&lt;li&gt;$\frac{d}{di} PV_{A}(i)|_{i_0} = \frac{d}{di} PV_{L}(i)|_{i_0}$&lt;/li&gt;&#xA;&lt;li&gt;$\frac{d^2}{di^2} PV_{A}(i)|_{i_0} &amp;gt; \frac{d^2}{di^2} PV_{L}|_{i_0}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Could we add any extra conditions so that the risk is reduced even more(e.g. could be look at $\frac{d^3}{di^3} PV_{A}(i)$ etc..)?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-23T18:57:16.000" Title="Immunization and Sensitivity Analysis" Tags="&lt;finance&gt;&lt;mathematical-modeling&gt;" />
  <row Id="15365" PostTypeId="1" AcceptedAnswerId="15369" CreationDate="2010-12-23T19:22:12.090" Score="0" ViewCount="65" Body="&lt;p&gt;I'm doing a physics course in my spare time and the problem set is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ball A is dropped from rest from a building of height H exactly as ball B is thrown up vertically from the ground. When they collide A has twice the speed of B. If the collision occurs at height h, what is h/H?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also gives a hint: &lt;em&gt;Write equations for heights $y_A, y_B$ and velocities $v_A, v_B$. What can you say about them at the time of the collision?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm stuck at the beginning. I'm trying to find a displacement function for ball A, and I realized that I need to find a velocity function first, and the only thing I know is that the ball is moving under gravity so then $a_A(t) = -g$ and then integrated this means $v_A(t) = -1/2gt^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good, except that now I got confused. Should I integrate the velocity function? It doesn't feel like I should, and even so I'm not really sure how - so I looked at the problem solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem solution says the displacement function is $y_A(t) = H - 1/2gt^2$ - which looks a lot like my velocity function! I am confused. The gravitational force &lt;em&gt;accelerates&lt;/em&gt; mass ($F = ma$ amirite), so how can this be?&lt;/p&gt;&#xA;" OwnerUserId="5085" LastEditorUserId="5085" LastEditorDisplayName="" LastEditDate="2010-12-23T19:38:40.583" LastActivityDate="2010-12-23T19:51:03.027" Title="Newtonian Physics: ball A is dropped, ball B is thrown - where do they coincide?" Tags="&lt;calculus&gt;&lt;physics&gt;" AnswerCount="2" />
  <row Id="15366" PostTypeId="1" AcceptedAnswerId="15377" CreationDate="2010-12-23T19:24:21.090" Score="16" ViewCount="145" Body="&lt;p&gt;This is prompted by &lt;a href=&quot;http://math.stackexchange.com/questions/15312/show-that-pi-i-j-v-i-v-j-le-kn2-for-1-le-v-1-v-2-v-n&quot;&gt;question 15312&lt;/a&gt;, but moved to the reals.  It must be solved already.  Pick n points $x_i \in [0,1]$ to maximize $\prod_{i &amp;lt; j} (x_i - x_j)$.  A little playing shows you don't want them evenly distributed-they need to push out to the ends.  With four points, Alpha says to use $\{0,\frac{1}{2}\pm\frac{1}{2\sqrt{5}},1\}$ and with five, $\{0,\frac{1}{2}-\frac{\sqrt{\frac{3}{7}}}{2},\frac{1}{2},\frac{1}{2}+\frac{\sqrt{\frac{3}{7}}}{2},1\}$&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-23T20:21:44.430" LastActivityDate="2010-12-23T21:01:08.773" Title="Spreading points in the unit interval to maximize the product of pairwise distances" Tags="&lt;real-analysis&gt;&lt;analysis&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="15367" PostTypeId="2" ParentId="639" CreationDate="2010-12-23T19:26:16.583" Score="-1" ViewCount="0" Body="&lt;p&gt;I would do it by changing my base. Instead of using binary for all the problems, I'd use base 3 for two poisoned bottles, base 4 for 3 poisoned bottles, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general problem, with &lt;em&gt;k&lt;/em&gt; poisoned bottles and &lt;em&gt;N&lt;/em&gt; total bottles, optimal number is:&#xA;roundup[log_base{&lt;em&gt;k&lt;/em&gt;+1}(&lt;em&gt;N&lt;/em&gt;) &lt;em&gt;k&lt;/em&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the two bottle case, &lt;em&gt;N&lt;/em&gt;=1000, the solution is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Label the 1000 bottles in base 3 from 0 to 999.&lt;/li&gt;&#xA;&lt;li&gt;Calculate log_base{3}(1000): 6.29 =&gt; we will have 7 places.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For i going from 0 to 6 (0th place being right-most)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Allocate a servant for bottles whose base 3 number has 1 in ith place.&lt;/li&gt;&#xA;&lt;li&gt;Allocate a servant for bottles whose base 3 number has 2 in ith place.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Repeat. You may not need a servant for step 2 if the highest number is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following that method, the final tally of servants required is 13, which is much better than the proposed 65.&lt;/p&gt;&#xA;" OwnerUserId="5086" LastActivityDate="2010-12-23T19:26:16.583" CommentCount="1" />
  <row Id="15368" PostTypeId="2" ParentId="15365" CreationDate="2010-12-23T19:49:18.970" Score="1" ViewCount="0" Body="&lt;p&gt;I see my fault: the integral of $a_A(t) = -g$ is $v_A(t) = -gt$, and therefore the displacement function is $y_A(t) = H - 1/2gt^2$.&lt;/p&gt;&#xA;" OwnerUserId="5085" LastActivityDate="2010-12-23T19:49:18.970" />
  <row Id="15369" PostTypeId="2" ParentId="15365" CreationDate="2010-12-23T19:51:03.027" Score="2" ViewCount="0" Body="&lt;p&gt;You double integrated to get the -(1/2)gt^2 !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate(-g) = -gt + C  (C=0 for starting at rest)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate(-gt) = -(1/2)gt^2 + C  (C=H for starting at top of building)&lt;/p&gt;&#xA;" OwnerUserId="5087" LastActivityDate="2010-12-23T19:51:03.027" />
  <row Id="15370" PostTypeId="2" ParentId="15332" CreationDate="2010-12-23T19:55:01.450" Score="7" ViewCount="0" Body="&lt;p&gt;It is possible to &lt;em&gt;express&lt;/em&gt; the finite-dimensionality in representation-theoretic terms: it is amounts to the fact that automorphic forms on $\mathrm{GL}_2(\mathbb A)$ form an &lt;em&gt;admissible&lt;/em&gt; representation of $\mathrm{GL}_2(\mathbb A)$.  (Actually, this statement is a little stronger, because it includes the Maass form case.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure that one can prove this purely representation-theoretically.  At least, it depends on analytic facts (or algebro-geometric equivalents, if one thinks in terms of Riemann--Roch), which have to be brought into play in one manner or another.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-23T21:29:30.217" LastActivityDate="2010-12-23T21:29:30.217" />
  <row Id="15371" PostTypeId="1" CreationDate="2010-12-23T19:56:16.960" Score="4" ViewCount="105" Body="&lt;p&gt;The hint for this problem is $a^{n+1} - 1 = (a + 1)(a^n - 1) - a(a^{n-1} - 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the problem is true because if you distribute the $a$ and the $-1$ the terms cancel out to equal the left side.  However, since it is telling me to use strong induction I am guessing there is more I am supposed to be doing.  On the hint I can see that it is a true statement, but I am not sure how to use that to prove the equation or how the right side of the hint relates to the right side of the problem.  Also, I do realize that in the case of the hint $n = 1$ would be the special case. &lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditorDisplayName="" LastEditDate="2010-12-23T20:36:08.143" LastActivityDate="2010-12-23T22:51:54.640" Title="Using the second principle of finite induction to prove $a^n -1 = (a-1)(a^{n-1} + a^{n-2} + ... + a + 1)$  for all $n \geq 1$" Tags="&lt;algebra-precalculus&gt;&lt;induction&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="15373" PostTypeId="1" CreationDate="2010-12-23T20:04:38.140" Score="1" ViewCount="108" Body="&lt;p&gt;I was reading a comment on MathOverflow and did not understand what was being suggested. I also don't understand Anton's response:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/18062#18062&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/17732/difference-between-measures-and-distributions/18062#18062&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially thought Regenbogen's comment was being applied to the derivative of the dirac distribution, i.e. he wanted to extend this continuous linear functional from $D$ to $C$. But I don't understand what dominating semi-norm (sublinear functional) he had in mind... And I don't understand Anton's response - How are the derivatives of the delta distribution not bounded linear functionals on $D$? They seem to have operator norm equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I thought perhaps he literally was referring to the measure mentioned earlier, i.e. a Radon measure acting as a distribution on $C$ and trivially extending to $D$... but this also seems to be a bounded linear functional on both $C$ and $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain what was being suggested, and how Anton's response makes sense?&lt;/p&gt;&#xA;" OwnerUserId="1260" LastEditorUserId="1260" LastEditorDisplayName="" LastEditDate="2010-12-24T16:29:38.740" LastActivityDate="2010-12-24T16:29:38.740" Title="Misapplication of the Hahn-Banach theorem" Tags="&lt;functional-analysis&gt;&lt;distribution-theory&gt;" CommentCount="3" />
  <row Id="15374" PostTypeId="2" ParentId="15354" CreationDate="2010-12-23T20:18:42.897" Score="7" ViewCount="0" Body="&lt;p&gt;You can use the product representation of $\Gamma(z)$, take logs and differentiate the resulting series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma(z) = \dfrac{e^{-\gamma z}}{z} \prod_{n=1}^{\infty} \left(1 + \dfrac{z}{n}\right)^{-1} \ e^{\frac{z}{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Polygamma_function&quot; rel=&quot;nofollow&quot;&gt;Polygamma Function&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Digamma_function&quot; rel=&quot;nofollow&quot;&gt;Digamma Function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T20:18:42.897" CommentCount="6" />
  <row Id="15375" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T20:29:07.940" Score="2" ViewCount="0" Body="&lt;p&gt;$\displaystyle a^{n+1} - 1 = (a + 1)(a^n + 1) - a(a^{n-1} - 1)$ is not right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try putting $a=1$, the right hand side is $\displaystyle 4$, while the left is $\displaystyle 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the hint was $\displaystyle a^{n+1} - 1 = (a-1)(a^n + 1) + a(a^{n-1} - 1)$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-23T20:29:07.940" CommentCount="1" />
  <row Id="15376" PostTypeId="2" ParentId="15335" CreationDate="2010-12-23T20:40:02.450" Score="6" ViewCount="0" Body="&lt;p&gt;You can also use complex numbers to make it mechanical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $\displaystyle \cos x = \dfrac{e^{ix} + e^{-ix}}{2}$ and just expand out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method makes finding integrals of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int P(\sin x, \cos x) \ \text{dx}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $P(x,y)$ is a polynomial in the variables $x,y$) purely mechanical, without having to do any clever algebra etc.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-23T21:00:35.077" LastActivityDate="2010-12-23T21:00:35.077" CommentCount="1" />
  <row Id="15377" PostTypeId="2" ParentId="15366" CreationDate="2010-12-23T20:46:50.477" Score="13" ViewCount="0" Body="&lt;p&gt;These points are known as Fekete points. A general Fekete problem is to maximize the product &#xA;$$\max_{z_1,...,z_n\in E}\prod\limits_{\quad 1\leq i &amp;lt; j \leq n}|z_i-z_j|$$&#xA;where $E\subset \mathbb C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case $E=[-1,1]$, there is a unique solution and the corresponding points  coincide with the zeros of $(1-x^2)P&amp;#39;_{n-1}(x)$, where $P_{n-1}$ is the Legendre polynomial of degree $n-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot give a precise reference at the moment, but this can be probably found in Szegő's book on orthogonal polynomials.  &lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-23T21:01:08.773" LastActivityDate="2010-12-23T21:01:08.773" CommentCount="10" />
  <row Id="15379" PostTypeId="2" ParentId="15334" CreationDate="2010-12-23T21:11:07.567" Score="2" ViewCount="0" Body="&lt;p&gt;Couldn't one just do a u-substitution with u=g-2v?&lt;/p&gt;&#xA;" OwnerUserId="1260" LastActivityDate="2010-12-23T21:11:07.567" CommentCount="1" />
  <row Id="15380" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T21:18:24.977" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ $ Put $\rm\ f(n) = a^n+a^{n-1}+\:\cdots\:+1\:.\ \ $  Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\ a^{n+1}-1\ = \ (a+1)\ (a^n-1) - a\ (a^{n-1}-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ }  =\ (a+1)\ ((a-1)\ f(n-1) - a\ (a-1)\ f(n-2))\quad $ by strong induction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ ((a+1)\ f(n-1)- a\ f(n-2))\quad$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ (\:f(n-1) + a\ (f(n-1)-f(n-2)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ \ldots $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ f(n)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-23T21:37:44.203" LastActivityDate="2010-12-23T21:37:44.203" />
  <row Id="15381" PostTypeId="2" ParentId="11492" CreationDate="2010-12-23T21:23:13.623" Score="1" ViewCount="0" Body="&lt;p&gt;Since the parameter $\sigma$ is positive, the quantity &#xA;$$&#xD;&#xA;-2\sigma \left ( rx^2 + y^2 + b(z-r)^2 - br^2 \right )&#xD;&#xA;$$&#xA;is negative if &#xA;$$&#xD;&#xA;rx^2 + y^2 + b(z-r)^2 &amp;gt; br^{2}.&#xD;&#xA;$$ &#xA;This inequality defines the exterior of an ellipsoid (call it $E_1$); note that the size of this ellipsoid is fixed by the parameters (that's a hint).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation&#xA;$$&#xD;&#xA;rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2 = C&#xD;&#xA;$$&#xA;defines a different ellipsoid, $E_2$, the size of which is determined by your choice of $C$ (another hint).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, remind yourself what it is that you want to show.  Typically, the goal is to show that there exists a $C$ such that $E_2$ defines a trapping region for the Lorenz equations, in which case it suffices to show that  $E_2$ can be made large enough so that it contains $E_1$.  There's really no additional calculation necessary to do this -- you just need to understand what you've done so far.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related (but different) question is to find an explicit lower bound on $C$ in terms of the parameters.  In this case, you can find bounds on each of $x$, $y$, and $z$ separately for points inside of $E_1$.  This will then give you a bound on the quantity&#xA;$$&#xD;&#xA;rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2&#xD;&#xA;$$&#xA;which then defines $C$.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2010-12-23T21:23:13.623" />
  <row Id="15382" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T21:24:41.750" Score="2" ViewCount="0" Body="&lt;p&gt;Sometimes the easiest way to figure out an induction argument like this is to prove a particular case.  I'll take care of proving $n = 3$, assuming you've already proven $n = 1$ and $n = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the hint, we have&#xA;$$a^{3} - 1 = (a + 1)(a^2 - 1) - a(a - 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the cases $n = 1$ and $n = 2$ hold, so we rewrite this as &#xA;$$a^{3} - 1 = (a + 1)(a - 1)(a + 1) - a(a - 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by factoring $(a - 1)$ on the right hand side, we have&#xA;$$a^3 - 1 = (a - 1)(a^2 + 2a + 1 - a) = (a - 1)(a^2 + a + 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is less interesting than the case $n = 4$; try to work that out on your own! After that, you should be able to work out a general $n + 1$ case.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-23T21:24:41.750" />
  <row Id="15383" PostTypeId="1" AcceptedAnswerId="15386" CreationDate="2010-12-23T21:27:21.000" Score="5" ViewCount="77" Body="&lt;p&gt;I have found the following ellipse representation $(x,y)=(x_0\cos(\theta+d/2),y_0\cos(\theta-d/2))$, $\theta \in [0,2\pi]$. This is a contour of bivariate normal distribution with uneven variances and correlation $\rho=\cos(d)$. I know that this is a rotated ellipse with centre $(0,0)$. How to find the lengths of major and minor axes and the angle between x-axis and major axis? &lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T11:53:52.650" LastActivityDate="2010-12-25T11:53:52.650" Title="Finding standard ellipse characteristics from specific ellipse parametrisation" Tags="&lt;geometry&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="2" />
  <row Id="15384" PostTypeId="1" CreationDate="2010-12-23T21:45:28.883" Score="2" ViewCount="67" Body="&lt;p&gt;Let $T$ be an invertible linear map from $\mathbb{R}^{3}$ to $\mathbb{R}^{3}$.&#xA;For any plane $V$, are there a linear map $S$ from&#xA;$\mathbb{R}^{3}$ to $\mathbb{R}^{3}$ and a plane $W$ such that&#xA;$\left\Vert S\left(y\right)\right\Vert_{3}\leq\sup_{x\in V,\left\Vert x\right\Vert _{3}=1}\left\Vert T\left(x\right)\right\Vert _{3}$&#xA;for all $y\in\mathbb{R}^{3}$ with $\left\Vert y\right\Vert _{3}=1$&#xA;and $S\left(x\right)=T\left(x\right)$ for all $x\in W$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4727" LastActivityDate="2010-12-23T21:45:28.883" Title=" Is there a linear map with norm dominated by a given linear and partially identical to the given one?" Tags="&lt;real-analysis&gt;&lt;linear-algebra&gt;&lt;analysis&gt;&lt;matrices&gt;&lt;functional-analysis&gt;" CommentCount="5" />
  <row Id="15385" PostTypeId="2" ParentId="15371" CreationDate="2010-12-23T21:52:08.817" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\ \ $ Show that $\rm\ (a^{n+1}-1)/(a-1)\ $ and $\rm\ a^n+a^{n-1}+\:\cdots\:+1\ $ are both solutions of the recurrence $\rm\ f(n+2)\ = (a+1)\ f(n+1) - a\ f(n)\ $ with identical initial conditions $\rm\ f(1),\ f(0)\:.\ \ \ $&lt;br&gt;&#xA;Now use strong induction to prove the &lt;strong&gt;uniqueness theorem&lt;/strong&gt; for solutions of such difference equations (which is very easy). This is the essence of the calculation in my other answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I've emphasized in &lt;a href=&quot;http://math.stackexchange.com/search?tab=newest&amp;amp;q=user:242%20uniqueness&quot;&gt;many posts here&lt;/a&gt; uniqueness theorems provide powerful tools for proving equalities. Luckily, here, the uniqueness theorem for difference (vs. differential) equations has an absolutely trivial proof by way of (strong) induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the corresponding first-order case is essentially &lt;a href=&quot;http://math.stackexchange.com/search?tab=newest&amp;amp;q=user:242%20telescopy&quot;&gt;&lt;strong&gt;telescopy&lt;/strong&gt;&lt;/a&gt;, also known as the fundamental theorem of difference calculus.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-23T22:51:54.640" LastActivityDate="2010-12-23T22:51:54.640" />
  <row Id="15386" PostTypeId="2" ParentId="15383" CreationDate="2010-12-23T22:10:40.057" Score="6" ViewCount="0" Body="&lt;p&gt;The semi-major axis will be the distance from the origin to the farthest point on the ellipse.  So $r^2=x_0^2\cos^2(\theta+d/2)+y_0^2\cos^2(\theta-d/2)$.  As $r^2$ is a monotonic function of $r$, you can maximize it instead, which simplifies things a bit.  The angle between the $X$ axis and the major axis will then be the arctangent of $y/x$ at this maximum point.  The semi-minor axis will be the distance from the origin to the closest point on the ellipse, and will be $\pi/2$ away from the semi-major axis.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-23T22:10:40.057" />
  <row Id="15387" PostTypeId="2" ParentId="15383" CreationDate="2010-12-23T22:36:39.350" Score="7" ViewCount="0" Body="&lt;p&gt;This formula rescales a standard ellipse $(\cos(\theta + d/2), \cos(\theta - d/2))$ (inscribed within the unit square) by the diagonal matrix $(x_0, y_0)$.  By symmetry, the values $\theta = 0$ and $\theta = \pi/2$ correspond to vertices of this standard ellipse, allowing us to find their coordinates (and thus the lengths of the semi-axes), whence we easily deduce its equation is $x^2 + y^2 - 2 \rho x y = 1 - \rho^2$.  Applying the diagonal matrix gives the conventional implicit form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{x}{x_0}\right)^2 + \left(\frac{y}{y_0}\right)^2 - 2 \rho \frac{x}{x_0} \frac{y}{y_0} = 1 - \rho^2 \text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you can &lt;a href=&quot;http://en.wikipedia.org/wiki/Ellipse&quot; rel=&quot;nofollow&quot;&gt;look up&lt;/a&gt; anything you want.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-12-23T22:36:39.350" CommentCount="2" />
  <row Id="15388" PostTypeId="1" AcceptedAnswerId="15390" CreationDate="2010-12-23T22:54:45.450" Score="3" ViewCount="118" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just started real analysis.  I don't have a background in proofs or logic, simply calculus.  So I'm trying to learn more about proofs--so forgive the basic question, please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you go about proving this theorem: If a and b are any numbers, then there is one and only one number x such that a + x=b.  This number is given by x=b+(-a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First part of the proof I understand--it's simple.  We simply use some axioms and do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a+b-a=b, which of course is true.  This is just true from plugging in the b+(-a) for x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how about the uniqueness issue?  In my text, it says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a+x)+(-a)=b+(-a)&#xA;x=b+(-a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this prove uniqueness?  Is this equation essentially saying that the ONLY possible value of x is expressed by b+(-a)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I have known to set this up as part of the proof if I was asked to prove something is unique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the basic question--have to start somewhere.&lt;/p&gt;&#xA;" OwnerUserId="5088" LastActivityDate="2010-12-24T19:56:17.983" Title="Basic question about proving uniqueness" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15390" PostTypeId="2" ParentId="15388" CreationDate="2010-12-23T23:10:02.263" Score="5" ViewCount="0" Body="&lt;p&gt;Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a + x = b \quad \text{and} \quad a + y = b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get&#xA;$$-a + (a + x) = -a + (a + y) \implies (-a + a) + x = (-a + a) + y $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies 0 + x = 0 + y \implies x = y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-23T23:10:02.263" CommentCount="5" />
  <row Id="15391" PostTypeId="1" AcceptedAnswerId="15393" CreationDate="2010-12-24T00:07:40.190" Score="3" ViewCount="58" Body="&lt;p&gt;Let $\kappa$ be a measurable cardinal, we say that $\mathcal{D}$ is a normal ultrafilter iff whenever $g\in\kappa^\kappa$ such that $g&amp;lt;_\mathcal{D} Id$, we have some $\alpha&amp;lt;\kappa$ such that $\{\gamma&amp;lt;\kappa | g(\gamma) = \alpha\}\in\mathcal{D}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\kappa$ a measurable cardinal and some $M$ which is a transitive collapse of an ultrapower by some $\kappa$-complete ultrafilter, $\mathcal{U}$, on $\kappa$, and $j_\mathcal{U}$ the canonical embedding of $V$ into $M$, we can reconstruct a normal ultrafilter by: $\mathcal{D} = \{X\subseteq\kappa | \kappa\in j_\mathcal{U}(X)\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given a set $A\in\mathcal{U}$ then $\kappa\in j_\mathcal{U}(A)$ iff $[Id]_\mathcal{U} E [C_A]_\mathcal{U}$ iff $\{\alpha&amp;lt;\kappa | \alpha\in A\}\in\mathcal{U}$. (where $E$ is the relation on the ultraproduct, and $C_x$ is the constant function giving $x$ for every $\alpha$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, clearly if $A\in\mathcal{U}$ then $\{\alpha&amp;lt;\kappa | \alpha\in A\}\in\mathcal{U}$ and therefore $\kappa\in j_\mathcal{U}(A)$ and therefore $A\in\mathcal{D}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that means the ultraproduct was given by a normal ultrafilter, even though we took some general ultrafilter, and I was told by my prof. that not every ultrafilter can be normalized through a permutation on $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where and what am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-24T01:19:21.650" Title="Normal ultrafilters on measurable cardinals" Tags="&lt;set-theory&gt;&lt;large-cardinals&gt;" AnswerCount="1" />
  <row Id="15393" PostTypeId="2" ParentId="15391" CreationDate="2010-12-24T01:19:21.650" Score="2" ViewCount="0" Body="&lt;p&gt;Hi Asaf, the problem with your argument is when you say that $\kappa\in j_{\mathcal U}(A)$ iff the class of the identity is a &quot;member&quot; of the class of the constantly-equal-to-$A$ function. This assumes that the identity represents $\kappa$, which is equivalent to the normality of ${\mathcal U}$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-24T01:19:21.650" CommentCount="2" />
  <row Id="15395" PostTypeId="2" ParentId="15251" CreationDate="2010-12-24T01:50:26.160" Score="3" ViewCount="0" Body="&lt;p&gt;I can prove it for the case when $Z = |X-Y|$ takes only integer values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q_i = P(Z=i)$ for $i=0,1,\dots$. Then, we need to show that $\frac{q_0+q_1}{q_0+q_1+q_2} \geq \frac{1}{3}$. This follows from the observation that $2q_0 \geq q_i$ for all $i$. This follows from Cauchy Schwarz inequality. Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}&#xD;&#xA;3(q_0+q_1) &amp;amp;\geq (q_0+q_1+q_2) \\&#xD;&#xA;2(q_0+q_1) &amp;amp;\geq q_2 \\&#xD;&#xA;\end{aligned}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is true since $2q_0 \geq q_2$. Thanks to iMath for this last observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $Z$ being real, I tried mimicking the proof above but the details don't quite work out. In this case, Cauchy-Schwarz still implies that $f_Z(z) \leq 2f_Z(0)$ for all $z$. However, the proof seems to need one more estimation along the lines of $\int_0^1 f_Z(z) dz \geq  f_Z(0)$.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="618" LastEditorDisplayName="" LastEditDate="2010-12-24T18:42:59.483" LastActivityDate="2010-12-24T18:42:59.483" CommentCount="3" />
  <row Id="15396" PostTypeId="1" CreationDate="2010-12-24T01:53:02.490" Score="1" ViewCount="41" Body="&lt;p&gt;I'm looking for an algorithm for the transitive reduction of a digraph (a DAG in fact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Transitive_reduction#Graph_algorithms_for_transitive_reduction&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; displays a formula R&lt;sup&gt;-&lt;/sup&gt; = R - (R &amp;#215; R&lt;sup&gt;+&lt;/sup&gt;) &amp;mdash; where R&lt;sup&gt;-&lt;/sup&gt; is the transitive reduction of R and R&lt;sup&gt;+&lt;/sup&gt; is the transitive closure of R &amp;mdash; but that Cartesian product doesn't form an edge (unless I'm missing something), so how can I take away from R?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say 3 depends on 2 and 1, and 2 depends on 1.  The edges of this graph G are (2, 1), (3, 1), and (3, 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm expecting the reduction to be G' = {(2, 1), (3, 2)}.&#xA;The transitive closure of G is G, right?&lt;/p&gt;&#xA;" OwnerUserId="623" LastActivityDate="2010-12-24T01:53:02.490" Title="Algorithm for the transitive reduction of a DAG" Tags="&lt;algorithms&gt;&lt;graph-theory&gt;" CommentCount="2" />
  <row Id="15399" PostTypeId="2" ParentId="15335" CreationDate="2010-12-24T03:46:38.150" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the &lt;em&gt;Half-angle&lt;/em&gt; formula. $$\cos{2x} = 2\cos^{2}{x}-1$$ and so you have $$1+\cos{x}= 1 + 2\cos^{2}\frac{x}{2} -1 =2\cos^{2}\frac{x}{2}$$ So the required integral is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\int (1+\cos{x})^{3} \rm{dx} &amp;amp;= \int 8 \cdot \cos^{6}\Bigl(\frac{x}{2}\Bigr) \ \rm{dx} \\ &amp;amp;= 8 \int \cos^{6}\Bigl(\frac{x}{2}\Bigr) \ \rm{dx}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2010-12-24T03:46:38.150" />
  <row Id="15400" PostTypeId="1" AcceptedAnswerId="15497" CreationDate="2010-12-24T03:53:27.317" Score="4" ViewCount="140" Body="&lt;p&gt;While combing around my notes looking for other possible examples for &lt;a href=&quot;http://math.stackexchange.com/questions/15260&quot;&gt;this question&lt;/a&gt;, I chanced upon another one of my unsolved problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cycloidal curves are curves generated by a circle rolling upon a plane or space curve. It's not too hard to derive the required parametric equations if the fixed curve is a plane curve, but I've had some trouble deriving the corresponding expression for space curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, here is the particular problem I was concerned with: consider a (cylindrical) helix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=a\cos\;t\\y&amp;amp;=a\sin\;t\\z&amp;amp;=ct\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and imagine a circle of radius $r$ whose plane is always perpendicular to the x-y plane rolling on the helix, starting at the point $(a,0,0)$ ($t=0$). Imagine a point in the plane of the circle at a distance $hr$ from the center. What are the parametric equations for the locus of the point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two obvious pieces of information I have are that &lt;strike&gt;the center of the circle also traces a helix, whose parametric equation differs from the original by a vertical shift of $r$&lt;/strike&gt; (per Tony, that was an erroneous assumption), and that the expression for the arclength of the helix, $s=\sqrt{a^2+c^2}t$, should figure into the final parametric equations. Otherwise, I'm not sure how to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one derive parametric equations for the &quot;helical cycloid&quot;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The physical model I had in mind was a screw ramp winding around a cylinder. Supposing that there was a car that needed to go to the top of the cylinder by driving on the ramp, and supposing that a spot is placed on one of the car's wheels, what are the equations for the locus of the spot?&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T09:33:22.773" LastActivityDate="2010-12-25T09:33:22.773" Title="A helical cycloid?" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="15401" PostTypeId="1" AcceptedAnswerId="15403" CreationDate="2010-12-24T03:56:38.483" Score="0" ViewCount="46" Body="&lt;p&gt;I'm developing an open-ended strategy game.  I am using the following formula to calculate damage (In PHP 'cause I don't know MathJaX!):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$rand = rand($a, $b) + $c;&#xA;$damage = $rand * sqrt(($d / 20) * $c));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a&lt;/code&gt;, &lt;code&gt;b&lt;/code&gt;, &lt;code&gt;c&lt;/code&gt;, and &lt;code&gt;d&lt;/code&gt; are all values that can be modified by the user over the course of play, either by buying a better item (&lt;code&gt;a&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt;), investing in the item (&lt;code&gt;c&lt;/code&gt;), or investing in their character &lt;code&gt;d&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do now is add a bit of randomness to the outcome of the equation.  Because the game is open ended:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a static value would become unnoticeable/negligible over time.&lt;/li&gt;&#xA;&lt;li&gt;a percentage based value would allow for too much noise over time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, I want to add a random value that is small at first, and grows with increased input, but has diminishing returns.  I'm sure I need some kind of logarithmic formula, but I'm not sure how to go about it!&lt;/p&gt;&#xA;" OwnerUserId="5092" LastActivityDate="2010-12-24T04:26:21.013" Title="How can I add random &quot;fuzz&quot; to this formula, so that the randomness range diminishes as values increase?" Tags="&lt;random-variable&gt;&lt;logarithms&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15402" PostTypeId="2" ParentId="15278" CreationDate="2010-12-24T04:25:08.033" Score="5" ViewCount="0" Body="&lt;p&gt;Dear Albert,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me suggest that you look at the book &quot;An introduction to topology and modern analysis&quot;, by Simmons.  It covers concepts of point set topology that you presumably already know, but also gives a fairly concise, but quite readable, introduction to abstract algebra.  It then brings these together in its final sections into a discussion of Banach algebras and related topics, culminating in a proof of the Gelfand--Naimark theorem.  [See below for a brief remark about this theorem, and the other theorems mentioned in the subsequent paragraphs.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you say that you are interested in &lt;em&gt;abstract&lt;/em&gt; harmonic analysis, the Gelfand--Naimark theorem is a good place to start.  (For example, it is not so far to go from there to the abstact form of Wiener's Tauberian theorem.)&#xA;Note: I am interpreting &lt;em&gt;abstract harmonic analysis&lt;/em&gt; to mean something like harmonic analysis on locally compact abelian groups (and related topics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simmons also has exercises, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was studying this stuff, the next place I went to after Simmons was Naimark's tome &lt;em&gt;Normed rings&lt;/em&gt;.  (There are various editions, and some of the later ones might be called &lt;em&gt;Normed algebras&lt;/em&gt; instead, if I'm not misremembering.  They are translated from Russian, so the slighly unusual, and changing, name may be an artefact of this; I'm not sure.  In any case, they are basically about the theory of Banach algebras and its applications to abstract harmonic analysis.)&#xA;This is a place where one can read about various group rings of topological groups, Haar measure, the general form of Wiener's Tauberian theorem, and other concepts of abstract harmonic analysis.  It is essentially too long to read from&#xA;start to finish, but in my experience one can dip into it in bits and pieces, and having a firm understanding of the material from Simmons helps a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naimark's book is a monograph, not a textbook as such, and although it has many historical comments and illustrative examples (although the examples are often at a theoretically fairly high level), I don't remember it as having exercises.  But in any case, I am not suggesting it as a first point of call, but as somewhere to go after you have some basics under your belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a book by Loomis, &lt;em&gt;An introduction to abstract harmonic analysis&lt;/em&gt;,&#xA;which also treats Haar measure, various group rings, and so on.  If I remember correctly it is  less condensed than Naimark and also less comprehensive.  My memory is that I preferred Naimark, but probably for idiosyncratic reasons.  I don't remember whether Loomis's book has exercises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the books I'm mentioning are probably out of print, so I'm also assuming that you have access to a university library or something similar.  (Any decent such library should have them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, some fundamental results that I would recommend aiming for, which combine algebra&#xA;and analysis nicely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Gelfand's generalization of Wiener's theorem (that if $f$ is a nowhere zero continuous periodic function whose Fourier series is absolutely convergent, than&#xA;the Fourier series of $1/f$ is also absolutely convergent).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Gelfand--Naimark theorem (identifying certain commutative Banach algebras with extra&#xA;structure as being algebras of continuous functions on compact topological space; it is a beautiful generalization of the classical spectral theorems for matrices).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The generalization of Wiener's Tauberian theorem to arbitrary commutative locally compact groups.  (Wiener's original theorem says that if $f$ is an $L^1$-function on the real line whose Fourier transform is nowhere zero, then&#xA;the translates of $f$ span a dense subspace of $L^1$.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More abstract, but basic to the previous example and lots of other things, is&#xA;the existence of Haar measure for any locally compact group.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As already mentioned,&#xA;the first two results are in Simmons, and are easier, but already involve a very nice interplay between analysis and algebra.  (The general framework is that of Banach algebras, which combine the analysis of Banach spaces with the&#xA;algebra of rings, ideals, and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second two results are in Naimark, and Haar measure is also in Loomis (and many other places) (and the general form of Wiener may be in Loomis too; I forget now).&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-24T14:02:12.033" LastActivityDate="2010-12-24T14:02:12.033" CommentCount="1" />
  <row Id="15403" PostTypeId="2" ParentId="15401" CreationDate="2010-12-24T04:26:21.013" Score="2" ViewCount="0" Body="&lt;p&gt;Let me expand upon my comment:&#xA;you can pick any positive increasing concave function $f$(damage) and produce a random disturbance in the range [-$f$(damage),+$f$(damage)]. That will give you a disturbance that increases with the damage, but it increases more and more slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of concave functions: log (to any base), n-th root for any integer n, arctan,... and of course you can scale them in any way you like. Moreover, any composition of concave functions is concave, so you can do e.g. s.th. like $log(\sqrt{x})$. Note that these functions behave very differently, as $x$ gets large: log falls below any $n$-root eventually, but is unbounded. arctan on the other hand asymptotically approaches $2\pi$, so for very large damage it will behave almost like a constant disturbance. You can influence when that happens by dilating, scaling and translating along the $x$-axis.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-24T04:26:21.013" />
  <row Id="15404" PostTypeId="2" ParentId="2949" CreationDate="2010-12-24T04:51:11.767" Score="1" ViewCount="0" Body="&lt;p&gt;The use of compactness to show existence of solutions to differential equations. It was not surprising as in unexpected, but in the sense that it opened so much possibility almost unreal. It felt like I was given an amazing super toy that can never be destructed.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T04:51:11.767" CommunityOwnedDate="2010-12-24T04:51:11.767" />
  <row Id="15405" PostTypeId="2" ParentId="15360" CreationDate="2010-12-24T04:55:42.630" Score="5" ViewCount="0" Body="&lt;p&gt;It isn't true in general, but the direction you used in your physics.SE answer is.  That is, if the sum converges, then the integral does too.  The decreasing hypothesis implies that the maximum value of $f$ on the cube $[n_1,n_1+1]\times\cdots\times[n_N,n_N+1]$ is $f(n_1,\ldots,n_N)$, so that the integral over that cube is less than or equal to $f(n_1,\ldots,n_N)$.  Adding up the integrals over all such cubes yields the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the other direction is that the function may drop off to zero in some directions rapidly enough to make the integral converge, while staying too big in another direction for the sum to converge.  For example, $N=2$, $f(x,y)=1$ if $x=1$, $f(x,y)=0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an additional &quot;shift&quot; you can go from integral convergence to sum convergence.  The decreasing hypothesis also implies that the minimum value of $f$ on the cube $[n_1,n_1+1]\times\cdots\times[n_N,n_N+1]$ is $f(n_1+1,\ldots,n_N+1)$, so that the integral over that cube is greater than or equal to $f(n_1+1,\ldots,n_N+1)$.  By adding up the integrals over all such cubes, this implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sum_{n_1=2}^\infty  \cdots \sum_{n_N=2}^\infty f(n_1,\ldots,n_N)&#xD;&#xA;\leq&#xD;&#xA;\int_1^\infty \cdots \int_1^\infty f(x_1,\ldots,x_N) dx_1 \dotsm dx_N.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-26T23:51:13.707" LastActivityDate="2010-12-26T23:51:13.707" />
  <row Id="15406" PostTypeId="1" CreationDate="2010-12-24T05:10:48.560" Score="1" ViewCount="89" Body="&lt;p&gt;How do we numerically integrate a rapidly decaying exponential function? A simple Gauss quadrature which is based on approximating the function by polynomial, I think will not work, since rapidly decaying exponential function will require more terms in the series. Is there a series approximation for the function?&lt;/p&gt;&#xA;" OwnerUserId="3892" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T00:55:32.450" LastActivityDate="2010-12-25T00:55:32.450" Title="Numerical integration - Gauss quadrature rule" Tags="&lt;integral&gt;&lt;numerical-methods&gt;&lt;approximation&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="15407" PostTypeId="2" ParentId="3273" CreationDate="2010-12-24T05:29:11.530" Score="2" ViewCount="0" Body="&lt;p&gt;Perhaps one should mention &lt;a href=&quot;http://en.wikipedia.org/wiki/Runge%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Runge's theorem&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Mergelyan%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Mergelyan's theorem&lt;/a&gt; which deal with approximation by rational functions and polynomials, respectively.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T05:29:11.530" />
  <row Id="15408" PostTypeId="2" ParentId="15400" CreationDate="2010-12-24T05:43:04.727" Score="2" ViewCount="0" Body="&lt;p&gt;I'm adjusting my answer, which as TonyK points out was based on an incorrect assumption.  Hopefully it's correct now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of the circle is not directly above the point of contact with the helix as my incorrect answer had asserted.  To see where it is, first picture the starting point at $(a,0,0)$, where the circle is parallel to the $y$-$z$ plane.  The tangent vector to the ellipse is in the direction of $(0,a,c)$, so the radius is perpendicular to this in the direction of $(0,-c,a)$.  So you have to add $\frac{r}{\sqrt{a^2+c^2}}(0,-c,a)$ to get to the center.  As the circle rolls along, the vector you add to get to the center will rotate around the $z$ axis to $\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$.  So the center of the circle is parametrized by $(a\cos(t),a\sin(t),ct)+\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arclength formula allows you to write down the rotation angle of the point about the center of the circle in terms of $t$ as $\frac{\sqrt{a^2+c^2}}{r}t$.  Let $b=\frac{\sqrt{a^2+c^2}}{r}$ from now on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we pretend for a minute that the point was just going to rotate about the center of the circle at this rate while the circle remains fixed, and simplify the problem for now by assuming that the point starts above the center of the circle, then the vector we would add to the center of the circle to parametrize the point would be $(0,hr\sin(bt),hr\cos(bt))$.  When looking toward the $y$-$z$ plane from the positive $x$ axis, this motion would appear clockwise, as it should because the circle will be starting off to the right from this perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the vector we actually add to the center will also be rotated by $t$ about the $z$-axis, so putting it all together (with minimal simplification) we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align*}&#xD;&#xA;x&amp;amp;=a\cos(t)+\frac{c}{b}\sin(t)-hr\sin(t)\sin(bt)\\&#xD;&#xA;y&amp;amp;=a\sin(t)-\frac{c}{b}\cos(t)+hr\cos(t)\sin(bt)\\&#xD;&#xA;z&amp;amp;=ct+\frac{a}{b}+hr\cos(bt).&#xD;&#xA;\end{align*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to adjust the starting point, you can change $bt$ to $b(t-t_0)$ for suitable $t_0$.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-24T08:51:02.613" LastActivityDate="2010-12-24T08:51:02.613" CommentCount="2" />
  <row Id="15409" PostTypeId="2" ParentId="9085" CreationDate="2010-12-24T05:45:48.790" Score="1" ViewCount="0" Body="&lt;p&gt;Maybe &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field&quot; rel=&quot;nofollow&quot;&gt;Lagrangian coordinates&lt;/a&gt; is what you are after.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T05:45:48.790" />
  <row Id="15411" PostTypeId="2" ParentId="14789" CreationDate="2010-12-24T06:41:12.523" Score="0" ViewCount="0" Body="&lt;p&gt;Actually, a straightforward componentwise analysis should do the job. First prove the quadratic rate for the fixed point iteration $x\leftarrow\phi(x)$, where $\phi:\mathbb{R}^n\to\mathbb{R}^n$ and the jacobian of $\phi$ is zero at the solution $\alpha=\phi(\alpha)$. Simply do what you do in 1 dimension, e.g., use Taylor's theorem with second order remainder. Then for the Newton method one has to substitute $\phi(x)=x-J(x)f(x)$ where $J$ is the jacobian of $f$, and show that the jacobian of such $\phi$ is zero at a solution $\alpha$ of $f(\alpha)=0$ if $J(\alpha)\neq0$.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastEditorUserId="2473" LastEditorDisplayName="" LastEditDate="2010-12-24T22:36:26.660" LastActivityDate="2010-12-24T22:36:26.660" />
  <row Id="15412" PostTypeId="2" ParentId="7554" CreationDate="2010-12-24T07:10:56.393" Score="0" ViewCount="0" Body="&lt;p&gt;In many cases you have a good idea of the limiting behavior of the function. For instance, you may have a conjecture and you want to test it. Then you can do some transformation on your function to a convenient form. For example, I have seen people plot $f(1/x)$ with $x\to0$ and guess the shape of $g(x)=f(1/x)$ from there. You can repeat this process to get more and more accurate pictures.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T07:10:56.393" />
  <row Id="15413" PostTypeId="2" ParentId="15406" CreationDate="2010-12-24T07:21:47.483" Score="1" ViewCount="0" Body="&lt;p&gt;Variable substitution to make $k=1$, and use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%E2%80%93Laguerre_quadrature&quot; rel=&quot;nofollow&quot;&gt;Gauss-Laguerre&lt;/a&gt;. For more advanced method, the &lt;a href=&quot;http://math.stackexchange.com/questions/2949/which-one-result-in-maths-has-surprised-you-the-most/2977#2977&quot;&gt;double exponential transformation&lt;/a&gt; maybe relevant.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-24T07:21:47.483" CommentCount="2" />
  <row Id="15414" PostTypeId="1" CreationDate="2010-12-24T07:28:41.770" Score="3" ViewCount="84" Body="&lt;p&gt;How do I show that $\text{Hom}(\mathbb{Z},G)$ and $G$ are isomorphic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method suggested by the book is to define a map $k$: $\text(\mathbb{Z},G) \to G$ by $f \to f(1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck on how exactly to show that it is an isomorphism. Could anyone shed some light into this? Is $f(1)$ an element of G?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For those who wanted to know, this is not homework)&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-24T07:44:01.500" LastActivityDate="2010-12-24T07:44:01.500" Title="How to show that for any abelian group $G$, $\text{Hom}(\mathbb{Z},G)$ is isomorphic to $G$." Tags="&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15415" PostTypeId="1" AcceptedAnswerId="15442" CreationDate="2010-12-24T07:36:24.667" Score="0" ViewCount="148" Body="&lt;p&gt;Prove that there is a largest integer $n$ such that $n \le x$ for any fixed rational $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about any fixed real number $x$?&lt;/p&gt;&#xA;" OwnerUserId="5035" LastEditorUserId="5035" LastEditorDisplayName="" LastEditDate="2010-12-24T13:22:07.747" LastActivityDate="2010-12-24T17:56:49.383" Title="Prove the existence of a largest integer less than or equal to a rational number" Tags="&lt;real-analysis&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="15416" PostTypeId="2" ParentId="15414" CreationDate="2010-12-24T07:37:49.923" Score="5" ViewCount="0" Body="&lt;p&gt;Any homomorphism $f$ from $\mathbb{Z}$ to $G$ is determined by its value at 1: indeed $f(n) = f(1)^n$ for any $n\in \mathbb{Z}$, since $f$ is a group homomorphism (I am using multiplicative notation for the group operation on $G$). So it is natural to consider the map suggested by the book: given $f\in \text{Hom}(\mathbb{Z},G)$, evaluate it at 1: $k:f\mapsto f(1)$. It sends $f$ to an element of $G$, namely $f(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check that this is a group homomorphism, you need to understand how $\text{Hom}(\mathbb{Z},G)$ is a group in the first place. That's very simple: the group operation is pointwise multiplication: the product of two homs $f$ and $g$ is given by $fg(n) = f(n)g(n)$ (to explain what $fg$ is you need to evaluate it on an arbitrary input, since you are describing a homomorphism). So now, you should have no difficulties verifying that $k$ respects the group structure.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-24T07:37:49.923" CommentCount="6" />
  <row Id="15417" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T07:49:12.760" Score="2" ViewCount="0" Body="&lt;p&gt;(Sorry to post this as an answer, but I can't comment yet). Do you mean that given a fixed rational number $x$, the set $\lbrace n: n\leq x, n \in \mathbb Z\rbrace$ has a maximum? (Hint: Use the well-ordering principle)&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-24T07:49:12.760" CommentCount="3" />
  <row Id="15418" PostTypeId="1" AcceptedAnswerId="15419" CreationDate="2010-12-24T08:04:50.660" Score="2" ViewCount="201" Body="&lt;p&gt;If we define $x$ = displacement, $v$ = velocity and $a$ = acceleration then I am used to the ideas that $a= \frac{dv}{dt} = \frac{d^2x}{dt^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I also understand $a=v \frac{dv}{dx}$. Can someone explain to me why this is?&lt;/p&gt;&#xA;" OwnerUserId="1699" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-28T15:11:15.143" LastActivityDate="2010-12-28T15:11:15.143" Title="Why does acceleration = $v\frac{dv}{dx}$" Tags="&lt;differential-equations&gt;&lt;physics&gt;&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15419" PostTypeId="2" ParentId="15418" CreationDate="2010-12-24T08:13:17.967" Score="8" ViewCount="0" Body="&lt;p&gt;Notice: $\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}$ by Chain rule. &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-24T08:13:17.967" />
  <row Id="15420" PostTypeId="1" CreationDate="2010-12-24T08:35:21.303" Score="0" ViewCount="31" Body="&lt;p&gt;Given set of points in 3D ( X = (x1, x2, x3), Y = (y1, y2, y3) ), how can I fit transformation from X to Y?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know this is called projective transformation.&#xA;Here is example of X and Y.&#xA;Blue and red lines in X are parallel, but they are not parallel in Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/YGEfS.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/LzjsM.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3532" LastActivityDate="2010-12-24T08:35:21.303" Title="Fitting projective transformation" Tags="&lt;linear-algebra&gt;&lt;transformation&gt;" CommentCount="3" />
  <row Id="15421" PostTypeId="1" CreationDate="2010-12-24T08:52:57.303" Score="0" ViewCount="59" Body="&lt;p&gt;Hello! I have this function, $e^{-x}$ bounded between 0 and 1500 and I have an approximation by Taylor Series of the same function bounded between 0 and 0.5. I would like to express my function $e^{-x}$ bounded between 0 and 15oo in terms of the last approximation I described, what is bounded between 0 and 0.5.&#xA;Does anyone know how I can do it?&#xA;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: This question is related with &lt;a href=&quot;http://math.stackexchange.com/questions/14206/development-of-a-specific-hardware-architecture-for-a-particular-algorithm&quot;&gt;Development of a hardware arquitecture for a particular algorithm.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4255" LastActivityDate="2010-12-24T16:14:42.817" Title="Taylor Series. Reusing an approximation of a function" Tags="&lt;functions&gt;&lt;approximation&gt;&lt;taylor-expansion&gt;&lt;mathematical-modeling&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15422" PostTypeId="2" ParentId="15400" CreationDate="2010-12-24T08:56:30.290" Score="3" ViewCount="0" Body="&lt;p&gt;The tangent to the helix at $P = (a \cos t, a \sin t, ct)$ is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = (-a \sin t, a \cos t, c)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = (\cos t, \sin t, 0)$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is horizontal, and perpendicular to $T$. The radius vector $R$ from $P$ to the centre $C$ of the circle is perpendicular to both of these, so it is proportional to  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H \wedge T = (c \sin t, -c \cos t, a)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has length $\sqrt{a^2 + c^2}$. So the centre of the circle is at $P + h H \wedge T$, with $h = r/\sqrt{a^2 + c^2}$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-24T09:23:04.850" LastActivityDate="2010-12-24T09:23:04.850" CommentCount="2" />
  <row Id="15423" PostTypeId="1" AcceptedAnswerId="15448" CreationDate="2010-12-24T08:57:57.547" Score="5" ViewCount="121" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Say we have $N$ [$3 \le N \le 100,000$] spheres indexed as $1,2,3,\cdots N$,all of them have identical weight apart from one.We have to determine which sphere it is (index) by using only the pair of scales.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could solve this problem by weighing repeatedly,but I am obviously interested weighing as minimum as possible so my question is what could be the optimal algorithm for this task? &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-24T09:12:51.533" LastActivityDate="2010-12-27T07:40:32.587" Title="Optimal algorithm for finding the odd spheres" Tags="&lt;algorithms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15424" PostTypeId="2" ParentId="15350" CreationDate="2010-12-24T09:07:14.513" Score="7" ViewCount="0" Body="&lt;p&gt;In the comments section to Willie Wong's answer, the following Dirichlet series came up: the Riemann $\zeta$-function, Dirichlet $L$-functions, and Ramanujans series $\sum_{n \geq 1}\tau(n) n^{-s}$, where $\tau(n)$ is the coefficient of $q^n$ in $\Delta(q) = q\prod_{n=1}^{\infty} (1-q^n)^{24}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First note that the $\zeta$-function is a special case of a Dirichlet $L$-function (it is the $L$-function of the trivial character).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is it that Dirichlet $L$-functions and Ramanujan's series have in common? Well, they are all automorphic $L$-functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An automorphic form (for the group $\mathrm{GL}_n$ over $\mathbb Q$; there are generalizations where $\mathbb Q$ is replaced by an arbitrary number field $F$ and  $\mathrm{GL}_n$ is replaced by an arbitrary reductive group, but to simplify the explanations, I will focus just on the simplest level of generality here) is a function on the product $\mathrm{GL}_n(\mathbb R)\times \mathrm{GL}_n(\mathbb Z/N\mathbb Z)$ for some integer $N \geq 1$ which&#xA;is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;invariant under the natural (diagonal) action of $\mathrm{GL}_n(\mathbb Z)$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;grows moderately at infinity with respect to the $\mathrm{GL}_n(\mathbb R)$-coordinates;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;satisfies a suitable differential equation in the $\mathrm{GL}_n(\mathbb R)$-coordinates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rather than explaining the generalities in more detail (they can be found in many places), I think it's better to illustrate them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. Dirichlet characters arise in the case $n = 1$: they are defined as functions&#xA;on $(\mathbb Z/N\mathbb Z)^{\times} =: \mathrm{GL}_1(\mathbb Z/N\mathbb Z)$,&#xA;and so we can make them into functions on $\mathrm{GL}_1(\mathbb R)\times&#xD;&#xA;\mathrm{GL}_1(\mathbb Z/N\mathbb Z)$ by defining them to be trivial on the&#xA;$\mathbb R^{\times}$-coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. If $f(\tau)$ is a holomorphic modular form of weight $k$ and level one (where $\tau$ is an upper half-plane variable as usual), we can make $f$ into a function on&#xA;$\mathrm{GL}_2(\mathbb R)$ by first identifying this matrix group with the collection of bases of $\mathbb R^2$, then identifying $\mathbb R^2$ with $\mathbb C$,&#xA;and then defining, for any $\mathbb R$-basis $\omega_1,\omega_2$ of $\mathbb C$,&#xA;$f(\omega_1,\omega_2) := \omega_2^{-k} f(\omega_1/\omega_2)$.  (This presumes&#xA;that $\omega_1/\omega_2$ is in the upper half-plane rather than the lower,&#xA;for simplicity.)  Thus we get a function of the required kind (with $N = 1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual modularity condition becomes invariance under $\mathbb GL_2(\mathbb Z)$.  The moderate growth condition becomes the condition that the Fourier expansion of $f$ involves only non-negative powers of $e^{2 \pi i \tau}$.  The differential equation is the Cauchy--Riemann equation expressing holomorphy of $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher level modular forms will involve values of $N$ that are $&amp;gt; 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.  Maass forms are similar to the preceding example, except that now the&#xA;differential equation expresses that a Maass form is an eigenvector of the Laplacian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any fixed $n$ and fixed $N$, we have Hecke operators acting on the space&#xA;of automorphic forms, labelled by primes $p$ not dividing $N$, and so we can&#xA;consider Hecke eigenforms.  In the case of Dirichlet characters, the fact that they are &lt;em&gt;characters&lt;/em&gt; of $(\mathbb Z/N\mathbb Z)^{\times}$ (rather than just arbitrary functions) can be reinterpreted as saying that they are Hecke eigenforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, Ramanujan's $\Delta$ is well-known to be a Hecke eigenform of weight $12$ and level $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an automorphic Hecke eigenform we can use the Hecke eigenvalues to make&#xA;an Euler product Dirichlet series, which will give Dirichlet $L$-functions&#xA;for Dirichlet characters, and Ramanujan's Dirichlet series for $\Delta$.&#xA;(In the Dirichlet character case, if a prime $p$ divides the conductor $N$, we just have a trivial factor in the Euler product for that prime; when $n &amp;gt; 1$,&#xA;and $N &amp;gt; 1$, it is a bit more of a battle to figure out what Euler factors&#xA;to put in at the primes dividing $p$, but it can be done.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, it is better to restrict to &lt;em&gt;cuspidal&lt;/em&gt; automorphic Hecke eigenforms.&#xA;Cuspidal is a vacuous condition when $n = 1$ (i.e. in that case we agree to call&#xA;everything cuspidal), and when $n &amp;gt; 1$ we replace &quot;moderate growth at infinity&quot; by &quot;rapid decay at infinity&quot;, suitably interpreted.  I'll assume that all my eigenforms are cuspidal&#xA;form now on.  (E.g. $\Delta$ is cuspidal.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way we get a natural class of $L$-functions which have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;meromorphic continuation to the whole complex plane, which is in fact&#xA;holomorphic with the sole exception of Riemann's $\zeta$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Functional equation with completely understood $\Gamma$-factors.  E.g. for a&#xA;weight $k$ modular form of level one, if the $p$the Hecke eigenvalue is $a_p$,&#xA;then the $L$-function is $\prod_p (1 - a_p p^{-s} + p^{k - 1 - 2s})^{-1},$ and the functional equation relates $s$ and $k - s$.  For $\Delta,$ I've already noted that $k = 12$.  (In general the functional equation relates the $L$-series of an automorphic eigenform with $L$-series of its &quot;complex conjugate&quot; suitably understood, just as in the case of Dirichlet characters that are not necessarily real values.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Conjecturally&lt;/em&gt;, they should all satisfy the analogue of RH, i.e. all non-trivial zeroes should line down the critical line, in the centre of the critical strip.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note incidentally that it is easy to change the apparent form of the functional equation.  E.g. if we make a change of variable $s \mapsto s + 11/2$ in Ramanujan's series, then the functional equation will become $s \mapsto 1 -s$&#xA;rather than $s \mapsto 12 -s $, and the critical line will be $\Re s = 1/2$, just as in the $\zeta$-function case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All cuspidal automorphic $L$-functions can be renormalized in a similar way, so that the symmetry of the functional equation is $s \mapsto 1 -s$.  This is called &lt;em&gt;unitary&lt;/em&gt; normalization, and is common in the automorphic forms literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to rescaling, there are only countably many automorphic eigenforms altogether (just because if we fix the level $N$ and (appropriately generalized version of) the weight the space of automorphic forms is finite dimensional) and so altogether we are talking about a very special class of just countably many Dirichlet series, but these seem to be the ones that naturally generalize $\zeta(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this general point of view is due to Langlands, and forms a part of the general Langlands program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point of view was given by Selberg, which focuses more on capturing the analytic properties necessary for getting good properties of a Dirichlet series, rather than beginning from a conceptual construction (as in the automorphic point of view).  Namely, he introduced the &lt;a href=&quot;http://en.wikipedia.org/wiki/Selberg_class&quot; rel=&quot;nofollow&quot;&gt;Selberg class&lt;/a&gt; of Dirichlet series.  Note that part of his axioms are an Euler product, analytic continuation, and functional equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sense is, though, that people expect the Selberg class of Dirichlet series to more-or-less coincide with the class of automorphic $L$-functions, so I think it is just two points of view on the same question: Langlands is showing how to &lt;em&gt;construct&lt;/em&gt; &quot;good&quot; Dirichlet series, and Selberg is writing down the &lt;em&gt;properties&lt;/em&gt; a &quot;good&quot; Dirichlet series should satisfy.  It turns out that &quot;good&quot; Dirichlet series are so special, though, that however you try to pick them out, you seem to end up with the same very special collection, namely the automorphic ones.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-24T09:13:10.720" LastActivityDate="2010-12-24T09:13:10.720" CommentCount="2" />
  <row Id="15425" PostTypeId="1" AcceptedAnswerId="15909" CreationDate="2010-12-24T09:08:16.957" Score="5" ViewCount="175" Body="&lt;p&gt;For an abelian variety $A_{/\mathbb{Q}}$, its volume $vol(A(\mathbb{R}))$ appears in the conjectured Birch Swinnerton-Dyer formula for the L-series at 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble in understanding the size of this volume, as a function of the (minimal) equations defining the variety. Hence the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have a hyperelliptic curve given by $$y^2 = x^{2g+1}+a_{2g}x^{2g}+\ldots +a_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for simplicity, assume that $a_i\approx M &amp;gt; 0$ for all $i$. What is a good crude estimate for the volume of the Jacobian?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: for an elliptic curve this is easy, since there is only one differential and hence only one simple integral.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2010-12-30T21:02:23.683" Title="Approximating the volume of the Jacobian of a hyperelliptic curve" Tags="&lt;number-theory&gt;&lt;algebraic-geometry&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15426" PostTypeId="1" AcceptedAnswerId="15428" CreationDate="2010-12-24T09:29:25.003" Score="4" ViewCount="118" Body="&lt;p&gt;Situation A: Once only, I toss 2 identical fair coins and don't look at the outcomes. A truthful observer looks at one of the coins and tells me that at least one of the coins is a head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situation B: Once only, I toss 2 identical fair coins and don't look at the outcomes. A truthful observer looks at both of the coins and tells me that at least one of the coins is a head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In A, what is the probability that there are 2 heads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In B, what is the probability that there are 2 heads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't both probabilities 1/2? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me refine the question. The agreement I have with the observer is this:&#xA;1) I will toss 2 identical coins.&lt;br&gt;&#xA;2) In situation A a third party will cover the coins with a cloth. The observer will look only at the outcome of the coin that comes to rest nearest him and report it to me. What is the probability that the second outcome will be the same as the first?&lt;br&gt;&#xA;3) In situation B the observer will look at both outcomes and report only the state (heads or tails) of the coin that came to rest nearest him. What is the probability that the second outcome will be the same as the first?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing 3) above to:&#xA;3) In situation B the observer will look at both outcomes, choose one of them, and truthfully report, &quot;There is at least one heads&quot;, or &quot;There is at least one tails&quot;, whichever is the case. What is the probability that the second outcome (that of the other coin) will be the same as the first?  &lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="1964" LastEditorDisplayName="" LastEditDate="2010-12-24T19:05:03.980" LastActivityDate="2010-12-25T08:08:41.083" Title="2 slightly different situations in which 2 coins are tossed. Does the knowledge of an observer effect the probabilities of the outcomes?" Tags="&lt;probability&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="15427" PostTypeId="2" ParentId="15426" CreationDate="2010-12-24T09:45:40.627" Score="2" ViewCount="0" Body="&lt;p&gt;A is easy.&lt;br&gt;&#xA;In B, if you know in advance that the observer is just going to let you know whether the number of heads is zero or non-zero, then the probablity is 1/3.&lt;br&gt;&#xA;If you don't know in advance what the observer is going to reveal, then the probability calculation involves the observer's strategy (what information the observer intends to reveal depending on the number of heads). So you can't calculate the probability. This is at the heart of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monty_Hall_problem&quot; rel=&quot;nofollow&quot;&gt;Monty Hall problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-24T09:45:40.627" CommentCount="1" />
  <row Id="15428" PostTypeId="2" ParentId="15426" CreationDate="2010-12-24T09:46:52.427" Score="7" ViewCount="0" Body="&lt;p&gt;No. There are four possibilities with two coins. For the time being identify one coin as X and one coin as Y. You have either &lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 1, Y = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 0, Y = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 1, Y = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 0, Y = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where 1 is heads and 0 is tails). A truthful observer who only looks at coin X and ignores coin Y will say: &quot;there is at least one head&quot; in cases 1 and 3. A truthful observer who only looks at coin Y and ignores coin X will say: &quot;there is at least one head&quot; in cases 1 and 2. In either of those situations, only one of the two cases have the other coin as a head. So the probability of two heads is 1/2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a truthful observer who looks at both coins will say: &quot;there is at least one head&quot; in cases 1, 2, and 3. Out of those only 1 case has two heads. So the probability of that is 1/3. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By choosing only one of the coins to look at, you are looking at the probability &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}\{ X = 1 | \mbox{ Observer chooses }Y\mbox{ and }Y = 1\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus the same with $X$ and $Y$ swapped. Since $X$ and $Y$ are independent, it is equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}\{X = 1\} \cdot \mathbb{P}\{\mbox{ Observer chooses } Y\} + \mathbb{P}\{Y = 1\} \cdot \mathbb{P}\{\mbox{ Observer chooses } X\} = 1/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where the observer sees both coins, you are looking at the conditional probability&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbb{P}\{ Y + X = 2 | Y + X \geq 1 \} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that the sum of the variables $Y$ and $X$ is 2 when we know their sum is at least 1. Clearly the conditioning is not independent of what you are testing! You can establish that the above is 1/3 by counting the total number of ways $A+B$ can be at least 1 as was done above.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-24T09:46:52.427" CommentCount="11" />
  <row Id="15429" PostTypeId="1" CreationDate="2010-12-24T10:28:28.640" Score="1" ViewCount="188" Body="&lt;p&gt;I am wondering whether there are any high resolution photos of Alexander Grothendieck available online. I did a search on Google Images, but most of the results are low quality and/or low resolution. &lt;a href=&quot;http://seven.ihes.fr/IMAGE/grothendieck.jpg&quot; rel=&quot;nofollow&quot;&gt;Here is a nice one&lt;/a&gt;, but I am wondering if there are any other good ones out there. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. Feel free to close this question if you think it's inappropriate... Obviously this is not really a mathematical question... I'm not so familiar with what the standards of appropriateness are here. Also, I guess this question should be community wiki but I don't see any option anywhere to make it such.)&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditorDisplayName="" LastEditDate="2010-12-24T10:34:11.277" LastActivityDate="2010-12-24T10:34:11.277" ClosedDate="2010-12-25T20:22:40.897" Title="High resolution photos of Grothendieck" Tags="&lt;algebraic-geometry&gt;" CommentCount="10" />
  <row Id="15430" PostTypeId="2" ParentId="15423" CreationDate="2010-12-24T10:40:35.967" Score="2" ViewCount="0" Body="&lt;p&gt;If $N = 3^n$ is a power of three, then we can do it $n+1$ weighings:&#xA;First, assume that the odd ball is heavier than the others (this is just to simplify the exposition; we will come back to this). Now label all the balls from $0$ to $N-1$, in ternary notation. For each ternary digit position, weigh all the balls with $0$ in that position against all the balls with $2$ in that position, leaving out the remaining balls (with 1 in that position). The result tells you (assuming that the odd ball is heavier) the ternary digit of the odd ball in that position: $0$ if the $0$'s outweigh the $2$'s; $2$ if the $2$'s outweigh the $0$'s; and $1$ if they are equal.&lt;br&gt;&#xA;So now you have the ternary expansion of the odd ball, assuming that it's heavy. If that assumption is wrong, and the odd ball is light, then the number of the odd ball is obtained by flipping all the $0$'s to $2$ and vice versa. To find out which of these two candidates is the odd ball, just weigh one of them against a ball that is known not to be odd.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't work if $N$ is not a power of three, because in general the number of balls with $0$ in a given position is not the same as the number of balls with $2$ in that position. So we can't weigh them against each other. But we can fix this by the simple device of splitting the balls into two equal heaps, numbering one heap upwards from $0$ and the other heap downwards from $3^n-1$, where $3^n$ is the smallest power of three that is $\ge N$. (If the number of balls is odd, label the left-over ball with $11...11$.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this procedure is optimal when $N$ is a power of three, but otherwise we can sometimes do better; for instance, if $N=4$, we only need two weighings to identify the odd ball (but we won't always know whether it is heavier or lighter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; This solution shows how to solve the problem in $n$ weighings for any $N \le 3^n$ when we know whether the odd ball is heavier or lighter. Now we can put this together with Moron's proposed solution (linked to in Moron's answer) to get a complete solution in $n$ weighings when $N \le (3^n-3)/2$. (This solution also tells us whether the odd ball is heavier or lighter.)&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2010-12-27T07:40:32.587" LastActivityDate="2010-12-27T07:40:32.587" CommentCount="3" />
  <row Id="15431" PostTypeId="1" CreationDate="2010-12-24T10:45:54.977" Score="4" ViewCount="85" Body="&lt;p&gt;Hi all, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a good introductory book  book for Markov chains and Markov processes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. &lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-24T10:57:55.760" LastActivityDate="2010-12-24T13:51:48.720" Title="Good introductory book for Markov processes" Tags="&lt;reference-request&gt;&lt;books&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15432" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T10:53:48.733" Score="1" ViewCount="0" Body="&lt;p&gt;First, as a clarification pointed out by Dactyl, you probably want a proof of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any rational (or real) $x$, there is a largest integer $n$ such that $n \leq x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reversal of words makes the claim false because it says that there is a fixed largest integer $n$ less than or equal to arbitrarily large $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for Dactyl's hint, you have to be careful when $x$ is negative because the well-ordering principle is for Natural numbers.  Specifically, not every nonempty subset of the integers has a smallest element (e.g., the integers themselves have no least element).  However, every nonempty subset of the integers that has a &lt;em&gt;lower bound&lt;/em&gt; has a least element, and you can prove this by induction on the absolute value of an integral lower bound.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, $n$ is typically used to denote a Natural number so maybe you're restricting yourself to nonnegative $x$ anyway.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit (in response to request for different proof):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\{n \in \mathbb{N}| |x| &amp;lt; n\}$.  The &lt;B&gt;Archimedean property&lt;/B&gt; tells us that for any Real number, we can find a Natural number that's larger so this set is non-empty.  Then by the well-ordering of the Natural numbers, there exists a least element in this set, call it $n_0$.  If $x \geq 0$, verify that $n_0 - 1$ is the desired integer, and if $x &amp;lt; 0$, verify that that $1 - n_0$ is the one you want if $x$ is an integer and $-n_0$ otherwise.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="4693" LastEditorDisplayName="" LastEditDate="2010-12-24T17:56:49.383" LastActivityDate="2010-12-24T17:56:49.383" />
  <row Id="15433" PostTypeId="2" ParentId="15431" CreationDate="2010-12-24T10:57:35.010" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.goodreads.com/book/show/6110201-an-introduction-to-markov-chain-analysis&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Markov Chain Analysis&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540234519&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Markov Processes&lt;/a&gt; would be a good start.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-24T10:57:35.010" />
  <row Id="15434" PostTypeId="1" AcceptedAnswerId="15481" CreationDate="2010-12-24T11:22:55.823" Score="2" ViewCount="106" Body="&lt;p&gt;This question is about Schauder bases in Banach spaces.&#xA;Suppose $(x_n)$ is a basic sequence in a Banach space $E$. Prove that $0$ is the only possible weak limit point of $\{x_n \in E \;\vert\; n \in \mathbb{N}\}$.&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-24T12:18:31.177" LastActivityDate="2010-12-25T05:12:45.337" Title="Limit point of basic sequence in Banach space" Tags="&lt;functional-analysis&gt;&lt;banach-spaces&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15435" PostTypeId="1" AcceptedAnswerId="15436" CreationDate="2010-12-24T11:25:07.757" Score="0" ViewCount="24" Body="&lt;p&gt;&lt;img src=&quot;http://i51.tinypic.com/2rfx91v.png&quot; alt=&quot;sketch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry for the draw)&#xA;I try to get the blue angle , i thought it was easy but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the curve is a Bézier curve.&#xA;the black dot shows the tangent of D&#xA;I know position of A B C D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to get the angle by addition of angle a and angle b &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the point F is found by multiplying AB per the time factor (between 0 and 1)&#xA;the point E is found using the y coordinate of B and x coordinate of A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i get a with Al-Kashi method with the triangle FBD.&#xA;i get b with Al-Kashi method too with the triangle EBA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is for rotation of an object who follow the curve&#xA;and i always get weird result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any better way to do it and if not what I'm doing wrong ?&#xA;thanks&lt;/p&gt;&#xA;" OwnerUserId="4279" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T11:31:55.720" LastActivityDate="2010-12-24T11:34:58.120" Title="getting angle between tangent of Bézier curve and an y axis" Tags="&lt;bezier-curve&gt;" AnswerCount="1" />
  <row Id="15436" PostTypeId="2" ParentId="15435" CreationDate="2010-12-24T11:34:58.120" Score="0" ViewCount="0" Body="&lt;p&gt;It's not too hard: if you know the slope $m$ of the tangent (assuming it's not vertical), then the angle it makes with the positive x-axis is $\arctan(m)$; thus, the angle it makes with the positive y-axis should be the complement of that, or $\frac{\pi}{2}-\arctan(m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better yet, since your Bézier curve is parametrically defined anyway, it would be a good idea to use the two-argument arctangent here to reckon the angle; I'll leave it to you to figure out the details.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-24T11:34:58.120" CommentCount="1" />
  <row Id="15437" PostTypeId="2" ParentId="15431" CreationDate="2010-12-24T11:35:15.347" Score="2" ViewCount="0" Body="&lt;p&gt;&quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0126848874&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Stochastic Modeling&lt;/a&gt;&quot; by Karlin and Taylor is a very good introduction to Stochastic processes in general. Bulk of the book is dedicated to Markov Chain. This book is more of applied Markov Chains than Theoretical development of Markov Chains. This book is one of my favorites especially when it comes to applied Stochastics.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-24T11:35:15.347" />
  <row Id="15438" PostTypeId="1" CreationDate="2010-12-24T12:55:58.103" Score="3" ViewCount="92" Body="&lt;p&gt;I want to make an origami of a sphere, so I planned to print some net of a pentakis icosahedron, but I have a image of another sphere with more polygons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.webkinesia.com/online/graphics/images/icos.gif&quot; alt=&quot;icos&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find the net of such model (I know it will be very fun to cut).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know if it has a name ?&lt;/p&gt;&#xA;" OwnerUserId="3709" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-24T13:04:27.980" LastActivityDate="2010-12-24T19:44:42.820" Title="How to make a sphere-ish shape with triangle faces ?" Tags="&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15441" PostTypeId="2" ParentId="15431" CreationDate="2010-12-24T13:44:57.200" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486453057&quot; rel=&quot;nofollow&quot;&gt;Theory of Markov Processes&lt;/a&gt;&lt;/em&gt; by Eugene Dynkin is a paperback published by Dover, so it has the advantage of being inexpensive. The author has made many contributions to the subject. Dynkin's lemma, the Dynkin diagram and the Dynkin system are named after him.&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditorDisplayName="" LastEditDate="2010-12-24T13:51:48.720" LastActivityDate="2010-12-24T13:51:48.720" />
  <row Id="15442" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T15:19:12.577" Score="3" ViewCount="0" Body="&lt;p&gt;The answer to your question depends on the set of facts you take for granted. In terms of elementary number theory I would argue as follows: Let $x=p/q$ with $q&amp;gt;0$. Then the $q$ consecutive numbers $p-q+1$, $p-q+2$, $\ldots$, $p$ form a complete residue system mod$\thinspace q$, so there is exactly one among them, say $r$, that is divisible by $q$. It follows that $n:=r/q$ is the largest integer $≤x$.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-24T15:19:12.577" CommentCount="2" />
  <row Id="15443" PostTypeId="1" CreationDate="2010-12-24T15:24:33.537" Score="10" ViewCount="229" Body="&lt;p&gt;This is a very basic question that I feel that I should know the answer to but haven't been able to think through clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In linear algebra, we learn that the basis of a finite dimensional vector space can be thought of as 'co-ordinates' of that space. And, we model what we intuitively understand as the euclidean plane, using the vector space $\mathbb{R^2}$ equipped with the standard inner product and metric etc. The underlying space is taken to be independent of the choice of basis, that is we understand that properties that are inherent to the space are those that will be invariant under change of basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\mathbb{R^2}$ comes with a canonical basis: this can be understood as saying that given any arbitrary two dimensional vector space and any basis, the vectors $v_i$ of the basis under the co-ordinate map maps to $e_i$. Since, we have also introduced inner products and thus a notion of parallel, the intuitive picture we now have of the co-odrinate grid is a criss-cross of lines. and the 'co-ordinates' are called, imaginatively, 'rectangular co-ordinates'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In school, we also learn about polar 'co-ordinates' of the plane. The associated picture is of concentric circles and rays fanning out of the origin.&#xA;However, these 'co-ordinates' do not fit within the 'Basic Linear Algebra' framework (since among other things, $0$ has no unique representation and the functions that change the variables are not linear). The one way of seeing the transformation of the 'rulings' of the plane is to consider $\mathbb{R^2}$ as $\mathbb{C}$ and the change as the map $\exp: \mathbb C \to \mathbb C$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the framework in which the notion of 'co-ordinates' subsumes both these pictures (likewise for cylindrical, spherical, etc in dimension(?) three). My second and connected question: is there a linear algebraic connection for using two numbers to represent points in polar co-ordinate, i.e. is it because the vector space dimension of $\mathbb{E^2}$ is two. My third question is: am I confusing different concepts of 'dimension' here?      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Thanks for all the replies. They are all great but given my continuing dissatisfaction, either I have not really understood the answers or haven't been able to communicate the question properly (or perhaps and quite likely, I don't know what I want to ask).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when we think of a (topological) manifold, we think of some object that is locally euclidean, i.e., we already have a 'handle' on $\mathbb{E^n}$ and want to use our knowledge of being able to do things, such as calculus, onto the new object . In my question, I am looking at ways we can 'handle' $\mathbb{E^2}$ itself, so invoking manifolds, seems a bit like putting the cart before the horse. I want to say something like this: The point of polar co-ordinates is to represent the plane by looking at it as $S^1\times \mathbb{R}$, where $S^1$ is a basic object like $\mathbb R$.&lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-25T23:16:33.320" LastActivityDate="2010-12-25T23:16:33.320" Title="What are polar co-ordinates?" Tags="&lt;calculus&gt;&lt;linear-algebra&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="15444" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T15:37:51.147" Score="10" ViewCount="0" Body="&lt;p&gt;Great observation; it is quite annoying that nobody bothers to explicitly tell you this.  The relevant concept here is the dimension of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Manifold&quot; rel=&quot;nofollow&quot;&gt;manifold&lt;/a&gt;, not the dimension of a vector space (although one is defined in terms of the other).  What &quot;coordinate&quot; means here is that you are specifying a certain isomorphism (say, differentiable) between $\mathbb{R}^n$ (or some nice submanifold thereof, for example restricting the values of some of the coordinates) and your manifold $M$, and you are using the points of $\mathbb{R}^n$ to parameterize points of $M$.  $M$ itself may be of the form $\mathbb{R}^m$, but the isomorphism need not be linear; it is an isomorphism of (say, smooth) manifolds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of requiring the isomorphism to be differentiable is that a differentiable isomorphism $\phi : M \to N$ between smooth manifolds induces an isomorphism in the ordinary vector space sense between their &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_space&quot; rel=&quot;nofollow&quot;&gt;tangent spaces&lt;/a&gt; at any point.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-24T15:37:51.147" CommentCount="4" />
  <row Id="15445" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T15:54:57.753" Score="4" ViewCount="0" Body="&lt;p&gt;Another way of looking at coordinate systems is they are a way of naming points.  There are many different systems which have been created because they make it easy to work on certain kinds of problem.  Polar coordinates make it easy to express things that happen around a center (compare $r=c$ to $\sqrt{x^2+y^2}=c$), but hard to express things in straight lines unless they pass through the origin and addition of vectors.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimension&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a discussion of various definitions of dimension.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-24T15:54:57.753" />
  <row Id="15446" PostTypeId="2" ParentId="15272" CreationDate="2010-12-24T15:56:42.387" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2005;task=show_msg;msg=2905.0001.0002.0001&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; has a proof that for any ordered space separable implies hereditarily separable.&#xA;And the Sorgenfrey line is a subspace of an ordered space (e.g. the double arrow).&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-24T15:56:42.387" />
  <row Id="15447" PostTypeId="2" ParentId="15421" CreationDate="2010-12-24T16:14:42.817" Score="3" ViewCount="0" Body="&lt;p&gt;Where is the Taylor series for (0,1500) centered?  If at zero, it is the same Taylor series as for (0,0.5).  But you really don't want a Taylor series for (0,1500)-it would take an enormous number of terms for any reasonable accuracy.  &lt;a href=&quot;http://www.nrbook.com/c/&quot; rel=&quot;nofollow&quot;&gt;Numerical Recipes&lt;/a&gt; (chapters 5 and 6) and all other numerical analysis books have suggestions for better models.  &lt;a href=&quot;http://people.math.sfu.ca/~cbm/aands/page_69.htm&quot; rel=&quot;nofollow&quot;&gt;Abramowitz and Stegun&lt;/a&gt; has a polynomial fit over (0,1) that is much more accurate than you need.  Given what you had in the previous question, you can't represent $\exp(-x)$ for $x$ much over 11 because you will underflow.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-24T16:14:42.817" CommentCount="3" />
  <row Id="15448" PostTypeId="2" ParentId="15423" CreationDate="2010-12-24T16:46:28.643" Score="4" ViewCount="0" Body="&lt;p&gt;This looks like a generalization of the classic $12$ ball problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to modify &lt;a href=&quot;http://www.cut-the-knot.org/blue/OddCoinProblems.shtml&quot; rel=&quot;nofollow&quot;&gt;Jack Wert's&lt;/a&gt; wonderful algorithm, (which was designed for the case when $N= \dfrac{3^m - 3}{2}$) to work for any $N$. I believe I had made an (incomplete) &lt;a href=&quot;http://stackoverflow.com/questions/3234790/algorithm-to-find-minimum-number-of-weightings-required-to-find-defective-ball-fr/3234846#3234846&quot;&gt;attempt&lt;/a&gt; when someone asked this on stackoverflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the numbers $\dfrac{3^m - 3}{2}$ are special, in the sense that they are the turning points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the variant of the problem where you are also required to tell if the odd sphere is heavier or lighter, for $\dfrac{3^m -3}{2} \lt N \le \dfrac{3^{m+1}-1}{2}$, the optimal number of weighings can be shown to be $m+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are only required to find the odd sphere and not necessarily figure out if it is heavier or lighter, the turning points are $\dfrac{3^m -3}{2} + 1$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-24T17:21:09.910" LastActivityDate="2010-12-24T17:21:09.910" />
  <row Id="15449" PostTypeId="2" ParentId="15251" CreationDate="2010-12-24T16:53:35.967" Score="2" ViewCount="0" Body="&lt;p&gt;Dinesh, you seem to have the answer (for integer distributions) - need to show&#xA;$3(q_0+q_1)\geq q_0+q_1+q_2$, ie, $2(q_0+q_1)\geq q_2$, which is true since $2q_0\geq q_i$ for any $i$.&lt;/p&gt;&#xA;" OwnerUserId="5031" LastEditorUserId="5031" LastEditorDisplayName="" LastEditDate="2010-12-24T17:08:28.507" LastActivityDate="2010-12-24T17:08:28.507" CommentCount="1" />
  <row Id="15450" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T17:08:38.200" Score="1" ViewCount="0" Body="&lt;p&gt;I think the key point is understanding the notion of &quot;howto parametrize points&quot; in a manifold, and in particular the &lt;em&gt;big question&lt;/em&gt; is &quot;How can I pass from delightful, amenable coordinates in $\mathbb{R}^n$ to those fancy 'things' with a lot of cosines, logarithms and so on?&quot;. &#xA;Well, the answer (IMO) is &quot;Because &lt;em&gt;I know&lt;/em&gt; how to do analysis nowhere but in open balls of $\mathbb {R}^n$&quot;. Studying the beginning of Mathematical Physics teached me that anytime I'm considering the speed of a motion over a (hopefully smooth) surface I'm considering a vector in the tangent space (plane) of that surface in a point: in this fashion &quot;local&quot; description of a motion is the description of a &quot;straight&quot; motion of a point $p$, with rules established, instant per instant, by the velocity vector $\vec v$ of the point $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be more precise. Fix a surface $\mathcal M$ in the euclidean space (if you don't know the precise definition of a 2-dimensional manifold don't worry and go on thinking to $\mathcal M$ as a sheet on which you can smoothly skate) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to mathematically capture the idea of $p$ to be &lt;em&gt;bonded to move on $\mathcal M$&lt;/em&gt;. So you define its motion to be not the entire three dimensional space $\mathbb R^3$ but only a precise subset of it, which is &lt;em&gt;parametrized&lt;/em&gt; by some diffeomorphism (if you don't know what a diffeomorphism is, then learn it on wikipedia, or visualize it as a &quot;highly differentiable homeomorphism&quot; ;) ) $f\colon U\to W\subseteq \mathcal M$, $U$ being an oper set of $\mathbb{R}^2$ and $W$ being an open set (with the subspace topology induced by $\mathbb R^3$) of $\mathcal M$, containing the point $p$. It means in a few words that you're taking a point $q=(q_1,q_2)$ in the affine space of $\mathbb{R}^2$ and you're smoothly sending it in a point $f(q_1,q_2)=(f_1(q),f_2(q),f_3(q))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this neighborhood of $p$, every happens &quot;as if you were on $\mathbb R^2$, upto go back with the inverse of $f$ on your starting $U$. A motion on $\mathcal M$ is totally captured by a motion on $U$, because you can think about a motion on $\mathcal M$ as a &quot;suitably deformed&quot; motion on $U$, where you know how to define the &lt;em&gt;position&lt;/em&gt; of a point, using coordinates, and other useful things like (piecewise) smoothness of a curve, tangent lines, et cetera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea to be 'bonden on $\mathcal M$' is finally captured by asking that, given the curve $\gamma\colon ]-\epsilon,\epsilon[\to \mathcal M\subseteq\mathbb{R}^3$ describing the motion of $p$ (i.e., for example $\gamma(0)=p$), the velocity vector $\gamma&amp;#39;(t)$ lies in $\mathbb{R}^3$ &lt;em&gt;but also&lt;/em&gt; in the tangent plane $T_p(\mathcal M)$, which can in fact be defined as the collection of &lt;em&gt;all&lt;/em&gt; possible tangent vectors to curves lying on $\mathcal M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now notice that the number '2' of independent coordinate you are using to uniquely describe the motion of $p$ on $\mathcal M$ is not at all a random number. It is the &lt;em&gt;dimension&lt;/em&gt; of the tangent plane in $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's find a basis of $T_p(\mathcal M)$. Nothing easier: do you remember the function $f\colon U\to W$? Brutally derive it, taking its differential $\text{d}f_q$, where I'm writing $q=f^{-1}(p)$, and you will get a very nice invertible matrix whose columns form a basis of $T_p(\mathcal M)$. Don't you like the canonical basis of $\mathbb{R}^2$, do you? Well, change it and see what's happen (the answer is: upto a linear automorphism of $T_p(\mathcal M)$ I'm doing nothing).&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-24T17:08:38.200" />
  <row Id="15451" PostTypeId="2" ParentId="7570" CreationDate="2010-12-24T17:23:18.287" Score="0" ViewCount="0" Body="&lt;p&gt;Is the limit of the $f_j$'s continuous? If so I think your statement is true. Let $g_N(x) = 1/N \ \sum f_j(x)$. Now if $f_j \to f$, then $g_N \to f$ as well (if each $f_j$ is within an epsilon of $f$, then $g_N$ is within an epsilon of $f$ as well; go far enough out in the sequence to make this happen). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say $x_n \to x$. Then the RHS is&#xA;$$ \lim_{N \to \infty} g_N (x) = f(x)$$, and the LHS is&#xA;$$ \lim_{n \to \infty} f(x_n) = f(x)$$ if $f$ is continuous.&lt;/p&gt;&#xA;" OwnerUserId="5100" LastActivityDate="2010-12-24T17:23:18.287" />
  <row Id="15452" PostTypeId="2" ParentId="15415" CreationDate="2010-12-24T17:28:56.560" Score="0" ViewCount="0" Body="&lt;p&gt;Since the problem is tagged &quot;real analysis&quot;, I'd just use the supremum of the set $\{ n \in \mathbf{Z} : n \le x \}$. Of course, you need to prove that this set is not empty and that the supremum is actually a maximum.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-24T17:28:56.560" />
  <row Id="15453" PostTypeId="1" AcceptedAnswerId="15454" CreationDate="2010-12-24T17:37:17.363" Score="5" ViewCount="199" Body="&lt;p&gt;I ran into this expression in a paper I was reading, and I'm confused about part of the meaning.  Here $u$ and $v$ are two polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u, v \in \mathbb{R}[x]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really familiar with usage of $[x]$ here, but if it means &quot;nearest integer&quot;, then isn't this expression equivalent to simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u, v \in \mathbb{Z}$$&lt;/p&gt;&#xA;" OwnerUserId="4724" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-25T00:49:26.447" LastActivityDate="2010-12-25T00:49:26.447" Title="Meaning of $\mathbb{R}[x]$" Tags="&lt;polynomials&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15454" PostTypeId="2" ParentId="15453" CreationDate="2010-12-24T17:40:08.187" Score="6" ViewCount="0" Body="&lt;p&gt;$\mathbb{R}[x]$ is the set of polynomials (with variable $x$) whose coefficients are taken from $\mathbb{R}$, the set of real numbers. It has got nothing to do with the greatest integer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, people talk about polynomials in $\mathbb{Q}[x]$, where coefficients are rational, or more generally, when the coefficients are taken from an arbitrary &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;Ring&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance of a usage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;: A real number $r$ is transcendental if and only if, for every $P \in \mathbb{Q}[x]$, we have that $P(r) \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Polynomial_ring&quot; rel=&quot;nofollow&quot;&gt;Polynomial Ring&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-24T17:54:20.633" LastActivityDate="2010-12-24T17:54:20.633" />
  <row Id="15455" PostTypeId="1" AcceptedAnswerId="15456" CreationDate="2010-12-24T18:12:33.777" Score="5" ViewCount="163" Body="&lt;p&gt;What does the $\ni$ (backwards element of) symbol mean?  It doesn't appear in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_symbols&quot; rel=&quot;nofollow&quot;&gt;Wikipedia list of mathematical symbols&lt;/a&gt;, and a Google search for &quot;backwards element of&quot; or &quot;backwards epsilon&quot; turns up contradictory (or unreliable) information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems it can mean both &quot;such that&quot;, or &quot;contains as an element&quot;.  Is this correct, and if so, which is the more common usage?&lt;/p&gt;&#xA;" OwnerUserId="4724" LastActivityDate="2010-12-24T21:44:04.473" Title="Backwards epsilon" Tags="&lt;notation&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="15456" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T18:21:13.547" Score="4" ViewCount="0" Body="&lt;p&gt;I believe the most common usage is &quot;such that&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to agree: &lt;a href=&quot;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this pdf here: &lt;a href=&quot;http://faculty.ssfs.org/~doschel/2008-2009%20School%20Year/Calculus/Math%20Symbols.pdf&quot; rel=&quot;nofollow&quot;&gt;http://faculty.ssfs.org/~doschel/2008-2009%20School%20Year/Calculus/Math%20Symbols.pdf&lt;/a&gt; seems to indicate a growing trend in using it as &quot;contains as an element&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, it was first used to mean &quot;such that&quot; (see the second pdf link above).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-24T18:43:10.837" LastActivityDate="2010-12-24T18:43:10.837" CommentCount="3" />
  <row Id="15457" PostTypeId="2" ParentId="15174" CreationDate="2010-12-24T18:29:15.017" Score="4" ViewCount="0" Body="&lt;p&gt;I'll describe the cube, the symmetries of the cube, and the generators of the symmetry of the cube.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The plus–minus–stars description of the faces of a cube:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The d-dimensional cube can be thought of as the set of points in &lt;b&gt;R&lt;/b&gt;&lt;sup&gt;d&lt;/sup&gt; with coordinates between −1 and +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its vertices are all sequences of plusses and minuses of length d.  For d=2 you get the four vertices { ++, +−, −+, −− }.  For d=3 you get the eight vertices { +++, ++−, …, −−+, −−−}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its edges are obtained from vertices by allowing one of the coordinates to vary.  For d=2 you get the four edges { ☆+, ☆−, +☆, −☆ } where for instance +☆ = { (1,y) : −1 ≤ y ≤ 1 } and ☆− = { (x,−1) : −1 ≤ x ≤ 1 }.  For d=3 you get the twelve edges { ☆++, ☆+−, …, −+☆, −−☆ }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its 2-faces are obtained from vertices by allowing two of the coordinates to vary.  For d=2 you get the single face { ☆☆ } where ☆☆ = { (x,y) : −1 ≤ x,y ≤ 1 }.  For d=3 you get the six faces { ☆☆+, ☆☆−, ☆+☆, ☆−☆, +☆☆, −☆☆ }.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its k-faces for k ≤ d are obtained from vertices by allowing k of the coordinates to vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some people the cube is defined by its k-faces for 0 ≤ k ≤ d.  A symmetry of &lt;b&gt;R&lt;/b&gt;&lt;sup&gt;d&lt;/sup&gt; is defined by what it does to an affine basis such as the all −1s vector combined with the vertices of the cube with exactly 1 coordinate with a +1 and d−1 coordinates with a −1.  So a symmetry is uniquely determined by where it sends the 0-faces (the vertices).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The symmetries of the cube as signed permutations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetry of the cube must take k-faces to k-faces.  In particular, it must take edges to edges.  Since it is determined by what it does to the vertices, we get two distinct operations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for each coordinate, we must send it to another coordinate&lt;/li&gt;&#xA;&lt;li&gt;for each coordinate, we can either reverse it or not&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For instance ☆++ can be sent to any of the other edges, like −☆+.  For the first point, we see that the first coordinate has been sent to the second coordinate (permuting the stars).  For the second point we have two possibilities based on what happens to the vertices inside ☆++: we must send { +++, −++ } to { −++, −−+ }, but we can do this in two distinct ways (negating the sign or not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first point, we get the action of the symmetry group on certain &quot;blocks&quot; gives rise to the symmetric group on d objects.  For the second point, we get that the action of the symmetry group on each block gives rise to the symmetric group on 2 objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for every coordinate we have to decide where to send it and whether to negate it.  The collection of all such decisions is called the group of &lt;a href=&quot;http://en.wikipedia.org/wiki/Signed_permutations&quot; rel=&quot;nofollow&quot;&gt;signed permutations&lt;/a&gt;, also known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperoctahedral_group&quot; rel=&quot;nofollow&quot;&gt;hyperoctahedral group&lt;/a&gt;.  Using the matrix or coordinate representation it is easy to check that all such signed permutations are in fact symmetries of the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for d=2 we could say &quot;switch the coordinates, then negate the first coordinate,&quot; which can be written simply as (x,y)↦(−y,x) in function notation, or in matrix notation: \begin{bmatrix}0&amp;amp;1\\-1&amp;amp;0\end{bmatrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More compactly, we could say [1,2]↦[-2,1] where we permute the coordinates and mark which coordinates get negated.  The action on the vertices is easy to describe: it sends [ ++, +−, −+, -- ] to [ −+, ++, −−, +− ].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, to describe such a symmetry we need to choose a permutation of the vertices (an element of the symmetric group on d objects), and then d choices of ± to decide whether to negate a coordinate or not.  The size of the group is thus d!⋅2&lt;sup&gt;d&lt;/sup&gt; and the structure of the group is the wreath product of Sym(d) acting on Sym(2), and the permutation structure of the group on the vertices is more specifically the product action.  In GAP this is written as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;WreathProductProductAction( SymmetricGroup(2), SymmetricGroup(d) );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The generators of the symmetry group on the vertices:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part of the symmetry group that just permutes the coordinates without doing any negating is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_group&quot; rel=&quot;nofollow&quot;&gt;symmetric group&lt;/a&gt; and has two common generating sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{ (1,2), (2,3), …, (d−1,d) }, the &lt;strong&gt;coxeter generators&lt;/strong&gt;, and&lt;/li&gt;&#xA;&lt;li&gt;{ (1,2,3,…,d), (1,2) }, the d-cycle generating set.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Coxeter generators are better for many theoretical reasons, and the d-cycle generating set is very short and good for computational reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part of the symmetry group can switch the first coordinate with any other coordinate, and so to finish generating the group it in fact suffices to merely switch the sign of one coordinate.  Thus we have the two generating sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{ (1,2), (2,3), …, (d−1,d), −d }, the coxeter generators, and&lt;/li&gt;&#xA;&lt;li&gt;{ (1,2,3,…,d), (1,2), −d }, a d-cycle generating set&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Presentation_of_a_group&quot; rel=&quot;nofollow&quot;&gt;presentation of the group&lt;/a&gt; on the Coxeter generators is very nice.  Call them g(1), g(2), … g(d−1), g(d).  Then we always have g(k)&lt;sup&gt;2&lt;/sup&gt; = 1 is the identity, g(i)⋅g(j) = g(j)⋅g(i) as long as 1 ≤ i ≤ d−2, and i+2 ≤ j ≤ d, that is, as long as i and j are not adjacent.  Furthermore, g(k)⋅g(k+1)⋅g(k) = g(k+1)⋅g(k)⋅g(k+1) for 1 ≤ k ≤ d−2, but for k = d−1 we have the longer: g(d−1)⋅g(d)⋅g(d−1)⋅g(d) = g(d)⋅g(d−1)⋅g(d)⋅g(d−1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you label the vertices from 1 to 2&lt;sup&gt;d&lt;/sup&gt;, then it is a little bit of a hassle to write down the generators, just because there are so many points.  I would probably just use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GeneratorsOfGroup( symmetry_group );&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;in GAP myself, but you could write them out if you wanted.&lt;/p&gt;&#xA;" OwnerUserId="583" LastActivityDate="2010-12-24T18:29:15.017" />
  <row Id="15458" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T18:33:11.233" Score="2" ViewCount="0" Body="&lt;p&gt;It is for example used for functions that are inconvenient to refer to with a symbol. E.g. if $G$ is a group, then the composition function can be referred to like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G \times G \owns (x,y) \mapsto xy \in G $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the LaTeX command for it is &lt;code&gt;\owns&lt;/code&gt; should also give a clue as to what it's used for...&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-24T18:33:11.233" CommentCount="1" />
  <row Id="15459" PostTypeId="2" ParentId="15443" CreationDate="2010-12-24T18:33:17.497" Score="2" ViewCount="0" Body="&lt;p&gt;Another question implicit in what you ask is why use other co-ordinate systems. One reason is that the form of many equations is dramatically simplified in other co-ordinate systems.  For example, the rectangular representation of the curve defined by the polar coordinate equation  r = 1 + cos(2(theata)) would make it much harder to understand than in its polar form. r= 3 is a circle of radius 3 in polar coordinates. For modeling of certain phenomena polar coordinates makes things much easier, including work involving orbit calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Orbit_equation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-24T18:33:17.497" />
  <row Id="15460" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T18:41:10.950" Score="7" ViewCount="0" Body="&lt;p&gt;The backwards epsilon notation for &quot;such that&quot; was introduced by Peano in 1898, e.g. from Jeff Miller's &lt;a href=&quot;http://jeff560.tripod.com/set.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Uses of Various Mathematical Symbols&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Such that&lt;/strong&gt;.  According to Julio González Cabillón, Peano introduced the backwards lower-case epsilon for &quot;such that&quot; in &quot;Formulaire de Mathematiques vol. II, #2&quot; (p. iv, 1898).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Peano introduced the backwards lower-case epsilon for &quot;such that&quot; in his 1889 &quot;Principles of arithmetic, presented by a new method,&quot; according van Heijenoort's From Frege to Gödel: A Source Book in Mathematical Logic, 1879--1931 [Judy Green].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-24T18:41:10.950" CommentCount="1" />
  <row Id="15461" PostTypeId="1" CreationDate="2010-12-24T19:09:40.130" Score="3" ViewCount="55" Body="&lt;p&gt;Hi,&#xA;I'm stuck on this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a &quot;truth-table&quot; (well, I don't know if it can be called truth table, if there aren't true/false values only):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; string |  a |  b&#xA;---------------------&#xA;      x |  1 |  0&#xA;      z |  1 |  0&#xA;     xx |  1 |  1&#xA;     xz |  1 |  1&#xA;     zx |  1 | -1&#xA;     zz |  1 | -1&#xA;    xxx |  0 |  1&#xA;    xxz |  0 |  1&#xA;    xzx |  2 |  1&#xA;    xzz |  2 |  1&#xA;    zxx |  2 | -1&#xA;    zxz |  2 | -1&#xA;    zzx |  0 | -1&#xA;    zzz |  0 | -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... the list goes on for string of any length, e.g. xxzxxzxxzzxzzxx -&gt; a = -4, b = -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; values from a given string, but the number of steps grows with the length of the string. I am hoping for some kind of better algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;em&gt;Edit: Here is my original algorithm:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rotation = 0 // 0 -&amp;gt; up, 1 -&amp;gt; right, 2 -&amp;gt; down, 3 -&amp;gt; left&#xA;pos_x = 0    // this is the &quot;b&quot;&#xA;pos_y = 0    // this is the &quot;a&quot;&#xA;&#xA;function rotate (n):&#xA;    rotation += n&#xA;    rotation %= 4 // result is positive integer, e.g. -1 % 4 = 3&#xA;&#xA;function forward (n):&#xA;    if n == 0: pos_y++&#xA;    if n == 1: pos_x++&#xA;    if n == 2: pos_y--&#xA;    if n == 3: pos_x--&#xA;&#xA;for char in string:&#xA;    forward()&#xA;    if char == &quot;x&quot;: rotate(1)  // to the right&#xA;    else:           rotate(-1) // to the left&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I can easily get the final rotation from a string:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function final_rotation (n):&#xA;    rot = (number of occurences of &quot;x&quot;) - (number of occurences of &quot;z&quot;)&#xA;    rot %= 4&#xA;    return rot&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But the pos_x and pos_y (or a and b)? No idea :/&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The Karnaugh-maps came to my mind - but I'm unsure how to deal with this, since the values of &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; aren't true/false, but integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it looks like it's ternary logic, because of the three possible states of string[position] - &lt;strong&gt;x, z, none&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand this, the last (rightmost) letter of a string doesn't have a say in what the &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt; values will be. But what does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is - how to find the algorithm for &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5102" LastEditorUserId="5102" LastEditorDisplayName="" LastEditDate="2010-12-24T19:57:01.010" LastActivityDate="2010-12-24T19:57:01.010" Title="Definition of functions based on &quot;fuzzy&quot; truth table" Tags="&lt;algorithms&gt;&lt;functions&gt;&lt;fuzzy-logic&gt;" CommentCount="7" />
  <row Id="15462" PostTypeId="2" ParentId="15261" CreationDate="2010-12-24T19:30:24.557" Score="3" ViewCount="0" Body="&lt;h3&gt;Introduction&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The following response attempts to address two aspects I perceive to be of interest in this question: one concerns whether there are some relatively rote or standard procedures for evaluating the convergence of certain kinds of improper multivariate integrals; the other is whether there is some simple intuition behind the subject.  To keep the discussion from becoming too abstract, I will use the double integral (introduced in the question) as a running example.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Synopsis&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Many integrals that are improper due to singular behavior of their integrand can be analyzed, both rigorously and intuitively, with a simple comparison.  The idea is that any singularity of the integrand that doesn't blow up too fast compared to the codimension of the manifold of singular points can be &quot;smoothed over&quot; by the integral, provided the domain of integration doesn't contact the singular points too &quot;tightly.&quot;  It remains to make these ideas precise.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Analysis&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When the domain of integration is relatively compact, as the one in the example is, the problems with convergence occur only at the possible singularities of the integrand within the closure of the domain, which in this case is the isolated point $(x,y) = (1,1)$.  However, it is evident that if the domain of integration were to be expanded, any zero of the curve $1 - x y = 0$ could be a singularity.  In general, many singularities occur this way: an integrand $f(\bf{x})$ is locally of the form $g(||h(\bf{x})||)$ where $h: \mathbb{R}^n \to \mathbb{R}^k$ and $g(r) = r^{-s} u(r)$ for some function $u$ that is bounded in some nonnegative neighborhood of $0$.  In this case $h(\bf{x}) = 1 - x y$ and $g(r) = r^{-1}$, so $s=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $h$ is differentiable at $\bf{0}$ with nonsingular derivative there, &lt;strong&gt;principles of Calculus&lt;/strong&gt; invite us to linearize the situation and &lt;strong&gt;geometric ideas&lt;/strong&gt; suggest a simple form for the linearization.  Specifically, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Implicit_function_theorem&quot; rel=&quot;nofollow&quot;&gt;Implicit Function Theorem&lt;/a&gt; guarantees that local coordinates can be found near such a singularity in which the derivative of $h$ is in the form $\left( \bf{1}_{k \times k} ; \bf{0}_{k \times n-k} \right)$.  The singularity itself can be translated to the origin where, to first order, the zeros of $h$ locally correspond with the vector subspace generated by the last $n-k$ coordinate axes.  The effect on the integral is to introduce a factor given by the determinant of the Jacobian, which is locally bounded and so does not affect the convergence.  In this example we can explicitly make these changes of coordinates by translating the singularity to $\bf{0}$, computing the gradient of $h$ there, and rotating it to point along the first axis.  This amounts to the change of variables $(u, v)$ = $(2 - x - y, y - x)$, in which $h(u, v) = u - \frac{1}{4}(u^2 - v^2)$, equal to $u$ to first order.  Within a small neighborhood of the origin, the domain of integration becomes a 90-degree wedge containing positive $(u,v)$ for which $u \le |v|$ and the zeros of $h$ coincide with the $v$ axis to first order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the case where the domain of integration locally contains an isolated point of the singular set $H$ (the zeros of $h$), which we translate to $\bf{0}$.  &lt;strong&gt;Estimate the integral near this singularity by adopting spherical coordinates there&lt;/strong&gt;.  To do this, we need a closer look at the domain of integration in spherical coordinates.  Any $\epsilon \gt 0$ determines the set of all possible unit direction vectors from $\bf{0}$ toward elements of the domain of integration that lie within a distance $\epsilon$ of $\bf{0}$.  In the example, this corresponds to the set of points on the unit circle with angles between $-\pi/4$ and $\pi/4$.  Suppose that for sufficiently small $\epsilon$ the closure (in the unit sphere) of this set of direction vectors does not include any of the tangent vectors of $H$ at $\bf{0}$.  Then an easy estimate shows that the angular part of the integral in spherical coordinates is bounded.  This reduces the task of integration to estimating the radial part.  Because the volume element in spherical coordinates $(\rho, \Omega)$ is $\rho^{n-1} d\rho d\Omega$, the radial integral is proportional to $\rho^{n-1} g(\rho) d \rho$.  &lt;strong&gt;This is the key calculation:&lt;/strong&gt; it shows how integration in $n$ dimensions can &quot;cancel&quot; a factor of $\rho^{1-n}$ in the integrand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're reduced to evaluating a 1D integral, improper at $0$, whose integrand behaves like  $\rho^{n-1} g(\rho)$.  By virtue of the assumptions about $g$, this is bounded above by a multiple of $\rho^{n-1-s}$.  &lt;em&gt;Provided&lt;/em&gt; $n-1-s \gt -1$, this converges (at a rate proportional to $\rho^{n-s}$).  In the example, $n=2$, $s=1$, so convergence is assured.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Generalizations&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When $n-1-s \le -1$, the behavior of the original integral depends on exactly how the domain &quot;pinches out&quot; as it approaches the singularity, so more detailed analysis is needed.  The spirit of the exercise doesn't change, though: with Calculus we linearize the situation (relying on simple estimates to take care of the second order remainder term), we use spherical coordinates centered at an isolated singularity in the domain (or more generally, cylindrical coordinates near a more complicated singularity) to perform the integration, and we estimate the rate of growth of the integral near the singularity in terms of a radial component and a contribution from a spherical angle.  The behavior of the integral over the spherical angle as we approach the singularity is determined by the shape of the domain near the singularity, so often some careful (and interesting) geometric analysis is needed.  This approach is characteristic of the proofs of many theorems in Several Complex Variables, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Edge-of-the-wedge_theorem&quot; rel=&quot;nofollow&quot;&gt;Edge of the Wedge Theorem&lt;/a&gt; (if I recall correctly--I'm reaching back to memories now a quarter century old).  One good reference is &lt;a href=&quot;http://books.google.com/books?id=uZZsh1w_RJcC&amp;amp;pg=PR7&amp;amp;source=gbs_selected_pages&amp;amp;cad=3#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Steven Krantz's book on Function Theory of SCV&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Summary&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The behavior of the integral of $(1-x y)^{-1}$ near $(1,1)$ can be analyzed by adopting polar coordinates $(\rho, \theta)$ centered at $(1,1)$, observing that the domain here forms a wedge whose point contacts the singular set $1 - x y = 0$ &quot;transversely,&quot; noticing that the radial behavior of the integrand is $\rho^{-1}$, remembering that the area element in polar coordinates has a $\rho^1 d \rho$ term, and noting that the integral of $\rho^{-1} \rho^1 d \rho$ converges at $0$.  Indeed, what originally looked like a singularity has entirely disappeared.  This procedure generalizes to multidimensional integrals subject to fairly mild restrictions on the nature of the singularities of their integrands and on the geometry of the domain of integration near those singularities.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-12-24T19:30:24.557" CommentCount="1" />
  <row Id="15463" PostTypeId="2" ParentId="15453" CreationDate="2010-12-24T19:32:31.650" Score="3" ViewCount="0" Body="&lt;p&gt;More generally, if $\rm\:R \subset S\:$ are rings and $\rm\:s\in S\:$ then $\rm\:R[s]\:$ denotes the &lt;strong&gt;ring-adjunction&lt;/strong&gt; of $\rm\:s\:$ to $\rm\:R\:,\:$ i.e. the smallest subring of $\rm\:S\:$ containing both $\rm\:R\:$ and $\rm\:s\:.\:$ The notation for the polynomial ring $\rm\:R[x]\:$ is  the special case where $\rm\:x\:$ is transcendental over $\rm\:R\ $ (an &quot;indeterminate&quot; in old-fashioned language),$\ $ i.e. $\rm\: x\:$ isn't a root of any polynomial with coefficients in $\rm\:R\:$. One may view $\rm\:R[x]\:$ as the adjunction of a &lt;em&gt;universal&lt;/em&gt; (or generic) element $\rm\:x\:$, in the sense that any other adjunction $\rm\:R[s]\:$ is a ring-image of $\rm\:R[x]\:$ under the evaluation homomomorphism $\rm\: x\to s\:.\ $ For example, if $\rm\:R \subset S\:$  are fields then $\rm\:R[s]\cong R[x]/(f(x))\:$ where $\rm\:f(x)\:$ is the minimal polynomial of $\rm\:s\:$ over $\rm\:R\:.\:$ Essentially this serves to faithfully ring-theoretically model $\rm\:s\:$ as a &quot;generic&quot; root $\rm\:x\:$ of the minimal polynomial $\rm\:f(x)\:$ for $\rm\:s\:.\:$ Polynomial rings may be characterized by the existence and uniqueness of such evaluation maps (&quot;universal mapping property&quot;), e.g. see any textbook on Universal Algebra, e.g. &lt;a href=&quot;http://math.berkeley.edu/~gbergman/245/index.html&quot; rel=&quot;nofollow&quot;&gt;Bergman.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-24T21:23:47.603" LastActivityDate="2010-12-24T21:23:47.603" />
  <row Id="15464" PostTypeId="2" ParentId="15438" CreationDate="2010-12-24T19:44:42.820" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a net of a buckyball, from &lt;a href=&quot;http://www.goldennumber.net/buckyball.htm&quot; rel=&quot;nofollow&quot;&gt;GoldenNumber.net&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8Bwto.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be possible to turn this into the kind of net you're looking for by replacing the pentagons and hexagons with 5 and 6 isosceles triangles (the heights of the triangles determine the &quot;elevation&quot; of the center vertex from the original pentagonal/hexagonal faces and thus affect the sphericality of the result). &lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-24T19:44:42.820" CommentCount="1" />
  <row Id="15465" PostTypeId="2" ParentId="15388" CreationDate="2010-12-24T19:56:17.983" Score="2" ViewCount="0" Body="&lt;p&gt;Simply use the (additive group) axioms to prove $\rm\ a + x = b\ \Rightarrow\ x = b + (-a)\:.\ $ That yields both existence &lt;em&gt;and&lt;/em&gt; uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in &lt;a href=&quot;http://groups.google.com/group/sci.math/browse_frm/thread/1bbf3ac81e6df7f9&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; on such a proof in Max Rosenlicht's &lt;em&gt;Introduction to Analysis&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-24T19:56:17.983" />
  <row Id="15467" PostTypeId="2" ParentId="15349" CreationDate="2010-12-24T21:36:32.173" Score="0" ViewCount="0" Body="&lt;p&gt;Use any algorithm for &lt;a href=&quot;http://en.wikipedia.org/wiki/Matching_%28graph_theory%29#Maximum_matchings_in_bipartite_graphs&quot; rel=&quot;nofollow&quot;&gt;Maximum Weighted Bipartite Matching&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-24T21:36:32.173" CommentCount="1" />
  <row Id="15468" PostTypeId="2" ParentId="15455" CreationDate="2010-12-24T21:44:04.473" Score="1" ViewCount="0" Body="&lt;p&gt;I have always used $\ni$ to represent &quot;such that.&quot; Many times people do not use this notation though; rather, they use s.t. or just say &quot;such that.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can see that Wikipedia uses the &quot;:&quot; notation, which is also correct; however, I usually only use that notation when in curly brackets (ie { }).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visit [1] to see an actual discussion of it in a list of mathematical symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a very good question, by the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-24T21:44:04.473" />
  <row Id="15469" PostTypeId="1" AcceptedAnswerId="15490" CreationDate="2010-12-24T22:00:01.693" Score="8" ViewCount="137" Body="&lt;p&gt;Does anyone know a combinatorial proof of the following identity, where $F_n$ is the $n$th Fibonacci number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n F_1 + (n-1)F_2 + \cdots + F_n = F_{n+4} - n - 3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not in the place I thought it most likely to appear: Benjamin and Quinn's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=FXLGzXwbwIAC&amp;amp;printsec=frontcover&amp;amp;dq=proofs+that+really+count&amp;amp;source=bl&amp;amp;ots=cLasNhVboM&amp;amp;sig=cgpcvX12VJy4hzDjlM8buMY1O1w&amp;amp;hl=en&amp;amp;ei=1RUVTY2vAY7EsAPklv3GAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCkQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Proofs That Really Count&lt;/a&gt;&lt;/em&gt;.  In fact, this may be a hard problem, as they say the similar identity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_1 + 2F_2 + \cdots + nF_n = (n+1)F_{n+2} - F_{n+4} +2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &quot;in need of a combinatorial proof.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, &lt;a href=&quot;http://books.google.com/books?id=n8Xx2oSlz4MC&amp;amp;pg=PA14&amp;amp;lpg=PA14&amp;amp;dq=benjamin+and+quinn+uncounted+fibonacci&amp;amp;source=bl&amp;amp;ots=UDfzehytLs&amp;amp;sig=mcWgV3BP2_Jyl5zi1fhL2EHqqfU&amp;amp;hl=en&amp;amp;ei=ThIVTcmsLo7ksQPcvMXIAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCIQ6AEwAg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (from Benjamin and Quinn's text) are several combinatorial interpretations of the Fibonacci numbers.&lt;/p&gt;&#xA;" OwnerUserId="2370" LastActivityDate="2010-12-25T06:16:12.707" Title="Combinatorial proof of a Fibonacci identity:   $n F_1 + (n-1)F_2 + \cdots + F_n = F_{n+4} - n - 3.$" Tags="&lt;combinatorics&gt;&lt;fibonacci-number&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="15470" PostTypeId="1" AcceptedAnswerId="15488" CreationDate="2010-12-24T22:39:09.373" Score="4" ViewCount="65" Body="&lt;p&gt;Let $f: X \to Y$ be a proper morphism of noetherian schemes. If the natural map $\mathcal{O}_Y \to f_*(\mathcal{O}_X)$ is an isomorphism, then a version of Zariski's main theorem states that the fibers $X_y, y \in Y$ are all connected. (The case listed as &quot;Zariski's main theorem&quot;  in Hartshorne is the case of $f$ birational and $Y$ normal.) This may be proved via the formal function theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fancier version of ZMT (EGA IV-8) is that a quasifinite separated morphism of finite presentation between quasicompact, quasiseparated schemes factors as the composite of an open immersion and a finite morphism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a direct way to deduce the connectedness theorem from the more general ZMT?&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-25T05:58:47.473" Title="How to get the connectedness theorem from the quasi-finite version of ZMT?" Tags="&lt;algebraic-geometry&gt;" AnswerCount="1" />
  <row Id="15471" PostTypeId="1" AcceptedAnswerId="15474" CreationDate="2010-12-24T22:46:51.917" Score="2" ViewCount="166" Body="&lt;p&gt;The sum of a finite number of rational numbers is of course a rational number, but the sum of an infinite number of rational numbers might be an irrational number.  Can someone give me some intuition why this sum might be irrational?  I just &quot;don't feel it.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2011-01-05T17:36:00.240" LastActivityDate="2011-01-05T17:36:00.240" Title="Sum of rational numbers" Tags="&lt;number-theory&gt;&lt;irrational-numbers&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="15472" PostTypeId="2" ParentId="15471" CreationDate="2010-12-24T22:55:13.000" Score="2" ViewCount="0" Body="&lt;p&gt;Any irrational number $x$ is the limit of a sequence of rational numbers $a_n$ (take for example the decimal expansion truncated after $n$ decimals, for $n=1,2,3,\dots$). Then&#xA;$$x = a_1 + (a_2-a_1) + (a_3-a_2) + \dots$$&#xA;is a sum of rational numbers, but it is irrational by construction.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-12-24T23:00:18.757" LastActivityDate="2010-12-24T23:00:18.757" />
  <row Id="15473" PostTypeId="2" ParentId="15471" CreationDate="2010-12-24T22:55:43.593" Score="1" ViewCount="0" Body="&lt;p&gt;An irrational number is a &quot;gap&quot; inside the rationals. The decimal expansion of any irrational number is an infinite sum converging to it. It gives better and better approximations to the irrational number, which is unfortunately just not &quot;there&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many more examples of such limiting constructions. Sometimes you get new objects and sometimes you don't. An example is the delta function which can be approximated using bona fide functions. So there are functions arbitrarily &quot;close&quot; (in some sense) to the delta function, but the delta function itself is just &quot;too good&quot; to be an actual function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, if these approximations are good enough, you can prove that the number is transcendental (Legendre's or Roth's theorem).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-24T22:55:43.593" />
  <row Id="15474" PostTypeId="2" ParentId="15471" CreationDate="2010-12-24T22:56:36.573" Score="11" ViewCount="0" Body="&lt;p&gt;Look at &lt;a href=&quot;http://mathforum.org/library/drmath/view/52829.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: $3.14159 \dots = 3 + \frac{1}{10} + \frac{4}{10^2} +\frac{1}{10^3} + \frac{5}{10^4}+ \frac{9}{10^5} + \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above expression is $\pi$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-24T22:56:36.573" />
  <row Id="15475" PostTypeId="2" ParentId="14828" CreationDate="2010-12-24T23:17:39.900" Score="1" ViewCount="0" Body="&lt;p&gt;In addition to the other answers, it's also noteworthy to know that there is a number system (more-or-less) integrating N, Z, Q and R (and &lt;em&gt;a lot&lt;/em&gt; more, but not C!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Surreal_number&quot; rel=&quot;nofollow&quot;&gt;surreal numbers&lt;/a&gt; take the basic idea from Dedekind cuts, assigning to each number a so-called &lt;em&gt;left set&lt;/em&gt; ('smaller') and &lt;em&gt;right set&lt;/em&gt; ('larger') of numbers (constraint to certain rules), bootstrapping the whole process from the empty set and ending up with the reals and weird infinitesimals like $\frac{1}{\sqrt{\omega - \pi}}$!&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2010-12-24T23:17:39.900" />
  <row Id="15476" PostTypeId="1" AcceptedAnswerId="15477" CreationDate="2010-12-25T00:06:36.170" Score="4" ViewCount="158" Body="&lt;p&gt;Let $0&amp;lt;a\le 1$ be fixed. Is the sequence $$a_N=\int_1^N x^{-a}e^{ix} dx$$ bounded?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T22:09:51.967" LastActivityDate="2010-12-27T22:09:51.967" Title="Is this a bounded sequence?" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;fourier-analysis&gt;" AnswerCount="2" />
  <row Id="15477" PostTypeId="2" ParentId="15476" CreationDate="2010-12-25T00:17:40.390" Score="5" ViewCount="0" Body="&lt;p&gt;Yes, and it converges to $\int_1^\infty x^{-a}e^{ix}dx$.  To see that this improper Riemann integral exists, you could break up the integrand into real and imaginary parts, break up the resulting real integrals over intervals of length $\pi$, and apply the alternating series test.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-25T00:17:40.390" />
  <row Id="15478" PostTypeId="2" ParentId="15476" CreationDate="2010-12-25T00:57:06.340" Score="7" ViewCount="0" Body="&lt;p&gt;I just thought of another answer. Using integration by parts, we get&#xA;$$\frac{1}{i}(N^{-a}e^{iN}-e^i)+\frac{a}{i}\int_1^N y^{-(1+a)}e^{iy} dy$$&#xA;The first term is bounded and the second term is dominated by the integral $\int_1^\infty y^{-(1+a)} dy$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-25T00:57:06.340" />
  <row Id="15479" PostTypeId="1" AcceptedAnswerId="15482" CreationDate="2010-12-25T01:57:02.287" Score="6" ViewCount="121" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are some good resources for a student who has taken algebraic and point-set topology and who wishes to learn how filters and ultrafilters are applied in topology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; I've seen in a number of papers where ultrafilters are briefly mentioned as tools which can be used to generalize an argument or something, but the few papers I've found which introduce filters mainly use them as as tools in Logic and mention topology only in passing.  The wikipedia page gives some good starting points, but I felt that it was more of a reference list than an introduction to the topic.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2010-12-25T05:45:42.473" Title="Introduction to Filters in Topology." Tags="&lt;topology&gt;&lt;online-resources&gt;" AnswerCount="2" />
  <row Id="15480" PostTypeId="1" AcceptedAnswerId="15483" CreationDate="2010-12-25T04:11:02.357" Score="3" ViewCount="50" Body="&lt;p&gt;This question is vaguely related to: &lt;a href=&quot;http://math.stackexchange.com/questions/14915/different-formulations-of-class-field-theory&quot;&gt;Different formulations of Class Field Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said there, I'm currently learning class field theory. For some motivation, I've also read a little about Langlands. Everything I've read about that program seems to be a generalization of the global formulation of CFT. Is the local language (and theorems) preserved as well? Are they obviously/with difficulty or/conjecturally equivalent?&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2010-12-25T05:06:38.233" Title="Does the Langland's program preserve CFT's distinction between local and global theories?" Tags="&lt;number-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15481" PostTypeId="2" ParentId="15434" CreationDate="2010-12-25T04:18:11.867" Score="1" ViewCount="0" Body="&lt;p&gt;Edit: I was initially assuming that $E$ is the closed linear span of the basic sequence, but then I realized that from your question there is no justification for me having done so.  Since my initial argument applied only to that case and I didn't know how to extend it, I did a Google search to find that one way to reduce to this case is to apply Mazur's theorem, and that a nice and succinct answer to your question is given as &lt;a href=&quot;http://books.google.com/books?id=FkjCulO5xg0C&amp;amp;lpg=PP1&amp;amp;pg=PA23#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Lemma 1.6.1&lt;/a&gt; in &lt;em&gt;Topics in Banach space theory&lt;/em&gt; by Albiac and Kalton.  (This is something I should have known, but I had apparently not thought enough about weak topologies.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, here's my original argument that applies after reducing to the case where $E$ is the closed linear span of $\{x_n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus every element $x$ of $E$ has a unique representation as an infinite sum $x=\sum_n a_n(x)x_n$, and the coefficient functionals $x\mapsto a_n(x)$ are bounded.  Let $B=\{x_n\}$ denote the basis, and suppose that $x$ is in $\overline{B}\setminus B$.  You want to show that $x=0$, which is equivalent to showing that $a_n(x)=0$ for all $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each $n$ and each $\varepsilon\gt0$, consider the subbasic neigborhood $U_{n,\varepsilon}$ of $x$ defined by  $U_{n,\varepsilon}= \{y:|a_n(y)-a_n(x)|\lt \epsilon \}$.  Suppose to get a contradiction that $x\neq 0$, so there exists a $k$ such that $a_k(x)\neq 0$.  Then $U_{k,\varepsilon}$ is disjoint from $B\setminus\{x_k\}$ for sufficiently small $\epsilon$, and since $x$ is in $\overline B$ you can deduce that $a_k(x)=1$.  Since $x\neq x_k$, there exists $m\neq k$ such that $a_m(x)\neq0$, and the same argument indicated above shows that $a_m(x)$ must be $1$.  Then $U_{k,1}\cap U_{m,1}$ is a neighborhood of $x$ that doesn't intersect $B$, a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-25T05:12:45.337" LastActivityDate="2010-12-25T05:12:45.337" />
  <row Id="15482" PostTypeId="2" ParentId="15479" CreationDate="2010-12-25T04:26:33.347" Score="3" ViewCount="0" Body="&lt;p&gt;Bourbaki's General Topology, Volume 1 devotes a chapter to filters. The notion of net and equivalence between filters and nets is developed (partly in a set of guided exercises) in Kelly's General Topology. &quot;The use of ultraproducts in Commutative algebra&quot; by Schoutens might interest you. &quot;The theory of ultrafilters&quot; by Comfort and Negrepontis is encyclopedic. Jech's &quot;Set theory&quot; has a chapter on ultrafilters. Sections 4 and 6 of Chang and Keisler's &quot;&quot;Model Theory&quot; is also worth a look. These are all available on the website library.nu.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditorDisplayName="" LastEditDate="2010-12-25T05:31:56.473" LastActivityDate="2010-12-25T05:31:56.473" />
  <row Id="15483" PostTypeId="2" ParentId="15480" CreationDate="2010-12-25T04:57:54.570" Score="5" ViewCount="0" Body="&lt;p&gt;Dear Nicole, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a local aspect to the Langlands program, known as the local Langlands correspondence.  In fact, Langlands conjectured the existence of such a correspondence for each local field $F$ and each reductive group $G$ over $F$.&#xA;He proved his local conjecture when $F$ is archimedean and $G$ is arbitrary.  The case when $G = \mathrm{GL}_n$ and $F$ is non-archimedean is now solved (by Laumon, Rapoport, and Stuhler in the function field case, and by Harris and Taylor in the case of $p$-adic fields).   There are many results known for other groups $G$ as well, but the full local conjectures for arbitrary $G$ are not yet settled, as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\mathrm{GL}_n$ case, the correspondence, roughly speaking, gives a bijeciton between (typically infinite dimensional!) irreducible representations of the group $\mathrm{GL}_n(F)$ and $n$-dimensional representations of the Galois group of $F$.  (So, unlike the abelian case, there is no isomorphism of groups, but rather a certain bijection between certain kinds of representations of rather different groups.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of general $G$ is more involved to state, and indeed, it is not so easy to find the precise conjecture in the literature.  In any case, it involves many complications, the most significant of which is probably so-called &lt;em&gt;endoscopy&lt;/em&gt;.  Note that Ngo won the Fields medal this summer for his work on this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as with local CFT, the motivations for the local conjecture come from the global theory.  On the one hand, representations of matrix groups over local fields arise naturally in the theory of automorphic forms (this is a generalization/reformulation of the classical theory of Hecke operators in the theory of modular forms), and, on the other hand, (at least certain) automorphic Hecke eigenforms are suppose to be related to representations of Galois groups of global fields by global reciprocity laws.  Restricting these global Galois representations to decomposition groups, one should recover the conjectural local correspondence.  (And as I noted in my answer to your earlier question, this is in fact the mechanism by which those local correspondences are normally constructed, in the cases when they can be constructed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'm not sure if I understand your last question about equivalence correctly, but if you are asking whether the existence of the global Langlands correspondence (whatever exactly that means) should follow from the local correspondence, the answer is no.  Consider the case of CFT: knowing all the local Artin maps lets you write down the global Artin map, but you still have to prove the global Artin reciprocity law.  Similarly, knowing the local Langlands correspondence allows one to write down a candidate for the global correspondence, but to prove that this candidate actually does the job is another, even more difficult, matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As one example: before the modularity theorem for elliptic curves was proved, people knew how to write down the &#xA;candidate $q$-expansion that should be the weight $2$ modular form attached to an elliptic curve over $\mathbb Q$; this was because the relevant local issues were all completely understood.  The problem was then to prove that this actually &lt;em&gt;was&lt;/em&gt; a weight $2$ modular form; this was a global issue, which was completely open until Wiles and Taylor, Breuil, Conrad and Diamond solved it.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-25T05:06:38.233" LastActivityDate="2010-12-25T05:06:38.233" />
  <row Id="15484" PostTypeId="1" CreationDate="2010-12-25T05:31:44.120" Score="1" ViewCount="146" Body="&lt;p&gt;Can you help me to prove this inequality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;|\sqrt{3} - m/n| \geq 1/(5n^2)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where m and n are integers.&#xA;Hint:$sqrt(3)$ is irrational.&lt;/p&gt;&#xA;" OwnerUserId="5049" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-25T06:56:39.637" LastActivityDate="2011-01-03T02:25:17.317" Title="Showing that $|\sqrt{3} - m/n| \geq 1/(5n^2)$" Tags="&lt;number-theory&gt;&lt;inequality&gt;" AnswerCount="1" />
  <row Id="15485" PostTypeId="2" ParentId="15479" CreationDate="2010-12-25T05:45:42.473" Score="0" ViewCount="0" Body="&lt;p&gt;As far as online resources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.uga.edu/~pete/convergence.pdf&quot; rel=&quot;nofollow&quot;&gt;Notes on convergence&lt;/a&gt; by Pete Clark,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://topologicalmusings.wordpress.com/2008/07/18/ultrafilter-convergence-compactness-and-the-spectrum-of-a-boolean-algebra/&quot; rel=&quot;nofollow&quot;&gt;Blog post&lt;/a&gt; by Todd Trimble,&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://terrytao.wordpress.com/tag/ultrafilters/&quot; rel=&quot;nofollow&quot;&gt;Blog posts&lt;/a&gt; by Terence Tao (see also the posts tagged ultralimit-analysis and nonstandard-analysis)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am also writing a &lt;a href=&quot;http://qchu.wordpress.com/tag/ultrafilters/&quot; rel=&quot;nofollow&quot;&gt;series of blog posts&lt;/a&gt; (unfinished) on the subject.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-25T05:45:42.473" />
  <row Id="15486" PostTypeId="1" AcceptedAnswerId="15492" CreationDate="2010-12-25T05:51:50.333" Score="2" ViewCount="344" Body="&lt;p&gt;How to understand the compactness in topology space in intuitive way? &lt;/p&gt;&#xA;" OwnerUserId="1080" LastActivityDate="2010-12-26T19:36:00.803" Title="How to understand compactness?" Tags="&lt;general-topology&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="15487" PostTypeId="2" ParentId="15471" CreationDate="2010-12-25T05:53:44.933" Score="5" ViewCount="0" Body="&lt;p&gt;The problem is psychological: you think of the &quot;infinite sum&quot; of rational numbers as an obvious, intuitive concept, but it's not.  It has a precise mathematical meaning, and that precise mathematical meaning only works if you allow the sums to be real numbers (which themselves have a precise mathematical meaning).  The definitions which allow these &quot;infinite sums&quot; to make sense are much less trivial than someone who's never worked through them would think.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-25T05:53:44.933" />
  <row Id="15488" PostTypeId="2" ParentId="15470" CreationDate="2010-12-25T05:58:47.473" Score="1" ViewCount="0" Body="&lt;p&gt;Dear Akhil,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure that this is directly possible.  If memory serves, Zariski's original version of his main theorem showed something like the following: if $f: X \to Y$ is birational with $Y$ normal, and if $y \in Y$ is a point where $f^{-1}$ is not defined (I think this is what Zariski calls a fundamental point), then each component of $f^{-1}(y)$ is positive dimensional.  &lt;em&gt;This&lt;/em&gt; result can be proved using Grothendieck's form of ZMT, I think: if $f^{-1}(y)$ contains an isolated point, we can choose a n.h. of this point such that the restriction of $f$ to this n.h. has finite fibres, hence is an isomorphism (here we are using Grothendieck's ZMT together with normality of $Y$), and so in fact&#xA;$f^{-1}$ can be defined at $y$ after all.  (This is quite possibly not quite correct, but hopefully is not totally bogus either, both in the argument and in the claim to some historical accuracy.  Also, I think that if you look e.g. in the commutative algebra part of the stacks project a version of ZMT along these lines is discussed; at least, there are results in which the hypothesis of an isolated point in the fibre plays a prominent role.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zariski's connectedness theorem came quite a bit later, in his monograph on formal functions, and the techniques were quite a bit more involved.  (They were a precursor to formal scheme techniques, in which one can complete in some directions but not in others, as opposed to earlier techniques with complete local rings, in which one completes in every direction around a point at once.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick glance over Mumford's discussion of ZMT in the Red Book (which is always a good place to go to for learning ZTM intuition) suggests that I'm not blundering here.  (Grothendieck's formulation of ZMT is what he calls version IV of ZMT, while the connectedness theorem is his version V, and he singles out version V as being &quot;more global&quot; than the other versions, and doesn't discuss any implications from the other versions to this one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[On the other hand, this contradicts the wikipedia entry, which suggests that Grothendieck's formulation does imply the connectedness theorem.  (But doesn't quite say how.)  There are plenty of people at Harvard you could ask for clarification on this point, of course ... .]&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-25T05:58:47.473" CommentCount="1" />
  <row Id="15489" PostTypeId="2" ParentId="12057" CreationDate="2010-12-25T06:01:03.390" Score="2" ViewCount="0" Body="&lt;p&gt;Simply repeatedly apply Euclid's lemma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad\quad\quad 1 = (pb+qa,qb)$&lt;br&gt;&#xA;$\rm\quad \iff\ 1 = (pb+qa,q) = (pb,q) = (b,q)\quad\ via\quad\ (p,q) = 1$&lt;br&gt;&#xA;$\quad\ $ and $\rm\ \ 1 = (pb+qa,b) = (qa,b) = (q,b)\quad\ via\quad\ (a,b) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\rm\ \ 1 = (pb+qa,qb)\ \iff\ 1 = (b,q)$&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-28T02:15:51.310" LastActivityDate="2010-12-28T02:15:51.310" />
  <row Id="15490" PostTypeId="2" ParentId="15469" CreationDate="2010-12-25T06:10:32.523" Score="7" ViewCount="0" Body="&lt;p&gt;Recall that $F_{n+1}$ is the number of ways to tile a board of length $n$ with tiles of length $1$ and $2$.  So $F_{n+4}$ is the number of ways to tile a board of length $n+3$ with tiles of length $1$ and $2$.  Note that $n+3$ such tilings use at most one tile of length $2$, so $F_{n+4} - (n+3)$ such tilings use at least two tiles of length $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given such a tiling, look at where the second-to-last tile of length $2$ is used.  The part after this tile is a tiling of some section of length $k+1$ where exactly one tile of length $2$ is used (which can be done in $k$ ways), and the part before this tile is a tiling of the remaining portion of length $n-k$ (which can be done in $F_{n-k+1}$ ways).  Sum over $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The bigger lesson to take away here is that convolution is much easier to deal with than Hadamard product.  Also, since the bijection I described above preserves the number of tiles of each type, the identity can be upgraded to an identity of $q$-Fibonacci numbers.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-25T06:16:12.707" LastActivityDate="2010-12-25T06:16:12.707" CommentCount="2" />
  <row Id="15491" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T06:40:58.547" Score="0" ViewCount="0" Body="&lt;p&gt;Look at as many examples of what sets are compact and which ones are not, then you get a feeling of what compactness tries to encapsulate. Then try to see why the examples of compact sets you looked at are compact and why the non compact examples are not. Human mind learns a lot by comparing and mixing stuff, so just compare mix, add repeat.  &lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-25T06:40:58.547" />
  <row Id="15492" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T06:52:11.477" Score="7" ViewCount="0" Body="&lt;p&gt;Maybe you should think about compactness, as something that takes local properties to global properties. For example, if $f:K\rightarrow \mathbb{R}$ is continuous, $K$ is compact, and $f(x)&amp;gt;t_x&amp;gt;0$ for all x, then you can find $t&amp;gt;0$ such that $f(x)&amp;gt;t&amp;gt;0$ for all x - so from $f(x)&amp;gt;t_x&amp;gt;0$ point wise, you know that $f&amp;gt;t&amp;gt;0$ as a function. (This is a simple consequence of Weierstrass theorem in $[a,b]\rightarrow \mathbb{R}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually we find some property that is true for every &quot;small&quot; enough open sets, then use compactness to reduce the case to finitely many open sets and use induction to show that the property is true for all of the space. This is at least how I understand compactness.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As the commenters below this message wrote (and I didn't emphasize enough), we usually use compactness to reduce infinite problems\conditions\restraints to a finite subset that cover the entire space, and then use some argument that works only for finite cases (like induction, taking max\min, take finite sums etc). In my example above, we wanted to find a minimum over all the lower bounds, but in the infinite case this is usually just an infimum (and can be zero), but when we reduce to a finite case there is a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way we can think of compactness as something that let us use some finite argument on infinite covers (and many times to transfer some property from the cover to the entire space). &lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-12-26T06:38:06.327" LastActivityDate="2010-12-26T06:38:06.327" CommentCount="6" />
  <row Id="15493" PostTypeId="2" ParentId="15484" CreationDate="2010-12-25T06:53:58.300" Score="3" ViewCount="0" Body="&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if $\displaystyle f(x) = x^2 -3$, then $$\left|f\left(\dfrac{m}{n}\right)\right| \ge \dfrac{1}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, I guess Vinod isn't interested in following up on the hints, here is almost a full proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any rational number $\displaystyle m/n$, we have that $\displaystyle |3 - \frac{m^2}{n^2}|  = \frac{|m^2 - 3n^2|}{n^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $\displaystyle \sqrt{3}$ is irrational, $\displaystyle |m^2 - 3n^2| \ge 1$ (it is an integer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\displaystyle |3 - \frac{m^2}{n^2} | \ge \frac{1}{n^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\displaystyle |\frac{m}{n} + \sqrt{3}| \le 5$, then we have that $\displaystyle |\frac{m}{n} - \sqrt{3}| \ge \frac{1}{5n^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case $\displaystyle |\frac{m}{n} + \sqrt{3}| \ge 5$, is easier, as then $\displaystyle \frac{m}{n}$ will be not be very close to $\displaystyle \sqrt{3}$.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2011-01-03T02:25:17.317" LastActivityDate="2011-01-03T02:25:17.317" CommentCount="4" />
  <row Id="15494" PostTypeId="2" ParentId="13624" CreationDate="2010-12-25T07:01:29.440" Score="2" ViewCount="0" Body="&lt;p&gt;For 1., clearly yes: if a cover has a finite refinement, then the sets it is a refinement of also cover $X$, and form a finite subcover.&#xA;For 2., see Nuno's comment on the other answer.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastActivityDate="2010-12-25T07:01:29.440" />
  <row Id="15495" PostTypeId="1" CreationDate="2010-12-25T07:32:59.570" Score="0" ViewCount="41" Body="&lt;p&gt;From example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chevalley basis for L consist of E, H, F in Book Lie Groups : A problem-oriented ... Harriet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E := matrix([[0,0,1],[0,0,-i],[-1,i,0]]);&#xA;X := matrix([[0,1,0],[-1,0,0],[0,0,0]]);&#xA;ad := E&amp;amp;*X - X&amp;amp;*E;&#xA;evalm(ad);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;above can not get the answer below in maple&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is&#xA;ad(E) := matrix([[0,-1,0],[0,0,-2],[0,0,0]]);   // How to get this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover for H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H := matrix([[1,0,0],[0,0,0],[0,0,-1]]);&#xA;Y := matrix([[0,0,1],[0,0,0],[-1,0,0]]);&#xA;ad := H&amp;amp;*Y - Y&amp;amp;*H;&#xA;evalm(ad);&lt;/p&gt;&#xA;" OwnerUserId="5070" LastActivityDate="2010-12-25T07:32:59.570" Title="How to calculate ad(X) for exp(t*ad(X))?" Tags="&lt;lie-groups&gt;" CommentCount="1" />
  <row Id="15496" PostTypeId="2" ParentId="15426" CreationDate="2010-12-25T08:08:41.083" Score="0" ViewCount="0" Body="&lt;p&gt;I'm posting this as a separate answer so as to stop bothering Willie with incessant comment notifications. Of course, the other answers that have been posted are completely correct, and I'm only trying to help you reconcile your intuitive reasoning with their results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In situation A, the information you obtain about the state of the coins is, &quot;The first coin is heads.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In situation B, the information you obtain is, &quot;At least one of the coins is heads.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not equivalent; the former is not implied by the latter, because in B you cannot conclude that the first coin is heads. In other words, the observer's statement &lt;em&gt;in conjunction with your knowledge of the observer's strategy&lt;/em&gt; enables you to make a stronger conclusion about the coins in situation A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish, I can clarify (a) how the observer's strategy affects things, as TonyK has also mentioned, and/or (b) why in situation B the probability is 1/3 and not 1/2 as you expect.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-25T08:08:41.083" CommentCount="16" />
  <row Id="15497" PostTypeId="2" ParentId="15400" CreationDate="2010-12-25T08:12:52.537" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sinx-and-cosx/1093#1093&quot;&gt;I like to think of helices on cylinders as images of lines under planar curling&lt;/a&gt;, so here's an approach incorporating that idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the circle rolling up a line drawn in the plane, with $P$ the point of tangency along the line and $Q$ a distinguished point on the circumference of the circle. (So, $Q$ traces out a cycloid in the plane.) Now, imagine the plane is made of thin paper, but the circle is made of stiff cardboard. If we curl the plane into a cylinder but the circle remains flat, the plane of the circle will be tangent to the cylinder along the (vertical) line passing through the point $P$. The point $P$ will follow a helix, and the point $Q$ --which lies in the plane of the circle-- traces out the helical cycloid in space. While the path that $P$ (and $Q$) takes is decidedly different after curling than before, the &lt;em&gt;displacement vector&lt;/em&gt; between $P$ and $Q$ in the plane of the circle is the same throughout. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before curling, if the circle of radius $r$ rolls along a horizontal track in the $uv$-plane, then we have $P=(r t, 0)$, and the point $Q$ traces out the standard cycloid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;u &amp;amp;= r ( t - \sin t ) \\&#xD;&#xA;v &amp;amp;= r ( 1 - \cos t )&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement vector from $P$ to $Q$ at time $t$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d := PQ = [m, n] = r[-\sin t, 1 - \cos t ] = - 2 r \sin\frac{t}{2} \; [\cos\frac{t}{2},-\sin\frac{t}{2}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotating the plane through an angle, say, $\theta := \rm{atan\frac{c}{a}}$, and writing $b$ for $\sqrt{a^2+c^2}$, the circle's point of tangency along the tilted track is given by ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_2 = (r t \cos\theta,r t \sin\theta) = \left( \frac{r t a}{b}, \frac{r t c}{b}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and the displacement vector, $d_2$, to the point $Q_2$ tracing out the tilted cycloid is given by ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;d_2 :&amp;amp;= [m \cos\theta - n \sin \theta, m \sin\theta + n \cos \theta ] \\&#xD;&#xA;&amp;amp;= \frac{1}{b} [a m - c n, c m + a n] \\&#xD;&#xA;&amp;amp;= -\frac{2r}{b}\sin\frac{t}{2}\left[\cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] \\&#xD;&#xA;\end{align}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the fun part: Curl the $(u,v)$-plane around a cylinder of radius $a$, such that the $uv$-origin aligns with the $xyz$-point $(a,0,0)$ and the $v$-axis runs parallel to the $z$ axis. The curled, tilted track will coincide with a helix. The horizontal distance travelled by $P_2$ in the plane becomes the length of horizontal circular &lt;em&gt;arc&lt;/em&gt; travelled by $P_3$ around the cylinder; upon dividing by the radius, $a$, of the cylinder, this length becomes the &lt;em&gt;angular&lt;/em&gt; &quot;distance&quot; --$s := \frac{rt}{b}$-- travelled by $P_3$; the vertical distances match. Therefore: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_3 = \left(a\cos s, a\sin s, \frac{rtc}{b} \right)=(a\cos s,a\sin s,c s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the displacement vector: The $u$ direction of the tangent plane coincides with the horizontal vector tangent to the cylinder at $P_3$; the $v$ direction coincides with the $z$ direction. Thus, the transformation from $uv$-coordinates to $xyz$-coordinates is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[1,0]\to[-\sin s, \cos s, 0]\hspace{0.5in}[0,1]\to[0,0,1]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image, $d_3$, of the displacement vector $d_2$, then, is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_3 = -\frac{2r}{b}\sin\frac{t}{2}\left[-\sin s \cos\left(\theta-\frac{t}{2}\right), \cos s \cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the path of $Q_3$, which traces the helical cycloid, is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;Q_3 = P_3 + d_3 = \left\{&#xD;&#xA;\begin{align}&#xD;&#xA;x &amp;amp;= a \cos s &amp;amp;+ \frac{2 r}{b} \sin s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\&#xD;&#xA;y &amp;amp;= a \sin s &amp;amp;- \frac{2 r}{b} \cos s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\&#xD;&#xA;z &amp;amp;= c s &amp;amp;- \frac{2 r}{b} \sin\frac{b s}{2r}\sin\left(\theta-\frac{bs}{2r}\right)&#xD;&#xA;\end{align}&#xD;&#xA;\right.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture with $a=c=1$ and $r=1/2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/vIRtM.jpg&quot; alt=&quot;Helical Cycloid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The above does not simply curl the planar cycloid around the cylinder. Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 = a^2 + \frac{4r^2}{b^2} \sin^2\frac{bs}{2r} \cos^2\left(\theta-\frac{bs}{2r}\right) \ge a^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that most of the helical cycloid lies outside the surface of the cylinder.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2010-12-25T08:12:52.537" />
  <row Id="15498" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T09:17:41.053" Score="3" ViewCount="0" Body="&lt;p&gt;As I mentioned in &lt;a href=&quot;http://math.stackexchange.com/questions/9978/real-analysis-continuity&quot;&gt;another question&lt;/a&gt;, I tend to intuit compactness as a sort of &quot;super closedness.&quot;  There are a few reasons for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hausdorff_space&quot; rel=&quot;nofollow&quot;&gt;Hausdorff spaces&lt;/a&gt;, every compact subset will automatically be closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, in &lt;a href=&quot;http://en.wikipedia.org/wiki/First-countable_space&quot; rel=&quot;nofollow&quot;&gt;first-countable&lt;/a&gt; spaces, one can think about &quot;closedness&quot; as &quot;closed under limit operations.&quot;  That is, a set $E$ is closed if and only if it contains the limits of all sequences in $E$.  By analogy, every &lt;em&gt;compact&lt;/em&gt; subset of a first-countable space is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sequential_compactness&quot; rel=&quot;nofollow&quot;&gt;sequentially compact&lt;/a&gt; -- that is, every sequence has a convergent subsequence.  (Note, however, that the converse of this is not generally true in first-countable spaces.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, in metric spaces, compactness is equivalent to being complete and totally bounded.  This can be viewed as a generalization of the Heine-Borel Theorem which classifies the compact sets of $\mathbb{R}^n$.  In such a scheme, completeness corresponds to closedness, and total boundedness to boundedness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that metric spaces are automatically Hausdorff and first-countable, so all of the above certainly applies to metric spaces.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For general topological spaces (without any assumption of &quot;Hausdorffness&quot; or first countability), one can also intuit compact topological spaces as those with relatively few open sets.  Indeed, if a set $X$ is compact under a given topology, then it will also be compact under any weaker (coarser) topology.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: To add to Prometheus' answer, I would also say that compactness is a property that plays especially well with continuity (&lt;a href=&quot;http://en.wikipedia.org/wiki/Extreme_value_theorem&quot; rel=&quot;nofollow&quot;&gt;Extreme-Value Theorem&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Heine-Cantor_theorem&quot; rel=&quot;nofollow&quot;&gt;Heine-Cantor Theorem&lt;/a&gt;) and convergence of functions (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dini%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Dini's Theorem&lt;/a&gt;).  As Prometheus said, this is in part because of its role in reducing global problems to local ones -- or, perhaps, infinite ones to finite ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm thinking particularly of the proof of Heine-Cantor which relies on the fact that the infimum of a finite number of positive distances is positive -- whereas an infinite number of positive distances may have an infimum of zero.)&lt;/p&gt;&#xA;" OwnerUserId="640" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-25T09:27:51.997" LastActivityDate="2010-12-25T09:27:51.997" CommentCount="2" />
  <row Id="15499" PostTypeId="1" CreationDate="2010-12-25T09:53:40.260" Score="4" ViewCount="77" Body="&lt;p&gt;I recently read that the group of diffeomorphisms of a smooth manifold which preserve some sort of geometric structure (e.g. Riemannian structure, conformal structure, etc.) frequently turn out to be a Lie group.  What are some examples of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read about the euclidean group $E(n)$ which are the isometries of $\mathbb{R}^n$, and also about the conformal automorphisms of the complex plane, upper half plane, and unit disc.  What others are there?  Do the groups Diff(M), Iso(M), etc. frequently turn out to be something recognizable?  What are the isometry groups or conformal groups of n-spheres, say, or some common 2-manifolds?&lt;/p&gt;&#xA;" OwnerUserId="640" LastActivityDate="2010-12-26T11:25:43.560" Title="Examples of automorphism groups of manifolds which turn out to be Lie groups" Tags="&lt;reference-request&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15500" PostTypeId="1" CreationDate="2010-12-25T09:54:26.847" Score="-3" ViewCount="86" Body="&lt;p&gt;Dear Professors, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for any &lt;em&gt;&lt;strong&gt;reference articles on the uses of the preposition &quot;of&quot;&lt;/strong&gt;&lt;/em&gt; that are applied most often in mathematics. I once read in the teacher's edition of Pre-algebra: An Accelerated Course by Dolciani that students should be taught, as early as possible, the many uses of the preposition &quot;of&quot; in mathematics, but I have been unable to find anything on this particular question/topic. &lt;em&gt;&lt;strong&gt;Could you please help me find something? (books, articles, etc.)&lt;/strong&gt;&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: The primary tag for this thread is a &lt;strong&gt;REFERENCE-REQUEST&lt;/strong&gt;. Reference-requests are for questions that the poster wants to seek for references &lt;strong&gt;(books, articles, etc.)&lt;/strong&gt; in a subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far the only references that I have found can be viewed at these links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.gsu.edu/~eslhpb/grammar/lecture_11/of.html&quot; rel=&quot;nofollow&quot;&gt;Professor Pat Byrd&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=L0PNx4i8G3UC&amp;amp;lpg=PR12&amp;amp;ots=GIyxSXUvi-&amp;amp;dq=Martin%2C%20Richard%20Milton%27s%20chapter%20X%20OF%20%27OF%27%20in%20Pragmatics%2C%20truth%20and%20language&amp;amp;pg=PA130#v=onepage&amp;amp;q&amp;amp;f=true&quot; rel=&quot;nofollow&quot;&gt;Richard Milton Martin, Pragmatics, truth, and language, chapter 10, Of 'Of'&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, they are at opposite ends of the spectrum with reference to my question about the uses of &quot;of&quot; most often applied in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yours respectfully.&lt;/p&gt;&#xA;" OwnerUserId="5030" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2010-12-25T10:37:22.787" LastActivityDate="2010-12-25T10:37:22.787" ClosedDate="2010-12-25T13:09:09.860" Title="Uses of the prepostion &quot;of&quot; applied most often in math. " Tags="&lt;reference-request&gt;" CommentCount="6" />
  <row Id="15501" PostTypeId="2" ParentId="15486" CreationDate="2010-12-25T10:05:29.183" Score="-1" ViewCount="0" Body="&lt;p&gt;For me, the compactness of a topological space means that it has &lt;em&gt;enough points&lt;/em&gt; to provide exact solutions to continuous equations. More precisely,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;compactness = Any equation that can be approximated by a consistent system of $\leq$ inequalities of continuous functions has a solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For instance, being a solution to the equation $x^2 = 2$ is equivalent to being a simultaneous solution to the infinite system of inequalities $\lbrace |x^2-2| \leq \frac 1 n \rbrace_{n\in\mathbb N}$. This system is consistent in the sense that every &lt;em&gt;finite&lt;/em&gt; subsystem has solutions; after all, we can approximate $\sqrt 2$ by rational numbers arbitrarily well. Then, it is the &lt;em&gt;compactness&lt;/em&gt; of bounded intervals of &lt;em&gt;real numbers&lt;/em&gt; that tells us that there exists a simultaneous solution of all of these inequalities, and hence of the equation itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, trying to encircle $\sqrt 2$ as a &lt;em&gt;rational number&lt;/em&gt; fails, you can enclose it by smaller and smaller intervals, but when you try to catch it with an infinite number of intervals, it &quot;falls through the cracks&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle discussed above is also known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_intersection_property&quot; rel=&quot;nofollow&quot;&gt;finite intersection property&lt;/a&gt;. The inequalities, or &quot;conditions&quot;, define a collection of closed sets and the exact solution we seek is contained in the intersection of infinitely many closed sets. If every finite intersection of these closed sets is non-empty, i.e. if every finite collection of conditions can be fulfilled, then one could expect that all of the conditions can be satisfied simultaneously. This is precisely what compactness ensures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Peano's theorem about the existence of solutions to ordinary differential equations relies on compactness to construct a good candidate.&lt;/li&gt;&#xA;&lt;li&gt;A commonly used technique for showing the existence of solution for partial differential equations is the calculus variations. In this context, the weak compactness of the unit ball in a Hilbert space is often used to get a weak solution to the PDE which is then upgraded to a strong solution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditorDisplayName="" LastEditDate="2010-12-26T19:36:00.803" LastActivityDate="2010-12-26T19:36:00.803" CommentCount="5" />
  <row Id="15502" PostTypeId="2" ParentId="9999" CreationDate="2010-12-25T11:30:59.483" Score="0" ViewCount="0" Body="&lt;p&gt;To answer the other direction of the question, i.e. to show that $F_n, L_n$ are fundamental for the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{2n}^2 - 5 F_{2n}^2 = 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_{2n+1}^2 - 5 F_{2n+1}^2 = -4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we do not need to invoke the full theory of Pell equations, algebraic extensions, continued fraction, etc.. since we have some concrete parameters at hand. Lets turn to the Pell equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^2 - a x^2 = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $(x_1, y_1)$ and $(x_2,y_2)$ be solutions of this equation. We see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(y_1^2 - a x_1^2)*(y_2^2 - a x_2^2) = y_1^2 * y_2^2 - a y_1^2 x_2^2 - a x_1^2 y_2^2 + a^2 x_1^2 * x_1^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (y_1 y_2)^2 + 2 a y_1 y_2 x_1 x_2 + (a x_1 x_2)^2 - a ((y_1 x_2)^2 + 2 y_1 x_2 x_1 y_2 +(x_1 y_2)^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= (y_1 y_2 + a x_1 x_2)^2 - a (y_1 x_2 + x_1 y_2)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we can go to a third solution $(x_3,y_3)$ from the given two solutions as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_3 := y_1 x_2 + x_1 y_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_3 := y_1 y_2 + a x_1 x_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call this operations $\times$. It spans an abelian group with the following unit and inverse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0,1) \times (x,y) = (x,y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-x,y) \times (x,y) = (0,1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group has a t-norm $|(x,y)|$ in $Q$. We define for $x&amp;gt;0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|(x,y)| := 1/x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can show that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|p \times q| \leq |p| |q|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse of the t-norm can be used to show that if $p_0$ is the smallest non-unit solution than any non-unit solution $q$ has the form $p_0^n = p_0 \times ... \times p_0$. Just observe that for $q\neq (0,1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q=p_0 \lor |p_0^{-1} \times q|^{-1} \leq {|q|^{-1} \over |p_0|^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $|p_0|^{-1} \neq 1$ this gives a decreasing terminating chain. Now exhibiting $p_0$ we should be done more or less.&lt;/p&gt;&#xA;" OwnerUserId="4414" LastEditorUserId="4414" LastEditorDisplayName="" LastEditDate="2010-12-25T11:40:52.717" LastActivityDate="2010-12-25T11:40:52.717" CommentCount="3" />
  <row Id="15503" PostTypeId="1" AcceptedAnswerId="15544" CreationDate="2010-12-25T12:47:42.630" Score="2" ViewCount="87" Body="&lt;p&gt;I have a functor $F\colon \mathbf{Rng}\to\mathbf{Grp}$, and a correspondence on objects which assigns to every group $F(R)$ a suitable subgroup $G_o(R)\subseteq F(R)$. Is there a way to turn $G$ into a functor, defining $G_o(R)\to G_o(S)$ via the maps I have between $F(R)$ and $F(S)$?&#xA;$$&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;F(R) &amp;amp;\to^{F(f)}&amp;amp; F(S) \\\&#xD;&#xA;\uparrow_{\iota_R} &amp;amp;&amp;amp; \uparrow_{\iota_S}\\\&#xD;&#xA;G_o(R) &amp;amp; &amp;amp;G_o(S)&#xD;&#xA;\end{array}&#xD;&#xA;$$&#xA;In this diagram vertical arrows are simply the existing injections. I thought to define $G_o(R)\to G_o(S)$ taking the obvious left inverse going from the copy of $G_o(-)$ into $F(-)$ to $G_o(-)$, (call $\pi_S$ this map, thn $G(R)\to G(S)$ is defined by $\pi_s\circ F(f)\circ \iota_R$) but I'm not sure this is going to work...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think you'll find it useful, $F$ is the functor which assigns to every ring its group of unities.&lt;/p&gt;&#xA;" OwnerUserId="685" LastActivityDate="2010-12-26T08:59:12.163" Title="Defining &quot;subfunctors&quot;" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15504" PostTypeId="1" AcceptedAnswerId="15506" CreationDate="2010-12-25T16:27:36.727" Score="3" ViewCount="78" Body="&lt;p&gt;In my course there's a paragraph: Taylor polynomial with Lagrange remainder,&lt;br&gt;&#xA;The paragraph starts with a theorem (I left out the constraints):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ( \exists \theta \in ]0,1[)(f(a +h) = T_{f,a,n}(a + h) + \frac{h^{n+1}}{(n+1)!}D^{n+1}f(a + \theta h)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $f$ an  $R-R$ function and $a$ and $a + h$ defined over an interval $I$.&lt;br&gt;&#xA;And the only other thing in the paragraph is a proof of the above theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I'm starting with the proof, could someone please explain the above theorem in &lt;em&gt;human language&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand what a Taylor polynomial is, and what it's good for, but I can't turn this into anything that makes sense...&lt;/p&gt;&#xA;" OwnerUserId="5109" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-25T17:08:51.090" LastActivityDate="2010-12-26T10:04:38.440" Title="Taylor polynomial with Lagrange remainder" Tags="&lt;calculus&gt;&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;taylor-expansion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15505" PostTypeId="1" AcceptedAnswerId="15599" CreationDate="2010-12-25T16:30:34.860" Score="4" ViewCount="185" Body="&lt;p&gt;Show that $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the LHS $\binom{n+r-1}{r}$ counts the number of ways of selecting $r$ objects from a set of size $n$ where order is not significant and repetitions are allowed. So you have $n$ people you form $r$ teams and select $r$ captains and select $(n-r)$ players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RHS divides up a team into 2 sets? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T20:16:39.867" LastActivityDate="2010-12-27T00:35:22.657" Title="Combinatorial Identity" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="15506" PostTypeId="2" ParentId="15504" CreationDate="2010-12-25T17:02:22.470" Score="2" ViewCount="0" Body="&lt;p&gt;I think what you've got there is the $n$th-degree Taylor polynomial for a function $f$ centered at $a$ and being evaluated at $x=a+h$.  Basically, the error term $\frac{h^{n+1}}{(n+1)!}D^{n+1}f(a + \theta h))$ looks like the last term in the $(n+1)$st degree Taylor polynomial would, except that instead of evaluating the $(n+1)$st derivative at $a$, it's evaluated somewhere between $a$ and $a+h$ ($\theta$ measures how much of the way from $a$ to $a+h$).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-25T17:02:22.470" />
  <row Id="15507" PostTypeId="1" CreationDate="2010-12-25T17:22:17.720" Score="9" ViewCount="209" Body="&lt;p&gt;We did in class $x^2+y^2$, which was easy, and we had for homework $2x^2+2xy+3y^2$, which I did (its values (if not square) must be divisible by form primes, or of the form x^2+5y^2, and clearly they can't all be the latter!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to test my method I took the middle coefficient down by one - and ran into trouble...&lt;/p&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2010-12-25T17:22:17.720" Title="Elementary proof that 2x^2+xy+3y^2 represents infinitely many primes?" Tags="&lt;number-theory&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="15508" PostTypeId="1" CreationDate="2010-12-25T17:57:51.353" Score="2" ViewCount="98" Body="&lt;p&gt;Hey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got a quick question on riemannian geometry. I'm not quite sure whether this is the right place to ask this question, since it might be a rather elementary one from a research point of view. Nontheless I think it might be too advanced for a typical homework question (which by the way it isn't) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $M^m$ be a manifold embedded into euclidean space $R^n$. Let L($\gamma$) be the &#xA;the lenght of a smooth curve $\gamma$: [0,1] --&gt; $R^n$ which is the number &#xA;$\int_0^1 \ |d/dt \gamma(t)| \ \mathrm{d}t$ . Define the distance function&#xA;d: M x M --&gt; [0,$\infty$) by d(p,q):=inf L ($\gamma$). The infimum is taken over all smooth paths connecting p and q. This formula defines a metric on M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wan't to show that &#xA;$\forall$ $p_0 \in M^m \exists U \subset M^m $ open neighborhood of $p_0$ such that (1-$\epsilon$) |p-q| &amp;lt; d(p,q) &amp;lt; (1+$\epsilon$) |p-q| .&#xA;Note that |.| denotes the usual euclidean distance on $M^m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we fix an arbitrary $p_0 \in M^m$. Without loss of generality (Translations and rotations preserves the length of a curve) we assume that &#xA;$p_0$ = 0 and $T_{p_{0}}M$ = $R^m$ x {0} .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the step that I don't understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there's a smooth function f: $\Omega$ --&gt; $R^{n-m}$ defined on an open neighborhood $\Omega \subset R^m$ of the origin such that &#xA;{ (x,y) $\in R^m x R^{n-m}$ | x $\in \Omega$ , y=f(x) } $\subset M^m$  , f(0)=0, df(0)=0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I don't really see where this is coming from. &#xA;It seems to me that it might be some combination of a characterisation of manifolds&#xA;and tangent spaces we introduced, which is the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M^m \subset R^n$, p $\in M^m$ &quot;smooth manifold&quot; , then&#xA;$\exists U \subset R^n$ and f: U --&gt; $R^{n-m}$ smooth such that p $\in$ U , df(q) surjective for all q $\in$ U $\cap$ M and U $\cap$ M = $f^{-1}$(0). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5110" LastActivityDate="2010-12-25T17:57:51.353" Title="Quick question on Riemannian geometry" Tags="&lt;differential-geometry&gt;&lt;differential&gt;" CommentCount="5" />
  <row Id="15509" PostTypeId="1" AcceptedAnswerId="15511" CreationDate="2010-12-25T18:31:45.457" Score="3" ViewCount="104" Body="&lt;p&gt;For $n \geq 4$, show that $\binom{\binom{n}{2}}{2} = 3 \binom{n}{3}+ 3 \binom{n}{4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS: So we have a set of $\binom{n}{2}$ elements, and we are choosing a $2$ element subset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: We are choosing a $3$ element subset and a $4$ element subset (each from a set of $n$ elements). But we multiply by $3$ by the multiplication principle for some reason. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T20:13:13.113" LastActivityDate="2010-12-25T20:13:13.113" Title="Combinatorial Proof" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" />
  <row Id="15510" PostTypeId="1" CreationDate="2010-12-25T18:39:37.920" Score="14" ViewCount="229" Body="&lt;p&gt;What is the shortest string $S$ over an alphabet of size $n$, such that every permutation of the alphabet is a substring of $S$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of this problem while reading a open problem on &lt;a href=&quot;http://garden.irmacs.sfu.ca/?q=op/smallest_universal_supersequence&quot; rel=&quot;nofollow&quot;&gt;shortest supersequence of all permutations&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="96" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-25T21:39:50.200" LastActivityDate="2010-12-26T06:59:19.260" Title="What is the shortest string that contains all permutations of an alphabet?" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="15511" PostTypeId="2" ParentId="15509" CreationDate="2010-12-25T18:50:31.297" Score="7" ViewCount="0" Body="&lt;p&gt;LHS: The $\binom{n}{2}$ is the number of pairs you can form of n distinct elements, so the LHS counts the number of ways to choose two distinct pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: Notice that you can choose two pairs that have a common element (but only one). If the two pairs are disjoint, then you need to choose four elements and then ask how you pair them. If the pairs have a common element, then you need to choose only three elements and then choose which is the common element.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="2125" LastEditorDisplayName="" LastEditDate="2010-12-25T19:58:56.420" LastActivityDate="2010-12-25T19:58:56.420" CommentCount="2" />
  <row Id="15512" PostTypeId="2" ParentId="15510" CreationDate="2010-12-25T19:10:20.697" Score="3" ViewCount="0" Body="&lt;p&gt;Here are some thoughts: given a permutation $abcde$, you can generate all its cyclic rotations with no additional cost: $abcdeabcd$. Then you'd want to capitalize of the largest number symbols, so we repeat the deepest symbol, in this case $a$, complete the permutation with $e$, and generate all cyclic rotations: $abcdeabcdaebcda$. So far we've generated all cyclic rotations of $abcde$ and $bcdae$. Notice that the first $n-1$ characters are rotated. We continue to generate all cyclic rotations of $cdabe,dabce$, and then we have to bring back one symbol shallower: $\ldots dabcedabcadebcad$. So we moved from $dabce$ to $bcade$. This corresponds to rotation of the middle $n-2$ characters, followed by the usual rotation of $n-1$ characters. Now we get back to the earlier operation (rotation of $n-1$), applying occasionally the new rotation, until we get stuck; we would then need to do a rotation of the middle $n-3$ characters, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would conjecture that the preceding scheme is optimal; the diligent reader can recursively calculate its length.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-25T19:10:20.697" />
  <row Id="15513" PostTypeId="2" ParentId="15510" CreationDate="2010-12-25T20:29:45.660" Score="5" ViewCount="0" Body="&lt;p&gt;[NB] This argument as stated is incorrect, and the error is pointed out by Yuval in the comments below. One can indeed use the De Bruijn graph to get all permutations, exactly as I described below, but as Yuval's comment implies, one will also get plenty of non-permutations. The length of the string $S$ is not what my argument says  it is, and the problem of the MINIMAL LENGTH sequence seems to require a different argument. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;This is very dangerous because all those undeserved points might cause someone to dismiss what might be an interesting and difficult question.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since there are $n!$ permutations of $n$ symbols, the string $S$ must have length&#xA;at least $n!+n-1$. That is, there must be $n!$ starting positions for a  permutation, and the last starting position must be succeeded by $n-1$ symbols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is in fact such a string, which is obtained from an Eulerian path in the De Bruijn graph $B(n,n)$. This is a directed graph whose vertices are all strings of $n-1$ distinct symbols, with  a directed edge from $u$ to $v$ if there is a symbol $s$ such that $v$ has the form $u_1s$, where $u_1$ is $u$ with the left-most symbol deleted. We label the edge with $s$. Every vertex has the same in-degree and out degree so the graph has an eulerian path. The labels of the edges on any such path, together with the string where the path ends, gives a minimal sequence of the kind you want. See this Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Bruijn_sequence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/De_Bruijn_sequence&lt;/a&gt; for terminology, references and a nice explanation. &lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditorDisplayName="" LastEditDate="2010-12-26T06:59:19.260" LastActivityDate="2010-12-26T06:59:19.260" CommentCount="5" />
  <row Id="15514" PostTypeId="1" AcceptedAnswerId="15517" CreationDate="2010-12-25T20:38:32.623" Score="3" ViewCount="171" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I solve the following derivative question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = \sin(\arctan x) + \tan(\arcsin x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-26T00:13:27.530" LastActivityDate="2010-12-26T04:03:54.567" Title="How can I find the derivative of $y = \sin(\arctan x) + \tan(\arcsin x)$?" Tags="&lt;calculus&gt;&lt;trigonometry&gt;" AnswerCount="4" />
  <row Id="15515" PostTypeId="1" CreationDate="2010-12-25T20:44:52.837" Score="2" ViewCount="66" Body="&lt;p&gt;Show that $\binom{2n}{2} = 2 \binom{n}{2}+n^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS: This is the number of pairs of $2n$ distinct elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: We can rewrite this as $2 \binom{n}{2}+ \binom{n}{1} \binom{n}{1}$. So you can divide up the $2n$ distinct elements into two subsets of $n$ elements. Then pick an element from each of these subsets to form a pair. Or you can just choose $2$ elements from the $n$ element subset (and multiply by $2$ since order matters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct idea? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-25T23:14:27.703" Title="Combinatorial Identity and Counting" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15516" PostTypeId="2" ParentId="15514" CreationDate="2010-12-25T20:53:40.040" Score="0" ViewCount="0" Body="&lt;p&gt;What's your intended result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A derivative, i.e. $y&amp;#39;$? Use the chain rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $x$? Use $cos^2(x) = 1 - sin^2(x)$ and $tan(x) = sin(x)/cos(x)$ to simplify&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-12-25T20:53:40.040" />
  <row Id="15517" PostTypeId="2" ParentId="15514" CreationDate="2010-12-25T20:57:28.200" Score="8" ViewCount="0" Body="&lt;p&gt;It's not clear to me where the derivative comes in to your question, but to simplify $\sin(\arctan x)+\tan(\arcsin x)$, I'd suggest the following: since arctangent and arcsine both result in angles in the first or fourth quadrants (between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$), draw two diagrams showing those two quadrants, one showing an angle labeled $\arctan x$, the other showing an angle labeled $\arcsin x$.  Draw right triangles to the $x$-axis in each diagram and label the legs in terms of $x$; for example, if the angle is $\arctan x$, that suggests that the leg opposite the angle is $x$ and the leg adjacent to the angle is $1$ (make sure that the labeling makes sense for positive and negative values of $x$).  Use these diagrams to find simpler algebraic expressions for $\sin(\arctan x)$ and $\tan(\arcsin x)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt;:  Here's a bit more detail on part of the problem.  If you're still stuck, please comment and let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by drawing an angle in the proper quadrant(s).  $\arctan x$ is an angle in the first or fourth quadrant, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/HWKLT.png&quot; alt=&quot;diagram 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\tan^{-1}(x)$ is another notation for $\arctan x$.)  Either the black (1st quadrant) or the gray (4th quadrant) triangle could be the proper triangle, depending on whether $x$ is positive or negative.  Since the angle is $\arctan x$, the tangent of that angle is $x$.  The easiest way to make that happen is for the leg opposite the angle to have length $x$ and the leg adjacent to the angle to have length $1$.  Note that if $x$ is negative, the gray triangle is the correct triangle and in the sense of directed-distances in the coordinate plane, that leg should have &quot;negative&quot; length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/JyHko.png&quot; alt=&quot;diagram 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the lengths of 2 sides of a right triangle, you can use the Pythagorean theorem to find the length of the third side.  In this case, the unknown side is the hypotenuse and the directed-distance length of the hypotenuse should always be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MjwgA.png&quot; alt=&quot;diagram 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all three sides of the triangle labeled, you can use the triangle to write an expression for $\sin(\arctan x)$, which is the sine of the marked angle, so the ratio of the length of the opposite leg to the length of the hypotenuse, $\frac{x}{\sqrt{x^2+1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The procedure for the $\tan(\arcsin x)$ term is similar, but the labeling of the sides of the triangle is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were working with, for example, arccosine, the two triangles would be in the first and second quadrants because the range of the arccosine function is $0\le\arccos x\le\pi$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-25T22:23:11.743" LastActivityDate="2010-12-25T22:23:11.743" />
  <row Id="15518" PostTypeId="2" ParentId="15514" CreationDate="2010-12-25T21:15:12.593" Score="0" ViewCount="0" Body="&lt;p&gt;@Isaac &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your answer.&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="3669" LastEditorDisplayName="" LastEditDate="2010-12-25T22:36:08.107" LastActivityDate="2010-12-25T22:36:08.107" CommentCount="1" />
  <row Id="15519" PostTypeId="2" ParentId="15162" CreationDate="2010-12-25T21:29:36.747" Score="2" ViewCount="0" Body="&lt;p&gt;Asaf's answer explains that there is no set of all cardinals, and that under the axiom of choice, the cardinals are well-ordered, and so it is impossible to have a homomorphism as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What remains is to see whether it is possible, in some models where the axiom of choice fails, to have a homomorphism $p$ as you suggest, with range a collection of sets, no two of which have the same cardinality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually possible. In fact, it holds in models where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_determinacy&quot; rel=&quot;nofollow&quot;&gt;axiom of determinacy&lt;/a&gt; is true, but the argument is sophisticated, and follows from recent work that started with results of Alexander Kechris and Scot Adams on Borel equivalence relations. The original reference is &quot;Linear algebraic groups and countable Borel equivalence relations&quot;, Journal of the American Mathematical Society, Vol 13 (4), (2000) 909-943. This paper is explicitly about Borel equivalence relations, but their arguments hold in more general contexts, assuming that all sets of reals are &quot;sufficiently nice&quot; in ways that the axiom of determinacy guarantees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benjamin Miller suggested the following general statement (I believe in this specific form the result is due to him):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\le$ is an analytic (i.e., $\Sigma^1_1$) partial order on a Hausdorff space $X$, then there is a sequence of countable analytic equivalence relations $(E_x)_{x \in X}$ such that $$ x &amp;lt; y \Longrightarrow | X/E_x | &amp;lt; | X/ E_x \sqcup X/ E_y | = | X/ E_x \times X/E_y | = | X/ E_y | $$ for all $x, y \in X$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;em&gt;analytic&lt;/em&gt; means the continuous image of a Borel set. Without choice, an inequality $|A|&amp;lt;|B|$ simply means that there is an injection from $A$ into $B$ but not one from $B$ into $A$. $A\sqcup B$ is the disjoint union of $A$ and $B$; this is the canonical set associated to the cardinal sum $|A|+|B|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why this solves the problem is that the ordering in the space $A/\sim$ is analytic in this sense, and so we can assign to each class $[a]$ the set $p([a])=X/E_{[a]}$ for $(E_{[a]})_{[a]\in A/\sim}$ a sequence of countable equivalence relations as granted by the statement above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What follows is Ben's (very high level) sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In all of these results, the key is the use of what we call &quot;rigidity arguments&quot;, which provide us with aperiodic countable Borel equivalence relations $F_x$ equipped with ergodic, invariant Borel probability measures $m_x$ with the property that for all distinct $x, y \in X$, the pair $(F_x, m_x)$ is $F_y$-ergodic (i.e., if $C$ is a set of full $m_x$-measure and $f$ is an $m_x$-measurable homomorphism from $F_x\upharpoonright C$ to $F_y$, then there is a subset $B$ of $C$ with $m_x(C \setminus B) = 0$ such that $f[B]$ is contained in a single $F_y$-class).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Let $E_x$ denote the disjoint union of the equivalence relations of the form $F_y$, for $y \le x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) We have that $x \le y$ implies that the identity map is a reduction of $E_x$ to $E_y$, i.e., it gives us an injection of $X/E_x$ into $X/E_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) If $x \le y$ is false, however, then any homomorphism from $E_x$ to $E_y$ would restrict down to a homomorphism from $F_x$ to $E_y$. Ergodicity of $m_x$ would then give a set $C$ of full $m_x$-measure and some $z \le y$ such that the map in question is a homomorphism from $F_x\upharpoonright C$ to $F_z$. As $x\ne z$, the map must therefore concentrate $m_x$-almost everywhere on a single $F_z$-class, and this implies that the original map could not have been a reduction. Thus $X/E_x$ cannot inject into $X/E_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) From the construction one can directly argue that (for any $x$) $X/E_x \times X/E_x$ is in bijection with $X/E_x$. From this, it is straightforward to check (using &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem&quot; rel=&quot;nofollow&quot;&gt;Schroeder-Bernstein&lt;/a&gt;) that if $x\le y$ then $X/E_x\sqcup X/E_y$, $X/E_x\times X/E_y$, and $X/E_y$ are all in bijection with one another. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-25T21:29:36.747" CommentCount="1" />
  <row Id="15520" PostTypeId="1" AcceptedAnswerId="15733" CreationDate="2010-12-25T21:40:11.670" Score="3" ViewCount="90" Body="&lt;p&gt;First, define a probability distribution with density:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\newcommand\bm\boldsymbol&#xA;\newcommand{\mbl}{\mathbf{l}}&#xA;f(\mbl \mid \bm \alpha)&#xA;= \frac{1}{B(\bm \alpha + \bm 1)}\exp\left(\mbl \cdot \bm\alpha\right)&#xA;\end{align*}&#xA;where&#xA;$\mathbf l \in (-\infty,0]^k, \sum_i\exp(l_i) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble finishing this integral (to calculate the convolution):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\newcommand{\mba}{\mathbf{a}}&#xA;\newcommand{\mbb}{\mathbf{b}}&#xA;\newcommand{\mbl}{\mathbf{l}}&#xA;\newcommand{\mbp}{\mathbf{p}}&#xA;\newcommand{\mbv}{\mathbf{v}}&#xA;\newcommand{\mbx}{\mathbf{x}}&#xA;f(\mbl \mid \mba, \mbb)&#xA;&amp;amp;= \int_{l_1}^0 \dotsi \int_{l_k}^0 f(\mbx \mid \mba)f(\mbl-\mbx \mid \mbb)d\mbx \newline&#xA;&amp;amp;\propto \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot \mba + (\mbl-\mbx) \cdot \mbb)d\mbx \newline&#xA;&amp;amp;= \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot (\mba - \mbb) + \mbl \cdot \mbb)d\mbx \newline&#xA;&amp;amp;= \prod_{i=1}^k\frac{\exp(a_il_i) - \exp(b_il_i)}{a_i-b_i} ???&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After looking at mpiktas's answer, the final line should have been&#xA;\begin{align*}&#xA;&amp;amp;= \prod_{i=1}^k\frac{\exp(b_il_i) - \exp(a_il_i)}{a_i-b_i}&#xA;\end{align*}&#xA;which ensures that the terms are all positive.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="774" LastEditorDisplayName="" LastEditDate="2010-12-29T12:03:57.223" LastActivityDate="2010-12-29T12:03:57.223" Title="Convolution of probability densities.  Having trouble integrating over vector" Tags="&lt;probability-theory&gt;&lt;integral&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15521" PostTypeId="1" AcceptedAnswerId="15540" CreationDate="2010-12-25T21:55:28.620" Score="4" ViewCount="97" Body="&lt;p&gt;I was trying to solve a problem similar to the &quot;how many ways are there to make change for a dollar&quot; problem.  I ran across a &lt;a href=&quot;http://www.maa.org/features/mathchat/mathchat_4_19_01.html&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; that said I could use a generating function similar to the one quoted below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer to our problem (293) is the&#xA;  coefficient of x^100 in the reciprocal&#xA;  of the following:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(1-x)(1-x^5)(1-x^10)(1-x^25)(1-x^50)(1-x^100)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I must be missing something as I can't figure out how they get from that to 293.  Any help on this would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5111" LastActivityDate="2010-12-26T09:59:54.467" Title="Making Change for a Dollar (and other number partitioning problems)" Tags="&lt;generating-functions&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="15522" PostTypeId="2" ParentId="15521" CreationDate="2010-12-25T22:09:29.000" Score="2" ViewCount="0" Body="&lt;p&gt;You &quot;just&quot; have to follow the prescription:  find the formal power series (no need to think about convergence) that is defined and check the number that multiplies x^100.  There's a reason I put just in quotes.  There is no obvious route to 293 that I can see.  Mathematica can do it with just one command, but I can't get Alpha to do it.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-25T22:09:29.000" />
  <row Id="15523" PostTypeId="2" ParentId="15521" CreationDate="2010-12-25T22:10:25.853" Score="1" ViewCount="0" Body="&lt;p&gt;I think you calculate $[x^{100}](1-x)^{100}(1-x^5)^{20}(1-x^{10})^{10}(1-x^{25})^4(1-x^{50})^2(1-x^{100})$, but that calculation seems to be brute force.&lt;/p&gt;&#xA;" OwnerUserId="774" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-26T00:16:17.030" LastActivityDate="2010-12-26T00:16:17.030" />
  <row Id="15524" PostTypeId="1" AcceptedAnswerId="15525" CreationDate="2010-12-25T22:36:48.340" Score="4" ViewCount="130" Body="&lt;p&gt;I'm not sure how well known this &quot;riddle&quot; is but here it goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 people go to a restaurant, each buy food worth 10.00.  When they're done, they give 30.00 to the waitress.  She gives the money to the manager, manager says they paid too much, gives 5.00 back to the waitress.  She comes back to the table, gives a dollar back to each person and then puts the remaining 2.00 in her pocket for tip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, each person has paid 9.00.  9.00 x 3 = 27.00.  Plus the 2.00 in the waitresses pocket is 29.00.  What happened to the 30th dollar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the issue with this way of calculating (since if you backward it works fine), that doesn't give us the correct result?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2010-12-25T23:24:29.567" Title="Riddle (simple arithmetic problem/illusion)" Tags="&lt;arithmetic&gt;&lt;riddle&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15525" PostTypeId="2" ParentId="15524" CreationDate="2010-12-25T22:50:53.427" Score="3" ViewCount="0" Body="&lt;p&gt;I am not sure how 'mathematical' the riddle part of this story is. :) Each guest paid 9 because together they paid 30, and got back 27. Of those 27, the manager got 25, and the waitress got the other 2 -- there is no sense/reason in adding 27 to 2 -- though you might have a gut feeling that you are headed in the 'right direction' of getting the initial 30 that way. But take a look -- there really is no extra dollar left, that's all it should be -- 25+2. In other words, each guest paid 9 dollars with the tip included into that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, possibly more insightful way to look at things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guests initially paid 30 dollars. The waitress returned them 3 dollars. So the guests ended up paying 27 (not thirty!). Of those 27 dollars, 2 dollars were pocketed as a tip by the waitress, so you could also say that they paid 8.333... each and then added a two dollar tip. I think it's a question of order of operations (and using it with respect to appropriate quantities) that could be confusing here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Punchline&lt;/strong&gt;: the waitress' 2 dollars were part of 27 dollars paid by the customers. They saved the other 3 dollars of the initial 30  because of what the manager said, so if you are wondering 'what happened to 30?' it's more of 27+3, or 25+2+3, where 25 is what they paid without a tip, 2 is the tip, and 3 is the amount they saved.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-25T23:24:29.567" LastActivityDate="2010-12-25T23:24:29.567" CommentCount="1" />
  <row Id="15526" PostTypeId="2" ParentId="9999" CreationDate="2010-12-25T23:04:14.250" Score="0" ViewCount="0" Body="&lt;p&gt;On another note, many people have mentioned the Binet formula. There is a nice way to see how this formula comes about. The Fibonacci sequence is a recurrence relation, so we can apply difference calculus to the sequence. In my advanced discrete mathematics course, we looked at the Fibonacci sequence, naturally. You can see the notes from the day we derived a formula for it[1]. It is on page 1 and page 2 of [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good text book to take a look at with regards to this is [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if I can help you anymore!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Kelley, W. &amp;amp; Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2010-12-25T23:04:14.250" />
  <row Id="15527" PostTypeId="2" ParentId="15515" CreationDate="2010-12-25T23:07:32.113" Score="0" ViewCount="0" Body="&lt;p&gt;$2n \choose 2$ is the sum of numbers from $1$ to $2n-1$, which is the sum of even numbers from $2$ to $2(n-1)$, which is $2{n\choose 2}$, and odd numbers from $1$ to $2n-1$. The latter is of course $n^2$.&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-25T23:07:32.113" />
  <row Id="15528" PostTypeId="2" ParentId="15514" CreationDate="2010-12-25T23:11:01.953" Score="1" ViewCount="0" Body="&lt;p&gt;First you want to simplify $\tan (\arcsin x)$ and $\sin(\arctan x)$. Let me show&#xA;you the latter one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\theta=\arctan x$. Then by definition $-\pi/2&amp;lt;\theta&amp;lt;\pi/2$ and $\tan\theta=x$.&#xA;Since $\sec \theta&amp;gt;0$ (because $-\pi/2&amp;lt;\theta&amp;lt;\pi/2$), and $$\sec^2\theta=1+\tan^2 \theta =1+x^2,$$ you get&#xA;$\sec \theta=\sqrt{1+x^2}$, so $\cos \theta =1/\sqrt{1+x^2},$ and &#xA;$$\sin\theta=\tan\theta \cos\theta =\frac{x}{\sqrt{1+x^2}},$$&#xA;i.e. $\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you can prove $\tan (\arcsin x)=\frac{x}{\sqrt{1-x^2}}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using quotient rule, you can prove that the derivative of $\frac{x}{\sqrt{1+x^2}}$&#xA;and $\frac{x}{\sqrt{1-x^2}}$ are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1+x^2)^{3/2}} \mbox{ and }\frac{1}{(1-x^2)^{3/2}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. Hence the derivative of  your function is $$\frac{1}{(1+x^2)^{3/2}}+\frac{1}{(1-x^2)^{3/2}}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf Edit.}$ Or you can use chain rule to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(\arctan x)\frac{1}{1+x^2}+\sec^2(\arcsin x)\frac{1}{\sqrt{1-x^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now simplify it.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-26T04:03:54.567" LastActivityDate="2010-12-26T04:03:54.567" />
  <row Id="15529" PostTypeId="2" ParentId="15515" CreationDate="2010-12-25T23:14:27.703" Score="1" ViewCount="0" Body="&lt;p&gt;Looks good to me.  Also you could just work numerically: ${{2n} \choose 2}=\frac{1}{2} \times 2n(2n-1)=n(2n-1)=2n^2-n=n(n-1)+n^2=2{n \choose 2}+n^2$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-25T23:14:27.703" CommentCount="1" />
  <row Id="15530" PostTypeId="1" CreationDate="2010-12-25T23:23:43.797" Score="-1" ViewCount="129" Body="&lt;p&gt;hi, i am trying to figure a algorithm problem that determines the two largest number of the a series of random 10 numbers. Thanks to your help.&lt;/p&gt;&#xA;" OwnerUserId="5112" LastActivityDate="2010-12-26T19:06:05.093" Title="algorithm of determines the two largest of a series" Tags="&lt;algorithms&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="15531" PostTypeId="2" ParentId="15530" CreationDate="2010-12-25T23:47:44.810" Score="5" ViewCount="0" Body="&lt;p&gt;Just go through the list and keep track of the two largest numbers, that is have two variables that are the the two largest elements so far while going through.&lt;/p&gt;&#xA;" OwnerUserId="5113" LastActivityDate="2010-12-25T23:47:44.810" CommentCount="2" />
  <row Id="15532" PostTypeId="2" ParentId="15530" CreationDate="2010-12-26T00:07:43.527" Score="3" ViewCount="0" Body="&lt;pre&gt;&lt;code&gt;list = [random, random, random, ..., random]&#xA;largest := (list[0] &amp;gt; list[1]) ? list[0] : list[1]&#xA;secondlargest := (list[0] &amp;lt;= list[1]) ? list[0] : list[1]&#xA;for each num in list:&#xA;    if num &amp;gt;= largest:&#xA;        secondlargest := largest&#xA;        largest := num&#xA;    elseif num &amp;gt;= secondlargest:&#xA;        secondlargest := num&#xA;return {largest, secondlargest}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You aren't going to be able to avoid iterating across the entire list because it is not sorted, and therefore you are going to have to search in $O(n)$ time. If the list is sorted, it becomes pretty much a non-problem, but there is no sense in sorting it if all you want is the two largest (if you have to do other things it might be worth it though) &lt;/p&gt;&#xA;" OwnerUserId="5038" LastActivityDate="2010-12-26T00:07:43.527" CommentCount="4" />
  <row Id="15533" PostTypeId="1" AcceptedAnswerId="15551" CreationDate="2010-12-26T00:11:33.337" Score="3" ViewCount="154" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the definition of a small class?  Is there no such thing as a set which contains other sets such as {{1,2},{1},{2}, 1, 2} (ie, is this really called a &quot;class&quot;)?  What are some prototypical examples of small classes?  What are some prototypical examples of large classes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; A few things I've been paging through (Rotman's homological algebra, a few papers which talk about categories, etc.) talk about small classes using definitions which are slightly confusing to me.  Most of them note that a &quot;class&quot; is a &quot;collection&quot; of &quot;sets&quot;, and that a small class is a class which is a set.  I'm not exactly sure where &quot;small&quot; ends and &quot;large&quot; begins in this respect.  See the specific questions below for what I mean by this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related Specific Questions:&lt;/strong&gt; If we use the notation {a,b,c} for a set with the elements 'a', 'b', and 'c', then is {a,b,c} a small class because it IS a set?  Is {{a,b,c}} a large class because it is a &quot;set which contains a set&quot;?  Is the set of real numbers a small class?  Is the set of natural numbers a small class?  Is the set of all functions from the real numbers to the real numbers a small class?  Is the class of all small classes a large class?&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2010-12-26T10:40:33.417" Title="Defining &quot;Small Classes.&quot;" Tags="&lt;set-theory&gt;&lt;examples&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15534" PostTypeId="2" ParentId="15530" CreationDate="2010-12-26T00:39:11.080" Score="1" ViewCount="0" Body="&lt;p&gt;It might be worth noting that the naive way to find the $k$ largest elements requires $O(nk)$ operations, since updating the list of $k$ largest elements could take $O(k)$ operations. However, if these are stored in a heap then we get the better complexity $O(n \log k)$; this method is practical (i.e. actually outperforms the naive algorithm) for moderately sized $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Debanjan's method shows how to solve this using $O(n + k\log k)$ operations, or $O(n)$ if we're not interested in the relative order of the $k$ largest elements. First you find the $k$th largest element using the classical $O(n)$ algorithm, and then you collect all elements at least as large as it is; if required, you then sort them.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-26T19:06:05.093" LastActivityDate="2010-12-26T19:06:05.093" CommentCount="1" />
  <row Id="15535" PostTypeId="2" ParentId="15533" CreationDate="2010-12-26T01:21:50.793" Score="2" ViewCount="0" Body="&lt;p&gt;As I understand it, the terms &quot;set&quot; and &quot;small class&quot; are equivalent.  Certainly a set may have other sets as elements, so {{1,2},{1},{2},1,2} is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally, a class is a collection of sets which can be &quot;unambiguously defined by a property that all its members share,&quot; to quote Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &lt;em&gt;formalize&lt;/em&gt; the notion of a large class -- that is, a class which is not small -- requires a choice of set theory that allows the formulation of such a concept.  The standard framework of the Zermelo-Fraenkel axioms (ZF) does &lt;em&gt;not&lt;/em&gt; have a formal notion of a large class: everything in consideration in that theory is a set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some standard examples of large classes are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the class of all sets&lt;/li&gt;&#xA;&lt;li&gt;the class of all groups&lt;/li&gt;&#xA;&lt;li&gt;the class of all topological spaces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are large classes because they are not small.  For example, the class of all sets is not itself a set: &lt;a href=&quot;http://en.wikipedia.org/wiki/Russel%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;Russell's paradox&lt;/a&gt; can be thought of as a proof of this statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your specific questions in light of the above: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;{a,b,c} is a small class, because it is a set.  &lt;/li&gt;&#xA;&lt;li&gt;{{a,b,c}} is a small class, because it is a set.&lt;/li&gt;&#xA;&lt;li&gt;Each of the set of real numbers and the set of natural numbers is a small class.&lt;/li&gt;&#xA;&lt;li&gt;The class of all small classes (i.e. the class of all sets) is a large class, as above.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5116" LastActivityDate="2010-12-26T01:21:50.793" />
  <row Id="15536" PostTypeId="1" AcceptedAnswerId="15539" CreationDate="2010-12-26T01:33:30.293" Score="2" ViewCount="54" Body="&lt;p&gt;How to input such system &#xA;&lt;img src=&quot;http://i.imgur.com/yFGkB.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;into maple so that it would solve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW we some part of system will be given to us,, which we do not know certently - may be some Ai's may be some  Bi's as constants. we do not know how much constants we will get. And we would need from given constants get a soltion for other a's and b's. in real life we will be provided with s as a constant but I hoped for general solution=) &lt;/p&gt;&#xA;" OwnerUserId="2065" LastEditorUserId="2065" LastEditorDisplayName="" LastEditDate="2010-12-26T13:36:44.297" LastActivityDate="2010-12-26T13:36:44.297" Title="Maple: How to input such system so that it would be solved?" Tags="&lt;computer-algebra-systems&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15537" PostTypeId="2" ParentId="14311" CreationDate="2010-12-26T01:45:17.763" Score="1" ViewCount="0" Body="&lt;p&gt;jerr18 ---&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the fact that my posted &quot;answer&quot; was clearly indicated to be a &quot;comment&quot; ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you do not understand what a proof is. As a writer: a proof is what makes me believe a result. As a reader whatever you wish to accept is a proof. A counterexample is what is needed to show a proof is defective. Sometimes a believer will use a comment to provide a counterexample. (I am not asking that you make me disbelieve my comment/proof by providing a counter example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question, I believe that the empty language is poly for any alphabet and it satisfies your requirement that every terminal appears an even number of times. I suppose your question should be reposed to conform to the effort that Yuval Filmus put forth.&lt;/p&gt;&#xA;" OwnerUserId="5055" LastActivityDate="2010-12-26T01:45:17.763" CommentCount="4" />
  <row Id="15538" PostTypeId="2" ParentId="15533" CreationDate="2010-12-26T01:50:07.037" Score="1" ViewCount="0" Body="&lt;p&gt;You already know the definition of a &quot;small class&quot;, it is what you gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be a class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then X is a set if and only there exists a class Y such that X is an element of Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let X be a class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then X is a set if and only if either/both of the following equivalent conditions hold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There exists an ordinal alpha such that X is an element of V_alpha.&lt;/li&gt;&#xA;&lt;li&gt;There exists an ordinal alpha such that X is a subclass of V_alpha.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(ordinal is defined at en.wikipedia.org/wiki/Ordinal_number,&#xA;&lt;br&gt;&#xA;V_alpha is defined at en.wikipedia.org/wiki/Von_neumann_universe)&lt;/p&gt;&#xA;" OwnerUserId="5114" LastActivityDate="2010-12-26T01:50:07.037" CommentCount="5" />
  <row Id="15539" PostTypeId="2" ParentId="15536" CreationDate="2010-12-26T03:01:22.477" Score="3" ViewCount="0" Body="&lt;p&gt;Apparently you want your solution be parametrized in $k,s$. I don't think Maple can&#xA;do that. What you need to do is to solve this linear system for some particular&#xA;$k, s$ and hopefully you can discern a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can solve linear systems using the linear algebra package: &gt;with(linalg);&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-26T03:01:22.477" />
  <row Id="15540" PostTypeId="2" ParentId="15521" CreationDate="2010-12-26T03:25:02.650" Score="3" ViewCount="0" Body="&lt;p&gt;You should be able to compute it using a Partial Fraction representation (involving complex numbers). For instance see this previous answer: &lt;a href=&quot;http://math.stackexchange.com/questions/12273/minimum-multi-subset-sum-to-a-target/12284#12284&quot;&gt;Minimum multi-subset sum to a target.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, this partial fraction expansion needs to be calculated only &lt;em&gt;one&lt;/em&gt; time. Once you have that, you can compute the way to make change for an arbitrary amount pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I doubt they really did that for finding the coefficient of $x^{100}$. It is probably quicker to just multiply out, ignoring the terms which would not contribute to the coefficient of $x^{100}$. Or you could try computing the partial fraction representation of only some of the terms and then multiply out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, if you are multiplying out to find the coefficient of $x^{100}$, it would be easier not to go to the reciprocal, which arises from considering an infinite number of terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need to multiply out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\sum_{j=0}^{100} x^j)\ (\sum_{j=0}^{20} x^{5j})\ (\sum_{j=0}^{10} x^{10j})\  (\sum_{j=0}^{4} x^{25j})\ (1 + x^{100})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which would amount to enumerating the different ways to make the change (and in fact is the the way we come up with the generating function in the first place).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could potentially do other things, like computing the $100^{th}$ derivative at $0$, or computing a contour integral of the generating function divided by $x^{100}$, but I doubt they went that route either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-26T03:25:02.650" CommentCount="1" />
  <row Id="15541" PostTypeId="1" AcceptedAnswerId="15581" CreationDate="2010-12-26T04:52:38.150" Score="4" ViewCount="56" Body="&lt;p&gt;This is a problem from Arnold's book on ODEs I cannot solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that if $f:U\to V$ is a diffeomorphism, then the Euclidean spaces with the domains $U$ and $V$ as subsets have the same dimension.&#xA;Hint. Use the implicit function theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-26T21:26:11.467" Title="invariance of dimension under diffeomorphism of real subspaces" Tags="&lt;real-analysis&gt;&lt;multivariable-calculus&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15542" PostTypeId="1" AcceptedAnswerId="15615" CreationDate="2010-12-26T05:14:51.390" Score="6" ViewCount="216" Body="&lt;p&gt;On page 18 of Rudin's &lt;em&gt;Real and Complex&lt;/em&gt; analysis he defines $0 \cdot \infty = 0$ and says that &quot;with this definition the commutative, associative, and distributive laws hold in $[0,\infty]$ without any restriction&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not clear to me is whether the quoted statement is a justification of the definition or just a consequence. Wouldn't the commutative, associative, and distributive laws also hold if we define $0 \cdot \infty = \infty$? &lt;/p&gt;&#xA;" OwnerUserId="1195" LastActivityDate="2010-12-27T05:22:21.563" Title="Arithmetic on $[0,\infty]$: is $0 \cdot \infty = 0$ the only reasonable choice?" Tags="&lt;real-analysis&gt;&lt;number-systems&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15543" PostTypeId="2" ParentId="11763" CreationDate="2010-12-26T06:40:51.440" Score="2" ViewCount="0" Body="&lt;p&gt;One way to think about this is to look at the chain level. Recall that the singular chains of a pair $C_n(X, Y)$ are sometimes defined as $C_n(X,Y) = C_n(X)/C_n(Y)$. So one way of thinking about this is that we are taking the space $X$ and attaching a cone to $Y$ so that any chain that lives in $Y$ can be shrunk to a point and forgotten about. Relative homology is then defined to be the relative cycles modulo the relative boundaries, i.e. we look at cycles AND chains living in $Y$, and we mod out not only by things that are boundaries, but things that are boundaries plus maybe a chain in $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This intuition works especially well for nice pairs $(X,Y)$ where $H_n(X,Y) = H_n(X/Y)$. In other words, we really are forgetting about all the stuff happening in $Y$ since we are shrinking it to a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wonderful treatment of all of this is given in Hatcher's book- he's very good about providing intuition.&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-12-26T06:40:51.440" />
  <row Id="15544" PostTypeId="2" ParentId="15503" CreationDate="2010-12-26T08:59:12.163" Score="1" ViewCount="0" Body="&lt;p&gt;Using only the things you have written, your question does not have an answer. In general there may not be any way to complete the commutative diagram you drew into a square (and in general there may not be any projection $\pi_S$, but this is a smaller problem). The basic problem is that you have only specified what $G_0$ does on objects, so as it stands $G_0$ is not even a functor (and in particular not a subfunctor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the example where $F$ is the group of units functor, as above. Let $G_0(R)$ be the whole group of units whenever $R$ is infinite, and the trivial group when $R$ is finite. Then you will see that there is in general no way to fill in the square above for e.g. the map $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately the condition that $G_0$ should be a subfunctor of $F$ means that you have very little choice in how you define $G_0(f)$, for a morphism $f$. The fact that the diagram above should be commutative &lt;em&gt;forces&lt;/em&gt; you to choose $G_0(f)$ to be equal to the restriction of $F(f)$ to $G_0(R)$. What you then need to check, in order for this to be well defined, is that this restriction always lands inside $G_0(S)$. (This fails in the example of the previous paragraph.) Whenever this condition holds, $G_0$ automatically becomes a functor using only that $F$ is a functor (check this).&lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2010-12-26T08:59:12.163" CommentCount="1" />
  <row Id="15545" PostTypeId="2" ParentId="15486" CreationDate="2010-12-26T08:59:27.530" Score="3" ViewCount="0" Body="&lt;p&gt;For nice enough spaces, e.g. metric spaces, compactness is equivalent to sequential compactness. I like to think of sequential compactness as saying that sequences of points can't run off to &quot;infinity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-12-26T08:59:27.530" CommentCount="1" />
  <row Id="15546" PostTypeId="2" ParentId="11763" CreationDate="2010-12-26T09:13:19.733" Score="3" ViewCount="0" Body="&lt;p&gt;See page 115 of Hatcher: Elements of $H_n(X,A)$ are represented by $n$-chains $\alpha \in C_n(X)$ such that $\partial \alpha \in C_{n-1}(A) \subset C_{n-1}(X)$. So you can think of an element of $H_n(X,A)$ as being an $n$-thing in $X$ whose boundary (an $(n-1)$-thing) lies in $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the relative homology groups fit into the long exact sequence&#xA;$$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \to \cdots,$$&#xA;you can think of them as giving a crude measurement of, e.g., the non-surjectiveness or non-injectiveness of the maps $H_n(A) \to H_n(X)$.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-12-26T09:13:19.733" />
  <row Id="15547" PostTypeId="2" ParentId="9186" CreationDate="2010-12-26T09:31:35.473" Score="2" ViewCount="0" Body="&lt;p&gt;I detect that you're really asking about Lie groups and Lie algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the theory of Lie groups and Lie algebras can be done algebraically or formally. Analysis only really comes in when you start to talk about convergence -- such as that of $\operatorname{exp} : \mathfrak{g} \to G$. If you don't care about convergence, you can still treat $\operatorname{exp}$ formally. Or, if $\mathfrak{g}$ is nice enough, say nilpotent, then everything works out -- you get convergence of $\operatorname{exp}$ for free.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2010-12-26T09:31:35.473" CommentCount="2" />
  <row Id="15548" PostTypeId="2" ParentId="15521" CreationDate="2010-12-26T09:59:54.467" Score="1" ViewCount="0" Body="&lt;p&gt;We can ease the calculation by noting that the number of ways of changing 100 equals the number of ways of representing the numbers less than or equal to $100$ as the sum of the numbers $5, 10, 25, 50$ and $100$, since the pennies can make up any remaining difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that all these number are divisible by $5$ we can conclude that the number of ways of representing $100$ in units of $1, 5, 10, 25, 50$ and $100$ is the sum of the coefficients up to and including the term in $x^{20}$ in the expansion of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{(1-x)(1-x^2)(1-x^5)(1-x^{10})(1-x^{20})} . $$&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-26T09:59:54.467" CommentCount="1" />
  <row Id="15549" PostTypeId="2" ParentId="15504" CreationDate="2010-12-26T10:04:38.440" Score="2" ViewCount="0" Body="&lt;p&gt;This formula is not meant to be a terribly complicated way to compute $f(a+h)$ (e.g. $\log(1+h)$) $exactly$, but it tells you the following: If you compute the value $T_{f,a,n}(a+h)$ (which is a simple polynomial value) instead you commit an error which is small because (i) $h^{n+1}$ might be very small, (ii) ${1\over(n+1)!}$ is very small and (iii) hopefully the $(n+1)^{\rm st}$ derivative of $f$ is bounded by a constant throughout, so that it doesn't matter where the point $a+\theta h$ guaranteed by the theorem lies.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-26T10:04:38.440" />
  <row Id="15550" PostTypeId="2" ParentId="15530" CreationDate="2010-12-26T10:35:54.740" Score="5" ViewCount="0" Body="&lt;p&gt;What you are looking for can be solved in $O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called finding the k-th order statistic.The Wikipedia entry of &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_algorithm&quot; rel=&quot;nofollow&quot;&gt;Selection algorithm&lt;/a&gt; provides some insight.But probably a better discussion is presented in &lt;strong&gt;Chapter 9, Medians and Other statistics from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0262032937&quot; rel=&quot;nofollow&quot;&gt;Introduction to Algorithms, 2nd Ed&lt;/strong&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a well described randomized $O(n)$ approach : &lt;a href=&quot;http://pine.cs.yale.edu/pinewiki/QuickSelect&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;QuickSelect&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-26T13:30:36.513" LastActivityDate="2010-12-26T13:30:36.513" CommentCount="2" />
  <row Id="15551" PostTypeId="2" ParentId="15533" CreationDate="2010-12-26T10:40:33.417" Score="1" ViewCount="0" Body="&lt;p&gt;The sets that we consider are actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Hereditary_set&quot; rel=&quot;nofollow&quot;&gt;hereditary sets&lt;/a&gt; meaning that all of their members are themselves hereditary sets.  What this informally means is that a hereditary set is a set of sets of sets of sets, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the empty set ($\emptyset$) is a hereditary set because it has no members so all of its members are vacuously hereditary sets.  Then $\{\emptyset\}$ is a hereditary set because its only member is a hereditary set.  Then $\{\emptyset, \{\emptyset\}\}$ is a hereditary set because both of its members are hereditary sets, and we can continue this process of tacking on the previous hereditary set as a member ad infinitum to get an infinite collection of hereditary sets.  This process describes how the Natural numbers are constructed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = \emptyset$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \{\emptyset\} = \{0\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = \{\emptyset, \{\emptyset\}\} = \{0, 1\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} = \{0, 1, 2\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vdots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n + 1 = \{0, 1, 2, \ldots, n\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the axiom of infinity, there exists an inductive set (set that contains $0$ and the successor of any of its members) so the class of Natural numbers is actually a hereditary set.  The integers can then be  constructed from the Natural numbers by considering equivalence classes of ordered pairs of Natural numbers where $(a, b)$ is equivalent to $(c, d)$ exactly when $a + d = b + c$ so $(a, b)$ will correspond to the integer $a - b$.  Ordered pairs of hereditary sets are also hereditary sets because $(a, b)$ can be formalized as $\{\{a\}, \{a, b\}\}$.  Consequently, the class of integers will also be a hereditary set.&#xA;The set of Rational numbers can then be defined as equivalence classes of ordered pairs of integers so it too will be a hereditary set.  The Real numbers, which can be constructed by considering select subsets of Rational numbers (or elements from the powerset of the Rational numbers), is also a hereditary set.  Once we formalize the Real numbers as a hereditary set, we can show that a function from the Reals into the Reals, which is a collection of ordered pairs of Reals, is a hereditary set.  Then we can show that the collection of all functions from the Reals into the Reals is a hereditary set.  In short, all of the natural objects would be &quot;small classes&quot; because they can be formed from the axioms of set theory (ZFC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that what determines whether a class is &quot;small&quot; or not is not whether it contains elements that look like sets such as $\{1, 2, 3\}$ or $\{a, b, c\}$ because we mostly restrict ourselves to hereditary sets anyway.  Instead, we have as a minimal requirement for smallness that the objects can either be provably constructible from the axioms of ZFC &lt;em&gt;or&lt;/em&gt; the assumption of their existence as sets does not lead to a contradiction in ZFC.  I include the first possibility here because we cannot prove (within ZFC) that the axioms of ZFC won't themselves lead to a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of what we would consider a large class is the class of all small classes or the class of all sets.  This is because such a class cannot be constructed from the axioms of ZFC, and it provably leads to a contradiction from ZFC (see Russell's Paradox from Brad's post).  In a nutshell, if we called the class of all sets $X$ and $X$ were a set, then $A = \{x \in X| x \notin X\}$ would have to be a set by the comprehension axiom.  But then $X \in X$ since $X$ is the class of all sets so $X \in A$ if and only if $X \notin A$, an immediate contradiction.  Other examples of &quot;large classes&quot; include the class of ordinals or as Ricky mentioned any object that does not fall into the $V$ hierarchy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing I should mention here is that you should always be aware of the context when considering whether an object is a large class (also known as proper class) or not.  Specifically, most of the natural objects that you consider will be small classes with respect to the set-theoretic universe.  However, sometimes classes are considered with respect to individual models.  A Grothendieck Universe for example, is a proper class if there are no inaccessible cardinals and we cannot prove that such a large cardinal is even possible from the ZFC axioms of set theory.  As another example, we sometimes do not allow ourselves use of all of the ZFC axioms, and the set of all even numbers for example would be a large class if we explicitly assumed that all sets were finite.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastActivityDate="2010-12-26T10:40:33.417" CommentCount="1" />
  <row Id="15552" PostTypeId="2" ParentId="15499" CreationDate="2010-12-26T11:25:43.560" Score="2" ViewCount="0" Body="&lt;p&gt;Dylan's right that diffeomorphism groups are usually infinite dimensional.  There's a theory of infinite dimensional Lie groups, I've heard, but I think it's a lot harder to work out things like convergence.  Worth noting, though, is that in this analogy, if $Diff(M)$ is a Lie group then its Lie algebra should be $Vect(M)$, the (infinite-dimensional) vector space of smooth vector fields on $M$.  The Lie algebra bracket is just the usual vector field bracket (possibly up to a sign? but if so, then it's only because of stupid clashing conventions).  Heuristically, to look for finite dimensional Lie groups we're going to want to put some more structure on $M$ and require that our diffeomorphisms preserve this.  If you're looking at things that aren't so homogeneous you get into the territory of $G$-bundles, which I think you should definitely look into if you don't already know about -- you don't need to know too much Lie theory to get the basics, and it's really cool (I think you'll like it too).  I don't have a reference offhand (please, anyone, post as a comment if you do) but I took an excellent class on them from Robert Bryant last year and a friend of mine is working on TeXing up the notes, so hopefully those will be available (at least in part) before too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, $Aut(M)/$isotopy is called the &lt;em&gt;mapping class group&lt;/em&gt;.  This can be written $MCG(M)=Aut(M)/Aut_0(M)$, where $Aut_0(M)$ is the connected component of the identity map, which if you think about it is all isotopies are anyways.  This has been extensively studied.  I don't remember much from a class I took with Jeff Brock a few years ago, but checking wikipedia reminds me that e.g. $MCG(\mathbb{T}^n)\simeq GL(n,\mathbb{Z})$.  It also says that the MCGs of surfaces have been extensively studied.  In any case, you can see that this is just the group $\pi_0(Aut(M))$, so assuming you understand $MCG(M)$ (!) then you only need to figure out $Aut_0(M)$ (!) to have the whole picture.  Which brings us back to the original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much, but like I said it's definitely easier (and finite-dimensionaler) if we restrict ourselves to manifolds with structure.  We can start by saying something simple: $Isom(S^n)=O(n+1)$.  One potentially fruitful way to think about this is to just start with a (matrix) group and find an invariant subspace of the (affine) space on which it acts, hopefully freely -- very often this should be a manifold, I think.  Otherwise things will probably be significantly more difficult.  If you put the flat metric on a torus and assume its fundamental domain is a square, then you should be able to look at isometries by what they do to your fundamental square sitting on a latticed plane.  To represent an isometry it'll need to be an isometry of the square, and to be a continuous map you'll need the four corners to be mapped to preimages of the same point under $\mathbb{R}^2\rightarrow \mathbb{T}^2$.  I'm pretty sure this ends up meaning $Isom_0(\mathbb{T}^2)\cong \mathbb{T}^2$.  But maybe that's not so surprising, since it's a group itself.  I guess all it's really telling you is that there are no isometries other than the &quot;multiply-by-$g$&quot; map for $g\in \mathbb{T}^2$.  The group $Conf(\mathbb{T}^2)$ should be pretty different, but I think you can't continuously dilate because that would eventually change the degree of your map (which is a discrete invariant), so a similar analysis should go through but you won't get any interesting &lt;em&gt;Lie&lt;/em&gt;-type behavior.  More interesting might be to try this all with the fundamental domain for an $n$-holed torus ($n\geq 2$) sitting on a latticed hyperbolic plane.  But it's late and I'm tired, so I'm not going to.  You should though!  If you do, let me know what you come up with.&lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2010-12-26T11:25:43.560" CommentCount="1" />
  <row Id="15553" PostTypeId="1" AcceptedAnswerId="15554" CreationDate="2010-12-26T12:33:56.790" Score="3" ViewCount="76" Body="&lt;p&gt;&lt;strong&gt;If $A$ and $B$ are two conformable matrices,and $AB = A$ and $BA = B$ then prove that A and B are idempotent matrices.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be trivial but I am not sure how to proceed on this.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-26T13:05:44.073" LastActivityDate="2010-12-26T13:05:44.073" Title="An elementary proof in matrix algebra" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;&lt;elementary&gt;" AnswerCount="1" />
  <row Id="15554" PostTypeId="2" ParentId="15553" CreationDate="2010-12-26T12:47:09.663" Score="5" ViewCount="0" Body="&lt;p&gt;$$Ax = ABx = A(BA)x = (AB)Ax = AAx$$&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2010-12-26T12:47:09.663" CommentCount="3" />
  <row Id="15555" PostTypeId="1" CreationDate="2010-12-26T13:16:35.977" Score="3" ViewCount="262" Body="&lt;p&gt;What is $\lim\limits{x\to 0}\frac{\lceil x\rceil}{x}$ ?  Here, $\lceil x \rceil$ is the ceiling function at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For left limit and right limit as $x\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in Advance .&lt;/p&gt;&#xA;" OwnerUserId="5122" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T01:46:25.560" LastActivityDate="2010-12-27T03:23:23.053" Title="what is $\lim\limits_{x\to 0} \lceil x\rceil/x$ ?" Tags="&lt;calculus&gt;&lt;limit&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="15556" PostTypeId="1" CreationDate="2010-12-26T13:36:00.910" Score="10" ViewCount="565" Body="&lt;p&gt;Some books say even numbers start from two but if you consider the number line concept, I think zero should be even because it is in between -1 and +1 (i.e in between 2 odd numbers). What is the real answer?&lt;/p&gt;&#xA;" OwnerUserId="5123" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-26T13:44:34.587" LastActivityDate="2010-12-27T21:38:08.560" Title="Is zero odd or even?" Tags="&lt;number-theory&gt;&lt;elementary&gt;&lt;math-history&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="15557" PostTypeId="2" ParentId="15556" CreationDate="2010-12-26T13:43:48.847" Score="7" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;YES!&lt;/strong&gt; zero is an even number. &lt;a href=&quot;http://mathforum.org/library/drmath/view/57132.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the Dr. Math's explanation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This seems to be a matter of confusion for many others around this planet,you may always like to ask  &lt;a href=&quot;http://www.google.com/search?btnG=1&amp;amp;pws=0&amp;amp;q=0+is+even+or+odd#hl=en&amp;amp;expIds=27642&amp;amp;sugexp=ldymls&amp;amp;xhr=t&amp;amp;q=Is+Zero+Even%2C+Odd%2C+or+Neither%3F&amp;amp;cp=30&amp;amp;qe=SXMgWmVybyBFdmVuLCBPZGQsIG9yIE5laXRoZXI_&amp;amp;qesig=Qog_VPHJ0jgC7-PI4nFBIQ&amp;amp;pkc=AFgZ2tlOVslOr2lA43pd92AsxK23pS2E9t-f3AFrd6jeF3PgpPmpcCSsSliwvaKIxTWjKuoevnHhEMGpNeprD9dQlSr05bBGkQ&amp;amp;pf=p&amp;amp;sclient=psy&amp;amp;source=hp&amp;amp;aq=f&amp;amp;aqi=&amp;amp;aql=&amp;amp;oq=Is+Zero+Even,+Odd,+or+Neither%3F&amp;amp;gs_rfai=&amp;amp;pbx=1&amp;amp;fp=b54c4bac999a8c85&quot; rel=&quot;nofollow&quot;&gt;google&lt;/a&gt; for your confusion.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditorDisplayName="" LastEditDate="2010-12-26T20:26:15.257" LastActivityDate="2010-12-26T20:26:15.257" CommentCount="12" />
  <row Id="15558" PostTypeId="2" ParentId="15555" CreationDate="2010-12-26T13:49:42.973" Score="8" ViewCount="0" Body="&lt;p&gt;Assuming that [x] is the floor of x, then look at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=f%5bx%5d+%3D%3D+Floor%5bx%5d%2Fx+&quot; rel=&quot;nofollow&quot;&gt;this graph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that [x] is the ceiling of x, then look at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=f%5bx%5d+%3D%3D+Ceiling%5bx%5d%2Fx+&quot; rel=&quot;nofollow&quot;&gt;this similar looking graph&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that [x] is the &quot;nearest integer&quot; function, then consider what the nearest integer is on the interval [-0.49, 0.49].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is something else, please specify.&lt;/p&gt;&#xA;" OwnerUserId="2959" LastEditorUserId="2959" LastEditorDisplayName="" LastEditDate="2010-12-26T17:23:04.557" LastActivityDate="2010-12-26T17:23:04.557" CommentCount="1" />
  <row Id="15559" PostTypeId="1" CreationDate="2010-12-26T14:26:57.530" Score="1" ViewCount="132" Body="&lt;p&gt;From wikipedia,&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_zeta_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Riemann_zeta_function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Furthermore, the fact that $\zeta(s) = \zeta(s^*)^*$ for all complex s ≠ 1 ($s^*$ indicating complex conjugation) implies that the zeros of the Riemann zeta function are symmetric about the real axis.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the zeros are symmetrical. But what about the other values of $\zeta(s)$?&#xA;My main aim is to find out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\zeta(s)$ symmetrical about the real axis for all $\Re(s) &amp;gt; 1$ ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T01:02:21.370" LastActivityDate="2010-12-27T01:02:21.370" Title="Is Riemann Zeta Function symmetrical about the real axis?" Tags="&lt;special-functions&gt;&lt;analytic-number-theory&gt;&lt;zeta-function&gt;" CommentCount="12" />
  <row Id="15560" PostTypeId="1" CreationDate="2010-12-26T14:51:09.213" Score="2" ViewCount="33" Body="&lt;p&gt;So I had a system&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#for given koefs &#xA;k:=3; n:=3; &#xA;#let us solve system:&#xA;koefSolution:= solve({&#xA; sum(a[i], i = 0 .. k) = 0,&#xA; sum(a[i], i = 0 .. k)-(sum(b[i], i = 0 .. k)) = 0,&#xA; sum(i^n*a[i], i = 0 .. k)-(sum(i^(n-1)*b[i], i = 0 .. k)) = 0&#xA;});&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I have a vector like &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;koefSolution := {  a[0] = 7*a[2]+26*a[3]-b[1]-4*b[2]-9*b[3], &#xA;                   a[1] = -8*a[2]-27*a[3]+b[1]+4*b[2]+9*b[3], &#xA;                   a[2] = a[2], &#xA;                   a[3] = a[3], &#xA;                   b[0] = -b[1]-b[2]-b[3], &#xA;                   b[1] = b[1], b[2] = b[2], &#xA;                   b[3] = b[3]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have &lt;code&gt;a[0]&lt;/code&gt; so I try &lt;code&gt;solve({koefSolution, a[0] = 1});&lt;/code&gt; why it does not solve my system for given &lt;code&gt;a[0]&lt;/code&gt;? ( main point here is to fill koefSolution with given a[] and b[] and optimize. So here I am trying to emulate analytic way of solving my system (It shall be capable to work for any given a[], b[], n, k... so that all students in our class would get good marks=))&lt;/p&gt;&#xA;" OwnerUserId="2065" LastActivityDate="2010-12-26T17:05:52.103" Title="How to solve such system with given parts of it? (maple)" Tags="&lt;computer-science&gt;&lt;computer-algebra-systems&gt;" AnswerCount="1" />
  <row Id="15561" PostTypeId="2" ParentId="15556" CreationDate="2010-12-26T15:15:56.757" Score="1" ViewCount="0" Body="&lt;p&gt;Which numbers are you using? Positive integer? In this case, you don't have to consider zero (that's why, perhaps, some books says that even numbers start from 2). If 0 is in your number set, then yes, it is divisible by 2.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2010-12-26T15:15:56.757" />
  <row Id="15562" PostTypeId="1" CreationDate="2010-12-26T15:49:24.040" Score="7" ViewCount="119" Body="&lt;p&gt;Last month, I was introduced to the card game SET. The game raises several interesting questions eg what's the probability that n randomly drawn cards contain k sets, is it possible to end with 12 (or 15) cards which contain no sets... I would like to present the solution to a question like this to undergrads, and through it introduce them to an area of mathematics. For example, as &lt;a href=&quot;http://math.stackexchange.com/questions/6190/a-question-related-to-the-card-game-set&quot;&gt;mentioned previously&lt;/a&gt;, there is the issue of maximal caps, which &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28game%29#Variations&quot; rel=&quot;nofollow&quot;&gt;Davis and MacLagan's paper&lt;/a&gt; relates to projective space and error-correcting codes. Can you give me any similar ideas? Perhaps relating to a variation of SET &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_%28game%29#Variations&quot; rel=&quot;nofollow&quot;&gt;listed here&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="782" LastEditorUserId="782" LastEditorDisplayName="" LastEditDate="2010-12-27T15:14:48.523" LastActivityDate="2010-12-27T15:14:48.523" Title="Mathematics related to the card game SET" Tags="&lt;combinatorics&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="15564" PostTypeId="1" AcceptedAnswerId="15569" CreationDate="2010-12-26T17:03:11.593" Score="2" ViewCount="150" Body="&lt;p&gt;I'm reading a book about Calculus on my own and am stuck at a problem, the problem is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two circles of radius 2 that have centers on the line x = 1 and pass through the origin. Find their equations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The equation for circle is $(x-h)^2 + (y-k)^2 = r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints will be really appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Here is what I did. I drew a triangle from the origin and applied the pythogoras theorem to find the perpendicular, the hypotenuse being 2 (radius = 2) and base 1 (because x = 1), the value of y-coordinate is $\sqrt3$. Can anyone confirm if this is correct?&lt;/p&gt;&#xA;" OwnerUserId="5125" LastEditorUserId="5125" LastEditorDisplayName="" LastEditDate="2010-12-27T03:38:22.137" LastActivityDate="2010-12-27T03:38:22.137" Title="Equation for a circle" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="15565" PostTypeId="2" ParentId="15560" CreationDate="2010-12-26T17:05:52.103" Score="1" ViewCount="0" Body="&lt;p&gt;In your solution, the free variables are $a[2],a[3],b[1],b[2],b[3]$. Since $a[0]$ is not&#xA;a free variable, setting it equal to 1 in your koefSolution does not yield a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is to set $a[0]=1$ in your original system, and solve for the rest&#xA;of the variables. I did that and I get:&#xA;$$ a[1]=-1-a[2]-a[3]$$&#xA;$$a[2]=a[2]$$&#xA;$$a[3]=a[3]$$&#xA;$$b[0]=1-7a[2]-26a[3]+3b[2]+8b[3]$$&#xA;$$b[1]=-1+7a[2]+26a[3]-4b[2]-9b[3]$$&#xA;$$b[2]=b[2]$$&#xA;$$b[3]=b[3]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free variables now are $a[2],a[3],b[2],b[3]$.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-26T17:05:52.103" />
  <row Id="15566" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T17:11:57.003" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt;: Find out what points $(x, y)$ with $x = 1$ also have distance $2$ to the origin. What would these points represent?&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="917" LastEditorDisplayName="" LastEditDate="2010-12-26T19:51:43.710" LastActivityDate="2010-12-26T19:51:43.710" CommentCount="4" />
  <row Id="15567" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T17:13:06.330" Score="3" ViewCount="0" Body="&lt;p&gt;Note that you can write down the equation of a circle if you know the co-ordinates of the center $(h,k)$ and its radius $r$. You already know that the two circles in question have radius $2$. It remains to figure out where their centers lie. You are told that their centers lie on the line $x=1$ which is a line parallel to the $Y$-axis. So you know the $x$-coordinate of the centers. The only thing that remains to be figured out are the $y$-coordinates of the two circles. To figure this out, you are given an additional information that both circles pass through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest drawing a picture of the Cartesian plane and of the line $x=1$ on it. You know the centers lie on this line and you know the centers have to be at a certain distance from the origin (why?). Given these two constraints, figure out what possible locations the centers can be at. If this isn't clear enough, post your work and indicate where you are getting stuck.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-12-26T17:13:06.330" CommentCount="2" />
  <row Id="15568" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T17:50:20.120" Score="0" ViewCount="0" Body="&lt;p&gt;Rephrasing Dario's hint: the center of each circle is a point of distance $2$ from the origin with $x=1$. If the center is the point $(x,y)$ then $x=1$, and so you can solve for $y$ (remember, the distance between $(x,y)$ and $(0,0)$ is $2$).&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-26T17:50:20.120" />
  <row Id="15569" PostTypeId="2" ParentId="15564" CreationDate="2010-12-26T18:01:54.617" Score="5" ViewCount="0" Body="&lt;p&gt;Hint: The distance between a point $P(h,k)=P(1,k)$ and the origin $O(0,0)$&#xA;is given by $\sqrt{1+k^{2}}$. This distance should be equal to $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/MrptX.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="752" LastEditorDisplayName="" LastEditDate="2010-12-26T21:17:10.877" LastActivityDate="2010-12-26T21:17:10.877" CommentCount="4" />
  <row Id="15570" PostTypeId="1" AcceptedAnswerId="15577" CreationDate="2010-12-26T18:50:17.767" Score="5" ViewCount="68" Body="&lt;p&gt;Show that $\displaystyle 1+ \sum\limits_{k=1}^{n} k \cdot k! = (n+1)!$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS: This is the number of permutations of an $n+1$ element set. We can rewrite this as $n!(n+1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS: It seems that the $k \cdot k!$ has a similar form to $(n+1)! = (n+1)n!$ Also we can write $1 = 0!$ I think you use the mulitplication principle is being used here (e.g. the permutations of a $k$ element set multiplied by $k$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a combinatorial proof is wanted (not an algebraic one).&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-26T20:01:50.983" Title="Permutation Identity and Sum" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15571" PostTypeId="1" CreationDate="2010-12-26T19:03:45.670" Score="8" ViewCount="724" Body="&lt;p&gt;I'm not going to prove something, this is just a question. From the first day which I went to University until now I had some root problems in some basic mathematical assumptions and concepts. Please help me to solve the most important of those problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example:&lt;/strong&gt; &lt;em&gt;&lt;strong&gt;infinity&lt;/em&gt;&lt;/strong&gt; ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is infinity ? Is that real ? Is infinity concept real ? I mean in real world. your answer is &quot;Yes&quot; ? What about &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeno%27s_paradoxes&quot; rel=&quot;nofollow&quot;&gt;Zeno's paradox&lt;/a&gt; ? I know that there are different solutions for this paradox ... but i don't talk about solutions. I'm saying &quot;this is logical&quot; because we accept some basic assumptions. But I can not understand it and compare it with real world. .... This problem assumes that there are infinite numbers between 2 numbers OR There are infinite points between two point. But in real world there is &quot;&lt;a href=&quot;http://iopscience.iop.org/0264-9381/20/15/101/fulltext&quot; rel=&quot;nofollow&quot;&gt;smallest distance&lt;/a&gt;&quot;. It's about &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_length&quot; rel=&quot;nofollow&quot;&gt;1.6*10^-35&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk more about my root problems: is number &quot;0&quot; real ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &lt;strong&gt;&quot;zero&quot;&lt;/strong&gt; ? You can say that &quot;there is 3 apple on the desk, if you take all of them there is 0 apple on the desk ? &quot; I still don't see number &quot;0&quot; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i know that they are tools which help us to solve different kind of &quot;real world&quot; problems ( I'm a software developer and I always use 0 and negative numbers ). But i can't see it in nature!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about &lt;a href=&quot;http://en.wikipedia.org/wiki/Empty_set&quot; rel=&quot;nofollow&quot;&gt;empty set&lt;/a&gt; ? I mean this --&gt; &lt;strong&gt;{}&lt;/strong&gt; ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is space between 2 bracket ... how can it be empty ? Is there any empty space in world ? for example a box which isn't any air in it ! It's an empty box ! ... no ... the element &quot;nothing&quot; is there ! &quot;nothing&quot; is a member of a set ... so there is no set with size &quot;zero&quot;. There is always a &quot;nothing&quot; or &quot;null&quot; element in there. Am I right ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk about other things which are panics in my mind!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about &lt;strong&gt;R set&lt;/strong&gt; ?!( &lt;strong&gt;real&lt;/strong&gt; numbers )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are millions of mathematical theorems which based on R set ! ... Somebody show me a set which is not countable in real world ? I mean in nature.  (I know &lt;a href=&quot;http://theoremoftheweek.wordpress.com/2010/03/18/theorem-21-the-real-numbers-are-uncountable/&quot; rel=&quot;nofollow&quot;&gt;Cantor's theorem&lt;/a&gt; , but it's based on &quot;infinity&quot; concept ! Which i can't understand it ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I know that this kind of mathematics help us to solve real world problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about logical relationships between those theorems or other things. They are logical if we base them on these concepts. I'm talking about concepts. Infinity concept, &quot;nothing&quot; concept. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if my question seems ridicules for someone !&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="5103" LastEditorDisplayName="" LastEditDate="2010-12-29T16:43:37.290" LastActivityDate="2010-12-29T18:16:25.023" CommunityOwnedDate="2010-12-29T16:43:37.290" Title="Do infinity and zero really exist ?" Tags="&lt;number-theory&gt;&lt;infinity&gt;&lt;number&gt;&lt;zeros&gt;" AnswerCount="6" CommentCount="19" FavoriteCount="4" />
  <row Id="15572" PostTypeId="2" ParentId="15505" CreationDate="2010-12-26T19:06:39.663" Score="-1" ViewCount="0" Body="&lt;p&gt;I don't claim to have a slick combinatorial proof, but I've upvoted the question in hopes that someone will provide one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would cop out and show equality as follows.  Dividing the LHS by the RHS and expanding the definition of the choose notation, we have&#xA;$$ (n-r)\frac{(n + r - 1)!}{r!(n-1)!} \frac{n!}{r!(n-r)!} \cdot \frac{1}{n} \frac{(2r)!(n-r-1)!}{(n+r-1)!} \frac{r!r!}{(2r)!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By cancelling appropriately, it follows that the LHS divided by the RHS is 1, so they're equal.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-26T19:06:39.663" />
  <row Id="15573" PostTypeId="2" ParentId="15571" CreationDate="2010-12-26T19:18:01.497" Score="6" ViewCount="0" Body="&lt;p&gt;It's not mathematics but rather physics which is based on these &quot;lies&quot;. In mathematics, we assume (if we're Platonists) that objects like the real numbers &quot;really exist&quot;, just not in the physical world, and then everything makes sense. Some people pretend that when they're doing mathematics, they're just combining axioms and derivation rules to prove theorems; these people have to take as an article of faith the fact that their chosen axiom system (e.g. ZFC) is consistent, otherwise all their toil makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When mathematics is applied to the real word, e.g. in physics, then often some approximations are made, like the fact that (at least in classical physics) a Cartesian coordinate system describes space, and the real line describes time. These approximations can be explained mathematically: taking discreteness into account in general will only slightly alter the results; but it complicates everything greatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some mathematics is not like that, for example when a poll is being taken, statisticians will calculate the standard deviation - this makes sense even if you don't believe in the real numbers; they are just a theoretical construct introduced to understand discrete phenomena. At a final count, everything reduces to finitistic reasoning, whose validity however rests on some unfounded belief in the consistency of some axiom system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not least, why do you oppose zero? If you take a ruler and mark a peg every inch, then you ask yourself &quot;how many pegs do I need to jump from 2in to 3in&quot;? The answer is $1$. Then &quot;how many pegs do I need to jump from 2in to 2in&quot;? The answer is zero. You can understand negative numbers this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, zero apples is the number of apples that remain after you've eaten all of them. And there are many more examples, in fact an entire book was written on the subject.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-26T19:18:01.497" CommentCount="5" />
  <row Id="15574" PostTypeId="2" ParentId="15571" CreationDate="2010-12-26T19:28:45.473" Score="7" ViewCount="0" Body="&lt;p&gt;It seems like your premise for mathematics being a 'lie' is that its theoretical background cannot be completely applied to the real world (e.g. having the smallest distance in the real world, but not in mathematics). If that's how you define a lie, sure, but mathematics is an abstract science and I am not sure why you think it's supposed to hold certain properties that are true in the real world -- abstraction is actually the whole point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solutions to Zeno's paradox have been offered, I am not sure what you mean by &quot;but i don't talk about solutions. I'm saying &quot;this is LOGICAL&quot;. But it's not TRUE.&quot; In fact, Zeno's paradox shows the &lt;em&gt;lie&lt;/em&gt; of the real world, as Zeno contested that motion does not exist and is an illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;You wanna tell me that &quot;there is 3 apple on the desk, if you take all of them there is 0 apple on the desk ? &quot; I still don't see number &quot;0&quot; ?&quot; -- true, but neither you see number 3 when there are 3 apples on the desk, correct? 0 is used to denote lack of objects, and it's definition is a tad more complicated than definition of other integers, and for a reason (it can be used differently in different branches of mathematics). Also, if you want a real-life definition of zero, think of not moving, not gaining weight, not making any progress -- all of this can essentially be seen as addition of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An empty set is a set (which, again, is an abstract concept) that has no elements. Careful with the words -- it's not the case that it has 'nothing' as an element, it's rather the case that it has no elements at all. Here, it comes down to semantics of the definition, &#xA;but you once again seem to be drawing parallels to real life and trying to find a corresponding entity. I do not quite see why that should be the approach -- we are talking about mathematics, not physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the set of Real numbers, again, same story. I do not know if you can find an uncountable set in the 'real world' (though I suppose you could contest that the set of points of any object is uncountably infinite, because as you keep zooming in you will be getting more and more particles, and it's our limitations that do not allow us to see past atoms). But that is not the point. Mathematics is based on some abstract ideas, and it's true that not all of them are necessarily reflected in nature. That does not, however, make mathematics based on 'lies.'&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-26T23:24:12.527" LastActivityDate="2010-12-26T23:24:12.527" CommentCount="2" />
  <row Id="15575" PostTypeId="1" AcceptedAnswerId="15576" CreationDate="2010-12-26T19:42:43.670" Score="4" ViewCount="197" Body="&lt;p&gt;Could someone tell me what the &quot;square of a graph&quot; $G^2$ is? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4654" LastActivityDate="2011-01-03T22:04:32.767" Title="Square of a graph" Tags="&lt;graph-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15576" PostTypeId="2" ParentId="15575" CreationDate="2010-12-26T19:56:26.160" Score="5" ViewCount="0" Body="&lt;p&gt;The square of a graph $G$ is obtained by starting with $G$, and adding edges between any two vertices whose distance in $G$ is $2$.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastEditorUserId="1508" LastEditorDisplayName="" LastEditDate="2011-01-03T22:04:32.767" LastActivityDate="2011-01-03T22:04:32.767" CommentCount="9" />
  <row Id="15577" PostTypeId="2" ParentId="15570" CreationDate="2010-12-26T20:01:50.983" Score="6" ViewCount="0" Body="&lt;p&gt;Here is a brief description of one combinatorial way to approach this. Suppose we are permuting $\{1,2,3,\ldots,n+1\}$. One permutation is $P=(1,2,3,\ldots,n+1)$. Any other permutation has a first position from the left which differs from $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $S_1$ be the set of permutations that first differs from $P$ in position $n$. Let $S_2$ be the set of permutations that first differs from $P$ in position $n-1$. Continue this until we get to $S_n$, which is the set of permutations that first differs from $P$ in position $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we count the size of $S_i$, we can convince ourselves that it is $(i+1)!-i! = i \cdot i!$. This is the part I'm skipping over. Also counting $P$ then gives us the left hand side.&lt;/p&gt;&#xA;" OwnerUserId="5124" LastActivityDate="2010-12-26T20:01:50.983" />
  <row Id="15578" PostTypeId="2" ParentId="15575" CreationDate="2010-12-26T20:11:31.347" Score="4" ViewCount="0" Body="&lt;p&gt;I would have thought that $G^2$ would either mean the box product of $G$ with itself, or the cross product of $G$ with itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definitions of these are as follows: If $G$ and $H$ are graphs with vertex sets $V_G$ and $V_H$, then the box product of $G$ and $H$ has vertex set $V_G \times V_H$ and has an edge from $(g_1, h_1)$ to $(g_2, h_2)$ if and only if &lt;em&gt;either&lt;/em&gt; (1) $g_1=g_2$ and there is an edge from $h_1$ to $h_2$ in $H$ or (2) $h_1=h_2$ and there is an edge from $g_1$ to $g_2$ in $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product of $G$ and $H$ has vertex set $V_G \times V_H$ and has an edge from $(g_1, h_1)$ to $(g_2, h_2)$ if and only if there is an edge from $g_1$ to $g_2$ in $G$ and an edge from $h_1$ to $h_2$ in $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since TonyK has found yet another definition, I would say that there is more than one thing $G^2$ can denote.&lt;/p&gt;&#xA;" OwnerUserId="448" LastActivityDate="2010-12-26T20:11:31.347" CommentCount="4" />
  <row Id="15580" PostTypeId="1" AcceptedAnswerId="15582" CreationDate="2010-12-26T21:24:20.987" Score="2" ViewCount="98" Body="&lt;p&gt;How can I calculate average distance from square center to points inside the square?&lt;/p&gt;&#xA;" OwnerUserId="5130" LastActivityDate="2010-12-26T21:40:15.637" Title="What is average distance from square center to some point?" Tags="&lt;geometry&gt;&lt;probability&gt;" AnswerCount="1" />
  <row Id="15581" PostTypeId="2" ParentId="15541" CreationDate="2010-12-26T21:26:11.467" Score="2" ViewCount="0" Body="&lt;p&gt;Thanks for the comments. After giving it more thought I was able to solve it on my own. Here's what I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $U\subset \mathbb{R}^{n+m},V\subset \mathbb{R}^m$ and $n&amp;gt;0$ (if $n&amp;lt;0$ just consider $f^{-1}$ instead of $f$ in what follows). We have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f&amp;#39;=\left(&#xD;&#xA;\begin{array}{cccccc}&#xD;&#xA; \frac{\partial f_1}{\partial x_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_1}{\partial x_n} &amp;amp; \frac{\partial f_1}{\partial y_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_1}{\partial y_m} \\&#xD;&#xA; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  \\&#xD;&#xA; \frac{\partial f_m}{\partial x_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_m}{\partial x_n} &amp;amp; \frac{\partial f_m}{\partial y_1} &amp;amp; \ldots  &amp;amp; \frac{\partial f_m}{\partial y_m}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} &amp;amp; \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(f^{-1}\right)&amp;#39;=\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA; \frac{\partial x_1}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial x_1}{\partial f_m} \\&#xD;&#xA; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  \\&#xD;&#xA; \frac{\partial x_n}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial x_n}{\partial f_m} \\&#xD;&#xA; \frac{\partial y_1}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial y_1}{\partial f_m} \\&#xD;&#xA; \ldots  &amp;amp; \ldots  &amp;amp; \ldots  \\&#xD;&#xA; \frac{\partial y_m}{\partial f_1} &amp;amp; \ldots  &amp;amp; \frac{\partial y_m}{\partial f_m}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{c}&#xD;&#xA; \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)} \\&#xD;&#xA; \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(f^{-1}\right)&amp;#39;\cdot f&amp;#39;=\left(&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} &amp;amp; \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)} \\&#xD;&#xA; \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} &amp;amp; \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=\left(&#xD;&#xA;\begin{array}{cc}&#xD;&#xA; I_{(n\times n)} &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; I_{(m\times m)}&#xD;&#xA;\end{array}&#xD;&#xA;\right)=I_{(n+m\times n+m)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}=I_{(m\times m)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Jacobian matrix $\frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}$ is not singular, we can apply the implicit function theorem. It follows that there is a function $g:A\to B$, where $A\subset \mathbb{R}^n,B\subset \mathbb{R}^m$ are open subsets and $A\times B\subset U$, such that $f(x,g(x))=\text{const}$ for all $x\in A$. It is then clear that $f$ cannot be bijective.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-26T21:26:11.467" CommentCount="1" />
  <row Id="15582" PostTypeId="2" ParentId="15580" CreationDate="2010-12-26T21:33:41.807" Score="3" ViewCount="0" Body="&lt;p&gt;For the average distance, what you want is the expected value of the distance from the center over the set of all points in the square, which is the sum of the probability of each point times the distance to that point.  The problem is (more or less) that there are a whole lot of points and each point has infinitesimal probability.  One way to deal with this is to use calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the square with sides of length 1 parallel to the axes and centered at the origin.  This square is the set of points $(x,y)$ with $-\frac{1}{2}\le x\le\frac{1}{2}$ and $-\frac{1}{2}\le y\le\frac{1}{2}$.  The total area of the square is 1.  The sum of the probabilities of each point (not dealing with distance yet) is $\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\;dx\;dy=1$.  Now, at each point, multiply the probability by the distance to get the average distance: $\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{x^2+y^2}\;dx\;dy$.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-12-26T21:40:15.637" LastActivityDate="2010-12-26T21:40:15.637" CommentCount="5" />
  <row Id="15583" PostTypeId="2" ParentId="11429" CreationDate="2010-12-26T21:56:21.903" Score="2" ViewCount="0" Body="&lt;p&gt;Conceptually, the way amortization of loan payments works is that the lender should come out even at the end of the loan when you account for future value of money based on the loan's interest rate.  If the lender loaned $A$ dollars at an interest rate per period $r$ over $n$ periods, the future value (at the end of the loan) of the money lent is $A(1+r)^n$.  If the payment is $p$, then the future value of the first payment is $p(1+r)^{n-1}$ (paid at the end of the first period), for the second is $p(1+r)^{n-2}$, for the $k$th payment is $p(1+r)^{n-k}$, up to the last payment which is paid at the end of the loan so its future value is $p(1+r)^0=p$.  If you take the sum of these future values of all the payments, it should be equal to the future value of the original loan.  This equation can be solved for $p$ in terms of the general $A$, $r$, and $n$ to get a generalized amortization formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on a period-to-period basis, the balance is defined by the initial balance $b_0=A$ and the recurrence relation $b_{k}=(1+r)b_{k-1}-p$.  That is, the change from one period to the next is to add the interest due and subtract the payment.  Knowing the parameters of the loan and the payment, you can use this to find the balance after a specific number of periods, as well as the breakdown of interest and principal in each payment.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-26T21:56:21.903" />
  <row Id="15584" PostTypeId="1" CreationDate="2010-12-26T22:17:49.077" Score="1" ViewCount="63" Body="&lt;p&gt;is integral of (1/x dx) the same as (dx/x) ??&lt;/p&gt;&#xA;" OwnerUserId="1755" LastActivityDate="2010-12-26T22:20:08.567" Title="small difference in integrals" Tags="&lt;integral&gt;" AnswerCount="2" />
  <row Id="15585" PostTypeId="2" ParentId="15584" CreationDate="2010-12-26T22:19:58.700" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, $\int \frac{dx}{x}$ is simply shorthand for $\int \frac1x \, dx$; they mean precisely the same thing.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2010-12-26T22:19:58.700" />
  <row Id="15586" PostTypeId="2" ParentId="15584" CreationDate="2010-12-26T22:20:08.567" Score="5" ViewCount="0" Body="&lt;p&gt;Yes. $\int \frac{1}{x} \mathrm dx = \int \frac{\mathrm dx}{x}$.&lt;/p&gt;&#xA;" OwnerUserId="3787" LastActivityDate="2010-12-26T22:20:08.567" />
  <row Id="15587" PostTypeId="1" AcceptedAnswerId="15589" CreationDate="2010-12-26T22:20:21.037" Score="-5" ViewCount="122" Body="&lt;p&gt;In &lt;a href=&quot;http://math.stackexchange.com/questions/15556/is-zero-odd-or-even&quot;&gt;Is zero odd or even?&lt;/a&gt; a simple question of $0 , -1$ being even or odd was raised and answered. The problem is domain of definition of even or odd numbers is over natural numbers. i.e. $1,2,3,\dots $. &#xA;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_(mathematics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Parity_(mathematics&lt;/a&gt;) Greeks did not consider 1 to be neither odd or even, as if one uses the definition odd number being $2k+1$ there is no $k$ in &lt;strong&gt;natural numbers&lt;/strong&gt; that will satisfy that $2k+1=1$.&lt;br&gt;&#xA;So calling 0,-1 even or odd is just abuse of the language. there are 2 more examples of extending the domain of definition and changing the vocabulary :&#xA;1.Gamma function, the $\Gamma \left( x + 1\right)$  coincides with factorial function over integers, but although the gamma function is also called factorial function, nobody will talk about $\frac{1}{2}!$ but $\Gamma$ of $\frac{-1}{2}$ is acceptable.&#xA;Another example :&lt;br&gt;&#xA;the series $1-1+1-1+\dots$ is divergent by the common definition of convergence, but using extended definition of convergence  &lt;a href=&quot;http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation&lt;/a&gt; it converges to $\frac{1}{2}$. Yet nobody will call it convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now after extending the definition of even or odd to integers, according to new definition 0 and -1 etc. must be included in either even or odd category by default, but not by the original definition over natural numbers. Now if nobody will hesitate to call 1,2,3 etc. even or odd, why 0,-1, etc. are not clear to be in which category they should be considered? One can define equipartition sets that the all the odd and even numbers will be included in one of the categories but neither of the new sets is set of even or odd numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are we to treat the sets $ \left\{{\dots, -2, 0, 2 , \dots}\right\} \text{ and }  \left\{{\dots, -3, -1, 1 , 3 , \dots}\right\}$ as even and odd numbers or are we to treat them equivalently as the even and odd numbers? remembering nobody will consider -1 apples as even or odd.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2010-12-26T22:40:20.137" LastActivityDate="2010-12-26T22:56:35.307" Title="Over extending domain of definitions" Tags="&lt;logic&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="15588" PostTypeId="2" ParentId="15556" CreationDate="2010-12-26T22:23:22.207" Score="3" ViewCount="0" Body="&lt;p&gt;This problem arose e.g. during the &lt;a href=&quot;http://uk.reuters.com/article/idUKNOA03972720080620&quot; rel=&quot;nofollow&quot;&gt;Beijing even-odd car ban for the 2008 Olympics&lt;/a&gt;, where cars with odd numbered licence plates were banned one day, then even the next day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice is between:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 is even and not odd,&lt;/li&gt;&#xA;&lt;li&gt;0 is odd and not even,&lt;/li&gt;&#xA;&lt;li&gt;0 is both even and odd,&lt;/li&gt;&#xA;&lt;li&gt;0 is neither even nor odd (like infinity or $\pi$) or&lt;/li&gt;&#xA;&lt;li&gt;0 is assigned a unique title (like how 1 called a &quot;unit&quot; -- neither prime nor composite).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a matter of definition, so while you could define 0 to be any of the above, it's best to choose the definition that will be the &lt;em&gt;most consistent&lt;/em&gt; with the usage of &quot;even&quot; and &quot;odd&quot; for numbers other than 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $W=\{2,4,6,\ldots\}$, $V=\{1,3,5,\ldots,\}$ and lets look at the properties of even and odd numbers on these sets that we are familiar with.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A number is either even or odd, and not both.&lt;/li&gt;&#xA;&lt;li&gt;If $w,x \in W$ then $w+x \in W$ (even + even = even).&lt;/li&gt;&#xA;&lt;li&gt;If $w \in W$ and $v \in V$ then $w+v \in V$ (even + odd = odd).&lt;/li&gt;&#xA;&lt;li&gt;If $y,v \in V$ then $y+v \in W$ (odd + odd = even).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;[and probably many others I've forgotten to write here]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would be desirable that whichever definition we choose for 0, it preserves the above properties.  Now lets say we let 0 be odd (the second and third cases listed above).  Then our definition is not consistent with these properties.  So, if we choose to define 0 as odd, we should have some substantial benefits to outweigh the losses.  On the other hand, defining 0 to be even and not odd is consistent with the above properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last two candidate definitions are essentially saying there is no consistent way of defining even or oddness to 0.  But in this case, there is -- 0 is even and not odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: We also have the property that elements of W are all divisible by 2, but whether or not 0 is divisible by 2 is another matter of definition, for which we should again apply the &quot;which is the most sensible definition&quot; concept.]&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-12-26T22:23:22.207" />
  <row Id="15589" PostTypeId="2" ParentId="15587" CreationDate="2010-12-26T22:55:04.750" Score="6" ViewCount="0" Body="&lt;p&gt;You seem to be proposing that it is somehow an error to use a word beyond the context in which it was originally defined.  This is a ludicrous &lt;em&gt;linguistic&lt;/em&gt; claim, let alone a mathematical one.  As the world and our understanding of it changes, the words we use to describe it must necessarily adapt to those changes.  To invent a new word every time we need to label a new concept would be endlessly tedious, and it would ignore the significant compressive power of metaphor.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-26T22:55:04.750" CommentCount="13" />
  <row Id="15590" PostTypeId="2" ParentId="15587" CreationDate="2010-12-26T22:56:35.307" Score="3" ViewCount="0" Body="&lt;p&gt;$\ldots,-4,-2,0,2,4,\ldots$ are even integers, $ 2,4,6, \ldots$ are even positive integers.&#xA;$\ldots,-5,-3,-1,1,3,5,\ldots$ are odd integers, $ 1,3,5, \ldots$ are odd positive integers.&#xA;What can be simpler than this?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-26T22:56:35.307" />
  <row Id="15591" PostTypeId="1" AcceptedAnswerId="15593" CreationDate="2010-12-26T23:01:32.640" Score="1" ViewCount="69" Body="&lt;p&gt;Suppose we have the following two identities: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\displaystyle \sum_{k=0}^{n} \binom{n}{k} = 2^n$&lt;/li&gt;&#xA;&lt;li&gt;$\displaystyle \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first says that the number of subsets of an $n$-set is $2^n$. The second says that the number of subsets of even size equals the number of subsets of odd size (of an $n$-set). Thus there are $2^{n-1}$ subsets of even length and $2^{n-1}$ subsets of odd length?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To combinatorially prove the second identity, let $A$ be a $k$-subset of $[n]$. Then note whether $k$ is odd or even? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-26T23:06:06.070" Title="Number of even and odd subsets" Tags="&lt;combinatorics&gt;&lt;binomial-coefficients&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15592" PostTypeId="1" AcceptedAnswerId="15594" CreationDate="2010-12-26T23:02:32.377" Score="4" ViewCount="163" Body="&lt;p&gt;I want to prove that the size (cardinality) of $\mathbb{N}$ is the same as the size of the set of all ordered triples of natural numbers ; just to be more clear that $|\mathbb{N}|=|\mathbb{N}\times\mathbb{N}\times\mathbb{N}|$.&lt;/p&gt;&#xA;" OwnerUserId="5131" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T00:29:43.257" LastActivityDate="2010-12-27T11:40:50.750" Title="function from $\mathbb{N}$ to $\mathbb{N}\times\mathbb{N}\times\mathbb{N}$" Tags="&lt;logic&gt;&lt;elementary-set-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="15593" PostTypeId="2" ParentId="15591" CreationDate="2010-12-26T23:06:06.070" Score="3" ViewCount="0" Body="&lt;p&gt;The second identity can be proven by an inclusion-exclusion argument, but maybe it's clearer this way: pick an arbitrary element of the $n$-element set.  Given a subset, if it contains the arbitrary element, remove it; otherwise, add it in.  (That is, take the symmetric difference.)  This defines a bijection from the set of subsets of odd size and the subsets of even size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalization with $2$ replaced by a larger integer is less trivial; see &lt;a href=&quot;http://math.stackexchange.com/questions/918/how-do-i-count-the-subsets-of-a-set-whose-number-of-elements-is-divisible-by-3-4&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell you are going through identities you know how to prove algebraically and trying to see how they can be proven combinatorially.  If you are interested in learning how to do this systematically you should start with Benjamin and Quinn's &lt;a href=&quot;http://books.google.com/books?id=n8Xx2oSlz4MC&amp;amp;pg=PA194&amp;amp;lpg=PA194&amp;amp;dq=art+benjamin+count&quot; rel=&quot;nofollow&quot;&gt;Proofs that Really Count&lt;/a&gt; and then, if you are really serious, Bergeron, Labelle, and Leroux's &lt;a href=&quot;http://books.google.com/books?id=83odtWY4eogC&amp;amp;printsec=frontcover&amp;amp;dq=bergeron+labelle+leroux&quot; rel=&quot;nofollow&quot;&gt;Combinatorial Species and Tree-like Structures&lt;/a&gt;.  There are also some blog posts and other material I could link you to if you want.  Asking a question on this more general topic certainly seems more efficient than what you're currently doing.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-26T23:06:06.070" CommentCount="2" />
  <row Id="15594" PostTypeId="2" ParentId="15592" CreationDate="2010-12-26T23:16:29.387" Score="6" ViewCount="0" Body="&lt;p&gt;Here is one way: I assume $0\in{\mathbb N}$. Given a triple $(a,b,c)$, assign to it the number $2^a(2(2^b(2c+1)-1)+1)-1$. The same idea shows that ${\mathbb N}$ has the same size as ${\mathbb N}^k$ for any positive integer $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can be easily modified and in fact gives a slightly easier formula if your convention is that the naturals begin with 1.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are of course many other ways. The idea I am using above is to first find a bijection between ${\mathbb N}\times{\mathbb N}$ and ${\mathbb N}$, and then use it to create the one you want. The bijection I used is $(a,b)\mapsto 2^a(2b+1)-1$. Another popular choice uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function&quot; rel=&quot;nofollow&quot;&gt;Cantor's pairing function&lt;/a&gt;, $\displaystyle (a,b)\mapsto \frac{(a+b)(a+b+1)}2 +b$. As a technical comment, there is actually an advantage to Cantor's choice, in that it is a polynomial, and it is easier to define (in a first-order way) in the usual structure of the natural numbers $({\mathbb N},+,\times,0,1,&amp;lt;)$. The choice I used requires exponentiation, and although this can also be defined in a first-order way, it is significantly less trivial (that exponentiation is first-order definable is the bulk of the argument for &lt;a href=&quot;http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem&quot; rel=&quot;nofollow&quot;&gt;Hilbert's 10th problem&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2010-12-26T23:24:26.197" LastActivityDate="2010-12-26T23:24:26.197" CommentCount="2" />
  <row Id="15595" PostTypeId="2" ParentId="15592" CreationDate="2010-12-26T23:16:44.393" Score="2" ViewCount="0" Body="&lt;p&gt;Do you know a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pairing_function&quot; rel=&quot;nofollow&quot;&gt;Pairing function&lt;/a&gt;?  Do you see how this can help?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-26T23:16:44.393" />
  <row Id="15596" PostTypeId="1" CreationDate="2010-12-26T23:53:24.907" Score="2" ViewCount="49" Body="&lt;p&gt;Say I have a time series, with actual measurements of a variable $a$ in different locations.  If I want to know the average of $p$ from time $t_1$ to $t_n$, I could say that $\overline{a_{p}}=\frac{\sum\limits_{t=1}^n a_{{p}_t}}{n}$, where $p$ is the coordinate of the point of measurement.  So far, so good.  Let's generalize and call this averaging function $m_p=f(a_p,t)$, where $f(a,t)$ is some deterministic function and $m$ is its result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say I want to calculate $m$ for a location $o$, where $o$ is a points for which I don't have actual measurements and $a_o$ was estimated through a an interpolating function $g(a,o)$.  My intuition tells me that in this case, we should first estimate every $a_{o_t}$ before applying $f(a,t)$. In other words, considering that $a_o = g(a,o)$, we get $m_o$ by doing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_o=f[g(a,o),t]=f(a_o,t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have seen so many papers that do the other way around.  They first calculate $f(a,t)$ for every known $a$ and then interpolate the results using the same g() even though there is no guarantee that $m$ behaves like $a$, or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_o=g[f(a,t),o]=g(m,o)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, they use the same logic when instead of $t$, there is some other variable $b$, i.e., it's not a time series, but $f(a,b)$ is a deterministic function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the models in those papers conceptually wrong?  If they are wrong, is there a techcnical term to call this kind of mistake?&lt;/p&gt;&#xA;" OwnerUserId="5133" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-26T23:59:01.643" LastActivityDate="2010-12-27T01:25:47.230" Title="Mean of interpolated data or interpolation of means in geostatistics" Tags="&lt;statistics&gt;" AnswerCount="1" />
  <row Id="15597" PostTypeId="1" AcceptedAnswerId="15600" CreationDate="2010-12-27T00:25:54.307" Score="5" ViewCount="124" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Infinity&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; on Wikipedia states the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering why is it the case for finite-dimensional spaces, but not for $\mathbb{R}^{\omega}$. Can't a PMI proof of a bijection between $[0;1]$ and &#xA;$\mathbb{R}^{\omega}$ be established, or would that proof be showing that one can establish a bijection between $[0;1]$ and $\mathbb{R}^n$ for any given $n\in \mathbb{R}/\mathbb{N}$ but not $\mathbb{R}^{\omega}$? Is there something special about infinitely-dimensional $R$ that makes it have more points than $\mathbb{R}$ or any segment on $\mathbb{R}$? Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5135" LastEditorDisplayName="" LastEditDate="2010-12-27T01:12:29.210" LastActivityDate="2010-12-27T01:12:29.210" Title="Is $[0;1]$ equinumerous to $\mathbb{R}^{\omega}$?" Tags="&lt;elementary-set-theory&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="15598" PostTypeId="2" ParentId="15571" CreationDate="2010-12-27T00:26:01.600" Score="14" ViewCount="0" Body="&lt;p&gt;Dear Michel,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your questions and struggles with these difficult concepts are completely understandable, but your presentation is terrible. You are unnecessarily confrontational and aggressive. Having difficulties with these concepts is completely all right. Assuming that during hundreds of years mathematicians conspired to build a make-believe world makes you a crackpot. Hence the series of down-votes.  I think the first thing you should do is to learn humility and present your ideas with respect. Don't assume that just because something does not make sense &lt;em&gt;to you&lt;/em&gt; it necessarily does not make sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, putting aside my initial discomfort with your attitude, let me try to say a few words towards resolving your issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;em&gt;infinity&lt;/em&gt; --- This is indeed something hard to grasp and there are grown mathematicians who are advocating similar views. They are a bit more sophisticated and a lot less arrogant though. As for &lt;em&gt;smallest distance&lt;/em&gt;, it has already been pointed out that that would be a statement in physics and not in math. On the other hand, if you believe in dividing integers by integers, you must realize that there are infinitely many numbers between any two numbers or infinitely many points between any two points, just divide the distance by $n$ where $n$ runs through $\{1,2,3,\dots\}$. But of course, you could try to argue that it does not make sense to talk about arbitrarily large numbers such as $2^{2011}$ as they have no practical applications. I can't say anything to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;em&gt;zero&lt;/em&gt; --- I always thought that whoever first invented zero and the empty set had to be a genius. These are indeed difficult concepts. However, you are confusing a few things here. You say &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...&quot;nothing&quot; is a member of a set...so there is no set with size &quot;zero&quot;. There is always a &quot;nothing&quot; or &quot;null&quot; element in there. Am I right ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, you are not right.  First off, what do you mean by &quot;nothing&quot;? The only sensible mathematical meaning of that is the &lt;em&gt;empty set&lt;/em&gt;. So, you are talking about a set that consist of one element, which is the empty set. That's different from the empty set that has no elements at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You wanna tell me that &quot;there is 3 apple on the desk, if you take all of them there is 0 apple on the desk ? &quot; I still don't see number &quot;0&quot; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you &lt;em&gt;are right&lt;/em&gt;! There is $0$ apple on the desk and you don't see the number $0$. There are many other things you don't see. For example you don't see &#xA;the dark side of the moon, but you would not doubt that it is there. You don't see the Sun at night but you know it exists even at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about negative numbers? Those don't exist either? So when you take out a loan and you owe the bank a ton of money, it is not real? If you accept negative numbers, you have to accept &lt;em&gt;zero&lt;/em&gt; as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;em&gt;uncountable sets&lt;/em&gt; --- I guess this is a more sophisticated version of your problem with &lt;em&gt;infinity&lt;/em&gt;. So, can we accept the existence of (non-zero ;) natural numbers? Say $1,2,3,\dots$? If so, then it is easy to show you an uncountable set: take the set that consists of all the subsets of $\{1,2,3,\dots\}$.&lt;/p&gt;&#xA;" OwnerUserId="4654" LastActivityDate="2010-12-27T00:26:01.600" CommentCount="2" />
  <row Id="15599" PostTypeId="2" ParentId="15505" CreationDate="2010-12-27T00:35:22.657" Score="3" ViewCount="0" Body="&lt;p&gt;Let $S$ be a set of $n+r-1$ elements. Both sides count the number of ways to select two disjoint sets $A,B\subseteq S$ of size $r$ and possibly an element $c\in S \setminus B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first observe that $(n-r)\binom{n}{r}=n\binom{n-1}{r}$ as both sides count the number of ways to form a team of size $r+1$ with a captain out of $n$ people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the above on the original LHS we get $n\binom{n-1}{r}\binom{n+r-1}{r}$, which corresponds to selecting $A$ ($r$ out of $n+r-1$), then $B$ ($r$ out of the remaining $n-1$) and $c$ ($n-1$ choices in $S\setminus B$ and one option of not choosing $c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RHS argument goes as follows: choose $2r$ elements for both $A$ and $B$, then choose $r$ of them to make $B$. Then, as before, there are $n$ options of choosing $c\in S\setminus B$ or none at all.&lt;/p&gt;&#xA;" OwnerUserId="5132" LastActivityDate="2010-12-27T00:35:22.657" />
  <row Id="15600" PostTypeId="2" ParentId="15597" CreationDate="2010-12-27T00:37:23.547" Score="10" ViewCount="0" Body="&lt;p&gt;Whoever told you that $[0, 1]$ is smaller than $\mathbb{R}^{\omega}$ is wrong; they both have the same cardinality as $\{ 0, 1 \}^{\omega} = \{ 0, 1 \}^{\omega \cdot \omega}$.  By the way, something about your post suggests to me that you are not convinced that this set has a rigorous definition, and it does: $\mathbb{R}^{\omega}$ is precisely the set of functions from $\omega$ to $\mathbb{R}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several other ways one might imagine putting infinitely many copies of $\mathbb{R}$ together.  The above is the infinite direct product; there is also the infinite direct &lt;em&gt;sum&lt;/em&gt; $\bigoplus_{n=1}^{\infty} \mathbb{R}$, which consists of all sequences of real numbers which are eventually zero; unlike the infinite direct product, the infinite direct sum is countable-dimensional.  And also one can consider the sequences $c_0$ of real numbers which converge to zero; this is not countable-dimensional but has nice topological properties.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-27T00:43:30.750" LastActivityDate="2010-12-27T00:43:30.750" CommentCount="5" />
  <row Id="15601" PostTypeId="2" ParentId="15597" CreationDate="2010-12-27T01:06:33.313" Score="4" ViewCount="0" Body="&lt;p&gt;The cleanest way of seeing the bijection from $\mathbb{R}^{\omega}$ to $\mathbb{R}$ is probably by taking advantage of your favorite bijection from $\mathbb{R}$ to $\left[0,1\right]$, then treating the numbers in that range as a sequence of decimals, 'stacking' them to get a $\omega\times\omega$ grid of digits, and using a diagonal sweep to compose them into a single decimal; so far instance, if the first three elements of your sequence of reals were .314159, .271828, and .161803, then take the first digit of the first, then the second digit of the first and the first digit of the second, then the third digit of the first, the second digit of the second, and the first digit of the first, continuing to interleave this way: .312471...  It's easy to convince yourself that this is a one-to-one bijection; you can go the other way by writing the digits of your single number one by one diagonally into a grid, and then taking the result as a series of reals.&lt;/p&gt;&#xA;" OwnerUserId="785" LastActivityDate="2010-12-27T01:06:33.313" CommentCount="4" />
  <row Id="15602" PostTypeId="1" AcceptedAnswerId="15604" CreationDate="2010-12-27T01:17:16.193" Score="15" ViewCount="339" Body="&lt;p&gt;I am trying to understand the notion of an orientable manifold.&lt;br&gt;&#xA;Let M be a smooth n-manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1}))&amp;gt; 0$ (where defined). My question is:&lt;br&gt;&#xA;Using this definition of orientation, how can one prove that the Möbius strip is not orientable?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T01:52:06.367" LastActivityDate="2010-12-28T23:09:31.700" Title="Why is the Möbius strip not orientable?" Tags="&lt;geometry&gt;&lt;topology&gt;&lt;differential-geometry&gt;&lt;algebraic-topology&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" />
  <row Id="15603" PostTypeId="2" ParentId="15596" CreationDate="2010-12-27T01:25:47.230" Score="2" ViewCount="0" Body="&lt;p&gt;If your interpolation scheme is linear in the input $a$'s (and all interpolation schemes I can think of are), then there is no difference between the mean of interpolated values and the interpolation of the means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your notation is a little confusing, so let me put it this way. Suppose you know the values $a_{i,j} = a(p_i,t_j)$ for $m$ points $p_i$ at $n$ times $t_j$. The interpolated value at a point $o$ at any time $t_j$ can typically be expressed as a weighted sum of the values at known points, $$a(o,t_j) := w_1 a_{1,j} + w_2 a_{2,j} + \cdots + w_m a_{m,j}.$$ The mean of these interpolated values is $$\bar a(o) = \frac{1}{n}\sum_j a(o,t_j).$$ If you expand this out, you'll find that it's equal to $$\begin{align}\bar a(o) &amp;amp;= w_1 \frac{1}{n}\sum_j a_{1,j} + w_2 \frac{1}{n}\sum_j a_{2,j} + \cdots + w_m \frac{1}{n}\sum_j a_{m,j} \\ &amp;amp;= w_1 \bar a_1 + w_2 \bar a_2 + \cdots + w_m \bar a_m,\end{align}$$ which amounts to interpolating the means. Presumably this is better from a computational perspective because if you want to find $\bar a(o)$ at many different $o$, you can keep the $\bar a_i$'s precomputed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your strategy of taking the mean of the interpolated values would be necessary if the sample points were moving. In that case, there's no sensible way to interpolate the means, because the interpolation weights would have been changing.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-27T01:25:47.230" CommentCount="4" />
  <row Id="15604" PostTypeId="2" ParentId="15602" CreationDate="2010-12-27T01:38:55.613" Score="15" ViewCount="0" Body="&lt;p&gt;If you had an orientation, you'd be able to define at each point $p$ a unit vector $n_p$ normal to the strip at $p$, in a way that the map $p\mapsto n_p$ is continuous. Moreover, this map is completely determined once you fix the value of $n_p$ for some specific $p$. (You have two possibilities, this uses a tangent plane at $p$, which is definable using a $(U_\alpha,\phi_\alpha)$ that covers $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the positivity condition you wrote gives you that the normal at any $p&amp;#39;$ is independent of the specific $(U_{\alpha&amp;#39;},\phi_{\alpha&amp;#39;})$ you may choose to use, and path connectedness gives you the uniqueness of the map. Now you simply check that if you follow a loop around the strip, the value of $n_p$ changes sign when you return to $p$, which of course is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is just a formalization of the intuitive argument.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-27T01:38:55.613" CommentCount="4" />
  <row Id="15605" PostTypeId="1" AcceptedAnswerId="15611" CreationDate="2010-12-27T01:57:47.227" Score="11" ViewCount="313" Body="&lt;p&gt;Let M be a smooth manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that for all $\alpha, \beta$, $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1})) &amp;gt; 0$ (where defined). I'm struggling to understand the reason this definition is made. My question is: What is the intuitive reason for this definition of an orientable manifold? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="640" LastEditorDisplayName="" LastEditDate="2010-12-27T03:45:48.947" LastActivityDate="2010-12-31T15:10:07.457" Title="How should I think about what it means for a manifold to be orientable? " Tags="&lt;geometry&gt;&lt;topology&gt;&lt;differential-topology&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="4" />
  <row Id="15606" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T02:32:20.457" Score="5" ViewCount="0" Body="&lt;p&gt;Get two intersecting charts $(U,\phi)=(U, (x^i))$, $(V,\psi)=(V, (y^i))$, with $p\in U\cap V$. Then there are two basis for the tangent space of $M$ at $p$: $\{ \frac{\partial}{\partial x^\alpha}\}$ and $\{\frac{\partial}{\partial y^\beta}\}$ . Intuitively, these two bases have the same orientation (in the vector space $T_pM$) if it is possible to go &quot;continuously&quot; from one to another. More precisely, they have $the$ $same$ orientation if the determinant of the matrix of change of basis from $\{ \frac{\partial}{\partial x^\alpha}\}$ to $\{\frac{\partial}{\partial y^\beta}\}$ is positive. If this determinant is negative, we will have to change the direction of one or more axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok. Given that, the question is: &quot;How to extend this concept of orientations from one tangent space to the entire manifold&quot;? That is where we define the $orientability$ of the manifold: a manifold will be $orientable$ if it admits an atlas such that, given two charts in this atlas with a nonempty intersection, at every point of this intersection the determinant of the matrix of change of basis from one coordinate basis to the other is positive (just like above). That is, the coordinate basis $\{ \frac{\partial}{\partial x^\alpha}\}$, $\{\frac{\partial}{\partial y^\beta}\}$ at a given point generated by any two charts of this &quot;orientable atlas&quot; have all the same orientation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix of change of basis between two coordinate basis is exactly the jacobian of the coordinate transformation.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2010-12-27T02:32:20.457" />
  <row Id="15607" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T02:34:12.000" Score="5" ViewCount="0" Body="&lt;p&gt;I am of the opinion that overthinking about &lt;em&gt;what things mean&lt;/em&gt; at the time one is first learning them can often be counterproductive. I like to quote, and fully agree with, D'Alambert, who famously said&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Allez en avant, et la foi vous viendra. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&quot;push on and faith will catch up with you&quot;, according to a translation I just googled for) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your concrete question: playing tetris, playing with your right hand and linear algebra should have convinced you that a real vector space has two orientations. Now, a manifold provides, among other things, a family of vector spaces with a piece of extra information: a way to move from one of them to others &quot;continuously&quot;. We say that the manifold is orientable if we can pick an orientation in each of them in a continuous way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I promise you that in not a long time, it'll make way more sense and you'll even find it difficult to explain it to others because it'll seem &lt;em&gt;intuitive&lt;/em&gt; to you!&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-27T02:34:12.000" CommentCount="7" />
  <row Id="15608" PostTypeId="2" ParentId="15555" CreationDate="2010-12-27T02:36:46.320" Score="2" ViewCount="0" Body="&lt;p&gt;What is the value of $\lceil x\rceil$ for small positive $x$? Think of $0&amp;lt;x&amp;lt;1$.&lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2010-12-27T02:36:46.320" />
  <row Id="15609" PostTypeId="2" ParentId="15555" CreationDate="2010-12-27T02:39:01.813" Score="0" ViewCount="0" Body="&lt;p&gt;$\lim \limits_{x \to 0} \frac{\lceil x \rceil}{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=lim+%28Ceiling%5bx%5d/x%29+as+x-%3E0&quot; rel=&quot;nofollow&quot;&gt;This is what the Alpha Wolf says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim \limits_{x \to 0^-} \frac{\lceil x \rceil}{x} = 0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lim \limits_{x \to 0^+} \frac{\lceil x \rceil}{x} = \infty$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two limits aren't equal. And what could this possible mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did we get this? Look &lt;a href=&quot;http://www.wolframalpha.com/input/?i=lim%20%28Ceiling%5bx%5d%29%20as%20x-%3E0&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim \limits_{x \to 0^-} \lceil x \rceil = 0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lim \limits_{x \to 0^+} \lceil x \rceil = 1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\lim \limits_{x \to 0^-} x = 0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\lim \limits_{x \to 0^+} x = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="2736" LastEditorDisplayName="" LastEditDate="2010-12-27T03:23:23.053" LastActivityDate="2010-12-27T03:23:23.053" CommentCount="4" />
  <row Id="15611" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T03:43:06.353" Score="6" ViewCount="0" Body="&lt;p&gt;I just want to flesh out something Ronaldo said &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Intuitively, these two bases have the same orientation (in the vector space $T_pM$) if it is possible to go &quot;continuously&quot; from one to another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To this end, consider the vector space $\mathbb{R}^n$.  Let $Gl = Gl_n(\mathbb{R})$ be the set of all invertible linear maps from $\mathbb{R}^n$ to itself and let $M = M_n(\mathbb{R})$ be the set of all linear maps (possibly noninvertible) from $\mathbb{R}^n$ to itself.  The space $M$ is naturally a vector space and is naturally isomorphic to $\mathbb{R}^{n^2}$.  Use this identification to topologize $M$.  The important thing about this topology is that polynomials in the entries of a matrix are continuous - this is all we'll use about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a map (the determinant) $det:M\rightarrow \mathbb{R}$, which is a continuous homomorphism (it's given as a polynomial in the entries of the matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can identify $Gl$ with $det^{-1}(\mathbb{R}-\{0\})$, i.e., $Gl$ is an open subset of $M = \mathbb{R}^{n^2}$.  Give $Gl$ the subspace topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we get to the main claim:  With this topology, $Gl$ has precisely two path components.  I'll prove this in a second.  In the mean time, notice that this makes what Ronaldo said precise:  If we start in the component of $Gl$ containing the identity matrix, then a path of elements in $Gl$, starting at the identity matrix, gives rise to a path of basis vectors starting with whatever choice of basis you've started with - it gives you a way to continuously go from one basis to another.  But you'll never be able to get to things of opposite orientation because that would require a path from one component of $Gl$ to another, which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'll establish the claim that $Gl$ has precisely two components.  First notice that is must have at least two components because $det:Gl\rightarrow \mathbb{R}^n-\{0\}$ is continuous, surjective, and the range has two components.  Let $X^+\subseteq Gl$ be those matrices of positive determinant and $X^-$ those of negative determinant, so we have $Gl = X^+\cup X^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I claim that $X^+$ and $X^-$ are homeomorphic.  To see this, let $A\in X^-$ be anything at all and consider the map $f:X^+\rightarrow X^-$ given by $f(B) = BA$.  Since matrix multiplication is given by polynomially in the entries of the matrices, $f$ is continuous.  It has inverse $f^{-1}(B) = BA^{-1}$, which is continuous for the same reason $f$ is.  Thus, $f$ is a homeomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the following maps and paths will be continuous for the same reason: they'll be given as polynomials in the entries of the matrices involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I just need to show $X^+$ is path connected.  So, let $A\in X^+$ be arbitrary.  I want to connect it to the identity matrix by a path.  Let $E_{ij}$ denote the matrix of all 0s with a 1 in the $ij$th slot.  Let $F_{ij}(t) = I + tE_{ij}$.  Notice that for $i\neq j$, $F_{ij}$ is triangular, so has determinant 1.  Left multiplying by $F_{ij}(t)$ corresponds to the row operation of adding $t$ times the $j$th row to the $i$th row, which doesn't change the determinant if $i\neq j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $F_{ij}(t_0)$ are just row operations, we know we can multiply by several of them (each with a different appropriate $t_0$) to bring $A$ to a diagonal form without changing the determinant.  By multiplying by $F_{ij}(t)$ and letting $t$ go from $0$ to $t_0$, we make a path out of this.  Doing this for all the $F_{ij}(t_0)$ simultaneously gives us a path from $A$ to a diagonal matrix where for all time, the determinant is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've now connected $A$ to a diagonal matrix $D$ by a path in $X^+$.  It remains to connect the diagonal matrix to the identity.  We know all the entries of $D$ are nonzero because $D$ has positive determinant.  If an entry $d$ of $D$ is positive, use the path $t + (1-t)d$ in that slot to continuously change $d$ to $1$, all the while keeping the determinant nonzero.  If $d$ is negative, using the same trick, we can change $d$ to $-1$.  At this point, the determinant must be $1$ - it's positive because we're in $X^+$ and it's a product of $\pm 1$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've now reduced $D$ to a new diagonal matrix $E$ whose entries are plus and minus 1s.  If all are positive, we're done, so assume we have a $-1$ somewhere.  Since $E$ is in $X^+$, there must be another entry which is $-1$.  For notation, let's assume one $-1$ is in the $i$th position and the other is in the $j$th position.  Think of the two $-1$s as $cos(t\pi)$ and think of the $ij$th entry and $ji$th entries as $sin(t\pi)$, all with $t=1$.  Now, let t go from 1 to 0.  One can easily check the determinant stays $1$, and this gives a path which changes the two $-1$s to $1$s.  Doing this finitely many times eliminates all the $-1$s and changes $E$ to the identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we've connected $A$ to the identity matrix by a path, so $X^+$ is connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Kudos to anyone who reads the whole thing - it ended up being WAY longer than I anticipated, but it was a good exercise for me to go through).&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-12-31T15:10:07.457" LastActivityDate="2010-12-31T15:10:07.457" CommentCount="2" />
  <row Id="15612" PostTypeId="1" CreationDate="2010-12-27T04:04:50.973" Score="1" ViewCount="137" Body="&lt;p&gt;Let $a,b$ be roots ($a\ne \pm b$) of a Lie algebra $g$ of type $X$, where $X$ can be classic or exceptional $(A,B,C,D,E,F,G)$. It is well known that the length of an $a$-string through $b$ is at most 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are all types of $g$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $a+b$ is a root and $a+2b$ is not a root?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $a+2b$ is a root and $a+3b$ is not a root?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $a+3b$ is a root?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastEditorUserId="2764" LastEditorDisplayName="" LastEditDate="2010-12-29T02:21:15.510" LastActivityDate="2010-12-29T02:21:15.510" Title="Length of root strings." Tags="&lt;lie-algebra&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="15614" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T04:41:21.350" Score="4" ViewCount="0" Body="&lt;p&gt;Suppose the universe has shape $M$.  Then $M$ is orientable if and only if you can write yourself a note, then begin a journey, and come back to always find that you can still read the note.  If $M$ is not orientable, it is possible to return from a journey to find that the note is written backwards and, indeed, that everything around you is written backwards, your friends' faces are all reversed, etc.  More seriously, the proteins in your body would have the wrong chirality and I think this would cause problems involving digestion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not known, to the best of my knowledge, whether our universe is orientable.  If you are interested in this more intuitive approach to topology and geometry before attacking its rigorous foundations, I recommend reading Weeks' &lt;a href=&quot;http://books.google.com/books?id=A8WBiUWy3SgC&amp;amp;printsec=frontcover&amp;amp;dq=the+shape+of+space&quot; rel=&quot;nofollow&quot;&gt;The Shape of Space&lt;/a&gt; and being very diligent about doing the exercises.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-27T04:41:21.350" CommentCount="4" />
  <row Id="15615" PostTypeId="2" ParentId="15542" CreationDate="2010-12-27T05:22:21.563" Score="6" ViewCount="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It may seem strange to define $0\cdot\infty=0$.  However, one verifies without difficulty that with this definition the &lt;em&gt;commutative, associative, and distributive laws hold in $[0,\infty]$ without any restriction&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way this is worded leads naturally to your question, as though Rudin were implying that this is the main justification for defining $0\cdot\infty$ in this way.  Rather, I see this as a bonus after making the convention consistent with what happens when integrating the $0$ function or integrating over a space of measure $0$, as KCd's comment indicates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask yourself what the possibilities are, you can start by supposing that $0\cdot\infty=x$ for some $x\in[0,\infty]$, and apply the distributive law to see that $x=2x$, so that $x=0$ or $x=\infty$.  You can then verify that with either convention the commutative, associative, and distributive laws will hold, so something more is needed to motivate the choice.  Such a choice will always depend on context, and in some cases it won't be a good idea to even define $0\cdot\infty$.  However, I am not aware of a mathematical context in which the convention $0\cdot\infty=\infty$ is useful.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-27T05:22:21.563" />
  <row Id="15616" PostTypeId="1" CreationDate="2010-12-27T05:45:41.520" Score="2" ViewCount="80" Body="&lt;p&gt;Suppose c is an isolated point in the domain D of a function f.&#xA;In the delta neighbourhood of c, does the function f have the value f(c).&lt;/p&gt;&#xA;" OwnerUserId="5049" LastActivityDate="2010-12-27T06:18:20.590" Title="Continuity of a function at an isolated point" Tags="&lt;real-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15617" PostTypeId="2" ParentId="15616" CreationDate="2010-12-27T06:10:56.420" Score="3" ViewCount="0" Body="&lt;p&gt;I see now that the comments above provide essentially the answer, with whatever definition of continuity you have. The following ties everything together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use as the definition of continuity $\lim_{x \rightarrow c}\,f(x) = f(c)$. Expand: For all $\varepsilon &amp;gt; 0$ there exists $\delta &amp;gt; 0$ such that whenever $0 &amp;lt; |x-c| &amp;lt; \delta$ it is true that $|f(x)-f(c)| &amp;lt; \varepsilon$. When $\delta$ is small enough, there are no points $x$ which work, so the part after the such that is &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuous_truth&quot; rel=&quot;nofollow&quot;&gt;vacuously true&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5124" LastActivityDate="2010-12-27T06:10:56.420" CommentCount="2" />
  <row Id="15618" PostTypeId="2" ParentId="15616" CreationDate="2010-12-27T06:18:20.590" Score="5" ViewCount="0" Body="&lt;p&gt;You can also see this is true using the topological definition of continuity at a point: a function is continuous at a point $f(x)$ if for any neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)$ is contained in $V$.  For an isolated point, you can take the neighborhood consisting of just the point $c$, so its image $f(c)$ will obviously be contained in $V$, as $V$ is a neighborhood of $f(c)$.&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2010-12-27T06:18:20.590" />
  <row Id="15619" PostTypeId="2" ParentId="15571" CreationDate="2010-12-27T07:06:29.017" Score="4" ViewCount="0" Body="&lt;p&gt;First of all please agree my apologies because of my attitude and the way I asked my question. I changed it again. I really didn't aggressive in reality just present my question in a bad way. But believe me or not, whenever I asked this kind of questions from my teachers they down voted me in real world. I should ask my question in more warily way. As Yuan commented in my question &quot;I'm not going to get any upvotes from christians.stackexchange.com if I ask them &quot;is Christianity based on lies?&quot;&quot;. BTW, I didn't sleep last night and thinking about my question and these answers and comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read all these 3 answers many times. And thank you. I think the real answer is combination of these 3 answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mathematics is different from Physics. We can't completely compare them with each other. Mathematics is an abstract science while Physics is natural science. So they are different types.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We should choose an axiomatic system before talking about a theorem. ( like ZFC ).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;About infinity: This is a part of Mathematics not Physics. We can imagine that there is a number bigger than any number. That number is Infinity ! ... I don't know if it exists in physics or not. For example &quot;is world infinity or not !?&quot;... &quot;Is there a new world inside a black hole or not!&quot; ... But in mathematics we can imagine it. And we can work with it without much problem. In many parts we see unbelievable theorems( in physics ) when we assume infinity is exists. But they are logical. So we can accept them. Infinity is just a tool. Nothing else I think!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;About Zero: zero is a concept. It's not real. We can't see it. But this concept is exists. Show me the word &quot;lie&quot;. What is a &quot;lie&quot; ? I can't see it. but this is a concept and exists. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Empty set: I can't imagine a set which don't have any element in Physics and reality. But I can imagine it in my mind. I can imagine &quot;nothing&quot;. it's a &lt;strong&gt;concept&lt;/strong&gt; again. so we call it &quot;empty set&quot; in mathematics. A set that there is nothing in there. I think I just playing with words in my question. nothing is not an element. it's a word. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Uncountable sets: If I can accept infinity. I can accept this too. So I have no word to say about.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think this answer is a combination of these answers. I found my answers I think. Thank you again for helping me get out of this confusion. But please let me know if this answer is wrong !&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="5103" LastEditorDisplayName="" LastEditDate="2010-12-27T17:18:28.257" LastActivityDate="2010-12-27T17:18:28.257" CommentCount="3" />
  <row Id="15620" PostTypeId="1" AcceptedAnswerId="15621" CreationDate="2010-12-27T07:32:31.437" Score="3" ViewCount="116" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, well-order is&#xA;defined as a strict total order on a set S&#xA;with the property that every&#xA;non-empty subset of S has a least&#xA;element in this ordering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order#Equivalent_formulations&quot; rel=&quot;nofollow&quot;&gt;later&lt;/a&gt;, well-order is&#xA;defined as a total order on S with&#xA;the property that every non-empty&#xA;subset of S has a least element in&#xA;this ordering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, a total order and&#xA;a strict total order are different.&#xA;One is not the other. So I was&#xA;wondering if well-order is defined&#xA;for total order or strict total&#xA;order or both? If for both, are they&#xA;equivalent in the sense that if a&#xA;total order is well-order, then its&#xA;corresponding strict total order is&#xA;also well-order? Vice versa?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;At the same &lt;a href=&quot;http://en.wikipedia.org/wiki/Well-order&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; page, it&#xA;also says &quot;a well-ordering is a&#xA;well-founded strict total order&quot;. As&#xA;I clicked into the definition of&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Well-founded_relation&quot; rel=&quot;nofollow&quot;&gt;Well-founded_relation&lt;/a&gt;, it says&#xA;&quot;a binary relation, R, is&#xA;well-founded (or wellfounded) on a&#xA;class X if and only if every&#xA;non-empty subset of X has a minimal&#xA;element with respect to R&quot;. As&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Minimal_element&quot; rel=&quot;nofollow&quot;&gt;minimal element&lt;/a&gt; is defined for&#xA;partial order not for strict total order, is it true that&#xA;well-founded order is a partial&#xA;order and not a strict total order? So the&#xA;aforementioned &quot;a well-ordering is a&#xA;well-founded strict total order&quot; is&#xA;not well-stated?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks and regards! &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="1281" LastEditorDisplayName="" LastEditDate="2010-12-27T07:38:59.953" LastActivityDate="2010-12-27T08:13:17.187" Title="questions about well-order" Tags="&lt;elementary-set-theory&gt;&lt;order-theory&gt;" AnswerCount="3" />
  <row Id="15621" PostTypeId="2" ParentId="15620" CreationDate="2010-12-27T08:02:38.707" Score="4" ViewCount="0" Body="&lt;p&gt;For the first question, taking strict and non-strict orders to be well-ordering is up to you and the way you use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in set theory it is generally easier to use strict ordering in the definition because it saves the trouble with $x\le y\wedge y\le x$, furthermore we want $\in$ to define some relations and it has to be strict by the axiom of foundation (also known as axiom of regularity in some parts of the globe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for minimal elements, they are not only for partial orders - but for any order. If $R$ is some relation on $A$ then $x$ is $R$-minimal if $\forall y (yRx \rightarrow y=x)$, note by the way that if $R$ is not reflexive then this is still true, but you could phrase it as $\forall y\neg(yRx)$ instead, which is clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way, in my opinion to understand deeply these choices of definitions is to study some set theory theorems about recursive definitions, transfinite inductions and the needed theorems for those. In these proofs it becomes very clear why one prefer strict relations over non-strict ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final remark, although strict and non-strict total orders are &quot;very different&quot;, they only differ by reflexivity which is some vacuous condition that you want to add when it's easier to have it - and you remove it when you find it easier to handle without it.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-27T08:02:38.707" />
  <row Id="15622" PostTypeId="2" ParentId="15620" CreationDate="2010-12-27T08:07:42.543" Score="4" ViewCount="0" Body="&lt;p&gt;&lt;br&gt;1. If a set is totally ordered, the strict total order is implied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of total order from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall a, b, c \in X$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $a ≤ b$ and $b ≤ a$ then $a = b$ (antisymmetry);&lt;br&gt;&#xA;  If $a ≤ b$ and $b ≤ c$ then $a ≤ c$ (transitivity);&lt;br&gt;&#xA;  $a ≤ b$ or $b ≤ a$ (totality).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Definition of strict total order from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\forall a, b \in X$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$a &amp;lt; b$ if and only if $a ≤ b$ and $a ≠ b$&lt;br&gt;&#xA;  $a &amp;lt; b$ if and only if not $b ≤ a$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you compare the definitions, you will see that there are no conditions under which the second set of rules is invalid, assuming the first set of rules. So, saying that a set is well-ordered if it is totally ordered makes sense, because total order implies existence of strict total order (and hence, well-ordering) on the given set.&lt;br&gt;&lt;br&gt;&#xA;2. I am fairly sure that one can use the term 'minimal element' with respect to totally ordered sets also.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-27T08:13:17.187" LastActivityDate="2010-12-27T08:13:17.187" CommentCount="4" />
  <row Id="15623" PostTypeId="2" ParentId="15620" CreationDate="2010-12-27T08:09:47.487" Score="7" ViewCount="0" Body="&lt;p&gt;If $\leq$ is a total order on a set $S$, then the new relation $&amp;lt;$ defined by $x &amp;lt; y$  iff ($x \leq y$ and $x \neq y$) is a strict total order on $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $&amp;lt;$ is a strict total order on a set $S$, then the new relation $\leq$ defined by &#xA;$x \leq y$  iff ($x &amp;lt; y$ or $x = y$) is a total order on $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, in a fairly evident way one can always exchange a total order for a strict total order and conversely.  So it doesn't really matter which definition is taken.  One can easily check that the definition of a well-order in one setting carries over to the definition of a well-order in the other setting.  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-27T08:09:47.487" CommentCount="3" />
  <row Id="15624" PostTypeId="1" CreationDate="2010-12-27T08:34:15.023" Score="4" ViewCount="146" Body="&lt;p&gt;I have a polygon in 2D (defined by a series of Vertex $V$ with coordinates). The polygon can be convex or concave. I have $n$ number of fix points I can put inside the polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, how can I distribute the fix points as uniformly as possible inside the polygon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation for this question is I want to create a mesh generator and I want all the triangular elements $E$ ( which is defined by a list of vertices $V$) to look good with no too small or too large angles. In order to control the granularity of the mesh I am thinking about using the number of fix points $n$ as the controlling parameter. The fix points are used to control the vertex of the triangular elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any algorithm for this?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="1489" LastEditorDisplayName="" LastEditDate="2010-12-27T23:09:31.777" LastActivityDate="2010-12-28T14:35:23.953" Title="Distribute a Fixed Number of Points &quot;Uniformly&quot; inside a Polygon" Tags="&lt;geometry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15625" PostTypeId="2" ParentId="15624" CreationDate="2010-12-27T09:05:53.300" Score="6" ViewCount="0" Body="&lt;p&gt;The simplest idea is to generate a sequence of points $(X_i,Y_i)$, where $X_i,Y_i$ are independent uniform random variables, in a rectangle containing the polygon (the minimal rectangle bounding the polygon). Then, you simply drop the points which fall outside the polygon. You continue until you get $n$ unrejected points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we could expect, this problem has been considered before by many people. You might find &lt;a href=&quot;http://www.gamedev.net/community/forums/topic.asp?topic_id=585566&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; very useful. The basic question is how fast/simple you want to generate the $n$ points. The method I described is most simple, but might be time consuming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: To add some mathematics to this post, let us justify the rejection method and triangulation approaches (though they might seem intuitively clear). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rejection method approach. Suppose that $(X,Y)$ is uniformly distributed in a rectangle $R$ bounding the polygon $M$, and that $S$ is an arbitrary square contained in $M$. Then,&#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S |(X,Y) \in M) = \frac{{{\rm P}((X,Y) \in S )}}{{{\rm P}((X,Y) \in M)}} = &#xD;&#xA;\frac{{{\rm area}(S) /{\rm area}(R)}}{{{\rm area}(M)/{\rm area}(R)}} = \frac{{\rm area}(S) }{{{\rm area}(M)}}.&#xD;&#xA;$$ &#xA;Hence, given that $(X,Y) \in M$, $(X,Y)$ is uniformly distributed in $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triangulation approach. Suppose that the polygon $M$ is partitioned into triangles $T_1,\ldots,T_m$, and set $p_i = {\rm area} (T_i)/ {\rm area}(M)$. Suppose that ${\rm P}((X,Y) \in T_i)=p_i$ and that, given $(X,Y) \in T_i$, $(X,Y)$ is uniformly distributed in $T_i$. Now, for fixed $k$, let $S \subset M$ an arbitrary square contained in $T_k$ (think of arbitrarily small squares). Then,&#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S) = \sum\limits_{i = 1}^m {{\rm P}((X,Y) \in S|(X,Y) \in T_i )p_i } = {\rm P}((X,Y) \in S|(X,Y) \in T_k )p_k.&#xD;&#xA;$$&#xA;Since, given that $(X,Y) \in T_k$, $(X,Y)$ is uniformly distributed in $T_k$, we have&#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S|(X,Y) \in T_k )  = \frac{{{\rm area}(S)}}{{{\rm area}(T_k )}}.&#xD;&#xA;$$&#xA;Hence, &#xA;$$&#xD;&#xA;{\rm P}((X,Y) \in S) = \frac{{{\rm area}(S)}}{{{\rm area}(T_k )}}\frac{{{\rm area}(T_k )}}{{{\rm area}(M)}} = \frac{{{\rm area}(S)}}{{{\rm area}(M)}}.&#xD;&#xA;$$&#xA;We conclude that $(X,Y)$ is uniformly distributed in $M$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-28T14:35:23.953" LastActivityDate="2010-12-28T14:35:23.953" CommentCount="6" />
  <row Id="15626" PostTypeId="2" ParentId="7867" CreationDate="2010-12-27T09:36:07.397" Score="0" ViewCount="0" Body="&lt;p&gt;Say you have $n$ data points $\{x_1,x_2,x_3,\ldots,x_n\}$. You have a range of values with which to compare them $[a,b]$. Then, as Ross suggests, I would also assign a score to this set of data points. I would base it on the standard deviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values that fall inside the interval get score zero, values $x_k$ that fall outside the interval get score $\min((x_k-a)^2,(x_k-b)^2)$. But if you prefer to take just absolute deviations like Ross, just adapt the formula. Then simply add all scores for all the data points to obtain one score for the entire set of data points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower the score, the closer the points are to the interval, the higher the farther they lie out. If you dispose of several data sets with different scores, you can define percentage deviations by looking at what percentage of data have a lower score than the data set you are examining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a small R script that illustrates the idea:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;n=1000;&#xA;size=10;&#xA;big&amp;lt;-matrix(rnorm(n,0,1),n%/%size,size);&#xA;liminf=-1;&#xA;limsup=1;&#xA;z=rep(0,length(big[,1]);&#xA;perc=z;&#xA;for (k in 1:length(big[,1]))&#xA;{ z[k]&amp;lt;-sum((big[k, big[k,]&amp;lt;liminf]-liminf)^2)+sum((big[k,big[k,]&amp;gt;limsup]-limsup)^2) ; }&#xA;mat&amp;lt;-matrix(c(1,2),1,2);&#xA;layout(mat);&#xA;hist(z,10,main=&quot;Score histogram&quot;);&#xA;for (k in 1:length(big[,1]))&#xA;{ perc[k]=length(z[z&amp;lt;z[k]])/length(z); }&#xA;plot(sort(z),sort(perc),main=&quot;Cumulative plot&quot;);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It makes a big matrix of 100 rows of each ten columns. Each row is to be interpreted as one data set. The data have been sampled from a random normal distribution, but you can adapt to any type of data of course. Then, I calculate the score of each row with respect to a chosen interval, here the interval $[-1,1]$. I plotted a score histogram as a bonus. After that, I compute for each score how many percent of the scores is smaller than that score. That's what the &lt;code&gt;perc&lt;/code&gt; is for. I also plotted the cumulative function for the scores based on the computed function &lt;code&gt;perc&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-27T13:09:44.917" LastActivityDate="2010-12-27T13:09:44.917" />
  <row Id="15627" PostTypeId="1" CreationDate="2010-12-27T09:38:34.543" Score="-2" ViewCount="224" Body="&lt;p&gt;Just solve it with your own mind without having a look to other answers !&lt;/p&gt;&#xA;" OwnerUserId="5139" LastActivityDate="2010-12-27T09:44:47.807" ClosedDate="2010-12-27T09:58:55.183" Title="1=5, 2=10, 3=15, 4=20 then 5=?" Tags="&lt;puzzle&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15628" PostTypeId="2" ParentId="15627" CreationDate="2010-12-27T09:44:47.807" Score="2" ViewCount="0" Body="&lt;p&gt;I see two ways you can look at this. One appears to be obvious (and is probably not something you are looking for, or rather something that you might expect us to say) -- $1=5$ means $1*5=5$, $2=10$ means $2*5=10$, and so on. Then, $5=?$ means $5*5=25$. Since this is too obvious, the answer could be $5=1$ (which is the very first equality that you provided). I think one could find some other possible answers, too. :)&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-27T09:44:47.807" CommentCount="5" />
  <row Id="15629" PostTypeId="1" AcceptedAnswerId="15633" CreationDate="2010-12-27T11:09:31.710" Score="2" ViewCount="292" Body="&lt;p&gt;How do I correctly represent the following pseudocode in math notation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1: Formula expanded.&#xA;EDIT2: Clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a,b) represents a line segment on a 1D line. a &amp;lt;= b for each segment. The division show below is done as per the following T-SQL code (which I suppose could be represented as a function in the formula?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input:  @a1 real, @b1 real, @an real, @bn real&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;DECLARE @Result real&#xA;&#xA;if @a1 &amp;lt;= @an begin&#xA;    SET @Result = @an - @b1&#xA;&#xA;    if @Result &amp;lt;= 0 RETURN 0&#xA;&#xA;    RETURN @Result / @an&#xA;end&#xA;&#xA;SET @Result = @a1 - @bn&#xA;&#xA;if @Result &amp;lt;= 0 RETURN 0&#xA;&#xA;RETURN @Result / @a1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if m = 1 then&#xA;  if (a,b)_1 intersects (a,b)_n then&#xA;    r = 1&#xA;  else if (a,b)_1 &amp;lt; (a,b)_n then&#xA;    r = (a,b)_1 / (a,b)_n&#xA;  else&#xA;    r = (a,b)_n / (a,b)_1&#xA;else if m = 2 then&#xA;  if (a,b)_1 intersects (a,b)_n then&#xA;    r = 1&#xA;  else if (a,b)_1 &amp;lt; (a,b)_n then&#xA;    r = (a,b)_1 / (a,b)_n&#xA;  else&#xA;    r = (a,b)_n / (a,b)_1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The m = 2 block is shown as being the same as the m = 1 one for simplicity's sake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The divisions are against the two points that are closets to each other, unless the segments intersect, at which point r = 1.&lt;/p&gt;&#xA;" OwnerUserId="4930" LastEditorUserId="4930" LastEditorDisplayName="" LastEditDate="2010-12-27T11:52:53.543" LastActivityDate="2010-12-28T06:44:31.907" Title="Representing IF ... THEN ... ELSE ... in math notation" Tags="&lt;computer-science&gt;&lt;notation&gt;&lt;elementary&gt;" AnswerCount="4" CommentCount="12" />
  <row Id="15630" PostTypeId="2" ParentId="15592" CreationDate="2010-12-27T11:40:50.750" Score="1" ViewCount="0" Body="&lt;p&gt;I will not give you an answer but a geometrical figure in proving what is equivalent to the naturals. You can think that the natural numbers is an elastic rope (infinite from one side )&#xA;and the natural numbers are marked points on this rope .If you want to prove that the integers have the same cardinality as $N^2$ you can think that $N^2$ is a lattice in $R^2$ and you have to create an 1-1 and surjective function or start passing the &quot;natural&quot; rope through the lattice such that every natural number fits on every vertex of the lattice .You can see then what is the function i.e. 1 goes to (0,1) 2 goes to (1,1) 3 goes to (1,0) 4 to (2,0) and you create a spiral round the (0,0).It is an 1-1 surjective function that sends n to (exercise).The same goes to 3 or n dimensions. Of course you can have different ways to use the elastic natural rope to create bijective functions (of course you must have a way to pass the rope through every vertex so usually you make a spiral in a way that grows biger and biger)  but you can have an image that $N$ have the same cardinality as $N^k$ or $Z^k$ for any k.  &lt;/p&gt;&#xA;" OwnerUserId="5010" LastActivityDate="2010-12-27T11:40:50.750" CommentCount="1" />
  <row Id="15631" PostTypeId="1" AcceptedAnswerId="15695" CreationDate="2010-12-27T11:54:30.867" Score="2" ViewCount="194" Body="&lt;p&gt;Hello, the problem is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all $x \in \mathbb{R}$, we assign a finite set $\Phi(x) \subset \mathbb{R} - \{ x \}$. We say that a set $S \subset \mathbb{R}$ is independent if for any $x,y \in S$, $x \notin \Phi(y)$ and $y \notin \Phi(x)$, that is, if $S \cap \Phi(S) = \emptyset$. Prove that there is an independent set $S$ with the cardinality of the continuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2926" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-27T12:20:19.963" LastActivityDate="2010-12-28T00:31:36.213" Title="Problem in set theory" Tags="&lt;set-theory&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="15632" PostTypeId="1" AcceptedAnswerId="15636" CreationDate="2010-12-27T12:40:31.217" Score="1" ViewCount="95" Body="&lt;p&gt;Given implicit function F(x,y) = 0, how can I find its asymptotes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  Sorry, my calculations were wrong. Here is correct function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(x,y)=\sqrt{(x-a)^2 + (y-b)^2} - \sqrt{(x-c)^2 + (y-d)^2}  - e$&lt;/p&gt;&#xA;" OwnerUserId="3532" LastEditorUserId="3532" LastEditorDisplayName="" LastEditDate="2010-12-27T16:04:28.563" LastActivityDate="2010-12-28T03:34:17.990" Title="finding hyperbola asymptotes" Tags="&lt;geometry&gt;&lt;analytic-geometry&gt;&lt;plane-curves&gt;&lt;conic-sections&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="15633" PostTypeId="2" ParentId="15629" CreationDate="2010-12-27T12:44:26.410" Score="6" ViewCount="0" Body="&lt;p&gt;In general, if you have &quot;If $\varphi$ then $\psi$, otherwise $\tau$&quot; you can write this as the following formula (or sentence, depending on $\varphi,\psi,\tau$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varphi\rightarrow\psi\wedge\neg\varphi\rightarrow\tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have several cases, you can either nest them (i.e. $\tau$ would be &quot;if second condition then, else ...&quot;) or if you can express them as $\varphi_1$ meaning only the first case holds, and none of the others as $(\varphi_1\rightarrow\psi_1)\wedge(\varphi_2\rightarrow\psi_2)\wedge\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a=b\rightarrow x=1)\wedge(a&amp;lt;b\rightarrow x=\frac{a}{b})\wedge(a&amp;gt;b\rightarrow x=\frac{b}{a})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have added $x$ as a variable, because writing just $\rightarrow 1$ seems very meaningless to me, you can of course replace $x$ by anything you'd like.&#xA;The idea, essentially is that you express &quot;IF ... THEN ... ELSE&quot; structures using the $\rightarrow$ (or $\implies$ sometimes) and I gave you in my original post the method of doing so.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-27T13:34:22.457" LastActivityDate="2010-12-27T13:34:22.457" CommentCount="9" />
  <row Id="15634" PostTypeId="1" CreationDate="2010-12-27T13:02:08.373" Score="6" ViewCount="99" Body="&lt;p&gt;Let $f$ be a function of period 1, Riemann integrable on [0,1]. Let $\xi_n$ be a sequence&#xA;which is equidistributed in $[0,1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Is it true that $$\frac{1}{N}\sum_{n=1}^N f(x+\xi_n)$$ converges to the constant $\int_0^1 f(y) dy$&#xA;for each $x\in \mathbb{R}$ as $N\to \infty$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) If so, is the convergence uniform over all $x$?&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T22:10:46.240" LastActivityDate="2010-12-28T01:41:03.847" Title="Equidistributed sequence and Riemann integrable function" Tags="&lt;analysis&gt;&lt;sequences-and-series&gt;&lt;integral&gt;&lt;fourier-analysis&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15635" PostTypeId="2" ParentId="15562" CreationDate="2010-12-27T13:18:12.697" Score="1" ViewCount="0" Body="&lt;p&gt;While I personally don't remember the mathematics, I can give you two leads which, if you follow it up, should give you some ideas. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Try contacting &lt;a href=&quot;http://www.math.wisc.edu/~ellenber/&quot; rel=&quot;nofollow&quot;&gt;Jordan Ellenberg&lt;/a&gt;. Several years ago he gave a talk to the Princeton University Undergraduate Math Club on precisely what you described in your question. He will definitely be able to give you some pointers. &lt;/li&gt;&#xA;&lt;li&gt;Also, try asking &lt;a href=&quot;http://www-math.mit.edu/people/profile.php?pid=1272&quot; rel=&quot;nofollow&quot;&gt;Aaron Potechin&lt;/a&gt;. he is an expert Set player and I remember seeing a pre-print of his on the Maximal Cap issue a few years ago, when he was still an undergrad. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-27T13:18:12.697" CommunityOwnedDate="2010-12-27T13:18:12.697" CommentCount="1" />
  <row Id="15636" PostTypeId="2" ParentId="15632" CreationDate="2010-12-27T13:26:35.290" Score="3" ViewCount="0" Body="&lt;p&gt;After rationalization, your implicit equation can be expressed in standard quadratic form as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$px^2+qxy+ry^2+sx+ty+u=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=4(a-c)^2-4e^2$$&lt;br&gt;&#xA;$$q=8(a-c)(b-d)$$&lt;br&gt;&#xA;$$r=4(b-d)^2-4e^2$$&lt;br&gt;&#xA;$$s=4(a+c)e^2-4(a-c)\left(a^2+b^2-c^2-d^2\right)$$&lt;br&gt;&#xA;$$t=4(b+d)e^2-4(b-d)\left(a^2+b^2-c^2-d^2\right)$$&lt;br&gt;&#xA;$$u=e^4-2\left(a^2+b^2+c^2+d^2\right)e^2+\left(a^2+b^2-c^2-d^2\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earlier version of my answer would now apply, but since Isaac has written a clearer way of doing it, I'll demonstrate the alternative &quot;dumb&quot; approach (&quot;dumb&quot; in the sense of not exploiting the structure of the problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, before you go through the trouble of doing these computations, you'd first want to ensure that the discriminant $\Delta=64e^2((a-c)^2+(b-d)^2-e^2)$ is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having done so, the idea is as follows: replace all instances of $y$ in the implicit equation $F(x,y)=0$ with $mx$, divide by $x^2$ (since the degree of the algebraic equation is 2), and let $x\to\infty$. This yields the quadratic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rm^2+qm+p=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which should be solved for $m$; the initial check of the discriminant would ensure that both roots of that quadratic in $m$ are real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $m_i$ be any of the two roots $m_1$ and $m_2$ of that quadratic, one now goes back to the original implicit equation, and replaces $y$ with $m_i x+k_i$. From this, note that the quadratic terms disappear, and only the linear terms remain. Divide by $x$, let $x\to\infty$, equate the remaining terms to 0, and solve for the appropriate value of $k_i$. If you do that, the expression you should get for $k_i$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_i=-\frac{tm_i+s}{2rm_i+q}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which $y=m_1 x+k_1$ and $y=m_2 x+k_2$ are your two asymptotes.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T03:34:17.990" LastActivityDate="2010-12-28T03:34:17.990" CommentCount="3" />
  <row Id="15637" PostTypeId="2" ParentId="15634" CreationDate="2010-12-27T13:34:51.860" Score="3" ViewCount="0" Body="&lt;p&gt;For question (a), see &lt;a href=&quot;http://en.wikipedia.org/wiki/Equidistributed_sequence&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; (the section &quot;Properties&quot;); for a proof, see &lt;a href=&quot;http://individual.utoronto.ca/hannigandaley/equidistribution.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-27T14:05:35.863" LastActivityDate="2010-12-27T14:05:35.863" CommentCount="2" />
  <row Id="15638" PostTypeId="1" AcceptedAnswerId="15639" CreationDate="2010-12-27T13:41:33.467" Score="5" ViewCount="210" Body="&lt;p&gt;Can you make a simple example of an uncountable ordinal? With &lt;i&gt;simple&lt;/i&gt; I mean that it is easy to prove that the ordinal is uncountable. I know that the set of all the countable ordinals is an uncountable ordinal, but the only proof that I know is quite complicated.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-28T02:33:55.830" LastActivityDate="2010-12-28T02:33:55.830" Title="Simple example of uncountable ordinal" Tags="&lt;set-theory&gt;&lt;cardinals&gt;&lt;ordinals&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="15639" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T13:49:05.897" Score="10" ViewCount="0" Body="&lt;p&gt;The proof that the set $\Omega$ of all countable ordinals is uncountable is not difficult. First, it's an ordinal. Next, if $\Omega$ were countable, then $\Omega + 1$ would also be a countable ordinal. Finally, it is impossible for $\Omega + 1$ to be in $\Omega$, because that would mean that $\Omega + 1 &amp;lt; \Omega$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastActivityDate="2010-12-27T13:49:05.897" CommentCount="10" />
  <row Id="15640" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T14:02:18.020" Score="4" ViewCount="0" Body="&lt;p&gt;The only two that I think have simple proofs are $\Omega$, as Carl says, and $2^{\aleph_0}$, the set of infinite binary sequences.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument&quot; rel=&quot;nofollow&quot;&gt;Cantor's diagonal argument&lt;/a&gt; proves this one uncountable.  Alternately, as all cardinals are ordinals, any cardinal greater than $\aleph_0$ is an uncountable ordinal just from the definition of countability.  That's simple to write, but probably not what you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-27T14:02:18.020" CommentCount="5" />
  <row Id="15641" PostTypeId="1" CreationDate="2010-12-27T14:07:04.457" Score="2" ViewCount="51" Body="&lt;p&gt;Given an abelian group $X$, let $F_n(X)$ denote the simplicial abelian group defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_n(X)_j=0$ for all $j&amp;lt;n$ and $F_n(X)_j=X$ for all $j\geq n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the appropriate zero and identity maps between them so the normalization (normalized moore complex) gives the appropriate homology (say, $d_i=0$ for every $i\geq 1$ (obviously for the parts higher than $n$) and $d_0=id_X$).  Then by the Hurewicz theorem, this simplicial abelian group seems like it should have homotopy concentrated in degree $n$ with $n$th component isomorphic to $X$ by computing the homology of the normalization.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this thing is a Kan complex (because it's a simplicial abelian group), isn't this a representative for the Eilenberg-Mac Lane space $\kappa(X,n)$?&lt;/p&gt;&#xA;" OwnerUserId="5142" LastActivityDate="2010-12-27T14:07:04.457" Title="What's stopping me from choosing the nth Eilenberg Mac Lane space to be the following simplicial abelian group?" Tags="&lt;homotopy-theory&gt;" CommentCount="3" />
  <row Id="15642" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T14:07:34.300" Score="5" ViewCount="0" Body="&lt;p&gt;The existence of an uncountable ordinal is established by the argument of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burali-Forti_paradox&quot; rel=&quot;nofollow&quot;&gt;Burali-Forti &quot;Paradox&quot;&lt;/a&gt;.  See for instance Section 1.4 &lt;a href=&quot;http://math.uga.edu/~pete/settheorypart3.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  (Yes, this is the same argument as in Carl Mummert's answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably the simplest way to show the existence of an uncountable ordinal.  As Section 1.2 of the notes linked to above shows, the usual operations of addition, multiplication and even exponentiation of ordinals do not produce uncountable ordinals from countable ones.  That this is not the case for exponentiation is a big difference between ordinal and cardinal arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-27T14:07:34.300" CommentCount="2" />
  <row Id="15643" PostTypeId="2" ParentId="15631" CreationDate="2010-12-27T14:58:33.393" Score="3" ViewCount="0" Body="&lt;p&gt;Here is a sketch of a proof (it uses axiom of choice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's introduce a equivalence relation on real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that two numbers $x$ and $y$ are &quot;equal&quot; of there exists a finite&#xA;set of reals $a_0 = x$, $a_1$, $a_2$, $\cdots$, $a_{n-1}$, $a_n = y$ such&#xA;that for any $i$ we have $a_i \in \Phi(a_{i+1})$ or $a_{i+1} \in \Phi(a_i)$.&#xA;Note that $n$ could be zero, so $x$ and $x$ are also considered equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that this equivalence relation is symmetric, reflexive and transitive.&#xA;Hence we can consider equivalence classes under this relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, every equivalence classes is countable (because $\Phi(x)$ is finite for all $x$).&#xA;Hence we have splitted reals into a set of disjoint countable sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that number of such sets is a continuum.&#xA;Now we can just take any element from each set to form $S$.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2010-12-27T14:58:33.393" CommentCount="1" />
  <row Id="15644" PostTypeId="1" AcceptedAnswerId="15645" CreationDate="2010-12-27T15:19:25.287" Score="0" ViewCount="170" Body="&lt;p&gt;If we only consider the natural numbers, is 3/2 undefined? If not what is the answer?&lt;/p&gt;&#xA;" OwnerUserId="5141" LastActivityDate="2010-12-28T01:56:53.580" Title="Is 3/2 undefined when only considering the natural numbers?" Tags="&lt;natural-number&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15645" PostTypeId="2" ParentId="15644" CreationDate="2010-12-27T15:22:27.863" Score="9" ViewCount="0" Body="&lt;p&gt;It depends on what you mean by $3/2$, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the usual intepretation where $3/2$ denotes the number such that when it is multiplied by $2$ gives you $3$, then, if you are restricting 'number' to mean 'natural number', then there is no such number.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-27T15:22:27.863" CommentCount="1" />
  <row Id="15646" PostTypeId="1" AcceptedAnswerId="15652" CreationDate="2010-12-27T15:24:09.367" Score="3" ViewCount="122" Body="&lt;h3&gt;Problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec r = r(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S: | \vec r  | = a$, with $\hat n$ outwards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec r_0 = \frac{3a}{5}(\hat x + \hat y - \hat z) = \frac{3a}{5}(1,1,-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F = k \frac{\vec r - \vec r_0}{ {| \vec r - \vec r_0 |}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_S \vec F \cdot \mathrm d \vec S $$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Solution (added after accepted answer)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Thanks to the accepted answer which confirmed that the explicit calculation on/in $S$ &lt;strong&gt;is&lt;/strong&gt; messy, and the suggestion of using translation invariance, I have noted down the solution below (let me know if you have further suggestions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argument is used which concludes that the integrand is zero inside the sphere $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Gauss' theorem: $\int_S \vec F \cdot \mathrm d \vec S = \int_V \nabla \cdot \vec F \mathrm dV $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the calculation is not easy to explicitly calculate in (or on) $S$, instead it is shown that $\nabla \cdot \vec F = 0$ exterior to $S_{\epsilon}$, which is a sphere containing the singularity, and since $S$ is exterior to $S_{\epsilon}$ the integrand must be $0$ and so $\int_V \nabla \cdot \vec F \mathrm dV = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{\epsilon}: |\vec r - \vec r_0 | = \epsilon, \epsilon &amp;gt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;translate the coordinate system so that $S_{\epsilon}$ is the origin in the translated system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec R = \vec r - \vec r_0 = R (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(\vec R) = k \frac{\vec R}{|\vec R|^3} = k \frac{1}{R^2} \hat R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{\epsilon}: |\vec R| = \epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \vec F (\vec R) = \frac{1}{R^2} \frac{\partial}{\partial R} \left(R^2 \vec F_R \right) = \frac{1}{R^2} \frac{\partial}{\partial R} \left(R^2 k \frac{1}{R^2} \right) = 0$$, ($k$ is a constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true for any value of $R$ except at the singularity ($\epsilon$ can be made as small as required), since $S$ does not contain the singularity, the integrand ($\nabla \cdot \vec F(\vec r)$ is 0 and the integral is therefore zero.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My question (updated)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am having difficulties in explicitly calculating the value of this integral. &#xA;Specifically, the divergence of the field becomes messy, I am not able to see how I can use the symmetry of $S$ due to $\vec r_0$. (When applying the Gauss' theorem, I am stuck in evaluating the divergence of the field). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to argue for that this integral is indeed zero (the field only has one singularity and it is exterior to the sphere $S$). However, I am not able to explicitly show (by calculation) that this integral is zero. Thankful for any help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that this is not homework, I am studying for an exam.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My question (original)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What is an easy way to calculate this integral? Any suggestions on the approaches below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies if this is due to lack of some basic knowledge (I am back studying after 2.5 years)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I understand the integral is zero (the point charge is located outside the sphere, anything flowing into the sphere will also flow out), however, I have issues with the calculation.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Calculate directly&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$\int_S \vec F \cdot \mathrm d \vec S = \int_0^{\pi} \mathrm d \theta \int_0^{2\pi} \mathrm d \phi r^2 \sin {\theta} \vec F_r \cdot \hat r = \int_0^{\pi} \mathrm d \theta \int_0^{2\pi} \mathrm d \phi r^2 \sin \theta k \frac{a - \frac{3a}{5} \left(  \sin \theta \cos \phi + \sin \theta \sin \phi - \cos \theta) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Using Gauss theorem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am tempted to use Gauss theorem, $\int_S \vec F \cdot \mathrm d \vec S = \int_V \nabla \cdot \vec F \mathrm dV $, in spherical coordinates. However $\vec F$ has components also in $\hat \theta$ and $\hat \phi$ (due to $\vec r_0$) and ${\left| \vec r - \vec r_0 \right|}^3$ is not that nice to derivate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F_r = k \frac{\vec r - \vec r_0}{ {\left| \vec r - \vec r_0 \right|}^3 } \cdot \hat r = k \frac{r - \frac{3a}{5} \left(  \sin \theta \cos \phi + \sin \theta \sin \phi - \cos \theta) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F_{\theta} = k \frac{\vec r - \vec r_0}{ {\left| \vec r - \vec r_0 \right|}^3 } \cdot \hat \theta = k \frac{ \frac{3a}{5} \left( \cos \theta \cos \phi + \cos \theta \sin \phi - \sin \theta) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec F_{\phi} = k \frac{\vec r - \vec r_0}{ {\left| \vec r - \vec r_0 \right|}^3 } \cdot \hat \phi = k \frac{ \frac{3a}{5} \left( - \sin \phi + \cos \phi ) \right)}{ {\left| \vec r - \vec r_0 \right|}^3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \vec F = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \vec F_r \right) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta F_{\theta} \right) + \frac{1}{r \sin \theta} \frac{\partial F_{\phi}}{\partial \phi}$$&lt;/p&gt;&#xA;" OwnerUserId="5143" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T01:16:24.663" LastActivityDate="2010-12-28T01:29:32.880" Title="Calculate surface integral of point charge located outside the surface" Tags="&lt;calculus&gt;&lt;vector-calculus&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="15647" PostTypeId="2" ParentId="15644" CreationDate="2010-12-27T15:36:32.723" Score="0" ViewCount="0" Body="&lt;p&gt;Kind'a similar analogy... $\sqrt{-1}$ is undefined when considering only real numbers. So mathematicians derived the whole complex number system to give that some meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its really defining the undefined that has progressed mathematics. So the answer to your question is, 3/2 is defined but only if you extend the natural numbers to a more generalized number system, like positive reals. :)&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2010-12-27T15:36:32.723" CommentCount="2" />
  <row Id="15648" PostTypeId="1" CreationDate="2010-12-27T15:48:02.487" Score="1" ViewCount="94" Body="&lt;p&gt;Show that $f_{n-1} + L_n = 2f_{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to find a $2$ to $1$ correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 1: Tilings an $n$-board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 2: Tiling of an $n-1$-board or tiling of an $n$-bracelet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to decompose a tiling of an $n$-board to a tiling of an $n-1$-board or a tiling of an $n-1$-bracelet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Proofs that Really Count&lt;/em&gt; by Art Benjamin and Jennifer Quinn&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2370" LastEditorDisplayName="" LastEditDate="2010-12-28T21:41:51.063" LastActivityDate="2010-12-28T21:41:51.063" Title="Lucas Numbers and Tilings" Tags="&lt;combinatorics&gt;&lt;fibonacci-number&gt;" CommentCount="4" />
  <row Id="15650" PostTypeId="1" CreationDate="2010-12-27T16:12:34.003" Score="2" ViewCount="127" Body="&lt;p&gt;Does f(n+1)/f(n) converge as n-&gt;infinity for f(n) defined by a linear recursion, for all linear recursions?&lt;/p&gt;&#xA;" OwnerUserId="5108" LastActivityDate="2010-12-28T10:03:44.447" Title="Linear Recursion.  Do ratio of consecutive terms converge for ALL linear recursions?" Tags="&lt;linear-algebra&gt;&lt;convergence&gt;&lt;recursive&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="15651" PostTypeId="2" ParentId="15632" CreationDate="2010-12-27T16:31:18.333" Score="3" ViewCount="0" Body="&lt;p&gt;The points $(a,b)$ and $(c,d)$ are the foci of the hyperbola and the hyperbola is the locus of points for which the difference of the distances to the foci is $|e|$, though because your $F(x,y)=0$ is only looking at the difference in one direction, your graph is probably only half of the hyperbola (the other half being $0=\sqrt{(x-a)^2+(y-b)^2}-\sqrt{(x-c)^2+(y-d)^2}+e$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymptotes of the hyperbola pass through the center, which is the midpoint of the segment joining the foci, $$\left(\frac{a+c}{2},\frac{b+d}{2}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f=\sqrt{(a-c)^2+(b-d)^2}$, the distance between the foci, and let $g=\sqrt{f^2-e^2}$.  If the foci had the same $x$-coordinate (that is, if the hyperbola were horizontally-oriented), the slopes of the asymptotes would be $\pm\frac{g}{e}$, so the acute angles the asymptotes make with the line through the foci are $\pm\arctan\frac{g}{e}$.  The acute angle the line through the foci makes with the $x$-axis is $\arctan\frac{b-d}{a-c}$, so the angles from the $x$-axis to the asymptotes are $\arctan\frac{b-d}{a-c}\pm\arctan\frac{g}{e}$, and the slopes of the asymptotes are $$\tan\left(\arctan\frac{b-d}{a-c}\pm\arctan\frac{g}{e}\right)=\frac{be-de+ag-cg}{ae-ce-bg+dg}\text{ or }\frac{be-de-ag+cg}{ae-ce+bg-dg}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a point and the slopes, you should be able to write equations of the asymptotes.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-27T16:31:18.333" CommentCount="1" />
  <row Id="15652" PostTypeId="2" ParentId="15646" CreationDate="2010-12-27T16:54:54.907" Score="3" ViewCount="0" Body="&lt;p&gt;(I don't know what you mean by $\hat x+\hat y -\hat z$, so I disregard the information about $r_0$.) Your field $\vec F$ is the gravitational field produced by a point mass at $r_0$. The divergence of this field is $\equiv0$ away from $r_0$. If the point $r_0$ is in the exterior of the sphere $S$ then it follows by Gauss' theorem that the integral in question is $0$. If $r_0$ lies in the interior of $S$ then you should draw a tiny sphere $S_\epsilon$ with center $r_0$ and apply Gauss' theorem to the region between $S_\epsilon$ and $S$. You are left with an integral over $S_\epsilon$ which you can almost do in your head.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-27T16:54:54.907" CommentCount="4" />
  <row Id="15653" PostTypeId="2" ParentId="15249" CreationDate="2010-12-27T17:03:42.800" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I'm adding my original comment as an answer at the OP's suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this notion of infinite set is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind-infinite_set&quot; rel=&quot;nofollow&quot;&gt;Dedekind infinite set&lt;/a&gt;. The Wikipedia article states some equivalent conditions under the section &lt;a href=&quot;http://en.wikipedia.org/wiki/Dedekind-infinite_set#Dedekind-infinite_sets_in_ZF&quot; rel=&quot;nofollow&quot;&gt;Dedekind infinite sets in ZF&lt;/a&gt; where it says explicitly that the Axiom of Choice is not required to prove their equivalence. In particular, one of the equivalent conditions for $A$ to be Dedekind infinite is that $A$ has a countably infinite subset.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-27T17:59:29.050" LastActivityDate="2010-12-27T17:59:29.050" CommentCount="3" />
  <row Id="15654" PostTypeId="2" ParentId="15605" CreationDate="2010-12-27T17:13:36.763" Score="4" ViewCount="0" Body="&lt;p&gt;I think it is very hard to get a better answer than any of the above. But there is a perspective that has not been mentioned, at least not explicitly. First, I would like to say that I happen to agree with Marianos answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body of your question is much more explicit than your title, so I will address the question in your title since it seems others have addressed the one in the body of your question. I like to think of orientations, or in fact most structures on a manifold, in terms of the tangent bundle on the manifold in question. The tangent bundle on a compact n dimensional manifold $M$ is classified (up to isomorphism) by a homotopy class of map $M \to BO(n)$. If we can lift this map over the map $BSO(n) \to BO(n)$ then the bundle is orientable, and hence so is the manifold. Here, the notion of orientability, our structure is really a fact about the tangent bundle. We can talk about almost complex structures or spin structures in a similar way. The map $BSO(n) \to BO(n)$ is induced by the inclusion $SO(n) \to O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to look at $\Lambda^n TM$, the top exterior power of the tangent bundle of M. Then an orientation is a nowhere zero continuous section of this bundle. Similarly, you can think of an orientation as a choice of generator of the top dimensional integral homology or cohomology. These are all probably less helpful than the above answers, but these are the ways I like to think about what an orientation is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please let me know if you want me to expand on some of these.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-27T17:13:36.763" CommentCount="1" />
  <row Id="15655" PostTypeId="2" ParentId="11763" CreationDate="2010-12-27T17:28:46.127" Score="2" ViewCount="0" Body="&lt;p&gt;First, let me say why all axiomatic treatments insist on talking about homology of pairs. One of the (most important/useful?) axioms is the Long exact sequence axiom, for any pair $A$ contained in $X$ we get the long exact sequence of homology groups. This is crucial for making computations. The concept of the homology of a pair is amazingly well behaved. Much better behaved than that of a quotient space, for homotopical purposes. So the best formulation of the axioms are in terms of pairs, because they require the least assumptions on the types of quotients you will allow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dylan alludes to, we really want to think of $(X,A)$ as the cofiber of the inclusion $A \to X$. This cofiber construction (quotient of the mapping cylinder, or rather the mapping cone) is a much better behaved notion. Its homotopy type depends only on the homotopy type of the inclusion. This is not the case for the standard quotient, unless the subspace is &quot;nice.&quot; For surfaces though, probably every subspace you are looking at is nice, so I think it would probably be okay if you just compute $H_*(X,A)$ as $H_*(X/A)$.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-12-27T17:28:46.127" />
  <row Id="15656" PostTypeId="1" CreationDate="2010-12-27T17:39:05.940" Score="24" ViewCount="454" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The volume of an $n$-dimensional ball of radius $1$ is given by the classical formula&#xA;$$V_n=\frac{\pi^{n/2}}{\Gamma(n/2+1)}.$$ &#xA;For small values of $n$, we have&#xA;$$V_1=2\qquad$$&#xA;$$V_2\approx 3.14$$&#xA;$$V_3\approx 4.18$$&#xA;$$V_4\approx 4.93$$&#xA;$$V_5\approx 5.26$$ &#xA;$$V_6\approx 5.16$$&#xA;$$V_7\approx 4.72$$&#xA;It is not difficult to prove that $V_n$ assumes its maximal value when $n=5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; Is there any non-analytic (i.e. geometric, probabilistic, combinatorial...) demonstration of this fact? What is so special about $n=5$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A similar question concerning the $n$-dimensional volume $S_n$ (&quot;surface area&quot;) of a unit $n$-sphere. Why is the maximum of $S_n$ attained at $n=7$ from a geometric point of view?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="723" LastActivityDate="2010-12-29T15:34:42.147" Title="Volumes of n-balls: what is so special about n=5?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="5" />
  <row Id="15657" PostTypeId="1" AcceptedAnswerId="15659" CreationDate="2010-12-27T17:40:30.437" Score="2" ViewCount="110" Body="&lt;p&gt;Given Kn define sequence S:  writing all vertices in order of traversing it by Euler circle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many distinct Hamilton cycles are in this sequence S?&lt;/p&gt;&#xA;" OwnerUserId="5145" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-27T22:12:12.293" LastActivityDate="2010-12-28T02:41:16.123" Title="Euler circle and Hamilton cycles" Tags="&lt;graph-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="15658" PostTypeId="2" ParentId="15638" CreationDate="2010-12-27T17:52:21.253" Score="5" ViewCount="0" Body="&lt;p&gt;Here is another way of arguing which is slightly different: Consider the set $X={\mathcal P}({\mathbb N}\times{\mathbb N})$ of all subsets of ${\mathbb N}^2$. Given $E\subseteq {\mathbb N}\times{\mathbb N}$, let $A_E$ be the set of all numbers that appear in either the domain or range of $E$ (this is sometimes called the &lt;em&gt;field&lt;/em&gt; of $E$). If it happens that $E$ is a well-ordering of $A_E$, let $\alpha_E$ be the unique ordinal that is order isomorphic to $(A_E,E)$. Otherwise, let $\alpha_E=0$. Then $\{\alpha_E\mid E\in X\}$ is a set (is the image of $X$ under the map $E\mapsto\alpha_E$) and it is obvious that it consists precisely of the countable ordinals. One easily sees then that it is itself an ordinal, and uncountable (since no ordinal belongs to itself). This is $\omega_1$.   &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-27T17:52:21.253" CommentCount="3" />
  <row Id="15659" PostTypeId="2" ParentId="15657" CreationDate="2010-12-27T18:02:30.173" Score="1" ViewCount="0" Body="&lt;p&gt;It can certainly vary.  As an Eulerean path uses all n(n-1)/2 edges and a Hamiltonian path uses n, you can't have more than (n-1)/2.  Can you have that many?  There aren't any Eulerean paths if n is even and &gt;2.  Depending upon your definition, I can define an Eulearean path that has no segment that is Hamiltonian &lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-28T01:38:01.077" LastActivityDate="2010-12-28T01:38:01.077" CommentCount="4" />
  <row Id="15660" PostTypeId="2" ParentId="15650" CreationDate="2010-12-27T18:10:26.130" Score="4" ViewCount="0" Body="&lt;p&gt;How about f(n+1)=f(n-1) started with f(0)=0, f(1)=1?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-27T18:10:26.130" />
  <row Id="15661" PostTypeId="2" ParentId="15656" CreationDate="2010-12-27T18:29:10.347" Score="1" ViewCount="0" Body="&lt;p&gt;The Gamma function has a minimum close to $5/2 +1$ for positive reals, and for your &quot;dimensionless quantity&quot; it's all that matters.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastEditorUserId="5105" LastEditorDisplayName="" LastEditDate="2010-12-27T18:36:35.583" LastActivityDate="2010-12-27T18:36:35.583" CommentCount="1" />
  <row Id="15662" PostTypeId="2" ParentId="15249" CreationDate="2010-12-27T18:40:52.170" Score="0" ViewCount="0" Body="&lt;p&gt;Surely a Dedekind infinite set trivially has an infinite proper subset: its image under the bijective function that is required for it to be Dedekind infinite.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2010-12-27T18:40:52.170" CommentCount="6" />
  <row Id="15663" PostTypeId="1" AcceptedAnswerId="15664" CreationDate="2010-12-27T19:04:24.600" Score="2" ViewCount="122" Body="&lt;p&gt;Let $f\colon\mathbb{R}^n\to\mathbb{R}$ be a function with $n\geq 2$, and define $S_{f}=\{(x_1,x_2, \ldots x_n) \in\mathbb{R}^n \mid f(x_1,x_2, \ldots x_n)=1\} = f^{-1}(1)$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $S_f$ path-connected when $f$ is $C^{\infty}$? If not, what about if $f \in \mathbb{Q}[x_1,\ldots,x_n]$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context&lt;/strong&gt;: I had to establish that the set of diagonal square matrices with $\det=1$ was path-connected, which I did, but it got me thinking on more general lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, please answer (if possible) such that one needs as little prerequisites as possible to understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;: Though Eric has already answered the question. There's a natural next question to ask: For what kinds of functions is this set path-connected? &lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-28T11:46:34.913" LastActivityDate="2010-12-28T11:46:34.913" Title="Show set is path-connected" Tags="&lt;topology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15664" PostTypeId="2" ParentId="15663" CreationDate="2010-12-27T19:10:16.633" Score="4" ViewCount="0" Body="&lt;p&gt;Not in general. Let $f:R \to R, f(x) = x^2 - 1$.  Then $S = \{\pm \sqrt{2}\}$ which is not path-connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: for the $n=2$ case you can take an equation defining a hyperbola, e.g. $x^2 - y^2$.  The set where this function is 1 has two connected components.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditorDisplayName="" LastEditDate="2010-12-27T19:22:33.943" LastActivityDate="2010-12-27T19:22:33.943" CommentCount="3" />
  <row Id="15666" PostTypeId="2" ParentId="15663" CreationDate="2010-12-27T19:25:14.877" Score="3" ViewCount="0" Body="&lt;p&gt;This properly belongs as a comment to Eric's answer, but I don't have enough reputation to post one.  More generally, defining $f:\mathbb{R}^n\to\mathbb{R}$ by $f(x_1,\ldots,x_n) = x_1^2-1$ gives $S$ with two connected components, the affine hyperplanes $\{\pm \sqrt 2\}\times\mathbb{R}^{n-1}$.&lt;/p&gt;&#xA;" OwnerUserId="5116" LastEditorUserId="5116" LastEditorDisplayName="" LastEditDate="2010-12-27T19:32:47.120" LastActivityDate="2010-12-27T19:32:47.120" CommentCount="2" />
  <row Id="15667" PostTypeId="1" AcceptedAnswerId="15673" CreationDate="2010-12-27T19:55:34.207" Score="2" ViewCount="68" Body="&lt;p&gt;Let $\sigma=(1\ 3),\ \tau=(2\ 4\ 5),\ \pi=(2\ 3\ 4) \in S_{5}$. Find $\pi\circ\tau\circ\sigma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the solution is $(1\ 4\ 5\ 3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i'm doing now is writing the permutations in the &quot;expanded&quot; form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example $\tau=\begin{pmatrix} &#xD;&#xA;  1\ 2\ 3\ 4\ 5\\ &#xD;&#xA;  1\ 5\ 3\ 2\ 4 &#xD;&#xA;\end{pmatrix}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then &quot;follow the numbers&quot; over the composition. This way, i find $(1\ 3\ 5\ 2)$, which is incorrect. What am i doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="5148" LastActivityDate="2010-12-27T20:41:27.627" Title="Can someone explain the algorithm for composition of cycles?" Tags="&lt;homework&gt;&lt;algorithms&gt;&lt;elementary&gt;&lt;permutations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15668" PostTypeId="1" AcceptedAnswerId="15693" CreationDate="2010-12-27T19:58:36.013" Score="7" ViewCount="192" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Axiom_of_choice&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; article, two examples are given which use/ do not use the axiom of choice. They are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Given an infinite pair of socks, one needs AC to pick one sock out of each pair.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given an infinite collection of pairs of shoes, one shoe can be specified without AC by choosing the left one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Aren't these equivalent examples (just with different objects)? Why can't one just choose the left sock in (i) (so that AC is not needed)? &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-27T21:23:02.523" LastActivityDate="2010-12-28T00:08:28.153" Title="Axiom of Choice Examples" Tags="&lt;intuition&gt;&lt;elementary-set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15669" PostTypeId="2" ParentId="15667" CreationDate="2010-12-27T20:00:18.510" Score="2" ViewCount="0" Body="&lt;p&gt;You're probably composing from the left to the right, instead of from the right to the left.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-27T20:00:18.510" CommentCount="2" />
  <row Id="15670" PostTypeId="2" ParentId="15556" CreationDate="2010-12-27T20:13:37.323" Score="3" ViewCount="0" Body="&lt;p&gt;Yes, the &quot;classification&quot; of naturals by their &lt;strong&gt;parity&lt;/strong&gt; (remainder modulo $2\:$) extends immediately to integers. Namely, &lt;em&gt;even&lt;/em&gt; integers are those divisible by $2$, i.e. $\rm\: n = 2\:m\equiv 0\ (mod\ 2)\:,\: $ and &lt;em&gt;odd&lt;/em&gt; integers are those leaving remainder $\:1\:$ when divided by $\:2\:,\ $ i.e. $\rm\ n = 2\:m + 1\equiv 1\ (mod\ 2)\:.\: $ The effectiveness of this parity classification arises from the fact that it is &lt;em&gt;compatible&lt;/em&gt; with integer arithmetic operations, i.e. if $\rm\ \bar{a}\ :=\ a\ (mod\ 2)\ $ then $\rm\ \overline{ a+b}\ =\ \bar a + \bar b,\ \ \overline{a\ b}\ =\ \bar a\ \bar b\:.\: $  Iterating, we infer that equalities between expressions composed of these integer operations (i.e. integer polynomial expressions) are &lt;em&gt;preserved&lt;/em&gt; by taking their images modulo $2$ (ditto mod $\rm\:m\:$ for any $\rm\:m\:,$ e.g mod $9$ reduction yields &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting%5Fout%5Fnines&quot; rel=&quot;nofollow&quot;&gt;casting out nines&lt;/a&gt;). In this way we can hope to understand integer equations by studying their images in the simpler (finite!) rings $\rm\: \mathbb Z/m\: $ of integers modulo $\rm\:m\:.\:$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if an integer coefficient polynomial has an integer root $\rm\ f(n) = 0\ $ then it remains a root modulo $2$, i.e. $\rm\ f(n)\equiv 0\ (mod\ 2)\:.\:$ So, contrapositively, if a polynomial has no roots modulo $2$ then it has no integer roots. This leads to the following simple &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PARITY ROOT TEST&lt;/strong&gt; $\ $ A polynomial $\rm\:P(x)\:$ with integer coefficients &#xA;has no integer roots if its constant coefficient and coefficient sum are both odd. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof&lt;/strong&gt; $\ $ The test  verifies that $\rm\  P(0) \equiv  P(1)\equiv 1\ \ (mod\ 2)\:,\ $ i.e. &#xA;that $\rm\:P(x)\:$ has no roots modulo $2$, hence no integer roots. $\quad$ &lt;strong&gt;QED&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. $\rm\:\ a\ X^2 + b\ X + c\ $  has no integer roots &#xA;if  $\rm\:c\:$  is odd and $\rm\:a,\:b\:$  have equal parity $\rm\:a\equiv b\ (mod\ 2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare the conciseness of this test to the messy reformulation that would result if we had to restrict it to positive integers. Then we could no longer represent polynomial equations in the normal form $\rm\:f(x) = 0\:$ but, rather, we would need to consider general equalities $\rm\:f(x) = g(x)\:$ where both polynomials have &lt;em&gt;positive&lt;/em&gt; coefficients. Now the test would be much messier - bifurcating into motley cases. Indeed, historically, before the acceptance of negative integers and zero, the formula for the solution of a quadratic equation was stated in such an analogous obfuscated way - involving many cases. But by extending the naturals to the ring of integers we are able to &lt;em&gt;unify&lt;/em&gt; what were previously motley separate cases into a single &lt;em&gt;universal&lt;/em&gt; method of solving a general quadratic equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous examples exist throughout history that help serve to motivate the reasons behind various number system enlargements. Studying mathematical history will help provide one with a much better appreciation of the motivations behind the successive enlargements of the notion of &quot;number systems&quot;, e.g. see&#xA;Kleiner: &lt;a href=&quot;http://retro.seals.ch/cntmng?type=pdf&amp;amp;rid=elemat-001:1998:53%3a:22&amp;amp;subp=hires&quot; rel=&quot;nofollow&quot;&gt;From numbers to rings: the early history of ring theory.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is but one of many examples where &quot;completing&quot; a structure in some manner serves to simplify its theory. Such ideas motivated many of the extensions of the classical number systems (as well as analogous geometrical and topological completions concept, e.g. adjoining points at $\infty$, projective closure, compactification, model completion, etc). For some interesting expositions on such methods see the &lt;a href=&quot;http://math.stackexchange.com/questions/3203/infinite-limits/3206#3206&quot;&gt;references here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; $\: $ Analogous remarks (on the power gained by normalizing equations to the form $\ldots = 0\:$) hold true more generally for any algebraic structure whose congruences are determined by ideals - so-called &lt;em&gt;ideal determined&lt;/em&gt; varieties, e.g. see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zps1nln5o.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; and see Gumm and Ursini: &lt;a href=&quot;http://www.springerlink.com/content/h2777414xl217t75/fulltext.pdf&quot; rel=&quot;nofollow&quot;&gt;Ideals in universal algebras&lt;/a&gt;. Without zero and negative numbers (additive inverses) we would not be able to rewrite expressions into such concise normal forms and we would not have available such powerful algorithms such as the Grobner basis algorithm, Hermite/Smith normal forms, etc. &lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-27T20:19:47.797" LastActivityDate="2010-12-27T20:19:47.797" />
  <row Id="15672" PostTypeId="2" ParentId="15668" CreationDate="2010-12-27T20:18:00.673" Score="3" ViewCount="0" Body="&lt;p&gt;The idea behind the axiom of choice is to tell you how to choose when you can't necessarily distinguish between the items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take out a pair of socks from the closet (or drawer, etc) you can't tell which one you had on your left foot and which on your right, or which one is this and which one is that. If you can, well... you need to remember to let someone else make sure you have matching socks :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea, formally, is that if you take a product of infinitely many non-empty sets then it is not empty, namely there is a function in the product which returns an element in each coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, given a set $I$ such that $\forall i\in I$ we have $A_i\not=\emptyset$ then there is $F\colon I\to\bigcup_{i\in I} A_i$ for which $F(i)\in A_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it, this is not a &quot;strange&quot; requirement when you are discussing mathematics, and it might come naturally in many places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for any two sets $A,B$ if there is $f\colon A\to B$ which is surjective then there is an injective $g\colon B\to A$. What are we doing? In a sense we choose one representative in each equivalence class of $a_1\sim a_2\iff f(a_1) = f(a_2)$. But if we have infinitely many equivalence classes - then we (usually) need to invoke some choice axiom (perhaps the axiom of choice, or countable choice, or dependent choice, etc...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, many times the use is not of the axiom of choice, but rather an equivalent principle called &quot;Zorn's Lemma&quot;, it states that if you have a partial order in which every chain is bounded from above - then there is some maximal element (that is no one is strictly above it in the order). Again not something that is unreasonable if you want infinitary processes to &quot;act&quot; similar to finitary ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&lt;br&gt;&#xA;After I gave it some extra thought, I came up with something that might clear up the things. There is a concept which is &quot;a definable element&quot;, that is that you can write some formula $\psi(x)$ such that $a$ is the only element (suppose in $A$) which satisfies the formula. If you want to choose from infinitely many sets then you need to be able and tell which element you have chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have at least one definable element in each set, suppose by some uniform formula $\varphi(x)$, then you can clearly choose the one element defined by $\varphi$. However if there are many definable elements in each set, or infinitely many formulae are needed - then you cannot express it simply, and then you must assume that you can do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as I remarked above, we simply want infinitary processes to behave well, like finitary ones.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-27T21:05:36.000" LastActivityDate="2010-12-27T21:05:36.000" />
  <row Id="15673" PostTypeId="2" ParentId="15667" CreationDate="2010-12-27T20:29:28.160" Score="2" ViewCount="0" Body="&lt;p&gt;When you read a composition of functions written in the usual notation for permutations, you must remember to read them from right to left. Thus, when you try to compute the composition you must start by looking successively at what does each permutation in the composition do to each integer from $1$ to $5$ (in this case), but from right to left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can start by writing &#xA;$$\pi \circ \tau \circ \sigma = \begin{pmatrix} 2 &amp;amp;3 &amp;amp;4 \end{pmatrix} \circ \begin{pmatrix} 2 &amp;amp;4 &amp;amp;5 \end{pmatrix} \circ \begin{pmatrix} 1 &amp;amp;3 \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you start with $1$. The rightmost permutation sends $1 \mapsto 3 $ and then the middle permutation sends $3 \mapsto 3$ and finally the leftmost permutation sends $3 \mapsto 4$. So in the end the total result is that the composition of the three of them sends $1 \mapsto 4$. In the same way you'll proceed for the other integers and in the end you'll get the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \pi \circ \tau \circ \sigma = \begin{pmatrix} 1 &amp;amp;2 &amp;amp;3 &amp;amp;4 &amp;amp;5 \\ 4 &amp;amp;2 &amp;amp;1 &amp;amp;5 &amp;amp;3  \end{pmatrix}$$&#xA;or in cycle notation $\pi \circ \tau \circ \sigma = \begin{pmatrix}  1 &amp;amp;4 &amp;amp;5 &amp;amp;3\end{pmatrix}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked doing the computation from left to right and the &quot;answer&quot; I got from doing that was $\pi \circ \tau \circ \sigma = \begin{pmatrix}  1 &amp;amp;3 &amp;amp;5 &amp;amp;2\end{pmatrix}$ which is what you originally had as your answer.&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="900" LastEditorDisplayName="" LastEditDate="2010-12-27T20:41:27.627" LastActivityDate="2010-12-27T20:41:27.627" CommentCount="1" />
  <row Id="15674" PostTypeId="2" ParentId="15650" CreationDate="2010-12-27T20:29:36.710" Score="3" ViewCount="0" Body="&lt;p&gt;The general solution for a (finite) linear recurrence is a function of the form $$f(n) = \sum_{i=1}^m P_i(n) \lambda_i^n,$$ where the $P_i$ are polynomials with possible complex coefficients, and the $\lambda_i$ can be complex as well (even if the sequence itself is real, for example a cyclic sequence of minimal period $&gt;2$); and vice versa. If the eigenvalue with maximal modulus is unique, say $\lambda$, then $f(n+1)/f(n) \rightarrow \lambda$. If all eigenvalues of maximal modulus are obtained from one of them by multiplying with (integral) roots of unity, then there is a periodic subsequence with this property (e.g. in Ross's example take every second term). Otherwise, the sequence will be pretty chaotic: for example $f(n) = \cos(n)$.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastActivityDate="2010-12-27T20:29:36.710" />
  <row Id="15675" PostTypeId="2" ParentId="15629" CreationDate="2010-12-27T20:48:10.887" Score="1" ViewCount="0" Body="&lt;p&gt;Or just $\begin{cases} a &amp;amp; b \\ c &amp;amp; d \end{cases}$&lt;/p&gt;&#xA;" OwnerUserId="5152" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-28T01:35:17.797" LastActivityDate="2010-12-28T01:35:17.797" CommentCount="3" />
  <row Id="15676" PostTypeId="1" CreationDate="2010-12-27T20:52:57.740" Score="4" ViewCount="130" Body="&lt;p&gt;Sorry for this uninteresting question but hopefully someone can provide some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to simplify the following expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}\displaystyle \frac{m}{m+v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a hint for a problem, but I don't know how to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: Would it bother you if I add the original problem?. Maybe some context is needed in order to solve this problem by using this hint.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-27T23:05:52.667" LastActivityDate="2010-12-27T23:05:52.667" Title="Simplification of sums" Tags="&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15677" PostTypeId="2" ParentId="15676" CreationDate="2010-12-27T20:57:09.370" Score="1" ViewCount="0" Body="&lt;p&gt;At a quick glance, it looks like there are some common factors to the two sums and that factoring is likely to help in simplifying.  Also, since the summations have the same indexing and range, they could be combined into a single summation.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-27T20:57:09.370" CommentCount="4" />
  <row Id="15678" PostTypeId="2" ParentId="15629" CreationDate="2010-12-27T21:01:21.120" Score="1" ViewCount="0" Body="&lt;p&gt;You could potentially convert it to a mathematical formula too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example say we had the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if (a &amp;lt; b) then c = 100 &#xA;else if (a &amp;gt; b) then c = 200&#xA;else c = 300.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c  = 300 \ (1 - \text{sgn}^2(a-b)) + \text{sgn}^2(a-b)(50 \ \text{sgn}(a-b) + 150)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\text{sgn}(x)$ is the sign of $x$, as defined here; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sign_function&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sign_function&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is defined as: 1 for positive, 0 for 0, and -1 for negative)&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-27T21:17:00.987" LastActivityDate="2010-12-27T21:17:00.987" />
  <row Id="15679" PostTypeId="2" ParentId="15676" CreationDate="2010-12-27T21:15:32.487" Score="1" ViewCount="0" Body="&lt;p&gt;I doubt it, the case $m=0$ would give a simpler formula for &lt;a href=&quot;http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind&quot; rel=&quot;nofollow&quot;&gt;Stirling numbers of the second kind&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n!\ S(r,n) = \sum_{v = 0}^{n} (-1)^{v} {n \choose v} (n-v)^{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you can write your expression in terms of these numbers...&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-27T21:15:32.487" CommentCount="4" />
  <row Id="15680" PostTypeId="2" ParentId="15556" CreationDate="2010-12-27T21:38:08.560" Score="2" ViewCount="0" Body="&lt;p&gt;The real answer depends on the definition, because there is math-history tag invoked there was a time that 1 was not considered an odd numbers, 0 and negative numbers for sure where not considered even or odd. Historically the concept was defined only for natural numbers. &#xA;These a days the set of all integers multiplied by 2 is considered the set of even numbers, i.e. $\dots,-4,-2,0,2,4,\dots$ n and all the integers not in that set are defined to be odd.&#xA;There is no real answer, it all depends on the definition, same way that the book is only dealing with natural numbers and not integers.&#xA;The concept was extended from naturals to integers, but there are uncountably many ways to define the even and odds beyond the natural numbers. Just make sure others know what definition you are using to label something even or odd. &lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-27T21:38:08.560" CommentCount="4" />
  <row Id="15681" PostTypeId="1" AcceptedAnswerId="15691" CreationDate="2010-12-27T21:49:49.817" Score="5" ViewCount="101" Body="&lt;p&gt;How many Borel-measurable functions from $\mathbb{R}$ to $\mathbb{R}$ are there? The motivation is from &lt;a href=&quot;http://mathoverflow.net/questions/32126/function-with-range-equal-to-whole-reals-on-every-open-set/49137#49137&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; answer of mine on MathOverflow.&lt;/p&gt;&#xA;" OwnerUserId="3078" LastEditorUserId="3078" LastEditorDisplayName="" LastEditDate="2010-12-27T22:02:43.790" LastActivityDate="2010-12-27T23:59:27.707" Title="How many Borel-measurable functions from R to R are there?" Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15682" PostTypeId="2" ParentId="13282" CreationDate="2010-12-27T22:55:51.737" Score="2" ViewCount="0" Body="&lt;p&gt;Other possibilities include using the conjugate gradient algorithm to check positive-definiteness. In theory, this method terminates after at most n iterations (n being the dimension of your matrix). In practice, it may have to run a bit longer. It is trivial to implement. You can also use a variant of the Lanczos method to estimate the smallest eigenvalue of your matrix (which is much easier than computing all eigenvalues!) Pick up a book on numerical linear algebra (check the SIAM collection.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate recall that such methods (and the Cholesky decomposition) check &lt;em&gt;numerical&lt;/em&gt; positive-definiteness. It is possible for the smallest eigenvalue of your matrix to be, say, 1.0e-16 and for cancellation errors due to finite-precision arithmetic to cause your Cholesky (or conjugate gradient) to break down.&lt;/p&gt;&#xA;" OwnerUserId="5155" LastActivityDate="2010-12-27T22:55:51.737" CommentCount="1" />
  <row Id="15683" PostTypeId="2" ParentId="15624" CreationDate="2010-12-27T23:06:00.453" Score="7" ViewCount="0" Body="&lt;p&gt;Just dropping points into the polygon according to a uniform distribution will not, unfortunately, generate very uniform solutions: with extremely high probability there will be large gaps and tight clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obtaining a highly uniform pattern within an arbitrary (convex) polygon is a difficult problem to solve; exact solutions are known only for special polygons and very small numbers of points.  As the number of points $n$ grows, they tend to fall into a hexagonal pattern.  However, like real crystals, these patterns can exhibit faults and fractures imposed by the polygon's boundary.  As a quick and dirty solution you might estimate the average spacing by dividing the polygon's area by $n$ and solving for the dimensions of a hexagon of that area, creating an hexagonal mesh of those dimensions, and performing some trial overlays of that mesh (translating it a bit in each trial) to find something that works well.  You are likely to run into trouble at places just inside the polygon's boundary, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One practical way to find solutions that really work is with &lt;a href=&quot;http://cat.inist.fr/?aModele=afficheN&amp;amp;cpsidt=2432316&quot; rel=&quot;nofollow&quot;&gt;spatial simulated annealing&lt;/a&gt;.  Although that's a computationally intensive and purely approximate approach, in practice you can get reasonable solutions for the intended application in an extremely short time (thousands of iterations rather than hundreds of thousands).&lt;/p&gt;&#xA;" OwnerUserId="1489" LastActivityDate="2010-12-27T23:06:00.453" CommentCount="3" />
  <row Id="15684" PostTypeId="1" CreationDate="2010-12-27T23:06:03.683" Score="1" ViewCount="120" Body="&lt;p&gt;Let $\tau \in \mathbb{S}^1$ be such that $\tau$ is not a root of unity. Let $E_\tau$ be the quotient space $S^1/\tau^{\mathbb Z}$. Consider it as a pointed space with basepoint the equivalence class of $1$. This is a path-connected space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fundamental group of this space?&lt;/p&gt;&#xA;" OwnerUserId="5156" LastActivityDate="2010-12-28T03:31:15.413" Title="Fundamental group of noncommutative torus" Tags="&lt;algebraic-topology&gt;&lt;noncommutative-geometry&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="15685" PostTypeId="2" ParentId="15624" CreationDate="2010-12-27T23:28:55.613" Score="4" ViewCount="0" Body="&lt;p&gt;As whuber mentions, using a uniform distribution for sampling will result in &quot;clumpiness&quot; of the points. One way of quantifying this &quot;clumpiness&quot; is through a measure called discrepancy. There are numerous &lt;a href=&quot;http://en.wikipedia.org/wiki/Low-discrepancy_sequence&quot; rel=&quot;nofollow&quot;&gt;low-discrepancy sequences&lt;/a&gt; out there. You may also want to look into the &lt;a href=&quot;http://people.csail.mit.edu/thouis/JonesPoissonPreprint.pdf&quot; rel=&quot;nofollow&quot;&gt;Poisson disk method&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2010-12-27T23:28:55.613" CommentCount="1" />
  <row Id="15686" PostTypeId="1" AcceptedAnswerId="15690" CreationDate="2010-12-27T23:29:24.397" Score="7" ViewCount="215" Body="&lt;p&gt;I have to explain the problem I am thinking first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions of $x^3+6x-20=0$ are $x = 2$, $-1+3i$ and $-1-3i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cubic formula for these solutions is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \sqrt[3]{10 + \sqrt{108}} + \sqrt[3]{10 + \sqrt{108}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First $\sqrt{108} \approx -10.392304845413$, $10.392304845413$ has two values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\sqrt[3]{10 + -10.392304845413}$ has three values $v_1\approx 0.3660 + 0.6339745i$, $v_2\approx -0.7320508$, $v_3\approx 0.3660254 - 0.6339745i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\sqrt[3]{10 + 10.392304845413}$ has three values $u_1\approx 2.7320508$, $u_2\approx -1.366 + 2.366$, $u_3\approx -1.366 - 2.366i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are nine possible values of $x$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$v_1 + u_1 = 3.098 + 0.63397i$&lt;/li&gt;&#xA;&lt;li&gt;$v_1 + u_2 = -1 + 3i$&lt;/li&gt;&#xA;&lt;li&gt;$v_1 + u_3 = -1 - 1.732i$&lt;/li&gt;&#xA;&lt;li&gt;$v_2 + u_1 = 2$&lt;/li&gt;&#xA;&lt;li&gt;$v_2 + u_2 = -2.098 + 2.366i$&lt;/li&gt;&#xA;&lt;li&gt;$v_2 + u_3 = -2.098 - 2.366i$&lt;/li&gt;&#xA;&lt;li&gt;$v_3 + u_1 = 3.098 - 0.63397i$&lt;/li&gt;&#xA;&lt;li&gt;$v_3 + u_2 = -1 + 1.732$&lt;/li&gt;&#xA;&lt;li&gt;$v_3 + u_3 = -1 - 3i$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but the correct ones are $v_1 + u_2$, $v_2 + u_1$, $v_3 + u_3$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; Here is a table of 3vu for Arturo Magidin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$3 v_1 u_1 = 3 + 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_1 u_2 = -6$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_1 u_3 = 3 - 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_2 u_1 = -6$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_2 u_2 = 3 - 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_2 u_3 = 3 + 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_3 u_1 = 3 - 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_3 u_2 = 3 + 5.196152i$&lt;/li&gt;&#xA;&lt;li&gt;$3 v_3 u_3 = -6$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the others values of $x$ (those which are not solutions)?&lt;/li&gt;&#xA;&lt;li&gt;Is there a better notation for taking roots that points out (for example) that the cube roots need to be synchronized?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-30T00:02:37.487" LastActivityDate="2010-12-30T00:38:26.410" Title="Different notations for roots?" Tags="&lt;algebra&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="15687" PostTypeId="1" AcceptedAnswerId="15688" CreationDate="2010-12-27T23:46:48.313" Score="3" ViewCount="166" Body="&lt;p&gt;Does anyone remember a movie about four mathematicians studying infinity, who ended up committing a suicide? It was aired online somewhere, and I can't find it (it's a documentary, if that helps).&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T02:13:26.547" LastActivityDate="2010-12-28T02:13:26.547" Title="Name of the math movie" Tags="&lt;reference-request&gt;&lt;popular-math&gt;" AnswerCount="2" />
  <row Id="15688" PostTypeId="2" ParentId="15687" CreationDate="2010-12-27T23:49:16.187" Score="7" ViewCount="0" Body="&lt;p&gt;Are you talking about &lt;strong&gt;Dangerous Knowledge&lt;/strong&gt; by BBC? Though it's not quite about four mathematicians (at least one of the people was physicist), and not all of them were studying infinity directly. That's the only one that comes to my mind as I watched it a couple of weeks ago.&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-27T23:49:16.187" CommentCount="1" />
  <row Id="15689" PostTypeId="2" ParentId="15687" CreationDate="2010-12-27T23:52:09.103" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Dangerous Knowledge&lt;/strong&gt; it is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cantor - Infinity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boltzmann - Entropy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Godel and Turing - Logic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://video.google.com/videoplay?docid=-5122859998068380459#&quot; rel=&quot;nofollow&quot;&gt;http://video.google.com/videoplay?docid=-5122859998068380459#&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-27T23:52:09.103" />
  <row Id="15690" PostTypeId="2" ParentId="15686" CreationDate="2010-12-27T23:58:25.780" Score="7" ViewCount="0" Body="&lt;p&gt;When applying Cardano's formulae, you must take specific branches of the square and cubic root functions. Also, note that the square root of 108 is justs a &lt;em&gt;single&lt;/em&gt; value; while $x^2 - 108=0$ has two (real) solutions, the expression $\sqrt{108}$ is a &lt;em&gt;single&lt;/em&gt; value: the nonnegative solution to $x^2-108=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, remember that if you have a depressed cubic, $x^3 + px + q = 0$, we introduce two variables $u$ and $v$ subject to the condition that $u+v = x$ and $3uv + p = 0$. This requires the product of $u$ and $v$ to be a real number. Eventually, you arrive at $u$ and $v$ being the cubic roots in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember that you are assuming that $uv$ is a real number. This is why you cannot just take all possible combinations with all possible roots. In other words, all those &quot;other values&quot; of $u+v$ are values that do not satisfy the system of equations you have:&#xA;\begin{align*}&#xA;(u+v)^3 + p(u+v) + q &amp;amp;=0\\&#xA;3uv + p &amp;amp;=0&#xA;\end{align*}&#xA;There is no general notation that tells you which branches of the complex-valued cubic and square root functions you must take, you just have to remember that the product of the two cubic roots has to be a real number. This is what cuts down from nine to three values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a standard way of listing all solutions: once you find &lt;strong&gt;one&lt;/strong&gt; value of $u$ and $v$ such that $uv$ is a real number, then the three solutions are given by $u+v$, $\omega u + \omega^2 v$, and $\omega^2u + \omega v$, where $\omega = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$ is a primitive cubic root of unity. (Notice that if $u$ is one value of the cubic root, then the other two values are $\omega u$ and $\omega^2u$, and likewise with $v$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;It seems that you are not quite aware of all the assumptions you are making about $u$ and $v$ when you use Cardano's formula. For one thing, if you review the derivation of the formula, say, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, you will see that the condition that $3uv+p=0$ is worked into the derivation. Here's the method: remember that $x=u+v$ is supposed to be a solution of&#xA;$$x^3 + px + q = 0.$$&#xA;Plugging in and doing a bit of work, we get:&#xA;\begin{align*}&#xA;0 &amp;amp;= (u+v)^3 + p(u+v) + q\\&#xA;  &amp;amp;= u^3 + 3u^2v + 3uv^2 + v^3 + p(u + v) + q\\&#xA;  &amp;amp;= u^3 + v^3 + 3uv(u+v) + p(u+v) + q\\&#xA;  &amp;amp;= u^3 + v^3 + (3uv+p)(u+v) + q.&#xA;\end{align*}&#xA;It is at this point that the extra condition $3uv+p=0$ is introduced, so that the equation reduces to $u^3+v^3 + q = 0$, Then $u^3+v^3 = -q$, and $(uv)^3 = -\frac{p^3}{27}$. Now remember that if $a$ and $b$ are two numbers, then $(z-a)(z-b) = z^2 - (a+b)z + ab$, so $a$ and $b$ are always the roots of the quadratic that has linear coefficient minus their sum, and constant coefficient their product. Applying this to $u^3$ and $v^3$, since we know their sum and their product, we deduce that $u^3$ and $v^3$ are the two roots of the quadratic:&#xA;$$z^2 +qz - \frac{p^3}{27}=0$$&#xA;and from this, using the quadratic formula, we obtain that&#xA;$$u^3 = -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}, \qquad v^3 = -\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}.$$&#xA;Therefore,&#xA;$$u = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},\qquad v = \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},$$&#xA;so that the solution $x$ is given by the formula you have:&#xA;$$x = u+v = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$&#xA;But the assumption that $3uv+p=0$ was instrumental in obtaining this formula. The very fact that you write that $x$ is equal to the sum of the two cubic roots &lt;strong&gt;already assumes&lt;/strong&gt; that the product of those two (complex-valued) cubic roots will be a real number, specifically, $-\frac{p}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While &lt;em&gt;you&lt;/em&gt; may perhaps not have made that assumption consciously, Cardano certainly did, and by using his formula you are also making the assumption, whether you were aware of it or not.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T00:38:26.410" LastActivityDate="2010-12-30T00:38:26.410" CommentCount="8" />
  <row Id="15691" PostTypeId="2" ParentId="15681" CreationDate="2010-12-27T23:59:27.707" Score="6" ViewCount="0" Body="&lt;p&gt;Sune, there are as many Borel-measurable functions as there are reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that a function is Borel-measurable iff the preimage of each open interval with rational end-points is a Borel set. Moreover, the function is completely determined by the sequence of pairs $(I,B)$ where $I$ varies over the intervals and $B$ is the preimage of $I$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are countably many possible $I$ and to each corresponds one of ${\mathfrak c}=|{\mathbb R}|$ many possible Borel sets. The total number of Borel functions is then bounded above by $|{\mathbb R}^{\mathbb N}|={\mathfrak c}$. Since each constant function is Borel-measurable, ${\mathfrak c}$ is also a lower bound.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-27T23:59:27.707" CommentCount="0" />
  <row Id="15692" PostTypeId="2" ParentId="15686" CreationDate="2010-12-28T00:01:20.040" Score="3" ViewCount="0" Body="&lt;p&gt;More of an extended comment than an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the way I've seen the radical notation used, it is mean to imply only the &quot;principal&quot; root, for some definition of &quot;principal&quot;—therein lies the trouble.  I have seen principal $n$th root defined as the $n$th root with the least non-negative argument or as the $n$th root with argument in $(-\frac{\pi}{n},\frac{\pi}{n}]$ (and, in fact, the latter is the only definition that I've seen implemented in calculators and computer algebra systems).  Most of the instances where I've seen the cubic formula (or quartic formula), it is intended for the first definition, though it is possible to write the formula (both) in such a way that takes the $n$th root only once and thus avoids the need for synchronizing the choice of principal root across multiple expressions (such as &lt;a href=&quot;http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations/786#786&quot;&gt;my answer here&lt;/a&gt; for quartics).&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-28T00:01:20.040" CommunityOwnedDate="2010-12-28T00:01:20.040" />
  <row Id="15693" PostTypeId="2" ParentId="15668" CreationDate="2010-12-28T00:08:28.153" Score="2" ViewCount="0" Body="&lt;p&gt;In both examples you are given an infinite family of sets of size 2, and a choice function picks an element of each family. In the case of the sets of shoes, each set comes with an ordering (left, right), and so we can &lt;em&gt;define&lt;/em&gt; a choice function explicitly. In the case of pairs of socks, this is not the case: Of course, given any pair, we can assign an ordering to it so we can select one of the two socks. However, there is no obvious way of &lt;em&gt;uniformly&lt;/em&gt; doing this for all pairs at the same time. This means (at least intuitively) that there is way of &lt;em&gt;defining&lt;/em&gt; a choice function. Its existence can only be granted by applying the axiom of choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several variants of this example. One that may be useful to think about is the following: One can show explicitly that if $A_n$ is a set of reals and $|A_n|=2$ for each $n\in{\mathbb N}$, then $\bigcup_n A_n$ is a (finite or infinite) countable set. However, it is consistent with the axioms of set theory except choice that there is a sequence $(A_n\mid n\in{\mathbb N})$ of sets, each $|A_n|=2$, and yet $\bigcup_n A_n$ is not countable. Although the construction of the model where this happens is technical, the point is that this formalizes the intuition that there is no &quot;explicit&quot; way of choosing a sock from each pair, simultaneously, and that any way of doing so is essentially non-constructive. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-28T00:08:28.153" CommentCount="1" />
  <row Id="15694" PostTypeId="2" ParentId="15646" CreationDate="2010-12-28T00:13:51.793" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: This reply assumes that the surface contains the singularity (&quot;source&quot;) at its center, so it does not actually pertain to this problem.  For the sake of clarity, I have now changed the surface from $S$ to $S&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a lot like an electrostatics problem, and it is probably trying to mimic one due the field's specific form.  I am going to treat this problem like it is so; someone please point out if what follows is incorrect since I am not absolutely sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field has rotational symmetry, and the $\vec{r_{0}}$ usually means the location of a &quot;source&quot; of the field.  Thus, no matter the location of the source, this field is still rotationally invariant.  Using that surface for the integral is called a &quot;Gaussian surface&quot; by physicists; it is a tool used to help solve a problem by exploiting its symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know you probably just want to solve the integral, so I will get to the point.  Your surface will be a sphere of radius $a$, and so will have a unit normal area vector $\hat{r}&amp;#39;$.  Let us write $\vec{r}&amp;#39;=\vec{r}-\vec{r_{0}}$ and $r=|\vec{r}|$.  If we rewrite the integrand, we get&#xA;$$k\frac{\vec{r}-\vec{r_{0}}}{|\vec{r}-\vec{r_{0}}|^{3}}=k\frac{\vec{r}&amp;#39;}{|\vec{r}&amp;#39;|^{3}}=k\frac{\hat{r}&amp;#39;}{|\vec{r}&amp;#39;|^{2}}~\text{since, for any vector }\vec{v},~\hat{v}=\frac{\vec{v}}{|\vec{v}|}.$$&#xA;The infinitesimal vector area element for the sphere will be&#xA;$$d\vec{S&amp;#39;}=\hat{r}&amp;#39;r&amp;#39;^{2}\sin\phi d\theta d\phi.$$&#xA;Thus, the integral becomes&#xA;$$\begin{aligned}&#xD;&#xA;\int\vec{F}\cdot d\vec{S&amp;#39;}&amp;amp;=\int k\left(\frac{\hat{r}&amp;#39;}{|\vec{r}&amp;#39;|^{2}}\right)\cdot\left(\hat{r}&amp;#39;r&amp;#39;^{2}\sin\phi d\theta d\phi\right) \\&#xD;&#xA;&amp;amp;=\int k\frac{r&amp;#39;^{2}}{r&amp;#39;^{2}}\hat{r}&amp;#39;^{2}\sin\phi d\theta d\phi \\&#xD;&#xA;&amp;amp;=k\int_{0}^{2\pi}d\theta\int_{0}^{\pi}\sin\phi d\phi \\&#xD;&#xA;&amp;amp;=4\pi k.&#xD;&#xA;\end{aligned}$$&#xA;Notice that the radius of the surface is irrelevant.  As long as it contains the &quot;source&quot; point in the center, the value of the integral will always be the same for this field.&lt;/p&gt;&#xA;" OwnerUserId="5153" LastEditorUserId="5153" LastEditorDisplayName="" LastEditDate="2010-12-28T01:29:32.880" LastActivityDate="2010-12-28T01:29:32.880" CommentCount="2" />
  <row Id="15695" PostTypeId="2" ParentId="15631" CreationDate="2010-12-28T00:31:36.213" Score="2" ViewCount="0" Body="&lt;p&gt;Charlie, this is a result of Sierpiński. It appears in &quot;    Sur un problème de la théorie des relations&quot;, Fundamenta Mathematicae, vol 28, (1937), 71-74. You can download the paper &lt;a href=&quot;http://matwbn.icm.edu.pl/tresc.php?wyd=1&amp;amp;tom=28&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general result of which this is a particular case is due to Hajnal (rather than &quot;independent&quot;, one usually talks of &quot;free&quot;). It appears in &quot;Proof of a conjecture of B. Ruziewicz&quot;, Fundamenta Mathematicae, vol 50, (1961), 123-128. It can be downloaded &lt;a href=&quot;http://matwbn.icm.edu.pl/tresc.php?wyd=1&amp;amp;tom=50&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you have difficulty understanding the arguments there, and I'll sketch the proof.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-28T00:31:36.213" />
  <row Id="15696" PostTypeId="1" AcceptedAnswerId="15699" CreationDate="2010-12-28T00:37:00.967" Score="1" ViewCount="49" Body="&lt;p&gt;I have an input array of line segments (a,b)_1 ... (a,b)_n. Let's call it &quot;S&quot; so (a,b)_S_1 ... (a,b)_S_n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another array of the same construct, let's call it &quot;I&quot;, so (a,b)_I_1 ... (a,b)_I_n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to mathematically express the shortest distance of each of I against S. In other words, the matrix formed from the product of I * S. To be very clear, each I is compared to each S. I want the minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know ||(a,b)_I_k-(a,b)_S_k|| denotes distance. How do I represent the matrix formed and indicate the Min of the cells?&lt;/p&gt;&#xA;" OwnerUserId="4930" LastActivityDate="2010-12-28T01:28:57.583" Title="Representing shortest distance of a matrix comparison" Tags="&lt;matrices&gt;&lt;elementary-set-theory&gt;&lt;notation&gt;&lt;elementary&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15698" PostTypeId="2" ParentId="58" CreationDate="2010-12-28T01:14:13.617" Score="1" ViewCount="0" Body="&lt;p&gt;I think that Benford's law is quite intuitive in &quot;real life&quot;, but it would indeed be more suprising if it would often hold in mathematics, and in fact it does, &lt;a href=&quot;http://arxiv.org/abs/1009.0780&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a nice related student paper.&lt;/p&gt;&#xA;" OwnerUserId="5105" LastActivityDate="2010-12-28T01:14:13.617" />
  <row Id="15699" PostTypeId="2" ParentId="15696" CreationDate="2010-12-28T01:28:57.583" Score="2" ViewCount="0" Body="&lt;p&gt;Denote $(a,b)_{S_i}$ by $S_i$ and $(a,b)_{I_k}$ by $I_k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance between $i^{th}$ line segment of $S$ and $j^{th}$ segment of $I$ is $|| (S_i,\  I_j) ||$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now denote the minimum by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{min}_{1 \le i \le n,1 \le j \le n} \{|| (S_i,\ I_j)||\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can denote the matrix by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{ij} = || (S_i,\ I_j)||$ and the min by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{min}_{1 \le i \le n,1 \le j \le n} \{ M_{ij} \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this help?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-28T01:28:57.583" CommentCount="9" />
  <row Id="15700" PostTypeId="2" ParentId="15686" CreationDate="2010-12-28T01:40:58.743" Score="6" ViewCount="0" Body="&lt;p&gt;Here's a pretty clean version of Cardano I use (adapted from Numerical Recipes). I assume here that the principal values for both the square and cube root (the result is in the interval $\left(-\frac{\pi}{n},\frac{\pi}{n}\right]$, which as Isaac says applies for most computing systems) are taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x^3+bx+c=0$, letting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q=\frac{b}{3}$ and $r=\frac{c}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=-\mathrm{sign}(r)\sqrt[3]{|r|+\sqrt{r^2+q^3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B=\begin{cases}\frac{q}{A}&amp;amp;\text{ if }A\neq0\\0&amp;amp;\text{ if }A=0\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the three roots can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A-B$$&lt;br&gt;&#xA;$$-\frac{A-B}{2}\pm i\frac{\sqrt{3}}{2}(A+B)$$&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-28T01:40:58.743" />
  <row Id="15701" PostTypeId="2" ParentId="15634" CreationDate="2010-12-28T01:41:03.847" Score="2" ViewCount="0" Body="&lt;p&gt;The answer to both questions (a) and (b) is yes, and is a consequence of the fact that Riemann integrable functions can be approximated from above and below by &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Step_function&amp;amp;oldid=396614578&quot; rel=&quot;nofollow&quot;&gt;step functions&lt;/a&gt;. The proof follows in a fairly straightforward manner once you realize that this allows you to reduce the problem to that of indicator functions of intervals. To be precise, if $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable then, for each fixed $\epsilon &amp;gt; 0$, there are step functions $g\le f\le h$ with $\int_0^1(h(y)-g(y))\,dy &amp;lt; \epsilon$.&#xA;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\frac1N\sum_{n=1}^Nf(\xi_n)-\int_0^1 f(y)\,dy&amp;amp;\le\frac1N\sum_{n=1}^Nh(\xi_n)-\int_0^1 h(y)\,dy+\int_0^1(h(y)-f(y))\,dy\\&#xD;&#xA;&amp;amp;\le\frac1N\sum_{n=1}^Nh(\xi_n)-\int_0^1 h(y)\,dy+\epsilon.&#xD;&#xA;\end{align}&#xD;&#xA;$$&#xA;Similarly, the reverse inequality holds with $g$ replacing $h$ and $-\epsilon$ replacing $\epsilon$ on the right hand side. So, convergence for step functions implies convergence for Riemann integrable functions. Also, replacing $\xi_n$ by $x+\xi_n$ (mod 1) above, the problem of uniform convergence over all $x$ is also reduced to that of step functions. Linearity further reduces the problem to that of indicator functions of intervals in $(0,1]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $f=1_A$ is an indicator function of an interval $A\subseteq(0,1]$ then the limit $\frac1N\sum_{n=1}^Nf(x+\xi_n)=\int_0^1f(y)\,dy$ follows by definition of equidistribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also obtain uniform convergence (in $x$) as follows: note that $y\mapsto f(x+y)=1_{\{y\in A-x{\rm\ (mod\ 1)}\}}$ is again an indicator function of an interval (taken mod 1). Convergence is uniform simultaneously on all intervals. For any $M &amp;gt; 0$ consider the finite collection of intervals $B_i=((i-1)/M,i/M]$. Then, any interval $A\subseteq(0,1]$ can be sandwiched between unions of the $B_i$. That is, $\cup_{i\in I}B_i\subseteq A\subseteq\cup_{j\in J}B_j$ for $I,J\subseteq\{1,2,\ldots,M\}$ with $I\subseteq J$ and $J\setminus I$ containing no more than two elements.&#xA;$$&#xD;&#xA;\frac1N\sum_{n=1}^N 1_A(\xi_n)-\int_0^1 1_A(y)\,dy\le\sum_{i\in J}\left(\frac1N\sum_{n=1}^N 1_{B_i}(\xi_n)-\int_0^1 1_{B_i}(y)\,dy\right) + 2/M.&#xD;&#xA;$$&#xA;The reverse inequality also holds if we make the sum run over $i\in I$ and replace $2/M$ by $-2/M$ on the right hand side. So,&#xA;$$&#xD;&#xA;\left\vert\frac1N\sum_{n=1}^N 1_A(\xi_n)-\int_0^1 1_A(y)\,dy\right\vert\le\sum_{i=1}^M\left\vert\frac1N\sum_{n=1}^N 1_{B_i}(\xi_n)-\int_0^1 1_{B_i}(y)\,dy\right\vert+2/M.&#xD;&#xA;$$&#xA;For any $\epsilon &amp;gt; 0$ we can choose $2/M &amp;lt; \epsilon/2$ and, using equidistribution, $N$ can be chosen large enough that the first term on the right hand side is less than $\epsilon/2$. So, the left hand side can be made less than $\epsilon$ by choosing $N$ large enough. This choice was independent of $A$ so, in fact, convergence is uniform on the set of all intervals $A$.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastActivityDate="2010-12-28T01:41:03.847" />
  <row Id="15702" PostTypeId="1" AcceptedAnswerId="15706" CreationDate="2010-12-28T01:43:23.033" Score="4" ViewCount="110" Body="&lt;p&gt;One way to prove that a field $K$ has no ideals except the entire field and the trivial ideal is to note the fact that every element $x$ has an inverse. By the definition of an ideal, if $x$ is in the ideal then $x^{-1}x$ is because $x^{-1} \in K$. But now we have that 1 is in the ideal, and so again by the definition of an ideal we have that every element is in the ideal. Therefore it is either the entire field or trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this works for any would-be ideal that has a unit; hence my question. I don't see how this coheres particularly with the idea that ideals are generalizations of things like &quot;multiple of $n$&quot;, or that we use them to form quotient rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain whether this has a deeper meaning or if it's not really important? I think it might have something to do with what is written in the &quot;motivation&quot; section in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article for ideals&lt;/a&gt; but I'm not really sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I do realize that not all subsets without a unit are ideals. Sorry for the confusion.&lt;/p&gt;&#xA;" OwnerUserId="2954" LastEditorUserId="2954" LastEditorDisplayName="" LastEditDate="2010-12-29T00:25:18.437" LastActivityDate="2010-12-29T00:28:35.997" Title="Interpretation of ideals as unitless subsets" Tags="&lt;ring-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="15703" PostTypeId="2" ParentId="15657" CreationDate="2010-12-28T01:45:40.660" Score="1" ViewCount="0" Body="&lt;p&gt;The question does not seem to be very precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;a href=&quot;http://books.google.com/books?id=PuL2KC_KCj0C&amp;amp;pg=PA21&quot; rel=&quot;nofollow&quot;&gt;well known&lt;/a&gt; that $K_{2k+1}$ can be decomposed into $k$ hamiltonian cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can find an Euler tour with exactly $\dfrac{n-1}{2}$ hamiltonian cycles for odd $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even $n$, we can do an &quot;incomplete tour&quot; with  $\dfrac{n}{2}-1$ (See link above) hamiltonian cycles.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-28T01:52:45.987" LastActivityDate="2010-12-28T01:52:45.987" CommentCount="2" />
  <row Id="15704" PostTypeId="2" ParentId="15644" CreationDate="2010-12-28T01:56:53.580" Score="1" ViewCount="0" Body="&lt;p&gt;One never says that the rational number 3/2 is &quot;undefined&quot; as an integer. Rather one simply says that it is nonintegral, or not an integer. On the other hand if 3/2 denotes the value of the (partial) division function on naturals $\rm\:\mathbb N^2\to \mathbb N$ then one may say that this &lt;em&gt;function&lt;/em&gt; is undefined at (3,2). &lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-28T01:56:53.580" />
  <row Id="15705" PostTypeId="2" ParentId="15702" CreationDate="2010-12-28T02:08:33.977" Score="3" ViewCount="0" Body="&lt;p&gt;Ideals are not &quot;unitless subsets&quot;, in the sense that there are many unitless subsets that aren't ideals. Proper ideals, indeed, contain no units, but they are also closed under addition, subtraction and multiplication by arbitrary ring elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why you're confused about the case of &quot;multiples of n&quot;. In the ring $\mathbb{Z}$ of integers, the proper ideals are precisely the multiples of some fixed integer. Indeed, there are many other &quot;unitless subsets&quot; in $\mathbb{Z}$, but they are not ideals. &lt;/p&gt;&#xA;" OwnerUserId="308" LastActivityDate="2010-12-28T02:08:33.977" />
  <row Id="15706" PostTypeId="2" ParentId="15702" CreationDate="2010-12-28T02:10:48.773" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, it's true that any ideal in a ring which contains a unit must be the whole ring. (Also: every left ideal that contains a left-invertible element must be the whole ring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are two items here. One is the notion of generalizing &quot;multiples of $n$&quot;,  and the second is the notion of making quotients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the first. The notion that ideals generalize things like &quot;all multiples of $n$&quot; in $\mathbb{Z}$ goes back all the way to Dedekind, who in fact came up with the term &quot;ideal&quot; (it was a counterpart to Kummer's &quot;ideal numbers&quot;; at the risk of touting my own horn, check out the paper with David McKinnon, &lt;a href=&quot;http://www.jstor.org/stable/30037491&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Gauss's Lemma for Number Fields&lt;/em&gt;&lt;/a&gt;, Amer. Math. Monthly &lt;strong&gt;112&lt;/strong&gt; no. 5 (2005) pp. 385-416).  Dedekind was looking at rings like $\mathbb{Z}[\sqrt{-5}]$, and he wanted something along the lines of unique factorization; Kummer had done this by introducing something called &quot;ideal numbers&quot;, which were like the primes for integers but which did not actually occur in the ring in order to explain things like $2\times 3 = (1+\sqrt{-5})(1-\sqrt{-5})$, even though none of $2$, $3$, $1+\sqrt{-5}$ and $1-\sqrt{-5}$ can be written as a product of two numbers (from $\mathbb{Z}[\sqrt{-5}]$), neither of which is $1$ or $-1$. The idea was that there were these &quot;ideal numbers&quot; $\alpha$, $\beta$, and $\gamma$, which had the property that $\alpha\beta=1+\sqrt{-5}$, $\alpha\gamma=1-\sqrt{-5}$, $\alpha^2 = 2$, and $\beta\gamma=3$. Instead of &quot;inventing&quot; these numbers, Dedekind considered a collection of all numbers which would be &quot;the multiples of $\alpha$&quot; &lt;em&gt;if&lt;/em&gt; $\alpha$ was actually an element of $\mathbb{Z}[\sqrt{-5}]$. A necessary and sufficient condition for a collection of elements of $\mathbb{Z}[\sqrt{-5}]$ to qualify as &quot;all multiples of $x$&quot; (where $x$ was either an ideal number of an actual number) was that the collection be nonempty, closed under sums and differences, and that if $a$ was in the collection of $r$ was in $\mathbb{Z}[\sqrt{-5}]$, then $ra$ was in the collection. They were &quot;ideals&quot; because they were playing the role of ideal numbers (here, &quot;ideal&quot; means &lt;em&gt;existing in fancy or imagination&lt;/em&gt;). This generalized to any ring contained in a finite extension of $\mathbb{Q}$ and containing all integral elements (roots of monic polynomials with integer coefficients), called the &quot;number field case&quot;; later Dedekind and Weber developed a similar theory in what is called the &quot;function field&quot; case. Later, when rings were abstracted and generalized from the number field and function field cases by Artin and by Noether, the conditions were extended but they were still based in the notion that the ideals corresponded to &quot;all multiples of&quot; an ideal number in the number field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is no problem with the fact that sometimes these collections are &lt;em&gt;everything&lt;/em&gt;: this happens whenever the collection satisfies the conditions and contains a unit. Some collections do, some don't. In fact, that is one way in which one can test to see if an element is a unit: if the ideal it generates is the whole ring, then it is a unit (this works in the case of commutative rings with unit, not in more general rings). This actually makes sense because &lt;em&gt;everything&lt;/em&gt; is a multiple of a unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For rings like $\mathbb{Z}$, and to some extent number rings as above, &lt;em&gt;every&lt;/em&gt; ideal consists exactly of &quot;all the multiples of $x$&quot; for some $x$ (though in the number ring, the $x$ may be an &quot;ideal number&quot;). For more general (commutative) rings (with unit), if you have a family of elements $x_1,\ldots,x_n$, then the elements of the ideal &lt;em&gt;generated&lt;/em&gt; by $x_1,\ldots,x_n$ are exactly those elements that are &quot;$R$-linear combinations&quot; of $x_1,\ldots,x_n$; that is, all elements that can be expressed as $r_1x_1+\cdots+r_nx_n$ for some $r_1,\ldots,r_n\in R$; this &lt;em&gt;includes&lt;/em&gt; all multiples of each $x_i$, as well as other elements. If one of the $x_i$ is a unit, then since everything is a multiple of a unit, then everything is in the ideal in question. There are other kinds of ideals (not &quot;finitely generated&quot;, as you start dealing with more and more complicated rings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on to the second question. As it happens, Dedekind also noted that the conditions on the set (nonempty, closed under sums and differences, and closed under products by elements of the ring) were exactly the condition one needed in order to be able to do &quot;congruence modulo&quot; the multiples of $x$, so he also called them &quot;modules&quot; (because you could do modular arithmetic with them). This is essentially the same as talking about quotients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one can get to ideals differently if you are interested in quotients. This is essentially the same thing I did in &lt;a href=&quot;http://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/14315#14315&quot;&gt;this answer&lt;/a&gt; about normal subgroups. Suppose you want to define an equivalence relation $\sim$ on the ring $R$ so that you can make $R/\sim$ into a ring using the operations $[a]+[b]=[a+b]$ and $[a][b]=[ab]$ (where $[a]$ is the equivalence class of $a$ under $\sim$). It turns out that the only way this can work is if $\sim$ is a subring of $R\times R$ (when we consider $\sim$ as a collection of ordered pairs of elements of $R$, hence a subset of $R\times R$), and that if you let $I$ be the collection of all elements of $R$ that are equivalent to $0$, then $I$ is an ideal of $R$ and the equivalence relation $\sim$ is $a\sim b\Longleftrightarrow a-b\in I$. You can try proving the theorems in the abovementioned answer for rings instead of groups, they are essentially the same arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this has no problem with regards to having units in the ideal: this is what happens if your equivalence relation is just &quot;everything is related to everything&quot;, or if your quotient is the one element ring; these are perfectly valid (if somewhat boring) equivalence relations and quotients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: in a commutative ring with $1$, an ideal is the whole ring &lt;strong&gt;if and only if&lt;/strong&gt; it contains a unit. If there are no units, then it is not the whole ring. A ring is said to be &lt;em&gt;simple&lt;/em&gt; if the only ideals are the zero ideal and the whole ring; for commutative rings, the only simple rings are the fields and the zero ring. In the noncommutative case there are simple rings that are not division rings: for example, the set of $n\times n$ matrices over any simple ring (for instance, any field) is simple for every $n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that while a &lt;em&gt;proper&lt;/em&gt; ideal necessarily does not contain a unit, it is not true that any subset that does not contain a unit is an ideal. The collection of all numbers that are either multiples of $2$ or of $3$ in $\mathbb{Z}$ contains no units, but is not an ideal (it is not even a subgroup). &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T00:28:35.997" LastActivityDate="2010-12-29T00:28:35.997" CommentCount="4" />
  <row Id="15707" PostTypeId="1" AcceptedAnswerId="15950" CreationDate="2010-12-28T02:36:26.713" Score="11" ViewCount="264" Body="&lt;p&gt;For which functions $f\colon [0,1] \to [0,1]$ is the function $g(x,y)=f(x)f(y)$ convex over $(x,y) \in [0,1]\times [0,1]$ ? Is there a nice characterization of such functions $f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious examples are exponentials of the form $e^{ax+b}$ and their convex combinations. Anything else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is a simple observation summarizing the status of this question so far. The class of such functions f includes all log-convex functions, and is included in the class of convex functions. So now, the question becomes: are there any functions $f$ that are not log-convex yet $g(x,y)=f(x)f(y)$ is convex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Jonas Meyer observed that, by setting $x=y$, the determinant of the hessian of $g(x,y)$ is positive if and only if $f$ is a log-convex. This resolves the problem for twice continuously differentiable $f$. Namely: if $f$ is $C^2$, then $g(x,y)$ is convex if and only if $f$ is log-convex.&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="5159" LastEditorDisplayName="" LastEditDate="2010-12-31T00:56:27.643" LastActivityDate="2010-12-31T01:34:20.630" Title="For what functions $f(x)$ is $f(x)f(y)$ convex?" Tags="&lt;convex-analysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" />
  <row Id="15708" PostTypeId="2" ParentId="15657" CreationDate="2010-12-28T02:41:16.123" Score="1" ViewCount="0" Body="&lt;p&gt;There aren't necessarily &lt;em&gt;any&lt;/em&gt; Hamiltonian cycles in your sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ross already pointed out, provided $n$ is even and $n &amp;gt; 2$, then all vertices have odd degree and no Eulerian circuit exists.  So, we need only worry about when $n$ is odd and $n &amp;gt; 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $K_5$ arranged as in the picture, an Eulerian circuit with no Hamiltonian cycles is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/StCdJ.png&quot; alt=&quot;$K_5$&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 2 \rightarrow 5 \rightarrow 4 \rightarrow 1 \rightarrow 5 \rightarrow 3 \rightarrow 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose $K_{n-2}$ has an Eulerian circuit with no Hamiltonian cycles.  First label the vertices of $K_n$ as $1, \ldots, n$.  Our Eulerian circuit with no Hamiltonian cycles for $K_n$ is given by first travelling $1 \rightarrow 2$, then travelling the Eulerian circuit with no Hamiltonian cycles for $K_{n-2}$ across the vertices $2, \ldots, n-1$, arriving back at $2$.  Then for $k = 2, 4, 6, \ldots, n - 3$, we travel $$k \rightarrow n \rightarrow k + 1 \rightarrow 1 \rightarrow k + 2.$$  Finally, we travel $n-1 \rightarrow n \rightarrow 1.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows from induction that $K_n$ contains an Eulerian circuit with no Hamiltonian cycles for all odd $n$.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-28T02:41:16.123" />
  <row Id="15709" PostTypeId="1" CreationDate="2010-12-28T02:54:33.603" Score="1" ViewCount="238" Body="&lt;p&gt;In &lt;em&gt;Conceptual Mathematics&lt;/em&gt; 1st edition, p. 325-236, there is a sketch of a proof, but I can't carry out the complete proof.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... This also follows from the appropriate universal mapping properties, which imply that the two composites satisfy properties that only the corresponding identity maps satisfy.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't figure this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me a clue?&lt;/p&gt;&#xA;" OwnerUserId="5158" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T02:56:12.853" LastActivityDate="2011-01-06T01:00:59.687" Title="A question in &quot;Conceptual Mathematics&quot;" Tags="&lt;category-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15710" PostTypeId="1" CreationDate="2010-12-28T03:20:49.937" Score="15" ViewCount="260" Body="&lt;p&gt;Why is it that $\mathbb{Q}$ cannot be homeomorphic to any complete metric space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly $\mathbb{Q}$ is not a complete metric space. But completeness is not a topological invariant, so why is the above statement true?&lt;/p&gt;&#xA;" OwnerUserId="5161" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T03:54:54.777" LastActivityDate="2010-12-28T16:26:06.617" Title="Why is it that $\mathbb{Q}$ cannot be homeomorphic to _any_ complete metric space?" Tags="&lt;topology&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="15711" PostTypeId="2" ParentId="15684" CreationDate="2010-12-28T03:31:15.413" Score="1" ViewCount="0" Body="&lt;p&gt;Can you see what the topology of the quotient $S^1/(\tau)$ is? Its open sets &quot;are&quot; the open sets in $S^1$ which are invariant under multiplication by $\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you see what the topology is, your question becomes easy.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-28T03:31:15.413" />
  <row Id="15712" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T03:31:16.643" Score="16" ViewCount="0" Body="&lt;p&gt;By the Baire Category Theorem, a space that is homeomorphic to a complete metric space must be a Baire Space: the intersection of a countable family of open dense sets must be dense. But $\mathbb{Q}$ does not have this property, because it is countable and no point is isolated: for each $q\in\mathbb{Q}$, let $\mathscr{O}_q = \mathbb{Q}\setminus\{q\}$. This is open (since $\{q\}$ is closed in $\mathbb{R}$, hence in the induced topology of $\mathbb{Q}$) and dense, since every open ball with center in $\mathbb{q}$ intersects $\mathscr{O}_q$. But &#xA;$$\bigcap_{q\in\mathbb{Q}} \mathscr{O}_q = \emptyset$$&#xA;is not dense. Hence $\mathbb{Q}$ is not a Baire space, hence cannot be homeomorphic to a complete metric space (or even to an open subset of a complete pseudometric space).&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T06:21:09.703" LastActivityDate="2010-12-28T06:21:09.703" CommentCount="5" />
  <row Id="15713" PostTypeId="1" CreationDate="2010-12-28T03:31:16.987" Score="1" ViewCount="110" Body="&lt;p&gt;I asked I question &lt;a href=&quot;http://math.stackexchange.com/questions/15676/simplification-of-sums&quot;&gt;here&lt;/a&gt; trying to obtain clarification about how to follow a hint. In spite of the fine answers I received there, the hint doesn't look very helpful. I'd like to know a hint for the following problem or a way to use the hint I already have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability $p_{m}(r,n)$ of finding &lt;em&gt;exactly&lt;/em&gt; $m$ cells empty placing $r$ balls into $n$ cells is $$p_{m}(r,n)=\frac{1}{n^{r}}\binom{n}{m}A(r,n-m)=\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}\left(1-\frac{m+\nu}{n}\right)^{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From such probability, conclude that the probability $x_{m}(r,n)$ of finding $m$ &lt;em&gt;or more cells&lt;/em&gt; empty equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}\left(1-\frac{m+\nu}{n}\right)^{r}\frac{m}{m+v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HINT: Evaluate $x_{m}(r,n)-p_{m}(r,n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without using the hint, I tried to find the pattern for $p_{m}$, $p_{m+1}$, etc with the goal of factoring each term and sum over something recognizable, but I couldn't find a way to factor the last term (for example, to factor $(n-m-1-\nu)^{r}$ into some terms including the original expression $(n-m-\nu)^{r}$...I don't think it is possible. Furthermore, I don't think that strategy is going to produce the answer I'm looking for.&lt;/p&gt;&#xA;" OwnerUserId="2569" LastEditorUserId="2569" LastEditorDisplayName="" LastEditDate="2010-12-28T19:14:07.677" LastActivityDate="2010-12-28T22:24:26.420" Title="Probability of finding $m$ or more cells empty" Tags="&lt;probability&gt;&lt;combinatorics&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="15714" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T03:33:51.197" Score="5" ViewCount="0" Body="&lt;p&gt;A complete metric space is not a countable union of nowhere dense sets, but $\mathbb{Q}$ is (each singleton is nowhere dense). Since nowhere denseness is a topological property, $\mathbb{Q}$ cannot be homeomorphic to a complete metric space.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-28T03:33:51.197" />
  <row Id="15715" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T03:57:18.440" Score="2" ViewCount="0" Body="&lt;p&gt;A subspace of a separable complete metric space is homeomorphic to a complete metric space if and only if it is a $G_\delta$. (I have a reference that says the direction relevant to your question is proved on page 197 of Bourbaki's &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=bQwhdmL6IjUC&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;General topology&lt;/a&gt;&lt;/em&gt;, volume 2, but I don't currently have access to that page.  The other direction is proved on &lt;a href=&quot;http://books.google.com/books?id=bQwhdmL6IjUC&amp;amp;lpg=PP1&amp;amp;pg=PA196#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 196&lt;/a&gt;.)  The set of rational numbers is not a $G_\delta$ in the real numbers.  I must admit that the only way I know how to prove that $\mathbb{Q}$ is not a $G_\delta$ is to appeal to Baire's theorem. (Every open set containing the rationals is dense, and the set of irrationals is a countable intersection of dense open sets, so if the set of rationals were a $G_\delta$, then the empty set (obtained by intersecting the sets of rational and irrational numbers) would be a countable intersection of dense open sets, contradicting Baire's theorem.) &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-28T04:03:03.253" LastActivityDate="2010-12-28T04:03:03.253" CommentCount="3" />
  <row Id="15716" PostTypeId="1" AcceptedAnswerId="15740" CreationDate="2010-12-28T05:13:27.780" Score="11" ViewCount="230" Body="&lt;p&gt;If $n$ is a natural number, let $\displaystyle \sigma_{11}(n) = \sum_{d \mid n} d^{11}$. &#xA;The modular form $\Delta$ is defined by $\displaystyle \Delta(q) = q \prod_{n=1}^{\infty}(1 - q^n)^{24}$.&#xA;Write $\tau(n)$ for the coefficient of $q^{n}$ in $\Delta(q)$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know why $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$. I think the proof may be somewhat difficult, so even just an outline of the argument would be much appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T05:48:22.940" LastActivityDate="2010-12-30T14:48:47.680" Title="Why is $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$?" Tags="&lt;number-theory&gt;&lt;algebraic-number-theory&gt;&lt;modular-forms&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="15717" PostTypeId="2" ParentId="15713" CreationDate="2010-12-28T06:04:57.103" Score="1" ViewCount="0" Body="&lt;p&gt;See the &lt;a href=&quot;http://probabilityandstats.wordpress.com/2010/04/04/a-formula-for-the-occupancy-problem/&quot; rel=&quot;nofollow&quot;&gt;following&lt;/a&gt; (Theorem 2). This gives the probability that exactly $m$ cells are empty. You could probably modify this to get the multiplicative factor $\frac{m}{m+v}$.&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-28T06:04:57.103" CommentCount="1" />
  <row Id="15718" PostTypeId="2" ParentId="15629" CreationDate="2010-12-28T06:21:44.337" Score="2" ViewCount="0" Body="&lt;p&gt;&#xA;How to implement If-then-else structures in propositional logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Example 1&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If P then&lt;br&gt;&#xA;&amp;nbsp; Q&lt;br&gt;&#xA;else&lt;br&gt;&#xA;&amp;nbsp; R&lt;br&gt;&#xA;end if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P -&amp;gt; Q) &amp;amp; (~P -&amp;gt; R)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Example 2&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If P then&lt;br&gt;&#xA;&amp;nbsp; Q&lt;br&gt;&#xA;else if R then&lt;br&gt;&#xA;&amp;nbsp; S&lt;br&gt;&#xA;else&lt;br&gt;&#xA;&amp;nbsp; T&lt;br&gt;&#xA;end if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P -&amp;gt; Q) &amp;amp; (~P &amp;amp; R -&amp;gt; S) &amp;amp; (~P &amp;amp; ~R -&amp;gt; T)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastEditorUserId="3515" LastEditorDisplayName="" LastEditDate="2010-12-28T06:44:31.907" LastActivityDate="2010-12-28T06:44:31.907" />
  <row Id="15719" PostTypeId="1" AcceptedAnswerId="15723" CreationDate="2010-12-28T06:46:39.920" Score="12" ViewCount="407" Body="&lt;p&gt;$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T06:56:57.080" LastActivityDate="2011-01-07T17:45:08.580" Title="How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$" Tags="&lt;calculus&gt;&lt;integral&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" />
  <row Id="15720" PostTypeId="1" CreationDate="2010-12-28T05:18:23.937" Score="3" ViewCount="117" Body="&lt;p&gt;I have a background in math, but I think the following construction appears often in physics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a term for a functional which maps one smooth function $f: R^n \rightarrow R$ to another $L(f):R \rightarrow R$ which represents the integral along each level set?  That is, &#xA;$L(f)(a) = \int_{f^{-1}(a)} d\mu $ &#xA;where the measure $d\mu$ might be the induced measure from the metric on $R^n$, or it might also depend on the original $f$ in some intelligent way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if the initial function is $f:R^2 \rightarrow R$ is defined $f(x)=x^2+y^2$ then $L(f)(a) = 2 \pi \sqrt{a}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the original function is $f:R \rightarrow R$ is defined $f(x)=x^2$, then we'd have $L(f)(a)=0$ for $a&amp;lt;0$, $1$ for $a=0$, and $2$ for $a&amp;gt;0$.  In both cases it might be more natural to integrate with a term derived from the value of $f&amp;#39;$ normal to the level set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose one could generalize to any smooth map $f:M \rightarrow M&amp;#39;$ from a Riemannian manifold to a another smooth manifold and get a new function that represents a computed volume of the preimage for each point:  $L(f):M&amp;#39; \rightarrow R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:  Is there a good way to compute $L(f)$ for some class of function $f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts would be appreciated.  It is reminiscent of (but I hope simpler than) stationary phase approximations and path integral calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stationary_phase_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stationary_phase_approximation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2010-12-30T19:23:13.310" LastActivityDate="2010-12-30T19:51:31.570" Title="Integrating over level sets of a function" Tags="&lt;calculus&gt;&lt;analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="15721" PostTypeId="1" CreationDate="2010-12-28T07:36:11.710" Score="8" ViewCount="219" Body="&lt;p&gt;How can I find solution for $x^3 \equiv 1 \pmod p$ ($p$ a prime) efficiently?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trivial root is $x_1 = 1$. I need to find other roots $x_2, x_3$.&lt;/p&gt;&#xA;" OwnerUserId="5164" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T07:44:05.557" LastActivityDate="2010-12-29T04:30:21.000" Title="Solve $x^3 \equiv 1 \pmod p$ for $x$" Tags="&lt;number-theory&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="15722" PostTypeId="2" ParentId="15721" CreationDate="2010-12-28T07:48:12.090" Score="5" ViewCount="0" Body="&lt;p&gt;First, note that if $p \equiv 2\text{ mod } 3$, then raising to the third power is a bijection on $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$, so that 1 is the only root. If $p \equiv 1\text{ mod } 3$, then the only way I can think of to find the roots is to find a primitive root modulo $p$ (i.e. a generator of the cyclic group $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$) and take it to the powers $(p-1)/3$ and $2(p-1)/3$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-28T07:48:12.090" CommentCount="4" />
  <row Id="15723" PostTypeId="2" ParentId="15719" CreationDate="2010-12-28T07:49:27.237" Score="28" ViewCount="0" Body="&lt;p&gt;It might not be elliptic after all... (unless I have made some mistake)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\displaystyle u = x -\frac{1}{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\displaystyle du = (1 + \frac{1}{x^2})dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $$\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} = -\frac{x^2(1 + 1/x^2)}{x(x-1/x)\sqrt{x^2(x^2 + 1/x^2)}} = -\frac{1 + 1/x^2}{(x-1/x)\sqrt{(x - 1/x)^2 + 2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= -\int \ \frac{\mathrm{du}}{u \sqrt{u^2 + 2}}$$&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-28T07:49:27.237" CommentCount="4" />
  <row Id="15724" PostTypeId="2" ParentId="15251" CreationDate="2010-12-28T08:04:38.377" Score="2" ViewCount="0" Body="&lt;p&gt;You may read the paper &quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.57.8591&quot; rel=&quot;nofollow&quot;&gt;123 theorem and its extensions&lt;/a&gt;&quot; by Noga Alon&lt;/p&gt;&#xA;" OwnerUserId="2481" LastActivityDate="2010-12-28T08:04:38.377" CommentCount="1" />
  <row Id="15725" PostTypeId="1" CreationDate="2010-12-28T08:58:46.910" Score="5" ViewCount="139" Body="&lt;p&gt;How do we solve the recurrence $T(n) = 2T(n/3) + n\log n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible to solve this recurrence by the Master method?&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-28T10:22:31.490" LastActivityDate="2010-12-28T21:07:41.663" Title="how to solve the recurrence $T(n) = 2T(n/3) + n\log n$" Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="15726" PostTypeId="2" ParentId="15725" CreationDate="2010-12-28T09:14:04.837" Score="3" ViewCount="0" Body="&lt;p&gt;Hint: It is possible to solve by &lt;a href=&quot;http://en.wikipedia.org/wiki/Master_theorem&quot; rel=&quot;nofollow&quot;&gt;Master theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more generic method is &lt;a href=&quot;http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method&quot; rel=&quot;nofollow&quot;&gt;Akra Bazzi&lt;/a&gt;, but you don't need that for this problem.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-28T09:14:04.837" CommentCount="4" />
  <row Id="15727" PostTypeId="2" ParentId="15716" CreationDate="2010-12-28T09:36:07.067" Score="10" ViewCount="0" Body="&lt;p&gt;The modular discriminant $\Delta (z) \in M_{12,0},$ the linear space of cusp forms of weight $12.$ Now dimension $M_{12,0} = 1$  (see, for example, &lt;a href=&quot;http://books.google.co.uk/books?id=lyqDSjLY6zEC&amp;amp;pg=PA118&amp;amp;lpg=PA118&amp;amp;dq=dimension+linear+space+cusp+form&amp;amp;source=bl&amp;amp;ots=xxacNSP9vX&amp;amp;sig=B0rUiRlyKk-7SiVYNGn9_KwILa8&amp;amp;hl=en&amp;amp;ei=eqcZTfmWMI-5hAfX79y3Dg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CDQQ6AEwBA#v=onepage&amp;amp;q=dimension%20linear%20space%20cusp%20form&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;T.M. Apostol: Moldular Functions and Dirichlet Series in Number Theory&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every modular form for the full modular group is a polynomial in $E_4$ and $E_6,$ where $E_{2k}$ are the Eisenstein series defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{2k}(z) = \frac{1}{2\zeta(2k)} \sum_{(m,n) \ne (0,0) } \frac{1}{(m+nz)^{2k}} \quad \textrm{for } Im(z) &amp;gt;0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the summation extends over all integral $m$ and $n$ not both equal to $0$. So if we use the Eisenstein series to construct cusp forms of weight $12,$ because $\textrm{dim } M_{12,0} = 1$ these will necessarily be constant multiples of $\Delta (z).$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{2k}(z) = 1 - \frac{4k}{B_{2k}} \sum_{n=1}^\infty \sigma_{2k-1}(n)e^{2 \pi i n z}&#xD;&#xA;\quad \textrm{for } Im(z) &amp;gt; 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma_k(n) = \sum_{d|n} d^k$ and the $B_{2k}$ are the Bernoulli numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can expect identities between the divisors functions $\sigma_{2k-1}(n)$ and $\tau(n)$ to exist. Ramanujan's paper &quot;On Certain Arithmetical Functions&quot; shows how to express&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{r,s}(x) = \sum_{n=1}^\infty n^r \phi_{s-r}(n)x^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for integer $r,s \ge 0$ and $ |x| &amp;lt; 1$ as polynomials in $E_2,E_4,$ and $E_6,$ from which many such identities follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To cite a random example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau(n) = n^2\sigma_7(n) - 540\sum_{k=1}^{n-1}k(n-k)\sigma_3(k)\sigma_3(n-k)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so $\tau(n) \equiv n^2\sigma_7(n) \textrm{ mod } 540.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: $B_{12} = -691/2730.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT3: For completeness, using Ramanujan's notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = 1 - 24 \sum_{n=1}^\infty \frac{nx^n}{1-x^n}$$&#xA;$$Q = 1 + 240 \sum_{n=1}^\infty \frac{n^3x^n}{1-x^n}$$&#xA;$$R = 1 - 504 \sum_{n=1}^\infty \frac{n^5x^n}{1-x^n},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the relevant entries from his paper &quot;On Certain Arithmetical Functions&quot; are $(3)$ and $(6)$ from his Table I, given below. (Note that $Q$ is $E_4$ and $R$ is $E_6.$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;1 - 24\phi_{0,1}(x) &amp;amp;= P \\&#xD;&#xA;1 + 240\phi_{0,3}(x) &amp;amp;= Q \\&#xD;&#xA;1 - 504\phi_{0,5}(x) &amp;amp;= R \quad (3) \\&#xD;&#xA;1 + 480\phi_{0,7}(x) &amp;amp;= Q^2 \\&#xD;&#xA;1 - 264\phi_{0,9}(x) &amp;amp;= QR \\&#xD;&#xA;691 + 65520\phi_{0,11}(x) &amp;amp;= 441Q^3+250R^2 \quad (6) \\&#xD;&#xA;1 - 24\phi_{0,13}(x) &amp;amp;= Q^2R \\&#xD;&#xA;3617 + 16320\phi_{0,15}(x) &amp;amp;= 1617Q^4+2000QR^2 \\&#xD;&#xA;43867 - 28728\phi_{0,17}(x) &amp;amp;= 38367Q^3R+5500R^3 \\&#xD;&#xA;174611 + 13200\phi_{0,19}(x) &amp;amp;= 53361Q^5+121250Q^2R^2 \\&#xD;&#xA;77683 - 552\phi_{0,21}(x) &amp;amp;= 57183Q^4R+20500QR^3 \\&#xD;&#xA;236364091 + 131040\phi_{0,23}(x) &amp;amp;= 49679091Q^6+176400000Q^3R^2 + 10285000R^4 \\&#xD;&#xA;657931 - 24\phi_{0,25}(x) &amp;amp;= 392931Q^5R+265000Q^2R^3 \\&#xD;&#xA;3392780147 + 6960\phi_{0,27}(x) &amp;amp;= 489693897Q^7+2507636250Q^4R^2 + 395450000QR^4 \\&#xD;&#xA;1723168255201 - 171864\phi_{0,29}(x) &amp;amp;= 815806500201Q^6R+88134070500Q^3R^3 + 26021050000R^5 \\&#xD;&#xA;7709321041217 + 32640\phi_{0,31}(x) &amp;amp;= 764412173217Q^8+5323905468000Q^5R^2 \\ &#xD;&#xA;&amp;amp;+ 1621003400000Q^2R^4&#xD;&#xA;&#xD;&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations $(3)$ and $(6)$, along with $1728 \sum_{n=1}^\infty \tau(n)x^n = Q^3 - R^2,$ give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$691+65520 \sum_{n=1}^\infty \sigma_{11}(n) x^n = 441 \times 1728 \sum_{n=1}^\infty \tau(n) x^n + 691R^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result follows by taking this equation modulo $691$ and noting that $566$ and $691$ are coprime. &lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2010-12-30T14:48:47.680" LastActivityDate="2010-12-30T14:48:47.680" CommentCount="3" />
  <row Id="15728" PostTypeId="2" ParentId="15650" CreationDate="2010-12-28T10:03:44.447" Score="1" ViewCount="0" Body="&lt;p&gt;As Ross's counterexample points out the answer, of course, is no in general.  However there are broad special cases where the answer is yes, which all more or less go back to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem&quot; rel=&quot;nofollow&quot;&gt;Perron-Frobenius theorem&lt;/a&gt;.  In particular if the recurrence is combinatorial in the sense that it counts the number of words of length $n$ in a regular language, then subject to some reasonable assumptions about the language the limit you are looking at will exist. &lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-28T10:03:44.447" />
  <row Id="15729" PostTypeId="2" ParentId="15719" CreationDate="2010-12-28T10:12:43.353" Score="12" ViewCount="0" Body="&lt;p&gt;Somewhat inspired by Moron's wonderful answer, I decided to see if a trigonometric solution would do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the substitution $x=\cot\left(\frac{\theta}{2}\right)$, $\mathrm dx=\frac{\mathrm d\theta}{\cos\;\theta-1}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\cot^4\frac{\theta}{2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\left(\frac{1+\cos\;\theta}{1-\cos\;\theta}\right)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac1{\sqrt{2}}\int \frac{\mathrm d\theta}{\cos\;\theta\sqrt{1+\cos^2\theta}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which integrates to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{2}}\tanh^{-1}\frac{\sin\;\theta}{\sqrt{1+\cos^2\theta}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undoing the substitution, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1{\sqrt{2}}\tanh^{-1}\left(x\sqrt{\frac{2}{x^4+1}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it is easy to verify that the derivative of this last expression gives the original integrand.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-28T10:12:43.353" />
  <row Id="15730" PostTypeId="1" CreationDate="2010-12-28T10:30:02.517" Score="3" ViewCount="71" Body="&lt;p&gt;Suppose A is a linear transformation from R^3 to R^3 and |det(A)| = 1. I know that A is volume preserving, but is it also area preserving? For example, if a and b are two vectors in R^3 that span a parallelogram, is the area of this parallelogram equal to the area of the paralellogram spanned by A(a) and A(b)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THank you!&lt;/p&gt;&#xA;" OwnerUserId="5165" LastActivityDate="2010-12-28T10:45:43.413" Title="Area preserving transformations" Tags="&lt;linear-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15731" PostTypeId="2" ParentId="15730" CreationDate="2010-12-28T10:45:43.413" Score="7" ViewCount="0" Body="&lt;p&gt;Matrices of the form $\begin{pmatrix}X &amp;amp; 0\\\\0 &amp;amp; \text{det}(X)^{-1}\end{pmatrix}$ with $X$ any invertible 2 by 2 matrix with determinant not equal to $\pm1$ give a host of counter examples: consider the action of such a matrix on a parallelogram in the subspace $\langle(1,0,0),(0,1,0)\rangle$.&lt;/p&gt;&#xA;" OwnerUserId="3212" LastActivityDate="2010-12-28T10:45:43.413" />
  <row Id="15732" PostTypeId="2" ParentId="15707" CreationDate="2010-12-28T11:43:23.683" Score="0" ViewCount="0" Body="&lt;p&gt;This may just be a silly observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, let $\alpha=(a_1,a_2)$ and $\beta=(b_1,b_2)$. If $f$ is convex and the sum of $g$ at the four corners of the square formed by $\alpha$ and $\beta$ is non-positive, then $g(t\alpha + (1-t)\beta) \leq tg(\alpha)+(1-t)g(\beta)$, where $t \in [0,1]$&lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-28T11:43:23.683" />
  <row Id="15733" PostTypeId="2" ParentId="15520" CreationDate="2010-12-28T11:55:21.640" Score="2" ViewCount="0" Body="&lt;p&gt;\begin{align}&#xA;\newcommand{\mba}{\mathbf{a}}&#xA;\newcommand{\mbb}{\mathbf{b}}&#xA;\newcommand{\mbl}{\mathbf{l}}&#xA;\newcommand{\mbp}{\mathbf{p}}&#xA;\newcommand{\mbv}{\mathbf{v}}&#xA;\newcommand{\mbx}{\mathbf{x}}&#xA;\int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot (\mba - \mbb) + \mbl \cdot \mbb)d\mbx &#xA;&amp;amp;=\exp(\mbl\cdot\mbb)\prod_{i=1}^k\int_{l_i}^0\exp(x_i(a_i-b_i))dx_i\\&#xA;&amp;amp;=\exp(\mbl\cdot\mbb)\prod_{i=1}^k\frac{1-\exp(l_i(a_i-b_i))}{a_i-b_i}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should fix the error in your calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a lot of questions still remain. What is $B(\alpha+1)$? And if your are trying to compute convolution, why isn't your integration domain $(-\infty,0]^k$?&lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-28T11:55:21.640" CommentCount="1" />
  <row Id="15734" PostTypeId="1" AcceptedAnswerId="15741" CreationDate="2010-12-28T12:01:23.910" Score="4" ViewCount="184" Body="&lt;p&gt;In&#xA;&lt;a href=&quot;http://math.stackexchange.com/questions/11945/limit-of-sqrt7-sqrt7-sqrt7-cdots&quot;&gt;Limit of $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$&lt;/a&gt;&#xA;Timothy Wagner gave a correct answer that was questioned for not having shown that the limit exists in the first place. My question is, are there any examples were a value of limit can be derived although the limit does not exist? Please note I am not questioning that limit must exist before it is exhibited but that, how a value for limit can be exhibited if the limit doesn't exist? Is that not a contradiction? On one hand we have a value for the limit and on the other hand the proof that it can't exist! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give an example were a more general form of convergence does not account for the calculated value. Otherwise it seems as if the value was hinting that the notion of convergence required adjustment and not the calculated value that required justification.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2010-12-28T13:52:48.827" LastActivityDate="2010-12-28T14:35:37.523" Title="Request for example of a value of limit of a sequence where sequence does not converge " Tags="&lt;analysis&gt;&lt;logic&gt;" AnswerCount="4" />
  <row Id="15735" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T12:06:55.683" Score="14" ViewCount="0" Body="&lt;p&gt;Yes. Consider the sequence $a_{n+1} = 2a_n + 1$, where $a_0 = 0$. If this sequence converged to some value $L$, then $(a_{n+1})$ would converge to the same value, so we would get $L = 2L + 1$, which gives $L = - 1$. But the sequence actually diverges to infinity.  I use this sort of example when I teach real analysis to drive home the point that we have to check for convergence before trying to compute the value of a limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method also comes up in with infinite series. Consider the series $1 + -1 + 1 + -1 + \cdots$, which is not convergent. If it &lt;em&gt;were&lt;/em&gt; convergent to a value $s$, then we would have $s = 1 + (-1 + 1 + -1 + \cdots)$ so $s = 1 - s$, which leads to $s = 1/2$. So we can compute a &quot;value&quot; even though the series diverges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact is actually useful in practice when we study generalized notions of convergence. For the moment, define a notion of series convergence to be &quot;nice&quot; if, whenever a series $\sum a_n$ converges, we have $\sum a_n = a_0 + \sum a_{n+1}$ and $\sum -a_n = -\sum a_n$. Then if $1 + -1 + 1 + -1 + \cdots$ converges under any &quot;nice&quot; notion of convergence, its limit will be $1/2$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-28T12:38:56.497" LastActivityDate="2010-12-28T12:38:56.497" CommentCount="17" />
  <row Id="15736" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T12:10:50.437" Score="2" ViewCount="0" Body="&lt;p&gt;What he showed was that &lt;em&gt;if&lt;/em&gt; a limit $L$ exists, then it has to satisfy the equation $L = \sqrt{7+L}$. This is typical of exercises early in a real analysis class where the only convergence theorem you have available is that bounded monotone sequences converge (this is easy to show given that the reals have the least upper bound property).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we have a sequence $(a_n)$ that is defined recursively by $a_1 = x$, $a_{n+1} = f(a_n)$ and is monotone and unbounded, then any fixed point of $f$ is going to be a candidate for a limit, but $(a_n)$ doesn't converge.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-28T12:22:38.487" LastActivityDate="2010-12-28T12:22:38.487" CommentCount="4" />
  <row Id="15737" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T12:13:44.277" Score="0" ViewCount="0" Body="&lt;p&gt;The point is often when we try to derive the limit, we invoke algebraic manipulations which would only work if it has already been established that a limit exists (say, by showing the sequence is cauchy). &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-28T12:13:44.277" CommentCount="7" />
  <row Id="15739" PostTypeId="1" AcceptedAnswerId="16899" CreationDate="2010-12-28T14:16:57.003" Score="0" ViewCount="231" Body="&lt;p&gt;I am referring to the Polyhedron Volume &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyhedron#Volume&quot; rel=&quot;nofollow&quot;&gt;formula here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac{\sum_{i=1}^k \vec{x_{i}} \cdot \hat{n_i} A_i}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, does the above formula applicable for the Polyhedron below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=155vFKEaDL3ZL8sy-zhil60zlL9wA07sJBOuZRHVvrKQ&amp;amp;w=960&amp;amp;h=720&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given faces for the polyhedron are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_1: n_1,n_2,n_3, n_4$$&#xA;$$f_2:n_2, n_3,n_8,n_5$$&#xA;$$f_3:n_1, n_4,n_6,n_7$$&#xA;$$f_4:n_6, n_8,n_5,n_7$$&#xA;$$f_5:n_1, n_2,n_5,n_7$$&#xA;$$f_6:n_6, n_8,n_3,n_4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for my application, the face $f_6$ may or may not intersect with the face $f_5$. Why is why I would prefer to apply the above formula mechanically, instead of having to determine the intersected line, and then compute the volume separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does, what would be the value of the above calculation?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$|v_1-v_2|$&lt;/li&gt;&#xA;&lt;li&gt;$|v_1|+|v_2|$&lt;/li&gt;&#xA;&lt;li&gt;None of the above?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there a proof showing why the answer is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Additional discussion on the validity of the above formula is &lt;a href=&quot;http://math.stackexchange.com/questions/16111/is-this-polyhedron-formula-correct&quot;&gt;shown here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Update the formula according to the wiki page. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2011-01-03T08:55:39.563" LastActivityDate="2011-01-09T16:53:15.887" Title="Does Divergence Theorem Polyhedron Volume Calculation Applicable for a Cut-Through Polygon?" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="15740" PostTypeId="2" ParentId="15716" CreationDate="2010-12-28T14:26:28.670" Score="8" ViewCount="0" Body="&lt;p&gt;The proof is not at all obvious if you begin simply with the formula&#xA;$$\Delta(q) = q \prod_{n=1}^{\infty} (1-q^n)^{24}.$$  However, as Derek Jennings explains in his answer, if you use the (absolutely crucial!) fact that $\Delta$ is a cusp form of weight twelve and level one, the proof is actually not very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Derek explains, the ring of modular forms of level one is generated by two $q$-expansions, namely&#xA;$$E_4 := 1 + 240 \sum_{n = 1}^{\infty} \sigma_3(n) q^n,$$&#xA;which has weight 4,&#xA;and&#xA;$$E_6 := 1 + 504 \sum_{n=1}^{\infty} \sigma_5(n) q^n,$$&#xA;which has weight 6.  (The coefficients $240$ and $-504$ come from Bernoulli numbers, as Derek explains, but we don't need that at the moment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we see that we can make two monomials of weight 12 from these, namely&#xA;$E_4^3$ and $E_6^2$.  How do we get $\Delta$?  Well, the constant term of $\Delta$ vanishes, while $E_4^3$ and $E_6^2$ have constant term $1$, so &#xA;$\Delta$ must be proportional to $E_4^3  - E_6^2$.  Since the coefficient&#xA;of $q$ in $\Delta$ is $1$ (i.e. $\tau(1) = 1$) while the coefficient&#xA;of $q$ in $E_4^3 - E_6^2$ is $1728,$ we find that &#xA;$$E_4^3 - E_6^2 = 1728 \Delta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is useful to note that we can also use $E_4^3$ (say) and $\Delta$ as a basis for the weight $12$ modular forms.  In fact, they are a very convenient basis, because if $f$ is any weight $12$ modular form, with constant term $a_0$, then we can subtract of $a_0 E_4^3$ to get rid of the constant term of $f$, and then&#xA;$f - a_0 E_4^3$ must be a mulitple of $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go further, we have to introduce another fact, also noted by Derek Jennings, namely that&#xA;there is a weight 12 modular form&#xA;$$E_{12} = 1 + \dfrac{65520}{691} \sum_{n=1}^{\infty} \sigma_{11}(n) q^n.$$&#xA;In fact, it is &#xA;(for me)&#xA;easier to work with&#xA;$$691 E_{12} = 691 + 65520\sum_{n=1}^{\infty} \sigma_{11}(n) q^n,$$&#xA;which has integer coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply the above procedure to write $691 E_{12}$ in terms of $E_4^3$ and $\Delta$, to find that&#xA;$$691 E_{12} = 691 E_4^3 + (65520 - 691\cdot 720) \Delta.$$ &#xA;Now all the $q$-expansions in this formula have integral coefficients, and so&#xA;what we find, looking at the coefficient of $q^n$, is that&#xA;$65520\sigma_{11}(n) \equiv 65520\tau(n)$ for each $n \geq 1$.&lt;br&gt;&#xA;Dividing by $65520$ (which is coprime to $691$) gives the desired formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way this is summarized is to say that&#xA;$$\dfrac{691}{65520} E_{12} = \dfrac{691}{65520} + \sum_{n=1}^{\infty} \sigma_{11}(n) q^n$$&#xA;is normalized (i.e. has coefficient of $q$ equal to $1$) and is a cuspform modulo $691$ (i.e. its constant term vanishes mod $691$).  This forces it (as we have just seen) to be congruent to $\Delta$ modulo $691$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me note that by far the best place to read about the theory of modular forms used here is Serre's beautiful book &lt;em&gt;A course in arithmetic&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: Since I was editing this anyway, let me add a cultural remark, namely that the study of congruences between Eisenstein series and cuspforms,&#xA;of which the congruence between $\dfrac{691}{65520} E_{12}$ and $\Delta$ considered above is the first example, is a central topic in modern number theory.  It lies at the heart of Mazur's determination of the possible torsion subgroups of elliptic curves over $\mathbb Q$, and is the basic method via which Ribet proved his &quot;converse to Herbrand's theorem'' result giving a criterion for non-triviality of various $p$-power subgroup of the class group of the cyclotomic field $\mathbb Q(\zeta_p)$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-28T19:50:59.233" LastActivityDate="2010-12-28T19:50:59.233" CommentCount="3" />
  <row Id="15741" PostTypeId="2" ParentId="15734" CreationDate="2010-12-28T14:35:37.523" Score="3" ViewCount="0" Body="&lt;p&gt;(This should be a comment under Carl's &lt;a href=&quot;http://math.stackexchange.com/questions/15734/request-for-example-of-a-value-of-limit-of-a-sequence-where-sequence-does-not-con/15735#15735&quot;&gt;answer&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Arjan: One very general way in which to formalize your idea of &quot;possible definitions of convergence&quot; is that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Banach_limit&quot; rel=&quot;nofollow&quot;&gt;Banach limits&lt;/a&gt;. Indeed, a Banach limit is essentially a convergence notion for bounded sequences satisfying a minimal set of reasonable conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Banach limits do exist---this was proved by Banach himself as a way to show off his Hahn-Banach theorem---and all Banach limits agree with the usual limit on sequences that converge. More interestingly, the set of bounded sequences on which &lt;em&gt;all&lt;/em&gt; Banach limits agree is strictly larger than the set of convergence sequences, and it was determined explicitly by [Lorentz, G. G. A contribution to the theory of divergent sequences. Acta Math. 80, (1948). 167--190. &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=MR0027868&quot; rel=&quot;nofollow&quot;&gt;MR0027868&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the result of Lorentz implies that not &lt;em&gt;all&lt;/em&gt; Banach limits agree on all sequences (in other words, that there is more than one Banach limit) It follows from this that there are divergent sequences which can be assigned at least &lt;em&gt;two&lt;/em&gt; different limits under different extensions of the notion of convergence.&lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-28T14:35:37.523" CommentCount="1" />
  <row Id="15742" PostTypeId="1" AcceptedAnswerId="15743" CreationDate="2010-12-28T14:45:32.910" Score="1" ViewCount="104" Body="&lt;p&gt;$$T (n) = 3T (n/5) + \log^2 n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am trying to solve this recurrence i have been told that it is master methods case 1 but how do i compare $\log^2n$ with $n^ {\log_5 3}$&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-28T15:13:36.017" LastActivityDate="2010-12-28T16:48:39.820" Title="stuck with master method." Tags="&lt;algorithms&gt;&lt;asymptotics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15743" PostTypeId="2" ParentId="15742" CreationDate="2010-12-28T15:34:40.163" Score="3" ViewCount="0" Body="&lt;p&gt;A rule of thumb is that polynomial functions always grow faster than logarithmic functions.  E.g. &lt;a href=&quot;http://www.cs.columbia.edu/~mihalis/cs4231/Handouts/Asymptotics.pdf&quot; rel=&quot;nofollow&quot;&gt;this lecture&lt;/a&gt;. So your polynomial function asymptotically dominates your log function, giving you case 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know why powers grow faster than logs, you can put the functions in a ratio and then evaluate $\lim_{n\to \infty}  \frac{f(n)}{g(n)}$.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2010-12-28T15:34:40.163" CommentCount="6" />
  <row Id="15744" PostTypeId="2" ParentId="15742" CreationDate="2010-12-28T15:42:38.337" Score="2" ViewCount="0" Body="&lt;p&gt;So we want to show that $T(n) = \Theta(n^{\log_{b}a})$ and $f(n) = O(n^{\log_{b}(a)-\epsilon})$. So $a=3, b=5, f(n) = \log^{2} n$, and $\log_{5} 3 = 0.6826$. So we check the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$f(n) = \log^{2} n = O(n^{0.6826- \epsilon})$. So choose any $\epsilon &amp;gt;0$. In other words, $\log^{2} n \in o(n^{0.6826- \epsilon})$. Note the derivative of $(\log n)^2 = \frac{2 \log n}{n}$ and the derivative of $n^{0.6826}$ is $\frac{0.6826}{n^{0.3174}}$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The result follows. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-28T15:53:35.853" LastActivityDate="2010-12-28T15:53:35.853" CommentCount="1" />
  <row Id="15746" PostTypeId="2" ParentId="15720" CreationDate="2010-12-28T15:53:17.547" Score="1" ViewCount="0" Body="&lt;p&gt;Using the Fourier transform representation of the Dirac measure on the level set, the volume&#xA;of the level set (which is the required integral according to the examples given in the question)&#xA;is given by (using the same notation as in the question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(f)(a) = \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{d}k \,exp(-ika)\, \int\,|\nabla f|\,exp(ikf)\,\mathrm{d}\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a Riemannian manifold, the gradient is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla f = \sum_{ij}g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two examples given in the question can be worked out according to this formula (the second example very easily, the first&#xA;one involves integrals of trigonometric functions of a squared angle, but it can also be &#xA;solved analytically).&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2010-12-28T15:53:17.547" />
  <row Id="15747" PostTypeId="2" ParentId="15710" CreationDate="2010-12-28T16:10:09.870" Score="5" ViewCount="0" Body="&lt;p&gt;There is another way to test whether a metric space is homeomorphic to a complete metric space (that is, whether the space is &lt;em&gt;completely metrizable&lt;/em&gt;: there is a complete metric that induces the same topology). This is known as the &lt;em&gt;Choquet game&lt;/em&gt; or the &lt;em&gt;strong Choquet game&lt;/em&gt;. It was introduced by Choquet in his &lt;em&gt;Lectures on analysis&lt;/em&gt; in 1969; a modern treatment can be found in &lt;em&gt;Classical descriptive set theory&lt;/em&gt; by Kechris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that each metric space is associated with a two-player game of perfect information, of countable length, such that the metric space is homeomorphic to a complete separable metric space if and only if the second player has a winning strategy in the game.  All the details and proofs can be found in Kechris' book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to show, once you have all the definitions, that the first player has a winning strategy on this game when it is played on $\mathbb{Q}$, which means the second player does not have a winning strategy, so $\mathbb{Q}$ is not completely metrizable. This gives an alternate way to prove that the rationals are not $G_\delta$ in the reals (cf. the answer by Jonas Meyer). First we prove that the rationals are not completely metrizable, by Choquet's method. Then we prove that a metric space is completely metrizable if and only if it is $G_\delta$ in its completion (actually, this is used in the proof of Choquet's result, so we had to prove it already). Since the completion of $\mathbb{Q}$ is $\mathbb{R}$, this shows that $\mathbb{Q}$ is not $G_\delta$ in $\mathbb{R}$. &lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorUserId="630" LastEditorDisplayName="" LastEditDate="2010-12-28T16:26:06.617" LastActivityDate="2010-12-28T16:26:06.617" />
  <row Id="15748" PostTypeId="2" ParentId="15742" CreationDate="2010-12-28T16:48:39.820" Score="2" ViewCount="0" Body="&lt;p&gt;This comment below makes explicit what the first answer already said, and hopefully removes the confusion regarding polylogs that seems to persist in the OP's mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, consider $$\lim_{n \to \infty} \frac{(\log n)^k}{n^\epsilon}$$ for any positive $\epsilon$ and any positive integer $k$ (integrality is not really necessary, but suffices for our purposes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&#xA;$$\lim_{n \to \infty} \frac{(\log n)^k}{n^\epsilon} = \lim_{n \to \infty} \underbrace{\frac{\log n}{n^{\epsilon/k}}\cdots \frac{\log n}{n^{\epsilon/k}}}_{k-\mathrm{times}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it all boils down to proving that $$\lim_{n \to \infty} \frac{\log n}{n^\gamma} = 0,$$&#xA;where $\gamma &amp;gt; 0$, and we already know how to prove this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I hope this provides the clarification the OP sought.&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-28T16:48:39.820" />
  <row Id="15749" PostTypeId="1" AcceptedAnswerId="15762" CreationDate="2010-12-28T16:49:11.707" Score="4" ViewCount="109" Body="&lt;p&gt;My question is on the convergence of the riemann sum, when the value spaces are square-integrable random variables. The convergence does depend on the evaluation point we choose, why is the case. here is some background to make this clearer&#xA;suppose $f\colon \Re \mapsto \Re $ is some continous function on $[a,b]$, the stieltjes integral of $f$ with respect to itself $f$ is $\int^{b}_{a} f(t)df(t)$&#xA;if we take a partition $ \Delta_n = \{t_0, t_1, \cdots, t_n \}$ of $[a,b]$ the riemmans sums is &#xA;$$ L_{n} = \sum^{n}_{i=1} f(t_{i-1})(f(t_{i})-f(t_{i-1})) $$&lt;br&gt;&#xA;Now if the limit exists say $\lim_{n\to\infty}L_{n}= A$ then if we choose the evaluation point $ t_{i}$ then the sum&lt;br&gt;&#xA;$$ R_{n} = \sum^{n}_{i=1} f(t_{i})(f(t_{i})-f(t_{i-1})) $$&lt;br&gt;&#xA;will also converge to $A$ so &#xA;$$\lim_{n\to\infty}L_{n} = \lim_{n\to\infty}R_{n}$$&#xA;Now we apply same idea for a stochastic integral here $W(t)$ is a wiener process and we wish to find&#xA; $$\int^{b}_{a}W(t)dW(t) $$ &#xA;$$ L_{n} = \sum^{n}_{i=1} W(t_{i-1})(W(t_{i})-W(t_{i-1})) $$&lt;br&gt;&#xA;$$ R_{n} = \sum^{n}_{i=1} W(t_{i})(W(t_{i})-W(t_{i-1})) $$&#xA;in $L^2$ norm the limits of  $L_{n}$ and $R_{n}$ exist but are different&#xA;$$\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = b-a $$&#xA;can someone explain why the limits are different ? If the limit exists which in this case it does. I would have expected  $\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = 0 $ in $L^2$ norm &lt;/p&gt;&#xA;" OwnerUserId="5176" LastActivityDate="2010-12-28T21:19:32.047" Title="stochastic integral &amp; stieltjes-integral" Tags="&lt;real-analysis&gt;&lt;stochastic-processes&gt;&lt;stochastic-integrals&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="15750" PostTypeId="1" AcceptedAnswerId="15763" CreationDate="2010-12-28T17:02:46.167" Score="12" ViewCount="221" Body="&lt;p&gt;After seeing &lt;a href=&quot;http://math.stackexchange.com/questions/15719/how-do-i-integrate-the-following-int-frac1x2-mathrm-dx1-x2-sqrt&quot;&gt;this&lt;/a&gt; recent question asking how to calculate the following integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int \frac{1 + x^2}{(1 - x^2) \sqrt{1 + x^4}} \, dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and some of the comments that suggested that it was an elliptic integral, I tried reading a little bit on the Wikipedia article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptic_integral&quot; rel=&quot;nofollow&quot;&gt;elliptic integrals&lt;/a&gt;.&#xA;It seems that the point is that most elliptic integrals cannot be expressed in terms of elementary functions. The Wikipedia article defines an elliptic integral as an integral of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int R \left( x, \sqrt{ P(x) } \right ) \, dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R(x, y)$ is a rational function and $P(x)$ is a polynomial of degree $3$ or $4$ with no repeated roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the article does mention in its introductory section that two exceptions in which the elliptic integrals can be expressed in terms of elementary functions are when the polynomial $P(x)$ has repeated roots or when the rational function $R(x, y)$ does not contain odd powers of $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example in question we have $P(x) = 1 + x^4$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(x, y) = \frac{1 + x^2}{(1 - x^2)y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so certainly it does not correspond to the two exceptions mentioned before. Thus I have a couple of questions about this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) What are the conditions for an elliptic integral (as defined in the Wikipedia article) to be expressible in terms of elementary functions? More specifically, are the two above cited conditions the only exceptions or are there any others which may explain why the above integral is expressible in terms of elementary functions?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) Depending on the answer to my first question, why is it that the above &quot;elliptic integral&quot; can be expressed in terms of elementary functions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: I'm not sure but I suppose that some conditions must be put on the rational function $R(x, y)$ so to avoid trivial cases, but I don't want to speculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="900" LastActivityDate="2010-12-30T04:47:52.143" Title="When is an elliptic integral expressible in terms of elementary functions?" Tags="&lt;calculus&gt;&lt;integral&gt;&lt;special-functions&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="15751" PostTypeId="2" ParentId="15749" CreationDate="2010-12-28T17:06:45.380" Score="0" ViewCount="0" Body="&lt;p&gt;Hint 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis of the explanation is in the different behaviour of the increments $f(t_i)-f(t_{i-1})$ and $W(t_i)-W(t_{i-1})$. The increments of the first are $O(t_i-t_{i-1})$ those of the second are $O(\sqrt{t_i-t_{i-1}})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint 2: $$R_n - L_n =\sum_{i=1}^n \left[ f(t_i)-f(t_{i-1}) \right]^2$$ and likewise for $W(t)$.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-28T17:21:34.067" LastActivityDate="2010-12-28T17:21:34.067" />
  <row Id="15752" PostTypeId="2" ParentId="15750" CreationDate="2010-12-28T17:11:51.023" Score="1" ViewCount="0" Body="&lt;p&gt;I think Hardy's monograph &lt;em&gt;The Integration of Functions of a Single Variable&lt;/em&gt; discusses this but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-28T17:11:51.023" CommentCount="1" />
  <row Id="15753" PostTypeId="2" ParentId="15721" CreationDate="2010-12-28T17:40:23.780" Score="5" ViewCount="0" Body="&lt;p&gt;Tonelli's algorithm for square-roots can easily be generalized to compute d'th roots in finite fields, e.g. see  Adleman; Manders; Miller: &lt;a href=&quot;http://www.cs.cmu.edu/~glmiller/Publications/AMM77.pdf&quot; rel=&quot;nofollow&quot;&gt;On taking roots in finite fields&lt;/a&gt;, and Bach; Shallit: &lt;a href=&quot;http://books.google.com/books?id=iJx1lP9ZcIkC&amp;amp;pg=PA160&quot; rel=&quot;nofollow&quot;&gt;Algorithmic number theory, section 7.3.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-28T17:40:23.780" />
  <row Id="15754" PostTypeId="2" ParentId="15749" CreationDate="2010-12-28T17:48:31.543" Score="1" ViewCount="0" Body="&lt;p&gt;First write &#xA;$$&#xD;&#xA;R_n  - L_n  = \sum\limits_{i = 1}^n {[W(t_i ) - W(t_{i - 1} )]^2 },&#xD;&#xA;$$&#xA;then consider &lt;a href=&quot;http://planetmath.org/encyclopedia/QuadraticVariationOfBrownianMotion.html&quot; rel=&quot;nofollow&quot;&gt;Quadratic variation of Brownian motion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-28T21:19:32.047" LastActivityDate="2010-12-28T21:19:32.047" CommentCount="1" />
  <row Id="15755" PostTypeId="2" ParentId="14506" CreationDate="2010-12-28T17:55:43.360" Score="2" ViewCount="0" Body="&lt;p&gt;I think there may be a simpler algorithm, which uses straightforward dynamic programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you correctly, you're interested in a BDD with $2n+1$ levels, the first and last containing the root and sink, respectively (we just delete all vertices from which $T$ is unreachable). Edges only go between adjacent levels, with the correct annotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's concentrate on the middle three levels $A,B,C$. Thus all $A \rightarrow B$ and $B \rightarrow C$ are labelled by $\pm x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is a node $c \in C$ reachable from a node $a \in A$? There must be some path going through $B$ with edges labeled by the same polarity of $x_n$. We can delete level $B$ and replace it by arrows going directly from $A$ to $C$; we label each arrow by the allowable assignments for $x_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is some path from the root to a vertex $c \in C$. What is its provenance? It must have passed through some vertex $a \in A$ via some edge $e$ going into $a$. We can shortcut the edge $e$ so that it goes directly to $c$, carrying both its original label and the new descriptive label concerning $x_n$. In this way, we can erase level $A$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having eliminated both levels $A,B$, we have removed $x_n$ from the BDD, and we can iterate our steps; having done this $n$ times, we will end up with a parametrization of &lt;i&gt;all&lt;/i&gt; solutions to the BDD.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Here is a run of the algorithm on your example. The graph is composed of five levels: $$L_0 = \{x_0\},\; L_1 = \{y_0,y_1\},\; L_2 = \{y_2\},\; L_3 = \{x_2,x_3\},\; L_4 = \{T\}.$$ There are 10 edges, which we list as tuples of from/to/label: $$(x_0,y_0,+x),(x_0,y_1,-x),(y_0,y_2,+y),(y_0,y_2,-y),(y_1,y_2,+y),$$ $$(y_2,x_2,+y),(y_2,x_3,-y),(x_2,T,-x),(x_3,T,-x),(x_3,T,+x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First iteration: the three middle levels are $L_1,L_2,L_3$. We start by eliminating $L_2$. Edges involving $y_2$ are removed, and the following edges are added: $$(y_0,x_2,+y),(y_0,x_3,-y),(y_1,x_2,+y).$$ Next we eliminate $L_1$. Edges involving $y_0,y_1$ are removed, and the following edges are added: $$(x_0,x_2,+x+y),(x_0,x_3,+x-y),(x_0,x_2,-x+y).$$ We are left with levels $L_0,L_3,L_4$ and the following 6 edges: $$(x_0,x_2,+x+y),(x_0,x_3,+x-y),(x_0,x_2,-x+y),(x_2,T,-x),(x_3,T,-x),(x_3,T,+x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second iteration: the three middle levels are $L_0,L_3,L_4$. We eliminate $L_3$. Edges involving $x_2,x_3$ are removed, and the following edges are added: $$(x_0,T,-x+y),(x_0,T,+x-y).$$ This gives us the answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Naively following the algorithm might give too many edges. However, notice that upon eliminating a variable $x_i$, we need only retain one assignment of the variables $x_{i+1},\ldots,x_n$ for each polarity of $x_i$. This way we won't get &lt;i&gt;all&lt;/i&gt; solutions, but we can decide whether there &lt;i&gt;is&lt;/i&gt; a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is easy to change the algorithm so that it maintains a succinct representation of all solutions. I leave that as an exercise for the reader.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditorDisplayName="" LastEditDate="2010-12-29T22:04:43.123" LastActivityDate="2010-12-29T22:04:43.123" CommentCount="2" />
  <row Id="15756" PostTypeId="1" CreationDate="2010-12-28T18:26:03.410" Score="8" ViewCount="311" Body="&lt;p&gt;I am interested in finding out about the current status of the planned book: The Triangle Book by John H. Conway and Steve Sigur. I understand that Steve Sigur died some time back. I got no reply from Prof. Conway; is there someone who knows the fate of this project. Is it abandoned or is it still under preparation? Here is some information about the planned &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1568811659&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2011-01-10T02:43:10.150" Title="Status of The Triangle Book" Tags="&lt;geometry&gt;&lt;books&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="15757" PostTypeId="1" AcceptedAnswerId="15761" CreationDate="2010-12-28T19:04:40.723" Score="2" ViewCount="67" Body="&lt;p&gt;Let's consider the sequence space $E =\mathbb R^{\mathbb N}$. If I believe in Choice, I have an isomorphism $E \simeq \mathbb R^{(\mathfrak c)}$ for some cardinal $\mathfrak c$. I further have some inequalities about $\mathfrak c$: first, $\mathfrak c = \dim \mathbb R^{\mathbb N} \leq |\mathbb R^{\mathbb N}| \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\times \aleph_0} = 2^{\aleph_0}$. Besides, $\mathbb R^{\mathbb N}$ is the dual space of the infinite-dimensional space $\mathbb R^{(\mathbb N)}$ so we have a strict inequality between their dimensions. To sum up, I get $\aleph_0 &amp;lt; \mathfrak c \leq 2^{\aleph_0}$. So, if I believe in the continuum hypothesis, I get $\mathfrak c = 2^{\aleph_0}$ and an isomorphism $\mathbb R^{\mathbb N} = \mathbb R^{(\mathbb R)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: as we have for a general set $A$ the isomorphism $(\mathbb R^{(A)})&amp;#39; \simeq \mathbb R^A$, an equivalent formulation of the question is: what's the dimension of the dual space of a $\aleph_0$-dimensional space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite ignorant about logic, so this raises a number of questions (if that's too much questions for the general rules of the website, please forgive me!)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What about the next step, that is the (algebraic) dual space of $\mathbb R^\mathbb N$? Its dimension should be $&amp;gt;2^{\aleph_0}$ but can we be more precise?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What if I do not suppose the continuum hypothesis? With Choice, it seems we've defined an increasing function $S$ on cardinals by the equation $\mathbb R^{\mathfrak c} (=(\mathbb R^{(\mathfrak c)})&amp;#39;) = \mathbb R^{(S(\mathfrak c))}$. Do we need (CH) to prove $S(\aleph_0) = 2^{\aleph_0}$? What is known about this function, depending on the various axioms about infinite dimensional sets? (I take it does not depend of the underlying field, but is it true?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What if I do not believe in Choice? I know the dimension terminology becomes quite useless, but can we still formulate some theorems about (non)existence of injective/surjective linear maps?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot, and my apologies for the imprecision of the questions. I hope it remains acceptable for MSE.&lt;/p&gt;&#xA;" OwnerUserId="5178" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T19:14:03.760" LastActivityDate="2010-12-28T20:00:28.947" Title="Dimension of the  sequence space and its dual, depending on status of (AC) and (CH)." Tags="&lt;linear-algebra&gt;&lt;set-theory&gt;&lt;axiom-of-choice&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="15758" PostTypeId="2" ParentId="15721" CreationDate="2010-12-28T19:04:56.743" Score="8" ViewCount="0" Body="&lt;p&gt;The integers modulo $p$ form a field. Since $x^3 - 1 = (x-1)(x^2+x+1)$, the problem is equivalent to solving $x^2+x+1\equiv 0 \pmod{p}$. For $p\neq 2$, the usual quadratic formula works; so you would need to find $y$ such that $y^2\equiv -3\pmod{p}$. If $p=2$, then $x^2+x+1=0$ has no solutions, and if $p=3$, then $x^3-1 = (x-1)^3$, so again there is no root other than $x=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the other primes, $p\gt 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using quadratic reciprocity, if $p\equiv 1\pmod{4}$, then $-1$ is a square modulo $p$ and we have:&#xA;$$\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right).$$&#xA;So if $p\equiv 1 \pmod{3}$ (hence $p\equiv 1 \pmod{12}$) then $-3$ is a square modulo $p$; if $p\equiv 2\pmod{3}$ (so $p\equiv 5\pmod{12}$), then $-3$ is not a square modulo $p$, so there is no other solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p\equiv 3\pmod{4}$, then $-1$ is not a square modulo $p$, and we have:&#xA;$$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right)$$&#xA;(since $\left(\frac{3}{p}\right) = -\left(\frac{p}{3}\right)$); again, if $p\equiv 1\pmod{3}$ (so $p\equiv 7\pmod{12}$) then $-3$ is a square modulo $p$; and if $p\equiv 2\pmod{3}$ (so $p\equiv 11\pmod{12}$) then $-3$ is not a square modulo $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, there are roots of $x^3-1$ modulo $p$ other than $x\equiv 1\pmod{p}$ if and only if $p\equiv 1\pmod{6}$. If $p=6k+1$ is such a prime, then the two roots of $x^3-1$ other than $x=1$ are precisely the roots of $x^2+x+1$, which are&#xA;$$\frac{-1\pm\sqrt{-3}}{2} = -3k(-1\pm a),$$&#xA;where $a$ is an integer such that $a^2\equiv -3\pmod{p}$. This means you still need to figure out the square root, which can be done using any of the methods suggested by Bill Dubuque, such as Tonelli's algorithm (say, in Shanks' version as described &lt;a href=&quot;http://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm#The_algorithm&quot; rel=&quot;nofollow&quot;&gt;in Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; As Alex Bartel notes in comments and in his own answer, one can avoid the use of quadratic reciprocity above. Since the units of $\mathbb{Z}/p\mathbb{Z}$ form a cyclic group of order $p-1$, there are nontrivial solutions to $x^3=1$ if and only if $3$ divides the order, that is, if and only if $p\equiv 1\pmod{3}$. From there, it is an easy jump to solutions if and only if $p\equiv 1 \pmod{6}$, since otherwise we would have $p\equiv 4\pmod{6}$ which is impossible since $p$ is prime. Once one establishes that, the quadratic formula can be applied safe in the knowledge that these primes must have $-3$ as a quadratic residue.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T04:30:21.000" LastActivityDate="2010-12-29T04:30:21.000" CommentCount="2" />
  <row Id="15759" PostTypeId="2" ParentId="12996" CreationDate="2010-12-28T19:35:11.343" Score="2" ViewCount="0" Body="&lt;p&gt;The answer is NO.&#xA;Take $g=sl(3)$ and $\Delta=\{a_1,a_2\}$. The unique choice for $\Delta&amp;#39;$ is $\{a_2\}$.&#xA;The result is clearly false in this scenery.&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-12-28T19:35:11.343" />
  <row Id="15760" PostTypeId="2" ParentId="15602" CreationDate="2010-12-28T19:39:26.457" Score="1" ViewCount="0" Body="&lt;p&gt;Let $M:=\{(x,y)|x\in\mathbb R, -1&amp;lt;y&amp;lt;1\}$ be an infinite strip and choose an  $L&amp;gt;0$. The equivalence relation $(x+L,-y)\sim(x,y)$ defines a Möbius strip $\hat M$. Let $\pi: M \to \hat M$ be the projection map. The Möbius strip  $\hat M$ inherits the differentiable structure from ${\mathbb R}^2$. We have to prove that $\hat M$ does not admit an atlas of the described kind which is compatible with the differentiable structure on $\hat M$. Assume that there is such an atlas $(U_\alpha,\phi_\alpha)_{\alpha\in I}$. We then define a function $\sigma:{\mathbb R}\to\{-1,1\}$ as follows: For given $x\in{\mathbb R}$ the point $\pi(x,0)$ is in $\hat M$, so there is an $\alpha\in I$ with $\pi(x,0)\in U_\alpha$. The map $f:=\phi_\alpha$ o $\pi$ is a diffeomorphism in a neighbourhood $V$ of $(x,0)$. Put $\sigma(x):={\rm sgn}\thinspace J_f(x,0)$, where $J_f$ denotes the Jacobian of $f$. One easily checks that $\sigma(\cdot)$ is well defined and is locally constant, whence it is constant on ${\mathbb R}$. On the other hand we have $f(x+L,y)\equiv f(x,-y)$ in $V$ which implies $\sigma(L)=-\sigma(0)$ $-$ a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2010-12-28T19:39:26.457" />
  <row Id="15761" PostTypeId="2" ParentId="15757" CreationDate="2010-12-28T20:00:28.947" Score="3" ViewCount="0" Body="&lt;p&gt;(Assuming the Axiom of Choice throughout) Suppose $\mathbf{V}$ is $\aleph_0$-dimensional over the field $F$. Then the cardinality of $\mathbf{V}^*$ is $|F|^{\aleph_0}$. If $|F|=\aleph_{\alpha}$, then as noted in &lt;a href=&quot;http://math.stackexchange.com/questions/14429/whats-the-cardinality-of-all-sequences-with-coefficients-in-an-infinite-set/14434#14434&quot;&gt;this answer&lt;/a&gt;, the cardinality is either $2^{\aleph_0}$, $\aleph_{\alpha}$, or $\aleph_{\gamma}^{\aleph_0}$, where $\aleph_{\gamma}\leq \aleph_{\alpha}$ has cofinality $\aleph_0$ and $\aleph_0\lt\aleph_{\alpha}=|F|$. You need to assume the Generalized Continuum Hypothesis to get more about the cardinality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, suppose the dimension of $\mathbf{V}^*$ is $d$. Then $\mathbf{V}^*$ is bijectable with the set of all almost-null functions $d\to F$; since $d$ is infinite, this is $d|F|=\max\{d,|F|\}=\max\{d,\aleph_{\alpha}\}$. So we must have $\max\{d,\aleph_{\alpha}\}=\aleph_{\alpha}^{\aleph_0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose $\mathbf{V}$ is an infinite dimensional vector space over $F$, and that $\kappa=\dim(\mathbf{V})\gt|F|=\lambda$. Then the dual has cardinality $\lambda^{\kappa}=2^{\kappa}\gt\kappa\gt|F|$, so the dimension will necessarily be $2^{\kappa}$. Thus your sequence of dimensions becomes $\kappa$, $2^{\kappa}$, $2^{2^{\kappa}}$, etc; that is, the $\beth$ function. I think that if $\dim(\mathbf{V})=|F|$, then you can't say precisely what $\dim(\mathbf{V}^*)$ is, other than that it is strictly larger than $|F|$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if at some point you get to a dimension which is strictly larger than the cardinality of the field, you will get a simple sequence, even without assuming the continuum hypothesis (or GCH). Unless the cardinality of the field is strongly inaccessible (or there is a strongly inaccessible cardinal between the dimension of $\mathbf{V}$ and the cardinality of $F$) you will eventually get there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know what happens without the assumption of the Axiom of Choice; if you assume that the field has a defined cardinality, and the vector space has given basis with a defined cardinality, then you can figure out the cardinality of the dual, but I don't know if you can find a basis for it in that case. The statement &quot;every vector space has a basis&quot; is equivalent to the Axiom of Choice.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-28T20:00:28.947" />
  <row Id="15762" PostTypeId="2" ParentId="15749" CreationDate="2010-12-28T20:11:12.890" Score="1" ViewCount="0" Body="&lt;p&gt;Limits of $R_n$ and $L_n$ coincide when Stieltjes integral exists. Existence of the Stieltjes integral does not follow from the existence of these limits. In general existence and definition of Stieltjes integral can be messy business as the Figure 2.1 in page 6 (page 10 of the ps file) of &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=4&amp;amp;ved=0CDsQFjAD&amp;amp;url=http%3A%2F%2Fwww.maphysto.dk%2Foldpages%2Fevents%2FLevyProc%2FCAC%2Fnotes%2Fchap1%2B2.ps.gz&amp;amp;rct=j&amp;amp;q=Young%20integral%20Dudley%20p-variation&amp;amp;ei=UkAaTf_HDYyw4Qa1j7mGAg&amp;amp;usg=AFQjCNF-ZJIQMvF_31i5A58zmUmsppJcyQ&quot; rel=&quot;nofollow&quot;&gt;this document can attest&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4742" LastActivityDate="2010-12-28T20:11:12.890" CommentCount="2" />
  <row Id="15763" PostTypeId="2" ParentId="15750" CreationDate="2010-12-28T20:20:58.987" Score="11" ViewCount="0" Body="&lt;p&gt;A consideration of &lt;a href=&quot;http://math.stackexchange.com/questions/15719/how-do-i-integrate-the-following-int-frac1x2-mathrm-dx1-x2-sqrt/15723#15723&quot;&gt;Moron's answer&lt;/a&gt; to the linked question&#xA;shows that there is a map from the elliptic curve &#xA;$y^2 = P(x)$ to the conic $v^2 = u^2 + 2$&#xA;given by&#xA;$$(x,y) \mapsto (x - \dfrac{1}{x}, \dfrac{y}{x}),$$&#xA;and the differential &#xA;$$\dfrac{1+x^2}{(1-x^2)\sqrt{1 + x^4}} \, dx$$ &#xA;on the elliptic curve&#xA;is the pull-back of the differential &#xA;$$\dfrac{1}{u v}\,du$$&#xA;on the conic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a conic has genus zero (i.e. it can be parameterized by a single variable,&#xA;using a classical ``$t$-substitution''), the integral of a differential on a conic can always be expressed via elementary functions.  Thus the same is true&#xA;of the integral of the original differential on the elliptic curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the general question is the same: if the differential in question can be pulled back from a map to a rational curve (i.e. a genus zero curve),&#xA;then the &quot;elliptic integral&quot; in question can be in fact integrated via elementary functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, any elliptic curve $y^2 = P(x)$ has a map to the the $x$-line given&#xA;by $(x,y) \mapsto x$.  So if the integral only involves rational functions of $x$ (which will be the case when $y$ appears to even powers, since we can always&#xA;substitute $P(x)$ for $y^2$) then it can be computed in elementary terms.  Also,&#xA;if $P(x)$ has repeated roots, then the curve $y^2 = P(x)$ itself is actually rational (it can be parameterized by a variation of the classical $t$-substiution for conics), and so any &quot;elliptic integral&quot; is actually elementarily integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  I have used some geomertic terminology here (&lt;em&gt;pull-back&lt;/em&gt;, &lt;em&gt;differential&lt;/em&gt;, &lt;em&gt;elliptic curve&lt;/em&gt;, &lt;em&gt;rational curve&lt;/em&gt;) because the modern point of view on this material is via algebraic geometry.  If some of this is unfamiliar, leave a comment and I (or someone else) can elaborate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: If we have a curve $C$ (which could be our elliptic curve $y^2 = P(x)$,&#xA;or our rational curve $v^2 = u^2 + 2$, or just the $x$-line, or ...) and if $\omega$ is a differential on $C$, then finding the indefinite integral of $\omega$ means finding some function $f$ on $C$ such that $\omega = df$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $\varphi: C&amp;#39; \to C$ is a map of curves, then $\varphi^* \omega&#xD;&#xA;= \varphi^* d f  = d (f\circ \varphi).$  So $f\circ \varphi$ is an indefinite&#xA;integral of the pulled back differential $\varphi^*\omega$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if $f$ is an elementary function of the coordinates on $C$,&#xA;and $\varphi$ is given by expressions which are elementary functions of the&#xA;coordinates, than the composite $f\circ \varphi$ will again be given by&#xA;elementary functions of the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what is happening in your example.&#xA;Explicitly, on our curve $v^2 = u^2 + u,$ we had for example the differential&#xA;$$\dfrac{1}{u v} \, du = \frac{1}{2 u^2 v}\, d (u^2 + 2) = \frac{1}{2(v^2-2)v}\, d(v^2) = \dfrac{1}{v^2-2}\, dv = \dfrac{1}{2\sqrt{2}}d\log\bigl( (v-\sqrt{2})/(v+\sqrt{2})\bigr).$$&#xA;Now pulling back this differential via our map $\varphi:(x,y)\mapsto (x-\dfrac{1}{x}, \dfrac{y}{x})$ we obtain&#xA;$$\dfrac{1 + x^2}{(1-x^2)y}\, dx = \dfrac{1}{2\sqrt{2}}d\log\bigl((y -&#xD;&#xA;\sqrt{2}x)/(y+ \sqrt{2}x) \bigr).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this example shows, &lt;em&gt;pulling back&lt;/em&gt; is just the theoretical terminology&#xA;for  &lt;em&gt;making a substitution&lt;/em&gt;, just as &lt;em&gt;a map of curves&lt;/em&gt; is just theoretical terminology for &lt;em&gt;a change of variables&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If memory serves, Miles Reid's wonderful book &lt;em&gt;Undergraduate algebraic geometry&lt;/em&gt; discusses some of this, and in particular gives some of the history of how the analytic theory of elliptic integrals turned into the &#xA;algebro-geometric theory of elliptic curves.  (If you don't know this book, don't be fooled by the title --- it is a great introduction to the subject for someone at any level, not just undergraduates!)  A much more detailed history can be found in Dieudonne's book on the history of algebraic geometry, but that book is probably not very readable unless you already have some feeling for algebraic geometry as a subject. &lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-28T23:16:20.330" LastActivityDate="2010-12-28T23:16:20.330" CommentCount="3" />
  <row Id="15764" PostTypeId="2" ParentId="15602" CreationDate="2010-12-28T20:26:33.150" Score="7" ViewCount="0" Body="&lt;p&gt;A simple answer that IMO is easy to justify using your definition of orientation goes like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;em&gt;any&lt;/em&gt; manifold $M$ and a point $p \in M$ there is a homomorphism $O : \pi_1(M,p) \to \mathbb Z_2$ and the idea is this: if $\phi : [0,1] \to M$ is a path such that $\phi(0)=\phi(1)=p$, given any basis for the tangent space to $M$ at $p$, $T_pM$ you can parallel transport that basis along the path, and you'll get a second basis for the tangent space at $\phi(1)=p$, $T_pM$.  And you can ask, is the change-of-basis map from your 1st to your 2nd basis for $T_pM$ orientation-preserving -- i.e. is the determinant of that linear transformation positive?  If it is, define $O(\phi)=0$, if the determinant is negative, define $O(\phi)=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact: the path-component of $p$ in the manifold $M$ is orientable if and only if $O$ is the zero function, $O=0$. You prove it by cutting your path $\phi$ into small segments and comparing orientations within charts -- the key analytical step is the intermediate value theorem, using that determinant is a continuous function of matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in this discussion &quot;parallel transport&quot; assumes a Riemann metric but you don't really need a Riemann metric for this argument to work.  The parallel transport of vectors along a path $\phi$ simply means continuously-varying vectors such that the vector corresponding to $t \in [0,1]$ is always tangent to the manifold, i.e. elements of $T_{\phi(t)} M$.  And of course if you're transporting $n$ vectors you demand that these $n$ vectors always make basis for $T_{\phi(t)}M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the case of the Moebius band, given any concrete model of the Moebius band you transport a basis along any path that goes once around the band and $O(\phi)=1$. &lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-12-28T23:09:31.700" LastActivityDate="2010-12-28T23:09:31.700" />
  <row Id="15765" PostTypeId="2" ParentId="15702" CreationDate="2010-12-28T20:36:43.267" Score="6" ViewCount="0" Body="&lt;p&gt;One way to think of an ideal as being the set of multiples of the (possibly non-existent) g.c.d of all the elements that contain it.  (I am thinking here of the case of a commutative ring with $1$, so that distinctions between left, right, and two-sided don't matter.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the integers, for example, any set of elements has a g.c.d. with all the  reasonable properties that you could want, and furthermore if $\{a_i\}_{i \in I}$ is a set of integers,&#xA;then the ideal generated by the $a_i$ is in fact principal, and its generator is a&#xA;g.c.d. of this collection.  In a more general ring, g.c.d.s don't necessarily exist, or even if they do, they don't have all the properties that they do in the integers.  So rather than trying to work with g.c.d.s, we can introduce ideals, which generally have better properties (or, rather, do many of the same jobs in more general rings that g.c.d.s and their set of multiples do in the context of the integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, one sees (a) why ideals are the natural kernels of quotient maps: think about the case of the integers, working {\em modulo} $n$ means setting all multiples of $n$ equal to zero; (b) why an ideal with a unit will be the trivial ideal (i.e. the whole ring): because the g.c.d. of any set containing a unit will have to be $1$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-28T20:36:43.267" />
  <row Id="15766" PostTypeId="1" AcceptedAnswerId="15768" CreationDate="2010-12-28T20:52:57.360" Score="4" ViewCount="89" Body="&lt;p&gt;From Rudin's &lt;em&gt;Principles of Mathematical Analysis&lt;/em&gt; (Chapter 2, Exercise 6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $E&amp;#39;$ be the set of all limit points of a set $E$. Prove that $E&amp;#39;$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I got it but my argument is a bit hand wavy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $x$ is a limit point of $E&amp;#39;$, then every neighborhood of $x$ contains some $y\in E&amp;#39;$, and every neighborhood of $y$ contains some $z\in E$. Therefore every neighborhood of $x$ contains some $z\in E$, and so $x$ is a limit point of $E$. Then $x\in E&amp;#39;$, so $E&amp;#39;$ is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that's bugging me is the leap from one neighborhood to another. Is this formally correct? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastActivityDate="2010-12-28T22:25:32.960" Title="Proving that the set of limit points of a set is closed." Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="15767" PostTypeId="2" ParentId="15725" CreationDate="2010-12-28T20:56:18.700" Score="0" ViewCount="0" Body="&lt;p&gt;From my understanding :&#xA;Since $$n^{log_3 2+0.4} = n^0.797$$ .(taking $\epsilon = 0.4$ in this case)&#xA;being smaller then $n \log(n)$  it qualifies for the case 3 . &#xA;Hence we check if ,&#xA;$$2\frac{n}{3} \log{n/3} \leq cn \log(n)$$&#xA;=$$\frac{2}{3} \log{n/3} \leq c  \log(n)$$&#xA;=$$ \frac{n^\frac{2c}{3}}{3} \leq \log(n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which i guess is true for positive values of c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am i doing anything wrong here ? or is everything okay :P&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="3545" LastEditorDisplayName="" LastEditDate="2010-12-28T21:07:41.663" LastActivityDate="2010-12-28T21:07:41.663" CommentCount="1" />
  <row Id="15768" PostTypeId="2" ParentId="15766" CreationDate="2010-12-28T21:01:37.447" Score="5" ViewCount="0" Body="&lt;p&gt;Your argument is correct, but incomplete: All you need to finish it is to ensure that you can find a neighborhood of $y$ contained in the neighborhood of $x$ that you began with (any will do, since all contain elements of $E$). Use the triangle inequality to find an appropriate radius for the neighborhood of $y$. &lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-28T21:01:37.447" CommentCount="1" />
  <row Id="15769" PostTypeId="1" AcceptedAnswerId="15770" CreationDate="2010-12-28T21:18:49.810" Score="3" ViewCount="162" Body="&lt;p&gt;How can I solve the recurrence &#xA;$$T (n) = T \left(\frac{n}{2} +\sqrt{n}\right) +\sqrt{6046}\ ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please don't just write the name of the method, as I just started learning this stuff and things are a little hard for me to figure out on my own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="3545" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-28T21:43:34.687" LastActivityDate="2010-12-29T09:03:47.497" Title="how to solve $T (n) = T \left(\frac{n}{2} +\sqrt{n}\right) +\sqrt{6046}$" Tags="&lt;asymptotics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="15770" PostTypeId="2" ParentId="15769" CreationDate="2010-12-28T21:35:43.473" Score="2" ViewCount="0" Body="&lt;p&gt;As you are interested in large $n, \sqrt{n}\lt \frac{n}{a}$ for any $a$.  So you can argue (assuming $T$ is increasing with $n$) that eventually $T(n) \leq T(\frac{2n}{3})+80$.  This fits the method you have been using.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-28T21:53:15.727" LastActivityDate="2010-12-28T21:53:15.727" CommentCount="8" />
  <row Id="15771" PostTypeId="2" ParentId="15766" CreationDate="2010-12-28T22:04:51.670" Score="0" ViewCount="0" Body="&lt;p&gt;An equivalent definition of a neighborhood of $x$ is that it is an open set containing $x$. If you adopt this definition, then your proof is perfect.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-28T22:04:51.670" CommentCount="2" />
  <row Id="15772" PostTypeId="2" ParentId="15769" CreationDate="2010-12-28T22:06:47.143" Score="1" ViewCount="0" Body="&lt;p&gt;I believe &lt;a href=&quot;http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method&quot; rel=&quot;nofollow&quot;&gt;Akra-Bazzi&lt;/a&gt; works for this and gives you $T(x) = \theta(\log x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recurrence you have satisfies the assumptions of Akra-Bazzi and we get $p=0$ (check out wiki for what $p$ is). $p=0$ implies $T(x) = \theta(\log x)$ in this case (as $g(x)$ is constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more elementary proof would involve showing $T(x) = \mathcal{O}(\log x)$ as in Ross's answer, then try to show $T(x) = \Omega(\log x)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is an explanation of applying Akra-Bazzi to this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Akra-Bazzi is used to solve recurrences of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T(x) = g(x) + \sum_{i=1}^{m} a_i T(b_i x + h_i(x)) \ \text{where}\  x &amp;gt; x_0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x$ is the variable, $a_i, b_i$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In you case, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(x) = \sqrt{6046} + T(0.5 x + \sqrt{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, this corresponds to $\displaystyle m=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumptions made by the theorem are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\displaystyle a_i \gt 0$. This is true in your case, as $\displaystyle a_1  = 1$. &lt;br/&gt;&#xA;2) $\displaystyle  0 \lt b_i \lt 1$. This is true in your case, as $\displaystyle b_1 = 0.5$. &lt;br/&gt;&#xA;3) $\displaystyle |g(x)| = \mathcal{O}(x^c)$ for some $c$. In our case $\displaystyle g(x) = \sqrt{6046}$ is constant. &lt;br/&gt;&#xA;4) $h_i(x) = \mathcal{O}(\frac{x}{\log^2 x})$. In our case $h_1(x) = \sqrt{x} = \mathcal{O}(\frac{x}{\log^2 x})$ as $\displaystyle \log^2 x \le K \sqrt{x}$ for sufficiently large $\displaystyle x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple more, but are trivial to verify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to apply Akra-Bazzi theorem, you need to find $\displaystyle p$ such that $\displaystyle \sum_{i=1}^{m} a_i (b_i)^p = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, we get $\displaystyle 0.5^p = 1$ and so $\displaystyle p = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we find $\displaystyle p$, by the Akra-Bazzi theorem we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(x) = \theta\left(x^p \left( 1 + \int_{1}^{x} \frac{g(t)}{t^{p+1}} \ \text{dt}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\displaystyle p = 0$ and $\displaystyle g(t)$ is constant, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(x) = \theta\left(1 + \int_{1}^{x} \frac{\sqrt{6046}}{t} \ \text{dt}\right) = \theta(\log x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also suggest you try to complete the elementary proof that $\displaystyle T(x) = \Omega(\log x)$. The relation $\displaystyle T(x) \leq T(2x/3) + 80$ only shows that $\displaystyle T(x) = \mathcal{O}(\log x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement $\displaystyle T(x) = \theta(\log x)$ is stronger. For instance, $\displaystyle T(x) = 10$ also satisifies $\displaystyle T(x) \leq T(2x/3) + 80$, but it won't satisfy your original recurrence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T07:21:28.000" LastActivityDate="2010-12-29T07:21:28.000" />
  <row Id="15773" PostTypeId="1" CreationDate="2010-12-28T22:16:41.610" Score="15" ViewCount="275" Body="&lt;p&gt;I have seen two proofs of the simplicity of $A_n,~ n \geq 5$ (Dummit &amp;amp; Foote, Hungerford). But, neither of them are such that they 'stick' to the head (at least my head). In a sense, I still do not have the feeling that I know why they are simple and why should it be 5 and not any other number (perhaps this is only because 3-cycles become conjugate in $A_n$ after $n$ becomes greater than 4).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the most illuminating proof of the simplicity of $A_n,~ n \geq 5$ that you know?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T01:12:54.370" LastActivityDate="2010-12-30T12:43:15.860" Title="Simplicity of $A_n$" Tags="&lt;group-theory&gt;&lt;big-list&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="15774" PostTypeId="2" ParentId="15713" CreationDate="2010-12-28T22:24:26.420" Score="0" ViewCount="0" Body="&lt;p&gt;To use Feller's original notation. Let $E_m(r,n)$ be the number of distributions leaving exactly $m$ cells empty, then we have $$p_m(r,n)=n^{-r}\binom{n}{k}A(r,n-m)$$&#xA;as the corresponding probability ($A$ is to be defined)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need compute the probability $x_m(r,n)$ of finding $m$ or more cells empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the probability that $m$ or more cells are empty is given by&#xA;$x_m = p_m + p_{m+1} + \cdots + p_{n}$. Now, observe the reason why the &lt;em&gt;hint&lt;/em&gt; was given: $x_{m+1} = x_m - p_m$ (i.e., $m+1$ or more cells being empty is the same as $\ge m$ cells being empty minus &lt;em&gt;exactly&lt;/em&gt; $m$ cells being empty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a sanity test, you could plug in the value of $x_m(r,n)$ given, and verify whether $x_{m}(r,n)-x_{m+1}(r,n)=p_m(r,n)$ holds (with the appropriate boundary cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper underlying reason for this hint is that Feller probably wants you to figure out an inclusion-exclusion type argument. If I get time, I will write out the details. But the above should be a sufficiently detailed hint?&lt;/p&gt;&#xA;" OwnerUserId="1709" LastActivityDate="2010-12-28T22:24:26.420" CommentCount="1" />
  <row Id="15776" PostTypeId="2" ParentId="15773" CreationDate="2010-12-28T22:37:53.007" Score="11" ViewCount="0" Body="&lt;p&gt;The one I like most goes roughly as follows (my reference is in French [Daniel Perrin, &lt;em&gt;Cours d'algèbre&lt;/em&gt;], but maybe it's the one in Jacobson's &lt;em&gt;Basic Algebra I&lt;/em&gt;) : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;you prove that A(5) is simple by considering the cardinal of the conjugacy classes and seeing that nothing can be a nontrivial normal subgroup (because no nontrivial union of conjugacy classes including {id} has a cardinal dividing 60). Actually, you don't have to know precisely the conjugacy classes in A(5).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, you consider a normal subgroup N in A(n), n &gt; 5 which is strictly larger than {id} and you prove (*) that it contains an element fixing at least n-5 points. The fact that A(5) is simple then gives that N contains every even permutation fixing the same n - 5 points. In particular, it contains a 3-cycle, and therefore contains all of A(n).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To prove (*), you consider a commutator [x,y], where x is nontrivial in your normal subgroup and y is a 3-cycle: by the very definition, it is the product of the 3-cycle and the conjugate of its inverse. So it's the product of two 3-cycles and has at the very least n-6 fixed points. But it's easy to see that you can chose the 3-cycle so that the commutator has n-5 fixed points (it is enough that the two 3-cycles have overlapping supports).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like this proof because it keeps the &quot;magical computation&quot; part to a minimum, that simply amounts to the fact that you have automatically knowledge about a commutator if you have knowledge about &lt;em&gt;one&lt;/em&gt; of his factors.&lt;/p&gt;&#xA;" OwnerUserId="5178" LastEditorUserId="5178" LastEditorDisplayName="" LastEditDate="2010-12-29T06:12:02.307" LastActivityDate="2010-12-29T06:12:02.307" />
  <row Id="15777" PostTypeId="1" AcceptedAnswerId="15785" CreationDate="2010-12-28T23:02:41.037" Score="6" ViewCount="66" Body="&lt;p&gt;I am aware that compactness fails on finite models, but the common counter-example uses models of arbitrary big finite size. So if we bound the size what results can we get? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we have an infinite set of sentences $\Sigma$, of a language of first order logic $\mathcal{L}$. If there is a natural number $n$ such that every finite $S\subset\Sigma$  has a model of size $\leq n$, then does $\Sigma$ have a finite model? How about a model of size $\leq n$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the language is finite or there is a  $k$ such that every function has at most $k$ arguments and every relation is at most $k$-place, we have finite many structures of size $\leq n$ of $\mathcal{L}$, let's call them $N_1,\ldots,N_m$. If for every $i\leq m$ there exists a finite $S_i\subset\Sigma$ such that $N_i$ doesn't satisfy $S_i$, then $\bigcup_{i\leq m}S_i$ wouldn't have a model of size $\leq n$, and thus we get that one of the models satisfy every sentence of $\Sigma$. Is this argument correct? And what about the general case, where we don't restrict the language whatsoever and the structures are infinite many?&lt;/p&gt;&#xA;" OwnerUserId="1772" LastActivityDate="2010-12-29T00:21:45.663" Title="Compactness on models with bounded finite size" Tags="&lt;logic&gt;&lt;model-theory&gt;" AnswerCount="2" />
  <row Id="15778" PostTypeId="1" AcceptedAnswerId="15781" CreationDate="2010-12-28T23:36:55.273" Score="9" ViewCount="119" Body="&lt;p&gt;This is probably just elementary, but I don't know how to do it. I would like to find the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x &amp;gt; 0$. Numerically, I have found that this minimum value lies between 0.365 and 0.37 (i.e., $x^{0.37} &amp;gt; \ln(x)$ for all $x &amp;gt; 0$, but $x^{0.365}$ is not). Is there any analytical way to find out exactly this minimum value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Based on the received answers, I finally came up with my own one as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the function $f(x) = x^a - \ln(x).$ This function is convex in $x$, and hence, achieves the unique minimum as $x^*$ such that $f&amp;#39;(x^*) = 0.$ Solving that equation yields $$f_{\mathrm{min}} = \min\limits_{x&amp;gt;0} f(x) = \frac{\ln(a)+1}{a}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by letting $f_{min} = 0$, we get the desired value $a^* = 1/e.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank everyone for the answers!&lt;/p&gt;&#xA;" OwnerUserId="1380" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T01:35:53.360" LastActivityDate="2010-12-29T01:35:53.360" Title="What is the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x &gt; 0$?" Tags="&lt;calculus&gt;&lt;real-analysis&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15779" PostTypeId="2" ParentId="15773" CreationDate="2010-12-28T23:43:23.773" Score="8" ViewCount="0" Body="&lt;p&gt;Here are the ingredients I think about:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$A_n$ is $n-2$-transitive on $\{1,2,\ldots,n\}$. That is, if $1\leq i_1\lt i_2\lt\cdots \lt i_{n-2}\leq n$ are any $n-2$ integers, and $j_1,\ldots,j_{n-2}$ are any $n-2$ distinct integers between $1$ and $n$, then there is an element of $A_n$ that maps $i_k$ to $j_k$ (in fact, one and only one element of $A_n$ that achieves this).  This is easy: just write out the corresponding permutation. If it is even, this gives you an element of $A_n$. If it is not even, then adding a transposition involving the elements that do not occur among the $j_k$ makes it even and does not change the image of the $i_k$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In particular, if $\sigma\in A_n$ fixes at least two elements, then every element with the same cycle structure to $\sigma$ is conjugate to $\sigma$ in $A_n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $N$ is a normal subgroup of $A_n$, and $\sigma\in N$ fixes at least two elements, then every element with the same cycle structure as $\sigma$ is in $N$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In particular, if $N$ is normal, contains a $3$-cycle, and $n\geq 5$, then $N$ contains all $3$-cycles; and since the $3$-cycles generate $A_n$ when $n\geq 3$, then $N=A_n$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $N$ is nontrivial and normal in $A_n$, and $n\geq 5$, then it contains a $3$-cycle: this involves a bit of manipulation, but the fact that $n\geq 5$ is as essential here as in (4) above: it gives you enough room to maneuver, room that you do not have in $A_4$ ($A_3$ and $A_2$ are also simple, but for silly reasons). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure if this particular line qualifies as &quot;illuminating&quot; in terms of giving you great insight into the structure of $A_n$; I feel they give me a good feel for what you can and cannot do in terms of playing with permutations (especially the transitivity, and the actual mechanics of point 5 above), and they remind me why $n\geq 5$ is important. &lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-28T23:43:23.773" />
  <row Id="15780" PostTypeId="2" ParentId="15778" CreationDate="2010-12-28T23:49:05.160" Score="5" ViewCount="0" Body="&lt;p&gt;Consider the minimum of the function $f(x)=x^{1/e}-\ln(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As a second step, verify that $x^a &amp;lt; \ln(x)$ at $x=(1/a)^{1/a}$, for any $0&amp;lt;a&amp;lt;1/e$.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-29T00:22:56.967" LastActivityDate="2010-12-29T00:22:56.967" CommentCount="3" />
  <row Id="15781" PostTypeId="2" ParentId="15778" CreationDate="2010-12-28T23:49:39.310" Score="7" ViewCount="0" Body="&lt;p&gt;$a = \max_{x&amp;gt;0} \frac{\ln(\ln(x))}{\ln(x)} = 1/e$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-28T23:49:39.310" CommentCount="1" />
  <row Id="15783" PostTypeId="1" AcceptedAnswerId="15790" CreationDate="2010-12-28T23:53:13.377" Score="2" ViewCount="91" Body="&lt;p&gt;I just asked this question on ask.metafilter, and it was suggested that I ask here.  Though I'm talking about coding something up, this question is about the math behind it, not the implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have done analysis in the past where we've computed an approximation for Gaussian curvature of a surface in Cartesian coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we've been doing for Cartesian (in MATLAB) is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[fu,fv] = gradient(Z)&#xA;&#xA;[fuu, fuv] = gradient(fu)&#xA;&#xA;[fvu,fvv] = gradient(fv)&#xA;&#xA;GC = (fuu*fvv - fuv*fuv)/(1 + fu^2 + fv^2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So now I have a surface that I'm modeling in cylindrical coordinates, and I can do the same thing as above for $r$ as a function of $\theta$ and $z$. The problem is that it's only taking into account the change in $r$, not the fact that there is curvature inherent in it being a cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking on &lt;a href=&quot;http://mathworld.wolfram.com/CylindricalCoordinates.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram&lt;/a&gt; (equations 27, 32 and 37 and thereabouts), it seems like there's a centripetal component that I don't know how to apply. Dividing by the (constant?) radius doesn't seem like it would work, so I think I'm missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated, either explaining how to modify these equations to work correctly, or some other approximation that has worked for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5185" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T01:41:39.213" LastActivityDate="2010-12-29T04:39:42.553" Title="How do I compute Gaussian curvature in cylindrical coordinates?" Tags="&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15784" PostTypeId="2" ParentId="15777" CreationDate="2010-12-29T00:10:30.420" Score="4" ViewCount="0" Body="&lt;p&gt;Your argument is correct. In fact, we can axiomatize any finite model, meaning that if ${\mathcal M}=(M,\dots)$ is a model (in a language of arbitrary size) then there is a set of sentences $\Sigma$ such that any model of $\Sigma$ is isomorphic to ${\mathcal M}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $M=\{a_1,\dots,a_n\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you say that the universe has size $|M|=n$, with a sentence $\tau$ such as &quot;there are $x_1,\dots,x_n$ which are pairwise different and such that each $y$ is equal to one of them.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each relational symbol $R$, consider the formula &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;there are $x_1,\dots,x_n$ pairwise different and such that $\bigwedge_{R^{\mathcal M}(a_{i_1},\dots,a_{i_k})}R(x_{i_1},\dots,x_{i_k})$ and such that $\bigwedge_{R^{\mathcal M}(a_{i_1},\dots,a_{i_k})\mbox{ fails}}\lnot R(x_{i_1},\dots,x_{i_k})$&quot;, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the first big conjunction ranges over all tuples of elements of $M$ that are in the interpretation of $R$, and the second runs over all tuples that are not in that interpretation. Given any model of this formula of size $n$, there is a bijection between this model and $M$ such that the interpretation of $R$ in this model is just the image under the bijection of the interpretation of $R$ in ${\mathcal M}$. Let $\phi_R$ be the formula such that the sentence we just wrote is $\exists \vec x\phi_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there is a sentence $\exists\vec x\psi_f$ describing completely each function $f^{\mathcal M}$, in the same sense as the sentence above completely describes $R^{\mathcal M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma$ consists of $\tau$, and the following formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any finitely many constant symbols $c_1,\dots,c_j$, any finitely many relational symbols $R_1,\dots,R_s$, and any finitely many function symbols $f_1,\dots,f_k$, say that $c_l^{\mathcal M}=a_{i_l}$ for each $l$, the sentence &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;$\exists \vec x(\bigwedge_{a=1}^s\phi_{R_a}\land\bigwedge_{b=1}^k\psi_{f_b}\land \bigwedge_{l=1}^j c_l=x_{i_l})$&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that if the language is finite, $\Sigma$ is finite as well. But the theory just described uniquely characterizes ${\mathcal M}$ even if the language is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is usually presented as a consequence of Beth’s deﬁnability theorem or of Svenonius’s theorem, see for example W. Hodges, Model theory. Cambridge University Press, (1993), Chapter 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I guess this says that there is no useful compactness result for finite models, since the only version that would hold would simply give us a model that we already knew we had.)&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-29T00:10:30.420" CommentCount="1" />
  <row Id="15785" PostTypeId="2" ParentId="15777" CreationDate="2010-12-29T00:15:34.650" Score="4" ViewCount="0" Body="&lt;p&gt;For the general case, without any restriction on the language, the Compactness theorem does hold for models having a specified finite size. To see this, argue as follows: suppose that every finite subtheory of $\Sigma$ has a model of finite size at most $n$. Let $\Sigma^+$ be the theory obtained by adding the axiom &quot;there are at most $n$ elements.&quot; Notice that every finite subtheory of $\Sigma^+$ is satisfiable. Thus, by the ordinary Compactness theorem, $\Sigma^+$ is satisfiable. Hence, $\Sigma$ has a model of size at most $n$, as desired.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditorDisplayName="" LastEditDate="2010-12-29T00:21:45.663" LastActivityDate="2010-12-29T00:21:45.663" />
  <row Id="15786" PostTypeId="2" ParentId="11095" CreationDate="2010-12-29T00:46:47.917" Score="0" ViewCount="0" Body="&lt;p&gt;How many different ways can &#xA;1000 be written as a sum of&#xA;two positive prime numbers ?&lt;/p&gt;&#xA;" OwnerUserId="5186" LastActivityDate="2010-12-29T00:46:47.917" CommentCount="1" />
  <row Id="15787" PostTypeId="1" AcceptedAnswerId="15789" CreationDate="2010-12-29T01:17:16.550" Score="8" ViewCount="239" Body="&lt;p&gt;I was wondering when in history did people start use the $\sqrt{}$ sign for square root, what did they use before, and why this curious nomenclature is adopted.&lt;/p&gt;&#xA;" OwnerUserId="3301" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T01:48:07.350" LastActivityDate="2010-12-29T01:48:07.350" Title="How did the square root get its shape?" Tags="&lt;notation&gt;&lt;math-history&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="15788" PostTypeId="2" ParentId="15787" CreationDate="2010-12-29T01:22:28.090" Score="2" ViewCount="0" Body="&lt;p&gt;If you're not looking for crazy amounts of information, there's always &lt;a href=&quot;http://en.wikipedia.org/wiki/Nth_root#History&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="2959" LastActivityDate="2010-12-29T01:22:28.090" />
  <row Id="15789" PostTypeId="2" ParentId="15787" CreationDate="2010-12-29T01:27:03.443" Score="9" ViewCount="0" Body="&lt;p&gt;You can find information about the history of the usage in Jeff Miller's page here: &lt;a href=&quot;http://jeff560.tripod.com/operation.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Uses of Symbols of Operation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoted:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Square root. The first use of a&#xA;  capital R with a diagonal line was in&#xA;  1220 by Leonardo of Pisa in Practica&#xA;  geometriae, where the symbol meant&#xA;  &quot;square root&quot; (Cajori vol. 1, page&#xA;  90).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The radical symbol first appeared in&#xA;  1525 in Die Coss by Christoff Rudolff&#xA;  (1499-1545). He used the symbol &lt;img src=&quot;http://i.imgur.com/xycFi.gif&quot; alt=&quot;alt text&quot;&gt;&#xA;  (without the vinculum) for square&#xA;  roots. He did not use indices to&#xA;  indicate higher roots, but instead&#xA;  modified the appearance of the radical&#xA;  symbol for higher roots.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is often suggested that the origin&#xA;  of the modern radical symbol is that&#xA;  it is an altered letter r, the first&#xA;  letter in the word radix. This is the&#xA;  opinion of Leonhard Euler in his&#xA;  Institutiones calculi differentialis&#xA;  (1775). However, Florian Cajori,&#xA;  author of A History of Mathematical&#xA;  Notations, argues against this theory.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In 1637 Rene Descartes used &lt;img src=&quot;http://i.imgur.com/JIlEB.gif&quot; alt=&quot;alt text&quot;&gt;, adding&#xA;  the vinculum to the radical symbol La&#xA;  Geometrie (Cajori vol. 1, page 375).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T01:32:04.923" LastActivityDate="2010-12-29T01:32:04.923" CommentCount="1" />
  <row Id="15790" PostTypeId="2" ParentId="15783" CreationDate="2010-12-29T01:38:51.117" Score="3" ViewCount="0" Body="&lt;p&gt;Your cylindrical coordinate surface $r(\theta,z)$ in Cartesian coordinates is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=r(\theta,z)\cos\;\theta\\y&amp;amp;=r(\theta,z)\sin\;\theta\\z&amp;amp;=z\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which now allows you to apply the usual Gaussian curvature formula. In particular, you should get the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=-\frac{r^3\frac{\partial^2 r}{\partial z^2}+r^2\left(\left(\frac{\partial^2 r}{\partial \theta\partial z}\right)^2-\frac{\partial^2 r}{\partial z^2}\frac{\partial^2 r}{\partial \theta^2}\right)+2r\frac{\partial r}{\partial \theta}\left(\frac{\partial^2 r}{\partial z^2}\frac{\partial r}{\partial \theta}-\frac{\partial r}{\partial z}\frac{\partial^2 r}{\partial \theta\partial z}\right)+\left(\frac{\partial r}{\partial \theta}\frac{\partial r}{\partial z}\right)^2}{\left(r^2+\left(r\frac{\partial r}{\partial z}\right)^2+\left(\frac{\partial r}{\partial \theta}\right)^2\right)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For completeness, if you have $z$ as a function of $r$ and $\theta$, your Cartesian parametrization is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=r\cos\;\theta\\y&amp;amp;=r\sin\;\theta\\z&amp;amp;=z(r,\theta)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the corresponding Gaussian curvature expression is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=\frac{r^2\frac{\partial^2 z}{\partial r^2}\left(\frac{\partial^2 z}{\partial \theta^2}+r\frac{\partial z}{\partial r}\right)-\left(\frac{\partial z}{\partial \theta}-r\frac{\partial^2 z}{\partial r\partial \theta}\right)^2}{\left(r^2\left(\left(\frac{\partial z}{\partial r}\right)^2+1\right)+\left(\frac{\partial z}{\partial \theta}\right)^2\right)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave the derivation of the Gaussian curvature expression for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}r&amp;amp;=f(u,v)\\\theta&amp;amp;=g(u,v)\\z&amp;amp;=h(u,v)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the interested reader.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-29T04:39:42.553" LastActivityDate="2010-12-29T04:39:42.553" CommentCount="0" />
  <row Id="15791" PostTypeId="3" CreationDate="2010-12-29T01:56:11.260" Score="0" ViewCount="0" Body="&lt;p&gt;Before asking a question on the site, please check if you can find your answer here: &lt;a href=&quot;http://jeff560.tripod.com/mathsym.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Uses of Various Mathematical Symbols&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T01:56:11.260" LastActivityDate="2010-12-29T01:56:11.260" />
  <row Id="15792" PostTypeId="3" CreationDate="2010-12-29T01:56:11.260" Score="0" ViewCount="0" Body="Meaning and history of mathematical symbols and notation." OwnerUserId="-1" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T01:56:11.260" LastActivityDate="2010-12-29T01:56:11.260" />
  <row Id="15793" PostTypeId="1" AcceptedAnswerId="15799" CreationDate="2010-12-29T04:03:51.913" Score="5" ViewCount="67" Body="&lt;p&gt;Let $\tau: E \to E$ be a measure-preserving transformation of the measure space $(E, \mathcal{E}, \mu)$, i.e. $\mu(\tau^{-1}(A)) = \mu(A)$ for all $A \in \mathcal{E}$. Let $\mathcal{E}_\tau = \{ A \in \mathcal{E} : \tau^{-1}(A) = A \}$. In my lecture notes, it is claimed that a measurable function $f$ is invariant (i.e. $f \circ \tau = f$) if and only if it is measurable with respect to $\mathcal{E}_\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is evidently false: Let $E = \{ 0, 1 \}$, and let $\mathcal{E} = \mathcal{P}(E)$ be the power set. Let $\mu$ be the uniform distribution. Let $\tau$ act on $E$ by transposing 0 and 1. Let $f: (E, \mathcal{E}) \to (E, \mathcal{E}_\tau)$ be the identity map on $E$. Clearly, $f$ is measurable and &lt;em&gt;not&lt;/em&gt; invariant. Yet it's also measurable with respect to $\mathcal{E}_\tau$: the preimage of every measurable set in its codomain is in $\mathcal{E}_\tau$, by construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I misinterpreted the claim? If not, is there a similar claim which &lt;em&gt;is&lt;/em&gt; true, e.g. by fixing the codomain of $f$?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2010-12-29T06:00:44.263" Title="Measure-preserving transformations and invariant functions" Tags="&lt;probability-theory&gt;&lt;measure-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15796" PostTypeId="2" ParentId="15571" CreationDate="2010-12-29T05:32:24.127" Score="1" ViewCount="0" Body="&lt;p&gt;As an indirect way of answering this question, I'd like to bring up an example from wave mechanics that I was told earlier this semester.  Physically, when an electromagnetic wave hits a perfect conductor, say from the left, the wave disappears to the right of the conductor.  What happens mathematically is the following: the charges emit a wave that is the exact mirror of the original one, cancelling it for all time and space to the right of the conductor.  Now, a natural question in this scenario is, are the two waves to the right of the conductor real, actually extending infinitely, or is there really no wave there any more?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is: the question does not make physical sense.  Physics is concerned with measurable quantities; you can ask &quot;what is the electric field to the right&quot;, and both interpretations of what's 'really' happening will give you the same (correct) answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, mathematics is built on abstractions.  The question of these abstractions being 'real' or not is not mathematical (nor logical), but philosophical; mathematics is the foundation of our technological society, which works quite well, which is the proof that mathematics is valid in the real world.&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2010-12-29T05:32:24.127" />
  <row Id="15797" PostTypeId="1" AcceptedAnswerId="15800" CreationDate="2010-12-29T05:43:47.080" Score="2" ViewCount="110" Body="&lt;p&gt;Does anyone know how to factorize the following expression: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4x^4+12x^{10/3} y^{2/3}+33x^{8/3} y^{4/3}+46x^2 y^2+33x^{4/3} y^{8/3}+12x^{2/3} y^{10/3}+4 y^4$$ &#xA;?&lt;/p&gt;&#xA;" OwnerUserId="5193" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T06:11:09.023" LastActivityDate="2010-12-29T06:15:15.727" Title="How to factorize  $4x^4+12x^{10/3} y^{2/3}+ \dots $?" Tags="&lt;algebra-precalculus&gt;&lt;polynomials&gt;&lt;symbolic-computation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15798" PostTypeId="2" ParentId="15656" CreationDate="2010-12-29T05:53:49.307" Score="1" ViewCount="0" Body="&lt;p&gt;If you compare the volume of the sphere to that of its enclosing hyper-cube you will find that this ratio continually diminishes.   The enclosing hyper-cube is 2 units in length per side if $R=1$.  Then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_1/2=1\qquad$$&#xA;$$V_2/4\approx 0.785$$&#xA;$$V_3/8\approx 0.5225$$&#xA;$$V_4/16\approx 0.308$$&#xA;$$V_5/32\approx 0.164$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this behavior is how we build hyper-spheres from low dimension to high dimensions.  Think for example extending $S_1$ to $S_2$.  We begin with a segment extending from $-1$ to $+1$ on the $x$ axis.  We build a 2 sphere by sweeping this sphere out along the $y$ axis using the scaling factor $\sqrt{1-y^2}$.  Compare this to the process of sweeping out the respective cube where the scale factor is $1$.  So not we only occupy approximately $3/4$ of the enclosing cube (i.e. square for $n=2$).  Likewise for $n=3$, we sweep the circle along the $z$ axis using the scaling factor, loosing event more volume compared to the cylinder if we had not scaled the circle as it was swept.  So as we extend $S_{n-1}$ to get $S_n$ we start with the diminished volume we have and loose even more as we sweep out into the $n^{th}$ dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be easier to explain with figures, however hopefully you can work through how this works for lower dimensions and extend to higher ones.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastEditorUserId="4726" LastEditorDisplayName="" LastEditDate="2010-12-29T15:34:42.147" LastActivityDate="2010-12-29T15:34:42.147" />
  <row Id="15799" PostTypeId="2" ParentId="15793" CreationDate="2010-12-29T06:00:44.263" Score="2" ViewCount="0" Body="&lt;p&gt;I suspect that &lt;em&gt;measurable function&lt;/em&gt; in the statement should be understood to be&#xA;&lt;em&gt;real-valued measurable function&lt;/em&gt;, or something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if each point $y$ in the codomain $Y$ of $f$ is measurable, and if $f$ is measuarble w.r.t. $\mathcal E_{\tau}$, then&#xA;we see that $f^{-1}(y)$ should be invariant under $\tau$, which is to say that&#xA;$f\circ\tau = f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your counter-example is that the points of $E$ are &lt;em&gt;not&lt;/em&gt; in&#xA;your particular $\mathcal E_{\tau}$, and so the preceding argument breaks down&#xA;(as you implicitly noted!).&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-29T06:00:44.263" CommentCount="1" />
  <row Id="15800" PostTypeId="2" ParentId="15797" CreationDate="2010-12-29T06:02:43.200" Score="7" ViewCount="0" Body="&lt;p&gt;It is possible, but likely messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First set $\displaystyle z = \left(\frac{x}{y}\right)^{2/3}$ and divide your expression by $\displaystyle y^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4z^6 + 12 z^5 + 33 z^4 + 46 z^3 + 33 z^2 + 12 z + 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now divide this by $\displaystyle z^3$ and set $\displaystyle t = z + 1/z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4(z^3 + 1/z^3) + 12(z^2 + 1/z^2) + 33(z+1/z) + 46$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 4(t^3 - 3t) + 12(t^2 - 2) + 33t + 46$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = 4t^3 + 12t^2 + 21t + 22$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a messy cubic, according to Wolfram Alpha. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note it is always possible to factorize a cubic in &quot;closed form&quot;, because of Cardano's method of finding the roots).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T06:15:15.727" LastActivityDate="2010-12-29T06:15:15.727" />
  <row Id="15801" PostTypeId="1" CreationDate="2010-12-29T06:07:48.687" Score="11" ViewCount="192" Body="&lt;p&gt;The (spherical) loxodrome, or the rhumb line, is the curve of constant bearing on the sphere; that is, it is the spherical curve that cuts the meridians of the sphere at a constant angle. A more picturesque way of putting it is that if one wants to travel from one point of a (spherical) globe to the antipodal point (say, from the North Pole to the South Pole) in a fixed direction, the path one would be taking on the globe would be a loxodrome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a unit sphere, the loxodrome that cuts the meridians at an angle $\varphi\in\left(0,90^\circ\right]$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}x&amp;amp;=\mathrm{sech}(t\cot\;\varphi)\cos\;t\\y&amp;amp;=\mathrm{sech}(t\cot\;\varphi)\sin\;t\\z&amp;amp;=\tanh(t\cot\;\varphi)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While playing around with loxodromes, I noted that for certain values of $\varphi$, one can orient two identical loxodromes such that they do not intersect (that is, one can position two ships such that if both take similar loxodromic paths, they can never collide). Here for instance are two loxodromes whose constant angle $\varphi$ is 60°, oriented such that they do not cross each other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/nB2WN.png&quot; alt=&quot;noncrossing 60° loxodromes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for the (extreme!) case of $\varphi=90^{\circ}$, the two loxodromes degenerate to great circles, and it is well known that two great circles must always intersect (at two antipodal points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less extreme, but seemingly difficult, would be the problem of positioning two 80° loxodromes such that they do not intersect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/huDYj.png&quot; alt=&quot;two 80° loxodromes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings me to my first question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) For what values of $\varphi$ does it become impossible to orient two&#xA;  loxodromes such that they do not cross each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For simplicity, one can of course fix one of the two loxodromes to go from the North Pole to the South Pole, and try to orient the other loxodrome so that it does not cross the fixed loxodrome.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That's the simpler version of my actual problem. Some experimentation seems to indicate that it is not possible to orient &lt;strong&gt;three&lt;/strong&gt; loxodromes such that they do not cross each other. So...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Is it true that for all (admissible) values of $\varphi$, one&#xA;  cannot position three loxodromes such that none of them cross each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've tried a bit of searching around to see if the problem has been previously considered, but I have not had any luck. Any pointers to the literature will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="498" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T16:38:30.860" LastActivityDate="2011-01-01T23:46:47.327" Title="On nonintersecting loxodromes" Tags="&lt;geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="15802" PostTypeId="2" ParentId="15686" CreationDate="2010-12-29T06:34:00.567" Score="0" ViewCount="0" Body="&lt;p&gt;Rewrite the equation as $$x = \omega^a \sqrt[3]{10 + \sqrt{108}} - \frac{6}{3 \omega^a \sqrt[3]{10 + \sqrt{108}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All three values of a = 0,1,2 produce a solution to the equation with no ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-30T00:11:40.473" LastActivityDate="2010-12-30T00:11:40.473" CommentCount="2" />
  <row Id="15803" PostTypeId="2" ParentId="15801" CreationDate="2010-12-29T07:43:22.107" Score="3" ViewCount="0" Body="&lt;p&gt;Not an answer, but an approach which might help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.ubc.ca/~israel/m103/mercator/mercator.html&quot; rel=&quot;nofollow&quot;&gt;Mercator projection&lt;/a&gt; is a map such that Loxodromes (corresponding to $0 \lt \varphi \lt 90^{\circ}$) get mapped to straight lines (and vice versa) and angles are preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An image of the map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/2tGEe.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you would need to pick parallel lines from the mercator map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem would be in choosing the correct antipodal points on the Mercator map and there could also be the problem of antipodal connecting lines possibly wrapping around the mercator map (which probably corresponds to your curvy $80^{\circ}$ loxodromes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another page which seems useful: &lt;a href=&quot;http://www.erikdeman.de/html/sail020m.htm&quot; rel=&quot;nofollow&quot;&gt;Notes on Loxodrome Calculations&lt;/a&gt; which I got from here: &lt;a href=&quot;http://www.erikdeman.de/html/sail022d.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.erikdeman.de/html/sail022d.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-29T18:27:06.833" LastActivityDate="2010-12-29T18:27:06.833" CommentCount="5" />
  <row Id="15807" PostTypeId="1" AcceptedAnswerId="15809" CreationDate="2010-12-29T09:03:24.013" Score="3" ViewCount="69" Body="&lt;p&gt;How to find the intersection between a set of two ranges of number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let me explain the question with an example,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{2,3} {3,8} would result to 0&#xA;{2,5} {3,8} would result to 2&#xA;{3,6} {3,8} would result to 3&#xA;{4,5} {3,8} would result to 1&#xA;{4,5} {3,8} would result to 1&#xA;{8,9} {3,8} would result to 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note:This could be easily done with series of if in a computer application, I am hungry for a mathematical solution!&lt;/p&gt;&#xA;" OwnerUserId="5197" LastEditorUserId="5197" LastEditorDisplayName="" LastEditDate="2010-12-29T09:39:18.257" LastActivityDate="2010-12-29T09:39:18.257" Title="What is the Shortest possible formula to find the intersection between a set of two ranges of number." Tags="&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="15808" PostTypeId="2" ParentId="15769" CreationDate="2010-12-29T09:03:47.497" Score="1" ViewCount="0" Body="&lt;p&gt;You can get precise asymptotics using $f(n)=n/2 \leq n/2 + \sqrt{n} \leq n/2 + \sqrt{n} + 1/2 = g(n)$.  Iterates of $f(n)$ and $g(n)$ have a closed form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulas for $f^k$ and $g^k$ (k'th iterates) imply that the number of iterations of $h(x)=x/2 + \sqrt{x}$ needed to bring $n &amp;gt; 4$ down into any interval $[0,C]$ with $C&amp;gt;4$, is $\log_2{n} + O(1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the recurrence, $T(n)$ increases by $\sqrt{6046}$ at each iteration, so the asymptotics are $T(n) = \sqrt{6046} \log_2(n) + O(1)$.&lt;/p&gt;&#xA;" OwnerUserId="467" LastActivityDate="2010-12-29T09:03:47.497" />
  <row Id="15809" PostTypeId="2" ParentId="15807" CreationDate="2010-12-29T09:18:25.430" Score="7" ViewCount="0" Body="&lt;p&gt;If I understood you correctly, you have $\{x,x+1,\ldots,x+n\}$ and $\{y,y+1,\ldots,y+k\}$ and granted $x,y$ are integers you want to find out the number of elements in the intersections (i.e. how many numbers appear in both sets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, take: $$\max\{0,\min\{y+k,x+n\}-\max\{x,y\}\}$$ and that should be it.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-12-29T09:18:25.430" CommentCount="3" />
  <row Id="15810" PostTypeId="1" CreationDate="2010-12-29T09:20:13.017" Score="6" ViewCount="98" Body="&lt;p&gt;The following problem is basically from Miranda's &quot;Algebraic Curves and Riemann Surfaces&quot;, which I am reading on my own; if there are any rules against posting textbook problems, my apologies!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ be a smooth projective curve defined by the homogeneous polynomial $F(x,y,z)=0$, with $\deg F = d \geq 3$. Let $f(x,y) = F(x,y,1)$. Show that if $p(u,v)$ is a polynomial of degree at most $d-3$, then $p(u,v) \frac{du}{\partial f/ \partial v}$ defines a holomorphic 1-form on the compact Riemann surface X. If $X$ is not smooth, but has nodes, then this form is a holomorphic 1-form on the resolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that this is a holomorphic $1$-form on the affine curve defined by $f$, since the charts are just projection to the $x$ or $y$ coordinate; in the former case the form is evidently holomorphic and in the latter case the form transforms to $p(u,v) \frac{dv}{\partial f/ \partial u}$. However, I'm a bit confused as to the computations involved in checking this on the other affine curves, and what extra argument is needed for the nodes case. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-29T18:04:34.833" Title="Differential Form on a Riemann Surface" Tags="&lt;riemann-surface&gt;&lt;differential-forms&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15811" PostTypeId="1" CreationDate="2010-12-29T11:11:54.013" Score="8" ViewCount="176" Body="&lt;p&gt;I've just started learning about rings. Rings are one additive abelian group strung together, through the associative law, with another structured operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't we continue stringing together operations in this manner (a multi-operation associative law)? Would what I'm thinking be encompassed by the ring definition through something I'm missing? If not, is this done and, if so, do useful objects come out of it? &lt;/p&gt;&#xA;" OwnerUserId="5198" LastActivityDate="2010-12-30T00:31:33.770" Title="Generalization of a ring?" Tags="&lt;algebra&gt;&lt;ring-theory&gt;&lt;definition&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="15812" PostTypeId="2" ParentId="15811" CreationDate="2010-12-29T11:19:41.587" Score="12" ViewCount="0" Body="&lt;p&gt;Absolutely!  Such things are studied in a few related disciplines encompassing &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;universal algebra&lt;/a&gt; and parts of &lt;a href=&quot;http://en.wikipedia.org/wiki/Category_theory&quot; rel=&quot;nofollow&quot;&gt;category theory&lt;/a&gt; (including what is called &quot;higher-dimensional algebra&quot;).  On the categorical side some keywords here include (algebras over an) &lt;a href=&quot;http://en.wikipedia.org/wiki/Operad_theory&quot; rel=&quot;nofollow&quot;&gt;operad&lt;/a&gt;, (algebras over a) &lt;a href=&quot;http://en.wikipedia.org/wiki/Monad_(category_theory)&quot; rel=&quot;nofollow&quot;&gt;monad&lt;/a&gt;, and (models of a) &lt;a href=&quot;http://en.wikipedia.org/wiki/Lawvere_theory&quot; rel=&quot;nofollow&quot;&gt;Lawvere theory&lt;/a&gt;.  More concretely there are also &lt;a href=&quot;http://en.wikipedia.org/wiki/Coalgebra&quot; rel=&quot;nofollow&quot;&gt;coalgebras&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Bialgebra&quot; rel=&quot;nofollow&quot;&gt;bialgebras&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf_algebra&quot; rel=&quot;nofollow&quot;&gt;Hopf algebras&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_algebra&quot; rel=&quot;nofollow&quot;&gt;Frobenius algebras&lt;/a&gt;, etc.  Ross Street's &lt;a href=&quot;http://books.google.com/books?id=u0slIpdBjYUC&amp;amp;printsec=frontcover&amp;amp;dq=ross+street+quantum+groups&quot; rel=&quot;nofollow&quot;&gt;Quantum Groups: a path to current algebra&lt;/a&gt; was written as an introduction to these more exotic types of algebra that people don't generally bother to tell you about as an undergrad.  Far from being esoteric generalizations of ordinary algebra, these structures turn up all over mathematics in the most unexpected places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of introduction, however, I don't think that what you've just described is a good way to get a sense of what a ring actually is.  The ring axioms capture the abstract properties that &lt;em&gt;symmetries of abelian groups&lt;/em&gt; satisfy, in the same way that the group axioms capture the abstract properties that &lt;em&gt;symmetries of sets&lt;/em&gt; satisfy.  The point here being that if $A$ is an abelian group and $f : A \to A$ some endomorphism of it, then there are two ways that one can combine such endomorphisms: one can compose them or add them.  This leads to the multiplication and addition operation, respectively, on the ring $\text{End}(A)$ of endomorphisms of $A$.  For example if $A$ is the &lt;em&gt;abelian group&lt;/em&gt; $\mathbb{Z}$, then $\text{End}(A)$ is the &lt;em&gt;commutative ring&lt;/em&gt; $\mathbb{Z}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a &quot;Cayley's theorem&quot; for rings which says that any abstract ring $R$ can be realized as endomorphisms of some abelian group: in fact $R$ is precisely the ring of endomorphisms of $R$ which respect &lt;em&gt;right&lt;/em&gt; multiplication, that is, which respect the right $R$-&lt;a href=&quot;http://en.wikipedia.org/wiki/Module_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;module&lt;/a&gt; structure on $R$.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-29T11:26:23.297" LastActivityDate="2010-12-29T11:26:23.297" />
  <row Id="15813" PostTypeId="2" ParentId="15811" CreationDate="2010-12-29T11:32:29.573" Score="3" ViewCount="0" Body="&lt;p&gt;The main point in a ring is not really its additive group or its multiplicative monoid: it's the distributive law, which ties the two operations. When you go on to more complicated structures, it's the laws that tie the operations that matter. See for instance the axioms of vector spaces and algebras. &lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-29T11:32:29.573" />
  <row Id="15814" PostTypeId="1" AcceptedAnswerId="15817" CreationDate="2010-12-29T11:37:54.543" Score="0" ViewCount="40" Body="&lt;p&gt;Hey,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a model which predicts the total final score (sum of the scores of both teams) of certain sports matches. I have checked the predictions of the model against the actual scores of matches for the last several years. The error in the predicted total score to the actual toal score is approximately normally distributed with mean = &lt;code&gt;-0.3&lt;/code&gt; and standard deviation = &lt;code&gt;13&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have a predicted total score &lt;code&gt;x&lt;/code&gt;, how can I determine the probability that the actual score will above or below a given score line &lt;code&gt;y&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say my model predicts that the total score will be &lt;code&gt;8&lt;/code&gt; and the score line is &lt;code&gt;5.5&lt;/code&gt;, what is the probability that the actual score will be &amp;lt;=5 and  what is the probability that the actual score will be &gt;5 (given that the errors in my prediction can be assumed to be normally distributed)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pete&lt;/p&gt;&#xA;" OwnerUserId="5199" LastActivityDate="2010-12-29T14:35:37.177" Title="Probability - Calculate chance total score will be over or under a score line" Tags="&lt;probability&gt;&lt;statistics&gt;&lt;normal-distribution&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15815" PostTypeId="1" AcceptedAnswerId="15822" CreationDate="2010-12-29T12:26:29.443" Score="3" ViewCount="81" Body="&lt;p&gt;I've asked &lt;a href=&quot;http://stackoverflow.com/questions/4526143/construct-polygons-out-of-union-of-many-polygons&quot;&gt;this question at SO&lt;/a&gt;, but only answer I got is a non-answer as far as I can tell, so I would like to try my luck here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I'm looking for a better-than-naive algorithm for the constructions of polygons out of union of many polygons, each with a list of Vertex $V$. The naive algorithm to find the union polygon(s) goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First take two polygons, union them,&#xA;  and take another polygon, union it&#xA;  with the union of the two polygons,&#xA;  and repeat this process until every&#xA;  single piece is considered. Then I will run&#xA;  through the union polygon list and&#xA;  check whether there are still some&#xA;  polygons can be combined, and I will&#xA;  repeat this step until a&#xA;  satisfactory result is achieved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a smarter algorithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this purpose, you can imagine each polygon as a jigsaw puzzle piece, when you complete them you will get a nice picture. But the catch is that a small portion ( say &amp;lt;5%) of the jigsaw is missing, and you are still require to form a picture as complete as possible; that's the polygon ( or polygons)-- maybe with holes-- that I want to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  I'm &lt;em&gt;not&lt;/em&gt; asking about how to union two polygons, but I am asking about--given that I know how to union two polygons--how to union $n$ number of (where $n&gt;&gt;2$) polygons in an efficient manner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,all the polygons can share edges, and some polygon's edge can be shared by one or many other polygon's edge. Polygons can't overlapped among themselves. &lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditorDisplayName="" LastEditDate="2010-12-29T14:04:39.620" LastActivityDate="2010-12-29T15:47:29.163" Title="How to Union Many Polygons Efficiently" Tags="&lt;computational-geometry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="15816" PostTypeId="2" ParentId="15164" CreationDate="2010-12-29T13:09:11.507" Score="1" ViewCount="0" Body="&lt;p&gt;The way to get the smaller angle spanned by $\mathbf e_1=(x_1,y_1)$ and $\mathbf e_2=(x_2,y_2)$ is through the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\min(|\arctan(x_1,y_1)-\arctan(x_2,y_2)|,2\pi-|\arctan(x_1,y_1)-\arctan(x_2,y_2)|)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\arctan(x,y)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atan2&quot; rel=&quot;nofollow&quot;&gt;two-argument arctangent&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-29T13:09:11.507" />
  <row Id="15817" PostTypeId="2" ParentId="15814" CreationDate="2010-12-29T13:49:18.407" Score="1" ViewCount="0" Body="&lt;p&gt;OK, sorry for misunderstanding your point. So, your model predicts $PV=8$ with errors $\epsilon \sim N(-0.3,13)$. Or, this is equivalent to saying that you expect the actual value to be $AV \sim N(8.3,13)$. ($PV=AV+\epsilon$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to compute the probability that $X\leq 5.5$ (or $x&gt;5.5$) under the hypothesis that your model is correct. This is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{P}(X\leq 5.5|X \sim N(8.3,13)) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what programming language or other computer program you use, but most math or statistics interpretative languages (Matlab, R, etc...) have the necessary commands inbuilt. So in R, you would use the command&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;pnorm(5.5,8.3,13)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which computes the cumulative normal distribution in point $5.5$ for mean and variance as given. For the probability bigger than $5.5$, you take &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1-pnorm(5.5,8.3,13)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you want the explicit formula for the computation, I can provide that too, but unless you want to make a lot of extra work implenting the functions yourself, I don't think that is really useful.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastEditorUserId="3567" LastEditorDisplayName="" LastEditDate="2010-12-29T14:35:37.177" LastActivityDate="2010-12-29T14:35:37.177" CommentCount="10" />
  <row Id="15818" PostTypeId="2" ParentId="15815" CreationDate="2010-12-29T14:02:10.790" Score="0" ViewCount="0" Body="&lt;p&gt;If the polygons either share exactly one edge or are disjoint then create a list of edges and the polygons they belong to and then remove each edge that has two polygons, joining those two polygons.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-29T14:02:10.790" CommentCount="4" />
  <row Id="15819" PostTypeId="1" AcceptedAnswerId="15821" CreationDate="2010-12-29T14:29:26.103" Score="4" ViewCount="113" Body="&lt;p&gt;From Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem&quot; rel=&quot;nofollow&quot;&gt;Weierstrass Factorization Theorem&lt;/a&gt;, I learned that every entire function can be represented as a product involving its zeroes. Examples are the sine and cosine function. The Riemann zeta function, however, is not entire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume the Riemann Hypothesis. Can $\zeta(s)$ be represented by an infinite product involving both its trivial zeroes at $\zeta(s)=-2n$ (for $n \in \mathbb{N}$) and its non-trivial zeroes at $\zeta(s)=\frac{1}{2} + i t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-29T14:46:08.647" LastActivityDate="2010-12-29T17:19:16.650" Title="Infinite product representation of a function in terms of its non-trivial zeroes?" Tags="&lt;functions&gt;&lt;analytic-number-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="15820" PostTypeId="2" ParentId="15819" CreationDate="2010-12-29T14:42:02.723" Score="3" ViewCount="0" Body="&lt;p&gt;(Would be a comment, but I don't have the reputation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Riemann zeta function is certainly not entire: it has a simple pole at s = 1. It is, however, meromorphic.&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2010-12-29T14:42:02.723" CommentCount="2" />
  <row Id="15821" PostTypeId="2" ParentId="15819" CreationDate="2010-12-29T15:19:36.520" Score="9" ViewCount="0" Body="&lt;p&gt;The standard approach to this problem is to write&#xA;$\xi(s) = \dfrac{s(s-1)}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s).$  This function is entire, and has zeroes precisely at the non-trivial zeroes of $\zeta(s)$.  It also has a slow enough rate of growth that it can be written as a product over its zeroes:&#xA;$$\xi(s) = \xi(0) \prod_{\rho}(1-\dfrac{s}{\rho}),$$&#xA;where the product is over zeroes of $\xi(s)$, i.e. over non-trivial zeroes of $\zeta(s)$.  (Here I am following more-or-less the notation in Edward's book &lt;em&gt;Riemann's zeta function&lt;/em&gt;, which is a good reference for these sort of things.)&#xA;[Edit: Also, to ensure convergence, the product should be taken over &quot;matching pairs&quot; of zeroes, i.e. the factors for a pair of zeroes of the form $\rho$ and $1-\rho$ should be combined.]&#xA;We can then write&#xA;$$\zeta(s) = \dfrac{2\xi(0)}{s(s-1)}\pi^{s/2}\dfrac{1}{\Gamma(s/2)}\prod_{\rho}(1-\dfrac{s}{\rho}).$$&#xA;If we now replace $\dfrac{1}{\Gamma(s/2)}$ by its Weierstrass product, we get&#xA;a product formula for $\zeta(s)$, namely&#xA;$$\zeta(s) = \dfrac{\xi(0)}{s-1} (\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1 +\dfrac{s}{2n})e^{-s/2n}&#xD;&#xA;\prod_{\rho}(1-\dfrac{s}{\rho}).$$&#xA;(Here $\gamma$ is Euler's constant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we can now compute $\xi(0)$, because we know that the value of $\zeta(s)$ at $s = 0$ is equal to $-1/2$.  We find that $\xi(0) = 1/2$,&#xA;and so &#xA;$$\zeta(s) = \dfrac{1}{2(s-1)}(\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1+\dfrac{s}{2n})e^{-s/2n}\prod_{\rho}(1-\dfrac{s}{\rho}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An added cultural remark: Riemann's &lt;a href=&quot;http://en.wikipedia.org/wiki/Explicit_formula&quot; rel=&quot;nofollow&quot;&gt;explicit formula&lt;/a&gt; for the prime counting function is obtained by taking a Fourier transform of the logarithm of this formula.  Combined with the fact that all the non-trivial zeroes $\rho$ have real part $&amp;lt; 1$ (proved by Hadamard and de la Vallee Poussin), this gives the prime number theorem.&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-29T17:19:16.650" LastActivityDate="2010-12-29T17:19:16.650" CommentCount="7" />
  <row Id="15822" PostTypeId="2" ParentId="15815" CreationDate="2010-12-29T15:47:29.163" Score="3" ViewCount="0" Body="&lt;p&gt;Likely the best method is to perform a simultaneous &lt;a href=&quot;http://en.wikipedia.org/wiki/Plane_sweep&quot; rel=&quot;nofollow&quot;&gt;plane sweep&lt;/a&gt; of all the polygons.&#xA;This is discussed in &lt;em&gt;The Algorithms Design Manual&lt;/em&gt; under &quot;&lt;a href=&quot;http://books.google.com/books?id=7XUSn0IKQEgC&amp;amp;pg=PA618&amp;amp;lpg=PA618&amp;amp;dq=plane+sweep+to+union+polygons&amp;amp;source=bl&amp;amp;ots=z6eFQRNNZd&amp;amp;sig=OYm-w6WA0jJQu3jtosuELrsafxA&amp;amp;hl=en&amp;amp;ei=7FUbTd-4CoWKlwed6MXhCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBMQ6AEwAA#v=onepage&amp;amp;q=plane%20sweep%20to%20union%20polygons&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Intersection Detection&lt;/a&gt;.&quot;&#xA;(Algorithms to find the intersection can be altered to instead find the union.)&#xA;It is also discussed in my textbook &lt;em&gt;&lt;a href=&quot;http://cs.smith.edu/~orourke/books/compgeom.html&quot; rel=&quot;nofollow&quot;&gt;Computational Geometry in C&lt;/a&gt;&lt;/em&gt;, Chapter 7, &quot;Intersection of NonConvex Polygons.&quot;  The time complexity is $O(n \log n + k)$ for $n$ total vertices and $k$ intersection points between edges of different polygons.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-29T15:47:29.163" />
  <row Id="15823" PostTypeId="1" AcceptedAnswerId="15826" CreationDate="2010-12-29T15:53:30.183" Score="2" ViewCount="123" Body="&lt;p&gt;Can $$\sin r\pi $$ be rational if $r$ is irrational? Either a direct or existence proof is fine.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-29T16:26:15.527" Title="Sine values being rational" Tags="&lt;number-theory&gt;&lt;algebra&gt;&lt;number&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="15824" PostTypeId="1" CreationDate="2010-12-29T16:14:58.213" Score="3" ViewCount="159" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Broadly I would like to know what is the connection between homogeneous binary quartic forms and elliptic curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the invariants on the first space under the action of $SL(2,\mathbb{C})$ are precisely the functions $g_2$ and $g_3$. Also given such a quartic form on $\mathbb{C}^2$, if one thinks its 0-set as the intersection of two conics in $\mathbb{C}^3$ then the determinant of the complex symmetric matrix representing the most general conic passing through this intersection is precisely the &quot;RHS&quot; of the elliptic curve equation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What does this determinant mean geometrically? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More specifically consider the $SL(2,\mathbb{C})$ action on $\mathbb{C}^2$ (spanned by $x$ and $y$) and define $U = x^2$, $V = 2xy$ and $W=y^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the action is by the matrix $ \left [ \begin{array}{c  c } &#xD;&#xA;                                     a &amp;amp; b \\&#xD;&#xA;                                     c &amp;amp; d \\&#xD;&#xA;                                                                 &#xD;&#xA;                           &#xD;&#xA;   \end{array}\right ]$ ($ad-bc=1$) of $SL(2,\mathbb{C})$ then the following transformation is affected on $\mathbb{C}^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left [  \begin{array}{c} &#xD;&#xA;                            x\\&#xD;&#xA;                            y\\&#xD;&#xA;                                                                 &#xD;&#xA;                          &#xD;&#xA;   \end{array} \right ]  \rightarrow \left [  \begin{array}{c} &#xD;&#xA;                            ax + by\\&#xD;&#xA;                            cx + dy\\&#xD;&#xA;                                                                 &#xD;&#xA;                           &#xD;&#xA;   \end{array} \right ] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turn implies the following transformation on $\mathbb{C}^3$ (defining the matrices, $X$ and $A$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = \left [  \begin{array}{c} &#xD;&#xA;                            U\\&#xD;&#xA;                            V\\&#xD;&#xA;                            W&#xD;&#xA;                                                                 &#xD;&#xA;                          &#xD;&#xA;   \end{array} \right ]  \rightarrow  A.X  = \left [  \begin{array}{c c c} &#xD;&#xA;                            a^2 &amp;amp; ab &amp;amp; b^2\\&#xD;&#xA;                            2ac &amp;amp; (ad+bc) &amp;amp; 2bd \\&#xD;&#xA;                            c^2 &amp;amp; cd &amp;amp; d^2\\&#xD;&#xA;                                                                 &#xD;&#xA;                           &#xD;&#xA;   \end{array} \right ] \left [  \begin{array}{c} &#xD;&#xA;                            U\\&#xD;&#xA;                            V\\&#xD;&#xA;                            W&#xD;&#xA;                                                                 &#xD;&#xA;                          &#xD;&#xA;   \end{array} \right ] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4UW-V^2$ being identically equal to $0$ is obviously kept invariant by the above transformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The invariance of $4UW-V^2$ can be thought of as $SL(2,\mathbb{C})$ preserving an inner-product on $\mathbb{C}^3$ with signature, $\left [ \begin{array}{c c c} &#xD;&#xA;                                     &amp;amp;   &amp;amp; 2 \\&#xD;&#xA;                                     &amp;amp; -1 &amp;amp;  \\&#xD;&#xA;                                     2 &amp;amp; &amp;amp;  \\&#xD;&#xA;                              &#xD;&#xA;                           &#xD;&#xA;   \end{array} \right ]$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But $A$ does not seem to be an orthogonal matrix though it satisfies the property of $det(A)=1$. Then why is it being said in the books that the $SL(2,\mathbb{C})$ action on $\mathbb{C}^2$ (spanned by $x$ and $y$) induces an action of $SO(3)$ on $\mathbb{C}^3$ (spanned by $U$, $V$ and $W$) with respect to the inner product $ \left [ \begin{array}{c  c c}  &#xD;&#xA;                                     &amp;amp;   &amp;amp; 2 \\&#xD;&#xA;                                     &amp;amp; -1 &amp;amp;  \\&#xD;&#xA;                                     2 &amp;amp; &amp;amp;  \\&#xD;&#xA;                              &#xD;&#xA;                           &#xD;&#xA;   \end{array}\right ]$ ? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-01-03T15:42:08.900" Title="Homogeneous binary quartic forms to elliptic curves " Tags="&lt;algebraic-geometry&gt;&lt;lie-groups&gt;" AnswerCount="1" />
  <row Id="15825" PostTypeId="1" CreationDate="2010-12-29T16:16:12.030" Score="2" ViewCount="80" Body="&lt;p&gt;Is there a good online community for discussing optimization models?  The ones I have found don't seem to have a critical mass for active discussions.&lt;/p&gt;&#xA;" OwnerUserId="5205" LastActivityDate="2010-12-29T16:16:12.030" Title="Mathematical Programming Community" Tags="&lt;linear-programming&gt;" CommentCount="2" />
  <row Id="15826" PostTypeId="2" ParentId="15823" CreationDate="2010-12-29T16:21:53.597" Score="7" ViewCount="0" Body="&lt;p&gt;As J. M. said, Niven's theorem does it.  There is some $r$ such that $\sin (r\pi) = \frac{1}{3}$  As $\sin (r\pi)$ is rational and not $0, \pm1, \pm \frac{1}{2}$, $r$ is not rational&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-29T16:21:53.597" />
  <row Id="15828" PostTypeId="2" ParentId="15811" CreationDate="2010-12-29T16:34:08.090" Score="6" ViewCount="0" Body="&lt;p&gt;Yes, any algebraic structure that has multiple operations will have laws like the distributive law that intertwine the operations. For otherwise the operations would not interact in any way and the structure could be studied as two independent structures with the non-interacting operations. For example, if we dropped the distributive law from the ring axioms then we'd simply have a set with a given abelian group structure and given monoid structure with no connection between the two structures. It is the distributive law that ties together these two structures and leads to the rich structure that is unique to rings - structure above and beyond the constituent structure of the additive group and multiplicative monoid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can observe the key role played by the distributive law even in the simplest results on rings. For example, consider the proof of the law of signs $\rm\ (-A)\:(-B) = A\:B\:.\ $ One simple proof is to observe that both terms are additive inverses of $\rm\ (-A)\:B\ $ hence they are equal by uniqueness of inverses. But to verify that they are inverse requires applying the distributive law. Similarly, any theorem that is truly ring-theoretic result $\:$ (i.e. $\:$ is not merely a result about abelian groups or monoids) must employ the distributive law in its proof (though perhaps obscured in some remote lemma).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogous remarks hold true for any algebraic structure with multiple operations, e.g. lattices with their intertwining absorption law $\rm\ X = X \vee (X \wedge Y)\ $ and it's dual, or distributive lattices (e.g. Boolean algebras) with their distributive law $\rm\ X \vee (Y \wedge Z) = (X\vee Y)\wedge (X\vee Z)\ $ and its dual or, more generally, the important modular law $\rm\ (X\wedge Y)\vee(Y\wedge Z) = Y \wedge ((X\wedge Y)\vee Z)\:.\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common properties of algebraic structures are studied in universal algebra (or general algebra). For example, one major theme is the study of the role played by properties of the lattices of congruences, e.g. congruence lattices of lattices are distributive, and congruence lattices of groups and rings are modular. These properties play  fundamental roles in the theories of these structures.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-30T00:31:33.770" LastActivityDate="2010-12-30T00:31:33.770" />
  <row Id="15829" PostTypeId="1" AcceptedAnswerId="15837" CreationDate="2010-12-29T17:27:18.553" Score="3" ViewCount="116" Body="&lt;p&gt;Let $p$ be a prime and $H$ a subgroup of a finite group $G$. Let $P$  be a p-sylow subgroup of G. Prove that there exists $g\in G$ such that $H\cap gPg^{-1}$ is  sylow subgroup of $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to do this, any hints? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; Originally it was unclear if the problem was for possibly infinite groups or just finite ones. However, since the definition of $p$-Sylow subgroup being used is that it is a $p$-subgroup such that the index and the order are relatively prime, the definition only applies to finite groups.&lt;/p&gt;&#xA;" OwnerUserId="5209" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:54:41.833" LastActivityDate="2010-12-30T16:59:51.297" Title="sylow subgroup of a subgroup" Tags="&lt;group-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="0" />
  <row Id="15830" PostTypeId="2" ParentId="15829" CreationDate="2010-12-29T17:32:39.393" Score="1" ViewCount="0" Body="&lt;p&gt;Well, we have $\vert P \vert =p^n$ where $\vert G \vert = p^n m$ and $(p,m)=1$.  What can be said about $\vert H \cap P \vert$ (or, for that matter, $\vert H \cap Q \vert$ for any Sylow p-subgroup $Q$ of $G$)?&lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="5137" LastEditorDisplayName="" LastEditDate="2010-12-29T17:38:37.937" LastActivityDate="2010-12-29T17:38:37.937" CommentCount="4" />
  <row Id="15831" PostTypeId="1" CreationDate="2010-12-29T17:36:27.577" Score="10" ViewCount="160" Body="&lt;p&gt;Some time ago when decomponsing the natural numbers, $\mathbb{N}$, in prime powes I noticed a pattern in their powers.  Taking, for example, the numbers $\lbrace 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 \rbrace$ and factorize them, we will get &#xA;$$\begin{align} 1&amp;amp;=2^0\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 2&amp;amp;=2^1\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\3&amp;amp;=2^0\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 4&amp;amp;=2^2\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 5&amp;amp;=2^0\times 3^0\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 6&amp;amp;=2^1\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 7&amp;amp;=2^0\times 3^0\times 5^0\times 7^1\times 11^0\times 13^0\times\ldots \\ 8&amp;amp;=2^3\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 9&amp;amp;=2^0\times 3^2\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 10&amp;amp;=2^1\times 3^0\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 11&amp;amp;=2^0\times 3^0\times 5^0\times 7^0\times 11^1\times 13^0\times\ldots \\ 12&amp;amp;=2^2\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 13&amp;amp;=2^0\times 3^0\times 5^0\times 7^0\times 11^0\times 13^1\times\ldots \\ 14&amp;amp;=2^1\times 3^0\times 5^0\times 7^1\times 11^0\times 13^0\times\ldots \\ 15&amp;amp;=2^0\times 3^1\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 16&amp;amp;=2^4\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\\end{align}$$&#xA;Now if we look at the powers of $2$ we will notice that they are $$\lbrace f_2(n)\rbrace=\lbrace 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4\rbrace$$ and for the powers of $3$ we have $$\lbrace f_3(n)\rbrace=\lbrace 0,0,1,0,0,1,0,0,2,0,0,1,0,0,1,0\rbrace$$&#xA;This, of course, is a well known fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then I wondered if there was a formula for $f_2(n)$ or $f_3(n)$ or $f_p(n)$, with $p\in \mathbb{P}$. It seemed impossible but I was able to devise the suitable formulas. They are&#xA;$$\displaystyle\begin{align} f_2(n)=\sum_{r=1}^{\infty}\frac{r}{{2^{r+1}}}\sum_{k=0}^{2^{r+1}-1}\cos\left( \frac{2k\pi(n+2^{r})}{2^{r+1}} \right)\end{align}$$&#xA;and for the general case we have&#xA;$$\displaystyle f_p(n)=\sum_{r=1}^{\infty}\frac{r}{p^{r+1}}\sum_{j=1}^{p-1}\left(\sum_{k=0}^{p^{r+1}-1}\cos\left( \frac{2k\pi(n+(p-j)p^{r})}{p^{r+1}} \right)\right)$$&#xA;If one cares to analyse the formula for $f_p(n)$ it can be concluded that it needs not to be restricted to the prime numbers, so that we have $f_m(n), m \in \mathbb{N}$ and similar  patterns for $\lbrace f_m(n)\rbrace$ will result. Now, the wonderfull thing is that we can express the &lt;em&gt;arithmetical divisor functions&lt;/em&gt; $\sigma_k(n)$ in terms of $f_m(n)$ as follows&#xA;$$\displaystyle \sigma_a(n)=1+\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{m^{a}}{m^{r+1}}\sum_{j=1}^{m-1}\left(\sum_{k=0}^{m^{r+1}-1}\cos\left( \frac{2k\pi(n+(m-j)m^{r})}{m^{r+1}} \right)\right)$$&#xA;And, if we consider the &lt;em&gt;divisor summatory function&lt;/em&gt;, $D(n)$, as&#xA;$$D(n)=\sum_{m \leq n}d(m)$$&#xA;with $$d(n)=\sigma_{0}(n)=\sum_{d|n}1$$&#xA;we can express $D(n)$ as &#xA;$$D(n)=\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{r}{m^{r+1}}\sum_{j=1}^{m-1}\left(\sum_{k=0}^{m^{r+1}-1}\cos\left( \frac{2k\pi(2^{n}+(m-j)m^{r})}{m^{r+1}} \right)\right)$$&#xA;Now, we know that, $d(n)$ and $D(n)$ are related to the Riemann zeta-function by&#xA;$$\zeta^{2}(z)=\sum_{n=1}^{\infty}\frac{d(n)}{n^{z}}$$&#xA;and &#xA;$$\zeta^{2}(z)=z\int_{1}^{\infty}\frac{D(x)}{x^{z+1}}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my questions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What can we say about the convergence of $f_m(z)$, $\sigma_a(z)$ and $D(z)$ with $z \in \mathbb{C}$? We can see that they converge for $z \in \mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;I think that $\sigma_a(z)$ and $D(z)$ are only curiosities and aren't interesting in the context of the Riemann zeta-function because they are hard to compute. What do you think? &lt;/li&gt;&#xA;&lt;li&gt;Are formulas $f_m(z)$, $\sigma_a(z)$ and $D(z)$ original. I think they are. I'd like to know if anyone has found something like this before. I've posted this as an answer to &lt;a href=&quot;http://mathoverflow.net/questions/29828/greatest-power-of-two-dividing-an-integer&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; post sometime ago.&lt;/li&gt;&#xA;&lt;li&gt;Finally, is this intereting enough to publish somewhere? I'm just an amateur...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To conclude I'd like to apologise for presenting all this formulas without showing how I got them but you can consider &lt;a href=&quot;http://math.stackexchange.com/questions/5574/a-triangular-representation-for-the-divisor-summatory-function-dx&quot;&gt;this&lt;/a&gt; previous post of mine and the question &lt;a href=&quot;http://mathoverflow.net/questions/29828/greatest-power-of-two-dividing-an-integer&quot; rel=&quot;nofollow&quot;&gt;Greatest power of two dividing an integer&lt;/a&gt;, &lt;a href=&quot;http://mathoverflow.net/questions/40427/difficult-infinite-sum&quot; rel=&quot;nofollow&quot;&gt;Difficult Infinite Sum&lt;/a&gt; and &lt;a href=&quot;http://fs.gallup.unm.edu/Atanassov-SomeProblems.pdf&quot; rel=&quot;nofollow&quot;&gt;On the 61-st, the 62-nd, and the 63-rd Smarandache's problem&lt;/a&gt; page 38.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now a challenge, can you present a formula for the &lt;em&gt;characteristic function of the prime numbers&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1747" LastEditorUserId="1747" LastEditorDisplayName="" LastEditDate="2011-01-05T09:24:02.660" LastActivityDate="2011-01-05T09:24:02.660" Title="Prime powers, patterns similar to $\lbrace 0,1,0,2,0,1,0,3\ldots \rbrace$ and formulas for $\sigma_k(n)$." Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;riemann-zeta&gt;" CommentCount="3" FavoriteCount="3" />
  <row Id="15832" PostTypeId="2" ParentId="12069" CreationDate="2010-12-29T18:03:44.410" Score="0" ViewCount="0" Body="&lt;p&gt;Hi, in searching the solution for a nearly problem, I have been falling in your interesting discussion. But there's something I doubt about Zarrax's answer. If the probablity density function is like your result, we can find out easily there's some values of alpha which makes the distribution function greater than 1, is it physically correct?&lt;/p&gt;&#xA;" OwnerUserId="5210" LastActivityDate="2010-12-29T18:03:44.410" CommentCount="1" />
  <row Id="15833" PostTypeId="2" ParentId="15810" CreationDate="2010-12-29T18:04:34.833" Score="2" ViewCount="0" Body="&lt;p&gt;You have two questions: how to change variables, and how to handle nodes.  The case of changing variables, for hyperelliptic curves without nodes, is in chapter III.5.5 of Shafarevich's Basic algebraic geometry.  The general case is on page 105 of Phillip Griffiths' Introduction to algebraic curves (China notes).  The discussion of the nodes there needs supplementing with the argument from page 98, i.e. a proof that the derivative vanishes simply on both the separate branches lying over the node, rather than at the point in the plane, as suggested there in a footnote.  Basically a fraction is holomorphic if the numerator vanishes as much as the denominator.  Here the equation for the curve vanishes twice at the node so the derivative in the denominator vanishes once and is canceled by the vanishing of the adjoint polynomial.  But you still have to finesse the point raised in Griffiths' footnote as mentioned above about the order of vanishing of the pullback to the normalization.  Warm up on some specific examples.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastActivityDate="2010-12-29T18:04:34.833" />
  <row Id="15834" PostTypeId="2" ParentId="15773" CreationDate="2010-12-29T18:15:52.130" Score="3" ViewCount="0" Body="&lt;p&gt;Since LIE's answer reduces one to the case of $A_5$, I like identifying $A_5$ with the icosahedral group and using the geometry of the icosahedron to make the calculations of the number of elements conjugate to those leaving vertices, edges, and faces, respectively fixed.  This is on page 20 of the notes for math 843-1 &lt;a href=&quot;http://www.math.uga.edu/~roy/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To identify Icos with $A_5$, note the 30 edges are partitioned into 5 subsets of size 6, by looking at the 5 sets of orthogonal triples in the 15 lines joining the centers of pairs of opposite edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then LIE's answer (and Arturo's) shows why $A_5$ is really the key case.  Of course the apparently ad hoc geometry used here is mirrored abstractly by the conjugacy classes of elements in the group.  I.e. every group acts on itself by conjugation, and sometimes we can represent that action by a geometric figure.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:27:45.773" LastActivityDate="2010-12-29T19:27:45.773" CommentCount="2" />
  <row Id="15835" PostTypeId="2" ParentId="15829" CreationDate="2010-12-29T18:16:20.763" Score="1" ViewCount="0" Body="&lt;p&gt;Let $P&amp;#39;$ be any Sylow $p$-subgroup of $H$. Then there is a maximal $p$-subgroup of $G$ containing it. What is it?&lt;/p&gt;&#xA;" OwnerUserId="5191" LastActivityDate="2010-12-29T18:16:20.763" CommentCount="2" />
  <row Id="15836" PostTypeId="2" ParentId="15571" CreationDate="2010-12-29T18:16:25.023" Score="3" ViewCount="0" Body="&lt;p&gt;Concerning infinity, here is my take. When we say, for example, that there is an infinity of natural numbers, we mean only that there is no largest number. Whenever you add 1 to a number, you will get another number that is greater still. Nothing very profound there. To say otherwise would greatly complicate even basic arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may even be possible to do away with the use of the word &quot;infinity&quot; (or any equivalent notion) in most if not all of mathematics. It may be nothing more than a convenient shorthand. When we say, for example, the limit of some expression as x tends to infinity, we could as easily have said, the limit as x increases without bound. The latter would have the advantage that there would be no confusion about there being some kind of endpoint called &quot;infinity.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2010-12-29T18:16:25.023" CommunityOwnedDate="2010-12-29T18:16:25.023" />
  <row Id="15837" PostTypeId="2" ParentId="15829" CreationDate="2010-12-29T18:32:24.993" Score="3" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The following works if $G$ is finite, but may fail in the infinite case; there are infinite groups in which there are $p$-Sylow subgroups $P$ and $P&amp;#39;$ for which no automorphism (inner or outer) of $G$ maps $P$ to $P&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $K$ be a $p$-Sylow subgroup of $H$ (we know it exists, though it may be trivial). Then $K$ is a $p$-subgroup of $G$ (even if $K={e}$) and by the Sylow Theorems, is contained in &lt;em&gt;some&lt;/em&gt; Sylow $p$-subgroup $Q$ of $G$. By the Sylow Theorems, $Q$ and $P$ are conjugate. Now just verify that $Q\cap H = K$. &lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:12:18.970" LastActivityDate="2010-12-29T19:12:18.970" CommentCount="9" />
  <row Id="15838" PostTypeId="1" CreationDate="2010-12-29T18:40:57.577" Score="0" ViewCount="103" Body="&lt;p&gt;Given the next sets: $A=\{1,2\}$ , $B=\mathbb{N}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Describe all the functions from $A$ to $B$ that are not injective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Describe all the functions from $B$ to $A$ that are not onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Prove or disprove the next:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) The next functions exist $f\colon A\to B$ and $g\colon B\to A$ if $g\circ f$ reversal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) the next functions exist $f\colon A\to B$ and $g\colon B\to A$ if $f\circ g$ reversal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need help here. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5214" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T19:53:59.400" LastActivityDate="2010-12-29T19:53:59.400" Title="Sets and functions" Tags="&lt;homework&gt;&lt;functions&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="15839" PostTypeId="2" ParentId="15709" CreationDate="2010-12-29T18:45:34.907" Score="3" ViewCount="0" Body="&lt;p&gt;A product $A\times B$ is a triple $(A,B,A\times B)$ equipped with maps $\pi_A\colon A\times B\to A$, and $\pi_B\colon A\times B\to B$ such that given any pair of maps $\psi_A\colon X\to A$ and $\psi_B\colon X\to B$, there is a &lt;strong&gt;unique&lt;/strong&gt; map $\psi\colon X\to A\times B$, such that $\psi_A=\pi_a\circ \psi$ and $\psi_B=\pi_B\circ \psi$. Hence for $X = A\times B$, (and some given maps) if the identity map $A\times B\to A\times B$ fulfills this role, then any other map also fulfilling it equals the identity map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. to prove that $fg$ is the identity, check that it does satisfy some property that is only satisfied by one unique map, and then check that the identity also satisfies it.  Examples are given in the algebra notes 843-4-5 on my website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, look on pages 8-9 of the notes 845-3, for exactly this proof of uniqueness of products.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2011-01-06T01:00:59.687" LastActivityDate="2011-01-06T01:00:59.687" CommentCount="3" />
  <row Id="15840" PostTypeId="2" ParentId="15838" CreationDate="2010-12-29T19:20:06.763" Score="2" ViewCount="0" Body="&lt;p&gt;The first two follow from the definitions, so I'm assuming you're having some difficulty with what injective (or one to one) and surjective (or onto) mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $f : A \rightarrow B$ be a function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our function $f$ is injective if, provided there is some $a$ in $A$ with $f(a) = b$, there is only one such $a$.  Put another way, everything in $B$ that is &quot;hit&quot; by $f$ is &quot;hit&quot; only one time.  This is why one-to-one is a good term:  for every output, there is one input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our function $f$ is surjective if, for every $b$ in $B$, there is some $a$ with $f(a) = b$.  Essentially, $f$ is surjective if everything in $B$ is &quot;hit&quot; by $f$.  This is why onto is a good term:  you can picture $B$ being covered by $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your part (a), think about functions from ${1,2}$ to $\mathbb{N}$ that do not have the property that &quot;for every output, there is one input.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your part (b), think about functions from $\mathbb{N}$ to ${1,2}$ that do not have the property that &quot;everything is hit&quot; by the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to reword part (c); it does not make sense as written.&lt;/p&gt;&#xA;" OwnerUserId="3537" LastActivityDate="2010-12-29T19:20:06.763" CommentCount="4" />
  <row Id="15841" PostTypeId="2" ParentId="14921" CreationDate="2010-12-29T19:35:09.690" Score="5" ViewCount="0" Body="&lt;p&gt;In John Stillwell's Mathematics and its History it is observed that Euler criticized Newton's classification for lacking a general principle, but that a closer examination of Newton/s work reveals one.  In fact his work gives a general classification into 5 types, depicted on &lt;a href=&quot;http://books.google.com/books?id=WNjRrqTm62QC&amp;amp;pg=PA112&quot; rel=&quot;nofollow&quot;&gt;page 112 of that book&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the difference in the real and complex cases, is the fact that the &quot;discriminant&quot; locus of singular cubics has real codimension one in the real case, hence separates the space of smooth cubics into distinct connected components, and this may be the types sought for.  By Ehresmann's theorem, a connected family of compact manifolds have the same topological type.  Thus curves on different connected components of the complement of the discriminant locus can have different homeomorphism type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least one (two?) of Newton's types is also singular and could represent the general (and special) point of the discriminant locus.  I am not an expert.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2011-01-01T00:52:56.590" LastActivityDate="2011-01-01T00:52:56.590" CommentCount="1" />
  <row Id="15842" PostTypeId="2" ParentId="15838" CreationDate="2010-12-29T19:53:34.627" Score="2" ViewCount="0" Body="&lt;p&gt;(a) Since $A$ has only two elements, the &lt;em&gt;only&lt;/em&gt; way for a function $f\colon A\to X$ (for &lt;em&gt;any&lt;/em&gt; set $X$) to &lt;em&gt;not&lt;/em&gt; be injective is if $f(1)=f(2)$. So you want all functions from $A$ to $B$ in which $f(1)=f(2)$. Now, while there are indeed &quot;lots of them&quot; (infinitely many) they all have a very easy description. &lt;em&gt;Hint:&lt;/em&gt; the word that describes these functions starts with &quot;con&quot; and ends with &quot;ants&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Again, since $A$ has only two elements, the &lt;em&gt;only&lt;/em&gt; way for a function $g\colon X\to A$ to &lt;em&gt;not&lt;/em&gt; be onto (surjective) is if either you never get $1$, or you never get $2$ as images of $g$. While there are many, many, functions from $B$ to $A$, there are very few in which this happens. &lt;em&gt;Hint&lt;/em&gt;: the word that describes theses functions is the same as in (a).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c-i) You want to know if you can define functions $A\stackrel{f}{\to} B\stackrel{g}{\to}A$ such that the composition is both one-to-one and onto. Not too hard, since there are only two functions from $A$ to $A$ that are both one-to-one and onto. See if you can come up with some easy functions $f$ and $g$ that do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c-ii) Here you want to know if you can find functions $B\stackrel{g}{\to}A\stackrel{f}{\to}B$ so that the composition is both one-to-one and onto. Again, this is not hard if you think about what these things mean. Can you have a one-to-one map from $B$ to $A$? Can you have an onto map from $A$ to $B$? Think about it. Note that these $f$ and $g$ don't have to be the same as in (c-i), they are only asking you if some such functions exist at all.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-29T19:53:34.627" />
  <row Id="15843" PostTypeId="1" CreationDate="2010-12-29T20:27:10.637" Score="3" ViewCount="122" Body="&lt;p&gt;Let $p_n$ be the n'th non-trivial zero of the Riemann zeta function. We define the &lt;em&gt;Zetor&lt;/em&gt; function (acronym of 'zeta' and 'zero') as follows: $$\zeta \rho (s) = \sum_{n=1}^{\infty} \frac{1}{(p_n)^s}. $$ &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which values of $s$ does $\zeta \rho (s)$ converge?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="1055" LastEditorDisplayName="" LastEditDate="2010-12-29T20:53:49.553" LastActivityDate="2010-12-29T20:53:49.553" Title="When does the &quot;Zetor function&quot; converge?" Tags="&lt;functions&gt;&lt;analytic-number-theory&gt;" CommentCount="7" />
  <row Id="15844" PostTypeId="1" AcceptedAnswerId="15845" CreationDate="2010-12-29T20:33:21.877" Score="4" ViewCount="69" Body="&lt;p&gt;I'm looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Intuitionistic_logic#Hilbert-style_calculus&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on Intuitionist logic&lt;/a&gt;, and I can't figure out how it would prove $(p \rightarrow p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it prove $(p \rightarrow p)$?  If yes, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, is there a correct (or reasonable, if there is no &quot;correct&quot;) axiomatization of intuitionistic logic available anywhere online?&lt;/p&gt;&#xA;" OwnerUserId="5114" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-29T20:40:45.007" LastActivityDate="2010-12-29T20:40:45.007" Title="How does (wikipedia's axiomatization of) intuitionistic logic prove  $p \rightarrow  p$  ?" Tags="&lt;logic&gt;&lt;constructive-mathematics&gt;" AnswerCount="1" />
  <row Id="15845" PostTypeId="2" ParentId="15844" CreationDate="2010-12-29T20:38:48.600" Score="7" ViewCount="0" Body="&lt;p&gt;&#xA;&lt;li&gt;p→(p→p) (THEN-1)&lt;/li&gt;&#xA;&lt;li&gt;p→((p→p)→p) (THEN-1)&lt;/li&gt;&#xA;&lt;li&gt;(p→((p→p)→p))→((p→(p→p))→(p→p)) (THEN-2)&lt;/li&gt;&#xA;&lt;li&gt;(p→(p→p))→(p→p) (MP from lines 2,3)&lt;/li&gt;&#xA;&lt;li&gt;(p→p) (MP from lines 1,4)&lt;/li&gt;&lt;/p&gt;&#xA;" OwnerUserId="5203" LastActivityDate="2010-12-29T20:38:48.600" />
  <row Id="15846" PostTypeId="1" AcceptedAnswerId="15848" CreationDate="2010-12-29T20:57:32.247" Score="4" ViewCount="127" Body="&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering today about why the word &lt;em&gt;differentiable&lt;/em&gt; is used for describing functions that have a derivative or are differentiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps because originally one considered &lt;em&gt;finite differences&lt;/em&gt;? But that seems somewhat not right, because &lt;em&gt;roughly speaking&lt;/em&gt; a derivative measures &lt;strong&gt;not&lt;/strong&gt; the difference $f(x+h)-f(x)$, but rather the ratio $(f(x+h)-f(x))/h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, could people here shed light on why we use &quot;differentiable&quot;? Any pointers to academic / historical / etymological explanations are also welcome. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1709" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T01:37:02.710" LastActivityDate="2010-12-30T01:37:02.710" Title="What is the origin of the term &quot;Differentiable&quot;?" Tags="&lt;reference-request&gt;&lt;terminology&gt;&lt;math-history&gt;" AnswerCount="1" />
  <row Id="15847" PostTypeId="2" ParentId="15773" CreationDate="2010-12-29T20:59:20.873" Score="2" ViewCount="0" Body="&lt;p&gt;The way to understand this theorem is to understand the following statement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Other than the even permutations, there's no way to choose a collection of cycle structures that's closed under multiplication.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To illustrate what I mean, I will prove the following theorem, which is related to but slightly easier than the simplicity of $A_n$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; &lt;em&gt;For $n \geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The main idea of the proof is to imagine what a normal subgroup of $S_n$ must look like:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Any normal subgroup of $S_n$ must be a union of conjugacy classes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But each conjugacy class of $S_n$ is simply the set of all permutations with a given cycle structure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore, a normal subgroup of $S_n$ must consist of a collection of possible cycle structures which is closed under multiplication.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, the conjugacy classes in $S_4$ are the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the identity&lt;/li&gt;&#xA;&lt;li&gt;transpositions $(\ast \;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;3-cycles $(\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;4-cycles $(\ast\;\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;Elements of the form $(\ast\;\ast)(\ast\;\ast)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any normal subgroup in $S_4$ would have to be the union of some of these, and would have to be closed under multiplication.  It's easy to check that the only possibilities are $A_4 = (1) \cup (3) \cup (5)$ and the 4-element subgroup $(1) \cup (5)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the conjugacy classes in $S_5$ are the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the identity&lt;/li&gt;&#xA;&lt;li&gt;transpositions $(\ast \;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;3-cycles $(\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;4-cycles $(\ast\;\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;5-cycles $(\ast\;\ast\;\ast\;\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;Elements of the form $(\ast\;\ast)(\ast\;\ast)$&lt;/li&gt;&#xA;&lt;li&gt;Elements of the form $(\ast\;\ast\;\ast)(\ast\;\ast)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's easy to check that the only finite union of these that is closed under multiplication is $A_5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, the rest of the proof just involves playing around with permutations and cycle structures.  In particular, it's not too hard to prove the following statement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For any nonidentity permutation $\alpha \in S_n$, there exists a permutation $\beta  \in S_n$ with the same cycle structure so that $\alpha\beta$ has the form $(\ast\;\ast\;\ast)$ or $(\ast\;\ast)(\ast\;\ast)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus any normal subgroup of $S_n$ must either contain all of the 3-cycles, or all elements of the form $(\ast\;\ast)(\ast\;\ast)$.  For $n \geq 5$, elements of these forms generate  all of $A_n$, and therefore $A_n$ is the only nontrivial normal subgroup of $S_n$.  (Since $A_n$ has index two, any nontrivial subgroup that contains $A_n$ must equal $A_n$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof that $A_n$ is simple is fairly similar, but complicated by the fact that the conjugacy classes in $A_n$ aren't precisely the cycle structures.  (Some cycle structures in $A_n$ have two corresponding conjugacy classes.)  Alternatively, it may be possible to prove that any normal subgroup of $A_n$ would also have to be normal in $S_n$, and then apply the above theorem to conclude that $A_n$ is simple.&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2010-12-29T20:59:20.873" CommentCount="1" />
  <row Id="15848" PostTypeId="2" ParentId="15846" CreationDate="2010-12-29T21:03:23.293" Score="9" ViewCount="0" Body="&lt;p&gt;From the &lt;a href=&quot;http://jeff560.tripod.com/mathword.html&quot; rel=&quot;nofollow&quot;&gt;Earliest Known Uses of Some of the Words of Mathematics&lt;/a&gt; webpage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;DIFFERENTIAL CALCULUS.&lt;/strong&gt; The term calculus differentialis was introduced by Leibniz in 1684 in Acta Eruditorum 3. Before introducing this term, he used the expression methodus tangentium directa (Struik, page 271). The OED has a nice quotation from Joseph Raphson’s Mathematical Dictionary of 1702: “A different way....passes....in France under the Name of Leibnitz's Differential Calculus, or Calculus of Differences.” &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="274" LastActivityDate="2010-12-29T21:03:23.293" CommentCount="1" />
  <row Id="15849" PostTypeId="1" AcceptedAnswerId="15851" CreationDate="2010-12-29T22:11:00.407" Score="2" ViewCount="87" Body="&lt;p&gt;I'm attempting to understand the proof of the statement that every finite domain $R$ is a division ring. For a fixed nonzero element $a \in R$, the left-translation map is injective since $R$ is a domain. Since $R$ is finite this map is also surjective which implies that 1 is in its image. Therefore we have the existence of an element $b \in R$ such that $ab=1$, right inverses exist for nonzero elements. By looking at the right-translation map, we get the existence of left inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a finite domain $R$ and $a,b,c \in R$ such that $ab=1$ and $ca=1$ how do we show that $b=c=a^{-1}$? How do we know that the nonzero elements are units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is as follows: The nonzero elements already form a multiplicative monoid since $R$ is a domain. But to form a group it is enough to demonstrate the existence of left (or right) inverses only by &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_group_theory#Alternative_Axioms&quot; rel=&quot;nofollow&quot;&gt;these alternate group axioms&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe we just need left inverses and an identity to get a division ring.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2010-12-30T17:20:22.103" Title="Equality of left and right inverses in a finite domain." Tags="&lt;abstract-algebra&gt;&lt;ring-theory&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="15850" PostTypeId="2" ParentId="15849" CreationDate="2010-12-29T22:16:53.870" Score="5" ViewCount="0" Body="&lt;p&gt;What happens if you mutliply the identity $ab = 1$ on the left by $c$?&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-29T22:16:53.870" CommentCount="1" />
  <row Id="15851" PostTypeId="2" ParentId="15849" CreationDate="2010-12-29T22:17:59.247" Score="4" ViewCount="0" Body="&lt;p&gt;In &lt;em&gt;any&lt;/em&gt; ring $R$, if $a$ has both a left and a right inverse, then the left and right inverses are the same and the element is a unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this in your case, just write this:&#xA;$$ b = 1\cdot b = (ca)b = c(ab) = c\cdot 1 = c.$$&#xA;Since $b=c$, then $ca = ba = ab = 1$, so $b=c=a^{-1}$, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the standard argument. It can be used, for example, to show that a function that has left and right inverses must be invertible (and the left and right inverses are the same). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also true that if you have a ring without zero divisors, which has a left identity $e\neq 0$ and every element has a left inverse, then the ring is a division ring. The argument is the one you give, essentially showing that the nonzero elements form a group under multiplication. But the fact that an element having both left and right inverses is in fact invertible is more basic than that.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-29T22:17:59.247" CommentCount="4" />
  <row Id="15852" PostTypeId="1" CreationDate="2010-12-29T22:44:00.737" Score="4" ViewCount="124" Body="&lt;p&gt;The word problem for finite groups is decidable. Is it obvious that this is true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm not entirely sure about what it means for the problem to be decidable (in this case---I think I understand what decidable means in general). I assume it means that we are given a fixed group &lt;em&gt;G&lt;/em&gt; (do we have to assume this), but is the generating set (the letters) also fixed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To decide the word problem for the group of symmetries of a square, with reflection $r$ and transposition $t$ as the generators, I would first find a canonical form for the group elements $1,r,r^2,r^3,t,tr,tr^2,tr^3$, and then note that using the relations $r^4=1$ and $rt=tr^3$ to show that any word can be reduced to something on my list. However, this seems like a lot of work, and it's not obvious to me what should be done in the case of an arbitrary finite group. &lt;/p&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="3656" LastEditorDisplayName="" LastEditDate="2010-12-30T00:27:59.003" LastActivityDate="2010-12-30T16:44:32.323" Title="The word problem for finite groups" Tags="&lt;group-theory&gt;&lt;computability&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="15853" PostTypeId="2" ParentId="15852" CreationDate="2010-12-29T22:51:25.437" Score="5" ViewCount="0" Body="&lt;p&gt;The sense in which I think this statement is usually meant is that you are given the entire multiplication table of the group.  Then it's obvious that the word problem is decidable.  You seem to be thinking about a different question: maybe you are given generators and relations and you are told that they define a finite group, then asked to solve the word problem.  This question, to me, is not obvious.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-12-29T22:51:25.437" CommentCount="10" />
  <row Id="15854" PostTypeId="2" ParentId="15849" CreationDate="2010-12-29T23:00:04.583" Score="2" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;HINT&lt;/strong&gt; $\:$ &lt;em&gt;Unify&lt;/em&gt; the rewrite rules $\rm\ ca\to 1,\  ab\to 1\ $ by creating a common generalization of their LHS,$\ $ viz. $\rm\ cab\:.\: $ Now apply the two rules to the unification $\rm\ cab\ $ to deduce that $\rm\ c = b\:.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same type of unification procedure is at the heart of procedures that generate normal-form rewriting algorithms for various algebraic structures, e.g. google the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knuth%E2%80%93Bendix_completion_algorithm&quot; rel=&quot;nofollow&quot;&gt;Knuth-Bendix completion algorithm.&lt;/a&gt; Many well-known algorithms may be considered as special cases of such, e.g. Gaussian elimination and its nonlinear generalizations such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis&quot; rel=&quot;nofollow&quot;&gt;Grobner basis algorithm,&lt;/a&gt; and also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Todd%E2%80%93Coxeter_algorithm&quot; rel=&quot;nofollow&quot;&gt;Todd-Coxeter algorithm&lt;/a&gt;  for enumerating cosets of group presented by generators and relations.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-30T17:20:22.103" LastActivityDate="2010-12-30T17:20:22.103" CommentCount="2" />
  <row Id="15855" PostTypeId="1" CreationDate="2010-12-29T23:05:02.523" Score="1" ViewCount="70" Body="&lt;p&gt;Let $X_1,...,X_n$ be some observations, and let $\theta$ be some parameter of the density function we want to estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, it is well known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l(\theta) = n^{-1} \sum_{i=1}^n \log f(X_i ; \theta)$ is called the average log-likelihood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $\mathbb{E}[\log f(X_1 ; \theta)]$ called? Meaning, &quot;the expected value of the likelihood&quot;? It can be called the cross-entropy, or perhaps it has other name, but it seems to me it should have a name relating it to the likelihood, such as &quot;population likelihood&quot; perhaps or something of that sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone knows? Wikipedia did not help much here.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditorDisplayName="" LastEditDate="2010-12-30T00:57:02.517" LastActivityDate="2010-12-31T03:56:26.060" Title="What is the &quot;population likelihood&quot; called?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15857" PostTypeId="1" AcceptedAnswerId="15864" CreationDate="2010-12-29T23:18:01.923" Score="3" ViewCount="143" Body="&lt;p&gt;Which of these two books should I read first? Is one of them more advanced than the other? I'm planning to use them to self-study geometry. &#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-30T18:49:15.130" Title="Coxeter's &quot;Geometry Revisited&quot; vs. &quot;Introduction to Geometry&quot;" Tags="&lt;geometry&gt;&lt;books&gt;&lt;self-learning&gt;" AnswerCount="5" />
  <row Id="15858" PostTypeId="2" ParentId="15852" CreationDate="2010-12-29T23:18:48.697" Score="7" ViewCount="0" Body="&lt;p&gt;One can prove quite simply a much more general result due to McKinsey, viz.&#xA;&lt;img src=&quot;http://i.imgur.com/lzh3L.jpg&quot; alt=&quot;alt text&quot;&gt;&#xA;&lt;img src=&quot;http://i.imgur.com/aHLrP.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-29T23:18:48.697" CommentCount="1" />
  <row Id="15859" PostTypeId="2" ParentId="15857" CreationDate="2010-12-29T23:26:08.530" Score="3" ViewCount="0" Body="&lt;p&gt;If you are talking about &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471504580&quot; rel=&quot;nofollow&quot;&gt;Introduction to Geometry&lt;/a&gt;&quot; by Coxeter and &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0883856190&quot; rel=&quot;nofollow&quot;&gt;Geometry Revisited&lt;/a&gt;&quot; by Coxeter and Greitzer, the consensus seems to be that both of them are pretty advanced, but &quot;Introduction to Geometry&quot; is significantly more so, while &quot;Geometry Revisited&quot; is closer to something 'right after high school geometry class,' so I guess you should start with the latter (assuming you do have some geometry knowledge already).&lt;/p&gt;&#xA;" OwnerUserId="3731" LastActivityDate="2010-12-29T23:26:08.530" />
  <row Id="15860" PostTypeId="1" AcceptedAnswerId="15863" CreationDate="2010-12-29T23:29:18.510" Score="2" ViewCount="66" Body="&lt;p&gt;In a population that grows, would a disparity in having a boy/girl probability cause the ratio of males to females tend to the same? e.g. if the probability of having boys was .49 and having a girl was .51 would that mean a growing population would tend to be 49% male and 51% female?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a non growing population?, would that mean finally one gender will be the only gender left after sufficient amount of time?&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2010-12-30T00:28:58.077" Title="Consequences of boy/girl probability disparity on population" Tags="&lt;probability&gt;" AnswerCount="2" />
  <row Id="15861" PostTypeId="2" ParentId="15860" CreationDate="2010-12-29T23:35:57.863" Score="1" ViewCount="0" Body="&lt;p&gt;Probability doesn't directly translate into the given proportions -- it actually provides you with the probability of the events happening. You could have .49 for girls and .51 for boys, yet have 99 to 1 ratio of girls to boys (which is &lt;strong&gt;unlikely&lt;/strong&gt;, but possible).&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-30T00:28:58.077" LastActivityDate="2010-12-30T00:28:58.077" />
  <row Id="15862" PostTypeId="2" ParentId="15857" CreationDate="2010-12-29T23:42:15.873" Score="4" ViewCount="0" Body="&lt;p&gt;Geometry Revisited is much more elementary, aimed at high-school teachers and bright high-school students or college frosh. For example, K. Strubecker wrote in his AMS Math Review&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tenor of the translation of Coxeter's beautiful tome Geometry revisited [Random House, New York, 1967] is in keeping with the objectives of the Klett Textbooks in Mathematics series which are intended to convey to freshmen and teachers of mathematics---via interesting representations---an approach to different aspects of mathematics, especially to geometry, that is kept as concrete as possible and so is applicable in schools. The volume contains six chapters which deal with the following topics: (1) Points and lines connected with a triangle; (2) some properties of circles; (3) collinearity and concurrence; (4) transformations; (5) an introduction to inversive geometry; (6) an introduction to projective geometry; The very lucid presentation takes the reader from elementary problems of plane Euclidean geometry to the fundamental concepts of non-Euclidean geometry, whose metric is briefly illustrated by the conformal model. Starting with simple geometric figures (triangle, lines, circle) and their properties, the volume advances to higher problems and figures in a manner that is convenient for the student and also whets his appetite. The always original developments use very simple tools (theorems of Ceva and Menelaos) and soon proceed to higher configurations (theorems of Pascal and Brianchon on the circle). The conics are obtained from circles as polar figures of the circles. The book is rich in remarkable facts and thereby is very effective in promoting the significance and the value of geometry in mathematical teaching, a promotion which is very necessary in view of today's predominance of set theory, analysis and algebra on the school and university level, and which deserves the skillful hand of distinguished scholars. An advantage in this recruiting endeavor is the high degree of visualizability of geometry, the easy comprehensibility of its problems and interesting theorems, and the challenge emanating from these problems to occupy oneself with their solutions. This purpose is also served by the numerous problems contained in the text whose solutions are listed at the end of the book. Many historical remarks are woven into the text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="242" LastActivityDate="2010-12-29T23:42:15.873" />
  <row Id="15863" PostTypeId="2" ParentId="15860" CreationDate="2010-12-29T23:57:57.240" Score="5" ViewCount="0" Body="&lt;p&gt;It's not possible for one gender to &quot;go extinct&quot;, as it were. That would happen if there were two different species whose rates of reproduction were different; that is, if girls produced only girls and boys produced only boys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, in each generation you expect 49% boys and 51% girls to be born. No matter what the initial population, once the first generation has died you will be left with an expected proportion of 49% males to 51% females. This holds whether or not the population is growing or fixed -- but only assuming the death rates for males and females are the same.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2010-12-29T23:57:57.240" />
  <row Id="15864" PostTypeId="2" ParentId="15857" CreationDate="2010-12-30T00:06:57.403" Score="4" ViewCount="0" Body="&lt;p&gt;If you are planning to self study and are just beginning geometry, I would recommend neither at this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometry Revisted is from a specific viewpoint: Visiting Elementary Geometry from the viewpoint of transformations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snapshot of the backcover of Geometry Revisited:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/KGQC9.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will miss out on some portions which are not covered by this book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to Geometry is an advanced book and presumes fairly high familiarity already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The books are excellent. I suggest you revisit Geometry Revisited &lt;em&gt;after&lt;/em&gt; you have done some amount of study and then perhaps take a shot at Introduction to Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having read it at some point myself, I can vouch for Geometry Revisited. It is an excellent book and ought be read at some point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your only choices to self learn were Geometry Revisited and Introduction to Geometry, then I would recommend Geometry Revisited.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T00:13:58.970" LastActivityDate="2010-12-30T00:13:58.970" />
  <row Id="15865" PostTypeId="1" CreationDate="2010-12-30T00:16:08.893" Score="2" ViewCount="129" Body="&lt;p&gt;For the solution of the cubic equation $x^3 + px + q = 0$ Cardano wrote it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this is ambiguous because it does not tell you which cube roots to match up. Why don't people write it this way today:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\frac{-p}{3\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is unambiguous.&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T01:31:25.427" LastActivityDate="2010-12-30T01:31:25.427" Title="Why not write the solutions of  a cubic this way?" Tags="&lt;algebra-precalculus&gt;&lt;algebra&gt;&lt;polynomials&gt;" CommentCount="6" />
  <row Id="15866" PostTypeId="1" AcceptedAnswerId="15873" CreationDate="2010-12-30T00:19:44.237" Score="7" ViewCount="123" Body="&lt;p&gt;I need to compute the intersection(s) between two circular arcs. Each arc is specified by its endpoints and their height. The height is the perpendicular distance from the chord connecting the endpoints to the middle of the arc. I use this representation because it is numerically robust for very slightly bent arcs, as well as straight line segments, for which the height is zero. In these cases, representing an arc using the center of its circle could lead to the center being far far away from the endpoints of the arc, and hence, numerically unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question at the highest level is how I would go about computing the intersection points, given that the centers of the circles of the arcs cannot necessarily be computed robustly. At a lower level, I am wondering if there is a parameterization of an arc using only the information I have stated above (which does not include the circle center). Of course, keep in mind numerical robustness is my principal concern here; otherwise I would just do the naive thing and compute the circle center for all non-linear arcs and hope for the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Formula for computing center of circle of arcs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the chord length is $2t$, and the height is $h$. The distance from the chord to the circle center is $c$, so that $r=h+c$. Then it follows that $c=(t^2-h^2)/2h$, which breaks down when $h$ is very small. Computing the location of the circle center is some simple vector arithmetic using the chord vector and its perpendicular.&lt;/p&gt;&#xA;" OwnerUserId="398" LastEditorUserId="398" LastEditorDisplayName="" LastEditDate="2010-12-30T01:26:05.417" LastActivityDate="2010-12-30T22:23:41.340" Title="arc-arc intersection, arcs specified by endpoints and height" Tags="&lt;geometry&gt;&lt;numerical-methods&gt;&lt;analytic-geometry&gt;&lt;computational-geometry&gt;&lt;intersection&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15867" PostTypeId="1" CreationDate="2010-12-30T00:21:19.133" Score="1" ViewCount="96" Body="&lt;p&gt;I am making a software that deals with employee trust. I am trying to make a math formula (no need to tell that I am bad in math :) ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the scenario:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Employee 1 trusts employee 2 = &amp;gt; 20% or 0.2 (average trust of employee 1 with employee 2)&#xA;Employee 2 deals with 10 customers =&amp;gt; 13%  (average - 10 customers trust on Employee 2)&#xA;Employee 1 also deals with 7 customers that customers belongs to Employee 2's 10 customers. =&amp;gt; 37% ( average 7 customers trust on employee 1)&#xA;&#xA;Employee 1 dealt with 10 customers for last 17 years.&#xA;Employee 2 dealt with 7 customers for last 1 years.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I want to calculate that how can we say that Employee 1 is much trust worthy than Employee 2? If I just see average than I can say Employee 1 but he only dealt with customers for 1 year. Now I want to make a generic formula to see which one is more trust worthy and I want to use all three values that are ( employee to employee, employee to customer, and number of year) to calculate average percentage. I want to increase or decrease an employee trust based on these values. Please remember I want to decrease as well for some employee I can not say that all employee or 90+ employee are trust worthy etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did:&#xA;I just simply plus employee to employee and employee to customer trust weight&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Employee 1 = 0.57&#xA;Employee 2 = 0.33&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't know what to do for years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current system just calculate employee to employee trust that is not good enough to trust itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not restricted to any programming language because I just want to make a generic formula. All ideas and suggestions are most welcome !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think my problem relates to any previously developed trust model then please do let me know (I already know web trust model but I don't know how to fit that in my situation)&lt;/p&gt;&#xA;" OwnerUserId="5223" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T01:07:25.573" LastActivityDate="2010-12-30T03:31:56.433" Title="Simple math formula to calculate average" Tags="&lt;mathematical-modeling&gt;" AnswerCount="1" />
  <row Id="15868" PostTypeId="1" CreationDate="2010-12-30T00:26:48.470" Score="2" ViewCount="93" Body="&lt;p&gt;[sorry for the bad english]&#xA;I am fond of astronomy and environment.&#xA;I wanna try to make a &quot;light pollution map&quot; but I haven't my satellites... so I use as approximation of light pollution the cities' population.&#xA;Let say we have for each city C citizens, each one spreads an average of X watt of electricity for lightning ( I have these data ).&#xA;Skip the units ( I need just a rough dimensionless &quot;light power&quot; ) :&#xA;city light power = C * X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a map, with many cities.&#xA;I know light power is inverse proportional to the distance^2.&#xA;I don't know about sky, air diffraction, clouds reflections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start from the simplest model. A flat terrain map.&#xA;N light sources, every one &#xA;   at position X(n), Y(n)&#xA;   has a specific &quot;total light power&quot; = C(n) * X(n) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a specific point of coordinates x,y wich is the light power, sum of all the cities light ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to calculate and plot, but it seems weird ( too far from some real satellite night shot ) and too slow to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help !&lt;/p&gt;&#xA;" OwnerUserId="5222" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T00:54:20.593" LastActivityDate="2010-12-31T18:21:27.527" Title="sum of N light sources" Tags="&lt;geometry&gt;&lt;physics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15869" PostTypeId="1" AcceptedAnswerId="15874" CreationDate="2010-12-30T00:27:07.223" Score="7" ViewCount="278" Body="&lt;p&gt;I've been reading On Numbers and Games and I noticed that Conway defines addition in his number system in terms of addition. Similarly in the analysis and logic books that I've read (I'm sure that this is not true of all such books) how addition works is assumed. From what I understand the traditional method of building the number system begins with the natural numbers (and zero)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0:=|\emptyset|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1:=|\{\emptyset\}|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2:=|\{\emptyset,\{\emptyset\}\}|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so forth. In this construction addition could(?) be defined as the disjoint union of the sets associated with the two numbers. Then the integers could be defined as the additive inverse and so forth. Is this the ideal way to do it though, is there a more elegant method?&lt;/p&gt;&#xA;" OwnerUserId="1930" LastActivityDate="2010-12-30T19:17:53.283" Title="How is addition defined?" Tags="&lt;logic&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="15870" PostTypeId="2" ParentId="15868" CreationDate="2010-12-30T00:51:30.757" Score="2" ViewCount="0" Body="&lt;p&gt;You may be confusing the energy distribution received by a satellite with the energy present at a place on the ground.  The  satellite is above the ground and records the amount of energy received as a function of direction (equivalently as a function of location on the ground) but doesn't add all the cities together.  So it would make a map with light at each city location measured as C*X/r^2.  This should match roughly with the photos you see, though different cities will have very different light/person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a location on the ground (in your model) you receive light from all other cities, again in the amount C*X/r^2 from each city, but there is no particular reason to expect this sum to match the satellite photos as the light is traveling along the ground, not up into space.  If you imagine a ring of cities in a circle, there would be a lot of light at the center, but the satellite would see nothing coming from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is surprising the calculation takes a long time.  How many locations are you calculating the power at?  Even if you have a few hundred cities in the list and are calculating at thousands of locations, that is a few million.  Did you get trapped by the curse of dimension and take a thousand points in each of the x and y directions, which would make hundreds of millions of calculations?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-30T00:51:30.757" />
  <row Id="15871" PostTypeId="2" ParentId="15869" CreationDate="2010-12-30T01:07:36.520" Score="8" ViewCount="0" Body="&lt;p&gt;Jacob, there are two ways of defining addition of natural numbers in set theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is the one you indicate: The sum of two numbers is the size of their disjoint union. In general, you define &lt;em&gt;cardinal addition&lt;/em&gt; this way: If $\kappa$ and $\lambda$ are cardinals (sizes of sets), say $\kappa=|A|$ and $\lambda=|B|$, then $\kappa+\lambda=|A\sqcup B|$, where $\sqcup$ denotes disjoint union. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way is to define &lt;em&gt;ordinal addition&lt;/em&gt;, considering natural numbers as ordered sets: $0=\emptyset$, $1=\{0\}$, $2=\{0,1\}$, etc, with $n=\{0,\dots,n-1\}$ ordered in the usual way, which actually corresponds to membership: $a&amp;lt;b$ iff $a\in b$. Here we identify two linearly ordered sets if they are order isomorphic. The relevant notion here is that of sum of ordered sets: If $(A,&amp;lt;)$ and $(B,\prec)$ are linear orders, their sum $A+B$ is the set $A\sqcup B$ ordered by $a$ is less than $b$ iff $a,b\in A$ and $a&amp;lt;b$, or $a,b\in B$ and $a\prec b$, or $a\in A$ and $b\in B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can check that these two notions coincide when $A$, $B$ are finite sets (i.e., when we look at $n+m$ for $n,m$ finite). One has to be careful, because the notions are different in general for infinite sets. For example, if $\aleph_0=|{\mathbb N}|$, then $\aleph_0+1=1+\aleph_0=\aleph_0$. However, if we think of $\aleph_0$ as the ordered set $\{0&amp;lt;1&amp;lt;\dots\}$, then $1+\aleph_0=\aleph_0$ (remember, we identify sets if they are order isomorphic). However, $\aleph_0+1$ is strictly larger than $\aleph_0$ as ordered sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a third option in the context of arithmetic: All we need is the notion of successor. The successor of 0 is 1, of 1 is 2, etc. Denote by $S(n)$ the successor of $n$ (in general, $S(n)=n+1$, of course). Note that &quot;successor of $n$&quot; is an order-theoretic notion: The smallest number larger than $n$. Then we define $n+m$ recursively: $n+0=n$, and $n+S(m)=S(n+m)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can go on, and define, for example, multiplication and exponentiation in these three ways as well. Cardinal multiplication is the size of the Cartesian product. Cardinal exponentiation $|A|^{|B|}$ is the size of the set of functions from $B$ to $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinal multiplication $\alpha\beta$ means: &quot;$\beta$ copies of $\alpha$&quot;. So $\aleph_02=\aleph_0+\aleph_0$ while $2\aleph_0=\aleph_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinal exponentiation is a bit trickier to define: For ordered sets $\alpha,\beta$ such that $\alpha$ has a minimum $0$, define $F(\alpha,\beta)$ as the set consisting of those functions $f:\beta\to\alpha$ such that there are only finitely many $\xi$ such that $f(\xi)\ne0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For functions $f,g$ in $F(\alpha,\beta)$ set $f\triangleleft g$ iff $f(\xi)&amp;lt;g(\xi)$ (in the order of $\alpha$) for $\xi$ largest (in the order of $\beta$) such that $f(\xi)\ne g(\xi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\alpha^\beta$ is defined as the order type of $(F(\alpha,\beta),\triangleleft)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, these notions coincide for finite numbers, but differ wildly for infinite sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recursive definitions are given by $n\cdot 0=0$ and $n\cdot S(m)=nm+n$, and $n^0=1$ (even if $n=0$) and $n^{S(m)}=n^m\cdot n$.&lt;/p&gt;&#xA;" OwnerUserId="462" LastActivityDate="2010-12-30T01:07:36.520" CommentCount="2" />
  <row Id="15873" PostTypeId="2" ParentId="15866" CreationDate="2010-12-30T01:11:11.620" Score="4" ViewCount="0" Body="&lt;p&gt;This is an interesting conundrum, and what I will suggest is just one possible approach, by no means a definitive answer.  Perhaps what you could do is compute Bézier curve segment approximating&#xA;your arcs, and then compute the intersection between Bézier segments.&#xA;To compute the Bézier segment for an arc,&#xA;you need the tangent vector at one endpoint (the tangent at the other endpoint is obtained by symmetry).&#xA;So this approach reduces to finding a tangent angle at one endpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the angle subtended by an arc at the circle center be $\theta$.&#xA;If the arc's chord length is $c$ and its height $h$, then, if I've calculated correctly,&#xA;$$\frac{\theta}{2} = \cos^{-1} \left( \frac{c^2 - 4h^2}{c^2+4h^2} \right) \;.$$&#xA;This avoids calculating the circle radius $r = (c^2 + 4h^2)/(8h)$.&#xA;Now from $\theta$ you could compute the needed tangent angle, and from there&#xA;a Bézier segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not analyzed this to see if it is indeed robust.&lt;/p&gt;&#xA;" OwnerUserId="237" LastActivityDate="2010-12-30T01:11:11.620" CommentCount="7" />
  <row Id="15874" PostTypeId="2" ParentId="15869" CreationDate="2010-12-30T01:25:06.027" Score="14" ViewCount="0" Body="&lt;p&gt;You don't have an a priori notion of cardinality, so you cannot really say things like &quot;$0:=|\emptyset|$&quot;. In fact, before you can define cardinals you usually define ordinals, and usually the definition of the naturals precedes the definition of the ordinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set-theoretic method begins by using the Axiom of Infinity, which states that there exists at least one inductive set; a set $X$ is &lt;em&gt;inductive&lt;/em&gt; if and only if (i) $\emptyset\in X$; and (ii) For all $x$, if $x\in X$, then $s(x) = x\cup\{x\}\in X$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let $X$ be any inductive set. Then we define&#xA;$$\mathbb{N} = \cap\{ A\subseteq X\mid \text{$A$ is inductive}\}.$$&#xA;One can then prove that $\mathbb{N}$ is well defined (it does not depend on the choice of $X$) and satisfies &quot;Peano's Axioms&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\emptyset \in\mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;If $x\in\mathbb{N}$, then $s(x) \in\mathbb{N}$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x\in\mathbb{N}$, $s(x)\neq\emptyset$.&lt;/li&gt;&#xA;&lt;li&gt;For all $x,y\in\mathbb{N}$, if $s(x)=s(y)$, then $x=y$.&lt;/li&gt;&#xA;&lt;li&gt;If $S\subseteq \mathbb{N}$ is inductive, then $S=\mathbb{N}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(In fact, 3 and 4 are just consequences of the definition of $s(x)$). We then define &quot;$0$&quot; to mean $\emptyset$, &quot;$1$&quot; to mean $s(0)$, &quot;$2$&quot; to mean $s(1)=s(s(0))$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can begin with Peano's Axioms. Here, there is a primitive notion called &quot;natural number&quot;, and a primitive symbol called $0$. We also have a primitive function $s$. The Peano Axioms would be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0$ is a natural number.&lt;/li&gt;&#xA;&lt;li&gt;If $n$ is a natural number, then $s(n)$ is a natural number.&lt;/li&gt;&#xA;&lt;li&gt;For all natural numbers $n$, $s(n)\neq 0$.&lt;/li&gt;&#xA;&lt;li&gt;For all natural numbers $n$ and $m$, if $s(n)=s(m)$, then $n=m$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Axiom Schema of Induction.&lt;/em&gt; If $\Phi$ is a predicate such that $\Phi(0)$ is true, and for all $n$, $\Phi(n)\Rightarrow \Phi(s(n))$, then for all natural numbers $k$, $\Phi(k)$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(You can also begin with $1$ instead of $0$; I use $0$ because it then parallels the set-theoretic construction). We then define &quot;$1$&quot; to mean $s(0)$; and &quot;$2$&quot; to mean $s(1)=s(s(0))$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then need the Recursion Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Recursion Theorem.&lt;/strong&gt; Given a set $X$, an element $a\in X$, and a function $f\colon X\to X$, there exists a unique function function $F\colon\mathbb{N}\to X$ such that $F(0)=a$ and $F(s(n)) = f(F(n))$ for all $n\in\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we have these definitions and theorem, we can start defining addition. Fix $n\in\mathbb{N}$. I'm going to define &quot;add $n$&quot;, $+_n\colon \mathbb{N}\to\mathbb{N}$ by letting&#xA;\begin{align*}&#xA;+_n(0) &amp;amp;= n,\\&#xA;+_n(s(m)) &amp;amp;= s(+_n(m)).&#xA;\end{align*}&#xA;Or, in usual notation,&#xA;\begin{align*}&#xA;n+0&amp;amp; = n,\\&#xA;n+s(m) &amp;amp;= s(n+m).&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these definitions, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $n\in\mathbb{N}$, $n+0=0+n=n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $S=\{n\in\mathbb{N}\mid n+0=0+n=n\}$. Note that $0\in S$, since $0+0 = 0$ by the definition of addition. Now assume that $k\in S$; that means that $k+0 = 0+k = k$. Then &#xA;$0+s(k) = s(0+k) = s(k)$ (first equality by the definition of addition with $0$, second by the induction hypothesis). And by the definition of addition with $s(k)$, we have $s(k)+0 = s(k)$. Therefore, $k\in S$ implies $s(k)\in S$. Thus, $S=\mathbb{N}$, as desired. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $n\in\mathbb{N}$, $s(n)=n+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $S=\{n\in\mathbb{N}\mid s(n)=n+1\}$. First, $0\in S$, since $s(0) = 1 = 0+1$, by the previous theorem. Assume that $k\in S$; that means that $s(k)=k+1$. Then $s(s(k)) = s(s(k)+0) = s(k)+s(0) = s(k)+1$. So $k\in S$ implies $s(k)\in S$, hence $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $\ell,n,m\in\mathbb{N}$, $\ell+(m+n) = (\ell+m)+n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Fix $\ell$ and $m$. Let $S=\{n\in\mathbb{N}\mid \ell+(m+n)=(\ell+m)+n\}$. We have $0\in S$, since&#xA;$$\ell + (m+0) = \ell + m = (\ell+m) + 0.$$&#xA;Now assume that $k\in S$; that means that $(\ell+m)+k = \ell+(m+k)$. We prove that $s(k)\in S$. We have:&#xA;$$(\ell+m)+s(k) = s((\ell+m)+k) = s(\ell+(m+k)) = \ell+s(m+k) = \ell+(m+s(k)).$$&#xA;Thus, if $k\in S$ then $s(k)\in S$. Hence, $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; For all $n\in\mathbb{N}$, $1+n = n+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $S=\{n\in\mathbb{N}\mid 1+n=n+1\}$. Then $0\in S$. Suppose that $k\in S$, so that $1+k = k+1 = s(k)$. Then we have:&#xA;$$1+s(k) = s(1+k) = s(k+1) = s(k+s(0)) = s(s(k+0)) = s(s(k)) = s(k)+1.$$&#xA;Thus, $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem.&lt;/strong&gt; For all $n,m\in\mathbb{N}$, $n+m=m+n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Fix $m$, and let $S=\{n\in\mathbb{N}\mid m+n=n+m\}$. First, $0\in S$, since $m+0=0+m$. Also, $1\in S$ by the previous lemma. Now assume that $k\in S$. Then $m+k=k+m$. To show that $s(k)\in S$, we have:&#xA;\begin{align*}&#xA;m+s(k) &amp;amp;= s(m+k) = s(k+m) = (k+m)+1 = k+(m+1) = k+(1+m)\\&#xA;       &amp;amp;= (k+1)+m = s(k)+m.&#xA;\end{align*}&#xA;Thus, $S=\mathbb{N}$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on. We can then define multiplication similarly, by fixing $n$ and defining&#xA;\begin{align*}&#xA;n\times 0 &amp;amp;= 0\\&#xA;n\times s(m) &amp;amp;= (n\times m) + n,&#xA;\end{align*}&#xA;and prove the usual properties of multiplication inductively. Then we can define exponentiation also recursively: fix $n$; then&#xA;\begin{align*}&#xA;n^0 &amp;amp; = 1\\&#xA;n^{s(m)} &amp;amp;= n^m\times n.&#xA;\end{align*}&#xA;We later define the order among the natural numbers by &#xA;$$a\leq b\Longleftrightarrow \exists n\in\mathbb{N}(a+n=b),$$&#xA;and prove the usual properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, we can construct $\mathbb{Z}$ from $\mathbb{N}$, $\mathbb{Q}$ from $\mathbb{Z}$, $\mathbb{R}$ from $\mathbb{Q}$, $\mathbb{C}$ from $\mathbb{R}$, etc. See for example my &lt;a href=&quot;http://math.stackexchange.com/questions/14828/set-theoretic-definition-of-numbers/14842#14842&quot;&gt;answer to this previous question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T19:17:53.283" LastActivityDate="2010-12-30T19:17:53.283" CommentCount="8" />
  <row Id="15875" PostTypeId="1" AcceptedAnswerId="15876" CreationDate="2010-12-30T02:00:23.237" Score="6" ViewCount="142" Body="&lt;p&gt;It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is &lt;em&gt;continuously&lt;/em&gt; differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then&lt;br&gt;&#xA;\begin{equation*}&#xA;u(x+h_1, y+h_2) - u(x,y) = \frac{du}{x} h_1 + \frac{du}{dy}h_2 + o(|h|)&#xA;\end{equation*}&#xA;and &#xA;\begin{equation*}&#xA;v(x+h_1, y+h_2) - v(x,y) = \frac{dv}{dx} h_1 + \frac{dv}{dy} h_2 + o(|h|).&#xA;\end{equation*}&#xA;Multiplying the second equation by $i$ and adding the two together gives &#xA;\begin{align*}&#xA;(u+iv)(z+h)-(u+iv)(z) &amp;amp;= \frac{du}{dx} h_1 + i \frac{dv}{dx} h_1 + \frac{du}{dy} h_2 + i \frac{dv}{dy} h_2 + o(|h|)\\&#xA; &amp;amp;= \left( \frac{du}{dx} + i \frac{dv}{dx} \right) (h_1+h_2i) + o(|h|).&#xA;\end{align*} &#xA;Now dividing by $h$ gives us the desired result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function?  &lt;/p&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T02:06:13.743" LastActivityDate="2010-12-30T02:37:18.010" Title="When is a function satisfying the Cauchy-Riemann equations holomorphic?" Tags="&lt;complex-analysis&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="15876" PostTypeId="2" ParentId="15875" CreationDate="2010-12-30T02:15:39.497" Score="6" ViewCount="0" Body="&lt;p&gt;See &lt;a href=&quot;http://www.jstor.org/pss/2321164&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-30T02:15:39.497" CommentCount="2" />
  <row Id="15877" PostTypeId="2" ParentId="15875" CreationDate="2010-12-30T02:23:21.447" Score="2" ViewCount="0" Body="&lt;p&gt;There's also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem&quot; rel=&quot;nofollow&quot;&gt;Looman–Menchoff theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T02:37:18.010" LastActivityDate="2010-12-30T02:37:18.010" CommentCount="3" />
  <row Id="15878" PostTypeId="1" AcceptedAnswerId="15879" CreationDate="2010-12-30T02:26:46.793" Score="5" ViewCount="136" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $f$ is differentiable at&#xA;  0, and that $f(0) = 0$. Prove that&#xA;  $f(x) = xg(x)$ for some function $g$&#xA;  which is continuous at 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a problem from Spivak's &lt;em&gt;Calculus&lt;/em&gt;, namely problem 27 of Chapter 10. (This is not homework, but rather self-study.) I am not sure how to go about this proof. The hint given in the text is to consider that $g(x)$ can be written as $f(x)/x$, but this puzzles me, because then continuity of $g$ at 0 says that $\lim_{x \to 0} g(x) = g(0) = f(0)/0 = 0/0$. &lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T02:32:02.060" LastActivityDate="2010-12-30T02:55:44.050" Title="Spivak's Calculus exercise. Chapter 10, Problem 27." Tags="&lt;calculus&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15879" PostTypeId="2" ParentId="15878" CreationDate="2010-12-30T02:28:06.137" Score="7" ViewCount="0" Body="&lt;p&gt;What is $\displaystyle \lim_{x \to 0} \frac{f(x) - f(0)}{x-0}$ ?&lt;/p&gt;&#xA;" OwnerUserId="1102" LastActivityDate="2010-12-30T02:28:06.137" CommentCount="4" />
  <row Id="15880" PostTypeId="2" ParentId="15878" CreationDate="2010-12-30T02:37:31.503" Score="0" ViewCount="0" Body="&lt;p&gt;Show that $f&amp;#39;(0) = g(0)$ from the the continuity of $g$. Note that $f&amp;#39;(0) = \lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h}$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-30T02:55:44.050" LastActivityDate="2010-12-30T02:55:44.050" />
  <row Id="15881" PostTypeId="1" CreationDate="2010-12-30T02:54:22.773" Score="10" ViewCount="180" Body="&lt;p&gt;(Motivation: I am going to be working with a high school student next week on long division, which is a subject I strongly dislike.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider:  $\frac{1110}{56}=19\frac{46}{56}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really a super easy problem, since once you realize $56*20=1120$ its trivial to write out $1110=56*19+46$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can work out the long division for yourself if you want; needless to say it makes an otherwise trivial problem into a tedious, multi-step process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long division is an &quot;effective procedure&quot;, in the sense that a &lt;a href=&quot;http://en.wikipedia.org/wiki/Turing_machine&quot; rel=&quot;nofollow&quot;&gt;Turing&lt;/a&gt; machine could do any division problem once it's given the instructions for the long division procedure.  To put it another way, an effective procedure is one for which given any problem of a specific type, I can apply this procedure systematically to this type of problem, and always arrive at a correct solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Are there other distinct effective procedures for doing division problems besides long division?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is there a way to measure how &lt;em&gt;efficient&lt;/em&gt; a given effective procedure is for doing division problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Does there exist an optimal effective procedure for division problems, in the sense that this procedure is the most efficient?&lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2010-12-30T03:13:39.970" LastActivityDate="2011-01-02T20:58:22.513" Title="Is long division the most optimal &quot;effective procedure&quot; for doing division problems?" Tags="&lt;arithmetic&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="15882" PostTypeId="2" ParentId="15881" CreationDate="2010-12-30T03:13:02.873" Score="3" ViewCount="0" Body="&lt;p&gt;1) yes, &lt;a href=&quot;http://en.wikipedia.org/wiki/Division_%28digital%29#Fast_division_methods&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Division_%28digital%29#Fast_division_methods&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) yes, &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_O_notation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Big_O_notation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) ?, This is an open problem.&lt;/p&gt;&#xA;" OwnerUserId="5114" LastActivityDate="2010-12-30T03:13:02.873" CommentCount="4" />
  <row Id="15883" PostTypeId="2" ParentId="15867" CreationDate="2010-12-30T03:31:56.433" Score="1" ViewCount="0" Body="&lt;p&gt;The solution to problems like this always starts with clear and detailed thinking about what data you have and what you want to do with it.  You seem to have 1) cross rating by employees of each other, 2) number of customers served, and 3)years of service.  One approach is to decide what weight you want to give each of these.  Say you want the cross rating to be 50% of the weight-you might scale it to a range of 0-50, scale the number of customers to 0-30 and scale experience to 0-20 and add up the points.  There are several issues here, however.  Maybe all your employees like each other and rate each other between 45 and 50.  Then it only contributes 5% of the grade, not 50.  Maybe one employee knows what you are doing and rates all the rest 0 so s/he will stand out.  Maybe one employee just grades harder than all the rest.  Is experience good, or do you want customers handled/year (in which case experience counts negatively)?  Maybe you want customers handled to &quot;decay&quot;-you get full credit for the ones this year, 80% for ones last year, and decreasing credit further in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are quite right this is independent of programming language-they can all do this, once we know what this is.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-30T03:31:56.433" CommentCount="3" />
  <row Id="15884" PostTypeId="1" AcceptedAnswerId="15886" CreationDate="2010-12-30T03:48:08.577" Score="5" ViewCount="51" Body="&lt;p&gt;I could swear I had a formula for this years ago in school, but I'm having trouble tracking it down. The problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 3 red balls and 3 black balls in a basket. I draw them out one at a time. How many different sequences of six balls can I get, i.e.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;rbrbrb&#xA;rrrbbb&#xA;bbbrrr&#xA;etc...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a general formula for n red and m black. This is not a homework, just my aging brain trying to recover an abstract formula.&lt;/p&gt;&#xA;" OwnerUserId="5227" LastActivityDate="2010-12-30T03:54:06.963" Title="Permutation with Duplicates" Tags="&lt;combinatorics&gt;&lt;permutations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="15885" PostTypeId="2" ParentId="15866" CreationDate="2010-12-30T03:50:53.597" Score="5" ViewCount="0" Body="&lt;p&gt;Have you considered finding the intersections using an implicit form for the circles, $$\frac{x^2}{r^2} + \frac{y^2}{r^2} + ax + by + c = 0?$$ This representation doesn't have any coefficients that diverge as the circle approaches a straight line. To find intersections, you'll have to solve a quadratic equation whose leading coefficient could be zero or arbitrarily close to it, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation#Alternative_formula&quot; rel=&quot;nofollow&quot;&gt;the alternative form of the quadratic formula&lt;/a&gt; should be able to deal with that robustly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll then have to do some jiggery-pokery to figure out whether the intersection points lie within the arcs. If the arc's bending angle is smaller than $\pi$, a projection onto the line joining the endpoints will suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Disclaimer: While all of this feels like it should work, I haven't analyzed it in any detail. Also, there could still be a problem when the circle is close to a line and you want the &lt;em&gt;longer&lt;/em&gt; arc. But I can't imagine that's a case that would turn up in any practical application.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; For a concrete example, here is the equation for a circular arc passing through the three points $(0,0)$, $(0.5, h)$, and $(1,0)$: $$\kappa^2 x^2 + \kappa^2 y^2 - \kappa^2 x - 2\eta y = 0,$$ where $$\begin{align}\kappa &amp;amp;= \frac{8h}{4h^2 + 1}, \\ \eta &amp;amp;= \frac{8h(4h^2-1)}{(4h^2+1)^2}.\end{align}$$ As you can see, the coefficients remain bounded as $h \to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt; Wait, that equation becomes trivial if $h = 0$, which is bad. We really want something like $x^2/r + y^2/r + ax + by + c,$  i.e. multiply the previous expression through by $r$. Then for the same example, our equation becomes $$\kappa x^2 + \kappa y^2 - \kappa x - 2\eta&amp;#39; y = 0,$$ where $\eta&amp;#39; = (4h^2-1)/(4h^2+1)$. Here are some explicit values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h = 1/2$: $$2 x^2 + 2 y^2 - 2 x = 0,$$ $h = 0.01$: $$0.07997 x^2 + 0.07997 y^2 - 0.07997 x + 1.998 y = 0,$$ $h = 0$: $$2 y = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, in this format, the linear terms will always be simply $-2(x_0/r)x$ and $-2(y_0/r)y$, where the center of the circle is at $(x_0,y_0)$. As the center goes to infinity but the endpoints remain fixed, these coefficients remain bounded and nonzero (i.e. not both zero).&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="856" LastEditorDisplayName="" LastEditDate="2010-12-30T22:23:41.340" LastActivityDate="2010-12-30T22:23:41.340" CommentCount="11" />
  <row Id="15886" PostTypeId="2" ParentId="15884" CreationDate="2010-12-30T03:54:06.963" Score="7" ViewCount="0" Body="&lt;p&gt;Place the $n+m$ balls in a row; the pick which $m$ you want to be red. You have $\binom{n+m}{m}=\frac{(n+m)!}{n!m!}$ possible ways of doing it, so that is the formula in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an alternative way of thinking about it: place the $n$ red in a row; now you need to decide where to insert the $m$ black. What you want to do is &lt;em&gt;choose&lt;/em&gt; the locations of the $m$ black balls; there are $n+1$ possible locations (before all the red balls, in the $n-1$ spaces between red balls, and after all the red balls). You want to pick those allowing repetitions, and without regard to the order in which you pick them (what matters is how many times you pick each gap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of ways in which you can select $r$ items from $k$ possibilities, allowing repetitions and without regard to order (combinations with repetitions) is $\binom{k+r-1}{r}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here you want to select $m$ out of $n+1$ possibilities, it gives $\binom{n+m}{m} = \frac{(n+m)!}{m!n!}$, same as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see a derivation of the formula, see for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Combinations#Number_of_combinations_with_repetition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on combinations&lt;/a&gt;. This is sometimes also called the &quot;stars and bars problem&quot;, so you can find a proof in the corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastActivityDate="2010-12-30T03:54:06.963" CommentCount="2" />
  <row Id="15887" PostTypeId="1" CreationDate="2010-12-30T04:26:28.060" Score="4" ViewCount="46" Body="&lt;p&gt;Can you please let me know if my &quot;proof&quot; is correct: ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ be a Hausdorff space and assume that each $y \in Y$ has a neighborhood $V$ such that $\overline{V}$ is regular. Prove $Y$ is regular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to show that $Y$ is regular we are going to show that for each $y \in Y$ and each open set $U$ which contains $y$ there exists an open neighborhood $W$ of $y$ such that $y \in W \subseteq \overline{W} \subseteq U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First observation: since regularity is hereditary then $V$ is regular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $y \in Y$ and $U$ a neighborhood of $y$. By assumption there exists a neighborhood $V$ of $y$ such that $\overline{V}$ is regular and hence $V$ is regular. Observe $U \cap V$ is a non-empty subset of $V$ and this set is open in $V$ as well. Since $V$ is regular we can find a non-empty subset $S$ of $V$ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S \subseteq \overline{S} \subseteq U \cap V \subseteq U$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since $V$ is open then $S$ is open in $X$ so the same set $S$ given by regularity of $V$ shows that $U$ is regular as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above incorrect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5229" LastActivityDate="2010-12-30T05:08:27.027" Title="Showing a space is regular by knowing each point has a nbd whose closure is regular" Tags="&lt;topology&gt;" AnswerCount="2" />
  <row Id="15888" PostTypeId="2" ParentId="15750" CreationDate="2010-12-30T04:47:52.143" Score="3" ViewCount="0" Body="&lt;p&gt;(This was intended to be yet another comment, but it got too big for the comment box.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I finally got a chance to look at the venerable &lt;a href=&quot;http://books.google.com/books?id=W6GV17SP0kQC&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Handbook of Elliptic Integrals for Engineers and Scientists&lt;/em&gt; by Byrd and Friedman&lt;/a&gt;, and there is indeed a short discussion there on pseudoelliptic integrals, which I shall quote in part:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One frequently encounters integrals that have all the appearances of being&#xA;  elliptic but which can ultimately be expressed solely in terms of&#xA;  elementary functions. These ... are called &lt;em&gt;pseudoelliptic integrals&lt;/em&gt;, and&#xA;  can always be evaluated as&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$v_0(t)+\sum_{j=1}^m b_j\ln\;v_j(t)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where the $b_j$ are constants and the $v_j(t)$ are algebraic functions. &lt;strong&gt;One&#xA;  need not, however, be too concerned at the outset about whether or not his&#xA;  integral at hand is genuinely elliptic, because the substitutions&#xA;  given in this book for the reduction and evaluation of elliptic integrals&#xA;  will lead to the right results (without additional labor) even though&#xA;  the integral might turn out to be elementary.&lt;/strong&gt; We give several important&#xA;  examples.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...the more general integral&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\int\frac{t\cdot R(t^2)}{\sqrt{at^4+bt^2+c}}\mathrm dt$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;($R(u)$ a rational function in $u$) ... reduces to integrals involving odd&#xA;  powers of Jacobi elliptic functions, and can thus always be ultimately&#xA;  evaluated in terms of elementary functions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A general case of a less obvious pseudoelliptic integral is&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\int \frac{R(t^2)}{\sqrt{(1-t^2)(1-k^2 t^2)}}\mathrm dt=\int \frac{R(\sin^2\theta)}{\sqrt{1-k^2\sin^2\theta}}\mathrm d\theta=\int R(\mathrm{sn}^2(u,k))\mathrm du$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;when... for all values of $t$ any one of the following relations holds:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\begin{cases}R(t^2)+R\left(\frac1{k^2 t^2}\right)&amp;amp;=0\\R(t^2)+R\left(\frac{1-k^2 t^2}{k^2(1-t^2)}\right)&amp;amp;=0\\R(t^2)+R\left(\frac{1-t^2}{1-k^2 t^2}\right)&amp;amp;=0\end{cases}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it's a handbook intended for use by nonspecialists, there's not much discussion of the conditions where an &quot;elliptic integral&quot; is in fact pseudoelliptic; I'd imagine advanced textbooks on the elliptic integrals would discuss the matter, but I have not been able to find any.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-30T04:47:52.143" CommentCount="2" />
  <row Id="15889" PostTypeId="2" ParentId="15887" CreationDate="2010-12-30T04:57:25.583" Score="1" ViewCount="0" Body="&lt;p&gt;There is a problem with your argument.  The closure of $S$ in $V$ might not be closed in $X$.  I can see why you went to $V$, because you wanted $S$ to be open.  However, if you work instead in $\overline V$, then the closure of $S$ will remain closed in $X$, and since $x$ is in the interior of $\overline V$, you'll still be able to find an open set that does what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to use the regularity of $\overline V$, and not just of $V$, because there are examples of nonregular spaces in which every point has a regular neighborhood, like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Hausdorff_manifold#Line_with_two_origins&quot; rel=&quot;nofollow&quot;&gt;line with 2 origins&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastActivityDate="2010-12-30T04:57:25.583" />
  <row Id="15890" PostTypeId="1" CreationDate="2010-12-30T05:00:13.047" Score="7" ViewCount="320" Body="&lt;p&gt;What books would you recommend to learn physics, being a a Math major, from classical mechanics, electricity, etc. to modern physics?&lt;/p&gt;&#xA;" OwnerUserId="1971" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-30T05:50:25.637" LastActivityDate="2011-01-02T19:22:30.967" Title="Books to learn physics, being a math major." Tags="&lt;soft-question&gt;&lt;reference-request&gt;&lt;physics&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="5" />
  <row Id="15891" PostTypeId="2" ParentId="15887" CreationDate="2010-12-30T05:08:27.027" Score="0" ViewCount="0" Body="&lt;p&gt;@Jonas Meyer: Thank you. I still have trouble and get confused using subspace topology of $\overline V$. Can you please help some more? (not homeowork, it is for fun).&lt;/p&gt;&#xA;" OwnerUserId="5230" LastActivityDate="2010-12-30T05:08:27.027" CommentCount="1" />
  <row Id="15892" PostTypeId="2" ParentId="15857" CreationDate="2010-12-30T05:28:39.890" Score="1" ViewCount="0" Body="&lt;p&gt;Thanks for your answers. Geometry Revisited it is. I already started to read it.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-30T05:28:39.890" CommentCount="1" />
  <row Id="15893" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T05:40:55.440" Score="2" ViewCount="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Kleppner/Kolenkow's  &lt;em&gt;An Introduction to Mechanics&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;David Morin's &lt;em&gt;Introduction to Classical Mechanics&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Purcell's &lt;em&gt;Electricity and Magnetism&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;A.P. French's &lt;em&gt;Principles of Modern Physics&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These books have more math than the typical introductory physics textbooks. But they also introduce the &quot;physicists&quot; way of thinking (e.g. how to gain physical intuition). &lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditorDisplayName="" LastEditDate="2010-12-30T05:47:38.430" LastActivityDate="2010-12-30T05:47:38.430" />
  <row Id="15894" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T05:45:38.117" Score="4" ViewCount="0" Body="&lt;p&gt;I will assume you are getting started - once you finish the equivalent of the first year of calculus-based physics, then you should dig in / branch out depending on your interests (optics, dynamics, modern physics - i.e. 20th century - E&amp;amp;M, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an overview of it all, I recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201021153&quot; rel=&quot;nofollow&quot;&gt;the Feynman Lectures on Physics&lt;/a&gt;, supplemented liberally with any of the solid textbooks (such as the following) and I suppose some Wikipedia.  I like the Feynman lectures because they are entertaining, conversational, and provide a great insight into physics.  Here are some example introductory texts - I can personally vouch for HLR.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471216437&quot; rel=&quot;nofollow&quot;&gt;Fundamentals of Physics - Halliday, Resnick, &amp;amp; Walker&lt;/a&gt;. Link is to current edition, but this is the book that many of us have used for decades, including myself at a small liberal arts school.  It's also the book used at Stanford.  I think you can use any edition from the last 10-20 years...&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534408427&quot; rel=&quot;nofollow&quot;&gt;Physics for Scientists and Engineers - Serway and Jewett&lt;/a&gt; (CalTech's intro book)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321500628&quot; rel=&quot;nofollow&quot;&gt;University Physics - Young, Hugh, Ford and Freedman&lt;/a&gt; (one of MIT's intro books)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Frankly, I recommend you go to the syllabus of the introductory, calculus-based physics from a school you respect and select that textbook as a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a physics major as an undergraduate, in addition to my studies in mathematics and engineering.  I still find that the first year's material is the most useful content, although I think that the concepts covered in introductory Modern Physics courses are really cool.  If you are interested, in addition to the Feynman lectures that I have read on the side, I was quite impressed with the text we used: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534493394&quot; rel=&quot;nofollow&quot;&gt;Modern Physics - Serway, Moses, Moyer&lt;/a&gt;.  While remaining very readable, it provides more of a textbook approach, whereas Feynman is a lecture style.&lt;/p&gt;&#xA;" OwnerUserId="4628" LastEditorUserId="4628" LastEditorDisplayName="" LastEditDate="2011-01-01T20:29:55.250" LastActivityDate="2011-01-01T20:29:55.250" CommentCount="4" />
  <row Id="15895" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T05:56:20.563" Score="5" ViewCount="0" Body="&lt;p&gt;As for mechanics I'd recommend:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stefan Banach - Mechanics (1951)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arnold VI - Mathematical Methods of Classical Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also recommend Griffiths books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Griffiths - Introduction to Electrodynamics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Griffiths - Introduction to Quantum Mechanics&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2010-12-30T05:56:20.563" CommentCount="1" />
  <row Id="15896" PostTypeId="1" CreationDate="2010-12-30T06:06:41.387" Score="3" ViewCount="47" Body="&lt;p&gt;I'm trying to figure out a generic way of determining a series of points on a Bézier curve where all the points are the same distance from their neighboring points. By distance I mean direct distance between the points not distance along the curve. See the image below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/8PXtm.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've written a program that will solve this in an iterative fashion, but I'm wondering if there is a direct solution I could use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My program starts by defining a circle of some initial radius (R) centered on the start point of the curve. It then intersects this circle with the curve to find the second point , which is R distance away from the start point). It then continues along the curve in this way finding points until it reaches the end of the curve. In most cases the distance between the last intersection point and the end point of the curve will not equal to R. The program then uses that difference to calculate a new value of R to try, and repeats the process.&lt;/p&gt;&#xA;" OwnerUserId="5029" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T06:18:47.933" LastActivityDate="2010-12-30T06:18:47.933" Title="Find points along a Bézier curve that are equal distance from one another" Tags="&lt;geometry&gt;&lt;bezier-curve&gt;" CommentCount="4" />
  <row Id="15897" PostTypeId="1" CreationDate="2010-12-30T06:16:48.620" Score="1" ViewCount="48" Body="&lt;p&gt;If we consider the unit square (i.e $[0,1] \times [0,1]$) with the lexicographic&#xA;order induced by ${\mathbb{R}}^{2}$, what are the limit points of the unit square?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is: the set of limit points is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ [0,1]^{2} \setminus ((0,1) \times \{1\}))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my answer correct? If not, how do you find them? I get confused with this topology.&lt;/p&gt;&#xA;" OwnerUserId="5230" LastActivityDate="2010-12-30T09:28:10.357" Title="Finding limit points in lexicographic order topology" Tags="&lt;topology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15898" PostTypeId="1" AcceptedAnswerId="16009" CreationDate="2010-12-30T06:28:40.873" Score="2" ViewCount="186" Body="&lt;p&gt;I define the function $D(\mathbf M)$ for square matrices by taking a maximal set of all triples  $(\mathbf v_i,k_i,n_i)_{i \in I}$ that satisfy $(\mathbf{M}-k\mathbf{I})^n\mathbf{v}_i = \mathbf{0}$ with the $\mathbf{v}_i$ linearly independent, and letting $$D(\mathbf M) = \prod_{i \in I} k_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf X=\begin{pmatrix}0&amp;amp;1&amp;amp;1\\1&amp;amp;0&amp;amp;1\\1&amp;amp;1&amp;amp;0\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the set:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$((-1\quad 0\quad 1)^T,-1,1)$&lt;/li&gt;&#xA;&lt;li&gt;$((-1\quad 1\quad 0)^T,-1,1)$&lt;/li&gt;&#xA;&lt;li&gt;$((1\quad 1\quad 1)^T,2,1)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so $D(\mathbf X) = -1 \cdot -1 \cdot 2 = 2$. It is equal to the determinant but I don't want to use that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How can it be shown that $D(\mathbf A\mathbf B) = D(\mathbf A)D(\mathbf B)$?&lt;/p&gt;&#xA;" OwnerUserId="4290" LastEditorUserId="4290" LastEditorDisplayName="" LastEditDate="2010-12-31T03:20:07.590" LastActivityDate="2010-12-31T19:45:35.830" Title="Proof of multiplicativity" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="16" />
  <row Id="15899" PostTypeId="1" AcceptedAnswerId="15900" CreationDate="2010-12-30T06:32:32.243" Score="1" ViewCount="239" Body="&lt;p&gt;What's the general form for $x^y$?&#xA;What's the specialized form for $x^1$ and $x^0$?&lt;/p&gt;&#xA;" OwnerUserId="1080" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T06:36:49.960" LastActivityDate="2011-01-03T03:00:47.787" Title="Square for $x^2$, Cube for $x^3$, Quartic for $x^4$, and what's for $x^1$?" Tags="&lt;algebra-precalculus&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="15900" PostTypeId="2" ParentId="15899" CreationDate="2010-12-30T07:40:34.513" Score="5" ViewCount="0" Body="&lt;p&gt;As J.M. and Rahul Narain pointed out in comments, $x^1$ is often referred to as &quot;linear&quot; and $x^0$ as &quot;constant.&quot;  Commonly, though certainly not universally, $x^y$ is a/the &quot;power&quot; with $x$ as the base and $y$ as the exponent.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-30T07:40:34.513" CommunityOwnedDate="2010-12-30T07:40:34.513" />
  <row Id="15901" PostTypeId="2" ParentId="15897" CreationDate="2010-12-30T09:28:10.357" Score="1" ViewCount="0" Body="&lt;p&gt;Every point $P = (a,b)$ on the square lies on some horizontal line segment $y = b$.  If $a \neq 0$ -- i.e., if $P$ is not the leftmost point on its horizontal line segment -- then you can build a sequence of points $(x_n,b)$ to the left of $a$ on that line segment converging to $P$.  If $a \neq 1$ -- i.e., if $P$ is not the rightmost point on its horizontal line segment -- then you can build a sequence of points $(x_n,b)$ to the right of $a$ on that line segment converging to $(a,b)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously $P$ can't be both the leftmost point &lt;em&gt;and&lt;/em&gt; the rightmost point, so the conclusion is...&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-30T09:28:10.357" />
  <row Id="15902" PostTypeId="1" AcceptedAnswerId="15910" CreationDate="2010-12-30T09:30:53.300" Score="14" ViewCount="441" Body="&lt;p&gt;This an interesting problem my friend has been working on for a while now (I just saw it an hour ago, but could not come up with anything substantial besides some PMI attempts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the full problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x_{1}, x_{2}, x_{3}, \cdots x_{y}$ be all the primes that are less than a given $n&amp;gt;1,n \in \mathbb{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that $$x_{1}x_{2}x_{3}\cdots x_{y} &amp;lt; 4^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas very much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-31T11:55:46.077" LastActivityDate="2011-01-02T09:46:20.843" Title="Show that product of primes, $\prod_{k=1}^{\pi(n)} p_k &lt; 4^n$ " Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="5" />
  <row Id="15903" PostTypeId="1" AcceptedAnswerId="15929" CreationDate="2010-12-30T09:44:14.633" Score="5" ViewCount="187" Body="&lt;p&gt;&lt;strong&gt;From &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0136020402&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Introductory Combinatorics&lt;/em&gt;&lt;/a&gt; by Richard Brualdi&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a chess master. He has 11 weeks to prepare for a competition so he decides that he will practice everyday by playing at least 1 game a day. To make sure that he's able to take a couple of breaks, he also decides that he wont play more than 12 games per week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how do you prove/dis-prove that there will be a succession of (consecutive) days in which he will play &lt;em&gt;k&lt;/em&gt; games, &lt;em&gt;k&lt;/em&gt; &gt;= 22?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I don't quite understand the use of the pigeonhole principle here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance! :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to clarify what I'm asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to prove that for &lt;strong&gt;every possible practice schedule&lt;/strong&gt; that the chess master makes, there will &lt;strong&gt;&lt;em&gt;always&lt;/em&gt;&lt;/strong&gt; be a consecutive sequence of days such that he plays &lt;strong&gt;exactly&lt;/strong&gt; &lt;em&gt;k&lt;/em&gt; games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, if &lt;em&gt;k&lt;/em&gt; = 22, how do you prove that no matter what his schedule is like, he will &lt;strong&gt;always&lt;/strong&gt; play &lt;strong&gt;exactly&lt;/strong&gt; &lt;em&gt;k&lt;/em&gt; games in &lt;em&gt;r&lt;/em&gt; days, 1 &amp;lt;= &lt;em&gt;r&lt;/em&gt; &amp;lt;= 77.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main question is: what is the general method for proving different values of &lt;em&gt;k&lt;/em&gt;? Are there values of &lt;em&gt;k&lt;/em&gt; (from 1 to 132 inclusive) that are impossible to obtain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that he does not necessarily have to play 132 games within 77 days!&#xA;Also, keep in mind the bounds of &lt;em&gt;k&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5166" LastEditorUserId="5166" LastEditorDisplayName="" LastEditDate="2010-12-30T17:28:56.323" LastActivityDate="2010-12-30T19:28:23.440" Title="Chess Master Problem" Tags="&lt;combinatorics&gt;&lt;pigeonhole-principle&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="15904" PostTypeId="2" ParentId="15902" CreationDate="2010-12-30T09:54:26.723" Score="2" ViewCount="0" Body="&lt;p&gt;How do I comment on here?  Must the primes be distinct?&lt;/p&gt;&#xA;" OwnerUserId="5237" LastActivityDate="2010-12-30T09:54:26.723" CommentCount="2" />
  <row Id="15905" PostTypeId="2" ParentId="15869" CreationDate="2010-12-30T10:02:32.090" Score="1" ViewCount="0" Body="&lt;p&gt;An extremely elegant way of defining numbers I want to mention does not use &lt;em&gt;sets&lt;/em&gt; but &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-calculus&quot; rel=&quot;nofollow&quot;&gt;lambda calculus&lt;/a&gt;, i.e. functions and function application. The system used there is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Church_encoding&quot; rel=&quot;nofollow&quot;&gt;Church encoding&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea goes the following: &lt;em&gt;Two&lt;/em&gt;, for example, means doing something for two times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, when we have some operation (a function) and a value, we apply this function twice on this value. In lambda notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 \equiv \lambda f\,x \mapsto f (f\,x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So generally, any number $n$ is defined as a function that takes another function and returns it's $n$th &lt;a href=&quot;http://en.wikipedia.org/wiki/Function_iteration&quot; rel=&quot;nofollow&quot;&gt;iterate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can simply define addition &lt;strong&gt;as a function composition&lt;/strong&gt;.&#xA;We first apply the function $n$ times, then $m$ times and thus we got a total of $n+m$ applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n + m \equiv \lambda f\,x \mapsto n f (m f x)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we end up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 + 3 \equiv \lambda f\,x \mapsto f (f (f (f (f\,x)))) \equiv 5 $$&lt;/p&gt;&#xA;" OwnerUserId="917" LastActivityDate="2010-12-30T10:02:32.090" />
  <row Id="15906" PostTypeId="2" ParentId="15903" CreationDate="2010-12-30T10:13:39.507" Score="1" ViewCount="0" Body="&lt;p&gt;Going clause-by-clause is a useful step to start things up. You know that he &lt;strong&gt;has&lt;/strong&gt; to play a game a day, so you know that the minimum amount of games per week will be 7. You also know that the maximum amount is 12. So, after distributing a game per every day of the week, you are left with 5 'extra' games. The way to maximize the number of games played per a day of the week, then, is to play all of those 5 extra games in the same day (in addition to 1 game that he &lt;strong&gt;has&lt;/strong&gt; to play). So that gives you 6 games per day. If you consider how many games one could play next day, then, it would be 1, as you can't have any 'extras' allotted to that day of the week, unless the next day is Monday, as that would be a &lt;strong&gt;new week&lt;/strong&gt; (and hence, a renewed quota). So, if you play 6 games on Sunday, and 6 games on Monday, you will max out the number of the games that can be played in two consecutive days (it will be 12). If by consecutive you meant two consecutive days, then the statement about $k\geq 22$ is disproved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the extra information (11 weeks) etc. I think you are looking for a different (not two) consecutive amount of days. Again, let's look at the minimum and maximum number of games he could play throughout the 11 weeks. 11 weeks = 77 days, so the minimum of the games he will play is 77, maximum -- 11*12 = 132. So, if you, again, were to distribute the games he &lt;strong&gt;has&lt;/strong&gt; to play, you get one per every day. If you are talking about &lt;strong&gt;any&lt;/strong&gt; amount of consecutive days, then let's take 22 -- since he has to play a game a day, he will clock up 22 games after 22 days. I think you are looking for something more interesting, which will require establishing a certain limit on the number of consecutive days (though even then, having to disprove that he will &lt;strong&gt;have&lt;/strong&gt; to play $\geq 22$ games in that span seems to be trivial as you can always (1) pick the minimum number of games (1 per day) to show that he won't &lt;strong&gt;have to&lt;/strong&gt; reach 22 for any number of successive days $&amp;lt;22$, or (2) use the same minimum-of-games-per-day rule to show that for any number of consecutive days $\geq 22$ he will have to play at least 22 games). Though I am sleepy, and might be missing something in the problem. :)&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="3731" LastEditorDisplayName="" LastEditDate="2010-12-30T10:20:53.947" LastActivityDate="2010-12-30T10:20:53.947" />
  <row Id="15907" PostTypeId="2" ParentId="15902" CreationDate="2010-12-30T10:15:55.980" Score="1" ViewCount="0" Body="&lt;p&gt;Note my comment above, also when you look a bit at the net you directly find proofs for this theorem. The easiest ones probably use induction, it is basically like this proof:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathrefresher.blogspot.com/2009/11/primorial.html&quot; rel=&quot;nofollow&quot;&gt;http://mathrefresher.blogspot.com/2009/11/primorial.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2010-12-30T10:15:55.980" />
  <row Id="15908" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T11:51:45.237" Score="4" ViewCount="0" Body="&lt;p&gt;If you want a solid and consistent approach to physics from the theoretical point of view (I assume that as a mathematician you don't want an experimentalist's point of view) then you can't go past the series by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lev_Landau#Landau_and_Lifshitz_Course_of_Theoretical_Physics&quot; rel=&quot;nofollow&quot;&gt;Landau and Lifshitz&lt;/a&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Course_of_Theoretical_Physics&quot; rel=&quot;nofollow&quot;&gt;&quot;Course of Theoretical Physics&quot;&lt;/a&gt; (&lt;a href=&quot;http://www.amazon.com/Course-Theoretical-Physics-All-Volumes/lm/RMVH4G2YKW3NH&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mechanics.&lt;/li&gt;&#xA;&lt;li&gt;The Classical Theory of Fields.&lt;/li&gt;&#xA;&lt;li&gt;Quantum Mechanics: Non-Relativistic Theory.&lt;/li&gt;&#xA;&lt;li&gt;Relativistic Quantum Theory.&lt;/li&gt;&#xA;&lt;li&gt;Statistical Physics.&lt;/li&gt;&#xA;&lt;li&gt;Fluid Mechanics.&lt;/li&gt;&#xA;&lt;li&gt;Theory of Elasticity.&lt;/li&gt;&#xA;&lt;li&gt;Electrodynamics of Continuous Media.&lt;/li&gt;&#xA;&lt;li&gt;Statistical Physics, Part 2.&lt;/li&gt;&#xA;&lt;li&gt;Physical Kinetics.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="954" LastActivityDate="2010-12-30T11:51:45.237" CommentCount="6" />
  <row Id="15909" PostTypeId="2" ParentId="15425" CreationDate="2010-12-30T12:01:40.040" Score="2" ViewCount="0" Body="&lt;p&gt;I assumed above that all coefficients are positive, but this isn't easiest to compute with, so apply the isomorphism $x \mapsto -x$, and now the coefficients are alternating. The volume is unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basis for the differentials is $\{\omega_i : 1\le i \le g\}$, where&#xA;$$\omega_i = x^{i-1}\frac{dx}{2y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $M$ large, there is a positive root $\alpha_1$ of size $M$, and the other $2g$ roots are small. We now need to figure out a basis for the real homology (loops fixed by complex conjugation). There are $g-1$ independent loops $\gamma_{j+1}$, the $j$-th loop being from $\alpha_{2j}$ to $\alpha_{2j+1}$ and back around the &quot;hole&quot; (for example, if all roots are real, these are simply the paths where the polynomial is positive). One more independent loop is from $\alpha_1$ to $\infty$ and back (being real, this is the easiest to understand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the jacobian of the curve is $|det((\int_{\gamma_j} \omega_i)_{ij})|$. So let's approximate these integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any $j&amp;gt;1$, since the roots involved are $O(1)$, the loop $\gamma_j$ is close to the origin. In this vicinity the root $\alpha_1$ contributes to $\int_{\gamma_j} \omega_i$ a factor of $M^{-1/2}$, and if we factor it out (recall that we only want a crude approximation), we are left with an integral close to the origin over a function that is relatively small. Hence $\int_{\gamma_j} \omega_i \approx M^{-1/2}$. Note that this is the same for any $i$ since close to the origin $x^{i-1}$ doesn't add much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $j=1$, the integral can be approximated as&#xA;$$\int_{\gamma_1} \omega_i = 2\int_M^\infty \frac{x^{i-1} dx}{2y} \approx \int_M^{2M} \frac{x^{i-1} dx}{\sqrt{x^{2g+1}+\cdot a_0}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small roots give each the denominator a factor of $M^{1/2}$, so we can approximate further to arrive at:&#xA;$$M^{-g} \int_M^{2M} \frac{x^{i-1} dx}{\sqrt{x-M}} \approx M^{1/2+i-1-g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have approximated all entries of the matrix. The top row's largest coefficient is of size $M^{-1/2}$, and hence the largest product of a generalised diagonal is $(M^{-1/2})^g$, since the entries in the other rows are also of size $M^{-1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crude approximation for the volume of the jacobian of a hyperelliptic curve given by a polynomial as above with positive coefficients of size $M$, a large number, is: $$Vol(J(\mathbb{R})) \approx M^{-g/2}$$&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditorDisplayName="" LastEditDate="2010-12-30T21:02:23.683" LastActivityDate="2010-12-30T21:02:23.683" />
  <row Id="15910" PostTypeId="2" ParentId="15902" CreationDate="2010-12-30T12:39:44.623" Score="34" ViewCount="0" Body="&lt;p&gt;I think the following argument is from Erdős: The binomial coefficient &#xA;$$ {n \choose {\lfloor n/2 \rfloor}} = \frac{n!}{{\lfloor n/2 \rfloor}!{\lceil n/2 \rceil}!}$$&#xA;is an integer. Any prime in the range ${\lceil n/2 \rceil}+1 \le p \le n$ will appear in the  numerator but not in the denominator, as a consequence $n \choose {\lfloor n/2 \rfloor}$ is divisible by the product of all the primes in the range ${\lceil n/2 \rceil}+1 \le p \le n$. On the other hand if $n$ is even then it will be the central term in the binomial expansion of $(1+1)^n$ so &#xA;    $$ {n \choose {\lfloor n/2 \rfloor}} \le 2^n $$ &#xA;and if $n$ is odd then $n \choose \lfloor n/2 \rfloor$ and $n \choose \lceil n/2 \rceil$ will be the two central terms of the binomial expansion of $(1+1)^n$, as they are equal we have&#xA;    $$ {n \choose {\lfloor n/2 \rfloor}} \le 2^{n-1} $$ &#xA;we apply this recursively to $n$, $\lceil n/2 \rceil, \lceil n/4 \rceil, \dots$, but&#xA; $$ { \lceil n/2 \rceil \choose \lfloor \lceil n/2 \rceil/2 \rfloor } =   &#xD;&#xA;    { \lceil n/2 \rceil \choose \lfloor n/4 \rfloor } \le 2^{\lfloor n/2 \rfloor } $$&#xA;using either of the two preceding inequalities depending on the parity of $\lceil n/2 \rceil$, by the same argument we get&#xA;  $${ \lceil n/4 \rceil \choose \lfloor \lceil n/4 \rceil/2 \rfloor } ={ \lceil n/4 \rceil \choose \lfloor n/8 \rfloor } \le 2^{\lfloor n/4 \rfloor }$$&#xA;and so on, so &#xA;$$ \begin{align}{n \choose {\lfloor n/2 \rfloor}}{{\lceil n/2 \rceil} \choose {\lfloor n/4 \rfloor}}{{\lceil n/4 \rceil} \choose {\lfloor n/8 \rfloor}}\cdots  &amp;amp;\le 2^n\cdot 2^{{\lfloor n/2 \rfloor}} \cdot 2^{{\lfloor n/4 \rfloor}} \cdots \\\\&#xD;&#xA; &amp;amp;\le 2^{n + {\lfloor n/2 \rfloor} + {\lfloor n/4 \rfloor} + \cdots } \\\\ &#xD;&#xA; &amp;amp;\le 2^{n(1 + 1/2 + 1/4 + \cdots) } = 2^{2n} = 4^n\end{align}$$&#xA;but the left hand side is divisible by all the primes up to $n$. So the product of any subset of these primes will also be bounded by $4^n$.&lt;/p&gt;&#xA;" OwnerUserId="3274" LastEditorUserId="3274" LastEditorDisplayName="" LastEditDate="2010-12-31T10:44:19.510" LastActivityDate="2010-12-31T10:44:19.510" CommentCount="6" />
  <row Id="15911" PostTypeId="2" ParentId="15773" CreationDate="2010-12-30T12:43:15.860" Score="1" ViewCount="0" Body="&lt;p&gt;A group-theoretic proof of the simplicity of $A_n$ for $n&amp;gt;4$ can be found in the books &lt;strong&gt;Finite Groups&lt;/strong&gt; by Michael Aschbacher or &lt;strong&gt;Theory of Finite Groups&lt;/strong&gt; by Hans Kurzweil and Bernd Stellmacher. Like the proof given by LIE one first proves that $A_5$ is simple, but then one proceeds by induction on $n$. The main difference is that the arguments are based on the $(n-2)$-transitivity of the natural action of $A_n$ and two more general, easily proved theorems for primitive permutation groups (recall that $2$-transitive implies primitive) instead of direct calculations with permutations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 1&lt;/strong&gt; A non-trivial normal subgroup of a primitive permutation group acts transitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The set $\omega\cdot N = \{\omega\cdot n\mid n \in N\}$ is a block for any $\omega \in \Omega$ and any normal subgroup $N$ of a transitive group on $\Omega$, hence $\omega\cdot N = \Omega$ by primitivity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theorem 2&lt;/strong&gt; The only $4$-transitive permutation group with a (proper) &lt;em&gt;regular&lt;/em&gt; normal subgroup is $S_4$ (in its natural action).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A subgroup $H$ of a permutation group $G$ is called &lt;em&gt;regular&lt;/em&gt; if the action restricted to $H$ is equivalent to the action of $H$ on itself by right multiplication. If additional $H$ is normal in $G$, then the action of a point stabilizer $G_\omega$ on $\Omega\setminus\{\omega\}$ is equivalent to the action of $G$ on $H\setminus\{1\}$ by conjugation. Take an elements $x \in H\setminus\{1\}$, an element $y \in H\setminus\{1, x, x^{-1}\}$ and set $z = xy$ to get three different elements of $H\setminus\{1\}$. By the $3$-transitivity of conjugation action of $G$ on $H\setminus\{1\}$ we conclude that $z&amp;#39;=x&amp;#39;y&amp;#39;$ holds for &lt;em&gt;any&lt;/em&gt; three different elements of $H\setminus\{1\}$ which easily implies $H \cong V_4 \cong Z_2\times Z_2$ and therefore $G \cong S_4$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simplicity of $A_5$ one observes that by Sylow's theorem any group of order $5$, $10$, $15$, $20$ or $30$ has a unique (hence characteristic) $5$-Sylow subgroup (For the case $|G| = 30$ observe that a group of order $30$ has a normal subgroup of index $2$, as an element of order $2$ is as product of $15$ transpositions not contained in the alternating group when considering the regular action of $G$ on itself. Then use the statement for groups of order $15$).&#xA;A proper normal subgroup $N$ of $A_5$ acts transitively on a $5$-element set by Theorem 1, hence its order is a multiple of $5$. Therefore $A_5$ has a unique $5$-Sylow subgroup which contradicts the fact that $A_5$ contains more than four $5$-cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the induction step assume $A_{n-1}$ to be simple for $n&amp;gt;5$. A normal subgroup $N \ne 1$ of $A_n$ is transitive by Theorem 1. As $N \cap A_{n-1}$ is normal in $A_{n-1}$, it is either $1$ or $A_{n-1}$ by induction. As $A_n$ is primitive, $A_{n-1}$ is as point stabilizer a maximal subgroup of $A_n$. Hence $N \cap A_{n-1} = 1$, and $N$ is a regular normal subgroup of $A_n$ contradicting Theorem 2.&lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2010-12-30T12:43:15.860" />
  <row Id="15912" PostTypeId="2" ParentId="15849" CreationDate="2010-12-30T13:17:01.220" Score="3" ViewCount="0" Body="&lt;p&gt;I thouht I would add an extra answer that goes in a slightly different direction than the rest.&#xA;An alternative proof that a finite domain is a division ring goes as follows:&#xA;Look at different powers of $a$ (ie, $a^n$ for various positive numbers $n$). Now, since the ring is finite, you will at some point get to some element you have already been at before, so you have $a^n = a^m$ with $n &amp;lt; m$. Now we can use that it is a domain, so this leads to $a^{m-n} = 1$ which means that not only does $a$ have an inverse. his inverse is in fact a power of $a$ (and then clearly it commutes with $a$ so it is also an inverse from the other side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is less elegant than the usual one, but I find it is nice to know that the inverse of an element in a finite ring is actually a power of the element itself.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastActivityDate="2010-12-30T13:17:01.220" />
  <row Id="15913" PostTypeId="1" CreationDate="2010-12-30T13:33:58.993" Score="0" ViewCount="84" Body="&lt;p&gt;What is the general advantage of a real sequence?&lt;/p&gt;&#xA;" OwnerUserId="5241" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-30T13:40:42.217" LastActivityDate="2010-12-30T13:40:42.217" ClosedDate="2010-12-30T17:00:55.820" Title="General advantage of a real sequence" Tags="&lt;real-analysis&gt;" CommentCount="5" />
  <row Id="15914" PostTypeId="2" ParentId="15881" CreationDate="2010-12-30T14:05:21.040" Score="4" ViewCount="0" Body="&lt;p&gt;Methods for division and the other basic arithmetic operations are explained in Knuth's ACP vol. 2, Seminumerical Algorithms.  For the sake of humans doing longhand division, perhaps the most relevant point is the discussion of trial divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you will recall, when faced with a problem like $1110/56$ (asked to find the quotient and remainder, which amounts to giving the fraction in mixed form), the basic instruction is to use the leading digit of the divisor to &quot;guess&quot; (estimate) the next digit of the quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often the case that this &quot;guess&quot; will be an overestimate, i.e. that the trial divisor proves to be too big for the divisor as a whole.  For example, here when trying to tell how many times the divisor as a whole $56$ will go into the leading digits $111$ of the dividend, we would &quot;guess&quot; $2$ times, because leading digit $5$ certainly goes into $11$ two times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turns out, frustratingly, to be a wrong guess in that $56$ times two gives $112$ (over the limit of $111$), and our trial divisor has to be reduced by one and checked again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knuth's explanation covers arbitrary (fixed positive integer) radix &lt;em&gt;b&lt;/em&gt; &amp;gt; 1.  Reading down to the radix ten (decimal) case, the point would be this.  If the divisor's leading digit is &quot;normalized&quot; to be greater than  or equal to half the radix (at least 5 in the decimal case, as was true in the example $1110/56$), then the trial divisor obtained in the usual way (estimating by using the leading digit of the divisor) will be at worst $2$ more than the correct next digit in the quotient.  [It is suggested that the desired normalization can be arranged by repeatedly doubling both numerator and denominator as necessary, until the leading digit of the denominator (divisor) reaches at least $b/2$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative method for getting the trial divisor is then explained that has a better worst case behavior.  If the next-to-leading digit of the divisor is at least $b/2$, then for the purpose of the trial divisor we round up the leading digit!  Now using that digit (or possible $b=10$ if the rounding takes us to the next &quot;place&quot; value) we will get an estimate for the next digit of the quotient that is off by at most $1$.  In the case where no rounding up took place, this means the trial divisor is either correct or too big by one, and in the case where we did round up, this means the trial divisor is either correct or too small by one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen a monograph where this alternative technique for longhand division was tried with a group of (if memory serves) high school math teachers, as a Master's Thesis in Education that someone did.  As it happens I'm in St. Augustine FL where I saw this monograph years ago in a used bookstore.  I'll go by there today and if it remains on the shelf post the reference details later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt;  I found the book (the bookstore itself had moved about five years ago!).  My memory of it was not perfect but here are some details:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;An Experimental Study of Two Methods of Long Division&lt;/em&gt; by Kenneth G. Fuller&lt;br&gt;&#xA;Bureau of Publications/Teachers College/Columbia University&#xA;New York, 1949 (hardback, 76 pages)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author refers in his acknowledgments to Prof. Clifford B. Upton, and in an early footnote refers to a paper by the same:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Making Long Division Automatic,&quot; &lt;em&gt;Tenth Yearbook, National Council of Teachers of Mathematics&lt;/em&gt;, 1935  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the percentages of correcting estimated quotients using a) a trial divisor that consists simply of a leading digit, and b) a trial divisor that is rounded up just when the following divisor digit is &lt;em&gt;greater than&lt;/em&gt; five.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this account Upton had shown method a) &quot;gives the correct quotient figure on the first trial in only 66.70 per cent of all cases.&quot;  With method b) one gets &quot;the correct quotient figure on the first trial in 80.43 per cent of all cases, a quotient figure requiring one correction in 19.29 per cent of all cases, and a quotient figure requiring two corrections in .28 per cent of all cases.&quot;  Method b) did not require &quot;more than two corrections&quot; in any circumstance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The study then proceeds to compare method b), the variant of our usual approach to trial divisors, and an &quot;experimental method&quot; that builds a table of multiples 1 through 9 of the actual divisor.  Fuller points out that the history of both methods can be followed back to Roberte Recorde's &lt;em&gt;Arithmetike&lt;/em&gt; (1579, orig. edition 1542).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building the table of multiples eliminates &quot;trial divisors&quot; and is certainly economical if there are to be several digits in the quotient; table construction can be done by alternate doubling and adding in the divisor.  Fuller's approach entailed checking the table of multiples with casting out 9's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuller's comparison of methods was carried out with three fifth grade class during the school year 1946-47 in Connecticut.  If I have read the statistical summary properly, the approach using a table of multiples proved more accurate (less error prone), though not in a statistically significant way except for certain longer problems (more digits in the divisor or quotient).  The table of multiples also took longer for the students to do, though this was in part because of extra checks involved on the answer and in part because of the modest sizes of problems (2-3 digits in the divisor, 2-4 digits in the quotient).&lt;/p&gt;&#xA;" OwnerUserId="3111" LastEditorUserId="3111" LastEditorDisplayName="" LastEditDate="2011-01-02T20:58:22.513" LastActivityDate="2011-01-02T20:58:22.513" CommentCount="2" />
  <row Id="15915" PostTypeId="1" AcceptedAnswerId="15916" CreationDate="2010-12-30T14:35:54.507" Score="1" ViewCount="68" Body="&lt;p&gt;In some course notes on &quot;evolution equations in probability spaces (etc)&quot; I have the following definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point $z \in \mathbb R^n$ is called the &lt;em&gt;approximate limit&lt;/em&gt; of $f$ at $x$ if for every $\epsilon &amp;gt; 0$ the set $\{y \in \Omega : |f(y) - z| \leq \epsilon\}$ has Lebesgue density 1 at $x$. Notation: $\tilde{f}(x) := f(y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand this definition, what exactly does it say intuitively? The given set also does not depend on $x$... is there something wrong?&lt;/p&gt;&#xA;" OwnerUserId="1120" LastActivityDate="2010-12-30T15:32:27.037" Title="Approximate limit." Tags="&lt;measure-theory&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="15916" PostTypeId="2" ParentId="15915" CreationDate="2010-12-30T15:04:37.843" Score="3" ViewCount="0" Body="&lt;p&gt;I don't know if this will help your intuition, but unwinding the definition a bit, this says that for all $\epsilon\gt0$, for all $\eta\gt0$, there exists $\delta\gt0$ such that in $B_\delta(x)$ you have $|f(y)-z|\leq\epsilon$ outside of a subset of measure less than $\eta\cdot m(B_\delta(x))$.  For a fixed $\epsilon$, letting $\eta$ go to zero makes the relative size of the set where $|f(y)-z|\gt\epsilon$ go to zero (by taking sufficiently small neighborhoods of $x$).  You get an ordinary limit when you don't have to include the exception of a small subset where $|f(y)-z|$ might be larger than $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest case, the limit may exist if you ignore the values of $f$ off a set of measure zero.  (E.g., take a continuous function but add $1$ to its values on a countable dense subset of the domain.)  In that case you can actually take $\eta=0$.  Perhaps more illustrative is the example of the function $f$ on $\mathbb{R}$ that is $1$ in an open interval of length $2^{-n}$ centered at $\frac{1}{n}$ for each positive integer $n$, and $0$ elsewhere.  Then the approximate limit of $f$ at $0$ is $0$, but for $\epsilon\lt1$ and $\delta\gt0$, the measure of the subset of $B_\delta(0)$ on which $|f(y)-0|\gt\epsilon$ is positive. &lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-30T15:32:27.037" LastActivityDate="2010-12-30T15:32:27.037" CommentCount="2" />
  <row Id="15917" PostTypeId="1" AcceptedAnswerId="15989" CreationDate="2010-12-30T15:08:38.103" Score="1" ViewCount="74" Body="&lt;p&gt;Hi, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}X=S\times(P_{FC}/P_{DC})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can be rewritten in terms of percentage change over a time period as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in}x=s+I_{FC}-I_{DC}=s-(I_{DC}-I_{FC})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I actually rewrite this? I think it has something to do with logarithms because the multiplication becomes an addition and the division becomes a subtraction, but I think taking the log on both sides won't give me the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,since $x$ is a percentage change over a time period, I figured its this must be true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{1in} x=\frac{X_{2}}{X_{1}}-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\frac{X_{2}}{X_{1}}-1$ is definitely not equal to $\log X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints would be much appreciated.&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="2973" LastEditorDisplayName="" LastEditDate="2010-12-31T01:58:40.873" LastActivityDate="2010-12-31T12:27:01.150" Title="Rewriting an equation in percentage terms" Tags="&lt;homework&gt;&lt;algebra&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="15918" PostTypeId="1" AcceptedAnswerId="15931" CreationDate="2010-12-30T15:32:04.977" Score="0" ViewCount="86" Body="&lt;p&gt;Hi I am wondering how best to explain this, I am working with WebGL (effectively OpenGL) and I have the ray cast from clicking in 3d space. I have the far and near values of the ray cast, the camera position and the target it is looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I work out where the ray cast starts and ends, effectively to draw a line to show the ray cast through 3d space. The most confusing part of this for me is the angles, subtracting the angles of the cameras face which are also in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Camera position would obviously be the start point, so I need to use that and using the direction the camera is facing, along with the near and far values of the ray to get the end point! I'm confused, please help.&lt;/p&gt;&#xA;" OwnerUserId="3849" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T21:41:31.307" LastActivityDate="2010-12-30T21:41:31.307" Title="Help with matrix mathematica" Tags="&lt;geometry&gt;&lt;matrices&gt;&lt;3d&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15919" PostTypeId="2" ParentId="15917" CreationDate="2010-12-30T16:06:11.530" Score="3" ViewCount="0" Body="&lt;p&gt;So $X = S \times (P_{FC}/P_{DC})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\log X = \log S + (\log  P_{FC}- \log P_{DC})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $\log X = \log S-(\log P_{DC}- \log P_{FC})$. &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-30T16:06:11.530" />
  <row Id="15920" PostTypeId="2" ParentId="15852" CreationDate="2010-12-30T16:44:32.323" Score="5" ViewCount="0" Body="&lt;p&gt;I would recommend Rotman's book &quot;An Introduction to the Theory of Groups&quot; for basic material on word problems of groups, and for a reasonably accessible proof of the Novikov-Boone theorem which asserts the existence of finitely presented groups with unsolvable word problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of the word problem of a group is generally applied to groups defined by a specific finite presentation, but it is not hard to prove that, for a group $G$, the solvability of the word problem of $G$ is independent of the given presentation of $G$, so it is customary simply to say that the word problem of $G$ is or is not solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots of familiar classes of groups are known to have solvable word problem, including finite groups, finitely presented residually finite groups (proof given above by Bill Dubuque), automatic groups, finitely generated nilpotent groups.&#xA;But, by  a recent result of Olga Kharlampovich, there exist finitely presented solvable groups of derived length $3$ with unsolvable word problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the word problem is semi-decidable for all groups given by a recursively enumerable presentation, in the sense that if you are given a word in the generators  that does represent the identity element of the group, then it is possible to prove that it does. You can do that naively by systematically trying all possibilities for expressing the word as a product of conjugates of defining relators, but there are more efficient and practical methods of attempting this by using rewriting systems or coset enumeration.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-30T16:44:32.323" />
  <row Id="15921" PostTypeId="2" ParentId="15903" CreationDate="2010-12-30T16:50:27.170" Score="2" ViewCount="0" Body="&lt;p&gt;He has $11$ weeks of training (or $77$ days). The minimum and maximum number of games for this period is $77$ and $11 \cdot 12=132$, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have distinct situations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) At any period of $k$ consecutive days he plays at least $k$ games. As well you asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The smallest period of $n$ days such that he plays $k=22$ games occur when he consider a week with a maximum number of games $12$ plus days after and before this week with a maximum possible number of games. Notice that in one day is possible to play $6$ games. So, take one day with $4$ games, one week with $12$ games, and one more day with $6$ games. You get then $22$ games in $9$ days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 1) and 2) we conclude that between $9$ and $k$ days we always can make the choice of $22$ games. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it is very useful!&lt;/p&gt;&#xA;" OwnerUserId="2764" LastActivityDate="2010-12-30T16:50:27.170" />
  <row Id="15922" PostTypeId="2" ParentId="15829" CreationDate="2010-12-30T16:59:51.297" Score="5" ViewCount="0" Body="&lt;p&gt;Let $G$ be the direct product of countably many copies of the dihedral group $D$ of order 6 (or, if you prefer, $D$ is the symmetric group $S_3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can construct a Sylow $2$-subgroup of $G$ by choosing Sylow $2$-subgroups of each of the direct factors of $G$, and taking their direct product. Since $D$ has three Sylow $2$-subgroups, $G$ has uncountably many Sylow $2$-subgroups, so they cannot all be conjugate in the countable group $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we let $P$ and $H$ be non-conjugate Sylow $2$-subgroups of $G$, then there is no $g \in G$ such that $H \cap gPg^{-1} \in {\rm Syl}_2(H)$.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2010-12-30T16:59:51.297" CommentCount="3" />
  <row Id="15923" PostTypeId="1" AcceptedAnswerId="15926" CreationDate="2010-12-30T17:17:04.223" Score="3" ViewCount="91" Body="&lt;p&gt;If $\kappa$ is a measurable cardinal, and $\mathcal{U}$ is a normal ultrafilter which is $\kappa$-complete then $\mathcal{U}$ extends the club filter (i.e. every club is a member of $\mathcal{U}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One result is that all the sets in the ultrafilter are stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;Now, given a stationary set $S$ such that its complement in $\kappa$ is also stationary (i.e. there is no club subset in $S$) we can choose either to include $S$ or $\kappa\setminus S$ in an extension of an ultrafilter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck in showing that I can have a normal ultrafilter with either $S$ or $\kappa\setminus S$ for every stationary $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have to prove something else, but it really reduces to this claim.)&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;In light of how the above statement is false, I will just give the original question:&lt;br&gt;&#xA;I have to show that there can be a normal ultrafilter such that the set of measurable cardinals below $\kappa$ is not in the ultrafilter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there aren't stationary many of them - then clearly it's true. And since the limit of $\omega$ measurable cardinals is singular (ergo non-measurable); the only nontrivial case is when there is a stationary set of measurable cardinals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer hits over partial solutions, and partial solutions over complete solutions. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-30T17:43:58.610" LastActivityDate="2011-01-01T17:05:54.090" Title="Normal ultrafilters and Stationary sets" Tags="&lt;homework&gt;&lt;set-theory&gt;&lt;large-cardinals&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15924" PostTypeId="2" ParentId="15868" CreationDate="2010-12-30T17:20:41.823" Score="2" ViewCount="0" Body="&lt;p&gt;I'm not sure what you mean by &quot;light power,&quot; but &lt;em&gt;intensity&lt;/em&gt; seems to be the measure you want (and is inversely proportional to the square of the distance.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensity at a point $(x,y)$ of a spherical light source originating at, say, $(x_0, y_0)$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \mathcal{P} / 4 \pi r^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal{P}$ is the power of the light source (what I think you've called $X$), and $r$ is the distance between $(x,y)$ and $(x_0, y_0)$. This gives the final expression, for all light sources $(x_n, y_n)$ with power $\mathcal{P}_n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \displaystyle \sum_{n} \frac{\mathcal{P}_n}{4 \pi ((x-x_n)^2 + (y-y_n)^2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Unless I've forgotten some physics and intensities don't sum like that). For the third dimension, just add the $z$ terms in the distance formula. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="5218" LastEditorUserId="5218" LastEditorDisplayName="" LastEditDate="2010-12-31T18:21:27.527" LastActivityDate="2010-12-31T18:21:27.527" CommentCount="2" />
  <row Id="15925" PostTypeId="1" AcceptedAnswerId="15936" CreationDate="2010-12-30T17:33:14.933" Score="3" ViewCount="72" Body="&lt;p&gt;What exactly is the variation of a function ? Is it a distace or an element of some space&#xA;The total of a real valued function $f\colon [0,t] \mapsto \Re $ is as below say $\pi = \{0=t_0,t_1,\cdots , t_n=t\} $ then the $p^{th}$ variation of $f$ is &#xA;$$ \lim_{\Vert \pi\Vert \to 0}V^p(f)=\sum_{i=1}^n \vert f(t_i)-f(t_{i-1})\vert^p $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say we have $g\colon [0,t] \mapsto \mathcal{X}$ where $\mathcal{x}$ is some linear, normed,Complete space. it seems natural to say variation of $g$ over $[0,t]$ is &#xA;$$ Z_n=\sum_{i=1}^n \Vert g(t_i)-g(t_{i-1})\Vert^p $$ where &#xA;$\Vert.\Vert$ is the norm of $\mathcal{X}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it exists what is the Total variation in this case is it $\lim_{n \to \infty}Z$ ie a value in $[0,\infty]$ or is it some $x$ such that $\Vert Z_n -  x \Vert \to 0 $&#xA;say for wiener process $W(s)$ it quadratic  variation over  $[0,t]$ is $t$ does it mean &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\Vert \pi\Vert \to 0}V^2(W)=\sum_{i=1}^n \vert W(t_i)-W(t_{i-1})\vert^2 $$ in $L^2(\Omega) $ norm or does it mean some $X \in L^2(\Omega)$ &lt;/p&gt;&#xA;" OwnerUserId="5176" LastActivityDate="2010-12-31T00:44:44.673" Title="Variation of a function" Tags="&lt;stochastic-processes&gt;" AnswerCount="1" />
  <row Id="15926" PostTypeId="2" ParentId="15923" CreationDate="2010-12-30T17:34:45.947" Score="3" ViewCount="0" Body="&lt;p&gt;Asaf, what you are saying is false. In $L[\mu]$, for example, there is a unique normal ultrafilter. (Unless I misunderstand. You are saying that for every $S$ stationary-costationary, you can find a normal $U$ with $S\in U$, right? Obviously, any normal $U$ satisfies that either $S$ or $\kappa\setminus S$ is in $U$, which is what you wrote.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit :&lt;/strong&gt; This is a nice homework problem. There are several different solutions, and you may want to study them afterward. Here is a way of thinking about this; it is perhaps not the most efficient, but it is very useful. Suppose $U$ on $\kappa$ is normal and concentrates on measurables. Then $\kappa$ is measurable in $M$, where $j:V\to M$ is given by $U$. So there is a normal $U&amp;#39;$ on $\kappa$ in $M$, and $U&amp;#39;$ really is normal (in $V$, not just in $M$). Let $k:V\to N$ be given by $U&amp;#39;$. what can you say about the sizes of $j(\kappa)$ vs $k(\kappa)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another approach: Use induction, use that you are at a measurable (so you have a normal $U$) and &quot;integrate&quot; the measures on small cardinals witnessing the claim. Check that the resulting measure is as you want. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me add a couple of details to the first approach (I know it is more elaborate than the other one, but the payoff is worth the effort): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $j:V\to M$ is the ultrapower embedding by a normal measure $U$ on $\kappa$, then ${}^\kappa M\subset M$ and from this it is easy to check that if $M\models$&quot;$U'$ is a normal measure on $\kappa$&quot;, then $U&amp;#39;$ is a normal measure on $\kappa$ in $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that $U$ concentrates on measurables. Since the identity represents $\kappa$ in $M$, it follows that $\kappa$ is measurable in $M$, and there is $U&amp;#39;$ as mentioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, we want to iterate this process: Form the embedding $k:V\to N$ given by $U&amp;#39;$, and if $U&amp;#39;$ concentrates on measurables, then we get $U&amp;#39;&amp;#39;$ in $N$ and form $l:V\to P$, etc. We would like this process to stop after finitely many times. For this, we want to associate to a normal measure $U$ on $\kappa$ an ordinal $\alpha_U$ in such a way that if $U&amp;#39;$ is in the ultrapower by $U$, then $\alpha_{U&amp;#39;}&amp;lt;\alpha_U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to do this is to set $\alpha_U=j(\kappa)$. If $U&amp;#39;\in M$, we can form in $M$ the ultrapower embedding by $U&amp;#39;$, call $k&amp;#39;$ the result, so $k&amp;#39;:M\to M&amp;#39;$, and recall we called $k:V\to N$ the ultrapower of $V$ by $U&amp;#39;$. The point is that $k&amp;#39;(\kappa)=k(\kappa)$, because ${}^\kappa M\subset M$, so all functions $f:\kappa\to\kappa$ are in $M$, and this and $U&amp;#39;$ is all we need to compute the value of the embedding at $\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $M\models$&quot;$j(\kappa)$ is inaccessible, while $k'(\kappa)&amp;lt;(2^\kappa)^+$&quot;, so $k&amp;#39;(\kappa)&amp;lt;j(\kappa)$, so $k(\kappa)&amp;lt;j(\kappa)$ or $\alpha_{U&amp;#39;}&amp;lt;\alpha_U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A posteriori&lt;/em&gt;, this argument shows that the relation $\prec$ on normal ultrafilters on $\kappa$ given by &quot;$U'\prec U$ iff $U&amp;#39;$ belongs to the ultrapower by $U$&quot; is well-founded, and so we can assign to $U$ the ordinal $o(U)$ given by its rank in this well-founded relation. This is Mitchell's ordering, and it is a very useful tool in large cardinal theory. &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="462" LastEditorDisplayName="" LastEditDate="2011-01-01T17:05:54.090" LastActivityDate="2011-01-01T17:05:54.090" CommentCount="6" />
  <row Id="15927" PostTypeId="1" CreationDate="2010-12-30T17:50:56.993" Score="0" ViewCount="72" Body="&lt;p&gt;Suppose that&#xA;$$u_1 = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0,\ldots,0\right)\quad\text{and}\quad&#xD;&#xA;v_1 = \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,0,\ldots,0\right)$$&#xA;and &#xA;$$u_l(n) = \sum_{k=0}^{2^{l-1}-1} u_1\left(n+ \frac{kN}{2^{l-1}}\right)\quad\text{and}\quad&#xD;&#xA;v_l(n) = \sum_{k=0}^{2^{l-1}-1} v_1\left(n+ \frac{kN}{2^{l-1}}\right).$$&#xA;How to show that &#xA;\begin{align*}&#xA;u_l(0)&amp;amp;=  \frac{1}{\sqrt{2}},\\&#xA;u_l(1)&amp;amp;=  \frac{1}{\sqrt{2}},\\&#xA;u_l(n)&amp;amp;=  0 &amp;amp;\quad&amp;amp;\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$,}\\&#xA;v_l(0)&amp;amp;=  \frac{1}{\sqrt{2}},\\&#xA;v_l(1)&amp;amp;=  -\frac{1}{\sqrt{2}},\\&#xA;v_l(n)&amp;amp;=  0&amp;amp;&amp;amp;\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$.}&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="2723" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-30T18:37:39.293" LastActivityDate="2010-12-30T18:37:39.293" Title="How to show that $u_l(0)=  \frac{1}{\sqrt{2}}$, $u_l(1)=  \frac{1}{\sqrt{2}}$ and $u_l(n)=  0$ for $2 \leq n \leq (\frac{N}{2^{l-1}}-1$ $\dots$" Tags="&lt;linear-algebra&gt;" CommentCount="5" />
  <row Id="15928" PostTypeId="1" AcceptedAnswerId="15930" CreationDate="2010-12-30T18:05:38.633" Score="6" ViewCount="127" Body="&lt;p&gt;Is there any rule for powers so that i can compare which one is greater without actually calculating? For example  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;54^53 and 53^54 &#xA;23^26 and 26^23&#xA;3^4 and 4^3 (very simple but how without actually calculating)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1515" LastEditorUserId="1515" LastEditorDisplayName="" LastEditDate="2010-12-30T18:18:18.330" LastActivityDate="2010-12-30T20:24:22.793" Title="How to Compare powers without calculating?" Tags="&lt;arithmetic&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="15929" PostTypeId="2" ParentId="15903" CreationDate="2010-12-30T18:14:23.317" Score="5" ViewCount="0" Body="&lt;p&gt;For $\displaystyle 1 \le k \le 24$ you can definitely show that the master must have played exactly $k$ games on some set of consecutive days, using pigeonhole principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the total number of games the master has played till the end of day $\displaystyle j$ is $\displaystyle g_j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the $\displaystyle 154$ numbers: $\displaystyle g_1, g_2, \dots, g_{77}, g_1 + k, g_2 + k, \dots, g_{77} + k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are a set of $\displaystyle 154$ numbers between $\displaystyle 1$ and $\displaystyle 132+k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k \lt 22$, two of these must be equal. Since $\displaystyle g_i \neq g_j$ (at least one game a day) we must have that $\displaystyle g_j + k = g_i$ for some $\displaystyle i,j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k=22$ we must have that the numbers are $\displaystyle 1,2, \dots, 154$, in which case, the first (and last) $\displaystyle 22$ days, the master must have played $\displaystyle 1$ game everyday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k=23$, we can assume $\displaystyle g_i \neq 23$, and since $g_i \ge 1$, we have $g_i + k \neq 23$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by an argument similar to above, we must have have $\displaystyle 154$ numbers taking all values in $\displaystyle 1, 2, \dots, 155$, except $\displaystyle 23$ and the master must have played $\displaystyle 1$ game each of the last $\displaystyle 23$ days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\displaystyle k=24$, you can show that the master must have played $\displaystyle 1$ game the first $\displaystyle 23$ days (after eliminating one of the numbers in $\displaystyle 133, 134 \dots$), then a big number of games the next, violating the $\displaystyle 12$ games per week restriction (this is where we actually used that restriction for a &lt;em&gt;specific&lt;/em&gt; week).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might be able to use this kind of argument to show for $\displaystyle k$ close to $\displaystyle 24$, but for larger $\displaystyle k$, I am guessing that you can find a set of games which will miss that (perhaps a computer search will help there).&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-30T19:28:23.440" LastActivityDate="2010-12-30T19:28:23.440" CommentCount="2" />
  <row Id="15930" PostTypeId="2" ParentId="15928" CreationDate="2010-12-30T18:24:07.313" Score="9" ViewCount="0" Body="&lt;p&gt;If $a\gt b\gt e , b^a\gt a^b$.  To see this, take logs.  You want to compare $a \ln b$ with $b \ln a$.  $\ln$ rises slowly, so the larger multiplier wins.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-30T18:24:07.313" CommentCount="8" />
  <row Id="15931" PostTypeId="2" ParentId="15918" CreationDate="2010-12-30T18:41:45.043" Score="0" ViewCount="0" Body="&lt;p&gt;Solved: I obtained the correct values from else where within the WebGL wrapper.&lt;/p&gt;&#xA;" OwnerUserId="3849" LastActivityDate="2010-12-30T18:41:45.043" CommentCount="2" />
  <row Id="15932" PostTypeId="2" ParentId="15857" CreationDate="2010-12-30T18:49:15.130" Score="1" ViewCount="0" Body="&lt;p&gt;Geometry Revisited has a much narrower domain of content than an Introduction to Geometry. If your goal is to get a sense of what different kinds of problems, techniques, and concepts geometry has evolved to deal with, Introduction to Geometry is a &quot;dated&quot; but somewhat comprehensive choice. (For example, the whole &quot;world&quot; of computational geometry came on the scene after Coexeter's book.) One typically does not read mathematics books from cover to cover and one can dabble in Introduction to Geometry to see some of geometry's many parts. If one has heavy sledding in places than looking at the wiki article of the topic causing trouble might help.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastActivityDate="2010-12-30T18:49:15.130" />
  <row Id="15933" PostTypeId="2" ParentId="12451" CreationDate="2010-12-30T18:52:01.277" Score="0" ViewCount="0" Body="&lt;p&gt;I am not expert, but my own limited view is this.  I said for thirty years I could see no interest in theorems like radon nikodym, having learned them very abstractly.  then once while teaching honors calculus i asked myself what the fundamental theorem of calculus should say for integrals of just Riemann integrable, not necessarily continuous functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After originally stating the wrong answer to my class, I learned (with help from an analyst friend who showed me a Cantor function) that the indefinite integral is characterized by being a function which has a derivative equal to the original integrand almost everywhere (wherever that integrand is continuous) and being also not just continuous but Lipschitz continuous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I realized at last the radon nikodym theorem is just the fundamental theorem of calculus for more general functions.  We may not realize the analogy from calc 1, because by considering only continuous integrands there we miss out on the singular part.  I.e. we forget to wonder why we are looking only at integrals of derivatives of C^1 (in particular Lipschitz) functions. So any application of FTC type is an application of RN, e.g. it lets you characterize constant functions (a.e?) by weak continuity and differentiability properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Fubini is of course repeated integration, so reduces any integral computation inductively to one of lower dimension (volumes by slicing).  E.g. to show some set has measure zero (as in Sard's theorem) you can do it inductively by showing most of the slices have lower dimensional measure zero.  (See Guillemin and Pollack, appendix, or Milnor's differential topology book.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my professors once suggested that virtually all problems in analysis are attacked by either dominated convergence or Fubini.  So if you see an exam problem that dominated convergence won't do, try Fubini.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer assumes you retain an interest in the topic even after your exam, unless you are at Harvard, where exams are perhaps still in january.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following up KCd's comment, you might also peruse old qualifying exams available on the websites at schools like Harvard and UGA.  Harvard's also has a few lists of typical questions of this nature: e.g. if every intersection of a certain subset S of the plane, with a line of slope 1 is countable, what can you say about the Lebesgue measure of S? Gosh, Harvard even has, with login capability, online copies of all exams since 1977, and paper copies in libraries of exams since 1836!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or consider a continuous weakly monotone increasing function f on the interval [0,1].  One knows that f is differentiable a.e. say with derivative g ≥ 0. If  g is integrable and G(x) is the integral of g from 0 to x, then to what extent does G determine f or g, if either?  When does G determine both f and g?  Give an example if possible where G does not determine f, respectively g.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2011-01-06T17:32:20.957" LastActivityDate="2011-01-06T17:32:20.957" CommentCount="3" />
  <row Id="15934" PostTypeId="2" ParentId="15890" CreationDate="2010-12-30T18:57:07.563" Score="5" ViewCount="0" Body="&lt;p&gt;Michael Spivak gave me a small volume he wrote on mechanics from the mathematicians perspective, and he is said to be preparing to release a longer book on physics explicitly for mathematicians.  In the meantime you might like Max Born's Atomic Physics.  I myself enjoyed, and learned from, at least looking at J.C.Maxwell's E&amp;amp;M.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2010-12-30T19:16:48.537" LastActivityDate="2010-12-30T19:16:48.537" CommentCount="4" />
  <row Id="15935" PostTypeId="2" ParentId="15720" CreationDate="2010-12-30T19:51:31.570" Score="0" ViewCount="0" Body="&lt;p&gt;If I remember correctly, what you are describing is close to the concept of a &lt;em&gt;distribution function&lt;/em&gt; (it is the name that I am slightly unsure about). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a measure space $(X,\mathcal{E},\mu)$, and a measurable function $f:X\to \mathbb{R}$, a concept often used in classical and harmonic analysis is the notion of the distribution function $\lambda_f:\mathbb{R}_+\to\mathbb{R}$ defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_f(a) = \mu\left( \{|f| &amp;lt; a\} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, $\lambda_f$ is returns the measure of the set on which $|f|$ is bounded by $a$. One can reformulate the $L^p$ norms of $f$ in terms of the distribution function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \|f\|_{L^p(X,\mu)} = p \int_0^\infty t^{p-1}\lambda_f(t) dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a device that is very useful for analytical estimates. Now, formally speaking the functional $L(f)$ you wrote down is very similar to the derivative $\lambda_f&amp;#39;$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another possibility that you may want to look into is the notion of pull-back distributions. If $f:M\to\mathbb{R}$ is a smooth map with non-vanishing gradient, then given any distribution (generalized function) $k$ on $\mathbb{R}$&#xA; you can define the pull-back distribution $f_*k$ on $M$. The definition of $f_*k$ depends closely on the notion of $L(f)$ you wrote down (see e.g. the introduction to distributions by Friedlander and Joshi), and is a generalization of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coarea_formula&quot; rel=&quot;nofollow&quot;&gt;co-area formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am not exactly sure how to answer your second question of &quot;is there a good way to compute&quot;. If you are asking whether given an analytic expression for the function $f$ whether you can have a &quot;closed form&quot; expression for $L(f)$, the answer is surely no, since the expression also depends on the chosen Riemannian metric/volume form/measure on $M$. If you are asking whether there are any convenient expressions to capture $L(f)$, I think you need to say a bit more about why you care about $L(f)$. You can certainly just write $L(f) = f_*\delta_a$ as the pull-back of the Dirac delta distribution on $\mathbb{R}$; whether that expression is a useful one for your purpose, I don't know.  &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2010-12-30T19:51:31.570" />
  <row Id="15936" PostTypeId="2" ParentId="15925" CreationDate="2010-12-30T20:21:57.850" Score="4" ViewCount="0" Body="&lt;p&gt;Consider the case where $W$ is a standard Brownian motion. Given a partition $\pi  = \{ 0 = t_0  &amp;lt; t_1  &amp;lt;  \cdots  &amp;lt; t_n  = t\}$ of the interval $[0,t]$, define&#xA;$$&#xD;&#xA;[W]^\pi   = \sum\limits_{i = 1}^n {[W(t_i ) - W(t_{i - 1} )]^2 }. &#xD;&#xA;$$&#xA;Let $|\pi | = \max _i (t_i  - t_{i - 1} )$ denote the mesh of the partition. As is well known, given a sequence of partitions $\pi_n$  with $|\pi_n| \to 0$ as $n \to \infty$, $[W]^{\pi _n }  \to t$ in the $L^2$ norm, that is ${\rm E}\big[([W]^{\pi _n }  - t)^2 \big] \to 0$. Hence, in particular, $[W]^{\pi _n }  \to t$ in probability, and thus, by definition, the quadratic variation of $W$ is given by $[W]_t = t$. The key point here is that the limit $t$ in the right-hand side of $[W]^{\pi _n }  \to t$ plays a role of a random variable, rather than of a constant. Of course, in this particular example, $t$ is also a constant, but in general the limit is a random variable, which is defined using convergence in probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A much simpler yet instructive example is given by the Poisson process $N$. It is readily seen that the variation of $N$ is given by&#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{|\pi _n | \to 0} \sum\limits_{i = 1}^n {|N(t_{i} ) - N(t_{i-1} )|} = N(t),&#xD;&#xA;$$&#xA;and the quadratic variation of $N$ is given by&#xA;$$&#xD;&#xA;\mathop {\lim }\limits_{|\pi _n | \to 0} \sum\limits_{i = 1}^n {[N(t_{i} ) - N(t_{i-1} )]^2} = N(t).&#xD;&#xA;$$&#xA;You may wish to consider different modes of convergence, as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-31T00:44:44.673" LastActivityDate="2010-12-31T00:44:44.673" CommentCount="4" />
  <row Id="15938" PostTypeId="2" ParentId="329" CreationDate="2010-12-30T21:05:27.587" Score="4" ViewCount="0" Body="&lt;p&gt;I recommend &lt;em&gt;Primes of the Form x&lt;sup&gt;2&lt;/sup&gt; + ny&lt;sup&gt;2&lt;/sup&gt;&lt;/em&gt;, by David Cox.  The question of which primes can be written as the sum of two squares was settled by Euler.  The more general question turns out to be much harder, and leads you to more advanced techniques in number theory like &lt;a href=&quot;http://en.wikipedia.org/wiki/Class_field_theory&quot; rel=&quot;nofollow&quot;&gt;class field theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_multiplication&quot; rel=&quot;nofollow&quot;&gt;elliptic curves with complex multiplication&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4880" LastEditorUserId="4880" LastEditorDisplayName="" LastEditDate="2010-12-30T23:45:33.203" LastActivityDate="2010-12-30T23:45:33.203" CommunityOwnedDate="2010-12-30T21:05:27.587" CommentCount="0" />
  <row Id="15939" PostTypeId="1" CreationDate="2010-12-30T21:06:42.593" Score="1" ViewCount="115" Body="&lt;p&gt;I have an $n\times p$  matrix $Z$ with $p\gt n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a diagonal matrix $A$ with positive entries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there is a known way to determine the&#xA;MP inverse of $A Z^T Z A$, if I know $A$ and the MP inverse of $Z^T Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5250" LastEditorUserId="498" LastEditorDisplayName="" LastEditDate="2010-12-31T00:36:05.407" LastActivityDate="2010-12-31T19:01:09.793" Title="Moore-Penrose inverse invariance" Tags="&lt;linear-algebra&gt;&lt;matrices&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15940" PostTypeId="1" CreationDate="2010-12-30T21:20:47.750" Score="2" ViewCount="29" Body="&lt;p&gt;In category theory, you see the idea of a class of epimorphisms being stable under pullback.  For example, in a regular category, the class of regular epimorphisms is closed under pullback.  Every place I've seen the notion of pullback-stability, it's always a part of a bigger definition, such as regular category, or Grothendieck topology.  Is there some bigger significance to the idea?  Is there a theory for pullback-stable classes of epimorphisms?&lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2010-12-30T21:20:47.750" Title="Pullback-stability for epimorphisms" Tags="&lt;category-theory&gt;" />
  <row Id="15941" PostTypeId="2" ParentId="5879" CreationDate="2010-12-30T21:26:24.213" Score="0" ViewCount="0" Body="&lt;p&gt;May I ask, how fast is this canonical labeling algorithm? (linear in n?)&#xA;I need to convert an adjacency matrix to canonical form but I don't know how :(&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;c.wilmer@gmail.com&lt;/p&gt;&#xA;" OwnerUserId="5251" LastActivityDate="2010-12-30T21:26:24.213" />
  <row Id="15942" PostTypeId="1" AcceptedAnswerId="15949" CreationDate="2010-12-30T22:15:16.153" Score="1" ViewCount="121" Body="&lt;p&gt;$f_n(x) :  [0,1] \rightarrow \bf R $,  and $f_n$ is $\frac{1}{n}$-periodic, $\max f_n(x) = n$, $\min f_n(x) = -n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $n$ and $v$ goes to infinity simultaneously, prove the convergence of&lt;br&gt;&#xA;$$\lim_{n,v \rightarrow \infty}  \int_0^1 f_n(x) e^{-i2\pi v x}\,dx $$&#xA;if $v,n$ are integers and $v/n$  is not an integer.&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2011-01-01T15:57:55.497" LastActivityDate="2011-01-01T15:57:55.497" Title="Convergence of  $\lim_{n,v \rightarrow \infty}  \int_0^1 f_n (x) e^{-i2\pi v x} \mbox{d} x $" Tags="&lt;fourier-analysis&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15943" PostTypeId="2" ParentId="12824" CreationDate="2010-12-30T22:53:30.267" Score="2" ViewCount="0" Body="&lt;p&gt;You can consider presentations for any algebraic system given by special constants, operations and equations.  For example, groups are given by the special constant 1 (for the multiplicative identity), the unary operation of inverse, x&lt;sup&gt;-1&lt;/sup&gt;, and the binary operation of multiplication, $xy$, the plus the equations that these must satisfy.  In that case, you can specify every object with such operations in terms of a set of generators, and the relations that the generators must satisfy.  The study of these kinds of constructions is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_algebra&quot; rel=&quot;nofollow&quot;&gt;universal algebra&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific case of commutative rings over a field, there is an explicit algorithm for working with presentations, known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis&quot; rel=&quot;nofollow&quot;&gt;Groebner basis&lt;/a&gt; algorithm.  It's widely implemented in symbolic algebra packages.  (In the case of commutative rings, you're basically just working with systems of polynomials.)&lt;/p&gt;&#xA;" OwnerUserId="4880" LastEditorUserId="4880" LastEditorDisplayName="" LastEditDate="2010-12-30T23:47:04.843" LastActivityDate="2010-12-30T23:47:04.843" />
  <row Id="15944" PostTypeId="2" ParentId="329" CreationDate="2010-12-30T23:07:44.343" Score="1" ViewCount="0" Body="&lt;p&gt;&lt;em&gt;Basic Number Theory&lt;/em&gt; by Andre Weil.  It's hard going and mind-blowing.&lt;/p&gt;&#xA;" OwnerUserId="4253" LastActivityDate="2010-12-30T23:07:44.343" CommunityOwnedDate="2010-12-30T23:07:44.343" />
  <row Id="15945" PostTypeId="2" ParentId="329" CreationDate="2010-12-30T23:08:08.963" Score="2" ViewCount="0" Body="&lt;p&gt;One of my colleagues, a number theorist, recommended the little book by van den eynden for beginners.  my favorite is by trygve nagell.  (I am a geometer.)  One of my friends, preparing for a PhD in arithmetic geometry?, started with the one recommended by Barry, Basic number theory.  As I recall it's for people who can handle Haar measure popping up on the first page of a &quot;basic&quot; book on number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also recommend Gauss's Disquisitiones Arithmeticae.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2011-01-05T20:01:58.207" LastActivityDate="2011-01-05T20:01:58.207" CommunityOwnedDate="2010-12-30T23:08:08.963" CommentCount="2" />
  <row Id="15946" PostTypeId="1" AcceptedAnswerId="15947" CreationDate="2010-12-30T23:59:17.993" Score="4" ViewCount="125" Body="&lt;p&gt;Is this series known to converge, and if so, what does it converge to (if known)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $p_n$ is prime number n, and $p_1 = 2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\limits_{n=1}^{\infty} \frac{1}{p_n}$$&lt;/p&gt;&#xA;" OwnerUserId="4576" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-31T00:31:45.397" LastActivityDate="2010-12-31T00:31:45.397" Title="Does the sum of reciprocals of primes converge?" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;prime-numbers&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="15947" PostTypeId="2" ParentId="15946" CreationDate="2010-12-31T00:02:46.237" Score="7" ViewCount="0" Body="&lt;p&gt;No, it does not converge. See this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges&quot; rel=&quot;nofollow&quot;&gt;Proof of divergence of sum of reciprocals of primes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact it is known that $$\sum_{p \le x} \frac{1}{p} = \log \log x + A + \mathcal{O}(\frac{1}{\log^2 x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;http://math.stackexchange.com/questions/12566/&quot;&gt;Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-31T00:12:54.347" LastActivityDate="2010-12-31T00:12:54.347" />
  <row Id="15948" PostTypeId="2" ParentId="15946" CreationDate="2010-12-31T00:08:37.593" Score="4" ViewCount="0" Body="&lt;p&gt;I would like to note that this implies that according the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem&quot; rel=&quot;nofollow&quot;&gt;Müntz-Szász Theorem&lt;/a&gt; that every continuous function in $[0,1]$ is a uniform limit of polynomials whose exponents are prime numbers!&lt;/p&gt;&#xA;" OwnerUserId="1120" LastEditorUserId="1120" LastEditorDisplayName="" LastEditDate="2010-12-31T00:15:05.557" LastActivityDate="2010-12-31T00:15:05.557" CommentCount="2" />
  <row Id="15949" PostTypeId="2" ParentId="15942" CreationDate="2010-12-31T00:28:29.597" Score="2" ViewCount="0" Body="&lt;p&gt;I'll assume you mean to take a limit in some sense as over double sequences indexed by natural numbers $n, \nu$ such that $\frac{\nu}{n}$ is not an integer. For $f_n$ an $\frac{1}{n}$-periodic function such that $\max f_n=-\min f_n =n$, let $g_n$ be defined by $f_n(x)=ng_n(nx)$. Then $g_n$ is $1$-periodic and has max and min equal to 1 and -1. Let $r=\frac{\nu}{n}$. Then you require that $r$ is not an integer.&#xA;Then writing the integral of interest in terms of $g_n$ and $r$ and $n$, you'll be very tempted to change variables and integrate from 0 to n. Do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using the periodicity of $g_n$, you'll want to break the integral into a sum of $n$ integrals. Finally, switch the order of summation and integration, be happy that you can look up or re-derive the formula for geometric sums, and be shocked (or not, but I was pleasantly surprised). &lt;/p&gt;&#xA;" OwnerUserId="5253" LastActivityDate="2010-12-31T00:28:29.597" CommentCount="3" />
  <row Id="15950" PostTypeId="2" ParentId="15707" CreationDate="2010-12-31T00:34:59.247" Score="4" ViewCount="0" Body="&lt;p&gt;Suppose $f$ is $C^2$.  First of all, because $g$ is convex in each variable, it follows that $f$ is convex, and hence $f&amp;#39;&amp;#39;\geq0$.  I did not initially have Slowsolver's insight that &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmically_convex_function&quot; rel=&quot;nofollow&quot;&gt;log convexity&lt;/a&gt; would be a criterion to look for, but naïvely checking for positivity of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hessian_matrix&quot; rel=&quot;nofollow&quot;&gt;Hessian&lt;/a&gt; of $g$ leads to the inequalities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;&amp;#39;(x)f(y)+f(x)f&amp;#39;&amp;#39;(y)\geq0$$&#xA;and&#xA;$$f&amp;#39;(x)^2f&amp;#39;(y)^2\leq f&amp;#39;&amp;#39;(x)f(x)f&amp;#39;&amp;#39;(y)f(y)$$&#xA;for all $x$ and $y$, coming from the fact that a real symmetric $2$-by-$2$ matrix is positive semidefinite if and only if its trace and determinant are nonnegative.  The first inequality follows from nonnegativity of $f$ and $f&amp;#39;&amp;#39;$.  The second inequality is equivalent to $f&amp;#39;^2\leq f&amp;#39;&amp;#39;f$.  To see the equivalence in one direction, just set $x=y$ and take square roots; in the other direction, multiply the inequalities at $x$ and $y$.  Since $\log(f)&amp;#39;&amp;#39;=\frac{f&amp;#39;&amp;#39;f-f&amp;#39;^2}{f^2}$, this condition is equivalent to $\log(f)&amp;#39;&amp;#39;\geq0$, meaning that $\log(f)$ is convex.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-31T01:34:20.630" LastActivityDate="2010-12-31T01:34:20.630" />
  <row Id="15952" PostTypeId="2" ParentId="12451" CreationDate="2010-12-31T01:01:22.093" Score="3" ViewCount="0" Body="&lt;p&gt;One basic example of what you can do with Fubini's theorem is to represent an integral of a function of a function in terms of its distribution function. For instance, there is the formula (for reasonable $\phi$, $f$ basically arbitrary but nonnegative)&#xA;$$ \int \phi \circ f d\mu = \int_t \mu(\{ f&amp;gt; t\}) \phi&amp;#39;(t) dt$$ which reduces questions about integrals of, for instance, $p$th powers to questions about the distribution function. This is (for example) how the boundedness of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function&quot; rel=&quot;nofollow&quot;&gt;Hardy-Littlewood maximal operator&lt;/a&gt;  on $L^p$ ($p&gt;1$) is proved: you get a bound on the distribution function of the maximal function by general methods and then do an interpolation process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove the formula above, as in Rudin, one can consider the collection $E$ of pairs $(x,t)$ such that $0 \leq t \leq f(x)$. This is a measurable subset of $X \times \mathbb{R}$ if $X$ is the initial measure space on which $f$ is defined and $\mathbb{R}$ has Lebesgue measure. &#xA;Then, one can write the second integral in the displayed equation &#xA;as $\int_t \phi&amp;#39;(t) dt \int_{x \in X} \chi_E(x,t) d \mu$ where $\chi_E$ denotes the characteristic function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rearranging this integral via Fubini's theorem allows one to integrate with respect to $t$ first, for each $x$; then $t$ goes from $0$ to $f(x)$, and one can see that this integral becomes $\int_x \int_{t=0}^{f(x)} \phi&amp;#39;(t) dt$, which is the right-hand-side of the displayed equation.&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-31T01:01:22.093" CommentCount="1" />
  <row Id="15954" PostTypeId="1" CreationDate="2010-12-31T01:34:02.283" Score="3" ViewCount="108" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral&quot;&gt;How can you prove that a function has no closed form integral?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;When they say that, e.g., Li(x) has no closed form (for some agreed upon definition of &quot;closed form&quot;), do they mean that it can be &lt;strong&gt;proved&lt;/strong&gt; that there isn't one, or just that none has been found yet.  Most authors seem to just assert that a given integral can't be represented in any simpler way, but I've been wondering if that's because they know this for a fact, or just that it's assumed that if there is one it would have been found by now.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a theorem, how would you go about proving it?&lt;/p&gt;&#xA;" OwnerUserId="5257" LastActivityDate="2010-12-31T01:34:02.283" ClosedDate="2010-12-31T02:09:58.803" Title="How do they know an integral has no closed form solution?" Tags="&lt;terminology&gt;&lt;closed-form&gt;" CommentCount="2" />
  <row Id="15955" PostTypeId="1" AcceptedAnswerId="15956" CreationDate="2010-12-31T02:34:26.537" Score="4" ViewCount="114" Body="&lt;p&gt;Let $f(x) = \sum_{i =0}^\infty a_i x^i$ be a power series which converges for all real $x$. Assume that $f(x)$ is not identically zero. I'm interested in the density of the zeros of $f(x)$. Let $Z$ be the set of zeros of $f(x)$. Which of the following claims about density of $Z$ are true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 1: $Z$ is nowhere dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 2: $Z$ is countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claim 3: For any $a,b \in \mathbb{R}$ , $Z \cap [a,b]$ is finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe (correct me if I'm wrong) that claim 3 implies the other two. I suspect all three claims are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the answers to these questions are well-known, though I was not able to find an obvious reference. Can anyone suggest a reference with a nice treatment of these questions?&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-31T04:48:20.067" LastActivityDate="2010-12-31T04:48:20.067" Title="Density of zeros of a power series over the reals" Tags="&lt;analysis&gt;&lt;power-series&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="15956" PostTypeId="2" ParentId="15955" CreationDate="2010-12-31T02:58:34.723" Score="6" ViewCount="0" Body="&lt;p&gt;Claim 3 is true, and it does imply the other 2.  It is enough to consider analytic functions, i.e. functions that have a power series expansion in some interval centered at each real number, which in particular holds if you have an everywhere convergent power series.  If $[a,b]$ had infinitely many zeros of $f$, then there would be a limit point $c$ of these zeros.  By continuity $f(c)=0$.  Let $n\gt0$ be the smallest positive integer such that $f^{(n)}(c)\neq0$ (using the assumption that $f$ is not identically $0$).  Then $f$ has power series expansion &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;f(x)&amp;amp;=\sum_{k=0}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k&#xD;&#xA;=\sum_{k=n}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k\\&#xD;&#xA;    &amp;amp;=(x-c)^n\sum_{k=n}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^{k-n}=(x-c)^ng(x),&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(x)$ is a continuous function such that $g(c)\neq0$, and hence $g(x)\neq0$ in some open interval $I$ containing $c$.  So the only zero of $f$ in $I$ is $c$, contradicting the fact that $c$ is a limit point of the set of zeros.  Hence, unless $f$ is identically zero, the limit point $c$ cannot exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference you can read any good text on complex analysis.  If you prefer to stick to the real case, there is the book &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=i4vw2STJl2QC&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;A primer of real analytic functions&lt;/a&gt;&lt;/em&gt; by Krantz and Parks.&lt;/p&gt;&#xA;" OwnerUserId="1424" LastEditorUserId="1424" LastEditorDisplayName="" LastEditDate="2010-12-31T03:09:23.310" LastActivityDate="2010-12-31T03:09:23.310" CommentCount="5" />
  <row Id="15957" PostTypeId="2" ParentId="15955" CreationDate="2010-12-31T02:59:16.647" Score="4" ViewCount="0" Body="&lt;p&gt;All three claims are true.  Claim 3 $\Rightarrow$ Claim 2 and Claim 1. The zeros of an analytic function are isolated. This implies Claim 3.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2010-12-31T02:59:16.647" CommentCount="4" />
  <row Id="15958" PostTypeId="1" AcceptedAnswerId="15959" CreationDate="2010-12-31T03:24:30.947" Score="1" ViewCount="81" Body="&lt;p&gt;On p.33 of Sergei Matveev's &quot;Lectures on Algebraic Topology&quot;, the &lt;em&gt;relative chain group&lt;/em&gt; of a pair $(K,L)$, where $L \subset K$ is a subcomplex, is defined as the free abelian group on simplices with interiors in $K \setminus L$. How is this different from just the chains on $K \setminus L$? How is this the same notion as $C_n(K)/C_n(L)$, i.e. chains in $K$ modded out by chains in $L$, which is another definition I have seen for the relative chain groups? &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-31T03:52:59.903" Title="The &quot;geometry&quot; behind relative homology" Tags="&lt;algebraic-topology&gt;" AnswerCount="1" />
  <row Id="15959" PostTypeId="2" ParentId="15958" CreationDate="2010-12-31T03:52:59.903" Score="3" ViewCount="0" Body="&lt;p&gt;The interior of the boundary of a simplex with interior in $K/L$ need not lie in $K/L$. As a result, the boundary map on the relative chain group $(K,L)$ (in your notation) has to be defined such that the (homological) boundary of a simplex includes only the pieces of the (geometric) boundary that don't lie in $L$. In this manner, we get the same thing as $C_*(K)/C_*(L)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, $K/L$ is not a simplicial complex, so it doesn't make sense to consider simplicial chains on $K/L$. It does make sense to consider singular chains. In that case $H_*(K,L)$ and $H_*(K-L)$ are generally different. (As a very simple example, take the index to be zero, and consider the case where $L$ disconnects the space $K$.)&lt;/p&gt;&#xA;" OwnerUserId="536" LastActivityDate="2010-12-31T03:52:59.903" CommentCount="1" />
  <row Id="15960" PostTypeId="2" ParentId="15855" CreationDate="2010-12-31T03:56:26.060" Score="4" ViewCount="0" Body="&lt;p&gt;slowsolver makes a very good point - the value of the parameter in the expectation is not explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one usually considers log-likelihood-ratios, rather than log-likelihoods - as in the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(\theta;\theta_0) = {\mathrm E}_{\theta_0}{\mathrm log}\frac{f(X|\theta)}{f(X|\theta_0)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is just a kullback-leibler divergence between the pdfs indexed by $\theta$ and $\theta_0$. assuming $\theta_0$ is the [fixed] true value of the parameter, this K-L number differs by a constant from the expected log-likelihood [taken wrt $\theta_0$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the significance of this K-L quantity is that it is negative if $\theta\ne\theta_0$ and is zero when $\theta=\theta_0$. [this follows directly from the concavity of log.] thus the K-L quantity, as a function of $\theta$, assumes its max at $\theta=\theta_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the average log-likelihood-ratio for the sample converges [for each $\theta$] to $K(\theta;\theta_0)$, one sees heuristically that the MLE of $\theta$ &lt;em&gt;ought&lt;/em&gt; to be near $\theta_0$ when $n$ is large. [this is a handwaving version of wald's proof that that the MLE is consistent to the true value of the parameter. a rigorous argument requires, &lt;em&gt;inter alia&lt;/em&gt;, continuity of $f$ in $\theta$ and conditions to insure that the K-L number stays bounded away from zero outside of a compact subset of $\theta$ [so that the MLE cannot escape to $\infty$] - as well as assorted other niceties.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the above behavior of the K-L number as a function of $\theta$  also underlies arguments that [maximum-]likelihood-ratio tests are consistent as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the expectation for the 'expected log-likelihood' is wrt some fixed pdf $f(x)$, which need not be one of the $f(\cdot|\theta)$ pdfs, then the value of $\theta$ maximizing the K-L quantity indicates the pdf in the model 'most like' $f(\cdot)$ [it has minimal K-L divergence from $f(\cdot)$]. when $f(\cdot)$ is not in the model [so that the model is incorrect], it is possible that the K-L function attains its max at more than one value of $\theta$. then the MLE can oscillate among the various maximizers of the K-L function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the expectation for the 'expected log-likelihood' is wrt $\theta$, rather than a fixed $\theta_0$ [or some other fixed $f(\cdot)$], i don't know offhand what interest there is in that quantity. [it is equivalent to considering $K(\theta,\theta_0)$ as a function of $\theta_0$, rather than as a function of $\theta$. but usually one wants the behavior of the entire average-log-likelihood[-ratio] function under the fixed true value $\theta_0$, not just its behavior at $\theta$ when $\theta$ is the true value.]&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2010-12-31T03:56:26.060" CommentCount="0" />
  <row Id="15962" PostTypeId="2" ParentId="370" CreationDate="2010-12-31T04:28:37.767" Score="1" ViewCount="0" Body="&lt;p&gt;as a young student, I enjoyed peter freyd's fun little book on abelian categories.  The nice thing about Freyd's book is it isn't boring, and it has little pieces of wisdom (opinion) such as the remark that categories are not really important, you just define them so you can define functors.  And in fact you just define functors so you can define natural transformations, the really interesting things.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you may disagree, but blunt debatable assertions (like this one) always make for more interesting reading.  Another provocative remark by this author is the observation that he himself seldom learnt math by reading books, but rather by talking to people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the nice link above I learned that Goldblatt also quotes a remark (which may have inspired Freyd's) by Eilenberg and Maclane that categories are entirely secondary to functors and natural transformations, on page 194 where he introduces these latter concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leinster's notes linked by Patrick, look nice - a bit like an introduction to Maclane's Categories for the working mathematician, chatty and full of debatable assertions, (many of which I disagree with, but enjoy thinking about).  He does not give much credit, but I believe the adjoint functor theorems he quotes without proof, (GAFT,...) may be due to Freyd.  Leinster's notes are easy reading and informative.&lt;/p&gt;&#xA;" OwnerUserId="5211" LastEditorUserId="5211" LastEditorDisplayName="" LastEditDate="2010-12-31T05:36:55.767" LastActivityDate="2010-12-31T05:36:55.767" CommunityOwnedDate="2010-12-31T04:28:37.767" />
  <row Id="15963" PostTypeId="1" AcceptedAnswerId="15973" CreationDate="2010-12-31T04:47:53.840" Score="9" ViewCount="204" Body="&lt;p&gt;Let $X$ and $Y$ be topological spaces. A function $f: X \rightarrow Y$ is defined as continuous if for each open set $U \subset Y$, $f^{-1}(U)$ is open in $X$. This definition makes sense to me when $X$ and $Y$ are metric spaces- it is equivalent to the usual $\epsilon-\delta$ definition. But why is this a good definition when $X$ and $Y$ are not metric spaces? &#xA;How should we think about this definition intuitively? &lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-12-31T15:43:53.757" LastActivityDate="2010-12-31T15:43:53.757" Title="What is the intuition for the point-set topology definition of continuity?" Tags="&lt;topology&gt;&lt;intuition&gt;" AnswerCount="6" CommentCount="1" />
  <row Id="15964" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T05:00:07.197" Score="3" ViewCount="0" Body="&lt;p&gt;You have it exactly right.  It works well when $X$ and $Y$ are metric spaces, and has proved useful in more general contexts.  When you think of open sets as points &quot;near&quot; another, this is the proper translation of the usual $\epsilon-\delta$ definition.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-31T05:00:07.197" />
  <row Id="15965" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T05:39:36.830" Score="6" ViewCount="0" Body="&lt;p&gt;Maybe it's just me, but I've never thought that the usual $\epsilon$-$\delta$ definition of continuity is intuitive at all. Why should a function be continuous at $x$ if every ball of radius $\epsilon$ around $f(x)$ contains the image under $f$ of a ball of radius $\delta$ around $x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, in metric spaces, I think of a function as continuous if it preserves limits, which can be intuitively (and generalizably) be phrased by saying that $f$ is continuous if and only if whenever $x$ is in the closure of a set $A$, then $f(x)$ is in the closure of the set $f(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Take a piece of paper and draw out the arguments of the next two paragraphs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the $\epsilon$-$\delta$ continuity implies 'closure' continuity, suppose that $f$ is not closure continuous at $x$, that is $x$ is in the closure of some set $A$ but $f(x)$ is not in the closure of $f(A)$. Then there exists an $\epsilon$-ball around $f(x)$ that does not intersect $f(A)$ even though every $\delta$-ball intersects $A$. Hence, some $\epsilon$-ball around $f(x)$ contains no image of a $\delta$-ball around $x$, and so $f$ is also not $\epsilon$-$\delta$ continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that 'closure' continuity implies $\epsilon$-$\delta$ continuity, suppose that $f$ is not $\epsilon$-$\delta$ continuous. Then there exists an $\epsilon$-ball around $f(x)$ that contains no image of a $\delta$-ball around $x$. In other words, the preimage of the $\epsilon$-ball around $f(x)$ contains no $\delta$-ball around $x$, so let $A$ be the collection of points that are not in the preimage of the $\epsilon$-ball. Then $x$ is in the closure of $A$ since any $\delta$-ball around $x$ has a point outside the preimage of the $\epsilon$-ball and hence in $A$, but $f(x)$ is not in the closure of $f(A)$ since the $\epsilon$-ball around $f(x)$ is disjoint from $f(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the cool thing to notice is that the above equivalence of definitions works perfectly fine if you replace $\delta$-balls and $\epsilon$-balls with open sets in the appropriate topological spaces, so really what you should care about is how to make sense of 'closure' continuity in a space that is not a metric space, and the answer is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kuratowski_closure_axioms&quot; rel=&quot;nofollow&quot;&gt;wiki:Kuratowski Closure Axioms&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also find useful the answers to &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets&quot; rel=&quot;nofollow&quot;&gt;this mathoverflow question&lt;/a&gt;, specifically &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/19156#19156&quot; rel=&quot;nofollow&quot;&gt;this one by sigfpe&lt;/a&gt; and &lt;a href=&quot;http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/19173#19173&quot; rel=&quot;nofollow&quot;&gt;this one by Vectornaught&lt;/a&gt;. The first one talks about how open sets can be thought of as rulers that try to measure imprecisely things in the vector space (but which doesn't explain why continuity it as it is)), while the second phrases the Kuratwoski Closure Axioms in terms of the intuitive notion of 'nearness' of points (which does account for continuity).&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2010-12-31T05:39:36.830" />
  <row Id="15966" PostTypeId="1" CreationDate="2010-12-31T05:51:02.327" Score="12" ViewCount="214" Body="&lt;p&gt;A positive integer is &lt;em&gt;$B$-smooth&lt;/em&gt; if and only if all of its prime divisors are less than or equal to a positive real $B$. For example, the $3$-smooth integers are of the form $2^{a} 3^{b}$ with non-negative exponents $a$ and $b$, and those integers less than or equal to $20$ are $\{1,2,3,4,6,8,9,12,16,18\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Ramanujan's first letter to G. H. Hardy, Ramanujan emphatically quotes (without proof) his result on the number of $3$-smooth integers less than or equal to $N &amp;gt; 1$,&#xA;\begin{eqnarray}&#xA;\frac{\log 2 N \ \log 3N}{2 \log 2 \ \log 3}.&#xA;\end{eqnarray}&#xA;This is an amazingly accurate approximation, as it differs from the exact value by less than 3 for the first $100,000$ or so integers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question&lt;/em&gt;: Knowing full well that Ramanujan rarely gave proofs of his own claims, I wonder if a proof of this particular estimate appears somewhere in the literature. Is this problem still open? If not, what is a reference discussing its proof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="2720" LastEditorDisplayName="" LastEditDate="2010-12-31T06:50:44.593" LastActivityDate="2010-12-31T23:18:45.257" Title="Ramanujan's First Letter to Hardy and the Number of $3$-Smooth Integers" Tags="&lt;number-theory&gt;&lt;real-analysis&gt;&lt;reference-request&gt;&lt;analytic-number-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="15967" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T05:58:19.797" Score="1" ViewCount="0" Body="&lt;p&gt;It's true that this definition generalizes that for metric spaces, but there are other generalizable definitions (e.g. takes convergent sequences to convergent sequences), and perhaps implicit in the OP's question is: why this particular definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course part of the answer is that it turns out to work well, but this is not too satisfying. The following are just a couple of things that just occurred to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, given a class of mathematical objects like a topological space (vector space, group, ring, etc.) it is natural to ask: what are the &quot;structure-preserving maps&quot; between such objects? Vector spaces are sets equipped with a scalar multiplication map; groups are sets equipped with a group operation, etc. and the notions of linear map and homomorphism are precisely defined to preserve this structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now with a topological space, of course, the structure comes as a set of &quot;open sets.&quot; Here the generalization from concepts of metric spaces is especially clear. Therefore, a continuous map, a structure-preserving map on a topological space, should be one that &quot;preserves open sets.&quot; At first you might think that such a map should take open sets to open sets (i.e. an open map), but examining the conditions on open sets shows that this is bad. The point is that if $f$ is a set map, then $f^{-1}$ is actually much &quot;nicer&quot; than $f$ in terms of how it commutes with unions, intersections, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I might make the following observation. A structure-preserving map from $X$ to a set $Y$ should prescribe a natural structure for $Y$. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-31T05:58:19.797" CommentCount="1" />
  <row Id="15968" PostTypeId="2" ParentId="15966" CreationDate="2010-12-31T06:29:33.437" Score="7" ViewCount="0" Body="&lt;p&gt;If $2^a3^b \lt N, a \ln 2 + b \ln 3 \lt \ln N$ or $a+\frac{b \ln 3}{\ln 2}\lt \frac{\ln N}{\ln 2}$.  For large $N$, we can ignore +1s, so to count the number less than $N$, we have $$\sum_{i=0}^\frac{\ln N}{\ln 3}\frac{\ln N-i\ln 3}{\ln 2}=\frac{(\ln N)^2}{\ln 2 \ln 3}-\frac{(\ln N)^2}{2\ln 2 \ln 3}$$ which is within a constant of the Ramanujan result.&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2010-12-31T06:29:33.437" CommentCount="0" />
  <row Id="15969" PostTypeId="2" ParentId="15966" CreationDate="2010-12-31T06:31:04.923" Score="12" ViewCount="0" Body="&lt;p&gt;Taking logs, one finds that $2^a 3^b \leq N$ if and only if &#xA;$a \log 2 + b \log 3 \leq \log N.$&#xA;So estimating the number of 3-smooth integers $\leq N$ is the same as estimating&#xA;the number of integer lattice points in the region&#xA;$a, b \geq 0, a \log 2 + b \log 3 \leq \log N.$&#xA;The standard estimate for the number integer lattice points in a convex region is the area of the region.  In our case the region is a triangle, of area equal to&#xA;$$\dfrac{1}{2}\dfrac{\log N}{\log 2}\dfrac{\log N}{\log 3},$$&#xA;which is close to the estimate you attribute to Ramanujan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, the estimate you wrote down is equal &#xA;to $$\dfrac{1}{2}\dfrac{\log N + \log 2}{\log 2}\dfrac{\log N + \log 3}{\log 3}&#xD;&#xA;= \dfrac{1}{2}(\dfrac{\log N}{\log 2} + 1)(\dfrac{\log N}{\log 3} + 1).$$&#xA;This probably improves the estimate coming from the area, by including correction terms coming from the integer points along the boundary of the region (in particular, the contributions coming from $a = 0$ or $b = 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The fact that these &quot;boundary&quot; contributions come with a multiplicity of $1/2$ seems similar to other contexts in which one approximates a discrete sum by an area, e.g. in Euler--MacLaurin summation, and perhaps thinking from the point of view would let one get a better handle on the precise estimates.)&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="221" LastEditorDisplayName="" LastEditDate="2010-12-31T06:41:32.640" LastActivityDate="2010-12-31T06:41:32.640" CommentCount="2" />
  <row Id="15970" PostTypeId="1" AcceptedAnswerId="15972" CreationDate="2010-12-31T06:58:11.373" Score="2" ViewCount="72" Body="&lt;p&gt;I am reading &lt;a href=&quot;http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec1.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Intuitively, an open set is a solid region minus its boundary. If we include the boundary, we get a closed set, which formally is defined as the complement of an open set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, question is if a &lt;code&gt;closed set&lt;/code&gt; includes interior points also then how can it be &lt;code&gt;complement&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know basic set theory.&#xA;Enlighten me! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="723" LastEditorDisplayName="" LastEditDate="2010-12-31T10:24:01.117" LastActivityDate="2010-12-31T10:24:01.117" Title="Does closed set contain only boundary points or interior points also?" Tags="&lt;topology&gt;&lt;elementary-set-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="15972" PostTypeId="2" ParentId="15970" CreationDate="2010-12-31T07:14:15.743" Score="2" ViewCount="0" Body="&lt;p&gt;A Closed set is by definition a set whose complement is an open set. Note that this also includes the possibility that a set is both open and closed, for example in a space with two connected components, each component is both open and closed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in what you have highlighted the complement of the solid region (inclusive of boundary) i.e. the whole space without the region, is open. Which, means that the solid region (inclusive of boundary) is closed.   &lt;/p&gt;&#xA;" OwnerUserId="4806" LastActivityDate="2010-12-31T07:14:15.743" CommentCount="7" />
  <row Id="15973" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T07:36:20.263" Score="7" ViewCount="0" Body="&lt;p&gt;One abstract way to think about continuity (in the sense that it generalizes to non-metric spaces) is that it is about &lt;em&gt;error.&lt;/em&gt;  A function $f : X \to Y$ is continuous at $x$ precisely when $f(x)$ can be &quot;effectively measured&quot; in the sense that, by measuring $x$ closely enough, we can measure $f(x)$ to any desired precision.  (In other words, the error in our measurement of $f(x)$ can be controlled.  &quot;Precision&quot; here means &quot;to within an arbitrary neighborhood of $f(x)$,&quot; so it does not depend on any metric notions.)  This is an abstract formulation of one of the most basic assumptions of science: that (most of) the quantities we try to measure ($f(x)$) depend continuously on the parameters of our experiments ($x$).  If they didn't, science would be effectively impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like thinking about limits, a function is continuous if and only if it preserves limits of &lt;a href=&quot;http://en.wikipedia.org/wiki/Filter_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;filters&lt;/a&gt; or, equivalently, &lt;a href=&quot;http://en.wikipedia.org/wiki/Net_(mathematics)&quot; rel=&quot;nofollow&quot;&gt;nets&lt;/a&gt;.  These are two ways to generalize converge of sequences to spaces which are not first-countable.  &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-31T10:37:20.663" LastActivityDate="2010-12-31T10:37:20.663" CommentCount="1" />
  <row Id="15974" PostTypeId="2" ParentId="15966" CreationDate="2010-12-31T07:47:10.873" Score="10" ViewCount="0" Body="&lt;p&gt;For any reference requirement related to Ramanujan, it is always a good idea to check the series of volumes, titled Ramanujan's notebooks, compiled and annotated by Bruce C. Berndt and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of Entry 15 in Ramanujan's second notebook, which is about numbers of the form $\displaystyle a^p b^q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference to this can be found here: &lt;a href=&quot;http://books.google.com/books?id=tX5UUCempn8C&amp;amp;pg=PA66&quot; rel=&quot;nofollow&quot;&gt;Ramanujan's notebooks, Volume 4&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I suggest you read page 66 onwards)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snapshot (from the google books link itself):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/OI2J1.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This volume talks about the &quot;proof&quot; Ramanujan gave (pages 68 and 69), provides references to Hardy's book (which apparently has a whole chapter on this) and also mentions a paper by Hardy and Littlewood which deals with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;De Bruijn has considered the number of $y$-smooth numbers in this paper: &lt;a href=&quot;http://alexandria.tue.nl/repository/freearticles/597499.pdf&quot; rel=&quot;nofollow&quot;&gt;On the number of positive integer less than x and free of prime factors greater than y&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of $y$-smooth numbers $\le x$ is apparently now known in literature as the DeBruijn function: $\psi(x,y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closely related function is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dickman_function&quot; rel=&quot;nofollow&quot;&gt;DeBruijn-Dickman&lt;/a&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a &lt;a href=&quot;http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1993__5_2/JTNB_1993__5_2_411_0/JTNB_1993__5_2_411_0.pdf&quot; rel=&quot;nofollow&quot;&gt;survey&lt;/a&gt; by Hildebrand and Tenenbaum which should be helpful.&lt;/p&gt;&#xA;" OwnerUserId="1102" LastEditorUserId="1102" LastEditorDisplayName="" LastEditDate="2010-12-31T23:18:45.257" LastActivityDate="2010-12-31T23:18:45.257" />
  <row Id="15975" PostTypeId="1" CreationDate="2010-12-31T08:08:25.343" Score="1" ViewCount="60" Body="&lt;p&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/13169/what-is-spec-mathbfcx-y-y2-x3-x2&quot;&gt;Here&lt;/a&gt; evgeniamerkulova explains intuition for formal completion very nicely. I was wondering if something could be said if the base ring were $\mathbb C$ instead of $\mathbb Z$.&lt;/p&gt;&#xA;" OwnerUserId="5260" LastActivityDate="2010-12-31T08:08:25.343" Title="Formal completion of a scheme at a point" Tags="&lt;algebraic-geometry&gt;" />
  <row Id="15976" PostTypeId="1" AcceptedAnswerId="16002" CreationDate="2010-12-31T08:33:21.347" Score="7" ViewCount="119" Body="&lt;p&gt;V.I. Arnold put the following question in his Mathematical trivium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an asymptotically stable equilibrium position become unstable in the Lyapunov sense under linearization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It puzzled me for a while, since my experience doesn't include such a pathological case - but I'm still not sure about this one.&lt;/p&gt;&#xA;" OwnerUserId="553" LastActivityDate="2011-01-01T10:34:57.810" Title="Lyapunov stability question (Arnold's trivium)" Tags="&lt;control-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="15977" PostTypeId="1" CreationDate="2010-12-31T08:44:26.170" Score="9" ViewCount="254" Body="&lt;p&gt;I am planning to write a little note detailing several proofs of Lagrange's theorem that every natural number can be written as the sum of four perfect squares. I know of three different proofs so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a completely elementary proof by descent. &lt;/li&gt;&#xA;&lt;li&gt;a proof via Minkowski's theorem and lattices. &lt;/li&gt;&#xA;&lt;li&gt;Jacobi's proof via modular forms. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anybody think of any more nice, relatively elementary proofs of this result? Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="1491" LastActivityDate="2010-12-31T15:49:01.310" Title="Proofs that every natural number is a sum of four squares" Tags="&lt;number-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="15978" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T09:00:26.523" Score="7" ViewCount="0" Body="&lt;p&gt;There's a proof in Herstein using quaternions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[See also Hardy and Wright's &lt;em&gt;An Introduction to the Theory of Numbers&lt;/em&gt;.  -- PLC.]&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-12-31T09:34:04.190" LastActivityDate="2010-12-31T09:34:04.190" />
  <row Id="15979" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T09:32:49.463" Score="5" ViewCount="0" Body="&lt;p&gt;One can easily deduce the four squares theorem from the Gauss-Legendre three squares theorem.  The latter is actually less elementary, but there is a very nice proof using Hasse-Minkowski and the Aubry-Davenport-Cassels Lemma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the approach taken in Serre's &lt;em&gt;A Course in Arithmetic&lt;/em&gt;, Appendix to Chapter IV.  Quadratic forms which satisfy the conclusion of the Aubry-Davenport-Cassels Lemma (&lt;strong&gt;ADC forms&lt;/strong&gt;) have been a topic of recent research of mine, so it is fair to say that this is currently my favorite proof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the quaternionic proof of Chris Card's answer (which is also a very important one) can be understood and, arguably, clarified in these terms.  The sum of four squares form does not itself satisfy the hypotheses of the ADC Lemma.  This is related to the fact that it is the norm form of the &lt;em&gt;nonmaximal order&lt;/em&gt; $\mathbb{Z}[i,j,k]$ in the Hamilton quaternion algebra $H = \left(\frac{-1,-1}{\mathbb{Q}} \right)$.  A maximal order containing it is $\mathbb{Z}[i,j,k,\frac{1+i+j+k}{2}]$.  The norm form on this order is the quadratic form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q = x^2 + y^2 + z^2 + w^2 + xw + yw + zw$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This form &lt;em&gt;does&lt;/em&gt; satisfy the hypothesis of the ADC Lemma (it is &lt;em&gt;Euclidean&lt;/em&gt; in my terminology), so it follows immediately from Hasse-Minkowski that it is universal, i.e., represents all positive integers.  Moreover the form $q$ is sufficiently closely &#xA;related to the sum of four squares form that a little elementary fiddling around shows that the sum of four squares form is universal as well.  For other examples of this phenomenon, see &lt;a href=&quot;http://opensiuc.lib.siu.edu/cgi/viewcontent.cgi?article=1113&amp;amp;context=math_articles&quot; rel=&quot;nofollow&quot;&gt;this recent preprint of R.W. Fitzgerald&lt;/a&gt; as well as some papers of J.I. Deutsch that are cited there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you dare, you can take a look at &lt;a href=&quot;http://www.math.uga.edu/~pete/ADCformsPart1b.pdf&quot; rel=&quot;nofollow&quot;&gt;my magnum opus on Euclidean forms and ADC forms&lt;/a&gt;.  &lt;em&gt;Caveat emptor&lt;/em&gt;: more than being unpolished, there are some unsanded sharp corners here.)  &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-31T09:32:49.463" />
  <row Id="15980" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T09:50:27.073" Score="4" ViewCount="0" Body="&lt;p&gt;Here's one to add to your list by Andrews, Ekhad and Zeilberger.&#xA;&lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/9206/9206203v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A short proof of Jacobi’s formula for the number of representations of an integer as the&#xA;sum of four squares.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1301" LastActivityDate="2010-12-31T09:50:27.073" />
  <row Id="15981" PostTypeId="1" AcceptedAnswerId="15982" CreationDate="2010-12-31T09:51:12.427" Score="4" ViewCount="147" Body="&lt;p&gt;all examples i have come across so far deal with norm defined on a vector space. Can norm only be defined on vector spaces&lt;/p&gt;&#xA;" OwnerUserId="5261" LastActivityDate="2010-12-31T18:31:19.203" Title="Can a norm only be defined on vector spaces ?" Tags="&lt;real-analysis&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="15982" PostTypeId="2" ParentId="15981" CreationDate="2010-12-31T10:07:33.680" Score="11" ViewCount="0" Body="&lt;p&gt;For instance, a norm can be defined on an abelian group: see Section 2.4 &lt;a href=&quot;http://math.uga.edu/~pete/8410Chapter2v2.pdf&quot; rel=&quot;nofollow&quot;&gt;of these notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is perhaps less commonly done, but one can define a norm on any group: see Exercise 2.17 in the aforelinked lecture notes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are many structures throughout mathematics that are called &quot;norms&quot;, and you can make more if you want!  But the above example is very closely related to the norm on a vector space: there is an induced metric, and so forth.  In fact a norm on a vector space in the usual sense gives a norm on the underlying additive group $(V,+)$.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-12-31T10:07:33.680" />
  <row Id="15983" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T10:17:26.600" Score="0" ViewCount="0" Body="&lt;p&gt;Suppose $x$ is a point in some topological space, then we can define $\mathcal{N}_x$ to be the set of (not necessarily open) neighbourhoods of $x$. Then &quot;$f$ is continuous at $x$&quot; is defined as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\forall V \in \mathcal{N}_{f(x)} \exists U \in \mathcal{N}_x : f(U) \subseteq V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or more colourfully: Whenever &quot;the enemy&quot; comes with a cleverly chosen and &quot;small&quot; neighbourhood of $f(x)$, we must be able to find a neighbourhood of $x$ that maps into said neighbourhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is nice about this is that it is (obviously?) the topological version of the $\varepsilon$-$\delta$ definition from $\mathbb{R}$ and metric spaces that we know (and love(?)), and it's relatively easy to prove that $f: X \to Y$ is continuous at $x$ for all $x \in X$ if and only if $f^{-1}(U)$ is open in $X$ for all open $U \subseteq Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to say is that I don't have much intuition for the &quot;preimage of open sets is open&quot; definition either, but it's not clear to me that you really need that. We take this as the definition because it's simple, it's entirely written in terms of the topologies of $X$ and $Y$ (i.e. the collections of their respective open sets) and easily shown to be equivalent to something which we do have an intuition about (assuming that one finds $\varepsilon$-$\delta$ intuitive, obviously).&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="1269" LastEditorDisplayName="" LastEditDate="2010-12-31T12:21:57.147" LastActivityDate="2010-12-31T12:21:57.147" />
  <row Id="15984" PostTypeId="1" AcceptedAnswerId="16001" CreationDate="2010-12-31T10:19:10.840" Score="7" ViewCount="242" Body="&lt;p&gt;I've had no luck with this one. None of the convergence tests pop into mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried looking at it in this form $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ and apply Dirichlets test. I know that $\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n} \to 0$ but not sure if it's decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding absolute convergence, I tried:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}|\geq \sin^2 n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1}{2}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}-\frac{1}{2}\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again I'm stuck with $\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$.&lt;br&gt;&#xA;Assuming it converges then I've shown that $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ doesn't converge absolutely.&lt;/p&gt;&#xA;" OwnerUserId="3237" LastEditorUserId="3237" LastEditorDisplayName="" LastEditDate="2010-12-31T11:50:35.817" LastActivityDate="2010-12-31T18:30:28.143" Title="Does $\sum_{n=1}^\infty\frac{\sin n}{n}(1+\frac{1}{2}+\cdots+\frac{1}{n})$ converge (absolutely)?" Tags="&lt;calculus&gt;&lt;sequences-and-series&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="15985" PostTypeId="1" AcceptedAnswerId="15986" CreationDate="2010-12-31T10:37:37.283" Score="1" ViewCount="107" Body="&lt;p&gt;Again, I am reading &lt;a href=&quot;http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec2.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&#xA;I am finding it a bit difficult to understand the definition of &lt;code&gt;n-dimensional smooth manifold&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{U_a; x^1_a, x^2_a, ..., x^n_a\}$ ----(1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the thing (1) a set? (I think it is not). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a tuple? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is $U_a$ a set or a set of sets? What is significance of the subscript $a$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like it very much if someone explains the definition easier to understand. With good examples.  &lt;/p&gt;&#xA;" OwnerUserId="705" LastEditorUserId="705" LastEditorDisplayName="" LastEditDate="2010-12-31T10:44:01.693" LastActivityDate="2011-01-01T10:33:00.080" Title="Help with definition of n-dimensional smooth manifold" Tags="&lt;topology&gt;&lt;differential-geometry&gt;" AnswerCount="2" />
  <row Id="15986" PostTypeId="2" ParentId="15985" CreationDate="2010-12-31T10:49:42.610" Score="4" ViewCount="0" Body="&lt;p&gt;This is there in the notes. $\{ U_a \}$ is an open cover of $M$. This is a set where each element $U_a$ is an open subset of $M$. The $a$ is used as an index for this set; so it would help to have put $a\in A$, say, where appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $x_a^i$ is a coordinate function on the open set $U_a$ (for given $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notes do give an example of the unit sphere which has an open cover consisting of two open subsets.&lt;/p&gt;&#xA;" OwnerUserId="638" LastActivityDate="2010-12-31T10:49:42.610" CommentCount="6" />
  <row Id="15987" PostTypeId="2" ParentId="15985" CreationDate="2010-12-31T10:57:54.193" Score="0" ViewCount="0" Body="&lt;p&gt;A n-dim manifold is local homeomorphism like a n-dim Euclidean Space.In the smooth situation,you can put the open set（just a common part） of R^n on the any part of the mainfold with the diffeomorphism,it keeps the full rank.&#xA; sphere is a good and basic example, you can white down the diffeomorphic mapping,but for many other figures that is complicated.&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="4706" LastEditorDisplayName="" LastEditDate="2011-01-01T10:33:00.080" LastActivityDate="2011-01-01T10:33:00.080" />
  <row Id="15988" PostTypeId="2" ParentId="15984" CreationDate="2010-12-31T11:05:51.343" Score="11" ViewCount="0" Body="&lt;p&gt;To show that the series does not converge absolutely, it suffices to show that $\sum\nolimits_{n = 1}^\infty  {|\frac{{\sin n}}{n}|}  = \infty $. For this purpose, see &lt;a href=&quot;http://www.math.niu.edu/~rusin/known-math/99/series&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;: the paragraph starting with &quot;It's not absolutely convergent&quot; as well as the one starting with &quot;This series exhibits rather irregular behaviour&quot; (two different approaches).&lt;/p&gt;&#xA;" OwnerUserId="2810" LastEditorUserId="2810" LastEditorDisplayName="" LastEditDate="2010-12-31T11:15:51.387" LastActivityDate="2010-12-31T11:15:51.387" />
  <row Id="15989" PostTypeId="2" ParentId="15917" CreationDate="2010-12-31T12:27:01.150" Score="0" ViewCount="0" Body="&lt;p&gt;The confusion stems from the fact that you do not say what quantities change. Say, if you mean that $X$ changes by a small amount $\Delta X$ then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\log (X+\Delta X) - \log X = \log (1+\frac{\Delta X}{X}) \approx \frac{\Delta X}{X} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $x$ is really the relative change in the quantity $X$ and likewise for the other quantities.&lt;/p&gt;&#xA;" OwnerUserId="3567" LastActivityDate="2010-12-31T12:27:01.150" />
  <row Id="15990" PostTypeId="1" CreationDate="2010-12-31T12:34:57.667" Score="0" ViewCount="102" Body="&lt;p&gt;In the computational topology literature, the reduction algorithm for computing the Smith normal form of a boundary matrix uses the notation $b_j &amp;gt; 1 \: \text{and}\: b_j|b_{j+1}$ in the context of the diagonal elements of the Smith matrix. Can anyone give me an idea for what it means?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastActivityDate="2010-12-31T12:34:57.667" Title="What does $b_i|b_{i+1}$ mean in this context?" Tags="&lt;linear-algebra&gt;&lt;abstract-algebra&gt;&lt;algebraic-topology&gt;" CommentCount="3" />
  <row Id="15991" PostTypeId="2" ParentId="15801" CreationDate="2010-12-31T12:42:38.370" Score="5" ViewCount="0" Body="&lt;p&gt;Updated answer to (2): Three non-intersecting $60^\circ$ loxodromes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/7igYI.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axes are coplanar and inclined at $120^\circ$ to each other. This image shows that symmetry better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ERO5C.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's the Mercator projection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/aQ3W3.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was to plot one loxodrome such that its Mercator projection is a (black) line through the origin. Then, I tilted the spherical loxodrome &quot;toward the camera&quot;; that is, I rotated the sphere about the horizontal axis to get new, curvy (red) projections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/9Sv5u.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From tilt-angles $108^\circ$ to $143^\circ$, the &quot;curve&quot; lies between parts of the &quot;line&quot;, indicating a range of red loxodromes that don't intersect the black one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/TWfR4.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a certain sub-range ($108^\circ$ to about $125^\circ$), a third (blue) non-intersecting loxodrome can be added by rotating the red one about the Mercator origin. Here's an image from the end of that range, where red and blue are tangent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/ZCKS8.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the end of the illustrated intro. Now for some equations ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with your parameterization of the loxodrome, then tilting via angle $\theta$, gives this parameterization of the Mercator projection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;u &amp;amp;= \rm{atan}\left( \frac{\sin t}{\cos t \cos \theta + \sinh\left(t \cot\phi\right) \sin \theta }\right) \\&#xD;&#xA;v &amp;amp;= \rm{atanh}\left( \frac{-\cos t \sin\theta + \sinh\left(t \cot\phi\right) \cos\theta }{\cosh\left(t \cot\phi\right)}\right)&#xD;&#xA;\end{align}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tilted loxodrome crosses into the range of (possible) non-intersection with the un-tilted loxodrome when the &quot;top&quot; of the outer loop about its tilted north pole meets the Mercator origin. (The nature of loxodromes guarantees that the two loxodromes will be tangent there.) The point on the loop corresponds to $t=\pi$, for which $u$ is already zero; for $v$ to vanish, we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = -\cos\pi \sin\theta + \sinh\left(\pi \cot\phi\right) \cos\theta = \sin\theta+\sinh\left(\pi \cot\phi \right) \cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\theta = -\sinh\left(\pi\cot\phi\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, appropriately adjusting the &quot;branch&quot; of $\rm{atan}$, the range begins at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_0 := \pi - \rm{atan}\left(\sinh\left(\pi\cot\phi\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of (possible) non-intersection ends when the loop around the tilted loxodrome's south pole brushes against the un-tilted loxodrome. This is when the point corresponding to $t=-\pi$ has $v = \pi\cot\phi$ (matching the upper-right point of the &quot;straight&quot; loxodrome projection). So, the range ends at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_1 := 2\;\rm{atan}\left( \sinh\left(\pi\cot\phi\right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I write &quot;range of &lt;em&gt;possible&lt;/em&gt; non-intersection&quot;, because that range collapses when $\theta_0 = \theta_1$. This gives us a critical loxodrome angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{*} = \rm{atan}\left(\frac{\pi}{ \rm{asinh}\left( \tan\frac{\pi}{3} \right) } \right) \approx 67.2565^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/c58Kn.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot arrive at two non-intersecting loxodromes with $\phi &amp;gt; \phi_{*}$ --in particular, with $\phi=80^\circ$-- by tilting one relative to the other in the way I've described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's $\phi = 80^\circ$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.imgur.com/PEfya.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, &quot;the way I've described&quot; lacks generality. In addition to &quot;vertical&quot; tilts, one should also consider &quot;lateral&quot; spins (horizontal shifts in the Mercator projection). I'll leave that, and a full investigation of the three-loxodrome scenario, as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="409" LastEditorDisplayName="" LastEditDate="2011-01-01T23:46:47.327" LastActivityDate="2011-01-01T23:46:47.327" CommentCount="2" />
  <row Id="15993" PostTypeId="2" ParentId="15902" CreationDate="2010-12-31T13:18:41.060" Score="3" ViewCount="0" Body="&lt;p&gt;I just like to point out that this argument is in Hardy and Wright, An Introduction to the Theory of Numbers, with the slight differences that they avoid the use of the floor and ceiling functions, and finish off (quite nicely, in my opinion) with induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll type it here, to save you looking it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorem: $\theta(n) &amp;lt; 2n \log 2$ for all $ n \ge 1,$ where $$\theta(x) = \log \prod_{p \le x} p.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M = { 2m+1 \choose m},$ an integer, which occurs twice in the expansion of $(1+1)^{2m+1}$ and so $2M &amp;lt; 2^{2m+1}.$ Thus $M&amp;lt; 2^{2m}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m+1 &amp;lt; p \le 2m+1,$ $p$ divides $M$ since it divides the numerator, but not the denominator, of $ { 2m+1 \choose m } = \frac{(2m+1)(2m)\ldots(m+2)}{m!}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \prod_{m+1 &amp;lt; p \le 2m+1} p \right) | M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta(2m+1) - \theta(m+1) = \sum_{m+1 &amp;lt; p \le 2m+1} \log p \le \log M &amp;lt; 2m \log 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem is clear for $n=1$ and $n=2,$ so suppose that it is true for all $n \le N-1.$ If $N$ is even, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta(N)= \theta(N-1) &amp;lt; 2(N-1) \log 2 &amp;lt; 2N \log 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $N$ is odd, $N=2m+1 $ say, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\theta(N)=\theta(2m+1) &amp;amp;=\theta(2m+1)-\theta(m+1)+\theta(m+1) \\&#xD;&#xA;&amp;amp;&amp;lt; 2m \log 2 +2(m+1) \log 2 \\&#xD;&#xA;&amp;amp;=2(2m+1) \log 2 = 2N \log 2,&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $m+1 &amp;lt; N.$ Hence the theorem is true for $n=N$ and the result follows by induction.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="1301" LastEditorDisplayName="" LastEditDate="2011-01-02T09:46:20.843" LastActivityDate="2011-01-02T09:46:20.843" CommentCount="2" />
  <row Id="15994" PostTypeId="2" ParentId="15981" CreationDate="2010-12-31T13:31:29.123" Score="2" ViewCount="0" Body="&lt;p&gt;The word &quot;norm&quot; is also used in connection with &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_domain&quot; rel=&quot;nofollow&quot;&gt; Euclidean domains (ring theory)&lt;/a&gt;, although that Wikipedia page tries to avoid confusion by preferring instead &quot;Euclidean function&quot;.  &lt;strong&gt;Example:&lt;/strong&gt;  The degree of a univariate polynomial with coefficients over a field gives a norm for that kind of ring to be a Euclidean domain.&lt;/p&gt;&#xA;" OwnerUserId="3111" LastActivityDate="2010-12-31T13:31:29.123" />
  <row Id="15995" PostTypeId="1" AcceptedAnswerId="16054" CreationDate="2010-12-31T13:48:48.927" Score="9" ViewCount="216" Body="&lt;p&gt;Interesting problem from a 2000 St. Petersburg school olympiad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There are 109 soldiers in a camp. Every night three of them go on watch patrol. Prove that it can be arranged so that after a while, every pair of soldiers has shared watch exactly three times.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm missing the key insight here.&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2011-01-01T20:11:45.513" Title="combinatorics olympiad problem - sort out a schedule" Tags="&lt;combinatorics&gt;&lt;contest-math&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15996" PostTypeId="2" ParentId="15963" CreationDate="2010-12-31T14:00:13.037" Score="1" ViewCount="0" Body="&lt;p&gt;Let me also add one little bit (perhaps this may even seem backwards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Topology is defined using sets satisfying some set of axioms, that we call Open sets. However, the collection of subsets we choose (respecting some set of axioms) to call a Topology on the space may vary with the underlying set. But, we can go from one such collection to another using a special class of maps. But, how do we define such maps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, properties that are to be 'intrinsic' to the Topology should not depend on this choice, so should be invariant across these maps. But, these properties (whatever they be) are defined via set operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps we want the assignment to be direct: i.e. $U$ is open and thus $f(U)$ should be open. But, notice that since the map is general: $f(A \cap B)$ is contained and not equal to $f(A) \cap f(B)$ etc. However, with the assignment $f^{-1}$ (which need not even be well-defined) all is fine with unions and intersections, i.e., we have equalities, so, the properties follow through almost trivially (see for example the proofs that continuous maps preserve compactness, connectedness etc.)  &lt;/p&gt;&#xA;" OwnerUserId="4806" LastEditorUserId="4806" LastEditorDisplayName="" LastEditDate="2010-12-31T14:16:30.637" LastActivityDate="2010-12-31T14:16:30.637" CommentCount="1" />
  <row Id="15997" PostTypeId="1" AcceptedAnswerId="15998" CreationDate="2010-12-31T14:09:13.373" Score="2" ViewCount="89" Body="&lt;p&gt;Hi, given a ring $R$, a &lt;em&gt;nilpotent&lt;/em&gt; ideal $I$ and a morphism $\phi$ of $R$-modules $M \to N$, such that $M/IM \to N/IN$ is an isomorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that this implies $\phi$ surjective, but what about injectivity?&lt;/p&gt;&#xA;" OwnerUserId="5262" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-31T18:03:56.400" LastActivityDate="2010-12-31T18:03:56.400" Title="$M/IM \to  N/IN$ an isomorphism $\Longrightarrow  M \to  N$ an  isomorphism?" Tags="&lt;commutative-algebra&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="15998" PostTypeId="2" ParentId="15997" CreationDate="2010-12-31T14:12:11.817" Score="4" ViewCount="0" Body="&lt;p&gt;The answer is no in general. E.g. if $M = R$ and $N = R/I$, then the natural surjection $M \to N$ will induce an isomorphism $M/I \to N/I$, which won't be&#xA;an isomorphism unless $I = 0$.&lt;/p&gt;&#xA;" OwnerUserId="221" LastActivityDate="2010-12-31T14:12:11.817" CommentCount="4" />
  <row Id="15999" PostTypeId="2" ParentId="15977" CreationDate="2010-12-31T14:52:48.957" Score="6" ViewCount="0" Body="&lt;p&gt;Perhaps the most beautiful solution is by way of Aubry's lemma - which employs a geometric variant of the Euclidean algorithm to turn a rational represention into an integral representation. This is the same technique that leads to the reflective generation of primitive Pythagorean triples and the associated &lt;a href=&quot;http://math.stackexchange.com/questions/5540/k2k12-being-a-perfect-square-for-infinitely-many-k/5567#5567&quot;&gt;ternary tree structure.&lt;/a&gt; Aubry's results are, in fact, very special cases of general results of Wall, Vinberg, Scharlau et al. on reflective lattices, i.e. arithmetic groups of isometries generated by reflections in hyperplanes. Generally reflections generate the orthogonal group of Lorentzian quadratic forms in dim &amp;lt; 10. See my &lt;a href=&quot;http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726&quot; rel=&quot;nofollow&quot;&gt;MO post here&lt;/a&gt; for further remarks and references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the results in this area are some of the most beautiful results in elementary number theory. Strangely enough, for all this beauty they appear to be little known. For example, over a decade ago when I mentioned to John Conway the connection between Aubry's work and Cassels and Pfister he was not aware of this (R.K. Guy told me that the presentation of the PPT ternary tree in their &lt;em&gt;Book of Numbers&lt;/em&gt; (1996) is based on a lecture he heard by an undergraduate, Richard Vogeler, at an MAA section at Brigham Young Univ. on 89-04-07.)  Also Pfister apparently was not aware of Aubry's work when he generalized Cassels result to arbitrary quadratic forms, founding the modern algebraic theory of quadratic forms (&quot;Pfister forms&quot;). Someday I hope to write something on the bizarre history of this beautiful circle of ideas so I would be grateful to hear from anyone who may know further details.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2010-12-31T15:49:01.310" LastActivityDate="2010-12-31T15:49:01.310" />
  <row Id="16000" PostTypeId="1" CreationDate="2010-12-31T15:00:33.767" Score="1" ViewCount="73" Body="&lt;p&gt;Excerpt from Waterhouse, 14.4 Structure of Finite Connected groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thm. Let $A$ represent a finite connected group scheme over a perfect field of characteristic $p$. Then $A$ has the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k[X_1, ..., X_n]/(X_1^{p^{e_1}}, \ldots, X_n^{p^{e_n}})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Let $I_A$ be the augmentation ideal of $A$. By connectedness, $I_A$ is nilpotent. [...] We assume therefore that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Hopf subalgebra $B = A^p$ is one of these truncated polynomial algebras, say with generators $x_i$ and relations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i^{q_i} = 0$. Choose $y_i \in A$ with $y_i^p = x_i$, and choose also a set $\{z_j\}$ in $A$ maximal with resprect to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the requirements that $z_j^p = 0$ and that the $z_j$ be linearly independent in $I_A/I_A^2$. Let $C = k[Y_i, Z_j]/(Y_i^&#xD;&#xA;&#xD;&#xA;{pq_i}, Z_j^p$ which maps in the obvious way to $A$. We claim that this map is an iso.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Embed $A^p$ in $C$ by $x_i \mapsto Y_i^p$. Then $C$ is a free $B$-module. By the main theorem, $A$ is also free over $A^p$. As in (14.2), then, it is enough to show that $C/I_BC \to A/I_BA$ is an iso.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the last sentence.&lt;/p&gt;&#xA;" OwnerUserId="5262" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2010-12-31T18:54:57.693" LastActivityDate="2010-12-31T18:54:57.693" Title="question regarding Waterhouse, affine group schemes" Tags="&lt;algebraic-geometry&gt;&lt;commutative-algebra&gt;&lt;algebraic-groups&gt;" />
  <row Id="16001" PostTypeId="2" ParentId="15984" CreationDate="2010-12-31T16:24:41.560" Score="10" ViewCount="0" Body="&lt;p&gt;The sequence $$a_n=\frac{1+\frac{1}{2}+\cdots +\frac{1}{n}}{n}$$ is decreasing because proving $a_{n+1}&amp;lt;a_n$ reduces to proving $$\frac{n}{n+1}&amp;lt;1+\frac{1}{2}+\cdots +\frac{1}{n}$$ which is clearly true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the sums $$s_k=\sum_{n=1}^k \sin n$$ are bounded. (This can be easily proved by writing $\sin n$ in its complex form and using finite geometric formula; in fact $|s_k|$ are bounded by $\frac{1}{|1-e^i|}+\frac{1}{|1-e^{-i}|}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore $a_n\to 0$ as $n\to \infty$ because $1+\frac{1}{2}+\cdots +\frac{1}{n}\approx \log n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the series converges by Dirichlet test. As Shai shows below, the convergence is not absolute.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastEditorUserId="3249" LastEditorDisplayName="" LastEditDate="2010-12-31T18:30:28.143" LastActivityDate="2010-12-31T18:30:28.143" />
  <row Id="16002" PostTypeId="2" ParentId="15976" CreationDate="2010-12-31T16:41:40.403" Score="4" ViewCount="0" Body="&lt;p&gt;For the system $dx/dt=y-x^3$, $dy/dt=-y^3$, the origin is an&#xA;asymptotically stable equilibrium. (It is even &lt;em&gt;globally&lt;/em&gt; attracting,&#xA;as can be seen by sketching the phase portrait with the help of the&#xA;nullclines $y=x^3$ and $y=0$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linearized system at the origin is $dx/dt=y$, $dy/dt=0$,&#xA;which is not Lyapunov stable: the solution starting at $(x_0,y_0)=(0,\epsilon)$&#xA;is $(x(t),y(t)) = (\epsilon t, \epsilon)$,&#xA;which goes to infinity if $\epsilon \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix corresponding to the linearized system is $\begin{pmatrix} 0 &amp;amp; 1 \\ 0 &amp;amp; 0 \end{pmatrix}$,&#xA;which has zero as a double eigenvalue.&#xA;If both eigenvalues have nonzero real part, then the linearized system determines the stability&#xA;of the original system. If the eigenvalues are $\pm c i$ (for some real $c \neq 0$), then&#xA;the linearized system is a neutral center (hence Lyapunov stable).&#xA;So the phenomenon in question can only occur if at least one eigenvalue is zero.&#xA;(I haven't really thought about what can or cannot happen in the case with one&#xA;zero eigenvalue and one real nonzero eigenvalue.)&lt;/p&gt;&#xA;" OwnerUserId="1242" LastEditorUserId="1242" LastEditorDisplayName="" LastEditDate="2011-01-01T10:34:57.810" LastActivityDate="2011-01-01T10:34:57.810" CommentCount="4" />
  <row Id="16003" PostTypeId="1" AcceptedAnswerId="16019" CreationDate="2010-12-31T16:58:59.917" Score="4" ViewCount="192" Body="&lt;p&gt;I may be over-complicating things, but something doesn't seem right (and I swear this isn't homework, I'm friggin 30 years old :P).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to see what it will cost me to make a TWO liter of Coca-Cola using coke syrup and CO2 (assume the water is free).  Assume CO2 costs 0.21usd per liter of water (25usd for 120 liters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coke syrup needs to be diluted at a &quot;5.4:1&quot; ratio with the carbonated water.&#xA;So how much will it cost me to make my own 2-liter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;syrup cost per liter = C / ((G * D) * 3.785)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;C is the cost of G gallons of coke syrup&lt;/li&gt;&#xA;&lt;li&gt;D is the dilution (5.4), and&lt;/li&gt;&#xA;&lt;li&gt;3.785 converts gallons to liters&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a 65.00usd box of 5 gallons of coke syrup, I get a cost of 0.64usd per liter.  Along with CO2 cost, that makes it 1.69usd for a 2-liter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That seems very, very costly, given that 2-liters in the store usually cost much less.  If I'm wrong, where am I going wrong?  If I'm right, why don't restaurants use 2-liters instead of fountain machines to increase their profits/lower their costs?  I know they charge a lot for a &lt;em&gt;glass&lt;/em&gt; of coke (and refills), not a &lt;em&gt;2-liter&lt;/em&gt;, but it still seems like it would make them more money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put that equation into an Excel sheet, and it seems like the syrup would have to be 30usd in order to get the cost down to 1.00usd/2-liter, and that doesn't even seem like much of a gain over buying 2-liter bottles from a grocery (especially with the equipment overhead costs).  I spoke to a restaurant manager, and &lt;em&gt;his&lt;/em&gt; cost is 56usd for a 5-gallon box, which puts the per 2-liter cost at $1.51 by my calculations, so that's when I suspected I was doing something wrong.&lt;/p&gt;&#xA;" OwnerUserId="5263" LastActivityDate="2010-12-31T23:27:02.890" Title="Ratios make me feel like an idiot - help me mix up some Coca-Cola" Tags="&lt;fractions&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="16004" PostTypeId="1" AcceptedAnswerId="16005" CreationDate="2010-12-31T17:27:39.497" Score="0" ViewCount="61" Body="&lt;p&gt;Let $(\Omega, \mathcal{F}, \textbf{P})$ be a probability space and $A,B, A_i$ events in $\mathcal{F}$. Prove the following properties of every probability measure: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Monotonicity: If $A \subseteq B$ then $\textbf{P}(A) \leq \textbf{P}(B)$.&lt;/li&gt;&#xA;&lt;li&gt;Sub-additivity: If $A \subseteq \bigcup_{i} A_i$ then $\textbf{P}(A) \leq \sum_{i} \textbf{P}(A_i)$. &lt;/li&gt;&#xA;&lt;li&gt;Continuity from Below: If $A_{i} \uparrow A$, then $\textbf{P}(A_i) \uparrow \textbf{P}(A)$. &lt;/li&gt;&#xA;&lt;li&gt;Continuity from Above: If $A_{i} \downarrow A$, then $\textbf{P}(A_i) \downarrow \textbf{P}(A)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the first, one could consider $\textbf{P}(B-A)$? For the second, one could consider $\textbf{P}(A_{i}-A)$ over all $i$? It seems like the third and fourth properties follow from the first two. In particular, if $A_{1} \subseteq A_2 \subseteq \dots$ and $\bigcup_{i} A_i = A$, then $\textbf{P}(A_1) \leq \textbf{P}(A_2) \leq \dots$ and $\sum_{i} \textbf{P}(A_i) = \textbf{P}(A)$.  &lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2010-12-31T18:04:06.467" Title="Properties of Probability Measure" Tags="&lt;probability-theory&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="16005" PostTypeId="2" ParentId="16004" CreationDate="2010-12-31T17:41:48.913" Score="5" ViewCount="0" Body="&lt;p&gt;Monotonicity follows from the countable additivity of measure. You need to recognize whenever $A \subseteq B$, we have $B = A \cup (B \cap A^{c})$. And note that $A$ and $B \cap A^{c}$ are disjoint sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-additivity is proved as follows. First prove that $\mathbf{P}(\displaystyle \cup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \mathbf{P}(A_i)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved by re-writing $\displaystyle \cup_{i=1}^{\infty} A_i$ as a disjoint union and then use the countable additivity of a measure to get the desired result. Now use monotonicity proved in the previous part to conclude the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What, continuity from below means is &quot;If we have a nested sequence of sets, i.e. $A_1 \subseteq A_2 \subseteq \ldots $ and $\cup_{i=1}^{\infty} A_i = A$, then we have continuity from below i.e. $\displaystyle \lim_{n \rightarrow \infty} \mathbf{P}(A_n) = \mathbf{P}(A)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick here again is to rewrite $\displaystyle \cup_{i=1}^{\infty} A_i$ as a disjoint union and massage around to get the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, continuity from above means, &quot;If we have a sequence of sets such that $A_{1} \supseteq A_{2} \supseteq \ldots $ and $\displaystyle \cap_{i=1}^{\infty} A_i = A$, then we have continuity from above i.e. $\displaystyle \lim_{n \rightarrow \infty} \mathbf{P}(A_n) = \mathbf{P}(A)$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from previous part when we consider the sequence of sets $B_i = A_1 \backslash A_i$ and noting that $\mathbb{P}(A_1) \leq 1 &amp;lt; \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: In general, all these properties except the last one hold true for any measure. The last property holds true if at-least one of the sets in the sequence we consider has a finite measure.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastEditorUserId="3302" LastEditorDisplayName="" LastEditDate="2010-12-31T18:04:06.467" LastActivityDate="2010-12-31T18:04:06.467" />
  <row Id="16007" PostTypeId="2" ParentId="15981" CreationDate="2010-12-31T18:31:19.203" Score="0" ViewCount="0" Body="&lt;p&gt;There are other kinds of norms, like field norms and norms of ideals. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Norm#In_mathematics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Norm#In_mathematics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2010-12-31T18:31:19.203" />
  <row Id="16008" PostTypeId="2" ParentId="15899" CreationDate="2010-12-31T18:51:44.450" Score="4" ViewCount="0" Body="&lt;p&gt;$x$ used to be called &lt;em&gt;latus&lt;/em&gt; (latin for &lt;em&gt;side&lt;/em&gt;---of a square, a cube)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in this, you &lt;em&gt;must&lt;/em&gt; read Cajori's &lt;em&gt;History of mathematical notations&lt;/em&gt;, which has a thorough discussion of the history of the notation and nomenclature of powers.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="274" LastEditorDisplayName="" LastEditDate="2011-01-03T03:00:47.787" LastActivityDate="2011-01-03T03:00:47.787" />
  <row Id="16009" PostTypeId="2" ParentId="15898" CreationDate="2010-12-31T18:59:33.047" Score="4" ViewCount="0" Body="&lt;p&gt;If I understand you correctly, you want to prove that, for a square matrix with complex coefficients (or more generally, any algebraically closed field) the product of the eigenvalues (with appropriate multiplicity) is a multiplicative function on the matrices, but &lt;em&gt;without&lt;/em&gt; using that the product of the eigenvalues is the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is pretty easy to show that the product of the Eigenvalues, with appropriate multiplicity, is the determinant, by using the Jordan canonical form. The determinant is invariant under conjugation, so the determinant of a matrix is the same as the determinant of its Jordan canonical form; this form is upper triangular, and the diagonal contains the eigenvalues, each with the appropriate algebraic multiplicity, yielding that the determinant is indeed the product of the eigenvalues; a Jordan canonical basis yields the triples you have. Once you have that the product is indeed the determinant, the conclusion follows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of obstacles to the project. First, we need to show that $D(\mathbf{M})$ is well-defined; that is, that it does not depend on the choice of maximal family of vectors. This can be achieved pretty much the way one shows that the generalized eigenspaces yield a decomposition of the vector space (see any book that covers the Jordan canonical form), but the simplest proofs I know will use the Cayley-Hamilton Theorem at some point, which involves the determinant. (The more complicated ones consider the vector space as a module over $\mathbb{C}[x]$ and use the structure theorem for modules over PIDs... don't ask.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But assuming this is taken for granted (that you can find a basis of generalized eigenvectors, so that your maximal family of triples will have $n$ elements, and that the number of times that each $k_i$ occurs is independent of the choice), then we can proceed along similar lines as the way we prove that the determinant is multiplicative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If there is vector $\mathbf{v}\neq 0$ such that $\mathbf{M}\mathbf{v}=\mathbf{0}$, then $D(\mathbf{M})=0$. Conversely, if $D(\mathbf{M})=0$, then there exists a vector $\mathbf{v}\neq 0$ such that $\mathbf{M}\mathbf{v}=\mathbf{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Immediate. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; $D(\mathbf{M})=0$ if and only if $\mathbf{M}$ is not invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; $\mathbf{M}$ is invertible if and only if it is of full rank. By the Rank-Nullity Theorem, this occurs if and only if the nullity of $\mathbf{M}$ is zero, which happens if and only if the nullspace is trivial, if and only if for all $\mathbf{v}$, $\mathbf{M}\mathbf{v}=\mathbf{0}\Rightarrow \mathbf{v}=\mathbf{0}$. The previous lemma then completes the proof. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; If $D(\mathbf{B})=0$, then $D(\mathbf{AB}) = 0 = D(\mathbf{A})D(\mathbf{B})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Let $\mathbf{v}\neq\mathbf{0}$ be such that $\mathbf{B}\mathbf{v}=\mathbf{0}$. Then $(\mathbf{AB})\mathbf{v}=\mathbf{0}$, so by previous result, $D(\mathbf{AB})=0$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposition.&lt;/strong&gt; If $D(\mathbf{A}) = 0$, then $D(\mathbf{AB}) = 0 = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If $D(\mathbf{B})=0$, then the result follows by the previous proposition. If $D(\mathbf{B})\neq 0$, then $B$ is invertible (as above). Since $D(\mathbf{A})=0$, there exists $\mathbf{v}\neq \mathbf{0}$ such that $\mathbf{A}\mathbf{v}=\mathbf{0}$. Then &#xA;$$(\mathbf{AB})(\mathbf{B}^{-1}\mathbf{v}) = \mathbf{A}\mathbf{v}=\mathbf{0} = 0(\mathbf{B}^{-1}\mathbf{v}),$$ &#xA;and since $\mathbf{B}^{-1}\mathbf{v}\neq\mathbf{0}$, then $D(\mathbf{AB}) = 0$, as desired. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If $\mathbf{B}$ is an elementary matrix, then  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If $\mathbf{B}$ is obtained from the identity by transposing two rows, then $D(\mathbf{B}) = -1$.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathbf{B}$ is obtained from the identity by adding a multiple of one row to another row, then $D(\mathbf{B}) = 1$.&lt;/li&gt;&#xA;&lt;li&gt;If $\mathbf{B}$ is obtained from the identity by multiplyig a row by $\lambda\neq 0 $, then $D(\mathbf{B}) = \lambda$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose we exchange rows $i$ and $j$. That means that $\mathbf{B}(e_i) = e_j$ and $\mathbf{B}(e_j) = e_i$, while $\mathbf{B}(e_k) = e_k$ if $k\neq i,j$. We get $n-2$ triples with $(e_k,1,1)$, $k\neq i,j$, and also the triples $(e_i+e_j,1,1)$ and $(e_i-e_j,-1,1)$. Then $D(\mathbf{B}) = 1^{n-1}(-1) = -1$, as claimed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose we add $r\neq 0$ times the $i$th row to the $j$th row. Then $\mathbf{B}(e_k) = e_k$ if $k\neq i$, and $\mathbf{B}(e_i) = e_i+re_j$. Take the $n-1$ triples $(e_k,1,1)$ with $k\neq i$, and the triple $(e_i,1,2)$. Note that this works, since $(\mathbf{B}-\mathbf{I})e_i = (e_i+re_j) - e_i = re_j$, and $(\mathbf{B}-\mathbf{I})(re_j) = re_j - re_j = \mathbf{0}$, so that $(\mathbf{B}-\mathbf{I})^2e_i = \mathbf{0}$, with $2$ minimal. Thus, $D(\mathbf{B}) = 1^n = 1$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we multiply the $i$th row by $\lambda$, then the $n-1$ triples $(e_j,1,1)$ with $j\neq i$, together with $(e_i,\lambda,1)$, make a maximal family and we have $D(\mathbf{B}) = 1^{n-1}\lambda = \lambda$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lemma.&lt;/strong&gt; If $\mathbf{A}$ and $\mathbf{B}$ are elementary matrices, then $D(\mathbf{AB}) = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sketch.&lt;/em&gt; There are several cases to consider, depending on which kind of elementary matrices they are and how they interact. For example, if $A$ is obtained from the identity by adding $r$ times the $i$th row to the $j$th row, and $B$ is obtained by multiplying the $\ell$th row by $\lambda$, then we must consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\ell\neq i,j$, then we can take the $n-3$ triples $(e_k,1,1)$ with $k\neq i,j,\ell$, the triple $(e_{\ell},\lambda,1)$, the triple $(e_j,1,1)$, and the triple $(e_i,1,2)$. Then $D(\mathbf{AB}) = 1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\ell = j$, then $\mathbf{AB}(e_k) = e_k$ if $k\neq i,j$; $\mathbf{AB}e_j = \lambda e_j$, and $\mathbf{AB}(e_i) = e_i+re_j$, so we can take the $n-2$ triples $(e_k,1,1)$ with $k\neq i,j$, together with $(e_j,\lambda,1)$ and $(e_i,1,2)$. Then $D(\mathbf{AB}) = 1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\ell=i$, then $\mathbf{AB}(e_k) = e_k$ if $k\neq i$, and $\mathbf{AB}(e_i) = \lambda e_i + \lambda re_j$. We can then take the $n-1$ triples $(e_k,1,1)$, $k\neq i$, and the triple $(e_i,\lambda,2)$, to get $D(\mathbf{AB})=1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Similar arguments hold for the other combinations of products of elementary matrices. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Corollary.&lt;/strong&gt; If $E_1,\ldots,E_k$ are elementary $n\times n$ matrices with coefficients in $\mathbb{C}$, then $D(E_1\cdots E_k) = D(E_1)\cdots D(E_k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; Induction on $k$. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THEOREM.&lt;/strong&gt; If $\mathbf{A}$ and $\mathbf{B}$ are $n\times n$ matrices with coefficients in $\mathbb{C}$, then $D(\mathbf{AB}) = D(\mathbf{A})D(\mathbf{B})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof.&lt;/em&gt; If either $D(\mathbf{A}) = 0$  or $D(\mathbf{B})=0$, then the result follows from previous proposition. If $D(\mathbf{A})\neq 0 \neq D(\mathbf{B})$, then each of $\mathbf{A}$ and $\mathbf{B}$ is invertible, hence a product of elementary matrices. The result now follows from the Corollary above. &lt;strong&gt;QED&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2010-12-31T19:45:35.830" LastActivityDate="2010-12-31T19:45:35.830" />
  <row Id="16010" PostTypeId="2" ParentId="15939" CreationDate="2010-12-31T19:01:09.793" Score="0" ViewCount="0" Body="&lt;p&gt;Yes, Z has full rank, and A is positive definite, but using the formula I do not&#xA;get how i can compute $(A Z^T Z A)^+ $ as a function of $(Z^T Z)^+ $ and $A$...&#xA;or am I doing something wrong?&lt;/p&gt;&#xA;" OwnerUserId="5267" LastActivityDate="2010-12-31T19:01:09.793" CommentCount="5" />
  <row Id="16011" PostTypeId="1" AcceptedAnswerId="16014" CreationDate="2010-12-31T19:50:40.350" Score="21" ViewCount="638" Body="&lt;p&gt;I don't know if this is a well know fact but I have observed that every number no matter how large that is equally divided by 9, will equal 9 if you add all the numbers it is made from until there is 1 digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick example of what I mean:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;9*99 = 891&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;8+9+1 = 18&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1+8 = 9&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This works even with really long numbers like 4376331&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? This doesn't work with any other number.&lt;/p&gt;&#xA;" OwnerUserId="5268" LastActivityDate="2011-01-01T16:30:55.763" Title="What makes 9 special?" Tags="&lt;number-theory&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="8" />
  <row Id="16012" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T19:57:29.690" Score="3" ViewCount="0" Body="&lt;p&gt;This property is based on the fact that 9 is one minus 10 which is the base of the number system you are working with because of this any power of 10 will equal 1 modulo 9.&lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-12-31T19:57:29.690" />
  <row Id="16013" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T19:58:04.493" Score="4" ViewCount="0" Body="&lt;p&gt;This follows from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisibility_rule&quot; rel=&quot;nofollow&quot;&gt;divisibility test&lt;/a&gt; for $9$ and from the fact that the sum of the digits of a natural number with more than one digit is strictly less than the number itself i.e. if $n$ is a natural number with more than one digit and if $s(n)$ denotes the sum of the digits of the natural number then $s(n)&amp;lt;n$. Your observation can be strengthened to note that if we keep repeatedly adding the digits, the final number we are left with is the remainder when divided by $9$.&lt;/p&gt;&#xA;" OwnerUserId="3302" LastActivityDate="2010-12-31T19:58:04.493" />
  <row Id="16014" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T19:58:20.250" Score="34" ViewCount="0" Body="&lt;p&gt;Not quite right, since $9\times 0 = 0$ and the digits don't add up to $9$; but otherwise correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it works is that we write numbers in base $10$, and when you divide $10$ by $9$, the remainder is $1$. Take a number, say, $184631$ (I just made it up). Remember what that really means:&#xA;$$184631 = 1 + 3\times 10 + 6\times 10^2 + 4\times 10^3 + 8\times 10^4 + 1\times 10^5.$$&#xA;The remainder when you divide any power of $10$ by $9$ is again just $1$, so adding the digits gives you a number that has the same remainder when dividing by $9$ as the original number does. Keep doing it until you get down to a single digit and you get the remainder of the original number when you divide by $9$, except that you get $9$ instead of $0$ if the number is a nonzero multiple of $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since every multiple of $9$ &lt;em&gt;is&lt;/em&gt; a multiple of $9$, you will always get $9$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you have a similar phenomenon with $3$ (a divisor of $9$), since adding the digits of a multiple of $3$ will always result in one of the one-digit multiples of $3$: $3$, $6$, or $9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we wrote in base $8$, instead of base $10$, then $7$ would have the property: if you write a number in base $8$ and add the digits (in base 8) until you get down to a single digit between $1$ and $7$, then the multiples of $7$ will always end yielding $7$, for precisely the same reason. And if we wrote in base $16$, then $15$ (or rather, F) would have the property. In general, if you write in base $b$, then $b-1$ has the property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting_out_nines&quot; rel=&quot;nofollow&quot;&gt;casting out nines&lt;/a&gt;, which in turn is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;nofollow&quot;&gt;modular arithmetic&lt;/a&gt;. It is what is behind many &lt;a href=&quot;http://en.wikipedia.org/wiki/Divisibility_tests&quot; rel=&quot;nofollow&quot;&gt;divisibility tests&lt;/a&gt; (e.g., for $2$, $3$, $5$, $9$, and $11$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coda.&lt;/strong&gt; This reminds me of an anecdote a professor of mine used to relate: a student once came to him telling him he had discovered a very easy way to test divisibility of any number $N$ by any number $b$: write $N$ in base $b$, and see if the last digit is $0$. I guess, equivalently, you could write $N$ in base $b+1$, and add the digits to see if you get $b$ at the end.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="742" LastEditorDisplayName="" LastEditDate="2011-01-01T01:50:18.590" LastActivityDate="2011-01-01T01:50:18.590" CommentCount="3" />
  <row Id="16015" PostTypeId="2" ParentId="16011" CreationDate="2010-12-31T20:55:42.317" Score="7" ViewCount="0" Body="&lt;p&gt;This is a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Casting_out_nines&quot; rel=&quot;nofollow&quot;&gt;casting out nines,&lt;/a&gt; which is best viewed in terms of modular arithmetic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rm\quad\quad\quad  mod\ 9:\ \ 10\equiv 1\ \Rightarrow\ P(10)\equiv P(1)\quad $ for all polynomials $\rm\ P(x) \in \mathbb Z[x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But radix notation has &lt;em&gt;polynomial&lt;/em&gt; form, e.g. $\rm\ 567 = P(10),\ \ P(x) = 5\: x^2 + 6\: x + 7\:,\ $ so the above says that, mod $9$,  an integer $\rm\ N = P(10)\ $ is congruent to $\rm\ P(1) = $ the sum $\rm\:S_N\:$ of its decimal digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yours is the special case $\rm\ N\equiv 0\ \ (mod\ 9)\ $ so the above implies that it remains $\equiv 0\ $ when you map $\rm\:N\:$ to its digit sum $\rm\:S_N\:$.  Since this map is strictly decreasing for $\rm\:N &amp;gt; 9\:,\ $ iterating it must eventually reach some nonzero $\rm\: N&amp;#39; \le 9\:.\ $ But since $\rm\ N&amp;#39;\equiv 0\ \ (mod\ 9)\ $ we conclude that $\rm\: N&amp;#39; = 9\:$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if modular arithmetic is unfamiliar then you may instead proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Factor Theorem&lt;/strong&gt; $\rm\;\Rightarrow\; X-1\ |\ P(X)-P(1)\ $ so $\rm\ X = 9\ \Rightarrow\ 9\ |\ P(10)-P(1) \;\:$ i.e. $\rm\; 9\ |\ N - S_N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous result holds true for any radix $\rm\:b\:$, i.e. one can cast $\rm\:b-1$'s in the same way, since $\rm\ P(b) \equiv P(1)\ \ (mod\ b-1)\:.\:$ Thus $\:9\:$ is &quot;special&quot; only in the way it relates to the radix $10\:$. One can similarly cast $11$'s by using $\rm\ P(10)\equiv P(-1)\ \ (mod\ 11)\:.\:$ One can devise countless &quot;divisibility tests&quot; by using modular reduction in this manner, e.g. see my &lt;a href=&quot;http://groups.google.com/groups?selm=y8zwsvvxdmy.fsf@nestle.csail.mit.edu&quot; rel=&quot;nofollow&quot;&gt;post here&lt;/a&gt; for casting out $91$'s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It deserves to be better known that one may also cast out nines to check rational arithmetic - as long as the fractions have denominator coprime to $3$, e.g. see Hilton; Pedersen: &lt;a href=&quot;http://www.jstor.org/stable/2689631&quot; rel=&quot;nofollow&quot;&gt;Casting out nines revisited&lt;/a&gt;. Analogous remarks hold true for any ring that has $\ \mathbb Z/9\ $ as an image - just as one can apply &lt;strong&gt;parity&lt;/strong&gt; arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. all rationals with odd denominator, or the Gaussian integers $\rm\:\mathbb Z[i]\:$, where the image $\rm\ \mathbb Z[i]/(2,i-1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\rm\ a+b\:i\ $ is even iff $\rm\ a\equiv b\ \ (mod\ 2)\ $, i.e. if $\rm\ a+b\:i\ $ maps to $\:0\:$ via the above isomorphism, which maps $\rm\ 2\to 0,\ i\to 1\:$.  These are but simple prototypical examples of problem-solving by way of &lt;strong&gt;modular reduction&lt;/strong&gt; - one of the keystones of abstract algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://math.stackexchange.com/questions/12846/using-fermats-little-theorem/12851#12851&quot;&gt;this post&lt;/a&gt; for &quot;casting out orders&quot; in cyclic groups, and see &lt;a href=&quot;http://groups.google.com/groups?selm=y8zoh73mzp8.fsf@nestle.ai.mit.edu&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; for an in-depth comparison of various inductive proofs of casting nines.&lt;/p&gt;&#xA;" OwnerUserId="242" LastEditorUserId="242" LastEditorDisplayName="" LastEditDate="2011-01-01T01:24:38.597" LastActivityDate="2011-01-01T01:24:38.597" />
  <row Id="16016" PostTypeId="1" AcceptedAnswerId="16017" CreationDate="2010-12-31T21:40:57.353" Score="2" ViewCount="131" Body="&lt;p&gt;This may be a simple question, but I can't make it out right now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let $M$ be a monoid, and $g_1, \ldots, g_n \in M$ be elements of finite order in $M$.  Is $\langle g_1, \ldots, g_n \rangle$ periodic (i.e, have all its elements of finite order)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The monoid I'm  interested in is the multiplicative monoid of $n \times n$ matrices over $\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5270" LastActivityDate="2011-01-01T05:37:27.103" Title="Is a monoid finitely generated by finite order elements periodic?" Tags="&lt;group-theory&gt;&lt;monoid&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="16017" PostTypeId="2" ParentId="16016" CreationDate="2010-12-31T21:53:33.383" Score="6" ViewCount="0" Body="&lt;p&gt;(Edit because I misread the question somewhat.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that this is false for an arbitrary monoid it suffices to exhibit a monoid generated by elements of finite order which has elements of infinite order.  Take, for example, $M = \langle a, b | a^2 = b^2 = 1 \rangle$.  In this monoid $ab$ has infinite order.  One can see this geometrically by constructing the following linear representation: let $a$ and $b$ act by reflection across two lines through the origin at an angle $\theta$ with respect to each other which is not a rational multiple of $\pi$.  (For this reason $M$ is known as the infinite dihedral group.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most important example of this phenomenon &quot;in nature&quot; is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_group&quot; rel=&quot;nofollow&quot;&gt;modular group&lt;/a&gt;, which has group presentation $\langle a, b | a^2 = b^3 = 1 \rangle$ and hence is generated by an element of order $2$ and an element of order $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about your particular $M$.  Does $\mathbb{N}$ contain $0$ for you?  (If not I am having trouble finding &lt;em&gt;any&lt;/em&gt; nontrivial elements of finite order.)&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="232" LastEditorDisplayName="" LastEditDate="2010-12-31T22:17:38.750" LastActivityDate="2010-12-31T22:17:38.750" CommentCount="6" />
  <row Id="16018" PostTypeId="1" CreationDate="2010-12-31T22:28:24.550" Score="4" ViewCount="62" Body="&lt;p&gt;Dear all,&#xA;I have systems like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A - \lambda B) X = F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where lambda is being updated inside a loop. I also have a limited number of eigenvectors of the matrix pair (A, B), say 40 eigenpair from a previous analysis step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could get the results with direct solution however due to repeated factorization of the operator matrix, I run into the O(n^3) wall. I was wondering if that could be possible to solve this system by some iterative methods. I, myself, tried some iterative methods, cg, minres, and others firstly in MATLAB, however cg like methods do not even converge due to the nature of the operator matrix since that can become indefinite in the loop, minres seems to converge hovewer that also does not converge in reasonable iteration counts. Also in these tries I had to use a PD preconditioner and do a factorization, which also induces other costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically are there some kind of a mathematical tricks for this kind of problem, where some limited number of eigenpairs are available from a previous analysis step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a poor engineer I tried all I know, so any further ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIA,&#xA;Umut&lt;/p&gt;&#xA;" OwnerUserId="5272" LastEditorUserId="1827" LastEditorDisplayName="" LastEditDate="2010-12-31T22:32:53.890" LastActivityDate="2010-12-31T22:32:53.890" Title="linear system solution, iterative vs direct" Tags="&lt;linear-algebra&gt;" CommentCount="1" />
  <row Id="16019" PostTypeId="2" ParentId="16003" CreationDate="2010-12-31T23:27:02.890" Score="3" ViewCount="0" Body="&lt;p&gt;With a $5.4:1$ ratio of carbonated water to syrup, assuming that the quantities are additive (mixing 5.4 units of carbonated water with 1 unit of syrup yields 6.4 units of the mixture), I get: $$5\text{ gallons of syrup}\times\frac{6.4\text{ gallons of mixture}}{1\text{ gallon of syrup}}\times\frac{3.785\text{ liters}}{1\text{ gallon}}\approx 121\text{ liters of mixture}.$$&#xA;At $\$65$ per box, $$\frac{\$65}{121\text{ liters}}\approx\$0.54\text{ per liter}.$$&#xA;This is all ignoring the cost of carbonating the water, because around me, a 2-liter bottle of Coke goes on sale for 99 cents with enough regularity that it already doesn't make sense to mix at home.&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2010-12-31T23:27:02.890" CommentCount="2" />
</posts>